I will grade problem 4 Dec 24. There is a solution for both (a) and (b) basically coinciding with one of Danny (without an obvious misprint): solution $u(x,t)=\phi (x-ct)+ \psi (x+ct)$ (thus does not working in higher dimensions where one based on energy integral works). Then
\begin{equation}
c^{-2}u_t^2 + u_x^2= \bigl( \phi'(x-ct)+\psi'(x+ct)\bigr)^2 + \bigl( \phi '(x-ct)-\psi'(x+ct)\bigr)^2=2 \bigl( \phi'(x-ct)\bigr)^2 + \bigl( \psi'(x+ct)\bigr)^2
\end{equation}
and integrating with respect to $x$ and making change of variables $x_{new}=x-ct$, $x_{new}=x+ct$ in the integrals containing $\phi'$ and $\psi'$ respectively we get
\begin{equation}
\int\bigl(c^{-2}u_t^2 + u_x^2\bigr)\,dx= 2 \int \bigl( \phi'^2 (x)+\psi'^2(x)\bigr)\,dx
\end{equation}
where r.h.e. does not depend on $t$. Here $\phi,\psi$ are related to $f,g$ in conditions by $f(x)= \phi(x)+\psi(x)$, $c^{-1}g(x)=-\phi'(x)+\psi'(x)$.
Meanwhile
\begin{equation}
c^{-2}u_t^2 - u_x^2= \bigl(- \phi '(x-ct)+\psi '(x+ct)\bigr)^2 + \bigl( \phi '(x-ct)+\psi'(x+ct)\bigr)^2=-4 \phi '(x-ct) \psi'(x+ct)
\end{equation}
and the r.h.e. is $0$ unless both factors are not $0$ which can happen only as $|x-ct|\le R, |x+ct|\le R\implies c|t|\le 2R$. Therefore as $|t|\ge 2c^{-1}R$ the r.h.e. is identically $0$ and integrating with respect to $x$ we get (b)
Remark. Obviously as either $u=\phi(x-ct)$ or $u=\psi (x+ct)$ we have $u_t^2=c^2u_x^2$ and $k(t)=p(t)$ for all $t$. The above proof is based on the observation that any solution with initial data supported in $\{|x|\le M\}$ after time $T=2c^{-1}M$ breaks into two waves $u_1=\phi(x-ct)$ and $u_2=\psi (x+ct)$which do not overlap. This is definitely not true for equation on the finite interval with energy preserving boundary conditions.
Remark. In higher dimensions (a) still holds for $u_{tt}-c^2\Delta u=0$
\begin{equation}
\int \bigl( |u_t|^2 + c^{-2}|\nabla u|^2\bigr)\,dV = \text{const}
\end{equation}
and (b) is replaced by
\begin{equation}
\int \bigl( |u_t|^2 - c^{-2}|\nabla u|^2\bigr)\,dV \to 0 \qquad \text{as}\quad t\to \pm \infty.
\end{equation}
While (a) is proven easily by an energy integral method, (b) becomes more subtle.
