Re: [femm] [A few questions]

[David Meeker <dmeeker@xxxxxxxxxxxxxxxxx>]

you se-ho wrote:

   To Dear Dr. Meeker and All:

How are you? I try to implement to ABC since I think it is more exact
compared to truncation of outer boundaries.

I am comparing the equation (46) and equation (21) in FEMM manual. I am
almost sure that mu_r disappeared because in outer spaced the domain should
be air or free space, so mu_r = 1. (I think even if domain is not air, it is
not difficult to include this term.) However, if I multiply mu_r and mu_0 in
equation (21) to both side, I think the term should be in the numerator. I
think equation (46) is correct from last e-mail. My guesses are

(1) mu_0 and mu_r coefficients should move to the 2nd term in equation (21).
or

(2) Equation (21) is correct, too. The equation (46) and (21) are different
books. Therefore, they are not consistent each other.

When applying the BC, I believe that I should apply the solution of equation
(45), which is A=r*exp(-(n/(mu*r)) on the boundary. By somehow, it started
from mixed boundary condition, but ended up with Dirichlet condition. Am I
Ok ?

Thanks.
Sincerely,
SE-HO YOU

The boundary condition that you want to apply on the outer edge to implement
the asymptotic boundary condition is (45):

dA/dr + (n/r) A = 0

but the format for the mixed boundary condition that falls "naturally" out of the finite element formulation is (21):

1/(mu_o*mu_r) dA/dn + c0 A + c1 = 0

If your outer boundary is circular, the normal direction is the same as the radial direction, i.e. dA/dr = dA/dn.  So, to put (45) in the same form as (21), just multiply both sides of (45) by 1/(mu_o*mu_r):

 1/(mu_o*mu_r) dA/dr +  1/(mu_o*mu_r) (n/r) A = 0

Now, just compare to (21) to get the right coefficient values:

c0 = n/(mu_o*mu_r*r)

c1=0

To get (46), I assumed that the material in the exterior region is air, where mu_r=1.  For an external region of air:

c0 = n/(mu_o*r)
c1=0

Except for some particular special cases, you'll want to use n=1, so things simplify even more to:

c0=1/(mu_o*r)
c1=0

Now, the asymptotic boundary condition is probably the most economical way of getting an outer boundary that mimics an "open" region.  However, since periodic boundary conditions have fairly recently been implemented, it is now possible apply the Kelvin transformation approach for open boundary problems, which is more accurate than the asymptotic boundary condition.  In theory, it exactly represents the contributions from the exterior region, i.e. it really is exactly solving the open boundary problem.  I've put in a new section that talks about applying the Kelvin transformation to 2-D planar problems in Appendix C3 of the new "rough draft" version.

Dave.
 

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