Elliptic Boundary Value Problems of Second Order in ...

Elliptic Boundary Value

Problems of Second Order in

Piecewise Smooth Domains

Mikhail BorsukUniversity of Warmia and Mazury

Olsztyn, PolandVladimir Kondratiev

Moscow State University, Russia

To the Centenary Anniversary of our teacherYa.B. Lopatinskiy

Contents

Introduction 1

Chapter 1. Preliminaries 91.1. List of symbols 91.2. Elementary inequalities 101.3. Domains with a conical point 121.4. The quasi-distance function rε and its properties 141.5. Function spaces 151.5.1 Lebesgue spaces 151.5.2 Regularization and Approximation by Smooth Functions 17

1.6. Hölder and Sobolev spaces 191.6.1 Notations and denitions 191.6.2 Sobolev imbedding theorems 21

1.7. Weighted Sobolev spaces 221.8. Spaces of Dini continuous functions 241.9. Some functional analysis 281.10. The Cauchy problem for a dierential inequality 291.11. Additional auxiliary results 321.11.1 Mean Value Theorem 321.11.2 Stampacchia's Lemma 321.11.3 Other assertions 331.11.4 The distance function 351.11.5 Extension Lemma 351.11.6 Dierence quotients 37

1.12. Notes 39

Chapter 2. Integral inequalities 412.1. The classical Hardy inequalities 412.2. The Poincaré inequality 442.3. The Wirtinger inequality: Dirichlet boundary condition 462.4. The Wirtinger inequality: Robin boundary condition 472.4.1 The eigenvalue problem 472.4.2 The Friedrichs - Wirtinger inequality 51

2.5. Hardy - Fridrichs - Wirtinger type inequalities 542.5.1 The Dirichlet boundary condition 542.5.2 The Robin boundary condition 61

2.6. Other auxiliary integral inequalities for n = 2 65

iii

iv CONTENTS

2.7. Notes 71

Chapter 3. The Laplace operator 733.1. Dini estimates of the generalized Newtonian potential 733.2. The equation with constant coecients. Green's function 833.3. The Laplace operator in weighted Sobolev spaces 873.4. Notes 87

Chapter 4. Strong solutions of the Dirichlet problem for linear equations 894.1. The Dirichlet problem in general domains 894.2. The Dirichlet problem in a conical domain 924.3. Estimates in weighted Sobolev spaces 924.4. The power modulus of continuity 1154.5. Lp− estimates 1184.6. Cλ− estimates 1214.7. Examples 1284.8. Higher regularity results 1304.9. Smoothness in a Dini - Liapunov region 1334.10. Unique solvability results 1444.11. Notes 151

Chapter 5. The Dirichlet problem for elliptic linear divergentequations in a nonsmooth domain 153

5.1. The best possible Hölder exponents for weak solutions 1535.1.1 Introduction 1535.1.2 The estimate of the weighted Dirichlet integral 1585.1.3 Local bound of a weak solution 1715.1.4 Example 1745.1.5 Hölder continuity of weak solutions 1755.1.6 Weak solutions of an elliptic inequality 177

5.2. Dini continuity of the rst derivatives of weak solutions 1815.2.1 Local Dini continuity near a boundary smooth portion 1815.2.2 Dini continuity near a conical point 1845.2.3 Global regularity and solvability 195

5.3. Notes 197

Chapter 6. The Dirichlet problem for semilinear equationsin a conical domain 201

6.1. The behavior of strong solutions for nondivergentequations near a conical point 201

6.1.1 The weighted integral estimates (0 < q ≤ 1) 2026.1.2 The estimate of the solution modulus (0 < q ≤ 1) 2056.1.3 The estimate of the solution modulus (q > 1) 210

6.2. The behavior of weak solutions for divergence equationsnear a conical point 216

6.3. Notes 223

CONTENTS v

Chapter 7. Strong solutions of the Dirichlet problem fornondivergence quasilinear equations 225

7.1. The Dirichlet problem in smooth domains 2257.2. The estimate of the Nirenberg type 2277.2.1 Introduction 2277.2.2 Formulation of the problem. The main result 2287.2.3 The Nirenberg estimate 2297.2.4 Behavior of the solution near a corner point

(weak smoothness) 2357.2.5 The weighted integral estimate 2377.2.6 Proof of Theorem 7.7 241

7.3. Estimates near a conical point 2457.3.1 Introduction 2457.3.2 The barrier function 2467.3.3 The weak smoothness of solutions 2477.3.4 Estimates in weighted spaces 2527.3.5 Lp− and pointwise estimates of the solution

and its gradient 2617.3.6 Higher regularity results 266

7.4. Solvability results 2727.5. Notes 275

Chapter 8. Weak solutions of the Dirichlet problem for ellipticquasilinear equations of divergence form 277

8.1. The Dirichlet problem in general domains 2778.2. The mLaplace operator with an absorption term 2808.2.1 Introduction 2808.2.2 Singular functions for the mLaplace operator and

the corresponding eigenvalue problem 2828.2.3 Eigenvalue problem for m−Laplacian in a bounded domain

on the unit sphere 2848.2.4 Integral estimates of solutions 2918.2.5 Estimates of solutions for singular right hand sides 299

8.3. Estimates of weak solutions near a conical point 3078.4. Integral estimates of second weak derivatives of solutions 3138.4.1 Local interior estimates 3138.4.2 Local estimates near a boundary smooth portion 3238.4.3 The local estimate near a conical point 328

8.5. Notes 330

Chapter 9. The behavior of weak solutions to the boundaryvalue problems for elliptic quasilinear equationswith triple degeneration in a neighborhoodof a boundary edge 331

9.1. Introduction. Assumptions. 331

vi CONTENTS

9.2. A weak comparison principle. The E. Hopf strong maximumprinciple 337

9.3. The boundedness of weak solutions 3439.4. The construction of the barrier function 3529.4.1 Properties of the function Φ(ω) 3539.4.2 About solutions of (9.1.3) - (9.1.4) 3549.4.3 About the solvability of (9.1.3) - (9.1.4) with ∀a0 > 0 358

9.5. The estimate of weak solutions in a neighborhood of a boundaryedge 363

9.6. Proof of the main Theorem 3709.7. Notes 373

Chapter 10. Sharp estimates of solutions to the Robin boundary valueproblem for elliptic non divergence second order equationsin a neighborhood of the conical point 375

10.1. Linear problem 37610.1.1 Formulation of the main result 37610.1.2 The Lieberman global and local maximum principle.

The comparison principle 37810.1.3 The barrier function. The preliminary estimate of

the solution modulus 38210.1.4 Global integral weighted estimate 38710.1.5 Local integral weighted estimates 40910.1.6 The power modulus of continuity at the conical point

for strong solutions 41710.1.7 Examples 419

10.2. Quasilinear problem 42210.2.1 Introduction 42210.2.2 The weak smoothness of the strong solution 42410.2.3 Integral weighted estimates 43210.2.4 The power modulus of continuity at the conical point

for strong solutions 45110.3. Notes 455

Bibliography 459

Notation Index 479

Index 481

Introduction

This book is devoted to the investigation of the behavior of weak orstrong solutions of the boundary values problems for the second order ellip-tic equations (linear and quasilinear) in the neighborhood of the boundarysingularities. Our main goal is to establish precise exponent of the solutiondecreasing rate and best possible conditions for this.

Since nowadays there exists a visible a full enough theory for linear el-liptic equations with partial derivatives, it has become possible to moveforward in the nonlinear equations analysis. Considerable success in thisdirection has been achieved particularly for the second order quasilinear el-liptic equations, due to the works of Schauder, Caccioppoli, Leray, etc. (see[43, 128, 209, 214 ]). They worked out a method that allows to prove exis-tence theorems given the appropriate a priori estimates. This method doesnot require preliminary construction of the fundamental solution and allowsapplying some functional analysis theorems instead of integral equation the-ory.

On the one hand, it proved possible to prove quite easily the solvabil-ity of boundary value problems for the second order quasilinear equations,given the Hölder coecients estimate of the solution rst derivatives of ap-propriate linear boundary problem, with a constant which depends only onthe maximum module of the problem coecients. Thus, there appeared anecessity of studying linear problems more deeply and giving them moreprecise estimates. The eorts of many mathematicians were directed to-wards this. L. Nirenberg [326] received the above mentioned estimate for atwo-dimensional nonselfadjoint equation, thanks to which it was possible toestablish the existence theorem of the Dirichlet problem for the second orderquasilinear elliptic equations under minimal assumptions on the smoothnessof the equation coecients. In the case of a multi-dimensional equation suchestimate was received by H. Cordes [84], however in the assumption thatthe equation complies with the condition (depending on the Euclidean spacedimension N > 2) stronger than that of the uniform ellipticity. On the otherhand, the attempts of receiving the above mentioned a priori estimate forthe second order general elliptic equations were not successful, as it turnedout that such an estimate in simply impossible.

Thus, to prove the classical solvability of boundary value problems for thesecond order quasilinear equations, it was necessary to create the methodswhich would allow receiving the estimates needed, directly for the non-linear

1

2 Introduction

problem itself. Such methods were created. The ideas of the new methodcould be already found in the works of S. Berstein and later De Giorgi andJ. Nash ([128, 214]) . O. Ladyzhenskaya and N. Ural'tseva improved anddeveloped this method. Then they published their well-known monograph[214], in which the method was worded and applied to dierent boundaryvalue problems. Their researches served as an incentive to create a wholeseries of their pupils' and other mathematicians' works (we note the works[206], [221] - [235], [364], [398]). All investigations mentioned above referto boundary value problems in suciently smooth domains. It should benoticed that denitive completing of these investigations cost a lot of eortof many mathematicians and over 30 years of work.

However, many problems of physics and technics lead to the necessity ofstudying boundary value problems in the domains with nonsmooth bound-ary. To the such domains, in particular, refer the domains which have onthe boundary a nite number of angular (N = 2) or conical (N > 2) points,edges etc. The 20-year-ago state of the theory of boundary value problem-s in non-smooth domains is described in detail in the well-known surveyof V.A. Kondrat'ev and O.A. Oleinik [175], in the book of A. Kufner andA.-M. Sandig [211] as well as in the monographs of V.G. Maz'ya and hiscolleagues [260, 196]. For this reason we will only dwell on its few aspectsand present in more detail the achievements of this theory.

Among the rst works studying the behavior of the boundary valueproblem solution in the neighbourhood of an angular boundary point forthe Laplace or Poisson equation, we can nd the works [77, 324, 400,125]. In the work [324] S. Nikol'skiy established the necessary and su-cient conditions of belonging to the Nikolskiy's space Hr

p of the Dirichletproblem solution for the Laplace equation. E. Volkov [400] described thenecessary and sucient conditions of belonging to the space Ck+α(G) (kis an integer, α ∈ (0, 1)) of the Dirichlet problem solution for the Pois-son equation 4u = f(x), x ∈ G, in the case, where G is a rectangle.V. Fufaev [125] considered the Poisson equation 4u = f(x), x ∈ G inthe domain G, where ∂G \ O is an innitely smooth curve, and in a cer-tain neighborhood of the point O the boundary ∂G consists of two seg-ments intersecting at an angle ω0. The smoothness of the Dirichlet prob-lem solution depends on the ω0 : the more small is angle ω0, the moresmooth is the solution (if f ∈ C∞(G).) There are exceptional values ω0,for which there are no obstacles for the smoothness. In particular, ifu∣∣∣∂G

= 0, f = 0 in a certain neighborhood of the point O and if πω0

is

an integer, then u ∈ C∞(G).Thus, the violation of the boundary smoothness condition leads to the

situation, when for the boundary value problem solution there appear sin-gularities in the neighborhood of the boundary irregular point. As we know,

Introduction 3

in the boundary value problems theory for elliptic equations in smooth do-mains, the situation is as follows: if the problem data is smooth enough,then the solution is also suciently smooth.

One of the rst works studying the general linear boundary value prob-lems for the domains with conical or angular points were V. Kondratiev'sfundamental works [158, 159] as well as papers of M. Birman & G. Skvortsov[47], G. Eskin [114, 115], Ya. Lopatinskiy [239], V. Maz'ya [247] - [250],[252, 291]. These works and extending their other works examine normalsolvability and regularity in the Sobolev weighted spaces of general linearelliptic problems in non-smooth domains under assumptions of sucientsmoothness of both the manifold ∂G \ O and the problem coecients. Thesolution is considered in special spaces of functions with the derivatives,which are summable with some power weight. These spaces well detect thebasic singularity of the solutions of such problems. It has also become clearthat the methods used for the analysis of boundary elliptic problems in s-mooth domains, are not applicable: in the regarded case it is impossible tostraighten the boundary by using a smooth transformation.

V. Kondratiev [158, 159] studied this problem in L2− Sobolev spaces,V. Maz'ya and B. Plamenevskiy [268] - [277] (see also [260] - [266], [279])extended the Kondratiev results to Lp− Sobolev and other spaces. There aremany other works concerning elliptic boundary value problems in nonsmoothdomains (see Bibliography).

The pioneering works in the studying elliptic boundary value problemsin nonsmooth domains for quasilinear equations was done by V. Maz'ya,I. Krol and B. Plamenevskiy [253, 255], [202] - [205], [267].

If we aimed at examining a nonlinear elliptic problem, we could come tothe necessity to clear up under which smoothness conditions for coecientsand right parts of linear problem, the solvability in appropriate functionalspaces and the necessary a priori estimates of the solution take place. Thisclearing up is dealt with in chapters 4 and 5. These chapters study the linearelliptic Dirichlet problem for the nondivergent form equation

Lu := aij(x)Diju(x) + ai(x)Diu(x) + a(x)u(x) = f(x), in G,u(x) = ϕ(x) on ∂G

(L)

and for the divergent form equation∂∂xi

(aij(x)uxj + ai(x)u) + bi(x)uxi + c(x)u =

= g(x) + ∂fj(x)∂xj

, x ∈ G;

u(x) = ϕ(x), x ∈ ∂G.(DL)

The questions of the smoothness of solutions in the neighborhood of anangular point for the linear nondivergence second order elliptic equationwere earlier studied in works [18] - [22]. There the authors assume that theequation coecients are Hölder - continuous . Our assumptions concerningto the smoothness of the coecients are minimally possible: the equation

4 Introduction

lieder coecients must be Dini - continuous at the conical point O, whereaslower ones can even grow (here we indicate the exact power growth order).In 4.7 we construct the examples which show that the Dini condition for theequation lieder coecients at the conical point, as well as the assumptionconcerning the lower equation coecients, are essential for the validity ofthe estimates derived in the chapters 4 -5. Otherwise in these estimates theexponent λ should be changed into λ − ε with ∀ε > 0. The fact that theexponent λ in these estimates cannot be increased either, is shown by thepartial solutions of the Laplace equation in the domain with the angular orconical point. In this sense the estimates of chapters 4 and 5 are the bestpossible.

The estimates obtained in 4.5, 4.6, 4.9 allow to formulate new exis-tence theorems of the linear Dirichlet problem solution. These theorems areformulated and proved in 4.10.

The regularity theory of strong solutions for this problem and its solv-ability in a smooth domain are well investigated [128, 215, 206, 209 ]. Butsuch theory in a nonsmooth domains is very small investigated. Existencetheorems obtained in 4.10 play a fundamental role further (chapter 7) dur-ing the considering the quasilinear problem

aij(x, u, ux)uxi,xj + a(x, u, ux) = 0, aij = aji, x ∈ Gu(x) = ϕ(x), x ∈ ∂G

(QL)

solvability. As mentioned above, to construct the theory of the Dirichletproblem solvability for quasilinear equations, the appropriate a priori es-timates of a nonlinear task itself are needed. Chapter 7 is dedicated toobtaining such estimates. The local Hölder estimate (near an angular orconical point) of the rst derivatives of solution is the central part in theseestimates. Although the results obtained in 7.2 are completely included inthe results of 7.3 (which is quite natural), the specic character of the planecase allowed us to single it out. Besides, the methods of obtaining estimatesare dierent in the cases where N = 2 and N > 2. We were interested indemonstrating the possibility of applying of the L. Nirenberg method for do-mains with an angular point. Thus it becomes possible to establish a basicestimate

|u(x)| ≤ c0|x|1+γ

with certain γ > 0. In the case of a conical point (N > 2) this method is notsuitable, as it is purely two-dimensional. To obtain the similar estimate inthis situation we resort to the barrier technique and apply the comparisonprinciple. Theorems of 7.4 also show that the (QL)− problem solutionshave the same regularity (at a conical point) as the (L)− problem solutions.

There is another fact which is worth pointing out. Known in lineartheory, the method of non-smooth domain approximation by the sequenceof smooth domains while examining nonlinear problems, is not applicable,because of the impossibility of the passage to the limit. We avoid this dif-culty by introducing a quasi-distance function rε(x). The introduction of

Introduction 5

such function allows us to work in the given domain, and then to providethe passage to the limit over ε → +0 (where rε(x) → r = |x|.) We use thesame method studying the problems (L) in Chapter 4 and (DL) in Chapter5.

In 7.3.6 we prove the theorems of the solution smoothness rise which areanalogous to the linear case. The results of 4.10 (concerning the solvabilityof the linear problem) and the estimates for solutions of the nonlinear prob-lem, given in 7.2, 7.3, permit to precede to examining the (QL)− problemsolvability in 7.4.

To sum up, in Chapters 4, 7 we have completely constructed the theoryof the rst boundary problem solvability for the second order non-divergentuniform elliptic equations in the domains with conical or angular points.

In Chapters 5, 8, 9 we consider the theory which refers to the equationsof divergent type. The history of research development of such equations ismuch richer, because of in this case it is possible to study weak solutionsof the problem, which is this way changed into the equivalent integral i-dentity with no second generalized derivatives of the sought function. Thedetailed history of studies of the linear problem can be found in surveys[90, 132, 170, 175 ]. We will only dwell on some of them. The exactsolution estimates near singularities on the boundary were obtained in theworks [395, 396] under the condition that lieder coecients of the equationsatisfy the Hölder condition, lowest coecients are missing: the rate of thesolution decrease in the neighborhood of a boundary point is characterizedby the function λ(%) (the latter is the least by the modulus eigenvalue ofthe Laplace - Beltrami operator in a domain Ω% on the sphere with zeroDirichlet data on ∂Ω%). In Chapter 5 we give results of works [168, 169].Under certain assumptions on the structure of the boundary of the domainin a neighborhood of the boundary point O and on the coecients of thelinear equation (DL), one obtains a power modulus of continuity at O fora weak solution of the Dirichlet problem vanishing at that point. Moreover,the exponent is the best possible for domains with the assumed boundarystructure in that neighborhood. The assumptions on the coecients of theequation are essential, as an example 5.1.4 shows.

In [21] A.Azzam and V.Kondrat'ev established the Hölder continuity ofthe rst derivatives of the (DL) problem weak solutions in the neighborhoodof an angular point, under the condition of the Hölder continuity of equa-tion coecients; in this case the Hölder exponent satises the inequalityα < π

ω0− 1. In 5.2 we generalize this result for the case of a conical point

and we also weaken the coecient smoothness requirements: they have tobe Dini continuous.

In Chapter 6 we study properties of strong and weak solutions of theDirichlet problem for semi-linear uniform elliptic second order equation in aneighborhood of conical boundary point.

6 Introduction

Let us refer to the quasilinear problem for the divergence form equation

Q(u, φ) ≡∫G

ai(x, u, ux)φxi + a(x, u, ux)φ dx = 0(DQL)

The regularity theory of weak solutions for this problem and its solvabili-ty in a smooth domain are well-known [128, 213, 214 ]; (see also [80, 82,127, 234]). The regularity theory of weak solutions for quasilinear ellipticequations of the arbitrary order and elliptic systems as well as their solv-ability in a smooth domain are investigated in the monographs [181]- [183],[357, 99, 40 ].

The rst works in the investigation of the solutions behavior for quasilin-ear elliptic equations in domains with angular and conical points were doneby V. Maz'ya, I. Krol and B. Plamenevskiy [204, 205, 202, 203, 267, 278 ].V. Maz'ya and I. Krol [202] - [205] have given estimates for the asymptoticbehavior near reentrant boundary points of the equation of the type (LPA)solutions. V. Maz'ya and B. Plamenevskiy [267, 278] have constructed theasymptotic solution of general quasilinear elliptic problem in a neighborhoodof angular or conical point.

Beginning from 1981, there appeared a cycle of the P. Tolksdorf [370]-[375], E. Miersemann [298] - [306] and M. Dobrowolski [97, 98] work-s, where they studied the behavior of the weak solutions to the (DQL)(see Chapter 8) in the neighborhood of an angular or conical boundarypoint. In [298] - [300] it is shown that a weak solution belongs to W 2 ∩C1+γ(G) with certain γ ∈ (0, 1), under the assumption that m = 2, ω0 ∈(0, π) ai(x, u, z), i = 1, . . . , N do not depend on x, u and the function a(x, u, z)does not depend on u, z. Some elaborations and generalizations for a widerclass of elliptic equations were made in [304]. In 8.1, chapter 8 of [132],P. Grisvard considered the problem (DQL) for N = 2, G is a convex poly-gon, ai(x, u, z) ≡ a(z)zi, i = 1, 2; a(x, u, z) ≡ f(x); here a(z) is a positivedecreasing function and a′′(z) is continuous. He proved the existence anduniqueness of the solution from the space W 2,m(G) ∩W 1,m

0 (G), if f(x) ∈C1+α(G), α ∈ (0, 1), 2 < m < 2

2−π/ω0, where ω0 is the measure of the

largest angle on the polygon boundary. In [370] - [373, 375], P. Tolks-dorf considered the problem (DQL) with ai(x, u, z) ≡ a(|z|2)zi + bi(z),i = 1, . . . , N ; a(x, u, z) ≡ f(x); under the following conditions:

ν(k + t)m−2 ≤ a(t2) ≤ µ(k + t)m−2; (ν − 12

)a(t) ≤ ta′(t) ≤ µa(t)

with some ν > 0, µ > 0, k ∈ [0, 1] and ∀t > 0. In addition, it is assumed that

limt→∞

ta′(t)a(t)

=m− 2

2> −1

2; lim|z|→∞

∣∣∣∣∂bj(z)∂zi

∣∣∣∣ · a−1(|z|2) = 0.

He obtained the upper- and lower- bounded estimates of the rate of thepositive weak solution decrease in the neighborhood of the boundary conicalpoint, which is characterized by the lowest module eigenvalue of the nonlinear

Introduction 7

eigenvalue problem (NEV P1) (see Chapter 8, 8.2.2). In Chapter 8 wegeneralize these results for a wider class of equations and analyze arbitrary(not only positive) weak solutions. It is also important to note here that theour estimates reinforce the established by O.Ladyzhenskaya and N.UraltsevaLipschiz-estimates of the (DQL) problem solution in the neighborhood ofthe boundary point, in the case when the boundary point is conical. In8.2 we establish the power weight estimates of weak solutions, similar tothe estimates in 7.3: in the latter the weight exponent is the best possible.The estimates of 8.2 allow to obtain the best possible estimates of the weaksolution module and its gradient. Finally, in 8.4 we estimate the secondgeneralized derivatives of weak solutions in the Sobolev weighted space withthe best weight exponent.

In Chapter 9 we investigate the behavior of weak solutions of the rstand mixed boundary value problems for the quasilinear second order ellipticequation with the triple degeneracy and singularity in the coecients in aneighborhood of the boundary edge. The coecients of our equation nearthe edge are close to coecients of the model equation

− d

dxi

(rτ |u|q|∇u|m−2uxi

)+ a0r

τ−mu|u|q+m−2 −

−µrτ |u|q−1|∇u|msgn u = f(x),(ME)

0 ≤ µ < 1, q ≥ 0, m > 1, a0 ≥ 0, τ ≥ m− 2.

The Chapter 10 is devoted to the investigation of the behavior of strongsolutions to the Robin boundary value problem for the second order ellipticequations (linear and quasilinear) in the neighborhood of a conical boundarypoint.

8 Introduction

CHAPTER 1

Preliminaries

1.1. List of symbols

Let us x some notations used in the whole book:

• [l] : the integral part of l (if l is not integer);• R the set of real numbers;• R+ the set of positive numbers;• RN the Ndimensional Euclidean space, N ≥ 2;• N the set of natural numbers;• N0 = N ∪ 0 the set of nonnegative integers;• x = (x1, . . . , xN ) an element of RN ;• O = (0, . . . , 0);• (r, ω) = (r, ω1, . . . , ωN−1) spherical coordinates in RN with pole Odened by

x1 = r cosω1,

x2 = r sinω1 cosω2,

. . . . . . . . . . . . . . .

xN−1 = r sinω1 sinω2 · · · sinωN−2 cosωN−1,

xN = r sinω1 sinω2 · · · sinωN−2 sinωN−1;

• SN−1 the unit sphere in RN ;• Br(x0) the open ball with radius r centred at x0;• Br(x0) the closed ball with radius r centred at x0;• ωN = 2πN/2

NΓ(N/2) the volume of the unit ball in RN ;• σN = NωN the area of the Ndimensional unit sphere;• RN+ the half-space x : xN > 0;• Σ the hyperplane x : xN = 0;• G a bounded domain in RN ;• dx volume element in RN ;• ds area element in RN−1;• dσ area element in RN−2;• ∂G the boundary of G, in what follows we shall assume that O ∈ ∂G;• d(x) := dist(x, ∂G);• n = (n1, . . . , nN ) exterior unit normal vector on ∂G;• G = G ∪ ∂G the closure of G;• measG the Lebesgue measure of G;

9

10 1 Preliminaries

• diamG the diameter of G;• K an open cone with vertex in O;• Ω := K ∩ SN−1;• C : the rotational cone x1 > r cos ω0

2 ;• ∂C : the lateral surface of C : x1 = r cos ω0

2 ;•⟨·, ·⟩ the scalar product of two vectors;

• Diu := ∂u∂xi

;• ∇u := (D1u, . . . ,DNu);• Diju := ∂2u

∂xi∂xj;

• D2u the Hessian of u;

• |∇u| := (N∑i=1

(Diu)2)1/2;

• |D2u| := (N∑

i,j=1(Diju)2)1/2;

• β = (β1, . . . , βN ), βi ∈ N0 an Ndimensional multi-index;• |β| := β1 + · · ·+ βN the length of the multi-index β;• Dβ

x = Dβ := ∂|β|

∂xβ11 ∂x

β22 ...∂x

βNN

a partial derivative of order |β|;

• ∂u∂n =

⟨∇u, n

⟩ the exterior normal derivative of u on ∂G;

• δji Kronecker's delta;• supp u : the support of u, the closure of the set on which u 6= 0;• c = c(∗, . . . , ∗) a constant depending only on the quantities appear-ing in the parentheses. The same letter c will be sometimes used todenote dierent constants depending on the same set of arguments.

1.2. Elementary inequalities

In this section we review some elementary inequalities (see e.g. [37, 141])which will be frequently used throughout this book.

Lemma 1.1. (Cauchy's Inequality) For a, b ≥ 0 and ε > 0, we have

ab ≤ ε

2a2 +

12εb2.(1.2.1)

Lemma 1.2. (Young's Inequality) For a, b ≥ 0, ε > 0 and p, q > 1with 1

p + 1q = 1, we have

ab ≤ 1p

(εa)p +1q

(b

ε

)q.(1.2.2)

Lemma 1.3. (Hölder's Inequality) Let ai, bi, i = 1, . . . , N, be nonnegative real numbers and p, q ∈ R with 1

p + 1q = 1, we have

N∑i=1

aibi ≤

(N∑i=1

api

)1/p( N∑i=1

bqi

)1/q

.(1.2.3)

1.2 Elementary inequalities 11

Lemma 1.4. (Theorem 41 [141]). Let a, b be nonnegative real numbersand m ≥ 1. Then

mam−1(a− b) ≥ am − bm ≥ mbm−1(a− b).(1.2.4)

Lemma 1.5. (Jensen's Inequality) (Theorem 65 [141], Lemma 1 [354]).Let ai, i = 1, . . . , N, be nonnegative real numbers and p > 0. Then

λN∑i=1

api ≤

(N∑i=1

ai

)p≤ Λ

N∑i=1

api ,(1.2.5)

where λ = min(1, Np−1) and Λ = max(1, Np−1).

Lemma 1.6. Let a, b ∈ R, m > 1. Then the familiar inequality

|b|m ≥ |a|m +m|a|m−2a(b− a).(1.2.6)

is just.

Proof. By Young's inequality (1.2.2) with ε = 1, p = m, q = mm−1 , we

have

m|a|m−2ab ≤ m|b| · |a|m−1 ≤ |b|m + (m− 1)|a|m =⇒ (1.2.6).

Lemma 1.7. For m > 1 the inequality holds

1∫0

|(1− t)z + tw|m−2dt ≥ c(m)(|z|+ |w|)m−2(1.2.7)

with the positive constant c(m).

Proof. This inequality is trivial, if 1 < m ≤ 2, and in this casec(m) = 1. Let m > 2. If |z| + |w| = 0, then this inequality holds withany c(m). Let now |z|+ |w| 6= 0. Setting

ζ =z

|z|+ |w|, η =

w

|z|+ |w|=⇒ |ζ|+ |η| = 1

we want proof the inequality1∫

0

|(1− t)ζ + tη|m−2dt ≥ c(m).

We consider the function

f(ζ, η) =

1∫0

|(1− t)ζ + tη|m−2dt

on the set E = (ζ, η) ∈ R2∣∣∣|ζ| + |η| = 1. This function is continuous on

E, since m > 2. The set E is nite-dimensional and bounded, and thereforeit is the compact set. By the Weierstrass Theorem, such function achieves

12 1 Preliminaries

the minimum on E in some point (ζ0, η0) ∈ E. It is clear that f(ζ0, η0) ≥ 0.Suppose that f(ζ0, η0) = 0. Then we have

|(1− t)ζ0 + tη0| = 0, ∀t ∈ [0, 1] =⇒ (ζ0, η0) = (0, 0) /∈ E.

Hence it follows that f(ζ0, η0) > 0 and therefore there is a positive constantc(m) such that the required inequality is fullled.

1.3. Domains with a conical point

Definition 1.8. Let G ⊂ RN be a bounded domain. We say that G hasa conical point in O if• O ∈ ∂G,• ∂G \ O is smooth,• G coincides in some neighbourhood of O with an open cone K,• ∂K ∩ SN−1 is smooth,• K is contained in a circular cone with the opening angle ω0 ∈ 0, 2π.

For a domain G which has a conical point at O ∈ ∂G we introduce thenotations:• Ω := K ∩ SN−1;• dΩ := area element of Ω;• Gba := G ∩ (r, ω) : 0 ≤ a < r < b, ω ∈ Ω− a layer in RN ;• Γba := ∂G ∩ (r, ω) : 0 ≤ a < r < b, ω ∈ ∂Ω− the lateral surface ofthe layer Gba;• Gd := G \Gd0;• Γd := ∂G \ Γd0;• Ωρ := Gd0 ∩ ∂B%(0), % ≤ d;• G(k) := G2−kd

2−(k+1)d, k = 0, 1, 2, . . . .

Let us recall some well known formulae related to spherical coordinates(r, ω1, . . . , ωN−1) centered at the conical point O :

dx = rN−1drdΩ,(1.3.1)

dΩρ = ρN−1dΩ,(1.3.2)

dΩ = J(ω)dω(1.3.3)

denotes the (N − 1)− dimensional area element of the unit sphere;

J(ω) = sinN−2 ω1 sinN−3 ω2 . . . sinωN−2,(1.3.4)

dω = dω1 . . . dωN−1,(1.3.5)

ds = rN−2drdσ(1.3.6)

denotes the (N − 1)− dimensional area element of the lateral surface of

the cone K, where dσ denotes the (N − 2)− dimensional area element on ∂Ω;

|∇u|2 =(∂u

∂r

)2

+1r2|∇ωu|2 ,(1.3.7)

1.3 Domains with a conical point 13

where |∇ωu| denotes the projection of the vector ∇u onto the tangent planeto the unit sphere at the point ω :

∇ωu =

1√q1

∂u

∂ω1, . . . ,

1√qN−1

∂u

∂ωN−1

,(1.3.8)

|∇ωu|2 =N−1∑i=1

1qi

(∂u

∂ωi

)2

,(1.3.9)

where q1 = 1, qi = (sinω1 · · · sinωi−1)2, i ≥ 2,

∆u =∂2u

∂r2+N − 1r

∂u

∂r+

1r2

∆ωu,(1.3.10)

∆ωu =1

J(ω)

N−1∑i=1

∂

∂ωi

(J(ω)qi· ∂u∂ωi

)=(1.3.11)

=N−1∑i=1

1qj sinN−i−1 ωi

∂

∂ωi

(sinN−i−1 ωi

∂u

∂ωi

)denotes the Beltrami-Laplace operator,

divωu =1

J(ω)

N−1∑i=1

∂

∂ωi

(J(ω)√qiu

).(1.3.12)

Lemma 1.9. Let α ∈ R and v(x) = rαu(x). Then

Div = αrα−2xiu+ rαDiu,

|∇v|2 ≤ c1

(α2r2α−2u2 + r2α|∇u|2

),

Dijv = rαDiju+ αrα−2 (xiDju+ xjDiu)

+(α2 − 2α)rα−4xixju+ αrα−2uδji ,

|D2v|2 ≤ c2

(r2α|D2u|2 + r2α−2|∇u|2 + r2α−4u2

)with constants c1, c2 > 0 depending only on α and N.

Lemma 1.10. Let there is d > 0 such that Gd0 be the convex rotationalcone with the vertex at O and the aperture ω0, thus

Γd0 =

(r, ω)

∣∣∣x21 = cot2 ω0

2

N∑i=2

x2i ; |ω1| =

ω0

2, ω0 ∈ (0, π)

.(1.3.13)

Then

xi cos(~n, xi)|Γd0 = 0, and cos(~n, x1)|Γd0 = − sinω0

2,(1.3.14)

Proof. By virtue of (1.3.13) we can rewrite the equation of Γd0 in thisway:

F (x) ≡ x21 − cot2 ω0

2

N∑i=2

x2i = 0.

14 1 Preliminaries

We use the formula cos(~n, xi) =∂F∂xi|∇F | , ∀i = 1, ..., N . Because of

∂F

∂x1= 2x1,

∂F

∂xi= −2 cot2 ω0

2xi, ∀i = 2, . . . , N,

then

xi cos(~n, xi)∣∣Γd0

=1|∇F |

xi∂F

∂xi

∣∣∣Γd0

=2|∇F |

(x2

1 − cot2 ω0

2

N∑i=2

x2i

)∣∣∣∣Γd0

= 0.

Because of

|∇F |2 =(∂F

∂x1

)2

+N∑i=2

(∂F

∂xi

)2

= 4(x2

1 + cot4 ω0

2

N∑i=2

x2i

)⇒

|∇F |2∣∣∣Γd0

= 4x21

(1 +

cos2 ω02

sin2 ω02

)=

4x21

sin2 ω02

,

then

cos(~n, x1)∣∣∣Γd0

= −2x1sin ω0

2

2x1= − sin

ω0

2, since ∠ (~n, x1) >

π

2.

1.4. The quasi-distance function rε and its properties

Let us assume as in Denition 1.8 that the cone K is contained in acircular cone K with the opening angle ω0. Furthermore, let us suppose thatthe axis of K coincides with (x1, 0, . . . , 0) : x1 > 0. In this case we denethe quasi-distance rε(x) as follows. We x the point Q = (−1, 0, ..., 0) ∈SN−1 \ Ω and consider the unit radius-vector ~l = OQ = −1, 0, ..., 0. Wedenote by ~r the radius-vector of the point x ∈ G and introduce the vec-tor ~rε = ~r − ε~l ∀ε > 0. Since ε~l /∈ Gd0 for all ε ∈]0, d[, it follows thatrε(x) = |~r − ε~l| 6= 0 for all x ∈ G. It is easy to see that rε(x) has thefollowing properties:

1. Lemma 1.11. ∃h > 0 such that:rε(x) ≥ hr and rε(x) ≥ hε, ∀x ∈ G, where

h =

1, if 0 < ω0 ≤ π ,sin ω0

2 , if π < ω0 < 2π.

Proof. From the denition of rε(x) we know that

r2ε = (x1 + ε)2 +

N∑i=2

x2i = (x1 + ε)2 + r2 − x2

1 = r2 + 2εx1 + ε2.

If 0 < ω0 ≤ π, we have x1 ≥ 0 and therefore we obtain eitherr2ε ≥ r2 ⇒ rε ≥ r or r2

ε ≥ ε2 ⇒ rε ≥ ε.

1.5 Function spaces 15

If x1 = r cosω ≤ 0, |ω| ∈ [π2 ,ω02 ], we obtain, by the Cauchy inequality:

either

|2εr cosω| ≤ r2 cos2 ω + ε2 ⇒ 2εr cosω ≥ −r2 cos2 ω − ε2 ⇒ rε ≥ r · sinω0

2or

|2εr cosω| ≤ ε2 cos2 ω + r2 ⇒ 2εr cosω ≥ −r2 − ε2 cos2 ω ⇒ rε ≥ ε · sinω0

2.

2. Corollary 1.12.

hr ≤ rε(x) ≤ r + ε ≤ 2hrε(x); ∀x ∈ G ∀ε > 0;

3. If x ∈ Gd, then rε(x) ≥ d2 for all ε ∈]0, d2 [,

4. limε→0+

rε(x) = r, for all x ∈ G.

5. |∇rε|2 = 1, and 4rε = N−1rε

.

Proof. Because ∂rε∂x1

= x1+εrε

and ∂rε∂xi

= xirε

(i ≥ 2), then

|∇rε|2 =N∑i=1

(∂rε∂xi

)2

=(x1+ε)2+

NPi=2

x2i

r2ε

= 1, and

4rε = ∂2rε∂x2

1+

N∑i=2

∂2rε∂x2i

= 1rε− (x1+ε)2

r3ε

+N∑i=2

(1rε− x2

ir3ε

)= N

rε− r2

εr3ε

=

= N−1rε

.

1.5. Function spaces

1.5.1. Lebesgue spaces. Let G be a domain in RN . For p ≥ 1 wedenote by Lp(G) be the space of Lebesgue integrable functions equippedwith the norm

‖u‖Lp(G) =

∫G

|u|pdx

1/p

.

Theorem 1.13. (Fubini's Theorem , see Theorem 9 11, Chapter III[100]). Let G1 ⊂ Rm1 , G2 ⊂ Rm2 and f ∈ L1(G1 ×G2). Then for almost allx ∈ G1 and y ∈ G2 the integrals∫

G1

f(x, y)dx and

∫G2

f(x, y)dy

exist. Moreover,∫G1×G2

f(x, y)dxdy =∫G1

∫G2

f(x, y)dy

dx =∫G2

∫G1

f(x, y)dx

dy.

16 1 Preliminaries

Theorem 1.14. (Hölder's Inequality , see Theorem 189 [141]). Letp, q > 1 with 1

p + 1q = 1 and u ∈ Lp(G), v ∈ Lq(G). Then∫

G

|uv|dx ≤ ‖u‖Lp(G)‖v‖Lq(G).(1.5.1)

If p = 1, then (1.5.1) is valid with q =∞.

Corollary 1.15. Let 1 ≤ p ≤ q and u ∈ Lp(G), v ∈ Lq(G). Then

‖u‖Lp(G) ≤ (measG)1/p−1/q‖v‖Lq(G).(1.5.2)

Corollary 1.16. (Interpolation inequality) Let 1 1. Then u+ v ∈ Lp(G) and

‖u+ v‖Lp(G) ≤ ‖u‖Lp(G) + ‖v‖Lp(G).(1.5.3)

Theorem 1.18. (Clarkson's Inequality , see 3.2, Chapter I [360]).Let u, v ∈ Lp(G). Then∥∥∥∥u+ v

2

∥∥∥∥pLp(G)

+∥∥∥∥u− v2

∥∥∥∥pLp(G)

≤ 12

(‖u‖pLp(G) + ‖v‖pLp(G)

), 2 ≤ p <∞;

∥∥∥u+ v

2

∥∥∥ pp−1

Lp(G)+∥∥∥∥u− v2

∥∥∥∥ pp−1

Lp(G)

≤(1

2‖u‖pLp(G) +

12‖v‖pLp(G)

) 1p−1

, 1 ≤ p ≤ 2.

Theorem 1.19. (Fatou's Theorem , see Theorem 19 6, Chapter III[100]). Let fk ∈ L1(G), k ∈ N, be a sequence of nonnegative functionsconvergent almost everywhere in G to the function f . Then∫

G

fdx ≤ sup∫G

fkdx.(1.5.4)

Lemma 1.20. [325, Lemma 1.3.8] Let G1 ⊂ Rm1 , G2 ⊂ Rm2 and f, fk ∈Lp(G1 ×G2), k = 1, 2, . . . , with 1 ≤ p ≤ ∞ and

limk→∞

‖f − fk‖Lp(G1×G2) = 0.

Then there is a subsequence fkl of fk such that

limkl→∞

‖f(y, z)− fkl(y, z)‖Lp(G2) = 0

holds for almost every y ∈ G1.

1.5 Function spaces 17

1.5.2. Regularization and Approximation by Smooth Function-s. Let us denote by Lploc(G) the linear space of all measurable functions whichare locally pintegrable in G, i.e. which are pintegrable on every compactsubset of G. Although Lploc(G) is not a normed spaces, it can be readilytopologized.

Definition 1.21. We say that a sequence um converges to u in thesense of Lploc(G) if um converges to u in Lp(G′) for each G′ ⊂⊂ G.

Let r = |x− y| for all x, y ∈ RN and h be any positive number. Further-more, let ψh(r) be a non-negative function in C∞(RN ) vanishing outside theball Bh(0) and satisfying

∫RN

ψh(r)dx = 1. Such a function is often called a

mollier. A typical example is the function ψh(r) given by

ψh(r) =

ch−N · exp

(h2

|r|2−h2

)for r < h, c = const > 0;

0 for |r| ≥ h,

where c is chosen so that∫ψh(r)dx = 1 and whose graph has the familiar

bell shape.

Definition 1.22. For L1loc(G) and h > 0, the regularization of u,

denoted by uh is then dened by the convolution

uh(x) =∫G

ψh(r)u(y)dy(1.5.5)

provided h < dist(x, ∂G).

It is clear that uh belongs to C∞(G′) for any G′ ⊂⊂ G provided h <dist(G′, ∂G). Furthermore, if u belongs to L1(G) and G is bounded, thenuh belongs to C∞(RN ) for arbitrary h > 0. As h tends to zero, the functionψh(r) tends to the Dirac delta distribution at the point x. The signicantfeature of regularization, which we partly explore now, is the sense in whichuh approximates u as h tends to zero. It turns out, roughly stated, that if ulies in a local space, then uh approximates u in the natural topology of thatspace.

Lemma 1.23. Let u ∈ C0(G). Then uh converges to u uniformly on anysubdomain G′ ⊂⊂ G.

Proof. We have

uh(x) =∫

|x−y|≤h

ψh(r)u(y)dy =∫|z|≤1

ψ1(|z|)|u(x− hz)|dz

18 1 Preliminaries

(putting z = x−y

h

). Hence if G′ ⊂⊂ G and 2h < dist(G′, ∂G),

supG′|u− uh| ≤ sup

x∈G′

∫|z|≤1

ψ1(|z|)|u(x)− u(x− hz)|dz ≤

≤ supx∈G′

sup|z|≤1|u(x)− u(x− hz)|.

Since u is uniformly continuous over the set Bh(G′) = x | dist(x,G′) < h,the sequence uh tends to u uniformly on G′.

Lemma 1.24. Let u ∈ Lploc(G)(Lp(G)

), p < ∞. Then uh converges to

u in the sense of Lploc(G)(Lp(G)

).

Proof. Using Hölder's inequality, we obtain from (1.5.5)

|uh(x)|p ≤∫|z|≤1

ψ1(|z|)|u(x− hz)|pdz

so that if G′ ⊂⊂ G and 2h < dist(G′, ∂G), then∫G′

|uh(x)|pdx ≤∫G′

∫|z|≤1

ψ1(|z|)|u(x− hz)|pdzdx =

=∫|z|≤1

ψ1(|z|)dz∫G′

|u(x− hz)|pdx ≤∫

Bh(G′)

|u|pdx,

where Bh(G′) = x : dist(x,G′) < h. Consequently‖uh‖Lp(G′) ≤ ‖u‖Lp(Bh(G′)).(1.5.6)

The proof can now be completed by an approximation based on Lemma 1.23.We choose ε > 0 together with a C0(G) function w satisfying

‖u− w‖Lp(Bh′ (G′)) ≤ ε

where 2h′ < dist(G′, ∂G). By virtue of Lemma 1.23, we have for sucientlysmall h that ‖w − wh‖Lp(G′) ≤ ε. Applying the estimate (1.5.6) to thedierence u− w we obtain

‖u− uh‖Lp(G′) ≤ ‖u− w‖Lp(G′) + ‖w − wh‖Lp(G′) + ‖uh − wh‖Lp(G′) ≤≤ 2ε+ ‖u− w‖Lp(Bh(G′)) ≤ 3ε

for small enough h ≤ h′. Hence uh converges to u in Lploc(G). The result foru ∈ Lp(G) can then be obtained by extending u to be zero outside G andapplying the result for Lploc(R

N ).

1.6 Hölder and Sobolev spaces 19

Lemma 1.25. (On the passage to the limit under the integralsymbol) [358, Theorem III.10] Let χ(x) ∈ L∞(G) and let χh(x) be theregularization of χ. Then for any u ∈ L1(G)

limh→0

∫G

χh(x)u(x)dx =∫G

χ(x)u(x)dx.

1.6. Hölder and Sobolev spaces

1.6.1. Notations and denitions. In this section G ⊂ RN is a bound-ed domain of the class C0,1. Let x0 ∈ RN be a point and f a function denedon G 3 x0. f is Hölder continuous with exponent α ∈ (0, 1) at x0 if the quan-tity

[f ]α;x0 = supx∈G

|f(x)− f(x0)||x− x0|α

is nite. [f ]α;x0 is said to be the α− Hölder coecient of f at x0 with respectto G.

f is uniformly Hölder continuous with exponent α ∈ (0, 1) in G if thequantity

[f ]α;G = supx,y∈Gx 6=y

|f(x)− f(y)||x− y|α

is nite.

• C l(G) : the Banach space of functions having all the derivatives of or-der at most l (if l=integer ≥ 0) and of order [l] (if l is non-integer) con-tinuous in G and whose [l]-th order partial derivatives are uniformlyHölder continuous with exponent l − [l] in G;|u|l;G is the norm of the element u ∈ C l(G); if l 6= [l] then

|u|l;G =[l]∑j=0

supG|Dju|+ sup

|α|=[l]supx,y∈Gx 6=y

|Dαu(x)−Dαu(y)||x− y|l−[l]

.

• C l0(G) : the set of functions in C l(G) with the compact support in G.• W k,p(G), 1 ≤ p <∞ : the Sobolev space equipped with the norm

‖u‖Wk,p(G) =

∫G

∑|β|≤k

|Dβu|pdx

1/p

,

• W k,p0 (G) is the closure of C∞0 (G) with respect to the norm ‖·‖Wk,p(G).

• W k,p(G \ O) = W k,p(G \Bε(0)), ∀ε > 0.• For p = 2 we use the notation

W k(G) ≡W k,2(G), W k0 (G) ≡W k,2

0 (G).

20 1 Preliminaries

Definition 1.26. Let us say that u ∈ W k,p(G) satisfy u ≤ 0 on ∂G inthe sense of traces, if its positive part u+ = maxu, 0 ∈ W k,p

0 (G). If u iscontinuous in a neighborhood of ∂G, then u satises u ≤ 0 on ∂G, if theinequality holds in the classical pointwise sense. Other denitions of theinequality at ∂G follow naturally. For example: u ≥ 0 on ∂G, if −u ≤ 0 on∂G; u = v on ∂G, if both u− v ≤ 0 and u− v ≥ 0 on ∂G;

sup∂G

u = infk|u ≤ k on ∂G, k ∈ R; inf∂G

u = − sup∂G

(−u).

• For Γ ⊆ ∂G and k ∈ 1, 2, . . . , the space W k− 1p,p(Γ) consists of traces

on Γ of functions from W k,p(G) and is equipped with the norm

‖ϕ‖Wk− 1

p ,p(Γ)= inf ‖Φ‖Wk,p(G),

where the inmum is taken over the set of all functions Φ ∈W k,p(G)such that Φ = ϕ on Γ in the sense of traces.For p = 2 we use the notation

W k−1/2(Γ) ≡W k− 12,2(Γ).

Theorem 1.27. [128, Theorem 7.28] (Interpolation inequality) LetG be a C1,1 domain in RN and u ∈W 2,p(G) with p ≥ 1. Then for all ε > 0

‖∇u‖Lp(G) ≤ ε‖u‖W 2,p(G) + cε−1‖u‖Lp(G)

with a constant c depending only on the domain G.

Theorem 1.28. [116, Section 4.3] (Trace Theorem) Let 1 ≤ p < ∞.There exists a bounded linear operator

T : W 1,p(G)→ Lp(∂G)

such that

Tu = u on ∂G

for all u ∈W 1,p(G) ∩ C0(G).

Henceforth, we will write simply u instead of Tu.

Theorem 1.29. (see e.g. (6.23), (6.24) Chapter I [212] or Lemma 6.36[234]). Let ∂G be piecewise smooth and u ∈ W 1,1(G). Then there is aconstant c > 0 which depends only on G such that∫

Γ

|u|ds ≤ c∫G

(|u|+ |∇u|) dx ∀Γ ⊆ ∂Ω(1.6.1)

If u ∈W 1,2(G), then∫∂G

v2ds ≤∫G

(δ|∇v|2 + cδv2)dx, ∀v(x) ∈W 1,2(G),∀δ > 0.(1.6.2)

1.6 Hölder and Sobolev spaces 21

If u ∈W 2,2(G), then∫∂G

(∂u

∂n

)2

ds ≤ c∫G

(2|∇u||D2u|+ |∇u|2

)dx.(1.6.3)

1.6.2. Sobolev imbedding theorems. We give the well known Sobo-lev inequalities and Kondrashov compactness results so-called the imbeddingtheorems (see [360], 1.4.5 - 1.4.6 [258], 7.7[128]).

Theorem 1.30. [409, Theorem 2.4.1], [128, Theorem 7.10] (Sobolevinequalities) Let G be a bounded open domain in RN and p > 1. Then

W 1,p0 (G) →

L

NpN−p (G) for p < N,

C0(G) for p > N.(1.6.4)

Furthermore, there exists a constant c = c(N, p) such that for all u ∈W 1,p

0 (G) we have

‖u‖LNp/(N−p)(G) ≤ c‖∇u‖Lp(G)(1.6.5)

for p < N and

supG|u| ≤ c(measG)1/N−1/p‖∇u‖Lp(G)(1.6.6)

for p > N.

The following Imbedding Theorems 1.31-1.34 was proved at rst bySobolev [360] and can be found with complete proofs in [311], [1, Theo-rem 5.4], [208, Sections 5.7,5.8] and [258, Section 1.4]. Let G be a C0,1

bounded domain in RN

Theorem 1.31. Let k ∈ N and p ∈ R with p ≥ 1, kp < N. Then theimbedding

W k,p(G) → Lq(G)(1.6.7)

is continuous for 1 ≤ q ≤ Np/(N − kp) and compact for 1 ≤ q < Np/(N −kp). If kp = N, then the imbedding (1.6.7) is continuous and compact forany q ≥ 1.

Theorem 1.32. Let k ∈ N0, m ∈ N and let p, q ∈ R with p, q ≥ 1. Ifkp < N, then the imbedding

Wm+k,p(G) →Wm,q(G)(1.6.8)

is continuous for any q ∈ R satisfying 1 ≤ q ≤ Np/(N − kp). If k = Np,then the imbedding (1.6.8) is continuous for any q ≥ 1.

Theorem 1.33. Let k,m ∈ N0 and p > 1. Then the imbedding

W k,p(G) → Cm+β(G)

is continuous if

(k −m− 1)p < N < (k −m)p and 0 < β ≤ k −m−N/p(1.6.9)

22 1 Preliminaries

and compact if the inequality in (1.6.9) is sharp. If (k −m− 1)p = N, thenthe imbedding is continuous for any β ∈ (0, 1).

Theorem 1.34. Let u ∈ W k,p(G) with k ∈ N, p ∈ R, kp > N andp > 1. Then u ∈ Cm(G) for 0 ≤ m < k−N/p and there exists a constant c,independent of u, such that

supx∈G|Dαu(x)| ≤ c‖u‖Wk,p(G)

for all |α| < k −N/p.

Theorem 1.35. Let G be lipschitzian domain and Ts ⊂ G be piecewiseCk−smooth s−dimensional manifold. Let k ≥ 1, p > 1, kp < N, N − kp <s ≤ N, 1 ≤ q ≤ q∗ = sp/(N − kp). Then the imbedding W k,p(G) → Lq(Ts)and the inequality

‖u‖Lq(Ts) ≤ c‖u‖Wk,p(G)(1.6.10)

hold. If q < q∗, then this imbedding is compact.

1.7. Weighted Sobolev spaces

Definition 1.36. For k ∈ N0, 1 < p < ∞ and α ∈ R we dene theweighted Sobolev space V k

p,α(G) as the closure of C∞0 (G \ 0) with respect tothe norm

‖u‖V kp,α(G) =

∫G

∑|β|≤k

rα+p(|β|−k)∣∣∣Dβu

∣∣∣p dx1/p

.

For Γ ⊆ ∂G and k ∈ 1, 2, . . . , the space V k−1/pp,α (Γ) consists of traces on Γ of

functions from V kp,α(G) and is equipped with the norm

‖u‖Vk−1/pp,α (Γ)

= inf ‖v‖V kp,α(G),

where the inmum is taken over the set of all functions v ∈ V kp,α(G) such

that v = u on Γ.For p = 2 we use the notations

Wkα(G) = V k

2,α(G), Wk−1/2α (Γ) = V

k−1/22,α (Γ).

Lemma 1.37. [159, 196] Let k′, k ∈ N with k′ ≤ k and

α− pk ≤ α′ − pk′.

Then V kp,α(G) is continuously imbedded in V k′

p,α′(G). Furthermore, the imbed-ding is compact if k′ < k and α− pk < α′ − pk′.

Lemma 1.38. [277, 319] Let (k−|γ|)p > N, then for every u ∈ V kp,α(G)

the following inequality is valid

|Dγu(x)| ≤ c|x|k−|γ|−(α+N)/p‖u‖V kp,α(G) ∀x ∈ Gd0

1.7 Weighted Sobolev spaces 23

with a constant c independent of u and some d > 0 depending only on G. Inparticular

V kp,α(G) → Cm(G)

for m < k − (α+N)/p.

Proof. Without loss of generality we can assume that G is a cone. Weintroduce new variables y = (y1, . . . , yN ) by x = yt with t > 0 and setv(y) := u(x). By the Sobolev Imbedding Theorem 1.34, we have

|Dγyv(y)| ≤ c

∑|δ|≤k

‖Dδyv‖Lp(G2

1) ∀y ∈ G21.

Returning back to the variables x

t|γ||Dγxu(x)| ≤ c

∑|δ|≤k

‖t|δ|−N/pDδxu‖Lp(G2t

t ) ∀x ∈ G2tt .

Multiplying both sides of this inequality by tN/p−k+α/p we obtain

t|γ|−k+(α+N)/p|Dγxu(x)| ≤ c

∑|δ|≤k

‖t|δ|−k+α/pDδxu‖Lp(G2t

t ) ∀x ∈ G2tt .

Therefore, because of t ≤ |x| ≤ 2t in G2tt , we have

|x||γ|−k+(α+N)/p|Dγxu(x)| ≤ c

∑|δ|≤k

‖|x||δ|−k+α/pDδxu‖Lp(G2t

t ) ∀x ∈ G2tt

with a constant c independent of t. Thus the assertion holds.

Lemma 1.39. Let k,m ∈ N0, β ∈ R with

(k −m− 1)p < N < (k −m)p and 0 < β ≤ k −m−N/p.Then for any u ∈ V k

p,α(G)∑|γ|=m

supx,y∈G,x 6=y

|Dγu(x)−Dγu(y)||x− y|β

≤ c|x|k−m−β−(α+N)/p‖u‖V kp,α(G) ∀x ∈ Gd0

with a constant c and some d > 0.

Proof. The proof is completely analogous to the proof of Lemma 1.38.

Lemma 1.40. [159, Lemma 1.1] Let u ∈ Wk−1/2α (Γd0). Then∫

Γd0

rα−2k+1u2(x)ds ≤ c‖u‖2Wk−1/2

α (Γd0).

Lemma 1.41. Let d > 0 and ρ ∈ (0, d). Then the inequality∫Γρ0

r3−N(∂u

∂n

)2

ds ≤ c1

∫Γρ0

r1−Nu2ds+ c2‖u‖2W

2

4−N (Gρ0)

is valid for all u ∈ W24−N(Gρ0) with constants c1, c2 independent of u.

24 1 Preliminaries

Proof. Let us rst recall that due to Theorem 1.29 we have∫Γρ0

(∂v

∂n

)2

ds ≤ c3

∫Gρ0

(r|D2v|2 +

1r|∇v|2

)dx

with a constant c3 > 0 depending only on Gd0. Setting v = r(3−N)/2u we have

∂v

∂n

∣∣∣∣Γρ0\O

= r(3−N)/2 ∂u

∂n

∣∣∣∣Γρ0\O

+3−N

2r(1−N)/2u ·

N∑i=1

xini

r.

Since

NPi=1

xini

r ≤ 1, therefore∫Γρ0

r3−N(∂u

∂n

)2

ds ≤ 2∫Γρ0

(∂v

∂n

)2

+(3−N)2

4r1−Nu2

⟨n,x

r

⟩2

ds

≤ c5

∫Gρ0

(r|D2v|2 + r−1|∇v|2

)ds+

(3−N)2

2

∫Γρ0

r1−Nu2ds.

The assertion then follows by Lemma 1.9.

1.8. Spaces of Dini continuous functions

Definition 1.42. The function A is called Dini continuous at zero ifthe integral

d∫0

A(t)tdt

is nite for some d > 0.

Definition 1.43. The function A is called an αfunction, 0 < α < 1,on (0, d], if t−αA(t) is monotonously decreasing on (0, d], i.e.

A(t) ≤ tατ−αA(τ), 0 < τ ≤ t ≤ d.(1.8.1)

In particular, setting t = cτ, c > 1, we have

A(cτ) ≤ cαA(τ), 0 < τ ≤ c−1d.(1.8.2)

If an αfunction A is Dini continuous at zero then we say that A is anαDini function. In that case we dene the function

B(t) =

t∫0

A(τ)τ

dτ.

1.8 Spaces of Dini continuous functions 25

Obviously, the function B is monotonously increasing and continuous on[0, d] and B(0) = 0.Integrating (1.8.1) over τ ∈ (0, t) we obtain

A(t) ≤ αB(t).(1.8.3)

Similarly, we derive from (1.8.1) the inequality

d∫δ

A(t)t2

dt =

d∫δ

tα−2A(t)tα

dt ≤ δ−αA(δ)

d∫δ

tα−2dt ≤ (1− α)−1A(δ)δ

and thus

(1.8.4) δ

d∫δ

A(t)t2

dt ≤ (1− α)−1A(δ) ≤ α(1− α)−1B(δ)

∀α ∈ (0, 1), 0 < δ < d.

Definition 1.44. The function B is called equivalent to A, writtenA ∼ B, if there exist positive constants C1 and C2 such that

C1A(t) ≤ B(t) ≤ C2A(t) ∀t ≥ 0.

Theorem 1.45. [113] A ∼ B if and only if

lim inft→0

A(2t)/A(t) > 1.(1.8.5)

Proof. At rst we remark that

2B(h) ≥ B(2h) =∫ 2h

0

A(t)tdt ≥

∫ 2h

h

A(t)tdt ≥ A(h) ln 2.

Therefore we must prove the equivalence of (1.8.5) to the inequality B(t) ≤CA(t).

The suciency: Let (1.8.5) be satised. Then there is a positive θ suchthat for suciently small t the inequality A(2t)

A(t) ≥ 1 + θ holds and therefore

A(2−kt) ≤ (1 + θ)−kA(t). Then

B(h) =h∫0

A(t)t dt =

∞∑k=0

2−kh∫2−k−1h

A(t)t dt ≤ ln 2

∞∑k=0

A(2−kh) ≤

≤ ln 2∞∑k=0

(1 + θ)−kA(h).

26 1 Preliminaries

The necessity: Let lim inft→+0

A(2t)/A(t) = 1. Then there is a sequence tn

such that A(2tn)A(tn) ≤ 1 + 1

n and we have

A(ntn)A(tn)

=A(ntn)

A((n− 1)tn)· A((n− 1)tn)A((n− 2)tn)

· · · A(2tn)A(tn)

≤

≤[A(2tn)A(tn)

]n−1

≤(

1 +1n

)n−1

≤ e.

Therefore

B(ntn) ≥ntn∫tn

A(t)tdt ≥ lnnA(tn) ≥ 1

elnnA(ntn),

andB(ntn)A(ntn)

≥ 1e

lnn, limn→∞

ntn = 0.

Thus A(t) and B(t) are not equivalent.

In some cases we shall consider functions A(t) such that

sup0<τ≤1

A(τt)A(τ)

≤ cA(t), t ∈ (0, d],(1.8.6)

with some constant c independent of t.Examples of α-Dini functions A(t) which satisfy (1.8.5), (1.8.6) with

c = 1 are:

tα, 0 ≤ t <∞;

tαln(1/t), t ∈ (0, d], d = min(e−1, e−1/α), e−1 < α < 1.

Definition 1.46. The Banach space C0,A(G) is the set of all boundedand continuous functions u on G ⊂ RN for which

[u]A;G = supx,y∈G,x 6=y

|u(x)− u(y)|A(|x− y|)

<∞.

It is equipped with the norm

‖u‖C0,A(G) = ‖u‖C0(G) + [u]A;G.

If k ≥ 1, then we denote by Ck,A(G) the subspace of Ck(G) consisting of allfunctions whose (k − 1)th order partial derivatives are uniformly Lipschitzcontinuous and each kth order derivative belongs to C0,A(G). It is a Banachspace equipped with the norm

‖u‖Ck,A(G) = ‖u‖Ck(G) +∑|β|=k

[Dβu]A;G.

1.8 Spaces of Dini continuous functions 27

Furthermore, let us introduce the following notation:

[u]A,x = supy∈G\x

|u(x)− u(y)|A(|x− y|)

.

Lemma 1.47. If A ∼ B, then [u]A ∼ [u]B.

Lemma 1.48. [128, p. 143, 6.7 (ii)] Let G be a bounded domain witha Lipschitz boundary ∂G. Then there are two positive constants L, %1 suchthat for any y ∈ G with dist(y, ∂G) ≤ %1 and any 0 < % ≤ %1 there existsx ∈ B%(y) such that B%/L(x) ⊂ G.

Theorem 1.49. [362, Inequality (10.1)] (Interpolation inequality)Let ∂G be Lipschitz. Then for any ε > 0 there exists a constant c = c(ε,G)such that for every u ∈ C1,A(G) the following inequality holds

N∑i=1

‖Diu‖C0(G) ≤ εN∑i=1

[Diu]A;G + c(ε,G)‖u‖C0(G).

Proof. Let L, % be given as in Lemma 1.48 and let ε > 0 be arbitrary.We choose % > 0 so small, that A(%(1 + 1/L)) ≤ ε. If dist(y, ∂G) > %1, thereare for every i ∈ 1, . . . N two points y1, y2 ∈ ∂B%(y) and y ∈ B%(y), suchthat

|Diu(y)| = 12%|u(y1)− u(y2)| ≤ 1

%‖u‖C0(G).

Thus

|Diu(y)| ≤ |Diu(y)|+ |Diu(y)−Diu(y)| ≤ 1%‖u‖C0(G) +A(%)[Diu]A;G.

If dist(y, ∂G) ≤ %1, there are y1, y2 ∈ ∂B%/L(x), y ∈ ∂B%/L(x) such that

|Diu(y)| = L

2%|u(y1)− u(y2)| ≤ L

%‖u‖C0(G).

Since |y − y| ≤ |y − x|+ |x− y| ≤ %(1 + 1/L) we conclude

|Diu(y)| ≤ |Diu(y)|+ |Diu(y)−Diu(y)| ≤ L

%‖u‖C0(G) +

+A(%(1 +1L

))[Diu]A;G,

which nally implies the statement.

Definition 1.50. We shall say that the boundary portion T ⊂ ∂G is ofclass C1,A if for each point x0 ∈ T there are a ball B = B(x0), a one-to-onemapping ψ of B onto a ball B′ and a constant K > 0 such that:

(i) B ∩ ∂G ⊂ T, ψ(B ∩G) ⊂ RN+ ;(ii) ψ(B ∩ ∂G) ⊂ Σ;(iii) ψ ∈ C1,A(B), ψ−1 ∈ C1,A(B′);(iv) ‖ψ‖C1,A(B) ≤ K, ‖ψ−1‖C1,A(B′) ≤ K.

28 1 Preliminaries

It is not dicult to see that for the dieomorphism ψ one has

K−1|ψ(x)− ψ(x′)| ≤ |x− x′| ≤ K|ψ(x)− ψ(x′)| ∀x, x′ ∈ B.(1.8.7)

Lemma 1.51. [362, Section 7] Let u, v ∈ C0,A(G). Then u · v ∈ C0,A(G)and

‖u · v‖C0,A(G) ≤ ‖u‖C0,A(G) · ‖v‖C0,A(G).

Lemma 1.52. [362, Section 7] Let A(t) be α-function on [0, d] and u ∈C0,A(B). Furthermore, let ψ : B′ → B be Lipschitz continuous with theLipschitz constant L. Then u ψ ∈ C0,A(B′) and

‖u ψ‖C0,A(B′) ≤ Lα‖u‖C0,A(B), L = max(1, L).(1.8.8)

Proof. Indeed, if x, y ∈ B′, |x− y| ≤ dL , then by (1.8.2)

|u(ψ(x))− u(ψ(y))| ≤ ‖u‖C0,A(B) · A(L|x− y|) ≤ ‖u‖C0,A(B) · Lα · A(|x− y|).

1.9. Some functional analysis

Definition 1.53. Let X,Y be Banach spaces. Then we denote byL(X,Y ) the linear space of all continuous linear mappings L : X → Y.

Theorem 1.54. [128, Theorem 5.2] (The method of continuity).Let X,Y be Banach spaces and L0, L1 ∈ L(X,Y ). Furthermore, let

Lt := (1− t)L0 + tL1 ∀t ∈ [0, 1]

and suppose that there exist a constant c such that

‖u‖X ≤ c‖Ltu‖Y ∀t ∈ [0, 1].

Then L1 maps X onto Y if and only if L0 maps X onto Y.

Theorem 1.55. [350] (Variational principle for the least positiveeigenvalue). Let H,V be Hilbert spaces with dense and compact imbeddings

V ⊂ H ⊂ V ′

and let A : V → V ′ be a continuous operator. We assume that the bilinearform

a(u, v) = (Au, v)His continuous and Vcoercive, i.e. there are constants c1 and c2 such that

|a(u, v)| ≤ c1‖u‖V ‖v‖V ,a(u, u) ≥ c2‖u‖2V

for all u, v ∈ V. Then the smallest eigenvalue ϑ of the eigenvalue problem

Au+ ϑu = 0

satises

ϑ = infv∈V

a(v, v)‖v‖2H

1.10 The Cauchy problem for a dierential inequality 29

Theorem 1.56. [128, Theorem 11.3] (The LeraySchauder Theo-rem). Let T be a compact mapping of a Banach space B into itself, andsuppose that there exist a constant M such that

‖x‖B < M

for all x ∈ B and σ ∈ [0, 1] satisfying x = σTx. Then T has a xed point.

1.10. The Cauchy problem for a dierential inequality

Theorem 1.57. Let V (%) be monotonically increasing, nonnegative dif-ferentiable function dened on [0, 2d] and satisfy the problem

V ′(ρ)− P(%)V (%) +N (ρ)V (2ρ) +Q(ρ) ≥ 0, 0 < ρ < d,

V (d) ≤ V0,(CP)

where P(%),N (%),Q(%) are nonnegative continuous functions dened on [0, 2d]and V0 is a constant. Then

(1.10.1) V (%) ≤ exp( d∫%

B(τ)dτ)

V0 exp(−

d∫%

P(τ)dτ)

+

+

d∫%

Q(τ) exp(−

τ∫%

P(σ)dσ)dτ

with

B(%) = N (%) exp( 2%∫%

P(σ)dσ).(1.10.2)

Proof. We dene functions

w(%) = V (%) exp( d∫%

P(σ)dσ)

(1.10.3)

R(%) = V0 +

d∫%

Q(τ) exp( d∫τ

P(σ)dσ)dτ(1.10.4)

30 1 Preliminaries

Multiplying the dierential inequality (CP ) by the integrating factor

exp(d∫%P(s)ds

)and integrating from ρ to d we get

V (d)− V (%) exp( d∫%

P(s)ds)

+

d∫%

N (τ) exp( d∫τ

P(s)ds)V (2τ)dτ +

+

d∫%

Q(τ) exp( d∫τ

P(s)ds)dτ ≥ 0.

Hence it follows that

w(%) ≤ R(%) +

d∫%

B(τ)w(2τ)dτ.(1.10.5)

Now we have

w(%)R(%)

≤ 1 +

d∫%

B(τ)w(2τ)R(2τ)

R(2τ)R(%)

dτ.(1.10.6)

Since R(2τ) ≤ R(%) for τ > %, then setting

z(%) =w(%)R(%)

(1.10.7)

we get

z(%) ≤ 1 +

d∫%

B(τ)z(2τ)dτ.(1.10.8)

Let us dene the function

Z(%) = 1 +

d∫%

B(τ)z(2τ)dτ.

The from (1.10.8) we have

z(%) ≤ Z(%)(1.10.9)

and

Z ′(%) = −B(%)z(2%) ≥ −B(%)Z(2%).

1.10 The Cauchy problem for a dierential inequality 31

Multiplying obtained dierential inequality by the integrating factor

exp(−

d∫%B(s)ds

)and using the equality

d

d%

[Z(%) exp

(−

d∫%

B(s)ds)]

= Z ′(%) exp(−

d∫%

B(s)ds)

+

+ B(%) exp(−

d∫%

B(s)ds)Z(%),

we have

d

d%

[Z(%) exp

(−

d∫%

B(s)ds)]≥ B(%) exp

(−

d∫%

B(s)ds)[Z(%)−Z(2%)

].

But

Z(2%) = 1 +

d∫2%

B(s)z(2s)ds ≤ 1 +

d∫%

B(s)z(2s)ds = Z(%).

Therefore

Z(%)−Z(2%) ≥ 0 ⇒ d

d%

[Z(%) exp

(−

d∫%

B(s)ds)]≥ 0.

Integrating from % to d hence we have

Z(%) exp(−

d∫%

B(s)ds)≤ Z(d) = 1 ⇒ Z(%) ≤ exp

( d∫%

B(s)ds)

Hence, by (1.10.9), we get

z(%) ≤ exp( d∫%

B(s)ds).(1.10.10)

Now, in virtue of (1.10.3), (1.10.7) and (1.10.10), nally we obtain

V (%) ≤ exp(−

d∫%

P(σ)dσ)R(%) exp

( d∫%

B(σ)dσ)

or with regard to (1.10.4) the desired estimate (1.10.1).

32 1 Preliminaries

1.11. Additional auxiliary results

1.11.1. Mean Value Theorem.

Theorem 1.58. Let f ∈ C0[a, b] with 0 ≤ a < b. Then there exist θ ∈(0, 1) and ξ ∈ (0, 1), such that

b∫a

f(x)dx ≥ (b− a)f((1− θ)a+ θb),

b∫a

f(x)dx ≤ (b− a)f((1− ξ)a+ ξb).

Proof. Let us assume thatb∫a

f(x)dx < (b− a)f((1− θ)a+ θb)

for all θ ∈ (0, 1). Integrating this inequality over θ ∈ (0, 1) we obtain thecontradiction

b∫a

f(x)dx < (b− a)

1∫0

f((1− θ)a+ θb)dθ =

b∫a

f(t)dt.

The other assertion is proved analogously.

1.11.2. Stampacchia's Lemma.

Lemma 1.59. (See Lemma 3.11 of [313], [363]). Let ϕ : [k0,∞)→ R bea nonnegative and nonincreasing function which satises

ϕ(h) ≤ C

(h− k)α[ϕ(k)]β for h > k > k0,(1.11.1)

where C,α, β are positive constants with β > 1. Then

ϕ(k0 + d) = 0,

where

dα = C |ϕ(k0)|β−1 2αβ/(β−1).

Proof. Dene the sequence

ks = k0 + d− d

2s, s = 1, 2, . . . .

From (1.11.1) follows that

ϕ(ks+1) ≤ C2(s+1)α

dα[ϕ(ks)]β , s = 1, 2, . . . .(1.11.2)

1.11 Additional auxiliary results 33

Let us prove by induction that

ϕ(ks) ≤ϕ(k0)2−sµ

, where µ =α

1− β< 0.(1.11.3)

For s = 0 the claim is trivial. Let us suppose that (1.11.3) is valid up to s.By (1.11.2) and the denition of dα if follows that

ϕ(ks+1) ≤ C 2(s+1)α

dα[ϕ(k0]β

2−sβµ≤ ϕ(k0)

2−(s+1)µ.(1.11.4)

Since the right hand side of (1.11.4) tends to zero as s→∞, we obtain

0 ≤ ϕ(k0 + d) ≤ ϕ(ks)→ 0.

1.11.3. Other assertions.

Lemma 1.60. (see Lemma 2.1 [78]). Let us consider the function

η(x) =

ex − 1, x ≥ 0−e−x + 1, x ≤ 0,

where κ > 0. Let a, b be positive constants, m > 1. If κ > (2b/a) +m, thenwe have

aη′(x)− b|η(x)| ≥ a

2ex ∀x ≥ 0;(1.11.5)

η(x) ≥[η( xm

)]m∀x ≥ 0.(1.11.6)

Moreover, there exist d ≥ 0 and M > 0 such that

η(x) ≤M[η( xm

)]m, η′(x) ≤M

[η( xm

)]m∀x ≥ d(1.11.7)

|η(x)| ≥ x ∀x ∈ R.(1.11.8)

Proof. The (1.11.5) is easy to show by direct calculation. By the de-nition, the (1.11.6) has the form(

exm − 1

)m≤ ex − 1, ∀x ≥ 0.(1.11.9)

We set for x ≥ 0 :

y = exm ≥ 1, f(y) = (y − 1)m + 1− ym.

Then

f ′(y) = m(y − 1)m−1 −mym−1 < 0,

hence it follows that f(y) is decreasing function, i.e. f(y) ≤ f(1), ∀y ≥ 1.Because of f(1) = 0, we get (1.11.9).

Further, the rst inequality from (1.11.7) has the form

ym − 1 ≤M(y − 1)m.(1.11.10)

34 1 Preliminaries

We consider the function g(y) = M(y− 1)m − ym + 1 and show that g(y) ≥0, ∀y ≥ y0 > 1. In fact, we have

g′(y) = Mm(y − 1)m−1 −mym−1 =⇒ g′(y0) = 0 for y0 =M

1m−1

M1

m−1 − 1> 1,

if we choose M > 1. In addition,

g′′(y) = Mm(m− 1)(y − 1)m−2 −m(m− 1)ym−2 =⇒

g′′(y0) = m(m− 1)ym−20

(M

1m−1 − 1

)> 0 =⇒

min g(y) = g(y0) = M(y0 − 1)m − ym0 + 1 = 1− M(M

1m−1 − 1

)m−1 ≥ 0,

since M > 1. Therefore

g(y) ≥ g(y0) > 0, ∀y ≥ y0 or for exm ≥ M

1m−1

M1

m−1 − 1=⇒

x ≥ d1 =m

κ

lnM

1m−1

M1

m−1 − 1,

i.e. the rst inequality from (1.11.7) is proved. Let us now prove the secondinequality from (1.11.7). We rewrite it in the form

M(y − 1)m ≥ κym.

Hence it follows:

M1m (y − 1) ≥ κ

1m (y − 1)y =⇒ y ≥ M

1m

M1m − κ

1m

,

if M > κ. The last inequality means that

exm ≥ M

1m

M1m − κ

1m

=⇒ x ≥ d2 =m

κ

lnM

1m

M1m − κ

1m

.

Thus, the (1.11.7) is proved, if we take M > κ; d = max(d1, d2).Finally, we prove the (1.11.8). From the denition we have

|η(x)| =

ex − 1, x ≥ 0e−x − 1, x ≤ 0.

It is sucient prove the inequality

ex − 1 ≥ x, x ≥ 0.

But it is obvious, because of the Taylor formula, since κ > 1.


1.11.4. The distance function. Let G be a domain in RN havingnon-empty boundary ∂G. The distance function d is dened by

d(x) = dist(x, ∂G).

Lemma 1.61. The distance function d is uniformly Lipschitz continuous.

Proof. For let x, y ∈ RN and let y∗ ∈ ∂G be such that |y − y∗| = d(y).Then

d(x) ≤ |x− y∗| ≤ |x− y|+ d(y)so that by interchanging x and y we have

|d(x)− d(y)| ≤ |x− y|.(1.11.11)

1.11.5. Extension Lemma.

Lemma 1.62. (See Lemma 3.9 [402]). Let D be a convex bounded set inRN , T ⊆ ∂D and f(x) ∈ C1,A(D), where A(t) is a non-decreasing function,

limt→+0

A(t) = 0, satisfying A(2t) ≤ 2A(t). Then there exists a function F (x)

with following properties:

1. F (x) ∈ C∞(D);

2. F (x) ∈ C1,A(D)

3. DαF (x) = Dαf(x), x ∈ T ; |α| ≤ 1;

4. |D2xxF (x)| ≤ Kd−1(x)A(d(x)),

where d(x) denotes the distance to T and K depends on N and A(t) only.

Proof. We shall use the concept of a partition of unity. Let us considera nite covering of D by a countable collection Dj of open sets Dj . Letζj be a locally nite partition of unity subordinate to this covering, i.e.

(i) ζk ∈ C∞0 (Dj) for some j = j(k);(ii) ζk ≥ 0,

∑ζk = 1 in D;

(iii) at each point of D there is a neighborhood in which only a nitenumber of the ζk are non-zero;

(iv)∑k

|Dαζk(x)| ≤ Cα(1 + d−α(x)), where Cα is independent of T ;

(ivv) there is a constant C independent of k and T such that

diam(supp ζk) ≤ Cd(x).

The proof of the existence of such a partition see e.g. the presentation ofWhitney's extension theorem in Hörmander (Lemma 3 [144]). Let x∗ ∈ Tbe a point satisfying d(x) = |x−x∗| and xk is any xed point in the supportof ζk. We write the Taylor expansion of f(x) at y

f(x) = P1(x, y) +R1(x, y), where

P1(x, y) = f(y) +∑j

(xj − yj)∂f

∂xj(y)

36 1 Preliminaries

and therefore, by mean value Lagrange's Theorem,

R1(x, y) = (f(x)− f(y))−∑j

(xj − yj)∂f

∂xj(y) =

=∑j

(xj − yj)(∂f

∂xj(y + θ(x− y))− ∂f

∂xj(y))

with some θ ∈ (0, 1). By assumptions about f, hence it follows

|R1(x, y)| ≤ |x− y|A(θ|x− y|) ≤ |x− y|A(|x− y|).(1.11.12)

SinceDxR1(x, y) = Dxf(x)−Dxf(y),

in the same way we get

|DxR1(x, y)| = |Dxf(x)−Dxf(y)| ≤ KA(|x− y|).(1.11.13)

Now we dene F (x) by

F (x) =

∑k

ζk(x)P1(x, xk) = P1(x, x∗) +∑k

ζk(x)(P1(x, xk)− P1(x, x∗)

),

x ∈ D \ T,f(x), x ∈ T.

Then we have

D2F (x) =∑k

D2ζk(x)(P1(x, xk)− P1(x, x∗)

)+

+ 2∑k

Dζk(x)(DxP1(x, xk)−DxP1(x, x∗)

).

But it is obviously

P1(x, xk)− P1(x, x∗) = R1(x, x∗)−R1(x, xk)

and therefore

D2F (x) =∑k

D2ζk(x)(R1(x, x∗)−R1(x, xk))

)+

+ 2∑k

Dζk(x)(DxR1(x, x∗)−DxR1(x, xk)

).

(1.11.14)

If x ∈ supp ζk, then by (ivv)

|x− xk| ≤ |x− x∗|+ |x∗ − xk| ≤ d(x) + Cd(x) = (1 + C)d(x).(1.11.15)

Therefore we obtain, by (iv) and by (1.11.12) - (1.11.13)

|D2F (x)| ≤ Kd−1(x)A(d(x))

and 4 is proved.


To prove 2, rst assume that

|x− y| ≤ 12d(x).(1.11.16)

By mean value Lagrange's Theorem

|DxF (x)−DxF (y)| ≤ |x− y| sup |D2xxF (z)|,

where the supremum is taken over those z for which |z − x| ≤ 12d(x). Then

using 4, it follows that

|DxF (x)−DxF (y)| ≤ KA(|x− y|).

On the other hand, if d(x) < 2|x− y|, we have

d(y) ≤ d(x) + |x− y| < 4|x− y|

and by the denition of F (x) and by (1.11.15):

|DxF (x)−Dxf(x∗)| ≤∑k

|ζk(x)||DxR1(x, x∗)−DxR1(x, xk)|+

+∑k

|Dζk(x)||R1(x, x∗)−R1(x, xk)| ≤ KA(d(x)) ≤

≤ KA(|x− y|).

Similarly,

|DxF (y)−Dxf(y∗)| ≤ KA(d(y)) ≤ cKA(|x− y|).

Since by assumption

|Dxf(x∗)−Dxf(y∗)| ≤ A(|x∗ − y∗|) ≤ KA(|x− y|),

the Lemma follows with the triangle inequality.

1.11.6. Dierence quotients. The investigation of dierential proper-ties of weak solutions to the boundary value problems may often be deducedthrough a consideration of their dierence quotients.

Definition 1.63. Let u ∈ Lm(G) and denote by ek (k=1, . . . N) theunit coordinate vector in the xk direction. The function

4hu(x) = 4hku(x) =

u(x+ hek)− u(x)h

, h 6= 0

is said to be the dierence quotients of u at x in the direction ek.

The following lemmas pertain to dierence quotients of functions inSobolev spaces.

Lemma 1.64. Let u ∈W 1,m(G). Then 4hu ∈ Lm(G′) for any G′ ⊂⊂ Gsatisfying h < dist(G′, ∂G), and we have

‖4hu‖Lm(G′) ≤ ‖Dku‖Lm(G).

38 1 Preliminaries

Proof. At rst, we suppose that u ∈ C1(G) ∩W 1,m(G). Then

4hu(x) =u(x+ hek)− u(x)

h=

=1h

h∫0

Dku(x1, . . . , xk−1, xk + ξ, xk+1, . . . , xN )dξ

so that by the Hölder inequality

|4hu(x)|m ≤ 1h

h∫0

|Dku(x1, . . . , xk−1, xk + ξ, xk+1, . . . , xN )|mdξ,

and hence setting Bh(G′) = x| dist(x,G′) < h∫G′

|4hu(x)|mdx ≤ 1h

h∫0

∫Bh(G′)

|Dku|mdxdξ ≤∫G

|Dku|mdx.

The extension to arbitrary functions in W 1,m(G) follows by a straight-forward approximation argument.

Lemma 1.65. Let u ∈ Lm(G), 1 < m < ∞, and suppose there exists aconstant K such that 4hu ∈ Lm(G′) and ‖4hu‖Lm(G′) ≤ K for all h > 0and G′ ⊂⊂ G satisfying h < dist(G′, ∂G). Then the weak derivative Dkuexists and satises ‖Dku‖Lm(G) ≤ K.

Proof. By the weak compactness of bounded sets in Lm(G′), thereexists a sequence hj tending to zero and a function v ∈ Lm(G) with‖v‖Lm(G) ≤ K satisfying

limhj→0

∫G

η4hjudx =∫G

ηvdx, ∀η ∈ C10 (G).

Now we haveThe summation by parts formula:∫

G

η4hjudx = −∫G

u4−hjηdx for hj < dist(supp η, ∂G).(1.11.17)

Hence

limhj→0

∫G

u4−hjηdx =∫G

uDkηdx =⇒

∫G

ηvdx = −∫G

uDkηdx,

whence v = Dku.

1.12 Notes 39

Lemma 1.66. Let u ∈W 1,m(G). Then

‖4hjk u(x)−Dku(x)‖Lm(G′) → 0, k = 1, . . . , N

for any sequence hj tending to zero and ∀G′ ⊂⊂ G. For some subsequencehjl functions 4

hjlk u(x) tend to Dku(x) a.e. in G.

Proof. For suciently small |hj | and a.e. x ∈ G′ we have

4hjk u(x)−Dku(x) =

1hj

1∫0

du(x+ thjek)dt

dt−Dku(x) =

=

1∫0

〈Dku(x+ thjek)−Dku(x)〉 dt

and therefore

‖4hjk u(x)−Dku(x)‖Lm(G′) ≤

1∫0

‖Dku(x+ thjek)−Dku(x)‖Lm(G′)dt.

But the right side in this inequality tends to zero as hj → 0, because ofDku(x) is continuous in the norm Lm(G′). Thus the rst part of Lemma isproved. The second part follows from properties of the space Lm.

1.12. Notes

The proof of the Cauchy, Young and Hölder inequalities 1.2 can befound in Chapter 1 [37] or in Chapter II [141]. The formulae (1.3.1) -(1.3.12) are proved in 2, Chapter 1 [307]. The proof of the Fubini andFatou Theorems see e.g. Theorem 9 11 and Theorem 19 6, Chapter III[100]. The proof of the integral inequalities 1.5 can be found in ChapterVI [141]. The Clarkson inequality is proved in Subsection 2 3, Chapter I[360]. The material in 1.8 is due to [74, 113]. The simplest version ofTheorem 1.57 in 1.10 goes back to G. Peano [328]; the special case wasformulated and proved by T. Gronwall [136] and S. Chaplygin [79]. Thecase N (%) ≡ 0 of this Theorem was considered in [168, 169]. The generalcase belongs [53, 54, 50].

40 1 Preliminaries

CHAPTER 2

Integral inequalities

2.1. The classical Hardy inequalities

Theorem 2.1. (The Hardy inequality, see Theorem 330 [141].)Let p > 1, s 6= 1 and

F (x) =

x∫0

f(ξ)dξ, if s > 1;

∞∫xf(ξ)dξ, if s < 1;

then ∫ ∞0

x−sF p(x)dx ≤(

p

|s− 1|

)p ∫ ∞0

x−s(xf)pdx(2.1.1)

The constant is the best.

We prove the partial case p = 2.

Theorem 2.2. Let f ∈ L2(0, d), d, β > 0 and F (x) =x∫0

yβ−12 f(y)dy.

Then

d∫0

x−2β−1F 2(x)dx ≤ 1β2

d∫0

f2(x)dx.(2.1.2)

Proof. Let 0 < δ < β. Then by Hölder's inequality (1.14)

|F (x)|2 ≤

∣∣∣∣∣∣x∫

0

yδf(y)yβ−δ−12dy

∣∣∣∣∣∣2

≤

≤x∫

0

y2δf2(y)dy

x∫0

y2β−2δ−1dy =

=1

2(β − δ)x2(β−δ)

x∫0

y2δf2(y)dy.

41

42 2 Integral inequalities

Therefore, by the Fubini Theorem 1.13,d∫

0

x−2β−1F 2(x)dx ≤ 12(β − δ)

d∫0

x−2δ−1

x∫0

y2δf2(y)dy

dx =

=1

2(β − δ)

d∫0

y2δf2(y)

d∫y

x−2δ−1dx

dy =

=1

2(β − δ)

d∫0

y2δf2(y)y−2δ − d−2δ

2δdy ≤

≤ 14δ(β − δ)

d∫0

f2(x)dx.

Noting, that 14δ(β−δ) becomes minimal for δ = 1

2β, we choose δ := 12β and

obtain the assertion.

Corollary 2.3. Let v ∈W 1,2(0, d), d > 0 with v(0) = 0. Thend∫

0

rN−5+αv2(r)dr ≤ 4(4−N − α)2

d∫0

rN−3+α

(∂v

∂r

)2

dr(2.1.3)

for α < 4−N provided that the integral on the right hand side is nite.

Proof. We apply Hardy's inequality (2.1.2) with F = v, β := 4−N−α2 ,

noting that F ′(r) = rβ−12 f(r) and therefore f2(r) = r1−2β

(∂v∂r

)2.

Remark 2.4. The constant in (2.1.3) is the best possible.

Corollary 2.5. If u ∈ C∞0 (Rn), α < 4−N and u(0) = 0, then∫RN

rα−4u2(x)dx ≤ 4(4−N − α)2

∫RN

rα−2|∇u(x)|2dx

provided that the integral on the right hand side is nite.

Proof. The assertion follows by integrating both sides of (2.1.3) over Ωfor large enough d and applying (1.3.7).

Corollary 2.6. If u ∈W 1,20 (G), α < 4−N, then∫

G

rα−4u2(x)dx ≤ 4(4−N − α)2

∫G

rα−2|∇u(x)|2dx,(2.1.4)

provided that the integral on the right hand side of (2.1.4) is nite.

Proof. The claim follows from Corollary 2.5 because C∞0 (G) is densein W 1,2

0 (G).

2.1 The classical Hardy inequalities 43

Note also another generalization of the Hardy inequality:

Theorem 2.7. The inequality∞∫

0

xα−p|f(x)|pdx ≤ pp

|α+ 1− p|p

∞∫0

xα|f ′(x)|pdx(2.1.5)

is true if p > 1, α 6= p − 1 and f(x) is absolutely continuous on [0,∞) andsatises the following boundary condition

f(0) = 0 when α < p− 1,lim

x→+∞f(x) = 0 when α > p− 1.

Lemma 2.8. Let 0 < ε < d, β > 0 and f ∈ L2(ε, d), F (x) =x∫εyβ−

12 f(y)dy.

Thend∫ε

x−2β−1F 2(x)dx ≤(

1β2

+1

2β2

εβ

dβ

) d∫ε

f2(x)dx.(2.1.6)

Proof. Let 0 < δ < β. Then by Hölder's inequality

|F (x)|2 ≤

∣∣∣∣∣∣x∫ε

yδf(y)yβ−δ−12dy

∣∣∣∣∣∣2

≤x∫ε

y2δf2(y)dy

x∫ε

y2β−2δ−1dy =

=1

2(β − δ)

[x2(β−δ) − ε2(β−δ)

] x∫ε

y2δf2(y)dy.

Therefore, by the Fubini Theorem,d∫ε

x−2β−1F 2(x)dx ≤

≤ 12(β − δ)

d∫ε

[x−2δ−1 − ε2(β−δ)x−2β−1

] x∫ε

y2δf2(y)dy

dx =

=1

2(β − δ)

d∫ε

y2δf2(y)

d∫y

[x−2δ−1 − ε2(β−δ)x−2β−1

]dx

dy =

=1

2(β − δ)

d∫ε

y2δf2(y)(y−2δ − d−2δ

2δ+ ε2(β−δ)d

−2β − y−2β

2β

)dy ≤

≤ 14(β − δ)

d∫ε

[1δ

+ε2(β−δ)

β· y

2δ

d2β

]f2(y)dy.


Now we choose δ := 12β and obtain the assertion.

Corollary 2.9. Let v ∈W 1,2(ε, d), d > 0 with v(ε) = 0. Then

(2.1.7)

d∫ε

rn−5+αv2(r)dr ≤

≤ 4(4− n− α)2

[1 +

12

(εd

) 4−n−α2

] d∫ε

rn−3+α

(∂v

∂r

)2

dr

for α < 4− n.

Proof. We apply the inequality (2.1.6) with F = v, β := 4−n−α2 , noting

that F ′(r) = rβ−12 f(r) and therefore f2(r) = r1−2β

(∂v∂r

)2.

2.2. The Poincaré inequality

Theorem 2.10. The Poincaré inequality for the domain in RN

(see e.g. (7.45) [128]).Let u ∈W 1(G) and G is bounded convex domain in Rn. Then

‖u− u‖2;G ≤ c(N)(diamG)N

(measS)1−1/N‖∇u‖2;G,(PI 1)

where

u =1

measS

∫S

u(x)dx,

and S is any measurable subset of G.

Theorem 2.11. The Poincaré inequality for the domain on thesphere (see e.g. Theorem 3.21 [143]).

Let u ∈W 1(Ω) and Ω is convex domain on the unit sphere SN−1. Then

‖u− uΩ‖2;Ω ≤ c(N,Ω)‖∇ωu‖2;Ω,(PI 2)

where

uΩ =1

measΩ

∫Ω

u(x)dΩ.

Also the following lemma is true.

Lemma 2.12. (see e.g. Lemma 7.16 [128]). Let u ∈W 1,1(G) and Gis bounded convex domain in RN . Then

|u− u| ≤ (diamG)N

N ·measS

∫G

|x− y|1−N |∇u(y)|dy a.e. in G,(2.2.1)

where

u =1

measS

∫S

u(x)dx,

2.2 The Poincaré inequality 45

and S is any measurable subset of G.

Now we shall prove the version of the Poincaré inequality.

Theorem 2.13. Let Gd0 be convex domain, Gd0 ⊂ G, G is bounded do-main in RN . Let u ∈ L2(G) and

∫Gd0

rα−2|∇u|2dx <∞, α ≤ 4−N. Then

∫G%0

rα−4|u− u|2dx ≤ c∫G%0

rα−2|∇u|2dx, ∀% ∈ (0, d),(2.2.2)

where

u =1

measG%%/2

∫G%%/2

u(y)dy(2.2.3)

and c > 0 depend only on N, d,measΩ.

Proof. Since α ≤ 4−N then from our assumption we have u ∈W 1(G).By density of C∞(G) ∩W 1(G) in W 1(G) we can consider u ∈ C1(G). Weuse Lemma 2.12, applying it for the domains G%%/2 and S = G%%/2. By thisLemma and the Hölder inequality, we have

(2.2.4) |u(x)− u|2 ≤ c( ∫G%%/2

|x− y|1−N |∇u(y)|dy)2

≤

≤ c∫

G%%/2

|x− y|1−N |∇u(y)|2dy∫

G%%/2

|x− y|1−Ndy =

=c

2% ·measΩ

∫G%%/2

|x− y|1−N |∇u(y)|2dy.

From (2.2.4) it follows

(2.2.5)∫

G%%/2

rα−4|u(x)− u|2dx ≤

≤ c

2% ·measΩ

∫G%%/2

rα−4

( ∫G%%/2

|x− y|1−N |∇u(y)|2dy)dx ≤

≤ c ·measΩ∫

G%%/2

rα−3

( ∫G%%/2

|x− y|1−N |∇u(y)|2dy)dx =

= c

∫G%%/2

|∇u(y)|2( ∫G%%/2

|x|α−3|x− y|1−Ndx)dy ≤ c

∫G%%/2

rα−2|∇u|2dx,


since∫G%%/2

|x|α−3|x− y|1−Ndx ≤ c%α−3

∫G%%/2

|x− y|1−Ndx = %α−3 · c2% ·measΩ ≤

≤ c%α−2.

Replacing in (2.2.5) % by 2−k% we can (2.2.5) rewrite so∫G(k)

rα−4|u− u|2dx ≤ c∫G(k)

rα−2|∇u|2dx, ∀% ∈ (0, d),

whence by summing over all k = 0, 1, · · · we get the required (2.2.2).

2.3. The Wirtinger inequality: Dirichlet boundary condition

Let Ω ⊂ SN−1 be bounded domain with smooth boundary ∂Ω. Weconsider the problem of the eigenvalues for the Laplace-Beltrami operator4ω on the unit sphere:

4ωu+ ϑu = 0, ω ∈ Ω,

u∣∣∣∂Ω

= 0,(EV P1)

which consists of the determination of all values ϑ (eigenvalues) for which(EV P1) has a non-zero weak solutions (eigenfunctions). In the following,we denote by ϑ the smallest positive eigenvalue of this problem.

Theorem 2.14. (The Wirtinger inequality) The following inequality

is valid for all u ∈W 1,20 (Ω)∫

Ω

u2(ω)dΩ ≤ 1ϑ

∫Ω

|∇ωu|2 dΩ.(2.3.1)

Proof. Let us consider the eigenvalue problem (EV P1) and denote by

a(u, v) :=∫Ω

⟨∇ωu,∇ωv

⟩dΩ

the bilinear form corresponding to the LaplaceBeltrami operator ∆ω.From Theorem 1.55 applied to the spaces V = W 1,2

0 (Ω), H = L2(Ω) follows,that the smallest positive eigenvalue ϑ of (EV P1) satises

ϑ = infv∈W 1,2

0 (Ω)

a(v, v)‖v‖22,Ω

Thus for all u ∈W 1,20 (Ω)∫

Ω

|∇ωu|2 dω = a(u, u) ≥ ϑ‖v‖2L2(Ω).

2.4 The Wirtinger inequality: Robin boundary condition 47

Remark 2.15. From the above proof follows that the constant in (2.3.1)is the best possible.

Now let θ(r) be the least eigenvalue of the Beltrami operator 4ω on Ωr

with Dirichlet condition on ∂Ωr. According to the variational principle ofeigenvalues (see Theorem 1.55 ) we have also

Theorem 2.16.∫Ωr

u2(ω)dΩr ≤1θ(r)

∫Ωr

|∇ωu|2 dΩr, ∀u ∈W 1,20 (Ωr).(2.3.2)

2.4. The Wirtinger inequality: Robin boundary condition

2.4.1. The eigenvalue problem. Let Ω ⊂ Sn−1 be a bounded domainwith smooth boundary ∂Ω. Let −→ν be the exterior normal to ∂Ω. Letγ(x) ∈ C0(∂Ω), γ(x) ≥ γ0 > 0. We consider the eigenvalue problem for theLaplace-Beltrami operator 4ω on the unit sphere:

4ωu+ ϑu = 0, ω ∈ Ω,∂u∂−→ν + γ(x)u

∣∣∣∂Ω

= 0,(EV P3)

which consists of the determination of all values ϑ (eigenvalues) for which(EV P3) has a non-zero weak solutions (eigenfunctions).

Definition 2.17. Function u is called a weak solution of the problem(EV P3) provided that u ∈W 1(Ω) and satises the integral identity∫

Ω

1qi

∂u

∂ωi

∂η

∂ωi− ϑuη

dΩ +

∫∂Ω

γ(x)uηdσ = 0(II)

for all η(x) ∈W 1(Ω).

Remark 2.18. The eigenvalue problem (EV P3) was studied in SectionVI [86] and in 2.5 [360]. We observe that ϑ = 0 is not an eigenvalue of(EV P3). In fact, setting in (II) η = u and ϑ = 0 we have∫

Ω

|∇ωu|2dΩ +∫∂Ω

γ(x)|u|2dσ = 0 =⇒ u ≡ 0.

Now, let us introduce the functionals on W 1(Ω):

F [u] =∫Ω

|∇ωu|2dΩ +∫∂Ω

γ(x)u2dσ, G[u] =∫Ω

u2dΩ,

H[u] =∫Ω

⟨|∇ωu|2 − ϑu2

⟩dΩ +

∫∂Ω

γ(x)u2dσ


and the corresponding to them bilinear forms

F (u, η) =∫Ω

1qi

∂u

∂ωi

∂η

∂ωidΩ +

∫∂Ω

γ(x)uηdσ,

G(u, η) =∫Ω

uηdΩ.

We dene yet the set

K =u ∈W 1(Ω)

∣∣∣ G[u] = 1.

Since K ⊂W 1(Ω), F [u] is bounded from below for u ∈ K. The greatest loverbound of F [u] for this family we denote by ϑ :

infu∈K

F [u] = ϑ

We formulate the following statement:

Theorem 2.19. (See Theorem of Subsection 4 2.5, p. 123 [360]).Let Ω ⊂ Sn−1 be a bounded domain with smooth boundary ∂Ω. Let γ(x) ∈C0(∂Ω), γ(x) ≥ γ0 > 0. There exist ϑ > 0 and a function u ∈ K such that

F (u, η)− ϑG(u, η) = 0 for arbitrary η ∈W 1(Ω).

In particular F [u] = ϑ. In addition, on Ω, u has continuous derivativesof arbitrary order and satises the equation 4ωu + ϑu = 0, ω ∈ Ω and theboundary condition of (EV P3) in the weak sense (for details see the Remarks2.20 below).

Proof. Because of F [v] is bounded from below for v ∈ K, there isϑ = inf

v∈KF [v]. Consider a sequence vk ⊂ K such that lim

k→∞F [vk] = ϑ (such

sequence exists by the denition of inmum). From K ⊂ W 1(Ω) it followsthat vk is bounded in W 1(Ω) and therefore compact in L2(Ω). Choosing asubsequence, we can assume that it is converging in L2(Ω). Furthermore,

‖vk − vl‖2L2(Ω) = G[vk − vl] < ε(2.4.1)

as soon as k, l > N(ε). From the obvious equality

G

[vk + vl

2

]=

12G[vk] +

12G[vl]−G

[vk − vl

2

]we obtain, using G[vk] = G[vl] = 1 and G[vk−vl2 ] < ε

ϑ , that

G

[vk + vl

2

]> 1− ε

ϑ

for big k, l. The functionals F [v] and G[v] are homogeneous quadratic func-tionals and therefore their ratio F [v]

G[v] does not change under the passage from


v to cv (c = const 6= 0) and hence

infv∈W 1(Ω)

F [v]G[v]

= infv∈K

F [v] = ϑ.

Therefore F [v] ≥ ϑG[v] for all v ∈ W 1(Ω). Since vk+vl2 ∈ W 1(Ω) together

with vk, vl ∈ K, then

F

[vk + vl

2

]≥ ϑG

[vk + vl

2

]> ϑ

(1− ε

ϑ

)= ϑ− ε, k, l > N(ε).

Then, taking k and l large enough that F [vk] < ϑ+ ε and F [vl] < ϑ+ ε, weobtain

F

[vk − vl

2

]=

12F [vk] +

12F [vl]− F

[vk + vl

2

]<

<12

(ϑ+ ε) +12

(ϑ+ ε)− (ϑ− ε) = 2ε.

Consequently,

F [vk − vl]→ 0, k, l→∞.(2.4.2)

From (8.2.10), (8.2.12) it follows that ‖vk−vl‖W 1(Ω) → 0, k, l→∞. Hencevk converges in W 1(Ω) and as result of the completeness of W 1(Ω) thereexists a limit function u ∈ W 1(Ω) such that ‖vk − u‖W 1(Ω) → 0, k → ∞.In addition,

|F [vk]− F [u]| =

∣∣∣∣∣∣∫Ω

(|∇ωvk|2 − |∇ωu|2

)dΩ +

∫∂Ω

γ(x)(v2k − u2)dσ

∣∣∣∣∣∣ =

=∣∣∣∫Ω

(∇ωvk −∇ωu)(∇ωvk +∇ωu)dΩ +

+∫∂Ω

γ(x)(vk − u)(vk + u)dσ∣∣∣ ≤

≤

∫Ω

|∇ω(vk − u)|2dΩ

1/2∫Ω

|∇ω(vk + u)|2dΩ

1/2

+

+

∫∂Ω

|vk − u|2dσ

1/2∫∂Ω

γ2(x)|vk + u|2dσ

1/2

→ 0, k →∞,

since by (1.6.2)∫∂Ω

|vk − u|2dσ

1/2

≤ C‖vk − u‖W 1(Ω) → 0, k →∞,


while the terms(∫

Ω

|∇ω(vk + u)|2dΩ)1/2

and(∫∂Ω

γ2(x)|vk + u|2dσ)1/2

are

bounded. Therefore we get

F [u] = limk→∞

F [vk] = ϑ.

Analogously one sees that G[u] = 1.Suppose now that η is some function from W 1(Ω). Consider the ratio

F [u+ µη]G[u+ µη]

=F [u] + 2µF (u, η) + µ2F [η]G[u] + 2µG(u, η) + µ2G[η]

.

It is a continuously dierentiable function of µ on some interval around thepoint µ = 0. This ratio has a minimum at µ = 0 equal to ϑ and by theFermat Theorem, we have(

F [u+ µη]G[u+ µη]

)′µ=0

=2F (u, η)G[u]− 2F [u]G(u, η)

G2[u]= 0,

which by virtue of F [u] = ϑ, G[u] = 1 gives

F (u, η)− ϑG(u, η) = 0, ∀η ∈W 1(Ω).

The rest part of our Theorem follows from the smoothness theory forelliptic boundary value problem in smooth domains (details see in 2.5 [360]).

Remark 2.20. Remarks about the eigenvalue problem (EV P3)(see Remarks on pp. 121 - 122 [360])

Consider a sequence of domains Ω′ lying in the interior of Ω and con-verging to Ω. Let the boundaries ∂Ω′ of these domains be piecewise con-tinuously dierentiable. The integral identity (II) from Denition 2.17 forη ∈W 1(Ω) has the form∫

Ω

1qi

∂u

∂ωi

∂η

∂ωi− ϑuη

dΩ +

∫∂Ω

γ(x)uηdσ = 0(2.4.3)

But∫Ω

1qi

∂u

∂ωi

∂η

∂ωi− ϑuη

dΩ = lim

Ω′→Ω

∫Ω′

1qi

∂u

∂ωi

∂η

∂ωi− ϑuη

dΩ =

= limΩ′→Ω

[−∫Ω′

η(4ωu+ ϑu) dΩ +∫∂Ω′

η∂u

∂−→νdσ

]= lim

Ω′→Ω

∫∂Ω′

η∂u

∂−→νdσ.

Thus, the equation (2.4.3) takes the form

limΩ′→Ω

∫∂Ω′

η∂u

∂−→νdσ +

∫∂Ω

γuη dσ = 0.(2.4.4)


If in addition ∂Ω′ → ∂Ω in the sense that not only the points of ∂Ω′ convergeto the points of ∂Ω but also the normals at these points converge to thecorresponding normals of ∂Ω, then∫

∂Ω

γuη dσ = limΩ′→Ω

∫∂Ω′

γuη dσ,

if we assume that γ is the value on ∂Ω of some function given on Ω.Then condition (2.4.4) takes the form

limΩ′→Ω

∫∂Ω′

(∂u

∂−→ν+ γu

)η dσ = 0.(2.4.5)

Thus, u satises the boundary condition of (EV P3) "in the weak sense."

Therefore, an eigenfunction of the problem (EV P3) will be dened to bea function u(x) 6≡ 0 satisfying equation in Ω for some ϑ and the boundarycondition in the sense of relation (2.4.5). The number ϑ is called the eigen-value corresponding to the eigenfunction u(x).

Theorem 2.19 proved implies the existence of an eigenfunction u corre-sponding to the eigenvalue ϑ in the sense indicated.

2.4.2. The Friedrichs - Wirtinger inequality. Now from the varia-tional principle we obtain

Theorem 2.21. Let ϑ be the smallest positive eigenvalue of problem(EV P3) (it exists according to Theorem 2.19). Let Ω ⊂ Sn−1 be a boundeddomain . Let u ∈W 1(Ω) and γ(x) ∈ C0(∂Ω), γ(x) ≥ γ0 > 0. Then

ϑ

∫Ω

u2(ω)dΩ ≤∫Ω

|∇ωu(ω)|2dΩ +∫∂Ω

γ(ω)u2(ω)dσ.(2.4.6)

Proof. Consider functionals F [u], G[u],H[u] onW 1(Ω) described above.We will nd the pair (ϑ, u) that gives the minimum of the functional F [u] onthe set K. For this we investigate the minimization of the quadratic func-tional H[u] on all functions u(ω), for which the integrals exist and whichsatisfy the boundary condition from (EV P3). The necessary condition ofexistence of the functional minimum is δH[u] = 0. By the calculation of therst variation δH we have

δH[u] = −2∫Ω

(4ωu+ ϑu)δudΩ + 2∫∂Ω

∂u

∂−→νδudσ + 2

∫∂Ω

γ(x)uδudσ.

Hence we obtain the Euler equation and the boundary condition that are our(EV P3). Backwards, let u(ω) be the solution of (EV P3). By Theorem 2.19,u ∈ C2(Ω). Therefore we can multiply both sides of the equation (EV P3)


by u and integrate over Ω using the Gauss-Ostrogradskiy formula:

0 =∫Ω

(u4ωu+ ϑu2)dΩ = ϑ

∫Ω

u2dΩ−∫Ω

|∇ωu|2dΩ+

+∫Ω

∂

∂ωi

(uJ

qi

∂u

∂ωi

)dω = ϑ

∫Ω

u2dΩ−∫Ω

|∇ωu|2dΩ +∫∂Ω

u∂u

∂−→νdσ =

= ϑ

∫Ω

u2dΩ−∫Ω

|∇ωu|2dΩ−∫∂Ω

γ(x)u2dσ =(by K)

ϑ− F [u]⇒

ϑ = F [u].

Consequently, the required minimum is the least eigenvalue of (EV P3).

The existence a function u ∈ K such that

F [u] ≤ F [v] for all v ∈ K(2.4.7)

was proved in Theorem 2.19.

Throughout what follows we work with the value

λ =2− n+

√(n− 2)2 + 4ϑ2

,(2.4.8)

where ϑ is the smallest positive eigenvalue of problem (EV P ).Therefore the Friedrichs - Wirtinger inequality will be written in the

following form

λ(λ+ n− 2)∫Ω

ψ2dΩ ≤∫Ω

|∇ωψ|2dΩ +∫∂Ω

γ(x)ψ2dσ,(2.4.9)

∀ψ ∈W 1(Ω), γ(x) ∈ C0(∂Ω), γ(x) ≥ γ0 > 0.

Example. Here we verify the existence of the least positive eigenvalue ofthe problem (EV P3) in the case N = 2. In this case (EV P3) has the form

ψ′′ + ϑψ = 0, ω ∈ (−ω02 ,

ω02 ),

±ψ′ + γ±ψ∣∣ω=±ω0

2= 0, γ± = const > 0.

(2.4.10)

Solving this problem we obtain that ϑ is a positive root of the transcedenceequation

tan(ω0

√ϑ) =

√ϑ (γ− + γ+)ϑ− γ+γ−

;(2.4.11)

ψ(ω) =√ϑ cos

[√ϑ(ω − ω0

2)]− γ+ sin

[√ϑ(ω − ω0

2)].(2.4.12)


The existence of the positive solution of (2.4.11) follows from the graphicmethod (see the gure below; see also Example I and Remark 10.29 fromSubsection 10.2.7).


2.5. Hardy - Fridrichs - Wirtinger type inequalities

2.5.1. The Dirichlet boundary condition. Let θ(r) be the leasteigenvalue of the Beltrami operator 4ω on Ωr with Dirichlet condition on∂Ωr and let a neighborhood Gd0 of the boundary point O satisfy the condi-tion:

θ(r) ≥ θ0 + θ1(r) ≥ θ2 > 0, r ∈ (0, d), whereθ0, θ2 are positive constants andθ1(r) is a Dini continuous at zero function:

limr→0

θ1(r) = 0,d∫0

|θ1(r)|r dr <∞.

(S)

This condition describes our very general assumptions on the structure ofthe boundary of the domain in a neighborhood of the boundary point O.

Theorem 2.22. (Generalized Hardy - Friedrichs - Wirtinger in-equality)Let U(d) =

∫Gd0

rα−2|∇u|2dx is nite and u(x) = 0 for x ∈ Γd0. Then

∫Gd0

rα−4|u|2dx ≤ H(λ,N, α)(

1 +|θ1(%)|θ2

)∫Gd0

rα−2|∇u|2dx(H-W)

with some % ∈ (0, d), whereH(λ,N, α) =

[ (4−N−α)2

4 + λ(λ+N − 2)]−1

,

λ = 12

(2−N +

√(N − 2)2 + 4θ0

)(*)

provided α ≤ 4−N.

Proof. Integrating (2.3.2) over r ∈ (0, d) we get∫Gd0

rα−4|u|2dx ≤∫Gd0

rα−2

θ(r)|∇ωu|2

r2dx ≤

∫Gd0

rα−2

θ0 + θ1(r)|∇ωu|2

r2dx,

but

1θ0 + θ1(r)

=1θ0

+( 1θ0 + θ1(r)

− 1θ0

)=

1θ0− θ1(r)θ0

(θ0 + θ1(r)

) ≤≤ 1θ0

+|θ1(r)|θ0θ2

2.5 Hardy - Fridrichs - Wirtinger type inequalities 55

because of the condition (S). Hence, applying the mean value theorem withregard to the continuous at zero of the function θ1(r) we obtain

θ0

∫Gd0


rα−2 |∇ωu|2

r2dx+

|θ1(%)|θ2

U(d)

for some % ∈ (0, d).Now we integrate the Hardy inequality over Ω and rewrite the result it

in the form (4−N − α2

)2∫Gd0


rα−2u2rdx, α ≤ 4−N.

Adding two last inequality and applying the formula for |∇u|2 and the de-nition of the value λ we get the required inequality (H-W).

Corollary 2.23. ∀δ > 0∃d > 0 such that∫Gd0

rα−4u2dx ≤(H(λ,N, α) + δ

) ∫Gd0

rα−2 |∇u|2 dx,

provided the integral on the right hand side is nite and u(x) = 0 for x ∈ Γd0in the sense of traces.

Proof. Because of the function θ1(%) is continuous at zero we establishthe statement.

For conical domains the following statements are true.

Corollary 2.24. Let∫Gd0

rα−2|∇u|2dx is nite and u vanish on Γd0 in

the sense of traces. Then∫Gd0

rα−4u2(x)dx ≤ 4(4−N − α)2

∫Gd0

rα−2

(∂u

∂r

)2

dx(2.5.1)

for α < 4−N and∫Gd0

rα−4u2(x)dx ≤ 1λ(λ+N − 2)

∫Gd0

rα−4 |∇ωv|2 dx(2.5.2)

for all α ∈ R.

Proof. Integrating both sides of Hardy's inequality (2.1.3) over Ω weobtain (2.5.1). The inequality (2.5.2) is derived similarly, by multiplyingthe generalized Wirtinger inequality (2.3.1) by rα+N−5 and integrating overr ∈ [0, d].


Theorem 2.25. Let u ∈W 1,2(Gd0) vanish on Γd0 in the W 1,2(Gd0) sense.Then ∫

Gd0

rα−4u2(x)dx ≤ H(λ,N, α)∫Gd0

rα−2 |∇u|2 dx,(2.5.3)

with H(λ,N, α) := [(4−N − α)2/4 + λ(λ+N − 2)]−1

for α ≤ 4−N, provided the integral on the right hand side of (2.5.3) is nite.

Proof. If α < 4−N, then the assertion follows by adding the inequali-ties (2.5.1), (2.5.2) and by taking into account the formula 1.3.7. If α = 4−N,then (2.5.3) coincides with (2.5.2).

Corollary 2.26. Let u ∈ W 1,2(G) with u|∂G = ϕ ∈ W

1/2α−2(∂Ω). Then

for every δ > 0 there exist a constant c = c(δ, λ,N, α) such that∫Gd0

rα−4u2(x)dx ≤ (1 + δ)H(λ,N, α)∫Gd0

rα−2 |∇u|2 dx(2.5.4)

+c(δ, λ,N, α)‖ϕ‖ W

1/2

α−2(Γd0),

for α ≤ 4 − N, provided the integral on the right hand side of (2.5.10) isnite.

Proof. Let Φ ∈ W

1α−2(Gd0) with Φ|Γd0 = ϕ on Γd0. Then the dierence

u− Φ satises the Generalized HardyWirtinger inequality∫Gd0

rα−4(u− Φ)2dx ≤ H(λ,N, α)∫Gd0

rα−2 |∇u−∇Φ|2 dx, α ≤ 4−N.

Applying Cauchy's inequality twice we obtain∫Gd0

rα−4u2dx =∫Gd0

rα−4((u− Φ)2 + 2uΦ− Φ2

)dx

≤ H(λ,N, α)∫Gd0

rα−2(|∇u|2 − 2

⟨∇u,∇Φ

⟩+ |∇Φ|2

)dx

+ε∫Gd0

rα−4u2dx+ ε−1

∫Gd0

rα−4Φ2dx

≤ H(λ,N, α)∫Gd0

rα−2((1 + δ1)|∇u|2 + (1 + δ−1

1 )|∇Φ|2))dx

+ε∫Gd0

rα−4u2dx+ ε−1

∫Gd0

rα−4Φ2dx


for all ε > 0, δ1 > 0. Thus the claim follows from the denition of the tracenorm, if we set δ = ε+δ1

1−ε .

Corollary 2.27. Let ε > 0 and u ∈W 1,2(Gε) with u(x) = 0 for x ∈ Γε.Then for α < 4−N we have∫

Gε

rα−4u2(x)dx ≤ c∫Gε

rα−2 |∇u|2 dx(2.5.5)

with a constant c = c(λ,N, α).

Let us denote by ζ : Gρ0 → [0, 1] a cuto function satisfying

ζ(r) =

1 for 0 ≤ r ≤ ρ/20 for r ≥ ρ

,

|ζ ′(r)| ≤ const · ρ−1 for 0 ≤ r ≤ ρ.

Corollary 2.28. Let u ∈ W 1,2(Gd0) vanish on Γd0. If α ≤ 4 − N andρ ∈ (0, d], then

(2.5.6)∫Gρ0

rα−4ζ2(r)u2(x)dx ≤ H(λ,N, α)∫Gρ0

rα−2(

(1 + δ)ζ2(r)|∇v|2 +

+ (1 + δ−1)(ζ ′)2(r)v2(x))dx

for all δ > 0.

Proof. The assertion follows directly by applying the generalized Hardy- Wirtinger inequality (2.5.3) to the product ζ(r)u(x) and the Cauchy in-equality with δ > 0.

Lemma 2.29. Let U(ρ) =∫Gρo

r2−N |∇u|2dx <∞, ρ ∈ (0, d). Then

∫Ω

(ρu∂u

∂r+N − 2

2u2

) ∣∣∣∣∣r=%

dΩ ≤ ρ

2λ+ θ1(%)h1(%)U ′(ρ),

where h1(%) ≤ 2√θ0+√θ2, % ∈ (0, d).

Proof. Writing U(ρ) in spherical coordinates and dierentiating by ρwe obtain

U ′(ρ) =∫Ω

(ρ

(∂u

∂r

)2

+1ρ|∇ωu|2

)∣∣∣∣∣r=%

dΩ.

Moreover, by Cauchy's inequality we have for all ε > 0

ρu∂u

∂r≤ ε

2u2 +

12ερ2

(∂u

∂r

)2

.


