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PDE Teaching for Spring, 2021 Revised on 2021-3-18 1 First order linear PDE with constant coe¢ cients. Assume u (x; y) is a C 1 function of two variables dened on some open set R 2 : If u (x; y) satises an equation of the form F (x; y; u; u x ;u y )=0; (x; y) 2 ; (1) we say it satises a rst order PDE on : Here F = F (x; y; z; p; q) is a given function dened on some open subset of R 3 : On the other hand, if we are given an equation of the form F (x; y; u; u x ;u y )=0; (2) where (x; y) lies on some open set R 2 ; and you want to solve the rst-order PDE, that means you want to nd a C 1 function u (x; y) dened at least on some small open set ~ such that (2) is satised for all (x; y) 2 ~ : In general, the solution u (x; y) you nd cannot be dened on the whole : If the function F (x; y; z; p; q) is linear in the three variables z; p; q; we say the PDE is linear, otherwise it is nonlinear. Therefore, a general rst-order linear PDE for a function u (x; y) of two variables has the form: a (x; y) u x (x; y)+ b (x; y) u y (x; y)+ c (x; y) u (x; y)= f (x; y) ; u = u (x; y) ; where a (x; y) ;b (x; y) ;c (x; y) ;f (x; y) are given functions. Solving a PDE is, in general, very di¢ cult, even for a rst-order linear PDE. In this elementary course, in most cases, we will focus on simple rst and second order linear PDE with constant coe¢ cients only. For simplicity of discussions, we will focus on a function u (x; y) with two vari- ables, but some theory of solution method can be easily generalized to a function u (x 1 ; ::: ;x n ) with n variables. 1.1 A simple rst order linear PDE with constant coe¢ cients. Consider the equation 3u x (x; y)+4u y (x; y)=0; u = u (x; y) (3) and we want to solve it. Without further conditions, the PDE has innitely many C 1 solutions, some of them are dened on whole R 2 ; some of them are dened on open subsets of R 2 only. Since this equation is linear with constant coe¢ cients, we can try to use a linear change of variables to convert it into an ODE and solve it. Let (s; t) be the new variables given by s t = J x y = ax + by cx + dy ; J = a b c d ; (4) where a; :::; d are constants to be determined later on. Under the change of variables, the function u (x; t) becomes a new function U (s; t) ; i.e. U (ax + by; cx + dy)= u (x; y) ; and by the chain rule (since we seek for a C 1 solution u (x; y) ; it is di/erentiable and the chain rule holds) we have u x (x; y)= U s (s; t) a + U t (s; t) c; u y (x; y)= U s (s; t) b + U t (s; t) d; (5) 1
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Page 1: 1 First order linear PDE with constant coe¢ cients.dhtsai/Lecture-notes-on-PDE-2021.pdfPDE Teaching for Spring, 2021 Revised on 2021-3-18 1 First order linear PDE with constant coe¢

PDE Teaching for Spring, 2021

Revised on 2021-3-18

1 First order linear PDE with constant coe¢ cients.

Assume u (x; y) is a C1 function of two variables de�ned on some open set � R2: If u (x; y)satis�es an equation of the form

F (x; y; u; ux; uy) = 0; (x; y) 2 ; (1)

we say it satis�es a �rst order PDE on : Here F = F (x; y; z; p; q) is a given function de�ned onsome open subset of � R3:On the other hand, if we are given an equation of the form

F (x; y; u; ux; uy) = 0; (2)

where (x; y) lies on some open set � R2; and you want to solve the �rst-order PDE, that meansyou want to �nd a C1 function u (x; y) de�ned at least on some small open set ~ � such that(2) is satis�ed for all (x; y) 2 ~: In general, the solution u (x; y) you �nd cannot be de�ned on thewhole :If the function F (x; y; z; p; q) is linear in the three variables z; p; q; we say the PDE is linear,

otherwise it is nonlinear. Therefore, a general �rst-order linear PDE for a function u (x; y) of twovariables has the form:

a (x; y)ux (x; y) + b (x; y)uy (x; y) + c (x; y)u (x; y) = f (x; y) ; u = u (x; y) ;

where a (x; y) ; b (x; y) ; c (x; y) ; f (x; y) are given functions.Solving a PDE is, in general, very di¢ cult, even for a �rst-order linear PDE. In this elementary

course, in most cases, we will focus on simple �rst and second order linear PDE with constantcoe¢ cients only. For simplicity of discussions, we will focus on a function u (x; y) with two vari-ables, but some theory of solution method can be easily generalized to a function u (x1; ::: ; xn) withn variables.

1.1 A simple �rst order linear PDE with constant coe¢ cients.

Consider the equation3ux (x; y) + 4uy (x; y) = 0; u = u (x; y) (3)

and we want to solve it. Without further conditions, the PDE has in�nitely many C1 solutions,some of them are de�ned on whole R2; some of them are de�ned on open subsets of R2 only. Sincethis equation is linear with constant coe¢ cients, we can try to use a linear change of variables toconvert it into an ODE and solve it. Let (s; t) be the new variables given by�

st

�= J

�xy

�=

�ax+ bycx+ dy

�; J =

�a bc d

�; (4)

where a; :::; d are constants to be determined later on. Under the change of variables, the functionu (x; t) becomes a new function U (s; t) ; i.e.

U (ax+ by; cx+ dy) = u (x; y) ;

and by the chain rule (since we seek for a C1 solution u (x; y) ; it is di¤erentiable and the chain ruleholds) we have

ux (x; y) = Us (s; t) a+ Ut (s; t) c; uy (x; y) = Us (s; t) b+ Ut (s; t) d; (5)

1

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which, in terms of matrix notation, is equivalent to the following gradient vector relation:�@u@x@u@y

�= JT

�@U@s@U@t

�; J =

�a bc d

�(6)

or the �rst-order di¤erential operator relation:�@@x@@y

�= JT

�@@s@@t

�; where

�st

�= J

�xy

�: (7)

For later use, we also need to know the second-order operator relation:8>>>><>>>>:

@2

@x2= @

@x

�@@x

�=�a @@s+ c @

@t

�(� � �) = a2 @

2

@s2+ 2ac @2

@s@t+ c2 @

2

@t2

@2

@x@y= @

@x

�@@y

�=�a @@s+ c @

@t

�(� � �) = ab @

2

@s2+ (ad+ bc) @2

@s@t+ cd @

2

@t2

@2

@y2= @

@y

�@@y

�=�b @@s+ d @

@t

�(� � �) = b2 @

2

@s2+ 2bd @2

@s@t+ d2 @

2

@t2;

(8)

i.e. we have @2

@x2@2

@x@y@2

@x@y@2

@y2

!=

�a cb d

��@2

@s2@2

@s@t@2

@s@t@2

@t2

��a bc d

�= JT

�@2

@s2@2

@s@t@2

@s@t@2

@t2

�J; (9)

or equivalently, the Hessain matrix relation: @2u@x2

@2u@x@y

@2u@x@y

@2u@y2

!= JT

�@2U@s2

@2U@s@t

@2U@s@t

@2U@t2

�J; J =

�a bc d

�: (10)

Back to the PDE, under the change of variables, the equation for U (s; t) becomes

0 = 3ux (x; y) + 4uy (x; y) = 3 [aUs (s; t) + cUt (s; t)] + 4 [bUs (s; t) + dUt (s; t)]

=�3a+ 4b| {z }�Us (s; t) + �3c+ 4d| {z }�Ut (s; t) : (11)

In terms of inner product in linear algebra, the above is the same as

0 = hru; vi =JTrU; v

�=DrU;

�JT�TvE= hrU; Jvi ; v =

�34

�; Jv =

�3a+ 4b3c+ 4d

�:

(12)If we choose a = 4; b = �3; we get (3c+ 4d)Ut (s; t) = 0 and if we require 3c + 4d 6= 0 (this isprecisely the change of variables condition det J = ad � bc 6= 0), we will get Ut (s; t) = 0 (this isjust an ODE with parameter s). After integration, we get

U (s; t) = F (s) ; (13)

where F (s) is an arbitrary C1 function de�ned on some open interval in R: Another way to see(13) is to note that for each �xed s; the function U (s; t) is independent of t due to Ut (s; t) = 0;hence U (s; t) is a constant C with respect to t: But this constant may depend on s; so we haveU (s; t) = C (s) ; which is exactly (13). Therefore, we conclude that any C1 solution of the PDE (3)must have the form:

u (x; y) = U (s; t) = F (s) = F (4x� 3y) (14)

for some C1 function F (s) de�ned on some open interval s 2 I:On the other hand, it is easy to check that for arbitrary C1 function F (s) ; s 2 I; the function

u (x; y) = F (4x� 3y) (15)

2

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is a solution of the PDE (3) de�ned on some open set � R2 ((x; y) 2 if and only if 4x�3y 2 I).Thus we can say that the "general solution" of the PDE is given by

u (x; y) = U (s; t) = F (s) = F (4x� 3y) (16)

for arbitrary C1 function F (s) : If we choose F (s) = s2; s 2 (�1;1) ; we get the solutionu (x; y) = (4x� 3y)2 and it is de�ned on R2; if F (s) =

ps; s 2 (0;1) ; then u (x; y) =

p4x� 3y

is de�ned only on the half-plane 4x� 3y > 0; :::; , etc.

De�nition 1.1 Any solution of equation (3) has constant value along each line of the form 4x�3y =�; where � is a constant. We call 4x� 3y = � a characteristic line of the equation (3). This canalso be seen from the fact that the gradient vector ru is everywhere perpendicular to the constantvector (3; 4) ; therefore its level curve must be a line.

Remark 1.2 Draw a picture for this ....

1.2 First order linear PDE with constant coe¢ cients; general discus-sions.

Let a; b; c be constants with a 6= 0 and b 6= 0 and f (x; y) be a given continuous function (de�nedon its "natural domain" � R2). Consider the PDE

aux (x; y) + buy (x; y) + cu (x; y) = f (x; y) ; (x; y) 2 � R2 (17)

and we want to �nd a C1 solution (or to �nd the general solution if possible) u (x; y) de�ned atleast on some open subset ~ of :

Remark 1.3 (Decomposing the general solution.) Note that if u (x; y) and v (x; y) are twosolutions of (17) de�ned on a common domain ~ � ; their di¤erence w (x; y) = u (x; y)� v (x; y)will satisfy the homogeneous equation on ~ :

awx (x; y) + bwy (x; y) + cw (x; y) = 0: (18)

By this, the general solution of (17) can be decomposed as a particular solution of the nonhomo-geneous equation (17) plus the general solution of the homogeneous equation (18).

Motivated by the method in the example in Section 1.1, we consider the linear change ofvariables: (

w = Ax+By

z = Cx+Dyi.e.

�wz

�=

�A BC D

��xy

�; AD �BC 6= 0 (19)

and denote the function corresponding to u (x; y) as U (w; z) (i.e. U (Ax+By;Cx+Dy) = u (x; y)). Bythe chain rule we have

aux (x; y) + buy (x; y) + cu (x; y)

= a [Uw (w; z)A+ Uz (w; z)C] + b [Uw (w; z)B + Uz (w; z)D] + cU (w; z)

=�aA+ bB| {z }�Uw (w; z) + �aC + bD| {z }�Uz (w; z) + cU (w; z) ; (20)

which indicates that the new equation for U (w; z) will become an ODE if and only if one of thefollowing is satis�ed (but not both):

either�ab

���AB

�= 0 or

�ab

���CD

�= 0: (21)

3

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By (21), we can choose the change of variables:(w = Ax+By = bx� ay

z = Cx+Dy = yi.e.,

�wz

�=

�b �a0 1

��xy

�; Jacobian = b 6= 0 (22)

and conclude the linear ODE for U (w; z) :

bUz (w; z) + cU (w; z)| {z } = F (w; z)

0@= f

�w + az

b; z

�| {z }

1A ; where U (w; z) = u

�w + az

b; z

�: (23)

where

u (x; y) = U (bx� ay; y) and U (w; z) = u

�w + az

b; z

�: (24)

Remark 1.4 Since here a 6= 0; b 6= 0; it is also OK if we use the change of variables w =bx� ay; z = x:

The ODE is de�ned on a domain � in wz-plane (� is the image of in xy-plane under thelinear map in (22); it is in one-one correspondence with ). By ODE theory we can solve (23) toget its general solution:

U (w; z) = e�cbz

�1

b

Zecbzf

�w + az

b; z

�dz + C (w)

�=e�

cbz

b

Zecbzf

�w + az

b; z

�| {z } dz + e�

cbzC (w) := I (w; z) + II (w; z) ; w = bx� ay; z = y;

(25)

where C (w) is an arbitrary C1 function of w (C (w) is "integration constant" for the integralRdz).We note that U (w; z) = I (w; z) + II (w; z) is the sum of a particular solution of bUz + cU = F

and the general solution of bUz + cU = 0: Moreover, we see that I (w; z) is de�ned on �:However, the domain of II (w; z) depends on how you choose C (w) : Back to u (x; y) ; wewill have

u (x; y) =

(a particular solution of aux + buy + cu = f (this is de�ned on )

+ the general solution of aux + buy + cu = 0 (this is given by e�cbyC (bx� ay) )

(26)and we note that the domain of e�

cbyC (bx� ay) depends on how you choose C (w) :

Note that (26) describes all possible solutions of (17) and is called the general solution of theequation (17).

Remark 1.5 (Important.) This is to explain why the integral functionRecbzf�w+azb; z�dz is de-

�ned on the domain � in wz-plane. Denote the integrand asH (w; z) and we look atRH (w; z) dz: For

each �xed w0 2 R with (w0; z0) 2 � for some z0; since � is an open set in wz-plane, the cross sectionfz 2 R : (w0; z) 2 �g is an open set O in R (a countable union of disjoint open intervals) and wenote that the inde�nite integral Z

H (w0; z) dz; z 2 O

is a well-de�ned function on O (since H (w; z) is a C1 function of (w; z) 2 �). One can use anexample to demonstrate this. We look at the ODE

ux (x; y) = exy: (27)

4

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The function exy is a C1 function on R2 and by integration we get

u (x; y) =

Zexydx =

( 1yexy; y 2 (�1; 0)

S(0;1) ; y 6= 0

x; y = 0;+ C (y) (28)

where C (y) is an arbitrary C1 function of y 2 R2: At this moment, it is not clear that we can �nda C1 particular solution of the ODE (27) de�ned on all R2: However, if we use the de�niteintegral

u (x; y) =

Z x

0

e�yd� =

( 1y(exy � 1) ; y 6= 0

x; y = 0;(denote this function as v (x; y) ) (29)

then we can obtain the following form of general solution

u (x; y) = v (x; y) + C (y) ; (x; y) 2 R2: (30)

One can see that the function v (x; y) is indeed a C1 function on R2 due to the following (writeexy as 1 + xy + x2y2

2!+ x3y3

3!+ � � ��):8>>>>>>>>>>><>>>>>>>>>>>:

(1) : limy!0exy�1y

= x;

(2) : vx (x; y) = exy; y 6= 0; vx (x; 0) = 1; limy!0 vx (x; y) = 1;

(3A) : vy (x; y) =�1y2(exy � 1) + x

yexy; y 6= 0;

(3B) : vy (x; 0) = limy!0v(x;y)�v(x;0)

y= limy!0

1y(exy�1)�x

y= x2

2;

(3C) : limy!0 vx (x; y) = limy!0

��1y2(exy � 1) + x

yexy�= limy!0

�xyexy�exy+1

y2

�= x2

2:

(31)

Remark 1.6 One can also express the general solution of aux + buy + cu = 0 as u (x; y) =e�

caxC (bx� ay) : Reason: multiply e� c

byC (bx� ay) by e�(bx�ay); where � = �c=ab.

Remark 1.7 (A trick to absorb the term cu:) For the homogeneous equation aux + buy + cu = 0;a 6= 0; b 6= 0; you can let v (x; y) = e

cbyu (x; y) ; then it will satisfy avx + bvy = 0: similarly, you

can let w (x; y) = ecaxu (x; y) ; then it will satisfy awx + bwy = 0: By this, we see that the general

solution of the homogeneous equation aux + buy + cu = 0 is

u (x; y) = e�caxC (bx� ay) or u (x; y) = e�

cbyC (bx� ay) ; (32)

where C (w) is arbitrary C1 functions of w:

De�nition 1.8 Any line L in xy-space of the form

bx� ay = �; � is a constant (33)

is called a characteristic line of the PDE (17). Characteristic line bx� ay = � correspond to thecoordinate line w = � in the (w; z) space. We use characteristic lines to do change of variables.

Remark 1.9 (Important.) The above says that the characteristic lines play an important rolein solving the PDE (17). They are the coordinate lines for the new variable (w; z) : Withoutusing the characteristic lines, we cannot convert the PDE into ODE.

5

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Remark 1.10 (Important.) Note that the integration function C (w) in (25) is constant alongeach characteristic line bx � ay = �: In particular, if c = 0 and f (x; y) = 0; then ODE (23)becomes bUz (w; z) = 0 and (25) becomes U (w; z) = C (w) and u (x; y) becomes

u (x; y) = C (bx� ay) ;

which is constant along each characteristic line. On the other hand, if c 6= 0 or f (x; y) is not a zerofunction, then u (x; y) is in general no longer a constant function along each characteristicline. However, if we know the value of u (x; y) at a point (x0; y0) 2 L (L is the characteristicline passing through (x0; y0) ; which has equation bx� ay = �0; �0 = bx0 � ay0), then the functionu (x; y) will satisfy an ODE on L (with respect to the parameter of the line L) and by solving theODE we can know all values of u (x; y) on L: More precisely, one can parametrize the line L as

L =

�(x; y) =

��0 + a�

b; �

�; � 2 (�1;1)

�; (x0; y0) =

��0 + a�0

b; �0

�and let

Q (�) = u

��0 + a�

b; �

�; Q (�0) = u (x0; y0) is known.

Then from the PDE (for simplicity of discussion, here we assume both u (x; y) and f (x; y) arede�ned on the whole plane R2)

aux (x; y) + buy (x; y) + cu (x; y) = f (x; y) ; 8 (x; y) 2 R2

we can derive

Q0 (�) =a

bux

��0 + a�

b; �

�+ uy

��0 + a�

b; �

�=1

bf

��0 + a�

b; �

�� c

bu

��0 + a�

b; �

�=1

bf

��0 + a�

b; �

�� c

bQ (�) ; 8 � 2 (�1;1) ;

and so we obtain the ODE along L (which is similar to (23)):

Q0 (�) +c

bQ (�) =

1

bf

��0 + a�

b; �

�; � 2 (�1;1) ; Q (�0) = u (x0; y0) is known. (34)

By ODE theory, one can know Q (�) for all � 2 (�1;1) : We can summarize the following:if u (x; y) satis�es the PDE (17) on open domain of R2; then it satis�es the ODE (34) on anycharacteristic line inside its domain.

Example 1.11 Find the general solution of the PDE

3ux � 2uy + u = x; u = u (x; y) : (35)

Solution:To make the change of variables and the ODE look better, we rewrite the equation

as �3ux + 2uy � u = �x with a = �3; b = 2; c = �1: According to the method, we introduce thechange of variables

w = bx� ay = 2x+ 3y; z = y; (36)

and the function u (x; y) becomes U (w; z) ; where by (23) we have

�3ux (x; y) + 2uy (x; y)� u (x; y) = 2Uz (w; z)� U (w; z) = �x = �w � 3z2

;

6

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and the ODE (in the variable z) for U (w; z) is

Uz (w; z)�1

2U (w; z) = �1

4(w � 3z) :

It has the solution

U (w; z) = ez2

�Ze�

z2

��14(w � 3z)

�dz + C (w)

= ez2

0@�w4

Ze�

z2dz| {z }+

3

4

Zze�

z2dz| {z }+C (w)

1A ;

where by Ze�

z2dz = �2e� z

2 ;

Zze�

z2dz = �2ze� z

2 � 4e� z2

we get

U (w; z) = ez2

��w4

��2e� z

2

�| {z }+34 ��2ze� z2 � 4e� z

2

�| {z }+C (w)�

=w

2� 32z � 3 + e z2C (w) :

As a result, the general solution of the original equation is

u (x; y) =2x+ 3y

2� 32y � 3 + e

y2C (2x+ 3y) = x� 3 + e

y2C (2x+ 3y) ;

where C (�) is an arbitrary C1 function. �

Remark 1.12 One can see that

u (x; y) =

(x� 3 (a particular solution of (35))

+ey2C (2x+ 3y) (the general solution of 3ux � 2uy + u = 0).

1.2.1 The physical meaning of the equation cux + ut = 0:

Let c > 0 be a constant. The equation cux + ut = 0; u = u (x; t) ; is called a simple transportequation. Here x 2 R represents space (one-dimensional) coordinate and t 2 R represents time.Assume we have a �uid (water, say), moving to the right (along a horizontal pipe with �xed crosssection in the positive x-direction) with constant speed c: Let u (x; t) be the concentration ofsome substance suspended in the water. The amount of the substance in the interval [0; b] at timet is given by

M =

Z b

0

u (x; t) dx:

A small time h > 0 later, the same substance (previously lies in the interval [0; b]) have moved tothe right by distance c � h: Hence we have

M =

Z b

0

u (x; t) dx =

Z b+ch

ch

u (x; t+ h) dx:

Note that the above identity is valid for all b > 0 and all h > 0: Di¤erentiation with respect to bgives

u (b; t) = u (b+ ch; t+ h) ; 8 b; h > 0

7

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(which says that u is constant in h 2 (0;1) along the ray (b+ ch; t+ h) for all h > 0) anddi¤erentiation with respect to h and letting h = 0 gives

0 = cux (b; t) + ut (b; t) ; 8 b > 0; t 2 R;

i.e. we have0 = cux (x; t) + ut (x; t) ; 8 x > 0; t 2 R:

Note that u (x; t) is constant along the line x� ct = � (it has a simple physical interpretation dueto the identity u (x; 0) = u (x+ ct; t) for any t 2 R). The general solution of the equation is givenby u (x; t) = F (x� ct) for arbitrary C1 function F (z) :

1.3 First order linear PDE with constant coe¢ cients plus side condi-tion.

Consider equation (17). We know that it has in�nitely many solutions if we do not impose anycondition on the solutions. The purpose of the extra "side condition" is to ensure that the solutionis unique. A general side condition has the form ujC = g; where C is a given curve in the planeR2; which intersects each characteristic line L "transversally" at exactly one point, andg is a given function on C: In most cases, we only consider the case when C = ~L is a line in R2: Nowthe side condition has the form

u (x;mx+ d) = g (x) ; 8 x 2 R; (37)

where m is the slope of the line ~L : y = mx+ d and d is some number. Here g (x) is a C1 functionof x (g (x) can also be a constant, say g (x) � 0): If the line ~L is vertical, then the side conditionhas the form

u (d; y) = g (y) ; 8 y 2 R: (38)

The general fact we know is the following:

Theorem 1.13 (Roughly speaking.) If the line ~L is not a characteristic line of (17), thenthe PDE with the side condition has a unique solution. If the line ~L is a characteristic line, thePDE with the side condition has either no solution or in�nitely many solutions.

Remark 1.14 (Important observation.) The key point is that if ~L is not a characteristicline, then it intersects each characteristic line L at exactly one point. This allows usto determine the integration function C (�) uniquely. Note that C (�) is constant along eachcharacteristic line L. See Remark 1.10.

We shall not make the above theorem precise. Instead, we will just look at some examples toconvince us the result.

Example 1.15 The equation 2ux (x; y)+5uy (x; y) = 0 has characteristic lines of the form 5x�2y =� (same as y = 5

2x+ �); � 2 R: Its general solution is given by

u (x; y) = C (5x� 2y) ; (39)

for arbitrary C1 function C (�) : Let L be the line y = 52x + 1; which is a characteristic line.

Consider the side condition problem(2ux (x; y) + 5uy (x; y) = 0

u�x; 5

2x+ 1

�= g (x) ; x 2 (�1;1)

(40)

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for some function g (x) : By (39), we have

u

�x;5

2x+ 1

�= C

�5x� 2

�5

2x+ 1

��= C (�2) (this is a constant !!): (41)

Therefore, if g (x) is not a constant function, then (40) has no solution at all. On the other hand,if g (x) is a constant function, say g (x) = 10; then any u (x; y) of the form u (x; y) = C (5x� 2y) isa solution as long as C satis�es C (�2) = 10: In such a case, we have in�nitely many solutions.Finally, if we replace the side condition by

u (x; 2x+ 7) = sinx; x 2 (�1;1) ; (42)

then the line y = 2x+ 7 is not a characteristic line and by (39) we can solve the equation

u (x; 2x+ 7) = C (5x� 2 (2x+ 7)) = C (x� 14) = sin x; x 2 (�1;1)

to get C (�) = sin (� + 14) ; � 2 (�1;1) : Therefore, the side condition problem (42) has theunique solution given by

u (x; y) = C (5x� 2y) = sin (5x� 2y + 14) ; (x; y) 2 R2: (43)

Example 1.16 In Example 1.11, the general solution of the equation

3ux � 2uy + u = x (44)

isu (x; y) = x� 3 + e

y2C (2x+ 3y) ; (45)

and if we put the side condition as

u

�x;�2

3x+ 2

�= g (x) ; x 2 (�1;1) ; (46)

for some g (x) ; then the line y = �23x+ 2 is a characteristic line, and we have

u

�x;�2

3x+ 2

�= x� 3 + C (6) e� 1

3x+1; x 2 (�1;1)

Therefore, unless g (x) has the form g (x) = x � 3 + ke�13x+1 for some constant k; we have no

solution satisfying the side condition. On the other hand, if g (x) is given by, say, g (x) = x �3 + 100e�

13x+1; there are in�nitely many solutions satisfying this side condition everywhere on

x 2 (�1;1) as long as we choose the function C (�) to satisfy C (6) = 100:

Example 1.17 Consider the same equation as in (44), but not with the condition

u (x; 4x� 2) = g (x) ; x 2 (�1;1)

for some g (x) : The line y = 4x� 2 is not a characteristic line, and by (45) we need to require

u (x; 4x� 2) = x� 3 + e2x�1C (14x� 6) = g (x) ; 8 x 2 (�1;1) ;

i.e.

C (14x� 6) = g (x)� x+ 3e2x�1

; x 2 (�1;1) ;

which implies (let � = 14x � 6; x = 114(� + 6)) the unique choice of the integration constant

function C (�) ; namely

C (�) =g�114(� + 6)

�� 1

14(� + 6) + 3

e214(�+6)�1

; � 2 (�1;1) :

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Example 1.18 Solve the equation(ux (x; y)� uy (x; y) + 2u (x; y) = 1

u (x; 0) = x2; x 2 (�1;1) :(47)

Solution:

By (25), one can check (or "guess") that the general solution of ux � uy + 2u = 1 is

u (x; y) =1

2+ e2yC (�x� y) (same as

1

2+ e2yC (x+ y) )

for arbitrary function C (�) : The side condition is prescribed on the x-axis, which is not a char-acteristic line. We need to choose C (�) so that

u (x; 0) =1

2+ C (x) = x2:

Hence we have C (x) = x2 � 1=2 and then

u (x; y) =1

2+ e2yC (x+ y) =

1

2+ e2y

�(x+ y)2 � 1

2

�; (x; y) 2 R2 (48)

is the unique solution of (47). �

Example 1.19 Solve the equation(ux (x; y) + 2uy (x; y)� 4u (x; y) = ex+y

u (x; 4x+ 2) = 0; x 2 (�1;1) :(49)

Solution:

By (25), one can check (or "guess") that the general solution of ux + 2uy � 4u = ex+y is

u (x; y) = �ex+y + e2yC (2x� y) :

By the side condition, we require (the line y = 4x+ 2 is not a characteristic line)

0 = u (x; 4x+ 2) = �e5x+2 + e8x+4C (�2x� 2) ;

i.e. we need to require C (�2x� 2) = e�3x�2: To get C (r) ; we let r = �2x� 2 and one can solvex in terms of r to get

�3x� 2 = �3�r + 2

�2

�� 2 = 3

2r + 1:

Hence C (r) = e32r+1 and the unique solution is

u (x; y) = �ex+y + e2ye32(2x�y)+1 = �ex+y + e3x+

12y+1; (x; y) 2 R2: (50)

In the next example, the side condition has the form ujC = g; where C is a curve in R2; not aline.

Example 1.20 Solve the equation(ux (x; y)� uy (x; y) + u (x; y) = 0

u (x; x3) = e�x (x+ x3) ; x 2 (�1;1) :(51)

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Solution:

Here u is speci�ed on the curve (unlike the previous examples, it is not a line) y = x3: Thiscurve intersects each characteristic line at exactly one point, which is good (otherwise,the solution may not exist). The general solution of ux � uy + u = 0 is given by u (x; y) =eyC (�x� y) (same as u (x; y) = eyC (x+ y)) for arbitrary function C (�) : Then we need to solve

u�x; x3

�= ex

3

C�x+ x3

�= e�x

�x+ x3

�and get

C�x+ x3

�=e�x (x+ x3)

ex3= e�(x+x

3) �x+ x3�: (52)

Hence C (r) = e�rr and the unique solution is

u (x; y) = eyC (x+ y) = ey (x+ y) e�(x+y) = (x+ y) e�x; (x; y) 2 R2: (53)

Remark 1.21 In the above example, the curve y = x3 intersects each characteristic line x + y =�; � 2 (�1;1) ; at exactly one point. With this, we have the unique solution u (x; y) satisfyingthe side condition. If this is not the case, the solution may not exist or may not be unique. Ifwe replace the side condition by

u�x; x2

�= e�x

�x+ x3

�; x 2 (�1;1) ;

then we note that the curve y = x2 intersects each characteristic line x+ y = �; � 2 (�1;1) ; ateither two points, or one point, or no intersection at all. Now the condition (52) becomes

C�x+ x2

�= e�(x+x

2) �x+ x3�; x 2 (�1;1) : (54)

One can check that it is "impossible" to �nd a function C (�) satisfying C (x+ x2) = e�(x+x2) (x+ x3) for

all x 2 (�1;1). Exercise: Show that it is impossible to �nd a C1 function C (�) de�ned

on � 2 [�1=4;1) satisfying C (x+ x2) = e�(x+x2) (x+ x3) for all x 2 (�1;1). Hint: look at

x = 1 and x = �2:

Remark 1.22 In general, if we want to solve an equation of the form C (h (x)) = g (x) ; then ifh0 (x) > 0 (or < 0) everywhere, one can solve x = h�1 (r) and get the function C (r) = g (h�1 (r)) :It satis�es C (h (x)) = g (h�1 (h (x))) = g (x) :

2 First order linear PDE with variable coe¢ cients.

In this section, we look at �rst order linear PDE for a 2-variable function u = u (x; y) with variablecoe¢ cients. The equation has the form

a (x; y)ux + b (x; y)uy| {z }+c (x; y)u = f (x; y) ; u = u (x; y) ; (55)

where the coe¢ cients a (x; y) ; b (x; y) ; c (x; y) ; f (x; y) are given C1 functions de�ned on somecommon open set � R2; with

a2 (x; y) + b2 (x; y) > 0 on : (56)

To solve (55) means to �nd a C1 function u (x; y) de�ned on some open subset ~ of whichsatis�es (55) on ~: The solution method will involve some ODE theory.

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Remark 2.1 The reason of assuming a (x; y) ; b (x; y) ; c (x; y) ; f (x; y) to be C1 functions on (not just C0 functions on ) is to ensure that the system of ODE (60) below has a unique solutionpassing through each point p 2 : This is because the vector �eld V (x; y) = (a (x; y) ; b (x; y)) is aC1 vector �eld.

Our purpose is to �nd a "general solution", which can encompass as many solutions as possible.Here, roughly speaking, "general" means that we have a solution formula and it contains anarbitrary integration function C (w) as in formula (25).Unlike the case of constant coe¢ cients (see equation (17)), where we can always decompose

the general solution u (x; y) into two parts, with the domain of the particular solution part being and the domain of the remaining part depending on how you choose the integration constantfunction C (w) ; it is not clear if we can do the same decomposition here.In this elementary course, we will be satis�ed if we can �nd some open subset ~ of

and you can �nd a general solution formula for the PDE (55) on ~:The following example says that the condition (56) is important for the existence of a solution

on :

Example 2.2 Consider the equation given by

xux (x; y) + yuy (x; y) = ex+y; (x; y) 2 = R2n f(0; 0)g ; x2 + y2 > 0 on :

and if we take = R2; then the condition (56) is not satis�ed due to x2 + y2 = 0 at (0; 0) (i.e.the PDE has a singularity at (0; 0)). One can see that "any" C1 solution u (x; y) cannot bede�ned near the origin (0; 0) : Otherwise, we will obtain the absurd identity

0 = 0ux (0; 0) + 0uy (0; 0) = e0+0 = 1:

Therefore, the natural domain for the PDE is = R2n f(0; 0)g :

Example 2.3 We now change the above equation as

xux (x; y) + yuy (x; y) = 0 (57)

and if we choose = R2n f(0; 0)g ; then (56) is satis�ed. Here we can �nd a solution de�ned on (say u (x; y) is a constant function). We can also �nd a non-constant solution de�ned on

(check the function u (x; y) = e�(yx)

2

for x 6= 0 and u (0; y) = 0 for y 6= 0). On the other hand, thesolution u (x; y) = y=x is de�ned only on a subset ~ of (n fy-axisg). Below, we will show youhow to �nd the general solution of (57) on some open subset ~ of :

We �rst look at the easier case

a (x; y)ux + b (x; y)uy = 0; u = u (x; y) ; (58)

where a (x; y) ; b (x; y) are given C1 functions on � R2 with a2 + b2 > 0 on and here we alsoassume that u (x; y) is a C1 solution de�ned on : One can rewrite the equation as

a (x; y)ux + b (x; y)uy = ru (x; y) ��a (x; y)b (x; y)

�= 0; (x; y) 2 : (59)

If we view the vector V (x; y) = (a (x; y) ; b (x; y)) as a C1 vector �eld on (the vector �eldV (x; y) has no equilibrium point on due to a2 + b2 > 0 on ), then (59) has a geomet-ric meaning. It says that if a C1 curve C : � (t) = (x (t) ; y (t)) 2 ; t 2 I (some open interval),satis�es the system of ODE (such a C1 curve is called an integral curve of the vector �eldV (x; y) on )

dx

dt= a (x; y) ;

dy

dt= b (x; y) ; 8 t 2 I; (60)

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then by the chain rule we have

d

dtu (x (t) ; y (t)) = ux (x (t) ; y (t))

dx

dt+ uy (x (t) ; y (t))

dy

dt= a (x (t) ; y (t))ux (x (t) ; y (t)) + b (x (t) ; y (t))uy (x (t) ; y (t)) = 0; 8 t 2 I:

Therefore, we can conclude:

Lemma 2.4 Let V (x; y) = (a (x; y) ; b (x; y)) be a C1 vector �eld on � R2 with a2 (x; y) +b2 (x; y) > 0 on : Then u (x; y) is a C1 solution of the homogeneous equation

a (x; y)ux + b (x; y)uy = 0; u = u (x; y) ; (x; y) 2 (61)

on if and only if for any C1 integral curve C : � (t) = (x (t) ; y (t)) ; t 2 I; of the ODE(60) lying on ; the function u (� (t)) ; t 2 I; is a constant function along the curve C:

Proof. The direction =) is clear. For the direction (=; we note that at any point p = (x0; y0) 2 (open set), there is a C1 integral curve C : � (t) = (x (t) ; y (t)) ; t 2 (�"; ") ; � (0) = p; lying on and passing through it. By the assumption we have

d

dtu (x (t) ; y (t)) = a (x (t) ; y (t))ux (x (t) ; y (t)) + b (x (t) ; y (t))uy (x (t) ; y (t)) = 0; 8 t 2 I;

which, at t = 0; implies a (p)ux (p) + b (p)uy (p) = 0: Since p 2 can be arbitrary, u (x; y) satis�esthe equation (61) on . �

De�nition 2.5 A curve C in the plane is called a characteristic curve of the PDE (58) (orthe more general PDE (55)) if at each point (x; y) 2 C; the vector V (x; y) = (a (x; y) ; b (x; y))is tangent to C at (x; y) : By a suitable parametrization (i.e., by solving the ODE (60)), acharacteristic curve C can be parametrized as C : � (t) = (x (t) ; y (t)) 2 ; t 2 I; wherex (t) and y (t) satisfy the system of ODE (60) on some interval I.

Remark 2.6 In the above lemma, the constant may be di¤erent on di¤erent integral curves. Thelemma says that an integral curve � (t) = (x (t) ; y (t)) ; t 2 I; is a level curve (in parametricform) of the function u (x; y) lying inside : Thus we have

integral curve = level curve = characteristic curve: (62)

Remark 2.7 (Important.) Roughly speaking, if we can know "all level curves of u (x; y) inside" (equivalent to "all solutions (x (t) ; y (t)) of the ODE (60) inside "), then one can determineu (x; y) uniquely on (as long as we can know the values of ujC ; where C is a curve on whichintersects each integral curve transversally at exactly one point). Such u (x; y) will be a solution ofthe PDE (58) on : This property is the same as that for the previous equation aux + buy = 0 withconstant coe¢ cients.

3 Finding general solution of the PDE (58).

Same as before, the characteristic curves play an important role in solving the PDE (58) (or themore general PDE (55)). We solve the ODE (60) �rst and obtain a family of solutions (x (t) ; y (t))with initial data (x (0) ; y (0)) 2 being arbitrary (each solution is de�ned on some maximal timeinterval I with 0 2 I). By ODE theory, solution curves with di¤erent initial datawill not intersectat all. Therefore, the domain can be viewed as the disjoint union of all integral curves in : Wethen try to convert (by deleting the time variable t or by the Inverse Function Theoremor by rewriting the system as dy=dx = b (x; y) =a (x; y) and solving it to get a relation

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between x and y) this family of ODE solutions (family of characteristic curves) (x (t) ; y (t)) intothe implicit form

h (x; y) = d (63)

where h (x; y) is a C1 function and d is an integration constant. In general, the functionh (x; y) is de�ned only on some open subset ~ of (however, it is also possible for h (x; y) to bede�ned on all ). As the constant d varies, the identity h (x; y) = d describes di¤erent characteristiccurves (in equation form, not in parametric form) of the equation a (x; y)ux + b (x; y)uy = 0 lyinginside ~:

Lemma 3.1 The above C1 function h (x; y) : ~! R is a solution of the PDE (58) on ~:

Proof. For each (x (0) ; y (0)) 2 ~; one can solve the ODE (60) to get an unique integral curve(x (t) ; y (t)) 2 ~ for t 2 I (some interval containing t = 0). Since we have h (x (t) ; y (t)) = d for allt 2 I; we have

@h

@x(x (t) ; y (t))

dx

dt(t) +

@h

@y(x (t) ; y (t))

dy

dt(t)

= a (x (t) ; y (t))@h

@x(x (t) ; y (t)) + b (x (t) ; y (t))

@h

@y(x (t) ; y (t)) = 0; 8 t 2 I:

In particular, at t = 0; we get

a (x (0) ; y (0))@h

@x(x (0) ; y (0)) + b (x (0) ; y (0))

@h

@y(x (0) ; y (0)) = 0: (64)

As (x (0) ; y (0)) 2 ~ can be arbitrary, we conclude

a (x; y)@h

@x(x; y) + b (x; y)

@h

@y(x; y) = 0; 8 (x; y) 2 ~: (65)

The result follows. �Remark 3.2 However, at this moment, the function h (x; y) is not yet a "general solu-tion" of the PDE on ~.

Remark 3.3 By the identity (65) and the assumption a2 (x; y)+ b2 (x; y) > 0 on ; we can see thatif @h

@x(x; y) 6= 0 on ~; we will have b (x; y) 6= 0 on ~ and similarly, if @h

@y(x; y) 6= 0 on ~; we will

have a (x; y) 6= 0 on ~. We will need this observation later on.

