apdes pan2008 - unknown

Chapter 1

Introduction to PDEs

1.1 Basic concepts and Examples

The di!erential equations are the class of equations involving derivatives of unknown func-tions. When the unknown function in the equation depends only on a single variable,the equation involves only the ordinary derivatives of the unknown function, we call it anordinary di!erential eqution, or in short ODE. Quite often, the unknown functions dependon several independent variables, and the equations involving the partial derivatives of theunknown functions, then the equations are called the partial di!erential equations, or inshort PDEs. In this course, the independent variables will always be formed by a timevariable t ! 0 and a space variable x " Rn, with n the spatial dimensions. In general, aPDE of the unknown function u(x1, x2, · · · , xn) takes the form

F (x, u, Du, ux1x1 , ux1x2 , · · · , uxnxn , · · · ) = 0, (1.1)

where x = (x1, x2, · · · , xn), Du = (ux1 , ux2 , · · · , uxn), and F is a function of the indepen-dent variable x and the unknown function u and finitely many partial derivatives of u. Theequation (1.1) is called m-th order, if the highest order of the derivatives of u in (1.1) ism. An m-th order PDE of u(x1, x2, · · · , xn) has a general form

F (x, u, Du, D2u, · · · , Dmu) = 0. (1.2)

Examples of Partial di!erential Equations:

(a) Heat equation ut = uxx

(b) Linear transport eqution ut + cux = 0, c " R

(c) Wave equation utt # c2uxx = 0, c > 0

(d) Burgers’ equation ut + uux = 0

(e) Viscous Burgers’ equation ut + uux = uxx

5

6 CHAPTER 1. INTRODUCTION TO PDES

(f) Laplace equation uxx + uyy + uzz = 0

We now consider the equation (1.2). In addition to its order, we also classify theequations by the relations between function F and u, as well as the derivatives of u. Whenthe dependence of F on u and its derivatives is linear, we say (1.2) is linear, otherwise wesay it is nonlinear. In the above examples, the (d) and (e) are nonlinear while the othersare linear. In the class of nonlinear PDEs, if F is linear with respect to the highest (m-th)derivatives of u, i.e., the coe!cients of the m-th order derivatives of u in (1.2) only dependon the independent variables x, u, and the derivatives of u up to (m # 1)-th order, wecall the equation is quasi-liner. In the quasi-linear case, if the coe!cients of the highestderivatives of u are functions of independent variable x only, the equation is then calledsemi-linear. The general linear PDE of m-th order takes the following form:

P (x, D)u = f(x),

where, P (x, D) is a general m-th order di"erential operator defined in (1.8) below. Theequation is called homogeneous if f(x) = 0.

We now introduce the concept of solution. Again, consider the equation (1.2), and werestrict ourself on the domain x " # $ Rn. u = !(x) is a solution of (1.2) on #, if !(x) andall of its derivatives appears in (1.2) are continuous in #, and after substituting it into theequation, it makes (1.2) an identity. Such a solution, we will refer it as a classical solution.

The following theorem gives an important properties for solutions to a linear PDE.

Theorem 1.1.1 (Principle of superposition) Let u1 and u2 be solutions to a givenhomogeneous linear PDE . Then for any constants " and µ,

"u1 + µu2

is also a solution of that equation.

1.2 Well-posed problems

Just as what happened in ODEs, a PDE, if solvable, often has many solutions. However, asmost equations have their physical or practical backgrounds, some conditions or constraintsare necessary to determine a unique solution to the realistic problem. In general, there aretwo classes of “side” conditions: the initial condition and the boundary conditions. Someproblems involve a mixture of initial and boundary conditions, called initial-boundary valueproblems.

We first start with the case of evolutionary PDEs, (or in which a time variable isinvolved). In this case, the initial conditions often involve the initial value of the unknownfunction, and the derivatives up to the next order of the highest time derivatives of theunknown function. The initial conditions (also called initial data) together with the PDEform an intial value problem, called Cauchy problem. The following example shows a typicalCauchy problem for a wave equation in one spatial dimension:

1.2. WELL-POSED PROBLEMS 7

Example 1.2.1 !utt # c2uxx = 0,

u(x, 0) = !(x), ut(x, 0) = #(x).(1.3)

When the problem is confined in a given domain, the constraints on the boundaryare often needed for the problem. There are many di"erent kinds of boundary conditionsdepending on the realistic applications. In this course, we will mainly discuss three typicalboundary conditions. The first one is to give the value of the unknown function itself atthe boundary of the domain. This type of boundary condition is called Dirichlet condition.The corresponding boundary value problem is thus called Dirichlet problem. The followingexample is a typical Dirichlet problem for a Poisson equation, describing the electronicfield, with u the electronic potential and $ the electronic density distribution.

Example 1.2.2 !$u = #4%$(x, y, z), (x, y, z) " # $ R3,

u(x, y, z) = !(x, y, z), (x, y, z) " &#.(1.4)

Here $ is the well-known Laplace operator. In Rn, it takes the form

$ =n"

i

&2i .

The second type is often called the Neumann condition, which assigns the normal derivativeof the unknow function at the boundary. The third type is the Robin condition, which givesthe nontrivial linear combination of the unknown function itself and its normal derivativeat the boundary. The next example shows a Neumann problem of Laplace equation in 3dimension.

Example 1.2.3 !$(um) = 0, (x, y, z) " # $ R3,

%u • ' = !(x, y, z), (x, y, z) " &#,

where ' is the outer unit normal to &#.

The following example is an initial-boundary value problem for heat equation involvingthe Robin boundary condition.

Example 1.2.4

#$%

$&

ut #$u = 0, (x, y, z) " # $ R3, t > 0,

u# 12%u • ' = !(x, y, z), (x, y, xz) " &#, t > 0,

u(x, y, z, 0) = u0(x, y, z), (x, y, z) " #.


As the problems we encounter often arise from real applications, we thus expect eachof our problem is uniquely solvable and the solution continuously depends on the relevantdata. In this case, we say our problem is well-posed. However, it is very delicate to tellwhether a problem is well-posed without any further techniques. The following example isdue to J. Hadamard.

Example 1.2.5 Consider the set of “initial-value” problems in the upper half-plane in R2,for n = 1, 2, · · · ,

#%

&$u = 0, y > 0,

u(x, 0) = 0, uy(x, 0) =sin(nx)

n, x " R,

(1.5)

and !$u = 0, y > 0,

u(x, 0) = 0, uy(x, 0) = 0, x " R.(1.6)

The problem (1.5) has a unique solution

un(x, y) =1

n2sinh(ny) sin(nx),

and the problem (1.6) has a unique soluton

u0(x, y) = 0.

We note that, as n &', the data of (1.5) tends uniformly to the data of (1.6). However,one has

limn!"

sup |un(x, y)# u0(x, y)| = ', y > 0

for each (x, y). Thus, arbitrarily small changes in the data lead to large changes in thesolution, this is called instability.

1.2.1 Characteristic and initial value problems

We see from last example that the Cauchy problem for a Laplace equation is not well-posed.This section is devoted to the study of how to set the initial value problem in a properway so that the initial data is compatible with the structure of the PDE. The notion ofcharacteristic plays an essential role in this context.

Let’s consider a rather general linear PDE of order m:

P (x, D)u = g(x), (1.7)

whereP (x, D) =

"

|!|=m

a!(x)D! +"

!<m

a!(x)D! + a(x). (1.8)

1.2. WELL-POSED PROBLEMS 9

Here x = (x0, x1, · · · , xn) " Rn+1, ( is a multi-index, ( = ((0, (1, · · · , (n), where each(i is a non-negative integer. |(| =

'(i, is the length of (. D = (D0, D1, · · · , Dn) is a

di"erential operator where Di = ""xi

, and

D! =&|!|

&x!00 · · · &x!n

n

.

We also define x! = x!00 · · ·x!n

n . We call

"

|!|=m

a!(x)D!

the principal part of the operator P (x, D).From the principle of superposition, we know that the solutions of (1.7) can be obtained

from a particular solution of (1.7) and a general solution of the following homogeneousequation:

P (x, D) = 0. (1.9)

Assume that equation (1.9) is defined in a neighborhood of a smooth n-dimensional surfaceS given by f(x) = 0. The initial value problem (or, Cauchy problem) for (1.9) consists ofassigning u and its derivatives of order ( (m# 1) on S, and it is required to solve (1.9) ina neighborhood of S. For instance, in Example 1.2.1, the initial surface is S = t = 0, andthe initial conditions are u(x, 0) = !(x), and ut(x, 0) = #(x), and we shall try to solve theequation

utt # c2uxx = 0

in the upper half plane {t > 0}.We now proceed to solve (1.9) as follows. First, we shall make proper change of co-

ordinates in a neighborhood of S to distinguish the time-like direction so that we knowin which direction we will solve our problem. This can be done by introducing the newindependent variables y0, y1, · · · , yn , where y1, · · · , yn are independent coordinates on S,and y0 = f(x); i.e., S corresponds to y0 = 0. Note that u is given on S, all the derivativesof u with respect to the new variables y1, · · · , yn on S.

By chain rule, we have, for ( = ((1, · · · , (n), |(| = m,

D!u =&!u

&x!00 · · · &x!n

n

=&mu

&ym0

(&y0

&x0

)!0

· · ·(

&y0

&xn

)!n

+ · · · ,

where the last dots represent derivatives of u with respect to y0 of orders < m, togetherwith derivatives of u with respect to yi, i = 1, · · · , n, and are thus all known quatities. Theequation (1.9) becomes

"

|!|=m

a!(x)

(&y0

&x0

)!0

· · ·(

&y0

&xn

)!n &mu

&ym0

+ · · · , (1.10)


where the dots are quantities known on S. Now it is clear that, if we want to solve this

equation for&mu

&ym0

, it is necessary and su!cient that

"

|!|=m

a!(x)

(&y0

&x0

)!0

· · ·(

&y0

&xn

)!n

)= 0,

or equivalently, that

"

|!|=m

a!(x)(%f(x))! ="

|!|=m

a!(x)

(&f(x)

&x0

)!0

· · ·(

&f(x)

&xn

)!n

)= 0,*x " S. (1.11)

When (1.11) fails to hold, the initial-value problem would be unreasonable, and in this casewe say S is a characteristic surface of P (x, D). Formally, we have the following defintion.

Definition 1.2.6 The surface S = {x : f(x) = 0} is said to be characteristic at a pointp " S for the operator P (x, D) defined in (1.8) if

"

|!|=m

a!(x)(%f(x))!|x=p = 0.

S is a characteristic surface for P (x, D) if it is characteristic at each point of S. Theequation "

|!|=m

a!(x))! = 0, (1.12)

with ) = ()0, )1, · · · , )n), is called the characteristic equation for the operator P (x, D) in(1.8).

By these terms, we know that a surface S is characteristic at p " S, for the operator(1.8), provided that the normal vector to S at p satisfies the characteristic equation (1.12).We remark that if f(x) = 0 is a characteristic surface for the operator (1.8), (1.11) showsthat the di"erential equation (1.9) imposes an additional restrictions on the data; namely,the known quantities, denoted by dots in (1.11) must vanish. Similar statement can bemade to the equation (1.7).

In the following, we assume ) is a unit normal vector given at a point of S, i.e.,

n"

k=0

)2k = 1. (1.13)

Now, we discuss several examples.

Example 1.2.7 The 3-dimensional wave equation:

ux0x0 #3"

k=1

uxkxk= 0.

1.3. CLASSIFICATIONS OF SECOND ORDER SEMILINEAR PDES 11

(Here, one usually denotes x0 as t) The characteristic equation is

)20 #

3"

k=1

)2k = 0,

which together with (1.13) gives )0 = ± 1#2. Therefore, a surface is characteristic for the

3-D wave equation if and only if its normal makes an angle of #4 with respect to the x0 axis.

Example 1.2.8 Consider the (n + 1)-dimensional Laplace equationn"

k=0

uxkxk= 0.

Here the characteristic equation readsn"

k=0

)2k = 0

which is incompatible with (1.13). Therefore, there are no (real) characteristics for Laplaceequation.

Example 1.2.9 Consider the following first-order linear equation

a(x, y)ux + b(x, y)uy = c(x, y)u + d(x, y).

The characteristic equation isa)0 + b)1 = 0.

Solving this together with (1.13) gives

()0, )1) = ± 1+a2 + b2

(b,#a).

Therefore, the characteristic curves are solutions of the following system!

x = a(x, y)

y = b(x, y).

This example is very useful in the next Chapter.

1.3 Classifications of second order semilinear PDEs

In this section, we further discuss the types of PDEs, for which the di"erent types ofequations often require di"erent methods to resolve. As the decisive coe!cients of quisa-linear PDEs depend on solutions, we will discuss semi-linear equations. Roughly speaking,the classifications of the semi-linear PDEs depend on how informations propagate. Ifthe information carried by solution propagates at a finite speed, we call it hyperbolic; ifthe information propagates at an infinite speed, we call it parabolic; if there is no (real)speed for the information to travel with, we call it elliptic. From the knowledge in theprevious section, we know the latter often links to the problem without reasonable time-like direction, or the problem is static.


1.3.1 Equations with several variables

To be precise, for n ! 2, we will now discuss only the second order semi-linear PDEs ofthe following form

n"

i,j=1

aij(x)uxixj + F (x, u, Du) = 0, (1.14)

where, x = (x1, x2, · · · , xn) " # with # an open subset in Rn, aij = aji as we expectuxixj = uxjxi . The linear principal part of this equation at a fixed point p " # is

n"

i,j=1

aij(p)uxixj(p), (1.15)

which corresponds to a quadratic form

n"

i,j=1

aij(p)*i*i = 0, ** = (*1, *2, · · · , *n) " Rn (1.16)

This is the left hand side of the characteristic equation of (1.14). By linear algebra, weknow that the numbers of positive, negative and zero (real) eigenvalues of the matrixA = (aij(p)) is invariant under the transformation of the form P T AP for any invertiblen, n matrix P . there is an orthogonal matrix O such that

OT AO = diag{"1, · · · , "n}, (1.17)

and the transformation * = Oy changes the quadratic form into diagonal form

Q(y) =n"

i=1

"iy2i . (1.18)

The way to transform Q(*) into its diagonal form is not unique. Upon re-scalings, we seethere is an invertible matrix U such that the transformation * = Uy changes Q(*) into itsstandard form

Q(y) =n"

i=1

+iy2i . (1.19)

Here, +i = 1, #1, or 0. (By Shur’s lemma of linear algebra, we know that the numbers ofpositive, negative and zero (real) eigenvalues of the matrix A = (aij(p)) is invariant underthe transformation of the form P T AP for any invertible n, n matrix P . ) If we fix sucha U , the transformation

y = P T x

will transfer the equation (1.14) into its standard form:

n"

i=1

+iuxixi + · · · = 0, (1.20)


where the dots terms contain at most first order derivatives.Now, we are able to give a precise classification for (1.14):

Definition 1.3.1 For (1.14) we have the following classifications:

• If all the eigenvalues of A have the same sign( all positive or all negative), we say(1.14) is elliptic at p.

• If A has zero eigenvalues, we call (1.14) parabolic at p.

• If (n # 1) eigenvalues of A have the same sign di!erent from the other one, we say(1.14) is hyperbolic at p. If both the number of positive eigenvalues and the numberof negative eigenvalues of A are greater than 1, and A had no zero eigenvalue, wecall (1.14) super hyperbolic.

Example 1.3.2 The following equation is super hyperbolic:

ux1x1 + ux2x2 # ux3x3 # ux4x4 = 0.

If the equation (1.14) is elliptic (or parabolic, or hyperbolic respectively) at each pointin #, we say (1.14) is elliptic (or parabolic, or hyperbolic respectively). If (1.14) is ofdi"erent types on di"erent points of #, we call (1.14) of mixed type.

Example 1.3.3 Tricomi equation

yuxx + uyy = 0,

is of mixed type on any region including points on the x-axis.

Unfortunately, when the independent variables are more than 2, there are examplesshowing that no matter how small the region # is, there does not exist a single change ofvariables such that the equation (1.14) is transfered into a single type. However, for onlytwo independent variables, under mild conditions on the coe!cients of the equation, it ispossible to make a change of variables such that the equation (1.13) was of the same typeon the whole region # (very small sometime).

1.3.2 The case of two variables

Consider the semi-linear PDE of the second order with independent variables x and y

a(x, y)uxx + 2b(x, y)uxy + c(x, y)uyy = F (x, y, u, ux, uy), (1.21)

where F , a, b and c are smooth functions with respect to their arguments. We also assumethat a, b and c are not all zero at any point on the working region #. According to theclassifications in last sub-section, equation (1.21) is classified as the sign of the followingdeterminant:

d := detA =

****

(a(x, y) b(x, y)b(x, y) c(x, y)

)**** = a(x, y)c(x, y)# b2(x, y). (1.22)

We then have the following cases:


• If d > 0, equation (1.21) is elliptic. A typical example of elliptic equations is theLaplace equation

uxx + uyy = 0,

where d = 1.

• If d < 0, equation (1.21) is hyperbolic. A typical example of hyperbolic equation isthe wave equation

utt # c2uxx = 0, c > 0,

where d = #c2.

• If d = 0 and this matrix A is not identically zero, the equation (1.21) is parabolic. Atypical example of parabolic equation is the Heat equation

ut # (uyy = 0, ( > 0.

We now describe how to transfer (1.21) into standard form. The linear principal partof (1.21) is

L0u = auxx + 2buxy + cuyy. (1.23)

For (x, y) " #, and any invertible smooth change of variables

* = *(x, y), , = ,(x, y),&(*, ,)

&(x, y))= 0, (1.24)

it is easy to compute that the linear principal part of (1.21) becomes

L0u = a$u$$ + 2b$u$% + c$u%%, (1.25)

where #$%

$&

a$ = a*2x + 2b*x*y + c*2

y

b$ = a*x,x + b(*x,y + *y,x) + c*y,y

c$ = a,2x + 2b,x,y + c,2

y .

(1.26)

Of course, we require the second order di"erentiability of * and ,. It is now clear that ifa$ = c$ = 0 and b$ )= 0, the equation (1.25) is of hyperbolic type, and it has a simple form

L0u = 2b$u$%,

which is called second standard form of hyperbolic equations. In this case, we know that* and , are solutions to the following equation

a!2x + 2b!x!y + c!2

y = 0. (1.27)

This is exact the characteristic equation of L0. !(x, y) = constant defines implicitly afamily of curves on xy-plane as y = y(x) (or, x = x(y) if necessary), which satisfies thefollowing ODE

a(dy

dx)2 # 2b

dy

dx+ c = 0. (1.28)


We note that (1.27) is a fully nonlinear PDE for ! while (1.28) is an (nonlinear) ODEwhich is easier to solve. As long as -%!- )= 0, these two equations are equivalent in thesense that !(x, y) = constant is an implicit solution to (1.28).

From the theory of classification in last sub-section and (1.27)–(1.28), we see that(1.28) will not give any real nontrivial solution to (1.27) in the elliptic region. While in thehyperbolic region, (1.28) gives two distinct families of real characteristic curves. Finally,in the parabolic set of points, (1.28) gives only one family of real characteristic curves.

We now show how to utilize the characteristic curves to transfer (1.21) into standardform. We will do so case by case.

First of all, assume that d < 0 in #, so (1.21) is hyperbolic. We solve (1.28) to obtaintwo distinct families of characteristic directions

dy

dx=

b ±+

b2 # ac

a. (1.29)

Integrating them, we find two families of characteristic curves

!1(x, y) = c1, !2(x, y) = c2.

When !2ix + !2

iy )= 0 for i = 1, 2, (1.29) implies that

&(!1, !2)

&(x, y))= 0.

Therefore, after the change of variables

* = !1(x, y), , = !2(x, y),

one obtains (1.25) with a$ = c$ = 0 and b$ )= 0. (1.21) becomes

u$% + F (*, ,, u, u$, u%) = 0,

which, after a further change of variables into s and t such that

* =1

2(s + t), , =

1

2(s# t),

becomes the standard form

uss # utt + F1(s, t, u, us, ut) = 0,

for some smooth F1.Secondly, we assume that d > 0 so that (1.21) is elliptic. We know that it is impossible

to obtain real-valued characteristic curves from (1.28). However, we are able to solve (1.28)for a complex-valued solution

!(x, y) = !1(x, y) + i!2(x, y) = constant,


where !1 and !2 are real functions, and i =+#1 is the imaginary unit. It is not hard to

show that if !x and !y don’t vanish at the same time, the following transformation

* = !1(x, y), , = !2(x, y),

satisfies&(!1, !2)

&(x, y))= 0.

Substituting * + i, into (1.27), separating the real and imaginary part, one discoversa$ = c$ )= 0 and b$ = 0. Therefore, (1.21) takes the following standard form

u$$ + u%% + G(*, ,, u, u$, u%) = 0,

for some smooth function G.Finally, we assume that d = 0 and thus (1.21) is parabolic. It is clear that ac = b2, and

one can assume a > 0 and c > 0 without loss of the generality since b )= 0. From (1.28) weobtain

dy

dx=

+c

a.

Solve this equation, we have the family !(x, y) = c3. Since -%*- )= 0, * = !(x, y) ).constant. Choose appropriate , = ,(x, y) such that

&(*, ,)

&(x, y))= 0.

For instance, if , = x, one has&(*, ,)

&(x, y)= #*y )= 0.

Otherwise, if *y = 0, equation (1.27) implies *x = 0, therefore * . constant, a contradic-tion. For some isolated points where *y = 0, one choose # to exclude these points.

