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.
$\def\d{\mathrm{d}}$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.