Wave Equation with Initial Conditions on Characteristic Curves I am trying to solve the initial value problem:
$$\begin{cases}
u_{tt}-u_{xx} =0\\
u|_{t = x^2/2} = x^3, \quad |x| \leq 1 \\
u_{t}|_{t = x^2 / 2} = 2x, \quad |x| \leq 1 \\
\end{cases}
$$
I'm unsure if we can use D'Alambert's formula in this problem, so what I have attempted is just using the general solution and going from there: $u(x, t) = f(x - t) + g(x + t)$, means (after plugging in initial conditions): $$f(x - \frac{x^2}{2}) + g(x + \frac{x^2}{2}) = x^3, \ \mathrm{and}$$ $$-f'(x - \frac{x^2}{2}) + g'(x + \frac{x^2}{2}) = 2x.$$
I am unsure where to go from here, or if my approach is correct. Could someone help me out?
 A: The general solution is $u(x,t)=f(x-t)+g(x+t)$
$u|_{t=\frac{x^2}{2}}=x^3$ :
$f\left(x-\dfrac{x^2}{2}\right)+g\left(x+\dfrac{x^2}{2}\right)=x^3......(1)$
$u_t(x,t)=f_t(x-t)+g_t(x+t)=g_x(x+t)-f_x(x-t)$
$u_t|_{t=\frac{x^2}{2}}=2x$ :
$g_x\left(x+\dfrac{x^2}{2}\right)-f_x\left(x-\dfrac{x^2}{2}\right)=2x$
$g\left(x+\dfrac{x^2}{2}\right)-f\left(x-\dfrac{x^2}{2}\right)=x^2+c......(2)$
$\therefore f\left(x-\dfrac{x^2}{2}\right)=\dfrac{x^3-x^2-c}{2}$ , $g\left(x+\dfrac{x^2}{2}\right)=\dfrac{x^3+x^2+c}{2}$
$f\left(-\dfrac{x^2-2x}{2}\right)=\dfrac{x^3-x^2-c}{2}$ , $g\left(\dfrac{x^2+2x}{2}\right)=\dfrac{x^3+x^2+c}{2}$
$f\left(-\dfrac{(x-1)^2-1}{2}\right)=\dfrac{x^3-x^2-c}{2}$ , $g\left(\dfrac{(x+1)^2-1}{2}\right)=\dfrac{x^3+x^2+c}{2}$
$f\left(-\dfrac{x^2-1}{2}\right)=\dfrac{(x+1)^3-(x+1)^2-c}{2}$ , $g\left(\dfrac{x^2-1}{2}\right)=\dfrac{(x-1)^3+(x-1)^2+c}{2}$
$f\left(-\dfrac{x^2-1}{2}\right)=\dfrac{x^3+2x^2+x-c}{2}$ , $g\left(\dfrac{x^2-1}{2}\right)=\dfrac{x^3-2x^2+x+c}{2}$
$f\left(-\dfrac{x^2-1}{2}\right)=\dfrac{x(x+1)^2-c}{2}$ , $g\left(\dfrac{x^2-1}{2}\right)=\dfrac{x(x-1)^2+c}{2}$
