Ockendon J., Howison S., Lacey A., Movchan A. Applied Partial Differential Equations

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178

ELLIPTIC EQUATIONS

By the Cauchy-Kowalevski theorem, there is a unique local solution to this Cauchy

problem if 9D and 9N are the transforms of arbitrary analytic functions, but the

solution only exists globally in y > 0 and has the required properties at infinity if

9D and 9N are related by (5.64). To interpret this result, one procedure is to note

that 9D is the convolution of ON and the Fourier inverse of -1/Iki. Unfortunately,

the latter is difficult to invert although, if we are daring, we can show that ik/IkI is

the Fourier transform of 1/ax (but we must use the arguments of §4.2.2 and then

wait for Exercise 5.21 to see this). This suggests that dgD/dx, whose transform is

-ikgD, is the convolution of gN and 1/7rx, which tempts us to write

d9D

_I /00 9N(f) dC

7r J 0 -x

(5.65)

providing we can make sense of the integral. A precisely similar argument for (5.64),

rewritten as

yields

9N(k) = ik

(kI)

9D(k),

00 d9D

9N (x)

a£x

(5.66)

but again we need to interpret the integral. Fortunately, we can revert to (5.60)

and integrate by parts to write it as

uD(x, y)

_ - J

tan-

£ - x

dgp

7r Y

for y > 0, where we must define the inverse tangent as lying between -ir/2 and

it/2. As y y 0,

tan-' (%

=2-£yx+o(y)

for

- x > 0 and

- z < 0, respectively. Hence, in order to find the limiting be-

haviour of UD (z, y) as y 10, we need to split the integral into lim,0

00 2+t

+ J 0O

)

so that, when y is small,

uD(x+y)=Em(- f

2 - E

dt dt

00 /

0" ( -2+y+... d£ dal

[ /

=9D(x)+EPVr

d9D

--x+...,

where

d9D

lim

X -f

d9D dt

(5.67)

/-00 d1

- x

!-+0 (,

dt t - z

EXPLICIT REPRESENTATIONS OF GREENS FUNCTIONS 179

is called a principal value integral. We can now differentiate with respect to y and

set y = 0 to find

F00

9N(x) =

1 PV

d9°

d ,

dt f - x

(5.68)

which tells us how to interpret the integrals above. The right-hand side of (5.68)

is called the Hilbert transform of dgD/d{.

5.6.1.3

Strips and rectangles

is a simple exercise to repeat the transform calculations above in the strip

0<y<1,-oo<x<oo,say, with

u(x,

y) = FX 9(f)H( - x, y) df,

(5.69)

ere now

e_ik= sinh ky

dk.

H(x, y) =

27t

sinh k

The function H is again the y derivative of the Green's function evaluated at rj = 1

and, by writing it as an infinite series of contributions from poles at k = in7r,

n = 0,±I,..., we can interpret the Green's function in terms of an infinitely

repeated series of images in y = 0 and y = 1.

It is sometimes possible to solve problems in semi-infinite strips by using a

different kind of image. Suppose that u is harmonic in the half-strip 0 < y < co,

0<x<1,andthat

u(x,0) = g(x)

for 0 < x < 1,

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

for y > 0,

together with boundedness conditions at (0, 0) and (1, 0). We simply extend the

problem periodically to -oo < x < oo with u odd in x; then (5.60) gives

u(x, y) _

f g(0

(

- 2n -C)2 + y2 - (x - (2n + 1) - {)2 +

y2) ,

=-x

(5.70)

which is much quicker than using Green's functions. We could also have proceeded

by separating the variables, to obtain a series which converges well for large y, in

contrast to (5.70), which works well for small y.

To those who find the prospect of similarly constructing series of images in

a rectangle daunting, the following technique for finding Green's functions will

come as a relief; in fact it works in any bounded domain for which we know the

eigenfunctions of the Laplacian. Suppose we want to solve the Dirichlet problem

for Laplace's equation in -a < x < a, -b < y < b, with, of course, appropriate

boundedness conditions at each vertex. As usual,

VG = 5(x - 06(y - r,),

(5.71)

180 ELLIPTIC EQUATIONS

with G = 0 on the boundary.73 What we now do is define the finite Fourier

transform of G, which is a generalisation of the Fourier series we discussed in §4.4.

