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Haoran Sun (haoransun@link.cuhk.edu.cn) Problem 1 (P135 Q15).
(a) Note that ∥cos(n + 1/2)x∥2 = π/2 on (0,π), then we have
Bn n
(b) The series converges for all x in (−2π,2π). −1 x ∈∈ (±−−π,π−) ∪
S(x) = 1 x ( 2π, π) (π,2π)
0 x = π
(c) Using Parseval’s equality, we have
Bn2∥Xn(x)∥2 = ∥ϕ(x)∥2 n=0
X
16 1 π
= π π (2n + 1)2 2 n=0
1 π
=
X (2n + 1)2 8
n=0 Problem 2 (P145 Q4).
(a) Let T(t)X(t) satisfies the boundary condition, then
T′(t) X′′(x)
= = −λ
kT X(x)
Assume that λ ≥ 0. If λ = 0, we have X0(x) = A + Bx. Else, let β2 = λ > 0, then we have the following form of the eigenfunctions.
X(x) = C cosβx + D sinβx
Applying the boundary condition, we can solve that
βnl βnl
tan =
2 2
Hence u(x,t) could be written in the form of
u e
(b) Suppose we can take the limit term by term, hence
lim e−
t→∞
= X 0 = 0
n=1
Consequently
lim u(x,t) = A + Bx t→∞
HW06 Haoran Sun

(c) Suppose λ = −β2 < 0. Note that
l
u(x,t)uxx(x,t)dx ≤ 0
However
l l l
u(x,t)uxx(x,t)dx = T(t)2ˆ X(x)X′′(x)dx = T(t)2β2ˆ X2(x)dx ≥ 0
0 0 0 where the contradiction occurs that λ = 0 in this case. Hence λ > 0.
(d) Since ⟨1,1⟩ = l, ⟨x,x⟩ = l3/3, then
1 l 3 l
A = ˆ ϕ(x)dx B = ˆ xϕ(x)dx l 0 l3 0
Problem 3 (P145 Q6). Suppose u is in the form of
1 ∞kt nπx
u(x,t) = A0 + X Ane cos
2l
n=1
where u(x,0) = ϕ(x) continuous on [0,l]. Claim. An bounded.
Proof. Since ϕ is continuous on [0,l], then ∥ϕ∥ bounded, hence
nπ nπ 2 nπ
|⟨ϕ,cos x⟩| ≤ ∥ϕ∥∥cos ∥ < ∞ ⇒ An = ⟨cos ,ϕ⟩ < M l l l l
Claim. The following series converges ∀t > 0
Annke−n2t
Proof. Note that ∀ϵ > 0, ∃N ∈ N s.t. ∀m > N we have < ϵ/M, then
|Annke−n2t| ≤ M nke−nt + ϵ
Hence the series converges.
According to the claims, ∃N ∈ N s.t. ∀x and ∀m > N we have
dk A e−n2l2π2kt cos nπx ≤ M e−n2l2π2kt < ϵ
k n l nm dx
which means the series converges uniformly with respect to x. Hence, using the theorem in the appendix, we have
d xk u(x,t) = d xk X n l l dxk Ane−n2l2π2kt cos nπx l dk dknπxdk n=1 n=1
∞ kkt nπx
= X Bnn e cos l
n=1
exists ∀k in t > 0.
HW06 Haoran Sun

Problem 4 (P145 Q11). Follow the same steps when proving the uniform convergence. Since the f′(x) piecewise continuous f′(x)X(x) also piecewise continuous, then it is integrable and hence
An = −n 1Bn′ Bn = n 1A′n
Applying Bessel’s inequality, we have the following series which is convergent
2 2
Therefore
∞
An cosnx + Bn sinnx ≤ X |Ann cosnxn| + |Bn sinnx|

|A | + |B |
|A′n| + |Bn′ |
1/2
|A′n||Bn′ |!
X X
Hence ∀ϵ > 0, we can choose N ∈ N s.t. ∀m > N and ∀x we have
f(x) − m An cosn nx2 + Bnn2sinnx
≤ M n = (|A′ | + |B′ | ) < ϵ
Hence the Fourier series converges uniformly.
Problem 5 (P160 Q5). Note that (x2+y2)/4+c is the solution of ∆u = 1. Since /4 satisfies the boundary condition, according to the uniqueness, the solution is u−
1/4.
Problem 6 (P160 Q11). Suppose there is a solution u, then
˚
D f dxdy dz = ˚D ∇ · ∇udxdy dz g dσ
Then if equality does not hold, there will be no solutions.
Problem 7of D, then ∆(P160 Q13)v ≤ 0. Note that. Let v = u+ϵ|x|2. Suppose v obtains its maximum in the interior domain
∆v = ∆u + 4nϵ > 0
where n is the dimension. This contradicts the assumption. Hence maxD v = max∂D v. Since D is bounded, we can also show that
maxv = maxu maxv = maxu
D D ∂D ∂D
Hence maxD u = max∂D u.