∇^2ψ(x,y) = -c*f(x,y) how to solve?
- From: Alex <alemns@xxxxxxxxxxx>
- Date: Thu, 10 Jul 2008 01:01:23 EDT
Dear All,
I am NOT familiar AT ALL with Matlab, but I've heard that would be easy (=less time-consuming, since I am not good at programming) to solve the following problem:
∇^2ψ(x,y) = -c*f(x,y) [Poisson]
where "c" is a constant.
bondary: ψ=0 for x=0, x=a, y=0, y=b
It's an electrostatic problem to be solved in a rectangular shape. I would like to have a graph of current as a function of x and y (arrow plot), as well as ψ and power (I*V)...the last two graphs like surface/contour...
((I have also heard that COMSOL would be a great option to solve such problem, but I don't have it)
My question is, should I need to decouple the above eq in first-order eqs? and....someone here would have, by accident, a code to solve this problem, or something similar so I could have some directions....any suggestions, hints...would be very helpful.
Thanks
.
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