[proofplan]
Differentiate the three terms in the energy. The kinetic term produces the pairing of $u_{tt}$ with $u_t$, the gradient term produces the pairing of $-\Delta u$ with $u_t$ after [integration by parts](/theorems/210), and the assumed chain rule for the potential term produces the pairing of $|u|^{p-1}u$ with $u_t$. These three terms add to the $L^2$ pairing of the [wave equation](/page/Wave%20Equation) with $u_t$, which is zero.
[/proofplan]
[step:Declare the localized integration by parts]
Fix $t\in I$. Let $J\subset I$ be a compact interval containing $t$ in its relative interior. By hypothesis there is $R_J>0$ such that $u(r,\cdot)$, $\nabla u(r,\cdot)$, and $\partial_tu(r,\cdot)$ vanish outside the Euclidean ball $B(0,R_J)\subset\mathbb R^n$ for every $r\in J$. Thus all spatial integrations below are effectively over a fixed bounded ball, and [integration by parts](/theorems/2098) has no boundary contribution.
Since $u(t,\cdot)\in H^2(\mathbb R^n)$ and $\partial_tu(t,\cdot)\in H^2(\mathbb R^n)$, the distributional Laplacian $\Delta u(t,\cdot)$ belongs to $L^2(\mathbb R^n)$ and the following identity is justified by the compact support and Sobolev integration by parts:
\begin{align*}
\int_{\mathbb R^n}\nabla u(t,x)\cdot \nabla \partial_tu(t,x)\,d\mathcal L^n(x)
=-\int_{\mathbb R^n}\Delta u(t,x)\partial_tu(t,x)\,d\mathcal L^n(x).
\end{align*}
[/step]
[step:Differentiate the kinetic and gradient terms]
Define the kinetic energy $K:I\to\mathbb R$ by
\begin{align*}
K(t):=\frac{1}{2}\int_{\mathbb R^n}|\partial_tu(t,x)|^2\,d\mathcal L^n(x).
\end{align*}
Because $u\in C^2(I;L^2(\mathbb R^n))$, the map $\partial_tu:I\to L^2(\mathbb R^n)$ is continuously differentiable with derivative $\partial_t^2u$. The Hilbert-space chain rule gives
\begin{align*}
K'(t)=\int_{\mathbb R^n}\partial_t^2u(t,x)\partial_tu(t,x)\,d\mathcal L^n(x).
\end{align*}
Define the gradient energy $G:I\to\mathbb R$ by
\begin{align*}
G(t):=\frac{1}{2}\int_{\mathbb R^n}|\nabla u(t,x)|^2\,d\mathcal L^n(x).
\end{align*}
Since $u\in C^1(I;H^2(\mathbb R^n))$, the map $\nabla u:I\to L^2(\mathbb R^n;\mathbb R^n)$ is continuously differentiable with derivative $\nabla\partial_tu$. Therefore
\begin{align*}
G'(t)=\int_{\mathbb R^n}\nabla u(t,x)\cdot\nabla\partial_tu(t,x)\,d\mathcal L^n(x).
\end{align*}
Using the integration by parts identity from the previous step,
\begin{align*}
G'(t)=-\int_{\mathbb R^n}\Delta u(t,x)\partial_tu(t,x)\,d\mathcal L^n(x).
\end{align*}
[guided]
The kinetic and gradient terms are ordinary differentiations in Hilbert spaces. For the kinetic term, the path $t\mapsto\partial_tu(t)$ lies in $C^1(I;L^2(\mathbb R^n))$, so differentiating the square norm gives
\begin{align*}
\frac{d}{dt}\left(\frac{1}{2}\|\partial_tu(t)\|_{L^2(\mathbb R^n)}^2\right)
=\langle \partial_t^2u(t),\partial_tu(t)\rangle_{L^2(\mathbb R^n)}.
\end{align*}
In integral notation this is
\begin{align*}
K'(t)=\int_{\mathbb R^n}\partial_t^2u(t,x)\partial_tu(t,x)\,d\mathcal L^n(x).
\end{align*}
For the gradient term, the path $t\mapsto\nabla u(t)$ lies in $C^1(I;L^2(\mathbb R^n;\mathbb R^n))$, so
\begin{align*}
\frac{d}{dt}\left(\frac{1}{2}\|\nabla u(t)\|_{L^2(\mathbb R^n)}^2\right)
=\int_{\mathbb R^n}\nabla u(t,x)\cdot\nabla\partial_tu(t,x)\,d\mathcal L^n(x).
\end{align*}
The compact-support hypothesis removes the boundary term in the spatial integration by parts, hence
\begin{align*}
\int_{\mathbb R^n}\nabla u(t,x)\cdot\nabla\partial_tu(t,x)\,d\mathcal L^n(x)
=-\int_{\mathbb R^n}\Delta u(t,x)\partial_tu(t,x)\,d\mathcal L^n(x).
\end{align*}
Thus the derivative of the sum of the quadratic terms is exactly the $L^2$ pairing of $\partial_t^2u(t)-\Delta u(t)$ with $\partial_tu(t)$.
[/guided]
[/step]
[step:Differentiate the potential term and use the equation]
Define the potential energy $P:I\to\mathbb R$ by
\begin{align*}
P(t):=\frac{1}{p+1}\int_{\mathbb R^n}|u(t,x)|^{p+1}\,d\mathcal L^n(x).
\end{align*}
The assumed differentiability of the $L^{p+1}$ term gives
\begin{align*}
P'(t)=\int_{\mathbb R^n}|u(t,x)|^{p-1}u(t,x)\partial_tu(t,x)\,d\mathcal L^n(x).
\end{align*}
Adding the identities for $K'(t)$, $G'(t)$, and $P'(t)$ yields
\begin{align*}
\frac{d}{dt}E[u(t),\partial_tu(t)]
=\int_{\mathbb R^n}\left(\partial_t^2u(t,x)-\Delta u(t,x)+|u(t,x)|^{p-1}u(t,x)\right)\partial_tu(t,x)\,d\mathcal L^n(x).
\end{align*}
The equation holds in $L^2(\mathbb R^n)$, while $\partial_tu(t,\cdot)\in L^2(\mathbb R^n)$, so the last integral is the $L^2$ [inner product](/page/Inner%20Product) of $0$ with $\partial_tu(t,\cdot)$. Hence
\begin{align*}
\frac{d}{dt}E[u(t),\partial_tu(t)]=0
\end{align*}
for every $t\in I$.
[/step]
[step:Conclude conservation on the interval]
The preceding step shows that the real-valued function
\begin{align*}
t\mapsto E[u(t),\partial_tu(t)]
\end{align*}
has derivative zero on $I$. Therefore it is constant on each connected component of $I$. Since $I$ is an interval, it is connected, and so for every $s,t\in I$,
\begin{align*}
E[u(t),\partial_tu(t)]=E[u(s),\partial_tu(s)].
\end{align*}
This proves the energy identity.
[/step]
