# Homework Help: Boundary Condition Problem

1. Nov 10, 2006

### Mazimillion

Hi, this question has been bugging me for weeks and any help would be greatly appreciated.

In lectures we derived a general expression for the potential distribution across the xy half plane (y>0) in terms of a known potential distribution along the boundary defined by the x-axis where the potential,phi, is described by Laplace's equation in 2-D. In the case where the potential along the boundary is: 0 for x greater than equal to a, 0 for x less then equal to -a, phi0 for -a<x<a, deduce an expression for the potential distribution throughout the half plane.

I think what is getting me are the 3 boundary conditions - i'm not sure exactly what i should be doing with them to find values for the coefficients.

i'm sorry that it's not very clear but if you have any ideas they would be greatly appreciated

thanks

2. Nov 10, 2006

### TMFKAN64

The solution to these things is always Fourier analysis, or something like it.

There are two boundary conditions here: the potential should go to zero as y gets large, and it should become your specified function of x when y is 0. Because we want the potential to die off, we have a decaying exponential in the y direction, and the general solution is a weighted sum of terms like like Ae^(-ky)e^(ikx). Since you can't restrict the value of k at all, this sum becomes an integral: Int A(k) e^(-ky)e^(ikx) dk. When y is 0, you know that the potential is your given f(x), so f(x) = Int A(k) e^(ikx) dx. So A(k) is just the Fourier transform of your boundary condition f(x)!

So, find A(k) by taking the Fourier transform of f(x), plug it into your original integral with e^(-ky), and you are done!

Last edited: Nov 10, 2006
3. Nov 10, 2006

### Mazimillion

Thanks for that - I've really made some progress. The only problem now is applying the boundary conditions on the x-axis. How do I apply both phi=0 for x<-a, x>a and phi=phi0 for -a<x<a?

I'm probably missing something really simple but i've been pulling my hair out for ages.

Thanks

4. Nov 10, 2006

### TMFKAN64

The integral I've mentioned is from k = negative infinity to positive infinity. f(x) is how I'm referring to your pulse between -a and a on the x axis.

The key is that A(k) and f(x) are a Fourier transform pair. Did they not discuss this in your class? So if f(x) = Int A(k) e^(ikx) dx, then A(k) = (1/2 pi) Int f(x) e^(-ikx) dx, where this integral is over all space. Since f(x) is zero except between -a and a, you can change the limits to there.

5. Nov 11, 2006

### Mazimillion

Got it now

Thanks a million, you've been a great help