I have encountered a peculiar problem in my mathematical physics test which is nothing I have seen before.
Solve $\Delta u(x) = 0$ inside a half-ball in $\mathbb R^3$, which is given by $|x| < R$, $x_3 > 0$, with the conditions $u|_{r=R} = 0$, $\frac{\partial u}{\partial x_3}|_{x_3=0} = q$ (a constant).
Separation of variables won't work here. I also thought I can separate the problem into two simpler ones: $$(1)\;\; \Delta w = 0, x_3 > 0, \frac{\partial w}{\partial x_3}|_{x_3=0} = q \implies w = qx_3 = qr\cos\theta$$ $$(2)\;\; \Delta v = 0, v|_{|x|=R} = -qR\cos\theta, \frac{\partial v}{\partial x_3}|_{x_3=0} = 0, |x| < R, x_3 > 0.$$
If we can solve (2), the solution should be $u = w + v$. Are there any ways to solve (2)? It looks easier, I tried reflection from half-ball to whole ball, but my knowledge of reflections in physics is very weak.
Any other ways to solve this problem are also of course welcome.
Let $u$ be a solution to $\Delta u = 0$ in $\{|x| < R, x_3 > 0\}$ satisfying the conditions $u|_{|x|=R} = 0$ and $\frac{\partial u}{\partial x_3}|_{x_3=0} = q$. Consider $U(x_1, x_2, x_3) = u(x_1, x_2, |x_3|)$ which is defined in the ball $\{|x| < R\}$.
Since $$\frac{\partial}{\partial x_3} [u(x_1,x_2,|x_3|)] = \frac{\partial u}{\partial x_3}(x_1,x_2,|x_3|) \cdot (2H(x_3)-1),$$ $$\frac{\partial^2}{\partial x_3^2} [u(x_1,x_2,|x_3|)] = \frac{\partial^2 u}{\partial x_3^2} \cdot (2H(x_3)-1)^2 + \frac{\partial u}{\partial x_3} \cdot 2\delta(x_3) = \frac{\partial^2 u}{\partial x_3^2} + \frac{\partial u}{\partial x_3}|_{x_3=0} 2\delta(x_3) = \frac{\partial^2 u}{\partial x_3^2} + 2q\delta(x_3),$$ we have a new Dirichlet problem which is $$\Delta U = \Delta u + 2q\delta(x_3) = 2q\delta(x_3);$$ $$U|_{|x|=R} = 0.$$
We are then, due to nature of $\delta(x_3)$, looking for a solution in the form $$U(x) = qx_3H(x_3) + f(x),$$ where $$\Delta f = 0, f|_{r=R} = -qx_3H(x_3) = -qR\cos\theta H(\cos\theta).$$
The problem for $f$ is standard and its solution has the form $$f(x) = \sum_{k \geq 0} A_k |x|^k P_k(\cos\theta),$$ where $P_k$ are the Legendre polynomials. For the boundary condition, putting $y = \cos\theta$ we have
$$\sum_{k \geq 0} A_k R^k P_k(y) = -q R y H(y).$$
The coefficients $A_k$ are computed using orthogonality $\int_{-1}^{1} dy P_k(y) P_m(y) = \delta_{km} \frac{2}{2m+1}:$
$$A_m = \frac{2m+1}{2R^m} \int_{0}^1 yP_m(y)dy.$$
$A_0 = \frac{1}{4}$, $A_1 = \frac{1}{2R}$. For further odd $m$ $A_m = 0$, while for even $m$ this is a lengthy computation with a lengthy answer. The computation can be done using $P_m(y) = \frac{1}{2^m m!}\frac{d^m}{dy^m} (y^2-1)^m$ and integration by parts. So the overall solution is
$$u = q|x|\cos\theta H(\cos\theta) + \frac{1}{4} + \frac{1}{2R}|x|\cos\theta + \sum_{k \geq 0} A_{2k} |x|^{2k} P_{2k}(\cos\theta)$$