The problem is from Evans PDE 2.5 problem 18 How to solve the equations?
Assume u solves the initial-value problem
$u_{tt}-\Delta u=0$ in $R^n \times (0,\infty)$
$u=0$, $u_t=h$ on $R^n\times \{t=0\}$
Show that $v:=u_t$ solves
$v_{tt}-\Delta v=0$ in $R^n \times (0,\infty)$
$v=h$, $\ v_t=0$ on $R^n\times \{t=0\}$
(This is Stokes' rule)
This is very straightforward. On $\mathbb{R}^n \times (0,\infty)$, your function $v$ satisfies $$v_{tt} - \Delta v = (\Delta u)_t - \Delta(u_t)$$ so you only need to see that $(\Delta u)_t = \Delta(u_t)$. For the second line, $v = h$ at $t=0$ is immediate from $u_t = h$ and for the last assertion, recall that $u=0$ initially so $\Delta u$ is also $0$ at $t=0$. This gives $v_t = u_{tt} = \Delta u = 0$ at $t=0$.
