Does this problem have a solution?
$$\begin{cases} \partial_t^2u(x,t)&=\partial_x^2u(x,t) \qquad x \in[-1,1] \quad t>0 \\ u(x,0)&=1-|x| \qquad \quad x \in[-1,1] \\ \partial_tu(x,0)&=0 \qquad \qquad \qquad t>0 \\ u(1,t)=u(-1,t)&=0 \end{cases} $$
If instead of $1-|x|$ were a $C^2$ function then I know from the theory that there is a classical solution which we can find e.g. by separation of variables.
$$\frac{\partial ^2 u}{\partial t^2}- \frac{\partial ^2 u}{\partial x^2}=0$$ $$\left(\frac{\partial}{\partial t}+\frac{\partial}{\partial x}\right) \left(\frac{\partial}{\partial t}-\frac{\partial}{\partial x}\right)u(x,t)=0$$
Hense, the general solution is : $$u=f(x+t)+g(x-t)$$ any derivable functions $f$ and $g$.
Condition : $\left( \frac{\partial u}{\partial t}\right)_{(x,0)} =0$ gives : $f'(x)+g'(x)=0$ , hense : $g=-f+$constant. $$u=f(x+t)-f(x-t)$$ Condition $u(x,0)=1-|x|$ leads to : $$f(x)-f(x)=1-|x|=0$$ This is impossible. So, either there is no solution, or there is somthing wrong in the wording of the problem.
