4Partial differential equations

IB Methods

4.4 Laplace’s equation in spherical polar coordinates

4.4.1 Laplace’s equation in spherical polar coordinates

We’ll often also be interested in taking

Ω = {(s, y, z) ∈ R

3

:

p

x

2

+ y

2

+ z

2

≤ a}.

Since our domain is spherical, it makes sense to use some coordinate system

with spherical symmetry. We use spherical coordinates (r, θ, φ), where

x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ.

The Laplacian is

∇

2

=

1

r

2

∂

∂r

r

2

∂

∂r

+

1

r

2

sin θ

∂

∂θ

sin θ

∂

∂θ

+

1

r

2

sin

2

θ

∂

2

∂φ

2

.

Similarly, the volume element is

dV = dx dy dz = r

2

sin θ dr dθ dφ.

In this course, we’ll only look at axisymmetric solutions, where

ψ

(

r, θ, φ

) =

ψ

(

r, θ

)

does not depend on φ.

We perform separation of variables, where we look for any set of solution of

∇

2

ψ = 0 inside Ω where ψ(r, θ) = R(r)Θ(θ).

Laplace’s equation then becomes

ψ

r

2

1

R

d

dr

(r

2

R

0

) +

1

Θ sin θ

d

dθ

sin θ

dΘ

dθ

= 0.

Similarly, since each term depends only on one variable, each must separately

be constant. So we have

d

dr

r

2

dR

dr

= λR,

d

dθ

sin θ

dΘ

dθ

= −λ sin θΘ.

Note that both these equations are eigenfunction equations for some Sturm-

Liouville operator and the Θ(

θ

) equation has a non-trivial weight function

w(θ) = sin θ.

4.4.2 Legendre Polynomials

The angular equation is known as the Legendre’s equation. It’s standard to set

x

=

cos θ

(which is not the Cartesian coordinate, but just a new variable). Then

we have

d

dθ

= −sin θ

d

dx

.

So the Legendre’s equation becomes

−sin θ

d

dx

sin θ(−sin θ)

dΘ

dx

+ λ sin θΘ = 0.

Equivalently, in terms of x, we have

d

dx

(1 − x

2

)

dΘ

dx

= −λΘ.

Note that since 0 ≤ θ ≤ π, the domain of x is −1 ≤ x ≤ 1.

This operator is a Sturm–Liouville operator with

p(x) = 1 − x

2

, q(x) = 0.

For the Sturm–Liouville operator to be self-adjoint, we had

(g, Lf) = (Lg, f) + [p(x)(g

∗

f

0

− g

0∗

f)]

1

−1

.

We want the boundary term to vanish. Since

p

(

x

) = 1

− x

2

vanishes at our

boundary

x

=

±

1, the Sturm-Liouville operator is self-adjoint provided our

function Θ(

x

) is regular at

x

=

±

1. Hence we look for a set of Legendre’s

equations inside (

−

1

,

1) that remains regular including at

x

=

±

1. We can try a

power series

Θ(x) =

∞

X

n=0

a

n

x

n

.

The Legendre’s equation becomes

(1 − x

2

)

∞

X

n=0

a

n

n(n − 1)x

n−2

− 2

∞

X

n=0

a

n

nx

n

+ λ

∞

X

n=0

a

n

x

n

= 0.

For this to hold for all

x ∈

(

−

1

,

1), the equation must hold for each coefficient

of x separately. So

a

n

(λ − n(n + 1)) + a

n−2

(n + 2)(n + 1) = 0.

This requires that

a

n+2

=

n(n + 1) − λ

(n + 2)(n + 1)

a

n

.

This equation relates

a

n+2

to

a

n

. So we can choose

a

0

and

a

1

freely. So we get

two linearly independents solutions Θ

0

(x) and Θ

1

(x), where they satisfy

Θ

0

(−x) = Θ

0

(x), Θ

1

(−x) = −Θ

1

(x).

In particular, we can expand our recurrence formula to obtain

Θ

0

(x) = a

0

1 −

λ

2

x

2

−

(6 − λ)λ

4!

x

4

+ ···

Θ

1

(x) = a

1

x +

(2 − λ)

3!

x

2

+

(12 − λ)(2 − λ)

5!

x

5

+ ···

.

We now consider the boundary conditions. We know that Θ(

x

) must be regular

(i.e. finite) at x = ±1. This seems innocent, However, we have

lim

n→∞

a

n+2

a

n

= 1.

This is fine if we are inside (

−

1

,

1), since the power series will still converge.

However, at

±

1, the ratio test is not conclusive and does not guarantee conver-

gence. In fact, more sophisticated convergence tests show that the infinite series

would diverge on the boundary! To avoid this problem, we need to choose

λ

such

that the series truncates. That is, if we set

λ

=

`

(

`

+ 1) for some

` ∈ N

, then

our power series will truncate. Then since we are left with a finite polynomial,

it of course converges.

In this case, the finiteness boundary condition fixes our possible values of

eigenvalues. This is how quantization occurs in quantum mechanics. In this case,

this process is why angular momentum is quantized.

