How solve the following PDE? I want to solve the following PDE:$$
\begin{cases}
u_{tt}-c^2u_{xx}=0, \quad x\in\mathbb{R},\ t\geq x\\
u(x,x)=φ(x), \quad x\in\mathbb{R}\\
u_t(x,x)=0, \quad x\in\mathbb{R}
\end{cases}
$$
where $φ:\mathbb{R} \to \mathbb{R}$, $φ\in C^1(\mathbb{R})$.
Thanks for your help.
 A: ﻿$\def\d{\mathrm{d}}$Case 1: $c^2 \neq 1$. Make substitution $(y, s) = (x - c^2 t, t - x)$, then$$
u_t = u_s - c^2 u_y, \quad u_{tt} = u_{ss} - 2c^2 u_{ys} + c^4 u_{yy}, \quad u_{xx} = u_{yy} - 2u_{ys} + u_{ss},
$$
and the equations become$$
\begin{cases}
(1 - c^2) u_{ss} - (c^2 - c^4) u_{yy} = 0, \quad y \in \mathbb{R},\ s > 0\\
u\bigr|_{s = 0} = φ\left( \dfrac{y}{1 - c^2} \right), \quad y \in \mathbb{R}\\
(u_s - c^2 u_y)\bigr|_{s = 0} = 0. \quad y \in \mathbb{R}
\end{cases} \tag{1}
$$
Because $u\bigr|_{s = 0} = φ\left( \dfrac{y}{1 - c^2} \right)$, then$$
u_y\bigr|_{s = 0} = \frac{\d}{\d y} \left( φ\left( \frac{y}{1 - c^2} \right) \right) = \frac{1}{1 - c^2} φ'\left( \frac{y}{1 - c^2} \right),
$$
and (1) becomes$$
\begin{cases}
u_{ss} - c^2 u_{yy} = 0, \quad y \in \mathbb{R},\ s > 0\\
u\bigr|_{s = 0} = φ\left( \dfrac{y}{1 - c^2} \right), \quad y \in \mathbb{R}\\
u_s\bigr|_{s = 0} = \dfrac{c^2}{1 - c^2} φ'\left( \dfrac{y}{1 - c^2} \right). \quad y \in \mathbb{R}
\end{cases} \tag{2}
$$
Thus\begin{align*}
u(y, s) &= \frac{1}{2} \left( φ\left( \frac{y + cs}{1 - c^2} \right) + φ\left( \frac{y - cs}{1 - c^2} \right) \right) + \frac{1}{2c} \int_{y - cs}^{y + cs} \dfrac{c^2}{1 - c^2} φ'\left( \dfrac{ξ}{1 - c^2} \right) \,\d ξ\\
&= \frac{1}{2} \left( φ\left( \frac{y + cs}{1 - c^2} \right) + φ\left( \frac{y - cs}{1 - c^2} \right) \right) + \frac{c}{2} \left( φ\left( \frac{y + cs}{1 - c^2} \right) - φ\left( \frac{y - cs}{1 - c^2} \right) \right)\\
&= \frac{1 + c}{2} φ\left( \frac{y + cs}{1 - c^2} \right) + \frac{1 - c}{2} φ\left( \frac{y - cs}{1 - c^2} \right),
\end{align*}
and$$
u(x, t) = \frac{1 + c}{2} φ\left( \frac{x + ct}{1 + c} \right) + \frac{1 - c}{2} φ\left( \frac{x - ct}{1 - c} \right).
$$
Case 2: $c^2 = 1$. Make substitution $(y, s) = (x, t - x)$, then$$
u_t = u_s, \quad u_{tt} = u_{ss}, \quad u_{xx} = u_{yy} - 2u_{ys} + u_{ss},
$$
and the equations become$$
\begin{cases}
2u_{ys} - u_{yy} = 0, \quad y \in \mathbb{R},\ s > 0\\
u\bigr|_{s = 0} = φ(y), \quad y \in \mathbb{R}\\
u_s\bigr|_{s = 0} = 0, \quad y \in \mathbb{R}
\end{cases} \tag{3}
$$
which implies$$
\begin{cases}
2u_s - u_y = η(s), \quad y \in \mathbb{R},\ s > 0\\
u\bigr|_{s = 0} = φ(y), \quad y \in \mathbb{R}
\end{cases} \tag{4}
$$
where $η(s)$ is a function to be determined. By the method of characteristics, the general solution to (4) is$$
u(y, s) = \frac{1}{2} \int_0^s η(ξ) \,\d ξ + φ\left( y + \frac{s}{2} \right),
$$
and plugging it back to (3) yields$$
0 = u_s\bigr|_{s = 0} = \frac{1}{2} η(s) + \frac{1}{2} φ(y) = 0. \quad \forall y \in \mathbb{R}
$$
For the existence of solutions to the original equations, $φ$ must be a constant (namely $-η(0)$) and$$
u(x, t) = \frac{1}{2} \int_0^{t - x} η(ξ) \,\d ξ - \frac{1}{2} η(0).
$$
It can be verified that $u(x, t)$ with the form above are indeed solutions.
