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Linear and Quasilinear Complex Equations of Hyperbolic and Mixed Type

Asian Mathematics Series A Series edited by Chung-Chun YangDepartment of Mathematics The Hong Kong University ofScience and Technology Hong Kong

Volume 1Dynamics of transcendental functionsXin-Hou Hua and Chung-Chun Yang

Volume 2Approximate methods and numerical analysis for elliptic complex equationsGuo Chun Wen

Volume 3Introduction to statistical methods in modern geneticsMark CK Yang

Volume 4Mathematical theory in periodic plane elasticityHai-Tao Cai and Jian-ke Lu

Volume 5Gamma lines On the geometry of real and complex functionsGrigor A Barsegian

Volume 6Linear and quasilinear complex equations of hyperbolic and mixed typeGuo Chun Wen

Guo Chun Wen School of Mathematical Sciences Peking University Beijing China

London and New York

First published 2002 by Taylor amp Francis 11 New Fetter Lane London EC4P 4EE

Simultaneously published in the USA and Canada by Taylor amp Francis Inc 29 West 35th Street New York NY 10001

Taylor amp Francis is an imprint of the Taylor amp Francis Group

copy 2002 Guo Chun Wen

All rights reserved No part of this book may be reprinted or reproducedor utilized in any form or by any electronicmechanical or other meansnow known or hereafter invented including photocopying and recording or in any information storage or retrieval system without permission inwriting from the publishers

Every effort has been made to ensure that the advice and information in this book is true and accurate at the time of going to press However neither the publisher nor the authors can accept any legal responsibility or liability for any errors or omissions that may be made In the case of drug administration any medical procedure or the use of technical equipment mentioned within this book you are strongly advised to consult the manufacturerrsquos guidelines

British Library Cataloguing in Publication DataA catalogue record for this book is available from the British Library

Library of Congress Cataloging in Publication DataA catalog record for this book has been requested

ISBN 0ndash415ndash26971ndash7

This edition published in the Taylor amp Francis e-Library 2006

ldquoTo purchase your own copy of this or any of Taylor amp Francis or Routledgersquoscollection of thousands of eBooks please go to wwweBookstoretandfcoukrdquo

ISBN 0-203-16658-2 Master e-book ISBN

ISBN 0-203-26135-6 (Adobe eReader Format)

(Print Edition)

Contents

Introduction to the series viiiPreface ix

Chapter IHyperbolic complex equations of first order 1

1 Hyperbolic complex functions and hyperbolic pseudoregular functions 1

2 Complex forms of linear and nonlinear hyperbolicsystems of first order equations 10

3 Boundary value problems of linear hyperbolic complex equations of first order 18

4 Boundary value problems of quasilinear hyperboliccomplex equations of first order 25

5 Hyperbolic mappings and quasi-hyperbolic mappings 35

Chapter IIHyperbolic complex equations of second order 39

1 Complex form of hyperbolic equations of second order 392 Oblique derivative problems for quasilinear hyperbolic

equations of second order 433 Oblique derivative problems for general quasilinear

hyperbolic equations of second order 504 Other oblique derivative problems for quasilinear

hyperbolic equations of second order 595 Oblique derivative problems for degenerate hyperbolic

equations of second order 66

vi Contents

Chapter IIINonlinear elliptic complex equations of first and second order 79

1 Generalizations of KeldychndashSedov formula for analytic functions 79

2 Representation and existence of solutions for elliptic complex equations of first order 90

3 Discontinuous oblique derivative problems for quasilinear elliptic equations of second order 95

4 Boundary value problems for degenerate ellipticequations of second order in a simply connected domain 108

Chapter IVFirst order complex equations of mixed type 119

1 The RiemannndashHilbert problem for simplest first order complex equation of mixed type 119

2 The RiemannndashHilbert problem for first order linear complex equations of mixed type 126

3 The RiemannndashHilbert problem for first orderquasilinear complex equations of mixed type 134

4 The RiemannndashHilbert problem for first order quasilinearequations of mixed type in general domains 138

5 The discontinuous RiemannndashHilbert problem for quasilinear mixed equations of first order 143

Chapter VSecond order linear equations of mixed type 157

1 Oblique derivative problems for simplest second order equation of mixed type 157

2 Oblique derivative problems for second order linear equations of mixed type 162

3 Discontinuous oblique derivative problems for secondorder linear equations of mixed type 171

4 The Frankl boundary value problem for second order linear equations of mixed type 177

5 Oblique derivative problems for second order degenerate equations of mixed type 194

Contents vii

Chapter VISecond order quasilinear equations of mixed type 200

1 Oblique derivative problems for second order quasilinear equations of mixed type 200

2 Oblique derivative problems for second order equations of mixed type in general domains 209

3 Discontinuous oblique derivative problems for second order quasilinear equations of mixed type 218

4 Oblique derivative problems for quasilinear equationsof mixed type in multiply connected domains 227

References 240Index 250

Introduction to the Series

The Asian Mathematics Series provides a forum to promote and reflect timelymathematical research and development from the Asian region and to providesuitable and pertinent reference on text books for researchers academics andgraduate students in Asian universities and research institutes as well as in theWest With the growing strength of Asian economic scientific and technologicaldevelopment there is a need more than ever before for teaching and researchmaterials written by leading Asian researchers or those who have worked in orvisited the Asian region particularly tailored to meet the growing demands ofstudents and researchers in that region Many leading mathematicians in Asiawere themselves trained in the West and their experience with Western methodswill make these books suitable not only for an Asian audience but also for theinternational mathematics community

The Asian Mathematics Series is founded with the aim to present significantcontributions from mathematicians written with an Asian audience in mind tothe mathematics community The series will cover all mathematical fields andtheir applications with volumes contributed to by international experts who havetaught or performed research in Asia The material will be at graduate level orabove The book series will consist mainly of monographs and lecture notes butconference proceedings of meetings and workshops held in the Asian region willalso be considered

Preface

In this book we mainly introduce first and second order complex equations ofhyperbolic and mixed (elliptic-hyperbolic) type in which various boundaryvalue problems for first and second order linear and quasilinear complex equationsof hyperbolic and mixed type are considered In order to obtain the results oncomplex equations of mixed type we need to first discuss some boundary valueproblems for elliptic and hyperbolic complex equations

In Chapters I and II the hyperbolic pseudoregular functions and quasi-hyperbolicmappings are introduced which are corresponding to pseudoanalytic functionsand quasiconformal mappings in the theory of elliptic complex equations On thebasis of hyperbolic notations the hyperbolic systems of first order equations andhyperbolic equations of second order with some conditions can be reduced tocomplex forms In addition several boundary value problems mainly the RiemannndashHilbert problem oblique derivative problems for some hyperbolic complex equationsof first and second order are discussed in detail

In Chapter III firstly the generalizations of the KeldychndashSedov formula foranalytic functions are given Moreover discontinuous boundary value problemsfor nonlinear elliptic complex equations of first and second order are discussedBesides some oblique derivative problems for degenerate elliptic equations ofsecond order are also introduced

In Chapter IV we mainly consider the discontinuous boundary value problemsfor first order linear and quasilinear complex equations of mixed type whichinclude the discontinuous Dirichlet problem and discontinuous RiemannndashHilbertproblem In the meantime we give some a priori estimates of solutions for theabove boundary value problems

For the classical dynamical equation of mixed type due to S A Chaplygin[17] the first really deep results were published by F Tricomi [77] 1) In ChaptersV and VI we consider oblique derivative boundary value problems for secondorder linear and quasilinear complex equations of mixed type by using a complexanalytic method in a special domain and in general domains which include theDirichlet problem (Tricomi problem) as a special case We mention that in thebooks [12] 1) 3) the author investigated the Dirichlet problem (Tricomi problem)for the simplest second order equation of mixed type ie uxx+sgnyuyy=0 in

x Preface

general domains by using the method of integral equations and a complicatedfunctional relation In the present book we use the uniqueness and existence ofsolutions of discontinuous RiemannndashHilbert boundary value problem for ellipticcomplex equations and other methods to obtain the solvability result of obliquederivative problems for more general equations and domains which includes theresults in [12] 1) 3) as special cases

Similarly to the book [86] 1) the considered complex equations and boundaryconditions in this volume are rather general and several methods are used Thereare two characteristics of this book one is that mixed complex equations are includedin the quasilinear case and boundary value conditions are almost considered inthe general oblique derivative case especially multiply connected domains areconsidered Another one is that complex analytic methods are used to investigatevarious problems about complex equations of hyperbolic and mixed type Wemention that some free boundary problems in gas dynamics and some problem inelasticity can be handled by using the results stated in this book

The great majority of the contents originates in investigations of the author andhis cooperative colleagues and many results are published here for the first timeAfter reading the book it can be seen that many questions about complexequations of mixed type remain for further investigations

The preparation of this book was supported by the National Natural ScienceFoundation of China The author would like to thank Prof H Begehr ProfW Tutschke and Mr Pi Wen Yang because they proposed some beneficialimproving opinions to the manuscript of this book

Beijing Guo Chun WenAugust 2001 Peking University

CHAPTER I

HYPERBOLIC COMPLEX EQUATIONS OFFIRST ORDER

In this chapter we first introduce hyperbolic numbers hyperbolic regular functionsand hyperbolic pseudoregular functions Next we transform the linear and non-linear hyperbolic systems of first order equations into complex forms Moreoverwe discuss boundary value problems for some hyperbolic complex equations of firstorder Finally we introduce the so-called hyperbolic mappings and quasihyperbolicmappings

1 Hyperbolic Complex Functions and HyperbolicPseudoregular Functions

11 Hyperbolic numbers and hyperbolic regular functions

First of all we introduce hyperbolic numbers and hyperbolic complex functions Theso-called hyperbolic number is z = x + jy where x y are two real numbers and j iscalled the hyperbolic unit such that j2 = 1 Denote

e1 = (1 + j)2 e2 = (1minus j)2 (11)

it is easy to see that

e1 + e2 = 1 ekel =

ek if k = l

0 if k = lk l = 1 2 (12)

and (e1 e2) will be called the hyperbolic element Moreover w = f(z) = u(x y) +jv(x y) is called a hyperbolic complex function where u(x y) v(x y) are two realfunctions of two real variables x y which are called the real part and imaginary partof w = f(z) and denote Rew = u(z) = u(x y) Imw = v(z) = v(x y) Obviously

z = x+ jy = microe1 + νe2 w = f(z) = u+ jv = ξe1 + ηe2 (13)

in which

micro = x+ y ν = x minus y x = (micro+ ν)2 y =(micro minus ν)2

ξ = u+ v η = u minus v u = (ξ + η)2 v = (ξ minus η)2

2 I Hyperbolic Equations of First Order

z = xminusjy will be called the conjugate number of z The absolute value of z is definedby |z| =

radic|x2 minus y2| and the hyperbolic model of z is defined by z= radic

x2 + y2The operations of addition subtraction and multiplication are the same with the realnumbers but j2 = 1 There exists the divisor of zero and denote by O = z |x2 = y2the set of divisors of zero and zero It is clear that z isin O if and only if |z| = 0 andz has an inversion

1z=

z

zz=

1x+ y

e1 +1

x minus ye2 =

1micro

e1 +1νe2

if and only if x + jy isin O and if the hyperbolic numbers z1 = micro1e1 + ν1e2 z2 =micro2e1 + ν2e2 isin O then

z1

z2= (micro1e1 + ν1e2)

(1micro2

e1 +1ν2

e2

)=

micro1

micro2e1 +

ν1

ν2e2

It is clear that |z1z2| = |z1||z2| but the triangle inequality is not true As for thehyperbolic model of z we have the triangle inequality z1+z2 le z1 + z2 and z1z2 le radic

2 z1 z2 In the following the limits of the hyperbolic number aredefined by the hyperbolic model The derivatives of a hyperbolic complex functionw = f(z) with respect to z and z are defined by

wz = (wx + jwy)2 wz = (wx minus jwy)2 (14)

respectively and then we have

wz = (wx minus jwy)2=[(ux minus vy) + j(vx minus uy)]2

= [(wx minus wy)e1 +(wx + wy)e2]2=wνe1 + wmicroe2

= [ξνe1 + ηνe2]e1+[ξmicroe1 + ηmicroe2]e2=ξνe1 + ηmicroe2

wz = [(ux + vy) + j(vx + uy)]2 = wmicroe1 + wνe2

= (ξe1 + ηe2)microe1 + (ξe1 + ηe2)νe2=ξmicroe1 + ηνe2

(15)

LetD be a domain in the (x y)-plane If u(x y) v(x y) are continuously differentiablein D then we say that the function w = f(z) is continuously differentiable in D andwe have the following result

Theorem 11 Suppose that the hyperbolic complex function w = f(z) is continu-ously differentiable Then the following three conditions are equivalent

(1) wz = 0 (16)

(2) ξν = 0 ηmicro = 0 (17)

(3) ux = vy vx = uy (18)

Proof From (15) it is easy to see that the conditions (1)(2) and (3) in Theorem11 are equivalent

1 Hyperbolic Complex Functions 3

The system of equations (18) is the simplest hyperbolic system of first orderequations which corresponds to the CauchyndashRiemann system in the theory of ellipticequations The continuously differentiable solution w = f(z) of the complex equation(16) in D is called a hyperbolic regular function in D

If the function w(z) is defined and continuous in the neighborhood of a point z0and the following limit exists and is finite

wprime(z0)= limzrarrz0

w(z)minusw(z0)z minus z0

= limmicrorarrmicro0νrarrν0

[ξ(z)minusξ(z0)

micro minus micro0e1 +

η(z)minusη(z0)ν minus ν0

e2

]

= [ξmicroe1 + ηνe2]|micro=micro0ν=ν0 = wz(z0)

then we say that w(z) possesses the derivative wprime(z0) at z0 From the above for-mula we see that w(z) possesses a derivative at z0 if and only if ξ(z) = Rew(z) +Imw(z) η(z) = Rew(z) minus Imw(z) possess derivatives at micro0 = x0 + y0 ν0 = x0 minus y0

respectively

Now we can define some elementary hyperbolic regular functions according toseries representations in their convergent domains as follows

zn = [microe1 + νe2]n = microne1 + νne2 =(x+ y)n + (x minus y)n

2+j(x+ y)n minus(x minus y)n

2

ez = 1 + z +z2

2+ middot middot middot+ zn

n+ middot middot middot = emicro e1 + eν e2 =

ex+y + exminusy

2+ j

ex+y minus exminusy

2

ln z = lnmicro e1 + ln ν e2 =ln(x+ y) + ln(x minus y)

2+ j

ln(x+ y)minus ln(x minus y)2

sin z = z minus z3

3+ middot middot middot+ (minus1)n z2n+1

(2n+ 1)+ middot middot middot = sinmicro e1 + sin ν e2

cos z = 1minus z2

2+ middot middot middot+ (minus1)n z2n

(2n)+ middot middot middot = cosmicro e1 + cos ν e2

tgz=sin z

cos z= (sinmicroe1 + sin νe2)

(1

cosmicroe1 +

1cos ν

e2

)= tgmicro e1 + tgν e2

ctgz=cos zsin z

=(cosmicroe1 + cos νe2)(

1sinmicro

e1+1

sin νe2

)=ctgmicro e1+ctgν e2

(1+z)α=1+αz+middot middot middot+α(αminus1) (αminusn+1)n

zn+middot middot middot=(1+micro)αe1+(1+ν)αe2

where n is a positive integer and α is a positive number Moreover we can define theseries expansion of hyperbolic regular functions and discuss its convergency

12 Hyperbolic continuous functions and their integrals

Suppose that w = f(z) = u(x y) + jv(x y) is any hyperbolic complex function in adomain D and possesses continuous partial derivatives of first order in D Then forany point z0 isin D we have

∆w = f primex(z0)∆x+ f prime

y(z0)∆y + ε(∆z)

where

∆w = f(z)minus f(z0) f primex(z0) = ux(z0) + jvx(z0) f prime

y(z0) = uy(z0) + jvy(z0)

and z = z0 +∆z ε is a function of ∆z and

lim∆zrarr0

ε(∆z) rarr 0

Suppose that C is a piecewise smooth curve in the domain D and w = f(z) =u+ jv = ξe1 + ηe2 is a continuous function in D Then the integral of f(z) along Cand D are defined byint

Cf(z)dz =

intC

udx+ vdy + j[int

Cvdx+ udy] =

intC[ξdmicroe1 + ηdνe2]int int

Df(z)dxdy =

int intD

udxdy + jint int

Dvdxdy

We easily obtain some properties of integrals of f(z) as follows

Theorem 12 (1) If f(z) g(z) are continuous functions in D and C is a piecewisesmooth curve in D thenint

C[f(z) + g(z)]dz =

intC

f(z)dz +int

Cg(z)dzint int

D[f(z) + g(z)]dxdy =

int intD

f(z)dxdy +int int

Dg(z)dxdy

(2) Under the conditions in (1) and denoting M1 = maxzisinC f(z) M2 = supzisinD

f(z) the length of C by l and the area of D by S then

int

Cf(z)dz le

radic2M1l

int int

Df(z)dxdy le

radic2M2S

(3) If C is a piecewise smooth closed curve and G is the finite domain bounded byC f(z) is continuously differentiable in G then we have Greenrsquos formulasint int

G[f(z)]zdxdy =

j

2

intC

f(z)dz

int intG[f(z)]zdxdy = minusj

2

intC

f(z)dz

(4) Under the conditions as in (3) of Theorem 11 and w = f(z) is a hyperbolicregular function in G then int

Cf(z)dz = 0

In the following we introduce the definition of hyperbolic pseudoregular functionsand prove some properties of hyperbolic pseudoregular functions

13 Hyperbolic pseudoregular functions and their properties

Let w(z) F (z) G(z) be continuous functions in a domain D and G(z) F (z) satisfythe conditions

ImF (z)G(z) = 0 in D (19)

Then for every point z0 isin D we can obtain a unique pair of real numbers δ0 and γ0such that

w(z0) = δ0F (z0) + γ0G(z0) (110)

SettingW (z) = w(z)minus δ0F (z)minus γ0G(z) (111)

it is easy to see thatW (z0) = 0 (112)

If the following limit exists and is finite

w(z0) = limzrarrz0

w(z)minus δ0F (z)minus γ0G(z)z minus z0

= limzrarrz0

W (z)minus W (z0)z minus z0

ie

w(z0)= [ReW (z0)+ImW (z0)]microe1+[ReW (z0)minusImW (z0)]νe2=W prime(z0)(113)

where micro = x + y ν = x minus y then we say that w(z) is a (FG)-derivative of w(z) atz0 In order to express the existence of (113) by partial differential equations wesuppose again that

Fz(z) Fz(z) Gz(z) and Gz(z) exist and are continuous (114)

in a neighborhood of z0 According to the definition of W (z) if Wz Wz exist then

Wz(z) = wz(z)minus δ0Fz(z)minus γ0Gz(z)

Wz(z) = wz(z)minus δ0Fz(z)minus γ0Gz(z)(115)

From (113) (115) and Theorem 11 we see that if w(z0) exists then Wz(z0) exists

Wz(z0) = w(z0) (116)

andWz(z0) = 0 (117)

and if wz(z) wz(z) are continuous in a neighborhood of z0 and (117) holds then wehave (113) and (116) Since

W (z) =

∣∣∣∣∣∣∣∣∣w(z) w(z0) w(z0)

F (z) F (z0) F (z0)

G(z) G(z0) G(z0)

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣F (z0) F (z0)

G(z0) G(z0)

∣∣∣∣∣∣ (118)

(117) can be rewritten as ∣∣∣∣∣∣∣∣∣wz(z0) w(z0) w(z0)

Fz(z0) F (z0) F (z0)

Gz(z0) G(z0) G(z0)

∣∣∣∣∣∣∣∣∣ = 0 (119)

If (113) exists then (116) can be written as

w(z0) =

∣∣∣∣∣∣∣∣∣wz(z0) w(z0) w(z0)

Fz(z0) F (z0) F (z0)

Gz(z0) G(z0) G(z0)

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣F (z0) F (z0)

G(z0) G(z0)

∣∣∣∣∣∣ (120)

Unfolding (119) and (120) respectively and arranging them we obtain

wz = aw + bw (121)

w = wz minus Aw minus Bw (122)

where

a = minus FGz minus FzG

F G minus FG b =

FGz minus FzG

FG minus FG

A = minus FGz minus FzG

F G minus FG B =

FGz minus FzG

FG minus FG

(123)

here a(z) b(z) A(z) and B(z) are called the characteristic coefficients of the gener-ating pair (FG) Obviously F = G = 0 and

Fz = aF + bF Gz = aG+ bG

Fz = AF +BF Gz = AG+BG

uniquely determine a b A and B Denote them by a(FG) b(FG) A(FG) and B(FG)

respectively

From the above discussion we see that if w(z0) exists then wz at z0 exists and(121) (122) are true If wz and wz(z) exist and are continuous in a neighborhoodz0 isin D and (121) holds at z0 then w(z0) exists and (122) is true

For any function w(z) if w(z) exists and is continuous in the domain D thenw(z) is called the first-class (FG) hyperbolic pseudoregular function or hyperbolicpseudoregular function for short It is clear that the following theorem holds

Theorem 13 w(z) is a hyperbolic pseudoregular function if and only if wz(z) andwz(z) exist and are continuous and (121) holds

By (19) it is easy to see that every function w(z) has a unique expression

w(z) = φ(z)F (z) + ψ(z)G(z) (124)

where φ(z) and ψ(z) are two real-valued functions Let

K(z) = φ(z) + jψ(z) (125)

Then we can give the following definition

If w(z) is the first-class (FG) hyperbolic pseudoregular complex function thenK(z) = φ(z) + jψ(z) is called the second-class (FG) hyperbolic pseudoregular func-tion

Theorem 14 K(z) = φ(z)+jψ(z) is a second-class (FG) hyperbolic pseudoregularfunction if and only if φ and ψ have continuous partial derivatives and

Fφz +Gψz = 0 (126)

Under this conditionw(z) = Fφz +Gψz (127)

holds where

φz = (e1φmicro + e2φν) = [(φx + φy)e1 + (φx minus φy)e2]2

φz = (e1φν + e2φmicro) = [(φx minus φy)e1 + (φx + φy)e2]2

ψz = (e1ψmicro + e2ψν) = [(ψx + ψy)e1 + (ψx minus ψy)e2]2

ψz = (e1ψν + e2ψmicro) = [(ψx minus ψy)e1 + (ψx + ψy)e2]2

(128)

Proof From

W (z) = [φ(z)minus φ(z0)]F (z) + [ψ(z)minus ψ(z0)]G(z)

it follows thatWz(z0) = F (z0)φz(z0) +G(z0)ψz(z0)

Wz(z0) = F (z0)φz(z0) +G(z0)ψz(z0)

Thus the proof can be immediately obtained

Setting

minusG

F= σ + jτ = (σ + τ)e1 + (σ minus τ)e2

minusF

G= σ + jτ = (σ + τ)e1 + (σ minus τ)e2

(129)

where σ τ σ and τ are real-valued functions and τ = 0 τ = 0 Hence (126) isequivalent to the system of equations

φx = σψx minus τψy φy = minusτψx + σψy (130)

If φ and ψ have continuous partial derivatives up to second order we find the deriva-tives with respect to x and y in (130) and then obtain

φxx minus φyy + δφx + γφy = 0

ψxx minus ψyy + δψx + γψy = 0(131)

where

δ =σy + τx

τ γ = minus σx + τy

τ

δ =σy + τx

τ γ = minusσx + τy

τ

(132)

In accordance with the following theorem the elimination is reasonable

Theorem 15 Let δ γ δ and γ be determined by (132) Then the real-valued func-tion φ (ψ) is the real (imaginary) part of a second-class hyperbolic pseudoregular func-tion if and only if it has continuous partial derivatives up to second order and satisfiesthe first (second) equation in (131)

Proof From the second formula in (131) we see that the function

φ(z) =int z

z0

[(σψx minus τψy)dx+ (minusτψx + σψy)dy] z0 z isin D

is single-valued and φ(z) ψ(z) satisfy system (130) The part of necessity can bederived from Theorem 23 Chapter II below

14 Existence of a generating pair (FG)

Theorem 16 Let a(z) and b(z) be two continuous functions in a bounded andclosed domain D = micro0 le micro le micro0+R1 ν0 le ν le ν0+R2 where R1 R2 are positiveconstants and denote z0 = micro0e1 + ν0e2 Then there exists a unique continuouslydifferentiable hyperbolic pseudoregular function w(z) satisfying the complex equation

wz = a(z)w(z) + b(z)w(z) (133)

and the boundary conditions

w(z) = c1(micro)e1 + c2(ν0)e2 when z isin L1

w(z) = c1(micro0)e1 + c2(ν)e2 when z isin L2(134)

where c1(micro) and c2(ν) are two real continuous functions on L1 L2 respectively L1 =micro0 le micro le micro0 +R1 ν = ν0 and L2 = micro = micro0 ν0 le ν le ν0 +R2The theorem is a special case of Theorems 33 and 34 below

Theorem 17 Let a(z) and b(z) be two continuous complex functions in the domainD as stated in Theorem 16 Then there exists a generating pair (FG) in D suchthat

a = a(FG) and b = b(FG) (135)

Proof Denote by F (z) and G(z) two solutions of the complex equation (133) sat-isfying the boundary conditions

w(z) = e1 + e2 = 1 when z isin L1 cup L2

andw(z) = e1 minus e2 = j when z isin L1 cup L2

respectively Then by Theorem 16 F (z) and G(z) have continuous partial deriva-tives and

Fz = aF + bF and Gz = aG+ bG

F = 1 G = j when z isin L1 cup L2

Hence a = a(FG) and b = b(FG) Whether ImF (z)G(z) = 0 in D remains to bediscussed

Theorem 18 Under the same conditions as in Theorem 17 and letting

b = minusa or b = a z isin D (136)

there exists a generating pair (FG) in D satisfying the complex equation (133) and

F (z) = 1 or G(z) = j in D (137)

Proof By the hypotheses in Theorem 17 and equation (136) there exists a uniquegenerating pair (FG) in D satisfying the conditions

Fz = a(F minus F ) or Gz = a(G+ G) z isin D

F (z) = 1 or G(z) = j when z isin L1 cup L2

Hence we have (137) The above results are similar to those in [9]1) (see [89])

2 Complex Forms of Linear and Nonlinear HyperbolicSystems of First Order Equations

In this section we transform linear and nonlinear hyperbolic systems of first orderequations into complex forms

21 Complex forms of linear hyperbolic systems of first order equations

We consider the linear hyperbolic system of first order partial differential equations⎧⎨⎩a11ux + a12uy + b11vx + b12vy = a1u+ b1v + c1

a21ux + a22uy + b21vx + b22vy = a2u+ b2v + c2(21)

where the coefficients akl bkl ak bk ck (k l = 1 2) are known functions in D in whichD is a bounded domain System (21) is called hyperbolic at a point in D if at thepoint the inequality

I = (K2 +K3)2 minus 4K1K4 gt 0 (22)

holds in which

K1 =∣∣∣∣∣a11 b11

a21 b21

∣∣∣∣∣ K2 =∣∣∣∣∣a11 b12

a21 b22

∣∣∣∣∣ K3 =∣∣∣∣∣a12 b11

a22 b21

∣∣∣∣∣ K4 =∣∣∣∣∣a12 b12

a22 b22

∣∣∣∣∣ If the inequality (22) at every point (x y) in D holds then (21) is called a hyperbolicsystem in D We can verify that (22) can be rewritten as

I = (K2 minus K3)2 minus 4K5K6 gt 0 (23)

where

K5 =∣∣∣∣∣a11 a12

a21 a22

∣∣∣∣∣ K6 =∣∣∣∣∣ b11 b12

b21 b22

∣∣∣∣∣ If the coefficients akl bkl (k l = 1 2) in D are bounded and the condition

I = (K2 +K3)2 minus 4K1K4 = (K2 minus K3)2 minus 4K5K6 ge I0 gt 0 (24)

holds in which I0 is a positive constant then (21) is called uniformly hyperbolic inD In the following we reduce system (21) to complex form

1) If K2 K3 are of same signs and K6 = 0 at the point (x y) isin D then we cansolve (21) for vy minusvx and obtain the system of first order equations⎧⎨⎩ vy = aux + buy + a0u+ b0v + f0

minusvx = dux + cuy + c0u+ d0v + g0(25)

where a b c d are known functions of akl bkl(k l = 1 2) and a0 b0 c0 d0 f0 g0 areknown functions of bkl ak bk ck(k l = 1 2) and

a = K1K6 b = K3K6 c = K4K6 d = K2K6

2 Complex Forms of Hyperbolic Systems 11

The condition (22) of hyperbolic type is reduced to the condition

∆ = I4K26 =

(b+ d)2

4minus ac gt 0 (26)

There is no harm in assuming that a minus c ge 0 because otherwise let y be replacedby minusy this requirement can be realized If a c are not of the same sign or one ofthem is equal to zero then minusac ge 0 bd ge 0 and may be such that a ge 0 minusc ge 0or a c are of same signs then we may assume that a gt 0 c gt 0 because otherwiseif v is replaced by minusv this requirement can be realized Moreover we can assumethat 0 lt c lt 1 otherwise setting v = hv herein h is a positive constant such thath ge c + 1 we have K4 = hK4 K6 = h2K6 and c = K4K6 = ch lt 1 bd ge 0Multiply the first formula of (25) by minusj and then subtract the second formula of(25) This gives

vx minus jvy = minus j(aux + buy + a0u+ b0v + f0)

minus dux minus cuy minus c0u minus d0v minus g0

Noting z = x+ jy w = u+ jv and using the relations⎧⎨⎩ux = (wz + wz + wz + wz)2 uy = j(minuswz + wz + wz minus wz)2

vx = j(wz minus wz + wz minus wz)2 vy = (minuswz minus wz + wz + wz)2

we getj(wz minus wz) = minus(aj + d)(wz + wz + wz + wz)2

minus(c+ bj)j(minuswz + wz + wz minus wz)2

+lower order terms

namely

(1 + q1)wz + q2wz = minusq2wz + (1minus q1)wz + lower order terms (27)

in whichq1 = [a minus c+ (d minus b)j]2 q2 = [a+ c+ (d+ b)j]2

Notingq0 = (1 + q1)(1 + q1)minus q2q2

= [(2 + a minus c)2 minus (d minus b)2 minus (a+ c)2 + (d+ b)2]4

= 1 + a minus c minus (d minus b)24 + (d+ b)24minus ac

= 1 + a minus c minus (d minus b)24 + ∆

= 1 + a minus c+ σ = 1 + a minus c+ bd minus ac

= (1 + a)(1minus c) + bd gt 0

where σ = ∆minus (b minus d)24 = bd minus ac ge 0 thus we can solve (27) for wz giving

wz minus Q1(z)wz minus Q2(z)wz = A1(z)w + A2(z)w + A3(z) (28)

in which

Q1(z) =minus2q2(z)q0(z)

Q2(z) =[q2q2 minus (q1 minus 1)(q1 + 1)]

q0

For the complex equation (28) if (a minus c)2 minus 4∆ ge 0 (1 + σ)2 minus 4∆ ge 0 ie

(K1 minus K4)2 minus (K2 minus K3)2 + 4K5K6 ge 0

(K26 +K2K3 minus K1K4)2 minus (K2 minus K3)2 + 4K5K6 ge 0

(29)

then we can prove

|Q1|+ |Q2| = |Q1Q1|12 + |Q2Q2|12 lt 1 (210)

where |Q1| = |Q1Q1|12 is the absolute value of Q1 In fact

|2q2| = |(a+ c)2 minus (d+ b)2|12 = |(a minus c)2 minus 4∆|12

|q2q2 minus (q1 minus 1)(q1 + 1)|= |(a minus c)24minus∆minus [a minus c+ (b minus d)j minus 2]

times[a minus c minus (d minus b)j + 2]4| = |(1 + σ)2 minus 4∆|12

(1 + σ)2 + (a minus c)2 + 2(1 + σ)(a minus c) = [1 + σ + (a minus c)]2 gt 0

(1 + σ)2(a minus c)2 + (4∆)2 minus 4∆(1 + σ)2 minus 4∆(a minus c)2

lt (4∆)2 + (1 + σ)2(a minus c)2 + 8∆(1 + σ)(a minus c)

[(a minus c)2 minus 4∆][(1 + σ)2 minus 4∆]

2[(a minus c)2 minus 4∆][(1 + σ)2 minus 4∆]12 lt 8∆ + 2(1 + σ)(a minus c)

(a minus c)2 minus 4∆ + (1 + σ)2 minus 4∆

+2[(a minus c)2 minus 4∆][(1 + σ)2 minus 4∆]12

lt (1 + σ + a minus c)2

and then|(a minus c)2 minus 4∆|12 + |(1 + σ)2 minus 4∆|12 lt 1 + σ + a minus c

thus we can derive (210)

2)K2 K3 at (x y) isin D have same signs K6 = 0 K5 = 0 by using similar methodswe can transform (21) into a complex equation in the form (28)

Now we discuss the case

3) K2 K3 are not of same signs K4 = 0 at the point (x y) isin D then we can solve(21) for uy vy and obtain the system of first order equations⎧⎨⎩ vy = aux + bvx + a0u+ b0v + f0

minusuy = dux + cvx + c0u+ d0v + g0(211)

where a b c d are known functions of akl bkl (k l = 1 2) and a0 b0 c0 d0 f0 g0 areknown functions of ak2 bk2 ak bk ck (k l = 1 2) and

a = K5K4 b = minusK3K4 c = K6K4 d = K2K4

∆ = I4K24 = (b+ d)24minus ac gt 0 (212)

Similarly to (25) multiply the second formula of (211) by j and then subtract thefirst formula of (211) we get

minusvy minus juy = wz minus wz = minus(a minus d j)ux minus (b minus c j)vx + lower order terms

= minus(a minus d j)(wz + wz + wz + wz)2

+(c minus b j)(wz minus wz + wz minus wz)2 + lower order terms

namely(1 + q1)wz + q2wz = (1minus q1)wz minus q2wz + lower order terms (213)

whereq1 =

[a minus c minus (d minus b)j]2

q2 =[a+ c minus (d+ b)j]

2

It is clear that

q0 = (1 + q1)(1 + q1)minus q2q2

=[(2 + a minus c)2 minus (d minus b)2 minus (a+ c)2 + (d+ b)2]

4

= (1 + a)(1minus c) + bd gt 0

thus we can solve (213) for wz ie

wz minus Q1(z)wz minus Q2(z)wz = A1(z)w + A2w + A3(z) (214)

in which

Q1(z) =[(1minus q1)(1 + q1) + |q2|2]

q0 Q2 =

minus2q2(z)q0

4) K2 K3 are not of same signs K4 = 0 K1 = 0 by using similar methods as in3) we can transform (21) into the complex equation of the form (214)

22 Complex forms of nonlinear hyperbolic systems of first orderequations

Next we consider the general nonlinear hyperbolic system of first order partial dif-ferential equations

Fk(x y u v ux uy vx vy) = 0 k = 1 2 (215)

where the real functions Fk(k = 1 2) are defined and continuous at every point (x y)in D and possess continuous partial derivatives in ux uy vx vy For system (215)its condition of hyperbolic type can be defined by the inequality (22) or (23) butin which

K1 =D(F1 F2)D(ux vx)

K2 =D(F1 F2)D(ux vy)

K3 =D(F1 F2)D(uy vx)

K4 =D(F1 F2)D(uy vy)

K5 =D(F1 F2)D(ux uy)

K6 =D(F1 F2)D(vx vy)

(216)

where Fkux Fkuy Fkvx Fkvy(k = 1 2) can be found as follows

Fkux =int 1

0Fktux(x y u v tux tuy tvx tvy)dt

Fkuy =int 1

0Fktuy(x y u v tux tuy tvx tvy)dt

Fkvx =int 1

0Fktvx(x y u v tux tuy tvx tvy)dt

Fkvy =int 1

0Fktvy(x y u v tux tuy tvx tvy)dt

(217)

By using the method in Subsection 21 for cases 1) K2 K3 are of same signs K5 orK6 = 0 2) K2 K3 are not of same signs K1 or K4 = 0 then system (215) can bereduced to the complex form

wz minus Q1wz minus Q2wz = A1w + A2w + A3 (218)

where z = x+ jy w = u+ jv and

Qk = Qk(z w wz wz) k = 1 2 Ak = Ak(z w wz wz) k = 1 2 3

In particular if (29) holds from the condition of hyperbolic type in (22) it followsthat (210) holds

Theorem 21 Let system (215) satisfy the condition of hyperbolic type as in (22)and the conditions from the existence theorem for implicit functions Then (215) issolvable with respect to wz and the corresponding hyperbolic complex equation of firstorder (218) can be obtained

As for the cases 3) K1 = K4 = 0 K2 K3 are not of same signs K5 = 0 or K6 = 0and 4) K5 = K6 = 0 K2 K3 are of same signs K1 = 0 or K4 = 0 we can transformthe quasilinear case of hyperbolic system (215) into the complex forms by using asimilar method in the next subsection

23 Complex forms of quasilinear hyperbolic systems of first orderequations

Finally we discuss the quasilinear hyperbolic system of first order partial differentialequations ⎧⎨⎩a11ux + a12uy + b11vx + b12vy = a1u+ b1v + c1

where the coefficients akl bkl(k l = 1 2) are known functions in (x y) isin D andak bk ck(k = 1 2) are known functions of (x y) isin D and u v isin IR The hyperbolicitycondition of (219) is the same as for system (21) ie for any point (x y) isin D theinequality

I = (K2 +K3)2 minus 4K1K4 = (K2 minus K3)2 minus 4K5K6 gt 0 (220)

holds in which

K1 =∣∣∣∣∣a11 b11

a21 b21

∣∣∣∣∣ K2 =∣∣∣∣∣a11 b12

a21 b22

∣∣∣∣∣ K3 =∣∣∣∣∣a12 b11

a22 b21

∣∣∣∣∣ K4 =

∣∣∣∣∣a12 b12

a22 b22

∣∣∣∣∣ K5 =∣∣∣∣∣a11 a12

a21 a22

∣∣∣∣∣ K6 =∣∣∣∣∣ b11 b12

b21 b22

∣∣∣∣∣ We first consider the case 1) K1 = K4 = 0 K2 K3 are not of same signs K6 = 0

at the point (x y) isin D From K1 = K4 = 0 there exist real constants λ micro such that

a11 = λb11 a21 = λb21 a12 = microb12 a22 = microb22

thusK2 = λK6 K3 = minusmicroK6 K5 = λmicroK6

and thenI = (K2 minus K3)2 minus 4K5K6 = [(λ+ micro)2 minus 4λmicro]K2

6

= (K2 +K3)2 minus 4K1K4 = (λ minus micro)2K26 gt 0

It is easy to see that λ = micro ie K2 = minusK3 in this case system (219) becomes theform ⎧⎨⎩ b11(λu+ v)x + b12(microu+ v)y = a1u+ b1v + c1

b21(λu+ v)x + b22(microu+ v)y = a2u+ b2v + c2(221)

Setting U = λu+ v V = microu+ v and noting∣∣∣∣∣∣Uu Uv

Vu Vv

∣∣∣∣∣∣ =∣∣∣∣∣∣λ 1

micro 1

∣∣∣∣∣∣ = λ minus micro = 0

andu =

U minus V

λ minus micro v =

microU minus λV

micro minus λ

system (221) can be written in the form⎧⎨⎩ b11Ux + b12Vy = aprime1U + bprime

1V + c1

b21Ux + b22Vy = aprime2U + bprime

2V + c2(222)

where

aprime1 =

a1 minus microb1

λ minus micro bprime

1 =minusa1 + λb1

λ minus micro aprime

2 =a2 minus microb2

2 =minusa2 + λb2

λ minus micro

thus ⎧⎨⎩Ux = [(aprime1b22 minus aprime

2b12)U + (bprime1b22 minus bprime

2b12)V + (c1b22 minus c2b12)]K6

Vy = [(aprime2b11 minus aprime

1b21)V + (c2b11 minus c1b21)]K6(223)

Subtracting the first equation from the second equation the complex equation ofW = U + jV

Wz +W z = A1(z W )W + A2(z W )W + A3(z W ) (224)

can be derived where A1 A2 A3 are known functions of bkl ak bk ck(k l = 1 2)

Moreover we consider system (219) with the condition 2) K5 = K6 = 0 K2 K3

are of same signs and K4 = 0 at the point (x y) isin D In this case due to K5 =K6 = 0 at the point (x y) isin D there exist real constants λ micro such that

a11 = λa12 a21 = λa22 b11 = microb12 b21 = microb22

thusK1 = λmicroK4 K2 = λK4 K3 = microK4

and then

I = (K2 +K3)2 minus 4K1K4 = [(λ+ micro)2 minus 4λmicro]K24 = (λ minus micro)2K2

4 gt 0

It is clear that if λ = micro ie K2 = K3 then system (219) can become the form⎧⎨⎩a12(λux + uy) + b12(microvx + vy) = a1u+ b1v + c1

a22(λux + uy) + b22(microvx + vy) = a2u+ b2v + c2(225)

Letξ =

x minus microy

λ minus micro η =

minusx+ λy

λ minus micro

it is easy to see that ∣∣∣∣∣∣ξx ξy

ηx ηy

∣∣∣∣∣∣ = 1(λ minus micro)2

∣∣∣∣∣∣1 minusmicro

minus1 λ

∣∣∣∣∣∣ = 1λ minus micro

= 0

and x = λξ + microη y = ξ + η Thus system (225) can be rewritten in the form⎧⎨⎩a12uξ + b12vη = a1u+ b1v + c1

a22uξ + b22vη = a2u+ b2v + c2(226)

This system can be solved for uξ vη namely⎧⎨⎩uξ = aprime1u+ bprime

1v + cprime1

vη = aprime2u+ bprime

2v + cprime2

(227)

whereaprime

1 = (a1b22 minus a2b12)K4 aprime2 = (a2a12 minus a1a22)K4

bprime1 = (b1b22 minus b2b12)K4 bprime

2 = (b2a12 minus b1a22)K4

cprime1 = (c1b22 minus c2b12)K4 cprime

2 = (c2a12 minus c1a22)K4

Denoting ζ = ξ + jη then system (227) can be written in the complex form

wζ + wζ = Aprime1(ζ w)w + Aprime

2(ζ w)w + Aprime3(ζ w) (228)

in which Aprime1 A

prime2 A

prime3 are known functions of ak2 bk2 ak bk ck(k = 1 2)

For 3) K1 = K4 = 0 K2 K3 are not of same signs and K5 = 0 and 4) K5 = K6 =0 K2 K3 are of same signs and K1 = 0 by using a similar method we can transform(219) into the complex equations (224) and (228) respectively We mention that itis possible that the case

b11 = λa11 b21 = λa21 b12 = microa12 b22 = microa22

occurs for 1) and 2) and

occurs for 3) and 4) then we can similarly discuss In addition if λ(x y) micro(x y) areknown functions of (x y) in D then in the left-hand sides of the two equations in(221) should be added b11λxu+ b12microyu and b21λxu+ b22microyu It is sufficient to modifythe coefficient of u and the system is still hyperbolic For 2)ndash4) we can similarlyhandle

The complex equations as stated in (224) and (228) can be written in the form

wz + wz = A(z w)w +B(z w)w + C(z w) (229)

which is equivalent to the system of first order equations

ux = au+ bv + f vy = cu+ dv + g (230)

where z = x + jy w = u + jv A = (a + jb minus c minus jd)2 B = (a minus jb minus c + jd)2C = f minus g Let

W (z) = ve1 + ue2 Z = xe1 + ye2 (231)where e1 = (1+j)2 e2 = (1minusj)2 From (230) we can obtain the complex equation

WZ = vye1 + uxe2 = A1W + A2W + A3 = F (ZW ) (232)

in which A1(ZW ) = de1 + ae2 A2(ZW ) = ce1 + be2 A3(ZW ) = ge1 + fe2 [92]

3 Boundary Value Problems of Linear Hyperbolic ComplexEquations of First Order

In this section we mainly discuss the RiemannndashHilbert boundary value problem forlinear hyperbolic complex equations of first order in a simply connected domain Wefirst give a representation of solutions for the above boundary value problem andthen prove the uniqueness and existence of solutions for the above problem by usingthe successive iteration

31 Formulation of the RiemannndashHilbert problem and uniqueness of itssolutions for simplest hyperbolic complex equations

Let D be a simply connected bounded domain in the x+ jy-plane with the boundaryΓ = L1 cup L2 cup L3 cup L4 where L1 = x = minusy 0 le x le R1 L2 = x = y + 2R1 R1 lex le R2 L3 = x = minusy minus 2R1 + 2R2 R = R2 minus R1 le x le R2 L4 = x = y 0 lex le R2 minus R1 and denote z0 = 0 z1 = (1 minus j)R1 z2 = R2 + j(R2 minus 2R1) z3 =(1 + j)(R2 minus R1) = (1 + j)R and L = L1 cup L4 where j is the hyperbolic unit Forconvenience we only discuss the case R2 ge 2R1 the other case can be discussed by asimilar method We consider the simplest hyperbolic complex equation of first order

wz = 0 in D (31)

The RiemannndashHilbert boundary value problem for the complex equation (31) maybe formulated as follows

Problem A Find a continuous solution w(z) of (31) in D satisfying the boundaryconditions

Re [λ(z)w(z)] = r(z) z isin L Im [λ(z0)w(z0)] = b1 (32)

where λ(z) = a(z) + jb(z) = 0 z isin L and λ(z) r(z) b1 satisfy the conditions

Cα[λ(z) L] = Cα[Reλ L] + Cα[Imλ L] le k0 Cα[r(z) L] le k2 (33)

|b1| le k2 maxzisinL1

1|a(z)minus b(z)| max

zisinL4

1|a(z) + b(z)| le k0 (34)

in which b1 is a real constant and α (0 lt α lt 1) k0 k2 are non-negative constants Inparticular when a(z) = 1 b(z) = 0 ie λ(z) = 1 z isin L Problem A is the Dirichletproblem (Problem D) whose boundary condition is

Re [w(z)] = r(z) z isin L Im [w(z0)] = b1 (35)

Problem A with the conditions r(z) = 0 z isin L b1 = 0 is called Problem A0

On the basis of Theorem 11 it is clear that the complex equation (31) can bereduced to the form

ξν = 0 ηmicro = 0 (micro ν) isin Q = 0 le micro le 2R 0 le ν le 2R1 (36)

3 Linear Hyperbolic Equations 19

where micro = x + y ν = x minus y ξ = u + v η = u minus v Hence the general solution ofsystem (36) can be expressed as

ξ = u+ v = f(micro) = f(x+ y) η = u minus v = g(ν) = g(x minus y) ie

u = [f(x+ y) + g(x minus y)]2 v = [f(x+ y)minus g(x minus y)]2(37)

in which f(t) g(t) are two arbitrary real continuous functions on [0 2R] [0 2R1]respectively From the boundary condition (32) we have

a(z)u(z) + b(z)v(z)=r(z) on L λ(z0)w(z0)=r(z0) + jb1 ie

[a((1minus j)x) + b((1minus j)x)]f(0) + [a((1minus j)x)minus b((1minus j)x)]

times g(2x) = 2r((1minus j)x) on [0 R1]

[a((1 + j)x) + b((1 + j)x)]f(2x) + [a((1 + j)x)minus b((1 + j)x)]

times g(0) = 2r((1 + j)x) on [0 R]

f(0)=u(0)+v(0)=r(0)+b1

a(0)+b(0) g(0)=u(0)minusv(0)=

r(0)minusb1

a(0)minusb(0)

(38)

The above formulas can be rewritten as

[a((1minus j)t2)+b((1minus j)t2)]f(0) + [a((1minus j)t2)minus b((1minus j)t2)]times g(t) = 2r((1minus j)t2) t isin [0 2R1]

[a((1 + j)t2) + b((1 + j)t2)]f(t) +[a((1 + j)t2)minus b((1 + j)t2)]times g(0) = 2r((1 + j)t2) t isin [0 2R] ie

f(x+ y) =2r((1 + j)(x+ y)2)

a((1 + j)(x+ y)2)+b((1 + j)(x+ y)2)(39)

minus [a((1+j)(x+y)2)minus b((1+j)(x+y)2)]g(0)a((1+j)(x+y)2) + b((1+j)(x+y)2)

0lex+yle2R

g(x minus y) =2r((1minus j)(x minus y)2)

a((1minus j)(x minus y)2)minus b((1minus j)(x minus y)2)

minus [a((1minusj)(xminusy)2)+b((1minusj)(xminusy)2)]f(0)a((1minusj)(xminusy)2)minusb((1minusj)(xminusy)2)

0lexminusyle2R1

Thus the solution w(z) of (31) can be expressed as

w(z) = f(x+ y)e1 + g(x minus y)e2

=12f(x+ y) + g(x minus y) + j[f(x+ y)minus g(x minus y)]

(310)

where f(x + y) g(x minus y) are as stated in (39) and f(0) g(0) are as stated in (38)It is not difficult to see that w(z) satisfies the estimate

Cα[w(z) D]leM1=M1(α k0 k2 D) Cα[w(z) D]leM2k2=M2(α k0 D)k2 (311)

where M1 M2 are two non-negative constants only dependent on α k0 k2 D andα k0 D respectively The above results can be written as a theorem

Theorem 31 Any solution w(z) of Problem A for the complex equation (31) pos-sesses the representation (310) which satisfies the estimate (311)

32 Uniqueness of solutions of the RiemannndashHilbert problem for linearhyperbolic complex equations

Now we discuss the linear case of the complex equation (232) namely

wz = A1(z)w + A2(z)w + A3(z) (312)

and suppose that the complex equation (312) satisfies the following conditions

Condition C Al(z) (l = 1 2 3) are continuous in D and satisfy

C[Al D] = C[ReAl D] + C[ImAl D] le k0 l = 1 2 C[A3 D] le k1 (313)

Due to w = u + jv = ξe1 + ηe2 wz = ξmicroe1 + ηνe2 wz = ξνe1 + ηmicroe2 from theformulas in Section 1 equation (312) can be rewritten in the form

ξνe1 + ηmicroe2 = [A(z)ξ +B(z)η + E(z)]e1

+[C(z)ξ +D(z)η + F (z)]e2 z isin D ie⎧⎨⎩ ξν = A(z)ξ +B(z)η + E(z)

ηmicro = C(z)ξ +D(z)η + F (z)z isin D

(314)

in which

A = ReA1 + ImA1 B = ReA2 + ImA2 C = ReA2 minus ImA2

D = ReA1 minus ImA1 E = ReA3 + ImA3 F = ReA3 minus ImA3

The boundary condition (32) can be reduced to

Re [λ(ξe1 + ηe2)] = r(z) Im [λ(ξe1 + ηe2)]|z=z0 = b1 (315)

where λ = (a + b)e1 + (a minus b)e2 Moreover the domain D is transformed into Q =0 le micro le 2R 0 le ν le 2R1 R = R2 minus R1 which is a rectangle and A B C D E F

are known functions of (micro ν) isin Q There is no harm in assuming that w(z0) = 0otherwise through the transformation

W (z) = w(z)minus [a(z0)minus jb(z0)][r(z0) + jb1]

[a2(z0)minus b2(z0)] (316)

the requirement can be realized For convenience sometimes we write z isin D or z isin Qand denote L1 = micro = 0 0 le ν le 2R1 L4 = 0 le micro le 2R ν = 0

Now we give a representation of solutions of Problem A for equation (312)

Theorem 32 If equation (312) satisfies Condition C then any solution w(z) ofProblem A for (312) can be expressed as

w(z) = w0(z) + Φ(z) + Ψ(z) in D

w0(z) = f(x+ y)e1 + g(x minus y)e2 Φ(z) = f(x+ y)e1 + g(x minus y)e2

Ψ(z) =int xminusy

0[Aξ +Bη + E]d(x minus y)e1 +

int x+y

0[Cξ +Dη + F ]d(x+ y)e2

(317)

where f(x+ y) g(x minus y) are as stated in (39) and f(x+ y) g(x minus y) are similar tof(x+ y) g(x minus y) in (39) but Φ(z) satisfies the boundary condition

Re [λ(z)Φ(z)]=minusRe [λ(z)Ψ(z)] z isin L Im [λ(z0)Φ(z0)]=minusIm [λ(z0)Ψ(z0)](318)

Proof It is not difficult to see that the functions w0(z) Φ(z) are solutions of thecomplex equation (31) in D which satisfy the boundary conditions (32) and (318)respectively and Ψ(z) satisfies the complex equation

[Ψ]z = [Aξ +Bη + E]e1 + [Cξ +Dη + F ]e2 (319)

and Φ(z) + Ψ(z) satisfies the boundary condition of Problem A0 Hence w(z) =w0(z)+Φ(z)+Ψ(z) satisfies the boundary condition (32) and is a solution of ProblemA for (312)

Theorem 33 Suppose that Condition C holds Then Problem A for the complexequation (312) has at most one solution

Proof Let w1(z) w2(z) be any two solutions of Problem A for (312) and substitutethem into equation (312) and the boundary condition (32) It is clear that w(z) =w1(z)minus w2(z) satisfies the homogeneous complex equation and boundary conditions

wz = A1w + A2w in D (320)

Re [λ(z)w(z)] = 0 if (x y) isin L w(z0) = 0 (321)

On the basis of Theorem 32 the function w(z) can be expressed in the form

w(z) = Φ(z) + Ψ(z)

Ψ(z) =int xminusy

0[Aξ +Bη]e1d(x minus y) +

int x+y

0[Cξ +Dη]e2d(x+ y)

(322)

Suppose w(z) equiv 0 in the neighborhood (sub D) of the point z0 = 0 We maychoose a sufficiently small positive number R0 lt 1 such that 8M3MR0 lt 1 whereM3 = maxC[A Q0] C[B Q0] C[CQ0] C[DQ0] M = 1+4k2

0(1+k20) is a positive

constant and m = C[w(z) Q0] gt 0 herein Q0 = 0 le micro le R0 cap 0 le ν le R0From (39)(310)(317)(318)(322) and Condition C we have

Ψ(z) le 8M3mR0 Φ(z) le 32M3k20(1 + k2

0)mR0

thus an absurd inequalitym le 8M3MmR0 lt m is derived It shows w(z) = 0 (x y) isinQ0 Moreover we extend along the positive directions of micro = x + y and ν = x minus ysuccessively and finally obtain w(z) = 0 for (x y) isin D ie w1(z)minus w2(z) = 0 in DThis proves the uniqueness of solutions of Problem A for (312)

33 Solvability of Problem A for linear hyperbolic complex equations offirst order

Theorem 34 If the complex equation (312) satisfies Condition C then ProblemA for (312) has a solution

Proof In order to find a solution w(z) of Problem A in D we can express w(z)in the form (317) In the following by using successive iteration we can find asolution of Problem A for the complex equation (312) First of all substitutingw0(z) = ξ0e1 + η0e2 of Problem A for (31) into the position of w = ξe1 + ηe2 in theright-hand side of (312) the function

w1(z) = w0(z) + Φ1(z) + Ψ1(z)

Ψ1(z) =int ν

0[Aξ0 +Bη0 + E]e1dν +

int micro

0[Cξ0 +Dη0 + F ]e2dmicro

(323)

can be determined where micro = x + y ν = x minus y Φ1(z) is a solution of (31) in Dsatisfying the boundary conditions

Re [λ(z)Φ1(z)] = minusRe [λ(z)Ψ1(z)] z isin L

Im [λ(z0)Φ1(z0)] = minusIm [λ(z0)Ψ1(z0)](324)

Thus from (323) (324) we have

w1(z)minus w0(z) = C[w1(z)minus w0(z) D] le 2M4M(4m+ 1)Rprime (325)

where M4 = maxzisinD(|A| |B| |C| |D| |E| |F |) m = w0 C(D) Rprime = max(2R1 2R)M = 1+4k2

0(1+ k20) is a positive constant as in the proof of Theorem 33 Moreover

we substitute w1(z) = w0(z)+Φ1(z)+Ψ1(z) and the corresponding functions ξ1(z) =Rew1(z)+Imw1(z) η1(z) = Rew1(z)minusImw1(z) into the positions of w(z) ξ(z) η(z)in (317) and similarly to (323)ndash(325) we can find the corresponding functionsΨ2(z)Φ2(z) in D and the function

w2(z) = w0(z) + Φ2(z) + Ψ2(z) in D

It is clear that the function w2(z)minus w1(z) satisfies the equality

w2(z)minus w1(z) = Φ2(z)minus Φ1(z) + Ψ2(z)minusΨ1(z)

= Φ2(z)minus Φ1(z) +int ν

0[A(ξ1 minus ξ0) +B(η1 minus η0)]e1dν

+int micro

0[C(ξ1 minus ξ0) +D(η1 minus η0)]e2dmicro

and

w2 minus w1 le [2M3M(4m+ 1)]2int Rprime

0RprimedRprime le [2M3M(4m+ 1)Rprime]2

2

where M3 is a constant as stated in the proof of Theorem 33 Thus we can find asequence of functions wn(z) satisfying

wn(z) = w0(z) + Φn(z) + Ψn(z)

Ψn(z) = +int ν

0[Aξn +Bηn + E]e1dν +

int micro

0[Cξn +Dηn + F ]e2dmicro

(326)

and wn(z)minus wnminus1(z) satisfies

wn(z)minuswnminus1(z)=Φn(z)minusΦnminus1(z) + Ψn(z)minusΨnminus1(z)

=Φn(z)minusΦnminus1(z) +int ν

0[A(ξnminus1minusξnminus2)

+B(ηnminus1minusηnminus2)]e1dν

+int micro

0[C(ξnminus1minusξnminus2)+D(ηnminus1minusηnminus2)]e2dmicro (327)

and then

wn minus wnminus1 le [2M3M(4m+ 1)]nint Rprime

0

Rprimenminus1

(n minus 1) dRprime le [2M3M(4m+ 1)Rprime]n

n (328)

From the above inequality it is seen that the sequence of functions wn(z) ie

wn(z) = w0(z) + [w1(z)minus w0(z)] + middot middot middot+ [wn(z)minus wnminus1(z)](n = 1 2 ) (329)

uniformly converges to a function wlowast(z) and wlowast(z) satisfies the equality

wlowast(z) = w0(z) + Φlowast(z) + Ψlowast(z)

Ψlowast(z) =int ν

0[Aξlowast +Bηlowast + E]e1dν +

int micro

0[Cξlowast +Dηlowast + F ]e2dmicro

(330)

It is easy to see that wlowast(z) satisfies equation (312) and the boundary condition (32)hence it is just a solution of Problem A for the complex equation (312) in the closeddomain D ([87]2)

34 Another boundary value problem for linear hyperbolic complexequations of first order

Now we introduce another boundary value problem for equation (312) in D with theboundary conditions

Re [λ(z)w(z)] = r(z) on L1 cup L5 Im [λ(z1)w(z1)] = b1 (331)

where L1=y=minusx0lexleRL5=y=(R+R1)[x(RminusR1)minus2R1R(R2minusR21)] R1

lexleR=R2minusR1R2 ge2R1 λ(z)=a(z)+jb(z)zisinL1λ(z)=a(z) +jb(z)=1+jzisinL5 and λ(z) r(z) b1 satisfy the conditions

Cα[λ(z) L1] le k0 Cα[r(z) L1cupL5] le k2 |b1| le k2maxzisinL1

1|a(z)minusb(z)| le k0 (332)

in which b1 is a real constant and α (0 lt α lt 1) k0 k2 are non-negative constantsThe boundary value problem is called Problem A1

On the basis of Theorem 11 it is clear that the complex equation (31) can bereduced to the form (36) in D The general solution of system (36) can be expressedas

w(z) = u(z) + jv(z)

= [u(z) + v(z)]e1 + [u(z)minus v(z)]e2

=f(x+y)e1+g(xminus y)e2

=12f(x+y)+g(xminusy)+j[f(x+y)minusg(xminusy)]

(333)

where f(t) (0 le t le 2R) g(t) (0 le t le 2R1) are two arbitrary real continuousfunctions Noting that the boundary condition (331) namely

a(z)u(z) + b(z)v(z)=r(z) on L1 cup L5 λ(z1)w(z1) =r(z1) + jb1 ie

[a((1minus j)x) + b((1minus j)x)]f(0) + [a((1minus j)x)minus b((1minus j)x)]g(2x)

= 2r((1minus j)x) on [0 R1] f(z1)=u(z1)+v(z1)=r(z1)+b1

a(z1)+b(z1)

Re [λ(z)w(z)]=u(z)+v(z)=r[(1+

R+R1

RminusR1j)

xminusj2RR1

RminusR1

]on [R1 R]

(334)

It is easy to see that the above formulas can be rewritten as

[a((1minusj)t2)+b((1minusj)t2)]f(0)+[a((1minusj)t2)minusb((1minusj)t2)]

timesg(t)=2r((1minusj)t2) tisin [0 2R1] f(x+y)=f[( 2R

RminusR1

)xminus 2RR1

RminusR1

]

f(t) = r[((1 + j)R minus (1minus j)R1)

t

2R+ (1minus j)R1

] t isin [0 2R]

4 Quasilinear Hyperbolic Equations 25

and then

0lexminusyle2R1

f(x+y) = r[((1+j)Rminus(1minusj)R1)x+y

2R+(1minusj)R1] 0lex+yle2R

(335)

Substitute the above function f(x+y) g(xminusy) into (333) the solution w(z) of (36)is obtained We are not difficult to see that w(z) satisfies the estimate

Cα[w(z) D]le M1 Cα[w(z) D] le M2k2 (336)

where M1 = M1(α k0 k2 D) M2 = M2(α k0 D) are two non-negative constants

Next we consider Problem A1 for equation (312) Similarly to before we canderive the representation of solutions w(z) of Problem A1 for (312) as stated in(317) where f(x + y) g(x minus y) possess the form (335) and L = L1 cup L2 z0 in theformula (318) should be replaced by L1 cup L5 z1 Moreover applying the successiveiteration the uniqueness and existence of solutions of Problem A1 for equation (312)can be proved but L z0 in the formulas (321) and (324) are replaced by L1 cupL5 z1We write the results as a theorem

Theorem 35 Suppose that equation (312) satisfies Condition C Then ProblemA1 for (312) has a unique solution w(z) which can be expressed in the form (317)where f(x+ y) g(x minus y) possess the form (335)

4 Boundary Value Problems of Quasilinear HyperbolicComplex Equations of First Order

In this section we mainly discuss the RiemannndashHilbert boundary value problem forquasilinear hyperbolic complex equations of first order in a simply connected domainWe first prove the uniqueness of solutions for the above boundary value problemand then give a priori estimates of solutions of the problem moreover by usingthe successive iteration the existence of solutions for the above problem is provedFinally we also discuss the solvability of the above boundary value problem in generaldomains

41 Uniqueness of solutions of the RiemannndashHilbert problem forquasilinear hyperbolic complex equations

In the subsection we first discuss the quasilinear hyperbolic complex equation

wz = F (z w) F = A1(z w)w + A2(z w)w + A3(z w) in D (41)

whereD is a simply connected bounded domain in the x+jy-plane with the boundaryΓ = L1 cup L2 cup L3 cup L4 as stated in Section 3

Suppose that the complex equation (41) satisfies the following conditions

Condition C

1) Al(z w) (l = 1 2 3) are continuous in z isin D for any continuous complexfunction w(z) and satisfy

2) For any continuous complex functions w1(z) w2(z) in D the equality

F (z w1)minusF (z w2)= A1(z w1 w2)(w1minusw2)+A2(z w1 w2)(w1minusw2) in D (43)

holds where C[Al D] le k0 l = 1 2 and k0 k1 are non-negative constants In particu-lar when (41) is a linear equation the condition (43) obviously holds

In order to give an a priori Cα(D)-estimate of solutions for Problem A we needthe following conditions For any hyperbolic numbers z1 z2(isin D) w1 w2 the abovefunctions satisfy

Al(z1 w1)minus Al(z2 w2) le k0[ z1 minus z2 α + w1 minus w2 ] l = 1 2

A3(z1 w1)minus A3(z2 w2) le k2[ z1 minus z2 α + w1 minus w2 ](44)

in which α(0 lt α lt 1) k0 k2 are non-negative constants

Similarly to (312) and (314) due to w = u + jv = ξe1 + ηe2 = ζ wz = ξmicroe1 +ηνe2 wz = ξνe1 + ηmicroe2 the quasilinear hyperbolic complex equation (41) can berewritten in the form

ξνe1 + ηmicroe2 = [A(z ζ)ξ +B(z ζ)η + E(z ζ)]e1

+[C(z ζ)ξ +D(z ζ)η + F (z ζ)]e2 z isin D ie⎧⎨⎩ ξν = A(z ζ)ξ +B(z ζ)η + E(z ζ)

ηmicro = C(z ζ)ξ +D(z ζ)η + F (z ζ)z isin D

(45)

in which

Obviously any solution of Problem A for equation (41) possesses the same rep-resentation (317) as stated in Theorem 32 In the following we prove the existenceand uniqueness of solutions for Problem A for (41) with Condition C

Theorem 41 If Condition C holds then Problem A for the quasilinear complexequation (41) has at most one solution

Proof Let w1(z) w2(z) be any two solutions of Problem A for (41) and substitutethem into equation (41) and boundary condition (32) By Condition C we see thatw(z) = w1(z) minus w2(z) satisfies the homogeneous complex equation and boundaryconditions

wz = A1w + A2w in D (46)

Re [λ(z)w(z)] = 0 z isin L Im [λ(z0)w(z0)] = 0 (47)

w(z) = Φ(z) + Ψ(z)

Ψ(z) =int xminusy

0[Aξ + Bη]e1d(x minus y) +

int x+y

0[Cξ + Dη]e2d(x+ y)

(48)

where the relation between the coefficients A B C D and A1 A2 is the same with thatbetween A B C D and A1 A2 in (45) Suppose w(z) equiv 0 in the neighborhood Q0(subD) of the point z0 = 0 we may choose a sufficiently small positive number R0 lt 1such that 8M5MR0 lt 1 where M5 = maxC[A Q0] C[B Q0] C[C Q0] C[DQ0]Similarly to the proof of Theorem 33 we can derive a contradiction Hence w1(z) =w2(z) in D

42 Solvability of Problem A for quasilinear hyperbolic complexequations

Theorem 42 If the quasilinear complex equation (41) satisfies Condition C thenProblem A for (41) has a solution

Proof Similarly to the proof of Theorem 34 we use the successive iteration tofind a solution of Problem A for the complex equation (41) Firstly substitutingw0(z) = ξ0e1 + η0e2 of Problem A for (31) into the position of w = ξe1 + ηe2 in theright-hand side of (41) the function

w1(z) = w0(z) + Φ1(z) + Ψ1(z)

Ψ1(z) =int ν

0[Aξ0 +Bη0 + E]e1dν +

int micro

(49)

Moreover we can find a sequence of functions wn(z) satisfyingwn(z) = w0(z) + Φn(z) + Ψn(z)

Ψn(z) =int ν

0[Aξnminus1 +Bηnminus1 + E]e1dν +

int micro

0[Cξnminus1 +Dηnminus1 + F ]e2dmicro

(411)

wn(z)minuswnminus1(z)=Φn(z)minusΦnminus1(z)+Ψn(z)minusΨnminus1(z)=Φn(z)minus Φnminus1(z)

+int ν

0[A(ξnminus1minusξnminus2)+B(ηnminus1minusηnminus2)]e1dν (412)

+int micro

0[C(ξnminus1minusξnminus2)+D(ηnminus1minusηnminus2)]e2dmicro

Denoting M5 = maxD(|A| |B| |C| |D|) we can obtain

0

Rprimenminus1

n (413)

in which m = w0 C(Q) Rprime = max(2R1 2R) M = 1 + 4k20(1 + k2

0) is a positiveconstant as in the proof of Theorem 33 The remained proof is identical with theproof of Theorem 34

43 A priori estimates of solutions of the RiemannndashHilbert problem forhyperbolic complex equations

We first give the boundedness estimate of solutions for Problem A

Theorem 43 If Condition C holds then any solution u(z) of Problem A for thehyperbolic equation (41) satisfies the estimates

C[w(z) D] le M6 C[w(z) D] le M7k (414)

in which M6 = M6(α k0 k1 k2 D) k = k1 + k2 M7 = M7(α k0 D) are non-negativeconstants

Proof On the basis of Theorems 41 and 42 we see that under Condition CProblem A for equation (41) has a unique solution w(z) which can be found by usingsuccessive iteration Due to the functions wn+1(z) minus wn(z) (n = 1 2 ) in D arecontinuous the limit function w(z) of the sequence wn(z) in D is also continuousand satisfies the estimate

C[w(z) D] leinfinsum

n=0

[2M5M(4m+ 1)Rprime]n

n= e2M5M(4m+1)Rprime

= M6 (415)

where Rprime = max(2R1 2R) This is the first estimate in (414) As for the second esti-mate in (414) if k = k1+k2 = 0 then it is true from Theorem 41 If k = k1+ k2 gt 0

let the solution w(z) of Problem A be substituted into (41) and (32) and dividingthem by k we obtain the equation and boundary conditions for w(z) = w(z)k

wz = A1w + A2w + A3k z isin D

Re [λ(z)w(z)] = r(z)k z isin D Im [λ(z0)w(z0)] = b1k(416)

Noting that A3k rk b1k satisfy the conditions

C[A3k D] le 1 C[rk L] le 1 |b1k| le 1

by using the method of deriving C[w(z) D] in (415) we can obtain the estimate

C[w(z) D] le M7 = M7(α k0 D)

From the above estimate the second estimate in (414) is immediately derived

Here we mention that in the proof of the estimate (414) we have not requiredthat the coefficients λ(z) r(z) of (32) satisfy a Holder (continuous) condition andonly require that they are continuous on L

Next we shall give the Cα(D)-estimates of solutions of Problem A for (41) andfirst discuss the linear hyperbolic complex equation (312) or (314)

Theorem 44 Suppose that the linear complex equation (312) satisfies the condi-tions (313) and (44) ie the coefficents of (312) satisfies the conditions

Cα[Al D] le k0 l = 1 2 Cα[A3 D] le k2 (417)

in which α(0 lt α lt 1) k0 k2 are non-negative constants Then the solution w(z) =w0(z) + Φ(z) + Ψ(z) satisfies the following estimates

Cα[w0(z) D] le M8 Cα[Ψ(z) D] le M8

Cα[Φ(z) D] le M8 Cα[w(z) D] le M8(418)

where w0(z) is a solution of (31) as stated in (310) M8 = M8(α k0 k1 k2 D) is anon-negative constant

Proof As stated before w0(z) is the function as in (310) which satisfies the esti-mate (311) namely the first estimate in (418) In order to prove that Ψ(z) =Ψ1(z) = Ψ1

1(z)e1 +Ψ21(z)e2 satisfies the second estimate in (418) from

Ψ11(z) =

int xminusy

0G1(z)d(x minus y) Ψ2

1(z) =int x+y

0G2(z)d(x+ y)

G1(z) = A(z)ξ +B(z)η + E(z) G2(z) = C(z)ξ +D(z)η + F (z)(419)

and (417) we see that Ψ11(z) = Ψ1

1(micro ν) Ψ21(z) = Ψ2

1(micro ν) in D with respect toν = x minus y micro = x+ y satisfy the estimates

Cα[Ψ11(middot ν) D] le M9R

prime Cα[Ψ21(micro middot) D] le M9R

prime (420)

respectively where Rprime = max(2R 2R1) M9 = M9(α k0 k1 k2 D) is a non-negativeconstant If we substitute the solution w0 = w0(z) = ξ0e1 + η0e2 of Problem A of(31) into the position of w = ξe1 + ηe2 in (419) and ξ0 = Rew0 + Imw0 η0 =Rew0 minus Imw0 from (417) and (311) we obtain

Cα[G1(micro middot) D] le M10 Cα[G2(middot ν) D] le M10

Cα[Ψ11(micro middot) D] le M10R

prime Cα[Ψ21(middot ν) D] le M10R

prime(421)

in which M10 = M10(α k0 k1 k2 D) is a non-negative constant Due to Φ(z) =Φ1(z) satisfies the complex equation (31) and boundary condition (318) and Φ1(z)possesses a representation similar to that in (317) the estimate

Cα[Φ1(z) D] le M11Rprime = RprimeM11(α k0 k1 k2 D) (422)

can be derived Thus setting w1(z) = w0(z)+Φ1(z)+Ψ1(z) w1(z) = w1(z)minusw0(z) itis clear that the functions w1

1(z) = Re w1(z)+ Im w1(z) w21(z) = Re w1(z) minusIm w1(z)

satisfy as functions of micro = x+ y ν = x minus y respectively the estimates

Cα[w11(middot ν) D] le M12R

prime Cα[w11(micro middot) D] le M12R

prime

Cα[w21(micro middot) D] le M12R

prime Cα[w21(middot ν) D] le M12R

prime(423)

where M12 = 2M13M(4m + 1) M = 1 + 4k20(1 + k0) m = Cα[w0 D] M13 =

maxD[|A| |B| |C| |D| |E| |F |] By using successive iteration we obtain thesequence of functions wn(z) (n = 1 2 ) and the corresponding functions w1

n =Re wn + Im wn w2

n = Re wn minus Im wn satisfying the estimates

Cα[w1n(middot ν) D] le

(M12Rprime)n

n Cα[w1

n(micro middot) D] le (M12Rprime)n

n

Cα[w2n(micro middot) D] le (M12R

prime)n

n Cα[w2

n(middot ν) D] le(M12R

prime)n

n

(424)

and denote by w(z) the limit function of wn(z) =sumn

m=0 wn(z) in D the correspondingfunctions w1 = Rew(z) + Imw(z) w2 = Rew(z)minus Imw(z) satisfy the estimates

Cα[w1(middot ν) D] le eM12Rprime Cα[w1(micro middot) D] le eM12Rprime

Cα[w2(micro middot) D] le eM12Rprime Cα[w2(middot ν) D] le eM12Rprime

Combining the first formula in (418) (420)ndash(424) and the above formulas the lastthree estimates in (418) are derived

Theorem 45 Let the quasilinear complex equation (41) satisfy Condition C and(44) Then the solution w(z) of Problem A for (41) satisfies the following estimates

Cα[w(z) D] le M14 Cα[w(z) D] le M15k (425)

where k = k1+k2 M14 = M14(α k0 k1 k2 D) M15 = M15(α k0 D) are non-negativeconstants

Proof According to the proof of Theorem 43 from the first formula in (425) thesecond formula in (425) is easily derived Hence we only prove the first formulain (425) Similarly to the proof of Theorem 44 we see that the function Ψ1(z) =Ψ1

1(micro ν)e1+Ψ21(micro ν)e2 still possesses the estimate (420) Noting that the coefficients

are the functions of z isin D and w and applying the condition (44) we can derivesimilar estimates as in (421) Hence we also obtain estimates similar to (422) and(423) and the constant M12 in (423) can be chosen as M12 = 2M13M(4m+1) m =Cα[w0(z) D] Thus the first estimate in (425) can be derived

Moreover according to the above methods we can obtain estimates for [Rew +Imw]ν [Rew minus Imw]micro analogous to those in (414) and (425)

44 The boundary value problem for quasilinear hyperbolic equations offirst order in general domains

In this subsection we shall generalize the domain D to general cases

1 The boundary L1 of the domain D is replaced by a curve Lprime1 and the boundary

of the domain Dprime is Lprime1 cup Lprime

2 cup L3 cup L4 where the parameter equations of the curvesLprime

1 Lprime2 are as follows

Lprime1 = γ1(x) + y = 0 0 le x le l Lprime

2 = x minus y = 2R1 l le x le R2 (426)

in which γ1(x) on 0 le x le l = 2R1 minus γ1(l) is continuous and γ1(0) = 0 γ1(x) gt 0on 0 lt x le l and γ1(x) is differentiable except at isolated points on 0 le x le l and1 + γprime

1(x) gt 0 By this condition the inverse function x = σ(ν) of x + γ1(x) = νcan be found and σprime(ν) = 1[1 + γprime

1(x)] hence the curve Lprime1 can be expressed by

x = σ(ν) = (micro+ ν)2 ie micro = 2σ(ν)minus ν 0 le ν le 2R1 We make a transformation

micro =2R[micro minus 2σ(ν) + ν]2R minus 2σ(ν) + ν

ν = ν 2σ(ν)minus ν le micro le 2R 0 le ν le 2R1 (427)

its inverse transformation is

micro =12R[2R minus 2σ(ν) + ν]micro+ 2σ(ν)minus ν ν = ν 0 le micro le 2R 0 le ν le 2R1 (428)

The transformation (427) can be expressed by⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

x =12(micro+ ν)

=2R(x+ y) + 2R(x minus y)minus (2R + x minus y)[2σ(x+ γ1(x))minus x minus γ1(x)]

4R minus 4σ(x+ γ1(x)) + 2x+ 2γ1(x)

y =12(micro minus ν)

=2R(x+ y)minus 2R(x minus y)minus (2R minus x+ y)[2σ(x+ γ1(x))minus x minus γ1(x)]

4R minus 4σ(x+ γ1(x)) + 2x+ 2γ1(x)

(429)

where γ1(x) = minusy and its inverse transformation is⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

x =12(micro+ ν) =

[2R minus 2σ(x+ γ1(x)) + x+ γ1(x)](x+ y)4R

+σ(x+ γ1(x))minus x+ γ1(x)minus x+ y

2

y =12(micro minus ν) =

+σ(x+ γ1(x))minus x+ γ1(x) + x minus y

2

(430)

Denote by z = x+ jy = f(z) z = x+ jy = fminus1(z) the transformation (429) and theinverse transformation (430) respectively In this case the system of equations andboundary conditions are

ξν = Aξ +Bη + E ηmicro = Cξ +Dη + F z isin Dprime (431)

Re [λ(z)w(z)] = r(z) z isin Lprime1 cup Lprime

2 Im [λ(z0)w(z0)] = b1 (432)

in which zprime0 = l minus jγ1(l) λ(z) r(z) b1 on Lprime

1 cup Lprime2 satisfy the conditions (33)(34)

Suppose system (431) in Dprime satisfies Condition C through the transformation (428)and ξν = ξν ηmicro = [2R minus 2σ(ν) + ν]ηmicro2R system (431) is reduced to

ξν = Aξ +Bη + E ηmicro = [2R minus 2σ(ν) + ν][Cξ +Dη + F ]

2R (433)

Moreover through the transformation (430) ie z = fminus1(z) the boundary condition(432) is reduced to

Re [λ(fminus1(z))w(fminus1(z))] = r(fminus1(z)) z isin L1 cup L2

Im [λ(fminus1(z0))w(fminus1(z0)] = b1(434)

in which z0 = f(z0) = 0 Therefore the boundary value problem (431)(432) istransformed into the boundary value problem (433)(434) On the basis of Theorems41 and 42 we see that the boundary value problem (433) (434) has a uniquesolution w(z) and then w[f(z)] is just a solution of the boundary value problem(431)(432) in Dprime

Theorem 46 If the complex equation (41) satisfies Condition C in the domainDprime with the boundary Lprime

1 cup Lprime2 cup L3 cup L4 where Lprime

1 Lprime2 are as stated in (426) then

Problem A with the boundary conditions

2 Im [λ(z0)w(z0)] = b1

has a unique solution w(z)

2 The boundary L1 L4 of the domain D are replaced by the two curves Lprimeprime1 Lprimeprime

4respectively and the boundary of the domain Dprimeprime is Lprimeprime

1 cup Lprimeprime2 cup Lprimeprime

3 cup Lprimeprime4 where the

parameter equations of the curves Lprimeprime1 Lprimeprime

2 Lprimeprime3 Lprimeprime

4 are as follows

Lprimeprime1 = γ1(x) + y = 0 0 le x le l1 Lprimeprime

2 = x minus y = 2R1 l1 le x le R2

Lprimeprime3 = x+ y = 2R l2 le x le R2 Lprimeprime

4 = minusγ4(x) + y = 0 0 le x le l2(435)

in which and γ1(0) = 0 γ4(R2) = 2R minus R2 γ1(x) gt 0 0 le x le l1 γ4(x) gt 0 0 lex le l2 γ1(x) on 0 le x le l1 γ4(x) on 0 le x le l2 are continuous and γ1(x) γ4(x)possess derivatives except at finite points on 0 le x le l1 0 le x le l2 respectively and1 + γprime

1(x) gt 0 1 + γprime4(x) gt 0 zprimeprime

1 = x minus jγ1(l1) isin L2 zprimeprime3 = x + jγ2(l2) isin L3 By the

conditions the inverse functions x = σ(ν) x = τ(micro) of x+ γ1(x) = ν x+ γ4(x) = microcan be found respectively namely

micro = 2σ(ν)minus ν 0 le ν le l1 + γ1(l1) ν = 2τ(micro)minus micro l2 + γ4(l2) le micro le 2R1 (436)

We make a transformation

micro = micro ν =2R1(ν minus 2τ(micro) + micro)2R1 minus 2τ(micro) + micro

0 le micro le 2R 2τ(micro)minus micro le micro le 2R1 (437)

micro = micro ν =2R1 minus 2τ(micro) + micro

2R1ν + 2τ(micro)minus ν 0 le micro le 2R 0 le ν le 2R1 (438)

Hence we have

x =12(micro+ ν) =

4R1x minus [2τ(x+ γ1(x))minus x minus γ1(x)](2R1 + x+ y)4R1 minus 4τ(x+ γ1(x)) + 2x+ 2γ1(x)

2R1y + [2τ(x+ γ1(x))minus x minus γ1(x)](2R1 minus x minus y)4R1 minus 4τ(x+ γ1(x)) + 2x+ 2γ1(x)

(439)

and its inverse transformation is

x =12(micro+ ν) =

4R1x+ [2τ(x+ γ1(x))minus x minus γ1(x)](2R1 minus x+ y)4R1

4R1y minus [2τ(x+ γ1(x))minus x minus γ1(x)](2R1 + x minus y)4R1

(440)

Denote by z = x + jy = g(z) z = x + jy = gminus1(z) the transformation (439) andits inverse transformation (440) respectively Through the transformation (438) wehave

(u+ v)ν =2R1 minus 2τ(micro) + micro

2R1(u+ v)ν (u minus v)micro = (u minus v)micro

system (431) in Dprimeprime is reduced to

ξν =2R1 minus 2τ(micro) + micro

2R1[Aξ +Bη + E] ηmicro = Cξ +Dη + F z isin Dprime (441)

where Dprime is a bounded domain with boundary Lprime1 cup Lprime

2 cup L3 cup L4 and Lprime1 = Lprimeprime

1 More-over through the transformation (440) the boundary condition (432) on Lprimeprime

1 cup Lprimeprime4

is reduced to

Re [λ(gminus1(z))w(gminus1(z))] = r(gminus1(z)) z isin Lprime1 cup L4

Im [λ(gminus1(z0))w(gminus1(z0)] = b1(442)

in which z0 = g(z0) = 0 Therefore the boundary value problem (431)(432) in Dprimeprime istransformed into the boundary value problem (441)(442) On the basis of Theorem46 we see that the boundary value problem (441)(442) has a unique solution w(z)and then w[g(z)] is just a solution of the boundary value problem (431)(432)

Theorem 47 If the complex equation (41) satisfies Condition C in the domainDprimeprime with the boundary Lprimeprime

1 cupLprimeprime2 cupLprimeprime

3 cupLprimeprime4 then Problem A with the boundary conditions

Re [λ(z)w(z)] = r(z) z isin Lprimeprime1 cup Lprimeprime

4 Im [λ(z0)w(z0)] = b1

has a unique solution w(z) where zprimeprime1 = x minus jγ1(l1) isin L2 z

primeprime3 = x+ jγ4(l2) isin L3

Now we give an example to illustrate the above results When R2 = 2R1 theboundary of the domain D is L1 = x = minusy 0 le x le R1 L2 = x = y + 2R1 R1 lex le R1 L3 = x = minusy + 2R1 R1 le x le 2R1 L4 = x = y 0 le x le R1 Wereplace L1 cup L4 by a left semi-circumference Lprimeprime

1 cup Lprimeprime4 with the center R1 and the

radius R1 namely

Lprimeprime1 = x minus y = ν y = minusγ1(x) = minus

radicR2

1 minus (x minus R1)2 0 le x le R1

Lprimeprime4 = x+ y = micro y = γ4(x) =

radicR2

where 1 + γprime1(x) gt 0 0 lt x le R1 1 + γprime

4(x) gt 0 0 le x lt R1 and

x = σ(ν) =R1 + ν minus

radicR2

1 + 2R1ν minus ν2

2

x = τ(micro) =R1 + micro minus

radicR2

1 + 2R1micro minus micro2

2

It is clear that according to the above method the domain D can be generalized toa general domainDprimeprime namely its boundary consists of the general curves Lprimeprime

1 Lprimeprime2 L

primeprime3 L

primeprime4

with some conditions which includes the circumference L = |z minus R1| = R1Finally we mention that some boundary value problems for equations (312) and

(41) with one of the boundary conditions

Re [λ(z)w(z)] = r(z) z isin L1 cup L2 Im [λ(z1)w(z1)] = b1

5 Quasi-hyperbolic Mappings 35

can be discussed where λ(z) r(z) b1 satisfy the conditions similar to those in(33)(34)(331) and Lj(j = 1 5) zj(j = 0 1 3) are as stated in Section 3

L6 = y =R2 minus 2R1

R2x 0 le x le R2 and λ(z) = 1 + j z isin L5 λ(z) = 1minus j z isin L6

For corresponding boundary value problems of hyperbolic systems of first ordercomplex equations whether there are similar results as before The problem needsto be investigated

5 Hyperbolic Mappings and Quasi-hyperbolic Mappings

Now we introduce the definitions of hyperbolic mappings and quasi-hyperbolic map-pings and prove some properties of quasi-hyperbolic mappings

51 Hyperbolic mappings

A so-called hyperbolic mapping in a domain D is a univalent mapping given by ahyperbolic continuously differentiable function w = f(z) = u + jv satisfying thesimplest hyperbolic system of first order equations

ux = vy vx = uy (51)

which maps D onto a domain G in the w-plane By Theorem 11 system (51) isequivalent to the system

ξν = 0 ηmicro = 0 (52)

where ξ = u+ v η = u minus v micro = x+ y ν = x minus y Noting∣∣∣∣∣microx νx

microy νy

∣∣∣∣∣ =∣∣∣∣∣ 1 1

1 minus1

∣∣∣∣∣ = minus2∣∣∣∣∣ ξu ηu

ξv ηv

∣∣∣∣∣ =∣∣∣∣∣ 1 1

1 minus1

∣∣∣∣∣ = minus2 (53)

if we find a homeomorphic solution of (52) then the solution

w = u+ jv = ξ[(micro+ ν)2 (micro minus ν)2]e1 + η[(micro+ ν)2 (micro minus ν)2]e2 (54)

of the corresponding system (51) is also homeomorphic In fact

ξ = ξ(micro) η = η(ν) (55)

is a homeomorphic solution of (52) if ξ(micro) and η(ν) are strictly monotonous continu-ous functions of micro(micro0 le micro le micro1) and ν(ν0 le ν le ν1) respectively When [ξ(micro) η(ν)]is univalent continuous in ∆ = micro0 le micro le micro1 ν0 le ν le ν1 and D(∆) is the closeddomain in the z = x + jy-plane corresponding to ∆ it is easy to see that[u(x y) v(x y)] is a homeomorphic solution of (51) in D(∆)

52 Quasi-hyperbolic mappings

In this subsection we first discuss the uniformly hyperbolic system in the complexform

wz = Q(z)wz Q(z) = a(z) + jb(z) (56)

where Q(z) is a continuous function satisfying the condition |Q(z)| le q0 lt 1 here q0

is a non-negative constant On the basis of the representation

wz = ξνe1 + ηmicroe2 wz = ξmicroe1 + ηνe2 Q = q1e1 + q2e2

from (56) it follows thatξν = q1ξmicro ηmicro = q2ην (57)

Due to Q = a + jb = q1e1 + q2e2 here q1 = a + b q2 = a minus b thus |Q|2 = |QQ| =|a2 minus b2| = |q1q2| le q2

0 lt 1 and a representation theorem of solutions for (56) can beobtained

Theorem 51 Let χ(z) be a homeomorphic solution of (56) in a domain D andw(z) be a solution of (56) in D Then w(z) can be expressed as

w(z) = Φ[χ(z)] (58)

where Φ(χ) is a hyperbolic regular function in the domain G = χ(D)

Proof Suppose that z(χ) is the inverse function of χ(z) we can find

w[z(χ)]χ = wzzχ + wz zχ = wz[zχ +Qzχ]

andχχ = 1 = χzzχ + χz zχ = χz[zχ +Qzχ]

χχ = 0 = χzzχ + χz zχ = χz[zχ +Qzχ]

From the above equalities we see χz = 0 zχ+Qzχ = 0 consequently Φ(χ) = w[z(χ)]satisfies

[Φ(χ)]χ = 0 χ isin G = χ(D)

This shows that Φ(χ) is a hyperbolic regular function in G = χ(D) therefore therepresentation (58) holds

Next we prove the existence of a homeomorphic solution of equation (56) withsome conditions for the coefficient of (56)

From (57) we see that the complex equation (56) can be written in the form

ξν = (a+ b)ξmicro ηmicro = (a minus b)ην (59)

Let ∆ = micro0 le micro le micro0 + R1 ν0 le ν le ν0 + R2 in which micro0 ν0 are two real num-bers and R1 R2 are two positive numbers if a b possess continuously differentiable

derivatives with respect to micro ν in ∆ then the solution w = ξe1+ ηe2 of (59) in ∆ isa homeomorphism provided that one of the following sets of conditions

b gt 0 minusb lt a lt b (510)

b lt 0 b lt a lt minusb (511)

holds and ξ and η are strictly monotonous continuous functions of micro (micro0 le micro lemicro0 +R1) and ν (ν0 le ν le ν0 +R2) respectively

Thus we have the following theorem

Theorem 52 Denote by D the corresponding domain of ∆ in the (x + jy)-planeand let w(z) be a continuous solution of (56) in D If (510) or (511) in D holdsξmicro(ξν) gt 0 and ηmicro (ην) gt 0 except some possible isolated points on micro (micro0 le micro lemicro0 +R1) (ν (ν0 le ν le ν0 +R2)) then the solution w(z) in D is a homeomorphism

In particular if the coefficient Q(z) = a + jb of the complex equation (56) is ahyperbolic constant which satisfies the condition

|Q(z)|2 = |QQ| = |a2 minus b2| = |q1q2| le q20 lt 1 z isin D

we make a nonsingular transformation

micro = minus(a+ b)σ + τ ν = σ minus (a minus b)τ (512)

Thus system (59) can be transformed into the system

ξσ = 0 ητ = 0 (σ τ) isin G (513)

where the domain G is the corresponding domain of ∆ under the transformation(512) According to the discussion of hyperbolic mappings we see that system (513)in G possesses a homeomorphic solution hence system (59) in ∆ has a homeomorphicsolution and then the complex equation (56) in D has a homeomorphic solution Theabove result can be written as a theorem

Theorem 53 Suppose that Q(z) = a+ jb is a hyperbolic constant and |Q(z)| lt 1Then the complex equation (56) in D has a homeomorphic solution

53 Other Quasi-hyperbolic mappings

Now we consider the hyperbolic complex equation

wz = Q(z)wz (514)

where Q(z) is a continuous function in D = 0 le x le R1 0 le y le R2 satisfyingthe condition |Q(z)| le q0 lt 1 If Q(z) in D is a hyperbolic regular function of z weintroduce a transformation of functions

W = w minus Qw ie w =W +Q(z)W1minus |Q(z)|2 (515)

Then (514) is reduced to the complex equation

Wz = 0 (516)

The solution W (z) of (516) in D is a hyperbolic regular function of z As statedbefore the complex equation (516) possesses a homeomorphic solution which realizesa hyperbolic mapping in D Moreover if Q(z) is a hyperbolic regular function in Dwe find the partial derivative with respect to z in (514) and obtain

wzz = Q(z)wzz ie wzz = Q(z)wzz (517)

from |Q(z)| lt 1 it follows that

(1minus |Q(z)|2)wzz = 0 ie wzz = 0 (518)

the solution of the above complex equation (518) is called a hyperbolic harmoniccomplex function

A hyperbolic harmonic complex function w(z) can be expressed as

w(z) = u(z) + jv(z) = φ(z) + φ(z) + ψ(z)minus ψ(z)

= φ(z) + ψ(z) + φ(z)minus ψ(z) = f(z) + g(z)

in which φ(z) ψ(z) are hyperbolic regular functions hence f(z) = φ(z)+ψ(z) g(z) =φ(z) minusψ(z) are hyperbolic regular functions This is a representation of hyperbolicharmonic functions through hyperbolic regular functions Hence in order to find ahyperbolic harmonic function it is sufficient to find solutions of the following twoboundary value problems with the boundary conditions

Re f(x) = Reφ0(x) Re f(jy) = Reφ1(y)

andIm g(x) = Imφ0(x) Im g(jy) = Imφ1(y)

respectively where φ0(x) φ1(y) are given hyperbolic complex functions on0 le x le R1 0 le y le R2 respectively and R1 R2 are two positive constants

At last we mention that the notations of hyperbolic numbers and hyperboliccomplex functions are mainly used in this and next chapters From Chapter III toChapter VI except in Section 5 Chapter V we do not use them

The references for this chapter are [5][9][12][19][26][29][32][34][38][44][51][59][68][74][80][83][85][87][89][92][97]

CHAPTER II

HYPERBOLIC COMPLEX EQUATIONS OFSECOND ORDER

In this chapter we mainly discuss oblique derivative boundary value problems forlinear and quasilinear hyperbolic equations of second order in a simply connecteddomain Firstly we transform some linear and nonlinear uniformly hyperbolic equa-tions of second order with certain conditions into complex forms give the uniquenesstheorem of solutions for the above boundary value problems Moreover by usingthe successive iteration the existence of solutions for several oblique derivative prob-lems is proved Finally we introduce some boundary value problems for degeneratehyperbolic equations of second order with certain conditions

1 Complex Form of Hyperbolic Equations of Second Order

This section deals with hyperbolic equations of second order in the plane domains wefirst transform some linear and nonlinear uniformly hyperbolic equations of secondorder with certain conditions into complex forms and then we state the conditionsof some hyperbolic complex equations of second order

11 Reduction of linear and nonlinear hyperbolic equations of secondorder

Let D be a bounded domain we consider the linear hyperbolic partial differentialequation of second order

auxx + 2buxy + cuyy + dux + euy + fu = g (11)

where the coefficients a b c d e f g are known continuous functions of (x y) isin Din which D is a bounded domain The condition of hyperbolic type for (11) is thatfor any point (x y) in D the inequality

I = ac minus b2 lt 0 a gt 0 (12)

holds If a b c are bounded in D and

I = ac minus b2 le I0 lt 0 a gt 0 (13)

40 II Hyperbolic Equations of Second Order

in D where I0 is a negative constant then equation (11) is called uniformly hyper-bolic in D Introduce the notations as follows

( )z =( )x + j( )y

2 ( )z =

( )x minus j( )y2

( )zz =( )xx minus ( )yy

4

( )zz =( )xx + ( )yy + 2j( )xy

4 ( )zz =

( )xx + ( )yy minus 2j( )xy

4

( )x = ( )z + ( )z ( )y = j[( )z minus ( )z] ( )xy = j[( )zz minus ( )zz]

( )xx = ( )zz + ( )zz + 2( )zz ( )yy = ( )zz + ( )zz minus 2( )zz

(14)

equation (11) can be written in the form

2(a minus c)uzz + (a+ c+ 2bj)uzz + (a+ c minus 2bj)uzz

+(d+ ej)uz + (d minus ej)uz + fu = g in D(15)

If a = c in D then equation (15) can be reduced to the complex form

uzz minus Re [Q(z)uzz + A1(z)uz]minus A2(z)u = A3(z) in D (16)

in which

Q =a+ c+ 2bj

a minus c A1 =

d+ ej

a minus c A2 =

f

2(a minus c) A3 =

g

2(a minus c)

If (a+ c)2 ge 4b2 then the conditions of hyperbolic type and uniformly hyperbolic aretransformed into

|Q(z)| lt 1 in D (17)

and|Q(z)| le q0 lt 1 in D (18)

respectively

For the nonlinear hyperbolic equation of second order

Φ(x y u ux uy uxx uxy uyy) = 0 in D (19)

from (14) we have Φ = F (z u uz uzz uzz) Under certain conditions equation (19)can be reduced to the real form

auxx + 2buxy + cuyy + dux + euy + fu = g in D (110)

and its complex form is as follows

a0uzz minus Re [quzz + a1uz]minus a2u = a3 in D (111)

1 Complex Form of Hyperbolic Equations 41

in which

a =int 1

0Φτuxx(x y u ux uy τuxx τuxy τuyy)dτ = a(x y u ux uy uxx uxy uyy)

2b =int 1

0Φτuxy(x y u ux uy τuxx τuxy τuyy)dτ = 2b(x y u ux uy uxx uxy uyy)

c =int 1

0Φτuyy(x y u ux uy τuxx τuxy τuyy)dτ = c(x y u ux uy uxx uxy uyy)

d =int 1

0Φτux(x y u τux τuy 0 0 0)dτ = d(x y u ux uy)

e =int 1

0Φτuy(x y u τux τuy 0 0 0)dτ = e(x y u ux uy)

f =int 1

0Φτu(x y τu 0 0 0 0 0)dτ = f(x y u)

g = minusΦ(x y 0 0 0 0 0 0) = g(x y)

and

a0 = 2(a minus c) =int 1

0Fτuzz(z u uz τuzz τuzz)dτ = a0(z u uz uzz uzz)

q=2(a+c+2bj)=minus2int 1

0Fτuzz(z u uz τuzz τuzz)dτ=q(z u uz uzz uzz)

a1 = 2(d+ ej) = minus2int 1

0Fτuz(z u τuz 0 0 0)dτ = a1(z u uz)

a2 = f = minusint 1

0Fτu(z τu 0 0 0 0)dτ = a2(z u)

a3 = minusF (z 0 0 0 0) = a3(z)

(112)

The condition of uniformly hyperbolic type for equation (110) is the same with (13)If a = c in D the complex equation (111) can be rewritten in the form

uzz minus Re [Quzz + A1uz]minus A2u = A3 in D (113)

whereQ = qa0 A1 = a1a0 A2 = a2a0 A3 = a3a0

are functions of z isin D u uz uzz uzz the condition of uniformly hyperbolic type for(113) is as stated in the form (18)

As stated in [12] 3) for the linear hyperbolic equation (11) or its complex form(16) if the coefficients a b c are sufficiently smooth through a nonsingular transfor-mation of z equation (11) can be reduced to the standard form

uxx minus uyy + dux + euy + fu = g (114)

or its complex form

uzz minus Re [A1(z)uz]minus A2(z)u = A3(z) (115)

12 Conditions of some hyperbolic equations of second order

Let D be a simply connected bounded domain with the boundary Γ = L1 cup L2 cupL3 cup L4 as stated in Chapter I where L1 = x = minusy 0 le x le R1 L2 = x =y + 2R1 R1 le x le R2 L3 = x = minusy minus 2R1 + 2R2 R2 minus R1 le x le R2 L4 = x =y 0 le x le R2 minus R1 and denote z0 = 0 z1 = (1minus j)R1 z2 = R2+ j(R2 minus 2R1) z3 =(1 + j)(R2 minus R1) L = L3 cup L4 here there is no harm in assuming that R2 ge 2R1In the following we mainly consider second order quasilinear hyperbolic equation inthe form

uzz minus Re [A1(z u uz)uz]minus A2(z u uz)u = A3(z u uz) (116)

whose coefficients satisfy the following conditions Condition C

1) Al(z u uz)(l = 1 2 3) are continuous in z isin D for all continuously differentiablefunctions u(z) in D and satisfy

2) For any continuously differentiable functions u1(z) u2(z) in D the equality

F (z u1 u1z)minus F (z u2 u2z) = Re [A1(u1 minus u2)z] + A2(u1 minus u2) in D (118)

holds whereC[Al(z u1 u2) D] le k0 l = 1 2 (119)

in (117)(119) k0 k1 are non-negative constants In particular when (116) is alinear equation from (117) it follows that the conditions (118) (119) hold

In order to give a priori estimates in Cα(D) of solutions for some boundary valueproblems we need to add the following conditions For any two real numbers u1 u2

and hyperbolic numbers z1 z2 isin D w1 w2 the above functions satisfy

Al(z1 u1 w1)minus Al(z2 u2 w2) le k0[ z1 minus z2 α + u1 minus u2 α + w1 minus w2 ] l = 1 2 A3(z1 u1 w1)minus A3(z2 u2 w2) le k1[ z1 minus z2 α + u1 minus u2 α + w1 minus w2 ]

(120)

where α (0 lt α lt 1) k0 k1 are non-negative constants

It is clear that (116) is the complex form of the following real equation of secondorder

uxx minus uyy = aux + buy + cu+ d in D (121)

in which a b c d are functions of (x y)(isin D) u ux uy(isin IR) and

A1 =a+ jb

2 A2 =

c

4 A3 =

d

4in D

Due to z = x+ jy = microe1 + νe2 w = uz = ξe1 + ηe2 and

wz =wx + jwy

2= ξmicroe1 + ηνe2 wz =

wx minus jwy

2= ξνe1 + ηmicroe2

the quasilinear hyperbolic equation (116) can be rewritten in the form

ξνe1 + ηmicroe2 = [A(z u w)ξ +B(z u w)η + C(z u w)u+D(z u w)]e1

+[A(z u w)ξ +B(z u w)η + C(z u w)u+D(z u w)]e2 ie⎧⎨⎩ ξν = A(z u w)ξ +B(z u w)η + C(z u w)u+D(z u w)

ηmicro = A(z u w)ξ + A(z u w)η + C(z u w)u+D(z u w)in D

(122)

in which

A =a+ b

4 B =

a minus b

4 C =

c

4 D =

d

4

In the following we mainly discuss the oblique derivative problem for linearhyperbolic equation (16) and quasilinear hyperbolic equation (116) in thesimply connected domain We first prove that there exists a unique solution of theboundary value problem and give a priori estimates of their solutions and then provethe solvability of the boundary value problem for general hyperbolic equations

2 Oblique Derivative Problems for Quasilinear HyperbolicEquations of Second Order

Here we first introduce the oblique derivative problem for quasilinear hyperbolic equa-tions of second order in a simply connected domain and give the representationtheorem of solutions for hyperbolic equations of second order

21 Formulation of the oblique derivative problem and the representationof solutions for hyperbolic equations

The oblique derivative problem for equation (116) may be formulated as follows

Problem P Find a continuously differentiable solution u(z) of (116) in D satisfyingthe boundary conditions

12

partu

partl= Re [λ(z)uz] = r(z) z isin L = L3 cup L4

u(0) = b0 Im [λ(z)uz]|z=z3 = b1

(21)

where l is a given vector at every point on L λ(z) = a(z) + jb(z) = cos(l x) +j cos(l y) z isin L b0 b1 are real constants and λ(z) r(z) b0 b1 satisfy the conditions

