Chapter 5

Fourier Expansions and Separation ofVariables in Rectangular Coordinates

Eigenfunctions come naturally in the treatment of vibrations in a conned space. Electro-magnetic eigenmodes arise in the same way as their mechanical counterpart, such as soundwaves in organ pipes. The basic equation for waves of speed c is

@2

@x2=

1

c2@2

@t2(5.1)

in one dimension. An eigenmode is sinusoidal in time, so that the wave equation becomes

@2

@x2+ k2 = 0; (5.2)

where c jkj is the angular frequency (usually called !). The solutions are plane wavesexp(ikx), with positive and negative k.

5.1 Convergence

If the waves are conned to a ring of length L, the eigenfunctions are simply plane waveswith k taking the special values (eigenvalues)

kn = 2n=L; n integer: (5.3)

The expansion

f(x) =1X

n=1Cn exp(iknx); (5.4)

with

Cn =Z L=2L=2

exp(iknx)f(x)dx=L; (5.5)

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is the basic Fourier expansion. It converges uniformly for any continuous periodic functionof period L. If f(x) is discontinuous or not periodic (i.e. if f(0) 6= f(L)), or worse if it issingular, the convergence is slow, not uniform, and likely to cause problems in numericalevaluations. At a simple jump of f from 0 to 1, for instance, any partial sum S(x) ofthe Fourier series will \overshoot" up to 1.179 and show some wiggles. The overshoot andwiggles occur over a narrower range as more terms are included in S, but never die out|thisis known as the \Gibbs phenomenon." One should be wary when using Fourier expansionsnear jumps and even near cusps. On the other hand for any square-integrable f there isconvergence in the mean, i.e. the mean square error

M =Z L=2L=2

jf(x) S(x)j2 dx (5.6)

can be made arbitrarily small. In practice, this means that most integrals involving f canalso be evaluated with arbitrarily small error.

5.2 Normalization

The eigenfunctions exp(iknx) are mutually orthogonal but are normalized to L:

Z L=2L=2

[exp(ikmx)] exp(iknx) dx =

Z L=2L=2

exp(2i(nm)x=L) dx = Lnm: (5.7)

This is not a problem, it simply entails that we must use dx=L in integrals such as (5.5). Onthe other hand, we may prefer to introduce normalized eigenfunctions

un(x) =1pL

exp(iknx) (5.8)

and obtain, in place of (5.4) and (5.5), the more symmetric pair

f(x) =1X

n=1anun(x); (5.9)

an =Z L=2L=2

un(x)f(x) dx: (5.10)

Often one deals with functions dened over all space, i.e. with the limit a ! 1.The limit can be tricky and sometimes it is better to keep a nite until the very end of thecalculation. But usually one can simply make the replacements kn ! k and

1Xn=1

! L2

Z 11

dk: (5.11)

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The most usual way to write the resulting Fourier-integral transform pair is

f(x) =Z 11

F (k) eikxdk

2; (5.12)

F (k) =Z 11

f(x) eikxdx; (5.13)

where f(k) corresponds to LCn of Eq. (5.5). Mathematicians (and Jackson, at times) havethe habit of using a factor 1=

p2 in front of each integral, instead of leaving the entire 1=2

attached to the k integral.

5.3 Other boundary conditions

In place of the basis functions exp(iknx) one can use cos(knx) and sin(knx) with kn 0.This serves only to produce messier formulas.

On the other hand, there are cases where the functions of interest are required tovanish at the boundaries (L=2; L=2) or (0; L). The relevant normalized eigenfunctions in(0; L) are

Un(x) =

s2

Lsin(qnx); (5.14)

withqn = n=L; n positive integer: (5.15)

The transform pair (5.9), (5.10) is still valid in this case and is in fact entirely generalfor expansions in orthonormal eigenfunctions, with appropriate modications. One obviousmodication is that for xed boundary conditions the sum over n runs from 1 to 1; whilefor periodic boundary conditions it runs from 1 to 1: The comments about convergenceapply to the general case as well.

