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.


1 Answer 1


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, 2017 at 18:00

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