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Laplace Equation on Spheres: Solutions and Legendre Polynomials, Study notes of Differential Equations and Transforms

These notes explore the laplace equation in spherical coordinates and its solutions invariant under rotations. The document derives the laplacian in spherical coordinates and uses the inner product of gradients to prove the result. The text then discusses finding harmonic functions on a sphere using separation of variables, leading to the legendre equation and the legendre polynomials as solutions.

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2021/2022

Uploaded on 09/27/2022

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Notes on the Laplace equation for spheres
§1. Laplacian in spherical coordinates
Let (r, φ, θ) be the spherical coordinates, related to the Cartesian coordinates by
x=rsin φcos θ, y =rsin φsin θ, z =rcos φ .
In polar coordinates, the Laplacian = 2
∂x2+2
∂y2+2
∂z2becomes
u=1
r2
∂r r2u
∂r +1
r2sin φ
∂φ sin φu
∂φ +1
r2sin2φ
2u
∂θ2(1)
Although it is possible to verify (1) directly using the chain rule, we give a simpler proof
using the formula
u· f=∂u
∂r
∂f
∂r +1
r2
∂u
∂φ
∂f
∂φ +1
r2sin2φ
∂u
∂θ
∂f
∂θ (2)
for the inner product of the gradients of two functions, which is quite easy to check. The
proof of (1) uses (2), the change of variable formula, and integration by parts, without much
calculation: For each smooth function f(x, y, z ) which vanishes outside of a bounded region
in R3, we have
RRR fu r2sin φ dr
=RRR fu dx dy dz change of variables
=RRR u· f dx dy dz integration by parts
=RRR ∂u
∂r
∂f
∂r +1
r2
∂u
∂φ
∂f
∂φ +1
r2sin2φ
∂u
∂θ
∂f
∂θ r2sin φ dr by (2)
=RRR fh
∂r r2∂u
∂r sin φ+
∂φ sin φ∂u
∂φ +2u
∂θ2idr integration by parts
=RRR fh1
r2
∂r r2∂u
∂r +1
r2sin φ
∂φ sin φ∂u
∂φ +1
r2sin2φ
2u
∂θ2ir2sin φ dr dr
Since the above equality holds for every smooth function fvanishing outside some bounded
region, (1) follows.
§2. Solutions invariant under rotations
How does one find a function u(r, φ θ) defined on a sphere 0 rRof radius Rsuch that
u= 0, u(R, φ, θ) = f(φ) (3)
where f(φ) is a (given) function on the boundary of the sphere r=R, expressed in spherical
coordinates, which is symmetric about the z-axis? The general procedure of separation of
variables says that one should first find harmonic functions of the form F(r)G(φ); these
“eigenfunctions” will be parametrized by some discrete parameters.
1
pf3

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Notes on the Laplace equation for spheres

§1. Laplacian in spherical coordinates

Let (r, φ, θ) be the spherical coordinates, related to the Cartesian coordinates by

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

In polar coordinates, the Laplacian ∆ = ∂

2 ∂x^2 +^

∂^2 ∂y^2 +^

∂^2 ∂z^2 becomes

∆ u =

r^2

∂r

r^2

∂u ∂r

r^2 sin φ

∂φ

sin φ

∂u ∂φ

r^2 sin^2 φ

∂^2 u ∂θ^2

Although it is possible to verify (1) directly using the chain rule, we give a simpler proof using the formula

∇u · ∇f =

∂u ∂r

∂f ∂r

r^2

∂u ∂φ

∂f ∂φ

r^2 sin^2 φ

∂u ∂θ

∂f ∂θ

for the inner product of the gradients of two functions, which is quite easy to check. The proof of (1) uses (2), the change of variable formula, and integration by parts, without much calculation: For each smooth function f (x, y, z) which vanishes outside of a bounded region in R^3 , we have ∫∫∫ f ∆u r^2 sin φ dr dφ dθ =

f ∆u dx dy dz change of variables = −

∇u · ∇f dx dy dz integration by parts = −

∫∫∫ (∂u ∂r

∂f ∂r +^

1 r^2

∂u ∂φ

∂f ∂φ +^

1 r^2 sin^2 φ

∂u ∂θ

∂f ∂θ

r^2 sin φ dr dφ dθ by (2)

f

[

∂ ∂r

r^2 ∂u∂r

sin φ + (^) ∂φ∂

sin φ (^) ∂φ∂u

  • ∂∂θ^2 u 2

]

dr dφ dθ integration by parts

f

[

1 r^2

∂ ∂r

r^2 ∂u∂r

  • (^) r (^2) sin^1 φ∂φ∂

sin φ ∂u∂φ

  • (^) r (^2) sin^12 φ∂

(^2) u ∂θ^2

]

r^2 sin φ dr dr dφ dθ

Since the above equality holds for every smooth function f vanishing outside some bounded region, (1) follows.

§2. Solutions invariant under rotations

How does one find a function u(r, φ θ) defined on a sphere 0 ≤ r ≤ R of radius R such that

∆u = 0, u(R, φ, θ) = f (φ) (3)

where f (φ) is a (given) function on the boundary of the sphere r = R, expressed in spherical coordinates, which is symmetric about the z-axis? The general procedure of separation of variables says that one should first find harmonic functions of the form F (r) G(φ); these “eigenfunctions” will be parametrized by some discrete parameters.

Separation of variables leads to the system of ordinary equations

d dr

r^2

dF dr

− k F (r) = 0, (4)

and 1 sin φ

d dφ

sin φ

dG dφ

  • k G(φ) = 0 (5)

for some k ∈ R. The equation (4) has solutions F (r) = rn, with n ∈ R≥ 0 and k = n(n + 1).

The equation (5) will look a lot better if we use the variable w = cos φ, with 0 ≤ w ≤ 1. Let H(w) be the function in w corresponding to G(φ), so that H(cos φ) = G(φ), then the equation (5) becomes the standard Legendre equation with parameter n:

d dw

(1 − w^2 )

dH dw

  • n(n + 1)H(w) = 0 (6)

Although one does not see any restriction on k at this stage other than n ≥ 0, it turns out that n has to be an integer, otherwise F (r)G(φ) cannot be a well-defined harmonic function on the sphere. This follows from a careful analysis of solutions of (6).

§3. Legendre polynomials

With n being a non-negative integer, the solutions of (6) that behave well at w = ±1 are all multiples of the Legendre polynomial Pn(w), defined by

Pn(w) =

2 n^ n!

dn dwn^

[(w^2 − 1)n] n ∈ Z≥ 0. (7)

Besides being the solution of the Legendre differential equation (6), the Legendre polynomials satisfy the orthogonality relations: ∫ (^1)

− 1

Pm(w)Pn(w) dw =

2 m + 1

δm,n (8)

where δm,n is equal to 0 if m 6 = n, and δm,n = 1 if m = n. To check (8), one uses integration by parts repeatedly. Another way to define the Legendre polynomials is a succinct expression of its generating function:

(1 − wt + t^2 )−^

(^12)

∑^ ∞

n=

tn^ Pn(w). (9)

We have P 0 (w) = 1, P 1 (w) = w, P 2 (w) = 3 w

(^2) − 1 2 ,^ P^3 (w) =^

1 2 (5w

(^3) − 3 w),etc. In case you like to

have a good reference for the special functions you encountered in this course, I recommend A Course of Modern Analysis by Whittaker a Watson, an absolute classic.