# Equipartition of energy

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?

• 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 $x=\infty$. Dec 26, 2014 at 7:44
• I thought $u_{tt}-u_{xx}=0$ $t\in(0,\infty)$ clearly implies its symmetry about time reversal. Dec 26, 2014 at 8:16
• 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. Dec 26, 2014 at 8:25
• Thanks for showing me a nice example. I completely agree that compact support is the one which makes $u_x=0$ at $x=\pm\infty$. Dec 26, 2014 at 8:40
• 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. Dec 26, 2014 at 8:44

(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 $t>M$, $k(t)=p(t)$.