Re: The relaxation method

[Ludwik Kowalski <kowalskiL@MAIL.MONTCLAIR.EDU>]

John Mallinckrodt wrote:

> On Fri, 1 Mar 2002, Ludwik Kowalski wrote:
>
> > I see that John's equipotentials are everywhere perpendicular
> > to the paper borders. How did you relax marginal cells (those
> > which have only three or two neighbors), John, to get this?
>
> Tempted as I may be to answer this question, John Denker has
> already proposed an alternate approach based on a model of good
> pedagogy that I am loathe to undermine.  Perhaps I will just
> provide the following hint which will allow you an alternate way
> of thinking about JD's "periodic boundary conditions".
>
> Laplace's equation is very well known to have unique solutions
> as long as appropriate boundary conditions are specified.
> Appropriate boundary conitions come in two forms:
>
> 1. "Dirichlet": Setting the value of the potential itself on the
> boundary.
>
> 2. "Neumann": Setting the normal derivative of the potential at
> the boundary.
>
> You are familiar with the Dirichlet case but are struggling with
> the Neumann case which is needed here.  So here's the hint: How
> would you specify the value of a marginal cell to insure that the
> normal derivative of the potential is zero at the boundary?

You forced every cell along the top margin to have the same
potential as the cell below it.
You forced every cell along the bottom margin to have the same
potential as the cell above it.
You forced every cell along the left margin to have the same
potential as the cell on its right.
You forced every cell along the right margin to have the same
potential as the cell on its left.
And you probably ignored corners.

Is that is what you did to get the exponential lines posted
this morning? I suppose it is OK to use Dirichlet and Neuman
specifications in the same problem (constant V for silver-coated
cells and dV/dn=0 along the margins). Is this correct?
Ludwik Kowalski