Underdetermined system of equations for infinite parallel plates? I am trying to mimic the separation of variable method to finding the electrostatic potential. For instance section 2.9 in Jackson 2nd edition.
Consider two infinite plates of potential $\pm V$ separated by a distance $d$

The only boundary conditions I can think of are written in the figure. I can't say anything about $V(x/y \rightarrow \pm \infty) =0$ since the plates are infinite.
But physically doesn't that mean the potential independent of x and y? Equipotentials lie parallel to the xy plane no?
Some progress,
$\phi$ is independent of x,y thus using Jackson's

$-\alpha^2=-\beta^2=0 \rightarrow \gamma^2=0$
And so
$Z=az+b.$
Applying boundary conditions 1 and 2 in the image drawn above, $b=-V$ and then $a=2V/d$. So then $\phi = 2V/d z -V$. But shouldn't $\phi$ go to zero as z goes to $\pm \infty$?
 A: Your conclusions are basically correct.  For a problem that is effectively one-dimensional (which this one is), all solutions to Laplace's equation are piecewise linear functions:  $\phi = a z + b$ for some constants $a$ and $b$.  However, if we include the sheets themselves in the region of $z$ we're solving for, then there will charge on the plates, and $\phi$ will satisfy Poisson's equation rather than Laplace's equation.
Instead, you need to solve Laplace's equation separately in each of the disconnected regions where $\rho = 0$, without assuming that the solutions "above" and "below" the infinite sheets necessarily have the same constants $a$ and $b$ as in the "interior" region.  Any discontinuities in the derivatives that arise at the boundaries can be interpreted as being due to the existence of surface charges, since
$$
\left. \frac{\partial V}{\partial z} \right|_\text{above} - \left. \frac{\partial V}{\partial z} \right|_\text{below} = \frac{\sigma}{\epsilon_0}.
$$
More broadly, it's also important to note that the reason you're not getting a solution for which $\phi \to 0$ as $z \to \pm \infty$ is simply that no such solution exists.  In fact, in 1D, the only solution for which $\phi$ doesn't diverge as $z \to \pm \infty$ is the solution in which $a = 0$.   In other words, if $\pm \infty$ is in your "region of interest", then the potential in the region of interest is a constant.
