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MATH 401: GREENS FUNCTIONS AND VARI-ATIONAL METHODS

The main goal of this course is to learn how to solve various PDE (= partialdifferential equation) problems arising in science. Since only very few special suchproblems have solutions we can write explicitly, solve here often means either (a)find an explicit approximation to the true solution, or (b) learn some importantqualitative propertiesof the solution.

A number of techniques are available for this. The method ofseparation of variables(together with Fourier series), perhaps familiar from earlier courses, unfortunatelyonly applies to very special problems. The related (but more general) method ofeigenfunction expansionis very useful and will be discussed in places in this course.

Integral transform techniques (such as the Fourier transformandLaplace transform)will also be touched on here. If time permits, we will also study some perturbationmethods, which are extremely useful if there is a small parameter somewhere in theproblem.

The bulk of this course, however, will be devoted to the method/notion ofGreensfunction(a function which, roughly speaking, expresses the effect of the data at onepoint on the solution at another point), and to variational methods(by which PDEproblems are solved by minimizing (or maximizing) some quantities). Variationalmethods are extremely powerful and even apply directly to some nonlinear problems(as well as linear ones). Greens functions are for linear problems, but in fact play a keyrole in nonlinear problems when they are treated as in some sense perturbationsof linear ones (which is frequently the only feasible treatment!).

We begin with Greens functions.

A. GREENS FUNCTIONS

I. INTRODUCTION

As an introductory example, consider the following initial-boundary value problemfor the inhomogeneous heat equation(HE) in one (space) dimension

u

t =

2u

x2 +f(x, t) 0< x < L, t >0 (HE)u(0, t) =u(L, t) = 0 (BC)u(x, 0) =u0(x) (IC)

which, physically, describes the temperature u(x, t) at time t and at point xalong arod 0xL of lengthL subject to (time- and space-varying) heat source f(x, t),and with the ends held fixed at temperature 0 (boundary condition (BC)) and withinitial (timet= 0) temperature distribution u0(x) (initial condition(IC)).

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This problem is easily solved by the (hopefully) familiar methods of separation ofvariables and eigenfunction expansion. Without going into details, the eigenfunctions

of the spatial part of the (homogeneous) PDE (namely d2

/dx2

) satisfying (BC) aresin(nx/L), n = 1, 2, 3, . . . (with corresponding eigenvaluesn22/L2), and so weseek a solution of the form

u(x, t) =

n=1

an(t) sin(nx/L).

We similarly expand the dataof the problem the source term f(x, t) and the initialcondition u0(x) in terms of these eigenfunctions; that is, as Fourier series:

f(x, t) =

n=1 fn(t)sin(nx/L), fn(t) = 2

L L

0

f(x, t) sin(nx/L)dx

u0(x) =

n=1

gnsin(nx/L), gn= 2

L

L0

u0(x)sin(nx/L)dx.

Plugging the expression for u(x, t) into the PDE (HE) and comparing coefficientsthen yields the family of ODE problems

an(t) + (n22/L2)an(t) =fn(t), an(0) =gn

which are easily solved (after all, they are first-order linear ODEs) by using an inte-grating factor, to get

an(t) =e(n22/L2)t gn+ t

0

e(n22/L2)sfn(s)ds

,

and hence the solution we sought:

u(x, t) = 2

L

n=1

e(n22/L2)t

L0

g(y) sin(ny/L)dy

+

t0

e(n22/L2)s

L0

f(y, s) sin(ny/L)dyds

sin(nx/L).

No problem. But it is instructive to re-write this expression by exchanging the orderof integration and summation (note we are not worrying here about the convergenceof the sum or the exchange of sum and integral suffice it to say that for reasonable(say continuous) functions g and fall our manipulations are justified and the sumconverges beautifully due to the decaying exponential) to obtain

u(x, t) =

L0

G(x, t; y, 0)u0(y)dy+

t0

L0

G(x, t; y, s)f(y, s)dyds (1)

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where

G(x, t; y, s) = 2

L

n=1 e(n22/L2)(ts) sin(ny/L)sin(nx/L).Expression (1) gives the solution as an integral (OK, 2 integrals) of the data (theinitial condition u0(x) and the source term f(x, t)) against the function G which iscalled, of course, theGreens functionfor our problem (precisely, for the heat equationon [0, L] with 0 boundary conditions).

Our computation above suggests a few observations about Greens functions:

if we can find the Greens function for a problem, we have effectively solved theproblem for anydata we just need to plug the data into an integral like (1)

a Greens function is a function of 2 sets of variables one set are the variablesof the solution (xand t above), the other set (y and sabove) gets integrated

one can think of a Greens function as giving the effect of the data at one point((y, s) above) on the solution at another point ((x, t))

the domain of a Greens function is determined by the original problem: in theabove example, the spatial variables x and y run over the interval [0, L] (therod), and the time variables satisfy 0 s t here the condition s treflects the fact that the solution at time t is only determined by the data at

previoustimes (not future times)

The first part of this course will be devoted to a systematic study of Greens functions,first for ODEs (where computations are generally easier), and then for PDEs, whereGreens functions really come into their own.

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II. GREENS FUNCTIONS FOR ODEs

1. An ODE boundary value problem

Consider the ODE (= ordinary differential equation) boundary value problem

Lu:= a0u+a1u

+a2u= f(x) x0< x < x1u(x0) =u(x1) = 0.

(2)

Here

L:= a0(x)d2

dx2+a1(x)

d

dx+a2(x)

is a (first-order, linear) differential operator. As motivation for problem (2), one canthink, for example, ofu(x) as giving the steady-state temperature along a rod [x0, x1]

with (non-uniform) thermal conductivity(x), subject to a heat sourcef(x) and withends held fixed at temperature 0, which leads to the problem

((x)u) = f, u(x0) =u(x1) = 0

of the form (2). Zero boundary conditions are the simplest, but later we will considerother boundary conditions (for example, if the ends of the rod are insulated, we shouldtakeu(x0) =u(x1) = 0).

We would like to solve (2) by finding a function G(x; z), the Greens function, sothat

u(x) = x1

x0

G(x; z)f(z)dz=: (Gx, f)

where we have introduced the notations

Gx(z) :=G(x; z)

(when we want to emphasize the dependence of G specifically on the variable z,thinking ofxas fixed), and

(g, f) :=

x1x0

g(z)f(z)dz ( inner product ).

Then since usolves Lu= f, we want

u(x) = (Gx, f) = (Gx, Lu).

Next we want to move the operator L over from u to Gx on the other side of theinner-product for which we need the notion ofadjoint.

Definition: The adjoint of the operator L is the operator L such that

(v,Lu) = (Lv, u) + boundary terms

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for allsmooth functions uand v.

The following example illustrates the adjoint, and explains what is meant by bound-

ary terms.Example: Let, as above, L = a0(x)

d2

dx2+a1(x)

ddx

+a2(x) acting on functions definedfor x0xx1. Then for two such (smooth) functions u(x) andv(x), integration byparts gives (check it!)

(v,Lu) =

x1x0

v[a0u+a1u+a2u]dx

=

x1x0

u[a0v+ (2a0 a1)v+ (a2+a0 a1)v]dx+ [a0vu a0vu+a1uv]

x1x0

= a0v+ (2a0

a1)v

+ (a2+a0

a1)v, u + [a0(vu

vu) + (a1

a0)uv]

x1

x0.

The terms after the integral (the ones evaluated at the endpoints x0andx1) are whatwe mean by boundary terms. Hence the adjoint is

L = a0d2

dx2+ (2a0 a1)

d

dx+ (a2+a

0 a1).

The differential operator L is of the same formas L, but with (in general) differentcoefficients.

An important class of operators are those which are equal to their adjoints.

Definition: An operator L is called (formally) self-adjointifL= L.

Example: Comparison ofL and L for our example L = a0 d2

dx2+ a1

ddx

+ a2 shows thatL is formally self-adjoint if and only ifa0=a1. Note that in this case

Lu= a0u+a0u

+a2u= (a0u)+a2u

which is an ordinary differential operator ofSturm-Liouvilletype.

Now we can return to our search for a Greens function for problem (2):

u(x) = (Gx, f) = (Gx, Lu) = (LGx, u) +BT

where we know from our computations above that the boundary terms are

BT = a0Gxu a0Gxu+a1Gxux1x0 = a0Gxux1x0where we used the zero boundary conditions u(x0) =u(x1) = 0 in problem (2). Wecan make the remaining boundary term disappear if we also imposethe boundaryconditions Gx(x0) =Gx(x1) = 0 on our Greens function G. Thus we are led to theproblem

u(x) = (LGx, u), Gx(x0) =Gx(x1) = 0

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for G.

So LGx should be a function g(z) which satisfies

u(x) = (g, u) = x1

x0

g(z)u(z)dz

for all (nice) functionsu. What kind of function is this? In fact, it is a (Dirac) deltafunction, which s not really a function at all! Rather, it is a generalized function, anotion we need to explore further before proceeding with Greens functions.

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2. Generalized functions (distributions).

The precise mathematical definition of a generalized function is:

Definition: A generalized function or distribution is a (continuous) linear func-tional acting on the space of test functions

C0 (R) ={ infinitely differentiable functions on R which vanish outside some interval }(an example of such a test function is

(x) =

e1/(a

2x2) a < x < a0 |x| a

for any a >0). That is, a distribution fmaps a test function to a real number

f() = (f, ) = f(x)(x)dxwhich it is useful to think ofas an integral (as in the parentheses above) hence thecommon notation (f, ) for f() but is not in general an integral (it cannot be since fis not in general an actual function!). Further, this map should be linear: fortest functions and and numbers and ,

f(+) =f() +f()

or (f,+) =(f, ) +(f, )

.

Some examples should help clarify.

Example:

1. If f(x) is a usual function (say, a piecewise continuous one), then it is alsoa distribution (after all, we wouldnt call them generalized functions if theydidnt include regular functions), which acts by integration,

f() = (f, ) =

f(x)(x)dx

which is indeed a linear operation. This is why we use inner-product (andsometimes even integral) notation for the action of a distribution when thedistribution is a real function, its action on test functionsisintegration.

2. The (Dirac) delta function denoted(z) (or more generally x(z) =(z x)for the delta function centred at a point x) isnota function, but a distribution,whose action on test functions is definedto be

(, ) :=(0)

=

(z)(z)dz

(x, ) :=(x)

=

(z x)(z)dz

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(where, again, the integral here is justnotation). That is,acts on test functionsby picking out their value at 0 (and xacts on test functions by picking out their

value at x).

