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I'm trying to prove uniqueness of solutions of the forced wave equation given standard boundary conditions in the Dirichlet, Neumann and Radiation cases.

$$ \frac{\partial^2u}{\partial t^2}=c^2\nabla^2u+s(\mathbf{r},t), \mathbf{r} \in \Omega $$

Let A denote the portion of $\partial \Omega$ on which we have a Dirichlet boundary condition.

Let B denote the portion of $\partial \Omega$ on which we have a Neumann boundary condition.

Let C denote the portion of $\partial \Omega$ on which we have a Radiation boundary condition.

Then we have boundary conditions

$$u(\mathbf{r},t) = f(\mathbf{r},t), \mathbf{r} \in A$$

$$\mathbf{\hat{n}} \centerdot \nabla u(\mathbf{r},t) = g(\mathbf{r},t), \mathbf{r} \in B$$

$$u(\mathbf{r},t)+ \alpha(\mathbf{r}) \mathbf{\hat{n}} \centerdot \nabla u(\mathbf{r},t) = h(\mathbf{r},t), \mathbf{r} \in C$$

and initial conditions $u(r,0) = u_0(r) $ and $ \frac{\partial u}{\partial t} (r,0) = v_0(r)$

I began with the standard technique of assuming the existence of two different solutions, $u_1$ and $u_2$, and their difference, $U = u_1-u_2$

I found that $U$ satisifies the unforced wave equation and that the boundary conditions become homogenous for U, and I believe that we were asked to make use of the energy functional to prove uniqueness. I believe I have a solution for the Dirichlet boundary conditions, by considering the time derivative of the energy functional and the initial conditions for $ \frac{\partial U}{\partial t} = 0 $

I cannot work out how to prove uniqueness for the Neumann and Radiation boundary regions, does that require a similar approach, or an entirely different one?

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In my opinion, the term radiating is better suited to diffusion equation; I prefer to think of that part of the boundary as being elastic: it resists the motion, but does not make it impossible. This means that the boundary does some nonzero work, and we should include it in the computation of energy.

Without the elastic part, the energy functional would be something like $E(t)=\frac12 \int_\Omega (U_t^2+c^2 |\nabla U|^2)$. With it, we need $$E(t)=\frac12 \int_\Omega (U_t^2+c^2 |\nabla U|^2) +\frac12 \int_C \alpha^{-1} c^2 U^2 $$

Differentiating with respect to $t$ and using the PDE yields $$E'(t) = \int_\Omega ( c^2U_t \nabla^2 U+c^2\nabla U \cdot \nabla U_t )+ c^2 \int_C \alpha^{-1} UU_t $$ Integrate the first term by parts: $$E'(t) = c^2 \int_{\partial\Omega}U_t U_\nu + \int_\Omega (-c^2 \nabla U_t \cdot \nabla U+c^2\nabla U \cdot \nabla U_t )+ c^2 \int_C \alpha^{-1} UU_t$$ where $U_\nu$ is the normal derivative. The integral over $\Omega$ vanishes, and it remains to show that $$U_t U_\nu + \alpha^{-1} U U_t \chi_C \le 0\quad \text{ on }\partial \Omega$$ And this is true thanks to the boundary conditions: check it on $A,B,C$ separately.

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