I'm working on an old problem from a previous exam.

Suppose $\alpha \geq 0$ and $T>0$ be given. Prove uniqueness of classical solutions to the initial boundary value problem

\begin{cases} u_t-\Delta u + \alpha u = f & \text{in } \Omega\times(0,T)\\ u=h & \text{on } \partial\Omega\times[0,T]\\ u=g& \text{on } \Omega\times\{t=0\} \end{cases}

by means of an energy method.

Through reading this post, I think that I should define



$$\frac{d}{dt}E(t)=\frac{d}{dt}\left(\int_{\Omega} u^2 dx\right)=2\int_{\Omega} u u_t dx=2\int_{\Omega}u(\Delta u -\alpha u)dx\le 2\int_{\Omega}u\Delta u dx \le 2\int_{\Omega}|u|^2dx$$ after which you can use Gronwall's inequality to get that $E(t)$ is identically zero.

Is this the correct approach for this problem? I'm not sure if I should do this differently because of the non-homogeneous term $f$.

Another approach is shown in this post where one substitutes $v(x,t) = e^{\alpha t}u(x,t)$ to get rid of the non-homogeneous term $f$.

  • $\begingroup$ Let $u_1$ and $u_2$ be two solutions to the problem. Then you can show that $v=u_2-u_1$ is the solution to the homogeneous problem, with zeros B.C.s and I.C.s, and is therefore identically zero $\endgroup$ – Dylan May 21 at 8:33
  • $\begingroup$ Do you need to define $v(x,t)=e^{-\alpha t} u(x,t)$ to show that $v=u_1-u_2$ is identically equal to 0? Can this be done without applying $v(x,t)=e^{-\alpha t}u(x,t)$ and using the fact that $v$ satisfies the above equation with 0 boundary conditions. So, $E_v(0)=0 \Rightarrow E_v(t) \leq E_v(0)=0 \Rightarrow v_t=0 \Rightarrow v=const=0$. Hence, $u_1=u_2$. $\endgroup$ – Axion004 May 21 at 15:07

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