Example 3.4 In the special case when the equation is aux + buy = 0; where a 6= 0; b 6= 0 areconstants, we solve the system of ODE

dx

dt= a;

dy

dt= b (66)

with initial condition (x0; y0) 2 R2 and obtain x (t) = at + x0; y (t) = bt + y0; t 2 (�1;1) : Bycancelling t; we get the equation

h (x; y) = bx� ay = d (d = bx0 � ay0) : (67)

One can see that the function h (x; y) = bx� ay is a solution of the PDE de�ned on R2. However,it is not the general solution. One can also rewrite the system as (view y as a function of x)

dy

dx=

dydtdxdt

=b

a(this is separable equation for y (x) ) (68)

and conclude y = (b=a)x+ � or useZady =

Zbdx; ay = bx+ �; � is integration constant, (69)

which also gives (67).

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Example 3.5 Consider the equation

xux + yuy = 0; u = u (x; y) ; (x; y) 2 = R2n f(0; 0)g : (70)

If we solve the system of ODE (60), we get x (t) = x0et and y (t) = y0e

t; t 2 (�1;1) ; where (x0; y0) 6=(0; 0) : The only way to cancel the parameter t is to look at the expression x=y (if y0 6= 0) or y=x (ifx0 6= 0). Therefore, we get the identity

x

y= d (missing x-axis), d is a constant

ory

x= d (missing y-axis), d is a constant

along each characteristic curve. Note that the function h (x; y) = x=y (or y=x) cannot be de�ned onthe whole due to the denominator. One can check that u (x; y) = x=y (or y=x) is a C1 solutionof the PDE (70) de�ned on n fx-axisg (n fy-axisg). Note that one can also rewrite the system ofODE (60) as

dy

dx=y

x(separable equation),

Z1

ydy =

Z1

xdx

and obtain the identityx

y= d (or

y

x= d).

To obtain "general solution" of the PDE (58), similar to the case of constant coe¢ cients, weintroduce the change of variables (in general, it is a nonlinear change of variables):(

w = h (x; y) ; (x; y) 2 ~

z = y;(71)

where we know that h (x; y) is a solution of the PDE a (x; y)ux+ b (x; y)uy = 0 on ~: By restrictingthe domain ~ to a smaller open subset if necessary, we may assume that the Jacobian of thischange of variables satis�es the regular condition:

det

�wx wyzx zy

�= det

�@h@x(x; y) @h

@y(x; y)

0 1

�=@h

@x(x; y) 6= 0 on ~: (72)

By the above assumption, we also have b (x; y) 6= 0 on ~ due to the condition (56). Now the functionu (x; y) becomes U (w; z) (i.e. U (h (x; y) ; y) = u (x; y) or U (w; z) = u (x (w; z) ; z)).

Remark 3.6 If @h@x= 0 somewhere on ~; we may have to restrict the domain of h (x; y) to a even

smaller domain ̂ � ~ so that @h@x6= 0 on ̂: Another alternative is to change the change of

variables as w = h (x; y) ; z = x and see if we have @h@y(x; y) 6= 0 on ~ or not.

Remark 3.7 Strictly speaking, by the inverse function theorem, even if we have @h@x(x; y) 6= 0 everywhere

on ~; the inverse function relation (x; y) ! (w; z) may be valid only on a small subdomain of~ (this is unlike the linear change of variables w = bx�ay; z = y before). However, by restrictingto an even smaller domain, we may assume that the inverse function relation (x; y) ! (w; z)is valid on ~ (on it we have @h

@x(x; y) 6= 0) and proceed to �nd the general form of the solution.

Finally, by the general form of the solution, we can �gure out the possible domain of the solution.

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By the chain rule, we get(ux (x; y) = Uw (w; z)

@w@x+ Uz (w; z)

@z@x= Uw (w; z)

@h@x

uy (x; y) = Uw (w; z)@w@y+ Uz (w; z)

@z@y= Uw (w; z)

@h@y+ Uz (w; z)

and, in terms of the new variables (w; z) ; the PDE for U (w; z) becomes:

0 = a (x; y)ux (x; y) + b (x; y)uy (x; y)

= a (x; y)Uw (w; z)@h

@x+ b (x; y)

�Uw (w; z)

@h

@y+ Uz (w; z)

�=

�a (x; y)

@h

@x(x; y) + b (x; y)

@h

@y(x; y)

�Uw (w; z) + b (x; y)Uz (w; z) = b (x; y)Uz (w; z) ; (73)

where in the above we have used the identity (65). The general solution solution of the ODE (73)is given by

U (w; z) = F (w) (74)

for arbitrary C1 function F (w) :

Remark 3.8 If we letw = h (x; y) ; z = x; (75)

then we have to assume that@h

@y6= 0 on ~:

Now the equation (73) becomes0 = a (x; y)Uz (w; z) (76)

and we get the same solution U (w; z) = F (w) (we now have a (x; y) 6= 0 on ~; see Remark 3.3).

Hence we get:

Theorem 3.9 Let a (x; y) ; b (x; y) be two C1 functions on a domain � R2 with a2 + b2 > 0 on: Consider the PDE a (x; y)ux+ b (x; y)uy = 0 on : There exists some open subset ~ of suchthat on ~ any C1 solution u (x; y) of the PDE has the form

u (x; y) = F (h (x; y)) ; (x; y) 2 ~; (77)

where F (�) is an arbitrary C1 function and the function h (x; y) ; which is de�ned on ~; comes fromsolving the ODE (60) on : Note that, in general, the open set ~ � can be very small.

Proof. By the discussion before Theorem 3.9, there exists some open subset ~ � such that on itthere is a C1 solution h (x; y) of the PDE, with @h

@x(x; y) 6= 0 on ~; and the change of variables (71)

is a bijection map between ~ and some open subset � in wz-space (which has a C1 inverse map).Moreover, on ~ we have the identity

a (x; y)ux (x; y) + b (x; y)uy (x; y) = b (x; y)Uz (w; z) ; (w; z) 2 �

for any C1 function u (x; y) on ~; where we also know that b (x; y) 6= 0 on ~: Therefore, if u (x; y)is a C1 solution on ~; we must have Uz (w; z) = 0 on � and the result follows. �

Remark 3.10 In the above lemma, the domain of the function F (�) will a¤ect the domain ofu (x; y) : However, if we choose F (�) a C1 function de�ned on R; then u (x; y) = F (h (x; y)) will bede�ned on ~:

16

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Remark 3.11 It is possible that there are di¤erent disjoint open sets ~1; ~2 of such that thefunctions h1 (x; y) on ~1 and h2 (x; y) on ~2 are di¤erent (and the integration constant functionsF and G on ~1 and ~2 can be independent to each other). However, in solving the PDE a (x; y)ux+b (x; y)uy = 0 on ; we are satis�ed as long as one can �nd one open set ~ of and the solutionformula (77) on ~:

Let us go back to the equation xux + yuy = 0 again:

Example 3.12 Find the general solution of the equation

xux + yuy = 0; u = u (x; y) ; (x; y) 2 = R2n f(0; 0)g : (78)

Solution:

We already know that h (x; y) = xyis a solution on n fx-axisg :We focus on the open half-plane

R2y+ = f(x; y) 2 R2 : y > 0g (or on R2y� = f(x; y) 2 R2 : y < 0g) and on it we do the change ofvariables (

w = xy2 (�1;1)

z = y 2 (0;1)same as

(x = wz; z 6= 0;

y = z;

which has Jacobian ���� 1y� xy2

0 1

���� = 1

y6= 0 on R2y+ :

We note that the above change of variables is a bijection map between the open set (�1;1)�(0;1) in wz-space and the open set (�1;1) � (0;1) in xy-space. By chain rule the equationon R2y+ becomes

0 = xux + yuy = x

�Uw (w; z)

@w

@x+ Uz (w; z)

@z

@x

�+ y

�Uw (w; z)

@w

@y+ Uz (w; z)

@z

@y

�= x

�Uw (w; z)

1

y

�+ y

�Uw (w; z)

�� x

y2

�+ Uz (w; z)

�= yUz; y > 0;

which gives the general solution U (w; z) = F (w) and so we get u (x; y) = F�xy

�for (x; y) 2

R2y+ and similarly on R2y� the general solution is given by u (x; y) = G

�xy

�for (x; y) 2 R2y� ; where

F (�) and G (�) are two arbitrary C1 functions de�ned on (�1;1) (in order for u (x; y) to havemaximal domain of de�nition).If we allow a solution u (x; y) to be unde�ned on x-axis (y = 0), then the above two inte-

gration constant functions F (�) and G (�) can be independent to each other. Note that eachintegral curve on R2y+ will not traverse into R

2y�, and vice versa. Therefore, there is no

discontinuity/inconsistency problem here even if F (�) and G (�) are di¤erent.However, it is possible to choose some special F and G so that u (x; y) can be de�ned across

the x-axis (more precisely, de�ned on the whole = R2n f(0; 0)g). One trivial solution is u (x; y) isa constant function; another less trivial one is

u (x; y) =

8>>>><>>>>:e�(

xy )

2

; y > 0; x 2 (�1;1)

0; y = 0; x 6= 0 2 (�1;1)

e�(xy )

2

; y < 0; x 2 (�1;1) :

(79)

We leave it to you to check that u 2 C1 () ; where = R2n f(0; 0)g : �

17

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Exercise 3.13 Show that the function u (x; y) given by (79) lies in the space C1 () ; where =R2n f(0; 0)g :

Remark 3.14 One can check that if u (x; y) is a C1 solution of xux+yuy = 0 de�ned on some openset � R2 containing the origin (0; 0) (singularity of the vector �eld), then it must be a constantsolution on :

Example 3.15 Find the general solution of the equation

yux � xuy = 0; u = u (x; y) ; (x; y) 2 = R2n f(0; 0)g : (80)

Solution:

The ODE for the equation is

dx

dt= y;

dy

dt= �x; dy

dx= �x

y(81)

which gives x00 (t)+x (t) = 0 and y00 (t)+y (t) = 0: The general solution (x (t) ; y (t)) of (81) is givenby

x (t) = A cos t+B sin t; y (t) = B cos t� A sin t; t 2 (�1;1); x (0) = A; y (0) = B;

where A; B are two arbitrary numbers with (A;B) 2 . The implicit form of the characteristiccurve (x (t) ; y (t)) is given by the equation

h (x; y) = x2 + y2 = A2 +B2 = d > 0 (82)

for arbitrary number d > 0 (when d = 0; the characteristic curve (x (t) ; y (t)) � (0; 0) is the equi-librium solution of the ODE). Each characteristic curve is a circle with radius

pd > 0; centered

at the origin.The function h (x; y) = x2 + y2 is de�ned on and is constant along each characteristic curve

lying in . Hence h (x; y) = x2 + y2 : ! R is a solution of (80). Moreover, it is de�ned on thewhole :Another way to �nd (82) is to rewrite the ODE system as (now we view y as a function of x)

dy

dx= �x

y(this is a separable equation; we get

Zydy = �

Zxdx) (83)

and solve it to get the implicit solution for y (x) ; i.e.

x2 + y2 = C > 0; C is a constant, (84)

which is the same as (82) (the explicit solution for (83) is given by y (x) =pC � x2).

Now we do the nonlinear change of variables(w = x2 + y2;

z = y:

The Jacobian of the change of variables is

det

�wx wyzx zy

�= 2x:

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Thus the change of variables is good either on the half-plane R2x+ or on the half-plane R2x� : On R

2x+ ;

the inverse function relation (x; y) ! (w; z) is globally valid with x =pw � z2; y = z: The

open domain in wz-space corresponding to R2x+ is

� =�(w; z) 2 R2 : w > z2

; � ! R2x+ is a bijection;

and the equation yux � xuy = 0 is equivalent to the equation xUz = 0: Similarly, on R2x� ; theinverse function relation (x; y) ! (w; z) is globally valid with x = �

pw � z2; y = z:

The general solution of the above ODE is U (w; z) = F (w) for arbitrary C1 function F (w) : Thuson R2x+ the general solution of the PDE is

u (x; y) = F�x2 + y2

�; x > 0; (85)

where F (w) is an arbitrary C1 function de�ned on open interval I � (0;1) : Similarly, on R2x� thegeneral solution of the PDE is

u (x; y) = G�x2 + y2

�; x < 0; (86)

where G (w) is an arbitrary C1 function de�ned on open interval J � (0;1) :Unlike the previous example, where each integral curve on R2y+ will not traverse into

R2y� (so F (�) on R2y+ and G (�) on R2y� can be independent), here integral curve on R2x+ willtraverse into R2x� : Hence we must choose F (�) = G (�) on (0;1) to make u (x; y) consistent onboth sides of y-axis and have larger domain (across both sides of y-axis). Hence anysolution u (x; y) of the equation on = R2n f(0; 0)g must have the form

u (x; y) = F�x2 + y2

�(87)

for arbitrary C1 function F (�) de�ned on (0;1) : In case F (w) has smaller domain (a; b) �(0;1) ; then u (x; y) has smaller domain�

(x; y) 2 R2 : a < x2 + y2 < b� :

We call (87) the general solution of the equation. Note that u (x; y) is constant along each char-acteristic curve (circle) x2 + y2 = C > 0. �

Remark 3.16 The function u (x; y) = x2+y2 is a solution of yux�xuy = 0 on = R2n f(0; 0)g : Infact, it also de�ned on the whole R2:

Example 3.17 Find the general solution of the equation

xux � yuy = 0; u = u (x; y) ; (x; y) 2 = R2n f(0; 0)g : (88)

Solution:

The system of ODE is

dx

dt= x;

dy

dt= �y; x (t) = c1e

t; y (t) = c2e�t

and we obtain the equation h (x; y) = xy = k (k 2 (�1;1) is a constant) on characteristic curves.Now we do the change of variables(

w = xy

z = ysame as

(x = w

z; z 6= 0

y = z:

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Its Jacobian is

det

�wx wyzx zy

�= y:

Thus the change of variables is good either onR2y+ = f(x; y) 2 R2 : y > 0g or onR2y� = f(x; y) 2 R2 : y < 0g :OnR2y+ the change of variables is a bijection map betweenR

2y+ and the open set� = f(w; z) 2 R2 : z > 0g

in wz-space and the equation xux � yuy = 0 on R2y+ is equivalent to the equation on �

0 = xux � yuy = x (Uwwx + Uzzx)� y (Uwwy + Uzzy) = �yUz; y > 0:

Therefore the general solution on R2y+ is given by

u (x; y) = F (xy) ; (x; y) 2 R2y+ ;

where F (�) is an arbitrary C1 function de�ned on (�1;1) and u (x; y) is de�ned on R2y+ : Similarly,on R2y� the general solution is given by

u (x; y) = G (xy) ; (x; y) 2 R2y�,

where G (�) is an arbitrary C1 function de�ned on (�1;1) : Unlike the example yux � xuy = 0,now each characteristic curve does not traverse from R2y+ to R

2y� ; and vice versa. If u (x; y) is not

de�ned on the x-axis (y = 0), then the function F and the function G in the above can be arbitraryand independent to each other. Compare with Example 3.12. �

Remark 3.18 If we choose z = x; the analysis is exactly the same except that the two opensets R2y+ and R

2y� become R

2x+ and R

2x� :

Remark 3.19 The function u (x; y) = xy is a solution of xux � yuy = 0 on = R2n f(0; 0)g : Infact, it also de�ned on the whole R2:

Example 3.20 Find the general solution of the equation

ux + yuy = 0; u = u (x; y) ; (x; y) 2 = R2: (89)

Note that the equation has no singularity on R2:

Solution:

The ODE has the form

dx

dt= 1;

dy

dt= y; x (t) = c1 + t; y (t) = c2e

t

ordy

dx=y

1;

Z1

ydy =

Zdx:

Each gives the implicit solution equation

h (x; y) = e�xy = k; k is an arbitrary constant.

Therefore we do the change of variables (now we choose z = x instead of z = y)(w = e�xy

z = xsame as

(x = z

y = ezw:

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and see that the Jacobian for this change of variables is

det

�wx wyzx zy

�= det

��e�xy e�x

1 0

�= �e�x 6= 0; (90)

which is everywhere nonzero (if we choose z = y; then the Jacobian is �e�xy; which is bad aty = 0). Thus this change of variables is globally good on R2: Now we have

0 = ux + yuy = (Uwwx + Uzzx) + y (Uwwy + Uzzy)

=��Uwe�xy + Uz

�+ yUwe

�x = Uz; U = U (w; z)

and so the general solution is

u (x; y) = U (w; z) = F (w) = F�e�xy

�;

where F (�) is an arbitrary C1 function and if F (�) is de�ned on (�1;1) ; then u (x; y) is de�nedon R2: The solution is constant along each characteristic curve e�xy = k: �

Until now we have only looked at equation of the simple form

a (x; y)ux + b (x; y)uy = 0; u = u (x; y) ; (x; y) 2 : (91)

In case we have the general form

a (x; y)ux + b (x; y)uy + c (x; y)u = f (x; y) ; u = u (x; y) ; (x; y) 2 ; (92)

the method is exactly the same. We can use the system of characteristic ODE (60) to �nd thefunction h (x; y) ; which satis�es

h (x; y) = d; (x; y) 2 ~ (93)

along each characteristic curve. After that we can use the change of variables(w = h (x; y)

z = y (or z = x)

to convert the PDE (92) into an ODE for U (w; z) : To see this, letA (w; z) ; B (w; z) ; C (w; z) ; U (w; z) ;F (w; z) be the functions corresponding to a (x; y) ; b (x; y) ; c (x; y) ; u (x; y) ; f (x; y) : Then by

a (x; y)ux + b (x; y)uy + c (x; y)u

= a (x; y)Uwhx + b (x; y) (Uwhy + Uz) + C (w; z)U

=

�a (x; y)hx + b (x; y)hy| {z }

�Uw +B (w; z)Uz + C (w; z)U = B (w; z)Uz + C (w; z)U;

we see that the PDE (92) for the function U (w; z) becomes an ODE of the form

B (w; z)Uz (w; z) + C (w; z)U (w; z) = F (w; z) ; (94)

which can be rewritten as

Uz (w; z) + P (w; z)U (w; z) = Q (w; z)| {z }; P (w; z) =C (w; z)

B (w; z); Q (w; z) =

F (w; z)

B (w; z): (95)

Note that (95) is a �rst order linear ODE (in the variable z) containing a parameterw; which can be solved.

21

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Example 3.21 Find the general solution of the equation

ux + yuy + u = ex+y; u = u (x; y) ; (x; y) 2 R2: (96)

Note that the equation has no singularity on R2:

Solution:

Similar to Example 3.20, we let w = e�xy; z = x; which is a global change of variables. Wehave

ux + yuy + u =��Uwe�xy + Uz

�+ y

�Uwe

�x�+ U

= Uz + U = ez+ezw; (w; z) 2 R2 (97)

which is an ODE for U (w; z) in the variable z with a parameter w; and get

U (w; z) = e�z�Z

e2z � eezwdz + F (w)

�: (98)

To �nd the integral, we let s = ez and get ds = ezdz (i.e. dz = 1sds)Z

e2z � eezwdz =Zs2 � esw 1

sds =

Zseswds =

Zsd

�1

wesw�

=s

wesw � 1

w

Zeswds =

s

wesw � 1

w2esw =

�ez

w� 1

w2

�ee

zw; s = ez

and so

U (w; z) = e�z��

ez

w� 1

w2

�ee

zw + F (w)

�; where w = e�xy; z = x:

We obtain the general solution

u (x; y) = e�x��

ex

e�xy� 1

(e�xy)2

�ey + F

�e�xy

��= ex+y

�1

y� 1

y2

�+ e�xF

�e�xy

�; y 6= 0 (99)

where F (�) is an arbitrary C1 function and if F (�) is de�ned on R; then u (x; y) is de�ned onR2nfy = 0g: �

Remark 3.22 (Important.) Since the ODE Uz + U = ez+ezw is de�ned on R2 (the function

ez+ezw is de�ned on R2), one should be able to �nd a solution u (x; y) de�ned on R2: We can use

de�nite integral to get

ezU (w; z) =

Z z

0

e2�+e�wd� + F (w) ; (100)

where by the inde�nite integral formulaZe2�ee

�wdz =

�e�

w� 1

w2

�ee

�w

we get Z z

0

e2�+e�wd� =

( �ez

w� 1

w2

�ee

zw ��1w� 1

w2

�ew; w 6= 0

12e2z � 1

2; w = 0:

(101)

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By (100) and (101), we conclude

U (w; z) = e�zV (w; z) + e�zF (w) ; w = e�xy; z = x: (102)

where

V (w; z) =

( �ez

w� 1

w2

�ee

zw ��1w� 1

w2

�ew; w 6= 0

12e2z � 1

2; w = 0:

(103)

We claim that the V (w; z) 2 C1 (R2) : To see this, it su¢ ces to check the continuity and derivativescontinuity across w = 0: By L�Hospital rule, we �rst have

limw!0

��ez

w� 1

w2

�ee

zw ��1

w� 1

w2

�ew�= lim

w!0

(ezw � 1) eezw � (w � 1) eww2

�0

0form

�= lim

w!0

ezeezw + (ezw � 1) eezwez � ew � (w � 1) ew

2w= lim

w!0

(ezw) eezwez � wew2w

= limw!0

(ez) eezwez � ew2

=1

2e2z � 1

2: (104)

Hence V (w; z) is continuous across w = 0: We leave the check of the rest as an exercise. From(102), we can obtain the general solution formula for u (x; y), which is slightly di¤erentfrom (99).

Recall the important fact that every solution of the PDE a (x; y)ux+ b (x; y)uy = 0 is constantalong the characteristic curves. That means if we know the value of u (x; y) at one point p 2 C (Cis a characteristic curve), then we know the values of u (x; y) at every point of C: This is also truefor the general nonhomogeneous equation

a (x; y)ux + b (x; y)uy + c (x; y)u = f (x; y) ; u = u (x; y) : (105)

More precisely, we have:

Lemma 3.23 Let u (x; y) be a C1 solution of (105) on some open set with (x0; y0) 2 : Let C :� (t) = (x (t) ; y (t)) 2 ; t 2 I; be the unique characteristic curve of (105) satisfying (x (0) ; y (0)) =(x0; y0) 2 C; t0 2 I: Then the function U (t) = u (x (t) ; y (t)) ; t 2 I; satis�es the ODE

U 0 (t) + C (t)U (t) = F (t) ; t 2 I; U (0) = u (x0; y0) ; (106)

where C (t) = c (x (t) ; y (t)) and F (t) = f (x (t) ; y (t)) : In particular, if we know the value of U (0) (i.e.if we know the value of u (x0; y0)), then we can know the value of U (t) on the whole interval I (bysolving the ODE (106)).

Proof. This is straightforward. We have

U 0 (t) =d

dtu (x (t) ; y (t)) = ux (x (t) ; y (t))x

0 (t) + uy (x (t) ; y (t)) y0 (t)

= a (x (t) ; y (t))ux (x (t) ; y (t)) + b (x (t) ; y (t))uy (x (t) ; y (t))

= �c (x (t) ; y (t))u (x (t) ; y (t)) + f (x (t) ; y (t)) = �C (t)U (t) + F (t) ; t 2 I:

Hence U (t) satis�es the ODE (106) on I and if we know U (0) ; U (t) can be expressed as

U (t) = e�R t0 C(�)d�

�Z t

0

�eR s0 C(�)d�F (s)

�ds+ U (0)

�; t 2 I:

The result follows. �

23

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3.1 First order linear PDE with variable coe¢ cients and side condition.

Due to Lemma 3.23, a "well-posed" side condition on a curve is that it intersects each charac-teristic curve C at exactly one point (so that we have exactly one initial condition U (t0) forthe ODE in Lemma 3.23 on each characteristic curve). Otherwise, the problem may haveno solution or in�nitely many solutions.

Example 3.24 Find the solution to the problem(yux � xuy = 0

u (s; s2) = s3; s 2 (0;1) :

Solution:

We already know that the general solution has the form u (x; y) = F (x2 + y2) for arbitrary C1

function F (�) : Since the curve := (s; s2) ; s 2 (0;1) ; intersects each characteristic curve x2+y2 =d > 0 at exactly one point, one should be able to �nd a unique F (�) satisfying the condition. Weneed to solve the equation

F�s2 + s4

�= s3; s 2 (0;1) :

Let r = s2 + s4 > 0; and let p = s2 > 0: We have p2 + p� r = 0 and so

p =�1�

p1 + 4r

2(the minus sign does not make sense).

Thus we have

F (r) = s3 = p3=2 =

��1 +

p1 + 4r

2

�3=2and the solution is given by

u (x; y) = F�x2 + y2

�=

�1 +

p1 + 4 (x2 + y2)

2

!3=2: (107)

The solution is de�ned on R2 and lies in the space C1 (R2)TC1 (R2n f(0; 0)g)): To see this, note

that we have

g (x; y) =�1 +

p1 + 4 (x2 + y2)

22 C1

�R2�; g (x; y) : R2 ! [0;1); g (0; 0) = 0

and the functionh (�) = �3=2 2 C1[0;1)

\C1 (0;1) ; h (0) = 0:

Therefore, chain rule implies

u (x; y) = h (g (x; y)) 2 C1�R2�\

C1�R2n f(0; 0)g

�; (108)

and we have

u�s; s2

�=

�1 +

p1 + 4 (s2 + s4)

2

!3=2=

��1 + (2s2 + 1)

2

�3=2= s3; 8 s 2 (0;1) :

24

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4 Second order linear PDE with constant coe¢ cients; clas-si�cation and canonical forms.

We now consider the linear second order equation with constant coe¢ cients, given by

auxx + 2buxy + cuyy + 2dux + 2euy + ku = f (x; y) ; u = u (x; y) (109)

where a; :::; k are all constants and f (x; y) is a given function de�ned on some open set � R2:Wewant to �nd a C2 function u (x; y) satisfying (109) on some open set (may be just a subset of ).Note that for a C2 function u (x; y) ; we have uxy (x; y) = uyx (x; y) on its domain.

Remark 4.1 Note that if u = u (x1; ::: ; xn) depends on n variables, the discussions below aresimilar. For convenience, we assume that u = u (x; y) depends only on 2 variables.

We can write (109) in the matrix form as

Trace

��a bb c

��uxx uxyuxy uyy

��+ 2

�de

���uxuy

�+ ku = f (x; y) ; (110)

which is helpful for us to understand the e¤ect of the change of variables. As we shall see soon,the "type" of the equation (109) is determined by the sign of the determinant of the coe¢ cientmatrix.

Lemma 4.2 Assume u (x; y) is a C2 function de�ned on some domain � R2: If we introduce thelinear change of variables given by

� = � (x; y) = Ax+By; � = � (x; y) = Cx+Dy; A; B; C; D are all const., (111)

i.e., ���

�= J

�xy

�; where J =

�A BC D

�; det J 6= 0; (112)

then the equation (110) for the function U (�; �) (where U (Ax+By;Cx+Dy) = u (x; y)) becomes

Tr

24J � a bb c

�JT| {z } �

�U�� U��U�� U��

�35+ (lower order terms) = F (�; �) ; (113)

where Tr (�) is the trace of a matrix.

Remark 4.3 Denote

M =

�a bb c

�; N = J

�a bb c

�JT ; det J 6= 0: (114)

We see that the matrix N is also symmetric. By a theorem from linear algebra, all eigenvaluesof both M and N are real. Moreover, the sign of detM and detN are the same due to detN =(det J)2 detM; where det J 6= 0: Finally, if J is an orthogonal matrix (i.e. JT = J�1), then bothM and N are similar and have the same eigenvalues. Our goal is to diagonalize M (i.e. make N tobe diagonal), which will reduce equation (109) into canonical form.

Proof. We have U (Ax+By;Cx+Dy) = u (x; y) and by (6) and (10), we have�uxuy

�= JT

�U�U�

�;

�uxx uxyuxy uyy

�= JT

�U�� U��U�� U��

�J; J =

�A BC D

�: (115)

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Thus the equation

Trace

��a bb c

��uxx uxyuxy uyy

��+ (lower order terms) = f (x; y)

becomes (note that for any two matrices M; N; we have the identity Tr (MN) = Tr (NM) inlinear algebra)

Trace

24� a bb c

�� JT

�U�� U��U�� U��

�J| {z }35+ (lower order terms)

= Trace

24J � a bb c

�JT| {z } �

�U�� U��U�� U��

�35+ (lower order terms) = F (�; �) (116)

The proof is done. �

De�nition 4.4 Since the matrix

M =

�a bb c

�is symmetric, it has two real eigenvalues �1 and �2: If both of them are positive (or negative),then we say the equation (109) is elliptic (this is equivalent to detM = ac � b2 > 0). If oneeigenvalue is positive and the other is negative, we say the equation is hyperbolic (this is equivalentto detM = ac� b2 < 0). If one eigenvalue is zero and the other is nonzero, we say the equation isparabolic (this is equivalent to detM = ac� b2 = 0). Note that by

det

�J

�a bb c

�JT�

| {z } = (det J)2 det�a bb c

�; where det J 6= 0;

we see that the type of the linear equation (110) is invariant under a linear change ofvariables.

Since the matrixM in (114) is symmetric, by linear algebra theory, we can �nd orthonormalbasis fv1; v2g (they are eigenvectors corresponding to �1; �2) such that

P TMP =

��1 00 �2

�; P T = P�1; (117)

where P is the orthogonal matrix P = (v1; v2) (v1; v2 are column vectors of P ). Assume thatv1 = (�; �) and v2 = (p; q) and let�

��

�= J

�xy

�=

�� �p q

��xy

�; J =

�� �p q

�= P T ; (118)

i.e. J = P T : We have

Trace

24J � a bb c

�JT| {z } �

�U�� U��U�� U��

�35 = Trace

24P T � a bb c

�P| {z } ��U�� U��U�� U��

�35= Trace

24� �1 00 �2

�| {z } �

�U�� U��U�� U��

�35 = �1U�� + �2U��| {z }; (119)

which will reduce the leading terms auxx + 2buxy + cuyy + � � � of the PDE (109) into canonicalform !!

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Remark 4.5 (Interesting observation ...) In case the original equation (109) is already aLaplace equation uxx+ uyy = 0; then the matrix M = I is the identity matrix and any orthogonalmatrix P will make

P TMP = P T IP = P TP =

�1 00 1

�; P = (v1; v2) :

With this, (113) becomes

Trace

24J � 1 00 1

�JT| {z } �

�U�� U��U�� U��

�35 ; J = P T ; JJT = P TP = I

= Trace

�JJT �

�U�� U��U�� U��

��= U�� + U�� = 0; (120)

which means that under any orthogonal linear change of variables (x; y) ! (�; �) ; the newequation for U (�; �) still has the same form

U�� + U�� = 0: (121)

The space of all 2 � 2 orthogonal matrices forms a group, which we denote it as O (2) : It isknown that the group O (2) is the set of all rotations about the origin O in R2 (they havedet = 1) together with all re�ections with respect to straight lines through the origin Oin R2 (they have det = �1). In terms of the polar coordinates (r; �) ; the �rst case correspondsto � ! � + �0 (�0 may be positive or negative) and the second case corresponds to � ! �� (say,re�ection with respect to the x-axis). Note that the Laplace equation under polar coordinates (r; �)has the form �

@2

@r2+1

r

@

@r+1

r2@2

@�2

�U (r; �) = 0

and one can see that if U (r; �) is a solution, so is U (r; � + �0) (rotation); the same for U (r;��) (re�ection).

By (119), we can conclude the following classi�cation result for equation (109):

Lemma 4.6 (Canonical form.) If the linear equation (110) is elliptic, then one can �nd asuitable linear change of variables (using eigenvalues and eigenvectors)

� = Ax+By; � = Cx+Dy; A; B; C; D are constants,

so that the equation for U (�; �) has the form

U�� + U�� + (lower order terms) = F (�; �) : (122)

For hyperbolic case, the equation has the form

U�� � U�� + (lower order terms) = F (�; �) ; (123)

and for parabolic case, the equation has the form

U�� + (lower order terms) = F (�; �) : (124)

Remark 4.7 The forms in (122), (123) and (124) are said to be in canonical forms. Anothercanonical form of the hyperbolic case is

U�� + (lower order terms) = F (�; �) : (125)

One can show that an equation of the form uxx � uyy = 0 can be converted into an equation of theform 4U�� = 0 (by the change of variables � = x+ y; � = x� y). Therefore, canonical form (123)and (125) are equivalent.

27

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Proof. For the elliptic case, by multiplying the equation by a minus sign if necessary, we mayassume �1 > 0; �2 > 0 (both are eigenvalues of the coe¢ cient matrix). By the change of variables(118), we can convert in into the form

�1U�� + �2U�� + (lower order terms) = 0; �1 > 0; �2 > 0:

If we let~� =

�p�1; ~� =

�p�2; (126)

then U (�; �) becomes ~U�~�; ~��with ~U

�~�; ~��= U

�p�1~�;p�2~��and we have

�1U�� + �2U�� + (lower order terms)

= ~U~�~�

�~�; ~��+ ~U~�~�

�~�; ~��+ (lower order terms) = ~F

�~�; ~��:

Thus we have arrived at the form (122). For the hyperbolic case, we have �1 > 0; �2 < 0: Thenwe replace (126) by

~� =�p�1; ~� =

�p��2

: (127)

For the parabolic case, we have �1 > 0; �2 = 0: Then we replace (126) by

~� =�p�1; ~� = �: (128)

The proof is done. �

Remark 4.8 (Important.) For parabolic case, in case there is NO U� term in (lower orderterms) of (124), we say the equation is degenerate. We will not look at a degenerate parabolicequation. It is actually a second order ODE in � of the form (view � as a parameter):

U�� + aU� + bU = F (�; �) ; a; b are constants. (129)

For an ODE of the above form, we can get rid of the middle term aU� by looking at the function

w = ea2�U| {z } (same as U = e�

a2�w). (130)

We have 8>>>><>>>>:bU = be�

a2�w

aU� = �a2

2e�

a2�w + ae�

a2�w�| {z }

U�� =a2

4e�

a2�w � ae�a

2�w�| {z }+e�a

2�w��

and the new equation for w (�; �) becomes

e�a2�

�w�� +

�b� a2

4

�w

�= F (�; �) ; i.e. w�� +

�b� a2

4

�w| {z } = e

a2�F (�; �) . (131)

One can use standard ODE method (variation of parameters method) to solve (131). From now on,we shall only focus on nondegenerate parabolic equation.

Motivated by Remark 4.8, one can go further to get rid of the �rst derivative terms in(122), (123) and (124). For simplicity, we just look at one example.

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Example 4.9 (This is for elliptic and hyperbolic equations.) Assume we have an ellipticequation in canonical form:

U�� + U�� + 3U� + 4U� + 5U = F (�; �) ; (132)

where we can write is as �U�� + 3U�| {z }

�+

�U�� + 4U�| {z }

�+ 5U = F (�; �)

Motivated by (129) and the function in (130), we let v (�; �) be the new function given by

v (�; �) = e32�+ 4

2�U (�; �) ; (133)

and compute

v� = e32�+ 4

2�

�3

2U + U�

�; v� = e

32�+ 4

2�

�4

2U + U�

�(134)

and

v�� = e32�+ 4

2�

�9

4U + 3U� + U��| {z }

�; v�� = e

32�+ 4

2�

�4U + 4U� + U��| {z }

�: (135)

Hence we obtain

v�� + v�� = e32�+ 4

2�

��U�� + 3U�| {z }

�+

�U�� + 4U�| {z }

�+

�9

4+ 4

�U

�and conclude

v�� + v�� �5

4v

= e32�+ 4

2� (U�� + 3U� + U�� + 4U� + 5U) = � (�; �) ; where � (�; �) = e

32�+ 4

2�F (�; �) : (136)

The new equation for v has no �rst derivatives terms. Note that, in general, one cannot choosea and b to get rid of the three terms 3U� +4U� +5U: Therefore, the term cv in Theorem 4.11 belowcannot be removed in general. The same for the hyperbolic equation.

Example 4.10 (This is for nondegenerate parabolic equations.) Assume we have the nonde-generate parabolic equation in canonical form:

U�� + 3U� + 4U� + 5U = F (�; �) : (137)

where we can write is as �U�� + 3U�| {z }

�+

�4U� + 5U| {z }

�= F (�; �) :

Now we let v (�; �) be the new function given by

v (�; �) = e32�+��U (�; �) ; � is a constant to be determined (138)

and compute 8<: v� = e32�+�� �

�32U + U�

�;

v� = e32�+�� � (�U + U�) (same as 4v� = e

32�+�� � (4�U + 4U�)

(139)

and

v�� = e32�+��

�9

4U + 3U� + U��| {z }

�; (140)

29

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where, since there is no U�� term in the original equation, we do not have to compute v��. By (139)and (140), we have

v�� + 4v� = e32�+��

�9

4U + 3U� + U��| {z }+4�U + 4U�

�: (141)

Therefore, if we choose � = 11=16; we will have

v�� + 4v� = e32�+ 11

16� (U�� + 3U� + 4U� + 5U) = � (�; �) ; where � (�; �) = e

32�+ 11

16�F (�; �) : (142)

Note that there is no v� term and v term in (142).

By the above two examples, we can improve Lemma 4.6 as:

Theorem 4.11 (Re�ned canonical form.) If the linear equation (110) is elliptic, then one can�nd a suitable linear change of variables (using eigenvalues, eigenvectors and scalings) andmultiply the solution by some exponential function so that, eventually, the equation has the form

v�� + v�� + cv = � (�; �) ; v = v (�; �) ; (143)

for some constant c 2 (�1;1) and some function � (�; �) : If the equation (110) is hyperbolic,the equation has the form

v�� � v�� + cv = � (�; �) ; v = v (�; �) ; (144)

for some constant c 2 (�1;1) and some function � (�; �) : If the equation (110) is parabolic andnondegenerate, the equation has the form

v�� + cv� = � (�; �) ; v = v (�; �) ; (145)

for some constant c 6= 0 and some function � (�; �) :

Remark 4.12 (Important.) The constant c in the elliptic case can be c > 0 or c = 0 or c <0: For c > 0; we can make it equal to 1 by doing the change of variables

~� =pc�; ~� =

pc�; ~v

�~�; ~��= v

��pc;�pc

�;

and for c < 0; we can make it equal to �1 by doing the change of variables

~� =p�c�; ~� =

p�c�; ~v

�~�; ~��= v

��p�c

;�p�c

�:

Thus in the elliptic case, we may simply assume c = 1 or 0 or �1: The constant c in the hyperboliccase can be c > 0 or c = 0 or c < 0: For c < 0; by switching the role of � and �; we may assume c >0 or c = 0: Hence for the hyperbolic case, eventually, we can simply assume c = 1 or 0: Finally,for the parabolic case, the constant c 6= 0 can be c > 0 or c < 0: So eventually we can simplyassume c = 1 or �1: However, since most parabolic equations come from physical phenomenoninvolving the behavior of some quantity v (�; �) depending on space and time. So � will representspace variable and � will represent time variable. In that case a model parabolic equationlooks like (assume � (�; �) = 0 for simplicity)

(1) : vt = vxx or (2) : vt = �vxx: (146)

We call (1) the "forward heat equation" (or just heat equation) and (2) the "backward heatequation". Since in reality, time cannot go backwards, so in a parabolic equation, we always fucuson the behavior of a solution v (x; t) as time goes forwards, i.e., as t is increasing. One canuse simple examples to see that, as time goes forwards, the heat equation (1) will make solutionbetter, while the backward heat equation (2) will make solution worse (look at e�t sin x and et sin xrespectively). Thus, as time goes forwards, equation (1) is well-posed, while (2) is ill-posed. In thiscourse, we will focus only on (1) (on the other hand, as time goes backwards, (1) will makesolution worse and (2) will make solution better...).