Generalized functions are so useful because we can perform on them many of theoperations we can perform on usual functions. We can

1. Differentiate them: iff is a usual differentiable function, then for a test function, by integration by parts,

(f, ) =

f(x)(x)dx=

f(x)(x)dx= (f, )

(there are no boundary terms becausevanishes outside of some interval). Now

iffis any distribution, these integrals make no sense, but we can use the abovecalculation as the definitionof how the distribution f acts on test functions:

(f, ) := (f, )and by iterating, we can differentiate f ntimes:

(f(n), ) = (f, (1)n(n)).Example:

(a) the derivative of a delta function:

(, ) = (, ) =(0)

(b) the derivative of the Heavyside function

H(x) :=

0 x01 x >0

(whichisa usual function, but is notdifferentiable in the classical sense atx= 0):

(H, ) = (H, ) = H(x)((x))dx) =0 (x)dx=(x)|x=x=0 =(0)

(since vanishes outside an interval). Hence

d

dxH(x) =(x).

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The fact that we can always differentiate a distribution is what makes them souseful for differential equations.

2. Multiply them by smooth functions: iff is a distribution and a(x) is a smooth(infinitely differentiable) function we define

(a(x)f, ) := (f, a(x))

(which makes sense, since a is again a test function). Note this definitionconincides with the usual one when f is a usual function.

3. Consider convergence of distributions: we say that a sequence{fj}j=1 of distri-butions converges to another distribution f if

limj

(fj, ) = (f, ) for all test functions.

This kind of convergence is called weak convergence.Example: Let(x) be a smooth, non-negative function with

(x)dx= 1, and

set j(x) := j(jx) for j = 1, 2, 3, . . .. Note that as j increases, the graphof j(x) is becoming both taller, and more concentrated near x = 0, whilemaintaining

j (x)dx= 1. In fact, we have

limj

j (x) = limj

j(jx) =(x)

in the weak sense it is a nice exercise to show this!

4. Compose them with invertible functions: let g : R R be a one-to-one andonto differentiable function, with g(x) > 0. If f is a usual function, then bychanging variables y = g(x) (so dy = g(x)dx), we have for the compositionf g(x) =f(g(x)),

(fg, ) =

f(g(x))(x)dx=

f(y)(g1(y))dy/g(g1(y)) = (f,

1

g g1 g1)

and so for fa distribution, we define

(f g, ) := (f, 1g g1 g

1).

Example: Composing the delta function with g(x) gives

((g(x)), ) = (, 1

g g1 g1) = (g1(0))g(g1(0)) ,

and in particular ifg(x) =cx (constant c >0)

((cx), ) =1

c(0) = (

1

c, )

and hence (cx) = 1c

(x).

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3. Greens functions for ODEs.Returning now to the ODE problem (2), we had concluded that we want our Greens

function G(x; z) =Gz(z) to satisfyu(x) = (LGx, u) u, Gx(x0) =Gx(x1) = 0.

After our discussion of generalized functions, then, we see that what we want is really

LGx(z) =(z x), Gx(x0) =Gx(x1) = 0.Notice that for z=x, we are simply solving LGx = 0. The strategy is to solve thisequation for x < z and for z > x, and then glue the two pieces together. Someexamples should help clarify.

Example: use the Greens function method to solve the problem

u =f(x), 0< x < L, u(0) =u(L) = 0. (3)(which could, of course, be solved simply by integrating twice).

First note that L = d2

dx2 = L (the operator is self-adjoint), so the problem for our

Greens function G(x; z) =Gx(z) is

Gx(z) =(z x), Gx(0) =Gx(L) = 0(here denotes d

dz). For z < xand z > x, we have simply Gx = 0, and so

Gx(z) =

Az+B 0z < xCz+D x < z L

The BC Gx(0) = 0 implies B = 0, and the BC Gx(L) = 0 implies D =LC, so wehave

Gx(z) =

Az 0z < x

C(z L) x < zLNow our task is to determine the remaining two unknown constants by using matchingconditionsto glue the two pieces together:

1. continuity: we demand that Gxbe continuous atz= x: Gx(x) =Gx(x+) (thenotation here is g(x) := lim0 g(x )). This yields Ax= C(x L).

2. jump condition: for any >0, integrating the equation Gx =(z x) betweenx

and x+ yields

Gx(x+) Gx(x ) =Gxx+

x= x+

xGx(z)dz=

x+x

(z x)dz= 1

and letting 0, we arrive atGx(x+) Gx(x) = 1.

This jump condition requires C A= 1.

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Solving the two equations for Aand Cyields C=x/L, A= (x L)/L, and so

G(x; z) =Gx(z) = z(x L)/L 0z < xx(z L)/L x < z Lwhich gives our solution of problem (3)

u(x) =

L0

G(x; z)f(z)dz=x L

L

x0

zf(z)dz+x

L

Lx

(z L)f(z)dz.

Remark:

1. whatever you may think of think of the derivation of this expression for thesolution, it is easy to check (by direct differentiation and fundamental theoremof calculus) that it is correct(assuming f is reasonable say, continuous).

2. notice the form of the Greens functionGx(z) (graph it!) it has a singularity(in the sense of being continuous, but not differentiable) at the point z= x. Thissingularity must be there, since differentiating G twice has to yield a deltafunction.

Example: use the Greens function method to solve the problem

x2u+ 2xu 2u= f(x), 0< x

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1. continuity Gx(x) =Gx(x+) implies Ax= C(x 1/x2)2. jump condition

1 =

x+x

(zx)dz= x+

x

(z2Gx)

2Gx]dz=z2Gx(z)x+

x=x2(Gx(x+)Gx(x))

impliesx2C(1 + 2/x3) x2A= 1.Solving the simultaneous linear equations for Aand Cyields

1 =C

(x2 + 2/x) x2(1 1/x3)= 3C/x = C=x/3, A= (x 1/x2)/3and so the Greens function is

G(x; z) =13 z(x 1/x2) 0z < xx(z 1/z2) x < z1and the corresponding solution formula is

u(x) =

10

G(x; z)f(z)dz=1

3(x 1/x2)

x0

zf(z)dz+1

3x

1x

(z 1/z2)f(z)dz.

Does this really solve (4) (supposing f is, say, continuous)? For 0 < x

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and we see that that the ODE is indeed solved. What about the BCs? Well, u(1) = 0obviously holds. As alluded to above, the BC at x = 0 is subtler. We see that

3 limx0+

u(x) = limx0+

1

x2 x

0

zf(z)dz limx0+

x 1

x

f(z)/z2dz.

Iff is smooth, we have f(z) =f(0) +O(z) for small z, so

3 limx0+

u(x) =f(0) f(0) =2f(0),

and so we require f(0) = 0 to genuinely satisfy the boundary condition at x= 0.

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4. Boundary conditions, and self-adjoint problems.

The only BCs we have seen so far have been homogeneous (i.e. 0) Dirichlet

(specifying thevalueof the function) ones; namelyu(x0) =u(x1) = 0. Lets make thismore general, first by considering an ODE problem with inhomogeneous DirichletBCs:

Lu:= a0u+a1u+a2u= f x0< x < x1u(x0) =u0, u(x1) =u1

. (5)

Recall that by integration by parts, for functions uand v,

(v,Lu) = (Lv, u) + [a0(vu vu) + (a1 a0)vu]

x1x0

.

Suppose we find a Greens function G(x; z) =Gx(z) solving the problem

LGx=(z x)Gx(x0) =Gx(x1) = 0

with the corresponding homogeneous(i.e. 0) BCs to the BCs in problem (5). Then

u(x) = (LGx, u) = (Gx, Lu) [a0Gxu]x1

x0= (Gx, f) [a0Gxu]

x1x0

=

x1x0

G(x; z)f(z)dz+a0(x0)Gx(x0)u0 a0(x1)Gx(x1)u1

a formula which gives the solution of problem (5) in terms of the Greens function

G(x; z), and the data (the source term f(x), and the boundary data u0 and u1).The other way to generalize boundary conditions is to include the value of the deriva-tive of u (as well as u itself) at the boundary (i.e. the endpoints of the interval).For example, ifu is the temperature along a rod [x0, x1] whose ends are insulated,we should impose the Neumann BCsu(x0) =u(x1) = 0 (no heat flux through theends).

In general, for the following discussion, think of imposing 2 boundary conditions,each of which is a linear combination ofu(x0), u

(x0), u(x1), and u(x1) equal to 0(homogeneous case) or some non-zero number (inhomogeneous case).

Definition:

1. Aproblem

Lu= fBCs on u

is called (essentially) self-adjointif

(a) L= L (so the operatoris self-adjoint), and

(b) (v,Lu) = (Lv,u) (i.e. with no boundary terms) whenever both u and vsatisfy thehomogeneousBCs corresponding to the BCs onuin the problem.

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Remark: As in the above example, the Greens function for a self-adjoint problemshould satisfy the homogeneous BCs corresponding to the BCs in the original

problem. More generally, the Greens function for a problem should satisfy:

2. The homogeneous adjoint boundary conditions for a problem

Lu= fBCs on u

are the BCs on v which guarantee that (v,Lu) = (Lv, u) (i.e. no boundaryterms) when usatisfies the homogeneous BCs corresponding to the BCs in theoriginal problem.

Remark:

1. these definitions are quite abstract it is better to see what is going on by doingsome specific examples

2. a problem can be non-self-adjoint even ifL= L, for example (see homework) u+q(x)u= f(x)

u(0) u(1) = 0, u(1) = 0

3. ifL=L, we can make Lu = fself-adjoint by multiplying by a function (again,see homework).

Example: (Sturm-Liouville problem)

Lu:= (p(x)u)+q(x)u= f(x) 0< x 0, and 0, 1, 0, 1 are numbers with 0, 0 not both0, and 1, 1not both0.

First notice that L = L (the operator is self-adjoint), and integration by parts (asusual) gives

(v,Lu) = (Lv,u) + [p(vu vu)]

1

0.

Ifuand v bothsatisfy the BCs in problem (6) (which are homogeneous) then

0[v(0)u(0) v(0)u(0)] =u(0)(0v(0)) v(0)(0u(0)) = 0

0[v(0)u(0) v(0)u(0)] =v(0)(0u(0)) u(0)(0v(0)) = 0

and so (since 00= 0),v(0)u(0) v(0)u(0) = 0. A similar computation shows thatv(1)u(1) v(1)u(1) = 0. Hence (v,Lu) = (Lv,u) (boundary terms disappear), andthe problemis, indeed, self-adjoint.

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Thus a Greens function G(x; z) =Gx(z) for problem (6) should satisfy

LGx = (p(z)Gx(z))+q(z)Gx(z) =(z x)0Gx(0) +0Gx(0) = 01Gx(1) +1G

x(1) = 0

.