Cα[λ(z) L] le k0 Cα[r(z) L] le k2 |b0| |b1| le k2

maxzisinL3

1|a(z)minus b(z)| maxzisinL4

1|a(z) + b(z)| le k0

(22)

in which α(0 lt α lt 1) k0 k2 are non-negative constants The above boundary valueproblem for (116) with A3(z u w) = 0 z isin D u isin IR w isin CI and r(z) = b0 = b1 =0 z isin L will be called Problem P0

By z = x+ jy = microe1+ νe2 w = uz = ξe1+ ηe2 the boundary condition (21) canbe reduced to

Re [λ(z)(ξe1 + ηe2)] = r(z) u(0) = b0 Im [λ(z)(ξe1 + ηe2)]|z=z3 = b1 (23)

where λ(z) = (a + b)e1 + (a minus b)e2 Moreover the domain D is transformed intoQ = 0 le micro le 2R 0 le ν le 2R1 R = R2 minus R1 which is a rectangle and A B C Dare known functions of (micro ν) and unknown continuous functions u w and they satisfythe condition

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

λ(z3)w(z3) = λ(z)[ξe1 + ηe2]|z=z3 = r(z3) + jb1 u(0) = b0

Re [λ(z)w(2Re1 + νe2)] = r(z)

if (x y) isin L3 = micro = 2R 0 le ν le 2R1

Re [λ(z)w(microe1 + 0e2)] = r(z)

if (x y) isin L4 = 0 le micro le 2R ν = 0

(24)

where λ(z) r(z) b0 b1 are as stated in (21) We can assume that w(z3) = 0otherwise through the transformation W (z) = w(z) minus [a(z3) minus jb(z3)][r(z3) + jb1][a2(z3)minus b2(z3)] the requirement can be realized

It is not difficult to see that the oblique derivative boundary value problem (Prob-lem P ) includes the Dirichlet boundary value problem (Problem D) as a special caseIn fact the boundary condition of Dirichlet problem (Problem D) for equation (121)is as follows

u(z) = φ(z) on L = L3 cup L4 (25)

We find the derivative with respect to the tangent direction s = (x∓jy)radic2 for (25)

in which ∓ are determined by L3 and L4 respectively it is clear that the followingequalities hold

Re [λ(z)uz] = r(z) z isin L Im [λ(z)uz]|z=z3 = b1 (26)

in which

λ(z) = a+ jb =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩1minus jradic2on L3

1 + jradic2on L4

r(z) =φxradic2on L = L3 cup L4

b+1 = Im

[1minus jradic2

uz(z3)]= minusφx + φx

2radic2

|z=z3minus0 = minusradic2φx|z=z3minus0

bminus1 = Im

[1 + jradic2

uz(z3)]=

radic2φx|z=z3+0 b0 = φ(0)

(27)

in which a = 1radic2 = b = minus1radic2 on L3 and a = 1

radic2 = minusb = minus1radic2 on L4

Noting that Problem P for (116) is equivalent to the RiemannndashHilbert problem(Problem A) for the complex equation of first order and boundary conditions

wz = Re [A1w] + A2u+ A3 in D (28)

Re [λ(z)w(z)] = r(z) z isin L Im [λ(z)w(z)]|z=z3 = b1 (29)

and the relationu(z) = 2Re

int z

0w(z)dz + b0 in D (210)

from Theorem 12 (3) Chapter I and equation (28) we see that

2ReintΓw(z)dz =

intΓw(z)dz +

intΓw(z)dz

= 2jint int

D[wz minus wz]dxdy = 4

int intDIm [wz]dxdy = 0

the above equality for any subdomain in D is also true hence the function determinedby the integral in (210) is independent of integral paths in D In this case wemay choose that the integral path is along two family of characteristic lines namelyfirst along one of characteristic lines x + y = micro (0 le micro le 2R) and then alongone of characteristic lines x minus y = ν (0 le ν le 2R1) for instance the value ofu(zlowast)(zlowast = xlowast + jylowast isin D ylowast le 0) can be obtained by the integral

u(zlowast) = 2Re[int

s1

w(z)dz +int

s2

w(z)dz]+ b0

where s1 = x+y = 0 0 le x le (xlowast minusylowast)2 s2 = xminusy = xlowast minusylowast (xlowast minusylowast)2 le x lexlowastminusylowast in which xlowastminusylowast is the intersection of the characteristic line xminusy = xlowastminusylowastthrough the point zlowast and real axis In particular when Aj = 0 j = 1 2 3 equation(116) becomes the simplest hyperbolic complex equation

uzz = 0 (211)

Problem P for (211) is equivalent to Problem A for the simplest hyperbolic complexequation of first order

wz = 0 in D (212)

with the boundary condition (29) and the relation (210) Hence similarly to Theorem31 Chapter I we can derive the representation and existence theorem of solutionsof Problem A for the simplest equation (212) namely

Theorem 21 Any solution u(z) of Problem P for the hyperbolic equation (211)can be expressed as (210) where w(z) is as follows

f(2R) = u(z3) + v(z3) =r((1 + j)R) + b1

a((1 + j)R) + b((1 + j)R)

g(0) = u(z3)minus v(z3) =r((1 + j)R)minus b1

a((1 + j)R)minus b((1 + j)R)

(213)

here f(x+ y) g(x minus y) possess the forms

g(x minus y)=2r((1minus j)(x minus y)2 + (1 + j)R)

a((1minus j)(x minus y)2 + (1 + j)R)minus b((1minus j)(x minus y)2 + (1 + j)R)

minus [a((1minusj)(xminusy)2+(1 + j)R)+b((1minus j)(x minus y)2 + (1 + j)R)]f(2R)a((1minus j)(x minus y)2 + (1 + j)R)minusb((1minus j)(x minus y)2 + (1 + j)R)

0 le x minus y le 2R1 (214)

f(x+ y)=2r((1+j)(x+y)2)minus[a((1 + j)(x+ y)2)minusb((1 + j)(x+ y)2)]g(0)

a((1 + j)(x+ y)2) + b((1 + j)(x+ y)2)

0 le x+ y le 2R

Moreover u(z) satisfies the estimate

C1α[u(z) D]leM1=M1(α k0 k2 D) C1

α[u(z) D]leM2k2=M2(α k0 D)k2 (215)

where M1 M2 are two non-negative constants only dependent on α k0 k2 D and αk0 D respectively

Proof Let the general solution w(z) = uz = 12f(x+ y) + g(x minus y) + j[f(x+ y)minus

g(x minus y)] of (212) be substituted in the boundary condition (21) we obtain

a(z)u(z) + b(z)v(z) = r(z) on L λ(z3)w(z3) = r(z3) + jb1 ie

[a((1minus j)x+ 2jR) + b((1minus j)x+ 2jR)]f(2R)+[a((1minus j)x+ 2jR)

minusb((1minus j)x+ 2jR)]g(2x minus 2R) = 2r((1minus j)x+ 2jR) on L3

[a((1 + j)x) + b((1 + j)x)]f(2x) + [a((1 + j)x)

minusb((1 + j)x)]g(0) = 2r((1 + j)x) on L4

f(2R)=u(z3)+v(z3)=r((1 + j)R)+b1

a((1 + j)R)+b((1 + j)R)

g(0)=u(z3)minusv(z3)=r((1 + j)R)minusb1

a((1 + j)R)minusb((1 + j)R)

and the above formulas can be rewritten as

[a((1minus j)t2 + (1 + j)R) + b((1minus j)t2 + (1 + j)R)]f(2R)

+ [a((1minus j)t2 + (1 + j)R)minus b((1minus j)t2 + (1 + j)R)]g(t)

= 2r((1minus j)t2 + (1 + j)R) t isin [0 2R1]

[a((1 + j)t2) + b((1 + j)t2)]f(t) + [a((1 + j)t2)

minusb((1 + j)t2)]g(0) = 2r((1 + j)t2) t isin [0 2R]thus the solution w(z) can be expressed as (213)(214) From the condition (22)and the relation (210) we see that the estimate (215) of u(z) for (211) is obviouslytrue

Next we give the representation of Problem P for the quasilinear equation (116)

Theorem 22 Under Condition C any solution u(z) of Problem P for the hyper-bolic equation (116) can be expressed as

u(z) = 2Reint z

0w(z)dz + b0 in D

w(z) = W (z) + Φ(z) + Ψ(z) in D

W (z) = f(micro)e1 + g(ν)e2 Φ(z) = f(micro)e1 + g(ν)e2

Ψ(z)=int ν

0[Aξ+Bη+Cu+D]e1dν+

int micro

2R[Aξ+Bη+Cu+D]e2dmicro

(216)

where f(micro) g(ν) are as stated in (214) and f(micro) g(ν) are similar to f(micro) g(ν) in(214) but r(z) b1 are replaced by minusRe [λ(z)Ψ(z)] minusIm [λ(z3)Ψ(z3)] namely

Re [λ(z)Φ(z)]=minusRe [λ(x)Ψ(x)] z isin L Im [λ(z3)Φ(z3)]=minusIm [λ(z3)Ψ(z3)](217)

Proof Let the solution u(z) of Problem P be substituted into the coefficients ofequation (116) Then the equation in this case can be seen as a linear hyperbolicequation (115) Due to Problem P is equivalent to the Problem A for the complexequation (28) with the relation (210) from Theorem 32 Chapter I it is not difficultto see that the function Ψ(z) satisfies the complex equation

[Ψ]z = [Aξ +Bη + Cu+D]e1 + [Aξ +Bη + Cu+D]e2 in D (218)

and Φ(z) = w(z) minus W (z) minus Ψ(z) satisfies the complex equation and the boundaryconditions

ξνe1 + ηmicroe2 = 0 (219)

Re [λ(z)(ξe1 + ηe2)] = minusRe [λ(z)Ψ(z)] on L

Im [λ(z)(ξe1 + ηe2)]|z=z3 = minusIm [λ(z3)Ψ(z3)](220)

By the representation of solutions of Problem A for (116) as stated in (317) Chap-ter I we can obtain the representation (216) of Problem P for (116)

22 Existence and uniqueness of solutions of Problem P for hyperbolicequations of second order

Theorem 23 If the complex equation (116) satisfies Condition C then ProblemP for (116) has a solution

Proof We consider the expression of u(z) in the form (216) In the following byusing successive iteration we shall find a solution of Problem P for equation (116)Firstly substitute

u0(z) = 2Reint z

0w0(z)dz + b0 w0(z) = W (z) = ξ0e1 + η0e2 (221)

into the position of u w = ξe1 + ηe2 in the right-hand side of (116) where w0(z) isthe same function with W (z) in (216) and satisfies the estimate (215) Moreoverwe have

u1(z) = 2Reint z

0w1(z)dz + b0 w1(z) = w0(z) + Φ1(z) + Ψ1(z)

Ψ1(z) =int xminusy

0[Aξ0 +Bη0 + Cu0 +D]e1d(x minus y)

+int x+y

2R[Aξ0 +Bη0 + Cu0 +D]e2d(x+ y)

(222)

from the first equality in (222) the estimate

C1[u0(z) D] le M3C[w0(z) D] + k2 (223)

can be derived where M3 = M3(D) From the third and second equalities in (222)we can obtain

C[Ψ1(z) D] le 2M4[(4 +M3)m+ k2 + 1]Rprime

C[Φ1(z) D] le 8M4k20(1 + k2

0)[(4 +M3)m+ k2 + 1]Rprime

C[w1(z)minus w0(z) D] le 2M4M [(4 +M3)m+ k2 + 1]Rprime

(224)

where M4 = maxD(|A| |B| |C| |D|) Rprime = max(2R1 2R) m = w0(z) C(D) M =1 + 4k2

0(1 + k20) is a positive constant Thus we can find a sequence of functions

wn(z) satisfying

un+1(z) = 2Reint z

0wn+1(z)dz + b0

wn+1(z) = w0(z) + Φn(z) +int ν

0[Aξn +Bηn + Cun +D]e1dν

+int micro

2R[Bηn + Aηn + Cun +D]e2dmicro

(225)

and then

wn minus wnminus1 le 2M4M [(4 +M3)m+ 1]n timesint Rprime

0

Rprimenminus1

(n minus 1) dRprime

le 2M4M [(4 +M3)m+ 1)]Rprimen

n

(226)

in D uniformly converges a function wlowast(z) and wlowast(z) satisfies the equality

wlowast(z) = ξlowaste1 + ηlowaste2

= w0(z) + Φlowast(z) +int ν

0[Aξlowast +Bηlowast + Culowast +D]e1dν

+int micro

2R[Aξlowast +Bηlowast + Culowast +D]e2dmicro

(228)

and the functionulowast(z) = 2Re

int z

0wlowast(z)dz + b0 (229)

is just a solution of Problem P for equation (116) in the closed domain D

Theorem 24 Suppose that Condition C holds Then Problem P for the complexequation (116) has at most one solution in D

Proof Let u1(z) u2(z) be any two solutions of Problem P for (116) we see thatu(z) = u1(z)minus u2(z) and w(z) = u1z(z)minus u2z(z) satisfies the homogeneous complexequation and boundary conditions

wz = Re [A1w] + A2u in D (230)

Re [λ(z)w(z)] = 0 on L Im [λ(z3)w(z3)] = 0 (231)

int z

0w(z)dz z isin D (232)

From Theorem 22 we see that the function w(z) can be expressed in the form

w(z) = Φ(z) + Ψ(z)

Ψ(z) =int ν

0[Aξ + Bη + Cu]e1dν +

int micro

2R[Aξ + Bη + Cu]e2dmicro

(233)

moreover from (232)C1[u(z) D] le M3C[w(z) D] (234)

can be obtained in which M3 = M3(D) is a non-negative constant By using theprocess of iteration similar to the proof of Theorem 23 we can get

w(z) = w1 minus w2 le 2M5M [(4 +M3)m+ 1]Rprimen

n

where M5 = maxD(|A| |B| |C|) Let n rarr infin it can be seen w(z) = w1(z)minus w2(z) =0 Ψ(z) = Φ(z) = 0 in D This proves the uniqueness of solutions of Problem P for(116)

3 Oblique Derivative Problems for General QuasilinearHyperbolic Equations of Second Order

In this section we first give a priori estimates in C1(D) of solutions for the obliquederivative problem moreover by using the estimates of solutions the existence ofsolutions for the above problem for general quasilinear equation is proved Finallywe discuss the oblique derivative problem for hyperbolic equations of second order ingeneral domains

31 A priori estimates of solutions of Problem P for hyperbolic equationsof second order

From Theorems 23 and 24 we see that under Condition C Problem P for equation(116) has a unique solution u(z) which can be found by using successive iterationNoting that wn+1(z)minuswn(z) satisfy the estimate (226) the limit w(z) of the sequenceof functions wn(z) satisfies the estimate

maxzisinD

w(z) = C[w(z) D] le e2M5M [(4+M3)m+1]Rprime= M6 (31)

and the solution u(z) of Problem P is as stated in (210) which satisfies the estimate

C1[u(z) D] le RlowastM6 + k2 = M7 (32)

where Rlowast = 2R1 + 2R Thus we have

3 General Hyperbolic Equations 51

Theorem 31 If Condition C holds then any solution u(z) of Problem P for thehyperbolic equation (116) satisfies the estimates

C1[u D] le M7 C1[u D] le M8k (33)

In the following we give the C1α(D)-estimates of solution u(z) for Problem P for

(116)

Theorem 32 If Condition C and (120) hold then any solution u(z) of ProblemP for the hyperbolic equation (116) satisfies the estimates

Cα[uz D] le M9 C1α[u D] le M10 C1

α[u D] le M11k (34)

in which k = k1 + k2 Mj = Mj(α k0 k1 k2 D) j = 9 10 M11 = M11(α k0 D) arenon-negative constants

Proof Similarly to Theorem 43 Chapter I it suffices to prove the first estimatein (34) Due to the solution u(z) of Problem P for (116) is found by the successiveiteration through the integral expression (216) we first choose the solution

u0(z) = 2Reint z

0w0(z)dz + b0 w0(z) = W (z) = ξ0e1 + η0e2 (35)

of Problem P for the equationuzz = 0 in D (36)

and substitute u0 w0 into the position of u w = ξe1 + ηe2 on the right-hand side of(116) where w0(z) is the same function with W (z) in (216) and w0(z) u0(z) satisfythe first estimates

Cα[w0 D] = Cα[Rew0D] + Cα[Imw0 D] le M12k2 C1α[u0 D] le M13k2 = M14 (37)

where Mj = Mj(α k0 D) j = 12 13 and then we have

u1(z) = 2Reint z

0w1(z)dz + b0 w1(z) = w0(z) + Φ1(z) + Ψ1(z)

Ψ1(z) = Ψ11(z)e1 +Ψ2

1(z)e2 Ψ11(z) =

int ν

0G1(z)dν

Ψ21(z) =

int micro

2RG2(z)dmicro G1(z) = G2(z) = Aξ0 +Bη0 + Cu0 +D

(38)

From (37) and the last three equalities in (38) it is not difficult to see thatΨ1

1(z) = Ψ11(micro ν) Ψ2

1(z) = Ψ21(micro ν) satisfy the Holder continuous estimates about

ν micro respectively namely

prime (39)

in which M15 = M15(α k0 k1 k2 D) there is no harm assuming that Rprime = max(2R12R) ge 1 By Condition C and (120) we can see that G1(z) Ψ1

1(z) G2(z) Ψ21(z)

about micro ν satisfy the the Holder conditions respectively ie

Cα[G1(micro middot) D] le M16Rprime Cα[Ψ1

1(micro middot) D] le M16Rprime

Cα[G2(middot ν) D] le M16Rprime Cα[Ψ2

1(middot ν) D] le M16Rprime

(310)

where M16 = M16(α k0 k1 k2 D) Moreover we can obtain the estimates of Ψ1(z)Φ1(z) as follows

Cα[Ψ1(z) D] le M17Rprime Cα[Φ1(z) D] le M17R

prime (311)

in which M17 = M17(α k0 k1 k2 D) Setting w11(z) = Re w1(z) + Im w1(z) w1

2(z) =Re w1(z) minus Im w1(z) w1(z) = w1(z) minus w0(z) and u1(z) = u1(z) minus u0(z) from (38)ndash(311) it follows that

Cα[w11(z) D] le M18R

prime Cα[w21(z) D] le M18R

prime

Cα[w1(z) D] le M18Rprime C1

α[u1(z) D] le M18Rprime

(312)

where M18 = M18(α k0 k1 k2 D) According to the successive iteration the esti-mates of functions w1

n(z) = Re wn(z) + Im wn(z) w2n(z) = Re wn(z) minus Im wn(z)

wn(z) = wn(z) minus wnminus1(z) and the corresponding function un(z) = un(z) minus unminus1(z)can be obtained namely

Cα[w1n(z) D] le

(M18Rprime)n

n Cα[w2

n(z) D] le(M18R

prime)n

n

Cα[wn(z) D] le (M18Rprime)n

n C1

α[un(z) D] le (M18Rprime)n

n

(313)

Therefore the sequences of functions

wn(z) = w0(z) +nsum

m=1wm(z) un(z) = u0(z) +

nsumm=1

um(z) (n = 1 2 )

uniformly converge to w(z) u(z) in D respectively and w(z) u(z) satisfy the esti-mates

Cα[w(z) D] le M9 = eM18Rprime C1

α[u(z) D] le M10 (314)

this is just the first estimate in (34)

32 The existence of solutions of Problem P for general hyperbolicequations of second order

Now we consider the general quasilinear equation of second order

uzz = F (z u uz) +G(z u uz)

F = Re [A1uz] + A2u+ A3

G = A4 uz σ +A5|u |τ z isin D

(315)

where F (z u uz) satisfies Condition C σ τ are positive constants and Aj(z u uz)(j = 4 5) satisfy the conditions in Condition C ie

C[Aj(z u uz) D] le k0 j = 4 5

and denote by Condition C prime the above conditions

Theorem 33 Let the complex equation (315) satisfy Condition C prime

(1) When 0 lt max(σ τ) lt 1 Problem P for (315) has a solution u(z) isin C11(D)

(2) When min(σ τ) gt 1 Problem P for (315) has a solution u(z) isin C11(D)

provided thatM19 = k1 + k2 + |b0|+ |b1| (316)

is sufficiently small

(3) In general the above solution of Problem P is not unique if 0ltmax(σ τ)lt1

Proof (1) Consider the algebraic equation for t

M8k1 + k2 + 2k0tσ + 2k0t

τ + |b1|+ |b0| = t (317)

where M8 is a non-negative constant in (33) it is not difficult to see that equation(317) has a unique solution t = M20 ge 0 Now we introduce a closed and convexsubsetBlowast in the Banach space C1(D) whose elements are the functions u(z) satisfyingthe conditions

u(z) isin C1(D) C1[u(z) D] le M20 (318)

We arbitrarily choose a function u0(z) isin Blowast for instance u0(z) = 0 and substitute itinto the position of u in coefficients of (315) and G(z u uz) from Theorems 23 and24 it is clear that problem P for

uzzminusRe [A1(z u0 u0z)uz]minusA2(z u0 u0z)uminusA3(z u0 u0z)=G(z u0 u0z) (319)

has a unique solution u1(z) isin Blowast By Theorem 31 we see that the solution u1(z)satisfies the estimate in (318) By using the successive iteration we obtain a sequenceof solutions um(z)(m = 1 2 ) isin Blowast of Problem P which satisfy the equations

um+1zz minus Re [A1(z um umz)um+1z]minus A2(z um umz)um+1

minusA3(z um umz) = G(z um umz) in D m = 1 2 (320)

and um+1(z) isin Blowast From (320) we see that um+1(z) = um+1(z)minus um(z) satisfies theequations and boundary conditions

um+1zz minus Re[A1um+1z]minus A2um+1

= G(z um umz)minus G(z umminus1 umminus1z) in D m = 1 2

Re [λ(z)(um+1(z)] = 0 on L Im [λ(z3)(um+1(z3)] = 0

(321)

Noting that C[G(z um umz)minus G(z umminus1 umminus1z) D] le 2k0M20 M20 is a solution ofthe algebraic equation (317) and according to Theorem 31

um+1 = C1[um+1 D] le M20 (322)

can be obtained Due to um+1 can be expressed as

um+1(z) = 2Reint z

0wm+1(z)dz wm+1(z) = w0(z) + Φm+1(z) + Ψm+1(z)

Ψm+1(z) =int xminusy

0[Aξm + Bηm + Cum + G]e1d(x minus y)

+int x+y

2R[Aξm + Bηm + Cum + G]e2d(x+ y)

(323)

in which the relation between A1 A2 G and A B C G is the same as that ofA1 A2 A3 and A B C D in Section 1 and G = G(z um umz) minus G(z umminus1 umminus1z)By using the method in the proof of Theorem 23 we can obtain

um+1 minus um = C1[um+1 D] le (M21Rprime)n

n

where M21 = 4M22M(M3 + 2)(2m0 + 1) Rprime = max(2R1 2R) m0 = w0(z) C(D)herein M22 = maxC[A Q] C[B Q] C[C Q] C[G Q] M = 1+ 4k2

0(1 + k20) From

the above inequality it is seen that the sequences of functions um(z) wm(z)ie

um(z)=u0(z)+[u1(z)minusu0(z)]+middot middot middot+[um(z)minusumminus1(z)](m=1 2 )

wm(z)=w0(z)+[w1(z)minusw0(z)]+middot middot middot+[wm(z)minuswmminus1(z)](m=1 2 )(324)

uniformly converge to the functions ulowast(z) wlowast(z) respectively and wlowast(z) satisfies theequality

wlowast(z) = w0(z) + Φlowast(z)

+int xminusy

0[Aσlowast +Bηlowast + Culowast +D]e1d(x minus y)

+int x+y

2R[Bσlowast + Aηlowast + Culowast +D]e2d(x+ y)

(325)

int z

is just a solution of Problem P for the nonlinear equation (315) in the closeddomain D

(2) Consider the algebraic equation for t

M8k1 + k2 + 2k0tσ + 2k0t

τ + |b0|+ |b1| = t (327)

it is not difficult to see that equation (327) has a solution t = M20 ge 0 providedthat M19 in (316) is small enough If there exist two solutions then we choosethe minimum of both as M20 Now we introduce a closed and convex subset Blowastof the Banach space C1(D) whose elements are of the functions u(z) satisfying theconditions

By using the same method in (1) we can find a solution u(z) isin Blowast of Problem P forequation (315) with min(σ τ) gt 1

(3) We can give an example to explain that there exist two solutions for equation(315) with σ = 0 τ = 12 namely the equation

uxx minus uyy = Au12 A = 8sgn(x2 minus y2) in D (329)

has two solutions u1(x y) = 0 and u2(x y) = (x2 minus y2)24 and they satisfy theboundary conditions

Re [λ(z)uz] = r(z) z isin L u(0) = b0 Im [λ(z)uz]|z=0 = b1 (330)

whereλ(z) = 1minus i z isin L1 λ(x) = 1 + i z isin L4

b0 = 0 r(z) = 0 z isin L = L1 cup L4 b1 = 0

33 The existence of solutions of Problem P for hyperbolic equations ofsecond order in general domains

1 The boundaries L3 L4 of the domain D are replaced by the curves Lprime3 L

prime4 and

the boundary of the domain Dprime is L1 cup L2 cup Lprime3 cup Lprime

4 where the parameter equationsof the curves Lprime

Lprime3 = x+ y = 2R l le x le R2 Lprime

4 = x+ y = micro y = γ1(x) 0 le x le l (331)

in which γ1(x) on 0 le x le l = γ1(l) is continuous and γ1(0) = 0 γ1(x) gt 0 on0 lt x le l and γ1(x) possesses the derivatives on 0 le x le l except some isolatedpoints and 1 + γprime

1(x) gt 0 By the condition we can find the inverse function x =τ(micro) = (micro+ ν)2 of x+ γ1(x) = micro and then ν = 2τ(micro)minus micro 0 le micro le 2R We make atransformation

0 le micro le 2R 2τ(micro)minus micro le ν le 2R1 (332)

2R1ν + 2τ(micro)minus micro 0 le micro le 2R 0 le ν le 2R1 (333)

Hence we have

x =12(micro+ ν) =

(334)

and

x =12(micro+ ν) =

(335)

Denote by z = x + jy = f(z) z = x + jy = fminus1(z) the transformations (334)and (335) respectively In this case setting w(z) = uz equation (116) in Dprime andboundary condition (21) on Lprime

3 cup Lprime4 can be rewritten in the form

ξν = Aξ +Bη + Cu+D ηmicro = Aξ +Bη + Cu+D z isin Dprime (336)

4 u(0) = b0 Im [λ(z3)w(z3)] = b1 (337)

in which z3 = l+ jγ1(l) λ(z) r(z) b1 on Lprime3 cupLprime

4 satisfy the condition (22) and u(z)and w(z) satisfy the relation

u(z) = 2Reint z

0w(z)dz + b0 in Dprime (338)

Suppose equation (116) in Dprime satisfies Condition C through the transformation(333) we have ξν = [2R1 minus 2τ(micro) + micro]ξν2R1 ηmicro = ηmicro system (336) is reduced to

ξν =[2R1 minus 2τ(micro) + micro][Aξ +Bη + Cu+D]

2R1 ηmicro = Aξ +Bη + Cu+D (339)

Moreover through the transformation (335) the boundary condition (337) on Lprime3cupLprime

4is reduced to

Re [λ(fminus1(z))w(fminus1(z))]=r(fminus1(z)) z isin L3 cup L4

in which z3 = f(z3) Therefore the boundary value problem (336)(337) is trans-formed into the boundary value problem (339)(340) On the basis of Theorems 22and 23 we see that the boundary value problem (339)(340) has a unique solutionw(z) and then w[f(z)] is just a solution of the boundary value problem (336)(337)

Theorem 34 If equation (116) satisfies Condition C in the domain Dprime with theboundary L1cupL2cupLprime

3cupLprime4 then Problem P with the boundary condition (337)(w = uz)

has a unique solution u(z) as stated in (338)

2 The boundaries L3 L4 of the domain D are replaced by two curves Lprimeprime3 Lprimeprime

4respectively and the boundary of the domain Dprimeprime is L1 cup L2 cup Lprimeprime

4 are as follows

Lprimeprime3 = x minus y = ν y = γ2(x) l le x le R2

Lprimeprime4 = x+ y = micro y = γ1(x) 0 le x le l

(341)

in which γ1(0) = 0 γ2(R2) = R2 minus 2R1 γ1(x) gt 0 0 le x le l γ2(x) gt 0 l le x leR2 γ1(x) on 0 le x le l γ2(x) on l le x le R2 are continuous and γ1(x) γ2(x) possessthe derivatives on 0 le x le l l le x le R2 except isolated points respectively and1+γprime

1(x) gt 0 1minusγprime2(x) gt 0 zprime

3 = l+jγ1(l) = l+jγ2(l) isin L3 (or L4) By the conditionswe can find the inverse functions x = τ(micro) x = σ(ν) of x+ γ1(x) = micro x minus γ2(x) = νrespectively namely

ν = 2τ(micro)minus micro 0 le micro le 2R micro = 2σ(ν)minus ν 0 le ν le l minus γ2(l) (342)

We first make a transformation

micro =2Rmicro

2σ(ν)minus ν ν = ν 0 le micro le 2σ(ν)minus ν 0 le ν le 2R1 (343)

micro =(2σ(ν)minus ν)micro

2R ν = ν 0 le micro le 2R 0 le ν le 2R1 (344)

The above transformation can be expressed by

x =12(micro+ ν) =

2R(x+ y) + (x minus y)[2σ(x minus γ2(x))minus x+ γ2(x)]2[2σ(x minus γ2(x))minus x+ γ2(x)]

2R(x+ y)minus (x minus y)[2σ(x minus γ2(x))minus x+ γ2(x)]2[2σ(x minus γ2(x))minus x+ γ2(x)]

(345)

x =12(micro+ ν) =

[2σ(x minus γ2(x))minus x+ γ2(x)](x+ y) + 2R(x minus y)4R

[2σ(x minus γ2(x))minus x+ γ2(x)](x+ y)minus 2R(x minus y)4R

(346)

Denote by z = x + jy = g(z) z = x + jy = gminus1(z) the transformation (345) andthe inverse transformation (346) respectively Through the transformation (344)we have

(u+ v)ν = (u+ v)ν (u minus v)micro =2σ(ν)minus ν

2R(u minus v)micro (347)

ξν = Aξ +Bη + Cu+D ηmicro =2σ(ν)minus ν

2R[Aξ +Bη + Cu+D] in Dprime (348)

Moreover through the transformation (346) the boundary condition (337) on Lprimeprime3cupLprimeprime

4is reduced to

Re [λ(gminus1(z))w(gminus1(z))] = r(gminus1(z)) z isin Lprime3 cup Lprime

4

in which z3 = g(zprime3) Besides the relation (338) is valid Therefore the boundary

value problem (336)(337) in Dprimeprime is transformed into the boundary value problem(348)(349) On the basis of Theorem 34 we see that the boundary value problem(348)(349) has a unique solution w(z) and then w[g(z)] is just a solution of theboundary value problem (336)(337) in Dprimeprime but we mention that the conditions ofcurve Lprime

3 Lprime4 through the transformation z = gminus1(z) must satisfy the conditions of

the curves in (331) For instance if z3 isin L3 γ1(x) ge x+2lminus2R on 2Rminus2l le x le lthen the above condition holds If z3 isin L4 γ2(x) ge 2l minusx on l le x le 2l then we cansimilarly discuss For other case it can be discussed by using the method as statedin Section 2 Chapter VI below

Theorem 35 If equation (116) satisfies Condition C in the domain Dprimeprime with theboundary L1 cup L2 cup Lprimeprime

3 cup Lprimeprime4 then Problem P with the boundary conditions

Re[λ(z)uz] = r(z) z isin Lprimeprime3 cup Lprimeprime

4 u(0) = b0 Im[λ(zprimeprime3 )uz(zprimeprime

3 )] = b1 (350)

has a unique solution u(z) as stated in (338) in Dprimeprime

By using the above method we can generalize the domain D to more generaldomain including the disk Dprimeprime = ||z minus R1|| lt R1 For the domain Dprimeprime we chooseR2 = 2R1 the boundary Lprimeprime of the domain Dprimeprime consists of Lprimeprime

1 Lprimeprime2 L

primeprime3 L

primeprime4 namely

Lprimeprime1 =

y = minusγ1(x) = minus

radicR2

Lprimeprime2 =

radicR2

1 minus (x minus R1)2 R1 le x le 2R1

Lprimeprime3 =

y = γ3(x) =

radicR2

Lprimeprime4 =

y = γ4(x) =

radicR2

where 1 + γprime1(x) gt 0 1 + γprime

4(x) gt 0 on 0 lt x le R1 1 minus γprime2(x) gt 0 1 minus γprime

3(x) gt 0 onR1 le x lt 2R1 The above curves can be rewritten as

Lprimeprime1 =

⎧⎨⎩x = σ1(ν) =R1 + ν minus

radicR2

1 + 2R1ν minus ν2

2

⎫⎬⎭

Lprimeprime2 =

⎧⎨⎩x = τ1(micro) =R1 + micro minus

radicR2

2

⎫⎬⎭

4 Other Oblique Derivative Problems 59

Lprimeprime3 =

⎧⎨⎩x = σ2(ν) =R1 + ν minus

radicR2

1 + 2R1ν minus ν2

2

⎫⎬⎭

Lprimeprime4 =

radicR2

2

⎫⎬⎭

where σ1(ν) σ2(ν) are the inverse functions of x + γ1(x) = ν x minus γ3(x) = ν on0 le ν le 2R1 and x = τ1(micro) x = τ2(micro) are the inverse functions of x minus γ2(x) =micro x+ γ4(x) = micro on 0 le micro le 2R1 respectively Through a translation we can discussthe unique solvability of corresponding boundary value problem for equation (116) inany disk ||z minus z0|| lt R where z0 is a hyperbolic number and R is a positive number

4 Other Oblique Derivative Problems for QuasilinearHyperbolic Equations of Second Order

In this section we discuss other oblique derivative problems for quasilinear hyperbolicequations Firstly the representation theorem of solutions for the above boundaryvalue problems is given moreover the uniqueness and existence of solutions for theabove problem are proved The results obtained include the corresponding result ofthe Dirichlet boundary value problem or the Darboux problem([12]3) as a specialcase

41 Formulation of other oblique derivative problems for quasilinearhyperbolic equations

We first state four other oblique derivative problems for equation (116) here thedomain D is the same as that in Section 1 but R2 = 2R1

Problem P1 Find a continuously differentiable solution u(z) of (116) in D satisfy-ing the boundary conditions

Re [λ(z)uz] = r(z) z isin L1 cup L2

u(0) = b0 Im [λ(z)uz]|z=z1 = b1(41)

where b0 b1 are real constants λ(z) = a(z) + jb(z) z isin L1 cup L2 and λ(z) r(z) b0 b1

satisfy the conditions

Cα[λ(z) L1 cup L2] le k0 Cα[r(z) L1 cup L2] le k2

|b0| |b1| le k2 maxzisinL1

zisinL2

1|a(z) + b(z)| le k0

(42)

If the boundary condition (41) is replaced by

Re [λ(z)uz] = r(z) z isin L1 Re [Λ(x)uz(x)] = R(x) x isin L0 = (0 R2)

u(0) = b0 Im [λ(z)uz]|z=z1 = b1(43)

where b0 b1 are real constants λ(z) = a(z) + jb(z) z isin L1 Λ(x) = 1 + j x isin L0and λ(z) r(z) R(x) b0 b1 satisfy the conditions

Cα[λ(z) L1] le k0 Cα[r(z) L1] le k2 |b0| |b1| le k2

Cα[R(x) L0] le k2 maxzisinL1

1|a(z)minus b(z)| le k0

(44)

in which α(0 lt α lt 1) k0 k2 are non-negative constants then the boundary valueproblem for (116) will be called Problem P2

If the boundary condition in (41) is replaced by

u(0) = b0 Im [λ(z)uz]|z=z1 = b1(45)

where b0 b1 are real constants λ(z) = a(z) + jb(z) z isin L2 Λ(x) = 1 minus j x isin L0and λ(z) r(z) R(x) b0 b1 satisfy the conditions

1|a(z) + b(z)| le k0

(46)

then the boundary value problem for (116) is called Problem P3 For Problem P2

and Problem P3 there is no harm in assuming that w(z1) = 0 otherwise throughthe transformation W (z) = w(z)minus [a(z1)minus jb(z1)][r(z1)+ jb1][a2(z1)minus b2(z1)] therequirement can be realized

u(x) = s(x) uy = R(x) x isin L0 = (0 R2) (47)

where s(z) R(x) satisfy the conditions

C1α[s(x) L0] le k2 Cα[R(x) L0] le k2 (48)

then the boundary value problem for (116) is called Problem P4

In the following we first discuss Problem P2 and Problem P3 for equation (211)

42 Representations of solutions and unique solvability of Problem P2

and Problem P3 for quasilinear hyperbolic equations

Similarly to Theorem 21 we can give the representation of solutions of Problem P2

and Problem P3 for equation (211) namely

Theorem 41 Any solution u(z) of Problem P2 for the hyperbolic equation (211)can be expressed as

u(z) = 2Reint z

0w(z)dz + b0 in D (49)

where w(z) is as follows

(410)

f(x+ y) = Re [(1 + j)uz(x+ y)] = R(x+ y) 0 le x+ y le 2R

g(xminusy)=2r((1minusj)(xminusy)2)minus[a((1minusj)(xminusy)2)+b((1minusj)(xminusy)2)]f(0)

0 le x minus y le 2R1

f(0) = u(z1) + v(z1) =r((1minus j)R1) + b1

a((1minus j)R1) + b((1minus j)R1)

(411)

As for Problem P3 for (211) its solution can be expressed as the forms (49) (410)but where f(x+ y) g(x minus y) possess the forms

f(x+ y) =2r((1+j)(x+y)2+(1minusj)R1)

a((1+j)(x+y)2+(1minusj)R1)minusb((1+j)(x+y)2+(1minusj)R1)

minus [a((1+j)(x+y)2+(1minusj)R1)minusb((1+j)(x+y)2+(1minusj)R1)]g(2R1)a((1 + j)(x+ y)2+(1minus j)R1)+b((1 + j)(x+ y)2+(1minus j)R1)

0 le x+ y le 2R (412)

g(2R1) = u(z1)minus v(z1) =r((1minus j)R1)minus b1

a((1minus j)R1)minus b((1minus j)R1)

g(x minus y) = Re [(1minus j)uz(x minus y)] = R(x minus y) 0 le x minus y le 2R1

Moreover the solution u(z) of Problem P2 and Problem P3 satisfies the estimate

C1α[u(z) D] le M23 C1

α[u(z) D] le M24k2 (413)

where M23 = M23(α k0 k2 D) M24 =M24(α k0 D) are two non-negative constants

Next we give the representation of solutions of Problem P2 and Problem P3 forthe quasilinear hyperbolic equation (116)

Theorem 42 Under Condition C any solution u(z) of Problem P2 for the hyper-bolic equation (116) can be expressed as

u(z) = 2Reint z

0w(z)dz + b0 w(z) = W (z) + Φ(z) + Ψ(z) in D

Ψ(z) =int ν

2R1

[Aξ +Bη + Cu+D]e1dν +int micro

0[Aξ +Bη + Cu+D]e2dmicro

(414)

where f(micro) g(ν) are as stated in (411) and f(micro) g(ν) are similar to f(micro) g(ν) in(411) but the functions r(z) R(x) b1 are replaced by the corresponding functionsminusRe [λ(z)Ψ(z)] minusRe [Λ(x)Ψ(x)] minusIm [λ(z1)Ψ(z1)] namely

Re [λ(z)Φ(z)] = minusRe [λ(z)Ψ(z)] z isin L1

Re [Λ(x)Φ(x)] = minusRe [Λ(x)Ψ(x)] x isin L0

Im [λ(z1)Φ(z1)] = minusIm [λ(z1)Ψ(z1)]

(415)

As to Problem P3 for (116) its solution u(z) possesses the expression (414) whereW (z) is a solution of Problem P3 for (211) and Φ(z) is also a solution of (211)satisfying the boundary conditions

(416)

Proof Let the solution u(z) of Problem P2 be substituted into the coefficients ofequation (116) Due to Problem P2 is equivalent to the Problem A2 for the complexequation (28) with the relation (210) where the boundary conditions are as follows

Re [λ(z)w(z)]=r(z) on L1 Re [Λ(x)w(x)]=R(x) on L0

Im [λ(z1)w(z1)] = b1(417)

According to Theorem 32 in Chapter I it is not difficult to see that the functionΨ(z) satisfies the complex equation

and noting that W (z) is the solution of Problem A2 for the complex equation (212)hence Φ(z) = w(z)minusW (z)minusΨ(z) = ξe1+ ηe2 satisfies the complex equation and theboundary conditions

ξmicroe1 + ηλe2 = 0 (419)

Re [λ(z)(ξe1 + ηe2)] = minusRe [λ(z)Ψ(z)] on L1

Re [Λ(x)(ξe1 + ηe2)] = minusRe [Λ(x)Ψ(x)] on L0

Im [λ(z)(ξe1 + ηe2)]|z=z1 = minusIm [λ(z1)Ψ(z1)]

(420)

The representation of solutions of Problem A2 for (116) is similar to (317) in ChapterI we can obtain the representation (414) of Problem P2 for (116) Similarly we canverify that the solution of Problem P3 for (116) possesses the representation (414)with the boundary condition (416)

By using the same method as stated in the proofs of Theorems 23 and 24 wecan prove the following theorem

Theorem 43 If the complex equation (116) satisfies Condition C then ProblemP2 (Problem P3) for (116) has a unique solution

43 Representations of solutions and unique solvability of Problem P4 forquasilinear hyperbolic equations

It is clear that the boundary condition (47) is equivalent to the following boundarycondition

ux = sprime(x) uy = R(x) x isin L0 = (0 R2) u(0) = s(0) ie

Re [(1 + j)uz(x)] = σ(x) Re [(1minus j)uz(x)] = τ(x) u(0) = r(0)(421)

in which

σ(x) =sprime(x) +R(x)

2 τ(x) =

sprime(x)minus R(x)2

(422)

for equation (211)

Theorem 44 Any solution u(z) of Problem P4 for the hyperbolic equation (211)can be expressed as (49) (410) where b0 = s(0) f(x+y) g(xminusy) possess the forms

f(x+ y) = σ(x+ y) =sprime(x+ y) +R(x+ y)

2 0 le x+ y le 2R

g(x minus y) = τ(x minus y) =sprime(x minus y)minus R(x minus y)

2 0 le x minus y le 2R1

(423)

and

f(0) =sprime(0) +R(0)

2 g(2R1) =

sprime(2R1)minus R(2R1)2

(424)

Moreover u(z) of Problem P4 satisfies the estimate (413)

Next we give the representation of Problem P4 for the quasilinear hyperbolicequation (116)

Theorem 45 Under Condition C any solution u(z) of Problem P4 for the hyper-bolic equation (116) can be expressed as (414) where b0 = s(0) W (z) is a solutionof Problem A4 for (212) satisfying the boundary condition (421) (W = uz) and Φ(z)is also a solution of (212) satisfying the boundary conditions

Re [(1+ j)Φ(x)] = minusRe [(1+ j)Ψ(x)] Re [(1minus j)Φ(x)] = minusRe [(1minus j)Ψ(x)] (425)

Proof Let the solution u(z) of Problem P4 be substituted into the coefficients ofequation (116) Due to Problem P4 is equivalent to the Problem A4 for the complexequation (28) with the relation (210) and the boundary conditions

Re [(1 + j)w(x)] = σ(z) Re [(1minus j)w(x)] = τ(x) x isin L0 (426)

according to Theorem 32 in Chapter I it can be seen that the function Ψ(z) satisfiesthe complex equation (418) and noting that W (z) is the solutions of Problem A4 forthe complex equation (212) ie (419) hence Φ(z) = w(z)minus W (z)minus Ψ(z) satisfiesequation (419) and boundary conditions

Re [(1 + j)(ξe1 + ηe2)] = minusRe [(1 + j)Ψ(x)]

Re [(1minus j)(ξe1 + ηe2)] = minusRe [(1minus j)Ψ(x)]x isin L0 (427)

Similarly to Theorem 42 the representation of solutions of Problem A4 for (28) issimilar to (317) in Chapter I we can obtain the representation (414) of Problem P4

for (116)

Theorem 46 If the complex equation (116) satisfies Condition C then ProblemP4 for (116) has a unique solution

Besides we can discuss the unique solvability of Problem P2 Problem P3 and Prob-lem P4 for general quasilinear hyperbolic equation (315) and generalize the aboveresults to the general domains Dprime with the conditions (331) and (341) respectively

Similarly to Problem P as in Section 2 we can discuss Problem P1 for equation(211) here the solution w(z) of equation (212) is as follows

f(0) = u(z1) + v(z1) =r((1minus j)R1) + b1

(428)

g(x minus y) =2r((1minusj)(xminusy)2)minus[a((1minusj)(xminusy)2)+b((1minusj)(xminusy)2)]f(0)

f(x+y)=2r((1+j)(x+y)2+(1minusj)R1)

minus [a((1+j)(x+y)2+(1minusj)R1)minusb((1+j)(x+y)2+(1minusj)R1)]g(2R1)a((1+j)(x+y)2+(1minusj)R1)+b((1+j)(x+y)2+(1minusj)R1)

0 le x+ y le 2R1

Moreover when we prove that Problem P1 of equation (116) has a unique solutionu(z) the integrals in (216)(222)(225)(228) and (233) possess the similar formsand the integral path in (221) can be chosen for instance the integral Ψ(z) in (216)is replaced by

Ψ(z) =int ν

2R1

0[Aξ +Bη + Cu+D]e1dmicro (430)

Now we explain that the Darboux problem is a special case of Problem P1 Theso-called Darboux problem is to find a solution u(z) for (116) such that u(z) satisfiesthe boundary conditions

u(x) = s(x) x isin L0 u(z) = φ(x) z = x+ jy isin L1 (431)

where s(x) φ(x) satisfy the conditions

C1[s(x) L0] le k2 C1[φ(x) L1] le k2 (432)

herein k2 is a non-negative constant [12]3) From (431) we have

u(x) = s(x) x isin L0 Re [(1minusj)uz]=φx(x) = φprime(x) z isin L1

u(0) = s(0) Im [(1minusj)uz]|z=z1minus0 = minusφprime(1) ie

u(x)=s(x) x isin L0 Re [λ(z)uz] = r(z) z isin L1

u(0) = b0 Im [λ(z)uz]|z=z1=b1

(433)

where we choose

λ(z)=(1minusj)radic2 r(z)=φprime(x)

radic2 z isin L1 b0 = s(0) b1=minusφprime(1)

radic2 (434)

it is easy to see that the boundary conditions (433)(434) possess the form of theboundary condition (41) This shows that the above Darboux problem is a specialcase of Problem P2

For more general hyperbolic equations of second order the corresponding boundaryvalue problems remain to be discussed

5 Oblique Derivative Problems for Degenerate HyperbolicEquations of Second Order

This section deals with the oblique derivative problem for the degenerate hyperbolicequation in a simply connected domain We first give the representation theorem ofsolutions of the oblique derivative problem for the hyperbolic equation and then byusing the method of successive iteration the existence and uniqueness of solutionsfor the above oblique derivative problem are proved

51 Formulation of the oblique derivative problem for degenerate hyper-bolic equations

It is known that the Chaplygin equation in the hyperbolic domain D possesses theform

K(y)uxx + uyy = 0 in D (51)

where K(y) possesses the first order continuous derivative K prime(y) and K prime(y) gt 0 ony1 lt y lt 0 K(0) = 0 and the domain D is a simply connected domain with theboundary L = L0 cup L1 cup L2 herein L0 = (0 2)

L1=x+int y

0

radicminusK(t)dt=0 xisin(0 1)

L2=

xminusint y

0

are two characteristic lines and z1 = x1 + jy1 = 1 + jy1 is the intersection point ofL1 and L2 In particular if K(y) = minus|y|m m is a positive constant thenint y

0

radicminusK(t)dt =

int y

0|t|m2dt = minus

int |y|

0d

2m+ 2

|t|(m+2)2 = minus 2m+ 2

|y|(m+2)2

In this case the two characteristic lines L1 L2 are as follows

L1 x minus 2m+ 2

|y|(m+2)2 = 0 L2 x+2

m+ 2|y|(m+2)2 = 2 ie

L1 y = minus(m+ 22

x)2(m+2) L2 y = minus[m+ 22

(2minus x)]2(m+2)

In this section we mainly consider the general Chaplygin equation of second order

K(y)uxx + uyy = dux + euy + fu+ g in D (52)

where DK(y) are as stated in (51) its complex form is the following equation ofsecond order

uzz = Re [Quzz + A1uz] + A2u+ A3 z isin D (53)

where

Q =K(y) + 1K(y)minus 1

A1 =d+ je

K(y)minus 1 A2 =

f

2(K(y)minus 1) A3 =

g

2(K(y)minus 1)

5 Degenerate Hyperbolic Equations 67

and assume that the coefficients Aj(z)(j = 1 2 3) satisfy Condition C It is clearthat equation (52) is a degenerate hyperbolic equation

The oblique derivative boundary value problem for equation (52) may be formu-lated as follows

Problem P1 Find a continuous solution u(z) of (53) in D which satisfies theboundary conditions

12

partu

u(0) = b0 Im [λ(z)uz]|z=z1 = b1

(54)

where l is a given vector at every point z isin L uz = [radic

minusK(y)ux + juy]2 umacrz =

[radic

minusK(y)ux minus juy]2 b0 b1 are real constants λ(z) = a(x) + jb(x) = cos(l x) +j cos(l y) z isin L and λ(z) r(z) b0 b1 satisfy the conditions

C1α[λ(z) Lj] le k0 C1

α[r(z) Lj] le k2 j = 1 2

zisinL2

1|a(z) + b(z)| le k0

(55)

in which α (12 lt α lt 1) k0 k2 are non-negative constants For convenience we canassume that uz(z1) = 0 ie r(z1) = 0 b1 = 0 otherwise we make a transformationuz minus [a(z1) minus jb(z1)][r(z1) + jb1][a2(z1) minus b2(z1)] the requirement can be realizedProblem P1 with the conditions A3(z) = 0 z isin D r(z) = 0 z isin L and b0 = b1 = 0will be called Problem P0

For the Dirichlet problem (Tricomi problem D) with the boundary condition

u(x) = φ(x) on L1 = AC =x = minus

int y

0

radicminusK(t)dt 0 le x le 1

L2 = BC =x = 2 +

int y

0

(56)

we find the derivative for (56) according to s = x on L = L1 cup L2 and obtain

us = ux + uyyx = ux minus (minusK(y))minus12uy = φprime(x) on L1

us = ux + uyyx = ux + (minusK(y))minus12uy = φprime(x) on L2 ie

(minusK(y))12U + V = (minusK(y))12φprime(x)2 = r(x) on L1

(minusK(y))12U minus V = (minusK(y))12φprime(x)2 = r(x) on L2 ie

Re [(1minus j)(U + jV )] = U minus V = r(x) on L1

Im [(1minus j)(U + jV )] = [minusU + V ]|z=z1minus0 = minusr(1minus 0)

Re [(1 + j)(U + jV )] = U + V = r(x) on L2

Im [(1 + j)(U + jV )] = [U + V ]|z=z1+0 = r(1 + 0)

whereU =

radicminusK(y)ux2 = U V = minusuy2 = minusV

a+ jb = 1minus j a = 1 = b = minus1 on L1

a+ jb = 1 + j a = 1 = minusb = minus1 on L2

From the above formulas we can write the complex forms of boundary conditions ofU + jV

Re [λ(z)(U + jV )] = r(z) z isin Lj (j = 1 2)

λ(z)=

⎧⎨⎩1minus j = a+ jb

1 + j = a minus jbr(x)=

⎧⎨⎩(minusK(y))12φprime(x)2 on L1

(minusK(y))12φprime(x)2 on L2

andu(z) = 2Re

int z

0(U minus jV )dz + φ(0) in D (57)

Hence Problem D is a special case of Problem P1

52 Unique solvability of Problem P for Chaplygin equation (51) in thehyperbolic domain D

In the subsection we discuss the Chaplygin equation (51) in the hyperbolic domainD where the arcs L1 = AC L2 = BC are the characteristics of (51) ie

x+int y

0

radicminusK(t)dt = 0 0 le x le 1 x minus

int y

0

radicminusK(t)dt = 2 1 le x le 2 (58)

Setting thatmicro = x+

int y

0

radicminusK(t)dt ν = x minus

int y

0

radicminusK(t)dt (59)

and thenmicro+ ν = 2x micro minus ν = 2

int y

0

radicminusK(t)dt

(micro minus ν)y = 2radic

minusK(y)radic

minusK(y) = (micro minus ν)y2

xmicro = 12 = xν ymicro = 12radic

minusK(y) = minusyν

(510)

hence we haveUx = Umicro + Uν Vy =

radicminusK(y)(Vmicro minus Vν)

Vx = Vmicro + Vν Uy =radic

minusK(y)(Umicro minus Uν)(511)

andK(y)Ux minus Vy = K(y)(Umicro + Uν)minus

radicminusK(y)(Vmicro minus Vν) = 0

Vx + Uy = Vmicro + Vν +radic

minusK(y)(Umicro minus Uν) = 0 in D(512)

where U = ux2 V = minusuy2 and U =radic

minusK(y)U V = minusV Noting that

(radicminusK(y)

)micro= minus1

2(minusK)minus12K prime(y)ymicro =

K prime

4K(radic

minusK(y))

ν= minusK prime4K (513)

we have(U minus V )micro =

radicminusK(y)Umicro + Vmicro +

K primeU4K

(U + V )ν =radic

minusK(y)Uν minus Vν minus K primeU4K

(514)

Moreover by (511) and (514) we obtain

minus2radic

minusK(y)(U + V )ν

=radic

minusK(y)[(U + V )micro minus (U + V )ν ]minusradic

minusK(y)[(U + V )micro + (U + V )ν ]

= minusradic

minusK(y)(U + V )x + (U + V )y

= minus12(minusK(y))minus12K prime(y)U =

K prime(y)2K(y)

U 2radic

minusK(y)(U minus V )micro

=radic

minusK(y)[(U minus V )micro + (U minus V )ν ]

+radic

minusK(y)[(U minus V )micro minus (U minus V )ν)]

=radic

minusK(y)(U minus V )x + (U minus V )y

K prime(y)2K(y)

U

(515)

Thus equation (515) can be written as a system of equations

(U+V )ν=minus(minusK)minus12 K prime(y)4K(y)

U (UminusV )micro=(minusK)minus12 K prime(y)4K(y)

U ie

Wmacrz=12[(minusK(y))12WxminusjWy]=A1W (z)+A2W (z) in ∆

(516)

in which W = U + jV A1(z)=A2(z)=minusjK prime8K in ∆ = 0 le micro le 2 0 le ν le 2and

u(z) = 2Reint z

0(U minus jV )dz + b0 in D (517)

where U minus jV = (minusK)minus12U + jV

In the following we first give the representation of solutions for the oblique deriva-tive problem (Problem P1) for system (516) inD For this we first discuss the systemof equations

(U + V )ν = 0 (U minus V )micro = 0 in D (518)

with the boundary condition12

partu

partl= Re [λ(z)(U + jV )] = r(z) z isin L

u(0) = b0 Im [λ(z)(U + iV )]|z=z1 = b1

(519)

in which λ(z) = a(z) + jb(z) on L1 cup L2 Similarly to Chapter I the solution ofProblem P1 for (518) can be expressed as

ξ = U + V = f(micro) η = U minus V = g(ν)

U(x y) =f(micro) + g(ν)

2 V (x y) =

f(micro)minus g(ν)2

ie W (z) =(1 + j)f(micro) + (1minus j)g(ν)

2

(520)

in which f(t) g(t) are two arbitrary real continuous functions on [0 2] For conve-nience denote by the functions a(x) b(x) r(x) of x the functions a(z) b(z) r(z) of zin (519) thus the first formula in (519) can be rewritten as

a(x)U(x y) + b(x)V (x y) = r(x) on L ie

[a(x)minusb(x)]f(x+y)+[a(x)+b(x)]g(xminusy)=2r(x) on L ie

[a(t2) + b(t2)]f(0) + [a(t2)minus b(t2)]g(t) = 2r(t2) t isin [0 2][a(t2+1)minusb(t2+1)]f(t)+[a(t2+1)+b(t2+1)]g(2)=2r(t2+1) t isin [0 2]

where

f(0) = U(0) + V (0) =r(1) + b1

a(1) + b(1) g(2) = U(2)minus V (2) =

r(1)minus b1

a(1)minus b(1)

Noting that the boundary conditions in (519) we can derive

U =12

f(micro) +

2r(ν2)minus (a(ν2) + b(ν2))f(0)a(ν2)minus b(ν2)

V =12

f(micro)minus 2r(ν2)minus (a(ν2) + b(ν2))f(0)

a(ν2)minus b(ν2)

or

U =12

g(ν) +

2r(micro2 + 1)minus (a(micro2 + 1)minus b(micro2 + 1))g(2)a(micro2 + 1) + b(micro2 + 1)

V =12

minusg(ν) +

(521)

if a(x)minus b(x) = 0 on [01] and a(x) + b(x) = 0 on [12] respectively From the aboveformulas it follows that

Re[(1+j)W (x)]= U+V =2r(x2+1)minus(a(x2+1)+b(x2+1))f(0)

a(x2+1)minusb(x2+1)

Re[(1minusj)W (x)]= UminusV =2r(x2)minus(a(x2)minusb(x2))g(2)

a(x2) + b(x2) x isin [0 2]

(522)

if a(x)minus b(x) = 0 on [01] and a(x) + b(x) = 0 on [12] respectively From (522)

W (z) =

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

12

[(1 + j)

+(1minus j)2r(ν2)minus (a(ν2) + b(ν2))f(0)

a(ν2)minus b(ν2)

] (523)

can be derived

Next we find the solution of Problem P1 for system (516) From (516) we have

U + V = minusint ν

2(minusK)minus12 K prime(y)

4K(y)Udν U minus V =

int micro

4K(y)Udmicro

W = U+jV =minus1+j

2

int ν

4K(y)Udν+

1minusj

2

int micro

4K(y)Udmicro

= minus1 + j

2

int ν

2

K prime(y)4K(y)

Udν +1minus j

2

int micro

0

K prime(y)4K(y)

Udmicro

the above last two integrals are along two characteristic lines s2 and s1 respectivelyBut according to the method in [66]1) if we denote by s1 the member of the familyof characteristic lines dxdy = minus

radicminusK(y) and by s2 the member of the family of

characteristic lines dxdy =radic

minusK(y) passing through the point P isin D and

ds1 =radic(dx)2 + (dy)2 =

radic1 +

(dy

dx

)2dx =

radic1minus K

minusKdx = minusradic

1minus Kdy

radic1 +

(dy

dx

)2dx =

radic1minus K

minusKdx =

radic1minus Kdy

(524)

then system (51) can be rewritten in the form

(U+V )s1 = (U+V )xxs1+(U+V )yys1

=1radic1minusK

[radicminusK(y)(U+V )xminus(U+V )y

]

=1

2radic1minusK

(minusK(y))minus12K prime(y)U=minus 12radic1minusK

K prime(y)K(y)

U

(U minusV )s2 = (U minusV )xxs2+(U minusV )yys2

=1radic1minusK

[radicminusK(y)(U minusV )x+(U minusV )y

]

= minus 12radic1minusK

(minusK(y))minus12K prime(y)U=1

2radic1minusK

K prime(y)K(y)

U

(525)

thus we obtain the system of integral equations

ξ = U + V = minusint s1

0

1

2radic(1minus K)

K prime(y)K(y)

Uds1 =int s1

0

12

K prime(y)K(y)

Udy

η = U minus V =int s2

0

1

2radic(1minus K)

K prime(y)K(y)

Uds2 =int s2

0

12

K prime(y)K(y)

Udy

(526)

where the integrals are along the two families of characteristics s1 and s2 respectivelySimilarly to next subsection the solution U+V UminusV can be obtained by the methodof successive iteration which can be expressed as

W (z) = U + jV = W (z) + Φ(z) + Ψ(z)

U+V =minusint s1

0

1

2radic(1minusK)

K prime(y)K(y)

Uds1 UminusV =int s2

0

1

2radic(1minusK)

K prime(y)K(y)

Uds2(527)

whereW (z) Φ(z) are the solutions of system (518) satisfying the boundary condition(519) respectively but the function r(z) b1 in the boundary condition of Φ(z) shouldbe replaced by minusRe [λ(z)Ψ(z)] on L1cupL2 and minusIm [λ(z1)Ψ(z1)] and then the function

u(z) = 2Reint z

0(U minus jV )dz + b0 = 2Re

int z

0[(minusK(y))minus12U + jV ]dz + b0 in D (528)

is just the solution of Problem P1 for (516) and the solution is unique Here wemention that firstly it suffices to consider the case of y1 le y le minusδ where δ is asufficiently small positive number and when we find out the solution of Problem P1

for equation (51) with the condition y1 le y le minusδ and then let δ rarr 0

Theorem 51 If the Chaplygin equation (51) in D satisfies the above conditionsthen Problem P1 for (51) in D has a unique solution

Finally we mention that the boundary condition of Problem P1

Re [λ(z)(U + jV )] = r(z) z isin L1 cup L2

can be replaced by

2U(x)=S(x)=radic

minusK[y(x)]ux ie u(x)=int x

0

S(x)dxradicminusK[y(x)]

+u(0)=s(x)

2V (x) = R(x) = uy on AB = L0 = (0 2)

(529)

where Cα[R(x) AB] le k2 lt infin C1α[s(x) AB] le k2

53 Unique solvability of the oblique derivative problem for degeneratehyperbolic equations

In this subsection we prove the uniqueness and existence of solutions of Problem P4

for the degenerate hyperbolic equation (52) the boundary condition of Problem P4

is as followsu(x) = s(x) uy(x) = R(x) on L0 = (0 2) (530)

where s(x) R(x) satisfy the condition C2α[s(x) L0] C2

α[R(x) L0] le k2 the aboveboundary value problem is also called the Cauchy problem for (52) Making a trans-formation of function

v(z) = u(z)minus yR(x)minus s(x) in D (531)

equation (52) and boundary condition (530) are reduced to the form

K(y)vxx + vyy = dvx + evy + fv +G

G = g + f(yR + s) + eR + d[yRprime(x) + sprime(x)]

minusK(y)[yRprimeprime(x) + sprimeprime(x)] in D

(532)

v(x) = 0 vy(x) = 0 on L0 (533)

Hence we may only discuss Cauchy problem (532)(533) and denote it by ProblemP4 again According to Subsection 43 Problem P4 for (532) is equivalent to theboundary value problem A for the hyperbolic system of first order equations therelation and the boundary conditions

ξs1 =2radicminusKradic1minus K

ξν =1

2radic1minus K

[( minusdradicminusKminus e minus 1

2K prime(y)K(y)

)ξ

+( minusdradicminusK

+ e minus 12

K prime(y)K(y)

)η minus fv minus G

]

ηs2 =2radicminusKradic1minus K

ηmicro =1

2radic1minus K

[( minusdradicminusKminus e+

12

K prime(y)K(y)

)ξ

+ e+12

K prime(y)K(y)

]

ξ = U + V η = U minus V vy = ξ minus η v(x) = 0

(534)

In particular if K(y) = minus|y|mh(y) m is a positive constant then

K prime(y)K(y)

=m|y|mminus1h(y)

K(y)minus |y|mhy

K(y)=

m

y+

hy

h

Integrating the hyperbolic system in (534) along the characteristics s1 s2 we obtainthe system of integral equations as follows

v(z) =int y

0(ξ minus η)dy in D

ξ(z) = minusint y

0[A1ξ +B1η + C1(ξ + η) +Dv + E]dy z isin s1

η(z) =int y

(535)

in this case

A1 = minuse

2minus hy

4h B1 =

e

2minus hy

4h A2 = minuse

2+

hy

4h B2 =

e

2+

hy

4h

C1 = minus12

dradicminusKminus m

4y C2 = minus1

2dradicminusK

+m

4y D = minusf

2 E = minusG

2

In the following we may only discuss the case ofK(y) = minus|y|mh(y) because otherwiseit can be similarly discussed In order to find a solution of the system of integralequations (535) we need to add the condition

limyrarr0

|y|d(x y)|y|m2 = 0 ie d(x y) asymp ε(y)|y|m2minus1 (536)

where ε(y) rarr 0 as y rarr 0 It is clear that for two characteristics s01 x = x1(y z0)

s02 x = x2(y z0) passing through P0 = z0 = x0 + jy0 isin D we have

|x1 minus x2| le 2|int y

0

radicminusKdy| le M |y|m2+1 for yprime lt y lt 0 (537)

for any z1 = x1 + jy isin s01 z2 = x2 + jy isin s0

2 where M(gt max[4radic

h(y)(m + 2) 1])is a positive constant Suppose that the coefficients of (535) possess continuouslydifferentiable with respect to x isin L0 and satisfy the condition

|Aj| |Ajx| |Bj| |Bjx| |yCj| |yCjx| |D| |Dx| |E||Ex| |1radich| |hyh| le M z isin D j = 1 2

(538)

According to the proof of Theorem 51 it is sufficient to find a solution of ProblemP4 for arbitrary segment minusδ le y le 0 where δ is a sufficiently small positive numberand choose a positive constant γ(lt 1) close to 1 such that the following inequalitieshold

3Mδ

2+[ε(y)M +m2]δm2

m+ 2lt γ

6δ2M2

m+ 6+8δMm+ 2

+2ε(y)M +m

m+ 2lt γ

(539)

Similar to [66]1) a solution of Problem P4 for (535) on minusδ lt y lt 0 can be foundFirstly let y0 isin (minusδ 0) and D0 be a domain bounded by the boundary y = 0 s0

1 s02

we choose v0 = 0 ξ0 = 0 η0 = 0 and substitute them into the corresponding positionsof v ξ η in the right-hand sides of (535) and obtain

ξ1(z)=minusint y

0[A1ξ0+B1η0+C1(ξ0+η0)+Dv0 + E]dy=minus

int y

0Edy z isin s0

1

η1(z)=int y

0[A2ξ0+B2η0+C2(ξ0+η0)+Dv0+E]dy=

int y

0Edy zisin s0

2

v1(z) = Reint y

0(ξ0 minus η0)dy = 0 in D0

(540)

By the successive iteration we find the sequences of functions vk ξk ηk whichsatisfy the relations

ξk+1(z) = minusint y

0[A1ξk +B1ηk + C1(ξk + ηk) +Dvk + E]dy z isin s0

1

ηk+1(z) =int y

2

vk+1(z) =int y

0(ξk minus ηk)dy in D0 k = 0 1 2

(541)

We can prove that vk ξk ηk in D0 satisfy the estimates

|vk(z)| |ξk(z)| |ηk(z)| le Mksum

j=0γj|y| |ξk(z) + ηk(z)|

j=0γj|y|m2+1

|vk+1(z)minus vk(z)| |ξk+1(z)minus ξk(z)| |ηk+1(z)minus ηk(z)| le Mγk|y||ξk+1(z) + ηk+1(z)minus ξk(z)minus ηk(z)| |vk+1(z1)minus vk+1(z2)

minusvk(z1)minus vk(z2)| |ξk+1(z1)minus ξk+1(z2)minus ξk(z1) + ξk(z2)||ηk+1(z1)minus ηk+1(z2)minus ηk(z1) + ηk(z2)| le Mγk|y|m2+1

(542)

In fact from (540) it follows that the first formula with k = 1 holds namely

|v1(z)|=0 le M |y| |ξ1(z)| le M |y| |η1(z)| le M |y|=Mγ0|y| le M1sum

j=0γj|y|

Moreover we get

|v1(z1)minusv1(z2)|=0|ξ1(z)+η1(z)|le|

int y

0[E(z1)minusE(z2)]dy| le2|

int y

0Ex[x1minusx2]dy|

le 4m+4

M2|y|m2+2 leMγ|y|m2+1 leM1sum

j=0γj|y|m2+1

|ξ1(z1)minusξ1(z2)| le |int y

0[E(xj(t z1)+jt)minusE(xj(t z2)+jt)]dt|

le |int y

0|Ex||xj(t z1)minus xj(t z2)|dy| le M |

int y

0|x1 minus x2|dy|

le M |int y

0M |y|m2+1dy| le Mγ|y|m2+1 le M

1sumj=0

γj|y|m2+1

|η1(z1)minusη1(z2)|= |int y

0[E(xj(t z1)+jt)minusE(xj(t z2)+jt)]dt| leM

1sumj=0

γj|y|m2+1

|v1(z)minus v0(z)| = |v1(z)| le Mγ|y| |ξ1(z)minus ξ0(z)| = |ξ1(z)| le Mγ|y||η1(z)minusη0(z)|= |η1(z)|leMγ|y| |v1(z1)minusv1(z2)minusv0(z1)minusv0(z2)|leMγ|y|m2+1

|ξ1(z)+η1(z)minusξ0(z)minusη0(z)|= |ξ1(z)+η1(z)|leγ|y|m2+1 leMγ|y|m2+1

|ξ1(z1)minusξ1(z2)minusξ0(z1)+ξ0(z2)|= |ξ1(z1)minusξ1(z2)| le Mγ|y|m2+1

|η1(z1)minusη1(z2)minusη0(z1)+η0(z2)|= |η1(z1)minusη1(z2)|leMγ|y|m2+1

In addition we use the inductive method namely suppose the estimates in (542) fork = n are valid then they are also valid for k = n + 1 In the following we onlygive the estimates of |vn+1(z)| |ξn+1(z)| |ξn+1 + ηn+1(z)| the other estimates can besimilarly given From (541) we have

|vn+1(z)| le |int y

0[ξnminusηn]dy|le2M |

int y

0

nsumj=0

γjydy|leMnsum

j=0γj|y|2 leM

n+1sumj=0

γj|y|

|ξn+1(z)| le |int y

0

⎡⎣(|A1|+|B1|+|D|)Mnsum

j=0γj|y|+|C1|M

nsumj=0

γj|y|m2+1+|E|⎤⎦ dy|

le M |int y

0

⎡⎣3M nsumj=0

γj|y|+(

ε(y)2|y|radich

+m

4|y|)

nsumj=0

γj|y|m2+1 + 1

⎤⎦ dy|

le M |y|⎧⎨⎩[32M |y|+

(|ε(y)|M +

m

2

) |y|m2

m+ 2

]nsum

j=0γj + 1

⎫⎬⎭ le Mn+1sumj=0

γj|y|

and

|ξn+1(z)+ηn+1(z)| le |int y

0

2sumj=1[A2(z2)ξn(z2)minusA1(z1)ξn(z1)+B2(z2)ηn(z2)

minusB1(z1)ηn(z1)] + C2(z2)(ξn(z2) + ηn(z2))

minusC1(z1)(ξn(z1) + ηn(z1))

+D(z2)vn(z2)minus D(z1)vn(z1) + E(z2)minus E(z1)]

dy|

Noting that

|D(z2)vn(z2)minus D(z1)vn(z1)| = |[D(z2)minus D(z1)]vn(z2) +D(z1)

times[vn(z2)minus vn(z1)]|

lensum

j=0γj|y||D(z2)minus D(z1)|+M2

nsumj=0

γj|y|m2+1

le M2nsum

j=0γj|y||x2minusx1|+M2

nsumj=0

γj|y|m2+1

le (M |y|+1)M2nsum

j=0γj|y|m2+1

and

|A2(z2)ξn(z2)minus A1(z1)ξn(z1) +B2(z2)ηn(z2)minus B1(z1)ηn(z1)|le|[A2(z2)minusA2(z1)]ξn(z2)+[A2(z1)minusA1(z1)]ξn(z2)+A1(z1)[ξn(z2)minusξn(z1)]

+[B2(z2)minusB2(z1)]ηn(z2)+B1(z1)[ηn(z2)minusηn(z1)]|+[B2(z1)minusB1(z1)]ηn(z2)

le 2M |y|[M |x1 minus x2|+ |y|m2]nsum

j=0γj +

∣∣∣∣∣hy

2h

∣∣∣∣∣ |ξn(z2)minus ηn(z2)|

le (2M |y|+ 3)M2nsum

j=0γj|y|m2+1

we get

|ξn+1(z) + ηn+1(z)| le |int y

0[(3M |y|+ 4)M2

nsumj=0

γj|y|m2+1

+(|C1|+ |C2|)Mnsum

j=0γj|y|m2+1 +M2|y|m2+1]dy|

le M |y|m2+1

[6M2y2

m+ 6+8M |y|m+ 2

+(|ε(y)|M +

m

2

)

times 2m+ 2

nsumj=0

γj +2M

m+ 4|y|⎤⎦

le Mn+1sumj=0

γj|y|m2+1

On the basis of the estimates (542) we can derive that vk ξk ηk in D0

uniformly converge to vlowast ξlowast ηlowast satisfying the system of integral equations

ξlowast(z) = minusint y

0[A1ξlowast +B1ηlowast + C1(ξlowast + ηlowast) +Dvlowast + E]dy z isin s1

ηlowast(z) =int y

vlowast(z) =int y

0(ξlowast minus ηlowast)dy in D0

and the function vlowast(z) satisfies equation (532) and boundary condition (533) henceulowast(z) = vlowast(z) + yR(x) + s(x) is a solution of Problem P4 for (52) From the abovediscussion we can see that the solution of Problem P4 for (52) in D is unique

Theorem 52 If the equation (52) in D satisfies the above conditions then ProblemP4 for (52) in D has a unique solution

Now we mention that if we denote

W (z)= U+jV = |y|m2UminusjV =12[|y|m2ux+juy]

W (z)= UminusjV = |y|m2U+jV =12[|y|m2uxminusjuy]

then W (z) = |y|m2UminusjV is a solution of the first order hyperbolic complex equation

Wmacrz = A1(z)W + A2(z)W + A3(z)u+ A4(z) in D

A1 = minus d

4|y|m2 + j

(m

8|y| minus e

4

) A3=minusf

4

A2 = minus d

4|y|m2 + j

(m

8|y| +e

4

) A4=minusg

4

(543)

and

u(z)=2Reint z

0uzdz=2Re

int z

0(UminusjV )d(x+jy) =2Re

int z

0(U+jV )d(xminusjy)

is a solution of equation (52) with K(y) = minus|y|mBy using the similar method we can prove the solvability of Problem P1 Problem

P2 and Problem P3 for equation (52) Moreover for general domain Dprime with non-characteristics boundary we can also discuss the solvability of Problem P1 ProblemP2 Problem P3 and Problem P4 for equation (52) Besides we can discuss thesolvability of corresponding boundary value problems for the hyperbolic equation inthe form

uxx +K(y)uyy = dux + euy + fu+ g in D (544)

under certain conditions where K(y) is as stated in (52)

The references for the chapter are [2][7][12][13][24][25][34][41][44][47][54][60][66][70][79][85][87][89][95]

CHAPTER III

NONLINEAR ELLIPTIC COMPLEXEQUATIONS OF FIRST ANDSECOND ORDER

In this chapter we discuss the representation and existence of solutions of discontinu-ous boundary value problems for nonlinear elliptic complex equations of first andsecond order which will be used in latter chapters

1 Generalizations of KeldychndashSedov Formula for AnalyticFunctions

It is known that the KeldychndashSedov formula gives the representation of solutions ofthe mixed boundary value problem for analytic functions in the upper half-plane (see[53]) But for many problems in mechanics and physics one needs a more generalformulas of solutions of the discontinuous RiemannndashHilbert boundary value problemfor analytic functions in the upper half-plane and other special domains In thissection we shall establish the representations of solutions of the general discontinuousboundary value problem for analytic functions in the upper half-plane and upper half-unit disk In the following sections and chapters we shall give applications to somenonlinear elliptic complex equations and quasilinear equations of mixed type

11 General discontinuous boundary value problem for analytic functionsin the upper half-plane

Let D be the upper half-plane and a(x) b(x) c(x) be known real functions on L =minusinfin lt x lt infin y = 0 where a(x) b(x) possess discontinuities of first kind at mdistinct points xj(j = 1 m minusinfin lt x1 lt middot middot middot lt xm lt infin) m is a positive integerand c(x) = O(|xminusxj|minusβj) in the neighborhood of xj(j = 1 2 m) on L herein βj(lt1 j = 1 2 m) are non-negative constants such that βj+γj lt 1 γj(j = 1 m)are as stated in (13) below Denote λ(x) = a(x)minus ib(x) and |a(x)|+ |b(x)| = 0 thereis no harm in assuming that |λ(x)| = 1 x isin Llowast = Lx1 xm Suppose thatλ(x) c(x) satisfy the conditions

λ(x) isin Cα(Lj) |x minus xj|βjc(x) isin Cα(Lj) j = 1 2 m (11)

80 III Elliptic Complex Equations

where Lj is the line segment from the point xjminus1 to xj on L x0 = xm Lj(j =1 2 m) do not include the end points L1 = x lt x1 cup x gt xm α(0 lt α lt 1)is a constant and the function λ(x) isin Cα(Linfin)(Linfin is a neighborhood of the point infin)is indicated as λ(1x) isin Cα(Llowast) here Llowast(sub L) is a neighborhood of the point x = 0

The discontinuous RiemannndashHilbert boundary value problem for analytic functionsin D may be formulated as follows

Problem A Find an analytic function Φ(z) = u(z)+iv(z) in D which is continuousin Dlowast = Dx1 x2 xm satisfying the boundary condition

Re [λ(x)Φ(x)] = au minus bv = c(x) z isin Llowast (12)

Problem A with the condition c(x) = 0 on Llowast is called Problem A0

Denote by λ(xj minus 0) and λ(xj + 0) the left limit and right limit of λ(x) as x rarrxj(j = 1 2 m) on L and

eiφj =λ(xjminus0)λ(xj+0)

γj=1πiln[λ(xjminus0)λ(xj+0)

]=

φj

πminusKj

Kj =[φj

π

]+ Jj Jj = 0 or 1 j = 1 m

(13)

in which 0 le γj lt 1 when Jj = 0 and minus1 lt γj lt 0 when Jj = 1 j = 1 m and

K =12(K1 + middot middot middot+Km) =

12

msumj=1

[φj

πminus γj

](14)

is called the index of Problem A and Problem A0 If λ(x) on L is continuous thenK = ∆Γ arg λ(x)2π is a unique integer If the function λ(x) on L is not continuouswe can choose Jj = 0 or 1 hence the index K is not unique We can require that thesolution Φ(z) satisfy the condition

Φ(z) = O(|z minus xj|minusδ) δ =

⎧⎨⎩βj + τ for γj ge 0 and γj lt 0 βj gt |γj||γj|+ τ for γj lt 0 βj le |γj| j = 1 m

(15)

in the neighborhood (sub D) of xj where τ (lt α) is an arbitrary small positive number

In order to find the solution of Problem A for analytic functions we first considerProblem A0 Making a transformation

Ψ(z) =Φ(z)Π(z)

Π(z) =mprod

j=1

(z minus xj

z + i

)γj

(16)

in which γj (j = 1 m) are as stated in (13) the boundary condition

Re [λ(x)Φ0(x)] = 0 x isin Llowast (17)

1 Generalizations of KeldychndashSedov Formula 81

of Problem A0 for analytic functions Φ0(z) is reduced to the boundary condition

Re [Λ(x)Ψ0(x)] = 0 Λ(x) = λ(x)Π(x)|Π(x)| x isin Llowast (18)

of Problem Alowast0 for analytic functions Ψ0(z) = Φ0(z)Π(z) Noting that

Λ(xj minus 0)Λ(xj + 0)

=λ(xj minus 0)λ(xj + 0)

(Π(xj minus 0)Π(xj + 0)

)=

λ(xj minus 0)λ(xj + 0)

eminusiπγj = plusmn1 (19)

the index of Λ(x) on L is

K =12π∆L arg Λ(x) =

12

msumj=1

[φj

πminus γj

]=12

msumj=1

Kj (110)

which is the same as the index of λ(x) on L If 2K is even provided that we changethe signs of Λ(x) on some line segments of Lj (j = 1 m) then the new functionΛlowast(x) on L is continuous its index is K too When 2K is odd we rewrite theboundary condition (18) in the form

Re[Λ(x)

x minus x0

x+ i

x minus x0Ψ0(x)

]= 0 x isin Llowast (111)

where x0(isin L) is a real number and x0 isin x1 xm thus similarly to before wechange the signs of Λ(x)(x minus x0)|x + i|(x + i)|x minus x0| on some line segments of Lsuch that the new function Λlowast(x) on L is continuous its index is Klowast = K minus 12Next we find an analytic function

Ψlowast(z) = i(

z minus i

z + i

)[K] (z minus x0

z + i

)eiS(z) in D (112)

which satisfies the homogeneous boundary condition

Re [Λlowast(x)Ψlowast(x)] = 0 x isin Llowast (113)

where [K] is the integer part of K S(z) is an analytic function in D satisfying theboundary condition

Re [S(x)] = arg

⎡⎣Λlowast(x)(

x minus i

x+ i

)[K](x minus x0

x+ i

)⎤⎦ x isin L Im [S(i)] = 0 (114)

Hence Problem Alowast0 for analytic functions possesses the solution

Ψ0(z) =

⎧⎨⎩Ψlowast(z) when 2K is even

(z minus x0)Ψlowast(z)(z + i) when 2K is odd(115)

and then Problem A0 for analytic functions has a non-trivial solution in the form

X(z) = Π(z)Ψ0(z) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩i(

z minus i

z + i

)K

Π(z)eiS(z) when 2K is even

i(

zminusi

z+i

)[K] zminusx0

z + iΠ(z)eiS(z) when 2K is odd

(116)

Take into account that X(z) has a zero of order [K] at the point z = i for K ge 0and a pole of order |[K]| at the point z = i for K lt 0 and a zero of order 1 at thepoint z = x0 when 2K is an odd integer moreover X(z) satisfies the homogeneousboundary condition (17) it is clear that iλ(x)X(x) is a real-valued function on L Letus divide the nonhomogeneous boundary condition (12) by iλ(x)X(x) and obtain

Re[Φ(x)iX(x)

]=

c(x)iλ(x)X(x)

=λ(x)c(x)iX(x)

x isin Llowast (117)

By using the Schwarz formula we get

Φ(z)iX(z)

=1πi

[int infin

minusinfinλ(t)c(t)

(t minus z)iX(t)dt+

Q(z)i

]

Φ(z) =X(z)πi

[int infin

minusinfinλ(t)c(t)

(t minus z)X(t)dt+Q(z)

]

(118)

If K ge 0 the function Q(z) possesses the form

Q(z) = i[K]sumj=0

[cj

(z minus i

z + i

)j

+ cj

(z minus i

z + i

)minusj]+

⎧⎪⎪⎨⎪⎪⎩0 when 2K is even

iclowastx0 + z

x0 minus z when 2K is odd

(119)

where clowast c0 are arbitrary real constants and cj (j=1 [K]) are arbitrary complexconstants from this we can see that the general solution Φ(z) includes 2K+1 arbitraryreal constants If 2K is odd we note (zminusx0)[(tminusz)(tminusx0)] = 1(tminusz)minus1(tminusx0)then the integral in (118) is understood as the difference of two integrals of Cauchytype If K lt 0 we have to take

Q(z) =

⎧⎪⎪⎨⎪⎪⎩iclowast = 0 when 2K is even

iclowastx0 + z

(120)

and require that the function in the square bracket of (118) has at least a zero point

of order |[K]| at z = i From

int infin

minusinfinλ(t)c(t)

(t minus z)X(t)dt+ iclowast

x0 + z

x0 minus z

=int infin

minusinfinλ(t)c(t)

(1minus(zminusi)(tminusi))(tminusi)X(t)dt+iclowast

1+(zminusi)(x0+i)1minus(zminusi)(x0minusi)

x0+i

x0minusi

=infinsum

j=0

int infin

minusinfinλ(t)c(t)(z minus i)j

(t minus i)j+1X(t)dt+ iclowast

[1 +

z minus i

x0 + i

]x0 + i

x0 minus i

infinsumj=0

(z minus i

x0 minus i

)j

=int infin

minusinfinλ(t)c(t)(tminusi)X(t)

dt+iclowastx0+i

x0minusi+

infinsumj=1

[int infin

minusinfinλ(t)c(t)

(tminusi)j+1X(t)dt+

2iclowastx0

(x0minusi)j+1

](zminusi)j

(121)

in the neighborhood of z = i this shows thatint infin

minusinfinλ(t)c(t)

X(t)(t minus i)jdt = 0 j = 1 minusK(= |[K]|) when 2K is even

int infin

minusinfinλ(t)c(t)

X(t)(t minus i)jdt+

2iclowastx0

(x0 minus i)j=0 j=2 [minusK]+1(= |[K]|)

when 2K is odd

(122)

then the function in the square bracket of (118) has a zero point of order |[K]| atz = i hence the function Φ(z) is analytic at z = i Besides when 2K is odd

clowast = ix0 minus i

x0 + i

int infin

minusinfinλ(t)c(t)

X(t)(t minus i)dt (123)

is a determined constant Therefore when K lt 0 Problem A has minus2K minus1 solvabilityconditions Thus we have the following theorem

Theorem 11 Problem A for analytic functions in D = Im z gt 0 has the follow-ing solvability result

(1) If the index K ge 0 the general solution Φ(z) of Problem A possesses the form(118) (119) which includes 2K + 1 arbitrary real constants

(2) If the index K lt 0 Problem A has minus2K minus 1 solvability conditions as stated in(122) When the conditions hold the solution of Problem A is given by the secondformula in (118) in particular

Φ(z)=X(z)(z minus i)|[K]|

πi

[int infin

minusinfinλ(t)c(t)

(t minus z)(t minus i)|[K]|X(t)dt+

2iclowastx0

(x0 minus z)(x0 minus i)|[K]|

](124)

in |z minus i| lt 1 where the constant clowast is determined as stated in (123)

Finally we mention that if x1 xm are first kind of discontinuities of c(x)and if γj gt 0 j = 1 2 m then the solution Φ(z) of Problem A is bounded in

Dlowast = Dx1 xm In general if γj le 0 (1 le j le m) the solution Φ(z) ofProblem A may not be bounded in the neighborhood of xj in Dlowast = Dx1 xmWe have

Φ(z) =

⎧⎨⎩O(|z minus xj|minusγj) if γj lt 0 Jj = 1

O(ln |z minus xj|) if γj = 0 Jj = 0(125)

in the neighborhood of xj on Dlowast but the integral

int z

iΦ(z)dz in D

is bounded In particular if m = 2n and

λ(x) =

⎧⎨⎩ 1 x isin (x2jminus1 x2j)

i x isin (x2j x2j+1)j = 1 n

and xj(j = 1 m = 2n) are first kind of discontinuous points of c(x) we canchoose γ2jminus1 = 12 K2jminus1 = 0 γ2j = minus12 K2j = 0 j = 1 n and thenthe index of the mixed boundary value problem is K = 0 In this case one canchoose Π(z) =

radicΠn

j=1(z minus x2jminus1)(z minus x2j) From the formula (118) with K = 0the KeldychndashSedov formula of the mixed boundary value problem for analytic func-tions in the upper half-plane is derived [53] If we chose γ2jminus1 = minus12 K2jminus1 =1 γ2j = minus12 K2j = 0 j = 1 n and the index of the mixed boundary valueproblem is K = n = m2 then the representation of solutions of the mixed bound-ary value problem for analytic functions can be written from (118) with K = nwhich includes 2K + 1 = m + 1 arbitrary real constants where the functionΠ(z) = 1

radicΠn

j=1(z minus x2jminus1)(z minus x2j)

12 The general discontinuous boundary value problem for analyticfunctions in the upper half-disk

Now we first introduce the general discontinuous RiemannndashHilbert problem (ProblemB) for analytic functions in the unit disk D = |z| lt 1 with the boundary conditions

Re [λ(z)Φ(z)] = au minus bv = c(z) Γ = |z| = 1 (126)

where λ(z) = a(z)minus ib(z) |λ(z)| = 1 on Γ and Z = z1 z2 zm are first kind ofdiscontinuous points of λ(z) on Γ and λ(z) c(z) satisfy the conditions

λ(z) isin Cα(Γj) |z minus zj|βjc(z) isin Cα(Γj) j = 1 2 m (127)

herein Γj is the arc from the point zjminus1 to zj on Γ and z0 = zm and Γj(j = 1 2 m)does not include the end points α(0 lt α lt 1) is a constant

Denote by λ(zj minus 0) and λ(zj + 0) the left limit and right limit of λ(z) as z rarrzj(j = 1 2 m) on Γ and

eiφj =λ(zj minus 0)λ(zj + 0)

γj =1πiln[λ(zj minus 0)λ(zj + 0)

]=

φj

πminus Kj

Kj =[φj

π

]+ Jj Jj = 0 or 1 j = 1 m

(128)

in which 0 le γj lt 1 when Jj = 0 and minus1 lt γj lt 0 when Jj = 1 j = 1 m Theindex K of Problem B is defined by (14) Let βj + γj lt 1 j = 1 m we requirethat the solution Φ(z) possesses the property

Φ(z) = O(|z minus zj|minusδ) δ =

(129)

in the neighborhood (sub Dlowast) of zj where τ(lt α) is an arbitrary small positive numberBy using a similar method as stated in Subsection 1 we can obtain the formula forsolutions of the boundary value problem

Theorem 12 Problem B for analytic functions in D = |z| lt 1 has the followingsolvability result

(1) If the index K ge 0 the general solution Φ(z) of Problem B possesses the form

Φ(z) =X(z)2πi

[intΓ

(t+ z)λ(t)c(t)(t minus z)tX(t)

dt+Q(z)] (130)

with

Q(z) = i[K]sumj=0(cjz

j + cjzminusj) +

iclowastz0 + z

z0 minus z when 2K is odd

(131)

where the constant clowast c0 are arbitrary real constants and cj (j = 1 [K]) arearbitrary complex constants which includes 2K + 1 arbitrary real constants

(2) If the index K lt 0 Problem B has minus2K minus 1 solvability conditions given by

intΓ

λ(t)c(t)X(t)tj

dt=0 j=1 minusK(= |[K]|) when 2K is even

intΓ

λ(t)c(t)X(t)tj

dt+iclowastzminusj+10 =0 j=1 [minusK]+1(= |[K]|) when 2K is odd

(132)When the conditions hold the solution of Problem B possesses the form

Φ(z) =X(z)z[K]

πi

[intΓ

λ(t)c(t)(t minus z)X(t)t|[K]| dt+

iclowast(z0 minus z)z|[K]|minus1

0

] (133)

where the constant clowast is determined via (132) as

clowast = iintΓ

λ(t)c(t)X(t)t

dt

In the above formula X(z) is a non-trivial solution of the homogeneous boundaryvalue problem (Problem B0) for analytic functions in the form

X(z) =

⎧⎨⎩ izKΠ(z)eiS(z) when 2K is even

iz[K](z minus z0)Π(z)eiS(z) when 2K is oddΠ(z) =

mprodj=1(z minus zj)γj (134)

in which S(z) is an analytic function in D satisfying the boundary conditions

Re [S(z)] = arg[Λlowast(z)z[K]] z isin Γ Im [S(0)] = 0

the function Λlowast(z) is similar to that in (113) [85]11)[86]1)

In addition through the conformal mapping from the upper half-unit disk D =|z| lt 1 Im z gt 0 onto the unit disk G = |ζ| lt 1 namely

ζ(z) = minusiz2 + 2iz + 1z2 minus 2iz + 1

z(ζ) =1

ζ + i

[1 + iζ minus

radic2(1minus ζ2)

]

we can obtain the result of the general discontinuous RiemannndashHilbert problem(Problem C) for analytic functions in the upper half-unit disk D = |z| lt 1 Im z gt0 namely

w(z) = Φ[ζ(z)] in D = |z| lt 1 Im z gt 0 (135)

is the solution of Problem C for analytic functions

In order to the requirement in latter chapters we give a well posed version (Prob-lem Bprime) of Problem B for analytic functions in D = |z| lt 1 namely we findan analytic function Φ(z) which is continuous in DZ and satisfies the boundarycondition (126) and the point conditions

Im [λ(zprimej)Φ(z

primej)] = bj j = 1 m (136)

where zprime1 z

primem (isin Z) are distinct points on Γ and bj (j = 1 m) are real con-

stants and we choose the index K = (m minus 1)2 of λ(z) on Γ = |z| = 1The homogeneous problem of Problem Bprime with the conditions c(z) = 0 on Γ andbj = 0 (j = 1 m) will be called Problem Bprime

0

Theorem 13 Problem Bprime for analytic functions in D = |z| lt 1 has a uniquesolution

Proof First of all we verify the uniqueness of solutions of Problem Bprime LetΦ1(z)Φ2(z) be two solutions of Problem Bprime for analytic functions Then the functionΦ(z) = Φ1(z) minus Φ2(z) is a solution of Problem Bprime

0 with the homogeneous boundaryconditions

Re [λ(z)Φ(z)]=0 on Γ=|z| = 1 Im [λ(zprimej)Φ(z

primej)]=0 j = 1 m (137)

According to the method in the proof of Theorem 11 or [8]2)[80]1) and [85]11) wesee that if Φ(z) equiv 0 in D then

m = 2K + 1 le 2ND +NΓ le 2K (138)

where ND NΓ are numbers of zero points in D and Γlowast = ΓZ respectively Thiscontradiction proves that Φ(z) equiv 0 ie Φ1(z) = Φ2(z) in D

Now we prove the existence of solutions of Problem Bprime for analytic functions Bythe representation (130) of the general solution of Problem B for analytic functionsit is easy to see that the general solution Φ(z) can be written as

Φ(z) = Φ0(z) +msum

j=1djΦj(z) (139)

where Φ0(z) is a special solution of Problem Bprime and Φj(z) (j = 1 m) are acomplete system of linearly independent solutions of Problem Bprime

0 and dj(j = 1 m)are arbitrary real constants In the following we prove that there exists a uniquesystem of real constants dprime

j(j = 1 m) such that |dprime1| + middot middot middot + |dprime

m| = 0 satisfyingthe equalities

msumj=1

dprimejΦj(zprime

j) = λ(zprimej)[c(z

primej) + ibj]minus Φ0(zprime

j) j = 1 m (140)

Then the analytic function Φ(z) = Φ0(z) +summ

j=1 dprimejΦj(z) satisfies the boundary con-

ditions (126) and (136) and thus is a solution of Problem Bprime According to thealgebraic theory it suffices to verify that the homogeneous system of algebraic sys-tem of equations (140) ie

Φlowast(zprimej) =

msumj=1

dprimejΦj(zprime

j) = 0 j = 1 m (141)

has no non-trivial solution Noting that the analytic function Φlowast(z) =summ

j=1 dprimejΦj(z)

is a solution of Problem Bprime0 from the uniqueness of solutions of Problem Bprime we see

that Φlowast(z) = 0 This proves the existence of solutions of Problem Bprime for analyticfunctions

Next we consider that D is the upper half-unit disk a(z) b(z) possess discon-tinuities of first kind at m distinct points z1 zm isin Γ cup L0 = Γprime = partD which arearranged according to the positive direction of partD Here Γ = |z| = 1 Im z gt 0L0 = minus1 le x le 1 y = 0 and z1 znminus1 isin Γ = |z| = 1 Im z gt 0 xn =minus1 xm = x0 = 1 isin L0 where n (lt m) m are positive integers and c(z) =O(|z minus zj|minusβj) in the neighborhood of zj (j = 1 2 m) on Γ in which βj(lt 1j = 1 2 m) are non-negative constants such that βj + γj lt 1 γj(j = 1 m)are as stated in (129) Denote λ(z) = a(z)minus ib(z) and |a(z)|+ |b(z)| = 0 there is noharm in assuming that |λ(z)| = 1 z isin Γprime = Γ cup L0 Suppose that λ(z) c(z) satisfyconditions again (127)

Problem C Find an analytic function Φ(z) = u(z)+iv(z) inD which is continuouson Dlowast = DZ satisfying the boundary condition

Re [λ(z)Φ(z)] = au minus bv = c(z) z isin Γlowast = ΓprimeZ (142)

here Z = z1 zm Problem C with the condition r(z) = 0 on Γlowast is calledProblem C0

The index K of Problem C is the same as stated in (14) We can require thatthe solution Φ(z) satisfies the condition (129)

In order to find the solution of Problem C for analytic functions it suffices tochoose a conformal mapping from the upper half-unit disk onto the upper half planeor the unit disk In the following we shall use the other method namely first find asolution of Problem A for analytic functions in D+ = Im z gt 0 with the boundarycondition

Re [λ(x)Φ(x)]=r(x) on L=(minusinfin infin)

r(x)=

⎧⎨⎩r(x) on L0=(minus1 1)c(x) on L1=(minusinfinltxltminus1) cup (1ltxltinfin)

(143)

in which λ(x) c(x) on L2 = (minusinfin lt x le minus1) cup (1 le x lt infin) are appropriatefunctions such that λ(x) |x minus xj|βjc(x) are piecewise Holder continuous functionsand continuous at the points x = minus1 1 and the index of λ(x) on L is K = 0 Forinstance setting

λ(x) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩λ(minus1 + 0) on (minusinfin minus1]λ(x) on (minus1 1)λ(16(x+ 3)minus 3) on [1 infin)

(144)

and denoting x2mminusj = 16(xj + 3)minus 3 j = n+ 1 m minus 1 we can determine thatthe index of above function λ(x) on L is K = 0 On the basis of Theorem 11 thesolution Ψ(z) of Problem A can be expressed in the form (118)(119) with K = 0and λ(z) c(z) are as stated in (127) Thus the function Φ(z) = Φ(z) minus Ψ(z) isanalytic in D and satisfies the boundary condition

Re [λ(z)Φ(z)] = r(z) =

⎧⎨⎩ r(z)minus Re [λ(z)Ψ(z)] z isin Γ

0 z isin L0(145)

Next similarly to Section 1 we make a transformation

Ψ(z) =Φ(z)Π(z)

Π(z) =2mminusnminus1prod

j=n+1j =m

(z minus xj

z + i

)γj

(146)

in which γj(j = n+1 mminus1 m+1 2mminusnminus1) are similar to those in (128)the boundary condition

Re [λ(z)Φ(z)] = 0 z isin L (147)

of Problem C0 for the analytic function Φ(z) is reduced to the boundary condition

Re [Λ(z)Ψ(z)] = 0 Λ(z) = λ(z)Π(z)|Π(z)| z isin L (148)

for the analytic function Ψ(z) = Φ(z)Π(z) Noting that

Λ(zj minus 0)Λ(zj + 0)

=λ(zj minus 0)λ(zj + 0)

(Π(zj minus 0)Π(zj + 0)

)=

λ(zj minus 0)λ(zj + 0)

eminusiπγj = plusmn1

j = n+ 1 m minus 1 m+ 1 2m minus n minus 1

(149)

the index of Λ(z) on L is the same as the index of λ(z) on L Due to 2K = 0 iseven provided that we change the sign of Λ(z) on some arcs Lj = (xjminus1 xj) (j =n + 1 2m minus n minus 1) Ln+1 = (minusinfin xn+1) cup (x2mminusnminus1 infin) then the new functionΛlowast(z) on Γ is continuous the index of λ(z) on L has not been changed Moreover wefind a solution of Problem A for analytic functions in Im z gt 0 with the boundaryconditions

Re [S(z)] = arg Λlowast(z) on L = (minusinfin infin) ImS(i) = 0 (150)

and denote Ψ(z) = Ψ(z)eminusiS(z)

Now we extend the analytic function Ψ(z) as follows

Φ(z) =

⎧⎪⎨⎪⎩Ψ(z) in D = |z| lt 1 Im z gt 0

minusΨ(z) in D = |z| lt 1 Im z lt 0(151)

It can be seen that the analytic function Φ(z) in |z| lt 1 satisfies the boundarycondition

Re [Λ(z)Φ(z)] = R(z) on |z| = 1 (152)

where

Λ(z)=

⎧⎨⎩λ(z)

λ(z)R(z)=

⎧⎨⎩r(z)eIm S(z) on Γ0=|z|=1 Im zgt0

minusr(z)eIm S(z) on Γ0=|z|=1 Im zlt0(153)

We can find the indexK prime of Λ(z) on |z| = 1 and by Theorem 12 the analytic functionΦ(z) in D with the boundary condition (152) can be found ie

Φ(z) = Φ(i1 + ζ

1minus ζ

)in D (154)

where Φ(z) is an analytic function as the function Φ(z) in (118) but in whichλ(z) c(z) K are replaced by λ[i(1 + ζ)(1minus ζ)] R[i(1 + ζ)(1minus ζ)] K prime respectivelyherein λ(z) R(z) are as stated in (153) It is clear that Φ(z) includes 2K prime + 1 arbi-trary real constants when K prime ge 0 and minus2K prime minus 1 solvability conditions when K prime lt 0

Thus the solution of Problem C for analytic functions in the upper half-unit disk Dis obtained ie

w(z) = Ψ(z) + Φ(z)Π(z)eiS(z) in D (155)

Theorem 14 When the index K ge 0 Problem C for analytic functions in D hasa solution in the form (155) including 2K + 1 arbitrary real constants and whenK lt 0 under minus2K minus 1 conditions Problem C for analytic functions possesses thesolution as stated in (155) Moreover the above solution of Problem C for analyticfunctions can be expressed by (135)

The KeldychndashSedov formula for analytic functions in the upper half-plane possessesimportant applications to the Tricomi problem for some equations of mixed type (see[12]1)3)) But more general boundary value problems for equations of mixed typecannot be solved by this formula Due to we have Theorems 11ndash14 such that theabove general problems can be solved In addition we can give the representation ofsolutions to the discontinuous RiemannndashHilbert boundary value problem for analyticfunctions in the zone domain D = 0 lt Im z lt 1 which can be used to solve someboundary value problems for nonlinear problems in mechanics

2 Representation and Existence of Solutions for EllipticComplex Equations of First Order

In this section we shall establish the representations for solutions of the generaldiscontinuous boundary value problem for elliptic complex equations of first orderin the upper half-unit disk Moreover we shall prove the existence of solutions fornonlinear elliptic complex equations of first order

21 Representation of solutions of the discontinuous RiemannndashHilbertproblem for elliptic complex equations in the upper half-unit disk

Let D be an upper half-unit disk with the boundary Γprime = Γcup L0 as stated in Section1 We consider the nonlinear uniformly elliptic systems of first order equations

Fj(x y u v ux vx uy vy) = 0 in D j = 1 2

Under certain conditions the system can be transformed into the complex form

wz = F (z w wz) F = Q1wz +Q2wz + A1w + A2w + A3 z isin D (21)

(see [86]1)) in which F (z w U) satisfy the following conditions

Condition C

1) Qj(z w U) Aj(z w) (j = 1 2) A3(z) are measurable in z isin D for all continu-ous functions w(z) in Dlowast = DZ and all measurable functions U(z) isin Lp0(Dlowast) and