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Proof. The proof is now obvious. We omit it. �

De�nition 4.13 Let v = v (�; �) : The equations v�� + v�� = 0; v�� � v�� = 0; v�� � v� = 0;are called Laplace equation (elliptic equation), wave equation (hyperbolic equation), andheat equation (nondegenerate parabolic equation), respectively.

Remark 4.14 In this course we shall focus on Laplace equation, wave equation and heatequation only.

5 General solutions of hyperbolic equations without lowerorder terms.

In this section, we look at equations of the form (note that (147) has NO lower order terms)

auxx + 2buxy + cuyy = 0; u = u (x; y) ; a; b; c are const., (147)

which can be written as

Trace

��a bb c

��uxx uxyuxy uyy

��= 0; det

�a bb c

�= ac� b2:

We want to �nd general C2 solution u = u (x; y) of (147) de�ned on R2: The canonical form of(147) is given by 8>><>>:

U�� + U�� = 0 (ac > b2; elliptic)

U�� � U�� = 0 or U�� = 0 (ac < b2; hyperbolic)

U�� = 0 (ac = b2; parabolic but degenerate).

(148)

Hence for hyperbolic and parabolic equations, we know how to solve them (but not forelliptic equation). The method is either by a change of variables (diagonalization method)or by a factorization method.

Lemma 5.1 Let A; B; C; D be constants with AD �BC 6= 0: Consider the equation

Aux +Buy = g (Dx� Cy) ; u = u (x; y) (149)

where g (�) is a given C0 function de�ned on R. Then the general solution of (149) is given by

u (x; y) = F (Bx� Ay) +G (Dx� Cy) ; (150)

where F (�) is an arbitrary C1 function de�ned on R and the C1 function G (�) satis�es

G0 (�) =g (�)

AD �BC ; 8 � 2 (�1;1) : (151)

In particular, if the function g (�) is arbitrary, then the general solution of (149) is given by (150),where both F (�) and G (�) are arbitrary C1 functions de�ned on R.

Remark 5.2 The case AD �BC = 0 will be discussed later on. See (178).

Remark 5.3 Also note that AD�BC 6= 0 implies that the two families of lines Bx�Ay = �; Dx�Cy = � are not parallel. As a consequence of this, the two terms F (Bx� Ay) ; G (Dx� Cy) in(156) are essentially di¤erent.

31

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Proof. First assume that B 6= 0: Let w = Bx � Ay; z = y; which is a global change of variablesdue to B 6= 0. The equation for U (w; z) is

Aux +Buy = AUwB +B [Uw (�A) + Uz]

= BUz = g (Dx� Cy) = g

�Dw + (AD �BC) z

B

�:

Hence we get

Uz =1

Bg

�Dw + (AD �BC) z

B

�and obtain

U (w; z) = F (w) +1

B

Zg

�Dw + (AD �BC) z

B

�dz = F (w) +G

�Dw + (AD �BC) z

B

�;

where G (�) satis�es (151). By this, the solution for u (x; y) is given by (150).

u (x; y) = F (Bx� Ay) +G

�D (Bx� Ay) + (AD �BC) y

B

�= F (Bx� Ay) +G (Dx� Cy) ;

which is (150).If B = 0; then we must have A 6= 0 due to AD�BC 6= 0: Now we let w = Bx�Ay; z = x:We

get

Aux +Buy = A [UwB + Uz] +BUw (�A)

= AUz = g (Dx� Cy) = g

�Cw + (AD �BC) z

A

�:

Hence we get

Uz =1

Ag

�Cw + (AD �BC) z

A

�and obtain

U (w; z) = F (w) +1

A

Zg

�Cw + (AD �BC) z

A

�dz = F (w) +G

�Cw + (AD �BC) z

A

�;

where now G (�) satis�es (151). Now we have

u (x; y) = F (Bx� Ay) +G

�C (Bx� Ay) + (AD �BC)x

A

�= F (Bx� Ay) +G (Dx� Cy) :

Again, this gives (150). �

Remark 5.4 (Interesting.) Another quick proof is to choose the change of variables w = Bx �Ay; z = Dx� Cy: It has nonzero Jacobian���� B �A

D �C

���� = AD �BC 6= 0:

Then we have

Aux +Buy = A [UwB + UzD] +B [Uw (�A) + Uz (�C)] = (AD �BC)Uz = g (z) ;

which gives

Uz =g (z)

AD �BC ; AD �BC 6= 0

and soU (w; z) = F (w) +G (z) = F (Bx� Ay) +G (Dx� Cy) ;

where G0 (z) = g(z)AD�BC :

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5.1 Solving hyperbolic equations; factorization method.

Lemma 5.5 Let a; b; c be three given constants with ac < b2: Then one can �nd constants A; B;C; D satisfying

AC = a; AD +BC = 2b; BD = c; AD �BC 6= 0: (152)

This means that, if ac < b2; we can factor the second order homogeneous polynomial ax2+2bxy+cy2

asax2 + 2bxy + cy2 = (Ax+By) (Cx+Dy) ; where AD �BC 6= 0: (153)

Remark 5.6 As long as A; B; C; D satisfy AC = a; AD +BC = 2b; BD = c; we have

4�ac� b2

�= 4ABCD � (AD +BC)2 = � (AD �BC)2 < 0:

Therefore, we automatically have AD �BC 6= 0.

Remark 5.7 If a = 1; we can choose A = C = 1; and then solve

B +D = 2b; BD = c

to getB = b�

pb2 � c; D =

c

b�pb2 � c

; for c 6= 0

andB = 2b; D = 0; for c = 0:

Remark 5.8 If a > 0; b = 0; c < 0; we can choose A = C =pa and B =

p�c; D = �

p�c:

Remark 5.9 If ac > b2; then (152) cannot be satis�ed. Therefore, the method in this section cannotbe used to elliptic equations.

Proof. (Read it yourself.) If a = 0; then by ac = 0 < b2; we must have b 6= 0: The numbers

A = 1; B =c

2b; C = 0; D = 2b;

satisfy (152). If c = 0; then we still have b 6= 0: The numbers

A =a

2b; B = 1; C = 2b; D = 0;

satisfy (152). If ac 6= 0; then b+pb2 � ac 6= 0 and b�

pb2 � ac 6= 0: The numbers

A = 1; B =b+pb2 � aca

; C = a; D =ac

b+pb2 � ac

satisfy (152). Finally, in each case we can see that AD �BC 6= 0: �

We now focus on the hyperbolic (ac� b2 < 0) case in (147). That is, the eigenvalues �1; �2 ofthe coe¢ cient matrix have di¤erent sign and we may assume �1 > 0; �2 < 0:

33

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Lemma 5.10 (Factorization method for hyperbolic equation.) Assume (147) is hyperbolic,i.e., ac < b2. Then one can decompose it as

auxx + 2buxy + cuyy =

�A@

@x+B

@

@y

���C@

@x+D

@

@y

�u

�= 0 (154)

for some constants A; B; C; D satisfying

AC = a; AD +BC = 2b; BD = c; AD �BC 6= 0: (155)

In particular, the general solution of (147) is given by

u (x; y) = F (Bx� Ay) +G (Dx� Cy) ; (x; y) 2 R2 (156)

for arbitrary C2 functions F (�) ; G (�) de�ned on R:

Proof. For a C2 function u; we have�A@

@x+B

@

@y

���C@

@x+D

@

@y

�u

�= A (Cux +Duy)x +B (Cux +Duy)y = ACuxx + (AD +BC)uxy +BDuyy:

Now by Lemma 5.5, there are numbers A; B; C; D satisfying

AC = a; AD +BC = 2b; BD = c; AD �BC 6= 0: (157)

Hence, for a C2 function u; it satis�es auxx + 2buxy + cuyy = 0 if and only if it satis�es�A@

@x+B

@

@y

���C@

@x+D

@

@y

�u

�= 0: (158)

We can �nd solutions of (158) by solving two �rst order PDE. Let v =�C @@x+D @

@y

�u: It

satis�es�A @@x+B @

@y

�v = 0: Hence v (x; y) = f (Bx� Ay) for arbitrary C1 function f and the

equation for u becomesCux +Duy = f (Bx� Ay) ; f 2 C1:

By Lemma 5.1, the general solution for u (x; y) is

u (x; y) = G (Dx� Cy) + F (Bx� Ay) ; (x; y) 2 R2;

where F; G are two arbitrary C2 functions de�ned on (�1;1) (since we want u (x; y) to be a C2solution, we must require F; G to be C2 functions). The proof is done. �

Remark 5.11 (Important.) Lemma 5.10 says that to solve the second order hyperbolic equation,it su¢ ces to solve two �rst order equations.

De�nition 5.12 We call the 2-parameter family of lines

Bx� Ay = �; Dx� Cy = �; AD �BC 6= 0

where �; � are arbitrary constants, the characteristic lines of the hyperbolic equation (154).

Example 5.13 Consider the second order linear equation in two variables:

uxx � 4uxy � 2uyy = 0; (x; y) 2 R2; u = u (x; y) : R2 ! R:

What is the type (elliptic, hyperbolic, or parabolic) of this equation ? Use factorization methodto �nd the general solution of the equation.

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Solution:

The equation has the form auxx + 2buxy + cuyy = 0; where a = 1; b = �2; c = �2 andac� b2 = �6 < 0: Therefore the equation is hyperbolic. We know that one can decompose it intotwo �rst order equations of the form (by Remark 5.7, we can choose A = C = 1)

uxx � 4uxy � 2uyy = (@x +B@y) (@x +D@y)u = uxx + (B +D)uxy +BDuyy:

Thus we solve B; D to satisfy the equation (note that now we have A = C = 1)

B +D = �4; BD = �2:We obtain (B;D) =

��2 +

p6;�2�

p6�or (B;D) =

��2�

p6;�2 +

p6�: Thus we choose (B;D) =�

�2 +p6;�2�

p6�and get

uxx � 4uxy � 2uyy =h@x +

��2 +

p6�@y

i h@x +

��2�

p6�@y

iu (159)

and the general solution is

u (x; y) = F (Bx� Ay) +G (Dx� Cy) = F���2 +

p6�x� y

�+G

���2�

p6�x� y

�= F

��p6� 2

�x� y

�+G

��p6 + 2

�x+ y

�; (160)

where F (z) : R! R and G (z) : R! R are two arbitrary C2 functions. �Remark 5.14 In terms of polynomial, the decomposition (159) is the same as

x2 � 4xy � 2y2 =�x+

��2 +

p6�y��

x+��2�

p6�y�: (161)

5.2 Solving hyperbolic equations; change of variables method.

Remark 5.15 By (164) below, we see that this method is essentially the same as the factorizationmethod. Computationally, you can just use the factorization method. Note that solving the equationU�� = 0 is the same as solving two �rst order equations. See equation (158) also.

We can also use a change of variables method to solve the equation

auxx + 2buxy + cuyy = 0; where ac < b2: (162)

Introduce the change of variables

� = Bx� Ay; � = Dx� Cy; Jacobian =

���� B �AD �C

���� = AD �BC; (163)

where A; B; C; D are constants satisfying

AC = a; AD +BC = 2b; BD = c; AD �BC 6= 0: (164)

Since AD�BC 6= 0; this change of variables is good on all R2: Let U (�; �) be the function u (x; y)in (�; �) variables. We have:

Lemma 5.16 (Change of variables method for hyperbolic equation.) Under the change ofvariables (163), (164), the equation for U (�; �) is given by

� (AD �BC)2 U�� = 0 (same as U�� = 0 since AD �BC 6= 0), (165)

which has general solution given by U (�; �) = F (�) + G (�) for arbitrary C2 functions (we wantU (�; �) to be a C2 function) F (�) and G (�) de�ned on R: As a consequence, the general solutionu (x; y) of (162) is

u (x; y) = F (Bx� Ay) +G (Dx� Cy) ; (x; y) 2 R2;for arbitrary C2 functions F (�) and G (�) de�ned on R:

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Proof. We have 8>>>>><>>>>>:

ux = BU� +DU�; uy = �AU� � CU�;

uxx = B2U�� + 2BDU�� +D2U��;

uxy = �ABU�� � (BC + AD)U�� � CDU��;

uyy = A2U�� + 2ACU�� + C2U��

and so

auxx + 2buxy + cuyy

=

(a [B2U�� + 2BDU�� +D2U��]

+2b [�ABU�� � (BC + AD)U�� � CDU��] + c [A2U�� + 2ACU�� + C2U��]

=

8><>:(aB2 � 2bAB + cA2)U��

+�a2BD � 2bBC � 2bAD + c2AC| {z }�U�� + �z }| {

aD2 � 2bCD + cC2�U��:

ByAC = a; AD +BC = 2b; BD = c; AD �BC 6= 0;

we have 8<:aB2 � 2bAB + cA2 = ACB2 � (AD +BC)AB +BDA2 = 0z }| {aD2 � 2bCD + cC2 = ACD2 � (AD +BC)CD +BDC2 = 0

and

a2BD � 2bBC � 2bAD + c2AC| {z }= 2ACBD � (AD +BC)BC � (AD +BC)AD +BD2AC

= 2ABCD �B2C2 � A2D2 = � (AD �BC)2 6= 0:

Hence the equation for U (�; �) is

� (AD �BC)2 U�� = 0 (same as U�� = 0).

Its general solution is U (�; �) = F (�)+G (�) for arbitrary C2 functions F (�) and G (�) de�ned onR: �

Example 5.17 Consider the second order linear equation in two variables:

uxx � 4uxy � 2uyy = 0; (x; y) 2 R2; u = u (x; y) : R2 ! R:

Use change of variables method to reduce it to canonical form (U�� � U�� = 0 or U�� = 0) andthen solve it.

Solution:

Recall that the numbers A; B; C; D satisfying

AC = a = 1; AD +BC = 2b = �4; BD = c = �2; (166)

are given by A = C = 1; B = �2 +p6; D = �2 �

p6: By Lemma 5.16, if we do the change of

variables (� = Bx� Ay =

��2 +

p6�x� y

� = Dx� Cy =��2�

p6�x� y;

(167)

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the new for U (�; �) is given by U�� (�; �) = 0: The general solution for U (�; �) is U (�; �) = F (�) +G (�) : Hence the general solution for u (x; y) is

u (x; y) = F (Bx� Ay) +G (Dx� Cy) = F���2 +

p6�x� y

�+G

���2�

p6�x� y

�= F

��p6� 2

�x� y

�+G

��p6 + 2

�x+ y

�; (168)

where F (z) : R! R and G (z) : R! R are two arbitrary C2 functions. �

5.3 Solving hyperbolic equations; diagonalization method (eigenvalue-eigenvector method).

Remark 5.18 Interesting question: Can you �nd the relation between this method and thefactorization method ?

Example 5.19 Consider the second order linear equation in two variables:

uxx � 4uxy � 2uyy = 0; (x; y) 2 R2; u = u (x; y) : R2 ! R:

Use diagonalization method to reduce it to canonical form (U�� � U�� = 0) and then solve it.

Solution:

One can write the equation as

Trace

��1 �2�2 �2

��uxx uxyuxy uyy

��= 0:

The eigenvalues of the coe¢ cient matrix are �1 = 2; �2 = �3 with corresponding orthonormaleigenvectors

v1 =1p5

��21

�; v2 =

1p5

�12

�: (169)

By (118) and (119), we introduce the change of variables

� = � 2p5x+

1p5y; � =

1p5x+

2p5y:

Then, in terms of the variables (�; �) ; we have

2U�� � 3U�� = 0:

Finally, we let ~� = 1p2�; ~� = 1p

3�; we have the �nal canonical form:

~U~�~� � ~U~�~� = 0 (which can be decomposed as�@

@ ~�+

@

@~�

���@

@ ~�� @

@~�

�~U

�= 0 )

and its general solution is

~U�~�; ~��= F

�~� � ~�

�+G

��~� � ~�

�(same as G

�~� + ~�

�),

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where F (z) and G (z) are two arbitrary C2 functions. Hence the general solution for u (x; y) is

u (x; y) =

8><>:F�1p2

�� 2p

5x+ 1p

5y�� 1p

3

�1p5x+ 2p

5y��

+G�1p2

�� 2p

5x+ 1p

5y�+ 1p

3

�1p5x+ 2p

5y�� (170)

=

8>>><>>>:F

��� 2p

10� 1p

15

�x+

�1p10� 2p

15

�y

�| {z }+G

��� 2p

10+ 1p

15

�x+

�1p10+ 2p

15

�y�:

By the identity� 2p

10� 1p

151p10� 2p

15

=2p15 +

p10

�p15 + 2

p10=p6 + 2

we can write F��� 2p

10� 1p

15

�x+

�1p10� 2p

15

�y

�| {z } as F

��p6 + 2

�x+ y

�| {z } and by the iden-

tity� 2p

10+ 1p

151p10+ 2p

15

=�2p15 +

p10p

15 + 2p10

= �p6 + 2

we can write G��� 2p

10+ 1p

15

�x+

�1p10+ 2p

15

�y�as G

���p6 + 2

�x+ y

�| {z }. Thus the general

solution can also be expressed as

u (x; y) = F��p

6 + 2�x+ y

�+G

���p6 + 2

�x+ y

�; (171)

which is the same as (160) and (168). �

Remark 5.20 (Interesting.) One can use eigenvalues and eigenvectors to express the generalsolution (170) as

u (x; y) =

8>>><>>>:F

�1p�1

��xy

�; v1

�� 1p

��2

��xy

�; v2

��+G

�1p�1

��xy

�; v1

�+ 1p

��2

��xy

�; v2

��;

(172)

where �1 = 2; �2 = �3 and v1; v2 are corresponding orthonormal eigenvectors given by (169).

6 General solutions of parabolic equations without lowerorder terms.

6.1 Solving parabolic equations; factorization method.

We now come to the parabolic case for the equation

auxx + 2buxy + cuyy = 0; where ac = b2: (173)

Here we may assume a; b; c are all nonzero (otherwise we are in a trivial case). As there is nolower order terms in (173), the parabolic equation is degenerate. Hence it is essentially an ODE.We have the following:

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Lemma 6.1 (Factorization method for parabolic equation.) Assume the equation (173) isparabolic, i.e. ac = b2 (with a 6= 0; b 6= 0; c 6= 0). Then one can decompose it as

auxx + 2buxy + cuyy =

�A@

@x+B

@

@y

���A@

@x+B

@

@y

�u

�= 0 (174)

for some constants A 6= 0; B 6= 0 (this is same as a 6= 0; c 6= 0): The general solution of (173)can be expressed as either one of the following two forms

(1) : u (x; y) = F (Bx� Ay) + x

AG (Bx� Ay) (175)

or(2) : u (x; y) = F (Bx� Ay) + y

BG (Bx� Ay) (176)

for arbitrary C2 functions F (z) ; G (z) de�ned on R:

Remark 6.2 If a = 0 (then b = 0; c 6= 0) or c = 0 (then b = 0; a 6= 0), then we are in a trivialcase. The equation now has the form cuyy = 0 or auxx = 0: We have not much to discuss.

Remark 6.3 The two forms in (175) and (176) are the same due to the identity

x

AG (Bx� Ay) = (Bx� Ay) + Ay

ABG (Bx� Ay) = y

B~G (Bx� Ay) : (177)

Proof. For a C2 function u; (174) is the same as

A (Aux +Buy)x +B (Aux +Buy)y = A2uxx + 2ABuxy +B2uyy = 0:

Hence we need to solveA2 = a; AB = b; B2 = c;

which is solvable due to ac�b2 = 0: If a > 0; b > 0; c > 0; then we can choose A =pa; B =

pc: If

a > 0; b < 0; c > 0; then we can choose A =pa; B = �

pc (if both a < 0 and c < 0; we can

simply multiply the equation by a minus sign).

Let w =�A @@x+B @

@y

�u: Then by

�A @@x+B @

@y

�w = 0; we see that w = G (Bx� Ay) for some

arbitrary C1 function G (z) de�ned on R: Next we solve�A@

@x+B

@

@y

�u (x; y) = G (Bx� Ay) : (178)

Since A 6= 0 and B 6= 0; we can do the change of variables w = Bx� Ay; z = x; to get

Aux +Buy = A [UwB + Uz] +BUw (�A) = AUz = G (w)| {z }and obtain the general solution for U (w; z) :

U (w; z) = F (w) +z

AG (w) ;

which givesu (x; y) = F (Bx� Ay) + x

AG (Bx� Ay) ; if A 6= 0;

where now F (z) ; G (z) are two arbitrary C2 functions de�ned on R: This is the form (175).Similarly, we can also do the change of variables w = Bx� Ay; z = y; to get

Aux +Buy = A (UwB) +B [Uw (�A) + Uz] = BUz = G (w)| {z }which gives the general solution of the form (176). �

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Remark 6.4 We can also do the change of variables w = Bx�Ay; z = Ax+By in (178) and get

Aux +Buy = A [UwB + UzA] +B [Uw (�A) + UzB] =�A2 +B2

�Uz = G (w)| {z } :

We now haveU (w; z) = F (w) +

z

A2 +B2G (w) ; (179)

which gives the symmetric form

u (x; y) = F (Bx� Ay) + Ax+By

A2 +B2G (Bx� Ay) : (180)

The three forms (175), (176) and (180) are all equivalent due to the following identities:

Ax+By

A2 +B2=

1

A2 +B2

��B (Bx� Ay)

A+(A2 +B2)x

A

�(181)

andAx+By

A2 +B2=

1

A2 +B2

�A (Bx� Ay)

B+(A2 +B2) y

B

�: (182)

6.2 Solving parabolic equations; diagonalization method.

What happens if we use diagonalization method to solve a parabolic equation (ac = b2) ? Onecan check that the coe¢ cient matrix of equation (173) has two eigenvalues �1 = a+ c; �2 = 0: Thecorresponding orthonormal eigenvectors are

v1 =1p

a2 + b2

�ab

�; v2 =

1pa2 + b2

�b�a

�:

By (119), under the change of variables

� =1p

a2 + b2(ax+ by) ; � =

1pa2 + b2

(bx� ay) ; (183)

equation (173) can be reduced to

Trace

��a+ c 00 0

���U�� U��U�� U��

��= 0:

Since a+ c 6= 0; the above is same asU�� = 0; (184)

which impliesU (�; �) = �g (�) + f (�) (185)

for arbitrary C2 functions f (�) and g (�) : The corresponding u (x; y) is given by

u (x; y) =1p

a2 + b2(ax+ by) g

�1p

a2 + b2(bx� ay)

�+ f

�1p

a2 + b2(bx� ay)

�= H (bx� ay) + ax+ byp

a2 + b2K (bx� ay) (186)

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for arbitrary C2 functions H (�) and K (�) : We can compare (186) with (180). Assume we are inthe case a > 0; b > 0; c > 0; then we have A =

pa; B =

pc and so b =

pac = AB: Hence

H (bx� ay) + ax+ bypa2 + b2

K (bx� ay)

= H�ABx� A2y

�+

A2x+ ABypA4 + A2B2

K�ABx� A2y

�= F (Bx� Ay) + Ax+By

A2 +B2G (Bx� Ay)

(187)

for another two arbitrary C2 functions F (�) and G (�) : Therefore, (186) is the same as (180).The check for the case a > 0; b < 0; c > 0 is similar. Therefore, both methods are actuallyequivalent.

7 Hyperbolic equations and parabolic equations of the form auxx+

2buxy + cuyy = 0 with initial conditions.

If we write the equation auxx+2buxy + cuyy = 0 as auxx+2buxt+ cutt = 0 and view t as time, thena pair of initial conditions (at time t = 0) of the form

u (x; 0) = � (x) ; ut (x; 0) = (x) ; x 2 (�1;1)

can determine the solution uniquely. Here � (x) and (x) are two given functions de�ned on(�1;1) : We will demonstrate this by direct computations.

Lemma 7.1 (Hyperbolic equation with initial conditions.) Let � 2 C2 (R) and 2 C1 (R) betwo given functions. Assume ac < b2 in equation (147) (we view y as time and denote it as t).Consider the hyperbolic equation with initial conditions (at time t = 0):( �

A @@x+B @

@t

� ��C @@x+D @

@t

�u�= 0; u = u (x; t)

u (x; 0) = � (x) ; ut (x; 0) = (x) ; x 2 (�1;1) ;(188)

where A; B; C; D are constants satisfying AD � BC 6= 0 and here we also assume thatB 6= 0 and D 6= 0 (they are the coe¢ cients of the operator @

@t). Then the initial value problem

(188) has a unique solution u (x; t) 2 C2 (R2). Moreover, the solution is given by

u (x; t) =AD

AD �BC��x� C

Dt

�� BC

AD �BC��x� A

Bt

�+

BD

AD �BC

Z x�CDt

x�ABt

(s) ds; (189)

where (x; t) 2 R2:

Remark 7.2 (2-parameter families of characteristic lines for hyperbolic equations.) Theassumption B 6= 0 and D 6= 0 is essential in the formula (189). There are 2-parameter familiesof characteristic lines for the hyperbolic equation�

A@

@x+B

@

@t

���C@

@x+D

@

@t

�u

�= 0;

namely the linesBx� At = const:; and Dx� Ct = const:: (190)

If we have B = 0 (then D 6= 0) or D = 0 (then B 6= 0), then the characteristic lines have the form

t = const:; and Dx� Ct = const:

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orBx� At = const:; and t = const::

By this, the initial conditions u (x; 0) = � (x) ; ut (x; 0) = (x) happen to occur on the charac-teristic line t = const: (i.e. t = 0). In general, we have either no solution to the initial valueproblem or in�nitely many solutions to the initial value problem. The same conclusion holds forthe second case. We will leave this as a homework problem for you to verify. Also see Remark 7.6.

Proof. The general solution of the equation is

u (x; t) = F (Bx� At) +G (Dx� Ct) (191)

for arbitrary C2 functions F (�) ; G (�) de�ned on R and we need to solve(F (Bx) +G (Dx) = � (x)

�AF 0 (Bx)� CG0 (Dx) = (x) :(192)

Di¤erentiate the �rst equation with respect to x to get the system of equations:(BF 0 (Bx) +DG0 (Dx) = �0 (x)

�AF 0 (Bx)� CG0 (Dx) = (x) ;

i.e. �B D�A �C

��F 0 (Bx)G0 (Dx)

�=

��0 (x) (x)

�:

Hence we get �F 0 (Bx)G0 (Dx)

�=

1

AD �BC

��C �DA B

���0 (x) (x)

�and so

F 0 (Bx) =�1

AD �BC (C�0 (x) +D (x)) ; G0 (Dx) =

1

AD �BC (A�0 (x) +B (x)) ;

which are the same as (note that we assume B 6= 0 and D 6= 0)( ddxF (Bx) = BF 0 (Bx) = �B

AD�BC (C�0 (x) +D (x)) ;

ddxG (Dx) = DG0 (Dx) = D

AD�BC (A�0 (x) +B (x)) :

(193)

If we let (x) be an antiderivative of (x) (it is not unique), we get(F (Bx) = �B

AD�BC (C� (x) +D(x)) + C1

G (Dx) = DAD�BC (A� (x) +B(x)) + C2

(194)

for some integration constants C1; C2: Now by the �rst equation of (192), we must have C1+C2 =0: Therefore, we conclude (note that we assume B 6= 0 and D 6= 0)

u (x; t)

= F (Bx� At) +G (Dx� Ct) = F

�B

�x� A

Bt

��+G

�D

�x� C

Dt

��=

( �BAD�BC

�C��x� A

Bt�+D

�x� A

Bt��

+ DAD�BC

�A��x� C

Dt�+B

�x� C

Dt��

=AD

AD �BC��x� C

Dt

�� BC

AD �BC��x� A

Bt

�+

BD

AD �BC

�x� C

Dt

��

�x� A

Bt

��(195)

42

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Note that the antiderivative (x) is not unique. However, the quantity

�x� C

Dt

��

�x� A

Bt

�(196)

is unique and is independent of the choice of the antiderivative (x) : For convenience, we canchoose (x) =

R x0 (s) ds and obtain

�x� C

Dt

��

�x� A

Bt

�=

Z x�CDt

x�ABt

(s) ds: (197)

Thus the solution formula for u (x; t) is given by

u (x; t) =AD

AD �BC��x� C

Dt

�� BC

AD �BC��x� A

Bt

�+

BD

AD �BC

Z x�CDt

x�ABt

(s) ds: (198)

The solution is de�ned on all (x; t) 2 R2: Since � 2 C2 (R) and 2 C1 (R) ; we have u (x; t) 2C2 (R� R) : The solution (198) satis�es the initial value problem (188). �

Remark 7.3 By direct check, one can see that (198) is indeed a solution of (188). First, thesolution u (x; t) has the form F (Bx� At) +G (Dx� Ct) : Hence it must be a solution. Second, wehave

u (x; 0)

=�B

AD �BC

�C� (x) +D

Z x

0

(s) ds

�+

D

AD �BC

�A� (x) +B

Z x

0

(s) ds

�= � (x)

and

ut (x; 0)

=�B

AD �BC

�C�0 (x)

��AB

�+D (x)

��AB

��+

D

AD �BC

�A�0 (x)

��CD

�+B (x)

��CD

��= (x) :

Thus u (x; t) given by (198) is indeed a solution of the equation satisfying the initial conditions.

To Be Continued

Remark 7.4 (Domain of dependence for hyperbolic equations.) By (198), the domainof dependence interval of the point (x0; t0) is the interval

�x0 � A

Bt0; x0 � C

Dt0�(or the inter-

val�x0 � C

Dt0; x0 � A

Bt0�) lying on the x-axis. Only the values of � (x) and (x) on the interval will

determine the value of u at the point (x0; t0) :

In case equation (188) is parabolic (i.e. A = C; B = D) and has the same initial conditions, wecan still solve it (note that now the equation is degenerate parabolic).

Lemma 7.5 (Parabolic equation with initial conditions.) Let � 2 C3 (R) and 2 C2 (R) betwo given functions. Assume ac = b2 in equation (147) (we view y as time and here we denote it ast). Consider the parabolic equation with initial conditions (at time t = 0):( �

A @@x+B @

@t

� ��A @@x+B @

@t

�u�= 0; u = u (x; t)

u (x; 0) = � (x) ; ut (x; 0) = (x) ; x 2 (�1;1) ;(199)

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where A; B are constants with A 6= 0 and B 6= 0 (same as in Lemma 6.1, here we assume A 6=0 and B 6= 0). Then the initial value problem (199) has a unique solution u (x; t) 2 C2 (R2).Moreover, the solution is given by

u (x; t) = �

�x� A

Bt

�+

t

B

�A�0

�x� A

Bt

�+B

�x� A

Bt

��; (200)

where (x; t) 2 R2:

Remark 7.6 (1-parameter family of characteristic lines for parabolic equations.) (Thisis a continuation of Remark 7.2.) In case B = 0 (of course, we must have A 6= 0), equationbecomes (

uxx (x; t) = 0;

u (x; 0) = � (x) ; ut (x; 0) = (x) ; x 2 (�1;1) ;(201)

and the general solution for u (x; t) is u (x; t) = A (t)x + B (t) for arbitrary functions of A (t) andB (t) ; t 2 (�1;1) : Now we need to require(

u (x; 0) = A (0) x+B (0) = � (x) ; x 2 (�1;1) ;

ut (x; 0) = A0 (0) x+B0 (0) = (x) ; x 2 (�1;1) :(202)

Clearly, in general, (202) has no solution. But if � (x) and (x) are of the form ax+b; cx+d; then(202) has in�nitely many solutions. Again, the reason for this is that the data is prescribedon the line t = 0; which is a characteristic line. On the other hand, if we have A = 0 (ofcourse, we must have B 6= 0), then now the characteristic line is x = const:; and equationbecomes (

utt (x; t) = 0;

u (x; 0) = � (x) ; ut (x; 0) = (x) ; x 2 (�1;1) ;(203)

and the general solution for u (x; t) is u (x; t) = A (x) t+B (x) for arbitrary functions of A (x) andB (x) ; x 2 (�1;1) : Now we need to require(

u (x; 0) = B (x) = � (x) ; x 2 (�1;1) ;

ut (x; 0) = A (x) = (x) ; x 2 (�1;1) :(204)

Hence the unique solution for the problem (203) is given by

u (x; t) = � (x) + t (x) ; (x; t) 2 R2: (205)

Note that now the line t = 0 is not a characteristic line.

Remark 7.7 (Interesting observation.) It is interesting to see that for the solution formula ofthe hyperbolic case in (189), i.e.

u (x; t) =AD

AD �BC��x� C

Dt

�� BC

AD �BC��x� A

Bt

�+

BD

AD �BC

Z x�CDt

x�ABt

(s) ds; (206)

if we �x A; B and let C ! A; D ! B; then the above formula will approach the formula (200) forthe parabolic case. To see this, we �rst note that

BD

AD �BC

Z x�CDt

x�ABt

(s) ds =

1

ABt� C

Dt

Z x�CDt

x�ABt

(s) ds

!t!

�x� A

Bt

�� t;

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which is the term tB

�� � �+B

�x� A

Bt��in (200). Also, note that

AD

AD �BC��x� C

Dt

�� BC

AD �BC��x� A

Bt

�=

( ADAD�BC�

�x� C

Dt�� BC

AD�BC��x� C

Dt�

+ BCAD�BC�

�x� C

Dt�� BC

AD�BC��x� A

Bt�

= �

�x� C

Dt

�+BC

��x� C

Dt�� �

�x� A

Bt�

AD �BC

= �

�x� C

Dt

�+BC

�0 (�)�ABt� C

Dt�

AD �BC (here we use mean value theorem)

! �

�x� A

Bt

�+BC

�0�x� A

Bt� �

ABt� C

Dt�

AD �BC = �

�x� A

Bt

�+

t

B

�A�0

�x� A

Bt

�+ � � �

�:

The proof is done.

Proof. If we choose the general solution to have the form

u (x; t) = F (Bx� At) + x

AG (Bx� At) ; A 6= 0; (207)

then by the initial conditions, we need to solve(F (Bx) + x

AG (Bx) = � (x)

�AF 0 (Bx)� xG0 (Bx) = (x) ; x 2 R;(208)

which gives (BF 0 (Bx) + 1

AG (Bx) + x

ABG0 (Bx) = �0 (x)

�AF 0 (Bx)� xG0 (Bx) = (x) ; x 2 R;and we �rst obtain G (Bx) = A�0 (x) + B (x) : Substituting it into the �rst identity of (208), wecan get

F (Bx) = � (x)� x

A[A�0 (x) +B (x)] ; G (Bx) = A�0 (x) +B (x) ; (209)

which gives(F (Bx� At) = F

�BBx�At

B

�= �

�Bx�AtB

�� 1

A

�Bx�AtB

� �A�0

�Bx�AtB

�+B

�Bx�AtB

��G (Bx� At) = G

�BBx�At

B

�= A�0

�Bx�AtB

�+B

�Bx�AtB

�and so

u (x; t) = F (Bx� At) + x

AG (Bx� At)

=

(��Bx�AtB

�� 1

A

�Bx�AtB

� �A�0

�Bx�AtB

�+B

�Bx�AtB

��+ xA

�A�0

�Bx�AtB

�+B

�Bx�AtB

��= �

�x� A

Bt

�+

t

B

�A�0

�x� A

Bt

�+B

�x� A

Bt

��; (x; t) 2 R2: (210)

Since we assume � 2 C3 (R) and 2 C2 (R) ; the solution u (x; t) given by (210) satis�es u (x; t) 2C2 (R� R) : �

Remark 7.8 By direct check, one can see that (210) is indeed a solution of (199).

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Remark 7.9 (Important.) Note that the value of the solution u at (x0; t0) depends only on theinitial data � and at the point x0 � A

Bt0:

Remark 7.10 If we choose the general solution to have the form

u (x; t) = F (Bx� At) + t

BG (Bx� At) ; (211)

then we have (F (Bx) = � (x)

�AF 0 (Bx) + 1BG (Bx) = (x) ; x 2 R;

which gives (BF 0 (Bx) = �0 (x)

�ABF 0 (Bx) +G (Bx) = B (x) ; x 2 R;and we conclude (

F (Bx) = � (x)

G (Bx) = B (x) + A�0 (x) :(212)

By the above we will get

u (x; t) = F (Bx� At) + t

BG (Bx� At) = F

�B

�x� A

Bt

��+

t

BG

�B

�x� A

Bt

��= �

�x� A

Bt

�+

t

B

�B

�x� A

Bt

�+ A�0

�x� A

Bt

��;

which is the same as in (210).

Example 7.11 Find the solution u (x; t) satisfying:

uxx + 2uxt + utt = 0; u (x; 0) = x2; ut (x; 0) = ex; x 2 R:Solution:

By (175) in Lemma 6.1 and the identity

uxx + 2uxt + utt = (@x + @t) [(@x + @t)u] = 0; A = B = 1;

we see that the general solution of this degenerate parabolic equation has the form

u (x; t) = F (x� t) + xG (x� t)for arbitrary C2 functions F (�) ; G (�) de�ned on R. Hence we need to solve(

F (x) + xG (x) = x2

�F 0 (x)� xG0 (x) = ex

and obtain (F 0 (x) +G (x) + xG0 (x) = 2x

�F 0 (x)� xG0 (x) = ex

and thenG (x) = 2x+ ex:

Next, by the identity F (x) + xG (x) = x2 we have

F (x) = x2 � xG (x) = x2 � x (2x+ ex) = �x2 � xex:The answer for u (x; t) is

u (x; t) = F (x� t) + xG (x� t)= � (x� t)2 � (x� t) ex�t + x

�2 (x� t) + ex�t

�= (x� t) (x+ t) + tex�t

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7.1 The wave equation with initial conditions.

In this section, we look at one-dimensional wave equation for the function u (x; t) ; given by

utt (x; t) = c2uxx (x; t) ; x 2 R; t 2 R: (213)

Under some assumptions (small amplitude, etc ...), the equation describes the motion of a vibratingstring, where u (x; t) represents the position of the string (I will not derive this in class). Herec > 0 is a constant given by

c =

sT

�; (214)

where T is the tension of the string and � is the density of the string (both are assumed to beconstants, not very realistic at all).

Remark 7.12 By a change of variable in time:

x = x; ~t = ct;

the function v�x; ~t�= u (x; t) = u

�x;

~tc

�will satisfy (213) with c = 1: Hence, the two equations

utt = c2uxx and utt = uxx are equivalent.

As (213) is a physical equation, it has initial conditions. They are the initial position andinitial velocity of the string, given by

u (x; 0) = � (x) ; ut (x; 0) = (x) ; x 2 R: (215)

Remark 7.13 (Useful motivation.) Roughly speaking, the wave equation comes from New-tonian mechanics, hence as long as the initial position and initial velocity are known, the wholeprocess of motion is uniquely determined.

With the above conditions, the solution u (x; t) satisfying (213) and (215) exists and is unique(which can be seen from (198)). Moreover, if we change � (x) and (x) a little bit, then thecorresponding solution will also change a little bit (we shall see this soon). In this sense, we saythat the problem (213) and (215) is well-posed.Since one can factorize the equation (213) as�

c@

@x+@

@t

���c@

@x� @

@t

�u

�= 0; u = u (x; t) ; (216)

by previous discussion we know the general solution of (213), de�ned on the whole space R2; isgiven by

u (x; t) = F (x+ ct) +G (x� ct) ; (x; t) 2 (�1;1)� (�1;1) (217)

for arbitrary C2 functions F; G de�ned on R:

De�nition 7.14 Any line of the form x+ ct = const: or x� ct = const: is called a characteristicline of the wave equation (213).

Remark 7.15 (The geometric meaning of the wave equation.) The solution u (x; t) givenby (??) consists of two traveling waves moving in opposite directions (positive and negative x-directions) with the same speed c (explain this and draw pictures). Moreover, since c =

pT=�; if the

tension T is large and the density � is small, then the traveling wave speed is large. This matcheswith physical observation.

