# Laplace equation in a rectangle; a non-symmetric solution

Consider Laplace's equation in a rectangle with specified boundary conditions. This problem is solved when $\epsilon_1 = \epsilon_2$ in the following link. $$\nabla \cdot \epsilon \nabla V=0$$

What if the $\epsilon_1$ and $\epsilon_2$ be different. We can still use the fact that at $x=a/2$, $V$ would be constant. However, its value is not $(V_1+V_2)/2$ since the symmetry is broken. Can we use the continuity of $dV/dx$ to find the value of V at $x=a/2$?

## 1 Answer

I think you can approach the problem in two similar manners. One that you mention and the other as follows:

Since your permittivity is a function of $(x,y)$, you can solve the equation assuming that $\epsilon = \epsilon(x,y)$, i.e., is not a constant anymore. The boundary conditions for your problem then become piecewise constant at the upper and lower boundaries (which is not trouble for the resulting integrals). With this method, you obtain $V = V(x,y)$ in the whole domain.

If you want me to elaborate some more, just tell me. Cheers!

• Could you elaborate more?
– Roy
Feb 6, 2015 at 16:55