For z=x, we have LGx= 0, so

Gx(z) =

c0w0(z) 0z < xc1w1(z) x < z1

where for j = 0, 1, wj denotes any fixed, non-zero solution of

Lwj = 0, jwj (j) +jwj (j) = 0

(which is an initial value problem for a second-order, linear, ODE hence there is aone-dimensional family of solutions), and c0,c1 are non-zero constants. Continuity ofG at ximplies

c0w0(x) =c1w1(x).

The jump condition at x is (check it!) p(x)[Gx(x+) Gx(x)] = 1, so

c1w1(x) c0w0(x) =

1

p(x).

Hence

c1[w0(x)w1(x) w1(x)w0(x)] =

w0(x)

p(x) ,

which involves the Wronskian

W=W[w0, w1](x) =w0(x)w1(x) w1(x)w0(x).

Recall that since w0 and w1 satisfyLw= 0,

p(x)W[w0, w1](x)

constant.

There are two possibilities:

1. IfW 0, then we cannot satisfy the equations for the coefficients above thereis no Greens function! In this case, w0 and w1 are linearly dependent, whichmeans thatw0 and w1 are both actually solutions of the homogeneous equationLu= 0 satisfying bothBCs in (6). Well discuss this case more later.

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2. Otherwise, W is non-zero everywhere in (0, 1) and so we have

c1=

w0(x)

p(x)W, c0=

w1(x)

p(x)W

and hence a Greens function

Gx(z) = 1

W[w0, w1]p(x)

w1(x)w0(z) 0z < xw0(x)w1(z) x < z1

and a solution to our Sturm-Liouville problem (6)

u(x) = 1

W[w0, w1]p(x)

w1(x)

x0

w0(z)f(z)dz+w0(x)

1x

w1(z)f(z)dz

.

Remark:

1. Using the variation of parametersprocedure from ODE theory would result inprecisely the same formula.

2. Notice that the Greens function here (and for all our examples so far) is sym-metricin its variables: Gx(z) =G(x; z) =G(z; x) =Gz(x). This is no accident rather, it is a general property of Greens functions for self-adjoint problems.

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5. Modified Greens functions.

We have seen that we run into trouble in constructing a Greens function for the

Sturm-Liouville problem Lu:= (p(x)u)+q(x)u= f(x) 0< x 0, q, andfsmooth functions, and0,1, 0, 1 numbers with0, 0 notboth0, and 1, 1 not both0) if the corresponding homogeneous problem ((7) withf0) has a non-trivial solution u:

Lu= 0, 0u(0) +0u(0) = 0, 1u(1) +1u

(1) = 0. (8)

In fact in this case, a simple integration by parts0 = (u,Lu) = (Lu,u) = (f, u)

leads to the solvability condition

(f, u) = 1

0

f(x)u(x)dx= 0 (9)

which the source termf mustsatisfy, to have any hope of a solution u. In fact:

Theorem: [Fredholm alternative (for the Sturm-Liouville problem)] Either

1. Problem (7) has exactly one solution; or,

2. There is a non-zero solution u of the corresponding homogeneous problem (8).In this case, problem (7) has a solution if and only if the solvability condition (9)holds (and the solution is not unique you can add any multiple ofu to getanother).

Proof:Weve already seen that if there is no suchu, we have a solution (constructed abovevia Greens function). Also, it is unique, since the difference of any two solutionswould be a solution of the homogeneous problem.

If there is a non-zero u, weve seen already that the solvability condition (9) isrequired. If it is satisfied, we will show below how to construct a solutionu using amodified Greens function.

So suppose now that u is a non-zero solution to (8), and that the solvability condi-tion (9) on f holds. A function G(x; z) = Gx(z) satisfying

LGx(z) =(z x) +c(x)u(z)same BCs on Gx as in (7)

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is called a modified Greens function. Notice that we can choose the constantc(x) here so that the solvability condition

0 = (Gx, Lu) = (LGx, u) = (x+c(x)u, u) =u(x) +c(x)(u, u)

holds namely c(x) =u(x)/(u, u). This allows the probem for Gx to be solved(though we wont do it here in general you can use the variation of parametersmethod to do it). Given such a G, if we have a solution uof (7), then

u(x) = (x, u) = (LGx cu, u) = (Gx, f) + (u, u)

(u, u)u

is a solution formula for u in terms off. Note that the constant in front ofu doesntmatter remember we can add anymultiple ofu and still have a solution. In factusingreciprocity(Gis symmetric inx and z), we can check this formula indeed solvesproblem (7):

Lu= (LxGz(x), f) = (z(x), f) + (c(z)u(x), f) = (x(z), f) u

(x)(u, u)

(u(z), f) =f

where we had to use (of course!) the solvability condition (9) on f. Similarly, theboundary conditions hold.

Hopefully an example will clarify the use of modified Greens functions.

Example: Solve u(x) =f(x), 0< x

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Continuity at x requires A= x+B. The jump condition Gx

x+

x = 1 already holds.(This leaves one free parameter, which is characteristic of modified Greens function

problems, since a multiple ofu (in this case a constant) can always be added.) So,assuming the solvability condition

10

f(x)dx = 0 holds, we arrive at a formula forthe general solution

u(x) = (Gx, f) +C

=12

10

z2f(z)dz+ (B+x)

x0

f(x)dz+B

1x

f(z)dz+

1x

zf(z)dz+C

=x

x0

f(z)dz+

1x

zf(z)dz+C

where we simplified using the solvability condition on f, and we absorbed 10 z2f(z)dzinto the general constant C. It is very easy to check that this expression solves ouroriginal problem (of course, we could have solved this particular problem just byintegrating).

Remark: So far in this section we discussed the issues of solvability conditions, modi-fied Greens functions, and so on, only for self-adjointproblems. For a general problem,the obstacle to solvability comes from considering the homogeneous adjoint problem:

Lu= 0hom. adj. BCs on u

.

If this problem has a non-trivial solution u, then the solvability condition forLu = fwith homogeneous BCs is

0 = (u, Lu) = (Lu,u) = (f, u).

One can also construct a modified Greens function in this case, but we wont go intoit here.

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6. Greens functions and eigenfunction expansion.

Consider again the Sturm-Liouville problem (with homogeneous BCs) Lu:= (p(x)u)+q(x)u= f(x) x0< x < x1

(BC)

0u(x0) +0u

(x0) = 01u(x1) +1u

(x1) = 0(10)

(p > 0, q, and f smooth functions). Recall that the Sturm-Liouville problem haseigenvalues

0 < 1< 2< 3 and corresponding eigenfunctions j(x) such that

Lj =j jj satisfies (BC)

which can be taken orthonormal:

(j, k) =jk =

1 j =k0 j=k .

Furthermore, the eigenfunctions are complete, meaning that any function g(x) on[x0, x1] can be expanded

g(x) =

j=0

cj j(x)

cj = (j, g) = Fourier coefficient.

To be more precise, this series does not (in general) converge at every point x in[x0, x1] (for example, it cannot converge at the endpoints ifg does not satisfy (BC)!), but rather converges in the L2-sense: provided

x1x0

g2(x)dx

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and try to find the coefficients cj(x). Using LGx = x, and applying the linearoperatorL to the expression for Gx, we find

x=LGx =

j=0

cj(x)(Lj )(z) =

j=0

cj(x)jj(z)

and so the coefficients cj (x)j of this expansion should satisfy

cj(x)j = (j, x) =j (x),

and hence we arrive at an expression for the Greens function as an eigenfunctionexpansion:

Gx(z) =

j=0 1j j(x)j(z).Remark:

1. The reciprocity (symmetry in x and z of G) is very clearly displayed by thisformula.

2. If, for some j, j = 0, this expression is ill-defined. Indeed, in this case, theeigenfunction j is a non-trivial solution to the homogeneous problem (i.e. it isa u), and we have already seen there is no (usual) Greens function in thiscase.

3. The corresponding solution formula for problem (10) is

u(x) = (Gx, f) =

x1x0

j=0

1

jj(x)j(z)f(z)dz=

j=0

1

jj(x)(j, f).

Notice that if j = 0 (for some j), the only way for this expression to makesense is if also (j, f) = 0 which is exactly the solvability condition on f wehave seen earlier.

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(one can think (yx) = (y1x1)(y2x2) in 2 dimensions, and (yx) =(y1 x1)(y2 x2)(y3 x3) in 3 dimensions).Supposing we can find a Greens function solving (12), we would like to derive aformula for the solution u of (11) in terms of G. To do this, we will use a simpleintegration by parts formula, for which we need to define the normal derivativeofa function uat the boundaryS= D:

u

n := n u

where ndenotes the outward unit normal vector to the curve/surface S.

Lemma 1 (Greens second identity) Let DR2 or3 be bounded by a smooth

curve/surfaceS. Then ifu1 andu2 are twice continuously differential functions onD,

D

(u1u2 u2u1) dV =

S

u1

u2n

u2 u1n

dS. (13)

Proof: subtracting the relations

(u1u2) =u1 u2+u1u2 (u2u1) =u1 u2+u2u1

and using the divergence theorem, we findD

(u1u2 u2u1) dV =

S

n (u1u2 u2u2)dS=

S

u1

u2n

u2 u1n

dS.

Takingu1 = u (solution of (11)) and u2=Gx (solution of (12)) in (13), we findD

(u(y)x(y) Gx(y)f(y))dy =

S

g(y)Gxn

dS(y)

or

u(x) = D

G(x; y)f(y)dy+ S

G(x; y)n(y)

g(y)dS(y) (14)

which is the solution formula we sought.

Free-space Greens functions. For a general region D, we havenochance of findingan explicit Greens function solving (12). One case we caneasily solve explicitly, isfor the whole space D= R2 or D= R3, where there is no boundary at all.

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R2: for a given xR2, we want to find Gx(y) solving Gx(y) =(y x). It isreasonable to try a Gwhich depends only on the distance from the singularity:

G= h(r), r:=|y x|. So for y=x (r >0), we need0 = G=

2

r2+

1

r

r

h(r) =

1

r(rh(r))

and so

h = c1/r = h= c1log r+c2for some constants c1 and c2. Including c2 is simply adding an overall constantto the Greens function, so we will drop it: c2 = 0. The constant c1is determinedby the condition Gx=x: for any >0, we have, by the divergence theorem,

1 = |yx|

x(y)dy= |yx|

Gx(y)dy = |yx|=

Gx(y) n dS(y)

=

|yx|=

c1y x

|y x|2 y x|y x|dS(y) =

c1

2 = 2c1

and so c1 = (2)1, and our expression for the two-dimensional free-space

Greens function is

GR2(x; y) = 1

2log |y x|.