Now, under such a transformation, (1.21) becomes standard form

u%% + G1(*, ,, u, u$, u%) = 0

for some smooth function G1.In the following, we show some examples.

Example 1.3.4 Discuss the type of the following equation

uxx + yuyy = 0 (1.30)

and change it into standard form.


Solution: We compute d = y. Therefore, this equation is elliptic in upper half planey > 0, is hyperbolic in lower half-plane y < 0, and is parabolic on x-axis. The ODEassociate to the characteristic equation is

(dy

dx)2 + y = 0.

Case 1: When y = 0, one substitute y = 0 into (1.30), one gets the standard form

uxx = 0.

The characteristic curve in this case is the integral curve ofdy

dx= 0, and that is x-axis

since y = 0.Case 2: In the hyperbolic region where y < 0, we solve the ODE and found

* = x + 2+#y = c1, , = x# 2

+#y = c2,

and thus the transformation

* = x + 2+#y, , = x# 2

+#y,

transfers, with some standard calculations, into standard form

u$% +1

2(* # ,)(u$ # u%) = 0, y < 0.

The characteristic curves in this case are two branches of the parabola y = #14(x # C)2,

where C is an arbitrary constant. The one with positive slope is corresponding to * =constant while the one with negative slope is corresponding to , = constant. Both branchesare tangent to x-axis.

Case 3: In the elliptic region y > 0, the ODE will become a pair of conjugate complexequations

dy

dx± i+

y = 0.

Solving the one with plus sign (you can choose either one), we obtain

x# 2i+

y = c.

The real and imaginary parts are

* = x, , = 2+

y,

which is our desired transformation. After some calculations, we arrive at

u$$ + u%% #1

,u% = 0, y > 0.


Example 1.3.5 Discuss the type of the following equation

yuxx + (x + y)uxy + xuyy = 0, (1.31)

and find the general solution when x )= y.

Solution: To solve the problem, we first compute

d = xy # 1

4(x + y)2 = #1

4(x# y)2 ( 0.

Therefore, when x = y, the equation(1.31) is parabolic; when x )= y, the equation (1.31) ishyperbolic. In the latter case, we substitute the parameters into (1.28) to find either

dy

dx= 1

ordy

dx=

x

y.

Therefore, the two families of characteristic curves are

y # x = c1, and y2 # x2 = c2.

Which tells us, if we perform the transformation

* = y # x, , = y2 # x2,

the Jacobian satisfies&(*, ,)

&(x, y)= 2(x# y) )= 0

in hyperbolic region. Furthermore, the equation (1.31) takes the second standard form

u$% +1

*u% = 0,

which is equivalent to(*u%)$ = 0.

Now, we integrate the above equation with respect to *,

*u% = f(,),

where f is any integrable function. Therefore,

u =1

*

,f(,) d, + g(*)

= g(y # x) +1

y # xh(y2 # x2),

where g and h are C2 functions. This formula gives the general solution of (1.31) on thehyperbolic region where x )= y.

1.4. PROBLEMS 19

1.4 Problems

Problem 1. For the following equations, identify which are linear, semi-linear, quasi-liner,or fully nonlinear, and their orders.

• (a) utt #$u = 0;

• (b) ut + uxxxx = 0;

• (c) div(|Du|p%2Du) = 0;

• (d) div

(Du

(1 + |Du|2)

)= 0;

• (e) ut + divF (u) = 0, F : R & Rn;

• (f) ut #$(u&) = 0;

• (g) ut #$u = f(u);

• (h)

!ut + u · Du#$u + Dp = 0,

divu = 0; u " R3, x " R3;

• (i) ut + divF (u) = 0, F : Rn & Rn&n.

Problem 2. Let # = {(x, y) : 0 < x < 1, 0 < y < 1}, and consider the following boundaryvalue problem #

$%

$&

uxx # uyy = 0, (x, y) " #,

u(x, 0) = f1(x), u(x, 1) = f2(x),

u(0, y) = g1(y), u(1, y) = g2(y).

where f1, f2, g1 and g2 are given functions. Is this problem well-posed?

Problem 3. Find the type of the following equation

3uxx # 2uxy + 2uyy # 2uyz + 3uzz + 5uy # ux + 10u = 0.

Transfer it into its standard form.

Problem 4. Identify the types of the following equations:

• (a) xuxx + 2yuxy + yuyy = 0;

• (b) uxx + (x# y)3uyy = 0;

• (c) yuxx + (x + y)uxy + xuyy = 0;


• (d) sin(x)uxx # 2 cos(x)uxy # (1 + sin(x))uyy = 0;

• (e) ezuxy # uxx = log(x2 + y2 + z2 + 1).

• (f) 7uxx # 10uxy # 22uyz + 7uyy # 16uxz # 5uzz = 0.

Problem 5. Transfer the following equations into standard form.

• (a) x2uxx + 2xyuxy + y2uyy = 0;

• (b) uxx + xyuyy = 0;

• (c) uxx # 2 cos(x)uxy # (3 + sin2(x))uyy # yuy = 0;

• (d) y2uxx # e#

2xxuxy + ux = 0, x > 0.

Problem 6. Determine the type of Tricomi equation

uxx + xuyy = 0,

and then transfer it into standard form.

Problem 7. Transfer the following equation

uxx + yuyy +1

2uy = 0

into standard form. Find the general solution.

Problem 8. Show that for any second order hyperbolic or elliptic PDEs with two variablesand constant coe!cients, one can always combine the change of variables (1.24) and thethe following transformation of unknown function

u = v exp("* + µ,)

to obtain a form ofv$$ ± v%% + cv = f.

Problem 9. Based on the problem 8, classify the following equations and transfer theminto a standard form without first order derivatives.

• (a) uxx + 4uxy + 3uyy + 3ux # uy + 2u = 0;

• (b) uxx + 2uxy + uyy + 5ux + 3uy + u = 0;

• (c) uxx # 6uxy + 12uyy + 4ux # u = sin(xy).

1.4. PROBLEMS 21

Problem 10. Make the change of unknown function u = v + w with v the new unknownfunctions, such that the following problems have homogeneous boundary conditions. Where(a) has Neumann boundary condition, (b) has Dirichlet boundary condition, while (c) hasone Neumann condition and one Robin condition.

(a)

#$%

$&

utt # c2uxx = 0, 0 < x < +', t > 0,

ux(0, t) = g(t), t ! 0,

u(x, 0) = !(x), ut(x, 0) = #(x), 0 ( x < +';

(b)

#$%

$&

utt # c2uxx = 0, 0 < x < l, t > 0,

u(0, t) = µ(t), u(l, t) = '(t), t ! 0,

u(x, 0) = !(x), ut(x, 0) = #(x), 0 ( x < l;

(c)

#$%

$&

utt # c2uxx = 0, 0 < x < l, t > 0,

#ux(0, t) = µ(t), ux(l, t) + u(l, t) = '(t), t ! 0,

u(x, 0) = !(x), ut(x, 0) = #(x), 0 ( x < l.

Problem 11. Determine w such that the change of unknown function u = vw changesthe equation

ut # uxx + aux + bu = f(x, t)

into

vt # vxx = f1(x, t).

Problem 12. Assume u is a solution of the heat equation

ut # a2uxx = 0,

with the formu(x, t) = u(

x+t).

Derive the ODE for u, and then solve the following problem

#$%

$&

ut # a2uxx = 0, 0 < x < +', t > 0,

u(0, t) = 0, t ! 0,

u(x, 0) = u0, 0 ( x < +',

where u0 is a constant.

Chapter 2

The first order quasi-linear PDEs

The first order quasi-linear PDEs have the following general form:

F (x, u, Du) = 0, (2.1)

where x = (x1, x2, · · · , x3) ! Rn, u = u(x), Du is the gradient of u. Such equation oftenappears in physical applications, such as the famous transport equation, Burgers’ equationand the Hamilton-Jacob equation

!ut + H(Du) = 0, (2.2)

where ! " 0. The distinguished feature of this class is that one can solve it through asystem of ODEs at least locally. Such a method is called method of characteristic. We willfirst introduce the general theory in two variables in first two sections, then we generalizeit into general cases. We then discuss in details on the semi-linear case in two variableswith some examples. The transport equation will be presented in fourth section with adirect approach.

2.1 Characteristic curves and integral surfaces

Consider the following first order quasi-linear PDE with two variables:

a(x, y, u)ux + b(x, y, u)uy = c(x, y, u). (2.3)

Here, u = u(x, y). (2.3) can be rewritten in the following form

(a, b, c) • (ux, uy,#1) = 0. (2.4)

Keeping this equation in mind, the vector field (a, b, c) (called characteristic direction)therefore plays an essential role in solving (2.3). On some interval I $ R, this vector fielddetermines the following parametric curves (called characteristic curves)

" : x = x(t), y = y(t), z = z(t), t ! I $ R,

23

24 CHAPTER 2. THE FIRST ORDER QUASI-LINEAR PDES

through the following equation

dx

a(x, y, z)=

dy

b(x, y, z)=

dz

c(x, y, z),

i.e. !""""#

""""$

dx

dt= a(x, y, z)

dy

dt= b(x, y, z)

dz

dt= c(x, y, z).

(2.5)

We call (2.5) the characteristic equation of (2.3). We see such curves, if exist, have thecharacteristic direction as tangent vectors. We remark that, from calculus, the integralsurface S : z = u(x, y), as the solution of (2.3), is tangent to the characteristic directioneverywhere. The following theorem states the relationship between " and S.

Theorem 2.1.1 If there is a point P = (x0, y0, z0) of the characteristic curve " lying onthe integral surface S : z = u(x, y), then " $ S.

Proof: Assume that

" : x = x(t), y = y(t), z = z(t), t ! I $ R,

is a solution of (2.5) such that for some parameter t = t0, it holds that

x0 = x(t0), y0 = y(t0), z0 = z(t0) = u(x0, y0).

Furthermore, we know thatP = (x0, y0, z0) ! S.

Definew = w(t) = z(t)# u(x(t), y(t)).

It is easy to check that w(t) is the solution to the following initial value problem!#

$

dw

dt= c(x, y, w + u)# uxa(x, y, w + u)# uyb(x, y, w + u),

w(t0) = 0.(2.6)

Clearly, w % 0 is a solution of (2.6) since z = u(x, y) is a solution of (2.3). Due toODE theory, we know that (2.6) has a unique solution. Therefore, w(t) % 0, namely,z(t) = u(x(t), y(t)), which means " $ S.

From the definition of S, we know that, for each point on S, there is a characteristiccurve passing through. Therefore, S is the union of all characteristic curves. Now, it is nothard to see from argument we made above that we did prove the following result.

2.2. CAUCHY PROBLEM 25

Theorem 2.1.2 The general solution of a first-order, quasi-linear PDE (2.3) is

f(#,$) = 0,

where f is an arbitrary function of #(x, y, u) and $(x, y, u) and # = constant = c1 and$ = constant = c2 are solution curves of

dx

a(x, y, u)=

dy

b(x, y, u)=

du

c(x, y, u),

or equivalently, equation (2.5) replacing z with u.

2.2 Cauchy problem

We now discuss the initial value problem for (2.3). We know from section 1.2 that, in orderto have the well-posed Cauchy problem, one should not prescribe the initial data on thecharacteristic curves. The following theorem confirms this observation.

Theorem 2.2.1 Suppose that x0(s), y0(s) and u0(s) are continuous di!erentiable func-tions of s in a closed interval 0 & s & 1 and that a, b and c are functions of x, y andu with continuous first order partial derivatives with respect to their arguments in somedomain D of (x, y, z)-space containing the initial curve

" : x = x0(s), y = y0(s), u = u0(s), (2.7)

where s ! [0, 1], and satisfying the condition%%%%

&a(x0(s), y0(s), u0(s)) b(x0(s), y0(s), u0(s))

x!0(s) y!0(s)

'%%%% '= 0. (2.8)

Then, there exists a unique solution u = u(x, y) of (2.3) in the neighborhood of C : x =x0(s), y = y0(s), s ! [0, 1], and the solution satisfies the initial condition

u0(s) = u(x0(s), y0(s)), s ! [0, 1]. (2.9)

Remark 2.2.2 The curve C, on which the Cauchy data is prescribed, is the projection ofthe initial curve " onto the (x, y)-plane. The condition (2.8) is precisely equivalent to thecurve C is not characteristic in the sense of section 1.2.

Roughly speaking, a proof of this theorem can be described as follows: One starts witha time t = 0 and solve the equation (2.5) with the initial data (2.7) which gives a uniquesolution

x = X(s, t), y = Y (s, t), z = Z(s, t), s ! [0, 1], t ! [0, T ], (2.10)

for some positive T , such that

(X(s, 0), Y (s, 0), Z(s, 0)) = (x0(s), y0(s), u0(s)).


Now, with the condition (2.8), one can solve (s, t) in terms of x and y to obtain

s = #(x, y), t = $(x, y).

Substituting these into z = Z(s, t), one has

u = Z(s, t) = Z(#(x, y), $(x, y)) % u(x, y),

the solution of the Cauchy problem. Of course, in general, the solutions exist only in aneighborhood of the initial curve ", and such a solution is call a local solution. The rigorousproof of this theorem involves the implicit function theorem and we shall not pursuit ithere.

2.3 General cases

We now generalize the theory we established in the previous two sections to the followinggeneral first order quasi-linear PDEs:

n(

k=1

ak(x1, x2, · · · , xn, u)uxk= c(x1, x2, · · · , xn, u), n " 2. (2.11)

The corresponding system of ODEs for characteristic curves reads!"#

"$

dxk

dt= ak(x1, x2, · · · , xn, z), k = 1, · · · , n,

dz

dt= c(x1, x2, · · · , xn, z).

(2.12)

The corresponding Cauchy problem is thus to find the integral surface of (2.11) z =u(x1, x2, · · · , xn) such that it pass through the following parametrized (n # 1) dimensioninitial surface S:

S :

)xk = fk(s1, s2, · · · , sn"1), k = 1, · · · , n,

z = u0(s1, s2, · · · , sn"1).(2.13)

To proceed, we could first find the characteristic curves passing through a point on S withparameter s = (s1, · · · , sn"1) to obtain the solution of (2.12)

)xk = Xk(s1, s2, · · · , sn"1, t), k = 1, · · · , n,

z = Z(s1, s2, · · · , sn"1, t),(2.14)

which satisfies the initial datum given in (2.13). Now, similar to (2.8), we propose thecondition

J = det

*

+++,

a1 a2 · · · an(s)!f1

!s1

!f2

!s1· · · !fn

!s1...

.... . .

...!f1

!sn!1

!f2

!sn!1· · · !fn

!sn!1

-

.../

%%%%%%%%%S

'= 0, (2.15)

2.4. SOME EXAMPLES 27

which is equivalent to that the initial surface is not characteristic. Under this condition(2.15), one can solve from (2.14) for s = (s1, s2, · · · , sn"1) and t in terms of x1, · · · , xn, andthen upon to a substitution, the solution u = Z(x1, · · · , xn) to (2.11) with initial condition(2.13) is obtained.

2.4 Some examples

In this section, we will show some examples on how to apply the method of characteristicestablished so far to solve some Cauchy problems on first order quasi-linear PDEs.

Example 2.4.1 Consider the semi-linear equation

a(x, y)ux + b(x, y)uu = f(x, y, u), (2.16)

where the functions a, b, and f are smooth functions. In this case, we could simply solvethe following system of ODEs

dx

dt= a(x, y),

dy

dt= b(x, y),

which determines a family of curves (x(t), y(t)) on (x, y)-plane. This family of curves areactually the projection of the characteristic curves on to (x, y)-plane. Then we solve theequation

dz

dt= f(x(t), y(t), z)

to obtain the characteristic curve ": (x(t), y(t), z(t)).

Example 2.4.2 We now solve the following Cauchy problem)

ut + vux = 0,

u(x, 0) = u0(x),(2.17)

where v > 0 is a constant, and u0 is a regular function. First of all, one easily verifies thatthe initial curve

" : t = 0, x = %, z = u0(x)

is not characteristic, as the determinant defined in (2.8) is 1.In view of Example 2.4.1, we solve the following

)x(s) = v,

t(s) = 1,(2.18)

where “ ˙ ” indicates the derivative with respect to the parameter s. By imposing theinitial conditions at s = 0, )

x(0) = x0,

t(0) = t0,


we determine a unique solution for (2.18) in the explicit form

)x(s) = vs + x0,

t(s) = s + t0.

Since the initial condition is assigned on t = 0, which implies (x0, t0) is a point of the curveand t0 = 0. So the characteristic curve may be rewritten, eliminating the parameter s, as

x(t) = vt + x0. (2.19)

This is a line in (x, t)-plane intersecting {t = 0} at x0 with the slope1

vor, equivalently

speed v.Now let #(t) = u(x(t), t) be a solution of (2.17). Along the characteristic curve (2.19)

we have:

#(t) = x(t)ux(x(t), t) + ut(x(t), t) = vux(x(t), t) + ut(x(t), t) = 0,

i.e., the solution of (2.17) is constant along the characteristics (2.19). So you can resolvethe ordinary di!erential equation for #:

#(t) = #(0),

i.e.u(vt + x0, t) = u(x(t), t) = u(x(0), 0) = u0(x0).

To determine the value of solution u at a generic point in (x, t)-plane, one simply determinesthe point x0 of intersection between the characteristic passing through (x, t) and the axis{t = 0}, that is, from x = vt + x0 to obtain

x0 = x# vt.

Ultimately, the solution of (2.17) is given by

u(x, t) = u0(x# vt).

Example 2.4.3 In this example, we solve the following Cauchy problem

!#

$x&u

&x+ 2y

&u

&u+

&u

&z= 3u,

u(x, y, 0) = #(x, y).(2.20)

On the initial surface

S : x = s1, y = s2, z = 0, w = #(s1, s2),

2.4. SOME EXAMPLES 29

we compute

J = det

*

+++,

x 2y 1&x

&s1

&y

&s1

&z

&s1&x

&s2

&y

&s2

&z

&s2

-

.../= det

*

,s1 2s2 11 0 00 1 0

-

/ = #1 '= 0.

Therefore, we expect the problem (2.20) has a unique solution near S. We solve the initialvalue problem !

"""#

"""$

dx

dt= x,

dy

dt= 2y,

dz

dt= 1,

dw

dt= 3w,

(x, y, z, w)|t=0 = (s1, s2, 0, #(s1, s2)),

to obtainx = s1e

t, y = s2e2t, z = t, w = #(s1, s2)e

3t.

From the first three equations in the above, we have

t = z, s1 = xe"z, s2 = ye"2z.

Therefore, the solution of the problem (2.20) is

w = u(x, y, z) = #(xe"z, ye"2z)e3z.

Example 2.4.4 Solve the following problem)

tut + xux = cu, x ! R, t " 0,

u(x, 1) = f(x),(2.21)

where c is a constant.For this problem, we note that the coe"cients (x, t, c) is singular at (0, 0, c) for c '= 0,

where the equation does not make sense except for u % 0. The initial curve is

" : x = s, t = 1, z = f(s),

which is not characteristic if s '= 0. Solving the system of ODEs

dx

d%= x,

dt

d%= t,

dz

d%= cz,

one obtains the characteristic curves

x(%, s) = se" , t(%, s) = e" , z(%, s) = f(s)ec" .

Therefore,

s =x

t, % = log t


and the solution isu(x, t) = z = f(

x

t)tc.

We note that, in the particular case where f % constant, we see from the above formulathat if c > 0 we have solution u(x, t) for all t " 0. On the other hand, if c < 0, we see thatu(x, t)() as t tends to zero. In the latter case, we say the solution blows up at t = 0.

We now use this example to explain what will happen if the initial curve is characteristic.Now, we fix c = 1, and choose the initial curve

"1 : x(0, %) = %, t(0, %) =%

!, z(0, %) =

'

!%

for s = 0, and ! '= 0, ' are arbitrary constants. One easily checks that both u(x, t) = #$x,

and u(x, t) = 't are solutions for the equation with the new initial data. We found infinitelymany solutions in this case.

Example 2.4.5 We now consider the following problem

)ut + uux = 0

u(x, 0) = u0(x).(2.22)

In view of example 2.4.1 and 2.4.2, one easily find the characteristic curves as the solutionof

dx

dt= u(x, t), x(0) = x0.

Also, we know that the solution u(x, t) is constant along each characteristic curves. There-fore, the characteristic curves are the family of straight lines

x(t) = x0 + tu0(x0).

So, the solution of u at each point (x, t) is thus determined by the characteristic line fromcertain x0 on {t = 0} passing through (x, t) where u(x, t) = u0(x0).

We remark that this process cannot go far in general. Suppose there are two pointa < b on x-axis such that u0(a) > u0(b). We know that the characteristic line issuing from(a, 0) will overtake the characteristic line from (b, 0) at

t =b# a

u0(a)# u0(b).

In this case, we could not assign a value for u(x, t) at this intersection point, which corre-sponds to two di!erent values. Actually, the solution blows up in this case in its derivative.More precisely, we carry out the following calculations. Let w = ux, which satisfies

wt + uwx = #w2.

2.5. LINEAR TRANSPORT EQUATION 31

Therefore, along the characteristic line, we have)

w = #w2

w(x0, 0) = u!0(x0).

It is now clear that

w(x0 + tu0(x0), t) =u!

0(x0)

1 + u!0(x0)t

.

Therefore, w tends to infinity in finite time if u!0(x0) < 0.

