1. Problem 5
Solve
2
u = 0 r ∈ B4
u = cos2 φ r ∈ ∂B4 .
This is a lot of work, first let u(r, θ, φ) = F (r ) G (θ ) H (φ) so we may seperate the DE.
In spherical coordinates
2 1 1 1
u= ∂r r 2 ∂r u + 2 ∂θ (sin θ∂θ u) + ∂2 u = 0.
r2 r sin θ r 2 sin2 θ φ
1 d dF 1 d dG 1 d2 H
0 = G (θ ) H (φ) r2 + F (r ) H ( φ ) sin θ + F (r ) G ( θ )
r2 dr dr r2 sin θ dθ dθ r2 sin2 θ dφ2
Dividing by u,
1 1 d dF 1 1 d dG 1 1 d2 H
0= r2 + sin θ + .
r2 F (r ) dr dr r2 sin θ G (θ ) dθ dθ r2 sin2 θ H (φ) dφ2
With some rearrangement,
1 d2 H 1 1 d dF 1 1 d dG
= −r2 sin2 θ r2 + sin θ .
H (φ) dφ2 r2 F (r ) dr dr r2 sin θ G (θ ) dθ dθ
Notice that when φ is fixed and r and θ vary the above equation holds even though the RHS is solely a
function of r, θ and the LHS is solely a function of φ. Hence,
1 d2 H
= − λ2
H (φ) dφ2
for some constant λ. We chose λ2 (the RHS is always negative then) to ensure that the solutions to the
above ODE are periodic, and the boundary conditions do not imply that H ≡ 0. Note, H (φ) = c1 exp(iλφ),
or if one desires; H (φ) = c1 sin(λφ) + c2 cos(λφ).
1 d2 H
Returning to the larger equation, we substitute H (φ) dφ2
= −λ2 so that
1 1 d dF 1 1 d dG λ2
0= r2 + sin θ − .
r2 F (r ) dr dr r 2 sin θ G ( θ ) dθ dθ r2 sin2 θ
Multiplying through by r2 and rearranging some more,
1 d dF λ2 1 1 d dG
r2 = 2
− sin θ .
F (r ) dr dr sin θ sin θ G (θ ) dθ dθ
We chose another seperation constant Ω to convert the above equation into two ODEs.
d dF
r2 − ΩF (r ) = 0 (i)
dr dr
d dG λ2
sin θ + G (θ ) − Ω sin θG (θ ) = 0 (ii)
dθ dθ sin θ
1

2. Equation (i) looks like an ODE for spherical Bessel functions at first glance, but is infact the so-called Euler
differential equation if we let Ω = l (l + 1). To solve, we apply the Frobenius method. Make the ansatz,
∞
F (r ) = ∑ an r n
n =0
with α fixed.
d dF d2 F dF
r2 = r2 + 2r
dr dr dr2 dr
and we compute the derivatives
∞
d2 F
= ∑ n ( n − 1 ) a n r n −2
dr2 n =0
∞
dF
= ∑ nan r n−1 .
dr n =0
Plugging into the ODE,
∞ ∞ ∞
∑ n ( n − 1) a n r n + 2 ∑ nan r n − l (l + 1) ∑ an r n = 0
n =0 n =0 n =0
∞
= ∑ an {n(n − 1) + 2n − l (l + 1)} rn .
n =0
∞
= ∑ an n2 + n − l ( l + 1) r n
n =0
∞
= ∑ an {n(n + 1) − l (l + 1)} rn
n =0
Now we make the observation that each term in this series is unique. As such, for such a series to terminate
to 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 = −(l + 1), l then the only possible way to rectify the situation is to let an = 0, hence all an for
n = l, −(l + 1) must be zero. Only two terms survive,
1
Fl (r ) = Al r l + Bl ,
r l +1
where l will be summed over in our final solution.
Equation (ii) is the associated Legendre DE which has solutions Plλ (cos φ) for λ = 0, ..., l. I think solving
(ii) is another application of the Frobenius method, I omit the derivation to save time; everyone knows this
ODE anyways.
Finally, our general solution is
∞ l
1
u(r, θ, φ) = ∑ ∑ Al r l + Bl
r l +1
Υλ (θ, φ)
l
l =0 λ=−l
where Υλ (θ, φ) = e−iλθ Plλ (cos φ) are the spherical harmonics familiar to anyone with a background in
l
quantum mechanics. We now impose our boundary condition. First note, there is an implied condition
that r → 0 causes singularities in our general solution. Thus, we take Bl = 0 ∀l.
∞ l
u(r = 4, φ) = ∑ ∑ Al 4l Plλ (cos φ)
l =0 λ=−l
Where we have absorbed the exponentials in θ into the Al s.
2