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 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

= 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
= 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)
dx
+ λ sin θΘ = 0.
Equivalently, in terms of x, we have
d
dx
(1 x
2
)
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.
(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
n2
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
n2
(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)
`k1
= (x
2
1)
`k1
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
(
xx
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
(
xr
)
Q
m
r=1
(
xx
r
)
2
. The first
few terms have roots outside (
1
,
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.
In our original equation
2
φ
= 0, we now set Θ(
θ
) =
P
`
(
cos θ
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).