| You need to specify the boundry conditions surrounding the grid of
resistors to get a unique solution. Once the boundry conditions are
specified, (and the problem is reduced somwhat via symmetry) a simple
relaxation method could be used to iteratively solve for the current
distribution. In partial differential equations, specifying the
value of a scalar potential on the boundry is called a Dirichlet
problem, and leads to what is called an elliptic boundry value
problem.
One approach to the "boundry at infinity" may be to model the current
flow as a diffusive process which approaches some steady state that
satisfies Kirchoffs laws. This would allow you to approach the final
solution in a mathematically satisfying way, as a well defined limit.
To be more specific, consider a small capacitor to be present at
each node of the grid and apply a step of current at the center, or
two opposite steps of current at selected grid points. Now you have
a large RC network, which will approach the solution you desire in the
steady state. And it is far closer to actual physical reality than
a capacitance free grid of resistors, which is never realisable in
practice.
- Jim
|
| > You need to specify the boundry conditions surrounding the grid of
> resistors to get a unique solution. Once the boundry conditions are
> specified, (and the problem is reduced somwhat via symmetry) a simple
> relaxation method could be used to iteratively solve for the current
> distribution. In partial differential equations, specifying the
> value of a scalar potential on the boundry is called a Dirichlet
> problem, and leads to what is called an elliptic boundry value
> problem.
What makes a sufficient set of boundary conditions ? Since the
grid is infinite, it seems that the condition is something like
V( infinity, infinity ) = 0. I'm not sure if this is meaningful.
The problem I see is that there are many V that should be able to
meet this criterion. It seems to me that the boundary conditions that
will determine a unique solution will have to be finite. That is,
it seems that the solution will become unique when given enough
( e.g. n) boundary value conditions of the form:
V(p,q) = k(m)
with p and q integers,
m natural, and
k(m) real for m between 1 and n.
> One approach to the "boundry at infinity" may be to model the current
> flow as a diffusive process which approaches some steady state that
> satisfies Kirchoffs laws. This would allow you to approach the final
> solution in a mathematically satisfying way, as a well defined limit.
Hmm. Sounds complicated. Maybe I should try to get the
solution for a finite grid (parametrised by something like the number
of resistors on a 'side' of the grid) and then let that number approach
infinity ?
> To be more specific, consider a small capacitor to be present at
> each node of the grid and apply a step of current at the center, or
> two opposite steps of current at selected grid points. Now you have
> a large RC network, which will approach the solution you desire in the
> steady state. And it is far closer to actual physical reality than
> a capacitance free grid of resistors, which is never realisable in
> practice.
It seems that I should be able to get the steady state solution without
having to solve for the voltage at each node for all time. In the steady
state, I think that the circuit reduces to just the resistors and that
the final state must depend only on the values of the resistors.
You mentioned something earlier about Tellegen's theorem. I hadn't heard
of that before. Could you tell what it is ?
Thanks for the help so far.
> - Jim
|
| Actually, the idea of capacitive loading is probably unnecessarily
complicated - it was just an attempt to make the problem more phyiscally
meaningful...
I forgot to mention Tellegen's theorem. It states that in a network,
SUM(v*i) = 0, and is actually valid for time varying and nonlinear
networks. It doesn't state anything fundamentally new and can be
derived from KVL and KCL, but surprisingly wasn't stated explicitly
until some time in the '50's, by B. Tellegen, in a Phillips Research
Reports paper. I can dig a copy out, I found a reprint of it once.
(Some of those early papers, like Nyquist's paper on regeneraton are
really worth reading.)
A sufficient set of boundry conditions is either the value of the
function along the boundry, or the normal of its gradient. You must
specify the whole boundry, either the value or gradient, or some
mixture, to have a unique solution. Boundries at infinity must be
handled by a limiting process using a finite boundry, though often a
simple handwave (boundry at infinity = 0) will work.
Here is one way of approaching the problem computationally:
By symmetry, we can consider only one quadrant of the grid. Define
mesh currents in each square, numbered I[i,j], and put a current
source of strength I at the origin. Now we have the boundry conditions
I[i,0] = -I[0,j] = I/2, and by further symmetry we clearly have
I[j,i] = -I[i,j], and I[i,i] = 0. Writing KVL around the resistors
surrounding one mesh square gives the relation on the mesh currents
I[i,j] = (1/4)*(I[i,j-1]+I[i+1,j]+I[i,j+1]+I[i-1,j]), and it is only
necessary to solve this along a 45 degree wedge due to symmetry.
Note that this relation, that each mesh current is the average of its
neighbors, is just the discrete form of Laplace's equation, so we can
estimate the character of the solution from normal electrostatics.
If we perform a conformal transformation of the 45 degree wedge via
ln(z), we find that the mesh currents along any ray are equal. In the
discrete case the mesh current at I[i,j] will be approximately
I[i,j] ~= (I/2)/(PI/2)*ATN(i,j). Near the origin, the discrete nature will
mess this up, but far away it is a fair approximation. This conforms to
intuition, since these mesh currents will approach each other far from the
origin, and sum to nearly zero along any resisitor.
To numerically solve this problem for the finite grid you can use
Gauss-Seidel iteration. Fix the values of I at the sides of the wedge,
and estimate the values along the circular border with the ATN approx
above, and initialize the interior points via the ATN approx as well.
Then just iteratively improve I[i,j] by settting it to the average of
its neighbors and repeat. This should converge to the solution because
the matrix of equations is diagonally dominant, and I think, positive
definite. You can accelerate the iteration with the usual tricks like
overrelaxation (make something like 1.5 times the change to any I
each iteratin) and Aitken's delta-squared extrapolation.
Now, to get an idea of the mesh currents near the origin for an infinite
boundry you can extrapolate to it. Assume you've calculated the
mesh currents for a number of wedge radii (say the sequence 24, 32, 48,
64, 96, etc) - you will find the final values near the origin to be
converging to something. The idea is to assume these mesh currents
have a power series expansion in h = 1/n (n = radius of wedge), and
by using several approximations above, eliminate the leading terms of
the expansion and extrapolate to the limit, where n -> infinity, h -> 0.
Naturally, you can use the previous estimates of a given grid size as
the starting values for the next size up to save time. This technique
will give surprisingly rapid convergence to the limiting values.
I don't think the mesh currents in the limiting case have any simple
relation in the discrete case - one way is to calculate them and find
out! I played with the problem a bit, hoping that the mesh currents
would fall out as something simple like Pascal's triangle, but no luck
so far... Perhaps the problem is related to certain problems in
crystallography where you need to sum up multidimensional series
over the crystal lattice.
- Jim
|
| Re .4 - yes, I was only pointing that out and quoting some known
results...
However, there is a difference, since the stated problem is not
to find an approximation but to solve a problem on a grid.
I tried iterating the solution for a few sizes of grid (16, 32, 64, 128)
with a simple minded averaging of neighboring mesh currents and
noticed that the currents near the origin converge to a limit with
a power series expansion of the form
x0 + x6*(1/n)^6 + x8*(1/n)^8 + ...
That is, the accuracy seems to go up by 64 for each doubling of the
size of the grid. I then printed the rational approximants to the
mesh current at [1,2], but it did not seem to be a simple rational
number, or an obvious quadratic. Too bad... the problem seems to have
no neat closed form solution.
- Jim
|