# Separation of variables and potential

1. Sep 19, 2015

### sayebms

1. The problem statement, all variables and given/known data
A potential satisfies $\nabla^2 Φ = 0$ in the 2d slab $-\inf < x < \inf$, $-b < y < b$, with boundary conditions $Φ(x, +b) = +V_s(x)$ on the top and $Φ(x, b) = -V_s(x)$ on the bottom, where

$V_s (x)= -V_0 for -a<x<0$
$V_s (x)=+V_0 for 0<x<a$
(a) what is $\Phi(x,y)$ in the interior?
2. Relevant equations
$\nabla^2 \phi=0$
$\nabla^2 = \frac{\partial ^2}{\partial x^2} + \frac{\partial ^2}{\partial y^2}$

3. The attempt at a solution
$\Phi = X(x)Y(y)$

then using separation of variables we arrive at the following:
$\frac{1}{X}\frac{\partial^2 X}{\partial x^2 } + \frac{1}{Y}\frac{\partial^2 Y}{\partial y^2 }=0$
$\frac{1}{Y}\frac{\partial^2 Y}{\partial y^2 }=-\alpha^2$ which gives us $Y(y)= Asin \alpha y +Bcos \alpha y$
$\frac{1}{X}\frac{\partial^2 X}{\partial x^2 }=\alpha ^2$ and solution of its is $X(x)=Ce^{\alpha x}+De^{-\alpha x}$
applying the periodic boundary conditions to the Y solution will give us :
$Y(b)=V_s =A sin \alpha b +B cos \alpha b$
$Y(-b)=-V_s =-A sin \alpha b +B cos \alpha b$
so summing these up will give B=0

my question is have I gotten it right till here and how I go about finding the $\alpha$ and the boundary conditions for $X(x)$? I appreciate any ideas or hints.

Last edited: Sep 19, 2015
2. Sep 19, 2015

### gabbagabbahey

Hi sayebms,

$Y(-b)=-V_s =-A sin \alpha b +B cos \alpha b$ does not match the boundary condition you gave in the problem statement.

Also, if the product of 2 numbers is zero, does that guarantee the the first number must be zero? If not, how do you justify claiming $2B cos \alpha b = 0$ means $B=0$?