(a) Solve the Darboux problem: $u_{tt}- u_{xx} = 0$, $t > \max\{-x,x\}$, $t \geq0$ \begin{equation*} u(x,t) = \begin{cases} \phi(t) &, \text{ si } x = t, t \geq0. \\ \psi(t) &, \text{ si } x=-t, t \geq 0. \end{cases} \end{equation*}
(b) Prove that the problem is well posed.
I'm trying to do this problem, but there are several things that I have not been able to understand well. In part (a) I'm not sure if the d'Alembert equation can be used directly for this homogeneous case, or perhaps another properly constructed solution is required from the problem itself. So, How can I solve this problem?
For part (b) we need to show that small changes in the initial conditions give rise to a small change in the solution.
For this, let $\epsilon>0$ and let $u_i$ be two solutions of the problem with initial conditions $f_i$, $g_i$, where $i = 1, 2$. Now, I need to show that, if $| f_1(x) − f_2(x)| < \delta $ and $|g_1(x) − g_2(x)| < \delta$ for all $x \in \mathbb{R}$, then for all $x \in \mathbb{R}$ and $0 \leq t \leq T$ we have $|u_1(x, t) − u_2(x, t)| < \epsilon$. How can I conclude this?
The characteristics flow from both boundaries to a point $(a,b)$ in the domain along lines of slope $\pm 1$. To solve for the two points they flow from, we can solve the slope equation
$$\frac{b-|x|}{a-x} = \pm 1 \implies x = \frac{a-b}{2}, \frac{a+b}{2}$$
Given the domain for the pde, $b>a$, the first solution is from the left boundary and the second solution is from the right. This means the solution to the problem is
$$u(x,t) = \psi\left(\frac{t-x}{2}\right) + \phi\left(\frac{t+x}{2}\right)$$
Can you solve the second part from here?