Wave equation: show eventually $\int_{\mathbb{R}}u_x^2 = \int_{\mathbb{R}}u_t^2$ Suppose $u$ solves the wave equation in $\mathbb{R}$ and has compactly supported initial data $f(x) = u(x,0)$ and $g(x)=u_t(x,0)$. Show that the "kinetic energy" $\int_{\mathbb{R}}u_t^2$ eventually equals the "potential energy" $\int_{\mathbb{R}}u_x^2$.
My attempt so far: When I expand $\int_{\mathbb{R}}u_t^2 - u_x^2$ using d'Alembert's formula,  I get $$\int_{\mathbb{R}}f'(x+t)f'(x-t)-g(x+t)g(x-t)\\ + f'(x+t)g(x+t) - f'(x-t)g(x-t)dx.$$
The first two terms will eventually be zero, because at least one of the two factors of each will be zero (since the initial data has compact support). I need to find a way to make the second two terms zero. I'm trying to do it by integration by parts, noting that the last two terms can be written as $$\left.f'g\right]_{x-t}^{x+t}$$ or $$\int_{x-t}^{x+t} f''g+g'f',$$
but this isn't getting me anywhere.
 A: Don't do integration by parts.  See that the third and fourth term are equal by using substitution.  Once you see it, you will see it is very easy.
A: By D'Alembert's formula:
\begin{alignat*}{2}
u(x,t) &= \frac{1}{2}\left[f(x+t)+f(x-t)\right]+\frac{1}{2}\int_{x-t}^{x+t}g(s)ds.
\end{alignat*}
Let $M$ be a positive constant such that 
\begin{alignat*}{2}
\text{supp}(f^{\prime}) \subseteq \left[ -M, M \right] \quad\text{and}\quad \text{supp}(g) \subseteq \left[ -M, M \right].
\end{alignat*}
First note:
For $t>M$,
\begin{alignat*}{2}
-M \le x-t \le M \Leftrightarrow 0 < t-M \le x \le M+t
\end{alignat*}
and
\begin{alignat*}{2}
-M \le x+t \le M \Leftrightarrow -t-M \le x \le M-t < 0.
\end{alignat*}
Second note:
\begin{alignat*}{2}
t > M \Rightarrow f(x+t) = g(x+t) = 0 \quad\forall x > 0
\end{alignat*}
Therefore,
\begin{alignat*}{2}
u_{t}^{2} &= \frac{1}{4}f^{\prime}(x-t)^{2}+\frac{1}{4}g(x-t)^{2}-\frac{1}{2}f^{\prime}(x-t)g(x-t) \\
&= u_{x}^{2}.  
\end{alignat*}
Similarly, $\forall x < 0$,
\begin{alignat*}{2}
u_{t}^{2} &= \frac{1}{4}f^{\prime}(x+t)^{2}+\frac{1}{4}g(x+t)^{2}-\frac{1}{2}f^{\prime}(x+t)g(x+t) \\
&= u_{x}^{2}.
\end{alignat*}
And for $x=0$,
\begin{alignat*}{2}
f^{\prime}(t) = f^{\prime}(-t) = g(t) = g(-t) = 0.
\end{alignat*}
Thus, 
\begin{alignat*}{2}
t &>M \Rightarrow \\
\int u_{t}^{2} &= \int u_{x}^{2}.
\end{alignat*}
