# Poisson's Equation (I think)

1. Mar 28, 2016

### joshmccraney

1. The problem statement, all variables and given/known data
$f_{xx}+f_{yy} = c$ where $c \neq 0$ is a constant.

2. Relevant equations
None

3. The attempt at a solution
I tried separating but obviously this fails since the constant term has mixed variables after separating. I can solve the homogenous equation where $c=0$ but was wondering what happens if $c \neq 0$.

I should say this isn't homework but I didn't know where else to put it. I didn't give any boundary conditions because I'm just curious of the technique to solve.

2. Mar 28, 2016

### LCKurtz

Here's the idea. Let's say you have $f_{xx}+f_{yy} = h(x)$ and at couple of NH boundary conditions $f(0,y) = a,~f(L,y) = b$ (a little more complicated than your example). You would start by letting $f(x,y) = u(x,y)+\Psi(x)$. Plugging that into the DE and boundary conditions gives$$u_{xx} +\Psi''(x) + u_{yy} = h(x)$$ $$u(0,y) + \Psi(0) = a,~u(L,y)+\Psi(L) = b$$
Now, if you can find $\Psi(x)$ satisfying $\Psi''(x) = h(x),~\Psi(0)=a,~\Psi(L) = b$ you would be left with a homogeneous system in $u$:$$u_{xx} + u_{yy} = 0, ~u(0,y) = u(L,y)=0$$And the $\Psi(x)$ equation can be solved by integrating twice giving two constants to take care of the $a$ and $b$. Then you solve the homogeneous $u$ system and your anwer is $f(x,y)=u(x,y)+\Psi(x)$.

3. Mar 28, 2016

### joshmccraney

Perfect, just what I was looking for!