# The energy method for $u_{tt}-du_t-u_{xx}=0, (0,1)\times(0,T)$

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

• 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. – EditPiAf Jan 23 '18 at 14:51
• The energy is, I think, $\frac{1}{2}(||\partial_tu||_{L^2[0,1]}^2+||\partial_xu||_{L^2[0,1]}^2)$ – 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.
• 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. – Robert Lewis Jan 23 '18 at 21:05
• @User124356 If $d> 0$ the above problem for $u_{tt} - u_{xx} = du_t$ is not well-posed. (+ See the comments to OP) – EditPiAf Dec 4 '20 at 14:59