If we consider the wave equation on the half line $\mathbb{R}_+$ such that
$u_{tt} -u_{xx}=0$ in $\mathbb{R_+}\times(0,\infty)$
$u(x,0)=g(x)$ and $u_t(x,0)=h(x)$ for $x\in \mathbb{R_+}$
$u(0,t)=0$ for $t \in \mathbb{R}_+$
with $g(0)=0$, $h(0)=0$,
the solution is given by
1) $u(x,t)=\frac{1}{2}[g(x+t)+g(x-t)]+\frac{1}{2}\int_{x-t}^{x+t}h(y)\,dy$ for $x\geq t \geq 0$
2) $u(x,t)=\frac{1}{2}[g(x+t)-g(t-x)]+\frac{1}{2}\int_{t-x}^{x+t}h(y)\,dy$ for $0\leq x \leq t$.
Why does Evans claim that $u$ does not belong to $C^2$ unless $g''(0)=0?$
I am not sure about how to proceed...certainly if I can show that $g$ and $\int_0^x h(t)\,dt$ are in $C^2$ then $u$ is also in $C^2$. And because we are unsure about the continuity of $g'$ and $g''$ at $0$ and of $h'$ at $0$ then it is worth investigating these functions at $0. But how do we calculate these?
Or do we return to how the solution was constructed which is via an odd reflection? If $\tilde{u},\tilde{g},\tilde{h}$ are the odd reflections of $u,g,h$ then the solution is given by
$\tilde{u}=\frac{1}{2}[\tilde{g}(x+t)-\tilde{g}(x-t)] + \frac{1}{2}\int_{x-t}^{x+t}\tilde{h}(y)\,dy$ so it suffices to investigate under what conditions $\tilde{g}$ and $\int_0^x \tilde{h}(y)\,dy$ are $C^2$?
Help very much appreciated...
