I was asked (in my PDE course) to solve for U(x,y) from the PDE: $$U_{xx}+U_{yy}=1$$ Over the region $r<a$, where $U$ vanishes at $r=a$. The first thing I did was to turn my Laplace Equation into its polar form: $$U_{rr}+\frac{1}{r}U_r+\frac{1}{r^2}U_{\theta \theta}=1$$ My PDE class likes to use a change of function method to solving these types of equations (instead of the traditional way learned from ODE where you solve for the homogeneous solution and add it to a particular solution of the inhomogeneous equation, although both methods give the same answer). In that case, I use $V=U-\frac{1}{4}r^2$ (as if my guess) so that when I solve for $U$ and plug it into my PDE, I end up with the homogeneous Laplace equation: $$V_{rr}+\frac{1}{r}V_r+\frac{1}{r^2}V_{\theta \theta}=0$$ I have seen that the solution is based on integrating the Poisson kernel as follows, as this was kind of derived in class (the Poisson kernel itself was derived but the integral and $\varphi$ stuff came from an analogy to the heat equation kernel): $$V(r,\theta)=\int_0^{2\pi}\frac{(a^2-r^2)*\varphi(\varphi_0)}{a^2+r^2-2ar\cos(\varphi_0-\theta)}\text{d}\varphi_0$$ However, I don't completely understand what the function $\varphi$ is supposed to represent (although I think I understand that $\varphi_0$ is a dummy variable that we integrate over). I would assume it is similar to what it represents in the heat equation kernel integral (not sure of the exact name of this), but those are the initial conditions of $U$, or $U(x,0)=\varphi(x)$. But that's where I'm stuck, as the Laplace equation doesn't take any initial conditions like the heat equation and instead has boundary conditions associated with it. How am I able to "apply" boundary conditions to my kernel integral?
I have tried to get a basic understanding of the Poisson kernel and its use in solving a Laplace equation online (via Googling of course), but it seems that my problem with the kernel is too simple and is overlooked in various explanations.
There is another Laplace equation problem in the problem set I am working on, but it is in 3 dimensions with more complicated boundary conditions, and I am hoping that an answer to this equation may help me to solve a harder problem on my own, although I may end up needing help for that problem too.
Thanks in advance for your help!
The original equation had the following boundary condition on the circle $r=a$ $$U(a \cos \varphi,a \sin \varphi)=0 ,$$ and hence $$V( a\cos \varphi,a\sin \varphi)=-\frac{a^2}{4}. $$ Thus your boundary data is $\varphi(\varphi_0) \equiv -a^2/4$.