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TRANSCRIPT
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).
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