It is a problem in Evan's PDE. I want to prove the smooth solution of the following PDE is zero: $$u_{tt}-du_t-u_{xx}=0, (0,1)\times(0,T) $$$u|_{x=0}=u|_{x=1}=u|_{t=0}=u_t|_{t=0}=0$.

The hint is to use the energy $\frac{1}{2}(||\partial_tu||_{L^2[0,1]}+||\partial_xu||_{L^2[0,1]})$, but when I differentiate the energy, I can only get the first and third term of PDE

  • $\begingroup$ I suppose that it comes from applying the energy method in problem 10 of chapter 7 (p. 448 in 2nd ed., 2010). Note the $+$ sign in $+d u_t$ (telegraph equation). It would be better to provide exact citation of the book, and attempts. $\endgroup$ – EditPiAf Jan 23 '18 at 14:51
  • 1
    $\begingroup$ The energy is, I think, $\frac{1}{2}(||\partial_tu||_{L^2[0,1]}^2+||\partial_xu||_{L^2[0,1]}^2)$ $\endgroup$ – Robert Lewis Jan 23 '18 at 21:04

Let us assume that the solution is sufficiently smooth and $d<0$. The time-derivative of the energy $E(t) = \frac{1}{2}\left( \|u_t\|_{L^2[0,1]}^2 + \|u_x\|_{L^2[0,1]}^2\right)$ writes $$ \begin{aligned} \frac{\text{d}}{\text{d}t}E(t) &= \int_0^1 \left( u_{tt}\, u_{t} + u_{xt}\, u_{x} \right) \text{d}x \\ &= \int_0^1 \left( u_{tt}\, u_{t} + u_{tx}\, u_{x} \right) \text{d}x \\ &= \int_0^1 u_{tt}\, u_{t}\, \text{d}x + \left[ u_{t}\, u_{x} \right]_0^1 - \int_0^1 u_{t}\, u_{xx}\, \text{d}x \\ &= \int_0^1 \left(u_{tt} - u_{xx}\right) u_{t}\, \text{d}x \\ &= d\int_0^1 (u_{t})^2\, \text{d}x \\ &\leq 0 \, . \end{aligned} $$ To show that the energy is decreasing, we have used successively the equality of mixed derivatives, integration by parts, the fact that $u$ is constant-in-time at the boundaries $x=0$ and $x=1$ of the domain, and the PDE itself. Since at $t=0$, the energy is $E(0) = 0$ and the energy is always positive, we have shown that the energy is equal to zero for all $t$, which means that $u$ is constant in time and space. Now, since it equals zero at the boundaries of the domain, it is necessary that $u$ is identically zero.

  • $\begingroup$ Nice job, endorsed. I noticed you use $\frac{1}{2}(||\partial_tu||_{L^2[0,1]}^2+||\partial_xu||_{L^2[0,1]}^2)$ for the energy; see my comment on the question itself. $\endgroup$ – Robert Lewis Jan 23 '18 at 21:05
  • $\begingroup$ My pleasure, sir! $\endgroup$ – Robert Lewis Jan 24 '18 at 6:48
  • $\begingroup$ @EditPiAf how do we solve this problem when d is non negative? $\endgroup$ – User124356 Dec 3 '20 at 23:46
  • $\begingroup$ @User124356 If $d> 0$ the above problem for $u_{tt} - u_{xx} = du_t$ is not well-posed. (+ See the comments to OP) $\endgroup$ – EditPiAf Dec 4 '20 at 14:59
  • $\begingroup$ @EditPiAf Yes. Thanks. I am trying to solve math.stackexchange.com/q/3933714/727808. Can you please suggest some hint? $\endgroup$ – User124356 Dec 7 '20 at 19:11

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.