There we expanded arbitrary functions in terms of complete sets of eigenfunctions

of the appropriate eigenvalue problems; here for an arbitrary function F(x, y) we

simply define

f °

IL = f f F(x, y) sin

minax

sin

6Wiry

dx dy, (5.72)

b o

where the kernel is just an eigenfunction of the Laplacian with zero Dirichlet

conditions, and corresponding eigenvalue -(m27r2/a2+n27r2/b2). Multiplying both

sides of (5.71) by this kernel and integrating gives

2 2

2 2l

Gmn = - sin

sin n6 / (

The inversion formula for (5.72) in this case is exactly that for Fourier series,

namely

G = 1 m7rx n7ry

mn sin

a sin b (5.73)

m.n

Of course, (5.72) can be generalised and used to obtain G in a general domain

provided that we know the eigenfunctions of the Laplacian in that domain. We

return to this question in §5.7.2.

5.6.2 Helmholtz' equation

Because of its importance in diffraction theory, we will accord Helmltoltz' equation

a special section. We will restrict ourselves to two dimensions for simplicity, and

we see that the Green's function for the whole plane satisfies

(02 + k2) G = b(x - t;).

(5.74)

By writing this in polar coordinates, we see that, instead of G being a combination

of log r and a constant as it was for Laplace's equation, it is now a combination of

the Bessel functions Jo(kr) and Yo(kr). Clearly, we need G to have a logarithmic

singularity with strength 1/27r as r = Ix - t j -+ 0, which, from known properties of

}o, demands that G =

However, we need to specify

the behaviour as r -+ oo more carefully before we can proceed. In particular, we

need to prescribe how much of the solution represents an `outgoing' wave as r - oo.

To do this we recall from Chapter 4 that waves in which u - f (r + aot) as r -+ 00

are incoming and those in which u - f (r - aot) are outgoing.74 Writing

731n fact, the problem for G is a good model for certain industrial processes involving current

flow in a conducting plate to which an electrode is attached at x

y = q; G is simply the

electric potential.

74The functions f (r ± aet) are not, of course, exact solutions of the wave equation except

in one space dimension, but as r -4 oo we expect arbitrary waves to become more and more

one-dimensional.

EXPLICIT REPRESENTATIONS OF GREEN'S FUNCTIONS 181

eik(rfaat)

and remembering the factor e-i7t in our derivation of (5.74), we obtain the moti-

vation for a radiation condition for Helmholtz' equation in the form

- ikG = o(G)

as r -+ oo, which is just a concise way of writing that G tends to a+ikr multiplied

by some function which may decay algebraically in r, rather than a-'kr multiplied

by such a function. In fact, it can be shown by using asymptotic methods (see

Exercise 8.11) that this radiation condition can be written more precisely as the

Sommerfeld radiation condition

lim r1'2 (G

- ikG f = 0.

r-oc

\\\

(5.75)

A quick glance at the asymptotic expansions of Jo and Yo [8] shows that the only

combination that satisfies (5.75) is a multiple of the Hankel function

Hou)(kr) = Jo(kr) +i}o(kr).

(5.76)

Thus

G=-4Ho1)(kIx -tl),

(5.77)

which, as expected from §5.1.5, is complex; it is still symmetric in x and t because

+ k2 is self-adjoint.

It would be sadistic to inflict on the reader a list of Green's functions for

Helmholtz' equation in the presence of boundaries, even though there are precious

few of them; one example is that of a rectangle with zero Dirichlet data, where we

can use a simple modification of (5.73), as long as -k2 is not an eigenvalue of the

Laplacian (see Exercise 5.16). However, we can point out that, although the pres-

ence of finite boundaries in an otherwise infinite domain requires that G should,

as usual, satisfy appropriate homogeneous boundary conditions on them, the ra-

diation condition (5.75) remains unaltered.75 We simply cite what can happen if

the geometry is simple enough. Suppose, for example, that

5r2 +r9r+r j02

+2u=0

7511 can be shown that the presence of finite boundaries merely introduces a 'directivity'

into the far-field so that

G-

A(el,...,em-i)eikr

r(m-t)/2 '

where m is the dimensionality and Os,...,O,,,_1 are polar angles (see Chapter 8); this makes

an interesting contrast with the far-field of the Green's function for Laplace's equation, which is

isotropic in general.

