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Assume the pde:

$$ u_{tt}(t,x) = c^2u_{xx}(t,x) + \sigma u_{txx}(t,x) -\mu u_{t}(t,x), \quad x \in [0,L], t>0 $$ $$ u_x(t,0) = u(t,L) = 0 $$ $$ u(0,x) = \phi(x), u_t(0,x) = \theta(x), x\in[0,L] $$ $$ \phi(L) = \theta(L) = 0, \phi'(0) = \theta '(0) = 0. $$ and the energy functional: $$ V(t) = \int_{0}^L\frac12u_t^2(t,x) + \frac{c^2}{2}u_x^2(t,x)dx $$

To prove uniqueness we'll assume $u_1$, $u_2$ are both solutions and then define $u$ as $u := u_1 - u_2$.

Then, we observe that $$u(0,x) = u_1(0,x) - u_2(0,x) = \phi(x) - \phi(x) \equiv 0$$ $$u_t(0,x) = u_{1,t}(0,x) - u_{2,t}(0,x) = \theta(x) - \theta(x) \equiv 0$$

So $u_x(0,x) = u_t(0,x) = 0$. Thus we have

$$ V(0) = \int_0^L 0 \, dx = 0 $$

Also, I have already shown that $V(t) \leq V(0)$ so

$$ V(t) \leq 0 $$

and since $V(t) \geq 0$, we have $V(t) \equiv 0$. Then since the integrand is non-negative:

$$ \frac12u_t^2(t,x) + \frac{c^2}{2}u_x^2(t,x) \equiv 0 \quad \quad (1) $$

Question:

Does $(1)$ guarantee that $u \equiv 0$ and why?

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    $\begingroup$ Is it because $u_t(t,x) = u_x(t,x) \equiv 0$ suggest $u(t,x)$ is a constant and due to the boundary conditions is identically $0$? $\endgroup$
    – Paris
    May 17 '20 at 7:46
  • $\begingroup$ Yes, that's right. You could write the argument up as an answer to your own question. $\endgroup$ May 19 '20 at 10:28
  • $\begingroup$ @RhysSteele Another quick question: $u = u_1 - u_2$ is itself a solution of the pde, right? (at least I assumed it was when using the functional). $\endgroup$
    – Paris
    May 19 '20 at 10:37
  • $\begingroup$ $u_1 - u_2$ solves the PDE with $0$ boundary conditions, yes. $\endgroup$ May 19 '20 at 10:44
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The equation

$$ \frac12u_t^2(t,x) + \frac{c^2}{2}u_x^2(t,x) = 0 $$

suggests that

$$ u_t(t,x) \equiv 0 \quad \text{and} \quad u_x(t,x) \equiv 0 $$

and thus $u(t,x) = \text{constant}$. Finally, the boundary condition $u(t,L) = 0$ implies that $$ u(t,x) \equiv 0 $$

and thus $u_1 \equiv u_2.$

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