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Without remembering the formula in Lemma 7.1, one can easily derive the solution formulasatisfying the conditions (215). We need to require(

u (x; 0) = F (x) +G (x) = � (x)

ut (x; 0) = cF 0 (x)� cG0 (x) = (x) :(218)

By this we obtain

F 0 (x) =c�0 (x) + (x)

2c; G0 (x) =

c�0 (x)� (x)2c

; (219)

and so

F (x) =� (x)

2+1

2c

Z x

0

(s) ds+ �; G (x) =� (x)

2� 1

2c

Z x

0

(s) ds+ " (220)

with suitable constants �; ": Here � + " = 0 by (218). Hence we get the unique solution given by

u (x; t) = F (x+ ct) +G (x� ct) = 1

2[� (x+ ct) + � (x� ct)] + 1

2c

Z x+ct

x�ct (s) ds: (221)

We can conclude the following:

Lemma 7.16 Assume � and in the initial conditions (215) satisfy � 2 C2 (R) and 2 C1 (R) : Thenthe function u (x; t) given by (221) is the unique C2 solution of the initial value problem(

utt (x; t) = c2uxx (x; t)

u (x; 0) = � (x) ; ut (x; 0) = (x) ; x 2 (�1;1)(222)

de�ned on the domain (x; t) 2 (�1;1)� [0;1) (or the domain (x; t) 2 (�1;1)� (�1;1)).

Remark 7.17 (Important.) We call (221) d�Alembert solution. It was due to him in 1746.Note that the function

1

2[� (x+ ct) + � (x� ct)] (223)

is an even function in t and the function

1

2c

Z x+ct

x�ct (s) ds (224)

is an odd function in t:

Remark 7.18 If � 2 Ck (R) and 2 Ck�1 (R) ; then u given by (221) represents a classical solutionu 2 Ck (R� [0;1)) of the initial value problem, but its regularity is not smoother (or worse) ingeneral. Thus the wave equation does not produce instantaneous smoothing of the initialdata as the heat equation does.

7.2 Domain of dependence and in�uence of the initial conditions forwave equation.

For convenience of discussion, we con�ne to nonnegative time t � 0 (this is not really essential). Recallthat the solution of the initial value problem (222) is

u (x; t) =1

2[� (x+ ct) + � (x� ct)] + 1

2c

Z x+ct

x�ct (s) ds; (x; t) 2 (�1;1)� [0;1): (225)

For each �xed time t = t0 2 [0;1) and �xed x0 2 (�1;1) ; we get an interval of the form[x0 � ct0; x0 + ct0] : The value of u (x0; t0) depends only on � at x0 � ct0; x0 + ct0; and on theinterval [x0 � ct0; x0 + ct0] : The values of � and outside the interval [x0 � ct0; x0 + ct0] will nota¤ect the value of u (x0; t0) :

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De�nition 7.19 The interval [x0 � ct0; x0 + ct0] lying on the x-axis is called the domain of de-pendence interval of the point (x0; t0) :

Remark 7.20 The above also says that for our wave equation "disturbances" or "sig-nals" only travel with speed c. To understand this more clearly, see the de�nition of "domainof in�uence".

In view of this, for each �xed x0 2 (�1;1) ; there is a region R � xt-plane (unboundedclosed set, lying on the upper half xt-plane) so that the values of u on this region can be a¤ectedby the values of � (x0) and (x0) : This region R � xt-plane is called the domain of in�uenceof the point (x0; 0) (equivalently, a point p 2 R if and only if its domain of dependence intervalcontains the point x0). The value of u at any point (x; t) lying out of of the region R is nota¤ected by the values of � (x0) and (x0) : The region R can be described as

R = f(x; t) 2 R� [0;1) : x� ct � x0 and x+ ct � x0g ; (226)

where the two half-lines x� ct = x0; t � 0; and x+ ct = x0; t � 0; intersect at the point (x0; 0) :

Remark 7.21 Draw a picture for the region R (or see Figure 1 in p. 39 of Strauss�s undergraduatePDE book).

Outside the domain of in�uence of the point (x0; 0) ; the value of u (x; t) is not a¤ected bythe values of � (x0) and (x0) : In view of this, we have the following obvious fact:

Lemma 7.22 For any given � (x) and (x) ; the domain of in�uence of the interval (lying on thex-axis) jxj � � is the region (lying on R� [0;1)) jxj � �+ct: In particular, if � (x) � (x) � 0 forjxj > � (i.e. both � (x) and (x) have compact support ), then u (x; t) � 0 on the regionjxj > � + ct:

Remark 7.23 Draw a picture for the above lemma.

Corollary 7.24 Assume both � (x) and (x) have compact support. Then u (x; t) given by (221)has compact support in x for each �xed t (however, the support of u (x; t) will become larger if t getslarger).

Example 7.25 Consider the wave equation with initial conditions:(utt (x; t) = c2uxx (x; t) ; c > 0

u (x; 0) = � (x) ; ut (x; 0) = (x) ; x 2 (�1;1) :

Assume that � = � 0 on the interval [�1� c; 1 + c] : Determine the region R � R� [0;1) suchthat u (x; t) � 0 on R:

Solution:

The region R is the trianglular-shaped region given by

R = f(x; t) : x� ct � �1� cg\f(x; t) : x+ ct � 1 + cg

\f(x; t) : t � 0g :

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7.3 Space-time separable solutions of the wave equation.

We look for certain special solutions of the wave equation utt (x; t) = c2uxx (x; t) with the separableform

u (x; t) = f (x) g (t) ;

where f (x) and g (t) are C2 functions de�ned on (�1;1) : Plug it into the equation to get

f (x) g00 (t) = c2f 00 (x) g (t)

and then we get the identityg00 (t)

c2g (t)=f 00 (x)

f (x):

The above identity cannot hold unless there is some constant K such that

g00 (t)

c2g (t)=f 00 (x)

f (x)= K; 8 x 2 domain of f; 8 t 2 domain of g:

There are three possibilities for the constant K : K = �2 > 0; K = 0; K = ��2 < 0: For the �rstcase, we have

g00 (t)

c2g (t)=f 00 (x)

f (x)= �2; � > 0;

which gives

g (t) = c1e�ct + c2e

��ct; f (x) = d1e�x + d2e

��x; x 2 (�1;1) ; t 2 (�1;1) ; (227)

where c1; c2; d1; d2 are integration constants. For the second case, we have

g00 (t)

c2g (t)=f 00 (x)

f (x)= 0;

which gives

g (t) = c1t+ c2; f (x) = d1x+ d2; x 2 (�1;1) ; t 2 (�1;1) : (228)

For the third case, we haveg00 (t)

c2g (t)=f 00 (x)

f (x)= ��2; � > 0;

which gives

g (t) = c1 cos (�ct)+c2 sin (�ct) ; f (x) = d1 cos (�x)+d2 sin (�x) ; x 2 (�1;1) ; t 2 (�1;1) :(229)

Thus we have:

Lemma 7.26 (Classi�cation of space-time separable solutions of the wave equation.)The following are the space-time separable solutions of the wave equation

u (x; t) =

8>><>>:�c1e

�ct + c2e��ct� �d1e�x + d2e

��x�(c1t+ c2) (d1x+ d2)

[c1 sin (�ct) + c2 cos (�ct)] [d1 sin (�x) + d2 cos (�x)] ; x 2 (�1;1) ; t 2 (�1;1)

and there are no others. Here � (� > 0), c1; c2; d1; d2 are all arbitrary constants.

Remark 7.27 (Important.) The space-time separable solutions are important if we want to useFourier series to express solutions of the wave equation.

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8 Lecture notes on 2019-3-28.

Example 8.1 For each n 2 N and L > 0; there is a solution, de�ned on R2; of the wave equationof the form (choose � = n�=L in the above)

un (x; t) =hAn sin

�n�cLt�+Bn cos

�n�cLt�isin�n�xL

�; An; Bn are const.,

which satis�es (un (0; t) = un (L; t) = 0; 8 t 2 [0;1)

un (x; 0) = Bn sin�n�xL

�; 8 x 2 [0; L] :

(230)

We call it a harmonic of the wave equation utt = c2uxx with �xed ends at x = 0 and x = L (byFourier series expansion applying to solutions of the wave equation with �xed ends, any such solutioncan be expressed as an in�nite sum of di¤erent harmonics). Note that a wave equation with�xed-end boundary condition is not enough to ensure a unique solution. One need one morecondition (say initial condition) to guarantee a unique solution. One can rewrite un (x; t) as

un (x; t) = Rn

hcos (�n) sin

�n�cLt�+ sin (�n) cos

�n�cLt�isin�n�xL

�= Rn sin

�n�cLt+ �n

�sin�n�xL

�; (231)

where Rn =pA2n +B2

n and the angle �n satis�es cos �n =AnRn; sin �n =

BnRn: Note that the value

of un (x; t) lies between �Rn and Rn: We call Rn the amplitude of un (x; t) and call �n the phaseof un (x; t) : The solution (231), de�ned on [0; L] ; is periodic in time with

un

�x; t+

2L

nc

�= un (x; t) ; 8 x 2 [0; L] ; t 2 R:

The time 2Lncis called the period of the solution.

8.1 Conservation of the total energy for wave equation.

Lemma 8.2 (Equipartition of energy for wave equations with compact support initialdata.) Assume both � (x) and (x) have compact support (this assumption is essential) in (222)and u 2 C2 (R� (�1;1)) solves the initial value problem (222). De�ne the kinetic energy andthe potential energy for the solution u (x; t) as

k (t) :=1

2

Z 1

�1u2t (x; t) dx and p (t) :=

1

2

Z 1

�1u2x (x; t) dx; t 2 (�1;1) : (232)

Then we have (1) : k (t) + c2p (t) is a constant for all time t 2 (�1;1). (2) : Moreover, fort 2 (�1;1) such that jtj is large enough, we have k (t) = c2p (t) : In particular, we have

k (t) = c2p (t) = const: (233)

when jtj is large enough.

Remark 8.3 By the conclusion (1) ; we see that the energy is equal to

k (t) + c2p (t) = k (0) + c2p (0)

=1

2

Z 1

�1u2t (x; 0) dx+

c2

2

Z 1

�1u2x (x; 0) dx =

1

2

Z 1

�1 2 (x) dx+

c2

2

Z 1

�1('0 (x))

2dx;

where we see that the improper integrals are actually proper integrals since � (x) and (x) havecompact support ...

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Proof. (1) : By Corollary 7.24, we know that u (x; t) given by (221) have compact support in x foreach �xed t 2 (�1;1) : Computing(

ut (x; t) =c2[�0 (x+ ct)� �0 (x� ct)] + 1

2[ (x+ ct) + (x� ct)]

ux (x; t) =12[�0 (x+ ct) + �0 (x� ct)] + 1

2c[ (x+ ct)� (x� ct)] ;

(234)

we can see that both ut (x; t) and ux (x; t) also have compact support in x for each �xed t 2(�1;1) : Hence the two improper integrals in (232) both converge (both are actually properintegrals).Hence, the di¤erentiation with respect to time can commute with the integral.That is, we have

d

dt

�k (t) + c2p (t)

�=

d

dt

�1

2

Z 1

�1u2t (x; t) dx+

c2

2

Z 1

�1u2x (x; t) dx

�=

Z 1

�1ut (x; t)utt (x; t) dx+ c2

Z 1

�1ux (x; t)uxt (x; t) dx

= c2Z 1

�1[ut (x; t)uxx (x; t) + ux (x; t)uxt (x; t)] dx

= c2Z 1

�1

@

@x[ut (x; t)ux (x; t)] dx = c2 (ux (x; t) � ut (x; t))

��x=1x=�1 = 0; 8 t 2 (�1;1) ;

where we have used the fact that both ut (x; t) and ux (x; t) also have compact support in x for each�xed t 2 (�1;1) :

Remark 8.4 If you have di¢ culty assuring the identity

d

dt

Z 1

�1u2t (x; t) dx = 2

Z 1

�1ut (x; t)utt (x; t) dx;

you can use de�nition and mean value theorem to see that

d

dt

Z 1

�1u2t (x; t) dx

= limh!0

Z 1

�1

u2t (x; t+ h)� u2t (x; t)h

dx = limh!0

Z 1

�12ut (x; t+ � (h))utt (x; t+ � (h)) dx;

where t + � (h) lies between t and t + h and in the above two limits the time t is �xed. As we willlet h! 0; we may assume that h 2 [�1; 1] and so t + � (h) 2 [t� 1; t+ 1] for all h 2 [�1; 1] : Onecan �nd a large number M > 0 so that

2ut (x; t+ � (h))utt (x; t+ � (h)) � 0 for all jxj �M and all h 2 [�1; 1] ;

which givesZ 1

�12ut (x; t+ � (h))utt (x; t+ � (h)) dx =

Z M

�M2ut (x; t+ � (h))utt (x; t+ � (h)) dx; 8 h 2 [�1; 1] :

Since the integrand 2ut (x; s)utt (x; s) is a continuous function in (x; s) 2 [�M;M ]�[t� 1; t+ 1] ;we have

lims!t

Z M

�M2ut (x; s)utt (x; s) dx =

Z M

�M2ut (x; t)utt (x; t) dx;

which is an elementary fact in Advanced Calculus. The above implies

limh!0

Z 1

�12ut (x; t+ � (h))utt (x; t+ � (h)) dx

= lims!t

Z M

�M2ut (x; s)utt (x; s) dx =

Z M

�M2ut (x; t)utt (x; t) dx =

Z 1

�12ut (x; t)utt (x; t) dx

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and we have

d

dt

Z 1

�1u2t (x; t) dx

= limh!0

Z 1

�12ut (x; t+ � (h))utt (x; t+ � (h)) dx =

Z 1

�12ut (x; t)utt (x; t) dx:

The proof is done.

For (2) ; by (234), we have

u2t (x; t) =

8>>>>>>>>>>><>>>>>>>>>>>:

c2

4

�(�0 (x+ ct))

2+ (�0 (x� ct))2| {z }�2�0 (x+ ct)�0 (x� ct)

�+14

� 2 (x+ ct) + 2 (x� ct)| {z }+2 (x+ ct) (x� ct)

+ c2

0B@ �0 (x+ ct) (x+ ct)| {z }+�0 (x+ ct) (x� ct)

��0 (x� ct) (x+ ct)� �0 (x� ct) (x� ct)| {z }1CA

and

c2u2x (x; t) =

8>>>>>>>>>>><>>>>>>>>>>>:

c2

4

�(�0 (x+ ct))

2+ (�0 (x� ct))2| {z }+2�0 (x+ ct)�0 (x� ct)

�+14

� 2 (x+ ct) + 2 (x� ct)| {z }�2 (x+ ct) (x� ct)

+ c2

0B@ �0 (x+ ct) (x+ ct)| {z }��0 (x+ ct) (x� ct)

+�0 (x� ct) (x+ ct)� �0 (x� ct) (x� ct)| {z }1CA :

By above, it su¢ ces to show thatZ 1

�1

c2

4[�2�0 (x+ ct)�0 (x� ct)] + 1

4[2 (x+ ct) (x� ct)]

+ c2[�0 (x+ ct) (x� ct)� �0 (x� ct) (x+ ct)]

!dx

=

Z 1

�1

c2

4[2�0 (x+ ct)�0 (x� ct)] + 1

4[�2 (x+ ct) (x� ct)]

+ c2[��0 (x+ ct) (x� ct) + �0 (x� ct) (x+ ct)]

!dx:

which is equivalent to showing thatZ 1

�1

�c2�0 (x+ ct)�0 (x� ct) + (x+ ct) (x� ct)

+c�0 (x+ ct) (x� ct)� c�0 (x� ct) (x+ ct)

!dx = 0: (235)

Since, for any x 2 R; the two points p = x+ ct; q = x� ct have distance 2c jtj. They both can notstay in the support of � and for large jtj for any x 2 R: Hence if jtj � 0 is large enough, we musthave

�0 (x+ ct)�0 (x� ct) = (x+ ct) (x� ct)= �0 (x+ ct) (x� ct) = �0 (x� ct) (x+ ct) = 0 for all x 2 (�1;1) :

The proof is done. �

Another energy property for the wave equation is the following:

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Lemma 8.5 (Omit this in class !!!) (Conservation of energy for wave equations with�xed-end condition.) Assume u 2 C2 (R2) is a solution of the wave equation utt (x; t) = c2uxx (x; t)and there is some L > 0 such that

u (0; t) = u (L; t) = 0; 8 t 2 (�1;1) ; (236)

i.e. u (x; t) satis�es "�xed-end" condition, then the total energy over the interval [0; L] :

1

2

�Z L

0

u2t (x; t) dx+ c2Z L

0

u2x (x; t) dx

�; t 2 (�1;1) (237)

is independent of time.

Proof. We �rst note that by (236) we have ut (0; t) = ut (L; t) = 0 for all t 2 (�1;1) : Compute

d

dt

�1

2

�Z L

0

u2t (x; t) dx+ c2Z L

0

u2x (x; t) dx

��=

Z L

0

ututtdx+ c2Z L

0

uxuxtdx = c2Z L

0

(utuxx + uxuxt) dx

= c2Z L

0

@

@x(utux) dx = c2 ut (L; t)| {z }ux (L; t)� c2 ut (0; t)| {z }ux (0; t) = 0; 8t 2 (�1;1) :

Hence the total energy over the interval [0; L] is independent of time. �

8.2 Nonhomogeneous wave equation.

We now consider the nonhomogeneous wave equation

utt (x; t)� c2uxx (x; t) = f (x; t) ; (x; t) 2 R2 (238)

with initial conditions

u (x; 0) = � (x) ; ut (x; 0) = (x) ; x 2 (�1;1) : (239)

Here the functions f (x; t) 2 C1 (R2) ; � (x) 2 C2 (R) ; (x) 2 C1 (R) are all given. By linearity, itsu¢ ces to look at the case when � (x) = (x) = 0 due to the following observation:

Lemma 8.6 If u (x; t) solves the problem(utt (x; t)� c2uxx (x; t) = f (x; t) ; (x; t) 2 R2

u (x; 0) = 0; ut (x; 0) = 0; x 2 R(240)

and v (x; t) solves the problem(vtt (x; t)� c2vxx (x; t) = 0; (x; t) 2 R2

v (x; 0) = � (x) ; vt (x; 0) = (x) ; x 2 R;(241)

then w (x; t) = u (x; t) + v (x; t) solves the problem(wtt (x; t)� c2wxx (x; t) = f (x; t) ; (x; t) 2 R2

w (x; 0) = � (x) ; wt (x; 0) = (x) ; x 2 R:(242)

Since we already know how to solve (241), it su¢ ces to focus on (240), which we change thenotation as: (

Utt (x; t)� c2Uxx (x; t) = f (x; t) ; (x; t) 2 R2

U (x; 0) = 0; Ut (x; 0) = 0; x 2 R;(243)

where f (x; t) is now assumed to be f (x; t) 2 C1 (R2) : We want to derive a solution formula forU (x; t) ; where U (x; t) 2 C2 (R2) :

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8.2.1 Motivation by Duhamel�s principle.

To understand the spirit of Duhamel�s principle, we �rst look at a simple case. Consider the ODEfor a time function u (t; s) with a parameter s 2 (�1;1) :8<:

du

dt(t; s) + ku (t; s) = 0; t 2 (s;1); k is a constant

u (s; s) = f (s) ; t = s;

(244)

where f (s) is a given continuous function de�ned on (�1;1) : We know its solution is given by

u (t; s) = e�k(t�s)f (s) ; 8 t 2 (s;1): (245)

Now if we integrate the above solution u (t; s) with respect to s from s = 0 to s = t 2(�1;1) ; we get

U (t) :=

Z t

0

u (t; s) ds =

Z t

0

e�k(t�s)f (s) ds; t 2 (�1;1) ; (246)

then U (t) will satisfy U (0) = 0 and

U 0 (t) = e�k(t�t)f (t) +

Z t

0

@

@t[e�k(t�s)f (s)]ds

= f (t)� kZ t

0

e�k(t�s)f (s) ds = f (t)� kU (t) ; t 2 (�1;1) ;

i.e., 8<:dU

dt(t) + kU (t) = f (t) ; t 2 (�1;1)

U (0) = 0:

(247)

That is: if we can solve the homogeneous problem (244) with initial condition f (s) at time t = s;then we can solve the nonhomogeneous problem (247) with initial condition 0 at time t = 0. Notethat the position of the function f in (244) has been switched in (247). This is called Duhamel�sprinciple.Let s 2 (�1;1) be a parameter (we can take it be �xed at �rst, but it can be arbitrary).

Assume u (x; t; s) solves the wave equation problem for each �xed s 2 R :(utt (x; t; s)� c2uxx (x; t; s) = 0 in (x; t) 2 R2;

u (x; s; s) = 0; ut (x; s; s) = f (x; s) ; x 2 R;(248)

where in (248), the "initial condition" is at time t = s: We observe the following:

Lemma 8.7 (Duhamel�s principle.) Let f (x; t) 2 C1 (R2) and let s 2 (�1;1) be a parameter.Assume u (x; t; s) is C2 in (x; t) 2 R2 and C1 in s 2 R; and solves the problem (248). Then if welet

U (x; t) =

Z t

0

u (x; t; s) ds; (249)

it will satisfy U (x; t) 2 C2 (R2) and solve the problem(Utt (x; t)� c2xxU (x; t) = f (x; t) ; (x; t) 2 R2

U (x; 0) = 0; Ut (x; 0) = 0; x 2 R:(250)

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Remark 8.8 (An useful derivative formula.) To understand the proof of Lemma 8.7, you needto mow the following derivative formula (assume � (x) ; � (x) : [a; b]! [c; d] are di¤erentiable withrespect to x 2 [a; b] and f (x; y) : [a; b]� [c; d]! R is a continuously di¤erentiable function):

d

dx

Z �(x)

�(x)

f (x; y) dy = f (x; � (x)) �0 (x)� f (x; � (x))�0 (x) +Z �(x)

�(x)

fx (x; y) dy: (251)

To see (251), we let the function F (�; x) be de�ned as

F (x; �) =

Z �

c

f (x; y) dy; x 2 [a; b] ; � 2 [c; d] :

In terms of F; we have the identityZ �(x)

�(x)

f (x; y) dy =

Z �(x)

c

f (x; y) dy �Z �(x)

c

f (x; y) dy = F (x; � (x))� F (x; � (x)) ; x 2 [a; b] :

(252)We also have

F� (x; �) = f (x; �) ; Fx (x; �) =@

@x

Z �

c

f (x; y) dy =

Z �

c

fx (x; y) dy: (253)

By the chain rule applied to (252) and use the identities in (253), we get

d

dx

Z �(x)

�(x)

f (x; y) dy =d

dx[F (x; � (x))� F (x; � (x))]

= Fx (x; � (x)) + F� (x; � (x)) �0 (x)� Fx (x; � (x))� F� (x; � (x))�0 (x)

=

Z �(x)

c

fx (x; y) dy + fx (x; � (x)) �0 (x)�

Z �(x)

c

fx (x; y) dy � fx (x; � (x))�0 (x)

= fx (x; � (x)) �0 (x)� fx (x; � (x))�0 (x) +

Z �(x)

�(x)

fx (x; y) dy; 8 x 2 [a; b] : (254)

You need to keep the identity (254) in mind.

Proof. We clearly have U (x; t) 2 C2 (R2) : We �rst note that U (x; 0) =R 00u (x; 0; s) ds =

0: Compute

Ut (x; t) =@

@t

Z t

0

u (x; t; s) ds = u (x; t; t) +

Z t

0

ut (x; t; s) ds =

Z t

0

ut (x; t; s) ds (255)

due to u (x; s; s) = 0: The above also gives Ut (x; 0) =R 00ut (x; 0; s) ds = 0:

Next, we compute 8>><>>:Ux (x; t) =

@@x

R t0u (x; t; s) ds =

R t0ux (x; t; s) ds

Uxx (x; t) =@@x

R t0ux (x; t; s) ds =

Z t

0

uxx (x; t; s) ds| {z }and

Utt (x; t) =@

@t

�Z t

0

ut (x; t; s) ds

�= ut (x; t; t) +

Z t

0

utt (x; t; s) ds

= f (x; t) +

Z t

0

utt (x; t; s) ds| {z }56

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due to ut (x; s; s) = f (x; s) : We conclude (note that utt (x; t; s)� c2uxx (x; t; s) = 0)

Utt (x; t)� c2xxU (x; t) = f (x; t) ; 8 (x; t) 2 R2:

The proof is done. �

Remark 8.9 Consider the problem(utt (x; t; s)� c2uxx (x; t; s) = 0 in R2;

u (x; s; s) = f (x; s) ; ut (x; s; s) = 0; x 2 R(256)

will not give us any good consequence.

By Lemma 8.7, it su¢ ces to look at problem (248). Note that if we are given an initial valueproblem of the form (

utt (x; t)� c2uxx (x; t) = 0 in (x; t) 2 R2;

u (x; t0) = 0; ut (x; t0) = f (x) ; x 2 R;(257)

then, by a change of variables v (x; t) = u (x; t+ t0) ; one can see that the solution of (257) is givenby

u (x; t) =1

2c

Z x+c(t�t0)

x�c(t�t0)f (�) d�: (258)

Therefore, the solution u (x; t; s) of (248) is given by the formula

u (x; t; s) =1

2c

Z x+c(t�s)

x�c(t�s)f (�; s) d�; (x; t) 2 R2; s 2 R: (259)

Note that the function u (x; t; s) given by (259) is at least C2 in (x; t) 2 R2 and C1 in s 2 R (dueto f (�; s) 2 C1 (R2)). By Lemma 8.7, the solution U (x; t) of the problem (250) is given by

U (x; t) =

Z t

0

u (x; t; s) ds =1

2c

Z t

0

Z x+c(t�s)

x�c(t�s)f (�; s) d�

!ds 2 C2

�R2�:

We can conclude the following:

Lemma 8.10 Assume f 2 C1 (R2) : Then the function U (x; t) given by

U (x; t) =1

2c

Z t

0

Z x+c(t�s)

x�c(t�s)f (�; s) d�

!ds 2 C2

�R2�; (x; t) 2 R2 (260)

is a C2 solution of the initial value problem(Utt (x; t)� c2Uxx (x; t) = f (x; t) ; (x; t) 2 R2

U (x; 0) = 0; Ut (x; 0) = 0; x 2 R(261)

on the domain (x; t) 2 R2: Moreover, the C2 (R2) solution of the problem (261) is uniquely given by(260).

Remark 8.11 One can give a direct check that U (x; t) given by (260) does satisfy (261). We have8<: Ux (x; t) =12c

R t0[f (x+ c (t� s) ; s)� f (x� c (t� s) ; s)] ds

Uxx (x; t) =12c

R t0[f 0 (x+ c (t� s) ; s)� f 0 (x� c (t� s) ; s)] ds

(262)

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and

Ut (x; t) =1

2c

(Z x+c(t�t)

x�c(t�t)f (�; t) d� +

Z t

0

[f (x+ c (t� s) ; s) c� f (x� c (t� s) ; s) (�c)] ds)

=1

2

Z t

0

[f (x+ c (t� s) ; s) + f (x� c (t� s) ; s)] ds

and by (262), we have

Utt (x; t) =

( 12[f (x+ c (t� t) ; t) + f (x� c (t� t) ; t)]

+12

R t0[f 0 (x+ c (t� s) ; s) c+ f 0 (x� c (t� s) ; s) (�c)] ds

= f (x; t) +c

2

Z t

0

[f 0 (x+ c (t� s) ; s)� f 0 (x� c (t� s) ; s)] ds = f (x; t) + c2Uxx (x; t) :

The check is done.

In the above lemma we assume f 2 C1 (R2) so that U 2 C2 (R2) : However, if f (x; t) dependsonly on x (or only on t), then it su¢ ces to assume that f 2 C0 (R) :

Corollary 8.12 Assume f (x; t) in (261) is a continuous function depending only on x; i.e.,f (x; t) = f (x) 2 C0 (R) : Let G (x) be a function satisfying G00 (x) = f (x) for all x 2 R; then thefunction U (x; t) in (260) has the form

U (x; t) = � 1

2c2[2G (x)�G (x+ ct)�G (x� ct)] ; (x; t) 2 R2; (263)

where U 2 C2 (R2) : Note that the solution U (x; t) depends on both space and time.

Remark 8.13 One can check that the solution (263) does not depend on the function G (x) as longas it satis�es G00 (x) = f (x) for all x 2 R: That is, if we replace G (x) by ~G (x) = G (x)+ax+ b forsome constants a; b; then we still have ~G00 (x) = f (x) : Using this ~G (x) in (263) will give the sameanswer U (x; t) :

Proof. The proof is straightforward. Let F (x) be such that F 0 (x) = f (x) for all x 2 R and G (x)be such that G0 (x) = F (x) for all x 2 R (and so G00 (x) = f (x)). We have

U (x; t) =1

2c

Z t

0

Z x+c(t�s)

x�c(t�s)f (�) d�

!ds =

1

2c

Z t

0

[F (x+ c (t� s))� F (x� c (t� s))] ds:

By

d

ds

��1cG (x+ c (t� s))

�= F (x+ c (t� s)) ; d

ds

�1

cG (x� c (t� s))

�= F (x� c (t� s)) ;

we get

U (x; t) =1

2c

��1cG (x+ c (t� s))

�����s=ts=0

� 1

2c

�1

cG (x� c (t� s))

�����s=ts=0

= � 1

2c2G (x) +

1

2c2G (x+ ct)� 1

2c2G (x) +

1

2c2G (x� ct)

= � 1

2c2[2G (x)�G (x+ ct)�G (x� ct)] ; (x; t) 2 R2:

The proof is done. �

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Corollary 8.14 Assume f (x; t) in (261) is a continuous function depending only on t; i.e.,f (x; t) = f (t) 2 C0 (R) : Then the function U (x; t) in (260) has the form

U (x; t) =

Z t

0

f (s) (t� s) ds 2 C2�R2�; 8 (x; t) 2 R2: (264)

Note that U (x; t) depends only on time t and is a C2 function of t 2 R: Note that the solutionU (x; t) depends only on time.

Remark 8.15 (Important observation !!) A remarkable thing is that for a continuous function f (t) 2C0 (R) ; the function U (t) :=

R t0f (s) (t� s) ds is a C2 function of t (not just a C1 function of

t). Note thatR t0f (s) (t� s) ds is a convolution-type integral and we have

d

dt

Z t

0

f (s) (t� s) ds =Z t

0

f (s) ds andd2

dt2

Z t

0

f (s) (t� s) ds = f (t) ; 8 t 2 R: (265)

On the other hand, since U (t) satis�es U 00 (t) = f (t) ; U (0) = 0; U 0 (0) = 0; we can also expressU (t) as the double integral

U (t) =

Z t

0

�Z s

0

f (�) d�

�ds; 8 t 2 R: (266)

Therefore, we have Z t

0

f (s) (t� s) ds =Z t

0

�Z s

0

f (�) d�

�ds; (267)

which can be veri�ed using the change of order of integration., i.e.Z t

0

�Z s

0

f (�) d�

�ds =

Z t

0

�Z t

f (�) ds

�d� =

Z t

0

f (�) (t� �) d� =Z t

0

f (s) (t� s) ds:

Therefore, the convolution-type integral is actually a double integral, which explains why wehave the higher regularity:

U (t) =

Z t

0

f (s) (t� s) ds 2 C2�R2�

for f (s) 2 C0 (R) : (268)

Proof. By (260), we have

U (x; t) =1

2c

Z t

0

Z x+c(t�s)

x�c(t�s)f (�; s) d�

!ds =

1

2c

Z t

0

Z x+c(t�s)

x�c(t�s)f (s) d�

!ds =

Z t

0

f (s) (t� s) ds:

We have8>><>>:Ut (x; t) = f (t) (t� t) +

R t0@@t(f (s) (t� s)) ds =

R t0f (s) ds; Utt (x; t) = f (t) ;

U (x; 0) = 0; Ut (x; 0) = 0;

Ux (x; t) = Uxx (x; t) = 0; Utt (x; t)� c2Uxx (x; t) = f (t) :

The main theorem in this section is the following:

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Theorem 8.16 Consider the following initial value problem for wave equation with a source:(utt (x; t)� c2uxx (x; t) = f (x; t) ; (x; t) 2 R2

u (x; 0) = � (x) ; ut (x; 0) = (x) ; x 2 R;(269)

where f 2 C1 (R2) ; � 2 C2 (R) ; 2 C1 (R) ; are given functions. Then the solution u (x; t) 2C2 (R2) of (269) is unique, de�ned on R2; and is given by the formula

u (x; t) =

8>><>>:1

2[� (x+ ct) + � (x� ct)] + 1

2c

R x+ctx�ct (s) ds

+1

2c

R t0

�R x+c(t�s)x�c(t�s) f (�; s) d�

�ds; (x; t) 2 R2:

(270)

Remark 8.17 (Important.) By the representation formula (270), we can conclude the following:The domain of dependence of the solution u at the point (x0; t0) ; t0 > 0; is given by a triangle4 in the (x; t) plane, where the triangle 4 is enclosed by the three lines:

line t = 0 (x-axis); line x = ct+ (x0 � ct0) ; line x = �ct+ (x0 + ct0) : (271)

Note that, when we look at the value u (x0; t0) ; the domain of integration in (270) for the functionf is precisely the triangle 4 described by (271).

To Be ContinuedThe above will be the coverage of the midterm exam on 4/18.

9 Lecture notes on 2019-4-1.

9.0.2 Change of variables method for nonhomogeneous wave equation. (OMIT THIS!!!!! )

The change of variables method (characteristic coordinates method) is more straightforward andnatural, and do not rely on the Duhamel�s principle.We �rst note the following simple fact:

Lemma 9.1 Let f (x; y) be a di¤erentiable function de�ned on R2 satisfying

f (�; �) = 0 and fx (�; �) = fy (�; �) ; 8 � 2 R: (272)

Then we must have fx (�; �) = fy (�; �) = 0 for all � 2 R: Conversely, if it satis�es

f (�; �) = 0 and fx (�; �) = 0; 8 � 2 R; (273)

then we must have fy (�; �) = 0 also for all � 2 R: Hence condition (272) is equivalent to condition(273).

Proof. Assume f (�; �) = 0 for all � 2 R: Then by the chain rule

0 =d

d�f (�; �) =

@f

@x(�; �)

d�

d�+@f

@y(�; �)

d�

d�= fx (�; �) + fy (�; �) :

The rest is clear. �

Recall that (see Section 5.2) when we solve the homogeneous wave equation utt� c2uxx = 0; weuse the change of variables

� = x+ ct; � = x� ct (274)

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to reduce the equation into the simple form (let ~u (�; �) = u (x; t))

�4c2~u�� (�; �) = 0

and obtain the general solution~u (�; �) = F (�) +G (�) : (275)

Hence the general C2 solution of utt � c2uxx = 0 is given by

u (x; t) = F (x+ ct) +G (x� ct)

for arbitrary C2 functions F (z) and G (z) de�ned on �1 < z <1:Following (274), the equation for ~u (�; �) becomes (we use the identity ~u�� = ~u�� since we assume

that u 2 C2 (R2) and so ~u 2 C2 (R2) also)

�4c2~u�� (�; �) = f

�� + �

2;� � �2c

�: (276)

To solve (276), we also need to consider the initial conditions u (x; 0) = ut (x; 0) = 0 for all x 2(�1;1) : In terms of ~u (�; �) ; it become

~u (�; �) = 0 and ~u� (�; �) = ~u� (�; �) ; 8 � 2 R (277)

and by Lemma 9.1, the initial conditions in (277) is equivalent to

(1) : ~u (�; �) = 0; (2) : ~u� (�; �) = 0; � 2 R: (278)

Thus the initial value problem (243) for ~u (�; �) becomes the following:(~u�� (�; �) = � 1

4c2f��+�2; ���

2c

�; 8 (�; �) 2 R2

~u (�; �) = 0; ~u� (�; �) = 0; 8 � 2 R:(279)

We �rst integrate the equation with respect to � to get ~u� (�; �) : We need to take the condi-tion ~u� (�; �) = 0 into consideration. In calculus, if we want to �nd H (�) satisfying H 0 (�) =g (�) ; H (�) = 0; then the answer is unique and is given by

H (�) =

Z �

g (�) d�:

Motivated by the above, we get (view ~u� (�; �) as H (�) ; where H (�) = 0)

~u� (�; �) = �1

4c2

Z �

f

�� + �

2;� � �2c

�d�; ~u� (�; �) = 0: (280)

Next, we integrate the above with respect to � to get ~u (�; �) and we need to take the condition~u (�; �) = 0 into consideration. We get (view ~u (�; �) as P (�) ; where P (�) = 0)

~u (�; �) =

Z �

�� 1

4c2

Z �

r

f

�r + �

2;r � �2c

�d�

�dr; (281)

which is the same as

~u (�; �) =1

4c2

Z �

�Z r

f

�r + �

2;r � �2c

�d�

�dr; 8 (�; �) 2 R2: (282)

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Therefore, we conclude that the unique solution of (279) is given by the formula (282). Going backto u (x; t) ; we get

u (x; t) =1

4c2

Z x+ct

x�ct

�Z r

x�ctf

�r + �

2;r � �2c

�d�

�dr:

If we do the change of variables (�; s) ! (r; �) by

� =r + �

2; s =

r � �2c

(same as r = � + cs; � = � � cs) ; (283)

we get

u (x; t) =1

4c2

Z x+ct

x�ct

�Z r

x�ctf

�r + �

2;r � �2c

�d�

�dr =

1

2c

Z t

0

Z x+c(t�s)

x�c(t�s)f (�; s) d�ds: (284)

We leave the veri�cation of the above identity to you as an exercise.

Remark 9.2 To verify (284), it is better to draw a picture for the change of variables (283).

10 Lecture notes on 2018-2-26.

Move the following stu¤ to back ........

The following divergence theorem (general version of the Fundamental Theorem of Calculus)will be needed often later on:

Theorem 10.1 (Divergence theorem.) Let be a C1 bounded domain in Rn andW : �! Rnis a vector �eld on � withW 2 C1 () \ C0

���. We have the identityZ

divWdx (volume integral in Rn) =Z@

W �Nd� (surface integral in Rn) (285)

where N is the unit outward normal to @: Here divW : ! R is the divergence of the vector�eldW and d� is the "surface measure" on @:

Remark 10.2 If n = 2; then d� means ds; where ds is the arc length di¤erential and the abovetheorem is the same as the familiar Green Theorem for plane region enclosed by a simple closedcurve �: More precisely, let C � R2 be a counterclockwise simple closed curve parametrized by� (t) = (x (t) ; y (t)) ; t 2 [a; b] ; where t is an arbitrary parameter (not necessarily the arc lengthparameter) and let � R2 be the open region enclosed by C with @ = C (@ means the boundaryof ). LetW 2 C1 () \ C0

���be a vector �eld on � given by

W (x; y) = (p (x; y) ; q (x; y)) ; (x; y) 2 �:

Now we �rst haveZ

divWdx (volume integral in R2) =ZZ

�@p

@x(x; y) +

@q

@y(x; y)

�dxdy (286)

and the unit outward normal N to (x (t) ; y (t)) 2 @; t 2 [a; b] ; is given by

N (x (t) ; y (t)) =

0@ y0 (t)q(x0 (t))2 + (y0 (t))2

;�x0 (t)q

(x0 (t))2 + (y0 (t))2

1A ; t 2 [a; b]

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and we also have

d� = ds =

q(x0 (t))2 + (y0 (t))2dt; t 2 [a; b] :

Hence we concludeZ@

W �Nd�

=

Z b

a

�(p (x (t) ; y (t)) ; q (x (t) ; y (t))) �

�y0(t)p

(x0(t))2+(y0(t))2; �x0(t)p

(x0(t))2+(y0(t))2

��q(x0 (t))2 + (y0 (t))2dt

=

Z b

a

[(p (x (t) ; y (t)) y0 (t)� q (x (t) ; y (t)))x0 (t)] dt =ZC

�qdx+ pdy (this is line integral).