182 ELLIPTIC EQUATIONS

in polar coordinates, with u = f (0) on r = 1. Then the solution in r > 1 that

satisfies (5.75) is

u(r, 0) _ t E (an cos n0 + bn sin n8) H,(,) (kr),00

n=0

and the solution in r < 1 is

u(r, 0) =

(en cos n0 + do sin nB) Jn (kr).

n=O

Here an, bn, cn and do are easily related to the Fourier coefficients of f (0). Also

J,(x), the Bessel function of the first kind of order n, satisfies

n +

(i_ TZ )in

= 0,

Jn(x) = 2nn (1 + O(x2))

as x -a 0,

while Y, (x), the Bessel function of the second kind, is equal to

J,,(x) cos nir - J_(x)

sin nir

(5.78)

for non-integral n, and to the limit of this expression for integral n; it is singular at

the origin (for more details see [8]); the combination

J. +iYn is known as a

Hankel function of order n. For future reference, we note that J,,(x) is proportional

to x" fo cos(x cos 0) sin'" 0 dB; this result is true even for n complex.

Lastly, we note that the situation becomes more complicated when the bound-

aries extend to infinity because the condition for outgoing waves may have to be

modified.

5.6.3

The modified Helmholtz equation

The modified Helmholtz equation,

(V2

- k2)u = 0,

has quite different properties from the ordinary Helmholtz equation, because in

an unbounded domain its solutions either grow or decay exponentially at infinity,

rather than algebraically. Its Green's function in R2 is the solution of

d2G

1 dG

dr2 +rdr

-kG=0

that decays at infinity and has the correct logarithmic behaviour at the origin,

namely G = -(1/2r)Ko(kr), where Ko(kr) is a modified Bessel function of the

third kind, equal to (br/2)H0("(ikr). It is left to the reader to show that the

Green's function in R3 is -e kr/47rr; note that neither of these Green's functions

is oscillatory.

TRANSFORMS 183

* 5.7

Green's functions, eigenfunction expansions and transforms

Results such as the series (5.73) suggest that we look more closely at the relation-

ships between Green's functions, eigenfunctions and transforms. They motivate

thinking of the Green's function G(x, t) for a self-adjoint operator C as a super-

position of eigenfunctions Oa(x) that satisfy Co,, = Aoa, and then regarding our

basic formula (5.52) as a transform formula in the sense of §4.4. Put another way,

the solution of Cu = f emerges as a superposition of 4a whose coefficients are easy

to find because

Oaf dx =

¢a Cu dx =

uLoa dx = A

u¢,\ dx.

Each elliptic problem can thus be solved by its customised transform, taken with

respect to its eigenfunctions 0a, and this procedure can be modified to cater for

non-self-adjoint operators. However, before we can describe the method we need

to review a few important results about eigenfunctions and eigenvalues of elliptic

operators.

5.7.1 Eigenvalues and eigenfunctions

Most of what we need to know about this large subject is a generalisation of

the Sturm-Liouville theory for two-point boundary value problems for self-adjoint

second-order ordinary differential equations [44]. In that theory, a principal result

is that the eigenvalues are real and the eigenfunctions are complete and can be

orthonormalised, those corresponding to different eigenvalues being orthogonal

automatically. The orthogonality results are proved in the same way as for matrices

but the completeness is more complicated. One relatively painless way to generalise

this procedure to a Dirichlet problem in more dimensions is to note that, if we

define the bilinear form a(u, v) of two suitable functions u and v by

u dxa(u, v) =

uLv dx = vL

for a self-adjoint elliptic operator C defined in a domain D, then the Rayleigh

quotient

a(u, u)

dx/

J (uI2 (Ix, where

u = 0 on OD,

is minimised" by the eigenfunction Oo corresponding to the lowest (smallest in

modulus) or principal eigenvalue of

Cu=Au, u=0 on OD.

Moreover, the (n + 1)th eigenfunction 4n can be found by carrying out the same

minimisation over test functions u that are orthogonal to their n predecessors.

761'his is an easy exercise in the calculus of variations (see Exercise 5.15).

184 ELLIPTIC EQUATIONS

Completeness can then be proved by using this minimisation property to show

that the remainder after n terms in a Fourier series type of expansion is bounded

by the inverse of the nth eigenvalue, in the mean-square sense (see [48]). This

procedure also works for Neumann and Robin problems, assuming the correct sign

in the latter boundary condition.