5.4 Higher dimensions

In 3 dimensions, the wave equation leads to the Helmholtz equation

r2 = k2 ; (5.16)which has plane wave solutions exp(ik x). Periodic boundary conditions on a box of sidesLx; Ly; Lz are not physically attainable in this case, but they are mathematically convenient.They lead to the set of orthogonal functions exp(ik x), with the j Cartesian component ofk given by

(k)j = 2nj=Lj; nj integer: (5.17)

The expansion of a function f(x) is

f(x) =1X

nx=1

1Xny=1

1Xnz=1

Cnx;ny ;nz exp(ik x); (5.18)

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with

Cnx;ny ;nz =Z Lx=2Lx=2

dx

Lx

Z Ly=2Ly=2

dy

Ly

Z Lz=2Lz=2

dz

Lzexp(ik x)f(x): (5.19)

In the continuum limit we obtain the Fourier integral representation

f(x) =ZF (k) eikx

d3k

(2)3; (5.20)

F (k) =Zf(x) eikxd3x; (5.21)

by the replacements1X

nx=1

1Xny=1

1Xnz=1

! V(2)3

Zd3k; (5.22)

V Cn ! f(k); (5.23)where V = LxLyLz is the volume of the \quantization box" and the integrals are over allspace. Similar formulae apply in two dimensions.

5.5 The Dirac delta function

This is a particularly important singular function. In one dimension, it has the generalexpression

(x x0) = Xn

un(x0)un(x) (5.24)

for any complete set of eigenfunctions Un. For the plane waves exp(iknx) this becomes

(x x0) = 1L

1Xn=1

exp [ikn(x x0)] ; (5.25)

and in the continuum limit

(x x0) =Zeik(xx

0) dk

2: (5.26)

The three-dimensional generalization is the usual (x) = (x)(y)(z), so that, for instance

(x x0) =Zeik(xx

0) d3k

(2)3: (5.27)

It is worth noting that the partial sum SN of (5.25) has a central peak that becomes higherand narrower for larger N , but also a lot of side wiggles that never go away:

SN(x) =1

a

ei2Nx=L ei2Nx=L1 ei2x=L =

sin [(2N + 1)x=L]

L sin(x=L): (5.28)

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5.6 Application to electrostatics

How do we use all this math in electrostatics, where in the simplest case we need solutionsof Laplaces equation, corresponding to k = 0 in the Helmholtz equation, with appropriateboundary conditions?

5.6.1 Rectangular box

Suppose the potential is assigned on the sides of an empty conducting rectangular box ofsides Lx; Ly; Lz. It is enough to consider the case where is not zero on one face only:the general case can be built by superposition. Suppose then that = V (x; y) on the facez = Lz, and vanishes on the other faces. We Fourier-expand in the dimensions x and y;cleverly picking eigenfunctions that vanish on the lateral faces:

(x; y; z) =Xnxny

sin(nxx=Lx) sin (nyy=Ly)Anxny(z): (5.29)

The expansion coecients are now functions of z, to be determined by using Laplacesequation and the remaining boundary conditions on the top and bottom faces.

From r2 = 0 for all x and y it follows that24nx

Lx

2 ny

Ly

!2+

@2

@z2

35Anxny(z) = 0; (5.30)

which has the solutions exp(nxnyz), with

nxny =

vuutnxLx

2+

ny

Ly

!2: (5.31)

To satisfy = 0 on the z = 0 face, we pick the combination

Anxny(z) = Anxny sinh(nxnyz): (5.32)

To satisfy the condition on the z = Lz face, which isXnxny

sin(nxx=Lx) sin (nyy=Ly)Anxny sinh(nxnyLz) = V (x; y); (5.33)

we use Fouriers theorem and nd

Anxny =4

LxLy sinh(nxnyLz)

Z Lx0

dxZ Ly0

dy V (x; y) sin(nxx=Lx) sin (nyy=Ly) :

(5.34)

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The complete solution was easily obtained because:

{ The starting equation (Laplaces equation in this case) is completely separablein the chosen variables x; y; z. That means that there are solutions of the formX(x)Y (y)Z(z). In our case the dierential operator is the sum of three separatepieces that depend only on x; y and z respectively. This is the simplest case ofseparability.

{ The boundary conditions are also separable in the same variables x; y; z.

Even partial separability may help. For instance, we may be able to expand in onedimension only, thus reducing the problem to a two-dimensional one.

We are still left with an innite double sum that must usually be evaluated numerically,with all the caveats about the behavior at discontinuities. It is worthwhile to look forapproximations that permit the series to be evaluated analytically near singular points.