(287)

By (286) and (287), we conclude the identityZZ

�@p

@x(x; y) +

@q

@y(x; y)

�dxdy =

ZC

�qdx+ pdy; (288)

which is exactly the Green Theorem. Note that the value of the line integralRC�qdx + pdy is

independent of parametrization.

Remark 10.3 If n = 3; then the surface measure d� means

d� =

s1 +

�@f

@x

�2+

�@f

@y

�2dxdy (289)

if the surface is represented by the graph of a function z = f (x; y) : Also, if the surface in R3 isgiven by the parametrization form

X (u; v) : (u; v) 2 U � R2 ! (x (u; v) ; y (u; v) ; z (u; v)) 2 R3;

thend� = jXu �Xvj dudv; where � is the cross product in R3: (290)

In case the surface in R3 is given by the equation

' (x; y; z) = 0

for some smooth function ' (x; y; z) : O � R3 ! R; then (assuming the surface can be expressed asz = f (x; y) for (x; y) 2 U � R2) we have

X (x; y) = (x; y; f (x; y)) : U � R2 ! R3

and by chain rule we have

d� = jXx �Xyj dxdy =q1 + f 2x + f 2ydxdy =

p'2x + '2y + '2zj'zj

dxdy: (291)

The �rst identity in (291) is clear, to see the second identity, note that by ' (x; y; f (x; y)) = 0 forall (x; y) 2 U � R2; we have

fx = �'x'z; fy = �

'y'z

(292)

and so q1 + f 2x + f 2y =

s1 +

�'x'z

�2+

�'y'z

�2=

p'2x + '2y + '2zj'zj

: (293)

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Proof. (Special case only.) We will give a proof of the theorem only for the case n = 2 andassuming that takes the simple form:

= f(x; y) : a � x � b; 0 � y � f (x) ; f (a) = f (b) = 0g (294)

where f (x) is a C1 function de�ned on [a; b] : Although may not be smooth at (a; f (a)) and(b; f (b)) ; divergence theorem still holds for such as shown below. Its boundary @ has two parts:the graph y = f (x) and the segment (x; 0) ; a � x � b: Call them @1 and @2 respectively. Wehave 8>><>>:

N at (x; f (x)) =(�f 0 (x) ; 1)q1 + (f 0 (x))2

N at (x; 0) = (0;�1) :WritingW (x; y) = (u (x; y) ; v (x; y)) ; the divergence theorem is equivalent toZZ

�@u

@x+@v

@y

�dxdy; divW =

@u

@x+@v

@y

=

Z@1

(u (x; y) ; v (x; y)) � (�f0 (x) ; 1)q

1 + (f 0 (x))2ds+

Z@2

(u (x; y) ; v (x; y)) � (0;�1) ds; (295)

where ds is the arc length di¤erential. Note that the boundary line integralR@ds in (295) has no

orientation. Clearly we have

y = f (x) ; ds =

q1 + (f 0 (x))2dx on @1

y = 0; ds = dx on @2

and so the RHS (right-hand side) of (295) becomesZ b

a

(u (x; f (x)) ; v (x; f (x))) � (�f 0 (x) ; 1) dx+Z b

a

(u (x; 0) ; v (x; 0)) � (0;�1) dx

=

Z b

a

[�u (x; f (x)) f 0 (x) + v (x; f (x))] dx�Z b

a

v (x; 0) dx

= �

z }| {Z b

a

u (x; f (x)) f 0 (x) dx+

Z b

a

[v (x; f (x))� v (x; 0)] dx| {z } : (296)

Also the LHS of (295) is

ZZ

�@u

@x+@v

@y

�dxdy =

z }| {Z b

a

Z f(x)

0

�@u

@x

�dydx+

Z b

a

Z f(x)

0

�@v

@y

�dydx| {z } : (297)

Now we have Z b

a

Z f(x)

0

�@v

@y

�dydx| {z } =

Z b

a

[v (x; f (x))� v (x; 0)] dx| {z } : (298)

Hence it su¢ ces to showz }| {Z b

a

u (x; f (x)) f 0 (x) dx+

Z b

a

Z f(x)

0

@u

@x(x; y) dydx = 0: (299)

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Note that the LHS (left-hand side) of (299) isZ b

a

u (x; f (x)) f 0 (x) +

Z f(x)

0

@u

@x(x; y) dy

!dx; (300)

where the integrand in (300) can be written as

u (x; f (x)) f 0 (x) +

Z f(x)

0

@u

@x(x; y) dy =

d

dx

Z f(x)

0

u (x; y) dy

!:

Hence we haveZ b

a

u (x; f (x)) f 0 (x) +

Z f(x)

0

@u

@x(x; y) dy

!dx

=

Z b

a

d

dx

Z f(x)

0

u (x; y) dy

!dx =

Z f(b)

0

u (b; y) dy �Z f(a)

0

u (a; y) dy = 0;

due to f (a) = f (b) = 0: The proof is done. �

10.1 The Laplace equation and its radial solutions; Green identities.

De�nition 10.4 Let u = u (x) = u (x1; ::: ; xn) 2 C2 () ; where � Rn is an open set. TheLaplace operator 4 acting on the function u : ! R is de�ned as

(4u) (x) = @2u

@x21(x) + � � �+ @2u

@x2n(x) ; x 2 : (301)

We call the function 4u : ! R the Laplacian of u: The equation

(4u) (x) = 0; 8 x 2 (302)

is called the Laplace equation of u on : This equation appears very often in physics. A C2 solutionu of (302), de�ned on ; is called a harmonic function on :

Example 10.5 The following functions

u (x; y) = x; y; ; x2 � y2; 2xy; ex cos y; ex sin y; x3 � 3xy2; (303)

are all harmonic functions de�ned on the whole plane R2: The functions

u (x; y) =x

x2 + y2;

�yx2 + y2

are harmonic functions de�ned on R2n f(0; 0)g :

De�nition 10.6 Let f (x) be a given continuous function on ; the equation

(4u) (x) = f (x) ; 8 x 2 (304)

is called the Poisson equation of u on :

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De�nition 10.7 Let V (x) : � Rn ! Rn be a C1 vector �eld on given by

V (x) = (V1 (x) ; ::: ; Vn (x)) ; Vi (x) : � Rn ! R is a C1 function.

Its divergence, denoted as (div V ) (x) (or just div V (x)): � Rn ! R; is a scalar function de�nedas

div V (x) =@V1@x1

(x) + � � �+ @Vn@xn

(x) ; x = (x1; ::: ; xn) 2 :

Note that div V (x) is a continuous function on :

Remark 10.8 By the above de�nition, we have the useful identity

(4u) (x) = div (ru) (x) ; x 2 ;

for any for any u 2 C2 () : By the divergence theorem, we haveZ

4udx =Z

div (ru) (x) dx =Z@

ru �Nd� =Z@

@u

@Nd� (surface integral in Rn) (305)

for a function u 2 C2 ()TC1���: Here d� is the surface measure on @:

Problem 10.9 Let u (x) : R2 ! R be a di¤erentiable function and let fv; wg � R2 be an ortho-normal basis at the origin O = (0; 0) 2 R2. Answer the following three questions:

1. Does the quantity

d

dt

����t=0

u (tv) +d

dt

����t=0

u (tw) (same as Dvu (O) +Dwu (O) ) (306)

depend on the choice of the orthonormal basis at the origin O = (0; 0) ? Give your reasons.

2. Does the quantityd2

dt2

����t=0

u (tv) +d2

dt2

����t=0

u (tw) (307)

depend on the choice of the orthonormal basis at the origin O = (0; 0) ? Give your reasons.Hint: �rst show that if v = (cos �; sin �) for some angle �; then we have

d2

dt2

����t=0

u (tv) =@2u

@x21(O) cos2 � + 2

@2u

@x1@x2(O) cos � sin � +

@2u

@x22(O) sin2 �;

then show thatd2

dt2

����t=0

u (tv) +d2

dt2

����t=0

u (tw) = 4u (O) :

3. Show that we have the following "average identities":

1

2�

Z 2�

0

�d

dt

����t=0

u (tv)

�d� = 0; v = (cos �; sin �) ; � 2 [0; 2�] (308)

and

1

2�

Z 2�

0

�d2

dt2

����t=0

u (tv)

�d� =

1

2(4u) (O) ; v = (cos �; sin �) ; � 2 [0; 2�] ; (309)

where (4u) (O) is the Laplacian of u at the origin.

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There are many useful consequence of divergence theorem. Among the most important are theGreen identities: Assume is a C1 bounded domain and u; v 2 C2 () \ C1

���: We have the

Green�s 1st identityZ

v4 udx+

Z

ru � rvdx =Z@

v@u

@Nd�; ru = gradient of u (310)

and Green�s 2nd identityZ

(v4 u� u4 v) dx =

Z@

�v@u

@N� u @v

@N

�d�: (311)

To see (310) we need to use the divergence theorem and the identity

div (vru) = v4 u+rv � ru: (312)

In particular, when u = v in (310), we get the identityZ

u4 udx+

Z

jruj2 dx =Z@

u@u

@Nd�; (313)

and when v � 1 in (310), we get the identityZ

4udx =Z@

@u

@Nd� =

Z@

ru �Nd�: (314)

Physically, the quantityR@ru �Nd� is called the �ux of the vector �eld ru across the boundary

of :

11 Lecture notes on 2018-3-1 and 2018-3-5.

We now look at the Laplace equation on Rn :

(4u) (x) = 0; x = (x1; ::: ; xn) 2 Rn; (315)

and we want to look for a special solution of the Laplace equation in Rn which is radial, i.e., it hasthe form

u (x) = v (r) ; r = jxj =qx21 + � � �+ x2n;

where v (r) is chosen so that it satis�es �u (x) = 0 for all x 2 Rn (at this moment, we do not knowif such function v (r) exists or not; we shall see that the function v (r) does exist, but cannot bede�ned at r = 0; which implies that the radial function u (x) cannot be de�ned at x = 0). Insteadof solving a PDE for u (x) ; we only have to solve an ODE for v (r) :By the chain rule, one can see that

@u

@xi(x) = v0 (r)

xir;

@2u

@x2i(x) = v00 (r)

x2ir2+ v0 (r)

�1

r� x2ir3

�; 1 � i � n;

which gives (sum over i = 1; 2; 3; :::; n)

�u (x) = v00 (r) +n� 1r

v0 (r) = 0; r > 0 (316)

and we obtain a second-order ODE for v (r) ; r 2 (0;1) : The general solution of (316) is

v (r) = v (jxj) = u (x) =

(Ar2�n +B; n > 2; r = jxj 2 (0;1)

A log r +B; n = 2; r = jxj 2 (0;1)(317)

where A; B are integration constants.

67

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Remark 11.1 Note that v (r) is not de�ned at r = 0: Hence the above radial solution u (x) is notde�ned at x = 0: We see that u 2 C1 (Rnn f0g) :

Remark 11.2 (Important.) Why does the Laplace equation have radial solutions? This is becausethe Laplace operator

4 = @2

@x21+ � � �+ @2

@x2n

has symmetry in it. If we change the Laplace operator into the non-symmetric form, say,

~4 = @2

@x21+ � � �+ @2

@x2n+

@

@x1= 4+ @

@x1(or other forms, say 4+ @2

@x21);

then if u (x) has the radial form v (r) ; r = jxj ; we have

~4u (x) = 4u (x) + @u

@x1(x) = v00 (r) +

n� 1r

v0 (r) + v0 (r)x1r;

which cannot produce a self-contained equation (ODE) for v (r) due to the term v0 (r) x1r: Thus for

the new "Laplace operator" ~4; it has no radial solution at all (except the trivial constant solutions).

Problem 11.3 Show that the general solution of the second order ODE

v00 (r) +n� 1r

v0 (r) = 0; r > 0

is given by (317). Hint: either let w (r) = v0 (r) and solve w (r) or multiply the equation by rn�1.

We can conclude the following:

Lemma 11.4 Consider the Laplace equation on Rn; given by

(4u) (x) = 0; x = (x1; ::: ; xn) � Rn: (318)

Its radial solutions (de�ned only on Rnn f0g) are given by

u (x) =

(A jxj2�n +B; n > 2; where x 2 Rnn f0g

A log jxj+B; n = 2; where x 2 R2n f0g(319)

and no others. Here A; B are two arbitrary constants.

De�nition 11.5 The radial functions u (x) given by (319) are called the fundamental solutionsof the Laplace equation (de�ned on Rnn f0g).

Remark 11.6 There are also many other non-radial solutions of (318) on Rn: For example, whenn = 2; all functions in (303) are non-radial harmonic functions. In fact, without further conditions,equation (318) has in�nitely many solutions.

In R3 we have (now we denote x 2 R3n f0g as (x; y; z) 2 R3n f0g) the radial solution of theLaplace equation:

u (x; y; z) =A

r+B =

Apx2 + y2 + z2

+B; (x; y; z) 2 R3n f0g

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and then

(ru) (x; y; z)

=A

x2 + y2 + z2

�xp

x2 + y2 + z2;

�ypx2 + y2 + z2

;�zp

x2 + y2 + z2

!

=A

j�!r j2���!rj�!r j (this is unit vector)

�; where �!r = (x; y; z) 2 R3n f0g : (320)

For suitable constant A; (320) describes the force �eld of the Newtonian earth gravity with pointmass at the origin. Also in R2; we have that radial solution of the Laplace equation:

u (x; y) = A log r +B = A logpx2 + y2 +B; (x; y) 2 R2n f0g : (321)

11.1 The application of Green identities to Dirichlet problem of Poissonequation.

Let � Rn be a bounded C1 domain and let f (x) ; h (x) be continuous functions on and@ respectively. The Dirichlet problem of Poisson equation has the form(

4u (x) = f (x) in � Rn

u (x) = h (x) on @:(322)

One can use (313) to show that (322) has a unique solution (we will not discuss the existence ofa solution here).

Lemma 11.7 (Uniqueness of solution.) Let u; v 2 C2 () \ C1���be two solutions of (322)

on : Then we must have u � v on �:

Proof. Set w = u� v 2 C2 () \ C1���: It satis�es(

4w (x) = 0 in � Rn

w (x) = 0 on @:(323)

By (313), we have the identityZ

w4 wdx+

Z

jrwj2 dx =Z@

w@w

@Nd�; (324)

which, together with (323), gives the identityZ

jrwj2 dx = 0; (325)

where we know that jrwj2 is a continuous function on with jrwj2 � 0 everywhere. Hence weconclude jrwj2 � 0 on and w (x) must be a constant function on : As w (x) = 0 on @; wemust have w (x) � 0 on �: The proof is done. �

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11.2 Elliptic equations and harmonic functions for Laplace operator.

De�nition 11.8 The Laplace equation on R2 is de�ned as

4u (x; y) = 0; (x; y) 2 R2; (326)

where the Laplace operator 4 acting on a C2 function u (x; y) : R2 ! R is

4u (x; y) = @2u

@x2(x; y) +

@2u

@y2(x; y) ; (x; y) 2 R2: (327)

Similarly, one can de�ne the Laplace equation on Rn for a C2 function u (x1; ::: ; xn) : Rn ! R: Onecan also de�ne the Laplace equation on a bounded domain (open connected set) � Rn:

De�nition 11.9 A C2 solution u of the Laplace equation 4u = 0 on a bounded domain � Rn iscalled a harmonic function on :

Remark 11.10 We will focus on n = 2 or n = 3 in this course.

According to the classi�cation, a Laplace equation on R2 is an elliptic equation. Moreover, onecan write it as

4u (x; y) = [div (ru)] (x; y) ; 8 (x; y) 2 R2,

whereru (x; y) =�@u@x(x; y) ; @u

@y(x; y)

�: This suggests that Laplace equation may have close relation

with the divergence theorem. It also has close relation with physical situation where the "vector�eld" (force, velocity, etc) of the physical system is given by the gradient of some potential functionu:

Remark 11.11 If we view y as time, then (327) becomes

@2u

@x2(x; t) +

@2u

@t2(x; t) = 0 on (x; t) 2 R2: (328)

With this, similar to the wave equation, one can look at the initial value problem(@2u@x2(x; t) + @2u

@t2(x; t) = 0 on (x; t) 2 R2;

u (x; 0) = � (x) ; ut (x; 0) = (x) ; x 2 (�1;1) :(329)

However, this problem is not well-posed (see exercise problem in Homework 12). Also, unlike thewave equation in which we can factor the wave equation utt � uxx = 0; now there is no generalsolution formula for elliptic equation (328).

11.3 The maximum principle for harmonic functions on bounded do-mains.

From now on, the letter will denote a bounded domain in R2 or R3:

Lemma 11.12 (Weak maximum principle.) Assume u 2 C2 ()TC0���. If4u � 0 (� 0) everywhere

in ; then

max�u = max

@u

�min�u = min

@u

�: (330)

Consequently for a harmonic function u 2 C2 ()TC0���; we have

min@

u � u (x) � max@

u, 8 x 2 �: (331)

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Remark 11.13 (A simple example.) Here is a simple example in the 1-dimensional case. Letf (x) = x2; x 2 (�1; 1) : It satis�es f 00 (x) = 2 > 0 everywhere in = (�1; 1) � R: The maximumvalue of f (x) on � is 1; attained at x = �1 2 @: Similarly, the function g (x) = �x2; x 2(�1; 1) satis�es g00 (x) = �2 < 0 everywhere in = (�1; 1) � R: The minimum value of g (x) on �is �1; attained at x = �1 2 @: Finally, the function h (x) = ax + b (a; b are any two numbers)is harmonic on = (�1; 1) : Its maximum value and minimum value attained at x = �1 2@ respectively. Unless h (x) is a constant function, otherwise, it is impossible for the harmonicfunction h (x) to attain its maximum value (or minimum value) at some x0 2 = (�1; 1) :

Proof. Assume �rst that 4u > 0 everywhere in : Then since u is continuous on � (compact set),there is some point p 2 � such that u (p) = max� u: If p 2 (interior point), we get

4u (p) = @2u

@x2(p) +

@2u

@y2(p) � 0;

a contradiction. Hence p must lie on the boundary of and so max� u � max@ u (note that wealways have max� u � max@ u).Next, assume that 4u � 0 everywhere in : Let

v (x; y) = u (x; y) + "�x2 + y2

�; (x; y) 2 �;

where " > 0 is a constant. We have

4v (x; y) = 4u (x; y) + 4" � 0 + 4" > 0 everywhere in :

Hence we havemax�u � max

�v = max

@v �

�max@

u�+ "max

�x2 + y2

�:

As " > 0 is arbitrary and is a bounded domain, we have, letting "! 0+; the following

max�u � max

@u:

On the other hand, we also have max� u � max@ u and so (330) is veri�ed. The proof of theminimum case is similar. �

Lemma 11.14 (Strong maximum principle.) Assume u 2 C2 ()TC0���is a harmonic

function and there exists a point p 2 such that u (p) = max� u or u (p) = min� u; then u must bea constant function on .

Proof. We will prove this later on. �

Let f 2 C0 () and h 2 C0 (@) : We can consider the following problem on bounded domain :

4u = f in ; u = h on @: (332)

This problem is well-posed and we have the following uniqueness property due to the maximumprinciple:

Lemma 11.15 The problem (332) has at most one solution u 2 C2 ()TC0���.

Proof. Assume there are two solutions u; v 2 C2 ()TC0���: Then the function w = u � v 2

C2 ()TC0���satis�es

4w = 0 in ; w = 0 on @: (333)

By the weak maximum principle, we have w � 0 in : Hence u � v in (and so on �). �

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11.4 Comparing the Laplace equation and the wave equation.

There are many striking di¤erences between the Laplace equation and the wave equation. One cannotice this from many aspects. For simplicity, we just look at one interesting example. For theconvenience of constructing a solution for the Laplace equation, here we take to be unbounded,with = R2 and consider the following two initial value problems (here for u (x; y) we view y astime and denote it as t):(

@2u@x2(x; t) + @2u

@t2(x; t) = 0; (x; t) 2 R2;

u (x; 0) = � (x) ; ut (x; 0) = (x) ; x 2 (�1;1) :(334)

and (@2u@t2(x; t) = @2u

@x2(x; t) ; (x; t) 2 R2;

u (x; 0) = � (x) ; ut (x; 0) = (x) ; x 2 (�1;1)(335)

We know that the equation in problem (335) is a wave equation and the ivp has a nice solutionformula given by:

u (x; t) =1

2[� (x+ ct) + � (x� ct)] + 1

2c

Z x+ct

x�ct (s) ds; 8 (x; t) 2 R2: (336)

From it we can see that the problem is well-posed (the solution exists, is unique, and depends"continuously" (in some function space) on the data � (x) ; (x).On the other hand, the equation in problem (334) is a Laplace equation and it is not well-

posed. To see this, in both examples we take � (x) = 0 and (x) = sinnxn; x 2 (�1;1) ;

where n 2 N is a positive integer. Clearly, the function

un (x; t) =1

2

Z x+t

x�t

sinns

nds =

1

2

��cosn (x+ t)

n2+cosn (x� t)

n2

�; (x; t) 2 R2 (337)

is a solution of the wave equation problem (335). On the other hand, one can verify that thefunction

un (x; t) =(sinhnt) (sinnx)

n2; (x; t) 2 R2 (338)

is a solution of Laplace equation problem (334). As n!1; the initial data for ut (x; 0) ; whichoscillates a lot, satis�es

limn!1

����sinnxn

���� = 0 uniformly in x 2 (�1;1) : (339)

For un (x; t) given by (337), it also satis�es

limn!1

un (x; t) = limn!1

1

2

��cosn (x+ t)

n2+cosn (x� t)

n2

�= 0 uniformly in (x; t) 2 R2: (340)

As for (338), we have

un

��

2n;1pn

�=

�sinhn

�1pn

�� �sinn

��2n

��n2

=sinhpn

n2=sinh �

�4=e� � e��2�4

!1 as n = �2 !1: (341)

By (341), we see that the solution to problem (334) does not depend on the initial data � (x) and (x)in a continuous way. Therefore, the initial value problem (334) is not well-posed. As n!1; wehave sinnx

n! 0 uniformly on x 2 R: So naturally, we expect that the solution un (x; t) should also be

small (in some sense). However, there exists a sequence (xn; tn) 2 R2 with limn!1 (xn; tn) = (0; 0)such that limn!1 un (xn; yn) =1:

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Remark 11.16 (Important.) On the other hand, if we take � R2 (the (x; t)-plane) to be a C1bounded domain and look at the following two Dirichlet problems (boundary value prob-lems): (

4u (x; t) = @2u@x2(x; t) + @2u

@t2(x; t) = f (x; t) in � R2

u (x; t) = h (x; t) on @(342)

and (@2u@x2(x; t)� @2u

@t2(x; t) = f (x; t) in � R2

u (x; t) = h (x; t) on @;(343)

where f (x; t) and h (x; t) are two continuous functions on and @ respectively (to be safe, maybetake � R2 to be a smooth bounded domain and f (x; t) and h (x; t) to be two smooth functionson � and @ respectively), then the problem (342) is well-posed and the problem (343) is NOTwell-posed. For example, take to be the unit disc in R2 and f = h � 0; then by Lemma 11.7,problem (342) has unique solution u (x; t) � 0; but problem (343) does not have unique solution. Thetwo functions (note that the general solution of the wave equation has the form F (x+ t)+G (x� t))

u (x; t) � 0; u (x; t) = (x+ t)2 + (x� t)2 � 2 (344)

are both solutions. From this, one can see that problem (343) is NOT well-posed.

Example 11.17 (Important.) For = R2; Laplace equation is also not well-posed for "Dirich-let problems (boundary value problems)" (not the above initial value problem with u (x; 0) =� (x) ; ut (x; 0) = (x)). Consider the equation(

4u (x; y) = 0 for y > 0

u (x; 0) = 0; x 2 R

and note thatuk (x; y) =

1

k(sin kx) eky; k 2 N

is a harmonic function in the upper half plane and uk (x; 0) = 1k(sin kx) ; which satis�es

supx2Rjuk (x; 0)� U (x; 0)j �

1

k! 0 as k !1;

where U (x; y) � 0: For large k; uk (x; 0) is even closer to U (x; 0) (note that uk (x; 0) oscillate a lotwhen k is large), but we have

supx2R; y2[0;")

juk (x; y)� U (x; y)j = supx2R; y2[0;")

����1k (sin kx) eky���� =1

for any " > 0:

11.5 Laplace equations are invariant under orthogonal linear transfor-mations.

Remark 11.18 Skip this section in class.

We have:

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Lemma 11.19 Let T : Rn ! Rn be an orthogonal linear transformation and let u (y) : Rn !R be any smooth function. Then

(4u) (Tx) = [4 (u � T )] (x) ; x 2 Rn: (345)

In particular, if u (y) is a harmonic function of y; i.e. 4yu (y) = 0; then the function ~u (x) =u(T (x)) is still a harmonic function of x; i.e. 4x~u (x) = 0:

Remark 11.20 See Remark 4.5 also.

Proof. Let y = Tx = Ax; where A = (aij) is an n � n orthogonal matrix ( ATA = I). Also letv (x) = u (y) = u (Tx) = u (Ax) : then the chain rule implies that

@v

@xi=

@u

@ys

@ys@xi

(sum over s) =@u

@ysasi

and@2v

@x2i=

�@

@xi

�@u

@ys

��asi =

�@2u

@ym@ysami

�asi (sum over s; m)

and then

[4 (u � T )] (x) = (4v) (x) =nXi=1

@2v

@x2i(x) =

nXi; m; s=1

@2u

@ym@ys(y) amiasi

=nX

m; s=1

"@2u

@ym@ys(y)

nXi=1

amiasi

!#=

nXi=1

@2u

@y2i(y) = (4u) (y) = (4u) (Tx) ;

where we have used the identity

nXi=1

amiasi = �ms =

(1; if m = s

0; if m 6= s:

Hence we get[4 (u � T )] (x) = (4u) (Tx) :

The proof is done. �

Remark 11.21 Draw a commutative diagram for the above lemma.

Remark 11.22 (Simple example.) If n = 1 and T : R! R is given by T (x) = ax; where a is aconstant and u (y) : R! R: Then

d2

dy2u (y)

����y=ax

= u00 (ax)

andd2

dx2(u (ax)) = u00 (ax) a2:

Therefore, if a = �1 (orthogonal), we have the identity.

Remark 11.23 This means that Laplace equation is invariant under orthogonal transformation.More precisely, if u (y) ; y = (y1; y2) 2 R2; satis�es

@2u

@y21(y1; y2) +

@2u

@y22(y1; y2) = 0 for all (y1; y2) 2 R2;

then the function v(x) := u (Ax) ; x = (x1; x2) 2 R2; where A is a 2 � 2 orthogonal matrix, alsosatis�es

@2v

@x21(x1; x2) +

@2v

@x22(x1; x2) = 0 for all (x1; x2) 2 R2:

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11.6 Laplace operator in polar coordinates (r; �) :

Remark 11.24 A major purpose of expressing Laplace operator in polar coordinates (r; �) is to �ndsome important special solutions, in particular, the radial solution U (r) and the angular solutionU (�) : In particular, we can use it to solve the Dirichlet problem of the Laplace equation on the discin R2 or on the ball in R3:

The polar coordinates (r; �) in R2 and the Euclidean coordinates (x; y) are related by

x = r cos �; y = r sin �; r > 0; 0 < � < 2�; (346)

where the change of variables is a di¤eomorphism between the following two open sets:

R2n f(x; 0) : x � 0g � xy-plane ! (r; �) 2 (0;1)� (0; 2�) � r�-plane. (347)

Let u (x; y) : R2 ! R be a C2 function on R2: Under the above change of variables u (x; y) becomesa function U (r; �) ; i.e.,

u (r cos �; r sin �) = U (r; �) :

What is the expression uxx (x; y) + uyy (x; y) under polar coordinates (r; �)? The answer is:

4u (x; y) =�@2

@x2+

@2

@y2

�u (x; y) =

�@2

@r2+1

r

@

@r+1

r2@2

@�2

�U (r; �) : (348)

Example 11.25 Let u (x; y) = x2y: Then U (r; �) = r3 cos2 � sin �: We have�@2

@x2+

@2

@y2

�u (x; y) = 2y:

On the other hand, we also have�@2

@r2+1

r

@

@r+1

r2@2

@�2

�U (r; �)

=

�@2

@r2+1

r

@

@r+1

r2@2

@�2

��r3 cos2 � sin �

�= 2r sin �:

Since y = r sin �; both sides of (348) are equal.

To derive (348), for the �rst derivatives, we have the relation:

@U

@r=

@

@r[u (r cos �; r sin �)] =

@u

@x

@x

@r+@u

@y

@y

@r

=@u

@xcos � +

@u

@ysin � =

1

r

�x@u

@x+ y

@u

@y

�(349)

and

@U

@�=

@

@�[u (r cos �; r sin �)] =

@u

@x

@x

@�+@u

@y

@y

@�

=@u

@x(�r sin �) + @u

@y(r cos �) = �y@u

@x+ x

@u

@y: (350)

We can the above as the system 8<:�r @@r

�U = x@u

@x+ y @u

@y�@@�

�U = �y @u

@x+ x@u

@y;

(351)

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or equivalently, the operator relation:8<: r @@r= x @

@x+ y @

@y;

@@�= �y @

@x+ x @

@y:

(352)

It says that the operator r @@ris roughly like @

@�: In the matrix form, we have�

r @@r@@�

�=

�x y�y x

��@@x@@y

�; (353)

and so �@@x@@y

�=

1

x2 + y2

�x �yy x

��r @@r@@�

�=

�cos � �1

rsin �

sin � 1rcos �

��@@r@@�

�: (354)

More precisely, the above gives

@u

@x= (cos �)

@U

@r��1

rsin �

�@U

@�;

@u

@y= (sin �)

@U

@r+

�1

rcos �

�@U

@�: (355)

In particular, we have two di¤erent ways to express the gradient vector of u :

ru (x; y) =�@u

@x;@u

@y

�=@u

@x(1; 0) +

@u

@y(0; 1)

=@U

@r(cos �; sin �) +

1

r

@U

@�(� sin �; cos �) : (356)

The above says that we can express the gradient vector ru (x; y) in terms of the orthonormal basis(cos �; sin �) ; (� sin �; cos �) :

Remark 11.26 Draw a picture for this.

Keep going and use (352) to get�r@

@r

�2U :=

�r@

@r

���r@

@r

�U

�=

�x@

@x+ y

@

@y

��x@u

@x+ y

@u

@y

�= x

@

@x

�x@u

@x+ y

@u

@y

�+ y

@

@y

�x@u

@x+ y

@u

@y

�= x2

@2u

@x2+ y2

@2u

@y2+ 2xy

@2u

@x@y+ x

@u

@x+ y

@u

@y: (357)

Similarly �@

@�

�2U =

@

@�

�@U

@�

�= �y @

@x

��y@u

@x+ x

@u

@y

�+ x

@

@y

��y@u

@x+ x

@u

@y

�= y2

@2u

@x2+ x2

@2u

@y2� 2xy @2u

@x@y� x@u

@x� y@u

@y: (358)

Add (357) and (358) to get the beautiful identity:�r@

@r

�2U +

�@

@�

�2U =

�x2 + y2

��@2u@x2

+@2u

@y2

�: (359)

Finally, one can simplify�r @@r

�2u as�

r@

@r

�2U =

�r@

@r

��r@U

@r

�= r2

@2U

@r2+ r

@U

@r(360)

and conclude the identity:

76

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Lemma 11.27 For any C2 function u (x; y) = u (r cos �; r sin �) = U (r; �) de�ned on R2; then onthe two open sets (347), we have the identity

�x2 + y2

��@2u@x2

+@2u

@y2

�=

�r@

@r

�2U +

�@

@�

�2U; (361)

which is the same as

@2U

@r2+1

r

@U

@r+1

r2@2U

@�2=@2u

@x2+@2u

@y2; r > 0; 0 < � < 2�: (362)

Remark 11.28 In particular, if u (x; y) = U (r) is a radial function, (362) becomes

U 00 (r) +1

rU 0 (r) =

@2u

@x2+@2u

@y2; r > 0; (363)

which matches with (316).

Example 11.29 If a function u (x; y) = U (r) ; r =px2 + y2; is radial, then by (361) we have�

r@

@r

�2U =

�x2 + y2

��@2u@x2

+@2u

@y2

�: (364)

Thus a radial harmonic function u (r) (de�ned on R2n f(0; 0)g) satis�es

@

@r

�r@U

@r

�= 0;

i.e.,

r@U

@r= const:

Henceu (x; y) = U (r) = a ln r + b = a log

px2 + y2 + b; (x; y) 6= (0; 0) : (365)

for some constants a; b: Note that u (x; y) is de�ned only on R2n f(0; 0)g and

lim(x;y)!(0;0)

u (x; y) =1 (if a > 0).

Its gradient vector is also in the radial direction, given by

ru (x; y) =�@u

@x(x; y) ;

@u

@y(x; y)

�= a

�x

x2 + y2;

y

x2 + y2

�; (x; y) 6= (0; 0) : (366)

Example 11.30 If a function u (x; y) = U (�) depends only on angle � 2 (0; 2�), then by (361) wehave �

@

@�

�2U =

�x2 + y2

��@2u@x2

+@2u

@y2

�: (367)

Thus an angular harmonic function U (�) (de�ned on R2n f(x; y) : x � 0g) satis�es

@U

@�= c; c is a constant.

i.e.,u (x; y) = U (�) = c� + d = c tan�1

y

x+ d (if x 6= 0) (368)

for some constants c; d: Its gradient vector is perpendicular to the radial direction, given by

ru (x; y) =�@u

@x;@u

@y

�= c

��y

x2 + y2;

x

x2 + y2

�; (x; y) 6= (0; 0) : (369)

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11.6.1 Di¤erent method.

By (354), we have the operator relation:

@

@x= (cos �)

@

@r��1

rsin �

�@

@�;

@

@y= (sin �)

@

@r+

�1

rcos �

�@

@�: (370)

Then we have

@2u

@x2= (cos �)

@

@r

�(cos �)

@U

@r��1

rsin �

�@U

@�

���1

rsin �

�@

@�

�(cos �)

@U

@r��1

rsin �

�@U

@�

=

8<: (cos2 �) @2U@r2

+ 1r2(cos � sin �) @U

@�� 1

r(cos � sin �) @

2U@r@�

+1rsin2 � @U

@r� 1

r(cos � sin �) @

2U@�@r

+ 1r2(cos � sin �) @U

@�+ 1

r2

�sin2 �

�@2U@�2

(371)

and

@2u

@y2= (sin �)

@

@r

�(sin �)

@U

@r+

�1

rcos �

�@U

@�

�+

�1

rcos �

�@

@�

�(sin �)

@U

@r+

�1

rcos �

�@U

@�

=

8<:�sin2 �

�@2U@r2� 1

r2(cos � sin �) @U

@�+ 1

r(cos � sin �) @

2U@r@�

+1rcos2 � @U

@r+ 1

r(cos � sin �) @

2U@�@r� 1

r2(cos � sin �) @U

@�+ 1

r2(cos2 �) @

2U@�2

:(372)

Adding (371) and (372) together, we get

@2u

@x2+@2u

@y2

=

8>>>>>>><>>>>>>>:

8<: (cos2 �) @2U@r2

+ 1r2(cos � sin �) @U

@�� 1

r(cos � sin �) @

2U@r@�

+1rsin2 � @U

@r� 1

r(cos � sin �) @

2U@�@r

+ 1r2(cos � sin �) @U

@�+ 1

r2

�sin2 �

�@2U@�2

+

8<:�sin2 �

�@2U@r2� 1

r2(cos � sin �) @U

@�+ 1

r(cos � sin �) @

2U@r@�

+1rcos2 � @U

@r+ 1

r(cos � sin �) @

2U@�@r� 1

r2(cos � sin �) @U

@�+ 1

r2(cos2 �) @

2U@�2

=@2U

@r2+1

r

@U

@r+1

r2@2U

@�2;

which gives the same formula.

11.7 Laplace operator in spherical coordinates (r; �; ') :

The sphere coordinates in R3 is given by (r; �; ') and its relation with respect to the Euclideancoordinates is

x = r sin' cos �; y = r sin' sin �; z = r cos'; x2 + y2 = r2 sin2 ';

where r > 0; � 2 (0; 2�) ; � 2 (0; �). By the above relation, we �rst have

r =px2 + y2 + z2;

@r

@x=x

r= sin' cos �;

@r

@y=y

r= sin' sin �;

@r

@z=z

r= cos'

and 8>>><>>>:� = tan�1 y

x;

@�@x= �y

x2+y2= �r sin' sin �

r2 sin2 '= � sin �

r sin';

@�@y= x

x2+y2= r sin' cos �

r2 sin2 '= cos �

r sin'

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and 8>>>>>>>>><>>>>>>>>>:

' = cos�1 zr;

@'@x= �1q

1�( zr )2

�zr2

@r@x= z

rpr2�z2

@r@x= r cos'

rpr2 sin2 '

sin' cos � = cos' cos �r

@'@y= r cos'

rpr2 sin2 '

sin' sin � = cos' sin �r

@'@z= �1q

1�( zr )2

�1r� z

r2@r@z

�= �1p

r2�z2 (1� cos2 ') = � sin2 'p

r2 sin2 '= � sin'

r:

With the above, we conclude the �rst order operator relation:8>>>>><>>>>>:

@@x= 1

r

n(sin' cos �)

�r @@r

�� (sin �)

�1

sin'@@�

�+ (cos' cos �) @

@'

o@@y= 1

r

n(sin' sin �)

�r @@r

�+ (cos �)

�1

sin'@@�

�+ (cos' sin �) @

@'

o@@z= 1

r

n(cos')

�r @@r

�� (sin') @

@'

o:

(373)

Based on (373), one can derive the identity (this can be very tedious !!):

4u (x; y; z) = @2U

@r2+2

r

@U

@r+

1

r2 sin2 '

@2U

@�2+

1

r2 sin'

@

@'

�sin'

@U

@'

�=@2U

@r2+2

r

@U

@r+

1

r2 sin2 '

"@2U

@�2+

�@

@'sin'

�2U

#: (374)

We will not prove this. The importance is that if u (x; y; z) = U (r) is a radial function only, thena radial harmonic function U (r) (de�ned on R3n f(0; 0; 0)g) satis�es

U 00 (r) +2

rU 0 (r) = 0 on (0;1) : (375)

Its solution is given by

U (r) =a

r+ b; r 2 (0;1) ; a; b are constants (376)

oru (x; y; z) =

apx2 + y2 + z2

+ b; (x; y; z) 2 R3n f(0; 0; 0)g ; (377)

which has a singularity at the origin O = (0; 0; 0) 2 R3 with

lim(x;y;z)!(0;0;0)

u (x; y; z) =1 (if a > 0).

12 Lecture notes on 2017-5-15.

12.1 Poisson integral formula; Dirichlet problem on a disc.

Let Ba (0) be the open disc centered at the origin (0; 0) with radius a > 0: We want to solve theDirichlet problem:(

4u (x; y) = 0 in (x; y) 2 Ba (0)

u (x; y) = h (x; y) on (x; y) 2 @Ba (0) (the boundary of Ba (0) ),(378)

where h (x; y) is a given continuous function de�ned on @Ba (0) : Due to the symmetry of thedomain and the rotational invariance of the Laplace operator, there is a solution formula for

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this problem. The solution lies in the space C2 (Ba (0))TC0�Ba (0)

�: Moreover, by the maximum

principle, the solution in the function space C2 (Ba (0))TC0�Ba (0)

�is unique.