Another result of great practical importance is that the lowest eigenfunction of

the Dirichlet eigenvalue problem has one sign. For second-order ordinary differen-

tial equations this is proved relatively easily by considering the way in which the

zeros of a function u(x) such that Cu = Au, u(O) = 0, `bunch up' near x = 0 as \

increases. However, this idea does not work for elliptic equations and, instead, we

set ourselves the task of showing that the zeros (or nodal lines) of the (n + 1)th

eigenfunction 0n divide D into no more than n + 1 subdomains. As explained in

detail in [12, 21], we assume the contrary and suppose there are m > n + 1 such

subdomains. Then, picking any n + 1 of these and denoting by ¢ni the restriction

of the eigenfunction to the ith of these n + 1 domains (i.e. ¢,, is 0n in the ith

subdomain and zero elsewhere), we can construct a test function

u =

ai0ni

i-0

that is normalised and also orthogonal to the preceding eigenfunctions

, ... ,

4'n_1. For this test function, the Rayleigh quotient becomes

at f OniLOni dx/

a?0ni dx = \n,

i=0

since Lq5ni = AnOni. This does not prove rigorously, but it does strongly suggest,

that u is the (n + 1)th eigenfunction; but we know it cannot be because u has

singularities on n nodal lines. We hope this contradiction will give the reader

enough confidence henceforth to assert that the principal eigenfunction has one

sign.

5.7.2 Green's functions and transforms

With this background, let us now embark on our discussion of Green's functions

and transforms. The series (5.73) for the Green's function of a rectangle suggests

the following result. Suppose the spectrum A. of a self-adjoint elliptic operator C

is discrete,77 that zero is not an eigenvalue,78 and that the real orthonormalised

eigenfunctions are 0n(x); then

G(x,

On (x,)0 (4).

(5.79)

This result can be derived by writing G as En cn(f)On(x) and observing that

LG = En cn(5)An4n(x) can only equal 6(x - r:) if c,n(t)Am =

we simply

"This usually means D is bounded, 8D is not too irregular and the operator has no unbounded

coefficients when written in canonical form.

78When zero is an eigenvalue, we must employ the devices used to introduce generalised Green's

functions in §5.5.

TRANSFORMS 185

multiply each side by 0,,,(x) and integrate. Moreover, either by using (5.79) with

A = 0 assumed not to be an eigenvalue, or by directly taking a finite Fourier

transform, the solution of Cu = f is

u(F,) =

(Lf )4n(x) dx f

)

(5.80)

/ An

This formula is fundamental to our discussion in the next section, but before we

put it into practice we need to make some remarks about normalisation and how

to cater for complex eigenfunctions.

We assume we have a self-adjoint operator G and boundary or boundedness

conditions such that the normalised eigenfunctions7 0a satisfying G¢.% = AOa are

complete (when it is convenient, we will use 0a to denote what we called O(x, k)

in (4.26), where A and k are related in a known way). This means that an arbi-

trary function f (x) can be expanded, either as a series Ea fa¢a(x), or an integral

f 1.\O.\ (x) dA, or a combination of both, depending on whether the spectrum is

discrete, continuous, or both, respectively. In any case, if the eigenfunctions are

orthonormal, we expect that

fA _ / dx; (5.81)

this is just the familiar Fourier series result in the discrete case and usually the

continuous spectrum result can be derived by taking a limit such as L -> 00

when the Fourier series is on the interval (-L, L). However, we must be careful

about what we mean by normalisation when we take this limit and, to illustrate

this important point, let us return to our Fourier transform discussion in §4.4.

Suppose we take C = d2/dx2, 0 < x < oo, and we seek eigenfunctions that vanish

at x = 0 and are bounded as x -+ oo. The only possibility is that A is a negative

number, in which case 0a is proportional to sin kx, k =

but what is the

normalisation? To answer this question, we specialise the Fourier transform and

its inverse to odd functions to show that

u(x) =

/ (f° u() sinkdI sin kx dk.

Now, this double integral is, formally,

2 00 00

u({) sinkCsinkxdkd

o 0

and so we can make the interpretation

sin k{ sin kx dk = 2 b(x - ). (5.82)

T91Ve denote the spectrum by the single scalar A even though a is parametrised by m numbers

when there are m independent variables x.

186 ELLIPTIC EQUATIONS

Thus the normalised eigenfunctions are s(x, k) = 217r sin kx and we have what

is called, to within a multiplicative constant, the Fourier sine transform. If we

define

/e(k) =

jf(x)5(xik)dx,

the inversion formula is

f (x) = f M fa (k)4(x, k)

dk.

