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The formula for the Laplacian
that gives the charge density in this second step is essentially the same
formula used in the relaxation algorithm in the previous step, so at the
very least there should be excellent consistency. Checking to see that
overall charge is conserved is a good diagnostic. See
http://www.monmouth.com/~jsd/physics/laplace.html
for more on this.
Since I set potential boundary conditions in my technique charge
is not conserved.
I don't see how you accommodated to azimuthal
symmetry in your relaxation calculation, John. It looks like you
have used the rectangular form of the Laplacian everywhere,
That will surely lead to funniness. Why do
you not use the cylindrical Laplacian? The calculation is still
two dimensional when you do so.
having progressed from
v.1.0 -> v.3.0 -> Excel 98 without having really ever read a
manual*.