2.5 Linear Transport Equation

The linear transport equation

ut + b • Du = 0, x ! Rn, t > 0 (2.23)

arises from many important applications, such as particle mechanics, kinetic theory. Hereb = (b1, b2, · · · , bn) is a constant vector. Of course, one can easily solve this PDE using thecharacteristic method. Here, we will use a direct approach instead. Actually, the (2.23)can be rewritten as

(Du, ut) • (b, 1) = 0,

which means the function u(x, t) is constant in the direction of (b, 1) on the (x, t)-space.Therefore, if we know the value of u at any point of the straight line through (x, t) withthe direction (b, 1) in (x, t)-space, we know the value of u(x, t).

Now let us consider the initial-value problem)

ut + b • Du = 0, x ! Rn, t > 0,

u(x, 0) = g(x), x ! Rn,(2.24)

where g(x) is a given function. Fix a point (x, t), the line through (x, t) in the directionofd (b, 1) is given in the parametric form with the parameter s as

(x + sb, t + s), s ! R.

The line hits the initial plane {t = 0} when s = #t at the point (x# bt, 0). We know thatu is constant on the line, we thus obtain

u(x, t) = g(x# tb), x ! Rn, t " 0. (2.25)

We see from the above that if (2.24) has a smooth solution, then it is given by (2.25).Conversely, if g is continuous di!erentiable, the (2.25) is the unique solution of (2.24). Wealso remark that the solution in the form of (2.25) reads as the information traveling in aconstant velocity b, such a solution is thus called traveling wave solution.


Remark 2.5.1 When g(x) is not C1, problem (2.24) does not admit a smooth solution.However, even when g is not continuous, the formula (2.25) does provide a reasonablecandidate for a solution. We may informally declare that (2.25) is a weak solution of(2.24). This makes sense even if g is discontinuous and so is u. This notion is very usefulin nonlinear PDEs.

Next, we consider the non-homogeneous problem

)ut + b • Du = f(x, t), x ! Rn, t > 0,

u(x, 0) = g(x), x ! Rn.(2.26)

As in the homogeneous case, we set

z(s) = u(x + sb, t + s), s ! R,

which satisfies

z(s) = Du(x + sb, t + s) • b + ut(x + sb, t + s) = f(x + sb, t + s).

Therefore,

u(x, t)# g(x# bt) = z(0)# z(#t) =

0 0

"t

z(s) ds

=

0 0

"t

f(x + sb, t + s) ds

=

0 t

0

f(x + (s# t)b, s) ds,

(2.27)

which gives the solution

u(x, t) = g(x# tb) +

0 t

0

f(x + (s# t)b, s) ds. (2.28)

2.6 Problems

Problem 1. Solve the following problems (x, y, z ! R, t > 0)

• (a) ut + uux = 1, u(x, 0) = h(x);

• (b) ux + uy = u, u(x, 0) = cos(x);

• (c) xuy # yux = u, u(x, 0) = h(x);

• (d) x2ux + y2uy = u2, u(x, y)|y=2x = 1;

• (e) xux + yuy + uz = u, u(x, y, 0) = h(x, y);

2.6. PROBLEMS 33

• (f)1n

k=1 xkuxk= 3u, u(x1, · · · , xn"1, 1) = h(x1, · · · , xn"1);

• (g) uux # uuy = u2 + (x + y)2, u(x, 0) = 1;

• (h) 2xyux + (x2 + y2)uy = 0, u = exp

&x

x# y

', on x + y = 1;

• (i) xux + yuy = u + 1, u(x, y) = x2 on y = x2.

Problem 2. Solve the Cauchy problem)

ut + b • Du + cu = 0, x ! Rn, t > 0,

u(x, 0) = g(x), x ! Rn.

where c ! R and b ! Rn are constant.

Problem 3. For the following problem!#

$ut + uux = 0

u(x, 0) =1

1 + x2,

find the solution and the time and the point where the solution blows up first.

Problem 4. If u is the C1 solution of the following equation

a(x, y)ux + b(x, y)uy = #u

on the closed unit disk # = {(x, y) : x2 + y2 & 1}. Suppose

a(x, y)x + b(x, y)y > 0, on x2 + y2 = 1,

prove that u % 0.

Problem 5. Show that the solution of the equation

yux # xuy = 0

containing the curve x2 + y2 = a2, u = y, does not exit.

Problem 6. Solve the following Cauchy problems:

• (a) x2ux # y2uy = 0, u( ex as y ();

• (b) yux + xuy = 0, u = sin(x) on x2 + y2 = 1;

• (c) #xux + yuy = 1, in 0 < x < y, u(x) = 2x on y = 3x;


• (d) 2xux + (x + 1)uy = y, in x > 0, u(1, y) = 2y;

• (e) xux # 2yuy = x2 + y2 in x > 0, y > 0, u = x2 on y = 1;

Problem 7. Show that u1 = ex, and u2 = e"y are solutions of the nonlinear equation

(ux + uy)2 # u2 = 0,

but that their sum (ex + e"y) is not a solution of the equation.

Problem 8. Solve the following equations:

• (a) (y + u)ux + (x + u)uy = x + y,

• (b) xu(u2 + xy)ux # yu(u2 + xy)uy = x4.

Problem 9. Solve the equationxzx + yzy = z,

and find the curves which satisfy the associated characteristic equations and intersect thehelix x2 + y2 = a2, z = b tan"1(y/x).

Problem 10. Find the family of curves which represent the general solution of the PDE

(2x# 4y + 3u)ux + (x# 2y # 3u)uy = #3(x# 2y).

Determine the particular member of the family which contains the line u = x and y = 0.

Problem 11. Find the solution of the equation

yux # 2xyuy = 2xu

with the condition u(0, y) = y3.

Problem 12. Find the solution surface of the equation

(u2 # y2)ux + xyuy + xu = 0, u = y = x, x > 0.

Chapter 3

Laplace Equation

Undoubtedly, the Laplace Equation is among the most important PDEs. Traditionally, wecall

!!u(x) = 0, x " " # Rn, n $ 2. (3.1)

the Laplace Equation, and its non-homogeneous companion

!!u(x) = f(x), x " " # Rn, n $ 2, (3.2)

the Poisson Equation.We recall that the Laplacian of u is

!u =n!

i!1

uxixi .

In both equations, x = (x1, · · · , xn) " " # Rn and the unknown is u : " % R. In (3.2),the function f : " % R is given fucntion. For later use, we have the following defintion:

Definition 3.0.1 If u " C2(") satisfies (3.1) in ", we call it a harmonic function. u "C2(") is called sub-harmonic (or super-harmonic) if it satisfies

!!u & 0($ 0).

Laplace equation and Poisson equations model many static physical fields with andwithout sources. They appear in the modeling of gravity, electric force, and the chemicalconcentration in equilibrium, and much more. Let V be any smooth subregion in ", thenet flux of u through !V is zero:

"

!V

F • " dS = 0,

where F is the flux density and " the unit outer normal of !V . By Divergence Theorem(see Theorem 3.1.1 below), we have

35

36 CHAPTER 3. LAPLACE EQUATION

"

V

' • F dx =

"

!V

F • " dS = 0,

and so' • F = 0, in ".

In many occasions, the flux F is proportional to the gradient Du, pointing in the oppositedirection. Therefore, one has

F = !aDu, a > 0. (3.3)

If u stands for the #$%

$&

chemical concentration

temperature

electrostatic potential,

the equation (3.3) is #$%

$&

Fick’s law of di#usion

Fourier’s law of heat conduction

Ohm’s law of electrical conduction.

Laplace equation also arises in the study of analytic functions and the probabilistic inves-tigation of Brownian motion.

3.1 The fundamental solution

From the form of Laplace equation, we see that all directions have the same weights inthe equation. Namely, the equation is invariant under any orthogonal transformation ofcoordinates. We thus seek to solve the equation explicitly by looking for the radial solutionwhich has the form

u(x) = v(r), r = |x|,where v is to be determined so that (3.1) holds. For i " {1, . . . , n},

uxi = v"(r)rxi = v"(r)xi

r, uxixi = v""(r)

x2i

r2+ v"(r)

'1

r! x2

i

r3

(.

One has

!u = v""(r) +n! 1

rv"(r) = 0.

If v" (= 0, we deduce

(log(v"))" =v""

v"=

1! n

rand hence

v"(r) =a

rn!1,

3.1. THE FUNDAMENTAL SOLUTION 37

for some constant a. Therefore, if r > 0, we find

v(r) =

#%

&b log r + c for n = 2

b

rn!2+ c for n $ 3,

where a and b are constants. We therefore discover the fundamental solution of Laplaceequation:

$(x) =

#$%

$&

! 1

2#log |x| for n = 2

1

n(n! 2)$(n)

1

|x|n!2for n $ 3,

(3.4)

where

$(n) =2#n/2

n%(n/2)

is the volume of the unite ball in Rn. The reason for particular choices of the constantswill be explained later.

From the derivation, we also have the following estimates

|D$(x)| & C

|x|n!1, |D2$(x)| & C

|x|n , (x (= 0), (3.5)

for some constant C > 0.

3.1.1 Green’s formula

In this subsection, we summarize some useful formulas due to Green. These formulasare convenient in the computations related to Laplacian. These are derived from theDivergence Theorem (or, Gauss formula):

Theorem 3.1.1 Let " be a bounded domain in Rn with smooth boundary !". Let " bethe unit outer normal of !". For any smooth vector field w " C1("), it holds that

"

!

' • w dx =

"

!!

w • " dS. (3.6)

Let " be a bounded domain in Rn with smooth boundary !". Let " be the unit outernormal of !". For u " C2("), one derives from the Divergence Theorem (letting w = Du)that "

!

!u dx =

"

!!

Du • " dS =

"

!!

!u

!"dS. (3.7)

Now, for u, v " C2("), by choosing w = uDv or w = vDu respectively in the DivergenceTheorem, we have


"

!

u!v dx =

"

!!

u!v

!"dS !

"

!

Du • Dv dx (3.8)

"

!

v!u dx =

"

!!

v!u

!"dS !

"

!

Dv • Du dx. (3.9)

We subtract the above two equations to get"

!

(u!v ! v!u) dx =

"

!!

(u!v

!"! v

!u

!") dS. (3.10)

Traditionally, (3.8) is called the first Green’s formula, while (3.10) is called the secondGreen’s formula.

3.1.2 Poisson Equation in Rn

From the construction, we know that the fundamental solution $(x) of Laplace equationis harmonic for x (= 0, so is $(x ! y) for x (= y. With this in mind, we will prove thefollowing

Theorem 3.1.2 If f " C20(Rn), then

u(x) =

"

Rn

$(x! y)f(y) dy " C(Rn) (3.11)

is a solution of the Poisson equation (3.2).

Proof. First of all, we have

u(x) =

"

Rn

$(x! y)f(y) dy =

"

Rn

$(y)f(x! y) dy,

henceu(x + hei)! u(x)

h=

"

Rn

$(y)

)f(x + hei ! y)! f(x! y)

h

*dy,

where h (= 0 and ei the unit vector in the direction of xi-axis. Note that

limh#0

f(x + hei ! y)! f(x! y)

h=

!f

!xi(x! y)

uniformly on Rn, and thus

!u

!xi(x) =

"

Rn

$(y)!f

!xi(x! y) dy, (i = 1, · · · , n). (3.12)

Similarly,!2u

!xi!xj(x) =

"

Rn

$(y)!2f

!xi!xj(x! y) dy, (i, j = 1, · · · , n). (3.13)

3.1. THE FUNDAMENTAL SOLUTION 39

We note that the integral above is continuous in the variable x, and thus u " C2(Rn).As $(x) is singular at x = 0, we have to pay substantial attention near the singularity.

Fix % > 0, let B(0, %) be the ball centered at 0 with radius %. We have

!u(x) =

"

B(0,")

$(y)!xf(x! y) dy +

"

Rn\B(0,")

$(y)!xf(x! y) dy

=: I" + J".

(3.14)

Now, it is clear that

I" & C)D2f)L!(Rn)

"

B(0,")

|$(y)| dy &+

C%2| log %| (n = 2)

C%2 (n = 3).(3.15)

On the other hand, we can estimate J" as following

J" =

"

Rn\B(0,")

$(y)!xf(x! y) dy

=

"

Rn\B(0,")

$(y)!yf(x! y) dy

= !"

Rn\B(0,")

Dy$(y) • Dyf(x! y) dy +

"

!B(0,")

$(y)!f

!v(x! y) dS(y)

=: K" + L",

(3.16)

where " is the inward unit normal along !B(0, %). We easily check

|L"| & C)Df)L!(Rn)

"

!B(0,")

|$(y)| dy &+

C%| log %| (n = 2)

C% (n = 3).(3.17)

It remains to compute K". In fact, we have

K" = !"

Rn\B(0,")

Dy$(y) • Dyf(x! y) dy

=

"

Rn\B(0,")

!y$(y)f(x! y) dy !"

!B(0,")

!$

!"(y)f(x! y) dS(y)

= !"

!B(0,")

!$

!"(y)f(x! y) dS(y).

We now note that

D$(y) =!1

n$(n)

y

|y|n , y (= 0,

and

" =!y

|y| = !y

%, on !B(0, %),


So,!$

!"(y) =

1

n$(n)%n!1, on !B(0, %).

We also note that n$(n)%n!1 is the surface area of !B(0, %), we have

K" = ! 1

n$(n)%n!1

"

!B(0,")

f(x! y) dS(y)

% !f(x), as % % 0.

(3.18)

We therefore conclude from (3.14)–(3.18) (letting % % 0) that

!!u = f.

Remark 3.1.3 Sometimes, we write

!!$ = &0, in Rn,

where &0 is the Dirac measure on Rn giving unit mass to the point 0. Formally, one cancomputes:

!!u(x) =

"

Rn

!!x$(x! y)f(y) dy

=

"

Rn

&xf(y) dy = f(x), (x " Rn).

3.1.3 Fundamental integral formulas

As we often solve the elliptic equations in bounded domains, we now carry out some usefulintegral formulas using the fundamental solution $(x).

Assume u " C2("). For any y " ", we choose ' > 0 suitably small so that the ballB#(y) centered at y with radius ' is inside ". On the region " \ B#(y), we substitute v(x)with $(x! y) in the second Green’s formula,

"

!!B!(y)

$(x! y)!xu dx =

"

!!

($!u

!"! u

!$

!") dSx +

"

!B!(y)

($!u

!"! u

!$

!") dSx. (3.19)

Similar to the computations carried out in (3.17)–(3.18), we have

|"

!B!(y)

$!u

!"dSx| = |$(')

"

!B!(y)

!u

!"dSx|

= |! $(')

"

B!(y)

!u dx|

&+

C'2| log '| (n = 2)

C'2 (n = 3).

% 0, as ' % 0;

3.2. PROPERTIES OF HARMONIC FUNCTIONS 41

and "

!B!(y)

u!$

!"dSx = !$"(')

"

!B!(y)

u dSx

=1

n$(n)'n!1

"

!B!(y)

u dSx

% u(y), as ' % 0.

Therefor, we obtain from (3.19) by letting ' % 0 that

u(y) = !"

!

$(x! y)!xu dx!"

!!

(u!$(x! y)

!"! $(x! y)

!u

!") dSx, *y " ". (3.20)

This formula is called Green’s representation of u(y).In particular, if u has compact support on ", then it holds

u(y) = !"

!

$(x! y)!xu dx, *y " ", u " C20("). (3.21)

If u is harmonic in ", we thus obtain the fundamental integral formula of harmonicfunctions

u(y) = !"

!!

(u!$(x! y)

!"! $(x! y)

!u

!") dSx, *y " ". (3.22)

We remark that, actually, for smooth harmonic function, (3.7) implies"

!!

!u

!"dS = 0. (3.23)

3.2 Properties of harmonic functions

In this section, we discuss some basic properties of harmonic functions. We now consider anopen bounded set " # Rn and suppose u is harmonic in ". Various interesting propertieswill be presented in orders.

3.2.1 Mean-value formulas

The mean-value formulas, which declare that u(x) equals both the average of u over thesphere !B(x, r) and the average of u over the whole ball B(x, r), as long as B(x, r) # ".It will play key roles in many important occasions.

To begin, we introduce the notion of mean value of u(x) over a domain ":

(u)! =1

|"|

"

!

u(x) dx. (3.24)

Theorem 3.2.1 If u " C2(") is harmonic, then

u(x) = (u)!B(x,r) = (u)B(x,r), (3.25)

for each ball B(x, r) # ".


Proof. By (3.22), using (3.23), one has

u(x) = !"

!B(x,r)

(u!$(x! y)

!"! $(x! y)

!u

!") dSy

= !$"(r)

"

!B(x,r)

u(y) dSy ! $(r)

"

!B(x,r)

!u

!"dSy

= !$"(r)

"

!B(x,r)

u(y) dSy

= (u)!B(x,r).

Now, for the mean-value on the ball, we have"

B(x,r)

u(y) dy =

" r

0

(

"

!B(x,s)

u(y) dSy) ds

= u(x)

" r

0

"

!B(x,s)

dSy ds = u(x) · (volume of B(x, r)).

This completes the proof.

Remark 3.2.2 If instead of harmonic, u " C2(") is sub-harmornic, then one has

u(x) & (u)!B(x,r), u(x) & (u)B(x,r). (3.26)

If u " C2(") is super-harmornic, then

u(x) $ (u)!B(x,r), u(x) $ (u)B(x,r). (3.27)

The converse of the mean-value formulas is also a true statement.

Theorem 3.2.3 If u " C2(") satisfies

u(x) = (u)!B(x,r)

for each ball B(x, r) # ", then u is harmonic.

Proof. Set((r) = (u)!B(x,r) = (u(x + rz))!B(0,1).

Then("(r) = (Du(x + rz) • z)!B(0,1)

=1

|!B(x, r)|

"

!B(x,r)

Du(y) • y ! x

rdSy

=1

|!B(x, r)|

"

!B(x,r)

!u

!"dSy

=r

n(!u)B(x,r)


If !u (= 0, there exists some ball B(x, r1) # " such that !u > 0 in B(x, r) (the other caseis similar). But

0 = ("(r1) =r1

n(!u)B(x,r1) > 0,

a contradiction.

3.2.2 Maximum principle and uniqueness

Assume that " is an open and bounded set in Rn. We first present

Theorem 3.2.4 (Strong maximum principle). Suppose u " C2(")+C(") is harmonic in", then

• (i) max!

u = max!!

u

• (ii) If " is connected and there exists a point x0 " " such that

u(x0) = max!

u,

then u is constant in ".

Remark 3.2.5 The first assertion in this theorem is the maximum principle and thesecond is the strong maximum principle. If one replaces u by !u, the similar assertionsare true when ”max” is replaced by ”min”.

Proof. Suppose there exists a point x0 " " with

u(x0) = M = max!

u.

Then for some r > 0, the ball B(x0, r) # ". By the mean-value formula on this ball, wehave

M = u(x0) = (u)B(x0,r) & M.

Here, the equality holds only if u , M in B(x0, r). Therefore, u(x) = M for any x "B(x, r). So, the set {x " "|u(x) = M} is both open and relatively closed in ", and thusequals " if " is connected. This proves assertion (ii). The first one follows immediately.

The strong maximum principle has many applications. A direct application is as follows:if " is connected and u " C2(") + C(") is a solution of

+!u = 0, in ",

u = g, on !",

where g $ 0, then u is positive everywhere in " if g is positive somewhere on !".An important application of the maximum principle is the uniqueness of solutions to

certain boundary value problems for Poisson equation.


Theorem 3.2.6 (Uniqueness). Let g " C(!"), f " C("). Then there exists at most onesolution u " C2(") + C(") of the boundary value problem

+!!u = f, in ",

u = g, on !".(3.28)

Proof. If u1 and u2 are two solutions, one applies the maximum principle to the har-monic functions w± = ±(u1!u2) which satisfies the Laplace equation with zero boundarycondition, therefore, w± , 0.

3.2.3 Harnack’s inequality

In this subsection, we will exploit an amazing averaging e#ect of Harmonic functions.Recall that we denote by V ## " if V # " is compact. The following theorem assert thatthe value of a non-negative harmonic function on " are all comparable within a compactsubset relative to ": it cannot be very large ( or very small) at any point in this subsetunless it is very large (or very small) everywhere. The idea is that since the compact sethas positive distance from !", there is room for the averaging e!ect of Laplace’s equationto occur.

Theorem 3.2.7 (Harnack’s inequality). For each connecterd open set V ## ", thereexists a positive constant C, depending only on V , such that

supV

u & C infV

u

for all nonnegative harmonic functions u in ".

Remark 3.2.8 The Harnack’s inequality in particular implies that, for any x and y " V ,it holds that

1

Cu(y) & u(x) & C u(y).

Proof. Let r = 14 dist(V, !"). Fixing any x " V , for any y " V such that |x! y| & r, we

haveu(x) = (u)B(x,2r)

$ 1

$(n)2nrn

"

B(y,r)

u dz

1

2n(u)B(y,r) =

1

2nu(y).

Therefore, we actually proved that

2!nu(y) & u(x) & 2nu(x),*x, y " V, |x! y| & r.


Since V is connected and V is compact, we can cover V by a chain of finitely many balls{Bi}N

i=1, each of which has radius r and Bi +Bi!1 (= -, for i = 2, · · · , N . Then,

2!nNu(y) & u(x) & 2nNu(y)

for all x, y " V .

3.2.4 Regularity

Now we prove that if u " C2(") is harmonic, then necessarily u " C$("). Thus harmonicfunctions are infinitely di!erentiable. This sort of assertion is called a regularity statements.It is interesting to see that the algebraic structure of Laplace equation leads to that all thepartial derivatives of u exist, even those which do not appear in the PDE.

