Solving Wave Equation with Initial Values

Question

I am trying to solve the wave equation:

$u_{tt}$ = $u_{xx}$

With initial values:

$u(x,0) =\begin{cases} x^3 - x, &\text{for }|x|\le 1,\ \\0, &\text{for }|x|\ge1\end{cases}$

$u_t(x,0) =\begin{cases} 1 - x^2, \text{for } |x|\le 1, \\ 0, \text{for } |x|\ge1 \end{cases}$

In the domain $D= \{(x,t) | - \infty < x<\infty, \space\space t>0\} $

I am using D'Alembert's Formula and after inputting the values in I have:

$u(x,t) = \frac{1}{2}[\frac{2}{3}(x+t)^3 + \frac{4}{3}(x-t)^3 - 2(x-t)] $

I am not sure where to go from here.

I think because the initial conditions are not differentiable at some points then there will be points in the solution that are not differentiable

So I'd need to consider the different cases of $|x+t|\le 1$?

I could be going down the complete wrong path so any help is appreciated.

Thanks

Answer 1

In fact the real result should be $u(x,t)=\dfrac{f(x+t)+f(x-t)+g(x+t)-g(x-t)}{2}$ , where $f(x)=\begin{cases}x^3-x&\text{for}~|x|\leq1\\0&\text{for}~|x|\geq1\end{cases}$ and $g(x)=\begin{cases}-\dfrac{2}{3}&\text{for}~x\leq-1\\x-\dfrac{x^3}{3}&\text{for}~|x|\leq1\\\dfrac{2}{3}&\text{for}~x\geq1\end{cases}$