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: $$$$\nabla^2V(r,\theta)=-\frac{\rho}{\epsilon_0}\tag{1}$$$$ for all points inside the hemisphere, and the usual laplace equation holds for points outside. I want to solve this PDE with the method of Green's functions. To wit, I want to find $$G(r,\theta)$$ such that: $$$$\nabla^2G(r-r',\theta-\theta')=\delta(r-r')\delta(\theta-\theta')\tag{2}$$$$ I'm having difficulty finding the appropriate Green's function. Because there isn't dependence on $$\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.

• Your question is unclear. Are you in three or in two spatial dimensions? What exactly do you mean by "northern hemisphere"? Is it the set of all points $(x,y,z)$ in $\mathbf{R}^3$ satisfying $x^2+y^2+z^2=R^2$ and $z \ge 0$ or do you mean $x^2+y^2+z^2 \le R^2$ and $z \ge 0$? Or is it something else? Is $\rho$ a surface charge density (dimension charge$/$area) or is it a charge density (dimension charge$/$volume) as your question suggests? Is it constant in the region where the charge is sitting? You refer to "books". Which books? Never seen a book where a scalar is replaced with with a vector. Commented Jan 6, 2023 at 7:56
• I'm sorry for the ambiguity. Please let me know if the question is any better. I feel like I'm generally lost on how to approach solving the PDE with Green's functions. I'm familiar with solving the Laplace equation with boundary conditions, but I just can't figure out how to approach continuous charge distributions. Commented Jan 7, 2023 at 4:13
• No problem, you are here to learn. In particular, how to ask precise questions containing all the relevant information. The question looks much better now. However, assuming that you are using spherical coordinates, your desired region for $\theta$ should probably read $0 \le \theta \le \pi/2$. (By the usual convention, the angle $\theta$ does not assume negative values.) Commented Jan 7, 2023 at 7:10
• woops my bad, yeah that was silly of me Commented Jan 7, 2023 at 11:03
• Comment to the post (v5): The RHS of eq. (2) is not the correct Dirac delta function. Commented Jan 7, 2023 at 11:35

1 Answer

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:

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

2. 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 $$\theta-\theta^\prime$$!

3. 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:

$$\quad \quad V_0(r, \theta) = \frac{\rho_0}{4 \pi} \int\limits_0^R d r^\prime r^{\prime 2} \int\limits_0^{\pi/2} d \theta^\prime \sin \theta^\prime G(r, \theta; r^\prime, \theta^\prime)$$.

• Thank you so much for the help! Just to be clear, are you finding green's function by first finding the potential as an integral, then reading off the appropriate green's function? I see why that's valid, but is there a way we can actually solve the PDE with the green's function in it. (ex. through fourier transforms) Commented Jan 8, 2023 at 4:42
• It depends, in particular on the symmetries of the problem. If you have translation invariance, Fourier transform is usually the method of choice as differential operators simply become multiplication operators and the problem to find the Green function can be reduced to an algebraic one. The main reason I took the detour via the potential in my answer was to make contact with concepts where I could expect you would be familiar with. By the way, a variant of this method can be used to find the Green function of $\Delta$ in $n$ dimensions if the one in $n+1$ dimensions is known. Commented Jan 8, 2023 at 6:34
• Oh I finished a course on introduction to mathematical physics (covers complex analysis, linear algebra (+tensors), ode methods, mde methods, pde methods, fourier analysis, laplace transforms, etc). Can you then provide a solution that actually solves the pde? I am familiar with the tools listed. Also, how do you do the variant problem you mentioned above? I'm fine with you just giving me sources to read on my own. Thanks Commented Jan 9, 2023 at 6:03
• I can recommend chapters 2,3 of the book "A. Alastuey, M. Clusel, M. Magro, P. Pujol, Physics and mathematical tools: methods and examples, World Scientific, 2016" (being an extension of an older edition in French). Commented Jan 9, 2023 at 10:16
• Thanks for the help! I'll look into it Commented Jan 10, 2023 at 2:08