Theorem 3.2.9 (Smoothness). If u " C(") satisfies the mean-value property (3.25) foreach ball B(x, r) # ", then u " C$(").

Before we proceed to prove this theorem, we first introduce an important tool. Define) " C$(Rn) by

)(x) =

+C exp{ 1

|x|2!1} if |x| < 1,

0 if |x| > 1,(3.29)

where C > 0 is chosen so that "

Rn

)(x) dx = 1.

Now, for each % > 0, set

)" = %!n)(x

%).

Such ) is called the standard mollifier, and )" " C$(Rn) statisfies"

Rn

)"(x) dx = 1, supp()") # B(0, %). (3.30)

The following lemma states some of the properties of mollifiers.

Lemma 3.2.10 If f : " % R is locally integrable, define

f " = )" * f =

"

!

)"(x! y)f(y) dy,

for x " "" = {x " "| dist(x, !") > %}. Then, it holds that

• (i) f " " C$("").

• (ii) f " % f , a.e. as % % 0.

• (iii) If f " C("), then f " % f uniformly on compact subsets of ".


We now give a proof to the regularity theorem using the mean-value formulas.Proof. Let ) be a standard mollifier, we will prove that u = u" on "". In fact, if x " "",then

u"(x) =

"

!

)"(x! y)u(y) dy

= %!n

"

B

(x, %))(|x! y|

%)u(y) dy

= %!n

" "

0

)(r

%)(

"

!B(x,r)

u(y) dS) dr

= %!nu(x)

" "

0

)(r

%)n$(n)rn!1 dr

= u(x)

"

B(0,")

)" dy = u(x).

Therefore, u " C$("") for any % > 0, and so u " C$(").

Remark 3.2.11 One should be careful that u is not assumed to have any regularity atboundary, even continuity.

A further applications of mean-value formulas will lead to the estimates on derivatives,for which we omit the proof.

Theorem 3.2.12 (Estimates on derivatives). Assume u is harmonic in ". Then it holdsthat

|D$u(x0)| &Ck

rn+k)u)L1(B(x0,r)) (3.31)

for each ball B(x0, r) # " and each multi-index $ of order |$| = k. Here, the Ck can bechosen as

C0 =1

$(n), Ck =

(2n+1nk)k

$(n), k = 1, · · · . (3.32)

The C$ regularity does not go to extreme of the harmonic functions. The followingtheorem states that harmonic functions are actually analytic.

Theorem 3.2.13 If u is harmonic in ", then u is analytic in ".

The next one confirms the analyticity of harmonic functions, which says that there areno nontrivial bounded harmonic functions on whole Rn.

Theorem 3.2.14 (Liouville’s Theorem). If u : Rn % R is harmonic and bounded, then uis constant.

3.3. GREEN’S FUNCTION 47

Proof. Fix x0 " Rn, r > 0, and apply Theorem 3.2.12 on B(x0, r),

|Du(x0)| &C1

rn+1)u)L1(B(x0,r))

& C1$(n)

r)u)L!(Rn) % 0, as r %..

Therefore, Du , 0, and so u is constant.

As a direct application of Liouville’s Theorem, we have

Theorem 3.2.15 (Representation formula). Let f " C20(Rn), n $ 3. Then any bounded

solution of!!u = f in Rn

has the form

u(x) =

"

Rn

$(x! y)f(y) dy + C, (x " Rn)

for some constant C.

Proof. Since $(x) % 0 as |x|%. for n $ 3,

u(x) =

"

Rn

$(x! y)f(y) dy

is a bounded solution of!!u = f in Rn.

If u is another bouned solution, w = u!u is bounded and harmonic, and thus is a constant.

Remark 3.2.16 When n = 2, $(x) is not bounded as |x|%., and so it is possible that"

Rn

$(x! y)f(y) dy

is not bounded as |x| % .. Therefore, the representation formula is not true in generalfor n = 2.

3.3 Green’s function

In last section, we obtained representation formula for problems on Rn. We now fix " tobe an bounded open domain in Rn with smooth boundary !". We will try to find a generalrepresentation formula for the solutions of the following boundary value problem

+!!u = f in ",

u = g on !".(3.33)


We intend to solve this problem through the Green’s representation formula (3.20). Werecall (3.20) below

u(x) = !"

!

$(x!y)!yu(y) dy!"

!!

(u(y)!$(x! y)

!"!$(x!y)

!u(y)

!") dSy, *x " ". (3.34)

It is clear that (3.34) permits us to solve the problem if we know how to deal with the term

of!u

!"on the boundary. Unfortunately, it is unknown to us. The idea is to introduce a

correction h(x, y) for each fixed x such that it solves the following boundary value problem.+

!!yh = 0 in ",

h = $(x! y) on !".(3.35)

By the Green’s formula, we have

!"

!

h(x, y)!yu(y) dy =

"

!!

(u(y)!h(x, y)

!"! h(x, y)

!u

!") dSy

=

"

!!

(u(y)!h(x, y)

!"! $(x! y)

!u

!") dSy.

(3.36)

We therefore arrived at the following definition.

Definition 3.3.1 Green’s function for the region " is

G(x, y) = $(x! y)! h(x, y), (x, y " ", x (= y).

With this notion, we add (3.36) to (3.34) to find

u(x) = !"

!

G(x, y)!yu(y) dy !"

!!

u(y)!G(x, y)

!"dSy, *x " ". (3.37)

where!G

!"(x, y) = 'yG(x, y) · "(y),

is the outer normal derivative of G with respect to variable y. The term!u

!"does not

appear in this formula (3.37). This means the correction h(x, y) is in its proper form.Now, suppose u(x) " C2(") solves the boundary value problem (3.33) for some contin-

uous functions f and g, using (3.37), we arrive at

Theorem 3.3.2 (Representation formula with Green’s function). If u(x) " C2(") solvesproblem (3.33), then

u(x) =

"

!

G(x, y)f(y) dy !"

!!

g(y)!G(x, y)

!"dSy, *x " ". (3.38)


Remark 3.3.3 Formally, (3.38) gives a nice formula to the solutions of the Poisson equa-tion with Dirichlet boundary condition if we know the Green’s function on the domain ",such a method is called Green’s method. However, for general domain, it is a di&cult taskto construct the Green’s function, for which requires to solve the problem (3.35). However,Green’s method is still significant in the following reasons:

• (i) For Poisson equation on a fixed domain, once we obtained the Green’s function,the existence of solutions to the problem (3.33) is given by the formula (3.38) for anycontinuous f and g.

• (ii) In the case when it is hard to find the solutions to (3.33), one can still use theformula (3.38) to discuss the certain behavior of the solutions.

• (iii) For some domains with simple geometry, explicit calculation of G is possible.The Dirichlet problem of Poisson equation on such domains are often important inthe applications.

• (iv) (3.38) transfers the (3.33) into a integral equation, which is convenient in certainoccasions even when the equation is semi-linear. The machinery of functional analysisis thus applied to obtain some interesting results.

Before going to specific examples, we discuss certain important properties of Green’sfunction.

First of all, it is clear that in ", when x (= y, G(x, y) is harmonic on x and on yeverywhere. Furthermore, G(x, y) % . as x % y at the order of |x ! y|n!2 if n > 2 andat the order of log |x! y| if n = 2.

Secondly, substituting u(x) = 1 into (3.37) we have"

!!

!G

!"dS = !1. (3.39)

Finally, we show that G(x, y) is symmetric in x and y.

Theorem 3.3.4 (Symmetry of Green’s function). For all x, y " ", x (= y, it holds that

G(y, x) = G(x, y).

Proof. Formally, we prove the theorem as following. For x (= y " ", $(|x! y|) is smoothon !", by definition, we know that

G(y, x) = $(x! y)! h(x, y), G(y, x) = $(y ! x)! h(y, x).

We note that $(x!y) = $(y!x) = $(|x!y|), and therefore, both h(x, y) and h(y, x) are theharmonic solutions of the same problem (3.35). By uniqueness we know h(x, y) = h(y, x)and therefore

G(y, x) = G(x, y), *x (= y " ".


3.3.1 Green’s function on a ball

We will construct Green’s function for the unit ball B(0, 1) using some reflection throughthe sphere !B(0, 1).

Definition 3.3.5 If x " Rn \ {0}, the point

x =x

|x|2

is called the point dual to x with respect to !B(0, 1). The mapping x % x is inversionthrough the unit sphere !B(0, 1).

We will use this inversion to construct Green’s function for the unit ball " = B(0, 1).Fix x " B(0, 1). We need to find a correction h(x, y) solving

+!!yh = 0 in B(0, 1),

h = $(x! y) on !B(0, 1),(3.40)

then the Green’s function reads

G(x, y) = $(x! y)! h(x, y).

We try to invert the singularity of $(x ! y) from x " B(0, 1) to x (" B(0, 1). Assumenow that n $ 3. The mapping

y % $(y ! x)

is harmonic for y (= x. Thus,y % |x|2!n$(y ! x)

is harmonic for y (= x, and so

h(x, y) = $(|x|(y ! x)) (3.41)

is harmonic in B(0, 1). Furthermore, if y " !B(0, 1) and x (= 0,

|x|2|y ! x|2 = |x|2'|y2 ! 2y · x

|x|2 +1

|x|2

(

= |x|2 ! 2y · x + 1 = |x! y|2.

Therefore, (|x||y ! x|)2!n = |x! y|2!n and so

h(x, y) = $(y ! x), *y " !B(0, 1). (3.42)

This verifies that h(x, y) is the one we were looking for.


Definition 3.3.6 Green’s function for the unit ball is

G(x, y) = $(y ! x)! $(|x|(y ! x)), x, y " B(0, 1), x (= y. (3.43)

We now solve the following boundary value problem+

!u = 0, in B(0, 1)

u = g, on !B(0, 1).(3.44)

By (3.38), we need to calculate!G

!"on the unit sphere. According to formula (3.43), for

y " !B(0, 1) one has!G

!yi=

!$(y ! x)

!yi! !$(|x|(y ! x))

!yi,

where!$(y ! x)

!yi=

!1

n$(n)

yi ! xi

|x! y|n ,

and

!$(|x|(y ! x))

!yi=

!1

n$(n)

yi|x|2 ! xi

(|x||y ! x|)n=

!1

n$(n)

yi|x|2 ! xi

|x! y|n , *y " !B(0, 1).

Therefore,!G

!"=

n!

i=1

yi!G

!yi(x, y)

=!1

n$(n)

1

|x! y|nn!

i=1

yi((yi ! xi)! yi|x|2 + xi)

=!1

n$(n)

1! |x|2

|x! y|n , *y " !B(0, 1).

Thus, we use (3.38) to yield the representation formula

u(x) =1! |x|2

n$(n)

"

!B(0,1)

g(y)

|x! y|n dSy. (3.45)

If now instead of (3.43), for r > 0, we want to solve the following boundary-valueproblem +

!u = 0, in B(0, r)

u = g, on !B(0, r).(3.46)

Then, u(x) = u(rx) solves (3.43) with g(x) = g(rx) replacing g in (3.43). After a directchange of variables, we obtain the Poisson’s formula

u(x) =r2 ! |x|2

n$(n)r

"

!B(0,r)

g(y)

|x! y|n dSy, *x " B(0, 1). (3.47)


The function

K(x, y) =r2 ! |x|2

n$(n)r

1

|x! y|n , x " B(0, r), y " !B(0, r), (3.48)

is called Poisson’s Kernel for the ball B(0, r).(3.47) was established under the assumption that (3.46) has a smooth solution. We

now prove that in fact (3.47) gives a solution.

Theorem 3.3.7 (Poisson’s formula for a ball). Assume g " C(!B(0, r)) and u is definedby (3.47). Then u is harmonic in B(0, r) and for each point x0 " !B(0, r),

limx#x0

u(x) = g(x0), x " B(0, r).

Proof. It is clear that K(x, y) $ 0 is harmonic when x (= y. Therefore, for x " B(0, r)and y " !B(0, r), we have

!u(x) =

"

!B(0,r)

!xK(x, y)g(y) dSy = 0.

To verify the boundary condition, we first remark that"

!B(0,r)

K(x, y) dSy = 1. (3.49)

Indeed, for each fixed y " !B(0, r), we denote x = +z where 0 & + < 1 and z " !B(0, r).By the mean-value formula for harmonic function, we have

1 = K(0, y)n$(n)rn!1 =

"

!B(0,r)

K(+z, y) dSz.

Now,

1 =

"

!B(0,r)

K(+z, y) dSz

=

"

!B(0,r)

K(+y, z) dSz

=

"

!B(0,r)

K(x, z) dSz.

Now, fix x0 " !B(0, r), % > 0. Choose & > 0 so small that

|g(y)! g(x0)| < %, if |y ! x0| < &, y " !B(0, r). (3.50)

Then, if |x! x0| < %2 , x " B(0, r), setting

V% = !B(0, r) +B(x0, &),


we compute

|u(x)! g(x0)| =

,,,,"

!B(0,r)

K(x, y)[g(y)! g(x0)] dSy

,,,,

&"

V"

K(x, y)|g(y)! g(x0)| dSy

+

"

!B(0,r)\V"

K(x, y)|g(y)! g(x0)| dSy

, I + J.

(3.51)

Now, (3.49)–(3.50) implies that

I & %

"

!B(0,r)

K(x, y) dSy = %.

For J , we note that if |x! x0| < %2 and |y ! x0| $ &, then

|y ! x| $ 1

2|y ! x0|.

Thus,

J & 2)g)L!

"

!B(0,r)\V"

K(x, y) dSy

& 2n+1)g)L!(r2 ! |x|2)n$(n)r

"

!B(0,r)\V"

|y ! x0|!n dSy

% 0, as x % x0.

Therefore, we could choose & > &" > 0 so small such that

|u(x)! g(x0)| < 2%, if |x! x0| < &".

This proves thatlim

x#x0

u(x) = g(x0), x " B(0, r).

3.3.2 Green’s function on half space

Again, we fix n $ 3. Let us consider the half space

Rn+ = {x = (x1, · · · , xn) " Rn|xn > 0}.

Although this region is unbounded, and so the calculations for Theorem 3.3.2 is notvalid. We will try to build Green’s function using the ideas developed so far. Later, wewill check directly that the derived representation formula gives the solution. We will alsouse the reflection idea about the boundary of the domain.


Definition 3.3.8 If x = (x1, · · · , xn!1, xn) " Rn+, its reflection in the plane !Rn

+ is thepoint

x = ((x1, · · · , xn!1,!xn).

We set

h(x, y) = $(y ! x) = $(y1 ! x1, · · · , yn!1 ! xn!1, yn + xn), (x, y " Rn+).

we note thath(x, y) = $(y ! x), if y " !Rn

+,

and hence +!h(x, y) = 0 in Rn

+

h(x, y) = $(y ! x) on !Rn+.

We thus has

Definition 3.3.9 Green’s function for the half-space Rn+ is

G(x, y) = $(y ! x)! $(y ! x), (x, y " Rn+, x (= y).

Clearly, if y " !Rn+,

!G

!"(x, y) = ! !G

!yn(x, y) =

2xn

n$(n)

1

|x! y|n .

Suppose u is a solution to the boundary-value problem+

!u = 0 in Rn+

u = g on !Rn+.

(3.52)

Then, formally, we expect from (3.38) that

u(x) =

"

!Rn+

K(x, y)g(y) dy, (x " Rn+) (3.53)

to be a representation formula for the solution. Here the function

K(x, y) =2xn

n$(n)

1

|x! y|n , (x, y " Rn+, x (= y), (3.54)

is the Poisson’s kernel for Rn+ and (3.53) is called the Poisson’s formula.

Similar to the case for a ball, we could verify directly that

Theorem 3.3.10 (Poisson’s formula for half-space). Assume g " C(Rn!1 + L$(Rn!1),and u is defined by (3.53). Then u is uniformly bounded harmonic function on Rn

+ and foreach x0 " !Rn

+, it holds that

limx#x0

u(x) = g(x0), x " Rn+.

3.4. HOPF’S MAXIMUM PRINCIPLE 55

3.4 Hopf’s maximum principle

We have put a lot of e#orts in the last section for the Dirichlet boundary value problemfor Poisson equation. The Green’s function method is particularly designed for this type ofproblems. The second boundary value problem, namely the Nuemann problem, is not wellstudied. We will establish the maximum principle for Nuemann problem in this section.

Theorem 3.4.1 (Hopf’s Lemma) Let B(y, R) # Rn (n $ 3), x0 " !B(y, R), u "C2(B(y, R)) + C1(B(y, R)) is sub-harmonic on B(y, R) such that

u(x0) > u(x), *x " B(y, R),

then!u

!"(x0) > 0,

where " is the outer unit normal of !B(y, R) at x0.

Proof. For ' " (0, R), and a positive parameter $ > 0, we define

v(x) = e!$r2 ! e!$R2, r = |x! y| > '.

Direct computation gives!v = e!$r2

(4$2r2 ! 2na).

Therefore, if we choose $ big enough, say $ = n# , then !v $ 0 on the region A =

B(y, R) \ B(y, '). We note that u(x) ! u(x0) < 0 on !B(y, '), there exists a % > 0 suchthat

w(x) = u(x)! u(x0) + %v(x) & 0, x " !B(y, ').

We note that v(x) = 0 on !B(y, R) and thus

w(x) & 0, x " !B(y, ').

We also note that w(x) is sub-harmonic on A, therefore the maximum principle for w onA implies that

w(x) & 0, *x " A.

But w(x0) = 0 and thus!u

!"(x0) $ 0,

that is!u

!"(x0) $ !%

!v

!"(x0) = !%v"(R) > 0.

We now introduce a concept concerning the structure of the boundary.


Definition 3.4.2 Let x0 " !", if there exists a ball B # " such that {x0} = B + ", wesay " satisfies the inner ball condition at x0. Meanwhile, we say Rn \ " satisfies the outerball condition at x0.

With this notion, based on Hopf’s lemma, we are able to prove the following

Theorem 3.4.3 (Hopf’s maximum principle) Assume u " C2(") + C1(") and !!u & 0(!!u $ 0), and x0 " !" such that

u(x0) > u(x) (u(x0) < u(x)), *x " ".

If " staisfies the inner ball condition at x0, then

!u

!"(x0) > 0 (

!u

!"(x0) < 0).

We now apply Hopf’s maximum principle to the Neumann problem. Consider+

!!u = f, in "!u!& = g, on !".

(3.55)

It is easy to see that the solution of the above Neumann problem, if exists, is not unique.For if u is one solution, then u + C for any constant C is another solution. However, wecould prove the following

Theorem 3.4.4 If " satisfies the inner ball condition at each boundary point, then thesolutions of Neumann problem (3.55) can only di!er by a constant.

Proof. Let u1 and u2 be two solutions to (3.55), then w = u1 ! u2 is the solution of

+!!w = f, in "!w!& = 0, on !".

Now, if w is not constant, by the maximum principle for harmonic function w, we know wattains its maximum at some x0 " !". According to Hopf’s maximum principle, we know

!w

!"(x0) > 0

which contradicts the boundary condition. So w is a constant.

So far, we proved the uniqueness and stability for Dirichlet problem by maximum prin-ciple, showed the relative uniqueness for Neumann problem via Hopf’s maximum principle.For the Dirichlet problem of Laplace equation on some special domains (such as a ball, andhalf space), we constructed Green’s functions and thus gave the existence for continuousboundary data. However, the solvability of Dirichlet and Neumann problems on generaldomains are not clear yer. These, however, will require the concept of weak solutions, forwhich the functional analysis will play central role. This topic will be presented later.

3.5. EXAMPLES 57

3.5 Examples

In the first two examples we construct the Green’s functions for the disk and upper halfplane in R2.

Example 3.5.1 Find the Green’s function for the unit disk

D(0, 1) = {(x1, x2)|x21 + x2

2 < 1} # R2.

Solution: We use the same idea as for n $ 3. Let ' =-

x21 + x2

2 < 1, $(x) be thefundamental solution. We recall that

$(x) = ! 1

2#log(|x|) = $(').

We will also use the inversion to invert the singularity out of the disk. Note for x = x|x|2

|x||y ! x| = |y ! x|, if |y| = 1.

The function $(|x|(y ! x)) is harmonic in y if x, y " D(0, 1) and

$(|x|(y ! x)) = $(x! y), for y " !D(0, 1).

Therefore, we have the Green’s function for the unit disk

G(x, y) = $(x! y)! $(|x|(y ! x)), x (= y " D(0, 1).

We now solve the Laplace equation on D(0, 1) with boundary data g(x). For this purpose,for |y| = 1, we compute

!G(y ! x)

!"= ! 1

2#

1! |x|2

|x! y|2 , *|y| = 1.

Therefore, we arrived at the Poisson’s formula

u(x) =1! |x|2

2#

"

!D(0,1)

g(y)

|x! y|2 d+y. (3.56)

We could now use polar coordinate to have another form of the above equation. Letx1 = ' cos(,), x2 = ' sin(,). and y1 = cos((), y2 = sin((), we thus have

u(', ,) =1

2#

" 2'

0

(1! '2)g(()

1! 2' cos((! ,) + '2d(. (3.57)

If the unit disk is replaced by a general disk D(0, r), (3.56) and (3.57) are replaced by

u(x) =r2 ! |x|2

2#r

"

!D(0,r)

g(y)

|x! y|2 d+y. (3.58)

u(', ,) =1

2#

" 2'

0

(r2 ! '2)g(()

r2 ! 2r' cos((! ,) + '2d(. (3.59)


Example 3.5.2 Find the Green’s function for the upper half plane

R2+ = {(x1, x2)|x2 > 0}.