2 Elliptic Equations of First Order 91

satisfyLp[Aj D] le k0 j = 1 2 Lp[A3 D] le k1 (22)

where Z = z1 zm Dlowast is any closed subset in D p0 p (2 lt p0 le p) k0 k1 arenon-negative constants

2) The above functions are continuous in w isin CI for almost every point z isin DU isin CI and Qj = 0 (j = 1 2) Aj = 0 (j = 1 2 3) for z isin D

3) The complex equation (21) satisfies the uniform ellipticity condition

|F (z w U1)minus F (z w U2)| le q0|U1 minus U2| (23)

for almost every point z isin D in which w U1 U2 isin CI and q0(lt 1) is a non-negativeconstant

Problem A The discontinuous RiemannndashHilbert boundary value problem for (21)is to find a continuous solution w(z) in Dlowast satisfying the boundary condition

Re [λ(z)w(z)] = c(z) z isin Γlowast = ΓprimeZ (24)

where λ(z) c(z) are as stated in Section 1 satisfying

Cα[λ(z)Γj] le k0 Cα[|z minus zj|βjc(z)Γj] le k2 j = 1 m (25)

herein α (12ltαlt1) k0 k2 are non-negative constants Assume that (βj + γj)β lt1 β = min(α 1 minus 2p0)2 γj βj(j = 1 m) are as stated in (128) (129)Problem A with A3(z) = 0 in D c(z) = 0 on Γlowast is called Problem A0 Theindex K of Problem A and Problem A0 is defined as in (14)

In order to prove the solvability of Problem A for the complex equation (21) weneed to give a representation theorem for Problem A

Theorem 21 Suppose that the complex equation (21) satisfies Condition C andw(z) is a solution of Problem A for (21) Then w(z) is representable by

w(z) = Φ[ζ(z)]eφ(z) + ψ(z) (26)

where ζ(z) is a homeomorphism in D which quasiconformally maps D onto the unitdisk G= |ζ| lt 1 with boundary L = |ζ| = 1 where ζ(minus1) = minus1 ζ(i) = iζ(1) = 1 Φ(ζ) is an analytic function in G ψ(z) φ(z) ζ(z) and its inverse functionz(ζ) satisfy the estimates

Cβ[ψ D] le k3 Cβ[φ D] le k3 Cβ[ζ(z) D] le k3 Cβ[z(ζ) G] le k3 (27)

Lp0 [|ψz|+ |ψz| D] le k3 Lp0 [|φz|+ |φz| D] le k3 (28)

Cβ[z(ζ) G] le k3 Lp0 [|χz|+ |χz| D] le k4 (29)

in which χ(z) is as stated in (214) below β = min(α 1 minus 2p0)2 p0 (2 lt p0 le p)kj = kj(q0 p0 k0 k1 D) (j = 3 4) are non-negative constantsMoreover if the

coefficients Qj(z) = 0 (j = 1 2) of the complex equation (21) in D then the rep-resentation (26) becomes the form

w(z) = Φ(z)eφ(z) + ψ(z) (210)

and if K lt 0 Φ(z) satisfies the estimate

Cδ[X(z)Φ(z) D] le M1 = M1(p0 β δ k D) (211)

in which

X(z)=mprod

j=1j =nm

|zminuszj|ηj |zminuszn|2ηn|zminuszm|2ηm ηj= |γj|+τ γj lt0 βj le|γj|

|βj|+τ other case(212)

Here γj (j = 1 m) are real constants as stated in (128) and δ τ (0 lt δ ltmin(β τ)) are sufficiently small positive constants and M1 is a non-negative con-stant

Proof We substitute the solution w(z) of ProblemA into the coefficients of equation(21) and consider the following system

ψz = Qψz + A1ψ + A2ψ + A3 Q =

Q1 +Q2wzwz for wz = 00 for wz = 0 or z isin D

φz = Qφz + A A =

⎧⎨⎩A1 + A2ww for w(z) = 00 for w(z) = 0 or z isin D

Wz = QWz W (z) = Φ[ζ(z)]

(213)

By using the continuity method and the principle of contracting mappings we canfind the solution

ψ(z) = Tf = minus 1π

int intD

f(ζ)ζ minus z

dσζ

φ(z) = Tg ζ(z) = Ψ[χ(z)] χ(z) = z + Th

(214)

of (213) where f(z) g(z) h(z) isin Lp02(D) 2 lt p0 le p χ(z) is a homeomorphism inD Ψ(χ) is a univalent analytic function which conformally maps E = χ(D) onto theunit disk G(see [85]11)) and Φ(ζ) is an analytic function in G We can verify thatψ(z) φ(z) ζ(z) satisfy the estimates (27) and (28) It remains to prove that z = z(ζ)satisfies the estimate (29) In fact we can find a homeomorphic solution of the lastequation in (213) in the form χ(z) = z + Th such that [χ(z)]z [χ(z)]z isin Lp0(D)[80]1)[85]9) Next we find a univalent analytic function ζ = Ψ(χ) which maps χ(D)onto G hence ζ = ζ(z) = Ψ[χ(z)] By the result on conformal mappings applyingthe method of Lemma 21 Chapter II in [86]1) we can prove that (29) is trueWhen Qj(z) = 0 in D j = 1 2 then we can choose χ(z) = z in (214) in this caseΦ[ζ(z)] can be replaced by the analytic function Φ(z) herein ζ(z)Ψ(z) are as statedin (214) it is clear that the representation (26) becomes the form (210) Thus theanalytic function Φ(z) satisfies the boundary conditions

Re [λ(z)eφ(z)Φ(z)] = c(z)minus Re [λ(z)ψ(z)] z isin Γlowast (215)

On the basis of Theorem 12 and the estimate (27) Φ(z) satisfies the estimate (211)

22 Existence of solutions of the discontinuous RiemannndashHilbert problemfor nonlinear complex equations in the upper half-unit disk

Theorem 22 Under the same conditions as in Theorem 21 the following state-ments hold

(1) If the index K ge 0 then Problem A for (21) is solvable and the generalsolution includes 2K + 1 arbitrary real constants

(2) If K lt 0 then Problem A has minus2K minus 1 solvability conditions

Proof Let us introduce a closed convex and bounded subset B1 in the Banachspace B = Lp0(D) times Lp0(D) times Lp0(D) whose elements are systems of functionsq = [Q(z) f(z) g(z)] with norms q = Lp0(Q D) + Lp0(f D) + Lp0(g D) whichsatisfy the condition

|Q(z)| le q0 lt 1 (z isin D) Lp0 [f(z) D] le k3 Lp0 [g(z) D] le k3 (216)

where q0 k3 are non-negative constants as stated in (23) and (27) Moreoverintroduce a closed and bounded subset B2 in B the elements of which are systemsof functions ω = [f(z) g(z) h(z)] satisfying the condition

Lp0 [f(z) D] le k4 Lp0 [g(z) D] le k4 |h(z)| le q0|1 + Πh| (217)

where Πh = minus 1π

intintD[h(ζ)(ζ minus z)2]dσζ

We arbitrarily select q = [Q(z) f(z) g(z)] isin B1 and using the principle of con-tracting mappings a unique solution h(z) isin Lp0(D) of the integral equation

h(z) = Q(z)[1 + Πh] (218)

can be found which satisfies the third inequality in (217) Moreover χ(z) = z+ This a homeomorphism in D Now we find a univalent analytic function ζ = Φ(χ)which maps χ(D) onto the unit disk G as stated in Theorem 21 Moreover we findan analytic function Ψ(ζ) in G satisfying the boundary condition in the form

Re [Λ(ζ)Φ(ζ)] = R(ζ) ζ isin L = ζ(Γ) (219)

in which ζ(z) = Ψ[χ(z)] z(ζ) is its inverse function ψ(z) = Tf φ(z) = Tg Λ(ζ) =λ[z(ζ)] exp[φ(z(ζ))] R(ζ) = r[z(ζ)] minus Re [λ[z(ζ)]ψ(z(ζ))] where Λ(ζ) R(ζ) on Lsatisfy conditions similar to λ(z) c(z) in (25) and the index of Λ(ζ) on L is K In thefollowing we first consider the case ofK ge 0On the basis of Theorem 12 we can findthe analytic function Φ(ζ) in the form (130) here 2K+1 arbitrary real constants canbe chosen Thus the function w(z) = Φ[ζ(z)]eφ(z) + ψ(z) is determined Afterwardswe find out the solution [f lowast(z) glowast(z) hlowast(z) Qlowast(z)] of the system of integral equations

f lowast(z)=F (z wΠf lowast)minusF (z w 0)+A1(z w)Tflowast+A2(z w)Tf lowast+A3(z w) (220)

Wglowast(z)=F (z w WΠglowast+Πf lowast)minusF (z wΠf lowast)+A1(z w)W+A2(z w)W (221)

S prime(χ)hlowast(z)eφ(z) = F [z w S prime(χ)(1 + Πhlowast)eφ(z) +WΠglowast +Πf lowast]

minusF (z wWΠglowast +Πf lowast)(222)

Qlowast(z) =hlowast(z)

[1 + Πhlowast] S prime(χ) = [Φ(Ψ(χ))]χ (223)

and denote by qlowast = E(q) the mapping from q = (Q f g) to qlowast = (Qlowast f lowast glowast)According to Theorem 21 from Chapter IV in [86]1) we can prove that qlowast = E(q)continuously maps B1 onto a compact subset in B1 On the basis of the Schauderfixed-point theorem there exists a system q = (Q f g) isin B1 such that q = E(q)Applying the above method from q = (Q f g) we can construct a functionw(z) = Φ[ζ(z)]eφ(z) + ψ(z) which is just a solution of Problem A for (21) Asfor the case of K lt 0 it can be similarly discussed but we first permit that thefunction Φ(ζ) satisfying the boundary condition (215) has a pole of order |[K]| atζ = 0 and find the solution of the nonlinear complex equation (21) in the formw(z) = Φ[ζ(z)]eφ(z) + ψ(z) From the representation we can derive the minus2K minus 1solvability conditions of Problem A for (21)

Besides we can discuss the solvability of the discontinuous RiemannndashHilbertboundary value problem for the complex equation (21) in the upper half-plane andthe zone domain For some problems in nonlinear mechanics as stated in [90] it canbe solved by the results in Theorem 22

23 The discontinuous RiemannndashHilbert problem for nonlinear complexequations in general domains

In this subsection let Dprime be a general simply connected domain with the boundaryΓprime = Γprime

1 cup Γprime2 herein Γ

prime1Γ

prime2 isin C1

α (0 lt α lt 1) and their intersection points zprime zprimeprime withthe inner angles α1π α2π(0 lt α1 α2 lt 1) respectively We discuss the nonlinearuniformly elliptic complex equation

wz = F (z w wz) F = Q1wz +Q2wz + A1w + A2w + A3 z isin Dprime (224)

in which F (z w U) satisfies Condition C in Dprime There exist m point Z = z1 =zprime znminus1 zn = zprimeprime zm on Γprime arranged according to the positive directionsuccessively Denote by Γprime

j the curve on Γprime from zjminus1 to zj j = 1 2 m and

Γprimej(j = 1 m) does not include the end points

Problem Aprime The discontinuous RiemannndashHilbert boundary value problem for (21)is to find a continuous solution w(z) in Dlowast = DprimeZ satisfying the boundary condition

Re [λ(z)w(z)] = c(z) x isin Γlowast = ΓprimeZ

Im [λ(zprimej)w(zprime

j)] = bj j = 1 m(225)

where zprimej bj(j = 1 m) are similar to those in (136) λ(z) c(z) bj(j = 1 m)

are given functions satisfying

Cα[λ(z)Γj] le k0 Cα[|z minus zj|βjc(z)Γj] le k2 |bj| le k2 j = 1 m (226)

3 Elliptic Equations of Second Order 95

herein α (12 lt α lt 1) k0 k2 are non-negative constants and assume that (βj +γj)β lt 1 β = min(α 1 minus 2p0)α0 βj(j = 1 m) are similar to those in (129)α0 = max(1α1 1α2 1) Problem A with A3(z) = 0 in D r(z) = 0 on Γprime andbj = 0 (j = 1 m) is called Problem Aprime

0 The index K = (m minus 1)2 of Problem Aand Problem A0 is defined as in (14)

In order to give the unique result of solutions of Problem Aprime for equation (224)we need to add one condition For any complex functions wj(z) isin C(Dlowast) Uj(z) isinLp0(Dprime)(j = 1 2 2 lt p0 le p) the following equality holds

F (z w1 U1)minus F (z w1 U2) = Q(U1 minus U2) + A(w1 minus w2) in Dprime (227)

in which |Q(z w1 w2 U1 U2)|le q0 A(z w1 w2 U1 U2)isinLp0(D) Especially if (224)is a linear equation then the condition (227) obviously is true

Applying a similar method as before we can prove the following theorem

Theorem 23 If the complex equation (224) in Dprime satisfies Condition C thenProblem Aprime for (224) is solvable If Condition C and the condition (227) hold thenthe solution of Problem Aprime is unique Moreover the solution w(z) can be expressed as(26)ndash(29) where β = min(α 1 minus 2p0)α0 If Qj(z) = 0 in D j = 1 2 in (224)then the representation (26) becomes the form

w(z) = Φ(z)eφ(z) + ψ(z) (228)

and w(z) satisfies the estimate

Cδ[X(z)w(z) D] le M1 = M1(p0 β δ k D) (229)

in which

X(z) =mprod

j=2j =1n

ηj = |γj|+ τ if γj lt 0 βj le |γj|

|βj|+ τ if γj ge 0 and γj lt 0 βj lt |γj|

(230)

here γj(j = 1 m) are real constants as stated in (128) δ τ (0 lt δ lt min(β τ))are sufficiently small positive constants and M1 is a non-negative constant

3 Discontinuous Oblique Derivative Problems for Quasilinear Elliptic Equations of Second Order

This section deals with the oblique derivative boundary value problems for quasi-linear elliptic equations of second order We first give the extremum principle andrepresentation of solutions for the above boundary value problem and then obtaina priori estimates of solutions of the above problem finally we prove the uniquenessand existence of solutions of the above problem

31 Formulation of the discontinuous oblique derivative problem forelliptic equations of second order

Let D be the upper half-unit disk as stated in Section 1 and Γprime = Γ cup L0 of D bethe boundary where Γ = |z| = 1 Im z ge 0 and L0 = (minus1 1) We consider thequasilinear uniformly elliptic equation of second order

where a b c d e f g are given functions of (x y) isin D and u ux uy isin IR Undercertain conditions equation (31) can be reduced to the the complex form

uzz=F (z u uz uzz) F =Re [Quzz+A1uz]+A2u+A3 in D (32)

where Q = Q(z u uz) Aj = Aj(z u uz) and

z = x+ iy uz =12[ux minus iuy] uz =

12[ux + iuy] uzz =

14[uxx + uyy]

Q(z) =minusa+ c minus 2bi

a+ c A1(z) =

minusd minus ei

a+ c A2(z) =

minusf

2(a+ c) A3(z) =

g

2(a+ c)

Suppose that equation (32) satisfies the following conditions

Condition C

1) Q(z u w) Aj(z u w) (j = 1 2 3) are continuous in u isin IR w isin CI for almostevery point z isin D u isin IRw isin CI and Q = 0 Aj = 0 (j = 1 2 3) for z isin D

2) The above functions are measurable in z isin D for all continuous functionsu(z) w(z) on Dlowast = DZ and satisfy

Lp[Aj(z u w) D]lek0 j=1 2 Lp[A3(z u w) D]lek1 A2(z u w)ge0 in D (33)

in which p0 p (2 lt p0 le p) k0 k1 are non-negative constants Z = minus1 13) Equation (32) satisfies the uniform ellipticity condition namely for any number

u isin IRw isin CI the inequality

|Q(z u w)| le q0 lt 1 (34)

for almost every point z isin D holds where q0 is a non-negative constant

The discontinuous oblique derivative boundary value problem for equation (32)may be formulated as follows

Problem P Find a continuously differentiable solution u(z) of (32) in Dlowast = DZwhich is continuous in D and satisfies the boundary conditions

12

partu

partν= Re [λ(z)uz] = r(z) z isin Γlowast = ΓprimeZ u(minus1) = b0 u(1) = b1 (35)

where Z = minus1 1 is the set of discontinuous points of λ(z) on Γlowast ν is a given vectorat every point on Γlowast λ(z) = a(x)+ ib(x) = cos(ν x)minus i cos(ν y) cos(ν n) ge 0 on ΓlowastIf cos(ν n) equiv 0 on Γlowast = ΓprimeZ then the condition u(1) = b1 can be canceled Heren is the outward normal vector at every point on Γlowast δ0(lt 1) is a constant b0 b1 arereal constants and λ(z) r(z) b0 b1 satisfy the conditions

Cα[λ(z)Γj] le k0 Cα[|z minus zj|βjr(z)Γj] le k2 j = 1 2 |b0| |b1| le k2 (36)

Herein α (12 lt α lt 1) k0 k2 are non-negative constants We assume that (βj +γj)β lt 1 β = min(α 1minus 2p0)2 βj(j = 1 m) are as stated in (127) ProblemP with A3(z) = 0 in D r(z) = 0 on Γ b0 = b1 = 0 is called Problem P0 Theindex of Problem P is K where K is defined as in (14) here we choose K = 0 andK = minus12 if cos(ν n) equiv 0 on Γlowast If A2(z) = 0 in D the last point condition in (35)can be replaced by

Im [λ(z)uz]|z=0 = b2 (37)

and we do not need the assumption cos(ν n) ge 0 on Γ where b2 is a real constantsatisfying the condition |b2| le k2 Then the boundary value problem for (32) will becalled Problem Q In the following we only discuss the case of K = 0 and the caseof K = minus12 can be similarly discussed

32 The representation theorem of Problem P for equation (32)

We first introduce a theorem

Theorem 31 Suppose that equation (32) satisfies Condition C Then there existtwo solutions ψ(z)Ψ(z) of the Dirichlet problem (Problem D) of (32) and its relatedhomogeneous equation

uzz minus Re [Q(z u uz)uzz + A1(z u uz)uz]minus A2(z u uz)u = 0 in D (38)

satisfying the boundary conditions

ψ(z) = 0 Ψ(z) = 1 on Γ (39)

respectively and ψ(z)Ψ(z) satisfy the estimates

C1β[ψ(z) D] le M2 C1

β[Ψ(z) D] le M2

Lp0 [ψzz D] le M3 Lp0 [Ψzz D] le M3 Ψ ge M4 gt 0 in D(310)

where β (0 lt β le α) Mj = Mj(q0 p0 β k0 k1 D) (j = 2 3 4) are non-negativeconstants

Proof We first assume that the coefficients Q = Aj = 0 (j = 1 2 3) of (32) inthe ε-neighborhood of z = minus1 1 ie Dε = |z plusmn 1| le ε Im z ge 0 ε gt 0 whereε = 1m (m is a positive integer) Introduce the transformation and its inversion

z(ζ) =1

ζ + i

[1 + iζ minus

radic2(1minus ζ2)

] (311)

The function ζ(z) maps D onto G = |ζ| lt 1 such that the boundary points minus1 0 1are mapped onto the points minus1 minusi 1 respectively Through the transformation equa-tion (32) is reduced to the equation

uζζ= |zprime(ζ)|2Re [Quζζ(zprime(ζ))2+(A1zprime(ζ)minusQzprimeprime(ζ)(zprime(ζ))3)uζ ]+A2u+A3 (312)

in G It is clear that equation (312) in G satisfies conditions similar to Condition CHence equation (312) and its related homogeneous equation

uζζ= |zprime(ζ)|2Re [Quζζ(zprime(ζ))2+(A1zprime(ζ)minusQzprimeprime(ζ)(zprime(ζ))3)uζ ]+A2u in G (313)

possess the solutions ψ(ζ) Ψ(ζ) satisfying the boundary conditions

ψ(ζ) = 0 Ψ(ζ) = 1 on L = ζ(Γ)

and ψ[ζ(z)]Ψ[ζ(z)] in D are the solutions of Problem D of (32)(38) satisfying theboundary condition (39) respectively and ψ(z)Ψ(z) satisfy the estimate (310) butthe constants Mj = Mj(q0 p0 β k0 k1 D ε) (j = 2 3 4) Now we consider

ψ(z) =

⎧⎨⎩ψ(z) in D

minusψ(z) in D = |z| lt 1 Im z lt 0(314)

It is not difficult to see that ψ(z) in ∆ = |z| lt 1 is a solution of the elliptic equationuzz minus Re [Quzz + A1uz]minus A2u = A3 in ∆ (315)

where the coefficients

Q=

⎧⎨⎩Q(z)

Q(z)A1=

⎧⎨⎩A1(z)

A1(z)A2=

⎧⎨⎩A2(z)

A2(z)A3=

⎧⎨⎩A3(z)

minusA3(z)in

⎧⎨⎩D

D

⎫⎬⎭

where D is the symmetrical domian of D with respect to the real axis It is clear thatthe coefficients in ∆ satisfy conditions similar to those from Condition C Obviouslythe solution ψ(z) satisfies the boundary condition ψ(z) = 0 on part∆ = |z| = 1Denote by ψm(z) the solution of equation (32) with Q = Aj = 0(j = 1 2 3) in theε = 1m-neighborhood of z = minus1 1 we can derive that the function ψm(z) in ∆satisfies estimates similar to ψ(z) in (310) where the constants Mj(j = 2 3) areindependent of ε = 1m Thus we can choose a subsequence of ψm(z) whichuniformly converges to ψlowast(z) and ψlowast(z) is just a solution of Problem D for theoriginal equation (32) in D Noting that the solution Ψ(z) = ψ(z)+ 1 of Problem Dfor equation (38) is equivalent to the solution ψ(z) of Problem D for the equation

with the boundary condition ψ(z) = 0 on Γ by using the same method we can provethat there exists a solution Ψ(z) of Problem D for (38) with the boundary conditionΨ(z) = 1 on Γ and the solution satisfies the estimates in (310)

Theorem 32 Suppose that equation (32) satisfies Condition C and u(z) is asolution of Problem P for (32) Then u(z) can be expressed as

u(z)=U(z)Ψ(z)+ψ(z) U(z)=2Reint z

0w(z)dz+b0 w(z)=Φ[ζ(z)]eφ(z) (317)

where ψ(z)Ψ(z) are as stated in Theorem 31 satisfying the estimate (310) ζ(z)is a homeomorphism in D which quasiconformally maps D onto the unit disk G=|ζ| lt 1 with boundary L where ζ(minus1) = minus1 ζ(1) = 1 ζ(i) = i Φ(ζ) is an analyticfunction in G φ(z) ζ(z) and its inverse function z(ζ) satisfy the estimates

Cβ[φ(z) D] le k3 Cβ[ζ(z) D] le k3

Cβ[z(ζ) G] le k3 Lp0 [|φz|+ |φz| D] le k3

Cβ[z(ζ) G] le k3 Lp0 [|χz|+ |χz| D] le k4

(318)

in which χ(z) is as stated in (214) β = min(α 1 minus 2p0)2 p0(2 lt p0 le p) kj =kj(q0 p0 k0 k1 D)(j = 3 4) are non-negative constants

Proof We substitute the solution u(z) of Problem P into the coefficients of equation(32) It is clear that (32) in this case can be seen as a linear equation Firstly onthe basis of Theorem 31 there exist two solutions ψ(z)Ψ(z) of Problem D of (32)and its homogeneous equation (38) satisfying the estimate (310) Thus the function

U(z) =u(z)minus ψ(z)

Ψ(z)in D (319)

is a solution of the equation

Uzz minus Re [QUzz + AUz] = 0 A = A1 minus 2(lnψ)z + 2Q(lnΨ)z in D (320)

and w(z) = Uz is a solution of the first order equation

wz =12[Qwz +Qwz + Aw + Aw] in D (321)

satisfying the boundary condition

12[partU

partν+ (lnΨ)νU ] = r(z)minus Re [λ(z)ψz] on Γlowast ie

Re [λ(z)Uz + (lnΨ)νU2] = r(z)minus Re [λ(z)ψz] on Γlowast(322)

By the following Lemma 33 we see that (lnΨ)ν gt 0 on Γlowast and similarly to Theorem21 the last formula in (317) can be derived and φ(z) ζ(z) and its inverse functionz(ζ) χ(z) satisfy the estimates (27)ndash(29)

Now we consider the linear homogeneous equation

uzz minus Re [Quzz + A1(z)uz]minus A2(z)u = 0 in D (323)

and give a lemma

Lemma 33 Let the equation (323) in D satisfy Condition C and u(z) be a con-tinuously differentiable solution of (323) in D If M = maxzisinD u(z) ge 0 then thereexists a point z0 isin partD such that u(z0) = M If z0 = x0 isin (minus1 1) and u(z) lt u(z0)in Dz0 then

partu

partl= lim

z(isinl)rarrz0

u(z0)minus u(z)|z minus z0| gt 0 (324)

where z (isin D) approaches z0 along a direction l such that cos(l y) gt 0

Proof From the result in Section 2 Chapter III [86]1) we see that the solutionu(z) in D attains its non-negative maximumM at a point z0 isin partD There is no harmin assuming that z0 is a boundary point of ∆ = |z| lt R because we can choose asubdomain(isin D) with smooth boundary and the boundary point z0 and then makea conformal mapping Thus this requirement can be realized By Theorem 31 wefind a continuously differentiable solution Ψ(z) of (323) in ∆ satisfying the boundarycondition Ψ(z) = 1 z isin part∆ = |z| = R and can derive that 0 lt Ψ(z) le 1 z isin ∆Due to V (z) = u(z)Ψ(z) is a solution of the following equation

LV = Vzz minus Re [A(z)Vz] = 0 A(z) = minus2(lnΨ)z + A1(z) in ∆ (325)

it is clear that V (z) lt V (z0) z isin ∆ and V (z) attains the maximum at the pointz0 Afterwards we find a continuously differentiable solution V (z) of (325) in ∆ =R2 le |z| le R satisfying the boundary condition

V (z) = 0 z isin part∆ V (z) = 1 |z| = R

2

It is easy to see that partV parts = 2Re [izVz] z isin part∆ and

partV

partn= 2Re

zVz

R z isin part∆

partV

partn= minus4Re zVz

R |z| = R

2

where s n are the tangent vector and outward normal vector on the boundary part∆Noting that W (z) = Vz satisfies the equation

Wz minus Re [A(z)W ] = 0 z isin ∆

and the boundary condition Re [izW (z)] = 0 z isin part∆ and the index of iz on theboundary part∆ equals to 0 hence W (z) has no zero point on part∆ thus partV partn =2Re [zW (z)R] lt 0 z isin part∆ The auxiliary function

V (z) = V (z)minus V (z0) + εV (z) z isin ∆

by selecting a sufficiently small positive number ε such that V (z) lt 0 on |z| = R2obviously satisfies V (z) le 0 z isin part∆ Due to LV = 0 z isin ∆ on the basis of themaximum principle we have

V (z) le 0 z isin part∆ ie V (z0)minus V (z) ge minusε[V (z0)minus V (z)] z isin ∆

Thus at the point z = z0 we have

partV

partnge minusε

partV

partngt 0

partu

partn= Ψ

partV

partn+ V

partΨpartn

ge minusεpartV

partn+ V

partΨpartn

gt 0

Moreover noting the condition cos(l n) gt 0 cos(l s) gt 0 partUparts = 0 at the point z0where s is the tangent vector at z0 it follows the inequality

partu

partl= cos(l n)

partu

partn+ cos(l s)

partu

partsgt 0 (326)

Theorem 34 If equation (32) satisfies Condition C and for any uj(z) isinC1(Dlowast) j = 1 2 uzz isin CI the following equality holds

F (z u1 u1z u1zz)minus F (z u2 u2z u2zz) = minusRe [Quzz+A1uz]minusA2u

where Lp[Aj D] lt infin j = 1 2 then the solution u(z) of Problem P is unique

Proof Suppose that there exist two solutions u1(z) u2(z) of Problem P for (32)it can be seen that u(z) = u1(z) minus u2(z) satisfies the homogeneous equation andboundary conditions

uzz = Re [Quzz + A1uz] + A2u in D

12

partu

partν= 0 z isin Γlowast u(minus1) = 0 u(1) = 0

(327)

If the maximum M = maxD u(z) gt 0 it is clear that the maximum point zlowast = minus1and 1 On the basis of Lemma 33 the maximum of u(z) cannot attain on (minus1 1)hence its maximum M attains at a point zlowast isin Γlowast If cos(ν n) gt 0 at zlowast from Lemma33 we get partupartν gt 0 at zlowast this contradicts the boundary condition in (327) ifcos(ν n) = 0 at zlowast denote by Γprime the longest curve of Γ including the point zlowast so thatcos(ν n) = 0 and u(z) = M on Γprime then there exists a point zprime isin ΓΓprime such that atzprime cos(ν n) gt 0 partupartn gt 0 cos(ν s) gt 0 (lt 0) partuparts ge 0 (le 0) hence (326) at zprime

holds it is impossible This shows zlowast isin Γ Hence maxD+ u(z) = 0 By the similarmethod we can prove minD+ u(z) = 0 Therefore u(z) = 0 u1(z) = u2(z) in D

Theorem 35 Suppose that equation (32) satisfies Condition C then the solutionu(z) of Problem P for (32) satisfies the estimates

C1δ [u(z) D]=Cβ[u(z) D]+Cδ[|X(z)|1βuz D]leM5

C1δ [u(z) D] le M6(k1 + k2)

(328)

in which β = min(α 1 minus 2p0) X(z) = |z + 1|2η1|z minus 1|2η2 M5 = M5(p0 β δ k D)M6 = M6(p0 β δ k0 D) are two non-negative constants

Proof We first verify that any solution u(z) of Problem P for (32) satisfies theestimate

S(u) = C[u(z) D] + C[|X(z)|1βuz D] le M7 = M7(p0 α k D) (329)

Otherwise if the above inequality is not true there exist sequences of coefficientsQm Am

j (j = 1 2 3) λm rm bmj (j = 1 2) satisfying the same conditions

of Q Aj(j = 1 2 3) λ r bj(j = 0 1) and Qm Amj (j = 1 2 3) weakly con-

verge in D to Q0 A0j(j = 1 2 3) and λm rm bm

j (j = 0 1) uniformly convergeon Γlowast to λ0 r0 b0

j(j = 0 1) respectively Let um is a solution of Problem P for(32) corresponding to Qm Am

j (j = 1 2 3) λm rm bmj (j = 0 1) but

maxD |um(z)| = Hm rarr infin as m rarr infin There is no harm in assuming that Hm ge 1Let Um = umHm It is clear that Um(z) is a solution of the boundary value problem

Umzz minus Re [QmUm

zz + Am1 Um

z ]minus Am2 Um =

Am3

Hm

partUm

partνm

=rm(z)Hm

z isin Γlowast Um(minus1) = bm0

Hm

Um(1) =bm1

Hm

From the conditions in the theorem we have

Lp[Am2 Um +

Am3

Hm

D] le M7 C[λjΓj] le M7

Cα

[|z minus zj|βj

rm(z)Hm

Γlowast]

le M7 j = 1 m

∣∣∣∣∣ bmj

Hm

∣∣∣∣∣ le M7 j = 0 1

whereM7=M7(q0 p0 α K D) is a non-negative constant According to the methodin the proof of Theorem 23 we denote

wm = Umz Um(z) = 2Re

int z

minus1wm(z)dz +

bm0

Hm

and can obtain that Um(z) satisfies the estimate

Cβ[Um(z) D] + Cδ[|X(z)|1βUmz D] le M8 (330)

in which M8 = M8 (q0 p0 δ α K D) δ (gt 0) are non-negative constants Hencefrom Um(z) and |X(z)|1βUm

z we can choose subsequences Umk(z) and|X(z)|1βUmk

z which uniformly converge to U0(z) and |X(z)|1βU0z in D respect-

ively and U0(z) is a solution of the following boundary value problem

U0zz = Re [Q

0U0zz + A0

1uz] + A02U

0 = 0 in D

partU0

partν= 0 on Γlowast U0(minus1) = 0 U0(1) = 0

By the result as stated before we see that the solution U0(z) = 0 However fromS(Um) = 1 the inequality S(U0) gt 0 can be derived Hence the estimate (329) istrue Moreover by using the method from S(Um) = 1 to (330) we can prove thefirst estimate in (328) The second estimate in (328) can be derived from the firstone

33 Existence of solutions of the discontinuous oblique derivative problemfor elliptic equations in the upper half-unit disk

Theorem 36 If equation (32) satisfies Condition C then Problem P for (32) issolvable

Proof Noting that the index K = 0 we introduce the boundary value problem Pt

for the linear elliptic equation with a parameter t(0 le t le 1)

Lu = uzz minus Re [Quzz + A1(z)uz] = G(z u) G = tA2(z)u+ A(z) (331)

for any A(z) isin Lp0(D) and the boundary condition (35) It is evident that whent = 1 A(z) = A3(z) Problem Pt is just Problem P When t = 0 the equation in(331) is

Lu=uzzminusRe [Quzz+A1uz]=A(z) ie wzminusRe [Qwz+A1w]=A(z) (332)

where w = uz By Theorem 37 below we see that Problem P for the first equation in(332) has a unique solution u0(z) which is just a solution of Problem P for equation(331) with t = 0 Suppose that when t = t0 (0 le t0 lt 1) Problem Pt0 is solvableie Problem Pt for (331) has a unique solution u(z) such that |X(z)|1βuz isin Cδ(D)We can find a neighborhood Tε = |t minus t0| lt ε 0 le t le 1 ε gt 0 of t0 such that forevery t isin Tε Problem Pt is solvable In fact Problem Pt can be written in the form

Luminust0[G(z u)minusG(z 0)]=(tminust0)[G(z u)minusG(z 0)]+A(z) z isin D (333)

and (35) Replacing u(z) in the right-hand side of (333) by a function u0(z) withthe condition |X(z)|1βu0z isin Cδ(D) especially by u0(z) = 0 it is obvious that theboundary value problem (333)(35) then has a unique solution u1(z) satisfying theconditions |X(z)|1βu1z isin Cδ(D) Using successive iteration we obtain a sequence ofsolutions un(z) satisfying the conditions |X(z)|1βunz isin Cδ(D)(n = 1 2 ) and

Lun+1minust0[G(z un+1)minusG(z 0)]=(tminust0)[G(z un)minusG(z 0)]+A(z) z isin D

Re [λ(z)un+1z] = r(z) z isin Γ un+1(minus1) = b0 un+1(1) = b1 n = 1 2

From the above formulas it follows that

L(un+1 minus un)z minus t0[G(z un+1)minus G(z un)]

= (t minus t0)[G(z un)minus G(z unminus1)] z isin D

Re [λ(z)(un+1z minus unz)] = 0 z isin Γ

un+1(minus1)minus un(minus1) = 0 un+1(1)minus un(1) = 0

(334)

Noting that

Lp[(t minus t0)(G(z un)minus G(z unminus1)) D] le |t minus t0|k0C1δ [un minus unminus1 D] (335)

where C1δ [un minus unminus1 D] = Cβ[un minus unminus1 D] + Cδ[|X(z)|1β(unz minus unminus1z) D] and

applying Theorem 35 we get

C1δ [un+1 minus un D] le |t minus t0|M6C

1δ [un minus unminus1 D] (336)

Choosing the constant ε so small that 2εM6 lt 1 it follows that

C1δ [un+1 minus un D] le C1

δ

un minus unminus1 D

2 (337)

and when n m ge N0 + 1 (N0 is a positive integer)

C1δ [un+1minusun D]le2minusN0

infinsumj=02minusjC1

δ [u1minusu0 D]le2minusN0+1C1δ [u1minusu0 D] (338)

Hence un(z) is a Cauchy sequence According to the completeness of the Banachspace C1

δ (D) there exists a function ulowast(z) isin C1δ (D) so that C1

δ [un minus ulowast D] rarr 0 forn rarr infin From (338) we can see that ulowast(z) is a solution of Problem Pt for everyt isin Tε = |t minus t0| le ε Because the constant ε is independent of t0 (0 le t0 lt 1)therefore from the solvability of Problem Pt when t = 0 we can derive the solvabilityof Problem Pt when t = ε 2ε [1ε] ε 1 In particular when t = 1 and A(z) =A3(z) Problem P1 for the linear case of equation (32) is solvable

Next we discuss the quasilinear equation (32) satisfying Condition C but we firstassume that the coefficients Q = 0 Aj(j = 1 2 3) = 0 in Dm = z isin D dist(zΓ) lt1m here m(ge 2) is a positive integer namely consider

uzz = Re [Qmuzz + Am1 uz] + Am

2 u+ Am3 in D (339)

where

Qm =

⎧⎨⎩Q(z u uz)

0Am

j =

⎧⎨⎩Aj(z u uz)

0in

⎧⎨⎩Dm = DDm

Dm

⎫⎬⎭ j = 1 2 3

Now we introduce a bounded closed and convex set BM in the Banach space B =C1

δ (D) any element of which satisfies the inequality

C1δ [u(z) D] le M5 (340)

where M5 is a non-negative constant as stated in (328) We are free to choose anarbitrary function U(z) isin BM and insert it into the coefficients of equation (339)It is clear that the equation can be seen as a linear equation hence there exists aunique solution u(z) of Problem P and by Theorem 35 we see u(z) isin BM Denoteby u(z) = S[U(z)] the mapping from U(z) isin BM to u(z) obviously u(z) = S[U(z)]maps BM onto a compact subset of itself It remains to verify that u(z) = S[U(z)]continuously maps the set BM onto a compact subset In fact we arbitrarily selecta sequence of functions Un(z) such that C1

δ [Un(z) minus U0(z) D] rarr 0 as n rarr infin

Setting un(z) = S[Un(z)] and subtracting u0(z) = S[U0(z)] from un(z) = S[Un(z)]we obtain the equation for un = un(z)minus u0(z)

unzminusRe [Qm(z Un Unz)unzz+Am1 (z Un Unz)unz]minusAm

2 (z Un Unz)un=Cn

Cn = Cn(z Un U0 u0) = Am3 minus Re [Qmu0zz + Am

1 u0z]minus Am2 u0

(341)

in which Qm = Qm(z Un Unz)minusQm(z U0 U0z) Amj = Am

j (z Un Unz)minusAmj (z U0 U0z)

j = 1 2 3 and the solution un(z) satisfies the homogeneous boundary conditions

Re [λ(z)uz] = 0 z isin Γlowast = ΓZ u(minus1) = 0 u(1) = 0 (342)

Noting that the function Cn = 0 in Dm according to the method in the formula(243) Chapter II [86]1) we can prove that

Lp[Cn D] rarr 0 as n rarr infin

On the basis of the second estimate in (328) we obtain

C1δ [un(z)minus u0(z) D] le M6Lp[Cn D] (343)

thus C1δ [un(z) minus u0(z) D] rarr 0 as n rarr infin This shows that u(z) = S[U(z)] in the

set BM is a continuous mapping Hence by the Schauder fixed-point theorem thereexists a function u(z) isin BM such that u(z) = S[u(z)] and the function u(z) is justa solution of Problem P for the quasilinear equation (339)

Finally we cancel the conditions the coefficients Q = 0 Aj (j = 1 2 3) = 0 inDm = z dist(zΓ) lt 1m Denote by um(z) a solution of Problem P for equation(339) By Theorem 35 we see that the solution satisfies the estimate (328) Hencefrom the sequence of solutions um(z) m = 2 3 we can choose a subsequenceumk(z) for convenience denote umk(z) by um(z) again which uniformly con-verges to a function u0(z) in D and u0(z) satisfies the boundary condition (35) ofProblem P At last we need to verify that the function u0(z) is a solution of equation(32) Construct a twice continuously differentiable function gn(z) as follows

gn(z) =

⎧⎨⎩ 1 z isin Dn = DDn

0 z isin D2n0 le gn(z) le 1 in DnD2n (344)

where n(ge 2) is a positive integer It is not difficult to see that the function umn (z) =

gn(z)um(z) is a solution of the following Dirichlet boundary value problem

umnzz minus Re [Qmum

nzz] = Cmn in D (345)

umn (z) = 0 on Γ (346)

where

Cmn =gn[Re (Am

1 umz )+Am

2 um]+um[gnzzminusRe (Qmgnzz)]+2Re [gnzumz minusQmgnzu

mz ] (347)

By using the method from the proof of Theorem 35 we can obtain the estimates ofum

n (z t) = um(z t) in Dn namely

C1β[u

mn Dn] le M9 um

n W 2p0

(Dn)le M10 (348)

where β = min(α 1 minus 2p0) 2 lt p0 le p Mj = Mj(q0 p0 α k0 k1 Mprimen gn Dn) j =

9 10 hereM primen = max1lemltinfin C10[um D2n] Hence from um

n (z) we can choose a sub-sequence unm(z) such that unm(z) unmz(z) uniformly converge to u0(z) u0z(z)and unmzz(z) unmzz(z) weakly converge to u0zz(z) u0zz(z) in Dn respectivelyFor instance we take n = 2 um

2 (z) = um(z) in D2 um2 (z) has a subsequence

um2(z) in D2 the limit function of which is u0(z) in D2 Next we take n = 3from um

3 (z) we can select a subsequence um3(z) in D3 the limit function is u0(z)in D3 Similarly from um

n (z)(n gt 3) we can choose a subsequence umn(z) inDn and the limit of which is u0(z) in Dn Finally from umn(z) in Dn we choosethe diagonal sequence umm(z) (m = 2 3 4 ) such that umm(z) ummz(z)uniformly converge to u0(z) u0z(z) and ummzz(z) ummzz(z) weakly converge tou0zz(z) u0zz(z) in any closed subset of D respectively the limit function u(z) = u0(z)is just a solution of equation (32) in D This completes the proof

Theorem 37 If equation (32) with A2(z) = 0 satisfies Condition C then ProblemQ for (32) has a unique solution

Proof By Theorem 23 we choose Dprime = Dn = m = 2 z1 = minus1 z2 = 1 and K = 0the second linear equation in (332) with A(z) = A3(z) has a unique solution w0(z)and the function

u0(z) = 2Reint z

minus1w0(z)dz + b0 (349)

is a solution of Problem Q for the first linear equation in (332) If u0(1) = bprime = b1then the solution is just a solution of Problem P for the linear equation (32) withA2(z) = 0 Otherwise u0(1) = bprime = b1 we find a solution u1(z) of Problem Q withthe boundary conditions

Re [λ(z)u1z] = 0 on Γ Im [λ(z)u1z]|z=0 = 1 u1(minus1) = 0

On the basis of Theorem 34 it is clear that u1(1) = 0 hence there exists a realconstant d = 0 such that b1 = bprime + du1(1) thus u(z) = u0(z) + du1(z) is just asolution of Problem P for the linear equation (32) with A2(z) = 0 As for thequasilinear equation (32) with A2 = 0 the existence of solutions of Problem Q andProblem P can be proved by the method as stated in the proof of last theorem

34 The discontinuous oblique derivative problem for elliptic equationsin general domains

In this subsection let Dprime be a general simply connected domain whose boundaryΓprime = Γprime

1 cupΓprime2 herein Γ

prime1Γ

prime2 isin C2

α(12 lt α lt 1) have two intersection points zprime zprimeprime with

the inner angles αprimeπ αprimeprimeπ (0 lt αprime αprimeprime lt 1) respectively We discuss the quasilinearuniformly elliptic equation

uzz = F (z u uz uzz) F = Re [Quzz + A1uz] + A1u+ A3 z isin Dprime (350)

in which F (z u uz uzz) satisfy Condition C in Dprime There are m points Z = z1 = zprime

znminus1 zn = zprimeprime zm on Γprime arranged according to the positive direction succes-sively Denote by Γprime

j the curve on Γprime from zjminus1 to zj j = 1 2 m z0 = zm and

Γprimej(j = 1 2 m) does not include the end points

Problem P prime The discontinuous oblique derivative boundary value problem for(350) is to find a continuous solution w(z) in Dlowast = DprimeZ satisfying the boundarycondition

12

partu

partν= Re [λ(z)uz] = c(z) z isin Γlowast = ΓprimeZ u(zj) = bj j = 1 m (351)

where cos(ν n) ge 0 λ(z) c(z) are given functions satisfying

herein α (12 lt α lt 1) k0 k2 are non-negative constants and assume that (βj +γj)β lt 1 β = min(α 1 minus 2p0)α0 βj(j = 1 m) are as stated in (127) α0 =max(1α1 1α2 1) Problem P prime with A3(z) = 0 inD r(z) = 0 on Γprime is called ProblemP prime

0 If cos(ν n) equiv 0 on each of Γj(j = 1 m) we choose the index K = m2 minus 1of Problem P prime which is defined as that in Subsection 23 If A2 = 0 in D the lastpoint conditions in (351) can be replaced by

u(zn) = bn Im [λ(z)uz]|zprimej= bj j = 1 n minus 1 n+ 1 m (353)

Here zprimej(isin Z j = 1 nminus 1 n+1 m) isin Γprime are distinct points and the condition

cos(ν n) ge 0 on Γprime can be canceled This boundary value problem is called ProblemQprime

Theorem 38 Let equation (350) in Dprime satisfy Condition C similar to before ThenProblem P prime and Problem Qprime for (350) are solvable and the solution u(z) can beexpressed by (317) but where β = min(α 1 minus 2p0)α0 Moreover if Q(z) = 0in D then the solution u(z) of equation (350) possesses the form in (317) wherew(z) = Φ(z)eφ(z) + ψ(z) and u(z) satisfies the estimate

C1δ [u D] = Cδ[u(z) D] + Cδ[X(z)w(z) D] le M11 = M11(p0 β δ k D) (354)

in which X(z) is given as

X(z) =mprod

j=2j =n

where ηj(j = 1 m) are as stated in (230) Besides the solution of Problem P prime

and Problem Qprime for (350) are unique if the following condition holds For any realfunctions uj(z) isin C1(Dlowast) Vj(z) isin Lp0(D)(j = 1 2) the equality

F (z u1 u1z V1)minusF (z u1 u1z V2) = Re [Q(V1minusV2)+A1(u1minusu2)z]+A2(u1minusu2) in Dprime

holds where |Q| le q0 in Dprime A1 A2 isin Lp0(Dprime)

Finally we mention that the above results can be generalized to the case of secondorder nonlinear elliptic equation in the form

uzz = F (z u uz uzz)

F = Re [Q(z u uz uzz) + A1(z u uz)uz] + A2(z u uz)u+ A3(z u uz) in D

satisfying the conditions similar to Condition C which is the complex form of thesecond order nonlinear elliptic equation

Φ(x y u ux uy uxx uxy uyy) = 0 in D

with certain conditions (see [86]1))

4 Boundary Value Problems for Degenerate EllipticEquations of Second Order in a Simply Connected Domain

This section deals with the oblique derivative problem for the degenerate ellipticequation of second order in a simply connected domain We first give a boundednessestimate of solutions of the oblique derivative problem for the equation and then byusing the principle of compactness the existence of solutions for the above obliquederivative problem is proved

41 Formulation of boundary value problems for degenerate ellipticequations

Let D be a simply connected domain with the boundary Γprime = Γ cup L0 where Γ isinC2

α (0 lt α lt 1) in the upper half plane with the end points minus1 1 and L0 = [minus1 1]and the inner angles of D at minus1 1 equal α1π α2π herein 0 lt α1 α2 lt 1 We considerthe elliptic equation of second order

Lu = ymuxx + uyy + a(x y)ux + b(x y)uy + c(x y)u = d(x y) in D (41)

here m is a positive number Its complex form is as follows

uzz minus Re [Q(z)uzz + A1(z)uz] + A2u = A3 in D (42)

4 Degenerate Elliptic Equations 109

where

Q(z) =1minus ym

1 + ym A1(z) = minus a+ bi

1 + ym A2(z) = minus c

2(1 + ym) A3(z) =

d

2(1 + ym)

Condition C

The coefficients Aj(z)(j = 1 2) are continuously differentiable in D and satisfy

C1α[Aj(z) D] le k0 j = 1 2 3 A2 = minus c

2(1 + ym)ge minus c

4ge 0 on D (43)

in which α(0 lt α lt 1) k0 are non-negative constants

The oblique derivative boundary value problem is as follows

Problem P In the domain D find a solution u(z) of equation (41) which iscontinuously differentiable in D and satisfies the boundary condition

lu=12

partu

partν+σ(z)u=φ(z) zisinΓ

partu

party=ψ(x) xisinL0

u(minus1)=b0 u(1)=b1

(44)

where ν is any unit vector at every point on Γ cos(ν n)ge 0 σ(z)ge σ0 gt 0 n is theunit outer normal at every point on Γ λ(z) = cos(ν x)minus i cos(ν y) and λ(z) φ(z)ψ(x) are known functions and b0 b1 are known constants satisfying the conditions

C1α[ηΓ] le k0 η = λ σ C1

α[φ L0] le k0 C1α[ψ L0] le k0 |b0| |b1| le k0 (45)

in which α (12 lt α lt 1) k0 σ0 are non-negative constants Problem P with theconditions A3(z) = 0 in D φ(z) = 0 on Γ ψ(z) = 0 on L0 and b0 = b1 = 0 is calledProblem P0 If cos(ν n) = 1 here n is a outward normal vector on Γ then Problem Pis the Neumann boundary value problem (Problem N) and if cos(ν n) gt 0 σ(z) = 0on Γ then Problem P is the regular oblique derivative problem ie third boundaryvalue problem (Problem O) in this case we choose σ(z) gt 0 on Γ If cos(ν n) = 0and σ(z) = 0 on Γ then from (44) we can derive

u(z) = 2Reint z

minus1uzdz + b0 = r(z) on Γ u(1) = b1 = 2Re

int 1

minus1uzdz + b0 (46)

In this case Problem P is called Problem D In the following there is no harm inassuming d(z) = 0 in (41)

42 A priori estimates of solutions for Problem P for (41)

First of all we give a lemma and then give a priori estimate of boundedness ofsolutions of Problem P for (41)

Lemma 41 Suppose that equation (41) or (42) satisfies Condition C and Lu ge0 (or Lu le 0) in D if the solution u(z) isin C2(D) cap C(D) of (41) attains its positivemaximum (or negative minimum ) at a point x0 isin (minus1 1) and maxΓ u(z) lt u(x0) (orminΓ u(z) gt u(x0)) on Γ then

limyrarr0

partu(x0 y)party

lt 0 (or limyrarr0

partu(x0 y)party

gt 0) (47)

if the limit exists

Proof Assume that the first inequality is not true namely

limyrarr0

partu(x0 y)party

= M prime ge 0 (48)

Obviously M prime = 0 Denote M = u(x0) B = maxD |b(z)| and by d the diameter ofD Thus there exists a small positive constant ε lt M such that maxΓ u(z) le M minus εMaking a function

v(z) =εu(z)

(MeBd minus εeBy)

we have

v(z)le ε(M minus ε)MeBdminusεeBd

ltεM

MeBdminusεon Γ v(x)lev(x0)=

εM

MeBdminusεon L0 (49)

Noting that Lu ge 0 the function v(x y) satisfies the inequality

ymvxx + vyy + a(x y)vx + b(x y)vy + c(x y)v ge 0 in D

where b = bminus2εBeBy(MeBd minusεeBy) c(x y) = cminusε(B+ b)BeBy(MeBd minusεeBy) le 0in D According to the above assumption we get

limyrarr0

partv(x0 y)party

=ε2BM

(MeBd minus ε)2gt 0

Hence v(x y) attains its maximum in D but from (49) it is impossible This provesthe first inequality in (47) Similarly we can prove the second inequality in (47)

Now we choose a positive constant η lt 1 and consider the equation

Lηu=(y+η)muxx+uyy+a(x y)ux+b(x y)uy+c(x y)u=d in D (410)

It is easy to see that (410) is a uniformly elliptic equation in D From Theorem36 we can derive that for every one of η = 1n gt 0 (n = 2 3 ) there exists asolution un(z) of Problem D for equation (410) In the following we shall give someestimates of the solution un(z)

Lemma 42 If Condition C holds then any solution un(z) of Problem P for (410)with d = 0 satisfies the estimate

C[un(z) D] le M12 = M12(α k0 D) (411)

where M12 is a non-negative constant

Proof We first discuss Problem D and choose two positive constants c1 c2 suchthat

c1 gec2+maxΓ

|r(z)|+maxD

ec2y c2 gtmaxL0

|ψ(x)|+maxD

|b|+2maxD

|d|+1

and make a transformation of function v(z) = c1 minus ec2y plusmn un(z) thus we have

Lηv le minusc2(c2 + b)ec2y + c(c1 minus ec2y) + 2maxD

|d| lt 0 z = x+ iy isin D

v gt 0 on Γ vy =partv

party= minusc2e

c2y plusmn ψ(z) lt 0 on L0(412)

by the extremum principle for elliptic equations the function v(z) cannot take thenegative minimum in D hence

v(z) = c1 minus ec2y plusmn un(z) ge 0 ie c1 ge ec2y ∓ un(z) in D (413)

hence |un(z)| le c1 minus ec2y le c1 = M12

For other case we introduce an auxiliary function v(z) = c1 minus ec2y plusmn un(z) wherec1 c2 are two positive constants satisfying the conditions

c2 gt maxD

|b(z)|+maxL0

|ψ(x)|+maxD

ec2y + 2maxD

|d|

c1 gt c2 +maxD

ec2y(1 +

c2

σ0

)+max

Γ

|φ(z)|σ0

(414)

We can verify that the function v(z) satisfies the conditions

Lηv lt 0 in D lv gt 0 on Γ vy lt 0 on L0 (415)

hence v(z) cannot attain the negative minimum in D Thus |un(z)| le c1 minus ec2y lec1 = M12 This completes the proof

Secondly from the sequence of solutions un(z) of Problem P for equation (410)we can choose a subsequence unk

(z) which uniformly converges to a solution ulowast(z)of (41) in any closed subset of DL In fact by Condition C and the estimate (411)we can derive the estimate of the solution un(z) as follows

C1β[un(z) D] le M13 = M13(β k0 D η) (416)

where η = 1n gt 0 and β (0 lt β le α) is a constant

Lemma 43 If Condition C holds then any solution un(z) of Problem P for (410)satisfies the estimate (416)

From the above lemma we can derive that the limit function ulowast(z) of unk(z)

satisfies the first boundary condition in (44) In order to prove that ulowast(z) satisfies

the second boundary condition in (44) we write the similar results in [24]1) as alemma

Lemma 44 Suppose that Condition C holds and 0 lt m lt 2 or m ge 2

a(x y) = O(ym2minus1+ε) ay = O(ym2minus2+ε) (417)

where ε is sufficiently small positive number Then any solution um(z) of Problem Pfor (410) with d = 0 satisfies the estimate

|uny| |((y + η)m+ε minus ηm+ε)u2nx| le M14 = M14(α k0 D) in Rn2δ0 (418)

where Rnδ = |x minus x0| lt ρ minus δ 0 lt y lt δl(Rnδ sub D) x0 isin (minus1 1) δ0 δ1 δ ρ(0 lt δ le 2δ0 lt ρ δ1 lt 1n) are small positive constants and M14 is a non-negativeconstant

Proof (1) First of all we prove the estimate

(uy)2 le [(y + η)m+ε1 minus ηm+ε1 ](ux)2 +M15 in Rnδ0 = Dlowast (419)

in which ε1(lt ε) M15 are non-negative constants f = f(x) = X4 = [ρ2 minus (x minusx0)2]4 g = g(x) = X2 = [ρ2 minus (x minus x0)2]2 are functions of x and F = ηm+ε1 minusY m+ε1 G = 1 minus Y ε1 H = minusY ε1 are functions of Y = y + η and introducing anauxiliary function

v(z) = f [F (ux)2 +G(uy)2] + gu2 +H in Dlowast (420)

if v(z) attains a positive maximum value at a point zlowast isin Dlowast then

v(z) gt 0 vx = vy = 0 Lη(v) = Lη(v) + cv le 0 at zlowast (421)

From (420) we get

vx = 2f [Fuxuxx +Guyuxy] + 2guux + f prime[F (ux)2 +G(uy)2] + gprimeu2 = 0

vy = 2f [Fuxuxy +Guyuyy] + 2guuy + f [F prime(ux)2 +Gprime(uy)2] +H prime = 0

Lη(v)=2f [FuxLη(ux)+GuyLη(uy)]+2guLη(u)+2fF [Y m(uxx)2+(uxy)2]

+2fG[Y m(uxy)2+(uyy)2]+2g[Y m(ux)2+(uy)2]+4Y mf prime[Fuxuxx+Guyuxy]

+4f [F primeuxuxy+Gprimeuyuyy]+Y mf primeprime[F (ux)2+G(uy)2]+f [F primeprime(ux)2+Gprimeprime(uy)2]

(422)

+4Y mgprimeuux+Y mgprimeprimeu2+H primeprime+af prime[F (ux)2+G(uy)2]

+agprimeu2+bf [F prime(ux)2+Gprime(uy)2]+bH prime+2cH

in which Y = y + η and from (410) we obtain

Lη(ux) = minus(axux + bxuy + cxu)

Lη(uy) = minus(mY mminus1uxx + ayux + byuy + cyu)

2fFY m(uxx)2 = 2fFY minusm(uyy + aux + buy + cu)2

(423)

and then we have

2fGuyLη(uy) =minusm

Y[2guuy + f [F prime(ux)2 +Gprime(uy)2] +H prime]minus 2mfF

Y(424)

timesuxuxy +2mfG

Yuy(aux + buy + cu)minus 2fGuy(ayux + byuy + cyu)

Substituting (423)(424) into Lη(v) it is not difficult to derive

1f

Lη(v) = 2(FY minusm +G)(Y m(uxy)2 + (uyy)2) + 2(2F prime minus mF

Y)uxuxy

+4[Gprimeuy + FY minusm(aux + buy + cu)]uyy minus 2Fux(axux + bxuy + cxu)

minus2Guy(ayux + byuy + cyu) +2mG

Yuy(aux + buy + cu)

+2FY minusm(aux + buy + cu)2 +[Y m

(f primeprime

fminus 2

f prime2

f 2

)+ a

f prime

f

](425)

times[F (ux)2 +G(uy)2] +(minusm

Y+ b)[F prime(ux)2 +Gprime(uy)2] + F primeprime(ux)2

+Gprimeprime(uy)2+2g

f[Y m(ux)2+(uy)2]+4Y m

(gprime

fminus f primeg

f 2

)uuxminus2mg

fYuuy

+[Y m

(gprimeprime

fminus 2

f primegprime

f 2

)+ a

gprime

f

]u2 +

H primeprime + (minusmY + b)H prime + 2cHf

Moreover by (411)(417)(420) and (425) we obtain

1f

Lη(v) = (2+o(1))[Y m(uxy)2+(uyy)2]+O

Y m+ε1minus1|uxuxy|

+Y mminus2+2ε+ε1|ux|2+(Y m2minus1+ε+ε1|ux|+Y ε1minus1|uy|+Y ε1)|uyy|+Y m2minus2+ε|uxuy|+Y minus1(uy)2+Y m2minus1+ε+ε1|ux|+Y minus1|uy|+Y ε1

+(

Y m

X2 +Y m2minus1+ε

X

)[Y m+ε1(ux)2+(uy)2]

+(1minusε1)(m+ε1)2(1+o(1))Y mminus2+ε1(ux)2

+ε1(m+1minusε1)(1+o(1))Y ε1minus2(uy)2

+2Y m

X2 (ux)2+2(uy)2

X2 +O

(Y m

X3 |ux|+Y minus1

X2 |uy|)

+ε1(m+1minusε1)(1+o(1))

X4 Y ε1minus2

(426)

When 0ltε1 ltmin(ε1) it is easy to see that the right-hand side of (426) is positivewhich contradicts (421) hence v(z) cannot have a positive maximum in Dlowast

On the basis of the estimate (411) we see that v(z) on the upper boundary |xminusx0|ltρy=1n of Dlowast is bounded and v(x)lefGψ2+gM2

12 on the lower boundary|xminusx0|ltρy=0 of Dlowast moreover v(x)lt0 on the left-hand side and right-hand side|xminusx0|=ρ0ltylt1n of Dlowast Thus the estimate (419) is derived

(2) Now we give the estimate

|((y+η)m+εminusηm+ε)u2nx|leM16 isin Rnδ0=Dlowast (427)

in which M16 is independent of η In fact we introduce the auxiliary function(420) where we choose that f=f(x)=X4=[(ρminusδ0)2minus(xminusx0)2]4g=g(x)=X2=[(ρminusδ0)2minus(xminusx0)2]2 and F =Y m+ε1 minusηm+ε1 G=Y ε2 H=Y ε3 herein Y =y+ηε2ε3

are positive constants satisfying 0lt2ε3 ltε2 leε12 If v(z) attains a positive maxi-mum value at a point zlowast isinDlowast then we have (421) Substituting (410)(420)(423)into (422) we get

Lη(v)f

= 2F [Y m(uxx)2+(uxy)2]+2G[Y m(uxy)2

+(uyy)2]+2g

f[Y m(ux)2+(uy)2]+Σ

(428)

in which

Σ = 4F primeuxuxy+2(2Gprime+

mG

Y

)uyuyy minus2Fux(axux+bxuy+cxu)

minus2Guy(ayux+byuy+cyu)+2mG

Yuy(aux+buy+cu)

+[Y m

(f primeprime

fminus2f

prime2

f 2

)+a

f prime

f

][F (ux)2+G(uy)2]+b[F prime(ux)2

+Gprime(uy)2]+F primeprime(ux)2+Gprimeprime(uy)2+4Y m

(gprime

fminus f primeg

f 2

)uux

+[Y m

(gprimeprime

fminus2f

primegprime

f 2

)+a

gprime

f

]u2+

H primeprime+bH prime+2cHf

(429)

From (411)(417)(420) and (429) it follows that

Σ = O

times|uxuy|+Y ε2minus2|uy|2+Y m+ε1|ux|+Y ε2minus1|uy|+(

Y m

X2 +Y m2minus1+ε

X

)

times(Y m+ε1|ux|2+Y ε2|uy|2)+Y m

X3 |ux|+Y ε3minus2

X4

geminusG(Y m|uxy|2+|uyy|2)

minus2 g

f(Y m|ux|2+|uy|2)+O

(Y m+ε1minus2|ux|2+Y ε2minus2|uy|2+ Y ε3minus2

X4

)

(430)

By (428)(430) if we can verify the following inequality

G[Y m(uxy)2+(uyy)2]+O

X4

)gt0 (431)

then the inequality Lη(v)gt0 Noting that F primeGprimeH prime are positive and from (411)(420) (422) we have

2|Fuxuxy+Guyuyy|geF prime|ux|2+Gprime|uy|2minus2 g

f|uuy|+H prime

f

geF prime|ux|2+ 1f(H primeminus u2

Gprime2 )geF prime|ux|2+1+o(1)f

H prime

Hence(F 2|ux|2+Y mG2|uy|2)(Y m|uxy|2+|uyy|2)=Y m(Fuxuxy+Guyuyy)2+(Y mGuyuxy minusFuxuyy)2

ge Y m

4(F prime|ux|2+1+o(1)

fH prime)2

(432)

By (419)(420)(432) we obtain

(F 2|ux|2+Y mG2|uy|2)[Y m|uxy|2+|uyy|2

+1G

O

X4

)]

ge Y m

4(mY m+ε1minus1|ux|2+ ε3+o(1)

X4 Y ε3minus1)2

+Y m+ε2(Y m+ε1+ε2|ux|2+Y ε2M15)Y minusε2O

(Y m+ε1minus2|ux|2+ Y ε2minus2

X4

)gt0

(433)

From (422) we see that uxuy cannot simultaneously be zero By (433) we have(431) such that Lη(v)gt0 holds This contradicts (421) Therefore v(z) cannotattain a positive maximum in Dlowast

On the basis of (411)(420) and the boundary condition (44) we see that vlefGψ2+gM2

12+H on the lower boundary of Dlowast is uniformly bounded Moreover thefunction v(z) is uniformly bounded on the upper left-hand and right-hand boundariesof Dlowast Thus the estimates in (418) are derived

Now we prove a theorem as follows

Theorem 45 Suppose that Condition C(417) hold and ax+cle0 in D Then anysolution un(z) of Problem P for (410) satisfies the estimate

|ux|leM17=M17(αk0D) in D (434)

where we assume that the inner angles αjπ(j=12) of D at z=minus11 satisfy the con-ditions 0ltαj(=12)lt1j=12 and M17 is a non-negative constant

Proof We find the derivative with respect to x to equation (410) and obtain

(y+η)muxxx+uxyy+a(xy)uxx+b(xy)uxy+[ax+c(xy)]ux=F (xy)

F =f(xy)minusbxuy minuscxu in D(435)

On the basis of Lemmas 41 and 44 we have

|F (xy)|= |f(xy)minusbxuy minuscxu|leM18 ltinfin in D

and equation (435) can be seen as a elliptic equation of ux and the solution ux

satisfies the boundary conditions

partu

parts=cos(sx)ux+cos(sy)uy=

partr(z)parts

on Γ (ux)y=ψprime(x) on L0 (436)

in which s is the tangent vector at every point Γ Noting that the angles αjπ(j=12)satisfy the conditions 0ltαj (=12)lt1j=12 it is easy to see that cos(sx) =0 atz=minus11 Thus the first boundary condition in (46) can be rewritten in the form

ux=R(z)=minuscos(sy)cos(sx)

uy+1

cos(sx)partr(z)parts

on Γ (437)

here R(z) is a bounded function in the neighborhood (subΓ) of z=minus11 hence by themethod in the proof of Lemma 42 we can prove that the estimate (434) holds Asfor cos(sx)=0 at z=minus1 or z=1 the problem remains to be solved

Theorem 46 Suppose that Condition C and (417) hold Then Problem P for(41) or (42) has a unique solution

Proof As stated before for a sequence of positive numbers η=1nn=23 wehave a sequence of solutions un(z) of the corresponding equations (410) withη=1n(n=23) which satisfy the estimate (416) hence from un(z) we canchoose a subsequence unk

(z) which converges to a solution u0(z) of (42) in DcupΓsatisfying the first boundary condition in (44) It remains to prove that u0(z) satisfiesthe other boundary condition in (44) For convenience we denote unk

(z) by u(z)x0 isany point in minus1ltx0 lt1 and give a small positive number β there exists a sufficientlysmall positive number δ such that |ψ(x)minusψ(x0)|ltβ when |xminusx0|ltδ Moreover weconsider an auxiliary function

v(z)=F (ux)2plusmnuy+G+f G=minusCyε2 minusτ ∓ψ(x0) f=minusC(xminusx0)2

F =

⎧⎨⎩Y m+1+ε2 minusηm+1+ε2 0ltmlt1

Y m+ε1 minus(m+ε1)ηmminus1+ε1Y +(m+ε1minus1)ηm+ε1 mge1(438)

where Y =y+ηη=1nε2(0ltε2 leε13) are positive constants and C is an undeter-mined positive constants We first prove that v(z) cannot attain its positive max-imum in Dlowast=|xminusx0|2+y2 ltσ2ygt0 Otherwise there exists a point zlowast such that

v(zlowast)=maxDlowast v(z)gt0 and then

vx=2Fuxuxxplusmnuxy+f prime=0 vy=2Fuxuxy plusmnuyy+F prime(ux)2+Gprime=0

Lη(v)=2FuxLη(ux)plusmnLη(uy)+2F [Y m(uxx)2+(uxy)2]+4F primeuxuxy

+[F primeprime+bF prime+cF ](ux)2plusmncuy+Y mf primeprime

+af prime+2cf+Gprimeprime+bGprime+2cG

(439)

and from (423) and (439) we obtain

plusmnLη(uy) = minusm

Y[2Fuxuxy+F prime(ux)2+Gprime]

plusmnm

Y(aux+buy+cu)∓(ayux+byuy+cyu)

(440)

Moreover by (423) and (439)ndash(440) we have

Lη(v) = minus2Fux(axux+bxuy+cxu)minus m

Y[2Fuxuxy+F prime(ux)2+Gprime

∓(aux+buy+cu)]∓(ayux+byuy+cyu)

+2F [Y m(uxx)2+(uxy)2]∓4F primeux(2Fuxuxx+f prime)

+[F primeprime+bF prime+cF ](ux)2plusmncuy+Y mf primeprime+af prime+2cf+Gprimeprime+bGprime+2cG

= 2FY m[uxx∓2F primeY minusm(ux)2]2+2F(uxy minus m

2Yux

)2

+[F primeprimeminus m

YF prime+bF prime+cF minus8F prime2Y minusmF (ux)2](ux)2

+[minus2F (bxuy+cxu)plusmn ma

Y∓ay ∓4F primef prime

]ux+Gprimeprimeminus m

YGprime+bGprime

+2cGminus(

m2

2Y 2+2ax

)F (ux)2plusmn m

Y(buy+cu)∓(byuy+cyu)

plusmncuy+Y mf primeprime+af prime+2cf

(441)

Choosing a sufficiently small positive number σ such that the domain Dlowast=|xminusx0|2+y2 ltσ2capygt0subDσltmin[ρminus2δ0δ1] where δ0δ1 are constants as stated inLemma 44 and |ψ(z)minusψ(x0)|ltτ we can obtain

Lη(v) ge ε2(m+1+ε2)(1+o(1))Y m+ε2minus1(ux)2+O( 1

Y

)|ux|

+Cε2(m+1minusε2)(1+o(1))yε2minus2 gt0 if 0ltmlt1(442)

and

Lη(v) ge (m+ε1)(m+ε1minus1)Y m+ε1minus2(ux)2+O(Y m2+εminus2)|ux|+Cε2(m+1minusε2)(1+o(1))Y ε2minus2 gt0 if mge1 ε=ε1

(443)

in which we use Lemmas 42 and 44 the conditions (417) (438) (441) and

F (ux)2=O(Y 2ε2) F prime(ux)2=O(Y minus1+2ε2) if mge1

It is clear that (442)(443) contradict (421) hence v(z) cannot attain a positivemaximum in Dlowast From (438) we get

v(z)=F (ux)2plusmn [uy minusψ(x0)]minusτ minusC[(xminusx0)2+yε2 ] (444)

Moreover it is easy to see that v(z)lt0 on the boundary of Dlowast provided that theconstant C is large enough Therefore v(z)le0 in Dlowast From F ge0 and (442)ndash(444)the inequality

plusmn[uyminusψ(x0)]minusτminusC[(xminusx0)2+Y ε2 ]le0 ie|uy minusψ(x0)|leτ+C[(xminusx0)2+Y ε2 ] in Dlowast

(445)

is derived Firstly let ηrarr0 and then let zrarrx0 τ rarr0 we obtain limpartupartyrarrψ(x0)Similarly we can verify limz(isinDlowast)rarrx0 uy=ψ(x0) when x0=minus11 Besides we can alsoprove that u(z)rarru(x0) as z(isinDlowast)rarrx0 when x0=minus11 This shows that the limitfunction u(z) of un(z) is a solution of Problem P for (41)

Now we prove the uniqueness of solutions of Problem P for (41) it suffices toverify that Problem P0 has no non-trivial solution Let u(z) be a solution of ProblemP0 for (41) with d=0 and u(z) equiv0 in D Similarly to the proof of Theorem 34we see that its maximum and minimum cannot attain in DcupΓ Moreover by usingLemma 41 we can prove that the maximum and minimum cannot attain at a pointin (minus11) Hence u(z)equiv0 in D

Finally we mention that for the degenerate elliptic equation

K(y)uxx+uyy=0 K(0)=0 K prime(y)gt0 in D (446)

which is similar to equation (41) satisfying Condition C and other conditions asbefore hence any solution u(z) of Problem P0 for (446) satisfies the estimates(411)(418) and (434) in D provided that the inner angles αjπ(j=12) of D atz=minus11 satisfy the conditions 0ltαj (=12)lt1j=12 Equation (446) is the Chap-lygin equation in elliptic domain Besides oblique derivative problems for the degen-erate elliptic equations of second order

uxx+ymuyy+a(xy)ux+b(xy)uy+c(xy)u=f(xy) in D

ym1uxx+ym2uyy+a(xy)ux+b(xy)uy+c(xy)u=f(xy) in D

needs to be considered where mm1m2 are non-negative constants

The references for this chapter are [3][6][11][15][18][23][24][30][33][38][39][40][46][48][50][53][58][60][65][67][76][78][80][81][82] [85][86][94][96][99]

CHAPTER IV

FIRST ORDER COMPLEX EQUATIONS OFMIXED TYPE

In this chapter we introduce the RiemannndashHilbert boundary value problem for firstorder complex equation of mixed (elliptic-hyperbolic) type in a simply connecteddomain We first prove uniqueness and existence of solutions for the above boundaryvalue problem and then give a priori estimates of solutions for the problem finallydiscuss the solvability of the above problem in general domains The results in thischapter will be used in the following chapters

1 The RiemannndashHilbert Problem for Simplest First OrderComplex Equation of Mixed Type

In this section we discuss the RiemannndashHilbert boundary value problem for the sim-plest mixed complex equation of first order in a simply connected domain Firstly weverify a unique theorem of solutions for the above boundary value problem Moreoverthe existence of solutions for the above problem is proved

11 Formulation of the RiemannndashHilbert problem for the simplestcomplex equation of mixed type

Let D be a simply connected bounded domain in the complex plane CI with theboundary partD = Γ cup L where Γ(sub y gt 0) isin Cα(0 lt α lt 1) with the end pointsz =0 2 and L=L1cupL2 L1=x=minusy 0 le x le 1

L2 = x = y + 2 1 le x le 2 Denote D+ =D cap y gt 0 Dminus = D cap y lt 0 and z1 = 1 minus iWithout loss of generality we may assume that Γ=|z minus 1|=1 yge0 otherwise through a conformalmapping this requirement can be realizedWe discuss the mixed system of first order equa-

tions

uxminusvy=0 vx+sgny uy=0 in D (11)

120 IV First Order Mixed Complex Equations

Its complex form is the following complex equation of first orderwz

wzlowast

= 0 in

D+

Dminus

(12)

where

w = u+ iv z = x+ iy wz =12[wx + iwy] wzlowast =

12[wx minus iwy]

Problem A Find a continuous solution w(z) of (12) in Dlowast = D(0 2cupxplusmn y =2 Im z le 0) or Dlowast = D(0 2 cup x plusmn y = 0 Im z le 0) satisfying the boundaryconditions

Re [λ(z)w(z)] = r(z) z isin Γ (13)

Re [λ(z)w(z)] = r(z) z isin Lj(j = 1 or 2) Im [λ(z1)w(z1)] = b1 (14)

where λ(z) = a(z) + ib(z) |λ(z)| = 1 z isin Γ cup Lj(j = 1 or 2) b1 is a real constantand λ(z) r(z) b1 satisfy the conditions

Cα[λ(z)Γ] le k0 Cα[r(z)Γ] le k2 |b1| le k2

Cα[λ(z) Lj] le k0 Cα[r(z) Lj] le k2 j = 1 or 2

a(z)minus b(z) = 0 on L1 or a(z) + b(z) = 0 on L2

(15)

in which α(0 lt α lt 1) k0 k2 are non-negative constants For convenience we mayassume that w(z1) = 0 otherwise through a transformation of the function w(z) minusλ(z1)[r(z1) + ib1] the requirement can be realized

This RiemannndashHilbert problem (Problem A) for (12) with r(z) = 0 z isin Γ cupL1 (or L2) and b1 = 0 will be called Problem A0 The number

K =12(K1 +K2) (16)

is called the index of Problem A and Problem A0 where

Kj =[φj

π

]+ Jj Jj = 0 or 1 eiφj =

λ(tj minus 0)λ(tj + 0)

γj =φj

πminus Kj j = 1 2 (17)

in which t1 = 2 t2 = 0 λ(t) = (1 minus i)radic2 on L0 = [0 2] or λ(t) = (1 + i)

radic2 on

L0 = [0 2] and λ(t1 minus0) = λ(t2+0) = exp(7πi4) or exp(πi4) Here we only discussthe case of K = (K1+K2)2 = minus12 on the boundary partD+ of D+ In order to ensurethat the solution w(z) of Problem A is continuous in the neighborhood(sub Dminus) of thepoint z = 0 or z = 2 we need to choose γ1 gt 0 or γ2 gt 0 respectively

1 Simplest Mixed Complex Equation 121

12 Uniqueness of solutions of the RiemannndashHilbert problem for thesimplest complex equation of mixed type

Theorem 11 Problem A for (12) has at most one solution

Proof Let w1(z) w2(z) be any two solutions of Problem A for (12) It is clear thatw(z) = w1(z)minus w2(z) is a solution of Problem A0 for (12) with boundary conditions

Re [λ(z)w(z)] = 0 z isin Γ (18)

Re [λ(z)w(z)] = 0 z isin Lj(j = 1 or 2) Im [λ(z1)w(z1)] = 0 (19)

Due to the complex equation (12) in Dminus can be reduced to the form

ξν = 0 ηmicro = 0 in Dminus (110)

where micro = x + y ν = x minus y ξ = u + v η = u minus v the general solution of system(110) can be expressed as

u(z) =f(x+ y) + g(x minus y)

2 v(z) =

f(x+ y)minus g(x minus y)2

(111)

in which f(t) g(t) are two arbitrary real continuously differentiable functions on [0 2]Noting the boundary condition (19) we have

au+bv = 0 on L1 or L2 [av minus bu]|z=z1=0 ie

[a((1minusi)x)+b((1minusi)x)]f(0)

+[a((1minusi)x)minusb((1minusi)x)]g(2x)=0 on [0 1] or

[a((1+i)xminus2i)+b((1+i)xminus2i)]f(2xminus2)+[a((1+i)xminus2i)minusb((1+i)xminus2i)]g(2)=0 on [1 2]

w(z1)=0 [u+v]|z1=f(0)=0 or [uminusv]|z1=g(2)=0

(112)

The second formula in (112) can be rewritten as

[a((1minusi)t2)+b((1minusi)t2)]f(0) +[a((1minusi)t2)minus b((1minus i)t2)]g(t) = 0

f(0) = g(t) = 0 or [a((1+i)t2+1minusi)+b((1minusi)t2+1minusi)]f(t)

+[a((1minusi)t2+1minusi)minusb((1minusi)t2+1minusi)]g(2)=0

g(t) = f(t) = 0 t isin [0 2]

(113)

Thus the solution (111) becomes

u(z) = v(z) =12f(x+ y) g(x minus y) = 0 or

u(z) = minusv(z) =12g(x minus y) f(x+ y) = 0

(114)

if a(z)minus b(z) = 0 on L1 or a(z) + b(z) = 0 on L2 respectively In particular we have

u(x) = v(x) =12f(x) x isin [0 2] or

u(x) = minusv(x) =12g(x) x isin [0 2]

(115)

Next due to f(0) = 0 or g(2) = 0 from (115) we can derive that

u(x)minus v(x) = 0 ie Re [(1 + i)w(x)] = 0 x isin [0 2] oru(x) + v(x) = 0 ie Re [(1minus i)w(x)] = 0 x isin [0 2]

(116)

Noting the index K = minus12 of Problem A for (12) in D+ and according to the resultin Section 1 Chapter III and [85]11)[86]1) we know that w(z) = 0 in D+ Thus

u(z) + v(z) = Re [(1minus i)w(z)] = f(x+ y) = 0 g(x minus y) = 0 or

u(z)minus v(z) = Re [(1 + i)w(z)] = g(x minus y) = 0 f(x+ y) = 0(117)

obviously

w(z) = u(z) + iv(z) = w1(z)minus w2(z) = 0 on Dminus (118)

This proves the uniqueness of solutions of Problem A for (12)

13 Existence of solutions of the RiemannndashHilbert problem for thesimplest complex equation of mixed type

Now we prove the existence of solutions of the RiemannndashHilbert problem (ProblemA) for (12)

Theorem 12 Problem A for (12) has a solution

Proof As stated before the general solution of (12) in Dminus can be expressed as

2 v(z) =

ie w(z) =(1 + i)f(x+ y) + (1minus i)g(x minus y)

2

(119)

in which f(t) g(t) are two arbitrary real continuously differentiable functions on [0 2]Taking into account the boundary condition (14) we have

au+ bv = r(x) on L1 or L2 ie

+ [a((1minusi)x)minusb((1minusi)x)]g(2x)

=2r((1minusi)x) on [0 1] f(0)

= [a(z1) + b(z1)]r(z1) + [a(z1)minus b(z1)]b1 or

[a((1+i)xminus2i)+b((1+i)xminus2i)]f(2xminus2)+[a((1+i)xminus2i)minusb((1+i)xminus2i)]g(2)

=2r((1+i)xminus2i) on [1 2] g(2)= [a(z1)minus b(z1)]r(z1)minus [a(z1) + b(z1)]b1

(120)

The second and third formulas in (120) can be rewritten as

[a((1minus i)t2)minus b((1minus i)t2)]g(t)

= 2r((1minus i)t2)minus [a((1minus i)t2) + b((1minus i)t2)]f(0) t isin [0 2] or[a((1+i)t2+1minusi)+b((1+i)t2+1minusi)]f(t)

=2r((1+i)t2+1minusi)

minus[a((1+i)t2+1minusi)minusb((1+i)t2+1minusi)]g(2) tisin [0 2]

(121)

thus the solution (119) possesses the form

u(z) =12f(x+ y) + g(x minus y) v(z) =

12f(x+ y)minus g(x minus y)

g(xminusy)=2r((1minusi)(xminusy)2)minus[a((1minusi)(xminusy)2)+b((1minusi)(xminusy)2)]f(0)

a((1minusi)(xminusy)2)minusb((1minusi)(xminusy)2)

u(z)=12g(xminusy)+f(x+ y) v(z)=

12minusg(xminusy)+f(x+ y) (122)

f(x+ y) =2r((1 + i)(x+ y)2 + 1minus i)

a((1 + i)(x+ y)2 + 1minus i) + b((1 + i)(x+ y)2 + 1minus i)

minus [a((1 + i)(x+ y)2 + 1minus i)minus b((1 + i)(x+ y)2 + 1minus i)]g(2)a((1 + i)(x+ y)2 + 1minus i) + b((1 + i)(x+ y)2 + 1minus i)

if a(z)minus b(z) = 0 on L1 or a(z) + b(z) = 0 on L2 respectively In particular we get

u(x) =12f(x) + g(x) v(x) =

12f(x)minus g(x)

g(x) =2r((1minus i)x2)minus [a((1minus i)x2) + b((1minus i)x2)]f(0)

a((1minus i)x2)minus b((1minus i)x2)

u(x) =12g(x) + f(x) v(x) =

12minusg(x) + f(x)

f(x)=2r((1+i)x2+1minusi)

a((1 + i)x2 + 1minus i) + b((1 + i)x2 + 1minus i)

minus [a((1+i)x2+1minusi)minusb((1+i)x2+1minusi)]g(2)a((1 + i)x2 + 1minus i) + b((1 + i)x2 + 1minus i)

x isin [0 2](123)

u(x)minusv(x)=2r((1minusi)x2)minus[a((1minusi)x2)+b((1minusi)x2)]f(0)

a((1minusi)x2)minusb((1minusi)x2) or

u(x)+v(x)=2r((1+i)x2+1minusi)

a((1+i)x2+1minusi)+b((1+i)x2+1minusi)

minus [a((1+i)x2+1minusi)minusb((1+i)x2+1minusi)]g(2)a((1+i)x2+1minusi)+b((1+i)x2+1minusi)

xisin [02]

f(0)=[a(1minusi)+b(1minusi)]r(1minusi)+[a(1minusi)minusb(1minusi)]b1 or

g(2)=[a(1minusi)minusb(1minusi)]r(1minusi)minus [a(1minusi)+b(1minusi)]b1

(124)

ieRe [(1+i)w(x)]=s(x)

s(x)=2r((1minusi)x2)minus[a((1minusi)x2)+b((1minusi)x2)]f(0)

xisin [0 2] or

Re [(1minusi)w(x)]=s(x)=2r((1+i)x2+1minusi)

a((1+i)x2+1minus i)minusb((1+i)x2+1minus i)

minus [a((1+i)x2+1minusi)minusb((1+i)x2+1minusi)]g(2)a((1+i)x2+1minus i)minusb((1+i)x2+1minus i)

xisin [0 2]

(125)

if a((1minus i)x)minus b((1minus i)x) = 0 on [01] or a((1 + i)x minus 2i) + b((1 + i)x minus 2i) = 0 on[12] respectively We introduce a conformal mapping ζ = ζ(z) from the domain D+

onto the upper half-plane G = Im ζ gt 0 such that the three points z = 0 1 2 mapto ζ = minus1 0 1 respectively it is not difficult to derive that the conformal mappingand its inverse mapping can be expressed by the elementary functions namely

ζ(z) =5(z minus 1)

(z minus 1)2 + 4 z(ζ) = 1 +

52ζ(1minus

radic1minus 16ζ225)

Denoting W (ζ) = w[z(ζ)] and

Λ(ζ) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

λ[z(ζ)] ζ isin Γ1 = ζ(Γ)

1minus iradic2

ζ isin Γ2 = minus1 le Re ζ le 1 Im ζ = 0 or

1 + iradic2

ζ isin Γ2 = minus1 le Re ζ le 1 Im ζ = 0

(126)

in which the points ζ1 = 1 ζ2 = minus1 are the discontinuous points of Λ(ζ) on partG =Im ζ = 0 from (16)(17) it can be seen that the index of Λ(ζ) on partG = Im ζ =0 is K = minus12 Hence according to the result of Theorem 11 Chapter III weknow that the discontinuous RiemannndashHilbert boundary value problem for analyticfunctions W (ζ) in G with the boundary condition

Re [Λ(z)W (ζ)] = R(ζ) =

⎧⎨⎩ r[z(ζ)] ζ isin Γ1

s[z(ζ)]radic2 ζ isin Γ2

(127)

has a unique solution W (ζ) in G as follows

W (ζ) =X(ζ)πi

[int infin

minusinfinΛ(t)R(t)(t minus ζ)X(t)

dt+ iclowast2 + ζ

2minus ζ

]in G (128)

and

X(ζ) = iζ minus 2ζ minus i

Π(ζ)eiS(ζ) Π(ζ) =(

ζ minus 1ζ + i

)γ1(

ζ + 1ζ + i

)γ2

clowast =2i+ 12 + i

int infin

minusinfinΛ(t)R(t)

X(t)(t minus i)dt

and S(ζ) is an analytic function in Im ζ gt 0 with the boundary condition

Re [S(t)] = arg[Λ1(t)(

t minus 2t+ i

)] on Im t = 0 Im [S(i)] = 0 (129)

where γj(j = 1 2) are as stated in (17) and Λ1(t) = λ(t)Π(t)(t minus 2)|x + i|[|Π(t)|times|t minus 2|(x + i)] and the boundedness of w(z) or boundedness of integral of thesolution w(z) in the neighborhood sub D0 2 of t1 = 2 and t2 = 0 is determined byJj = 0 γj gt 0 or Jj = 0 γj = 0 and Jj = 1(j = 1 2) respectively Hence Problem Afor (12) has a solution w(z) in the form

w(z)=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

W [ζ(z)] z isin D+0 2

12

(1 + i)f(x+ y) + (1minus i)

times2r((1minusi)(xminusy)2)minus[a((1minusi)(xminusy)2)+b ((1minusi)(xminusy)2)]f(0)a((1minus i)(x minus y)2)minus b ((1minus i)(x minus y)2)

or

12

(1minusi)g(xminusy)+

2(1+i)r((1+i)(x+y)2+1minusi)a((1+i)(x+y)2+1minusi)+b ((1+i)(x+y)2+1minusi)

minus [a((1 + i)(x+ y)2 + 1minus i) + b ((1 + i)(x+ y)2 + 1minus i)]g(2)a((1+i)(x+y)2+1minusi)+b ((1+i)(x+y)2+1minusi)

z = x+ iy isin Dminus0 2(130)

in which f(0) g(2) are as stated in (124) W (ζ) in D+0 2 is as stated in (128)and from (123) we derive that

f(x+ y) = u(x+ y) + v(x+ y) = Re [(1minus i)W (ζ(x+ y))]

g(x minus y) = u(x minus y)minus v(x minus y) = Re [(1 + i)W (ζ(x minus y))](131)

where W [ζ(x+ y)] and W [ζ(x minus y)] are the values of W [ζ(z)] on 0 le z = x+ y le 2and 0 le z = x minus y le 2 respectively

From the foregoing representation of the solution w(z) of Problem A for (12) andthe mapping ζ(z) we can derive that w(z) satisfies the estimate

Cβ[w(z)X(z) D+] + Cβ[wplusmn(z)Y plusmn(z) Dminus] le M1 (132)

in which X(z) = Π2j=1|z minus tj|2|γj |+δ Y plusmn(z) = |xplusmn y minus tj|2|γj |+δ wplusmn(z) = Rew plusmn Imw

β(0 lt β lt δ) δ are sufficiently small positive constants and M1 = M1(β k0 k2 D)is a non-negative constant [85]15)

Finally we mention that if the index K is an arbitrary even integer or 2K isan arbitrary odd integer the above RiemannndashHilbert problem for (12) can be con-sidered but in general the boundary value problem for K le minus1 have some solvabilityconditions or its solution for K ge 0 is not unique

2 The RiemannndashHilbert Problem for First Order LinearComplex Equations of Mixed Type

In this section we discuss the RiemannndashHilbert boundary value problem for first orderlinear complex equations of mixed (elliptic-hyperbolic) type in a simply connecteddomain Firstly we give the representation theorem and prove the uniqueness ofsolutions for the above boundary value problem secondly by using the method ofsuccessive iteration the existence of solutions for the above problem is proved

21 Formulation of RiemannndashHilbert problem of first order complexequations of mixed type

Let D be a simply connected bounded domain in the complex plane CI with theboundary partD = Γcup L where Γ L = L1 cup L2 D

+ = D cap y gt 0 Dminus = D cap y lt 0and z1 = 1minus i are as stated in Section 1

We discuss the first order linear system of mixed (elliptic-hyperbolic) type equa-tions ⎧⎨⎩ux minus vy = au+ bv + f

vx + sgny uy = cu+ dv + gz = x+ iy isin D (21)

2 Linear Mixed Complex Equations 127

in which a b c d f g are functions of (x y)(isin D) its complex form is the followingcomplex equation of first order

wz

wzlowast

= F (z w) F = A1(z)w + A2(z)w + A3(z) in

D+

Dminus

(22)

where

12[wx minus iwy]

A1 =a minus ib+ ic+ d

4 A2 =

a+ ib+ ic minus d

4 A3 =

f + ig

2

Condition C

Aj(z) (j = 1 2 3) are measurable in z isin D+ and continuous in Dminus in Dlowast =D(0 2 cup x plusmn y = 2 Im z le 0) or Dlowast = D(0 2 cup x plusmn y = 0 Im z le 0) andsatisfy

Lp[Aj D+] le k0 j = 1 2 Lp[A3 D+] le k1 (23)

C[Aj Dminus] le k0 j = 1 2 C[A3 Dminus] le k1 (24)

where p (gt 2) k0 k1 are non-negative constants

22 The representation and uniqueness of solutions of the RiemannndashHilbert problem for mixed complex equations

We first introduce a lemma which is a special case of Theorem 21 Chapter III

Lemma 21 Suppose that the complex equation (22) satisfies Condition C Thenany solution of Problem A for (22) in D+ with the boundary conditions (13) and

Re [λ(x)w(x)] = s(x) λ(x) = 1minus i or 1 + i x isin L0 Cα[s(x) L0] le k3 (25)

can be expressed asw(z) = Φ(z)eφ(z) + ψ(z) z isin D+ (26)

where Im [φ(z)] = 0 z isin L0 = (0 2) and φ(z) ψ(z) satisfies the estimates

Cβ[φD+] + Lp0 [φz D+] le M2 Cβ[ψD+] + Lp0 [ψz D+] le M2 (27)

in which k3 β (0 lt β le α) p0 (2 lt p0 le 2) M2 = M2(p0 β k D+) are non-negativeconstants k = (k0 k1 k2 k3) Φ(z) is analytic in D+ and w(z) satisfies the estimate

Cβ[w(z)X(z) D+] le M3(k1 + k2 + k3) (28)

in which

X(z) = |z minus t1|η1|z minus t2|η2 ηj =2|γj|+ δ if γj lt 0

δ γj ge 0j = 1 2 (29)

here γj(j = 1 2) are real constants as stated in (17) and δ is a sufficiently smallpositive constant and M3 = M3(p0 β k0 D

+) is a non-negative constant

Theorem 22 If the complex equation (22) satisfies Condition C in D then anysolution of Problem A with the boundary conditions (13) (14) for (22) can beexpressed as

w(z) = w0(z) +W (z) (210)

where w0(z) is a solution of Problem A for the complex equation (12) and W (z)possesses the form

W (z) = w(z)minus w0(z) w(z) = Φ(z)eφ(z) + ψ(z) in D+

φ(z) = φ0(z) + Tg = φ0(z)minus 1π

int intD+

g(ζ)ζ minus z

dσζ ψ(z) = Tf in D+

W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν

2g1(z)dνe1 +

int micro

0g2(z)dmicroe2 in z isin Dminus

(211)

in which φ(z) = 0 on L0 e1 = (1 + i)2 e2 = (1minus i)2 micro = x+ y ν = x minus y φ0(z)is an analytic function in D+ and

g(z) =

A1 + A2ww w(z) = 00 w(z) = 0

f = A1ψ + A2ψ + A3 in D+

g1(z) = Aξ +Bη + E g2(z) = Cξ +Dη + F in Dminus

(212)

where ξ = Rew+Imw η = RewminusImw A = ReA1+ImA1 B = ReA2+ImA2 C =ReA2 minus ImA2 D = ReA1 minus ImA1 E = ReA3 + ImA3 F = ReA3 minus ImA3 andφ(z) ψ(z) satisfy the estimates

Cβ[φ(z) D+] + Lp0 [φz D+] le M4 Cβ[ψ(z) D+] + Lp0 [ψz D+] le M4 (213)

where M4 = M4(p0 β k D+) is a non-negative constant Φ(z) is analytic in D+ andΦ(z) is a solution of equation (12) in Dminus satisfying the boundary conditions

Re [λ(z)(eφ(z)Φ(z) + ψ(z))] = r(z) z isin Γ

Re [λ(x)(Φ(x)eφ(x) + ψ(x))] = s(x) x isin L0

Re [λ(x)Φ(x)] = Re [λ(x)(W (x)minusΨ(x))] z isin L0

Re [λ(z)Φ(z)] = minusRe [λ(z)Ψ(z)] z isin L1 or L2

(214)

Moreover the solution w0(z) of Problem A for (12) satisfies the estimate (132)namely

Cβ[w0(z)X(z) D+] + Cβ[wplusmn0 (z)Y

plusmn(z) Dminus] le M5(k1 + k2) (215)

where wplusmn0 (z) = Rew0(z)plusmn Imw0(z) Y plusmn(z) =

prod2j=1 |x plusmn y minus tj|ηj j = 1 2 X(z) ηj =

2|γj| + δ (j = 1 2) β are as stated in (132) and M5 = M5(p0 β k0 D) is a non-negative constant

Proof Let the solution w(z) be substituted in the position of w in the complex equa-tion (22) and (212) thus the functions g1(z) g2(z) and Ψ(z) in Dminus in (211)(212)can be determined Moreover we can find the solution Φ(z) of (12) with the boundarycondition (214) where

s(x)=

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

2r((1minus i)x2)minus 2R(1minus i)x2)a((1minus i)x2)minus b((1minus i)x2)

+Re [λ(x)Ψ(x)] x isin L0 or

2r((1+i)x2+1minusi)minus2R((1+i)x2+1minusi)a((1 + i)x2 + 1minus i) + b((1 + i)x2 + 1minus i)

+ Re [λ(x)Ψ(x)] x isin L0

(216)here and later R(z) = Re [λ(z)Ψ(z)] on L1 or L2 thus

w(z) = w0(z) +W (z) =

⎧⎨⎩ Φ(z)eφ(z) + ψ(z) in D+

w0(z) + Φ(z) + Ψ(z) in Dminus(217)

is the solution of Problem A for the complex equation

wz

wzlowast

= A1w + A2w + A3 in

⎧⎨⎩D+

Dminus

⎫⎬⎭ (218)

which can be expressed as in (210) and (211)

23 The unique solvability of the RiemannndashHilbert problem for firstorder complex equations of mixed type

Theorem 23 Let the mixed complex equation (22) satisfy Condition C ThenProblem A for (22) has a solution in D

Proof In order to find a solution w(z) of Problem A in D we express w(z) in theform (210)ndash(212) In the following we shall find a solution of Problem A by usingthe successive iteration First of all denoting the solution w0(z) = (ξ0e1 + η0e2) ofProblem A for (12) and substituting them into the positions of w = (ξe1 + ηe2)in the right-hand side of (22) similarly to (210)ndash(212) we have the correspondingfunctions g0(z) f0(z) in D+ and the functions

W1(z) = w1(z)minus w0(z) w1(z) = Φ1(z)eφ1(z) + ψ1(z)

φ1(z) = φ0(z)minus 1π

int intD+

g0(ζ)ζ minus z

dσζ ψ1(z) = Tf0 in D+

w1(z) = w0(z) +W1(z) W1(z) = Φ1(z) + Ψ1(z)

Ψ1(z)=int ν

2[Aξ0+Bη0+E]e1dν+

int micro

0[Cξ0+Dη0+F ]e2dmicro in Dminus

(219)

can be determined where micro = x + y ν = x minus y and the solution w0(z) satisfies theestimate (215) ie

plusmn(z) Dminus] le M6 = M5(k1 + k2) (220)

where βX(z) Y plusmn(z) are as stated in (215) Moreover we find an analytic functionΦ1(z) in D+ and a solution Φ1(z) of (12) in Dminus satisfying the boundary conditions

Re [λ(z)(eφ1(z)Φ1(z) + ψ1(z))] = r(z) z isin Γ

Re [λ(x)(Φ1(x)eφ1(x) + ψ1(x))] = s1(x) x isin L0

Re [λ(x)Φ1(x)] = minusRe [λ(x)Ψ1(x)] z isin L0

Re [λ(z)Φ1(z)] = minusRe [λ(z)Ψ1(z)] z isin L1 or L2

Im [λ(z1)Φ1(z1)] = minusIm [λ(z1)Ψ1(z1)]

(221)

in which

s1(x)=

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

2r((1minusi)x2)minus2R1((1minusi)x2)a((1minusi)x2)minusb((1minusi)x2)

+Re [λ(x)Ψ1(x)] x isin L0 or

2r((1+i)x2+1minusi)minus2R1((1+i)x2+1minusi)a((1+i)x2 + 1minus i) + b((1 + i)x2 + 1minus i)

+Re [λ(x)Ψ1(x)] xisinL0

here and later R1(z) = Re [λ(z)Ψ1(z)] on L1 or L2 and

w1(z) = w0(z) +W1(z) =

⎧⎨⎩ Φ1(z)eφ1(z) + ψ1(z) in D+

w0(z) + Φ1(z) + Ψ1(z) in Dminus(222)

satisfies the estimate

Cβ[w1(z)X(z) D+] + C[wplusmn1 (z)Y

plusmn(z) Dminus] le M7 = M7(p0 β k D) (223)

where φ1(z) ψ1(z) Φ1(z) are similar to the functions in Theorem 22 Furthermore wesubstitute w1(z) = w0(z) +W1(z) and the corresponding functions w+

1 (z) = ξ1(z) =Rew1(z)+Imw(z) wminus

1 (z) = η1(z) = Rew1(z)minusImw(z) into the positions of w ξ η in(211)(212) and similarly to (219)ndash(222) we can find the corresponding functions

φ2(z) ψ2(z) Φ2(z) in D+ and Ψ2(z)Φ2(z) and W2(z) = Φ2(z) + Ψ2(z) in Dminus andthe function

w2(z) = w0(z) +W2(z) =

⎧⎨⎩ Φ2(z)eφ2(z) + ψ2(z) in D+

w0(z) + Φ2(z) + Ψ2(z) in Dminus(224)

satisfies a similar estimate of the form (223) Thus there exists a sequence of functionswn(z) as follows

wn(z) = w0(z) +Wn(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Φn(z)eφn(z) + ψn(z) in D+

w0(z) + Φn(z) + Ψn(z) in Dminus

Ψn(z)=int ν

2[Aξnminus1+Bηnminus1+E]e1dν

+int micro

0[Cξnminus1+Dηnminus1+F ]e2dmicro in Dminus

(225)

and then

|[wplusmn1 (z)minuswplusmn

0 (z)]Yplusmn(z)|le|Φplusmn

1 (z)Yplusmn(z)|+

radic2[|Y +(z)

int ν

2[Aξ0+Bη0+E]e1dν|

+|Y minus(z)int micro

0[Cξ0 +Dη0 + F ]e2dmicro|

]le 2M8M(4m+ 1)Rprime in Dminus

(226)where m = maxC[w+

0 (z)Y +(z) Dminus] + C[wminus0 (z)Y minus(z) Dminus] M8 = maxzisinDminus(|A|

|B| |C| |D| |E| |F |) Rprime = 2 M = 1 + 4k20(1 + k2

0) M5 is a constant as stated in(220) It is clear that wn(z)minus wnminus1(z) satisfies

wn(z)minuswnminus1(z)=Φn(z)minusΦnminus1(z)+int ν

2[A(ξnminusξnminus1)+B(ηnminusηnminus1)]e1dν

+int micro

0[C(ξn minus ξnminus1) +D(ηn minus ηnminus1)]e2dmicro in Dminus

(227)

where n = 1 2 From the above equality we can obtain

|[wplusmnn minus wplusmn

nminus1]Yplusmn(z)| le [2M8M(4m+ 1)]n

timesint Rprime

0

Rprimenminus1

n in Dminus

(228)

and then we can see that the sequence of functions wplusmnn (z)Y

plusmn(z) ie

wplusmnn (z)Y

plusmn(z)=wplusmn0 (z)+[w

plusmn1 (z)minuswplusmn

0 (z)]+middot middot middot+[wplusmnn (z)minuswplusmn

nminus1(z)]Y plusmn(z) (229)

(n=1 2 ) in Dminus uniformly converge to functions wplusmnlowast (z)Y

plusmn(z) and wlowast(z) satisfiesthe equality

Ψlowast(z) =int ν

2[Aξlowast +Bηlowast + E]e1dν

+int micro

0[Cξlowast +Dηlowast + F ]e2dmicro in Dminus

(230)

where ξlowast = Rewlowast + Imwlowast η = Rewlowast minus Imwlowast and wlowast(z) satisfies the estimate

C[wplusmnlowast (z)Y

plusmn(z) Dminus] le e2M8M(4m+1)Rprime (231)

Moreover we can find a sequence of functions wn(z)(wn(z) = Φn(z)eφn(z) + ψn(z))in D+ and Φn(z) is an analytic function in D+ satisfying the boundary conditions

Re [λ(z)(Φn(z)eφn(z) + ψn(z))] = r(z) z isin Γ

Re [λ(x)(Φn(x)eφn(x) + ψn(x))] = s(x) x isin L0(232)

in which

sn(x)=

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

2r((1minusi)x2)minus2Rn((1minusi)x2)a((1minusi)x2)minusb((1minusi)x2)