Thus, we obtain by the (2.3.2)∫Ω

(ρu∂u

∂r+N − 2

2u2

)dΩ ≤ ε+N − 2

2

∫Ω

u2dΩ +ρ2

2ε

∫Ω

(∂u

∂r

)2

dΩ ≤

≤ ε+N − 22θ(%)

∫Ω

|∇ωu|2dΩ +ρ2

2ε

∫Ω

(∂u

∂r

)2

dΩ =

=ρ2

2

∫Ω

1ε

(∂u

∂r

)2

+ε+N − 2%2θ(%)

|∇ωu|2dΩ.

Let us choose ε from the equality

1ε

=ε+N − 2θ(%)

=⇒ ε =12

(2−N +

√(N − 2)2 + 4θ(%)

).

Then we get ∫Ω

(ρu∂u

∂r+N − 2

2u2

)dΩ ≤ %2

2ε

∫Ω

|∇u|2dΩ,

or in virtue of the condition (S)

∫Ω

(ρu∂u

∂r+N − 2

2u2

) ∣∣∣∣∣r=%

dΩ ≤ %

2−N +√

(N − 2)2 + 4[θ0 + θ1(%)]U ′(%).

Because of (*) by elementary calculation we have(2−N +

√(N − 2)2 + 4[θ0 + θ1(%)]

)− 2λ =

=(2−N +

√(N − 2)2 + 4[θ0 + θ1(%)]

)−(2−N +

√(N − 2)2 + 4θ0

)=

= θ1(%)h1(%),

where

h1(%) =4√

(N − 2)2 + 4[θ0 + θ1(%)] +√

(N − 2)2 + 4θ0

≤ 2√θ0 +

√θ2.

Hence follows the statement of Lemma.

Corollary 2.30. Let Gd0 be the conical domain andU(ρ) =

∫Gρo

r2−N |∇u|2dx <∞, ρ ∈ (0, d).. Then

∫Ω

(ρu∂u

∂r+N − 2

2u2

) ∣∣∣∣∣r=%

dΩ ≤ ρ

2λU ′(ρ).


Proof. Writing U(ρ) in spherical coordinates and dierentiating by ρwe obtain

U ′(ρ) =∫Ω

(ρ

(∂u

∂r

)2

+1ρ|∇ωu|2

)∣∣∣∣∣r=%

dΩ.

Moreover, by Cauchy's inequality we have for all ε > 0

ρu∂u

∂r≤ ε

2u2 +

12ερ2

(∂u

∂r

)2

.

Thus, choosing ε = λ we obtain by the generalized Wirtinger's inequality(2.3.1) with (2.4.8):∫

Ω

(ρu∂u

∂r+N − 2

2u2

) ∣∣∣∣∣r=%

dΩ ≤ ε+N − 22

∫Ω

u2dΩ +ρ2

2ε

∫Ω

(∂u

∂r

)2

dΩ

≤ ε+N − 22λ(λ+N − 2)

∫Ω

|∇ωu|2dΩ +ρ2

2ε

∫Ω

(∂u

∂r

)2

dΩ =ρ

2λU ′(ρ).

Let us assume that the cone K is contained in a circular cone K withthe opening angle ω0 and that the axis of K coincides with (x1, 0, . . . , 0) :x1 > 0. We dene the vector l ∈ RN by l = (−1, 0, . . . , 0).

Lemma 2.31. Let v ∈ C0(Gdε) ∩ W 1(Gdε), v(ε) = 0; γ(x) ≥ γ0 > 0.Then for any ε > 0∫

Gdε

rα−4v2dx ≤ Hε(λ, n, α)∫Gdε

rα−2|∇v|2dx,(2.5.7)

(2.5.8) Hε(λ, n, α) =

=1

λ(λ+ n− 2) + 14(4− n− α)2 · 1

1+ 12( εd)

4−n−α2

, α ≤ 4− n;

in addition, it is obvious that

Hε(λ, n, α) = H(λ, n, α) +O(ε) =⇒(2.5.9)

limε→+0

Hε(λ, n, α) = H(λ, n, α),

where H(λ, n, α) is determined by (2.5.3).

Proof. By Theorem 2.21 the inequality (2.4.9) holds. Multiplying it byrn−5+α and integrating over r ∈ (ε, d) we obtain (2.5.7) for α = 4 − n. If


α < 4−n we consider the inequality (2.1.7) and integrate it over Ω; then wehave

14

(4− n− α)2 · 1

1 + 12

(εd

) 4−n−α2

∫Gdε

rα−4v2dx ≤∫Gdε

rα−2v2rdx.

Adding this inequality with above for α = 4 − n (see (2.5.2) for Gdε) andusing the formula |∇u|2 =

(∂u∂r

)2+ 1

r2 |∇ωu|2 , we get the desired.

Lemma 2.32. Let v ∈ C0(G) ∩W 1(G), v(0) = 0; γ(x) ≥ γ0 > 0. Thenfor any ε > 0 ∫

Gd0

rα−4ε v2dx ≤ Hε(λ, n, α)

∫Gd0

rα−2ε |∇v|2dx,(2.5.10)

where Hε(λ, n, α) is determined by (2.5.8).

Proof. We perform the change of variables yi = xi − εli, i = 1, . . . , nand use the inequality (2.5.7) - (2.5.8):∫

Gd0

rα−4ε v2(x)dx =

∫Gdεε

|y|α−4v2(y + εl)dy ≤

≤ Hε(λ, n, α)∫Gdεε

|y|α−2|∇yv(y + εl)|2dy = Hε(λ, n, α)∫Gd0

rα−2ε |∇v|2dx.

Lemma 2.33. Let u ∈W 1,2(Gd0) with u(x) = 0 for x ∈ Γd0. Then∫Gd0

rα−2ε r−2u2(x)dx ≤

(3h

)2−α· 1λ(λ+N − 2)

∫Gd0

rα−2ε |∇u|2dx(2.5.11)

for all α ∈ R, where

h =

1, if 0 < ω0 ≤ π ,sin ω0

2 , if π < ω0 < 2π..

Proof. We consider the Wirtinger inequality (2.3.1), multiply bothsides of this inequality by (2−kd + ε)α−2rN−3 with ε > 0, taking into ac-count that

2−k−1d+ ε < r + ε < 2−kd+ ε in G(k),

and integrate over r ∈ (2−k−1d, 2−kd) :∫G(k)

r−2(2−kd+ ε)α−2u2dx ≤ 1λ(λ+N − 2)

∫G(k)

((2−kd+ ε)α−2|∇u|2

)dx.


Since rε ≤ r + ε < 2−kd+ ε in G(k) and α ≤ 2, we obtain∫G(k)

r−2(2−kd+ ε)α−2u2dx ≤ 1λ(λ+N − 2)

∫G(k)

rα−2ε |∇u|2dx.

On the other hand, by Lemma 1.11, in G(k)

2−kd+ ε = 2 · 2−k−1d+ ε < 2r + ε ≤ 3hrε ⇒

(2−kd+ ε)α−2 ≥(

3h

)α−2

rα−2ε .

Hence it follows(3h

)α−2 ∫G(k)

r−2rα−2ε u2dx ≤ 1

λ(λ+N − 2)

∫G(k)


Summing up this inequalities for k = 0, 1, 2, . . . , we nally obtain the re-quired (2.5.11).

Theorem 2.34. Let G be a unbounded domain. Let u ∈ W 1(G) vanishfor |x| > R≫ 1. Then∫

G

rα−4u2(x)dx ≤ H(λ,N, α)∫G

rα−2 |∇u|2 dx,(2.5.12)

with H(λ,N, α) := [(4−N − α)2/4 + λ(λ+N − 2)]−1

for α ≥ 4 − N, provided the integral on the right hand side of (2.5.12) isnite.

Proof. Similarly to the Theorem 2.25, if we apply the Theorem 2.7with p = 2 and α replied by α+ n− 3.

2.5.2. The Robin boundary condition.

Theorem 2.35. The Hardy - Friedrichs - Wirtinger inequalityLet u ∈ C0(G) ∩W 1(G) and γ(x) ∈ C0(∂G \ O), γ(x) ≥ γ0 > 0. Then

∫Gd0

rα−4u2dx ≤ H(λ, n, α)

∫Gd0

rα−2|∇u|2dx+∫Γd0

rα−3γ(x)u2(x)ds

,

(2.5.13)

H(λ, n, α) =1

λ(λ+ n− 2) + 14(4− n− α)2

, α ≤ 4− n

provided that integrals on the right are nite.


Proof. By Theorem 2.21 the inequality (2.4.9) holds. Multiplying it byrn−5+α and integrating over r ∈ (0, d) we obtain

(2.5.14)∫Gd0

rα−4u2dx ≤ 1λ(λ+ n− 2)

∫Gd0

rα−2 1r2|∇ωu|2dx+

+1

λ(λ+ n− 2)

∫Γd0

rα−3γ(x)u2(x)ds, ∀α ≤ 4− n.

Hence (2.5.13) follows for α = 4−n. Now, let α < 4−n. We shall show thatu(0) = 0. In fact, from the representation u(0) = u(x) −

(u(x) − u(0)

)by

the Cauchy inequality we have12 |u(0)|2 ≤ |u(x)|2 + |u(x)− u(0)|2. Putting v(x) = u(x)− u(0) we obtain

12|u(0)|2

∫Gd0

rα−4dx ≤∫Gd0

rα−4u2(x)dx+∫Gd0

rα−4|v|2dx <∞(2.5.15)

(the rst integral from the right is nite by (2.5.14) and the second integralis also nite, in virtue of Corollary 2.6). Since

∫Gd0

rα−4dx = measΩ

d∫0

rα+n−5dr =∞,

by α + n − 4 < 0, the assumption u(0) 6= 0 contradicts (2.5.15). Thusu(0) = 0.

Therefore we can use the Corollary 2.6:∫Gd0

rα−4u2dx ≤ 4|4− n− α|2

∫Gd0

rα−2u2rdx.(2.5.16)

Adding the inequalities (2.5.14), (2.5.16) and using the formula|∇u|2 =

(∂u∂r

)2+ 1

r2 |∇ωu|2 , we get the desired (2.5.13).

Lemma 2.36. Let Gd0 be the conical domain and

V (ρ) =∫G%o

r2−n|∇v|2dx+∫Γ%0

r1−nγ(x)v2(x)ds <∞, % ∈ (0, d).

Then ∫Ω

(%v∂v

∂r+n− 2

2v2

) ∣∣∣∣∣r=%

dΩ ≤ %

2λV ′(%).


Proof. Writing V (%) in spherical coordinates

V (%) =

%∫0

r2−n

∫Ω

|∇v|2dΩ

rn−1dr +

%∫0

r1−n

∫∂Ω

γ(x)|v|2dσ

rn−2dr =

=

%∫0

r

∫Ω

|∇v|2dΩ

dr +

%∫0

1r

∫∂Ω

γ(x)|v|2dσ

dr

and dierentiating with respect to % we obtain

V ′(%) =∫Ω

(%

(∂v

∂r

)2

+1%|∇ωv|2

)∣∣∣∣∣r=%

dΩ +1%

∫∂Ω

γ(%, ω)v2(%, ω)dσ.

Moreover, by Cauchy's inequality, we have for all ε > 0

ρv∂v

∂r≤ ε

2v2 +

12ερ2

(∂v

∂r

)2

.

Thus choosing ε = λ we obtain, by the Friedrichs - Wirtinger inequality(2.4.9),

∫Ω

(%v∂v

∂r+n− 2

2v2

) ∣∣∣∣∣r=%

dΩ ≤ %2

2ε

∫Ω

(∂v

∂r

)2

dΩ +

+ε+ n− 2

2

∫Ω

v2dΩ ≤ %2

2ε

∫Ω

(∂v

∂r

)2

dΩ +

+ε+ n− 2

2λ(λ+ n− 2)

∫Ω

|∇ωv|2dΩ +∫∂Ω

γ(x)v2(x)dσ

=%

2λV ′(%).

Lemma 2.37. Let v ∈ C0(Gdε) ∩ W 1(Gdε), v(ε) = 0; γ(x) ≥ γ0 > 0.Then for any ε > 0

(2.5.17)∫Gdε

rα−4v2dx ≤

≤ Hε(λ, n, α)

∫Gdε

rα−2|∇v|2dx+∫Γdε

rα−3γ(x)v2(x)ds

,


(2.5.18) Hε(λ, n, α) =

=1

λ(λ+ n− 2) + 14(4− n− α)2 · 1

1+ 12( εd)

4−n−α2

, α ≤ 4− n;

in addition, it is obvious that

Hε(λ, n, α) = H(λ, n, α) +O(ε) =⇒(2.5.19)

limε→+0

Hε(λ, n, α) = H(λ, n, α),

where H(λ, n, α) is determined by (2.5.13).

Proof. By Theorem 2.21 the inequality (2.4.9) holds. Multiplying it byrn−5+α and integrating over r ∈ (ε, d) we obtain (2.5.17) for α = 4 − n. Ifα < 4−n we consider the inequality (2.1.7) and integrate it over Ω; then wehave

14

(4− n− α)2 · 1

1 + 12

(εd

) 4−n−α2

∫Gdε

rα−4v2dx ≤∫Gdε

rα−2v2rdx.

Adding this inequality with above for α = 4 − n (see (2.5.14) for Gdε) andusing the formula |∇u|2 =

(∂u∂r

)2+ 1

r2 |∇ωu|2 , we get the desired result.

Lemma 2.38. Let v ∈ C0(G) ∩W 1(G), v(0) = 0; γ(x) ≥ γ0 > 0. Thenfor any ε > 0

(2.5.20)∫Gd0

rα−4ε v2dx ≤

≤ Hε(λ, n, α)

∫Gd0

rα−2ε |∇v|2dx+

∫Γd0

rα−3ε γ(x)v2(x)ds

,

where Hε(λ, n, α) is determined by (2.5.18).

2.6 Other auxiliary integral inequalities for n = 2 65

Proof. We perform the change of variables yi = xi − εli, i = 1, . . . , nand use the inequality (2.5.17) with (2.5.18):∫

Gd0

rα−4ε v2(x)dx =

∫Gdεε

|y|α−4v2(y + εl)dy ≤

≤ Hε(λ, n, α)

∫Gdεε

|y|α−2|∇yv(y + εl)|2dy +∫

Γdεε

|y|α−3γ(y + εl)v2(y + εl)ds

=

= Hε(λ, n, α)

∫Gd0


∫Γd0

rα−3ε γ(x)v2(x)ds

.

2.6. Other auxiliary integral inequalities for n = 2

In the following lemmata we assume that n = 2 and we denote by ζ acuto function dened in the previous section.

Lemma 2.39. Let u ∈W 2,20 (G). Then∫

Gρ0

r−αε ζ2(r)|∇u|4dx ≤ c(maxx∈Gρ0

|u(x)|)2

∫Gρ0

(r−αε ζ2|D2u|2 +

+ α2r−2−αε ζ2|∇u|2 + r−αε (ζ ′)2|∇u|2

)dx

for all α ≥ 0, ρ ∈ (0, d) with a suciently small d > 0.

Proof. Taking into account that ζ(r)u(x) = 0 on ∂Gρ0, we obtain bypartial integration∫Gρ0

r−αε ζ2(r)|∇u|4dx =∫Gρ0

r−αε ζ2(r)|∇u|2⟨∇u,∇u

⟩dx =

= −∫Gρ0

u(x)N∑i=1

Di

(r−αε ζ2(r)|∇u|2Diu

)dx = −

∫Gρ0

u(x)(r−αε ζ2|∇u|2∆u+

+ 2r−αε ζ2N∑

i,j=1

DiuDjuDiju− αr−1−αε ζ2|∇u|2

⟨∇u, x− εl

rε

⟩+

+ 2r−αε ζζ ′|∇u|2⟨∇u, x

r

⟩)dx.


Therefore∫Gρ0

r−αε ζ2(r)|∇u|4dx ≤∫Gρ0

|u(x)|(αr−1−α

ε ζ2|∇u|3 +

+ 4r−αε ζ2|∇u|2|D2u|+ 2r−αε ζ|ζ ′||∇u|3)dx.

Applying the Cauchy inequality with σ > 0 we get∫Gρ0

r−αε ζ2(r)|∇u|4dx ≤ supx∈Gρ0

|u(x)|∫Gρ0

(σ

2r−αε ζ2|∇u|4 +

γ2

2σr−α−2ε ζ2|∇u|2|

+3σr−αε ζ2|∇u|4 +2σr−αε ζ2|D2u|2 +

1σr−γε |ζ ′|2|∇u|2

)dx.

Choosing σ = (7 supx∈Gρ0 |u(x)|)−1 we obtain the assertion.

Lemma 2.40. Let u ∈W 2,20 (G). Then for all ε > 0, α > 0∫

G

r2rα−2ε |∇u|4dx ≤ c(sup

x∈G|u(x)|)2

∫Gd

(|∇u|2 + |D2u|2

)dx

+ 4( supx∈Gd0

|u(x)|)2

∫Gd0

((2 + (α− 2))2rα−2

ε |∇u|2 + rα−2ε r2|D2u|2

)dx

with a constant c depending only on α and d.

Proof. Taking into account that v vanishes on ∂G, we obtain by partialintegration and Cauchy's inequality∫G

r2rα−2ε |∇u|4dx =

∫G

r2rα−2ε |∇u|2

N∑i=1

DiuDiudx =

= −∫G

u

N∑i=1

Di

(r2rα−2

ε |∇u|2Diu)dx =

= −∫G

u(2rα−2ε |∇u|2

⟨∇u, x

⟩+ (α− 2)r2rα−2

ε |∇u|2⟨∇u, x− εl

⟩+

+2r2rα−2ε

N∑i,j=1

DiuDjuDiju+ r2rα−2ε |∇u|2

N∑i=1

Diiu

dx ≤

≤∫G

|u|(r2rα−2

ε |∇u|2|D2v|+ 2rrα−2ε |∇u|3 + |α− 2|r2rα−3

ε |∇u|3)dx ≤


≤∫G

|u|(r2rα−2

ε (12δ|D2u|2 +

δ

2|∇u|4) +

32δr2rα−2

ε |∇u|4 +

+1δrα−2ε |∇u|2 +

(α− 2)2

2δr2rα−4

ε |∇u|2)dx ≤ 2δ sup

x∈Gd0|u|∫Gd0

r2rα−2ε |∇u|4dx+

+12δ

supx∈Gd0

|u|∫Gd0

(r2rα−2

ε |D2u|2 + (2 + (α− 2)2)rα−2ε |∇u|2

)dx+

+ 2δ1 supx∈G|u|∫Gd

r2rα−2ε |∇u|4dx+

12δ1

supx∈G|u|c(d, α)

∫Gd

(|∇u|2 + |D2u|2

)dx

for all δ, δ1 > 0. Setting δ = 1/(4 supx∈Gd0 |u|) and δ1 = 1/(4 supx∈G |u|), weobtain the assertion.

Lemma 2.41. Let u ∈W 2,20 (G) and

w =(D2uD12u−D1uD22u

D1uD12u−D2uD11u

).(2.6.1)

Then there exists a constant K ≥ 0 such that

∫∂G

r2rα−2ε

⟨w, n

⟩dσ ≤ K

∫∂G\(Γd0∪O)

r2rα−2ε

(∂u

∂n

)2

dσ.(2.6.2)


Proof. To evaluate the boundary integral we decompose ∂G into ∂G =Γd0 ∪ 0 ∪ Γ and take into consideration that v vanishes on ∂G. At rst weverify that

⟨w, n

⟩= 0 on Γd0. We write Γd0 = Γd1,0 ∪ Γd2,0 (see the gure).

Now we have:∂u

∂x1

∣∣∣Γd1,0

= 0,∂2u

∂x21

∣∣∣Γd1,0

= 0, n1

∣∣∣Γd1,0

= 0, n2

∣∣∣Γd1,0

= −1 ⇒⟨w, n

⟩= 0 on Γd1,0.

Furthern1

∣∣∣Γd2,0

= cos(π

2+ ω0

)= − sinω0; n2

∣∣∣Γd2,0

= cosω0

Let us perform the rotation of axes about the origin O, through an angleω0 :

x′1 = x1 cosω0 + x2 sinω0,

x′2 = −x1 sinω0 + x2 cosω0.

Then ∂∂x1

= cosω0∂∂x′1− sinω0

∂∂x′2

,∂∂x2

= sinω0∂∂x′1

+ cosω0∂∂x′2

.


Therefore

ux1x1 = cos2 ω0ux′1x′1 − sin(2ω0)ux′1x′2 + sin2 ω0ux′2x′2 ;

ux1x2 =12

sin(2ω0)ux′1x′1 + cos(2ω0)ux′1x′2 −12

sin(2ω0)ux′2x′2 ;

ux2x2 = sin2 ω0ux′1x′1 + sin(2ω0)ux′1x′2 + cos2 ω0ux′2x′2 ;

Since∂u

∂x′1

∣∣∣Γd2,0

= 0,∂2u

∂x′21

∣∣∣Γd2,0

= 0,

then we obtain

⟨w, n

⟩∣∣∣Γd2,0

= − sinω0

cosω0ux′2

(cos(2ω0)ux′1x′2 −

12

sin(2ω0)ux′2x′2

)+

+ sinω0ux′2

(sin(2ω0)ux′1x′2 + cos2 ω0ux′2x′2

)+

+ cosω0

− sinω0ux′2

(cos(2ω0)ux′1x′2 −

12

sin(2ω0)ux′2x′2

)−

− cosω0ux′2

(− sin(2ω0)ux′1x′2 + sin2 ω0ux′2x′2

)≡

≡ 0.

Now we calculate⟨w, n

⟩∣∣∣Γ. We suppose that Γ is a smooth curve. The last

means that there is a coordinate system (y1, y2) centered at x0 ∈ Γ such thatthe positive y2−axis is parallel to the outward normal −→n to Γ at x0 and theequation of the portion of Γ has a form

y2 = ψ(y1); ψ′′(y1) ≤ K, K ≥ 0,

where the number K can choose independent of x0. Let us perform thetransformation of coordinates

yi = cik(xk − x0

k

), i = 1, 2,

where (cik) is the orthogonal matrix. In particular, we have

n1

∣∣∣Γ= c21, n2

∣∣∣Γ= c22,

∂∂x1

= c11∂∂y1

+ c21∂∂y2

,∂∂x2

= c12∂∂y1

+ c22∂∂y2

.

Hence it follows

ux1x1 = c211uy1y1 + 2c11c21uy1y2 + c2

21uy2y2 ;

ux1x2 = c11c12uy1y1 + (c21c12 + c11c22)uy1y2 + c21c22uy2y2 ;

ux2x2 = c212uy1y1 + 2c12c22uy1y2 + c2

22uy2y2


Because of u∣∣∣Γ= 0, we have u (y1, ψ(y1)) = 0 near x0. If we dierentiate this

equality, then we getuy1 + uy2ψ

′(y1) = 0,uy1y1 + 2uy1y2ψ

′(y1) + uy2y2ψ′2(y1) + uy2ψ

′′(y1) = 0.

But ψ′(y1)∣∣∣x0

= 0, therefore

uy1

∣∣∣x0

= 0; uy1y1

∣∣∣x0

= −∂u∂n

∣∣∣x0

·ψ′′(y1).

In addition,

ux1(x0) = c21∂u

∂n

∣∣∣x0

; ux2(x0) = c22∂u

∂n

∣∣∣x0

Now we can obtain

⟨w, n

⟩∣∣∣x0∈Γ

= c21∂u

∂n

c22

(−c11c12

∂u

∂nψ′′(y1) + (c11c22 + c21c12)uy1y2 +

+ c21c22uy2y2

)−

− c21

(−c2

12

∂u

∂nψ′′(y1) + 2c12c22uy1y2 + c2

22uy2y2

)+

+ c22∂u

∂n

c21

(−c11c12

∂u

∂nψ′′(y1) + (c11c22 + c21c12)uy1y2 +

+ c21c22uy2y2

)−

− c22

(−c2

11

∂u

∂nψ′′(y1) + 2c11c21uy1y2 + c2

21uy2y2

)=

= ψ′′(y1)(∂u

∂n

)2

in virtue of det (cik) = 1. Thus from above calculations we get the desired(2.6.2).

Lemma 2.42. Let u ∈W 2,20 (G). Then for all γ, ε > 0 and all α ∈ R

Jαε[u] ≡∫G

r2rα−2ε ((D12u)2 −D11uD22u)dx ≤(2.6.3)

≤ γ∫Gd0

r2rα−2ε |D2u|2dx+ c1(α, γ, h)

∫Gd0

rα−2ε |∇u|2dx+

+c2

∫Gd

(|∇u|2 + |D2u|2

)dx

2.7 Notes 71

with a constant c2 depending only on α, γ, d,diamG and measG.

Proof. Since G is a strictly Lipschitz domain and the set of all C∞0 (G)functions is dense inW 2,2

0 (G), it suces to prove (2.6.3) for smooth functions.In order to estimate Jαε[u] we integrate it by parts, once with respect to x1

and once with respect to x2 and add the resulting equations. As a result weobtain

(2.6.4) 2Jαε[u] =∫∂G

r2rα−2ε

⟨w, n

⟩dσ −

−∫G

r2rα−4ε

⟨2x+ (α− 2)(x− εl), w

⟩dx

with w dened by (2.6.1). The boundary integral in (2.6.4) we evaluate byLemma 2.41 ∫

∂G

r2rα−2ε

⟨w, n

⟩dσ ≤ K

∫Γ

r2rα−2ε

(∂u

∂n

)2

dσ.

By properties of the function rε and Theorem 1.29, we obtain∫∂G

r2rα−2ε

⟨w, n

⟩dσ ≤ cK

∫Gd

(|D2u|2 + |∇u|2)dx ∀ε > 0.(2.6.5)

The domain integral in (2.6.4) is estimated using (1.2.5) and the Cauchyinequality with ∀γ > 0:

(2.6.6) (2− α)∫G

rα−3ε r2

⟨x− εlrε

, w⟩dx+ 2

∫Grα−2ε |

⟨x,w

⟩|dx

≤ γ∫G

r2rα−2ε |D2u|2dx+ c2(α, γ, h)

∫G


The desired assertion then follows from (2.6.5) and (2.6.6).

2.7. Notes

The classical Hardy inequality was rst proved by G. Hardy [ 141]. Thevarious extensions of this inequality as well the proof of Theorem 2.7 can befound in [359, 107]. We have followed the account 2.2 of [128]. For otherversions of the Poincaré inequality, see [141]. The one-dimensional Wirtingerinequality is given and proved in Chapter VII [141]. The variational principlefor the Dirichlet boundary condition is given more detail in 4.1 [141]. Thematerial in 2.4.2, 2.5.2 is new. Subsection 2.6 is based on the ideas of [213](see there Lemma 4.5, Chapter II and 8, Chapter III).


CHAPTER 3

The Laplace operator

3.1. Dini estimates of the generalized Newtonian potential

We shall consider the Dirichlet problem for the Poisson equation4v = G +

n∑j=1

DjF j , x ∈ G,

(PE)v(x) = 0, x ∈ ∂G.

Let Γ(x− y) be the normalized fundamental solution of Laplace's equation.The following estimates are known (see e.g. (2.12), (2.14)[128]):

|Γ(x− y)| = 1N(N − 2)ωN

|x− y|2−N , N ≥ 3,

|DiΓ(x− y)| ≤ 1NωN

|x− y|1−N ,

(3.1.1)

|DijΓ(x− y)| ≤ 1ωN|x− y|−N ,

|DβΓ(x− y)| ≤ C(N, β)|x− y|2−N−β.

We dene the functions

z(x) =∫G

Γ(x− y)G(y)dy, w(x) = Dj

∫G

Γ(x− y)F j(y)dy,(3.1.2)

assuming that the functions G(x) and F j(x), j = 1, . . . , N are integrable onG. The function z(x) is called the Newtonian potential with density functionG(x), and w(x) is called the generalized Newtonian potential with densityfunction divF . We now give estimates for these potentials. In the followingthe D operator is always taken with respect to the x variable.

Lemma 3.1. Let ∂G ∈ C1,A,G ∈ Lp(G), p > N,F j ∈ C0,A(G), j =1, . . . , N, where A is an α− function Dini continuous at zero.

73

74 3 The Laplace operator

Then z ∈ C1(RN ), w ∈ C2(G) and for any x ∈ G

Diz(x) =∫G

DiΓ(x− y)G(y)dy,(3.1.3)

Diw(x) =∫G0

DijΓ(x− y)(F j(y)−F j(x)

)dy −(3.1.4)

−F j(x)∫∂G0

DiΓ(x− y)νj(y)dyσ

(i=1, . . . , N); here G0 is any domain containing G for which the Gaussdivergence theorem holds and F j are extended to vanish outside G.

Proof. By virtue of the estimate (3.1.1) for DiΓ, the functions

vi(x) =∫G

DiΓ(x− y)G(y)dy, i = 1, . . . , N

are well dened. To show that vi = Diz, we x a function ζ ∈ C1(R)satisfying

0 ≤ ζ ≤ 1, 0 ≤ ζ ′ ≤ 2, ζ(t) = 0 for t ≤ 1, ζ(t) = 1 for t ≥ 2

and dene for ε > 0

zε(x) =∫G

Γ(x− y)ζ(|x− y|ε

)G(y)dy.

Clearly, zε(x) ∈ C1(RN ) and

vi(x)−Dizε(x) =∫

|x−y|≤2ε

Di

(1− ζ

(|x− y|ε

))Γ(x− y)

G(y)dy

so that

|vi(x)−Dizε(x)| ≤ sup |G|∫

|x−y|≤2ε

(|DiΓ|+

2ε|Γ|)dy ≤

≤ sup |G| ·

2NεN−2 for N > 2,4ε(1 + | lg 2ε|) for N = 2.

Consequently, zε and Dizε converge uniformly in compact subsets of RN toz and vi respectively as ε→ 0. Hence, z ∈ C1(RN ) and Diz = vi.

3.1 Dini estimates of the generalized Newtonian potential 75

By virtue of the estimate (3.1.1) for DijΓ, and the Dini continuous of F jthe functions

ui(x) =∫G0


)dy −

−F j(x)∫∂G0

DiΓ(x− y)νj(y)dyσ, i = 1, . . . , N

are well dened. Let us dene for ε > 0

vε(x) =∫G

DiΓ(x− y)ζ(|x− y|ε

)F j(y)dy.

Clearly, vε(x) ∈ C1(G) and dierentiating, we obtain

n∑j=1

Djvε(x) =∫G

Dj

(DiΓ(x− y)ζ

(|x− y|ε

))F j(y)dy =

= F j(x)∫G0

Dj

(DiΓ(x− y)ζ

(|x− y|ε

))dy +

+∫G

Dj

(DiΓ(x− y)ζ

(|x− y|ε

))(F j(y)−F j(x)

)dy =

=∫G0

Dj

(DiΓ(x− y)ζ

(|x− y|ε

))(F j(y)−F j(x)

)dy −

−F j(x)∫∂G0

DiΓ(x− y)νj(y)dyσ

provided ε is suciently small. Hence, by subtraction

|ui(x)−n∑j=1

Djvε(x)| =

=

∣∣∣∣∣∫

|x−y|≤2ε

Dj

(1− ζ

(|x− y|ε

))DiΓ(x− y)

(F j(y)−F j(x)

)dy

∣∣∣∣∣ ≤≤ [F j ]A;x ·

∫|x−y|≤2ε

(|DijΓ|+

2ε|DiΓ|

)A(|x− y|)dy ≤

≤ C(N,G)

2ε∫0

A(t)tdt ·

N∑j=1

[F j ]A;x


provided 2ε < dist(x, ∂G). Consequentlyn∑j=1

Djvε(x) converges to ui uni-

formly on compact subsets of G as ε→ 0, and since vε converges uniformlyto vi = Diz in G, we obtain w ∈ C2(G) and ui = Diw. This completes theproof of Lemma 3.1.

Let B1 = BR(x0), B2 = B2R(x0) be concentric balls in RN and z(x), w(x)be Newtonian potentials in B2.

Lemma 3.2. Suppose G ∈ Lp(B2), p > N/2, and F j ∈ L∞(B2), j =1, . . . , N. Then

|z|0;B1 ≤ c(p)R2/p′ ln1/p′(1

2R)‖G‖p;B2 , N = 2;

(3.1.5)

|z|0;B1 ≤ c(p,N)R2−N+N/p′‖G‖p;B2 , N ≥ 3;

|w|0;B1 ≤ 2RN∑j=1

|F j |0;B2 .(3.1.6)

Proof. The estimates follow from inequalities (3.1.1), Hölder's inequal-ity for integrals and Lemma 3.1.

Lemma 3.3. Let G ∈ Lp(B2), p > N,F j ∈ C0,A(B2), j = 1, . . . , N,where A is an α− function Dini continuous at zero. Thenz, w ∈ C1,B(B1) and

‖z‖1,B;B1 ≤ c(p,N,R,A−1(2R)

)‖G‖p;B2 ,(3.1.7)

‖w‖1,B;B1 ≤ c(p,N,R, α,A−1(2R),B(2R)

) N∑j=1

‖F j‖0,A;B2 .(3.1.8)

Proof. Let x, x ∈ B1 and G = B2. By formulas (3.1.3), (3.1.4), takinginto account (3.1.1) and Hölder's inequality for integrals and setting|x− y| = t, y − x = tω, dy = tN−1dtdΩ, we have

|Diz| ≤ (NωN )−1

∫B2

|x− y|1−N |G(y)|dy ≤

≤ (NωN )−1‖G‖p;B2

∫B2

|x− y|(1−N)p′dy1/p′

=(3.1.9)

=p− 1p−N

(2R)(p−N)/(p−1)‖G‖p;B2 ;


|Diw(x)| ≤ (NωN )−1R1−NN∑j=1

|F j(x)|∫∂B2

dyσ +

+ω−1N

N∑j=1

[F j ]A,x ·∫B2

A(x− y)|x− y|N

dy ≤ 2N−1N∑j=1

|F j(x)|+(3.1.10)

+NN∑j=1

[F j ]A,x ·2R∫0

A(t)tdt ≤ c(N)B(2R)

N∑j=1

(|F j(x)|+

N∑j=1

[F j ]A,x).

Taking into account (3.1.3) we obtain by subtraction

|Diz(x)−Diz(x)| ≤∫B2

|DiΓ(x− y)−DiΓ(x− y)| · |G(y)|dy.

We set δ = |x − x|, ξ = 12(x − x) and represent B2 = Bδ(ξ) ∪ B2 \ Bδ(ξ).

Then ∫Bδ(ξ)

|DiΓ(x− y)−DiΓ(x− y)| · |G(y)|dy ≤

≤∫

Bδ(ξ)

|DiΓ(x− y)| · |G(y)|dy +∫

Bδ(ξ)

|DiΓ(x− y)| · |G(y)|dy ≤

≤ (NωN )−1 ∫Bδ(ξ)

|x− y|1−N |G(y)|dy +∫

Bδ(ξ)

|x− y|1−N |G(y)|dy≤

≤ 2(NωN )−1

∫B3δ/2(x)

|x− y|1−N |G(y)|dy ≤(3.1.11)

≤ 2(NωN )−1‖G‖p;B2

( ∫B3δ/2(x)

|x− y|(1−N)p′dy)1/p′

≤

≤ 2(NωN )−1/p‖G‖p;B2

(3δ2

)1−N/pN + (1−N)p′−1/p′ ≤

≤ 2(NωN )−1/p(2R)1−N/p

N + (1−N)p′−1/p′· A(|x− x|)A(2R)

‖G‖p;B2

(here we take into account that δα ≤ (2R)αA(δ)/A(2R) for all α > 0 by(1.8.1), since δ ≤ 2R). Similarly,


∫B2\Bδ(ξ)

|DiΓ(x− y)−DiΓ(x− y)| · |G(y)|dy ≤

≤ |x− x|∫

B2\Bδ(ξ)

|DDiΓ(x− y)| · |G(y)|dy

(for some x between x and x)

≤ δω−1N

∫|y−ξ|≥δ

|x− y|−N |G(y)|dy ≤ 2Nδω−1N

∫|y−ξ|≥δ

|ξ − y|−N |G(y)|dy

(3.1.12)

(since |y − ξ| ≤ 2|y − x|)

≤ 2Nδω−1N ‖G‖p;B2

( ∫|y−ξ|≥δ

|ξ − y|−np′dy)1/p′

≤ 2Nδ1−N/pω−1/pN (p− 1)1/p′‖G‖p;B2

≤ 2N (2R)1−N/pω−1/pN (p− 1)1/p′ · A(|x− x|)

A(2R)‖G‖p;B2 .

From (3.1.11) and (3.1.12), taking into account (1.8.3), we obtain:

(3.1.13) |Diz(x)−Diz(x)| ≤ c(N, p,R)A−1(2R)‖G‖p;B2A(|x− x|)≤ c(N, p,R)A−1(2R)‖G‖p;B2B(|x− x|), ∀x, x ∈ B1.

The rst from the required estimates (3.1.7) follows from the inequalities(3.1.5) and (3.1.13).

Now we derive the estimate (3.1.8). By (3.1.4) for x, x ∈ B1 we have

Diw(x)−Diw(x) =N∑j=1

(F j(x)J1j +

(F j(x)−F j(x)

)J2j

)

+J3 + J4 +N∑j=1

(F j(x)−F j(x)

)J5j + J6,(3.1.14)

where

J1j =∫∂B2

(DiΓ(x− y)−DiΓ(x− y)

)νj(y)dyσ,

J2j =∫∂B2

DiΓ(x− y)νj(y)dyσ,

J3 =∫

Bδ(ξ)

DijΓ(x− y)(F j(x)−F j(y)

)dy,


J4 =∫

Bδ(ξ)


)dy,

J5j =∫

B2\Bδ(ξ)

DijΓ(x− y)dy,

J6 =∫

B2\Bδ(ξ)

(DijΓ(x− y)−DijΓ(x− y)

)(F j(x)−F j(y)

)dy

We estimate these integrals:

|J1j | ≤ |x− x|∫∂B2

|DDiΓ(x− y)|dyσ

(for some point x between x and x)

≤ |x− x|Nω−1N

∫∂B2

|x− y)|−Ndyσ ≤ N22N−1|x− x|R−1

(since |x− y)| ≥ R for y ∈ ∂B2)

≤ N22N−1A(|x− x|)R−1δ/A(δ) ≤ N22NA(|x− x|)/A(2R)

(since δ = |x− x| ≤ 2R and δ/A(δ) ≤ 2R/A(2R) by (1.8.1))

≤ N22NαB(δ)/A(2R) (by (1.8.3)).

Next,

|J2j | ≤ 2N−1,

|J3| ≤ ω−1N [F j ]A,x

∫Bδ(ξ)

|x− y)|−NA(|x− y|)dy

≤ ω−1N [F j ]A,x

∫B3δ/2(x)

|x− y)|−NA(|x− y|)dy

= N [F j ]A,x

3δ/2∫0

A(t)tdt ≤ N

(32

)α[F j ]A,xB(δ) (by (1.8.2)).

By analogy with the estimate for J3 we obtain

|J4| ≤ N(3

2

)α[F j ]A,xB(δ).

By (3.1.1) it is obvious

|J5j | ≤ 2N .


At last

|J6| ≤ |x− x|∫

B2\Bδ(ξ)

|DDijΓ(x− y)| · |F j(x)−F j(y)|dy

(for some point x between x and x)

≤ |x− x|c(N)∫

|y−ξ|≥δ

|x− y)|−N−1 · |F j(x)−F j(y)|dy

≤ c(N)δ[F j ]A,x∫

|y−ξ|≥δ

|x− y)|−N−1A(|x− y|)dy

≤ c(N)δ[F j ]A,x∫

|y−ξ|≥δ

|ξ − y)|−N−1A(

32|ξ − y|

)dy

(since |x− y| ≤ 32|ξ − y| ≤ 3|x− y)|)

≤ c(N)ωNδ(

32

)α[F j ]A,x

R∫δ

t−2A(t)dt

(since A

(32t

)≤(

32

)αA(t) by (1.8.2)

)≤ c(N)ωN

α

1− α

(32

)α[F j ]A,xB(δ) by (1.8.4).

Now from (3.1.14) and the above estimates we obtain

|Diw(x)−Diw(x)| ≤ c(N,α)N∑j=1

(|F j(x)|A−1(2R) +

+[F j ]A,x + [F j ]A,x)B(|x− x|) ∀x, x ∈ B1.(3.1.15)

Finally, from (3.1.10) and (3.1.15) it follows that w(x) ∈ C1,B(B1) and theestimate (3.1.8) holds. Lemma 3.3 is proved.

Now we can assert a C1,B interior estimate:

Lemma 3.4. Let G be a domain in RN , and let v(x) ∈ C1,B(G) be ageneralized solution of Poisson's equation (PE) with G ∈ L N

1−α(G),F j ∈

C0,A(G), where A is an α− function satisfying the Dini condition at zero.Then for any two concentric balls B1 = BR(x0), B2 = B2R(x0) ⊂⊂ G wehave

‖v‖1,B;B1 ≤ c(|v|0;B2 + ‖G‖ N

1−α ;B2+

N∑j=1

‖F j‖0,A;B2

),(3.1.16)

where c = c(N,R, α,A−1(2R),B(2R)

).


Proof. It is easily shown that the Newtonian potential, given by

V (x) =∫G

Γ(x− y)G(y)dy +∫G

DjΓ(x− y)F j(y)dy

is a weak solution of the equation from (PE). We can write

v(x) = V (x) + v(x), x ∈ B2,(3.1.17)

where v(x) is harmonic in B2. By Lemma 3.3, we have

‖V ‖1,B;B1 ≤ c(‖G‖ N

1−α ;B2+

N∑j=1

‖F j‖0,A;B2

),(3.1.18)


). By Theorem 2.10 [128] we obtain:

(3.1.19) ‖v‖1,B;B1 ≤ |v|1,B1 +N∑i=1

supx,y∈B1x6=y

|Div(x)−Div(y)|B(|x− y|)

≤

≤ |v|1,B1 + supx∈B1

|D2v| · supx,y∈B1x 6=y

|x− y|B(|x− y|)

≤

≤ c1

(R,A−1(2R)

)|v|2,B1 ≤ c2

(R,A−1(2R)

)|v|0,B2 ≤ c2

(|v|0,B2 + |V |0,B2

)≤

≤ c3

(|v|0,B2 + ‖G‖ N

1−α ;B2+

N∑j=1

‖F j‖0,B2

)in virtue of Lemma 3.2. From (3.1.17) - (3.1.19) it follows the desired esti-mate (3.1.16).

Corresponding boundary estimates can be derived in a similar way. Letus rst derive the appropriate extension of the estimate for the generalizedNewtonian potential w(x) with density function divF .

Lemma 3.5. Let F j ∈ C0,A(B+2 ) (j = 1, . . . , N). Then w ∈ C1,B(B+

1 )and

‖w‖1,B;B+1≤ c(p,N,R, α,A−1(2R),B(2R)

) N∑j=1

‖F j‖0,A;B+2.(3.1.20)

Proof. We assume that B2 intersects Σ since otherwise the result isalready contained in Lemma 3.3. The representation (3.1.4) holds for Diw(x)with G0 = B+

2 . If either i or j 6= N , then the portion of the boundary integral∫∂B+

2 ∩Σ

DiΓ(x− y)νj(y)dyσ =∫

∂B+2 ∩Σ

DjΓ(x− y)νi(y)dyσ = 0

since νi or νj = 0 on Σ. The estimates in Lemma 3.3 for Diw(x) (i orj 6= N) then proceed exactly as before with B2 replaced by B+

2 , Bδ(ξ)replaced by Bδ(ξ) ∩ B+

2 and ∂B2 replaced by ∂B+2 \ Σ. Finally DNNw can


be estimated from the equation of the problem (PE) and the estimates onDkkw for k = 1, . . . , N − 1.

Theorem 3.6. Let v(x) ∈ C0(B+2 ) be a generalized solution of equation

(PE) in B+2 with G ∈ L N

1−α(B+

2 ),F j ∈ C0,A(B+2 ) (j = 1, . . . , N), where A

is an α− function satisfying the Dini condition at zero, and let v = 0 on

B2 ∩ Σ. Then v ∈ C1,B(B+1 ), and

‖v‖1,B;B+1≤ c(|v|0;B+

2+ ‖G‖ N

1−α ;B+2

+N∑j=1

‖F j‖0,A;B+2

),(3.1.21)


).

Proof. We use the method of reection. Let x′ = (x1, . . . , xN−1), x∗ =(x′,−xN ) and dene

F i∗(x) =

F i(x) if xN ≥ 0,F i(x∗) if xN ≤ 0

(i = 1, . . . , N).

We assume that B2 intersects Σ; otherwise Lemma 3.4 implies (3.1.21). Weset B−2 = x ∈ RN

∣∣x∗ ∈ B+2 and D = B+

2 ∪B−2 ∪ (B2 ∩ Σ). Then F i∗(x) ∈

C0,A(D) and

‖F i∗‖0,A;D ≤ 2‖F i‖0,A;B+2

; (i = 1, . . . , N).

Let

G(x, y) = Γ(x− y)− Γ(x− y∗) = Γ(x− y)− Γ(x∗ − y)

denote the Green's function of the half-space RN+ , and consider

w(x) = −∫B+

2

DyG(x, y)−→F (y)dy, Dy = (Dy1 , . . . , DyN ).

For each i = 1, . . . , N let wi(x) denote the component of w given by

wi(x) =∫B+

2

DyiΓ(x− y)F i(y)dy +∫B+

2

DyiΓ(x∗ − y)F i(y)dy.

We can see that w(x) and wi(x) vanish on B2 ∩ Σ. Noting that

∫B+

2

Γ(x∗ − y)F i(y)dy =∫B−2

Γ(x− y)F i∗(y)dy, (i = 1, . . . , N − 1),

3.2 The equation with constant coecients. Green's function 83

we obtain

(3.1.22) wi(x) = Di

[2∫B+

2

Γ(x− y)F i(y)dy −∫D

Γ(x− y)F i∗(y)dy],

(i = 1, . . . , N − 1).

And when i = N , since∫B+

2

DyNΓ(x∗ − y)FN (y)dy =∫B−2

DyNΓ(x− y)FN∗ (y)dy,

we have

wN (x) = DN

∫D

Γ(x− y)FN∗ (y)dy.(3.1.23)

Letting

w∗i (x) = −Di

∫D

Γ(x− y)F i∗(y)dy, (i = 1, . . . , N),

we have by Lemma 3.3

‖w∗‖1,B;B+1≤ c(p,N,R, α,A−1(2R),B(2R)

) N∑j=1

‖F j∗‖0,A;D ≤

≤ 2c(p,N,R, α,A−1(2R),B(2R)

) N∑j=1

‖F j‖0,A;B+2.

Combining this with Lemma 3.5, we obtain

‖w‖1,B;B+1≤ c(p,N,R, α,A−1(2R),B(2R)

) N∑j=1

‖F j‖0,A;B+2.(3.1.24)

Now let v(x) = v(x)−V (x), where V (x) is the Newtonian potential fromLemma 3.4. Then v(x) is harmonic in B+

2 and v(x) = 0 on Σ. By Schwarzreection principle v(x) may be extended to a harmonic function in B2 andhence the estimate (3.1.21) follows from the interior derivative estimate forharmonic functions by Theorem 2.10 [128] (see the proof of Lemma 3.4).

3.2. The equation with constant coecients. Green's function

LetL0 ≡ aij0 Dij , aij0 = aji0

be a dierential operator with constant coecients aij0 satisfying

ν|ξ|2 ≤ aij0 ξiξj ≤ µ|ξ|2, ∀ξ ∈ RN

for positive constants ν, µ and let det(aij0

)= 1.


Definition 3.7. The Green's function of the rst kind of the operator L0

for the domain G is the function G(x, y) satisfying the following properties

• L0G(x, y) = δ(x− y), x ∈ G, where δ(x− y) is the Dirac function;• G(x, y) = 0, x ∈ ∂G.

For the properties, the existence and the construction of Green's func-tions in detail see e.g. 5.1 [43], chapter I [310]. We note following state-ments:

Lemma 3.8. Let G(x, y) be the Green function of L0 in RN+ , N ≥ 3.Then G(x, y) satises the following inequalities:

G(x, y) ≤

|x− y|2−N ,CyN |x− y|1−N ,CxNyN |x− y|−N ;

|DiG(x, y)| ≤

C|x− y|1−N ,CyN |x− y|−N ;

|DijG(x, y)| ≤

C|x− y|−N ,CyN |x− y|−1−N ,

where C depends only on ν, µ,N.

Proof. Let A be the matrix (aij0 ) and T be a constant matrix whichdenes a nonsingular linear transformation x′ = xT from R

N onto RN .Letting v(x′) = v(xT) one veries easily that

aij0 Dijv(x) = aij0 Dij v(x′),

where A = TtAT, Tt = T transpose. For suitable orthogonal matrix T, Ais a diagonal matrix Λ whose diagonal elements are the eigenvalues λ1, . . . , λNof A. If Q = TΛ−1/2, where Λ−1/2 = [λ−1/2

i δji ], then the transformationx′ = xQ takes L0v = 4′v(x′), i.e. L0 is transformed into the Laplace op-erator. By a further rotation we may assume that Q takes the half-spacexN > 0 onto the half-space x′N > 0.

Since the orthogonal matrix T preserves length, we have:

Λ−1/2|x| ≤ |x′| = |xQ| ≤ λ−1/2|x|;λ = minλ1, . . . , λN = ν; Λ = maxλ1, . . . , λN = µ.

It follows that G(x′, y′) = G(xQ, yQ) is the Green function of the Laplaceoperator in the half-space x′N > 0.

The corresponding inequalities for G(x′, y′) are well known, since weknow G(x′, y′) explicitly (see e.g. 2.4 [128] or 8, 10 Chapter I [310]).Here C depends on N only . Now required inequalities follow easily, sincethe dilation of distance is bounded above and below with µ and ν.

3.2 The equation with constant coecients. Green's function 85

In the same way we can prove the next Lemma (here we use the explicit formof the Green function for a ball,- see.e.g. 2.5 [128],- and a homothety).

Lemma 3.9. Let G(x, y) be the Green function of L0 for the ball B%(0).Then G(x, y) satises the following inequalities:

G(x, y) ≤ C|x− y|2−N , |∇xG(x, y)| ≤ C|x− y|1−N , for x, y ∈ B%(0);∣∣∣ ∂∂yi∇xG(x, y)

∣∣∣ ≤ C%−N , ∣∣∣ ∂2

∂yi∂yj∇xG(x, y)

∣∣∣ ≤ C%−N−1,

for y ∈ B%/2(0), |x| = %, N ≥ 3,

where C depends only on ν, µ,N.

Finally, we note the Green representation formula

u(y) =∫∂G

u(x)∂G(x, y)∂νx

dsx +∫G

G(x, y)L0udx,(3.2.1)

where G(x, y) is the Green function of the operator L0 in G and ∂∂ν denotes

the conormal derivative, i.e. the derivative with direction cosines aij0 nj ,i = 1, · · · , N. It is well known that this formula is valid in a Dini - Liapunovregion (see Chapter I [310]).

Now we establish a necessary preliminary result that extends Lemma 3.4and Theorem 3.6 from Poisson's equation to other elliptic equations withconstant coecients. We state these extensions in the following theorem:

Theorem 3.10. In the equation

L0u ≡ aij0 Diju = G(x) +∂F j(x)∂xj

, aij0 = aji0 , x ∈ G(ECC )

let A = (aij0 ) be a constant matrix such that

ν|ξ|2 ≤ aij0 ξiξj ≤ µ|ξ|2, ∀ξ ∈ RN

for positive constants ν, µ.(a) Let G be a domain in RN , and let v(x) ∈ C1,B(G) be a generalized so-

lution of equation (ECC) with G ∈ L N1−α

(G), F j ∈ C0,A(G), (j = 1, . . . , N)where A is an α− function satisfying the Dini condition at zero. Then forany two concentric balls B1 = BR(x0), B2 = B2R(x0) ⊂⊂ G we have

‖v‖1,B;B1 ≤ c(|v|0;B2 + ‖G‖ N

1−α ;B2+

N∑j=1

‖F j‖0,A;B2

),(3.2.2)

where c = c(N,R, α, ν, µ,A−1(2R),B(2R)

).

(b) Let v(x) ∈ C0(B+2 ) be a generalized solution of equation

L0v = G(x) + ∂Fj(x)∂xj

in B+2 with G ∈ L N

1−α(B+

2 ),F j ∈ C0,A(B+2 ),


(j = 1, . . . , N), where A is an α− function satisfying the Dini condition

at zero, and let v = 0 on B2 ∩ Σ. Then v ∈ C1,B(B+1 ), and

‖v‖1,B;B+1≤ c(|v|0;B+

2+ ‖G‖ N

1−α ;B+2

+N∑j=1

‖F j‖0,A;B+2

),(3.2.3)

where c = c(N,R, α, ν, µ,A−1(2R),B(2R)

).

Proof. Let T be a constant matrix which denes a nonsingular lineartransformation y = xT from R

N onto RN . Letting v(y) = v(xT) one verieseasily that

aij0 Dijv(x) = aij0 Dij v(y),

where A = TtAT, Tt = T transpose. For suitable orthogonal matrix T, Ais a diagonal matrix Λ whose diagonal elements are the eigenvalues λ1, . . . , λNof A. If Q = TΛ−1/2, where Λ−1/2 = [λ−1/2

i δji ], then the transformation

y = xQ takes L0v = G(x) + ∂Fj(x)∂xj

into the Poisson equation 4v(y) =

G(y) + qkj∂ eFj(y)∂yk

. By a further rotation we may assume that Q takes thehalf-space xN > 0 onto the half-space yN > 0.

Since the orthogonal matrix T preserves length, we have:

Λ−1/2|x| ≤ |y| = |xQ| ≤ λ−1/2|x|;λ = minλ1, . . . , λN = ν; Λ = maxλ1, . . . , λN = µ.

It follows that if B(y0) is the image of B(x0) under the transformation y =xQ then the norms ‖ • ‖k,A dened on B and B are equivalent, i.e. thesenorms are related by the inequality

c−1‖v‖k,A;B ≤ ‖v‖k,A; eB ≤ c‖v‖k,A;B; k = 0, 1,

where c = c(k,N, ν, µ).Similarly if B+(y0) with boundary portion σ on yN = 0 is the image of

B+(x0) with a boundary portion σ on Σ, the norms ‖ • ‖k,A dened on B+

and B+ are equivalent, i.e. these norms are related by the inequality

c−1‖v‖k,A;B+∪σ ≤ ‖v‖k,A; eB+∪eσ ≤ c‖v‖k,A;B+∪σ; k = 0, 1,

where c = c(k,N, ν, µ).To prove part (a) of our Theorem we apply Lemma 3.4 in B(y0) to obtain

‖v‖1,B;B1 ≤ c‖v‖1,B; eB1≤ C

(|v|

0; eB2+ ‖G‖ N

1−α ; eB2+

N∑j=1

‖F j‖0,A; eB2

)≤

≤ C(|v|0;B2 + ‖G‖ N

1−α ;B2+

N∑j=1

‖F j‖0,A;B2

),

which is the desired conclusion (3.2.2).

3.3 The Laplace operator in weighted Sobolev spaces. 3.4 Notes 87

Part (b) of our Theorem is proved in the same way, using Theorem3.6.

3.3. The Laplace operator in weighted Sobolev spaces

Let G be a conical domain. We consider the Dirichlet problem for thePoisson equation

∆u = f in G,u = g on ∂G.

(DPE)

It is known from the classical paper by Kondrate'v [159] that the behav-ior of solutions of (DPE) is controlled by the eigenvalues of the eigenvalueproblem (EV P1) for the LaplaceBeltrami operator ∆ω.

Theorem 3.11. (See Theorem 4.1 [272], Theorem 2.6.5 [197]).Let p ∈ (1,∞), k ∈ N with k ≥ 2 and α ∈ R. Let λ be dened by (2.4.8) withthe smallest positive eigenvalue ϑ of (EVP1). Then the Dirichlet problem(DPE) has a unique solution u ∈ V k

p,α(G) for all f ∈ V k−2p,α (G),

g ∈ V k−1/pp,α (∂G) if and only if

−λ+ 2−N < k − (α+N)/p < λ.

In this case the following apriori estimate is valid

‖u‖V kp,α(G) ≤ c‖f‖V k−2

p,α (G) + ‖g‖Vk−1/pp,α (∂G)

.

3.4. Notes

Section 3.1 is a modication of Chapter 4 [128]: we replaced Höldercontinuity by Dini continuity.

Discussions of boundary value problems for the Laplacian in nonsmoothdomains can be found in a number of works (see e.g., [2, 9, 90, 88, 112,115, 125, 126, 132, 134, 159, 175, 196, 197, 211, 239, 240, 246,275, 245, 319, 324, 329, 344, 347, 353, 395, 400, 404, 405, 406 ]).Theorem 3.11 was established for the rst time in the work [159] for p = 2.V.G. Maz'ya and B.A. Plamenevsky [272] extended this result to the case1 < p < ∞. For details we refer to [197] (in particular, see Notes 1.5, 2.7there).

Other boundary value problems for the Laplace equation or for gener-al second order elliptic equations and systems with constant coecients innonsmooth domain have been studied in many works: W. Zajaczkowski andV. Solonnikov [406] - Neumann problem in a domain with edges, P. Grisvard[132], M. Dauge [91], N. Wigley [404, 405] - Neumann and mixed problemon curvilinear polyhedra, L. Stupelis - Neumann problem in a plane angle,N. Grachev and V. Maz'ya [130, 131] - Neumann problem in a polyhedralcone, Y. Saito - the limiting equation for Neumann Laplacians on shrink-ing domains [348], V. Maz'ya and J. Rossmann [288] - [290] - the mixedproblem in a polyhedral domain. J. Banasiak [30] investigated the elliptic


transmission problem for Laplacian in plane domains with curvilinear poly-gons as its boundaries. New elliptic regularity results for polyhedral Laplaceinterface problems for anisotropic materials are established by V. Maz'ya,J. Elschner, J. Rehberg and G. Schmidt [259].

CHAPTER 4

Strong solutions of the Dirichlet problem for linear

equations

4.1. The Dirichlet problem in general domains

Let G ⊂ RN be a bounded domain. We consider the following Dirichletproblem

Lu := aij(x)Diju(x) + ai(x)Diu(x) + a(x)u(x) = f(x), in G,u(x) = ϕ(x) on ∂G,

(L)

where the coecients aij(x) = aji(x) and satisfy the uniform ellipticity con-dition

ν|ξ|2 ≤ aij(x)ξiξj ≤ µ|ξ|2 ∀ξ ∈ RN , x ∈ Gwith the ellipticity constants ν, µ > 0.

Let us recall some well known facts about W 2,p(G) solutions of thisproblem.

Theorem 4.1. (Unique solvability) [128, Theorem 9.30 and the re-mark in the end of 9.5].

Let G satisfy an exterior cone condition at every boundary point and letbe given p ≥ N. Let• aij ∈ C0(G) ∩ L∞(G), ai ∈ Lq(G), a ∈ Lp(G), i, j = 1, . . . , N,where q > N, if p = N, and q = N, if p > N ;• ∀x ∈ G : a(x) ≤ 0,• f ∈ Lp(G), ϕ ∈ C0(∂G).Then the boundary value problem (L) has a unique solution

u ∈W 2,ploc (G) ∩ C0(G).

Theorem 4.2. [128, Theorem 9.1] (Alexandrov's Maximum Prin-

ciple) Let u ∈ W 2,Nloc (G) ∩ C0(G) satisfy the boundary value problem (L).

Furthermore let

•(

N∑i=1|ai|2

)1/2

, f ∈ LN (G),

• ∀x ∈ G : a(x) ≤ 0.Then

supGu ≤ sup

∂Gu+ + c‖f‖LN (G),

89

90 4 Strong solutions of the Dirichlet problem for linear equations

where c depends only on N, ν, diamG and

∥∥∥∥( N∑i=1|ai|2

)1/2∥∥∥∥LN (G)

.

Theorem 4.3. The E. Hopf strong maximum principle (see The-orems 9.6, 3.5 [128]).

Let L be elliptic in the domain G and ai(x), i = 1, . . . , N ;a(x) ∈ L∞loc(G), a(x) ≤ 0. If u ∈ W 2,N

loc (G) satises L[u] ≥ 0(≤ 0) in G,then u cannot achieve a nonnegative maximum (non-positive minimum) inG unless it is a constant.

Applying the Alexandrov Maximum Principle to the dierence of twofunctions we obtain the following Comparison Principle.

Theorem 4.4. (Comparison principle) Let L be elliptic in G, let(N∑i=1|ai|2

)1/2

, f ∈ LN (G), ∀x ∈ G : a(x) ≤ 0 and u, v ∈W 2,Nloc (G) ∩ C0(G)

with Lu ≥ Lv in G and u ≤ v on ∂G. Then u ≤ v throughout G.

Theorem 4.5. [128, Theorem 9.26], [380] (Local maximum prin-ciple) Let G be a bounded domain with subdomains T,G′ such that T ⊂G′ ⊂ G and suppose that ai ∈ Lq(G), q > N and a ∈ LN (G). Letu ∈ W 2,N (G) ∩ C0(G) satisfy Lu ≥ f in G and u ≤ 0 on T ∩ ∂G wheref ∈ LN (G′). Then for any p > 0, we have

supTu ≤ c

‖f‖LN (G′) + ‖u‖Lp(G′)

,

where the constant c depends only on N,µ, ν, p, ‖ai‖q,G′ , ‖a‖N,G′ , T,G′, G.

Theorem 4.6. [128, Theorem 9.13] (Lpestimate) Let G be a bound-ed domain in RN and T ⊂ ∂G be of the class C1,1. Furthermore, let u ∈W 2,p(G), 1 N if p ≤ N and q = p if p > N,• a ∈ Lr(G), where r > N/2 if p ≤ N/2 and r = p if p > N/2.

Then, for any domain G′ ⊂⊂ G ∪ T,

‖u‖W 2,p(G′) ≤ c(‖u‖Lp(G) + ‖f‖Lp(G)

),(4.1.1)

where c depends only on N, p, ν, µ, T,G′, G, the moduli of continuity of the

coecients aij on G′ and on

∥∥∥∥( N∑i=1|ai|2

)1/2∥∥∥∥Lq(G)

, ‖a‖Lr(G).

Theorem 4.7. [4, Theorem 15.2] Let G be a bounded domain of classCk with k ∈ N, k ≥ 2 and suppose that the coecients of the operator Lbelong to Ck−2(G) and have Ck−2− norms bounded by K. Let u be a solution

4.1 The Dirichlet problem in general domains 91

of (L) with f ∈ W k−2,p(G) and ϕ ∈ W k−1/p,p(∂G). Then u ∈ W k,p(G) andthe following estimate is valid

‖u‖Wk,p(G) ≤ c‖f‖Wk−2,p(G) + ‖ϕ‖Wk−1/p,p(∂G) + ‖u‖Lp(G)

,

where c depends only on ν, µ,K, k, p, the domain G, and the modulus ofcontinuity of the leading coecients of L.

By use of a suitable cuto function we obtain the following localizedversion of the above theorem.

Theorem 4.8. [4, Theorem 15.3] Let G be a bounded domain of classCk with subdomains T,G′ such that T ⊂ G′ ⊂ G. We suppose that thecoecients of the operator L belong to Ck−2(G) with k ∈ N, k ≥ 2. Let u be

a solution of (L) with f ∈ W k−2,p(G′) and ϕ ∈ W k−1/p,p(∂G′ ∩ ∂G). Thenu ∈W k,p(T ) and the following estimate is valid

‖u‖Wk,p(T ) ≤ c‖f‖Wk−2,p(G′) + ‖ϕ‖Wk−1/p,p(∂G′∩∂G) + ‖u‖Lp(G′)

.

The more strong result is valid for the case N = 2; it is the Bernsteinestimate (see in detail 19 Chapter III, the inequality (19.20) [214]):

Theorem 4.9. Let G ⊂ R2 be a bounded domain and G′ ⊂⊂ G\O be anysubdomain with a W 2,p, p > 2 boundary portion T = (∂G′ ∩ ∂G) ⊂ ∂G \ O.Let u ∈ W 2(G) be a strong solution of the equation aij(x)Diju(x) = f(x)in G′ with u = 0 on T in the sense of W 1(G). Let the equation satises theuniform ellipticity condition with the ellipticity constants ν, µ. Then, for anysubdomain G′′ ⊂⊂ G′ ∪ T :

‖u‖2W 2(G′′) ≤ C∫G′

(u2 + f2

)dx,

where C depends on ν, µ, p, T,G′′, G′.

At last, we cite one Theorem about local gradient bound for uniformlyelliptic equations in general form with two variables.

Theorem 4.10. [213, Theorem 17.4], [214, Theorem 19.4].Let G ⊂ R2 be a bounded domain and G′ ⊂⊂ G \ O be any subdomain witha W 2,p, p > 2 boundary portion T = (∂G′ ∩ ∂G) ⊂ ∂G \ O. Let u ∈W 2(G′)be a strong solution of the problem (L) in G′, where L is uniformly elliptic,satisfying the inequality

‖ai(x), a(x), f(x)‖Lp(G′) ≤ µ1; ‖ϕ(x)‖W 2,p(G′) ≤ µ1.

Then for any subdomain G′′ ⊂⊂ G′∪T there is a constant M1 > 0 dependingonly on ν, µ, µ1, p, ‖u‖2,G′ , ‖ϕ‖W 2,p(G′) and G

′, G′′, T such that

supG′′|∇u| ≤M1.


4.2. The Dirichlet problem in a conical domain

In the following part of this chapter we denote by G ⊂ RN a boundeddomain with a conical point in O as described in Section 1.3.

Definition 4.11. A (strong) solution of the Dirichlet problem (L) inG is a function u ∈ W 2(Gε) ∩ C0(G), ∀ε > 0 which satises the equationsLu = f for almost all x ∈ G and the boundary condition u = ϕ for allx ∈ ∂G.

In the following we will always assume that the coecients aij(x), ai(x)and a(x) satisfy the following conditions:

A1) Uniform ellipticity condition:

ν|ξ|2 ≤ aij(x)ξiξj ≤ µ|ξ|2 ∀ξ ∈ RN , x ∈ G

with some ν, µ > 0.A2) aij(0) = δji .

A3) aij ∈ C0(G), ai ∈ Lp(G), a ∈ Lp/2(G), p > N.A4) There exists a monotonically increasing nonnegative function A such

that N∑i,j=1

|aij(x)− aij(y)|21/2

≤ A(|x− y|),

|x|

(N∑i=1

ai2(x)

)1/2

+ |x|2|a(x)| ≤ A(|x|)

for x, y ∈ G.

Remark 4.12. The Assumption A4) guarantees that the coecients ai

and a are bounded on G \Bε(0) for every ε > 0.

4.3. Estimates in weighted Sobolev spaces

Theorem 4.13. Let u be a solution of (L) and let λ be the smallestpositive eigenvalue of (EV P1). Suppose that

limr→+0

A(r) = 0(4.3.1)

and that f ∈ W0α(G), ϕ ∈ W3/2

α (∂G) ∩ C0(∂G), where

4−N − 2λ < α ≤ 2.(4.3.2)

Then u ∈ W2α(G) and

‖u‖ W

2

α(G)≤ c

(‖u‖L2(G) + ‖f‖

W0

α(G)+ ‖ϕ‖

W3/2

α (∂G)

),

4.3 Estimates in weighted Sobolev spaces 93

where c > 0 depends only on ν, µ, α, λ, N, maxx∈GA(|x|), G. Furthermore, if

N < 4, there exists real constant c2 independent of u, such that

|u(x)| ≤ c2|x|(4−N−α)/2, ∀x ∈ Gd0(4.3.3)

for some d > 0.

Proof. Let Φ ∈ W2α(G)∩C0(G) be an arbitrary extension of the bound-

ary function ϕ into G. The function v = u−Φ then satises the homogeneousDirichlet problem

aij(x)Dijv(x) + ai(x)Div(x) + a(x)v(x) = F (x) in G,v(x) = 0 on ∂G,

(L)0

where

F (x) = f(x)−(aij(x)DijΦ(x) + ai(x)DiΦ(x) + a(x)Φ(x)

).(4.3.4)

Since aij(0) = δji , we have

(4.3.5) ∆v(x) = F (x)−(aij(x)− aij(0)

)Dijv(x)−− ai(x)Div(x)− a(x)v(x) in G.

Case I: 4−N ≤ α ≤ 2.Integrating by parts we show that∫Gε

rα−2v∆vdx = −εα−2

∫Ωε

v∂v

∂rdΩε −

∫Gε

⟨∇v,∇rα−2v

⟩dx =

= −εα−2

∫Ωε

v∂v

∂rdΩε −

∫Gε

rα−2|∇v|2dx+

+ (2− α)∫Gε

rα−4v⟨x,∇v

⟩dx.

Integrating again by parts we obtain∫Gε

rα−4v⟨x,∇v

⟩dx =

12

∫Gε

⟨rα−4x,∇v2

⟩dx

= −12εα−3

∫Ωε

v2dΩε −12

∫Gε

v2N∑i=1

Di(rα−4xi)dx

= −12εα−3

∫Ωε

v2dΩε −N + α− 4

2

∫Gε

rα−4v2dx

becauseN∑i=1

Di(rα−4xi) = Nrα−4 + (α− 4)rα−5N∑i=1

x2i

r= (N + α− 4)rα−4.


Thus, multiplying both sides of (L)0 by rα−2v(x) and integrating overGε, we obtain

(4.3.6) εα−2

∫Ωε

v∂v

∂rdΩε +

∫Gε

rα−2|∇v|2dx+2− α

2εα−3

∫Ωε

v2dΩε +

+2− α

2(N + α− 4)

∫Gε

rα−4v2dx =

=∫Gε

rα−2v(−F (x) +

(aij(x)− aij(0)

)Dijv(x) +

+ ai(x)Div(x) + a(x)v(x))dx.

Let us estimate in the above equation the integrals over Ωε. To this endwe consider the function

M(ε) = maxx∈Ωε

|v(x)|.

Since v ∈ C0(G) and v = 0 on ∂G, we have

limε→+0

M(ε) = 0.

Lemma 4.14.

limε→+0

εα−2

∫Ωε

v∂v

∂rdΩε = 0, ∀α ∈ [4−N, 2].(4.3.7)

Proof. We consider the set G2εε and we have Ωε ⊂ ∂G2ε

ε . Now we usethe inequality (1.6.1) ∫

Ωε

|w|dΩε ≤ c∫G2εε

(|w|+ |∇w|)dx.

Setting w = v ∂v∂r we nd

|w|+ |∇w| ≤ c(r2v2xx + |∇v|2 + r−2v2)

. Therefore we get∫Ωε

∣∣∣∣v∂v∂r∣∣∣∣ dΩε ≤ c

∫G2εε

(r2v2xx + |∇v|2 + r−2v2)dx.(4.3.8)

Let us now consider the sets G5/2εε/2 and G2ε

ε ⊂ G5/2εε/2 and new variables

x′ dened by x = εx′. Then the function w(x′) = v(εx′) satises in G21/4 the

equation

aij(εx′)∂2w

∂x′i∂x′j

+ εai(εx′)∂w

∂x′i+ ε2a(εx′)w = ε2F (εx′).(4.3.9)


Applying the L2estimate (4.1.1) for the solution w in G21/4 we get∫

G21

(|D′2w|2 + |∇′w|2

)dx′ ≤ c

∫G

5/21/2

(ε4F 2(εx′) + w2

)dx′,(4.3.10)

where c > 0 depends only on ν, µ,G, and maxx′∈G5/2

1/2

A(|x′|); here:

|D′2w|2 =N∑

i,j=1

∣∣∣∣∣ ∂2w

∂x′i∂x′j

∣∣∣∣∣2

, |∇′w|2 =N∑i=1

∣∣∣∣ ∂w∂x′i∣∣∣∣2 .

Returning to the variable x, we obtain∫G2εε

(r2|D2v|2 + |∇v|2 + r−2v2

)dx ≤ c

∫G

5/2εε/2

(r2F 2 + r−2v2

)dx.(4.3.11)

By the Mean Value Theorem 1.58 with regard to v ∈ C0(G) we have

∫G

5/2εε/2

r−2v2dx =

5/2ε∫ε/2

rN−3

∫Ω

v2(r, ω)dΩdr

≤ 2ε(θ1ε)N−3

∫Ω

v2(θ1ε, ω)dΩ

≤ 2εN−2θN−31 M2(θ1ε) meas Ω

(4.3.12)

for some 12 < θ1 <

52 .

From (4.3.8), (4.3.11) and (4.3.12) we obtain

(4.3.13)∫Ωε

∣∣∣∣v∂v∂r∣∣∣∣ dΩε ≤ c1ε

N−2M2(ε) + c2

∫G

5/2εε/2

r2F 2dx ≤

≤ c1εN−2M2(ε) + c3ε

2−α∫

G5/2εε/2

rαF 2dx, ∀α ≤ 2.

Hence we obtain as well

εα−2

∫Ωε

∣∣∣∣v∂v∂r∣∣∣∣ dΩε ≤ c1ε

α+N−4M2(ε) + c3

∫G

5/2εε/2

rαF 2dx, ∀α ≤ 2.(4.3.14)


By the assumption A4) and hypotheses of our Theorem we have that F ∈W

0α(G), hence

limε→+0

∫G

5/2εε/2

rαF 2dx = 0(4.3.15)

and thus from (4.3.14) with regard to that v(0) = 0 we deduce the validityof statement (4.3.7) of our Lemma.

Further, we get by the Cauchy inequality∫Gε

rα−2v(x)F (x)dx =∫Gε

(rα/2−2v(x))(rα/2F (x))dx

≤ δ

2

∫Gε

rα−4v2dx+12δ

∫Gε

rαF 2(x)dx(4.3.16)

for arbitrary δ > 0. Applying Assumption A4) together with the Hölder andthe Cauchy inequality

(4.3.17) rα−2v((aij(x)− aij(0)

)Dijv(x) + ai(x)Div(x) + a(x)v(x)

)≤ A(r)

((r

α2 |D2v|)(r

α2−2v) + rα−2|∇v|(r−1v) + rα−4v2

)≤ A(r)

(rα|D2v|2 + rα−2|∇v|2 + 2rα−4v2

).

Finally, from (4.3.6) - (4.3.17) we obtain

(4.3.18)∫Gε

rα−2|∇v|2dx+2− α

2(N + α− 4)

∫Gε

rα−4v2dx

≤ εα−2

∫Ωε

v∂v

∂rdΩε +

δ

2

∫Gε

rα−4v2dx+

+12δ

∫Gε

rαF 2(x)dx+∫Gε

A(|x|)(rα|D2v|2 + rα−2|∇v|2 + 2rα−4v2

)dx

for all δ > 0.

Let us now estimate the last integral in (4.3.18). Due to the assumption(4.3.1) we have

∀δ > 0 ∃d > 0 such that A(r) < δ for all 0 < r < d.(4.3.19)

Let 4ε < d. From (4.3.11), (4.3.12) follows that∫G3εε

rα|D2v|2dx ≤ c4εα+N−4 + c3

∫G

7/2εε/2

rαF 2dx,


and consequently

∫Gε

A(r)rα|D2v|2dx =∫G3εε

A(r)rα|D2v|2dx+∫Gd3ε

A(r)rα|D2v|2dx+

+∫Gd

A(r)rα|D2v|2dx ≤ c4A(3ε)εα+N−4 +

+ c3A(3ε)∫

G7/2εε/2

rαF 2dx+ δ

∫G2dε

(rαF 2(x) + rα−4v2)dx

+c5 maxr∈[d,diamG]

A(r)∫Gd

|D2v|2dx

for all δ > 0 and 0 < ε < d/4. Here, c5 depends only on α, d and diamG.

Applying all this estimates to the inequality (4.3.18) we obtain

(4.3.20)∫Gε

rα−2|∇v|2dx+2− α

2(N + α− 4)

∫Gε

rα−4v2dx

≤ εα−2

∫Ωε

v∂v

∂rdΩε + c6A(3ε)

(εα−N−4 +

∫G

7/2εε/2

rαF 2dx

)+

+ δ

∫Gε

(rα−2|∇v|2 + rα−4v2)dx+ c7

∫Gd

|D2v|2dx+ c8

∫G

rαF 2(x)dx

for all δ > 0 and 0 < ε < d/4.Finally, we apply Theorem 4.6 to the solution v of (L)0 in Gd

∫Gd

|D2v|2dx ≤ c9

∫Gd/2

(v2 + f2 + |aijDijΦ + aiDiΦ + aΦ|2

)dx

≤ c9

∫Gd/2

(v2 + f2)dx+ c10‖ϕ‖2W 3/2,2(∂G\Γd/20 )

.(4.3.21)

Furthermore, if (2−α)(N +α−4) = 0, then we apply the inequality (2.5.2).


Now, let δ > 0 be small enough and d > 0 chosen according to (4.3.19).Then we obtain from (4.3.20) the estimate∫Gε

(rα|D2v|2 + rα−2|∇v|2 + rα−4v2

)dx ≤ εα−2

∫Ωε

v∂v

∂rdΩε +

+ c11A(3ε)(εα+N−4 +

∫G

7/2εε/2

rαF 2dx

)+ c12

(‖v‖2L2(G) + ‖f‖2

W0

α(G)+

+ ‖ϕ‖2W

3/2

α (∂G)

)where the constants c11 and c12 do not depend on ε. Letting ε→ +0, applyingLemma 4.14 and noting that

‖u‖ W

2

α(G)≤ ‖v‖

W2

α(G)+ ‖ϕ‖

W3/2

α (∂G)

we obtain the assertion of our Theorem in the case I.

Case II: 4−N − 2λ < α < 4−N.Due to the embedding theorem (see Lemma 1.37) we have

f ∈ W04−N (G), ϕ ∈ W3/2

4−N (∂G) ∩ C0(∂G).

Therefore, by case I, u ∈ W24−N (G) and∫

G

(r4−N |D2u|2 + r2−N |∇u|2 + r−Nu2

)dx ≤ const.

According to (4.3.10) with % = 2−kd, k = 0, 1, 2, . . . , we have∫G

3/21/2

(|D′2w|2 + |∇′w|2

)dx′ ≤ c13

∫G2

1/4

(2−4kd4F 2(x′2−kd) + w2

)dx′.

Multiplying both sides of this inequality by (2−kd+ε)α−2 with ε > 0, takinginto account that

2−k−1d+ ε < r + ε < 2−kd+ ε in G(k)

and returning to the variables x we obtain∫G(k)

r2(r + ε)α−2|D2v|dx ≤ c13

∫G(k−1)∪Gk∪G(k+1)

(r2(r + ε)α−2F 2 + r−2(r + ε)α−2v2

)dx.

Since rε ≤ r + ε ≤ 2hrε in G with h dened as in Lemma 1.11, we obtain

(4.3.22)∫G(k)

r2rα−2ε |D2v|dx ≤ c14

∫G(k−1)∪Gk∪G(k+1)

(r2rα−2

ε F 2 +

+ r−2rα−2ε v2

)dx.


Summing up the inequalities (4.3.22) for k = 0, 1, 2, . . . , we nally obtain∫Gd0

r2rα−2ε |D2v|dx ≤ c14

∫G2d

0

(rαF 2 + r−2rα−2

ε v2)dx,(4.3.23)

since α ≤ 2 and rε ≥ hr.Let us return back to the equation (L)0. Multiplying its both sides by

rα−2ε v and integrating by parts twice we obtain (compare with case I)

(4.3.24)∫G

rα−2ε |∇v|2dx =

2− α2

(4−N − α)∫G

rα−4ε v2dx+

+∫G

rα−2ε v

((aij(x)− aij(0)

)Dijv(x) + ai(x)Div(x) + a(x)v(x)

)dx−

−∫G

rα−2ε vF (x)dx.

By assumption A4) we obtain with the help of the Cauchy and the Hölderinequalities and the properties of the quasidistance rε

rα−2ε v

N∑i,j=1

(aij(x)− aij(0)

)Dijv(x) +

N∑i=1

ai(x)Div(x) + a(x)v(x)

≤ c(h)A(r)

(rα−2ε r2|D2v|2 + rα−2

ε |∇v|2 + rα−2ε r−2v2

)and

rα−2ε vF (x) ≤ δ

2rα−2ε r−2v2 + c(δ, h)rαF 2, ∀δ > 0.

Decomposing G into G = Gd0 ∪Gd, we then obtain from (4.3.24)∫G


2− α2

(4−N − α)∫G

rα−4ε v2dx

+c(h)A(d)∫Gd0

(rα−2ε r2|D2v|2 + rα−2

ε |∇v|2 + rα−2ε r−2v2

)dx

+δ

2

∫Gd0

rα−2ε r−2v2dx+ c15

∫Gd

(|D2v|2 + v2

)dx

+c(δ, h)∫G

rαF 2(x)dx =: J1 + J2 + J3 + J4 + J5

with an arbitrary δ > 0. Let us further estimate the right hand side of thisinequality.


By the inequality (2.5.7) - (2.5.9),

J1 ≤2− α

2(4−N − α)Hε(λ,N, α)

∫G


Thus

Cε(λ,N, α)∫G

rα−2ε |∇v|2dx ≤ J2 + J3 + J4 + J5,

where

Cε(λ,N, α) = 1− 2− α2

(4−N − α)Hε(λ,N, α).

The integrals J2, J3, J4 and J5 can be estimated using (4.3.23), (4.3.21)and Lemma 2.33 . In this way we obtain

C(λ,N, α)∫G

rα−2ε |∇v|2dx ≤ c16 [A(d) + δ +O(ε)]

∫G

rα−2ε |∇v|2dx

+ c17

(‖v‖L2(G) + ‖f‖

W0

α(G)+ ‖ϕ‖

W3/2

α (∂G)

),

where

C(λ,N, α) = 1− 2− α2

(4−N − α)H(λ,N, α) > 0

due to assumption (4.3.2). Choosing δ > 0 and d > 0 small enough andpassing to the limits as ε→ 0, by the Fatou Theorem we obtain the assertion,if we recall (4.3.23).

The estimate (4.3.3) follow directly from Lemma 1.38.

Remark 4.15. On the belonging of weak solutions to W 2(G). Sup-pose that all assumptions of Theorem 4.13 are fullled with

aij(x) = δji , x ∈ G, ∀i, j = 1, . . . , N ; f ∈ L2(G), ϕ ∈ W3/2(∂G)∩C0(∂G).

We want study the regularity of a weak solution u ∈ W 1(G). The followingstatement is valid:

Proposition 4.16. A weak solution u ∈ W 1(G) belongs to W 2(G), ifeither• N ≥ 4;

or• N = 2 and 0 < ω0 < π;

or• N = 3 and the domain G is convex;

or• N = 3 and Ω ⊂ Ω0 = (ϑ, ϕ)

∣∣0 < |ϑ| < ϑ0; 0 < ϕ < 2π, where ϑ0 isthe smallest positive root of the Legendre function P 1

2(cosϑ).


Proof. We apply Theorem 4.13 with α = 0. Since λ > 0, then forN ≥ 4, α = 0 the assumption (4.3.2) of Theorem 4.13 is fullled andtherefore we have

u ∈ W20(G) =⇒ u ∈W 2(G).(4.3.25)

If N = 2 and 0 < ω0 < π, then the assumption (4.3.2) with α = 0 of Theorem4.13 is fullled too, since in this case we have that λ > 1. Therefore we haveagain (4.3.25).

Let now N = 3. If G is a convex domain, then it is well known (see e.g.Theorem 3 2 chapter VI [86]) that λ > 1. Then the assumption (4.3.2) withα = 0 of Theorem 4.13 is fullled and therefore (4.3.25) is valid. Let G ⊂ R3

be any domain and denote by Ω0 ⊂ S2 the domain, in which the problem(EV P1) is solvable for λ = 1

2 :4ωψ + 12

(1 + 1

2

)ψ = 0, ω ∈ Ω0,

ψ∣∣∣∂Ω0

= 0.

Right now the assumption (4.3.2) with α = 0 of Theorem 4.13 is fullled, ifλ > 1

2 . Again in virtue of the monotony Theorem 3 2 chapter VI [86] wehave Ω ⊂ Ω0. Let us reduce to the eigenvalue problem above. We shall lookfor the particular solution in the form ψ = ψ(ϑ). Then ψ(ϑ) is a solution ofthe Sturm - Liouville problem

1sinϑ ·

ddϑ

(sinϑdψdϑ

)+ 3

4ψ = 0, |ϑ| ≤ ϑ0,

ψ(−ϑ0) = ψ(ϑ0) = 0.

A solution of the equation of this problem is the Legendre function of rstgenus ψ(ϑ) = P1/2(cosϑ). This function has precisely one root on the interval(0, π) (see e.g. example 39, page 158 [401]); we denote it by ϑ0.

Theorem 4.17. Let u(x) be a strong solution of problem (L) and as-sumptions A1) - A4) are satised with A(r) Dini-continuous at zero. Sup-pose

ϕ(x) ∈W

3/2

4−N−2λ(∂G),

(4.3.26)∫∫G

r4−N−2λH−1(r)f2(x)dx+∫∂G

r1−N−2λH−1(r)ϕ2(x)dσ <∞,

where H(r) is a Dini-continuous at zero, monotone increasing function, λ isthe smallest positive eigenvalue of problem (EV P1) with (2.4.8).


Then u(x) ∈W

2

4−N (G) and

(4.3.27) ‖u‖2W

2

4−N (G%0)≤ C%2λ

(‖u‖22,G +

∫∫G

r4−N−2λH−1(r)f2(x)dx+

+∫∂G

r1−N−2λH−1(r)ϕ2(x)dσ + ‖ϕ‖2W

3/2

4−N−2λ(∂G)

), 0 < % < d,

where the constant C > 0 depends only on ν, µ, d,A(d),H(d), N, λ, measG,

and on the quantitiesd∫0

A(r)r dr,

d∫0

H(r)r dr.

Proof. Since u ∈ W

24−N (G) due to Theorem 4.13, it remains to prove

(4.3.27). Let

U(%) :=∫G%0

r2−N |∇u|2dx.

We write the equation (L) in the form

∆u(x) = f(x)−(aij(x)− aij(0)

)Diju(x)− ai(x)Diu(x)− a(x)u(x),

multiply both sides by r2−Nu and integrate over G%0, % ∈ (0, d). As a resultwe obtain

(4.3.28) U(%) =∫Γ%0

r2−Nϕ(x)∂u

∂ndσ +

∫Ω

(%u∂u

∂r+N − 2

2u2

)dΩ

+∫G%0

r2−Nu(x)((aij(x)− aij(0)

)Diju(x) +

+ ai(x)Diu(x) + a(x)u(x)− f(x))dx.

We will estimate each integral on the right hand side of this equation fromabove. From Lemma 1.41 and Lemma 1.40 it follows by Cauchy's inequality∫Γ%0

r2−Nϕ(x)∂u

∂ndσ =

∫Γ%0

(√H(r)r(3−N)/2 ∂u

∂n

)·(H−1/2(r)r(1−N)/2ϕ(x)

)dσ

≤ H(%)2

∫Γ%0

r3−N(∂u

∂n

)2

dσ +12

∫Γ%0

H−1(r)r1−Nϕ2(x)dσ(4.3.29)

≤ c1H(%)‖u‖2W

2

4−N (G%0)+ c1

∫Γ%0

H−1(r)r1−Nϕ2(x)dσ.


Moreover, as in the proof of Theorem 4.13 we have∫G%0


)Diju(x) + ai(x)Diu(x) + a(x)u(x)

)dx ≤

≤ A(%)∫G%0

(r4−N |D2u|2 + r2−N |∇u|2 + 2r−Nu2

)dx(4.3.30)

and∫G%0

r2−Nu(x)f(x)dx =∫G%0

(√H(r)r−N/2u(x)

)·(H−1/2(r)r2−N/2f(x)

)dx

≤ H(%)2

∫G%0

r−Nu2(x)dx+12

∫G%0

H−1(r)r4−Nf2(x)dx.(4.3.31)

Therefore, using (4.3.29)(4.3.31) and Corollary 2.30, (2.5.10) from Corollary2.26 we obtain from (4.3.28) the inequality

U(%) ≤ %

2λU ′(%) + ε(%)

∫G%0

r4−N |D2u|2dx+ δ(%)U(%) + F(%),

where

(4.3.32)

ε(%) = A(%) + c1H(%),δ(%) = c2(λ,N)(A(%) +H(%)),F(%) = c1

∫Γ%0

H−1(r)r1−Nϕ2(x)dσ+

+12

∫G%0

H−1(r)r4−Nf2(x)dx+

+c2(λ,N)(A(%) +H(%)

)‖ϕ‖2

W3/2

4−N (Γ%0).

Let us now estimate∫G%0

r4−N |D2u|2dx. To this end we consider again the

estimate (4.3.11) with ε replaced by 2−k%. Summing up this inequalities fork = 0, 1, . . . , we obtain∫G%0

r4−N |D2u|2dx ≤ c3

∫G2%

0

(r4−NF 2(x) + r−Nu2

)dx+ c4‖ϕ‖2

W3/2

4−N (Γ2%0 ).

Inserting the denition (4.3.4) of F and applying (2.5.2) we then obtain

(4.3.33)∫G%0

r4−N |D2u|2dx ≤ c5

(U(2%) + ‖f‖2

W0

4−N (G2%0 )

+

+ ‖ϕ‖2W

3/2

4−N (Γ2%0 )

), 0 < % < d


and therefore

(4.3.34) U(%) ≤ %

2λU ′(%) + c5ε(%)U(2%) + δ(%)U(%) + F(%) +

+ c5ε(%)(‖f‖2

W0

4−N (G2%0 )

+ ‖ϕ‖2W

3/2

4−N (Γ2%0 )

).

Moreover we have the initial condition (see the proof of Theorem 4.13):

U(d) =∫Gd0

r2−N |∇u|2dx ≤ c(‖u‖2L2(G) + ‖f‖2

W0

4−N (G)+ ‖ϕ‖2

W3/2

4−N (∂G)

)≡ V0.

From (4.3.34) we obtain the dierential inequality (CP ) from 1.10 with

P(%) = 2λ

%

(1− δ(%)

);

N (%) = 2λ% c5ε(%);

Q(%) = 2λ% F(%) + c6

ε(%)%

(‖f‖2

W0

4−N (G2%0 )

+ ‖ϕ‖2W

3/2

4−N (Γ2%0 )

).

(4.3.35)

Now we apply Theorem 1.57. For this we have:

2%∫%

P(s)ds = 2λ ln 2− 2λ

2%∫%

δ(s)sds ≤ 2λ ln 2⇒

exp( 2%∫%

P(s)ds)

= 22λ exp(−2λ

2%∫%

δ(s)sds

)≤ 22λ.(4.3.36)

Further,

B(%) = N (%) exp( 2%∫%

P(s)ds)≤ 22λ 2λ

%c5ε(%);⇒

d∫%

B(τ)dτ ≤ 2λ22λc5

d∫0

ε(τ)τdτ.(4.3.37)


In addition

d∫%

P(s)ds = 2λ lnd

%− 2λ

d∫%

δ(s)sds⇒

exp(−

d∫%

P(s)ds)≤(%

d

)2λ

exp(

2λ

d∫0

δ(s)sds

)≤ c7

(%

d

)2λ

.

(4.3.38)

exp(−∫%

τP(s)ds)≤(%

τ

)2λ

exp(

2λ

d∫0

δ(s)sds

)≤ c7

(%

τ

)2λ

,

0 < % < τ < d.

Now by Theorem 1.57 from (1.10.1) by virtue of (4.3.38), and (4.3.37) weobtain:

U(%) ≤ c8%2λ

V0 +

d∫%

τ−2λQ(τ)dτ,(4.3.39)

where c8 is a positive constant depending only on N,λ,d∫0

A(s)+H(s)s ds. We

have now to estimated∫%τ−2λQ(τ)dτ. For this we recall (4.3.35) and therefore

we obtain

(4.3.40)

d∫%

τ−2λQ(τ)dτ ≤ 2λ

d∫%

τ−2λ−1F(τ)dτ +

+ c6ε(d)

d∫%

τ−2λ−1

(‖f‖2

W0

4−N (G2τ0 )

+ ‖ϕ‖2W

3/2

4−N (Γ2τ0 )

)dτ

Now, by changing the order of integration in virtue of the Fubini Theoremin the integral

d∫%

τ−2λ−1( τ∫

0

rαK(r)dr)dτ =

%∫0

rαK(r)( d∫%

τ−2λ−1dτ)dr +

+

d∫%

rαK(r)( d∫r

τ−2λ−1dτ)dr =

%−2λ − d−2λ

2λ

%∫0

rαK(r)dr +


+1

2λ

d∫%

rαK(r)(r−2λ − d−2λ

)dr ≤ 1

2λ

%∫0

rα%−2λK(r)dr +

+1

2λ

d∫%

rα−2λK(r)dr ≤ 12λ

d∫0

rα−2λK(r)dr

we nd

1)

d∫%

τ−2λ−1

(∫∫Gτ0

r4−NH−1(r)f2(x)dx)dτ ≤

≤ 12λ

∫∫Gd0

r4−N−2λH−1(r)f2(x)dx.

2)

d∫%

τ−2λ−1

(∫Γτ0

r1−NH−1(r)ϕ2(x)dσ)dτ ≤

≤ 12λ

∫Γd0

r1−N−2λH−1(r)ϕ2(x)dσ.

In the same way we nd

3)

d∫%

τ−2λ−1‖ϕ‖2W

3/2

4−N (Γ2τ0 )

dτ ≤ 12λ‖ϕ‖2

W3/2

4−N−2λ(Γ2d0 )

.

4)

d∫%

τ−2λ−1‖f‖2W

0

4−N (G2τ0 )dτ ≤ 1

2λ‖f‖2

W0

4−N−2λ(G2d0 )

Hence and from (4.3.39), (4.3.35) it follows:

(4.3.41) U(%) ≤ C%2λ

(‖u‖22,G +

∫∫G

r4−N−2λH−1(r)f2(x)dx+

+ ‖ϕ‖2W

3/2

4−N−2λ(∂G)

+∫∂G

r1−N−2λH−1(r)ϕ2(x)dσ

), 0 < % < d.

At last we apply (4.3.33) and deduce from (4.3.41) the validity of (4.3.27).

Theorem 4.18. Let u(x) be a strong solution of problem (L) and as-sumptions A1) - A4) are satised with A(r) Dini-continuous at zero. Sup-pose

f ∈W

0

4−N (G) ϕ(x) ∈W

3/2

4−N (∂G) ∩ C0(∂G)


and there exist real numbers s > 0, ks ≥ 0 such that

ks =: sup%>0

%−s(‖f‖

W0

4−N (G%0)+ ‖ϕ‖

W3/2

4−N (Γ%0)

).(4.3.42)

Then there are d ∈ (0, 1e ) and a constant C > 0 depends only on ν,

µ, d,A(d), N, s, λ,measG, and on the quantityd∫0

A(r)r dr, such that ∀% ∈ (0, d)

(4.3.43) ‖u‖ W

2

4−N (G%0)≤ C

(‖u‖2,G + ‖f‖

W0

4−N (G)+ ‖ϕ‖

W3/2

4−N (∂G)+ ks

)×

×

%λ, if s > λ,

%λ ln3/2(

1%

), if s = λ,

%s, if s < λ.

Proof. We consider the function v = u−Φ as a solution of homogeneousproblem (L)0 in the form (4.3.5) with (4.3.4). Multiplying both sides of(4.3.5) by r2−Nv and integrating over G%0, we obtain

(4.3.44)∫G%0

r2−Nv4vdx = −∫G%0

(r2−N(aij(x)− aij(0)

)vvxixj +

+ r2−Naivxiv + r2−Na(x)v2

)dx+

∫G%0

r2−NvF (x)dx

Integrating by parts twice we show that∫G%0

r2−Nv4vdx =∫Ω

(%v∂v

∂r+N − 2

2v2)dΩ−

∫G%0

r2−N |∇v|2dx(4.3.45)

We dene

V (%) :=∫G%0

r2−N |∇v|2dx.

Because of A4), Corollary 2.30, (2.5.3) and the Cauchy inequality we obtainfor ∀δ > 0

(4.3.46) V (%) ≤ %

2λV ′(%) + cA(%)

∫G%0

r4−Nv2xxdx+

+ cA(%)V (%) +δ

2V (%) +

12δ

(‖f‖2

W0

4−N (G2%0 )

+ ‖ϕ‖2W

3/2

4−N (Γ2%0 )

).


If we take into account (4.3.42), by (4.3.33) we get:

(4.3.47) V (%) ≤ %

2λV ′(%) + c1A(%)V (2%) + c2

(A(%) + δ

)V (%) +

+ c31δk2s%

2s, ∀δ > 0, 0 < % < d.

1) s > λ

Choosing 2λc2δ = %ε, ∀ε > 0 we obtain from (4.3.47) the problem (CP )1.10 with

P(%) =2λ%− 2λc2

A(%)%− %ε−1; N (%) = 2λc1

A(%)%

; Q(%) = k2sc4%

2s−1−ε.

Now we haved∫%

P(τ)dτ = 2λ lnd

%− 2λc2

d∫%

A(τ)τ

dτ − dε − %ε

ε⇒

exp( 2%∫%

P(τ)dτ)≤ 22λ;

d∫%

B(τ)dτ ≤ 22λ+1λc1

d∫0

A(τ)τ

dτ ;

exp(−

d∫%

P(τ)dτ)≤(%d

)2λexp(

2λc2

d∫0

A(τ)τ

dτ

)exp(ε−1dε

)= c5

(%d

)2λ,

if we recall (1.10.2).In this case we have as well:

d∫%

Q(τ) exp(−

τ∫%

P(σ)dσ)dτ ≤ k2

sc4c5%2λ

d∫%

τ2s−2λ−ε−1dτ ≤ k2sc6%

2λ,

since s > λ.Now we apply Theorem 1.57: then from (1.10.1) by virtue of deduced

inequalities and with regard to (4.3.33) we obtain the rst statement of(4.3.43).

2) s < λ

In this case we have from (4.3.47) the problem (CP ) 1.10 with

P(%) =2λ(1− δ)

%− 2λc2

A(%)%

; N (%) = 2λc1A(%)%

;

Q(%) = k2sc8δ

−1%2s−1, ∀δ > 0.


Now similarly the case 1) we have :

exp( 2%∫%

P(τ)dτ)≤ 22λ(1−δ);

d∫%

B(τ)dτ ≤ 22λ+1c1

d∫0

A(τ)τ

dτ

exp(−

d∫%

P(τ)dτ)≤(%d

)2λ(1−δ)exp(

2λc2

d∫0

A(τ)τ

dτ

)= c9

(%d

)2λ(1−δ),

if we recall (1.10.2).In this case we have as well:

d∫%

Q(τ) exp(−

τ∫%

P(σ)dσ)dτ ≤ k2

sc10δ−1%2λ(1−δ)

d∫%

τ2s−2λ(1−δ)−1dτ ≤

≤ k2sc11%

2s,

if we choose δ ∈ (0, λ−sλ ).Now we apply Theorem 1.57: then from (1.10.1) by virtue of deduced

inequalities we obtain

V (%) ≤ c12

(V0%

2λ(1−δ) + k2s%

2s)≤ c13(V0 + k2

s)%2s,

because of chosen δ. Taking into account of (4.3.33) we deduced the thirdstatement of (4.3.43).

3) s = λ

As in the proof of Theorem 4.17 we consider the function U(%) satisfyingthe equation (4.3.28). We will estimate each integral on the right hand sideof this equation from above. From Lemma 1.41 and Lemma 1.40 it followsby the Hölder inequality for integrals∫Γ%0

r2−Nϕ(x)∂u

∂ndσ =

∫Γ%0

(r(3−N)/2 ∂u

∂n

)·(r(1−N)/2ϕ(x)

)dσ ≤

≤

∫Γ%0

r3−N(∂u∂n

)2dσ

1/2

·

∫Γ%0

r1−Nϕ2(x)dσ

1/2

≤(4.3.48)

≤ c1‖ϕ‖ W

3/2

4−N (G%0)‖u‖

W2

4−N (G%0)+ c2‖ϕ‖2

W3/2

4−N (G%0)

≤

≤ c1ks%λ‖u‖

W2

4−N (G%0)+ c2k

2s%

2λ


in virtue of the assumption (4.3.42). In the same way:∫G%0

r2−Nu(x)f(x)dx =∫G%0

(r−N/2u(x)

)·(r2−N/2f(x)

)dx ≤(4.3.49)

≤ cU1/2(%)‖f‖ W

0

4−N (G%0)≤ cks%λU1/2(%).

Moreover, as in the proof of Theorem 4.13 we have

(4.3.50)∫G%0


)Diju(x) + ai(x)Diu(x) +

+ a(x)u(x))dx ≤ A(%)‖u‖2

W2

4−N (G%0).

Therefore, using (4.3.48)(4.3.50) and Corollary 2.30 we obtain from (4.3.28)the inequality

U(%) ≤ %

2λU ′(%) +A(%)‖u‖2

W2

4−N (G%0)+ c1ks%

λ‖u‖ W

2

4−N (G%0)+

+ ks%λU1/2(%) + c2k

2s%

2λ.

Now we apply the inequality (4.3.33); then as result we obtain:

(4.3.51) U(%) ≤ %

2λU ′(%) +

(A(%) + δ(%)

)U(2%) +A(%)U(%) +

+ c3k2sδ−1(%)%2λ, ∀δ(%) > 0.

Moreover we have the initial condition (see the proof of Theorem 4.13):

U(d) =∫Gd0

r2−N |∇u|2dx ≤ c(‖u‖2L2(G) + ‖f‖2

W0

4−N (G)+ ‖ϕ‖2

W3/2

4−N (∂G)

)≡ V0.

From (4.3.47) we obtain the dierential inequality (CP ) from 1.10 with

P(%) =2λ%− 2λ

A(%)%

; N (%) = 2λA(%) + δ(%)

%;

Q(%) = 2c3λk2sδ−1(%)%2λ−1, ∀δ(%) > 0.

We choose

δ(%) =1

λ22λ+1 ln(ed%

) , 0 < % < d,


where e is the Euler number. Since according to the assumption of TheoremA(%) is Dini - continuous at zero, then we have:

exp( 2%∫%

P(τ)dτ)≤ 22λ;

exp( d∫%

B(τ)dτ)≤ exp

C(λ)

d∫0

A(τ)τ

dτ

ln(ed%

);

−d∫%

P(τ)dτ ≤ ln(%d

)2λ+ 2λ

d∫0

A(τ)τ

dτ ⇒

exp(−

d∫%

P(τ)dτ)≤(%d

)2λexp

C(λ)

d∫0

A(τ)τ

dτ

if we recall (1.10.2). In this case we have as well:

d∫%

Q(τ) exp(−

τ∫%

P(σ)dσ)dτ ≤ k2

sC(λ)%2λ

d∫%

ln(edτ

)dττ≤

≤ k2sC(λ)%2λ ln2

(ed%

).

Now we apply Theorem 1.57: then from (1.10.1) by virtue of deducedinequalities we obtain

U(%) ≤ C(V0 + k2s)%

2λ ln3 1%, 0 < % < d <

1e.(4.3.52)

Taking into account of (4.3.33) we deduced the second statement of (4.3.43).

Both the following Theorems and examples from Section 4.7 show thatassumptions about the smoothness of the coecients of (L), i.e. Dini-continuity at zero of the function A(r) from the hypothesis A4) aboveTheorems 4.17 and 4.18 are essential for their validity.

Theorem 4.19. Let u(x) be a strong solution of problem (L) and as-sumptions A1) - A4) are satised with A(r), which is a continuous at zerofunction, but not Dini-continuous at zero. Suppose

f(x) ∈W

0

4−N (G), ϕ(x) ∈W

3/2

4−N (∂G) ∩ C0(∂G)

and there exist real numbers s > 0, ks ≥ 0 such that

ks =: sup%>0

%−s(‖f‖

W0

4−N (G%0)+ ‖ϕ‖

W3/2

4−N (Γ%0)

).(4.3.53)


Then for ∀ε > 0 there are d ∈ (0, 1) and a constant Cε > 0 depends onlyon ν, µ, d, s,N, ε, λ,meas G, such that ∀% ∈ (0, d)

(4.3.54) ‖u‖ W

2

4−N (G%0)≤ Cε

(‖u‖2,G + ‖f‖

W0

4−N (G)+ ‖g‖

W1/2

4−N (∂G)+ ks

)×

×

%λ−ε, if s > λ,

%s−ε, if s ≤ λ.

Proof. As above in Theorem 4.18 we nd (4.3.47), from which by theCauchy inequality we get the problem (CP ) 1.10 with

P(%) =2λ%

(1− δ

2− C8A(%)

), ∀δ > 0; N (%) = 2λC8

A(%)%

;

Q(%) = k2sC20%

2s−1.

Therefore we have:

−d∫%

P(τ)dτ = 2λ(1− δ

2) ln

%

d+ 2λC8

d∫%

A(τ)τ

dτ.

Now we apply the mean value theorem for integrals:

d∫%

A(τ)τ

dτ ≤ A(d) lnd

%

and choose d > 0 by continuity of A(r) so that 2C8A(d) < δ. Thus we obtain

exp(−

d∫%

P(τ)dτ)≤(%d

)2λ(1−δ), ∀δ > 0

Similarly we have

exp(−

τ∫%

P(σ)dσ)≤(%τ

)2λ(1−δ), ∀δ > 0.

Further it is obviously

2%∫%

P(τ)dτ ≤ 2λ ln 2


and with regard to (1.10.2)

d∫%

B(τ)dτ ≤ 2λ22λC8

d∫%

A(τ)τ

dτ ≤ 2λ22λC8A(d) lnd

%≤ δλ22λ ln

d

%⇒

exp( d∫%

B(τ)dτ)≤(%d

)−δλ22λ

, ∀δ > 0.

Hence by (1.10.1) of Theorem 1.57 we deduce

(4.3.55) U(%) ≤(%d

)−δλ22λV0

(%d

)2λ(1−δ)+

+

d∫%

Q(τ) exp(−

τ∫%

P(σ)dσ)dτ

, ∀δ > 0.

Now we estimate the last integral:

d∫%

Q(τ) exp(−

τ∫%

P(σ)dσ)dτ

≤ k2

sC20%2λ(1−δ)

d∫%

τ2s−2λ(1−δ)−1dτ =

= k2sC20%

2λ(1−δ)d2s−2λ(1−δ) − %2s−2λ(1−δ)

2s− 2λ(1− δ)≤ k2

sC21

%2λ(1−δ), if s ≥ λ%2s, if s < λ

(4.3.56)

(in this connection we choose δ > 0 so that δ 6= λ−sλ ).

From (4.3.55) - (4.3.56) and because of (4.3.33) it follows the desiredestimate (4.3.54).

We can correct Theorem 4.19 in the case s = λ, if A(r) ∼ 1ln 1r

.

Theorem 4.20. Let u(x) be a strong solution of problem (L) and as-sumptions A1) - A4) are satised with A(r) ∼ 1

ln 1r

, A(0) = 0. Suppose

f(x) ∈W

0

4−N (G), ϕ(x) ∈W

3/2

4−N (∂G)

and there exist real number kλ ≥ 0 such that

kλ =: sup%>0

%−λ(‖f‖

W0

4−N (G%0)+ ‖ϕ‖

W3/2

4−N (Γ%0)

).(4.3.57)


Then there are d ∈ (0, 1e ) and constants C > 0, c > 0 depends only on

ν, µ, d,N, λ,measG, such that

(4.3.58) ‖u‖ W

2

4−N (G%0)≤ C

(‖u‖2,G + ‖f‖

W0

4−N (G)+ ‖ϕ‖

W3/2

4−N (∂G)+ kλ

)×

× %λ lnc+1 1%, 0 < % < d.

Proof. As above in Theorem 4.18 we obtain the problem (CP ) 1.10with

P(%) =2λ%− 2λc5

A(%)%

; N (%) = 2λc5A(%)%

+ λδ(%)%

;

Q(%) = c(λ)k2λ

(1 + δ−1(%)

)%2λ−1.

We choose

δ(%) =1

2λ ln(ed%

) , 0 < % < d,

where e is the Euler number. Since according to the assumption of TheoremA(%) ∼ δ(%) for suitable small d > 0, then we have:

exp( 2%∫%

P(τ)dτ)≤ 22λ; exp

( d∫%

B(τ)dτ)≤ C(d, λ) lnc(λ)

(ed%

);

−d∫%

P(τ)dτ ≤ ln(%d

)2λ+ 2λc5

d∫%

dτ

τ ln(edτ

) = ln(%d

)2λ+ 2λc5 ln ln

(ed%

)⇒

exp(−

d∫%

P(τ)dτ)≤(%d

)2λlnc(ed%

),

if we recall (1.10.2). In this case we have as well:

d∫%

Q(τ) exp(−

τ∫%

P(σ)dσ)dτ ≤ k2

λC(λ)%2λ

d∫%

1 + δ−1(τ)τ

lnc(eτ%

)dτ ≤

≤ k2λC(λ)%2λ lnc+2

(ed%

).

Now we apply Theorem 1.57: then from (1.10.1) by virtue of deducedinequalities we obtain

U(%) ≤ C25(V0 + k2λ)%2λ ln2c+2 1

%, 0 < % < d <

1e.(4.3.59)

From (4.3.59) and because of (4.3.33) it follows the desired estimate (4.3.58).

4.4 The power modulus of continuity 115

4.4. The power modulus of continuity

Theorem 4.21. Let u ∈W 2,N (G)∩C0(G) be a strong solution of prob-lem (L) and assumptions A1) - A4) are satised with A(r) Dini-continuousat zero. Suppose, in addition,

ai ∈ Lp(G), p > N ; a ∈ LN (G), f ∈ LN (G) ∩W

0

4−N (G),

ϕ(x) ∈W

3/2

4−N (∂G) ∩ V 2−1/NN,0 (∂G) ∩ Cλ(∂G)

and there exist real numbers s > 0, ks ≥ 0, k ≥ 0 such that

ks =: sup%>0

%−s(‖f‖

W0

4−N (G%0)+ ‖ϕ‖

W3/2

4−N (Γ%0)

);(4.4.1)

k =: sup%>0

%1−s(‖f‖

N ;G2%%/4

+ ‖ϕ‖V

2−1/NN,0 (Γ2%

%/4)

).(4.4.2)

Then there are d ∈ (0, 1e ) and a constant C > 0 depends only on ν, µ, d, s,

N, λ,meas G and on the quantityd∫0

A(r)r dr, such that ∀x ∈ Gd0

(4.4.3) |u(x)| ≤ C(‖u‖2,G + ‖f‖

W0

4−N (G)+ ‖ϕ‖

W3/2

4−N (∂G)+

+ |ϕ|Cλ(∂G) + ks + k)×

|x|λ, if s > λ,

|x|λ ln3/2(

1|x|

), if s = λ,

|x|s, if s < λ.

Proof. Let the functions Φ, v and F be dened as in the proof of The-orem 4.13. We remark that Φ(0) = 0 due to Lemma 1.38.

Let us introduce the function

ψ(%) =

%λ, if s > λ,

%λ ln3/2(

1%

), if s = λ,

%s, if s < λ,

(4.4.4)

for 0 < % < d and consider two sets G2%%/4 and G%%/2 ⊂ G

2%%/4, % > 0. We make

transformation x = %x′; v(%x′) = ψ(%)w(x′). The function w(x′) satisesthe problem

aij(%x′)wx′ix′j + %ai(%x′)wx′i + %2a(%x′)w = %2

ψ(%)F (%x′), x′ ∈ G21/4

w(x′) = 0, x′ ∈ Γ21/4,


where

%2

ψ(%)F (%x′) =

%2

ψ(%)f(%x′)− 1

ψ(%)

(aij(%x′)Φx′ix

′j

+ %ai(%x′)Φx′i+

+ %2a(%x′)Φ(%x′))≤ %2

ψ(%)|f |+ µ

ψ(%)|Φx′x′ |+

A(%)ψ(%)

(|∇′Φ|+ |Φ|).

Let us now rstly notice that∫G2

1/4

(∣∣∣∣%( N∑i=1

|ai(%x′)|2)1/2∣∣∣∣p +

(%2|a(%x′)|

)N)dx′ ≤

≤ c(N, p)∫

G2%%/4

Ap(r) +AN (r)rN

dx ≤ c(N, p)AN−1(d) meas Ω

2d∫0

A(r)r

dr

and

%2

ψ(%)‖F (%x′)‖LN (G2

1/4) ≤ c(µ,A(d))

%

ψ(%)

(‖f‖

LN (G2%%/4

)+ ‖ϕ‖

V2−1/NN,0 (Γ2%

%/4)

)≤ kc(µ,A(d))

%s

ψ(%)≤ k · const(µ,A(d), s, λ, d),(4.4.5)

because of (4.4.2) and (4.4.4). We apply now Theorem 4.5 (Local MaximumPrinciple), because of proved there estimates we have:

(4.4.6) supG1

1/2

|w(x′)| ≤ C(N, ν, µ)(∫∫

G21/4

w2dx′) 1

2

+

+%2

ψ(%)

(∫∫G2

1/4

|F |Ndx′) 1N.

Returning back to the variable x and the function v(x) by Theorem 4.18with (4.4.4), we obtain:

(4.4.7)∫∫G2

1/4

w2dx′ =1

ψ2(%)

∫∫G2%%/4

r−Nv2dx ≤

≤ C(‖u‖2,G + ‖f‖

W0

4−N (G)+ ‖ϕ‖

W3/2

4−N (∂G)+ ks

)2;

Because of ϕ ∈ Cλ(∂G) we then obtain

|u(x)| ≤ |v(x)|+ |Φ(x)| ≤ |v|+ |Φ(x)− Φ(0)| ≤ |v|+ |x|λ|ϕ|λ,∂G

4.4 The power modulus of continuity 117

Hence and from (4.4.5), (4.4.6) and (4.4.7) it follows:

supG%%/2

|u(x)| ≤ C(‖u‖2,G + ‖f‖

W0

4−N (G)+ ‖ϕ‖

W3/2

4−N (∂G)+

+ |ϕ|λ,∂G + ks + k)ψ(%).

Putting now |x| = 23% we obtain nally the desired estimate (4.4.3).

Theorem 4.22. Let u(x) be a strong solution of problem (L) and as-sumptions A1) - A4) are satised with A(r), which is a continuous at zerofunction, but not Dini-continuous at zero. Suppose in addition


0

4−N (G),

ϕ ∈W

3/2

4−N (∂G) ∩ V 2−1/NN,0 (∂G) ∩ Cλ(∂G)

and there exist real numbers s > 0, ks ≥ 0, k ≥ 0 such that

ks =: sup%>0

%−s(‖f‖

W0

4−N (G%0)+ ‖ϕ‖

W3/2

4−N (Γ%0)

);(4.4.8)

k =: sup%>0

%1−s(‖f‖

N ;G2%%/4

+ ‖ϕ‖V

2−1/NN,0 (Γ2%

%/4)

).(4.4.9)

Then ∀ε > 0 there are d ∈ (0, 1) and a constant Cε > 0 depends only onν, µ, d, s, ε,N, λ,meas G and on A(diamG), such that ∀x ∈ Gd0

(4.4.10) |u(x)| ≤ Cε(‖u‖2,G + ‖f‖

W0

4−N (G)+ ‖ϕ‖

W3/2

4−N (∂G)+

+ |ϕ|λ,∂G + ks + k)×

|x|λ−ε, if s > λ,

|x|s−ε, if s ≤ λ.

Proof. We repeat verbatim the proof of Theorem 4.21 by taking

ψ(%) =


%s−ε, if s ≤ λ,

and applying Theorem 4.19.

Theorem 4.23. Let u(x) be a strong solution of problem (L)and assump-tions A1) - A4) are satised with A(r) ∼ 1

ln 1r

, A(0) = 0. Suppose in addition


0

4−N (G),

ϕ(x) ∈W

3/2

4−N (∂G) ∩ V 2−1/NN,0 (∂G) ∩ Cλ(∂G)


and there exist real numbers kλ ≥ 0, k ≥ 0 such that

kλ =: sup%>0

%−λ(‖f‖

W0

4−N (G%0)+ ‖ϕ‖

W3/2

4−N (Γ%0)

);(4.4.11)

k =: sup%>0

%1−λ(‖f‖

N ;G2%%/4

+ ‖ϕ‖V

2−1/NN,0 (Γ2%

%/4)

).(4.4.12)

Then there are d ∈ (0, 1e ) and constants C > 0, c > 0 depends only on

ν, µ, d,N, λ,meas G and on A(diamG), such that ∀x ∈ Gd0

(4.4.13) |u(x)| ≤ C(‖u‖2,G + ‖f‖

W0

4−N (G)+ ‖ϕ‖

W3/2

4−N (∂G)+

+ |ϕ|λ,∂G + kλ + k)× |x|λ lnc+1 1

|x|.

Proof. We repeat verbatim the proof of Theorem 4.21 by taking

ψ(%) = %λ lnc+1 1%


4.5. Lp− estimates

In this and next Sections we establish the exact smoothness of strongsolutions of (L).

Let u be a strong solution of (L), where p > N , and let

f ∈ Lp(G), ϕ ∈W 2−1/p,p(∂G).

Let us consider the sets G2%%/4 and G%%/2 ⊂ G

2%%/4 and new variables x′ dened

by x = %x′. Then the function z(x′) = v(%x′) = v(x) satises in G21/4 the

problem

(4.5.1)

aij(%x′) ∂2z

∂x′i∂x′j

+ %ai(%x′) ∂z∂x′i

+ %2a(%x′)z = %2f(%x′)−

−(aij(%x′) ∂2Φ

∂x′i∂x′j

+ %ai(%x′) ∂Φ∂x′i

+ %2a(%x′)Φ), x′ ∈ G2

1/4

w(x′) = 0, x′ ∈ Γ21/4,

where the functions Φ, v be dened as in the proof of Theorem 4.13.

By the Sobolev Imbedding Theorems 1.33 and 1.34 we have

supx′,y′∈G1

1/2

x′ 6=y′

|z(x′)− z(y′)||x′ − y′|β

≤ c‖z‖W 2,N (G11/2

), ∀β ∈ (0, 1);

(4.5.2)

supx′∈G1

1/2

|∇′z(x′)|+ supx′,y′∈G1

1/2

x′ 6=y′

|∇′z(x′)−∇′z(y′)||x′ − y′|1−N/p

≤ c‖z‖W 2,p(G11/2

), p > N.

4.5 Lp− estimates 119

By the local Lp apriori estimate (4.6) for solutions of (4.5.1) we obtain

(4.5.3) ‖z‖W 2,p(G11/2

) ≤ c(N, ν, µ,A(2))‖z‖Lp(G2

1/4) + %2‖f‖Lp(G2

1/4) +

+ ‖ϕ‖W 2−1/p,p(∂G21/4

)

.