Solution. Similar to subsection 3.3.2, we choose

G(x, y) = $(y ! x)! $(y ! x)

wherex = (x1,!x2), for x = (x1, x2) " R2

+.

Therefore

G(x, y) = ! 1

2#log

.-(y1 ! x1)2 + (y2 ! x2)2

-(y1 ! x1)2 + (y2 + x2)2

/. (3.60)

The corresponding Poisson’s formula for Laplace equation with boundary data g(x) onR2

+ is

u(x) =1

#

" $

!$

x2g(y1)-(y1 ! x1)2 + x2

2

dy1 (3.61)

Example 3.5.3 For a > 0, D(0, a) is the disk centered at the origin on R2. Solve thefollowing boundary value problem

#$%

$&

!!u = 0, in D(0, a),

u(a, () = g(() =

+1, 0 < ( < #,

0, # < ( < 2#.

Solution: We could use (3.59) to solve this problem:

u(', ,) =1

2#

" 2'

0

(a2 ! '2)g(()

a2 ! 2a' cos((! ,) + '2d(

=a2 ! '2

2#

" '

0

1

(a2 + '2)! 2a' cos((! ,)d(.

Set c = a2 + '2, d = !2a' and - = (! ,, we reduce the above equation into

u(', ,) =a2 ! '2

2#

" '!(

!(

1

c + d cos(-)d-.

Except for the singular points - = ±#, we have that

F (-) =2/

c2 ! d2arctan

./c2 ! d2 tan( )

2)

c + d

/,

3.5. EXAMPLES 59

is the anti-derivative of1

c + d cos(-).

We note that ( " (0, #) and , " [0, 2#], therefore, if , " (0, #), - " (!#, #), and so

u(', ,) = lim"#0

.a2 ! '2

2#

2

a2 ! '2arctan

./c2 ! d2 tan(*!(

2 )

c + d

//,,,,,

*='!"

*="

=1

#arctan(

a + '

a! 'cot(

,

2)) +

1

#arctan(

a + '

a! 'tan(

,

2))

However, if , " (#, 2#), then - " (!2#, 0) which contains - = !#. We have to split theintegral at - = !#. Therefore, we have

u(', ,) = lim"#0

'1

#arctan

'a + '

a! 'tan(

(! ,

2)

((,,,,*=(!'!"

*="

+ lim"#0

'1

#arctan

'a + '

a! 'tan(

(! ,

2)

((,,,,*='!"

*=(!'+"

=1

#

#

2+

1

#arctan(

a + '

a! 'tan(

,

2))

+1

#arctan(

a + '

a! 'cot(

,

2)) +

1

#

#

2

= 1 +1

#arctan(

a + '

a! 'cot(

,

2)) +

1

#arctan(

a + '

a! 'tan(

,

2)).

One easily verify the boundary condition by taking limit ' % a in the above two cases.

The above example shows that using Poisson’s formula could lead to tedious calcula-tions. We show in the following couple examples some specific tricks in two and threedimensions.

Example 3.5.4 Solve the following Dirichlet problem

+!!u = 0, in D(0, 1)

u(', ,) = A sin2 , + B cos2 ,, on ' = 1,

where x = (x1, x2) = (' cos ,, ' sin ,) and A and B are constants.

Solution. It is easy to check that 'n sin(n,) and 'n cos(n,) are harmonic functions on R2,and

A sin2 , + B cos2 , =A + B

2+

B ! A

2cos(2,),

therefore, we find the solution

u(', ,) =A + B

2+

B ! A

2'2 cos(2,).


Example 3.5.5 Find a bounded solution to the following Dirichlet problem outside a unitball in R3:

+!!u = 0, r > 1,

u|r=1 = 25+4x2

,

where r = |x|.Solution. We know that the function

u(x) =1

|x! x0|is a harmonic function out of the unit ball if x0 " B(0, 1). We try to see if we could choosea x0 so that the above function satisfies the boundary condition.

Since 25+4x2

= (54 + x2)!1, we need to find x0 such that

5

4+ x2 = |x! x0|2 = 1! 2x · x0 + x2

0

which is equivalent to

x01 = x03 = 0, x02 = !1

2.

And such a point is inside the unit ball. Therefore,

u(x) =10

x21 + (x2 + 1

2)2 + x2

3

.

Example 3.5.6 Let " be the triangle on R2 with vertices (!1, 0), (1, 0) and (0,/

3). Solvethe following Dirichlet problem

+!!u = 2, in "

u = 0, on !".

Solution. We first observe the equations for the sides of the triangle are

y = 0, y +/

3x!/

3 = 0, y !/

3x!/

3 = 0.

We thus guess that the solution has the following form

u(x, y) = cy(y +/

3x!/

3)(y !/

3x!/

3)

with c the constant to be determined. Clearly, the boundary condition is fulfilled. A directcalculation gives

!!u = 4/

3c = 2,

and so c =%

36 and the solution is

u(x, y) =

/3

6y(y +

/3x!

/3)(y !

/3x!

/3).

3.5. EXAMPLES 61

The following example is the famous Hadamard’s three circles theorem.

Example 3.5.7 Let D be a annular region on R2 centered at the origin. The outer circlehas radius R1 and the inner one has radius R1, u(x, y) is a sub-harmonic function on D.Set

M(r) = maxx2+y2=r2

u(x, y), R1 < r1 < r < r2 < R2,

then

M(r) &M(r1) log( r2

r ) + M(r2) log( rr1

)

log( r2r1

).

Solution. For r (= 0, we define

((r) = a + b log r

where a and b are chosen such that

((r1) = M(r1), ((r2) = M(r2).

Therefore, we find

((r) =M(r1) log( r2

r ) + M(r2) log( rr1

)

log( r2r1

).

Consider now

v(x, y) = u(x, y)! ((-

x2 + y2),

which satisfies +!!v & 0, if r1 < r < r2

v & 0, if r = r1 or r = r2.

By the maximum principle, we know that

v & 0, if r1 < r < r2.

Therefore,

u(x, y) & ((r), if r1 < r < r2.

This implies that

M(r) & ((r), if r1 < r < r2.


3.6 Problems

Problem 1. Show that the Laplace operator takes the following form under cylindricalcoordination (r, ,, z):

!u =1

r

!

!r(r

!u

!r) +

1

r2

!2u

!,2+

!2u

!z2.

Problem 2. Show that the Laplace operator takes the following form under sphericalcoordinate (r, ,, ().

!u =1

r2

!

!r(r2!u

!r) +

1

r2 sin2 ,

)!

!,(sin ,

!u

!,) +

!2u

!(2

*.

Problem 3. Prove the following functions are harmonic.

• (a) x3 ! 3xy2, and 3x2y ! y3.

• (b) sh(ny)sin(nx), sh(ny)cos(nx), ch(ny)sin(nx), and ch(ny)cos(nx).

• (c) sh(x)(ch(x) + cos(y))!1 and sin(y)(ch(x) + cos(y))!1.

Problem 4. Prove the following functions are harmonic in the polar coordinate.

• (a) ln(r), and ,.

• (b) rn cos(n,) and rn sin(n,).

• (c) r ln(r) cos(,)! r, sin(,) and r ln(r) sin(,) + r, cos(,).

Problem 5. Find the Green’s function for the first quadrant of R2, namely the domain

" = {(x, y) " R2|x > 0, y > 0}.

Problem 6. Find the Green’s function for the upper half ball B+(0, r) in R3.

Problem 7. Find the Green’s function for the first octant in R3, namely the domain

" = {(x, y, z) " R3|x > 0, y > 0, z > 0}.

Problem 8. Find the Green’s function for the unit square in R2, namely the domain

" = {(x, y) " R2|1 > x > 0, 1 > y > 0}.

Problem 9. Let D(0, r) is the disk on R2 with boundary C. For each of the followingboundary conditions, find the function u so that it is harmonic on D(0, r).

3.6. PROBLEMS 63

• (a) u|C = A cos(().

• (b) u|C = A + B sin(().

Problem 10. Solve the following Dirichlet problem+

uxx + uyy + uzz = 0, x2 + y2 + z2 < 1,

u(r, ,, ()|r=1 = 3 cos(2,) + 1,

where (r, ,, () is the spherical coordinate.

Problem 11. Let B be a unit ball in Rn (n $ 2), and u is the smooth solution of thefollowing problem +

!!u = f in B

u = g, on !B.

Prove that there exists a constant C, depending only on n, such that

maxB

|u| & C(max!B

|g| + max!B

|f |).

Problem 12. Assume u is harmonic, prove the following statements

• (a) If ( : R % R is a smooth convex function, then v = ((u) is sub-harmonic.

• (b) Prove v = |Du|2 is sub-harmonic.

Problem 13. Use Poisson’s formula for the ball to prove

rn!2 r ! |x|(r + |x|)n!1

u(0) & u(x) & rn!2 r + |x|(r ! |x|)n!1

u(0),

where u(x) is harmonic for x " B(0, r) # Rn with n $ 3. This is an explicit form ofHarnack’s inequality.

Problem 14. Let u be the solution of+

!!u = 0, in Rn+

u = g on !Rn+

given by the Poisson’s formula for the half-space. Assume g is bounded and g(x) = |x| forx " !Rn

+, |x| & 1. Show Du is not bounded near x = 0.

Problem 15. Let "+ # Rn+ and T = !"+ + !Rn

+ is a non-empty open set. Assumeu " C("+) is harmonic in "+, with u = 0 on T . Set


v(x) =

+u(x1, · · · , xn!1, xn), xn > 0

!u(x1, · · · , xn!1,!xn), xn < 0.

Prove that v(x) is harmonic on "+0T 0"! where "! is the reflection of "+ about xn = 0.This result is called Schwarz reflection theorem.

Problem 16. Using example to show that the maximum principle is not valid for

uxx + uyy + cu = 0, c > 0.

Problem 17. Find the Green’s function for the following wedge:

" = {(', ,, z) : ' > 0, 0 < , <#

4, z " R}.

Problem 18. Find the Green’s function for a domain between two parallel planes:

" = {(x, y, z) : 0 < z < 1}.

Chapter 4

Heat equation

Let ! be an open set in Rn, we will study the Heat equation

ut ! µ"u = 0, x " !, t > 0 (4.1)

and the non-homogeneous Heat equation

ut ! µ"u = f(x, t), x " !, t > 0 (4.2)

with appropriate initial boundary conditions. The constant µ > 0 models the heat di#u-sion.

A guiding principle is that any assertion about harmonic functions yields a analogousstatement about solutions of the heat equation. However, we will follow a slight di#erentapproach to show di#erent techniques.

4.1 Physical derivation

Assume ! was occupied by the homogeneous media without heat source inside. Let u(x, t)be the temperature at x and time t, J(x, t) is the heat flux. For each sub-region G withsmooth boundary !G and unit outer normal ",

!

!G

J · " dS

is the total heat flow out of G per unit time. According to Fourier’s law,

J = !k#xu

with k > 0 the constant of heat conductivity. Therefore, we update the last equation by

!!

!G

k!u

!"dS.

65

66 CHAPTER 4. HEAT EQUATION

On the other hand, the first principle of thermodynamics says that the absolute tempera-ture u(x, t) is proportional to the heat, therefore, we know

!

G

CV ut dx = !!

!G

k!u

!"dS,

where CV > 0 is the specific heat constant. Now, by divergence theorem, one has!

G

ut ! µ"u dx = 0

with µ = kCV

. This leads to heat equation. The f in non-homogeneous heat equationmodels the interior heat source.

4.2 Fundamental solution

4.2.1 Derivation

We observe that the Heat Equation is invariant under the transformation:

x $ #x, t $ #t, for # " R.

Therefore, the ratio |x|2t is an important variable for heat equation, and we shall look for a

solution of the form

u(x, t) = v(r2

t), r = |x|, t > 0. (4.3)

Such class of solutions are called self-similar solution.A quicker approach is to find a solution u of the special structure

u(x, t) = t!"v(|x|%

t).

Let y = x"t, and we substitute it into (4.1) to obtain

$v +1

2y · Dv + "v = 0 (4.4)

which reduces into

$w +1

2rw# + w## +

n! 1

rw# = 0

for w(r) = v(|y|) where r = |y|. Now, $ = n2 is the magic number so that the equation

becomes

(rn!1w#)# +1

2(rnw)# = 0.

Hence,

rn!1w# +1

2rnw = 0

4.3. FOURIER TRANSFORM AND INTIAL VALUE PROBLEM 67

if we require w and w# tends to 0 if r $&. Therefore,

w = be!r2

4

for some constant b. Therefore,

u(x, t) = bt!n2 e!

|x|24t

solves (4.1).This motivates the following deinition:

Definition 4.2.1 The function

E(x, t) =

"(4%t)!

n2 e!

|x|24t , (x " Rn, t > 0)

0 (x " Rn, t > 0),

is called the fundamental solution of the heat equation.

We note that E(x, t) is smooth except the singular point (0, 0). In the next section,the fundamental solution is naturally derived when we apply Fourier transform to solvethe initial value problem.

4.3 Fourier transform and intial value problem

4.3.1 Fourier transform

Let function f(x) and g(x) defined on x " Rn are continuously di#erentiable and absolutelyintegrable, then the Fourier transform of f(x) is defined as

F(f)(&) = f(&) = (2%)!n2

!

Rnx

f(x)e!ix·# dx (4.5)

and the inverse Fourier transform of g(&) is

F!1(g)(x) = g(x) = (2%)!n2

!

Rn!

f(x)eix·# d&. (4.6)

We remark that the Fourier transform and its inverse are naturally generalized into thesquare integrable function space L2(Rn). We recall that for a function u(x) " L2(Rn),

'u'L2(Rn) =

#!

Rn

u2(x) dx

$ 12

< &.

We also recall that

f ( g =

!

Rn

f(y)g(x! y) dy.

We list some properties of Fourier transform in the following theorem.


Theorem 4.3.1 (Properties of Fourier Transform). Assume f, g " L2(Rn).

• (i) 'f'L2(Rn) = 'f'L2(Rn) = 'f'L2(Rn);

• (ii)

!

Rn

fg dx =

!

Rn

f ¯g d&,

• (iii) For each multi-index $ such that D"f " L2(Rn), %D"f = (i&)"f .

• (iv) !(f ( g) = (2%)n2 f g.

• (v) f = ˇf.

Example 4.3.2 For x " R, find the Fourier transform for f(x) = e!a|x|.

Solution.

f(&) =1%2%

!

Re!a|x|e!ix# dx

=1%2%

!

Re!a|x|(cos(x&)! isin(x&)) dx

= 21%2%

! $

0

eaxcos(x&) dx

=1%2%

2a

&2 + a2.

Example 4.3.3 For & " Rn and t > 0, find the inverse Fourier transform for

f(&) = (2%)!n2 e!|#|2t.

Solution.

f(x) = (2%)!n

!

Rn

e!|#|2teix·# d&

=n&

k=1

#1

2%

! $

!$e!t#2

k+ixk#k d&k

$

=n&

k=1

#1

%

! $

0

e!t#2kcos(xk&k) d&k

$

=n&

k=1

I(xk).

By Euler’s formula ! $

0

e!y2dy =

%%

2,

we know that

I(0) =1

2%

%t.


Di!erentiating I(xk) once, and using integration by parts, one has

I # +xk

2tI = 0,

therefore,

I(xk) =1

2%

%te!

x2k

4t .

Finally, we obtain

f(x) = F!1((2%)!n2 e!|#|2t) = (4%t)!

n2 e!

|x|24t = E(x, t).

4.3.2 Initial Value Problem

We now employ the Fourier transform to solve the following Cauchy problem for heatequation "

ut !"u = 0

u(x, 0) = '(x).(4.7)

We perform Fourier transform in x, and denote

u(&, t) = (2%)!n2

!

Rn

u(x, t)e!ix·# dx

'(&) = (2%)!n2

!

Rn

'(x)e!ix·# dx,

we thus arrive at an initial value problem for the ODE about u(&, t)'(

)

du(&, t)

dt+ |&|2u(&, t) = 0

u(&, 0) = '(&).(4.8)

Clearly, one hasu(&, t) = '(&)e!|#|2t,

therefore, we perform inverse Fourier transform

u(x, t) = F!1(u(&, t))

= F!1('(&)e!|#|2t)

= F!1('(&)) ( F!1((2%)!n2 e!|#|2t)

= E(x, t) ( '(x)

=

!

Rn

E(x! y, t)'(y) dy.

(4.9)

This also gives a natural way to derive the fundamental solution E(x, t). We list someproperties here for later applications:


Theorem 4.3.4 (Properties of Fundamental solution) Let E(x, t) be the fundamental so-lution of Heat equation.

• (i) For )x, y " Rn and t > 0, E(x! y, t) > 0.

• (ii) For )x, y " Rn and t > 0, (!t !")E(x! y, t) = 0.

• (iii) For )x " Rn and t > 0,

!

Rn

E(x! y, t) dy = 1.

• (iv) For any ( > 0 and )x " Rn, limt%0+

!

|y!x|>$

E(x! y, t) dy = 0.

Proof. By the expression of E(x, t), the first two properties are obvious. For (iii), usingsubstitution y = x +

%4tz, we compute!

Rn

E(x! y, t) dy = (4%t)!n2

!

Rn

e!|x!y|2

4t dy

= (%)!n2

!

Rn

e!|z|2 dz

= 1.

Using the same substitution, we can prove (iv) as following

limt%0+

!

|y!x|>$

E(x! y, t) dy

= limt%0+

(%)!n2

!

|z|> ""4t

e!|z|2 dz = 0.

We remark that the formula (4.9) gives a formal solution to the intial value problem(4.7), where we assumed that the existence of Fourier transform of initial function '(x),and we used the inverse transformation, this often requires '(x) to be C1 and absolutelyintegrable. However, under much weaker conditions on '(x), we are able to prove that(4.9) does give the classical solution to (4.7).

Theorem 4.3.5 If '(x) " C(Rn) and there exist constants M > 0 and A > 0 such that

|'(x)| * MeA|x|2 , )x " Rn,

thenu(x, t) = E(x, t) ( '(x)

is a C$ solution of (4.7) on the region

! = {(x, t)|x " Rn, 0 < t * T}, for T <1

4A.


Proof. 1. We first show the continuity of u(x, t). For any constants a > 0 and t0 " (0, T ),we define the set

V = {(x, t)||x| * a, t0 * t * T}.

For (x, t) " V , we see that

|u(x, t)| * M

!

Rn

E(x! y, t)eA|y|2 dy

* cM

!

Rn

exp{A|y|2 + A|x! y|2} dy,

where c = (4%t0)!n2 and A = ! 1

4T . We note that

A|y|2 + A|x! y|2 = (A + A)|y ! A

A + Ax|2 +

AA

A + A|x|2.

Therefore, for A + A < 0, i.e., T < 14A , we have

|u(x, t)| * MeAA

A+A|x|2

!

Rn

exp{(A + A)|y ! A

A + Ax|2} dy

* cM(!%

A + A)!

n2 e

AAA+A

|x|2 .

Therefore, we know that the integral in (4.9) converges uniformly and absolutely on V ,and so u(x, t) is continuous on V . For a > 0 and t0 > 0 arbitrary, u(x, t) is continuous on!.

2. We now show that u(x, t) " C$(!) and it satisfies the Heat equation. This isbecause E(x ! y, t) is infinitely di#erentiable, with uniformly bounded derivatives, on V .For each multi-index $ = ($0, $1, · · · , $n),

D"u(x, t) =

!

Rn

'(y)D"E(x! y, t) dy.

One can estimate this integral in a similar manner as in step 1 on V to prove its absoluteand uniform convergence on V . Therefore, one proves that u(x, t) " C$(!). We now usethe property of E(x, t) to show that

(!t !")u(x, t) =

!

Rn

'(y)(!t !")E(x! y, t) dy = 0.

3. Finally, we prove that u(x, t) verifies the initial condition, namely, we shall prove forany x0 " Rn,

lim(x,t)%(x0,0+)

u(x, t) = '(x0).


Letting v0(x) = '(x)! '(x0), using the property of E(x, t), one has

u(x, t) =

!

Rn

'(x0)E(x! y, t) dy +

!

Rn

v0(y)E(x! y, t) dy

= '(x0) +

!

Rn

v0(y)E(x! y, t) dy.

Since v0(x) is continuous and v0(x0) = 0, for any fixed ) > 0 there exists ( such that when|x! x0| < (, |v0(x)| < ). For |x! x0| > (, there exists B()) > 0 large enough such that

|v0(x)| * )eB(%)|x!x0|2 .

Therefore, for any x " Rn, one has

|!

Rn

v0(y)E(x! y, t) dy|

*!

B

(x0, ()|v0(y)|E(x! y, t) dy +

!

Rn\B(x0,$)

|v0(y)|E(x! y, t) dy

* ) +)

(4%t)n2

!

Rn

exp{B())|y|2 ! |x! y|2

4t} dy

* ) +)

(1! 4Bt)n2e

B|x|21!4Bt ,

where we have chosen t so small such that 1 ! 4Bt > 0. Now, when t is su$cient small,we see that |u(x, t)! '(x0)| < 2). Therefore, we verified that

lim(x,t)%(x0,0+)

u(x, t) = '(x0).

Remark 4.3.6 Some remarks are in order.

• 1. By the property of the fundamental solution E(x, t), it is easy to see that

u(x, t) =

!

Rn

E(x! y, t)'(y) dy * supy&Rn

'(y).

Similarly, on finds the bound from below. Therefore, if '(x) is bounded, then

infy&Rn

'(y) * u(x, t) * supy&Rn

'(y) (4.10)

• 2. From the theorem, it is obvious that the smaller A is, the larger existence time.If '(x) is bounded, A = 0, then the solution is global in time.