+Re [λ(x)Ψn(x))] x isin L0 or

2r((1+i)x2+1minusi)minus2Rn((1minusi)x2 +1minusi)a((1+i)x2+1minusi)+b((1+i)x2+1minusi)

+Re [λ(x)Ψn(x)] xisinL0

(233)here and later Rn(z) = Re [λ(z)Ψn(z)] on L1 or L2 From (231) it follows that

Cβ[sn(x)X(x) L0] le 2k2k0 +[2M8M(4m+ 1)Rprime]n

n = M9 (234)

and then the estimate

Cβ[wn(z)X(z) D+] le M3(k1 + k2 +M9) (235)

thus from wn(z)X(x) we can choose a subsequence which uniformly converge afunction wlowast(z)X(z) in D+ Combining (231) and (235) it is obvious that thesolution wlowast(z) of Problem A for (22) in D satisfies the estimate

Cβ[wlowast(z)X(z) D+] + C[wplusmnlowast (z)Y

Theorem 24 Suppose that the complex equation (22) satisfies Condition C ThenProblem A for (22) has at most one solution in D

Proof Let w1(z) w2(z) be any two solutions of Problem A for (22) By ConditionC we see that w(z) = w1(z)minusw2(z) satisfies the homogeneous complex equation andboundary conditions

Lw = A1w + A2w in D (237)

Re [λ(z)w(z)] = 0 z isin Γ Re [λ(x)w(x)] = s(x) x isin L0

Re [λ(z)w(z)] = 0 z isin L1 or L2 Re [λ(z1)w(z1)] = 0(238)

From Theorem 22 the solution w(z) can be expressed in the form

w(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Φ(z)eφ(z) φ(z) = T g in D+

g(z) =

⎧⎨⎩A1 + A2ww w(z) = 0 z isin D+

0 w(z) = 0 z isin D+

Φ(z) + Ψ(z)

Ψ(z) =int ν

2[Aξ +Bη]e1dν +

int micro

0[Cξ +Dη]e2dmicro in Dminus

(239)

where Φ(z) is analytic in D+ and Φ(z) is a solution of (12) in Dminus satisfying theboundary condition (214) but ψ(z) = 0 z isin D+ r(z) = 0 z isin Γ and

s(x)=

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

minus2R((1minusi)x2)a((1minusi)x2)minus b((1minusi)x2)

minus2R((1minus i)x2 + 1minus i)a((1 + i)x2 + 1minus i) + b((1 + i)x2 + 1minus i)

+Re [λ(x)Ψ(x)] x isin L0

By using the method in the proof of Theorem 23 we can derive that

|wplusmn(z)Y plusmn(z)| le [2M8M(4m+ 1)Rprime]n

n in Dminus (240)

Let n rarr infin we get wplusmn(z) = 0 ie w(z) = w1(z) minus w2(z) = 0 Ψ(z) = Φ(z) = 0 inDminus Noting that w(z) = Φ(z)eφ(z) satisfies the boundary conditions in (238) we seethat the analytic function Φ(z) in D+ satisfies the boundary conditions

Re [λ(z)eφ(z)Φ(z)] = 0 z isin Γ Re [λ(x)Φ(x)] = 0 x isin L0 (241)

and the index of the boundary value problem (241) is K = minus12 hence Φ(z) = 0 inD+ and then w(z) = Φ(z)eφ(z) = 0 in D+ namely w(z) = w1(z)minus w2(z) = 0 in D+This proves the uniqueness of solutions of Problem A for (22)

From Theorems 23 and 24 we see that under Condition C Problem A forequation (22) has a unique solution w(z) which can be found by using successiveiteration and w(z) of Problem A satisfies the estimates

Cβ[w(z)X(z) D+] le M11 C[wplusmn(z)Y plusmn(z) Dminus] le M12 (242)

where wplusmn(z) = Rew(z)plusmn Imw(z) X(z) Y plusmn(z) are as stated in (132) and β(0 lt βlt δ) Mj = Mj(p0 β k D) (j = 11 12) are non-negative constants k = (k0 k1 k2)Moreover we can derive the following theorem

Theorem 25 Suppose that equation (22) satisfies Condition C Then any solutionw(z) of Problem A for (22) satisfies the estimates

Cβ[w(z)X(z) D+] le M13(k1 + k2) C[wplusmn(z)Y plusmn(z) Dminus] le M14(k1 + k2) (243)

in which Mj = Mj(p0 β k0 D)(j = 13 14) are non-negative constants

Proof When k1 + k2 = 0 from Theorem 23 it is easy to see that (243) holds Ifk1 + k2 gt 0 then it is seen that the function W (z) = w(z)(k1 + k2) is a solution ofthe homogeneous boundary value problem

Lw = F (z w)k Fk = A1W + A2W + A3k in D

Re [λ(z)W (z)] = r(z)k z isin Γ Im [λ(z1)W (z1)] = b1k

Re [λ(z)W (z)] = r(z)k z isin Lj j = 1 or 2

where Lp[A3kD+] le 1 C[A3kDminus] le 1 Cα[r(z)kΓ] le 1 Cα[r(z)k Lj] le1 j = 1 or 2 |b1k| le 1 On the basic of the estimate (242) we can obtain theestimates

where Mj = Mj(p0 β k0 D) (j = 13 14) are non-negative constants From (244) itfollows the estimate (243)

From the estimates (243)(244) we can see the regularity of solutions of ProblemA for (22) In the next section we shall give the Holder estimate of solutions ofProblem A for first order quasilinear complex equation of mixed type with the morerestrictive conditions than Condition C which includes the linear complex equation(22) as a special case

3 The RiemannndashHilbert Problem for First Order QuasilinearComplex Equations of Mixed Type

In this section we discuss the RiemannndashHilbert boundary value problem for firstorder quasilinear complex equations of mixed (elliptic-hyperbolic) type in a simplyconnected domain We first give the representation theorem and prove the uniquenessof solutions for the above boundary value problem and then by using the successiveiteration the existence of solutions for the above problem is proved

31 Representation and uniqueness of solutions of RiemannndashHilbertproblem for first order quasilinear complex equations of mixed type

Let D be a simply connected bounded domain as stated in Subsection 21 We discussthe quasilinear mixed (elliptic-hyperbolic) system of first order equations

⎧⎨⎩ux minus vy = au+ bv + f

3 Quasilinear Mixed Complex Equations 135

in which a b c d f g are functions of (x y) (isin D) u v (isin IR) its complex form isthe following complex equation of first order

wz

wzlowast

= F (z w) F = A1w + A2w + A3 in

D+

Dminus

(32)

where Aj = Aj(z w) j = 1 2 3 and the relations between Aj (j = 1 2 3) anda b c d f g are the same as those in (22)

Condition C

1) Aj(z w) (j = 1 2 3) are continuous in w isin CI for almost every point z isin D+and are measurable in z isin D+ and continuous on Dminus for all continuous functionsw(z) in Dlowast = D(0 2cupxplusmny = 2 y le 0) or Dlowast = D(0 2cupxplusmny = 0 y le 0)and satisfy

Lp[Aj D+] le k0 j = 1 2 Lp[A3 D+] le k1

C[Aj Dminus] le k0 j = 1 2 C[A3 Dminus] le k1(33)

2) For any continuous functions w1(z) w2(z) on Dlowast the following equality holds

F (z w1)minus F (z w2) = A1(z w1 w2)(w1 minus w2) + A2(z w1 w2)(w1 minus w2) in D (34)

whereLp[Aj D+] le k0 C[Aj Dminus] le k0 j = 1 2 (35)

in (33)(35) p (gt 2) k0 k1 are non-negative constants In particular when (32) isa linear equation (22) the condition (34) is obviously valid

The boundary conditions of RiemannndashHilbert problem for the complex equation(32) are as stated in (13)(14) Let the solution w(z) of Problem A be substitutedin the coefficients of (32) Then the equation can be viewed as a linear equation(22) Hence we have the same representation theorems as Lemma 21 and Theorem22

Theorem 31 Suppose that the quasilinear complex equation (32) satisfies Condi-tion C Then Problem A for (32) has a unique solution in D

Proof We first prove the uniqueness of the solution of Problem A for (32) Letw1(z) w2(z) be any two solutions of Problem A for (32) By Condition C we seethat w(z) = w1(z)minusw2(z) satisfies the homogeneous complex equation and boundaryconditions

wz = A1w + A2w in D (36)

Re [λ(z)w(z)] = 0 z isin L1 or L2 Re [λ(z1)w(z1)] = 0 (37)

where the conditions on the coefficients Aj(j = 1 2) are the same as in the proof ofTheorem 24 for the linear equation (22) Besides the remaining proof is the samein the proof of Theorems 23 and 24

Next noting the conditions (33)(34) by using the same method the existenceof solutions of Problem A for (32) can be proved and any solution w(z) of ProblemA for (32) satisfies the estimate (243)

In order to give the Holder estimate of solutions for (32) we need to add thefollowing condition

3) For any complex numbers z1 z2(isin D) w1 w2 the above functions satisfy

|Aj(z1 w1)minus Aj(z2 w2)| le k0[|z1 minus z2|α + |w1 minus w2|] j = 1 2

On the basis of the results of Theorem 44 in Chapter I and Theorem 23 inChapter III we can derive the following theorem

Theorem 32 Let the quasilinear complex equation (32) satisfy Condition C and(38) Then any solution w(z) of Problem A for (32) satisfies the following estimates

Cδ[X(z)w(z) D+] le M15 Cδ[Y plusmn(z)wplusmn(z) Dminus] le M16 (39)

in which wplusmn(z) = Rew(z)plusmn Imw(z) and

X(z)=2prod

j=1|z minus tj|ηj Y plusmn(x)=

2prodj=1

|xplusmn y minus tj|ηj ηj=2|γj|+2δ if γj lt0

2δ γj ge 0(310)

here γj(j = 1 2) are real constants as stated in (17) and δ is a sufficiently smallpositive constant and Mj = Mj(p0 β k D) (j = 15 16) are non-negative constantsk = (k0 k1 k2)

32 Existence of solutions of Problem A for general first order complexequations of mixed type

Now we consider the general quasilinear mixed complex equation of first order

Lw =

wz

wzlowast

= F (z wz) +G(z w) z isin

D+

Dminus

F = A1w + A2w + A3 G = A4 |w |σ z isin D

(311)

in which F (z w) satisfies Condition C σ is a positive constant and A4(z w) satisfiesthe same conditions as Aj(j = 1 2) where the main condition is

C[A4(z w) D] le k0 (312)

and denote the above conditions by Condition C prime

Theorem 33 Let the mixed complex equation (311) satisfy Condition C prime

(1) When 0 lt σ lt 1 Problem A for (311) has a solution w(z)

(2) When σ gt 1 Problem A for (311) has a solution w(z) provided that

M17 = k1 + k2 + |b1| (313)

(M13 +M14)k1 + k2 + 2k0tσ + |b1| = t (314)

in which M13 M14 are constants stated in (243) It is not difficult to see that theequation (314) has a unique solution t = M18 ge 0 Now we introduce a closed andconvex subset Blowast of the Banach space C(D) whose elements are the function w(z)satisfying the condition

C[w(z)X(z) D+] + C[wplusmn(z)Y plusmn(z) Dminus] le M18 (315)

We arbitrarily choose a function w0(z) isin Blowast for instance w0(z) = 0 and substitute itinto the position of w in the coefficients of (311) and G(z w) From Theorem 31 itis clear that problem A for

Lw minus A1(z w0)w minus A2(z w0)w minus A3(z w0) = G(z w0) (316)

has a unique solution w1(z) By (243) we see that the solution w1(z) satisfies theestimate in (315) By using successive iteration we obtain a sequence of solutionswm(z)(m = 1 2 ) of Problem A which satisfy the equations

Lwm+1 minus A1(z wm)wm+1z minus A2(z wm)wm+1

+A3(z wm) = G(z wm) in D m = 1 2 (317)

and wm+1(z)X(z) isin Blowast m = 1 2 From (317) we see that wm+1(z) = wm+1(z)minuswm(z) satisfies the complex equation and boundary conditions

Lwm+1minusA1wm+1minusA2wm+1=G G(z)=G(z wm)minusG(z wmminus1) in D

Re [λ(z)wm+1(z)]=0 on Γ cup Lj j=1 or 2 Im [λ(z1)wm+1(z1)]=0(318)

where m=1 2 Noting that C[X(z)G(z) D] le 2k0M18 M18 is a solution of thealgebraic equation (314) and according to the proof of Theorem 23

C[wm+1X(z) D+] + C[wplusmnm+1(z)Y

plusmn(z) Dminus] le M18 (319)

can be obtained The function wm+1 can be expressed as

wm+1(z) = w0(z) + Φm+1(z) + Ψm+1(z)

2[Aξm+1 + Bηm+1 + E]e1d(x minus y)

+int x+y

0[Cξm+1 + Dηm+1 + F ]e2d(x+ y) in Dminus

(320)

in which the relation between A B C D E F and A1 A2 G is the same as that ofA B C D E F and A1 A2 A3 in (212) By using the method from the proof ofTheorem 25 we can obtain

plusmn(z) Dminus] le (M20Rprime)m

m

where M20 = 2M19M(M5+1)(4m+1) Rprime = 2 m = C[wplusmn0 (z)Y plusmn(z) D] herein M19 =

maxC[AQ] C[BQ] C[CQ] C[DQ] C[EQ] C[F Q] M =1 + 4k20(1 + k2

0) Fromthe above inequality it is seen that the sequence of functions wm(z)X(z) ie

wplusmnm(z)Y

0 (z)]+middot middot middot+[wplusmnm(z)minuswplusmn

mminus1(z)]Y plusmn(z) (321)

(m = 1 2 ) uniformly converge to wplusmnlowast (z)Y

plusmn(z) and similarly to (230) the corres-ponding function wlowast(z) satisfies the equality

Ψlowast(z) =int xminusy

2[Aξlowast +Bηlowast + E]e1d(x minus y)

+int x+y

0[Cξlowast +Dηlowast + F ]e2d(x+ y) in Dminus

(322)

and the function wlowast(z) is just a solution of Problem A for the quasilinear equation(311) in the closure of the domain D

(2) Consider the algebraic equation

(M13 +M14)k1 + k2 + 2k0tσ + |b1| = t (323)

for t It is not difficult to see that equation (323) has a solution t = M18 ge 0provided that M17 in (313) is small enough Now we introduce a closed and convexsubset Blowast of the Banach space C(D) whose elements are the functions w(z) satisfyingthe conditions

By using the same method as in (1) we can find a solution u(z) isin Blowast of Problem Afor equation (311) with σ gt 1

4 The RiemannndashHilbert Problem for First Order QuasilinearEquations of Mixed type in General Domains

This section deals with the RiemannndashHilbert boundary value problem for quasilinearfirst order equations of mixed (elliptic-hyperbolic) type in general domains

4 Mixed Equations in General Domains 139

41 Formulation of the oblique derivative problem for second orderequations of mixed type in general domains

Let D be a simply connected bounded domain D in the complex plane CI with theboundary partD = Γ cup L where Γ L are as stated in Section 1 Now we consider thedomain Dprime with the boundary Γ cup Lprime

1 cup Lprime2 where the parameter equations of the

curves Lprime1 Lprime

2 are as follows

2 = x minus y = 2 l le x le 2 (41)

in which γ1(x) on 0 le x le l = γ1(l) + 2 is continuous and γ1(0) = 0 γ1(x) gt 0on 0 le x le l and γ1(x) is differentiable on 0 le x le l except finitely many pointsand 1 + γprime

1(x) gt 0 Denote Dprime+ = Dprime cap y gt 0 = D+ Dprimeminus = Dprime cap y lt 0and zprime

1 = l minus iγ1(l) Here we mention that in [12]1)3) the author assumes that thederivative of γ(x) satisfies γprime

1(x) gt 0 on 0 le x le l and other conditions

We consider the first order quasilinear complex equation of mixed type as statedin (32) in Dprime and assume that (32) satisfies Condition C in Dprime

The oblique derivative boundary value problemfor equation (32) may be formulated as follows

Problem Aprime Find a continuous solution w(z)of (32) in Dlowast = D0 Lprime

2 which satisfies theboundary conditions

Re [λ(z)w(z)] = r(z) z isin Lprime1

Im [λ(z)uz]|z=zprime1= b1

(43)

where λ(z) = a(x)+ ib(x) and |λ(z)| = 1 on ΓcupLprime1 and b0 b1 are real constants and

λ(z) r(z) b0 b1 satisfy the conditions

Cα[λ(z) Lprime1] le k0 Cα[r(z) Lprime

1] le k2 maxzisinLprime

1

1|a(x)minus b(x)| le k0

(44)

in which α (12 lt α lt 1) k0 k2 are non-negative constants The boundary valueproblem for equation (32) with A3(z u uz) = 0 z isin D u isin IR uz isin CI r(z) =0 z isin Γ cup Lprime

1 and b0 = b1 = 0 will be called Problem Aprime0 The number

K =12(K1 +K2) (45)

is called the index of Problem Aprime and Problem Aprime0 as stated in Section 1 Similarly

we only discuss the case of K = minus12 on partD+ because in this case the solution ofProblem Aprime is unique Besides we choose γ1 gt 0 In the following we first discussthe domain Dprime and then discuss another general domain Dprimeprime

42 The existence of solutions of Problem A for first order equations ofmixed type in general domains

1 By the conditions in (41) the inverse function x = σ(ν) of x + γ1(x) = ν =x minus y can be found and σprime(ν) = 1[1 + γprime

1(x)] Hence the curve Lprime1 can be expressed

by x = σ(ν) = (micro + ν)2 ie micro = 2σ(ν) minus ν 0 le ν le l + γ1(l) We make atransformation

micro=2[microminus2σ(ν)+ν

2minus2σ(ν)+ν

] ν=ν 2σ(ν)minusν lemicrole2 0leν le2 (46)

where micro ν are real variables its inverse transformation is

micro = [2minus 2σ(ν) + ν]micro2 + 2σ(ν)minus ν ν = ν 0 le micro le 2 0 le ν le 2 (47)

It is not difficult to see that the transformation in (46) maps the domain Dprimeminus ontoDminus The transformation (46) and its inverse transformation (47) can be rewrittenas ⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

x =12(micro+ ν) =

4x minus (2 + x minus y)[2σ(x+ γ1(x))minus x minus γ1(x)]4minus 4σ(x+ γ1(x)) + 2x+ 2γ1(x)

4y minus (2minus x+ y)[2σ(x+ γ1(x))minus x minus γ1(x)]4minus 4σ(x+ γ1(x)) + 2x+ 2γ1(x)

(48)

and ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

x =12(micro+ ν) =

[2minus 2σ(x+ γ1(x)) + x+ γ1(x)](x+ y)4

2

[2minus 2σ(x+ γ1(x)) + x+ γ1(x)](x+ y)4

2

(49)

Denote by z = x+ jy = f(z) z = x+ jy = fminus1(z) the transformation (48) and theinverse transformation (49) respectively In this case the system of equations is

ξν = Aξ +Bη + E ηmicro = Cξ +Dη + F z isin Dprimeminus (410)

which is another form of (32) in Dprimeminus Suppose that (32) in Dprime satisfies ConditionC through the transformation (47) we obtain ξν = ξν ηmicro = [2minus 2σ(ν) + ν]ηmicro2 inDprimeminus and then

ξν = Aξ +Bη + E

ηmicro =[2minus 2σ(ν) + ν][Cξ +Dη + F ]

2

in Dminus (411)

and through the transformation (48) the boundary condition (43) is reduced to

Re [λ(fminus1(z))w(fminus1(z))] = r(fminus1(z)) z isin L1

in which z1 = f(zprime1) Therefore the boundary value problem (410)(43) is trans-

formed into the boundary value problem (411)(412) ie the corresponding Prob-lem A in D On the basis of Theorem 31 we see that the boundary value problem(32)(in D+)(411)(42)(412) has a unique solution w(z) and w(z) is just a solutionof Problem A for (32) in Dprime with the boundary conditions (42)(43)

Theorem 41 If the mixed equation (32) in Dprime satisfies Condition C in the domainDprime with the boundary Γ cup Lprime

1 cup Lprime2 where Lprime

1 Lprime2 are as stated in (41) then Problem

Aprime for (32) with the boundary conditions (42) (43) has a unique solution w(z)

2 Next let the domain Dprimeprime be a simply connected domain with the boundaryΓ cup Lprimeprime

1 cup Lprimeprime2 where Γ is as stated before and

Lprimeprime1 = γ1(x) + y = 0 0 le x le l Lprimeprime

2 = γ2(x) + y = 0 l le x le 2 (413)

in which γ1(0) = 0 γ2(2) = 0 γ1(x) gt 0 0 le x le l γ2(x) gt 0 l le x le 2 γ1(x) on0 le x le l γ2(x) on l le x le 2 are continuous and differentiable except at isolatedpoints and 1 + γprime

1(x) gt 0 1 minus γprime2(x) gt 0 Denote Dprimeprime+ = Dprimeprime cap y gt 0 = D+ and

Dprimeprimeminus = Dprimeprime cap y lt 0 and zprimeprime1 = l minus iγ1(l) = l minus iγ2(l) We consider the Riemannndash

Hilbert problem (Problem Aprime) for equation (32) in Dprimeprime with the boundary conditions(42) and

Re [λ(z)w(z)] = r(z) z isin Lprimeprime2 Im [λ(zprimeprime

1 )w(zprimeprime1 )] = b1 (414)

where zprimeprime1 = l minus iγ1(l) = l minus iγ2(l) and λ(z) r(z) satisfy the corresponding condition

Cα[λ(z)Γ] le k0 Cα[r(z)Γ] le k2 |b1| le k2 Cα[λ(z) Lprimeprime2] le k0

Cα[r(z) Lprimeprime2] le k2 max

zisinLprimeprime1

1|a(x)minus b(x)| max

zisinLprimeprime2

1|a(x) + b(x)| le k0

(415)

in which α (12 lt α lt 1) k0 k2 are non-negative constants By the conditions in(413) the inverse function x = τ(micro) of x minus γ2(x) = micro can be found namely

ν = 2τ(micro)minus micro 0 le micro le 2 (416)

micro = micro ν =2ν

2τ(micro)minus micro 0 le micro le 2 0 le ν le 2τ(micro)minus micro (417)

micro = micro = x+ y

ν =[2τ(micro)minus micro]ν

2

=[2τ(x minus γ2(x))minus x+ γ2(x)](x minus y)

2 0 le micro le 2 0 le ν le 2

(418)

Hence we have

x =12(micro+ ν) =

2(x minus y) + (x+ y)[2τ(x minus γ2(x))minus x+ γ2(x)]2[2τ(x minus γ2(x))minus x+ γ2(x)]

minus2(x minus y) + (x+ y)[2τ(x minus γ2(x))minus x+ γ2(x)]2[2τ(x minus γ2(x))minus x+ γ2(x)]

x =12(micro+ ν) =

14[(2τ(x minus γ2(x))minus x+ γ2(x))(x minus y) + 2(x+ y)]

14[(minus2τ(x minus γ2(x)) + x minus γ2(x))(x minus y) + 2(x+ y)]

(419)

Denote by z = x+ jy = g(z) z = x+ jy = gminus1(z) the transformation (418) and itsinverse transformation in (419) respectively Through the transformation (418) weobtain

(u+ v)ν = [τ(micro)minus micro2](u+ v)ν (u minus v)micro = (u minus v)micro in Dprimeminus (420)

System (410) in Dprimeprimeminus is reduced to

ξν = [τ(micro)minus micro2][Aξ +Bη + E]

ηmicro = Cξ +Dη + Fin Dprimeminus (421)

Moreover through the transformation (419) the boundary condition (414) on Lprimeprime2 is

reduced to

Re [λ(gminus1(z))w(gminus1(z))] = r[gminus1(z)] z isin Lprime2

Im [λ(gminus1(zprime1))w(gminus1(zprime

1)] = b1(422)

in which zprime1 = zprimeprime

1 = g(zprimeprime1 ) Therefore the boundary value problem (410)(414) is

transformed into the boundary value problem (421)(422) According to the methodin the proof of Theorem 41 we can see that the boundary value problem (32) (inD+) (421) (42) (422) has a unique solution w(z) and then w(z) is a solution of theboundary value problem (32)(42)(414) But we mention that through the trans-formation (417) or (419) the boundaries Lprimeprime

1 Lprimeprime2 are reduced to Lprime

1 Lprime2 respectively

such that Lprime1 L

prime2 satisfy the condition as stated in (41) In fact if the intersection zprimeprime

1of Lprimeprime

1 and Lprimeprime2 belongs to L2 and γ1(x) ge 2(1minus l) + x 2minus 2l le x le l then the above

requirement can be satisfied If zprimeprime1 isin L1 = x + y = 0 γ2(x) ge 2l minus x l le x le 2l

then we can proceed similarly

Theorem 42 If the mixed equation (32) satisfies Condition C in the domain Dprimeprime

with the boundary Γcup Lprimeprime1 cup Lprimeprime

2 where Lprimeprime1 Lprimeprime

2 are as stated in (413) then Problem Aprime

for (32) (42) (414) in Dprimeprime has a unique solution w(z)

5 Discontinuous RiemannndashHilbert Problem 143

5 The Discontinuous RiemannndashHilbert Problem forQuasilinear Mixed Equations of First Order

This section deals with the discontinuous RiemannndashHilbert problem for quasilinearmixed (elliptic-hyperbolic) complex equations of first order in a simply connecteddomain Firstly we give the representation theorem and prove the uniqueness ofsolutions for the above boundary value problem Afterwards by using the method ofsuccessive iteration the existence of solutions for the above problem is proved

51 Formulation of the discontinuous RiemannndashHilbert problem forcomplex equations of mixed type

Let D be a simply connected domain with the boundary Γ cup L1 cup L2 as stated asbefore where D+ = |z minus 1| lt 1 Im z gt 0 We discuss the first order quasilinearcomplex equations of mixed type as stated in (32) with Condition C

In order to introduce the discontinuous Riemann-Hilbert boundary value problemfor the complex equation (32) let the functions a(z) b(z) possess discontinuities offirst kind at m minus 1 distinct points z1 z2 zmminus1 isin Γ which are arranged accordingto the positive direction of Γ and Z = z0 = 2 z1 zm = 0 cup x plusmn y = 0 x plusmn y =2 y le 0 wherem is a positive integer and r(z) = O(|zminuszj|minusβj) in the neighborhoodof zj(z = 0 1 m) on Γ in which βj(j = 0 1 m) are sufficiently small positivenumbers Denote λ(z) = a(z) + ib(z) and |a(z)| + |b(z)| = 0 there is no harmin assuming that |λ(z)| = 1 z isin Γlowast = ΓZ Suppose that λ(z) r(z) satisfy theconditions

λ(z) isin Cα(Γj) |z minus zj|βjr(z) isin Cα(Γj) j = 0 1 m (51)

herein Γj is an arc from the point zjminus1 to zj on Γ and Γj(j = 1 m) does notinclude the end points α(0 lt α lt 1) is a constant

Problem Alowast Find a continuous solution w(z) of (32) in Dlowast = DZ and w(z) onZ maybe become infinite of an order lower than unity which satisfies the boundaryconditions

Re [λ(z)w(z)] = r(z) z isin Lj (j = 1 or 2) Im [λ(z)w(z)]|z=z1 = b1 (53)

where b1 are real constants λ(z) = a(x)+ ib(x)(|λ(z)| = 1) z isin Γcup Lj(j = 1 or 2)and λ(z) r(z) b1 satisfy the conditions

Cα[λ(z)Γj] le k0 Cα[r(z)Γj] le k2 |b1| le k2

maxzisinL1

1|a(z)minus b(z)| le k0 or max

zisinL2

1|a(z) + b(z)| le k0

(54)

in which α (12 lt α lt 1) k0 k2 are non-negative constants The above discontinuousRiemannndashHilbert boundary value problem for (32) is called Problem Alowast

Denote by λ(zjminus0) and λ(zj+0) the left limit and right limit of λ(z) as z rarr zj (j =0 1 m) on Γ and

]=

φj

πminus Kj

Kj =[φj

π

]+ Jj Jj = 0 or 1 j = 0 1 m

(55)

in which zm = 0 z0 = 2 λ(z) = exp(minusiπ4) on L0 = (0 2) and λ(z0 minus 0) = λ(zn +0) = exp(minusiπ4) or λ(z) = exp(iπ4) on L0 and λ(z0 minus 0) = λ(zn + 0) = exp(iπ4)and 0 le γj lt 1 when Jj = 0 and minus1 lt Jj lt 0 when Jj = 1 j = 0 1 m and

K =12(K0 +K2 + middot middot middot+Km) =

msumj=0

[φj

2πminus γj

2

](56)

is called the index of Problem Alowast Now the function λ(z) on ΓcupL0 is not continuouswe can choose Jj = 0 or 1 (0 le j le m) hence the index K is not unique Here wechoose the index K = minus12 Let βj + γj lt 1 j = 0 1 m We can require thatthe solution u(z) satisfy the conditions

uz=O(|zminuszj|minusδ) δ=

⎧⎨⎩βj+τ for γj ge0 and γj lt0 βj gt |γj||γj|+τ for γj lt0 βj le|γj| j=1 m

(57)

in the neighborhood of zj in Dlowast where τ(lt α) is a small positive number

For Problem Alowast of the quasilinear complex equation (32) we can prove that thereexists a unique solution by using a similar method as stated in the last section

Next we discuss the more general discontinuous RiemannndashHilbert problem Asstated before denote L = L1 cup L2 L1 = x = minusy 0 le x le 1 L2 = x = y + 2 1 lex le 2 and D+ = D cap y gt 0 Dminus = D cap y lt 0 Here there are n pointsE1 = a1 E2 = a2 En = an on the segment AB = (0 2) = L0 where a0 = 0 lta1 lt a2 lt middot middot middot lt an lt an+1 = 2 and denote by A = A0 = 0 A1 = (1minus i)a12 A2 =(1minusi)a22 An = (1minusi)an2 An+1 = C = 1minusi and B1 = 1minusi+(1+i)a12 B2 =1 minus i + (1 + i)a22 Bn = 1 minus i + (1 + i)an2 B = Bn+1 = 2 on the segmentsAC CB respectively Moreover we denote Dminus

1 = Dminus capcup[n2]j=0 (a2j le xminusy le a2j+1)

Dminus2 = Dminuscapcup[(n+1)2]

j=1 (a2jminus1 le x+y le a2j) and Dminus2j+1 = Dminuscapa2j le xminusy le a2j+1

j = 0 1 [n2] Dminus2j = Dminus cap a2jminus1 le x + y le a2j j = 1 [(n + 1)2] and

Dminuslowast = DminusZ Z = cupn+1

j=0 (x plusmn y = aj y le 0) Dlowast = D+ cup Dminuslowast

Problem Alowast Find a continuous solution w(z) of (32) in Dlowast = DZ whereZ = z0 z1 zm a1 an cup x plusmn y = aj y le 0 j = 1 n and the abovesolution w(z) satisfies the boundary conditions (52) and

Re [λ(z)w(z)]=r(z)

zisinL3=sum[n2]

j=0 A2jA2j+1

Re [λ(z)w(z)]=r(z)

zisinL4=sum[(n+1)2]

j=1 B2jminus1B2j

Im [λ(z)w(z)]|z=A2j+1=c2j+1

j=0 1 [n2]

Im [λ(z)w(z)]|z=B2jminus1=c2j

j=1 [(n+ 1)2]

(58)

where cj(j = 1 n+ 1) are real constants λ(z) = a(x) + ib(x) |λ(z)| = 1 z isin Γand λ(z) r(z) cj(j = 1 n+ 1) satisfy the conditions

Cα[λ(z)Γ] le k0 Cα[r(z)Γ] le k2 |cj| le k2 j = 0 1 n+ 1

Cα[λ(z) Lj] le k0 Cα[r(z) Lj] le k2 j = 3 4

maxzisinL3

1|a(x)minus b(x)| and max

zisinL4

1|a(x) + b(x)| le k0

(59)

where α (12 lt α lt 1) k0 k2 are non-negative constants The above discontinuousRiemannndashHilbert boundary value problem for (32) is called Problem Alowast

Denote by λ(tj minus 0) and λ(tj +0) the left limit and right limit of λ(z) as z rarr tj =zj(j = 0 1 m zm+k = ak k = 1 n zn+m+1 = 2) on Γ cup L0 (L0 = (0 2)) and

eiφj =λ(tj minus 0)λ(tj + 0)

γj =1πiln(

)=

φj

πminus Kj

Kj =[φj

π

]+ Jj Jj = 0 or 1 j = 0 1 m+ n

(510)

in which [a] is the largest integer not exceeding the real number a λ(z) = exp(minusiπ4)on Lprime

1 = AB cap Dminus1 and λ(a2j + 0) = λ(a2j+1 minus 0) = exp(minusiπ4) j = 0 1 [n2]

and λ(z) = exp(iπ4) on Lprime2 = AB cap Dminus

2 and λ(a2jminus1 + 0) = λ(a2j minus 0) = exp(iπ4)j = 1 [(n + 1)2] and 0 le γj lt 1 when Jj = 0 and minus1 lt γj lt 0 whenJj = 1 j = 0 1 m+ n and

K =12(K0 +K1 + middot middot middot+Km+n) =

m+nsumj=0

(φj

2πminus γj

2

)(511)

is called the index of Problem Alowast and Problem Alowast0 We can require that the solution

w(z) in D+ satisfy the conditions

w(z) = O(|z minus zj|minusτ ) τ = γprimej + δ j = 0 1 m+ n (512)

in the neighborhood of zj(0 le j le m + n) in D+ where γprimej = max(0 minusγj) (j =

1 m minus 1 m + 1 m + n) γprimem = max(0 minus2γm) γprime

0 = max(0 minus2γ0) and γj(j =0 1 m+n) are real constants in (510) δ is a sufficiently small positive numberand choose the index K = minus12 Now we explain that in the closed domain Dminusthe functions u + v u minus v corresponding to the solution w(z) in the neighborhoodsof the 2n + 2 characteristic lines Z0 = x + y = 0 x minus y = 2 x plusmn y = aj(j =m + 1 m + n) y le 0 may be not bounded if γj le 0(j = m m + n + 1)Hence if we require that u + v u minus v in DminusZ0 is bounded then it needs to chooseγj gt 0 (j = 0 1 m+ n+ 1)

52 Representation of solutions for the discontinuous RiemannndashHilbertproblem

We first introduce a lemma

Lemma 51 Suppose that the complex equation (32) satisfies Condition C Thenthere exists a solution of Problem Alowast for (32) in D+ with the boundary conditions(52) and

Re [λ(z)w(z)]|z=x = s(x) Cβ[s(x) Lprimej] le k3 j = 1 2

λ(x) =

⎧⎨⎩ 1minus i on Lprime1 = Dminus

1 cap AB

1 + i on Lprime2 = Dminus

2 cap AB

(513)

Cβ[w(z)X(z) D+] le M21(k1 + k2 + k3) (514)

in which k3 is a non-negative constant s(x) is as stated in the form (525) belowX(z) = Πm+n

j=0 |z minus zj|γprimej+δ herein γprime

j = max(0 minusγj)(j = 1 m minus 1 m + 1 m +n) γprime

0 = max(0 minus2γ0) γprimem = max(0 minus2γm) and γj(j = 0 1 m + n) are real

constants in (510) β (0 lt β lt δ) δ are sufficiently small positive numbers andM21 = M21(p0 β k0 D

By using the method as in the proofs of Theorems 21ndash23 Chapter III Theorem12 and Lemma 21 we can prove the lemma

Theorem 52 Problem Alowast for equation (12) in D has a unique solution w(z)

Proof First of all similarly to Theorem 12 the solution w(z) = u(z) + iv(z) ofequation (12) in Dminus can be expressed as (119) According to the proof of Theorem12 we can obtain f(x + y) on Lprime

1 = Dminus1 cap AB and g(x minus y) on Lprime

2 = Dminus2 cap AB in

the form

g(x)=k(x)=2r((1minusi)x2)minus[a((1minusi)x2)+b ((1minusi)x2)]h2j

a((1minus i)x2)minus b ((1minus i)x2)

on L2j+1 = Dminus2j+1 cap AB

f(x)=h(x)=2r((1 + i)x2 + 1minus i)

minus [a((1+i)x2+1minusi)minusb ((1+i)x2+1minusi)]k2jminus1

a((1+i)x2+1minusi)+b ((1+i)x2+1minusi)

on L2j = Dminus2j cap AB

(515)

where Dminusj (j = 1 2 2n+ 1) are as stated in Subsection 51 and

h2j = Re [λ(A2j+1)(r(A2j+1) + ic2j+1)] + Im [λ(A2j+1)(r(A2j+1) + ic2j+1)]

on L2j+1 j = 0 1 [n2]

k2jminus1 = Re [λ(B2jminus1)(r(B2jminus1) + ic2j)]minus Im [λ(B2jminus1)(r(B2jminus1) + ic2j)]

on L2j j = 1 [(n+ 1)2]

where L2j+1 = Dminus2j+1 cap AB j = 0 1 [n2] L2j = Dminus

2j cap AB j = 1 [(n + 1)2]By using Theorem 22 Chapter III choosing an appropriate index K = minus12 thereexists a unique solution w(z) of Problem Alowast in D+ with the boundary conditions(52) and

Re [λ(x)w(x)] =

⎧⎨⎩k(x)

h(x)λ(x) =

⎧⎪⎨⎪⎩1minus i on Lprime

1 = Dminus1 cap AB

2 cap AB

and denote

Re [λ(x)w(x)] =

⎧⎨⎩h(x) on Lprime1

k(x) on Lprime2

Cβ[X(x)k(x) Lprime1] le k2 Cβ[X(x)h(x) Lprime

2] le k2

(516)

herein β(0 lt β le α lt 1) k2 are non-negative constants

Next we find a solution w(z) of Problem Alowast for (12) in Dminus with the boundaryconditions

Re [λ(z)w(z)] = r(z) on L3 cup L4

Im [λ(z)w(z)]|z=A2j+1 = c2j+1 j = 0 1 [n2]

Im [λ(z)w(z)]|z=B2jminus1 = c2j j = 1 [(n+ 1)2]

(517)

and (516) where cj(j = 1 n + 1) are as stated in (58) By the result and themethod in Chapters I and II we can find the solution of Problem Alowast for (12) in Dminus

1in the form

w(z) = w(z) + λ(A2j+1)[r(A2j+1) + ic2j+1]

w(z) =12[(1 + i)f2j+1(x+ y) + (1minus i)g2j+1(x minus y)]

f2j+1(x+ y) = Re [λ(x+ y)w(x+ y)] g2j+1(x minus y)

=2r((1minus i)(x minus y)2)

a((1minus i)(x minus y)2)minus b((1minus i)(x minus y)2)in Dminus

2j+1

j = 0 1 [n2]

w(z) = w(z) + λ(B2jminus1)[r(B2jminus1) + ic2j]

w(z)=12[(1 + i)f2j(x+ y)+(1minus i)g2j(x minus y)] f2j(x+ y)

= h(x+ y)=2r((1+i)(x+y)2+1minusi)

a((1+i)(x+y)2+1minusi)+b ((1+i)(x+y)2+1minusi)

g2j(x minus y) = Re [λ(x minus y)w(x minus y)] in Dminus2j j = 1

[n+ 12

]

Furthermore from the above solution we can find the solution of Problem Alowast for (12)in DminusDminus

1 cup Dminus2 and the solution w(z) of Problem Alowast for (12) in Dminus possesses

the form

w(z) =12[(1 + i)f(x+ y) + (1minus i)g(x minus y)]

f(x+ y) = Re [(1minus i)w(x+ y)] in DminusDminus2

g(x minus y) = k(x minus y) in Dminus1

f(x+ y) = h(x+ y) in Dminus2

g(x minus y) = Re [(1 + i)w(x minus y)] in DminusDminus1

(518)

where k(x) h(x) are as stated in (515) and w(z) is the solution of Problem Alowast for(12) with the boundary conditions (52)(516)

Theorem 53 Let the complex equation (32) satisfy Condition C Then any solu-tion of Problem Alowast for (32) can be expressed as

w(z) = w0(z) +W (z) in D (519)

where w0(z) is a solution of Problem Alowast for equation (12) and W (z) possesses theform

W (z) = w(z)minus w0(z) in D w(z) = Φ(z)eφ(z) + ψ(z)

int intD+

g(ζ)ζ minus z

W (z) = Φ(z) + Ψ(z) in Dminus

Ψ(z) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

int ν

a2j+1

g1(z)dνe1+int micro

0g2(z)dmicroe2 in Dminus

2j+1 j=0 1 [n2]

int ν

2g1(z)dνe1+

int micro

a2jminus1

g2(z)dmicroe2 in Dminus2j j=1 [(n+1)2]

(520)

in which e1 = (1 + i)2 e2 = (1 minus i)2 micro = x + y ν = x minus y φ0(z) is an analyticfunction in D+ Im [φ(z)] = 0 on L0 and

g(z) =

⎧⎨⎩A1 + A2w(w) w(z) = 00 w(z) = 0 z isin D+

f = A1Tf + A2Tf + A3 in D+

(521)

where ξ = Rew + Imw η = Rew minus Imw and φ(z) ψ(z) satisfy the estimates

Cβ[φ(z) D+] + Lp0 [φz D+] le M22

Cβ[ψ(z) D+] + Lp0 [ψz D+] le M22(522)

where p0 (0 lt p0 le p) M22 = M22 (p0 α k D+) are non-negative constants Φ(z) isanalytic in D+ and Φ(z) is a solution of equation (12) in Dminus satisfying the boundaryconditions

Re [λ(z)eφ(z)Φ(z)] = r(z)minus Re [λ(z)ψ(z)] z isin Γ

Re [λ(z)Φ(z)] = minusRe [λ(z)Ψ(z)] z isin L3 cup L4

Im [λ(z)(Φ(z) + Ψ(z))]|z=A2j+1 = 0 j = 0 1 [n2]

Im [λ(z)(Φ(z) + Ψ(z))]|z=B2jminus1 = 0 j = 1 [(n+ 1)2]

(523)

Moreover the solution w0(z) of Problem Alowast for (12) satisfies the estimate in the form

Cβ[X(z)w0(z) D+] + C[wplusmn0 (micro ν)Y plusmn(micro ν) Dminus] le M23(k1 + k2) (524)

in which X(z) = Πm+nj=0 |z minus zj|γprime

j+δ Y plusmn(z) = Y plusmn(micro ν) = Πm+nj=0 |x plusmn y minus zj|γprime

j+δwplusmn

0 (micro ν) = Rew0(z)plusmn Imw0(z) w0(z) = w0(micro ν) micro = x + y ν = x minus yγprime

j(j = 0 1 m + n) are as stated in (514) M23 = M23 (p0 β k0 D) is a non-negative constant

Proof Let the solution w(z) of Problem Alowast be substituted into the complex equa-tion (32) and the solution w0(z) = ξ0e1 + η0e2 of Problem Alowast for equation (12)be substituted in the position of w in (521) Thus the functions f(z) g(z) in D+

and g1(z) g2(z) and Ψ(z) in Dminus in (520)(521) can be determined Moreover byTheorem 52 we can find an analytic function Φ(z) in D+ and a solution Φ(z) of(12) in Dminus with the boundary conditions (523) where

s(x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

2r((1minus i)x2)minus 2R((1minus i)x2)a((1minus i)x2)minus b((1minus i)x2)

minus [a((1minus i)x2) + b((1minus i)x2)]h2j

a((1minus i)x2)minus b((1minus i)x2)+ Re [λ(x)Ψ(x)]

x isin (a2j a2j+1) j = 0 1 [n2]

minus [a((1 + i)x2 + 1minus i)minus b((1 + i)x2 + 1minus i)]k2jminus1

+Re [λ(x)Ψ(x)] x isin (a2jminus1 a2j) j = 1 [(n+ 1)2]

(525)

in which the real constants h2j(j = 0 1 [n2]) k2jminus1(j = 1 [(n + 1)2]) are asstated in (515) thus

w(z) = w0(z) +W (z) =

⎧⎨⎩ Φ(z)φ(z) + ψ(z) in D+

w0(z) + Φ(z) + Ψ(z) in Dminus

is the solution of Problem Alowast for the complex equation

wz

wzlowast

= A1w + A2w + A3 in

⎧⎨⎩D+

Dminus

⎫⎬⎭ (526)

and the solution w0(z) of Problem Alowast for (12) satisfies the estimate (524)

53 Existence and uniqueness of solutions of the RiemannndashHilbertproblem for (32)

Theorem 54 Suppose that the complex equation (32) satisfies Condition C ThenProblem Alowast for (32) is solvable

Proof In order to find a solution w(z) of Problem Alowast in D we express w(z) inthe form (519)ndash(521) On the basis of Theorem 52 we see that Problem Alowast for(12) has a unique solution w0(z)(= ξ0e1+η0e2) and substitute it into the position ofw = ξe1 + ηe2 in the right-hand side of (32) Similarly to (219) from (519)ndash(521)

we obtain the corresponding functions g0(z) f0(z) in D+ g10(z) g

20(z) in Dminus and the

functions

φ1(z)= φ0(z)minus 1π

int intD+

g0(ζ)ζ minus z

dσζ ψ(z)=Tf0 in D+

Ψ1(z)=

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

int ν

a2j+1

g10(z)dνe1+

int micro

2j+1 j=0 1 [n2]

int ν

2g10(z)dνe1+

int micro

a2jminus1

g20(z)dmicroe2 in Dminus

2j j=1 [(n+1)2]

(527)

Moreover by Theorem 52 we can find an analytic function Φ(z) inD+ and a solutionΦ1(z) of (12) in Dminus satisfying the boundary conditions

Re [λ(z)(Φ1(z)eφ1(z) + ψ1(z))] = r(z) z isin Γ

Re [λ(x)(Φ1(x) + ψ1(x))] = s(z) x isin L0

Re [λ(x)Φ1(x)] = Re [λ(x)(W1(x)minusΨ1(x))] z isin L0

Re [λ(z)Φ1(z)] = minusRe [λ(z)Ψ1(z)] z isin L3 cup L4

Im [λ(z)(Φ1(z) + Ψ1(z))]|z=A2j+1 = 0 j = 0 1 [n2]

Im [λ(z)(Φ1(z) + Ψ1(z))]|z=B2jminus1 = 0 j = 1 [(n+ 1)2]

(529)

where

s1(x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

2r((1minus i)x2)minus 2R1((1minus i)x2)a((1minus i)x2)minus b ((1minus i)x2)

minus [a((1minus i)x2) + b ((1minus i)x2)]h2j

a((1minus i)x2)minus b ((1minus i)x2)+ Re [λ(x)Ψ1(x)]

x isin (a2j a2j+1) j = 0 1 [n2]

2r((1+i)x2+1minusi)minus2R1((1+i)x2+1minusi)a((1 + i)x2 + 1minus i) + b ((1 + i)x2 + 1minus i)

minus [a((1 + i)x2 + 1minus i)minus b ((1 + i)x2 + 1minus i)]k2jminus1

a((1 + i)x2 + 1minus i) + b ((1 + i)x2 + 1minus i)

+Re [λ(x)Ψ1(x)] x isin (a2jminus1 a2j) j = 1 [(n+ 1)2]

in which the real constants h2j k2jminus1 are as stated in (515) and

w1(z) = w0(z) +W1(z) = w0(z) + Φ1(z) + Ψ1(z) in Dminus (530)

Cβ[w1(z)X(z) D+]+C[wplusmn1 (micro ν)Y plusmn(micro ν) Dminus]leM24=M24(p0 β k Dminus) (531)

here wplusmn1 (micro ν) = Rew1(micro ν) plusmn Imw1(micro ν) Y plusmn(micro ν) X(z) Y plusmn(z) are as stated in

(524) Furthermore we substitute w1(z) = w0(z)+W1(z) and corresponding functionsw1(z) ξ1(z) = Rew1(z) + Imw1(z) η1(z) = Rew1(z) minus Imw1(z) into the positionsw(z) ξ(z) η(z) in (520) (521) and similarly to (527)ndash(530) we can find thecorresponding functions φ2(z) ψ2(z) Φ2(z) in D+ and Ψ2(z)Φ2(z) and W2(z) =Φ2(z) + Ψ2(z) in Dminus The function

w2(z) = w0(z) +W2(z) =

⎧⎨⎩ Φ2(z)eφ2(z) + ψ2(z) in D+

w0(z) + Φ2(z) + Ψ2(z) in Dminus(532)

satisfies the similar estimate in the form (531) Thus there exists a sequence offunctions wn(z) as follows

wn(z) = w0(z) +Wn(z) =

⎧⎨⎩ Φn(z)eφn(z) + ψn(z) in D+

Ψn(z)=

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

int ν

a2j+1

g1nminus1(z)e1dν+

int micro

0g2

nminus1(z)e2dmicro in Dminus2j+1 j=0 1 [n2]

int ν

2g1

nminus1(z)e1dν+int micro

a2jminus1

g2nminus1(z)e2dmicro in Dminus

2j j=1 [(n+1)2]

g1nminus1(z)=Aξnminus1+Bηnminus1+E g2

nminus1(z)=Cξnminus1+Dηnminus1 + F in Dminus

(533)

and then

|[wplusmn1 (micro ν)minus wplusmn

0 (micro ν)]Y plusmn(micro ν)| le |Φplusmn1 (micro ν)Y plusmn(micro ν)|

+radic2

|Y minus(micro ν)|[max

1lejlen+1|int ν

a2j+1

g10(z)e1dν|+ |

int ν

2g20(z)e1dν|

]

+|Y +(micro ν)|[|int micro

0g10(z)e2dmicro|+ max

1lejlen+1|int micro

a2jminus1

g20(z)e2dmicro|

]

le 2M25M(4m+ 1)Rprime in Dminus

(534)

wherem = C[w+0 (micro ν)Y +(micro ν) Dminus]+C[wminus

0 (micro ν)Y minus(micro ν) Dminus] M = 1+ 4k20(1+k2

0)

Rprime = 2 M25 = maxzisinDminus(|A| |B| |C| |D|) It is clear that wn(z)minus wnminus1(z) satisfies

wn(z)minus wnminus1(z)

= Φn(z)minus Φnminus1(z) +

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int ν

a2j+1

[g1nminus1minusg1

nminus2]e1dν +int micro

0[g2

nminus1minusg2nminus2]e2dmicro in Dminus

2j+1

j=0 1 [n2]int ν

2[g1

nminus1minusg1nminus2]e1dν+

int micro

a2jminus1

[g2nminus1minusg2

nminus2]e2dmicro inDminus2j

j=1 [(n+1)2]

(535)

Moreover we can find the solution w(z) of Problem Alowast for (32) in the setDminus(cup[n2]

j=0 Dminus2j+1) cup (cup[(n+1)2]

j=1 Dminus2j) = DminusDminus

1 cup Dminus2 From the above result

nminus1]Yplusmn| le [2M25M(4m+ 1)]n

int Rprime

0

Rprimenminus1

(n minus 1) dRprime

le [2M25M(4m+ 1)Rprime]n

n in Dminus

(536)

can be obtained and then we see that the sequence of functions wplusmnn (micro ν)Y plusmn(micro ν)

ie

wplusmnn (micro ν)Y plusmn(micro ν)=wplusmn

0 +[wplusmn1 minuswplusmn

0 ]+ +[wplusmnn minuswplusmn

nminus1]Y plusmn(micro ν)(n=1 2 ) (537)

in Dminus uniformly converge to wplusmnlowast (micro ν)Y plusmn(micro ν) and wlowast(z) = [w+(micro ν) + wminus(micro ν)

minusi(w+(micro ν)minus wminus(micro ν))]2 satisfies the equality

wlowast(z)=w0(z)+Φlowast(z)+Ψlowast(z)

Ψlowast(z)=

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

int ν

a2j+1

g1lowast(z)e1dν+

int micro

0g2

lowast(z)e2dmicro inDminus2j+1 j=01[n2]

int ν

2g1

lowast(z)e1dν+int micro

a2jminus1

g2lowast(z)e2dmicro inDminus

2j j=1[(n+1)2]

g1lowast(z)=Aξlowast+Bηlowast+Eg2

lowast(z)=Cξlowast+Dηlowast+F inDminus

(538)

and the corresponding function ulowast(z) is just a solution of Problem Alowast for equation(32) in the domain Dminus and wlowast(z) satisfies the estimate

C[wplusmnlowast (micro ν)Y plusmn(micro ν) Dminus] le M26 = e4M25M(2m+1)Rprime

(539)

In addition we can find a sequence of functions wn(z)(wn(z) = Φn(z)eφn(z)+ψn(z))in D+ and Φn(z) is an analytic function in D+ satisfying the boundary conditions

in which

sn(x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

2r((1minus i)x2)minus 2Rn((1minus i)x2)a((1minus i)x2)minus b((1minus i)x2)

minus [a((1minus i)x2) + b((1minus i)x2)]h2jradic2[a((1minus i)x2)minus b((1minus i)x2)]

+ Re [λ(x)Ψn(x)]

x isin (a2j a2j+1) j = 0 1 [n2]

2r((1+i)x2+1minusi)minus2Rn((1+i)x2+1minusi)a((1 + i)x2 + 1minus i) + b((1 + i)x2 + 1minus i)

+Re [λ(x)Ψn(x)] x isin (a2jminus1 a2j) j = 1 [(n+ 1)2]

(541)

in which the real constants h2j k2jminus1 are as stated in (515) From (531)(539) itfollows that

Cβ[X(x)sn(x) L0] le M27 = M27(p0 β k D) (542)

Cβ[wn(z)X(z) D+] le M21(k1 + k2 +M27) (543)

Thus from wn(z)X(x) we can choose a subsequence which uniformly converges toa function wlowast(z)X(z) in D+ Combining (543) and (539) it is obvious that thesolution wlowast(z) of Problem Alowast for (32) in D satisfies the estimate

Theorem 55 If the complex equation (32) satisfies Condition C then ProblemAlowast for (32) has at most one solution in D

Proof Let w1(z) w2(z) be any two solutions of Problem Alowast for (32) By ConditionC we see that w(z) = w1(z)minus w2(z) satisfies the homogeneous complex equation

wz

wzlowast

= A1w + A2w in

⎧⎨⎩D+

Dminus

⎫⎬⎭ (545)

and boundary conditions

Re [λ(z)w(z)] = 0 z isin Γ

Re [λ(x)w(x)] = s(x) x isin L0

Re [λ(z)w(z)] = 0 z isin L3 cup L4

Im [λ(z)w(z)]|z=A2j+1 = 0 j = 0 1 [n2]

Im [λ(z)w(z)]|z=B2jminus1 = 0 j = 1 [(n+ 1)2]

(546)

in which

s(x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

+Re [λ(x)Ψ(x)] x isin (a2j a2j+1) j = 0 1 [n2]

(547)

w(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

g(z) =

⎧⎨⎩A1 + A2ww w(z) = 0 z isin D+

Φ(z) + Ψ(z)

Ψ(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int ν

a2j+1

[Aξ + Bη]e1dν +int micro

0[Cξ + Dη]e2dmicro

in Dminus2j+1 j = 0 1 [n2]int ν

2[Aξ + Bη]e1dν +

int micro

a2jminus1

[Cξ + Dη]e2dmicro

in Dminus2j j = 1 [(n+ 1)2]

(548)

By using the method in the proof of Theorem 54 we can get that

n in Dminus (549)

Let n rarr infin we get wplusmn(z) = 0 ie w(z) = w1(z) minus w2(z) = 0 and then Φ(z) =Ψ(z) = 0 in Dminus thus s(x) = 0 on L0 Noting that w(z) = Φ(z)eφ(z) satisfies theboundary conditions in (546) we see that the analytic function Φ(z) in D+ satisfiesthe boundary conditions

and the index of the boundary value problem (550) is K = minus12 hence Φ(z) = 0 inD+ and then w(z) = Φ(z)eφ(z) = 0 in D+ namely w(z) = w1(z)minus w2(z) = 0 in D+This proves the uniqueness of solutions of Problem Alowast for (32)

From Theorems 54 and 55 we see that under Condition C Problem Alowast forequation (32) has a unique solution w(z) which can be found by using successiveiteration and the solution w(z) satisfies the estimate (544) ie

Cβ[w(z)X(z) D+] + C[wplusmn(micro ν)Y plusmn(micro ν) Dminus] le M29 (551)

where k = (k0 k1 k2) M29 = M29 (p0 β k D) is a non-negative constant Moreoverwe have

Theorem 56 Suppose that equation (32) satisfies Condition C Then any solutionw(z) of Problem Alowast for (32) satisfies the estimates (544) and

Cβ[w(z)X(z) D+] + C[w(z) Dminus] le M30 (k1 + k2) (552)

where X(z) Y (z) are as stated in (524) respectively and M30 = M30 (p0 β δ k0 D)is a non-negative constant

From the estimates (551) and (552) we can see the regularity of solutions ofProblem Alowast for (32)

Finally we mention that if the index K is an arbitrary even integer or 2K is anarbitrary odd integer the above Problem Alowast for (32) can be considered But ingeneral Problem Alowast for (32) with K le minus1 has minus2K minus 1 solvability conditions orwhen K ge 0 its general solution includes 2K + 1 arbitrary real conditions

For more general first order complex equations of mixed type the discontinuousRiemannndashHilbert boundary value problem remains to be discussed

The references for this chapter are [3][8][12][16][20][25][35][36][42][44][52][55][60][63][73][75][83][85][95][98]

CHAPTER V

SECOND ORDER LINEAR EQUATIONS OFMIXED TYPE

In this chapter we discuss the oblique derivative boundary value problem for sec-ond order linear equation of mixed (elliptic-hyperbolic) type in a simply connecteddomain We first prove uniqueness and existence of solutions for the above bound-ary value problem and then give a priori estimates of solutions for the problemfinally discuss the existence of solutions for the above problem in general domainsIn books [12]1)3) the author investigated the Dirichlet problem (Tricomi problem)for the mixed equation of second order ie uxx + sgny uyy = 0 In [69] theauthor discussed the Tricomi problem for the generalized Lavrentprimeev-Bitsadze equa-tion uxx + sgny uyy + Aux + Buy + Cu = 0 ie uξη + auξ + buη + cu = 0 with theconditions a ge 0 aξ+abminus c ge 0 c ge 0 in the hyperbolic domain In this section wecancel the above assumption in [69] and obtain the solvability result on the discon-tinuous Poincare problem which includes the corresponding results in [12]1)3)[69]as special cases

1 Oblique Derivative Problems for Simplest Second OrderEquation of Mixed Type

In this section we introduce the oblique derivative boundary value problem for sim-plest mixed equation of second order in a simply connected domain and verify theuniqueness and existence of solutions for the above boundary value problem

Let D be a simply connected bounded domain in the complex plane CI with theboundary partD = Γ cup L where Γ(sub y gt 0) isin C2

α(0 lt α lt 1) with the end pointsz = 0 2 and L = L1 cup L2 L1 = x = minusy 0 le x le 1 L2 = x = y + 2 1 le x le 2and denote D+ = D cap y gt 0 Dminus = D cap y lt 0 and z1 = 1 minus i We mayassume that Γ = |z minus 1| = 1 y ge 0 otherwise through a conformal mapping therequirement can be realized

11 The oblique derivative problem for simplest second order equationof mixed type

In A V Bitsadzersquos books [12]1)3) the author discussed the solvability of sev-eral boundary value problems including the Dirichlet problem or Tricomi problem

158 V Second Order Linear Mixed Equations

(Problem D or Problem T ) for the second order equation of mixed type

uxx + sgny uyy = 0 in D (11)

the equation is sondashcalled Lavrentprimeev-Bitsadze equation its complex form is as followsuzz

uzlowastzlowast

= 0 in

D+

Dminus

(12)

whereuzlowast = uz wzlowast =

12[wx minus iwy]

Now we formulate the oblique derivative boundary value problem as follows

Problem P Find a continuously differentiable solution u(z) of (12) in Dlowast =D0 x minus y = 2 or Dlowast = Dx + y = 0 2 which is continuous in D and sat-isfies the boundary conditions

12

partu

partl= Re [λ(z)uz] = r(z) z isin Γ u(0) = b0 u(2) = b2 (13)

12

partu

partl= Re [λ(z)uz] = r(z) z isin Lj (j = 1 or 2) Im [λ(z)uz]|z=z1 = b1 (14)

where l is a given vector at every point on Γcup Lj (j = 1 or 2) λ(z) = a(x) + ib(x) =cos(l x) minus i cos(l y) if z isin Γ and λ(z) = a(z) + ib(z) = cos(l x) + i cos(l y) ifz isin Lj (j = 1 or 2) b0 b1 are real constants and λ(z) r(z) b0 b1 b2 satisfy theconditions

Cα[λ(z)Γ]lek0 Cα[r(z)Γ]lek2 Cα[λ(z)Lj]lek0 Cα[r(z)Lj]lek2 j=1 or 2

cos(l n)ge0 on Γ |bj|lek2 j=0 1 2 maxzisinL1

1|a(z)minusb(z)| or maxzisinL2

1|a(z)+b(z)| lek0

(15)

in which n is the outward normal vector at every point on Γ α (12 lt α lt 1) k0 k2

are non-negative constants For convenience we may assume that uz(z1) = 0 other-wise through a transformation of function Uz(z) = uz(z) minus λ(z1)[r(z1) + ib1] therequirement can be realized

The boundary value problem for (12) with r(z) = 0 z isin Γ cup Lj (j = 1 or 2) andb0 = b1 = b2 = 0 will be called Problem P0 The number

K =12(K1 +K2) (16)

is called the index of Problem P and Problem P0 where

Kj =[φj

π

]+Jj Jj = 0 or 1 eiφj =

γj =φj

πminusKj j = 1 2 (17)

1 Simplest Mixed Equation of Second Order 159

in which t1 = 2 t2 = 0 λ(t) = eiπ4 on L0 = (0 2) and λ(t1 minus 0) = λ(t2 + 0) =exp(iπ4) or λ(t) = ei7π4 on L0 and λ(t1 minus 0) = λ(t2 + 0) = exp(i7π4) Here wechoose K = 0 or K = minus12 on the boundary partD+ of D+ if cos(ν n) equiv 0 on Γand the condition u(2) = b2 can be canceled In this case the solution of ProblemP for (12) is unique In order to ensure that the solution u(z) of Problem P iscontinuously differentiable in Dlowast we need to choose γ1 gt 0 If we require that thesolution u(z) in D is only continuous it is suffices to choose minus2γ1 lt 1 minus2γ2 lt 1Problem P in this case still includes the Dirichlet problem as a special case If theboundary condition Re [λ(z)uz] = r(z) on Lj(j = 1 or 2) in (14) is replaced byRe [λ(z)uz] = r(z) on Lj(j = 1 or 2) then Problem P does not include the aboveDirichlet problem (Problem D) as a special case

Setting that uz = w(z) it is clear that Problem P for (12) is equivalent to theRiemannndashHilbert boundary value problem (Problem A) for the first order complexequation of mixed type

wz

wzlowast

= 0 in

D+

Dminus

(18)

with the boundary conditions

Re [λ(z)w(z)] = r(z) z isin Γ u(2) = b2

Re [λ(z)w(z)] = r(z) z isin Lj (j = 1 or 2) Im[λ(z1)w(z1)] = b1(19)

int z

0w(z)dz + b0 in D (110)

in which the integral path in Dminus is as stated in Chapter II On the basis of the resultin Section 1 Chapter IV we can find a solution w(z) of Problem A for the mixedcomplex equation (18) as stated in (130) Chapter IV but the function λ(x) in theintegral formula in D+ should be replaced by λ(x) on L0 = (0 2) the function w(z)in Dminus should be replaced by w(z) in the second formula in (130) Chapter IV Hencewe have the following theorem

Theorem 11 Problem P for the mixed equation (12) has a unique solution in theform (110) where

w(z) = w(z) + λ(z1)[r(z1)minus ib1]

w(z) =

⎧⎪⎪⎨⎪⎪⎩W [ζ(z)] z isin D+0 2

12(1minus i)f(x+ y) +

12(1 + i)g(x minus y)

g(x minus y) =2r((1minus i)(x minus y)2)

a((1minus i)(x minus y)2)minus b((1minus i)(x minus y)2)

minus [a((1minusi)(xminusy)2)+b((1minusi)(xminusy)2)]f(0)a((1minus i)(x minus y)2)minus b((1minus i)(x minus y)2)

or

w(z) =12(1 + i)g(x minus y) +

12(1minus i)f(x+ y)

f(x+ y) =2r((1 + i)(x+ y)2 + 1minus i)

a((1 + i)(x+ y)2 + 1minus i) + b((1 + i)(x+y)2+1minusi)

minus [a((1+i)(x+y)2+1minusi)minusb((1+i)(x+y)2+1minusi)]g(2)a((1+i)(x+y)2+1minusi)+b((1+i)(x+y)2+1minusi)

z = x+ iy isin Dminus0 2

(111)

in which f(0) = [a(z1) + b(z1)]r(z1) + [a(z1) minus b(z1)]b1 g(2) = [a(z1) minus b(z1)]r(z1) minus[a(z1)+ b(z1)]b1 W (ζ) on D+0 2 is as stated in (128) Chapter IV but where thefunction λ(z) on L0 is as stated in (17) and λ(z) r(z) b1 are as stated in (13) (14)Moreover the functions

f(x+ y) = U(x+ y 0)minus V (x+ y 0) = Re [(1 + i)W (ζ(x+ y))]

g(x minus y) = U(x minus y 0) + V (x minus y 0) = Re [(1minus i)W (ζ(x minus y))](112)

where U = ux2 V = minusuy2 W [ζ(x+ y)] and W [ζ(xminus y)] are the values of W [ζ(z)]on 0 le z = x+ y le 2 and 0 le z = x minus y le 2 respectively

From the above representation of the solution u(z) of Problem P for (12) we canderive that u(z) satisfies the estimate

Cβ[u(z) D] + Cβ[w(z)X(z) D+] + C[wplusmn(z)Y plusmn(z) Dminus] le M1k2 (113)

in which k2 are as stated in (15) wplusmn(z) = Rew ∓ Imw and

X(z) =2prod

j=1|z minus tj|ηj Y plusmn(z) =

2prodj=1

|x plusmn y minus tj|ηj

ηj =2|γj|+ δ if γj lt 0

δ γj ge 0j = 1 2

(114)

here γj (j = 1 2) are real constants as stated in (17) and β δ (β lt δ) are sufficientlysmall positive constant andM1 = M1(p0 β k0 D

+) is a non-negative constant Fromthe estimate (113) we can see the regularity of solutions of Problem P for (12)

Finally we mention that if the index K is an arbitrary even integer or 2K is anarbitrary odd integer the above oblique derivative problem for (11) or (12) can beconsidered but in general these boundary value problems for K le minus1 have somesolvability conditions or their solutions for K ge 0 are not unique

12 The Dirichlet boundary value problem for simplest second orderequation of mixed type

The Dirichlet problem (Problem D or Problem T ) for (12) is to find a solution of(12) with the boundary conditions

u(z) = φ(z) on Γ cup Lj(j = 1 or 2) (115)

where φ(z) satisfies the condition

C1[φ(z)Γ cup Lj] le k2 j = 1 or 2 (116)

In the following we shall explain that Problem D is a special case of Problem P Infact denote w = uz in D Problem D for the mixed equation (12) is equivalent toProblem A for the mixed equation (18) with the boundary condition (19) and therelation (110) in which

λ(z) = a+ ib =

⎧⎪⎪⎨⎪⎪⎩i(z minus 1) θ = arg(z minus 1) on Γ

1minus iradic2on L1 or

1 + iradic2on L2

r(z) =

⎧⎪⎪⎨⎪⎪⎩φθ on Γ

φxradic2on L1 or

φxradic2on L2

b1 = Im[1 + iradic2

uz(z1 minus 0)]=

φx minus φxradic2

|z=z1minus0 = 0 or

b1 = Im[1minus iradic2

uz(z1 + 0)]= 0 b0 = φ(0)

(117)

in which a = 1radic2 = b = minus1radic2 on L1 or a = 1

As for the index K = minus12 of Problem D on partD+ we can derive as followsAccording to (117) the boundary conditions of Problem D in D+ possess the form

Re [i(z minus 1)w(z)] = r(z) = φθ on Γ

Re[1minus iradic2

w(x)]= s(x) =

φprime((1minus i)x2)radic2

x isin [0 2] or

Re[1 + iradic2

w(x)]= s(x) =

φprime((1 + i)x2 + 1minus i)radic2

x isin [0 2]

it is clear that the possible discontinuous points of λ(z) on partD+ are t1 = 2 t2 = 0and

λ(t1 + 0) = e3πi2 λ(t2 minus 0) = eπi2

λ(t1 minus 0) = λ(t2 + 0) = eπi4 or λ(t1 minus 0) = λ(t2 + 0) = e7πi4

λ(t1 minus 0)λ(t1 + 0)

= eminus5πi4 = eiφ1 λ(t2 minus 0)λ(t2 + 0)

= eπi4 = eiφ2 or

= eπi4 = eiφ1 λ(t2 minus 0)λ(t2 + 0)

= eminus5πi4 = eiφ2

In order to insure the uniqueness of solutions of Problem D we choose that

minus1 lt γ1 =φ1

πminus K1 = minus5

4minus (minus1) = minus1

4lt 0

0 le γ2 =φ2

πminus K2 =

14

lt 1 or

0 le γ1 =φ1

πminus K1 =

14

lt 1

minus1 lt γ2 =φ2

πminus K2 = minus5

4lt 0

thusK1 = minus1 K2 = 0 K =

K1 +K2

2= minus1

2 or

K1 = 0 K2 = minus1 K =K1 +K2

2= minus1

2

In this case the unique solution w(z) is continuous in Dlowast = D0 x minus y = 2 orDlowast = Dx + y = 0 2 for the first case w(z) in the neighborhood of t2 = 0 isbounded and w(z) in the neighborhood of t1 = 2 possesses the singularity in theform |z minus 2|minus12 and its integral (110) is bounded for the second case w(z) in theneighborhood of t1 = 2 is bounded and w(z) in the neighborhood of t2 = 0 possessesthe singularity of |z|minus12 and its integral is bounded If we require that the solutionw(z) = uz is bounded in D+0 2 then it suffices to choose the index K = minus1 inthis case the problem has one solvability condition

From Theorem 11 it follows that the following theorem holds

Theorem 12 Problem D for the mixed equation (12) has a unique continuoussolution in D as stated in (110) where λ(z) r(z) b1 are as stated in (117) and W (ζ)in D+0 2 is as stated in (128) Chapter IV but in which λ(x) = (1 + i)

radic2 or

(1minus i)radic2 on L0 and f(x+ y) g(x minus y) are as stated in (112) [85]15)

2 Oblique Derivative Problems for Second Order LinearEquations of Mixed Type

In this section we mainly discuss the uniqueness and existence of solutions for secondorder linear equations of mixed type

21 Formulation of the oblique derivative problem for mixed equationsof second order

Let D be a simply connected bounded domain D in the complex plane CI with theboundary partD = Γcup L as stated in Section 1 We consider the linear mixed equation

2 Oblique Derivative Problems 163

of second orderuxx + sgny uyy = aux + buy + cu+ d in D (21)

where a b c d are functions of z(isin D) its complex form is the following equation ofsecond order ⎧⎨⎩uzz = Re [A1(z)uz] + A2(z)u+ A3(z) in D+

uzlowastzlowast = Re [A1(z)uz] + A2(z)u+ A3(z) in Dminus(22)

where

12[ux + iuy] uzz =

14[uxx + uyy]

uzlowast =12[ux + iuy] = uz uzzlowast =

12[(uz)x minus i(uz)y] =

14[uxx minus uyy]

A1 =a+ ib

2 A2 =

c

4 A3 =

d

4in D

Suppose that equation (22) satisfies the following conditions Condition C

The coefficients Aj(z) (j = 1 2 3) in (22) are measurable in z isin D+ and continu-ous in Dminus and satisfy

Lp[Aj D+] le k0 j = 1 2 Lp[A3 D+] le k1 A2 ge 0 in D+ (23)

where p (gt 2) k0 k1 are non-negative constants If the condition (24) is replaced by

Cα[Aj Dminus] le k0 j = 1 2 Cα[A3 Dminus] le k1

in which α(0 lt α lt 1) is a real constant then the conditions will be called ConditionC prime

The oblique derivative boundary value problem (Problem P ) for equation (22) isto find a continuously differentiable solution u(z) of (22) in Dlowast = D0 x minus y = 2or Dlowast = Dx + y = 0 2 which is continuous in D and satisfies the boundaryconditions (13) and (14) in which b0 b2 is a real constant satisfying the condition|b0| |b2| le k2 The index K is defined as stated in Section 1 now we discuss the case

K =12(K1 +K2) = 0 (25)

where

Kj =[φj

π

γj =φj

in which t1 = 2 t2 = 0 λ(t) = eiπ4 on L0 = (0 2) and λ(t1 minus 0) = λ(t2 + 0) =exp(iπ4) or λ(t) = ei7π4 on L0 and λ(t1 minus0) = λ(t2+0) = exp(i7π4) We mention

that if A2 = 0 in D or cos(l n) equiv 0 on Γ then we do not need the point conditionu(2) = b2 in (13) and only choose the index K = minus12 Because if cos(l n) equiv 0 onΓ from the boundary condition (13) we can determine the value u(2) by the valueu(0) namely

u(2)=2Reint 2

0uzdz+u(0)=2

int 2

0Re [i(z minus 1)uz]dθ+b0=2

int 0

πr(z)dθ+b0

in which λ(z) = i(z minus 1) θ = arg(z minus 1) on Γ In brief we choose that

K=

⎧⎪⎪⎨⎪⎪⎩0

minus12the point conditions are

⎧⎨⎩u(0)=b0 u(2)=b2

u(0) = b0

⎫⎬⎭ if

⎧⎨⎩ cos(l n) equiv0cos(l n)equiv0on Γ

In order to ensure that the solution u(z) of Problem P is continuously differentiablein Dlowast we need to choose γ1 gt 0 or γ2 gt 0 If we only require that the solution iscontinuous it suffices to choose minus2γ2 lt 1 minus2γ1 lt 1 respectively In the followingwe shall only discuss the case K = 0 and the case K = minus12 can be similarlydiscussed

22 The representation and uniqueness of solutions for the obliquederivative problem for (22)

Now we give the representation theorems of solutions for equation (22)

Theorem 21 Let equation (22) satisfy Condition C in D+ u(z) be a continuoussolution of (22) in D+ and continuously differentiable in D+

lowast = D+0 2 Thenu(z) can be expressed as

u(z) = U(z)Ψ(z) + ψ(z) in D+

U(z) = 2Reint z

0w(z)dz + b0 w(z) = Φ(z)eφ(z) in D+

(27)

where ψ(z) Ψ(z) are the solutions of equation (22) in D+ and

uzz minus Re [A1uz]minus A2u = 0 in D+ (28)

respectively and satisfy the boundary conditions

ψ(z) = 0 Ψ(z) = 1 on Γ cup L0 (29)

where ψ(z) Ψ(z) satisfies the estimates

C1β[ψD+] le M2 ψ W 2

p0(D+)le M2 (210)

C1β[Ψ D+] le M3 Ψ W 2

p0(D+)le M3Ψ(z) ge M4 gt 0 z isin D+ (211)

in which β (0 lt β le α) p0 (2 lt p0 le p) Mj = Mj(p0 β k D) (j = 2 3 4) arenon-negative constants k = (k0 k1 k2) Moreover U(z) is a solution of the equation

Uzz minus Re [AUz] = 0 A = minus2(lnΨ)z + A1 in D+ (212)

where Im [φ(z)] = 0 z isin L0 = (0 2) and φ(z) satisfies the estimate

Cβ[φD+] + Lp0 [φz D+] le M5 (213)

in which β(0 lt β le α) M5 = M5 (p0 β k0 D) are two non-negative constantsΦ(z) is analytic in D+ If u(z) is a solution of (22) in D+ satisfying the boundaryconditions (13) and

Re [λ(z)uz]|z=x = s(x) λ(x) = 1 + i or 1minus i x isin L0 Cβ[s(x) L0] le k3 (214)

then the following estimate holds

Cβ[u(z) D+] + Cβ[uzX(z) D+] le M6(k1 + k2 + k3) (215)

in which k3 is a non-negative constant s(x) can be seen as stated in the form (223)below X(z) is as stated in (114) and M6 = M6(p0 β k0 D

+) is a non-negativeconstant

Proof According to the method in the proof of Theorem 31 Chapter III theequations (22)(28) in D+ have the solutions ψ(z) Ψ(z) respectively which satisfythe boundary condition (29) and the estimates (210)(211) Setting that

Ψ(z) (216)

it is clear that U(z) is a solution of equation (212) which can be expressed the secondformula in (27) where φ(z) satisfies the estimate (213) and Φ(z) is an analyticfunction in D+ If s(x) in (214) is a known function then the boundary valueproblem (22)(13)(214) has a unique solution u(z) as stated in the form (27)which satisfies the estimate (215)

Theorem 22 Suppose that the equation (22) satisfies Condition C Then anysolution of Problem P for (22) can be expressed as

u(z) = 2Reint z

0w(z)dz + b0 w(z) = w0(z) +W (z) (217)

where w0(z) is a solution of Problem A for the complex equation (18) with the bound-ary conditions (13) (14)(w0(z) = u0z) and W (z) possesses the form

W (z) = w(z)minus w0(z) in D w(z) = Φ(z)eφ(z) + ψ(z) in D+

int intD+

g(ζ)ζ minus z

W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν

2g1(z)dνe1 +

int micro

(218)

in which e1 = (1 + i)2 e2 = (1 minus i)2 micro = x + y l = x minus y φ0(z) is an analyticfunction in D+ such that Im [φ(x)] = 0 on L0 and

g(z)=

A12+A1w(2w) w(z) =00 w(z)=0 zisinD+

f(z)=Re[A1φz]+A2u+A3 in Dminus

g1(z)=g2(z)=Aξ+Bη+Cu+D ξ=Rew+Imw η=RewminusImw

A=ReA1+ImA1

2 B=

ReA1minusImA1

2 C=A2D=A3 in Dminus

(219)

where Φ(z) and Φ(z) are the solutions of equation (18) in D+ and Dminus respectivelysatisfying the boundary conditions

Re [λ(z)(Φ(z)eφ(z) + ψ(z))] = r(z) z isin Γ

(220)

where λ(x) = 1 + i or 1 minus i x isin L0 Moreover by Theorem 12 Chapter IV thesolution w0(z) of Problem A for (18) and u0(z) satisfy the estimate in the form

Cβ[u0(z) D]+Cβ[w0(z)X(z) D+]+Cβ[wplusmn0 (z)Y

plusmn(z) Dminus]leM7(k1+k2) (221)

in which wplusmn(z) = Rew(z)∓ Imw(z) X(z) Y plusmn(z) are as stated in (114)

u0(z) = 2Reint z

0w0(z)dz + b0 (222)

and M7 = M7(p0 β k0 D) is a non-negative constant From (222) it follows that

Cβ[u0(z) D] le M8Cβ[w0(z)X(z) D+] + Cβ[wplusmn0 (z)Y

plusmn(z) Dminus+ k2

where M8 = M8(D) is a non-negative constant

Proof Let u(z) be a solution of Problem P for equation (22) and w(z) = uzu(z) be substituted in the positions of w u in (219) thus the functions g(z)f(z) g1(z) g2(z) and ψ(z) φ(z) in D+ and Ψ(z) in Dminus in (218)(219) can bedetermined Moreover we can find the solution Φ(z) in D+ and Φ(z) in Dminus of (18)with the boundary conditions (220) where

s(x)=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

2r((1minusi)x2)minus2R((1minusi)x2)a((1minus i)x2)minus b((1minus i)x2)

+Re [λ(x)Ψ(x)] or

+Re [λ(x)Ψ(x)] on L0

(223)

here and later R(z) = Re [λ(z)Ψ(z)] on L1 or L2 thus

w(z) = w0(z) +W (z) =

⎧⎪⎨⎪⎩Φ(z)φ(z) + ψ(z) in D+

wz

wzlowast

= Re [A1w] + A2u+ A3 in

⎧⎨⎩D+

Dminus

⎫⎬⎭ (224)

which can be expressed as the second formula in (217) and u(z) is a solution ofProblem P for (22) as stated in the first formula in (217)

Theorem 23 If equation (22) satisfies Condition C then Problem P for (22) hasat most one solution in D

Proof Let u1(z) u2(z) be any two solutions of Problem P for (22) By ConditionC we see that u(z) = u1(z)minusu2(z) and w(z) = uz satisfies the homogeneous equationand boundary condition

wz

wzlowast

= Re [A1w] + A2u in

D+

Dminus

(225)

Re [λ(z)w(z)] = 0 z isin Γ u(0) = 0 u(2) = 0

Re [λ(z)w(z)] = 0 z isin Lj(j = 1 or 2) Im [λ(z1)w(z1)] = 0(226)

w(z) =

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

Φ(z)eφ(z) + ψ(z) ψ(z) = Tf φ(z) = φ0(z) + T g in D+

w0(z) + Φ(z) + Ψ(z)

Ψ(z) =int ν

2[Aξ+Bη+Cu]e1dν+

int micro

0[Aξ+Bη+Cu]e2dmicro in Dminus

(227)

where g(z) is as stated in (219) Φ(z) in D+ is an analytic function and Φ(z) is asolution of (18) in Dminus satisfying the boundary condition (220) φ(z) ψ(z) possessthe similar properties as φ(z) ψz(z) in Theorem 21 If A2 = 0 in D+ then ψ(z) = 0Besides the functions Φ(z) Φ(z) satisfy the boundary conditions⎧⎨⎩Re [λ(x)Φ(x)] = s(x)

Re [λ(x)Φ(x)] = Re [λ(x)(W (x)minusΨ(x))]on L0 (228)

where s(x) is as stated in (223) From (217) with b0 = 0 we can obtain

C[u(z) D] le M8C[w(z)X(z) D+] + C[wplusmn0 (z)Y

plusmn(z) Dminus (229)

By using the method of iteration the estimate

C[w(z) Dminus] le [2M9M(4m+ 1)Rprime]n

n(230)

can be derived where M9 = maxC[A Dminus] C[BDminus] C[CDminus] M = 1+ 4k20(1 +

k20) and m = C[w(z) Dminus] gt 0 Let n rarr infin from (229) it follows that w(z) = 0 in

Dminus and Ψ(z) = 0 Φ(z) = 0 z isin Dminus Thus the solution u(z) = 2Reint z0 w(z)dz is

the solution of equation (28) with the boundary conditions

Re [λ(z)uz]=0 on Γ Re [λ(x)uz(x)]=0 on L0=(0 2) u(0)=0 u(2)=0(231)

in which λ(x) = 1 + i or 1 minus i x isin L0 Similarly to the proof of Theorem 34Chapter III we can obtain u(z) = 0 on D+ This shows the uniqueness of solutionsof Problem P for (22)

23 The solvability of the oblique derivative problem for (22)

Theorem 24 Suppose that the mixed equation (22) satisfies Condition C ThenProblem P for (22) has a solution in D

Proof It is clear that Problem P for (22) is equivalent to Problem A for thecomplex equation of first order and boundary conditions

wz

wzlowast

= F F = Re [A1w] + A2u+ A3 in

⎧⎨⎩D+

Dminus

⎫⎬⎭ (232)

Re [λ(z)w(z)] = r(z) z isin Γ

Re [λ(z)w(z)] = r(z) z isin Lj(j = 1 or 2) Im [λ(z1)w(z1)] = b1(233)

and the relation (217) From (217) it follows that

C[u(z) D] le M8[C(w(z)X(z) D+) + C(wplusmn(z)Y plusmn(z) Dminus)] + k2 (234)

where X(z) Y plusmn(z) wplusmn(z) are as stated in (114) respectively M8 = M8(D) is anon-negative constant In the following by using successive iteration we shall find asolution of Problem A for the complex equation (232) inD Firstly denoting the solu-tion w0(z)(= ξ0e1+η0e2) of Problem A for (18) and u0(z) in (217) and substitutingthem into the position of w = (ξe1+ ηe2) u(z) in the right-hand side of (232) simi-larly to (218)(219) we have the corresponding functions f1(z) g1(z) g1

2(z) g12(z)

and

w1(z) = Φ1(z)eφ1(z) + ψ1(z) in D+

φ1(z)= φ0(z)+Tg1= φ0(z)minus 1π

int intD+

g1(ζ)ζminusz

dσζ ψ1(z)=Tf1 in D+

W1(z)=Φ(z)+Ψ(z) Ψ(z)=int ν

2g11(z)dνe1+

int micro

(235)

where micro = x + y ν = x minus y where Φ1(z) is a solution of (18) in Dminus satisfying theboundary conditions

Re [λ(x)Φ1(x)] = Re [λ(x)(W1(z)minusΨ1(x))] z isin L0

(236)

andw1(z) = w0(z) +W1(z) = w0(z) + Φ1(z) + Ψ1(z) in Dminus (237)

Cβ[w(z)X(z) D+] + C[wplusmn1 (z)Y

plusmn(z) Dminus] le M10 = M10(p0 β k Dminus) (238)

Furthermore we substitute w1(z) = w0(z)+W1(z) and corresponding functions w1(z)ξ1(z) = w+(z) = Rew1(z)minusImw1(z) η1(z) = wminus(z) = Rew1(z)+Imw1(z) u1(z) intothe positions w(z) ξ(z) η(z) u(z) in (218)(219) and similarly to (235)ndash(237)we can find the corresponding functions ψ2(z) φ2(z) Φ2(z) in D+ Ψ2(z)Φ2(z) andW2(z) = Φ2(z) + Ψ2(z) in Dminus and the function

w2(z) = Φ2(z)eφ2(z) + ψ2(z) in D+

w2(z) = w0(z) +W2(z) = w0(z) + Φ2(z) + Ψ2(z) in Dminus(239)

satisfies the similar estimate in the form (238) Thus there exists a sequence offunctions wn(z) and

wn(z) = Φn(z)eφn(z) + ψn(z) in D+

wn(z) = w0(z) +Wn(z) = w0(z) + Φn(z) + Ψn(z)

Ψn(z)=int ν

2g1

n(z)e1dν +int micro

0g2

n(z)e2dmicro in Dminus

(240)

and then

|[wplusmn1 (z)minus wplusmn

0 (z)]Y plusmn(z)| le |Φplusmn1 (z)Y plusmn(z)|

+radic2[|Y +(z)

int ν

2[Aξ0 +Bη0 + Cu0 +D]e1dν|

0[Aξ0 +Bη0 + Cu0 +D]e2dmicro|] le 2M11M(4m+ 1)Rprime in Dminus

(241)

where M11 = maxzisinDminus(|A| |B| |C| |D|) m = C[w0(z)X(z) Dminus] Rprime = 2 M = 1 +4k2

0(1 + k20) It is clear that wn(z)minus wnminus1(z) satisfies

= Φn(z)minus Φnminus1(z) +int ν

2[A(ξn minus ξnminus1) +B(ηn minus ηnminus1) + C(un minus unminus1)]e1dν

+int micro

0[A(ξn minus ξnminus1) +B(ηn minus ηnminus1) + C(un minus unminus1)]e2dmicro in Dminus

(242)

where n = 1 2 From the above equality the estimate

nminus1]Y plusmn(z)| le [2M11M(4m+ 1)]n timesint Rprime

0

Rprimenminus1

(n minus 1) dRprime

n in Dminus

(243)

can be obtained and then we can see that the sequence of functions wplusmnn (z)Y

plusmn(z)ie

wplusmnn (z)Y

plusmn(z) = wplusmn0 (z) + [w

plusmn1 (z)minus wplusmn

0 (z)] + middot middot middot+ [wplusmnn (z)minus wplusmn

(n = 1 2 ) in Dminus uniformly converge to wplusmnlowast (z)Y

plusmn(z) and wlowast(z) = [w+lowast (z) +

wminuslowast (z)minus i(w+

lowast (z)minus wminuslowast (z))]2 satisfies the equality

Ψlowast(z) =int ν

2[Aξlowast+Bηlowast+Culowast+D]e1dν+

int micro

0[Aξlowast+Bηlowast+Culowast+D]e2dmicro in Dminus

(245)

and the corresponding function ulowast(z) is just a solution of Problem P for equation(22) in the domain Dminus and wlowast(z) satisfies the estimate

In the meantime we can obtain the estimate

Cβ[wn(z)X(z) D+] le M12 = M12(p0 β k D) (247)

hence from the sequence wn(z) we can choose a subsequence which uniformlyconverges to wlowast(z) in D+ and wlowast(z) satisfies the same estimate (247) Combining(246) and (247) it is obvious that the solution wlowast(z) = uz of Problem A for (22)in D satisfies the estimate

plusmn(z) Dminus] le M13 = M13(p0 β k D)

where M13 is a non-negative constant Moreover the function u(z) in (217) is asolution of Problem P for (22) where w(z) = wlowast(z)

From Theorems 23 and 24 we see that under Condition C Problem A forequation (232) has a unique solution w(z) which can be found by using successiveiteration and the corresponding solution u(z) of Problem P satisfies the estimates

Cβ[u(z) D+] + Cβ[uzX(z) D+] le M14

C[u(z) D] + C[uplusmnz Y plusmn(z) D] le M15

(248)

where X(z) Y plusmn(z) is as stated in (114) and Mj = Mj (p0 β k D) (j = 14 15)are non-negative constants k = (k0 k1 k2) Moreover we can derive the followingtheorem

3 Discontinuous Oblique Derivative Problems 171

Theorem 25 Suppose that equation (22) satisfies Condition C Then any solutionu(z) of Problem P for (22) satisfies the estimates

Cβ[u(z) D+] + Cβ[uzX(z) D+] le M16(k1 + k2)

C[u(z) Dminus] + C[uplusmnz Y plusmn(z) Dminus] le M17(k1 + k2)

(249)

in which Mj = Mj(p0 β k0 D) (j = 16 17) are non-negative constants

From the estimates (248)(249) we can see that the regularity of solutions ofProblem P for (22) (see [85]15))

3 Discontinuous Oblique Derivative Problems for SecondOrder Linear Equations of Mixed Type

This section deals with an application of method of integral equations to second orderequations of mixed type We mainly discuss the discontinuous Poincare boundaryvalue problem for second order linear equation of mixed (elliptic-hyperbolic) type iethe generalized Lavrentprimeev-Bitsadze equation with weak conditions by the methodof integral equations We first give the representation of solutions for the aboveboundary value problem and then give the solvability conditions of the above problemby the Fredholm theorem for integral equations

31 Formulation of the discontinuous Poincare problem for mixedequations of second order

Let D be a simply connected bounded domain D in the complex plane CI with theboundary partD = Γ cup L as stated in Section 1 We consider the second order linearequation of mixed type (21) and its complex form (22) with Condition C prime

In order to introduce the discontinuous Poincare boundary value problem forequation (22) let the functions a(z) b(z) possess the discontinuities of first kind atm+ 2 distinct points z0 = 2 z1 zm+1 = 0 isin Γ and Z = z0 z1 zm+1 whichare arranged according to the positive direction of Γ where m is a positive integerand r(z) = O(|z minus zj|minusβj) in the neighborhood of zj(j = 0 1 m + 1) on Γ inwhich βj(j = 0 1 m+1) are small positive numbers Denote λ(z) = a(x)+ ib(x)and |a(x)|+ |b(x)| = 0 there is no harm in assuming that |λ(z)| = 1 z isin Γlowast = ΓZSuppose that λ(z) r(z) satisfy the conditions

λ(z) isin Cα(Γj) |z minus zj|βjr(z) isin Cα(Γj) j = 0 1 m+ 1 (31)

herein Γj is an arc from the point zjminus1 to zj on Γ and zm+1 = 0 and Γj(j =0 1 m+ 1) does not include the end points and α (0 lt α lt 1) is a constant

Problem Q Find a continuously differentiable solution u(z) of (22) in Dlowast =DZ(Z = 0 x minus y = 2 y le 0 or Z = x+ y = 0 y le 0 2) which is continuous in

D and satisfies the boundary conditions

12

partu

partl+εσ(z)u=Re [λ(z)uz]+εσ(z)u=r(z)+Y (z)h(z) zisinΓ u(0)=b0 (32)

12

partu

partl= Re [λ(z)uz] = r(z) z isin L1 or L2 Im [λ(z)uz]|z=z1 = b1 (33)

where l is a vector at every point on Γ cup Lj (j = 1 or 2) z1 = 1 minus i b0 b1 are realconstants λ(z) = a(x)+ib(x) = cos(l x)minusi cos(l y) z isin Γ and λ(z) = a(x)+ib(x) =cos(l x) + i cos(l y) z isin Lj (j = 1 or 2) and λ(z) r(z) b0 b1 satisfy the conditions

Cα[λ(z)Γ] le k0 Cα[σ(z)Γ] le k0 Cα[r(z)Γ] le k2 |b0| |b1| le k2

Cα[λ(z) Lj] le k0 Cα[σ(z) Lj] le k0 Cα[r(z) Lj] le k2 j = 1 or 2

maxzisinL1

1|a(x)minus b(x)| or max

zisinL2

1|a(x) + b(x)| le k0

(34)

in which α (12 lt α lt 1) k0 k2 are non-negative constants ε is a real parameterBesides the functions Y (z) h(z) are as follows

Y (z)=ηm+1prodj=0

|z minus zj|γj |zminuszlowast|l z isin Γlowast h(z)=

⎧⎨⎩ 0zisinΓ ifK geminus12hjηj(z) zisinΓj ifKltminus12

(35)

in which Γj(j = 0 1 m) are arcs on Γlowast = ΓZ and ΓjcapΓk = φ j = k hj isin J (J =φ if K ge minus12 J = 1 2K prime minus 1 if K lt minus12 K prime = [|K| + 12]) are unknownreal constants to be determined appropriately herein h1 = 0 l = 1 if 2K is oddzlowast(isin Z) isin Γlowast is any fixed point and l = 0 if 2K is even Γj(j = 1 2K prime minus 1) arenon-degenerate mutually disjointed arcs on Γ and Γj cap Z = φ j = 1 2K prime minus 1ηj(z) is a positive continuous function on the interior point set of Γj such thatηj(z) = 0 on ΓΓj and

Cα[ηj(z)Γ] le k0 j = 1 2K prime minus 1 (36)

and η = 1 or minus1 on Γj (0 le j le m + 1Γm+1 = (0 2)) as stated in [93] Theabove discontinuous Poincare boundary value problem for (22) is called Problem QProblem Q for (22) with A3(z) = 0 z isin D r(z) = 0 z isin Γ cup Lj (j = 1 or 2) andb0 = b1 = 0 will be called Problem Q0

Denote by λ(zj minus 0) and λ(zj + 0) the left limit and right limit of λ(z) as z rarrzj (0 le j le m+ 1) on Γ cup L0 and

γj =1πiln

=φj

πminus Kj

Kj =[φj

π

]+ Jj Jj = 0 or 1 j = 0 1 m+ 1

(37)

in which zm+1 = 0 z0 = 2 λ(z) = eiπ4 on L0 = (0 2) and λ(z0 minus0) = λ(zm+1+0) =exp(iπ4) or λ(z) = eminusiπ4 on L0 and λ(z0 minus 0) = λ(zm+1 + 0) = exp(minusiπ4) and0 le γj lt 1 when Jj = 0 and minus1 lt Jj lt 0 when Jj = 1 j = 0 1 m+ 1 and

K =12(K0 +K2 + middot middot middot+Km+1) =

m+1sumj=0

(φj

2πminus γj

2

)(38)

is called the index of Problem Q and Problem Q0 Let βj+γj lt 1 j = 0 1 m+1we can require that the solution u(z) satisfy the condition uz = O(|z minus zj|minusδj) in theneighborhood of zj (j = 0 1 m+ 1) in Dlowast where

τj=

⎧⎨⎩βj+τ for γj ge0 and γj lt0 βj gt |γj||γj|+τ for γj lt0 βj le|γj|

δj=

⎧⎨⎩ 2τj j=0 m+1

τj j=1 m(39)

and τ δ(lt τ) are small positive numbers In order to ensure that the solution u(z)of Problem Q is continuously differentiable in Dlowast we need to choose γ1 gt 0 or γ2 gt 0respectively

32 The representation and solvability of the oblique derivative problemfor (22)

Now we write a representation theorem of solutions for equation (22) which is similarto Theorem 22

Theorem 31 If equation (22) satisfies Condition C prime and ε = 0 A2 ge 0 in D+then any solution of Problem Q for (22) can be expressed as

u(z) = 2Reint z

0w(z)dz + c0 w(z) = w0(z) +W (z) (310)

where w0(z) is a solution of Problem A for equation (18) with the boundary conditions

Re [λ(z)w(z)] = r(z) + Y (z)h(z) z isin Γ

Re [λ(z)w(z)] = r(z) z isin Lj (j = 1 or 2) Im [λ(z)w(z)]|z=z1 = b1(311)

and W (z) possesses the form

W (z) = w(z)minus w0(z) W (z) = Φ(z)eφ(z) + ψ(z)

int intD+

g(ζ)ζ minus z

W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν

2g1(z)dνe1 +

int micro

(312)

in which e1 =1 + i

2 e2 =

1minus i

2 micro = x+ y ν = x minus y and

g(z)=A12+A1w(2w) w(z) = 00 w(z) = 0 z isin D+

f(z)=Re [A1φz]+A2u+A3 in D+

g1(z)=g2(z)=Aξ+Bη+Cu+D ξ=RewminusImw η=Rew+Imw

A=ReA1+ImA1

2 B=

ReA1minusImA1

2 C=A2 D=A3 in Dminus

(313)

where φ0(z) is an analytic function in D+ such that Im [φ(x)] = 0 on L0 = (0 2) andΦ(z)Φ(z) are the solutions of the equation (18) in D+ Dminus respectively satisfyingthe boundary conditions

(314)

where

s(x)=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

+Re [λ(x)Ψ(x)] or

(315)

in which s(x) can be written similar to (29) Moreover from Theorem 11 if theindex K le minus12 the solution u0(z)(w0(z) = u0z(z)) of Problem Q for (12) satisfiesthe estimate in the form

Cδ[u0(z) D] + Cδ[w0(z)X(z) D+] + C[wplusmn0 (z)Y

in which δ is a small positive constant p0 (2 lt p0 le p) M18 = M18 (p0 δ k0 D) aretwo non-negative constants

wplusmn0 (z)=Rew0(z)∓ Imw0(z) X(z)=Πm+1

j=0 |zminustj|ηj Y plusmn(z)=prod2

j=1 |x plusmn yminustj|ηj

ηj = 2|γj|+ δ j = 0 m+ 1 ηj = |γj|+ δ j = 1 m

andu0(z) = 2Re

int z

0w0(z)dz + c0 (317)

In order to prove the solvability of ProblemQ for (22) denote w = uz and considerthe equivalent boundary value problem (Problem B) for the mixed complex equation⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

wz minus Re [A1(z)w] = εA2(z)u+ A3(z) z isin D+

wzlowast minus Re [A1(z)w] = A3(z) z isin Dminus

u(z) = 2Reint z

0w(z)dz + b0

(318)

Re [λ(z)w] = r(z)minus εσ(z)u+ Y (z)h(z) z isin Γ

Re [λ(z)uz] = r(z) z isin Lj (j = 1 or 2) Im [λ(z)uz]|z=z1 = b1(319)

where b0 b1 are real constants are as stated in (32)(33) According to the method inSection 5 Chapter IV we can find the general solution of Problem B1 for the mixedcomplex equation ⎧⎨⎩wz minus Re [A1(z)w] = A3(z) z isin D+

wzlowast minus Re [A1(z)w] = A3(z) z isin Dminus(320)

which can be expressed as

w(z) = w0(z) +2K+1sumk=1

ckwk(z) (322)

in which w0(z) is a special solution of Problem B1 and wk(z)(k = 1 2K + 1K ge 0) is the complete system of linear independent solutions for the homogeneousproblem of Problem B1 Moreover denote by H2u the solution of Problem B2 for thecomplex equation ⎧⎨⎩wz minus Re [A1(z)w] = A2(z)u z isin D+

wzlowast minus Re [A1(z)w] = A2(z)u z isin Dminus(323)

Re [λ(z)w(z)] = minusσ(z)u+ Y (z)h(z) z isin Γ

Re [λ(z)w(z)] = 0 z isin Lj (j = 1 or 2) Im [λ(z)w(z)]|z=z1 = 0(324)

and the point conditions

Im [λ(aj)w(aj)] = 0 j isin J =

⎧⎨⎩ 1 2K + 1 K ge 0

φ K lt 0(325)

where aj isin ΓZ are the distinct points It is easy to see that H2 is a boundedoperator from u(z) isin C1(D) (ie C(u D)+C(X(z)uz D+) + C(Y plusmn(z)uplusmn

z Dminus) lt infin)to w(z) isin Cδ(D) (ie Cδ(u D) + Cδ(X(z)w(z) D+) + Cδ(Y plusmn(z)wplusmn(z) Dminus) lt infin)herein X(z) Y plusmn(z) are functions as stated in (316) Furthermore denote

u(z) = H1w + c0 = 2Reint z

0w(z)dz + c0 (326)

where c0 is arbitrary real constant It is clear that H1 is a bounded operator fromX(z)w(z) isin Cδ(D) to u(z) isin C1(D) On the basis of Theorem 31 the function w(z)can be expressed as an integral From (326) and w(z) = w(z)+ εH2u we can obtaina nonhomogeneous integral equation (K ge 0)

u minus εH1H2u = H1w(z) + c0 +2K+1sumk=1

ckH1wk(z) (327)

Due to H1H2 is a completely continuous operator in C1(D) we can use the Fredholmtheorem for the integral equation (327) Denote by

εj(j = 1 2 ) 0 lt |ε1| le |ε2| le middot middot middot le |εn| le |εn+1| le middot middot middot (328)

are the discrete eigenvalues for the homogeneous integral equation

u minus εH1H2u = 0 (329)

Noting that Problem Q for the complex equation (22) with ε = 0 is solvable hence|ε1| gt 0 In the following we first discuss the case of K ge 0 If ε = εj(j =1 2 ) ie it is not an eigenvalue of the homogeneous integral equation (329)then the nonhomogeneous integral equation (327) has a solution u(z) and the generalsolution of Problem Q includes 2K + 2 arbitrary real constants If ε is an eigenvalueof rank q as stated in (328) applying the Fredholm theorem we obtain the solv-ability conditions for nonhomogeneous integral equation (327) there is a system ofq algebraic equations to determine the 2K + 2 arbitrary real constants setting thats is the rank of the corresponding coefficients matrix and s le min(q 2K +2) we candetermine s equalities in the q algebraic equations hence Problem Q for (22) hasq minus s solvability conditions When these conditions hold then the general solutionof Problem Q includes 2K + 2 + q minus s arbitrary real constants As for the case ofK lt 0 it can be similarly discussed Thus we can write the above result as in thefollowing theorem

Theorem 32 Suppose that the linear mixed equation (22) satisfies Condition C primeIf ε = εj (j = 1 2 ) where εj(j = 1 2 ) are the eigenvalues of the homogeneousintegral equation (329) Then

