Problem 5

In spherical coordinates (φ latititude, θ azimuth).

∇2u =1r2 ∂r

(r2∂ru

)+

1r2 sin φ

∂φ

(sin φ∂φu

)+

1r2 sin2 φ

∂2θu = 0.

inside of the region B.Apply seperation of variables as usual; u = F(r)G(θ)H(φ)

∇2u = GH1r2

ddr

(r2 dF

dr

)+ FG

1r2 sin φ

ddφ

(sinφ

dHdφ

)+ FH

1r2 sin2 φ

d2Gdθ2 = 0

0 =1

r2Fddr

(r2 dF

dr

)+

1r2 sin φH

ddφ

(sin φ

dHdφ

)+

1r2 sin2 φG

d2Gdθ2 .

First we seperate the equation in θ ,

−r2 sin2 φ

1

r2Fddr

(r2 dF

dr

)+

1r2 sin φH

ddφ

(sin φ

dHdφ

)= −k2 =

1G

d2Gdθ2

and conclude that Gk (θ) = Ake±ikθ where k ∈ Z.Subtituting−k2 into the original equation,

0 =1

r2Fddr

(r2 dF

dr

)+

1r2 sin φH

ddφ

(sin φ

dHdφ

)− k2

r2 sin2 φ.

Multiplying through by r2 and rearranging,

1F(r)

ddr

(r2 dF

dr

)=

k2

sin2 φ− 1

sin φ

1H(φ)

ddφ

(sin φ

dHdφ

).

We choose another seperation constant Ω to convert the above equation into two ODEs.

ddr

(r2 dF

dr

)−ΩF(r) = 0 (i)

ddφ

(sin φ

dHdφ

)+

λk2

sin φH(θφ)−Ω sin φH(φ) = 0 (ii)

Equation (i) looks like an ODE for spherical Bessel functions at first glance, but is infact the so-called Eulerdifferential equation if we let Ω = l(l + 1). To solve, we apply the Frobenius method. Make the ansatz,

F(r) =∞

∑n=0

anrn

with α fixed.

ddr

(r2 dF

dr

)= r2 d2F

dr2 + 2rdFdr

and we compute the derivatives

d2Fdr2 =

∞

∑n=0

n(n− 1)anrn−2

1

dFdr

=∞

∑n=0

nanrn−1.

Plugging into the ODE,

∞

∑n=0

n(n− 1)anrn + 2∞

∑n=0

nanrn − l(l + 1)∞

∑n=0

anrn = 0

=∞

∑n=0

an n(n− 1) + 2n− l(l + 1) rn.

=∞

∑n=0

an

n2 + n− l(l + 1)

rn

=∞

∑n=0

an n(n + 1)− l(l + 1) rn

Now we make the observation that each term in this series is unique. As such, for such a series to terminateto zero either an = 0 ∀n∈ Z+ (which leads to obvious trivialities) OR n(n + 1)− l(l + 1) = 0.n(n + 1) = l(l + 1) and n = −(l + 1), l.If n 6= −(l + 1), l then the only possible way to rectify the situation is to let an = 0, hence all an forn 6= l, −(l + 1) must be zero. Only two terms survive,

Fl(r) = Alrl + Bl1

rl+1 ,

where l will be summed over in our final solution.Equation (ii) is the associated Legendre DE which has solutions Pk

l (cos φ) for k = 0, ..., l. I think solving(ii) is another application of the Frobenius method, I omit the derivation to save time; everyone knows thisODE anyways.Finally, our general solution is

u(r, θ, φ) =∞

∑l=0

l

∑k=−l

(Alrl + Bl

1rl+1

)Υk

l (θ, φ)

where Υkl (θ, φ) = e−ikθ Pk

l (cos φ) are the spherical harmonics familiar to anyone with a background inquantum mechanics. We now impose our boundary condition. First note, there is an implied conditionthat r → 0 causes singularities in our general solution. Thus, we take Bl = 0 ∀l. Second, assuming∇2u = 0holds on the boundary, we have a much simpler PDE;

∇2u =1r2 ∂r

(r2∂ru

)+

1r2 sin φ

∂φ

(sin φ∂φu

)= 0.

on ∂B. The simplification comes from the fact that ∂θu = 0 on the boundary by symmetry. Hence, thesolution on the boundary is of the form

u(φ) =∞

∑l=0

Al4l Pl (cos φ) = cos2 φ.

To determine the A′ls we must expand cos2 φ in a basis of Legendre polynomials and exploit some ortho-gonality properties of Legendre polynomials.

2

cos2 φ =∞

∑l=0

Λl Pl (cos φ)

∫ π

0sin φ cos2 φPj(cos φ)dφ =

∫ π

0

∞

∑l=0

Λl sin φPl (cos φ) Pj(cos φ)dφ

Let v = cos φ,

∫ 1

−1v2Pj(v)dv =

∫ 1

−1

∞

∑l=0

Λl Pl (v) Pj(v)dv =∞

∑l=0

Λl

∫ 1

−1Pl (v) Pj(v)dv =

∞

∑l=0

Λl2

2l + 1δl,j

Λj =2j + 1

2

∫ 1

−1v2Pj(v)dv.

Also,

∫ 1

−1

∞

∑l=0

Al4l Pl (cos φ) Pj (cos φ) sin φdφ = Aj4j 22j + 1

and we wantAj4j = Λj

whence

Aj =2j + 12(4j)

∫ 1

−1v2Pj(v)dv.

Our general solution to the entire problem is then;

u(r, θ, φ) =∞

∑l=0

l

∑k=0

(2l + 12(4l)

∫ 1

−1v2Pl(v)dv

)rlΥk

l (θ, φ).

3