R3: we play the same game in three dimensions. Try G

x(y) =h(r),r :=

|y

x|,

so for r >0

0 = Gx=

2

r2+

2

r

r

h(r) =

1

r2(r2h(r))

and so

h = c1/r2 = h=c1/r+c2and again we set c2 = 0. Constant c1 is determined by

1 =

|yx|

x(y)dy =

|yx|

Gx(y)dy=|yx|=

Gx(y) ndS(y)

= |yx|=

c1 y x|y x|3 y x|y x|dS(y) =

c12

42 = 4c1

and so c1 = (4)1, and our expression for the three-dimensional free-space

Greens function is

GR3(x; y) = 1

4|y x| .

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Knowing the free-space Greens functions, we are lead immediately to solution for-mulas for the Poisson equation u(x) =f(x) in R2 and R2:

u(x) = 12 R2log |x y|f(y)dy R2 14

R3

f(y)|xy| R

3 . (15)

The following theorem makes precise the claim that these formulas indeed solve Pois-sons equation:

Theorem: Letf(x) be a twice continuously differentiable function on R2 (respectivelyR3) with compact support (i.e. it vanishes outside of some ball). Then u(x) definedby (15) is a twice continuously differentiable function, and u= f(x) for all x.

Proof: well do the R3 case (the R2 case is analogous). Notice first that by the changeof variable yx y in the integral,

u(x) = 14

R3

f(y)|x y|dy=

14

R3

f(x y)|y| dy.

If ej denotes the j -th standard basis vector, then

u(x+hej) u(x)h

= 14

R3

f(x+hej y) f(x y)

h

dy

|y| ,

and since f(x+hejy)f(xy)

h f

xj(x y) ash0 uniformly in y (since f is continu-

ously differentiable and compactly supported),

u

xj(x) = lim

h0

u(x+hej) u(x)h

=

1

4 R3f

xj(x

y)

dy

|y|.

Similarly,2u

xjxk(x) =1

4

R3

2f

xjxk(x y) dy|y|

which is a continuous function ofx (again, because fis twice continuously differen-tiable and compactly supported) - hence uis twice continuously differentiable. Now,to find u we want to integrate by parts twice. But the Greens function 1/|y| isnot twice continuously differentiable (indeed not even continuous) at y = 0, so to dothis legally, we remove a small neighbourhood of the origin: for >0,

u(x) = 1

4 |y|> f(x y) dy|y| 14 |y| f(x y) dy|y| =: A + B .Now,

|B| 14

maxx

|f(x)||y|

dy

|y| = 1

4max

x|f(x)|

0

rdr

= 1

4max

x|f(x)|

2

20 as 0

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(f is bounded because it is continuous and compactly supported). For the otherterm, we apply Greens second identity on the domain{|y| > }, and use the factthat (1/|y|) = 0 for|y|> 0 to find

A= 1

4

|y|=

n

1

|y|f(x y) 1

|y|

nf(x y)

dS(y) =:C+ D.

For the second term,

|D| 14

max |f||y|=

1

|y

|dS(y) =

1

4max |f|1

42 0 as 0.

Finally, using n

1|y| =

1|y|2 ,

C= 1

4

|y|=

1

|y|2 f(x y)dS(y) = 1

42

|y|=

f(x y)dS(y) =f(x y)

for some y with|y| = , by the mean-value theorem for integrals. So as 0,y

0, and so sending

0 in all terms yields u(x) =f(x), as required.

Remark:

1. for the above theorem to hold true, we dont really need f to be compactlysupported (just with enough spatial decay so the the integral defining u makessense), or twice continuously differentiable (actually, fjust needs to be a littlebetter than continuous although just continuous is not quite enough)

2. the free space Greens functions are not just useful for solving Poissons equationon R2 or R3, but in fact play a key role in boundary value problems as we willsoon see

3. a physical interpretation: the three-dimensional free-space Greens function(4|x y|)1 is the electrostatic potential generated by a point charge locatedat x (i.e. the Coulomb potential).

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2. The method of images.

Example: Solve the boundary value problem for Laplaces equation on the 1/2-plane

{(x1, x2)|x2>0}by the Greens function method: u= 0 x2>0

u(x1, 0) =g(x1) x2= 0 . (16)

The corresponding Greens function problem is: for xwith x2 > 0, Gx(y) =x(y) y2>0

Gx(y)0 y2= 0 .

We know that the two-dimensional free-space Greens function Gf(x; y) = log |yx|/2 will generate the right delta function at x, but most certainly does notsatisfythe boundary conditions. The idea of the method of images is to balanceGf(x, y)with other copies ofGf with singularities at different points in hopes of satisfyingthe boundary conditions. For this problem, the geometry suggests placing an imagecharge (the name comes from the interpretation of free-space Greens functions aspotentials generated by point electric charges) at x := (x1, x2), the reflection ofxabout the x1 axis. So we set

Gx(y) :=Gf(x; y) Gf(x; y) = 1

2(log |y x| log |y x|) .

Notice thaty2 = 0 = |y x|=|y x| = Gx(y) = 0

so we have satisfied the boundary conditions. But have we satisfied the right PDE?Yes, since for y with y2 > 0,

Gx(y) =(y x) (y x) =(y x)since the singularity introduced at x lies outside the domain, and so does not con-tribute. So we have our Greens function. To write the corresponding solution for-mula, we need to compute its normal derivative on the boundary y2 = 0: for y withy2= 0 (using

|y

x

|=

|y

x

|),

Gx(y) = 12|y x|2 (y x (y x)) =

1

2|y x|2 (x x)

and since the outward unit normal on the boundary is n= (0, 1),

nGx(y) =

1

2|y x|2 (x2 x2) = x2

((y1 x1)2 +x22).

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Hence our formula for the solution of problem (16) is

u(x) =

x2 g(y1)(y1 x1)2 +x22 dy1.

Now, does this expression really yield a solution to problem (16)? For x2 > 0, thefunction x2/(x

21+x

22) is harmonic:

x2

x21+x22

=

x1

2x1x2(x21+x

22)

2+

x2

x21 x22(x21+x

22)

2

=2x2(x21+x22) + 8x21x2 2x2(x21+x22) 4x2(x21 x22)

(x21+x22)

3 = 0,

and hence so isx2/((y1

x1)2 +x22) for anyy1. So supposing the functiong is bounded

and continuous, we may differentiate under the integral sign to conclude that u(x) =0 for x2>0. What about the boundary conditions? Since

(s

2 + 1)1ds= , andchanging variables y1 = x2s, we have for fixed x1, and for x2 > 0,

|u(x1, x2) g(x1)|= 1

1

s2 + 1[g(x1+x2s) g(x1)] ds

4max |g|

M

ds

s2 + 1+

1

MM

1

1 +s2|g(x1+x2s) g(x1)|ds.

Let > 0 be given, and then choose Mlarge enough so that the first term above is< /2. Then since a continuous function on a compact set is uniformly continuous,we may choose x2 small enough so that the second term above is < /2. Thus wehave shown that

limx20

u(x1, x2) =g(x1)

i.e., that the boundary conditions in problem (16) are also satisfied.

In practice, the method of images can only be used to compute the Greens functionfor very special, highly symmetric geometries. Examples include:

half-plane, quarter-plane, half-space, octant, etc.

disks, balls

some combinations of the above, such as a 1/2-disk

Example: Solve the boundary value problem for Laplaces equation in the 3D ball ofradius a{x R3 | |x|< a} using the method of images:

u= 0 |x|< au= g |x|= a . (17)

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Proof (for theR3 case): take x= 0 in (18) to find

u(x) =

a

4 |y|=a g(y)|y|3dS(y) = 14a2 |y|=a g(y)dS(y)as required.

Solvability conditions and modified Greens functions.

Example: Consider the problem u= f inD R2

un

=g on D (19)

with so-called Neumann boundary conditions. Notice that the corresponding

homogeneous problem u = 0 in D R2u

n = 0 on D

has the non-trivial solution u(x)1. As a result, there is a solvability condition onthe datafandg in problem (19), which can be derived using Greens second identity:

0 =

D

uudx=

D

uudx+

D

(u

nu uu

n)dS

and since u/n0, we arrive at the solvability condition

D f(x)dx= D g(x)dS(x).Problem (19) will not have a Greens function, but will have a modified Greensfunction, satisfying

Gx(y) =(y x) +C yD

nGx(y) = 0 yD

for some constant C, which can be determined using the divergence theorem:

1 =

D

(y x)dy=

D

(Gx C)dy=

D

nGxdS C|D|=C|D|

(where|D|denotes the area/volume ofD), and henceC=1/|D|. If we can find themodified Greens function, we can construct the family ofsolutions of problem (19)(assuming the solvability condition holds), as

u(x) =

D

Gx(y)f(y)dy

D

Gx(y)g(y)dS(y) +A

for any constant A.

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3. Greens functions for Laplacian: some general theory.

So far we have mainly discussed methods for explicit computation of Greens func-

tions and solutions for certain problems involving the Laplacian operator. For generalregions, of course, explicit computations are not possible. Nevertheless, some generaltheory tells us that Greens functions exist, and that the corresponding solution for-mulas do yield solutions to the problems of interest. We describe some of this theoryhere.

Let D be a region in R2 or R3 bounded by a smooth curve/surface S = D. LetGf(x; y) denote the free-space Greens function (i.e. log |xy|/(2) for R2 and(4|y x|)1 for R3).Definition: a (Dirichlet) Greens function G(x; y) for the operator and the regionD is a function continuous for x, y

D:= D

S, x

=y, satisfying

1. the function Hx(y) :=Gx(y) Gfx(y) is harmonic (i.e. yHx(y) = 0) for yD2. Gx(y)0 for yS

Remark:

the first property implies that Gx solves Gx=x, and also implies that Gx(y)is a smooth function on D\{x} (since Gfx is, and since harmonic functions aresmooth).

the terminology Dirichlet Greens function refers to the Dirichlet boundaryconditions specifying the value of the function (rather than, say, its normalderivative) on the boundary.

Properties of Greens functions:

1. Existence: a Greens function as above exists. (The proof of this fact calls forsome mathematics a little beyond the level of this course eg., some functionalanalysis so we will have to simply take it as given.)

2. Uniqueness: the Greens function is unique in particular, we may refer tothe Greens function (rather than a Greens function).

