Description

Haoran Sun (haoransun@link.cuhk.edu.cn)
Problem 1 (P160 Q7). The solution is in the form of
c1 1 2
u(r,θ,ϕ) = + c2 + r r 6
Applying the boundary condition where u(a,θ,ϕ) = u(b,θ,ϕ) = 0, we can solve c1 and c2
ab b2 − a2 1 1 b3 − a3 1 2 2 c1 = = ab(a + b) c2 = = (a + ab + b )
6 b − a 6 6 b − a 6
Problem 2 (P165 Q6). Let X(x)Y (y)Z(z) be a solution, then
X′′ Y ′′ Z′′
∆u = 0 ⇒ + + = 0
X Y Z
set boundary conditions
ux(0,y,z) = ux(1,y,z) = uy(x,0,z) = uy(x,1,z) = uz(x,y,0) = 0
We have
Xm(x) = cosmπx, Yn(y) = cosnπy
Hence
Z ′′ = (m2 + n2)π2, Z′(0) = Z′(1) = 0 ⇒ Z(z) = Acosh√m2 + n2πz Z
Let the solution be in the form
1 1 ∞∞
u(x,y,z) = A00 + ∑ Am0 cosmπxcoshmπz + ∑ A0n cosnπy coshmπz
4 2
m=1n=1
Amn cosmπ cosnπy cosh√m2 + n2πz m=1n=1
where
Amn dxdy cosmπxcosnπyg(x,y)
Problem 3 (P172 Q2). Since
u 2 n=1
we have
u 2 n=1
3
⇒ A0 = 2, An = 0, B1 = , B2 = ··· = 0
a
which means u(r,θ) = 1 + 3rsinθ/a.
1 HW07 Haoran Sun

Problem 4 (P176 Q4). Let
u 2 n=1 an 2π
Anˆ dϕh(ϕ)cosnϕ
0
an 2π
Bnˆ dϕh(ϕ)sinnϕ
0
Therefore
u dϕ h
dϕ h
dϕ h
dϕ h neinφ + (a/r)−ne−inφ]
1 ˆ 2π [ −aeiφ −ae−iφ ]
= dϕ h(ϕ) 1 + +
2π 0 r aeiφ r ae−iφ
r2 − a2
dϕ h(ϕ) r2 + a2 − 2arcos(θ − ϕ)
Problem 5 (P176 Q13). In this case, the eigenfunction Θ(θ) and R(r) have the form
nπ(θ − α) βnπα βnπ
Θn(θ) = sin Rn(r) = Anr − + Bnr− −α
β − α
Therefore
u Bnr
nπ(θ−α)
Anb β−α h(θ)]
Bn b h(θ)]
2