Returning back to the variables x, from (4.5.2), (4.5.3) it follows:

(4.5.4) supx,y∈G%

%/2

x 6=y

|v(x)− v(y)||x− y|β

≤ c%−β%−1‖v‖

LN (G2%%/4

)+ ‖f‖

V 0N,N (G2%

%/4)

+

+ ‖ϕ‖V

2−1/NN,N (Γ2%

%/4)

, ∀β ∈ (0, 1);

(4.5.5) supG%%/2

|∇v| ≤ c%−1%−N/p‖v‖

Lp(G2%%/4

)+ ‖f‖

V 0p,2p−N (G2%

%/4)

+

+ ‖ϕ‖V

2−1/pp,2p−N (Γ2%

%/4)

;

(4.5.6) supx,y∈G%

%/2

x 6=y

|∇v(x)−∇v(y)||x− y|1−N/p

≤ c%Np−2

%−N/p‖v‖Lp(G2%

%/4)

+

+ ‖f‖V 0p,2p−N (G2%

%/4)

+ ‖ϕ‖V

2−1/pp,2p−N (Γ2%

%/4)

.

Moreover, if we rewrite the inequality (4.5.3) in the equivalent form∫G1

1/2

(|D′2z|p + |∇′z|p + |z|p

)dx′ ≤ c

∫G2

1/4

(|z|p + r2p|f |p +

+|D′2Φ|p + |∇′Φ|p + |Φ|p)dx′,

multiply both sides of this inequality by %α−2p and return to the variablesx, then we obtain ∫

G%%/2

(rα|D2v|p + rα−p|∇v|p + rα−2p|v|p

)dx ≤

≤ c∫

G2%%/4

(rα−2p|v|p + rα|f |p + rα|D2Φ|p + rα−p|∇Φ|p + rα−2p|Φ|p

)dx

and consequently

(4.5.7) ‖v‖V 2p,α(G%

%/2) ≤ c

‖v‖

V 0p,α−2p(G2%

%/4)

+ ‖f‖V 0p,α(G2%

%/4)

+

+ ‖ϕ‖V

2−1/pp,α (Γ2%

%/4)

.


Theorem 4.24. Let u be a strong solution of the boundary value problem(L) and let the assumptions of Theorem 4.21 be satised. Furthermore, wesuppose that

f ∈ V 0p,α(G), ϕ ∈ V 2−1/p

p,α (∂G), p > N

with

α >

(2− λ)p−N, if s > λ,

(2− s)p−N, if s ≤ λand there is some constant k2 ≥ 0 such that

k2 =: sup%>0

‖f‖V 0p,α(G%

%/2) + ‖ϕ‖

V2−1/pp,α (Γ%

%/2)

χ(%),(4.5.8)

where

χ(%) ≡

%λ−2+α+N

p , if s > λ,

%λ−2+α+N

p ln3/2 1% , if s = λ,

%s−2+α+N

p , if s < λ

(4.5.9)

for all suciently small % > 0.Then u ∈ V 2

p,α(G) and the estimate

‖u‖V 2p,α(G%0) ≤ cχ(%)(4.5.10)

holds with c independent of u.

Proof. The statement of Theorem follows from (4.5.7), since

‖v‖V 0p,α−2p(G2%

%/4)

=( ∫G2%%/4

rα−2p|v|pdx)1/p

≤in virtue of (4.4.3), (4.4.4)

≤ cψ(%)( 2%∫%/4

rα−2p+N−1dr

)1/p

≤ c%−2+α+Np ψ(%) = cχ(%).

Hence and from (4.5.7), (4.5.8) replacing % by 2−k%, we have

‖u‖V 2p,α−2p(G(k)) ≤ cχ(2−k%).

By summing this inequalities over all k = 0, 1, · · · we obtain desired assertion.

In similar way we prove following theorems.


f ∈ V 0p,α(G), ϕ ∈ V 2−1/p

p,α (∂G), p > N

4.6 Cλ− estimates 121

with

α >

(2− λ)p−N, if s > λ,

(2− s)p−N, if s ≤ λand there is some constant k2 ≥ 0 such that

‖f‖V 0p,α(G%

%/2) + ‖ϕ‖

V2−1/pp,α (Γ%

%/2)≤ k2

%λ−2−ε+α+N

p , if s > λ,

%s−2−ε+α+N

p , if s ≤ λ(4.5.11)

for all suciently small % > 0 and ∀ε > 0.

Then u ∈ V 2p,α(G) and the estimate

‖u‖V 2p,α(G%0) ≤ cε

%λ−2−ε+α+N

p , if s > λ,

%s−2−ε+α+N

p , if s ≤ λ(4.5.12)

holds with cε independent of u.


f ∈ V 0p,α(G), ϕ ∈ V 2−1/p

p,α (∂G), p > N, α > (2− λ)p−N

and there is some constant k2 ≥ 0 such that

‖f‖V 0p,α(G%

%/2) + ‖ϕ‖

V2−1/pp,α (Γ%

%/2)≤ k2%

λ−2+α+Np lnc+1 1

%(4.5.13)

for all suciently small % > 0, where c is dened by Theorem 4.23.

Then u ∈ V 2p,α(G) and the estimate

‖u‖V 2p,α(G%0) ≤ C%

λ−2+α+Np lnc+1 1

%(4.5.14)

holds with C independent of u.

4.6. Cλ− estimates

Let known be that

|v(x)| ≤ c0ψ(|x|), x ∈ Gd0.

Then we have

%−1‖v‖LN (G2%

%/4)≤ c1ψ(%);(4.6.1)

%−Np ‖v‖

Lp(G2%%/4

)≤ c2ψ(%);(4.6.2)

%−2‖v‖Lp(G2%

%/4)≤ c3%

Np−2ψ(%);(4.6.3)


Theorem 4.27. Let u be a strong solution of the boundary value prob-lem (L) and let the assumptions of Theorem 4.21 be satised. Let λ = 1.

Then

u ∈

Cβ(G), ∀β ∈ (0, 1) if s ≥ 1,Cs(G), if 0 < s < 1.

(4.6.4)

Proof. From (4.5.4) and (4.6.1), (4.4.2) it follows

supx,y∈G%

%/2

x 6=y

|v(x)− v(y)||x− y|β

≤ c%−βψ(%), ∀β ∈ (0, 1),(4.6.5)

where in our case ψ(%) is dened by (4.4.4). By Theorem 4.21 hence follows

supx,y∈G%

%/2

x 6=y

|v(x)− v(y)||x− y|β

≤ c

%1−β, if s > 1,%1−β−ε, if s = 1,%s−β, if s < 1,

(4.6.6)

∀ε > 0, ∀β ∈ (0, 1).By denition of the set G%%/2 we have |x − y| ≤ 2% and therefore from

(4.6.6) it follows:

(4.6.7) |v(x)− v(y)| ≤ c|x− y|β

%1−β, if s > 1,%1−β−ε, if s = 1,%s−β, if s < 1,

≤

≤ c

|x− y|β , if s ≥ 1,|x− y|s, if s < 1

∀β ∈ (0, 1), ∀x, y ∈ G%%/2.

If |x− y| ≥ % = |x|, then from Theorem 4.21 it follows:

(4.6.8)|v(x)− v(y)||x− y|β

≤ 2|v(x)||x− y|−β ≤ 2cψ(%)%−β ≤

≤ c

%1−β, if s > 1,%1−β−ε, if s = 1,%s−β, if s < 1,

≤ const,

if we choose β = s for 0 < s < 1. This together with ϕ ∈ Cλ proves ourTheorem.

By repeating verbatim the proof of previous Theorem we obtain the nextTheorems.



Then

u ∈

Cβ(G), ∀β ∈ (0, 1) if s ≥ 1,Cs−ε(G), ∀ε > 0, if 0 < s < 1.

(4.6.9)

Proof. In this case we have (4.6.5) with ψ(%) from Theorem 4.22, i.e.

supx,y∈G%

%/2

x6=y

|v(x)− v(y)||x− y|β

≤ c

%1−β−ε, if s ≥ 1,%s−β−ε, if s < 1,

(4.6.10)

∀ε > 0, ∀β ∈ (0, 1).Hence follows our statement, if we choose ε = 1 − β for s ≥ 1 and

β = s− ε for 0 < s < 1.


Then

u ∈ C1−ε(G), ∀ε > 0.(4.6.11)

Proof. In this case we have (4.6.5) with ψ(%) from Theorem 4.23, i.e.

supx,y∈G%

%/2

x 6=y

|v(x)− v(y)||x− y|β

≤ c%1−β−ε, ∀ε > 0, ∀β ∈ (0, 1).(4.6.12)

Hence follows our statement, if we choose β = 1− ε.

Theorem 4.30. Let u be a strong solution of the boundary value problem(L) and let the assumptions of Theorem 4.21 be satised. Let 0 < λ < 1.

Then

u ∈

Cλ(G), if s > λ,

Cλ−ε(G), ∀ε > 0, if s = λ,

Cs(G), if 0 < s < λ.

(4.6.13)

Proof. By Theorem 4.21 from (4.6.5) it follows

supx,y∈G%

%/2

x 6=y

|v(x)− v(y)||x− y|β

≤ c

%λ−β , if s > λ,

%λ−β−ε, if s = λ,

%s−β , if s < λ,

(4.6.14)

∀ε > 0, ∀β ∈ (0, 1).Putting

β =

λ, if s > λ,

λ− ε, if s = λ,

s, if 0 < s < λ,

we obtain the required assertion.



Then

u ∈

Cλ−ε(G), ∀ε > 0, if s ≥ λ,Cs−ε(G), if 0 < s < λ.

(4.6.15)


supx,y∈G%

%/2

x 6=y

|v(x)− v(y)||x− y|β

≤ c

%λ−β−ε, if s ≥ λ,%s−β−ε, if s < λ,

(4.6.16)

∀ε > 0, ∀β ∈ (0, 1).Putting

β =

λ− ε, if s ≥ λ,s− ε, if 0 < s < λ,

we obtain the required assertion.


Then

u ∈ Cλ−ε(G), ∀ε > 0.(4.6.17)


supx,y∈G%

%/2

x 6=y

|v(x)− v(y)||x− y|β

≤ c%λ−β−ε(4.6.18)

∀ε > 0, ∀β ∈ (0, 1).Putting

β = λ− ε,we obtain the required assertion.

Now, let be fullledAssumption A5) There exists some constant k ≥ 0 such that

k =: sup%>0

‖f‖V 0p,2p−N (G2%

%/4)

+ ‖ϕ‖V

2−1/pp,2p−N (Γ2%

%/4)

ψ(%), p > N.A5)

Then from (4.6.2), (4.5.5) and (4.5.6) we obtain

supG%%/2

|∇v| ≤ c%−1ψ(%),(4.6.19)

supx,y∈G%

%/2

x 6=y

|∇v(x)−∇v(y)||x− y|1−N/p

≤ c%Np−2ψ(%),(4.6.20)


Theorem 4.33. Let u be a strong solution of the boundary value problem(L) and let the assumptions of Theorem 4.21 and A5) with ψ(%) dened by(4.4.4) be satised. Then is true the next estimate

|∇u(x)| ≤ c

|x|λ−1, if s > λ,

|x|λ−1 ln3/2 1|x| , if s = λ,

|x|s−1, if s < λ.

(4.6.21)

Moreover,

• 1) if λ ≥ 2− Np , then

u ∈

C

2−Np (G), if s > λ,

C2−N

p−ε(G), ∀ε > 0, if s = λ,

Cs−λ+2−N

p (G), if λ− 1 + Np ≤ s < λ.

• 2) if 1 < λ ≤ 2− Np , then

u ∈

Cλ(G), if s > λ,

Cλ−ε(G), ∀ε > 0, if s = λ,

Cs(G), if 1 ≤ s < λ.

Proof. From (4.6.19), (4.6.20) with (4.4.4) we obtain:

supG%%/2

|∇v| ≤ c

%λ−1, if s > λ,

%λ−1 ln3/2 1% , if s = λ,

%s−1, if s < λ,

(4.6.22)

supx,y∈G%

%/2

x 6=y

|∇v(x)−∇v(y)||x− y|1−N/p

≤ c

%Np−2+λ

, if s > λ,

%Np−2+λ−ε

, ∀ε > 0, if s = λ,

%Np−2+s

, if s < λ.

(4.6.23)

Putting |x| = 23% we obtain from (4.6.22) the (4.6.21).

Now we set

κ =

0, if s > λ,

−ε, if s = λ,

s− λ, if λ− 1 + Np ≤ s < λ.

Let us consider the rst case λ ≥ 2− Np . If x, y ∈ G

%%/2, then |x−y| ≤ 2%

and therefore %κ ≤ c|x− y|κ, since κ ≤ 0. Then from (4.6.23) we get

|∇v(x)−∇v(y)| ≤ c

|x− y|1−N/p, if s > λ,

|x− y|1−N/p−ε, ∀ε > 0, if s = λ,

|x− y|1−N/p+s−λ, if s < λ.


If x, y ∈ G and |x− y| ≥ % = |x|, then by (4.6.22) we get

|∇v(x)−∇v(y)||x− y|1−N/p+κ

≤ 2|∇v||x− y|N/p−1−κ ≤ c%λ−1+κ%Np−1−κ =

= c%Np−2+λ ≤ const;

we have taken into account that in the considered case 1 − N/p + κ ≥ 0.Thus the case 1) of our Theorem is proved.

Now we consider the second case 1 < λ ≤ 2− Np . If x, y ∈ G%%/2, then

|x−y| ≤ 2% and therefore %κ ≤ c|x−y|κ, since κ ≤ 0. Then from (4.6.23)we get

|∇v(x)−∇v(y)| ≤ c|x− y|1−N/p%Np−2+λ+κ ≤ c|x− y|λ−1+κ.


|∇v(x)−∇v(y)||x− y|λ−1+κ

≤ 2|∇v||x− y|1−λ−κ ≤ c%λ−1+κ|x− y|1−λ−κ ≤ const;

we have taken into account that in the considered case 1− λ− κ ≤ 0. Thusthe case 2) of our Theorem as well is proved.

Theorem 4.34. Let u be a strong solution of the boundary value problem(L) and let the assumptions of Theorem 4.22 and A5) with ψ(%) dened byTheorem 4.22 be satised. Then is true the next estimate

|∇u(x)| ≤ c

|x|λ−1−ε, if s ≥ λ,|x|s−1−ε, if s < λ.

∀ε > 0.(4.6.24)

Moreover,

• 1) if λ ≥ 2− Np , then

u ∈

C

2−Np−ε(G), if s ≥ λ,

Cs−λ+2−N

p−ε(G), if λ− 1 + N

p < s < λ.∀ε > 0;

• 2) if 1 < λ ≤ 2− Np , then

u ∈

Cλ−ε(G), if s ≥ λ,Cs−ε(G), if 1 < s < λ.

∀ε > 0


Proof. From (4.6.19), (4.6.20) with ψ(%) from Theorem 4.22 we obtain:

supG%%/2

|∇v| ≤ c

%λ−1−ε, if s ≥ λ,%s−1−ε, if s < λ,

(4.6.25)

supx,y∈G%

%/2

x 6=y

|∇v(x)−∇v(y)||x− y|1−N/p

≤ c

%Np−2+λ−ε

, if s = λ,

%Np−2+s−ε

, if s < λ.(4.6.26)

∀ε > 0. Putting |x| = 23% we obtain from (4.6.25) the (4.6.24).

Let us consider the rst case λ ≥ 2− Np . If x, y ∈ G

%%/2, then |x−y| ≤ 2%

and therefore %−ε ≤ c|x− y|−ε. Then from (4.6.26) we get

|∇v(x)−∇v(y)| ≤ c

|x− y|1−N/p−ε, if s ≥ λ,|x− y|1−N/p+s−λ−ε, if s < λ.

∀ε > 0.

If x, y ∈ G and |x− y| ≥ % = |x|, then by (4.6.25) we get• 1) for s ≥ λ :

|∇v(x)−∇v(y)||x− y|1−N/p−ε

≤ 2|∇v||x− y|N/p−1+ε ≤ c%λ−1−ε|x− y|N/p−1+ε ≤

≤ c|x− y|Np−2+λ ≤ const.

• 2) for N/p− 1 + λ < s < λ :

|∇v(x)−∇v(y)||x− y|1−N/p−ε+s−λ

≤ 2|∇v||x− y|N/p−1+ε−s+λ ≤

≤ c%s−1−ε|x− y|N/p−1+ε+λ−s ≤ c%Np−2+λ ≤ const.

Thus the case 1) of our Theorem is proved.

Now we consider the second case 1 < λ ≤ 2− Np . For this we dene

κ =

−ε, if s ≥ λ,s− λ− ε, if 1 < s < λ.

If x, y ∈ G%%/2, then |x−y| ≤ 2% and therefore %κ ≤ c|x−y|κ, since κ < 0.Then from (4.6.26) we get

|∇v(x)−∇v(y)| ≤ c|x− y|1−N/p%Np−2+λ+κ ≤ c|x− y|λ−1+κ.


|∇v(x)−∇v(y)||x− y|λ−1+κ

≤ 2|∇v||x− y|1−λ−κ ≤ c%λ−1+κ|x− y|1−λ−κ ≤ const;


we have taken into account that in the considered case 1− λ− κ < 0. Thusthe case 2) of our Theorem as well is proved.

At last, in the same way we prove

Theorem 4.35. Let u be a strong solution of the boundary value problem(L) and let the assumptions of Theorem 4.23 and A5) with ψ(%) dened byTheorem 4.23 be satised. Then is true the next estimate

|∇u(x)| ≤ C|x|λ−1 lnc+1 1|x|.(4.6.27)

Moreover,

• 1) if λ ≥ 2− Np , then u ∈ C

2−Np−ε(G), ∀ε > 0;

• 2) if 1 < λ ≤ 2− Np , then u ∈ C

λ−ε(G), ∀ε > 0.

4.7. Examples

Let us present some examples which demonstrate that the assumptionson the coecients of the operator L are essential for the validity of Theoremsof Sections 4.4 - 4.6.

Let N = 2, let the domain G lie inside the sector

G∞0 = (r, ω)∣∣0 < r <∞, 0 < ω < ω0, 0 < ω ≤ 2π

and suppose that O ∈ ∂G and in some neighborhood Gd0 of O the boundary∂G coincides with the sides ω = 0 and ω = ω0 of the sector G∞0 . In our casethe least eigenvalue of (EVP1) is λ = π

ω0.

Example 4.36. Let us consider the function

u(r, ω) = rλ(

ln1r

)(λ−1)/(λ+1)

sin(λω), λ =π

ω0

in

Gd0 := x ∈ R2 : 0 < r < d, 0 < ω < ω0.It satises the equation

N∑i,j=1

aij(x)Diju = 0 in Gd0

with

a11(x) = 1− 2λ+ 1

x22

r2 ln(1/r),

a12(x) = a21(x) =2

λ+ 1x1x2

r2 ln(1/r),

a22(x) = 1− 2λ+ 1

x21

r2 ln(1/r),

aij(0) = δji

4.7 Examples 129

and the boundary conditions

u = 0 on Γd0.

If d < e−2, then the equation is uniformly elliptic with ellipticity constants

ν = 1− 2ln(1/d)

and µ = 1.

Furthermore,

A(r) =2

(λ+ 1) ln(1/r),

d∫0

A(r)r

dr = +∞,

i.e. the leading coecients of the equation are continuous but not Dinicontinuous at zero. From the explicit form of the solution u we have

|u(x)| ≤ c|x|λ−ε, ‖u‖ W

2

2(G%0)≤ c%λ−ε(4.7.1)

for all ε > 0. This example shows that it is not possible to replace λ −ε in (4.7.1) by λ without additional assumptions regarding the continuitymodulus of the leading coecients of the equation at zero.

Example 4.37. Let Gd0 be dened as in the previous example and let

u(x) = rλ ln(1r

) sin(λω), λ =π

ω0.

The function u satises∆u+ 2λ

r2 ln(1/r)u = 0 in Gd0,

u = 0 on Γd0.

Here

A(r) =2λ

ln (1/r),

d∫0

A(r)r

dr = +∞.

Thus the assumptions about the lower order coecients are essential, too.

Example 4.38. The function

u(x) = rλ ln(1r

) sin(λω), λ =π

ω0.

satises ∆u = f := −2λrλ−2 sin(λω) in Gd0,u = 0 on Γd0.

Here, all assumptions on the coecients are satised but

‖f‖ W

0

2(G%0)≤ c%s

with s = λ. This veries the importance of conditions of our Theorems.


4.8. Higher regularity results

Now we begin the study of the higher regularity of the strong solutionsof the problem (L). This smoothness depends on the value λ.

Theorem 4.39. Let p, α ∈ R, k ∈ N satisfy p ≥ 1, k ≥ 2. Let u ∈W 2,N (G) ∩ C0(G) be a strong solution of the boundary value problem (L)and assumptions of Theorem 4.21 with s > λ are satised. Suppose, inaddition, that there are derivatives Dlaij , Dlai, Dla, |l| ≤ k−2 and numbersµl ≥ 0 such that N∑

i,j=1

|Dlaij(x)|21/2

+ |x|l+1

(N∑i=1

|Dlai(x)|2)1/2

+ |x|l+2|Dla(x)| ≤ µl;

x ∈ G; |l| = 1, 2, . . . , k − 2.

If f ∈ V k−2p,α (G), ϕ ∈ V k−1/p

p,α (∂G) and

‖f‖V k−2p,α (G2%

%/4)

+ ‖ϕ‖Vk−1/pp,α (Γ2%

%/4)≤ k1%

λ−k+α+Np ,(4.8.1)

where

α > p(k − λ)−N,(4.8.2)

then there are numbers c > 0, d > 0 such that u ∈ V kp,α(Gd0) and the following

estimate is valid

‖u‖V kp,α(G%0) ≤ c%λ−k+α+N

p , % ∈ (0, d).(4.8.3)

Proof. Let us consider two sets G2%%/4 and G

%%/2 ⊂ G

2%%/4, % > 0. We make

transformation x = %x′; u(%x′) = %λw(x′). The function w(x′) satises theproblem

aij(%x′)wx′ix′j + %ai(%x′)wx′i + %2a(%x′)w = %2−λf(%x′),

x′ ∈ G21/4;

w(x′) = %−λϕ(%x′), x′ ∈ Γ21/4.

(L)′

By Theorem 4.7 we have

(4.8.4) ‖w‖Wk,p(G11/2

) ≤ Ck(‖w‖Lp(G2

1/4) + %2−λ‖f‖Lp(G2

1/4) +

+ %−λ‖ϕ‖Wk−1/p,p(Γ21/4

)

),

where Ck does not depend on w and depends only on G,N, p, ν, µ andmaxx∈G2%

%/4

A(|x|). Returning to the variables x, u, multiplying both sides of this

4.8 Higher regularity results 131

inequality by %α+Np−k and noting that %/4 ≤ r ≤ 2% in G2%

%/4 we obtain

(4.8.5) ‖u‖V kp,α(G%%/2

) ≤ C‖f‖

V k−2p,α (G2%

%/4)

+ ‖ϕ‖Vk−1/pp,α (Γ2%

%/4)

+

+ ‖u‖V 0p,α−kp(G2%

%/4)

.

By Theorem 4.21 we have |u(x)| ≤ c0|x|λ, therefore

‖u‖pV 0p,α−kp(G2%

%/4)

=∫

G2%%/4

%α−kp|u(x)|pdx ≤ cp0∫

G2%%/4

%α−kp+λpdx ≤

≤ cp0measΩ · %α+N+p(λ−k).

Hence and from (4.8.6) with regard to (4.8.1) it follows

‖u‖V kp,α(G%%/2

) ≤ C%λ−k+α+N

p , % ∈ (0, d).(4.8.6)

Replacing % in the above inequality by 2−m% and summing up the resultinginequalities for every m = 0, 1, 2, . . . , we obtain

‖u‖V kp,α(G%0) ≤ C%λ−k+α+N

p

∞∑m=0

2−m(λ−k+α+Np

).

By (4.8.2), the numerical series from the right converges. Thus the estimate(4.8.3) is proved.

Theorem 4.40. Let u be a strong solution of (L). Suppose that the con-ditions of Theorem 4.21 with s > λ and Theorem 4.39 are satised. Let, inaddition,

k − 1 < λ ≤ k − N

p, k ≥ 2, p > N.(4.8.7)

Then u ∈ Cλ(Gd0) and there are nonnegative numbers Cl such that

|Dlu(x)| ≤ Cl|x|λ−|l| ∀x ∈ Gd0; |l| = 0, 1, . . . , k − 1(4.8.8)

for some d > 0. If λ = k − 1, p = N, then u ∈ Cλ−ε(Gd0), ∀ε > 0.

Proof. We consider the function w(x′) as a solution of the problem (L)′

in the domain G21/4. By the Sobolev imbedding Theorem 1.33,

W k,p(G) → Ck−1+β(G), 0 < β ≤ 1− N

p

and, in addition,

(4.8.9)∑|l|≤k−1

supx′∈G1

1/2

|Dlx′w(x′)|+ sup

x′,y′∈G11/2

x′ 6=y′

|Dk−1x′ w(x′)−Dk−1

y′ w(y′)||x′ − y′|1−N/p

≤

≤ c‖w‖Wk,p(G11/2

)


with a constant c independent of u and dened only by N, p and the domainG. Returning back to the variables x, u we have for ∀% ∈ (0, d) :

supx∈G%

%/2

|Dlu(x)| ≤ Cl%k−|l|−N+αp ‖u‖V kp,α(G%

%/2), |l| = 0, 1, . . . , k − 1

(4.8.10)

supx,y∈G%

%/2

x 6=y

|Dk−1u(x)−Dk−1u(y)||x− y|1−N/p

≤ c%−αp ‖u‖V kp,α(G%

%/2).

Since %/2 ≤ r = |x| ≤ % for x ∈ G%%/2, by (4.8.6), from 4.8.10) it follows

|Dlu(x)| ≤ Cl|x|λ−|l|, |l| = 0, 1, . . . , k − 1; x ∈ Gd0;(4.8.11)

supx,y∈G%

%/2

x 6=y

|Dk−1u(x)−Dk−1u(y)||x− y|1−N/p

≤ c%λ−k+Np .(4.8.12)

Now from (4.8.12) for τ = λ− k + Np ≤ 0 we have

|Dk−1u(x)−Dk−1u(y)| ≤ c%τ |x− y|λ−k+1−τ ∀x, y ∈ G%%/2.(4.8.13)

Since τ ≤ 0, we have

|x− y|τ ≥ (2%)τ ∀x, y ∈ G%%/2and therefore from (4.8.13) it follows

|Dk−1u(x)−Dk−1u(y)| ≤ c2−τ |x− y|λ−k+1 ∀x, y ∈ G%%/2.(4.8.14)

The inequality (4.8.14) together with the (4.8.11) leads to the assertion u ∈Cλ(Gd0), if the (4.8.7) is fullled.

Let now λ = k − 1, p = N. Then, by the Sobolev imbedding Theorem1.33, we have

(4.8.15) supx′,y′∈G1

1/2

x′ 6=y′

|Dk−2x′ w(x′)−Dk−2

y′ w(y′)||x′ − y′|β

≤ c‖w‖Wk,p(G11/2

),

∀β ∈ (0, 1); k ≥ 2.

Returning back to the variables x, u and considering the inequality (4.8.6),we have for ∀% ∈ (0, d) :

supx,y∈G%

%/2

x 6=y

|Dk−2u(x)−Dk−2u(y)||x− y|β

≤ c%2−β−α+Np ‖u‖V kp,α(G%

%/2) ≤

≤ c%λ−k+2−β = c%1−β , ∀β ∈ (0, 1); k ≥ 2.(4.8.16)

4.9 Smoothness in a Dini - Liapunov region 133

The inequality (4.8.16) for β = 1 − ε, ∀ε > 0 together with the (4.8.11)for |l| = 0, 1, . . . , k − 2 means u ∈ Cλ−ε(Gd0), ∀ε > 0. Thus the assertionfollows.

4.9. Smoothness in a Dini - Liapunov region

In this Section we shall study strong solutions u ∈ W 2,ploc (G) ∩W 1,p(G),

p > N of (L) in a Dini - Liapunov region G. We follow some results inK.-O. Widman [402], [403].

Definition 4.41. A Dini - Liapunov surface is a closed, bounded(N − 1)− dimensional surface S satisfying the following conditions:

• At every point of S there is a uniquely dened tangent (hyper-) plane,and thus also a normal.• There exists a Dini function A(r) such that if θ is the angle betweentwo normals, and r is the distance between their foot points, then theinequality θ ≤ A(r) holds.• There is a constant % > 0 such that if Ω% is a sphere with radius %and center x0 ∈ S, then a line parallel to the normal at x0 meets S atmost once inside Ω%.

A Dini - Liapunov surface is called a Liapunov surface, if A(r) = crγ ,γ ∈ (0, 1). Dini - Liapunov and Liapunov regions are regions the boundaryof which are Dini - Liapunov and Liapunov surfaces respectively.

For the properties of Liapunov regions see Günter [139]. In particularwe note that a Dini - Liapunov domain belongs to C1,A.

Since some minor complications arise from the logarithmic singularity ofthe fundamental solutions of the elliptic equation with constant coecientsin the case N = 2, we will concentrate on domains in RN with N ≥ 3.

We note that it is well known, that the rst derivatives of u are continuousfunctions which are locally absolutely continuous on all straight lines parallelto one of the coordinate axis except those issuing from a set of (N − 1)−dimensional Lebesgue measure zero on the orthogonal hyperplane.

Further we will always suppose that the following Assumptions on theequation (L) satisfy:• A1)−A4) with det

(aij)

= 1, which is no further restriction.W4) There exists a α− Dini function A such that N∑

i,j=1


≤ A(|x− y|), ∀x, y ∈ G.

(N∑i=1

|ai(x)|2)1/2

+ |a(x)|+ |f(x)| ≤ Kdλ−2(x),W5)

where λ ∈ (1, 2) and by d(x) is denoted the distance from x to ∂G.


Theorem 4.42. Let G be a bounded Liapunov domain in RN with a

Cλ, 1 < λ < 2 boundary portion T ⊂ ∂G. Let u(x) be a strong solution ofthe problem (L) with ϕ(x) ∈ Cλ(∂G). Suppose the coecients of the equationin (L) satisfy assumptions W1) - W5).

Then u ∈ Cλ(G′) for any domain G′ ⊂⊂ G ∪ T and

|u|λ;G′∪T ≤ c(N,T,G, ν, µ,K, k, d′)(|u|0;G + ‖f‖p;G + |ϕ|λ;∂G

),(4.9.1)

where d′ = dist(G′, ∂G \ T ), k = maxi,j=1,...,N

‖aij , ‖0,A;G

, N 0 and center x0 ∈ T such that a line parallel to the normal at x0

intersects T at most once inside S%. We can choose % > 0 so small that anytwo normals issuing from points of T inside S% form an angle less than π

4 ,say. It will be no restriction to assume that x0 = O and that the positivexN− axis is along the inner normal of T at x0. Then, inside S% the surfaceT is described by

xN = h(x′) ∈ Cλ(|x′| < 12%+ ε); x′ = (x1, . . . , xN−1).

Now we use Extension Lemma 1.62 to extend the function xN − h(x′) fromT into G. We denote this extension by H(x). Since ∂H

∂xN= 1 on T we can

consider the connected region G′ that is a connected component of the set

x∣∣|x′| < 1

2%,

∂H

∂xN>

12, H > 0

which has T as portion of its boundary. By Extension Lemma 1.62 H hasthe following properties in G′ :

1. H(x) ∈ C∞(G′);

2. H(x) ∈ Cλ(G′);

3. K1[xN − h(x′)] ≤ H(x) ≤ K2[xN − h(x′)], =⇒d(x) ≥ K3H(x), K1,K2,K3 > 0, (see also 2 [232]);

4. |D2xxH(x)| ≤ Kdλ−2(x);

5. H(x) is strictly monotonic considered as a function of xN

for each x′, |x′| < 12%.

From 3 follows

Corollary 4.43.

d(x) ≥ 12K3|x|, x ∈ G′.


Proof. In fact, we have:

d(x) ≥ K3H(x) = K3 (H(x)−H(x0)) = K3|∇H| · |x| ≥12K3|x|.

Similarly, let Φ(x) be an extension of the boundary function ϕ(x) from Tinto G. By Extension Lemma 1.62 Φ has the following properties in G′ :

1. Φ(x) ∈ C∞(G′);

2. Φ(x) ∈ Cλ(G′)

3. |D2xxΦ(x)| ≤ Kdλ−2(x).

Now we atten the boundary portion T . Let us consider the dieomorphismψ that is given in the following way:

yk = xk; k = 1, . . . , N − 1,yN = H(x).

The mapping y = ψ(x), x ∈ G′, is one-one and maps G′ onto a region D′

which contains the set y∣∣|y′| < 1

2%, 0 < yN < τ for some τ > 0, in such away that T and |y′| < 1

2% correspond.Let us consider the problem (L) for the function v = u−Φ. The function

v then satises the homogeneous Dirichlet problemaij(x)Dijv(x) + ai(x)Div(x) + a(x)v(x) = F (x) in G,v(x) = 0 on ∂G,

(L)0

where

F (x) = f(x)−(aij(x)DijΦ(x) + ai(x)DiΦ(x) + a(x)Φ(x)

).(4.9.2)

Under the mapping y = ψ(x), let v(y) = v(x). Since

vxi =∂ψk∂xi

vyk , vxixj =∂ψk∂xi

∂ψm∂xj

vykym +∂2ψk∂xi∂xj

vyk ,

it follows from (L)0 that v(y) is a strong solution in D′ of the problemaij(y)Dij v(y) + ai(y)Div(y) + a(y)v(y) = F (y) in D′,v(y′) = 0 on |y′| ≤ 1

2%,(L)0

where

F (y) = f(y)−(aij(y)DijΦ(y) + ai(y)DiΦ(y) + a(y)Φ(y)

),

aij(y) = akm(x)∂ψi∂xk

∂ψj∂xm

, ai(y) = ak(x)∂ψi∂xk

,

a(y) = a(x), f(y) = f(x), Φ(y) = Φ(x),

x = ψ−1(y).

(4.9.3)


It is not dicult to observe that the conditions on coecients of the equationand on the portion T are invariant under maps of class C1,A. Further by theellipticity condition:

aij(y)ξiξj = akm(x)∂(ξiyi)∂xk

∂(ξj∂yj)∂xm

≥

≥ νN∑k=1

( ∂

∂xk

( N∑i=1

ξiyi))2

=

= ν

N∑k=1

(N∑i=1

ξi∂yi∂xk

)2

= ν

N∑k=1

(ξk + ξN

∂yN∂xk

)2=

= ν

(ξ2 + 2ξ2

N − 2ξNN−1∑k=1

ξk∂h

∂xk+ ξ2

N

[1 +

N−1∑k=1

(∂h

∂xk

)2]).

But by the Cauchy inequality with ∀ε > 0:

2ξN∂h

∂xkξk ≤ εξ2

N

(∂h

∂xk

)2

+1εξ2k,

therefore from the previous inequality it follows that

aij(y)ξiξj ≥ ν(

1− 1ε

)ξ′

2 + (1− ε)ξ2N

N−1∑k=1

(∂h

∂xk

)2

+ 4ξ2N

=

= ν(

1− 1ε

)ξ′

2 + ξ2N

[4 + (1− ε)|∇h|2

]≥

≥ ν(

1− 1ε

)ξ′

2 + ξ2N

[4 + (1− ε)K2

], ∀ε > 1.

(4.9.4)

Now we show that there is ε > 1 such that

1− 1ε

= 4 + (1− ε)K2

For this we solve the equation

K2ε2 − (3 +K2)ε− 1 = 0

and obtain

ε =12

+3

2K2+

√14

+10

4K2+

94K4

.

Hence we see that ε > 1 and we have also:

1− 1ε

=8

K2 + 5 +√K4 + 10K2 + 9

.


Thus from (4.9.4) it follows nally that

aij(y)ξiξj ≥ νc(K)ξ2,

(4.9.5)

c(K) =8

K2 + 5 +√K4 + 10K2 + 9

.

Now we rewrite the problem (L)0 in the formL0v ≡ aij0 Dij v = G(y), y ∈ D′

v(y′) = 0 on |y′| ≤ 12%,

where

aij0 = aij(0);

G(y) = f(y)−(aij(y)DijΦ(y) + ai(y)DiΦ(y) + a(y)Φ(y)

)−

−((aij(y)− aij(0)

)Dij v(y) + ai(y)Div(y) + a(y)v(y)

),

(4.9.6)

and we can apply to this problem Theorem 3.10:

(4.9.7) |v|λ,D′′′ ≤ c(|v|0,D′′ + ‖G‖p,D′′

); N < p <

N

2− λ,

∀D′′′ ⊂ D′′ ⊂ D′.

Noting that dx = |J |dy, where J = D(ψ1,...,ψN )D(x1,...,xN ) is Jacobian of the trans-

formation ψ(x) and J = ∂H∂xN

≥ 12 , further, from assumptions A1) -A4),

W4) - W5) and (4.9.3), (4.9.6) - (4.9.7) with regard for above propertiesof H(x),Φ(x) it follows that

|v|λ,G′′′ ≤ c1|v|0,G′′ + c(p, µ,K)

∫G′′

⟨Ap(d(x))|vxx|p + |f |p +

+ dp(λ−2)(x)(|∇v|p + |v|p + |ϕ|λ + 1)⟩dx

1p

;

N < p <N

2− λ, ∀G′′′ ⊂ G′′ ⊂ G′

(4.9.8)

(here G′′′ = ψ−1(D′′′), G′′ = ψ−1(D′′)).Now we apply Lp− estimate (Theorem 4.6) to the solution of (L)0:∫

D′′

|vyy|pdy ≤ c∫D′

(|v|p + |F |p

)dy,(4.9.9)


where c depends on N, ν, µ,K, λ,A, D′, |ψ|λ, |ψ−1|λ withK from the assump-tion W5). Considering the property 4 of H, from (4.9.9) it follows

∫G′′

|vxx|pdx ≤ c1(|H|1)∫D′′

|vyy|pdy +

+ c2(|ψ−1|1)∫G′′

dp(λ−2)(x)|∇v|pdx ≤

≤ cc1

∫D′

(|v|p + |F |p

)dy +

+ c2

∫G′′

dp(λ−2)(x)|∇v|pdx ≤

≤ c∫G′

dp(λ−2)(x) (|∇v|p + |v|p + |ϕ|λ + 1) +

+ |f |pdx

(4.9.10)

in virtue of the property 3 of Φ and the assumption W5).Thus from (4.9.8) and (4.9.10) we obtain

(4.9.11) |v|λ,G′′ ≤ c1|v|0,G′ + c

∫G′

dp(λ−2)(x)(|∇v|p + |v|p + |ϕ|λ +

+ 1)

+ |f |pdx

1p

; N < p <N

2− λ, ∀G′′ ⊂ G′ ⊂ G ∪ T.

Step 2.Let x0 ∈ G, x∗0 ∈ ∂G be arbitrary points. Put d = 1

4d(x0). We rewritethe equation (L)0 in the form

aij(x∗0)Dijv = F +(aij(x∗0)− aij(x)

)Dijv,(L)∗0

where, by the assumption W5) and the properties of Φ,

|F| ≤ c(µ,K)dλ−2(x)(1 + |v|+ |∇v|).(4.9.12)

Let G(x, y) be the Green function of the operator aij(x∗0)Dij in the ballB%(0). Then according to the Green representation formula (3.2.1), almost


everywhere

(4.9.13) v(y) =∫

∂B%(x0)

v(x)∂G(x, y)∂νx

dsx +

+∫

B%(x0)

G(x, y)F +

(aij(x∗0)− aij(x)

)Dijv

dx ≡

≡ J1(y) + J2(y), % ∈ [2d, 3d].

Remark 4.44. We observe that the Green representation is valid be-cause v and Div are absolutely continuous on almost every line parallel toone of the coordinate axis, and thus partial integration is allowed.

Now using Lemma 3.9 with the Hölder inequality,

|DkJ1(y)|p ≤

C%−N ∫∂B%(x0)

|v|dsx

p

≤ Cd1−p−N∫

∂B%(x0)

|v|pdsx

from which follows

∫Bd(x0)

dp(λ−2)(y)|DkJ1(y)|pdy ≤ Cdp(λ−3)+1

∫∂B%(x0)

|v|pdsx,(4.9.14)

if we take into account that

d(y) ≤ |d(y)− d(x0)|+ d(x0) ≤ d+ 4d = 5d;

d(y) ≥ d(x0)− |y − x0| ≥ 4d− d = 3d;and therefore

3d ≤ d(y) ≤ 5d.(4.9.15)


Similarly, by Lemma 3.9 and the Hölder inequality,

|DkJ2(y)|p =

∣∣∣∣∣∣∣∫

B%(x0)

DkG(x, y)F +


)Dijv

dx

∣∣∣∣∣∣∣p

dx ≤

≤ C

∫B3d(x0)

|x− y|1−N∣∣F +


)Dijv

∣∣ dxp

dx =

= C( ∫B3d(x0)

|x− y|p−1p

(α−N) ×

× |x− y|1−N−p−1p

(α−N) ∣∣F +(aij(x∗0)− aij(x)

)Dijv

∣∣ dx)pdx ≤≤ C

∫B3d(x0)

|x− y|α−Ndx

p−1

×

×∫

B3d(x0)

|x− y|p(1−α)+α−N ∣∣F +(aij(x∗0)− aij(x)

)Dijv

∣∣p dx,∀α ∈ (0, 1)

or ∫Bd(x0)

dp(λ−2)(y)|DkJ2(y)|pdy ≤

≤ Cdp(λ−1)

∫B3d(x0)

∣∣F +(aij(x∗0)− aij(x)

)Dijv

∣∣p dx.Hence and from (4.9.13), (4.9.14) we have

(4.9.16)∫

Bd(x0)

dp(λ−2)(y)|∇v(y)|pdy ≤ Cdp(λ−3)+1

∫∂B%(x0)

|v|pdsx +

+ Cdp(λ−1)

∫B3d(x0)

∣∣F +(aij(x∗0)− aij(x)

)Dijv

∣∣p dx.Now we take into account that d(x0) ≤ d(x) + |x − x0| and therefore inB3d(x0) hold

d =14d(x0) ≤ 1

4d(x) +

34d =⇒ d ≤ d(x).


Therefore, integrating (4.9.16) with respect to % from 2d to 3d, we get theinequality

(4.9.17)∫

Bd(x0)

dp(λ−2)(x)|∇v(x)|pdx ≤

≤ C∫

B3d(x0)

dp(λ−3)(x)|v|p + dp(λ−1)(x)

(|F|p +

+ |(aij(x)− aij(x∗0)

)Dijv|p

)dx.

Finally, from the inequalities (4.9.10), (4.9.12), (4.9.17) follows the inequality∫Bd(x0)

dp(λ−2)(x)|∇v(x)|pdx ≤ c1

∫B4d(x0)

dp(2λ−3)(x)|∇v(x)|pdx+

+c2

∫B4d(x0)

dp(λ−3)(x)|v|p + dp(2λ−3)(x)

dx.(4.9.18)

Step 3.It is well known (see, for example, 2.2.2 [197]), that the smallest positive

eigenvalue ϑ of the problem (EV P1) for (N − 1)− dimensional sphere orhalf-sphere is equal to N − 1 and therefore, by the formula (2.4.8), thecorresponding value λ = 1. To the problem (L)0 we apply Theorem 4.21 in(N − 1)− dimensional sphere with s = λ > 1; as a result we obtain

|v(x)| ≤ c0d(x), x ∈ B4d(x0).

Therefore from (4.9.18) we get∫Bd(x0)

dp(λ−2)(x)|∇v(x)|pdx ≤ c1

∫B4d(x0)

dp(2λ−3)(x)|∇v(x)|pdx+

+ c2

∫B4d(x0)

dp(λ−2)(x)dx.(4.9.19)

Now consider the region G′t dened by

G′t = x ∈ G′| d(x) > t,where d(x) is the boundary distance function of G while dt(x) will be thatof Gt. We apply the following Lemma on the covering :

Lemma 4.45. (See Lemma 3.1 [402], 1.2.1 [258]). Let G be any boundedopen domain in RN and let B be the set of balls B = B 1

4d(x)(x) with center

x and radius 14d(x), d(x) being the distance from x to ∂G. Then there exists

a denumerable sequence of balls B(k)∞1 , B(k) = B 14d(x(k))(x

(k)) with the

properties:

•⋃B(k) = G;


• every point of G is inside at most C(N) of balls B′(k)∞1 , B′(k) =

B 34d(x(k))(x

(k)); C(N) depends only on N .

Let us choose a covering B(k)∞1 of G′t. Assuming the centers of theballs in the covering to be x(k)∞1 , dene x(k)∗ as one the points satisfyingx(k)∗ ∈ ∂G∩∂G′, |x(k)−x(k)∗| = d(x(k)). Then apply the estimate (4.9.19)for each k with x0 = x(k) and x∗0 = x(k)∗. Since

C1dk ≤ dt(x) ≤ C2dk for |x− x(k)| ≤ 4dk where dk =14dt(x(k)),

we get∫Bdk (x(k))

dp(λ−2)t (x)|∇v(x)|pdx ≤ c1

∫B4dk

(x(k))

dp(2λ−3)t |∇v|pdx+

+ c2

∫B4dk

(x(k))

dp(λ−2)t (x)dx.

Now, summing these inequalities over all k, we have

(4.9.20)∫G′t

dp(λ−2)t (x)|∇v(x)|pdx ≤ c1

∫G′t


+ c2

∫G′t

dp(λ−2)t (x)dx.

Since c1 does not depend on t and λ > 1, we can nd some t′ which isindependent of t, and is such that

c1dp(λ−1)(x) <

12,

if d(x) < t′. Then, if t < 12 t′,

c1

∫G′t

dp(2λ−3)t |∇v|pdx ≤ 1

2

∫G′t∩d(x)≤t′

dp(λ−2)t (x)|∇v(x)|pdx+

+ c1

∫G′t∩d(x)>t′

dp(2λ−3)t |∇v|pdx.

Hence and from (4.9.20) it follows∫G′t

dp(λ−2)t (x)|∇v(x)|pdx ≤ C

∫G′t∩d(x)>t′


+ C

∫G′t

dp(λ−2)t (x)dx.

(4.9.21)


It should be noted that the second integral does not depend on t, but dependson |∇v| and t′; in fact, since t < 1

2 t′ we have G′t ∩ d(x) > t′ = G′t′ and

dt(x) = d(x)− t > t′ − 12 t′ = 1

2 t′ on this set. We note that G′t′ ⊂ G′t ⊂ G′.

Finally, from (4.9.21) we get∫G′t

dp(λ−2)t (x)|∇v(x)|pdx ≤ C(λ, p, diamG)

∫G′t′

|∇v(x)|pdx+

+ C

∫G′t

dp(λ−2)t (x)dx,

(4.9.22)

since λ > 1.Now we apply the Lp− estimate (Theorem 4.6) to the solution of (L)0:∫

G′t′

|∇v(x)|pdx ≤ c∫G′t

(|v|p + |f |p +Kd

p(λ−2)t

)dx,(4.9.23)

where c depends on N, ν, µ, |ϕ|λ, λ,A, G′ with K from the property 3 of Φ.Then from (4.9.22), (4.9.23) we have∫

G′t

dp(λ−2)t (x)|∇v(x)|pdx ≤ c

∫G′t

(|v|p + |f |p + d

p(λ−2)t

)dx(4.9.24)

Step 4.Let x0 ∈ T and N < p < N

2−λ , 1 < λ < 2. Then, by Corollary 4.43, weget ∫

Bd(x0)

dp(λ−2)(x)dx ≤ cd∫

0

rp(λ−2)+N−1dr ≤ const.

Performing a covering of G′ by the spheres with centers x0 ∈ T we get hencethat ∫

G′

dp(λ−2)(x)dx ≤ C <∞.(4.9.25)

Similarly, by setting ρ(x) = d(x)− t, we obtain∫G′t

dp(λ−2)t (x)dx =

∫G′∩ρ(x)>0

ρp(λ−2)(x)dx ≤ C <∞.(4.9.26)

Hence, if we put t = tk and let k →∞, (4.9.24), (4.9.26) imply that (4.9.11)is nite, with Fatou's Lemma. The theorem is proved.


4.10. Unique solvability results

In this section we investigate the existence of solutions in weighted Sobo-lev spaces for the boundary value problem (L) under minimal assumptions onthe smoothness of the coecients. Let λ be the smallest positive eigenvalueof (EV P1) with (2.4.8).

Theorem 4.46. Let p ∈ (1,∞), α, β ∈ R with

−λ+ 2−N < 2− (β +N)/p ≤ 2− (α+N)/p < λ.

Furthermore, let us assume that

|x|(α−β)/pA(|x|)→ 0 as |x| → 0(4.10.1)

and suppose that assumptions A1) - A4) are fullled. If u ∈ V 2p,β(G) is a so-

lution of the boundary value problem (L) with f ∈ V 0p,α(G), ϕ ∈ V 2−1/p

p,α (∂G)then u ∈ V 2

p,α(G) and the following apriori estimates are valid

‖u‖V 2p,α(G) ≤ c

‖f‖V 0

p,α(G) + ‖ϕ‖V

2−1/pp,α (∂G)

+ ‖u‖V 0p,α(G)

(4.10.2)

with a constant c > 0 which depends only on ν, µ, α,N,A(diamG) and themoduli of continuity of aij .

Proof. We write the equation Lu = f in the form

∆u(x) = f(x)−

−((

aij(x)− δji) ∂2u

∂xi∂xj(x) + ai(x)

∂u

∂xi(x) + a(x)u(x)

).

(4.10.3)

Due to Theorem 3.11 we then have

‖u‖V 2p,α(G) ≤ c2

‖∆u‖V 0

p,α(G) + ‖ϕ‖V

2−1/pp,α (∂G)

.(4.10.4)

Estimating the V 0p,αnorm of the right hand side of (4.10.11) we obtain from

the condition A4)

‖∆u‖pV 0p,α(G)

≤ c3

(‖f‖p

V 0p,α(G)

+∫G

Ap(|x|)rα(|D2u|p +

+ r−p|∇u|p + r−2p|u|p)dx

)with c3 depending only on p and N.

Decomposing the domain G into G = Gd0 ∪Gd we then obtain

(4.10.5) ‖∆u‖V 0p,α(G) ≤ c4

(‖f‖V 0

p,α(G) +

+ supx∈(0,d)

|x|(α−β)/pA(|x|)‖u‖V 2p,β(Gd0) + sup

x∈GA(|x|)‖u‖W 2,p(Gd)

)

4.10 Unique solvability results 145

with c4 depending only on N, p and d. Since all terms on the right hand sideof (4.10.5) are nite, we conclude that u ∈ V 2

p,α(G).Furthermore, from the local Lp apriori estimates (see Theorem 4.6)

applied to the solution u of (L) we have

(4.10.6) ‖u‖W 2,p(Gd) ≤ c5

(‖f‖Lp(Gd/2) + ‖ϕ‖W 2−1/p,p(Γd/2) +

+ ‖u‖Lp(Gd/2)

)≤ c6

(‖f‖V 0

p,α(G) + ‖ϕ‖V

2−1/pp,α (∂G)

+

+ ‖u‖V 0p,α(G)

)with c6 depending only on N, p, ν, µ,G, d, α, the moduli of continuity of thecoecients aij on Gd and on

∥∥∥( N∑i=1

|ai|2)1/2∥∥∥

LN (G), ‖a‖Lp(G).

Combining the estimates (4.10.4)(4.10.6) and taking the continuity ofthe imbedding

V 2p,α(G) → V 2

p,β(G)

into account we arrive at

‖u‖V 2p,α(G) ≤ c7 sup

|x|∈(0,d)|x|(α−β)/pA(|x|)‖u‖V 2

p,α(Gd0)

+c8

(‖f‖V 0

p,α(G) + ‖ϕ‖V

2−1/pp,α (∂G)

+ ‖u‖V 0p,α(G)

).

Choosing d small enough and applying the condition (4.10.1) we obtain

‖u‖V 2p,α(G) ≤ c9

(‖f‖V 0

p,α(G) + ‖ϕ‖V

2−1/pp,α (∂G)

+ ‖u‖V 0p,α(G)

).(4.10.7)

Theorem 4.47. Let p ∈ (1,+∞), α ∈ R with

−λ+ 2−N < 2− α+N

p< λ

and let u ∈ V 2p,α(G) be a strong solution of (L) with f ∈ V 0

p,α(G) and ϕ ∈V

2−1/pp,α (∂G). If u is the only solution in the space V 2

p,α(G) then followinga-priori estimate is valid

‖u‖V 2p,α(G) ≤ c

(‖f‖V 0

p,α(G) + ‖ϕ‖V

2−1/pp,α (∂G)

).(4.10.8)

Proof. Due to Theorem 4.46 we have

‖u‖V 2p,α(G) ≤ c

(‖Lu‖V 0

p,α(G) + ‖u‖V

2−1/pp,α (∂G)

+ ‖u‖V 0p,α(G)

).


Let us suppose that (4.10.8) is not valid. Then there exists a sequenceuj∞j=1 ⊂ V 2

p,α(G) such that

‖uj‖V 2p,α(G) ≥ j

(‖Luj‖V 0

p,α(G) + ‖uj‖V 2−1/pp,α (∂G)

+ ‖uj‖V 0p,α(G)

).

After the normalization ‖uj‖V 2p,α(G) = 1 we obtain

‖Luj‖V 0p,α(G) + ‖uj‖V 2−1/p

p,α (∂G)+ ‖uj‖V 0

p,α(G) ≤1j.

Since the imbedding V 2p,α(G) → V 0

p,α(G) is compact, there exists a subse-quence uj′∞j′=1 such that

uj′ → u∗ in V 0p,α(G) for some u∗ ∈ V 0

p,α(G).

Moreover, we have

‖ui′ − uj′‖V 2p,α(G) ≤ c

(‖Lui′ − Luj′‖V 0

p,α(G) +

+ ‖ui′ − uj′‖V 2−1/pp,α (∂G)

+ ‖ui′ − uj′‖V 0p,α(G)

).

Thus uj′∞j′=1 is a Cauchy sequence in V2p,α(G). Consequently u∗ belongs to

V 2p,α(G) and is a nontrivial solution of the boundary value problem (L) withf ≡ 0, ϕ ≡ 0, in contradiction to the unique solvability assumption.

Theorem 4.48. Let p ≥ N,α ∈ R and suppose that assumptions A1) -A4) are fullled and that a(x) ≤ 0 for all x ∈ G. Then the Dirichlet problem(L) has a unique solution u ∈ V 2

p,α(G) for all f ∈ V 0p,α(G) ∩ Lp(G), ϕ ∈

V2−1/pp,α (∂G) if and only if

0 < 2− (α+N)/p < λ.

In this case the following apriori estimate is valid

‖u‖V 2p,α(G) ≤ c

‖f‖V 0

p,α(G) + ‖ϕ‖V

2−1/pp,α (∂G)

.(4.10.9)

Proof. We prove the existence of a solution by the method of continuity(see Theorem 1.54). We consider the family of boundary value problemsdepending on the parameter t ∈ [0, 1] :

Ltu := tLu+ (1− t)∆u = f in G,u = ϕ on ∂G.

(L)t

The operator Lt is uniformly elliptic with the ellipticity constants

µt = max1, µ, νt = max1, ν

and is continuous if considered between the Banach spaces

Lt : V 2p,α(G)→ V 0

p,α(G)× V 2−1/pp,α (∂G)


Let us denote by ut a solution of the boundary value problem (L)t for t ∈[0, 1]. We will show that

‖ut‖V 2p,α(G) ≤ c1

‖f‖V 0

p,α(G) + ‖ϕ‖V

2−1/pp,α (∂G)

∀t ∈ [0, 1](4.10.10)

with a constant c1 independent of t, ut and f, ϕ. To this end we write theequation Ltut = f in the form

(4.10.11) ∆ut(x) = f(x)− t((aij(x)− aij(0)

)Dijut(x) + ai(x)Diut(x) +

+ a(x)ut(x)).

Due to Theorem 3.11 we then have

‖ut‖V 2p,α(G) ≤ c2

‖∆ut‖V 0

p,α(G) + ‖ϕ‖V

2−1/pp,α (∂G)

.(4.10.12)

Estimating the V 0p,αnorm of the right hand side of (4.10.11) we obtain from

the condition A4)

‖∆ut‖pV 0p,α(G)

≤ c3

(‖f‖p

V 0p,α(G)

+∫G

Ap(|x|)rα(|D2ut|p + r−p|∇ut|p +

+ r−2p|ut|p)dx)

with c3 depending only on p and N.Decomposing the domain G into G = Gd0 ∪Gd we then obtain

(4.10.13) ‖∆ut‖V 0p,α(G) ≤ c4

(‖f‖V 0

p,α(G) +A(d)‖ut‖V 2p,α(Gd0) +

+ supx∈GA(|x|)‖ut‖W 2,p(Gd)

)with c4 depending only on N, p and d. Furthermore, from the Lp estimate(see Theorem 4.6) applied to the solution ut of (L)t we have

(4.10.14) ‖ut‖W 2,p(Gd) ≤ c5

(‖f‖Lp(Gd/2) + ‖ϕ‖W 2−1/p,p(Γd/2) +

+ ‖ut‖Lp(Gd/2)

)≤ c6

(‖f‖V 0

p,α(G) + ‖ϕ‖V

2−1/pp,α (∂G)

+

+ ‖ut‖V 0p,α−2p−1(G)

)with c5 depending only on N, p, ν, µ,G, d, the continuity moduli of the coef-cients aij on Gd and on

∥∥∥( N∑i=1

|ai|2)1/2∥∥∥

Lp(G), ‖a‖Lp/2(G), p > N.


Combining the estimates (4.10.12)(4.10.14) we arrive at

‖ut‖V 2p,0(G) ≤ c2c4A(d)‖ut‖V 2

p,α(Gd0) + c7

(‖f‖V 0

p,α(G) + ‖ϕ‖V

2−1/pp,α (∂G)

+

+ ‖ut‖V 0p,α−2p−1(G)

).

If we choose d small enough, then

c2c4A(d) ≤ 1/2

due to the continuity of the function A. Therefore,

(4.10.15) ‖ut‖V 2p,α(G) ≤ 2c7

(‖f‖V 0

p,α(G) + ‖ϕ‖V

2−1/pp,α (∂G)

+

+ ‖ut‖V 0p,α−2p−1(G)

).

We remark that according to Lemma 1.38 we have V 2p,α(G) → C0(G) and

ϕ ∈ C0(∂G) for 0 < 2 − (α + N)/p. Thus the boundary value problem (L)can have at most one solution in the space V 2

p,α(G) due to Theorem 4.1. Dueto Lemma 1.37 the imbedding

V 2p,α(G) → V 0

p,α−2p−1(G)

is compact and we can apply the standard compactness argument (see The-orem 4.47) in order to get rid of the ‖ut‖V 0

p,α−2p−1(G) term on the right handside of (4.10.15). Thus

‖ut‖V 2p,0(G) ≤ c11

(‖f‖V 0

p,α(G) + ‖ϕ‖V

2−1/pp,α (∂G)

).

Since the boundary value problem (L)t is uniquely solvable for t = 0 due toTheorem 3.11 we conclude from Theorem 1.54 that (L)t is uniquely solvablefor t = 1, too.

Theorem 4.49. Let Γd ∈ C1,1 with some d > 0. Let λ ∈ (1, 2) and thenumbers are given q ≥ N

2−λ , N ≤ p < q and α ∈ R satisfying the inequality

0 < 2− (α+N)/p < λ.

Suppose that assumptions A1) - A4) are fullled with A(r) Dini-continuousat zero and, in addition,

A6) a ∈ LN (G) and a(x) ≤ 0 for all x ∈ G;A7) f ∈ V 0

q,α(G)∩Lq(G), ϕ ∈ V 2−1/qq,α (∂G)∩W 2−1/q,q(∂G) and there exist

real numbers s > λ, k1 ≥ 0, k2 ≥ 0, k3 ≥ 0 such that

k1 =: sup%>0

%−s(‖f‖

W0

4−N (G%0)+ ‖ϕ‖

W3/2

4−N (Γ%0)

)+

+ sup%>0

%1−s(‖f‖

N ;G2%%/4

+ ‖ϕ‖V

2−1/NN,0 (Γ2%

%/4)

),

k2 =: sup%>0

%2−λ−α+N

q

(‖f‖

V 0q,α(G2%

%/4)

+ ‖ϕ‖V

2−1/qq,α (Γ2%

%/4)

),


W5)

(N∑i=1

|ai(x)|2)1/2

+ |a(x)|+ |f(x)| ≤ k3dλ−2(x), x ∈ Gε, ∀ε > 0,

where d(x) is the distance from x to ∂G.

Then the problem (L) has a unique solution

u ∈W 2,qloc (G) ∩ V 2

p,α(G) ∩ Cλ(G)

and the following a-priori estimate is valid

‖u‖Cλ(G) ≤ K(4.10.16)

with the constant K independent of u and dened only by N, q, ν, µ, λ, s, k1,

k2, k3, ‖f‖Lq(G),maxx∈GA(|x|), ‖ϕ‖

V2−1/qq,α (∂G)

,d∫0

A(t)t dt and the domain G.

Proof. In virtue of Theorem 4.1 the problem (L) has a unique solutionu ∈W 2,q

loc (G)∩C0(G). Using the Hölder inequality with s = qp > 1, s′ = q

q−pwe obtain

∫G

rα|f |pdx =∫G

rα/s|f |p · rα/s′dx ≤

∫G

rα|f |qdx

p/q

·

∫G

rαdx

(q−p)/q

≤ C‖f‖pV 0q,α(G)

,

since α+N > p(2− λ) > 0. Now it is easy to verify that all assumptions ofTheorem 4.48 are fullled and therefore according to this Theoremu ∈ V 2

p,α(G) and the estimate (4.10.9) is true.Now let us prove u ∈ Cλ(G) and the estimate (4.10.16). For this we

apply the local estimates of 4.4, 4.6, 4.9. We consider the partition ofunity

1 =∑k

ζk(x), where ζk(x) ∈ C∞0 (Gj),⋃j

Gj = G.

Let Φ ∈ V 2q,α(G)∩C0(G) be an arbitrary extension of the boundary function

ϕ into G. The function v = u − Φ then satises the homogeneous Dirichletproblem:

Lv = F in G,v = 0 on ∂G.

(L)0

with F (x) determined by (4.3.4). Setting vk(x) = ζk(x)v(x) we have

Lvk(x) = Fk(x) ≡ ζk(x)F (x) + 2aij(x)ζkxjvxi +(aij(x)ζkxixj +

+ ai(x)ζkxi)v(x).

(4.10.17)


At rst we consider such ζk(x) the support of which intersects with thed− vicinity of the origin O. The assumptions of our theorem guarantee thefullment of all conditions of Theorems 4.21, 4.33 and therefore we have:

|Fk(x)| ≤ ck(|F (x)|+ |∇v(x)|+ |x|−1|v(x)|

)≤

≤ ck(|F (x)|+ |x|λ−1

)≤ ck

(|f(x)|+ |x|λ−1 +

+ |Φxx|+A(|x|)|x|

|∇Φ|+ A(|x|)|x|2

|Φ|), x ∈ Gd0,

(4.10.18)

if we recall (4.3.4). Now we verify that we can apply to the solutions of(4.10.17) Theorems 4.21, 4.33, too. In fact, by (4.10.18) and the assumptionA7), we obtain:

∫G%0

r4−NF 2k (x)dx ≤ 2c2

k

∫G%0

(r4−NF 2(x) + r2λ+2−N

)dx ≤ 2c2

kk21%

2s +

+ 2c2k

measΩ2λ+ 2

%2λ+2 ≤ Ck%2es s > λ, % ∈ (0, d).

Similarly

‖Fk‖N ;G%%/2≤ ck

(‖f‖

N ;G2%%/4

+ ‖ϕ‖V

2−1/NN,0 (Γ2%

%/4)

)+

+

∫G%0

rN(λ−1)dx

1/N

≤ ckk1%s−1 + ck%

λ;

hence the (4.4.2) follows with s > λ and Fk ∈ LN (G), since s > λ > 1. Weveried the conditions of Theorem 4.21.

Further,

∫G%%/2

r2q−N |Fk(x)|qdx ≤ cqk∫

G%%/2

(r2q−N |f(x)|q + r2q−N |Φxx|N + rq−N |∇Φ|q

+ r−N |Φ|q + rq(λ+1)−N)dx ≤

≤ cqk%2q−N−α

∫G%%/2

(rα|f |q + rα|Φxx|q + rα−q|∇Φ|q +

+ rα−2q|Φ|q)dx+ Ck%

q(λ+1) ≤

≤ ckkq2%qλ, % ∈ (0, d)

because of the Assumption A7). By this the fullment of the AssumptionA5) is veried and therefore all conditions of Theorem 4.33 are fullled.

4.11 Notes 151

Finally, on the basis of the Alexandrov Maximum Principle (see Theorem4.2) we have

M0 = supG|u| ≤ sup

∂G|ϕ|+ c‖f‖LN (G).

Thus, by Theorems 4.21, 4.33, we get vk(x) ∈ Cλ(Gd0) and

‖vk‖Cλ(Gd0)≤ Kk.(4.10.19)

Now let us consider such ζk(x) the support of which intersects with theΓd with some d > 0. In this case we can apply the Widman local estimates(see 4.9) near the smooth piece of the boundary of G. In particular, byTheorem 4.42 with regard to the Assumption W5), we obtain

‖Fk(x)‖ ≤ Ckdλ−1(x).(4.10.20)

The inequality (4.10.20) and the assumption W5) allow to apply Theorem4.42 to the equation (4.10.17), too. Therefore we can conclude that

‖vk(x)‖Cλ(G

jkd )≤ Cjk .(4.10.21)

Finally, if the support of ζk(x) belongs strictly to the angular domain G,since u ∈ W 2,q

loc (G), q ≥ N2−λ , by the Sobolev imbedding Theorem, we have

that vk(x) ∈ Cλ(G′k), ∀G′k ⊂⊂ G and in virtue of Theorem 4.7 for k = 2the estimate

‖vk(x)‖Cλ(G′k) ≤ C‖vk‖W 2,q(G′k) ≤ Ck(4.10.22)

holds.

Since v(x) =∑k

vk(x) and this sum is nite, from the estimates (4.10.19),

(4.10.21), (4.10.22) it follows that v ∈ Cλ(G) and the validity of (4.10.16).Thus our Theorem is proved.

Since the Widman results (4.9) are true for the Liapunov domains, in thisway the following Theorem is proved:

Theorem 4.50. Let Γd ∈ Cλ with some d > 0. Let the assumptionsof Theorem 4.49 be fullled. Then the problem (L) has a unique solution

u ∈W 2,qloc (G) ∩ Cλ(G) and the estimate (4.10.16) holds.

4.11. Notes

The behavior of the problem (L)- solutions near a conical point wasstudied in the case of the Hölder continuity coecients in [16] - [19], [395,396]. Our presentation of the results of this Chapter follows [53, 56, 57,58, 63, 66]. These results were generalized in [366, 50] on linear ellipticequations whose coecients may degenerate near a conical boundary point.Theorem 4.48 was known earlier in two cases: either when the problem (L)equation is the Poisson equation [397] or when G is a cone, but the lowestequation coecients are more smooth (theorem 2.2 [187]). Theorems 4.49


and 4.50 are new because without our new estimates from 4.5, 4.6 as wellas the Widman estimates from 4.9 they could not be proved. Moreover, inthese theorems we weaken the smoothness requirement on the surface ∂G\O.In Theorem 4.49 these requirements allow to "straighten" a locally smoothpiece of surface; in Theorem 4.50 the surface ∂G \ O can be the Liapunovsurface, because in such a domain the Widman results (4.9) are correct,and we use them in the neighborhood of a smooth piece of ∂G \ O.

Other boundary value problems (the Neumann problem, mixed problem)for general elliptic second order equations in nonsmooth domains have beenstudied by A. Azzam [20], A. Azzam and E. Kreyszig [22, 23], G. Lieberman[227], V. Chernetskiy [81].

CHAPTER 5

The Dirichlet problem for elliptic linear divergent

equations in a nonsmooth domain

5.1. The best possible Hölder exponents for weak solutions

5.1.1. Introduction. In this Section, the behavior of weak solutions ofthe Dirichlet problem for a second order elliptic equation in a neighborhoodof a boundary point is studied. Under certain assumptions on the structureof the domain boundary in a neighborhood of the boundary point O and onthe equation coecients, one obtains a power modulus of continuity at Ofor a generalized solution of the Dirichlet problem vanishing at that point.Moreover, the exponent is the best possible for domains with the assumedboundary structure in that neighborhood. The assumptions on the equationcoecients are essential, as the example in 5.1.4 shows.

Next, it is shown, with the help of the previous results on the continuitymodulus at boundary points of the domain, that a weak solution of theDirichlet problem in a domain G belongs to a Hölder space Cλ in the closeddomain G, the exponent λ being determined by the structure of the domainboundary and being the best possible for the class of domains in question.

We consider weak solutions of the Dirichlet problem for the linear uni-formly elliptic second order equation of the divergent form

∂∂xi


= g(x) + ∂fj(x)∂xj

, x ∈ G;

u(x) = ϕ(x), x ∈ ∂G(DL)

(summation over repeated indices from 1 to N is understood), where G ⊂ RNis a bounded domain with the boundary ∂G.

At rst, we describe our very general assumptions on the structure ofthe domain boundary in a neighborhood of the boundary point O. Namely,we denote by θ(r) the least eigenvalue of the Beltrami operator 4ω on Ωr

with the Dirichlet condition on ∂Ωr. According to the variational theoryof eigenvalues (the analog of the Wirtinger inequality: see (2.3.2) Theorem2.15), we have ∫

Ωr

u2(ω)dΩr ≤1θ(r)

∫Ωr

|∇ωu|2 dΩr, ∀u ∈W 1,20 (Ωr).(5.1.1)

153

154 5 Divergent equations in a non-smooth domain

Assumption I.

θ(r) ≥ θ0 + θ1(r) ≥ θ2 > 0, whereθ0, θ2 are positive constants and

θ1(r) is a Dini continuous at zero function:

limr→0

θ1(r) = 0,

d∫0

|θ1(r)|r

dr <∞.

Assumption II.

• (i) Uniform ellipticity condition:

ν|ξ|2 ≤N∑

i,j=1

aij(x)ξiξj ≤ µ|ξ|2 ∀ξ ∈ RN , x ∈ G

with some ν, µ > 0.• (ii) aij(0) = δji .

• (iii) aij(x) ∈ C0(G), (i, j = 1, . . . , N); ai(x), bi(x) ∈ Lp(G), (i =1, . . . , N); c(x) ∈ Lp/2(G), p > N .• (v) There exists a monotonically increasing nonnegative function A

such that N∑i,j=1

|aij(x)− aij(0)|21/2

+ |x|

(N∑i=1

|ai(x)|2 +N∑i=1

|bi(x)|2)1/2

+

+ |x|2|c(x)| ≤ A(|x|) ∀x ∈ G.

• (iv) g(x), f i(x) (i = 1, . . . , N) ∈ L2(G), ϕ(x) ∈W12 (∂G).

Definition 5.1. The function u(x) is called a weak solution of theproblem (DL) provided that u(x) − Φ(x) ∈ W 1

0 (G) and satises the inte-gral identity∫

G

aij(x)uxjηxi + ai(x)uηxi − bi(x)uxiη − c(x)uη

dx =

=∫G

f i(x)ηxi − g(x)η

dx(II)

for all η(x) ∈W 10 (G).

Lemma 5.2. Let u(x) be a weak solution of (DL). For

∀v(x) ∈ V :=v ∈W 1(G%0) | v(x) = 0, x ∈ Γ%0

the equality

5.1 The best possible Hölder exponents for weak solutions 155

∫G%0

(aij(x)uxj + ai(x)u− f i(x)

)vxi +

(g(x)− bi(x)uxi − c(x)u

)vdx =

=∫Ω%


)v(x) cos(r, xi)dΩ%(5.1.2)

holds for a.e. % ∈ (0, d).

Proof. By u(x) ∈W 10 (G) and because of

∫G%0

|∇u|2dx =

%∫0

d%

∫Ω%

|∇u(r, ω)|2dΩr,

from the Fubini Theorem follows that the function

V (r) =∫Ωr

|∇u(r, ω)|2dΩr(5.1.3)

is determined and nite for a.e. all r ∈ (0, d). We consider the function

J(%) ≡∫Ω%


)v(x) cos(r, xi)dΩ(5.1.4)

for a.e. % ∈ (0, d) ∀v ∈ V . By virtue of ellipticity condition and assumptionson the equation coecients we have

aij(x)uxj cos(r, xi) ≤ µ|∇u|;ai(x)u cos(r, xi) ≤ r−1A(r)|u|,

therefore using the Cauchy inequality, we get

J(%) ≤(1 + µ%N−1 +A(%)%N−2

) ∫Ω

(|∇u|2 + u2 + v2)dΩ.(5.1.5)

Since the integral (5.1.3) is nite for a.e. all r ∈ (0, d) from (5.1.5) followsthat the function J(%) is determined and nite for a.e. all % ∈ (0, d).

Now let χ%(x) be the characteristic function of the set G%0 and (χ%)h bethe regularization of χ (see 1.5.2, chapter 1)

(χ%)h(x) =∫G

ψh(|x− y|)χ%(y)dy(5.1.6)


where ψh(|x− y|) is the mollier. It is well known that the regularization isa innity dierentiable function in the whole of the space and

∂(χ%)h(x)∂xi

=∫G

χ%(y)∂ψh(|x− y|)

∂xidy =

= −∫G

χ%(y)∂ψh(|x− y|)

∂yidy (i = 1, . . . , N).(5.1.7)

Let us take a function v ∈ W 10 (G), and set η(x) = (χ%)h(x)v(x) in the

integral identity (II). It is easily seen that such function η(x) is admissibleand moreover,

∂η

∂xi= (χ%)h(x)

∂v

∂xi− v(x)

∫G

χ%(y)∂ψh(|x− y|)

∂yidy.

Denoting by

A(x) ≡(aij(x)uxj + ai(x)u− f i(x)

)vxi +


)v(x)

from (II) follows:

(5.1.8)∫G

A(x)(χ%)h(x)dx =∫G


)v(x)×

×

∫G

χ%(y)∂ψh(|x− y|)

∂yidy

dx = (by the Fubuini Theorem)

=∫G

χ%(y)

∫G


)v(x)

∂ψh(|x− y|)∂yi

dx

dy =

(by the Theorem about dierentiability of the integral)

=∫G

χ%(y)

∂

∂yi

∫G


)v(x)ψh(|x− y|)dx

dy =

(by denition of the regularization)

=∫G

χ%(y)

∂

∂yi

((aijuxj + aiu− f i

)v)h(y)

dy =

=∫G%0

∂

∂yi


)v)h(y)dy =

=∫∂G%0


)v)h(y) cos(−→n , yi)dyσ =


=∫Ω%

(aijuxj + aiu− f i

)v(x) cos(r, xi)dΩ% +

+∫∂G%0

((aij(x)uxj + ai(x)u− f i(x)

)v(x)

)h

(y)−

−(aij(y)uxj (y) + ai(y)u(y)− f i(y)

)v(y)

cos(−→n , yi)dyσ

in virtue of ∂G%0 = Γ%0 ∪ Ω% and v(x)∣∣∣Γ%0

= 0.

Now we show that A(x) ∈ L1(G). First of all because of the assumptionson coecients,

|A(x)| ≤(µ|∇u|+ |x|−1A(|x|)|u|+ |−→f |

)|∇v|+

+(|g|+ |x|−1A(|x|)|∇u|+ |x|−2A(|x|)|u|

)|v|.

Using the Cauchy inequality, we have

|A(x)| ≤ c(µ)(|∇u|2 + |∇v|2 + v2 + g2 + |−→f |2

)+

+A(|x|)(|∇u|2 + |∇v|2 + |x|−2(u2 + v2)

).

Further, we apply the inequality (2.5.2) (see Corollary 2.23)∫G%0

|x|−2u2(x)dx ≤ C∫G%0

|∇u|2dx

Because of this inequality and the above bounds the belonging A(x) ∈ L1(G)is obvious.

Now in virtue of Lemma 1.21 we can then obtain

(5.1.9) limh→0

∫G

A(x)(χ%)h(x)dx =∫G

A(x)χ%(x)dx =∫G%0

A(x)dx =

=∫G%0

((aij(x)uxj + ai(x)u− f i(x)

)vxi +


)v(x)

)dx

Next, setting Ai(x) ≡ aij(x)uxj + ai(x)u − f i(x) we have Ai(x)v(x) ∈L1(G) (i = 1, . . . , N) and in virtue of Lemma 1.20

limh→0‖(Ai(x)v(x)

)h−Ai(x)v(x))‖L1(G%0) = 0, (i = 1, . . . , N).(5.1.10)


Representing G%0 = (0, %) × Ω%, because of Lemma 1.16, we obtain from(5.1.10) that for some subsequence hn

(5.1.11) limhn→0

‖(Ai(x)v(x)

)hn−Ai(x)v(x))‖L1(Ω%) = 0 a.e. % ∈ (0, d)

(i = 1, . . . , N).

Similarly, representing G%0 = Γ%0 × (−ω0, ω0), because of the same Lemma1.16, we obtain from (5.1.10) that for some subsequence hm

(5.1.12) limhm→0

‖(Ai(x)v(x)

)hmn−Ai(x)v(x))‖L1(Γ%0) = 0

a.e.ω1 ∈ (−ω0, ω0), (i = 1, . . . , N).

Thus, performing in (5.1.8) the passage to the limit over h→ 0 by (5.1.9) -(5.1.12) we get the required equality. Lemma 5.2 is proved.

5.1.2. The estimate of the weighted Dirichlet integral. Settingv = u− Φ we obtain that v(x) satises the integral identity∫

G

aij(x)vxjηxi + ai(x)vηxi − bi(x)vxiη − c(x)vη

dx =

=∫G

F i(x)ηxi − G(x)η

dx (II)0

for all η(x) ∈W 1,20 (G),where

F i(x) = f i(x)− aij(x)DjΦ− ai(x)Φ(x) (i = 1, . . . , N),

G(x) = g(x)− bi(x)DiΦ− c(x)Φ(x).(5.1.13)

At rst, we will obtain a global estimate for the weighted Dirichlet inte-gral.

Theorem 5.3. Let u(x) be a weak solution of the problem (DL) andsuppose that assumptions I, II are satised with a continuous at zero functionA(r). Let us assume, in addition, that

g ∈ W0α(G), f ∈ W0

α−2(G), ϕ ∈ W1/2α−2(∂G),(5.1.14)

where 4−N − 2λ < α ≤ 2

λ = 12

(2−N +

√(N − 2)2 + 4θ0

)(*)

Then we have u(x) ∈ W1α−2(G) and

‖u‖ W

1

α−2(G)≤ C

‖u‖W 1,2(G) + ‖g‖

W0

α(G)+

+ ‖f‖ W

0

α−2(G)+ ‖ϕ‖

W1/2

α−2(∂G)

,

(5.1.15)


where C > 0 is the constant dependent only on α, λ, ω0, N, µ,G and indepen-dent of u.

Proof. Replacing u by v = u − Φ and setting η(x) = rα−2ε v(x), with

regard to

ηxi = rα−2ε vxi + (α− 2)rα−3

ε

xi − εlirε

v(x)

we obtain

(5.1.16)∫G


2− α2

∫G

rα−4ε (xi − εli)(v2)xidx+

+ (2− α)∫G

((aij(x)− aij(0)

)vxj + ai(x)v + F i(x)

)rα−3ε

xi − εlirε

v(x)dx−

−∫G

((aij(x)− aij(0)

)vxj + ai(x)v + F i(x)

)rα−2ε vxidx+

+∫G

(bi(x)vxi + c(x)v − G(x)

)rα−2ε v(x)dx

We transform the rst integral on the right:

∫G

rα−4ε (xi − εli)(v2)xidx = −

∫G

v2 ∂

∂xi

(rα−4ε (xi − εli)

)dx,

because of v ∈W 1,20 (G). By elementary calculation

∂

∂xi

(rα−4ε (xi − εli)

)= Nrα−4

ε + (α− 4)(xi − εli)rα−5ε

xi − εlirε

=

= (N + α− 4)rα−4ε ,

we obtain

(5.1.17)2− α

2

∫G

rα−4ε (xi − εli)(v2)xidx =

=(2− α)(4−N − α)

2

∫G

rα−4ε v2dx.


We estimate the other integrals on the right by using our assumptions and(5.1.13):

|(aij(x)− aij(0)

)vxj + ai(x)v + F i(x)| ≤

≤ A(r)|∇v|+A(r)r−1(|v|+ |Φ|) + µ|∇Φ|+ |f |;|bi(x)vxi + c(x)v − G(x)| ≤

≤ A(r)r−1(|∇v|+ |∇Φ|) +A(r)r−2(|v|+ |Φ|) + |g|.

(5.1.18)

Now from (5.1.16), (5.1.17) it follows that

(5.1.19)∫G

rα−2ε |∇v|2dx ≤ (2− α)(4−N − α)

2

∫G

rα−4ε v2dx+

+c(N,α)∫G

rα−2ε A(r)

⟨r−1|∇v|(|v|+ |Φ|)+r−1|v||∇Φ|+r−2(v2 + |v||Φ|)+

+ |∇v|2+⟩

+ rα−2ε (µ|∇Φ||∇v|+ |v||g|) + rα−3

ε A(r)(|v||∇v|+

+ r−1v2 + r−1|v||Φ|) + µrα−3ε |v||∇Φ|

dx.

Further, we estimate by the Cauchy inequality with ∀δ > 0:

r−1|∇v||v| ≤ 12|∇v|2 +

12r−2|v|2;

r−1|∇v||Φ| ≤ 12|∇v|2 +

12r−2|Φ|2;

r−1|∇Φ||v| ≤ 12|∇Φ|2 +

12r−2|v|2;

r−2|v||Φ| ≤ 12r−2|v|2 +

12r−2|Φ|2;

µ|∇v||∇Φ| ≤ δ

2|∇v|2 +

µ2

2δ|∇Φ|2;

|g||v| = (r−1|v|)(r|g|) ≤ δ

2r−2|v|2 +

12δr2|g|2;

r−1ε |∇v||v| ≤

12|∇v|2 +

12r−2ε |v|2;

|v||Φ| ≤ 12|v|2 +

12|Φ|2;

µr−1ε |∇Φ||v| ≤ δ

2r−2ε |v|2 +

µ2

2δ|∇Φ|2.

(5.1.20)


As a result from (5.1.19) we obtain

(5.1.21)∫G

rα−2ε |∇v|2dx ≤ (2− α)(4−N − α)

2

∫G

rα−4ε v2dx+

+ c(N,α, µ)∫G

rα−2ε A(r)|∇v|2 + rα−2

ε r−2A(r)|v|2 +

+ rα−2ε A(r)(|∇Φ|2 + r−2|Φ|2) + rα−4

ε A(r)|v|2 + rα−3ε r−1A(r)|v|2 +

+ rα−3ε r−1A(r)|Φ|2 + rα−2

ε |∇Φ|2 +δ

2rα−2ε |∇v|2 +

+δ

2rα−2ε r−2|v|2 + rα−2

ε r2|g|2dx.

Now we apply the inequality (2.5.7) - (2.5.9) to the rst integral from theright side; because of the condition (*) of our Theorem we have

C(λ,N, α) = 1− (2− α)(4−N − α)2

H(λ,N, α) > 0.

Therefore we can write the inequality (5.1.21) in the following way

C(λ,N, α)∫G

rα−2ε |∇v|2dx ≤ c0 [A(d) + δ +O(ε)]

∫G


+ c1(N,α, µ)∫G

A(r)rα−2ε r−2|v|2 + rα−4

ε |v|2dx+ δ

∫G

(rα−2ε r−2|v|2

)dx+

+ c2(N,α, µ, ω0)∫G

(rα−4|Φ|2 + rα−2|∇Φ|2 + rα|g|2

)dx, ∀δ > 0

(here we use property 1) of the function rε(x)). We apply now Lemmas 2.30,2.31 and choose δ > 0 from the condition(

1 +1

λ(λ+N − 2)

)δ =

12C(λ,N, α).

As a result we obtain∫G

rα−2ε |∇v|2dx ≤ c(N,α, µλ, ω0)

∫G

[A(r) +O(ε)] rα−2

ε |∇v|2 + |v|2 +

+ rα−2(|∇Φ|2 + r−2|Φ|2) + rα|g|2dx.

We now write the representation G = Gd0 ∪ Gd and choose d > 0 so smallthat

A(d)c(N,α, µλ, ω0) < 1

(this is possible because of the continuity at zero of A(r)).


Thus, nally we obtain

∫G

rα−2ε |∇u|2dx ≤ O(ε)

∫G


+c(N,α, µλ, ω0)∫G

u2+|∇u|2+rα−2|∇Φ|2+rα−4|Φ|2+rα|g|2

)dx, ∀ε > 0.

Passaging to the limit when ε → +0 by the Fatou Theorem we have therequired estimate (5.1.15).

We pass now to the derivation of the local estimate for the weightedDirichlet integral. For this together with Assumptions I, II we make yetthe following

Assumptions III.

• (ivv) the function A(r) satises the Dini condition at zero, i.e.d∫0

A(r)r dr <∞;

• (w)∫G

r4−Nn−2λH−1(r)g2(x)dx <∞;∫G

r4−N−2λH−1(r)ϕ2(x)dx <∞;

∫G

r2−N−2λH−1(r)(

N∑i=1|f i(x)|2dx+ |∇Φ|2

)<∞,

where H(r) is a continuous, monotone increasing, Dini continuous atzero function.

Theorem 5.4. Let u(x) be a weak solution of (DL) and suppose thatassumptions I, II, III are satised. Then there exist positive constants d,C1,independent of u, g, fi, ϕ, such that

∫G%0

r2−N |∇u|2dx ≤ C1%2λ

∫G2d

0

|u(x)|2 + |∇u|2 + g2(x) +

N∑i=1

|f i(x)|2 +

+|Φ|2 + |∇Φ|2 + r4−N−2λH−1(r)g2(x) + r−N−2λH−1(r)|Φ|2 +

r2−N−2λH−1(r)|∇Φ|2 + r2−N−2λH−1(r)N∑i=1

|f i(x)|2dx,(5.1.22)

ρ ∈ (0, d).

Proof. By the above proved Theorem 5.3 we have that u(x) ∈ W1α−2(G).

Therefore we can apply Lemma 5.2 and take the function r2−n(u(x)−Φ(x))

as v(x) in the equal (5.1.2). Now replacing u by v = u − Φ as a result we


obtain

∫G%0

(aij(x)vxj + ai(x)v −F i(x)

)(r2−Nvxi + (2−N)r−Nxiv

)+

+(G(x)− bi(x)vxi − c(x)v

)r2−Nv

dx =

= %

∫Ω

(aij(x)vxj + ai(x)v −F i(x)

)v(x) cos(r, xi)dΩ.

Hence we have

(5.1.23)∫G%0

r2−N |∇v|2dx =N − 2

2

∫G%0

r−Nxi∂v2

∂xidx + %

∫Ω

v∂v

∂rdΩ +

+∫G%0

(aij(x)− aij(0)

)((N − 2)r−Nvxivxj − r2−Nvxivxj

)+

+ (N − 2)r−Nxiai(x)v2 + r2−Nv(bi(x)vxi + c(x)v − G

)+

+ r2−Nvai(x)vxi + r2−NF i(x)vxi + (2−N)r−NvxiF i(x)dx+

+ %

∫Ω

(aij(x)− aij(0)

)vvxj + ai(x)v2 − vF i(x)

cos(r, xi)dΩ.

The rst integral from the right we transform in the following way:

∫G%0

r−Nxi∂v2

∂xidx =

∫Ω%

r−Nv2xi cos(r, xi)dΩ% −

−∫G%0

v2(Nr−N −Nxir−N−1xi

r

)dx =

∫Ω

v2dΩ.


Therefore we can rewrite (5.1.23) in this way:

(5.1.24)∫G%0

r2−N |∇v|2dx =∫Ω

(%v∂v

∂r+N − 2

2v2

)dΩ +

+∫G%0

(aij(x)− aij(0)

)((N − 2)r−Nvxivxj − r2−Nvxivxj

)+

+ (N − 2)r−Nxiai(x)v2 + r2−Nv(bi(x)vxi + c(x)v − G

)+

+ r2−Nvai(x)vxi + r2−NF i(x)vxi + (2−N)r−NvxiF i(x)dx+

+ %

∫Ω

(aij(x)− aij(0)

)vvxj + ai(x)v2 − vF i(x)

cos(r, xi)dΩ.

We set V (ρ) =∫Gρo

r2−N |∇v|2dx and estimate every integral from the right

side. The rst integral is estimated by Lemma 2.29. We estimate otherintegrals from the right side by using our assumptions and (5.1.18), (5.1.20)as well as

r2−N |v||g| =(√H(r)r−

N2 |v|

)(√H−1(r)r2−N

2 |g|)≤ 1

2H(r)r−N |v|2 +

+12H−1(r)r4−N |g|2;

r2−N |vxiF i(x)| ≤ 12H(r)r2−N |∇v|2 +

12H−1(r)r2−N |F|2 ≤

≤ 12H(r)r2−N |∇v|2 +

12H−1(r)r2−N

(|f |2 + µ2|∇Φ|2 + r−2A2(r)|Φ|2

).

Then we obtain

(5.1.25) V (%) ≤ ρ

2λ+ θ1(%)h1(%)V ′(ρ) + c(N,µ)

∫G%0

A(r)r2−N |∇v|2 +

+A(r)r−Nv2 +A(r)r2−N |∇Φ|2 +A(r)r−NΦ2 +H(r)r2−N |∇v|2 +

+H(r)r−Nv2 +H−1(r)r2−N(r2g2 + |f |2 +µ2|∇Φ|2 + r−2A2(r)|Φ|2

)dx+

+%2−N∫Ω%

A(%)|v||∇v|+%−1A(%)v2+|v|

(|f |+µ|∇Φ|+%−1A(%)|Φ|

)dΩ%,

where 0 ≤ h1(%) ≤ 2√θ0+√θ2. To estimate the last integral from the right


side we apply the Cauchy inequality and the inequality (H-W):

(5.1.26) %2−N∫Ω%

A(%)|v||∇v|+ %−1A(%)v2 + |v|

(|f |+ µ|∇Φ|+

+ %−1A(%)|Φ|)dΩ% ≤

(A(%) +H(%)

) ∫Ω

(%2|∇v|2 + v2 + Φ2

)dΩ +

+%2H−1(%)∫Ω

(|∇Φ|2 + f2

)dΩ ≤ c(N,λ, θ2)

(A(%) +H(%)

)%V ′(%) +F1(%),

where

F1(%) = %2H−1(%)∫Ω

(|∇Φ|2 + f2

)dΩ +A(%)

∫Ω

|Φ|2dΩ(5.1.27)

Thus, from (5.1.25) - (5.1.27) we obtain

V (%) ≤ ρ

2λ+ θ1(%)h1(%)V ′(ρ) + c1(N,µ, λ, θ2)

(A(%) +H(%)

)V (%) +

+ c2(N,µ, λ, θ2)(A(%) +H(%)

)%V ′(%) + F1(%) + F2(%),

where

F2(%) =

%∫0

K(r)dr

K(r) =∫Ωr

H−1(r)r4−Ng2 +H−1(r)r2−N |f |2 +(5.1.28)

+H−1(r)r2−N |∇Φ|2 + r−NH−1(r)|Φ|2dΩr.

Finally, setting

P(%) =2λ+ θ1(%)h1(%)

%·

1− c1

(A(%) +H(%)

)1 +

(2λ+ θ1(%)h1(%)

)c2

(A(%) +H(%)

) ;

(5.1.29)

Q(%) =2λ+ θ1(%)h1(%)

%· F1(%) + F2(%)

1 +(2λ+ θ1(%)h1(%)

)c2

(A(%) +H(%)

) ,as result we get the dierential inequality (CP) 1.10 with N (%) = B(%) ≡ 0

V ′(%) ≥ P(%)V (%)−Q(%), % ∈ (0, d).(5.1.30)

It is easy to verify that

P(%) =2λ%

+δ(%)%,


where δ(%) satises the Dini condition at zero. Therefore we have

d∫%

P(s)ds = ln(d

%

)2λ

+

d∫%

δ(s)sds,

From this it follows that

(d

%

)2λ

≤ exp( d∫%

P(s)ds)≤(d

%

)2λd∫

0

δ(s)sds, ∀% ∈ (0, d).(5.1.31)

Now because of Theorem 1.52 we obtain

V (%) ≤ V (d) exp(−

d∫%

P(s)ds)

+

d∫%

Q(τ) exp(−

τ∫%

P(s)ds)dτ,(5.1.32)

and in virtue of (5.1.31) hence we have

V (%) ≤ C%2λ

(V (d) +

d∫%

τ−2λQ(τ)dτ),(5.1.33)

where C > 0 is a constant independent of v.Now we estimate the last integral. Because of (5.1.29) we get

d∫%

τ−2λQ(τ)dτ ≤ c3

d∫%

τ−2λ−1F1(τ)dτ + c4

d∫%

τ−2λ−1F2(τ)dτ

From (5.1.27) it follows that

(5.1.34)

d∫%

τ−2λ−1F1(τ)dτ ≤d∫%

τ−2λ+1H−1(τ)∫Ω

(|∇Φ|2 + f2

)dΩdτ +

+

d∫%

τ−2λ−1A(τ)∫Ω

|Φ|2dΩdτ ≤

≤∫Gd0

r2−2λ−NH−1(r)

(f2 + |∇Φ|2

)+ r−2λ−NH−1(r)Φ2

dx.


Further, because of (5.1.28) we change the order of integration and obtain

d∫%

τ−2λ−1( τ∫

0

K(r)dr)dτ =

%∫0

K(r)dr

d∫%

τ−2λ−1dτ +

+

d∫%

K(r)dr

d∫r

τ−2λ−1dτ =1

2λ

%∫0

K(r)(%−2λ − d−2λ

)dr +

+1

2λ

d∫%

K(r)(r−2λ − d−2λ

)dr ≤ 1

2λ

d∫0

r−2λK(r)dr.

Hence in virtue of (5.1.28) it follows that

(5.1.35)

d∫%

τ−2λ−1F2(τ)dτ ≤ 12λ

∫Gd0

r4−2λ−NH−1(r)g2(x) +

+ r−2λ−NH−1(r)Φ2 + r2−2λ−NH−1(r)(f2 + |∇Φ|2

)dx.

From (5.1.33) - (5.1.35) together with Theorem 5.3 follows the required(5.1.22). Theorem 5.4 is proved.

Theorem 5.5. Let u(x) be a weak solution of the problem (DL) withϕ ≡ 0 and suppose that assumptions I, II are satised with a continuous atzero function A(r), but not Dini-continuous. Let us assume, in addition,that

g ∈ W04−N−2λ(G), f i ∈ W0

2−N−2λ(G) i = 1, . . . , N.(5.1.36)

Then for every ε > 0 there exist positive constants d, cε, independent ofu, g, fi, such that

(5.1.37)

∫G%0

r2−N |∇u|2dx ≤ cε%2λ(1−ε)∫G

|u(x)|2 + |∇u|2 +

+ r4−N−2λg2(x) + r2−N−2λf2(x)dx,ρ ∈ (0, d), ∀ε > 0.


Proof. Similar to (5.1.24) we get from (DL)

(5.1.38)∫G%0

r2−N |∇u|2dx =∫Ω

(%u∂u

∂r+N − 2

2u2

)dΩ +

+∫G%0

(aij(x)− aij(0)

)((N − 2)r−Nuxiuxj − r2−Nuxiuxj

)+

+ (N − 2)r−Nxiai(x)u2 + r2−Nu(bi(x)uxi + c(x)u− g(x)

)+

+ r2−Nuai(x)uxi + r2−Nf i(x)uxi + (2−N)r−Nuxif i(x)dx+

+ %

∫Ω

(aij(x)− aij(0)

)uuxj + ai(x)u2 − uf i(x)

cos(r, xi)dΩ.

We set U(ρ) =∫Gρo

r2−N |∇u|2dx and estimate every integral from the right

side. The rst integral is estimated by Lemma 2.29.The other integrals fromthe right side we estimate by using our assumptions and (5.1.18) as well

r2−N |u||g| ≤ δ

2r−Nu2 +

12δr4−N |g|2,

r2−N |uxif i(x)| ≤ δ

2r2−N |∇u|2 +

12δr2−N |f |2,

∀δ > 0.

Thus we obtain

(5.1.39) U(%) ≤ c(N,µ)∫G%0

(A(r) + δ

)(r−Nu2 + r2−N |∇u|2

)+

+12δ

(r2−N |f |2 + r4−N |g|2

)dx+

ρ

2λU ′(ρ) +

+ %2−N∫Ω%

A(%)|u||∇u|+ %−1A(%)u2 + |u||f |

dΩ%, ∀δ > 0.


As above in (5.1.26), to estimate the last integral from the right side weapply the Cauchy inequality and the Wirtinger inequality:

(5.1.40) %2−N∫Ω%

(A(%)|u||∇u|+ %−1A(%)u2 + |u||f |

)dΩ% ≤

≤ A(%)∫Ω

(%2|∇u|2 + u2

)∣∣∣r=%

dΩ +δ

2

∫Ω

u2∣∣∣r=%

dΩ +%2

2δ

∫Ω

f2dΩ ≤

≤ c(A(%) + δ

)%U ′(%) +

%2

2δ

∫Ω

f2dΩ, ∀δ > 0.

Thus, from (5.1.39) - (5.1.40) we obtain

(5.1.41) U(%) ≤( ρ

2λ+ δ%

)U ′(ρ) +

(cA(%) +

δ

2

)U(%) + F1(%) + F2(%),

∀δ > 0, % ∈ (0, d),

where

F1(%) =%2

2δ

∫Ω

f2dΩ, F2(%) =12δ

%∫0

K(r)dr,

(5.1.42)

K(r) =∫Ωr

r4−N |g|2 + r2−N |f |2

dΩr.

Finally, since A(r) is continuous at zero and A(%) ≤ A(d), % ∈ (0, d), wecan choose ∀δ > 0 such d > 0 that cA(d) < δ

2 . Therefore we can rewrite(5.1.41) in this way:

(5.1.43) U(%) ≤ ρ

2λ(1 + δ)U ′(ρ) + δU(%) + F1(%) + F2(%),

∀δ > 0, % ∈ (0, d),

Setting now

P(%) =2λ%· 1− δ

1 + δ, B(%) ≡ 0,

(5.1.44)

Q(%) =2λ

1 + δ· F1(%) + F2(%)

%,


as a result we get the dierential inequality (CP ) 1.10. Now, putting ε =2δ

1+δ by calculating, we have:

exp(−

d∫%

P(s)ds)

=(%

d

)2λ(1−ε), ∀% ∈ (0, d).(5.1.45)

Now, because of Theorem 1.52, we obtain

U(%) ≤ c%2λ(1−ε)(U(d) +

d∫%

τ−2λ(1−ε)Q(τ)dτ),(5.1.46)

where c > 0 is a constant independent of u.Now we estimate the last integral. Because of (5.1.44) we get

(5.1.47)

d∫%

τ−2λ(1−ε)Q(τ)dτ =2λ

1 + δ

d∫%

τ−2λ(1−ε)−1F1(τ)dτ +

+2λ

1 + δ

d∫%

τ−2λ(1−ε)−1F2(τ)dτ.

From (5.1.42) it follows that

(5.1.48)

d∫%

τ−2λ(1−ε)−1F1(τ)dτ =12δ

d∫%

τ−2λ(1−ε)+1

∫Ω

f2dΩdτ =

=12δ

∫Gd%

r2−N−2λ(1−ε)|f |2dx ≤ c∫Gd0

r2−2λ−Nf2dx.

Further, because of (5.1.42) we change the order of integration and obtain

d∫%

τ−2λ(1−ε)−1( τ∫

0

K(r)dr)dτ =

%∫0

K(r)dr

d∫%

τ−2λ(1−ε)−1dτ +

+

d∫%

K(r)dr

d∫r

τ−2λ(1−ε)−1dτ =1

2λ(1− ε)

%∫0

K(r)(%−2λ(1−ε)−d−2λ(1−ε))dr+

+1

2λ(1− ε)

d∫%

K(r)(r−2λ(1−ε) − d−2λ(1−ε))dr ≤ cε d∫

0

r−2λK(r)dr.


Hence in virtue of (5.1.42) it follows that

(5.1.49)

d∫%

τ−2λ(1−ε)−1F2(τ)dτ ≤ cε∫Gd0

r4−2λ−Ng2(x) + r2−2λ−Nf2

dx.

From (5.1.46) - (5.1.49), together with Theorem 5.3, follows the required(5.1.37). Theorem 5.5 is proved.

5.1.3. Local bound of a weak solution. We pass now to the estab-lishing of the local (near the singular boundary point) bound for a weaksolution of the problem (DL).

Theorem 5.6. Let u(x) be a weak solution of the problem (DL) andsuppose that assumptions I, II, III are satised. Let us assume, in addition,that g(x) ∈ Lp(G) for some p > N/2, f i(x) ∈ Lq(G), (i = 1, . . . , N) forsome q > N , Φ ∈W 1,s(G), s = max(2p, q) > N and∫

G

r2p−N−pλ|g(x)|pdx <∞;N∑i=1

∫G

rq−N−qλ|f i(x)|qdx <∞;

∫G

(r−N−sλ|Φ|s + rs−N−sλ|∇Φ|s

)dx <∞.

(5.1.50)

Then there exist positive constants d, c, independent of u, g, fi, ϕ, such that

|u(x)| ≤ c|x|λ(∫

G

(|u|2 + |∇u|2 + g2(x) +

+r4−N−2λH−1(r)g2(x) + r2−N−2λH−1(r)N∑i=1

|f i(x)|2 +

+r−N−2λH−1(r)|Φ|2 + r2−N−2λH−1(r)|∇Φ|2)dx

1/2

+(5.1.51)

+∫G

r2p−N−pλ|g(x)|pdx1/p

+∫G

N∑1

rq−N−qλ|f i(x)|qdx1/q

+

+(∫G


)dx

)2/s), x ∈ Gd0.

Proof. At rst we refer to well-known local estimate at the boundary(see e.g. 8.10 [128]).

Lemma 5.7. Let the (i), (iii) of assumptions II are satised and supposethat F i(x) ∈ Lq(G), (i = 1, . . . , N); G(x) ∈ Lp(G)) for some q > N, p > N

2 .


Then if v(x) ∈W 10 (G) is a solution of the problem (II)0, we have

(5.1.52) supG′′|v(x)|2 ≤ C

∫G′

v2dx+(∫G′

|G|pdx)2/p

+

+N∑i=1

(∫G′

|F i|qdx)2/q

, ∀G′′ ⊂ G′ ⊂ G,

where C = const(N, ν, µ, q, p, dist(G′′, G′)).

We make the change of variables x = %x′. Then the function v(x′) =u(%x′)− Φ(%x′) satises the problem:

∂∂x′i

(aij(%x′)vx′j + %ai(%x′)v) + %bi(%x′)vx′i + %2c(%x′)v =

= %2G(%x′) + %∂Fj(%x′)∂x′j

, x′ ∈ G21/4,

v(x′) = 0, x′ ∈ Γ21/4

(5.1.53)

in the domain G21/4, where

G(%x′) = g(%x′)− %−1bi(%x′)Φx′i− c(%x′)Φ;

F i(%x′) = f i(%x′)− %−1aij(%x′)Φx′j− ai(%x′)Φ

and therefore, because of the (i) and (v) of assumptions II,

%2|G(%x′)| ≤ %2|g(%x′)|+ c(|∇′Φ|+ |Φ|);%|F i(%x′)| ≤ %|f i(%x′)|+ c(|∇′Φ|+ |Φ|).

(5.1.54)

From an estimate of the type (5.1.52) for (5.1.53) and the domains G′′ = G11/2

and G′ = G21/4 we obtain

supG1

1/2

|v(x′)|2 ≤ C ∫G2

1/4

v2dx′ + %4

( ∫G2

1/4

|G|pdx′)2/p

+

+ %2N∑i=1

( ∫G2

1/4

|F i|qdx′)2/q

.

Hence, by (5.1.54) and the Hölder inequality, we have

supG1

1/2

|v(x′)|2 ≤ C ∫G2

1/4

v2dx′ + %4

( ∫G2

1/4

|g|pdx′)2/p

+

+ %2N∑i=1

( ∫G2

1/4

|f i|qdx′)2/q

+( ∫G2

1/4

(|Φ|s + |∇′Φ|s

)dx′)2/s

.


Now, returning again to the variables x, we nd that

supG%%/2

|v(x)|2 ≤ C ∫G2%%/4

r−Nv2dx+( ∫G2%%/4

r2p−N |g|pdx)2/p

+

+N∑i=1

( ∫G2%%/4

rq−N |f i|qdx)2/q

+( ∫G2%%/4

(r−N |Φ|s + rs−N |∇Φ|s

)dx

)2/s.

We apply the inequality (H-W) to the rst integral from the right side:

(5.1.55) supG%%/2

|v(x)|2 ≤ C ∫G2%%/4

r2−N |∇v|2dx+( ∫G2%%/4

r2p−N |g|pdx)2/p

+

+N∑i=1

( ∫G2%%/4

rq−N |f i|qdx)2/q

+( ∫G2%%/4

(r−N |Φ|s + rs−N |∇Φ|s

)dx

)2/s.

Now, because of the bound (5.1.22) from Theorem 5.4 as well as the Sobolevimbedding theorem,

maxG%%/2

|Φ| ≤ C(N, s)‖∇Φ‖Ls(G%%/2), s > N ;

from (5.1.55) it follows that

|u(x)|2 ≤ C1%2λ

∫G2d

0

|u(x)|2 + |∇u|2 + g2(x) +

N∑i=1

|f i(x)|2 +

+ |Φ|2 + |∇Φ|2 + r4−N−2λH−1(r)g2(x) + r−N−2λH−1(r)|Φ|2 +

+ r2−N−2λH−1(r)|∇Φ|2 + r2−N−2λH−1(r)N∑i=1

|f i(x)|2dx+

+ C2%2λ(∫

G2d0

r2p−N−pλ|g|pdx)2/p

+N∑i=1

(∫G2d

0

rq−N−qλ|f i|qdx)2/q

+

+(∫G2d

0


)dx

)2/s, ∀x ∈ G%%/2.

Setting now |x| = 23% hence we obtain the required estimate (5.1.51). Thus

Theorem 5.6 is proved.

In a similar way, using the bound (5.1.37) and Theorem 5.5 instead of(5.1.22) and Theorem 5.4, we get the following


Theorem 5.8. Let u(x) be a weak solution of the problem (DL) withϕ ≡ 0 and suppose that assumptions I, II are satised with a continuous atzero function A(r), but not Dini-continuous. Let us assume, in addition,that g(x) ∈ Lp(G) for some p > N/2, f i(x) ∈ Lq(G), (i = 1, . . . , N) forsome q > N and∫

G

r2p−N−pλ|g(x)|pdx <∞;N∑i=1

∫G

rq−N−qλ|f i(x)|qdx <∞.(5.1.56)

Then for every ε > 0 there exist positive constants d, cε, independent ofu, g, fi, such that

|u(x)| ≤ cε|x|λ−ε(∫

G

(|u(x)|2 + |∇u|2 + r4−N−2λg2(x) +

+ r2−N−2λ|f |2(x))dx

1/2

+∫G

r2p−N−pλ|g(x)|pdx1/p

+

+∫G

N∑1

rq−N−qλ|f i(x)|qdx1/q

), x ∈ Gd0, ∀ε > 0.

(5.1.57)

5.1.4. Example. We provide an example to show that the assumption(v) is essential for the validity of the estimates (5.1.22) and (5.1.51).

Let N = 2, let the domain G lie inside the sector

G∞0 = (r, ω)∣∣0 < r <∞, 0 < ω < ω0, 0 < ω ≤ 2π

and suppose that O ∈ ∂G and in some neighborhood Gd0 of O the boundary∂G coincides with the sides ω = 0 and ω = ω0 of the sector G∞0 . In ourcase the least eigenvalue of (EVP1) is λ = π

ω0. We consider Example 4.36 of

Section 4.7 and rewrite it in the form (DL):

a11(x) = 1− 2λ+ 1

· x22

r2 ln(1/r),

a12(x) = a21(x) =2

λ+ 1· x1x2

r2 ln(1/r),

a22(x) = 1− 2λ+ 1

· x21

r2 ln(1/r),(5.1.58)

aij(0) = δji , i, j = 1, 2;

b1(x) = −1rA(r) cosω ; b2(x) = −1

rA(r) sinω,

a1(x) = a2(x) = c(x) = g(x) = f1(x) = f2(x) = ϕ(x) ≡ 0,


where

A(r) =2

(λ+ 1) ln(1/r), =⇒

d∫0

A(r)r

dr = +∞.

Clearly, the equation (5.1.58) is uniformly elliptic in Gd0 for 0 < d < e−2 withthe ellipticity constants

ν = 1− 2ln(1/d)

and µ = 1.

The equation (5.1.58) has a particular solution of the form

u(r, ω) = rλ(

ln1r

)(λ−1)/(λ+1)

sin(λω), λ =π

ω0,

that satises the boundary conditions

u = 0 on Γd0.

This solution is continuous in G, and easy verify that it belongs to W 1(G).Clearly, this solution does not satisfy (5.1.51), and therefore not (5.1.22),since (5.1.22) implies (5.1.51).

5.1.5. Hölder continuity of weak solutions. We shall now assumethat aij(x), i, j = 1, . . . , N are continuous in G and satisfy a Dini conditionon ∂G, i.e. there exists a continuous function A(t) such that

|aij(x)− aij(y)| ≤ A(|x− y|)

for any points x ∈ ∂G and y ∈ G, with1∫0

A(t)t dt <∞. Let O be any point

on ∂G.We place the origin at O and perform a linear change of independentvariables such that aij(O) = δji , where a

ij(O) is the coecient of ∂2u∂x′i∂x

′jin

the equation of (DL), written in terms of the new variables x′. As in theIntroduction, we dene a function θ(r) for the point O and shall supposethat Assumptions I are satised for all points O ∈ ∂G, where θ0, θ1, θ2 donot depend on O. For the point O we construct integrals in the variables x′

of the form (5.1.50) and (w) from Assumptions III and assume that they arebounded by constants independent of O.

Theorem 5.9. Let u(x) be a weak solution of the problem (DL) andsuppose that assumptions I, II, III (indicated above) are satised. Let usassume, in addition, that g(x), f i(x) ∈ Lp(G), (i = 1, . . . , N); Φ(x) ∈W 1,2p

for some p ≥ N1−λ , λ < 1, where λ is dened by (5.3.1). Suppose that (5.1.50)

is fullled.Then u ∈ Cλ(G). If λ = 1 and g, f i ∈ L∞(G), (i = 1, . . . , N), then

u ∈ Cλ−ε(G) for ∀ε > 0.


Proof. We consider an arbitrary pair of points x, y ∈ G. Let

max(d(x), d(y)) < 2|x− y|.

By virtue of (5.1.51) of Theorem 5.6, in this case we have

|u(x)− u(y)||x− y|λ

≤ 2|u(x)|dλ(x)

+2|u(y)|dλ(y)

≤ C1,

where C1 is the positive constant.Consider the case 2|x − y| < d(x) = %. We make a change of variables

x− x = %x′. Then the function v(x′) = u(x+ %x′)− Φ(x+ %x′) satises inthe domain G2

0 the problem

∂∂x′i

(aij(x+ %x′)vx′j + %ai(x+ %x′)v) + %bi(x+ %x′)vx′i + %2c(x+ %x′)v =

= %2G(x+ %x′) + %∂Fj(x+%x′)∂x′j

, x′ ∈ G20;

v(x′) = 0, x′ ∈ Γ20,

where

G(x+ %x′) = g(x+ %x′)− %−1bi(x+ %x′)Φx′i− c(x+ %x′)Φ;

F i(x+ %x′) = f i(x+ %x′)− %−1aij(x+ %x′)Φx′j− ai(x+ %x′)Φ.

(5.1.59)

This problem satises the ellipticity condition (i) with the same constantsν, µ and its coecients are uniformly bounded in virtue of the condition (v),since G is a bounded domain. On the basis of Theorem 15.3' in [4], we have∫

G10

|∇′v|pdx′ ≤ C2

∫G2

0

(|v|p + %2p|G|p + %p

N∑i=1

|F i|p)dx′,(5.1.60)

where the constant C2 does not depend on v. Because of conditions (i), (v),from (5.1.60) and (5.1.59) it follows that

(5.1.61)∫G1

0

|∇′v|pdx′ ≤ C3

∫G2

0

(|v|p + %2p|g|p + %p

N∑i=1

|f i|p +

+ |∇′Φ|p + |Φ|p)dx′,

where the constant C3 does not depend on v. Since according to Theorem 5.6the function u(x) is bounded in G, and by our assumptions about g, f i,Φ, itfollows from (5.1.61) that v ∈ W 1,p(G1

0), where p ≥ N1−λ . From the Sobolev

Imbedding Theorem 1.33 it follows that u ∈ Cλ(G), if λ < 1. We therefore


have

(5.1.62) |u(x)− u(y)|p = |v(0)− v(y)|p ≤

≤ C4|y|pλ∫G2

0

(|v|p + %2p|g|p + %p

N∑i=1

|f i|p + |∇′Φ|p + |Φ|p)dx′ ≤

≤ C4%−pλ|x−y|pλ·

∫G∩|x−x|<%

(|v|p+%2p|g|p+%p

N∑i=1

|f i|p+|∇′Φ|p+|Φ|p)%−Ndx;

|v(x)| ≤ C5|x− x∗|λ ≤ C5(2%)λ for x ∈ G ∩B%(x0),

where C4, C5 = const, y = %−1(y − x) and x∗ is a point of ∂G such thatd(x) = |x− x∗|. From (5.1.62) we have

|u(x)− u(y)| ≤ C6|x− y|λ, C6 = const.

If λ = 1, then according to the Sobolev Imbedding Theorem 1.33 v(x′) ∈C1−ε, where ε = const > 0, and therefore u(x) ∈ C1−ε. This proves ourTheorem.

5.1.6. Weak solutions of an elliptic inequality. In this subsectionwe consider the properties of weak solutions of an elliptic inequality:

∂∂xi

(aij(x)uxj + ai(x)u) + bi(x)uxi + c(x)u ≤≤ g(x) + ∂fj(x)

∂xj, x ∈ G ⊂ K;

u(x) = 0, x ∈ ∂G \ O.(IDL)

Definition 5.10. The function u(x) is called a weak solution of theproblem (IDL) provided that u(x) ∈ W 1(Gε), ∀ε > 0 and satises theintegral inequality∫

G

aij(x)uxjηxi + ai(x)uηxi − bi(x)uxiη − c(x)uη

dx ≤

≤∫G

f i(x)ηxi − g(x)η

dx(II*)

whatever η ≥ 0 may be, η(x) ∈W 1(G) and has a support compact in G.

Theorem 5.11. Let u(x) be a weak solution of (IDL) in G, let G ⊂ Kbe a bounded domain, and suppose that assumptions II are satised. Let usassume, in addition

• u > 0 in G,•∫G

rα−2|∇u|2dx <∞ at 2 ≤ α < N + 2λ,

• g ∈ W0α(G), f ∈ W0

α−2(G).


There exists δ > 0 such that A(|x|) ≤ δ, x ∈ K, where δ depends only onα,K, then ∫

G

(rα−2|∇u|2 + rα−4u2

)dx ≤ c

∫G

(rα−2|f |2 + rαg2

)dx,(5.1.63)

where c > 0 is independent of u, g, f i or G.

Proof. We may redene the functions u, η beyond G as having a zerovalue. Let us assume that aij ≡ δji beyond G. Then from the inequality (II*)it follows

(5.1.64)∫K

uxiηxidx ≤∫K

(aij(0)− aij(x)

)uxjηxi − ai(x)uηxi +

+ bi(x)uxiη + c(x)uη + f i(x)ηxi − g(x)ηdx.

Let us set δ = maxGA(|x|) and let us consider a function

ϑ(t) ∈ C∞(R1), ϑ(t) ≥ 0,

ϑ(t) ≡

0 for t < 1,1 for t > 2.

Now let us consider the function

η(x) = rα−2ϑε(r)u(x) where ϑε(r) = ϑ(rε

).

The function η(x) can be taken as a probe function in (5.1.64), becauseu∣∣∣Γε

= 0. By calculating, we obtain

ηxi = rα−2ϑε(r)uxi + (α− 2)rα−4ϑε(r)xiu(x) +1εϑ′(rε

)rα−3xiu(x).

Now from (5.1.64) with this probe function it follows that

(5.1.65)∫K

⟨rα−2ϑε(r)|∇u|2 + (α− 2)rα−4ϑε(r)xiuuxi +

+1εrα−3ϑ′

(rε

)xiuuxi

⟩dx ≤

∫K

(aij(0)− aij(x)

)[rα−2ϑε(r)uxiuxj +

+ (α− 2)rα−4ϑε(r)xiuuxj +1εrα−3ϑ′

(rε

)xiuuxj

]−

− ai(x)[rα−2ϑε(r)uuxi + (α− 2)rα−4ϑε(r)xiu2 +

1εrα−3ϑ′

(rε

)xiu

2]

+

+ bi(x)rα−2ϑε(r)uuxi + c(x)rα−2ϑε(r)u2 + rα−2ϑε(r)uxifi(x) +

+(α−2)rα−4ϑε(r)xiuf i(x)+1εrα−3ϑ′

(rε

)xiuf

i(x)−rα−2ϑε(r)ug(x)dx.


If we observe that

ϑε(r) = ϑ(rε

)=

0 for r < ε,

1 for r > 2ε,⇒ ϑ′ε(r)

= 0 for r < ε and r > 2ε,6= 0 for ε < r < 2ε,

then we obtain

1): ∫K

rα−2ϑε(r)|∇u|2dx =∫Gε

rα−2ϑε(r)|∇u|2dx;

2):

(α− 2)∫K

rα−4ϑε(r)xiuuxi =α− 2

2

∫Gε

rα−4ϑε(r)xi∂u2

∂xi=

=2− α

2ε

∫G2εε

rα−3ϑ′(rε

)u2dx+

(2− α)(N + α− 4)2

∫Gε

rα−4ϑε(r)u2dx;

3):

1ε

∫K

rα−3ϑ′(rε

)xiuuxi = − 1

2ε2

∫G2εε

rα−2ϑ′′(rε

)u2dx−

− N + α− 32ε

∫G2εε

rα−3ϑ′(rε

)u2dx

(here we integrated by parts).