The resulting polynomial solutions

P

`

(

x

) are called Legendre polynomials.

For example, we have

P

0

(x) = 1

P

1

(x) = x

P

2

(x) =

1

2

(3x

2

− 1)

P

3

(x) =

1

2

(5x

3

− 3x),

where the overall normalization is chosen to fix P

`

(1) = 1.

It turns out that

P

`

(x) =

1

2

`

`!

d

`

dx

`

(x

2

− 1)

`

.

The constants in front are just to ensure normalization. We now check that this

indeed gives the desired normalization:

P

`

(1) =

1

2

`

d

`

dx

`

[(x − 1)

`

(x + 1)

`

]

x=1

=

1

2

`

`!

`!(x + 1)

`

+ (x − 1)(stuff)

x=1

= 1

We have previously shown that the eigenfunctions of Sturm-Liouville operators

are orthogonal. Let’s see that directly now for this operator. The weight function

is w(x) = 1.

We first prove a short lemma: for 0 ≤ k ≤ `, we have

d

k

dx

k

(x

2

− 1)

`

= Q

`,k

(x

2

− 1)

`−k

for some degree k polynomial Q

`,k

(x).

We show this by induction. This is trivially true when k = 0. We have

d

k+1

dx

k+1

(x

2

+ 1)

`

=

d

dx

Q

`,k

(x)(x

2

− 1)

`−k

= Q

0

`,k

(x

2

− 1)

`−k

+ 2(` − k)Q

`,k

x(x

2

− 1)

`−k−1

= (x

2

− 1)

`−k−1

h

Q

0

`,k

(x

2

− 1) − 2(` − k)Q

`,k

x

i

.

Since

Q

`,k

has degree

k

,

Q

0

`,k

has degree

k −

1. So the right bunch of stuff has

degree k + 1.

Now we can show orthogonality. We have

(P

`

, P

`

0

) =

Z

1

−1

P

`

(x)P

`

0

(x) dx

=

1

2

`

`!

Z

1

−1

d

`

dx

`

(x

2

− 1)

`

P

`

0

(x) dx

=

1

2

`

`!

d

`−1

dx

`−1

(x

2

− 1)

2

P

`

0

(x)

−

1

2

`

`!

Z

1

−1

d

`−1

dx

`−1

(x

2

− 1)

`

dP

`

0

dx

dx

= −

1

2

`

`!

Z

1

−1

d

`−1

dx

`−1

(x

2

− 1)

`

dP

`

0

dx

dx.

Note that the first term disappears since the (

` −

1)

th

derivative of (

x

2

−

1)

`

still

has a factor of

x

2

−

1. So integration by parts allows us to transfer the derivative

from x

2

− 1 to P

`

0

.

Now if

` 6

=

`

0

, we can wlog assume

`

0

< `

. Then we can integrate by parts

`

0

times until we get the `

0th

derivative of P

`

0

, which is zero.

In fact, we can show that

(P

`

, P

`

0

) =

2δ

``

0

2` + 1

.

hence the P

`

(x) are orthogonal polynomials.

By the fundamental theorem of algebra, we know that

P

`

(

x

) has

`

roots. In

fact, they are always real, and lie in (−1, 1).

Suppose not. Suppose only

m < `

roots lie in (

−

1

,

1). Then let

Q

m

(

x

) =

Q

m

r=1

(

x−x

r

), where

{x

1

, x

2

, ··· , x

m

}

are these

m

roots. Consider the polynomial

P

`

(

x

)

Q

m

(

x

). If we factorize this, we get

Q

`

r=m+1

(

x−r

)

Q

m

r=1

(

x−x

r

)

2

. The first

few terms have roots outside (

−

1

,

1), and hence do not change sign in (

−

1

,

1).

The latter terms are always non-negative. Hence for some appropriate sign, we

have

±

Z

1

−1

P

`

(x)Q

m

(x) dx > 0.

However, we can expand

Q

m

(x) =

m

X

r=1

q

r

P

r

(x),

but (P

`

, P

r

) = 0 for all r < `. This is a contradiction.

4.4.3 Solution to radial part

In our original equation

∇

2

φ

= 0, we now set Θ(

θ

) =

P

`

(

cos θ

). The radial

equation becomes

(r

2

R

0

)

0

= `(` + 1)R.

Trying a solution of the form

R

(

r

) =

r

α

, we find that

α

(

α

+ 1) =

`

(

`

+ 1). So

the solution is α = ` or −(` + 1). Hence

φ(r, θ) =

X

a

`

r

`

+

b

`

r

`+1

P

`

(cos θ)

is the general solution to the Laplace equation

∇

2

φ

= 0 on the domain. In we

want regularity at r = 0, then we need b

`

= 0 for all `.

The a

`

are fixed if we require φ(a, θ) = f(θ) on ∂Ω, i.e. when r = 1.

We expand

f(θ) =

∞

X

`=0

F

`

P

`

(cos θ),

where

F

`

= (P

`

, f) =

Z

π

0

P

`

(cos θ)f(θ) d(cos θ).

So

φ(r, θ) =

∞

X

`=0

F

`

r

a

α

P

`

(cos θ).