Description

Haoran Sun (haoransun@link.cuhk.edu.cn)
Problem 1 (P52 Q3). Using the formula

Problem 2 (P53 Q14). Suppose 1 = 4̸ kta, we have
(1)
(2)
For t ∈ (0,1/(4ak)), we have 1 − 4kat > 0, then

This upper bound makes sense ∀x ∈ Ω, which means u(x,t) is meaningful.
Ceax2, according to equation 2, we can see that u(x,t) = ∞ when t > 1/(4ak), a eky (k > 0) term appears in the integral term, which means the integral not exists (goes to infinity). Problem 3 (P60 Q3). Define
ϕ˜(x) = ϕ(|x|)
Then

Problem 4 (P66 Q3). Perform the odd extension to both ϕ(x) and ψ(x) to ϕ˜(x) and ψ˜(x), then we get

Plugin ϕ(x) = f(x), ψ(x) = cf′(x), then ∀x ≥ 0, we have two cases
1. x > ct ≥ 0, we get
HW03 Haoran Sun

2. 0 ≤ x ≤ ct, we get

Problem 5 (P67 Q10). Perform odd extension of ϕ and ψ on (0,π) and perform even extension on whole R, we get
ϕ˜(x) = cosx, ψ˜(x) = 0
Then the solution of this wave equation is

using the fact that c = 3.
Problem 6 (P79 Q2). Using the formula that

we have

Problem 7 (P79 Q14). Define a new function v(x,t) = u(x,t) − xk(t), then we have
on x ∈ (0,∞). Then we can perform even extension on ϕ, ψ and f. Applying the formula

When 0 ≤ ct ≤ x, the three parts red r, blud b, and orange o equals to
HW03 Haoran Sun

o = −xk(t) + xk(0) + txk′(0)
Then v(x,t) = −xk(t), which means u(x,t) = 0 when 0 ≤ ct ≤ x.
When ct > x ≥ 0, the red part r equals to

The blue part b equals to

The orange part o equals to

Therefore we have

for ct > x > 0.