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Consider the Neumann problem $$\Delta u = c, \qquad x \in \Omega = B(\mathbf{x},1) \subseteq \mathbb{R}^2$$ equipped with the boundary condition $\nabla u \cdot \mathbf{n} = 1 \quad \forall x \in \partial \Omega$ where $c$ is a constant.

Find all $c$ for which there exists a solution and find all these solutions.

The boundary condition suggests usage of the Green theorem: $$\int_\Omega \varphi \Delta u = - \int \nabla \varphi \nabla u + \int_{\partial \Omega} \varphi \nabla u \cdot \mathbf{n}$$

Which results in $$c \int_\Omega\varphi = - \int \nabla \varphi \nabla u + \int_{\partial \Omega} \varphi$$ If I would take $\varphi = u$ then this results in $$\|\nabla u\|^2_{L^2} = \int_{\partial \Omega} u - c \int_\Omega u$$

But how could I continue? I don't really see how to reach a answer to the question. I guess I would look for a upper bound of $c$?

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WLOG we may assume $\bf x = 0$, i.e. $\Omega$ is the unit disk centred at $0$. Since everything is rotationally invariant, if there is a solution there is one that has radial symmetry. The radially symmetric solutions of $\Delta u = c$ are $u(r,\theta) = f(r) = c r^2/4 + a \ln(r) + b$, and for this to be continuous at $r=0$ we need $a=0$. Then we want $f'(1) = c/2 = 1$. So we need $c=2$, in which case $u(r,\theta) = r^2/2 + b$ is a solution. Now all you have to do is show that these are all the solutions.

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  • $\begingroup$ I have already shown that the problem $\Delta u = f$ in $\Omega$, $\nabla u \cdot \mathbf{n} = g$ on $\Gamma$ has a unique solution up to a constant. Which would imply here $u(r,\theta) = \frac{r^2}{2} + b$ to be unique up to constant which is taken care of through the $b$. I guess this would be valid :) $\endgroup$ – dietervdf Jan 29 '17 at 23:58

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