Solving real life problem of gravity using Laplace's equation in polar coordinates

Question

Suppose we have an empty space: then the gravitational potential will be equal for all points to 1 constant.

I want to solve the Laplace equation for such space in polar coordinates.

Let the gravitational potential over space be described by Laplace's equation $$ G_{\rho\rho}+\frac{1}{\rho}G_{\rho}+\frac{1}{\rho^{2}}G_{\theta\theta}=0 $$ with boundary states $G(\rho,0) = G(\rho,\pi)=A$ and $G(a,\theta) = A$, $|G(\rho,\theta)|<\infty $.

Questions

1. Since the Laplace equation in polar coordinates is linear with respect to $G(\rho,\theta)$, is its general solution equal to $$ G = G_{1}+G_{2}+G_{3} $$ where $G_{i}$ satisfies the $i$-th boundary condition while all other ones are set to $0$?

2. On the problems for determining the $G_{i}$ component of the solution: if the condition set to $0$ is $G_{i}(a,\theta)$ does this implies that this components equals a constant $C$?

EDIT

I understand from my comments that my question isnt clear so I will try to clarify some things in this section.
Let $$ G_{\rho\rho}+\frac{1}{\rho}G_{\rho}+\frac{1}{\rho^{2}}G_{\theta\theta}=0 $$ with boundary conditions $G(\rho,0) = G(\rho,\pi)=A$ and $G(a,\theta) = A$, $|G(\rho,\theta)|<\infty $.

We have 3 inhomogenous boundary conditions: can we evaluate $G$ indirectly as a linear combination of $G_{1}+G_{2}+G_{3}$?

Here

• $G_{1}$ is the solution of the boundary value problem $$G_{1\rho\rho}+\frac{1}{\rho}G_{1\rho}+\frac{1}{\rho^{2}}G_{1\theta\theta}=0$$ with boundary conditions $G_{1}(\rho,0) =A, G_{1}(\rho,\pi)=0$ and $G_{1}(a,\theta) = 0$, $|G_{1}(\rho,\theta)|<\infty$,

• $G_{2}$ is the solution of the boundary value problem $$ G_{2\rho\rho}+\frac{1}{\rho}G_{2\rho}+\frac{1}{\rho^{2}}G_{2\theta\theta}=0 $$ with boundary conditions $G_{2}(\rho,0) =0, G_{2}(\rho,\pi)=A$ and $G_{2}(a,\theta) = 0$, $|G_{2}(\rho,\theta)|<\infty $ and finally

• $G_{3}$ is the solution of this boundary value problem: $G_{3\rho\rho}+\frac{1}{\rho}G_{3\rho}+\frac{1}{\rho^{2}}G_{3\theta\theta}=0$ with boundary conditions $G_{3}(\rho,0) =0, G_{3}(\rho,\pi)=0$ and $G_{3}(a,\theta) = A$, $|G_{3}(\rho,\theta)|<\infty $.

And in the case of the components $G_{1},G_{2}$ where the boundary condition is $G_{1}(a,\theta)=G_{2}(a,\theta)=0$ isn't the solution a constant $C$?

Answer 1

• In OP your domain is the upper half disk $$ \{(\rho,\theta)\,:0\le\rho\le a\,,0\le \theta \le \pi\}=\{(x,y)\,:x^2+y^2\le a^2\,,y\ge 0\} $$ with radius $a\,.$

• Your boundary condition is that $G$ equals the constant $A$ on the boundary \begin{align}&\{\rho=a,0\le\theta\le\pi\}\cup\{0\le\rho\le a,\theta\in\{0,\pi\}\}\\ &=\{x^2+y^2=a^2\,,y\ge 0\}\cup\{-a\le x\le a\,,y=0\}\,.\end{align}

• It does not matter in which coordinates you express this. The unique solution is always trivial: $G\equiv A$ which follows from the maximum principle by which a harmonic function attains its maximum and minimum on the boundary: hence $\max G=\min G=A\,.$