4Partial differential equations

IB Methods

4.3 Separation of variables

Unfortunately, the use of complex variables is very special to the case where

Ω ⊆ R

2

. In higher dimensions, we proceed differently.

We let Ω =

{

(

x, y, z

)

∈ R

3

: 0

≤ x ≤ a,

0

≤ y ≤ b, z ≥

0

}

. We want the

following boundary conditions:

ψ(0, y, z) = ψ(a, y, z) = 0

ψ(x, 0, z) = ψ(x, b, z) = 0

lim

z→∞

ψ(x, y, z) = 0

ψ(x, y, 0) = f(x, y),

where

f

is a given function. In other words, we want our function to be

f

when

z = 0 and vanish at the boundaries.

The first step is to look for a solution of ∇

2

ψ(x, y, z) = 0 of the form

ψ(x, y, z) = X(x)Y (y)Z(z).

Then we have

0 = ∇

2

ψ = Y ZX

00

+ XZY

00

+ XY Z

00

= XY Z

X

00

X

+

Y

00

Y

+

Z

00

Z

.

As long as ψ 6= 0, we can divide through by ψ and obtain

X

00

X

+

Y

00

Y

+

Z

00

Z

= 0.

The key point is each term depends on only one of the variables (

x, y, z

). If we

vary, say,

x

, then

Y

00

Y

+

Z

00

Z

does not change. For the total sum to be 0,

X

00

X

must

be constant. So each term has to be separately constant. We can thus write

X

00

= −λX, Y

00

= −µY, Z

00

= (λ + µ)Z.

The signs before λ and µ are there just to make our life easier later on.

We can solve these to obtain

X = a sin

√

λx + b cos

√

λx,

Y = c sin

√

λy + d cos

√

λx,

Z = g exp(

p

λ + µz) + h exp(−

p

λ + µz).

We now impose the homogeneous boundary conditions, i.e. the conditions that

ψ vanishes at the walls and at infinity.

– At x = 0, we need ψ(0, y, z) = 0. So b = 0.

– At x = a, we need ψ(a, y, z) = 0. So λ =

nπ

a

2

.

– At y = 0, we need ψ(x, 0, z) = 0. So d = 0.

– At y = b, we need ψ(x, b, z) = 0. So µ =

mπ

b

2

.

– As z → ∞, ψ(x, y, z) → 0. So g = 0.

Here we have n, m ∈ N.

So we found a solution

ψ(x, y, z) = A

n,m

sin

nπx

a

sin

mπy

b

exp(−s

n,m

z),

where A

n,m

is an arbitrary constant, and

s

2

n,m

=

n

2

a

2

+

m

2

b

2

π

2

.

This obeys the homogeneous boundary conditions for any

n, m ∈ N

but not the

inhomogeneous condition at z = 0.

By linearity, the general solution obeying the homogeneous boundary condi-

tions is

ψ(x, y, z) =

∞

X

n,m=1

A

n,m

sin

nπx

a

sin

mπy

b

exp(−s

n,m

z)

The final step is to impose the inhomogeneous boundary condition

ψ

(

x, y,

0) =

f(x, y). In other words, we need

∞

X

n,m=1

A

n,m

sin

nπx

a

sin

mπy

b

= f(x, y). (†)

The objective is thus to find the A

n,m

coefficients.

We can use the orthogonality relation we’ve previously had:

Z

a

0

sin

kπx

a

sin

nπx

a

dx = δ

k,n

a

2

.

So we multiply (†) by sin

kπx

a

and integrate:

∞

X

n,m=1

A

n,m

Z

a

0

sin

kπx

a

sin

nπx

a

dx sin

mπy

b

=

Z

a

0

sin

kπx

a

f(x, y) dx.

Using the orthogonality relation, we have

a

2

∞

X

m=1

A

k,m

sin

mπy

b

=

Z

a

0

sin

kπx

a

f(x, y) dx.

We can perform the same trick again, and obtain

ab

4

A

k,j

=

Z

[0,a]×[0,b]

sin

kπx

a

sin

jπy

b

f(x, y) dx dy. (∗)

So we found the solution

ψ(x, y, z) =

∞

X

n,m=1

A

n,m

sin

nπx

a

sin

mπy

b

exp(−s

n,m

z)

where

A

m,n

is given by (

∗

). Since we have shown that there is a unique solution

to Laplace’s equation obeying Dirichlet boundary conditions, we’re done.

Note that if we imposed a boundary condition at finite

z

, say 0

≤ z ≤ c

,

then both sets of exponential

exp

(

±

√

λ + µz

) would have contributed. Similarly,

if

ψ

does not vanish at the other boundaries, then the

cos

terms would also

contribute.

In general, to actually find our

ψ

, we have to do the horrible integral for

A

m,n

, and this is not always easy (since integrals are hard). We will just look at

a simple example.

Example. Suppose f(x, y) = 1. Then

A

m,n

=

4

ab

Z

a

0

sin

nπx

a

dx

Z

b

0

sin

mπy

b

dy =

(

16

π

2

mn

n, m both odd

0 otherwise

Hence we have

ψ(x, y, z) =

16

π

2

X

n,m odd

1

nm

sin

nπx

a

sin

mπy

b

exp(−s

m,n

z).