Equipartition of energy in the wave equation 

(i) Fix $t$. The derivatives of $k(t)$ and $p(t)$ are $$k_t(t) = \frac{\partial}{\partial t} \frac{1}{2} \int_{-\infty}^{\infty} u_t^2 (x,t) \, dx = \int_{-\infty}^{\infty} u_t \cdot u_{tt} \, dx,$$ \[ p_t(t) = \frac{\partial}{\partial t} \frac{1}{2} \int_{-\infty}^{\infty} u_x^2 (x,t) \, dx = \int_{-\infty}^{\infty} u_x \cdot u_{xt} \, dx \] \[= - \int_{-\infty}^{\infty} \Delta u \cdot u_t \, dx +\left[ u_t u_x \right]_{x=-\infty}^{x=\infty} \hspace{0.5cm} \] \[= - \int_{-\infty}^{\infty} \Delta u \cdot u_t \, dx \hspace{0.5cm}\] where we have used integration by parts for which the boundary terms drop due to the finite propagation speed of waves, i.e. $u(x,t)=0$ for $t$ large enough on every compact subset of $\mathbb{R}^n$.

Hence $$k_t(t)+p_t(t)=\int_{-\infty}^{\infty} u_t (u_{tt}- \Delta u) \, dx = 0,$$

which implies that $k(t)+p(t)$ is constant in $t$. 

(ii) By D'Alembert's formula, 
$$u(x,t)=\frac{1}{2} \left[ g(x+t) +g(x-t) \right] +\frac{1}{2} \int_{x-t}^{x+t} h(y) \, dy.$$

Thus  $$u_t=\frac{1}{2} \left[ g'(x+t) -g'(x-t) \right] + \frac{1}{2} \left[ h(x+t) +h(x-t)\right],$$ and $$u_x=\frac{1}{2} \left[ g'(x+t) +g'(x-t) \right] + \frac{1}{2} \left[ h(x+t) -h(x-t)\right].$$
This implies
\[ k(t)-p(t) = \frac{1}{2} \int_{-\infty}^{\infty} \left( \frac{1}{2} \left[ g'(x+t) -g'(x-t) \right] + \frac{1}{2} \left[ h(x+t) +h(x-t)\right] \right)^2 \, dx\]
\[- \frac{1}{2} \int_{-\infty}^{\infty} \left( \frac{1}{2} \left[ g'(x+t) +g'(x-t) \right] + \frac{1}{2} \left[ h(x+t) -h(x-t)\right] \right) ^2  dx.\]

Let $L>0$ such that $\text{supp}(g'), \text{supp}(h)\subseteq [-L,L]$. We  show that the above integral is zero by showing $u_t^2=u_x^2$. 

Case 1: If $x+t > L$, then  $x>L-t$, and hence $g'(x+t)=h(x+t)=0 \Rightarrow u_t^2=u_x^2$.

Case 2: If $x-t<-L$,  $x<t-L$, and hence $g'(x-t)=h(x-t)=0 \Rightarrow u_t^2=u_x^2$. 

Case 3: If $L-t \geq x \geq t-L$,  is not satisfied for any x when $t>L$. So this case does not happen for large $t$.

Therefore, $k(t)=p(t)$ for large enough $t$.