Solvability of Neumann problem Let $D$ be the domain in $\mathbb{R}^2$ such that $x^2 + y^2 <4$ , and $\alpha, \beta$ and $ \gamma $ be real constants. For which values of these three constants there is no solution for:$$\Delta u =0 , (x,y)\in D$$
$$\partial_n u= \alpha x^2 + \beta y + \gamma, (x,y) \in \partial D$$
 How to solve the equation for the values of these three constants which the solution exists?
Thanks.
 A: In polar coordinates $(x,y) = (r\cos\phi,r\sin\phi)$ the general solution (which is finite at $r=0$) is
$$ u(r,\phi) = A_0 + \sum_{n=1}^\infty r^n \big[A_n\cos(n\phi) + B_n\sin(n\phi)\big] $$
where $D = \{ 0<r < 2\}$. The gradient is
$$ u_r(r,\phi) = \alpha r^2\cos^2\phi + \beta r\sin(\phi) + \gamma = \frac{\alpha r^2}{2}\big[\cos(2\phi) + 1\big] + \beta r \sin(\phi) + \gamma $$
Evaluated at the boundary:
$$ u_r(2,\phi) = 2\alpha\cos(2\phi) + 2\beta \sin(\phi) + (2\alpha + \gamma) $$
You also have, from the general solution
$$ u_r(2,\phi) = \sum_{n=1}^\infty n2^{n-1}\big[A_n\cos(n\phi) + B_n\sin(\phi)\big] $$
Comparing coefficients gives
\begin{cases} 
2\alpha + \gamma = 0 \\
B_1 = 2\beta \\
4A_2 = 2\alpha 
\end{cases}
The first condition is required for the solution to exist.
A: The Neumann problem has a solution iff $\int_{S}{fd\sigma=0}$  , where $S$,  is the circle with center the origin and radius $2$ and $f(x,y)=\alpha x^2 + \beta y + \gamma$ . A parametrization for this circle is $(2\cos{t}, 2\sin{t})$. Now if we take the above mentioned line - integral we get the desired condition: $2a+\gamma=0$.
Edit The proposition above is
