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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

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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