5.6.2 Rectangular trough

This is a two-dimensional problem: nothing depends on z. The boundary conditions arethat vanishes at x = 0 and x = L, as well as at y = 1, and it has an assigned value V (x)at y = 0 (Fig. 2.10). The solution is

(x; y) =1X

n=1

An sin(nx=L) exp(ny=L); (5.35)

with

An =2

L

Z L0V (x) sin(nx=L) dx: (5.36)

For instance, if V (x) is constant, we have

An =4V

nfor odd n; (5.37)

and 0 otherwise. In simple cases like this the series can be summed. As in most two-dimensional problems, it is convenient to work with the complex variable

z = x+ iy (5.38)

instead of x and y. In our case we note that

=4V

Im

Xn odd

1

nexp(inz=a): (5.39)

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We put Z = eiz=L and recall that

ln(1 + Z) =1X

n=1

1

n(1)n+1Zn;

ln(1 Z) =1X

n=1

1

nZn; (5.40)

so that by subtracting,

ln1 + Z

1 Z = 2X

n odd

1

nZn: (5.41)

Then

=2V

Im

ln

1 + Z

1 Z

=2V

arctan

2 ImZ

1 jZj2!: (5.42)

With ImZ = ey=L sin(x=L) and jZj2 = e2y=L; we obtain nally

=2V

arctan

"sin(x=L)

sinh(y=L)

#: (5.43)

Plots of the equipotentials and eld lines are shown in the gure.

x=0 x=L

F=0 F=0

F= V

In the corner near the origin, V (1 2=); where = arctan(y=x) as usual.The eld is azimuthal, E = 2V=: The energy density diverges as 1=

2 and the totalenergy for > 0 also diverges as 0 ! 0; although only logarithmically. More generally, itcosts innite energy to join two plates at any angle when there is a potential dierence Vbetween them. In practice, a small gap must be left and the energy of the adjoining plates

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goes like (V=)2 lnw per unit length of the common edge, where w the width of the gap.This logarithmic divergence also occurs for adjoining edges that are not straight.

Jackson has many problems and examples of adjoining plates and shells at dierentpotentials. Such problems are not very realistic, for the reason discussed above.

5.6.3 Polar coordinates

Some problems in electrostatics are best solved in polar coordinates (; ). Laplaces equationin polar coordinates is

r2 = 1

@

@

@

@

!+

1

2@2

@2= 0: (5.44)

Lets assume that the solution is separable in these coordinates, so that (; ) = R()Q().Substituting this into Laplaces equation, and multiplying by 2=RQ, we obtain

R

d

d

dR

d

!+

1

Q

d2Q

d2= 0: (5.45)

The rst term is only a function of and the second term is only a function of , so theymust both be equal to a constant which we will call 2. Therefore,

1

Q

d2Q

d2= 2; (5.46)

with the solutionQ() = A cos() +B sin(): (5.47)

The radial function R() is a solution of

R

d

d

dR

d

! 2 = 0: (5.48)

To solve this, seek solutions of the form R() = a. Substituting into Eq. (5.6.3), we ndsolutions for = . We must treat = 0 separately; in this case Q() = A0 + B0, andR() = a0 + b0 ln(=R), with R some scale factor. Therefore, our general solution is

(; ) = [a0 + b0 ln(=R)](A0 +B0) + (a + b

)(A cos +B sin ): (5.49)

This result is quite general. Suppose that we now permit the full range of , so that (; +2) = (; ); in this case B0 = 0, and = n, with n a non-negative integer:

(; ) = a0 + b0 ln(=R) +1X

n=1

an

n + bnn (An cosn+Bn sinn) : (5.50)

As an example, lets consider a grounded conducting cylinder placed in a uniformelectric eld perpendicular to its axis. For !1, to produce a uniform eld we must have(; ) = E0 cos; the solution is then of the form

(; ) = a0 E0 cos+1X

n=1

n (An cosn+Bn sinn) : (5.51)

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On the cylinder we have ( = a; ) = 0, so that a0 = 0 and A1 = E0a2, with all of the

other Ans and Bns zero. The solution is therefore

(; ) = E0 a

2

!cos; (5.52)

which we also obtained using the method of images.

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