Suppose i have this poisson equation
$u_{xx}+u_{yy}=x\\ x^2+y^2<3\\ u(\sqrt 3, \theta)=e^{\theta}$
and i've changed it into polar world like this (if i'm not false),
$\dfrac{\partial^2 u}{\partial r^2}+\dfrac{1}{r}\dfrac{\partial u}{\partial r}+\dfrac{1}{r^2}\dfrac{\partial^2 u}{\partial \theta^2}=r\cos (\theta)\\ r\lt \sqrt 3\\ u(\sqrt 3, \theta)=e^{\theta}$
I want to use green function. But i still don't understand what is the green function and how do i solve this. Please help me to solve this with step-by step. Actually i have done with this problem with separation variable's method. But i just want to know more about solving this with the green function. Some theories said that this has similarities with variation of parameters, and some say about integral forms with G functions or some kind of dirac delta. (I don't know, it confused me)
I have watched youtube, read many books, journals, but they are just talk about theory without example. So it looks like i got nothing from all of those.
Thanks for advance!
Let $ \vec p=\left<x, y\right>$ and $\vec s=\left<s_x, x_y\right>$. Then for the Cartesian Laplacian, Green's function is $G(\vec p, \vec s)=\frac{1}{2\pi}\ln(||\vec p-\vec s||)$ or $G(x, y, x', y')=\frac{1}{4\pi}\ln((x-x')^2+(y-y')^2)$ as can be found in the table of Green's functions listed here.
The solution to the Laplacean $\nabla^2u=f$ on $\Omega$ for Dirichlet boundary conditions $u=g$ on $\partial\Omega$ is
$$ u(x) = \int_\Omega G(x,y)f(y)\,dy + \int_{\partial\Omega}\frac{\partial G(x,y)}{\partial n}g(y)\,dy $$
So the solution to your problem is $$ u(\vec p)=\int\int_\Omega s_x\frac{\ln(||\vec p-\vec s||)}{2\pi}ds_x ds_y+\int_{\partial \Omega}e^\theta \frac{1}{2\pi||\vec p-\vec s||}ds $$
or
$$ u(x, y)=\int_0^{2\pi}\int_0^\sqrt{3} r\sin\theta\frac{\ln((x-r\sin\theta)^2+(y-r\cos\theta)^2)}{4\pi}r dr d\theta +\int_0^{2\pi}e^\theta \frac{(x-\sqrt{3}\sin\theta)\sin\theta+(y-\sqrt{3}\cos\theta)\cos\theta}{2\pi((x-\sqrt{3}\sin\theta)^2+(y-\sqrt{3}\cos\theta)^2)}d\theta $$
Where $\Omega$ is the circle of radius $\sqrt{3}$, and $\partial \Omega$ is the circle's perimeter. This is how to use Green's function.