[proofplan]
We prove the identity by passing from the [wave equation](/page/Wave%20Equation) to the two characteristic derivatives $F=\partial_tu+c\partial_xu$ and $G=\partial_tu-c\partial_xu$. The wave equation implies that $F$ and $G$ are transported rigidly along opposite characteristic families, so their $L^2$ norms are invariant under translation. Finally, the energy is exactly one quarter of the sum of these two squared $L^2$ norms, which gives conservation of $E(t)$.
[/proofplan]
[step:Define the characteristic derivatives and prove their transport equations]
Define the functions
\begin{align*}
F:[0,T]\times\mathbb{R} &\to \mathbb{R}
\end{align*}
by
\begin{align*}
F(t,x)=\partial_tu(t,x)+c\partial_xu(t,x),
\end{align*}
and
\begin{align*}
G:[0,T]\times\mathbb{R} &\to \mathbb{R}
\end{align*}
by
\begin{align*}
G(t,x)=\partial_tu(t,x)-c\partial_xu(t,x).
\end{align*}
Since $u\in C^2([0,T]\times\mathbb{R})$, both $F$ and $G$ are $C^1$ functions on $[0,T]\times\mathbb{R}$. For every $(t,x)\in[0,T]\times\mathbb{R}$,
\begin{align*}
\partial_tF(t,x)-c\partial_xF(t,x)
=
\partial_{tt}u(t,x)-c^2\partial_{xx}u(t,x)
=
0.
\end{align*}
Similarly,
\begin{align*}
\partial_tG(t,x)+c\partial_xG(t,x)
=
\partial_{tt}u(t,x)-c^2\partial_{xx}u(t,x)
=
0.
\end{align*}
[guided]
The quantities $F$ and $G$ are chosen because the wave operator factors into first-order transport operators. Define
\begin{align*}
F:[0,T]\times\mathbb{R} &\to \mathbb{R}
\end{align*}
by
\begin{align*}
F(t,x)=\partial_tu(t,x)+c\partial_xu(t,x),
\end{align*}
and define
\begin{align*}
G:[0,T]\times\mathbb{R} &\to \mathbb{R}
\end{align*}
by
\begin{align*}
G(t,x)=\partial_tu(t,x)-c\partial_xu(t,x).
\end{align*}
Because $u$ is $C^2$, the first derivatives $\partial_tu$ and $\partial_xu$ are $C^1$, hence $F$ and $G$ are $C^1$.
Now compute the directional derivative of $F$ along the transport operator $\partial_t-c\partial_x$:
\begin{align*}
\partial_tF(t,x)-c\partial_xF(t,x)=\partial_{tt}u(t,x)+c\partial_{xt}u(t,x)-c\partial_{tx}u(t,x)-c^2\partial_{xx}u(t,x).
\end{align*}
Since $u\in C^2$, the mixed partial derivatives agree:
\begin{align*}
\partial_{xt}u(t,x)=\partial_{tx}u(t,x).
\end{align*}
Therefore the middle terms cancel and
\begin{align*}
\partial_tF(t,x)-c\partial_xF(t,x)
=
\partial_{tt}u(t,x)-c^2\partial_{xx}u(t,x)
=
0.
\end{align*}
The same calculation for $G$ with the transport operator $\partial_t+c\partial_x$ gives
\begin{align*}
\partial_tG(t,x)+c\partial_xG(t,x)
=
\partial_{tt}u(t,x)-c^2\partial_{xx}u(t,x)
=
0.
\end{align*}
Thus $F$ is transported with speed $-c$, while $G$ is transported with speed $c$.
[/guided]
[/step]
[step:Recover the characteristic derivatives from their initial values]
Fix $t\in[0,T]$ and $x\in\mathbb{R}$. Define
\begin{align*}
\phi_F:[0,t] \to \mathbb{R}
\end{align*}
by
\begin{align*}
\phi_F(s)=F(s,x+c(t-s)).
\end{align*}
By the chain rule,
\begin{align*}
\phi_F'(s)=\partial_tF(s,x+c(t-s))-c\partial_xF(s,x+c(t-s))=0.
\end{align*}
Hence $\phi_F$ is constant on $[0,t]$, and evaluating at $s=t$ and $s=0$ gives
\begin{align*}
F(t,x)=F(0,x+ct).
\end{align*}
Similarly define
\begin{align*}
\phi_G:[0,t] \to \mathbb{R}
\end{align*}
by
\begin{align*}
\phi_G(s)=G(s,x-c(t-s)).
\end{align*}
Then
\begin{align*}
\phi_G'(s)=\partial_tG(s,x-c(t-s))+c\partial_xG(s,x-c(t-s))=0,
\end{align*}
so
\begin{align*}
G(t,x)=G(0,x-ct).
\end{align*}
[/step]
[step:Use translation invariance of Lebesgue measure to conserve the two $L^2$ norms]
For each $t\in[0,T]$, the functions $F(t,\cdot)$ and $G(t,\cdot)$ belong to $L^2(\mathbb{R})$ because they are linear combinations of $\partial_tu(t,\cdot)$ and $\partial_xu(t,\cdot)$. Using the identities from the previous step and the substitutions $y=x+ct$ and $z=x-ct$, for which $d\mathcal{L}^1(y)=d\mathcal{L}^1(x)$ and $d\mathcal{L}^1(z)=d\mathcal{L}^1(x)$, we obtain
\begin{align*}
\int_{\mathbb{R}} |F(t,x)|^2\,d\mathcal{L}^1(x)=\int_{\mathbb{R}} |F(0,y)|^2\,d\mathcal{L}^1(y)
\end{align*}
and
\begin{align*}
\int_{\mathbb{R}} |G(t,x)|^2\,d\mathcal{L}^1(x)=\int_{\mathbb{R}} |G(0,z)|^2\,d\mathcal{L}^1(z).
\end{align*}
[/step]
[step:Rewrite the energy as the sum of the characteristic energies]
For every $(t,x)\in[0,T]\times\mathbb{R}$, expanding the squares gives
\begin{align*}
|F(t,x)|^2+|G(t,x)|^2=|\partial_tu(t,x)+c\partial_xu(t,x)|^2+|\partial_tu(t,x)-c\partial_xu(t,x)|^2.
\end{align*}
The cross terms cancel, so
\begin{align*}
|F(t,x)|^2+|G(t,x)|^2=2(\partial_tu(t,x))^2+2c^2(\partial_xu(t,x))^2.
\end{align*}
Therefore
\begin{align*}
E(t)=\frac{1}{4}\int_{\mathbb{R}}\left(|F(t,x)|^2+|G(t,x)|^2\right)\,d\mathcal{L}^1(x).
\end{align*}
Using the two conserved $L^2$ norms from the previous step,
\begin{align*}
E(t)=\frac{1}{4}\int_{\mathbb{R}}\left(|F(0,x)|^2+|G(0,x)|^2\right)\,d\mathcal{L}^1(x)=E(0).
\end{align*}
Since $t\in[0,T]$ was arbitrary, $E(t)=E(0)$ for every $t\in[0,T]$.
[/step]
