Consider $\Delta u =f(x) , x \in \Omega $ and $\nabla u\cdot n +\alpha u = g(x) , x\in \partial\Omega $, where $n$ is outward normal. Can anyone give me a hind how to find sufficient condition on $\alpha$ so that the solution is unique. Thanks
Condition to have unique solution.
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0@RahulNarain : thank you , i didn't know that . – 2012-07-03
2 Answers
I think the condition is $\alpha \geq 1,$ which you need for coercivity of the bilinear form, which in turn can be got by using the fact that$\lVert \nabla u \rVert_{L^2(\Omega)}^2 + \lVert u \rVert^2_{L^2(\partial\Omega)} \geq C\lVert u \rVert_{H^1(\Omega)}^2$ for every $u \in H^1(\Omega).$
Added: If you multiply by a test function $v \in H^1(\Omega)$ and IBP, you get $\int_\Omega{\nabla u \nabla v} - \int_{\partial\Omega}{v\nabla u \cdot \nu} = \int_\Omega{-fv}$ where $\nu$ is the normal. Plugging in your boundary condition and moving the term involving $g$ on the other side: $\int_\Omega{\nabla u \nabla v} + \alpha\int_{\partial\Omega}{uv} = \int_{\partial\Omega}{gv} - \int_\Omega{fv}$ So your bilinear form $b(u,v)$ (for Lax-Milgram) is the LHS. For coercivity, we need $b(v,v) \geq C_1\lVert v \rVert^2_{H^1(\Omega)}$ for some $C_1$. We have $b(v,v) = \lVert \nabla v \rVert^2_{L^2(\Omega)} + \alpha \lVert v \rVert^2_{L^2(\partial\Omega)},$ which implies coercivity if $\alpha \geq 1$ because you can use the statement I gave at the top of this post.
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0@Court I would like to add to Court's solution. A sharper condition is \alpha > 0. Just note that $ b(v,v) \ge \min \{1, \alpha \} (\| \nabla v \|_{L^2 (\Omega)} + \| v \|_{L^2(\partial \Omega)} \ge \min \{1, \alpha \} C \| u \|_{H^1(\Omega)}$ – 2014-01-05
I think that this paper can be very useful and essentially contains the answer.