What Are the Steps to Solve a Boundary Condition Problem with Three Conditions?

[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

 

[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!

 

[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

 

[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.

 

[Mazimillion]

Got it now

Thanks a million, you've been a great help