As the domain Ba (0) is a disc, it is natural to use polar coordinates (r; �) instead of the Euclideancoordinates (x; y) : The Laplace equation under polar coordinates (r; �) is given by

@2u

@r2+1

r

@u

@r+1

r2@2u

@�2= 0; u = u (r; �) ; 0 < r < a; 0 < � < 2�: (379)

and we we want to solve it. If we multiply the equation by r2; we get

r2@2u

@r2+ r

@u

@r+@2u

@�2= 0; u = u (r; �) ; 0 < r < a; 0 < � < 2�: (380)

We then try a special form of solution (separable solution), namely u (r; �) = f (r) g (�) for somefunctions f (r) and g (�) ; where g (�) is required to be a periodic function with period 2� sincethe solution u (x; y) is de�ned on the whole disc Ba (0) :

Remark 12.1 The advantage of equation (380) over equation (379) is that r = 0 is allowed in(380). For r 6= 0; the two equations are equivalent.

Plug u (r; �) = f (r) g (�) into (380) to get

r2f 00 (r) g (�) + rf 0 (r) g (�) + f (r) g00 (�) = 0; 0 < r < a; 0 < � < 2�: (381)

One can write (381) as the separate form:

r2f 00 (r) + rf 0 (r)

f (r)= �g

00 (�)

g (�);

which implies thatr2f 00 (r) + rf 0 (r)

f (r)= �g

00 (�)

g (�)= � (382)

for some constant � 2 R: (382) gives the system of ODEs:(g00 (�) + �g (�) = 0; 8 � 2 (0; 2�):

r2f 00 (r) + rf 0 (r)� �f (r) = 0; 8 r 2 (0; a) :(383)

Since we are looking for a solution u (r; �) = f (r) g (�) de�ned on the whole disc Ba (0) ; we mustrequire g (�) to be periodic in � and so we have g (0) = g (2�) : This impose a sign condition onthe constant �; which is � � 0 (note that for � < 0 there is NO periodic solution g (�) satisfyingg00 (�) + �g (�) = 0).

Case 1: � > 0: In this case, we must have � = n2 for some n 2 N (otherwise, we will not haveg (0) = g (2�)). The general solution of g00 (�) + n2g (�) = 0 is

g (�) = C cosn� +D sinn�; � 2 [0; 2�] ; (384)

for some constants C; D: As for f (r) we get

r2f 00 (r) + rf 0 (r)� n2f (r) = 0; r 2 (0; a) ; (385)

which is an Euler equation (you have learned this from elementary ODE course). We recall thefollowing result: Consider a second order linear equation of the form

r2y00 (r) + �ry0 (r) + �y (r) = 0; r 2 (0;1) ; �; � are constants.

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One can use change of variables to convert it into a linear equation with constant coe¢ cients.Let x = ln r; r 2 (0;1) ; (same as r = ex); where x 2 (�1;1) will be the new variable. Thefunction y (r) will become a function ~y (x) : That is, ~y (x) = y (r) : The equation for ~y (x) is

d2~y

dx2+ (�� 1) d~y

dx+ �~y (x) = 0; x 2 (�1;1) :

By the above, the general solution for the ODE (385) is (check it yourself)

f (r) = Arn +B

rn; r 2 (0; a) ; for some constants A; B:

Thus for each n 2 N; the function

u (r; �) =

�Arn +

B

rn

�(C cosn� +D sinn�) ; (386)

will be a solution of the equation (380) de�ned on r 2 (0; a) ; � 2 [0; 2�] :

Case 2: � = 0: In this case, g (�) must be a constant and we have

r2f 00 (r) + rf 0 (r) = 0 (same as�r@

@r

�2f (r) = 0).

The general solution for f (r) is

f (r) = A log r +B; r > 0; for some constants A; B:

Thus the functionu (r; �) = A log r +B; r 2 (0; a) ; (387)

is a (radial) solution of the equation (380) de�ned on r 2 (0; a) ; � 2 [0; 2�] :

Remark 12.2 Until now, we can conclude the following: For any n 2 N and any constantsA; B; C; D; the following functions

u (r; �) =

( �Arn + B

rn

�(C cosn� +D sinn�) ; r 2 (0; a) ; � 2 [0; 2�]

A log r +B; r 2 (0; a)(388)

are solutions to equation (380) on r 2 (0; a) (or r 2 (0;1)), � 2 [0; 2�] : The correspondingfunctions u (x; y) are harmonic on Ba (0) n f(0; 0)g (or on R2n f(0; 0)g).

Remark 12.3 (Important observation.) If we take � < 0; say � = ��2 for some � > 0; we get(g00 (�)� �2g (�) = 0; 8 � 2 (0; 2�) :

r2f 00 (r) + rf 0 (r) + �2f (r) = 0; 8 r 2 (0; a) :

The general solution of the above is given by(g (�) = c1e

�� + c2e���; 8 � 2 (0; 2�)

f (r) = C1 cos (� ln r) + C2 sin (� ln r) ; 8 r 2 (0; a) ;

for integration constants c1; c2; C1; C2: Hence we get the following harmonic functions for any� > 0 :

cos (� ln r) � e��; sin (� ln r) � e��; cos (� ln r) � e���; sin (� ln r) � e���; (389)

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de�ned on (r; �) 2 (0; a) � (0; 2�) (or on (r; �) 2 (0;1) � (0; 2�)). On the other hand, if we take� > 0; say � = �2 for some � > 0; we get(

g00 (�) + �2g (�) = 0; 8 � 2 (0; 2�) :

r2f 00 (r) + rf 0 (r)� �2f (r) = 0; 8 r 2 (0; a) :

Then the general solutions for g (�) and f (r) are:(g (�) = c1 cos (��) + c2 sin (��) ; 8 � 2 (0; 2�)

f (r) = C1e� ln r + C2e

�� ln r = C1r� + C2r

��; 8 r 2 (0; a) ;

where we note that the function g (�) = c1 cos (��) + c2 sin (��) is not necessarily 2�-periodic unless� = n 2 N: In such a case, we obtain the following harmonic functions for any � > 0 :

r� cos (��) ; r� sin (��) ; r�� cos (��) ; r�� sin (��) ; (390)

de�ned on (r; �) 2 (0; a)� (0; 2�) (or on (r; �) 2 (0;1)� (0; 2�)). Finally, for � = 0; we get(g (�) = c1� + c2; 8 � 2 (0; 2�) :

f (r) = C1 ln r + C2; 8 r 2 (0; a) ;

and we get the following harmonic functions:

1; ln r; �; ln r � �; (391)

de�ned on (r; �) 2 (0; a)� (0; 2�) (or on (r; �) 2 (0;1)� (0; 2�)).

Finally, since we want u (r; �) to be well-de�ned at r = 0; we need to throw away the terms Brn

and A log r in (388). Thus for each n 2 N and any constants A0; An; Bn; the following functions

un (r; �) =

(rn (An cosn� +Bn sinn�) ; n 2 N;

A0;(392)

are harmonic functions de�ned on r 2 (0; a) (or r 2 (0;1)), � 2 [0; 2�] : They are solutions of theequation (380) on r 2 (0; a) (or r 2 (0;1)), � 2 [0; 2�] : Each corresponding function un (x; y) areharmonic on Ba (0) n f(0; 0)g (or on R2n f(0; 0)g).In terms of the Euclidean coordinates (x; y) ; we have8>>>>><>>>>>:

r cos � = x; r sin � = y;

r2 cos 2� = x2 � y2; r2 sin 2� = 2xy;

r3 cos 3� = x3 � 3xy2; r3 sin 3� = �y3 + 3x2y;

r4 cos 4� = x4 � 6x2y2 + y4; r4 sin 4� = 4x3y � 4xy3;

... , etc. Thus un (x; y) is, in fact, a harmonic polynomial (with homogeneous degree n). More-over, it is smooth on the whole plane R2 and satis�es

@2un@x2

+@2un@y2

= 0 at (0; 0) : (393)

Therefore, each un (x; y) is harmonic on the whole plane R2:

To take care of the boundary condition u = h on @Ba (0) in the Dirichlet problem (378), we�rst express h on @Ba (0) as the form h = h (�) ; � 2 [0; 2�] ; and recall the following result fromFourier series theory:

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Lemma 12.4 (Fourier series result.) Assume h (�) is a 2�-periodic C1 function de�ned onR: then the following series

a02+

1Xn=1

an cosn� + bn sinn�; � 2 [0; 2�] (394)

converges absolutely and uniformly to h (�) on [0; 2�] : Here8>>><>>>:an =

1

Z 2�

0

h (�) cos (n�) d�; n = 0; 1; 2; 3; :::;

bn =1

Z 2�

0

h (�) sin (n�) d�; n = 1; 2; 3; ::::

(395)

The series (394) is called the Fourier series of the function h:

Proof. Omit. �

We now assume that h (�) is a 2�-periodic C1 function and see what we can do. We nowconsider the sum of un (r; �) for all n = 0; 1; 2; 3; :::; and get a function of the form

u (r; �) =A02+

1Xn=1

rn (An cosn� +Bn sinn�) ; (r; �) 2 [0; a)� [0; 2�] ; (396)

where now the constants A0; An; Bn are still arbitrary.

Remark 12.5 Since A0=2 and each rn (An cosn� +Bn sinn�) is harmonic on (r; �) 2 [0; a) �[0; 2�] (also on [0;1)� [0; 2�]), we "expect" that the series (396) converges to a harmonic functionon [0; a) � [0; 2�] : Moreover, if we choose the coe¢ cients A0; An and Bn suitably, we can makeu = h on @Ba (0). Both expectations can be ful�lled. See below for details.

To satisfy the boundary condition we need to require

u (a; �) =A02+

1Xn=1

an (An cosn� +Bn sinn�) = h (�) ; 8 � 2 [0; 2�] : (397)

Hence we must require8>><>>:A0 =

1

R 2�0h (') d'

An =1

�anR 2�0h (') cos (n') d'; Bn =

1

�anR 2�0h (') sin (n') d'; n 2 N:

(398)

Remark 12.6 It is impossible to have u (a; �) = h (�) for arbitrary h (�) if we only consider �nitesum in (397).

By the above, we arrive at a series of the form

u (r; �) =1

2�

Z 2�

0

h (') d'+

1Xn=1

rn

�an

Z 2�

0

h (') [cosn' cosn� + sinn' sinn�] d'

=1

2�

Z 2�

0

h (') d'+1

2�

1Xn=1

Z 2�

0

2�ra

�ncosn (� � ') � h (') d'; (399)

where (r; �) 2 [0; a)� [0; 2�] : To go on, we need the following fact:

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Lemma 12.7 The series

1 + 2

1Xn=1

�ra

�ncosn (� � ') (400)

converges absolutely on (r; �; ') 2 [0; a)� [0; 2�]� [0; 2�] and uniformly on (r; �; ') 2 [0; a� "]�[0; 2�]� [0; 2�] for any small " > 0: Moreover, we have

@k

@rk

1 + 2

1Xn=1

�ra

�ncosn (� � ')

!= 2

1Xn=1

@k

@rk

h�ra

�ncosn (� � ')

i(401)

and@k

@�k

1 + 2

1Xn=1

�ra

�ncosn (� � ')

!= 2

1Xn=1

@k

@�k

h�ra

�ncosn (� � ')

i(402)

on (r; �; ') 2 [0; a) � [0; 2�] � [0; 2�] for all k 2 N: For each k 2 N; the series on the right handside of (401) and (402) also converge absolutely on [0; a) � [0; 2�] � [0; 2�] and uniformly on[0; a� "]� [0; 2�]� [0; 2�] : Finally, we have the identity

1 + 2

1Xn=1

�ra

�ncosn (� � ') = a2 � r2

a2 � 2ar cos (� � ') + r2; 8 (r; �; ') 2 [0; a)� [0; 2�]� [0; 2�] :

(403)In particular, the function

a2 � r2a2 � 2ar cos (� � ') + r2

=a2 � jxj2

jx� zj2; (404)

where x = (r cos �; r sin �) 2 Ba (0) ; z = (a cos'; a sin') 2 @Ba (0) ; is harmonic on Ba (0) foreach �xed z 2 @Ba (0) (i.e. for each �xed ').

Remark 12.8 (Be careful.) Note that the series (400) diverges when r = a for any �; ': It hasthe form

1 + 21Xn=1

cosn (� � ')

and limn!1 cosn (� � ') does not converge to zero for any values of �; ':

Proof. This is a consequence of standard Series Theory. To prove (403), you may have to useEuler�s formula for complex number z = rei�; r 2 [0;1); � 2 [0; 2�] ; which is

zn = [r (cos � + i sin �)]n = rn (cosn� + i sinn�) ; 8 n 2 N:

Then look at the series

1 + 21Xn=1

�ra

�n[cosn (� � ') + i sinn (� � ')]

and use the identity

1 +

1Xn=1

zn =1

1� z ; 8 z 2 C with jzj < 1:

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By Lemma 12.7, one can moveP1

n=1 into the integral sign and (399) becomes

u (r; �) =1

2�

Z 2�

0

(1 + 2

1Xn=1

�ra

�ncosn (� � ')

)h (') d'

=a2 � r22�

Z 2�

0

h (')

a2 � 2ar cos (� � ') + r2d'; (r; �) 2 [0; a)� [0; 2�] : (405)

We call (405) the Poisson Integral Formula.

Remark 12.9 (Useful observation.) In case we have h (�) = cosn� or sinn�; n 2 N; then wedo not have to do computation since we have (for h (�) = cosn�)

u (r; �) =a2 � r22�

Z 2�

0

cosn'

a2 � 2ar cos (� � ') + r2d'

=

Z 2�

0

1

2�

(1 + 2

1Xm=1

�ra

�mcosm (� � ')

)!cosn'd'

=1

�ra

�n Z 2�

0

cosn (� � ') cosn'd'

=1

�ra

�n Z 2�

0

[cosn� cosn'+ sinn� sinn'] cosn'd'

=1

�ra

�ncosn� �

Z 2�

0

cos2 n'd' =�ra

�ncosn�:

Thus the function u (r; �) =�ra

�ncosn� is harmonic in Ba (0) with u (a; �) = cosn� = h (�) on

@Ba (0) :

Remark 12.10 One can also see, just from the formula (405), that one can di¤erentiate (withrespect to r or �) under the integral sign since a2�2ar cos (� � ')+ r2 > 0 for all (r; �; ') 2 [0; a)�[0; 2�]�[0; 2�] : The integral is proper and the integrand is a smooth function of (r; �) 2 [0; a)�[0; 2�] :This will also imply that u (r; �) is a C1 function on (r; �) 2 [0; a) � [0; 2�] : That is, harmonicfunctions are automatically C1 functions.

One can also write u (r; �) in a more geometric way as

u (x) =a2 � jxj2

2�a

Z 2�

0

h (')

jx� zj2ad' =

a2 � jxj2

2�a

Zjzj=a

h (z)

jx� zj2ds; x 2 Ba (0) ; (406)

wherex = (r cos �; r sin �) 2 Ba (0) ; z = (a cos'; a sin') 2 @Ba (0) ;

and the integral on the right hand side of (406) is the line integral with respect to arc lengthparameter s on @Ba (0) and we know that ds = ad':

13 Lecture notes on 2017-5-22.

The Poisson Integral Formula (??) is motivated by the fact that h 2 C1 (so that we can applythe Fourier series theory). However, to solve the Dirichlet problem (378), it su¢ ces to assumethat h 2 C0 (i.e. h is a continuous function). We have:

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Theorem 13.1 Let h be a continuous function on @Ba (0) and let

u (x) =a2 � jxj2

2�a

Zjzj=a

h (z)

jx� zj2ds; x = (x; y) 2 Ba (0) ; (407)

Then u (x) = u (x; y) is harmonic in Ba (0) and for each �xed p 2 @Ba (0) we have

limx!p; x2Ba(0)

u (x) = h (p) : (408)

That is, we have u 2 C2 (Ba (0))TC0�Ba (0)

�.

Remark 13.2 The integral in (407) is a proper integral.

Proof. In the integral (407), we have jx� zj 6= 0 for each x 2 Ba (0) and z 2 @Ba (0) : Thus theintegral is a regular integral (not an improper integral) and one can di¤erentiate (with respect tox or y) under the integral sign. Thus u (x) = u (x; y) is harmonic in Ba (0).Next we note that

a2 � jxj2

2�a

Zjzj=a

1

jx� zj2ds

=a2 � r22�

Z 2�

0

1

a2 � 2ar cos (� � ') + r2d'

=1

2�

Z 2�

0

1 + 2

1Xn=1

�ra

�ncosn (� � ')

!d' = 1; 8 x 2 Ba (0) ; (409)

where in the above we have used the identityZ 2�

0

2

1Xn=1

�ra

�ncosn (� � ')

!d' = 2

1Xn=1

��ra

�n Z 2�

0

cosn (� � ') d'�= 2

1Xn=1

0 = 0:

Hence

ju (x)� h (p)j

=

�����a2 � jxj22�a

Zjzj=a

h (z)� h (p)jx� zj2

ds

����� � a2 � jxj2

2�a

Zjzj=a

jh (z)� h (p)jjx� zj2

ds:

Since h is continuous at p; for any " > 0 there exists a small arc C (p) � @Ba (0) centered at pwith length 2� such that

jh (z)� h (p)j < " if z 2 C (p) :Now

ju (x)� h (p)j �

8><>:a2�jxj22�a

Rz2C(p)

jh(z)�h(p)jjx�zj2 ds

+a2�jxj22�a

Rz2@Ba(0)nC(p)

jh(z)�h(p)jjx�zj2 ds

� "+a2 � jxj2

2�a

Zz2@Ba(0)nC(p)

jh (z)� h (p)jjx� zj2

ds:

Now if x 2 Ba (0) and x! p 2 @Ba (0) ; we will have

a2 � jxj2

2�a! 0

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and Zz2@Ba(0)nC(p)

jh (z)� h (p)jjx� zj2

ds �Zz2@Ba(0)nC(p)

jh (x)� h (p)j��2

�2 ds � 2M��2

�2 � 2�awhere M = supz2@Ba(0) jh (z)j : Hence

limx!p; x2Ba(0)

ju (x)� h (p)j � "+ limx!p; x2Ba(0)

a2 � jxj2

2�a� 2M�

�2

�2 � 2�a!= "

But since " > 0 is arbitrary, we obtain limx!p; x2Ba(0) u (x) = h (p) : �

We can summarize the following:

Theorem 13.3 (Solution of the Dirichlet problem (378).) Consider the Dirichlet problem forthe Laplace equation on Ba (0):(

4u (x; y) = 0 in (x; y) 2 Ba (0)

u (x; y) = h (x; y) on (x; y) 2 @Ba (0) (the boundary of Ba (0) ),(410)

where h (x; y) is a given continuous function de�ned on @Ba (0) : The solution in the space

C2 (Ba (0))\

C0�Ba (0)

�is unique and is given by

u (x) =

8><>:a2 � jxj2

2�a

Zjzj=a

h (z)

jx� zj2ds; x 2 Ba (0)

h (x) ; x 2 @Ba (0) :(411)

Remark 13.4 (Important observation. ) By the representation formula (411), we see that u (x)is actually a C1 function in Ba (0) and one can di¤erentiate under the integral sign.

Example 13.5 Assume h (x; y) : @Ba (0) ! R is an odd (even) continuous function with respectto y (i.e. in the odd case we have h (x;�y) = �h (x; y) for (x; y) 2 @Ba (0) and in the even case wehave h (x;�y) = h (x; y) for (x; y) 2 @Ba (0)). Then by the Poisson Integral Formula, the solutionu (x; y) for (

4u (x; y) = 0 in (x; y) 2 Ba (0)

u (x; y) = h (x; y) on (x; y) 2 @Ba (0) (the boundary of Ba (0) ),(412)

is also an odd (even) function of y: In particular, for the odd case, we have u (x; 0) = 0 on thex-axis inside the open disc Ba (0) : To see this, note that for any � 2 [0; 2�] we have

u (r;��) = a2 � r22�

Z 2�

0

h (')

a2 � 2ar cos (�� � ') + r2d'

=a2 � r22�

Z 2�

0

h (')

a2 � 2ar cos (� + ') + r2d'

=a2 � r22�

Z 0

�2�

h (��)a2 � 2ar cos (� � �) + r2

d�; ' = ��

=a2 � r22�

Z 2�

0

h (��)a2 � 2ar cos (� � �) + r2

d�

=a2 � r22�

Z 2�

0

�h (�)a2 � 2ar cos (� � �) + r2

d� = �u (r; �) :

The proof for the even case is similar.

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Example 13.6 (Strong maximum principle for harmonic functions on the disc Ba (0) :)Assume that u (x) 2 C2 (Ba (0))

TC0�Ba (0)

�; x = (x; y) 2 R2; is harmonic on Ba (0). Let

M = maxx2Ba(0) u (x) and m = minx2Ba(0) u (x) : If there exists x0 2 Ba (0) such that u (x0) =

M or u (x0) = m; then u must be a constant function on Ba (0). To see this, by the PoissonIntegral Formula, we have

u (x) =

8><>:a2 � jxj2

2�a

Zjzj=a

h (z)

jx� zj2ds; x = (x; y) 2 Ba (0)

h (x) ; x = (x; y) 2 @Ba (0) ;

where h (x) �M for all x 2 @Ba (0) : By the identity (see (409) or use the Poisson Integral Formulato the constant harmonic function M)

0 = u (x0)�M =a2 � jx0j2

2�a

Zjzj=a

h (z)�Mjx0 � zj2

ds; where h (z)�M � 0 on @Ba (0) ;

we must have h (z) =M on @Ba (0) : By the weak maximum principle, for a harmonic functionu (x) 2 C2 (Ba (0))

TC0�Ba (0)

�we have the estimate:

min@Ba(0)

u � u (x) � max@Ba(0)

u; 8 x 2 Ba (0): (413)

Hence u (x) must be a constant function on Ba (0):

13.1 Mean value property and the maximum principle.

Theorem 13.7 (Mean value property.) Let u (x; y) be harmonic in any open disc Ba (0) �R2 centered at the origin (0; 0) with radius a > 0 and u 2 C2 (Ba (0))

TC0�Ba (0)

�. Then we have

u (0) =1

2�a

Zjzj=a

u (z) ds; (414)

that is, the value of u at the center of the disc is equal to its average on the circumference jzj = a:The integral in (414) is the line integral with respect to arc length parameter s0 on @Ba (0) ; ds =ad'; ' 2 [0; 2�] :

Remark 13.8 If the disc is Ba (x0) � R2; then we have

u (x0) =1

2�a

Zjz�x0j=a

u (z) ds:

Remark 13.9 There is also a mean value property for harmonic functions u 2 C2 (Ba (0))TC0�Ba (0)

�de�ned

on an open disc Ba (0) � Rn:

Proof. This is obvious by the representation formula (407) if we choose x = 0. �By Theorem 13.7, we know that for any open set � R2; if u 2 C2 () is harmonic on ; then

for any open disc Ba (x0) �� ; a > 0; we have the mean value formula

u (x0) =1

2�a

Zjx�x0j=a

u (x) ds (line integral on the circle @Ba (x0) ): (415)

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Lemma 13.10 (Another form of mean value property.) The mean value property of harmonicfunctions also implies the following

u (x0) =1

�a2

ZBa(x0)

u (x) dx (double integral on the disc Ba (x0) ) (416)

for any Ba (x0) �� ; a > 0:

Proof. By (415), we have

ru (x0) =1

2�

Zjx�x0j=r

u (x) ds =1

2�

Z 2�

0

u (x0 + r (cos �; sin �)) � rd�

for any Br (x0) �� ; r > 0; and if we integrate with respect to the radius r from 0 to a; we getZ a

0

ru (x0) dr =1

2�

Z a

0

�Z 2�

0

u (x0 + r (cos �; sin �)) � rd��dr

and soa2

2u (x0) =

1

2�

Z 2�

0

Z a

0

u (x0 + r (cos �; sin �)) � rdrd�;

i.e.,

u (x0) =1

�a2

ZBa(x0)

u (x) dx:

The proof is done. �

14 Lecture notes on 2017-6-5.

14.1 Gradient estimate and Liouville theorem for harmonic functions.

Lemma 14.1 (Derivatives estimate of harmonic functions.) Assume u (x; y) 2 C2 () isharmonic on � R2: Then we have����@u@x (x0; y0)

���� �or ����@u@y (x0; y0)����� � 2a max

@Ba(x0;y0)juj (417)

as long as Ba (x0; y0) �� :

Proof. First note that u 2 C1 () (note that a C2 harmonic function is automatically C1; this isan exercise problem). In particular, the function @u

@x

�or @u

@y

�is also a harmonic function on : Hence

by the mean value formula we have

@u

@x(x0; y0) =

1

�a2

ZBa(x0;y0)

@u

@x(x; y) dxdy;

where by the divergence theorem, we knowZBa(x0;y0)

@u

@x(x; y) dxdy

=

ZBa(x0;y0)

div (u (x; y) ; 0) dxdy =

Z@Ba(x0;y0)

h(u (x; y) ; 0) ; Nout (x; y)i ds:

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Therefore����@u@x (x0; y0)����

=

���� 1�a2ZBa(x0;y0)

@u

@x(x; y) dxdy

���� = ���� 1�a2Z@Ba(x0;y0)

h(u (x; y) ; 0) ; Nout (x; y)i ds����

� 1

�a2

Z@Ba(x0;y0)

jh(u (x; y) ; 0) ; Nout (x; y)ij ds �2�a

�a2max

@Ba(x0;y0)j(u (x; y) ; 0)j = 2

amax

@Ba(x0;y0)juj :

(418)

The proof is done. �

We also have:

Corollary 14.2 (Derivatives estimate of nonnegative harmonic functions.) Assume u (x; y) 2C2 () is harmonic on � R2 with u � 0 everywhere in : Then we have����@u@x (x0; y0)

���� �or ����@u@y (x0; y0)����� � 2au (x0; y0) (419)

as long as Ba (x0; y0) �� :Proof. In (418), we have����@u@x (x0; y0)

���� � 1

�a2

Z@Ba(x0;y0)

jh(u (x; y) ; 0) ; Nout (x; y)ij ds

and since u � 0 everywhere in ; we havejh(u (x; y) ; 0) ; Nout (x; y)ij � u (x; y) ; 8 (x; y) 2 @Ba (x0; y0) :

By this, we get����@u@x (x0; y0)���� � 1

�a2

Z@Ba(x0;y0)

u (x; y) ds =1

�a22�a � u (x0; y0) =

2

au (x0; y0) :

The proof is done. �Example 14.3 Let � R2 be a domain. Assume u (x; y) 2 C2 () is harmonic on � R2 withu � �13 everywhere in : Assume Ba (x0; y0) �� and u (x0; y0) = �5 (here Ba (x0; y0) is an opendisc in R2 centered at (x0; y0) with radius a > 0). Then we have the following derivatives estimate:����@u@x (x0; y0)

���� �or ����@u@y (x0; y0)����� � :

Solution: The function v (x; y) = u (x; y) + 13 is harmonic on � R2 with v � 0 everywhere in: Hence we have����@u@x (x0; y0)

���� = ����@v@x (x0; y0)���� � 2av (x0; y0) = 2

a(u (x0; y0) + 13) =

2

a(�5 + 13) = 16

a:

Theorem 14.4 (Liouville theorem of harmonic functions.) If u is harmonic on R2 and isbounded either above or below on R2; then u must be a constant function.

Remark 14.5 Note that here u (x; y) is de�ned on the whole space R2:

Proof. Without loss of generality, assume u is bounded below (if u is bounded above, we can lookat �u). By adding a large constant, we may assume u � 0 on Rn: Apply (419) to u (x; y) with = R2; a!1; to get

@u

@x(x0; y0) =

@u

@y(x0; y0) � 0; 8 (x0; y0) 2 R2:

As the point (x0; y0) is arbitrary, we are done. �

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14.2 The maximum principle.

One can use the mean value property to prove the following:

Lemma 14.6 (Strong maximum principle for harmonic functions.) Let � R2 be abounded domain (open and connected) in R2. Assume u 2 C2 ()

TC0���is a harmonic function

and there exists a point p 2 such that u (p) = max� u or u (p) = min� u; then u must be aconstant function on .

Remark 14.7 Draw a picture for this for 4u � 0; 4u � 0; 4u = 0; in the 1-dimensional case.

Proof. We �rst prove the case u (p) = max� u (call this value M). Since is an open set, one can�nd an open disc Ba (p) � for some a > 0: By Example 13.6, we know that u (x) = M for allx 2 Ba (p) : Hence the set

D = fx 2 : u (x) =Mg � is both open and closed in : It is clear that D � is open in : To see that D � is alsoclosed in ; let qn 2 D be a sequence in D with limn!1 qn = q� 2 (i.e. the sequence qn 2 D hasa limit point q� 2 ), then since u 2 C0

���; we have

M = limn!1

u (qn) = u�limn!1

qn

�= u (q�) ; q� 2 ;

which implies that q� 2 D: By de�nition, this means that D � is closed in : As is connected,the only set which is both open and closed in is either the empty set � or the whole set : SinceD is not empty (because p 2 D), we must have D = : That is, u �M on all of :The proof for the minimum case is similar since one can apply the above argument to the

harmonic function �u: �

15 Notes on p. 170: Harmonic functions are automaticallyC1 functions.

Lemma 15.1 Let � R2 be any open set and u 2 C2 () is harmonic. Then we must haveu 2 C1 () :

Proof. For any x0 2 one can �nd a small open disc Ba (x0) � and we have the identity

u (x) =a2 � jxj2

2�a

Zjx0�x0j=a

u (x0)

jx� x0j2ds0; 8 x = (x; y) 2 Ba (x0) : (420)

Since the integrand is di¤erentiable with respect to any x 2 Ba (x0) for any order and jx� x0j 6= 0 forall x0 2 @Ba (x0) ; one can di¤erentiate under the integral sign (the integral (420) is a regular integralover compact domain). Therefore, u 2 C1 (Ba (x0)) and so u 2 C1 () : �

16 Notes on p. 174: Dirichlet problem for an annulus.

Let Aa;b (0) be the open annulus centered at the origin (0; 0) with outer and inner radii b > a > 0:Wewant to solve the equation 8>><>>:

4u (x; y) = 0 in (x; y) 2 Aa;b (0) ;

u = g (�) for x2 + y2 = a2;

u = h (�) for x2 + y2 = b2:

(421)

where g (�) and h (�) are two given continuous functions de�ned on � 2 [0; 2�] :The following fact is important:

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Lemma 16.1 There exists a unique solution u 2 C2 (Aa;b (0))TC0�Aa;b (0)

�for the above Dirich-

let problem.

Proof. We will omit this. �

Again, we use polar coordinates (r; �) and �nd all possible separable functions which are har-monic in Aa;b (0) : They are:

1; log r; rn cosn�; rn sinn�; r�n cosn�; r�n sinn�; n 2 N:

Hence we consider a series of the form

u (r; �) =1

2(A0 +B0 log r) +

1Xn=1

h�Anr

n + ~Anr�n�cosn� +

�Bnr

n + ~Bnr�n�sinn�

i; (422)

where the coe¢ cients A0; An; ~An; B0; Bn; ~Bn are to be determined. We need to solve the 2 �2 system of equations:8><>:

12(A0 +B0 log a) +

P1n=1

h�Ana

n + ~Ana�n�cosn� +

�Bna

n + ~Bna�n�sinn�

i= g (�) ;

12(A0 +B0 log b) +

P1n=1

h�Anb

n + ~Anb�n�cosn� +

�Bnb

n + ~Bnb�n�sinn�

i= h (�) ;

for all � 2 [0; 2�] : Motivated by the theory of Fourier series, we need to require8<:12(A0 +B0 log a) =

12�

R 2�0g (�) d�

12(A0 +B0 log b) =

12�

R 2�0h (�) d�

and 8<: Anan + ~Ana

�n = 1�

R 2�0g (�) cos (n�) d�

Anbn + ~Anb

�n = 1�

R 2�0h (�) cos (n�) d�

and 8<: Bnan + ~Bna

�n = 1�

R 2�0g (�) sin (n�) d�

Bnbn + ~Bnb

�n = 1�

R 2�0h (�) sin (n�) d�:

One can �nd all of the coe¢ cients A0; An; ~An; B0; Bn; ~Bn uniquely. Moreover, the series (422)

will converge and represent a harmonic function in the space C2 (Aa;b (0))TC0�Aa;b (0)

�:We will

not prove this.

17 Notes on p. 174: Dirichlet problem for the exterior ofa circle.

In this case, we consider the problem on the exterior of a circle:8>><>>:4u (x; y) = 0 for x2 + y2 > a2;

u = h (�) for x2 + y2 = a2;

u is bounded as x2 + y2 !1;

(423)

where h (�) is a continuous function de�ned on � 2 [0; 2�] :

The bounded condition in (423) is necessary due to the following:

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Lemma 17.1 Without the bounded condition in (423), the solution is not unique.

Proof. We can just look at an example. Consider the problem(4u (x; y) = 0 for x2 + y2 > 1;

u = 1 for x2 + y2 = 1:

It has two solutions u � 1 and u � 1 + log r; where the second solution is not bounded asx2 + y2 !1: �Lemma 17.2 There is a unique solution u 2 C2 (a)

TC0��a�for the above Dirichlet problem

(423) (with the condition u is bounded as x2+y2 !1), where a = Ba (0)C= f(x; y) 2 R2 : x2 + y2 > a2g.

Proof. Without loss of generality, we may assume that a = 1 and h (�) � 0: Consider the Kelvintransformation ~u of u; which is de�ned as

~u (x; y) = u (p; q) ; (x; y) 2�(x; y) 2 R2 : 0 < x2 + y2 < 1

= B1 (0) n f(0; 0)g ; (424)

where

(p; q) =

�x

x2 + y2;

y

x2 + y2

�=� xr2;y

r2

�; 0 < r =

px2 + y2 < 1:

By the identities

@p

@x=y2 � x2r4

;@q

@x=�2xyr4

;@p

@y=�2xyr4

;@q

@y=x2 � y2r4

;

and (~uxx = upp (px)

2 + 2upqpxqx + uqq (qx)2

~uyy = upp (py)2 + 2upqpyqy + uqq (qy)

2 ; upp + uqq = 0

it is easy to see that ~u (x; y) is a bounded harmonic function de�ned on B1 (0) n f(0; 0)g with ~u 2C2 (B1 (0) n f(0; 0)g) and continuous up to @B1 (0) ; and

~u (x; y) = u (x; y) � 0 on @B1 (0) :

We claim that ~u (x; y) � 0 onB1 (0) n f(0; 0)g : For each �xed 0 < r < 1; letMr = max@Br(0) j~u (x; y)j :Weknow thatMr is a bounded function for 0 < r < 1: By the maximum principle, we have (note thatlogpx2 + y2 is harmonic on R2n f(0; 0)g)

� Mr

log rlogpx2 + y2 � ~u (x; y) � Mr

log rlogpx2 + y2; 8 (x; y) 2 B1 (0) nBr (0) :

Now we let r ! 0+ in the above inequalities to get ~u (x; y) = 0 (note that limr!0+Mr

log r= 0) for

each �xed (x; y) 2 B1 (0) n f(0; 0)g : The proof is done. �Remark 17.3 Draw a picture for the Kelvin transformation.

Remark 17.4 Assume that u (r; �) is harmonic on (r; �) 2 (1;1)� [0; 2�] : Using polar coordinates(r; �) ; the Kelvin transformation becomes

~u (r; �) = u

�1

r; �

�= u (s; �) ; s =

1

r> 1; 0 < r < 1:

Then one can use chain rule to show that ~u (r; �) satis�es the equation

@2~u

@r2+1

r

@~u

@r+1

r2@2~u

@�2= 0 on (r; �) 2 (0; 1)� [0; 2�] (425)

due to the identity

@2u

@s2+1

s

@u

@s+1

s2@2u

@�2= 0 on (s; �) 2 (1;1)� [0; 2�] :

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Example 17.5 (Interesting fact.) Another related fact is the following: Let the upper half-planexn � 0 be denoted as Rn+: Assume u (x1; :::; xn) 2 C2

�Rn+�T

C��Rn+�T

L1�Rn+�(i.e., bounded on

the upper half-plane) with (4u = 0 in Rn+

u = 0 on @Rn+:(426)

Then u � 0 (for the proof, see Schwarz Re�ection Principle section in elliptic-pde.tex). Withoutthe bounded condition, there is an unbounded solution u (x1; ::: ; xn) = xn: Hence we do not haveuniqueness

17.1 Finding solution using series method.

Motivated by the method in the Poisson Integral Formula, we now consider a series for the solution u (r; �) of(423) of the form

u (r; �) =A02+

1Xn=1

r�n (An cosn� +Bn sinn�) ; (r; �) 2 (a;1)� [0; 2�] ; (427)

where we know that each r�n cosn� and r�n sinn�; n 2 N; is harmonic and bounded on the region(r; �) 2 (a;1)� [0; 2�] :

Remark 17.6 Note that we have thrown away the terms rn cosn� and rn sinn�:

By the boundary condition, we hope that

u (a; �) =A02+

1Xn=1

a�n (An cosn� +Bn sinn�) = h (�) ; � 2 [0; 2�] :

With this, we choose

A0 =1

Z 2�

0

h (') d'

and

An =an

Z 2�

0

h (') cos (n') d'; Bn =an

Z 2�

0

h (') sin (n') d'; n 2 N:

Thus we arrive at a series of the form

u (r; �) =1

2�

Z 2�

0

h (') d'+1Xn=1

an

�rn

Z 2�

0

h (') [cosn' cosn� + sinn' sinn�] d'

=1

2�

Z 2�

0

h (') d'+1

2�

1Xn=1

Z 2�

0

2�ar

�ncosn (� � ') � h (') d' (428)

where (r; �) 2 (a;1)� [0; 2�] : To go on, we need the following:

Lemma 17.7 The series

1 + 21Xn=1

�ar

�ncosn (� � ') (429)

converges absolutely on (r; �; ') 2 (a;1) � [0; 2�] � [0; 2�] and uniformly on (r; �; ') 2 [a +";1)� [0; 2�]� [0; 2�] for any small " > 0: Moreover, we have

@k

@rk

1 + 2

1Xn=1

�ar

�ncosn (� � ')

!= 2

1Xn=1

@k

@rk

h�ar

�ncosn (� � ')

i(430)

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and@k

@�k

1 + 2

1Xn=1

�ar

�ncosn (� � ')

!= 2

1Xn=1

@k

@�k

h�ar

�ncosn (� � ')

i(431)

on (r; �; ') 2 (a;1) � [0; 2�] � [0; 2�] for all k 2 N: For each k 2 N; the series on the right handside of (430) and (431) also converge absolutely on (a;1) � [0; 2�] � [0; 2�] and uniformly on[a+ ";1)� [0; 2�]� [0; 2�] : Finally, we have the identity

1 + 21Xn=1

�ar

�ncosn (� � ') = r2 � a2

a2 � 2ar cos (� � ') + r2; 8 (r; �; ') 2 (a;1)� [0; 2�]� [0; 2�] :

(432)In particular, the function

r2 � a2a2 � 2ar cos (� � ') + r2

=jxj2 � a2

jx� x0j2; x 2 Ba (0)

C(433)

where x = (r cos �; r sin �) ; r > a; x0 = (a cos'; a sin') 2 @a; is harmonic on a = Ba (0)Cfor

each �xed x0 2 @a.

