Description

Haoran Sun (haoransun@link.cuhk.edu.cn)
Problem 1 (P38 Q2). Using the formula for the wave equation

Problem 2 (P38 Q5). Sketch:

Problem 3 (P38 Q7). Since ϕ and ψ are odd functions, then

Therefore u(−x,t) = −u(x,t), u an odd function in x.
Problem 4 (P38 Q9). Setting
ξ = x − t
η = 4x + t
we have
∂x = ∂ξ + 4∂η
∂t = −∂ξ + ∂η
Then

Then the general solution of the equation ∂ξ∂ηu = 0 would be
u(x,t) = F(ξ) + G(η) = F(x − t) + G(4x + t)
Then
ϕ(x) = F(x) + G(4x) ψ(x) = −F′(x) + 4G′(4x) ⇒ Ψ(x) = −F(x) + G(4x)
where Ψ(x) is any function that Ψ′(x) = ψ(x), then

According to the boundary condition u(x,0) = x2, ut(x,0) = ex, we have

Problem 5 (P41 Q1). Using the conservation law, we know that

Then

by the first vanish theorem. Then we can solve that u(x,t) = k for some constant k. Since ϕ(x) = u(x,0) = 0, we have u(x,t) = u(x,0) = 0.
Problem 6 (P46 Q4).
1. Since the maximum of u is on ∂pΩT , then

Also, since u not a constant function, then ∀(x,t) ∈ ΩT ∂pΩT
u(x,t) < maxu(x,t) = 1
Similarly, according to the minimum principle, one can show that minu(x,t) = 0 and thus
u(x,t) > 0 ∀(x,t) ∈ ΩT ∂pΩT Finally we have 0 < u(x,t) < 1 on ΩT ∂pΩT = (0,1) × (0,∞).
2. Define v(x,t) = u(1 − x,t).
Claim. vt = vxx
Proof. Since
Then
We can also show that u and v have the same boundary condition where v(x,0) = u(1 − x,0) = 4x(1 − x), v(0,t) = u(1,t) = u(0,t) = v(1,t) = 0
Then, according to the uniqueness of diffusion equation, u(x,t) = v(x,t) on ΩT .
3. The energy method shows that with u(x,0) = 0, ut(0,t) = ut(0,l) = 0, we have
E = −κZ ux2 dx ≤ 0 d
dt Ω
Since 0 < u < 1 in ΩT ∂pΩT , ∀t ∃x ∈ (0,1) s.t. ux(x,t) = 0̸ . Therefore ∀t E = −κZ ux2 dx < 0 d
dt Ω
Problem 7 (P46 Q6). To prove the comparison principle, we can prove an equivalent statement:
if u is a solution to diffusion equation with u ≥ 0 on ∂pΩT , then we have u ≥ 0 on ΩT . Proof. According to the minimum principle

then u ≥ 0 on ΩT .
Therefore ∀u,v solution to the diffusion equation with same boundary condition and u ≤ v, we know that w = v −u a solution to the diffusion equation with zero boundary condition. According to our claim, on Ω = [0,l] × [0,∞)
w ≥ 0 ⇒ v − u ≥ 0
Problem 8.
Claim. The function v(x,t) = u(x,t) − ϵt with ϵ > 0 takes its maximum on ∂pΩT .
Proof. Prove by contradiction. Suppose ∃(xϵ,tϵ) ∈ ΩT ∂pΩT s.t.
maxv(x,t) = v(xϵ,tϵ)
ΩT
Then, then we have
vt(xϵ,tϵ − xx ϵ ϵ
vt(xϵ,tϵ) − κvxx(xϵ,tϵ) = ut(xϵ,tϵ) − κuxx(xϵ,tϵ) − ϵ ≤−ϵ < 0
Then we have contradiction. Hence v takes its maximum on some point (xϵ,tϵ) ∈ ∂pΩT .
Since ∂pΩT compact, we can pick a sequence of ϵn s.t. (xϵn,tϵn) → (x0,t0) ∈ ∂pΩT as ϵn → 0+
(which was proved in the class,every bounded sequence has a converged subsequence). Using the fact thatu takes its maximum on (x0,y0) ∈ ∂pΩT . limϵ→0+ maxv = maxu