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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.

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2 Answers 2

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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.

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  • $\begingroup$ I'm having trouble seeing intuitively how they could be equal. If $x+t$ is a point where $f'$ and $g$ are nonzero, and $x-t$ is outside the support of $f$ and $g$, then they are unequal, no? $\endgroup$
    – Eric Auld
    Nov 16, 2013 at 4:01
  • $\begingroup$ Oh, perhaps you're saying the integrals are equal, excuse me. Yes, you're right, it is quite easy. $\endgroup$
    – Eric Auld
    Nov 16, 2013 at 4:03
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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*}

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