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?
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 $u_1 \equiv u_2.$