Green's Function Method for Poisson's Equation: Uniformly Charged Hemisphere I wish to solve for the potential of a uniformly charged "northern" hemisphere ($r\leq R $ and $0\leq\theta\leq \frac{\pi}{2}$ and $0\leq\phi\leq 2\pi$) of uniform volume charge density $\rho$. Due to symmetry, the potential will only depend on $r$ and $\theta$. The equivalent PDE problem goes as follows:
\begin{equation}
\nabla^2V(r,\theta)=-\frac{\rho}{\epsilon_0}\tag{1}
\end{equation}
for all points inside the hemisphere, and the usual laplace equation holds for points outside. I want to solve this PDE with the method of Green's functions. To wit, I want to find $G(r,\theta)$ such that:
\begin{equation}
\nabla^2G(r-r',\theta-\theta')=\delta(r-r')\delta(\theta-\theta')\tag{2}
\end{equation}
I'm having difficulty finding the appropriate Green's function. Because there isn't dependence on $\phi$, am I solving for the Green's function in 3D space that's only dependent on two variables? Is that possible? At the end, I'd be doing a volume integral to find the solution right? I believe I'm generally lost on how to approach solving the PDE in this manner.
EDIT: I realized I've misunderstood the books I've been reading (currently using Riley, Hobson, Bence).  The book shows the Green's function for radial dependence, but I can't find literature on finding the appropraite Green's function when the nature of the problem requires radial and angular dependence.
 A: We start with the well known formula
$\Delta \frac{1}{|\vec{x}-\vec{x}^\prime|} = - 4 \pi \,\delta^{(3)}(\vec{x}-\vec{x}^\prime)$,
where the Laplace operator $\Delta = \sum\limits_{i=1}^3 \partial^2 /\partial x_i^2$ acts on the unprimed variables $x_i$. As a consequence,
$V(\vec{x})= \frac{1}{4 \pi} \int d^3 x^\prime \frac{\rho(\vec{x}^\prime)}{|\vec{x}-\vec{x}^\prime|}$
is a solution of the equation $\Delta V(\vec{x}) = -\rho(\vec{x})$, where $\rho(\vec{x})$ is an arbitrary charge density (no symmetry used yet and $\varepsilon_0=1$ for simplicity).
In the next step, we specialize to the case of a charge density being invariant under rotations around the $3$-axis. Thus, in spherical coordinates ($x_1 = r \sin \theta \cos \varphi, x_2=r\sin \theta \sin \varphi, x_3=r \cos \theta)$, the charge density is now independent of the angle $\varphi$. Inserting this charge density in the formula for the potential, we obtain
$V(r, \theta, \varphi) = \frac{1}{4 \pi} \! \int\limits_0^\infty \! dr^\prime r^{\prime \, 2} \!\int\limits_0^\pi \! d \theta^\prime \sin \theta^\prime \rho(r^\prime, \theta^\prime)\!\int\limits_0^{2 \pi} \! d \varphi^\prime [r^2+r^{\prime \, 2}-2 r r^\prime (\sin \theta \sin \theta^\prime \cos(\varphi^\prime \!- \! \varphi) + \cos \theta \cos \theta^\prime)]^{-1/2}$.
Performing the variable transformation $\psi = \varphi^\prime-\varphi$ and exploiting the periodicity properties of the cosine in the last integral, we find the $\varphi$-independent expression
$G(r,\theta;r^\prime, \theta^\prime)= \int\limits_0^{2 \pi} d \psi \, [r^2+r^{\prime \, 2}-2 r r^\prime (\sin \theta \sin \theta^\prime \cos \psi+\cos \theta \cos \theta^\prime)]^{-1/2}   $.
As to be expected, also the potential is thus independent of $\varphi$ and we find the final formula
$V(r, \theta) = \frac{1}{4 \pi} \int\limits_0^\infty d r^\prime r^{\prime \, 2} \int\limits_0^\pi d \theta^\prime \sin \theta^\prime \, G(r,\theta;r^\prime, \theta^\prime) \, \rho(r^\prime, \theta^\prime)$
for the potential generated by a charge distribution with the aforementioned symmetry properties. Comparing with $\Delta V = -\rho$, we see that the function  $G(r, \theta; r^\prime, \theta^\prime)$ satisfies the equation
$\Delta G(r, \theta; r^\prime, \theta^\prime) = \left(\frac{1}{r^2} \frac{\partial}{\partial r} r^2 \frac{\partial}{\partial r}+\frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \sin \theta \frac{\partial}{\partial \theta} \right) G(r, \theta; r^\prime, \theta^\prime) = -4 \pi \frac{\delta(r-r^\prime) \delta(\theta-\theta^{ \prime})}{r^{\prime \, 2} \sin \theta^{ \prime}}$
and can be regarded as the desired Green function for the problem under investigation.
Remarks:

*

*The integral defining $G(r, \theta; r^\prime, \theta^\prime)$ can be expressed in terms of two complete elliptic integrals of the first kind.


*For the transformation of the three-dimensional delta function in cartesian coordinates into the corresponding expression in spherical (or other) coordinates, the Jacobian must be taken into account! This is the reason for the occurence of the factor $1/r^{\prime 2} \sin \theta^\prime$ in the last equation. This is also the reason, why your equ. (2) for the Green function cannot be correct. It is also obvious for physical reasons that the Green function cannot depend only on the coordinate differences $r-r^\prime$ and $\theta-\theta^\prime$!


*The solution of your problem for the charge density $\rho(r, \theta) = \rho_0 \Theta(R-r)  \Theta(\theta) \Theta(\pi/2-\theta)$ with constant $\rho_0$ is now straightforward:
$\quad \quad V_0(r, \theta) = \frac{\rho_0}{4 \pi} \int\limits_0^R d r^\prime r^{\prime 2} \int\limits_0^{\pi/2} d \theta^\prime \sin \theta^\prime G(r, \theta; r^\prime, \theta^\prime)$.