Further we estimate the integrals on the right hand side of (5.1.65) usingthe Cauchy inequality and taking into account Assumptions II. As a resultwe get:∫

Gε

rα−2ϑ(rε

)|∇u|2dx ≤ (2− α)(4−N − α)

2

∫Gε

rα−4ϑ(rε

)u2dx+

+c1

∫G2εε

(rα−2|∇u|2 + rα−4u2

)dx+ c2

∫Gε

A(r)ϑ(rε

)(rα−2|∇u|2 + rα−4u2

)dx+

(5.1.66)

+c3

∫Gε

ϑ(rε

)[σ(rα−2|∇u|2 + rα−4u2

)+

1σ

(rα−2|f |2 + rαg2

)]dx+

+c4

∫G2εε

ϑ′(rε

)(rα−2|f |2 + rα−4u2

)dx, ∀σ > 0.


Since all necessary integrals there exist (by the assumptions of our Theorem),we may ε tend to zero; then we obtain

(5.1.67)∫G

rα−2|∇u|2dx ≤ (2− α)(4−N − α)2

∫G

rα−4u2dx+

+ (c2δ + c3σ)∫G

(rα−2|∇u|2 + rα−4u2

)dx+

+c3

σ

(rα−2|f |2 + rαg2

)]dx, ∀σ > 0

(here we took into account that A(r) ≤ δ in G by denition of δ).

Now we apply the Hardy - Wirtinger inequality (see Theorem 2.34) forunbounded cone that is true at α ≥ 4 − N. Since by the condition of ourTheorem

2 ≤ α < N + 2λ,then it is easy to verify that

(2− α)(4−N − α)2

H(λ,N, α) < 1,

where H(λ,N, α) is from (2.5.12). Therefore from (5.1.67) it follows that

(5.1.68) C(λ,N, α)∫G

rα−2|∇u|2dx ≤ (c2δ + c3σ)∫G

(rα−2|∇u|2 +

+ rα−4u2)dx+

c3

σ

(rα−2|f |2 + rαg2

)]dx, ∀σ > 0

with

C(λ,N, α) = 1− (2− α)(4−N − α)2

H(λ,N, α) > 0.

Now we require that

δ =C(λ,N, α)

2c2(5.1.69)

and choose a constant σ so that c3σ = 14C(λ,N, α). Then from (5.1.68) we

obtain the required inequality (5.1.63).

5.2 Dini continuity of the rst derivatives of weak solutions 181

5.2. Dini continuity of the rst derivatives of weak solutions

We consider weak solutions to the Dirichlet problem (DL) in a boundeddomain G ⊂ RN with boundary ∂G that is a Dini-Lapunov surface contain-ing the origin O as a conical point. The last means that ∂G \O is a smoothmanifold but near O the domain G is dieomorc to a cone.

5.2.1. Local Dini continuity near a boundary smooth portion.

Theorem 5.12. Let A be an α− Dini function (0 < α < 1) satisfyingthe condition (1.8.5). Let G be a domain in RN with a C1,A boundary portionT ⊂ ∂G. Let u(x) ∈ W 1(G) be a weak solution of the problem (DL) withϕ(x) ∈ C1,A(∂G) Suppose the coecients of the equation in (DL) satisfy theconditions

aij(x)ξiξj ≥ ν|ξ|2, ∀x ∈ G, ξ ∈ RN ; ν = const > 0;

aij , ai, f i ∈ C0,A(G) (i, j = 1, . . . , N),

bi, c ∈ L∞(G), g ∈ LN/(1−α)(G).

Then u ∈ C1,B(G ∪ T ) and for every G′ ⊂⊂ G ∪ T

‖u‖1,B;G′ ≤ c(N,T, ν, k, d′)(|u|0;G + ‖g‖ N

1−α ;G

+N∑i=1

‖f i‖0,A;G + ‖ϕ‖1,A;∂G

),

(5.2.1)

where d′ = dist(G′, ∂G \ T ) and k = maxi,j=1,...,N

‖aij , ai‖0,A;G, |bi, c|0;G

.

Proof. At rst we atten the boundary portion T . By the denition ofa C1,A domain, at each point x0 ∈ T there is a neighborhood B of x0 anda C1,A dieomorphism ψ that atten the boundary in B. Let B%(x0) ⊂⊂B and set γ = B%(x0) ∩ G, γ = ψ(γ); τ = B%(x0) ∩ T ⊂ ∂γ andτ = ψ(τ) ⊂ ∂γ (τ is a hyperplane portion of ∂γ).Under the mapping y =ψ(x), let v(y) = v(x), η(y) = η(x). Since

vxi =∂ψk∂xi

vyk , dx = |J |dy,

where J = D(ψ1,...,ψN )D(x1,...,xN ) is a Jacobian of the transformation ψ(x), it follows

from (II)0 that∫eγ

⟨aij(y)vyj + ai(y)v − F i(y)

⟩ηyi +

+⟨G(y)− bi(y)vyi − c(y)v(y)

⟩η(y)

|J |dy = 0 (II)0


for all η(y) ∈W 1,20 (γ), where

aij(y) = akm(x)∂ψi∂xk

∂ψj∂xm

, ai(y) = ak(x)∂ψi∂xk

,

bi(y) = bk(x)∂ψi∂xk

, c(y) = c(x),

F i(y) = Fk(x)∂ψi∂xk

, G(y) = G(x).

It is not dicult to observe that conditions on coecients of the equationand on the portion T are invariant under maps of class C1,A. Indeed, let usconsider the dieomorphism ψ that is given in the following way:

yk = xk − x0k; k = 1, . . . , N − 1

yN = xN − h(x′), x′ = (x1, . . . , xN−1)

where xN = h(x′) is the equation of the surface τ and h ∈ C1,A(τ). In virtueof the property (iv) of ψ it is easy to seen that |∇h| ≤ K. We have also that|J | = 1. Further by the ellipticity condition:

aij(y)ξiξj = akm(x)∂(ξiyi)∂xk

∂(ξj∂yj)∂xm

≥

≥ νN∑k=1

( ∂

∂xk

( N∑i=1

ξiyi))2

=

= νN∑k=1

(N∑i=1

ξi∂yi∂xk

)2

= νN∑k=1

(ξk + ξN

∂yN∂xk

)2=

= ν

(ξ2 + 2ξ2

N − 2ξNN−1∑k=1

ξk∂h

∂xk+ ξ2

N

[1 +

N−1∑k=1

(∂h

∂xk

)2]).

But by the Cauchy inequality with ∀ε > 0:

2ξN∂h

∂xkξk ≤ εξ2

N

(∂h

∂xk

)2

+1εξ2k,

therefore from the previous inequality follows

aij(y)ξiξj ≥ ν(

1− 1ε

)ξ′

2 + (1− ε)ξ2N

N−1∑k=1

(∂h

∂xk

)2

+ 4ξ2N

=

= ν(

1− 1ε

)ξ′

2 + ξ2N

[4 + (1− ε)|∇h|2

]≥

≥ ν(

1− 1ε

)ξ′

2 + ξ2N

[4 + (1− ε)K2

], ∀ε > 1.

(5.2.2)

Now we show that there is ε > 1 such that

1− 1ε

= 4 + (1− ε)K2


For this we solve the equation

K2ε2 − (3 +K2)ε− 1 = 0

and obtain

ε =12

+3

2K2+

√14

+10

4K2+

94K4

.

Hence we see that ε > 1 and we have also:

1− 1ε

=8

K2 + 5 +√K4 + 10K2 + 9

.

Thus from (5.2.2) follows nally

aij(y)ξiξj ≥ νc(K)ξ2,

(5.2.3)

c(K) =8

K2 + 5 +√K4 + 10K2 + 9

.

Therefore after the preliminary attening of the portion T by means of adieomorphism ψ ∈ C1,A it is sucient prove the theorem in the case T ⊂ Σ.We use the perturbation method. We freeze the leading coecients aij(x)at x0 ∈ G∪ T by setting aij(x0) = aij0 and rewrite the equation (DL) in theform of the Poisson equation (PE) for the function v(x) = u(x)−ϕ(x) with

G(x) = g(x)− bi(x)(Div +Diϕ)− c(x)(v(x) + ϕ(x)

),(5.2.4)

F i(x) =(aij(x0)− aij(x)

)Djv − aij(x)Djϕ

− ai(x)(v(x) + ϕ(x)

)+ f i(x), (i = 1, . . . , N).

(5.2.5)

Now we can apply Theorem 3.6 and thus we obtain the desired assertion ofour Theorem. In this connection we use following estimates for functions(5.2.4), (5.2.5):

(5.2.6) ||G|| N1−α ;B+

2≤ ||g|| N

1−α ;B+2

+ k

( N∑i=1

|Div|0;B+2

+ |v|0;B+2

+

+N∑i=1

|Diϕ|0;B+2

+ |ϕ|0;B+2

)≤ ||g|| N

1−α ;B+2

+

+ k

(εN∑i=1

|Div|0,A;B+2

+ cε|v|0;B+2

+ |ϕ|1;B+2

)(by (1.11.6)),

(5.2.7)N∑i=1

||F i||0,A;B+2≤ nkA(2R)||∇v||0,A;B+

2+ k

N∑i=1

|Div|0;B+2

+

+ c(k)(|v|0;B+

2+ ||ϕ||1,A;B+

2

)+

N∑i=1

||f i||0,A;B+2.


Taking into account once more the interpolation inequality (Theorem 1.49)and the condition (1.8.5) that ensures the equivalence [. . . ]A ∼ [. . . ]B, from(5.2.6)-(5.2.7) we nally obtain the inequality

(5.2.8) ||G|| N1−α ;B+

2+

N∑i=1

||F i||0,A;B+2≤

≤ k(ε+NA(2R)

)||v||1,B;B+

2+ cε(k)

(|v|0;B+

2+ ||ϕ||1,A;B+

2

)+

+N∑i=1

||f i||0,A;B+2

+ ||g|| N1−α ;B+

2∀ε > 0.

Since A(t) is the continuous function, choosing ε,R > 0 suciently small weobtain the desired assertion and the estimate (5.2.1) in a standard way from(3.2.3), and (5.2.7), (5.2.8).

5.2.2. Dini-continuity near a conical point. We consider the prob-lem (DL) under following assumptions :

(i) ∂G is a Dini-Lyapunov surface and contains the conical point O;(ii) the uniform ellipticity holds

νξ2 ≤ aij(x)ξiξj ≤ µξ2, ∀x ∈ G, ξRN ,ν, µ = const > 0; aij(0) = δji , (i, j = 1, . . . , N);

(iii) aij(x), ai(x) ∈ C0,A(G), (i, j = 1, . . . , N), where A(t) is an α −Dini function on (0, d], α ∈ (0, 1), satisfying the conditions (1.8.5) -(1.8.6) and also

sup0<%<1

%λ−1

A(%)≤ const,(5.2.9)

|x|

(N∑i=1

|bi(x)|2)1/2

+ |x|2|c(x)| ≤ A(|x|);

(iv) g(x) ∈ L N1−α

(G), ϕ(x) ∈ C1,A(∂G), f j(x) ∈ C0,A(G), j = 1, . . . , N ;

(v)

∫G

r4−N−2λH−1(r)g2(x)dx <∞;

∫G

r2−N−2λH−1(r)( N∑i=1

|f i|2 + |∇Φ|2 + r−2Φ2)dx <∞,

where H(t) is a continuous monotone increasing function satisfyingthe Dini condition at t = 0.


Theorem 5.13. Let u(x) be the generalized solution of (DL) and supposeassumptions (i) - (v) are satised. Then there exist d > 0 and a constantc > 0 independent of u(x) and dened only by parameters and norms of thegiven functions appearing in assumptions (i)-(v), such that

(5.2.10) |u(x)| ≤ c|x|A(|x|)

(||g|| N

1−α ;G +N∑i=1

||f i||0,A;G + ||ϕ||1,A;∂G +

+∫G

(r4−N−2λH−1(r)g2(x) + r2−N−2λH−1(r)

N∑i=1

|f i(x)|2 +

+ r2−N−2λH−1(r)|∇Φ|2 + |u|2 + |∇u|2)dx

1/2), ∀x ∈ Gd0;

(5.2.11) |∇u(x)| ≤ cA(|x|)

(||g|| N

1−α ;G +N∑i=1

||f i||0,A;G + ||ϕ||1,A;∂G +

+∫G

(r4−N−2λH−1(r)g2(x) + r2−N−2λH−1(r)

N∑i=1

|f i(x)|2 +

+ r2−N−2λH−1(r)|∇Φ|2 + |u|2 + |∇u|2)dx

1/2), ∀x ∈ Gd0.

Proof. We use the Kondrat'ev method of layers: we move away fromthe conical point of % > 0 and work in G2%

%/4; after the change of variables

x = ρx′ the layer G2%%/4; takes the position of a xed domain G2

1/4 withsmooth boundary.

Step 1. We consider a solution u(x) in the domain G2d0 with some po-

sitive d << 1; then u(x) is a weak solution in G2d0 of the problem:

∂∂xi


= g(x) + ∂fj(x)∂xj

, x ∈ G2d0 ; (DL)0,2d

u(x) = ϕ(x), x ∈ Γ2d0 ⊂ ∂G2d

0 .

Wemake the change of variables x = %x′ and function v(x′) = %−1A−1(%)u(%x′),% ∈ (0, d), 0 < d << 1. Then the function v(x′) satises in the domain G2

1/4

the problem:∂∂x′i

(aij(%x′)vx′j + %ai(%x′)v) + %bi(%x′)vx′i + %2c(%x′)v =

= %A−1(%)g(%x′) +A−1(%)∂fj(%x′)∂x′j

, x′ ∈ G21/4;

v(x′) = %−1A−1(%)ϕ(%x′), x′ ∈ Γ21/4.


To solve this problem we use Theorem 5.12 about the local Dini-continuityof the rst derivatives for weak solutions of the problem (DL). We checkthe possibility of using this theorem. Since under assumption (ii), A(t) ismonotone increasing function, % ∈ (0, d), 0 < d << 1, from the inequality%−1|x− y| ≥ |x− y| it follows that

A(|x′ − y′|) = A(%−1|x− y|) ≥ A(|x− y|)

and by (iii) we have∑i,j

‖ai,j(%x′)‖0,A;G21/4

+ %∑i

‖ai(%x′)‖0,A;G21/4≤

≤∑i,j

‖ai,j(x)‖0,A;G2%

%/4+ d

∑i

‖ai(x)‖0,A;G2%

%/4<∞.

Further, let Φ(x) be a regularity preserving extension of the boundary func-tion ϕ(x) into a domain Gdε ∀ε (such an extension exists; see e.g. the Lemma6.38 [128]). Since ϕ(x) ∈ C1,A(∂G) we have

‖Φ‖1,A;G2%

%/4≤ c(G)‖ϕ‖

1,A;Γ2%%/4≤ const.

By denition of the norm in C1,A we obtain

supx,y∈G2%

%/4

x 6=y

|∇Φ(x)−∇Φ(y)|A(|x− y|)

≤ ‖Φ‖1,A;G2%

%/4≤ c(G)‖ϕ‖

1,A;Γ2%%/4.(5.2.12)

Now we show that by (v) and by the smoothness of ϕ(x)

|ϕ(x)| ≤ c|x|A(|x|), |∇Φ(x)| ≤ cA(|x|), ∀x ∈ G2%%/4.(5.2.13)

Indeed from the equality

ϕ(x)− ϕ(0) =

1∫0

d

dtΦ(τx)dτ = xi

1∫0

∂Φ(τx)∂τxi

dτ

by Hölder's inequality we have:

|ϕ(x)− ϕ(0)| ≤ r|∇Φ|.(5.2.14)

From (iv) it follows that

(5.2.15)∫G%0

(r2−N |∇Φ|2 + r−N |ϕ|2)dx =∫G%0

(r2−N−2λH−1(r)|∇Φ|2 +

+ (r−N−2λH−1(r)|ϕ|2)(r2λH(r)

)dx ≤ const · %2λH(%).

Since |ϕ(0)| ≤ |ϕ(x)|+ |ϕ(x)− ϕ(0)|, by (5.2.14)we obtain

|ϕ(0)| ≤ |ϕ(x)|+ r|∇Φ|.


Squaring both sides of last inequality, multiplying by r−N and integratingover G%0 we obtain

|ϕ(0)|2∫G%0

(r−Ndx ≤ 2∫G%0

(r2−N |∇Φ|2 + r−N |ϕ|2)dx <∞(5.2.16)

by (5.2.15). Since∫G%0

r−Ndx = mesΩ%∫0

drr = ∞, the assumption ϕ(0) 6= 0

contradicts (5.2.16). Thus ϕ(0) = 0 Then from (5.2.12) we have

|∇Φ(x)−∇Φ(y)| ≤ constA(|x− y|)‖ϕ‖1,A;Γ2%

%/4, ∀x, y ∈ G2%

%/4,

|∇Φ(y)| ≤ |∇Φ(x)−∇Φ(y)|+ |∇Φ(x)| ≤≤ cA(|x− y|)‖ϕ‖

1,A;Γ2%%/4

+ |∇Φ(x)|

Hence considering y to be xed in G2%%/4 and x as variable, we get

|∇Φ(y)|2∫

G2%%/4

r2−Ndx ≤ 2c2‖ϕ‖1,A;Γ2%

%/4

∫G2%%/4

r2−NA2(|x− y|)dx+

+ 2∫

G2%%/4

r2−N |∇Φ(x)|2dx

or by (5.2.15)

%2|∇Φ(y)|2 ≤ c(mesΩ, k1)(%2A2(%) + %2λH(%)

), ∀y ∈ G2%

%/4.

Hence the assumption (5.2.9) yields the second inequality of (10.2.85). Nowthe rst inequality of (10.2.85) follows from (5.2.14) and ϕ(0) = 0. Thus(10.2.85) is proved.

Now we obtain:

(5.2.17) %−1A−1(%)‖ϕ(%x′)‖1,A;Γ21/4≤ c%−1A−1(%)‖Φ(%x′)‖1,A;G2

1/4=

= c%−1A−1(%)

supx′∈G2

1/4

|Φ(%x′)|+ supx′∈G2

1/4

|∇′Φ(%x′)|+

+ supx′,y′∈G2

1/4

x′ 6=y′

|∇′Φ(%x′)−∇′Φ(%y′)|A(|x′ − y′|)

≤ cA−1(%) sup

x,y∈G2%%/4

x 6=y

|∇Φ(x)−∇Φ(y)|A(%−1|x− y|)

+

+ c1 = c1 + cA−1(%) sup0<t<4%

A(t)A(%−1t)

· [∇Φ]0,A;G2%

%/4≤ const, ∀% ∈ (0, d),

by (10.2.85), since by (1.8.6)

sup0<t<4%

A(t)A(%−1t)

= sup0<τ<4

A(τ%)A(τ)

≤ cA(%).


In the same way we have:

(5.2.18) A−1(%)‖f j(%x′)‖0,A;G21/4

= A−1(%)(|f j(x)|

0,A;G2%%/4

+

+ supx,y∈G2%

%/4

x 6=y

|f j(x)− f j(y)|A(%−1|x− y|)

).

Since f j ∈ C0,A(G), we get

|f j(x)− f j(y)| ≤ cjA(|x− y|), ∀x, y ∈ G2%%/4,(5.2.19)

(5.2.20)∫G%0

r2−N |f j(x)|2dx =∫G%0

(r2−N−2λH−1(r)|f j(x)|2

)(r2λH(r)

)dx ≤

≤ const · %2λH(%)

by (v). Now let y be xed in G2%%/4. Then

|f j(y)| ≤ |f j(x)|+ |f j(x)− f j(y)| ≤ |f j(x)|+ cjA(|x− y|)

Hence

|f j(y)|2∫

G2%%/4

r2−Ndx ≤ 2c2j

∫G2%%/4

r2−NA2(|x− y|)dx+ 2∫

G2%%/4

r2−N |f j(x)|2dx.

Calculations and (5.2.20) give

%2|f j(y)|2 ≤ c(cj , k1,mesΩ)(%2A2(%) + %2λH(%)

)∀y ∈ G2%

%/4.

Hence by the assumption (5.2.9) it follows that

|f j(x)| ≤ cjA(%) ∀x ∈ G2%%/4, j = 1, . . . , N(5.2.21)

Further, in the same way as in the proof of (5.2.17),

(5.2.22) supx,y∈G2%

%/4

x 6=y

|f j(x)− f j(y)|A(%−1|x− y|)

≤ [f j ]0,A;G2%

%/4sup

0<t<4%

A(t)A(%−1t)

≤

≤ cA(%)[f j ]0,A;G2%

%/4.

Now from (5.2.18), (5.2.21) and (5.2.22) we obtain

A−1(%)N∑j=1

‖f j(%x′)‖0,A;G21/4≤ const.(5.2.23)


It remains to verify the niteness of %A−1(%)‖g(%x′)‖ N1−α ;G2

1/4. We have

%A−1(%)( ∫G2

1/4

|g(%x′)|N

1−αdx′) 1−α

N =

= %αA−1(%)( ∫G2%%/4

|g(x)|N

1−αdx) 1−α

N ≤

≤ dαA−1(d)( ∫G2%%/4

|g(x)|N

1−αdx) 1−α

N ≤ const ∀% ∈ (0, d),

by the condition (1.8.1). Thus the conditions of Theorem 5.12 are satised.By this theorem we have:

‖v‖1,B;G11/2≤(|v|0;G2

1/4+ %−1A−1(%)‖ϕ(%x′)‖1,A;Γ2

1/4+

+%A−1(%)‖g(%x′)‖ N1−α ;G2

1/4+A−1(%)

N∑j=1

‖f j(%x′)‖0,A;G21/4

)×(5.2.24)

×c(N, ν,G, max

i,j=1,...,N

(‖ai,j(%x′)‖0,A;G2

1/4, %‖ai(%x′)‖0,A;G2

1/4

),A(2%)

),

∀% ∈ (0, d).

Step 2. To estimate |v|0;G21/4

we use the local estimate at the boundary

of the maximum of the modulus of a solution (Theorem 8.25 [128]). Wecheck the assumptions of this theorem. To this end, we set

z(x′) = v(x′)− %−1A−1(%)Φ(%x′)

and write the problem for the function z(x′) :∂∂x′i

(aij(%x′)zx′j + %ai(%x′)z) + %bi(%x′)zx′i + %2c(%x′)z =

= G(x′) + ∂F j(x′)∂x′j

, x′ ∈ G21/4;

z(x′) = 0, x′ ∈ Γ21/4.

where

G(x′) ≡ %A−1(%)g(%x′)−A−1(%)bi(%x′)Φx′i(%x′)−

− %A−1(%)c(%x′)Φ(%x′),(5.2.25)

F i(x′) ≡ A−1(%)f i(%x′)− %−1A−1(%)aij(%x′)Φx′j(%x′)−

−A−1(%)ai(%x′)Φ(%x′), (i = 1, . . . , N).(5.2.26)


At rst we verify the necessary smoothness of coecients (see the remark inthe end of §8.10 [128]). Let q > N ; we have:

∫G2

1/4

|%ai(%x′)|qdx′ = %q−N∫

G2%%/4

|ai(x)|qdx ≤

≤ c(G)dq‖ai‖q0,A;G, ∀% ∈ (0, d).

(5.2.27)

By (iii) we also obtain

∫G2

1/4

|%bi(%x′)|qdx′ = %q−N∫

G2%%/4

|bi(x)|qdx ≤ 4q%−N∫

G2%%/4

|rbi(x)|qdx ≤

≤ 4q%−N∫

G2%%/4

Aq(r)dx ≤ 2N+2q

∫G2%%/4

r−NAq(r)dx =(5.2.28)

= 2N+2qmesΩ

2%∫%/4

Aq(r)r

dr ≤ 2N+2qmesΩ · Aq−1(2d)

2d∫0

A(r)r

dr,

∫G2

1/4

|%2c(%x′)|q/2dx′ = %q−N∫

G2%%/4

|c(x)|q/2dx ≤ 4q%−N∫

G2%%/4

|r2c(x)|q/2dx ≤

≤ 2N+2q

∫G2%%/4

r−NAq/2(r)dx ≤ 2N+2qmesΩ · Aq−2

2 (2d)

2d∫0

A(r)r

dr,(5.2.29)

∀% ∈ (0, d).

In the same way from (5.2.25) we get

%A−1(%)‖G(x′)‖q/2;G21/4

= %A−1(%)( ∫G2%%/4

|g(x)|q/2 +

+( N∑i=1

|bi(x)|)q/2|∇Φ|q/2 + |c(x)|q/2|Φ(x)|q/2

%−Ndx

)2/q.(5.2.30)


By (iv) setting q = N/(1− α) > N and applying Holder's inequality for theintegrals we obtain

%A−1(%)( ∫G2%%/4

%−N |g(x)|q/2dx)2/q

≤

≤ cραA−1(%)( ∫G2%%/4

%−N/2|g(x)|q/2dx)2/q

≤(5.2.31)

≤ cραA−1(%)‖g‖q;G2%

%/4(mesΩ ln 8)1/q ≤ c(d, α, q,mesΩ,A(d))‖g‖

q;G2%%/4,

since by (1.8.1), %αA−1(%) ≤ dαA−1(d) ∀% ∈ (0, d). Similarly

%A−1(%)( ∫G2%%/4

r−N( N∑

i=1

|bi(x)|)q/2|∇Φ|q/2 + |c(x)|q/2|Φ(x)|q/2

dx)2/q

≤

≤ c(mesΩ)2/q‖ϕ‖1,A;Γ2%

%/4A

q−22 (%)

2%∫%/4

A(r)r

dr.(5.2.32)

From (5.2.30)-(5.2.32) we obtain

‖G(x′)‖q/2;G21/4≤ c(d, α, q,mesΩ,A(d),

2%∫%/4

A(r)r

dr

)×

×(‖g‖

q;G2%%/4

+ ‖ϕ‖1,A;Γ2%

%/4

), q =

N

1− α> N.(5.2.33)

Finally, in the same way from (5.2.26) we have

N∑i=1

∫G2

1/4

|F i(x′)|qdx′ ≤ c(N, q,G, max

j=1,...,N

( N∑i=1

‖ai,j‖q0,A;G ,

N∑i=1

‖ai‖q0,A;G

))×

×∫

G2%%/4

r−NA−q(r)( N∑i=1

|f i(x)|q + |∇Φ|q + |Φ(x)|q)dx.(5.2.34)

It follows from (10.2.85) as %→ +0 that |∇Φ(0)| = 0. Therefore

|∇Φ(x)| = |∇Φ(x)−∇Φ(0)| ≤ A(|x|)‖ϕ‖1,A;Γ2%

%/4, ∀x ∈ G2%

%/4,

and hence we have

|Φ(x)| ≤ r|∇Φ(x)| ≤ |x|A(|x|)‖ϕ‖1,A;Γ2%

%/4, ∀x ∈ G2%

%/4.


Similarly it follows from (5.2.21) as % → +0 that f j(0) = 0, ∀j = 1, . . . , N.Therefore we have ∀x ∈ G2%

%/4

|f j(x)| = |f j(x)− f j(0)| ≤ A(r)[f j ]0,A;G2%

%/4.

Consequently, estimating the right side of (5.2.34) and taking into accountthe inequalities obtained, we have

N∑i=1

‖F i‖q;G21/4≤ c(N, q,G, max

j=1,...,N

( N∑i=1

‖ai,j‖0,A;G ,

N∑i=1

‖ai‖0,A;G

))×

×mesΩ ·( N∑i=1

‖f i‖0,A;G2%

%/4+ ‖ϕ‖

1,A;Γ2%%/4

).(5.2.35)

So all conditions of Theorem 8.25 [128] are satised. By this theorem weget

supx′∈G1

1/2

|z(x′)| ≤ c(‖z‖2;G2

1/4+ ‖G‖ N

2(1−α);G2

1/4+

N∑i=1

‖F i‖ N1−α ;G2

1/4

)≤

≤ c(‖z‖2;G2

1/4+ ‖g‖ N

1−α ;G2%%/4

+N∑i=1

‖f i‖0,A;G2%

%/4+ ‖ϕ‖

1,A;Γ2%%/4

).(5.2.36)

Setting w(x) = u(x)− ϕ(x) we have for w(x) the problem∂∂xi

(aij(x)wxj + ai(x)w) + bi(x)wxi + c(x)w =

= G(x) + ∂F j(x)∂xj

, x ∈ G2d0 ;

w(x) = 0, x ∈ Γ2d0 ⊂ ∂G2d

0 ,

where

G(x) = g(x)− bi(x)Φxi − c(x)Φ(x),

F i(x) = f i(x)− aij(x)Φxj − ai(x)Φ(x).

Moreover by assumptions (i), (ii)

|aij(x)− δji | ≤ ‖ai,j‖0,A;GA(|x|), x ∈ G.

In virtue of the estimate (5.1.22) of Theorem 5.4 there is a constant c > 0independent on w,G, F i such that

(5.2.37)∫G%0

r2−N |∇w|2dx ≤ c%2λ

∫G2

0d

|w(x)|2 + |∇w|2 +G2(x) +

+N∑i=1

|F i(x)|2 + r4−N−2λH−1(r)G2(x) + r2−N−2λH−1(r)N∑i=1

|F i(x)|2dx,

ρ ∈ (0, d).


Our assumptions guarantee that the integral on the right side is nite. Sincez(x′) = %−1A−1(%)w(%x′) we obtain from (5.2.37)

(5.2.38)∫

G21/4

|∇′z|2dx′ ≤ 2N−2%−2A−2(%)∫

G2%%/4

r2−N |∇w|2dx ≤

≤ c%2λ−2A−2(%)∫G

|w(x)|2 + |∇w|2 +G2(x) +

N∑i=1

|F i(x)|2 +

+ r4−N−2λH−1(r)G2(x) + r2−N−2λH−1(r)N∑i=1

|F i(x)|2dx, ρ ∈ (0, d).

By assumptions (i)-(iv) we have

|G(x)|2 ≤ c|g|2 +A2(r)(r−2|∇Φ|2 + r−4Φ2)

,

N∑i=1

|F i(x)|2 ≤ c N∑i=1

|f i(x)|2 +

+ maxi,j=1,...,N

(‖ai,j‖0,A;G , ‖ai‖0,A;G

)(|∇Φ|2 + Φ2)

.

(5.2.39)

Applying now the Friedrichs inequality and taking into account (5.2.9), weobtain from (5.2.38), (5.2.39)

(5.2.40) ‖z‖22;G21/4≤ c‖∇′z‖22;G2

1/4≤ c%2λ−2A−2(%)

∫G

|w(x)|2 +

+ |∇w|2 + g2(x) +N∑i=1

|f i(x)|2 + |∇Φ|2 + Φ2 + r4−N−2λH−1(r)g2(x) +

+ r2−N−2λH−1(r)N∑i=1

|f i(x)|2 + r2−N−2λH−1(r)|∇Φ|2 +

+ r−2A2(r)|∇Φ|2dx ≤ const

||g||2N

1−α ;G+

N∑i=1

||f i||20,A;G + ||ϕ||21,A;∂G +

+∫G

(r4−N−2λH−1(r)g2(x) + r2−N−2λH−1(r)

N∑i=1

|f i(x)|2 +

+ r2−N−2λH−1(r)|∇Φ|2 + |w|2 + |∇w|2)dx


by assumptions (iii)-(v). By the denition of z(x′), inequalities (5.2.36),(5.2.40) and assumptions (i)-(v) we nally obtain

(5.2.41) |v|0;G21/4≤ |z|0;G2

1/4+ %−1A−1(%)|ϕ|0;Γ2

1/4≤

≤ c

(||g|| N

1−α ;G +N∑i=1

||f i||0,A;G + ||ϕ||1,A;∂G +

+∫G

(r4−N−2λH−1(r)g2(x) + r2−N−2λH−1(r)

N∑i=1

|f i(x)|2 +

+ r2−N−2λH−1(r)|∇Φ|2 + |w|2 + |∇w|2)dx

1/2).

Step 3. Returning to the variables x, u(x), we now obtain from inequal-ities (5.2.24), (5.2.41)

(5.2.42) %−1A−1(%) supx∈G%

%/2

|u(x)|+A−1(%) supx∈G%

%/2

|∇u(x)|+

+ supx,y∈G%

%/2

x 6=y

|∇u(x)−∇u(y)|A(%)B|x− y|)

≤ c

(||g|| N

1−α ;G +N∑i=1

||f i||0,A;G + ||ϕ||1,A;∂G +

+∫G

(r4−N−2λH−1(r)g2(x) + r2−N−2λH−1(r)

N∑i=1

|f i(x)|2 +

+ r2−N−2λH−1(r)|∇Φ|2 + |u|2 + |∇u|2)dx

1/2).

Setting |x| = 2%/3 we deduce from (5.2.42) the validity of (5.2.10), (5.2.11).This completes the proof of Theorem 5.13.

Remark 5.14. As an example of A(r), that satises all the conditionsof Theorem 5.13, besides the function rα, one may take A(r) = rα ln(1/r),provided λ ≥ 1 + α. In the case of A(r) = rα the result of [21] follows fromTheorem 5.13 for a single equation and the estimate (5.2.10) coincides with(5.1.51).


5.2.3. Global regularity and solvability.

Theorem 5.15. indexDini gradient continuity!global!conical domain LetA be an α -Dini function (0 < α < 1) that satises the conditions (1.8.5) -(1.8.6), (5.2.9). Let G \ O be a domain of class C1,A and O ∈ ∂G be aconical point of G. Suppose that the assumptions (i)-(iv) are valid and

(vi)∫G

(c(x)η − ai(x)Diη

)dx ≤ 0, ∀η ≥ 0, η ∈ C1

0 (G).

Then the generalized problem (DL) has a unique solution u ∈ C1,A(G) andwe have the estimate

‖u‖1,A;G ≤ c

(||g|| N

1−α ;G +N∑i=1

||f i||0,A;G + ||ϕ||1,A;∂G +

+∫G

(r4−N−2λH−1(r)g2(x) + r2−N−2λH−1(r)

N∑i=1

|f i(x)|2 +

+r2−N−2λH−1(r)|∇Φ|2)dx

1/2).

(5.2.43)

Proof. The inequality (5.2.42) implies that

|∇u(x)−∇u(y)| ≤ cB(|x− y|)

(||g|| N

1−α ;G + ||ϕ||1,A;∂G +

+N∑i=1

||f i||0,A;G +∫G

(|u|2 + |∇u|2 + r4−N−2λH−1(r)g2(x) +

+r2−N−2λH−1(r)N∑i=1

|f i(x)|2 + r2−N−2λH−1(r)|∇Φ|2)dx

1/2)

∀x, y ∈ G%%/2, % ∈ (0, d).

(5.2.44)

From (5.2.42), (5.2.44) we now infer that u ∈ C1,B(Gd0). Indeed, let x, y ∈ Gd0and % ∈ (0, d). If x, y ∈ G%%/2 then (5.2.44) holds. If |x − y| > % = |x| thenby (5.2.42) we obtain

|∇u(x)−∇u(y)|B(|x− y|)

≤ 2cA(|x|)B−1(|x|)

(||g|| N

1−α ;G + ||ϕ||1,A;∂G +


+N∑i=1

||f i||0,A;G +∫G

(|u|2 + |∇u|2 + r4−N−2λH−1(r)g2(x) +

+ r2−N−2λH−1(r)N∑i=1

|f i(x)|2 + r2−N−2λH−1(r)|∇Φ|2)dx

1/2)≤

≤ 2cα

(||g|| N

1−α ;G + ||ϕ||1,A;∂G +N∑i=1

||f i||0,A;G +

+∫G

(|u|2 + |∇u|2 + r4−N−2λH−1(r)g2(x) +

+ r2−N−2λH−1(r)N∑i=1

|f i(x)|2 + r2−N−2λH−1(r)|∇Φ|2)dx

1/2)

in view of (1.8.3). Because of the condition (1.8.5) for the equivalence of Aand B, we derive u ∈ C1,B(Gd0) and the estimate

‖u‖1,A;Gd0≤ c

(||g|| N

1−α ;G +N∑i=1

||f i||0,A;G + ||ϕ||1,A;∂G +

+∫G

(|u|2 + |∇u|2 + r4−N−2λH−1(r)g2(x) +

+r2−N−2λH−1(r)N∑i=1

|f i(x)|2 + r2−N−2λH−1(r)|∇Φ|2)dx

1/2).

(5.2.45)

following from the above arguments.By means of a partition of unity, from the bounds (5.2.1) of Theorem

5.12 and (5.2.45) we derive

‖u‖1,A;G ≤ c

(||g|| N

1−α ;G +N∑i=1

||f i||0,A;G + ||ϕ||1,A;∂G +

+|u|0;G +∫G

(|u|2 + |∇u|2 + r4−N−2λH−1(r)g2(x) +

+r2−N−2λH−1(r)N∑i=1

|f i(x)|2 + r2−N−2λH−1(r)|∇Φ|2)dx

1/2).

(5.2.46)

By the assumption (vi) that guarantees the uniqueness of the solution forthe problem (DL), we have the bound (see Corollary 8.7 [128])∫

G

(|u|2 + |∇u|2

)dx ≤ C

∫G

(g2 +

N∑i=1

|f i(x)|2 + |∇Φ|2 + Φ2)dx,

5.3 Notes 197

which together with the global boundedness of weak solution (Theorem 8.16[128]) and the bound (5.2.46), leads to the desired estimate (5.2.43).

Finally, the global estimate (5.2.43) leads to the assertion on the uniquesolvability in C1,A(G) . This is proved by an approximation argument (seee.g. the proof of Theorem 8.34 [128]).

Remark 5.16. The conclusion of Theorem 5.15 is best possible. This isshown for the function A(r) = rα, λ ≥ 1 + α, α ∈ (0, 1) in [169] (see alsoexamples in Section 4.7 of the Chapter 4).

5.3. Notes

The best possible Hölder exponents for weak solutions was rst obtainedin [168, 169]. There the method of non-smooth domain approximation bythe sequence of smooth domains was used. We apply here the quasi-distancefunction rε(x). The introduction of such function allows us to work in thegiven domain, and then to provide the passage to the limit over ε → +0(where rε(x)→ r = |x|.)

The Lp−regularity of the (DL) in the cone was studied in [83], and inthe domains with angles - in [246]. Finally, let us point yet at two works.In [8] Alkhutov and Kondrat'ev proved the single-valued solvability in thespace W 1,p

0 (G) of the (DL) in arbitrary convex bounded domain G assumingonly the continuity in G of the lieder coecients .

Hölder estimates for the rst derivatives of generalized solutions to theproblem (DL) are well known in the case, if the leading coecients aij(x) ofthe equation are Hölder continuous (see e.g. 8.11 [128] for smooth domainsand [21] for the domain with angular point). Here we derive Dini-estimatesfor the rst derivatives of generalized solutions of the problem (DL) in a do-main with conical boundary point under minimal condition on the smooth-ness of the leading coecients (Dini-continuity). The presentation of Section5.2 follows [64]. It should be noted that the interior Dini-continuity of therst and second derivatives of generalized solutions to the problem (DL) wasinvestigated in [74],[221] under the condition of Dini-continuity of the rstderivatives of the leading coecients.

Recently, V. Kozlov and V. Maz'ya [193, 194] derived an asymptoticformula near a point O at the smooth boundary of a new type for weaksolutions of the Dirichlet problem for elliptic equations of arbitrary order.We formulate an idea of their results for the linear uniformly elliptic secondorder equation:

∂∂xi

(aij(x)uxj = g(x), x ∈ G;u(x) = 0, x ∈ ∂G,

where G ⊂ RN is a bounded domain with smooth boundary ∂G. It is

assumed that aij(x) are measurable and bounded complex-valued functions,u(x) has a nite Dirichlet integral and g = 0 in a certain neighborhood Gd0


of the origin O. In addition, let there exists a constant symmetric matrixA0 =

(aij0

)with positive denite real part such that the function

A(r) := supx∈Gr0

‖A(x)−A0‖

is suciently small for r < d, where A =(aij). Let us dene the function

Q(x) =< (A(x)−A)n, n > −N < A−1(A(x)−A)n, x >< n, x >< A−1x, x >−1

σN (detA)1/2 < A−1x, x >N/2,

where n is the exterior unit normal at O. The following asymptotic formulaholds:

u(x) = exp

− ∫Gd0\G

|x|0

Q(y)dy +O

d∫|x|

A2(r)r

dr

×

×

Cd(x) +O

|x|2−ε d∫|x|

A(r)r2−ε dr

+O

(|x|2−ε

),

(5.3.1)

where C = const and ε is a small positive number. The sharp two-sidedestimate for the Hölder exponent of u at the origin may be derived from(5.3.1).

They establish also the following criterion: under the conditiond∫0

A2(r)r dr <∞ all solutions u are Lipschitz at the origin if and only if

lim infr→+0

∫Gd0\Gr0

<Q(x)dx > −∞.(5.3.2)

Needless to say, this new one-sided restriction (5.3.2) is weaker that theclassical Dini condition at the origin.

We point to the work [259] yet. In this work it is studied Lq−regularityof weak solutions of the Dirichlet problem for linear elliptic second orderequation in the divergent form with piecewise constant leading coecientsin a Lipschitz polyhedron.

Other boundary value problems (the Neumann problem, mixed problem)for elliptic variational equations in smooth, convex or nonsmooth domainshave been studied by V. Adolfsson, D. Jerison [3] studied Lp− integrabilityof the second order derivatives for the Neumann problem in convex domains,J. Banasiak [27] - [29], BR:89 J. Banasiak & G.F. Roach [31, 32] consid-ered the mixed boundary value problem of Dirichlet oblique-derivative typein plane domains with piecewise dierentiable boundary, K. Gröger [135] es-tablished a W 1,p−estimate for solutions to mixed boundary value problems,P. Shi & S. Wright [355] investigated the higher integrability of the gradi-ent in linear elasticity, M.K.V. Murthy & G. Stampacchia [314] considered

5.3 Notes 199

a variational inequality with mixed boundary conditions, W. Zajaczkowskiand V. Solonnikov [406] investigated the Neumann problem in a domainwith edges.


CHAPTER 6

The Dirichlet problem for semilinear equations

in a conical domain

6.1. The behavior of strong solutions for nondivergentequations near a conical point

In this Section we study the properties of strong solutions of the Dirichletproblem for nondivergent semilinear uniformly elliptic second order equationsin a neighborhood of a conical boundary point:

Lu := aij(x)Diju(x) + ai(x)Diu(x) + a(x)u(x) == g(u) + f(x) in G,

g(u) = a0(x)u|u|q−1, q > 0;u(x) = 0 on ∂G \ O.

(SL)

Let G ⊂ RN be a bounded domain with a conical point in O as described inSection 1.3 of chapter 1. We shall assume that Gd0 is a convex cone for smalld > 0.

Definition 6.1. By a strong solution of the Dirichlet problem (SL) inG we mean a function u ∈ W 2(G) ∩ C0(G \ O) which satises the equationof (SL) for almost all x ∈ G and the boundary condition for all x ∈ ∂G \O.

In the following we will always supposeAssumptions:

a) the uniform ellipticity condition:

ν|ξ|2 ≤ aij(x)ξiξj ≤ µ|ξ|2 ∀ξ ∈ Rn, x ∈ G

with some ν, µ > 0; aij(0) = δji ;aa) aij ∈ C0(G), ai ∈ Lp(G) and a ∈ Lp/2(G) with some p > N ;aaa) there exists a monotonically increasing nonnegative continuous at zero

function A(r), A(0) = 0 such that for x, y ∈ G N∑i,j=1


≤ A(|x− y|),

|x|

(N∑i=1

ai2(x)

)1/2

+ |x|2|a(x)| ≤ A(|x|);

201

202 6 The Dirichlet problem for semilinear equations in a conical domain

b) a0(x) is a nonnegative measurable in G function;c) there exist real numbers k1 ≥ 0 and β > −1 such that

|f(x)| ≤ k1|x|β.

6.1.1. The weighted integral estimates (0 < q ≤ 1). Now we provecertain weighted integral estimates of strong solutions of (SL). Here thefunction a0(x) can be unbounded.

Theorem 6.2. Let u be a strong solution of (SL) and the conditions a)- aaa), b) are satised. Suppose that a0(x) ∈ V 0

2/(1−q);4/(1−q)−N (G),

f(x) ∈ W04−N (G), 0 < q < 1.

Then u(x) ∈ W24−N (G) and there is a positive constant c, determined by

ν, µ, q,N,maxx∈GA(|x|), G such that

∫G

(r4−N |D2u|2 + r2−N |Du|2 + r−N |u|2 + a0(x)r2−N |u|1+q

)dx ≤

(6.1.1)

≤ c∫G

(u2 + r4−Nf2(x) + 2 + a

2/(1−q)0 (x)r4/(1−q)−N)dx.

Proof. We multiply both parts of the equation of (SL) by r2−Nε u(x)

and integrate over the domain (G). Similar to the theorem 4.13 proof fromChapter 4 we have∫

G

r2−Nε |∇u|2dx+

∫G

a0(x)r2−Nε |u|1+qdx ≤

≤ c(h)A(d)∫Gd0

(r2−Nε r2|D2u|2 + r2−N

ε |∇u|2 + r2−Nε r−2u2

)dx+(6.1.2)

+c(d)∫Gd

(|D2u|2 + u2

)dx+

12

∫G

r4−Nε f2(x)dx, ∀ε > 0, d > 0.

By the layers method based on the local L2− Schauder's estimate, we derivethe inequality (see the derivation of (4.3.23)

(6.1.3)∫Gd0

r2−Nε r2|D2u|2dx ≤ c

∫G2d

0

(r2−Nε r−2|u|2 + r4−N

ε a20(x)|u|2q +

+ r4−Nε f2(x)

)dx, ∀ε > 0, d > 0,

6.1 Strong solutions for nondivergent equations 203

where c is a constant depending only on ν, µ, q,N,maxx∈GA(|x|), G. Taking into

account that q < 1 by Young's inequality we have

(6.1.4) r4−Nε a2

0(x)|u|2q =(r−Nqε |u|2q

)(r4−N+Nqε a2

0(x))≤ σr−Nε |u|2 +

+ c(σ, q)a2/(1−q)0 (x)r4/(1−q)−N

ε , ∀σ > 0.

Now the estimate (6.1.1) sought for follows from (6.1.2) - (6.1.4) under propersmall d > 0 with the help of the same arguments as during the completionof the Theorem 4.13 proof.

Theorem 6.3. Let u be a strong solution of (SL) and the conditionsa) - c) with A(r) that is Dini-continuous at zero are satised. In addition,

suppose f(x) ∈ LN (G) ∩ W04−N (G), a0(x) ∈ V 0

2/(1−q);4/(1−q)−N (G) and there

is a constant k2 ≥ 0 such that

‖a0‖1/(1−q)V 02/(1−q),4/(1−q)−N (G%0)

≤ k2%2+β, % ∈ (0, d).(6.1.5)

Then there are positive constants c and d ∈ (0, e−1) such that for % ∈(0, d)

‖u‖ W

2

4−N (G%0)≤ c(‖u‖L2(G) + ‖f‖

W0

4−N (G)+ ‖a0‖1/(1−q)V 0

2/(1−q);4/(1−q)−N (G)+ k1 +

+k2

)%λ, if β + 2 > λ,

%λ ln3/2(

1%

), if β + 2 = λ,

%β+2, if β + 2 < λ.

(6.1.6)

Proof. At rst, because of Theorem 6.2, we have u(x) ∈ W

24−N (G).

Now we introduce the function

U(%) =∫G%0

r2−N |Du|2dx, % ∈ (0, d)

and multiply both parts of (SL) by r2−Nu(x) and integrate the obtainedequality over the domain G%0, % ∈ (0, d). As a result, similarly to Theorems4.18 we obtain

U(%) +∫G%0

a0(x)r2−N |u|1+qdx ≤ %

2λU ′(%) + cA(%)U(2%) +

+ cA(%)∫G2%

0

(a

2/(1−q)0 (x)r4/(1−q)−N + r4−Nf2(x)

)dx+

+ cA(%)U(%) +∫G%0

r2−N |u||f(x)|dx;

(6.1.7)


for this we used the inequalities (6.1.3), (6.1.4) with ε = 0, σ = 1.

From the hypothesis (c) we have∫G2%

0

r4−Nf2(x)dx ≤ k21measΩ

2(β + 2)(2%)4+2β(6.1.8)

and as well apply the Cauchy and Poincaré inequalities

(6.1.9)∫G%0

r2−N |u||f(x)|dx ≤ k1

∫G%0

rβ+2−N |u|dx =

+ k1

∫G%0

(r−N/2|u|

)rβ+2−N/2dx ≤ cδU(%) + cδ−1k2

1%2β+4.

From (6.1.5), (6.1.7) - (6.1.9) nally we obtain the dierential inequality

(6.1.10) U(%) ≤ %

2λU ′(%) + c1A(%)U(2%) + c2

(A(%) + δ

)U(%) +

+ c3δ−1(k2

1 + k22

)%2β+4, ∀δ > 0, 0 < % < d.

Moreover, by Theorem 6.2, we have the initial condition

(6.1.11) U0 ≡ U(d) =∫Gd0

r2−N |∇u|2dx ≤

≤ c(‖u‖2L2(G) + ‖f‖2

W0

4−N (G)+ ‖a0‖2/(1−q)V 0

2/(1−q);4/(1−q)−N (G)

).

The dierential inequality (6.1.10) with initial condition is the same type as(4.3.47) with s = β + 2 or (4.3.51), if β + 2 = λ. Repeating verbatim theinvestigation of these inequalities in the proof of Theorem 4.18 we obtain

U(%) ≤ c(‖u‖2L2(G) + ‖f‖2

W0

4−N (G)+ ‖a0‖2/(1−q)V 0

2/(1−q);4/(1−q)−N (G)+ k2

1 + k22

)×

×

%2λ, if β + 2 > λ,

%λ ln3(

1%

), if β + 2 = λ,

%2(β+2), if β + 2 < λ.

(6.1.12)

From (6.1.3) - (6.1.4) passing to the limits as ε→ 0 we obtain

(6.1.13)∫G%0

r4−N |D2u|2dx ≤ c∫G2%

0

(r−N |u|2 + a

2/(1−q)0 (x)r4/(1−q)−N +

+ r4−Nf2(x))dx, 0 < % < d.

Now taking into account the inequality (H-W) from (6.1.12), (6.1.13) it fol-lows the desired (6.1.6).


Theorem 6.4. Let u ∈ W 2,N (G) be a strong solution of (SL) and theconditions a) - c) with A(r) that is Dini-continuous at zero are satised. Inaddition, suppose

a(x) ≤ 0, f(x) ∈ LN (G) ∩ W04−N (G), a0(x) ∈ V 0

N/(1−q);2qN/(1−q)(G).

Then there is a positive constant c such that

‖u‖V 2N,0(G) ≤ c

(‖a0‖1/(1−q)V 0

N/(1−q);2qN/(1−q)(G)+ ‖f‖LN (G)

).(6.1.14)

Proof. By Theorem 4.48, there exists the unique solution u ∈ V 2N,0(G)

of the linear problem Lu = F (x), x ∈ G,u(x) = 0, x ∈ ∂G \ O

provided λ > 1, F ∈ LN (G) and

‖u‖V 2N,0(G) ≤ c‖F‖N,G,(6.1.15)

where c > 0 depends only on ν, µ,N,maxx∈GA(|x|), ‖ai‖p,G, ‖a‖p/2,G,

p > N and the domain G. The condition λ > 1 is fullled by the convexityof Gd0. From (6.1.15) with F (x) = f(x) + a0(x)u|u|q−1 using the inequality(1.2.5) we obtain:

(6.1.16)∫G

(|D2u|N + r−N |Du|N + r−2N |u|N

)dx ≤

≤ 2N−1

∫G

(|a0(x)|N |u|qN + |f(x)|N

)dx.

Using Young's inequality and taking into account q ∈ (0, 1) we have

(6.1.17) |a0(x)|N |u|qN =(r−2qN |u|qN

) (r2qN |a0(x)|N

)≤

≤ εr−2N |u|N + εq/(q−1)r2qN/(1−q)|a0(x)|N/(1−q), ∀ε > 0.

By the choice ε = 2−N from (6.1.16) - (6.1.17) it follows the desired (6.1.14).

6.1.2. The estimate of the solution modulus (0 < q ≤ 1). Now wewant to deduce the estimate of our solution modulus in the case (0 < q ≤ 1).To that end we introduce the function

ψ(%) =

%λ if, λ < β + 2;%λ ln3/2 1

% if, λ = β + 2;%β+2 if, λ > β + 2.

(6.1.18)


Theorem 6.5. Let u(x) ∈ W 2,N (G) be a strong solution of (SL) andthe conditions a) - c) with A(r) that is Dini-continuous at zero are satised.

In addition, suppose a(x) ≤ 0, a(x) ∈ LN (G), f(x) ∈ LN (G) ∩ W04−N (G),

a0(x) ∈ LN/(1−q)(G) ∩ V 02/(1−q);4/(1−q)−N (G) together with (6.1.5) and there

exists a nonnegative constant k0 such that

‖a0(x)‖L

N1−q (G2%

%/4)≤ k0%

1−2qψq(%).(6.1.19)

Then there are positive constants c0, d independent of u such that thefollowing estimates are held:

1) |u(x)| ≤ c0|x|2

1−q , x ∈ Gd0, if λ > β + 2, 0 < q < 1− 2λ

;

2) |u(x)| ≤ c0|x|λ, x ∈ Gd0, if λ < β + 2, 1− 2λ≤ q ≤ 1;

3) |u(x)| ≤ c0|x|β+2, x ∈ Gd0, if λ > β + 2, 1− 2λ≤ q ≤ 1;

4) |u(x)| ≤ c0|x|λ ln1|x|, x ∈ Gd0, if λ > β + 2, 1− 2

λ≤ q ≤ 1.

Proof. Let us perform the variables substitution x = %x′, u(%x′) =ψ(%)v(x′) in the problem (SL). Let G′ be the image of the domain G undertransformation of coordinates xi = %x′i; i = 1, · · · , N. As a result we inferthat v(x′) is a solution of the problem

(SL)′

aij(%x′)vx′ix′j + %ai(%x′)vx′i + %2a(%x′)v = %2

ψ(%)f(%x′)+

+%2ψq−1(%)a0(%x′)v|v|q−1, x′ ∈ G′,v(x′) = 0, x′ ∈ ∂G′.

We apply now Theorem 4.5 (Local Maximum Principle):

(6.1.20) supx′∈G1

1/2

|v(x′)| ≤ c ∫

G21/4

v2(x′)dx′

1/2

+%2

ψ(%)‖f‖LN (G2

1/4) +

+ %2ψq−1(%)

∫G2

1/4

|a0(%x′)|N |v|qN )dx′

1/N

.

By the inequality (6.1.17), we have

(6.1.21) |a0(x)|N |v|qN ≤ εqN/(q−1)|x′|2qN/(1−q)|a0(x)|N/(1−q) +

+ εN |x′|−2N |v|N , ∀ε > 0.


From (6.1.20) - (6.1.21) we get

supx′∈G1

1/2

|v(x′)| ≤ c

∫G2

1/4

v2(x′)dx′

1/2

+ c%2

ψ(%)‖f‖LN (G2

1/4) +

+ %2ψq−1(%)ε

∫G2

1/4

|x′|−2N |v|Ndx′

1/N

+

+ cεqq−1

∫G2

1/4

|x′|2qN1−q |a0|

N1−q dx′

1/N

, ∀ε > 0.

(6.1.22)

Now we estimate each term on the right in (6.1.22)

•

∫G2

1/4

v2(x′)dx′

1/2

≤ 2n/2ψ−1(%)

∫G2%%/4

r−Nu2(x)dx

1/2

≤ const,

in virtue of (6.1.6);

• %2ψ−1(%)‖f‖LN (G21/4

) ≤ 2%ψ−1(%)

∫G2%%/4

|f |Ndx

1/N

≤ ck1%β+2ψ−1(%) ≤ const; here we apply the hypothesis c) and the

denition (6.1.18) of ψ(%);

•

∫G2

1/4

|x′|−2N |v|Ndx′1/N

≤ 24

∫G2%%/4

%−N |v|Ndx

1/N

≤

≤ 26%

∫G2%%/4

r−2N |v|Ndx

1/N

≤ 26 %ψ(%)

(∫G

r−2N |u|Ndx)1/N

;

•

∫G2

1/4

|x′|2qN1−q |a0|

N1−q dx′

1/N

≤ c(q,N)%−1

∫G2%%/4

|a0|N

1−q dx

1/N

;

In (6.1.22) we choose ε = ψ(%)% ; then, because of (6.1.16), from two latter

estimates we obtain:

ε

∫G2

1/4

|x′|−2N |v|Ndx′

1/N

≤ const(6.1.23)


and

(6.1.24) εqq−1

∫G2

1/4

|x′|2qN1−q |a0|

N1−q dx′

1/N

≤

≤ c(

%

ψ(%)

) q1−q 1

%

∫G2%%/4

|a0|N

1−q dx

1/N

≤ const,

in virtue of the hypothesis (6.1.19). The obtained estimates result for (6.1.22):

supx′∈G1

1/2

|v(x′)| ≤ c(1 + %2ψq−1(%)

).(6.1.25)

Now we show that for all interesting cases of our Theorem,

%2ψq−1(%) <∞,∀% > 0(6.1.26)

is true.

1) β + 2 < λ ⇒ ψ(%) = %β+2. In this case we have:

%2ψq−1(%) = %q(β+2)−β <∞,∀% > 0,

if β + 2 ≤ 21−q . Choosing the best exponent β + 2 = 2

1−q < λ we getthe rst statement of our Theorem. In fact, from (6.1.25) - (6.1.26)we have

|v(x′)| ≤M ′0 = const ∀x′ ∈ G11/2.

Returning to former variables hence it follows

|u(x)| ≤M ′0ψ(%) = M ′0%2

1−q , ∀x ∈ G%%/2, % ∈ (0, d).

Setting |x| = 23% hence follows the required assertion.

2) β + 2 > λ ⇒ ψ(%) = %λ. In this case we have:

%2ψq−1(%) = %2+λ(q−1) <∞,∀% > 0,

if 1 − 2λ ≤ q ≤ 1. Repeating verbatim stated above arguments as in

the rst case we get the second statement of our Theorem.

3) β + 2 < λ ⇒ ψ(%) = %β+2. In this case we have:

%2ψq−1(%) = %q(β+2)−β = %2q%β(q−1) ≤ %2q%(λ−2)(q−1) <∞,∀% > 0,



the rst case we get the third statement of our Theorem.

4) β + 2 = λ ⇒ ψ(%) = %λ ln3/2 1% . In this case we have:

%2ψq−1(%) = %2+λ(q−1) ln

32 (q−1) 1

% <∞,∀% > 0,


the rst case we get the fourth statement of our Theorem.

Now we go on to the deduction of some corollaries from Theorem 6.5.

Lemma 6.6. Let a(x) ≥ a0 = const > 0. and the hypotheses a), aaa)are satised. There are positive numbers η, %, determined only by ν, µ, q,a0, N such that, if u(x) is a strong solution of the equation or (SL) withf(x) ≡ 0 and 0 < q < 1 in the ball B%(0) and |u(x)| < η, x ∈ ∂B%(0), thenu(0) = 0.

Proof. Let s > 21−q . We set R(x) = |x|s. Then

LR(x)− a0(x)Rq(x) = srs−2∑

i

aii(x) + (s− 2)aij(x)xixj

r2+ xia

i(x) +

+1sa(x)r2

− a0(x)rsq ≤

≤ srs−2(µ(s+N − 2) +A(r)

)− a0r

sq.

By the continuity of A(r) at zero, there exists d > 0 such that A(r) < 1 assoon as r < d. Therefore we obtain

LR(x)− a0(x)Rq(x) ≤ srs−2(µ(s+N − 2) + 1

)− a0r

sq < 0,

provided r < d and rs−2−sq < a0

s(

1+µ(s+N−2)) . So

LR(x)− a0(x)Rq(x) < 0 provided % = min

d;

a0

s(1 + µ(s+N − 2)

) .By the Maximum Principle (see below Theorem 6.8), |u| < R provided thatη < %s, hence u(0) = 0.

Theorem 6.7. Let u(x) ∈W 2,N (G) be a strong solution of (SL) and theconditions a) - c) with A(r) that is Dini-continuous at zero are satised. Letλ > β + 2, 0 < q < 1 − 2

λ . In addition, suppose that f(x) ≡ 0,a0(x) ∈ LN/(1−q)(G) ∩ V 0

2/(1−q);4/(1−q)−N (G) together with (6.1.5),


a(x) ≤ 0, a(x) ∈ LN (G), a0(x) ≥ a0 = const > 0, and there exist a nonne-gative constant k0 such that

‖a0(x)‖L

N1−q (G2%

%/4)≤ k0%

1+βq.(6.1.27)

Then there is a positive constant d independent of u such thatu(x) ≡ 0, x ∈ Gd0.

Proof. Let c0, d > 0 are chosen according to Theorem 6.4 and such thatGd0 ⊂ G. Let x0 ∈ Gd0. We make the transform x − x0 = %x′,

u(x) = %2

1−q v(x′). The function v(x′) is a solution of (SL)′ with f ≡ 0 and,by Lemma 6.6, we have v(0) = 0 provided |v(x′)| < η for |x′| = R with somepositive R, η. Hence u(x0) = 0 provided |u(x)| < η%

21−q for |x − x0| = R%.

But the latter condition is satised, in virtue of the assertion 1) of Theorem6.5, if we set η = c0, R = 2. Thus we get u(x0) = 0. Since any x0 ∈ Gd0 weobtain the assertion of our Theorem.

6.1.3. The estimate of the solution modulus (q > 1). Let us recallthe well known Comparison Principle.

Theorem 6.8. (Comparison principle) Suppose D ∈ RN is a bound-ed domain, L is elliptic in D, a(x) ≤ 0 in D. Let us dene the functiong(x, u) with the properties:

g(x, u1) ≥ g(x, u2) for u1 ≥ u2.

Let u, v ∈W 2,Nloc (D) ∩ C0(D) satisfy the inequalities

Lu ≤ g(x, u), Lv ≥ g(x, v) in D.

Then

u ≥ v on ∂D ⇒ u ≥ v throughout D.

Proof. Let w = u− v. We have:

Lw = Lu− Lv ≤ g(x, u)− g(x, v) ≤ 0

on D− = x ∈ D | w(x) < 0 and w ≥ 0 on ∂D. From the Alexandrovmaximum principle (Theorem 4.2) we get

w(x) ≥ infD−

w(x) = inf∂D−

w(x) = 0, ∀x ∈ D.

Now we consider the case q > 1. For this at rst we study the C0 ∩W 2,Nloc −

solutions of dierential inequality in RN :

signu · Lu− a0(x)|u|q ≥ −k,(DI)

In this connection we suppose:


(*) L is the uniformly elliptic operator with the ellipticity constants ν, µ,(ν ≤ µ) and with bounded coecients(

N∑i=1

|ai(x)|2)1/2

+ |a(x)| ≤ m, a(x) ≤ 0,

a0(x) ≥ a0 > kN q−1, ∀x ∈ G,

where m,a0, k are nonnegative constants.

We derive as a preliminary the next statements.

Lemma 6.9. Let L satisfy (*). There are a bounded domain D ⊂ RNcontaining the origin O and a positive function U(x) dened in D such that

LU − a0Uq ≤ −k, x ∈ D,

U(0) = 1, limx→∂D

U(x) =∞.(6.1.28)

Proof. We rst set U(x) =N∑i=1

y(xi), where y(t) is a positive solution

of the Cauchy problemµy′′(t) +m|y′(t)| − a0y

q = − kN ,

y(0) = 1N , y

′(0) = 0.(6.1.29)

By setting y′ = p(y) we get

t =

y∫1/N

dη

p(η).(6.1.30)

The function p(y) is a solution of the Cauchy problemµpp′ +m|p| − a0y

q = − kN ,

p(

1N

)= 0.

(6.1.31)

Now we apply the Hardy theorem (see, for example, Theorem 3 5, chapterV [38]). By virtue of this theorem, any positive solution of (6.1.31) fulllsthe asymptotic relation

p(η) ∼ ηκ as η → +∞, κ ∈ R.(6.1.32)

Now we calculate the quantity κ. From (6.1.31) we inferµηκp′(η) +mηκ ∼ a0η

q as η → +∞ or

p′(η) ∼ 1µ

(a0ηq−κ −m) as η → +∞.(6.1.33)

Integrating the relation (6.1.33) with regard to (6.1.32) we nd

p(η) ∼ 1µ

(a0ηq−κ+1

q − κ+ 1−mη) as η → +∞.(6.1.34)


From (6.1.32) and (6.1.34) it follows that κ = q − κ + 1 ≥ 1,or κ = q+1

2 , q ≥ 1 ⇒∞∫

1/N

dη

p(η)∼

∞∫1/N

dη

ηq+1

2

<∞, if q > 1.(6.1.35)

From (6.1.30) and (6.1.35) it follows that

y(t)→∞ as t→∞∫

1/N

dη

p(η)<∞.

Now we remark that, because of (*),

y′′(0) =1µ

(a0

N q− k

N

)> 0

and consequently, by the continuity of y′′, we have y′′(t) > 0 in a certainneighborhood of zero. Therefore, returning now to U(x) we have

Uxi = y′(xi), Uxixj = δji y′′(xi),

LU ≡N∑i=1

aii(x)y′′(xi) +N∑i=1

ai(x)y′(xi) + a(x)N∑i=1

y(xi) ≤

≤ µN∑i=1

y′′(xi) +m

N∑i=1

|y′(xi)| =N∑i=1

(a0y

q(xi)−k

N

)≤ a0U

q − k,

if we recall (6.1.29) and use the Jensen inequality (1.2.5). Thus we proved(6.1.28) as well as our Lemma.

Lemma 6.10. There exists R0 > 0 such that in the ball BR0(0) there is nosolution of the inequality (DI), satisfying the condition|u(0)| > 1.

Proof. For R0 we take any number R such that BR(0) ⊃⊃ D, whereD is the domain constructed in Lemma 6.9. Let u be a positive solution of(DI) in BR0(0) with u(0) > 1. We dene in D the function w = u − U,where U is the function constructed in Lemma 6.9. By Lemma 6.9, thefunction w has the following properties: lim

x→∂Dw(x) = −∞, w(0) > 0. We

set D+ = x ∈ D | w(x) > 0. Since O ∈ D+ we have D+ 6= ∅. Now weapply the comparison principle (Theorem 6.8) to w in D:

Lu ≥ a0uq − k ≡ g(u) in D,

LU ≤ a0Uq − k = g(U) in D,

u < U on ∂D.

From the comparison principle it follows that w < 0 in D, and hence w < 0in D+ . We get a contradiction with the denition of D+.


Lemma 6.11. If u(x) is a strong solution of the inequality (DI) in BR(x0)such that |u(x0)| > h, then

R ≤ R0h1−q

2 ,(6.1.36)

where R0 depends only on ν, µ, q, a0, N.

Proof. We make the change of variables x− x0 = h1−q

2 x′ and u = hv.The function v satises (DI), and |v(0)| > 1. Hence, by Lemma 6.10,v(x′) is dened in a ball of radius not exceeding R0, i.e. in the ball|x′| < Rh

q−12 ≤ R0 ⇒ R ≤ R0h

1−q2 .

Corollary 6.12. Let G be a bounded domain containing the origin O.Let u(x) be a strong solution of inequality (DI) in G \ O. Then

|u(x)| ≤ c|x|2

1−q ,(6.1.37)

where c > 0 is a constant depending on ν, µ, q, a0, N.

Now we are estimating the modulus of a strong solution of (SL). At rstwe derive an auxiliary estimate.

Lemma 6.13. Let u(x) be a strong solution of (SL) and the conditionsa) - c) are satised. In addition, suppose a0(x) ≥ a0 = const > 0. Thenthere are d > 0, c0 > 0 such that the inequality

|u(x)| ≤ c0|x|2

1−q , x ∈ Gd0.(6.1.38)

holds.

Proof. Let us perform the substitution of variables x = %x′, u(%x′) =hv(x′), h > 0 in the problem (SL). The function v(x′) is a solution in thedomain G1

1/2 of the problem

(6.1.39)

L′v ≡ aij(%x′)vx′ix′j + %ai(%x′)vx′i + %2a(%x′)v =

= %2hq−1a0(%x′)v|v|q−1 + %2h−1f(%x′), x′ ∈ G11/2,

v(x′) = 0, x′ ∈ Γ11/2.

Now we choose h > 0 so

%2hq−1 = 1(6.1.40)

Because of a0(x) ≥ a0 > 0 and the assumption c), from (6.1.39) - (6.1.40) itfollows that

signv · L′v − a0|v|q ≥ %2qq−1 f(%x′)signv ≥ −k1%

β+ 2qq−1(6.1.41)

But β > −1, q > 1, therefore β + 2qq−1 > q+1

q−1 > 0. Hence for

0 < % ≤ d < 1 we have %β+ 2qq−1 ≤ %

q+1q−1 ≤ d

q+1q−1 . Now from (6.1.41) we

obtain

signv · L′v − a0|v|q ≥ −k1dq+1q−1 .(6.1.42)


By setting k = k1dq+1q−1 , we see from (6.1.42) that for a small positive d,

namely

0 < d <

(a0

k1N q−1

) q−1q+1

,(6.1.43)

the following inequalities

signv · L′v − a0|v|q ≥ −k, a0 > kN q−1(6.1.44)

hold. This allows us to apply Corollary 6.12 and we obtain

|v(x′)| ≤M ′0, x′ ∈ G11/2,

where M ′0 > 0 is a constant depending only on ν, µ, q, a0, N, supx∈GA(|x|). Re-

turning to the former variables we get

|u(x)| ≤M ′0%2

1−q , x ∈ G%%/2, % ∈ (0, d).(6.1.45)

Taking |x| = 2%3 nally we arrive to the desired inequality (6.1.38).

Lemma 6.14. Let L be a linear elliptic operator with the conditions a) -aaa). Then for ∀γ ∈ (−λ−N + 1, λ− 1) there exist a number d > 0 and the

function w ∈ C0(Gd0) ∩ C2(Gd0) with the following properties:

Lw ≤ −(λ+N + γ − 1)(λ− γ − 1)2λ(λ+N − 2)

rγ−1, x ∈ Gd0,(6.1.46)

0 ≤ w(x) ≤ c|x|1+γ , x ∈ Gd0,(6.1.47) w(x) > 0, x ∈ Γd0,w(x) ≥ %γ+1

λ(λ+N−2) , x ∈ Ω%, 0 < % ≤ d,(6.1.48)

where c depends only on λ, γ,N,Ω.

Proof. Let us consider in the domain Ω ⊂ SN−1 the auxiliary Dirichletproblem for the Beltrami - Laplace operator

4ωψ + (1 + γ)(N − 1 + γ)ψ = −1, ω ∈ Ω,ψ(ω) = 0, ω ∈ ∂Ω.

It is well known (see Subsection 3 2, chapter 7 [201]) that this problem hasthe unique solution having the properties

ψ ∈ C2(Ω) ∩ C0(Ω), ψ > 0 in Ω, ‖ψ‖C2(Ω) ≤ c(γ,N,Ω)

provided the inequality

(1 + γ)(N − 1 + γ) < λ(λ+N − 2)(6.1.49)

is satised. The solutions of the latter inequality are the numbers

γ ∈ (−λ−N + 1, λ− 1).


Now we dene the function

w(x) = |x|1+γ

(ψ(ω) +

1λ(λ+N − 2)

).(6.1.50)

By direct calculating we get

4w = −(λ+N + γ − 1)(λ− γ − 1)λ(λ+N − 2)

|x|γ−1.(6.1.51)

Now, by the assumptions a) - aaa), we have

Lw = 4w +(aij(x)− aij(0)

)Dijw(x) + ai(x)Diw(x) + a(x)w(x) ≤

≤ 4w + cA(r)(|D2w|+ r−1|Dw|+ r−2|w|) ≤

≤ rγ−1

(cA(r)− (λ+N + γ − 1)(λ− γ − 1)

λ(λ+N − 2)

),

where c > 0 depend only on λ, γ,N,Ω. By the continuity of A(r) at zero, wend d > 0 such that

A(r) ≤ (λ+N + γ − 1)(λ− γ − 1)2cλ(λ+N − 2)

, r ∈ (0, d).

By this, (6.1.46) is proved. The other properties of w are trivial.

Definition 6.15. The above constructed function w we shall call thebarrier function.

Theorem 6.16. Let u(x) be a strong solution of (SL) and the conditionsa) - c) are satised. In addition, suppose 0 < a0 ≤ a(x) ≤ a1, where a0, a1

are positive constants.Then for ∀ε > 0 there are positive constants cε, d independent of u such

that the following estimate holds

|u(x)| ≤ cε|x|λ−ε, x ∈ Gd0,(6.1.52)

if β + 2 ≥ λ > 1, q > 1 +2

λ+N − 2.

Proof. Since |a0(x)| ≤ a1 then from (SL), in virtue of (6.1.38) and theassumption c), it follows

Lu ≥ −a1|u|q − k1rβ ≥ −cq0a1r

2q1−q − k1r

β.(6.1.53)

Set

γ − 1 =2q

1− q∈ (−λ−N,λ− 2).(6.1.54)

It is easily seen that such a number γ satises Lemma 6.14 about the barrierfunction. Let

B ≥ 2λ(λ+N − 2)(k1 + a1cq0)

(λ+N − 1 + γ)(λ− 1− γ).(6.1.55)


Now from (6.1.53), (6.1.46), (6.1.54) taking into account β ≥ λ− 2 > 2q1−q it

follows that

L(Bw) ≤ Lu, x ∈ Gdε , ∀ε > 0.(6.1.56)

Moreover, from the properties of the barrier function it follows that

u(x) = 0 < w(x), x ∈ Γdε , ∀ε > 0,(6.1.57)

Bw(x) ≥ B

λ(λ+N − 2)|x|

21−q ≥ c0|x|

21−q ≥ u(x),(6.1.58)

x ∈ Ω%, 0 < % ≤ d, if B ≥ c0λ(λ+N − 2).

Thus, if the number B > 0 satises (6.1.55), (6.1.58), then it is proved thatL(Bw) ≤ Lu in Gdε ,u(x) ≤ Bw(x) on ∂Gdε .

By the comparison principle (Theorem 4.4), hence we obtain

u(x) ≤ Bw(x), x ∈ Gdε , ∀ε ∈ (0, d).

Similarly u(x) is estimated from below. Thus we get

|u(x)| ≤ c|x|1+γ , x ∈ Gd0 \ O,where γ satises (6.1.54); in particular, we can choose 1 + γ = λ − ε,∀ε > 0, that gives us the estimate sought for. Our Theorem is proved.

6.2. The behavior of weak solutions for divergence equationsnear a conical point

Here we study the properties of weak solutions of the Dirichlet problemfor the divergence semilinear uniformly elliptic second order equation in aneighborhood of conical boundary point:

Lu := ∂∂xi

(aij(x)uxj + ai(x)u) + bi(x)uxi + c(x)u == a0(x)u|u|q−1, q > 0, x ∈ Gε;

u(x) = 0 on Γε, ∀ε > 0.(DSL)

Definition 6.17. The function u(x) ∈ W 1(Gε) ∩ L∞(Gε) is called aweak solution of the problem (DSL) provided that it satises the integralidentity ∫

G

aij(x)uxjηxi + ai(x)uηxi − bi(x)uxiη − c(x)uη +

+a0(x)u|u|q−1ηdx = 0(III)

for all η(x) ∈W 1(G), which has a support compact in G.

In the following we will always supposeAssumptions:

i) G ⊂ K is bounded domain;

6.2 Weak solutions for divergence equations 217

a) the uniform ellipticity condition:

ν|ξ|2 ≤ aij(x)ξiξj ≤ µ|ξ|2 ∀ξ ∈ Rn, x ∈ G

with some ν, µ > 0; aij(0) = δji ;aa) aij(x) ∈ C0(G), (i, j = 1, . . . , N); ai(x), bi(x) ∈ Lp(G),

(i = 1, . . . , N); c(x) ∈ Lp/2(G), p > N ;aaa) there exists a monotonically increasing nonnegative continuous at zero

function A(r), A(0) = 0 such that for all x ∈ G

N∑i,j=1

|aij(x)− aij(0)|21/2

+ |x|

(N∑i=1

ai2(x)

)1/2

+ |x|2|a(x)| ≤ A(|x|);

b) 0 ≤ a0(x) ≤ a0 = const is a nonnegative measurable in G function;c) for all η(x) ∈W 1(G) which has a support compact in G∫

G

(c(x)η − ai(x)Diη

)dx ≤ 0.

Now we derive a bound of the weak solution of (DSL) modulus. Let λ bethe smallest positive eigenvalue of (EV PI) with (2.4.8).

Theorem 6.18. Let u be a weak solution of (DSL) and the conditionsi),a) - c) are satised. Suppose that∫

G

rα|Du|2dx <∞ at some α ∈ [2, 2λ+N.)

Then ∀ε > 0, there is a positive constant cε, determined only by ν, µ, q,N,maxx∈GA(|x|), G such that

|u(x)| ≤ cε|x|λ−ε.(6.2.1)

Proof. Let v ∈W 1(Gd0) be a weak solution of the linear problemLv = 0 in Gd0,

v∣∣∣Ωd

= u+, v∣∣∣Γd0

= 0,

where u+ is the positive part of u. The constant d > 0 we choose so thatGd0 ⊂ G. Such v exists and is unique. By Theorem 5.8, we obtain

|v(x)| ≤ cε|x|λ−ε.(6.2.2)

Let us show that

u(x) ≤ v(x).(6.2.3)


Suppose the contrary is true, i.e. we have u(x) > v(x) in a domain D ⊂ Gd0.Then L(u− v) ≥ 0 in D,∫

G

rα|D(u− v)|2dx <∞ ∀α ∈ [2, 2λ+N)(6.2.4)

is satised. In fact, in D we obtain:

L(u− v) = a0(x)u|u|q−1 > a0(x)v|v|q−1 ≥ 0,

since, by weak maximum principle, v ≥ 0 in Gd0. Moreover, by Theorem 5.5,it is easily seen that∫

Gd0

rα|Dv|2dx <∞ ∀α ∈ [2, 2λ+N)

and therefore (6.2.4) is veried. From (6.2.4) and Theorem 5.11 it followsthat u = v. Thus, u satises (6.2.1) too.

Theorem 6.18 is a simple extension of well known results of the linearequation theory to (DL). It should be noted that we cannot take u > 0 in(6.2.1) without additional restrictions. The following theorem is only validfor solutions of nonlinear equations. Note that the behavior of u(x) in theneighborhood of the vertex of the cone is not restricted a priori in the the-orem, which is mandatory in the theory of linear problems. It is usuallyrequired in linear problems that either the Dirichlet integral be limited orthe solution be continuous.

Theorem 6.19. If a0(x) ≥ a0 = const > 0, x ∈ G, q > 1,

21− q

> 2−N − λ,(6.2.5)

then inequality (6.2.1) is satised.

Proof. We state the assertion established in [?]. Let q > 1, a0(x) ≥a0 > 0, and u(x) be a solution of (DSL) in some domain G 3 O, which isinside the unit sphere |x|<1 and vanishes in that part of ∂G which is strictlyinside the sphere. Then

|u(0)| ≤ C1,

∫|x|<1/2, x∈G

|∇u|2dx ≤ C2,(6.2.6)

where the constants C1 and C2 are only dependent on the elliptic constantsof (DSL) [assumption a)] and on a0 and q. If we change the variables so thatx = ρx′, u = hv, and ρ−2 = hq−1, which retains the structure of (DSL),then we obtain the following assertion from (6.2.6).


Let u(x) be a solution of (DSL) in domain G2%%/2 and vanishes in Γ2%

%/2.Then ∫

G3%/23%/4

|∇u|2dx ≤ C2%2· 1+q

1−q+N.(6.2.7)

According to (6.2.7) ∫G

rα|∇u|2dx <∞.(6.2.8)

if

α+ 2 · 1 + q

1− q+N > 0(6.2.9)

Since, in view of the condition of Theorem 6.19,

2 · 1 + q

1− q+N = 2

( 21− q

− 1 +N)> 2−N − 2λ,

we can choose an α < 2λ+N − 2 which satises (6.2.9). In this case (6.2.8)is satised and we can use Theorem 6.18. This is the proof of the inequality(6.2.1).

Theorem 6.20. If 0 < q < 1, a0(x) ≥ a0 = const > 0,2

1−q < λ, u(x) ∈ W 12 (G), then u(x) ≡ 0 in some neighborhood of the vertex

of the cone K.

Proof. The following statement was proved in [?]. Let G 3 O be in theunit sphere, let u(x) be the solution of (DSL), and let u(x) = 0 in that partof ∂G which is strictly inside the unit sphere. There exists a B = const > 0which depends only on q, ν, µ and on a0. If |u(x)| ≤ B at |x| = 1, x ∈ G,then u(0) = 0. The constant B does not depend on either u or the structureof domain G. Thus, using the transform x = %x′, u = hv, %−2 = hq−1, wereadily obtain the following statement.

Let u(x) be the solution of (DSL) in the part of the domain G 3 Olying inside the sphere |x| < 2%, and vanishes in that part of ∂G, inside thesphere. If

|u(x)| ≤ B%2

1−q(6.2.10)

for |x| = 2%, x ∈ G, then u(x) ≡ 0 for |x| < %.If the conditions of Theorem 6.20 are satised, we will obtain (6.2.1), by

applying Theorem 6.18. Inequality (6.2.1) yields (6.2.10) at small %. Hence,u(x) ≡ 0 if |x| < %.

Note that if the condition (6.2.5) of Theorem 6.19 is not satised, then(DSL) has unbounded solutions in the neighborhood of x = 0. We willnow prove it provided ai(x) = bi(x) = c(x) ≡ 0. We state some assertionsabout the characteristics of the solutions of linear elliptic equations in conic


domains which we shall use [159].

1. Let K be a cone in Rn. We consider the boundary value problemN∑

i,j=1

∂∂xiaij(x) ∂u∂xj =

N∑i=1

∂f i

∂xi+ f0(x), x ∈ K,

u(x) = 0, x ∈ ∂K.(6.2.11)

Let β be such that β2−(N−2)2

4 is not an eigenvalue of (EVP1). Thereexists a δ > 0, which depends on K and β, such that if

|aij(x)− δji | ≤ δ, x ∈ K,(6.2.12) ∫

K

|x|β+2f20dx+

∫K

N∑i=1

|x|β|f i|2dx <∞,

then there exists a unique solution of (6.2.11) such that

(6.2.13)∫K

|x|β|∇u|2dx+∫K

|x|β−2u2dx ≤ C∫K

|x|β+2f20dx+

+ C

∫K

N∑i=1

|x|β |f i|2dx.

This statement was proved for aij(x) ≡ δji in [159]. It follows fromthe Banach theorem on the invertibility of the sum of an invertibleoperator and the operator which is small by the norm.

2. (see [159]) Suppose that K is a cone in RN , limx→0

aij(x) = δji , the

numbers β1 and β2 are such that the interval [β21−(N−2)2

4 ,β2

2−(N−2)2

4 ]has no points from spectrum of (EVP1), and u(x) is a solution of(6.2.11),

N∑i=1

∫K

|x|β1 |f i|2dx+N∑i=1

∫K

|x|β2 |f i|2dx+

+∫K

|x|β1+2f20dx+

∫K

|x|β2+2f20dx <∞,∫

K

|x|β2 |∇u|2dx+∫K

|x|β2−2u2dx <∞.

Then ∫K1

0

|x|β1 |∇u|2dx+∫K1

0

|x|β1−2u2dx <∞.(6.2.14)


Let us consider (6.2.11) where f i ≡ 0, i = 0, 1, ..., N . We will show that ifδ in (6.2.12) is small, aij(x) ≡ δji at |x| > R1, then (6.2.11) has a nontrivialsolution. Let Γ0(x) = |x|2−N−λΦ(ω), where Φ(ω) > 0 in K is the eigenfunc-tion of (EVP1) corresponding to ϑ. We seek Γ(x), the solution of (6.2.11),in the form of Γ(x) = Γ0(x)− V (x), where V (x) is a solution of

(6.2.15)

N∑i,j=1

∂∂xi

(aij(x) ∂V∂xj

)=

N∑i,j=1

∂∂xi

(aij(x)− δji )∂Γ0∂xj

=

=N∑i=1

∂∂xiF i(x), x ∈ K,

V (x) = 0 x ∈ ∂K.

Note thatN∑i=1

∫K

|F i|2|x|βdx <∞,(6.2.16)

if β > N − 2 + 2λ. We x β so that N − 2 + 2λ < β < −N + 2 − 2λ1−(K),

where λ1−(K) = 1

2

(2−N −

√(N − 2)2 + 4ϑ2

), ϑ2 is the smallest eigenvalue

of (EVP1) which is larger than ϑ. It follows from (6.2.13) that according tothe condition of (6.2.12) there exists a V (x), a solution of (6.2.15), such that∫

K

|x|β−2V 2dx+∫K

|x|β|∇V |2dx ≤N∑i=1

∫K

|x|β|F i|2dx.(6.2.17)

We will now discuss some characteristics of V (x) and Γ(x). It follows fromthe classical estimates of the solutions of the elliptic dierential equationsthat

V 2(x) ≤ Cλ−N∫

K2λλ/2

V 2dx, if |x| = λ.(6.2.18)

From this and (6.2.17) we have

V 2(x) ≤ C1|x|−N+2−β = o(|x|4−N−2λ) as x→∞.

Besides, |Γ0(x)| ≤ C|x|2−N−λ → 0 as |x| → ∞. Hence, Γ → 0 as x → ∞.Since Γ0(x) = Φ(ω)|x|2−N−λ and V (x) = o(|x|2−N−λ), then Γ 6≡ 0. Notethat ∫

K10

|x|α|∇Γ|2dx+∫K1

0

|x|α−2Γ2dx =∞(6.2.19)

at any α such that α < N − 2 + 2λ. Otherwise we would have|Γ(x)| ≤ Cε|x|λ−ε, i.e., Γ(x) → 0 as x → 0. This is impossible, in view


of the maximum principle. Finally, from (6.2.14) we have∫K1

0

Γ2|x|s−2dx+∫K1

0

|x|s|∇Γ|2dx <∞(6.2.20)

regardless of s > N −2+2λ. According to (6.2.20) and (6.2.18) we also havethat |Γ(x)| ≤ Cε|x|2−N−λ−ε for any ε > 0.

Using Γ(x), we construct an unbounded solution of (DSL) providedai(x) = bi(x) = c(x) ≡ 0. Suppose that d1 is so small that for x ∈ Kd1

0

(6.2.12) is satised for some β > N − 2 + 2λ. We change aij(x) at |x| > d1,taking them equal to δji . We constructed Γ(x), a solution of (6.2.11) atf i ≡ 0, that is unbounded in the neighborhood of x = 0 and satises (6.2.20).Suppose that Γ(x)→ +∞ along a sequence xm → 0.

Let Γk(x) be a solution of (DSL) in the domain Gk : x ∈ K,2−k < |x| < dk, k = 1, 2, ..., such that

u∣∣∣∂K∩Gk

= 0, u∣∣∣|x|=d1

= Γ, u∣∣∣|x|=2−k

= Γ.(6.2.21)

Then (DSL) has a solution satisfying (6.2.21), and it is unique. It can beconstructed by the variational method. Let us consider z(x) = −Γ(x) +Γk(x). This function is a solution of

(6.2.22)

L(z) = a0(x)|Γk|q−1Γk = a0(x) |Γk|

q−1Γk−|Γk|q−1ΓΓk−Γ (Γk − Γ)+

+a0(x)|Γk|q−1Γ, x ∈ Kd1

2−k,

z∣∣∣∂K

d12−k

= 0.

It follows from Theorem 5.11 that for any α < 2λ+N − 2,∫Kd12−k

|x|α|∇z|2dx+∫

Kd12−k

|x|α−2z2dx ≤

(6.2.23)

≤ C∫Kd10

|x|α+2|Γ(x)|2qdx ≤ CCε∫Kd10

|x|α+2q(2−N−λ)+2−εdx ≤ A,

if

α+ 2 + 2q(2−N − λ)− ε+N > 0.(6.2.24)

We can choose α such that (6.2.24) is valid and α < N − 2 + 2λ, since inthis case 2 > (1− q)(2−N − λ). The constant A on the right-hand side of(6.2.23) is not dependent on k. Let us consider the boundary value problemin K:

L(Z) = |a∗0(x)||Γ(x)|q = f0(x), x ∈ K,Z = 0, x ∈ ∂K,

(6.2.25)

6.3 Notes 223

where

a∗0(x) =

a0(x) at |x| < d1,

0 for |x| > d1.

If α satised (6.2.24) and α < N − 2 + 2λ, then (6.2.22) has a solution suchthat (6.2.25) holds and∫

K

|x|α|∇Z|2dx+∫K

|x|α−2Z2dx <∞.(6.2.26)

It follows from (6.2.26) and (6.2.18) that Z(x) → 0 as |x| → ∞. In view ofTheorem 5.11, Z(x) < 0 in K. From this and (6.2.22) we have

|Z(x)| ≤ |Z(x)|.(6.2.27)

This implies that qk(x) = a0(x) |Γk|q−1Γk−|Γ|q−1Γ

Γk−Γ is uniformly bounded with

respect to k in each domain Kd1d0, d0 > 0. Hence, the functions Z(x) form a

sequence which contains a subsequence compact in the sense of the topologyof inform convergence in each subdomain Kd1

d0. Let Z0(x) be its limit. It

follows from (6.2.24) that Z0(x) satises (6.2.26). Thus, u(x) = Γ(x)−z0(x)is the solution of (DSL) with ai(x) = bi(x) = c(x) ≡ 0 in Kd1 . Accordingto (6.2.19) and (6.2.23)∫

K

|x|q−2u2dx =∞ for any α < N − 2 + 2λ,

satisfying (6.2.24). It implies that u(x) is the solution of (DSL) withai(x) = bi(x) = c(x) ≡ 0 which is unbounded in any neighborhood of theorigin.

6.3. Notes

The properties of the (SL) solutions in a neighborhood of an isolatedsingular point were studied in [170, 171]. Positive solutions of singular valueproblems for the semilinear equations in smooth domains was investigatedalso in [172, 173, 174 ]. The solutions smoothness of some superlinearelliptic equations was investigated by S. Pohozhaev [334, 336, 337 ].

The results of Section 6.1 was established in [62] and of Section 6.2 - in[163].

We point out other problems, which are not investigated here. M. Marcusand L. Veron studied [242, 243, 244 ] the uniqueness and expansion proper-ties of the positive solutions of the equation 4u+hu−kup = 0 in nonsmoothdomain G, subject to the condition u(x)→∞, when dist(x, ∂G)→ 0, whereh, k are continuous functions in G, k > 0 and p > 1. They proved that thesolution is unique, when ∂G has the local graph property. They obtainedthe asymptotic behavior of solutions, when ∂G has a singularity of conicalor wedge-like type; if ∂G has a re-entrant cuspidal singularity then the rate


of blow-up may not be of the same order as in the previous more regularcases.

Many other problems for elliptic semilinear equations was studied byL. Veron together with colleagues in works [45, 46, 118, 119, 129, 179,339, 389, 391, 394, 340 ] as well as by other authors [33, 36, 51, 106,109, 110, 178, 327, 367 ].

Semilinear degenerate elliptic equations and axially symmetric problemswere considered by J. Below and H. Kaul [39].

CHAPTER 7

Strong solutions of the Dirichlet problem for

nondivergence quasilinear equations

7.1. The Dirichlet problem in smooth domains

Let G ⊂ RN be a bounded domain with a smooth boundary ∂G. Weconsider the Dirichlet problem:

aij(x, u, ux)uxi,xj + a(x, u, ux) = 0, aij = aji, x ∈ Gu(x) = ϕ(x), x ∈ ∂G

(QL)

(summation from 1 to N is assumed over repeated indices). The valueM0 = max

x∈G|u(x)| is assumed to be known.

Remark 7.1. For the nding of M0 see for example 10.2 [128].

Let us dene the set M =

(x, u, z)∣∣x ∈ G, u ∈ R, z ∈ RN .With regard

to the equation of the problem (QL) we assume that on the set M thefollowing conditions are satised:(A) Caratheodory for the functions a(x, u, z), aij(x, u, z) ∈ CAR,

(i, j = 1, . . . , N); that is:(i) for ∀u, z the functions a(x, u, z), aij(x, u, z) (i, j = 1, . . . , N) are

measurable on G as the functions of variable x;(ii) for almost all x ∈ G functions a(x, u, z), aij(x, u, z)

(i, j = 1, . . . , N) are continuous with respect to u, z;(B) the uniform ellipticity : there exist positive constants ν, µ indepen-

dent of u, z and such that

νξ2 ≤ aij(x, u, z)ξiξj ≤ µξ2, ∀ξ ∈ RN ;

(C) there exist a number µ1 and functions b(x), f(x) ∈ Lq,loc(G), q ≥ Nindependent of u, z such that:

|a(x, u, z)| ≤ µ1|z|2 + b(x)|z|+ f(x).

Let us recall some well known facts about W 2,p(G), p ≥ N solutions ofthis problem.

Definition 7.2. A bounded open set T ⊂ ∂G is said to be of type(A) if there exist two positive constants %0 and θ0 such that for every ballBr(x0), x0 ∈ T with radius r ≤ %0 and every connected component Gr,i ofthe intersection Br(x0)∩G the inequality measGr,i ≤ (1−θ0)measBr holds.

225

226 7 Strong solutions of the Dirichlet problem for nondivergence quasilinear equations

Theorem 7.3. (See 2 [215].) Let u ∈ W 2,Nloc (G) ∩ C0(G) be a strong

solution of (QL) and suppose that assumptions (A) - (C) are satised. LetG be of type (A) and ϕ ∈ Cβ(G), β ∈ (0, 1). Then u ∈ Cα(G), α ∈ (0, 1)and |u(x) − ϕ(x)|α,G ≤ Mα, where α is determined by N, ν, µ, β, θ0, G and

Mα depends on the same values and also on µ1,M0, ‖b‖N , ‖f‖N , |ϕ|β,∂G.

Theorem 7.4. (See Theorem 2.1 [217].) Let u ∈ W 2,Nloc (G) ∩ C0(G)

be a strong solution of (QL) and suppose that assumptions (A) - (C) aresatised. Let T ⊂ ∂G be a piece of class W 2,q, q > N . Then there is aconstant c > 0 depending only on N, ν, µ, µ1, q, ‖b‖q,M0 and the domain G

such that, if ϕ∣∣∣T

= 0, then |∇u|0,T ≤ c (1 + ‖f‖q) .

Yet let us introduce a set:

M(u) ≡

(x, u, z)∣∣x ∈ G, u = u(x), z = ∇u(x)

.

We assume in addition that in the neighborhood of the set M(u) is fullledthe following condition(D) the functions aij(x, u, z), (i, j = 1, . . . , N) have weak rst order deriva-

tives over all its own arguments and there exist the nonnegative con-stants µ0, µ2, µ3, k2 and the functions g(x), h(x) ∈ Lq,loc(G \ O), q >N independent of u, z such that:

N∑i,j,k=1

∣∣∣∣∂aij(x, u, z)∂zk− ∂aik(x, u, z)

∂zj

∣∣∣∣ ≤ µ0

(1 + |z|2

)−1/2 ;

N∑i,j=1

∣∣∣∣∣N∑k=1

(∂aij(x, u, z)

∂uz2k −

∂akj(x, u, z)∂u

zkzi +

+∂aij(x, u, z)

∂xkzk −

∂akj(x, u, z)∂xk

zi

)∣∣∣∣∣ ≤ (1 + |z|2)1/2 (µ2|z|+ g(x)) ;

‖g(x)‖q,G%%/2≤ k2%

N/q−1+γ , % ∈ (0, d∗);N∑

i,j=1

∣∣∣∣∣∂aij(x, u, z)∂u

∣∣∣∣∣2

+N∑k=1

∣∣∣∣∣∂aij(x, u, z)∂xk

∣∣∣∣∣2

1/2

≤ h(x);

N∑

i,j,k=1

∣∣∣∣∂aij(x, u, z)∂zk

∣∣∣∣2

1/2

≤ µ3,

where γ is a number from the estimate (7.3.1).

Theorem 7.5. (See Theorems 4.1, 4.3 [215].) Let G be a bounded do-main in RN with a W 2,q− boundary portion T ⊂ ∂G. Let u ∈ C0(G) ∩C1(G′)∩W 2,q

loc (G \O), q > N be a strong solution of (QL) and suppose that

assumptions (A)− (D) are satised. Let ϕ ∈ C1+α(∂G), α ∈ (0, 1). Then

7.2 The estimate of the Nirenberg type 227

there are the constants M1 > 0, γ ∈ (0, 1) depending only on N, ν, µ, µ0, µ1, µ2,µ3, q, α, ‖f‖q, ‖b‖q, ‖g‖q, ‖h‖q, ‖ϕ‖C1+α(∂G),M0 and the domain G such that

for ∀G′ ⊂⊂ (G ∪ T ) the inequality

‖u‖C1+γ(G′) ≤M1

holds.

7.2. The estimate of the Nirenberg type

7.2.1. Introduction. Until recently the problem of the solution smooth-ness to the boundary value problems for the second order quasilinear ellipticequations of nondivergence form remained open. An exception is Nirenberg'spaper [326], in which this problem was investigated for equations with twoindependent variables in a bounded plane domain with a smooth boundary.In the last decade, thanks to the eorts of many mathematicians, rst of allO.A. Ladyzhenskaya and N.N.Ural'tseva (see their survey [ 215], 1986), thisproblem has received a denitive solution for equations in multidimensionaldomains bounded by a suciently smooth boundary. As concerns the equa-tions in domains with a piecewise smooth boundary, only the investigation[89] of I. I. Danilyuk is known (we emphasize that here we are talking ofelliptic nonlinear and nondivergence equations). There the solvability of theDirichlet problem is proved for a two-dimensional equation in the Sobolevspace W 2,p for p > 2 and suciently close to 2.

In the present section we investigate the behavior of solutions of theDirichlet problem for a uniformly elliptic quasilinear equation of second or-der of nondivergence form near a corner point of the boundary of a boundedplane domain. It is here assumed that the coecients of the equation sat-isfy minimal conditions of smoothness and coordinated growth (no higherthan quadratic) modulo the gradient of the unknown function. We rst ex-tend to domains whose boundary contains a corner point and to equationswith an unbounded right side the method of Nirenberg [326] for estimatingthe Hölder constant of the rst derivatives of solutions. The weighted L2-estimate of the second derivatives of a solution obtained in this manner (wecall it the Nirenberg estimate) and the Sobolev imbedding theorems make itpossible to estimate the maximum of the modulus of a solution and its gradi-ent and thus establish power rate of decay (temporarily with a small positiveexponent) of a solution in a neighborhood of a corner point. Using the weaksmoothness of a solution established in 7.2.4, in 7.2.5 we rene the Niren-berg estimate and obtain a weighted integral estimate with best-possibleexponent of the weight. While for the Nirenberg estimate boundedness ofthe leading coecients of the equation was sucient, it is now necessary torequire their continuity. The estimate of 7.2.5 makes it possible to obtainsharp estimates of the modulus of a solution and of its gradient as well as


weighted Lp-estimates of the second derivatives, and to prove Hölder conti-nuity of the rst derivatives of a solution with best-possible Hölder exponent.

7.2.2. Formulation of the problem. The main result. Let G ⊂ R2

be a bounded domain with boundary ∂G which is assumed to be a Jordancurve smooth everywhere except at a point O ∈ ∂G; in some neighborhoodof the point O the boundary ∂G consists of two segments intersecting at anangle ω0 ∈ (0, π). We place the origin of a rectangular coordinate system(x1, x2) at the point O. Let (r, ω) be a polar coordinate system with poleat O. We direct the abscissa of the rectangular coordinate system along theray ω = 0 on which one the segments of ∂G lies, and we situate the ordinateaxis so that the second segment of ∂G lying on the ray ω = ω0 lies in theupper half-plane x2 > 0. For any numbers d > a ≥ 0 we set

Gda = G ∩ (r, ω)| a < r < d; 0 < ω < ω0.

(we henceforth assume that d is a suciently small positive number);

Γd1,a = (r, 0)| a < r < d; Γd2,a = (r, ω0)| a < r < d;

Γda = Γd1,a ∪ Γdd,a; Sd = (d, ω)| 0 ≤ ω ≤ ω0.

Definition 7.6. A strong solution of problem (QL) is a function u ∈W 2(G) satisfying the equation of the problem for almost all x ∈ G andthe boundary condition u − Φ ∈ W 2

0 (G) with any Φ ∈ W 2(G) such thatΦ(x) = ϕ(x), x ∈ ∂G.

The main result of this section is the proof of the following theorem.

Theorem 7.7. Suppose u ∈ W 2(G) is a solution of problem (QL) with

aij(0, 0, 0) = δji = the Kronecker symbol (i, j,= 1, 2), conditions (A)− (C)are satised, and the following quantities are known:

M0 = maxx∈G|u(x)|, M1 = ess sup

x∈G|∇u(x)|.(7.2.1)

Suppose the functions aij(x, u, ux) (i, j = 1, 2) are Dini-continuous at thepoint (0, 0, 0),

b2, f ∈ V 0p,α(G), ϕ ∈ Cπ/ω0(∂G) ∩ W3/2

0 (∂G) ∩ V 2−1/pp,α (∂G),

p > 2 α > p(2− π/ω0)− 2, 0 < ω0 < π,

and there exist numbers k1, k2 > 0 and s > π/ω0 such that for all ρ ∈ (0, d)the following inequalities hold:

||b2||2,Gρ0 + ||f ||2,Gρ0 + ||ϕ|| W

3/2

0

(Γρ0) ≤ k1ρs−1(7.2.2)

||b2||V 0p,α(Gρ

ρ/2) + ||f ||V 0

p,α(Gρρ/2

) + ||ϕ||V

2−1/ρp,α (Γρ

ρ/2)≤ k2ρ

π/ω0−2+α+2p .(7.2.3)

Then the following assertions are true:


1) u ∈ W22(G), and

||u|| W

2

2(Gρ0)≤ cρπ/ω0 , 0 < ρ < d;(7.2.4)

2) for 0 < ρ < d

|u(x)| ≤ c1|x|π/ω0 , x ∈ Gρ0,(7.2.5)

|∇u(x)| ≤ c2|x|π/ω0−1, x ∈ Gρ0;(7.2.6)

3) u ∈ V 2p,α(G), and

||u||V 2p,α(Gρ0) ≤ c3ρ

π/ω0−2+α+2p , 0 < ρ < d;(7.2.7)

4) if p ≥ 22−π/ω0

with π/2 < ω0 < π, then u ∈ Cπ/ω0(G).

7.2.3. The Nirenberg estimate. Let Φ(x) be any extension of theboundary function ϕ(x) into the domain G. The change of function v(x) =u(x) − Φ(x) reduces the inhomogeneous problem (QL) to the homogenousproblem

aij(x, v + Φ, vx + Φ(x)) · vxixj = F (x, v, vx), x ∈ G,v(x) = 0, x ∈ ∂G,

(QL)0

(7.2.8) F (x, v, vx) ≡ −aij(x, v + Φ, vx + Φ(x) · Φxixj −− a(x, v + Φ, vx + Φ(x)),

where by assumptions (b) and (c) the following condition is satised:

(7.2.9) |F (x, v, p)| ≤ 2µ1|p|2 + b(x) · |p|+ 2µ|Φxx|+ 2µ1 · |∇Φ|2 + b(x) · |∇Φ|+ f(x).

By a solution of problem (QL)0 we mean a function v ∈ W

20(G) satisfying

the equation of the problem almost everywhere in G. By Theorem 7.3, suchsolution is Hölder continuous in G, and there exists γ0 ∈ (0, 1), dependingon ν−1, µ, and ω0, such that

|v(x)| ≤ c0 · |x|γ0(7.2.10)

with a positive constant c0 depending on ν−1, µ, µ1, ω0,M0, ‖b2‖2,G, ‖f‖2,G,and ‖ϕ‖W 3/2(∂G).

Theorem 7.8. (cf. [326]; see also [213], Chapter IX, 6). Supposeu(x) ∈ W 2(G) is a solution of problem (QL), assumptions (A)− (C) aresatised, and the quantities (7.2.1) are known. Then there exists a con-stant γ, determined by the quantities γ, µ, µ1, ω0, c0, γ0, d, and satisfying theinequality

0 < γ < 2min(γ0;π/ω0 − 1) = γ∗,(7.2.11)


such that if f, b2 ∈ W0−γ∗(G) and ϕ ∈ C1(∂G)∩ W3/2

−γ∗(∂G), then u ∈ W2−γ(G),

and

‖u‖ W

2

−γ(Gρ/20 )≤ c(d), 0 < ρ ≤ d.(7.2.12)

Proof. First of all, we consider the expression

V ≡ aijvxi,xj(vx1x1

a22+vx2x2

a11

)and write it in the form

V =(a11

a22v2x1x1

+2a12

a22vx1x1vx1x2 + v2

x1x2

)+

+(a22

a11v2x2x2

+2a12

a11vx2x2vx1x2 + v2

x1x2

)+

+ 2(vx1x1vx2x2 − v2

x1x2

).

Because of (QL)0, the uniform ellipticity condition (B), hence it follows

(7.2.13)ν

µ

(v2x1x1

+ 2v2x1x2

+ v2x2x2

)≤

≤ 2(v2x1x2− vx1x1vx2x2

)+ F

(vx1x1

a22+vx2x2

a11

).

Now, let ζ(r) be a cut-o function for the domain G%0, % ∈ (0, d) :

ζ(r) =

1 0 ≤ r ≤ %/2,0 r ≥ %,

(7.2.14)

0 ≤ ζ(r) ≤ 1; |ζ ′(r)| ≤ c%−1; 0 ≤ r ≤ %.

Multiplying both sides of (7.2.13) by r−γε ζ2(r) and integrating over G%0, wehave

ν

2µ·∫Gρ0

r−γε v2xxζ

2(r)dx ≤ J (1)ε (ρ) + J (2)

ε (ρ),(7.2.15)

where

J (1)ε (ρ) =

∫Gρ0

r−γε ζ2(r)(v2x1x2− vx1x2vx1x2)dx,(7.2.16)

J (2)ε (ρ) =

12ν

∫Gρ0

r−γε ζ2(r) (|vx1x1 |+ |vx2x2 |) · |F (x, v, vx)|dx.(7.2.17)

Repeating the computations made in the proof of Lemma 2.41, we obtain

J (1)ε (ρ) = J (11)

ε (ρ) + J (12)ε (ρ) + J (13)

ε (ρ),(7.2.18)


J (11)ε (ρ) =

γ

2

∫Gρ0

r−γ−2ε · ζ2(r)xiwi(x)dx,(7.2.19)

J (12)ε (ρ) = ε

γ

2·∫Gρ0

r−γ−2ε · ζ2(r)w2(x)dx(7.2.20)

J (13)ε (ρ) = −

∫Gρ0

r−γ · ζζ ′(r)xirwi(x)dx,(7.2.21)

where the wi(x) are dened by (2.6.1), by virtue of which

xiwi(x) = vx2 ·∂vx1

∂ω− vx1 ·

∂vx2

∂ω,(7.2.22)

and therefore (7.2.19) can be rewritten in the form

J (11)ε (ρ) =

γ

2·

ρ∫0

r−γ−2ε · rζ2(r)dr

ω0∫0

(vx2 ·

∂vx1

∂ω− vx1 ·

∂vx2

∂ω

)dω.(7.2.23)

To estimate the integral J11ε we perform the transformation of coordinates.

From the rectangular system (x1, x2) we go over to an ane system (y1, y2):we place the axis OY1 along the axis OX1 (along the ray ω = 0,) and wedirect the axis OY2 along the ray ω = ω0. We then have

vx2

∂vx1

∂ω− vx1

∂vx2

∂ω=

1sinω0

(vy2

∂vy1

∂ω− vy1

∂vy2

∂ω

)Further, by the boundary condition v(x) = 0, x ∈ ∂G, we have

vy1(r, 0) = 0, vy2(r, ω0) = 0,

and hence

(7.2.24) J (11)ε (ρ) =

γ

2sinω0·

ρ∫0

r−γ−2ε rζ2(r)dr ·

ω0∫0

[vy2(r, ω)−

− vy2(r, ω0)]∂vy1

∂ω− [vy1(r, ω)− vy1(r, 0)] · ∂vy2

∂ω

dω.

By the Hölder inequality for integrals

(7.2.25) |vy1(r, ω)− vy1(r, 0)|2 =

∣∣∣∣∣ω∫

0

∂vy1(r, θ)∂θ

dθ

∣∣∣∣∣2

≤ ω ·ω∫

0

∣∣∣∣∣∂vy1(r, θ)∂θ

∣∣∣∣∣2

dθ

≤ ωr2 ·ω∫

0

|∇vy1(r, θ)|2dθ,


and, similarly,

|vy2(r, ω0)− vy2(r, ω)|2 ≤ (ω0 − ω) · r2 ·ω0∫0

|∇vy2(r, θ)|2dθ.(7.2.26)

We estimate the integrals in (7.2.24) by Cauchy's inequality and considerthe estimates (7.2.25) and (7.2.26). As a result, we obtain

J (11)ε (ρ) ≤ γ

4sinω0·

ρ∫0

r−γ−2ε rζ2(r)

ω0∫0

[(∂vy1

∂ω

)2+(∂vy2

∂ω

)2]dω +

+r2 ·ω0∫0

[ω ·

ω∫0

|∇vy1(r, θ)|2dθ + (ω0 − ω)

ω0∫ω

|∇vy2(r, θ)|2dθ

]dω

dr ≤

≤ 2 + ω20

8sinω0γ ·∫Gρ0

r−γε ζ2(r)v2xxdx(7.2.27)

by property 2) of the function rε(x). By property 3) of this function andCauchy's inequality we can estimate the integral (7.2.20) as follows:

J (12)ε (ρ) ≤ γ

2

∫Gρ0

[r−γε ζ2(r)v2xx + r−γ−2

ε · ζ2(r)|∇v|2]dx(7.2.28)

Finally, applying Cauchy's inequality with ∀δ > 0 and considering (2.6.1),we estimate the integral (7.2.21):

J (13)ε (ρ) ≤

∫Gρ0

[4δr−γε · ζ2(r)v2

xx +1δr−γε ζ ′

2(r) · |∇v|2]dx.(7.2.29)

Thus, from the representation (7.2.18) and estimates (7.2.27) - (7.2.29) weobtain

(7.2.30) J (1)ε (ρ) ≤

(4δ +

γ

2+ γ

2 + ω20

8sinω0

)∫Gρ0

r−γε ζ2(r)v2xxdx+

+γ

2

∫Gρ0

r−γ−2ε ζ2(r)|∇v|2dx+

1δ

∫Gρ0

r−γε ζ ′2(r)|∇v|2dx, ∀δ > 0.


We now turn to the estimation of the integral (7.2.17). Using Cauchy'sinequality, considering the condition (7.2.9), and applying Lemma 2.39 to-gether with the inequality (7.2.10), we obtain

J (2)ε (ρ) ≤ 1

2µ[δ + cδ−1(1 + 8µ2

1)c20d

2γ0] ∫Gρ0


+c

δµ(4µ2

1 +12

)c20d

2γ0

∫Gρ0

(r−γ−2ε · ζ2(r) + r−γε · ζ ′

2(r))|∇v|2dx+

+c(δ−1, µ, µ1,M0)∫Gρ0

[r−γε ζ2(r)

(b4(x) + f2(x) + Φ2

xx

)+

+(r−γ−2ε · ζ2(r) + r−γε · ζ ′

2(r))|∇Φ|2

]dx, ∀δ > 0, 0 < ρ ≤ d.

(7.2.31)

Lemma 7.9. Under the conditions of Theorem 7.8 we have∫Gρ0

r−γ−2ε · ζ2(r)|∇v|2dx ≤ c(µ, µ1, c0, γ0, γ, ω0, d)×

×ρ2γ0−γ−2 +

∫Gρ0


∫Gρ0

[r−γε ζ2(r)

(b4(x) + f2(x) +

+Φ2xx

)+ r−γ−2

ε · ζ2(r)|∇Φ|2]dx,

0 < ρ ≤ d.

(7.2.32)

Proof. We multiply the equality (QL)0 by r−γ−2ε ζ2(r)v(x) and inte-

grate it over the domain Gρ0. Integrating by parts, we have∫Gρ0

r−γ−2ε ζ2(r)|∇v|2dx =

(2 + γ)2

2

∫Gρ0

r−γ−4ε · ζ2(r)v2dx−

−(γ + 2)∫Gρ0

r−γ−4ε · ζζ ′(r)xi

r(xi − εli)v2dx− 2

∫Gρ0

r−γ−2ε · ζζ ′(r)xi

rvvxidx+

+∫Gρ0

[aij(x, v + Φ, vx + Φx)− aij(0, 0, 0)] vxixjvr−γ−2ε ζ2(r)dx+

+∫Gρ0

[aij(x, v + Φ, vx + Φx)Φxixj +(7.2.33)

+a(x, v + Φ, vx + Φx)]r−γ−2ε ζ2(r)v(x)dx.

We estimate the integrals on the right using Cauchy's inequality, assumptions(B), (C), and the estimate (7.2.10) - the Hölder continuity of v(x). As a


result, from (7.2.33) we obtain[(π

ω0

)2

−

(2 + γ

2

)2]·∫Gρ0

r−γ−2ε ζ2(r)|∇v|2dx ≤ c1(c0, γ, µ1)(δ + dγ0)×

×∫Gρ0

r−γ−2ε ζ2(r)|∇v|2dx+ c2(µ, γ, ω0, δ

−1) ·∫Gρ0

[r−γε · ζ2(r)v2

xx +

+r−γ−2ε ζ ′

2(r)v2]dx+ c3(µ, µ1, γ0, ω0, d, γ, δ

−1)×(7.2.34)

×∫Gρ0

[r−γε · ζ2(r)(Φ2

xx + b4(x) + f2(x)) + r−γ−2ε · ζ2(r) · |∇Φ|2

]dx;

0 < ρ ≤ d, ∀δ > 0.

Further, by property 2) of the function rε(x) and the properties of the func-tion ζ(r) with consideration of the inequality (7.2.11) we have∫

Gρ0

r−γ−2ε ζ ′

2(r)v2dx ≤ (cc0)2 · ρ−2 ·∫Gρ0

r2γ0−γ−2dx =

=ω0 · (cc0)2

2γ0 − γ· ρ2γ0−γ−2.

(7.2.35)

Since by (7.2.11) the left side of the (7.2.34) contains a strictly positiveconstant factor, choosing the quantities δ, d > 0 suciently small, we obtainthe desired inequality (7.2.32).

Returning to the inequality (7.2.15), on the basis of (7.2.30)-(7.2.32) andthe choice of the quantities γ, δ, d > 0 as suciently small, we obtain

(7.2.36)∫Gρ0

r−γε ζ2(r)v2xxdx ≤ c(ν, µ, µ1, γ0, γ, c0, ω0,M0, d)

ρ2γ0−γ−2 +

+∫Gρ0

r−γε ζ ′2(r)|∇v|2dx+

∫Gρ0

[r−γε ζ2(r)

(b4(x) + f2(x) + Φ2

xx

)+

+(r−γ−2ε ζ2(r) + r−γε ζ ′

2(r))· |∇Φ|2

]dx

, 0 < ρ ≤ d.

Finally, noting that by the hypotheses of theorem the quantities (7.2.1) areknown, in analogy to (7.2.35) we have∫

Gρ0

r−γε ζ ′2(r)|∇u|2dx ≤ c2M2

1ρ−2 ·

∫Gρ0

r−γdx =ω2

0c2M2

1

2− γ%−γ .


Therefore, by the properties of the functions rε and ζ(r) the inequality(7.2.36) gives∫Gρ0

r−γε ζ2(r)u2xxdx ≤ c(ν, µ, µ1,M0, γ0, γ, c0, d, ω0)×

×M2

1ρ−γ + ρ2γ0−γ−2 +

∫Gρ0

[r−γ∗ζ2(r)(b4(x) + f2(x) + Φ2

xx) +

+ r−γ∗−2 · ζ2(r)|∇Φ|2]dx

,

where 0 < ρ ≤ d. Since the right side does not depend on ε, passing to thelimit as ε→ +0 in it, we nally obtain∫Gρ/20

r−γu2xxdx ≤

∫Gd0

r−γζ2(r)u2xxdx ≤ c(ν, µ, µ1,M0,M1, γ0, γ, c0, d, ω0)×

×

d−γ + d2γ0−γ−2 + ||b2||2

W0

−γ∗ (Gd0)+ ||f ||2

W−γ∗ (Gd0)+ ||ϕ||2

W3/2

−γ∗ (Γd0)

The assertion of the theorem and the estimate (7.2.12) follow from this esti-mate.

Corollary 7.10. Suppose the hypotheses of Theorem 7.8 are satisedexcept for the niteness of M1. Then∫

Gρ/20

u2xxdx ≤ c(ν, µ, µ1,M0, γ0, γ, c0, d, ω0)×

×

ρ−2

∫Gρ0

|∇u|2dx+∫Gρ0

[b4(x) + f2(x)]dx+ ‖ϕ‖2W

3/2

0 (Γρ0)

(7.2.37)

Proof. This follows from (7.2.15), the Hardy-Wirtinger inequality, andestimates (7.2.30), (7.2.31), and (7.2.34) for γ = 0 and suciently smallδ, d > 0 with consideration of the properties of the functions rε and ξ(r).

7.2.4. Behavior of the solution near a corner point (weaksmoothness). In this Subsection we establish power decay of a solution ofthe homogeneous problem (QL)0 near a corner point.

Theorem 7.11. Suppose the conditions of Theorem 7.8 are satised, andlet γ > 0 be the number determined by this theorem. Suppose, moreover, that

f, b2 ∈ V 0ρ,−γ(G), ϕ ∈ C1(∂G) ∩ V 2−1/ρ

p,−γ (∂G), p > 2,

and

||f ||V 0p,−γ(Gρ

ρ/2) + ||b2||V 0

p,−γ(Gρρ/2

) + ||ϕ||V

2−1/ρp,−γ (Γρ

ρ/2)≤ k1ρ

2/ρ−1.(7.2.38)


Then

|v(x)| ≤ c1 · |x|1+γ/2(7.2.39)

|∇v(x)| ≤ c2 · |x|γ/2.(7.2.40)

Proof. The inequality (7.2.39) follows from the imbedding theorem(Lemma 1.38), because of Theorem 7.8.

To estimate the modulus of the gradient of the solution in the ring G11/2

we consider the function

z(x′) = v(ρx′) · ρ−1−γ/2,(7.2.41)

assuming that v ≡ 0 outside G. In G11/2 this function satises

aij(x′)zxijxij = F (x′),

aij(x′) ≡ aij(ρx′, u(ρx′), ρ−1ux′(ρx′)),(7.2.42)

F (x′) ≡ −ρ1−γ/2 · a(ρx′, u(ρx′), ρ−1ux′(ρx′))− ρ−1−γ/2aij(x′)Φx′jx′j(ρx′),

where, by assumptions (B), (C),

(7.2.43) |F (x′)| ≤ (µ1 +12

)ρ−1−γ/2 · |∇′u|2 + ρ1−γ/2(f + b2) +

+ µρ−1−γ/2|Φx′x′ |.

