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

Solution 1:

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.