3. Symmetry (reciprocity): G(x; y) =G(y; x).

4. Relation to solution of BVP for Poisson: let fbe a smooth function onD, and let g be a smooth function on S. Then u solves the boundary valueproblem for Poissons equation

u= f Du= g D

(20)

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if and only if

u(x) = D G(x; y)f(y)dy+ D Gx(y)n g(y)dS(y). (21)Some (sketches of) proofs:

Symmetry: a formal (i.e. cheating) argument giving the symmetry is:

G(x; y) G(y; x) =

D

(G(x; z)y(z) G(y; z)x(z)) dz

=

D

(Gx(z) Gy(z) Gy(z)Gx(z)) dz

= S

Gx(z) Gy(z)n(z)

Gy(z) Gx(z)n(z)

dS(z) = 0.This is not a rigorous argument, because we applied Greens second identityto functions (i.e. the Greens functions) which are NOT twice continuouslydifferentiable (indeed, not even continuous). To make the argument rigorous,try applying this identity on the domain D with balls of (small) radius aroundthe points x and y removed the Greens functions are smooth on this newdomain. Then take 0. This is left as an exercise.

Representation of solution of Poisson: suppose u is twice continuously differen-tiable onD, continuous on

D:= D S, and satisfies (20). We will show that uis given by formula (21). Assume first that f= 0, so that u is harmonic in D.

We claim then that

u(x) =

S

u

nGfx Gfx

nu

dS. (22)

Assuming this for a moment, applying Greens second identity to the harmonicfunctions u and Hx := Gx Gfx in D, we arrive at

0 =

Su

Hxn

Hx un

dS

which we add to (22) to arrive at

u(x) =

S

u

Gxn

Gx un

dS=

S

Gxn

gdS

as required. Now lets prove (22) (we will just do the three-dimensional casehere, for simplicity). Let D be D with a ball of (small) radius about x

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removed, and apply Greens second identity to u and Gfx onD (both functionsare smooth and harmonic here) to find

0 =

D

u

Gfxn

Gfxu

n

dS

=

D

u

Gfxn

Gfxu

n

dS

|yx|=

y x|y x|

u

y x4|y x|3

1

4|y x|u

dS.

Now the second term in the second integral is bounded by

1

4max |u| 42 = (max |u|)0 as 0,

while the first term is, in the limit as

0,

lim0

1

42

|yx|=

u(y)dS(y) =u(x)

(by the mean value theorem for integrals), and we recover (22), as desired.Finally, suppose f= 0 in (20). Let

w(x) :=

D

G(x; y)f(y)dy.

Notice that for x S, G(x; y) = 0 (reciprocity), and so w(x) = 0. Just as wedid for the Poisson equation in R

3

, we can show that w= f, and so

(u w) = u f=f f= 0,and we may write

u(x) w(x) =

D

Gxn

[g(y) w(y)]dS(y) =

D

Gxn

g(y)dS(y)

and we recover (21) as needed.

Formula (21) solves the Poisson BVP: it remains to show (21) really does solvethe BVP (20). We already argued that the first term in (21) (what we calledw above) solves w= fand has zero boundary conditions. It remains to showthat the boundary integral is harmonic, and has boundary values g. We wontdo this here the proof can be found in rigorous PDE texts.

Uniqueness of the Greens function: we will prove this using the maximumprinciple in the next section.

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4. The maximum principle.

Let D be a (open, connected) region in Rn, bounded by a smooth surface D.

Theorem: [maximum principle for harmonic functions]. Let u be a functionwhich is harmonic in D (u= 0) and continuous on D= D D. Then

1. uattains its maximum and minimum values on the boundary D

2. ifu also attains its maximum or minimum value at an interior point ofD, itmust be a constant function

Remark:

1. umust attain its max. and min. values somewhereon D, since it is a continuousfunction on a closed, bounded set

2. the maximum part of the theorem extends tosubharmonicfunctions (u0),and the minimum part to superharmonicfunctions (u0).

Well give two proofs of the maximum principle.

Proof #1: suppose u does, in fact attain its maximum (say) at an interior pointx0D. Then for any r >0 such that the ball Br of radius r about x0 lies in D, themean-value property of harmonic functions implies that

u(x0) = 1

|Br

| Bru dSu(x0)

since u(x0) is the maximum value ofu, and so we must have equality above, and umust be equal to u(x0) everywhere on Br. Hence u is constant on Br. Repeatingthis argument, we can fill out all ofD, and conclude u is constant on D. (A slickargument for this: the set where u(x) =u(x0) is both closed and open in D, hence isall ofD, since D is connected.)

Our second proof does not rely so heavily on the mean-value property (at least for thefirst statement of the maximum principle), and hence generalizes to other Laplacian-like (a.k.a elliptic) operators, for which the mean-value property does not hold.

Proof # 2: again, suppose u were to have a maximum (say) at an interior point

x0 D. Then we know from vector calculus that u(x0) 0, which is almost acontradiction to u= 0, but not quite (because it israther than0,

v(x) :=u(x) +|x|2.Notice that

v=u+ 2x, u= u+ 2n= 2n >0

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(n is the space dimension), and so v really cannot have a maximum at an interiorpoint ofD. That means that for any xD,

u(x)v(x)< maxyD

v(y) = maxyD

[u(y) +|y|2]maxyD

u(y) + maxyD

|y|2

and now letting 0, we seeu(x)max

yDu(y),

which is the first statement of the maximum principle (sometimes called the weakmaximum principle. For the second statement (sometimes called thestrong maximumprinciple), we can use the mean value property as before.

Consequences of the maximum principle:

Uniqueness of solutions of the Dirichlet problem for the Poisson equation:Theorem: Letfandg be continuous functions on D and D respectively. Thereis at most one function u which is twice continuously differentiable in D, con-tinuous on D, and solves

u= f Du= g D

.

Proof: If u1(x) and u2(x) are both solutions, then their difference w(x) :=u1(x)

u2(x) solves w= 0 D

w= 0 D .

The maximum principle says thatw attains its maximum and minimum on theboundaryD. Sincew0 onD, we must havew0 inD, and henceu1u2in D .

Uniqueness of the (Dirichlet) Greens function:Theorem: There is at most one Dirichlet Greens function for a given region D.

Proof: if G(x; y) is a (Dirichlet) Greens function for D, then the functionHx(y) :=G(x; y) Gf(x; y) (whereGf denotes the free-space Greens function)solves the problem

Hx(y) = 0 yDHx(y) =Gf(x; y) yD

and so is unique, by the previous theorem.

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5. Greens functions by eigenfunction expansion.

We will just do one example.

Example: Find the Dirichlet Greens function in the infinite wedge of angle , de-scribed in polar coordinates (r, ) by 0, r0.So given a point x in the wedge, we are looking for a function Gx solving Gx =xinside the wedge, and Gx = 0 on the rays = 0 and =. It is natural to work inpolar coordinates. Let (r, ) be the polar coordinates ofx. Then the delta functioncentred at x is written in polar coordinates as 1

r(r r)( ) (the factor 1/r is

there since integrating in polar coordinates, dy becomes rdrd check it). Also, theLaplacian in polar coordinates is

=

2

r2 +

1

r

r +

1

r2

2

2

and so we are looking for G(r, ) which solves Grr +

1r

Gr+ 1r2

G = 1r

(r r)( )G(r, 0) =G(r, ) = 0

The eigenfunctions of 2

2(the angular part of the Laplacian) which satisfy zero bound-

ary conditions at = 0 and = are sin(n/), n = 1, 2, 3, . . ., so we will try tofind G in the form of an eigenfunction expansion

G(r, ) = n=1

cn(r) sin(n/)

(in fact, a Fourier sine series since the eigenfunctions are sines), and our job is tofind the coefficients cn(r).

Applying term-by-term to the expansion, we find

n=1

cn+

1

rcn

n22

2 cn

sin(n/) =

1

r(r r)( ),

and integrating (in ) both sides against sin(k/) yields

ck +1

rck

k22

2 ck =

2

1

r(rr)

0

sin(k/)()d= 2r

sin(k/)(rr).

For r=r , this isck+

1

rck

k 22

2 ck = 0

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an equation of Euler type whose solutions are of the form r b, where plugging r b inyields

0 =b(b 1) +b k2

2

2 = (b k/)(b+k/),so b=k/. Thus we find

ck(r) =

Ark/ +Brk/ 0< r < r

Crk/ +Drk/ r < r

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Now setting z1 :=ei(), z2 :=ei(+

), noting that|z/1 |=|z/2 |= /

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we arrive at a representation formula for u(x, t), xV, 0t < T,

u(x, t) = T0 V G(x, t; y, )f(y, )dyd+ V G(x, t; y, 0)u0(y)dy

T0

V

G

ng(y, )dS(y)d.

(25)

Note that the above causality condition implies that the solution at time t shouldnot depend on any of its values at later times i.e. we are solving forward in time.

Problem (24) for the Greens function looks a little odd. It is a backwards heatequation, which would be nasty, except that is also solved backwards in time (startingfrom time = t and going down to = 0). So, to straighten it out, it is useful tochange the time variable from to :=t . Problem (24) then becomes

G

DG= (y x)()G= 0 fory VG= 0 for

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2. Free-space Greens function for the heat equation.

We will find the free-space Greens function for the heat equation by using theFourier

transform, so let us first recall the definition and some key properties of it.Definition: Let fbe an integrable function on Rn (that means

Rn

|f(x)|dx 0,

e

a|x|2

2 () =an/2e||2

2a .

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It is property 4 which makes the Fourier transform so useful for differential equations it converts differential equations into algebraic ones.

Property 2 is deep, and difficult to prove. See an analysis textbook for this. Theother properties are easy to show. We will do 4 and 7 two that we will use shortly.

Proof of property 4: integrating by parts,fxj

() = (2)n/2Rn

eixf

xj(x)dx=(2)n/2

Rn

(ij)eixf(x)dx= ijf().

And for the other one,

i

jf() =i

j(2)n/2

Rn

eixf(x)dx= i(2)n/2Rn

(ixj)eixf(x)dx= (xjf(x))().

Proof of property 7: the higher-dimensional cases follow easily from the case n = 1,so well do that one. Completing the square,

ea2|x|2() = (2)1/2

eixea2x

2

dx= e2

2a (2)1/2

ea2 (x+i/a)

2

dx.

The last integral is the integral of the entire complex function f(z) = eaz2/2 along

the contour z= x+i/a, < x

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as needed.

Now lets return to the problem of finding the free-space Greens function for the heat

equation. That is, solving (26) when V = Rn. Let G(x, t; , ) denote the Fouriertransform ofG(x, t; y, ) in the variable y. Property 4 of the Fourier transform showsthat in the variable y corresponds to multiplication by||2. Also, note that

x(y)() = (2)n/2

Rn

eiy(y x)dy = (2)n/2eix,

and so we have to solve

G+D||2G= (2)n/2eix().