4 Frankl Boundary Value problem 177

(1) When K ge 0 Problem Q for (22) is solvable and the general solution u(z)of Problem Q for (22) includes 2K + 2 arbitrary real constants

(2) When K lt0 Problem Q for (22) has minus2Kminus1minuss solvability conditions sle1If ε is an eigenvalue of homogeneous integral equation (329) with the rank q

(3) When K ge 0 Problem Q for (22) has q minus s solvability conditions and s lemin (q 2K + 2)

(4) When K lt 0 Problem Q for (22) has minus2K minus 1 + q minus s solvability conditionsand s le min (minus2K minus 1 + q 1 + q)

Moreover we can derive the solvability result of Problem P for equation (22) withthe boundary condition (32) in which h(z) = 0

4 The Frankl Boundary Value Problem for Second OrderLinear Equations of Mixed Type

This section deals with the Frankl boundary value problem for linear second orderequations of mixed (elliptic-hyperbolic) type ie for generalized Lavrentprimeev-Bitsadzeequations We first give representation formula and prove uniqueness of solutions forthe above boundary value problem moreover we obtain a priori estimates of solutionsfinally by the method of parameter extension the existence of solutions is proved Inthe books [12]1)3) the Frankl problem was discussed for the special mixed equationsof second order uxx+sgny uyy = 0 In the book [73] the Frankl problem was discussedfor the mixed equation with parabolic degeneracy sgny|y|muxx + uyy = 0 which is amathematical model of problem of gas dynamics There the existence of solutions ofFrankl problem was proved by using the method of integral equations In this sectionwe will not use this method We are proving the solvability of the Frankl problemfor generalized linear Lavrentprimeev-Bitsadze equations generalizing the correspondingresult from [12]1)3)

41 Formulation of the Frankl problem for second order equations ofmixed type

Let D be a simply connected bounded domain in the complex plane CI with theboundary partD = Γ cup AprimeA cup AprimeC cup CB where Γ(sub x gt 0 y gt 0) isin C2

micro(0 lt micro lt 1)with the end points A = i and B = a AprimeA = x = 0 minus1 le y le 1 AprimeC = x minus y =1 x gt 0 y lt 0 is the characteristic line and CB = 1 le x le a y = 0 and denoteD+ = D cap y gt 0 Dminus = D cap y lt 0 Without loss of generality we may assumethat Γ = x2a2 + y2 = 1 x gt 0 y gt 0 otherwise through a conformal mappingfrom D+ onto the domain Dprime+ = x2a2 + y2 lt 1 x gt 0 y gt 0 such that threeboundary points i 0 1 are not changed then the above requirement can be realized

Frankl Problem Find a continuously differentiable solution u(z) of equation(22) in Dlowast = D1 a i minusi x + y = 0 which is continuous in D and satisfies theboundary conditions

u = ψ1(s) on Γ (41)

u = ψ2(x) on CB (42)

partu

partx= 0 on AprimeA (43)

u(iy)minus u(minusiy) = φ(y) minus1 le y le 1 (44)

Here ψ1(s) ψ2(x) φ(y) are given real-valuedfunctions satisfying the conditions

C1α[ψ1(s)S]lek2 C1

α[ψ2(x)CB]lek2

C1α[φ(y)A

primeA]lek2 ψ1(0)=ψ2(a)(45)

in which S = 0 le s le l s is the arc length parameter on Γ normalized suchthat s = 0 at the point B l is the length of Γ and α (0 lt α lt 1) k2 are non-negative constants The above boundary value problem is called Problem F and thecorresponding homogeneous problem is called Problem F0

LetU =

12ux V = minus1

2uy W = U + iV in D (46)

then equation (22) can be written as the complex equationWz

W zlowast

= Re [A1W ] + A2u+ A3 in

⎧⎨⎩D+

Dminus

⎫⎬⎭

u(z) = 2Reint z

aW (z)dz + ψ1(0)

(47)

If A1 = A2 = A3 = 0 in D then it is clear that

U(x y) =12ux =

12[f(x+ y) + g(x minus y)]

minusV (x y) =12uy =

12[f(x+ y)minus g(x minus y)]

(48)

in Dminus From the boundary conditions (41)ndash(43) it follows that

U(0 y) =12

partu

partx= 0 minusV (0 y) =

12[u(0 y)]y =

12[u(0 minusy)]y +

12φprime(y)

= V (0 minusy) +12φprime(y) = minusF (y) +

12φprime(y)

F (y) = minusV (0 minusy) minus1 le y le 0

(49)

and then

U(0 y) =12[f(y) + g(minusy)] = 0 minus1 le y le 0

minusV (0 y) =12[f(y)minus g(minusy)] = minusF (y) +

12φprime(y) minus1 le y le 0

U(0 y) + V (0 y) = g(minusy) = F (y)minus 12φprime(y) minus1 le y le 0

U(0 y)minus V (0 y) = f(y) = minusF (y) +12φprime(y) minus1 le y le 0

f(y) = minusg(minusy) f(y) = g(minusy)minus 2F (y) + φprime(y) minus1 le y le 0 ie

f(y minus x) = minusg(x minus y) f(y minus x) = g(x minus y)minus 2F (y minus x) + φprime(y minus x)

U(x y) + V (x y) = g(x minus y) = F (y minus x)minus 12φprime(y minus x) 0 le x minus y le 1

U(x y)minus V (x y) = f(x+ y) = minusg(minusx minus y)

= minusF (x+ y) +12φprime(x+ y) 0 le minusx minus y le 1

(410)

Hence

U(x y) =12[f(x+ y)minus f(y minus x)] 0 le x minus y le 1

minusV (x y)=12[f(x+y)minusf(yminusx)]minusF (y minus x)+

12φprime(yminusx) 0lexminusyle1

U(x 0) + V (x 0) = g(x) = F (minusx)minus 12φprime(minusx) 0 le x le 1

(411)

In particular we have

U(x 0) =12[f(x)minus f(minusx)] =

12[f(x) + F (minusx)minus 1

2φprime(minusx)]

minusV (x 0) =12[f(x)minus f(minusx)]minus F (minusx) +

12φprime(minusx)

=12[f(x)minus F (minusx) +

12φprime(minusx)] on OC

(412)

The boundary conditions of the Frankl problem are

partu

partl=2Re [λ(z)W (z)] = r(z) z isin Γ cup CB u(a) = b0 = ψ1(0)

U(0 y) =12

partu

partx= r(0 y) = Re [λ(iy)W (iy)] = 0 minus1 le y le 1

(413)

Re [λ(x)W (x)] = r(x)=1radic2[F (minusx)minus 1

2φprime(minusx)] xisinL0=(0 1) (414)

in which l is the tangent vector on the boundary Γ and

λ(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

cos(l x)minus i cos(l y)

1

1 + iradic2

1

r(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

ψprime1(s) on Γ = BA

0 on AO

1radic2[F (minusx)minus 1

ψprime2(x) on CB

We shall prove the solvability of the Frankl problem for equation (22) by using themethods of parameter extension and symmetry extension

We can choose the index K = minus12 of λ(z) on the boundary partD+ of D+ In factdue to the boundary condition

Re [λ(z)W (z)] =12Re [λ(z)(ux minus iuy)] = r(z) on partD+ = AO cup OB cup BA (415)

and λ(z) = 1 on AO cup CB λ(z) = exp(iπ4) on OC λ(z) = cos(l x) minus i cos(l y)on Γ denote t1 = 0 t2 = 1 t3 = a t4 = i it is seen λ(a + 0) = exp(i3π2) andλ(i minus 0) = exp(iπ) we have

Kj=[φj

π

λ(tjminus0)λ(tj+0)

γj=φj

πminusKj j=1 4

eiφ1=λ(t1minus0)λ(t1+0)

=ei0

eiπ4 = eminusiπ4 0ltγ1=φ1

πminusK1=minus1

4minusK1 lt

34

lt1

=eiπ4

ei0 = eiπ4 minus1ltγ2=φ2

πminusK2=

14minusK2=minus3

4lt0

eiφ3 =λ(t3 minus 0)λ(t3 + 0)

=ei0

ei3π2 0 le γ3 =φ3

πminus K3 = minus3

2minus K3 =

12

lt 1

=eiπ

ei0 = eiπ 0 le γ4 =φ4

πminus K4 = 1minus K4 = 0 lt 1

(416)

here [b] is the largest integer not exceeding the real number b we choose K1 = minus1K3 = minus2 K2 = K4 = 1 Under these conditions the index K of λ(z) on theboundary partD+ of D+ is just as follows

K =12(K1 +K2 +K3 +K4) = minus1

2 (417)

Noting that U(0 y) = 0 on AprimeA we can extend W (z) onto the reflected domainD of D about the segment AprimeA In fact we introduce the function

W (z) =

⎧⎨⎩W (z) in D

minusW (minusz) on D(418)

this function W (z) is a solution of the equation

⎧⎨⎩ Wz

W zlowast

⎫⎬⎭ = Re [A1W ] + A2u+ A3 in

⎧⎨⎩D

D

⎫⎬⎭u(z) = 2Re

int z

1W (z)dz + ψ1(0)

(419)

2Re[λ(z)W (z)]=r(z) zisinΓcupCBcupΓcupBC u(a)=b0=ψ1(0)=u(minusa)

Re[λ(x)W (x)]=r(x) xisinL2=(01)cup(minus10)(420)

in which

A1 =

⎧⎨⎩A1(z)

minusA1(minusz)A2 =

⎧⎨⎩A2(z)

A2(minusz)A3 =

⎧⎨⎩A3(z) in D

A3(minusz) in D+ cup Dminus(421)

and

λ(z) =

⎧⎨⎩λ(z)

λ(minusz)r(z) =

⎧⎨⎩ r(z) Γ cup CB

minusr(minusz) Γ cup BC

λ(z) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩1 + iradic2

1minus iradic2

r(z) =

⎧⎨⎩r(z) on OC = (0 1)

minusr(minusz) on CO = (minus1 0)

(422)

herein Γ BC = (minusa minus1) CO and AB are the reflected curves of Γ CB OC and BAabout the imaginary axis respectively We choose the index of the function λ(z) onthe boundary part(D+cupD+cupAO) of the elliptic domain D+cupD+cupAO asK = minus12 Infact noting that λ(z) = 1 on CBcupBC λ(z) = exp(iπ4) on OC λ(z) = exp(minusiπ4)on CO we denote t1 = 0 t2 = 1 t3 = a t4 = i t5 = minusa t6 = minus1 it is seenλ(a+ 0) = exp(i3π2) λ(i minus 0) = λ(i+ 0) = exp(iπ) λ(minusa minus 0) = exp(iπ2) hencewe have

Kj=[φj

π

]+Jj Jj=0 or 1 ejφj =

γj=φj

πminusKj j=1 6

=eminusiπ4

eiπ4 =eminusiπ2 0ltγ1=φ1

πminusK1=minus1

2minusK1=

12

lt1

=eiπ4

ei0 =eiπ4 minus1ltγ2=φ2

πminusK2=

14minusK2=minus3

4lt0

=ei0

ei3π2 =eminusi3π2 0ltγ3=φ3

πminusK3=minus3

2minusK3=

12

lt1

=eiπ

eiπ= ei0 0 le γ4 =

φ4

eiφ5=λ(t5 minus 0)λ(t5 + 0)

=eiπ2

ei0 =eiπ2 0 lt γ5=φ5

πminus K5=

12

minus K5 =12

lt 1

=ei0

eminusiπ4 =eiπ4 minus1ltγ6=φ6

πminusK6=

14minusK6=minus3

4lt0

(423)

If we choose K1 = minus1 K2 = K6 = 1 K3 = minus2 K4 = K5 = 0 the index K of λ(z) isjust

K =12(K1 +K2 + middot middot middot+K6) = minus1

2 (424)

We can discuss the solvability of the corresponding boundary value problem(419) (420) and then derive the existence of solutions of the Frankl problemfor equation (22)

42 Representation and a priori estimates of solutions to the Franklproblem for (22)

First of all similarly to Lemma 21 we can prove the following theorem

Theorem 41 Let equation (22) satisfy Condition C in D+ and u(z) be a continu-ous solution of (22) in D+

lowast = D+0 1 a i Then u(z) can be expressed as

u(z) = U(z)Ψ(z) + ψ(z) in D+

U(z) = 2Reint z

aw(z)dz + b0 w(z) = Φ(z)eφ(z) in D+

(425)

Here ψ(z) Ψ(z) are the solutions of equation (22) in D+ and

ψ(z) = 0 Ψ(z) = 1 on Γ cup L

partψ(z)partx

= 0partΨ(z)

partx= 0 on AO

(427)

where L = (0 a) They satisfy the estimates

C1γ [X(z)ψ(z) D+] le M19 X(z)ψ(z) W 2

p0(D+)le M19 (428)

C1γ [X(z)Ψ(z) D+]leM20X(z)Ψ(z)W 2

p0(D+)leM20Ψ(z)geM21 gt0 zisinD+ (429)

in which X(z) = |x + y minus t1|η1prod4

j=2 |z minus tj|ηj ηj = maxminusγj + δ δ j = 1 2 3 4herein tj γj(j = 1 2 3 4) are as stated in (416) δ γ(γ lt δ) are small positiveconstants p0 (2 lt p0 le p) Mj = Mj (p0 γ k D) (j = 19 20 21) are non-negativeconstants k = (k0 k1 k2) Moreover U(z) is a solution of the equation

where Im [φ(z)] = 0 z isin partD+ Re [φ(0)] = 0 and φ(z) satisfies the estimate

Cβ[φ(z) D+] + Lp0 [φz D+] le M22 (431)

in which β (0 lt β le α) M22 = M22 (p0 α k0 D+) are two non-negative constants

Φ(z) is analytic in D+ If u(z) is a solution of Problem F then W (z) = uz satisfiesthe boundary conditions

Re [λ(z)W (z)] = r(z) on Γ cup AO u(a) = b0 = ψ1(0) (432)

Re [λ(x)W (x)] = r(x) λ(x) =

⎧⎪⎨⎪⎩1 + iradic2on L0 = (0 1)

1 on L1 = (1 a)(433)

Theorem 42 Suppose that equation (22) satisfies Condition C Then any solutionof the Frankl problem for (22) can be expressed as

u(z) = 2Reint z

aW (z)dz + b0 b0 = ψ1(0) (434)

Here W (z) is a solution of the equation

Wz

Wzlowast

= Re [A1W ] + A2u+ A3 in

D+

Dminus

(435)

satisfying the boundary conditions (413) minus (414) (W (z) = uz) and W (z) possessesthe form

W (z) = Φ(z)eφ(z) + ψ(z)

int intD+

g(ζ)ζ minus z

W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν

1g1(z)dνe1 +

int micro

(436)

in which e1 = (1 + i)2 e2 = (1 minus i)2 micro = x + y ν = x minus y φ0(z) is an analyticfunction in D+ such that Im φ(x) = 0 on Γ cup AO cup L and

g(z)=A12+A1W(2W ) W (z) = 00 W (z)=0

in D+

f(z) = Re [A1uz] + A2u+ A3 f(z) = Re [A1φz] + A2u+ A3 in D+

g1(z)=g2(z)=Aξ+Bη+Cu+D ξ=ReW+ImW η=ReW minusImW

A=ReA1+ImA1

2 B=

ReA1minusImA1

2 C=A2 D= A3 in Dminus

(437)

where Φ(z) in D+ and Φ(z) in Dminus are solutions of the equationWz

W zlowast

= 0 in

D+

Dminus

(438)

Re [eφ(z)Φ(z)] = minusRe [ψ(z)] z = iy isin AO

Re [λ(x)Φ(x)eφ(x)] = r(x)minus Re [λ(x)ψ(x)] x isin L0 = (0 1)

Re [λ(x)(Φ(x) + Ψ(x))] = r(x) = Re [λ(x)W (x)] x isin L0 = (0 1)

Re [Φ(x)] = minusRe [Ψ(x)] x isin OAprime u(a) = b0 = ψ1(0)

(439)

where λ(x) on L = (0 a) is as stated in (415)

Proof Let u(z) be a solution of the Frankl problem for equation (22) andW (z) = uz u(z) be substituted in the positions of w u in (437) Thus the functionsg(z) f(z) g1(z) g2(z) and ψ(z) φ(z) in D+ and Ψ(z) in Dminus in (436)(437) aredetermined Moreover we can find the solution Φ(z) in D+ and Φ(z) in Dminus of (438)with the boundary condition (439) where r(z) as stated in (415) namely

r(z) = H(F φ) z isin Γ cup AO cup L (440)

thus

W (z) =

⎧⎪⎨⎪⎩Φ(z)eφ(z) + ψ(z) in D+

Φ(z) + Ψ(z) in Dminus(441)

is the solution of Problem A for the complex equation (435) with the boundaryconditions (413)(414) which can be expressed as in (436) and u(z) is a solutionof the Frankl problem for (22) as stated in (434)

Next we discuss the uniqueness of solutions of the Frankl problem for (22)

Theorem 43 Suppose that the mixed equation (22) satisfies Condition C Thenthe Frankl problem for (22) has at most one solution u(z) isin C(D) cap C1(D)

Proof We consider equation (22) in D+ As stated before if u1(z) u2(z) are twosolutions of the Frankl problem for (22) then u(z) = u1(z) minus u2(z) is a solution ofthe homogeneous equation⎧⎨⎩uzz

uzzlowast

⎫⎬⎭=Re [A1uz]+A2u in

⎧⎨⎩D+

Dminus

⎫⎬⎭

u(z) = 2Reint z

aW (z)dz W (z) = uz in D

u(z)=U(z)Ψ(z) Uz=Φ(z)eφ(z) in D+

W (z) = Φ0(z) + Φ(z) + Ψ(z) in Dminus

(442)

in which

Ψ(z) =int ν

1g1(z)dνe1 +

int micro

0g2(z)dmicroe2 g1(z) = g2(z) = Aξ +Bη + Cu in Dminus

and Ψ(z) Φ(z) φ(z)Φ(z) are similar to those in Theorem 42 and Φ(z)Φ(z) Φ0(z)are solutions of equation (438) in D+ and Dminus respectively satisfying the conditions

2Re [λ(z)Φ(z)]=r(z)=0 on ΓcupAOcupCB U(a)=0

Re [λ(x)Φ(x)] =12[f(x) + F (minusx)] on L0 = (0 1)

Φ(x) = Φ(x) Φ0(x) = uz(x)minusΨ(x)minus Uz(x)eminusφ(x) on L0

Re [Φ0(z)]=Re [Ψ(z)] Im [Φ0(z)]=Im [uzminusΨ(z)minusUz(z)eminusφ(z)] on AprimeO

(443)

According to Theorem 41 the solution U(z) = 2Reint za Φ(z)dz of equation (12)

U(z) = S(z) = 0 on Γ cup CB

U(x) = 2int x

aΦ(x)dx = 2

int x

aUz(x)eminusφ(x)dx =

int x

0Φ(x)dx minus

int a

0Φ(x)dx

=S(x)=int x

0[f(x)+F (minusx)]dx=g(x)+

12U(ix) on (0 1)

(444)

where g(x) =int x0 f(x)dx

int x0 F (minusx)dx = U(ix)2 Besides the harmonic function U(z)

in D+ satisfies the boundary condition

partU(z)partx

= 0 on AO (445)

Moreover there exists a conjugate harmonic function V (z) in D+ such that V (0) = 0From the above last formula we can derive that V (iy) =

int y0 Vydy =

int y0 Uxdy = 0 on

AO By the Cauchy theorem we haveintpartD+

[U(z) + iV (z)]2dz = 0

= minusint a

1[V (x)]2dx minus

intΓ[V (z)]2

(dx

ds+ i

dy

ds

)ds (446)

+iint 0

1[U(iy)]2dy +

int 1

0[U2(x)minus V 2(x) + 2i U(x)V (x)]dx

Due to the continuity of Uy on (01) V (x) =int x0 Vxdx = minus int x

0 Uydx = 2int x0 V (x)dx =

minus int x0 [f(x) minus F (minusx)]dx = minusg(x) + U(ix)2 is obtained From the imaginary part in

(446) and the above formula it is clear thatintΓ[V (z)]2

party

partsds+

int 1

0[U(iy)]2dy + 2

int 1

0[g(x)]2 minus 1

4[U(ix)]2dx = 0 (447)

Hence we getU(iy) = 0 on AO g(x) = 0 on OC

and then f(x) = gprime(x) = 0 F (minusx) = [U(ix)]x2 = 0 on OC Due to the functionr(z) = S(z) = 0 on partD+ in (443)(444) and the index K = minus12 hence Φ(z) = 0 inD+ and then the solution u(z) of the homogeneous Frankl problem for (442) in D+

satisfies u(z) = u1(z)minus u2(z) = 0 Moreover we can derive u(z) = u1(z)minus u2(z) = 0in Dminus This proves the uniqueness of solutions for the Frankl problem for (22) in D

Finally we give an a priori estimate of solutions to the Frankl problem for equation(22) From the estimate we can see the singular behavior of uz at the discontinuityset Z = 1 a i minusi x + y = 0 It becomes infinity of an order not exceeding 34 atz = 1 infinite of order not exceeding a small positive number δ at the points i minusiand uz is bounded at the point set a x+y = 0 In fact we can prove that z = i minusiare removable singular points In [12]3) the author pointed out that uz can becomeinfinity of an order less than 1

Theorem 44 Suppose that equation (22) satisfies Condition C in D and the func-tion r(z) in (414) is H(F φ) especially

r(x) = H(F φ) =1radic2[F (minusx)minus 1

2φprime(minusx)] x isin L0 = (0 1) (448)

Then any solution u(z) of the Frankl problem for equation (22) in D+ satisfies theestimate

C1γ [u D+] = Cγ[u(z) D+] + C[uzX(z) D+] le M23(k1 + k2) (449)

where X(z) is as stated in (429) ie

X(z) = |x+ y minus t1|η1

4prodj=2

|z minus tj|ηj ηj = maxminusγj + δ δ j = 1 2 3 4 (450)

and M23 = M23(p0 γ δ k0 D) is a non-negative constant

Proof On the basis of the uniqueness of solutions of the Frankl problem for (22)in Theorem 43 and the results in [12]3) by using reductio ad absurdum we canderive the estimate (449) In fact from (483)(484) in the proof of Theorem 46below we see that the function [Wn+1 un+1] (Wn+1(z) = Wn+1(z)minusWn(z) un+1(z) =un+1(z)minus un(z)) is a solution of the boundary value problem

[Wn+1]z minus t0G(z un+1 Wn+1) = (t minus t0)G(z un Wn) z isin D+

Re [λ(z)Wn+1(z)] = 0 z isin Γ cup AO cup CB

Re [λ(z)Wn+1(z)]minus t0H(Fn+1 0) = (t minus t0)H(Fn 0)] on L0 = (0 1)

un+1(z) =int z

aWn+1(z)dz z isin D+

(451)

where G(z u W ) = Re [A1W ] + A2u + A3 G(z un wn) = G(z un+1 wn+1) minus G(zun wn) Fn+1 = Fn+1 minus Fn On the basis of Theorem 41 the solution Wn+1 of theboundary value problem (451) can be expressed as

un+1(z) = Un+1(z)Ψn+1(z) + ψn+1(z) in D+

Un+1(z)=2Reint z

1Φn+1(z)dz Un+1z=Φn+1(z)eφn+1(z) in D+

Wn+1(z) = Φ0n+1(z) + Φn+1(z) + Ψn+1(z) in Dminus

(452)

where Φn+1(z)Φ0n+1(z) are the solutions of equation (438) in Dminus Ψn+1(z) is a

solution of the equation in (451) in Dminus ψn+1(z) Ψn+1(z) are the solutions ofthe equation in (451) and its homogeneous complex equation in D+ satisfying theboundary conditions

ψn+1(z) = 1 Ψn+1(z) = 1 on Γ cup L

partψn+1(z)partx

= 0partΨn+1(z)

partx= 0 on AO

(453)

According to the proof of Theorem 43 we see that the function Un+1(z) satisfies theboundary conditions

Un+1(z) = S(z) = 0 on Γ cup CB Un+1(a) = 0

Un+1(x)=Reint x

aΦn+1xdx=S(x)

S(x)=int x

0[f(x)+Fn+1(minusx)]dx minus 1

2

int x

0φprime

n+1(minusx)dx

= g(x) +12Un+1(ix) +

12φn+1(minusx) on (0 1)

(454)

where

g(x)=int x

0f(x)dx

int x

0Fn+1(minusx)dx=

Un+1(ix)2

Un+1(minusx)=minusint x

0φprime

n+1(minusx)dx

Besides we can see that the harmonic function Un+1(z) in D+ satisfies the boundarycondition

partUn+1(z)partx

= 0 on AO (455)

Moreover there exists a conjugate harmonic function Vn+1(z) in D+ such thatVn+1(0) = 0 We shall verify that

limnrarrinfinmax

D+

|X(x)Un+1x| = 0 limnrarrinfin

int 1

0[Un+1(iy)]2dy = 0 (456)

Suppose that limnrarrinfinint 10 |Un+1(iy)|dy = C gt 0 due to

intΓ[Vn+1(z)]2

party

partsds+

int 1

0[Un+1(iy)]2dy

+2int 1

0

[g(x)]2 minus 1

4[t0Un+1(ix) + (t minus t0)Un(ix)]2

dx = 0

(457)

0 |Un+1(iy)|2dy2 for n = nk rarr infin then similarly to (447) from (457) we canderive that

Un+1(iy) = 0 on AO g(x) = 0 on OC (458)

This contradiction proves thatint 10 [Un+1(ix)]2dx = 0

int 10 [Un+1(ix)]2dx = 0 and Un+1 =

Un+1(z) minus Un(z) = 0 in D+ for n ge N0 where N0 is a sufficiently large positivenumber Hence un+1 = un+1(z) minus un(z) = ψn+1(z) minus ψn(z) in D+ for n ge N0Similarly to the proof of the first estimate in (428) we can obtain

C1γ [X(z)un+1(z) D+] le M24|t minus t0|C1

γ [X(z)un(z) D+] (459)

in whichM24 = M24(p γ δ k0 D+) is a non-negative constant Choosing the constant

ε so small that εM24 le 12 and |t minus t0| le ε it follows that

C1γ [un+1 D+] le εM24|t minus t0|C1

γ [un D+] le 12C1

γ [un D+]

C1γ [un+1 D+]le2minusn+N0

infinsumj=N0

2minusjC1γ [u1 minus u0 D+]le2minusn+N0+1C1

γ [u1minusu0 D+]

for n gt N0 Therefore there exists a continuous function ulowast(z) on D+ such that

ulowast(z) =infinsum

j=0un+1 =

infinsumj=0[un+1 minus un(z)]

From the estimate ofsumn

j=0[uj+1(z)minus uj(z)] = un+1(z)minus u0(z) in D+ the estimate

C1γ [un+1 D] = Cγ[un+1 D+] + C[un+1zXD+] le M25 (460)

can be derived where M25 = M25(p0 γ δ k0 D+) is a non-negative constant More-

over we can derive a similar estimate of ulowast(z) in D+ and Dminus which gives the estimate(449)

43 The solvability of the Frankl problem for (22)

Theorem 45 Suppose that the mixed equation (22) satisfies Condition C andA1(z) = A2(z) = 0 in D ie⎧⎨⎩uzz = A3(z) z isin D+

uzzlowast = A3(z) z isin Dminus(461)

Then the Frankl problem for (461) has a solution in D

Proof It is clear that the Frankl problem for (461) is equivalent to the followingProblem A for the complex equation of first order and boundary conditions

Wz

W zlowast

= A3(z) in

⎧⎨⎩D+

Dminus

⎫⎬⎭ (462)

Re [λ(z)W (z)] = r(z) = H(F φ) z isin Γ cup AO cup L

Re [λ(z)W (z)] = r(z) = 0 λ(z) = 1 z isin OAprime(463)

int z

aW (z)dz + b0 in D (464)

in which λ(z) r(z) are as stated in (415) and (440)

In order to find a solution W (z) of Problem A in D we can express W (z) in theform (434)ndash(437) In the following by using the method of parameter extensionwe shall find a solution of Problem A for the complex equation (462) We considerequation (462) and the boundary conditions with the parameter t isin [0 1]

Re [λ(z)W (z)] = tH(F φ) +R(z) on partD+ = Γ cup AO cup L (465)

in which H(F φ) on partD+ = Γ cup AO cup L is as stated in (448) and R(z)X(z) isinCγ(partD+) this problem is called Problem Ft

When t = 0 the unique solution of Problem F0 for the complex equation (461)can be found by a method given in Section 1 and its solution [W0(z) u0(z)] can be

expressed as

u0(z) = 2Reint z

aW0(z)dz + b0 W0(z) = W (z) in D b0 = ψ1(0)

W (z) = Φ(z) + ψ(z) ψ(z) = TA3 = minus 1π

int intD+

A3(ζ)ζminusz

dσζ in D+

W (z)=Φ(z)+Ψ(z) Ψ(z)=int ν

1A3(z)e1dν+

int micro

0A3(z)e2dmicro in Dminus

(466)

where Φ(z) is an analytic function in D+ and Φ(z) is a solution of (438) in Dminussatisfying the boundary conditions

Re [λ(z)W (z)] = R(z) z isin Γ cup L

Re [λ(z)(Φ(z) + Ψ(z))] = R(z) λ(z) = 1 z = iy isin AO

Re [λ(z)(Φ(z) + Ψ(z))] = R(z) λ(z) = 1 z = iy isin OAprime

Re [λ(x)(Φ(x)+Ψ(x))]=R(x)=Re [λ(x)W (x)] z=x isin OC u(a)=b0

(467)

Suppose that when t = t0 (0 le t0 lt 1) Problem Ft0 is solvable ie Problem Ft0

for (462) has a solution [W0(z) u0(z)] (u0(z) isin C1γ(D)) We can find a neighborhood

Tε = |t minus t0| le ε 0 le t le 1(0 lt ε lt 1) of t0 such that for every t isin Tε Problem Ft

is solvable In fact Problem Ft can be written in the formWz

Wzlowast

= A3(z) in

⎧⎨⎩D+

Dminus

⎫⎬⎭

Re [λ(z)W (z)]minus t0H(F φ) = (t minus t0)H(F φ) +R(z) on partD+

(468)

ReplacingW (z) u(z) in the right-hand sides of (468) by a functionW0(z) isin Cγ(partD+)and the corresponding function u0(z) in (466) ie u0(z) isin C1

γ(D) especially by thesolution [W0(z) u0(z)] of Problem F0 it is obvious that the boundary value problemfor such an equation in (468) then has a solution [W1(z) u1(z)] u1(z) isin C1

γ(partD)Using successive iteration we obtain a sequence of solutions [Wn(z) un(z)] un(z) isinC1

γ(D) n = 1 2 which satisfy the equations and boundary conditionsWn+1z

Wn+1zlowast

= A3(z) in

⎧⎨⎩D+

Dminus

⎫⎬⎭

Re [λ(z)Wn+1(z)]minus t0H(Fn+1 φ) = (t minus t0)H(Fn φ) +R(z) on partD+

Re [λ(z)Wn+1(z)] = 0 z isin OAprime

(469)

[Wn+1 minus Wn]z = 0 z isin D

Re [λ(z)(Wn+1(z)minusWn(z))]minust0[H(Fn+1minusFn 0)]=(tminust0)[H(FnminusFnminus10)](470)

Noting that

|t minus t0|Cγ[XH(Φn minus Φnminus1 0) L0] le |t minus t0|Cγ[X(Φn minus Φnminus1) L0] (471)

and applying Theorem 44 we have

C1γ [un+1 minus un D+] le M26C

1γ [Φn minus Φnminus1 D+] (472)

where M26 = M26(p0 γ δ k0 D+) Choosing the constant ε so small that εM24 le 12

and |t minus t0| le ε it follows that

C1γ [un+1 minus un D+] le εM26C

1γ [un minus unminus1 D+] le 1

2C1

γ [un minus unminus1 D+] (473)

C1γ [un+1 minus un D+] le 2minusN0

γ [u1 minus u0 D+] le 2minusN0+1C1γ [u1 minus u0 D+] (474)

γ(D+) there exists a function ulowast(z) isin C1γ(D+) and Wlowast(z) = ulowastz(z) so that

C1γ [un minus ulowast D+] = Cγ[un minus ulowast D+] + C[X(Wn minus Wlowast) D+] rarr 0 as n rarr infin We cansee that [Wlowast(z) ulowast(z)] is a solution of Problem Ft for every t isin Tε = |t minus t0| leε Because the constant ε is independent of t0 (0 le t0 lt 1) therefore from thesolvability of Problem F0 when t0 = 0 we can derive the solvability of Problem Ft

when t = ε 2ε [1ε]ε 1 In particular when t = 1 and R(z) = 0 Problem F1 iethe Frankl problem for (461) in D+ is solvable

As for the solution [W (z) u(z)] in Dminus it can be obtained by (410)(411) and themethod in Chapters I and II namely

u(z) = 2Reint z

aW (z)dz + b0 on Dminus b0 = ψ1(0)

W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν

1A3(z)e1dν +

int micro

0A3(z)e2dmicro

Φ(z) =12[(1 + i)f(x+ y) + (1minus i)g(x minus y)]

f(x+ y) = Re [(1minus i)(W (x+ y)minusΨ(x+ y))]

g(x minus y) = Re [(1 + i)(W (x minus y)minusΨ(x minus y))] z isin Dminus cap x+ y ge 0

(475)

where W (x+ y) W (x minus y) are the values on 0 le z = x+ y le 1 0 le x minus y le 1 of thesolution W (z) of Problem F for (461) in D+ and Ψ(x+ y) Ψ(x minus y) are the valueson 0 le z = x + y le 1 0 le x minus y le 1 of Ψ(z) respectively Moreover the functionW (z) in Dminus cap x+ y le 0 can be obtained by (475)(418) In fact from (475) wehave found the function W (z) on OC prime = x + y = 0 0 le x le 12 by (418) weobtain the function W (z) = minusW (minusz) on OC primeprime = x minus y = 0 minus12 le x le 0 and

denote σ(x) = Re [(1 minus i)Ψ(z)] on OC prime τ(x) = Re [(1 + i)Ψ(z)] on OC primeprime Hence thesolution u(z) in Dminus cap x+ y le 0 is as follows

u(z) = 2Reint z

0W (z)dz + u(0) Ψ(z) =

int ν

0A3(z)e1dν +

int micro

0A3(z)e2dmicro

W (z) =12[(1minus i)f(x+ y) + (1 + i)g(x minus y)] + Ψ(z)

f(x+ y) = τ((x+ y)2) + ReW (0) + ImW (0)

g(x minus y) = σ((x minus y)2) + ReW (0)minus ImW (0)z isin Dminus cap x+ y le 0

(476)

in which Φ(z) and Ψ(z) are the functions from (475) Furthermore we can provethat the solution u(z) satisfies the boundary conditions (41)ndash(44) This completesthe proof

Theorem 46 Suppose that the mixed equation (22) satisfies Condition C Thenthe Frankl problem for (22) has a solution in D

Proof Similarly to the proof of Theorem 45 we see that the Frankl problem for(22) is equivalent to Problem A for first order complex equation and boundary con-ditions

Wz

Wzlowast

= G G = G(z u W ) = Re [A1W ] + A2u+ A3 in

⎧⎨⎩D+

Dminus

⎫⎬⎭ (477)

Re [λ(z)W (z)] = r(z) = H(F φ) z isin Γ cup AO cup L (478)

and the relation (464) in which r(z) = H(F φ) on z isin partD+ = Γ cup AO cup L is asstated in (463)

In order to find a solution W (z) of Problem A in D we can express W (z) in theform (434)ndash(437) In the following by using the method of parameter extension asolution of Problem A for the complex equation (477) will be found We considerthe equation and boundary conditions with the parameter t isin [0 1]

Wz = tG+K(z) G = G(z u W ) = Re [A1W ] + A2u+ A3 in D+ (479)

where K(z) isin Lp(D+) and R(z)X(z) isin Cγ(partD+) This problem is called ProblemFt

When t = 0 the complex equation (479) becomes the equation

Wz = K(z) in D+ (481)

From Theorem 45 we can find the unique solution of Problem F0 for (479) Supposethat when t = t0 (0 le t0 lt 1) Problem Ft0 is solvable ie Problem Ft0 for (479) hasa solution [W0(z) u0(z)] (u0 isin C1

γ(D)) We can find a neighborhood Tε = |t minus t0| le

ε 0 le t le 1(0 lt ε lt 1) of t0 such that for every t isin Tε Problem Ft is solvable Infact Problem Ft can be written in the form

Wz minus t0G(z u W ) = (t minus t0)G(z u W ) +K(z) in D+

Re [λ(z)W (z)]minus t0H(F φ) = (t minus t0)H(F φ) +R(z) on partD+(482)

γ(D) especially by thesolution [W0(z) u0(z)] of Problem F0 it is obvious that the boundary value problemfor such equation in (482) then has a solution [W1(z) u1(z)] u(z) isin C1

γ(partD) Usingsuccessive iteration we obtain a sequence of solutions [Wn(z) un(z)] un(z) isin C1

γ(D)n = 1 2 which satisfy the equations and boundary conditions

Wn+1zminust0G(z un+1 Wn+1) = (t minus t0)G(z un Wn) +K(z) in D+ (483)

Re [λ(z)Wn+1(z)]minust0H(Fn+1 φ)=(t minus t0)H(Fn φ)+R(z) on partD+ (484)

[Wn+1 minus Wn]z minus t0[G(z un+1 Wn+1)minus G(z un Wn)]

= (t minus t0)[G(z un Wn)minus G(z unminus1 Wnminus1)] z isin D+

Re [λ(z)(Wn+1(z)minus Wn(z))]minus t0[H(Fn+1 minus Fn 0)]

= (t minus t0)[H(Fn minus Fnminus1 0)] z isin L0

(485)

Noting that

Lp[(tminust0)(G(z un Wn)minusG(z unminus1 Wnminus1)) D+] le 2k0|tminust0|C1γ [unminusunminus1 D+]

|t minus t0|Cγ[XH(Fn minus Fnminus1 0) L0] le |t minus t0|Cγ[X(Fn minus Fnminus1) L0](486)

and according to the method in the proof of Theorem 44 we can obtain

C1γ [un+1 minus un D+] le M27[2k0 + 1]C1

whereM27 = M27(p0 γ δ k0 D+) Choosing the constant ε so small that εM27

(2k0 + 1) le 12 and |t minus t0| le ε it follows that

C1γ [un+1 minus un D+] le εM27(2k0+1)C1

γ [un minus unminus1 D+] le 12C1

γ [u1 minus u0 D+] le 2minusN0+1C1γ [u1 minus u0 D+]

C1γ [un minus ulowast D+] = Cγ[un minus ulowast D+] + C[X(Wn minus Wlowast) D+] rarr 0 as n rarr infin We cansee that [Wlowast(z) ulowast(z)] is a solution of Problem Ft for every t isin Tε = |t minus t0| le εBecause the constant ε is independent of t0 (0 le t0 lt 1) therefore from the solvabilityof Problem Ft0 when t0 = 0 we can derive the solvability of Problem Ft whent = ε 2ε [1ε] ε 1 In particular when t = 1 and K(z) = 0 R(z) = 0 ProblemF1 ie the Frankl problem for (22) in D+ is solvable

The existence of the solution [W (z) u(z)] of Problem F for (22) in Dminus can beobtained by the method in Chapters I and II

5 Oblique Derivative Problems for Second Order DegenerateEquations of Mixed Type

In this section we discuss the oblique derivative problem for second order degenerateequations of mixed type in a simply connected domain We first give the represen-tation of solutions of the boundary value problem for the equations and then provethe uniqueness of solutions for the problem Moreover we introduce the possibilityto prove the existence of the above oblique derivative problem

51 Formulation of oblique derivative problems for degenerate equationsof mixed type

α(0 lt α lt 1) with the endpoints z = 0 2 and L = L1 cup L2 L1 = x +

int y0

radicminusK(t)dt = 0 x isin (0 1) L2 =

x minus int y0

radicminusK(t)dt = 2 x isin (1 2) and z1 = x1 + jy1 = 1 + jy1 is the intersection

point of L1 and L2 In this section we use the hyperbolic numbers Denote D+ =D cap y gt 0 Dminus = D cap y lt 0 We may assume that Γ = |z minus 1| = 1 y ge 0 andconsider the linear degenerate mixed equation of second order

Lu = K(y)uxx + uyy = dux + euy + fu+ g in D (51)

where K(y) possesses the first order continuous derivatives K prime(y) and K prime(y) gt 0 ony = 0 K(0) = 0 The following degenerate mixed equation is a special case

Lu = sgny|y|m uxx + uyy = dux + euy + fu+ g in D (52)

where m is a positive constant d e f g are functions of z(isin D) Similarly to (543)Chapter II we denote W (z)= UminusiV =ym2U+iV =[ym2uxminusiuy]2 Wmacrz=[ym2Wx+iWy]2 in D+ and W (z)= U+jV = |y|m2UminusjV =[|y|m2ux+juy]2 Wmacrz=[|y|m2WxminusjWy]2 in Dminus then equation (52) in D can be reduced to the form

5 Degenerate Mixed Equations 195

⎧⎨⎩ Wmacrz

Wmacrz

⎫⎬⎭ = A1(z)W + A2(z)W + A3(z)u+ A4(z) in

D+

Dminus

A1 =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩im

8y+

d

4ym2 +ie

4=

d

4ym2 + i

(m

8y+

e

4

)

jm

8|y| +minusd

4|y|m2 minus je

4=

minusd

4|y|m2 + j

(m

8|y|minuse

4

) in

⎧⎨⎩D+

Dminus

⎫⎬⎭

A2 =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩d

4ym2+ i

(m

8yminus e

4

)

minusd

4|y|m2+j

(m

8|y|+e

4

)

A3=

⎧⎪⎪⎪⎨⎪⎪⎪⎩f

4

minusf

4

A4=

⎧⎪⎪⎨⎪⎪⎩g

4

minusg

4

in

⎧⎨⎩D+

Dminus

⎫⎬⎭

(53)

and

u(z)=

⎧⎪⎪⎨⎪⎪⎩2Re

int z

0uzdz+u(0) in D+

2Reint z

0(UminusjV )d(x+jy)+u(0) in Dminus

is a solution of equation (52)

The coefficients Aj(z) (j = 1 2 3) in (52) are continuous in D+ and continuousin Dminus and satisfy

C[Aj D+] le k0 j = 1 2 C[A3 D+] le k1 A2 ge 0 in D+

where p (gt 2) k0 k1 are non-negative constants If the above conditions is replaced by

C1α[Aj Dplusmn] le k0 j = 1 2 C1

α[A3 Dplusmn] le k1 (55)

in which α (0 lt α lt 1) is a real constant then the conditions will be called ConditionC prime

Problem P Find a continuously differentiable solution u(z) of (52) in Dlowast =D0 L2 which is continuous in D and satisfies the boundary conditions

lu=partu

partl=2Re [λ(z)uz]=r(z) zisinΓ u(0)=b0 u(2)=b2 (56)

Re [λ(z)umacrz] = r(z) z isin L1 Im [λ(z)umacrz]|z=z1 = b1 (57)

where uz = [radicminusKux + iuy]2 λ(z) = a(x) + ib(x) = cos(l x) minus i cos(l y) if z isin Γ

and λ(z) = a(z) + jb(z) if z isin L1 b0 b1 b2 are real constants and λ(z)(|λ(z)| =1) r(z) b0 b1 b2 satisfy the conditions

Cα[λ(z)Γ]lek0 Cα[r(z)Γ]lek2 Cα[λ(z)L1]lek0 Cα[r(z)L1]lek2

cos(l n) ge 0 on Γ |b0| |b1| |b2|lek2 maxzisinL1

(58)

are non-negative constants For convenience we may assume that uz(z1) = 0 other-wise through a transformation of function Uz(z) = uz(z)minusλ(z1) [r(z1)+jb1][a2(z1)+b2(z1)] the requirement can be realized If cos(l n) = 0 on Γ where n is theoutward normal vector on Γ then the problem is called Problem D in whichu(z) = 2Re

int z0 uzdz + b0 = φ(z) on Γ

Problem P for (52) with A3(z) = 0 z isin D r(z) = 0 z isin Γcup Lj (j = 1 or 2) andb0 = b1 = 0 will be called Problem P0 The number

K =12(K1 +K2)

Kj =[φj

π

]+ Jj Jj = 0or 1 eiφj =

γj =φj

in which t1 = 2 t2 = 0 λ(t) = eiπ4 on L0 = (0 2) and λ(t1 minus 0) = λ(t2 + 0) =exp(iπ4) Here we choose K = 0 or K = minus12 on the boundary partD+ of D+ ifcos(ν n) equiv 0 on Γ and the condition u(2) = b2 can be canceled In fact if cos(l n) equiv 0on Γ from the boundary condition (56) we can determine the value u(2) by the valueu(0) namely

u(2)=2Reint 2

0uzdz+u(0)=2

int 2

0Re [i(zminus1)uz]dθ+b0=2

int 0

πr(z)dθ+b0 (510)

in which λ(z) = i(zminus1) θ = arg(zminus1) on Γ In order to ensure that the solution u(z)of Problem P is continuously differentiable in Dlowast we need to choose γ1 gt 0 If werequire that the solution is only continuous it suffices to choose minus2γ2 lt 1 minus2γ1 lt 1respectively In the following we shall only discuss the case K = 0 and the caseK = minus12 can be similarly discussed Problem P in this case still includes theDirichlet problem (Problem D) as a special case

52 Representation and uniqueness of solutions of oblique derivativeproblem for degenerate equations of mixed type

Now we give the representation theorem of solutions for equation (52)

Theorem 51 Suppose that the equation (52) satisfies Condition C prime Then anysolution of Problem P for (52) can be expressed as

u(z) = 2Reint z

0w(z)dz + b0 w(z) = w0(z) +W (z) (511)

where w0(z) is a solution of Problem A for the complex equation

Wmacrz = 0 in D (512)

with the boundary conditions (56) (57) (w0(z) = u0z on Γ w0(z) = u0z on L1) andW (z) in Dminus possesses the form

W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν

2g1(z)dνe1 +

int micro

0g2(z)dmicroe2 in Dminus (513)

in which e1 = (1 + j)2 e2 = (1 minus j)2 micro = x minus 2|y|m2+1(m + 2) ν = x +2|y|m2+1(m+ 2)

g1(z)= A1ξ+B1η+Cu+D ξ=Rew+Imw η=RewminusImw

g2(z)= A2ξ+B2η+Cu+D C= minus f

4|y|m2 D= minus g

4|y|m2

A1=1

4|y|m2

[m

2yminus d

|y|m2 minuse

] B1=

14|y|m2

[m

2yminus d

|y|m2+e

]

A2=1

4|y|m2

[m

2|y|minusd

|y|m2 minuse

] B2=

14|y|m2

[m

2|y|minusd

|y|m2+e

]in Dminus

(514)

Φ(z) is the solutions of equation (52) and w(z) in D+ and Φ(z) in Dminus satisfy theboundary conditions

Re [λ(z)w(z)] = r(z) z isin Γ Re [λ(x)w(x)] = s(x) x isin L0

Re [λ(x)Φ(x)] = Re [λ(x)(W (z)minus Φ(x))] x isin L0 = (0 2)

Re [λ(z)(Φ(z) + Ψ(z))] = 0 z isin L1 Im [λ(z1)(Φ(z1) + Ψ(z1))] = 0

(515)

where λ(x) = 1 + i x isin L0 Moreover by Section 5 Chapter II we see that w0(z) isa solution of Problem A for equation (512) and

u0(z) = 2Reint z

0w0(z)dz + b0 in D (516)

Proof Let u(z) be a solution of Problem P for equation (52) and w(z) = uz u(z)be substituted in the positions of w u in (513) (514) thus the functions g1(z) g2(z)and Ψ(z) in Dminus in (513)(514) can be determined Moreover we can find the solutionΦ(z) in Dminus of (512) with the boundary condition (515) where s(x) on L0 is afunction of λ(z) r(z)Ψ(z) thus

w(z) = w0(z) + Φ(z) + Ψ(z) in Dminus (517)

is the solution of Problem A in Dminus for equation (52) which can be expressed as thesecond formula in (511) and u(z) is a solution of Problem P for (52) as stated inthe first formula in (511)

Theorem 52 Suppose that equation (52) satisfies Condition C prime Then ProblemP for (52) has at most one solution in D

Proof Let u1(z) u2(z) be any two solutions of Problem P for (52) It is easy tosee that u(z) = u1(z)minus u2(z) and W (z) = uz satisfy the homogeneous equation andboundary conditions

Wz

= A1W + A2W + A3u in

D+

Dminus

(518)

Re [λ(z)W (z)] = 0 z isin Γ u(0) = 0 u(2) = 0

Re [λ(z)W (z)] = 0 z isin L1 Im [λ(z1)W (z1)] = 0(519)

where W (z) = uz in D+ According to the method as stated in Section 5 ChapterII the solution W (z) in the hyperbolic domain Dminus can be expressed in the form

W (z) = Φ(z) + Ψ(z)

Ψ(z) =int ν

2[Aξ+Bη+Cu]e1dν+

int micro

(520)

where Φ(z) is a solution of (512) in Dminus satisfying the boundary condition (515)Similarly to the method in Section 5 Chapter II Ψ(z) = 0 Φ(z) = 0 W (z) = 0z isin Dminus can be derived Thus the solution u(z) = 2Re

int z0 w(z)dz is the solution of

the homogeneous equation of (52) with homogeneous boundary conditions of (56)and (57)

2Re [λ(z)uz]=0 on Γ Re [λ(x)uz(x)]=0 on L0 u(0)=0 u(2)=0 (521)

in which λ(x) = 1 + i x isin L0 = (0 2)

Now we verify that the above solution u(z) equiv 0 in D+ If u(z) equiv 0 in D+noting that u(z) satisfies the boundary condition (521) and similarly to the proof ofTheorem 34 Chapter III we see that its maximum and minimum cannot attain inD+ cup Γ Hence u(z) attains its maximum and minimum at a point zlowast = xlowast isin L0 =(0 2) By using Lemma 41 Chapter III we can derive that ux(xlowast) = 0 uy(xlowast) = 0and then

Re [λ(x)uz(xlowast)] =12

[radicminusK(y)ux(xlowast) + uy(xlowast)

]= 0

this contradicts the second equality in (521) Thus u(z) equiv 0 in D+ This completesthe proof

53 Solvabilty problem of oblique derivative problems for degenerateequations of mixed type

From the above discussion we see in order to prove the existence of solutions ofProblem P for equation (52) the main problem is to find a solution of the oblique

derivative problem for the degenerate elliptic equation of second order ie equation(52) in elliptic domain D+ and the oblique derivative boundary conditions is (56)and

Re [λ(x)uz(x)]=s(x) on L0=(0 2) ie

12

[radicminusK(y)ux + uy

]= s(x) on L0

(522)

which is more general than the case as stated in Section 4 Chapter III We try to solvethe problem by using the method of integral equations or the method of auxiliaryfunctions which will be discussed in detail in our other publishers

CHAPTER VI

SECOND ORDER QUASILINEAR EQUATIONSOF MIXED TYPE

This chapter deals with several oblique derivative boundary value problems for sec-ond order quasilinear equations of mixed (elliptic-hyperbolic) type We shall dis-cuss oblique derivative boundary value problems and discontinuous oblique derivativeboundary value problems for second order quasilinear equations of mixed (elliptic-hyperbolic) type Moreover we shall discuss oblique derivative boundary value prob-lems for general second order quasilinear equations of mixed (elliptic-hyperbolic) typeand the boundary value problems in multiply connected domains In the meantimewe shall give a priori estimates of solutions for above oblique derivative boundaryvalue problems

1 Oblique Derivative Problems for Second Order QuasilinearEquations of Mixed Type

In this section we first give the representation of solutions for the oblique derivativeboundary value problem and then prove the uniqueness and existence of solutionsof the problem and give a priori estimates of solutions of the above problem Finallywe prove the solvability of oblique derivative problems for general quasilinear secondorder equations of mixed type

11 Formulation of the oblique derivative problem for second orderequations of mixed type

Let D be a simply connected bounded domain D in the complex plane CI as statedin Chapter V We consider the second order quasilinear equation of mixed type

uxx + sgny uyy = aux + buy + cu+ d in D (11)

where a b c d are functions of z(isin D) u ux uy (isin IR) its complex form is the fol-lowing complex equation of second order

Luz=

uzz

uzzlowast

=F (z u uz) F =Re [A1uz]+A2u+A3 in

D+

Dminus

(12)

where Aj = Aj(z u uz) j = 1 2 3 and

uzz =14[uxx + uyy] uzzlowast =

14[uxx minus uyy]

A1 =a+ ib

2 A2 =

c

4 A3 =

d

4in D

Suppose that the equation (12) satisfies the following conditions namely

Condition C

1) Aj(z u uz) (j = 1 2 3) are continuous in u isin IR uz isin CI for almost everypoint z isin D+ and measurable in z isin D+ and continuous in Dminus for all continuouslydifferentiable functions u(z) in Dlowast = D0 xminusy = 2 or Dlowast = Dx+y = 0 2 andsatisfy

Lp[Aj D+] le k0 j = 1 2 Lp[A3 D+] le k1 A2 ge 0 in D+

2) For any continuously differentiable functions u1(z) u2(z) in Dlowast the equality

holds where Aj = Aj(z u1 u2) (j = 1 2) satisfy the conditions

Lp[Aj D+] le k0 C[Aj Dminus] le k0 j = 1 2 (15)

in (13)(15) p (gt 2) k0 k1 are non-negative constants In particular when (12) isa linear equation the condition (14) obviously holds

Problem P The oblique derivative boundary value problem for equation (12) isto find a continuously differentiable solution u(z) of (12) in Dlowast = D0 x minus y = 2which is continuous in D and satisfies the boundary conditions

12

partu

partl= Re [λ(z)uz] = r(z) z isin Γ

12

partu

partl= Re [λ(z)uz] = r(z) z isin L1

Im [λ(z)uz]z=z1 = b1 u(0) = b0 u(2) = b2

(16)

where l is a given vector at every point on Γ cup L1 λ(z) = a(x) + ib(x) = cos(l x)∓i cos(l y) and ∓ are determined by z isin Γ and z isin L1 respectively b0 b1 b2 are realconstants and λ(z) r(z) b0 b1 b2 satisfy the conditions

Cα[λ(z)Γ]lek0 Cα[r(z)Γ]lek2 cos(ln)ge0onΓ |b0||b1||b2|lek2

Cα[λ(z)L1]lek0 Cα[r(z)L1]lek2 maxzisinL1 [1|a(x)minusb(x)|]lek0(17)

are non-negative constants For convenience we may assume that w(z1) = 0

202 VI Second Order Quasilinear Mixed Equations

otherwise through a transformation of function W (z) = w(z) minus λ(z1)[r(z1) minus ib1]the requirement can be realized Here we mention that if A2(z) = 0 in D1we can cancel the assumption cos(l n) ge 0 on Γ and if the boundary conditionRe [λ(z)uz] = r(z) z isin L1 is replaced by Re [λ(z)uz] = r(z) z isin L1 then ProblemP does not include the Dirichlet problem (Tricomi problem) as a special case

The boundary value problem for (12) with A3(z u uz) = 0 z isin D u isin IR uz isinCIr(z) = 0 z isin partD and b0 = b2 = b1 = 0 will be called Problem P0 The number

K =12(K1 +K2) (18)

Kj =[φj

π

γj =φj

in which [a] is the largest integer not exceeding the real number a t1 = 2 t2 = 0λ(t) = exp(iπ4) on L0 and λ(t1minus0) = λ(t2+0) = exp(iπ4) here we only discuss thecase of K=0 on partD+ if cos(l n) equiv 0 on Γ or K=minus12 if cos(l n) equiv 0 on Γ becausein this case the last point condition in (16) can be eliminated and the solution ofProblem P is unique In order to ensure that the solution u(z) of Problem P in Dlowast iscontinuously differentiable we need to choose γ1 gt 0 If we require that the solutionof Problem P in D is only continuous it suffices to choose minus2γ1 lt 1 minus2γ2 lt 1

Besides if A2 = 0 in D the last condition in (16) is replaced by

Im [λ(z)uz]|z=z2 = b2 (110)

where the integral path is along two family of characteristic lines similar to thatin (210) Chapter II z2(= 0 2) isin Γ and b2 is a real constant with the condition|b2| le k2 and here the condition cos(l n) ge 0 is canceled then the boundary valueproblem for (12) will be called Problem Q

12 The existence and uniqueness of solutions for the oblique derivativeproblem for (12)

Similarly to Section 2 Chapter V we can prove the following results

Lemma 11 Let equation (12) satisfy Condition C Then any solution of ProblemP for (12) can be expressed as

u(z) = 2Reint z

0w(z)dz + b0 w(z) = w0(z) +W (z) in D (111)

where the integral path in Dminus is the same as in Chapter II and w0(z) is a solutionof Problem A for the equation

Lw =

wz

wzlowast

= 0 in

D+

Dminus

(112)

with the boundary condition (16) (w0(z) = u0z) and W (z) possesses the form

int intD+

g(ζ)ζ minus z

W (z)=Φ(z) + Ψ(z) Ψ(z)=int ν

2g1(z)dνe1 +

int micro

(113)

in which Im [φ(z)] = 0 on L0 = (0 2) e1 = (1 + i)2 e2 = (1 minus i)2 micro = x + yν = x minus y φ0(z) is an analytic function in D+ and continuous in D+ such thatIm [φ(x)] = 0 on L0 and

g(z)=

A12+A1w(2w)w(z) =00w(z)=0 zisinD+

f(z)=Re[A1φz]+A2u+A3 inD+

A=ReA1+ImA1

2 B=

ReA1minusImA1

2 C=A2 D=A3 inDminus

(114)

where Φ(z) is an analytic function in D+ and Φ(z) is a solution of equation (112)in Dminus satisfying the boundary conditions

(115)

in which λ(x) = 1 + i x isin L0 = (0 2) and s(x) is as stated in (223) Chapter VMoreover by Theorem 11 Chapter V the solution w0(z) of Problem A for (112) andu0(z) satisfy the estimate in the form

Cβ[u0(z) D]+Cβ[w0(z)X(z) D+]+Cβ[wplusmn0 (micro ν)Y plusmn(micro ν) Dminus]leM1(k1+k2) (116)

where

X(z) =2prod

j=1

|z minus tj|ηj Y plusmn(z) = Y plusmn(micro ν) = [|ν minus 2||micro minus 2|]ηj

ηj =

2|γj|+ δ γj lt 0

δ γj ge 0j = 1 2

(117)

herein wplusmn0 (micro ν) = Rew0(z) ∓ Imw0(z) w0(z) = w0(micro ν) micro = x + y ν = x minus y

and γ1 γ2 are the real constants in (19) β(lt δ) δ are sufficiently small positiveconstants and

u0(z) = 2Reint z

0w0(z)dz + b0 in D (118)

where p0(2 lt p0 le p) M1 = M1(p0 β k0 D) are non-negative constants

Theorem 12 Suppose that equation (12) satisfies Condition C Then Problem Pfor (12) has a unique solution u(z) in D

Theorem 13 Suppose that the equation (12) satisfies Condition C Then anysolution u(z) of Problem P for (12) satisfies the estimates

C1β[u D+] = Cβ[u(z) D+] + Cβ[uzX(z) D+] le M2

C1[u Dminus] = Cβ[u(z) Dminus] + C[uplusmnz (micro ν)Y plusmn(micro ν) Dminus] le M3

C1β[u D+] le M4(k1 + k2) C1[u Dminus] le M4(k1 + k2)

(119)

where X(z) Y plusmn(micro ν) are stated in (117) and Mj = Mj (p0 β k0 D) (j = 2 3 4) arenon-negative constants

13 C1α(D)-estimate of solutions of Problem P for second order equations

of mixed type

Now we give the C1α(D)-estimate of solutions u(z) for Problem P for (12) but it

needs to assume the following conditions For any real numbers u1 u2 and complexnumbers w1 w2 we have

|Aj(z1 u1 w1)minusAj(z2 u2 w2)|lek0[|z1minusz2|α+|u1minusu2|α+|w1minusw2|] j=1 2

Theorem 14 If Condition C and (120) hold then any solution u(z) of ProblemP for equation (12) in Dminus satisfies the estimates

C1β[u Dminus]=Cβ[u Dminus]+Cβ[uplusmn

z (micro ν)Y plusmn(micro ν) Dminus]

le M5 C1β[u Dminus] le M6(k1 + k2)

(121)

in which uplusmnz (micro ν) = Reuz ∓ Imuz β (0 lt β le α) M5 = M5(p0 β k D) M6 =

M6(p0 β k0 D) are non-negative constants k = (k0 k1k2)

Proof Similarly to Theorem 13 it suffices to prove the first estimate in (121)Due to the solution u(z) of Problem P for (12) is found by the successive iterationthrough the integral expressions (111) (113) and (114) we first choose the solutionof Problem A of (112) in the form (118) ie

u0(z) = 2Reint z

0w0(z)dz + b0 w0(z) = ξ0(z)e1 + η0(z)e2 in D (122)

and substitute them into the positions of u0 w0 in the right-hand side of (114) wecan obtain Ψ1(z) w1(z) u1(z) as stated in (111)ndash(114) Denote

u1(z) = 2Reint z

0w1(z)dz + b0 w1(z) = w0(z) + Φ1(z) + Ψ1(z)

Ψ11(z) =

int ν

2G1(z)dν G1(z) = Aξ0 +Bη0 + Cu0 +D

Ψ21(z) =

int micro

0G2(z)dmicro G2(z) = Aξ0 +Bη0 + Cu0 +D

(123)

from the last two equalities in (123) it is not difficult to see that G1(z) =G1(micro ν) Ψ1

1(z) = Ψ11(micro ν) andG2(z) = G2(micro ν) Ψ2

1(z) = Ψ21(micro ν) satisfy the Holder

estimates about ν micro respectively namely

Cβ[G1(middot ν) Dminus] le M7 Cβ[Ψ11(middot ν) Dminus] le M7R

Cβ[G2(micro middot) Dminus] le M8 Cβ[Ψ21(micro middot) Dminus] le M8R

(124)

where Mj = Mj(p0 β k D) (j = 7 8) and R = 2 Moreover from (123) we canderive that Ψ1

1(micro ν) Ψ21(micro ν) about micro ν satisfy the Holder conditions respectively

namelyCβ[Ψ1

1(micro middot) Dminus] le M9R Cβ[Ψ21(middot ν) Dminus] le M9R (125)

where M9 = M9(p0 β k D) Besides we can obtain the estimate of Φ1(z) ie

Cβ[Φ1(z) Dminus] le M10R = M10(p0 β k D)R (126)

in which Φ1(z) satisfies equation (112) and boundary condition of Problem P butin which the function Ψ(z) is replaced by Ψ1(z) Setting w1(z) = w0(z) + Φ1(z) +Ψ1(z) and by the first formula in (123) we can find the function u1(z) from w1(z)Furthermore from (125)(126) we can derive that the functions wplusmn

1 (z) = wplusmn1 (micro ν) =

Re w1(z) ∓ Im w1(z) (w1(z) = w1(z) minus w0(z)) and u1(z) = u1(z) minus u0(z) satisfy theestimates

Cβ[wplusmn1 (micro ν)Y plusmn(micro ν) Dminus] le M11RCβ[u1(z) Dminus] le M11R (127)

where M11 = M11(p0 β k D) Thus according to the successive iteration we canobtain the estimates of functions wplusmn

n (z) = wplusmnn (micro ν) = Re wn(z)∓ Im wn(z) (wn(z) =

wn(z)minuswnminus1(z)) and the corresponding function un(z) = un(z)minusunminus1(z) satisfy theestimates

Cβ[wplusmnn (micro ν)Y plusmn(micro ν) Dminus] le (M11R)n

n Cβ[un(z) Dminus] le (M11R)n

n (128)

wn(z) =nsum

m=1

wm(z) + w0(z) un(z) =nsum

m=1

um(z) + u0(z) n = 1 2 (129)

uniformly converge to w(z) u(z) in any close subset ofDlowast respectively and w(z) u(z)satisfy the estimates

Cβ[wplusmn(micro ν)Y plusmn(micro ν) Dminus] le eM11R Cβ[u(z) Dminus] le M5 (130)

From the estimates (119) and (121) we can see the regularity of solutions of Prob-lem P for (12) Moreover it is easy to see that the derivatives [w+(micro ν)]ν [wminus(micro ν)]microsatisfy the estimates similar to those in (130)

As for Problem Q for (12) we can similarly discuss its unique solvability

14 The solvability for the oblique derivative problem for general secondorder quasilinear equations of mixed type

Luz =

uzz

uzzlowast

= F (z u uz) +G(z u uz) z isin

D+

Dminus

F = Re [A1uz] + A2u+ A3 G = A4|uz|σ + A5|u|τ z isin D

(131)

where F (z u uz) satisfies Condition C σ τ are positive constants and Aj(z u uz)(j = 4 5) satisfy the conditions in Condition C the main conditions of which are

Lp[Aj(z u uz) D+] le k0 C[Aj(z u uz) Dminus] le k0 j = 4 5

(1) When 0 lt max(σ τ) lt 1 Problem P for (131) has a solution u(z) isin C(D)

(2) When min(σ τ) gt 1 Problem P for (131) has a solution u(z) isin C(D)provided that

M12 = k1 + k2 + |b0|+ |b1| (132)

(3) When min(σ τ) gt 1 Problem P for the equation

Luz =

uzz

uzzlowast

= F (z u uz) + εG(z u uz) z isin

D+

Dminus

(133)

has a solution u(z) isin C(D) provided that the positive number ε in (133) is appro-priately small where the functions F (z u uz) G(z u uz) are as stated in (131)

Proof (1) Consider the algebraic equation

M4k1 + k2 + 2k0tσ + 2k0t

τ + |b0|+ |b1| = t (134)

for t where M4 is the constant stated in (119) It is not difficult to see that equation(134) has a unique solution t = M13 ge 0 Now we introduce a bounded closed andconvex subset Blowast of the Banach space B = C1(D) whose elements are the functionsu(z) satisfying the condition

C1[u(z) D]=Cβ[u(z) D]+Cβ[uzX(z) D+]+C[uplusmnz (micro ν)Y plusmn(micro ν) Dminus]leM13 (135)

We arbitrarily choose a function u0(z) isin B for instance u0(z) = 0 and substitute itinto the position of u in the coefficients of (131) and G(z u uz) From Theorem 12it is clear that problem P for

Luz minus Re [A1(z u0 u0z)uz]minus A2(z u0 u0z)u minus A3(z u0 u0z) = G(z u0 u0z) (136)

has a unique solution u1(z) From Theorem 14 we see that the solution u1(z) satisfiesthe estimate in (135) By using successive iteration we obtain a sequence of solutionsum(z) (m = 1 2 ) isin Blowast of Problem P which satisfy the equations

Lum+1z minus Re [A1(z um umz)um+1z]minus A2(z um umz)um+1

+A3(z um umz) = G(z um umz) in D m = 1 2 (137)

and um+1(z) isin Blowast From (137) we see that and um+1(z) = um+1(z)minusum(z) satisfiesthe equations and boundary conditions

Lum+1zminusRe[A1um+1z]minusA2um+1=G(zumumz)minusG(zumminus1umminus1z) in D

Re[λ(z)um+1z]=0 on Γ Re[λ(z)um+1z]=0 on L1 Im[λ(z)um+1z]|z=z1=0(138)

in which m = 1 2 Noting that C[G(z um umz) minus G(z umminus1 umminus1z) D] le2k0M13 M13 is a solution of the algebraic equation (134) and according to The-orem 13 the estimate

um+1 = C1[um+1 D] le M14 = M14(p0 β k0 D) (139)

um+1(z) = 2Reint z

0wm+1(z)dz in D wm+1(z) = Φm+1(z) + Ψm+1(z)

2[Aξm+1 + Bηm+1 + Cum+1 + D]e1d(x minus y)

+int x+y

0[Aξm+1 + Bηm+1 + Cum+1 + D]e2d(x+ y) in Dminus

(140)

in which Φm+1(z) Ψm+1(z) are similar to the functions Φ(z) Ψ(z) in (113) the rela-tion between A1 A2 G and A B C D is the same as that of A1 A2 A3 and A B C Din (114) and G = G(z um umz)minus G(z umminus1 umminus1z) By using the method from theproof of Theorem 13 we can obtain

um+1 minus um = C1[um+1 D] le (M14Rprime)m

m

where M14 = 2M4(M15 + 1)MR(4m0 + 1) R = 2 m0 = w0(z)X(z) C(D) hereinM15 = maxC[A Q] C[B Q] C[C Q] C[DQ] M = 1 + 4k2

0(1 + k20) From the

above inequality we see that the sequence of functions um(z) ieum(z) = u0(z) + [u1(z)minus u0(z)] + middot middot middot+ [um(z)minus umminus1(z)] m = 1 2 (141)

uniformly converges to a function ulowast(z) and wlowast(z) = ulowastz satisfies the equality

wlowast(z) = w0(z) + Φlowast(z) + Ψlowast(z) in Dminus

Ψlowast(z) = +int xminusy

2[Aξlowast +Bηlowast + Culowast +D]e1d(x minus y)

+int x+y

0[Aξlowast +Bηlowast + Culowast +D]e2d(x+ y) in Dminus

(142)

int z

0wlowast(z)dz + b0 in D (143)

is just a solution of Problem P for the general quasilinear equation (131) in D

M4k1 + k2 + 2k0tσ + 2k0t

τ + |b0|+ |b1| = t (144)

for t It is not difficult to see that the equation (144) has a solution t = M13 ge 0provided that the positive constant M12 in (132) is small enough Now we introducea bounded closed and convex subset Blowast of the Banach space C1(D) whose elementsare the functions u(z) satisfying the condition

By using the same method as in (1) we can find a solution u(z) isin Blowast of Problem Pfor equation (131) with min(σ τ) gt 1

(3) There is no harm in assuming that k1 k2 in (13)(17) are positive constantswe introduce a bounded closed and convex subset Bprime of the Banach space C1(D)whose elements are the functions u(z) satisfying the condition

C1[u(z) D] le (M4 + 1)(2k1 + k2) (146)

where M4 is a constant as stated in (119) and we can choose an appropriately smallpositive number ε such that C[εG(z u uz) D] le k1 Moreover we are free to choosea function u0(z) isin Bprime for instance u0(z) = 0 and substitute it into the position of uin the coefficients of (133) and G(z u uz) From Theorem 12 it is seen that thereexists a unique solution of Problem P for

Lu minus Re [A1(z u0 u0z)uz]minus A2(z u0 u0z)u minus A3(z u0 u0z) = G(z u0 u0z)

and u1(z) isin Bprime Thus similarly to the proof in (1) by the successive iteration asolution of Problem P for equation (133) can be obtained

By using a similar method as before we can discuss the solvability of ProblemP and the corresponding Problem Q for equation (12) or (131) with the boundaryconditions

Re [λ(z)uz] = r(z) z isin Γ u(0) = b0 u(2) = b2

Re [λ(z)uz] = r(z) z isin L2 Im [λ(z)uz]|z=z1 = b1

in which the coefficients λ(z) r(z) b0 b1 b2 satisfy the condition (17) but where theconditions Cα[λ(z) L1] le k0 Cα[r(z) L1] le k2 maxzisinL1 [1|a(x) minus b(x)|] le k0 arereplaced by Cα[λ(z) L2] le k0 Cα[r(z) L2] le k2 maxzisinL2 [1|a(x) + b(x)|] le k0 andin (19) the condition λ(t) = eiπ4 on L0 = (0 2) and λ(t1minus0) = λ(t2+0) = exp(iπ4)is replaced by λ(t) = eminusiπ4 on L0 = (0 2) and λ(t1 minus 0) = λ(t2 + 0) = exp(minusiπ4)Besides the setDlowast = D0 xminusy = 2 in Condition C is replaced byDlowast = Dx+y =0 2 if the constant γ2 gt 0 in (19)

2 Oblique Derivative Problems for Second Order Equationsof Mixed Type in General Domains

This section deals with oblique derivative boundary value problem for secondorder quasilinear equations of mixed (elliptic-hyperbolic) type in general domainsWe prove the uniqueness and existence of solutions of the above problem In refs[12]1)3) the author discussed the Dirichlet problem (Tricomi problem) for secondorder equations of mixed type uxx + sgny uyy = 0 by using the method of integralequations and a complicated functional relation In the present section by usinga new method the solvability result of oblique derivative problem for more generaldomains is obtained

21 Oblique derivative problem for second order equations of mixed typein another domain

Let D be a simply connected bounded domain D in the complex plane CI with theboundary partD = Γ cup L where Γ(sub y gt 0) isin C2

micro (0 lt micro lt 1) with the end pointsz = 0 2 and L = L1 cup L2 cup L3 cup L4 and

L1 = x+ y = 0 0 le x le a2 L2 = x minus y = a a2 le x le a

L3 = x+ y = a a le x le 1 + a2 L4 = x minus y = 2 1 + a2 le x le 2(21)

where a (0 lt a lt 2) is a constant Denote D+ = D cap y gt 0 and Dminus = D cap y lt0 Dminus

1 = Dminuscapxminusy lt a andDminus2 = Dminuscapx+y gt a Without loss of generality we

may assume that Γ = |z minus 1| = 1 y ge 0 otherwise through a conformal mappingthis requirement can be realized

We consider the quasilinear second order mixed equation (12) and assume that(12) satisfies Condition C in D here Dminus is as stated before Problem P for(12)

inD is to find a continuously differentiablesolution of (12) in Dlowast = D0 a 2 sat-isfying the boundary conditions

12

partu

partl=Re[λ(z)uz]=r(z) zisinΓ

u(0)=b0 u(a)=b1 u(2)=b2 (22)12

partu

partl=Re[λ(z)uz]=r(z) zisinL1cupL4

Im[λ(z)uz]|z=zj=bj+2 j=12

where l is a given vector at every point on Γ cup L1 cup L4 z1 = (1 minus i)a2 z2 =(1 + a2) + i(1 minus a2) λ(z) = a(x) + ib(x) = cos(l x) minus i cos(l y) z isin Γ andλ(z) = a(x) + ib(x) = cos(l x) + i cos(l y) z isin L1 cup L4 bj(j = 0 1 4) are realconstants and λ(z) r(z) bj(j = 0 1 4) satisfy the conditions

Cα[λ(z)Γ] le k0 Cα[r(z)Γ] le k2 Cα[λ(z) Lj] le k0

Cα[r(z) Lj] le k2 j = 1 4 |bj| le k2 j = 0 1 4

cos(l n)ge0 on Γ maxzisinL1

1|a(x)minusb(x)| max

zisinL4

1|a(x)+b(x)| lek0

(23)

are non-negative constants The number

K =12(K1 +K2 +K3) (24)

is called the index of Problem P on the boundary partD+ of D+ where

Kj=[φj

π

]+Jj Jj=0or1 eiφj =

λ(tj minus0)λ(tj+0)

γj=φj

πminusKj j=123 (25)

in which [b] is the largest integer not exceeding the real number b t1 = 2 t2 = 0t3 = a λ(t) = eiπ4 on (0 a) and λ(t3 minus 0) = λ(t2+0) = exp(iπ4) and λ(t) = eminusiπ4

on (a 2) and λ(t1 minus 0) = λ(t3 + 0) = exp(minusiπ4) Here we only discuss the caseK = 12 or K = 0 if cos(l n) = 0 on Γ because in this case the solution of ProblemP is unique and includes the Dirichlet problem (Tricomi problem) as a special caseWe mention that if the boundary condition Re [λ(z)uz] = r(z) z isin Lj (j = 1 4) isreplaced by

Re [λ(z)uz] = r(z) z isin Lj (j = 1 4)

then Problem P does not include the Dirichlet problem (Tricomi problem) as a spe-cial case In order to ensure that the solution u(z) of Problem P is continuouslydifferentiable in Dlowast we need to choose γ1 gt 0 γ2 gt 0 and can select γ3 = 12 Ifwe only require that the solution u(z) in D is continuous it is sufficient to chooseminus2γ1 lt 1 minus2γ2 lt 1 minusγ3 lt 1

Besides we consider the oblique derivative problem (Problem Q) for equation(12) with A2 = 0 and the boundary condition (22) but the last point conditionsu(a) = b1 u(2) = b2 in (22) is replaced by

Im [λ(z)uz]|zprimej= cj j = 1 2 (26)

in which zprimej(j = 1 2) isin Γlowast = Γ0 2 are two points and c1 c2 are real constants and

|c1| |c2| le k2

Similarly to Section 1 we can give a representation theorem of solutions of ProblemP for equation (12) in which the functions Ψ(z) in (113) λ(x) s(x) on L0 in (115)X(z) Y plusmn(micro ν) in (117) are replaced by

Ψ(z) =

⎧⎪⎪⎨⎪⎪⎩int ν

a

g1(z)dνe1 +int micro

0g2(z)dmicroe2 z isin Dminus

1 int ν

2g1(z)dνe1 +

int micro

a

g2(z)dmicroe2 z isin Dminus2

(27)

λ(x) =

1 + i x isin Lprime

0 = (0 a)

1minus i x isin Lprimeprime0 = (a 2)

(28)

s(x)=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

2r((1minusi)x2)minus2Re[λ((1minusi)x2)Ψ((1minusi)x2)a((1minusi)x2)minusb((1minusi)x2)

+Re[λ(x)Ψ(x)]

minus [a((1minusi)x2)+b((1minusi)x2)]f(0)a((1minusi)x2)minusb((1minusi)x2)

xisin(0a)

2r((1+i)x2+1minusi)minus2Re[λ((1+j)x2+1minusi)Φ((1+j)x2+1minusi)]a((1+i)x2+1minusi)+b((1+i)x2+1minusi)

minusa((1+i)x2+1minusi)minusb((1+i)x2+1minusi)g(2)minush(x)a((1+i)x2+1minusi)+b((1+i)x2+1minusi)

h(x)=Re[λ(x)Ψ(x)]

times[a((1+i)x2+1minusi)+b((1+i)x2+1minusi)]2xisin(a2)

f(0)=[a(z1)+b(z1)]r(z1)+[a(z1)minusb(z1)]b3

g(2)=[a(z2)minusb(z2)]r(z2)minus [a(z2)+b(z2)]b4

(29)

X(z) =3prod

j=1

|z minus tj|ηj Y plusmn(z) = Y plusmn(micro ν) =3prod

j=1

[|micro minus tj||ν minus tj|]ηj

ηj =

2max(minusγj 0) + δ j = 1 2

max(minusγj 0) + δ j = 3

(210)

respectively δ is a sufficiently small positive constant besides L1 and the point z1 in(115) should be replaced by L1 cup L4 and z1 = (1minus i)a2 z2 = 1 + a2 + (1minus a2)i

Now we first prove the unique solvability of Problem Q for equation (12)

Theorem 21 If the mixed equation (12) in the domain D satisfies Condition Cthen Problem Q for (12) has a unique solution u(z) as stated in the form

u(z) = 2Reint z

0w(z)dz + b0 in D (211)

wherew(z) = Φ(z)eφ(z) + ψ(z) in D+

int intD+

g(ζ)ζ minus z

w(z) = w0(z) + Φ(z) + Ψ(z) in Dminus

Ψ(z) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩int ν

a

1 int ν

2g1(z)dνe1 +

int micro

a

(212)

where Im [φ(z)] = 0 on L0 = (0 2) e1 = (1 + i)2 e2 = (1 minus i)2 micro = x + yν = x minus y φ0(z) is an analytic function in D+ and continuous in D+ such thatIm [φ(z)] = 0 on L0 and f(z) g(z) g1(z) g2(z) are as stated in (114) and Φ(z) isan analytic function in D+ and Φ(z) is a solution of equation (112) satisfying theboundary conditions the first conditions and

Im [λ(zj)Φ(zj)] = minusIm [λ(zj)Ψ(zj)] j = 1 2

Im [λ(zprimej)Φ(z

primej)] = cj j = 1 2

(213)

in which λ(x) s(x) are as stated in (28)(29)

Proof By using a similar method as stated in Section 2 Chapter V we can proveTheorem 21 provided that L1 or L2 in the boundary conditions Section 2 ChapterV is replaced by L1 cup L4 the point conditions Im [λ(z1)w(z1)] = b1 in Section 2Chapter V is replaced by Im [λ(zj)w(zj)] = bj j = 1 2 and so on the formula (214)Chapter V is replaced by

Re [λ(z)uz] = Re [λ(z)w(z)] = s(x)

λ(x) =

1 + i x isin Lprime

0 = (0 a)

Cβ[s(x) Lprime0] + Cβ[s(x) Lprimeprime

0] le k3

Theorem 22 Suppose that the equation (12) satisfies Condition C Then ProblemP for (12) has a unique solution u(z) and the solution u(z) satisfies the estimates

C1β[u(z) D+] = Cβ[u(z) D+] + Cβ[uzX(z) D+] le M15

C1[u(z) Dminus] = Cβ[u(z) Dminus] + C[uplusmnz Y plusmn(z) Dminus] le M15

C1β[u(z) D+]leM16(k1+k2) C1[u(z) Dminus]leM16(k1+k2)

(214)

where

X(z)=3prod

j=1

|zminustj|ηj Y plusmn(z)=3prod

j=1

|xplusmnyminustj|ηj ηj=

2max(minusγj0)+δ j=12

max(minusγj0)+δ j=3(215)

herein t1 = 2 t2 = 0 t3 = a γ1 γ2 γ3 are real constants in (25) β(lt δ) δ aresufficiently small positive constants and M15 = M15(p0 β k D) M16 = M16(p0 βk0 D) are non-negative constants k = (k0 k1 k2)

Proof First of all we prove the uniqueness of solutions of Problem P for (12)Suppose that there exist two solutions u1(z) u2(z) of Problem P for (12) By Condi-tion C we can see that u(z) = u1(z)minusu2(z) and w(z) = uz satisfies the homogeneousequation and boundary conditions

wz

wzlowast

= f f = Re [A1w] + A2u in

D+

Dminus

(216)

Re [λ(z)w(z)] = r(z) z isin Γ u(0) = 0 u(a) = 0

u(2) = 0 Re [λ(z)w(z)] = 0 z isin L1 cup L4(217)

and (211) By using the method of proofs in Theorems 23 and 24 Chapter Vw(z) = uz = 0 u(z) = 0 in Dminus can be derived Thus we have

2Re[1minus iradic2

uz

]=

partu

partl= 0 on (0 a) 2Re

[1 + iradic2

uz

]=

partu

partl= 0 on (a 2) (218)

it is clear that (1minus i)radic2 = cos(l x) minus i cos(l y) = exp(minusiπ4) on (0 a) and

(1 + i)radic2 = cos(l x) + i cos(l y) = exp(iπ4) on (a 2) On the basis of the max-

imum principle of solutions for (12) with A3 = 0 in D+ if maxD+ u(z) gt 0 then itsmaximum M attains at a point zlowast isin Γ cup L0 obviously zlowast = 0 a and 2 and we canprove zlowast isin Γ by the method as stated in the proof of Theorem 23 Chapter V More-over it is not difficult to prove that if zlowast isin L0 then partupartl = 0 at zlowast This contradicts(217) Thus maxD+ u(z) = 0 By the similar method we can prove minD+ u(z) = 0Hence u(z) = 0 u1(z) = u2(z) in D+

Secondly we first prove the existence of solutions of Problem P for the linearequation (12) with A2 = 0 ie

uzz = Re [A1uz] + A3 z isin D+

uzzlowast = Re [A1uz] + A3 z isin Dminus(219)

By Theorem 21 we can prove the solvability of Problem P for (219) In fact ifu0(a) = b1 u0(2) = b2 then the solution u0(z) is just a solution of Problem P for(12) Otherwise u0(a) = cprime

1 = b1 or u0(2) = cprime2 = b2 we find a solution u2(z) of

Problem P for (219) with the boundary conditions

Re [λ(z)ukz] = 0 z isin Γ uk(0) = 0

Re [λ(z)ukz] = 0 z isin L1 cup L4 k = 1 2

Im [λ(z)ukz]|z=zprimej= δkj k j = 1 2

(220)

in which δ11 = δ22 = 1 δ12 = δ21 = 0 It is clear that

J =

∣∣∣∣∣u1(a) u2(a)

u1(2) u2(2)

∣∣∣∣∣ = 0 (221)

Because otherwise there exist two real constants d1 d2 (|d1| + |d2| = 0) such thatd1u1(z) + d2u2(z) equiv 0 in D and

d1u1(a) + d2u2(a) = 0 d1u1(2) + d2u2(2) = 0 (222)

According to the proof of uniqueness as before we can derive d1u1(z) + d2u2(z) equiv 0in D the contradiction proves J = 0 Hence there exist two real constants d1 d2such that

d1u1(a) + d2u2(a) = cprime1 minus b1 d1u1(2) + d2u2(2) = cprime

2 minus b2 (223)

thus the functionu(z) = u0(z)minus d1u1(z)minus d2u2(z) in D (224)

is just a solution of Problem P for equation (12) in the linear case Moreover we canobtain that the solution u(z) of Problem P for (12) satisfies the estimates in (214)we can rewrite in the form

C1[u D] = Cβ[u(z) D] + Cβ[uzX(z) D+]

+C[uplusmnz (micro ν)Y plusmn(micro ν) Dminus] le M17

C1[u D] le M18(k1 + k2)

(225)

where X(z) Y plusmn(micro ν) are as stated in (210) and M17 = M17(p0 β k D) M18 =M18(p0 β k0 D) are non-negative constants k = (k0 k1 k2) By using the estimateand method of parameter extension the existence of solutions of Problem P forquasilinear equation (12) can be proved

22 Oblique derivative problem for second order equations of mixed typein general domains

Now we consider the domain Dprime with the boundary partDprime = Γ cup Lprime where Lprime =Lprime

1 cup Lprime2 cup Lprime

3 cup Lprime4 and the parameter equations of the four curves Lprime

1 Lprime2 Lprime

3 Lprime4 are

Lprime1 = γ1(x) + y = 0 0 le x le l1 Lprime

2 = x minus y = a l1 le x le a

Lprime3 = x+ y = a a le x le l2 Lprime

4 = γ2(x) + y = 0 l2 le x le 2(226)

in which γ1(0) = 0 γ2(2) = 0 γ1(x)gt 0 on 0le x le l1 = γ(l1)+a γ2(x)gt 0 on l2 =minusγ(l2)+a le x le 2 γ1(x) on 0 le x le l1 γ2(x) on l2 le x le 2 are continuous andγ1(x) γ2(x) are differentiable on 0le x le l1 l2 le x le 2 except some isolated pointsand 1 + γprime

1(x)gt 0 on 0 le x le l1 1 minus γprime2(x)gt 0 on l2 le x le 2 Denote Dprime+ =Dprimecap

y gt 0=D+ Dprimeminus=Dprime cap y lt 0 Dprimeminus1 =Dprimeminus cap x lt a and Dprimeminus

2 =Dprimeminus cap x gt aHere we mention that in [12]1)3) the author assumed 0 lt minusγprime

1(x) lt 1 on 0 le x le l1and some other conditions

We consider the quasilinear second order equation of mixed (elliptic-hyperbolic)type (12) in Dprime Assume that equation (12) satisfies Condition C but the hyper-bolic domain Dminus is replaced by Dprimeminus

Problem P prime The oblique derivative problem for equation (12) is to find a con-tinuously differentiable solution of (12) in Dlowast = Dprime0 a 2 for (12) satisfying theboundary conditions

12

partu

partl= Re [λ(z)uz] = r(z) z isin Γ u(0) = b0 u(a) = b1 u(2) = b2

12

partu

partl=Re [λ(z)uz]=r(z) zisinLprime

1cupLprime4 Im [λ(z)uz]|z=zj

=bj+2 j=12(227)

Here l is a given vector at every point on Γ cup Lprime1 cup Lprime

4 z1 = l1 minus iγ1(l1) z2 =l2 minus iγ2(l2) λ(z) = a(x) + ib(x) = cos(l x) minus i cos(l y) z isin Γ and λ(z) = a(x) +ib(x) = cos(l x) + i cos(l y) z isin Lprime

1 cup Lprime4 bj(j = 0 1 4) are real constants and

λ(z) r(z) bj(j = 0 1 4) satisfy the conditions

Cα[λ(z)Γ] le k0 Cα[r(z)Γ] le k2

Cα[λ(z) Lprimej] le k0 Cα[r(z) Lprime

j] le k2 j = 1 4

cos(l n) ge 0 on Γ |bj| le k2 0 le j le 4

maxzisinLprime

1

zisinLprime4

1|a(x) + b(x)| le k0

(228)

are non-negative constants In particular if Lprimej = Lj(j = 1 2 3 4) then Problem P prime

in this case is called Problem P

In the following we discuss the domain Dprime with the boundary ΓcupLprime1cupLprime

2cupLprime3cupLprime

4where Lprime

1 Lprime2 Lprime

3 Lprime4 are as stated in (226) and γ1(x) γ2(x) satisfy the conditions

1 + γprime1(x) gt 0 on 0 le x le l1 and 1minus γprime

2(x) gt 0 on l2 le x le 2 By the conditions the

inverse functions x = σ(ν) x = τ(micro) of x+ γ1(x) = ν = xminus y xminus γ2(x) = micro = x+ ycan be found respectively namely

micro = 2σ(ν)minus ν 0 le ν le a ν = 2τ(micro)minus micro a le micro le 2 (229)

They are other expressions for the curves Lprime1 Lprime

4 Now we make a transformationin Dprimeminus

micro =a[micro minus 2σ(ν) + ν]a minus 2σ(ν) + ν

ν = ν 2σ(ν)minus ν le micro le a 0 le ν le a

micro=micro ν=a[2τ(micro)minusmicrominus2]+(2minusa)ν

2τ(micro)minus micro minus a alemicrole2 aleν le2τ(micro)minusmicro

(230)

in which micro ν are variables If (micro ν) isin Lprime1 Lprime

2 Lprime3 Lprime

4 then (micro ν) isin L1 L2 L3 L4

respectively The inverse transformation of (230) is

micro =1a[a minus 2σ(ν) + ν]micro+ 2σ(ν)minus ν

=1a[a minus 2σ(x+ γ1(x)) + x+ γ1(x)](x+y)+2σ(x+γ1(x))minusxminusγ1(x)

ν= ν= xminusy 0 le micro le a 0 le ν le a micro = micro = x+ y

ν =1

2minus a[(2τ(micro)minus micro)(ν minus a)minus a(ν minus 2)]

=12minusa

[(2τ(xminusγ2(x))minusx+γ2(x))(xminusyminusa)minusa(xminusyminus2)]

a le microle2 ale ν le2

(231)

It is not difficult to see that the transformations in (231) map the domains Dprimeminus1 Dprimeminus

2onto the domains Dminus

1 Dminus2 respectively Moreover we have

x =12(micro+ ν) =

2ax minus (a+ x minus y)[2σ(x+ γ1(x))minus x minus γ1(x)]2a minus 4σ(x+ γ1(x)) + 2x+ 2γ1(x)

2ay minus (a minus x+ y)[2σ(x+ γ1(x))minus x minus γ1(x)]2a minus 4σ(x+ γ1(x)) + 2x+ 2γ1(x)

x =12(micro+ ν) =

12a[a minus 2σ(x+ γ1(x)) + x+ γ1(x)](x+ y)

+σ(x+ γ1(x))minus 12(x+ γ1(x)minus x+ y)

+σ(x+ γ1(x))minus 12(x+ γ1(x) + x minus y)

(232)

and

x=12(micro+ν) =

[2τ(xminusγ2(x))minusx+γ2(x)](x+y+a)+2(xminusy)minus2a(1+x)2[2τ(xminusγ2(x))minusx+γ2(x)minusa]

y=12(microminusν) =

[2τ(xminusγ2(x))minusx+γ2(x)](x+yminusa)minus2(xminusy)+2a(1minusy)2[2τ(xminusγ2(x))minusx+γ2(x)minusa]

x=12(micro+ν) =

12(2minusa)

[2τ(xminusγ2(x))minusx+γ2(x))(xminus yminusa)

+2(x+ y+a)minus2ax]

12(2minusa)

[2τ(xminusγ2(x))minusx+γ2(x))(minusx+ y+a)

+2(x+ yminusa)minus2ay]

(233)

Denote by z = x + iy = f(z) z = x + iy = g(z) and z = x + iy = fminus1(z) z =x+iy = gminus1(z) the transformations and their inverse transformations in (232) (233)respectively Through the transformation (230) we have

(U+V )ν=(U+V )ν (UminusV )micro=1a[aminus2σ(ν)+ν](UminusV )micro in Dprimeminus

1

(U + V )ν=2τ(micro)minus micro minus a

2minus a(U + V )ν (U minus V )micro=(U minus V )micro in Dprimeminus

2

(234)

Equation (12) in Dprimeminus can be rewritten in the form

ξν = Aξ +Bη + Cu+D ηmicro = Aξ +Bη + Cu+D z isin Dprimeminus (235)

where ξ = U + V = (ux minus uy)2 η = U minus V = (ux+ uy)2 under the transformation(234) it is clear that system (235) in Dprimeminus is reduced to

ξν=Aξ+Bη+ Cu+D ηmicro=1a[aminus2σ(ν)+ν][Aξ+Bη+Cu+D] in Dminus

1

ξν =2τ(micro)minusmicrominusa

2minus a[Aξ+Bη+Cu+D] ηmicro = Aξ+Bη+Cu+D in Dminus

2

(236)

Moreover through the transformations (232)(233) the boundary condition (223)on Lprime

1 cup Lprime4 is reduced to

Re [λ(fminus1(z))w(fminus1(z))]=r[fminus1(z)] zisinL1 Im [λ(fminus1(z3))w(fminus1(z3))]=b1

Re [λ(gminus1(z))w(gminus1(z))]=r[gminus1(z)] z isin L4 Im [λ(gminus1(z4))w(gminus1(z4))]=b2(237)

in which z3 = f(z3) z4 = g(z4) Therefore the boundary value problem (12) (inD+) (235) (227) (26) is transformed into the boundary value problem (12) (236)(227) (237) According to the proof of Theorem 21 we see that the boundary value

problem (12) (236) (227) (237) has a unique solution w(z) and then w[z(z)] is asolution of the boundary value problem (12)(22) (w = uz) in Dprimeminus and the function

u(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

2Reint z

0w(z)dz + b0 in D+

2Reint z

a

w[f(z)]dz + u(a) in Dprimeminus1

2Reint z

2w[g(z)]dz + u(2) in Dprimeminus

2

(238)

is just a solution of Problem P for (12) in Dprime where u(a) = b1 u(2) = b2

Theorem 23 If the mixed equation (12) in the domain Dprime satisfies Condition Cthen Problem P prime for (12) with the boundary condition (22) has a unique solutionu(z) as stated in (238) where z1 = l1 minus iγ1(l1) z2 = l2 minus iγ2(l2)

By using the above method and the method in Section 4 Chapter IV we candiscuss the unique solvability of Problem P prime for equation (12) in some more generaldomains Dprimeprime including the domain Dprimeprime = |z minus 1| lt 1 Im z ge 0 cup |z minus a2| lta24 Im z lt 0 cup |z minus 1minus a2| lt (2minus a)24 Im z lt 0

3 Discontinuous Oblique Derivative Problems for SecondOrder Quasilinear Equations of Mixed Type

This section deals with discontinuous oblique derivative problems for quasilinear sec-ond order equations of mixed (elliptic-hyperbolic) type in a simply connected domainFirstly we give a representation theorem and prove the uniqueness of the solutionfor the above boundary value problem and then by using the method of successiveiteration the existence of solutions for the above problem is proved

31 Formulation of discontinuous oblique derivative problems for secondorder equations of mixed type

Let D be a simply connected domain with the boundary ΓcupL1 cupL2 as stated beforewhere D+ = |zminus1| lt 1 Im z gt 0 We discuss the second order quasilinear complexequations of mixed type as stated in (12) with Condition C In order to introducethe discontinuous oblique derivative boundary value problem for equation (12) letthe functions a(z) b(z) possess discontinuities of first kind at m minus 1 distinct pointsz1 z2 zmminus1 isin Γ which are arranged according to the positive direction of Γ andZ = z0 = 2 z1 zm = 0 cup x + y = 0 x minus y = 2 Im z le 0 where m is apositive integer and r(z) = O(|z minus zj|minusβj) in the neighborhood of zj(j = 0 1 m)on Γ in which βj(j = 0 1 m) are sufficiently small positive numbers Denoteλ(z) = a(z) + ib(z) and |a(z)| + |b(z)| = 0 There is no harm in assuming that

|λ(z)| = 1 z isin Γlowast = ΓZ Suppose that λ(z) r(z) satisfy the conditions

λ(z) isin Cα(Γj) |z minus zj|βjr(z) isin Cα(Γj) j = 1 m (31)

herein Γj is the arc from the point zjminus1 to zj on Γ and z0 = 2 and Γj(j = 1 m)does not include the end points α (0 lt α lt 1) is a constant Besides there existn points E1 = a1 E2 = a2 En = an on the segment AB = L0 = (0 2) andE0 = 0 En+1 = 2 where a0 = 0 lt a1 lt a2 lt middot middot middot lt an lt an+1 = 2 Denote byA = A0 = 0 A1 = (1minus i)a12 A2 = (1minus i)a22 An = (1minus i)an2 An+1 = C =1minus i and B1 = 1minus i + (1 + i)a12 B2 = 1minus i + (1 + i)a22 Bn = 1minus i + (1 +i)an2 Bn+1 = B = 2 on the segments AC CB respectively Moreover we denoteDminus

1 = Dminuscapcup[n2]j=0 (a2j le xminusy le a2j+1) Dminus

2 = Dminuscapcup[(n+1)2]j=1 (a2jminus1 le x+y le a2j)

and Dminus2j+1 = Dminus cap a2j le x minus y le a2j+1 j = 0 1 [n2] Dminus

2j = Dminus cap a2jminus1 lex + y le a2j j = 1 [(n + 1)2] and Dminus

lowast = Dminuscupn+1j=0 (x plusmn y = aj y le 0)

Dlowast = D+ cup Dminuslowast

Problem P prime Find a continuous solution u(z) of (12) in D which is continuouslydifferentiable in Dlowast = D+ cup Dminus

lowast and satisfies the boundary conditions

12

partu

12

partu

partl= 2Re [λ(z)uz] = r(z) z isin L3 =

[n2]sumj=0

A2jA2j+1

12

partu

[(n+1)2]sumj=1

B2jminus1B2j

(32)

Im [λ(z)uz]|z=A2j+1 = c2j+1 j = 0 1 [n2]

Im [λ(z)uz]|z=B2jminus1 = c2j j = 1 [(n+ 1)2]

u(zj) = bj j = 0 1 m u(aj) = bm+j j = 1 n

(33)

where l is a vector at every point on Γ cup L3 cup L4 bj(j = 0 1 m + n)cj(j = 0 1 n + 1 c0 = b0) are real constants λ(z) = a(x) + ib(x) =cos(l x)minus i cos(l y) z isin Γ λ(z) = a(x) + ib(x) = cos(l x) + i cos(l y) z isin L3 cup L4and λ(z) r(z) cj(j = 0 1 n+ 1) satisfy the conditions

Cα[λ(z) Lj]lek0 Cα[r(z) Lj]lek2 j=34 |bj|lek2 j=0 1 m+n

cos(l n) ge 0 on Γ maxzisinL3

1|a(x)minus b(x)| le k0 max

zisinL4

1|a(x) + b(x)| le k0

(34)

where n is the outward normal vector at every point on Γ α (12 lt α lt 1) k0 k2 arenon-negative constants The above discontinuous oblique derivative boundary valueproblem for (12) is called Problem P prime Problem P prime for (12) with A3(z u uz) = 0z isin D r(z) = 0 z isin Γ cup L3 cup L4 bj = 0(j = 0 1 m + n) and cj = 0(j = 0 1 n+1) will be called Problem P prime

0 Moreover we give the same definitionsas in (510) (511) Chapter IV but choose K = (m+nminus1)2 or K = (m+n)2minus1if cos(ν n) equiv 0 on Γ and the condition u(zm) = bm can be canceled Besides werequire that the solution u(z) in D+ satisfies the conditions

uz = O(|z minus zj|minusδ) δ =

βj + τ for γj ge 0 and γj lt 0 βj gt |γj||γj|+ τ for γj lt 0 βj le |γj| j = 0 1 m+ n

(35)in the neighborhood of zj (0 le j le m) aj (1 le j le n) in D+ where γprime

j =max (0 minusγj) (j = 0 1 m + n) γprime

0 = max (0 minus2γ0) γprimem = max (0 minus2γm) and

γj (j = 0 1 m + n) are real constants as stated in (510) Chapter IV δ is asufficiently small positive number Now we explain that in the closed domain Dminusthe derivatives ux+uy ux minusuy of the solution u(z) in the neighborhoods of the 2n+2characteristic lines Z prime = x+ y = 0 x minus y = 2 x plusmn y = aj (j = 1 n) y le 0 maybe not bounded if γj le 0(j = 0 1 n+1) Hence if we require that the derivativeuz of u(z) in DminusZ prime is bounded then we need to choose γj gt 0 (j = 0 1 n+ 1)If we only require that the solution u(z) is continuous in D it suffices to chooseminus2γ0 lt 1 minus2γm lt 1 minusγj lt 1 (j = 1 m minus 1 m+ 1 m+ n)

Furthermore we need to introduce another oblique derivative boundary valueproblem

Problem Qprime If A2(z) = 0 in D we find a continuously differentiable solution u(z) of(12) inDlowast which is continuous in D and satisfies the boundary conditions (32)(33)but the point conditions in (33) are replaced by

u(2) = b0 = d0 Im [λ(z)uz]|z=zprimej= dj j = 1 m+ n (36)

where zprimej(isin Z) isin Γ(j = 0 1 m+n) are distinct points dj(j=0 1 m+n) are

real constants satisfying the conditions |dj| le k2 j = 0 1 m + n but we do notassume cos(ν n) ge 0 on each Γj(j = 1 m)

32 Representations of solutions for the oblique derivative problem for(12)

First of all we give the representation of solutions of Problem Qprime for the equationuzz

uzzlowast

= 0 in

D+

Dminus

(37)

It is clear that Problem Qprime for (37) is equivalent to the following boundary valueproblem (Problem Aprime) for the first order complex equation

Lw =

wz

wzlowast

= 0 in

D+

Dminus

(38)

Re [λ(z)w(z)] = r(z) on Γ

Im [λ(z)w(z)]|z=zprimej= bj j = 1 m+ n

Im [λ(z)w(z)]|z=A2j+1 = c2j+1 j = 0 1 [n2]

(39)

and the relation

u(z) = 2Reint z

2w(z)dz + b0 (310)

where the integral path is appropriately chosen Thus from Theorem 52 ChapterIV we can derive the following theorem

