[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*}[/step]

[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: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]