For >0, this is G+ D||2G= 0, an ODE which is easily solved to find

G= C eD||2,

whereC=C(x,t,) can be found by a jump condition:

1 =

0+0

()d= (2)n/2eix 0+

0

G+D||2G

= (2)n/2eixG

=0+=0= (2)

n/2eixC

where we used the fact that G is bounded, so that the second term in the integral

contributes nothing, and the causality condition that G = 0 for < 0. Thus wearrive at

G= (2)n/2eixeD||2.

Finally, then, inverting the Fourier transform, and using properties 7 and 6, we findfor >0,

G(x, t; y, ) = (4D)n/2e|yx|2

4D

and so, changing back to =t , our free-space Greens function (also called thefundamental solution) of the heat equation in Rn is

G(x, t; y, ) = (4D(t ))n/2e |yx|24D(t) < t0 > t

.

We obtain a solution formula for the heat equation in Rn, u

t Du= f(x, t), x Rn, t >0

u(x, 0) =u0(x), x Rn , (27)

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by substituting our expression for Ginto (25):

u(x, t) = t0

V(4D(t))n/2e

|yx|2

4D(t) f(y, )dy d+(4Dt)n/2 V

e |yx|

2

4Dt u0(y)dy.

(28)It is not hard to prove the following (though we will not do it here):

Theorem: Suppose f(x, t) is continuously differentiable and bounded on Rn [0, T],and u0(x) is continuous on R

n. Then for 0< t < T, u(x, t) given by formula (28) iscontinuously differentiable in t and twice continuously differentiable in x, and solvesthe heat equation in (27). Furthermore, for any x Rn, limt0 u(x, t) =u0(x).Based on our free-space solution, we can infer a couple of important properties ofdiffusion:

Instantaneous smoothing: in the absence of sources, solutions of the heatequation become instantaneously smooth (even if the initial data is not). Inthe free-space case, this property is reflected in the fact that the fundamentalsolution (4Dt)n/2e|x|

2/4Dt is infinitely differentiable for t >0 (though it is adelta function at t= 0!), and in the second integral of (28) (the one containingthe intial datau0(x)), derivatives inxandtwill fall on the fundamental solution.

Infinite propagation speed: suppose f 0 (no sources), and u0(x) 0 ispositive in a ball of radius 1 about the origin, and vanishes outside a ball of

radius 2. Then for any t >0, and any x Rn

,

u(x, t) = (4Dt)n/2Rn

e|yx|2/(4Dt)u0(x)dx >0.

That is, the solution becomes positive at al lpoints in space instantaneously fort >0 hence the initial, localized disturbance is propagated with infinite speed.

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Just as for the Laplace/Poisson equation, an immediate consequence of the maximumprinciple is the uniquenessof solutions of the initial-boundary-value problem for the

heat equation.

As usual let Vbe a bounded (open, connected) domain in Rn. Fix any T > 0. Letf(x, t), g(x, t), andu0(x) be continuous functions (on V (0, T), V [0, T], andV,respectively), and consider the problem

ut

Du= f(x, t), xV , 0< tTu(x, t) =g(x, t), x V, 0tTu(x, 0) =u0(x), xV

. (29)

Theorem: There is at most one function u(x, t) which is continuous on V[0, T],continuously differentiable (once int, twice inx) inV(0, T], and solves problem (29).

Proof: if there are 2 solutions, their difference w(x, t) satisfies

wt

Dw= 0, xV, t >0w(x, t) = 0, x V, t >0w(x, 0) = 0, xV

.

Applying the (parabolic) maximum principle, we conclude that the max. and min.values ofw on the cylinder are 0. Hence w0.

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4. Methods of images and eigenfunction expansion.

Notation: to make the writing easier, we will often denote derivatives using subscripts,

eg. ut= ut , uxx = 2

ux2 , etc.

Example: (diffusion on the 1/2-line). Solve ut uxx = f(x, t) x >0, t >0u(0, t) = 0 t >0u(x, 0) =u0(x) x >0 (30)by finding the Greens function.

Recalling our notation = t , the problem for the Greens function G(x, t; y, )is: for x >0, t >0, G Gyy =(y x)() y >0, 0G|y=0 = 0G= 0 0,

G(x, t; y, ) :=Gf(x, t; y, ) Gf(x, t; y, ) = 14

e

(yx)2

4 e (y+x)2

4

.

ThenG Gyy =(y x)() (y+x)() =(y x)()

(since x >0 and y >0), and furthermore

G|y=0= 1

4 e y24 e y24 = 0so we are in business! Replacing byt , our Greens function is, for < t(recallit is zero for > t),

G(x, t; y, ) = 14(t )

e

(yx)2

4(t) e (y+x)2

4(t)

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and the resulting solution formula for our half-line problem (30) is

u(x, t) = t0

01

4(t ) e (yx)24(t) e (y+x)24(t) f(y, )dy d+

14t

0

e

(yx)2

4t e (y+x)2

4t

u0(y)dy.

It can be checked rigorously that for reasonable functions fand u0(precisely: boundedand continuous, withfalso continuously differentiable), this formula gives a continu-ously differentiable (once in t twice inx) functionu which solves (30) but we wontdo it here.

Example: (diffusion on a rod). Solve ut uxx = 0 0< x < L, t >0u(0, t) =u(L, t) = 0 t >0u(x, 0) =u0(x) 0xL

(31)

by finding the Greens function.

The method of images will not work so well here, so we try an eigenfunction expansioninstead. The eigenfunctions of the spatial part of the differential operator (d2/dx2)with zero BCs at x = 0 and x = L are sin(nx/L), n = 1, 2, 3, . . ., and so we seekour Greens function in the form

G(x, t; y, ) =

n=1

gn(x, t; )sin(ny/L)

for >0. We require G Gyy =(y x)(), thus

n=1

(gn)+

n22

L2 gn

sin(ny/L) =(y x)()

and so

(gn)+n22

L2 gn =

2

L

L0

sin(ny/L)(y x)()dy = 2L

sin(nx/L)().

For >0, we have (gn)+ (n22/L2)gn= 0, and so

gn = C e(n22/L2)

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and we determine the constant C from a jump condition:

1 = 0+

0 ()d=

L

2 sin(nx/L) 0+

0 (gn)+ (n22/L2)gn d=

L

2 sin(nx/L)gn|=0+=0=

L

2sin(nx/L)C

(where we used G = 0 for < 0). Hence C = 2sin(nx/L)/L, and we have anexpression for our Greens function

G= 2

L

n=1

e(n22/L2) sin(nx/L) sin(ny/L)

(which indeed satisfies the zero boundary conditions at y = 0 and y = L). Thecorresponding formula for the solution of problem (31) is

u(x, t) = 2

L

n=1

e(n22t/L2) sin(nx/L)

L0

sin(ny/L)u0(y)dy.

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5. Greens function for the 1D wave equation.

Suppose we measure the displacement u(x, t) from equilibrium, at each point x in a

bounded regionV in Rn (representing an elastic string (n= 1), membrane (n= 2), orsolid (n= 3)) , and at each time t. The string/membrane/solid is subjected to forcesf(x, t), and at its boundary, it is held at fixed displacement g(x, t) (xV). At timet= 0, at eachxV, the initial displacement is u0(x), and the initial velocity is v0(x).Assuming the displacements are small, the following initial-boundary value problemfor thewave equationis a reasonable description of the displacementu(x, t) at latertimes t >0:

wave equation:

2ut2

c2u= f(x, t), xV, t >0boundary value: u(x, t) =g(x, t), x V, t >0initial conditions: u(x, 0) =u0(x),

ut

(x, 0) =v0(x), x

V

, (32)

wherec >0 is the wave speed. The wave equation also describes the propagation ofother waves, such as sound and light.

Just as for the heat equation, we would like to solve problems like this one via Greensfunctions. We will jump right in, by looking immediately for the free-space Greensfunctions (or fundamental solutions) in dimensions one, two, and three (unlikefor the heat equation, the form of the Greens function depends significantly on thedimension).

1D free-space Greens function for the wave equation:As with the heat equation, the Greens function should be a function of two sets

of space-time variables: x, t; y, . And as for the heat equation, it is sometimesconvenient to work with := t , rather than. Hence, forx, t, y, R, we seekG(x, t; y, ) solving

G c2Gyy =(y x)(), < y

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which is easily verified by integration by parts. Thus if we let G(x, t; y, s) be theLaplace transform ofG in the variable , and useL() = 1, and the causality condi-tion, we find s2G c2Gyy =(y x).Solving this ODE (in y) to the left and right ofx, and imposing the conditions thatG decay as y , and that is be continuous at y= x, yields

G= Aesc|yx|.

As usual, the remaining constant A is determined by the jump condition:

1 = x+

x

(y

x)dy=

x+

x s2G

c2Gyy dx=c

2Gy

|

y=x+y=x= 2scA

henceA= (2sc)1, and

G(x; y, s) = 1

2c

e(|yx|/c)s

s .

Now for anyr0, the Laplace transform of the Heavyside function

H(t) :=

0 t

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Integrating by parts, we have, for any T > t,

u(x, t) = T0

u(y, )(y x)( t)dyd= T

0 u(y, )(G c2Gyy )dyd

=

T0

(u c2uyy)Gdyd+

(uG uG)|=T=0 dy

=

(v0(y)G(x, t; y, 0) u0(y)G(x, t; y, 0))dy

where we used the causality condition GG 0 for > t. Now, since H = ,

G=G =12c

( 1c|y x|) =1

2[(y (x+ct)) +(y (x ct))] ,

and so our solution to (34) is

u(x, t) = 1

2c[u0(x+ct) +u0(x ct)] + 1

2c

x+ctxct

v0(y)dy,

which is known as DAlemberts formula.

Remark:

Notice that DAlemberts formula represents the sum of two waves, one movingto the left with speed c, one moving to the right with speed c:

u(x, t) =f+(x+ct) +f(x ct), f(z) :=12

u0(z) 12c

z0

v0(z)dz.

It is simple to check DAlemberts formula does, in fact, solve problem (34).Indeed, iff is anytwice differentiable function, then f(x ct) solves the waveequation: [

t2

x2]f(x ct) = c2f c2f = 0. It is also easy to check the

initial data are satisfied. Hence, assuming u0 is twice differentiable, and v0 isonce differentiable, we have solved problem (34).

Finite speed of propagation: The solution at space-time point (x, t) dependsonly on the initial data (u0andv0) in the interval [x ct,x + ct] (draw a graph!) that is, signals propagate with speed at most c.

Sidenote: DAlemberts formula makes sense even ifu0 and v0 are not differ-entiable (just continuous) though then the PDE doesnt hold in the classicalsense, only in the sense of distributions.

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6. The wave equation in higher dimensions.