To the equation (7.2.42) we apply Theorem 4.10 regarding the boundednessof the modulus of the gradient of a solution inside the domain and near asmooth portion of the boundary:

ess supG1

1/2

|∇′z| ≤M ′1(7.2.44)

where M ′1 is determined only by ν, µ, µ1, ω0, and the integrals

∫G1

1/2

z2dx′,

( ∫G1

1/2

|F (x′)|pdx′)1/ρ

, p > 2.

We shall verify the niteness of these integrals. We have∫G1

1/2

z2dx′ ≤∫

Gρρ/2

r−γ−4 · v2dx ≤∫Gd0

r−γ−4v2dx ≤ c(d)


by Theorem 7.8. Further, by (7.2.43) and the assumption (7.2.38) of thetheorem we have( ∫

G11/2

|F (x′)|ρdx′)1/p

≤ c(µ, µ1, p)

∫Gρρ/2

rp(1−γ2

)−2 ·[|∇u|2p + |Φxx|p +

+ b2p + fp]dx

1/p

≤ c(µ, µ1, p)

ω0M

21d

1−γ/2

[p · (1− γ/2)]1/p+ k1

.

Returning to the function v(x), from (7.2.41) and (7.2.44) we obtain

|∇v(x)| ≤M ′1ργ/2, x ∈ Gρρ2∩Gd0.

Setting |x| = 2ρ/3, from this we obtain (7.2.40). Theorem 7.11 is proved.

7.2.5. The weighted integral estimate. We can rene the Niren-berg estimate on the basis of the weak smoothness of a solution establishedin Subsection 7.2.4. This renement is possible due to the requirement ofcontinuity of the leading coecients of the equation.

Theorem 7.12. Suppose u ∈W 2(G) is a solution of problem (QL) andthe assumptions of Theorem 7.11 are satised. Suppose the functionsaij(x, u, z) (i, j = 1, 2) are continuous at the point (0, 0, 0). If, in addition,

b2, f ∈ W0α(G), ϕ ∈ W3/2

α (∂G), and

2− 2π/ω0 < α ≤ 2,(K)

then u ∈ W2α(G), and

(7.2.45) ||u|| W

2

α(G)≤ c ·

(||u||2,G + ||f ||

W0

α(G)+ ||b2||

W0

α(G)+

+ ||ϕ|| W

3/2

α (∂G)

),

where c > 0 is a constant depending only on ν−1, µ, µ1, α, ω0,M0,M1, γ0, c0,meas G and diam G and also on k1, p, c1 and c2 of (7.2.38)-(7.2.40).

Proof. Let rε(x) be the function dened in 1.4 of Chapter 1. Wemultiply both sides of (QL)0 by rα−2

ε (x)v(x) and integrate over the domainG, using the condition (B) and integration by parts:∫

G


(2− α)2

2

∫G

rα−4ε v2dx+

(7.2.46)

+∫G

rα−2ε (x)v(x)

[aij(x, u, ux)− aij(0, 0, 0)]vxixj − F (x, v, vx)

dx.


We decompose G into two subdomains Gd0 and Gd, in each of which we ob-tain an upper bound for integral on the right side of (7.2.46).

Estimates in Gd0. By the continuity of aij(x, u, z) at the point (0, 0, 0)assumed in the theorem, for any δ > 0 there exists d0(δ) > 0 such that(

2∑i,j=1

|aij(x, u, ux)− aij(0, 0, 0)|2)1/2

< δ,(7.2.47)

provided that

|x|+ |u(x)|+ |∇u(x)| < d0.(7.2.48)

The smoothness of the boundary function ϕ(x) assumed in Theorem 7.11makes it possible to conclude, by Lemma 1.38, that ϕ(0) = 0 and|∇Φ(0)| = 0. Therefore, by (7.2.39) and (7.2.40) of Theorem 7.11 we have,for any x ∈ Gd0,

|x|+ |u(x)|+ |∇u(x)| ≤ |x|+ |v(x)|+ |∇v(x)|+ |Φ(x)− ϕ(0)|+

+|∇Φ(x)−∇Φ(0)| ≤ d+ c1d1+γ/2 + c2d

γ/2 +12d0,

and hence (7.2.48) is ensured because of the sucient smallness of d > 0.With the Cauchy inequality we now estimate the integral∫

Gd0

rα−2ε (x)v(x) [aij(x, u, ux)− aij(0, 0, 0)] vxixjdx ≤

(7.2.49)

≤ δ

2

∫Gd0

(r2rα−2

ε v2xx + rα−2

ε

v2

r2

)dx, ∀δ > 0.

Further, by the condition (7.2.9) with the help of Cauchy's inequality and(7.2.39) we obtain∫

Gd0

rα−2ε (x)v(x)F (x, v, vx)dx ≤

(12

+ 2µ1

)c1d

1+γ/2

∫Gd0


+(

12

+ 2µ1

)M0d

α−2

∫Gd

|∇v|2dx+32δ

∫G

rα−2ε

v2

r2dx+

(7.2.50)

+12δ

∫G

r2rα−2ε

(b4(x) + f2(x)

)dx+

+c(µ, µ1,M0, δ−1)

∫G

(r2rα−2ε Φ2

xx + rα−2ε |∇Φ|2)dx, ∀δ > 0.


Estimates in Gd. By condition (B) and properties of the function rε(x)we have

(7.2.51)∫Gd

rα−2ε v(x)[aij(x, u, ux)− aij(0, 0, 0)]vxixjdx ≤

≤ µ+ 12

dα−2

∫Gd

(v2xx + v2

)dx.

On the basis of (7.2.49) - (7.2.51) and the inequality (2.5.7) - (2.5.9) from(7.2.46) we now obtain for ∀δ > 0

(7.2.52)(π/ω0)2 − ((2− α)/2)2

(π/ω0)2 + ((2− α)/2)2·∫G

rα−2ε · |∇v|2dx ≤

≤[2δ ·H(α, ω0) +

(12

+ 2µ1

)c1d

1+γ/2 +O(ε)] ∫G


+δ

2·∫Gd0

r2rα−2ε ·v2

xxdx+ c(µ, µ1,M0, δ−1) ·

∫G

(rα−2ε r2Φ2

xx+ rα−2ε · |∇Φ|2)dx+

+12δ

∫G

r2rα−2ε

(b4(x) + f2(x)

)dx+ c(µ, µ1,M0)dα−2

∫Gd

(v2xx + v2)dx.

To estimate the integral with second derivatives in (7.2.52) we apply themethod of S. N. Bernstein (see, for example, [214], Chapter III, 19). Werewrite the problem (QL)0 in the form

∆v = F , x ∈ G; ,v(x) = 0, x ∈ ∂G;

(7.2.53)

where

F ≡ −[aij(x, v + Φ, vx + Φx)− aij(0, 0, 0)

]vxixj + F (x, v, vx).(7.2.54)

We multiply (7.2.53) rst by r2rα−2ε · vx1x1 and next by r2rα−2

ε · vx2x2 , andadd the equalities thus obtained; we integrate the result over G:∫

G

r2rα−2ε · v2

xxdx =∫G

r2rα−2ε ·∆v · Fdx+ 2Jα,ε[v],(7.2.55)

where the last term is dened by (2.6.3) and for it Lemma 2.42 holds. Weestimate the rst term on the right in (7.2.55) on the basis of (7.2.53), (7.2.47)


and conditions (B), (7.2.9):

(7.2.56)∫G

r2rα−2ε · 4v · Fdx ≤ δ1

∫G

r2rα−2ε · v2

xxdx+

+ c1

∫G

r2rα−2ε

(|∇v|4 + b4(x) + f2(x) + Φ2

xx + |∇Φ|4)

+

+ c(d, α)(1 + µ)∫Gd

v2xxdx, ∀δ1 > 0.

From (7.2.55), (7.2.56) on the basis of Lemmas 2.39, 2.40, 2.42 and withconsideration of (7.2.10) ,- the Hölder continuity of v(x),- since d, δ1 > 0were chosen suciently small, we obtain∫

G

r2rα−2ε v2

xxdx ≤ c(α,M0, µ,measG, diamG)∫Gd

(v2xx + v2)dx+

+ c(M0, µ, µ1)∫G

[rα(b4(x) + f2(x) + Φ2

xx

)+

+ rα−2|∇Φ|2]dx+ c(α,M0)

∫G


(7.2.57)

From Theorem 4.9 and Lemma 2.39 with α = 0 it follows the followingLemma:

Lemma 7.13.

(7.2.58)∫Gd

v2xxdx ≤ c(ν−1, µ, µ1, c0, γ0)×

×∫Gd

[v2(x) + b4(x) + f2(x) + Φ2xx + |∇Φ|2 + Φ2]dx.

From (7.2.52), (7.2.57), (7.2.58), and the condition (K) of the theorem,by choosing δ, and d suciently small we nally obtain∫

G

(r2rα−2ε u2

xx + rα−2ε · |∇u|2 + rα−4

ε · u2)dx ≤ O(ε)∫G


+ c(α, ω0, ν−1, µ, µ1,M0, c0, c1, γ0, γ, diamG,measG)×

×∫G

[u2 + rα(b4(x) + f2(x) + Φ2xx) + rα−2 · |∇Φ|2 + rα−4 ·Φ2]dx ∀ε > 0.

Passing to the limit as ε→ +0, we establish Theorem 7.12 and (7.2.45).


7.2.6. Proof of Theorem 7.7.

Proof. That u belongs to the space W22(G) follows from Theorem 7.12

for α = 2, so that to prove assertion 1) we need to prove (7.2.4). For thiswe multiply both sides of (QL)0 by v(x) and integrate over the domain Gρ0,0 < ρ < d. Setting

V (ρ) =∫Gρ0

|∇v|2dx(7.2.59)

we obtain

V (ρ) = ρ

ω0∫0

v∂v

∂rdω +

∫Gρ0

v(x)[aij(x, u, ux)− aij(0, 0, 0)]vxixj −

− v(x) · F (x, v, vx)dx.

(7.2.60)

We shall obtain an upper bound for each integral on the right. For the rstintegral we have Corollary 2.30:

ρ

ω0∫0

v∂v

∂rdω ≤ ρω0

2π· V ′(ρ).(7.2.61)

Condition (7.2.3) of the theorem ensures that (7.2.38) is satised, and henceestimates (7.2.39) and (7.2.3) of Theorem 7.11 are valid. On the basis of theseestimates and the assumed Dini continuity of the functions aij(x, u, ux) at(0, 0, 0) and the smoothness of the boundary function ϕ(x) it is not hard toestablish the existence of a positive, monotonically increasing function δ(r),continuous on [0, d], which satises a Dini condition at zero and is such that(

2∑i,j=1

|aij(x, u(x), ux(x))− aij(0, 0, 0)|2)1/2

≤ δ(ρ), |x| < ρ.

In fact, by Dini-continuity, we have:

(2∑

i,j=1

|aij(x, u(x), ux(x))− aij(0, 0, 0)|2)1/2

≤ A(|x|+ |u(x)|+ |∇u(x)|),

where A(t) satises the Dini condition at zero, i.e.d∫0

A(t)t < ∞. But from

estimates (7.2.39) and (7.2.3) it follows

|x|+ |u(x)|+ |∇u(x)| ≤ |x|+ c1|x|1+γ/2 + c2|x|γ/2 ≤ c|x|γ/2.

Hence we obtain

A(|x|+ |u(x)|+ |∇u(x)|) ≤ A(crγ/2) ≡ δ(r),


where1∫

0

δ(r)rdr =

1∫0

A(crγ/2)r

dr =2γ

c∫0

A(t)tdt <∞.

Therefore, by the Cauchy inequality we obtain∫Gρ0

v(x) · [aij(x, u, ux)− aij(0, 0, 0)]vxixjdx ≤

(7.2.62)

≤ 12δ(ρ) ·

∫gρ0

(r2 · v2xx + r−2 · v2)dx.

Finally, on the basis of (7.2.8) and by the condition (7.2.9) and the Höldercontinuity of the function v (inequality(7.2.10)) we obtain∫

Gρ0

v(x) · F (x, v, xx)dx ≤

≤(1

2+ 2µ1

)c0ρ

γ0V (ρ) + (1 + µ)ρδ ·∫Gρ0

r−2v2dx(7.2.63)

+c(µ, µ1,M0)ρ−δ ·∫Gρ0

[r2(Φ2xx + b4(x) + f2(x)

)+ |∇Φ|2

]dx, ∀δ > 0.

From (7.2.60) on the basis of (7.2.61)-(7.2.63), the Hardy-Wirtinger inequa-lity for α = 2, and the estimate (7.2.37) of Corollary 7.10 it now follows thatthe function V (ρ) satises the dierential inequality (CP ) from 1.10 of thePreliminaries, in which

P(ρ) =2πω0ρ− 2c0

π

ω0(12

+ 2µ1)ργ0−1 +ω0

πρ−1δ(ρ) + 2

ω0

π(1 + µ)ρδ−1,

N (ρ) =π

ω0ρ−1δ(ρ)c(ν, µ, µ1,M0, c0, γ0, ω0),

Q(ρ) =2πω0kρ2s−1−δ k = k2

1 · c(µ, µ1,M0) · (1 + 22s),(7.2.64)

V0 = V (d) ≤M21 ·measG

∀δ ∈(

0, 2s− 2π

ω0

)(here k1 and s are dened in condition (7.2.2)). According to Theorem 1.57the estimate (1.10.1) holds, which together with (7.2.37) leads to the desiredestimate (7.2.4).

The estimate (7.2.5) now follows from the imbedding theorem (Lemma1.38) and (7.2.4).


To prove the remaining assertions of Theorem 7.7 we apply the methodof rings and arguments analogous to those in the proof of Theorem 7.11.We perform the coordinate transformation x = ρx′. In G1

1/2 the functionz(x′) = ρ−λv(ρx′) satises (7.2.42) and (7.2.43) with γ replaced by 2(λ− 1).

By the Sobolev-Kondrashov theorems on imbedding of function spaceswe have

( ∫G1

1/2

|∇′z|pdx′)1/p

≤ c ·

[ ∫G1

1/2

(z2x′x′ + z2)dx′

]1/2

, ∀p > 2.(7.2.65)

(7.2.66) supx′∈G7/8

5/8

|∇′z|+ supx′,y′∈G7/8

5/8

x 6=y′

|∇′z(x′)−∇′z(y′)||x′ − y′|1−2/p

≤

≤ c‖z‖W 2,p(G

7/85/8

), ∀p > 2.

We consider (7.2.42) as a linear equation whose leading coecients are Dinicontinuous. By Theorem 10.17- the Lp-estimates for the solution inside thedomain and near a smooth portion of the boundary we have

||z||W 2,p(G

7/85/8

)≤ c(ν, µ, δ(1))

[ ∫G1

1/2

(|F (x′)|p + |z|p)dx′]1/p

,

∀p > 1, G7/85/8 ⊂ G

11/2.

(7.2.67)

Returning to the variable x and the function v(x), from (7.2.64)-(7.2.66) by(7.2.43) and (7.2.4) we obtain

( ∫Gρρ/2

|∇v|pdx

)1/p

≤ c · ρπ/ω0+2/p−1, ∀p ≥ 2,(7.2.68)

and also, considering the smoothness of the boundary function,

supx∈G7ρ/8

5ρ/8

|∇v(x)|+ ρ1−2/p · supx,y∈G7ρ/8

5ρ/8

x 6=y

|∇v(x)−∇v(y)||x− y|1−2/p

≤


(7.2.69) ≤ c(ν, µ, µ1, p, δ(ρ))

ρ1−2/p

( ∫Gρρ/2

|∇v|2pdx)1/p

+

+( ∫Gρρ/2

r−p−2|v|pdx)1/p

+

+ ρ1− 2+α

p ·⟨ ∫Gρρ/2

[rα(fp + b2p + |Φxx|p

)+ rα−p|∇Φ|p

]dx⟩1/p

,

(7.2.70) ||v||V 2p,α(G

7ρ/85ρ/8

)≤ c(ν, µ, µ1, δ(ρ))

[ ∫Gρρ/2

(rα|∇v|2p +

+ rα−2p|v|p)dx

]1/p

+⟨ ∫Gρρ/2

[rα · (fp + b2p +

+ |Φxx|p) + rα−p · |∇Φ|p]dx⟩1/p

.

We now note that by the estimate (7.2.5), already proved,

(7.2.71)∫Gρ0

r−p−2|v|pdx ≤ cp1 ·∫Gρ0

r(λ−1)p−2dx =

= ω0cp1 ·

ρ(λ−1)p

(λ− 1)p, λ = π/ω0 > 1,

∫Gρ0

rα−2p|v|pdx ≤ cp1 ·∫Gρ0

rα+(λ−2)pdx =cp1ω0

2 + p(λ− 2)ρ2+α+p(λ−2),(7.2.72)

if λ− 2 + 2+αp > 0. In addition, we have also

ρ1−2/p( ∫Gρρ/2

|∇v|2pdx)1/p

≤ %M21 ≤M2

1%λ−1,(7.2.73)

and ( ∫Gρρ/2

rα|∇v|2pdx)1/p

≤ cM21%

α+2p ≤ cM2

1%λ−2+α+2

p ,(7.2.74)

since 1 < λ < 2.

7.3 Estimates near the conical point 245

From (7.2.69), on the basis of (7.2.71), (7.2.73)and (7.2.3), we obtain(7.2.6). This concludes the proof of assertion 2) of Theorem 7.7. The asser-tion 3) and (7.2.7) follow in exactly the same way from (7.2.70) on the basisof (7.2.72), (7.2.74), and assumption (7.2.3). Finally, suppose the conditionsof assertion 4) are satised. Returning to (7.2.69), by (7.2.71), (7.2.73), and(7.2.3) we have

|∇v(x)−∇v(y)| ≤ cρ|x− y|π/ω0−1−,

∀x, y ∈ G7ρ/85ρ/8, κ =

π

ω0− 2 +

2p≤ 0.

(7.2.75)

By the denition of the sets G7ρ/85ρ/8 we have |x − y| ≥ (7

4ρ), since κ ≤ 0.Therefore, from (7.2.75) we obtain

|∇v(x)−∇v(y)| ≤ c · |x− y|π/ω0−1, ∀x, y ∈ G7ρ/85ρ/8,

whence assertion 4) follows. Theorem 7.7 is proved.

7.3. Estimates near a conical point

7.3.1. Introduction. In 7.2 we have investigated the behavior of strongsolutions to the Dirichlet problem for uniform elliptic quasi-linear second or-der equation of non-divergent form near an angular point of the boundary ofa plane bounded domain. There in particular it's proved that the rst orderderivatives of the strong solution are Hölder continuous with the exponentπω0−1, if π2 < ω0 < π and this exponent is the best possible (ω0 is an angle of

intersection of segments of the domain boundary in the angular point). Two-dimensionality of the domain was stipulated by Nirenberg's method whichwe applied to obtain the estimate

|u(x)| ≤ c0|x|1+γ(7.3.1)

with a certain γ ∈ (0, 1) in the neighborhood of an angular point. Otherresults of 7.2 don't depend upon two-dimensionality of the domain andmay be obtained by the methods presented in 7.2 in the multidimensionalcase. First we build the barrier function and with the aid of the ComparisonPrinciple establish the estimate (7.3.1) with a certain now small γ > 0.Then, by the Kondratiev layers method, basing on results of Ladyzhenskaya- Uraltseva - Lieberman [215, 217, 221 ] and on the estimate (7.3.1) weestablish the estimate

|∇u(x)| ≤ c1|x|γ .(7.3.2)

On the basis of (7.3.1), (7.3.2) we prove the integral weight estimates for thesecond generalized derivatives of the solution with the best weight exponent.These estimates allow to obtain exact estimates of the solution's moduli andits gradient and weight Lq− estimates of the second generalized derivativesof the solution and as well as to prove the Hölder continuity of the rstderivatives of the solution with the best Hölder exponent.


Definition 7.14. A strong solution of the problem (QL) is the function

u(x) ∈W 2,qloc (G \ O) ∩ C0(G), q ≥ N

satisfying the equation of the problem for almost all x ∈ G and the boundarycondition of the problem for all x ∈ ∂G. The value M0 = max

x∈G|u(x)| is

assumed to be known.

We shall further assume throughout that the below conditions aresatised(S) for ∀ε0 > 0 there exists d0 > 0 such that

Gd00 =

x ∈ G

∣∣arccos(xNr

)<π

2− ε0

⇔ Gd0

0 ⊂ xN ≥ 0 ⇒ λ > 1

(J) ϕ(x) ∈W 2− 1q,q(∂G), q ≥ N ; aij(x, u, z) ∈W 1,q(M), q > N ;

there exist a number β > −1 and nonnegative number k1 such that:

b(x) + f(x) ≤ k1|x|β,(7.3.3)

where functions b(x), f(x) are from the condition (C).

7.3.2. The barrier function. Let G0 = G∞0 be a innite cone andG0 ⊂ xN ≥ 0, Γ0 it lateral surface. We consider the second order linearoperator

L0 = aij(x)∂2

∂xi∂xj; aij(x) = aji(x), x ∈ G0,

νξ2 ≤ aij(x)ξiξj ≤ µξ2 ∀x ∈ G0,∀ξ ∈ RN ; ν, µ = const > 0;

Lemma 7.15. (About the existence of the barrier function). There exista number h > 0, determined only by G0, a number γ0 and a functionw(x) ∈ C1(G0)∩C2(G0) depending only on G0 and ellipticity constants ν, µof the operator L0 such that ∀γ ∈ (0, γ0]:

L0w(x) ≤ −νh2|x|γ−1, x ∈ G0;(7.3.4)

0 ≤ w(x) ≤ |x|1+γ ; |∇w(x)| ≤ 2(1 + h2)1/2|x|γ , x ∈ G0.(7.3.5)

Proof. We set x′ = (x1, ..xN−2), x = xN−1, y = xN . In the half-space y ≥ 0 we consider a cone K with the vertex O such that K ⊃ G0 (itis possible since G0 ⊂ y ≥ 0). Let ∂K be the lateral surface of K and theequation of ∂K∩ (xOy) is y = ±hx, so that inside K the inequality y > h|x|is true. We consider the function

w(x′;x, y) = (y2 − h2x2)yγ−1, γ ∈ R.(7.3.6)

Renaming the operator L0 coecients: aN−1,N−1 = a,aN−1,N = b, aN,N = c. we get:

L0w = awxx + 2bwxy + cwyy;

νη2 ≤ aη21 + 2bη1η2 + cη2

2 ≤ µη2;

η2 = η21 + η2

2 ∀η1, η2 ∈ R.(7.3.7)


We calculate the operator L0 on the function (7.3.6):

L0w = −h2yγ−1ϕ(γ), t = x/y, |t| < 1/h,

ϕ(γ) = 2(a− 2bt+ ct2)− (3ct2 − 4bt+ ch−2)γ − c(h−2 − t2)γ2;(7.3.8)

Since, by (7.3.7), ϕ(0) = 2(a−2bt+ct2) > 2ν and ϕ(γ) is the square function,then it is obvious that there is a number γ0 > 0, depending only on ν, µ, hsuch that ϕ(γ) > ν for γ ∈ [0, γ0]. From (7.3.6) and (7.3.8) we now obtainall the statements of our Lemma.

7.3.3. The weak smoothness of solutions. The above constructedfunction and the Comparison Principle (see Theorem 4.4) allow to estimateu(x) in the neighborhood of conical point. Without loss of generality weassume ϕ(x) ≡ 0.

Theorem 7.16. Let u(x) be a solution of (QL) and satisfy the conditions(S), (A), (B), (C) on the set M(u). Then there exist nonnegative numbersd < d0, γ dened only by values ν, µ,N, k1, β, γ0, d0,M0 and the domain Gsuch that in Gd0 the estimate (7.3.1) with a constant c0, independent of u(x)and dened only by the values ν, µ,N, k1, β, γ0, d0,M0 and the domain G,holds.

Proof. We consider the linear elliptic operator:

L = aij(x)∂2

∂xi∂xj+ ai(x)

∂

∂xi, x ∈ G;

aij(x) = aij(x, u(x), ux(x)); ai(x) = b(x)|∇u(x)|−1uxi(x),

where we assume ai(x) = 0, i = 1, . . . , N in such points x, for which|∇u(x)| = 0. Let us introduce the auxiliary function:

v(x) = −1 + exp(ν−1µ1u(x))(7.3.9)

Then we get:

Lv(x) ≡ ν−1µ1

(aij(x)uxixj + ν−1µ1a

ij(x)uxiuxj + b(x)|∇u(x)|)×

× exp(ν−1µ1u(x)) = ν−1µ1 (b(x)|∇u(x)| − a(x, u(x), ux(x)) +

+ ν−1µ1aij(x)uxiuxj exp

(ν−1µ1u(x)

)≥ −ν−1µ1f(x) exp(ν−1µ1M0)

in virtue of the assumptions (B), (C). By the condition (7.3.3), now weobtain

Lv(x) ≥ −ν−1µ1k1rβ exp(ν−1µ1M0), x ∈ Gd0.(7.3.10)

Let γ0 be the number dened by the barrier function Lemma and the numberγ satises the inequality:

0 < γ ≤ min(γ0, β + 1).(7.3.11)


We calculate the operator L for the barrier function (7.3.6)

Lw(x′;x, y) = −h2yγ−1ϕ(γ)+|∇u|−1b

((h2(1−γ)x2yγ−2+(1+γ)yγ

)∂u∂x−

− 2h2xyγ−1)∂u

∂y

)≤ −νh2yγ−1 + 2(1 + h)byγ , ∀(x′;x, y) ∈ G0.

Returning to the previous denotes and considering the inequality (7.3.3), weget:

Lw(x) ≤(−νh2 + 2(1 + h)k1d

1+β)rγ−1, x ∈ Gd0.

Now let the number d ∈ (0, d0) satisfy the inequality:

d ≤(

νh2

4k1(1 + h)

)1/(1+β)

.(7.3.12)

Then nally we have

Lw(x) ≤ −12νh2rγ−1, x ∈ Gd0.(7.3.13)

Now let us dene a number A :

A ≥ 2k1µ1ν−2h−2d1+β−γ exp(M0µ1/ν).(7.3.14)

Then from (7.3.10) and (7.3.13) with regard to (7.3.11) it follows

L(Aw(x)) ≤ Lv(x), x ∈ Gd0.(7.3.15)

In addition, from (7.3.5) and (7.3.9) it follows

Aw(x) ≥ 0 = v(x), x ∈ Γd0.(7.3.16)

Now we compare the functions v and w on Ωd. In virtue of the assumption(S) we have on the set G ∩ r = d ∩ xN−1OxN that

xN−1 = d sinϑ, xN = d cosϑ, |ϑ| < π/2− ε0, d ≤ d0,

and there is a cone K ⊃ G0 such that 0 < h < tan ε0 (see the proof of thebarrier function Lemma). From (7.3.6) it follows:

w∣∣∣r=d≥ d1+γ(sin ε0)γ−1(sin2 ε0 − h2 cos2 ε0) > 0.(7.3.17)

On the other hand, by Theorem 7.3 |u(x)| ≤ Mα|x|α, where α ∈ (0, 1) isdened by ν−1, µ,N and the domain G, but Mα is dened by the samevalues and M0, k1, β, d0. Therefore, by the well-known inequality et − 1 ≤t/(1− t), t < 1, we have

v(x)∣∣∣r=d≤ −1 + exp(ν−1µ1Mαd

α) ≤ 2ν−1µ1Mαdα,(7.3.18)

if d is so small that holds

d ≤ (2µ1Mαν−1)−1/α.(7.3.19)


Choosing a number A so large that the following inequality

A ≥ 2ν−1µ1Mαdα−1−γ(sin ε0)1−γ(sin2 ε0 − h2 cos2 ε0)−1(7.3.20)

would be satised, from (7.3.17),(7.3.18) it follows

Aw(x) ≥ v(x), x ∈ Ωd.(7.3.21)

Thus, if d ∈ (0, d0) is chosen according to (7.3.12), (7.3.19), the numberγ is chosen according to (7.3.11), and A is chosen according to (7.3.14),(7.3.20), then from (7.3.15), (7.3.16), (7.3.21) we obtain

Lv(x) ≥ L(Aw(x)), x ∈ Gd0; v(x) ≤ Aw(x), x ∈ ∂Gd0.In this case, because of the Comparison Principle (see Theorem 4.4), wehave v(x) ≤ Aw(x), x ∈ Gd0. Returning to the function u(x), from (7.3.9)we obtain

u(x) = νµ−11 ln(1 + v(x)) ≤ νµ−1

1 ln(1 +Aw(x)) ≤ Aνµ−11 w(x), x ∈ Gd0.

In the analogous way the inequality u(x) ≥ −Aνµ−11 w(x), x ∈ Gd0 is proved,

if we consider v(x) = 1 − exp(−ν−1µ1u(x)) as an auxiliary function. By(7.3.5), the proof of our Theorem is completed.

Basing on the layer method and the assumption (D), we can now provea gradient bound for solutions near a conical point.

Theorem 7.17. Let u(x) be a strong solution of the problem (QL),q > N and the assumptions (S), (A)− (J) on the set M(u) are fullled.Then in the domain Gd0, 0 < d ≤ min(d0, d) the estimate (7.3.2) is true witha constant c1, depending only on ν−1, µ, µ0, µ1, µ2, u, q, β, γ,κ1, κ2,M0 andthe domain G.

Proof. Let us consider in the layer G11/2 the function v(x′) = ρ−1−γu(ρx′),

taking u ≡ 0 outside G. Let us perform the change of variables x = ρx′ inthe equation (QL). The function v(x′) satises the equation

aij(x′)vxixj = F (x′), x′ ∈ G11/2(QL)′

where

aij(x′) ≡ aij(ρx′, ρ1+γv(x′), ργvx′(x)),

F (x′) ≡ −ρ1−γa(ρx′, ρ1+γv(x′), ργvx′(x′)).

By the assumption (D),

vraimaxG1

1/2

|∇′v| ≤M ′1,(7.3.22)

where M ′1 is determined only by ν, µ, µ1, k1, c0,M0, β, γ,N, q. Returning toformer variables from (7.3.22) we obtain

|∇u(x)| ≤M ′1ργ , x ∈ Gρρ/2.

Taking |x| = 2ρ/3, we arrive to the sought estimate (7.3.2). The Theoremis proved.


Let us now establish a "weak" solution smoothness of the problem (QL)in the neighborhood of a conical point.

Theorem 7.18. Let u(x) be a strong solution of (QL), q > N and theassumptions (S), (A)− (J) are fullled. Let γ0 be the number dened by the

barrier function Lemma. Then u(x) ∈ G1+γ(Gd0) for some d ∈ (0,min(d0, d))and ∀γ ∈ (0, γ∗], where γ∗ = min(γ0;β + 1; 1−N/q).

Proof. Let a number d ∈ (0,min(d0, d)) be xed so that the estimates(7.3.1), (7.3.2) are satised according to Theorems 7.16, 7.17. Let us considerin the layer G1

1/2 the equation of (QL)′ for the function v(x′) = ρ−1−γu(ρx′).By the Sobolev - Kondrashov imbedding Theorem 1.33

supx′,y′∈G1

1/2

x′ 6=y′

|∇′v(x′)−∇′v(y′)||x′ − y′|1−N/q

≤ c(N, q,G)‖v‖2,q;G11/2

; q > N.(7.3.23)

Let us verify that for the solution v(x′) we can apply Theorem 4.6 about Lq

- estimate inside a domain and near a smooth boundary portion. In fact,by the assumption (A), the functions aij(x, u, z) are continuous on the setM(u), i.e. for ∀ε > 0 there exists such η(ε) that

|aij(x, u(x), ux(x))− aij(y, u(y), ux(y))| < ε,

as soon as

|x− y|+ |u(x)− u(y)|+ |ux(x)− ux(y)| < η(ε), ∀x, y ∈ Gρρ/2, ρ ∈ (0, d).

The assumption (D) guaranteed the existence of the local a-priori estimateinside the domain Gρρ/2 and near a smooth portion of the boundary Γρρ/2 :there exist the number κ > 0 and the number M1 > 0 such that

|u(x)− u(y)|+ |∇u(x)−∇u(y)| ≤M1|x− y|e, ∀x, y ∈ Gρρ/2, ρ ∈ (0, d).

Then the functions aij(x′) are continuous in G11/2 and consequently are u-

niform continuous. It means that ∀ε > 0 there exists δ > 0 (we choose thenumber δ such that: δd + M1(δd)e < η) such that |aij(x′) − aij(y′)| < ε,

if only |x′ − y′| < δ, ∀x′, y′ ∈ G11/2. We see, that the assumptions of the

theorem about the local Lq a-priori estimate for the (QL)′ are satised. Bythis theorem, we have:

‖v‖q2,q:G1

1/2

≤ c4

∫G2

1/4

(|v|q + ρq(1−γ)|a(ρx′, ρ1+γv, ργux′)|q

)dx′(7.3.24)


with a constant c4, independent of v and a, and being determined only byN, ν, µ, µ1, γ, β, k1, q,M0,M1, d0, d. The estimate (7.3.1) gives rise to

(7.3.25)∫

G21/4

|v|qdx′ =∫

G2%%/4

%−q(1+γ)|u(x)|q%−Ndx ≤ cq,γmesΩ2ρ∫

ρ/4

dr

r≤

≤ cq,γmesΩ ln 8.

In the analogous way, by the assumption (C) together with the inequality(7.3.3) and the estimate (7.3.2), we obtain

(7.3.26)∫

G21/4

ρq(1−γ)|a(ρx′, ρ1+γv, ργvx′)|qdx′ ≤ ρq(1−γ)−N∫

G2ρρ/4

(µ1|∇u|2 +

+ b(x)|∇u|+ f(x))qdx ≤ 2N3q−1ρq(1−γ)mesΩ

2ρ∫Gρ/4

(µq1c

2q1 r

2qγ−1 +

+ (k1c1)qrq(β+γ)−1 + kq1rqβ−1

)dr ≤ c(N, q, γ, β, µ1, c1, k1),

since 0 < γ ≤ 1 + β. From(7.3.24)-(7.3.26) it follows

‖v‖2,q;G11/2≤ c(N, ν, µ, µ1, γ, β, k1, q,M0,M1, c0, c1).(7.3.27)

Now from (7.3.23) and (7.3.27) we obtain

supx′,y′∈G1

1/2

x′ 6=y′

|∇′v(x′)−∇′v(y′)||x′ − y′|1−N/q

≤ c5, q > N,(7.3.28)

where c5 = c(N, ν, µ, µ1, γ, β, k1, q,M0,M1, c0, c1, G).Returning to the variables x, u, we get

supx,y∈Gρ

ρ/2

x 6=y

|∇u(x)−∇u(y)||x− y|1−N/q

≤ c5ργ−1+N/q, q > N, ρ ∈ (0, d).(7.3.29)

Now let us recall that by the assumptions of our Theorem q ≥ N/(1 − γ).Let us put τ = γ − 1 +N/q ≤ 0, then from (7.3.29) it follows

|∇u(x)−∇u(y)| ≤ c5ρτ |x− y|γ−τ ∀x, y ∈ Gρρ/2, ρ ∈ (0, d).

By the denition of the set Gρρ/2, |x − y| ≤ 2ρ and consequently|x− y|τ ≥ (2ρ)τ , since τ ≤ 0.. Therefore:

supx,y∈Gρ

ρ/2

x 6=y

|∇v(x)−∇v(y)||x− y|γ

≤ 2−γc5, ρ ∈ (0, d).(7.3.30)


Now, let x, y ∈ Gd0 and ∀ρ ∈ (0, d). If x, y ∈ Gρρ/2, then (7.3.30) is fullled.If |x− y| > ρ = |x|, then, by the estimate (7.3.2), we have

|∇u(x)−∇u(y)||x− y|γ

≤ 2ρ−γ |∇u(x)| ≤ 2c1.

From here and (7.3.30) we conclude that

supx,y∈Gd0x 6=y

|∇u(x)−∇u(y)||x− y|γ

≤ const.

This inequality together with the estimates (7.3.1), (7.3.2) means u(x) ∈C1+γ(Gd0). Our Theorem is proved.

7.3.4. Estimates in weighted spaces. On the basis of the estimatesof 7.3.3 let us now derive the weighted integral estimates of the weak secondorder derivatives of strong solutions and establish the best-possible weightedexponent; for the simplicity we take ϕ(x) ≡ 0.

Theorem 7.19. Let u(x) be a strong solution of the problem (QL),q > N and the assumptions (S), (A)− (J) on the set M(u) are fullled. Inaddition, suppose

aij(0, 0, 0) = δji , (i, j = 1..N).

Then there exist positive numbers d, c2, independent of u(x), such that, ifb(x), f(x) ∈ V 0

2,α(G) and

4−N − 2λ < α ≤ 2,(7.3.31)

then u(x) ∈ V 22,α(Gd/20 ) and the estimate

(7.3.32)∫

Gd/20

(rαu2xx + rα−2|∇u|2 + rα−4u2)dx ≤ c2

∫G

2/d0

(u2 + |∇u|2 +

+ rα(b2(x) + f2(x)))dx,

is true, where d and c2 are dened by the values N, ν, µ, µ1, γ, β, k1, q, d0, d,M0,M1, λ, α and the domain G.

Proof. 1. 2−N ≤ a ≤ 2.

In this case, by the estimates (7.3.1), (7.3.2),∫Gd0

(rα−2|∇u|2 + rα−4u2)dx ≤ c(α,N, γ)dα+N−2+2γ .(7.3.33)


Now we get over to obtaining of the weighted estimate of the second orderweak derivatives of the solution. Let us x d ∈ (0,min(d, d0)] and considerthe sets G(k), k = 0, 1, 2, .... Let us perform the change of variables

x = (2−kd)x′, u((2−kd)x′) = v(x′)

in the equation of the problem (QL). As a result the domain G(k) of thespace (x1, . . . , xN ) transforms at the domain G1

1/2 of the space (x′1, .., x′N ),

and the equation takes the form

aij(x′)vx′ix′j = F (x′), aij(x′) ≡ aij((2−kd)x′, v(x′), d−12kvx′),

F (x′) ≡ −(2−kd)2a((2−kd)x′, v(x′), d−12kvx′).

To its solution let us apply the L2- estimate inside the domain and near asmooth boundary portion (the possibility to us this estimate is substantiatedunder the proof of Theorem 7.18; see the inequality (7.3.24)):∫

G11/2

v2x′x′dx

′ ≤ c4

∫G2

1/4

(v2(x′) + F 2(x′)

)dx′(7.3.34)

where the constant c4 is independent of v and F and is determined only bythe quantities pointed in (7.3.24). In the inequality (7.3.34) we return toformer variables and taking into account the denition of the sets G(k) :∫

G(k)

rαu2xxdx ≤ c4

∫G(k−1)∪G(k)∪G(k+1)

(rα−4u2 + rαa2(x, u, ux)

)dx.(7.3.35)

We sum the inequalities(7.3.35) over k = 0, 1, .., [log2(d/ε)] ∀ε ∈ (0, d) andwe get: ∫

Gdε

rαu2xxdx ≤ c4

∫G2dε/4

(rα−4u2 + rαa2(x, u, ux)

)dx.(7.3.36)

Taking into consideration the niteness of the integral (7.3.33) and becauseof the assumption (C) and the estimate (7.3.2), from (7.3.36) it follows:

(7.3.37)∫Gdε

rαu2xxdx ≤ c4c(γ, d, c1)

∫G2d

0

(rα−4u2 + rαf2(x) + rαb2(x) +

+ rα−2|∇u|2)dx ∀ε > 0,


where c4 is independent of ε. Therefore, by the Fatou Theorem, in (7.3.37)one can perform the passage to the limit over ε→ +0 and as a result we get

(7.3.38)∫Gd0

rαu2xxdx ≤

≤ c4

∫G2d

0

(rα−4u2 + rαf2(x) + rαb2(x) + rα−2|∇u|2

)dx.

The inequality (7.3.38) together with (7.3.33) means that u(x) ∈ V 22,α(Gd0).

We are coming now to the derivation of the estimate (7.3.32). Let ζ(r)be the cut-o function on the segment [0, d] :

ζ(r) ∈ C2[0, d], ζ(r) ≡ 1, if r ∈ [0, d/2], 0 ≤ ζ(r) ≤ 1, if r ∈ [d/2, d];

ζ ≡ 0, if r ≥ d; ζ(d) = ζ ′(d) = 0.

We multiply both parts of the problem (QL) equation by ζ2(r)rα−2u(x)and integrate over the domain Gd0. Taking into account the assumptionaij(0, 0, 0) = δji , twice integrating by parts we obtain

(7.3.39)∫Gd0

ζ2(r)rα−2|∇u|2dx+2− α

2(N + α− 4)

∫Gd0

ζ2(r)rα−4u2(x)dx =

=∫

Gdd/2

((N + 2α− 5)ζζ ′rα−3 + ζζ ′′rα−2 + ζ ′2rα−2)u2(x)dx+

+∫Gd0

ζ2(r)rα−2u(x)(aij(x, u, ux)− aij(0, 0, 0)uxixj + a(x, u, ux)

)dx

From the assumptions (A), (D), (J), by the Sobolev imbedding Theorem, itfollows that aij(x, u, z) are continuous at any point (x, u, z) ∈ M(u)

(i, j = 1, . . . , N) and in particular at the point (0, 0, 0). This means thatfor ∀δ > 0 there exists dδ > 0 such that

|aij(x, u(x), ux(x))− aij(0, 0, 0)| < δ(7.3.40)

as soon as

|x|+ |u(x)|+ |∇u(x)| < dδ.(7.3.41)

By the estimates (7.3.1), (7.3.2),

|x|+ |u(x)|+ |∇u(x)| ≤ d+ c0d1+γ + c1d

γ ∀x ∈ Gd0.(7.3.42)

Let us now choose d > 0, maybe more smaller than before, such that theinequality

d+ c0d1+γ + c1d

γ ≤ dδ(7.3.43)


would be fullled. Then the inequality (7.3.40) is fullled and therefore, bythe Cauchy inequality and the (7.3.38), we get

(7.3.44)∫Gd0

ζ2(r)rα−2aij(x, u, ux)− aij(0, 0, 0)uuxixjdx ≤ δ∫Gd0

ζ2(r)×

× rα−2|u||uxx|dx ≤δ

2

∫Gd0

(rαu2xx + rα−4u2)dx ≤

≤ δ

2(1 + c4)

∫G

2/d0

(rα−4u2 + rαf2(x) + rαb2(x) + rα−2|∇u|2)dx ∀δ > 0

Further, by the assumption (C), the estimates (7.3.1), (7.3.2) and the Cauchyinequality ,

(7.3.45)∫Gd0

ζ2(r)rα−2u(x)a(x, u, ux)dx ≤ µ1c0d1+γ

∫Gd0

ζ2(r)rα−2|∇u|2dx+

+12

(c1dγ + δ)

∫Gd0

ζ2(r)rα−4u2(x)dx+12c1d

γ

∫Gd0

ζ2(r)rαb2(x)dx+

+12δ

∫Gd0

ζ2(r)rαf2(x)dx, ∀δ > 0.

From (7.3.39), (7.3.44), (7.3.45) it follows

(7.3.46)∫Gd0

ζ2(r)rα−2|∇u|2dx+2− α

2(N + α− 4)

∫Gd0

ζ2(r)rα−4u2(x)dx ≤

≤ cδ(δ + dγ)∫

Gd/20

(rα−2 + |∇u|2 + rα−4u2)dx+ c8

∫G2d

0

rα(b2 + f2)dx+

+ c7

∫G2dd/2

(|∇u|2 + u2)dx ∀δ > 0 where c6 = c(µ1, c0, c1, c4),

c7 = c(µ1, c0, c1, c4, N, α, γ, d), c8 = c(δ, γ, c1, c4, d).


If N + α− 4 ≤ 0, then let us also use the inequality (2.5.3). As a result wehave

(7.3.47) C(λ,N, α)∫

Gd/20

rα−2|∇u|2dx ≤ c9(δ + dγ)∫

Gd/20

rα−2|∇u|2dx+

+ c10

∫G2d

0

(|∇u|2 + u2 + rα(b2 + f2))dx ∀δ > 0,

where

C(λ,N, α) = 1− 2− α2

(4−N − α)H(λ, α,N) > 0 (by (7.3.31)),

c9 = c(µ1, c0, c1, c4, N, α, λ), c10 = c(µ1, c0, c1, c4, N, α, γ, d, δ).

Now we choose the numbers δ and d such that:

δ =14c−1

9 C(λ,N, α),(7.3.48)

c9dγ ≤ 1

4C(λ,N, α)(7.3.49)

Then from (7.3.47) we nally obtain the inequality∫Gd/20

rα−2|∇u|2dx ≤ 2c10

C(λ, α,N)

∫G2d

0

(|∇u|2 + u2 + rα(b2 + f2))dx,(7.3.50)

being true only for such d ∈ (0,min d0, d), that (7.3.49) and (7.3.43) are ful-lled for dδ, being determined by the continuity of aij(x, u, z) at (0, 0, 0) forδ from the equality (7.3.48). The inequality (7.3.50) together with (7.3.38)and (2.5.3) leads us to the desired (7.3.32).

2. 4−N − 2λ < α < 2−N .

By the assumption (J), we have b(x), f(x) ∈ W

02−N (Gd00), consequently,

u(x) ∈ W22−N (Gd/20 ), i.e.∫

Gd/20

(r2−Nu2xx + r−N |∇u|2 + r−N−2u2)dx <∞,(7.3.51)

which was proved in the case 1).Now we use the function rε(x) dened in 1.4. We consider again the in-

equality (7.3.34). We multiply both parts of this inequality by(2−kd + ε)α−2 ∀ε > 0 and take into account that in G(k):2−k−1d + ε < r + ε < 2−kd + ε. Then returning to former variables we


get∫G(k)

r2(r + ε)α−2u2xxdx ≤ c4

∫G(k−1)∪G(k)∪G(k+1)

(r−2(r + ε)α−2u2 +

+ (r + ε)αa2(x, u, ux))dx.

Hence, by the Corollary 1.12, it follows

∫G(k)

rα−2ε u2

xxdx ≤ c4

∫G(k−1)∪G(k)∪G(k+1)

(r−2rα−2ε u2 + rαε a

2(x, u, ux))dx.

Summing this inequalities over all k = 0, 1, 2..., we obtain∫Gd0

rα−2ε u2

xxdx ≤ c4

∫G2d

0

(r−2rα−2ε u2 + rαε a

2(x, u, ux))dx.(7.3.52)

Let us now multiply both parts of the problem (QL) equation byζ2(r)rα−2

ε u(x) and integrate over Gd0; twice having applied the formula ofintegration by parts; as a result we have

(7.3.53)∫Gd0

ζ2(r)rα−2ε |∇u|2dx =

2− α2

(4−N − α)∫Gd0

ζ2(r)rα−4ε u2(x)dx+

+∫

Gdd/2

u2(x)(2(α− 2)ζζ ′(xi − εli)xirrα−4ε +Nζζ ′r−1rα−2

ε + ζ ′2rα−2ε +

+ ζζ ′′rα−2ε )dx+

∫Gd0

ζ2(r)rα−2ε u(x)

((aij(x, u, ux)−

− aij(0, 0, 0))uxixj + a(x, u, ux))dx.

Let d ∈ (0,min(d, d0)] be so small that (7.3.43) is fullled, and consequently(7.3.40) is fullled too. Then, by the Cauchy inequality,

(7.3.54)∫Gd0

ζ2(r)rα−2ε (aij(x, u, ux)− aij(0, 0, 0))uuxixjdx ≤

≤ δ∫Gd0

ζ2(r)rα−2ε (r|uxx|)(r−1|u|)dx ≤

≤ δ

2

∫Gd0

(ζ2(r)r2rα−2

ε u2xx + ζ2(r)r−2rα−2

ε u2)dx, ∀δ > 0.


Similarly, by the assumption (C) in view of the estimates (7.3.1), (7.3.2),

(7.3.55)∫Gd0

ζ2(r)rα−2ε u(x)a(x, u, ux)dx ≤

≤ c10µ1d1+γ

∫Gd0

ζ2(r)rα−2ε |∇u|2dx+

12

(c1dγ + δ)

∫Gd0

ζ2(r)rα−4ε u2dx+

+12c1d

γ

∫Gd0

ζ2(r)rαε b2(x)dx+

12δ

∫Gd0

ζ2(r)rαε f2(x)dx, ∀δ > 0.

From (7.3.52) - (7.3.55) with regard to the properties of rε(x) (see 1.4) andζ(r) it follows

(7.3.56)∫

Gd/20

rα−2ε |∇u|2dx ≤ c12(µ1, c1, c4, γ, d, δ)

∫G2d

0

rαε (x)(b2 + f2)dx+

+(1

2(δ + c1d

γ) + (2− α)(4−N − α)) ∫Gd/20

rα−4ε u2dx+

+ c11(µ1, c0, c1, c4, γ)d2γ

∫Gd/20


δ

2(1 + c4)

∫G2d

0

r−2rα−2ε u2dx.

The rst and third integrals on the right we estimate with the aid of (2.5.7)-(2.5.9), but for the bound of the latter integral on the right we use Lemma1.11 about rε(x) and take into account the negativity of α; as a result from(7.3.56) we obtain

(7.3.57) C(λ,N, α)∫

Gd/20

rα−2ε |∇u|2dx ≤

(12

(δ + c1dγ)H(λ, α,N) +

+ c11d2γ +

δ(1 + c4)2λ(λ+N − 2)

+O(ε)) ∫Gd/20


+ c14

∫G2d

0

(u2 + |∇u|2 + rαf2(x) + rαb2(x))dx, ∀δ > 0

with C(N,λ, α) being the same as in (7.3.47). Let us now choose δ and d asthe following

δ =12C(λ, α,N)

(H(λ, α,N) +

1 + c4

λ(λ+N − 2)

)−1,(7.3.58)

12c1d

γH(λ, α,N) + c11d2γ ≤ 1

4C(λ, α,N).(7.3.59)


Then from (7.3.57) it follows

(7.3.60)∫

Gd/20

rα−2ε |∇u|2dx ≤ O(ε) ·

∫Gd/20


+2c14

C(λ, α,N)

∫G2d

0

(u2 + |∇u|2 + rαf2(x) + rαb2(x))dx, ∀ε > 0.

Finally, from (7.3.52), (2.5.7)-(2.5.9)with regard to the assumption (C) andthe estimates (7.3.1), (7.3.2), because of (7.3.60), we have:

(7.3.61)∫

Gd/20

(r2rα−2ε u2

xx + rα−2ε |∇u|2 + rα−4

ε u2)dx ≤

≤ O(ε)·∫

Gd/20

rα−2ε |∇u|2dx+c2

∫G2d

0

(u2+|∇u|2+rαf2(x)+rαb2(x))dx, ∀ε > 0,

where

c2 = c(N,λ, α, ν, µ, µ1, γ, β, k1, q,M0, d0, d)

and independent of ε. The inequality (7.3.61) holds for that d ∈ (0,min(d0, d)]for which (7.3.59), (7.3.43) are fullled with dδ, being determined by the con-tinuity of aij(x, u, z) at (0, 0, 0) for δ, being assigned by (7.3.58). Performingin the inequality (7.3.61) the passage to the limit over ε→ +0, by the FatouTheorem, we get the desired estimate (7.3.32).

Remark 7.20. By the continuity of the equation leading coecients atthe point (0, 0, 0) and the estimates (7.3.1), (7.3.2), the condition aij(0, 0, 0) =δji , (i, j = 1..N) of our Theorem is not implied to be restrictive, since thereexists the orthogonal transformation of coordinates, which transforms anelliptic equation with leading coecients, frozen at the point, to canonicalform, whose main part is dened by Laplacian.

Theorem 7.21. Let u(x) be a strong solution of the problem (QL),q > N and the hypotheses of Theorem 7.19 are satised. In addition, supposethat β > λ− 2. Then there exist positive numbers d and c15 independent ofu(x) and being dened only by the quantities from hypotheses (B)− (J) andby G such that u(x) ∈ W2

4−N (Gd/20 ) and the inequality

|u| W

2

4−N (Gρ0)≤ c15ρ

λ, ρ ∈ (0,d

2)(7.3.62)

holds.


Proof. The belonging of u(x) to W 24−N (Gd/20 ) follows from Theorem

7.19, therefore it is required to prove only the estimate (7.3.62). We set

U(ρ) ≡∫Gρ0

r2−N |∇u|2dx(7.3.63)

Let us multiply both parts of the (QL) equation by r2−Nu(x) and integrateover the domain Gρ0, ρ ∈ (0, d2):

(7.3.64) U(ρ) =∫Ω

(ρu(ρ, ω)∂u

∂r

∣∣∣r=ρ

+N − 2

2u2(ρ, ω))dω +

+∫Gρ0

u(x)r2−N(aij(x, u, ux)− aij(0, 0, 0)uxixj + a(x, u, ux)

)dx.

Let us upper estimate every integral on the right. The rst integral estimatesby Corollary 2.30. By the assumption (C) and the Cauchy inequality withδ = ρε, ∀ε > 0 with regard to (7.3.1), (7.3.2), we have

(7.3.65)∫Gρ0

r2−Nu(x)a(x, u, ux)dx ≤ µ1c0ρ1+γV (ρ) +

12c1ρ

γ

∫Gρ0

(r−Nu2 +

+ r4−Nb2(x))dx+

12

∫Gρ0

(ρεr−Nu2 + ρ−εr4−Nf2(x)

)dx.

Let us also apply the inequality (2.5.3) with α = 4−N and also (7.3.3):

(7.3.66)∫Gρ0

r2−Nu(x)a(x, u, ux)dx ≤(µ1c0ρ

1+γ +12H(λ,N, 4−N)×

× (ρε+c1ργ))U(ρ)+(4λ)−1(1+c1)k2

1measΩρ2s−ε, ∀ε > 0, s = β+2 > λ.

Further aij(x, u, z)− aij(0, 0, 0) = (aij(0, 0, z)− aij(0, 0, 0)) ++ (aij(x, u, z)− aij(0, 0, z)) . From the assumption (J), by the Sobolevimbedding Theorem, taking into account the estimates (7.3.1), (7.3.2), wehave (

N∑i,j=1

|aij(x, u(x), ux(x))− aij(0, 0, 0)|2)1/2

≤ δ(ρ), |x| < ρ

δ(ρ) = c(N, q, c0, c1, γ, d)ργ , ρ ∈ (0,d

2),

0 < γ ≤ γ∗ = min(γ0; 1 + β; 1− N

q).


Therefore applying the Cauchy inequality, the (7.3.38), the inequality (2.5.3)with α = 4−N , and the condition (7.3.3) we get∫

Gρ0

r2−Nu(x)(aij(x, u, ux)− aij(0, 0, 0))uxixjdx ≤

≤ 12δ(ρ)

∫Gρ0

(r4−Nu2xx + r−Nu2)dx ≤ 1

2H(λ,N, 4−N)δ(ρ)U(ρ) +

(7.3.67)

+c4

2(H(λ,N, 4−N) + 1)δ(ρ)U(2ρ) +

k21c4

4λmeasΩδ(ρ)(2ρ)2s.

From (7.3.64) basing upon Corollary 2.30, (7.3.66)-(7.3.67) we conclude thatU(ρ) satises the inequality for the Cauchy problem (CP ) with

P(%) =2λ%− c

(%γ−1 + %ε−1

); N (%) = c%γ−1;

Q(%) = ck21

(%2s+γ−1 + %2s−ε−1

), s > λ, ∀ε ∈ (0, 2(s− λ));

V0 =∫Gd0

r4−N |∇u|2dx ≤ c1measΩ2(2 + γ)

d2(γ+2).

According to the Theorem 1.57 the estimate (1.10.1) holds, which leads tothe estimate U(ρ) ≤ cρ2λ. This estimate together with (7.3.37) and (2.5.3)gives the desired estimate (7.3.62).

7.3.5. Lp− and pointwise estimates of the solution and its gra-dient. Let us make precise the exponent γ in the estimates (7.3.1), (7.3.2)and the Hölder exponent for the rst order weak derivatives of the strongsolution in the neighborhood of conical point O. We recall that ϕ(x) ≡ 0.

Theorem 7.22. Let u(x) be a strong solution of the problem (QL),q > N and it is known the value M0 = max

x∈G|u(x)|. Let the assumptions (S),

(A)− (J) be fullled with β > λ − 2 > −1. Then there exist nonnegativenumbers d ≤ d∗ = min(d, d) and c0, c1, c2, c3, independent of u(x) and beingdened only by quantities N,λ, ν, µ, µ1, β, k1, q,M0,M1, d0, d, and G, suchthat the following assertions hold:

1. |u(x)| ≤ c0|x|λ; |∇u(x)| ≤ c1|x|λ−1, x ∈ Gd/20 ;

2. u(x) ∈ W24−N (Gd/20 ) and ||u||

W2

4−N (Gρ0)≤ c2ρ

λ, 0 < ρ < d/2;

3. if α+ q(λ− 2) +N > 0, then u(x) ∈ V 2q,α(Gd/20 ) and

‖u‖V 2q,α(Gρ0) ≤ c3ρ

λ−2+α+Nq , 0 < ρ < d/2;

4. if 1 < λ < 2, q ≥ N2−λ , then u(x) ∈ Cλ(Gd/20 ).


Proof. The assertion 2) is proved in Theorem 7.21. To prof the remain-ing assertions we consider the sets Gρρ/2 and G

2ρρ/4 ⊃ G

ρρ/2. Let us perform the

transformation of coordinates x = ρx′ in the equation of (QL). The functionv(x′) = ρ−λu(ρx′) satises in G2

1/4 the equation (QL)′ for γ = λ − 1. Thelocal boundary Lq− estimate, Theorem 4.6 seems to be applicable to thesolution v(x′) (the justication of the possibility of its application see in theproof of Theorem 7.18):

|v|q2,q;G1

1/2

≤ c4

∫G2

1/4

(|v|q + ρ(2−λ)q|a(ρx′, ρλv, ρλ−1vx)|q)dx′, ∀q > 1(7.3.68)

with the constant c4 independent of v and a.Let at rst 2 ≤ N < 4. By the estimate (7.3.62) we have:

‖v‖22,2;G11/2≤ c(N)ρ−2λ

∫Gρρ/2

(r4−Nu2xx + r2−N |∇u|2 + r−Nu2)dx ≤ c(N)c2

15

Therefore from the Sobolev imbedding theorem it follows

supx′∈G1

1/2

|v(x′)| ≤ c(N, q)‖v‖2,2;G11/2≤ c(N, q)c15 = c0 or

|u(x)| ≤ c0ρλ, x ∈ Gρρ/2.

Putting |x| = 34ρ hence we obtain the rst bound of statement 1) of our

theorem. The second bound of that assertion follows from Theorem 7.17having been considered under γ = λ− 1.

Let now N ≥ 4. In this case let us apply the local maximum principle(see Theorem 4.5):

(7.3.69) supx′∈G1

1/2

|v(x′)| ≤ c(N, ν−1, µ)(‖v‖2,G2

1/4+

+ ρ2−λ‖a(ρx′, ρλv, ρλ−1vx′)‖N,G21/4

).

Let us estimate from above the addends of the right part of (7.3.69). Therst addend is estimated as well as above (see (7.3.62))

|v|22,G21/4≤ 2Nρ−2λ

∫G2ρρ/4

r−Nu2dx ≤ 2Nc215(7.3.70)


By the assumption (C) in view of (7.3.2), we have:

(7.3.71)∫

G21/4

|a(ρx′, ρλv, ρλ−1vx′)|Ndx ≤13

6N∫

G2ρρ/4

(µN1 |∇u|2N + fN (x) +

+ bN (x)|∇u|N)r−Ndx ≤ 1

36N

∫G2ρρ/4

(µN1 (r2−N |∇u|2)(r−2|∇u|2N−2) +

+ (r2−N |∇u|2)(kN1 rβN−2|∇u|N−2) + kN1 r

βN−N)dx ≤

≤ 13

6N(µN1 c

2N−21 ρ2γ(N−1)−2 + kN1 c

N−21 ργ(N−2)+βN−2

) ∫G2ρρ/4

r2−N |∇u|2dx+

+ (3βN)−1(6k)NmeasΩ(2βN − 2−βN )ρβN , ρ ∈ (0,d

2)

Hence with regard to (7.3.62) we obtain

(7.3.72) ρ2−λ|a(ρx′, ρλv, ρλ−1vx′)|N,G21/4≤ c16ρ

2−λ+2(λ−1)/N+2γ(N−1) +

+ c17ρ(2−λ+β)+2(λ−1)/N+γ(N−2)/N + c18ρ

β+2−λ, ∀ρ ∈ (0,d

2)

From (7.3.69), (7.3.70), and (7.3.72) with regard to β ≥ λ− 2 we get

supx′∈G1

1/2

|v(x′)| ≤ c19 + c20ρ2−λ+2(λ−1)/N+2γ(N−1)/N(7.3.73)

Let us remind that λ > 1 and γ > 0 and is determined by Theorem 7.16.As well as in the case 2 ≤ N < 4, for the validity of assertion 1) of our

theorem it is suciently to derive the bound

supx′∈G1

1/2

|v(x′)| ≤M ′0 = const(7.3.74)

Let us show, that repeating a nite number of times the procedure ofderiving of (7.3.73) with dierent exponents γ, it is possible to deduce theestimate (7.3.74). So let the exponent of ρ in (7.3.73) be negative (otherwise(7.3.73) means (7.3.74)). From (7.3.73) we have:

|u(x)| ≤ c21|x|2+2(λ−1)/N(7.3.75)

and from here, by Theorem 7.17 with γ = γ1

γ1 = 1 +2N

(λ− 1)(7.3.76)

we get also the inequality

|∇u(x)| ≤ c22|x|γ1 .(7.3.77)


Let us repeat the procedure of the deduction of (7.3.71), (7.3.72) havingapplied the inequality (7.3.77) instead of (7.3.2) (i.e. replacing γ on γ1); asa result we get

supx′∈G1

1/2

|v(x′)| ≤ c19 + c20ρ2−λ+2(λ−1)/N+2γ1(N−1)/N(7.3.78)

If in this inequality the exponent of ρ is negative, then putting

γ2 = 1 +2N

(λ− 1) +2(N − 1)

Nγ1(7.3.79)

we rst obtain by Theorem 7.17 the inequality

|∇u(x)| ≤ c22|x|γ2(7.3.80)

and next repeating the procedure pointed above as well as the bound

supx′∈G1

1/2

|v(x′)| ≤ c19 + c20ρ2−λ+2(λ−1)/N+2γ2(N−1)/N .(7.3.81)

Let us set

t =2(N − 1)

N≥ 3

2, ∀N ≥ 4(7.3.82)

and consider the numerical sequence γk :γ1 is determined by the equality (7.3.76),γ2 = (1 + t)γ1,γ3 = (1 + t+ t2)γ2,. . . . . . . . . . . . . . . . . . . . . . . . . . .γk+1 = (1 + t+ · · ·+ tk)γ1 = tk+1−1

t−1 , k = 0, 1, · · · ,Repeating the expounded procedure k times we get

supx′∈G1

1/2

|v(x′)| ≤ c19 + c20ρ1−λ+γk+1 , ρ ∈ (0, d/2).(7.3.83)

Let us show that for ∀N ≥ 4 we can nd such an integer k that

1− λ+ γk+1 ≥ 0.(7.3.84)

In fact, from the denition of the numerical sequence γk and (7.3.76) itfollows

1− λ+ γk+1 =tk+1 − 1t− 1

+λ− 1

N(t− 1)

(2tk+1 − 2−Nt+N

).

The rst addend on the right is positive. For the second addend from (7.3.82)it follows

2tk+1 − 2−Nt+N = 2k+2

(1− 1

N

)k+1

−N ≥ 0,

if (2N − 2N

)k+1

≥ N

2or k + 1 ≥

ln N2

ln 2N−2N

.


Hence we obtain the validity of (7.3.84) for

k =

[ln N

2

ln 2N−2N

], ∀N ≥ 4,

where [a] is the integral part of a. Thus statement 1) is proved.

Now let us refer the proof of statement 3) of our theorem. Multiplyingboth sides of the inequality (7.3.68) by %α−2q and returning to the variablesx, u we rewrite the inequality obtained in such way replacing ρ by 2−kρ andnext sum all inequalities over k = 0, 1, . . . . As result we have

‖u‖qV 2q,α(G%0)

≤ c4

∫G2%

0

(rα|a(x, u, ux)|q + rα−2q|u|q

), q > 1.(7.3.85)

Taking into account the assumption (C) and the bounds from assertion 1)proved above we obtain

|a(x, u, ux)|q ≤ C(µ1, k1, q,N)(|∇u|2q + rβq|∇u|q + rβq

)≤

≤ C(r2q(λ−1) + rq(β+λ−1) + rβq

)Hence and from (7.3.85) it follows


≤ CmeasΩ

2%∫0

(rα+q(λ−2) + rα+2q(λ−1) + rα+q(β+λ−1) + rα+βq

)dr.

Since β > λ− 2 and therefore rqβ < rq(λ−2), nally we establish


≤ C%α+N+q(λ−2),(7.3.86)

provided α+N + q(λ− 2) > 0. The latter means the required statement 3).

Finally, let us prove statement 4). By the Sobolev - Kondrashov imbed-ding Theorem 1.33

supx′,y′∈G1

1/2

x′ 6=y′

|∇′v(x′)−∇′v(y′)||x′ − y′|1−N/q

≤ c(N, q,G)‖v‖2,q;G11/2

; q > N.

Returning to the variables x, u, we get

supx,y∈Gρ

ρ/2

x 6=y

|∇u(x)−∇u(y)||x− y|1−N/q

≤ C‖u‖qV 2q,0(G

%/42% )≤ Cρλ−2+N/q,

q > N, ρ ∈ (0, d),

in virtue of (7.3.86). Verbatim repeating the proof of Theorem 7.18 withγ = λ − 1, provided N + q(λ − 2) ≤ 0, we get the validity of statement4).


7.3.6. Higher regularity results. In this subsection we examine thequestion of a smoothness rise of the Dirichlet problem solutions for the ellipticsecond order non-divergence quasi-linear equations near the conical bound-ary point. Let us consider the strong solution from W 2,q(G) ∩ C1+γ(G) of(QL). As well as in the linear case the solution smoothness in the quasilinearcase depends on the quantity λ determining value of the cone solid angle ina neighborhood of the point O.

Let us dene the set

MM0,M1 = (x, u, z)∣∣x ∈ G, |u| ≤M0, |z| ≤M1

As for the equation of the problem (QL) we assume that the followingconditions are satised on the set MM0,M1 :(E) the uniform ellipticity: there exist the positive constants ν, µ such that

for ∀(x, y, z) ∈MM0,M1 ,∀ξ ∈ RN

νξ2 ≤ aij(x, u, z)ξiξj < µξ2; aij(0, 0, 0) = δji , i, j = 1, . . . , N ;

(F) aij(x, u, z) ∈ Cm(MM0,M1) (i, j = 1, ..., N) for some integerm ≥ 1 and the partial derivatives of the functions aij(x, u, z) overall their arguments up to the order m are bounded on MM0,M1 ;

(G) there exist generalized partial derivatives of the function a(x, u, z) overall their arguments up to the order m ≥ 1, nonnegative functions fl(x)and the numbers µl, kl (l = 1, ...,m) such that the following inequalities

|Dl1uD

l2x a(x, u, z)| ≤ µ1|z|2 + fl(x); 1 ≤ l1 + l2 ≤ m;(7.3.87)

|Dl1uD

l2x D

l3z a(x, u, z)| ≤ µ1|z|+ fl(x); 0 ≤ l1 + l2 ≤ m− 1;(7.3.88)

|Dl1uD

l2x D

l3z a(x, u, z)| ≤ µl; 2 ≤ l1 + l2 + l3 ≤ m,(7.3.89)

where

fl(x) ≤ kl|x|λ−2−l, f0 ≤ k0|x|β, β > λ− 2(7.3.90)

are fullled.

Theorem 7.23. Let λ > 1, p > N be given and let the integer m satisfythe condition

1 ≤ m < λ− 2 +N/p.(7.3.91)

Let the assumptions (A)− (G) be satised and let the functionu(x) ∈ V 2

p,0(G) be a solution of the problem (QL) with

M0 = max|u(x)| : x ∈ G, M1 = max|∇u(x)| : x ∈ G.

Moreover, let ϕ(x) ∈ V m+2−1/pp,0 (∂G) ∩ V 3/2

2,4−N (∂G) and there exist the non-

negative numbers k′0, k′1, ..., k

′m and s > l such that the inequalities

(7.3.92) ||ϕ||V

3/22,4−NΓρ0

≤ k′0ρs, ||ϕ||V

2−1/p+mp,0 Γρ

ρ/2

≤ k′mρλ−2−m+N/p,

ρ ∈ (0, d)


hold. Then u(x) ∈ V m+2p,0 (G) and there exist the numbers d ∈ (0, d) and

Cm > 0 such that

||u||Vm+2p,0 Gρ0

≤ Cmρλ−2−m+N/p, ρ ∈ (0, d),(7.3.93)

where Cm is determined only by the quantities taking part in the assumptionsof the theorem and by G.

Proof. We apply the usual iteration procedure over m. Let m = 1. Letus consider the equation of (QL) in the domain Gρρ/2, ρ ∈ (0, d). The lateralsurface Γρρ/2 of Gρρ/2 is unboundedly smooth, because Gd0 is a convex cone.By denition of smooth domains, for every point x0 ∈ Γρρ/2 there exists aneighborhood Γ ⊂ Γρρ/2 of this point and a dieomorphism χ from C2+m

rectifying the boundary in Γ. Let D ⊂ Gρρ/2 be such that Γ ⊂ D. Let usperform the transformation y = χ(x) = (χ1(x), ..., χ(x)) and let χ(D) =D′, χ(Γ) = Γ′ ⊂ ∂D′, (Γ′ is a plane portion of the boundaryD′), v(y) =u(χ−1(y)). In this case χ, χ−1 ∈ C2+m and Jacobian |∇χ| 6= 0. Besides, onecan suppose the norms in C2+m of transformations χ determining the localrepresentation of the boundary Γρρ/2 to be uniformly bounded with respectto x0 ∈ Γρρ/2. In the new variables the equation of (QL) takes the form:

Aij(y, v, vy)vyiyj +A(y, v, vy) = 0, y ∈ D′,(QL)′

where

A(y, v, vy) = a(x, u, ux) + aij(x, u, ux)vyk∂2χk∂xi∂xj

(7.3.94)

Aij(y, v, vy) = akl(x, u, ux)∂χi∂xk

∂χj∂xl

,

Let us notice that in D′ by condition (E)

κ1νξ2 ≤ Aijξiξj ≤ κ2µξ

2,(7.3.95)

where κ1 = inf|∇χ(x)|2 > 0 : x ∈ D′;κ2 = sup|∇χ(x)|2 > 0 : x ∈ D′.The coordinate system can be chosen such that the positive axis yN wouldbe parallel to the normal toward Γ′ and the axes y1, ..., yN−1 parallel therays at plane Γ′. Let ek be the xed coordinate vectors (k = 1, ..., N − 1).For suciently small |h| we dene the dierence quotients

vk(y;h) =1hv(y)− v(y1, ..., yk−1, yk − h, yk+1, ..., yN ), k = 1, ..., N − 1.

We set:

yt = ty + (1− t)(y − hek); vt(y) = tv(y) + (1− t)v(y − hek).

Then the function w(y) ≡ vk(y, h) satises the linear equation

aij(y)wyiyj + ai(y)wyi + a(y)w = f(y), y ∈ D′,(L)


where

aij(y) = Aij(y, v(y), vy(y));

ai(y) = vypyl(y − h)∫ 1

0

∂Apl(yt, vt, vty)∂vtyi

dt+∫ 1

0

∂A(yt, vt, vty)∂vtyi

dt,

a(y) = vypyl(y − h)∫ 1

0

∂Apl(yt, vt, vty)∂vt

dt+∫ 1

0

∂A(yt, vt, vty)∂vt

dt,

−f(y) = vypyl(y − h)∫ 1

0

∂Apl(yt, vt, vty)∂ytk

dt+∫ 1

0

∂A(yt, vt, vty)∂ytk

dt,

k = 1, ...N − 1. Since the directories ek(k = 1, ...N − 1) are parallel to thetangent plane to Γ, we have w|Γ′ = ψk(y, h), y ∈ Γ′, ψ(y) = ϕ(χ−1y). Letus apply the local Lp - estimate near smooth boundary portion (Theorem4.6) to the solution w(y). Let us verify the fulllment of all conditions ofthe above estimate. (7.3.95) implies the fulllment of the uniform ellipticitycondition for the (L) equation. Since our solution u(x) ∈ C1+γ(G), thehypothesis (E) guarantees continuity of the coecients aij(y) in D′. Since,by the assumptions of our theorem uxx ∈ Lp(D′), p > N, then by assertions1) and 3) of Theorem 7.22 in view of the assumptions (F), (G) we have:

∥∥∥( N∑i=1

|ai(y)|2) 1

2

∥∥∥N,D′

≤ C(|χ|2,D′

)(µ1|uxx|N,G3ρ/2

0

+

+ ‖µ+ (µ1 + µ1)|∇u|+ f0(x)‖N,Gρρ/2

)≤

≤ C(|χ|2,D′ , N, p, k0, µ, µ1, µ1

)(ρ+ ρλ + ρλ−2+N/p + ρλ−1

)≤ const,

by the inequality (7.3.91). Similarly:

||a||p,D′ ≤ C(|χ|2,D′

)(||µ1|∇u|2 + µ1|∇u|+ f0(x)||p,Gρ

ρ/2+ µ1||uxx||p,G3ρ/2

0

)≤

≤ C(|χ|2,D′ , N, p, k0, µ1, µ1

)(ρ2(λ−1)+N/p + ρλ−2+N/p + ρλ−1+N/p

)≤ const;

‖f‖p,D′ ≤ C(|χ|3,D′

)(||µ1|∇u|2 + µ1|∇u|+ f1(x)||p,Gρ

ρ/2+ µ1‖uxx‖p,G3ρ/2

0

)≤

≤ C(|χ|2,D′ , N, p, k0, µ1, µ1

)%λ−3+N/p.

So the local Lp - estimate for the solutions of (L)′ gives us the inequality:

||w||2,p;D” ≤ const(||w||p,D′ + ||f ||p;D + ||ϕk(y − h)||2−1/p,p;Γ′),(7.3.96)

∀D” b D′ ∪ Γ′,

where const is independent of w, f, , ϕk, h and dewpends only on k, p, ν,µ,κ1,κ2 and the moduli of continuity of the coecients aij(y) on D′; the


latter are estimated in the following way:

|aij(y1)− aij(y2)|D′ = |Aij(y1, v(y1), vy(y1))−Aij(y2, v(y2), vy(y2))| =

=∣∣∣akl(x1, u(x1), ux(x1))

∂χi(x1)∂xk

∂χj(x1)∂xl

− akl(x2, u(x2), ux(x2))∂χi(x2)∂xk

×

×∂χj(x2)∂xl

∣∣∣ ≤≤ |akl(x1, u(x1), ux(x1))− akl(x2, u(x2), ux(x2))| · |∇χ|2 +

+µ∣∣∣∂χi(x1)∂xk

∂χj(x1)∂xl

− ∂χi(x2)∂xk

∂χj(x2)∂xl

∣∣∣ ≤ µκ1/22 |χ|2,D|x1 − x2|+

+κ2µ1

(|x1 − x2|+ |u(x1)− u(x2)|+ |∇u(x1)−∇u(x2)|

)≤

≤ 2ρ(κ2µ1 + µκ

1/22 |χ|2,D + c1ρ

γ)

+ C(2ρ)γ

by u(x) ∈ C1+γ(G). Further, we have by the denition of w(y):

||w||p,D′ =∣∣∣∣∣∣v(y)− v(y − hek)

h

∣∣∣∣∣∣p,D′≤ C(|χ−1|1)||∇u(x)||p,Gρ

ρ/2(7.3.97)

Analogously we obtain:

||ϕk(y, h)||2−1/p,p;Γ′ ≤ C(|χ−1|1)||ϕ(x)||3−1/p,p;Γρρ/2,(7.3.98)

and nally

(7.3.99) ||f ||p,D′ ≤ C(|χ|2,D′)(∣∣∣∣µ1|∇u|2 + µ1|∇u|+ f1(x)

∣∣∣∣p,Gρ

ρ/2

+

+ µ1||uxx||p,G3ρ/20

)Now from (7.3.96) - (7.3.99) we obtain the inequality:∣∣∣∣∣∣v(y)− v(y − hek)

h

∣∣∣∣∣∣2,p;D′

≤ const(||ϕ(x)||3−1/p,p;Γρ

ρ/2+

+∣∣∣∣|∇u|2∣∣∣∣

p,Gρρ/2

+ ||∇u(x)||p,Gρρ/2

+ ||f1(x)||p,Gρρ/2

+ ||uxx||p,G3ρ/20

)≤

≤ const%λ−3+N/p,

where const on the right is independent of h. This fact allows us to concludeon the basis of Fatou's theorem that there exists vyh ∈W 2,p(D′′) and performpassage to the limit h→ 0:

(7.3.100)∣∣∣∣∣∣ ∂v∂yk

∣∣∣∣∣∣2,p;D′′

≤ const(||ϕ(x)||3−1/p,p;Γρ

ρ/2+ ||uxx||p,G3ρ/2

0

+

+∣∣∣∣|∇u(x)|2

∣∣∣∣p,Gρ

ρ/2

+ ||∇u(x)||p,Gρρ/2

+ ||f1(x)||p,Gρρ/2

)≤

≤ const%λ−3+N/p, k = 1, ..., N − 1.


We consider again the equation (QL)′ and dierentiate it over yN :

(7.3.101) ANN (y, v, vy)vyNyNyN = −N−1∑k=1

AkNvykyNyN +N−1∑i,j=1

AijvyiyiyN +

+∂Aij∂vyl

vyiyjvylyN +∂Aij∂v

vyiyjvyN +∂Aij∂yN

vyiyj +∂Aij∂vyl

vylyN +∂A

∂yN+∂A

∂vvyN

,

y ∈ D′,

where

ANN = akl(y, v, vy)∂χN∂xk

∂χN∂xl

≥ ν|∇χN (x)|2 ≥ κ1ν.

Since u(x) ∈ W 2,p(Gρρ/2), vyk ∈ W 2,p(D′′), 1 ≤ k ≤ N − 1 then from(7.3.101) we obtain v(y) ∈ W 3,p(D′′), by the assumptions (F), (G). Thenby Sobolev's imbedding theorems 1.32, 1.34 we can derive:

1. if p > 2N then v(y) ∈ C2(D′′) and in this case |v|2;D′′ ≤ c||v||3,p/2;D′′ .2. if N p; inparticular v(y) ∈W 2,2p(D′′) for p ≥ 3N/2.

By the above statements and equation (7.3.101), we obtain v(y) ∈W 3,p(D′′)and therefore u(x) ∈ W 3,p(G7p/8

5p/8), if p ≥ 3N/2. Now we need to examineonly p ∈ (N, 3N/2). From above we have v(y) ∈W 2,q1(D′′) and by (7.3.101)v(y) ∈W 3,q1/2(D′′). Let us use again imbedding

W 3,q ⊂W 2,q∗, q∗ =Nq

N − q, q < N ;

as a result

v(y) ∈W 2,q2(D′′), q2 = Nq1/(2N − q1) = Np/(4N − 3p), if N < p < 4N/3,

and v(y) ∈ C2(D′′), if p ≥ 4N/3.

We repeat that procedure s times:

v(y) ∈W 3,qs/2(D′′) ∩W 2,qs(D′′), qs =Np

N2s − (2s − 1)p,(7.3.102)

if N < p < N/(1 − 2−s). We choose an integer number s ≥ 1 in such waythat qs ≥ 2p. Solving that inequality we obtain s = [log2((2p−N)/(p−N))],where [a] is the integral part of a. Thus from (7.3.102) we nd

u(x) ∈W 3,ρ(G7ρ/85ρ/8) ∩W 2,2p(G7ρ/8

5ρ/8), ∀ρ ∈ (0, d).


We proceed to derivation of the estimate (7.3.93) under m = 1. From(7.3.101), by (7.3.100), we have

(7.3.103)(∫D′′

∫|vyNyNyN |

pdy)1/p

≤ (νχ1)−1µχ2

N−1∑k=1

∣∣∣∣∣∣ ∂v∂yk

∣∣∣∣∣∣2,p;D′′

+

+(µ+ (µ1 + µ1)|∇yv|D′′ + |f0(y)|D′′

)||vyy||p,D′′ +

+∣∣∣∣µ1|∇yv|3 + (µ1 + µ1)|∇yv|2 + (µ1 + f0(y))|∇yv|+ f1(y)

∣∣∣∣p,D′′

+

+ µ1||vyy||2p,D′′ + µ1||vyy||p,D′′(1 + |∇yv|D′′)C(|χ|2,D′′)

From (7.3.100), (7.3.103) in the variables x, u(x) taking into account thehypothesis (G) we obtain

||u||3,p;G

7ρ/85ρ/8

≤ C(|χ|3,Gρ

ρ/2, ν, µ,N, p,κ1,κ2, µ1, µ1

)(||uxx||p,G3ρ/2

0

+

+∣∣∣∣|∇u|3 + |∇u|2 + |∇u|(1 + f0(x)) + |f1(x)|+ u2

xx

∣∣∣∣p,Gρ

ρ/2

+

+(1 + |f0(x) +∇u|Gρρ/2

)||uxx||p,Gρρ/2

+ ||ϕ||3−1/p,p,Γρρ/2

).

From here basing on Theorem 7.22 for ρ ∈ (0, d)

||u||3,p;G

7ρ/85ρ/8

≤ Cρλ−3+N/p,(7.3.104)

C = C(|χ|3,Gρ

ρ/2, ν, µ, p,κ1,κ2, µ1, µ1, λ, k0, k1, d, c, c3

).

Replacing in (7.3.104) ρ by 2−kρ, summing the inequalities obtained over allk = 0, 1, 2, ... and taking into account (7.3.91) under m = 1 we come to thesought estimate (7.3.93) under m = 1.

Repeating such procedure by induction we conclude the validity of theassertions of Theorem 7.23.

Theorem 7.24. Let all assumptions of Theorem 7.23 excepting of (7.3.91)be fullled. If m ≥ 0 is the integer and

m+ 1 < λ ≤ m+ 2− N

p, p > N,(7.3.105)

then u(x) ∈ Cλ(G). In addition, there exist constants ck, (k = 0, ...,m + 1)independent of u(x) such that

|∇ku(x)| ≤ Ck|x|λ−k, x ∈ Gd0, k = 0, ...,m+ 1.(7.3.106)

If λ = m+ 1, p ≥ N then u ∈ Cλ−ε(G), ∀ε > 0.

Proof. Let the function v(x′) = ρ−λu(ρx′) be a solution in the layerG1

1/2 of (QL)′. Verbally repeating the proof of Theorem 7.22 and using theresults of Theorem 7.23 we obtain all assertions of Theorem 7.24.


7.4. Solvability results

Let us include the problem (QL) to a family of one-parametric problemsfor t ∈ [0, 1]

aij(x, u, ux)uxi,xj + ta(x, u, ux) = 0, x ∈ Gu(x) = tϕ(x), x ∈ ∂G

(QL)t

With regard to the problem (QL) we assume the hypotheses (S), (A)− (J)to be satised. In addition, suppose(M) for every solution ut(x) of the problem (QL)t the value

M0 = supx∈G|ut(x)|, ∀t ∈ [0, 1] is known;

(K) ϕ(x) ∈ W324−N (∂G)∩V

2− 1q

q,0 (∂G), q ≥ N ; there exist nonnegative num-bers k3, k4, k5 and s > λ such that

b(x) + f(x) + |Φxx(x)| ≤ k3dλ−2(x), x ∈ Gε, ∀ε > 0;

‖ϕ‖ W

324−N (Γ%0)

≤ k4%s, ‖ϕ‖

V2− 1

qq,0 (Γ%

%/2)≤ k5%

λ−2+Nq , % ∈ (0, d).

Theorem 7.25. Let Γd ∈W 2,p and the assumptions (S), (A)− (J), (M),

(K) under q = p > N be fullled. If either λ ≥ 2 or 1 < λ < 2, N λ−2, q > N

2−λ be giv-

en numbers, and let Γd ∈W 2,p. Suppose the hypotheses (S), (A)− (J), (M), (K)are fullled. Then the problem (QL)t has at least one solution

ut(x) ∈W 2,qloc (G) ∩ V 2

p,0(G) ∩ Cλ(G)

for ∀t ∈ [0, 1].

Proof. We rst shall establish that for some γ ∈ (0, 1) and all t ∈ [0, 1]every solution ut(x) ∈W 2,q

loc (G) ∩ C0(G) satises the inequality

|ut(x)|1+γ,G ≤ K(7.4.1)

with a constant K being independent of ut(x) and t. Let us represent G =Gd0 ∪Gd with some positive suciently small d. From Theorem 7.18 we con-clude that under given assumptions there exist such positive d and γ0 thatut(x) ∈ C1+γ(Gd0) and the estimate (7.4.1) holds with ∀γ ∈ (0, γ∗], whereγ∗ = min(γ0;β+ 1; 1−N/q). The membership ut(x) ∈ C1+γ(Gd) and corre-sponding apriori estimate follow from the assumption (D) (local estimatesnear a smooth boundary portion have been established in [215, 217, 221 ],but in strictly contained subdomain follows, by the Sobolev - Kondrashovimbedding theorem 1.33. In such way the membership ut(x) ∈ C1+γ(G) andthe apriori estimate (7.4.1) are established.

7.4 Solvability results 273

The bound (7.4.1) allows to apply the Leray - Schauder xed point the-orem 1.56. To apply this theorem we x γ ∈ (0, 1) and consider the Banach s-pace B = C1+γ(G) for Theorem 7.25 or B = C1+γ

0 (G) =

=v ∈ C1+γ(G)

∣∣∣v(0) = |∇v| = 0

for Theorem 7.26. Let us dene theoperator T, by letting ut = tTv, as the unique solution from the spaceV 2p,0(G) (Theorem 7.25) or W 2,q

loc (G) ∩ V 2p,0(G) ∩ Cλ(G) (Theorem 7.26) for

any v ∈ B of the linear problemaij(x)uxixj = At(x), x ∈ G,u(x) = tϕ(x), x ∈ ∂G.

(L)t

where aij(x) = aij(x, v(x), vx(x)), At(x) = −ta(x, v(x), vx(x)). It exists byTheorem 4.48 (Theorem 4.49). In fact, it is not dicult to verify that allhypotheses of these Theorems are fullled. In particular, by the assumption(A), aij(x, v(x), vx(x)) ∈ W 1,p(M), p > N and therefore by the imbeddingtheorem aij(x) ∈ C1−N/p(G). In addition, for ut(x) the bound (4.10.9) holds.In virtue of the assumption (C) it has the form

(7.4.2) ‖ut‖V 2p,0(G) ≤ c

(µ1|∇v|2 + |∇v|‖b(x)‖p,G +

+ ‖f‖p,G + ‖ϕ‖V

2−1/pp,0 (∂G)

), ∀t ∈ [0, 1]

It is clear that the solvability of the problem (QL)t in the correspondingspace is equivalent to the solvability of the equation ut = tTv in the Banachspace B. Now we verify that all hypotheses of the Leray - Schauder xedpoint theorem 1.56 are fullled. This theorem guarantees the existence of ax point of the map T.

At rst, we verify that T is the compact mapping of the space B ontoitself. From the bound (7.4.2) it follows that the operator T maps bounded inB sets into bounded sets of the space V 2

p,0(G), and they are precompact setsin C1+γ(G), if γ < 1−N

p . Thus T is the compact mapping. Now we verify thatT is the continuous mapping onto B. Let the sequence vk(x) ⊂ B convergeto v(x) ∈ B.We set uk(x) = Tvk(x). By stated above, uk(x) ⊂ V 2

p,0(G). It iswell known that in the space V 2

p,0(G) every bounded set is weakly compact.We leave the notation uk(x) for a weak convergent subsequence and denotethe weakly limit by lim

k→∞uk(x) = u(x) ∈ V 2

p,0(G). The last means

limk→∞

∫G

g(x)Dαuk(x)dx =∫G

g(x)Dαu(x)dx, |α| ≤ 2,

(7.4.3)

∀g(x) ∈ Lp′(G);1p

+1p′

= 1.

Since now it is obvious that

aijk (x)(uk)xixj −Ak(x) ∈ Lp(G),


where

aijk (x) = aij(x, vk(x), vkx(x)), Ak(x) = −a(x, vk(x), vkx(x)),

then we prove that

limk→∞

∫G

g(x)(aijk (x)(uk)xixj −Ak(x)

)dx =

(7.4.4)

=∫G

g(x)(aij(x)uxixj −A(x)

)dx, ∀g(x) ∈ Lp′(G).

In fact, by the continuity of aij(x, v(x), vx(x)) on M and, because ofvk(x)→ v(x) in C1+γ(G), we have

(7.4.5) limk→∞

aijk (x) = limk→∞

aij(x, vk(x), vkx(x)) =

= aij

(x, lim

k→∞vk(x), lim

k→∞vkx(x)

)= aij(x).

Similarly we verify that limk→∞

Ak(x) = A(x). Now for ∀g(x) ∈ Lp′(G) we

obtain

limk→∞

∫G

g(x)(aijk (x)(uk)xixj − aij(x)uxixj

)dx ≤

(7.4.6)

≤ supx∈G|aijk (x)− aij(x)| · ‖ukxx‖p,G‖g‖p′,G +

∫G

(ukxixj − uxixj

)aij(x)g(x)dx.

Since the equation of the problem (QL) is uniformly elliptic then aij(x)g(x) ∈Lp′(G) and by (7.4.3) we get that the last summand in (7.4.6) tends to zero

as k →∞. By the proved above aijk (x) ∈ C1−Np (G), therefore by the Arzela

Theorem the limit (7.4.5) is uniform, consequently

limk→∞

supx∈G|aijk (x)− aij(x)| = 0.

In addition, vk(x) is uniformly bounded in B, hence by the bound (7.4.2)we obtain that ‖ukxx‖p,G ≤ const for ∀k. Hence, the rst summand in (7.4.6)tends to zero as k →∞ too. In the same way we verify that

limk→∞

∫G

g(x) (Ak(x)−A(x)) dx = 0.

Thus the equality (7.4.4) is proved. Since uk = Tvk, then the left side of(7.4.4) is equal to zero and hence∫

G

g(x)(aij(x)uxixj −A(x)

)dx = 0, ∀g(x) ∈ Lp′(G).

7.5 Notes 275

Hence it follows that aij(x)uxixj = A(x) for almost all x ∈ G. Further,uk(x) = ϕ(x), x ∈ ∂G and, by uk(x) ⊂ V 2

p,0(G) because of the imbeddingtheorem, uk(x) ⊂ C1+γ(G), 0 < γ ≤ 1− N

p . Therefore

u(x)∣∣∣∂G

= limk→∞

uk(x)∣∣∣∂G

= ϕ(x).

Thus we proved the equality u = Tv. But then we have

limk→∞

Tvk(x) = limk→∞

uk(x) = u(x) = Tv(x) = T

(limk→∞

vk(x)),

that is T is the continuous mapping. All hypotheses of the Leray - SchauderTheorem are veried and Theorem 7.25 is proved.

Theorem 7.26 is proved in the same way. Let us turn our attention tosome details only. We consider in the space B the bounded set

VK =v ∈ C1+γ

0 (G)∣∣∣|v(x)|1+γ,G ≤ K

.

In this case we apply Theorem 4.49 with α = 0 for the solvability of thelinear problem (L)t. We must verify only the assumption A7) of Theorem4.49. For this by our assumption (J) we get:∫G%0

r4−NA2t (x)dx ≤

∫G%0

r4−Na2(x, v, vx)dx ≤ c∫G%0

r4−N(µ2

1K4 +K2b2(x) +

+f2(x))dx ≤ c

∫G%0

(r4−N + r4−N+2β

)dx ≤

≤ cmeas Ω(

14%4 +

14 + 2β

%2β+4

)≤ C%2s, s > λ; ∀t ∈ [0, 1];

∫G%%/2

|At(x)|qdx ≤ c∫

G%%/2

(µq1K

2q +Kqbq(x) + f q(x))dx ≤

≤ cmeas Ω(%N + %qβ+N

)≤ C%N+(λ−2)q, ∀t ∈ [0, 1].

7.5. Notes

The condition (D) can be replaced by any other condition which guar-antee the existence of a priori estimate

|u|1+γ;G′ ≤M1, γ ∈ (0, 1)

for any smooth subdomain G′ ⊂⊂ G \ O (see [84, 221, 326, 128 ]).The results of Chapter 7 refer to the problem (QL) with its equation as

non-divergent. Such problems in non-smooth domains, have not almost beenstudied before. Only the research of I.I. Danilyuk [89] is known here. In this


work, using the methods of complex variable function theory and integralequations, the author proved the solvability in the space W 2,2+ε(G), ε > 0is suciently small, G ⊂ R2 and contains angular points. However, as we'llsee below (7.2), the requirements for these problems in this work are toohigh and the number ε > 0 is not precise. The formulated Theorem 7.7 from7.2.2 shows:

0 < ε < 2 · π/ω0 − 12− π/ω0

, ifπ

2< ω0 < π.

The results of Sections 7.2 -7.4 were rst established in [54, 55, 57, 58, 59,60, 61, 63]. We follow these articles.

One of the rst investigations of the solutions behavior in a neighbor-hood of the boundary without assumption it smoothness and convexity forquasilinear elliptic equation with two independent variables was done byN. Fandyushina [121].

N. Trudinger [379] has established a necessary and sucient conditionon boundary data for the solvability of the Dirichlet problem for a quasilinearelliptic equation aij(ux)uxixj = 0.

Solutions to some other quasilinear equations in nonsmooth domainswere studied in [10, 101, 292, 293, 330, 331, 332, 333, 407 ].

The results of this chapter was generalized in [366] on quasilinear ellipticequations whose coecients may degenerate near a conical boundary point,namely, the uniform ellipticity condition on the set M has a form

ν|x|τξ2 ≤ aij(x, u, z)ξiξj ≤ µ|x|τξ2, ∀ξ ∈ RN , 0 < τ ≤ 1;

lim|x|→+0

|x|−τaij(x, u, z) = δji .

CHAPTER 8

Weak solutions of the Dirichlet problem for elliptic

quasilinear equations of divergence form

In this chapter we investigate the behavior of weak solutions to theDirichlet problem for uniformly elliptic quasilinear equations of divergenceform in a neighborhood of a boundary conical point. We consider weak solu-tionsu ∈W 1,m(G) ∩ Lp(G), m > 1 of the dierential equation

Q(u, φ) ≡∫G

ai(x, u, ux)φxi + a(x, u, ux)φ dx = 0(DQL)

for all φ(x) ∈W 1,m0 (G)∩Lp(G). We suppose that Q is elliptic in G, namely

there are positive constants ν, µ such that

ν|z|m−2|ξ|2 ≤ ∂ai(x, u, z)∂zj

ξiξj ≤ µ|z|m−2|ξ|2, m > 1;(E)

∀(x, u, z) ∈ G× R× RN , ∀ξ ∈ RN .

8.1. The Dirichlet problem in general domains

Theorem 8.1. Maximum principle (see Theorem 10.9 10.5 [128]).Let u ∈ W 1,m(G), m > 1 be a weak solution of (DQL) and suppose that Qsatises the structure conditions

(i) ai(x, u, z)zi ≥ ν|z|m − g(x);(ii) a(x, u, z)sign z ≥ −µ2|z|m−1 − f(x),

where ν, µ2 = const > 0, and f(x), g(x) ∈ Lp/m(G) are nonnegative measur-able functions. Then we have the estimate

supx∈G|u(x)| ≤ C

(‖f‖p/m,G + ‖g‖p/m,G

)where C = C(N,m, ν, µ2, p,measG).

Theorem 8.2. The weak Harnack inequality (see Theorem 1.1 [378]).Let u ∈ W 1,m(G), m > 1 be a weak nonnegative solution of (DQL) and

suppose that Q satises the structure conditions

(i) ai(x, u, z)zi ≥ ν|z|m − µ1um;

(ii)

√N∑i=1

(ai(x, u, z))2 ≤ µ2|z|m−1 + µ3u

m−1;

277

278 8 Weak solutions of the Dirichlet problem for divergence form quasilinear equations

(iii) |a(x, u, z)| ≤ µ4|z|m−1 + µ5um−1,

with ν, µ2 > 0; µ1, µ3, µ4, µ5 ≥ 0. Then for any ball B3R ⊂ G, there holds

‖u‖Lp(B2R) ≤ CRNp infBR

u(x),(8.1.1)

where C depends only on m,N, p, ν, µ1, µ2, µ3, µ4, µ5 and p ∈ (0, (m−1)NN−m ) if

m < N or p ∈ (0,∞) if m ≥ N.Theorem 8.3. Hölder continuity of weak solutions (see Theorems

2.1 - 2.2 2, chapter IX [213]).Let G be of type (A) (see Denition 7.2). Let u ∈ W 1,m(G) ∩ L∞(G),

m > 1 with vrai maxG|u| = M0 < ∞ be a weak solution of the (DQL) and

suppose that following assumptions are satised

(a) ai(x, u, z)zi ≥ ν|z|m − g(x);

(b)

√N∑i=1

(ai(x, u, z))2 ≤ µ1|z|m−1 + ϕ1(x);

(c) |a(x, u, z)| ≤ µ1|z|m + ϕ2(x),where 1 < m ≤ N, and ϕi(x) are nonnegative and

‖g(x)‖Lp/m(G), ‖ϕ1(x)‖Lp/(m−1)(G), ‖ϕ2(x)‖Lp/m(G) ≤ const, p > N.

Then u(x) is Hölder-continuous in G.

Remark 8.4. We observe that the condition (a) follows from the ellip-ticity condition (E) and the condition (b). In fact, we have

ai(x, u, z)zi = zizj

1∫0

∂ai(x, u, z)∂zj

∣∣∣ez=tz

dt+ ziai(x, u, 0) ≥

≥ ν|z|21∫

0

tm−2|z|m−2dt− ziai(x, u, 0) ≥ ν

m− 1|z|m − ϕ1(x)|z| ≥

≥(

ν

m− 1− ε)|z|m − cεϕm

′1 (x), ∀ε > 0

in virtue of the Young inequality.

Theorem 8.5. Existence Theorem (see Theorem 9.2 9, chapter IV[214]).

Let 1 < m < N, 1 ≤ p < ∞. Let the functions ai(x, u, z), a(x, u, z) becontinuous with respect to u, z and satisfy the conditions

(i) Q(u, φ) is coercive, i.e.

Q(u, u) ≥ h(‖u‖W 1,m

0 (G)∩Lp(G))− c1 for ∀u ∈W 1,m

0 (G) ∩ Lp(G),

where c1 > 0, and h(t) is a continuous positive function such thatlimt→∞

h(t) =∞;

8.1 The Dirichlet problem in general domains 279

(ii) |ai(x, u, z)| ≤ µ|z|m−1 + µ|u|ep/m′ + ϕ1(x), ϕ1(x) ∈ Lm′(G);

(iii) |a(x, u, z)| ≤ µ|z|m/ep′ + µ|u|ep−1 + ϕ2(x), ϕ2(x) ∈ Lep′(G);with p 0,nondecreasing function.

Then the problem (DQL) has at least one weak solution from

W 1,m0 (G) ∩ Lp(G).

Remark 8.6. If the functions ai(x, u, z) are dierentiable with respectto z, then the condition (iv) follows from the ellipticity condition (E). In fact,let the ellipticity condition (E) be satised. Then considering two cases: 1)m ≥ 2 and 2) 1 < m < 2, we obtain

1) m ≥ 2:

(ai(x, u, z)− ai(x, u, w)) (zi − wi) =

= (zi − wi)1∫

0

d

dtai (x, u, w + t(z − w)) dt =

=

1∫0

∂ai (x, u, w + t(z − w))∂zj

dt · (zi − wi)(zj − wj) ≥

≥ ν|z − w|21∫

0

|w + t(z − w)|m−2dt ≥ νc(m)|z − w|m

in virtue of Lemma 1.7 and m ≥ 2.

2) 1 < m < 2: We have again

(ai(x, u, z)− ai(x, u, w)) (zi − wi) ≥ ν|z − w|21∫

0

|w + t(z − w)|m−2dt

But

|w + t(z − w)| ≤ |w|+ t|z − w| ⇒ |w + t(z − w)|m−2 ≥ (|w|+ t|z − w|)m−2


and therefore1∫

0

|w + t(z − w)|m−2dt ≥1∫

0

(|w|+ t|z − w|)m−2dt =

=1

|z − w|

|w|+|z−w|∫|z|

τm−2dτ =1

m− 1(|w|+ |z − w|)m−1 − |w|m−1

|z − w|.


(ai(x, u, z)− ai(x, u, w)) (zi − wi) ≥

≥ ν|z − w|m− 1

(|w|+ |z − w|)m−1 − |w|m−1

.

It is easy to verify that in both cases the function ψ(ζ) satises the conditionsof (iv).

Theorem 8.7. Hölder continuity of the rst derivatives of weaksolutions (see Theorem 1 [225]).

Let µ,M0 be positive constants. Let G be a bounded domain in RN withC1+α, α ∈ (0, 1] boundary. Let u(x) be a bounded weak solution of (DQL)with |u| ≤M0. Suppose (DQL) satises the ellipticity condition (E) and thestructure conditions√√√√ N∑

i=1

|ai(x, u, z)− ai(y, v, z)|2 ≤ µ(1 + |z|)m−1 (|x− y|α + |u− v|α) ;

|a(x, u, z)| ≤ µ(1 + |z|)m

for all (x, u, z) ∈ ∂G× [−M0,M0]× RN and all (y, v) ∈ G× [−M0,M0].Then there is a positive constant γ = γ(α, ν−1µ,m,N) such that

u ∈ C1+γ(G). Moreover we have

‖u‖C1+γ(G) ≤ C(α, ν−1µ,M0,m,N).

8.2. The mLaplace operator with an absorption term

8.2.1. Introduction. We consider the Dirichlet problem∆mu := −div (|∇u|m−2∇u) = −a0(x)u|u|q−1 + f(x) in G,u(x) = 0 on ∂G \ 0,

(LPA)

where 1 < m < ∞, q > 0 and a0(x) ≥ 0, f(x) are measurable functions inG.

Definition 8.8. A function u is called a generalized solution of(LPA), if u ∈W 1,m(Gε) ∩ Lq+1(Gε) ∀ε > 0 and it satises∫

G

|∇u|m−2⟨∇u,∇η

⟩+ a0(x)u|u|q−1η − fη dx = 0(II)

8.2 The mLaplace operator with an absorption term 281

for any η ∈W 1,m(G)∩Lq+1(G) having a compact support in G and u(x) = 0on Γε for all ε > 0 in the sense of traces.

Definition 8.9. A function u is called a weak solution of (LPA), ifu ∈W 1,m

0 (G) ∩ Lq+1(G) and satises (II) for all η ∈W 1,m0 (G) ∩ Lq+1(G).

Let us denote

ai(z) := |z|m−2zi.(8.2.1)

We verify that the ellipticity condition (E) is satised with

µ =

m− 1 for m ≥ 21 for 1 < m ≤ 2

, ν =

1 for m ≥ 2m− 1 for 1 < m ≤ 2.

(8.2.2)

Theorem 8.10. Weak comparison principle. Let u, v ∈ W 1,m(G)satisfy ∆mu ≤ ∆mv in the weak sense, i.e.∫

G

(ai(∇u)− ai(∇v)) ηxidx ≤ 0

for all nonnegative η ∈W 1,m0 (G) and let

u ≤ v on ∂G.

Then

u ≤ v in G.

Proof. Since u− v ≤ 0 on ∂G, we may set

η = max(u− v, 0).

By the ellipticity condition (E) and by Remark 8.6, we have∫G

(ai(∇u)− ai(∇v)) (uxi − vxi)dx ≥∫G

ψ (|∇(u− v)|) dx > 0,

because ψ(ζ) is a continuous, positive for ζ > 0, nondecreasing function.Hence, by standard arguments, we obtain the required assertion.

Theorem 8.11. Let u(x) be a bounded weak solution of (LPA) with

|u(x)| ≤ M0. Suppose that a0(x), f(x) ∈ Lp/m(G), p > N. If f(x) ≥ 0in G, then u(x) ≥ 0 in G.

Proof. Choose η = u− = max−u(x), 0 as a test function in theintegral identity (II). We obtain:∫

G

⟨|∇u−|m + a0(x)|u−|q+1 + f(x)u−

⟩dx = 0 =⇒

∫G

|∇u−|mdx+∫G

a0(x)|u−|q+1dx = −∫G

f(x)u−dx ≤ 0,


since u− ≥ 0. By Theorem 8.3, u(x) is continuous in G. Due to a0(x) ≥ 0and u

∣∣∂G

= 0 we get u−(x) = 0 in G and therefore u(x) ≥ 0 in G.

8.2.2. Singular functions for the mLaplace operator andthe corresponding eigenvalue problem. The rst eigenvalue problemwhich characterizes the singular behavior of the solutions of (LPA) can bederived by inserting in 4mv = 0 the function of the form v = rλφ(ω) whichleads to the nonlinear eigenvalue problem

D(λ, φ) = 0 in Ω, φ = 0 on ∂Ω,(NEV P1)

where

D(λ, φ) = −divω(λ2φ2 + |∇ωφ|2)m−2

2 ∇ωφ −

− λλ(m− 1) +N −m(λ2φ2 + |∇ωφ|2)m−2

2 φ.(D)

We formulate the Tolksdorf result:

Theorem 8.12. [371, 372]. There exists a solution (λ0, φ) ∈ R+ ×C∞(Ω) of (NEV P1) such that

λ0 > max

0,m−Nm− 1

, φ > 0 in Ω, φ2 + |∇ωφ|2 > 0 in Ω.(8.2.3)

Remark 8.13. In the case N = 2, by direct calculation (see (9.4.14)),we can obtain

λ0 =

m+(2−)(m−2)+(1−)

√m2−(2−)(m−2)2

2(m−1)(2−) , if ω0 < 2π;

m−1m , if ω0 = 2π,

(8.2.4)

where κ = ω0π .

In order to construct a barrier function which can be used in the weakcomparison principle, we prove a solvability property of the operator D as-sociated to the eigenvalue problem (NEV P1).

Theorem 8.14. For 0 ≤ λ < λ0 there exists a solution φ of the problem

D(λ, φ) = 1 in Ω, φ = 0 on ∂Ω,(8.2.5)

with φ > 0 in Ω.

This theorem will be proved in a sequence of lemmas. In the proofs ofthese lemmas we frequently use the fact that every solution (λ, φ) of (8.2.5)corresponds to a solution of

∆m(rλφ) = r(λ−1)(m−1)−1 in Gd0,

which, by local regularity of the PseudoLaplace equation, implies that φ ∈Cβ(Ω) ∩W 1+ε,m

0 (Ω) for β, ε > 0.

Lemma 8.15. The problem (8.2.5) is solvable for all 0 ≤ λ < λ0.


Proof. We prove that Fredholm's alternative holds for (8.2.5) in thesense that if (8.2.5) is not solvable then λ is an eigenvalue of D. For thispurpose, we choose a suciently large α ∈ R such that the problem

D(λ, φ) + α|φ|m−2φ = g in Ω, φ = 0 on ∂Ω

is uniquely solvable for all g ∈ H−1,m′(Ω), 1m + 1

m′ = 1, and denote thesolution operator by φ = Φg. By the regularity of D, Φ: Cβ(Ω)→ Cβ(Ω) isa compact operator for a β > 0. Moreover, Φ is homogeneous of degree 1

m−1 .

The problem D(λ, φ) = f in Ω, φ = 0 on ∂Ω, is then equivalent to

φ− αFφ = Φf,(8.2.6)

where Fφ = Φ(|φ|m−2φ) is compact and homogeneous of degree 1. Theoperator Id− αF is studied on the unit ball

B1 = φ ∈ Cβ(Ω) : ||φ||Cβ ≤ 1

If 0 /∈ (Id−αF )(∂B1) then K. Borsuk's theorem states that (8.2.6) is solvablefor suciently small f. Since (8.2.6) is equivalent to D(λ, φ) = f and D(λ, ·)is homogeneous of degree m− 1 we can solve D(λ, φ) = f for all f.

Lemma 8.16. Let (λ, φ) be a solution of (8.2.5). Then φ(ω) 6= 0 for allω ∈ Ω.

Proof. Let K = (r, ω) : 1 < r < 2, ω ∈ Ω. If (λ, φ) is a solution of(8.2.5) then v = rλφ(ω) solves

∆mv = r(λ−1)(m−1)−1 in K, v = 0 on (1, 2)× ∂Ω,(8.2.7)

v = crφ for r = 1, 2.(8.2.8)

Assume that φ(ω0) = 0 for ω0 ∈ Ω.We apply the weak comparison principleon the domain K using the function v. It follows that every solution of

∆mu = f in K, u = v on ∂K,

with f ∈ C∞0 (K), satises u(r, ω0) ≤ 0 which is a contradiction.

Lemma 8.17. For suciently small λ ≥ 0, the solution of (8.2.5) is u-nique and satises φ > 0 in Ω.

Proof. The operator D(0, ·) is strictly monotone on W 1,m0 (Ω). Hence,

the problem (8.2.5) is uniquely solvable and the comparison principle impliesφ > 0 in Ω. Since D(λ, ·) is continuous in λ, the conclusion also holds forsuciently small λ ≥ 0.

Lemma 8.18. There exists a constant c = c(λ1) such that ||φ||1,m ≤ cfor all solutions (λ, φ) of (8.2.5) satisfying 0 ≤ λ ≤ λ1 < λ0.

Proof. Assuming the converse we obtain a sequence (λi, φi) solving(8.2.5) with

λi → λ, ||φi||1,m →∞.


For the normalized functions

φi =φi

||φi||1,m

we obtain that D(λi, φi)→ 0 in W−1,m′(Ω) and, by regularity, ||φi||1+ε,m ≤c. Hence, we can extract a subsequence φik such that φik → φ in W 1,m

0 (Ω)and D(λ, φ) = 0 with ||φ||1,m = 1. This contradicts the fact that there is noeigenvalue of D in the interval [0, λ1).

Proof of Theorem 8.14. Lemma 8.18 implies a kind of continuity ofthe solutions (λ, φ) in the following sense. If λi → λ with 0 ≤ λi, λ < λ0,then there exists a subsequence φik such that

φik → φ in Cα(Ω),

where (λ, φ) is a solution of (8.2.5). Hence, by Lemmas 8.16 and 8.17 thereexists a solution (λ, φ) with φ > 0 in Ω for all 0 ≤ λ < λ0.

8.2.3. Eigenvalue problem for m−Laplacian in a bounded do-main on the unit sphere. For technical reasons we consider eigenvalueproblem for m−Laplacian in a bounded domain Ω on the unit sphere SN−1.

−divω(|∇ωψ|m−2∇ωψ) = µ|ψ|m−2ψ in Ω,ψ = 0 on ∂Ω.

(NEV P2)

Definition 8.19. We say that µ is an eigenvalue, if there exists a con-tinuous function ψ ∈W 1,m

0 (Ω), ψ 6≡ 0 such that∫Ω

|∇ωψ|m−2 1

qi

∂ψ

∂ωi

∂η

∂ωi− µ|ψ|m−2ψη

dΩ = 0(II2)

whenever η(x) ∈ W 1,m0 (Ω). The function ψ is called a weak eigenfunction

(a weak solution of the problem (NEV P2)).

We characterize the rst eigenvalue µ(m) of (NEV P2) by

µ(m) = infψ∈W 1,m

0 (Ω)ψ 6=0

∫Ω

|∇ωψ|mdΩ∫Ω

|ψ|mdΩ.(8.2.9)

Theorem 8.20. There exists a solution (µ, ψ) of (NEVP2) with µ > 0and ψ > 0 in Ω. Furthermore, the following Wirtinger's inequality holds∫

Ω

|ψ|m dω ≤ 1µ(m)

∫Ω

|∇ωψ|m dω, ∀ψ ∈W 1,m0 (Ω)(Wm)

with a sharp constant 1µ(m) .


Proof. Let us introduce the functionals on W 1,m(Ω):

F [u] =∫Ω

|∇ωu|mdΩ, G[u] =∫Ω

|u|mdΩ,

H[u] =∫Ω

⟨|∇ωu|m − µ|u|m

⟩dΩ

and the corresponding forms

F (u, η) =∫Ω

|∇ωu|m−2 1qi

∂u

∂ωi

∂η

∂ωidΩ, G(u, η) =

∫Ω

|u|m−2uηdΩ.

Now, we dene the set

K =u ∈W 1,m

0 (Ω)∣∣∣ G[u] = 1

.

Since K ⊂ W 1,m0 (Ω), F [u] is bounded from below for u ∈ K. The greatest

lower bound of F [u] for this family we denote by µ :

infu∈K

F [u] = µ.

Since F [v] is bounded from below for v ∈ K, there isµ = inf

v∈KF [v]. Consider a sequence vk ⊂ K such that lim

k→∞F [vk] = µ

(such a sequence exists by the denition of inmum). From K ⊂ W 1,m0 (Ω)

it follows that vk is bounded in W 1,m0 (Ω) and therefore compact in Lm(Ω).

Choosing a subsequence we can assume that it is converging in Lm(Ω). Fur-thermore,

‖vk − vl‖mLm(Ω) = G[vk − vl] < ε(8.2.10)

as soon as k, l > N(ε). Now we use Lemma 1.6:∣∣∣∣vk + vl2

∣∣∣∣m ≥ |vk|m +m

2|vk|m−2vk(vl − vk), m > 1.

We integrate this inequality over Ω∫Ω

∣∣∣∣vk + vl2

∣∣∣∣m dΩ ≥∫Ω

|vk|mdΩ +m

2

∫Ω

|vk|m−2vk(vl − vk)dΩ.

Further, by the Young inequality (1.2.2) with p = mm−1 , q = m, we have

∣∣∣m2vk|vk|m−2(vl − vk)

∣∣∣ ≤ m

2|vk|m−1|vl − vk| ≤

m− 12

δmm−1 |vk|m +

+1

2δm|vl − vk|m, ∀δ > 0.


This yields that∫Ω

∣∣∣∣vk + vl2

∣∣∣∣m dΩ ≥(

1− m− 12

δmm−1

)∫Ω

|vk|mdΩ−

− 12δm

∫Ω

|vl − vk|mdΩ, ∀δ > 0.

This implies that

G

[vk + vl

2

]≥(

1− m− 12

δmm−1

)G[vk]−

12δm

G[vl − vk], ∀δ > 0.

By using G[vk] = G[vl] = 1 and G[vl − vk] < ε1 we obtain

G

[vk + vl

2

]> 1− m− 1

2δ

mm−1 − ε1

2δm, ∀δ, ε1 > 0

for big k, l. Now we choose δm = εm−1m

1 . By setting ε = mµ2 ε

1m1 we get

G

[vk + vl

2

]> 1− ε

µ(8.2.11)

for big k, l. The functionals F [v] and G[v] are homogeneous functionals andtherefore their ratio F [v]

G[v] does not change under the passage from v to cv(c = const 6= 0) and hence

infv∈W 1(Ω)

F [v]G[v]

= infv∈K

F [v] = µ.

Therefore F [v] ≥ µG[v] for all v ∈W 1,m0 (Ω). Since vk+vl

2 ∈W 1,m0 (Ω) together

with vk, vl ∈ K, then

F

[vk + vl

2

]≥ µG

[vk + vl

2

]> µ

(1− ε

µ

)= µ− ε, k, l > N(ε).

Let us take k and l large enough so that F [vk] < µ + ε and F [vl] < µ + ε.We apply Clarkson's inequalities (Theorem 1.18)• 1) m ≥ 2

F

[vl − vk

2

]≤ 1

2F [vl] +

12F [vk]− F

[vl + vk

2

]<

< µ+ ε− (µ− ε) = 2ε

• 2) 1 < m ≤ 2

F1

m−1

[vl − vk

2

]≤(

12F [vk] +

12F [vl]

) 1m−1

−

− F1

m−1

[vk + vl

2

]< (µ+ ε)

1m−1 − (µ− ε)

1m−1 <

<2ε

m− 1(µ+ ε)

2−mm−1


by Lemma 1.4. Consequently,

F [vk − vl]→ 0, k, l→∞.(8.2.12)

From (8.2.10), (8.2.12) it follows that ‖vk − vl‖W 1,m0 (Ω)

→ 0, k, l → ∞.Hence vk converges in W 1,m

0 (Ω) and as a result of the completeness ofW 1,m

0 (Ω) there exists a limit function u ∈W 1,m0 (Ω) such that

‖vk − u‖W 1,m0 (Ω)

→ 0, k →∞.

In addition, again by Lemma 1.4 and the Hölder inequality

|F [vk]− F [u]| =

∣∣∣∣∣∣∫Ω

(|∇ωvk|m − |∇ωu|m) dΩ

∣∣∣∣∣∣ ≤≤ m

∫Ω

|∇vk|m−1|∇ω(vk − u)|dΩ ≤

≤ m

∫Ω

|∇ω(vk − u)|mdΩ

1/m∫Ω

|∇ωvk|mdΩ

(m−1)/m

→ 0,

k →∞, since vk ∈W 1,m0 . Therefore we get

F [u] = limk→∞

F [vk] = µ.

Analogously one sees that G[u] = 1.Suppose now that η is some function from W 1,m

0 (Ω). Consider the ratioF [u+εη]G[u+εη] . It is a continuously dierentiable function of ε on some intervalaround the point ε = 0. This ratio has a minimum at ε = 0 equal to µ andby the Fermat Theorem, we have(

F [u+ εη]G[u+ εη]

)′ε=0

= mF (u, η)G[u]− F [u]G(u, η)

G2[u]= 0,

which by virtue of F [u] = µ, G[u] = 1 gives

F (u, η)− µG(u, η) = 0, ∀η ∈W 1,m0 (Ω).

Further, if u is an eigenfunction of µ, then it follows from the formula(8.2.9) that |u| is one also. But then, by the weak Harnack inequality, The-orem 8.2, either |u| > 0 in the whole domain G or u ≡ 0 (the latter casebeing excluded for eigenfunctions). By continuity, either u or −u is positivein the whole domain G. Indeed, suppose that u = 0 at some point x0 ∈ G.Let B3R(x0) be a ball with so small R that B3R ⊂ G. Then inf

BRu(x) = 0,

so in turn ‖u‖Lp(B2R) = 0 by (8.1.1). That is, u = 0 in B2R. Chaining thengives the conclusion u ≡ 0 in G, thus proving the theorem.