• 3. It also clear that even if '(x) is replaced by measurable function, while otherconditons keep the same, u(x, t) given in (4.9) is still the C$ solution to (4.7), andverifies the initial conditions at all the continuous points of '.

• 4. If '(x) + 0 and ' ,- 0, the u(x, t) > 0 for any x " Rn and t > 0. This is called theinfinite propogation speed for disturbances. If the initial temperature is nonnegativeand positive somewhere, the temperature is positive everywhere at any later time.


4.3.3 Nonhomogeneous Problem

We now derive a general formula for the following initial-value problem"

ut !"u = f(x, t) x " Rn, t > 0

u(x, 0) = 0.(4.11)

For this purpose, we note that

u(x, t; s) =

!

Rn

E(x! y, t! s)f(y, s) dy

solves the following initial value poblem"

ut(x, t; s)!"xu(x, t; s) = 0, x " Rn, t s,

u(x, s; s) = f(x, s).(4.12)

Of course, u(x, t; s) does not solve (4.11), however, the Duhamel’s Principle allows us tobuild a solution of (4.11) from the solutions of (4.12), by integrating with respect to s.More precisely, consider

u(x, t) =

! t

0

u(x, t; s) ds

=

! t

0

!

Rn

E(x! y, t! s)f(s, y) dyds, x " Rn, t > 0.

(4.13)

The following Theorem confirms that (4.13) gives the solution to (4.11).

Theorem 4.3.7 Assume f " C21((0,&) . Rn) has compact support. Deinfe u by (4.13),

then

• (i) u " C21((0,&). Rn),

• (ii) ut !"u = f(x, t),

• (iii) For each x0 " Rn, lim(x,t)%(x0,0+)

u(x, t) = 0.

Proof.1. We first change variables to write

u(x, t) =

! t

0

!

Rn

E(y, s)f(x! y, t! s) dyds.

We note that E(y, s) is smooth near s = t > 0 and f has compact support, then

ut(x, t) =

! t

0

!

Rn

E(y, s)ft(x! y, t! s) dyds

+

!

Rn

E(y, s)f(x! y, 0) dy,


uxi(x, t) =

! t

0

!

Rn

E(y, s)fxi(x! y, t! s) dyds,

uxixj(x, t) =

! t

0

!

Rn

E(y, s)fxixj(x! y, t! s) dyds.

Thus, u(x, t) " C21((0,&). Rn).

2. We not compute

ut !"u

=

! t

0

!

Rn

E(y, s)[(!t !"x)f(x! y, t! s)] dyds +

!

Rn

E(y, s)f(x! y, 0) dy

= I% + J% + K,

where,

I% =

! t

%

!

Rn

E(y, s)[(!!s !"y)f(x! y, t! s)] dyds,

J% =

! %

0

!

Rn

E(y, s)[(!!s !"y)f(x! y, t! s)] dyds,

K =

!

Rn

E(y, s)f(x! y, 0) dy.

Now,

|J%| * ('ft'L# + 'D2f'L#)

! %

0

!

Rn

E(y, s) dyds * C).

Integrating by parts, we find

I% =

! t

%

!

Rn

[(!!s !"y)E(y, s)]f(x! y, t! s)] dyds

+

!

Rn

E(y, ))f(x! y, t! )) dy !!

Rn

E(y, s)f(x! y, 0) dy.

=

!

Rn

E(y, ))f(x! y, t! )) dy !K,

Therefore, we conclude that

ut !"u = lim%%0

!

Rn

E(y, ))f(x! y, t! )) dy

= f(x, t), x " Rn, t > 0.

Finally, we note that'u(x, t)'L# * t'f'L# ,

which tends to zero as t $ 0+. This concludes the proof.

4.4. MAXIMUM PRINCIPLE AND APPLICATIONS 75

By linear superposition principle, we know that

u(x, t) =

!

Rn

E(x! y, t)'(y) dy +

! t

0

!

Rn

E(x! y, t! s)f(s, y) dyds (4.14)

solves "ut !"u = f(x, t) x " Rn, t > 0

u(x, 0) = '(x),(4.15)

under conditions on ' and f as above.

4.4 Maximum Principle and applications

4.4.1 Maximum Principle

Assume ! / Rn is open, bounded set. We first introduce the following concept.

Definition 4.4.1 Fix a time T > 0, the parabolic cylinder

!T = !. (0, T ].

The parabolic boundary of !T is%T = !T \ !T .

We remark that the parabolic interior of !T contains the top ! . {t = T}. Theparabolic boundary %T comprises the bottom, the vertical sides !! . [0, T ], but not thetop.

We now state the maximum principle.

Theorem 4.4.2 (Maximum Principle) Assume u(x, t) " C21(!T ) solves the heat equation

in !T , thenmax!T

u = max"T

u.

Proof. Assume that u(x, t) does not attains it maximum at %T but at a point (x', t') " !T .Namely, there exists m < M such that

max!T

u = u(x', t') = M > m = max"T

u.

Define

v(x, t) = u(x, t) +M !m

2nd2|x! x'|2,

where d = 2 max dist{x', !!}. We see that v(x', t') = M and

v(x, t)|"T < m +M !m

2n< M.


Therefore, v(x, t) attains its maximum at some (x1, t1) " !T . We observe that at thispoint (x1, t1),

"v(x1, t1) * 0, vt(x1, t1) + 0,

and sovt !"v + 0, at (x1, t1).

But on the other hand, direct computation gives

vt !"v = !M !m

d2< 0,

a contradiction. Therefore,max!T

u = max"T

u.

If one replaces u with !u, one actually obtains that

max!T

|u| = max"T

|u|.

The following comparison principle is a direct consequence of the above maximumprinciple.

Theorem 4.4.3 (Comparison Principle) Let u1, u2 " C21(!T ) be two solutions of the heat

equation. Ifu1 * u2, on %T ,

thenu1 * u2, on !T .

In order to restore the strong maximum principle, we introduce the parabolic mean-value formula. we introduce the following concept.

Definition 4.4.4 For fixed x " Rn, t > 0 and r > 0, we define

V (x, t; r) =

*(y, s) " Rn+1

+ |s * t, E(x! y), t! s) + 1

rn

+.

We remark that the boundary of V (x, t; r) is a level set of E(x ! y, t ! s), and the point(x, t) is at the center of the top. With the help of this notion, we have

Theorem 4.4.5 (Mean-value property for heat equation). Let u " C21(!T ) solve the heat

equation. Then

u(x, t) =1

4r2

!!

V (x,t;s)

u(y, s)|x! y|2

(t! s)2dyds, (4.16)

for each V (x, t; r) / !T .

With the help of this theorem, a similar argument as for Laplace equation, one canprove

4.4. MAXIMUM PRINCIPLE AND APPLICATIONS 77

Theorem 4.4.6 (Strong maximum principle). Let u " C21(!T ) solve the heat equation. If

! is connected and there exists a point (x0, t0) " !T such that

u(x0, t0) = max!T

u,

then u is constant in !t0.

4.4.2 Uniqueness and Stability

The first application is to the following initial boundary value problem:"

ut !"u = f(x, t), (x, t) " !T ,

u|"T = '(x, t).(4.17)

Theorem 4.4.7 The solution of (4.17) is unique and continuously depends on the initialboundary data.

Proof. If ui(i = 1, 2) are solutions of"

uit !"ui = f(x, t), (x, t) " !T ,

ui|"T = 'i(x, t),

then w = u1 ! u2 solves"

wt !"w = f(x, t), (x, t) " !T ,

w|"T = '1 ! '2.

By maximum principle, one has

max!T

|w| = max!T

|'1 ! &2|.

This shows the continuous dependence. In particular, if '1 = '2, w - 0, one finds theuniqueness.

We now turn to the initial value problem. For this purpose we first establish themaximum principle for the Cauchy problem (4.7). From Theorem 4.3.5, we know that if'(x) " C(Rn) satisfies the growth condition

|'(x)| * MeA|x|2 ,

then the formula (4.9) u(x, t) = E(x, t) ( '(x) solves (4.7) for any x " Rn, 0 * t * T(T < 1

4T ). Furthermore, from the proof of Theorem 4.3.5, we know that there exist M1 > 0and A1 + 0, such that

|u(x, t)| * M1eA1|x|2 , (x, t) " Rn . [0, T ].

This motivates the following maximum principle.


Theorem 4.4.8 (Maximum principle of Cauchy problem). Let u " C21(Rn . (0, T ]) 0

C(Rn . [0, T ]) solves (4.7), and satisfies the growth estimate

|u(x, t)| * MeA|x|2 , (x, t) " Rn . [0, T ], (4.18)

for constants M > 0, A + 0. Then

supRn([0,T ]

u = supRn

'.

Proof. We first assume4AT < 1 (4.19)

and so there is ) > 0 that4A(T + )) < 1.

Fix y " Rn, µ > 0, we define

v(x, t) = u(x, t)! µ

(T + )! t)n2e

|x!y|24(T+#!t) . (4.20)

Clearly, it holdsvt !"v = 0, in Rn . (0, T ].

Fix r > 0, define !T = B(y, r). (0, T ]. By maximum principle,

max!T

v = max"T

v.

We not check the value of v at %T . First of all, if x " Rn,

v(x, 0) = u(x, 0)! µ

(T + ))n2e

|x!y|24(T+#)

* u(x, 0) = '(x).

(4.21)

On the sides where |x! y| = r, t " [0, T ], we have

v(x, 0) = u(x, 0)! µ

(T + )! t)n2e

r2

4(T+#!t)

* MeA(|y|+r)2 ! µ

(T + ))n2e

r2

4(T+#) .(4.22)

We now know 14(T+%) = A + ( for some small ( > 0. Thus we have

v(x, t) * MeA(|y|+r)2 ! µ(4(A + ()n2 e(A+$)r2 * sup

Rn',

for r su$ciently large. We thus conclude that

v(x, t) * supRn

',

4.5. EXAMPLES 79

for all x " Rn and t * T as long as 4AT < 1. Let µ $ 0, we showed that

u(x, t) * supRn

', if 4AT < 1.

For general case where (4.19) fails, we can repeat the above procedure on [0, 18A ], [ 1

8A , 28A ],

· · · . Thus the proof of the theorem is complete.Using this maximum principle, one easily prove the following uniqueness result.

Theorem 4.4.9 (Uniqueness for Cauchy problem). Let '(x) " C(Rn), f " C(Rn.[0, T ]).Then there exists at most one solution u " C2

1(Rn . (0, T ]) 0 C(Rn . [0, T ]) of (4.15) ,satisfying the growth estimate

|u(x, t)| * MeA|x|2 , (x, t) " Rn . [0, T ], (4.23)

for constants M > 0, A + 0.

Now, a natural question arises: Is there any other solutions to (4.15) if we don’t requirethe growth restriction (4.23). The answer is YES. This will be explained in the examples.

4.5 Examples

The first example of A. N. Tychonov is to answer the uniqueness problem for the Cauchyproblem without growth restriction (4.23).

Example 4.5.1 There are infinitely many solutions to the initial value problem"

ut ! uxx = 0, x " R, t > 0,

u(x, 0) = 0,

without the growth restriction (4.23).

Solution. For some g(t) " C$(R), we define

u(x, t) =$,

k=0

dkg(t)

dtkx2k

(2k)!, x " R, t " R. (4.24)

If the series converge in a nice way, we have

ut =$,

k=0

dk+1g(t)

dtk+1

x2k

2k!,

and

uxx =$,

k=1

dkg(t)

dtkx2k!2

(2k ! 2)!=

$,

k=0

dk+1g(t)

dtk+1

x2k

2k!.


Therefore, we see u(x, t) solves the heat equation. Now, we choose

g(t) =

"exp{!t!"}, $ > 1, t > 0

0, t * 0.

It is clear that g(t) is analytic except for t = 0. It remains to verify that u(x, t) attainsthe initial data when t $ 0+. For this purpose, we need to compute the derivatives ofg(t). Due to complex analysis, the Cauchy integral formula, we have

dkg(t)

dtk=

k!

2%i

!

"

e!z!$

|z ! t|k+1dz,

where the path % is chosen as the circle: |z ! t| = *t for * " (0, 1). For Re(z) > 0, z" isdefined as its principal value. Now, for some # " R, the point on % is described as

z = t + *tei& = t(1 + *ei&), Re(!z!") = !t"Re(1 + ei&)!".

For small *, we have

Re(1 + ei&)!" >1

2, and so

Re(!z!") < !1

2t!", |d

kg(t)

dtk| * k!

(*t)ke!

12t$ .

Note that k!(2k)! < 1

k! , we have

|u(x, t)| *$,

k=0

|x|2k

k!(*t)ke!

12t$ = exp{1

t(|x|2

*! 1

2t1!")}.

Now, it is clear that on each interval [x1, x2], when t $ 0+, u(x, t) $ 0 uniformly. Thisshows that (4.24) determines a solution (called Tychonov’s solution) for each $ > 1.

The second example is due to E Rothe showing that the backward heat equation isill-posed.

Example 4.5.2 Consider the initial value problem"

ut ! uxx = 0

u(x, 0) = '(x), x " R.

If '(x) = 0, we see u(x, t) = 0 is a solution. Now, if

'(x) = # sin(x

#), # > 0,

4.5. EXAMPLES 81

thenu(x, t) = #e!

t%2 sin(

x

#)

is a solution. We see when # is su$ciently small, then '(x) is arbitrarily close to 0. Thisis true when t > 0; but for t < 0, it is not. So the solution is not stable about the initialdata if t < 0.

In what follows, we show some tricks in solving initial boundary value problem of heatequation in one space dimension.

Example 4.5.3 Consider the heat conduction problem on a finite bar

'-(

-)

ut ! uxx = 0, t > 0, x " (0, l),

u(x, 0) = '(x), x " (0, l),

u(0, t) = u(l, t) = 0, t + 0,

(4.25)

where '(x) " C[0, l] such that '(0) = '(l) = 0.

Solution. If u(x, t) is a solution, since u(0, t) = 0, we could extend u(x, t) as an oddfunction in x into (!l, 0). Then we further extend u(x, t) as a periodic function in x withperiod 2l to whole real line. This will transfer the problem into a Cauchy problem

"ut ! uxx = 0, t > 0, x " R,

u(x, 0) = &(x), x " R,

where&(x) = '(x), x " (0, l)

&(x) = !'(!x), x " (!l, 0)

&(x + 2l) = &(x), x " R.

Since '(x) " C[0, l], &(x) satisfies all conditions for existence and uniqueness, and therefore

u(x, t) =

! $

!$&(y)E(x! y, t) dy (4.26)

We know that &(x) is odd in x, so

u(0, t) =

! $

!$&(y)E(y, t) dy = 0.

Also, &(l ! x) is odd in x, so

u(l, t) =

! $

!$&(l ! y)E(y, t) dy = 0.


Therefore, u(x, t) determined by the formula (4.26) is a solution of (4.25). We could rewrite(4.26) as

u(x, t) =

! l

0

'(y)G(x, y, t) dy

where

G(x, y, t) =1

2%

%t

$,

n=!$

#exp{!x! y ! 2nl)2

4t}! exp{!x + y ! 2nl)2

4t}$

.

Example 4.5.4 We solve the above problem (4.25) again by the method of separation ofvariables, called Fourier method.

Solution. Assumingu(x, t) = X(x)T (t),

one getsT #

T=

X ##

X= !#

for # the constant ratio as the first term depends only on t while the second depends onlyon x. Therefore, we obtained two equations

T # + #T = 0, X ## + #X = 0. (4.27)

For X, it satisfies the following two points boundary value problem

"X ## + #X = 0, 0 < x < l,

X(0) = X(l) = 0.(4.28)

This is a typical Sturm-Liouville type eigenvalue problem. We now derive the generalmethod for such type of problems.

First of all, if # < 0, we have the general solution of X as

X(x) = Ae"!&x + Be!

"!&x,

which gives zero solution by the boundary condition.If # = 0, X ## = 0, which with the boundary condition, give zero solution.Now, if # = k2 > 0 for k > 0, we have the general solution of X is

X(x) = A cos(kx) + B sin(kx),

where A = 0 as X(0) = 0, and B sin(kl) = 0 as X(l) = 0. If B = 0, we get zero solutionagain. If B ,= 0, we have

k =n%

l, or, #n = (

n%

l)2, n = 1, 2, · · · (4.29)

4.5. EXAMPLES 83

We remark that n < 0 does not give new values of #. We thus obtained the non-zerosolutions

Xn(x) = B sin(n%

lx), n = 1, 2, · · · (4.30)

We call #n the eigenvalue and Xn the corresponding eigenfunction. We now substitute the#n into the equation of T to obtain

Tn = ane!&nt. (4.31)

We thus obtained the formal solution of the heat equation with the boundary data

u(x, t) =$,

n=1

ane!(n&

l )2t sin(n%

lx). (4.32)

The constant B is absorbed by an, which will be determined by the initial condition,

'(x) =$,

n=1

an sin(n%

lx),

and therefore,

an =2

l

! l

0

'(x) sin(n%

lx) dx.

The equation (4.32) gives a formal solution, we need to prove the convergent propertiesfor this series. As '(x) " C([0, l]), there exists a constant K > 0 such that |an| * K forall n. Now, for any t1 > t0 > 0, on the domain [0, l]. [t0, t1], we have

|ane!(n&

l )2t sin(n%

lx)| * Ke!(n&

l )2t.

Therefore, the series (4.32) converges uniformly and absolutely. Therefore, u(x, t) is con-tinuous in this domain and thus is continuous for x " [0, l], t > 0. Furthermore, it verifiesthe boundary conditions. Similarly, one show the continuous di#erentiability in t and inx. Therefore, (4.32) defines the solution to problem (4.25).

Example 4.5.5 Assume there is a infinitely long, heat conductive, thin rod (x " R).Assume the rod is made of two di#erent materials on the part of x < 0 and on the partof x > 0 respectively. At the transition point x = 0, the heat flux must be continuous onboth side. We denote the left temperature by u(x, t) and the right temperature by v(x, t).Therefore, we need to solve

'--------(

--------)

ut ! µuxx = 0, x < 0, t > 0,

vt ! "vxx = 0, x > 0, t > 0,

u(x, 0) = '(x), x * 0,

v(x, 0) = +(x), x + 0,

u(0, t) = v(0, t), t + 0,

,ux(0, t) = !vx(0, t), t + 0,

(4.33)


where, '(0) = +(0), µ and " are head conduction constants, and , and ! are the heattransfer constants.

Solution. We assume this problem has solution (u(x, t), v(x, t)). We define

w(x, t) = au(!x, t) + bv(

."

µx, t), x + 0 (4.34)

where a and b are constants to be determined. We easily verify that w satisfies the equation

wt ! µwxx = 0, x > 0, t > 0.

We now introduce a function

u'(x, t) =

"u(x, t), x < 0,

w(x, t), x > 0.

It is clear that u' satisfies u't !µu'xx = 0 on R \ {0}. Now, we require that u'(x, t) satisfiesthe following compatibility condition at x = 0,

"u(0, t) = w(0, t) = au(0, t) + bv(0, t),

ux(0, t) = wx(0, t) = !aux(0, t) + b/

'µvx(0, t).

(4.35)

By the conditions of (4.33), the above conditions implies that

"u(0, t) = au(0, t) + bu(0, t),

!ux(0, t) = !a!ux(0, t) + b,/

'µux(0, t),

(4.36)

which gives

a + b = 1, !a! + b,

."

µ= !.

Thus,

a = 1! b, b =2!

! + ,/

'µ

.

We now substitute a and b into (4.34) to recover the initial data for u' on x + 0:

u'(x, 0) = a'(!x) + b+(

."

µx), x > 0.

We now define

''(x) =

"'(x), x < 0,

a'(!x) + b+(/

'µx), x + 0.

4.6. PROBLEMS 85

We thus obtained a Cauchy problem for u'

"u't ! µu'xx = 0, x " R, t > 0

u'(x, 0) = ''(x), x " R.(4.37)

Now, set y = x"µ , and then the solution of (4.37) is given by

u'(x, t) =1%2%t

! $

!$''(%

µz)e!( x"

µ!z)2

4t dz, x " R.

4.6 Problems

Problem 1. Use Fourier transform method to give a solution formula to the followingCauchy problem

"ut ! uxx + xu = 0, x " R, t > 0,

u(x, 0) = '(x).

Problem 2. Use Fourier transform method to solve

'(

)

uxx + uyy = 0, x " R, y > 0,

u(x, 0) = '(x), lim%x2+y2%$

u = 0. (4.38)

Problem 3. Assume a static temperature distribution u(x, y) on upper half plane satisfiesthe constraints

u(x, 0) =

"1, if |x| * a,

0, if |x| > a.

Prove that

u(x, y) =1

%(arctan(

a + x

y) + arctan(

a! x

y)).

Problem 4. Solve the Cauchy problem of 1-d heat equation ut!a2uxx = 0, (x " R, t > 0)with the following initial conditions:

• (a) u(x, 0) = sin(x);

• (b) u(x, 0) = x2 + 1.


Problem 5. Use extension method to solve the following problem

'-(

-)

ut ! a2uxx = 0, x > 0, t > 0,

u(x, 0) = '(x), x > 0

u(0, t) = 0, t + 0; '(0) = 0.

Problem 6. Use extension method to solve the following problem

'-(

-)

ut ! uxx = 0, x > 0, t > 0,

u(x, 0) = '(x), x > 0

ux(0, t) = 0, t > 0; '(0) = '#(0) = 0.