In practice, most people work with 7r/218(k) and retain the factor 2/a in the

inversion; the same remark about normalisation applies to the Fourier transform.

The lesson to be learnt from this discussion is that, if we define fA by (5.81),

we should always be able to use the inversion formula

f (x) = f

fa0a dA,

(5.83)

as long as we are prepared to use results like (5.82) to define the normalisation.

We must also be careful in defining the range of integration, which is very problem

dependent and leads to a sum over a discrete set of A when the spectrum is discrete.

In such a case, f, is a superposition of delta functions and (5.83) is a generalised

Fourier series.

We finally remind readers of the symmetry of (5.81) and (5.83) compared to the

Fourier transform formulae i(k) = f f (x)e'kx dx and f (x) = (1/27x) f f (k)e`'ks dk.

This reflects the fact that (5.81) depends on 0,\ being real. Since we use the complex

eigenfunctions e'k; for the Fourier transform, all inner products for such problems

need to be defined for Hermitian operators, as explained in §4.4. Thus we need to

define

(u, v) = f u(x)v(x)dx, (5.84)

which means that, in (5.81), the Fourier coefficients are with respect to the con-

jugate eigenfunctions. When the eigenfunctions are complex, the crucial result on

which the theory of transforms depends, and of which (5.82) is a special case, is

f ¢a(x)Q.%(x') dA = d(x - x'),

(5.85)

where the integration is taken over whatever range is used in (5.81).

5.8

Transform solutions of elliptic problems

As happens all too often with very general theories, we soon encounter hideous

technical complications when we try to apply the ideas of §5.7 to specific elliptic

problems, even in two dimensions. This is simply because of the difficulty of iden-

tifying and characterising the eigenvalues and eigenfumctions of partial differential

TRANSFORM SOLUTIONS OF ELLIPTIC PROBLEMS 187

operators in any useful way. Fortunately, there is a widely occurring class of prob-

lems where we can decompose these eigenfunctions into those of ordinary differen-

tial operators, and we will henceforth restrict our attention to this class. They are

characterised by the property that the variables are separable, so that the partial

differential operator can be manipulated into the form C(x, y) = C1(x) + £2(y),

where Cl and G2 only involve coefficients and differentials in x and y, respec-

tively. For simplicity, we will also restrict attention to cases where the original

differential equation is homogeneous, i.e. has zero right-hand side; when the right-

hand side is non-zero it is possible to proceed by expanding it in terms of the

eigenfunctions of C, but this often leads to considerable technical difficulties. As-

suming separability and homogeneity, the representation of u as a double series

u = E cmnomn(x, y), where Co.. = Amncmn, simplifies to u = E caOa(x),ia(y),

where C j Oa (x) = A O.% (x) and C2+ba (v) = -\

(y). For homogeneous problems in

which separability is not possible, all that we can say is that the solution can be

represented as a series in $mn, but an explicit formula for Omn is rarely available.

Let us now turn to some examples.

5.8.1

Laplace's equation with cylindrical symmetry: Hankel transforms

Suppose we wish to solve Laplace's equation in cylindrical polars,

02U 1 9U I 02U 02U

r2 + r 8r + T2_

a_2 2 + 8z2

0 (5.86)

U 8

in z > 0 with u(r, 8, 0) = g(r), so that the solution is independent of 6, and with

whatever decay rate is needed at infinity to ensure uniqueness. We recall that, to

find the solution (5.60) of the half-plane problem in two dimensions, it was easy to

take a Fourier transform in x and solve the resulting ordinary differential equations

in y. This was because the eigenfunctions of d2/dx2 are so easy to find. When we

separate the variables here we find

ar2 +

r = AmA.

Hence we have no problem with the z derivatives in (5.86), which lead to eigen-

functions e'`4, but to take a transform in r we need to consider the spectrum

of d2/dr2 + (1/r) d/dr, which is not a self-adjoint operator. Hence, as in §4.5.2,

for the function 0,\(r) that describes the dependence of u on r, we consider not

the eigenvalue problem

d +

r ddr = Ana

but, instead, the self-adjoint problem

(5 87).

(rdd ! ) = Aro

The factor r on the right-hand side introduces a slight complication into the ar-

gument in §5.7, namely that a weight function has to be introduced into the inner