Theorem 31 The boundary value problem Qprime for (37) in D has a unique continu-ous solution u(z) as stated in (310) where the solution w(z) of Problem Aprime for (38)in Dminus possesses the form

w(z) =12[(1minus i)f(x+ y) + (1 + i)g(x minus y)]

f(x+ y) = Re [(1 + i)w(x+ y)] in DminusDminus2

g(x minus y) = Re [(1minus i)w(x minus y)] in DminusDminus1

(311)

herein w(x + y)(0 le x + y le 2) w(x minus y)(0 le x minus y le 2) are values of the solutionw(z) of Problem Aprime for (38) in D+ with the first boundary condition in (39) and theboundary condition

Re [λ(x)w(x)] =

⎧⎨⎩k(x) on Lprime1 = Dminus

1 cap AB

h(x) on Lprime2 = Dminus

2 cap AB(312)

in which k(x) h(x) can be expressed as

k(x)=2r((1minusi)x2)minus[a((1minusi)x2)+b((1minusi)x2)]h2j

h(x)=2r((1 + i)x2 + 1minus i)

minus [a((1+i)x2+1minusi)minusb((1+i)x2+1minusi)]k2jminus1

(313)

where Dminusj (j = 1 2 2n+ 1) are as stated before and

on L2j+1 j = 0 1 [n2]

on L2j j = 1 [(n+ 1)2]

2j cap AB j = 1 [(n+ 1)2]

Next we give the representation theorem of solutions of Problem Qprime for equation(12)

Theorem 32 Suppose that equation (12) satisfies Condition C Then any solutionof Problem Qprime for (12) can be expressed as

u(z) = 2Reint z

2w(z)dz + c0 w(z) = w0(z) +W (z) in D (314)

where w0(z) is a solution of Problem Aprime for the complex equation (38) with the bound-ary condition (32) (36) (w0(z) = u0z) and w(z) possesses the form

w(z)=W (z)+w0(z) inD w(z)=Φ(z)eφ(z)+ψ(z)

φ(z)=φ0(z)+Tg=φ0(z)minus 1π

int intD+

g(ζ)ζminusz

dσζ ψ(z)=Tf in D+

W (z)=Φ(z)+Ψ(z) in Dminus

Ψ(z)=

⎧⎪⎪⎪⎨⎪⎪⎪⎩

int ν

a2j+1

2j+1 j=01[n2]

int ν

2g1(z)dνe1+

int micro

a2jminus1

g2(z)dmicroe2 in Dminus2j j=1[(n+1)2]

(315)

in which e1 = (1 + i)2 e2 = (1 minus i)2 micro = x + y ν = x minus y φ0(z) is an analyticfunction in D+ Im [φ(z)] = 0 on Lprime = (0 2) and

g(z)=

f(z)=Re[A1φz]+A2u+A3

A=ReA1+ImA1

2 B=

ReA1minusImA1

2 C=A2D=A3 in D+

(316)

where Φ(z) is an analytic function in D+ and Φ(z) is a solution of the equation (38)in Dminus satisfying the boundary conditions

Re [λ(x)(Φ(x)eφ(x) + ψ(x))] = s(x) x isin L0 = (0 2)

Im [λ(z)(Φ(z)eφ(z) + ψ(z))|z=zprimej= bj j = 1 m+ n

Im [λ(z)(Φ(z) + Ψ(z))]|z=A2j+1 = 0 j = 0 1 [n2]

(317)

where s(x) can be written as in (319) below Moreover the solution u0(z) of ProblemQprime for (37) the estimate

C1[u0(z) Dminus] = Cβ[u0(z) Dminus] + C[wplusmn0 (micro ν)Y plusmn(micro ν) Dminus] le M19(k1 + k2) (318)

in which Y plusmn(z) = Y plusmn(micro ν) = Πn+1j=0 |x plusmn y minus aj|γprime

j+δ wplusmn0 (micro ν) = Rew0(z)∓ Imw0(z)

w0(z) = w0(micro ν) micro = x+y ν = xminusy are as stated in (524) Chapter IV and u0(z) isas stated in the first formula of (314) where w(z) = w0(z) M19 = M19(p0 β k0 D)is a non-negative constant

Proof Let u(z) be a solution of Problem Qprime for equation (12) and w(z) = uzu(z) be substituted in the positions of w u in (316) Thus the functions g(z) f(z)ψ(z) φ(z) in D+ and g1(z) g2(z)Ψ(z) in Dminus in (315)(316) can be determinedMoreover we can find the solution Φ(z) in D+ and Φ(z) in Dminus of (38) with theboundary condition (317) where

s(x)=2r((1minusi)x2)minus [a((1minusi)x2)+b((1minusi)x2)]h2j

a((1minusi)x2)minusb((1minusi)x2)

xisin(a2ja2j+1) j=01[n2]

s(x)=2r((1+i)x2+1minusi)minus[a((1+i)x2+1minusi)minusb((1+i)x2+1minusi)]k2jminus1

xisin(a2jminus1a2j) j=1[(n+1)2]

(319)

in which the real constants hj(j = 0 1 n) are of the form

h2j = Re [λ(A2j+1)(r(A2j+1) + ic2j+1) + Im [λ(A2j+1)(r(A2j+1) + ic2j+1)]

on L2j+1 j = 0 1 [n2]

k2jminus1 = Re [λ(B2jminus1)(r(B2jminus1) + ic2j)minus Im [λ(B2jminus1)(r(B2jminus1) + ic2j)]

on L2j j = 1 [(n+ 1)2]

where L2j+1 = Dminus2j+1capAB j = 0 1 [n2] L2j = Dminus

2jcapAB j = 1 [(n+1)2]Thus

w(z) = w0(z) +W (z) =

⎧⎨⎩ Φ(z)φ(z) + ψ(z) in D+

is the solution of Problem Aprime for the complex equationwz

wzlowast

= Re [A1w] + A2u+ A3 in

D+

Dminus

(320)

and u(z) is a solution of Problem Qprime as stated in (314)

33 Unique solvability for the discontinuous oblique derivative problemfor (12)

Theorem 33 Suppose that equation (12) satisfies Condition C Then Problem Qprime

for (12) has a unique solution in D

Proof The proof is similar to the proof of Theorems 23 and 24 Chapter V but theboundary condition on Lj(j = 1 or 2) and the point condition in which are modifiedFor instance the boundary condition on Lj(j = 1 or 2) and the point condition in(14) Chapter V are replaced by that on L3 cup L4 and Im [λ(z)w(z)]|z=A2j+1 = 0j = 0 1 [n2] Im [λ(z)w(z)]|z=B2jminus1 = 0 j = 1 [(n + 1)2] respectively theintegral in (218) Chapter V is replaced by

Ψ1(z)=

⎧⎪⎪⎪⎨⎪⎪⎪⎩int ν

a2j+1

g0(z)e1dν+int micro

0g0(z)e2dmicro in Dminus

2j+1 j=0 1 [n2]int ν

2g0(z)e1dν+

int micro

a2jminus1

g0(z)e2dmicro in Dminus2j j=1 [(n+1)2]

g0(z) = Aξ0 +Bη0 + Cu0 +D in Dminus

(321)

and so on and the characteristic lines through the points z1 = 1 minus i are re-placed by the characteristic lines through the points A2j+1(j = 0 1 [n2]) B2jminus1

(j = 1 [(n+ 1)2]

Moreover we can obtain the estimates of solutions of Problem Qprime for (12)

Theorem 34 Suppose that equation (12) satisfies Condition C Then any solutionu(z) of Problem Qprime for (12) satisfies the estimates

C1[u(z) Dminus] = Cβ[u(z) Dminus] + C[wplusmn(micro ν)Y plusmn(micro ν) Dminus] le M20

C1β[u(z) D+] = Cβ[u(z) D+] + Cβ[uzX(z) D+] le M21(k1 + k2)

C1[u(z) Dminus] le M21(k1 + k2) = M22

(322)

where

X(z) =m+nprodj=0

|z minus aj|γprimej+δ Y plusmn(z) = Y plusmn(micro ν) =

n+1prodj=0

|x plusmn y minus aj|γprimej+δ

wplusmn0 (micro ν)=Rew0(z)∓Imw0(z) w0(z)=w0(micro ν) micro=x+y ν=xminusy

(323)

in which γprimej = max (0 minusγj) (j = 1 mminus1 m+1 m+n) γprime

0 = max (0 minus2γ0) γprimem =

max (0 minus2γm) and γj (j = 0 1 m + n) are real constants as stated beforeβ (0 lt β lt δ) δ are sufficiently small positive numbers and k = (k0 k1 k2) M20 =M20(p0 β k D) M21 = M21(p0 β δ k0 D) are two non-negative constants

From the estimate (322) we can see the regularity of solutions of Problem Qprime for(12)

Next we consider the oblique derivative problem(Problem P prime) for the equation(12)

Theorem 35 Suppose that the mixed equation (12) satisfies Condition C ThenProblem P prime for (12) has a solution in D

Proof First of all we prove the uniqueness of solutions of Problem P prime for (12)Suppose that there exist two solutions of Problem P prime for (12) By Condition Cit can be seen that u(z) = u1(z) minus u2(z) and w(z) = uz satisfy the homogeneousequation and boundary conditions

wz

wzlowast

D+

Dminus

u(zj) = 0 j = 0 1 m u(aj) = 0 j = 1 n

Re [λ(z)w(z)]=0 z isin Γ Re [λ(z)w(z)]=0 zisinL3cupL4

Im [λ(z)w(z)]|z=A2j+1 = 0 j = 0 1 [n2]

(324)

By using the method of proof in Theorem 33 w(z) = uz = 0 u(z) = 0 in Dminus canbe verified Thus we have

2Re [λ(x)uz] =partu

partl= 0 on Lprime = (0 2)

it is clear that λ(x) = cos(l x)minusi cos(l y) = exp(minusiπ4) on Lprime1 and λ(x) = cos(l x)minus

i cos(l y) = exp(iπ4) on Lprime2 On the basis of the maximum principle of solutions for

the equationuzz = Re [A1uz] + A2u z isin D+ (325)

if maxD+ u(z) gt 0 then its maximum attains at a point zlowast isin Γ cup Lprime obviouslyzlowast = zj (j = 0 1 m) aj (j = 1 n) and we can prove zlowast isin Γ by the methodas stated in the proof of Theorem 34 Chapter III Moreover it is not difficult toprove that if zlowast isin Lprime then partupartl = 0 at zlowast Hence maxD+ u(z) = 0 By the similarmethod we can prove minD+ u(z) = 0 Hence u(z) = 0 u1(z) = u2(z) in D

Secondly we first prove the existence of solutions of Problem P prime for equation(12) in the linear case From Theorem 33 it can be seen that Problem Qprime for(12) has a solution ulowast(z) in D if ulowast(zj) = bj j = 0 1 m ulowast(aj) = bm+j j =1 n then the solution ulowast(z) is just a solution of Problem P prime for (12) Otherwise[ulowast(aprime

1) ulowast(aprime

m+n)] = [dlowast1 d

lowastn+m] in which aprime

j = zj j = 1 m aprimej = ajminusm j =

m+1 m+n we find m+n solutions u1(z) um+n(z) of Problem Qprime for (325)with the boundary conditions

Re [λ(z)ukz] = 0 z isin Γ Re [λ(z)ukz] = 0 z isin L3 cup L4

uk(2) = 0 Im [λ(z)ukz]|z=zprimej= δjk j k = 1 m+ n

Im [λ(z)ukz]|z=A2j+1 = 0 j = 0 1 [n2]

Im [λ(z)ukz]|z=B2jminus1 = 0 j = 1 [(n+ 1)2]

(326)

It is obvious that U(z) =summ+n

k=1 uk(z) equiv 0 in D moreover we can verify that

J =

∣∣∣∣∣∣∣∣∣u1(aprime

1) middot middot middot um+n(aprime1)

u1(aprimem+n) middot middot middot um+n(aprime

m+n)

∣∣∣∣∣∣∣∣∣ = 0thus there exist m + n real constants cprime

1 cprime2 middot middot middot cprime

m+n which are not equal to zerosuch that

cprime1u1(aprime

k) + cprime2u2(aprime

k) + middot middot middot+ cprimem+num+n(aprime

k) = dlowastk minus bk k = 1 middot middot middot m+ n

thus the function

u(z) = ulowast(z)minusm+nsumk=1

cprimekuk(z) in D

is just a solution of Problem P prime for the linear equation (12) with A2 = 0 in D Inaddition we can derive that the solution u(z) of Problem P prime for (12) satisfies theestimates similar to (322) Afterwards we consider the equation with the parametert isin [0 1]

Luz=

uzz

uzzlowast

==Re [A1uz]+ t[A2u+A3] + A(z) in

D+

Dminus

(327)

4 Problems in Multiply Connected Domains 227

where A(z) is any function in D satisfying the condition C[A(z)X(z) D+] +C[Aplusmn(micro ν)Y plusmn(micro ν) Dminus] lt infin By using the method of parameter extension namelywhen t = 0 we see that Problem P prime for such equation has a unique solution bythe above discussion Moreover assuming that when t = t0 isin (0 1] Problem P prime forequation (327) has a solution then we can prove that there exists a small positiveconstant ε such that for any t isin |tminust0| le ε t isin [0 1] Problem P prime for such equation(327) has a solution Thus we can derive that there exists a solution u(z) of ProblemP prime for equation (327) with t = 1 especially when A(z) = 0 in D ie Problem P prime forequation (12) has a solution u(z) This completes the proof

4 Oblique Derivative Problems for Quasilinear Equations ofMixed Type in Multiply Connected Domains

In this section we discuss the oblique derivative boundary value problems for quasilin-ear second order equations of mixed (elliptic-hyperbolic) type in multiply connecteddomains We first give a representation of solutions for the above boundary valueproblem and then prove the uniqueness and existence of solutions of the above prob-lem and give a priori estimates of solutions of the above problem In the book [9]2)the author proposed the Dirichlet boundary value problem (Tricomi problem) for sec-ond order equations of mixed type in multiply connected domains In [12] 1)3) theauthor only discussed the Dirichlet problem (Problem T2) for the Lavrentprimeev-Bitsadzeequation of mixed (elliptic-hyperbolic) type uxx + sgny uyy = 0 in a special doublyconnected domain Up to now we have not seen that other authors have solved itin multiply connected domains In this section we try to discuss the oblique deriva-tive problem for quasilinear equations of mixed type in multiply connected domainswhich includes the Dirichlet problem (Problem T2) as a special case

41 Formulation of the oblique derivative problem for second order equa-tions of mixed type

Let D be an N -connected bounded domain D in the complex plane CI with theboundary partD = Γ cup L where Γ =

sumNj=1 Γj isin C2

α(0 lt α lt 1) in y gt 0 withthe end points z = a1 = 0 b1 a2 b2 aN bN = 2 and L = cup2N

j=1Lj L1 = x =minusy 0 le x le 1 L2 = x = minusy + b1 b1 le x le b1 + (a2 minus b1)2 L3 = x =y+ a2 b1+(a2 minus b1)2 le x le a2 L4 = x = minusy+ b2 b2 le x le b2+(a3 minus b2)2 L2Nminus1 = x = y + aN bNminus1 + (aN minus bNminus1)2 le x le aN L2N = x = y + 2 1 lex le 2 in which a1 = 0 lt b1 lt a2 lt b2 lt middot middot middot lt aN lt bN = 2 and denoteD+ = D cap y gt 0 Dminus = D cap y lt 0 Dminus

1 = Dminus capx+ y lt b1 Dminus2 = Dminus cap b1 lt

x + y lt a2 Dminus3 = Dminus cap a2 lt x + y lt b2 Dminus

2Nminus2 = Dminus cap bNminus1 lt x + y ltaN Dminus

2Nminus1 = DminuscapaN lt x+y and z1 = 1minusi z2 = b1+(a2minusb1)(1minusi)2 zN =bNminus1+ (aN minus bNminus1)(1minus i)2 We assume that the inner angles πα2jminus1 πα2j of D+

at the points z = aj bj(j = 1 N) are greater than zero and less than π

We consider the quasilinear second orderequation of mixed type (11) and its com-plex form (12) with Condition C

The oblique derivative boundary valueproblem for equation (12) may be formu-lated as follows

Problem P primeprime Find a continuous solutionu(z) of equation (12) in D which is con-tinuously differentiable in Dlowast = DZ andsatisfies the boundary conditions

12

partu

12

partu

partl= Re [λ(z)uz] = r(z) z isin Lprime (41)

Im [λ(z)uz]|z=zj=cj j=1 N u(aj)=dj u(bj)=dN+j j=1 N

(42)where Z = xplusmny = aj xplusmny = bj j = 1 N y le 0 Lprime = cupN

j=1L2jminus1 l is a vectorat every point on ΓcupLprime λ(z) = a(x)+ ib(x) = cos(l x)∓ i cos(l y) z isin ΓcupLprime ∓ aredetermined by z isin Γ and Lprime respectively cj dj dN+j(j = 1 N) are real constantsand λ(z) r(z) cj dj dN+j(j = 1 N) satisfy the conditions

Cα[λ(z)Γ]lek0 Cα[r(z)Γ]lek2 Cα[λ(z) Lprime]lek0 Cα[r(z) Lprime]lek2

cos(l n) ge 0 on Γ |cj| |dj| |dN+j| le k2 j = 1 N

maxzisinL1

zisinLprimeprime

1|a(x) + b(x)| le k0

(43)

in which n is the outward normal vector on Γ Lprimeprime = cupNj=2L2jminus1 α (12 lt α lt 1)

k0 k2 are non-negative constants Here we mention that if A2 = 0 in D+ then we cancancel the condition cos(l n) ge 0 on Γ and if the boundary condition Re [λ(z)uz] =r(z) z isin Lprime is replaced by Re [λ(z)uz] = r(z) z isin Lprime then Problem P primeprime does notinclude the Dirichlet problem (Tricomi problem)

The boundary value problem for (12) with A3(z u uz) = 0 z isin D u isin IR uz isin CIr(z) = 0 z isin Γ cup Lprime and cj = 0 (j = 0 1 N) and dj = 0 (j = 1 2N) will becalled Problem P primeprime

0 The number

K =12(K1 +K2 + middot middot middot+K2N) (44)

is called the index of Problem P primeprime and Problem P primeprime0 on the boundary partD+ of D+

where

Kj=[φj

π

γj=φj

πminusKj j=1N (45)

in which [a] is the largest integer not exceeding the real number a and t1 = a1 = 0t2 = b1 t3 = a2 t4 = b2 t2Nminus1 = aN t2N = bN = 2 and

λ(t) = eiπ4 on lj = (aj bj) λ(t2jminus1 + 0) = λ(t2j minus 0) = eiπ4 j = 1 N

If cos(l n) equiv 0 on each of Γj (j = 1 N) then we select the index K = N minus 1 onpartD+ If cos(l n) equiv 0 on Γj (j = 1 N) then we select the index K = N2 minus 1on partD+ and the last N point conditions in (42) can be eliminated In this caseProblem P includes the Dirichlet problem (Tricomi problem) as a special case Nowwe explain that in the closed domain Dminus the derivative ux plusmn uy of the solution u(z)in the neighborhoods of 4N characteristic lines x plusmn y = aj x plusmn y = bj(j = 1 N)may not be bounded if γjαj le 0(j = 1 2N) Hence if we require that the solutionu(z) in DminusZ is bounded where Z = x + y = aj x + y = bj x minus y = aj x minus y =bj y le 0 (j = 1 N) then it needs to choose γj gt 0 (j = 1 2N) hereinγj (j = 1 2N) are as stated in (45) If we require that solution u(z) is onlycontinuous in D it suffices to choose minusγjαj lt 1 (j = 1 2N)

Moreover we need to introduce another oblique derivative boundary value prob-lem

Problem Qprimeprime If A2 = 0 in D one has to find a continuously differentiable solutionu(z) of (12) in Dlowast which is continuous in D and satisfies the boundary conditions(41)(42) but the last N point conditions are replaced by

Im [λ(z)uz]|z=zprimej= dprime

j j = 1 N (46)

where zprimej(j = 1 N minus 1) are distinct points such that zprime

j isin Γ zprimej isin Z (j =

1 N minus 1) and dprimej (j = 1 N) are real constants satisfying the conditions

|dprimej| le k2 j = 1 N In the case the condition cos(l n) ge 0 on Γ can be can-

celed and we choose the index K = N minus 1

42 Representation and uniqueness of solutions for the oblique derivativeproblem for (12)

Now we give representation theorems of solutions for equation (12)

Theorem 41 Suppose that equation (12) satisfies Condition C Then any solutionof Problem P primeprime for (12) can be expressed as

u(z) = 2Reint z

0w(z)dz + d1 w(z) = w0(z) +W (z) (47)

where w0(z) is a solution of Problem A for the equation

wz

wzlowast

= 0 in

D+

Dminus

(48)

with the boundary conditions (41) and (42)(w0(z) = u0z) and W (z) possesses theform

W (z) = w(z)minus w0(z) w(z) = Φ(z)eφ(z) + ψ(z)

int intD+

g(ζ)ζ minus z

Ψ(z) =

⎧⎪⎪⎨⎪⎪⎩int micro

0g2(z)dmicroe2 +

int ν

2g1(z)dνe1 in Dminus

2jminus1 j = 1 2 Nint micro

0g2(z)dmicroe2 +

int ν

aj+1

g1(z)dνe1 in Dminus2j j = 1 2 N minus 1

(49)

in which Im [φ(z)] = 0 on L0 = cupNj=1lj lj = (aj bj) j = 1 N e1 = (1 + i)2

e2 = (1minus i)2 micro = x+ y ν = x minus y φ0(z) is an analytic function in D+ and

g(z)=

A12+A1w(2w) w(z) =00 w(z) = 0 z isin D+

g1(z) = g2(z) = Aξ +Bη + Cu+D ξ = Rew + Imw η = Rew minus Imw

A =ReA1 + ImA1

2 B =

ReA1 minus ImA1

2 C = A2 D = A3 in Dminus

(410)where Φ(z) is analytic in D+ and Φ(z) is a solution of equation (48) in Dminus satisfyingthe boundary conditions

Re [λ(z)Φ(z)] = minusRe [λ(z)Ψ(z)] z isin Lprime

Im [λ(zj)Φ(zj)] = minusIm [λ(zj)Ψ(zj)] j = 1 N

(411)

where λ(x) = 1+i on L0 and s(x) is as stated in (414) below Moreover the solutionu0(z) of Problem P primeprime for (48) in Dminus satisfies the estimate in the form

Cβ[u0(z) D] + Cβ[X(z)w(z) D] + C[wplusmn0 (micro ν)Y plusmn(micro ν) Dminus] le M23(k1 + k2) (412)

in which X(z) = Π2Nj=1[|zminus tj||γj |αj+δ Y plusmn(z) = Y plusmn(micro ν) = Π2N

j=1[|microminus tj||ν minus tj|]|γj |αj+δδ β (0 lt β lt δ) γj (j = 1 2N) are as stated in (45) wplusmn

0 (micro ν) = Rew0(z) ∓Imw0(z) w0(z) = w0(micro ν) micro = x+ y ν = x minus y and

u0(z) = 2Reint z

0w0(z)dz + d1 (413)

and M23 = M23(p0 β k0 D) is a non-negative constant

Proof Let u(z) be a solution of Problem P primeprime for equation (12) and w(z) = uzu(z) be substituted in the positions of w u in (410) thus the functions g(z) f(z)ψ(z) φ(z) in D+ and g1(z) g2(z)Ψ(z) in Dminus in (49)(410) can be determinedMoreover we can find the solution Φ(z) in D+ and Φ(z) in Dminus of (48) with theboundary conditions (411) where

s(x)=2r((1minusi)x2)minus2Re [λ((1minusi)x2)Ψ((1minusi)x2)]minush(x)

a((1minus i)x2)minus b((1minus i)x2)+Re [λ(x)Ψ(x)] (414)

on L0 in which

h(x)=[a

((1minusi)x2

)+b

((1minusi)x2

)][Re (λ(z1)(r(z1)+ic1))+Im (λ(z1)(r(z1)+ic1))]

and

Ψ(x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int ν

2g1(z)dνe1 =

int ν

2g1((1minus i)ν2)dνe1 z = x+ iy = (1minus i)x isin L1int micro

0g2(z)dmicroe2 =

int micro

0g2((1 + i)micro2 + (1minus i)aj2))dmicroe2

z = x+ iy = (1 + i)x minus aji isin L2jminus1 j = 2 Nint micro

0g2(z)dmicroe2=

int micro

0g2((1+i)micro2+1minusi))dmicroe2 z=(1+i)xminus2iisinL2N

Thus

w(z) = w0(z) +W (z) =

⎧⎨⎩ Φ(z)φ(z) + ψ(z) in D+

is the solution of Problem A for the complex equationwz

wzlowast

= Re [A1w] + A2u+ A3 in

D+

Dminus

(415)

and u(z) is a solution of Problem P for (12) as stated in the first formula in (47)

Theorem 42 Suppose that equation (12) satisfies Condition C Then ProblemP primeprime for (12) has at most one solution in D

Proof Let u1(z) u2(z) be any two solutions of Problem P primeprime for (12) It is clearthat u(z) = u1(z) minus u2(z) and w(z) = uz satisfies the homogeneous equation andboundary conditions

wz

wzlowast

= Re [A1w] + A2u in

D+

Dminus

Re [λ(z)w(z)] = 0 z isin Γ Re [λ(x)w(x)] = s(x) x isin L0 u(0) = 0

Re [λ(z)w(z)] = 0 z isin Lprime Im [λ(zj)w(zj)] = 0 j = 1 N

(416)

in which

s(x)=minus2Re [λ((1minus i)x2)Ψ((1minus i)x2)]

a((1minus i)x2)minus b((1minus i)x2)+ Re [λ(x)Ψ(x)] on L0 (417)

w(z)=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Φeφ(z)+ψ(z) ψ(z)=Tf φ(z)= φ0(z)+T g in D+

Φ(z) + Ψ(z) in Dminus

Ψ(z)=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int micro

0[Aξ+Bη+Cu]edmicroe2+

int ν

2[Aξ+Bη+Cu]dνe1 in Dminus

2jminus1

j = 1 2 Nint micro

0[Aξ+Bη+Cu]edmicroe2 +

int ν

aj+1

[Aξ+Bη+Cu]dνe1 in Dminus2j

j = 1 2 N minus 1

(418)

where g(z) is as stated in (410) Φ(z) in D+ is an analytic function and Φ(z) is asolution of (48) in Dminus satisfying the boundary condition (411) in which W (z) =w(z) If A2 = 0 in D+ then ψ(z) = 0 besides the functions Φ(z) Φ(z) satisfy theboundary conditions⎧⎨⎩Re [λ(x)Φ(x)] = s(x)

Re [λ(x)Φ(x)] = Re [λ(x)(W (x)minusΨ(x))]x isin L0 (419)

where s(x) is as stated before Noting that

C[u(z) Dminus] le M24C[X(z)w(z) D+] + C[wplusmn(z)Y plusmn(z) Dminus]and applying the method of iteration we can get

|wplusmn(z)Y plusmn(z)| le [2NM25M((4 +M24)m+ 1)Rprime]n

n in Dminus

where M24 = M24(D) M25 = maxDminus [|A| |B| |C|] M = 1 + 4k20(1 + k2

0) m =C[w(z) Dminus] Rprime = 2 Let n rarr infin we can derive wplusmn(z) = 0 ie w(z) =w1(z) minus w2(z) = 0 u(z) = 0Ψ(z) = Φ(z) = 0 in Dminus and s(x) = 0 on L0Besides noting that the solution u(z) of the equation

uzz = Re [A1uz] + A2u in D+ (420)

12

partu

partl=Re[λ(z)uz]=0 zisinΓcupL0 u(aj)=0 u(bj)=0 j=1N (421)

and the index of the above boundary value problem is K = N minus 1 on the basis ofTheorem 37 Chapter III we see that u(z) = 0 in D+ This proves the uniquenessof solutions of Problem P primeprime for (12) in D As for the general equation (12) we canprove the uniqueness of solutions of Problem P primeprime by the extremum principle for ellipticequations of second order by the method in the proof of Theorem 34 Chapter III

43 The solvability for the oblique derivative problem for (12)

First of all we prove the existence of solutions of Problem P primeprime for equation (37) inD It is obvious that Problem P primeprime for equation (37) is equivalent to the followingboundary value problem (Problem Aprimeprime) for (38) with the boundary conditions

Re [λ(z)w(z)] = r(z) on Γ Re [λ(z)w(z)] = r(z) on Lprime

Im [λ(z)w(z)]|z=zj= cj j = 1 N

u(aj) = dj u(bj) = dN+j j = 1 N

(422)

and the relation

u(z) = 2Reint z

0w(z)dz + d1 (423)

Similarly to the method in the proof of Theorem 31 we can get the following theorem

Theorem 43 Problem P primeprime for (37) in D has a unique continuous solution u(z)

Proof From the second and third boundary conditions in (422) we can obtain thefollowing conditions

Re [λ(x)w(x)] = k(x) on L0

k(x) =2r((1minus i)x2)minus [a((1minusi)x2)+b((1minusi)x2)]h

a((1minus i)x2)minus b((1minus i)x2)on L0

h = [Re (λ(z1)(r(z1)+ic1))+Im (λ(z1)(r(z1)+ic1))]

(424)

According to the method in the proof of Theorem 22 Chapter III we can find asolution w(z) of (38) in D+ with the first boundary condition in (422) and (424)Thus we can find the solution of Problem P primeprime for (37) in D as stated in (423) andthe solution w(z) of Problem Aprimeprime for (38) in Dminus

1 possesses the form

w(z) = w(z) + λ(z1)[r(z1)minus ic1]

f(x+ y) = Re [(1 + i)w(x+ y)]

a((1minusi)(xminusy)2)minus b((1minusi)(xminusy)2)in Dminus

1 0 b1

Similarly we can write the solution of Problem P primeprime in Dminusj (j = 2 3 2N minus 1) as

w(z)= w(z)+λ(zj)[r(zj)minusicj]

w(z)=12[(1minusi)fj(x+y)+(1+i)gj(xminusy)] in Dminus

j j=2N

f2j(x+y)=2r((1+i)(x+y)2+(1minusi)aj+12)

a((1+i)(x+y)+(1minusi)aj+12)minusb((1+i)(x+y)2+(1minusi)aj+12)

g2j(xminusy)=2r((1minusi)(xminusy)2)

a((1minusi)(xminusy)2)+b((1minusi)(xminusy)2)in Dminus

2j j=12Nminus1

f2jminus1(x+y)=Re[(1minusi)w(x+y)]

g2jminus1(xminusy)=2r((1minusi)(xminusy)2)

a((1minusi)(xminusy)2)minusb((1minusi)(xminusy)2)in Dminus

2jminus1 j=2N

(425)in which Dminus

j (j = 2 3 2N minus 1) are as stated before

Theorem 44 Suppose that the mixed equation (12) satisfies Condition C ThenProblem P primeprime for (12) in D has a solution

Proof It is clear that Problem P primeprime for (12) is equivalent to Problem Aprimeprime for thecomplex equation of first order and boundary conditions

wz

wzlowast

= Re [A1w] + A2u+ A3 in

D+

Dminus

(426)

Re [λ(z)w(z)] = r(z) z isin Γ Re [λ(z)w(z)] = r(z) z isin Lprime

Im [λ(zj)w(zj)] = cj u(aj)=dj u(bj)=dN+j j=1 N(427)

and the relation (423) In order to find a solution w(z) of Problem Aprimeprime for (426)in D we express w(z) in the form (49)ndash(410) and use the successive iterationFirst of all denoting the solution w0(z) of Problem Aprimeprime for (426) and substitutingw0(z)(= ξ0e1 + η0e2) and the corresponding function u0(z) into the positions of w(z)(= ξe1 + ηe2) u(z) in the right hand side of (426)(49) and (410) thus the corres-ponding functions g0(z) f0(z) and the functions

int intD+

g0(ζ)ζ minus z

w1(z) = w0(z) +W1(z) W1(z) = Φ1(z) + Ψ1(z) (428)g0(z) = Aξ0 +Bη0 + Cu0 +D in Dminus

Ψ1(z) =

0g0(z)dmicroe2 +

int ν

0g0(z)dmicroe2 +

int ν

aj+1

can be determined where micro = x+ y ν = x minus y and the solution w0(z) = u0z u0(z)satisfies the estimate (412) Moreover we find a solution Φ1(z)Φ1(z) of (48) satis-fying the boundary conditions

Re [λ(z)Φ1(z)] = minusRe [λ(z)Ψ1(z)] z isin Lprime

Im [λ(zj)Φ1(zj)] = minusIm [λ(zj)Ψ1(zj)] j = 1 N

(429)

where λ(x) = 1 + i on L0 and the function

s1(x)=2r((1minusi)x2)minus2Re [λ((1minusi)x2)Ψ1((1minusi)x2)]minush(x)

+Re [λ(x)Ψ1(x)]

h(x) =[a

((1minus i)x

2

)+ b

((1minus i)x

2

)][Re (λ(z1)(r(z1) + ic1))

+Im (λ(z1)(r(z1) + ic1))] on L0

(430)

in which w1(z) satisfies the estimate

Cβ[u1(z) D] + Cβ[X(z)w1(z) D+] + C[wplusmn1 (micro ν)Y plusmn(micro ν) Dminus] le M26 (431)

here wplusmn1 (micro ν) = Rew1(micro ν) ∓ Imw1(micro ν) X(z) Y plusmn(micro ν) is as stated in (412)

M26 = M26(p0 β k D) is a non-negative constant Thus we can obtain a sequence offunctions wn(z) and

wn(z) = w0(z) +Wn(z) = w0(z) + Φn(z) + Ψn(z) in Dminus

Ψn(z) =

0gnminus1(z)dmicroe2 +

int ν

2gnminus1(z)dνe1 in Dminus

int ν

aj+1

gnminus1(z)dνe1 in Dminus2j j = 1 2 N minus 1

gnminus1(z) = Bξnminus1 + Aηnminus1 + Cunminus1 +D in Dminus(432)

and then|[wplusmn

1 (micro ν)minus wplusmn0 (micro ν)]Y plusmn(micro ν)|

le |Φplusmn1 (micro ν)Y plusmn(micro ν)|+radic

2|Y +(micro ν)| |int micro

0g0(z)e2dmicro|

+|Y minus(micro ν)|[|int ν

2g0(z)e1dν|+

Nminus1sumj=1

|int micro

aj+1

g0(z)e1dν|]

le 2NM27M((4 +M24)m+ 1)Rprime in Dminus

(433)

wherem = C[w+0 (micro ν)Xplusmn(micro ν) Dminus]+C[wminus

0 (micro ν)Y plusmn(micro ν) Dminus] M = 1+4k20(1+k2

0)Rprime = 2 M27 = maxzisinDminus(|A| |B| |C| |D|) It is clear that wn(z)minus wnminus1(z) satisfies

wn(z)minuswnminus1(z)=Φn(z)minusΦnminus1(z)+Ψn(z)minusΨnminus1(z) inDminus

Ψn(z)minusΨnminus1(z)=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int micro

0[gn(z)minusgnminus1(z)]dmicroe2+

int ν

2[gn(z)minusgnminus1(z)]dνe1 inDminus

2jminus1

j=12Nint micro

int ν

aj+1

[gn(z)minusgnminus1(z)]dνe1 inDminus2j

j=12N minus1(434)

where n = 1 2 From the above equality

|[wplusmnn (micro ν)minus wplusmn

nminus1(micro ν)]Y plusmn(micro ν)|

le [2NM27M((4 +M24)m+ 1)]nint Rprime

0

Rprimenminus1

(n minus 1) dRprime

le [2NM27M((4 +M24)m+ 1)Rprime]n

n in Dminus

(435)

can be obtained and we can see that the sequences of functions wplusmnn (micro ν)Y plusmn(micro ν)

ie

wplusmnn (micro ν)Y plusmn(micro ν) = wplusmn

0 + [wplusmn1 minus wplusmn

0 ] + middot middot middot+ [wplusmnn minus wplusmn

nminus1]Y plusmn(micro ν) (436)

(n = 1 2 ) inDminus uniformly converge to wplusmnlowast (micro ν)Xplusmn(micro ν) and wlowast(z) = [w+(micro ν)+

wminus(micro ν) minusi(w+(micro ν)minus wminus(micro ν))]2 satisfies the equality

wlowast(z)=w0(z)+Φlowast(z)+Ψlowast(z) in Dminus

Ψlowast(z)=

⎧⎪⎪⎪⎨⎪⎪⎪⎩int micro

0glowast(z)dmicroe2+

int ν

2glowast(z)dνe1 in Dminus

int ν

aj+1

glowast(z)dνe1 in Dminus2j j = 1 2 N minus 1

glowast(z) = Bξlowast + Aηlowast + Culowast +D in Dminus

(437)

and the corresponding function ulowast(z) is just a solution of Problem P primeprime for equation(12) in the domain Dminus and wlowast(z) satisfies the estimate

C[wplusmnlowast (micro ν)Y plusmn(micro ν) Dminus] le M28 = e2NM27M((4+M24)m+1)Rprime

(438)

Besides we see that the solution wlowast(z) = uz of Problem Aprimeprime for (426) and the cor-responding function ulowast(z) in D satisfy the estimate

Cβ[ulowast(z) D] + Cβ[wlowast(z)X(z) D+] + C[wplusmnlowast (micro ν)Y plusmn(micro ν) Dminus] le M29 (439)

where M29 = M29(p0 α k D) is a non-negative constant Moreover the function u(z)in (423) is a solution of Problem P primeprime for (12)

From the proof of Theorem 44 we can obtain the estimates of any solution u(z)of Problem P primeprime and the corresponding function w(z) = uz

Theorem 45 Suppose that equation (12) satisfies Condition C Then any solutionu(z) of Problem P primeprime for (12) satisfies the estimates

C1[u(z) D] = Cβ[u(z) D] + Cβ[X(z)w(z) D+]

+C[wplusmn(micro ν)Y plusmn(micro ν) Dminus] le M30 C1β[u(z) D] le M31(k1 + k2)

(440)

where

X(z) =2Nprodj=1

|z minus tj||γj |αj+δ Y plusmn(z) = Y plusmn(micro ν) =2Nprodj=1

[|micro minus tj||ν minus tj|]|γj |αj+δ (441)

herein γj(j = 1 2N) are real constants in (45) β(0 lt β lt δ) δ are suffi-ciently small positive constants and k = (k0 k1 k2) M30 = M30(p0 β k D) M31 =M31(p0 β δ k0 D) are two non-negative constants

Next we consider the oblique derivative problem (Problem Qprimeprime) for the equa-tion (12)

Theorem 46 Suppose that the mixed equation (12) with A2 = 0 satisfies ConditionC Then its Problem Qprimeprime has a solution in D

Proof It is clear that Problem Qprimeprime is equivalent to the following boundary valueproblem

wz

wzlowast

= Re [A1w(z)] + A3 in

D+

Dminus

(442)

Im [λ(zprimej)w(z

primej)] = dprime

j j = 1 N(443)

Noting that w(z) satisfies the second boundary condition in (411) namely

Re [λ(x)w(z)] = s(x)s(x) = Re [λ(x)Ψ(x)] (444)

+2r((1minusi)x2)minus2Re [λ((1minus i)x2)Ψ((1minusi)x2)]minush(x)

a((1minusi)x2)minusb((1minusi)x2)on L0

and the index K = N minus 1 similarly to the proof of Theorems 23 and 24 ChapterV we can find a unique solution w(z) of the boundary value problem (442)ndash(444)and the function u(z) in (423) is just a solution of Problem Qprimeprime in D

Finally we mention that the above result includes the Dirichlet problem (Tricomiproblem) as a special case In fact if Γ1 = |z minus 1| = 1 Γj = |z minus aj| = Rjaj = bjminus1 +(aj minus bjminus1)2 Rj = (aj minus bjminus1)2 j = 2 N R1 = 1 the boundarycondition of the Dirichlet problem is

u(z) = φ(x) on Γ cup Lprime (445)

which can be rewritten as

Re [λ(z)w(z)] = r(z) on Γ Im [λ(zj)w(zj)] = cj j = 1 N

Re [λ(z)w(z)] = r(z) on Lprime u(z) = 2Reint z

0w(z)dz + d0 in D

(446)

in which d0 = φ(0)

λ(z) = a+ ib =

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

i(z minus 1) θ = arg(z minus 1) on Γ1

i(zminusaj)Rj θ=arg(zminusaj) on Γj j=2 N

(1 + i)radic2 on L1

(1minus i)radic2 on Lprimeprime = cupN

j=2L2jminus1

(447)

and

r(z) =

⎧⎪⎨⎪⎩φθRj on Γj j = 1 N

φxradic2on Lprime = cupN

j=1L2jminus1

c1=Im[1 + iradic2

uz(z1 minus 0)]=

φx minus φxradic2

|z=z1minus0=0

cj = Im[1minus iradic2

uz(zj + 0)]= 0 j = 2 N

(448)

radic2 = minusb = minus1radic2 on Lprimeprime

If the index of Problem D on partD+ is K = N2 minus 1 we can argue as followsAccording to (446)(447) the boundary conditions of Problem D in D+ possess theform

Re [i(z minus 1)w(z)] = r(z) = φθ z isin Γ

Re[1minus iradic2

w(x)]= s(x) =

x isin lj = (aj bj) j = 1 N

It is clear that the possible points of discontinuity of λ(z) on partD+ are t1 = a1 =0 t2 = b1 t3 = a2 t4 = b2 t2Nminus1 = aN t2N = bN = 2 and

λ(a1 minus 0) = λ(bj + 0) = eiπ2 j = 1 2 N minus 1

λ(bN + 0) = λ(aj minus 0) = e3iπ2 j = 2 N

λ(aj + 0) = λ(bj minus 0) = eiπ4 j = 1 N

= eiπ4 = eiφ1 0 lt γ1 =φ1

πminus K1 =

14

minus 0 =14

lt 1

λ(t2N minus 0)λ(t2N + 0)

=eminus5iπ4=eiφ2N minus1ltγ2N=φ2N

πminusK2N=minus5

4minus(minus1)=minus1

4lt0

= e5iπ4 = eiφj 0ltγj=φj

πminus Kj=

54minus1= 1

4lt1 j=3 5 2Nminus1

=eminusiπ4=eiφj minus1ltγj=φj

πminus Kj=minus1

4=minus1

4lt0 j=2 4 2Nminus2

Thus K1 = K2 = K4 = middot middot middot = K2Nminus2 = 0 K3 = K5 = middot middot middot = K2Nminus1 = 1 K2N = minus1in the case where the index of Problem D on partD+ is

K =12(K1 +K2 + middot middot middot+K2N) =

N

2minus 1

Hence Problem D for (12) has a unique continuous solution u(z) in D If we requirethat the derivative w(z) = uz of the solution u(z) is bounded in DZ it suffices toreplace K2 = K4 = middot middot middot = K2Nminus2 = minus1 K2N = minus2 and then the index K = minus1 Inthis case the problem has one solvability condition

The oblique derivative problems for the Chaplygin equation of second order

K(y)uxx + uyy = 0 K(0) = 0 K prime(y) gt 0 in D

where D is a multiply connected domain was proposed by L Bers in [9]2) but theproblem was still not solved For more general second order quasilinear degenerateequations of mixed type and second order mixed equations in higher dimensionaldomains many boundary value problems need to be investigated and solved

The references for this chapter are [4][6][12][14][27][31][37][39][45][56][61][64][71][72][73][85][88][90][98]

References

[1] R Adams Sobolev spaces Academic Press New York 1975

[2] S Agmon L Nirenberg and M H Protter A maximum principle for a class ofhyperbolic equations and applications to equations of mixed ellipticndashhyperbolictype Comm Pure Appl Math 6 (1953) 455ndash470

[3] H Begehr 1) Boundary value problems for mixed kind systems of firstorder partial differential equations 3 Roumanian-Finnish Seminar on ComplexAnalysis Bucharest 1976 Lecture Notes in Math 743 Springer-Verlag Berlin1979 600ndash6142) Complex analytic methods for partial differential equations World ScientificSingapore 1994

[4] H Begehr and A Dzhureav An introduction to several complex variables andpartial differential equations Addison Wesley Longman Harlow 1997

[5] H Begehr and R P Gilbert 1) Pseudohyperanalytic functions Complex Vari-ables Theory Appl 9 (1988) 343ndash3572) Transformations transmutations and kernel functions I II Longman Har-low 1992 1993

[6] H Begehr and A Jeffrey 1) Partial differential equations with complex analysisLongman Harlow 19922) Partial differential equations with real analysis Longman Harlow 1992

[7] H Begehr and Wei Lin A mixed-contact boundary problem in orthotropic elas-ticity Partial Differential Equations with Real Analysis Longman Harlow1992 219ndash239

[8] H Begehr and Guo-chun Wen 1) The discontinuous oblique derivative problemfor nonlinear elliptic systems of first order Rev Roumaine Math Pures Appl33 (1988) 7ndash192) A priori estimates for the discontinuous oblique derivative problem for ellipticsystems Math Nachr 142(1989) 307ndash3363) Nonlinear elliptic boundary value problems and their applications AddisonWesley Longman Harlow 1996

[9] L Bers 1) Theory of pseudoanalytic functions New York University New York19532) Mathematical aspects of subsonic and transonic gas dynamics Wiley NewYork 1958

References 241

[10] L Bers F John and M Schechter Partial differential equations IntersciencePubl New York 1964

[11] L Bers and L Nirenberg 1) On a representation theorem for linear ellipticsystems with discontinuous coefficients and its application Conv Intern EqLin Derivate Partiali Trieste Cremonense Roma 1954 111ndash1402) On linear and nonlinear elliptic boundary value problems in the plane ConvIntern Eq Lin Derivate Partiali Trieste Cremonense Roma 1954 141ndash167

[12] A V Bitsadze 1) Differential equations of Mixed type Mac Millan Co NewYork 19642) Boundary value problems for elliptic equations of second order NaukaMoscow 1966 (Russian) Engl Transl North Holland Publ Co Amsterdam19683) Some classes of partial differential equations Gordon and Breach New York19884) Partial Differential Equations World Scientific Singapore 1994

[13] A V Bitsadze and A N Nakhushev Theory of degenerating hyperbolic equa-tions Dokl Akad Nauk SSSR 204 (1972) 1289ndash1291 (Russian)

[14] B Bojarski 1)Generalized solutions of a system of differential equations of firstorder and elliptic type with discontinuous coefficients Mat Sb N S 43(85)(1957) 451ndash563 (Russian)2) Subsonic flow of compressible fluid Arch Mech Stos 18 (1966) 497ndash520Mathematical Problems in Fluid Mechanics Polon Acad Sci Warsaw 19679ndash32

[15] B Bojarski and T Iwaniec Quasiconformal mappings and non-linear ellipticequations in two variables I II Bull Acad Polon Sci Ser Sci Math AstrPhys 22 (1974) 473ndash478 479ndash484

[16] F Brackx R Delanghe and F Sommen Clifford analysis Pitman London1982

[17] S A Chaplygin Gas jets Complete Works Moscow-Leningrad Vol 2 1933

[18] D Colton 1) Partial differential equations in the complex domain PitmanLondon 19762) Analytic theory of partial differential equations Pitman London 1980

[19] R Courant and H Hilbert Methods of mathematical physics II IntersciencePubl New York 1962

[20] I I Daniljuk 1) Nonregular boundary value problems in the plane Izdat NaukaMoscow 1975 (Russian)2) Selected works Naukova Dumka Kiev 1995 (Russian etc)

242 References

[21] Ju V Devingtalprime Existence and uniqueness of the solution of the Frankl prob-lem Uspehi Mat Nauk 14 (1959) no1 (85) 177ndash182 (Russian)

[22] A Dzhuraev 1) Methods of singular integral equations Nauk SSSR Moscow1987 (Russian) Engl transl Longman Harlow 19922) Degenerate and other problems Longman Harlow 1992

[23] Zheng-zhong Ding The general boundary value problem for a class of semilinearsecond order degenerate elliptic equations Acta Math Sinica 27 (1984) 177ndash191(Chinese)

[24] Guang-chang Dong 1) Boundary value problems for degenerate elliptic partialdifferential equations Sci Sinica 13 (1964) 697ndash7082) Nonlinear second order partial differential equations Amer Math SocProvidence RI 1991

[25] Guang-chang Dong and Min-you Chi Influence of Tricomirsquos Mathematical workin China Mixed Type Equations BSB Teubner Leipzig 90 1986 105ndash111

[26] A Douglis and L Nirenberg Interior estimates for elliptic systems of partialdifferential equations Comm Pure Appl Math 8 (1953) 503ndash538

[27] Ai-nong FangQuasiconformal mappings and the theory of functions for systemsof nonlinear elliptic partial differential equations of first order Acta MathSinica 23 (1980) 280ndash292 (Chinese)

[28] F I Frankl 1) On the problems of Chaplygin for mixed subsonic and supersonicflows Izv Akad Nuak SSSR Ser Mat 9 (1945) 121ndash1432) Two gas-dynamical applications of the Lavrentprimeev-Bitsadze boundary valueproblem Vest Leningrad Univ Ser Mat Meh Astronom 6 (1951) 3ndash7 (Rus-sian)3) Gas flows past profiles with a local supersonic zone ending in a direct shockwave Prikl Mat Meh 20 (1956) 196ndash202 (Russian)4) Selected works on gas dynamics Nauka Moscow 1973 (Russian)

[29] A Friedman Variational principles and free boundary problems Wiley NewYork 1982

[30] F D Gakhov Boundary value problems Fizmatgiz Moscow 1963 (Russian)Pergamon Oxford 1966

[31] D Gilbarg and N S Trudinger Elliptic partial differential equations of secondorder Springer-Verlag Berlin 1977

[32] R P Gilbert 1) Function theoretic methods in partial differential equationsAcademic Press New York 19692) Plane ellipticity and related problems Amer Math Soc Providence RI1981

References 243

[33] R P Gilbert and J L Buchanan First order elliptic systems A functiontheoretic approach Academic Press New York 1983

[34] R P Gilbert and G N Hile 1) Generalized hypercomplex function theoryTrans Amer Math Soc 195 (1974) 1ndash292) Degenerate elliptic systems whose coefficients matrix has a group inverseComplex Variables Theory Appl 1 (1982) 61ndash88

[35] R P Gilbert and Wei Lin 1) Algorithms for generalized Cauchy kernels Com-plex Variables Theory Appl 2 (1983) 103ndash1242) Function theoretic solutions to problems of orthotropic elasticity J Elasticity15 (1985) 143ndash154

[36] R P Gilbert and Guo-chun Wen 1) Free boundary problems occurring in planarfluid dynamics Nonlinear Analysis 13 (1989) 285ndash3032)Two free boundary problems occurring in planar filtrations Nonlinear Analy-sis 21 (1993) 859ndash868

[37] Chao-hao Gu and Jia-xing Hong Some developments of the theory of partial dif-ferential equations of mixed typeMixed Type Equations BSB Teubner Leipzig90 1986 120ndash135

[38] W Haack and W Wendland Lectures on Pfaffian and partial differential equa-tions Pergamon Press Oxford 1972

[39] G N Hile and M H Protter 1) Maximum principles for a class of first orderelliptical systems J Diff Eq 24(1) (1977) 136ndash1512) Properties of overdetermined first order elliptic systems Arch Rat MechAnal 66 (1977) 267ndash293

[40] Zong-yi Hou Dirichlet problem for a class of linear elliptic second order equa-tions with parabolic degeneracy on the boundary of the domain Sci Record (NS) 2 (1958) 244ndash249 (Chinese)

[41] L Hormander Linear partial differential operators Springer-Verlag Berlin1964

[42] G C Hsiao and W Wendland A finite element method for some integral equa-tions of the first kind J Math Anal Appl 58 (1977) 449ndash481

[43] Loo-Keng Hua 1) On Lavrentprimeevrsquos partial differential equation of the mixedtype Sci Sinica 13 (1964) 1755ndash1762 (Chinese)2) A talk starting from the unit circle Science Press Beijing 1977 (Chinese)

[44] Loo-keng Hua Wei Lin and Tzu-chien Wu Second order systems of partialdifferential equations in the plane Pitman London 1985

244 References

[45] T Iwaniec Quasiconformal mapping problem for general nonlinear systems ofpartial differential equations Symposia Math 18 Acad Press London 1976501ndash517

[46] T Iwaniec and A Mamourian On the first order nonlinear differential systemswith degeneration of ellipticity Proc Second Finnish-Polish Summer School inComplex Analysis (Jyvaskyla 1983) Univ Jyvaskyla 28 (1984) 41ndash52

[47] Xin-hua Ji and De-quan Chen 1) The equations of mixed type of elliptic andhyperbolic in the nndashdimensional real projective space Acta Math Sinica 23(1980) 908ndash9212) The non-homogeneous equations of mixed type in the real projective planerMixed Type Equations BSB Teubner Leipzig 90 1986 280ndash300

[48] M V Keldych On certain cases of degeneration of equations of elliptic type onthe boundary of a domain Dokl Akad Nauk SSSR(NS) 77 (1951) 181ndash183(Russian)

[49] A G Kuzprimemin Equations of mixed type with non-classical behavior of charac-teristics Mixed Type Equations BSB Teubner Leipzig 90 1986 180ndash194

[50] O A Ladyshenskaja and N N Uraltseva Linear and quasilinear elliptic equa-tions Academic Press New York 1968

[51] E Lanckau and W Tutschke Complex analysis Methods trends and applica-tions Akademie-Verlag Berlin 1983

[52] M A Lavrentprimeev and A V Bitsadze The problem of equations of mixed typeDokl AN SSSR 1950 Vol 70 373ndash376

[53] M A Lavrentprimeev and B V Shabat Methods of function theory of a complexvariable GITTL Moscow 1958 (Russian)

[54] J Leray Hyperbolic differential equations Princeton Univ Press 1954

[55] J Leray and J Schauder Topologie et equations fonczionelles Ann Sci EcoleNorm Sup 51 (1934) 45ndash78 YMH 1 (1946) 71ndash95 (Russian)

[56] Ming-de Li and Yu-chun Qin On boundary value problems for singular ellipticequations Hangzhoudaxue Xuebao (Ziran Kexue) 1980 no2 1ndash8

[57] Jian-bing Lin On some problems of Frankl Vestnik Leningrad Univ 16 (1961)no13 28ndash34 (Russian)

[58] Chien-ke Lu 1) Boundary value problems for analytic functions World Scien-tific Singapore 19932) Complex variable methods in plane elasticity World Scientific Singapore1995

References 245

[59] L G Mikhailov A new class of singular integral equations and its applicationsto differential equations with singular coefficients Wolters-Noordhoff Gronin-gen 1970

[60] C Miranda Partial differential equations of elliptic type Springer-VerlagBerlin 1970

[61] V N Monakhov 1)Transformations of multiply connected domains by thesolutions of nonlinear L-elliptic systems of equations Dokl Akad Nauk SSSR220 (1975) 520ndash523 (Russian)2) Boundary value problems with free boundaries for elliptic systems AmerMath Soc Providence RI 1983

[62] N I Mushelishvili 1) Singular integral equations Noordhoff Groningen 19532) Some basic problems of the mathematical theory of elasticity Nauka Moscow1946 (Russian) Noordhoff Groningen 1953

[63] J Naas and W Tutschke Some probabilistic aspects in partial complex differ-ential equations Complex Analysis and its Applications Akad Nauk SSSRIzd Nauka Moscow 1978 409ndash412

[64] L Nirenberg 1) On nonlinear elliptic partial differential equations and Holdercontinuity Comm Pure Appl Math 6 (1953) 103ndash1562) An application of generalized degree to a class of nonlinear problems CollAnalyse Fonct Liege 1970 Vander Louvain 1971 57ndash74

[65] O A Oleinik On equations of elliptic type degenerating on the boundary of aregion Dokl Akad Nauk SSSR (NS) 87 (1952) 885ndash888

[66] M H Protter 1) The Cauchy problem for a hyperbolic second order equationCan J Math 6 (1954) 542ndash5532) An existence theorem for the generalized Tricomi problem Duke Math J 21(1954) 1ndash7

[67] M H Protter and H F Weinberger Maximum principles in differential equa-tions Prentice-Hall Englewood Cliffs N J 1967

[68] Deqian Pu Function-theoretic process of hyperbolic equations Integral Equa-tions and Boundary Value Problems World Scientific Singapore 1991161ndash169

[69] S P Pulprimekin The Tricomi problem for the generalized Lavrentprimeev-Bitsadze equa-tion Dokl Akad Nuak SSSR 118 (1958) 38ndash41 (Russian)

[70] J M Rassias 1) Mixed type equations BSB Teubner Leipzig 90 19862) Levture notes on mixed type partial differential equations World ScientificSingapore 1990

246 References

[71] M S Salakhitdinov and B Islomov The Tricomi problem for the general linearequation of mixed type with a nonsmooth line of degeneracy Soviet Math Dokl34 (1987) 133ndash136

[72] M Schneider The existence of generalized solutions for a class of quasilinearequations of mixed type J Math Anal Appl 107(1985) 425ndash445

[73] M M Smirnov Equations of mixed type Amer Math Soc Providence RI1978

[74] S L Sobolev Applications of functional analysis in mathematical physicsAmer Math Soc Providence RI 1963

[75] He-sheng Sun 1) The problems of the rigidity of the surfaces with mixed Gausscurvature and boundary value problems for the equations of mixed type Proc1980 Beijing Sym Diff Geom Diff Eq Beijing 1982 1441ndash14502) Tricomi problem for nonlinear equation of mixed type Sci in China (SeriesA) 35 (1992) 14ndash20

[76] Mao-ying Tian The general boundary value problem for nonlinear degenerateelliptic equations of second order in the plane Integral Equations and BoundaryValue Problems World Scientific Singapore 1991 197ndash203

[77] F Tricomi 1) Sulle equazioni lineari alle derivate parziali di 2o ordine di tipomisto atti Accad Naz Lincei Mem Cl Sci Fis Nat (5) 14 (1923) 133ndash247(1924)2) Repertorium der Theorie der Differentialgleichungen SpringerndashVerlagBerlin 1968

[78] N S Trudinger Nonlinear oblique boundary value problems for nonlinear ellip-tic equations Trans Amer Math Soc 295 (1986) 509ndash546

[79] W Tutschke 1) The Riemann-Hilbert problem for nonlinear systems of differ-ential equations in the plane Complex Analysis and its Applications AkadNauk SSSR Izd Nauka Moscow 1978 537ndash5422) Boundary value problems for generalized analytic functions for several com-plex variables Ann Pol Math 39 (1981) 227ndash238

[80] I N Vekua 1) Generalized analytic functions Pergamon Oxford 19622) New methods for solving elliptic equationsNorth-Holland Publ Amsterdam1967

[81] N D Vvedenskaya On a boundary problem for equations of elliptic type degen-erating on the boundary of a region Dokl Akad Nauk SSSR(NS) 91 (1953)711ndash714 (Russian)

[82] Chung-fang Wang Dirichlet problems for singular elliptic equations Hang-zhoudaxue Xuebao (Ziran Kexue) 1978 no2 19ndash32 (Chinese)

References 247

[83] W Warschawski On differentiability at the boundary in conformal mappingProc Amer Math Soc 12 (1961) 614ndash620

[84] E Wegert Nonlinear boundary value problems for homeomorphic functions andsingular integral equations Akademie-Verlag Berlin 1992

[85] Guo-chun Wen 1) Modified Dirichlet problem and quasiconformal mappings fornonlinear elliptic systems of first order Kexue Tongbao (A monthly J Sci) 25(1980) 449ndash4532) The Riemann-Hilbert boundary value problem for nonlinear elliptic systemsof first order in the plane Acta Math Sinica 23 (1980) 244ndash255 (Chinese)3) The singular case of the Riemann-Hilbert boundary value problem Beijing-daxue Xuebao (Acta Sci Natur Univ Peki) 1981 no4 1ndash14 (Chinese)4) The mixed boundary value problem for nonlinear elliptic equations of secondorder in the plane Proc 1980 Beijing Sym Diff Geom Diff Eq Beijing 19821543ndash15575) On a representation theorem of solutions and mixed boundary value problemsfor second order nonlinear elliptic equations with unbounded measurable coeffi-cients Acta Math Sinica 26 (1983) 533ndash537 (Chinese)6) Oblique derivative boundary value problems for nonlinear elliptic systems ofsecond order Scientia Sinica Ser A 26 (1983) 113ndash1247) Some nonlinear boundary value problems for nonlinear elliptic equations ofsecond order in the plane Complex Variables Theory Appl 4 (1985) 189ndash2108) Applications of complex analysis to nonlinear elliptic systems of partial dif-ferential equations Analytic Functions of one Complex Variable Amer MathSoc Providence RI 1985 217ndash2349) Linear and nonlinear elliptic complex equations Shanghai Science TechnPubl Shanghai 1986 (Chinese)10) Some boundary value problems for nonlinear degenerate elliptic complexequations Lectures on Complex Analysis World Scientific Singapore 1988265ndash28111) Conformal mappings and boundary value problems Amer Math SocProvidence RI 199212) Function theoretic methods for partial differential equations and theirapplications and computations Advan Math 22 (1993) 391ndash40213) Nonlinear irregular oblique derivative problems for fully nonlinear ellipticequations Acta Math Sci 15 (1995) 82ndash9014) A free boundary problem in axisymmetric filtration with two fluids ActaMath Appl Sinica 1998 139ndash14315) Oblique derivative problems for linear second order equations of mixed typeScience in China 41 (1998) 346ndash35616) Approximate methods and numerical analysis for elliptic complex equationsGordon and Breach Science Publishers Amsterdam 199917) Linear and nonlinear parabolic complex equations World Scientific Singa-pore 1999

248 References

18) Nonlinear partial differential complex equations Science Press Beijing 1999(Chinese)

[86] Guo-chun Wen and H Begehr 1) Boundary value problems for elliptic equa-tions and systems Longman Harlow 19902) Existence of solutions of Frankl problem for general Lavrentprimeev-Bitsadze equa-tions Revue Roumaine Math Pure Appl 45(2000) 141ndash160

[87] Guo-chun Wen and Shi-xiang Kang 1) The Dirichlet boundary value problemfor ultra-hyperbolic systems of first order J Sichuan Normal Univ (NaturSci) 1992 no1 32ndash43 (Chinese)2) The Riemann-Hilbert boundary value problem of linear hyperbolic complexequations of first order J Sichuan Normal Univ (Natur Sci) 21 (1998)609ndash614

[88] Guo-chun Wen and Ping-qian Li The weak solution of Riemann-Hilbert prob-lems for elliptic complex equations of first order Appl Anal 45 (1992) 209ndash227

[89] Guo-chun Wen and Zhao-fu Luo Hyperbolic complex functions and hyperbolicpseudoregular functions J Ningxia Univ (Natur Sci) 19 (1998) no1 12ndash18

[90] Guo-chun Wen Chung-wei Tai and Mao-ying Tain Function Theoretic methodsof free boundary problems and their applications to mechanics Higher Educa-tion Press Beijing 1995 (Chinese)

[91] Guo-chun Wen and Mao-ying Tain 1) Solutions for elliptic equations of secondorder in the whole plane J Math 2 (1982) no1 23ndash36 (Chinese)2) Oblique derivative problems for quasilinear equations of mixed type in generaldomains I Progress in Natural Science 4 (1999) no1 85ndash953) Oblique derivative problems for nonlinear equations of mixed type in generaldomains Comm in Nonlinear Sci amp Numer Simu 34 (1998) 148ndash151

[92] Guo-chun Wen and Wen-sui Wu The complex form of hyperbolic systems offirst order equations in the plane J Sichuan Normal University (Natur Sci)1994 no2 92ndash94 (Chinese)

[93] Guo-chun Wen and C C Yang 1)Some discontinuous boundary value problemsfor nonlinear elliptic systems of first order in the plane Complex VariablesTheory Appl 25 (1994) 217ndash2262) On general boundary value problems for nonlinear elliptic equations of secondorder in a multiply connected domain Acta Applicandae Mathematicae 43(1996) 169ndash189

[94] Guo-chun Wen Guang-wu Yang and Sha Huang Generalized analytic functionsand their generalizations Hebei Education Press Hebei 1989 (Chinese)

[95] Guo-chun Wen and Zhen Zhao Integral equations and boundary value problemsWorld Scientific Singapore 1991

References 249

[96] W Wendland 1) On the imbedding method for semilinear first order ellipticsystems and related finite element methods Continuation methods AcademicPress New York 1978 277-3362) Elliptic systems in the plane Pitman London 1979

[97] Xi-jin Xiang Pan-complex functions and its applications in mathematics andphysics Northest Normal University Press Changchun 1988 (Chinese)

[98] C C Yang G C Wen K Y Li and Y M Chiang Complex analysis and itsapplications Longman Harlow 1993

[99] Guo-chun Wen and Wei-ping Yin Applications of functions of one complexvariable Capital Normal University Press 2000 (Chinese)

Index

algebraic equation 53 54 87 137 138176 206 207 208

algebraic theory 87analytic function 79ndash93 99 125

128 130 132 133 149 150 151153 155 165ndash167 174 184 190202 212 223 228 230 232

a priori estimate 25 28 42 43 50 109119 157 177 182 186 200 227

auxiliary function 100 111 112 114116 199

AV Bitsadze 157

Banach space 53 55 93 104 137 138191 193 207 208

boundary condition 9 18ndash2427 29 30 32 34 38 43ndash46 48 4953 55ndash59 60ndash65 67 68 70 7273 78 80ndash82 84 86 88 8991ndash100 101 103 105ndash107 109111 112 115 116 120 121 122125 127ndash130 132 133 135 137139 140 141ndash144 146ndash148 149150 151 153ndash155 158 159 160161 163ndash169 172ndash175 177ndash185187 189 190 192 193 195ndash199201ndash203 205 207 209ndash215217ndash220 223 224ndash226 228ndash235237 238

boundary value problem 1 18 24ndash2531 32 34 35 38 39 42ndash44 5658ndash60 65 67 72 73 78 84 90 97102 103 107 108 119 126 133134 139 141ndash143 155 157 158160 165 171 175 177 178 182187 190 193 194 195 200 201

217 219 220 227 228 232 237238 239

boundedness 28 108 109 125

Cα(D)-estimate 26 29 42C1(D)-estimate 50C1

α(D)-estimate 51 204Cauchy problem 73CauchyndashRiemann system 3Cauchy sequence 104 191 193Cauchy theorem 186Chaplygin equation 66 68 72 118

239characteristic coefficient 6characteristic lines 45 66 71 177 202

224 229characteristics 68 72 74 78closed and convex set 53 55 93 104

137 138 207 208coefficients matrix 176compact subset 94 104completely continuous operator 176completeness 104 191 193complete system of linearly

independent solutions 87 175complex equation 1 3 8 9 12 13

16ndash24 27 29 32 34 36ndash38 4547ndash49 53 62ndash64 90 91 94 95120 121 127ndash129 132 135 137143 146 148 150 154 165 167168 175 176 178 184 187 189192 197 200 206 218 221 222234

complex equation of mixed(elliptic-hyperbolic) type 119126 129 134 136 156 159

Index 251

Condition C 21 22 25ndash28 30 32 3442 47ndash53 56 58 62ndash64 67 90 9194ndash101 104 106ndash112 116 118127ndash129 132ndash136 139ndash143 146148 150 154 155 156 163ndash165167 168 170 171 182 183 185186 189 192 195 201 202 204206 209 212 215 218 222 224225 227 228 229 231 234 237

Condition C prime 53 163 173 176 195196 198 206

condition of hyperbolic type 10 11 1314 39 40

condition of uniformly hyperbolic type41

conformal mapping 86 92 100 124157 177 209

conjugate harmonic function185 188

continuity method 92continuous function 70 172 188continuously differentiable 2ndash4 36 74

108 109 159 164 173 196 202210 219 226

continuously differentiable function 842 105 121 122 201

continuously differentiable solution 343 59 96 158 163 171 177 195201 220 229

continuous mapping 105continuous solution 37 107 120 139

144 162 219 228convergency 3convergent domain 3

Darboux problem 59 65degenerate elliptic equation 108 118

199degenerate equation of mixed type

194 196 199degenerate hyperbolic equation 39

66 73degenerate mixed equation 194diagonal sequence 106

Dirichlet (boundary value) problem18 44 59 67 97 105 157 159160 196 202 209 210 227 229238

discontinuities of first kind 79 83 143171 218

discontinuity 186discontinuous boundary value problem

79discontinuous oblique derivative

(boundary value) problem 95 96103 106 107 171 200 218 219224

discontinuous Poincare (boundaryvalue) problem 157 171 172

discontinuous point 84 97 125 161discontinuous RiemannndashHilbert

problem 79 80 90 91 93 94125 143 144 145 146 156

discrete eigenvalue 176divisor of zero 2doubly connected domain 227

eigenvalue 176 177elementary function 124elliptic (complex) equation 3 10 96

98 103 106 108 111 116 232elliptic domain 118 181 199equation of mixed type 89 90 139

140 158 171 200 204 209 214218 227 228

estimates of functions 52 205estimates of solutions 28 108 110

111 136 224 235 237existence and uniqueness of solutions

27 48 66 150 202existence of solutions 25 39 52 55 79

87 90 93 103 106 108 119 122126 134 143 157 177 182 193198 213 218 226 233

expression of curve 216expression of solutions 48 50 62expression theorem 46extremum principle 95 232

252 Index

(F G)-derivative 5(F G)-hyperbolic pseudoregular

complex function 7finite kind of discontinuous points 84Frankl (boundary value) problem 177

179 180 182ndash184 186 188 189191 193 194

Fredholm theorem 171 176

gas dynamics 177general boundary value problem 90general Chaplygin equation 66general discontinuous boundary value

problem 79 84 90general discontinuous

RiemannndashHilbert problem 84 86144

general domain 31 34 50 55 58 6478 94 106 119 138ndash140 157 209214 218

general hyperbolic equation 43 52 65general LavrentprimeevndashBitsadze equation

157 171 177general nonlinear hyperbolic system

14general quasilinear equation of mixed

type 200 206 208general quasilinear hyperbolic

equation 50 52 64general quasilinear mixed equation

136general solution 24 46 82 83 85 87

93 121 122 156 175 176generating pair 6 8 9Greenrsquos formula 4

harmonic function 185 188higher dimensional domain 238Holder continuous condition 29

51 52 205Holder continuous estimate 51 204Holder continuous function 88Holder estimate 134 136Holder function 88homeomorphic (solution) 35ndash37 92

homeomorphism 37 91ndash93 99homogeneous boundary condition 81

82 86 101 105homogeneous boundary value problem

86 134 178 198homogeneous (complex) equations 21

27 49 97 101 132 154 187 195homogeneous equation 98 99 185

198 213 225 231homogeneous Frankl problem 186homogeneous integral equation 176

177homogeneous problem 175 198homogeneous system 87hyperbolic 10 17hyperbolic (complex) equation 1 14

28 37 39 55 63hyperbolic complex functions 1ndash3 38hyperbolic constant 37hyperbolic continuous function 4hyperbolic continuously differentiable

function 35hyperbolic domain 66 68 157 198

212hyperbolic element 1hyperbolic equation 28 39 42 43

46ndash48 50 51 61 62 65 78hyperbolic harmonic complex function

38hyperbolicity condition 15hyperbolic mapping 1 35 37hyperbolic model 2hyperbolic number 1 2 26 38 42 59hyperbolic pseudoregular functions 1

5 7 8hyperbolic regular functions 1 3 5

36ndash38hyperbolic system 1 10 14 35 73hyperbolic unit 1

implicit function 14index 80 81 84 88 90 93 95 97 100

103 107 120 125 126 139 144145 156 158 160ndash164 180 181

Index 253

186 196 202 228 229 232 237238 239

integral equation 93 171 176integral expression 51 204integral formula 159integral of Cauchy type 82integral path 45 65 159 202inverse function 31 32 36 55 57 59

91 93 99 140 141 216inverse mapping 124inverse transformation 31ndash33 55ndash57

140 141 216inversion 97

KeldychndashSedov formula 79 84 90

LavrentprimeevndashBitsadze equation 157 227L Bers 239linear and nonlinear hyperbolic

complex equation 10 20 22 39linear and quasilinear hyperbolic

complex equation 39linear and quasilinear hyperbolic

complex system 9linear complex equation 29 134linear complex equation of mixed type

126linear degenerate mixed equation 194linear elliptic equation 103linear equation 42 95 99 104 106

135 200 226linear equation of mixed type 157 162

171 172linear homogeneous equation 99linear hyperbolic complex equation 18

25 29 39 41 43 47linear hyperbolic system 10linear mixed equation 162 176linear system of mixed type 126

mathematical model 177maximum 100 101 110 118 198 213

226maximum point 101maximum principle 100 211 213 226mechanics 79

method of integral equation 171 177209

method of iteration 168 193 232method of parameter extension 177

180 189 192 214 227method of successive iteration 66 72

143minimum 55 118 198mixed boundary value problem 79 84mixed (complex) equation 127 129

137 141 142 162 165 167 168171 175 176 177 184 189 192212 218 224 237 239

mixed equation with parabolicdegeneracy 177

mixed system 119monotonous continuous function 37multiply connected domain 200 227

238 239

negative minimum 110 111Neumann boundary value problem 109non-degenerate mutually disjointed

arcs 172nonhomogeneous integral equation

176nonlinear boundary condition 82nonlinear complex equation 94nonlinear elliptic (complex) equation

79 90 108nonlinear equation 54nonlinear hyperbolic equation 40nonlinear hyperbolic system 14nonlinear mechanics 90 94nonlinear uniformly elliptic system

90non-singular transformation 37 41non-trivial solution 82 86 87 118

oblique derivative (boundary value)problem 39 43 44 50 59 66 6769 73 95 108 109 139 157 158160 162ndash164 168 173 194ndash196198ndash202 206 209 211 214 215219 224 225 228 233 237 239

254 Index

parameter equation 31 33 55 57 139214

partial differential equation 5 10 1415 39

piecewise smooth curve 4physics 79point condition 86 97 107 164 176

202 211 212 220 224 229positive maximum 110 113 115 118principle of compactness 108principle of contracting mapping 92

93principle of extremum 111Problem A 18ndash23 26ndash30 32 34 45

46ndash48 80 83 88 89 91ndash95120ndash122 126 127 129 132ndash138140 141 159 165ndash168 170 173184 189 192 197 202 203 204229 231 233

Problem A0 18 21 80 82 120 121Problem A1 24 25Problem A2 62Problem A4 64Problem Aprime 94 95 139ndash142 219ndash222Problem Aprime

0 95 139Problem Aprimeprime 233 234 237Problem Alowast 144ndash150 153ndash156Problem Alowast

0 81 91 95 145Problem Alowast 143 144Problem B 84ndash86 175Problem B0 86Problem B1 175Problem B2 175Problem Bprime 86 87Problem Bprime

0 86 87Problem C 86 88 90Problem C0 88Problem C prime 136 137 139 171Problem D 18 44 67 68 97ndash99 109

110 111 158 159 160 161 162196 238 239

Problem F 178 183 191 193Problem F0 178 189 190 192Problem F1 194

Problem Ft 189 190 192 194Problem Ft0 189 190 192 193Problem N 109Problem O 109Problem P 43ndash56 58 64 68 96 97

99 101ndash106 109ndash111 116 118158ndash161 163 165ndash168 170 171195ndash198 201ndash204 205 207ndash215218 229 230 231

Problem P0 44 67 97 109 118 158196 197 202

Problem P1 59 64ndash72 78 104Problem P2 60ndash63 64 65 78Problem P3 60ndash63 64 78Problem P4 60 63 64 73 74 78Problem Pt 103 104Problem Pt0 103Problem P 209Problem P prime 107 204 215 218 219

225Problem P prime

0 107 220 223ndash225Problem P primeprime 228ndash229 231ndash234 236

237Problem P primeprime

0 228Problem Q 97 106 171ndash177 202

206 211 212Problem Q0 172 173Problem Q 209Problem Qprime 107 108 220ndash226Problem Qprimeprime 229 237Problem T 158Problem T2 227Process of iteration 50

quasi-hyperbolic mapping 1 35 37quasilinear (complex) equation 27 30

34 135 138 139 144quasilinear (complex) equation of

mixed type 79 134 143 199 200209 215 218 227 228

quasilinear degenerate equation ofmixed type 239

quasilinear elliptic equation 95quasilinear equation 104ndash106 214

Index 255

quasilinear hyperbolic equation31 42 43 47 59ndash61 63

quasilinear hyperbolic system 15quasilinear mixed (complex) equation

136 143 209quasilinear mixed system 134quasilinear uniformly elliptic equation

96 107

rank 176 177reductio ad absurdum 187reflected domain 180regularity of solution 134 156 160

171 206 225representation 20 25 30 35 36 38

43 46ndash48 63 64 79 87 90ndash92 94134 160 182 198 229

representation of solutions 18 20 2548 61 63 64 69 79 84 90 95146 171 200 220 227

representation theorem 36 43 5966 91 97 126 134 135 143164 173 196 211 218 222

RiemannndashHilbert (boundary value)problem 18 20 25 28 45119ndash122 125 126 127 129 134135 138 141 150 159

removable singular point 185

Schauder fixed-point theorem 94 105Schwarz formula 82sequence 28 106 116sequence of coefficients 102sequence of functions 23 28 30 49

50 52 54 75 104 131 132 138152 153 169 170 205 207 208235 236

sequence of solutions 53 103 105 111116 137 190 193

series expansion 3simplest complex equation of mixed

type 119 121 122simplest equation of mixed type 157

160

simplest hyperbolic (complex)equation 18 45 46

simplest hyperbolic system 3 35simplest mixed equation 157simply connected domain 18 25 42

43 66 106 108 119 126 134 139141 143 157 194 200 208 218

singularity 162solvability 22 25 27 43 78 83 91

94 104 119 157 160 168 173175 177 180 182 189 191 194200 206 209 211 213 231

solvability condition 83 89 94 126156 160 162 171 176 177 239

subsequence 102 105 106 111 116132 154 170

successive iteration 18 22 25 27 2830 39 48 50ndash53 75 103 126 129133 134 137 150 155 168 170190 193 204 205 207 208 218234

symmetry extension 180system (of equations) 32 35 37 69

71 72 140system of integral equations 72 74

77 93

third boundary value problem 109transformation 21 31ndash34 37 44

55ndash58 60 67 80 88 97 111120 140 141 142 158 196202 216 217

triangle inequality 2Tricomi problem 67 90 157 202

209 210 213 227 229 236

uniformly bounded 115uniformly converge 23 49 52 54 77

98 102 106 111 131 132 138153 154 170 205 208 236

uniformly elliptic system 90uniformly elliptic equation 110uniformly hyperbolic 10 40uniformly hyperbolic system 36

256 Index

unique continuous solution 162221 233 239

uniqueness and existence of solutions18 25 59 72 95 119 157 162200 209 227

uniqueness of solutions 20 22 2587 118 121 122 126 127133ndash134 143 150 154 162 164168 177 184 186 187 194 196202 206 213 225 229 233

uniqueness theorem 39 119 232unique solution 25 28 32 34 43 50

53 56 58 63 64 65 72 78 8693 103 104 106 116 125 133135 137 141 142 144 147 150155 159 162 165 189 192 204207 208 212 217 224 227 238

unique solvability 59 61 63 64 6873 129 206 211 218 224

unique system 87unit disk 84 86 92 99univalent analytic function 92 93univalent continuous 35univalent mapping 35upper half-plane 79 84 90 94 124upper half-(unit) disk 79 84 86 88

90 93 96 103

weakly converge 102 106well posed version 86

zone domain 90 94

Book Cover
HALF-TITLE
SERIES TITLE
TITLE
COPYRIGHT
CONTENTS
INTRODUCTION TO THE SERIES
PREFACE
CHAPTER I HYPERBOLICCOMPLEXEQUATIONSOFFIRSTORDER
CHAPTER II HYPERBOLICCOMPLEXEQUATIONSOFSECONDORDER
CHAPTER III NONLINEARELLIPTICCOMPLEXEQUATIONSOFFIRSTANDSECONDORDER
CHAPTER IV FIRSTORDERCOMPLEXEQUATIONSOFMIXEDTYPE
CHAPTER V SECONDORDERLINEAREQUATIONSOFMIXEDTYPE
CHAPTER VI SECONDORDERQUASILINEAREQUATIONSOFMIXEDTYPE
REFERENCES
INDEX

Page 2: Linear and quasilinear complex equations of hyperbolic and ...webéducation.com/wp-content/uploads/2018/09/Linear...and hyperbolic pseudoregular functions. Next we transform the linear

London and New York

(Print Edition)

Contents

vi Contents

Contents vii

Preface

x Preface

CHAPTER I

e1 = (1 + j)2 e2 = (1minus j)2 (11)

e1 + e2 = 1 ekel =

ek if k = l

0 if k = lk l = 1 2 (12)

in which

1z=

z

zz=

1x+ y

e1 +1

x minus ye2 =

1micro

e1 +1νe2

z1

(1micro2

e1 +1ν2

e2

)=

micro1

micro2e1 +

ν1

ν2e2

(15)

(1) wz = 0 (16)

(2) ξν = 0 ηmicro = 0 (17)

(3) ux = vy vx = uy (18)

[ξ(z)minusξ(z0)

e2

]

respectively

2

ez = 1 + z +z2

ex+y + exminusy

2+ j

ex+y minus exminusy

2

2+ j

sin z = z minus z3

cos z = 1minus z2

tgz=sin z

(1

cosmicroe1 +

1cos ν

e2

ctgz=cos zsin z

1sinmicro

e1+1

sin νe2

y(z0)∆y + ε(∆z)

where

y(z0) = uy(z0) + jvy(z0)

lim∆zrarr0

ε(∆z) rarr 0

Cf(z)dz =

intC

udx+ vdy + j[int

Cvdx+ udy] =

Df(z)dxdy =

int intD

udxdy + jint int

Dvdxdy

C[f(z) + g(z)]dz =

intC

f(z)dz +int

Cg(z)dzint int

int intD

f(z)dxdy +int int

Dg(z)dxdy

int

Cf(z)dz le

radic2M1l

int int

Df(z)dxdy le

radic2M2S

G[f(z)]zdxdy =

j

2

intC

f(z)dz

2

intC

f(z)dz

Cf(z)dz = 0

ImF (z)G(z) = 0 in D (19)

w(z0) = δ0F (z0) + γ0G(z0) (110)

w(z0) = limzrarrz0

= limzrarrz0

ie

Wz(z0) = w(z0) (116)

andWz(z0) = 0 (117)

W (z) =

∣∣∣∣∣∣∣∣∣w(z) w(z0) w(z0)

F (z) F (z0) F (z0)

G(z) G(z0) G(z0)

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣F (z0) F (z0)

G(z0) G(z0)

∣∣∣∣∣∣ (118)

Fz(z0) F (z0) F (z0)

Gz(z0) G(z0) G(z0)

∣∣∣∣∣∣∣∣∣ = 0 (119)

w(z0) =

∣∣∣∣∣∣∣∣∣wz(z0) w(z0) w(z0)

Fz(z0) F (z0) F (z0)

Gz(z0) G(z0) G(z0)

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣F (z0) F (z0)

G(z0) G(z0)

∣∣∣∣∣∣ (120)

wz = aw + bw (121)

where

F G minus FG b =

FGz minus FzG

FG minus FG

F G minus FG B =

FGz minus FzG

FG minus FG

(123)

respectively

w(z) = φ(z)F (z) + ψ(z)G(z) (124)

K(z) = φ(z) + jψ(z) (125)

Fφz +Gψz = 0 (126)

holds where

(128)

Proof From

Wz(z0) = F (z0)φz(z0) +G(z0)ψz(z0)

Setting

minusG

minusF

(129)

where

δ =σy + τx

τ

δ =σy + τx

τ

(132)

φ(z) =int z

z0

wz = a(z)w(z) + b(z)w(z) (133)

a = a(FG) and b = b(FG) (135)

F (z) = 1 or G(z) = j in D (137)

I = (K2 +K3)2 minus 4K1K4 gt 0 (22)

holds in which

K1 =∣∣∣∣∣a11 b11

a21 b21

∣∣∣∣∣ K2 =∣∣∣∣∣a11 b12

a21 b22

∣∣∣∣∣ K3 =∣∣∣∣∣a12 b11

a22 b21

∣∣∣∣∣ K4 =∣∣∣∣∣a12 b12

a22 b22

where

K5 =∣∣∣∣∣a11 a12

a21 a22

∣∣∣∣∣ K6 =∣∣∣∣∣ b11 b12

b21 b22

a = K1K6 b = K3K6 c = K4K6 d = K2K6

∆ = I4K26 =

(b+ d)2

4minus ac gt 0 (26)

+lower order terms

namely

= (1 + a)(1minus c) + bd gt 0

in which

q0

(29)

then we can prove

|Q1|+ |Q2| = |Q1Q1|12 + |Q2Q2|12 lt 1 (210)

∆ = I4K24 = (b+ d)24minus ac gt 0 (212)

whereq1 =

2

It is clear that

q0 = (1 + q1)(1 + q1)minus q2q2

4

= (1 + a)(1minus c) + bd gt 0

in which

Q1(z) =[(1minus q1)(1 + q1) + |q2|2]

q0 Q2 =

minus2q2(z)q0

(216)

Fkux =int 1

Fkuy =int 1

Fkvx =int 1

Fkvy =int 1

(217)

holds in which

K1 =∣∣∣∣∣a11 b11

a21 b21

∣∣∣∣∣ K2 =∣∣∣∣∣a11 b12

a21 b22

∣∣∣∣∣ K3 =∣∣∣∣∣a12 b11

a22 b21

∣∣∣∣∣ K4 =

∣∣∣∣∣a12 b12

a22 b22

∣∣∣∣∣ K5 =∣∣∣∣∣a11 a12

a21 a22

∣∣∣∣∣ K6 =∣∣∣∣∣ b11 b12

b21 b22

6

Vu Vv

∣∣∣∣∣∣ =∣∣∣∣∣∣λ 1

micro 1

andu =

U minus V

λ minus micro v =

microU minus λV

micro minus λ

1V + c1

2V + c2(222)

where

aprime1 =

a1 minus microb1

1 =minusa1 + λb1

2 =a2 minus microb2

2 =minusa2 + λb2

λ minus micro

2b12)V + (c1b22 minus c2b12)]K6

1b21)V + (c2b11 minus c1b21)]K6(223)

and then

4 gt 0

Letξ =

x minus microy

λ minus micro η =

minusx+ λy

λ minus micro

ηx ηy

minus1 λ

= 0

a22uξ + b22vη = a2u+ b2v + c2(226)

1v + cprime1

2v + cprime2

(227)

whereaprime

2(ζ w)w + Aprime3(ζ w) (228)

in which Aprime1 A

prime2 A

ux = au+ bv + f vy = cu+ dv + g (230)

wz = 0 in D (31)

zisinL4

1|a(z) + b(z)| le k0 (34)

times g(0) = 2r((1 + j)x) on [0 R]

f(0)=u(0)+v(0)=r(0)+b1

a(0)+b(0) g(0)=u(0)minusv(0)=

r(0)minusb1

a(0)minusb(0)

(38)

f(x+ y) =2r((1 + j)(x+ y)2)

a((1 + j)(x+ y)2)+b((1 + j)(x+ y)2)(39)

0lex+yle2R

0lexminusyle2R1

(310)

wz = A1(z)w + A2(z)w + A3(z) (312)

(314)

in which

[a2(z0)minus b2(z0)] (316)

w(z) = w0(z) + Φ(z) + Ψ(z) in D

Ψ(z) =int xminusy

int x+y

0[Cξ +Dη + F ]d(x+ y)e2

(317)

[Ψ]z = [Aξ +Bη + E]e1 + [Cξ +Dη + F ]e2 (319)

wz = A1w + A2w in D (320)

w(z) = Φ(z) + Ψ(z)

Ψ(z) =int xminusy

int x+y

(322)

Ψ(z) le 8M3mR0 Φ(z) le 32M3k20(1 + k2

0)mR0

w1(z) = w0(z) + Φ1(z) + Ψ1(z)

Ψ1(z) =int ν

0[Aξ0 +Bη0 + E]e1dν +

int micro

(323)

w2(z) = w0(z) + Φ2(z) + Ψ2(z) in D

+int micro

and

2

wn(z) = w0(z) + Φn(z) + Ψn(z)

Ψn(z) = +int ν

int micro

(326)

+int micro

and then

0

Rprimenminus1

n (328)

Ψlowast(z) =int ν

int micro

(330)

w(z) = u(z) + jv(z)

(333)

a(z1)+b(z1)

Re [λ(z)w(z)]=u(z)+v(z)=r[(1+

R+R1

RminusR1j)

xminusj2RR1

RminusR1

]on [R1 R]

(334)

RminusR1

)xminus 2RR1

RminusR1

]

t

2R+ (1minus j)R1

] t isin [0 2R]

and then

0lexminusyle2R1

(335)

Condition C

(45)

in which

wz = A1w + A2w in D (46)

w(z) = Φ(z) + Ψ(z)

Ψ(z) =int xminusy

int x+y

(48)

w1(z) = w0(z) + Φ1(z) + Ψ1(z)

Ψ1(z) =int ν

0[Aξ0 +Bη0 + E]e1dν +

int micro

(49)

Ψn(z) =int ν

int micro

(411)

+int ν

+int micro

0

Rprimenminus1

n (413)

n=0

= M6 (415)

C[w(z) D] le M7 = M7(α k0 D)

Ψ11(z) =

int xminusy

1(z) =int x+y

0G2(z)d(x+ y)

prime (420)

prime(421)

prime

prime(423)

n =Re wn + Im wn w2

(M12Rprime)n

n Cα[w1

n

prime)n

n Cα[w2

prime)n

n

(424)

x =12(micro+ ν)

4R minus 4σ(x+ γ1(x)) + 2x+ 2γ1(x)

(429)

x =12(micro+ ν) =

2

(430)

2 Im [λ(z0)w(z0)] = b1 (432)

2R (433)

2 Im [λ(z0)w(z0)] = b1

4 are as follows

4 = minusγ4(x) + y = 0 0 le x le l2(435)

Hence we have

x =12(micro+ ν) =

(439)

x =12(micro+ ν) =

(440)

1 cup Lprimeprime4

is reduced to

4 Im [λ(z0)w(z0)] = b1

radius R1 namely

radicR2

1 + 2R1ν minus ν2

2

radicR2

2

1 Lprimeprime2 L

primeprime3 L

primeprime4

ξν = 0 ηmicro = 0 (52)

microy νy

∣∣∣∣∣ =∣∣∣∣∣ 1 1

1 minus1

∣∣∣∣∣ = minus2∣∣∣∣∣ ξu ηu

ξv ηv

∣∣∣∣∣ =∣∣∣∣∣ 1 1

1 minus1

∣∣∣∣∣ = minus2 (53)

ξ = ξ(micro) η = η(ν) (55)

wz = Q(z)wz Q(z) = a(z) + jb(z) (56)

w(z) = Φ[χ(z)] (58)

ξσ = 0 ητ = 0 (σ τ) isin G (513)

wz = Q(z)wz (514)

Wz = 0 (516)

CHAPTER II

( )z =( )x + j( )y

2 ( )z =

( )x minus j( )y2

4

( )zz =( )xx + ( )yy + 2j( )xy

4 ( )zz =

4

(14)

in which

Q =a+ c+ 2bj

a minus c A1 =

d+ ej

a minus c A2 =

f

2(a minus c) A3 =

g

2(a minus c)

|Q(z)| lt 1 in D (17)

respectively

in which

a =int 1

2b =int 1

c =int 1

d =int 1

e =int 1

f =int 1

0Φτu(x y τu 0 0 0 0 0)dτ = f(x y u)

g = minusΦ(x y 0 0 0 0 0 0) = g(x y)

and

a2 = f = minusint 1

0Fτu(z τu 0 0 0 0)dτ = a2(z u)

a3 = minusF (z 0 0 0 0) = a3(z)

(112)

or its complex form

(120)

A1 =a+ jb

2 A2 =

c

4 A3 =

d

4in D

wz =wx + jwy

wx minus jwy

(122)

in which

A =a+ b

4 B =

a minus b

4 C =

c

4 D =

d

4

12

partu

u(0) = b0 Im [λ(z)uz]|z=z3 = b1

(21)

maxzisinL3

1|a(z) + b(z)| le k0

(22)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Re [λ(z)w(2Re1 + νe2)] = r(z)

(24)

u(z) = φ(z) on L = L3 cup L4 (25)

in which

λ(z) = a+ jb =

1 + jradic2on L4

b+1 = Im

[1minus jradic2

2radic2

bminus1 = Im

[1 + jradic2

uz(z3)]=

(27)

wz = Re [A1w] + A2u+ A3 in D (28)

int z

0w(z)dz + b0 in D (210)

2ReintΓw(z)dz =

intΓw(z)dz +

intΓw(z)dz

= 2jint int

s1

w(z)dz +int

s2

w(z)dz]+ b0

uzz = 0 (211)

wz = 0 in D (212)

f(2R) = u(z3) + v(z3) =r((1 + j)R) + b1

a((1 + j)R) + b((1 + j)R)

a((1 + j)R)minus b((1 + j)R)

(213)

a((1 + j)(x+ y)2) + b((1 + j)(x+ y)2)

0 le x+ y le 2R

α[u(z) D]leM2k2=M2(α k0 D)k2 (215)

[a((1 + j)x) + b((1 + j)x)]f(2x) + [a((1 + j)x)

f(2R)=u(z3)+v(z3)=r((1 + j)R)+b1

a((1 + j)R)+b((1 + j)R)

[a((1 + j)t2) + b((1 + j)t2)]f(t) + [a((1 + j)t2)

u(z) = 2Reint z

0w(z)dz + b0 in D

w(z) = W (z) + Φ(z) + Ψ(z) in D

Ψ(z)=int ν

int micro

(216)

u0(z) = 2Reint z

0w0(z)dz + b0 w0(z) = W (z) = ξ0e1 + η0e2 (221)

u1(z) = 2Reint z

0w1(z)dz + b0 w1(z) = w0(z) + Φ1(z) + Ψ1(z)

Ψ1(z) =int xminusy

+int x+y

(222)

C1[u0(z) D] le M3C[w0(z) D] + k2 (223)

C[Φ1(z) D] le 8M4k20(1 + k2

0)[(4 +M3)m+ k2 + 1]Rprime

(224)

wn(z) satisfying

un+1(z) = 2Reint z

0wn+1(z)dz + b0

+int micro

(225)

and then

0

Rprimenminus1

(n minus 1) dRprime

n

(226)

+int micro

(228)

int z

wz = Re [A1w] + A2u in D (230)

int z

w(z) = Φ(z) + Ψ(z)

Ψ(z) =int ν

int micro

(233)

n

maxzisinD

C1[u D] le M7 C1[u D] le M8k (33)

(116)

α[u D] le M11k (34)

u0(z) = 2Reint z

0w0(z)dz + b0 w0(z) = W (z) = ξ0e1 + η0e2 (35)

u1(z) = 2Reint z

0w1(z)dz + b0 w1(z) = w0(z) + Φ1(z) + Ψ1(z)

Ψ1(z) = Ψ11(z)e1 +Ψ2

1(z)e2 Ψ11(z) =

int ν

0G1(z)dν

Ψ21(z) =

int micro

(38)

prime (39)

1(z) G2(z) Ψ21(z)

(310)

prime (311)

prime

(312)

Cα[w1n(z) D] le

(M18Rprime)n

n Cα[w2

n(z) D] le(M18R

prime)n

n

n C1

n

(313)

wn(z) = w0(z) +nsum

m=1wm(z) un(z) = u0(z) +

nsumm=1

um(z) (n = 1 2 )

α[u(z) D] le M10 (314)

F = Re [A1uz] + A2u+ A3

(315)

M8k1 + k2 + 2k0tσ + 2k0t

τ + |b1|+ |b0| = t (317)

(321)

um+1 = C1[um+1 D] le M20 (322)

um+1(z) = 2Reint z

+int x+y

(323)

n

0(1 + k20) From

+int xminusy

+int x+y

(325)

int z

M8k1 + k2 + 2k0tσ + 2k0t

τ + |b0|+ |b1| = t (327)

prime4 and

Hence we have

x =12(micro+ ν) =

(334)

and

x =12(micro+ ν) =

(335)

4 u(0) = b0 Im [λ(z3)w(z3)] = b1 (337)

u(z) = 2Reint z

4is reduced to

4 are as follows

(341)

micro =2Rmicro

x =12(micro+ ν) =

(345)

x =12(micro+ ν) =

(346)

4is reduced to

4

3 )] = b1 (350)

1 Lprimeprime2 L

primeprime3 L

primeprime4 namely

Lprimeprime1 =

radicR2

Lprimeprime2 =

radicR2

Lprimeprime3 =

y = γ3(x) =

radicR2

Lprimeprime4 =

y = γ4(x) =

radicR2

Lprimeprime1 =

⎧⎨⎩x = σ1(ν) =R1 + ν minus

radicR2

1 + 2R1ν minus ν2

2

⎫⎬⎭

Lprimeprime2 =

radicR2

2

⎫⎬⎭

Lprimeprime3 =

⎧⎨⎩x = σ2(ν) =R1 + ν minus

radicR2

1 + 2R1ν minus ν2

2

⎫⎬⎭

Lprimeprime4 =

radicR2

2

⎫⎬⎭

u(0) = b0 Im [λ(z)uz]|z=z1 = b1(41)

zisinL2

1|a(z) + b(z)| le k0

(42)

u(0) = b0 Im [λ(z)uz]|z=z1 = b1(43)

(44)

u(0) = b0 Im [λ(z)uz]|z=z1 = b1(45)

1|a(z) + b(z)| le k0

(46)

u(z) = 2Reint z

0w(z)dz + b0 in D (49)

(410)

f(0) = u(z1) + v(z1) =r((1minus j)R1) + b1

(411)

f(x+ y) =2r((1+j)(x+y)2+(1minusj)R1)

0 le x+ y le 2R (412)

α[u(z) D] le M24k2 (413)

u(z) = 2Reint z

Ψ(z) =int ν

2R1

(414)

(415)

(416)

Im [λ(z1)w(z1)] = b1(417)

(420)

in which

2 τ(x) =

(422)

for equation (211)

2 0 le x+ y le 2R

(423)

and

f(0) =sprime(0) +R(0)

2 g(2R1) =

(424)

for (116)

f(0) = u(z1) + v(z1) =r((1minus j)R1) + b1

(428)

0 le x+ y le 2R1

Ψ(z) =int ν

2R1

C1[s(x) L0] le k2 C1[φ(x) L1] le k2 (432)

u(0) = b0 Im [λ(z)uz]|z=z1=b1

(433)

where we choose

radic2 (434)

L1=x+int y

0

L2=

xminusint y

0

radicminusK(t)dt =

int y

0|t|m2dt = minus

int |y|

0d

2m+ 2

|t|(m+2)2 = minus 2m+ 2

|y|(m+2)2

L1 x minus 2m+ 2

|y|(m+2)2 = 0 L2 x+2

m+ 2|y|(m+2)2 = 2 ie

L1 y = minus(m+ 22

x)2(m+2) L2 y = minus[m+ 22

(2minus x)]2(m+2)

where

A1 =d+ je

K(y)minus 1 A2 =

f

2(K(y)minus 1) A3 =

g

2(K(y)minus 1)

12

partu

u(0) = b0 Im [λ(z)uz]|z=z1 = b1

(54)

[radic

α[r(z) Lj] le k2 j = 1 2

zisinL2

1|a(z) + b(z)| le k0

(55)

int y

0

L2 = BC =x = 2 +

int y

0

(56)

Re [(1 + j)(U + jV )] = U + V = r(x) on L2

Im [(1 + j)(U + jV )] = [U + V ]|z=z1+0 = r(1 + 0)

whereU =

λ(z)=

andu(z) = 2Re

int z

x+int y

0

int y

0

int y

0

int y

0

int y

0

radicminusK(t)dt

minusK(y)radic

(510)

(radicminusK(y)

)micro= minus1

K prime

4K(radic

minusK(y))

K primeU4K

(U + V )ν =radic

(514)

minus2radic

minusK(y)(U + V )ν

=radic

= minusradic

K prime(y)2K(y)

U 2radic

=radic

+radic

=radic

K prime(y)2K(y)

U

(515)

U ie

(516)

u(z) = 2Reint z

partu

u(0) = b0 Im [λ(z)(U + iV )]|z=z1 = b1

(519)

2 V (x y) =

2

(520)

where

f(0) = U(0) + V (0) =r(1) + b1

a(1) + b(1) g(2) = U(2)minus V (2) =

r(1)minus b1

a(1)minus b(1)

U =12

f(micro) +

V =12

a(ν2)minus b(ν2)

or

U =12

g(ν) +

V =12

minusg(ν) +

(521)

a(x2+1)minusb(x2+1)

a(x2) + b(x2) x isin [0 2]

(522)

W (z) =

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

12

[(1 + j)

a(ν2)minus b(ν2)

] (523)

can be derived

U + V = minusint ν

int micro

4K(y)Udmicro

W = U+jV =minus1+j

2

int ν

4K(y)Udν+

1minusj

2

int micro

4K(y)Udmicro

= minus1 + j

2

int ν

2

K prime(y)4K(y)

Udν +1minus j

2

int micro

0

K prime(y)4K(y)

Udmicro

radic1 +

(dy

dx

)2dx =

radic1minus K

1minus Kdy

radic1 +

(dy

dx

)2dx =

radic1minus K

minusKdx =

radic1minus Kdy

(524)

=1radic1minusK

]

=1

2radic1minusK

K prime(y)K(y)

U

=1radic1minusK

]

2radic1minusK

K prime(y)K(y)

U

(525)

0

1

2radic(1minus K)

K prime(y)K(y)

Uds1 =int s1

0

12

K prime(y)K(y)

Udy

0

1

2radic(1minus K)

K prime(y)K(y)

Uds2 =int s2

0

12

K prime(y)K(y)

Udy

(526)

W (z) = U + jV = W (z) + Φ(z) + Ψ(z)

U+V =minusint s1

0

1

2radic(1minusK)

K prime(y)K(y)

0

1

2radic(1minusK)

K prime(y)K(y)

Uds2(527)

u(z) = 2Reint z

int z

can be replaced by

2U(x)=S(x)=radic

0

+u(0)=s(x)

2V (x) = R(x) = uy on AB = L0 = (0 2)

(529)

(532)

v(x) = 0 vy(x) = 0 on L0 (533)

ξν =1

2radic1minus K

2K prime(y)K(y)

)ξ

+ e minus 12

K prime(y)K(y)

]

ηmicro =1

2radic1minus K

12

K prime(y)K(y)

)ξ

+ e+12

K prime(y)K(y)

]

(534)

K prime(y)K(y)

=m|y|mminus1h(y)

K(y)minus |y|mhy

K(y)=

m

y+

hy

h

v(z) =int y

ξ(z) = minusint y

η(z) =int y

(535)

in this case

A1 = minuse

2minus hy

4h B1 =

e

2minus hy

4h A2 = minuse

2+

hy

4h B2 =

e

2+

hy

4h

C1 = minus12

dradicminusKminus m

4y C2 = minus1

2dradicminusK

+m

4y D = minusf

2 E = minusG

2

limyrarr0

0

(538)