2D free-space Greens function for the wave equation:We look for a Greens function depending on the spatial variable y only throughr:=|y x|, leading to the problem

G c2

Grr+ 1r

Grr

= ()(y x), r >0, >0G0, 0, the ODE

Grr +1

rGrs

2

c2G= 0

is solved by G = AK0((s/c)r), where K0(z) is the modified Bessel function of order0 which solves

z2K0 (z) +zK0(z) z2K0(z) = 0

K0(z)log zas z0K0(z) (const) ezz as z

.

The constant A can be determined using the divergence theorem: denoting the diskof radius about xbyB,

1 = lim0+

B

(y x)dy=c2 lim0+

B

Gdy =c2 lim0+

r=

G

ndS

=c2Asc

lim0+

r=

c

srdS=2c2A

so A=(2c2)1, andG= 1

2c2K0

sc

r

.

It turns out we can invert the Laplace transform on the Bessel function explicitly. In

fact, for b >0, we have

L

1t2 b2 H(t b)

=K0(bs),

and so

G= 1

2c21

2 r2/c2 H( r/c)

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and replacing=t ,r =|y x|, our two-dimensional free-space Greens functionfor the wave equation is

G(x, t; y, ) = 1

2c2

(t )2 1c2

|y x|2H

t 1

c|y x| .

Remark: As with the 1D wave equation, the fact that the Greens function is sup-ported inside the cone{ |y x| c(t )}shows that signals propagate with speedno greater than c. Notice also, as in the 1D case, signals do not propagate sharplyin 2D, since for fixed x, y, , the Greens function is non-zero for times t beyond+ 1

c|y x| (though it does decay like 1/t).

It is left as an exercise to use this expression for the free-space Greens function toshow that the solution of the Cauchy problem

utt=c2u x R2, t >0

u(x, 0) =u0(x), ut(x, 0) =v0(x) ,

for the 2D wave equation, is

u(x, t) = 1

2c2

|yx|ct

v0(y)t2 1

c2|y x|2

dy+

t

|yx|ct

u0(y)t2 1

c2|y x|2

dy

,sometimes known as Poissons formula.

3D free-space Greens function for the wave equation:We again look for a Greens function depending on the spatial variabley only throughr:=|y x|, leading to the problem

G c2

Grr+ 2r

Grr

= ()(y x), r >0, >0G0, 0, the ODE

Grr +2

rGrs

2

c2G= 0

is solved by (weve seen this before in a homework problem)

G= Ae

sc

r

r .

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B. VARIATIONAL METHODS

I. EIGENVALUE PROBLEMSLet D Rn be a bounded domain, and consider the following variable-coefficientgeneralizations of the heat equation

r(x)ut = [p(x)u] q(x)u xD, t >0un

+ bu= 0 x D (35)

and the wave equation r(x)utt= [p(x)u] q(x)u xD, t >0

un

+bu= 0 xD (36)

for (smooth) functionsp(x)> 0 (the spatially variable diffusion rate in the first case,

wave speed in the second), r(x)> 0, and q(x) (a source of heat loss in the first case,friction in the second), and constant b0 (notice the boundary conditions are a mixof Dirichlet and Neumann for the moment).

Suppose for some 0, we seek solutions of (35) of the separated-variables formu(x, t) = et(x), or solutions of (36) of the (oscillatory in time) form u(x, t) =ei

t(x). Then in both cases, as an easy substitution will show, we arrive at thefollowing problem for (x), involving only the variable x:

L(x) := [p(x)] +q(x)= r(x) in Dn

+b= 0 on D . (37)

Problem (37) is an eigenvalue problem for the (differential) operator L. If thereis a non-zero solution (x) for some R, we call an eigenfunction, and thecorrespondingeigenvalue.

We will be studying the eigenvalue problem (37) for the next few lectures.

1. Basic properties of eigenvalues and eigenfunctions

We begin by listing the fundamental properties of the eigenvalues and eigenfunctionsof problem (37). Our basic assumptions: D is a smooth, bounded, connected regionin Rn; p, q, and r are smooth functions on D, with p(x)> 0 and r(x)> 0.

1. The eigenvalues form an infinite sequence, tending to plus infinity:

123. . . , j asj .The multiplicityof an eigenvalue is the dimension of the subspace of eigen-functions corresponding to that is, the maximum number of linearly in-dependent solutions ofL = r. It is always finite. Convention: we willincorporate multiplicity in our ordered list of eigenvalues above by repeatingeach eigenvalue according to its multiplicity.

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2. Eigenfunctions corresponding to different eigenvalues are orthogonal in thesense:

j=k = (j, k)r := D

j(x)k(x)r(x)dx= 0.

This means that by normalizing, we may (and will!) assume that the eigenfunc-tions j corresponding to the eigenvalues j form an orthonormal set:

(j, k)r =

D

j(x)k(x)r(x)dx = jk =

1 j=k0 j=k

(for eigenvalues of multiplicity greater than 1, we can choose an orthonormalset of eigenfunctions using the Gram-Schmidt orthogonalization procedure).

3. The eigenfunctions form acompleteset. This means that any function uwhichis continuous on D can be expanded in terms of the eigenfunctions:

u(x) =

j=1

cjj(x), cj = (j, u)r =

D

j(x)u(x)r(x)dx

in the sense

limNu(x) N

j=1

cjj(x)= 0where denotes the L2 norm, defined by

f2 :=

D

(f(x))2dx.

4. Ifq(x)0, then 10 (and hence j 0 for all j) and 1 = 0 only ifq0,b= 0, and 1 = constant.

Proofs of properties 1 (existence of eigenfunctions) and 3 (completeness of eigenfunc-tions) are beyond the scope of this course, although we will sketch some ideas a bitlater on. Properties 2 and 4 are easy to prove, and we will do that here.

First, notice that L is self-adjoint. That is, ifu and v are smooth functions on Dboth satisfying the boundary conditions: u

n+bu = v

n+bv = 0 on D, then using

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the divergence theorem,

(v, Lu) (Lv, u) = D

[v(x)(Lu)(x) u(x)(Lv)(x)] dx=

D

[v (pu) +qvu+u (pv) quv] dx

=

D

(p(uv vu))dx=

D

p

u

v

n v u

pn

dS

=b

D

[p(uv+uv)] dS= 0.

Proof of property 2, orthogonality of eigenfunctions with different eigenvalues: using

Lj = jrj, Lk = krk, the fact that j and k satisfy the BCs, and the self-adjointness ofL, we have

(j k)(j, k)r = (jrj, k) (j, krk)r = (Lj, k) (j, Lk) = 0.

Hence ifj=k, then (j, k)r = 0. To prove property 4, we first derive a simple, but important identity: ifu is a smoothfunction satisfying the BCs, then

(u, Lu) = D u(x) [ (p(x)u(x)) +q(x)u(x)] dx=

D

(u(x)p(x)u(x)) +p(x)|u(x)|2 +q(x)u2(x) dx=

D

p(x)|u(x)|2 +q(x)u2(x) dx

D

p(x)u(x)u

n(x)dS(x)

=

D

p(x)|u(x)|2 +q(x)u2(x) dx+b

D

p(x)u2(x)dS(x).

(38)

Proof of property 4, positivity of first eigenvalue for positiveq: apply the above iden-tity to the eigenfunction 1 corresponding to the first eigenvalue 1:

1=1(1, 1)r = (1, 1r1) = (1, L1)

=

D

p(x)|1(x)|2 +q(x)21(x)

dx+b

D

21(x)dS(x) 0

since p > 0, b 0, and assuming q 0. And the only way to get 1 = 0 is if 1 constant , q0, and b= 0.

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2. The energy, and variational principles for eigenvalues.

The identity (38) turns out to be a very important and useful one. We will specializefor now to the two most common boundary conditions:

Neumann BCs: b= 0 i.e. un

= 0 on D

Dirichlet BCs: b= i.e. u= 0 on D.In both of these cases, the boundary integral in (38) vanishes. Lets give this quantitya name.

Definition: Given the operatorL(i.e. the functionspand q), the(Dirichlet) energyof a function uon D is

E(u) := D p(x)|u(x)|2 +q(x)u2(x) dx.Now supposeuis any (smooth) function satisfying the boundary conditions (Dirichletor Neumann). By completeness of the eigenfunctions, we may write

u(x) =

j=1

cjj(x),

and notice that

(u, u)r = j=1

cjj,

k=1

ckkr

=

j,k=1

cjck(j, k)r =

j,k=1

cj ckjk =

j=1

c2j .

Now, as computed in (38),

E(u) = (u, Lu) =

j,k=1

cjck(j, Lk) =

j,k=1

cjck(j, krk) =

j,k=1

cjckk(j , k)r

=

j,k=1

cjckkjk =

j=1

kc2k.

And since k1 for all k,E(u)1

j=1

c2k =1(u, u)r,

or

1 E(u)(u, u)r

,

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and we see that the only way to get equality here, is if the only non-zero coefficientscj are those corresponding to the lowest eigenvalue (i.e. j > 1 = cj = 0).That is, equality only holds if and only ifu is an eigenfunction corresponding to thelowest eigenvalue 1. This observation gives us our first example of a variationalprinciple:

Theorem: [variational principle for the first eigenvalue]: the lowest eigenvalueof the operator L (with Dirichlet or Neumann BCs) is given by

1= minu satisfying BCs

E(u)

(u, u)r= min

u satisfying BCs

D

[p(x)|u(x)|2 +q(x)u2(x)] dxD

u2(x)r(x)dx .

Remark: The quantity E(u)(u, u)r

is sometimes called the Rayleigh quotient.

Here is a simple example of how a variational principle like this might be used.

Example: Find an upper bound for the lowest eigenvalue (with r1) ofL= d2dx2

+xon the interval [0, 1] with Dirichlet (zero) BCs at the endpoints.

We have D = [0, 1], r(x) 1, p(x) 1, q(x) = x. According to the variationalprinciple, anyfunctionu(x) on [0, 1] which is zero at the endpoints, gives us an upperbound on 1: 1E(u)/(u, u).

1. One simple such choice is u(x) =x(1 x) =x x2. Then u(x) = 1 2x, andwe compute

(u, u) = 1

0

(x x2)2dx= 1

0

(x2 2x3 + x4)dx= (x3/3 x4/2 + x5/5)|10 = 1

30

and

E(u) =

10

(1 2x)2 +x(x x2)2 dx

=

10

(x5 2x4 +x3) + 4x2 4x+ 1 dx= 1

3+

60

and hence

1 E(u)(u, u)

= 10 +1

2.