Problem 7. For constant d, solve the following problem

'-(

-)

ut ! uxx = 0, x > 0, t > 0,

u(x, 0) = '(x), x > 0

ux(0, t) + du(0, t) = 0, t > 0.

Problem 8. For constant - > 0, use separation of variables method to solve

'-(

-)

ut ! uxx = 0, 0 < x < l, t > 0,

u(x, 0) = '(x), 0 * x * l

u(0, t) = 0, ux(l, t) + -u(l, t) = 0, t + 0.

Problem 9. Prove the solution of

'-(

-)

ut ! uxx = 0, x " R, t > 0,

u(x, 0) = 1, x > 0,

u(x, 0) = !1, x < 0,

isu(x, t) = h(

x

2%

t),

where, h is the error function

h(x) =2%%

! x

0

e!t2 dt.

Problem 10. If u1(x, t) and u2(y, t) are respectively the solutions of the problems"

u1t ! u1xx = 0, x " R, t > 0,

u1(x, 0) = '1(x),

4.6. PROBLEMS 87

and "u2t ! u2yy = 0, y " R, t > 0,

u2(y, 0) = '2(y).

Prove the function u(x, y, t) = u1u2 is the solution of"

ut ! (uxx + uyy) = 0, (x, y) " R2, t > 0,

u(x, y, 0) = '1(x)'2(y).

Problem 11. Derive the solution formula for the following problem

'-(

-)

ut ! (uxx + uyy) = 0, (x, y) " R2, t > 0,

u(x, y, 0) =n,

i=1

$i(x).i(y).

Problem 12. Let v(x, t) be the solution of"

vt ! a2vxx = 0, x > 0, t > 0,

v(x, 0) = 0, v(0, t) = 1.

Prove the Duhamel integral

u(x, t) =!

!t

! t

0

v(x, t! /)g(/) d/

solves "ut ! a2uxx = 0, x > 0, t > 0,

u(x, 0) = 0, u(0, t) = g(t).

Problem 13. Prove the weak maximum principle: Let ! be a bounded open set in Rn,and !T is the parabolic cylinder. If u(x, t) " C2

1(!T ) satisfies

ut !"u * 0,

thenmax!T

u = max"T

u

where %T is the parabolic boundary. If * is replaced by +, and the max is repalced bymin, then the statement is also valid.

Problem 14. Let u(x, t) " C21(!T ) be the solution of

"ut !"u = f(x, t), (x, t) " !T ,

u(x, t)|"T = '(x, t).


DefineF = sup

QT

|f |, B = sup"T

|'|.

Prove thatmax!T

u * FT + B.

Problem 15. Assume u solves ut !"u = 0, prove the following statements

• (a) If ' : R $ R is a smooth convex function, then v = '(u) satisfies

vt !"v * 0.

• (b) Prove v = |Du|2 + u2t also satisfies the above inequality.

Problem 16. Define the following parabolic di#erential operator

Lu = ut ! a2uxx + b(x, t)ux + c(x, t)u.

Assume c(x, t) > 0, if u(x, t) " C21(!T ) satisfies

Lu * 0, in !T ,

thenmax!T

u * max"T

u+,

where u+(x, t) = max{u(x, t), 0)}.

Problem 17. Consider the same operator L as in Problem 16. Assume for some constantc0 > 0, c(x, t) > !c0, and u(x, t) " C2

1(!T ) satisfies

Lu * 0, in !T ,

if max"T

u * 0, then

max!T

u * 0.

Chapter 5

Wave equation

In this chapter, we discuss the wave equation

utt ! a2!u = f, (5.1)

where a > 0 is a constant. We will discover that solutions of the wave equation behave ina di"erent way comparing with the solutions of Laplace’s equation or the heat equation.

5.1 Physical derivation

In physics or mechanics, the wave equation serves as a simplified model for the oscillationson a vibrating string (n = 1), membrane (n = 2), or elastic solid (n = 3). In these models,u(x, t) is the displacement of the pointed mass in certain directions of the point x at timet " 0. In (5.1), f models the external force.

For instance, we assume that an elastic solid occupied a region # in R3 without externalforce. For any subregion G # #, !F (x, t) is the contact force density acting on G throughthe boundary "G. Normalize the mass density to be unity. Let # be the unit outer normalvector of "G. The acceleration within G is

d2

dt2

!

G

u dx =

!

G

utt dx,

while the net contact force is

!!

!G

!F · # dS,

then, by Newton’s law, one has!

G

utt dx = !!

!G

!F · # dS =

!

G

$ · !F dx

where we have applied the divergence theorem. Therefore, we conclude that

utt = !$ · !F .

89

90 CHAPTER 5. WAVE EQUATION

For elastic body, !F = !F ($u), so

utt +$ · !F ($u) = 0.

In the case of small oscillations, |$u| is very small, and so !F ($u) % !a2$u, therefore,

utt ! a2!u = 0.

We remark that, from the physical interpretation, it is mathematically appropriate tospecify two initial conditions, on the displacement u and the velocity ut at t = 0. This willappears often in the rest of this chapter.

5.2 Solution by Spherical means

Unlike the heat equation or Laplace equation, we will present an elegant method to solve(5.1) first for n = 1 directly and then for n " 2 by the method of spherical means.

5.2.1 d’Alembert’s Formula, n = 1

We first start with the Cauchy problem for the wave equation in one space dimension.Consider "

utt ! uxx = 0, in R& (0,')

u(x, 0) = g(x), ut(x, 0) = h(x), x ( R,(5.2)

where g and h are given functions.From the characteristic method we showed in Chapter 1, we know that the general

solution of the 1-D wave equation takes the following form

u(x, t) = F (x + t) + G(x! t)

for any C2 functions F and G. We now determine the F and G through the initial data.It turns out that

F (x) + G(x) = g(x), F !(x)!G!(x) = h(x), (5.3)

which implies

F !(x) =1

2(g!(x) + h(x)), G!(x) =

1

2(g!(x)! h(x)). (5.4)

Therefore, one derives the following d’Alembert’s formula:

u(x, t) =1

2[g(x + t) + g(x! t)] +

1

2

! x+t

x"t

h(y) dy. (5.5)

We thus proved the following Theorem.

Theorem 5.2.1 Assume g(x) ( C2(R) and h(x) ( C1(R). Then u(x, t) defined byd’Alembert’s formula (5.5) is the unique solution of (5.2).

5.2. SOLUTION BY SPHERICAL MEANS 91

In the following example, we apply the d’Alembert’s formula to an initial boundaryvalue problem using the reflection method.

Example 5.2.2 Consider#$%

$&

utt ! uxx = 0, in R+ & (0,')

u(x, 0) = g(x), ut(x, 0) = h(x), x ( R+,

u(0, t) = 0, t > 0

(5.6)

where g and h are given functions such that g(0) = h(0) = 0.We perform the following odd reflection.

u(x, t) =

"u(x, t), x " 0, t " 0

!u(!x, t), x ) 0, t " 0,

g(x) =

"g(x), x " 0,

!g(!x), x ) 0,

h(x, t) =

"h(x), x " 0

!h(!x), x ) 0.

(5.7)

It is clear that "utt ! uxx = 0, in R& (0,')

u(x, 0) = g(x), ut(x, 0) = h(x), x ( R,(5.8)

Hence, the d’Alembert’s formula gives

u(x, t) =1

2[g(x + t) + g(x! t)] +

1

2

! x+t

x"t

h(y) dy. (5.9)

Finally, we transform this expression into the region for x " 0 and t " 0:

u(x, t) =

#$$%

$$&

1

2[g(x + t) + g(x! t)] +

1

2

! x+t

x"t

h(y) dy, if x " t " 0,

1

2[g(x + t)! g(t! x)] +

1

2

! x+t

"x+t

h(y) dy, if 0 ) x ) t.(5.10)

5.2.2 Spherical means

For n " 2, u ( Cm(Rn & [0,')) (m " 2) solves the initial problem"

utt !!u = 0, in Rn & (0,')

u(x, 0) = g(x), ut(x, 0) = h(x), x ( Rn,(5.11)


From the physical experience on the wave propagation in R3, it appears that the wavepropagates spherically. This motivates the approach of spherical means. We will first studythe average of u over certain spheres. These averages, as functions of the time t and theradius r, solve the Euler-Poisson-Darboux equation, which could be transfered into theone-dimensional wave equation for which we know how to solve.

Definition 5.2.3 Let x ( Rn, t > 0, r > 0. Define

U(x; r, t) = (u(y, t))!B(x,r) =1

|"B(x, r)|

!

!B(x,r)

u(y, t) dSy. (5.12)

Similarly, define

G(x; r) = (g(x))!B(x,r), H(x; r) = (h(x))!B(x,r).

Lemma 5.2.4 Fix x ( Rn, and let u solves (5.11), then U ( Cm(R+ & [0,')) solves theinitial value problem for Euler-Poisson-Darboux equation

#%

&Utt ! Urr !

n! 1

rUr = 0, in R+ & (0')

U(x; r, 0) = G(x; r), Ut(x; r, 0) = H(x; r)., in R+. (5.13)

Proof. For r > 0, one has

Ur(x; r, t) =r

n(!u(y, t))B(x,r). (5.14)

Therefore, one has Ur(x; r, t) * 0 as r * 0+. Then we di"erentiate (5.14) to have

Urr = (!u)!B(x,r) + (1

n! 1)(!u)B(x,r),

which implies that

limr#0+

Urr =1

n!u.

Similar calculations show the regularity of U . Now, by the wave equation, one has

(rn"1Ur)r =1

n$(n)

!

!B(x,r)

utt(y, t) dSy

= rn"1(utt)!B(x,r) = rn"1Utt.

This completes the proof of the theorem.


5.2.3 Kirchho! ’s formula, n = 3

We now take n = 3 and suppose u ( C2(R3 & [0,')) solves the initial value problem(5.11), and U , G, H defined in the Definition 5.2.3. The magic transform is

U = rU, G = rG, H = rH, (5.15)

which solves the following problem#$%

$&

Utt ! Urr = 0, in R+ & (0,')

U(x; r, 0) = G(x; r), Ut(x; r, 0) = H(x; r), r ( R+,

U(x; 0, t) = 0, t > 0.

(5.16)

From the Example 5.2.2, we know that for 0 ) r ) t,

U(x; r, r) =1

2[G(x; r + t)! G(x; t! r)] +

1

2

! r+t

"r+t

H(y) dy, (5.17)

We note that for continuous function u, one has

limr#0+

U(x; r, t) = u(x, t).

Therefore,

u(x, t) = limr#0+

U(x; r, t)

r

= limr#0+

[G(x; r + t)! G(x; t! r)

2r+

1

2r

! r+t

"r+t

H(y) dy

= G!(x; t) + H(x; t).

Hence one reaches the Kirchho"’s formula

u(x, t) ="

"t(t(g)!B(x,t)) + t(h)!B(x,t). (5.18)

A further calculation gives the more explicit form

u(x, t) =1

|"B(x, t)|

!

!B(x,t)

[th(y) + g(y) + Dg(y) · (y ! x)] dSy. (5.19)

5.2.4 Poisson’s formula, n = 2

Unlike the 3D case, when n = 2, the Euler-Poisson-Darboux equation cannot be transferredinto the one-dimensional wave equation. We will apply the method of descent introducedby Hadamard. The idea is to take the initial value problem (5.11) for n = 2 and regard itas a problem for n = 3 where the third spatial variable x3 does not appear.


To this purpose, we assume u ( C2(R2 & [0,')) solves (5.11) for n = 2, and we write

u(x1, x2, x3, t) = u(x1, x2, t). (5.20)

Therefore, forg(x1, x2, x3) = g(x1, x2), h(x1, x2, x3) = h(x1, x2),

we have "utt !!u = 0, in R3 & (0,'),

u = g, ut = h, on R3 & {t = 0}.(5.21)

Denote by x = (x1, x2) ( R2, and x = (x1, x2, 0) ( R3, we know from the Kirchho"’sformula (5.18) that

u(x, t) = u(x, t) ="

"t(t(g)!B(x,t)) + t(h)!B(x,t), (5.22)

where B(x, t) is the ball in R3 centered at x with radius t > 0. For y ( B(x, t) and%(y) = (t2 ! |y ! x|2) 1

2 , we note that

(g)!B(x,t) =1

4&t2

!

!B(x,t)

g dS

=2

4&t2

!

B(x,t)

g(y)(1 + |D%(y)|2) 12 dy.

Since (1 + |D%(y)|2) 12 = t(t2 ! |y ! x|2) 1

2 , we thus have

(g)!B(x,t) =2

2&t

!

B(x,t)

g(y)

(t2 ! |y ! x|2) 12

dy

=t

2

'g(y)

(t2 ! |y ! x|2) 12

(

B(x,t)

.

Further simplification gives the following Poisson’s formula:

u(x, t) =1

2

'tg(y) + t2h(y) + tDg(y) · (y ! x)

(t2 ! |y ! x|2) 12

(

B(x,t)

(5.23)

for x ( R2, t > 0.

5.2.5 Further generalization

We now generalized the previous results to any dimensions for n " 3. The followingidentities can be proved by induction.


Lemma 5.2.5 Let ' : R * R be Ck+1. Then for k = 1, 2, · · · , the following identitieshold

• (d2

dr2)(

1

r

d

dr)k"1(r2k"1'(r)) = (

1

r

d

dr)k(r2k d'(r)

dr).

• (1

r

d

dr)k"1(r2k"1'(r)) =

k"1)

j=0

(kj rj+1dj'(r)

drj, where (k

j are constants independent of '.

• (k0 = 1 · 3 · 5 · · · (2k ! 1).

Now, we assume that n " 3 is an odd integer and set n = 2k + 1. Then the followingtransformation #

$%

$&

U(r, t) = (1r

!!r )

k"1(r2k"1U(x; r, t))

G(r) = (1r

!!r )

k"1(r2k"1G(x; r))

H(r) = (1r

!!r )

k"1(r2k"1H(x; r)),

(5.24)

converts the Euler-Poisson-Darboux equation into the one-dimensional wave equation:

#$%

$&

Utt ! Urr = 0, in R+ & (0,')

U(x; r, 0) = G(x; r), Ut(x; r, 0) = H(x; r), r ( R+,

U(x; 0, t) = 0, t > 0.

(5.25)

Solution for odd n. Follow the similar steps in the case of n = 3, we have for n = 2k +1and %n = 1 · 3 · 5 · · · (n! 2) the following representation formula:

u(x, t) =1

%n

*("

"t)(

1

t

"

"t)

n!32 (tn"2(g)!B(x,t)) + (

1

t

d

dt)

n!32 (tn"2(h)!B(x,t))

+. (5.26)

Solution for even n. Then, the method of descent will give the results for even n. Foreven n, and %n = 2 · 4 · 5 · · · (n! 2) · n, we have the following solution formula

u(x, t) =

1

%n

,("

"t)(

1

t

"

"t)

n!22 (tn(

g(y)

(t2 ! |y ! x|2) 12

)B(x,t))

-

+1

%n

,(1

t

d

dt)

n!22 (tn(

g(y)

(t2 ! |y ! x|2) 12

)B(x,t))

-.

(5.27)


5.3 Nonhomogeneous problem

We now study the initial value problem for the nonhomogeneous wave equation

"utt !!u = f, in Rn & (0,')

u(x, 0) = 0, ut(x, 0) = 0, x ( Rn.(5.28)

By Duhamel’s principle, let u(x, t; s) be the solutions of

"utt(·; s)!!u(·; s) = 0, in Rn & (0,')

u(x, 0) = 0, ut(x, 0) = f(·, s), on Rn & {t = s},(5.29)

we set

u(x, t) =

! t

0

u(x, t; s) ds, (x ( Rn, t " 0). (5.30)

The following theorem asserts this gives the solution of (5.28).

Theorem 5.3.1 For n " 2 and f ( C [n2 ]+1(Rn & [0,')), u(x, t) defined in (5.30) is a

solution of (5.28).

Proof. Through the solutions constructed in the last section, u(·, ·; s) ( C2(Rn & [0,').Then, we compute

ut(x, t) = u(x, t; t) +

! t

0

ut(x, t; s) ds =

! t

0

ut(x, t; s) ds,

utt(x, t) = ut(x, t; t) +

! t

0

utt(x, t; s) ds

= f(x, t) +

! t

0

u + tt(x, t; s) ds.

Furthermore

!u(x, t) =

! t

0

!u(x, t; s) ds =

! t

0

utt(x, t; s) ds.

Therefore,utt(x, t) = !u(x, t) = f(x, t), (x ( Rn, t > 0).

Clearly, u(x, 0) = ut(x, 0) = 0 for any x ( Rn.

Example 5.3.2 When n = 3, Kirchho"’s formula implies

u(x, t, ; s) = (t! s)(f(y, s)!B(x,t"s).

5.4. ENERGY METHOD 97

Therefore,

u(x, t) =

! t

0

(t! s)(f(y, s)!B(x,t"s) ds

=1

4&

! t

0

!

!B(x,t"s)

f(y, s)

t! sdSds

=1

4&

! t

0

!

!B(x,r)

f(y, t! r)

rdSdr.

Hence, one obtains the solution of (5.28) for n = 3

u(x, t) =1

4&

!

B(x,t)

f(y, t! |y ! x|)|y ! x| dy, (x ( R3, t " 0) (5.31)

wheref(y, t! |y ! x|)

4&|y ! x|is called retarded potential.

5.4 Energy method

In this section, we introduce the energy method, which is a powerful tool in the theory ofpartial di"erential equations. Let # # Rn be a bounded, open set with a smooth boundary"#. Define #T = #& (0, T ], $T = #T \ #T , where T > 0.

5.4.1 Wave equation

We first consider the following initial boundary value problem

#$%

$&

utt !!u = f, in #T ,

u = g, on $T ,

ut = h, on #& {t = 0}.(5.32)

We now prove the uniqueness for the above problem using energy method.

Theorem 5.4.1 There exists at most one solution u ( C2(#T ) for (5.32).

Proof: Let u and u be two solutions of (5.32), then w = u! u soloves

#$%

$&

wtt !!w = 0, in #T ,

w = 0, on $T ,

wt = 0, on #& {t = 0}.


Define the energy

e(t) =1

2

!

!

(w2t + |Dw|2)(x, t) dx, (0 ) t ) T ).

It is clear that

e!(t) =

!

!

wtwtt + Dw · Dwt dx

=

!

!

wt(wtt !!w) dx = 0.

Therefore, e(t) = e(0) = 0 for all t ( [0, T ], and so wt + 0 and Dw + 0 in #T . By theinitial boundary conditions, we know that w + 0 and thus u = u in #T .

5.4.2 Domain of dependence

Energy method is also useful to study some local behavior of solutions. We will illustratethe domain of dependence of solutions to the wave equation and the finite propagationspeed properties.

Now, suppose u ( C2 solves

utt !!u = 0, in Rn & (0,').

Fix x0 ( Rn, t0 > 0. We define the cone

C = {(x, t)|0 ) t ) t0, |x! x0| ) t0 ! t}.

We first prove the following finite propagation speed property, which says that anydisturbance originating outside B(x0, t0) has no e"ect on the solution within C.

Theorem 5.4.2 If u + ut + 0 on B(x0, t0), then u + 0 within the cone C.

Proof: Define the local energy

e(t) =1

2

!

B(x0,t0"t)

(u2t + |Du|2)(x, t) dx, t ( [0, t0].

We now compute

e!(t) =

!

B(x0,t=0"t)

ututt + Du · Dut dx! 1

2

!

!B(x0,t0"t)

(u2t + |Du|2) dS

=

!

B(x0,t0"t)

ut(utt !!u) dx! 1

2

!

!B(x0,t0"t)

(u2t + |Du|2 ! 2ut

"u

"#) dS

= !1

2

!

!B(x0,t0"t)

(u2t + |Du|2 ! 2ut

"u

"#) dS

5.4. ENERGY METHOD 99

We now observe that, by Cauchy-Schwartz inequality,

|ut"u

"#| ) 1

2u2

t +1

2|Du|2.

Therefore, we havee!(t) ) 0

which means e(t) ) e(0) = 0 for all t ( [0, t0]. Thus ut + 0 and Du + 0 and so u + 0within the cone C.

The bottom of the cone C is B(x0, t0) on the initial hyper-plane, is called the domainof dependence for the point (x0, t0).

It is now clear that Theorem 5.4.2 also implies the uniqueness for the following Cauchyproblem "

utt !!u = f, in Rn,

u = g, ut = h, on Rn & {t = 0}.(5.33)

5.4.3 Energy method for Heat equation

We now study the large time asymptotic behavior of the solutions to the following initialboundary value problem for heat equation.

#$%

$&

ut !!u = 0, in #& (0,'),

u = 0, on "#,

u = g, on #& {t = 0}.(5.34)

Theorem 5.4.3 The solution of (5.34) converges to zero exponentially fast in time.

Proof: Let u be the solution of the problem (5.34), we define the energy

e(t) =

!

!

u2 dx.

Now we compute

e!(t) = 2

!

!

uut dx

= 2

!

!

u!u dx

= !2

!

!

|Du|2 dx.

However, by Poincare inequality, one knows that there exists a positive constant C = C(#)such that

C

!

!

u2 dx )!

!

|Du|2 dx,


for any u ( H10 (#). Therefore, we have

e!(t) + 2Ce(t) ) 0,

which impliese(t) ) e(0)e"2Ct, ,t > 0.

Therefore, one sees that e(t) * 0 as t * ' and thus u converges to zero in energy normexponentially fast in time.

5.5 Initial boundary value problem: n = 1

For wave equation in one spatial dimension, there are several approaches to solve the initialboundary value problem, including the characteristic method, reflection method and theFourier method (separation of variables). We will focus on the method of separation ofvariables.

