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Solutions to Exercises

4
2
1 2
Exercise 129. (e x+t/4 − e x−t ) + x + t .
5 4

2
2
Exercise 134. u(x, t) = x + t + 3t /2.

Exercise 136. D’Alembert’s formula implies

1 1
P(x, t) = [f(x + 4t) + f(x − 4t)] + [H(x + 4t) − H(x − 4t)],
2 8


x |x| < 1,

x
R
where H(x) = g(s)ds. Hence H(x) = 1 x > 1, Notice that at x = 10 :
0 0

−1 x < −1.

1
f(10+4t) = 0, f(10−4t) ≤ 10, |H(t)| ≤ 1, t > 0. Therefore, P(10, t) ≤ 5+ =
4
21 < 6, and the structure will not collapse.
4

Exercise 138. Unfortunately, this problem is no-trivial. (a) The solution is not
3
−1
classical when x ± 2t = −1, 0, 1, 2, 3. (b) u(1, 1) = 1/3 + e − e /2 − e /2.

R 1
Exercise 140. u(x, t) = v(x, t)dx + f(t) = [sin(x − t) − sin(x + t)] + f(t),
2
where f(t) is an arbitrary function.

Exercise 142. (0, 0) and (1, T).

√ √
Exercise 143. 2u(x, t) = Erf((x + l)/2 kt) − Erf((x − l)/2 kt).

Exercise 145. e 3x+9kt .

2
Exercise 147. x + 2kt.

Exercise 149. v(x, t) satisfies the diffusion equation with initial condition φ(x).

R ∞ (x−V t−z) 2 dz
Exercise 150. u(x, t) = exp[− ]φ(z) √ .
−∞ 4kt 4πkt

√
Exercise 152. 1 − Erf[x/ 4kt].

2
2
Exercise 153. w(x, t) = (4πkt) −1/2 R ∞ (e −(x−y) /4kt + e −(x+y) /4kt )φ(y)dy.
0

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