Let the initial value problem for the wave equation:

$u_{tt}-u_{xx}=0\;\;\forall x \in \mathbb R\;\;,t\gt 0\\ u(x,0)=u_t(x,0)=0\;\;,\vert x \vert \gt R\;\;\text{for}\;R\gt 0$

Prove that: $\int_{\mathbb R} u^2\;dx\;=c_1t+c_2\;\;,t \gt R\text{ and }c_1,c_2 \text{ constants}$.

The professor also gave us some hints:

  1. Prove the differential identity: $\;2v(v_{tt}-v_{xx})\;=\;2({v_x}^2-{v_t}^2)\;+\;(v^2)_{tt}\;-\;2(vv_x)_x\; $ where $v\equiv v(x,t)$ is a smooth function.
  2. Use the above identity and the equipartition of energy.

Well, I proved the given identity and then I wrote:

$\int_{\mathbb R} 2u(u_{tt}-u_{xx})\;dx\;=\int_{\mathbb R} 2({u_x}^2-{u_t}^2)\;+\;(u^2)_{tt}\;-\;2(uu_x)_x\;dx\;\Rightarrow \int_{\mathbb R} 2({u_x}^2-{u_t}^2)\;+\;(u^2)_{tt}\;-\;2(uu_x)_x\;dx\;=0\;$

Now, $\int_{\mathbb R} ({u_x}^2-{u_t}^2)\; dx\;=0\;$ for $\;t\gt R\;$ by the equipartition of energy.

If I could get rid of the term $2(uu_x)_x\;$ then $\int_{\mathbb R} (u^2)_{tt} dx \;$ could be written as $(\int_{\mathbb R} u^2\;\;dx)_{tt}\;$ and the exercise is complete.

At this point I've been stuck! I don't know how to procced..I would appreciate any help. Hints and other solutions are of course welcome.

Thanks in advance.


By the fundamental theorem of calculus, the integral of the derivative of a function is given by the difference of the values of that function. The function at hand is $uu_x$, which vanishes for all large $x$. Hence, $$\int_{\mathbb{R}} (uu_x)_x = 0$$ which completes the proof that $\left(\int_{\mathbb{R}} u^2\right)_{tt} = 0$.

  • $\begingroup$ Well that was quite easy! I've got exams tomorrow and I am in panic, I guess :) Thanks a lot for your time! $\endgroup$ Jan 22 '17 at 18:00

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