3Mδ

2+[ε(y)M +m2]δm2

m+ 2lt γ

6δ2M2

m+ 6+8δMm+ 2

+2ε(y)M +m

m+ 2lt γ

(539)

1 s02

ξ1(z)=minusint y

int y

0Edy z isin s0

1

η1(z)=int y

0[A2ξ0+B2η0+C2(ξ0+η0)+Dv0+E]dy=

int y

0Edy zisin s0

2

v1(z) = Reint y

(540)

1

ηk+1(z) =int y

2

vk+1(z) =int y

(541)

j=0γj|y|m2+1

(542)

j=0γj|y|

Moreover we get

int y

0Ex[x1minusx2]dy|

le 4m+4

j=0γj|y|m2+1

le |int y

int y

0|x1 minus x2|dy|

le M |int y

1sumj=0

γj|y|m2+1

1sumj=0

γj|y|m2+1

|vn+1(z)| le |int y

int y

0

nsumj=0

γjydy|leMnsum

j=0γj|y|2 leM

n+1sumj=0

γj|y|

0

⎡⎣(|A1|+|B1|+|D|)Mnsum

j=0γj|y|+|C1|M

nsumj=0

γj|y|m2+1+|E|⎤⎦ dy|

le M |int y

0

⎡⎣3M nsumj=0

γj|y|+(

ε(y)2|y|radich

+m

4|y|)

nsumj=0

γj|y|m2+1 + 1

⎤⎦ dy|

le M |y|⎧⎨⎩[32M |y|+

(|ε(y)|M +

m

2

) |y|m2

m+ 2

]nsum

j=0γj + 1

γj|y|

and

0

dy|

Noting that

lensum

nsumj=0

γj|y|m2+1

le M2nsum

nsumj=0

γj|y|m2+1

le (M |y|+1)M2nsum

j=0γj|y|m2+1

and

j=0γj +

∣∣∣∣∣hy

2h

j=0γj|y|m2+1

we get

0[(3M |y|+ 4)M2

nsumj=0

γj|y|m2+1

+(|C1|+ |C2|)Mnsum

j=0γj|y|m2+1 +M2|y|m2+1]dy|

le M |y|m2+1

[6M2y2

m+ 6+8M |y|m+ 2

+(|ε(y)|M +

m

2

)

times 2m+ 2

nsumj=0

γj +2M

m+ 4|y|⎤⎦

le Mn+1sumj=0

γj|y|m2+1

ηlowast(z) =int y

vlowast(z) =int y

A1 = minus d

4|y|m2 + j

(m

8|y| minus e

4

) A3=minusf

4

A2 = minus d

4|y|m2 + j

(m

8|y| +e

4

) A4=minusg

4

(543)

and

u(z)=2Reint z

0uzdz=2Re

int z

0(U+jV )d(xminusjy)

CHAPTER III

]=

φj

πminusKj

Kj =[φj

π

]+ Jj Jj = 0 or 1 j = 1 m

(13)

12

msumj=1

[φj

πminus γj

](14)

(15)

Ψ(z) =Φ(z)Π(z)

Π(z) =mprod

j=1

(z minus xj

z + i

)γj

(16)

)=

12

msumj=1

[φj

πminus γj

]=12

msumj=1

Kj (110)

Re[Λ(x)

x minus x0

x+ i

x minus x0Ψ0(x)

Ψlowast(z) = i(

z minus i

z + i

)[K] (z minus x0

z + i

)eiS(z) in D (112)

Re [S(x)] = arg

⎡⎣Λlowast(x)(

x minus i

x+ i

)[K](x minus x0

x+ i

)⎤⎦ x isin L Im [S(i)] = 0 (114)

Ψ0(z) =

X(z) = Π(z)Ψ0(z) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩i(

z minus i

z + i

)K

i(

zminusi

z+i

)[K] zminusx0

(116)

Re[Φ(x)iX(x)

]=

c(x)iλ(x)X(x)

=λ(x)c(x)iX(x)

Φ(z)iX(z)

=1πi

[int infin

minusinfinλ(t)c(t)

(t minus z)iX(t)dt+

Q(z)i

]

Φ(z) =X(z)πi

[int infin

minusinfinλ(t)c(t)

]

(118)

Q(z) = i[K]sumj=0

[cj

(z minus i

z + i

)j

+ cj

(z minus i

z + i

)minusj]+

iclowastx0 + z

(119)

Q(z) =

iclowastx0 + z

(120)

int infin

minusinfinλ(t)c(t)

x0 + z

x0 minus z

=int infin

minusinfinλ(t)c(t)

x0+i

x0minusi

=infinsum

j=0

int infin

[1 +

z minus i

x0 + i

]x0 + i

x0 minus i

infinsumj=0

(z minus i

x0 minus i

)j

=int infin

dt+iclowastx0+i

x0minusi+

infinsumj=1

[int infin

minusinfinλ(t)c(t)

(tminusi)j+1X(t)dt+

2iclowastx0

(x0minusi)j+1

](zminusi)j

(121)

minusinfinλ(t)c(t)

int infin

minusinfinλ(t)c(t)

X(t)(t minus i)jdt+

2iclowastx0

when 2K is odd

(122)

x0 + i

int infin

minusinfinλ(t)c(t)

πi

[int infin

minusinfinλ(t)c(t)

2iclowastx0

](124)

Φ(z) =

int z

iΦ(z)dz in D

λ(x) =

radicΠn

]=

φj

πminus Kj

Kj =[φj

π

]+ Jj Jj = 0 or 1 j = 1 m

(128)

(129)

Φ(z) =X(z)2πi

[intΓ

dt+Q(z)] (130)

with

j + cjzminusj) +

iclowastz0 + z

(131)

intΓ

λ(t)c(t)X(t)tj

intΓ

λ(t)c(t)X(t)tj

Φ(z) =X(z)z[K]

πi

[intΓ

0

] (133)

clowast = iintΓ

λ(t)c(t)X(t)t

dt

X(z) =

z(ζ) =1

ζ + i

[1 + iζ minus

radic2(1minus ζ2)

]

Im [λ(zprimej)Φ(z

primej)] = bj j = 1 m (136)

where zprime1 z

0

primej)]=0 j = 1 m (137)

m = 2K + 1 le 2ND +NΓ le 2K (138)

j=1djΦj(z) (139)

msumj=1

dprimejΦj(zprime

j) j = 1 m (140)

Φlowast(zprimej) =

msumj=1

dprimejΦj(zprime

j) = 0 j = 1 m (141)

j=1 dprimejΦj(z)

r(x)=

(143)

λ(x) =

(144)

Re [λ(z)Φ(z)] = r(z) =

0 z isin L0(145)

Ψ(z) =Φ(z)Π(z)

j=n+1j =m

(z minus xj

z + i

)γj

(146)

Re [λ(z)Φ(z)] = 0 z isin L (147)

)=

(149)

Φ(z) =

⎧⎪⎨⎪⎩Ψ(z) in D = |z| lt 1 Im z gt 0

Re [Λ(z)Φ(z)] = R(z) on |z| = 1 (152)

where

Λ(z)=

⎧⎨⎩λ(z)

λ(z)R(z)=

Φ(z) = Φ(i1 + ζ

1minus ζ

)in D (154)

Condition C

w(z) = Φ[ζ(z)]eφ(z) + ψ(z) (26)

w(z) = Φ(z)eφ(z) + ψ(z) (210)

in which

X(z)=mprod

j=1j =nm

ψz = Qψz + A1ψ + A2ψ + A3 Q =

φz = Qφz + A A =

(213)

int intD

f(ζ)ζ minus z

dσζ

φ(z) = Tg ζ(z) = Ψ[χ(z)] χ(z) = z + Th

(214)

h(z) = Q(z)[1 + Πh] (218)

prime1Γ

prime2 isin C1

j)] = bj j = 1 m(225)

w(z) = Φ(z)eφ(z) + ψ(z) (228)

in which

X(z) =mprod

j=2j =1n

(230)

12[ux + iuy] uzz =

14[uxx + uyy]

a+ c A1(z) =

minusd minus ei

a+ c A2(z) =

minusf

2(a+ c) A3(z) =

g

2(a+ c)

Condition C

|Q(z u w)| le q0 lt 1 (34)

12

partu

Im [λ(z)uz]|z=0 = b2 (37)

ψ(z) = 0 Ψ(z) = 1 on Γ (39)

β[Ψ(z) D] le M2

z(ζ) =1

ζ + i

[1 + iζ minus

radic2(1minus ζ2)

] (311)

ψ(ζ) = 0 Ψ(ζ) = 1 on L = ζ(Γ)

ψ(z) =

⎧⎨⎩ψ(z) in D

Q=

⎧⎨⎩Q(z)

Q(z)A1=

⎧⎨⎩A1(z)

A1(z)A2=

⎧⎨⎩A2(z)

A2(z)A3=

⎧⎨⎩A3(z)

minusA3(z)in

⎧⎨⎩D

D

⎫⎬⎭

0w(z)dz+b0 w(z)=Φ[ζ(z)]eφ(z) (317)

(318)

Ψ(z)in D (319)

12[partU

and give a lemma

partu

partl= lim

z(isinl)rarrz0

2

partV

partn= 2Re

zVz

R z isin part∆

partV

partn= minus4Re zVz

R |z| = R

2

partV

partnge minusε

partV

partngt 0

partu

partn= Ψ

partV

partn+ V

partΨpartn

ge minusεpartV

partn+ V

partΨpartn

gt 0

partu

partl= cos(l n)

partu

partn+ cos(l s)

partu

partsgt 0 (326)

12

partu

(327)

C1δ [u(z) D] le M6(k1 + k2)

(328)

j (j = 1 2 3) λm rm bmj (j = 0 1) but

Umzz minus Re [QmUm

zz + Am1 Um

z ]minus Am2 Um =

Am3

Hm

partUm

partνm

=rm(z)Hm

Hm

Um(1) =bm1

Hm

Lp[Am2 Um +

Am3

Hm

Cα

[|z minus zj|βj

rm(z)Hm

Γlowast]

le M7 j = 1 m

∣∣∣∣∣ bmj

Hm

∣∣∣∣∣ le M7 j = 0 1

int z

minus1wm(z)dz +

bm0

Hm

U0zz = Re [Q

0U0zz + A0

1uz] + A02U

0 = 0 in D

partU0

(334)

Noting that

δ

un minus unminus1 D

2 (337)

2 u+ Am3 in D (339)

where

Qm =

⎧⎨⎩Q(z u uz)

0Am

j =

⎧⎨⎩Aj(z u uz)

0in

⎧⎨⎩Dm = DDm

Dm

⎫⎬⎭ j = 1 2 3

C1δ [u(z) D] le M5 (340)

2 (z Un Unz)un=Cn

1 u0z]minus Am2 u0

(341)

gn(z) =

umn (z) = 0 on Γ (346)

where

Cmn =gn[Re (Am

1 umz )+Am

mz ] (347)

C1β[u

mn Dn] le M9 um

n W 2p0

(Dn)le M10 (348)

u0(z) = 2Reint z

prime1Γ

prime2 isin C2

12

partu

X(z) =mprod

j=2j =n

where

Q(z) =1minus ym

2(1 + ym) A3(z) =

d

2(1 + ym)

Condition C

2(1 + ym)ge minus c

4ge 0 on D (43)

lu=12

partu

party=ψ(x) xisinL0

(44)

u(z) = 2Reint z

int 1

limyrarr0

partu(x0 y)party

lt 0 (or limyrarr0

partu(x0 y)party

gt 0) (47)

if the limit exists

limyrarr0

partu(x0 y)party

= M prime ge 0 (48)

v(z) =εu(z)

(MeBd minus εeBy)

we have

ltεM

εM

limyrarr0

partv(x0 y)party

=ε2BM

C[un(z) D] le M12 = M12(α k0 D) (411)

c1 gec2+maxΓ

|r(z)|+maxD

ec2y c2 gtmaxL0

|ψ(x)|+maxD

|b|+2maxD

|d|+1

party= minusc2e

c2 gt maxD

|b(z)|+maxL0

|ψ(x)|+maxD

ec2y + 2maxD

|d|

c1 gt c2 +maxD

ec2y(1 +

c2

σ0

)+max

Γ

|φ(z)|σ0

(414)

C1β[un(z) D] le M13 = M13(β k0 D η) (416)

From (420) we get

(422)

(423)

and then we have

Y(424)

timesuxuxy +2mfG

1f

Y)uxuxy

Yuy(aux + buy + cu)

(f primeprime

fminus 2

f prime2

f 2

)+ a

f prime

f

](425)

(gprime

fminus f primeg

f 2

)uuxminus2mg

fYuuy

+[Y m

(gprimeprime

fminus 2

f primegprime

f 2

)+ a

gprime

f

]u2 +

1f

Lη(v) = (2+o(1))[Y m(uxy)2+(uyy)2]+O

+(

Y m

X2 +Y m2minus1+ε

X

)[Y m+ε1(ux)2+(uy)2]

+2Y m

X2 (ux)2+2(uy)2

X2 +O

(Y m

X3 |ux|+Y minus1

X2 |uy|)

+ε1(m+1minusε1)(1+o(1))

X4 Y ε1minus2

(426)

Lη(v)f

+(uyy)2]+2g

(428)

in which

mG

Y

Yuy(aux+buy+cu)

+[Y m

(f primeprime

fminus2f

prime2

f 2

)+a

f prime

f

(gprime

fminus f primeg

f 2

)uux

+[Y m

(gprimeprime

fminus2f

primegprime

f 2

)+a

gprime

f

]u2+

(429)

Σ = O

Y m

X2 +Y m2minus1+ε

X

)

X3 |ux|+Y ε3minus2

X4

minus2 g

f(Y m|ux|2+|uy|2)+O

X4

)

(430)

X4

)gt0 (431)

f|uuy|+H prime

f

H prime

ge Y m

fH prime)2

(432)

By (419)(420)(432) we obtain

+1G

O

X4

)]

ge Y m

X4 Y ε3minus1)2

X4

)gt0

(433)

partu

partr(z)parts

uy+1

on Γ (437)

F =

(439)

plusmnm

(440)

2Yux

)2

YGprime+bGprime

+2cGminus(

m2

2Y 2+2ax

)F (ux)2plusmn m

(441)

Y

)|ux|

and

(443)

(445)

CHAPTER IV

tions

wzlowast

= 0 in

D+

Dminus

(12)

where

12[wx minus iwy]

(15)

K =12(K1 +K2) (16)

Kj =[φj

π

γj =φj

radic2 on

Re [λ(z)w(z)] = 0 z isin Γ (18)

2 v(z) =

(111)

(112)

g(t) = f(t) = 0 t isin [0 2]

(113)

(114)

u(x) = v(x) =12f(x) x isin [0 2] or

(115)

(116)

obviously

2 v(z) =

2

(119)

=2r((1minusi)x) on [0 1] f(0)

(120)

(121)

f(x+ y) =2r((1 + i)(x+ y)2 + 1minus i)

u(x) =12f(x) + g(x) v(x) =

12f(x)minus g(x)

u(x) =12g(x) + f(x) v(x) =

12minusg(x) + f(x)

x isin [0 2](123)

xisin [02]

(124)

xisin [0 2] or

xisin [0 2]

(125)

ζ(z) =5(z minus 1)

(z minus 1)2 + 4 z(ζ) = 1 +

52ζ(1minus

Λ(ζ) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

1minus iradic2

1 + iradic2

(126)

Re [Λ(z)W (ζ)] = R(ζ) =

(127)

W (ζ) =X(ζ)πi

[int infin

dt+ iclowast2 + ζ

2minus ζ

]in G (128)

and

ζ minus 1ζ + i

)γ1(

ζ + 1ζ + i

)γ2

clowast =2i+ 12 + i

int infin

minusinfinΛ(t)R(t)

X(t)(t minus i)dt

t minus 2t+ i

)] on Im t = 0 Im [S(i)] = 0 (129)

w(z)=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

12

(1 + i)f(x+ y) + (1minus i)

or

12

wz

wzlowast

= F (z w) F = A1(z)w + A2(z)w + A3(z) in

D+

Dminus

(22)

where

12[wx minus iwy]

4 A2 =

a+ ib+ ic minus d

4 A3 =

f + ig

2

Condition C

Cβ[w(z)X(z) D+] le M3(k1 + k2 + k3) (28)

in which

δ γj ge 0j = 1 2 (29)

w(z) = w0(z) +W (z) (210)

int intD+

g(ζ)ζ minus z

W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν

2g1(z)dνe1 +

int micro

(211)

g(z) =

A1 + A2ww w(z) = 00 w(z) = 0

f = A1ψ + A2ψ + A3 in D+

(212)

(214)

s(x)=

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

w(z) = w0(z) +W (z) =

⎧⎨⎩ Φ(z)eφ(z) + ψ(z) in D+

wz

wzlowast

= A1w + A2w + A3 in

⎧⎨⎩D+

Dminus

⎫⎬⎭ (218)

int intD+

g0(ζ)ζ minus z

w1(z) = w0(z) +W1(z) W1(z) = Φ1(z) + Ψ1(z)

Ψ1(z)=int ν

int micro

(219)

(221)

in which

s1(x)=

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

w1(z) = w0(z) +W1(z) =

⎧⎨⎩ Φ1(z)eφ1(z) + ψ1(z) in D+

w0(z) + Φ1(z) + Ψ1(z) in Dminus(222)

w2(z) = w0(z) +W2(z) =

⎧⎨⎩ Φ2(z)eφ2(z) + ψ2(z) in D+

w0(z) + Φ2(z) + Ψ2(z) in Dminus(224)

wn(z) = w0(z) +Wn(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Ψn(z)=int ν

+int micro

(225)

and then

1 (z)Yplusmn(z)|+

radic2[|Y +(z)

int ν

|B| |C| |D| |E| |F |) Rprime = 2 M = 1 + 4k20(1 + k2

+int micro

(227)

timesint Rprime

0

Rprimenminus1

n in Dminus

(228)

plusmn(z) ie

wplusmnn (z)Y

Ψlowast(z) =int ν

+int micro

(230)

in which

sn(x)=

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

n = M9 (234)

Cβ[wn(z)X(z) D+] le M3(k1 + k2 +M9) (235)

Lw = A1w + A2w in D (237)

w(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

g(z) =

⎧⎨⎩A1 + A2ww w(z) = 0 z isin D+

Φ(z) + Ψ(z)

Ψ(z) =int ν

2[Aξ +Bη]e1dν +

int micro

(239)

s(x)=

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

n in Dminus (240)

wz

wzlowast

= F (z w) F = A1w + A2w + A3 in

D+

Dminus

(32)

Condition C

wz = A1w + A2w in D (36)

X(z)=2prod

2prodj=1

2δ γj ge 0(310)

Lw =

wz

wzlowast

D+

Dminus

(311)

C[A4(z w) D] le k0 (312)

M17 = k1 + k2 + |b1| (313)

(M13 +M14)k1 + k2 + 2k0tσ + |b1| = t (314)

+A3(z wm) = G(z wm) in D m = 1 2 (317)

wm+1(z) = w0(z) + Φm+1(z) + Ψm+1(z)

+int x+y

(320)

m

wplusmnm(z)Y

+int x+y

(322)

(M13 +M14)k1 + k2 + 2k0tσ + |b1| = t (323)

2 are as follows

2 = x minus y = 2 l le x le 2 (41)

(43)

1

(44)

K =12(K1 +K2) (45)

2minus2σ(ν)+ν

x =12(micro+ ν) =

(48)

and ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

x =12(micro+ ν) =

[2minus 2σ(x+ γ1(x)) + x+ γ1(x)](x+ y)4

2

[2minus 2σ(x+ γ1(x)) + x+ γ1(x)](x+ y)4

2

(49)

ξν = Aξ +Bη + E

2

in Dminus (411)

2 = γ2(x) + y = 0 l le x le 2 (413)

zisinLprimeprime1

zisinLprimeprime2

1|a(x) + b(x)| le k0

(415)

2

(418)

Hence we have

x =12(micro+ ν) =

(419)

reduced to

1)] = b1(422)

such that Lprime1 L

1of Lprimeprime

maxzisinL1

zisinL2

1|a(z) + b(z)| le k0

(54)

]=

φj

πminus Kj

Kj =[φj

π

]+ Jj Jj = 0 or 1 j = 0 1 m

(55)

msumj=0

[φj

2πminus γj

2

](56)

(57)

Re [λ(z)w(z)]=r(z)

zisinL3=sum[n2]

j=0 A2jA2j+1

Re [λ(z)w(z)]=r(z)

zisinL4=sum[(n+1)2]

j=1 B2jminus1B2j

j=0 1 [n2]

j=1 [(n+ 1)2]

(58)

maxzisinL3

zisinL4

1|a(x) + b(x)| le k0

(59)

γj =1πiln(

)=

φj

πminus Kj

Kj =[φj

π

]+ Jj Jj = 0 or 1 j = 0 1 m+ n

(510)

m+nsumj=0

(φj

2πminus γj

2

)(511)

λ(x) =

1 cap AB

2 cap AB

(513)

Cβ[w(z)X(z) D+] le M21(k1 + k2 + k3) (514)

the form

f(x)=h(x)=2r((1 + i)x2 + 1minus i)

(515)

on L2j+1 j = 0 1 [n2]

on L2j j = 1 [(n+ 1)2]

Re [λ(x)w(x)] =

⎧⎨⎩k(x)

h(x)λ(x) =

1 = Dminus1 cap AB

2 cap AB

and denote

Re [λ(x)w(x)] =

k(x) on Lprime2

2] le k2

(516)

Im [λ(z)w(z)]|z=A2j+1 = c2j+1 j = 0 1 [n2]

(517)

1in the form

w(z) = w(z) + λ(A2j+1)[r(A2j+1) + ic2j+1]

2j+1

j = 0 1 [n2]

= h(x+ y)=2r((1+i)(x+y)2+1minusi)

[n+ 12

]

the form

(518)

w(z) = w0(z) +W (z) in D (519)

int intD+

g(ζ)ζ minus z

Ψ(z) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

int ν

a2j+1

2j+1 j=0 1 [n2]

int ν

2g1(z)dνe1+

int micro

a2jminus1

(520)

g(z) =

⎧⎨⎩A1 + A2w(w) w(z) = 00 w(z) = 0 z isin D+

(521)

Cβ[ψ(z) D+] + Lp0 [ψz D+] le M22(522)

Im [λ(z)(Φ(z) + Ψ(z))]|z=A2j+1 = 0 j = 0 1 [n2]

(523)

j+δwplusmn

s(x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

x isin (a2j a2j+1) j = 0 1 [n2]

(525)

w(z) = w0(z) +W (z) =

⎧⎨⎩ Φ(z)φ(z) + ψ(z) in D+

wz

wzlowast

= A1w + A2w + A3 in

⎧⎨⎩D+

Dminus

⎫⎬⎭ (526)

functions

int intD+

g0(ζ)ζ minus z

Ψ1(z)=

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

int ν

a2j+1

g10(z)dνe1+

int micro

2j+1 j=0 1 [n2]

int ν

2g10(z)dνe1+

int micro

a2jminus1

2j j=1 [(n+1)2]

(527)

Im [λ(z)(Φ1(z) + Ψ1(z))]|z=A2j+1 = 0 j = 0 1 [n2]

(529)

where

s1(x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

x isin (a2j a2j+1) j = 0 1 [n2]

w2(z) = w0(z) +W2(z) =

⎧⎨⎩ Φ2(z)eφ2(z) + ψ2(z) in D+

w0(z) + Φ2(z) + Ψ2(z) in Dminus(532)

wn(z) = w0(z) +Wn(z) =

Ψn(z)=

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

int ν

a2j+1

g1nminus1(z)e1dν+

int micro

0g2

int ν

2g1

a2jminus1

2j j=1 [(n+1)2]

(533)

and then

+radic2

1lejlen+1|int ν

a2j+1

g10(z)e1dν|+ |

int ν

2g20(z)e1dν|

]

1lejlen+1|int micro

a2jminus1

g20(z)e2dmicro|

]

(534)

0)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int ν

a2j+1

[g1nminus1minusg1

0[g2

2j+1

j=0 1 [n2]int ν

2[g1

int micro

a2jminus1

[g2nminus1minusg2

j=1 [(n+1)2]

(535)

int Rprime

0

Rprimenminus1

(n minus 1) dRprime

n in Dminus

(536)

ie

Ψlowast(z)=

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

int ν

a2j+1

g1lowast(z)e1dν+

int micro

0g2

int ν

2g1

a2jminus1

2j j=1[(n+1)2]

(538)

(539)

in which

sn(x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

+ Re [λ(x)Ψn(x)]

x isin (a2j a2j+1) j = 0 1 [n2]

(541)

Cβ[wn(z)X(z) D+] le M21(k1 + k2 +M27) (543)

wz

wzlowast

= A1w + A2w in

⎧⎨⎩D+

Dminus

⎫⎬⎭ (545)

Im [λ(z)w(z)]|z=A2j+1 = 0 j = 0 1 [n2]

(546)

in which

s(x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(547)

w(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

g(z) =

⎧⎨⎩A1 + A2ww w(z) = 0 z isin D+

Φ(z) + Ψ(z)

Ψ(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int ν

a2j+1

2[Aξ + Bη]e1dν +

int micro

a2jminus1

[Cξ + Dη]e2dmicro

(548)

n in Dminus (549)

CHAPTER V

uzlowastzlowast

= 0 in

D+

Dminus

(12)

12[wx minus iwy]

12

partu

12

partu

1|a(z)+b(z)| lek0

(15)

K =12(K1 +K2) (16)

Kj =[φj

π

γj =φj

wz

wzlowast

= 0 in

D+

Dminus

(18)

int z

0w(z)dz + b0 in D (110)

w(z) =

⎧⎪⎪⎨⎪⎪⎩W [ζ(z)] z isin D+0 2

or

w(z) =12(1 + i)g(x minus y) +

12(1minus i)f(x+ y)

f(x+ y) =2r((1 + i)(x+ y)2 + 1minus i)

(111)

X(z) =2prod

2prodj=1

δ γj ge 0j = 1 2

(114)

λ(z) = a+ ib =

1 + iradic2on L2

r(z) =

⎧⎪⎪⎨⎪⎪⎩φθ on Γ

φxradic2on L1 or

φxradic2on L2

b1 = Im[1 + iradic2

uz(z1 minus 0)]=

φx minus φxradic2

|z=z1minus0 = 0 or

uz(z1 + 0)]= 0 b0 = φ(0)

(117)

Re[1minus iradic2

w(x)]= s(x) =

x isin [0 2] or

Re[1 + iradic2

w(x)]= s(x) =

x isin [0 2]

= eπi4 = eiφ2 or

minus1 lt γ1 =φ1

πminus K1 = minus5

4lt 0

0 le γ2 =φ2

πminus K2 =

14

lt 1 or

0 le γ1 =φ1

πminus K1 =

14

lt 1

minus1 lt γ2 =φ2

πminus K2 = minus5

4lt 0

K1 +K2

2= minus1

2 or

K1 = 0 K2 = minus1 K =K1 +K2

2= minus1

2

radic2 or

where

12[ux + iuy] uzz =

14[uxx + uyy]

14[uxx minus uyy]

A1 =a+ ib

2 A2 =

c

4 A3 =

d

4in D

K =12(K1 +K2) = 0 (25)

where

Kj =[φj

π

γj =φj

u(2)=2Reint 2

0uzdz+u(0)=2

int 2

int 0

πr(z)dθ+b0

K=

⎧⎪⎪⎨⎪⎪⎩0

⎧⎨⎩u(0)=b0 u(2)=b2

u(0) = b0

⎫⎬⎭ if

u(z) = U(z)Ψ(z) + ψ(z) in D+

U(z) = 2Reint z

(27)

ψ(z) = 0 Ψ(z) = 1 on Γ cup L0 (29)

p0(D+)le M2 (210)

Cβ[φD+] + Lp0 [φz D+] le M5 (213)

Ψ(z) (216)

u(z) = 2Reint z

0w(z)dz + b0 w(z) = w0(z) +W (z) (217)

int intD+

g(ζ)ζ minus z

W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν

2g1(z)dνe1 +

int micro

(218)

g(z)=

A=ReA1+ImA1

2 B=

ReA1minusImA1

(219)

(220)

u0(z) = 2Reint z

0w0(z)dz + b0 (222)

s(x)=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

+Re [λ(x)Ψ(x)] or

(223)

w(z) = w0(z) +W (z) =

⎧⎪⎨⎪⎩Φ(z)φ(z) + ψ(z) in D+

wz

wzlowast

= Re [A1w] + A2u+ A3 in

⎧⎨⎩D+

Dminus

⎫⎬⎭ (224)

wz

wzlowast

= Re [A1w] + A2u in

D+

Dminus

(225)

Re [λ(z)w(z)] = 0 z isin Γ u(0) = 0 u(2) = 0

w(z) =

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

w0(z) + Φ(z) + Ψ(z)

Ψ(z) =int ν

2[Aξ+Bη+Cu]e1dν+

int micro

(227)

n(230)

wz

wzlowast

= F F = Re [A1w] + A2u+ A3 in

⎧⎨⎩D+

Dminus

⎫⎬⎭ (232)

2(z) g12(z)

and

w1(z) = Φ1(z)eφ1(z) + ψ1(z) in D+

int intD+

g1(ζ)ζminusz

2g11(z)dνe1+

int micro

(235)

(236)

w2(z) = Φ2(z)eφ2(z) + ψ2(z) in D+

Ψn(z)=int ν

2g1

0g2

(240)

and then

+radic2[|Y +(z)

int ν

(241)

+int micro

(242)

0

Rprimenminus1

(n minus 1) dRprime

n in Dminus

(243)

plusmn(z)ie

wplusmnn (z)Y

Ψlowast(z) =int ν

int micro

(245)

(248)

(249)

12

partu

12

partu

maxzisinL1

zisinL2

1|a(x) + b(x)| le k0

(34)

Y (z)=ηm+1prodj=0

(35)

γj =1πiln

=φj

πminus Kj

Kj =[φj

π

]+ Jj Jj = 0 or 1 j = 0 1 m+ 1

(37)

m+1sumj=0

(φj

2πminus γj

2

)(38)

τj=

δj=

⎧⎨⎩ 2τj j=0 m+1

τj j=1 m(39)

u(z) = 2Reint z

0w(z)dz + c0 w(z) = w0(z) +W (z) (310)

int intD+

g(ζ)ζ minus z

W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν

2g1(z)dνe1 +

int micro

(312)

in which e1 =1 + i

2 e2 =

1minus i

A=ReA1+ImA1

2 B=

ReA1minusImA1

(313)

(314)

where

s(x)=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

+Re [λ(x)Ψ(x)] or

(315)

ηj = 2|γj|+ δ j = 0 m+ 1 ηj = |γj|+ δ j = 1 m

andu0(z) = 2Re

int z

0w0(z)dz + c0 (317)

u(z) = 2Reint z

0w(z)dz + b0

(318)

w(z) = w0(z) +2K+1sumk=1

ckwk(z) (322)

⎧⎨⎩ 1 2K + 1 K ge 0

φ K lt 0(325)

0w(z)dz + c0 (326)

ckH1wk(z) (327)

u = ψ1(s) on Γ (41)

u = ψ2(x) on CB (42)

partu

α[ψ2(x)CB]lek2

C1α[φ(y)A

LetU =

12ux V = minus1

2uy W = U + iV in D (46)

W zlowast

= Re [A1W ] + A2u+ A3 in

⎧⎨⎩D+

Dminus

⎫⎬⎭

u(z) = 2Reint z

aW (z)dz + ψ1(0)

(47)

U(x y) =12ux =

(48)

U(0 y) =12

partu

12[u(0 y)]y =

12[u(0 minusy)]y +

12φprime(y)

(49)

and then

(410)

Hence

(411)

2φprime(minusx)]

12φprime(minusx)

(412)

partu

U(0 y) =12

partu

(413)

λ(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

1

1 + iradic2

1

r(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

0 on AO

ψprime2(x) on CB

Kj=[φj

π

γj=φj

πminusKj j=1 4

=ei0

πminusK1=minus1

4minusK1 lt

34

lt1

=eiπ4

πminusK2=

14minusK2=minus3

4lt0

=ei0

πminus K3 = minus3

2minus K3 =

12

lt 1

=eiπ

(416)

K =12(K1 +K2 +K3 +K4) = minus1

2 (417)

W (z) =

⎧⎨⎩W (z) in D

⎧⎨⎩ Wz

W zlowast

⎫⎬⎭ = Re [A1W ] + A2u+ A3 in

⎧⎨⎩D

D

⎫⎬⎭u(z) = 2Re

int z

1W (z)dz + ψ1(0)

(419)

in which

A1 =

⎧⎨⎩A1(z)

minusA1(minusz)A2 =

⎧⎨⎩A2(z)

A2(minusz)A3 =

⎧⎨⎩A3(z) in D

and

λ(z) =

⎧⎨⎩λ(z)

λ(minusz)r(z) =

λ(z) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩1 + iradic2

1minus iradic2

r(z) =

⎧⎨⎩r(z) on OC = (0 1)

(422)

Kj=[φj

π

γj=φj

πminusKj j=1 6

=eminusiπ4

πminusK1=minus1

2minusK1=

12

lt1

=eiπ4

πminusK2=

14minusK2=minus3

4lt0

=ei0

πminusK3=minus3

2minusK3=

12

lt1

=eiπ

φ4

=eiπ2

πminus K5=

12

minus K5 =12

lt 1

=ei0

πminusK6=

14minusK6=minus3

4lt0

(423)

2 (424)

u(z) = U(z)Ψ(z) + ψ(z) in D+

U(z) = 2Reint z

(425)

ψ(z) = 0 Ψ(z) = 1 on Γ cup L

partψ(z)partx

= 0partΨ(z)

partx= 0 on AO

(427)

p0(D+)le M19 (428)

Cβ[φ(z) D+] + Lp0 [φz D+] le M22 (431)

Re [λ(x)W (x)] = r(x) λ(x) =

⎧⎪⎨⎪⎩1 + iradic2on L0 = (0 1)

1 on L1 = (1 a)(433)

u(z) = 2Reint z

aW (z)dz + b0 b0 = ψ1(0) (434)

Wz

Wzlowast

= Re [A1W ] + A2u+ A3 in

D+

Dminus

(435)

W (z) = Φ(z)eφ(z) + ψ(z)

int intD+

g(ζ)ζ minus z

W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν

1g1(z)dνe1 +

int micro

(436)

g(z)=A12+A1W(2W ) W (z) = 00 W (z)=0

in D+

A=ReA1+ImA1

2 B=

ReA1minusImA1

(437)

W zlowast

= 0 in

D+

Dminus

(438)

(439)

thus

W (z) =

⎧⎪⎨⎪⎩Φ(z)eφ(z) + ψ(z) in D+

uzzlowast

⎧⎨⎩D+

Dminus

⎫⎬⎭

u(z) = 2Reint z

(442)

in which

Ψ(z) =int ν

1g1(z)dνe1 +

int micro

(443)

U(x) = 2int x

aΦ(x)dx = 2

int x

0Φ(x)dx minus

int a

0Φ(x)dx

=S(x)=int x

12U(ix) on (0 1)

(444)

partU(z)partx

= 0 on AO (445)

int y0 Vydy =

int y0 Uxdy = 0 on

[U(z) + iV (z)]2dz = 0

= minusint a

1[V (x)]2dx minus

intΓ[V (z)]2

(dx

ds+ i

dy

ds

)ds (446)

+iint 0

1[U(iy)]2dy +

int 1

party

partsds+

int 1

0[U(iy)]2dy + 2

int 1

0[g(x)]2 minus 1

4[U(ix)]2dx = 0 (447)

4prodj=2

un+1(z) =int z

(451)

Un+1(z)=2Reint z

(452)

ψn+1(z) = 1 Ψn+1(z) = 1 on Γ cup L

partψn+1(z)partx

= 0partΨn+1(z)

partx= 0 on AO

(453)

Un+1(x)=Reint x

aΦn+1xdx=S(x)

S(x)=int x

2

int x

0φprime

n+1(minusx)dx

= g(x) +12Un+1(ix) +

(454)

where

g(x)=int x

0f(x)dx

int x

0Fn+1(minusx)dx=

Un+1(ix)2

0φprime

n+1(minusx)dx

partUn+1(z)partx

= 0 on AO (455)

limnrarrinfinmax

D+

int 1

0[Un+1(iy)]2dy = 0 (456)

intΓ[Vn+1(z)]2

party

partsds+

int 1

0[Un+1(iy)]2dy

+2int 1

0

[g(x)]2 minus 1

dx = 0

(457)

Un+1(iy) = 0 on AO g(x) = 0 on OC (458)

γ [X(z)un(z) D+] (459)

γ [un D+] le 12C1

γ [un D+]

infinsumj=N0

γ [u1minusu0 D+]

j=0un+1 =

Wz

W zlowast

= A3(z) in

⎧⎨⎩D+

Dminus

⎫⎬⎭ (462)

int z

aW (z)dz + b0 in D (464)

expressed as

u0(z) = 2Reint z

int intD+

A3(ζ)ζminusz

dσζ in D+

1A3(z)e1dν+

int micro

(466)

(467)

Wzlowast

= A3(z) in

⎧⎨⎩D+

Dminus

⎫⎬⎭

(468)

Wn+1zlowast

= A3(z) in

⎧⎨⎩D+

Dminus

⎫⎬⎭

(469)

Noting that

2C1

u(z) = 2Reint z

W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν

1A3(z)e1dν +

int micro

0A3(z)e2dmicro

(475)

u(z) = 2Reint z

0W (z)dz + u(0) Ψ(z) =

int ν

0A3(z)e1dν +

int micro

0A3(z)e2dmicro

f(x+ y) = τ((x+ y)2) + ReW (0) + ImW (0)

(476)

Wz

Wzlowast

⎧⎨⎩D+

Dminus

⎫⎬⎭ (477)

Wz = K(z) in D+ (481)

(485)

Noting that

int y0

x minus int y0

⎧⎨⎩ Wmacrz

Wmacrz

⎫⎬⎭ = A1(z)W + A2(z)W + A3(z)u+ A4(z) in

D+

Dminus

A1 =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩im

8y+

d

4ym2 +ie

4=

d

4ym2 + i

(m

8y+

e

4

)

jm

8|y| +minusd

4|y|m2 minus je

4=

minusd

4|y|m2 + j

(m

8|y|minuse

4

) in

⎧⎨⎩D+

Dminus

⎫⎬⎭

A2 =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩d

4ym2+ i

(m

8yminus e

4

)

minusd

4|y|m2+j

(m

8|y|+e

4

)

A3=

⎧⎪⎪⎪⎨⎪⎪⎪⎩f

4

minusf

4

A4=

⎧⎪⎪⎨⎪⎪⎩g

4

minusg

4

in

⎧⎨⎩D+

Dminus

⎫⎬⎭

(53)

and

u(z)=

⎧⎪⎪⎨⎪⎪⎩2Re

int z

0uzdz+u(0) in D+

2Reint z

lu=partu

(58)

K =12(K1 +K2)

Kj =[φj

π

γj =φj

u(2)=2Reint 2

0uzdz+u(0)=2

int 2

int 0

πr(z)dθ+b0 (510)

u(z) = 2Reint z

0w(z)dz + b0 w(z) = w0(z) +W (z) (511)

W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν

2g1(z)dνe1 +

int micro

4|y|m2 D= minus g

4|y|m2

A1=1

4|y|m2

[m

2yminus d

|y|m2 minuse

] B1=

14|y|m2

[m

2yminus d

|y|m2+e

]

A2=1

4|y|m2

[m

2|y|minusd

|y|m2 minuse

] B2=

14|y|m2

[m

2|y|minusd

|y|m2+e

]in Dminus

(514)

(515)

u0(z) = 2Reint z

0w0(z)dz + b0 in D (516)

Wz

D+

Dminus

(518)

Re [λ(z)W (z)] = 0 z isin Γ u(0) = 0 u(2) = 0

W (z) = Φ(z) + Ψ(z)

Ψ(z) =int ν

2[Aξ+Bη+Cu]e1dν+

int micro

(520)

]= 0

12

]= s(x) on L0

(522)

CHAPTER VI

Luz=

uzz

uzzlowast

D+

Dminus

(12)

14[uxx minus uyy]

A1 =a+ ib

2 A2 =

c

4 A3 =

d

4in D

Condition C

12

partu

12

partu

Im [λ(z)uz]z=z1 = b1 u(0) = b0 u(2) = b2

(16)

K =12(K1 +K2) (18)

Kj =[φj

π

γj =φj

Im [λ(z)uz]|z=z2 = b2 (110)

u(z) = 2Reint z

0w(z)dz + b0 w(z) = w0(z) +W (z) in D (111)

Lw =

wz

wzlowast

= 0 in

D+

Dminus

(112)

int intD+

g(ζ)ζ minus z

2g1(z)dνe1 +

int micro

(113)

g(z)=

A=ReA1+ImA1

2 B=

ReA1minusImA1

(114)

(115)

where

X(z) =2prod

j=1

ηj =

2|γj|+ δ γj lt 0

δ γj ge 0j = 1 2

(117)

u0(z) = 2Reint z

0w0(z)dz + b0 in D (118)

(119)

of mixed type

(121)

u0(z) = 2Reint z

u1(z) = 2Reint z

0w1(z)dz + b0 w1(z) = w0(z) + Φ1(z) + Ψ1(z)

Ψ11(z) =

int ν

2G1(z)dν G1(z) = Aξ0 +Bη0 + Cu0 +D

Ψ21(z) =

int micro

(123)

(124)

namelyCβ[Ψ1

n (128)

wn(z) =nsum

m=1

um(z) + u0(z) n = 1 2 (129)

Luz =

uzz

uzzlowast

D+

Dminus

(131)

M12 = k1 + k2 + |b0|+ |b1| (132)

Luz =

uzz

uzzlowast

D+

Dminus

(133)

M4k1 + k2 + 2k0tσ + 2k0t

τ + |b0|+ |b1| = t (134)

um+1 = C1[um+1 D] le M14 = M14(p0 β k0 D) (139)

um+1(z) = 2Reint z

+int x+y

(140)

m

0(1 + k20) From the

+int x+y

(142)

int z

M4k1 + k2 + 2k0tσ + 2k0t

τ + |b0|+ |b1| = t (144)

C1[u(z) D] le (M4 + 1)(2k1 + k2) (146)

12

partu

u(0)=b0 u(a)=b1 u(2)=b2 (22)12

partu

zisinL4

1|a(x)+b(x)| lek0

(23)

K =12(K1 +K2 +K3) (24)

Kj=[φj

π

γj=φj

|c1| |c2| le k2

Ψ(z) =

⎧⎪⎪⎨⎪⎪⎩int ν

a

1 int ν

2g1(z)dνe1 +

int micro

a

(27)

λ(x) =

1 + i x isin Lprime

0 = (0 a)

(28)

s(x)=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

+Re[λ(x)Ψ(x)]

xisin(0a)

h(x)=Re[λ(x)Ψ(x)]

(29)

X(z) =3prod

j=1

ηj =

(210)

u(z) = 2Reint z

0w(z)dz + b0 in D (211)

int intD+

g(ζ)ζ minus z

Ψ(z) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩int ν

a

1 int ν

2g1(z)dνe1 +

int micro

a

(212)

Im [λ(zprimej)Φ(z

(213)

λ(x) =

1 + i x isin Lprime

0 = (0 a)

0] le k3

(214)

where

X(z)=3prod

j=1

wz

wzlowast

D+

Dminus

(216)

2Re[1minus iradic2

uz

]=

partu

[1 + iradic2

uz

]=

partu

partl= 0 on (a 2) (218)

(220)

J =

∣∣∣∣∣u1(a) u2(a)

u1(2) u2(2)

∣∣∣∣∣ = 0 (221)

d1u1(a) + d2u2(a) = 0 d1u1(2) + d2u2(2) = 0 (222)

2 minus b2 (223)

C1[u D] le M18(k1 + k2)

(225)

1 Lprime2 Lprime

3 Lprime4 are

4 = γ2(x) + y = 0 l2 le x le 2(226)

12

partu

12

partu

=bj+2 j=12(227)

j] le k2 j = 1 4

maxzisinLprime

1

zisinLprime4

1|a(x) + b(x)| le k0

(228)

4where Lprime

1 Lprime2 Lprime

(230)

2 Lprime3 Lprime

ν =1

=12minusa

(231)

x =12(micro+ ν) =

(232)

and

x=12(micro+ν) =

12(2minusa)

+2(x+ y+a)minus2ax]

12(2minusa)

(233)

1

2

(234)

1

2

(236)

u(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

2Reint z

0w(z)dz + b0 in D+

2Reint z

a

2Reint z

2

(238)

12

partu

12

partu

[n2]sumj=0

A2jA2j+1

12

partu

[(n+1)2]sumj=1

B2jminus1B2j

(32)

Im [λ(z)uz]|z=A2j+1 = c2j+1 j = 0 1 [n2]

(33)

zisinL4

1|a(x) + b(x)| le k0

(34)

uzzlowast

= 0 in

D+

Dminus

(37)

Lw =

wz

wzlowast

= 0 in

D+

Dminus

(38)

Im [λ(z)w(z)]|z=A2j+1 = c2j+1 j = 0 1 [n2]

(39)

and the relation

u(z) = 2Reint z

2w(z)dz + b0 (310)

(311)

Re [λ(x)w(x)] =

1 cap AB

2 cap AB(312)

h(x)=2r((1 + i)x2 + 1minus i)

(313)

on L2j+1 j = 0 1 [n2]

on L2j j = 1 [(n+ 1)2]

2j cap AB j = 1 [(n+ 1)2]

u(z) = 2Reint z

2w(z)dz + c0 w(z) = w0(z) +W (z) in D (314)

int intD+

g(ζ)ζminusz

Ψ(z)=

⎧⎪⎪⎪⎨⎪⎪⎪⎩

int ν

a2j+1

2j+1 j=01[n2]

int ν

2g1(z)dνe1+

int micro

a2jminus1

(315)

g(z)=

A=ReA1+ImA1

2 B=

ReA1minusImA1

2 C=A2D=A3 in D+

(316)

Im [λ(z)(Φ(z) + Ψ(z))]|z=A2j+1 = 0 j = 0 1 [n2]

(317)

(319)

on L2j+1 j = 0 1 [n2]

on L2j j = 1 [(n+ 1)2]

w(z) = w0(z) +W (z) =

⎧⎨⎩ Φ(z)φ(z) + ψ(z) in D+

wzlowast

= Re [A1w] + A2u+ A3 in

D+

Dminus

(320)

Ψ1(z)=

⎧⎪⎪⎪⎨⎪⎪⎪⎩int ν

a2j+1

2j+1 j=0 1 [n2]int ν

2g0(z)e1dν+

int micro

a2jminus1

(321)

(j = 1 [(n+ 1)2]

(322)

where

X(z) =m+nprodj=0

n+1prodj=0

(323)

wz

wzlowast

D+

Dminus

u(zj) = 0 j = 0 1 m u(aj) = 0 j = 1 n

Im [λ(z)w(z)]|z=A2j+1 = 0 j = 0 1 [n2]

(324)

1) ulowast(aprime

m+n)] = [dlowast1 d

Im [λ(z)ukz]|z=A2j+1 = 0 j = 0 1 [n2]

(326)

J =

∣∣∣∣∣∣∣∣∣u1(aprime

m+n)

cprime1u1(aprime

thus the function

cprimekuk(z) in D

Luz=

uzz

uzzlowast

D+

Dminus

(327)

sumNj=1 Γj isin C2

12

partu

12

partu

maxzisinL1

zisinLprimeprime

1|a(x) + b(x)| le k0

(43)

0 The number

where

Kj=[φj

π

γj=φj

πminusKj j=1N (45)

j j = 1 N (46)

u(z) = 2Reint z

0w(z)dz + d1 w(z) = w0(z) +W (z) (47)

wz

wzlowast

= 0 in

D+

Dminus

(48)

int intD+

g(ζ)ζ minus z

Ψ(z) =

0g2(z)dmicroe2 +

int ν

0g2(z)dmicroe2 +

int ν

aj+1

(49)

g(z)=

A12+A1w(2w) w(z) =00 w(z) = 0 z isin D+

A =ReA1 + ImA1

2 B =

ReA1 minus ImA1

(411)

u0(z) = 2Reint z

0w0(z)dz + d1 (413)

on L0 in which

h(x)=[a

((1minusi)x2

)+b

((1minusi)x2

and

Ψ(x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int ν

2g1(z)dνe1 =

int ν

0g2(z)dmicroe2 =

int micro

0g2(z)dmicroe2=

int micro

Thus

w(z) = w0(z) +W (z) =

⎧⎨⎩ Φ(z)φ(z) + ψ(z) in D+

wzlowast

= Re [A1w] + A2u+ A3 in

D+

Dminus

(415)

wz

wzlowast

= Re [A1w] + A2u in

D+

Dminus

(416)

in which

w(z)=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Ψ(z)=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int micro

int ν

2jminus1

j = 1 2 Nint micro

int ν

aj+1

j = 1 2 N minus 1

(418)

n in Dminus

uzz = Re [A1uz] + A2u in D+ (420)

12

partu

(422)

and the relation

u(z) = 2Reint z

0w(z)dz + d1 (423)

(424)

f(x+ y) = Re [(1 + i)w(x+ y)]

1 0 b1

j j=2N

2j j=12Nminus1

2jminus1 j=2N

wz

wzlowast

= Re [A1w] + A2u+ A3 in

D+

Dminus

(426)

int intD+

g0(ζ)ζ minus z

Ψ1(z) =

0g0(z)dmicroe2 +

int ν

0g0(z)dmicroe2 +

int ν

aj+1

(429)

+Re [λ(x)Ψ1(x)]

h(x) =[a

((1minus i)x

2

)+ b

((1minus i)x

2

)][Re (λ(z1)(r(z1) + ic1))

+Im (λ(z1)(r(z1) + ic1))] on L0

(430)

Ψn(z) =

int ν

aj+1

and then|[wplusmn

0g0(z)e2dmicro|

2g0(z)e1dν|+

Nminus1sumj=1

|int micro

aj+1

g0(z)e1dν|]

(433)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int micro

int ν

2jminus1

j=12Nint micro

int ν

aj+1

j=12N minus1(434)

0

Rprimenminus1

(n minus 1) dRprime

n in Dminus

(435)

ie

Ψlowast(z)=

int ν

aj+1

(437)

(438)

(440)

where

X(z) =2Nprodj=1

wz

wzlowast

= Re [A1w(z)] + A3 in

D+

Dminus

(442)

Im [λ(zprimej)w(z

primej)] = dprime

j j = 1 N(443)

0w(z)dz + d0 in D

(446)

in which d0 = φ(0)

λ(z) = a+ ib =

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

(1 + i)radic2 on L1

j=2L2jminus1

(447)

and

r(z) =

j=1L2jminus1

c1=Im[1 + iradic2

uz(z1 minus 0)]=

φx minus φxradic2

|z=z1minus0=0

uz(zj + 0)]= 0 j = 2 N

(448)

Re[1minus iradic2

w(x)]= s(x) =

πminus K1 =

14

minus 0 =14

lt 1

πminusK2N=minus5

4lt0

πminus Kj=

54minus1= 1

4lt1 j=3 5 2Nminus1

πminus Kj=minus1

4=minus1

4lt0 j=2 4 2Nminus2

N

2minus 1

References

References 241

242 References

References 243

244 References

References 245

246 References

References 247

248 References

References 249

Index

128 130 132 133 149 150 151153 155 165ndash167 174 184 190202 212 223 228 230 232

AV Bitsadze 157

Banach space 53 55 93 104 137 138191 193 207 208

217 219 220 227 228 232 237238 239

Index 251

108 109 159 164 173 196 202210 219 226

problem 79 80 90 91 93 94125 143 144 145 146 156

140 158 171 200 204 209 214218 227 228

87 90 93 103 106 108 119 122126 134 143 157 177 182 193198 213 218 226 233

252 Index

179 180 182ndash184 186 188 189191 193 194

103 107 120 125 126 139 144145 156 158 160ndash164 180 181

Index 253

186 196 202 228 229 232 237238 239

140 141 216inversion 97

137 141 142 162 165 167 168171 175 176 177 184 189 192212 218 224 237 239

238 239

254 Index

110 111 158 159 160 161 162196 238 239

Problem P0 44 67 97 109 118 158196 197 202

225Problem P prime

0 228Problem Q 97 106 171ndash177 202

mixed type 79 134 143 199 200209 215 218 227 228

Index 255

96 107

171 206 225representation 20 25 30 35 36 38

43 46ndash48 63 64 79 87 90ndash92 94134 160 182 198 229

50 52 54 75 104 131 132 138152 153 169 170 205 207 208235 236

160

43 66 106 108 119 126 134 139141 143 157 194 200 208 218

94 104 119 157 160 168 173175 177 180 182 189 191 194200 206 209 211 213 231

subsequence 102 105 106 111 116132 154 170

77 93

55ndash58 60 67 80 88 97 111120 140 141 142 158 196202 216 217

209 210 213 227 229 236

98 102 106 111 131 132 138153 154 170 205 208 236

256 Index

53 56 58 63 64 65 72 78 8693 103 104 106 116 125 133135 137 141 142 144 147 150155 159 162 165 189 192 204207 208 212 217 224 227 238

90 93 96 103

zone domain 90 94

Book Cover
HALF-TITLE
SERIES TITLE
TITLE
COPYRIGHT
CONTENTS
PREFACE
REFERENCES
INDEX

Page 3: Linear and quasilinear complex equations of hyperbolic and ...webéducation.com/wp-content/uploads/2018/09/Linear...and hyperbolic pseudoregular functions. Next we transform the linear

London and New York

(Print Edition)

Contents

vi Contents

Contents vii

Preface

x Preface

CHAPTER I

e1 = (1 + j)2 e2 = (1minus j)2 (11)

e1 + e2 = 1 ekel =

ek if k = l

0 if k = lk l = 1 2 (12)

in which

1z=

z

zz=

1x+ y

e1 +1

x minus ye2 =

1micro

e1 +1νe2

z1

(1micro2

e1 +1ν2

e2

)=

micro1

micro2e1 +

ν1

ν2e2

(15)

(1) wz = 0 (16)

(2) ξν = 0 ηmicro = 0 (17)

(3) ux = vy vx = uy (18)

[ξ(z)minusξ(z0)

e2

]

respectively

2

ez = 1 + z +z2

ex+y + exminusy

2+ j

ex+y minus exminusy

2

2+ j

sin z = z minus z3

cos z = 1minus z2

tgz=sin z

(1

cosmicroe1 +

1cos ν

e2

ctgz=cos zsin z

1sinmicro

e1+1

sin νe2

y(z0)∆y + ε(∆z)

where

y(z0) = uy(z0) + jvy(z0)

lim∆zrarr0

ε(∆z) rarr 0

Cf(z)dz =

intC

udx+ vdy + j[int

Cvdx+ udy] =

Df(z)dxdy =

int intD

udxdy + jint int

Dvdxdy

C[f(z) + g(z)]dz =

intC

f(z)dz +int

Cg(z)dzint int

int intD

f(z)dxdy +int int

Dg(z)dxdy

int

Cf(z)dz le

radic2M1l

int int

Df(z)dxdy le

radic2M2S

G[f(z)]zdxdy =

j

2

intC

f(z)dz

2

intC

f(z)dz

Cf(z)dz = 0

ImF (z)G(z) = 0 in D (19)

w(z0) = δ0F (z0) + γ0G(z0) (110)

w(z0) = limzrarrz0

= limzrarrz0

ie

Wz(z0) = w(z0) (116)

andWz(z0) = 0 (117)

W (z) =

∣∣∣∣∣∣∣∣∣w(z) w(z0) w(z0)

F (z) F (z0) F (z0)

G(z) G(z0) G(z0)

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣F (z0) F (z0)

G(z0) G(z0)

∣∣∣∣∣∣ (118)

Fz(z0) F (z0) F (z0)

Gz(z0) G(z0) G(z0)

∣∣∣∣∣∣∣∣∣ = 0 (119)

w(z0) =

∣∣∣∣∣∣∣∣∣wz(z0) w(z0) w(z0)

Fz(z0) F (z0) F (z0)

Gz(z0) G(z0) G(z0)

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣F (z0) F (z0)

G(z0) G(z0)

∣∣∣∣∣∣ (120)

wz = aw + bw (121)

where

F G minus FG b =

FGz minus FzG

FG minus FG

F G minus FG B =

FGz minus FzG

FG minus FG

(123)

respectively

w(z) = φ(z)F (z) + ψ(z)G(z) (124)

K(z) = φ(z) + jψ(z) (125)

Fφz +Gψz = 0 (126)

holds where

(128)

Proof From

Wz(z0) = F (z0)φz(z0) +G(z0)ψz(z0)

Setting

minusG

minusF

(129)

where

δ =σy + τx

τ

δ =σy + τx

τ

(132)

φ(z) =int z

z0

wz = a(z)w(z) + b(z)w(z) (133)

a = a(FG) and b = b(FG) (135)

F (z) = 1 or G(z) = j in D (137)

I = (K2 +K3)2 minus 4K1K4 gt 0 (22)

holds in which

K1 =∣∣∣∣∣a11 b11

a21 b21

∣∣∣∣∣ K2 =∣∣∣∣∣a11 b12

a21 b22

∣∣∣∣∣ K3 =∣∣∣∣∣a12 b11

a22 b21

∣∣∣∣∣ K4 =∣∣∣∣∣a12 b12

a22 b22

where

K5 =∣∣∣∣∣a11 a12

a21 a22

∣∣∣∣∣ K6 =∣∣∣∣∣ b11 b12

b21 b22

a = K1K6 b = K3K6 c = K4K6 d = K2K6

∆ = I4K26 =

(b+ d)2

4minus ac gt 0 (26)

+lower order terms

namely

= (1 + a)(1minus c) + bd gt 0

in which

q0

(29)

then we can prove

|Q1|+ |Q2| = |Q1Q1|12 + |Q2Q2|12 lt 1 (210)

∆ = I4K24 = (b+ d)24minus ac gt 0 (212)

whereq1 =

2

It is clear that

q0 = (1 + q1)(1 + q1)minus q2q2

4

= (1 + a)(1minus c) + bd gt 0

in which

Q1(z) =[(1minus q1)(1 + q1) + |q2|2]

q0 Q2 =

minus2q2(z)q0

(216)

Fkux =int 1

Fkuy =int 1

Fkvx =int 1

Fkvy =int 1

(217)

holds in which

K1 =∣∣∣∣∣a11 b11

a21 b21

∣∣∣∣∣ K2 =∣∣∣∣∣a11 b12

a21 b22

∣∣∣∣∣ K3 =∣∣∣∣∣a12 b11

a22 b21

∣∣∣∣∣ K4 =

∣∣∣∣∣a12 b12

a22 b22

∣∣∣∣∣ K5 =∣∣∣∣∣a11 a12

a21 a22

∣∣∣∣∣ K6 =∣∣∣∣∣ b11 b12

b21 b22

6

Vu Vv

∣∣∣∣∣∣ =∣∣∣∣∣∣λ 1

micro 1

andu =

U minus V

λ minus micro v =

microU minus λV

micro minus λ

1V + c1

2V + c2(222)

where

aprime1 =

a1 minus microb1

1 =minusa1 + λb1

2 =a2 minus microb2

2 =minusa2 + λb2

λ minus micro

2b12)V + (c1b22 minus c2b12)]K6

1b21)V + (c2b11 minus c1b21)]K6(223)

and then

4 gt 0

Letξ =

x minus microy

λ minus micro η =

minusx+ λy

λ minus micro

ηx ηy

minus1 λ

= 0

a22uξ + b22vη = a2u+ b2v + c2(226)

1v + cprime1

2v + cprime2

(227)

whereaprime

2(ζ w)w + Aprime3(ζ w) (228)

in which Aprime1 A

prime2 A

ux = au+ bv + f vy = cu+ dv + g (230)

wz = 0 in D (31)

zisinL4

1|a(z) + b(z)| le k0 (34)

times g(0) = 2r((1 + j)x) on [0 R]

f(0)=u(0)+v(0)=r(0)+b1

a(0)+b(0) g(0)=u(0)minusv(0)=

r(0)minusb1

a(0)minusb(0)

(38)

f(x+ y) =2r((1 + j)(x+ y)2)

a((1 + j)(x+ y)2)+b((1 + j)(x+ y)2)(39)

0lex+yle2R

0lexminusyle2R1

(310)

wz = A1(z)w + A2(z)w + A3(z) (312)

(314)

in which

[a2(z0)minus b2(z0)] (316)

w(z) = w0(z) + Φ(z) + Ψ(z) in D

Ψ(z) =int xminusy

int x+y

0[Cξ +Dη + F ]d(x+ y)e2

(317)

[Ψ]z = [Aξ +Bη + E]e1 + [Cξ +Dη + F ]e2 (319)

wz = A1w + A2w in D (320)

w(z) = Φ(z) + Ψ(z)

Ψ(z) =int xminusy

int x+y

(322)

Ψ(z) le 8M3mR0 Φ(z) le 32M3k20(1 + k2

0)mR0

w1(z) = w0(z) + Φ1(z) + Ψ1(z)

Ψ1(z) =int ν

0[Aξ0 +Bη0 + E]e1dν +

int micro

(323)

w2(z) = w0(z) + Φ2(z) + Ψ2(z) in D

+int micro

and

2

wn(z) = w0(z) + Φn(z) + Ψn(z)

Ψn(z) = +int ν

int micro

(326)

+int micro

and then

0

Rprimenminus1

n (328)

Ψlowast(z) =int ν

int micro

(330)

w(z) = u(z) + jv(z)

(333)

a(z1)+b(z1)

Re [λ(z)w(z)]=u(z)+v(z)=r[(1+

R+R1

RminusR1j)

xminusj2RR1

RminusR1

]on [R1 R]

(334)

RminusR1

)xminus 2RR1

RminusR1

]

t

2R+ (1minus j)R1

] t isin [0 2R]

and then

0lexminusyle2R1

(335)

Condition C

(45)

in which

wz = A1w + A2w in D (46)

w(z) = Φ(z) + Ψ(z)

Ψ(z) =int xminusy

int x+y

(48)

w1(z) = w0(z) + Φ1(z) + Ψ1(z)

Ψ1(z) =int ν

0[Aξ0 +Bη0 + E]e1dν +

int micro

(49)

Ψn(z) =int ν

int micro

(411)

+int ν

+int micro

0

Rprimenminus1

n (413)

n=0

= M6 (415)

C[w(z) D] le M7 = M7(α k0 D)

Ψ11(z) =

int xminusy

1(z) =int x+y

0G2(z)d(x+ y)

prime (420)

prime(421)

prime

prime(423)

n =Re wn + Im wn w2

(M12Rprime)n

n Cα[w1

n

prime)n

n Cα[w2

prime)n

n

(424)

x =12(micro+ ν)

4R minus 4σ(x+ γ1(x)) + 2x+ 2γ1(x)

(429)

x =12(micro+ ν) =

2

(430)

2 Im [λ(z0)w(z0)] = b1 (432)

2R (433)

2 Im [λ(z0)w(z0)] = b1

4 are as follows

4 = minusγ4(x) + y = 0 0 le x le l2(435)

Hence we have

x =12(micro+ ν) =

(439)

x =12(micro+ ν) =

(440)

1 cup Lprimeprime4

is reduced to

4 Im [λ(z0)w(z0)] = b1

radius R1 namely

radicR2

1 + 2R1ν minus ν2

2

radicR2

2

1 Lprimeprime2 L

primeprime3 L

primeprime4

ξν = 0 ηmicro = 0 (52)

microy νy

∣∣∣∣∣ =∣∣∣∣∣ 1 1

1 minus1

∣∣∣∣∣ = minus2∣∣∣∣∣ ξu ηu

ξv ηv

∣∣∣∣∣ =∣∣∣∣∣ 1 1

1 minus1

∣∣∣∣∣ = minus2 (53)

ξ = ξ(micro) η = η(ν) (55)

wz = Q(z)wz Q(z) = a(z) + jb(z) (56)

w(z) = Φ[χ(z)] (58)

ξσ = 0 ητ = 0 (σ τ) isin G (513)

wz = Q(z)wz (514)

Wz = 0 (516)

CHAPTER II

( )z =( )x + j( )y

2 ( )z =

( )x minus j( )y2

4

( )zz =( )xx + ( )yy + 2j( )xy

4 ( )zz =

4

(14)

in which

Q =a+ c+ 2bj

a minus c A1 =

d+ ej

a minus c A2 =

f

2(a minus c) A3 =

g

2(a minus c)

|Q(z)| lt 1 in D (17)

respectively

in which

a =int 1

2b =int 1

c =int 1

d =int 1

e =int 1

f =int 1

0Φτu(x y τu 0 0 0 0 0)dτ = f(x y u)

g = minusΦ(x y 0 0 0 0 0 0) = g(x y)

and

a2 = f = minusint 1

0Fτu(z τu 0 0 0 0)dτ = a2(z u)

a3 = minusF (z 0 0 0 0) = a3(z)

(112)

or its complex form

(120)

A1 =a+ jb

2 A2 =

c

4 A3 =

d

4in D

wz =wx + jwy

wx minus jwy

(122)

in which

A =a+ b

4 B =

a minus b

4 C =

c

4 D =

d

4

12

partu

u(0) = b0 Im [λ(z)uz]|z=z3 = b1

(21)

maxzisinL3

1|a(z) + b(z)| le k0

(22)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Re [λ(z)w(2Re1 + νe2)] = r(z)

(24)

u(z) = φ(z) on L = L3 cup L4 (25)

in which

λ(z) = a+ jb =

1 + jradic2on L4

b+1 = Im

[1minus jradic2

2radic2

bminus1 = Im

[1 + jradic2

uz(z3)]=

(27)

wz = Re [A1w] + A2u+ A3 in D (28)

int z

0w(z)dz + b0 in D (210)

2ReintΓw(z)dz =

intΓw(z)dz +

intΓw(z)dz

= 2jint int

s1

w(z)dz +int

s2

w(z)dz]+ b0

uzz = 0 (211)

wz = 0 in D (212)

f(2R) = u(z3) + v(z3) =r((1 + j)R) + b1

a((1 + j)R) + b((1 + j)R)

a((1 + j)R)minus b((1 + j)R)

(213)

a((1 + j)(x+ y)2) + b((1 + j)(x+ y)2)

0 le x+ y le 2R

α[u(z) D]leM2k2=M2(α k0 D)k2 (215)

[a((1 + j)x) + b((1 + j)x)]f(2x) + [a((1 + j)x)

f(2R)=u(z3)+v(z3)=r((1 + j)R)+b1

a((1 + j)R)+b((1 + j)R)

[a((1 + j)t2) + b((1 + j)t2)]f(t) + [a((1 + j)t2)

u(z) = 2Reint z

0w(z)dz + b0 in D

w(z) = W (z) + Φ(z) + Ψ(z) in D

Ψ(z)=int ν

int micro

(216)

u0(z) = 2Reint z

0w0(z)dz + b0 w0(z) = W (z) = ξ0e1 + η0e2 (221)

u1(z) = 2Reint z

0w1(z)dz + b0 w1(z) = w0(z) + Φ1(z) + Ψ1(z)

Ψ1(z) =int xminusy

+int x+y

(222)

C1[u0(z) D] le M3C[w0(z) D] + k2 (223)

C[Φ1(z) D] le 8M4k20(1 + k2

0)[(4 +M3)m+ k2 + 1]Rprime

(224)

wn(z) satisfying

un+1(z) = 2Reint z

0wn+1(z)dz + b0

+int micro

(225)

and then

0

Rprimenminus1

(n minus 1) dRprime

n

(226)

+int micro

(228)

int z

wz = Re [A1w] + A2u in D (230)

int z

w(z) = Φ(z) + Ψ(z)

Ψ(z) =int ν

int micro

(233)

n

maxzisinD

C1[u D] le M7 C1[u D] le M8k (33)

(116)

α[u D] le M11k (34)

u0(z) = 2Reint z

0w0(z)dz + b0 w0(z) = W (z) = ξ0e1 + η0e2 (35)

u1(z) = 2Reint z

0w1(z)dz + b0 w1(z) = w0(z) + Φ1(z) + Ψ1(z)

Ψ1(z) = Ψ11(z)e1 +Ψ2

1(z)e2 Ψ11(z) =

int ν

0G1(z)dν

Ψ21(z) =

int micro

(38)

prime (39)

1(z) G2(z) Ψ21(z)

(310)

prime (311)

prime

(312)

Cα[w1n(z) D] le

(M18Rprime)n

n Cα[w2

n(z) D] le(M18R

prime)n

n

n C1

n

(313)

wn(z) = w0(z) +nsum

m=1wm(z) un(z) = u0(z) +

nsumm=1

um(z) (n = 1 2 )

α[u(z) D] le M10 (314)

F = Re [A1uz] + A2u+ A3

(315)

M8k1 + k2 + 2k0tσ + 2k0t

τ + |b1|+ |b0| = t (317)

(321)

um+1 = C1[um+1 D] le M20 (322)

um+1(z) = 2Reint z

+int x+y

(323)

n

0(1 + k20) From

+int xminusy

+int x+y

(325)

int z

M8k1 + k2 + 2k0tσ + 2k0t

τ + |b0|+ |b1| = t (327)

prime4 and

Hence we have

x =12(micro+ ν) =

(334)

and

x =12(micro+ ν) =

(335)

4 u(0) = b0 Im [λ(z3)w(z3)] = b1 (337)

u(z) = 2Reint z

4is reduced to

4 are as follows

(341)

micro =2Rmicro

x =12(micro+ ν) =

(345)

x =12(micro+ ν) =

(346)

4is reduced to

4

3 )] = b1 (350)

1 Lprimeprime2 L

primeprime3 L

primeprime4 namely

Lprimeprime1 =

radicR2

Lprimeprime2 =

radicR2

Lprimeprime3 =

y = γ3(x) =

radicR2

Lprimeprime4 =

y = γ4(x) =

radicR2

Lprimeprime1 =

⎧⎨⎩x = σ1(ν) =R1 + ν minus

radicR2

1 + 2R1ν minus ν2

2

⎫⎬⎭

Lprimeprime2 =

radicR2

2

⎫⎬⎭

Lprimeprime3 =

⎧⎨⎩x = σ2(ν) =R1 + ν minus

radicR2

1 + 2R1ν minus ν2

2

⎫⎬⎭

Lprimeprime4 =

radicR2

2

⎫⎬⎭

u(0) = b0 Im [λ(z)uz]|z=z1 = b1(41)

zisinL2

1|a(z) + b(z)| le k0

(42)

u(0) = b0 Im [λ(z)uz]|z=z1 = b1(43)

(44)

u(0) = b0 Im [λ(z)uz]|z=z1 = b1(45)

1|a(z) + b(z)| le k0

(46)

u(z) = 2Reint z

0w(z)dz + b0 in D (49)

(410)

f(0) = u(z1) + v(z1) =r((1minus j)R1) + b1

(411)

f(x+ y) =2r((1+j)(x+y)2+(1minusj)R1)

0 le x+ y le 2R (412)

α[u(z) D] le M24k2 (413)

u(z) = 2Reint z

Ψ(z) =int ν

2R1

(414)

(415)

(416)

Im [λ(z1)w(z1)] = b1(417)

(420)

in which

2 τ(x) =

(422)

for equation (211)

2 0 le x+ y le 2R

(423)

and

f(0) =sprime(0) +R(0)

2 g(2R1) =

(424)

for (116)

f(0) = u(z1) + v(z1) =r((1minus j)R1) + b1

(428)

0 le x+ y le 2R1

Ψ(z) =int ν

2R1

C1[s(x) L0] le k2 C1[φ(x) L1] le k2 (432)

u(0) = b0 Im [λ(z)uz]|z=z1=b1

(433)

where we choose

radic2 (434)

L1=x+int y

0

L2=

xminusint y

0

radicminusK(t)dt =

int y

0|t|m2dt = minus

int |y|

0d

2m+ 2

|t|(m+2)2 = minus 2m+ 2

|y|(m+2)2

L1 x minus 2m+ 2

|y|(m+2)2 = 0 L2 x+2

m+ 2|y|(m+2)2 = 2 ie

L1 y = minus(m+ 22

x)2(m+2) L2 y = minus[m+ 22

(2minus x)]2(m+2)

where

A1 =d+ je

K(y)minus 1 A2 =

f

2(K(y)minus 1) A3 =

g

2(K(y)minus 1)

12

partu

u(0) = b0 Im [λ(z)uz]|z=z1 = b1

(54)

[radic

α[r(z) Lj] le k2 j = 1 2

zisinL2

1|a(z) + b(z)| le k0

(55)

int y

0

L2 = BC =x = 2 +

int y

0

(56)

Re [(1 + j)(U + jV )] = U + V = r(x) on L2

Im [(1 + j)(U + jV )] = [U + V ]|z=z1+0 = r(1 + 0)

whereU =

λ(z)=

andu(z) = 2Re

int z

x+int y

0

int y

0

int y

0

int y

0

int y

0

radicminusK(t)dt

minusK(y)radic

(510)

(radicminusK(y)

)micro= minus1

K prime

4K(radic

minusK(y))

K primeU4K

(U + V )ν =radic

(514)

minus2radic

minusK(y)(U + V )ν

=radic

= minusradic

K prime(y)2K(y)

U 2radic

=radic

+radic

=radic

K prime(y)2K(y)

U

(515)

U ie

(516)

u(z) = 2Reint z

partu

u(0) = b0 Im [λ(z)(U + iV )]|z=z1 = b1

(519)

2 V (x y) =

2

(520)

where

f(0) = U(0) + V (0) =r(1) + b1

a(1) + b(1) g(2) = U(2)minus V (2) =

r(1)minus b1

a(1)minus b(1)

U =12

f(micro) +

V =12

a(ν2)minus b(ν2)

or

U =12

g(ν) +

V =12

minusg(ν) +

(521)

a(x2+1)minusb(x2+1)

a(x2) + b(x2) x isin [0 2]

(522)

W (z) =

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

12

[(1 + j)

a(ν2)minus b(ν2)

] (523)

can be derived

U + V = minusint ν

int micro

4K(y)Udmicro

W = U+jV =minus1+j

2

int ν

4K(y)Udν+

1minusj

2

int micro

4K(y)Udmicro

= minus1 + j

2

int ν

2

K prime(y)4K(y)

Udν +1minus j

2

int micro

0

K prime(y)4K(y)

Udmicro

radic1 +

(dy

dx

)2dx =

radic1minus K

1minus Kdy

radic1 +

(dy

dx

)2dx =

radic1minus K

minusKdx =

radic1minus Kdy

(524)

=1radic1minusK

]

=1

2radic1minusK

K prime(y)K(y)

U

=1radic1minusK

]

2radic1minusK

K prime(y)K(y)

U

(525)

0

1

2radic(1minus K)

K prime(y)K(y)

Uds1 =int s1

0

12

K prime(y)K(y)

Udy

0

1

2radic(1minus K)

K prime(y)K(y)

Uds2 =int s2

0

12

K prime(y)K(y)

Udy

(526)

W (z) = U + jV = W (z) + Φ(z) + Ψ(z)

U+V =minusint s1

0

1

2radic(1minusK)

K prime(y)K(y)

0

1

2radic(1minusK)

K prime(y)K(y)

Uds2(527)

u(z) = 2Reint z

int z

can be replaced by

2U(x)=S(x)=radic

0

+u(0)=s(x)

2V (x) = R(x) = uy on AB = L0 = (0 2)

(529)

(532)

v(x) = 0 vy(x) = 0 on L0 (533)

ξν =1

2radic1minus K

2K prime(y)K(y)

)ξ

+ e minus 12

K prime(y)K(y)

]

ηmicro =1

2radic1minus K

12

K prime(y)K(y)

)ξ

+ e+12

K prime(y)K(y)

]

(534)

K prime(y)K(y)

=m|y|mminus1h(y)

K(y)minus |y|mhy

K(y)=

m

y+

hy

h

v(z) =int y

ξ(z) = minusint y

η(z) =int y

(535)

in this case

A1 = minuse

2minus hy

4h B1 =

e

2minus hy

4h A2 = minuse

2+

hy

4h B2 =

e

2+

hy

4h

C1 = minus12

dradicminusKminus m

4y C2 = minus1

2dradicminusK

+m

4y D = minusf

2 E = minusG

2

limyrarr0

0

(538)

3Mδ

2+[ε(y)M +m2]δm2

m+ 2lt γ

6δ2M2

m+ 6+8δMm+ 2

+2ε(y)M +m

m+ 2lt γ

(539)

1 s02

ξ1(z)=minusint y

int y

0Edy z isin s0

1

η1(z)=int y

0[A2ξ0+B2η0+C2(ξ0+η0)+Dv0+E]dy=

int y

0Edy zisin s0

2

v1(z) = Reint y

(540)

1

ηk+1(z) =int y

2

vk+1(z) =int y

(541)

j=0γj|y|m2+1

(542)

j=0γj|y|

Moreover we get

int y

0Ex[x1minusx2]dy|

le 4m+4

j=0γj|y|m2+1

le |int y

int y

0|x1 minus x2|dy|

le M |int y

1sumj=0

γj|y|m2+1

1sumj=0

γj|y|m2+1

|vn+1(z)| le |int y

int y

0

nsumj=0

γjydy|leMnsum

j=0γj|y|2 leM

n+1sumj=0

γj|y|

0

⎡⎣(|A1|+|B1|+|D|)Mnsum

j=0γj|y|+|C1|M

nsumj=0

γj|y|m2+1+|E|⎤⎦ dy|

le M |int y

0

⎡⎣3M nsumj=0

γj|y|+(

ε(y)2|y|radich

+m

4|y|)

nsumj=0

γj|y|m2+1 + 1

⎤⎦ dy|

le M |y|⎧⎨⎩[32M |y|+

(|ε(y)|M +

m

2

) |y|m2

m+ 2

]nsum

j=0γj + 1

γj|y|

and

0

dy|

Noting that

lensum

nsumj=0

γj|y|m2+1

le M2nsum

nsumj=0

γj|y|m2+1

le (M |y|+1)M2nsum

j=0γj|y|m2+1

and

j=0γj +

∣∣∣∣∣hy

2h

j=0γj|y|m2+1

we get

0[(3M |y|+ 4)M2

nsumj=0

γj|y|m2+1

+(|C1|+ |C2|)Mnsum

j=0γj|y|m2+1 +M2|y|m2+1]dy|

le M |y|m2+1

[6M2y2

m+ 6+8M |y|m+ 2

+(|ε(y)|M +

m

2

)

times 2m+ 2

nsumj=0

γj +2M

m+ 4|y|⎤⎦

le Mn+1sumj=0

γj|y|m2+1

ηlowast(z) =int y

vlowast(z) =int y

A1 = minus d

4|y|m2 + j

(m

8|y| minus e

4

) A3=minusf

4

A2 = minus d

4|y|m2 + j

(m

8|y| +e

4

) A4=minusg

4

(543)

and

u(z)=2Reint z

0uzdz=2Re

int z

0(U+jV )d(xminusjy)

CHAPTER III

]=

φj

πminusKj

Kj =[φj

π

]+ Jj Jj = 0 or 1 j = 1 m

(13)

12

msumj=1

[φj

πminus γj

](14)

(15)

Ψ(z) =Φ(z)Π(z)

Π(z) =mprod

j=1

(z minus xj

z + i

)γj

(16)

)=

12

msumj=1

[φj

πminus γj

]=12

msumj=1

Kj (110)

Re[Λ(x)

x minus x0

x+ i

x minus x0Ψ0(x)

Ψlowast(z) = i(

z minus i

z + i

)[K] (z minus x0

z + i

)eiS(z) in D (112)

Re [S(x)] = arg

⎡⎣Λlowast(x)(

x minus i

x+ i

)[K](x minus x0

x+ i

)⎤⎦ x isin L Im [S(i)] = 0 (114)

Ψ0(z) =

X(z) = Π(z)Ψ0(z) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩i(

z minus i

z + i

)K

i(

zminusi

z+i

)[K] zminusx0

(116)

Re[Φ(x)iX(x)

]=

c(x)iλ(x)X(x)

=λ(x)c(x)iX(x)

Φ(z)iX(z)

=1πi

[int infin

minusinfinλ(t)c(t)

(t minus z)iX(t)dt+

Q(z)i

]

Φ(z) =X(z)πi

[int infin

minusinfinλ(t)c(t)

]

(118)

Q(z) = i[K]sumj=0

[cj

(z minus i

z + i

)j

+ cj

(z minus i

z + i

)minusj]+

iclowastx0 + z

(119)

Q(z) =

iclowastx0 + z

(120)

int infin

minusinfinλ(t)c(t)

x0 + z

x0 minus z

=int infin

minusinfinλ(t)c(t)

x0+i

x0minusi

=infinsum

j=0

int infin

[1 +

z minus i

x0 + i

]x0 + i

x0 minus i

infinsumj=0

(z minus i

x0 minus i

)j

=int infin

dt+iclowastx0+i

x0minusi+

infinsumj=1

[int infin

minusinfinλ(t)c(t)

(tminusi)j+1X(t)dt+

2iclowastx0

(x0minusi)j+1

](zminusi)j

(121)

minusinfinλ(t)c(t)

int infin

minusinfinλ(t)c(t)

X(t)(t minus i)jdt+

2iclowastx0

when 2K is odd

(122)

x0 + i

int infin

minusinfinλ(t)c(t)

πi

[int infin

minusinfinλ(t)c(t)

2iclowastx0

](124)

Φ(z) =

int z

iΦ(z)dz in D

λ(x) =

radicΠn

]=

φj

πminus Kj

Kj =[φj

π

]+ Jj Jj = 0 or 1 j = 1 m

(128)

(129)

Φ(z) =X(z)2πi

[intΓ

dt+Q(z)] (130)

with

j + cjzminusj) +

iclowastz0 + z

(131)

intΓ

λ(t)c(t)X(t)tj

intΓ

λ(t)c(t)X(t)tj

Φ(z) =X(z)z[K]

πi

[intΓ

0

] (133)

clowast = iintΓ

λ(t)c(t)X(t)t

dt

X(z) =

z(ζ) =1

ζ + i

[1 + iζ minus

radic2(1minus ζ2)

]

Im [λ(zprimej)Φ(z

primej)] = bj j = 1 m (136)

where zprime1 z

0

primej)]=0 j = 1 m (137)

m = 2K + 1 le 2ND +NΓ le 2K (138)

j=1djΦj(z) (139)

msumj=1

dprimejΦj(zprime

j) j = 1 m (140)

Φlowast(zprimej) =

msumj=1

dprimejΦj(zprime

j) = 0 j = 1 m (141)

j=1 dprimejΦj(z)

r(x)=

(143)

λ(x) =

(144)

Re [λ(z)Φ(z)] = r(z) =

0 z isin L0(145)

Ψ(z) =Φ(z)Π(z)

j=n+1j =m

(z minus xj

z + i

)γj

(146)

Re [λ(z)Φ(z)] = 0 z isin L (147)

)=

(149)

Φ(z) =

⎧⎪⎨⎪⎩Ψ(z) in D = |z| lt 1 Im z gt 0

Re [Λ(z)Φ(z)] = R(z) on |z| = 1 (152)

where

Λ(z)=

⎧⎨⎩λ(z)

λ(z)R(z)=

Φ(z) = Φ(i1 + ζ

1minus ζ

)in D (154)

Condition C

w(z) = Φ[ζ(z)]eφ(z) + ψ(z) (26)

w(z) = Φ(z)eφ(z) + ψ(z) (210)

in which

X(z)=mprod

j=1j =nm

ψz = Qψz + A1ψ + A2ψ + A3 Q =

φz = Qφz + A A =

(213)

int intD

f(ζ)ζ minus z

dσζ

φ(z) = Tg ζ(z) = Ψ[χ(z)] χ(z) = z + Th

(214)

h(z) = Q(z)[1 + Πh] (218)

prime1Γ

prime2 isin C1

j)] = bj j = 1 m(225)

w(z) = Φ(z)eφ(z) + ψ(z) (228)

in which

X(z) =mprod

j=2j =1n

(230)

12[ux + iuy] uzz =

14[uxx + uyy]

a+ c A1(z) =

minusd minus ei

a+ c A2(z) =

minusf

2(a+ c) A3(z) =

g

2(a+ c)

Condition C

|Q(z u w)| le q0 lt 1 (34)

12

partu

Im [λ(z)uz]|z=0 = b2 (37)

ψ(z) = 0 Ψ(z) = 1 on Γ (39)

β[Ψ(z) D] le M2

z(ζ) =1

ζ + i

[1 + iζ minus

radic2(1minus ζ2)

] (311)

ψ(ζ) = 0 Ψ(ζ) = 1 on L = ζ(Γ)

ψ(z) =

⎧⎨⎩ψ(z) in D

Q=

⎧⎨⎩Q(z)

Q(z)A1=

⎧⎨⎩A1(z)

A1(z)A2=

⎧⎨⎩A2(z)

A2(z)A3=

⎧⎨⎩A3(z)

minusA3(z)in

⎧⎨⎩D

D

⎫⎬⎭

0w(z)dz+b0 w(z)=Φ[ζ(z)]eφ(z) (317)

(318)

Ψ(z)in D (319)

12[partU

and give a lemma

partu

partl= lim

z(isinl)rarrz0

2

partV

partn= 2Re

zVz

R z isin part∆

partV

partn= minus4Re zVz

R |z| = R

2

partV

partnge minusε

partV

partngt 0

partu

partn= Ψ

partV

partn+ V

partΨpartn

ge minusεpartV

partn+ V

partΨpartn

gt 0

partu

partl= cos(l n)

partu

partn+ cos(l s)

partu

partsgt 0 (326)

12

partu

(327)

C1δ [u(z) D] le M6(k1 + k2)

(328)

j (j = 1 2 3) λm rm bmj (j = 0 1) but

Umzz minus Re [QmUm

zz + Am1 Um

z ]minus Am2 Um =

Am3

Hm

partUm

partνm

=rm(z)Hm

Hm

Um(1) =bm1

Hm

Lp[Am2 Um +

Am3

Hm

Cα

[|z minus zj|βj

rm(z)Hm

Γlowast]

le M7 j = 1 m

∣∣∣∣∣ bmj

Hm

∣∣∣∣∣ le M7 j = 0 1

int z

minus1wm(z)dz +

bm0

Hm

U0zz = Re [Q

0U0zz + A0

1uz] + A02U

0 = 0 in D

partU0

(334)

Noting that

δ

un minus unminus1 D

2 (337)

2 u+ Am3 in D (339)

where

Qm =

⎧⎨⎩Q(z u uz)

0Am

j =

⎧⎨⎩Aj(z u uz)

0in

⎧⎨⎩Dm = DDm

Dm

⎫⎬⎭ j = 1 2 3

C1δ [u(z) D] le M5 (340)

2 (z Un Unz)un=Cn

1 u0z]minus Am2 u0

(341)

gn(z) =

umn (z) = 0 on Γ (346)

where

Cmn =gn[Re (Am

1 umz )+Am

mz ] (347)

C1β[u

mn Dn] le M9 um

n W 2p0

(Dn)le M10 (348)

u0(z) = 2Reint z

prime1Γ

prime2 isin C2

12

partu

X(z) =mprod

j=2j =n

where

Q(z) =1minus ym

2(1 + ym) A3(z) =

d

2(1 + ym)

Condition C

2(1 + ym)ge minus c

4ge 0 on D (43)

lu=12

partu

party=ψ(x) xisinL0

(44)

u(z) = 2Reint z

int 1

limyrarr0

partu(x0 y)party

lt 0 (or limyrarr0

partu(x0 y)party

gt 0) (47)

if the limit exists

limyrarr0

partu(x0 y)party

= M prime ge 0 (48)

v(z) =εu(z)

(MeBd minus εeBy)

we have

ltεM

εM

limyrarr0

partv(x0 y)party

=ε2BM

C[un(z) D] le M12 = M12(α k0 D) (411)

c1 gec2+maxΓ

|r(z)|+maxD

ec2y c2 gtmaxL0

|ψ(x)|+maxD

|b|+2maxD

|d|+1

party= minusc2e

c2 gt maxD

|b(z)|+maxL0

|ψ(x)|+maxD

ec2y + 2maxD

|d|

c1 gt c2 +maxD

ec2y(1 +

c2

σ0

)+max

Γ

|φ(z)|σ0

(414)

C1β[un(z) D] le M13 = M13(β k0 D η) (416)

From (420) we get

(422)

(423)

and then we have

Y(424)

timesuxuxy +2mfG

1f

Y)uxuxy

Yuy(aux + buy + cu)

(f primeprime

fminus 2

f prime2

f 2

)+ a

f prime

f

](425)

(gprime

fminus f primeg

f 2

)uuxminus2mg

fYuuy

+[Y m

(gprimeprime

fminus 2

f primegprime

f 2

)+ a

gprime

f

]u2 +

1f

Lη(v) = (2+o(1))[Y m(uxy)2+(uyy)2]+O

+(

Y m

X2 +Y m2minus1+ε

X

)[Y m+ε1(ux)2+(uy)2]

+2Y m

X2 (ux)2+2(uy)2

X2 +O

(Y m

X3 |ux|+Y minus1

X2 |uy|)

+ε1(m+1minusε1)(1+o(1))

X4 Y ε1minus2

(426)

Lη(v)f

+(uyy)2]+2g

(428)

in which

mG

Y

Yuy(aux+buy+cu)

+[Y m

(f primeprime

fminus2f

prime2

f 2

)+a

f prime

f

(gprime

fminus f primeg

f 2

)uux

+[Y m

(gprimeprime

fminus2f

primegprime

f 2

)+a

gprime

f

]u2+

(429)

Σ = O

Y m

X2 +Y m2minus1+ε

X

)

X3 |ux|+Y ε3minus2

X4

minus2 g

f(Y m|ux|2+|uy|2)+O

X4

)

(430)

X4

)gt0 (431)

f|uuy|+H prime

f

H prime

ge Y m

fH prime)2

(432)

By (419)(420)(432) we obtain

+1G

O

X4

)]

ge Y m

X4 Y ε3minus1)2

X4

)gt0

(433)

partu

partr(z)parts

uy+1

on Γ (437)

F =

(439)

plusmnm

(440)

2Yux

)2

YGprime+bGprime

+2cGminus(

m2

2Y 2+2ax

)F (ux)2plusmn m

(441)

Y

)|ux|

and

(443)

(445)

CHAPTER IV

tions

wzlowast

= 0 in

D+

Dminus

(12)

where

12[wx minus iwy]

(15)

K =12(K1 +K2) (16)

Kj =[φj

π

γj =φj

radic2 on

Re [λ(z)w(z)] = 0 z isin Γ (18)

2 v(z) =

(111)

(112)

g(t) = f(t) = 0 t isin [0 2]

(113)

(114)

u(x) = v(x) =12f(x) x isin [0 2] or

(115)

(116)

obviously

2 v(z) =

2

(119)

=2r((1minusi)x) on [0 1] f(0)

(120)

(121)

f(x+ y) =2r((1 + i)(x+ y)2 + 1minus i)

u(x) =12f(x) + g(x) v(x) =

12f(x)minus g(x)

u(x) =12g(x) + f(x) v(x) =

12minusg(x) + f(x)

x isin [0 2](123)

xisin [02]

(124)

xisin [0 2] or

xisin [0 2]

(125)

ζ(z) =5(z minus 1)

(z minus 1)2 + 4 z(ζ) = 1 +

52ζ(1minus

Λ(ζ) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

1minus iradic2

1 + iradic2

(126)

Re [Λ(z)W (ζ)] = R(ζ) =

(127)

W (ζ) =X(ζ)πi

[int infin

dt+ iclowast2 + ζ

2minus ζ

]in G (128)

and

ζ minus 1ζ + i

)γ1(

ζ + 1ζ + i

)γ2

clowast =2i+ 12 + i

int infin

minusinfinΛ(t)R(t)

X(t)(t minus i)dt

t minus 2t+ i

)] on Im t = 0 Im [S(i)] = 0 (129)

w(z)=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

12

(1 + i)f(x+ y) + (1minus i)

or

12

wz

wzlowast

= F (z w) F = A1(z)w + A2(z)w + A3(z) in

D+

Dminus

(22)

where

12[wx minus iwy]

4 A2 =

a+ ib+ ic minus d

4 A3 =

f + ig

2

Condition C

Cβ[w(z)X(z) D+] le M3(k1 + k2 + k3) (28)

in which

δ γj ge 0j = 1 2 (29)

w(z) = w0(z) +W (z) (210)

int intD+

g(ζ)ζ minus z

W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν

2g1(z)dνe1 +

int micro

(211)

g(z) =

A1 + A2ww w(z) = 00 w(z) = 0

f = A1ψ + A2ψ + A3 in D+

(212)

(214)

s(x)=

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

w(z) = w0(z) +W (z) =

⎧⎨⎩ Φ(z)eφ(z) + ψ(z) in D+

wz

wzlowast

= A1w + A2w + A3 in

⎧⎨⎩D+

Dminus

⎫⎬⎭ (218)

int intD+

g0(ζ)ζ minus z

w1(z) = w0(z) +W1(z) W1(z) = Φ1(z) + Ψ1(z)

Ψ1(z)=int ν

int micro

(219)

(221)

in which

s1(x)=

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

w1(z) = w0(z) +W1(z) =

⎧⎨⎩ Φ1(z)eφ1(z) + ψ1(z) in D+

w0(z) + Φ1(z) + Ψ1(z) in Dminus(222)

w2(z) = w0(z) +W2(z) =

⎧⎨⎩ Φ2(z)eφ2(z) + ψ2(z) in D+

w0(z) + Φ2(z) + Ψ2(z) in Dminus(224)

wn(z) = w0(z) +Wn(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Ψn(z)=int ν

+int micro

(225)

and then

1 (z)Yplusmn(z)|+

radic2[|Y +(z)

int ν

|B| |C| |D| |E| |F |) Rprime = 2 M = 1 + 4k20(1 + k2

+int micro

(227)

timesint Rprime

0

Rprimenminus1

n in Dminus

(228)

plusmn(z) ie

wplusmnn (z)Y

Ψlowast(z) =int ν

+int micro

(230)

in which

sn(x)=

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

n = M9 (234)

Cβ[wn(z)X(z) D+] le M3(k1 + k2 +M9) (235)

Lw = A1w + A2w in D (237)

w(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

g(z) =

⎧⎨⎩A1 + A2ww w(z) = 0 z isin D+

Φ(z) + Ψ(z)

Ψ(z) =int ν

2[Aξ +Bη]e1dν +

int micro

(239)

s(x)=

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

n in Dminus (240)

wz

wzlowast

= F (z w) F = A1w + A2w + A3 in

D+

Dminus

(32)

Condition C

wz = A1w + A2w in D (36)

X(z)=2prod

2prodj=1

2δ γj ge 0(310)

Lw =

wz

wzlowast

D+

Dminus

(311)

C[A4(z w) D] le k0 (312)

M17 = k1 + k2 + |b1| (313)

(M13 +M14)k1 + k2 + 2k0tσ + |b1| = t (314)

+A3(z wm) = G(z wm) in D m = 1 2 (317)

wm+1(z) = w0(z) + Φm+1(z) + Ψm+1(z)

+int x+y

(320)

m

wplusmnm(z)Y

+int x+y

(322)

(M13 +M14)k1 + k2 + 2k0tσ + |b1| = t (323)

2 are as follows

2 = x minus y = 2 l le x le 2 (41)

(43)

1

(44)

K =12(K1 +K2) (45)

2minus2σ(ν)+ν

x =12(micro+ ν) =

(48)

and ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

x =12(micro+ ν) =

[2minus 2σ(x+ γ1(x)) + x+ γ1(x)](x+ y)4

2

[2minus 2σ(x+ γ1(x)) + x+ γ1(x)](x+ y)4

2

(49)

ξν = Aξ +Bη + E

2

in Dminus (411)

2 = γ2(x) + y = 0 l le x le 2 (413)

zisinLprimeprime1

zisinLprimeprime2

1|a(x) + b(x)| le k0

(415)

2

(418)

Hence we have

x =12(micro+ ν) =

(419)

reduced to

1)] = b1(422)

such that Lprime1 L

1of Lprimeprime

maxzisinL1

zisinL2

1|a(z) + b(z)| le k0

(54)

]=

φj

πminus Kj

Kj =[φj

π

]+ Jj Jj = 0 or 1 j = 0 1 m

(55)

msumj=0

[φj

2πminus γj

2

](56)

(57)

Re [λ(z)w(z)]=r(z)

zisinL3=sum[n2]

j=0 A2jA2j+1

Re [λ(z)w(z)]=r(z)

zisinL4=sum[(n+1)2]

j=1 B2jminus1B2j

j=0 1 [n2]

j=1 [(n+ 1)2]

(58)

maxzisinL3

zisinL4

1|a(x) + b(x)| le k0

(59)

γj =1πiln(

)=

φj

πminus Kj

Kj =[φj

π

]+ Jj Jj = 0 or 1 j = 0 1 m+ n

(510)

m+nsumj=0

(φj

2πminus γj

2

)(511)

λ(x) =

1 cap AB

2 cap AB

(513)

Cβ[w(z)X(z) D+] le M21(k1 + k2 + k3) (514)

the form

f(x)=h(x)=2r((1 + i)x2 + 1minus i)

(515)

on L2j+1 j = 0 1 [n2]

on L2j j = 1 [(n+ 1)2]

Re [λ(x)w(x)] =

⎧⎨⎩k(x)

h(x)λ(x) =

1 = Dminus1 cap AB

2 cap AB

and denote

Re [λ(x)w(x)] =

k(x) on Lprime2

2] le k2

(516)

Im [λ(z)w(z)]|z=A2j+1 = c2j+1 j = 0 1 [n2]

(517)

1in the form

w(z) = w(z) + λ(A2j+1)[r(A2j+1) + ic2j+1]

2j+1

j = 0 1 [n2]

= h(x+ y)=2r((1+i)(x+y)2+1minusi)

[n+ 12

]

the form

(518)

w(z) = w0(z) +W (z) in D (519)

int intD+

g(ζ)ζ minus z

Ψ(z) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

int ν

a2j+1

2j+1 j=0 1 [n2]

int ν

2g1(z)dνe1+

int micro

a2jminus1

(520)

g(z) =

⎧⎨⎩A1 + A2w(w) w(z) = 00 w(z) = 0 z isin D+

(521)

Cβ[ψ(z) D+] + Lp0 [ψz D+] le M22(522)

Im [λ(z)(Φ(z) + Ψ(z))]|z=A2j+1 = 0 j = 0 1 [n2]

(523)

j+δwplusmn

s(x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

x isin (a2j a2j+1) j = 0 1 [n2]

(525)

w(z) = w0(z) +W (z) =

⎧⎨⎩ Φ(z)φ(z) + ψ(z) in D+

wz

wzlowast

= A1w + A2w + A3 in

⎧⎨⎩D+

Dminus

⎫⎬⎭ (526)

functions

int intD+

g0(ζ)ζ minus z

Ψ1(z)=

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

int ν

a2j+1

g10(z)dνe1+

int micro

2j+1 j=0 1 [n2]

int ν

2g10(z)dνe1+

int micro

a2jminus1

2j j=1 [(n+1)2]

(527)

Im [λ(z)(Φ1(z) + Ψ1(z))]|z=A2j+1 = 0 j = 0 1 [n2]

(529)

where

s1(x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

x isin (a2j a2j+1) j = 0 1 [n2]

w2(z) = w0(z) +W2(z) =

⎧⎨⎩ Φ2(z)eφ2(z) + ψ2(z) in D+

w0(z) + Φ2(z) + Ψ2(z) in Dminus(532)

wn(z) = w0(z) +Wn(z) =

Ψn(z)=

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

int ν

a2j+1

g1nminus1(z)e1dν+

int micro

0g2

int ν

2g1

a2jminus1

2j j=1 [(n+1)2]

(533)

and then

+radic2

1lejlen+1|int ν

a2j+1

g10(z)e1dν|+ |

int ν

2g20(z)e1dν|

]

1lejlen+1|int micro

a2jminus1

g20(z)e2dmicro|

]

(534)

0)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int ν

a2j+1

[g1nminus1minusg1

0[g2

2j+1

j=0 1 [n2]int ν

2[g1

int micro

a2jminus1

[g2nminus1minusg2

j=1 [(n+1)2]

(535)

int Rprime

0

Rprimenminus1

(n minus 1) dRprime

n in Dminus

(536)

ie

Ψlowast(z)=

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

int ν

a2j+1

g1lowast(z)e1dν+

int micro

0g2

int ν

2g1

a2jminus1

2j j=1[(n+1)2]

(538)

(539)

in which

sn(x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

+ Re [λ(x)Ψn(x)]

x isin (a2j a2j+1) j = 0 1 [n2]

(541)

Cβ[wn(z)X(z) D+] le M21(k1 + k2 +M27) (543)

wz

wzlowast

= A1w + A2w in

⎧⎨⎩D+

Dminus

⎫⎬⎭ (545)

Im [λ(z)w(z)]|z=A2j+1 = 0 j = 0 1 [n2]

(546)

in which

s(x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(547)

w(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

g(z) =

⎧⎨⎩A1 + A2ww w(z) = 0 z isin D+

Φ(z) + Ψ(z)

Ψ(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int ν

a2j+1

2[Aξ + Bη]e1dν +

int micro

a2jminus1

[Cξ + Dη]e2dmicro

(548)

n in Dminus (549)

CHAPTER V

uzlowastzlowast

= 0 in

D+

Dminus

(12)

12[wx minus iwy]

12

partu

12

partu

1|a(z)+b(z)| lek0

(15)

K =12(K1 +K2) (16)

Kj =[φj

π

γj =φj

wz

wzlowast

= 0 in

D+

Dminus

(18)

int z

0w(z)dz + b0 in D (110)

w(z) =

⎧⎪⎪⎨⎪⎪⎩W [ζ(z)] z isin D+0 2

or

w(z) =12(1 + i)g(x minus y) +

12(1minus i)f(x+ y)

f(x+ y) =2r((1 + i)(x+ y)2 + 1minus i)

(111)

X(z) =2prod

2prodj=1

δ γj ge 0j = 1 2

(114)

λ(z) = a+ ib =

1 + iradic2on L2

r(z) =

⎧⎪⎪⎨⎪⎪⎩φθ on Γ

φxradic2on L1 or

φxradic2on L2

b1 = Im[1 + iradic2

uz(z1 minus 0)]=

φx minus φxradic2

|z=z1minus0 = 0 or

uz(z1 + 0)]= 0 b0 = φ(0)

(117)

Re[1minus iradic2

w(x)]= s(x) =

x isin [0 2] or

Re[1 + iradic2

w(x)]= s(x) =

x isin [0 2]

= eπi4 = eiφ2 or

minus1 lt γ1 =φ1

πminus K1 = minus5

4lt 0

0 le γ2 =φ2

πminus K2 =

14

lt 1 or

0 le γ1 =φ1

πminus K1 =

14

lt 1

minus1 lt γ2 =φ2

πminus K2 = minus5

4lt 0

K1 +K2

2= minus1

2 or

K1 = 0 K2 = minus1 K =K1 +K2

2= minus1

2

radic2 or

where

12[ux + iuy] uzz =

14[uxx + uyy]

