5
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ 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 $x=\infty$. $\endgroup$ – Math.StackExchange Dec 26 '14 at 7:44
  • $\begingroup$ I thought $u_{tt}-u_{xx}=0$ $t\in(0,\infty)$ clearly implies its symmetry about time reversal. $\endgroup$ – Math.StackExchange Dec 26 '14 at 8:16
  • 1
    $\begingroup$ 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. $\endgroup$ – Math.StackExchange Dec 26 '14 at 8:25
  • 1
    $\begingroup$ Thanks for showing me a nice example. I completely agree that compact support is the one which makes $u_x=0$ at $x=\pm\infty$. $\endgroup$ – Math.StackExchange Dec 26 '14 at 8:40
  • 1
    $\begingroup$ 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 '14 at 8:44
4
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.