Differentiate the energy and use the <chain rule>:
$$
\frac{dE}{dt}
=\int_{\mathbb R^3}
\left(u_tu_{tt}+\nabla u\cdot\nabla u_t+|u|^{p-1}uu_t\right)dx.
$$
After <integration by parts>, the middle term is $-\int(\Delta u)u_t$. Therefore
$$
\frac{dE}{dt}
=\int u_t\left(u_{tt}-\Delta u+u|u|^{p-1}\right)dx=0
$$
by the <defocusing semilinear wave equation>. Thus the total energy is conserved.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 154 / 2 / 2 / Solution

{title2=$u_{tt}-\Delta u+u|u|^{p-1}=0$}

The defocusing power-type wave equation has the conserved nonnegative energy
$$
E(u)=\int\left(\frac{|u_t|^2+|\nabla u|^2}{2}+\frac{|u|^{p+1}}{p+1}\right)dx.
$$


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