14[uxx minus uyy]

A1 =a+ ib

2 A2 =

c

4 A3 =

d

4in D

K =12(K1 +K2) = 0 (25)

where

Kj =[φj

π

γj =φj

u(2)=2Reint 2

0uzdz+u(0)=2

int 2

int 0

πr(z)dθ+b0

K=

⎧⎪⎪⎨⎪⎪⎩0

⎧⎨⎩u(0)=b0 u(2)=b2

u(0) = b0

⎫⎬⎭ if

u(z) = U(z)Ψ(z) + ψ(z) in D+

U(z) = 2Reint z

(27)

ψ(z) = 0 Ψ(z) = 1 on Γ cup L0 (29)

p0(D+)le M2 (210)

Cβ[φD+] + Lp0 [φz D+] le M5 (213)

Ψ(z) (216)

u(z) = 2Reint z

0w(z)dz + b0 w(z) = w0(z) +W (z) (217)

int intD+

g(ζ)ζ minus z

W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν

2g1(z)dνe1 +

int micro

(218)

g(z)=

A=ReA1+ImA1

2 B=

ReA1minusImA1

(219)

(220)

u0(z) = 2Reint z

0w0(z)dz + b0 (222)

s(x)=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

+Re [λ(x)Ψ(x)] or

(223)

w(z) = w0(z) +W (z) =

⎧⎪⎨⎪⎩Φ(z)φ(z) + ψ(z) in D+

wz

wzlowast

= Re [A1w] + A2u+ A3 in

⎧⎨⎩D+

Dminus

⎫⎬⎭ (224)

wz

wzlowast

= Re [A1w] + A2u in

D+

Dminus

(225)

Re [λ(z)w(z)] = 0 z isin Γ u(0) = 0 u(2) = 0

w(z) =

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

w0(z) + Φ(z) + Ψ(z)

Ψ(z) =int ν

2[Aξ+Bη+Cu]e1dν+

int micro

(227)

n(230)

wz

wzlowast

= F F = Re [A1w] + A2u+ A3 in

⎧⎨⎩D+

Dminus

⎫⎬⎭ (232)

2(z) g12(z)

and

w1(z) = Φ1(z)eφ1(z) + ψ1(z) in D+

int intD+

g1(ζ)ζminusz

2g11(z)dνe1+

int micro

(235)

(236)

w2(z) = Φ2(z)eφ2(z) + ψ2(z) in D+

Ψn(z)=int ν

2g1

0g2

(240)

and then

+radic2[|Y +(z)

int ν

(241)

+int micro

(242)

0

Rprimenminus1

(n minus 1) dRprime

n in Dminus

(243)

plusmn(z)ie

wplusmnn (z)Y

Ψlowast(z) =int ν

int micro

(245)

(248)

(249)

12

partu

12

partu

maxzisinL1

zisinL2

1|a(x) + b(x)| le k0

(34)

Y (z)=ηm+1prodj=0

(35)

γj =1πiln

=φj

πminus Kj

Kj =[φj

π

]+ Jj Jj = 0 or 1 j = 0 1 m+ 1

(37)

m+1sumj=0

(φj

2πminus γj

2

)(38)

τj=

δj=

⎧⎨⎩ 2τj j=0 m+1

τj j=1 m(39)

u(z) = 2Reint z

0w(z)dz + c0 w(z) = w0(z) +W (z) (310)

int intD+

g(ζ)ζ minus z

W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν

2g1(z)dνe1 +

int micro

(312)

in which e1 =1 + i

2 e2 =

1minus i

A=ReA1+ImA1

2 B=

ReA1minusImA1

(313)

(314)

where

s(x)=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

+Re [λ(x)Ψ(x)] or

(315)

ηj = 2|γj|+ δ j = 0 m+ 1 ηj = |γj|+ δ j = 1 m

andu0(z) = 2Re

int z

0w0(z)dz + c0 (317)

u(z) = 2Reint z

0w(z)dz + b0

(318)

w(z) = w0(z) +2K+1sumk=1

ckwk(z) (322)

⎧⎨⎩ 1 2K + 1 K ge 0

φ K lt 0(325)

0w(z)dz + c0 (326)

ckH1wk(z) (327)

u = ψ1(s) on Γ (41)

u = ψ2(x) on CB (42)

partu

α[ψ2(x)CB]lek2

C1α[φ(y)A

LetU =

12ux V = minus1

2uy W = U + iV in D (46)

W zlowast

= Re [A1W ] + A2u+ A3 in

⎧⎨⎩D+

Dminus

⎫⎬⎭

u(z) = 2Reint z

aW (z)dz + ψ1(0)

(47)

U(x y) =12ux =

(48)

U(0 y) =12

partu

12[u(0 y)]y =

12[u(0 minusy)]y +

12φprime(y)

(49)

and then

(410)

Hence

(411)

2φprime(minusx)]

12φprime(minusx)

(412)

partu

U(0 y) =12

partu

(413)

λ(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

1

1 + iradic2

1

r(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

0 on AO

ψprime2(x) on CB

Kj=[φj

π

γj=φj

πminusKj j=1 4

=ei0

πminusK1=minus1

4minusK1 lt

34

lt1

=eiπ4

πminusK2=

14minusK2=minus3

4lt0

=ei0

πminus K3 = minus3

2minus K3 =

12

lt 1

=eiπ

(416)

K =12(K1 +K2 +K3 +K4) = minus1

2 (417)

W (z) =

⎧⎨⎩W (z) in D

⎧⎨⎩ Wz

W zlowast

⎫⎬⎭ = Re [A1W ] + A2u+ A3 in

⎧⎨⎩D

D

⎫⎬⎭u(z) = 2Re

int z

1W (z)dz + ψ1(0)

(419)

in which

A1 =

⎧⎨⎩A1(z)

minusA1(minusz)A2 =

⎧⎨⎩A2(z)

A2(minusz)A3 =

⎧⎨⎩A3(z) in D

and

λ(z) =

⎧⎨⎩λ(z)

λ(minusz)r(z) =

λ(z) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩1 + iradic2

1minus iradic2

r(z) =

⎧⎨⎩r(z) on OC = (0 1)

(422)

Kj=[φj

π

γj=φj

πminusKj j=1 6

=eminusiπ4

πminusK1=minus1

2minusK1=

12

lt1

=eiπ4

πminusK2=

14minusK2=minus3

4lt0

=ei0

πminusK3=minus3

2minusK3=

12

lt1

=eiπ

φ4

=eiπ2

πminus K5=

12

minus K5 =12

lt 1

=ei0

πminusK6=

14minusK6=minus3

4lt0

(423)

2 (424)

u(z) = U(z)Ψ(z) + ψ(z) in D+

U(z) = 2Reint z

(425)

ψ(z) = 0 Ψ(z) = 1 on Γ cup L

partψ(z)partx

= 0partΨ(z)

partx= 0 on AO

(427)

p0(D+)le M19 (428)

Cβ[φ(z) D+] + Lp0 [φz D+] le M22 (431)

Re [λ(x)W (x)] = r(x) λ(x) =

⎧⎪⎨⎪⎩1 + iradic2on L0 = (0 1)

1 on L1 = (1 a)(433)

u(z) = 2Reint z

aW (z)dz + b0 b0 = ψ1(0) (434)

Wz

Wzlowast

= Re [A1W ] + A2u+ A3 in

D+

Dminus

(435)

W (z) = Φ(z)eφ(z) + ψ(z)

int intD+

g(ζ)ζ minus z

W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν

1g1(z)dνe1 +

int micro

(436)

g(z)=A12+A1W(2W ) W (z) = 00 W (z)=0

in D+

A=ReA1+ImA1

2 B=

ReA1minusImA1

(437)

W zlowast

= 0 in

D+

Dminus

(438)

(439)

thus

W (z) =

⎧⎪⎨⎪⎩Φ(z)eφ(z) + ψ(z) in D+

uzzlowast

⎧⎨⎩D+

Dminus

⎫⎬⎭

u(z) = 2Reint z

(442)

in which

Ψ(z) =int ν

1g1(z)dνe1 +

int micro

(443)

U(x) = 2int x

aΦ(x)dx = 2

int x

0Φ(x)dx minus

int a

0Φ(x)dx

=S(x)=int x

12U(ix) on (0 1)

(444)

partU(z)partx

= 0 on AO (445)

int y0 Vydy =

int y0 Uxdy = 0 on

[U(z) + iV (z)]2dz = 0

= minusint a

1[V (x)]2dx minus

intΓ[V (z)]2

(dx

ds+ i

dy

ds

)ds (446)

+iint 0

1[U(iy)]2dy +

int 1

party

partsds+

int 1

0[U(iy)]2dy + 2

int 1

0[g(x)]2 minus 1

4[U(ix)]2dx = 0 (447)

4prodj=2

un+1(z) =int z

(451)

Un+1(z)=2Reint z

(452)

ψn+1(z) = 1 Ψn+1(z) = 1 on Γ cup L

partψn+1(z)partx

= 0partΨn+1(z)

partx= 0 on AO

(453)

Un+1(x)=Reint x

aΦn+1xdx=S(x)

S(x)=int x

2

int x

0φprime

n+1(minusx)dx

= g(x) +12Un+1(ix) +

(454)

where

g(x)=int x

0f(x)dx

int x

0Fn+1(minusx)dx=

Un+1(ix)2

0φprime

n+1(minusx)dx

partUn+1(z)partx

= 0 on AO (455)

limnrarrinfinmax

D+

int 1

0[Un+1(iy)]2dy = 0 (456)

intΓ[Vn+1(z)]2

party

partsds+

int 1

0[Un+1(iy)]2dy

+2int 1

0

[g(x)]2 minus 1

dx = 0

(457)

Un+1(iy) = 0 on AO g(x) = 0 on OC (458)

γ [X(z)un(z) D+] (459)

γ [un D+] le 12C1

γ [un D+]

infinsumj=N0

γ [u1minusu0 D+]

j=0un+1 =

Wz

W zlowast

= A3(z) in

⎧⎨⎩D+

Dminus

⎫⎬⎭ (462)

int z

aW (z)dz + b0 in D (464)

expressed as

u0(z) = 2Reint z

int intD+

A3(ζ)ζminusz

dσζ in D+

1A3(z)e1dν+

int micro

(466)

(467)

Wzlowast

= A3(z) in

⎧⎨⎩D+

Dminus

⎫⎬⎭

(468)

Wn+1zlowast

= A3(z) in

⎧⎨⎩D+

Dminus

⎫⎬⎭

(469)

Noting that

2C1

u(z) = 2Reint z

W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν

1A3(z)e1dν +

int micro

0A3(z)e2dmicro

(475)

u(z) = 2Reint z

0W (z)dz + u(0) Ψ(z) =

int ν

0A3(z)e1dν +

int micro

0A3(z)e2dmicro

f(x+ y) = τ((x+ y)2) + ReW (0) + ImW (0)

(476)

Wz

Wzlowast

⎧⎨⎩D+

Dminus

⎫⎬⎭ (477)

Wz = K(z) in D+ (481)

(485)

Noting that

int y0

x minus int y0

⎧⎨⎩ Wmacrz

Wmacrz

⎫⎬⎭ = A1(z)W + A2(z)W + A3(z)u+ A4(z) in

D+

Dminus

A1 =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩im

8y+

d

4ym2 +ie

4=

d

4ym2 + i

(m

8y+

e

4

)

jm

8|y| +minusd

4|y|m2 minus je

4=

minusd

4|y|m2 + j

(m

8|y|minuse

4

) in

⎧⎨⎩D+

Dminus

⎫⎬⎭

A2 =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩d

4ym2+ i

(m

8yminus e

4

)

minusd

4|y|m2+j

(m

8|y|+e

4

)

A3=

⎧⎪⎪⎪⎨⎪⎪⎪⎩f

4

minusf

4

A4=

⎧⎪⎪⎨⎪⎪⎩g

4

minusg

4

in

⎧⎨⎩D+

Dminus

⎫⎬⎭

(53)

and

u(z)=

⎧⎪⎪⎨⎪⎪⎩2Re

int z

0uzdz+u(0) in D+

2Reint z

lu=partu

(58)

K =12(K1 +K2)

Kj =[φj

π

γj =φj

u(2)=2Reint 2

0uzdz+u(0)=2

int 2

int 0

πr(z)dθ+b0 (510)

u(z) = 2Reint z

0w(z)dz + b0 w(z) = w0(z) +W (z) (511)

W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν

2g1(z)dνe1 +

int micro

4|y|m2 D= minus g

4|y|m2

A1=1

4|y|m2

[m

2yminus d

|y|m2 minuse

] B1=

14|y|m2

[m

2yminus d

|y|m2+e

]

A2=1

4|y|m2

[m

2|y|minusd

|y|m2 minuse

] B2=

14|y|m2

[m

2|y|minusd

|y|m2+e

]in Dminus

(514)

(515)

u0(z) = 2Reint z

0w0(z)dz + b0 in D (516)

Wz

D+

Dminus

(518)

Re [λ(z)W (z)] = 0 z isin Γ u(0) = 0 u(2) = 0

W (z) = Φ(z) + Ψ(z)

Ψ(z) =int ν

2[Aξ+Bη+Cu]e1dν+

int micro

(520)

]= 0

12

]= s(x) on L0

(522)

CHAPTER VI

Luz=

uzz

uzzlowast

D+

Dminus

(12)

14[uxx minus uyy]

A1 =a+ ib

2 A2 =

c

4 A3 =

d

4in D

Condition C

12

partu

12

partu

Im [λ(z)uz]z=z1 = b1 u(0) = b0 u(2) = b2

(16)

K =12(K1 +K2) (18)

Kj =[φj

π

γj =φj

Im [λ(z)uz]|z=z2 = b2 (110)

u(z) = 2Reint z

0w(z)dz + b0 w(z) = w0(z) +W (z) in D (111)

Lw =

wz

wzlowast

= 0 in

D+

Dminus

(112)

int intD+

g(ζ)ζ minus z

2g1(z)dνe1 +

int micro

(113)

g(z)=

A=ReA1+ImA1

2 B=

ReA1minusImA1

(114)

(115)

where

X(z) =2prod

j=1

ηj =

2|γj|+ δ γj lt 0

δ γj ge 0j = 1 2

(117)

u0(z) = 2Reint z

0w0(z)dz + b0 in D (118)

(119)

of mixed type

(121)

u0(z) = 2Reint z

u1(z) = 2Reint z

0w1(z)dz + b0 w1(z) = w0(z) + Φ1(z) + Ψ1(z)

Ψ11(z) =

int ν

2G1(z)dν G1(z) = Aξ0 +Bη0 + Cu0 +D

Ψ21(z) =

int micro

(123)

(124)

namelyCβ[Ψ1

n (128)

wn(z) =nsum

m=1

um(z) + u0(z) n = 1 2 (129)

Luz =

uzz

uzzlowast

D+

Dminus

(131)

M12 = k1 + k2 + |b0|+ |b1| (132)

Luz =

uzz

uzzlowast

D+

Dminus

(133)

M4k1 + k2 + 2k0tσ + 2k0t

τ + |b0|+ |b1| = t (134)

um+1 = C1[um+1 D] le M14 = M14(p0 β k0 D) (139)

um+1(z) = 2Reint z

+int x+y

(140)

m

0(1 + k20) From the

+int x+y

(142)

int z

M4k1 + k2 + 2k0tσ + 2k0t

τ + |b0|+ |b1| = t (144)

C1[u(z) D] le (M4 + 1)(2k1 + k2) (146)

12

partu

u(0)=b0 u(a)=b1 u(2)=b2 (22)12

partu

zisinL4

1|a(x)+b(x)| lek0

(23)

K =12(K1 +K2 +K3) (24)

Kj=[φj

π

γj=φj

|c1| |c2| le k2

Ψ(z) =

⎧⎪⎪⎨⎪⎪⎩int ν

a

1 int ν

2g1(z)dνe1 +

int micro

a

(27)

λ(x) =

1 + i x isin Lprime

0 = (0 a)

(28)

s(x)=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

+Re[λ(x)Ψ(x)]

xisin(0a)

h(x)=Re[λ(x)Ψ(x)]

(29)

X(z) =3prod

j=1

ηj =

(210)

u(z) = 2Reint z

0w(z)dz + b0 in D (211)

int intD+

g(ζ)ζ minus z

Ψ(z) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩int ν

a

1 int ν

2g1(z)dνe1 +

int micro

a

(212)

Im [λ(zprimej)Φ(z

(213)

λ(x) =

1 + i x isin Lprime

0 = (0 a)

0] le k3

(214)

where

X(z)=3prod

j=1

wz

wzlowast

D+

Dminus

(216)

2Re[1minus iradic2

uz

]=

partu

[1 + iradic2

uz

]=

partu

partl= 0 on (a 2) (218)

(220)

J =

∣∣∣∣∣u1(a) u2(a)

u1(2) u2(2)

∣∣∣∣∣ = 0 (221)

d1u1(a) + d2u2(a) = 0 d1u1(2) + d2u2(2) = 0 (222)

2 minus b2 (223)

C1[u D] le M18(k1 + k2)

(225)

1 Lprime2 Lprime

3 Lprime4 are

4 = γ2(x) + y = 0 l2 le x le 2(226)

12

partu

12

partu

=bj+2 j=12(227)

j] le k2 j = 1 4

maxzisinLprime

1

zisinLprime4

1|a(x) + b(x)| le k0

(228)

4where Lprime

1 Lprime2 Lprime

(230)

2 Lprime3 Lprime

ν =1

=12minusa

(231)

x =12(micro+ ν) =

(232)

and

x=12(micro+ν) =

12(2minusa)

+2(x+ y+a)minus2ax]

12(2minusa)

(233)

1

2

(234)

1

2

(236)

u(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

2Reint z

0w(z)dz + b0 in D+

2Reint z

a

2Reint z

2

(238)

12

partu

12

partu

[n2]sumj=0

A2jA2j+1

12

partu

[(n+1)2]sumj=1

B2jminus1B2j

(32)

Im [λ(z)uz]|z=A2j+1 = c2j+1 j = 0 1 [n2]

(33)

zisinL4

1|a(x) + b(x)| le k0

(34)

uzzlowast

= 0 in

D+

Dminus

(37)

Lw =

wz

wzlowast

= 0 in

D+

Dminus

(38)

Im [λ(z)w(z)]|z=A2j+1 = c2j+1 j = 0 1 [n2]

(39)

and the relation

u(z) = 2Reint z

2w(z)dz + b0 (310)

(311)

Re [λ(x)w(x)] =

1 cap AB

2 cap AB(312)

h(x)=2r((1 + i)x2 + 1minus i)

(313)

on L2j+1 j = 0 1 [n2]

on L2j j = 1 [(n+ 1)2]

2j cap AB j = 1 [(n+ 1)2]

u(z) = 2Reint z

2w(z)dz + c0 w(z) = w0(z) +W (z) in D (314)

int intD+

g(ζ)ζminusz

Ψ(z)=

⎧⎪⎪⎪⎨⎪⎪⎪⎩

int ν

a2j+1

2j+1 j=01[n2]

int ν

2g1(z)dνe1+

int micro

a2jminus1

(315)

g(z)=

A=ReA1+ImA1

2 B=

ReA1minusImA1

2 C=A2D=A3 in D+

(316)

Im [λ(z)(Φ(z) + Ψ(z))]|z=A2j+1 = 0 j = 0 1 [n2]

(317)

(319)

on L2j+1 j = 0 1 [n2]

on L2j j = 1 [(n+ 1)2]

w(z) = w0(z) +W (z) =

⎧⎨⎩ Φ(z)φ(z) + ψ(z) in D+

wzlowast

= Re [A1w] + A2u+ A3 in

D+

Dminus

(320)

Ψ1(z)=

⎧⎪⎪⎪⎨⎪⎪⎪⎩int ν

a2j+1

2j+1 j=0 1 [n2]int ν

2g0(z)e1dν+

int micro

a2jminus1

(321)

(j = 1 [(n+ 1)2]

(322)

where

X(z) =m+nprodj=0

n+1prodj=0

(323)

wz

wzlowast

D+

Dminus

u(zj) = 0 j = 0 1 m u(aj) = 0 j = 1 n

Im [λ(z)w(z)]|z=A2j+1 = 0 j = 0 1 [n2]

(324)

1) ulowast(aprime

m+n)] = [dlowast1 d

Im [λ(z)ukz]|z=A2j+1 = 0 j = 0 1 [n2]

(326)

J =

∣∣∣∣∣∣∣∣∣u1(aprime

m+n)

cprime1u1(aprime

thus the function

cprimekuk(z) in D

Luz=

uzz

uzzlowast

D+

Dminus

(327)

sumNj=1 Γj isin C2

12

partu

12

partu

maxzisinL1

zisinLprimeprime

1|a(x) + b(x)| le k0

(43)

0 The number

where

Kj=[φj

π

γj=φj

πminusKj j=1N (45)

j j = 1 N (46)

u(z) = 2Reint z

0w(z)dz + d1 w(z) = w0(z) +W (z) (47)

wz

wzlowast

= 0 in

D+

Dminus

(48)

int intD+

g(ζ)ζ minus z

Ψ(z) =

0g2(z)dmicroe2 +

int ν

0g2(z)dmicroe2 +

int ν

aj+1

(49)

g(z)=

A12+A1w(2w) w(z) =00 w(z) = 0 z isin D+

A =ReA1 + ImA1

2 B =

ReA1 minus ImA1

(411)

u0(z) = 2Reint z

0w0(z)dz + d1 (413)

on L0 in which

h(x)=[a

((1minusi)x2

)+b

((1minusi)x2

and

Ψ(x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int ν

2g1(z)dνe1 =

int ν

0g2(z)dmicroe2 =

int micro

0g2(z)dmicroe2=

int micro

Thus

w(z) = w0(z) +W (z) =

⎧⎨⎩ Φ(z)φ(z) + ψ(z) in D+

wzlowast

= Re [A1w] + A2u+ A3 in

D+

Dminus

(415)

wz

wzlowast

= Re [A1w] + A2u in

D+

Dminus

(416)

in which

w(z)=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Ψ(z)=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int micro

int ν

2jminus1

j = 1 2 Nint micro

int ν

aj+1

j = 1 2 N minus 1

(418)

n in Dminus

uzz = Re [A1uz] + A2u in D+ (420)

12

partu

(422)

and the relation

u(z) = 2Reint z

0w(z)dz + d1 (423)

(424)

f(x+ y) = Re [(1 + i)w(x+ y)]

1 0 b1

j j=2N

2j j=12Nminus1

2jminus1 j=2N

wz

wzlowast

= Re [A1w] + A2u+ A3 in

D+

Dminus

(426)

int intD+

g0(ζ)ζ minus z

Ψ1(z) =

0g0(z)dmicroe2 +

int ν

0g0(z)dmicroe2 +

int ν

aj+1

(429)

+Re [λ(x)Ψ1(x)]

h(x) =[a

((1minus i)x

2

)+ b

((1minus i)x

2

)][Re (λ(z1)(r(z1) + ic1))

+Im (λ(z1)(r(z1) + ic1))] on L0

(430)

Ψn(z) =

int ν

aj+1

and then|[wplusmn

0g0(z)e2dmicro|

2g0(z)e1dν|+

Nminus1sumj=1

|int micro

aj+1

g0(z)e1dν|]

(433)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int micro

int ν

2jminus1

j=12Nint micro

int ν

aj+1

j=12N minus1(434)

0

Rprimenminus1

(n minus 1) dRprime

n in Dminus

(435)

ie

Ψlowast(z)=

int ν

aj+1

(437)

(438)

(440)

where

X(z) =2Nprodj=1

wz

wzlowast

= Re [A1w(z)] + A3 in

D+

Dminus

(442)

Im [λ(zprimej)w(z

primej)] = dprime

j j = 1 N(443)

0w(z)dz + d0 in D

(446)

in which d0 = φ(0)

λ(z) = a+ ib =

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

(1 + i)radic2 on L1

j=2L2jminus1

(447)

and

r(z) =

j=1L2jminus1

c1=Im[1 + iradic2

uz(z1 minus 0)]=

φx minus φxradic2

|z=z1minus0=0

uz(zj + 0)]= 0 j = 2 N

(448)

Re[1minus iradic2

w(x)]= s(x) =

πminus K1 =

14

minus 0 =14

lt 1

πminusK2N=minus5

4lt0

πminus Kj=

54minus1= 1

4lt1 j=3 5 2Nminus1

πminus Kj=minus1

4=minus1

4lt0 j=2 4 2Nminus2

N

2minus 1

References

References 241

242 References

References 243

244 References

References 245

246 References

References 247

248 References

References 249

Index

128 130 132 133 149 150 151153 155 165ndash167 174 184 190202 212 223 228 230 232

AV Bitsadze 157

Banach space 53 55 93 104 137 138191 193 207 208

217 219 220 227 228 232 237238 239

Index 251

108 109 159 164 173 196 202210 219 226

problem 79 80 90 91 93 94125 143 144 145 146 156

140 158 171 200 204 209 214218 227 228

87 90 93 103 106 108 119 122126 134 143 157 177 182 193198 213 218 226 233

252 Index

179 180 182ndash184 186 188 189191 193 194

103 107 120 125 126 139 144145 156 158 160ndash164 180 181

Index 253

186 196 202 228 229 232 237238 239

140 141 216inversion 97

137 141 142 162 165 167 168171 175 176 177 184 189 192212 218 224 237 239

238 239

254 Index

110 111 158 159 160 161 162196 238 239

Problem P0 44 67 97 109 118 158196 197 202

225Problem P prime

0 228Problem Q 97 106 171ndash177 202

mixed type 79 134 143 199 200209 215 218 227 228

Index 255

96 107

171 206 225representation 20 25 30 35 36 38

43 46ndash48 63 64 79 87 90ndash92 94134 160 182 198 229

50 52 54 75 104 131 132 138152 153 169 170 205 207 208235 236

160

43 66 106 108 119 126 134 139141 143 157 194 200 208 218

94 104 119 157 160 168 173175 177 180 182 189 191 194200 206 209 211 213 231

subsequence 102 105 106 111 116132 154 170

77 93

55ndash58 60 67 80 88 97 111120 140 141 142 158 196202 216 217

209 210 213 227 229 236

98 102 106 111 131 132 138153 154 170 205 208 236

256 Index

53 56 58 63 64 65 72 78 8693 103 104 106 116 125 133135 137 141 142 144 147 150155 159 162 165 189 192 204207 208 212 217 224 227 238

90 93 96 103

zone domain 90 94

Book Cover
HALF-TITLE
SERIES TITLE
TITLE
COPYRIGHT
CONTENTS
PREFACE
REFERENCES
INDEX

Page 4: Linear and quasilinear complex equations of hyperbolic and ...webéducation.com/wp-content/uploads/2018/09/Linear...and hyperbolic pseudoregular functions. Next we transform the linear

London and New York

(Print Edition)

Contents

vi Contents

Contents vii

Preface

x Preface

CHAPTER I

e1 = (1 + j)2 e2 = (1minus j)2 (11)

e1 + e2 = 1 ekel =

ek if k = l

0 if k = lk l = 1 2 (12)

in which

1z=

z

zz=

1x+ y

e1 +1

x minus ye2 =

1micro

e1 +1νe2

z1

(1micro2

e1 +1ν2

e2

)=

micro1

micro2e1 +

ν1

ν2e2

(15)

(1) wz = 0 (16)

(2) ξν = 0 ηmicro = 0 (17)

(3) ux = vy vx = uy (18)

[ξ(z)minusξ(z0)

e2

]

respectively

2

ez = 1 + z +z2

ex+y + exminusy

2+ j

ex+y minus exminusy

2

2+ j

sin z = z minus z3

cos z = 1minus z2

tgz=sin z

(1

cosmicroe1 +

1cos ν

e2

ctgz=cos zsin z

1sinmicro

e1+1

sin νe2

y(z0)∆y + ε(∆z)

where

y(z0) = uy(z0) + jvy(z0)

lim∆zrarr0

ε(∆z) rarr 0

Cf(z)dz =

intC

udx+ vdy + j[int

Cvdx+ udy] =

Df(z)dxdy =

int intD

udxdy + jint int

Dvdxdy

C[f(z) + g(z)]dz =

intC

f(z)dz +int

Cg(z)dzint int

int intD

f(z)dxdy +int int

Dg(z)dxdy

int

Cf(z)dz le

radic2M1l

int int

Df(z)dxdy le

radic2M2S

G[f(z)]zdxdy =

j

2

intC

f(z)dz

2

intC

f(z)dz

Cf(z)dz = 0

ImF (z)G(z) = 0 in D (19)

w(z0) = δ0F (z0) + γ0G(z0) (110)

w(z0) = limzrarrz0

= limzrarrz0

ie

Wz(z0) = w(z0) (116)

andWz(z0) = 0 (117)

W (z) =

∣∣∣∣∣∣∣∣∣w(z) w(z0) w(z0)

F (z) F (z0) F (z0)

G(z) G(z0) G(z0)

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣F (z0) F (z0)

G(z0) G(z0)

∣∣∣∣∣∣ (118)

Fz(z0) F (z0) F (z0)

Gz(z0) G(z0) G(z0)

∣∣∣∣∣∣∣∣∣ = 0 (119)

w(z0) =

∣∣∣∣∣∣∣∣∣wz(z0) w(z0) w(z0)

Fz(z0) F (z0) F (z0)

Gz(z0) G(z0) G(z0)

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣F (z0) F (z0)

G(z0) G(z0)

∣∣∣∣∣∣ (120)

wz = aw + bw (121)

where

F G minus FG b =

FGz minus FzG

FG minus FG

F G minus FG B =

FGz minus FzG

FG minus FG

(123)

respectively

w(z) = φ(z)F (z) + ψ(z)G(z) (124)

K(z) = φ(z) + jψ(z) (125)

Fφz +Gψz = 0 (126)

holds where

(128)

Proof From

Wz(z0) = F (z0)φz(z0) +G(z0)ψz(z0)

Setting

minusG

minusF

(129)

where

δ =σy + τx

τ

δ =σy + τx

τ

(132)

φ(z) =int z

z0

wz = a(z)w(z) + b(z)w(z) (133)

a = a(FG) and b = b(FG) (135)

F (z) = 1 or G(z) = j in D (137)

I = (K2 +K3)2 minus 4K1K4 gt 0 (22)

holds in which

K1 =∣∣∣∣∣a11 b11

a21 b21

∣∣∣∣∣ K2 =∣∣∣∣∣a11 b12

a21 b22

∣∣∣∣∣ K3 =∣∣∣∣∣a12 b11

a22 b21

∣∣∣∣∣ K4 =∣∣∣∣∣a12 b12

a22 b22

where

K5 =∣∣∣∣∣a11 a12

a21 a22

∣∣∣∣∣ K6 =∣∣∣∣∣ b11 b12

b21 b22

a = K1K6 b = K3K6 c = K4K6 d = K2K6

∆ = I4K26 =

(b+ d)2

4minus ac gt 0 (26)

+lower order terms

namely

= (1 + a)(1minus c) + bd gt 0

in which

q0

(29)

then we can prove

|Q1|+ |Q2| = |Q1Q1|12 + |Q2Q2|12 lt 1 (210)

∆ = I4K24 = (b+ d)24minus ac gt 0 (212)

whereq1 =

2

It is clear that

q0 = (1 + q1)(1 + q1)minus q2q2

4

= (1 + a)(1minus c) + bd gt 0

in which

Q1(z) =[(1minus q1)(1 + q1) + |q2|2]

q0 Q2 =

minus2q2(z)q0

(216)

Fkux =int 1

Fkuy =int 1

Fkvx =int 1

Fkvy =int 1

(217)

holds in which

K1 =∣∣∣∣∣a11 b11

a21 b21

∣∣∣∣∣ K2 =∣∣∣∣∣a11 b12

a21 b22

∣∣∣∣∣ K3 =∣∣∣∣∣a12 b11

a22 b21

∣∣∣∣∣ K4 =

∣∣∣∣∣a12 b12

a22 b22

∣∣∣∣∣ K5 =∣∣∣∣∣a11 a12

a21 a22

∣∣∣∣∣ K6 =∣∣∣∣∣ b11 b12

b21 b22

6

Vu Vv

∣∣∣∣∣∣ =∣∣∣∣∣∣λ 1

micro 1

andu =

U minus V

λ minus micro v =

microU minus λV

micro minus λ

1V + c1

2V + c2(222)

where

aprime1 =

a1 minus microb1

1 =minusa1 + λb1

2 =a2 minus microb2

2 =minusa2 + λb2

λ minus micro

2b12)V + (c1b22 minus c2b12)]K6

1b21)V + (c2b11 minus c1b21)]K6(223)

and then

4 gt 0

Letξ =

x minus microy

λ minus micro η =

minusx+ λy

λ minus micro

ηx ηy

minus1 λ

= 0

a22uξ + b22vη = a2u+ b2v + c2(226)

1v + cprime1

2v + cprime2

(227)

whereaprime

2(ζ w)w + Aprime3(ζ w) (228)

in which Aprime1 A

prime2 A

ux = au+ bv + f vy = cu+ dv + g (230)

wz = 0 in D (31)

zisinL4

1|a(z) + b(z)| le k0 (34)

times g(0) = 2r((1 + j)x) on [0 R]

f(0)=u(0)+v(0)=r(0)+b1

a(0)+b(0) g(0)=u(0)minusv(0)=

r(0)minusb1

a(0)minusb(0)

(38)

f(x+ y) =2r((1 + j)(x+ y)2)

a((1 + j)(x+ y)2)+b((1 + j)(x+ y)2)(39)

0lex+yle2R

0lexminusyle2R1

(310)

wz = A1(z)w + A2(z)w + A3(z) (312)

(314)

in which

[a2(z0)minus b2(z0)] (316)

w(z) = w0(z) + Φ(z) + Ψ(z) in D

Ψ(z) =int xminusy

int x+y

0[Cξ +Dη + F ]d(x+ y)e2

(317)

[Ψ]z = [Aξ +Bη + E]e1 + [Cξ +Dη + F ]e2 (319)

wz = A1w + A2w in D (320)

w(z) = Φ(z) + Ψ(z)

Ψ(z) =int xminusy

int x+y

(322)

Ψ(z) le 8M3mR0 Φ(z) le 32M3k20(1 + k2

0)mR0

w1(z) = w0(z) + Φ1(z) + Ψ1(z)

Ψ1(z) =int ν

0[Aξ0 +Bη0 + E]e1dν +

int micro

(323)

w2(z) = w0(z) + Φ2(z) + Ψ2(z) in D

+int micro

and

2

wn(z) = w0(z) + Φn(z) + Ψn(z)

Ψn(z) = +int ν

int micro

(326)

+int micro

and then

0

Rprimenminus1

n (328)

Ψlowast(z) =int ν

int micro

(330)

w(z) = u(z) + jv(z)

(333)

a(z1)+b(z1)

Re [λ(z)w(z)]=u(z)+v(z)=r[(1+

R+R1

RminusR1j)

xminusj2RR1

RminusR1

]on [R1 R]

(334)

RminusR1

)xminus 2RR1

RminusR1

]

t

2R+ (1minus j)R1

] t isin [0 2R]

and then

0lexminusyle2R1

(335)

Condition C

(45)

in which

wz = A1w + A2w in D (46)

w(z) = Φ(z) + Ψ(z)

Ψ(z) =int xminusy

int x+y

(48)

w1(z) = w0(z) + Φ1(z) + Ψ1(z)

Ψ1(z) =int ν

0[Aξ0 +Bη0 + E]e1dν +

int micro

(49)

Ψn(z) =int ν

int micro

(411)

+int ν

+int micro

0

Rprimenminus1

n (413)

n=0

= M6 (415)

C[w(z) D] le M7 = M7(α k0 D)

Ψ11(z) =

int xminusy

1(z) =int x+y

0G2(z)d(x+ y)

prime (420)

prime(421)

prime

prime(423)

n =Re wn + Im wn w2

(M12Rprime)n

n Cα[w1

n

prime)n

n Cα[w2

prime)n

n

(424)

x =12(micro+ ν)

4R minus 4σ(x+ γ1(x)) + 2x+ 2γ1(x)

(429)

x =12(micro+ ν) =

2

(430)

2 Im [λ(z0)w(z0)] = b1 (432)

2R (433)

2 Im [λ(z0)w(z0)] = b1

4 are as follows

4 = minusγ4(x) + y = 0 0 le x le l2(435)

Hence we have

x =12(micro+ ν) =

(439)

x =12(micro+ ν) =

(440)

1 cup Lprimeprime4

is reduced to

4 Im [λ(z0)w(z0)] = b1

radius R1 namely

radicR2

1 + 2R1ν minus ν2

2

radicR2

2

1 Lprimeprime2 L

primeprime3 L

primeprime4

ξν = 0 ηmicro = 0 (52)

microy νy

∣∣∣∣∣ =∣∣∣∣∣ 1 1

1 minus1

∣∣∣∣∣ = minus2∣∣∣∣∣ ξu ηu

ξv ηv

∣∣∣∣∣ =∣∣∣∣∣ 1 1

1 minus1

∣∣∣∣∣ = minus2 (53)

ξ = ξ(micro) η = η(ν) (55)

wz = Q(z)wz Q(z) = a(z) + jb(z) (56)

w(z) = Φ[χ(z)] (58)

ξσ = 0 ητ = 0 (σ τ) isin G (513)

wz = Q(z)wz (514)

Wz = 0 (516)

CHAPTER II

( )z =( )x + j( )y

2 ( )z =

( )x minus j( )y2

4

( )zz =( )xx + ( )yy + 2j( )xy

4 ( )zz =

4

(14)

in which

Q =a+ c+ 2bj

a minus c A1 =

d+ ej

a minus c A2 =

f

2(a minus c) A3 =

g

2(a minus c)

|Q(z)| lt 1 in D (17)

respectively

in which

a =int 1

2b =int 1

c =int 1

d =int 1

e =int 1

f =int 1

0Φτu(x y τu 0 0 0 0 0)dτ = f(x y u)

g = minusΦ(x y 0 0 0 0 0 0) = g(x y)

and

a2 = f = minusint 1

0Fτu(z τu 0 0 0 0)dτ = a2(z u)

a3 = minusF (z 0 0 0 0) = a3(z)

(112)

or its complex form

(120)

A1 =a+ jb

2 A2 =

c

4 A3 =

d

4in D

wz =wx + jwy

wx minus jwy

(122)

in which

A =a+ b

4 B =

a minus b

4 C =

c

4 D =

d

4

12

partu

u(0) = b0 Im [λ(z)uz]|z=z3 = b1

(21)

maxzisinL3

1|a(z) + b(z)| le k0

(22)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Re [λ(z)w(2Re1 + νe2)] = r(z)

(24)

u(z) = φ(z) on L = L3 cup L4 (25)

in which

λ(z) = a+ jb =

1 + jradic2on L4

b+1 = Im

[1minus jradic2

2radic2

bminus1 = Im

[1 + jradic2

uz(z3)]=

(27)

wz = Re [A1w] + A2u+ A3 in D (28)

int z

0w(z)dz + b0 in D (210)

2ReintΓw(z)dz =

intΓw(z)dz +

intΓw(z)dz

= 2jint int

s1

w(z)dz +int

s2

w(z)dz]+ b0

uzz = 0 (211)

wz = 0 in D (212)

f(2R) = u(z3) + v(z3) =r((1 + j)R) + b1

a((1 + j)R) + b((1 + j)R)

a((1 + j)R)minus b((1 + j)R)

(213)

a((1 + j)(x+ y)2) + b((1 + j)(x+ y)2)

0 le x+ y le 2R

α[u(z) D]leM2k2=M2(α k0 D)k2 (215)

[a((1 + j)x) + b((1 + j)x)]f(2x) + [a((1 + j)x)

f(2R)=u(z3)+v(z3)=r((1 + j)R)+b1

a((1 + j)R)+b((1 + j)R)

[a((1 + j)t2) + b((1 + j)t2)]f(t) + [a((1 + j)t2)

u(z) = 2Reint z

0w(z)dz + b0 in D

w(z) = W (z) + Φ(z) + Ψ(z) in D

Ψ(z)=int ν

int micro

(216)

u0(z) = 2Reint z

0w0(z)dz + b0 w0(z) = W (z) = ξ0e1 + η0e2 (221)

u1(z) = 2Reint z

0w1(z)dz + b0 w1(z) = w0(z) + Φ1(z) + Ψ1(z)

Ψ1(z) =int xminusy

+int x+y

(222)

C1[u0(z) D] le M3C[w0(z) D] + k2 (223)

C[Φ1(z) D] le 8M4k20(1 + k2

0)[(4 +M3)m+ k2 + 1]Rprime

(224)

wn(z) satisfying

un+1(z) = 2Reint z

0wn+1(z)dz + b0

+int micro

(225)

and then

0

Rprimenminus1

(n minus 1) dRprime

n

(226)

+int micro

(228)

int z

wz = Re [A1w] + A2u in D (230)

int z

w(z) = Φ(z) + Ψ(z)

Ψ(z) =int ν

int micro

(233)

n

maxzisinD

C1[u D] le M7 C1[u D] le M8k (33)

(116)

α[u D] le M11k (34)

u0(z) = 2Reint z

0w0(z)dz + b0 w0(z) = W (z) = ξ0e1 + η0e2 (35)

u1(z) = 2Reint z

0w1(z)dz + b0 w1(z) = w0(z) + Φ1(z) + Ψ1(z)

Ψ1(z) = Ψ11(z)e1 +Ψ2

1(z)e2 Ψ11(z) =

int ν

0G1(z)dν

Ψ21(z) =

int micro

(38)

prime (39)

1(z) G2(z) Ψ21(z)

(310)

prime (311)

prime

(312)

Cα[w1n(z) D] le

(M18Rprime)n

n Cα[w2

n(z) D] le(M18R

prime)n

n

n C1

n

(313)

wn(z) = w0(z) +nsum

m=1wm(z) un(z) = u0(z) +

nsumm=1

um(z) (n = 1 2 )

α[u(z) D] le M10 (314)

F = Re [A1uz] + A2u+ A3

(315)

M8k1 + k2 + 2k0tσ + 2k0t

τ + |b1|+ |b0| = t (317)

(321)

um+1 = C1[um+1 D] le M20 (322)

um+1(z) = 2Reint z

+int x+y

(323)

n

0(1 + k20) From

+int xminusy

+int x+y

(325)

int z

M8k1 + k2 + 2k0tσ + 2k0t

τ + |b0|+ |b1| = t (327)

prime4 and

Hence we have

x =12(micro+ ν) =

(334)

and

x =12(micro+ ν) =

(335)

4 u(0) = b0 Im [λ(z3)w(z3)] = b1 (337)

u(z) = 2Reint z

4is reduced to

4 are as follows

(341)

micro =2Rmicro

x =12(micro+ ν) =

(345)

x =12(micro+ ν) =

(346)

4is reduced to

4

3 )] = b1 (350)

1 Lprimeprime2 L

primeprime3 L

primeprime4 namely

Lprimeprime1 =

radicR2

Lprimeprime2 =

radicR2

Lprimeprime3 =

y = γ3(x) =

radicR2

Lprimeprime4 =

y = γ4(x) =

radicR2

Lprimeprime1 =

⎧⎨⎩x = σ1(ν) =R1 + ν minus

radicR2

1 + 2R1ν minus ν2

2

⎫⎬⎭

Lprimeprime2 =

radicR2

2

⎫⎬⎭

Lprimeprime3 =

⎧⎨⎩x = σ2(ν) =R1 + ν minus

radicR2

1 + 2R1ν minus ν2

2

⎫⎬⎭

Lprimeprime4 =

radicR2

2

⎫⎬⎭

u(0) = b0 Im [λ(z)uz]|z=z1 = b1(41)

zisinL2

1|a(z) + b(z)| le k0

(42)

u(0) = b0 Im [λ(z)uz]|z=z1 = b1(43)

(44)

u(0) = b0 Im [λ(z)uz]|z=z1 = b1(45)

1|a(z) + b(z)| le k0

(46)

u(z) = 2Reint z

0w(z)dz + b0 in D (49)

(410)

f(0) = u(z1) + v(z1) =r((1minus j)R1) + b1

(411)

f(x+ y) =2r((1+j)(x+y)2+(1minusj)R1)

0 le x+ y le 2R (412)

α[u(z) D] le M24k2 (413)

u(z) = 2Reint z

Ψ(z) =int ν

2R1

(414)

(415)

(416)

Im [λ(z1)w(z1)] = b1(417)

(420)

in which

2 τ(x) =

(422)

for equation (211)

2 0 le x+ y le 2R

(423)

and

f(0) =sprime(0) +R(0)

2 g(2R1) =

(424)

for (116)

f(0) = u(z1) + v(z1) =r((1minus j)R1) + b1

(428)

0 le x+ y le 2R1

Ψ(z) =int ν

2R1

C1[s(x) L0] le k2 C1[φ(x) L1] le k2 (432)

u(0) = b0 Im [λ(z)uz]|z=z1=b1

(433)

where we choose

radic2 (434)

L1=x+int y

0

L2=

xminusint y

0

radicminusK(t)dt =

int y

0|t|m2dt = minus

int |y|

0d

2m+ 2

|t|(m+2)2 = minus 2m+ 2

|y|(m+2)2

L1 x minus 2m+ 2

|y|(m+2)2 = 0 L2 x+2

m+ 2|y|(m+2)2 = 2 ie

L1 y = minus(m+ 22

x)2(m+2) L2 y = minus[m+ 22

(2minus x)]2(m+2)

where

A1 =d+ je

K(y)minus 1 A2 =

f

2(K(y)minus 1) A3 =

g

2(K(y)minus 1)

12

partu

u(0) = b0 Im [λ(z)uz]|z=z1 = b1

(54)

[radic

α[r(z) Lj] le k2 j = 1 2

zisinL2

1|a(z) + b(z)| le k0

(55)

int y

0

L2 = BC =x = 2 +

int y

0

(56)

Re [(1 + j)(U + jV )] = U + V = r(x) on L2

Im [(1 + j)(U + jV )] = [U + V ]|z=z1+0 = r(1 + 0)

whereU =

λ(z)=

andu(z) = 2Re

int z

x+int y

0

int y

0

int y

0

int y

0

int y

0

radicminusK(t)dt

minusK(y)radic

(510)

(radicminusK(y)

)micro= minus1

K prime

4K(radic

minusK(y))

K primeU4K

(U + V )ν =radic

(514)

minus2radic

minusK(y)(U + V )ν

=radic

= minusradic

K prime(y)2K(y)

U 2radic

=radic

+radic

=radic

K prime(y)2K(y)

U

(515)

U ie

(516)

u(z) = 2Reint z

partu

u(0) = b0 Im [λ(z)(U + iV )]|z=z1 = b1

(519)

2 V (x y) =

2

(520)

where

f(0) = U(0) + V (0) =r(1) + b1

a(1) + b(1) g(2) = U(2)minus V (2) =

r(1)minus b1

a(1)minus b(1)

U =12

f(micro) +

V =12

a(ν2)minus b(ν2)

or

U =12

g(ν) +

V =12

minusg(ν) +

(521)

a(x2+1)minusb(x2+1)

a(x2) + b(x2) x isin [0 2]

(522)

W (z) =

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

12

[(1 + j)

a(ν2)minus b(ν2)

] (523)

can be derived

U + V = minusint ν

int micro

4K(y)Udmicro

W = U+jV =minus1+j

2

int ν

4K(y)Udν+

1minusj

2

int micro

4K(y)Udmicro

= minus1 + j

2

int ν

2

K prime(y)4K(y)

Udν +1minus j

2

int micro

0

K prime(y)4K(y)

Udmicro

radic1 +

(dy

dx

)2dx =

radic1minus K

1minus Kdy

radic1 +

(dy

dx

)2dx =

radic1minus K

minusKdx =

radic1minus Kdy

(524)

=1radic1minusK

]

=1

2radic1minusK

K prime(y)K(y)

U

=1radic1minusK

]

2radic1minusK

K prime(y)K(y)

U

(525)

0

1

2radic(1minus K)

K prime(y)K(y)

Uds1 =int s1

0

12

K prime(y)K(y)

Udy

0

1

2radic(1minus K)

K prime(y)K(y)

Uds2 =int s2

0

12

K prime(y)K(y)

Udy

(526)

W (z) = U + jV = W (z) + Φ(z) + Ψ(z)

U+V =minusint s1

0

1

2radic(1minusK)

K prime(y)K(y)

0

1

2radic(1minusK)

K prime(y)K(y)

Uds2(527)

u(z) = 2Reint z

int z

can be replaced by

2U(x)=S(x)=radic

0

+u(0)=s(x)

2V (x) = R(x) = uy on AB = L0 = (0 2)

(529)

(532)

v(x) = 0 vy(x) = 0 on L0 (533)

ξν =1

2radic1minus K

2K prime(y)K(y)

)ξ

+ e minus 12

K prime(y)K(y)

]

ηmicro =1

2radic1minus K

12

K prime(y)K(y)

)ξ

+ e+12

K prime(y)K(y)

]

(534)

K prime(y)K(y)

=m|y|mminus1h(y)

K(y)minus |y|mhy

K(y)=

m

y+

hy

h

v(z) =int y

ξ(z) = minusint y

η(z) =int y

(535)

in this case

A1 = minuse

2minus hy

4h B1 =

e

2minus hy

4h A2 = minuse

2+

hy

4h B2 =

e

2+

hy

4h

C1 = minus12

dradicminusKminus m

4y C2 = minus1

2dradicminusK

+m

4y D = minusf

2 E = minusG

2

limyrarr0

0

(538)

3Mδ

2+[ε(y)M +m2]δm2

m+ 2lt γ

6δ2M2

m+ 6+8δMm+ 2

+2ε(y)M +m

m+ 2lt γ

(539)

1 s02

ξ1(z)=minusint y

int y

0Edy z isin s0

1

η1(z)=int y

0[A2ξ0+B2η0+C2(ξ0+η0)+Dv0+E]dy=

int y

0Edy zisin s0

2

v1(z) = Reint y

(540)

1

ηk+1(z) =int y

2

vk+1(z) =int y

(541)

j=0γj|y|m2+1

(542)

j=0γj|y|

Moreover we get

int y

0Ex[x1minusx2]dy|

le 4m+4

j=0γj|y|m2+1

le |int y

int y

0|x1 minus x2|dy|

le M |int y

1sumj=0

γj|y|m2+1

1sumj=0

γj|y|m2+1

|vn+1(z)| le |int y

int y

0

nsumj=0

γjydy|leMnsum

j=0γj|y|2 leM

n+1sumj=0

γj|y|

0

⎡⎣(|A1|+|B1|+|D|)Mnsum

j=0γj|y|+|C1|M

nsumj=0

γj|y|m2+1+|E|⎤⎦ dy|

le M |int y

0

⎡⎣3M nsumj=0

γj|y|+(

ε(y)2|y|radich

+m

4|y|)

nsumj=0

γj|y|m2+1 + 1

⎤⎦ dy|

le M |y|⎧⎨⎩[32M |y|+

(|ε(y)|M +

m

2

) |y|m2

m+ 2

]nsum

j=0γj + 1

γj|y|

and

0

dy|

Noting that

lensum

nsumj=0

γj|y|m2+1

le M2nsum

nsumj=0

γj|y|m2+1

le (M |y|+1)M2nsum

j=0γj|y|m2+1

and

j=0γj +

∣∣∣∣∣hy

2h

j=0γj|y|m2+1

we get

0[(3M |y|+ 4)M2

nsumj=0

γj|y|m2+1

+(|C1|+ |C2|)Mnsum

j=0γj|y|m2+1 +M2|y|m2+1]dy|

le M |y|m2+1

[6M2y2

m+ 6+8M |y|m+ 2

+(|ε(y)|M +

m

2

)

times 2m+ 2

nsumj=0

γj +2M

m+ 4|y|⎤⎦

le Mn+1sumj=0

γj|y|m2+1

ηlowast(z) =int y

vlowast(z) =int y

A1 = minus d

4|y|m2 + j

(m

8|y| minus e

4

) A3=minusf

4

A2 = minus d

4|y|m2 + j

(m

8|y| +e

4

) A4=minusg

4

(543)

and

u(z)=2Reint z

0uzdz=2Re

int z

0(U+jV )d(xminusjy)

CHAPTER III

]=

φj

πminusKj

Kj =[φj

π

]+ Jj Jj = 0 or 1 j = 1 m

(13)

12

msumj=1

[φj

πminus γj

](14)

(15)

Ψ(z) =Φ(z)Π(z)

Π(z) =mprod

j=1

(z minus xj

z + i

)γj

(16)

)=

12

msumj=1

[φj

πminus γj

]=12

msumj=1

Kj (110)

Re[Λ(x)

x minus x0

x+ i

x minus x0Ψ0(x)

Ψlowast(z) = i(

z minus i

z + i

)[K] (z minus x0

z + i

)eiS(z) in D (112)

Re [S(x)] = arg

⎡⎣Λlowast(x)(

x minus i

x+ i

)[K](x minus x0

x+ i

)⎤⎦ x isin L Im [S(i)] = 0 (114)

Ψ0(z) =

X(z) = Π(z)Ψ0(z) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩i(

z minus i

z + i

)K

i(

zminusi

z+i

)[K] zminusx0

(116)

Re[Φ(x)iX(x)

]=

c(x)iλ(x)X(x)

=λ(x)c(x)iX(x)

Φ(z)iX(z)

=1πi

[int infin

minusinfinλ(t)c(t)

(t minus z)iX(t)dt+

Q(z)i

]

Φ(z) =X(z)πi

[int infin

minusinfinλ(t)c(t)

]

(118)

Q(z) = i[K]sumj=0

[cj

(z minus i

z + i

)j

+ cj

(z minus i

z + i

)minusj]+

iclowastx0 + z

(119)

Q(z) =

iclowastx0 + z

(120)

int infin

minusinfinλ(t)c(t)

x0 + z

x0 minus z

=int infin

minusinfinλ(t)c(t)

x0+i

x0minusi

=infinsum

j=0

int infin

[1 +

z minus i

x0 + i

]x0 + i

x0 minus i

infinsumj=0

(z minus i

x0 minus i

)j

=int infin

dt+iclowastx0+i

x0minusi+

infinsumj=1

[int infin

minusinfinλ(t)c(t)

(tminusi)j+1X(t)dt+

2iclowastx0

(x0minusi)j+1

](zminusi)j

(121)

minusinfinλ(t)c(t)

int infin

minusinfinλ(t)c(t)

X(t)(t minus i)jdt+

2iclowastx0

when 2K is odd

(122)

x0 + i

int infin

minusinfinλ(t)c(t)

πi

[int infin

minusinfinλ(t)c(t)

2iclowastx0

](124)

Φ(z) =

int z

iΦ(z)dz in D

λ(x) =

radicΠn

]=

φj

πminus Kj

Kj =[φj

π

]+ Jj Jj = 0 or 1 j = 1 m

(128)

(129)

Φ(z) =X(z)2πi

[intΓ

dt+Q(z)] (130)

with

j + cjzminusj) +

iclowastz0 + z

(131)

intΓ

λ(t)c(t)X(t)tj

intΓ

λ(t)c(t)X(t)tj

Φ(z) =X(z)z[K]

πi

[intΓ

0

] (133)

clowast = iintΓ

λ(t)c(t)X(t)t

dt

X(z) =

z(ζ) =1

ζ + i

[1 + iζ minus

radic2(1minus ζ2)

]

Im [λ(zprimej)Φ(z

primej)] = bj j = 1 m (136)

where zprime1 z

0

primej)]=0 j = 1 m (137)

m = 2K + 1 le 2ND +NΓ le 2K (138)

j=1djΦj(z) (139)

msumj=1

dprimejΦj(zprime

j) j = 1 m (140)

Φlowast(zprimej) =

msumj=1

dprimejΦj(zprime

j) = 0 j = 1 m (141)

j=1 dprimejΦj(z)

r(x)=

(143)

λ(x) =

(144)

Re [λ(z)Φ(z)] = r(z) =

0 z isin L0(145)

Ψ(z) =Φ(z)Π(z)

j=n+1j =m

(z minus xj

z + i

)γj

(146)

Re [λ(z)Φ(z)] = 0 z isin L (147)

)=

(149)

Φ(z) =

⎧⎪⎨⎪⎩Ψ(z) in D = |z| lt 1 Im z gt 0

Re [Λ(z)Φ(z)] = R(z) on |z| = 1 (152)

where

Λ(z)=

⎧⎨⎩λ(z)

λ(z)R(z)=

Φ(z) = Φ(i1 + ζ

1minus ζ

)in D (154)

Condition C

w(z) = Φ[ζ(z)]eφ(z) + ψ(z) (26)

w(z) = Φ(z)eφ(z) + ψ(z) (210)

in which

X(z)=mprod

j=1j =nm

ψz = Qψz + A1ψ + A2ψ + A3 Q =

φz = Qφz + A A =

(213)

int intD

f(ζ)ζ minus z

dσζ

φ(z) = Tg ζ(z) = Ψ[χ(z)] χ(z) = z + Th

(214)

h(z) = Q(z)[1 + Πh] (218)

prime1Γ

prime2 isin C1

j)] = bj j = 1 m(225)

w(z) = Φ(z)eφ(z) + ψ(z) (228)

in which

X(z) =mprod

j=2j =1n

(230)

12[ux + iuy] uzz =

14[uxx + uyy]

a+ c A1(z) =

minusd minus ei

a+ c A2(z) =

minusf

2(a+ c) A3(z) =

g

2(a+ c)

Condition C

|Q(z u w)| le q0 lt 1 (34)

12

partu

Im [λ(z)uz]|z=0 = b2 (37)

ψ(z) = 0 Ψ(z) = 1 on Γ (39)

β[Ψ(z) D] le M2

z(ζ) =1

ζ + i

[1 + iζ minus

radic2(1minus ζ2)

] (311)

ψ(ζ) = 0 Ψ(ζ) = 1 on L = ζ(Γ)

ψ(z) =

⎧⎨⎩ψ(z) in D

Q=

⎧⎨⎩Q(z)

Q(z)A1=

⎧⎨⎩A1(z)

A1(z)A2=

⎧⎨⎩A2(z)

A2(z)A3=

⎧⎨⎩A3(z)

minusA3(z)in

⎧⎨⎩D

D

⎫⎬⎭

0w(z)dz+b0 w(z)=Φ[ζ(z)]eφ(z) (317)

(318)

Ψ(z)in D (319)

12[partU

and give a lemma

partu

partl= lim

z(isinl)rarrz0

2

partV

partn= 2Re

zVz

R z isin part∆

partV

partn= minus4Re zVz

R |z| = R

2

partV

partnge minusε

partV

partngt 0

partu

partn= Ψ

partV

partn+ V

partΨpartn

ge minusεpartV

partn+ V

partΨpartn

gt 0

partu

partl= cos(l n)

partu

partn+ cos(l s)

partu

partsgt 0 (326)

12

partu

(327)

C1δ [u(z) D] le M6(k1 + k2)

(328)

j (j = 1 2 3) λm rm bmj (j = 0 1) but

Umzz minus Re [QmUm

zz + Am1 Um

z ]minus Am2 Um =

Am3

Hm

partUm

partνm

=rm(z)Hm

Hm

Um(1) =bm1

Hm

Lp[Am2 Um +

Am3

Hm

Cα

[|z minus zj|βj

rm(z)Hm

Γlowast]

le M7 j = 1 m

∣∣∣∣∣ bmj

Hm

∣∣∣∣∣ le M7 j = 0 1

int z

minus1wm(z)dz +

bm0

Hm

U0zz = Re [Q

0U0zz + A0

1uz] + A02U

0 = 0 in D

partU0

(334)

Noting that

δ

un minus unminus1 D

2 (337)

2 u+ Am3 in D (339)

where

Qm =

⎧⎨⎩Q(z u uz)

0Am

j =

⎧⎨⎩Aj(z u uz)

0in

⎧⎨⎩Dm = DDm

Dm

⎫⎬⎭ j = 1 2 3

C1δ [u(z) D] le M5 (340)

2 (z Un Unz)un=Cn

1 u0z]minus Am2 u0

(341)

gn(z) =

umn (z) = 0 on Γ (346)

where

Cmn =gn[Re (Am

1 umz )+Am

mz ] (347)

C1β[u

mn Dn] le M9 um

n W 2p0

(Dn)le M10 (348)

u0(z) = 2Reint z

prime1Γ

prime2 isin C2

12

partu

X(z) =mprod

j=2j =n

where

Q(z) =1minus ym

2(1 + ym) A3(z) =

d

2(1 + ym)

Condition C

2(1 + ym)ge minus c

4ge 0 on D (43)

lu=12

partu

party=ψ(x) xisinL0

(44)

u(z) = 2Reint z

int 1

limyrarr0

partu(x0 y)party

lt 0 (or limyrarr0

partu(x0 y)party

gt 0) (47)

if the limit exists

limyrarr0

partu(x0 y)party

= M prime ge 0 (48)

v(z) =εu(z)

(MeBd minus εeBy)

we have

ltεM

εM

limyrarr0

partv(x0 y)party

=ε2BM

C[un(z) D] le M12 = M12(α k0 D) (411)

c1 gec2+maxΓ

|r(z)|+maxD

ec2y c2 gtmaxL0

|ψ(x)|+maxD

|b|+2maxD

|d|+1

party= minusc2e

c2 gt maxD

|b(z)|+maxL0

|ψ(x)|+maxD

ec2y + 2maxD

|d|

c1 gt c2 +maxD

ec2y(1 +

c2

σ0

)+max

Γ

|φ(z)|σ0

(414)

C1β[un(z) D] le M13 = M13(β k0 D η) (416)

From (420) we get

(422)

(423)

and then we have

Y(424)

timesuxuxy +2mfG

1f

Y)uxuxy

Yuy(aux + buy + cu)

(f primeprime

fminus 2

f prime2

f 2

)+ a

f prime

f

](425)

(gprime

fminus f primeg

f 2

)uuxminus2mg

fYuuy

+[Y m

(gprimeprime

fminus 2

f primegprime

f 2

)+ a

gprime

f

]u2 +

1f

Lη(v) = (2+o(1))[Y m(uxy)2+(uyy)2]+O

+(

Y m

X2 +Y m2minus1+ε

X

)[Y m+ε1(ux)2+(uy)2]

+2Y m

X2 (ux)2+2(uy)2

X2 +O

(Y m

X3 |ux|+Y minus1

X2 |uy|)

+ε1(m+1minusε1)(1+o(1))

X4 Y ε1minus2

(426)

Lη(v)f

+(uyy)2]+2g

(428)

in which

mG

Y

Yuy(aux+buy+cu)

+[Y m

(f primeprime

fminus2f

prime2

f 2

)+a

f prime

f

(gprime

fminus f primeg

f 2

)uux

+[Y m

(gprimeprime

fminus2f

primegprime

f 2

)+a

gprime

f

]u2+

(429)

Σ = O

Y m

X2 +Y m2minus1+ε

X

)

X3 |ux|+Y ε3minus2

X4

minus2 g

f(Y m|ux|2+|uy|2)+O

X4

)

(430)

X4

)gt0 (431)

f|uuy|+H prime

f

H prime

ge Y m

fH prime)2

(432)

By (419)(420)(432) we obtain

+1G

O

X4

)]

ge Y m

X4 Y ε3minus1)2

X4

)gt0

(433)

partu

partr(z)parts

uy+1

on Γ (437)

F =

(439)

plusmnm

(440)

2Yux

)2

YGprime+bGprime

+2cGminus(

m2

2Y 2+2ax

)F (ux)2plusmn m

(441)

Y

)|ux|

and

(443)

(445)

CHAPTER IV

tions

wzlowast

= 0 in

D+

Dminus

(12)

where

12[wx minus iwy]

(15)

K =12(K1 +K2) (16)

Kj =[φj

π

γj =φj

radic2 on

Re [λ(z)w(z)] = 0 z isin Γ (18)

2 v(z) =

(111)

(112)

g(t) = f(t) = 0 t isin [0 2]

(113)

(114)

u(x) = v(x) =12f(x) x isin [0 2] or

(115)

(116)

obviously

2 v(z) =

2

(119)

=2r((1minusi)x) on [0 1] f(0)

(120)

(121)

f(x+ y) =2r((1 + i)(x+ y)2 + 1minus i)

u(x) =12f(x) + g(x) v(x) =

12f(x)minus g(x)

u(x) =12g(x) + f(x) v(x) =

12minusg(x) + f(x)

x isin [0 2](123)

xisin [02]

(124)

xisin [0 2] or

xisin [0 2]

(125)

ζ(z) =5(z minus 1)

(z minus 1)2 + 4 z(ζ) = 1 +

52ζ(1minus

Λ(ζ) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

1minus iradic2

1 + iradic2

(126)

Re [Λ(z)W (ζ)] = R(ζ) =

(127)

W (ζ) =X(ζ)πi

[int infin

dt+ iclowast2 + ζ

2minus ζ

]in G (128)

and

ζ minus 1ζ + i

)γ1(

ζ + 1ζ + i

)γ2

clowast =2i+ 12 + i

int infin

minusinfinΛ(t)R(t)

X(t)(t minus i)dt

t minus 2t+ i

)] on Im t = 0 Im [S(i)] = 0 (129)

w(z)=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

12

(1 + i)f(x+ y) + (1minus i)

or

12

wz

wzlowast

= F (z w) F = A1(z)w + A2(z)w + A3(z) in

D+

Dminus

(22)

where

12[wx minus iwy]

4 A2 =

a+ ib+ ic minus d

4 A3 =

f + ig

2

Condition C

Cβ[w(z)X(z) D+] le M3(k1 + k2 + k3) (28)

in which

δ γj ge 0j = 1 2 (29)

w(z) = w0(z) +W (z) (210)

int intD+

g(ζ)ζ minus z

W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν

2g1(z)dνe1 +

int micro

(211)

g(z) =

A1 + A2ww w(z) = 00 w(z) = 0

f = A1ψ + A2ψ + A3 in D+

(212)

(214)

s(x)=

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

w(z) = w0(z) +W (z) =

⎧⎨⎩ Φ(z)eφ(z) + ψ(z) in D+

wz

wzlowast

= A1w + A2w + A3 in

⎧⎨⎩D+

Dminus

⎫⎬⎭ (218)

int intD+

g0(ζ)ζ minus z

w1(z) = w0(z) +W1(z) W1(z) = Φ1(z) + Ψ1(z)

Ψ1(z)=int ν

int micro

(219)

(221)

in which

s1(x)=

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

w1(z) = w0(z) +W1(z) =

⎧⎨⎩ Φ1(z)eφ1(z) + ψ1(z) in D+

w0(z) + Φ1(z) + Ψ1(z) in Dminus(222)

w2(z) = w0(z) +W2(z) =

⎧⎨⎩ Φ2(z)eφ2(z) + ψ2(z) in D+

w0(z) + Φ2(z) + Ψ2(z) in Dminus(224)

wn(z) = w0(z) +Wn(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Ψn(z)=int ν

+int micro

(225)

and then

1 (z)Yplusmn(z)|+

radic2[|Y +(z)

int ν

|B| |C| |D| |E| |F |) Rprime = 2 M = 1 + 4k20(1 + k2

+int micro

(227)

timesint Rprime

0

Rprimenminus1

n in Dminus

(228)

plusmn(z) ie

wplusmnn (z)Y

Ψlowast(z) =int ν

+int micro

(230)

in which

sn(x)=

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

n = M9 (234)

Cβ[wn(z)X(z) D+] le M3(k1 + k2 +M9) (235)

Lw = A1w + A2w in D (237)

w(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

g(z) =

⎧⎨⎩A1 + A2ww w(z) = 0 z isin D+

Φ(z) + Ψ(z)

Ψ(z) =int ν

2[Aξ +Bη]e1dν +

int micro

(239)

s(x)=

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

n in Dminus (240)

wz

wzlowast

= F (z w) F = A1w + A2w + A3 in

D+

Dminus

(32)

Condition C

wz = A1w + A2w in D (36)

X(z)=2prod

2prodj=1

2δ γj ge 0(310)

Lw =

wz

wzlowast

D+

Dminus

(311)

C[A4(z w) D] le k0 (312)

M17 = k1 + k2 + |b1| (313)

(M13 +M14)k1 + k2 + 2k0tσ + |b1| = t (314)

+A3(z wm) = G(z wm) in D m = 1 2 (317)

wm+1(z) = w0(z) + Φm+1(z) + Ψm+1(z)

+int x+y

(320)

m

wplusmnm(z)Y

+int x+y

(322)

(M13 +M14)k1 + k2 + 2k0tσ + |b1| = t (323)

2 are as follows

2 = x minus y = 2 l le x le 2 (41)

(43)

1

(44)

K =12(K1 +K2) (45)

2minus2σ(ν)+ν

x =12(micro+ ν) =

(48)

and ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

x =12(micro+ ν) =

[2minus 2σ(x+ γ1(x)) + x+ γ1(x)](x+ y)4

2

[2minus 2σ(x+ γ1(x)) + x+ γ1(x)](x+ y)4

2

(49)

ξν = Aξ +Bη + E

2

in Dminus (411)

2 = γ2(x) + y = 0 l le x le 2 (413)

zisinLprimeprime1

zisinLprimeprime2

1|a(x) + b(x)| le k0

(415)

2

(418)

Hence we have

x =12(micro+ ν) =

(419)

reduced to

1)] = b1(422)

such that Lprime1 L

1of Lprimeprime

maxzisinL1

zisinL2

1|a(z) + b(z)| le k0

(54)

]=

φj

πminus Kj

Kj =[φj

π

]+ Jj Jj = 0 or 1 j = 0 1 m

(55)

msumj=0

[φj

2πminus γj

2

](56)

(57)

Re [λ(z)w(z)]=r(z)

zisinL3=sum[n2]

j=0 A2jA2j+1

Re [λ(z)w(z)]=r(z)

zisinL4=sum[(n+1)2]

j=1 B2jminus1B2j

j=0 1 [n2]

j=1 [(n+ 1)2]

(58)

maxzisinL3

zisinL4

1|a(x) + b(x)| le k0

(59)

γj =1πiln(

)=

φj

πminus Kj

Kj =[φj

π

]+ Jj Jj = 0 or 1 j = 0 1 m+ n

(510)

m+nsumj=0

(φj

2πminus γj

2

)(511)

λ(x) =

1 cap AB

2 cap AB

(513)

Cβ[w(z)X(z) D+] le M21(k1 + k2 + k3) (514)

the form

f(x)=h(x)=2r((1 + i)x2 + 1minus i)

(515)

on L2j+1 j = 0 1 [n2]

on L2j j = 1 [(n+ 1)2]

Re [λ(x)w(x)] =

⎧⎨⎩k(x)

h(x)λ(x) =

1 = Dminus1 cap AB

2 cap AB

and denote

Re [λ(x)w(x)] =

k(x) on Lprime2

2] le k2

(516)

Im [λ(z)w(z)]|z=A2j+1 = c2j+1 j = 0 1 [n2]

(517)

1in the form

w(z) = w(z) + λ(A2j+1)[r(A2j+1) + ic2j+1]

2j+1

j = 0 1 [n2]

= h(x+ y)=2r((1+i)(x+y)2+1minusi)

[n+ 12

]

the form

(518)

w(z) = w0(z) +W (z) in D (519)

int intD+

g(ζ)ζ minus z

Ψ(z) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

int ν

a2j+1

2j+1 j=0 1 [n2]

int ν

2g1(z)dνe1+

int micro

a2jminus1

(520)

g(z) =

⎧⎨⎩A1 + A2w(w) w(z) = 00 w(z) = 0 z isin D+

(521)

Cβ[ψ(z) D+] + Lp0 [ψz D+] le M22(522)

Im [λ(z)(Φ(z) + Ψ(z))]|z=A2j+1 = 0 j = 0 1 [n2]

(523)

j+δwplusmn

s(x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

x isin (a2j a2j+1) j = 0 1 [n2]

(525)

w(z) = w0(z) +W (z) =

⎧⎨⎩ Φ(z)φ(z) + ψ(z) in D+

wz

wzlowast

= A1w + A2w + A3 in

⎧⎨⎩D+

Dminus

⎫⎬⎭ (526)

functions

int intD+

g0(ζ)ζ minus z

Ψ1(z)=

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

int ν

a2j+1

g10(z)dνe1+

int micro

2j+1 j=0 1 [n2]

int ν

2g10(z)dνe1+

int micro

a2jminus1

2j j=1 [(n+1)2]

(527)

Im [λ(z)(Φ1(z) + Ψ1(z))]|z=A2j+1 = 0 j = 0 1 [n2]

(529)

where

s1(x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

x isin (a2j a2j+1) j = 0 1 [n2]

w2(z) = w0(z) +W2(z) =

⎧⎨⎩ Φ2(z)eφ2(z) + ψ2(z) in D+

w0(z) + Φ2(z) + Ψ2(z) in Dminus(532)

wn(z) = w0(z) +Wn(z) =

Ψn(z)=

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

int ν

a2j+1

g1nminus1(z)e1dν+

int micro

0g2

int ν

2g1

a2jminus1

2j j=1 [(n+1)2]

(533)

and then

+radic2

1lejlen+1|int ν

a2j+1

g10(z)e1dν|+ |

int ν

2g20(z)e1dν|

]

1lejlen+1|int micro

a2jminus1

g20(z)e2dmicro|

]

(534)

0)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int ν

a2j+1

[g1nminus1minusg1

0[g2

2j+1

j=0 1 [n2]int ν

2[g1

int micro

a2jminus1

[g2nminus1minusg2

j=1 [(n+1)2]

(535)

int Rprime

0

Rprimenminus1

(n minus 1) dRprime

n in Dminus

(536)

ie

Ψlowast(z)=

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

int ν

a2j+1

g1lowast(z)e1dν+

int micro

0g2

int ν

2g1

a2jminus1

2j j=1[(n+1)2]

(538)

(539)

in which

sn(x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

+ Re [λ(x)Ψn(x)]

x isin (a2j a2j+1) j = 0 1 [n2]

(541)

Cβ[wn(z)X(z) D+] le M21(k1 + k2 +M27) (543)

wz

wzlowast

= A1w + A2w in

⎧⎨⎩D+

Dminus

⎫⎬⎭ (545)

Im [λ(z)w(z)]|z=A2j+1 = 0 j = 0 1 [n2]

(546)

in which

s(x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(547)

w(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

g(z) =

⎧⎨⎩A1 + A2ww w(z) = 0 z isin D+

Φ(z) + Ψ(z)

Ψ(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int ν

a2j+1

2[Aξ + Bη]e1dν +

int micro

a2jminus1

[Cξ + Dη]e2dmicro

(548)

n in Dminus (549)

CHAPTER V

uzlowastzlowast

= 0 in

D+

Dminus

(12)

12[wx minus iwy]

12

partu

12

partu

1|a(z)+b(z)| lek0

(15)

K =12(K1 +K2) (16)

Kj =[φj

π

γj =φj

wz

wzlowast

= 0 in

D+

Dminus

(18)

int z

0w(z)dz + b0 in D (110)

w(z) =

⎧⎪⎪⎨⎪⎪⎩W [ζ(z)] z isin D+0 2

or

w(z) =12(1 + i)g(x minus y) +

12(1minus i)f(x+ y)

f(x+ y) =2r((1 + i)(x+ y)2 + 1minus i)

(111)

X(z) =2prod

2prodj=1

δ γj ge 0j = 1 2

(114)

λ(z) = a+ ib =

1 + iradic2on L2

r(z) =

⎧⎪⎪⎨⎪⎪⎩φθ on Γ

φxradic2on L1 or

φxradic2on L2

b1 = Im[1 + iradic2

uz(z1 minus 0)]=

φx minus φxradic2

|z=z1minus0 = 0 or

uz(z1 + 0)]= 0 b0 = φ(0)

(117)

Re[1minus iradic2

w(x)]= s(x) =

x isin [0 2] or

Re[1 + iradic2

w(x)]= s(x) =

x isin [0 2]

= eπi4 = eiφ2 or

minus1 lt γ1 =φ1

πminus K1 = minus5

4lt 0

0 le γ2 =φ2

πminus K2 =

14

lt 1 or

0 le γ1 =φ1

πminus K1 =

14

lt 1

minus1 lt γ2 =φ2

πminus K2 = minus5

4lt 0

K1 +K2

2= minus1

2 or

K1 = 0 K2 = minus1 K =K1 +K2

2= minus1

2

radic2 or

where

12[ux + iuy] uzz =

14[uxx + uyy]

14[uxx minus uyy]

A1 =a+ ib

2 A2 =

c

4 A3 =

d

4in D

K =12(K1 +K2) = 0 (25)

where

Kj =[φj

π

γj =φj

u(2)=2Reint 2

0uzdz+u(0)=2

int 2

int 0

πr(z)dθ+b0

K=

⎧⎪⎪⎨⎪⎪⎩0

⎧⎨⎩u(0)=b0 u(2)=b2

u(0) = b0

⎫⎬⎭ if

u(z) = U(z)Ψ(z) + ψ(z) in D+

U(z) = 2Reint z

(27)

ψ(z) = 0 Ψ(z) = 1 on Γ cup L0 (29)

p0(D+)le M2 (210)

Cβ[φD+] + Lp0 [φz D+] le M5 (213)

Ψ(z) (216)

u(z) = 2Reint z

0w(z)dz + b0 w(z) = w0(z) +W (z) (217)

int intD+

g(ζ)ζ minus z

W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν

2g1(z)dνe1 +

int micro

(218)

g(z)=

A=ReA1+ImA1

2 B=

ReA1minusImA1

(219)

(220)

u0(z) = 2Reint z

0w0(z)dz + b0 (222)

s(x)=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

+Re [λ(x)Ψ(x)] or

(223)

w(z) = w0(z) +W (z) =

⎧⎪⎨⎪⎩Φ(z)φ(z) + ψ(z) in D+

wz

wzlowast

= Re [A1w] + A2u+ A3 in

⎧⎨⎩D+

Dminus

⎫⎬⎭ (224)

wz

wzlowast

= Re [A1w] + A2u in

D+

Dminus

(225)

Re [λ(z)w(z)] = 0 z isin Γ u(0) = 0 u(2) = 0

w(z) =

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

w0(z) + Φ(z) + Ψ(z)

Ψ(z) =int ν

2[Aξ+Bη+Cu]e1dν+

int micro

(227)

n(230)

wz

wzlowast

= F F = Re [A1w] + A2u+ A3 in

⎧⎨⎩D+

Dminus

⎫⎬⎭ (232)

2(z) g12(z)

and

w1(z) = Φ1(z)eφ1(z) + ψ1(z) in D+

int intD+

g1(ζ)ζminusz

2g11(z)dνe1+

int micro

(235)

(236)

w2(z) = Φ2(z)eφ2(z) + ψ2(z) in D+

Ψn(z)=int ν

2g1

0g2

(240)

and then

+radic2[|Y +(z)

int ν

(241)

+int micro

(242)

0

Rprimenminus1

(n minus 1) dRprime

n in Dminus

(243)

plusmn(z)ie

wplusmnn (z)Y

Ψlowast(z) =int ν

int micro

(245)

(248)

(249)

12

partu

12

partu

maxzisinL1

zisinL2

1|a(x) + b(x)| le k0

(34)

Y (z)=ηm+1prodj=0

(35)

γj =1πiln

=φj

πminus Kj

Kj =[φj

π

]+ Jj Jj = 0 or 1 j = 0 1 m+ 1

(37)

m+1sumj=0

(φj

2πminus γj

2

)(38)

τj=

δj=

⎧⎨⎩ 2τj j=0 m+1

τj j=1 m(39)

u(z) = 2Reint z

0w(z)dz + c0 w(z) = w0(z) +W (z) (310)

int intD+

g(ζ)ζ minus z

W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν

2g1(z)dνe1 +

int micro

(312)

in which e1 =1 + i

2 e2 =

1minus i

A=ReA1+ImA1

2 B=

ReA1minusImA1

(313)

(314)

where

s(x)=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

+Re [λ(x)Ψ(x)] or

(315)

ηj = 2|γj|+ δ j = 0 m+ 1 ηj = |γj|+ δ j = 1 m

andu0(z) = 2Re

int z

0w0(z)dz + c0 (317)

u(z) = 2Reint z

0w(z)dz + b0

(318)

w(z) = w0(z) +2K+1sumk=1

ckwk(z) (322)

⎧⎨⎩ 1 2K + 1 K ge 0

φ K lt 0(325)

0w(z)dz + c0 (326)

ckH1wk(z) (327)

u = ψ1(s) on Γ (41)

u = ψ2(x) on CB (42)

partu

α[ψ2(x)CB]lek2

C1α[φ(y)A

LetU =

12ux V = minus1

2uy W = U + iV in D (46)

W zlowast

= Re [A1W ] + A2u+ A3 in

⎧⎨⎩D+

Dminus

⎫⎬⎭

u(z) = 2Reint z

aW (z)dz + ψ1(0)

(47)

U(x y) =12ux =

(48)

U(0 y) =12

partu

12[u(0 y)]y =

12[u(0 minusy)]y +

12φprime(y)

(49)

and then

(410)

Hence

(411)

2φprime(minusx)]

12φprime(minusx)

(412)

partu

U(0 y) =12

partu

(413)

λ(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

1

1 + iradic2

1

r(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

0 on AO

ψprime2(x) on CB

Kj=[φj

π

γj=φj

πminusKj j=1 4

=ei0

πminusK1=minus1

4minusK1 lt

34

lt1

=eiπ4

πminusK2=

14minusK2=minus3

4lt0

=ei0

πminus K3 = minus3

2minus K3 =

12

lt 1

=eiπ

(416)

K =12(K1 +K2 +K3 +K4) = minus1

2 (417)

W (z) =

⎧⎨⎩W (z) in D

⎧⎨⎩ Wz

W zlowast

⎫⎬⎭ = Re [A1W ] + A2u+ A3 in

⎧⎨⎩D

D

⎫⎬⎭u(z) = 2Re

int z

1W (z)dz + ψ1(0)

(419)

in which

A1 =

⎧⎨⎩A1(z)

minusA1(minusz)A2 =

⎧⎨⎩A2(z)

A2(minusz)A3 =

⎧⎨⎩A3(z) in D

and

λ(z) =

⎧⎨⎩λ(z)

λ(minusz)r(z) =

λ(z) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩1 + iradic2

1minus iradic2

r(z) =

⎧⎨⎩r(z) on OC = (0 1)

(422)

Kj=[φj

π

γj=φj

πminusKj j=1 6

=eminusiπ4

πminusK1=minus1

2minusK1=

12

lt1

=eiπ4

πminusK2=

14minusK2=minus3

4lt0

=ei0

πminusK3=minus3

2minusK3=

12

lt1

=eiπ

φ4

=eiπ2

πminus K5=

12

minus K5 =12

lt 1

=ei0

πminusK6=

14minusK6=minus3

4lt0

(423)

2 (424)

u(z) = U(z)Ψ(z) + ψ(z) in D+

U(z) = 2Reint z

(425)

ψ(z) = 0 Ψ(z) = 1 on Γ cup L

partψ(z)partx

= 0partΨ(z)

partx= 0 on AO

(427)

p0(D+)le M19 (428)

Cβ[φ(z) D+] + Lp0 [φz D+] le M22 (431)

Re [λ(x)W (x)] = r(x) λ(x) =

⎧⎪⎨⎪⎩1 + iradic2on L0 = (0 1)

1 on L1 = (1 a)(433)

u(z) = 2Reint z

aW (z)dz + b0 b0 = ψ1(0) (434)

Wz

Wzlowast

= Re [A1W ] + A2u+ A3 in

D+

Dminus

(435)

W (z) = Φ(z)eφ(z) + ψ(z)

int intD+

g(ζ)ζ minus z

W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν

1g1(z)dνe1 +

int micro

(436)

g(z)=A12+A1W(2W ) W (z) = 00 W (z)=0

in D+

A=ReA1+ImA1

2 B=

ReA1minusImA1

(437)

W zlowast

= 0 in

D+

Dminus

(438)

(439)

thus

W (z) =

⎧⎪⎨⎪⎩Φ(z)eφ(z) + ψ(z) in D+

uzzlowast

⎧⎨⎩D+

Dminus

⎫⎬⎭

u(z) = 2Reint z

(442)

in which

Ψ(z) =int ν

1g1(z)dνe1 +

int micro

(443)

U(x) = 2int x

aΦ(x)dx = 2

int x

0Φ(x)dx minus

int a

0Φ(x)dx

=S(x)=int x

12U(ix) on (0 1)

(444)

partU(z)partx

= 0 on AO (445)

int y0 Vydy =

int y0 Uxdy = 0 on

[U(z) + iV (z)]2dz = 0

= minusint a

1[V (x)]2dx minus

intΓ[V (z)]2

(dx

ds+ i

dy

ds

)ds (446)

+iint 0

1[U(iy)]2dy +

int 1

party

partsds+

int 1

0[U(iy)]2dy + 2

int 1

0[g(x)]2 minus 1

4[U(ix)]2dx = 0 (447)

4prodj=2

un+1(z) =int z

(451)

Un+1(z)=2Reint z

(452)

ψn+1(z) = 1 Ψn+1(z) = 1 on Γ cup L

partψn+1(z)partx

= 0partΨn+1(z)

partx= 0 on AO

(453)

Un+1(x)=Reint x

aΦn+1xdx=S(x)

S(x)=int x

2

int x

0φprime

n+1(minusx)dx

= g(x) +12Un+1(ix) +

(454)

where

g(x)=int x

0f(x)dx

int x

0Fn+1(minusx)dx=

Un+1(ix)2

0φprime

n+1(minusx)dx

partUn+1(z)partx

= 0 on AO (455)

limnrarrinfinmax

D+

int 1

0[Un+1(iy)]2dy = 0 (456)

intΓ[Vn+1(z)]2

party

partsds+

int 1

0[Un+1(iy)]2dy

+2int 1

0

[g(x)]2 minus 1

dx = 0

(457)

Un+1(iy) = 0 on AO g(x) = 0 on OC (458)

γ [X(z)un(z) D+] (459)

γ [un D+] le 12C1

γ [un D+]

infinsumj=N0

γ [u1minusu0 D+]

j=0un+1 =

Wz

W zlowast

= A3(z) in

⎧⎨⎩D+

Dminus

⎫⎬⎭ (462)

int z

aW (z)dz + b0 in D (464)

expressed as

u0(z) = 2Reint z

int intD+

A3(ζ)ζminusz

dσζ in D+

1A3(z)e1dν+

int micro

(466)

(467)

Wzlowast

= A3(z) in

⎧⎨⎩D+

Dminus

⎫⎬⎭

(468)

Wn+1zlowast

= A3(z) in

⎧⎨⎩D+

Dminus

⎫⎬⎭

(469)

Noting that

2C1

u(z) = 2Reint z

W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν

1A3(z)e1dν +

int micro

0A3(z)e2dmicro

(475)

u(z) = 2Reint z

0W (z)dz + u(0) Ψ(z) =

int ν

0A3(z)e1dν +

int micro

0A3(z)e2dmicro

f(x+ y) = τ((x+ y)2) + ReW (0) + ImW (0)

(476)

Wz

Wzlowast

⎧⎨⎩D+

Dminus

⎫⎬⎭ (477)

Wz = K(z) in D+ (481)

(485)

Noting that

int y0

x minus int y0

⎧⎨⎩ Wmacrz

Wmacrz

⎫⎬⎭ = A1(z)W + A2(z)W + A3(z)u+ A4(z) in

D+

Dminus

A1 =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩im

8y+

d

4ym2 +ie

4=

d

4ym2 + i

(m

8y+

e

4

)

jm

8|y| +minusd

4|y|m2 minus je

4=

minusd

4|y|m2 + j

(m

8|y|minuse

4

) in

⎧⎨⎩D+

Dminus

⎫⎬⎭

A2 =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩d

4ym2+ i

(m

8yminus e

4

)

minusd

4|y|m2+j

(m

8|y|+e

4

)

A3=

⎧⎪⎪⎪⎨⎪⎪⎪⎩f

4

minusf

4

A4=

⎧⎪⎪⎨⎪⎪⎩g

4

minusg

4

in

⎧⎨⎩D+

Dminus

⎫⎬⎭

(53)

and

u(z)=

⎧⎪⎪⎨⎪⎪⎩2Re

int z

0uzdz+u(0) in D+

2Reint z

lu=partu

(58)

K =12(K1 +K2)

Kj =[φj

π

γj =φj

u(2)=2Reint 2

0uzdz+u(0)=2

int 2

int 0

πr(z)dθ+b0 (510)

u(z) = 2Reint z

0w(z)dz + b0 w(z) = w0(z) +W (z) (511)

W (z) = Φ(z) + Ψ(z) Ψ(z) =int ν

2g1(z)dνe1 +

int micro

4|y|m2 D= minus g

4|y|m2

A1=1

4|y|m2

[m

2yminus d

|y|m2 minuse

] B1=

14|y|m2

[m

2yminus d

|y|m2+e

]

A2=1

4|y|m2

[m

2|y|minusd

|y|m2 minuse

] B2=

14|y|m2

[m

2|y|minusd

|y|m2+e

]in Dminus

(514)

(515)

u0(z) = 2Reint z

0w0(z)dz + b0 in D (516)

Wz

D+

Dminus

(518)

Re [λ(z)W (z)] = 0 z isin Γ u(0) = 0 u(2) = 0

W (z) = Φ(z) + Ψ(z)

Ψ(z) =int ν

2[Aξ+Bη+Cu]e1dν+

int micro

(520)

]= 0

12

]= s(x) on L0

(522)

CHAPTER VI

Luz=

uzz

uzzlowast

D+

Dminus

(12)

14[uxx minus uyy]

A1 =a+ ib

2 A2 =

c

4 A3 =

d

4in D

Condition C

12

partu

12

partu

Im [λ(z)uz]z=z1 = b1 u(0) = b0 u(2) = b2

(16)

K =12(K1 +K2) (18)

Kj =[φj

π

γj =φj

Im [λ(z)uz]|z=z2 = b2 (110)

u(z) = 2Reint z

0w(z)dz + b0 w(z) = w0(z) +W (z) in D (111)

Lw =

wz

wzlowast

= 0 in

D+

Dminus

(112)

int intD+

g(ζ)ζ minus z

2g1(z)dνe1 +

int micro

(113)

g(z)=

A=ReA1+ImA1

2 B=

ReA1minusImA1

(114)

(115)

where

X(z) =2prod

j=1

ηj =

2|γj|+ δ γj lt 0

δ γj ge 0j = 1 2

(117)

u0(z) = 2Reint z

0w0(z)dz + b0 in D (118)

(119)

of mixed type

(121)

u0(z) = 2Reint z

u1(z) = 2Reint z

0w1(z)dz + b0 w1(z) = w0(z) + Φ1(z) + Ψ1(z)

Ψ11(z) =

int ν

2G1(z)dν G1(z) = Aξ0 +Bη0 + Cu0 +D

Ψ21(z) =

int micro

(123)

(124)

namelyCβ[Ψ1

n (128)

wn(z) =nsum

m=1

um(z) + u0(z) n = 1 2 (129)

Luz =

uzz

uzzlowast

D+

Dminus

(131)

M12 = k1 + k2 + |b0|+ |b1| (132)

Luz =

uzz

uzzlowast

D+

Dminus

(133)

M4k1 + k2 + 2k0tσ + 2k0t

τ + |b0|+ |b1| = t (134)

um+1 = C1[um+1 D] le M14 = M14(p0 β k0 D) (139)

um+1(z) = 2Reint z

+int x+y

(140)

m

0(1 + k20) From the

+int x+y

(142)

int z

M4k1 + k2 + 2k0tσ + 2k0t

τ + |b0|+ |b1| = t (144)

C1[u(z) D] le (M4 + 1)(2k1 + k2) (146)

12

partu

u(0)=b0 u(a)=b1 u(2)=b2 (22)12

partu

zisinL4

1|a(x)+b(x)| lek0

(23)

K =12(K1 +K2 +K3) (24)

Kj=[φj

π

γj=φj

|c1| |c2| le k2

Ψ(z) =

⎧⎪⎪⎨⎪⎪⎩int ν

a

1 int ν

2g1(z)dνe1 +

int micro

a

(27)

λ(x) =

1 + i x isin Lprime

0 = (0 a)

(28)

s(x)=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

+Re[λ(x)Ψ(x)]

xisin(0a)

h(x)=Re[λ(x)Ψ(x)]

(29)

X(z) =3prod

j=1

ηj =

(210)

u(z) = 2Reint z

0w(z)dz + b0 in D (211)

int intD+

g(ζ)ζ minus z

Ψ(z) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩int ν

a

1 int ν

2g1(z)dνe1 +

int micro

a

(212)

Im [λ(zprimej)Φ(z

(213)

λ(x) =

1 + i x isin Lprime

0 = (0 a)

0] le k3

(214)

where

X(z)=3prod

j=1

wz

wzlowast

D+

Dminus

(216)

2Re[1minus iradic2

uz

]=

partu

[1 + iradic2

uz

]=

partu

partl= 0 on (a 2) (218)

(220)

J =

∣∣∣∣∣u1(a) u2(a)

u1(2) u2(2)

∣∣∣∣∣ = 0 (221)

d1u1(a) + d2u2(a) = 0 d1u1(2) + d2u2(2) = 0 (222)

2 minus b2 (223)

C1[u D] le M18(k1 + k2)

(225)

1 Lprime2 Lprime

3 Lprime4 are

4 = γ2(x) + y = 0 l2 le x le 2(226)

12

partu

12

partu

=bj+2 j=12(227)

j] le k2 j = 1 4

maxzisinLprime

1

zisinLprime4

1|a(x) + b(x)| le k0

(228)

4where Lprime

1 Lprime2 Lprime

(230)

2 Lprime3 Lprime

ν =1

=12minusa

(231)

x =12(micro+ ν) =

(232)

and

x=12(micro+ν) =

12(2minusa)

+2(x+ y+a)minus2ax]

12(2minusa)

(233)

1

2

(234)

1

2

(236)

u(z) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

2Reint z

0w(z)dz + b0 in D+

2Reint z

a

2Reint z

2

(238)

12

partu

12

partu

[n2]sumj=0

A2jA2j+1

12

partu

[(n+1)2]sumj=1

B2jminus1B2j

(32)

Im [λ(z)uz]|z=A2j+1 = c2j+1 j = 0 1 [n2]

(33)

zisinL4

1|a(x) + b(x)| le k0

(34)

uzzlowast

= 0 in

D+

Dminus

(37)

Lw =

wz

wzlowast

= 0 in

D+

Dminus

(38)

Im [λ(z)w(z)]|z=A2j+1 = c2j+1 j = 0 1 [n2]

(39)

and the relation

u(z) = 2Reint z

2w(z)dz + b0 (310)

(311)

Re [λ(x)w(x)] =

1 cap AB

2 cap AB(312)

h(x)=2r((1 + i)x2 + 1minus i)

(313)

on L2j+1 j = 0 1 [n2]

on L2j j = 1 [(n+ 1)2]

2j cap AB j = 1 [(n+ 1)2]

u(z) = 2Reint z

2w(z)dz + c0 w(z) = w0(z) +W (z) in D (314)

int intD+

g(ζ)ζminusz

Ψ(z)=

⎧⎪⎪⎪⎨⎪⎪⎪⎩

int ν

a2j+1

2j+1 j=01[n2]

int ν

2g1(z)dνe1+

int micro

a2jminus1

(315)

g(z)=

A=ReA1+ImA1

2 B=

ReA1minusImA1

2 C=A2D=A3 in D+

(316)

Im [λ(z)(Φ(z) + Ψ(z))]|z=A2j+1 = 0 j = 0 1 [n2]

(317)

(319)

on L2j+1 j = 0 1 [n2]

on L2j j = 1 [(n+ 1)2]

w(z) = w0(z) +W (z) =

⎧⎨⎩ Φ(z)φ(z) + ψ(z) in D+

wzlowast

= Re [A1w] + A2u+ A3 in

D+

Dminus

(320)

Ψ1(z)=

⎧⎪⎪⎪⎨⎪⎪⎪⎩int ν

a2j+1

2j+1 j=0 1 [n2]int ν

2g0(z)e1dν+

int micro

a2jminus1

(321)

(j = 1 [(n+ 1)2]

(322)

where

X(z) =m+nprodj=0

n+1prodj=0

(323)

wz

wzlowast

D+

Dminus

u(zj) = 0 j = 0 1 m u(aj) = 0 j = 1 n

Im [λ(z)w(z)]|z=A2j+1 = 0 j = 0 1 [n2]

(324)

1) ulowast(aprime

m+n)] = [dlowast1 d

Im [λ(z)ukz]|z=A2j+1 = 0 j = 0 1 [n2]

(326)

J =

∣∣∣∣∣∣∣∣∣u1(aprime

m+n)

cprime1u1(aprime

thus the function

cprimekuk(z) in D

Luz=

uzz

uzzlowast

D+

Dminus

(327)

sumNj=1 Γj isin C2

12

partu

12

partu

maxzisinL1

zisinLprimeprime

1|a(x) + b(x)| le k0

(43)

0 The number

where

Kj=[φj

π

γj=φj

πminusKj j=1N (45)

j j = 1 N (46)

u(z) = 2Reint z

0w(z)dz + d1 w(z) = w0(z) +W (z) (47)

wz

wzlowast

= 0 in

D+

Dminus

(48)

int intD+

g(ζ)ζ minus z

Ψ(z) =

0g2(z)dmicroe2 +

int ν

0g2(z)dmicroe2 +

int ν

aj+1

(49)

g(z)=

A12+A1w(2w) w(z) =00 w(z) = 0 z isin D+

A =ReA1 + ImA1

2 B =

ReA1 minus ImA1

(411)

u0(z) = 2Reint z

0w0(z)dz + d1 (413)

on L0 in which

h(x)=[a

((1minusi)x2

)+b

((1minusi)x2

and

Ψ(x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int ν

2g1(z)dνe1 =

int ν

0g2(z)dmicroe2 =

int micro

0g2(z)dmicroe2=

int micro

Thus

w(z) = w0(z) +W (z) =

⎧⎨⎩ Φ(z)φ(z) + ψ(z) in D+

wzlowast

= Re [A1w] + A2u+ A3 in

D+

Dminus

(415)

wz

wzlowast

= Re [A1w] + A2u in

D+

Dminus

(416)

in which

w(z)=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Ψ(z)=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int micro

int ν

2jminus1

j = 1 2 Nint micro

int ν

aj+1

j = 1 2 N minus 1

(418)

n in Dminus

uzz = Re [A1uz] + A2u in D+ (420)

12

partu

(422)

and the relation

u(z) = 2Reint z

0w(z)dz + d1 (423)

(424)

f(x+ y) = Re [(1 + i)w(x+ y)]

1 0 b1

j j=2N

2j j=12Nminus1

2jminus1 j=2N

wz

wzlowast

= Re [A1w] + A2u+ A3 in

D+

Dminus

(426)

int intD+

g0(ζ)ζ minus z

Ψ1(z) =

0g0(z)dmicroe2 +

int ν

0g0(z)dmicroe2 +

int ν

aj+1

(429)

+Re [λ(x)Ψ1(x)]

h(x) =[a

((1minus i)x

2

)+ b

((1minus i)x

2

)][Re (λ(z1)(r(z1) + ic1))

+Im (λ(z1)(r(z1) + ic1))] on L0

(430)

Ψn(z) =

int ν

aj+1

and then|[wplusmn

0g0(z)e2dmicro|

2g0(z)e1dν|+

Nminus1sumj=1

|int micro

aj+1

g0(z)e1dν|]

(433)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int micro

int ν

2jminus1

j=12Nint micro

int ν

aj+1

j=12N minus1(434)

0

Rprimenminus1

(n minus 1) dRprime

n in Dminus

(435)

ie

Ψlowast(z)=

int ν

aj+1

(437)

(438)

(440)

where

X(z) =2Nprodj=1

wz

wzlowast

= Re [A1w(z)] + A3 in

D+

Dminus

(442)

Im [λ(zprimej)w(z

primej)] = dprime

j j = 1 N(443)

0w(z)dz + d0 in D

(446)

in which d0 = φ(0)

λ(z) = a+ ib =

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

(1 + i)radic2 on L1

j=2L2jminus1

(447)

and

r(z) =

j=1L2jminus1

c1=Im[1 + iradic2

uz(z1 minus 0)]=

φx minus φxradic2

|z=z1minus0=0

uz(zj + 0)]= 0 j = 2 N

(448)

Re[1minus iradic2

w(x)]= s(x) =

πminus K1 =

14

minus 0 =14

lt 1

πminusK2N=minus5

4lt0

πminus Kj=

54minus1= 1

4lt1 j=3 5 2Nminus1

πminus Kj=minus1

4=minus1

4lt0 j=2 4 2Nminus2

N

2minus 1

References

References 241

242 References

References 243

244 References

References 245

246 References

References 247

248 References

References 249

Index

128 130 132 133 149 150 151153 155 165ndash167 174 184 190202 212 223 228 230 232

AV Bitsadze 157

Banach space 53 55 93 104 137 138191 193 207 208

217 219 220 227 228 232 237238 239

Index 251

108 109 159 164 173 196 202210 219 226

problem 79 80 90 91 93 94125 143 144 145 146 156

140 158 171 200 204 209 214218 227 228

87 90 93 103 106 108 119 122126 134 143 157 177 182 193198 213 218 226 233

252 Index

179 180 182ndash184 186 188 189191 193 194

103 107 120 125 126 139 144145 156 158 160ndash164 180 181

Index 253

186 196 202 228 229 232 237238 239

140 141 216inversion 97

137 141 142 162 165 167 168171 175 176 177 184 189 192212 218 224 237 239

238 239

254 Index

110 111 158 159 160 161 162196 238 239

Problem P0 44 67 97 109 118 158196 197 202

225Problem P prime

0 228Problem Q 97 106 171ndash177 202

mixed type 79 134 143 199 200209 215 218 227 228

Index 255

96 107

171 206 225representation 20 25 30 35 36 38

43 46ndash48 63 64 79 87 90ndash92 94134 160 182 198 229

50 52 54 75 104 131 132 138152 153 169 170 205 207 208235 236

160

43 66 106 108 119 126 134 139141 143 157 194 200 208 218

94 104 119 157 160 168 173175 177 180 182 189 191 194200 206 209 211 213 231

subsequence 102 105 106 111 116132 154 170

77 93

55ndash58 60 67 80 88 97 111120 140 141 142 158 196202 216 217

209 210 213 227 229 236

98 102 106 111 131 132 138153 154 170 205 208 236

256 Index

53 56 58 63 64 65 72 78 8693 103 104 106 116 125 133135 137 141 142 144 147 150155 159 162 165 189 192 204207 208 212 217 224 227 238

90 93 96 103

zone domain 90 94

Book Cover
HALF-TITLE
SERIES TITLE
TITLE
COPYRIGHT
CONTENTS
PREFACE
REFERENCES
INDEX

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Linear and quasilinear complex equations of hyperbolic and...

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