Solutions of a Poisson equation on an ellipsoid, with Neumann boundary condition. I want to find the number of solutions of the following problem:
Fix $n \in \mathbb{N}$, with $n \geq 2$. Define the domain $\Omega$ as the ellipsoid
$$\Omega = \left\{x = (x_1,...,x_n) \in \mathbb{R}^n : \sum_{k=1}^n \frac{x_k^2}{\mu_i^2}<1\right\},$$
for positive constants $\mu_i$.
Consider the equation
$$\begin{cases}\Delta u = f & \text{in }\Omega\\
\displaystyle \frac{\partial u}{\partial \nu} = 0 & \text{in } \partial \Omega\end{cases}$$
where $\nu$ is the (unitary) exterior normal to the boundary $\partial \Omega$.
So, the question is: how many solutions $u\in C^2(\bar\Omega)$ does this problem have? Zero? A unique solution? Multiple solutions?
Thank you in advance for your answers!
 A: Excuse me, it was an easy consequence of the divergence theorem.
Assume we have a solution. Thus, it is necessary that
$$0 = \int_{\partial \Omega} \nabla u \cdot n dS = \int_\Omega \nabla \cdot \nabla u dV = \int_\Omega \Delta u dV.$$
We can change variables here, $y_i = x_i/\mu_i$, being the Jacobian of the transformation J with image $\Omega$:
$$J = \begin{vmatrix} \mu_i & & 0 \\
& \ddots & \\
0& & \mu_n \end{vmatrix} = \prod \mu_i = C > 0.$$
So, we have:
$$ 0 = \int_{B(0,1)} (1-y_1 \mu_1) J dy_1 ... dy_n .$$
Change to polar coordinates, following the conventions here, to obtain
\begin{eqnarray*}
 0 &=& C\int_0^1 \int_0^{2\pi} \int_0^\pi ... \int_0^\pi  (1 - r \cos(\phi_1)) r^{n-1}\sin^{n-2}(\phi_1)\sin^{n-3}(\phi_2)\cdots \sin(\phi_{n-2})\,
\,d\phi_1 \, d\phi_2\cdots d\phi_{n-1}\, dr \\
&=& C' \int_0^1 \int_0^{\pi}  (1 - r \cos(\phi_1)) r^{n-1}\sin^{n-2}(\phi_1) \, d\phi_1 \, dr,
\end{eqnarray*}
for a new constant C' > 0.
By Fubini's theorem,
$$  \int_0^1 \int_0^{\pi}  (1 - r \cos(\phi_1)) r^{n-1}\sin^{n-2}(\phi_1) \, d\phi_1 \, dr = \int_0^1 (r^{n-1} \underbrace{\int_0^\pi \sin^{n-2}(\phi_1) d\phi_1}_{>0} - r^n \underbrace{\int_0^1 cos(\phi_1) \sin^{n-2} (\phi_1) d\phi_1 }_{= 0}) dr$$
We arrive in this way to a contradiction. So, there is no solution.
