How can Poisson's equation be solved when the constant term is nonzero?

[member 428835]

Homework Statement

$f_{xx}+f_{yy} = c$ where $c \neq 0$ is a constant.

Homework Equations

None

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.

 

[LCKurtz]

Homework Statement

$f_{xx}+f_{yy} = c$ where $c \neq 0$ is a constant.

Homework Equations

None

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.

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 $f(x,y)=u(x,y)+\Psi(x)$.

 

[member 428835]

Perfect, just what I was looking for!