Description

Haoran Sun (haoransun@link.cuhk.edu.cn)
Problem 1 (P184 Q5). Consider any solutions satisfies the boundary condition w and u where
∆u = 0. Let v = w − u, hence ∂v/∂n = 0. Then
E[w] = E[u + v]
n
= E[u] + E[v] + ˚ dx ∇u · ∇v − ‹ dσ v∇u · n
D ∂D
= E[u] + E[v] + ‹ dσ v∇u · n − ˚ dx v∆u − ‹ dσ v∇u · n
∂D D ∂D
= E[u] + E[v]
Since
1 2 1 2
E[v] = ˚ dx |∇v| − ‹ dσ v∇v · n = ˚ dx |∇v| ≥ 0
2 D 2 D
Hence E[w] − E[u] = E[v] ≥ 0.
Problem 2 (P184 Q7). Define the operation
(∇u1,∇u2) = ˚ dx ∇u1 · ∇u2
D
Let
w˜(x) = w0 + c1w1 + ··· + cnwn
let c0 = 1, hence
E[w˜] = (∇w,˜ ∇w˜) = ∑cicj(∇ui,∇uj)
ij
To minimize the energy, we should have ∂E[w˜]/∂ci = 0, which means
∂E[w˜]
= 2ci(∇wi,∇wi) + 2∑cj(∇wi,∇wj) = 2∑cj(∇wi,∇wj) = 0
∂ci i̸=j j
Hence we have a linear system with n unknowns and n equations
∑
n cj(∇wi,∇wj) = −(∇wi,∇w0)
j=1
for i = 1,…,n.
Problem 3 (P187 Q2). Since
1
dx ∇ · ∇ϕ(x)
D r
HW08 Haoran Sun

1 1 1 1 1
= − ‹ dσ ∇ϕ(x) · n + ‹ dσϕ(x)∇ · n − ˚ dx
4π ∂D r 4π ∂D r 4π D r
Let Dϵ = BR(0) Bϵ(0) for some R > 0 large (ϕ vanish) and ϵ > 0 small, since ∆1/r = 0, then
dx n
Dϵ

for x1,x2 ∈ ∂Bϵ(0) according to the mean value theorem. Hence, easy to show that
1 1 1
lim − ˚ ϵ dx ∆ϕ(x) = − ˚ dx ϵ→0 4π D r 4π D r
Problem 4 (P196 Q1). Since G′′(x) = 0, we have G(x) = Ax + b except x0. Then we have
G(x) =Ax + B x ∈ (0,x0)
Cx + D x ∈ (x0,l)
Applying the boundary and continuity condition, we have
B = 0
Cl + D = 0
Ax0 + B = Cx0 + D
Note that H(x) = G(x) + |x − x0|/2 differentiable at x0
H ⇒ A − 1/2 = C + 1/2
Hence we have four equations, easy to solve that
l − x0 −x0
A = B = 0 C = D = x0 x0 l
Hence
G(x,x0) = {ll−−llxx0xx0 xx ∈∈ (0(x,x0,l0))
Problem 5 (P196 Q6).
(a) The Green’s function for the half plane is
G(x,x | − where x .
HW08 Haoran Sun

(b) The solution is
∞
u(x0) = ˆ dσ h(x)(x,x0) = −ˆ dx h
∂D ∂n −∞ ∂y
where
G[(x,y),(x
(c) Plugin h(x) = 1, we get
− ∞ 1 0 − y0 0 + y0 u(x0,y0) = ˆ dx 2
−∞
1 ∞
= ˆ dx
π (x
−∞