Now we shall prove the inequality (Wm). Consider the described abovefunctionals F [u], G[u],H[u] on W 1,m

0 (Ω). We will nd the pair (µ, u) that


gives the minimum of the functional F [u] on the set K. For this we investi-gate the minimization of the functional H[u] on all functions u(ω), for whichthe integral exists and which satisfy the boundary condition from (NEV P2).The necessary condition of existence of the functional minimum is δH[u] = 0.By the calculation (with the help of formulas from Section 1.3) of the rstvariation δH we have

δH[u] = δ

∫Ω

[N−1∑i=1

1qi

(∂u

∂ωi

)2]m

2

− µ(u2)m2

dΩ

=

= −m∫Ω

δu ·N−1∑i=1

∂

∂ωi

(J(ω)qi· |∇ωu|m−2 ∂u

∂ωi

)dω −

−mµ∫Ω

δu · |u|m−2udΩ =

= −m∫Ω

δu · 1J(ω)

N−1∑i=1

∂

∂ωi


∂ωi

)+

+ µu|u|m−2dΩ =

= −m∫Ω

δu ·divω(|∇ωu|m−2∇ωu) + µ|u|m−2u

dΩ =⇒ (NEVP2).

Backwards, let u(ω) be the solution of (NEV P2). We multiply both sidesof the equation (NEV P2) by u and integrate over Ω, using the Gauss-Ostrogradskiy formula:

0 =∫Ω

u · divω

(|∇ωu|m−2∇ωu

)+ µ|u|m

dΩ =

= µ

∫Ω

|u|mdΩ +

+∫Ω

u ·N−1∑i=1

∂

∂ωi


∂ωi

)dω =

= µ

∫Ω

|u|mdΩ−

−N−1∑i=1

J(ω)qi|∇ωu|m−2

(∂u

∂ωi

)2

dω =


=∫Ω

(µ|u|m − |∇ωu|m) dΩ = µG[u]− F [u] =

=(by K)

µ− F [u]⇒ µ = F [u],

consequently, the required minimum is the least eigenvalue of (NEV P2).

The existence of a function u ∈ K such that

F [u] ≤ F [v] for all v ∈ Khas been proved above.

The one-dimensional Wirtinger inequality.Now we consider the case N = 2 and thus let Ω =

[−ω0

2 ,ω02

]be an arc

on the unit circle. Then our eigenvalue problem is(|ψ′|m−2ψ′

)′ + µψ|ψ|m−2 = 0, ω ∈(−ω0

2 ,ω02

); m > 1,

ψ(±ω0

2

)= 0.

The Wirtinger inequality in this case take the following formω02∫

−ω02

|ψ|mdω ≤ 1µ(m)

ω02∫

−ω02

|ψ′|mdω, ∀ψ ∈W 1,m0

(−ω0

2,ω0

2

).

We want to calculate the sharp constant µ(m). First of all, we note that thesolutions of our eigenvalue problem are determined uniquely up to a scalarmultiple. We consider the solution normed by the condition ψ(0) = 1. Inaddition, it is easy to see that ψ(−ω) = ψ(ω) and therefore ψ′(0) = 0. Thuswe can suppose

0 ≤ ψ(ω) ≤ 1.This we shall take into consideration for the solution of the problem.

Rewriting the equation in the form

(m− 1)|ψ′|m−2ψ′′

+ µψ|ψ|m−2 = 0

and solving it direct by the preset parameter method we obtain

|ψ′|m =µ

m− 1(1− ψm) .

By integrating, from this equation it follows

± m

√µ

m− 1· ω =

1∫ψ

dtm√

1− tm.

Taking into account the boundary condition we get

m

√µ

m− 1· ω0

2=

1∫0

dtm√

1− tm.


Let Γ(x) be a gamma-function and the beta-function B(x, y) = Γ(x)·Γ(y)Γ(x+y) .

Then we have (see, e.g., formula (16) 1.5.1, Chapter 1 [34]):

1∫0

dtm√

1− tm=

1mB

(1m, 1− 1

m

)=

Γ(

1m

)· Γ(1− 1

m

)mΓ(1)

=

=1m

Γ(

1m

)· Γ(

1− 1m

)=

π

m sin(πm

) ;

here we used the formula

Γ(z) · Γ(1− z) =π

sin(πz), Rez > 0.

Thus we get

µ(m) = (m− 1)

(2ω0· π

m sin(πm

))m , ∀m > 1

Hence, in particular, we have the well-known result

µ(2) =(π

ω0

)2

.

At last, we calculate µ(1) = limm→1+0

µ(m). For this we rewrite obtained

result above in this way

µ1m (m− 1)−

1m =

2ω0· 1m· Γ(

1m

)· Γ(

1− 1m

).

We multiply this equality by (m− 1) and use the formula zΓ(z) = Γ(1 + z) :

µ1m (m− 1)1− 1

m =2ω0· Γ(

1m

)·(

1− 1m

)· Γ(

1− 1m

)=

=2ω0· Γ(

1m

)· Γ(

2− 1m

)→ 2

ω0as m→ 1 + 0,

since Γ(1) = 1. On the other hand, by limx→+0

xx = 1, we have

limm→1+0

µ1m (m− 1)1− 1

m = µ. Hence it follows that

µ(1) =2ω0.


The last leads to the Wirtinger inequality for the case m = 1 :

ω02∫

−ω02

|ψ|dω ≤ ω0

2

ω02∫

−ω02

|ψ′|dω, ∀ψ ∈W 1,10

(−ω0

2,ω0

2

).

8.2.4. Integral estimates of solutions. The aim of this section is topresent integral estimates for the solutions of (LPA). Moreover, the weakcomparison principle is not used in the proof, so that it may be applied alsoto the case of elliptic systems.

Theorem 8.21. Let a0 ∈ L mm−1−q

(G), if 0 < q < m − 1 and 0 < a0 ≤a0(x) ≤ a1 (a0, a1− const.), if q ≥ m−1. Let f ∈ V 0

mm−1

,2(G). Then the weak

solution u of the problem (LPA) belongs to V 1m,0(G) the inequality∫

G

(|∇u|m + r−m|u|m + a0(x)|u|1+q

)dx ≤ c(N,G)

∫G

|rf |mm−1dx(8.2.13)

holds.

Proof. Let us consider the function

Θ ∈ C∞(R), Θ(t) ≥ 0, Θ(t) =

0, t < 1;1, t > 2.

(8.2.14)

Inserting η(x) = u(x)Θ(|x|ε

)with ε > 0 into the integral identity (II) we

obtain

(8.2.15)∫G

(|∇u|m + a0(x)|u|1+q

)Θ(|x|ε

)dx

≤ c1ε−1

∫G2εε

|u||∇u|m−1dx+∫G

|u||f |Θ(|x|ε

)dx.

By Young's inequality and (Wm) we get

(8.2.16) ε−1

∫G2εε

|u||∇u|m−1dx

≤ c2ε−1

∫G2εε

(r|∇u|m + r1−m|u|m

)dx ≤ c3(µ0)

∫G2εε

|∇u|mdx.


From (8.2.15) and (8.2.16) it follows that:

(8.2.17)∫G

(|∇u|m + a0(x)|u|1+q

)Θ(|x|ε

)dx

≤ c3

∫G2εε

|∇u|mdx+∫G

|u||f |Θ(|x|ε

)dx.

Passing to the limit as ε→ 0 and applying the Young inequality to the lastintegral on the right hand side of (8.2.17), we obtain the assertion.

Corollary 8.22. Let m > N. Under the suppositions of Theorem 8.21,a weak solution u(x) of (LPA) is bounded and Hölder-continuous in G.

Proof. This follows from Theorem 8.21 in view of the embedding the-orem

|u(x)| ≤ c0|x|1−Nm

∫G

|rf |mm−1

1m

, x ∈ G

We set µ0 = µ(N) and observe that µ0 = µ0(Ω) is the smallest positiveeigenvalue of (NEV P2) for m = N.

Theorem 8.23. Let m = N and let the following condition be satised∫Gρ0

|rf |N/(N−1) dx ≤ cρκ.

Let χ0 = 2√µ0

(1+µ0)(N−2)/N . Then for any weak solution of (LPA) the bound∫Gρ0

|∇u|N dx ≤

≤ c(N,µ0,Ω)

(ρ/d)χ0 , if χ0 < κ,

(ρ/d)χ0 lnNN−1 (d/ρ), if χ0 = κ, ρ ∈ (0, d),

(ρ/d)κ, if χ0 > κ

(8.2.18)

is satised.

Remark 8.24. It is well known that if m = N = 2, then µ0 = λ20 = π2

ω20,

where ω0 is the quantity of the angle with the vertex 0. In this case theassertion of the Theorem was proved in chapter 5 (see Theorems 5.4, 5.5).

The proof of the theorem will be carried out due to the following lemma.


Lemma 8.25. Let 2 ≤ m ≤ N. For any function u ∈ W 1,m0 (G) with

∇u(ρ, ·) ∈ Lm(Ω) we have∫Ω

ρuur +

N −m2

u2

|∇u|m−2 dω ≤ ρ2

χ

∫Ω

|∇u|m dω,(8.2.19)

where

χ =m−N +

√4µ+ (N −m)2

(1 + µ)(m−2)/m.(8.2.20)

Proof. From the Cauchy inequality we obtain

ρuur +N −m

2u2 ≤ ε+N −m

2u2 +

12ερ2u2

r , ε > 0,

and hence∫Ω

ρuur +

N −m2

u2

|∇u|m−2 dω ≤ ρ2

∫Ω

ε+N −m2

(uρ

)2+

+12εu2rBigr|∇u|m−2 dω =: A

The right hand side is estimated by Young's inequality(u

ρ

)2

|∇u|m−2 ≤ m− 2m

δ−2/(m−2)|∇u|m +2mδ

(|u|ρ

)m,

u2r |∇u|m−2 ≤ m− 2

mδ−2/(m−2)|∇u|m +

2mδ |ur|m, ∀δ > 0,

which implies by Wirtinger's inequality (Wm)

A ≤ ρ2

∫Ω

m− 2

2mδ−2/(m−2)

(ε+N −m+

1ε

)|∇u|m +

+2δm

(ε+N −m

2

(|u|ρ

)m+

12ε|ur|m

)dω ≤

≤ ρ2

∫Ω

m− 2

2mδ−2/(m−2)

(ε+N −m+

1ε

)|∇u|m +

+δ

m

(ε+N −m

µ

∣∣∣∣∇ωuρ∣∣∣∣m +

1ε|ur|m

)dω

We choose ε > 0 such that ε+N−mµ = 1

ε , which gives

ε =12

(m−N +

√(N −m)2 + 4µ

),


and hence, ∫Ω

ρuur +

N −m2

u2

|∇u|m−2 dω ≤

≤ ρ2

mε

(m− 2

2δ−2(m−2)(µ+ 1) + δ

)∫Ω

|∇u|m dω.

The lemma is proved by choosing δ = (1 + µ)(m−2)/m.

Remark 8.26. For m = N = 2 the constant χ is sharp.

Proof of Theorem 8.23.. Let

V (ρ) =∫Gρ0

|∇u|N dx.

From (LPA) it follows that

V (ρ) +∫Gρ0

a0(x)|u|1+q dx = ρN−2

∫Ω

ρuur|∇u|N−2 dω +∫Gρ0

ufdx.

In view of

V ′(ρ) = ρN−1

∫Ω

|∇u|N dω,

we obtain from Lemma 8.25

V (ρ) ≤ ρ

χ0V ′(ρ) +

∫Gρ0

|uf | dx.

The second term of the right hand side can be estimated by the condition ofthe Theorem and Wirtinger's inequality (Wm),

∫Gρ0

|uf | dx ≤

∫Gρ0

r−N |u|N dx

1/N ∫

Gρ0

|rf |N/(N−1) dx

(N−1)/N

≤

≤ cρκN−1N V

1N (ρ).

Thus we get the dierential inequality for V (ρ) :

V (ρ) ≤ ρ

χ0V ′(ρ) + cρκ

N−1N V

1N (ρ)

In view of Theorem 8.21, as an initial condition for this dierential inequality,we can use

V (d) ≤∫G

|∇u|N dx ≤ c∫G

|rf |N/(N−1) dx ≡ V0.


By puttingW (ρ) = VN−1N (ρ), we obtain the dierential inequality for W (ρ) :

W (ρ) ≤ NN−1

ρχ0W ′(ρ) + cρκ

N−1N , 0 < ρ < d

W (d) = VN−1N

0 .

Solving the Cauchy problem for the corresponding equation, we get

W ∗(ρ) =(ρd

)χ0N−1N

(V

NN−1

0 +

+ κχ0

N−1N ln d

ρ , if χ0 = κ,

dN−1N

(κ−χ0)−ρN−1N

(κ−χ0)

κ−χ0, if χ0 6= κ

.

It is well known that the solution of the dierential inequality can be estimat-ed by the solution W ∗(ρ) of the corresponding equation: W (ρ) ≤W ∗(ρ) andhence we obtain nally the required estimate. Theorem 8.23 is proved.

Lemma 8.27. Let q > m− 1, a0(x) ≥ a0 > 0, (a0 - const). Let

|f(x)| ≤ f1|x|β, x ∈ Gd0, where

β > −1 if m > N,

β > −m if m ≤ N.

Then for any generalized solution u(x) of (LPA) the inequality

||u||p;Gρρ/2≤ c(a0,m,N, p, q, f1)ρ

Np− mq−m+1 ∀p > m(8.2.21)

holds.

Proof. We consider the cut-o function

ζ(r) =

0, r ∈

[0, ρ4]∪ [2ρ,∞) ,

1, r ∈[ρ

2 , ρ]

;

0 ≤ ζ(r) ≤ 1, |∇ζ| ≤ cρ−1, r ∈[ρ

4,ρ

2

]∪ [ρ, 2ρ].

By putting in (II)

η(x) = |u|tsgnu · ζs(|x|) ∀t ≥ 1, s > 0,

we obtain

t

∫G2ρρ/4

ζs(r)(|u|t−1|∇u|m + a0(x)|u|t+q

)dx ≤(8.2.22)

≤ s∫

G2ρρ/4

|u|t|∇u|m−1ζs−1|∇ζ|dx+∫

G2ρρ/4

|u|t|f |ζsdx.


By the Young inequality

s|u|t|∇u|m−1ζs−1|∇ζ| ≤ m− 1m

εmm−1 |u|t−1|∇u|mζs +

+1mε−msm|u|t+m−1|∇ζ|mζs−m, ∀ε > 0,

choosing ε =(

tmm−1

)m−1m and taking into account that ∇ζ = O(ρ−1), from

(2.13) we get

a0

∫G2ρρ/4

|u|t+qζsdx ≤ c(m)sm

tm−1

∫G2ρρ/4

r−m|u|t+m−1ζs−mdx+(8.2.23)

+∫

G2ρρ/4

|u|t|f |ζsdx.

Applying the Hölder inequality to integrals with p = t+qt+m−1 > 1,

p′ = t+qq−m+1 we obtain ∫

G2ρρ/4

r−m(|u|t+m−1ζs−m)dx ≤(8.2.24)

≤

∫G2ρρ/4

r−mp′dx

1/p′ ∫

G2ρρ/4

|u|t+qζ(s−m) t+qt+m−1dx

t+m−1t+q

.

Let us now choose s = m(t+q)q−m+1 ; then from (8.2.23), (8.2.24) it follows that

a0

∫G2ρρ/4

|u|t+qζsdx ≤∫

G2ρρ/4

|u|t|f |ζsdx+(8.2.25)

+c1(m,N, t, q)sm

tm−1ρN(q−m+1)

t+q−m

∫G2ρρ/4

|u|t+qζsdx

t+m−1t+q

.

We estimate the rst right hand side term in (8.2.25) by the Hölder inequality

∫G2ρρ/4

|u|t|f |ζsdx ≤

∫G2ρρ/4

|u|t+qζsdx

tt+q ∫G2ρρ/4

|f |t+qq ζsdx

qt+q

.


Then from (8.2.25) we obtain

a0

∫G2ρρ/4

|u|t+qζsdx

qt+q

≤

∫G2ρρ/4

|f |t+qq ζsdx

qt+q

+(8.2.26)

+c1(m,N, t, q)sm

tm−1ρN(q−m+1)

t+q−m

∫G2ρρ/4

|u|t+qζsdx

m−1t+q

.

Again, by the Young inequality, taking into account that q > m− 1, we get

c1(m,N, t, q)sm

tm−1ρN(q−m+1)

t+q−m

∫G2ρρ/4

|u|t+qζsdx

m−1t+q

≤(8.2.27)

≤ a0

2

∫G2ρρ/4

|u|t+qζsdx

qt+q

+ c2(m,N, t, q, a0)ρnqt+q− mqq−m+1 .

Now by setting p = t + q > 1 + q > m, from (8.2.26), (8.2.27) we arrive atthe inequality (8.2.21) sought for.

Lemma 8.28. Suppose the conditions of the Lemma 8.27 hold. Let u(x)be any generalized solution of (LPA). Then the inequality∫

Gρρ/2

(|∇u|m + |u|1+q

)dx ≤ c(a0,m,N, q, f1)ρN−

(1+q)m1+q−m(8.2.28)

is valid.

Proof. Let us consider the inequality (8.2.22) with t = 1 and ∀s > 0∫G2ρρ/4

|∇u|mζsdx+ a0

∫G2ρρ/4

|u|1+qζsdx ≤(8.2.29)

≤ cs∫

G2ρρ/4

r−1|u||∇u|m−1ζs−1dx+∫

G2ρρ/4

|u||f |ζsdx.


By estimating the rst right side term in (8.2.29) with the help of the Younginequality, we have

12

∫G2ρρ/4

|∇u|mζsdx+ a0

∫G2ρρ/4

|u|1+qζsdx ≤(8.2.30)

≤ c(m)sm∫

G2ρρ/4

r−m|u|mζs−mdx+∫

G2ρρ/4

|u||f |ζsdx.

By using the Young inequality once again with p = 1+qm , p′ = 1+q

1+q−m and∀δ > 0 :

c(m)smr−m(|u|mζs−m) ≤ δ|u|1+qζ(s−m) 1+qm +

+c(δ,m, s)r−m1+q

1+q−m ;

we set

s =(1 + q)m1 + q −m

.(8.2.31)

As a result, from (8.2.30) we get

12

∫G2ρρ/4

|∇u|mζsdx+ a0

∫G2ρρ/4

|u|1+qζsdx ≤ δ∫

G2ρρ/4

|u|1+qζsdx+

+ c(δ,m, q)∫

G2ρρ/4

r−sdx+∫

G2ρρ/4

|u||f |ζsdx ∀δ > 0.

Hence, by choosing δ = a02 , we obtain

(8.2.32)∫

G2ρρ/4

(|∇u|m + |u|1+q

)ζsdx ≤ c(a0,m, q,N)ρN−s +

+ ε

∫G2ρρ/4

r−m|u|mζsdx+ cε

∫G2ρρ/4

(r|f |)mm−1 ζs(r)dx, ∀ε > 0.

Taking into account the inequality (Wm) and choosing ε > 0 properly, from(8.2.31), (8.2.32) we get the inequality (8.2.28) sought for. This completesthe proof of Lemma 8.28.

Corollary 8.29. Let q > mNN−m − 1, 1 < m < N and the hypothesis of

Lemma 8.27 about the functions a0(x), f(x) holds. Then for any generalized


solution u(x) of (LPA) the inequality∫Gρ0

(|∇u|m + r−m|u|m + |u|1+q

)dx ≤ c(a0, N,m, q, f1, d),(8.2.33)

∀ρ ∈ (0, d)

is valid.

Proof. By replacing ρ with 2−kρ (k = 0, 1, 2, . . . ) in (8.2.28) and sum-ming the received inequalities over all k, we obtain (8.2.33).

8.2.5. Estimates of solutions for singular right hand sides. Westate two results of M. Dobrowolski (Theorems 1, 2 [98]). Let λ0 be theleast positive eigenvalue and φ(ω) be the corresponding eigenfunction of(NEV P1) (see (8.2.3)).

Theorem 8.30. Let u ∈W 1,m(G) be a weak solution of the problem∆mu = f(x), x ∈ Gd0,u(x) = g(x), x ∈ Ωd,

u(x) = 0, x ∈ Γd0.(PL)0

Assume that g(x) ∈ C1(Ωd) and

|f(x)| ≤ f1|x|β with f1 ≥ 0, β > λ0(m− 1)−m.(8.2.34)

Then

|u(x)| ≤ c0|x|λ0 , |∇u(x)| ≤ c1|x|λ0−1, x ∈ Gd0.

Theorem 8.31. Assume that 0 ≤ f(x) ≤ f1|x|β with β > λ0(m−1)−mand a0(x) ≡ 0. Then each nonvanishing weak solution of (LPA) admits thesingular expansion

u(r, ω) = krλ0φ(ω) + v(x)

with k > 0 and

|v(x)| ≤ c|x|λ0+δ, |∇v| ≤ c|x|λ0+δ−1, |vxx| ≤ c|x|λ0+δ−2,

where the maximum δ > 0 depends on β and the eigenvalue problem (NEV P1).

The proof of these results is based on the weak comparison principlefor the Pseudo-Laplace operator. Here we shall prove the estimates of themodulus of generalized and weak solutions of (LPA) with a0 ≥ 0. Let d > 0be a small xed number. We also suppose that

|f(x)| ≤ f1|x|β, β > −Np

(8.2.35)

with some p > Nm .


Observe that a function v = rαφ(ω) is a weak solution v ∈W 1,m0 , if φ(ω)

is suciently smooth and

α >m−Nm

.(8.2.36)

Since ∆mv ∼ rα(m−1)−m and the right-hand side of (LPA)

−a0(x)v|v|q−1 + f(x) ∼ rαq + rβ,

hence we obtain that

rα(m−1)−m ∼ rαq + rβ.(8.2.37)

These arguments suggest the following theorems to us.

Theorem 8.32. Let u(x) be a weak solution of (LPA). Let 1 < m < N,q > 0 be given. Let a0(x) ≥ a0 > 0 (a0 is a constant) and let f(x) ∈Lp(G), p > N

m . Then there exists the constant M0 > 0, depending only on‖f(x)‖Lp(G), measG,N,m, q, p, a0, such that

||u||L∞(G) ≤M0.

Proof. Let us introduce the set A(k) = x ∈ G, |u(x)| > k and letχA(k) be a characteristic function of the set A(k). We note that A(k + d) ⊆A(k) ∀d > 0. By setting φ(x) = η((|u| − k)+)χA(k) · sgnu in (II), whereη is dened by Lemma 1.60 from Preliminaries and k ≥ k0 (without lossof generality we can assume that k0 ≥ 1), on the strength of the Theoremassumptions we get the inequality:

(8.2.38)∫

A(k)

|∇u|mη′((|u| − k)+)dx+ a0

∫A(k)

|u|qη((|u| − k)+)dx ≤

≤∫

A(k)

|f(x)|η((|u| − k)+)dx.

Now we dene the function wk(x) := η

((|u| − k)+

m

). By the denition of

η(x) (see Lemma 1.60 from Preliminaries):

e κ(|u|−k)+ |∇u|m =(mκ

)m|∇wk|m, κ > 0

and by the choice of κ > m according to Lemma 1.60, using (1.11.5) -(1.11.7), from (8.2.38) we obtain

(8.2.39)12

(m

κ

)m ∫A(k)

|∇wk|mdx+ a0kq0

∫A(k)

|wk|mdx ≤

≤ c7M

∫A(k+d)

|f(x)||wk|mdx+ c8e κd⟨ ∫A(k)\A(k+d)

|f(x)|dx⟩.


By the assumptions of the Theorem we have that f(x) ∈ Lp(G), p > Nm .

Then by the Hölder inequality for integrals with the exponents p and p′

(1p + 1

p′ = 1) :

∫A(k+d)

|f ||wk|mdx ≤ ‖f(x)‖Lp(G)

∫A(k)

|wk|mp′dx

1p′

.(8.2.40)

Letting m# = mNN−m , from the interpolation inequality (see Lemma 1.16) for

Lp−norms we obtain:

∫A(k)

|wk|mp′dx

1p′

≤

∫A(k)

|wk|mdx

θ

·

∫A(k)

|wk|m#dx

(1−θ)mm#

with θ ∈ (0, 1), which is dened by the equality

1p′

= θ +(1− θ)mm#

=⇒ θ = 1− N

pm.

Thus from (8.2.40) we get:

∫A(k+d)

|f ||wk|mdx ≤ ‖f(x)‖Lp(G)

∫A(k)

|wk|mdx

θ

×

×

∫A(k)

|wk|m#dx

(1−θ)mm#

.

(8.2.41)

By using the Young inequality with the exponents 1θ and

1(1−θ) , from (8.2.41)

we obtain

∫A(k+d)

|f ||wk|mdx ≤θ ‖f(x)‖

1θ

Lp(G)

ε1/θ

∫A(k)

|wk|mdx+

+ ε1

(1−θ) (1− θ)

∫A(k)

|wk|m#dx

m

m#

, ∀ε > 0.

(8.2.42)


It follows from (8.2.39), (8.2.42) that:

(8.2.43)12

(m

κ

)m ∫A(k)

|∇wk|mdx+ a0kq0

∫A(k)

|wk|mdx ≤

≤ c9ε−1/θ

∫A(k)

|wk|mdx+ c10ε1

(1−θ)

∫A(k)

|wk|m#dx

m

m#

+

+ c11

∫A(k)

|f(x)|dx, ∀ε > 0,

where

c9 = θMc7 ‖f(x)‖1θ

Lp(G) ;

c10 = (1− θ)Mc7;

c11 = c8eκd.

Now we use the Sobolev imbedding Theorem 1.30. Then from (8.2.43)we get:

(8.2.44)12

(m

c1κ

)m ∫A(k)

|wk|m#dx

m

m#

+ a0kq0

∫A(k)

|wk|mdx ≤

≤ c9ε−1/θ

∫A(k)

|wk|mdx+ c10ε1

(1−θ)

∫A(k)

|wk|m#dx

m

m#

+

+ c11

∫A(k)

|f(x)|dx, ∀ε > 0,

Now, we can choose ε in order to have

c10ε1

1−θ =14

(m

c1κ

)m(8.2.45)

and k0 such that

c9ε− 1θ = a0k

q0(8.2.46)

We obtain that from (8.2.44) it results: ∫A(k)

|wk|m#dx

m

m#

≤ c12

∫A(k)

|f(x)|dx ∀k ≥ k0.(8.2.47)


At last, by Young's inequality we get:∫A(k)

|f(x)|dx ≤ ‖f(x)‖Lp(G) meas 1− 1pA(k).

Therefore from (8.2.47) it follows that ∫A(k)

|wk|m#

m

m#

≤ c12 ‖f(x)‖Lp(G) meas1− 1

pA(k).(8.2.48)

Let now l > k > k0. By (1.11.8) of Preliminaries and the denition ofthe function wk(x) : |wk| ≥ 1

m(|u| − k)+, and therefore∫A(l)

|wk|m#dx ≥

(l − km

)m#

mesA(l).

From (8.2.48) it now follows that:

meas A(l) ≤(

m

l − k

)m# ∫A(k)

|wk|m#dx ≤(8.2.49)

≤(

m

l − k

)m# (c12 ‖f(x)‖Lp(G)

)m#

m

measm#

m

1− 1

p

A(k),

∀l > k ≥ k0.

Now we setψ(k) = mes A(k).

Then from (8.2.49) it follows that

ψ(l) ≤ c13

(m

l − k

)m#

[ψ(k)]m#

m

1− 1

p

.(8.2.50)

From the denition of m# and the assumption p > Nm we note that

γ =m#

m

(1− 1

p

)> 1.

Then from (8.2.50) we get

ψ(l) ≤ c19

(l − k)m# ψγ(k) ∀l > k ≥ k0

and therefore we have, according to Lemma 1.59 of Preliminaries, that ψ(k0+δ) = 0 with δ depending only on the quantities in the formulation of Theorem8.32. This means that |u(x)| < k0 + δ for almost all x ∈ G. Theorem 8.32 isproved.


Corollary 8.33. Let 1 < m < N, q > mNN−m − 1, β > −N

s with some

s > Nm be given numbers. Let a0(x) ≥ a0 > 0 (a0 - const) and |f(x)| ≤ f1|x|β.

Suppose

a0(x), f(x) ∈ Lp/m(G), p > N.

Then any generalized solution u(x) of (LPA) is Hölder-continuous in G.

Proof. This assertion follows from Theorem 8.32 and Theorem 8.3 ac-cording to the inequality (8.2.33).

Theorem 8.34. Let 1 < m < N, q > m − 1 be given. Let 0 < a0 ≤a0(x) ≤ a1, (a0, a1 const) and let (8.2.35) is satised with some β ≥ 0.Let u(x) be any generalized solution of (LPA). If, in addition,

λ0 <β +m

m− 1, q >

mN

N −m− 1,(8.2.51)

then

|u(x)| ≤ c0|x|λ0 , x ∈ Gd0.(8.2.52)

Proof. First we apply Lemma 8.27. From the inequality (8.2.21) underp→∞ the estimate follows

|u(x)| ≤ c|x|m

m−1−q .(8.2.53)

Hence, in view of (8.2.36) the second inequality (8.2.51) is justied. Now weconsider the auxiliary problem

∆mv = f1|x|β, x ∈ Gd0,v(x) = u+(x), x ∈ Ωd,

v(x) = 0, x ∈ Γd0

(8.2.54)

with some d > 0, f1 ≥ 0, where u+(x) is the positive part of u(x).Under the assumptions of our Theorem, by the existence Theorem 8.5,

there is a weak solution of the auxiliary problem (8.2.54). Further, by The-orem 8.7, we have that u(x) ∈ C1+γ(Gdd/2). Then, in view of Theorem 8.30,we have

0 ≤ v(x) ≤ c0|x|λ0 , |∇v| ≤ c|x|λ0−1, x ∈ Gd0.(8.2.55)

We wish to prove that

u(x) ≤ v(x), x ∈ Gd0,(8.2.56)

by this, the Theorem will be proved. To do this, we apply the proof bycontradiction. We suppose that u(x) > v(x) on some set D ⊂ Gd0 is fullled. By Corollary 8.33, the set D is a domain. From (LPA) and (8.2.54) wehave

∆mu ≤ f(x) ≤ f1|x|β = ∆mv, ∀x ∈ D,i.e. ∫

D

(|∇u|m−2uxi − |∇v|m−2vxi

)ηxidx ≤ 0(8.2.57)


for ∀η(x) ∈W 1,m0 (D) ∩ Lq+1(D), η(x) ≥ 0. We put

w = u− v, ut = tu+ (1− t)v ∀t ∈ [0, 1],

aij(x) =

1∫0

∂ai(utx)∂utxj

dt,

where ai(z) are dened by (8.2.1). Then from (8.2.57) we obtain∫D

aij(x)wxjηxidx ≤ 0(8.2.58)

for ∀η(x) ∈ W 1,m0 (D) ∩ Lq+1(D), η(x) ≥ 0. We remind that the ellipticity

condition (E) - (8.2.2) holds. Thus, the function w(x) > 0 in D and satisesthe integral inequality (8.2.58). Further, by the conditions of the Theorem,the inequality (8.2.33) holds and in particular∫

D

(|∇u|m + r−m|u|m

)dx ≤ const.(8.2.59)

The same inequality is true for the function v(x) : really, (8.2.59) for v(x)follows from (8.2.55), if we take into account (8.2.3) and m < N. But nowwe can state the validity of the inequality∫

D

(|∇w|m + r−m|w|m

)dx ≤ const.(8.2.60)

This circumstance makes it possible to put in (8.2.58) the function η(x) =w(x)Θ

(|x|ε

)with Θ(t), dened by (8.2.14). As a result we obtain

∫D

Θ(|x|ε

)|∇w|2

1∫0

|∇ut|m−2dt

dx ≤

≤ c∫

D∩G2εε

r−1w|∇w|

1∫0

|∇ut|m−2dt

dx ≤(8.2.61)

≤ c∫

D∩G2εε

(|∇w|m + r−mwm + |∇v|m

)dx,

(by the Young inequality). In view of (8.2.55) and (8.2.60) the right handintegral is uniformly bounded over ε > 0. Therefore, it is possible to makethe passage to the limit over ε→ 0 that implies∫

D

|∇w|2 1∫

0

|∇ut|m−2dt

dx ≤ 0.(8.2.62)


By the continuity of w(x) and in view of w(x) = 0, x ∈ ∂D, from (8.2.64)we get w(x) ≡ 0 ∀x ∈ D. The contradiction to our assumption w(x) > 0∀x ∈ D is nished By this, (8.2.56) and the assertion of Theorem 8.34 areproved.

Lemma 8.35. Let u(x) be a weak solution of the problem (LPA). Iff(x) ≥ 0 for a.e. x ∈ G then u(x) ≥ 0 a.e. in G.

Proof. We dene

G− = x ∈ G∣∣ u(x) < 0.

Choose η = max−u(x), 0 as a test function in the integral identity (II).We obtain: ∫

G−

⟨|∇u|m + a0(x)|u|q+1

⟩dx =

∫G−

f(x)u(x)dx ≤ 0.

Hence it follows that u(x) = 0, x ∈ G−. Thus u(x) ≥ 0 a.e. in G.

Theorem 8.36. Let 1 < m < N, q > 0 be given. Let a0(x) ≥ a0 > 0(a0 is a constant) and let (8.2.35) be satised. Let u(x) be a weak boundedsolution of (LPA) with sup

G|u(x)| = M0. Suppose, in addition,

f(x) ≥ 0; a0(x) ≤M−q0 f(x) a.e. inG.

The following assertion holds: if λ0 <β+mm−1 , then

0 ≤ u(x) ≤ c0|x|λ0 , x ∈ Gd0.(8.2.63)

Proof. From the equation of (LPA) we have

∆mu = F (x), F (x) ≡ f(x)− a0(x)u|u|q−1.

By Lemma 8.35, u ≥ 0. Therefore, in view of our assumptions, we get that0 ≤ F (x) ≤ f1|x|β. By the assumption on λ0, β, the conditions of Theorem8.31 are satised. By this Theorem we get (8.2.63).

Theorem 8.37. Let 1 < m < N, q > 0 be given. Let a0(x) ≥ a0 > 0(a0 is a constant) and let (8.2.35) be satised. Let u(x) be a weak solutionof (LPA).

The following assertion holds: if λ0 >β+mm−1 , then

|u(x)| ≤ c0|x|β+mm−1 , x ∈ Gd0.(8.2.64)

Proof. By Theorem 8.32 we verify that u(x) is a bounded function. Weset λ = m+β

m−1 . By the conditions of our Theorem,

0 < λ < λ0.

We takev(x) = A|x|λφ(ω)

8.3 Estimates of weak solutions near a conical point 307

as the barrier functions, where ∀A > 0 and (λ, φ) is a solution of (8.2.5); itexists in view of Theorem 8.14. In this connection

∆mv = Am−1|x|λ(m−1)−m, x ∈ Gd0,v(x) = Adλφ(ω) ≥ 0, x ∈ Ωd,

v(x) = 0, x ∈ Γd0.

By the function φ(ω) properties (see Theorem 8.14 and Lemma 8.18) it iseasy to verify that

0 ≤ v(x) ≤ cA|x|λ,∫G

(|∇v|m + r−m|v|m

)dx ≤ const.

Wishing to prove that u(x) ≤ v(x), x ∈ Gd0 (by this, the assertion of theTheorem will be proved), we suppose by contradiction that on some setD ⊂ Gd0 the inequality u(x) > v(x) is satised. Since u(x) is bounded in G,then by Theorem 8.3 it is Hölder-continuous. This implies that the set D isa domain. Further, we have for x ∈ D:

∆mu(x) ≤ f(x) ≤ f1|x|β = f1|x|λ(m−1)−m ≤

≤ Am−1|x|λ(m−1)−m = ∆mv(x), if A ≥ f1

m−1

1 .

Moreover, (8.2.59) is valid by Theorem 8.21. In fact, for this it obviouslysuces to show that

∫G

|rf |mm−1dx is nite. Because of (8.2.35) we have

∫G

|rf |mm−1dx ≤ f

mm−1

1

d∫0

rmm−1

(β+1)+N−1dr <∞,

if mm−1(β + 1) +N > 0. But by (8.2.35) and since N > m we obtain

m

m− 1(β+1)+N >

m

m− 1(1− N

p)+N >

m

m− 1(1−m)+N = N −m > 0.

Now we repeat the arguments of the proof of Theorem 8.34 word for wordand obtain the required assertion of Theorem 8.37.

8.3. Estimates of weak solutions near a conical point

In this Section we investigate the behavior of the weak solutions of the(DQL) near a conical point. Let λ0 be the least positive eigenvalue of theproblem (EV P1) (see Theorem 8.12). Let us introduce the number

q =(1− t)(m− 1)

t, 0 < t ≤ 1.

Concerning the equation of the (DQL) we make the following


Assumptions:

the functions ai(x, u, z), a(x, u, z) are continuously dierentiable with re-

spect to the x, u, z variables in Md,M0 = Gd0 × [−M0,M0]×RN and satisfy:

E) ν|u|q|z|m−2|ξ|2 ≤ ∂ai(x,u,z)∂zj

ξiξj ≤ µ|u|q|z|m−2|ξ|2, ∀ξ ∈ RN \ 0;

1)

√N∑i=1

∣∣∣∣∂a(x,u,z)∂zi

∣∣∣∣2 ≤ µ|u|q−1|z|m−1;

2) ∂a(x,u,z)∂u ≥ ν|u|q−2|z|m;

3)

∣∣∣∣∂ai(x,u,z)∂zj− |u|q|z|m−4

(δji |z|2 + (m− 2)zizj

)∣∣∣∣ ≤≤ c1(r)rβ+m−λ0(m−1)|u|q|z|m−2 + c2(r)rβ+2−λ0 |u|

1−tt ;

4)

∣∣∣∣∂ai(x,u,z)∂xi

∣∣∣∣+ |a(x, u, z)| ≤ c3(r)rβ−m(λ0−1)|u|m(1−t)

t |z|m+

+c4(r)|u|βtλ0 + c5(r)rβ,

where ν, µ > 0, β > (m − 1)λ0 − m are constants, ci(r) are nonnegative,continuous at zero functions with ci(0) = 0; i = 1, . . . , 5.

At rst, we transform our problem (DQL) into such problem in whichthe leading coecients are independent of u explicit.

Lemma 8.38. Let us make the change of function

u = v|v|t−1; 0 < t ≤ 1.(8.3.1)

Suppose that

u∂ai(x, u, z)

∂u=

1− tt· ∂ai(x, u, z)

∂zjzj ; i = 1, . . . , N.(U)

Then the problem (DQL) takes the form:

Qt(v, φ) ≡∫G

⟨Ai(x, vx)φxi +A(x, v, vx)φ

⟩dx = 0(8.3.2)

for all φ(x) ∈W 1,m0 (G) ∩ L∞(G), where

Ai(x, ζ) ≡ ai(x, v|v|t−1, t|v|t−1ζ),

(8.3.3)

A(x, v, ζ) ≡ a(x, v|v|t−1, t|v|t−1ζ).


Proof. In fact, by calculating, from (8.3.1) - (8.3.3) it follows that

dAidv

=∂ai(x, u, z)

∂u· t|v|t−1 +

∂ai(x, u, z)∂zj

· t(t− 1)|v|t−2sign v · ζj =

=∂ai(x, u, z)

∂u· t|v|t−1 +

∂ai(x, u, z)∂zj

· (t− 1)|v|t−2sign v · |v|1−tzj =

=1v

(tu∂ai(x, u, z)

∂u+ (t− 1)

∂ai(x, u, z)∂zj

zj

)= 0,

that means the required statement.

Remark 8.39. It is easy to see that we can taket = 1, if dai(x,u,z)

du = 0,t = 1− ε, ∀ε ∈ (0, 1), if dai(x,u,z)

du 6= 0.(8.3.4)

The change (8.3.1) transforms our assumptions into the following:

(E) ν|ζ|m−2|ξ|2 ≤ ∂Ai(x,ζ)∂ζj

ξiξj ≤ µ|ζ|m−2|ξ|2, ∀ξ ∈ RN \ 0;

1)

√N∑i=1

∣∣∣∣∂A(x,v,ζ)∂ζi

∣∣∣∣2 ≤ µ|v|−1|ζ|m−1;

2) ∂A(x,v,ζ)∂v ≥ νtm+1|v|−2|ζ|m;

3)∣∣∣∣∂Ai(x,ζ)∂ζj

− tm−1|ζ|m−4(δji |ζ|2 + (m− 2)ζiζj

)∣∣∣∣ ≤≤ c1(r)rβ+m−λ0(m−1)|ζ|m−2 + c2(r)rβ+2−λ0 ;

4)∣∣∣∣∂Ai(x,ζ)∂xi

∣∣∣∣+ |A(x, v, ζ)| ≤ c3(r)rβ−m(λ0−1)|ζ|m+c4(r)|v|βλ0 +c5(r)rβ.

The main statement of this Section is presented by the following theorems.

Theorem 8.40. Let u(x) ∈ W 1,m(G) ∩ L∞(G), 1 < m < N be aweak solution of the (DQL). Suppose that the assumptions E), (U),1)− 4)are fullled. Then there exists a constant c0 > 0, depending only on theparameters and norms of functions occuring in the assumptions, such that

|u(x)| ≤ c0|x|tλ0 .(8.3.5)

Proof. Making the transformation (8.3.1) in the problem (DQL) tothe equation Qt(v, φ) = 0 we shall estimate the function v(x) under theassumptions (E), 1)− 4). At rst, for some d > 0 we consider the auxiliary


problem ∆mw = f1|x|β, x ∈ Gd0,w(x) = v+(x), x ∈ Ωd,

w(x) = 0, x ∈ Γd0,(8.3.6)

where v+(x) is the positive part of v(x) and the constants

f1 ≥ 0, β > (m− 1)λ0 −m.Under the assumptions of our Theorem, by the existence Theorem 8.5, thereis a weak solution w(x) of the auxiliary problem (8.3.6). Further, by Theorem8.7, we have that v(x) ∈ C1+γ(Gdd/2). Then, in view of Theorem 8.30, wehave

0 ≤ w(x) ≤ c0|x|λ0 , |∇w| ≤ c1|x|λ0−1,

(8.3.7)

|wxx| ≤ c2|x|λ0−2, x ∈ Gd0.

Now let φ ∈ L∞(Gd0) ∩W 1,m0 (Gd0) be any nonnegative function. For the

operator Qt, that is dened by (8.3.2), applying the assumptions 3)− 4) andestimates (8.3.7) we obtain:

Qt(w, φ) =∫Gd0

⟨Ai(x,wx)φxi +A(x,w,wx)φ

⟩dx =

=∫Gd0

φ(x)⟨− d

dxiAi(x,wx) +A(x,w,wx)

⟩dx =

=∫Gd0

φ(x)⟨− d

dxi

(Ai(x,wx)− tm−1|∇w|m−2wxi

)+ f1r

β +

+A(x,w,wx)⟩dx =

∫Gd0

φ(x)⟨f1r

β − ∂Ai(x,wx)∂xi

+A(x,w,wx)−

−[∂Ai(x,wx)

∂wxj− tm−1|∇w|m−4

(δji |∇w|

2 +

+ (m− 2)wxiwxj]wxixj

⟩dx ≥

≥∫Gd0

φ(x)⟨f1r

β −∣∣∣∣∂Ai(x,wx)

∂xi

∣∣∣∣− |A(x,w,wx)| −


−∣∣∣∂Ai(x,wx)

∂wxj− tm−1|∇w|m−4

(δji |∇w|

2 +

+ (m− 2)wxiwxj∣∣∣ · |wxx|⟩dx ≥

≥∫Gd0

φ(x)⟨f1r

β − c3(r)rβ−m(λ0−1)|∇w|m − c4(r)|w|βλ0 − c5(r)rβ −

− c1(r)rβ+m−λ0(m−1)|∇w|m−2|wxx| − c2(r)rβ+2−λ0 |wxx|⟩dx ≥

≥∫Gd0

φ(x)rβ⟨f1 −

5∑i=1

ci(r)⟩dx.

Hence, choosing d > 0 by the continuity of ci(r), (i = 1, . . . , 5) such small

that5∑i=1

ci(r) ≤ 12f1 we get

Qt(w, φ) ≥ 12f1

∫Gd0

φ(x)rβ ≥ 0.

Thus, from (8.3.2) and (8.3.6) we get:Q(w, φ) ≥ 0 = Q(v, φ) ∀φ ≥ 0 in Gd0;w(x) ≥ v(x), x ∈ ∂Gd0.

Besides that, one can readily verify that all the other conditions of the com-parison principle (Theorem 9.6) are fullled; by this principle we get

v(x) ≤ wε(x), ∀x ∈ Gd0.

Similarly one can prove that

v(x) ≥ −w(x), ∀x ∈ Gd0.

Thus, nally, we obtain

|v(x)| ≤ w(x) ≤ c0|x|λ0 , ∀x ∈ Gd0.

Returning to the old variables, in virtue of (8.3.1) we get the required esti-mate (8.3.5). Our Theorem is proved.

Theorem 8.41. Let u(x) ∈W 1,m(G)∩L∞(G), 1 < m < N be a weaksolution of the (DQL). Suppose that the assumptions E), (U),1)− 4) are


fullled. Suppose, in addition,

√√√√ N∑i=1

|ai(x, u, z)− ai(y, v, z)|2 ≤ µ(1 + |z|)m−1 (|x− y|α + |u− v|α)

for all (x, u, z) ∈ ∂G× [−M0,M0]× RN and all (y, v) ∈ G× [−M0,M0].Then there exists a constant c1 > 0, depending only on the the parameters

and norms of the functions occurring in the assumptions, such that

|∇u(x)| ≤ c1|x|tλ0−1.(8.3.8)

Proof. Let us consider in the layer G11/2 the function v(x′) = %−tλ0u(%x′),

taking u ≡ 0 outside G. Let us perform in the equation (DQL) the changeof variables x = %x′. The function v(x′) satises the equation

∫G1

1/2

ai(x′, v, vx′)φx′i + a(x′, v, vx′)φ

dx′ = 0,

∀φ(x′) ∈W 1,m0 (G1

1/2) ∩ L∞(G11/2);

ai(x′, v, vx′) ≡ ai(%x′, %tλ0v, %tλ0−1vx′),a(x′, v, vx′) ≡ %a(%x′, %tλ0v, %tλ0−1vx′).

(DQL)′

In virtue of the assumptions of our Theorem, we can apply the LiebermanTheorem 8.7:

supG1

1/2

|∇′v| ≤M ′1,

where M ′1 > 0 is determined only by t, λ0, α, ν, µ,N,G and c0 from (8.3.5).Hence, returning to the function u(x) we get

|∇u(x)| ≤M ′1%tλ0−1, x ∈ G%%/2.

Letting |x| = 23%, we obtain the desired inequality (8.3.8).

Corollary 8.42. From Remark 8.39 it follows that the estimates (8.3.5),(8.3.8) can be rewritten in the following form

|u(x)| ≤ c

|x|λ0 , if dai(x,u,z)

du = 0,|x|λ0−ε, ∀ε ∈ (0, 1), if dai(x,u,z)

du 6= 0.(8.3.9)

|∇u(x)| ≤ c

|x|λ0−1, if dai(x,u,z)

du = 0,|x|λ0−1−ε, ∀ε ∈ (0, 1), if dai(x,u,z)

du 6= 0.(8.3.10)

8.4 Integral estimates of second weak derivatives of solutions 313

8.4. Integral estimates of second weak derivatives of solutions

In this Section we will derive a priori estimates of second derivatives (interms of the Sobolev weighted norm) of solutions to the (DQL) in a neighbor-hood of a conical boundary point. We give an example which demonstratesthat the estimates obtained are exact.

We dene the set M = G×R×RN and we will suppose that the ellipticitycondition (E) and the following assumptions are fullled: there exist anumber µ > 0 and nonnegative functions

f(x) ∈ L2(G) ∩ L(m+2)/m(G) ∩ Lp/m(G),

g(x) ∈ L2(m+2)/m(G) ∩ L(m+2)/(m−1)(G) ∩ Lp/(m−1)(G),p > N

such that

(A) ai(x, u, z), a(x, u, z) ∈ C1(M), i = 1, . . . , N ;

(B) |a(x, u, z)|+∣∣∣∣ N∑i=1

∂ai(x,u,z)∂xi

∣∣∣∣ ≤ µ|z|m + f(x)|z|m−2

2 ;

(C)

√N∑i=1

∣∣∣∂ai(x,u,z)∂u

∣∣∣2 +

√N∑i=1

∣∣∣∂a(x,u,z)∂zi

∣∣∣2 ≤ µ|z|m−1 + g(x)|z|m−2

2 ;

(D) |z| ·

√N∑i=1|ai(x, u, z)− ai(y, v, z)|2 + |a(x, u, z)− a(y, v, z)| ≤

≤ µ|z|m (|x− y|+ |u− v|) , ∀x, y ∈ G, ∀u, v ∈ R;

(F)

∣∣∣∣ N∑i=1

∂ai(x,u,0)∂xi

∣∣∣∣ ≤ f(x);

√N∑i=1

∣∣∣∂ai(x,u,0)∂u

∣∣∣2 ≤ g(x).

We make transformation x = %x′. Let v(x′) = u(%x′) and G′ be the image ofG under this transformation. Let d > 0 be so small that, if % ∈ (0, d), thenG2

1/4 ⊂ G′. Further, our problem (DQL) takes the form

∫G′


dx′ = 0,

∀φ(x′) ∈W 1,m0 (G′) ∩ L∞(G′);

ai(x′, v, vx′) ≡ ai(%x′, v, %−1vx′),a(x′, v, vx′) ≡ %a(%x′, v, %−1vx′).

(DQL)′

At rst we establish the strong interior estimate.

8.4.1. Local interior estimates. In this subsection we derive localinterior integral estimates of weak solutions of the problem (DQL).


Theorem 8.43. Let u(x) be a bounded weak solution of the problem(DQL). Let us assume that the hypotheses (A), (B), (C), (D), (E), (F) are

fullled on the set M. Let ∀G ⊂⊂ G%%/4 ⊂ G. Then there exists the integral∫eG

(|∇u|m+2 + |∇u|m−2u2

xx

)dx and we have the estimate

∫eG

(|∇u|m+2 + |∇u|m−2u2

xx

)dx ≤

(8.4.1)

≤ C∫eG

(ρ−2|∇u|m + f2 + f

m+2m + g

m+2m−1 + g

2(m+2)m

)dx.

Proof. Let the image of G be G′ ⊂⊂ G21/4 ⊂ G′. For all x′0 ∈ G′ and

all σ such that 0 < σ < dist (G′, ∂G21/4) we take

φ(x′) = 4−hk(ζ2(x′)4h

kv(x′))

as the test function in the (DQL)′, where ζ(x′) ∈ C∞0(B2σ(x′0)

)is a cuto

function such that

ζ(x′) = 1 in Bσ(x′0), 0 ≤ ζ(x′) ≤ 1, |∇′ζ| ≤ cσ−1 in B2σ(x′0).

Then for suciently small |h| ≤ σ, summing formula (1.11.17) by parts, weobtain

(8.4.2)∫

B2σ(x′0)

4hk ai(x

′, v, vx′)(ζ2∂4h

kv(x′)∂x′i

+ 2ζζx′i4hkv(x′)

)+

+4hkv(x′)ζ24h

k a(x′, v, vx′)

dx′ = 0,

where

4hk ai(x

′, v, vx′) = aij(x′)∂4h

kv(x′)∂x′j

+ ai(x′),

4hk a(x′, v, vx′) = bj(x′)

∂4hkv(x′)∂x′j

+ b(x′)


with

aij(x′) ≡1∫

0

∂ai(x′, v, vtx′)∂vt

x′j

dt; bj(x′) ≡1∫

0

∂a(x′, v, vtx′)∂vt

x′j

dt;

ai(x′) ≡ ai (x′ + hek, v(x′ + hek), vx′(x′ + hek))− ai (x′, v(x′), vx′(x′ + hek))h

,

b(x′) ≡ a (x′ + hek, v(x′ + hek), vx′(x′ + hek))− a (x′, v(x′), vx′(x′ + hek))h

,

vt(x′) = (1− t)v(x′) + tv(x′ + hek).

Thus hence we get (for brevity we denote B2σ = B2σ(x′0)):∫B2σ

aij(x′)∂4h

kv(x′)∂x′i


ζ2dx′ ≤∫B2σ

∣∣∣∣∣aij(x′)∂4hkv(x′)∂x′j

2ζζx′i4hkv(x′)

∣∣∣∣∣+

+∣∣∣∣ai(x′)∂4h

kv(x′)∂x′i

ζ2

∣∣∣∣+∣∣∣ai(x′)2ζζx′i4h

kv(x′)∣∣∣+(8.4.3)

+

∣∣∣∣∣bj(x′)∂4hkv(x′)∂x′j

4hkv(x′)ζ2

∣∣∣∣∣+∣∣∣b(x′)4h

kv(x′)ζ2∣∣∣dx′.

Letting

Pk(x′) ≡ |∇′v(x′)|+ |∇′v(x′ + hek)|,(8.4.4)

by assumptions (C), (D), (E), and applying Lemma 1.7 we have

aij(x′)∂4h

kv(x′)∂x′i


≥ νc(m)%1−mPm−2k (x′)|∇′4h

kv(x′)|2;

|aij(x′)| ≤ µ

m− 1%1−mPm−2

k (x′), (i, j = 1, . . . , N);

|ai(x′)| ≤ µ%1−mPm−1k (x′)

(1 + |4h

kv(x′|), (i = 1, . . . , N);

|bj(x′)| ≤ %1−m⟨µPm−1

k (x′) + %m2 g(%x′)P

m−22

k

⟩, (j = 1, . . . , N);

|b(x′)| ≤ µ%1−mPmk (x′)(

1 + |4hkv(x′|

).

(8.4.5)


Now from (8.4.3) - (8.4.5) it follows that

∫B2σ

Pm−2k (x′)|∇′4h

kv(x′)|2ζ2dx′ ≤

≤ c(ν, µ,m)∫B2σ

(Pm−2k (x′)|∇′4h

kv(x′)||4hkv(x′)|ζ|∇′ζ|+

Pm−1k (x′)|∇′4h

kv(x′)|ζ2 + Pm−1k (x′)|∇′4h

kv(x′)||4hkv(x′|ζ2 +(8.4.6)

+Pm−2

2k (x′)|∇′4h

kv(x′)||4hkv(x′|%

m2 g(%x′)ζ2 +

+Pm−1k (x′)|4h

kv(x′)|ζ|∇′ζ|+ Pm−1k (x′)|4h

kv(x′)|2ζ|∇′ζ|+

+Pmk (x′)|4hkv(x′|ζ2 + Pmk (x′)|4h

kv(x′|2ζ2

)dx′.

Now we estimate each term on the right using the Cauchy inequality with∀ε > 0 :

|∇′4hkv(x′)||4h

kv(x′)|ζ|∇′ζ| ≤ ε

2|∇′4h

kv(x′)|2ζ2 +

+12ε|4h

kv(x′)|2|∇′ζ|2;

Pm−1k (x′)|∇′4h

kv(x′)| ≤ ε

2Pm−2k |∇′4h

kv(x′)|2 +12εPmk ;

Pm−1k (x′)|∇′4h

kv(x′)||4hkv(x′| ≤ ε

2Pm−2k |∇′4h

kv(x′)|2 +

+12εPmk |4h

kv(x′)|2;

Pm−2

2k (x′)|∇′4h

kv(x′)||4hkv(x′|%

m2 g(%x′) ≤ ε

2Pm−2k |∇′4h

kv(x′)|2 +

+12ε|4h

kv(x′)|2%mg2(%x′)

Pm−1k (x′)|4h

kv(x′)|ζ|∇′ζ| ≤ 12Pm−2k (x′)|4h

kv(x′)|2ζ2 +

+12Pmk |∇′ζ|2;

Pm−1k (x′)|4h

kv(x′)|2ζ|∇′ζ| ≤ 12Pmk (x′)|4h

kv(x′)|2ζ2 +

+12Pm−2k |4h

kv(x′)|2|∇′ζ|2;

Pmk (x′)|4hkv(x′|ζ2 ≤ 1

2Pmk (x′)|4h

kv(x′)|2ζ2 +12Pmk ζ

2.


Choosing ε > 0 in an appropriate way we get from (8.4.6)∫B2σ

Pm−2k (x′)|∇′4h

kv(x′)|2ζ2dx′ ≤ c(ν, µ,m)∫B2σ

Pmk |4h

kv(x′)|2ζ2 +

+(Pm−2k (x′)|4h

kv(x′)|2 + Pmk (x′)) (ζ2 + |∇′ζ|2

)+

+|4hkv(x′)|2%mg2(%x′)ζ2

dx′.(8.4.7)

In order to estimate the integral∫B2σ

Pmk |4hkv(x′)|2ζ2dx′ we take

φ(x′) =(v(x′)− v(x′0)

)ζ2(x′)|4h

kv(x′)|2

as the test function in the (DQL)′:

(8.4.8)∫B2σ

ai(x′, v, vx′)

((v(x′)− v(x′0)

)ζ2(x′)|4h

kv(x′)|2)x′i

+

+ a(x′, v, vx′)(v(x′)− v(x′0)

)ζ2(x′)|4h

kv(x′)|2dx′ = 0

Now we use the representation

ai(x′, v, z) = aij(x′, v, z)zj + ai(x′, v, 0)

(8.4.9)

aij(x′, v, z) =

1∫0

∂ai(x′, v, τz)∂(τzj)

dτ, (i, j = 1, . . . N).

Therefore from (8.4.8) it follows that

(8.4.10)∫B2σ

aij(x′, v, vx′)vx′ivx′j |4hkv(x′)|2ζ2(x′)dx′ =

= −∫B2σ

a(x′, v, vx′)

(v(x′)− v(x′0)

)|4h

kv(x′)|2ζ2(x′) +

+ 2aij(x′, v, vx′)(v(x′)− v(x′0)

)vx′j

(|4h

kv(x′)|2ζζx′i +

+4hkv(x′)

∂(4hkv(x′))∂x′i

ζ2)

+

+ ai(x′, v, 0)((v(x′)− v(x′0)

)ζ2(x′)|4h

kv(x′)|2)x′i

dx′.


In the last term on the right we integrate by parts and so obtain

(8.4.11)∫B2σ

aij(x′, v, vx′)vx′ivx′j |4hkv(x′)|2ζ2(x′)dx′ =

= −∫B2σ

(v(x′)− v(x′0)

)a(x′, v, vx′)|4h

kv(x′)|2ζ2(x′) +

+ 2aij(x′, v, vx′)vx′j

(|4h

kv(x′)|2ζζx′i +4hkv(x′)

∂(4hkv(x′))∂x′i

ζ2)−

−(∂ai(x′, v, 0)

∂x′i+∂ai(x′, v, 0)

∂vvx′i

)ζ2(x′)|4h

kv(x′)|2dx′.

After a simple computation, using assumptions (B), (E), (F) and taking intoaccount 0 < % < d < 1, we obtain from (8.4.11)∫B2σ

|∇′v|m|4hkv|2ζ2dx′ ≤ c(ν, µ,m)

∫B2σ

|v(x′)− v(x′0)|

|∇′v|m|4h

kv|2ζ2 +

+|∇′v|m−1|4hkv|2ζ|∇′ζ|+ |∇′(4h

kv)||∇′v|m−1|4hkv|ζ2 + %mf(%x′)|4h

kv|2ζ2 +

+%1+m2 f(%x′)|∇′v|

m−22 |4h

kv|2ζ2 + %m−1g(%x′)|∇′v||4hkv|2ζ2

dx′.(8.4.12)

Taking into consideration Remark 8.4 we observe that all hypotheses of The-orem 8.3 about Hölder continuity of weak solutions are fullled and conclude

|v(x′)− v(x′0)| ≤ cσα, x′ ∈ B2σ(x′0), α ∈ (0, 1).

Moreover, we use the Cauchy inequality

|∇′v|m−1ζ|∇′ζ| ≤ 12|∇′v|mζ2 +

12|∇′v|m−2|∇′ζ|2;

|∇′(4hkv)||∇′v|m−1|4h

kv| ≤12|∇′v|m−2|∇′(4h

kv)|2 + |∇′v|m|4hkv|2.

Hence and from (8.4.12) it follows that∫B2σ

|∇′v|m|4hkv|2ζ2dx′ ≤ c(ν, µ,m)σα

∫B2σ

|∇′v|m|4h

kv|2ζ2 +

+|∇′v|m−2|4hkv|2|∇′ζ|2 + |∇′(4h

kv)|2|∇′v|m−2ζ2 + %mf(%x′)|4hkv|2ζ2 +

+%1+m2 f(%x′)|∇′v|

m−22 |4h

kv|2ζ2 + %m−1g(%x′)|∇′v||4hkv|2ζ2

dx′.(8.4.13)

Now we consider the function w(x′) = v(x′ + hek). It is easy to observethat this function is the bounded weak solution of the equation

− d

dx′iai(x′ + hek, w(x′), wx′(x′)

)+ a

(x′ + hek, w(x′), wx′(x′)

)= 0.


Then we write the corresponding integral identity with the test function

φ(x′) =(w(x′)− w(x′0)

)ζ2(x′)|4h

kv(x′)|2

and repeat verbatim the deduction of (8.4.13). As a result we get

∫B2σ

|∇′w|m|4hkv|2ζ2dx′ ≤ c(ν, µ,m)σα

∫B2σ

|∇′w|m|4h

kv|2ζ2 +

+|∇′w|m−2|4hkv|2|∇′ζ|2 + |∇′(4h

kv)|2|∇′w|m−2ζ2 +

+%mf(%(x′ + hek)

)|4h

kv|2ζ2 + %m−1g(%(x′ + hek)

)|∇′w||4h

kv|2ζ2 +

+%1+m2 f(%(x′ + hek)

)|∇′w|

m−22 |4h

kv|2ζ2dx′.(8.4.14)

Let us sum the estimates (8.4.13) - (8.4.14), applying the inequality (1.2.5)of Lemma 1.5; then recalling the notation (8.4.4) we have

∫B2σ

Pmk (x′)|4hkv|2ζ2dx′ ≤ c(ν, µ,m)σα

∫B2σ

Pmk (x′)|4h

kv|2ζ2 +

+Pm−2k (x′)|4h

kv|2|∇′ζ|2 + |∇′(4hkv)|2Pm−2

k (x′)ζ2 +

+%m⟨f(%x′) + f

(%(x′ + hek)

)⟩|4h

kv|2ζ2 +

+%1+m2⟨f(%x′) + f

(%(x′ + hek)

)⟩Pm−2

2k (x′)|4h

kv|2ζ2 +

+%m−1⟨g(%x′) + g

(%(x′ + hek)

)⟩Pk(x′)|4h

kv|2ζ2dx′.

Choosing, if it is necessary, σ ∈ (0, dist (G, ∂G21/4)) smaller such that c(ν, µ,m)σα ≤

12 hence we obtain

∫B2σ

Pmk (x′)|4hkv|2ζ2dx′ ≤ c(ν, µ,m)σα

∫B2σ

|∇′(4h

kv)|2Pm−2k (x′)ζ2 +

+Pm−2k (x′)|4h

kv|2|∇′ζ|2 + %m⟨f(%x′) + f

(%(x′ + hek)

)⟩|4h

kv|2ζ2 +

+%1+m2⟨f(%x′) + f

(%(x′ + hek)

)⟩Pm−2

2k (x′)|4h

kv|2ζ2 +

+%m−1⟨g(%x′) + g

(%(x′ + hek)

)⟩Pk(x′)|4h

kv|2ζ2dx′.(8.4.15)


In the same way from (8.4.7), (8.4.15)∫B2σ

Pm−2k (x′)|∇′4h

kv(x′)|2ζ2dx′ ≤ c(ν, µ,m)∫B2σ

(Pm−2k (x′)|4h

kv|2 +

+Pmk (x′)) (ζ2 + |∇′ζ|2

)+ %m

⟨f(%x′) + f

(%(x′ + hek)

)⟩|4h

kv|2ζ2 +

+%1+m2⟨f(%x′) + f

(%(x′ + hek)

)⟩Pm−2

2k (x′)|4h

kv|2ζ2 +

+%m−1⟨g(%x′) + g

(%(x′ + hek)

)⟩Pk(x′)|4h

kv|2ζ2 +

+%mg2(%x′)|4hkv(x′)|2ζ2

dx′.(8.4.16)

From (8.4.15) - (8.4.16) follows the estimate∫B2σ

(Pmk (x′)|4h

kv|2 + Pm−2k (x′)|∇′4h

kv(x′)|2)ζ2dx′ ≤


(Pm−2k (x′)|4h

kv|2 + Pmk (x′)) (ζ2 + |∇′ζ|2

)+

+%m⟨f(%x′) + f

(%(x′ + hek)

)⟩|4h

kv|2ζ2 + %mg2(%x′)|4hkv(x′)|2ζ2 +

+%1+m2⟨f(%x′) + f

(%(x′ + hek)

)⟩Pm−2

2k (x′)|4h

kv|2ζ2 +

+%m−1⟨g(%x′) + g

(%(x′ + hek)

)⟩Pk(x′)|4h

kv|2ζ2

dx′.(8.4.17)

Further, by the Young inequality, we have for ∀ε > 0

%m−1g(%x′)Pk(x′)|4hkv|2 ≤

ε

mPmk (x′)|4h

kv|2 +

+m− 1m

ε1

1−m%mgmm−1 (%x′)|4h

kv|2.

Then choosing ε > 0 from the equality 2εm c(ν, µ,m) = 1

2 we can rewrite(8.4.17) in the following way:∫

B2σ

(Pmk (x′)|4h

kv|2 + Pm−2k (x′)|∇′4h

kv(x′)|2)ζ2dx′ ≤


(Pm−2k (x′)|4h

kv|2 + Pmk (x′)) (ζ2 + |∇′ζ|2

)+

+%1+m2⟨f(%x′) + f

(%(x′ + hek)

)⟩Pm−2

2k (x′)|4h

kv|2ζ2 +

+%m⟨f(%x′) + f

(%(x′ + hek)

)+ g2(%x′) +

+gmm−1 (%x′) + g

mm−1

(%(x′ + hek)

)⟩|4h

kv|2ζ2

dx′.(8.4.18)


Again, by the Young inequality, we have for ∀δ > 0

%mf(%x′)|4hkv|2 ≤

2δm+ 2

|4hkv|m+2 +

+m

m+ 2δ−

2m%m+2f

m+2m (%x′),

%1+m2 f(%x′)P

m−22

k (x′)|4hkv|2 ≤

δ

2Pm−2k |4h

kv|4 +12δ%m+2f2(%x′),

%mg2(%x′)|4hkv|2 ≤

2δm+ 2

|4hkv|m+2 +

+m

m+ 2δ−

2m%m+2g

2(m+2)m (%x′),

%mgmm−1 (%x′)|4h

kv|2 ≤2δ

m+ 2|4h

kv|m+2 +

+m

m+ 2δ−

2m%m+2g

m+2m−1 (%x′)

and therefore from (8.4.18) it follows that∫B2σ

(Pmk (x′)|4h

kv|2 + Pm−2k (x′)|∇′4h

kv(x′)|2)ζ2dx′ ≤


δ(|4h

kv|m+2 + Pm−2k (x′)|4h

kv|4)

+

+(Pm−2k (x′)|4h

kv|2 + Pmk (x′)) (ζ2 + |∇′ζ|2

)+

+δ−1%m+2⟨f2(%x′) + f2

(%(x′ + hek)

)⟩ζ2 +

+δ−2m%m+2

⟨fm+2m (%x′) + f

m+2m(%(x′ + hek)

)+

+gm+2m−1 (%x′) + g

m+2m−1 (%x′) + g

2(m+2)m (%(x′)

⟩ζ2

dx′.(8.4.19)

If δ > 0 is suciently small, then hence we get∫B2σ

Pm−2k (x′)|∇′4h

kv(x′)|2 +⟨Pmk |4h

kv|2 −

−c(ν, µ,m)δ(|4h

kv|m+2 + Pm−2k (x′)|4h

kv|4)⟩

ζ2dx′ ≤


(Pm−2k (x′)|4h

kv|2 + Pmk (x′)) (ζ2 + |∇′ζ|2

)+

+δ−1%m+2⟨f2(%x′) + f2

(%(x′ + hek)

)⟩ζ2 +


+δ−2m%m+2

⟨fm+2m (%x′) + f

m+2m(%(x′ + hek)

)+

+gm+2m−1 (%x′) + g

m+2m−1 (%x′) + g

2(m+2)m (%(x′)

⟩ζ2

dx′.(8.4.20)

Now we verify that

limh→0

∫B2σ

(Pm−2k (x′)|4h

kv|2 + Pmk (x′)) (ζ2 + |∇′ζ|2

)dx′ =

=(2m−2 + 2m

) ∫B2σ

|∇′v|m(ζ2 + |∇′ζ|2

)dx′.

In fact, in virtue of Lemma 1.66, 4hkv converges to Dku in the norm Lm and

a.e., and, by the Egorov Theorem, almost uniformly. Analogously, the almostuniform convergence of f (%(x′ + hek)) , g (%(x′ + hek)) to f(%x′), g(%x′) re-spectively is veried.

Thus, we can apply the Fatou Theorem and perform the passage to thelimit in the (8.4.20) as h→ 0 :

∫B2σ

([2m − c(ν, µ,m) · δ(1 + 2m−2)

]|∇′v|m

∣∣∣∣ ∂v∂x′k∣∣∣∣2 + |∇′v|m−2

∣∣∣∣∂|∇′v|∂x′k

∣∣∣∣2)×

×ζ2(x′)dx′ ≤ c1

∫B2σ

|∇′v|m(ζ2 + |∇′ζ|2

)dx′ +

+c2%m+2

∫B2σ

⟨f2(%x′) + f

m+2m (%x′) + g

2(m+2)m (%(x′) + g

m+2m−1 (%x′)

⟩dx′,

k = 1, . . . , N.

Let us now choose δ > 0 from the equality c(ν, µ,m) ·δ = 2m−1

1+2m−2 ; then hencewe get

∫B2σ

(|∇′v|m

∣∣∣∣ ∂v∂x′k∣∣∣∣2 + |∇′v|m−2

∣∣∣∣∂|∇′v|∂x′k

∣∣∣∣2)ζ2(x′)dx′ ≤(8.4.21)

≤ c1

∫B2σ

|∇′v|m(ζ2 + |∇′ζ|2

)dx′ +

+c2%m+2

∫B2σ

⟨f2(%x′) + f

m+2m (%x′) + g

2(m+2)m (%(x′) + g

m+2m−1 (%x′)

⟩dx′,

k = 1, . . . , N.


After summing up over all k = 1, . . . , N , by the properties of the functionζ(x′), we establish∫

Bσ

(|∇′v|m+2 + |∇′v|m−2|vx′x′ |2

)dx′ ≤ c1

∫B2σ

|∇′v|mdx′ +

+c2%m+2

∫B2σ

⟨f2(%x′) + f

m+2m (%x′) + g

m+2m−1 (%x′) + g

2(m+2)m (%(x′)

⟩dx′.

(8.4.22)

By the covering argument we obtain∫eG′

(|∇′v|m+2 + |∇′v|m−2|vx′x′ |2

)dx′ ≤ c1

∫eG′

|∇′v|mdx′ +

+c2%m+2

∫eG′

⟨f2(%x′) + f

m+2m (%x′) + g

m+2m−1 (%x′) + g

2(m+2)m (%(x′)

⟩dx′.

(8.4.23)

Returning to previous variables x, u we get the desired (8.4.1).

8.4.2. Local estimates near a boundary smooth portion. In thissubsection we derive local integral estimates near a boundary smooth portionof weak solutions of the problem (DQL). Let x′0 ∈ Γ′ ⊂ Γ2

1/4 and let U′(x′0) ⊂

G21/4 be a neighborhood of x′0. Since our assumption on the boundary of G is

such that ∂G \O is smooth, then there exists a dieomorphism: U ′(x′0) −→B+

2σ(x′0), which attens the boundary i.e. maps Γ′ onto Σ2σ ⊂ x′N = 0being a plane part of ∂B+

2σ(x′0).So we may suppose that G′ = B+

2σ(x′0) in the (DQL)′, that takes theform

∫B+

2σ(x′0)


dx′ = 0,

∀φ(x′) ∈W 1,m0 (B+

2σ(x′0)) ∩ L∞(B+2σ(x′0));

ai(x′, v, vx′) ≡ ai(%x′, v, %−1vx′),a(x′, v, vx′) ≡ %a(%x′, v, %−1vx′).

(DQL)′0

We denote U(x0) as the preimage of U ′(x′0) under the transformation x =%x′. It is obvious that U(x0) ⊂ G2%

%/4.

Theorem 8.44. Let u(x) be a weak bounded solution of the problem(DQL). Let us assume that the hypotheses (A), (B), (C), (D), (E), (F) are


fullled on the set M. Let ∀G ⊂ G2%%/4 ⊂ G. Then we have the estimate∫

eG

(|∇u|m+2 + |∇u|m−2u2

xx

)dx ≤

(8.4.24)

≤ C∫eG

(ρ−2|∇u|m + f2 + f

m+2m + g

m+2m−1 + g

2(m+2)m

)dx.

Proof. Repeating verbatim the procedure of the deduction of the esti-mate (8.4.21), we establish∫

B+2σ

(|∇′v|m

∣∣∣∣ ∂v∂x′k∣∣∣∣2 + |∇′v|m−2

∣∣∣∣∂|∇′v|∂x′k

∣∣∣∣2)ζ2(x′)dx′ ≤

≤ c1

∫B+

2σ

|∇′v|m(ζ2 + |∇′ζ|2

)dx′ +(8.4.25)

+c2%m+2

∫B+

2σ

⟨f2(%x′) + f

m+2m (%x′) + g

2(m+2)m (%(x′) + g

m+2m−1 (%x′)

⟩dx′,

k = 1, . . . , N − 1.

It remains only to consider the case k = N. For this, by Theorem 8.43, usingthe covering argument we can easily establish that

φ(x′)ai(x′, v, v′x) ∈W 1,10 (G′), ∀φ(x′) ∈W 1,m

0 (G′)∩L∞(G), ∀G′ ⊂⊂ G′, i = 1, . . . N.

Therefore we have from (DQL)′:

− d

dx′iai(x′, v, v′x) + a(x′, v, v′x) = 0 a.e. x′ ∈ G′.

Then we obtain

∂aN (x′, v, vx′)∂vx′N

vx′Nx′N

= a(x′, v, vx′)−N−1∑i=1

∂ai(x′, v, vx′)∂vx′j

vx′ix′j −

− ∂ai(x′, v, vx′)∂v

vx′i −∂ai(x′, v, vx′)

∂x′i.

Hence, in virtue of assumptions (B), (C), (E), follows the inequality

|∇′v|m−2∣∣∣vx′Nx′N ∣∣∣ ≤ c(ν, µ,m)

|∇′v|m−2

N−1∑i=1

∣∣∣vx′ix′j ∣∣∣+ |∇′v|m +

+ %1+m2 f(%x′)|∇′v|

m−22 + %

m2 g(%x′)|∇′v|

m2

.


From this, using the Young inequality, it is easy to obtain the inequality

(8.4.26) |∇′v|m−2∣∣∣vx′Nx′N ∣∣∣2 ≤ c(ν, µ,m)

|∇′v|m−2

N−1∑i=1

∣∣∣vx′ix′j ∣∣∣2 +

+ |∇′v|m+2 + %m+2f2(%x′) + %m+2g2(m+2)m (%x′)

.

Let ζ(x′) ∈ C∞0(B+

2σ(x′0))be a cuto function such that

ζ(x′) = 1 in B+σ (x′0), 0 ≤ ζ(x′) ≤ 1, |∇′ζ| ≤ cσ−1 in B+

2σ(x′0).

Let us now multiply both sides of this inequality by ζ2(x′) and integrate overB+

2σ(x′0); as a result we deduce∫B+

2σ

|∇′v|m−2∣∣∣vx′Nx′N ∣∣∣2 ζ2(x′)dx′ ≤ c(ν, µ,m)

∫B+

2σ

|∇′v|m−2

N−1∑i=1

∣∣∣vx′ix′j ∣∣∣2 ζ2(x′) +

+|∇′v|m+2ζ2(x′) + %m+2f2(%x′) + %m+2g2(m+2)m (%x′)

dx′.(8.4.27)

We estimate the rst term on the right in (8.4.27) by means of (8.4.25):∫B+

2σ

|∇′v|m−2∣∣∣vx′Nx′N ∣∣∣2 ζ2(x′)dx′ ≤ c1

∫B+

2σ

⟨|∇′v|m+2ζ2(x′) +

+|∇′v|m(ζ2 + |∇′ζ|2

)⟩dx′ + c2%

m+2

∫B+

2σ

⟨f2(%x′) +

+fm+2m (%x′) + g

2(m+2)m (%(x′) + g

m+2m−1 (%x′)

⟩dx′.(8.4.28)

Summing (8.4.25) and (8.4.28) we obtain∫B+

2σ

|∇′v|m−2 |vx′x′ |2 ζ2(x′)dx′ ≤ c1

∫B+

2σ

⟨|∇′v|m+2ζ2(x′) +

+|∇′v|m(ζ2 + |∇′ζ|2

)⟩dx′ + c2%

m+2

∫B+

2σ

⟨f2(%x′) +

+fm+2m (%x′) + g

2(m+2)m (%(x′) + g

m+2m−1 (%x′)

⟩dx′.(8.4.29)

We embark on the estimating of∫B+

2σ

|∇′v|m+2ζ2(x′)dx′. Because of (8.4.25),

it is sucient to estimate the integral∫B+

2σ

|∇′v|mv2x′Nζ2(x′)dx′. For this we

turn again to the (DQL)′0 and we take

φ(x′) =(v(x′)− v(x′0)

)v2x′Nζ2(x′)


as a test function. Then, in virtue of the representation (8.4.9), we have

∫B+

2σ

aij(x′, v, vx′)vx′ivx′jv2x′Nζ2(x′)dx′ =

= −∫B+

2σ

(v(x′)− v(x′0)

) ⟨aij(x′, v, vx′)vx′j

(2vx′N vx′ix′N ζ

2(x′) +

+2ζ(x′)ζx′iv2x′N

)+ a(x′, v, vx′)v2

x′Nζ2(x′)

⟩dx′ −

−∫B+

2σ

ai(x′, v, 0)⟨(v(x′)− v(x′0)

)v2x′Nζ2(x′)

⟩x′i

dx′.

Hence, by integrating by parts in the last term on the right and so obtainand because of the assumption (E), follows the inequality

ν

m− 1%1−m

∫B+

2σ

|∇′v|mv2x′Nζ2(x′)dx′ ≤

≤∫B+

2σ

∣∣v(x′)− v(x′0)∣∣ ⟨2|aij(x′, v, vx′)|∇′v||vx′N ||vx′x′ |ζ

2(x′) +

+2|aij(x′, v, vx′)|∇′v|ζ(x′)|∇′ζ|v2x′N

+ |a(x′, v, vx′)|v2x′Nζ2(x′) +

+∣∣∣∣∂ai(x′, v, 0)

∂x′i+∂ai(x′, v, 0)

∂vvx′i

∣∣∣∣ v2x′Nζ2(x′)

⟩dx′.

Now we observe again that all hypotheses of Theorem 8.3 about Höldercontinuity of weak solutions are fullled and conclude

|v(x′)− v(x′0)| ≤ cσα, x′ ∈ B+2σ(x′0), α ∈ (0, 1).

Therefore, by assumptions (B), (E), (F), we get

∫B+

2σ

|∇′v|mv2x′Nζ2(x′)dx′ ≤ c(ν, µ,m)σα

∫B+

2σ

⟨|∇′v|m−1|vx′N ||vx′x′ |ζ

2(x′) +

+|∇′v|m−1ζ(x′)|∇′ζ|v2x′N

+ |∇′v|mv2x′Nζ2(x′) + %mf(%x′)v2

x′Nζ2(x′) +

+%1+m2 f(%x′)|∇′v|

m−22 v2

x′Nζ2(x′) + %m−1g(%x′)|∇′v|v2

x′Nζ2(x′)

⟩dx′.

(8.4.30)


Let us apply again the Cauchy - Young inequalities:

|∇′v|m−1|vx′N ||vx′x′ | ≤12|∇′v|m−2|vx′x′ |2 +

12|∇′v|mv2

x′N,

|∇′v|m−1ζ(x′)|∇′ζ|v2x′N≤ 1

2|∇′v|mv2

x′Nζ2 +

12|∇′v|m|∇′ζ|2,

%mf(%x′)v2x′N≤ 2m+ 2

|vx′N |m+2 +

m

m+ 2%m+2f

m+2m (%x′) ≤

≤ 2m+ 2

|∇′v|mv2x′N

+m

m+ 2%m+2f

m+2m (%x′),

%1+m2 f(%x′)|∇′v|

m−22 v2

x′N≤ 1

2|∇′v|mv2

x′N+

12%m+2f2(%x′),

%m−1g(%x′)|∇′v|v2x′N≤ 1m|∇′v|mv2

x′N+m− 1m

%mgmm−1 (%x′)v2

x′N≤

≤ 3m+ 2

|∇′v|mv2x′N

+m− 1m+ 2

%m+2gm+2)m (%x′).

Hence and from (8.4.30) we nally obtain∫B+

2σ

|∇′v|mv2x′Nζ2(x′)dx′ ≤ c1σ

α

∫B+

2σ

⟨|∇′v|m−2v2

x′x′ζ2(x′) +

+|∇′v|mv2x′Nζ2(x′) + |∇′v|m|∇′ζ|2

⟩dx′ + c2%

m+2

∫B+

2σ

⟨f2(%x′) +

+fm+2m (%x′) + g

m+2m (%x′)

⟩ζ2(x′)dx′.(8.4.31)

Combining (8.4.25), (8.4.29), (8.4.31), choosing σ suciently small and usingthe properties of ζ(x′), we get∫

B+σ

(|∇′v|m+2 + |∇′v|m−2 |vx′x′ |2

)dx′ ≤ c1

∫B+

2σ

|∇′v|mdx′ +

(8.4.32)

+c2%m+2

∫B+

2σ

⟨f2(%x′) + f

m+2m (%x′) + g

m+2m−1 (%x′) + g

2(m+2)m (%(x′)

⟩dx′.

By the covering argument, returning to the previous variables x, u we getthe desired (8.4.24).

From Theorems 8.43 - 8.44 follows immediately the following theorem:

Theorem 8.45. Let u(x) be a weak bounded solution of the problem(DQL). Let us assume that the hypotheses (A), (B), (C), (D), (E), (F) are


fullled on the set M. Then we have the estimate∫G2%%

(|∇u|m+2 + |∇u|m−2u2

xx

)dx ≤

(8.4.33)

≤ C∫

G4%%/2

(r−2|∇u|m + f2 + f

m+2m + g

m+2m−1 + g

2(m+2)m

)dx, ∀ρ ∈ (0, d).

8.4.3. The local estimate near a conical point.

Theorem 8.46. Let u(x) be a weak bounded solution of the problem(DQL). Let λ0 be the least eigenvalue of the problem (NEV P1) (it is de-termined by Theorem 8.12) and t ∈ (0, 1] be the number that is determinedby (8.3.4). Let us assume that the hypotheses (U),1)− 4) from Section 8.3and (A), (B), (C), (D), (E), (F) are fullled. In addition, suppose∫

G4%%/2

(f2 + f

m+2m + g

m+2m−1 + g

2(m+2)m

)dx ≤ K%N−2+m(tλ0−1).(8.4.34)

If γ > 2−N −m(tλ0 − 1), then we have the estimate

(8.4.35)∫G%0

(rγ |∇u|m−2u2

xx + rγ−2|∇u|m + rγ−2−m|u|m)dx ≤

≤ C%γ+N−2+m(tλ0−1), ∀% ∈ (0, d).

Proof. By Theorem 8.45 together with (8.4.34) we have∫G2%%

rγ |∇u|m−2u2xxdx ≤ C

∫G4%%/2

rγ−2|∇u|mdx+K%γ+N−2+m(tλ0−1).(8.4.36)

Let us now apply Theorems 8.40, 8.41; according to the estimates (8.3.5),(8.3.8) and (8.4.36) we obtain

(8.4.37)∫G2%%

(rγ |∇u|m−2u2


≤ C%γ+N−2+m(tλ0−1), ∀% ∈ (0, d).


Let us dene the sequence %k = 21−k%. We rewrite the inequality (8.4.37)replacing % by %k in it:∫

G2%k%k

(rγ |∇u|m−2u2


≤ C2(1−k)%, ∀% ∈ (0, d),(8.4.38)

κ = γ +N − 2 +m(tλ0 − 1) > 0.

Summing the inequalities (8.4.38) over all k = 1, 2, . . . we have∫G2%

0

(rγ |∇u|m−2u2


≤ C%∞∑k=1

2(1−k) =C

1− 2−%,

since κ > 0.

Example.Let us look at the problem∆mu := −div (|∇u|m−2∇u) = 0 in G0,

u∣∣∣ω=± 1

2ω0

= 0,

where m > 1 and

G0 =x = (r, ω)

∣∣∣0 < r <∞, |ω| ≤ ω0

2

, ω0 ∈ (0, 2π)

is the plane angle. We use the results of Chapter 9. In Subsection 9.4 ofChapter 9 we constructed the solution of our problem in the form

w(x) = rλΦ(ω), ω ∈ [−ω0

2,ω0

2], λ > 0

with Φ(ω) ≥ 0 and λ = λ0, determined by (8.2.4). By the properties ofΦ(ω), established in Subsection 9.4, it is not dicult to deduce the followingestimates:

0 < u(x) ≤ rλ0 , |∇u| ≤ c1rλ0−1, |uxx| ≤ c2r

λ0−2.(8.4.39)

Now we can establish the condition of the niteness of the integral∫G%0

(rγ |∇u|m−2u2

xx + rγ−2|∇u|m + rγ−2−m|u|m)dx.

From the estimates (8.4.39) it follows that the integral above is nite, if

the integral%∫0

rγ−1+m(λ0−1)dr is convergent, that holds under the condition

γ > (1− λ0)m. This shows that the statement of Theorem 8.46 is precise.


8.5. Notes

The properties of weak solutions of the (LPA) in the neighborhood ofisolated singularities have been studied by many authors (see e.g. [155, 390]and the literature cited therein). We point out the great cycle of the L. Veronworks [381] - [394].

The behavior of solutions near a conical boundary points is treated onlyin special cases: in [372, 98, 59 ] for a0(x) ≡ 0, in [52] for bounded solutionsand m = 2. In this Chapter we extended these results to the more generalquasilinear case m 6= 2.

The problem (NEV P1) was studied by P.Tolksdorf [371, 372, 373, 375 ]and a more detailed analysis is carried out by Aronsson [11], Krol [202, 203]and 9.5.2, Chapter 9.

The solvability property of the operator D associated with the eigenvalueproblem (NEV P1), Theorem 8.14, as proved here is due to M. Dobrowolski[98, 68].

There is a number of works relating to the estimation of the rst eigen-value of the m− Laplacian in a Riemannian manifold, i.e of the problem(NEV P2) (see, e.g., [220, 408, 368, 153 ]). Apropos the one-dimensionalWirtinger inequality see also Theorems 256, 257 [141].

The other L∞− estimates of weak solutions of the problem (DQL) canbe found in 10.5, Chapter 10 [128] and in 7, Chapter IV [214].

Integral estimates of second weak derivatives of the (DQL) weak so-lutions in smooth domains were established in [213, 214, 215, 398 ]. InSection 8.4 we make these estimates more precise in the case of smooth do-mains as well as establish new estimates for nonsmooth domains; here wefollow [57, 70].

G. Savaré [351] obtained recently the certain new regularity results forsolutions of the Dirichlet and Neumann problems to some linear and quasi-linear elliptic equation of the variational structure in the lipschitz domains.M. Fuchs & Li Gongbao [124] established L∞−bound for weak solutions ofthe Dirichlet problem for the quasilinear elliptic equation on Orlicz - Sobolevspaces. S. Knobloch [156] considered the Neumann problem for (DQL) ina plane domain with corners.

CHAPTER 9

The behavior of weak solutions to the boundary

value problems for elliptic quasilinear equations

with triple degeneration in a neighborhood

of a boundary edge

9.1. Introduction. Assumptions.

This chapter is devoted to the estimate of weak solutions to the boundaryvalue problems for elliptic quasilinear degenerate second order equations.We investigate the behavior of weak solutions of the rst and mixed boundaryvalue problems for quasilinear elliptic equation of the second order with tripledegeneracy and singularity in the coecients in a neighborhood of singularboundary point.

Let G be a domain in RN , N ≥ 3, bounded by (N − 1)− dimensionalmanifold ∂G and let Γ1,Γ2 be open nonempty submanifolds of ∂G, possessingthe following properties: Γ1∩Γ2 = ∅, ∂G = Γ1∪Γ2, where Γ1∩Γ2 is smooth(N − 2)− dimensional submanifold that contains an edge Γ0 ⊆ Γ1 ∩ Γ2. Wealso x a partition of 0, 1, 2 into two subsets N and D. The union of theΓj with j ∈ D is going to be the part of the boundary where we considera Dirichlet boundary condition, but with j ∈ N is going to be the partof the boundary where we consider rst order boundary conditions: eitherNeumann or the third BVP. In what follows we suppose 0, 1 ∈ D. If 2 ∈ D,then our problem is Dirichlet problem; if 2 ∈ N , then our problem is mixedBVP.

We derive almost exact estimate of the weak solution in a neighbor-hood of an edge of the boundary for the problem

− d

dxiai(x, u, ux) + a0a(x, u, ux) + b(x, u, ux) = f(x),

x ∈ G; a0 ≥ 0u(x) = 0, x ∈ ∂G, if 2 ∈ D and x ∈ ∂G \ Γ2, if 2 ∈ N ;ai(x, u, ux)ni(x) + σ(x, u) = g(x), x ∈ Γ2, if 2 ∈ N

(BVP)

(summation over repeated indices from 1 to N is understood); here: ni(x),i = 1, . . . , N are components of the unit outward normal to Γ2.

For x = (x1, . . . , xN ) let us dene cylindrical coordinates (x, r, ω):

x = (x1, . . . , xN−2), r =√x2N−1 + x2

N , ω = arctgxN−1

xN.

331

332 9 Elliptic quasilinear equations with triple degeneration in a domain with an edge

For suciently small number d > 0 we also dene the sets:

Gd0 = G ∩ (x, r, ω)∣∣ x ∈ RN−2, 0 < r < d, ω ∈ (−ω0/2, ω0/2); ω0 ∈ (0, 2π);

Γdj = Γj ∩Gd0 ⊂ ∂Gd0, j = 0, 1, 2;

Ωd = G ∩ (x, r, ω)∣∣ x ∈ RN−2, r = d, ω ∈ [−ω0/2, ω0/2] ⊂ ∂Gd0.

We shall assume the following:• ∂G \ Γ0 is smooth submanifold in RN ;• there exists a number d > 0 such that

Γd0 = (x, 0, 0)| |x| < d ⊂ Γ0

is the straight edge with the center in the origin;• Gd0 is locally dieomorphic to the dihedral cone

Dd = (r, ω)∣∣ 0 < r < d, ω ∈ (−ω0/2, ω0/2) × RN−2; 0 < ω0 < 2π;

thus we assume that Gd0 ⊂ G and, consequently, the domain G is a"wedge" in some vicinity of the edge.• ω |Γ1= −ω0/2; ω |Γ2= ω0/2.

Let C0(G) be the set of continuous functions on G and let Lm(G) andW k,m(G),m > 1 be the usual Lebesgue and Sobolev spaces respectively.By N1

m,q(ν, ν0, G) we shall denote a set of functions u(x) ∈ L∞(G) havingrst weak derivatives with the nite integral∫

G

(ν(x)|u|q|∇u|m + ν0(x)|u|q+m

)dx <∞, q ≥ 0, m > 1,(9.1.1)

where ν0(x) and ν(x) are two nonnegative measurable in G functions suchthat

ν−10 (x) ∈ Lt(G), ν−1(x) ∈ Lt(G); ν0(x) ∈ Ls(G),

1s

+1t<m

N;

(9.1.2)

1 +1t< m < N

(1 +

1t

), t > max

(N,

N

m− 1), N > m > 1.

If X(G) is one of the above spaces, then by X(G,Γ) ∀Γ ⊆ ∂G we denote asubset of functions u(x) ∈ X(G) vanishing on Γ in the sense of traces. Nowwe dene the space V :

V :=

N1m,q(ν, ν0, G, ∂G), if BVP is Dirichlet problem;

N1m,q(ν, ν0, G, ∂G \ Γ2), if BVP is mixed problem.

We set also: V0 is V for q = 0. Let us dene for ∀ε ≥ 0 the number

θε :=

12(ω0 + ε), if BVP is Dirichlet problem;ω0 + ε, if BVP is mixed problem,

9.1 Introduction. Assumptions. 333

and let λ be the least positive number satisfying+∞∫0

[(m− 1)y2 + λ2

](y2 + λ2)

m−42 dy

(m− 1 + q + µ)(y2 + λ2)m2 + λ(2−m+ τ)(y2 + λ2)

m−22 − a0

=

= θ0,(9.1.3)

λm(q +m− 1 + µ) + λm−1(2−m+ τ) > a0.(9.1.4)

We shall use the following notation: (|u| − k)+ := max (|u| − k; 0).Concerning the equation of (BV P ) we make the following

Assumptions:

Let 1 < m < N, l > N, q ≥ 0, 0 ≤ µ < 1 be given numbers and letα(x), α0(x), b0(x) be nonnegative functions .

1) f(x), α(x), α0(x), b0(x), g(x) are measurable functions such that:

ν−10 (x)

(α0(x) + b0(x) + f(x)

)∈ Lp(G); α(x) ∈ Lm′(G); g(x) ∈ Lα(Γ2);

1p<m

N− 1t− 1s, α >

N − 1m− 1− N

t

,1m

+1m′

= 1;

ai(x, u, ξ), i = 1, . . . , N ; a(x, u, ξ), b(x, u, ξ), σ(x, u) are Caratheodory func-tions: G× R× N→ R, possessing the properties:

2) ai(x, u, ξ)ξi ≥ ν(x)|u|q|ξ|m − α0(x); a(x, u, ξ)u ≥ ν0(x)|u|q+m;σ(x, u) · sign u ≥ 0;

3) |b(x, u, ξ)| ≤ µν(x)|u|q−1|ξ|m + b0(x);

4)

√n∑i=1

a2i (x, u, ξ) ≤ ν(x)|u|q|ξ|m−1 + ν

1m (x)ν1/m′

0 (x)|u|q+m−1 +

+ α(x)ν1m (x);

5) |a(x, u, ξ)| ≤ ν1m′ (x)ν1/m

0 (x)|u|q|ξ|m−1 + ν0(x)|u|q+m−1 ++ α(x)ν1/m

0 (x)|u|q;

6)∫Γ2

|σ(x, u)|ds <∞ ∀u ∈ L∞(G ∪ Γ2).

In addition suppose that: the functions ai(x, u, ξ), a(x, u, ξ), b(x, u, ξ), σ(x, u)are continuously dierentiable with respect to the x, u, ξ variables in Md,M0 =Gd0 × [−M0,M0]× RN and satisfy in Md,M0 :

7) (m− 1)u∂ai(x,u,ξ)∂u = q ∂ai(x,u,ξ)∂ξjξj ; i = 1, . . . , N ;


8) ∂ai(x,u,ξ)∂ξj

pipj ≥ γm,qν(x)|u|q|ξ|m−2p2, ∀p ∈ RN \ 0;

9)

√N∑i=1

∣∣∣∣∂b(x,u,ξ)∂ξi

∣∣∣∣2 ≤ ν(x)|u|q−1|ξ|m−1;

10) ∂b(x,u,ξ)∂u ≥ ν(x)|u|q−2|ξ|m; ∂a(x,u,ξ)

∂u ≥ γm,qν0(x)|u|q+m−2;∂σ(x,u)∂u ≥ 0;

11)

√N∑i=1

∣∣∣∣ai(x, u, ξ)− rτ |u|q|ξ|m−2ξi

∣∣∣∣2 ≤ c1(r)rτ |u|q|ξ|m−1 + ψ1(r);

12)

∣∣∣∣∂ai(x,u,ξ)∂ξj−rτ |u|q|ξ|m−4

(δji |ξ|2 +(m−2)ξiξj

)∣∣∣∣ ≤ c2(r)rτ |u|q|ξ|m−2+

+ c2(r)ψ2(r)|u|q

m−1 ;

13)

∣∣∣∣∂ai(x,u,ξ)∂xi− τrτ−2|u|q|ξ|m−2xiξi

∣∣∣∣ ≤ c3(r)rτ−1|u|q|ξ|m−1 + ψ3(r);

14)

∣∣∣∣a(x, u, ξ) − rτ−mu|u|q+m−2

∣∣∣∣ +∣∣∣∣b(x, u, ξ) + µrτu|u|q−2|ξ|m

∣∣∣∣ ≤≤ c4(r)r|u|q|ξ|m−1 + |u|qψ4(r),

where γm,q > 0, ci(r) are nonnegative, continuous at zero functions withci(0) = 0; in addition to that let there exist numbers ki ≥ 0, such thatψi(r) ≤ kirβi , i = 1, ..., 4;

β1 =l(N − 1)−N(m− 1)

l(N −m)τ − 2

l(m− 1) + λ(q +m− 1);

β2 = τ −m+ 2 + λ(q +m− 1)m− 2m− 1

;

β3 = τ −m+ λ(q +m− 1);(9.1.5)

β4 =(l −m)Nl(N −m)

τ − 2lm+ λ(m− 1),

κ =l(N −m+ 1)−N

l(N −m)τ − 2

l+ ε, ∀ε > 0.

Remark 9.1. Our assumptions 11) - 14) essentially mean that the co-ecients of the (BV P ) near the edge Γ0 are close to coecients of model

9.1 Introduction. Assumptions. 335

equation

− d

dxi

(rτ |u|q|∇u|m−2uxi

)+ a0r

τ−mu|u|q+m−2 −

−µrτ |u|q−1|∇u|msign u = f(x),(ME)

0 ≤ µ < 1, q ≥ 0, m > 1, a0 ≥ 0, τ ≥ m− 2.

Definition 9.2. Function u(x) is called a weak solution of (BV P ) pro-vided that u(x) ∈ V and satises the integral identity

(II )∫G

ai(x, u, ux)φxi + a0a(x, u, ux)φ+ b(x, u, ux)φ dx =

=∫G

f(x)φdx+∫Γ2

g(x)− σ(x, u)φds

for all φ(x) ∈ V

One can easy verify that assumptions 1) - 6) together with (9.1.2) guar-antee the correctness of such denition.

We need the following auxiliary statements:

Lemma 9.3. Let m# denote the number associated to m by the relation

1m#

=1m

(1 +

1t

)− 1N

(9.1.6)

and suppose that assumption (9.1.2) holds. Then there exist constantsc1 > 0, c2 > 0, c3 > 0 depending only on meas G,ω0, N,m, t, ||ν−1

0 ||Lt(G),

||ν−1||Lt(G) such that∫G

ν0(x)|v|mdx ≤ c1

∫G

ν(x)|∇v|mdx,(9.1.7)

∫G

|v|m#dx

m

m#

≤ c2

∫G

(ν0(x)|v|m + ν(x)|∇v|m) dx(9.1.8)

for any v(x) ∈ V0 and also∫G

ν0(x)|u|q+mdx ≤ c3

∫G

ν(x)|u|q|∇u|mdx,(9.1.9)

for any u(x) ∈ V.


Proof. The proof for (9.1.7) had been given either in 1.5 [99] or in thestatements 3.2 - 3.5 [313]. The inequality (9.1.9) is obtained from (9.1.7),by performing in the latter the change of function:

u = v|v|σ−1, σ =m

q +m.

Now we prove the inequality (9.1.8) following the Theorem 3.1 [313]. Weshall deduce the inequality (9.1.8) from the corresponding ones for the imbed-ding Sobolev Theorem 1.31, namely if 1 < m < N then

‖v‖L

mNN−m (G)

≤ C‖v‖W 1,m(G), ∀v ∈W 1,m(G).(9.1.10)

If we put 1 = 1 + 1

t then we have from (9.1.2)

1 < mκ < N and κ +κ

t= 1.

Now, by using the Hölder integral inequality with p = 1 , p

′ = 11− , we

obtain

‖v‖Lm(G) =

∫G

|v|mν0 (x)ν−0 (x)dx

1m

≤

≤ ‖ν−10 (x)‖

1m

Lt(G) ·

∫G

ν0(x)|v|mdx

1m

.

(9.1.11)

Similarly,

‖∇v‖Lm(G) =

∫G

|∇v|mν(x)ν−(x)dx

1m

≤

≤ ‖ν−1(x)‖1m

Lt(G) ·

∫G

ν(x)|∇v|mdx

1m

.

(9.1.12)

We consider now the inequality (9.1.10) replacing m by mκ (in this connec-tion we verify that Nm

N−m = m#); then we obtain

‖v‖Lm

#(G)≤ C

(‖v‖Lm(G) + ‖∇v‖Lm(G)

)Hence and from (9.1.11), (9.1.12) it follows the required inequality (9.1.8).

Lemma 9.4. There exists a constant c4 > 0 depending on N,m, t,G,Γ2

such that for any v(x) ∈ N1m,0(ν, ν0, G, ∂G \ Γ2)(∫

Γ2

|v|α∗ds) 1α∗

≤ c4

∫G

(ν0(x)|v|m + ν(x)|∇v|m) dx 1m,(9.1.13)

9.2 A weak comparison principle. Strong maximum principle of E.Hopf 337

where

α∗ =m(N − 1)N −m+ N

t

.(9.1.14)

Proof. By the theorem of trace for Sobolev spaces (Theorem 1.35), wehave

‖v‖Lα∗ (Γ2) ≤ c‖v‖W 1,m(G)

with α∗ from (9.1.14). Hence and from the inequalities (9.1.11), (9.1.12) itfollows the desired inequality (9.1.13).

Corollary 9.5. (From Lemmas 9.3, 9.4).

(9.1.15)

∫G

|v|m#dx

m

m#

+(∫

Γ2

|v|α∗ds) mα∗

≤

≤ c5

∫G

(ν0(x)|v|m + ν(x)|∇v|m) dx

for any v(x) ∈ N1m,0(ν, ν0, G, ∂G\Γ2), where the constant c5 > 0 depends on

N,m, t,G,Γ2, ||ν−10 ||Lt(G), ||ν−1||Lt(G).

The main statement of this chapter is in the following theorem.Main Theorem Let u(x) be a weak solution to (BVP) and let λ be least

positive solution of (9.1.3)-(9.1.4). Suppose that the assumptions (9.1.2) and1) - 14) with m ≥ 2 are fullled. Let there are nonnegative constants f1, g1

such that

|f(x)| ≤ f1rτ−m+λ(q+m−1), x ∈ Gd0;

(9.1.16)

|g(x)| ≤ g1rτ−m+1+λ(q+m−1), x ∈ Γd2.

Then ∀ε > 0 there exists a constant cε > 0, depending only on the parametersand norms of functions occuring in the assumptions, such that

|u(x)| ≤ cεrλ−ε.(9.1.17)

9.2. A weak comparison principle. The E. Hopf strong maximumprinciple

Now we shall prove a weak comparison principles for quasilinear equationwhich extend corresponding results in chapter 10, Theorem 10.7 [128] andin chapter 3, Lemma 3.1 [372] (see also [295]).

Let Ω ⊂ RN be a bounded domain with lipschitzian boundary ∂Ω =

∂1Ω ∪ ∂2Ω. We consider the second order quasilinear degenerate operator Q


of the form:

Q(v, φ) ≡∫Ω

⟨Ai(x, vx)φxi +A(x, v)φ+ B(x, v, vx)φ−

− f(x)φ⟩dx+

∫∂2Ω

⟨Σ(x, v)− g(x)

⟩φds

(9.2.1)

for v ∈ N1m,0(ν, ν0,Ω, ∂Ω\∂2Ω) and for all nonnegative φ ∈ N1

m,0(ν, ν0,Ω, ∂Ω\∂2Ω) under following assumptions:

the functions f(x), g(x) are summable on Ω and ∂2Ω respectively; thefunctions Ai(x, η),A(x, v),B(x, v, η),Σ(x, v) are Caratheodory, continuouslydierentiable with respect to the v, η variables in M = Ω×R×RN and satisfyin M the inequalities:

(i) ∂Ai(x,η)∂ηj

pipj ≥ γmν(x)|η|m−2p2, ∀p ∈ RN \ 0;

(ii)

√N∑i=1

∣∣∣∣∂B(x,v,η)∂ηi

∣∣∣∣2 ≤ ν(x)|v|−1|η|m−1;

(iii) ∂B(x,v,η)∂v ≥ ν(x)|v|−2|η|m; ∂A(x,v)

∂v ≥ γmν0(x)|v|m−2; ∂Σ(x,v)∂v ≥ 0;

here: m > 1; γm > 0; ν0(x), ν(x) are the functions dened by (9.1.2) .

Theorem 9.6. Let operator Q satisfy assumptions (i) - (iii). Let thefunctions v, w ∈ N1

m,0(ν, ν0,Ω, ∂Ω \ ∂2Ω) satisfy the inequality

Q(v, φ) ≤ Q(w, φ)(9.2.2)

for all non-negative φ ∈ N1m,0(ν, ν0,Ω, ∂Ω \ ∂2Ω) and also the inequality

v(x) ≤ w(x), on ∂Ω \ ∂2Ω(9.2.3)

holds in the weak sense. Then

v(x) ≤ w(x), a.e. in Ω.(9.2.4)

Proof. Let us dene

z = v − w; vt = tv + (1− t)w, t ∈ [0, 1].


Then we have: 0 ≥ Q(v, φ)−Q(w, φ) =

=∫Ω

⟨φxizxj

∫ 1

0

∂Ai(x, vtx)∂vtxj

dt+ zφ

∫ 1

0

∂A(x, vt)∂vt

dt+

+φzxi

∫ 1

0

∂B(x, vt, vtx)∂vtxi

dt+ φz

∫ 1

0

∂B(x, vt, vtx)∂vt

dt

⟩dx+

+∫∂2Ω

φz

∫ 1

0

∂Σ(x, vt)∂vt

dtds

(9.2.5)

for all non-negative φ ∈ N1m,0(ν, ν0,Ω, ∂Ω \ ∂2Ω).

Now let k ≥ 1 be any odd number. We dene the set

Ω+ := x ∈ Ω | v(x) > w(x).As the test function in the integral inequality (9.2.2) we choose

φ = max(v − w)k, 0.

By assumptions (i) - iii) then we obtain

kγm

∫Ω+

ν(x)zk−1(∫ 1

0|∇vt|m−2dt

)|∇z|2dx+

γm

∫Ω+

ν0(x)zk+1(∫ 1

0|vt|m−2dt

)dx+

∫Ω+

ν(x)zk+1(∫ 1

0|vt|−2|∇vt|mdt

)dx ≤

(9.2.6)

≤∫

Ω+

ν(x)zk(∫ 1

0|vt|−1|∇vt|m−1dt

)|∇z|dx.

Now we use the Cauchy inequality

zk|∇z||vt|−1|∇vt|m−1 =

(|vt|−1z

k+12 |∇vt|m/2

)·

(zk−1

2 |∇z||∇vt|m/2−1

)

≤ ε

2|vt|−2zk+1|∇vt|m +

12εzk−1|∇z|2|∇vt|m−2, ∀ε > 0.

Hence, taking ε = 2, we obtain from (9.2.6)(kγm −

14) ∫Ω+

ν(x)zk−1|∇z|2(∫ 1

0|∇vt|m−2dt

)dx ≤ 0.(9.2.7)

Now choosing the odd number k ≥ max(

1; 12γm

)in view of z(x) ≡ 0 on a.e.

∂Ω+, we get from (9.2.7) z(x) ≡ 0 in a.e.Ω+. We have nished with thecontradiction to our denition of the set Ω+. By this the (9.2.4) is proved.


Remark 9.7. The operator Q, generated by the model equation (ME)with q = 0, satisfy all assumption (i) - (iii). In fact, we have for this case:

ν(x) = rτ , ν0(x) = a0rτ−m, Ai(x, η) = ν(x)|η|m−2ηi,

A(x, v) = ν0(x)v|v|m−2, B(x, v, η) = −µν(x)v−1|η|m.

Therefore

ν−1(x)∂Ai(x, η)∂ηj

= δji |η|m−2 + (m− 2)|η|m−4ηiηj

and hence

ν−1(x)∂Ai(x, η)∂ηj

pipj = |η|m−2|p|2 + (m− 2)|η|m−4(piηi)2 ≥ γm|η|m−2|p|2,

where

γm =

1, if m ≥ 2;m− 1, if 1 < m ≤ 2,

i.e. (i) holds.Further,

∂B(x, v, η)∂ηi

= −µmν(x)v−1|η|m−2ηi

and hence (ii) holds. At last

∂A(x, v)∂v

= (m− 1)ν0(x)|v|m−2,∂B(x, v, η)

∂v= µν(x)|v|−2|η|m,

and therefore (iii) holds as well.

Now we want prove the strong Hopf maximum principle (cf. 3.2 [372].)In addition to (i)-(iii) we shall suppose

(v) |η|·∣∣∣∣ N∑i=1

∂Ai(x, η)∂ηi

∣∣∣∣+ ∣∣∣∣ N∑i=1

∂Ai(x, η)∂xi

∣∣∣∣+ ∣∣B(x, v, η)+µν(x)v−1|η|m∣∣ ≤

≤ γmν(x)|η|m−1

for some non-negative constants γm, µ.

Lemma 9.8. Let Bd(y) be an open ball of radius d > 0 centered at y,

contained in Ω ⊂ RN and v(x) ∈ N1m,0(ν, ν0, Bd(y))∩C1

(Bd(y)

)be a solution

of

Q0(v, φ) ≡∫

Bd(y)

⟨Ai(x, vx)φxi + B(x, v, vx)φ

⟩dx = 0(9.2.8)


for all nonnegative φ ∈ L∞(Bd(y)) ∩ W 1,m(Bd(y), ∂Bd(y)

). Suppose that

assumptions (i)− (v) are fullled. Assume that

v(x) > 0, x ∈ Bd(y), v(x0) = 0 for some x0 ∈ ∂Bd(y).(9.2.9)

Then

|∇v(x0)| 6= 0.(9.2.10)

Proof. We consider the annular region

R = Bd(y) \Bd/2(y) =x∣∣ d

2< |x− y| < d

and the function

w(x) = e−σ|x−y|2 − e−σd2

, x ∈ R, σ > 0.

Direct calculation gives:

0 ≤ w(x) ≤ e−σ|x−y|2 ;(9.2.11)

wxi = −2σ(xi − yi)e−σ|x−y|2; |∇w| = 2σ|x− y|e−σ|x−y|2 ;(9.2.12)

wxixj =⟨4σ2(xi − yi)(xj − yj)− 2σδji

⟩e−σ|x−y|

2;(9.2.13)

L(εw) ≡ −dAi(x, εwx)dxi

+ B(x, εw, εwx) =

= −ε∂Ai(x, εwx)∂(εwxj )

wxixj −∂Ai(x, εwx)

∂xi+ B(x, εw, εwx) =

= −4εσ2e−σ|x−y|2 ∂Ai(x, εwx)

∂(εwxj )(xi − yi)(xj − yj) +

+ 2εσe−σ|x−y|2 ∂Ai(x, εwx)

∂(εwxi)− ∂Ai(x, εwx)

∂xi+ B(x, εw, εwx), ∀ε > 0.

By assumptions (i), (v) hence it follows that

L(εw) ≤ −εm−1ν(x)|∇w|m−2e−σ|x−y|2 ·⟨4γm|x− y|2σ2 −

(9.2.14)

−2γmσ − 4|x− y|γmσ⟩− µεm−1ν(x)w−1|∇w|m, ∀ε > 0.

Now we observe by (9.2.11), (9.2.12) that

|∇w|w

> 2σ|x− y|(9.2.15)

and therefore we have from (9.2.14) in region R:

L(εw) ≤ −εm−1ν(x)|∇w|m−2e−σ|x−y|2 ·⟨(γm + µ)d2σ2 − 2(1 + 2d)γmσ

⟩,

ε > 0.

If we choose σ ≥ 2(1+2d)γm(γm+µ)d2 , then we obtain

L(εw) ≤ 0 in R, ∀ε > 0.(9.2.16)


Since v > 0 on ∂Bd/2(y) there is a constant ε > 0 for which v − εw ≥ 0 on∂Bd/2(y). This inequality is also satised on ∂Bd(y) where w = 0. By virtueof (9.2.16) we have

Q0(εw, φ) =∫

Bd(y)

φL(εw)dx ≤ 0 = Q0(v, φ).

Thus we obtained: Q0(v, φ) ≥ Q0(εw, φ) in R;v ≥ εw on ∂R.

(9.2.17)

By weak comparison principle (Theorem 9.6) from (9.2.17) it follows that

v ≥ εw throughout R.(9.2.18)

Since x0 ∈ ∂Bd(y) and w(x0) = 0 now we have:

v(x)− v(x0)|x− x0|

≥ εw(x)− w(x0)|x− x0|

and therefore

|∇v(x0)| ≥ ε|∇w(x0)| = 2εσde−σd2> 0, Q.E.D.

Theorem 9.9. (Strong maximum principle of E.Hopf). Assumethat Ω is connected and v(x) ∈ N1

m,0(ν, ν0,Ω) ∩ C1(Ω) is non-negative weaksolution of ∫

Ω

⟨Ai(x, vx)φxi + B(x, v, vx)φ

⟩dx = 0

for all nonnegative φ ∈ L∞(Ω) ∩W 1,m(Ω, ∂Ω). Assume that v(x) 6≡ 0. Sup-pose that assumptions (i)− (v) are fullled. Then

v(x) > 0, x ∈ Ω(9.2.19)

Proof. Assume that v(x0) = 0 for some x0 ∈ Ω. Then, we can nd aball Bd(y) ⊂ Ω, satisfying the hypotheses of Lemma 9.8, i.e. x0 ∈ ∂Bd(y.)By this Lemma we have that |∇v(x0)| 6= 0. But 0 = v(x0) = inf

x∈Ωv(x) and

therefore |∇v(x0)| = 0. This, however, is a contradiction. Therefore, theconclusion of Theorem must be true.

Lemma 9.10. Let u(x) be a weak solution of (BVP) and let the assump-tions 2), 3) with α0(x) ≡ 0, b0(x) ≡ 0 be fullled. If in addition

f(x) ≥ 0, g(x) ≥ 0 for a.e. x ∈ G

then u(x) ≥ 0 a.e. in G.

9.3 The boundedness of weak solutions 343

Proof. Choose φ = u− = max−u(x), 0 as a test function in theintegral identity (II). We obtain:

∫G

⟨ai(x,−u−,−u−x )(−u−xi) + a0 · a(x,−u−,−u−x )(−u−) +

+ b(x,−u−,−u−x )(−u−) + f(x)u−⟩dx =

= −∫Γ2

⟨(−u−)σ(x,−u−) + g(x)u−

⟩ds.

By virtue of assumptions 2), 3)

(1− µ)∫G

ν(x)|u−|q|∇u−|mdx+ a0

∫G

ν0(x)|u−|q+mdx ≤

≤ −∫G

fu−dx−∫Γ2

⟨(−u−)σ(x,−u−) + g(x)u−

⟩ds ≤ 0,

since u− ≥ 0. Due to µ < 1, a0 ≥ 0 and u∣∣∂G\Γ2

= 0 we get u−(x) = 0 a.e. inG, i.e. u(x) ≥ 0 a.e. in G.

9.3. The boundedness of weak solutions

The goal of this section is to derive L∞(G)− a priori estimate of theweak solution to problem (BVP). The main statement of this section is inthe following theorem.

Theorem 9.11. Let u(x) be a weak solution of (BVP) and assumptions(9.1.2), 1) - 3) hold. Then there exists the constant M0 > 0, depending onlyon ||g||Lα(Γ2), ||ν−1(x), ν−1

0 (x)||Lt(G),∥∥ν−1

0 (x)(α0(x) + b0(x) + |f(x)|

)∥∥Lp(G)

,

meas G,ω0, N,m, µ, q, p, t, s, a0, such that

||u||L∞(G) ≤M0.

Proof. Let us introduce the set A(k) = x ∈ G, |u(x)| > k and letχA(k) be a characteristic function of the set A(k). We note that A(k + d) ⊆A(k) ∀d > 0. By setting φ(x) = η((|u|−k)+)χA(k) · signu in (II), where η isdened by Lemma 1.60 and k ≥ k0 (without loss of generality we can assume


k0 ≥ 1), on the strength of the assumptions 2)- 3) we get the inequality:

∫A(k)

ν(x)|u|q|∇u|mη′((|u| − k)+)dx+

+a0

∫A(k)

ν0(x)|u|q+m−1η((|u| − k)+)dx+

+∫

Γ2∩A(k)

σ(x, u)(sign u)η((|u| − k)+)ds ≤

≤ µ∫

A(k)

ν(x)|∇u|m|u|q−1η((|u| − k)+)dx+(9.3.1)

+∫

A(k)

(b0(x) + |f(x)|

)η((|u| − k)+)dx+

+∫

A(k)

α0(x)η′((|u| − k)+)dx+∫

Γ2∩A(k)

|g(x)|η((|u| − k)+)ds.

Now we dene the function wk(x) := η

((|u| − k)+

m

). By (1.11.7) from

Lemma 1.60 we have

(9.3.2)∫

Γ2∩A(k)

|g(x)|η((|u| − k)+)ds ≤M ·∫

Γ2∩A(k+d)

|g(x)||wk|mds+

+ e d ·∫

Γ2∩A(k+d)\A(k)

|g(x)|ds

Now we apply Lemma 9.4. In virtue of Hölder's inequality and (9.1.13) -(9.1.14) we get:

∫Γ2∩A(k+d)

|g(x)||wk|mds ≤

(∫Γ2∩A(k+d)

|wk|α∗ds

) mα∗

· ||g||L N−1

m−1−Nt

(Γ2) ≤

≤ c4||g||L N−1

m−1−Nt

(Γ2) ·∫

A(k)

(ν(x)|∇wk|m + ν0(x)|wk|m

)dx.


Then by assumptions 2) from (9.3.1) - (9.3.2) it follows that∫A(k)

ν(x)|u|q|∇u|m⟨η′((|u| − k)+)− µη((|u| − k)+)

⟩dx ≤

≤Mc4||g||L N−1

m−1−Nt

(Γ2) ·∫

A(k)


)dx+

+∫

A(k)

⟨α0(x)η′((|u| − k)+) + (b0 + |f |)η((|u| − k)+)

⟩dx+(9.3.3)

+e d ·∫

Γ2∩A(k)

|g(x)|ds.

