Question

Solve the Cauchy problem on a semi-infinite domain:
u_tt - 4u_xx = 0 on 0<x<infinity, t>0
u(x,0) = xe^(-x)
u_t(x,0) = 0
u(0,t) = 0

Answer 1

Solve the Cauchy problem on a semi-infinite domain:

$$U_{tt} - 4U_{xx} = 0$$ $$U(x, 0) = x e^{-x} = \varphi(x)$$ $$U_t(x, 0) = 0 = \psi(x)$$ $$U(0, t) = 0 = \mu(t)$$ $$0 \leq x < +\infty; t \geq 0$$

Solution:

a) $x - 2t \geq 0$

$$U(x, t) = \frac{\varphi(x - 2t) + \varphi(x + 2t)}{2} + \frac{1}{4} \int_{x - 2t}^{x + 2t} \psi(y) \, dy = \frac{(x - 2t) e^{2t - x} + (x + 2t) e^{x + 2t}}{2}$$

b) $x - 2t < 0$

$$U(x, t) = \frac{\varphi(x + 2t) - \varphi(2t - x)}{2} + \frac{1}{4} \int_{2t - x}^{x + 2t} \psi(y) \, dy = \frac{(x + 2t) e^{-(x + 2t)} - (2t - x) e^{x - 2t}}{2}$$

Answer:

$$U(x, t) = \begin{cases} \frac{(x - 2t) e^{2t - x} + (x + 2t) e^{x + 2t}}{2}, & \text{when } x \geq 2t \\ \frac{(x + 2t) e^{-(x + 2t)} - (2t - x) e^{x - 2t}}{2}, & \text{when } x < 2t \end{cases}$$