4Partial differential equations

IB Methods

4.6 Laplace’s equation in cylindrical coordinates

We let

Ω = {(r, θ, z) ∈ R

3

: r ≤ a, z ≥ 0}.

In cylindrical polars, we have

∇

2

φ =

1

r

∂

∂r

r

∂φ

∂r

+

1

r

2

∂

2

φ

∂θ

2

+

∂

2

φ

∂z

2

= 0.

We look for a solution that is regular inside Ω and obeys the boundary conditions

φ(a, θ, z) = 0

φ(r, θ, 0) = f(r, θ)

lim

z→∞

φ(r, θ, z) = 0,

where f is some fixed function.

Again, we start by separation of variables. We try

φ(r, θ, z) = R(r)Θ(θ)Z(z).

Then we get

1

rR

(rR

0

)

0

+

1

r

2

Θ

00

Θ

+

Z

00

Z

= 0.

So we immediately know that

Z

00

= µZ

We now replace Z

00

/Z by µ, and multiply by r

2

to obtain

r

R

(rR

0

)

0

+

Θ

00

Θ

+ µr

2

= 0.

So we know that

Θ

00

= −λΘ.

Then we are left with

r

2

R

00

+ rR

0

+ (µr

2

− λ)R = 0.

Since we require periodicity in θ, we have

Θ(θ) = a

n

sin nθ + b

n

cos nθ, λ = n

2

, n ∈ N.

Since we want the solution to decay as z → ∞, we have

Z(z) = c

µ

e

−

√

µz

The radial equation is of Sturm-Liouville type since it can be written as

d

dr

r

dR

dr

−

n

2

r

R = −µrR.

Here we have

p(r) = r, q(r) = −

n

2

r

, w(r) = r.

Introducing x = r

√

µ, we can rewrite this as

x

2

d

2

R

dx

2

+ x

dR

dx

+ (x

2

− n

2

)R = 0.

This is Bessel’s equation. Note that this is actually a whole family of differential

equations, one for each

n

. The

n

is not the eigenvalue we are trying to figure out

in this equation. It is already fixed by the Θ equation. The actual eigenvalue we

are finding out is µ, not n.

Since Bessel’s equations are second-order, there are two independent solution

for each n, namely J

n

(x) and Y

n

(x). These are called Bessel functions of order

n

of the first (

J

n

) or second (

Y

n

) kind. We will not study these functions’

properties, but just state some useful properties of them.

The

J

n

’s are all regular at the origin. In particular, as

x →

0

J

n

(

x

)

∼ x

n

.

These look like decaying sine waves, but the zeroes are not regularly spaced.

On the other hand, the

Y

n

’s are similar but singular at the origin. As

x →

0,

Y

0

(x) ∼ ln x, while Y

n

(x) ∼ x

−n

.

Now we can write our separable solution as

φ(r, θ, z) = (a

n

sin nθ + b

n

cos nθ)e

−

√

µz

[c

µ,n

J

n

(r

√

µ) + d

µ,n

Y

n

(r,

√

µ)].

Since we want regularity at

r

= 0, we don’t want to have the

Y

n

terms. So

d

µ,n

= 0.

We now impose our first boundary condition

φ

(

a, θ, z

) = 0. This demands

J

n

(a

√

µ) = 0. So we must have

√

µ =

k

ni

a

,

where

k

ni

is the

i

th root of

J

n

(

x

) (since there is not much pattern to these roots,

this is the best description we can give!).

So our general solution obeying the homogeneous conditions is

φ(r, θ, z) =

∞

X

n=0

X

i∈roots

(A

ni

sin nθ + B

ni

cos nθ) exp

−

k

ni

a

z

J

n

k

ni

r

a

.

We finally impose the inhomogeneous boundary condition

φ

(

r, θ,

0) =

f

(

r, θ

). To

do this, we use the orthogonality condition

Z

a

0

J

n

k

nj

r

a

J

n

k

ni

r

a

r dr =

a

2

2

δ

ij

[J

0

n

(k

ni

)]

2

,

which is proved in the example sheets. Note that we have a weight function

r

here. Also, remember this is the orthogonality relation for different roots of

Bessel’s functions of the same order. It does not relate Bessel’s functions of

different orders.

We now integrate our general solution with respect to cos mθ to obtain

1

π

Z

π

−π

f(r, θ) cos mθ dθ =

X

i∈roots

J

m

k

mi

r

a

.

Now we just have Bessel’s function for a single order

j

. So we can use the

orthogonality relation for the Bessel’s functions to obtain

B

mj

=

1

[J

0

m

(k

mj

)]

2

πa

2

Z

a

0

Z

π

−π

cos mθJ

m

k

mj

r

a

f(r, θ)r dr dθ.

How can we possibly perform this integral? We don’t know how

J

m

looks like,

what the roots

k

mj

are, and we multiply all these complicated things together

and integrate! Often, we just don’t. We ask our computers to do this numerically

for us. There are indeed some rare cases where we are indeed able to get an

explicit solution, but these are rare.