5.5.1 Homogeneous equation

The Fourier method is motivated from the physical fact: vibrations can be decomposedinto simple oscillations according to its frequency. Therefore, one can try to look for thesolution of the form

un(x, t) = Xn(x)Tn(t), n = 1, 2, · · · ,

then, determine the constants in the following formula

u(x, t) =$)

n=1

CnXn(x)Tn(t),

to solve the problem.We now show this idea using the following example.

Example 5.5.1 Solve the following problem

#$%

$&

utt ! a2uxx = 0, 0 < x < l, t > 0,

u(0, t) = 0, u(l, t) = 0, t " 0,

u(x, 0) = f(x), ut(x, 0) = g(x), 0 ) x ) l,

(5.35)

where f and g are C1 satisfying the compatibility condition

f(0) = f(l) = 0, g(0) = g(l) = 0.

Solution: The Fourier method consists the following steps.

5.5. INITIAL BOUNDARY VALUE PROBLEM: N = 1 101

(i) Separation of variables. We first look for the solution of the form

u(x, t) = X(x)T (t) -= 0

satisfying the boundary conditions. This leads to

T !!(t)

a2T (t)=

X !!(x)

X(x)= !), X(x)T (t) -= 0.

It is clear that ) is a constant. We thus have"

T !!(t) + )a2T (t) = 0, t > 0

X !!(x) + )X(x) = 0, 0 < x < l.(5.36)

For u(x, t) to satisfy the boundary condition, one requires X(0) = X(l) = 0.

(ii) Solve the eigenvalue problem. We now solve the following eigenvalue problem

"X !!(x) + )X(x) = 0, 0 < x < l

X(0) = X(l) = 0.(5.37)

where the ) is called the eigenvalue if it gives a non-zero solution X"(x) which is calledthe associated eigenfunction. There are three cases concerning with the sign of ).

(a) If ) < 0, the general solution is

X(x) = Ae%""x + Be"

%""x,

where A and B will be determined through the boundary conditions. In deed,

X(0) = A + B = 0

X(l) = Ae%""l + Be"

%""l = 0.

Therefore, A = B = 0 and ) < 0 is not the eigenvalue.

(b) If ) = 0, then X(x) = A+Bx which is identically zero with the boundary conditionsX(0) = 0 = X(l). So, ) = 0 is not an eigenvalue either.

(c) We now consider ) = k2 > 0. The general solution takes the form

X(x) = A cos(kx) + B sin(kx).

Using the boundary condition X(0) = 0 = X(l), one has A = 0 and B sin(kl) = 0. Sincewe need the non-zero solution, B -= 0 and

k =n&

l, )n = (

n&

l)2, n = 1, 2, · · · (5.38)


This gives the eigenvalues for problem (5.37) and the corresponding eigenfunctions are

Xn(x) = sin(n&

lx), n = 1, 2, · · · (5.39)

We now substitute )n into the first equation in (5.36),

Tn(t) = cn cos(an&

lt) + dn sin(

an&

lt), n = 1, 2, · · ·

Therefore, we found

un(x, t) = Xn(x)Tn(t) = [cn cos(an&

lt) + dn sin(

an&

lt)] sin(

n&

lx), n = 1, 2, · · · (5.40)

where cn and dn are arbitrary constants to be determined later.

(iii) In this step, we hope to find the solution of the initial boundary value problem(5.35) through linear combination of un(x, t). Let

u(x, t) =$)

n=1

[cn cos(an&

lt) + dn sin(

an&

lt)] sin(

n&

lx). (5.41)

Using the initial conditions, we require

u(x, 0) =$)

n=1

cn sin(n&

lx) = f(x),

ut(x, 0) =$)

n=1

dnan&

lsin(

n&

lx) = g(x).

(5.42)

From the conditions of f and g and we require they are C1, then

cn =2

l

! l

0

f(x) sin(n&

lx) dx,

dn =2

an&

! l

0

g(x) sin(n&

lx) dx.

(5.43)

We therefore obtained the formal solution of the problem (5.35). One could further verifythat the series solution has good convergent properties to be a classical solution if f andg are C4([0, l]) and f(0) = f !!(0) = g(0) = f(l) = f !!(l) = g(l) = 0.

We see from this example that as long as the corresponding eigenvalue problem iswell solved, we are able to find at least the formal solution of the initial value problemfor homogeneous wave equation in one space dimension. The eigenvalue problem will bediscussed in great details at the end of this chapter. For the formal solution to be classicalsolution, it often requires very restrictive properties on the data, this however could beresolved by the notion of weak solution.


5.5.2 Non-homogeneous equation

In this section, we further show that the eigenfunctions have further applications to solvethe initial boundary value problem for the non-homogeneous equation. This is so-calledexpansion about eigenfunctions.

Again, we will illustrate our ideas using specific example.

Example 5.5.2 Solve the following problem

#$%

$&

utt ! a2uxx = f(x, t), 0 < x < l, t > 0,

u(0, t) = 0, u(l, t) = 0, t " 0,

u(x, 0) = 0, ut(x, 0) = 0, 0 ) x ) l,

(5.44)

Solution: Unless f(x, t) has certain specific structure, u(x, t) = X(x)T (t) is not a solutionof the non-homogeneous wave equation. However, motivated by the constant variationmethod of ODEs, we seek the solution of the following form

u(x, t) =$)

n=1

Tn(t)Xn(x) (5.45)

where Xn(x) are the eigenfunctions for the corresponding homogeneous wave equationunder the same boundary conditions. In this case,

Xn(x) = sin(n&

lx), n = 1, 2, · · · .

We hope that the solution could be expanded into the series of Xn(x) with variant coef-ficients Tn(t). Clearly, (5.45) satisfies the boundary conditions. We will determine Tn(t)through the equation and the initial conditions. Therefore, Tn(t) satisfies

)

n=1

[T !!(t) + (an&

l)2Tn(t)] sin(

n&

lx) = f(x, t),

u(x, 0) =$)

n=1

Tn(0) sin(n&

lx) = 0

ut(x, 0) =$)

n=1

T !n(0) sin(n&

lx) = 0.

(5.46)

Setting

fn(t) =l

2

! l

0

f(x, t) sin(n&

lx) dx, n = 1, 2, · · · ,

one has "T !!n (t) + (an#

l )2Tn(t) = fn(t)

Tn(0) = T !n(0) = 0, n = 1, 2, · · · .


Hence, we found

Tn(t) =l

an&

! t

0

fn(*) sin(an&

l(t! *)) d*, n = 1, 2, · · · .

We finally obtain the formal solution of the problem (5.44)

u(x, t) =$)

n=1

[l

an&

! t

0

fn(*) sin(an&

l(t! *)) d* ] sin(

n&

lx). (5.47)

For general initial boundary value problem, one can apply the linear superpositionprinciple and some basic techniques to transfer the non-homogeneous boundary conditionsto homogeneous ones. We omit the details.

5.5.3 Sturm-Liouville Theory

In this subsection, we briefly discuss the theory of Sturm-Liouville for eigenvalue problem.Consider the initial boundary value problem

#$%

$&

Lu = 0, x ( (a, b), t > 0,

u(x, 0) = '(x), ut(x, 0) = +(x), x ( [a, b]

$1u(a, t) + $2ux(a, t) = 0, (1u(b, t) + (2ux(b, t) = 0, t " 0,

(5.48)

where

Lu = A(t)utt + C(x)uxx + D(t)ut + E(x)ux + (F1(t) + F2(x))u.

We assume that $i and (i are constants for i = 1, 2 such that

$21 + $2

2 -= 0, (21 + (2

2 -= 0.

Suppose A(t) " A0 > 0, C(x) ) C0 < 0, A0 and C0 are constants. All other coe%cientsare continuous. We also assume F2(x) > 0.

If we want to perform the method of separation of variables, u(x, t) = X(x)T (t), for )the possible eigenvalue, we need to solve the equation of T (t)

AT !!(t) + DT !(t) + F1T + )T = 0 (5.49)

and the eigenvalue problem

#$%

$&

CX !! + EX ! + F2X ! )X = 0,

$1X(a) + $2X !(a) = 0,

(1X(b) + (2X !(b) = 0.

(5.50)


In order to solve this eigenvalue problem, we first convert it into the self-adjoint form.Multiplying the equation by

S = ! 1

Cexp{

! x

0

E

Cdx},

the equation became[p(x)X !]! ! q(x)X + )SX = 0, (5.51)

where

p(x) = !SC " exp{! x

0

E

C0dx} = p0 > 0,

q(x) = SF2 > 0, S(x) > 0.

Therefore, (5.50) has been changed into the standard form of Sturm-Liouville eigenvalueproblem #

$%

$&

[p(x)X !]! ! q(x)X + )SX = 0

$1X(a) + $2X !(a) = 0,

(1X(b) + (2X !(b) = 0.

(5.52)

We first list some properties of the eigenfunctions.

Theorem 5.5.3 Let X1 and X2 are eigenfunctions corresponding to the same eigenvalue), then X1 = CX2 for certain nonzero constant C.

Theorem 5.5.4 If X1 and X2 are eigenfunctions corresponding to eigenvalues )1 and )2

respectively, then ! b

a

SX1X2 dx = 0.

We now introduce the working space for the Sturm-Liouville theory. For 0 < S(x) (C([a, b]), we first define

L2S + {y(x) ( L1

loc([a, b]) :

! b

a

S(x)y2(x) dx < '},

equipped with the inner product

(y1, y2)S +! b

a

S(x)y1(x)y2(x) dx, ,y1, y2 ( L2S.

For the functions from C10([a, b]) (C1 functions vanishing at endpoints), we introduce an

inner product

(y1, y2)H +! b

a

[p(x)y!1y!2 + q(x)y1y2] dx, ,y1, y2 ( C1

0([a, b]), (5.53)


where p(x) and q(x) are continuous functions such that there is a constant p0 > 0 andp(x) > p0, q(x) > 0 for any x ( [a, b].

(5.53) defines a norm . ·.H , we denote H0,1p,q for the closure space of C1

0([a, b]) under thenorm . · .H . This is a Hilbert space.

We now use the Dirichlet boundary condition as the example to illustrate the procedure.Consider now

"[p(x)X !]! ! q(x)X + )SX = 0

X(a) = X(b) = 0.(5.54)

Definition 5.5.5 X ( H0,1p,q is said to be a weak solution of (5.54) if it satisfies

(X, y)H = )(X, y)S

for any y ( C10 [a, b].

Consider the functional

J(X) =(X, X)H

(X, X)S, ,X ( H0,1

p,q . (5.55)

Theorem 5.5.6 There exits 0 -= X ( H0,1p,q such that

J(X) = infz&H0,1

p,q

J(z).

Now, we set K1 = H0,1p,q , we call

)1 = inf0 '=X&K1

J(X)

the first eigenvalue of (5.54), and the function 0 -= X1 ( K1 such that

(X1, X1)H

(X1, X1)S= )1

the eigenfunction corresponding to )1.

Now, we setK2 = {X ( K1|(X, X1)S = 0}.

Similarly, we define)2 = inf

X&K2

J(X)

as the second eigenvalue and 0 -= X2 ( K2 such that

(X2, X2)H

(X2, X2)S= )2

5.6. PROBLEMS 107

the eigenfunction corresponding to )2. Inductively, we set

Kn = {X ( K1|(X, X1)S = (X, X2)S = · · · = (X, Xn"1)S = 0},

and the n-th eigenvalue)n = inf

X&Kn

J(X)

and 0 -= Xn ( Kn such that(Xn, Xn)H

(Xn, Xn)S= )n

the eigenfunction corresponding to )n.

Theorem 5.5.7 Let )n be eigenvalues of (5.54) and Xn the corresponding eigenfunctionsof )n. Then

• 0 < )1 < )2 < · · · < )n < · · ·

• )n *' as n *'.

• {Xn}$n=1 form an orthogonal basis of L2S.

We therefore solved the Sturm-Liouville problem (5.54). The general case can be treatedin a similar way.

5.6 Problems

Problem 1. Let u ( C2(R& [0,')) be the solution of

"utt ! uxx = 0, x ( R, t > 0,

u(x, 0) = '(x), ut(x, 0) = +(x),

where '(x) and +(x) have compact support. Define

K(t) =1

2

! +$

"$u2

t (x, t) dx

P (t) =1

2

! +$

"$u2

x(x, t) dx.

Prove the following

• (a). K(t) + P (t)=constant;

• (b). When t is su%ciently large, K(t) = P (t).


Problem 2. Solve the following initial value problem

#$%

$&

utt ! uxx = 0, x ( R, t > ax,

u|t=ax = '(x), x ( R,

ut|t=ax = +(x), x ( R,

where a -= ±1. If the initial data are given on x ( [b1, b2], then on what region can youdetermine the solution?

Problem 3. Prove the initial value problem

"utt ! uxx = 6(x + t), x ( R, t > x,

u|t=x = 0, ut|t=x = +(x), x ( R,

has solutions if and only if +(x)!3x2 = constat. When the solutions exist, it is not unique.Why does this problem behave di"erently from problem 2?

Problem 4. Solve the following problems

• (a)

"utt ! a2uxx = 0, x ( R, t > 0,

u|x"at=0 = '(x), u|x+at=0 = +(x), '(0) = +(0);

• (b)

"utt ! a2uxx = 0, x ( R, t > 0,

u|t=0 = '(x), u|x"at=0 = +(x), '(0) = +(0);

• (c)

"uxx + 2 cos(x)uxy ! sin2(x)uyy ! sin(x)uy = 0,

u|y=sin(x) = '(x), uy|y=sin(x) = +(x), x, y ( R;

• (d)

"uxx + yuxy + 1

2uy = 0, x ( R, y < 0

u|y=0 = '(x), uy|y=0 < ';

• (e)

"y2uyy ! x2uxx = 0, x ( R, y > 1,

u|y=1 = f(x), uy|y=1 = g(x).

Problem 5. Let h be a nonzero constant, F and G be C2 functions. Prove that

u(x, t) =1

h! x(F (x! at) + G(x + at))

is the general solution of

[(1! x

h)2ux]x =

1

a2(1! x

h)2utt.

5.6. PROBLEMS 109

Find the solution of this equation with the following initial conditions

u(x, 0) = '(x), ut(x, 0) = +(x).

Problem 6. If w(x, t; *) is the solution of

(A)

"wtt ! a2wxx = 0, x ( R, t > *,

w(x, * ; *) = 0, wt(x, * ; *) = f(x, *),

prove that the function

u(x, t) =

! t

0

w(x, t; *) d*

is the solution of

(B)

"utt ! a2uxx = f(x, t), x ( R, t > 0,

u(x, 0) = 0, ut(x, 0) = 0,

This is the so-called Duhamel principle.

Problem 7. Use the separation of variable method to solve the following problems

• (a)

#$%

$&

utt ! a2uxx = 0, 0 < x < l, t > 0,

u(x, 0) = x2 ! 2lx, ut(x, 0) = 0,

u(0, t) = ux(l, t) = 0, t " 0;

• (b)

#$$$$$$%

$$$$$$&

utt ! a2uxx = 0, 0 < x < l, t > 0,

u(x, 0) =

"hxc , x ( [0, c]

h l"xl"c , x ( (c, l].

ut(x, 0) = 0,

u(0, t) = u(l, t) = 0, t " 0;

Problem 8. Solve the following initial boundary value problems.

• (a)

#$%

$&

utt ! a2uxx = Ax, 0 < x < l, t > 0,

u(x, 0) = ut(x, 0) = 0, 0 ) x ) l,

u(0, t) = u(l, t) = 0, t " 0;

• (b)

#$%

$&

utt ! a2uxx = 0, 0 < x < l, t > 0,

u(x, 0) = ut(x, 0) = 0, 0 ) x ) l,

u(0, t) = 0, u(l, t) = A(sin(,t)! ,t), t " 0;


• (c)

#$%

$&

utt ! a2uxx = bx, 0 < x < l, t > 0,

u(x, 0) = ut(x, 0) = 0, 0 ) x ) l,

u(0, t) = 0, u(l, t) = Bt, t " 0.

Problem 9. Let x = r cos('), y = r sin(') be the coordinates in R2. Solve the followingproblem "

uxx + uyy = 0, 0 < r < R, 0 < ' < $,

u|$=0 = u|$=% = 0, u|r=R = f(x, y).

Problem 10. Solve the initial boundary value problem#$%

$&

utt + a2uxxxx = 0, 0 < x < l, t > 0,

u(x, 0) = x(x! l), ut(x, 0) = 0, 0 ) x ) l,

u(0, t) = u(l, t) = uxx(0, t) = uxx(l, t) = 0, t " 0.

Problem 11. Solve the initial boundary value problem"

ut = a2uxx ! b2u, 0 < x < l, t > 0,

u(x, 0) = u0, u(0, t) = 0, ux(l, t) + hu(l, t) = 0.

Problem 12. Solve the initial boundary value problem#$$$%

$$$&

ut = c2(uxx + uyy), 0 < x < a, 0 < y < b, t > 0,

u(x, y, 0) = A,

u(0, y, t) = 0 = ux(a, y, t),

uy(x, 0, t) = u(x, b, t) = 0.

Problem 13. Assume the eigenvalue problem#$$$$$$%

$$$$$$&

[p(x)X !(x)]! ! q(x)X(x) + )-(x)X(x) = 0,

p " p0 > 0, - > -0 > 0, q " 0, x ( (0, l),

A0X(0) + B0X !(0) = 0,

A1X(l) + B1X !(l) = 0,

A20 + B2

0 -= 0, A21 + B2

1 -= 0,

has eigenvalue. Where Ai and Bi (i = 0, 1) are constants, B0 ) 0, A0 " 0, A1 " 0, andB1 " 0. Prove that the eigenvalue is positive. Furthermore, if X1 and X2 correspond todi"erent eigenvalues )1 and )2, then

! l

0

-(x)X1(x)X2(x) dx = 0.

5.6. PROBLEMS 111

Problem 14. Using the energy function

E(t) =1

2

! l

0

(ku2x + -u2

t + qu2) dx

to prove that the problem

#$%

$&

-(x)utt = (k(x)ux)x ! q(x)u, 0 < x < l, t > 0,

u(x, 0) = ut(x, 0) = 0,

u(0, t) = u(l, t) = 0, t > 0,

has the only solution u + 0. Here k(x) " k0 > 0, q(x) " 0, -(x) " -0 > 0, k0 and -0 aretwo constants.

Problem 15. If w(x, t; *) is the solution of

(C)

#$%

$&

wtt ! a2wxx = 0, 0 < x < l, t > *,

w|x=0 = w|x=l = 0, t > *

w(x, * ; *) = 0, wt(x, * ; *) = f(x, *), 0 ) x ) l,

prove that the function

u(x, t) =

! t

0

w(x, t; *) d*

is the solution of

(D)

#$%

$&

utt ! a2uxx = f(x, t), 0 < x < l, t > 0,

u(0, t) = u(l, t) = 0, t " 0,

u(x, 0) = 0, ut(x, 0) = 0, 0 < x < l,

This is the Duhamel principle.

Problem 16. Assume !E = (E1, E2, E3) and !B = (B1, B2, B3) are solution of the Maxwellsystem #

$%

$&

!Et = $& !B!Bt = !$& !E

$ · !B = $ · !E = 0.

Prove that if u = Ei or u = Bi for i = 1, 2, 3, the utt !!u = 0.

Problem 17. Use the Kirchho" formula to solve"

utt = a2(uxx + uyy + uzz), (x, y, z) ( R3, t > 0

u|t=0 = x3 + y2z, ut|t=0 = 0.


Problem 18. Assume u(x, t) is the solution for the following problem"

utt ! a2!u = 0, x ( R3, t > 0

u(x, 0) = '(x), ut(x, 0) = +(x).

If '(x), +(x) ( C$c (R3), prove there is a constant C such that

|u(x, t)| ) C

t, x ( R3, t > 0.

Problem 19. Use the Poisson’s formula to solve"utt = a2(uxx + uyy), (x, y) ( R2, t > 0

u|t=0 = x2(x + y), ut|t=0 = 0.

Problem 20. Solve the following initial value problem"

utt = a2(uxx + uyy) + c2u, (x, y) ( R2, t > 0

u|t=0 = '(x, y), ut|t=0 = +(x, y).

(Hint: choose v(x, y, z) = ecza u(x, y), apply Kirchho" formula to v.)

Problem 21. Let u(x, y, t) be the solution of"

utt ! 4(uxx + uyy) = 0u, (x, y) ( R2, t > 0

u|t=0 = '(x, y), ut|t=0 = +(x, y).

Here

'(x, y) = +(x) =

"0, (x, y) ( #,

1, (x, y) ( R2 \ #,

where # = {(x, y)||x| ) 1, |y| ) 1}. Determine the region where u(x, y, t) + 0 for t > 0.

Problem 22. Solve the following problem"

utt !!u = 2(y ! t), (x, y, z) ( R3, t > 0

u|t=0 = 0, ut|t=0 = x2 + yz.

(Hint: Decompose the problem into two using the linear superposition principle.)

Problem 23. A vibrating string under frictional force satisfies the following equation

utt ! a2uxx ! cut = 0, c > 0.

Prove the energy of this equation decreasing in time. With this fact, prove the uniquenessof the following problem

#$%

$&

utt ! a2uxx ! cut = f(x, t), 0 < x < l, t > 0,

u(0, t) = u(l, t) = 0, t " 0,

u(x, 0) = '(x), ut(x, 0) = +(x), 0 ) x ) l.

apdes pan2008 - unknown

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