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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

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    what do you mean by $\nabla\cdot n$? Do you mean $\nabla u\cdot n$ instead?2012-07-03
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    @Paul : thanks for pointing out.2012-07-03
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    For future reference, you should write the operators as `\Delta` and `\nabla` instead of `\triangle` and `\triangledown`.2012-07-03
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    @RahulNarain : thank you , i didn't know that .2012-07-03

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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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    Sir , can you tell me how can i get the condition on $\alpha$ using the fact that u have given ?2012-07-04
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    @Theorem I updated the answer.2012-07-04
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    @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
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I think that this paper can be very useful and essentially contains the answer.