By the denition of η(x) (see Lemma 1.60) and wk(x):

e (|u|−k)+ |∇u|m =(mκ

)m|∇wk|m, κ > 0

and by the choice of κ > m+ 2µ according to Lemma 1.60, using (1.11.5) -(1.11.7), from (9.3.3) we obtain

(9.3.4)12

(mκ

)mkq0

∫A(k)

ν(x)|∇wk|mdx ≤ c7M

∫A(k+d)

h(x)|wk|mdx+

+Mc4||g||L N−1

m−1−Nt

(Γ2) ·∫

A(k)


)dx+

+ c8e d⟨ ∫A(k)\A(k+d)

h(x)dx+∫

Γ2∩A(k)

|g(x)|ds⟩,

where

h(x) = α0(x) + b0(x) + |f(x)|.(9.3.5)

Now, by (9.1.7) from (9.3.4) it follows that

(9.3.6) (kq0 − c9)∫

A(k)

ν(x)|∇wk|mdx ≤ c10

∫A(k+d)

h(x)|wk|mdx+

+ c11e d⟨ ∫A(k)\A(k+d)

h(x)dx+∫

Γ2∩A(k)

|g(x)|ds⟩,


where

(9.3.7) c9 = 2(κm

)m(1 + c1)Mc4||g||L N−1

m−1−Nt

(Γ2); c10 = 2(κm

)mMc7;

c11 = 2(κm

)mc8.

By assumptions 1) we get that ν−10 (x)h(x) ∈ Lp(G), where p is such that

1p <

mN −

1t −

1s . By Hölder's inequality with exponents p and p′ (1

p + 1p′ = 1):

∫A(k+d)

h|wk|mdx ≤∥∥ν−1

0 (x)h(x)∥∥Lp(G)

∫A(k)

νp′

0 (x)|wk|mp′dx

1p′

.(9.3.8)

From the inequality 1p <

mN −

1t −

1s it follows that mp′ < m#, where m# is

dened in (9.1.6). Let j be a real number such that mp′ < j < m#. Fromthe interpolation inequality

∫A(k)

νp′

0 (x)|wk|mp′dx

1p′

≤

≤

∫A(k)

ν0(x)|wk|mdx

θ

·

∫A(k)

νj/m0 (x)|wk|jdx

(1−θ)m

j

with θ ∈ (0, 1), which is dened by the equality 1mp′ = θ

m + 1−θj , on the

strength of Hölder's inequality with exponents m#

j and m#

m#−j , from (9.3.8)we get:

∫A(k+d)

h|wk|mdx ≤ c12

∫A(k)

ν0(x)|wk|mdx

θ

×

×

∫A(k)

|wk|m#dx

(1−θ)mm#

,

c12 =∥∥ν−1

0 (x)h(x)∥∥Lp(G)

‖ν0(x)‖1−θLs(G) ,

(9.3.9)

provided we choose j = smm#

sm+m# ∈ (mp′,m#) in virtue of (9.1.6) and (9.1.2).By using the Young inequality with exponents 1

θ and 1(1−θ) , from (9.2.9) we


obtain

∫A(k+d)

h|wk|mdx ≤c13

ε1/θ

∫A(k)

ν0(x)|wk|mdx+

+ε1

(1−θ) (1− θ)

∫A(k)

|wk|m#dx

m

m#

,

c13 = θ∥∥ν−1

0 (x)h(x)∥∥ 1θ

Lp(G)‖ν0(x)‖

1−θθ

Ls(G) , ∀ε > 0.

(9.3.10)

It follows from (9.3.6), (9.3.10) that:

(kq0 − c9)∫

A(k)

ν(x)|∇wk|mdx ≤ c14ε−1/θ

∫A(k)

ν0(x)|wk|mdx+

+c16

⟨ ∫A(k)

h(x)dx+∫

Γ2∩A(k)

|g(x)|ds⟩

+(9.3.11)

+c15ε1

(1−θ)

∫A(k)

|wk|m#dx

m

m#

,

where ∀ε > 0, c14 = c13c10, c15 = (1− θ)c10, c16 = c11e d.Further, from (9.3.11), by (9.1.7), we get:

(kq0 − c9 − c1c14ε− 1θ )∫

A(k)

ν(x)|∇wk|mdx ≤ c15ε1

(1−θ)

∫A(k)

|wk|m#dx

m

m#

+

+c16

⟨ ∫A(k)

h(x)dx+∫

Γ2∩A(k)

|g(x)|ds⟩, ∀ε > 0, ∀k ≥ k0.(9.3.12)

Let us choose

c1c14ε

− 1θ = 1

2kq0 ⇒ ε = (2c1c14)θk−qθ0 ;

kq0 ≥ 4c9.(9.3.13)


By virtue of (9.1.15) we obtain:

( 14c5

kq0 − c15ε1

(1−θ)) ∫

A(k)

|wk|m#dx

m

m#

+

(∫Γ2∩A(k)

|wk|α∗ds

) mα∗≤

≤ c16

⟨ ∫A(k)

h(x)dx+∫

Γ2∩A(k)

|g(x)|ds⟩,(9.3.14)

if we choose

18c5

kq0 ≥ c15ε1

(1−θ) ;(9.3.15)

so by (9.3.13), (9.3.15) we choose:

k0 ≥ max

1; (8c5c15)1−θq (2c1c14)

θq ; (4c9)

1q

(9.3.16)

therefore from (9.3.15) it follows that

(9.3.17)

∫A(k)

|wk|m#dx

m

m#

+

(∫Γ2∩A(k)

|wk|α∗ds

) mα∗

≤

≤ c17

⟨ ∫A(k)

h(x)dx+∫

Γ2∩A(k)

|g(x)|ds⟩∀k ≥ k0,

where

c17 = max

4θc−θ1 cθ5c−θ14 c

θ−115 c16; 2c5c

−19 c16; 8c5c16

.

At last, by Young's inequality with exponents p, s, 11− 1

p− 1s

, we get:

∫A(k)

h(x)dx ≤∥∥ν−1

0 (x)h(x)∥∥Lp(G)

‖ν0(x)‖Ls(G) meas 1− 1p− 1sA(k).

In just the same way∫Γ2∩A(k)

|g(x)|ds ≤ ||g||Lα(Γ2) ·[meas(Γ2 ∩A(k))

] 1α′ ,

1α

+1α′

= 1.


Therefore from (9.3.17) it follows that ∫A(k)

|wk|m#

m

m#

+

(∫Γ2∩A(k)

|wk|α∗ds

) mα∗

≤

≤ 17

⟨∥∥ν−10 (x)h(x)

∥∥Lp(G)

‖ν0(x)‖Ls(G) meas 1− 1p− 1sA(k) +(9.3.18)

+||g||Lα(Γ2) ·[meas(Γ2 ∩A(k))

] 1α′⟩,

where 1− 1p −

1s > 0 in virtue of (9.1.2) and assumptions 1).

Let now l > k > k0. By (1.11.8) and the denition of the function wk(x) :|wk| ≥ 1

m(|u| − k)+, and therefore∫A(l)

|wk|m#dx ≥

(l − km

)m#

·meas A(l);

∫Γ2∩A(l)

|wk|α∗ds ≥

(l − km

)α∗·meas (Γ2 ∩A(l)).

From (9.3.18) it now follows that:

meas A(l) +[meas(Γ2 ∩A(l))

]m#

α∗ ≤(9.3.19)

≤(

m

l − k

)m#

·

∫A(k)

|wk|m#dx+

(∫Γ2∩A(k)

|wk|α∗ds

)m#

α∗≤

≤ 12

(m

l − k

)m#

· (2c17)m#

m

(∥∥ν−10 (x)h(x)

∥∥Lp(G)

‖ν0(x)‖Ls(G) +

+||g||Lα(Γ2)

)m#

m

×

meas

m#

m

1− 1

p− 1s

A(k) +[meas (Γ2 ∩A(k))

]m#

mα′

,

∀l > k ≥ k0.

Now we set

ψ(k) = meas A(k) +[meas(Γ2 ∩A(k))

]m#

α∗ .

Then from (9.3.19) it follows that

ψ(l) ≤ c18

(m

l − k

)m#

·

⟨[ψ(k)]

m#

m

1− 1

p− 1s

+ [ψ(k)]α∗mα′

⟩.(9.3.20)

Relying on (9.1.2), (9.1.6), (9.1.14) and assumptions 1) we note that

γ = min

m#

m

(1− 1

p− 1s

);α∗

ma′

> 1.


Then from (9.3.20) we get

ψ(l) ≤ c19

(l − k)m# ψγ(k) ∀l > k ≥ k0

and therefore we have, because of Lemma 1.59, that ψ(k0 + δ) = 0 withδ depending only on quantities in the formulation of Theorem 9.11. Thismeans that |u(x)| < k0 +δ for almost all x ∈ G. Theorem 9.11 is proved.

To complete this section let us derive some a priori integral estimates ofsolutions.

Theorem 9.12. Let u(x) be a weak solution of (BVP) and assumptions(9.1.2), 1) - 3) hold. Let us suppose in addition that∫

G

ν1

1−q−m0 (x)

(b0(x) + |f(x)|

) q+mq+m−1 <∞, g(x) ∈ Lm+q

q(Γ2).

Then the inequality

(9.3.21)∫G

(ν(x)|u|q|∇u|m + ν0(x)|u|q+m

)dx ≤

≤ C

∫G

ν1

1−q−m0 (x)

(b0(x) + |f(x)|

) q+mq+m−1dx+

∥∥ν−10 (x)α0(x)

∥∥p‖ν0(x)‖s +

+∥∥ν−1(x)

∥∥ 1m−1

Lt(G) +∥∥ν−1

0 (x)∥∥ 1m−1

Lt(G) +∫Γ2

|g(x)|m+qq ds

holds, where C > 0 is a constant depending only on N, m, q, µ, a0, measG.

Proof. By setting in (II) φ = u we get, in virtue of assumptions 2) -3):

(9.3.22) (1− µ)∫G

ν(x)|u|q|∇u|mdx+ a0

∫G

ν0(x)|u|q+mdx ≤

≤∫G

α0(x)dx+∫G

(b0(x) + |f(x)|

)|u(x)|dx+

∫Γ2

|g(x)||u(x)|ds.

By the Young inequality with p = q +m; p′ = q+mq+m−1 ∀ε > 0 :(

b0(x) + |f(x)|)|u(x)| =

(ν

1q+m

0 (x)|u(x)|)(

ν−1q+m

0

(b0(x) + |f(x)|

))≤

≤ εν0(x)|u|q+m + cεν1

1−q−m0 (x)

(b0(x) + |f(x)|

) q+mq+m−1 ;(9.3.23)

|g(x)||u(x)| ≤ ε|u(x)|m+qm + cε|g(x)|

m+qq .


Further, by Lemma 1.29 and by Young's inequality:∫Γ2

|u(x)|m+qm ds ≤ c6

∫G

(|u(x)|

m+qm +

m+ q

m|u(x)|

qm |∇u|

)dx ≤

≤ m+ q

mc6

∫G

⟨(ν

1m (x)|u|

qm |∇u|

)ν−

1m (x) +

(ν

1m0 (x)|u|

q+mm)ν− 1m

0 (x)⟩dx ≤

≤ m+ q

mc6

∫G

⟨ 1m

(ν(x)|u|q|∇u|m + ν0(x)|u|q+m

)+(9.3.24)

+1m′(ν−

m′m (x) + ν

−m′

m0 (x)

)⟩dx.

In addition we have:∫G

α0(x)dx ≤∥∥ν−1

0 (x)α0(x)∥∥p‖ν0(x)‖s ‖1‖1− 1

p− 1s

;

(9.3.25)∫G

ν−m′m (x)dx =

∫G

(ν−1(x)

) 1m−1 (x)dx ≤

∥∥ν−1(x)∥∥ 1m−1

Lt(G) · (measG)1− 1t(m−1) ,

where, by (9.1.2): t(m− 1) > 1.From (9.3.22) - (9.3.25) it follows that

(1− µ)∫G

ν(x)|u|q|∇u|mdx+ a0

∫G

ν0(x)|u|q+mdx ≤

≤ ε1

∫G

ν(x)|u|q|∇u|mdx+ ε2

∫G

ν0(x)|u|q+mdx+(9.3.26)

+c(ε1, ε2,m, q,N, t,measG)

∫Γ2

|g(x)|m+qq ds+

∥∥ν−1(x)∥∥ 1m−1

Lt(G) +

+∥∥ν−1

0 (x)∥∥ 1m−1

Lt(G) +∫G

ν1

1−q−m0 (x)

(b0(x) + |f(x)|

) q+mq+m−1dx+

+∥∥ν−1

0 (x)α0(x)∥∥p‖ν0(x)‖s

,

Now, if a0 > 0, then we choose ε1 = 1−µ2 , ε2 = a0

2 ; and if a0 = 0, then wetake advantage of (9.1.9) and choose ε1 = ε2c3 = 1−µ

4 . For both cases, from(9.3.26) we obtain the required (9.3.21). Theorem 3.2 is proved.


9.4. The construction of the barrier function

Let us set

ν(x) = rτ , ν0(x) = rτ−m, τ ≥ m− 2; m ≥ 2.

In this section, in N− dimensional innite dihedral cone

G0 =x = (x, r, ω)

∣∣ x ∈ RN−2, 0 < r <∞, −ω0

2< ω <

ω0

2, ω0 ∈ (0, 2π)

with the edge Γ0 =

(x, 0, 0)

∣∣ x ∈ RN−2, that contains the origin, and

lateral faces

Γ1 =

(x, r,−ω0

2)∣∣ x ∈ RN−2, 0 < r <∞

;

Γ2 =

(x, r,+ω0

2)∣∣ x ∈ RN−2, 0 < r <∞

we shall consider only the homogeneous boundary value problem

Lw ≡ − d

dxi

(rτ |w|q|Nw|m−2wxi

)+ a0r

τ−mw|w|q+m−2−

−µrτw|w|q−2|Nw|m = 0, x ∈ G0,

(BVP)0

w(x) = 0, x ∈ Γ0 ∪ Γ1 ∪ Γ2; for Dirichlet problem;w(x) = 0, x ∈ Γ0 ∪ Γ1; ∂w

∂n = 0, x ∈ Γ2 for mixed problem;a0 ≥ 0, 0 ≤ µ < 1, q ≥ 0, m ≥ 2, τ ≥ m− 2

and construct the function that will be the barrier for the non-homogeneousproblem. We shall seek for the solution of the problem (BVP)0 as:

w(x) = rλΦ(ω), ω ∈ [−ω0

2,ω0

2], λ > 0(9.4.1)

with Φ(ω) ≥ 0 and λ satisfying (9.1.3)-(9.1.4). By substituting the func-tion (9.4.1) in (BVP)0 and calculating in the cylindrical coordinates, for thefunction Φ(ω) we get the following Sturm - Liouvill boundary problem:

d

dω

[(λ2Φ2 + Φ′2

)m−22 |Φ|qΦ′

]+

+λ[λ(q +m− 1)−m+ 2 + τ ]Φ|Φ|q(λ2Φ2 + Φ′2

)m−22 =

= a0Φ|Φ|q+m−2 − µΦ|Φ|q−2(λ2Φ2 + Φ′2

)m2, ω ∈ (−ω0/2, ω0/2),

(StL)Φ(−ω0/2) = Φ(ω0/2) = 0 for Dirichlet problem;Φ(−ω0/2) = Φ′(ω0/2) = 0 for mixed problem.

9.4 The construction of the barrier function 353

By setting Φ′/Φ = y, we arrive at the Cauchy problem for y(ω) :

[(m− 1)y2 + λ2

](y2 + λ2)

m−42 y′ + (m− 1 + q + µ)(y2 + λ2)

m2 +

+λ(2−m+ τ)(y2 + λ2)m−2

2 = a0, ω ∈ (−ω0/2, ω0/2),(CPE)

y(0) = 0 for Dirichlet problem;y(ω0/2) = 0 for mixed problem.

From the equation of (CPE) we get:

−[(m− 1)y2 + λ2

](y2 + λ2)

m−42 y′ =

= (m− 1 + q + µ)(y2 + λ2)m2 + λ(2−m+ τ)(y2 + λ2)

m−22 − a0 =

= (y2 + λ2)m−2

2[(m− 1 + q + µ)(y2 + λ2) + λ(2−m+ τ)

]− a0 ≥

(9.4.2)

≥ (y2 + λ2)m−2

2 [λ2(m− 1 + q + µ) + λ(2−m+ τ)]− a0 ≥≥ λm(m− 1 + q + µ) + λm−1(2−m+ τ)− a0 > 0

by virtue of (9.1.4). Thus, it is proved that y′(ω) < 0, ω ∈ (−ω0/2, ω0/2).Therefore y(ω) decreases on the interval (−ω0/2, ω0/2).

9.4.1. Properties of the function Φ(ω). We turn in detail our atten-tion to the properties of the function Φ(ω). The case of Dirichlet problem seeas well [72]. First of all, we note that the solutions of (StL) are determineduniquely up to a scalar multiple. We consider the solution normed by thecondition

1 =

Φ(0) for Dirichlet problem;Φ(ω0

2 ) for mixed problem.(9.4.3)

We rewrite the equation of (StL) in the following form

−Φ[(m− 1)Φ′2 + λ2Φ2

] (λ2Φ2 + Φ′2

)m−42 Φ′′ = (q + µ)

(λ2Φ2 + Φ′2

)m2 +

+Φ2(λ2Φ2 + Φ′2

)m−42λ[λ(m− 1)−m+ 2 + τ ]

(λ2Φ2 + Φ′2

)+

+(m− 2)λ2Φ′2− a0Φm.(9.4.4)

Now, since m ≥ 2 from (9.4.4) it follows that

(9.4.5) − Φ[(m− 1)Φ′2 + λ2Φ2

] (λ2Φ2 + Φ′2

)m−42 Φ′′ ≥ −a0Φm +

+(λ2Φ2 + Φ′2

)m−22

(q+µ)(λ2Φ2 + Φ′2

)+λ[λ(m−1)−m+2+τ ]Φ2

≥

≥ Φm

(q + µ+m− 1)λm + (2−m+ τ)λm−1 − a0

> 0

(here we take into account that (q + µ + m − 1)λ2 + (2 −m + τ)λ > 0 by(9.1.4)).


Summarizing the above we obtain the following properties of functionΦ(ω) :

Φ(ω) ≥ 0, Φ′′(ω) < 0 ∀ω ∈ (−ω0/2, ω0/2).(9.4.6)

Corollary 9.13.

max[−ω0/2,ω0/2]

Φ(ω) = 1 ⇒ 0 ≤ Φ(ω) ≤ 1 ∀ω ∈ [−ω0/2, ω0/2].(9.4.7)

Let us proceed with the problem of (CPE) solvability. Rewriting theequation of (CPE) in the form resolved with respect to the derivative y′ =g(y, ) we observe that by (9.4.2) g(y) 6= 0 ∀y ∈ R. Moreover, g(y) andg′(y) being rational functions with non-zero denominators are continuousfunctions. By the theory of ordinary dierential equations the Chauchyproblem (CPE) is uniquely solvable in the interval

(−ω0

2 ,ω02

]. By integrating

(StL) - (CPE) we obtain

Φ(ω) = exp

ω∫0

y(ξ)dξ for Dirichlet problem;ω∫

ω0/2

y(ξ)dξ for mixed problem.(9.4.8)

y∫0

[(m− 1)z2 + λ2](z2 + λ2)m−4

2 dz

(m− 1 + q + µ)(z2 + λ2)m2 + λ(2−m+ τ)(z2 + λ2)

m−22 − a0

=

=

−ω for Dirichlet problem;ω02 − ω for mixed problem.

(9.4.9)

Hence, in particular, we get from (9.1.3) that limω→−ω0

2+0y(ω) = +∞. The last

allows to prove the solvability of the eigenvalue problem (StL) The expression(9.1.3) yields the equation for the sharp nding of the exponent λ in (9.4.1).

9.4.2. About solutions of (9.1.3) - (9.1.4). We may calculate ex-plicit the exponent λ for m = 2 or a0 = 0. In fact, integrating the (9.1.3) weobtain:

m = 2.

λ =

√τ2 + (π/θ0)2 + 4a0(1 + q + µ)− τ

2(1 + q + µ).(9.4.10)

a0 = 0, m 6= 2.

We denote the value λ in this case by λ0 :


(9.4.11)λ0(m− 2)(m− 1 + q + µ) + (m− 1)(2−m+ τ)√

(m− 1 + q + µ)2λ20 + (m− 1 + q + µ)(2−m+ τ)λ0

=

= (m− 2)(1− κ) + κτ,

where κ = 2θ0π . Hence we get the quadratic equation whence it follows

λ0 =

=

m(m−2)+[(1−)(m−2)+τ ]√m2+(2−m+τ)[(m−2)(2−)+τ ]

2(m−1+q+µ)[(m−2)(2−)+τ ] +

+ m−2−τ2(m−1+q+µ) , if θ0 < π;

m(m−2)−4τ(τ+2−m)+(2τ+2−m)√m2+4τ(τ+2−m)

8τ(m−1+q+µ) , if θ0 = π.

(9.4.12)

It is easily to see that λ0 > 0. From (9.4.12) we have for θ0 = π2 also

λ0 =m− 2− τ

2(m− 1 + q + µ)+m(m− 2) + τ

√τ2 + 4(m− 1)

2κ(m− 1 + q + µ)(m− 2 + τ.(9.4.13)

Now from (9.4.12) - (9.4.13) we deduce following special cases of value λ0 :

τ = 0

λ0 =

m+(2−)(m−2)+(1−)

√m2−(2−)(m−2)2

2(m−1+q+µ)(2−) , if θ0 < π;

(m−1)2

m(m−1+q+µ) , if θ0 = π.

(9.4.14)

Proof. We prove the second equality of (9.4.14). Applying the Taylorformula

√1± t = 1± 1

2 t+ o(t) for t→ 0, from (9.4.12) we obtain

λ0

∣∣∣=2

=m(m− 2)− 4τ2 + 4(m− 2)τ +m(2τ + 2−m)

√1 + 4τ(τ+2−m)

m2

8τ(m− 1 + q + µ)=

=m(m− 2)− 4τ2 + 4(m− 2)τ +m(2τ + 2−m)

[1 + 2τ(τ+2−m)

m2 + o(τ)]

8τ(m− 1 + q + µ)=

=m(−2τ + 3m− 4) + (τ + 2−m)(2τ + 2−m)

4m(m− 1 + q + µ)+o(τ)τ

;

hence it follows

λ0

∣∣∣=2τ=0

= limτ→0

λ0

∣∣∣=2

=(m− 1)2

m(m− 1 + q + µ), Q.E.D.


Similarly, on the other hand, from the rst equal of (9.4.14) we have

λ0

∣∣∣τ=0

=1

2κ(m− 1 + q + µ)

κ(m− 2) +m1 + (1− κ)

√1− (2−)(m−2)2

m2

2− κ

=

=1

2κ(m− 1 + q + µ)

κ(m− 2) +m− κ(1− κ)(m− 2)2

2m+o(2− κ)

2− κ

;

hence it follows

λ0

∣∣∣=2τ=0

= lim→2−0

λ0

∣∣∣τ=0

=(m− 1)2

m(m− 1 + q + µ),

Q.E.D.

τ = m− 2From (9.4.12) immediately it follows

λ0 =mπ

4θ0(m− 1 + q + µ), θ0 ≤ π.

ω0 → 0

We want investigate the behavior of λ0 for κ → 0. For this we rewrite(9.4.12) in the next way:

λ0 =m(m− 2) + [m− 2 + κ(τ −m+ 2)]m

√1 + 2(2−m+τ)+2(m−2)(2−m+τ)

m2

2κ2(m− 1 + q + µ)(τ −m+ 2) + 4κ(m− 2)(m− 1 + q + µ)+

+m− 2− τ

2(m− 1 + q + µ)=

=m(m− 2) + [m− 2 + κ(τ −m+ 2)]

m+

2m(2−m+ τ)(κ + 2m− 4)

2κ2(m− 1 + q + µ)(τ −m+ 2) + 4κ(m− 2)(m− 1 + q + µ)+

+m− 2− τ

2(m− 1 + q + µ)+o(κ)κ

=m(

1− (τ−m+2)2(m−2) + o(κ)

)2κ(m− 1 + q + µ)

+

+m− 2− τ

2(m− 1 + q + µ)+o(κ)κ

+

+m(τ −m+ 2) + 1

2m(2−m+ τ)(κ + 2m− 4)[κ(τ −m+ 2) +m− 2]2κ(m− 1 + q + µ)(τ −m+ 2) + 4(m− 2)(m− 1 + q + µ)

=

=m

2(m− 1 + q + µ)· 1κ

+O(1);


hence we get nally

λ0 =mπ

4(m− 1 + q + µ)· 1θ0

+O(1) for θ0 → 0.

That coincides with the Krol result for the Pseudo-Laplacian (q = µ = τ == a0 = 0), see p. 145 [203].

m→ +∞

We want investigate the behavior of λ0 for m→ +∞. For this we rewrite(9.4.12) in the next way:

1) if κ < 2,

λ0 =m− 2− τ

2(m− 1 + q + µ)+

m2 − 2m2κ(2− κ)m2 +O(m)

+

+[(1− κ)m+ 2κ − 2 + κτ ]

√(1− κ)2m2 +O(m)

2κ(2− κ)m2 +O(m)=

=m− 2− τ

2(m− 1 + q + µ)+

[1 + (1− κ)|1− κ|]m2 +O(m3/2)2κ(2− κ)m2

;

hence it follows

limm→+∞

λ0 =12

+1 + (1− κ)|1− κ|

2κ(2− κ)=

1

(2−) , if κ ≤ 1,

1, if 1 ≤ κ < 2;

2) if κ = 2,

λ0 =m2 − 2m+ 4mτ − 4τ2 − 8τ

8τ(m− 1 + q + µ)+

+(2τ + 2−m)m

√1 + 4τ(τ+2−m)

m2

8τ(m− 1 + q + µ)=

=m2 − 2m(1− 2τ)− 4τ(τ + 2)

8τ(m− 1 + q + µ)−

−[m2 − 2m(τ + 1)]

(1 + 2τ(τ+2−m)

m2 + o( 1m2 ))

8τ(m− 1 + q + µ)=

=8mτ +O( 1

m)8mτ +O(1)

=⇒ lim→2−0m→+∞

λ0 = 1.


Thus nally we have

limm→+∞

λ0 =

π2

4θ0(π−θ0) , if 0 < θ0 ≤ π2 ;

1, if π2 ≤ θ0 ≤ π.

That coincides with the Aronsson result for the Pseudo-Laplacian (q = µ == τ = a0 = 0), see [11].

9.4.3. About the solvability of (9.1.3) - (9.1.4) with ∀a0 > 0. Weset

F(λ, a0, ω0) = −θ0 +

(9.4.15)

+

+∞∫0

[(m− 1)y2 + λ2](y2 + λ2)m−4

2 dy

(m− 1 + q + µ)(y2 + λ2)m2 + λ(2−m+ τ)(y2 + λ2)

m−22 − a0

.

By making the substitution: y = tλ, t ∈ (0,+∞) we obtain:

F(λ, a0, ω0) = −θ0 +

+∞∫0

Λ(λ, a0, t)dt,

where

Λ(λ, a0, t) ≡(9.4.16)

≡ [(m− 1)t2 + 1](t2 + 1)m−4

2

λ(m− 1 + q + µ)(t2 + 1)m2 + (2−m+ τ)(t2 + 1)

m−22 − a0λ1−m

.

Then the equation (9.1.3) takes the form

F(λ, a0, ω0) = 0.(9.4.17)

According to the above, we have:

F(λ0, 0, ω0) = 0.(9.4.18)

The direct calculations yield:

∂Λ∂λ

= −[(m− 1)t2 + 1](t2 + 1)m−4

2 ×

(9.4.19)

× (m− 1 + q + µ)(t2 + 1)m2 + a0(m− 1)/λm[

λ(m− 1 + q + µ)(t2 + 1)m2 + (2−m+ τ)(t2 + 1)

m−22 − a0λ1−m

]2 < 0

∀ t, λ, a0;


∂Λ∂a0

=λ1−m[(m− 1)t2 + 1](t2 + 1)

m−42[

λ(m− 1 + q + µ)(t2 + 1)m2 + (2−m+ τ)(t2 + 1)

m−22 − a0λ1−m

]2(9.4.20)

> 0 ∀t, λ, a0.

Therefore, we can apply the theorem about implicit functions: in a certainneighborhood of the point (λ0, 0) the equation (9.4.17) (and so the equa-tion (9.1.3) as well) determines λ = λ(a0, ω0) as a single-valued continuousfunction of a0, depending continuously on the parameter ω0 and having con-tinuous partial derivatives ∂λ

∂a0, ∂λ∂ω0

. Now, we analyze the properties of λ asthe function λ(a0, ω0). First of all, from (9.4.17) we get:

∂F∂λ

∂λ

∂a0+∂F∂a0

= 0,∂F∂λ

∂λ

∂ω0+∂F∂ω0

= 0;

hence, it follows that

∂λ

∂a0= −

(∂F∂a0

)(∂F∂λ

) ;∂λ

∂ω0= −

(∂F∂ω0

)(∂F∂λ

) .(9.4.21)

But, on the strength of (9.4.19), (9.4.20) we have:

∂F∂a0

=

+∞∫0

∂Λ∂a0

dt > 0,∂F∂λ

=

+∞∫0

∂Λ∂λ

dt < 0,∂F∂θ0

= −1 ∀(λ, a0);

(9.4.22)

∂F∂ω0

=∂F∂θ0· dθ0

dω0=

−1

2 , if BVP is Dirichlet problem;−1, if BVP is mixed problem,

From (9.4.21) - (9.4.22) we get:

∂λ

∂a0> 0;

∂λ

∂ω0< 0 ∀a0.(9.4.23)

So, the function λ(a0, ω0) increases with respect to a0 and decreases withrespect to ω0. Applying the analytic continuation method, we obtain thesolvability of the equation (9.1.3) ∀a0.

Corollary 9.14.

λ = λ(a0, ω0) ≥ λ0 > 0 ∀a0 ≥ 0.


Further, multiplying the equation of (StL) by Φ(ω) and integrating overthe interval (−ω0

2 ,ω02 ), we get:

(9.4.24) (1− µ)

ω02∫

−ω02

|Φ|q(λ2Φ2 + Φ′2)m−2

2 Φ′2dω = −a0

ω02∫

−ω02

|Φ|q+mdω +

+ [λ2(m− 1 + q + µ) + λ(2−m+ τ)]

ω02∫

−ω02

|Φ|q+2(λ2Φ2 + Φ′2)m−2

2 dω ≥

≥⟨λm(m− 1 + q + µ) + λm−1(2−m+ τ)− a0

⟩ ω02∫

−ω02

|Φ|q+mdω > 0,

by virtue of (9.1.4).

Lemma 9.15. There takes place the inequalityω02∫

−ω02

|Φ|q|Φ′|mdω ≤ c(q, µ,m, τ, λ)

ω02∫

−ω02

|Φ|q+mdω.(9.4.25)

Proof. From (9.4.24), by Young's inequality with p = mm−2 , p

′ = m2 , it

follows that:

(1− µ)

ω02∫

−ω02

|Φ|q|Φ′|mdω ≤

≤ [λ2(m− 1 + q + µ) + λ(2−m+ τ)]

ω02∫

−ω02

|Φ|qΦ2(λ2Φ2 + Φ′2)m−2

2 dω ≤

≤ ε

ω02∫

−ω02

|Φ|q(λ2Φ2 + Φ′2)m2 dω + cε

ω02∫

−ω02

|Φ|q+mdω ≤

≤ ε

ω02∫

−ω02

|Φ|q|Φ′|mdω + cε

ω02∫

−ω02

|Φ|q+mdω ∀ε > 0,

since m ≥ 2. Choosing ε = 1−µ2 , we obtain the required (9.4.25).

Lemma 9.16. Let the assumption (9.1.4) holds, and in addition

q + µ < 1.(9.4.26)


Thenω02∫

−ω02

|Φ′|mdω ≤ c(q, µ,m, τ, λ, ω0).(9.4.27)

Proof. Dividing the equation of (StL) by Φ|Φ|q−2, we get

Φd

dω

[(λ2Φ2 + Φ′2

)m−22 Φ′

]+λ[λ(q+m−1)−m+2+τ ]Φ2

(λ2Φ2 + Φ′2

)m−22

+ qΦ′2(λ2Φ2 + Φ′2

)m−22 = a0|Φ|m − µ

(λ2Φ2 + Φ′2

)m2.

On integrating the obtained equality:

(9.4.28) (1− q − µ)

ω02∫

−ω02

(λ2Φ2 + Φ′2)m2 dω + a0

ω02∫

−ω02

|Φ|mdω =

= λ(λm+ 2−m+ τ)

ω02∫

−ω02

Φ2(λ2Φ2 + Φ′2)m−2

2 dω.

Since q+µ < 1, λ(λm+2−m+τ) > 0 and m ≥ 2 we shall get the required(9.4.27), if we apply the Young inequality with p = m

m−2 , p′ = m

2 , ∀ε > 0.Finally, if there were q + µ ≥ 1, then from (9.4.28) we would get

(q−1+µ)λm

ω02∫

−ω02

|Φ|mdω+λm−1(λm+2−m+τ)

ω02∫

−ω02

|Φ|mdω ≤ a0

ω02∫

−ω02

|Φ|mdω,

which would contradict (9.1.4), by virtue of Φ 6≡ 0. The Lemma is proved.

(9.4.1), (9.1.3) give us the function w = rλΦ(ω), which will be a barrierfor our boundary problem (BVP).

Lemma 9.17. Let ζ(r) ∈ C∞0 [0, d]. Then

ζ(r)w(x) ∈ N1m,q(r

τ , rτ−m, Gd0, Gd0 \ Γd2).

If (9.1.4) and (9.4.26) hold, then

ζ(r)w(x) ∈ N1m,0(rτ , rτ−m, Gd0, G

d0 \ Γd2).

Proof. At rst, we observe that w ∈ L∞(Gd0) since λ > 0. Now we shallprove that

Iq[w] ≡∫Gd0

(rτ−m|w|q+m + rτ |w|q|∇w|m

)dx <∞.(9.4.29)


The direct calculations give:

|∇w|m = rm(λ−1)(λ2Φ2 + Φ′2)m2 .(9.4.30)

Therefore

Iq[w] =∫Gd0

rτ+m(λ−1)+qλΦm+q(ω)+rτ+m(λ−1)+qλ|Φ|q(λ2Φ2+Φ′2)

m2

dx ≤

≤ c(λ,m)

d∫0

rτ+(m+q)λ−m+1dr

ω02∫

−ω02

(|Φ|q|Φ′|m + |Φ|q+m

)dω.

It is clear that, by virtue of Lemma 9.15, Iq[w] is nite. To prove the secondassertion of Lemma we have to demonstrate that

I[w] ≡∫Gd0

(rτ−m|w|m + rτ |∇w|m

)dx <∞.(9.4.31)

We have again:

I[w] =∫Gd0

rτ+m(λ−1)(λ2Φ2 + Φ′2)

m2 + rτ+m(λ−1)Φm(ω)

dx ≤

≤ c(λ,m)

d∫0

rτ+mλ+1−mdr

ω02∫

−ω02

(λ2Φ2 + Φ′2)

m2 + Φm

dω.

I[w] is nite by Lemma 9.16. Thus,

I[w] ≤ c(m,λ,N, q, µ, ω0, d).

Lemma 9.17 is proved.

Example 9.18. Let m = 2 and we shall consider the boundary valueproblem (BVP)0 for the equation

d

dxi(rτ |w|qwxi) = a0r

τ−2w|w|q − µrτw|w|q−2|∇w|2, x ∈ G0,

(9.4.32)a0 ≥ 0, 0 ≤ µ < 1, q ≥ 0, τ ≥ 0.

From (9.4.8), (9.1.3) it follows that the solution of our problem is the function

w(r, ω) = rλ ×

cos

11+q+µ

(πωω0

)for Dirichlet problem;

cos1

1+q+µ

(πω2ω0− π

4

)for mixed problem.

(9.4.33)

9.5 The estimate of weak solutions in a neighborhood of a boundary edge 363

where

λ =

√τ2 + (π/θ0)2 + 4a0(1 + q + µ)− τ

2(1 + q + µ)

(see (9.4.10)). It is easy to check that for such λ the inequality (9.1.4) isfullled.

By calculating Φ′(ω) one can readily see that all the properties of thefunction Φ(ω) hold. Moreover, we have:

ω02∫

−ω02

Φ′2(ω)dω =π

(1 + q + µ)2ω0·

Γ(32)Γ( 1−q−µ

2(1+q+µ)

)Γ(2+q+µ

1+q+µ

) ×

×

1, if BVP is Dirichlet problem;14 , if BVP is mixed problem,

(9.4.34)

provided q + µ < 1. This integral is nonconvergent , if q + µ ≥ 1. At thesame time ∀q > 0 we have:

ω02∫

−ω02

|Φ(ω)|qΦ′2(ω)dω =π

(1 + q + µ)2ω0·

Γ(32)Γ( 1−µ

2(1+q+µ)

)Γ(2+ 3

2q+µ

1+q+µ

) ×

×

1, if BVP is Dirichlet problem;14 , if BVP is mixed problem,

(9.4.35)

since µ < 1. This completely agrees with Lemmas 9.15 - 9.17, sinceω02∫

−ω02

|Φ(ω)|q+2dω =ω0√π·

Γ( q+ 3+µ

2(1+q+µ)

)Γ(2+ 3

2q+µ

1+q+µ

) .This demonstrates that w(x) ∈ N1

2,0(rτ , rτ−m, Gd0), if q + µ < 1, andw(x) /∈ N1

2,0(rτ , rτ−m, Gd0), if q + µ ≥ 1; for the latter case we havew(x) ∈ N1

2,q(rτ , rτ−m, Gd0).

9.5. The estimate of weak solutions in a neighborhood of aboundary edge

In this section we derive almost exact estimate of the weak solution of(BVP) in a neighborhood of a boundary edge. For our purpose we are goingto apply the comparison principle (see Theorem 9.6) and use the barrierfunction constructed in 9.4. It is easy to verify by assumptions 8) - 10)that all assumptions (i) - (iii) of the comparison principle (see 9.2) arefullled.


To begin with let us make some transformations. At rst we introducethe change of function

u = v|v|t−1; t =m− 1

q +m− 1.(9.5.1)

By virtue of the assumption 7) as a result the problem (BV P ) takes theform:

Q(v, φ) ≡∫G

⟨Ai(x, vx)φxi + a0A(x, v)φ+ B(x, v, vx)φ−(II)

−f(x)φ⟩dx+

∫Γ2

⟨Σ(x, v)− g(x)

⟩φds = 0

for v(x) ∈ V0 and any φ(x) ∈ V0, where

Ai(x, η) ≡ ai(x, v|v|t−1, t|v|t−1η), A(x, v) ≡ a(x, v|v|t−1, t|v|t−1η),

B(x, v, η) ≡ b(x, v|v|t−1, t|v|t−1η), Σ(x, v) ≡ σ(x, v|v|t−1)

and by assumptions 11) - 14):

11)

√N∑i=1

∣∣∣∣Ai(x, η)− tm−1rτ |η|m−2ηi

∣∣∣∣2 ≤ c1(r)rτ |η|m−1 + ψ1(r);

12)∣∣∣∣∂Ai(x,η)

∂ηj− tm−1rτ |η|m−4

(δji |η|2 + (m− 2)ηiηj

)∣∣∣∣ ≤≤ c2(r)rτ |η|m−2 + c2(r)ψ2(r);

13)∣∣∣∣∂Ai(x,η)

∂xi− τtm−1rτ−2|η|m−2xiηi

∣∣∣∣ ≤ c3(r)rτ−1|η|m−1 + ψ3(r);

14)∣∣∣∣A(x, v)− rτ−mv|v|m−2

∣∣∣∣+∣∣∣∣B(x, v, η) + µtmrτv−1|η|m

∣∣∣∣ ≤≤ c4(r)r|η|m−1 + |v|tqψ4(r).

Remark 9.19. Our assumptions 11) - 14) essentially mean that theoperator of the problem (BVP) is approximated near the edge Γ0 by theoperator of the problem for the (ME). Furthermore, by the assumption 7)coecients ai(x, u, ux) i = 1, ..., N after the substitution (9.5.1) do not de-pend on v explicit. For instance, the model equation (ME) satises theseassumptions; in fact, after the substitution (9.5.1) the (ME) takes the form:

L0v(x) ≡ −tm−1 d

dxi

(rτ |∇v|m−2vxi

)+ a0r

τ−mv|v|m−2 − µtmrτv−1|∇v|m =

= f(x), x ∈ G.


We shall make additional studies. Let us set

a0 = t1−ma0; µ = tµ; λ =1tλ; Φ(ω) = Φ

1t (ω),(9.5.2)

where t = m−1q+m−1 .

Now the function

w = rλΦ(ω)(9.5.3)

will play the role of the barrier. By (StL) - (CPE), (9.1.3) one can easilycheck that (λ,Φ(ω)) is a solution of the problem

d

dω

[(λ

2Φ2 + Φ′2)m−2

2 Φ′]

+ µ 1Φ

(λ

2Φ2 + Φ′2)m

2 =

= a0Φ|Φ|m−2 − λ[λ(m− 1)−m+ 2 + τ ]Φ(λ

2Φ2 + Φ′2)m−2

2,

ω ∈ (−ω0/2, ω0/2),(NEVP)

Φ(−ω0/2) = Φ(ω0/2) = 0 for Dirichlet problem;Φ(−ω0/2) = Φ′(ω0/2) = 0 for mixed problem.

+∞∫0

[(m− 1)y2 + λ

2]

(y2 + λ2)

m−42 dy

(m− 1 + µ)(y2 + λ2)

m2 + λ(2−m+ τ)(y2 + λ

2)m−2

2 − a0

= θ0.

It is evident, that the properties of (λ,Φ) established in 9.4.1 - 9.4.2 alsoremain valid for the (λ,Φ(ω)). In particular, (9.1.4) takes the form:

Pm(λ) ≡ (m− 1 + µ)λm + (2−m+ τ)λm−1 − a0 > 0.(9.5.4)

We consider the perturbation of the problem (NEVP). Namely, ∀ε ∈ (0, 2π−ω0) on the segment [−ω0+ε

2 , ω0+ε2 ] we dene the problem for (λε,Φε) :

d

dω

[(λ2εΦ

2ε + Φ

′2ε

)m−22 Φ′ε

]+ µ 1

Φε

(λ2εΦ

2ε + Φ

′2ε

)m2 =

= (a0 − ε)Φε|Φε|m−2 − λε[λε(m− 1)−m+ 2 + τ ]Φε

(λ2εΦ

2ε + Φ

′2ε

)m−22,

ω ∈ (−ω0+ε2 , ω0+ε

2 ),(NEVP)ε

Φε(−ω0+ε2 ) = Φε(ω0+ε

2 ) = 0 for Dirichlet problem;Φε(−ω0+ε

2 ) = Φ′ε(ω0+ε

2 ) = 0 for mixed problem;

+∞∫0

[(m− 1)y2 + λ2

ε

](y2 + λ2

ε)m−4

2 dy

(m− 1 + µ)(y2 + λ2ε)

m2 + λε(2−m+ τ)(y2 + λ2

ε)m−2

2 + ε− a0

= θε;

Pm(λε) + ε > 0.


The problem (NEVP)ε is obtained from the problem (NEVP) by replacingin the latter ω0 by ω0 + ε and a0 by a0 − ε. In virtue of monotonicity of thefunction λ(ω0, a0) , established in 9.4.2 (see. (9.4.23)), we get

0 < λε < λ, limε→+0

λε = λ.(9.5.5)

We denote by λ0 the value of λ for a0 = 0. It clearly follows from (9.4.12)

that λ0 = λ0

∣∣∣∣q=0;µ=µ

. In just the same way as in 9.4.2 we calculate that

λ > λ0. From (9.5.5) it follows that

0 <12λ0 < λε < λ(9.5.6)

for suciently small ε > 0.Next we shall consider separately the case of Dirichlet problem and the

case of mixed boundary value problem.

Dirichlet problem.

Lemma 9.20. There exists ε∗ > 0 such that

Φε

(ω0

2

)≥ ε

ω0 + ε, ∀ε ∈ (0, ε∗).(9.5.7)

Proof. We turn to the (9.5.4): Pm(λ) > 0. Since Pm(λ) is a polyno-mial, by continuity, there exists a δ∗− neighborhood of λ, in which (9.5.4)is satised as before, i.e. there exists δ∗ > 0 such that Pm(λ) > 0 for∀λ

∣∣ |λ − λ| < δ∗. We choose the number δ∗ > 0 in the such way; inparticular the inequality

Pm(λ− δ) > 0 ∀δ ∈ (0, δ∗)

holds. We recall that λ solves (NEVP). By virtue of (9.5.5), now for everyδ ∈ (0, δ∗) we can put

λε = λ− δand solve (NEVP)ε together with this λε with respect to ε; let ε(δ) > 0 beobtained solution. Since (9.5.5) is true,

limδ→+0

ε(δ) = +0.

Thus we have the sequence of problems (NEVP)ε with respect to

(λε,Φε(ω)) ∀ε∣∣ 0 < ε < min(ε(δ);π − ω0) = ε∗(δ), ∀δ ∈ (0, δ∗).(9.5.8)

We consider Φε(ω) with ∀ε from (9.5.8). In the same way as (9.4.5) we verifythat

Φ′′ε(ω) < 0, ∀ω ∈[−ω0 + ε

2,ω0 + ε

2

].


But this inequality means that the function Φε(ω) is convex up on [−ω0+ε2 , ω0+ε

2 ],i.e.

Φε(α1ω1 + α2ω2) ≥ α1Φε(ω1) + α2Φε(ω2), ∀ω1, ω2 ∈[−ω0 + ε

2,ω0 + ε

2

];

α1 ≥ 0, α2 ≥ 0∣∣α1 + α2 = 1.

We put

α1 =ω0

ε+ ω0, α2 =

ε

ε+ ω0; ω1 =

ε+ ω0

2, ω2 = 0.

By (NEVP)ε we get

Φε

(ω0

2

)≥ ε

ω0 + εΦε(0) =

ε

ω0 + ε,

q.e.d. Lemma is proved.

Corollary 9.21.ε

ω0 + ε≤ Φε(ω) ≤ 1, ∀ω ∈ [−ω0/2, ω0/2]; ∀ε ∈ (0, ε∗).(9.5.9)

Mixed problem.

Lemma 9.22. There exists ε∗ > 0 such that

Φε

(−ω0

2

)≥ ε

2(ω0 + ε)∀ε ∈ (0, ε∗).(9.5.10)

Proof. We turn back to (9.5.4): Pm(λ) > 0. Since Pm(λ) is a polyno-mial, then, by the continuity, there exists a δ∗− neighborhood of the pointλ, in which (9.5.4) remains valid, i.e. there exists δ∗ > 0 such that

Pm(λ) > 0 ∀λ∣∣ |λ− λ| < δ∗.

We choose the number δ∗ > 0 to guarantee this; particulary, the inequality

Pm(λ− δ) > 0 ∀δ ∈ (0, δ∗)(9.5.11)

holds. Let us recall that λ is a solution of (NEVP). By (9.5.5) we can nowput for every δ ∈ (0, δ∗)

λε = λ− δ(9.5.12)

and solve (NEVP)ε together with λε with respect to ε; let ε(δ) > 0 be theobtained solution. Since (9.5.5) holds, then

limδ→+0

ε(δ) = +0.

Thus, we get the sequence of the problems (NEVP)ε with respect to

(9.5.13) (λε,Φε(ω)) ∀ε∣∣ 0 < ε < min(ε(δ); 2π − ω0) = ε∗(δ),

∀δ ∈ (0, δ∗).


We consider Φε(ω) from (9.5.13). In the same way as in (9.4.5) we verifythat

Φ′′ε(ω) < 0 ∀ω ∈(−ω0 + ε

2,ω0 + ε

2

).

This inequality means that the function Φε(ω) is convex on the segment[−ω0+ε

2 , ω0+ε2 ], i.e.

Φε(α1ω1 + α2ω2) ≥ α1Φε(ω1) + α2Φε(ω2) ∀ω1, ω2 ∈[−ω0 + ε

2,ω0 + ε

2

];

α1 ≥ 0, α2 ≥ 0∣∣α1 + α2 = 1.

We put

α1 =ω0 + 1

2ε

ε+ ω0, α2 =

ε

2(ε+ ω0); ω1 = −ε+ ω0

2, ω2 =

ε+ ω0

2.

By (NEVP)ε we get

Φε

(−ω0

2

)≥ ε

2(ω0 + ε)Φε

(ω0 + ε

2

)=

ε

2(ω0 + ε).

The Lemma is proved.

Corollary 9.23.ε

2(ω0 + ε)≤ Φε(ω) ≤ 1 ∀ω ∈ [−ω0/2, ω0/2]; ∀ε ∈ (0, ε∗).(9.5.14)

Lemma 9.24. For any ε > 0 the inequalities:

0 < Φ′ε(ω) ≤ 2ε−1, ω ∈[−ω0

2,ω0

2

];(9.5.15)

−C(q, µ, τ,m, λ, ω0)ε−3 ≤ Φ′′ε < 0, ω ∈

[−ω0

2,ω0

2

](9.5.16)

hold, where C(q, µ, τ,m, λ, ω0) > 0.

Proof. From (NEVP)ε, (9.5.10), (9.5.14) we have:

ε

2(ε+ ω0)≤ Φε(ω) ≤ 1, Φ′ε(ω) ≥ 0, Φ′′ε(ω) < 0,(9.5.17)

∀ω ∈(−ω0 + ε

2,ω0 + ε

2

); Φε

(−ω0 + ε

2

)= 0, ∀ε > 0.

Hence it follows that Φ′ε(ω) decreases on(−ω0 + ε

2,ω0 + ε

2

). By the La-

grange mean value theorem, we have:

Φε

(−ω0 + ε

2

)− Φε

(−ω0

2

)= −ε

2Φ′ε(ω) ⇒ εΦ′ε(ω) ≤ 2


with some ω ∈(−ω0 + ε

2,−ω0

2

). Hence, by decreasing of Φ′ε(ω) we get

(9.5.15). From the equation of (NEVP)ε for Φε it follows that

−Φ′′ε =

1

Φε

[(m− 1)Φ′ε

2 + λ2εΦ2

ε

] (λ2εΦ2

ε + Φ′ε2)m−4

2

µ(λ2εΦ

2ε + Φ′ε

2)m

2 +

+Φ2ε

(λ2εΦ

2ε + Φ′ε

2)m−4

2

λε[λε(m− 1)−m+ 2 + τ ]

(λ2εΦ

2ε + Φ′ε

2)

+

+(m− 2)λ2εΦ′ε2

+ (ε− a0)Φmε

(9.5.18)

and therefore by virtue of (9.5.6), (9.5.15) and (9.5.17)

−Φ′′ε (ω) ≤

[λ2ε(2m− 3 + µ) + (2−m+ τ)λε + ελ2−m

ε

]Φε + µΦ−1

ε Φ′ε2 ≤

≤ C(q, µ, τ,m, λ, ω0)ε−3.

Lemma 9.25. There exists a positive constant c0 = c0(m, q, µ, τ, ω0) suchthat

Φ′ε(ω0

2)≥ c0ε

m+3, 0 < ε 1.(9.5.19)

Proof. By the Lagrange mean value theorem in virtue of (NEVP)ε wehave:

Φ′ε

(ω0

2

)= Φ′ε

(ω0

2

)− Φ′ε

(ω0 + ε

2

)= −ε

2Φ′′ε(ω)(9.5.20)

with some ω ∈(ω0

2,ω0 + ε

2

). From the equation (9.5.18) it follows that

−Φ′′εΦε

[(m− 1)Φ′ε

2 + λ2εΦ

2ε

] (λ2εΦ

2ε + Φ′ε

2)m−4

2 ≥

≥ Φmε

⟨[(µ+m− 1)λmε + (2 + τ −m)λm−1

ε − a0

]+ ε

⟩> εΦm

ε ,(9.5.21)

by (9.5.11), (9.5.12) and (9.5.4).

Since Φ′ε(ω) is decreasing continuous function and Φ′ε

(ω0 + ε

2

)= 0, then

for suciently small ε > 0 we can assert that 0 < Φ′ε(ω) < 1,

ω ∈(ω0

2,ω0 + ε

2

). Therefore we obtain:


1) if m ≥ 4, then[(m− 1)Φ′ε

2 + λ2εΦ

2ε

] (λ2εΦ

2ε + Φ′ε

2)m−4

2 ≤[(m− 1)Φ′ε

2 + λ2εΦ

2ε

]m−22 ≤

≤(m− 1 + λ2

ε

)m−22

≤(m− 1 + λ

2)m−2

2

, ω ∈(ω0

2,ω0 + ε

2

),

by (9.5.5); hence from (9.5.14), (9.5.21) it follows that

−Φ′′ε (ω) >

(m− 1 + λ

2) 2−m

2

εΦm−1ε ≥

(m− 1 + λ

2) 2−m

2 1[2(1 + ω0)]m−1

εm,

and, in virtue of (9.5.20), the required (9.5.19) is proved;2) if 2 ≤ m < 4, then from (9.5.21), by (9.5.6):

−Φ′′ε (ω) > εΦm−1

ε

(λ2εΦ

2ε + Φ′ε

2) 4−m

2

(m− 1)Φ′ε2 + λ2

εΦ2ε

> ελ4−mε Φ3

ε

1

λ2 +m− 1

>

> ε

(λ0

2

)4−m 1

8(λ2 +m− 1)(1 + ω0)3ε3, ω ∈

(ω0

2,ω0 + ε

2

)and, by virtue of (9.5.20), we again obtain (9.5.19).

9.6. Proof of the main Theorem

Proof. Let (λε,Φε(ω)) be a solution of the problem (NEVP)ε with xedε ∈ (0, ε∗), where ε∗ is determined by (9.5.13). We dene the function whichwe shall use as barrier function. Namely, let us consider the function

wε(x, r, ω) = ArλεΦε(ω), ∀x ∈ RN−2, r ∈ [0, d], ω ∈ [−ω0/2, ω0/2],

where A > 0 is a number to be chosen below. Let us apply the comparisonprinciple (Theorem 9.6) to the problem (II), comparing its solution v(x) withbarrier function wε(x) in the domain Gd0. The direct calculations demonstratethat:

L0wε(x, r, ω) = (At)m−1r(m−1)λε−m+τ

− d

dω

[(λ2εΦ

2ε + Φ′2ε

)m−22 Φ′ε

]−

− λε[λε(m− 1)−m+ 2 + τ ]Φε

(λ2εΦ

2ε + Φ′2ε

)m−22 + a0Φm−1

ε −

− µ 1Φε

(λ2εΦ

2ε + Φ′2ε

)m2

= εr(m−1)λε−m+τ

(tAΦε(ω)

)m−1.

By virtue of (9.5.1), (NEVP)ε and by (9.5.14), we obtain:

L0wε(x, r, ω) ≥ εm[

A(m− 1)2θε(q +m− 1)

]m−1

r(m−1)λε−m+τ ;(9.6.1)

9.6 Proof of the main Theorem 371

further,

wε(x)∣∣Ωd≥ Aε

2θεdλ,(9.6.2)

in virtue of (9.5.5), (9.5.14); in addition,

wε(x) ≥ 0 = v(x), x ∈ ∂Gd0 \(Ωd ∪ Γd2

).(9.6.3)

At last it is not dicult to calculate:

c0Aεm+1rλε−1 ≤ |∇wε| ≤ c1Aε

−1rλε−1; |∇2wε| ≤ c2Aε−3rλε−2,(9.6.4)

if we take into account (9.5.15), (9.5.16) from Lemma 9.24 and (9.5.19) fromLemma 9.25.

Now let φ ∈ L∞(Gd0)∩W 1,m(Gd0, ∂Gd0\Γd2) be any nonnegative function.

For the operator Q that is dened by (II) we obtain:

Q(wε, φ) =∫Gd0

φ(x)⟨− d

dxiAi(x,∇wε) + a0A(x,wε) + B(x,wε,∇wε)−

− f(x)⟩dx+

∫Γd2

φ(x)⟨Ai(x,∇wε)ni(x) + Σ(x,wε)− g(x)

⟩ds

and hence, by the denition of the operator L0:

Q(wε, φ) =∫Gd0

φ(x)

⟨− d

dxi

(Ai(x,∇wε)− tm−1rτ |∇wε|m−2wεxi

)

+ L0wε(x) + a0

(A(x,wε)− rτ−m|wε|m−2wε

)+

+

(B(x,wε,∇wε) + µtmrτ |∇wε|mw−1

ε

)− f(x)

⟩dx+

+∫Γd2

φ(x)

⟨Σ(x,wε)− g(x) + tm−1rτ |∇wε|m−2∂wε

∂n+

+(Ai(x,∇wε)− tm−1rτ |∇wε|m−2wεxi

)ni(x)

⟩ds.

(9.6.5)

By the assumption 2),

Σ(x,wε) = σ(x,wtε) ≥ 0, since wε(x) ≥ 0.(9.6.6)

Further,

∂wε∂n

∣∣∣∣∣Γd2

=1r

∂wε∂ω

∣∣∣∣ω=

ω02

= Arλε−1Φ′ε(ω0

2)≥ c0Aε

m+3rλε−1,(9.6.7)


by Lemma 9.25. Therefore, in virtue of (9.6.1), (9.6.6) - (9.6.7) from (9.6.5)it follows that

Q(wε, φ) ≥∫Gd0

φ(x)

⟨εm[

A(m− 1)2(ω0 + ε)(q +m− 1)

]m−1

r(m−1)λε−m+τ

−N∑

i,j=1

∣∣wεxixj ∣∣ ·∣∣∣∣∣∂Ai(x,∇wε)∂wεxj

− tm−1rτ |∇wε|m−4 ×

×(δji |∇wep|

2 + (m− 2)wεxiwεxj)∣∣∣∣∣−

−

∣∣∣∣∣∂Ai(x,∇wε)∂xi− τtm−1rτ−2|∇wε|m−2xiwεxi

∣∣∣∣∣−− a0

∣∣∣A(x,wε)− rτ−m|wε|m−2wε

∣∣∣−∣∣∣∣∣B(x,wε,∇wε) + µtmrτ |∇wε|mw−1ε

∣∣∣∣∣− |f(x)|

⟩dx+

+∫Γd2

φ(x)

⟨c0At

m−1εm+3rτ+λε−1|∇wε|m−2 −

−

√√√√ N∑i=1

∣∣∣∣Ai(x,∇wε)− tm−1rτ |∇wε|m−2wεxi

∣∣∣∣2 −− |g(x)|

⟩ds.

(9.6.8)

Now, taking into account the assumptions (9.1.16), 11)−14), (9.1.5) and theinequalities (9.6.4), from (9.6.8) we get:

Q(wε, φ) ≥ C1(m, q, ε, A, ω0)∫Gd0

φ(x)r(m−1)λε−m+τ⟨

1− c2(r)r(m−2)(λ−λε) −

− c(r)− (f1 + k3)r(m−1)(λ−λε) − k4rt(m−1)(λ−λε)

⟩dx+

+ C2(m, q, µ, τ, ε, A, ω0)∫Γd2

φ(x)rτ+(m−1)(λε−1)⟨

1− c1(r)−

− (k1 + g1)r(m−1)(λ−λε)

⟩ds,

9.7 Notes 373

where c(r) = c2(r) + c3(r) + c4(r). Fixing A > 0 and ε > 0, we can choosed > 0 so small (because of the continuity of the functions c1(r), c2(r), c3(r),c4(r) at zero) that

Q(wε, φ) ≥ 0, ∀φ ≥ 0.(9.6.9)

Further, by Theorem 9.11

v(x)∣∣∣∣Ωd

≤M1/t0 ,

therefore, by (9.6.2)

wε

∣∣∣∣Ωd

≥ Aε

2θεdλ ≥M1/t

0 ≥ v(x)∣∣∣∣Ωd

,(9.6.10)

provided that A > 0 is chosen suciently large:

A ≥ 2θεε

(M0

dλ

) q+m−1m−1

.(9.6.11)

Thus, from (9.6.9), (9.6.3), (9.6.10) and (II) we get:Q(wε, φ) ≥ 0 = Q(v, φ) ∀φ ≥ 0 in Gd0;wε(x) ≥ v(x), x ∈ ∂Gd0 \ Γd2.

Besides that, one can readily verify that all the other conditions of the com-parison principle (Theorem 9.6) are fullled; by this principle we get

v(x) ≤ wε(x), ∀x ∈ Gd0.Similarly one can prove that

v(x) ≥ −wε(x), ∀x ∈ Gd0.Thus, nally, we obtain

|v(x)| ≤ wε(x) ≤ Arλε , ∀x ∈ Gd0.On returning to the old variables, in virtue of (9.5.1) we get the requiredestimate (9.1.17). The main Theorem is proved.

9.7. Notes

The presentation of this Chapter follows [67]. Boundary value problemsin the smooth domains for quasilinear degenerate elliptic second order equa-tions have been intensively studied recently (see [6, 26, 49, 75, 78, 87, 99,135, 142, 145, 218, 219, 349 ] etc., and the vast bibliography there). Lessstudied are the problems of this kind in the domains with non-smooth bound-ary. In the paper [210], existence results for quasilinear degenerate ellipticboundary value problems are considered. In the paper [238] are examinedthe well posedness and regularity of the solution of degenerate quasilinearelliptic equations arising from bimaterial problems in elastic-plastic mechan-ics in lipschitzian domain. The papers [58], [59], [68], [98], [372] are de-voted to the study of weak solutions behavior for the special cases of the


(BV P ) equation in a neighborhood of conical boundary point. In [72] theDirichlet problem has been studied in a domain with edge on the bound-ary for the model equation (ME). In [132] the properties of the solutionsof (BV P ) for the Laplace operator has been investigated in a plain domainwith a polygonal boundary (see there Chapter 4). Such studies are importan-t for numerical solving of the boundary value problems (see, for ex., [97]).The Hölder continuity of weak solutions to the Dirichlet problem for thedegenerate elliptic linear and quasilinear divergence equations was proved inSection 3 [117] (linear equation) and in 2 [26] (quasilinear equation withm = 2.)

Recently, C. Ebmeyer and J. Frehse [103, 104] considered mixed bound-ary value problems for quasilinear elliptic equations and systems of the di-vergent form in a polyhedron. They proved W s,2, s < 3

2 -regularity and Lp−properties of the rst and the second derivatives of a solution.

CHAPTER 10

Sharp estimates of solutions to the Robin boundary

value problem for elliptic non divergence second

order equations in a neighborhood of the conical

point

The present Chapter is devoted to investigating of the behavior of strongsolutions to the the Robin boundary value problem for the second orderelliptic equations (linear and quasilinear) in the neighborhood of a conicalboundary point.

Let G ⊂ RN , N ≥ 2 be a bounded domain with the boundary ∂G thatis a smooth surface everywhere except at the origin O ∈ ∂G and near thepoint O it is a convex conical surface with its vertex at O. We consider theelliptic value problems:

L[u] ≡ aij(x)uxixj + ai(x)uxi + a(x)u = f(x);aij = aji, x ∈ G,

B[u] ≡ ∂u∂~n + 1

|x|γ(x)u = g(x), x ∈ ∂G \ O.(LRP)

and aij(x, u, ux)uxixj + a(x, u, ux) = 0, aij = aji, x ∈ G,∂u∂~n + 1

|x|γ(x)u = g(x), x ∈ ∂G \ O.(QLRP)

(summation over repeated indices from 1 to N is understood), ~n denotes theunite outward normal to ∂G \O. We obtain the best possible estimatesof the strong solutions of these problems near a conical boundary point.

A principal new feature here is the consideration of equations with coef-cients whose smoothness isthe minimal possible! Our examples demon-strate this fact. The exact solution estimates near singularities on the bound-ary are obtained under the condition that leading coecients of the equationsatisfy the Dini condition, the lowest coecients can increase: the rate ofthe solution decrease in the neighborhood of a conical point is characterizedby the smallest eigenvalue ϑ of the Laplace - Beltrami operator in a domainΩ on the unit sphere (see (EV P3) 2.4.1).

Let us refer to the problem (QLRP ). We obtain the best possibleestimates of the strong solutions of the problem (QLRP ) near a conical

375


boundary point. Our theorems also show that the quasilinear problem solu-tions have the same regularity (near a conical point) as the linear problemsolutions.

10.1. Linear problem

10.1.1. Formulation of the main result.

Definition 10.1. A strong solution of the problem (LRP ) is a function

u(x) ∈W 2,Nloc (G)∩W 2(Gε)∩C0(G) that for each ε > 0 satises the equation

for almost all x ∈ Gε and the boundary condition in the sense of traces onΓε.

We assume the existence d > 0 such that Gd0 is the convex rotationalcone with the vertex at O and the aperture ω0 ∈ (π2 , π) (see (1.3.13)). Re-garding the equation we assume that the following conditions are satised:

(a) The condition of the uniform ellipticity:

νξ2 ≤ aij(x)ξiξj ≤ µξ2, ∀x ∈ G, ∀ξ ∈ RN ;

ν, µ = const > 0, and aij(0) = δji (the Kronecker symbol) ;

(b) aij ∈ C0(G), ai ∈ Lp(G), p > N, a, f ∈ LN (G); for them theinequalities( N∑

i,j=1

|aij(x)− aij(y)|2) 1

2

≤ A(|x− y|);

|x|( N∑i=1

|ai(x)|2) 1

2

+ |x|2|a(x)| ≤ A(|x|)

hold for x, y ∈ G, where A(r) is a monotonically increasing, nonneg-ative function, continuous at 0 , A(0) = 0;

(c) there exist numbers f1 ≥ 0, g1 ≥ 0, s > 1, β ≥ s − 2, γ0 > tan ω02

such that

|f(x)| ≤ f1|x|β , |g(x)| ≤ g1|x|s−1, γ(x) ≥ γ0,

γ(x) ∈ L∞(∂G) ∩ C1(∂G \ O);(d) a(x) ≤ 0 in G.

We denote M0 = maxx∈G|u(x)| (see Proposition 10.11). Our main results

are the following theorems. Let

λ =2−N +

√(N − 2)2 + 4ϑ2

,

where ϑ is the smallest positive eigenvalue of the problem (EV P3) (see Sub-section 2.4.1).

10.1 Linear problem 377

Theorem 10.2. Let u be a strong solution of the problem (LRP ) and as-sumptions (a) - (d) are satised with A(r) Dini-continuous at zero. Suppose,in addition, that

g(x) ∈ W124−N (∂G), as well as a(x) ∈ W0

4−N (G), γ(x) ∈ W122−N (∂G), if u(0) 6= 0

and there exist numbers

ks =: sup%>0

%−s(‖f‖

W0

4−N (G%0)+ ‖g‖

W1/2

4−N (Γ%0)+

+|u(0)|(‖a‖

W0

4−N (G%0)+ ‖γ‖

W1/2

2−N (Γ%0)

)),

(10.1.1)

κs =: sup%>0

%1−s(‖f‖N,G%

%/2+ |u(0)|‖a‖N,G%

%/2

).

Then there are d ∈ (0, 1) and a constant C > 0 depending only on ν, µ, d, s,

N, λ, γ0, ‖γ‖C1(∂G\O),meas G and on the quantityd∫0

A(r)r dr such that

∀x ∈ Gd0

|u(x)− u(0)| ≤ C(|u|0,G + ‖f‖

W0

4−N (G)+ ‖g‖

W1/2

4−N (∂G)+ g1 +

+ks + κs + |u(0)|(1 + ‖a‖

W0

4−N (G)+ ‖γ‖

W1/2

2−N (∂G)

))×

×

|x|λ, if s > λ,

|x|λ ln3/2(

1|x|

), if s = λ,

|x|s, if s < λ.

(10.1.2)

If, in addition, there is a number

τs =: sup%>0

%−s(‖f + u(0)a‖V 0

p,2p−N (G%%/2

) + ‖g‖V

1− 1p

p,2p−N (Γ%%/2

)+

|u(0)|‖γ‖V

1− 1p

p,p−N (Γ%%/2

)

),(10.1.3)

then

|∇u(x)| ≤ C(|u|0,G + τs + ‖f‖

W0

4−N (G)+ ‖g‖

W1/2

4−N (∂G)+ g1 +

+ks + κs + |u(0)|(1 + ‖a‖

W0

4−N (G)+ ‖γ‖

W1/2

2−N (∂G)

))×

×

|x|λ−1, if s > λ,

|x|λ−1 ln3/2(

1|x|

), if s = λ,

|x|s−1, if s < λ.

(10.1.4)

378 10 Robin boundary value problem in a nonsmooth domain

Theorem 10.3. Let u be a strong solution of the problem (LRP ) andthe assumptions of Theorem 10.2 are satised with A(r) that is a functioncontinuous at zero but not Dini-continuous at zero. Then there are d ∈ (0, 1)and for each ε > 0 a constant Cε > 0 depending only on ε, ν, µ, d, s,N, λ, γ0,‖γ‖C1(∂G\O),meas G and on A(diamG) such that ∀x ∈ Gd0

|u(x)− u(0)| ≤ Cε(|u|0,G + ‖f‖

W0

4−N (G)+ ‖g‖

W1/2

4−N (∂G)+ g1 +

+|u(0)|(1 + ‖a‖

W0

4−N (G)+ ‖γ‖

W1/2

2−N (∂G)

)+ ks + κs

)×

×

|x|λ−ε, if s > λ,

|x|s−ε, if s ≤ λ

(10.1.5)

and

|∇u(x)| ≤ Cε(|u|0,G + ‖f‖

W0

4−N (G)+ ‖g‖

W1/2

4−N (∂G)+ g1 +

+|u(0)|(1 + ‖a‖

W0

4−N (G)+ ‖γ‖

W1/2

2−N (∂G)

)+ ks + κs + τs

)×

×

|x|λ−1−ε, if s > λ,

|x|s−1−ε, if s ≤ λ.

(10.1.6)

Theorem 10.4. Let u be a strong solution of the problem (LRP ) and theassumptions of Theorem 10.2 are satised with s ≥ λ, A(r) ln 1

r ≤ const,r > 0 and A(0) = 0. Then there are d ∈ (0, 1) and the constants C > 0, c > 0depending only on ν, µ, d,N, λ, γ0, ‖γ‖C1(∂G\O),meas G and on A(diamG)such that ∀x ∈ Gd0

(10.1.7) |u(x)− u(0)| ≤ C(|u|0,G + ‖f‖

W0

4−N (G)+ ‖g‖

W1/2

4−N (∂G)+ g1 +

+ |u(0)|(1 + ‖a‖

W0

4−N (G)+ ‖γ‖

W1/2

2−N (∂G)

)+ ks + κs

)|x|λ lnc+1 1

|x|

and

(10.1.8) |∇u(x)| ≤ C(|u|0,G + ‖f‖

W0

4−N (G)+ ‖g‖

W1/2

4−N (∂G)+ g1 +

+ |u(0)|(1 + ‖a‖

W0

4−N (G)+ ‖γ‖

W1/2

2−N (∂G)

)+ ks + κs + τs

)|x|λ−1 lnc+1 1

|x|.

10.1.2. The Lieberman global and local maximum principle.The comparison principle.

Definition 10.5. Let the domain G be at least Lipschitz. A vector−→β

is said to point into G at x0 ∈ ∂G if there is a positive constant t0 such thatx0 + t

−→β ∈ G for 0 < t < t0. A vector eld

−→β , dened on some subset T of

∂G, points into G if−→β (x0) points into G at x0 for all x0 ∈ T.


In this Section we consider the linear elliptic oblique derivative problemL[u] ≡ aij(x)uxixj + ai(x)uxi + a(x)u = f(x);

aij = aji, x ∈ G,B0[u] ≡ βi(x) ∂u∂xi + γ(x)u = g(x), x ∈ ∂G.

(OP)

Definition 10.6. It is said that the operator B0 (or the vector eld−→β )

is oblique at a point x0 ∈ ∂G if there is a coordinate system (x1, x′) =

(x1, . . . , xN ) centered at x0 such that−→β (x0) is parallel to the positive

x1−axis and if there is a Lipschitz function χ dened on some (N − 1)−dimensional ball Bd(x0) such that

G ∩Bd(x0) = x ∈ RN∣∣x1 > χ(x′), |x| < d.

Definition 10.7. It is said that a vector eld−→β = (β1, β′) dened in a

neighborhood of some x0 ∈ ∂G has modulus of obliqueness δ near x0 if, forany ε > 0, there is a coordinate system such that

G ∩Bd(x0) = x ∈ RN∣∣x1 > χ(x′), |x| < d

with a Lipschitz function χ such that

sup |∇χ| sup|β′||β1|≤ δ + ε.

Here β′ = (β2, . . . , βN ).

Definition 10.8. Let x = (x1, x′) be a point in RN and

−→β be a vector

eld such that

<−→β ,−→n >=

N∑i=1

βi cos(−→n , xi) < 0.

We assume that there are positive constants d, h,m1 and ε < 1 such that

Gd0 = x ∈ RN∣∣∣x1 > h|x′|, |x| < d ⊂ G,

|β′| ≤ m1β1 onΓd0, where hm1 ≤ 1− ε.

By Denition 10.7, this means that the modulus of obliqueness at x0 ∈Γd0 is less than 1 .

Remark 10.9. In the denitions above the vector eld−→β can have dis-

continuities and ∂G allows to be piecewise. In this connection see also [224].

Remark 10.10. For the convex rotational cone Gd0 with the vertex atO, the aperture ω0 ∈ (0, π) and the vector

−→β = −−→n on Γd0 we have (see


Lemma 1.10), by (1.3.13) - (1.3.14):

h = cotω0

2, β1 = sin

ω0

2,

β′2 =

N∑i=2

(βi)2 =cot4 ω0

2 sin2 ω02

x21

n∑i=2

x2i = cos2 ω0

2⇒

|β′| = cosω0

2≤ m1 sin

ω0

2⇒ h ≤ m1.

Hence it follows that the modulus of obliqueness at x0 ∈ Γd0 is less than 1, if

h = cotω0

2< 1 ⇒ ω0 >

π

2.

Proposition 10.11. The global maximum principle (see Lemma1.1 [222], Proposition 2.1 [231]; see as well as [230]).

Let G be a bounded domain in RN with the C1− boundary ∂G \ O andGd0 be a convex rotational cone with vertex at O and the aperture ω0 ∈ (π2 , π).Let u(x) be a strong solution of the problem (LRP ). Suppose the operator Lis uniformly elliptic with the ellipticity constants 0 < ν ≤ µ, ai(x), f(x) ∈LN (G), g(x) ∈ L∞(∂G), a(x) ≤ 0 in G, γ(x) ≥ γ0 > 0 on ∂G. Then

maxx∈G|u(x)| ≤ C

(‖g‖L∞(∂G) + ‖f‖LN (G)

),

where C = C(ν, γ0, N, diamG, ‖ai‖LN (G)).

Remark 10.12. We observe that the vector −−→n points into G if G is abounded domain in RN with the C1− boundary ∂G \O and Gd0 be a convexrotational cone Gd0 with vertex at O and the aperture ω0 ∈ (π2 , π).

Proposition 10.13. The strong maximum principle (see Corollary3.2 [231]).

Let G be a bounded domain in RN with the C1− boundary ∂G\O and Gd0be a convex rotational cone with vertex at O and the aperture ω0 ∈ (π2 , π).Suppose u(x) ∈ C0(G) has nonnegative maximum at some x0 ∈ Γd0, andsuppose there is a positive constant d such that u ∈ W 2,N

loc (Gd0). Suppose theoperator L is uniformly elliptic with the ellipticity constants 0 < ν ≤ µ,ai(x), a(x) ∈ LN (Gd0), a(x) ≤ 0 in Gd0, as well γ(x) ∈ L∞(Γd0), γ(x) ≥ γ0 >0 on Γd0. If

L[u] ≥ 0 in Gd0, B[u] ≤ 0 on Γd0,(10.1.9)

then u is constant in Gd0.

Proposition 10.14. The local maximum principle (see Theorem3.3 [222], Theorem 4.3 [231]; see as well as [230]).

Let the hypotheses of Proposition 10.11 hold. In addition, suppose ai(x) ∈Lp(G), p > N ; a(x) ∈ LN (G). Then for any q > 0 and σ ∈ (0, 1), we have


supGσR0

|u(x)| ≤ C

1

measGR0

∫GR0

|u|qdx

1/q

+R(‖f‖LN (GR0 ) + ‖g‖L∞(∂G)

) ,

where C = C(ν, µ, γ0, N, p,R,G, ‖ai‖Lp(G), ‖a‖LN (G)).

Proposition 10.15. The maximum principle.Let G be a bounded domain in RN with the C1− boundary ∂G \ O and

Gd0 be a convex rotational cone with vertex at O and the aperture ω0 ∈ (π2 , π).Let u(x) be a strong solution of the problem

L[u] = f(x) in Gd0,

B[u] = g(x) on Γd0,u = h(x) on Ωd ∪ O

and suppose the operator L is uniformly elliptic with the ellipticity con-stants 0 < ν ≤ µ, ai(x), a(x) ∈ L∞loc(G

d0), a(x) ≤ 0 in Gd0, as well

γ(x) ∈ L∞(Γd0), g(x) ∈ L∞(Γd0), h(x) ∈ L∞(Ωd ∪ O) γ(x) ≥ γ0 > 0 onΓd0. In addition, suppose that the functions w1(x), w2(x) can be found, whichsatises the inequalities:

L[w1] ≤ f(x) in Gd0,

B[w1] ≥ g(x) on Γd0,w1 ≥ h(x) on Ωd ∪ O

and L[w2] ≥ f(x) in Gd0,

B[w2] ≤ g(x) on Γd0,w2 ≤ h(x) on Ωd ∪ O

respectively. Then the solution u satises the inequalities:

w2(x) ≤ u(x) ≤ w1(x) in Gd0.

Proof. Under such circumstance, the function v = u− w1 satises thethree inequalities:

L[v] ≥ 0 in Gd0,B[v] ≤ 0 on Γd0,v ≤ 0 on Ωd ∪ O.

According to the E. Hopf strong maximum principle, Theorem 4.3 , if v is notidentically constant, it can only have a nonnegative maximum at a point onthe boundary. By Proposition 10.13, v cannot have a nonnegative maximumon Γd0 unless it is a constant. Thus v can only have a nonnegative maximumon Ωd ∪ O and therefore we conclude that v ≤ 0 in Gd0. To obtain a lowerbound we consider the function v = w2 − u and reasoning in the same wayas we did for w1.


Proposition 10.16. The comparison principle.Let Gd0 be a convex rotational cone with vertex at O and the aperture ω0 ∈

(π2 , π). Let L be uniformly elliptic in Gd0 with the ellipticity constants 0 <

ν ≤ µ, ai(x), a(x) ∈ L∞loc(Gd0), a(x) ≤ 0 in Gd0. Let γ(x) ∈ L∞(Γd0), γ(x) ≥γ0 > 0 on Γd0. Suppose that v and w are functions in W 2,N

loc (Gd0) ∩ C0(Gd0)satisfying

L[w(x)] ≤ L[v(x)], x ∈ Gd0;B[w(x)] ≥ B[v(x)], x ∈ Γd0;w(x) ≥ v(x), x ∈ Ωd ∪ O.

(10.1.10)

Then v(x) ≤ w(x) in Gd0.

Proof. This Proposition is the direct consequence of Proposition 10.15.

Theorem 10.17. Lp−estimate of solutions of the elliptic obliqueproblem in the smooth domain (see Theorem 15.3 of [4]).

Let G be a domain in RN with a C2 boundary portion T ⊂ ∂G. LetL be uniformly elliptic in G with the ellipticity constants 0 < ν ≤ µ andu ∈W 2,p(G), p > 1 be a strong solution of the problem

L[u] = f in G,

B[u] = g on T

in the weak sense, where

(i) aij(x), ai(x), a(x) ∈ C0(G); γ(x) ∈ C1(T );(ii) f(x) ∈ Lp(G), g(x) ∈W 1− 1

p,p(T ).

Then, for any domain G′ ⊂⊂ G ∪ T we have

‖u‖W 2,p(G′) ≤ C(‖u‖Lp(G) + ‖f‖Lp(G) + ‖g‖

W1− 1

p ,p(T ),

)where the constant C is independent of u and depends only on N, p, ν, µ,T,G′, G, ‖ai(x)‖C0(G), ‖a(x)‖C0(G), ‖γ(x)‖C1(T ) and the moduli of continuity

of the coecients aij(x) on G′.

10.1.3. The barrier function. The preliminary estimate of thesolution modulus. Let Gd0 be a convex rotational cone with a solid angleω0 ∈ (0, π) and the lateral surface Γd0 such that Gd0 ⊂ x1 ≥ 0. Let us denethe linear elliptic operator:

L0 ≡ aij(x)∂2

∂xi∂xj; aij(x) = aji(x), x ∈ Gd0;

νξ2 ≤ aij(x)ξiξj ≤ µξ2,∀x ∈ Gd0,∀ξ ∈ RN ; ν, µ = const > 0

and the boundary operator:

B ≡ ∂

∂−→n+

1|x|γ(x), γ(x) ≥ γ0 > 0, x ∈ Γd0.


Lemma 10.18. (Existence of the barrier function).Fix the numbers γ0 > tan ω0

2 , δ > 0, g1 ≥ 0, d ∈ (0, 1). There exist h > 0depending only on ω0, the number κ0 ∈ (0, γ0 cot ω0

2 −1), a number B > 0 and

a function w(x) ∈ C1(G0) ∩ C2(G0) that depend only on ω0, the ellipticityconstants ν, µ of the operator L0 and the quantities γ0, δ, g1, such that forany κ ∈ (0; min(δ,κ0)) the following hold:

L0[w(x)] ≤ −νh2|x|−1; x ∈ Gd0;(10.1.11)

B[w(x)] ≥ g1|x|δ; x ∈ Γd0 \ O;(10.1.12)

0 ≤ w(x) ≤ C0(κ0, B, ω0)|x|+1; x ∈ Gd0;(10.1.13)

|∇w(x)| ≤ C1(κ0, B, ω0)|x|; x ∈ Gd0.(10.1.14)

Proof. Let (x, y, x′) ∈ RN , where x = x1, y = x2, x′ = (x3, ..., xN ). In

x1 ≥ 0 we consider the cone K with the vertex in O, such that K ⊃ Gd0(we recall that Gd0 ⊂ x1 ≥ 0). Let ∂K be the lateral surface of K and let∂K ∩yOx = Γ± be x = ±hy, where h = cot ω0

2 , 0 < ω0 < π, such that in theinterior of K the inequality x > h|y| holds. We shall consider the function:

w(x; y, x′) ≡ x−1(x2 − h2y2) +Bx+1,

(10.1.15)

with some κ ∈ (0; 1), B > 0.

Let the coecients of the operator L0 be: a2,2 = a, a1,2 = b, a1,1 = c. Thenwe have:

L0w = awyy + 2bwxy + cwxx;(10.1.16)

νη2 ≤ aη21 + 2bη1η2 + cη2

2 ≤ µη2;

(10.1.17)

η2 = η21 + η2

2; ∀η1, η2 ∈ R.

Let us calculate the operator L0 on the function (10.1.15). For t = yx , |t| <

1h

we obtain:

L0w = −h2x−1φ(κ),

where

φ(κ) = 2a− 4bt+ 4btκ − ch−1(1 +B)(κ2 + κ) + ct2κ2 − 3ct2κ + 2ct2 =

= c(t2 − h−2(1 +B))κ2 + (4bt− ch−2(1 +B)− 3ct2)κ + 2(ct2 − 2bt+ a);

c(t2 − h−2(1 +B)) = c

(y2

x2− 1 +B

h2

)≤ −c B

h2< 0.

Because of (10.1.17), we have φ(0) = 2(ct2− 2bt+ a) ≥ 2ν and since φ(κ) isa square function there exists the number κ0 > 0 depending only on ν, µ, h


such that φ(κ) ≥ ν for κ ∈ [0;κ0]. Therefore we obtain (10.1.11).Now, let us notice that

Γ± : x = ±hy, h = cotω0

2, 0 < ω0 < π.(10.1.18)

Then we have

on Γ+ :

x = r cos ω0

2 ,

y = r sin ω02

∠(~n, x) = π

2 + ω02 ,

∠(~n, y) = ω02

on Γ− :

x = r cos ω0

2 ,

y = −r sin ω02

∠(~n, x) = π

2 + ω02 ,

∠(~n, y) = π + ω02

sinω0

2=

1√1 + h2

, cosω0

2=

h√1 + h2

.

Therefore we obtain:

wx = (1 + κ)x(1 +B)− (κ − 1)h2y2x−2 ⇒ wx∣∣Γ±

= [2 +B(1 + κ)]x,

(10.1.19)

wy = −2h2yx−1 ⇒ wy∣∣Γ±

= ∓2hx.

Because of

∂w

∂~n

∣∣∣∣∣Γ±

= wx cos∠(~n, x)∣∣∣∣Γ±

+ wy cos∠(~n, y)∣∣∣∣Γ±

and (10.1.19), we get:

∂w

∂~n

∣∣∣∣∣Γ±

= −r h

(1 + h2)+1

2

[2(1 + h2) +B(1 + κ)].

Hence it follows that:

B[w]∣∣∣Γ±≥ h(

1 + h2)+1

2

r[Bhγ0 −B(1 + κ)− 2(1 + h2)

].

Since h > 1γ0

for κ ≤ κ0 we obtain:

B[w]∣∣∣Γd±≥ h0r(

1 + h2)0+1

2

[B(hγ0 − 1− κ0)− 2(1 + h2)

]≥ g1r

δ, 0 < r < d < 1

if we choose:

κ ≤ δ ⇒ r ≥ rδ;(10.1.20)

B ≥g1

(1 + h2

)0+12

h0+ 2(1 + h2)

· 1hγ0 − 1− κ0


(it should be pointed out that we can choose (if it is necessary) κ0 so smallthat κ0 < hγ0 − 1).

Now we'll show (10.1.13). Let us rewrite the function (10.1.15) in spher-ical coordinates. Recalling that h = cot ω0

2 we obtain:

w(x; y, x′) = (1 +B)(r cosω)1+ − h2r2 sin2 ω(r cosω)−1 =

= r1+ cos−1 ω

(B cos2 ω +

χ(ω)sin2 ω0

2

), ∀ω ∈

[−ω0

2;ω0

2],

where

χ(ω) = sin(ω0

2− ω

)· sin

(ω0

2+ ω

).

We nd χ′(ω) = − sin 2ω and χ′(ω) = 0 for ω = 0. Now we see thatχ′′(0) = −2 cos 0 = −2 < 0. In this way we have

maxω∈[−ω0/2,ω0/2]

χ(ω) = χ(0) = sin2 ω0

2

and therefore:

w(x; y, x′) ≤ r1+ cos−1 ω(B cos2 ω + 1) ≤ r1+ cos+1 ω

(B +

1cos2 ω

)≤

≤ r1+(B +

1cos2 ω

).

Hence (10.1.13) follows. Finally, (10.1.14) follows in virtue of (10.1.19).

Now we can estimate |u(x)| for (LRP ) in the neighborhood of a conicalpoint.

Theorem 10.19. Let u(x) be a strong solution of the problem (LRP )and satisfy assumptions (a)-(d). Then there exist numbers d ∈ (0, 1) andκ > 0 depending only on ν, µ,N,κ0, ω0, f1, β, γ0, s, g1,M0 and the domain Gsuch that

|u(x)− u(0)| ≤ C0|x|+1, x ∈ Gd0,(10.1.21)

where the positive constant C0 depends only on ν, µ,N, f1, g1, β, s, γ0,M0 andthe domain G, and does not depend on u(x).

Proof. Without loss of generality we may suppose that u(0) ≥ 0. Letus take the barrier function w(x) dened by (10.1.15) with κ ∈ (0,κ0) andthe function v(x) = u(x)− u(0). For them we shall show:

L(Aw(x)) ≤ Lv(x), x ∈ Gd0;B[Aw(x)] ≥ B[v(x)], x ∈ Γd0;Aw(x) ≥ v(x), x ∈ Ωd ∪ O.

(10.1.22)


Let us calculate the operator L on these functions. Because of Lemma 10.18and the assumptions (b), (d), we obtain:

Lv(x) = Lu(x)− a(x)u(0) = f(x)− a(x)u(0) ≥ f(x) ≥ −f1rβ ;

Lw(x) ≤ L0w + ai(x)wxi ≤ −νh2r−1 +A(r)r

C1r ≤ −1

2νh2r0−1.

By the continuity of A(r), d > 0 has been chosen so small that

C1A(r) ≤ C1A(d) ≤ 12νh2 for r ≤ d.(10.1.23)

Since 0 < κ < κ0, hence it follows that

L[Aw(x)] ≤ −12Aνh2r0−1 ≤ Lv(x), x ∈ Gd0,

if numbers κ0, A are chosen such that

κ0 ≤ β + 1, A ≥ 2f1

νh2.(10.1.24)

From (10.1.12) we get:

B[Aw]∣∣∣Γd±≥ Ag1r

δ.(10.1.25)

Let us calculate B[v] on Γd±. If A ≥ 1 and 0 < δ ≤ s− 1 then

B[v(x)] =∂u

∂~n+

1|x|γ(x) (u(x)− u(0)) = g(x)− 1

|x|γ(x)u(0)

≤ g(x) ≤ g1rs−1 ≤ g1r

δ ≤ B[Aw], x ∈ Γd±

(10.1.26)

by (10.1.25).Now we compare v(x) and w(x) on Ωd. Since x2 ≥ h2y2 in K, from

(10.1.15) we have

w(x)∣∣∣r=d≥ B|x|1+

∣∣∣r=d

= Bd1+ cos+1 ω0

2.(10.1.27)

On the other hand

v(x)∣∣∣Ωd

= (u(x)− u(0))∣∣∣Ωd≤M0(10.1.28)

and therefore from (10.1.27)-(10.1.28), in virtue of (10.1.20), we obtain:

Aw(x)∣∣∣Ωd≥ ABd1+ cos+1 ω0

2≥ A

g1

(1 + h2

)0+12

h0+ 2(1 + h2)

×

× 1hγ0 − 1− κ0

d1+0h1+0(1 + h2)−1+0

2 ≥

≥M0 ≥ v∣∣∣Ωd,


where A is made great enough to satisfy

A ≥ M0(hγ0 − 1− κ0)

hd1+0

[g1 + 2h0

(1 + h2

) 1−02

] .(10.1.29)

Thus, if we choose the small number d > 0 according to (10.1.23) and largenumbers B > 0, A ≥ 1 according to (10.1.20), (10.1.24), (10.1.29), we providethe validity of (10.1.22).

Therefore the functions v(x), Aw(x) satisfy the comparison principle,Proposition 10.16, and we have:

u(x)− u(0) ≤ Aw(x), x ∈ Gd0.(10.1.30)

Similarly, we derive the estimate

u(x)− u(0) ≥ −Aw(x),

if we consider an auxiliary function v(x) = u(0)− u(x). Theorem is proved,in virtue of (10.1.13).

10.1.4. Global integral weighted estimate.

Theorem 10.20. Let u(x) be a strong solution of the problem (LRP ).Let assumptions (a) - (c) be satised. Suppose, in addition, that g(x) ∈W

12α(∂G), where

4−N < α < 2.(10.1.31)

Then u(x) ∈W 2α(G) and

‖u‖ W 2α(G)

≤ C(‖u‖2,G + ‖f‖

W 0α(G)

+ ‖g‖ W

12

α (∂G)

),(10.1.32)

where the constant C > 0 depends only on ν, µ, α,N, ‖ai‖p,G, i = 1, . . . , N ;‖a‖N,G, γ0, ‖γ‖C1(∂G\O), the moduli of continuity of the coecients aij andthe domain G.

Proof. Since aij(0) = δji , we have

(10.1.33) ∆u(x) = f(x)−(aij(x)− aij(0)

)Diju(x)− ai(x)Diu(x)−

− a(x)u(x) in G.


Integrating by parts, using the Gauss - Ostrogradskiy formula, we showthat ∫

Gε

rα−2u∆udx = −εα−2

∫Ωε

u∂u

∂rdΩε +

∫Γε

rα−2u∂u

∂−→nds−

−∫Gε

⟨∇u,∇

(rα−2u

)⟩dx = −εα−2

∫Ωε

u∂u

∂rdΩε +

+∫Γε

rα−2u∂u

∂−→nds−

∫Gε

rα−2|∇u|2dx+

+ (2− α)∫Gε

rα−4u⟨x,∇u

⟩dx.

Integrating again by parts we obtain∫Gε

rα−4u⟨x,∇u

⟩dx =

12

∫Gε

⟨rα−4x,∇u2

⟩dx−

− 12εα−3

∫Ωε

u2dΩε +12

∫Γε

rα−4u2xi cos(−→n , xi)ds−

− 12

∫Gε

u2N∑i=1

Di(rα−4xi)dx = −12εα−3

∫Ωε

u2dΩε +

+∫Γd

rα−3u2 ∂r

∂−→nds− N + α− 4

2

∫Gε

rα−4u2dx,

because of

N∑i=1

Di(rα−4xi) = Nrα−4 + (α− 4)rα−5N∑i=1

x2i

r= (N + α− 4)rα−4

and (1.3.14) of Lemma 1.10

Thus, multiplying both sides of (10.1.33) by rα−2u(x) and integratingover Gε, because of the boundary condition of the (LRP ), we obtain

∫Gε

rα−2|∇u|2dx+2− α

2εα−3

∫Ωε

u2dΩε +∫Γε

rα−3γ(x)u2ds+(10.1.34)


+2− α

2(N +α− 4)

∫Gε

rα−4u2dx = −εα−2

∫Ωε

u∂u

∂rdΩε +

∫Γε

rα−2g(x)uds+

+2− α

2

∫Γd

rα−3u2 ∂r

∂−→nds+

∫Gε

rα−2u(−f(x) +

(aij(x)− aij(0)

)Diju(x) +

+ ai(x)Diu(x) + a(x)u(x))dx.

Let us estimate the integral over Ωε in the above equality. To end this weconsider the function

M(ε) = maxx∈Ωε

|u(x)|.

Lemma 10.21.

limε→+0

εα−2

∫Ωε

u∂u

∂rdΩε = 0, ∀α ∈ (4−N, 2].(10.1.35)

Proof. We consider the set G2εε . We have Ωε ⊂ ∂G2ε

ε . Now we use theinequality (1.6.1) ∫

Ωε

|w|dΩε ≤ c∫G2εε

(|w|+ |∇w|)dx.

Setting w = u∂u∂r we nd |w|+ |∇w| ≤ c(r2u2xx + |∇u|2 + r−2u2). Therefore

we get ∫Ωε

∣∣∣∣u∂u∂r∣∣∣∣ dΩε ≤ c

∫G2εε

(r2u2xx + |∇u|2 + r−2u2)dx.(10.1.36)

Let us now consider the sets G5ε/2ε/2 and G2ε

ε ⊂ G5ε/2ε/2 and new variables

x′ dened by x = εx′. Then the function w(x′) = u(εx′) satises in G5/21/2 the

problem aij(εx′) ∂2w

∂x′i∂x′j

+ εai(εx′) ∂w∂x′i

+ ε2a(εx′)w = ε2f(εx′),

x′ ∈ G5/21/2,

∂w∂~n′ + 1

|x′|γ(εx′)w = εg(εx′), x′ ∈ Γ5/21/2.

(LPR)′

Because of the interior and near a smooth portion of the boundary L2−estimate,Theorem 10.17, for the equation (LPR)′ solution we have:∫G2

1

(w2x′x′ + |∇′w|2

)dx′ ≤ C1

∫G

5/21/2

(ε4f2 + w2

)dx′ +

+ C2ε2 inf

∫G

5/21/2

(|∇′G|2 + |G|2)dx′,


where inmum is taken over all G such that G∣∣Γ

5/21/2

= g and the constants

C1, C2 > 0 depend only on ν, µ, maxx′,y′∈G5/2

1/2

A(|x′ − y′|), ‖γ‖C1(Γ

5/21/2

)and the

domain G.

Returning to the variable x, we obtain

(10.1.37)∫G2εε

(r2|D2u|2 + |∇u|2 + r−2u2

)dx ≤ c

∫G

5ε/2ε/2

(r2f2 + r−2u2

)dx+

+ C2 inf∫

G5ε/2ε/2

(r2|∇G|2 + |G|2)dx.

By the Mean Value Theorem 1.58 with regard to u ∈ C0(G), we have

∫G

5ε/2ε/2

r−2u2dx =

5ε/2∫ε/2

rN−3

∫Ω

u2(r, ω)dΩdr

≤ 2ε(θ1ε)N−3

∫Ω

u2(θ1ε, ω)dΩ

≤ 2εN−2θN−31 M2(θ1ε) meas Ω

(10.1.38)

for some 12 < θ1 <

52 . From (10.1.36), (10.1.37) and (10.1.38) we obtain

(10.1.39)∫Ωε

∣∣∣∣u∂u∂r∣∣∣∣ dΩε ≤ c1ε

N−2M2(ε) + c2

∫G

5ε/2ε/2

r2f2dx+

+ C2 inf∫

G5ε/2ε/2

(r2|∇G|2 + |G|2)dx ≤ c1εN−2M2(ε) +

+ c3ε2−α

∫G

5ε/2ε/2

rαf2 + rα|∇G|2 + rα−2|G|2

dx, ∀α ≤ 2.

Also we have

(10.1.40) εα−2

∫Ωε

∣∣∣∣u∂u∂r∣∣∣∣ dΩε ≤ c1ε

α+N−4M2(ε) +

+ c3

∫G

5ε/2ε/2

rαf2 + rα|∇G|2 + rα−2|G|2

dx, ∀α ≤ 2.


By the hypotheses of our Theorem, we have f ∈ W

0α(G), g(x) ∈

W12α(∂G),

hence

limε→+0

∫G

5ε/2ε/2

rαf2 + rα|∇G|2 + rα−2|G|2

dx = 0.(10.1.41)

Because u ∈ C0(G) and 4−N < α ≤ 2, from (10.1.40) - (10.1.41), we deducethe validity of the statement (10.1.35) of our Lemma.

Now we estimate each integral from the right hand side of (10.1.34):

1) ∫Γd

rα−3u2 ∂r

∂−→nds ≤ dα−3

∫Γd

u2ds since r ≥ d, α ≤ 2;

hence, applying (1.6.2), we get∫Γd

rα−3u2 ∂r

∂−→nds ≤ δdα−3

∫Gd

|∇u|2dx+ cδ

∫Gd

|u|2dx; ∀δ > 0;(10.1.42)

2) using the Cauchy inequality we obtain∫Γε

rα−2|u||g|ds =∫Γε

(rα−1

21√γ(x)

|g|)(rα−3

2

√γ(x)|u|

)ds ≤

≤ δ

2

∫Γε

rα−3γ(x)u2ds+1

2δγ0

∫∂G

rα−1g2ds; ∀δ > 0;(10.1.43)

3) we get, by the Cauchy inequality,∫Gε

rα−2u(x)f(x)dx =∫Gε

(rα/2−2u(x))(rα/2f(x))dx

≤ δ

2

∫Gε

rα−4u2dx+12δ

∫Gε

rαf2(x)dx, ∀δ > 0;(10.1.44)

4) applying assumption b) together with the Cauchy inequality we obtain

(10.1.45) rα−2u((aij(x)− aij(0)

)Diju(x) + ai(x)Diu(x) + a(x)u(x)

)≤ A(r)

((r

α2 |D2u|)(r

α2−2u) + rα−2|∇u|(r−1u) + rα−4u2

)≤ A(r)

(rα|D2u|2 + rα−2|∇u|2 + 2rα−4u2

).


Finally, by (10.1.42) - (10.1.45), from (10.1.34) we obtain

(10.1.46)∫Gε

rα−2|∇u|2dx+2− α

2(N + α− 4)

∫Gε

rα−4u2dx+

+12

∫Γε

rα−3γ(x)u2ds ≤ εα−2

∫Ωε

u∂u

∂rdΩε + δ

∫Gε

rα−4|u|2dx+

+2− α

2c

∫Gd

(|∇u|2 + |u|2

)dx+ cδ

∫G

rαf2(x)dx+1

2γ0

∫∂G

rα−1g2ds+

+∫Gε

A(|x|)(rα|D2u|2 + rα−2|∇u|2 + 2rα−4u2

)dx

for ∀δ > 0.

Let us now estimate the last integral in (10.1.46). Due to assumption b)we have

∀δ > 0 ∃d > 0 such that A(r) < δ for all 0 < r < d.(10.1.47)

Let 2ε < d. From (10.1.37), (10.1.38) it follows that

(10.1.48)∫G2εε

rα|D2u|2dx ≤ cεα−2

∫G2εε

r2|D2u|2dx ≤ cεα+N−4M2(ε) +

+ c

∫G

5ε/2ε/2

(rαf2 + rα|∇G|2 + rα−2|G|2

)dx,

and consequently∫Gε

A(r)rα|D2u|2dx =∫G2εε

A(r)rα|D2u|2dx+∫Gd2ε

A(r)rα|D2u|2dx+

+∫Gd

A(r)rα|D2u|2dx ≤ cA(2ε)∫

G5ε/2ε/2

(rαf2 + rα|∇G|2 + rα−2|G|2

)dx+

+δ∫G2dε

(rαf2(x) + rα|∇G|2 + rα−2|G|2 + rα−4u2

)dx+

+cA(2ε)εα+N−4 + c maxr∈[d,diamG]

A(r)∫Gd

|D2u|2dx(10.1.49)

for ∀δ > 0 and 0 < ε < d/2. Here c does not depend on ε.


Applying all these estimates to the inequality (10.1.46) we obtain

(10.1.50)∫Gε

rα−2|∇u|2dx+2− α

2(N + α− 4)

∫Gε

rα−4u2dx ≤

≤ cA(2ε)(εα+N−4 +

∫G

5ε/2ε/2

(rαf2 + rα|∇G|2 + rα−2|G|2

)dx

)+

+ δ

∫Gε

(rα−2|∇u|2 + rα−4u2)dx+ c

∫Gd

(|D2u|2 + |∇u|2 + u2

)dx+

+ c

∫G

(rαf2 + rα|∇G|2 + rα−2|G|2

)dx+ c

∫∂G

rα−1g2ds+ εα−2

∫Ωε

u∂u

∂rdΩε

for ∀δ > 0 and 0 < ε < d/2.Finally, we apply L2− estimate, Theorem 10.17, to the solution u of the

(LRP ) in Gd∫Gd

(|D2u|2 + |∇u|2

)dx ≤ c

∫Gd/2

(u2 + f2

)dx+ c‖g‖2

W 1/2,2(Γd/2).(10.1.51)

Now we use the inequality∫Γd0

rα−1g2(x)ds ≤ C‖g‖2W

1/2

α (Γd0)

(10.1.52)

(see Lemma 1.40). Then from (10.1.50), (10.1.51) and (10.1.52) we obtain

(10.1.53)∫Gε

rα−2|∇u|2dx+2− α

2(N + α− 4)

∫Gε

rα−4u2dx ≤

≤ εα−2

∫Ωε

u∂u

∂rdΩε+cA(2ε)

(εα+N−4+

∫G

5ε/2ε/2

(rαf2+rα|∇G|2+rα−2|G|2

)dx

)

+ δ

∫Gε

(rα−2|∇u|2 + rα−4u2)dx+ c(‖u‖22,G + ‖f‖2

W 0α(G)

+ ‖g‖2W

12

α (∂G)

)

for ∀δ > 0 and 0 < ε < d/2.


Now, since 4 − N < α < 2, we can choose δ = min(

12 ; (2−α)(N+α−4)

4

).

Then

(10.1.54) cα,N

∫Gε

(rα−2|∇u|2 + rα−4u2)dx ≤ εα−2

∫Ωε

u∂u

∂rdΩε +

+ cA(2ε)( ∫G

5ε/2ε/2

(rαf2 + rα|∇G|2 + rα−2|G|2

)dx+ εα+N−4

)+

+ c(‖u‖22,G + ‖f‖2

W 0α(G)

+ ‖g‖2W

12

α (∂G)

)We observe that the constant c in (10.1.54) does not depend on ε. Thereforewe can perform the passage to the limit as ε → +0 by the Fatou theorem:indeed, we apply Lemma 10.21, (10.1.41) and use the continuity of A(r) andA(0) = 0. Thus, we get

(10.1.55)∫G

(rα−2|∇u|2 + rα−4u2)dx ≤ c(‖u‖22,G + ‖f‖2

W 0α(G)

+

+ ‖g‖2W

12

α (∂G)

).

Now from (10.1.37) we obtain

(10.1.56)∫G2εε

rα|D2u|2dx ≤ c∫

G5ε/2ε/2

(rαf2 + rα−4u2

)dx+

+ C2 inf∫

G5ε/2ε/2

(rα|∇G|2 + rα−2|G|2)dx.

Let ε = 2−kd, (k = 0, 1, 2, ...) and let us sum the obtained inequalities overall k. Then we have:

(10.1.57)∫Gd0

(rαu2

xx + rα−2|∇u|2)dx ≤ C3

∫G2d

0

rα−4u2dx+

+ C4‖g‖2W

1/2

α (Γ2d0 )

+ C5‖f‖2W

0

α(G2d0 ).

From (10.1.55), (10.1.57) and (10.1.51) we deduce the validity of our Theo-rem.

Theorem 10.22. Let u(x) be a strong solution of the problem (LRP ).Let N ≥ 3 and assumptions (a) - (c) be satised. Suppose, in addition, that

g(x) ∈ W122 (∂G). Then u(x) ∈ W2

2(G) and

‖u‖ W 2

2 (G)≤ C

(|u|0,G + ‖f‖

W 02 (G)

+ ‖g‖ W

12

2 (∂G)

),(10.1.58)


where the constant C > 0 depends only on ν, µ,N, ‖ai‖p,G, i = 1, . . . , N ;‖a‖N,G, γ0, ‖γ‖C1(∂G\O), the moduli of continuity of the coecients aij andthe domain G.

Proof. We repeat verbatim the proof of Theorem 10.20 with α = 2.Then from (10.1.53) and (10.1.40) we have∫

Gε

|∇u|2dx ≤ c4

∫G

5ε/2ε/2

r2f2 + r2|∇G|2 + |G|2

dx+

+ cA(2ε)εN−2 + δ1

∫Gε

|∇u|2dx+

+ δ2

∫Gε

r−2u2dx+ c1εN−2M2(ε) +

+ c(‖u‖22,G + ‖f‖2

W 02 (G)

+ ‖g‖2W

12

2 (∂G)

)(10.1.59)

for any δ1 > 0, δ2 > 0 and 0 < ε < d/2. Now, since N ≥ 3 we can estimate∫Gd0

r−2u2dx ≤ |u|20,GmeasΩ∫ d

0rN−3dr <

dN−2

N − 2measΩ|u|20,G.

Therefore, for δ1 = 12 it follows from (10.1.59) that

12

∫Gε

|∇u|2dx ≤ c4

∫G

5ε/2ε/2

r2f2 + r2|∇G|2 + |G|2

dx+

+ c1εN−2M2(ε) + cA(2ε)εN−2 +

+ c(|u|20,G + ‖f‖2

W 02 (G)

+ ‖g‖2W

12

2 (∂G)

)(10.1.60)

for any ε ∈ (0, d/2). Performing the passage to the limit as ε → +0 by theFatou theorem we deduce from this the validity of our Theorem.

Now we consider α = 4 − N, N ≥ 2. In order to do this, we turn toTheorem 10.19, based on Lemma 10.18 about the existence of the barrierfunction.

Theorem 10.23. Let u be a strong solution of the problem (LRP ). Let

assumptions (a) - (d) be satised. Suppose, in addition, that g(x) ∈ W124−N (∂G)

and a(x) ∈ W04−N (G), γ(x) ∈ W

122−N (∂G), if u(0) 6= 0.


Then(u(x)− u(0)

)∈ W2

4−N (G) and

(10.1.61)

∫∂G

r1−Nγ(x)|u(x)− u(0)|2ds

12

+ ‖u(x)− u(0)‖ W

2

4−N (G)≤

≤ C(|u|0,G + ‖f‖

W0

4−N (G)+ |u(0)| ·

(1 + ‖γ‖

W1/2

2−N (∂G)+

+ ‖a‖ W

0

4−N (G)

)+ ‖g‖

W1/2

4−N (∂G)

),

where the constant C > 0 depends only on ν, µ,N, ‖ai‖p,G, i = 1, . . . , N ;‖a‖N,G, γ0, ‖γ‖C1(∂G\O), the moduli of continuity of the coecients aij andthe domain G.

Proof. Setting v(x) = u(x)− u(0) we have v ∈ C0(G), v(0) = 0 and vis a strong solution of the problem

(LRP )0

aij(x)vxixj + ai(x)vxi + a(x)v = f(x)− a(x)u(0) ≡

≡ f0(x), x ∈ G;∂v∂~n + 1

|x|γ(x)v = g(x)− 1|x|γ(x)u(0) ≡ g0(x), x ∈ ∂G \ O.

We repeat verbatim the arguments of the proof of Theorem 10.20with α = 4−N . Then from (10.1.34) with regard to a(x) ≤ 0 we have

(10.1.62)∫Gε

r2−N |∇v|2dx+∫Γε

r1−Nγ(x)v2ds ≤ ε2−N

∣∣∣∣∣∣∫Ωε

v∂v

∂rdΩε

∣∣∣∣∣∣+

+∫Γε

r2−Ng(x)vds+ |u(0)|∫Γε

r1−Nγ(x)|v|ds+N − 2

2

∫Γd

r1−Nv2 ∂r

∂−→nds+

+∫Gε

r2−Nv(−f(x)+u(0)a(x)+

(aij(x)− aij(0)

)Dijv(x)+ai(x)Div(x)

)dx.

We estimate each term of (10.1.62). First (10.1.40) has the form

(10.1.63) ε2−N∫Ωε

∣∣∣∣v∂v∂r∣∣∣∣ dΩε ≤ c1 max

x∈Ωε|u(x)− u(0)|2 +

+ c3

∫G

5ε/2ε/2

r4−Nf2 + r4−N |∇G|2 + r2−N |G|2

dx+

+ c4u2(0)

∫G

5ε/2ε/2

r4−Na2(x) + r2−N |∇γ|2 + r−Nγ2(x)

dx.


By the hypotheses of our Theorem, we get

limε→+0

ε2−N∫Ωε

v∂v

∂rdΩε = 0.(10.1.64)

Using the Cauchy inequality we get

(10.1.65) |u(0)|∫Γε

r1−Nγ(x)|v|ds ≤ δ

2

∫Γε

r1−Nγ(x)|v|2ds+

+12δ|u(0)|2

∫Γε

r1−Nγ(x)ds, ∀δ > 0.

Since γ(x) ≥ γ0 > 0 and because of (10.1.52),

∫Γε

r1−Nγ(x)ds ≤ 1γ0

∫Γε

r1−Nγ2(x)ds ≤ c

γ0‖γ‖2

W1/2

2−N (∂G).(10.1.66)

From (10.1.62), (10.1.65) (with δ = 1) and (10.1.66) it follows that

(10.1.67)∫Gε

r2−N |∇v|2dx+12

∫Γε

r1−Nγ(x)v2ds ≤ ε2−N

∣∣∣∣∣∣∫Ωε

v∂v

∂rdΩε

∣∣∣∣∣∣+

+∫Γε

r2−Ng(x)vds+ |u(0)|2 c

2γ0‖γ‖2

W1/2

2−N (∂G)+N − 2

2

∫Γd

r1−Nv2 ∂r

∂−→nds+

+∫Gε

r2−Nv(−f(x)+u(0)a(x)+

(aij(x)− aij(0)

)Dijv(x)+ai(x)Div(x)

)dx.

Taking into account the estimates (10.1.42), (10.1.43) (with δ = 1), (10.1.44),(10.1.45), (10.1.49), (10.1.51), (10.1.52) we obtain

12

∫Γε

r1−Nγ(x)|v|2ds+∫Gε

r2−N |∇v|2dx ≤ ε2−N

∣∣∣∣∣∣∫Ωε

v∂v

∂rdΩε

∣∣∣∣∣∣+(10.1.68)


+ c|u(0)|2(‖γ‖2

W1/2

2−N (∂G)+ ‖a‖2

W0

4−N (G)

)+

+ cA(2ε)(

1 +∫

G5ε/2ε/2

(r4−Nf2 +

+ r4−N |∇G|2 + r2−N |G|2)dx

)+

+ δ1

∫Gε

r2−N |∇v|2dx+ δ2

∫Gε

r−Nv2dx+

+ c(‖v‖22,G + ‖f‖2

W0

4−N (G)+ ‖g‖2

W124−N (∂G)

)for any δ1, δ2 > 0 and 0 < ε < d/2.

Finally, we apply Theorem 10.19. The assumptions of our Theoremguarantee the fullment of all suppositions of this Theorem. Therefore wecan estimate∫

Gd0

r−Nv2dx ≤ C20 ·measΩ

∫ d

0r2+1dr ≤ cd2+2, κ > 0; =⇒

∫G

r−Nv2dx <∞.(10.1.69)

Now choosing δ1 = 12 , because of (10.1.69), we may perform the passage to

the limit as ε → +0 by the Fatou Theorem in (10.1.68). By (10.1.64), weget

(10.1.70)∫G

r2−N |∇v|2dx+∫∂G

r1−Nγ(x)v2ds ≤

≤ δ∫G

r−Nu2dx+ c1

(‖u‖22,G + ‖f‖2

W0

4−N (G)+

+ ‖g‖2W

124−N (∂G)

)+ c2u

2(0)(‖a‖2

W0

4−N (G)+ ‖γ‖2

W1/2

2−N (∂G)

)for any δ > 0.

From (10.1.69) - (10.1.70) it follows that v ∈ W12−N (G); moreover,

v(0) = 0. This makes possible to apply the Hardy -Friedrichs - Wirtingerinequality (2.5.13). Therefore choosing appropriatly small δ > 0 we deduce


from (10.1.70) the inequality∫G

(r2−N |∇v|2 + r−Nv2

)dx+ +

∫∂G

r1−Nγ(x)v2ds ≤ c1

(‖v‖22,G +

(10.1.71)

+‖f‖2W

0

4−N (G)+ ‖g‖2

W124−N (∂G)

)+ c2u

2(0)(‖a‖2

W0

4−N (G)+ ‖γ‖2

W1/2

2−N (∂G)

).

Finally, putting in (10.1.57) α = 4 − N and replacing f by f0 and g by g0

from the (LRP )0 we obtain∫Gd0

r4−Nv2xxdx ≤ C3

∫G2d

0

r−Nv2dx+ C4‖g‖2W

1/2

α (Γ2d0 )

+

+ C5‖f‖2W

0

α(G2d0 )

+ C6u2(0)

(‖a‖2

W0

4−N (G)+

+ ‖γ‖2W

1/2

2−N (∂G)

).

(10.1.72)

From (10.1.71), (10.1.72) follows the desired estimate (10.1.61).

Theorem 10.24. Let u be a strong solution of the problem (LRP) andλ be as above (see (2.4.8)) . Let assumptions (a) - (d) with β > λ − 2 be

satised. Suppose, in addition, that g(x) ∈ W12α(∂G), where

4−N − 2λ < α < 4−N

and a(x) ∈ W0α(G), γ(x) ∈ W

12α−2(∂G), if u(0) 6= 0.

Then(u(x)− u(0)

)∈ W2

α(G) and

(10.1.73)

∫∂G

rα−3γ(x)(u(x)− u(0)

)2ds

12

+ ‖u(x)− u(0)‖ W

2

α(G)≤

≤ C(|u|0,G + ‖f‖

W0

α(G)+ ‖g‖

W1/2

α (∂G)+

+ |u(0)|(

1 + ‖a‖ W

0

α(G)+ ‖γ‖

W1/2

α−2(G)

)),

where the constant C > 0 depends only on ν, µ, λ, α,N, ‖ai‖p,G, i = 1, . . . , N ;‖a‖N,G, γ0, ‖γ‖C1(∂G\O), the moduli of continuity of the coecients aij andthe domain G.

Proof. We consider the function v(x) = u(x)−u(0) which satises theproblem (LRP )0 and multiply both sides of the equation of the (LRP )0 by


rα−2ε v(x) and integrate over G; we obtain:∫

G

rα−2ε v4vdx =

∫G

rα−2ε v

f(x)− a(x)u(0)−

−〈(aij(x)− aij(0))vxixj + ai(x)vxi + a(x)v〉, ∀ε > 0.

(10.1.74)

We transform the integral from the left in (10.1.74) by the Gauss-Ostrogradskiyformula:

(10.1.75)∫G

rα−2ε v4vdx =

∫∂G

rα−2ε v

∂v

∂~nds−

∫G


+2− α

2

∫G

rα−3ε

∂v2

∂xi

∂rε∂xi

dx.

Because of the boundary condition of the (LRP )0, we obtain

(10.1.76)∫G

rα−2ε v4vdx =

2− α2

∫G

rα−3ε

∂v2

∂xi

∂rε∂xi

dx−

−∫G


∫∂G

rα−2ε v

g(x)− 1

rγ(x)u(0)− 1

rγ(x)v

ds, ∀ε > 0.

Now we transform the second integral from the right in (10.1.76). For thiswe use the Gauss-Ostrogradskiy formula once more:

(10.1.77)∫G

rα−3ε

∂rε∂xi

∂v2

∂xidx =

∫∂G

rα−3ε v2 ∂rε

∂xicos (~n, xi) ds−

−∫G

v2 ∂

∂xi

(rα−3ε

∂rε∂xi

)dx.

Because of ∂rε∂x1

= x1+εrε

, ∂rε∂xi= xi

rε(i ≥ 2), ∂G = Γd0 ∪ Γd and by (1.3.14), we

obtain:

(10.1.78)∫∂G

rα−3ε v2 ∂rε

∂xicos (~n, xi) ds = −ε sin

ω0

2

∫Γd0

rα−4ε v2ds+

+∫Γd

rα−3ε v2∂rε

∂~nds.

However, by the fourth property of rε, we have:

−∫G

v2 ∂

∂xi

(rα−3ε

∂rε∂xi

)dx = (4−N − α)

∫G

rα−4ε v2dx.(10.1.79)


From (10.1.77)-(10.1.79) it follows that

(10.1.80)2− α

2

∫G

rα−3ε

∂v2

∂xi

∂rε∂xi

dx =2− α

2

∫Γd

rα−3ε v2∂rε

∂~nds−

− 2− α2

ε sinω0

2

∫Γd0

rα−4ε v2ds+

(2− α) (4−N − α)2

∫G

rα−4ε v2dx, ∀ε > 0.