Remark 17.8 Same as before, the series (429) diverges for any �; '; when r = a:

By the above lemma, (428) can be written as

u (r; �)

=1

2�

Z 2�

0

(1 + 2

1Xn=1

�ar

�ncosn (� � ')

)h (') d'

=r2 � a22�

Z 2�

0

h (')

a2 � 2ar cos (� � ') + r2d'; (r; �) 2 (a;1)� [0; 2�] : (434)

We still call (434) the Poisson Integral Formula. One can rewrite it as

u (x) =jxj2 � a22�a

Zjx0j=a

h (x0)

jx� x0j2ds0; x 2 a; (435)

where the integral is a line integral with respect to arc length parameter s0 on jx0j = a (i.e., @a).Same as before, we can conclude the following:

Theorem 17.9 Consider the Dirichlet problem for the Laplace equation on the exterior of a circle:8>><>>:4u (x; y) = 0 for (x; y) 2 a;

u = h (�) for (x; y) 2 @a;

u is bounded as x2 + y2 !1;

(436)

where h (�) is a continuous function de�ned on � 2 [0; 2�] : The solution in the space

C2 (a)\

C0��a�; a = Ba (0)

C

is unique and is given by

u (x) =

8><>:jxj2 � a22�a

Zjx0j=a

h (x0)

jx� x0j2ds0; x 2 a

h (x) ; x 2 @a:(437)

For each �xed p 2 @a we havelim

x!p; x2au (x) = h (p) : (438)

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Remark 17.10 When h � 1; we actually haveZ 2�

0

1

a2 � 2ar cos (� � ') + r2d' =

2�

jr2 � a2j (439)

as long as a; r > 0 with a 6= r: The proof is left to you as an exercise. Note that the denominator isgiven by jx� x0j2 ; where x = (r cos �; r sin �) ; x0 = (a cos'; a sin') ; a 6= r: So no matter r > a orr < a; we always have Z 2�

0

1

a2 � 2ar cos (� � ') + r2d' > 0:

Thus the answer makes sense. In particular, when r = a; the integral diverges.

The following asymptotic behavior is interesting on its own.

Corollary 17.11 Assume u (r; �) satis�es (436) (with the condition u is bounded as x2+y2 !1).Then we have the limit

limr!1

u (r; �) =1

2�

Z 2�

0

h (') d'; (440)

where the convergence is uniform with respect to � 2 [0; 2�] :

Remark 17.12 In fact, (440) is consistent with the mean value formula.

Remark 17.13 (Important.) Note that to have (440), we need the condition that u is boundedas x2+ y2 !1: This is because in the series expansion of u (r; �) ; we have thrown away the termsrn cosn� and rn sinn�: See Remark 17.6.

Proof. By the Poisson Integral Formula (304), we have

limr!1

u (r; �) = limr!1

1

2�

Z 2�

0

r2 � a2a2 � 2ar cos (� � ') + r2

h (') d' =1

2�

Z 2�

0

h (') d':

The proof is done. �

Corollary 17.14 Let u 2 C2 (a)TC0��a�; a = Ba (0)

C; be the solution of the Dirichlet problem8>><>>:

4u (x; y) = 0 for (x; y) 2 a;

u = h (�) for (x; y) 2 @a

u is bounded as x2 + y2 !1:

(441)

Then its Kelvin transformation ~u; which is de�ned as

~u (x; y) = u

�a2x

x2 + y2;

a2y

x2 + y2

�; (x; y) 2 Ba (0) n f(0; 0)g ; (442)

is a harmonic function de�ned on a: Moreover, if we de�ne ~u at (0; 0) as

~u (0; 0) =1

2�

Z 2�

0

h (') d'; (443)

then ~u 2 C2 (Ba (0))TC0�Ba (0)

�is a harmonic function on Ba (0) with ~u = h on @Ba (0) :

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Proof. Same as before, we know that ~u 2 C2 (Ba (0) n f(0; 0)g) is harmonic on Ba (0) n f(0; 0)gwith ~u = u = h on @Ba (0) and continuous up to @Ba (0) ; Moreover, by (440), we know that

lim(x;y)!(0;0)

~u (x; y) =1

2�

Z 2�

0

h (') d':

Hence if we de�ne ~u (0; 0) as in (443), it is continuous at (0; 0) : Thus ~u 2 C0�Ba (0)

�:

Next let v 2 C2 (Ba (0))TC0�Ba (0)

�be the unique harmonic function on Ba (0) with v = h on

@Ba (0) : Consider the di¤erence w = ~u� v: Then w = 0 on @Ba (0) and

w 2 C2 (Ba (0) n f(0; 0)g)\

C0�Ba (0)

�is harmonic and bounded on Ba (0) n f(0; 0)g. By the proof in Lemma 17.2, we know that w � 0on Ba (0) n f(0; 0)g. As w is continuous at (0; 0) ; we have w � 0 on Ba (0): Hence ~u � v on Ba (0).The proof is done. �

Finally, we have:

Corollary 17.15 Assume u (r; �) 2 C2 (a)TC0��a�satis�es (436). Then we have mean value

formula1

2�

Z 2�

0

u (r; �) d� =1

2�

Z 2�

0

h (') d' for all r 2 (a;1) : (444)

Proof. By Corollary 17.14 and the mean value property, we have

~u (0; 0) =1

2�

Z 2�

0

~u (r cos �; r sin �) d� =1

2�

Z 2�

0

u

�a2

rcos �;

a2

rsin �

�d�

for all r 2 (0; a) : Thus we have

1

2�

Z 2�

0

h (') d' =1

2�

Z 2�

0

u (r; �) d�; 8 r 2 (a;1) :

The proof is done. �

Remark 17.16 One can also use the following argument: we haveZ 2�

0

u (r; �) d�

=1

2�

Z 2�

0

(Z 2�

0

1 + 2

1Xn=1

�ar

�ncosn (� � ')

!h (') d'

)d�

=1

2�

Z 2�

0

(Z 2�

0

1 + 2

1Xn=1

�ar

�ncosn (� � ')

!d�

)h (') d'; r 2 (a;1)

where (Z 2�

0

1 + 2

1Xn=1

�ar

�ncosn (� � ')

!d�

)= 2�

and so the integral1

2�

Z 2�

0

u (r; �) d� =1

2�

Z 2�

0

h (') d'

is independent of r 2 (a;1) :

To Be Continued

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17.2 The heat equation.

17.2.1 The fundamental solution.

The one-dimensional heat equation (nondegenerate parabolic equation) has the form

ut = kuxx; u = u (x; t) ; k > 0 is a constant, x 2 R (445)

For higher dimensional heat equation, it has the form

ut = k4 u; u = u (x1; ::: ; xn; t) ; k > 0 is a constant, (x1; ::: ; xn) 2 Rn

4 = @2

@x21+ � � �+ @2

@x2nis the Laplace operator in Rn:

Without loss of generality, we may assume that the constant k = 1: Moreover, we focus only on theone-dimensional case in this course.Unlike the wave equation utt = uxx, the heat equation ut = uxx is much more di¢ cult to solve. It

is not di¢ cult to guess some special solutions, like u (x; t) = x; 2t+x2; x3+6xt; et+x; et�x; e�t cosx; e�t sin x; et coshx; et sinh x; etc(note that et coshx and et sinh x are linear combinations of et+x and et�x). All are de�ned onR�R: One can check that the only space-time separable solutions of the heat equation are "essen-tially" of the form

u (x; t) = et+x; et�x; e�t cosx; e�t sin x

and no others. Also note that the solutions u (x; t) = x; 2t+ x2; x3 + 6xt are polynomial solutionswith u (x; 0) = x; x2; x3: There is a formula for a polynomial solution with u (x; 0) = xn for anyn 2 N:A major di¤erence between the wave equation and the heat equation is the following: For wave

equation, if u (x; t) is a solution, so is the function u (x;�t) ; but for the heat equation, if u (x; t)is a solution, the function u (x;�t) is no longer a solution (unless u (x; t) = x). Another importantproperty of the heat equation is that if u (x; t) is a solution of the heat equation, so is the function~u (x; t) = �u (�x; �2t) for any constants � 6= 0 and � 6= 0 (for the wave equation, if u (x; t) is asolution, so is the function ~u (x; t) = �u (�x; �t) for any constants � 6= 0 and � 6= 0):

18 Lecture notes on 2018-4-16.

18.0.2 The heat equation and the di¤usion; motivation from two simple physicalexamples.

Example 18.1 (Fick�s law.) See Example 4 in p. 14 of Strauss�s book.

Example 18.2 (Fourier�s law.) See Example 5 in p. 16 of Strauss�s book.

If we do not impose any "side condition" on the heat equation ut = uxx; then on R2 it hasin�nitely many solutions. To get unique solution (we hope so), we focus on the following initialvalue problem for the heat equation:(

ut (x; t) = uxx (x; t) ; (x; t) 2 R� (0;1)

u (x; 0) = f (x) ; x 2 R:(446)

Here f (x) is a given continuous function on R and we want (hope) to �nd a "unique" solutionu (x; t) 2 C2 (R� (0;1))

TC0 (R� [0;1)) : Note that, unlike the wave equation, here we do not

need the condition ut (x; 0) = g (x) for the heat equation. This is due to physical phenomenon

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(heat equation is not a mechanical equation coming from Newton�s law) and also due to the factthat if u (x; t) is C2 up to t = 0 (with f 2 C2 (R)), then we also have

ut (x; 0) = uxx (x; 0) = f 00 (x) ; x 2 R;

i.e. the condition ut (x; 0) = g (x) is automatically implied by the condition u (x; 0) = f (x) : Onthe other hand, for the case of wave equation, we can not determine ut (x; 0) from the conditionu (x; 0) (but we can determine utt (x; 0) from the condition u (x; 0) due to the identity utt (x; 0) =uxx (x; 0)).Unfortunately, the initial value problem (446) still has in�nitely many solutions (this

is unlike the wave equation, which has a unique solution once we know u (x; 0) and ut (x; 0)) unlesswe impose condition on the behavior of solution u (x; t) for large jxj. From now on, we willfocus only on some "special solution" of (446), which is the one that is closest to the physicalreality.

19 Lecture notes on 2018-4-19.

19.1 Fundamental solution of the heat equation and its properties.

Recall that for the Laplace equation (4u) (x) = 0 in Rn there is a radial solution (with asingularity at the origin of Rn; i.e. x = 0) of the form

u (x) =

(A jxj2�n +B; n > 2; where x 2 Rnn f0g

A log jxj+B; n = 2; where x 2 R2n f0g ;

where A; B are arbitrary constants. It plays an important role in the theory of Laplace equation.For the heat equation ut = 4u, we also have a radial solution (with a singularity at the origin

of Rn+1; i.e. (x; t) = (0; 0))), which plays an important role in the theory of parabolic equation.

Lemma 19.1 Let

K (x; t) =

8><>:1

(4�t)n=2e�

jxj24t ; x 2 Rn; t > 0

0; x 2 Rn; t � 0:(447)

Then K (x; t) 2 C1 (Rn+1n f(0; 0)g) (the point (0; 0) means x = 0 2 Rn; t = 0 2 R), is radial inx; and it satis�es the heat equation @tu = 4u in Rn+1n f(0; 0)g : The function K (x; t) is called thefundamental solution of the heat equation.

Remark 19.2 See advanced PDE book for the motivation of the solution (447).

Remark 19.3 The only singularity ofK in (447) is at the origin (0; 0) : Away from (0; 0) ; K (x; t) issmooth and satis�es the heat equation. We call it the fundamental solution of the heat equation.

Proof. We will prove the lemma for n = 1 only. The proof for n > 1 is similar. Since

K (0; t) =

8><>:1p4�t

; t > 0

0; t � 0;

we see that K (x; t) is not continuous at (0; 0) : Moreover, we have

limt!0+

K (0; t) = limt!0+

1p4�t

=1:

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To check that K (x; t) 2 C1 (R2n f(0; 0)g) ; it su¢ ces to look at the behavior of K (x; t) on theset S = f(x; 0) 2 R2 : x 6= 0g : By the limit

limt!0+

�1

t�e�

�t

�= 0; 8 const. �; � > 0;

one can check that K (x; t) is C1 at any point of S: Computing

@

@x

�1p4�t

e�x2

4t

�=

1p4�t

�� x2t

�e�

x2

4t ; t > 0

@2

@x2

�1p4�t

e�x2

4t

�=

1p4�t

�� 12t+x2

4t2

�e�

x2

4t ; t > 0

@

@t

�1p4�t

e�x2

4t

�=

1p4�

��12t�3=2

�e�

x2

4t +1p4�t

e�x2

4tx2

4t2; t > 0 (448)

we see that K (x; t) satis�es the heat equation on R � (0;1) : Clearly it also satis�es the heatequation on R� (�1; 0) : At any point (x0; 0) 2 S; x0 6= 0; we have Kxx (x0; 0) = 0: Also note that

limt!0�

K (x0; t)�K (x0; 0)t

= 0 (x0 6= 0)

and

limt!0+

K (x0; t)�K (x0; 0)t

= limt!0+

�1

t

1p4�t

e�x204t

�= 0 (x0 6= 0; x20 > 0),

and so we have Kt (x0; 0) = 0: The proof is done. �

Remark 19.4 (Another proof.) We can also verify that K (x; t) =�1=p4�t�e�x

2=4t satis�es theheat equation ut = uxx on (x; t) 2 R� (0;1) by the following quick method: We �rst claim that apositive function u (x; t) > 0 satis�es the heat equation ut = uxx on some open set � R2 if andonly if the function v (x; t) = log u (x; t) satis�es the equation

vt (x; t) = vxx (x; t) + v2x (x; t) ; (x; t) 2 :

To see this, assume that u > 0 satis�es the heat equation on and let v = log u. Then we have

vt =utu=uxxu; vx =

uxu; vxx =

uxxu��uxu

�2:

Hence we get vt = vxx + v2x: On the other hand, if v satis�es vt = vxx + v2x on ; then the functionu = ev satis�es

ut = evvt = ev�vxx + v2x

�; ux = evvx; uxx = evvxx + evv2x

and so ut = uxx: Now let

v = log

�1p4�t

e�x2

4t

�= �1

2log 4� � 1

2log t� x2

4t:

We have

vt = �1

2t+x2

4t2; vx = �

x

2t; vxx = �

1

2t

and so vt = vxx+v2x: Hence the function

�1=p4�t�e�x

2=4t satis�es the heat equation on R�(0;1) :

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19.1.1 Basic properties of the fundamental solution of the heat equation.

We list some important properties of the fundamental solution. For convenience, we only focus onthe case n = 1: For each property, it is also valid for the case n > 1 and the proof is similar.

Lemma 19.5 For any �xed " > 0; we have

limt!0+

K (x; t) = 0 uniformly in the region fx 2 R : jxj � "g : (449)

Alsolimjxj!1

K (x; t) = 0 uniformly in the region t 2 (0;1) (or t 2 (�1;1) ). (450)

Remark 19.6 We also have

limt!1

K (x; t) = 0 uniformly in x 2 (�1;1) . (451)

This is easy due to

jK (x; t)j =���� 1p4�t

e�x2

4t

���� � 1p4�t

for all x 2 (�1;1) ; t > 0: (452)

Remark 19.7 Explain the de�nition of "converges uniformly".

Remark 19.8 Draw a picture for K (x; t) with t! 0+:

Proof. For (449), we have for t > 0 the inequality

0 < K (x; t) =1p4�t

e�x2

4t � 1p4�t

e�"2

4t ; 8 jxj � "

and the conclusion follows. For (450), for �xed x 2 R; x 6= 0; the maximum value of the positivefunction

1p4�t

e�x2

4t ; t 2 (0;1)

over t 2 (0;1) ; is attained at the point t = x2=2 with maximum value equal to

1

jxjp2�e�

12 :

To see this, one can look at (448). Hence the result follows. �

Lemma 19.9 For each �xed t > 0; we haveZ 1

�1K (x; t) dx =

Z 1

�1

1p4�t

e�x2

4t dx = 1; t > 0: (453)

Moreover, the convergence of the integral is uniform with respect to t 2 (0; T ) for any �xed T > 0:

Remark 19.10 Draw a picture for K (x; t) showing the propertyR1�1K (x; t) dx = 1 for all t > 0:

Remark 19.11 Similarly, we have:Z 1

�1

1p4�t

e�(x�y)24t dy = 1; 8 x 2 R; t > 0 (454)

and Z 1

�1

1p4�t

e�(x�y)24t dx = 1; 8 y 2 R; t > 0: (455)

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Proof. We �rst recall the following improper integral identity from calculus:Z 1

�1e�s

2

ds =p�: (456)

By a change of variables (let s = �y + �), we haveZ 1

�1e�(�y+�)

2

dy =

p�

�; 8 � 2 R; � > 0: (457)

Letting x =p4ts; we obtainZ 1

�1

1p4�t

e�x2

4t dx =

Z 1

�1

1p4�t

e�s2p4tds =

1p�

Z 1

�1e�s

2

ds = 1:

Next, let T > 0 be a �xed time. For any " > 0; then there exists a large M > 0 (M depends onlyon " and T ) such that for all t 2 (0; T ) we have the estimate (again, let x =

p4ts)

0 <

Z 1

M

1p4�t

e�x2

4t dx =1p�

Z 1

Mp4t

e�s2

ds <1p�

Z 1

Mp4T

e�s2

ds < "; 8 t 2 (0; T ) :

The same result holds for the integralR �M�1

1p4�te�

x2

4t dx: Therefore, the convergence of the integralis uniform with respect to t 2 (0; T ) for any �xed T > 0: �

20 Lecture notes on 2018-4-26.

Lemma 20.1 For any �xed � > 0; we have

limt!0+

Zjy�xj>�

1p4�t

e�(x�y)24t dy = 0 uniformly in x 2 R; (458)

which means that the values of the fundamental solution K (x� y; t) (view it as a function of y withparameter x) concentrate around x as t! 0+:

Remark 20.2 For �xed � > 0; the above means that: for any " > 0; there exists ~" > 0 (dependingonly on ") such that if 0 < t < ~"; we have����Z

jy�xj>�

1p4�t

e�(x�y)24t dy

���� < " for all x 2 R:

Proof. Let y = x+p4ts: Then

limt!0+

Zjy�xj>�

1p4�t

e�(x�y)24t dy = lim

t!0+1p�

Zjsj>�=

p4t

e�s2

ds = 0: (459)

Note that the right hand side of (459) does not depend on x 2 R: Hence we have convergence tozero uniformly in x 2 R: The proof is done. �

Lemma 20.3 Let � (x) be a bounded function de�ned on (�1;1) and is continuous at x =x0: Then we have

lim(x;t)!(x0;0+)

Z 1

�1

1p4�t

e�(x�y)24t � (y) dy = � (x0) : (460)

In particular, we also have

limt!0+

Z 1

�1

1p4�t

e�(x0�y)

2

4t � (y) dy = � (x0) : (461)

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Remark 20.4 The above two limits have di¤erent meaning.

Proof. Letu (x; t) =

Z 1

�1

1p4�t

e�(x�y)24t � (y) dy; (x; t) 2 (�1;1)� (0;1) : (462)

For any " > 0; we choose � > 0 such that j� (y)� � (x0)j < " if jy � x0j < 2�: Let M = supR j�j : Ifjx� x0j < �; then

ju (x; t)� � (x0)j =���� 1p4�t

Z 1

�1e�

(x�y)24t (� (y)� � (x0)) dy

����� 1p

4�t

�Zjy�xj<�

e�(x�y)24t j� (y)� � (x0)j dy +

Zjy�xj��

e�(x�y)24t j� (y)� � (x0)j dy

�� 1p

4�t

�Zjy�x0j<2�

e�(x�y)24t j� (y)� � (x0)j dy + 2M

Zjy�xj��

e�jx�yj24t dy

�� "

Zjy�x0j<2�

1p4�t

e�(x�y)24t dy + 2M

Zjy�xj��

1p4�t

e�(x�y)24t dy

which, by (453) and (458), implies

ju (x; t)� � (x0)j � "+ 2M";

if t > 0 is small enough and jx� x0j < �. Hence we have

lim(x;t)!(x0;0+)

u (x; t) = � (x0)

and (460) is proved. (461) is a consequence of (460). �Lemma 20.5 Let � (y) be a continuous bounded function de�ned on (�1;1) : Then we have�

@�+�

@t�@x�

Z 1

�1

1p4�t

e�(x�y)24t � (y) dy

�(x; t) =

Z 1

�1

��@�+�

@t�@x�1p4�t

e�(x�y)24t

�(x; t; y)

�� (y) dy

(463)for all (x; t) 2 (�1;1)� (0;1) and all �; � 2 N

Sf0g :

Proof. For any �xed �; � 2 NSf0g and �xed (x0; t0) 2 (�1;1)� (0;1) the function�

@�+�

@t�@x�1p4�t

e�(x�y)24t

�(x0; t0; y) ; y 2 (�1;1)

decays exponentially in the variable y as jyj ! 1: Moreover, one can check that the integralZ 1

�1

��@�+�

@t�@x�1p4�t

e�(x�y)24t

�(x; t; y)

�� (y) dy; (x; t) is near (x0; t0)

converges uniformly for all (x; t) in some neighborhood of (x0; t0) : By standard theory in ad-vanced calculs, we have the identity (463). As the point (x0; t0) 2 (�1;1) � (0;1) is arbitrary,the proof is done. �

20.1 The maximum principle.

Assume that u (x; t) is a solution of the heat equation ut = uxx on (�1;1) � (�1;1) : Themaximum principle of the di¤usion equation says that (roughly speaking), for �xed time t0; wehave ut (x0; t0) � 0 if u (x0; t0) has a local maximum at x = x0 (so the value of u (x0; t0) will"decrease" at the moment t = t0); and ut (x0; t) � 0 if u (x0; t) has a local minimum at x = x0 (sothe value of u (x0; t0) will "increase" at the moment t = t0) (draw a picture for this). This is calledthe maximum principle of the heat equation. It matches with the physical phenomenon thatheat goes from hot points to cold points and vice versa.The maximum principle on the unbounded domain x 2 (�1;1) is more di¢ cult to describe.

We will discuss the maximum principle on bounded domains only.

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20.1.1 The maximum principle on bounded domains.

Let UT � R2 be the set given by

UT =�(x; t) 2 R2 : 0 < x < `; 0 < t � T

; `; T > 0 (464)

and assume that u = u (x; t) 2 C2 (UT )TC0��UT�satis�es the heat equation ut = uxx on UT (note

that the segment (x; T ) ; 0 < x < `; is included). Note that since u is continuous on the compactset �UT ; it has global maximum and minimum on �UT :

Remark 20.6 Explain the meaning of u 2 C2 (UT ) :

The maximum principle says the following:

Lemma 20.7 (Weak maximum principle for heat equation.) Assume u 2 C2 (UT )TC0��UT�

satis�es the heat equation ut = uxx on UT : Then we have

max�UT

u = max�T

u; (465)

where �T := �UT � UT ; which is called the parabolic boundary of UT :

Proof. Assume v 2 C2 (UT )TC0��UT�is a function such that

vxx (x; t)� vt (x; t) > 0 in UT : (466)

Then since v 2 C0��UT�; there is a point (x0; t0) 2 �UT such that v (x0; t0) = max �UT v: If (x0; t0) 2

UT with t < T; then from calculus we know that

vx (x0; t0) = 0; vxx (x0; t0) � 0; vt (x0; t0) = 0: (467)

This contradicts vxx � vt > 0 in UT :If (x; t) 2 UT with t = T; then we replace vt (x0; t0) = 0 by vt (x0; t0) � 0 in (467) and get the

same contradiction. Thus the point (x0; t0) must lie on the parabolic boundary of UT andcannot lie on UT (for v (x; t) satisfying the di¤erential inequality (466)). In such a case we have

max�UT

v = max�T

v (call this value M), where vxx (x; t)� vt (x; t) > 0 in UT ; (468)

and moreover, v (x; t) cannot attain the value M on UT :Now let v (x; t) = u (x; t)+"x2 (" > 0 is a small constant), where u 2 C2 (UT )

TC0��UT�satis�es

the heat equation on UT . We now have

vxx (x; t)� vt (x; t) = uxx (x; t) + 2"� ut (x; t) = 2" > 0 in UT :

By the above discussion, we know that

max�UT

v = max�T

v = max�T

�u (x; t) + "x2

���max�T

u (x; t)

�+ "`2

and by u (x; t) = v (x; t)� "x2 � v (x; t) ; we get

max�UT

u � max�UT

v ��max�T

u (x; t)

�+ "`2: (469)

As " > 0 is arbitrary, letting "! 0+ (note that here ` is �nite), we obtainmax �UT u � max�T u (x; t) :Onthe other hand, we also have max �UT u � max�T u (x; t) : Hence max �UT u = max�T u (x; t) : �

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Remark 20.8 (Be careful.) In the above proof, we do not exclude the possibility that the maximumof u (x; t) (note that ut = uxx) can also be attained at some point in UT : For example, when thesolution u (x; t) is a constant, then this can happen. However, this is the only case that can happen(this is the strong maximum principle).

Problem 20.9 Instead of using v (x; t) = u (x; t) + "x2; now use the function v (x; t) = u (x; t) �"t; " > 0; and repeat the same argument of proof.

We also have the following minimum principle:

Corollary 20.10 (Weak minimum principle for heat equation.) Assume u 2 C2 (UT )TC0��UT�

satis�es the heat equation ut = uxx on UT : Then we have

min�UT

u = min�T

u: (470)

Remark 20.11 Again, here we do not exclude the possibility that the minimum can be attained atsome point in UT :

Proof. The proof for the minimum case is similar by looking at �u (it also satis�es the heatequation) and the identity

max�UT(�u) = max

�T(�u)

becomes�min

�UTu = �min

�Tu:

Corollary 20.12 Assume u 2 C2 (UT )TC0��UT�satis�es the heat equation ut = uxx on UT and u �

0 on the parabolic boundary �T ; then u � 0 on �UT :

Proof. This is a consequence of the maximum-minimum principle. �

Example 20.13 (Give this as an homework problem ....) Let u (x; t) be one of the followingfunctions:

t+x2

2; et+x; et�x; e�t cosx; e�t sin x; et coshx; et sinh x; (x; t) 2 R2:

They all satisfy the heat equation ut = uxx (note that et coshx and et sinh x are linear combinationsof et+x and et�x). Let UT = (0; 1)� (0; T ]: We have8>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>:

(1) : max �UT

�t+ x2

2

�= T + 1

2; attained at (1; T ) 2 �T := �UT � UT

(2) : max �UT (et+x) = eT+1; attained at (1; T ) 2 �T

(3) : max �UT (et�x) = eT�0; attained at (0; T ) 2 �T

(4) : max �UT (e�t cosx) = e�0 cos 0 = 1; attained at (0; 0) 2 �T

(5) : max �UT (e�t sin x) = e�0 sin 1 = sin 1; attained at (1; 0) 2 �T

(6) : max �UT (et coshx) = eT cosh 1; attained at (1; T ) 2 �T

(7) : max �UT (et sinh x) = eT sinh 1; attained at (1; T ) 2 �T :

From the above, we see that each solution attains its maximum point on the parabolic boundary �T :Also note that the maximum can be attained at any corner point (there are four of them) of �T :

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One can also use Energy Method (integral method) to prove the following (without using themaximum principle):

Lemma 20.14 Assume u 2 C2��UT�satis�es the heat equation ut = uxx on UT and u � 0 on the

parabolic boundary �T ; then u � 0 on �UT :

Proof. Let E (t) ; 0 � t � T; be the quantity

E (t) =1

2

Z `

0

u2 (x; t) dx � 0; 0 � t � T:

Then E (t) is a di¤erentiable function on [0; T ] ; E (0) = 0; and we have

dE

dt(t) =

Z `

0

u (x; t)ut (x; t) dx =

Z `

0

u (x; t)uxx (x; t) dx

=

Z `

0

��d

dx[u (x; t)ux (x; t)]

�� (ux (x; t))2

�dx

= [u (x; t)ux (x; t)]jx=`x=0 �Z `

0

(ux (x; t))2 dx = �

Z `

0

(ux (x; t))2 dx � 0: (471)

Hence we have0 � E (t) � E (0) = 0; 8 t 2 [0; T ]: (472)

Thus E (t) = 0 for all time t 2 [0; T ] and so u � 0 on �UT : �

21 Lecture notes on 2018-4-30.

To Be Continued

21.1 Uniqueness and stability of the heat equation on bounded do-mains.

21.1.1 Uniqueness.

Consider the Dirichlet problem for the heat equation:8><>:ut (x; t)� uxx (x; t) = f (x; t) for 0 < x < `; t > 0

u (x; 0) = � (x) ; 0 � x � `u (0; t) = g (t) ; u (`; t) = h (t) ; t � 0;

(473)

where we assume that(f 2 C0 ((0; `)� (0;1)) ; � 2 C0 [0; `] ; g; h 2 C0[0;1)

� (0) = g (0) ; � (`) = h (0) :(474)

Lemma 21.1 Assume u1; u2 2 C2 ((0; `)� (0;1))TC0 ([0; `]� [0;1)) satis�es the Dirichlet prob-

lem (473). Then we must have u1 (x; t) � u2 (x; t) on [0; `]� [0;1):

Proof. For any T > 0; apply the maximum principle to the function w = u1 � u2 on the domainUT given by (464) to get the conclusion. �

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21.1.2 Stability.

There are two types of stability for the heat equation. One is due to the maximum principle; theother is due to energy estimate.

Lemma 21.2 Assume u1; u2 2 C2 ((0; `)� (0;1))TC0 ([0; `]� [0;1)) satis�es the Dirichlet prob-

lem (473) with

u1 (x; 0) = �1 (x) 2 C0 [0; `] ; u2 (x; 0) = �2 (x) 2 C0 [0; `] :

Then we havemax

(x;t)2[0;`]�[0;1)ju1 (x; t)� u2 (x; t)j = max

x2[0;`]j�1 (x)� �2 (x)j : (475)

Remark 21.3 Since we assume that u1; u2 2 C2 ((0; `)� (0;1))TC0 ([0; `]� [0;1)) ; we must

have�1 (0) = g (0) ; �1 (`) = h (0)

and�2 (0) = g (0) ; �2 (`) = h (0) :

Proof. For any T > 0; apply the maximum principle to the function w = u1 � u2 on the domainUT given by (464) to get

max(x;t)2[0;`]�[0;T ]

ju1 (x; t)� u2 (x; t)j = maxx2[0;`]

j�1 (x)� �2 (x)j :

As T > 0 is arbitrary, we get the conclusion. �

Lemma 21.4 Assume u1; u2 2 C2 ([0; `]� [0;1)) satis�es the Dirichlet problem (473) with

u1 (x; 0) = �1 (x) 2 C0 [0; `] ; u2 (x; 0) = �2 (x) 2 C0 [0; `] :

Then we haveZ `

0

ju1 (x; t)� u2 (x; t)j2 dx �Z `

0

j�1 (x)� �2 (x)j2 dx; 8 t 2 [0;1): (476)

Proof. Let w = u1 � u2 on the domain UT : Similar to (471) and (472), we now have E 0 (t) �0 on [0;1) and so

0 � E (t) =1

2

Z `

0

w2 (x; t) dx � E (0) =1

2

Z `

0

j�1 (x)� �2 (x)j2 dx; 8 t 2 [0;1):

As T > 0 is arbitrary, we get the conclusion. �

22 Lecture notes on 2017-4-10 and 2017-4-13.

22.1 Some calculus results (this will be covered by TA in recitationclass).

To study the solution of a heat equation de�ned on the whole space x 2 R we need some propertieson the change of order in di¤erentiation and improper integrals.Let I � R be an arbitrary interval and consider the improper integralZ 1

�1f (x; y) dy; x 2 I; (477)

107

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where we assume f (x; y) is a continuous function on its domain I � (�1;1). If both integralsZ 1

0

f (x; y) dy;

Z 0

�1f (x; y) dy

converge for each x 2 I; then we say (477) converges on I

De�nition 22.1 Assume (477) converges on I: We say it converges uniformly on I if for any" > 0; there exists some number M > 0 (depending only on ") such that����Z

fjxj�Mgf (x; y) dy

���� = ����Z 1

�1f (x; y) dy �

Z M

�Mf (x; y) dy

���� < "; 8 x 2 I: (478)

Remark 22.2 (478) implies����Z 1

M

f (x; y) dy +

Z �M

�1f (x; y) dy

���� < "; 8 x 2 I: (479)

Remark 22.3 The same de�nition can be applied to the improper integralZ 1

0

f (x; y) dy; x 2 I:

Lemma 22.4 (Weierstrass M-test.) Let I � R be an interval and f 2 C0 (I � [0;1)) : Assume thereexists a function M (y) : [0;1)! R such that

R10jM (y)j dy converges and we also have

jf (x; y)j � jM (y)j ; 8 x 2 I; y 2 [0;1)

then Z 1

0

f (x; y) dy

converges uniformly on x 2 I:

Proof. Since the integralR10jM (y)j dy is �nite, there exist real numbersA > 0 such that

R1AjM (y)j dy <

": Hence we have����Z 1

A

f (x; y) dy

���� � Z 1

A

jf (x; y)j dy �Z 1

A

jM (y)j dy < "; 8 x 2 I:

By de�nition, the integralR10f (x; y) dy converges uniformly on x 2 I: �

Example 22.5 Consider the continuous function f (x; y) = e�xy on (0;1)� [0;1): The integral

F (x) :=

Z 1

0

f (x; y) dy =

Z 1

0

e�xydy =1

x; x 2 (0;1)

converges uniformly to 1=x on the interval [c;1) for any �xed constant c > 0: However, it does notconverge uniformly to 1=x on the interval (0; c) : To see this, note that for any �xed R > 0 we have����Z 1

R

e�xydy

���� = ����Z R

0

e�xydy � 1x

���� = e�xR

x; x 2 (0; c) ; where lim

x!0+e�xR

x=1:

On the other hand, one can see that the integralZ 1

0

e�(x+1)ydy =1

x+ 1; x 2 (0;1)

does converges uniformly to 1= (x+ 1) on the interval (0;1) :

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Lemma 22.6 Let I � R be an arbitrary open interval. Suppose f 2 C0 (I � (�1;1)) andR1�1 f (x; y) dy converges uniformly on I; then the function

F (x) =

Z 1

�1f (x; y) dy; x 2 I (480)

is continuous on I:

Remark 22.7 The uniform convergence condition in Lemma 22.6 is necessary. However, it is pos-sible that

R1�1 f (x; y) dy does not converge uniformly on I; but the function F (x) is still continuous

on I: For example:R10e�xydy = 1

xon I = (0;1) :

Proof. Fix x0 2 I and for x1 2 I close to x0; we have (note that I is open) for any �xedM > 0; thefollowing:

F (x1)� F (x0)

=

Zfjyj�Mg

[f (x1; y)� f (x0; y)] dy +Zfjyj�Mg

[f (x1; y)� f (x0; y)] dy| {z } := I + II|{z};where we know that I ! 0 if x1 ! x0 (since the integral

Rfjxj�Mg is proper and we can use a known

result). For II; by the de�nition of uniform convergence, for given " > 0; we have

II =

������Zfjyj�Mg

[f (x1; y)� f (x0; y)] dy| {z }������

�����Zfjyj�Mg

f (x1; y) dy

����+ ����Zfjyj�Mg

f (x0; y) dy

���� < 2�; 8 x1; x0 2 I

if M is large enough (this M depends only on "). By the above, if M > 0 is large enough we havethe estimate

jF (x1)� F (x0)j � jIj+ 2"; 8 x1; x0 2 Iand so

limx1!x0

jF (x1)� F (x0)j � limx1!x0

jIj+ 2" = 2":

As " > 0 is arbitrary, we have limx1!x0 jF (x1)� F (x0)j = 0: The proof is done. �

Example 22.8 Let f (x; y) = xe�xy 2 C0 ([0;1)� [0;1)) : We �rst note that the function

F (x) :=

Z 1

0

xe�xydy =

(1; x 2 (0;1)

0; x = 0

is not continuous on [0;1): We have����Z 1

R

xe�xydy

���� = ����Z R

0

xe�xydy � F (x)���� =

(e�xR; x 2 (0;1)

0; x = 0:

The integralR10xe�xydy does not converge uniformly on [0; c) due to the limit limx!0 e

�xR = 1 forany R > 0; but it converges uniformly on [c;1) for any �xed constant c > 0. The limiting functionis not continuous on [0;1) due to this non-uniform convergence.

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Lemma 22.9 Let I � R be an arbitrary open interval. Suppose f 2 C0 (I � (�1;1)) andR1�1 f (x; y) dy converges (does not have to be uniform) on I: If

@f@x2 C0 (I � (�1;1)) andR1

�1@f@x(x; y) dy converges uniformly on I; then F (x) =

R1�1 f (x; y) dy; x 2 I; is di¤erentiable

on I with

F 0 (x) =

Z 1

�1

@f

@x(x; y) dy; 8 x 2 I: (481)

In particular, the function F (x) =R1�1 f (x; y) dy is continuous on I (since a di¤erentiable function

is continuous).

Remark 22.10 The uniform convergence condition in Lemma 22.9 is necessary.

Remark 22.11 Note that the function F 0 (x) is continuous on I due to Lemma 22.6. Hence F (x)is a C1 function on I:

Proof. First by Lemma 22.6, we know that the function

G (x) :=

Z 1

�1

@f

@x(x; y) dy; x 2 I

is continuous on I: We have

F 0 (x)�Z 1

�1

@f

@x(x; y) dy

= limh!0

1

h

�Z 1

�1[f (x+ h; y)� f (x; y)] dy

��Z 1

�1

@f

@x(x; y) dy

and by mean value theorem the above is equal to

limh!0

Z 1

�1

�@f

@x(s (x+ h) + (1� s)x; y)� @f

@x(x; y)

�dy; s 2 (0; 1) :

Since the function G (x) is continuous on I; the above limit is zero and we obtain the identity(481). �

23 Lecture notes on 2017-4-17.

23.1 Heat equation on the whole line with initial condition.

We now consider the following initial value problem for heat equation de�ned on the whole line:(ut = uxx; x 2 (�1;1) ; t 2 (0;1)

u (x; 0) = � (x) ; x 2 (�1;1) :(482)

Here � (x) is a given continuous function on (�1;1) and we want to �nd a solution

u (x; t) 2 C2 ((�1;1)� (0;1))\

C0 ((�1;1)� [0;1)) (483)

satisfying (482).

Remark 23.1 Unlike wave equation (which is a mechanical equation governed by Newton�s lawF = ma), the heat equation with initial conditions of the form

u (x; 0) = � (x) ; ut (x; 0) = (x) ; x 2 (�1;1) (484)

is not well-posed (for arbitrary � (x) and (x)). As we will see soon, it has too many conditions.

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Lemma 23.2 Let � (y) be a continuous bounded function de�ned on y 2 (�1;1) : Then theintegral

u (x; t) :=

Z 1

�1

1p4�t

e�(x�y)24t � (y) dy; (x; t) 2 (�1;1)� (0;1) (485)

converges for each (x; t) 2 (�1;1)� (0;1) : Moreover, we have u 2 C1 ((�1;1)� (0;1)) and

ut (x; t) = uxx (x; t) ; 8 (x; t) 2 (�1;1)� (0;1) : (486)

We also have the identity

@m+n

@tm@xn

�Z 1

�1

1p4�t

e�(x�y)24t � (y) dy

�=

Z 1

�1

@m+nu

@tm@xn

�1p4�t

e�(x�y)24t

�� (y) dy (487)

for all (x; t) 2 (�1;1)� (0;1) ; for any m; n 2 NSf0g :

Remark 23.3 The idea to show (486) is: one can di¤erentiate under the integral sign.

