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I have the following Neumann problem, and I don't know what I'm missing:

Let $\Omega = D(0,1)\setminus\{0\}$. I want to find all $f \in C^2(\Omega)$, such that

$$\Delta_{\Bbb R^2} f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = r \cos\varphi = g(r,\varphi) \qquad (x,y) \in \Omega$$

with

$$\frac{\partial f}{\partial r} (1,\varphi) = 0.$$

So, what I did was to convert the Laplacian to polar coordinates and decompose $f(r,\varphi) = \sum_{n\in\Bbb Z} f_n(r) \cdot e^{in\varphi}$ as well as $g(r,\varphi) = \sum_{n\in\Bbb Z} g_n(r) \cdot e^{in\varphi}$ into its respective fourier series in order to get an equation for the coefficients $f_n(r)$ and $g_n(r)$. Obviously $g_n = 0$ except $g_{\pm 1} = \frac{r}{2}$. So I get

\begin{align} \text{for} \quad n \in \{-1,1\}&: \qquad \frac{1}{r} \frac{d}{dr} \left( r \frac{df_{\pm 1}(r)}{dr} \right) - \frac{1}{r^2} f_{\pm 1}(r) = \frac{r}{2} \\ \text{for} \quad n \notin \{-1,1\}, \,\, n \in \Bbb Z&: \qquad \frac{1}{r} \frac{d}{dr} \left( r \frac{df_{n}(r)}{dr} \right) - \frac{n^2}{r^2} f_{n}(r) = 0 \end{align}

Using the ansatz $f_n = c_n r^\alpha$, where $\alpha \in \Bbb R$, I get

\begin{align} \text{for} \quad n \in \{-1,1\}&: \qquad f_{\pm 1}(r) = \frac{1}{16} r^3 \\ \text{for} \quad n \notin \{-1,1\}, \,\, n \in \Bbb Z&: \qquad f_n(r) = c_n r^n \end{align}

and hence

$$f(r,\varphi) = \sum_{n\in\Bbb Z} f_n(r) \cdot e^{in\varphi} = \frac{1}{8} r^3 \cos \varphi + \sum_{n\in\Bbb Z \setminus \{-1,1\}} c_n r^n e^{in\varphi} \overset{\color{red}{?}}{=} \frac{1}{8} r^3 \cos \varphi + \frac{a_0}{2} + \sum_{\color{red}{n \geq 2}} r^n (A_n \cos n\varphi + B_n \sin n\varphi) \qquad a_0 = \text{const.}$$

The Neumann condition yields

$$\frac{\partial f}{\partial r} (1,\varphi) = {3 \over 8} \cos \varphi + \sum_{\color{red}{n \geq 2}} n (A_n \cos n\varphi + B_n \sin n\varphi) \overset{!}{=} 0$$

The crucial part is the summation index $n$. If I would have $n \geq 1$ it would immediately follow that $B_n = 0$ and $A_n = 0 \,\, \forall n$ except $A_1 = - {3 \over 8}$, which would imply

$$f(r,\varphi) = \frac{1}{8} r^3 \cos \varphi + \frac{a_0}{2} - \frac{3}{8} r \cos \varphi \qquad a_0 = \text{const.},$$

which solves the above stated problem. But is this really the only $f \in C^2(\Omega)$ solving the problem? With $n \geq 2$ the Neumann condition seems to have no solution. What am I missing? I suspect, that it should probably be $n \geq 1$, but I don't see how to get there. Or am I totally wrong? Any help would be appreciated.

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  • $\begingroup$ First, in the inhomogeneous Neumann problem, you must always be careful about the question of existence. There is a simple criterion which is made clear from the weak formulation: $0 = \int_\Omega \Delta u v dx = \int_{\partial \Omega} v \nabla u \cdot n dS - \int_\Omega \nabla u \cdot \nabla v dx = \int_{\partial \Omega} v g dS - \int_\Omega \nabla u \cdot \nabla v dx$. If you consider the test function $v \equiv 1$ then you see that you must have $\int_{\partial \Omega} g dS = 0$. This does occur in your case. $\endgroup$ – Ian Dec 6 '15 at 1:24
  • $\begingroup$ Given that, one can show using the Lax-Milgram theorem that the solution to the Neumann problem is unique up to a constant. So then the only concern is about detecting a solution which is singular at the origin. $\endgroup$ – Ian Dec 6 '15 at 1:25

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