From (10.1.74), (10.1.75) and (10.1.80) with regard to a(x) ≤ 0 we obtainthe following equality:

(10.1.81)∫G

rα−2ε |∇v|2dx+ ε

2− α2

sinω0

2

∫Γd0

rα−4ε v2ds+

+∫∂G

rα−2ε

1rγ(x)v2ds =

(2− α) (4− α−N)2

∫G

rα−4ε v2dx+

+∫G

rα−2ε v

(aij(x)− aij(0)

)vxixj + ai(x)vxi + a(x)u(0)− f(x)

dx+

+2− α

2

∫Γd

rα−3ε v2∂rε

∂~nds+

∫∂G

rα−2ε vg(x)ds−u(0)

∫∂G

rα−2ε

1rγ(x)vds, ∀ε > 0.

Now we estimate the integral over Γd. Because of on Γd : rε ≥ hr ≥hd⇒ (α− 3) ln rε ≤ (α− 3) ln(hd), since α < 2, we have rα−3

ε |Γd ≤ (hd)α−3

and therefore:2− α

2

∫Γd

rα−3ε v2∂rε

∂~nds ≤ 2− α

2(hd)α−3

∫Γd

v2ds.(10.1.82)

By (1.6.2), we obtain:∫Γd

v2ds ≤ Cδ∫Gd

v2dx+ δ

∫Gd

|∇v|2dx, ∀δ > 0.(10.1.83)

By the Cauchy inequality,

vg =(r

12

1√γ(x)

|g|)(r−

12

√γ(x)|v|

)≤ δ

2r−1γ(x)v2 +

12δγ0

rg2, ∀δ > 0;

taking into account property 1) of rε we obtain

(10.1.84)∫∂G

rα−2ε |v||g|dσ ≤ δ

2

∫∂G

rα−2ε

1rγ(x)v2ds+

+1

2δγ0hα−2

∫∂G

rα−1g2ds, ∀δ > 0.


From assumptions (a)− (b) we have

maxΓd0

|aij(x)− aij(0)| ≤ A(d), and maxΓd|aij(x)− aij(0)| ≤ 1 + µ.

From this, by the Cauchy inequality and assumption (b), we obtain:

(10.1.85)∫Gd0

rα−2ε v

(aij(x)− aij(0)

)vxixj + |x|ai(x)r−1vxi

dx ≤

≤ A(d)C1(N)∫Gd0

(r2rα−2

ε v2xx + rα−2

ε |∇v|2 + r−2rα−2ε v2

)dx.

Similarly, we have:

(10.1.86)∫Gd

rα−2ε v

(aij(x)− aij(0))vxixj + ai(x)vxi

dx ≤

≤ C2(N, diamG)(hd)α−2

∫Gd

(v2xx + |∇v|2

)dx.

Further, from the Cauchy inequality we obtain:

(10.1.87)∫G

rα−2ε vf(x)dx ≤ δ

2

∫G

r−2rα−2ε v2dx+

+12δ

∫G

r2rα−2ε f2(x)dx, ∀δ > 0;

(10.1.88) |u(0)|∫∂G

rα−2ε r−1γ(x)|v|ds ≤ δ

2

∫∂G

rα−2ε r−1γ(x)|v|2ds+

+12δ|u(0)|2

∫∂G

rα−2ε r−1γ(x)ds, ∀δ > 0;

(10.1.89)∫G

rα−2ε v(x)u(0)a(x)dx ≤ δ

2

∫G


+12δ|u(0)|2

∫G

r2rα−2ε a2(x)dx, ∀δ > 0.


As a result from (10.1.81)- (10.1.89) we obtain with ∀δ > 0:

∫G


∫∂G

r−1rα−2ε γ(x)v2ds ≤

≤ (2− α) (4− α−N)2

∫G

rα−4ε v2dx+

+ δ

∫∂G

r−1rα−2ε γ(x)v2ds+

+1

2γ0hα−2

∫∂G

rα−1g2ds+

+A(d)C3 (δ, λ,N)∫Gd0

(r2rα−2

ε v2xx + rα−2

ε |∇v|2 +

+ r−2rα−2ε v2

)dx+ Cδ

∫G

r2rα−2ε f2(x)dx+

+ δ

∫G

(rα−2ε |∇v|2 + r−2rα−2

ε v2)dx+

+ C4 (α, h, d, diamG)∫Gd

(v2xx + |∇v|2

)dx+

+ Cδ|u(0)|2(∫∂G

rα−2ε r−1γ(x)ds+

+∫G

r2rα−2ε a2(x)dx

).

(10.1.90)

Now we consider two sets G2ρρ/4 and Gρρ/2 ⊂ G2ρ

ρ/4, ρ > 0. We make thecoordinate transformation x = ρx′. The function z(x′) = v(ρx′) in G2

1/4

satises the equation

aij(ρx′)zx′ix′j + ρai(ρx′)zx′i + ρ2a(ρx′)z = ρ2f(ρx′),

x′ ∈ G21/4

∂z∂~n′ + 1

|x′|γ(ρx′)z = ρg(ρx′), x′ ∈ Γ21/4.

(LRP )′′


Because of interior and near a smooth portion of the boundary L2−estimates,Theorem 10.17, for the equation of the (LRP )′′ solution we have:∫

G11/2

(z2x′x′ + |∇′z|2

)dx′ ≤ C5

∫G2

1/4

(ρ4f2 + z2

)dx′ +

+ C6%2 inf

∫G2

1/4

(|∇′G|2 + |G|2)dx′,

where inmum is taken over all G such that G∣∣Γ2

1/4

= g and the constants

C5, C6 > 0 depend only on ν, µ, maxx′∈G2

1/4

A(|x′|), ‖γ‖C1(Γ21/4

) and the domain

G. Multiplying both sides of this inequality by (%+ ε)α−2 and returning tothe variable x, we obtain:∫Gρρ/2

⟨%2(%+ ε)α−2v2

xx + (%+ ε)α−2|∇v|2⟩dx ≤ C5

∫G2ρρ/4

⟨(%2(%+ ε)α−2f2 +

+ %−2(%+ ε)α−2v2

⟩dx+C6(%+ ε)α−2 inf

∫G2%%/4

(%2|∇G|2 + |G|2)dx, ∀ε > 0.

Now, in the domain Gρρ/2, we have:

%

2< r < %⇒ r < % < 2r ⇒ %+ ε < 2r + ε ≤ 3

hrε by the property 1) of rε

⇒ (%+ ε)α−2 ≥ (3h−1)α−2rα−2ε , since α < 2.

Similarly in the domain G2ρρ/4 we have:

%

4< r < 2%⇒ 1

2r < % < 4r ⇒ %+ ε ≥ 1

2r + ε >

12

(r + ε) ≥ 12rε ⇒

(%+ ε)α−2 ≤ 22−αrα−2ε , since α < 2.

Thus we obtain∫Gρρ/2

(r2rα−2ε v2

xx + rα−2ε |∇v|2)dx ≤

≤ C7(h, α)C5

∫G2ρρ/4

(rαf2 + r−2rα−2ε v2)dx+

+ C6 inf∫

G2%%/4

(rα|∇G|2 + rα−2|G|2)dx, ∀ε > 0.


Let ρ = 2−kd, (k = 0, 1, 2, ...) and let us sum the obtained inequalities overall k. Then we have:

(10.1.91)∫Gd0

(r2rα−2

ε v2xx + rα−2

ε |∇v|2)dx ≤ C8

∫G2d

0


+ C9‖g‖2W

1/2α (Γ2d

0 )+ C10‖f‖2

W 0α(G2d

0 ), ∀ε > 0.

Finally, we use once more the interior and near a smooth portion of theboundary L2− estimate for the equation (L) solution. We obtain analogous-ly:

(10.1.92)∫Gd

(v2xx + |∇v|2

)dx ≤ C11

∫Gd/2

(f2 + v2

)dx+

+ C12‖g‖2W

1/2

(Γd/2)

≤ C(d, diamG)(‖f‖2

W0

α

(Gd/2) + ‖g‖2W

1/2

α (Γd/2)

)+

+ C11

∫Gd/2

v2 dx.

Since α < 2 and by the property 1) of rε we have rα−2ε ≤ rα−2 and therefore

with regard to (10.1.52) we get

(10.1.93)∫∂G

rα−2ε r−1γ(x)ds ≤

∫∂G

rα−3γ(x)ds ≤ 1γ0

∫∂G

rα−3γ2(x)ds ≤

≤ c

γ0‖γ‖2

W1/2

α−2(G);

From (10.1.90)-(10.1.93) we obtain:

(10.1.94)∫∂G


∫G

(r2rα−2ε v2

xx + rα−2ε |∇v|2)dx ≤

≤ (2− α) (4− α−N)2

∫G

rα−4ε v2dx+ δ

∫∂G


+(A(d) + δ

)C13 (d, λ,N)

∫G

(rα−2ε |∇v|2 + r−2rα−2

ε v2)dx+

+ C14(α, d, h, δ, γ0diamG)(‖v‖22,G + ‖f‖2

W0

α(G)+ ‖g‖2

W1/2

α (∂G)

+

+ |u(0)|2(‖a‖2

W0

α(G)+ ‖γ‖2

W1/2

α−2(G)

)), ∀ε > 0.

By our assumptions of the Theorem, in virtue of the obvious embeddingW

0α(G) →

W0β(G), W1/2

α (∂G) → W

1/2β (∂G), ∀β ≥ α,


we obtain

g(x) ∈ W124−N (∂G), a(x) ∈ W0

4−N (G), γ(x) ∈ W122−N (∂G),

and therefore by Theorem 10.23 v(x) ∈ W

24−N (G). But then we can apply

Theorem 2.19 and according to (2.4.9) we have

∫Ω

v2(r, ω)dΩ ≤ 1λ(λ+N − 2)

∫Ω

|∇ωv(r, ω)|2dΩ +∫∂Ω

γ(r, ω)v2ds,

for a.e. r ∈ (0, d).

Multiplying both sides of this inequality by (%+ ε)α−2rN−3 and integratingover r ∈ (%2 , %) we obtain

∫G%%/2

(%+ ε)α−2r−2v2dx ≤ 1λ(λ+N − 2)

∫G%%/2

(%+ ε)α−2|∇v|2dx+

+∫

Γ%%/2

r−1(%+ ε)α−2γ(x)v2ds

, ∀ε > 0

or since %+ ε ∼ rε

∫G%%/2

rα−2ε r−2v2dx ≤ 1

λ(λ+N − 2)

∫G%%/2


+∫

Γ%%/2

r−1rα−2ε γ(x)v2ds

, ∀ε > 0.

Letting ρ = 2−kd, (k = 0, 1, 2, ...) and summing the obtained inequalitiesover all k we get:

(10.1.95)∫Gd0

rα−2ε r−2v2dx ≤ 1

λ(λ+N − 2)

∫Gd0


+∫Γd0


, ∀ε > 0.


Therefore from (10.1.94), (10.1.95) it follows that

(10.1.96)∫∂G


∫G

(r2rα−2ε v2

xx + rα−2ε |∇v|2)dx ≤

≤ (2− α) (4− α−N)2

∫G

rα−4ε v2dx+

+(A(d) + δ

)C15 (d, λ,N)

∫G


∫∂G


+

+ C14(α, d, h, δ, γ0, diamG)(‖v‖22,G + ‖f‖2

W0

α(G)+ ‖g‖2

W1/2

α (∂G)

+

+ |u(0)|2(‖a‖2

W0

α(G)+ ‖γ‖2

W1/2

α−2(G)

)).

Finally, we use Lemma 2.38 and take into account that rε ≥ r, because ofthe convexity of Gd0. Then from (10.1.96) we get:

∫∂G


∫G

(r2rα−2ε v2

xx + rα−2ε |∇v|2)dx ≤

≤ 2 (2− α) (4− α−N)(4−N − α)2 + 4λ(λ+N − 2)

∫G


∫∂G


+

+[C15

(A(d) + δ

)+O(ε)

]∫G


∫∂G


+

+C14(α, d, h, δ, γ0, diamG)(‖v‖22,G + ‖f‖2

W0

α(G)+ ‖g‖2

W1/2

α (∂G)

+

+|u(0)|2(‖a‖2

W0

α(G)+ ‖γ‖2

W1/2

α−2(G)

)), ∀ε > 0.(10.1.97)

In our case, by 4−N − 2λ < α < 4−N, we have

2 (2− α) (4− α−N)(4−N − α)2 + 4λ(λ+N − 2)

< 1


and therefore we can rewrite (10.1.97) in the form(1− 2 (2− α) (4− α−N)

(4−N − α)2 + 4λ(λ+N − 2)

)∫G

rα−2ε |∇v|2)dx+

+∫∂G


+∫G

r2rα−2ε v2

xxdx ≤

≤[C15

(A(d) + δ

)+O(ε)

]∫G


∫∂G


+

+ C14(α, d, h, δ, γ0, diamG)(‖v‖22,G + ‖f‖2

W0

α(G)+ ‖g‖2

W1/2

α (∂G)

+

+ |u(0)|2(‖a‖2

W0

α(G)+ ‖γ‖2

W1/2

α−2(G)

)).

In this case we choose

δ =1

4C15

(1− 2 (2− α) (4− α−N)

(4−N − α)2 + 4λ(λ+N − 2)

)and next d > 0 such that, by the continuity of A(r) at zero,

C15A(d) ≤ 14

(1− 2 (2− α) (4− α−N)

(4−N − α)2 + 4λ(λ+N − 2)

).

Thus we have∫G

r2rα−2ε v2

xxdx+ (1−O(ε))

∫G


∫∂G


≤≤ C16(α, d, h, δ, γ0, diamG)

(‖v‖22,G + ‖f‖2

W0

α(G)+ ‖g‖2

W1/2

α (∂G)

+

+|u(0)|2(‖a‖2

W0

α(G)+ ‖γ‖2

W1/2

α−2(G)

)), ∀ε > 0.(10.1.98)

We observe that the right hand side of (10.1.98) does not depend on ε.Therefore we can perform the passage to the limit as ε → +0 by the FatouTheorem. Hence it follows that

(10.1.99)∫∂G

rα−3γ(x)v2ds+∫G

(rαv2xx + rα−2|∇v|2)dx ≤

≤ C16(α, d, h, δ, γ0, diamG)(‖v‖22,G + ‖f‖2

W0

α(G)+ ‖g‖2

W1/2

α (∂G)

+

+ |u(0)|2(‖a‖2

W0

α(G)+ ‖γ‖2

W1/2

α−2(G)

)).


Now, by the Hardy -Friedrichs - Wirtinger inequality (2.5.13), from (10.1.99)we get the desired estimate (10.1.73).

10.1.5. Local integral weighted estimates.

Theorem 10.25. Let u(x) be a strong solution of the problem (LRP)and assumptions (a)- (d) be satised for A(r) being Dini-continuous at zero.Suppose, in addition, that

g(x) ∈ W1/24−N (∂G) and

a(x) ∈ W04−N (G), γ(x) ∈ W1/2

2−N (∂G), if u(0) 6= 0,

and there is ks from (10.1.1).

Then (u(x)− u(0)) ∈ W

24−N (G) and there are d ∈ (0, 1) and a constant

C > 0 depending only on ν, µ, d,A(d), N, s, λ, γ0, g1, ‖γ‖C1(∂G\O),measG and

on the quantityd∫0

A(r)r dr, such that ∀% ∈ (0, d)

‖u(x)− u(0)‖ W

2

4−N (G%0)≤ C

(|u|0,G + ‖f‖

W0

4−N (G)+ ‖g‖

W1/2

4−N (∂G)+

(10.1.100)

+|u(0)|(1 + ‖a‖

W0

4−N (G)+ ‖γ‖

W1/2

2−N (∂G)

)+ ks

)%λ, if s > λ,

%λ ln3/2(

1%

), if s = λ,

%s, if s < λ.

Proof. From Theorem 10.23 it follows that v(x) = u(x)− u(0) belongsto W2

4−N (G) , so it is enough to prove the estimate (10.1.100). We set

V (ρ) ≡∫Gρ0

r2−N |∇v|2dx+∫Γ%0

r1−Nγ(x)v2ds(10.1.101)

and multiply both sides of the (L)0 equation by r2−Nv(x) and integrate overthe domain Gρ0, 0 < ρ < d. As the result we obtain

V (%) =∫Ω

(%v∂v

∂r+N − 2

2v2

)dΩ +

∫Γ%0

r2−Nvgds− u(0)∫Γ%0

r1−Nvγ(x)ds+

(10.1.102)

+∫G%0

r2−Nv

(aij(x)− aij(0)

)vxixj + ai(x)vxi + a(x)v − f(x) + u(0)a(x)

dx.

We shall obtain an upper bound for each integral on the right. According toLemma 2.36, we estimate the rst integral.


Lemma 10.26.∫G%0

r4−Nv2xxdx ≤ C1

(ν, µ,N, d,A(d), g1, ‖γ‖C1(∂G\O)

)(V (2%) +

+‖f‖2W

0

4−N (G2%0 )

+ ‖g‖2W

1/2

4−N (Γ2%0 )

+(10.1.103)

+u2(0)(‖a‖2

W0

4−N (G2%0 )

+ ‖γ‖2W

1/2

2−N (Γ2%0 )

)).

Proof. The proof is analogous to reasoning deriving (10.1.57).

Now we estimate the second integral in (10.1.102). By the Cauchy in-equality and Lemma 1.40:

(10.1.104)∫Γ%0

r2−N |v||g|ds ≤ δ

2

∫Γ%0

r1−Nγ(x)v2ds+

+C2

δγ0‖g‖2

W1/2

4−N (Γ%0), ∀δ > 0.

By (10.1.65)-(10.1.66), we obtain

(10.1.105) |u(0)|∫Γ%0

r1−Nγ(x)|v|ds ≤ δ

2

∫Γ%0

r1−Nγ(x)|v|2ds+

+ |u(0)|2 C2

δγ0‖γ‖2

W1/2

2−N (Γ%0), ∀δ > 0.

To estimate the last integral in (10.1.102) we use the Cauchy inequality,(2.5.13) with α = 4−N and with the assumption (b) regarding the equationcoecients. We get

(10.1.106)∫G%0

r2−Nv

(aij(x)− aij(0)

)vxixj + ai(x)vxi + a(x)v

dx ≤

≤ A(%)∫G%0

r4−Nv2xxdx+A(%)C2 (λ,N)V (%)

and

(10.1.107)∫G%0

r2−N |v(x)|| − f(x) + u(0)a(x)|dx ≤ δ

2

∫G%0

r−Nv2(x)dx+

+1δ

∫G%0

r4−N (f2(x) + |u(0)|2a2(x))dx ≤ δ

2H(λ,N, 4−N)V (%) +

+1δk2s%

2s, ∀δ > 0,


because of the supposition (10.1.1). By Lemma 2.36 and (10.1.103) -(10.1.107), from (10.1.102) we get the dierential inequality

(10.1.108) V (%) ≤ %

2λV ′(%) + C1A(%)V (2%) + C4 (δ +A(%))V (%) +

+ C5δ−1k2

s%2s, ∀δ > 0, 0 < % < d.

We adjoin the initial condition V (d) ≤ V0 to it. By Theorem 10.23 forα = 4−N we have

(10.1.109) V (d) =∫Gd0

r2−N |∇u|2dx+∫Γd0

r1−Nγ(x)v2ds ≤

≤ C(|u|20,G + ‖f‖2

W0

4−N (G)+ ‖g‖2

W1/2

4−N (∂G)+

+ |u(0)|2 ·(‖γ‖2

W1/2

2−N (∂G)+ ‖a‖2

W0

4−N (G)

))≡ V0.

1) s > λ

Setting δ = %ε we obtain, from (10.1.108), the problem (CP ) with

P(%) =2λ%− C7

(A(%)%

+ %ε−1

); N (%) = 2λC1

A(%)%

;

Q(%) = k2sC6%

2s−1−ε, ∀ε > 0.

Now we have, by (1.10.2),d∫%

P(τ)dτ = 2λ lnd

%− C7

d∫%

A(τ)τ

dτ +dε − %ε

ε

⇒exp( 2%∫%

P(τ)dτ)≤ 22λ;

d∫%

B(τ)dτ ≤ 22λ+1λC1

d∫0

A(τ)τ

dτ ;

exp(−

d∫%

P(τ)dτ)≤(%d

)2λexp(C7

d∫0

A(τ)τ

dτ

)exp(C7ε

−1dε)

=

= C8

(%d

)2λ.

In this case we also have

d∫%

Q(τ) exp(−

τ∫%

P(σ)dσ)dτ ≤ k2

sC9%2λ

d∫%

τ2s−2λ−ε−1dτ ≤ k2sC10%

2λ,

since s > λ.


Now we apply Theorem 1.57: then from (1.10.1), by virtue of the deducedinequalities and with regard to (10.1.103), we obtain the rst statement of(10.1.100).

2) s = λ

Taking in (10.1.108) any function δ(%) > 0 instead of δ > 0 we obtainthe problem (CP ) with

P(%) =2λ(1− δ(%))

%− C7

A(%)%

; N (%) = 2λC1A(%)%

;

Q(%) = k2sC6δ

−1(%)%2λ−1.

We choose

δ(%) =1

2λ ln(ed%

) , 0 < % < d,

where e is the Euler number. Then we obtain

exp( 2%∫%

P(τ)dτ)≤ 22λ;

d∫%


d∫0

A(τ)τ

dτ ;

−d∫%

P(τ)dτ ≤ ln(%d

)2λ+

d∫%

dτ

τ ln(edτ

) + C7

d∫0

A(τ)dτ =

= ln(%d

)2λ+ ln ln

(ed%

)+ C7

d∫0

A(τ)dτ ⇒

exp(−

d∫%

P(τ)dτ)≤(%d

)2λln(ed%

)exp(C7

d∫0

A(τ)τ

dτ

),

because of (1.10.2). In this case we also have

d∫%

Q(τ) exp(−

τ∫%

P(σ)dσ)dτ ≤ k2

sC11%2λ

d∫%

δ−1(τ)τ−1 ln(edτ

)dτ ≤

≤ k2sC12%

2λ ln3(ed%

).

Now we apply Theorem 1.57: from (1.10.1), by virtue of the deducedinequalities, we obtain

V (%) ≤ C17(V0 + k2s)%

2λ ln3 1%, 0 < % < d <

1e.

Taking into account (10.1.103), we obtain the second statement of (10.1.100).


3) 0 < s < λ

From (10.1.108) we obtain the problem (CP ) with

P(%) =2λ(1− δ)

%− C7

A(%)%

; N (%) = 2λC1A(%)%

; Q(%) = k2sC6δ

−1%2s−1, ∀δ > 0.

Now similar to case 1) we have :

exp( 2%∫%

P(τ)dτ)≤ 22λ(1−δ);

d∫%


d∫0

A(τ)τ

dτ

exp(−

d∫%

P(τ)dτ)≤(%d

)2λ(1−δ)exp(C7

d∫0

A(τ)τ

dτ

)= C13

(%d

)2λ(1−δ),

because of (1.10.2).In this case we also have

d∫%

Q(τ) exp(−

τ∫%

P(σ)dσ)dτ ≤ k2

sC16δ−1%2λ(1−δ)

d∫%

τ2s−2λ(1−δ)−1dτ ≤

≤ k2sC14%

2s,

if we choose δ ∈ (0, λ−sλ ).Now we apply Theorem 1.57: then from (1.10.1), by virtue of the deduced

inequalities, we obtain

V (%) ≤ C15

(V0%

2λ(1−δ) + k2s%

2s)≤ C16(V0 + k2

s)%2s,

because of chosen δ.Taking into account (10.1.103), we deduce the third statement of (10.1.100).

Theorems 10.27 and 10.28 together with examples from Subsection 10.2.7show that the assumptions about the smoothness of the coecients of (L)in Theorem 10.25 (i.e. Dini-continuity of the function A(r) at zero fromthe hypothesis (b)) are essential for their validity.

Theorem 10.27. Let u(x) be a strong solution of the problem (LRP ) andthe assumptions of Theorem 10.25 be satised with A(r) that is a continuousat zero , but not Dini-continuous at zero. Then there are d ∈ (0, 1) and foreach ε > 0 a constant Cε > 0 depending only on ε, ν, µ, d, s,N, λ, γ0, ‖γ‖C1(∂G\O),


g1,measG, such that ∀% ∈ (0, d)

‖u(x)− u(0)‖ W

2

4−N (G%0)≤ C

(|u|0,G + ‖f‖

W0

4−N (G)+ ‖g‖

W1/2

4−N (∂G)+

(10.1.110)

+|u(0)|(1 + ‖a‖

W0

4−N (G)+ ‖γ‖

W1/2

2−N (∂G)

)+ ks

)%λ−ε, if s > λ,

%s−ε, if s ≤ λ.

Proof. As above in Theorem 10.25, we get the problem (CP ): (10.1.108)- (10.1.109) with

P(%) =2λ%

(1− δ

2− C7A(%)

), ∀δ > 0; N (%) = 2λC1

A(%)%

;

Q(%) = k2sC17%

2s−1.

Therefore we have:

−d∫%

P(τ)dτ = 2λ(1− δ

2) ln

%

d+ 2λC7

d∫%

A(τ)τ

dτ.

Now we apply the mean value theorem for integrals:

d∫%

A(τ)τ

dτ ≤ A(d) lnd

%

and choose d > 0 by continuity of A(r) so that 2C7A(d) < δ. Thus we obtain

exp(−

d∫%

P(τ)dτ)≤(%d

)2λ(1−δ), ∀δ > 0

Similarly we have

exp(−

τ∫%

P(σ)dσ)≤(%τ

)2λ(1−δ), ∀δ > 0.

Further it is obvious that

2%∫%

P(τ)dτ ≤ 2λ ln 2


and with regard to (1.10.2)

d∫%

B(τ)dτ ≤ 2λ22λC7

d∫%

A(τ)τ

dτ ≤ 2λ22λC7A(d) lnd

%≤ δλ22λ ln

d

%⇒

exp( d∫%

B(τ)dτ)≤(%d

)−δλ22λ

, ∀δ > 0.

Hence, by (1.10.1) of Theorem 1.57, we have

(10.1.111) V (%) ≤(%d

)−δλ22λV0

(%d

)2λ(1−δ)+

+

d∫%

Q(τ) exp(−

τ∫%

P(σ)dσ)dτ

, ∀δ > 0.

Now we estimate the last integral:

d∫%

Q(τ) exp(−

τ∫%

P(σ)dσ)dτ ≤ k2

sC17%2λ(1−δ)

d∫%

τ2s−2λ(1−δ)−1dτ =

= k2sC17%

2λ(1−δ)d2s−2λ(1−δ) − %2s−2λ(1−δ)

2s− 2λ(1− δ)≤ k2

sC18

%2λ(1−δ), if s ≥ λ%2s, if 0 < s < λ

(10.1.112)

(we choose δ > 0 so that δ 6= λ−sλ ).

From (10.1.111) - (10.1.112) and because of (10.1.103) Lemma 10.26follows the desired estimate (10.1.110).

We can improve Theorem 10.27 in the case s ≥ λ, if A(r) ln 1r ≤ const.

Theorem 10.28. Let u(x) be a strong solution of the problem (LRP) andthe assumptions of Theorem 10.25 be satised with s ≥ λ and A(r) ln 1

r ≤const,A(0) = 0. Then there are d ∈ (0, 1) and the constants C > 0, c > 0depending only on ν, µ, d,N, λ, γ0, g1, ‖γ‖C1(∂G\O),measG, such that

‖u(x)− u(0)‖ W

2

4−N (G%0)≤ C

(|u|0,G + ‖f‖

W0

4−N (G)+ ‖g‖

W1/2

4−N (∂G)+

(10.1.113)

+|u(0)|(1 + ‖a‖

W0

4−N (G)+ ‖γ‖

W1/2

2−N (∂G)

)+ kλ

)%λ lnc+1 1

%, 0 < % < d.

Proof. As above in Theorem 10.25, we get the problem (CP ): (10.1.108)- (10.1.109). Taking in (10.1.108) any function δ(%) > 0 instead of δ > 0 we


obtain the problem (CP ) with

P(%) =2λ(1− δ(%))

%− C7

A(%)%

; N (%) = 2λC1A(%)%

;

Q(%) = k2sC6δ

−1(%)%2s−1.

We choose

δ(%) =1

2λ ln(ed%

) , 0 < % < d,

where e is the Euler number. Because of (1.10.2), since A(%) ≤ Cδ(%), wehave:

exp( 2%∫%

P(τ)dτ)≤ 22λ; exp

( d∫%

B(τ)dτ)≤ lnc

(ed%

), c > 0;

−d∫%

P(τ)dτ ≤ ln(%d

)2λ+ c

d∫%

dτ

τ ln(edτ

) = ln(%d

)2λ+ c ln ln

(ed%

)⇒

exp(−

d∫%

P(τ)dτ)≤(%d

)2λlnc(ed%

)for suitable small d > 0.. In this case we also have

d∫%

Q(τ) exp(−

τ∫%

P(σ)dσ)dτ ≤ k2

sC19%2λ

d∫%

δ−1(τ)τ2(s−λ) lnc(eτ%

)dττ≤

≤ k2sC20%

2λ lnc+2(ed%

),

because s ≥ λ.Now we apply Theorem 1.57: then from (1.10.1), by virtue of the deduced

inequalities, we obtain

V (%) ≤ C21(V0 + k2s)%

2λ ln2c+2 1%, 0 < % < d <

1e.(10.1.114)

From (10.1.114) and because of (10.1.103) the desired estimate (10.1.113)follows.


10.1.6. The power modulus of continuity at the conical pointfor strong solutions. In this Section we prove Theorems 10.2, 10.3, 10.4.

Proof of Theorem 10.2.

We dene the functions v(x) = u(x)− u(0) and

ψ(%) =

%λ, if s > λ,

%λ ln3/2(

1%

), if s = λ,

%s, if s < λ,

(10.1.115)

for 0 < % < d and consider two sets G2%%/4 and G%%/2 ⊂ G2%

%/4, % > 0. Wemake transformation x = %x′; v(%x′) = ψ(%)z(x′). Because of (LRP )0, thefunction z(x′) satises the problem

aij(%x′)zx′ix′j + %ai(%x′)zx′i + %2a(%x′)z =

= %2

ψ(%)

(f(%x′)− u(0)a(%x′)

), x′ ∈ G2

1/4

∂z∂~n′ + 1

|x′|γ(%x′)z = %ψ(%)

(g(%x′)− u(0)γ(%x′)

%|x′|

)] ≤

≤ %ψ(%)g(%x′), x′ ∈ Γ2

1/4,

(LRP )′0

since without loss of generality we can suppose that u(0) ≥ 0.We apply nowProposition 10.14. Because of the estimates proved there, we have

supG1

1/2

|z(x′)| ≤ C( ∫

G21/4

z2dx′) 1

2

+%

ψ(%)supG2

1/4

|g(%x′)|+

+ρ2

ψ(%)

( ∫G2

1/4

|f(%x′)− u(0)a(%x′)|Ndx′) 1N,

(10.1.116)

where the constant C > 0 depends only on

∥∥∥∥∥(N∑1|ai|2

)1/2∥∥∥∥∥Lq(G2

1/4)

, ‖a‖LN (G21/4

),

M0, g1, γ0, ‖γ‖L∞(∂G), N, ν, diamG,ω0 andd∫0

A(r)r dr, sup

14<%<2

ρψ(%) . Returning to

the variable x and the function u(x), by Theorem 10.25 with (10.1.115), weobtain:

(10.1.117)∫

G21/4

z2dx′ =1

ψ2(%)

∫G2%%/4

%−N |u(x)− u(0)|2dx ≤ C(

(|u|0,G +

+‖f‖ W

0

4−N (G)+‖g‖

W1/2

4−N (∂G)+|u(0)|

(1+‖a‖

W0

4−N (G)+‖γ‖

W1/2

2−N (∂G)

)+ks

)2;


( ∫G2

1/4

|f(%x′)− u(0)a(%x′)|ndx′) 1N

= %−1

( ∫G2%%/4

|f(x)− u(0)a(x)|Ndx) 1N

⇒ %2

ψ(%)

( ∫G2

1/4

|f(x)− u(0)a(x)|Ndx′) 1N

≤ ‖f + u(0)a‖N,G2%

%/4

%

ψ(%)≤

≤ κs%s

ψ(%)≤ const(N, s, λ, d) · κs(10.1.118)

by our assumptions. From (10.1.116), (10.1.117) and (10.1.118) we get:

(10.1.119) supGρρ/2

|u(x)− u(0)| ≤ C(|u|0,G + ‖f‖

W0

4−N (G)+ ‖g‖

W1/2

4−N (∂G)+

+ g1 + |u(0)|(1 + ‖a‖

W0

4−N (G)+ ‖γ‖

W1/2

2−N (∂G)

)+ ks + κs

)ψ(%).

Putting now |x| = 23% we nally obtain the desired estimate (10.1.2).

By the Sobolev Imbedding Theorems we have

supx′∈G1

1/2

|∇′z(x′)| ≤ c‖z‖W 2,p(G11/2

), p > N.(10.1.120)

By the local Lp apriori estimate, Theorem 10.17, for the solution of theequation of the (LRP )′0 inside the domain and near a smooth portion of theboundary we have:

(10.1.121) ‖z‖W 2,p(G11/2

) ≤ c(N, ν, µ,A(2))

%2

ψ(%)‖f + u(0)a‖Lp(G2

1/4) +

+%

ψ(%)

∥∥∥g + u(0)γ(%x′)%|x′|

∥∥∥W 1−1/p,p(Γ2

1/4)

+ ‖z‖Lp(G21/4

)

.

Returning back to the variables x, from (10.1.120), (10.1.121) it follows that

supG%%/2

|∇v| ≤ c%−1%−N/p‖v‖

Lp(G2%%/4

)+ %2−N/p‖f + u(0)a‖

p,G2%%/4

+

+ %2−N/p‖g‖V

1−1/pp,0 (Γ2%

%/4)

+ %1−N/p|u(0)|‖γ‖V

1−1/pp,0 (Γ2%

%/4)

or

(10.1.122) supG%%/2

|∇v| ≤ c%−1|v|

0,G2%%/4

+ |u(0)|‖a‖V 0p,2p−N (G2%

%/4)

+

+ ‖f‖V 0p,2p−N (G2%

%/4)

+ ‖g‖V

1−1/pp,2p−N (Γ2%

%/4)

+ |u(0)|‖γ‖V

1−1/pp,p−N (Γ2%

%/4)

.

Because of (10.1.119), (10.1.2) and by the assumption (10.1.3), from(10.1.122) we get the required (10.1.4).


Proof of Theorem 10.3

We repeat verbatim the proof of Theorem 10.2 taking

ψ(%) =


%s−ε, if s ≤ λ,


Proof of Theorem 10.4

We repeat verbatim the proof of Theorem 10.2 taking

ψ(%) = %λ lnc+1 1%


10.1.7. Examples. We present the examples that show that the con-ditions of Theorems 10.2 - 10.4 (in particular the Dini condition for thefunction A(r) in condition (b) at the point O in Theorem 10.2) are essentialfor their validity. Suppose N = 2, the domain G lies inside the corner

G0 = (r, ω) |r > 0; −ω0

2< ω <

ω0

2, ω0 ∈]0, π[,

O ∈ ∂G and in some neighborhood of O the boundary ∂G coincides withthe sides of the corner ω = −ω0

2 and ω = ω02 . We denote

Γ± = (r, ω) | r > 0; ω = ±ω0

2

and we putγ(x)

∣∣∣ω=±ω0

2

= γ± = const > 0.

I. (See Example of Subsection 2.4.2)4u = 0, x ∈ G0;(∂u∂n + 1

rγ±u

)∣∣∣∣Γ±

= 0.

We verify that the function u(r, ω) = rλψ(ω) is a solution of ourproblem, if λ2 is the least positive eigenvalue of the problem

ψ′′

+ λ2ψ = 0, ω ∈(−ω0

2 ,ω02

)(±ψ′ + γ±ψ)

∣∣∣∣ω=±ω0

2

= 0

and ψ(ω) is a regular eigenfunction associated with λ2. Precisely λ isdened from the transcedence equation

tan(λω0) =λ(γ+ + γ−)λ2 − γ+γ−

.(10.1.123)


Then we nd the eigenfunction

ψ(ω) = λ cos[λ(ω − ω0

2)]− γ+ sin

[λ(ω − ω0

2)].(10.1.124)

The existence of the positive solution of (10.1.123) may be deducedby the graphic method. This example shows that the exponent λ in(10.1.2) cannot be increased.

Remark 10.29. In order to have λ > 1 we show that the conditionγ(x) ≥ γ0 > tan ω0

2 from the assumption (c) of our Theorems isjustied. In fact, we rewrite the equation (10.1.123) in the equivalentform

λ =1ω0

(arctan

γ+

λ+ arctan

γ−λ

).(10.1.125)


1 < λ <1ω0

(arctan γ+ + arctan γ−) ⇒

ω0 < arctanγ+ + γ−1− γ+γ−

, provided γ+γ− < 1(10.1.126)

has to be fullled. But our condition from the assumption (c) meansthat γ± ≥ γ0 > tan ω0

2 . Hence we obtain

γ+ + γ−1− γ+γ−

≥ 2γ0

1− γ20

>2 tan ω0

2

1− tan2 ω02

= tanω0, ω0 <π

2.

Thus we get (10.1.126). In the case γ± ≥ γ0 > tan ω02 ≥ 1 for ω0 ∈

[π2 , π) the inequality λ > 1 is fullled a fortiori, because of the propertyof the monotonic increase of the eigenvalues together with the increaseof γ(x) (see for example Theorem 6 2, chapter VI [86]).

II. The function

u(r, ω) = rλ

(ln

1r

)λ−1λ+1

ψ(ω)

with λ and ψ(ω) dened by (10.1.123) - (10.1.124) is a solution of theproblem

N∑i,j=1

aij(x)uxixj = 0, x ∈ G0,(∂u∂n + 1

rγ±u

)∣∣∣∣Γ±

= 0, γ± > 0


in the corner G0, where

a11(x) = 1− 2λ+ 1

· x22

r2 ln 1/r, r > 0;

a12(x) = a21(x) =2

λ+ 1· x1x2

r2 ln 1/r, r > 0;

a22(x) = 1− 2λ+ 1

· x21

r2 ln 1/r, r > 0;

aij(0) = δji , (i, j = 1, 2).

In the domain Gd0, d < e−2 the equation is uniformly elliptic withellipticity constants µ = 1 and ν = 1 − 2

ln(1/d) . Further, A(r) =2

λ+1 ln−1(

1r

), i.e., the function A(r) does not satisfy the Dini condition

at zero. Moreover, aij(x) are continuous at the point O. This exampleshows that the condition of Theorem 10.2 about Dini-continuity of theleading coecients of the (LRP ) are essential, and it illustrates theprecision of the assumptions of Theorem 10.4 as well.

III. The functionu(r, ω) = rλ ln

1rψ(ω)

λ and ψ(ω) dened by (10.1.123) - (10.1.124) is a solution of theproblem

4u+ 2λr2 ln 1

r

u = 0, x ∈ G0,(∂u∂n + 1

rγ±u

)∣∣∣∣Γ±

= 0, γ± > 0

in the corner G0. This example shows that the assumptions of The-orem 10.4 on the lowest coecients of the (LRP ) are precise andessential.

IV. The functionu(r, ω) = rλ ln

1rψ(ω)

with λ and ψ(ω) dened by (10.1.123) - (10.1.124) is a solution of theproblem

4u = −2λrλ−2ψ(ω), x ∈ G0,(∂u∂n + 1

rγ±u

)∣∣∣∣Γ±

= 0, γ± > 0

in the corner G0. All assumptions of Theorem 10.3 are fullled withs = λ. This example shows the precision of the assumptions for theright hand side of the (LRP ) in Theorem 10.2.


10.2. Quasilinear problem

10.2.1. Introduction. In this Section we consider the elliptic valueproblem (QLRP ). We obtain the best possible estimates of the problem(QLRP ) strong solutions near a conical boundary point. The analogousresults were established in Chapter 7 for the Dirichlet problem.

Definition 10.30. A strong solution of the problem (QLRP) is a func-

tion u(x) ∈ C0(G)∩W 1(G)∩W 2,qloc (G \O), q ≥ N that satises the equation

for almost all x ∈ G, and the boundary condition in the sense of traces on∂G \ O.

We assume that M0 = maxx∈G|u(x)| is known.

Let us recall some known facts about W 2,ploc (G)−solutions (p > N) of the

quasilinear oblique derivative problem in smooth domains.

Theorem 10.31. Local gradient bound estimate (see Theorems 13.13,13.14 [234]).

Let G′ ⊂⊂ G \ O be any subdomain with a C2 boundary portionT = (∂G′ ∩ ∂G) ⊂ ∂G \ O. Let u ∈ W 2,p(G′) ∩ C1(T ), p > N be a strongsolution of the problem

aij(x, u, ux)uxi,xj + a(x, u, ux) = 0, x ∈ G′,∂u∂~n + 1

|x|γ(x)u = g(x), x ∈ T

with |u| ≤M0. Suppose that

aij(x, u, z), a(x, u, z) ∈ C1(G′ × [−M0,M0]× RN ),

γ(x), g(x) ∈ C1(T )

and there are positive constants ν, µ, µ1,K such that aij(x, u, z), a(x, u, z)satisfy

νξ2 ≤ aij(x, u, z)ξiξj ≤ µξ2, ∀ξ ∈ RN ;

|z|2∣∣∣∣∂aij∂zk

∣∣∣∣+ |z|∣∣∣∣∂aij∂u

∣∣∣∣+∣∣∣∣∂aij∂xk

∣∣∣∣ ≤ µ1|z|;

|a(x, u, z)| ≤ µ1

(1 + |z|2

);

|z|2∣∣∣∣∂a∂u

∣∣∣∣+ |z|2∣∣∣∣ ∂a∂zk

∣∣∣∣+∣∣∣∣ ∂a∂xk

∣∣∣∣ ≤ µ1|z|3

for |z| ≥ K. Then for any subdomain G′′ ⊂⊂ G′ ∪ T there is a constantM1 > 0 depending only on N, ν, µ, µ1, ‖γ‖C1(T ), ‖g‖C1(T ),M0,K and G′, G′′, Tsuch that

supG′′|∇u| ≤M1.

10.2 The quasilinear problem 423

Theorem 10.32. Local Hölder gradient estimate (see Lemma 2.3[233]).

Let G′ ⊂⊂ G \ O be any subdomain with a C2 boundary portionT = (∂G′ ∩ ∂G) ⊂ ∂G \ O. Let u ∈ W 2,p(G′) ∩ C1(T ), p > N be a strongsolution of the problem

aij(x, u, ux)uxi,xj + a(x, u, ux) = 0, x ∈ G′,∂u∂~n + 1

|x|γ(x)u = g(x), x ∈ T

with |u| ≤M0, |∇u| ≤M1. Suppose that

aij(x, u, z), a(x, u, z) ∈ C1(G′ × [−M0,M0]× [−M1,M1]),

γ(x), g(x) ∈ C1(T )

and there are positive constants ν, µ, µ1 such that aij(x, u, z), a(x, u, z) satisfy

νξ2 ≤ aij(x, u, z)ξiξj ≤ µξ2, ∀ξ ∈ RN ;∣∣∣∣∂aij∂zk

∣∣∣∣+∣∣∣∣∂aij∂u

∣∣∣∣+∣∣∣∣∂aij∂xk

∣∣∣∣+ |a(x, u, z)| ≤ µ1

for |u| ≤M0, |∇u| ≤M1. Then for any subdomain G′′ ⊂⊂ G′ ∪ T there areconstants C > 0, κ ∈ (0, 1) depending only on N, ν, µ, µ1, ‖γ‖C1(T ),

‖g‖C1(T ),M0,M1 and G′, G′′, T such that

|u|1+e,G′′ ≤ C.

We assume the existence d > 0 such that Gd0 is the convex rotationalcone with the vertex at O and the aperture ω0 ∈ (π2 , π) (see (1.3.13)). LetM = (x, u, z)|x ∈ G, u ∈ R, z ∈ RN. Regarding the equation we assumethat the following conditions are satised on M:

(A) aij(x, u, z) ∈W 1,q(M), q > N ; γ(x) ∈ L∞(∂G) ∩ C1(∂G \ O);the condition of Caratheodory: functions

a(x, u, z),∂a(x, u, z)

∂u∈ CAR,

that is:(i) they are measurable on G as functions of variable x for ∀u, z;(ii) they are continuous with respect to u, z for almost all x ∈ G;

(B) the condition of uniform ellipticity:

νξ2 ≤ aij(x, u, z)ξiξj ≤ µξ2,

∀ξ ∈ RN , x ∈ G, u ∈ R, z ∈ RN ; ν, µ = const > 0;

(C) ∂a(x,u,z)∂u ≤ 0;

424 10 The Robin boundary value problem in a nonsmooth domain

(D) there exist numbers β > −1, γ0 > tan ω02 , γ1 ≥ γ0, positive constants

δ, µ1, k1, g0 ≥ 0 and nonnegative functionsb(x), f(x) ∈ Lqloc(G \ O), q ≥ N such that on M the inequalities:

|a(x, u, z)|+∣∣∣∂a(x, u, z)

∂u

∣∣∣ ≤ µ1|z|2 + b(x)|z|+ f(x),

b(x) + f(x) ≤ k1|x|β, |g(x)| ≤ g0|x|δ;γ0 ≤ γ(x) ≤ γ1

hold;(E) the problem (QLRP ) coecients satisfy such conditions that guaran-

tee u ∈ C1+e(G′) and the existence the local a-priori estimate

|u|1+e,G′ ≤M1, κ ∈ (0, 1)

for any smooth G′ ⊂⊂ G \ O (see Theorems 10.31, 10.32).

Proposition 10.33. The local maximum principle (see Theorem3.3 [222], Theorem 4.3 [231]; see as well as [230]).

Let G be a bounded domain in RN with the C1− boundary ∂G \ O andGd0 be a convex rotational cone with vertex at O and the aperture ω0 ∈ (π2 , π).Let u(x) be a strong solution of the problem (QLRP ) with |u| ≤ M0. Sup-pose the conditions (A), (B), (C) are satised. In addition, suppose thatthere are nonnegative number µ1 and nonnegative functions b(x) ∈ Ls(G),s > N, f(x) ∈ LN (G), such that:

|a(x, u, z)| ≤ µ1|z|2 + b(x)|z|+ f(x).

Suppose nally that g ∈ L∞(∂G).Then for any q > 0 and σ ∈ (0, 1) there is a constant

C = C(ν, µ, µ1,M0, γ0, ω0, N, p,R,G, ‖b‖Ls(G), ‖f‖LN (G)) such that

supGσR0

|u(x)| ≤ C

1

measGR0

∫GR0

|u|qdx

1/q

+R(‖f‖LN (GR0 ) + ‖g‖L∞(∂G)

) .

10.2.2. The weak smoothness of the strong solution. First weestimate |u(x)| for (QLRP ) in the neighborhood of a conical point. For thiswe use the barrier function, constructed in Lemma 10.18, Subsection 10.1.2.

Theorem 10.34. Let u(x) be a strong solution of problem (QLRP) andassumptions (A)-(D) be satised. Then there exist the numbers d > 0 andκ > 0 depending only on ν, µ, µ1, N,κ0, ω0, k1, β, δ, γ0, g0,M0 and the domainG such that

|u(x)− u(0)| ≤ C0|x|+1, x ∈ Gd0,(10.2.1)

where the positive constant C0 does not depend on u but depends only onν, µ, µ1, g0, N,k1, β, γ0,M0 and the domain G.


Proof. Let us take the linear elliptic operator:

L ≡ aij(x)∂2

∂xi∂xj+ ai(x)

∂

∂xi, x ∈ G;

aij(x) = aij(x, u(x), ux(x)); ai(x) = b(x)|∇u(x)|−1uxi(x).(10.2.2)

Here we suppose that ai(x) = 0, i = 1, ..., N in such points x, where|∇u(x)| = 0. Let us take the barrier function (10.1.15) and dene the auxil-iary function v(x) as follow:

v(x) = −1 + exp(ν−1µ1(u(x)− u(0))).(10.2.3)

For them we shall show thatL(Aw(x)) ≤ Lv(x), x ∈ Gd0;B[Aw(x)] ≥ B[v(x)], x ∈ Γd0 \ O;Aw(x) ≥ v(x), x ∈ Ωd ∪ O.

(10.2.4)

Let us calculate operator L on the function (10.2.3). We obtain:

Lv(x) ≡ ν−1µ1(aij(x)uxixj + ν−1µ1aij(x)uxiuxj + b(x)|∇u(x)|)×

× exp(ν−1µ1(u(x)− u(0))) = ν−1µ1

[b(x)|∇u(x)| − a(x, u(x), ux(x))

+ aij(x)uxiuxj]

exp(ν−1µ1(u(x)− u(0))) ≥≥ −ν−1µ1f(x) exp(2ν−1µ1M0)

in virtue of assumptions (C) and (D). Because f(x) ≤ k1rβ , we obtain:

Lv(x) ≥ −ν−1µ1k1rβ exp(2ν−1µ1M0), x ∈ Gd0.(10.2.5)

Let us calculate the operator L on the barrier function (10.1.15). Letthe number κ0 be such that Lemma 10.18 holds and suppose κ satises theinequality:

0 < κ ≤ min(δ,κ0, β + 1).

By (10.1.11) and (10.2.2), we obtain

Lw = L0w + b(x)uxi|∇u(x)|

wxi ≤ −νh2|x|−1 + b(x)|∇w| ≤

≤ −νh2|x|−1 + b(x)|x|√

2 + 4h2 +B(1 + κ0)2.

Because of b(x) ≤ k1rβ, we get in Gd0:

Lw ≤ r−1(− νh2 + dβ+1k1

√2 + 4h2 +B(1 + κ0)2

)≤ −1

2νh2r−1,


if

k1

√2 + 4h2 +B(1 + κ0)2dβ+1 ≤ 1

2νh2 ⇒

d ≤( νh2

2k1

√2 + 4h2 +B(1 + κ0)2

) 11+β

.(10.2.6)

Hence, in virtue of (10.2.5) it follows that

L[Aw(x)] ≤ Lv(x), x ∈ Gd0,

if we dene a number A such that

A ≥ 2µ1k1ν−2h−2d1−0+β exp(2ν−1µ1M0).(10.2.7)

From (10.1.12) we get:

B[Aw]∣∣∣Γd±≥ Ag0r

δ(10.2.8)

Let us calculate the operator B on the dened by (10.2.3) function v(x):

B[v(x)] ≡ ∂v

∂~n+

1|x|γ(x)v(x), x ∈ Γd± \ O.

By the (QLRP ) boundary condition, we have

∂v

∂~n

∣∣∣Γd±\O

= ν−1µ1 exp(ν−1µ1(u(x)− u(0)))∂u

∂~n

∣∣∣Γd±\O

=

= ν−1µ1 exp(ν−1µ1(u(x)− u(0)))⟨g(x)− γ(x)

|x|u(x)

⟩.

(10.2.9)

Using (10.2.9) and our assumptions, we calculate:

B[v]∣∣∣Γd±\O

≤ ν−1µ1 exp(2ν−1µ1M0)⟨g0r

δ − γ(x)|x|

u(x)⟩

+γ(x)|x|

v(x) ≤

(10.2.10)

≤ exp(2ν−1µ1M0)g0rδ +

γ(x)|x|

[v − (1 + v)ν−1µ1u(x)

]; v > −1.

Because of (10.2.3), we have:

exp(ν−1µ1(u(x)− u(0))) = v + 1⇒ ν−1µ1(u(x)− u(0)) = ln(1 + v)⇒⇒ (1 + v)ν−1µ1u(x) = (1 + v) ln(1 + v) + (1 + v)ν−1µ1u(0),

and therefore from (10.2.10) we obtain, if only u(0) ≥ 0:

B[v]∣∣∣Γd±\O

≤ exp(2ν−1µ1M0)g0rδ +

γ(x)|x|

[v − (1 + v) ln(1 + v)

]≤

≤ exp(2ν−1µ1M0)g0rδ; v > −1.(10.2.11)

Indeed, if we denote f(v) = v − (1 + v) ln(1 + v); v > −1 we get f ′(v) == − ln(1 + v); f ′′(v) = − 1

1+v . We see that f ′(v) = 0 ⇔ v = 0 and


f ′′(0) == −1 < 0. Then we obtain:

maxv>−1

f(v) = f(0) = 0 =⇒ (10.2.11).

Taking into account (10.2.8) and (10.2.11), we choose:

A ≥ ν−1µ1g−10 exp(2ν−1µ1M0)(10.2.12)

and we obtain

B[Aw] ≥ B[v] on Γd± \ O.If u(0) < 0, we have to take instead of the function v(x), dened by (10.2.3),the function

z(x) := 1− exp(−ν−1µ1(u(x)− u(0))).(10.2.13)

Now we compare v(x) and w(x) on Ωd. Since x2 ≥ h2y2 in K, from(10.1.15) we have

w(x)∣∣∣r=d≥ B|x|1+

∣∣∣r=d

= Bd1+ cos+1 ω0

2.(10.2.14)

On the other hand

v(x)∣∣∣Ωd

= −1 + exp(ν−1µ1(u(x)− u(0)))∣∣∣Ωd≤

≤ −1 + exp(2ν−1µ1M0)(10.2.15)

and therefore from (10.2.14)-(10.2.15), in virtue of (10.1.20), we obtain:

Aw(x)∣∣∣Ωd≥ ABd1+ cos+1 ω0

2≥ A

g0

(1 + h2

)0+12

h0+ 2(1 + h2)

×

× 1hγ0 − 1− κ0

d1+0h1+0(1 + h2)−1+0

2 ≥

≥ exp(2ν−1µ1M0)− 1 ≥ v∣∣∣Ωd,

if we choose A enough great

A ≥ [exp(2ν−1µ1M0)− 1](hγ0 − 1− κ0)

hd1+0

[g0 + 2h0

(1 + h2

) 1−02

] .(10.2.16)

Thus, if we choose small number d > 0 according to (10.2.6) and large num-bers A,B according to (10.1.20), (10.2.7), (10.2.12), (10.2.16), we providethe validity of (10.2.4).

Therefore the functions (10.1.15) and (10.2.3) satisfy the comparisonprinciple, Proposition 10.16, and, by it, we have:

v(x) ≤ Aw(x), x ∈ Gd0.(10.2.17)

Returning to the function u(x) from (10.2.3), on the basis (10.2.17), we have

u(x)− u(0) = νµ−11 ln(v(x) + 1) ≤ νµ−1

1 ln(Aw(x) + 1) ≤ νµ−11 Aw(x).


Similarly, we derive the estimate

u(x)− u(0) ≥ −νµ−11 Aw(x),

if we consider an auxiliary function (10.2.13). In virtue of (10.1.13), thetheorem is proved

Now we'll estimate the gradient modulus of the problem (QLRP ) solu-tion near a conical point.

Theorem 10.35. Let u(x) be a strong solution of the problem (QLRP ),q > N and suppose that assumptions (A)-(E) are satised. Let κ > 0 be anumber, dened by Theorem 10.34. Then there exists the number d > 0 suchthat

|∇u(x)| < C1|x|, x ∈ Gd0,(10.2.18)

where the constant C1 does not depend on u, but depends only on ν, µ,N, k1, β, g0, γ0,M1

and the domain G.

Proof. Let us consider the set Gρρ/2 ⊂ G, 0 < ρ < d. We make thetransformation x = ρx′; v(x′) = ρ−1−u(ρx′). The function v(x′) satisesthe problem:

aij(x′)vx′ix′j = F (x′), x′ ∈ G11/2,

∂v∂~n′ + 1

|x′|γ(ρx′)v(x′) = ρ−g(ρx′), x ∈ Γ11/2,

(QLRP )′

where

aij(x′) ≡ aij(ρx′, ρ1+v(x′), ρvx′(x′)),

F (x′) ≡ −ρ1−a(ρx′, ρ1+v(x′), ρvx′(x′)).

We apply now the assumption (E):

maxx′∈G1

1/2

|∇′v(x′)| ≤M ′1.(10.2.19)

Returning to the variable x and the function u(x) we obtain from (10.2.19):

|∇u(x)| ≤M1ρ, x ∈ Gρρ/2, 0 < ρ < d.

Putting now |x| = 23ρ we obtain the desired estimate (10.2.18)

Corollary 10.36. Let u(x) be a strong solution of the problem (QLRP ),q > N and suppose that assumptions (A)-(E) are satised. Then u(0) = 0and therefore the inequality (10.2.1) take a form

|u(x)| ≤ C0|x|+1, x ∈ Gd0.(10.2.20)

Proof. From the problem boundary condition it follows that

γ(x)u(x) = |x|g(x)− |x|∂u∂n, x ∈ ∂G \ O.

By the assumption (D) and the estimate (10.2.18), we obtain

γ0|u(x)| ≤ γ(x)|u(x)| ≤ |x||g(x)|+ |x||∇u| ≤ g0|x|δ+1 + C1|x|+1.


By tending |x| → 0 , because of the continuity of u(x), we get γ(0)|u(0)| = 0,whence taking into account γ0 > 0 we nd u(0) = 0.

Theorem 10.37. Let u(x) be a strong solution of the problem (QLRP ),q > N and suppose that assumptions (A)-(E) are satised. Let κ0 be thenumber from Lemma about the barrier function. In addition, let g(x) ∈

V1− 1

qq,q (∂G) and

‖g‖V

1− 1q

q,q (Γ2ρρ/4

)≤ Cρ+N

q .(10.2.21)

Then u(x) ∈ C1+(Gd0), 0 < κ ≤ min(κ,κ0, β+1, 1−Nq ) for some d ∈ (0, 1).

Proof. Let d > 0 be a number such that estimates (10.1.21) and(10.2.18) are satised. Let us consider the set Gρρ/2 ⊂ G; 0 < ρ < d. Wemake the transformation x = ρx′; v(x′) = ρ−1−u(ρx′), where κ > 0 is de-ned by Theorem 10.34. The function v(x′) satises the problem (QLRP )′.By the Sobolev Embedding Theorem, we have:

supx′, y′ ∈ G1

1/2

x′ 6= y′

|∇′v(x′)−∇′v(y′)|

|x′ − y′|1−Nq

≤ C(N, q,G)‖v‖W 2,q(G11/2

);(10.2.22)

q > N.We shall verify that the local interior and near a smooth boundary por-

tion Lq a-priori estimate (Theorem 4.8) for the solution of the (QLRP )′ equa-tion holds. On the basis assumption (E) we have that functions aij(x, u, z)are continuous on M: ∀ε > 0 there exist such η that

|aij(x, u(x), ux(x))− aij(y, u(y), ux(y))| < ε,

if only

|x− y|+ |u(x)− u(y)|+ |ux(x)− ux(y)| < η, ∀x, y ∈ Gρρ/2, ρ ∈ (0, d).

The assumption (E) guarantees the existence of the local interior and neara smooth boundary portion apriori C1+e - estimate: there exist a numberκ > 0 and a number M1 > 0 such that

|u(x)− u(y)|+ |∇u(x)−∇u(y)| ≤M1|x− y|e, ∀x, y ∈ Gρρ/2, ρ ∈ (0, d).

Then functions aij(x′) are uniformly continuous in G11/2. It means that

∀ε > 0 exists δ > 0 (we choose the number δ such that: δd+M1(δd)e < η)such that |aij(x′)− aij(y′)| < ε, if only |x′ − y′| < δ, ∀x′, y′ ∈ G1

1/2. We seethat assumptions of Theorem 10.17 about local Lq− a-priori estimate for the


(QLRP )′ are satised. By this theorem, we have:

‖v‖qW 2,q(G1

1/2)≤ C3

∫G2

1/4

⟨|v|q + ρq(1−)|a(ρx′, ρ1+v(x′), ρvx′(x))|q

⟩dx′ +

+C4 inf∫

G21/4

(|∇′G|q + |G|q)dx′ · ρ−q,(10.2.23)

where the constants C3, C4 do not depend on v. Returning to the variable xand using the estimate (10.1.21) we obtain:

∫G2

1/4

|v|qdx′ =∫G2ρρ4

ρ−q(1+)|u(x)|qρ−Ndx ≤

≤ Cq0mesΩC(q,κ)

2ρ∫ρ4

dr

r= Cq0mesΩC(q,κ) ln 8.

(10.2.24)

Similarly, by the assumption (D), the estimate (10.2.18) and the inequality

( N∑i=1

ci

)t≤ N t−1

N∑i=1

cti; ∀ci > 0, t ≥ 1,

we have: ∫G2

1/4

ρq(1−)|a(ρx′, ρ1+v(x′), ρvx′(x))|qdx′ ≤

≤ ρq(1−)−N∫G2ρρ4

(µ1|∇u|2 + b(x)|∇u|+ f(x)

)qdx ≤

≤ 2N3q−1ρq(1−)mesΩ

2ρ∫ρ4

(µq1C

2q1 r2q−1 + (k1C1)qrq(β+)−1 + kq1r

qβ−1)dr ≤

(10.2.25)

≤ C(N, q,κ, β, µ1, C1, k1),


if only 0 < κ ≤ 1 + β. Because of the assumption (10.2.21) of our theorem,we have:

(10.2.26)∫

G21/4

(|∇′G|q + |G|q)dx′ · ρ−q =∫G2ρρ4

(ρq|∇G|q + |G|q)ρ−q−Ndx ≤

≤ C∫G2ρρ4

(rq|∇G|q + |G|q

)dx · ρ−q−N ≤ const.

In virtue of (10.2.23), we obtain from (10.2.24)-(10.2.26)

‖v‖qW 2,q(G1

1/2)≤ C.(10.2.27)

From (10.2.22) and (10.2.27) we have

supx′, y′ ∈ G1

1/2

x′ 6= y′

|∇′v(x′)−∇′v(y′)|

|x′ − y′|1−Nq

≤ C5; q > N(10.2.28)

Returning to the variable x and the function u we have:

supx, y ∈ Gρρ/2x 6= y

|∇u(x)−∇u(y)|

|x− y|1−Nq

≤ C5ρ−1+N

q , q > N, ρ ∈ (0, d).(10.2.29)

Let us recall that from assumptions of our theorem we have κ ≤ κ∗ ≤ 1− Nq .

From this we obtain q ≥ N1− .We take τ = κ−1+N

q ≤ 0. Then from (10.2.29)it follows that

|∇u(x)−∇u(y)| ≤ C5ρτ |x− y|−τ , ∀x, y ∈ Gρρ/2, ρ ∈ (0, d)(10.2.30)

Because x, y ∈ Gρρ/2, then |x− y| ≤ 2ρ and, because τ ≤ 0, |x− y|τ ≥ (2ρ)τ .That's way we obtain:

|∇u(x)−∇u(y)| ≤ C52−τ |x− y|, ∀x, y ∈ Gρρ/2, ρ ∈ (0, d)⇒

supx, y ∈ Gρρ/2x 6= y

|∇u(x)−∇u(y)||x− y|

≤ C52−τ ρ ∈ (0, d).(10.2.31)

Let now x, y ∈ Gd0. If x, y ∈ Gρρ/2,∀ρ ∈ (0, d) we have (10.2.31). If |x− y| >

ρ = |x| then, because of (10.2.18), we have:

|∇u(x)−∇u(y)||x− y|

≤ 2ρ−|∇u(x)| ≤ 2C1.


From this and (10.2.31) it follows that

supx, y ∈ Gd0x 6= y

|∇u(x)−∇u(y)||x− y|

≤ const.(10.2.32)

Because of (10.2.32), (10.2.1) and (10.2.18), we get thatu ∈ C1+(Gd0).

10.2.3. Integral weighted estimates. On the ground of the obtainedin Subsection 10.2.2 estimates we shall deduce integral weighted estimatesof second order generalized derivatives of a strong solution and establish thebest possible exponent of the weight.

Theorem 10.38. Let u(x) be a solution of the problem (QLRP), q > Nand let λ be dened by (2.4.8) for (EVP3). Suppose that assumptions (A)-(E) are satised. In addition, suppose that

(AA) aij(0, u(0), 0) = δji (i, j = 1, ..., N) - the Kronecker symbol.Then there exist the numbers d,C > 0, which do not depend on u, such

that, if b(x), f(x) ∈ W0α(G), g(x) ∈ W1/2

α (∂G), γ(x) ∈ W1/2α−2(∂G) for

4−N − 2λ < α ≤ 2,(10.2.33)

then u(x) ∈ W2α(Gd/20 ) and

(10.2.34)∫

Gd/20

(rαu2

xx + rα−2|∇u|2 + rα−4|u(x)|2)dx ≤ C

|u|20 +

+ |g‖2W

1/2

α (Γ2d0 )

+∫G2d

0

(|∇u|2 + rα(b2(x) + f2(x))

)dx+ 1

.

Proof. We break the proof into three steps:

Step 1. 4−N < α ≤ 2, N ≥ 3

By (10.2.18), we obtain:

(10.2.35)∫Gd0

(rα−2|∇u|2 + rα−4u2

)dx ≤ CmeasΩ

d∫0

rα−2+2 · rN−1dr +

+M20measΩ

d∫0

rα−4+N−1dr ≤ C(α,N,M0,measΩ,κ)(dα+N−2+2 +

+ dα+N−4)≤ const.


By assumption (A), we have aij(x, u, z) ∈ W 1,q(M), q > N and, by theembedding Theorem, aij(x, u, z), i, j = 1, ..., N are uniformly continuous onM. Therefore for ∀δ > 0 exists dδ > 0 such that

|aij(x, u(x), ux(x))− aij(y, u(y), ux(y))| < δ(10.2.36)

if only

|x− y|+ |u(x)− u(y)|+ |ux(x)− ux(y)| < dδ.(10.2.37)

By (10.2.1), (10.2.18) and (10.2.32) we get:

(10.2.38) |x− y|+ |u(x)− u(y)|+ |ux(x)− ux(y)| < d+ C0d1+ + C1d

,

∀x ∈ Gd0.

Now we choose d > 0 such that the inequality

d+ C0d1+ + C1d

≤ dδ(10.2.39)

holds. For such d we may guarantee (10.2.36) in Gd0.Now we shall estimate second derivatives of the problem (QLRP ) so-

lution. We make the transformation x = %x′, u(%x′) = v(x′). Then(x1, ..., xN ) ∈ G%%/2 → G1

1/2 3 (x′1, ..., x′N ) and the function v(x′) satises

the problem: aij(x′)vx′ix′j = F (x′), x′ ∈ G1

1/2,∂v∂~n′ + 1

|x′|γ(%x′)v(x′) = %g((%x′), x ∈ Γ11/2,

(QLRP )′′

where

aij(x′) ≡ aij(%x′, v(x′), %−1vx′(x′)),

(10.2.40)

F (x′) ≡ −%2a(%x′, v(x′), %−1vx′(x′)).

Because of (10.2.36), we can apply Theorem 10.17 about interior and near asmooth portion of the boundary L2−estimate to the solution of the (QLRP )′′

equation:

(10.2.41)∫

G11/2

(v2x′x′ + |∇′v|2)dx′ ≤ C4

∫G2

1/4

(v2(x′) + F 2(x′) +

+ %2(|∇′G|2 + G2)dx′,

where the constant C4 does not depend on v, F, g and it is dened by ν, µ,‖γ(x)‖C1(Γ2

1/4), moduli of continuity of aij(x′) and G2

1/4. Returning to the


variable x and the function u(x) in (10.2.41) we obtain:

(10.2.42)∫

G%%/2

rαu2xxdx ≤ C4

∫G2%%/4

(rα−4u2 + rαa2(x, u, ux) + rα|∇G|2 +

+ rα−2G2)dx.

Putting in (10.2.42) % = 2−kd and summing over k = 0, 1, ..., log2(d/ε)∀ε ∈ (0, d) we get

(10.2.43)∫Gdε

rαu2xxdx ≤ C4

∫G2dε/4

(rα−4u2 + rαa2(x, u, ux) +

+ rα|∇G|2 + rα−2G2)dx

By the assumption (D) and (10.2.18) with regard to (10.2.35), we have:

(10.2.44)∫Gdε

rαu2xxdx ≤ C4

∫G2d

0

(rα−4u2 + rαf2(x) + rαb2(x) +

+ rα−2|∇u|2 + rα|∇G|2 + rα−2G2)dx, ∀ε > 0,

where the constant C4 does not depend on ε. Therefore we can perform thepassage to the limit as ε→ +0, by the Fatou theorem, and we get:

(10.2.45)∫Gd0

rαu2xxdx ≤ C4

∫G2d

0

(rα−4u2 + rαf2(x) + rαb2(x) +

+ rα−2|∇u|2)dx+ ‖g‖2

W1/2

α (Γ2d0 )

.

On the basis of the inequalities (10.2.35) and (10.2.45), we have u(x) ∈W

2α(Gd0). Now we shall prove (10.2.34). Let ζ(r) ∈ C2[0, d] be cut o function

such that

ζ(r) ≡ 1 for r ∈ [0, d/2]; 0 ≤ ζ(r) ≤ 1 for r ∈ [d/2, d];

ζ(r) ≡ 0 for r > d; ζ(d) = ζ ′(d) = 0;

|ζ ′(r)| ≤ C

rfor r ∈ [d/2, d].

We multiply both sides of the (QLRP ) equation by ζ2(r)rα−2u(x) and inte-grate over Gd0. We get:

(10.2.46)∫Gd0

ζ2(r)rα−2u4 udx = −∫Gd0

ζ2(r)rα−2u(aij(x, u, ux)−

− aij(0, u(0), 0))uxixj + a(x, u, ux)dx


We apply the Gauss- Ostrogradskiy formula:∫Gd0

ζ2(r)rα−2u4 udx =∫∂Gd0

ζ2(r)rα−2u∂u

∂~nds−

−∫Gd0

ζ2(r)rα−2|∇u|2dx−

−∫Gd0

ζ(r)ζ ′(r)xirα−3∂u2

∂xidx+

+2− α

2

∫Gd0

ζ2(r)xirα−4∂u2

∂xidx.

(10.2.47)

Because of the (QLRP ) boundary condition and by properties of ζ(r), weobtain: ∫

Gd0

ζ2(r)rα−2u4 udx = −∫Gd0

ζ2(r)rα−2|∇u|2dx−

−∫Gd0


∂xidx+

+2− α

2

∫Gd0


∂xidx+

+∫Γd0

ζ2(r)rα−2ug(x)− 1rγ(x)uds.

(10.2.48)

Now we calculate the second and the third integral from the right in (10.2.48).For this we use the Gauss-Ostrogradskiy formula once more:

(10.2.49)∫Gd0


∂xidx =

∫∂Gd0

ζ(r)ζ ′(r)rα−3u2xi cos(~n, xi)ds−

−∫Gd0

u2 ∂

∂xi

(ζ(r)ζ ′(r)xirα−3

)dx =

∫Ωd∪Γd0

ζ(r)ζ ′(r)rα−3u2xi cos(~n, xi)ds−

−∫Gd0

u2⟨ζ ′

2(r)rα−2 + ζ(r)ζ ′′(r)rα−2 + (α+N − 3)ζ(r)ζ ′(r)rα−3⟩dx


and

(10.2.50)∫Gd0


∂xidx =

∫∂Gd0

ζ2(r)rα−4u2xi cos(~n, xi)ds−

−∫Gd0

u2 ∂

∂xi〈ζ2(r)xirα−4〉dx =

∫Ωd∪Γd0

ζ2(r)rα−4u2xi cos(~n, xi)ds−

−∫Gd0

u2⟨

2ζ(r)ζ ′(r)rα−3 + (α+N − 4)ζ2(r)rα−4⟩dx.

Because ζ(r)∣∣∣Ωd

= 0, ζ ′(r)∣∣∣Γd/20

= 0 and xi cos(~n, xi)∣∣∣Γd0

= 0, from

(10.2.46)- (10.2.50) it follows that

(10.2.51)∫Gd0

ζ2(r)rα−2|∇u|2dx+(2− α)(α+N − 4)

2

∫Gd0

ζ2(r)rα−4u2dx+

+∫Γd0

ζ2(r)rα−3γ(x)u2ds =∫Γd0

ζ2(r)rα−2ug(x)ds+

+∫

Gdd/2

u2⟨

(2α+N − 5)ζ(r)ζ ′(r)rα−3 + ζ(r)ζ ′′(r)rα−2 + (ζ ′(r))2rα−2⟩dx+

+∫Gd0

ζ2(r)rα−2u(aij(x, u, ux)− aij(0, u(0), 0))uxixj + a(x, u, ux)dx.

Now, by the Cauchy inequality, we estimate the rst integral from the right:

∫Γd0

ζ2(r)rα−2|u||g|ds =∫Γd0

ζ2(r)(rα−1

21√γ(x)

|g|)(rα−3

2

√γ(x)|u|

)ds ≤

≤ 12

∫Γd0

ζ2(r)rα−3γ(x)u2ds+1

2γ0

∫∂G

rα−1g2ds.(10.2.52)


Choosing adequate δ in (10.2.52) we obtain from (10.2.51) the estimate

(10.2.53)∫Gd0

ζ2(r)rα−2|∇u|2dx+(2− α)(α+N − 4)

2

∫Gd0

ζ2(r)rα−4u2dx+

+12

∫Γd0

ζ2(r)rα−3γ(x)u2ds ≤∫

Gdd/2

u2〈(2α+N − 5)ζ(r)ζ ′(r)rα−3 +

+ ζ(r)ζ ′′(r)rα−2 + (ζ ′(r))2rα−2〉dx+∫Gd0

ζ2(r)rα−2ua(x, u, ux)dx+

+∫Gd0

ζ2(r)rα−2uuxixj (aij(x, u, ux)− aij(0, u(0), 0))dx+

+∫Gd0

ζ2(r)rα−2ua(x, u, ux)dx+1

2γ0

∫∂G

rα−1g2ds.

Using the Cauchy inequality, (10.2.36) and (10.2.45) we obtain:

(10.2.54)∫Gd0

ζ2(r)rα−2uuxixj (aij(x, u, ux)− aij(0, u(0), 0))dx ≤

≤ δ∫Gd0

rα−2|uxx||u|dx ≤δ

2

∫Gd0

⟨rα|uxx|2 + rα−4u2

⟩dx ≤

≤ δC5

∫G2d

0

⟨rα−4u2 + rαf2(x) + rαb2(x) + rα−2|∇u|2

⟩dx+

+1

2γ0

∫∂G

rα−1g2ds, ∀δ > 0.

From the assumption (D) and by (10.2.1), (10.2.18) we get:

rα−2ua(x, u, ux) ≤ rα−2|u|⟨µ1|∇u|2 + b(x)|∇u|+ f(x)

⟩≤

≤ |∇u|⟨µ1r

α−2|u||∇u|+ rα−2b(x)|u|⟩

+ rα−2|u||f | ≤

≤ C1d⟨rα−2|∇u|2 + rα−4u2 + rαb2(x)

⟩+δ

2rα−4u2 +

+12δrαf2, ∀δ > 0


and therefore:

(10.2.55)∫Gd0

ζ2(r)rα−2ua(x, u, ux)dx ≤ Cd∫Gd0

ζ2(r)rα−2|∇u|2dx+

+(Cd +

δ

2

)∫Gd0

ζ2(r)rα−4u2(x)dx+ Cd∫Gd0

ζ2(r)rαb2(x)dx+

+12δ

∫Gd0

ζ2(r)rαf2(x)dx, ∀δ > 0.

From (10.2.53) and (10.2.54), (10.2.55) it follows that∫Gd/20

rα−2|∇u|2dx+(2− α)(α+N − 4)

2

∫Gd/20

rα−4u2dx ≤

≤ C5(δ + d)∫

Gd/20

〈rα−2|∇u|2 + rα−4u2〉dx+

+C6

∫G2d

0

rα(b2(x) + f2(x))dx+

+C7

∫G2dd/2

(|∇u|2 + u2)dx+1

2γ0

∫∂G

rα−1g2ds, ∀δ > 0.

(10.2.56)

In our case N +α− 4 > 0. If α < 2 then we choose d, δ appropriate positivesmall and obtain

(10.2.57)∫

Gd/20

rα−2|∇u|2dx ≤ C|u|20 +

12γ0

∫∂G

rα−1g2ds+

+∫G2d

0

|∇u|2 + rα(b2(x) + f2(x))

dx.

If α = 2 then again for appropriate positive small d, δ and, because of(10.2.35), we get the validity of (10.2.57). Now we use Lemma 1.40. From(10.2.45), (10.2.57) with regard to (10.2.35) follows (10.2.34).


Step 2. α = 4−N, N ≥ 2.

Because of (10.2.1) and (10.2.18), we obtain:

(10.2.58)∫Gd0

(r2−N |∇u|2 + r−N |u(x)|2

)dx ≤

≤ CmeasΩd∫

0

r2−N+2 · rN−1dr ≤ Cd2+2 ≤ const.

Hence it follows that u ∈ W

12−N (G). We repeat verbatim the arguments of

the deduction of (10.2.45) and (10.2.56) for α = 4−N ; we obtain

(10.2.59)∫Gd0

r4−Nu2xxdx ≤ C4

‖g‖2

W1/2

4−N (Γ2d0 )

+∫G2d

0

(r−Nu2 + r4−Nf2(x) +

+ r4−Nb2(x) + r2−N |∇u|2)dx

;

(10.2.60)12

∫Γd/20

r1−Nγ(x)u2(x)ds+∫

Gd/20

r2−N |∇u|2dx ≤

≤ C5(δ+d)∫

Gd/20

〈r2−N |∇u|2 + r−Nu2〉dx+C6

∫G2d

0

r4−N (b2(x) + f2(x))dx+

+ C7

∫G2dd/2

(|∇u|2 + u2)dx+ C8‖g‖2W

1/2

4−N (Γ2d0 ), ∀δ > 0.

Since u ∈ W12−N (G) we can apply the Hardy - Friedrichs - Wirtinger inequal-

ity (2.5.13) for α = 4−N. Then from (10.2.59) and (10.2.60) we obtain againthe validity of (10.2.34).

Step 3. 4−N − 2λ < α < 4−N .

From the assumption (D) it follows that b(x), f(x) ∈ W

04−N (Gd0). In the

second step we proved that u(x) ∈ W24−N (Gd/20 ); it means∫

Gd/20

(r4−Nu2

xx + r2−N |∇u|2 + r−N |u(x)|2)dx <∞.(10.2.61)

In this step we use the quasi-distance rε(x) =

√(x1 + ε)2 +

N∑i=2x2i (see 1.4).


Similarly to (10.2.41) we obtain

(10.2.62)∫

G11/2

(u2x′x′ + |∇′u|2)dx′ ≤ C4

∫G2

1/4

(u2(x′) + %4a2(%x′, u, %−1ux′) +

+ %2(|∇′G|2 + G2))dx′,

We put % = 2−kd and notice that

2−k−1d+ ε < r + ε < 2−kd+ ε

in G(k) and

2−k−2d+ ε < r + ε < 2−k+1d+ ε

in G(k−1) ∪ G(k) ∪ G(k+1). We multiply the inequality (10.2.62) by (2−kd +ε)α−2. Returning to the variable x, we obtain:

(10.2.63)∫G(k)

(r2(r + ε)α−2u2

xx + (r + ε)α−2|∇u|2)dx ≤

≤ C∫

G(k−1)∪G(k)∪G(k+1)

(r−2(r + ε)α−2u2 + r2(r + ε)α−2a2(x, u, ux) +

+ r2(r + ε)α−2|∇G|2 + (r + ε)α−2G2)dx, ∀ε > 0

and, in virtue of property 2) for rε(x), we have:

(10.2.64)∫G(k)

(r2rα−2

ε u2xx + rα−2

ε |∇u|2)dx ≤

≤ C∫

G(k−1)∪G(k)∪G(k+1)

(r−2rα−2

ε u2 + r2rα−2ε a2(x, u, ux) +

+ r2rα−2ε |∇G|2 + rα−2

ε G2)dx, ∀ε > 0.

Whence, by summing over all k = 0, 1, 2, ..., we get:

(10.2.65)∫Gd0

(r2rα−2

ε u2xx + rα−2

ε |∇u|2)dx ≤

∫G2d

0

(r−2rα−2

ε u2 +

+ r2rα−2ε a2(x, u, ux) + r2rα−2

ε |∇G|2 + rα−2ε G2

)dx, ∀ε > 0.


Now we multiply both sides of the (QLRP )0 equation by ζ2(r)rα−2ε u(x) and

integrate over Gd0; we obtain:

(10.2.66)∫Gd0

ζ2(r)rα−2ε u4 udx = −

∫Gd0

ζ2(r)rα−2ε u

a(x, u, ux) +

+ (aij(x, u, ux)− aij(0, u(0), 0))uxixjdx, ∀ε > 0.

Using the Gauss - Ostrogradskiy formula in the integral from the left andthe (QLRP )0 boundary condition we obtain:∫

Gd0

ζ2(r)rα−2ε u4 udx = −

∫Gd0

ζ2(r)rα−2ε |∇u|2dx−

−∫Gd0

ζ(r)ζ ′(r)rα−2ε

xir

∂u2

∂xidx+

+2− α

2

∫Gd0

ζ2(r)rα−3ε

∂rε∂xi

∂u2

∂xidx+

+∫Γd0

ζ2(r)rα−2ε u

g(x)− 1

rγ(x)u

ds.

(10.2.67)

From (10.2.66), (10.2.67) it follows that

(10.2.68)∫Gd0


∫Γd0

ζ2(r)rα−2ε u2 1

rγ(x)ds =

=∫Gd0

∂u2

∂xi

2− α2

ζ2(r)rα−3ε

∂rε∂xi− ζ(r)ζ ′(r)rα−2

ε

xir

dx+

+∫Gd0

ζ2(r)rα−2ε u

(aij(x, u, ux)− aij(0, u(0), 0))uxixj + a(x, u, ux)

dx+

+∫Γd0

ζ2(r)rα−2ε ug(x)ds.

We shall estimate the rst integral from the right. For this we use the Gauss- Ostrogradskiy formula once more and take into account property 5) of rε(x)and Lemma 1.10

xi cos(~n, xi)∣∣∣Γd0

= 0, cos(~n, x1)∣∣∣Γd0

= − sinω0

2, ζ(d) = 0.


As a result we obtain:

(10.2.69)∫Gd0

∂u2

∂xi

2− α2

ζ2(r)rα−3ε

∂rε∂xi− ζζ ′rα−2

ε

xir

=

= −2− α2

ε sinω0

2·∫Γd0

ζ2(r)rα−4ε u2ds+

α− 22

∫Gd0

u2 ∂

∂xi

(ζ2(r)rα−3

ε

∂rε∂xi

)dx+

+∫Gd0

u2 ∂

∂xi

(ζζ ′rα−2

ε

xir

)dx = −2− α

2ε sin

ω0

2·∫Γd0


+(α− 2)(α+N − 4)

2

∫Gd0


+2(α−2)∫Gd0

ζζ ′rα−4ε

(r + ε

x1

r

)u2dx+

∫Gd0

u2rα−2ε

(ζ ′

2 + ζζ ′′ +N − 1r

ζζ ′)dx.

Finally, from (10.2.68), (10.2.69) we get:

∫Gd0


2− α2

ε sinω0

2·∫Γd0


+∫Γd0


rγ(x)ds =

(2− α)(4− α−N)2

∫Gd0


+2(α−2)∫Gd0

ζζ ′rα−4ε

(r + ε

x1

r

)u2dx+

∫Gd0

u2rα−2ε

(ζ ′

2 + ζζ ′′ +N − 1r

ζζ ′)dx

+∫Γd0

ζ2(r)rα−2ε ug(x)ds+

∫Gd0

ζ2(r)rα−2ε u

a(x, u, ux) +


Recalling properties of the function ζ(r) hence follows

∫Gd0


∫Γd0


rγ(x)ds ≤(10.2.70)


≤ (2− α)(4− α−N)2

∫Gd0

ζ2(r)rα−4ε u2dx+ c

∫Gdd/2

u2dx+

+∫Γd0

ζ2(r)rα−2ε |u||g(x)|ds+

∫Gd0

ζ2(r)rα−2ε u

a(x, u, ux) +


Let d > 0 be such that (10.2.39) and (10.2.36) hold. Using the Cauchyinequality we obtain:

(10.2.71)∫Gd0

ζ2(r)rα−2ε (aij(x, u, ux)− aij(0, u(0), 0))vvxixjdx ≤

≤ δ∫Gd0

ζ2(r)rα−2ε (r|uxx|)(r−1|u|)dx ≤

≤ δ

2

∫Gd0

⟨ζ2(r)r2rα−2

ε u2xx + ζ2(r)r−2rα−2

ε u2⟩dx, ∀δ > 0

In addition, by the assumption (D) and estimates (10.2.1), (10.2.18), we get:

(10.2.72)∫Gd0

ζ2(r)rα−2ε |u||a(x, u, ux)|dx ≤

≤ 12

(C1d + δ)

∫Gd0

ζ2(r)rα−2ε r−2u2dx+

12δ

∫Gd0

ζ2(r)rα−2ε r2f2dx+

+ µ1C0d1+

∫Gd0


12C1d

∫Gd0

ζ2(r)rα−2ε r2b2dx, ∀δ > 0.

Because of the property of rε(x), we have rε ≥ r. From α ≤ 2 it follows thatrα−2ε ≤ rα−2. We know also that b(x), f(x) ∈ W0

α(G) and therefore:

12δ

∫Gd0

ζ2(r)rα−2ε r2f2dx ≤ 1

2δ

∫G

rαf2dx;

(10.2.73)12C1d

∫Gd0

ζ2(r)rα−2ε r2b2dx ≤ 1

2C1d

∫G

rαb2dx.

By the Cauchy inequality with regard to γ(x) ≥ γ0 > 0,

|u||g| =(r1/2 1√

γ(x)|g|)(r−1/2

√γ(x)|u|

)≤ δ

2r−1γ(x)u2 +

12δγ0

rg2, ∀δ > 0.


Taking into account the rst property of rε we obtain

(10.2.74)∫Γd0

ζ2(r)rα−2ε ug(x)ds ≤ δ

2

∫Γd0

ζ2(r)rα−2ε

1rγ(x)u2ds+

+1

2δγ0

∫Γd0

rα−1g2ds, ∀δ > 0.

From (10.2.70)-(10.2.74) it follows that∫Gd/20

rα−2ε |∇u|2dx+ (1− δ)

∫Γd/20

r−1rα−2ε γ(x)ds ≤

≤ δ

2

∫Gd0

r2rα−2ε u2

xxdx+(2− α)(4−N − α)

2

∫Gd/20

rα−4ε u2dx+

+12

(C1d + 2δ)

∫Gd/20

rα−2ε r−2u2dx+ Cd1+

∫Gd/20

rα−2ε |∇u|2dx+(10.2.75)

+C∫G2d

0

(u2 + |∇u|2)dx+ C

∫G2d

0

rα(b2 + f2)dx+

+1

2δγ0

∫Γd0

rα−1g2ds, ∀δ > 0.

Taking into account (10.2.65), (10.2.75) and choosing suciently small δ > 0we get: ∫

Gd/20

(r2rα−2ε u2

xx + rα−2ε |∇u|2)dx+

∫Γd/20

r−1rα−2ε γ(x)ds ≤

≤ (2− α)(4−N − α)2

∫Gd/20

rα−4ε u2dx+ C(d + δ)

∫Gd/20

rα−2ε r−2u2dx+

+C∫G2d

0

(u2 + |∇u|2 + rα(b2 + f2) + rα|∇G|2 + rα−2G2

)dx+(10.2.76)

+1

2δγ0

∫Γd0

rα−1g2ds+ Cd2∫

Gd/20

rα−2ε |∇u|2dx, ∀δ > 0, ∀ε > 0.


Since by (10.2.61) u(x) ∈ W24−N (Gd/20 ), we can apply Theorem 2.21 and then

we have (see the inequality (2.4.9))

∫Ω

u2(r, ω)dΩ ≤ 1λ(λ+N − 2)

∫Ω

|∇ωu(r, ω)|2dΩ +∫∂Ω

γ(x)u2(x)dσ,

for a.e. r ∈ (0, d).Multiplying both sides of this inequality by (%+ε)α−2rN−3

and integrating over r ∈ (%2 , %) we obtain

∫G%%/2

(%+ ε)α−2r−2u2dx ≤ 1λ(λ+N − 2)

∫G%%/2

(%+ ε)α−2|∇u|2dx+

+1

λ(λ+N − 2)

∫Γ%%/2

r−1(%+ ε)α−2γ(x)u2ds, ∀ε > 0

or since %+ ε ∼ rε

∫G%%/2

rα−2ε r−2u2dx ≤ 1

λ(λ+N − 2)

∫G%%/2


+∫

Γ%%/2

r−1rα−2ε γ(x)u2ds

, ∀ε > 0.

Letting ρ = 2−kd, (k = 0, 1, 2, ...) and summing obtained inequalities over kwe get:

(10.2.77)∫Gd0

rα−2ε r−2u2dx ≤ 1

λ(λ+N − 2)

∫Gd0


+∫Γd0

r−1rα−2ε γ(x)u2ds

, ∀ε > 0.


In addition, Lemma 2.38 holds. Therefore from (10.2.76), in virtue of (10.2.77)and (2.5.20) together with (2.5.18), (2.5.19), we obtain

K(λ,N, α)

∫

Gd/20


∫Γd/20

r−1rα−2ε γ(x)u2(x)ds

+

+∫

Gd/20

r2rα−2ε u2

xxdx ≤ O(ε)

∫

Gd/20


∫Γd/20


+

+1

2δγ0

∫Γd0

rα−1g2ds+ C

∫G2d

0

(u2 + |∇u|2 +

(10.2.78)

+rα(b2 + f2) + rα|∇G|2 + rα−2G2)dx+

+C(d + δ)

∫

Gd/20


∫Γd/20


, ∀δ > 0, ∀ε > 0,

where

K(λ,N, α) = 1− (2− α)(4−N − α)2

H(λ,N, α) =

= 1− 2(2− α)(4−N − α)(4−N − α)2 + 4λ(λ+N − 2)

> 0,

because of 4−N − 2λ < α < 4−N. We choose δ = K(λ,N,α)4C and d > 0 such

that d ≤ K(λ,N,α)4C ; as a result we get

∫Gd/20

r2rα−2ε u2

xx + (1−O(ε))

∫

Gd/20


∫Γd/20


≤≤ C

∫G2d

0

(u2 + |∇u|2 + rα(b2 + f2) + rα|∇G|2 + rα−2G2

)dx+

+C∫Γd0

rα−1g2ds, ∀ε > 0.(10.2.79)

We observe that the right side does not depend on ε. Therefore we canperform the passage to the limit as ε → +0, by the Fatou Theorem. Hence


we get:

(10.2.80)∫

Gd/20

(rαu2

xx + rα−2|∇u|2)dx ≤ C

∫G2d

0

(u2 + |∇u|2 +

+ rα(b2 + f2) + rα|∇G|2 + rα−2G2)dx+ C

∫Γd0

rα−1g2ds.

Finally, using Lemma1.40 we obtain the desired estimate (10.2.34).

Theorem 10.39. Let u(x) be a solution of the problem (QLRP), q > Nand let λ be dened by (2.4.8) for (EVP3). Suppose that assumptions (A)-

(E) are satised for β > λ−2. Suppose, in addition, that g(x) ∈W

1/2

4−N (∂G)and there is

supρ>0

ρ−λ−ε‖g‖ W

1/2

4−N (Γρ0):= k2, ∀ε > 0.(10.2.81)

Then there exist numbers d,C > 0 not depending on u such that

u(x) ∈ W24−N (Gd/20 ) and

‖u(x)‖ W

2

4−N (Gρ0)≤ C

(‖u‖W 1(G) + k1 + k2

)ρλ, ρ ∈ (0,

d

2).(10.2.82)

Proof. The belonging u(x) ∈ W24−N (Gd/20 ) follows from Theorem 10.38.

So it is enough to derive the estimate (10.2.82). We set

V (ρ) ≡∫Gρ0

r2−N |∇u|2dx+∫Γρ0

r1−Nγ(x)u2ds(10.2.83)

and multiply both sides of the (QLRP )0 equation by r2−Nu(x) and integrateover Gρ0, ρ ∈ (0, d2). As a result we obtain :

V (ρ) =∫Ω

(ρu∂u

∂r+N − 2

2u2)dω +

∫Γρ0

r2−Nugds+

+∫Gρ0

u(x)r2−N

(aij(x, u, ux)− aij(0, u(0), 0))uxixj +

+ a(x, u, ux)dx, ρ ∈ (0,

d

2).

(10.2.84)

We shall obtain an upper bound for each integral on the right. First of them,we use Lemma 2.36:∫

Ω

(%u∂u

∂r+N − 2

2u2

)dΩ ≤ %

2λV ′(%).(10.2.85)


We estimate the second integral in (10.2.84); by the Cauchy inequality withregard to Lemma1.40, we get:

(10.2.86)∫Γρ0

r2−Nugds =∫Γρ0

(r

1−N2 γ1/2(x)u(x)

)(r

3−N2 γ−1/2(x)g(x)

)ds ≤

≤ δ

2

∫Γρ0

r1−Nγ(x)u2(x)ds+1

2δγ0

∫Γρ0

r3−Ng2(x)ds ≤

≤ δ

2

∫Γρ0

r1−Nγ(x)u2(x)ds+ C‖g(x)‖2W

1/2

4−N (Γρ0)

, ∀δ > 0.

To estimate the third integral in (10.2.84) we use the assumption (A) :

aij(x, u, z) ∈W 1,q(M), q > N ⇒ aij(x, u, z) ∈ Cδ(M), 0 < δ < 1− N

q,

(i, j = 1, ..., N),

by the embedding Theorem. The last together with (10.2.38) means that

|aij(x, u, ux)− aij(0, u(0), 0)| ≤ δ(%), |x| ≤ %,

where

δ(%) ∼ %δ, δ ∈(

0, 1− N

q

).(10.2.87)

Therefore, by the Cauchy and Hardy- Friedrichs - Wirtinger (2.5.13) inequa-lities, we obtain

(10.2.88)∫Gρ0

r2−N |u(x)||uxixj | |aij(x, u, ux)− aij(0, u(0), 0)| dx ≤

≤ δ(ρ)∫Gρ0

r4−N |uxx|2dx+ Cδ(ρ)V (ρ).

We apply the inequality (10.2.59) and once more the Hardy - Friedrichs -Wirtinger inequality (2.5.13); then from (10.2.88) we get

(10.2.89)∫Gρ0

r2−N |u(x)||uxx| |aij(x, u, ux)− aij(0, u(0), 0)| dx ≤

≤ Cδ(ρ)V (ρ) + V (2ρ) + ‖f‖2

W0

4−N (G2ρ0 )

+ ‖b‖2W

0

4−N (G2ρ0 )

+

+ ‖g‖2W

1/2

4−N (Γ2ρ0 )

.


Similarly to (10.2.55), considering the Hardy - Friedrichs - Wirtinger inequal-ity (2.5.13), we obtain

(10.2.90)∫Gρ0

r2−Nu(x)a(x, u, ux)dx ≤ C

(ρ + δ)V (ρ) +

+ ρ∫Gρ0

r4−Nb2(x))dx+

12δ

∫Gρ0

r4−Nf2(x)dx, ∀δ > 0.

From (10.2.84), in virtue of (10.2.86)-(10.2.90) for δ = %ε, ∀ε > 0, it followsthat

V (%) ≤ %

2λV ′(%) + Cδ(%)V (2%) + C (δ(%) + % + %ε)V (%) +

+ C‖b‖2

W0

4−N (G2%0 )

+ %−ε‖f‖2W

0

4−N (G2%0 )

+ ‖g‖2W

1/2

4−N (Γ2ρ0 )

.

(10.2.91)

Hence it follows the Cauchy problem for dierential inequality:V ′(ρ)− P(%)V (%) +N (ρ)V (2ρ) +Q(ρ) ≥ 0, 0 < ρ < d,

V (d) ≤ V0,(CP )

where

P(ρ) =2λρ− C

(δ(ρ)ρ

+ ρ−1 + ρε−1

);(10.2.92)

N (ρ) = Cδ(ρ)ρ

;(10.2.93)

(10.2.94) Q(ρ) = C‖b‖2

W0

4−N (G2%0 )

+ %−ε‖f‖2W

0

4−N (G2%0 )

+

+ ‖g‖2W

1/2

4−N (Γ2ρ0 )

ρ−1.

We adjoin to it the initial condition V (d) ≤ V0. By Theorem 10.38 forα = 4−N,

V (d) =∫Gd0

r2−N |∇u|2dx+∫Γd0

r1−Nγ(x)u2ds ≤ C

|u|20 +

+∫G2d

0

(|∇u|2 + r4−N (b2(x) + f2(x))

)dx+

+ ‖g‖2W

1/2

4−N (Γ2d0 )

+ 1

≡ V0.

(10.2.95)


By Theorem 1.57,

(10.2.96) V (%) ≤ exp( d∫%

B(τ)dτ)

V0 exp(−

d∫%

P(τ)dτ)

+

+

d∫%

Q(τ) exp(−

τ∫%

P(σ)dσ)dτ

with

B(%) = N (%) exp( 2%∫%

P(σ)dσ).

Now, by means of simple calculations, from (10.2.92), (10.2.93) with regardto (10.2.87) we have:

exp(−

d∫ρ

P(τ)dτ)≤ C

(ρd

)2λ;

d∫ρ

B(τ)dτ ≤ C = const.(10.2.97)

In addition,

(10.2.98) Q(τ) exp(−

τ∫ρ

P(σ)dσ)≤ C%2λτ−2λ−1

‖b‖2

W0

4−N (G2τ0 )

+

+ τ−ε‖f‖2W

0

4−N (G2τ0 )

+ ‖g‖2W

1/2

4−N (Γ2τ0 )

, ∀ε > 0.

Let us recall the assumption (D) :

‖b‖2W

0

4−N (G2τ0 )≤ ck2

1τ2β+4; τ−ε‖f‖2

W0

4−N (G2τ0 )≤ ck2

1τ2β+4−ε.

Since β > λ− 2, we can put β = λ− 2 + ε, ∀ε > 0. Therefore we get:

‖b‖2W

0

4−N (G2τ0 )

+ τ−ε‖f‖2W

0

4−N (G2τ0 )≤ ck2

1τ2λ+ε, ∀ε > 0.(10.2.99)

From (10.2.98), (10.2.99) with regard to (10.2.81) we obtain

d∫%

Q(τ) exp(−

τ∫ρ

P(σ)dσ)dτ ≤ C(k2

1 + k22)%2λ.(10.2.100)

Finally, from (1.10.1), by (10.2.95), (10.2.97),(10.2.100), it follows that

V (ρ) ≤ C(N,λ, d,κ)⟨‖u‖2W 1,2(G) + k2

1 + k22

⟩%2λ.(10.2.101)

At last, from (10.2.59) and (10.2.101) we deduce the validity of (10.2.82).


10.2.4. The power modulus of continuity at the conical pointfor strong solutions. Now we shall precise the exponent κ in the esti-mates (10.2.1) and (10.2.18) and we prove the Hölder continuity of the rstderivatives of the strong solutions in the neighborhood of a conical point.

Theorem 10.40. Let λ > 1 be dened by (2.4.8) for the problem (EV P3)and let u(x) be the problem (QLRP ) strong solution, q > N. Suppose theassumptions(A), (AA), (C)-(E) are satised for β > λ−2 > −1, δ > λ−1 > 0. Supposethat functions g(x), γ(x) satisfy conditions of Theorem 10.39. Then thereexist numbers d > 0, C0, C1 not depending on u(x), but depending only onN,λ, ν, µ, µ1, β, k1, k2, q, g0,M0,M1 and the domain G, such that

1) |u(x)| ≤ C0|x|λ; |∇u(x)| ≤ C1|x|λ−1, x ∈ Gd/20 .

In addition, if g(x) ∈ V 1−1/qq,α (∂G) and

‖g(x)‖V

1−1/qq,α (Γ%0)

≤ C%λ−2+α+Nq , 0 < ρ < d/2(10.2.102)

then there exist numbers d > 0, C2, not depending on u(x) but only onN,λ, ν, µ, µ1, β, k1, k2, q, g0,M0,M1 and the domain G, such that:

2) if α+ q(λ− 2) +N > 0 then u(x) ∈ V 2q,α(G) and

‖u(x)‖V 2q,α(Gρ0) ≤ C2ρ

λ−2+N+αq ; 0 < ρ < d/2;

3) if 1 < λ < 2, q > N2−λ then u(x) ∈ Cλ(Gd/20 ).

Proof. Let us consider the sets Gρρ/2 and G2ρρ/4 ⊃ G

ρρ/2, 0 < ρ < d/2. We

make the transformation x = ρx′; w(x′) = ρ−λu(ρx′). The function w(x′)satises the problem

aij(x′)wx′ix′j = F (x′), x′ ∈ G11/2,

∂w∂~n′ + 1

|x′|γ(ρx′)w(x′) = %1−λg(ρx′), x ∈ Γ11/2,

(QLRP )′0

where

aij(x′) ≡ aij(ρx′, u, ρλ−1wx′(x′)),

F (x′) ≡ −ρ2−λa(ρx′, ρλw(x′), ρλ−1wx′(x′)).

The Lq− estimate (10.2.23) is satised for the function w(x′) (see proof ofTheorem 10.37):

(10.2.103) ‖w‖qW 2,q(G1

1/2)≤ C3

∫G2

1/4

⟨|w|q + ρqλ|∇′w|2q + %q|b|q|∇′w|q +

+ ρq(2−λ)|f |q⟩dx′ + C4ρ

q(1−λ)

∫G2

1/4

(|∇′g|q + |g|q)dx′,


where the constants C3, C4 do not depend on w.For the rst we consider the case 2 ≤ N < 4. By the Sobolev Imbedding

Theorem, we have:

supx′∈G1

1/2

|w(x′)| ≤ C‖w‖2,2;G11/2.(10.2.104)

Returning to the variable x and because of the estimate (10.2.82) of Theorem10.39, we get:

‖w‖22,2;G11/2

=∫

G11/2

(|wx′x′ |2 + |∇′w|2 + w2

)dx′ ≤

≤ C(N)ρ−2λ

∫G1

1/2

(r4−N |uxx|2 + r2−N |∇u|2 +

+ r−Nu2)dx ≤ C.

(10.2.105)

From (10.2.104), (10.2.105) it follows that

supx′∈G1

1/2

|w(x′)| ≤ C0,

and returning to the variable x we get:

|u(x)| ≤ C0ρλ; x ∈ Gρρ/2.

Putting now |x| = 23ρ we obtain the rst estimate of 1) of our theorem.

Let now N ≥ 4. We apply the Lieberman local maximum principle,Proposition 10.33; then, by the condition (D), we have:

(10.2.106) supx′∈G1

1/2

w(x′) ≤ C

( ∫G2

1/4

w2dx′) 1

2 + ρ1−λ+δg0 +

+ ρ2−λ( ∫G2

1/4

|a(ρx′, ρλw(x′), ρλ−1wx′)|Ndx′) 1N

.

We shall upper bound estimate each integral from the right hand side of(10.2.106). First integral we estimate by (10.2.82) of Theorem 10.39:∫

G21/4

w2dx′ ≤ ρ−2λ

∫G2ρρ/4

r−Nu2dx ≤ C.(10.2.107)


Because of the assumption (D) and the estimate (10.2.18), from (??) itfollows that

(10.2.108)∫

G21/4

|a(ρx′, ρλw(x′), ρλ−1wx′)|Ndx′ ≤ c(N)∫

G2ρρ/4

(µN1 |∇u|2N +

+ bN (x)|∇u|N + fN (x))r−Ndx ≤ c(N)

∫G2ρρ/4

(µN1 (r2−N |∇u|2)×

× (r−2|∇u|2N−2) + (r2−N |∇u|2) · (kN1 rβN−2|∇u|N−2) + kN1 rβN−N

)dx ≤

≤ c(N)(µN1 C

2N−21 ρ2(N−1)−2+kN1 C

N−21 ρ(N−2)+βN−2

) ∫G2ρρ/4

r2−N |∇u|2dx+

+ c(N)(β)−1(k1)N measΩ(2βN − 2−2βN

)ρβN , 0 < ρ < d/2.

Because of (10.2.82), hence we obtain:

(10.2.109) ρ2−λ

( ∫G2

1/4

|a(ρx′, ρλw(x′), ρλ−1wx′)|Ndx′) 1

N

≤

≤ C(ρ2−λ+

2(λ−1)N

+2(N−1)

N +ρ2−λ+β+2(λ−1)N

+(N−2)

N +ρβ+2−λ), ∀ρ ∈ (0,

d

2).

We recall that β > λ− 2, δ ≥ λ− 1. Hence and from (10.2.106),(10.2.107),(10.2.109) it follows that

supx′∈G1

1/2

w(x′) ≤ C1 + C2ρ2−λ+

2(λ−1)N

+2(N−1)

N .(10.2.110)

We recall as well as that λ > 1 and κ > 0. To prove the validity of 1) (likeas in the rst case) is enough to obtain the estimate:

supx′∈G1

1/2

w(x′) ≤ const.(10.2.111)

We shall show that the repetition by the nite time of the procedure of thereceiving of the (10.2.110) for various κ can lead to the estimate (10.2.111).

Let the exponent of ρ in (10.2.110) be negative (otherwise the (10.2.110)means the (10.2.111)). Returning to the function u(x) in (10.2.110) andputting |x| = 2

3ρ we obtain:

u(x) ≤ C|x|2+2(λ−1)N ,(10.2.112)

and hence, by Theorem 10.35 for κ = κ1,

κ1 = 1 +2(λ− 1)

N,(10.2.113)


we get:

|∇u(x)| ≤ C|x|1 .(10.2.114)

Let us repeat the procedure of the receiving of inequalities (10.2.109) and(10.2.110), applying the estimate (10.2.114) instead of the (10.2.18) (i.e.changing κ for κ1); as a result we obtain:

supx′∈G1

1/2

w(x′) ≤ C1 + C2ρ2−λ+

2(λ−1)N

+21(N−1)

N .(10.2.115)

If the exponent of ρ in (10.2.115) is negative, then putting

κ2 = 1 +2(λ− 1)

N+

2(N − 1)N

κ1,(10.2.116)

by Theorem 10.35 for κ = κ2, we get:

|∇u(x)| ≤ C19|x|2 ,(10.2.117)

and next repeating above procedure we get the estimate:

supx′∈G1

1/2

w(x′) ≤ C1 + C2ρ2−λ+

2(λ−1)N

+22(N−1)

N .(10.2.118)

Letting

t =2(N − 1)

N≥ 3

2∀N ≥ 4,(10.2.119)

we consider the number sequence κk:

κ1 dened by (10.2.113);

κ2 = κ1(1 + t);

κ3 = κ2(1 + t+ t2);· · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·

κk+1 = κ1(1 + t+ ...+ tk) = κ1tk+1 − 1t− 1

; k = 0, 1, ....(10.2.120)

Repeating the stated process k times we obtain estimates:

(10.2.121) supx′∈G1

1/2

w(x′) ≤ C1 + C2ρ1−λ+k+1 , 0 < ρ < d/2;

k = 0, 1, ....

Now we shall show that ∀N ≥ 4 exists integer k such that:

1− λ+ κk+1 ≥ 0.(10.2.122)

From (10.2.113) and (10.2.120) we have:

1− λ+ κk+1 =tk+1 − 1t− 1

+λ− 1

N(t− 1)(2tk+1 − 2−Nt+N).

10.3 Notes 455

The rst term on the right is positive, by (10.2.119). For the second termfrom (10.2.119) it follows that

2tk+1 − 2−Nt+N = 2k+2(1− 1/N)k+1 −N ≥ 0

if (2N − 2)/Nk+1 ≥ N/2. Hence we get that (10.2.122) holds if

k + 1 ≥ln N

2

ln 2N−2N

.

Choosing k =

[ln N

2

ln 2N−2N

], where [a] is the integer part of a, we guarantee

(10.2.122) ∀N ≥ 4. By this, the validity of 1) of our theorem is proved.

The validity of the second estimate we get from Theorem 10.35 forκ = λ− 1.

Now we shall prove the validity of 2). Returning to the variable x andthe function u(x) in (10.2.103) we have:∫G%%/2

(|uxx|q + %−q|∇u|q + %−2q|u|q

)dx ≤ C4

∫G2%%

(%−2q|u|q + |∇u|2q +

+ |b|q|∇u|q + |f |q + |∇g|q + %−q|g|q)dx.

Multiplying this inequality by %α, replacing % by 2−k% and summing over allk = 0, 1, ... we obtain:


≤ C4

∫G2%

0

(rα−2q|u|q + rα|∇u|2q + rα|b|q|∇u|q +

+ rα|f |q + rα|∇g|q + rα−q|g|q)dx; ∀q > 1.

Using estimates from 1), by the assumption (D) and the assumption (10.2.102)of our theorem, taking into consideration β > λ− 2 > −1 we get:

‖u‖V 2q,0(Gρ0) ≤ C%

λ−2+α+Nq(10.2.123)

if only α +N + (λ− 2)q > 0 . From (10.2.123) we obtain the validity of 2)of our theorem.

Finally, repeating verbatim the proof of Theorem 10.37 for κ = λ − 1,we obtain the validity of 3) of our theorem.

10.3. Notes

Many mathematicians have considered the third boundary value prob-lem. The oblique derivative problem for elliptic equations in non smooth do-mains investigated M.Faierman [120], M.Garroni, V.A.Solon-


nikov and M.Vivaldi [126],P. Grisvard [132], G. Liberman [222, 223, 229,231, 224], H.Reisman [347] etc.

P. Grisvard investigated (Chapter 4 [132]) the properties of the se-cond weak derivatives of the weak solutions of the oblique problem for theLaplace operator in a plane domain with a polygonal boundary. He estab-lished W 2,p− a priori estimates for such solutions and conditions, when suchestimates hold.

M. Dauge and S. Nicaise [93] investigated oblique derivative and inter-face problems associated to the Laplace operator on a polygon: they obtainedindex formulae, a calculus of the dimension of the kernel, an expansion ofthe weak solutions into regular and singular parts and formulae for the co-ecients of the singularities in such expansions

M. Faierman [120] extended the P. Grisvard results to the elliptic oper-ator of the form

L = −N∑i=1

aii(x)D2i +

N∑i=1

ai(x)Di + a(x),

in a N− dimensional rectangle.H.Reisman [347] considered elliptic boundary value problems for the

equation from (L) with innitely dierentiable coecients in a bounded do-main Ω ⊂ RN (N ≥ 3) with non smooth boundary that has dihedral edges.He considered boundary conditions that are an oblique derivative on one sideof the edge and an oblique derivative or a Dirichlet condition on the otherside of the edge. The main results in his work are uniqueness, existence andregularity theorems for such problems in weighted Sobolev spaces.

M.G.Garroni, V.A.Solonnikov and M.A.Vivaldi [126] considered the fol-lowing elliptic boundary value problem for the Poisson equation on the in-nite angle: −∆u+ su = f(x), x ∈ dϑ,(

∂u∂−→n + hi

∂u∂r

)∣∣∣γi

= ϕi(r), i = 0, 1,

where dϑ ⊂ R2 is the innite angle of opening ϑ ∈ (0, 2π] with sides γ0 andγ1 given by

γ0 = 0 ≤ x1 <∞, x2 = 0,

γ1 = x1 = r cosϑ, x2 = r sinϑ, 0 ≤ r =√x2

1 + x22 <∞

in a Cartesian coordinate system x1, x2; −→n is the exterior normal to γi; h0

and h1 are given real constants; s is a complex parameter with Re s ≡ a2 ≥ 0.Authors obtained estimates of the solution of the above problem, whichare uniform with respect to s in weighted Sobolev spaces introduced byV.A.Kondrat'ev for the investigation of elliptic boundary value problems indomains with angular and conical points at the boundary. In this spaces the

10.3 Notes 457

distance |x| from the origin, with an appropriate exponent, is the weight.The spaces, in which the solution exists, depend on the sign of h0 + h1.

At last, oblique derivative problems in Lipschitz domains was investigat-ed by G. Lieberman [222, 223, 229, 231, 224 ]. He studied the problemof the existence and the regularity of solutions in Lipschitz domains for el-liptic equations with Hölder continuous coecients. He proved [222, 231]the local and global maximum principle (see Propositions 10.11, 10.14) forthe oblique derivative problem for general second order linear and quasi-linear elliptic equations in arbitrary Lipschitz domains. Without makingany continuity assumptions on the known functions, he derived the Harnackand Hölder estimates for strong solutions near the boundary of the domain.As well as he bounded the maximum of the solution modulus in terms anappropriate norm and the known functions.

An important element in the study of elliptic equations is the modulus ofcontinuity estimate for the gradient of the solutions. Usually this modulus ofthe continuity estimate is in fact a Hölder estimate, so it is often referred toas a Hölder gradient estimate. For elliptic nonlinear oblique boundary valueproblem in a smooth domain, the Hölder gradient estimate was rst provedby G. Lieberman [233, 234] and by Lieberman - Trudinger [235].

M. Dauge and S. Nicaise [93] investigated the oblique derivative andinterface problems on polygonal domains.


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10 Notation Index 479

Notation Index

: C l(G), 19: C0,A, 26: Ck,A(G), 26: C l0(G), 19: G(k), 12: Gba, 12: Gd, 12: Lp(G), 15: Lploc(G), 17: Lu, 89: M0, 225: M1, 228: V k

p,α(G), 22

: Vk−1/pp,α (Γ), 22

: W k,p(G), 19: W k,p(G \ O), 19: W k,p

0 (G), 19: W k−1/2(Γ), 20: W

k− 1p,p(Γ), 20

: W k(G), 19: W k

0 (G), 19: X(G,Γ), 332: [u]A;G, 26: [f ]α;G, 19: ∆ωu, 13: ∆m, 280: Γ(x− y), 73: Γba, 12: Γd, 12: Ω, 12: Ωρ, 12: Φ(ω), 354:Wkα(G), 22

:Wk−1/2α (Γ), 22

: χ0, 292: λ, 52, 333: λ0, 282, 355358: B, 375: L, 352: D(λ, φ), 282: G(x, y), 85

: M, 225, 423: M(u), 226: N1

m,q(ν, ν0, G), 332: µ(m), 284: µ0, 292: ∇ωu, 13: ω0, 12, 13: θ(r), 47: θε, 332: 4hu(x), 37: ϑ, 4652: ai(z), 281: d(x), 35: dΩ, 12: rε(x), 14: uh, 17: divω, 13

480 10 Notation Index

Index

LpestimatesDirichlet problemlocal boundary, 88global, 88local boundary, 89, 117, 317local interior, 307near a conical point, 118119, 286291

oblique problemlocal boundary, 376

αDini function, 22

apriori estimatesglobal elliptic inequalityDirichlet problem, 174

global in weighted Sobolev spaceDirichlet problem, 90, 156, 196, 231,285

Robin problem, 381, 388, 389, 393local in weighted Sobolev spaceDirichlet problem, 99, 104, 109, 111,160, 164, 197, 222, 224, 246, 253,255, 292, 322

Robin problem, 403, 407, 409, 426,441

power modulus of continuityfor linear Dirichlet problem, 113, 115for linear Robin problem, 371, 372for pseudolaplacian, 293for quasilinear Dirichlet problem, 223,255, 298, 300, 303, 331

for quasilinear Robin problem, 445for semilinear equation, 199, 204

barrier function, 209, 240, 346, 377Beltrami-Laplace operator, 11Bernstein estimate, 89

Caratheodory's function, 219Cauchy's Inequality, 8Clarkson's Inequality, 14comparison principleDirichlet problem

m-Laplacian, 275linear equation, 88semilinear equation, 204

mixed problem, 332, 376conical point, 10convex rotational cone, 11

dierence quotients, 35dierential inequality (CP ), 27Dini - Liapunov region, 131Dini - Liapunov surface, 131Dini continuity, 22, 52Dini gradient continuitylocalnear a conical point, 181smooth boundary, 178

Dirichlet problemfor the Poisson equationin a conical domain, 85in a smooth domain, 71

distance function, 33

eigenvalue, 4445eigenvalue problemfor the Laplace-Beltrami operatorDirichlet boundary condition, 44Robin boundary condition, 45

elliptic inequality, 174elliptic operator, 271equation(ECC), 83(ME), 329

extension lemma, 33

Fatou's Theorem, 14xed point LeraySchauder Theorem, 27Fubini's Theorem, 13fundamental solution, 71

generalized solution, 274global bound of a weak solutionquasilinear Dirichlet problem, 294quasilinear mixed problem, 337

481

482 Index

global Dini regularityfor divergent equationsconical domain, 191

gradient boundlocal for linear Dirichlet problem, 89local for linear Robin problem, 371,372

local for quasilinear Dirichlet problem,220, 223, 255, 293, 305

local for quasilinear Robin problem, 445Green's functionof a ball, 83of a domain, 82of the half-space, 80, 82

Hölder continuity, 17generalized solution, 298weak solution, 172, 286quasilinear equation, 272

Hölder estimatelinear Dirichlet problem, 120122quasilinear Dirichlet problem, 220

Hölder gradient estimatelinear Dirichlet problem, 123, 124, 126quasilinear Dirichlet problem, 220, 274

Hölder's Inequality, 8, 14Hardy inequality, 3941Hardy-Friedrichs-Wirtingertype inequalityDirichlet boundary condition, 5259Robin boundary condition, 59

higher regularitylinear problem, 128, 129quasilinear equation, 260, 265

inequality for boundary and domain in-tegrals, 1821, 55, 56, 60

Interpolation inequality, 14, 18, 25

Jensen's Inequality, 9

linear problemsmoothness in a Dini - Liapunov re-gion, 132

local bound of a strong solutionsemilinear problem, 200, 209

local bound of a weak solutionlinear problem, 168, 171semilinear problem, 211

maximum principlefor quasilinear equations, 271local, 88mixed problem

global, 375of Alexandrov, 87Robin problemglobal, 374local, 374, 418strong, 374

strong of Hopf, 88, 336method of continuity, 26, 144Minkowski's Inequality, 14mixed problem, 325modulus of obliqueness, 373

Newtonian potential, 71

oblique operator, 373

Poincaré inequality, 42, 43problem

(BV P )0, 346(StL), 346(BVP), 325(CPE), 347(DL), 151(DPE), 85(DQL), 271(DSL), 210(EVP1), 44(EVP3), 45(IDL), 174(L), 87(LPA), 274(LRP), 369(NEVP1), 276(NEVP2), 278(PE), 71(QL), 219(QLRP), 369(SL), 195

quasi-distance function, 12

regularization of a function, 15

semilinear equationdivergent, 210nondivergent, 195

set of type (A), 219Sobolev imbedding theorems, 19space of Dini continuous functions, 24Stampacchia's Lemma, 30strong solution, 90, 195, 222, 240, 370,

416

the Bernstein method, 233

Index 483

unbounded solutions, 213, 216uniform ellipticity condition, 87, 90, 152,

181, 195, 211, 219unique solvabilitylinear problemconical domain, 142, 144, 146, 149smooth domain, 87

variational principle, 26, 45

weak eigenfunction, 44, 45, 278weak Harnack inequality, 271weak solution, 152, 210, 271, 275, 329weighted Sobolev space, 20Wirtinger inequalityDirichlet boundary condition, 4445Robin boundary condition, 50

Young's Inequality, 8

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