# Uniqueness of PDE via energy functional

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?

• 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 $0$? May 17, 2020 at 7:46
• Yes, that's right. You could write the argument up as an answer to your own question. May 19, 2020 at 10:28
• @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). May 19, 2020 at 10:37
• $u_1 - u_2$ solves the PDE with $0$ boundary conditions, yes. May 19, 2020 at 10:44

## 1 Answer

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 $$u_1 \equiv u_2.$$