Let $u$ solve the initial-value problem or the wave equation in one dimension: $$\begin{cases}u_{tt}-u_{xx}=0 & \text{in } \mathbb{R} \times (0,\infty) \\ u = g, u_t = h & \text{on } \mathbb{R} \times \{t=0\}. \end{cases}$$ Suppose $g,h$ have compact support. The kinetic energy is $k(t) := \frac 12 \int_{-\infty}^\infty u_t^2(x,t) \, dx$ and the potential energy is $p(t) := \frac 12 \int_{-\infty}^\infty u_x^2(x,t) \, dx$. Prove

(a) $k(t)+p(t)$ is constant in $t$,

(b) $k(t)=p(t)$ for all large enough times $t$.

This is Chapter 2, Exercise 24 of PDE Evans, 2nd edition.

I am only doing part (a) right now; my work is shown below:

Define $$e(t):=k(t)+p(t)=\int_{-\infty}^\infty u_t^2+u_x^2 \, dx.$$ Then $$e'(t)=\frac 12 \int_{-\infty}^\infty 2u_tu_{tt}+2u_{x}u_{xt} \, dx= \int_{-\infty}^\infty u_tu_{tt}+u_{x}u_{xt} \, dx.$$

Now, I want to get $e'(t)=0$ so that $e(t)$ is constant. How can I go about doing this? I do know that I can use $u_{tt}-u_{xx}=0$, if $t > 0$.

With $t>0$ into mind, do I have to consider the $t=0$ case separately? Or can I treat both cases together as $t \ge 0$, since $g$ and $h$ have compact support?

  • 4
    $\begingroup$ How about $u_{t}u_{tt}+u_xu_{xt}=u_{t}u_{xx}+u_xu_{xt}=\dfrac{d(u_xu_t)}{dx}$? I think you can use the boundary condition that wave function and its time or space derivative is 0 at $x=\infty$. $\endgroup$ Dec 26, 2014 at 7:44
  • $\begingroup$ I thought $u_{tt}-u_{xx}=0$ $t\in(0,\infty)$ clearly implies its symmetry about time reversal. $\endgroup$ Dec 26, 2014 at 8:16
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    $\begingroup$ Wave equation's space derivative should be always continuous except at the point where potential is infinity. So, I don't think we should pay a special attention to the point at $x=0$ in calculating the integral. Since the derivative is continuous, $\int_{-\infty}^{\infty} \dfrac{d(u_xu_t)}{dx}dx=[u_xu_t]^{\infty}_{-\infty}$. Since the space derivative of wave function at $x=\pm\infty$ is 0 (otherwise the wave function isn't integrable), the integral is 0. $\endgroup$ Dec 26, 2014 at 8:25
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    $\begingroup$ Thanks for showing me a nice example. I completely agree that compact support is the one which makes $u_x=0$ at $x=\pm\infty$. $\endgroup$ Dec 26, 2014 at 8:40
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    $\begingroup$ And thanks to you also for your help, especially in the initial comment, with your observations of using $u_{xx}=u_{tt}$ and the product rule. $\endgroup$
    – Cookie
    Dec 26, 2014 at 8:44

1 Answer 1


(a) We define$$ e(t)\equiv k(t)+p(t)=\frac12\int_{-\infty}^\infty \left( u_t^2+u_x^2\right)\,dx. $$ Since $g,h$ have compact supports, we have that\begin{align*} \frac{d}{dt}e(t)&=\frac12\int_{-\infty}^\infty 2u_tu_{tt}+2u_xu_{xt}\,dx\\ &=\int_{-\infty}^\infty u_tu_{tt}\,dx-\int_{-\infty}^\infty u_{xx}u_t\,dx\\ &=\int_{-\infty}^\infty u_t(u_{tt}-u_{xx})\,dx=0. \end{align*} Hence, $e(t)\equiv e(0)$ and so $k(t)+p(t)$ is constant in $t$.

(b) By d'Alembert's formula, we have$$ u(x,t)\frac12 (g(x+t)+g(x-t))+\frac12 \int_{x-t}^{x+t} h(y)\,dy. $$ Thus\begin{align*} u_t&=\frac12 (g'(x+t)-g'(x-t))+\frac12(h(x+t)+h(x-t)),\\ u_x&=\frac12 (g'(x+t)+g'(x-t))+\frac12(h(x+t)-h(x-t)) \end{align*} We assume that there exists a positive constant $M$ so that $[-M,M]\supseteq supp(g')$ and $[-M,M]\supseteq supp(h)$. Note that for a fixed $t>M$,$$ -M\leq x-t\leq M\Leftrightarrow 0<t-M\leq x\leq t+M$$and $$-M\leq x+t \leq M\Leftrightarrow -t-M\leq x\leq -t+M<0.$$ Thus, when $t>M$,

$\,\,\,\,\,$(i) $0<t-M\leq x\leq t+M$. Then,$$ h(x+t)=g(x+t)=0 $$and so$$ u_t^2=\frac14 g'(x-t)^2+\frac14 h(x-t)^2-\frac12 g'(x-t)h(x-t)=u_x^2. $$

$\,\,\,\,\,$(ii) $-t-M\leq x\leq -t+M<0$. Then,$$ u_t^2=\frac14 g'(x+t)^2+\frac14 h(x+t)^2+\frac12 g'(x+t)h(x+t)=u_x^2. $$

$\,\,\,\,\,$(iii) Otherwise,$$ g'(x+t)=g'(x-t)=h(x+t)=h(x-t)=0. $$ Hence, combining all the cases, it follows that, when $t>M$, $k(t)=p(t)$.


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