2. A little more thought will give us a better (i.e. smaller) upper bound, supposing

is small. For = 0, we know the (Dirichlet) eigenfunctions of d2dx2

are sin(kx),k = 1, 2, . . ., with eigenvalues 2k2. Lets use the lowest = 0 eigenfunction

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u(x) =

2sin(x) (notice we have normalized it so (u, u) = 1) as our testfunction in the variational principle:

1 E(u)(u, u)

= 1

0

(

2 cos(x))2 +x(

2sin(x))2

dx

=2 + 2

10

x sin2(x)dx= 2 +1

2,

which is indeed better (since 2 < 10), and in particular becomes exact in thelimit0.

We can easily generalize the above variational principle to the higher eigenvalues.Suppose u is a function on D satisfying the boundary conditions, and which is or-

thogonal to the first n 1 eigenfunctions:(u, j)r = 0 j = 1, 2, . . . , n 1.

Then c1 = c2 = = cn1= 0, so

u(x) =

j=n

cnj(x), (u, u)r =

j=n

c2j

and

E(u) =

j=n jc2jn

j=n c2jn(u, u)rwith equality if and only ifcj = 0 for all j with j > n i.e., if and only ifu is aneigenfunction with eigenvalue n. Hence:

Theorem: [variational principle for higher eigenvalues]: the n-th eigenvalue ofthe operator L (with Dirichlet or Neumann BCs) is given by

n= minu satisfying BCs,(u,1)r==(u,n1)r=0

E(u)

(u, u)r.

Remark: Our variational principles apply also for the more general boundary condi-

tions un +bu= 0 on D, provided we use the full form of the energy from (38):

E(u) =

D

(p|u|2 +qu2) +b

D

pu2.

In practice, the variational principle above for higher eigenvalues (n > 1) is not souseful, since we typically do not know what the eigenfunctions 1, . . . , n1 are! A

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more useful version is the following, in which we minimize over functions orthogonalto anyset ofn 1 functions (not necessarily the eigenfunctions), and then maximizeover that set.Theorem: [Courant max-min principle]

n= maxfunctions m1,...,mn1

min

u satisfying BC,(u,mj)r=0

E(u)

(u, u)r

Proof: let m1, m2, . . . , mn1 by any functions on D. We will build a function u,satisfying the BCs, and which is orthogonal to all the mk, of the form

u(x) =n

j=1 ajj(x).That is, we want

0 = (u, mk)r =n

j=1

aj (j , mk)r, k= 1, 2, . . . , n 1.

This is n 1 linear equations in the n variables a1, a2, . . . , an. Since there are morevariables than equations, linear algebra tells us there is a non-zero solution (in fact,at least a one-parameter family of solutions). Now as above,

E(u) = (u, Lu) =

nj=1

ja2j nn

j=1

a2j =n(u, u)r,

hence

minu satisfying BC,(u,mj)r=0

E(u)

(u, u)rn.

But since this is true for anychoice ofm1, . . . , mn1, it is true for the maximum overall such choices:

maxfunctions m1,...,mn1

minu satisfying BC,(u,mj)r=0

E(u)

(u, u)r n.

It remains to show that the maximum actually equals n that is, for some choice ofm1, . . . , mn1, and for anyuwhich is orthogonal to all themj ,E(u)/(u, u)rn. Forthis, we take mj =j for j= 1, . . . , n 1. Then any uorthogonal to these functionscan be written

u(x) =

j=n

cjj(x),

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and, as above

E(u) =

j=n jc2j n

j=n c2j =n(u, u)ras required. .

Important remark: It turns out that in the case of Neumann BCs, the variationalprinciples for eigenvalues discussed above (including the max-min principle) remaintrue if the minimization is done over alltest functions u i.e. we do not have toimpose the Neumann BCs on the functions u inserted into the Rayleighquotient. In the case of Dirichlet BCs, however, the test functions u do indeed needto vanish on the boundary. We will return to this point when we discuss the calculusof variations.

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3. Bounds on eigenvalues.

We fix here Dirichlet (zero) BCs, and, as always, denote by n the n-th (Dirichlet)eigenvalue ofL= p(x) +q(x) (with respect to r(x)).Bounds on Coefficients

Suppose 0< pminp(x)pmaxqminq(x)qmax0< rminr(x)rmaxin our domain D. Denote by n,min the n-th eigenvalue of

(pmin) +qmin= n,minrmax in D

= 0 on D ,

and byn,max the n-th eigenvalue of (pmax) +qmax= n,maxrmin inD= 0 on D

.

Theorem:n,minnn,max.

Proof: notice that for any function uon D,

E(u) = D p|u|2 +qu2 dx D pmax|u|2 +qmaxu2 dx=: Emax(u)and

(u, u)r =

D

u2r dx

D

u2rmindx= (u, u)rmin,

so thatE(u)

(u, u)r Emax(u)

(u, u)rmin.

Also notice that for any functions m1, . . . , mn1 on D,

(u, mj)r = 0 if and only if (u, mj)rmin = 0 where mj(x) :=r(x)

rminmj(x).

So by the min-max principle,

n= maxm1,...,mn1

min{u=0D ,(u,mj)r=0}

E(u)

(u, u)r

maxm1,...,mn1

min{u=0 D ,(u,mj)rmin=0}

Emax(u)

(u, u)rmin=n,max.

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The other inequality is really the same thing.

Example: Find upper and lower bounds for the n-th eigenvalue ofL =

d2

dx2+ x2 on

[0, 1] with Dirichlet (zero) BCs. Here 0.So here D= [0, 1], p1, r1, q(x) =x2. Lets use the easy bounds

qmin:= 0q(x) =: qmaxon [0, 1]. Son is sandwiched between the n-th (Dirichlet) eigenvalue of d2dx2 , andthat of d2

dx2+. That is

2n2 n2n2 +.Notice that we get a better (i.e. smaller) upper bound for the first eigenvalue by

taking sin(x) (the first Dirichlet eigenfunction of

d2

dx2 ) as a test function in the

variational principle:

1 E(sin(x))(sin(x), sin(x))

=

10

2 cos2(x) +x2 sin2(x)

dx1

0 sin2(x)dx

=2/2 +/6

1/2 =2+

1

3.

Bounds on geometry

Now lets fix the coefficientsp, q, and r , and make the dependence on the domain Dexplicit; that is, denote byn(D) then-th eigenvalue of our operatorL =p+ qon D with Dirichlet BCs.

Theorem:DD = n(D)n(D).

That is, the smaller the domain, the larger the eigenvalue.

Proof: Again, well use the max-min principle. Let m1, . . . , mn1 be any given func-tions on D. Notice that ifu is a function on D which vanishes on D and satisfies(with integrals restricted to D), (m1, u)r,D = = (mn1, u)r,D = 0 (i.e. u is anadmissible test function for the Rayleigh quotient in the min-max principle for D),then its extension

u(x) := u(x) xD

0 xD\Dis a function onD, vanishing on the boundary D, and satisfying (with integrals nowover D) (m1, u)r,D = = (mn1, u)r,D = 0 (i.e. u is an admissible test functionfor the Rayleigh quotient in the min-max principle for D). Further (denoting regionof integration with a subscript),

ED(u) =ED(u), (u,u)r,D= (u, u)r,D.

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So

min{u=0 D, (mj ,u)r=0}E(u)

(u,u)r min{u0 in D\D, (mj ,u)r=0}E(u)

(u,u)r (minimizing over smaller set)

= min{u=0 D, (mj ,u)r=0}

E(u)

(u, u)r

max{fns. m1,...,mn1 on D}

min{u=0 D, (mj ,u)r=0}

E(u)

(u, u)r=n(D).

Since this inequality holds for all choices ofm1, . . . , mn1 (functions onD), maximiz-ing over all such choices yields n(D)n(D), as required. Example: Find upper and lower bounds for the Dirichlet eigenvalues of

on the

following 2D domain, which is the union of 2 rectangles: D := ( [0, a][0, b] )( [a, a +h] [c, c+g] ) with a,b,c,h,g positive, and 0< c < c+g < b(draw it!).Well, we have

[0, a] [0, b] =:DminDDmax:= [0, a +h] [0, b].By separation of variables, we know the Dirichlet eigenvalues of in a rect-angle [0, a][0, b] are 2(n2/a2 +m2/b2), m, n = 1, 2, 3, . . . (with eigenfunctionssin(nx1/a) sin(mx2/b)). It is a bit annoying, though, to notate them. One way isto denote

k(a, b) :=k th element of the set{n2/a2 +m2/b2 |n, m= 1, 2, 3, . . .}ordered by size, and counting multiplicity (so, eg., 1(1, 1) = 1,2(1, 1) = 5,3(1, 1) =5, 4(1, 1) = 8, etc.). In this notation,n([0, a] [0, b]) =2n(a, b), and so we canconclude for the original problem

2n(a+h, b)n2n(a, b).In particular, for the first eigenvalue,

2 1(a+h)2 + 1b212 1a2 + 1b2 .One more observation, using the geometric bounds established in this section. Asusual, Let D Rn be a bounded domain, and let k(D) denote the k-th Dirichleteigenvalue of the operator L = p(x) +q(x) on D. Letpmin, qmin, and rmaxbe constants such that 0 < pmin p(x), qmin < q(x), and r(x) rmax. Since Dis bounded, it fits inside an n-dimensional (hyper-) cube C of (sufficiently large)

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side-length a. Separation of values shows easily that the Dirichlet eigenvalues (withrespect to rmax) ofLmin:= pmin +qmin on Care

pmin2

rmaxa2(k21+ +k2n) +

qminrmax

, k1, . . . , kn= 1, 2, 3, . . . .

In particular, the eigenvalues go off to infinity: min,k(C) ask . Then oureigenvalue bounds above imply that the Dirichlet eigenvalues k(D) ofL on D satisfy

k(D)min,k(C) , k ,

and so,

limk k(D) =.This was a general property of the eigenvalues we listed at the beginning which oureigenvalue bounds (based, in turn, on variational principles) give a nice slick proof of.

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II. CALCULUS OF VARIATIONS

1. Euler-Lagrange equations.We begin with an example.

Example: Let D Rn be a bounded domain. Let q(x) and g(x) be given smoothfunctions on D and D respectively. Among all functions u(x) on D (which aresufficiently smooth say, twice continuously differentiable) and satisfy u(x) = g(x)for x D, find the one which minimizes the functional

I(u) :=

D

1

2|u(x)|2 +q(x)u(x)

dx.

Well, supposeI(u) is minimized by a functionu (withu=g on D). Consider nowa one-parameter familyof functions nearby u:

u(x) :=u(x) +w(x)

wherew is some fixed function. In order foru to be in our allowed class of functions(twice continuously differentiable and equal to g on the boundary), we require thatw also be twice continuously differentiable, and w= 0 on D. Key observation:

u minimizes I(u) = I(u) is minimized at = 0.Since I(u) is a function of just the single variable , we are in the realm of simple

calculus, and we know