H: Laplace’s Equation in Spherical Coordinates
Consider
r 2 ∇2 u =
∂
∂r
r2 ∂u
+
∂r
1
∂
sin(φ) ∂φ
∂u
sin(φ) ∂φ
+
1
∂2u
sin2 (φ) ∂θ2
=0
(1)
for r < 1, 0 ≤ φ ≤ π, 0 ≤ θ < 2π
u(1, θ, φ) = g(θ, φ) for 0 ≤ φ ≤ π, 0 ≤ θ < 2π
You can view this as a steady-state heat conduction problem with specified heat distribution maintained on the outer boundary of the unit sphere.
Since the equation seems a little intimidating, let us only consider some special cases.
Case 1: g(θ, φ) = γ = constant
Then, since
depend on θ and φ, neither does u, so (1) becomes
g does not
du
d
2 du
r dr = 0 ⇒ dr = ra2 ⇒ u = − ar + b = b = γ,
dr
where we have invoked the boundedness condition at r = 0 by setting a = 0.
Remark: Constants and constant/r are the only solutions (“potentials”) that
depend only on the radial distance from the origin; u = 1/r is called the
Newton potential in physics.
Case 2: g = g(φ) only
Then (1) becomes

∂

r2 ∂u
+
 ∂r
∂r


1
∂
sin(φ) ∂φ
∂u
sin(φ) ∂φ
= 0 r < 1, 0 ≤ φ ≤ π
0≤φ≤π
u(1, φ) = g(φ)
Let u(r, φ) = R(r)Φ(φ). Then
1 d
1
d
dΦ
2 dR
r
=−
sin(φ)
=λ,
R dr
dr
Φ sin(φ) dφ
dφ
which gives
d
dr
d2 R
dR
2 dR
r
− λR = r2 2 + 2r
− λR = 0
dr
dr
dr
1
(2)
again giving us a Cauchy-Euler type equation to solve on 0 ≤ r < 1. Also,
dΦ
d
sin(φ)
+ λ sin(φ)Φ = 0 for 0 ≤ φ ≤ π
(3)
dφ
dφ
2
For (2), R = rα gives
√ characteristic equation α + α − λ = 0 with solutions 2α = −1 ± 1 + 4λ. For (3) we put it in a more standard form
d
d
by letting Φ(φ) = P (x), where x = cos(φ). Hence, dφ
= − sin(φ) dx
, so
2
dP
d
sin(φ) dx sin (φ) dx + λ sin(φ)P = 0, or
d
2 dP
(1 − x )
+ λP = 0 for −1 < x < 1 .
(4)
dx
dx
This is Legendre’s equation. It is also a singular Sturm-Liouville equation.
It can be shown through a study of series solutions that the only bounded
solutions are when λ = n(n+1), with n = 0, 1, 2, . . ., and in fact the solutions
to
d
2 dP
(1 − x )
+ n(n + 1)P = 0
(5)
dx
dx
are polynomials Pn (x) called the Legendre polynomials. A scaling has
been adopted that has become convention so that
P0 (x) ≡ 1 , P1 (x) = x .
0.1
(6)
Some Properties of Legendre Polynomials
1. Pure recurrence relation: The Pn0 s are related by
nPn (x) = (2n − 1)xPn−1 (x) − (n − 1)Pn−2 (x) n ≥ 2
So, given (6), we have, for example,
3
1
5
3
35
15
3
P2 (x) = x2 − , P3 (x) = x3 − x, P4 (x) = x4 − x2 + , (7)
2
2
2
2
8
4
8
63 5 35 3 15
P5 (x) = x − x + x ,
8
4
8
1
231x6 − 315x4 + 104x2 − 5 ,
P6 (x) =
16
etc.
2
n
d
2
n
2. Rodrigue’s formula: Pn (x) = 2n1n! dx
n [(x − 1) ].
This can be used to obtain various properties, including the important
orthogonality relation, that is
3. Orthogonality relation:
Z 1
0
Pn (x)Pm (x)dx =
n 6= m
n=m
2
2m+1
−1
In spherical coordinates this gives us
Z π
0
Pn (cos φ)Pm (cos φ) sin φ dφ =
2
2m+1
0
n 6= m
n=m
4. Note that Pn (x) is an odd function of x if n = odd, and an even function
if n is even: Pn (−x) = (−1)n Pn (x).
0.2
Solution to Laplace’s Equation in the Ball
Now, if λ = n(n + 1), from the characteristic equation for the R equation,
α2 + α − λ = 0, roots are α = n, −(n + 1), so R(r) = arn + br−(n+1) .
For boundedness of R at r = 0, set b = 0. Thus, we have the modes
un (r, φ) = rn Pn (cos φ). Therefore,
u(r, φ) =
∞
X
An rn Pn (cos φ) .
n=0
For the boundary condition, using (6),(7),
∞
X
A2
A3
2
3
g(φ) =
An Pn (cos φ) = A0 + A1 x +
(3x − 1) +
(5x − 3x) + . . . ,
2
2
n=0
where x = cos φ. Hence, integrating both sides and using the orthogonality
relation, we have
Z π
Z π
∞
X
g(φ)Pm (cos φ) sin φ dφ =
An
Pn (cos φ)Pm (cos φ) sin φ dφ
0
0
n=0
=
∞
X
n=0
Z
1
An
Pn (x)Pm (x)dx =
−1
3
2Am
2m + 1
which implies
2m + 1
Am =
2
Example: Consider
Z
π
g(φ)Pm (cos φ) sin φ dφ for m ≥ 0 .
0
∇2 u = 0
r < 1, 0 ≤ φ ≤ π
u(1, φ) = cos(3φ)
Thus, in the unit ball, u does not depend on θ (so on latitude lines u =
constant, the constant only depending on the value of φ). Note that
cos(3φ) = cos((2 + 1)φ) = cos(2φ) cos(φ) − sin(2φ) sin(φ)
= (2 cos2 (φ) − 1) cos(φ) − 2 sin2 (φ) cos(φ)
= 2 cos3 (φ) − cos(φ) − 2(cos(φ) − cos3 (φ))
= 4 cos3 (φ) − 3 cos(φ) = 4x3 − 3x .
But P3 (x) = 25 x3 − 32 x, so 4x3 − 3x = 58 25 x3 − 32 x − 35 x, so that cos(3φ) =
8
P (cos φ) − 53 P1 (cos φ) = g(φ). Therefore,
5 3
u(r, φ) =
∞
X
An rn Pn (cos φ)
n=0
= A0 + A1 rP1 (cos φ) + A2 r2 P2 (cos φ) + A3 r3 P3 (cos φ)
3
8
= − rP1 (cos φ) + r3 P3 (cos φ) .
5
5
Exercise: Write the solution u(r, φ) to
2
∇ u=0
r < 1, 0 ≤ φ ≤ π
u(1, φ) = cos(4φ)
in an expansion in Legrendre polynomials.
4