In the method of separation of variables, we need homogeneous BCs.
For the elliptic pde with inhomogeneous BCs:
$u_{xx}+u_{yy}=0$, with $0<x<a$ and $0<y<b$.
With $u(x=0,y)=0$ and $u(x=a,y)=A$
And $u(x,y=0)=0$ and $u(x,y=b)=B$
We want to turn the inhom BCs into hom BCs so we need that $u(x=0,y)=0$ and $u(x=a,y)=A$ to both equal zero right?
But why don't we worry about the other conditions? $u(x,y=0)=0$ and $u(x,y=b)=B$? Are these not BCs? I know what initial conditions are which is when time equals zero. But in this, there is no time so how do we know which are BCs and which are not?
If you set $u=0$ on all $4$ edges, then you can tell me the unique solution right away, right?
When you're performing separation of variables, a two-variable problem results in one separation parameter. A three-variable problem results in two separation parameters. Etc. A pair of homogeneous conditions in one variable determines the possible values of one of the separation parameters. So you can only have one pair of homogeneous conditions when you have two variables. You can only have two pairs of homogeneous conditions when you have three variables. Etc.