Show Uniqueness of Solution for Boundary Value Problem

Question

Let $G \subseteq R^n$ be a simple, connected and bounded region with smooth boundary and let $f : \overline G \to \mathbb R$, $g : \partial G \to \mathbb R$ be continuous. Show that the following boundary value problem $\Delta u = f$ and $$ \frac{\partial u}{\partial n} + \alpha u \Big|_{\partial G} = g, \quad \alpha > 0 $$ has at most one solution.

Any hints how to show this, most of the time I can show uniqueness by applying the maximum principle, but here I do not see how to use it?

Answer 1

Assume there are two solutions, $u_1,u_2$, satisfying the bvp. Let $w=u_1-u_2$, then $w$ satisfies $$\Delta w=0$$ on $G$, and $$\partial w/\partial n+\alpha w|_{\partial G}=0$$ now multiply by $w$ and integrate by parts: $$0=\int_G w\Delta w dx =\int_G\nabla\cdot(w\nabla w)-\nabla w\cdot \nabla w dx=\int_{\partial G}w\nabla w\cdot n dS-\int_G\nabla w\cdot \nabla wdx$$

$$=-(\int_{\partial G}\alpha w^2 dS+\int_G\nabla w\cdot \nabla wdx)$$

thus $\nabla w=0$ in $G\Rightarrow w=C$ in $G$.

but also we have $w=0$ on $\partial G$, so by the continuity of $w$, we have $$0=w=u_1-u_2$$

Thus $u_1=u_2$.

Answer 2

Assume $g=0$. Then $$ \begin{align} 0=\int_{\Omega} u\nabla^{2}u\,dV & = \int_{\Omega}(\nabla\cdot(u\nabla u)-|\nabla u|^{2})dV \\ & = \int_{\partial\Omega}u\frac{\partial u}{\partial n}\,dS-\int_{\Omega}|\nabla u|^{2}\,dV \\ & = -\alpha\int_{\partial\Omega}u^{2}dS-\int_{\Omega}|\nabla u|^{2}\,dV. \end{align} $$ NOTE: The reason I assumed $g=0$ was because the difference of two solutions will be a solution of your problem with $g=0$, and that's what you want to show is $0$.