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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When $d>0$ the uniqueness still follows. For the energy $E(t):=\frac12( \|u_t\|_{L^2}^2+\|u_x\|_{L^2}^2)$, we get $$E'(t)= d\int_0^1(u_t)^2~dx\leq d\left(\int_0^1(u_t)^2+(u_x)^2~dx\right)=2d E(t).$$ Then multiplying $e^{-2dt}$ on both sides, we find $$\frac{d}{dt}\left(e^{-2dt}E(t)\right)\leq 0.$$ That means $e^{-2dt}E(t)$ is decreasing in time, but $E(0)=0$ forces $e^{-2dt}E(t)=0$ for $0<t<T$.
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.