Proof. Since � is a bounded function on R, for each �xed (x0; t0) 2 (�1;1)� (0;1) ; the integral(let y = x�

p4ts) Z 1

�1

1p4�t

e�(x�y)24t � (y) dy =

1p�

Z 1

�1e�s

2

��x�p4ts�ds (488)

converges uniformly for all (x; t) lying on the open rectangle R centered at (x0; t0) :

R =

�(x; t) : x0 � 1 < x < x0 + 1;

t02< t <

3t02

�:

By

@

@x

�1p4�t

e�(x�y)24t

�=

1p4�t

��x� y

2t

�e�

(x�y)24t ;

@

@t

�1p4�t

e�(x�y)24t

�=

@2

@x2

�1p4�t

e�(x�y)24t

�=

1p4�t

� 12t+(x� y)2

4t2

!e�

(x�y)24t ;

the integral Z 1

�1

1p4�t

��x� y

2t

�e�

(x�y)24t � (y) dy;

which is equal to

1p�

Z 1

�1

�1ptse�s

2

���x�p4ts�ds

�� Cp

tmax j�j

Z 1

�1jsj e�s2ds

�; (489)

and the integral Z 1

�1

1p4�t

� 12t+(x� y)2

4t2

!e�

(x�y)24t � (y) dy;

which is equal to1p�

Z 1

�1

�� 12te�s

2

+1

ts2e�s

2

���x�p4ts�ds; (490)

both also converges uniformly for all (x; t) 2 R: By Lemma 22.9, we have on the rectangle R thefollowing identities:

@u

@t=

@

@t

Z 1

�1

1p4�t

e�(x�y)24t � (y) dy =

Z 1

�1

@

@t

�1p4�t

e�(x�y)24t

�� (y) dy

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and@2u

@x2=

@2

@x2

Z 1

�1

1p4�t

e�(x�y)24t � (y) dy =

Z 1

�1

@2

@x2

�1p4�t

e�(x�y)24t

�� (y) dy:

This implies ut = uxx on the rectangle R and so on (x; t) 2 (�1;1) � (0;1) : Note thatut; ux and uxx are all continuous functions on R due to uniform convergence and so on (�1;1)�(0;1) and so u 2 C2 (R) : Moreover, one can check that for any m; n 2 N

Sf0g we have

@m+nu

@tm@xn=

Z 1

�1

@m+nu

@tm@xn

�1p4�t

e�(x�y)24t

�� (y) dy (491)

and the integral in (491) also converges uniformly for all (x; t) 2 R. By Lemma 22.9 again,we have u 2 C1 (R) : Since the point (x0; t0) 2 (�1;1) � (0;1) is arbitrary, we have u 2C1 ((�1;1)� (0;1)) : The lemma is proved.

We now can conclude the main theorem for the initial value problem of the heat equation onthe real line:

Theorem 23.4 Assume � (x) is a continuous bounded function de�ned on (�1;1). Then thefunction

u (x; t) =

8<:R1�1

1p4�te�

(x�y)24t � (y) dy; x 2 (�1;1) ; t 2 (0;1)

� (x) ; t = 0(492)

belongs to C1 (R� (0;1))TC0 (R� [0;1)) (continuous up to t = 0) and satis�es the initial

value problem (ut (x; t) = uxx (x; t) ; x 2 (�1;1) ; t 2 (0;1)

u (x; 0) = � (x) ; x 2 (�1;1) :(493)

Remark 23.5 (Important.) As long as t > 0; u (x; t) becomes a smooth function even if theinitial data � (x) is only a continuous function. We call this a smoothing e¤ect of the heat equation.This is unlike the wave equation, which has no smoothing e¤ect.

Corollary 23.6 (The maximum principle.) The solution u (x; t) given by (492), where � (x) isa continuous bounded function de�ned on (�1;1) ; satis�es the maximum principle:

infR� � u (x; t) � sup

R� for all x 2 (�1;1) ; t 2 (0;1) : (494)

Proof. We have

u (x; t) =

Z 1

�1

1p4�t

e�(x�y)24t � (y)|{z} dy �

Z 1

�1

1p4�t

e�(x�y)24t sup

R�| {z } dy

= supR� �Z 1

�1

1p4�t

e�(x�y)24t dy = sup

R�

and similarly u (x; t) � infR �: �

Corollary 23.7 (In�nite speed of propagation of the heat equation.) Let � (x) be a contin-uous bounded function de�ned on (�1;1) : Assume � (x) � 0 everywhere, has compact support,and � 6� 0. Then the solution u (x; t) given by (492) satis�es

u (x; t) > 0; 8 x 2 (�1;1) ; t 2 (0;1) ; (495)

i.e., as long as time is positive, u (x; t) is positive everywhere no matter how large jxj is (that is whywe say the equation has in�nite speed of propagation).

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Remark 23.8 This is di¤erent from the wave equation. The function u (x; t) = � (x� t) satis�esthe wave equation utt = uxx with u (x; 0) = � (x) : However, for t > 0; u (x; t) = 0 if jxj > 0 is largeenough.

Proof. Since � is not a zero function, we have � (x0) > 0 for some x0 2 (�1;1) : By continuity,� > 0 on (x0 � "; x0 + ") for some " > 0: Now at any (x; t) 2 (�1;1)� (0;1) ; we have

u (x; t) =

Z 1

�1

1p4�t

e�(x�y)24t � (y) dy �

Z x0+"

x0�"

1p4�t

e�(x�y)24t � (y) dy > 0:

The proof is done. �

Lemma 23.9 (Tonelli�s theorem.) Let � (x; y) be a continuous "nonnegative" function de-�ned on R2 = (�1;1)� (�1;1) : Then the �niteness of any one of the following three integrals:Z 1

�1

�Z 1

�1� (x; y) dx

�dy;

Z 1

�1

�Z 1

�1� (x; y) dy

�dx;

ZZR2� (x; y) dxdy

implies that of the other two. Moreover, their values are all equal.

Remark 23.10 The condition � (x; y) � 0 on R2 is essential.

Proof. We omit it. �

24 Lecture notes on 2017-????.

24.0.1 Polynomial solutions of the 1-dimensional heat equation.

Consider the 1-dimensional heat equation ut (x; t)� uxx (x; t) = 0 with initial data

u (x; 0) = p0 (x) ; x 2 (�1;1) ; (496)

where p0 (x) is a polynomial de�ned on x 2 (�1;1) with degree n 2 N: We try to look for aspace-time polynomial solution u (x; t) of the heat equation of the form

u (x; t) = p0 (x) + p1 (x) t+ p2 (x) t2 + p3 (x) t

3 + � � �;

where each pi (x) is also a polynomial in x 2 (�1;1) : We compute

ut (x; t) = p1 (x) + 2p2 (x) t+ 3p3 (x) t2 + � � �

anduxx (x; t) = p000 (x) + p001 (x) t+ p002 (x) t

2 + p003 (x) t3 + � � �;

and by comparing the coe¢ cient functions, we require8>>>>>>>>><>>>>>>>>>:

p1 (x) = p000 (x) ;

p2 (x) =12p001 (x) =

12p00000 (x) ;

p3 (x) =13p002 (x) =

13!p(6)0 (x) ;

� � �pk (x) =

13p002 (x) =

1k!p(2k)0 (x) ;

� � �:

(497)

Since p0 (x) is a polynomial with �nite degree n 2 N; the above process will stop at some k:Moreover, wesee that all of the other polynomials p1 (x) ; p2 (x) ; p3 (x) ; ::::; are all uniquely determined by

113

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p0 (x) ; where we have u (x; 0) = p0 (x) ; i.e. p0 (x) is the initial condition of the heat equa-tion. This is an important property of the heat equation ut = uxx de�ned on the entire spacex 2 (�1;1) ; i.e. the solution is uniquely determined by the initial condition (undersome suitable reasonable assumption on the behavior of the solution at x = �1).We look at some simple examples.

Example 24.1 Take p0 (x) = x: Then p1 (x) = p000 (x) = 0 and so on. The function u (x; t) = x isa polynomial solution.

Example 24.2 Take p0 (x) = x2: Then p1 (x) = p000 (x) = 2 and p2 (x) = 0 and so on. The function

u (x; t) = p0 (x) + p1 (x) t = x2 + 2t (498)

is a polynomial solution.

Example 24.3 Take p0 (x) = x3: Then p1 (x) = p000 (x) = 6x and p2 (x) = 0 and so on. The function

u (x; t) = p0 (x) + p1 (x) t = x3 + 6xt (499)

is a polynomial solution.

Example 24.4 Take p0 (x) = x4: Then p1 (x) = p000 (x) = 12x2 and p2 (x) = 12 and p3 (x) = 0 and

so on. The function

u (x; t) = p0 (x) + p1 (x) t+ p2 (x) t2 = x4 + 12x2t+ 12t2 (500)

is a polynomial solution.

Example 24.5 Take p0 (x) = x5: Then p1 (x) = p000 (x) = 20x3 and p2 (x) = 60x and p3 (x) = 0 and

so on. The function

u (x; t) = p0 (x) + p1 (x) t+ p2 (x) t2 = x5 + 20x3t+ 60xt2 (501)

is a polynomial solution.

Remark 24.6 In all of the above examples, note that t is like x2 (so that each term has the samedegree !!).

Remark 24.7 (Important.) If we use the representation formula

u (x; t) =1p4�t

Z 1

�1e�

(x�y)24t p0 (y) dy; t > 0; (502)

we will get the same answer. Moreover, we can allow t 2 (�1;1) : Note that the integral (502)converges for any polynomial p0 (y) : Moreover, di¤erentiation can move into the integral sign.

25 Lecture notes on 2017-4-20.

Corollary 25.1 (Conservation of total energy.) Let � (x) be a continuous bounded functionde�ned on (�1;1) (� (x) may not be nonnegative). Assume

R1�1 j� (x)j dx converges. Then the

solution u (x; t) given by (492) satis�esZ 1

�1u (x; t) dx =

Z 1

�1� (x) dx; 8 t 2 (0;1) : (503)

This means that the total energy (heat) is conserved.

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Proof. For each x 2 (�1;1) ; let �+ (x) = max f� (x) ; 0g � 0 and �� (x) = �min f� (x) ; 0g �0: Then we have

� (x) = �+ (x)� �� (x) ; j� (x)j = �+ (x) + �� (x) ; 8 x 2 (�1;1) :

The convergence ofR1�1 j� (x)j dx implies that of

R1�1 �

+ (x) dx andR1�1 �

� (x) dx: Also, since � (x)is a bounded function, for each �xed (x; t) 2 (�1;1)� (0;1) ; the three improper integralsZ 1

�1

1p4�t

e�(x�y)24t � (y) dy;

Z 1

�1

1p4�t

e�(x�y)24t �+ (y) dy;

Z 1

�1

1p4�t

e�(x�y)24t �� (y) dy

all converge. Now we haveZ 1

�1u (x; t) dx =

Z 1

�1

�Z 1

�1

1p4�t

e�(x�y)24t � (y) dy

�dx

=

Z 1

�1

�Z 1

�1

1p4�t

e�(x�y)24t �+ (y) dy �

Z 1

�1

1p4�t

e�(x�y)24t �� (y) dy

�dx (504)

and by Lemma 23.9, we haveZ 1

�1

�Z 1

�1

1p4�t

e�(x�y)24t �+ (y) dy

�dx

=

Z 1

�1

24�+ (y)Z 1

�1

1p4�t

e�(x�y)24t dx| {z }

35 dy = Z 1

�1�+ (y) dy <1

and similarly Z 1

�1

�Z 1

�1

1p4�t

e�(x�y)24t �� (y) dy

�dx =

Z 1

�1�� (y) dy <1:

Therefore, the two iterated integrals in (504) converge and we concludeZ 1

�1u (x; t) dx =

Z 1

�1�+ (y) dy �

Z 1

�1�� (y) dy =

Z 1

�1� (x) dx; 8 t 2 (0;1) :

The proof is done. �

25.1 Discontinuous bounded initial data.

What happens if the initial condition � (x) is a bounded function de�ned on (�1;1) but discon-tinuous somewhere (here we assume that � (x) is discontinuous only at a �nite number of points)?We have:

Lemma 25.2 Let � (x) be a bounded function de�ned on (�1;1) and at x = x0 it is discon-tinuous and satis�es

limx!x+0

� (x) = A; limx!x�0

� (x) = B; where A 6= B (505)

Then the function

u (x; t) =

Z 1

�1

1p4�t

e�(x�y)24t � (y) dy; x 2 (�1;1) ; t 2 (0;1)

lies in the space u 2 C1 ((�1;1)� (0;1)) and satis�es the heat equation

ut (x; t) = uxx (x; t) ; 8 (x; t) 2 (�1;1)� (0;1) (506)

with

limt!0+

u (x0; t) = limt!0+

Z 1

�1

1p4�t

e�(x0�y)

2

4t � (y) dy =A+B

2: (507)

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Remark 25.3 (Be careful.) In general, the limit

lim(x;t)!(x0;0+)

u (x; t) (508)

does not exist (see Example 25.4 below). On the other hand, if � (x) is continuous at x = x0; thenwe have

lim(x;t)!(x0;0+)

u (x; t) = limt!0+

u (x0; t) = � (x0) : (509)

Proof. It su¢ ces to verify (507). Let M = supR j�j and let

u (x; t) =

Z 1

�1

1p4�t

e�(x�y)24t � (y) dy; (x; t) 2 (�1;1)� (0;1) : (510)

Then (let y = x0 +p4ts)

u (x0; t) =1p�

Z 1

0

e�s2

��x0 +

p4ts�ds+

1p�

Z 0

�1e�s

2

��x0 +

p4ts�ds: (511)

For any " > 0; there exists � > 0 such that if x 2 (x0; x0 + �) ; then j� (x)� Aj < ": Hence the �rstintegral in (511) satis�es���� 1p�

Z 1

0

e�s2

��x0 +

p4ts�ds� A

2

���� = ���� 1p�Z 1

0

e�s2h��x0 +

p4ts�� A

ids

����� 1p

Z �=p4t

0

e�s2�����x0 +p4ts�� A��� ds+ 1p

Z 1

�=p4t

e�s2�����x0 +p4ts�� A��� ds

� "

2+ 2M � 1p

Z 1

�=p4t

e�s2

ds

and so

limt!0+

1p�

Z 1

0

e�s2

��x0 +

p4ts�ds =

A

2:

Similarly, we have

limt!0+

1p�

Z 0

�1e�s

2

��x0 +

p4ts�ds =

B

2:

The proof is done. �

Example 25.4 Let

� (x) =

(1; x > 0

0; x < 0;; � (x) is not continuous at x = 0:

It is a bounded function. De�ne the function

u (x; t) =

Z 1

�1

1p4�t

e�(x�y)24t � (y) dy =

Z 1

0

1p4�t

e�(x�y)24t dy

and let y = x�p4ts to get (we will get the same result if we let y = x+

p4ts)

u (x; t) =1p�

Z x=p4t

�1e�s

2

ds =1p�

Z 0

�1+

Z x=p4t

0

!e�s

2

ds

=1

2+

1p�

Z x=p4t

0

e�s2

ds; (x; t) 2 (�1;1)� (0;1) :

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We note that8>>>>>>>>>>>>><>>>>>>>>>>>>>:

u (x; t) 2 C1 ((�1;1)� (0;1)) and ut = uxx on (�1;1)� (0;1)

lim(x;t)!(x0;0+)

u (x; t) = 1; if x0 > 0

lim(x;t)!(x0;0+)

u (x; t) = 0; if x0 < 0

limt!0+

u (0; t) = 12= 1+0

2;

lim(x;t)!(0;0+)

u (x; t) = lim(x;t)!(0;0+)

�12+ 1p

R x=p4t0

e�s2ds�does not exist.

Example 25.5 Let the initial data � (x) be

� (x) =

(e�x; x 2 (0;1)

0; x 2 (�1; 0) :

� is bounded but not continuous at x = 0: Now we have

u (x; t) =

Z 1

�1

1p4�t

e�(x�y)24t � (y) dy

=

Z 1

0

1p4�t

e�x2�2xy+y2+4ty

4t dy =

Z 1

0

1p4�t

e�[y+(2t�x)]2

4t et�xdy (let y = x� 2t+p4ts)

=1p�et�x

Z 1

2t�xp4t

e�s2

ds; (x; t) 2 (�1;1)� (0;1) :

It satis�es ut (x; t) = uxx (x; t) on (�1;1)� (0;1) and8>>>>>>>><>>>>>>>>:

lim(x;t)!(x0;0+)

�1p�et�x

R12t�xp4t

e�s2ds

�= 1p

�e�x0

R1�1 e

�s2ds = e�x0 ; if x0 > 0

lim(x;t)!(x0;0+)

�1p�et�x

R12t�xp4t

e�s2ds

�= 1p

�e�x0

R11 e�s

2ds = 0; if x0 < 0

limt!0+

u (0; t) = limt!0+

�1p�etR1

2tp4t

e�s2ds

�= 1p

R10e�s

2ds = 1

2:

26 Lecture notes on 2017-4-24.

26.1 Some unbounded initial data.

For some special unbounded continuous initial data � (x) ; we can still use the integral (510) to �ndthe answer. We look at two examples.

Example 26.1 For some special initial data � (x) ; one can evaluate the integralZ 1

�1

1p4�t

e�(x�y)24t � (y) dy (512)

explicitly and it can be well-de�ned at t = 0 (or even for t < 0): For example, when � (x) =e�x; x 2 (�1;1) ; which is continuous but not bounded on (�1;1) : However, the integral in(510) still converges. To see this, we haveZ 1

�1

1p4�t

e�(x�y)24t � (y) dy =

Z 1

�1

1p4�t

e�x2�2xy+y2+4ty

4t dy;

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where by

x2 � 2xy + y2 + 4ty = y2 + 2 (2t� x) y + x2 = [y + (2t� x)]2 � 4t (t� x)

we havee�

x2�2xy+y2+4ty4t = e�

[y+(2t�x)]24t et�x:

Now let y = x� 2t+p4ts to get

u (x; t) =1p�et�x

Z 1

�1e�

[y+(2t�x)]24t ds =

1p�et�x

Z 1

�1e�s

2

ds = et�x: (513)

The function u (x; t) = et�x is de�ned on (�1;1)�(�1;1) with ut (x; t) = uxx (x; t) on (�1;1)�(0;1) and

u (x; 0) = � (x) = e�x; x 2 (�1;1) :However, we need t > 0 in the integral (513) due to the term 1=

p4�t:

Example 26.2 In the case � (x) = Me�jxj2

; where M > 0 and � > 0 are constants, we can �ndu (x; t) explicitly by (510). For simplicity, let M = 1 and compute

1p4�t

Z 1

�1e�

(x�y)24t � (y) dy =

1p4�t

Z 1

�1e�

(x�y)2�4t�y24t dy

where by

� (x� y)2 � 4t�y24t

= �(1� 4�t) y2 � 2xy + x2

4t= �

�p1� 4�ty � xp

1�4�t

�2+ x2 �

�xp1�4�t

�24t

= � p

1� 4�ty � xp1�4�t

2pt

!2+

�x2

1� 4�t

we get

1p4�t

Z 1

�1e�

(x�y)2�4t�y24t dy =

1p4�t

e�x2

1�4�t

Z 1

�1exp

��p

1� 4�t2pt

y � x

2ptp1� 4�t

�2!dy

and use the identity Z 1

�1e�(�y+�)

2

dy =

p�

�; 8 � 2 R; � > 0

we �nd1p4�t

Z 1

�1e�

(x�y)24t � (y) dy =

1p4�t

e�x2

1�4�t2ptp�p

1� 4�t=

1p1� 4�t

e�x2

1�4�t : (514)

Note that (514) (and the above calculation) is valid only if we restrict t 2 (0; 1=4�): Moreover

lim(x;t)!(x0;0+)

1p1� 4�t

e�x2

1�4�t = e�x20 = � (x0) ; 8 x0 2 (�1;1) :

Note that the explicit solution

u (x; t) =1p

1� 4�te

�x2

1�4�t (515)

is de�ned on (�1;1)� (�1; 1=4�) with ut (x; t) = uxx (x; t) on (�1;1)� (�1; 1=4�) and

u (x; 0) = � (x) = e�x2

; x 2 (�1;1) :

However, we need t > 0 in the integral (514) due to the term 1=p4�t: We call (515) a backward

fundamental solution of the heat equation.

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26.2 The behavior of the solution when jxj is large.Lemma 26.3 Assume � (x) is a continuous bounded function de�ned on (�1;1) with

limjxj!1

� (x) = 0 (516)

and u (x; t) is given by

u (x; t) =

Z 1

�1

1p4�t

e�(x�y)24t � (y) dy; x 2 (�1;1) ; t 2 (0;1) : (517)

Then for each t > 0 it satis�eslimjxj!1

u (x; t) = 0: (518)

Remark 26.4 The condition (516) is necessary. Otherwise, we can take the example � (x) �1; which is continuous and bounded but it does not satisfy (516). Then the solution u (x; t) � 1 doesnot satisfy (518).

Proof. Let M = supR j�j : We have for each �xed t > 0 that

ju (x; t)j �Z 1

�1

1p4�t

e�(x�y)24t j� (y)j dy

� max[x��;x+�]

j�j �Z x+�

x��

1p4�t

e�(x�y)24t dy +M

ZR�[x��;x+�]

1p4�t

e�(x�y)24t dy

� max[x��;x+�]

j�j+ 2Mp4�t

Z 1

e�r2

4t dr

for any x 2 (�1;1) and any � > 0: Note thatZ 1

e�r2

4t rn�1dr ! 0 as � !1

and so we can choose � large such that

ju (x; t)j � max[x��;x+�]

j�j+ ": (519)

Finally for such chosen �; we know thatmax[x��;x+�] j�j ! 0 if jxj ! 1:Hence we get limjxj!1 u (x; t) =0: �

26.3 The relation between the heat equation and the wave equation.

Lemma 26.5 Assume that v (x; t) is a bounded function and satis�es the wave equation vtt =vxx on R2: Then the function

u (x; t) =1p4�t

Z 1

�1e�

s2

4t v (x; s) ds; x 2 R; t 2 (0;1) ; (520)

satis�es the heat equation on R� (0;1) with

lim(�;t)!(x;0+)

u (�; t) = v (x; 0) for all x 2 (�1;1) :

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Proof. The function v (x; t) has the form v (x; t) = F (x+ t) + G (x� t) for some C2 boundedfunction F (�) ; G (�) de�ned on R and so

u (x; t) =1p4�t

Z 1

�1e�

s2

4t v (x; s) ds

=1p4�t

Z 1

�1e�

s2

4t [F (x+ s) +G (x� s)] ds := I + II:

We note that

I =1p4�t

Z 1

�1e�

s2

4tF (x+ s) ds =1p4�t

Z 1

�1e�

(x�y)24t F (y) dy (let y = x+ s)

and

II =1p4�t

Z 1

�1e�

s2

4tG (x� s) ds = 1p4�t

Z 1

�1e�

(x�y)24t G (y) dy (let y = x� s).

Hence we have ut = uxx on R � (0;1) and lim(�;t)!(x;0+) u (�; t) = F (x) + G (x) = v (x; 0) forall x 2 R: The proof is done. �

27 Lecture notes on 2017-4-27 and 2017-5-1.

27.1 Heat equation with a source.

Let f (x; t) be a function de�ned on (�1;1)�(0;1) :We now consider the following heat equationwith a source: (

ut � uxx = f (x; t) ; (x; t) 2 (�1;1)� (0;1)

u (x; 0) = 0; x 2 (�1;1) ;(521)

where the initial condition is understood in the following way:

lim(�;t)!(x;0+)

u (�; t) = 0; 8 x 2 (�1;1) : (522)

Remark 27.1 If we know how to solve (521), then we know how to solve the more general problem(ut � uxx = f (x; t) ; (x; t) 2 (�1;1)� (0;1)

u (x; 0) = � (x) ; x 2 (�1;1) :(523)

In this case the solution u (x; t) of (523) is given by

u (x; t) = v (x; t) + w (x; t) ;

where v (x; t) solves (521) and w (x; t) solves(ut � uxx = 0; (x; t) 2 (�1;1)� (0;1)

u (x; 0) = � (x) ; x 2 (�1;1) :(524)

There is a solution formula for the problem (521) if we impose some condition on the functionf (x; t) : We assume that(

f (x; t) 2 C2 ((�1;1)� [0;1))

jf (x; t) ; fx (x; t) ; fxx (x; t) ; ft (x; t)j �M for all (x; t) 2 (�1;1)� [0;1)(525)

for some constant M:

Remark 27.2 Note that for heat equation, it is not well-posed for t < 0: Hence we only have tofocus on t � 0:

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27.1.1 Motivation by Duhamel�s principle.

First example. Consider the ODE for a time function u (t; s) with a parameter s 2 [0;1) :8<:du

dt(t; s) + ku (t; s) = 0; t 2 (s;1); k is a constant

u (s; s) = f (s) ; t = s:

(526)

We know its solution is given by

u (t; s) = e�k(t�s)f (s) ; 8 t 2 [s;1): (527)

Now if we integrate the above solution u (t; s) with respect to s from s = 0 to s = t; we get

U (t) :=

Z t

0

u (t; s) ds =

Z t

0

e�k(t�s)f (s) ds; t > 0; (528)

then U (t) will satisfy U (0) = 0 and

U 0 (t) = e�k(t�t)f (t) +

Z t

0

@

@t[e�k(t�s)f (s)]ds

= f (t)� kZ t

0

e�k(t�s)f (s) ds = f (t)� kU (t) ; t > 0;

i.e., 8<:dU

dt(t) + kU (t) = f (t) ; t 2 (0;1)

U (0) = 0:

(529)

That is: if we can solve the homogeneous problem (526), then we can solve the nonhomogeneousproblem (529). Note that the position of the function f in (526) has been switched in (529). Thisis called Duhamel�s principle.

Second example. Recall that the fundamental solution of the heat equation is given by

K (x; t) =

8><>:1p4�t

e�x2

4t ; (x; t) 2 (�1;1)� (0;1)

0; x 2 (�1;1) ; t � 0:

Let s � 0 be a parameter and let u (x; t; s) be a solution to the initial value problem(ut (x; t; s)� uxx (x; t; s) = 0 in (�1;1)� (s;1)

u (x; s; s) = f (x; s) on (�1;1)� ft = sg :(530)

Clearly (530) has a solution given by the integral

u (x; t; s) =

Z 1

�1K (x� y; t� s) f (y; s) dy; (x; t) 2 (�1;1)� (s;1) ; s � 0:

Duhamel�s principle says that the function (for each �xed t > 0; we integrate s from 0 to t)

U (x; t) =

Z t

0

u (x; t; s) ds =

Z t

0

Z 1

�1K (x� y; t� s) f (y; s) dyds; (x; t) 2 (�1;1)� (0;1)

(531)

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will be a solution to the problem (521). Formally we can easily see that U (x; t) =R t0u (x; t; s) ds

satis�es (521) since

Ut (x; t) = u (x; t; t) +

Z t

0

ut (x; t; s) ds = f (x; t) +

Z t

0

ut (x; t; s) ds

Uxx (x; t) =

Z t

0

uxx (x; t; s) ds =

Z t

0

ut (x; t; s) ds

and so (Ut (x; t)� Uxx (x; t) = f (x; t) ; (x; t) 2 (�1;1)� (0;1)

U (x; 0) =R 00u (x; 0; s) ds = 0; x 2 (�1;1) :

(532)

27.1.2 The solution of the heat equation with a source.

We will prove that, under the assumption (525), the above Duhamel�s principle can be veri�edrigorously. We have:

Theorem 27.3 De�ne u (x; t) as

u (x; t) =

Z t

0

Z 1

�1K (x� y; t� s) f (y; s) dyds

=

Z t

0

Z 1

�1

1p4� (t� s)

e�(x�y)24(t�s) f (y; s) dyds; (x; t) 2 (�1;1)� (0;1) : (533)

Then under assumption (525) we have

(1) : u 2 C2 ((�1;1)� (0;1))

(2) : ut (x; t)� uxx (x; t) = f (x; t) in (�1;1)� (0;1)

(3) : limt!0+

u (x; t) = 0 uniformly in x 2 (�1;1) :(534)

Remark 27.4 The uniform convergence (3) in (534) also implies

lim(x;t)!(x0;0+)

u (x; t) = 0; 8 x0 2 (�1;1) : (535)

Remark 27.5 (Important.) We can check that the integralZ t

0

Z 1

�1

1p4� (t� s)

e�(x�y)24(t�s) f (y; s) dyds; 0 < e�

(x�y)24(t�s) � 1

converges for each (x; t) 2 (�1;1)� (0;1) due to jf (y; s)j �M and the identityZ 1

�1

1p4� (t� s)

e�(x�y)24(t�s) dy = 1 for each s 2 (0; t) : (536)

In fact, we have Z t

0

Z 1

�1

����� 1p4� (t� s)

e�(x�y)24(t�s) f (y; s)

����� dyds�Z t

0

M �

Z 1

�1

1p4� (t� s)

e�(x�y)24(t�s) dy

!ds �

Z t

0

Mds =Mt: (537)

Hence, the double integral (533) converges. Moreover, due to (537), the improper integralR t0is

actually more like a proper integral.

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Remark 27.6 Note that

@

@tK (x� y; t� s) = 1p

4� (t� s)

� 1

2 (t� s) +(x� y)2

4 (t� s)2

!e�

(x�y)24(t�s)

and then for �xed (x; t) 2 (�1;1)� (0;1) we haveZ t

0

Z 1

�1

�@

@tK (x� y; t� s)

�f (y; s) dyds

=

Z t

0

Z 1

�1

"1p

4� (t� s)

� 1

2 (t� s) +(x� y)2

4 (t� s)2

!e�

(x�y)24(t�s)

#f (y; s) dy

!ds: (538)

The double integral (538) is an improper integral. In the �rst integral the domain of integrationis unbounded, while in the second integral, the integrand blows up when s is close to t: Withoutputting extra assumption on f (x; t) ; it is not clear whether we can commute di¤erentiation @t (or@x; @xx) and the double integral

R t0

R1�1 :

Proof. Since K (x; t) has a singularity at (0; 0) and the integral is in space-time regioncontaining the singularity (0; 0), we can not do di¤erentiation under the integral signdirectly (compare with the proof of the homogeneous initial value problem).First, one can do change of variables to get

u (x; t) =

Z t

0

Z 1

�1K (x� y; t� s) f (y; s) dyds

=

Z t

0

Z 1

�1K (y; s) f (x� y; t� s) dyds; (x; t) 2 (�1;1)� (0;1) :

With this, when we do di¤erentiation on u (x; t) ; it will hit on the function f (x� y; t� s) insteadof on the fundamental solution K (x� y; t� s) :Recall that our main assumption on f is the following:(

f (x; t) 2 C2 ((�1;1)� [0;1))

jf (x; t) ; fx (x; t) ; fxx (x; t) ; ft (x; t)j �M for all (x; t) 2 (�1;1)� [0;1)(539)

for some constant M: By (539), the function

P (x; t; s) :=

Z 1

�1K (y; s) f (x� y; t� s) dy

=

Z 1

�1

1p4�s

e�y2

4s f (x� y; t� s) dy; (x; t; s) 2 (�1;1)� (0;1)� (0; t]

satis�es the following:

(1) : P (x; t; s) is continuous with respect to (x; t; s) on (�1;1)� (0;1)� (0; t] with

jP (x; t; s)j =����Z 1

�1K (y; s) f (x� y; t� s) dy

���� �M; 8 (x; t; s) 2 (�1;1)�(0;1)�(0; t]: (540)

(2) : P (x; t; s) is di¤erentiable with respect to t 2 (0;1) on (�1;1)� (0;1)� (0; t] with

Pt (x; t; s) =

Z 1

�1K (y; s) ft (x� y; t� s) dy;

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andjPt (x; t; s)j �M; 8 (x; t; s) 2 (�1;1)� (0;1)� (0; t]: (541)

Moreover, Pt (x; t; s) is continuous on (�1;1)� (0;1)� (0; t]:

(3) : P (x; t; s) is di¤erentiable with respect to x 2 (�1;1) (once and twice) on (�1;1) �(0;1)� (0; t] with

Px (x; t; s) =

Z 1

�1K (y; s) fx (x� y; t� s) dy; Pxx (x; t; s) =

Z 1

�1K (y; s) fxx (x� y; t� s) dy

(542)and

jPx (x; t; s)j �M; jPxx (x; t; s)j �M; 8 (x; t; s) 2 (�1;1)� (0;1)� (0; t]: (543)

Moreover, Px (x; t; s) and Pxx (x; t; s) are both continuous on (�1;1)� (0;1)� (0; t]:

(4) : Due to (540), (541) and (543), the four improper integrals (with respect to s 2 (0; t])Z t

0

P (x; t; s) ds;

Z t

0

Pt (x; t; s) ds;

Z t

0

Px (x; t; s) ds;

Z t

0

Pxx (x; t; s) ds (544)

are actually like proper integrals, even if they look like improper integrals when s � 0 due tothe function K (y; s) : Also note that when s � t > 0; every function behaves well since K (y; s) issmooth at s = t and f (x� y; t� s) is C2 up to s = t:

(5) : By (4) ; we now have for (x; t) 2 (�1;1)� (0;1) the following identities (one can useLemma 27.7 below to prove (545)):8<:

@@t

R t0P (x; t; s) ds = P (x; t; t) +

R t0Pt (x; t; s) ds

@@x

R t0P (x; t; s) ds =

R t0Px (x; t; s) ds;

@2

@x2

R t0P (x; t; s) ds =

R t0Pxx (x; t; s) ds:

(545)

By (5) ; we have for (x; t) 2 (�1;1)� (0;1) :

ut (x; t) =

Z 1

�1K (y; t) f (x� y; 0) dy +

Z t

0

Z 1

�1K (y; s) ft (x� y; t� s) dyds (546)

and

ux (x; t) =

Z t

0

Z 1

�1K (y; s) fx (x� y; t� s) dyds (547)

and

uxx (x; t) =

Z t

0

Z 1

�1K (y; s) fxx (x� y; t� s) dyds: (548)

In particular, we see that u 2 C2 ((�1;1)� (0;1)) since f 2 C2 ((�1;1)� [0;1)) :Compute

ut (x; t)� uxx (x; t)

=

8<:R1�1K (y; t) f (x� y; 0) dy

+R t0

R1�1K (y; s) (ft � fxx) (x� y; t� s) dyds

=

8>>><>>>:R "0

R1�1K (y; s) (�fs � fyy) (x� y; t� s) dyds

+R t"

R1�1K (y; s) (�fs � fyy) (x� y; t� s) dyds

+R1�1K (y; t) f (x� y; 0) dy

:= I" + J" + II:

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Now

jI"j � 2MZ "

0

Z 1

�1K (y; s) dyds = 2M" (549)

and by the assumption (539) and integration by parts (with respect to dy), we �nd

J" =

Z t

"

Z 1

�1�Kyy (y; s) f (x� y; t� s) dyds�

Z t

"

Z 1

�1K (y; s) fs (x� y; t� s) dyds

=

8<:R t"

R1�1 [�Kyy (y; s) +Ks (y; s)] f (x� y; t� s) dyds

�R t"

R1�1 @s [K (y; s) f (x� y; t� s)] dyds

=

Z 1

�1K (y; ") � f (x� y; t� ") dy �

Z 1

�1K (y; t) f (x� y; 0) dy

=

Z 1

�1K (y; ") f (x� y; t� ") dy � II (550)

since (Ks �Kyy) (y; s) = 0 (the principle is: we must use the equation for the fundamental solutionsomewhere !!): We conclude that

ut (x; t)� uxx (x; t) = I" +

Z 1

�1K (y; ") f (x� y; t� ") dy: (551)

By (549) and letting "! 0; we obtain

ut (x; t)� uxx (x; t) = f (x; t) in (�1;1)� (0;1) : (552)

Finally note that for t > 0

supx2(�1;1)

ju (x; t)j = supx2(�1;1)

����Z t

0

Z 1

�1K (y; s) f (x� y; t� s) dyds

���� ; t > 0

�M

Z t

0

Z 1

�1K (y; s) dyds =Mt! 0 as t! 0: (553)

The proof is done. �

Some calculus fact again. We recall the following calculus fact: Let a < b; c < d be four�nite constants. If f (x; y) and @f

@x(x; y) are both in C0 ([a; b]� [c; d]) ; then the function F (x) =R d

cf (x; y) dy is di¤erentiable with respect to x 2 [a; b] and satis�es

dF

dx(x) =

d

dx

�Z d

c

f (x; y) dy

�=

Z d

c

@f

@x(x; y) dy for all x 2 [a; b] : (554)

In particular, dFdx(x) is continuous on [a; b] ; i.e. F (x) 2 C1 [a; b] : Now a slight variant of the above

fact is:

Lemma 27.7 If f (x; y) and @f@x(x; y) are both in C0 ((a; b)� (c; d))

TL1 ((a; b)� (c; d)) ; then the

function F (x) is di¤erentiable with respect to x 2 (a; b) and satis�es (554). Moreover, we haveF (x) 2 C1 (a; b) :

Remark 27.8 A function g (x; y) lies in the space L1 ((a; b)� (c; d)) means that g (x; y) is abounded function de�ned on (a; b)� (c; d) :

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Proof. Since both f (x; y) and @f@x(x; y) are bounded functions on (a; b)� (c; d) (denote the bound

of jf (x; y)j and��@f@x(x; y)

�� by M), the two integralsZ d

c

f (x; y) dy;

Z d

c

@f

@x(x; y) dy

converge for each �xed x 2 (a; b) : For �xed x 2 (a; b) ; choose number x0 > 0 such that the compactinterval [x; x+ x0] � (a; b) : For small h > 0 with [x; x+ h] � [x; x+ x0] ; by the mean value theoremwe have

F (x+ h)� F (x)h

�Z d

c

@f

@x(x; y) dy

=

Z d

c

�f (x+ h; y)� f (x; y)

h� @f

@x(x; y)

�dy =

Z d

c

�@f

@x(��; y)�

@f

@x(x; y)

�dy;

where �� is a number depending on h; x; y with �� 2 [x; x+ h] : Hence����Z d

c

�f (x+ h; y)� f (x; y)

h� @f

@x(x; y)

�dy

�����Z d

c

����@f@x (��; y)� @f

@x(x; y)

���� dy := I + II + III;

where

I =

Z c+~"

c

����@f@x (��; y)� @f

@x(x; y)

���� dy; II =

Z d�~"

c+~"

����@f@x (��; y)� @f

@x(x; y)

���� dyIII =

Z d

d�~"

����@f@x (��; y)� @f

@x(x; y)

���� dy:Here ~" > 0 is some small positive number and we use it to decompose the open interval (c; d) into(c; c+ ~")

S[c+ ~"; d� ~"]

S(d� ~"; d) :

As the function @f@xis continuous on the compact interval [x; x+ x0]�[c+ ~"; d� ~"] ; by continuous

function theory, it is uniformly continuous on [x; x+ x0]� [c+ ~"; d� ~"] : For given " > 0; thereexists some � > 0; such that as long as the above h > 0 satis�es 0 < h < �; we have

II =

Z d�~"

c+~"

����@f@x (��; y)� @f

@x(x; y)

���� dy � Z d�~"

c+~"

"dy = " (d� c� 2~") :

We also haveI � 2M ~"; III � 2M ~"

and soI + II + III � " (d� c� 2~") + 4M ~";

which implies

limh!0+

F (x+ h)� F (x)h

=

Z d

c

@f

@x(x; y) dy; x 2 (a; b)

Similarly we have

limh!0�

F (x+ h)� F (x)h

=

Z d

c

@f

@x(x; y) dy; x 2 (a; b) :

Thus the function F (x) is di¤erentiable with respect to x 2 (a; b) and satis�es

dF

dx(x) =

d

dx

�Z d

c

f (x; y) dy

�=

Z d

c

@f

@x(x; y) dy; 8 x 2 (a; b) :

By the same argument, one can see thatR dc@f@x(x; y) dy is a continuous function on (a; b) : Hence

F (x) 2 C1 (a; b) and the proof is done. �

126


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