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Integral Equations
Stan Richardson

The Condenser Problem

An example discussed by Fox and Goodwin [3] in an early paper concerned with the numerical solution of integral equations is the equation derived by Love [4] to describe the electrostatic field produced by a condenser consisting of two parallel circular plates. This is

when dimensionless variables are taken so that the plates have unit radius. Here is the distance between the plates. This is the relevant equation when the potentials of the plates are equal in magnitude but opposite in sign; if these are equal in both magnitude and sign, the first sign on the right-hand side becomes a sign.

Of particular interest in this problem is the capacitance of the condenser, which is the total charge on one of the plates, given in this nondimensional formulation by

Beginning again with

we now use

and

Here we have taken advantage of the fact that the solution is an even function of , so we need not compute entries in values for , and used a construction that ensures that our approximation is also an even function. There has also been a modest increase in the degree of the InterpolatingPolynomial employed.

We will judge convergence by the estimate we obtain for the capacitance. In fact, most numerical work actually quotes values for the capacitance multiplied by the factor . This is convenient because a circular disc of unit radius on its own has a capacitance of , so the difference between the value of this quantity and 1 is a measure of the interference effect between the two discs. This scaled capacitance tends to 1 as . We therefore introduce

Fox and Goodwin [3] carried out their calculations for , and we will begin with this value, too.

A little experimentation with both the interpolation used and the number of iterations performed now tells us that we have convergence to 6 significant figures with

Fox and Goodwin [3] do not evaluate the capacitance, but this was calculated for , both from the figures given by Fox and Goodwin and independently, by Cooke [5], who gives the value 1.8208 for the scaled capacitance.

With this solution in hand, we can plot the equipotentials round the condenser. This was done manually by Love [6], but Mathematica makes the task somewhat easier. With cylindrical polar coordinates chosen so that a plate at potential lies in the plane and one at potential is in the plane , both in the region , the potential is independent of and is given by

In fact, with approxsoln as a polynomial, Mathematica can perform the integrations here analytically, but a numerical approach seems preferable. We have also incorporated the known values of the potential on the plates and the plane into this definition.

We can economize on the effort required to produce a reasonable sketch by exploiting the symmetry and plotting initially only in the first quadrant, where and . We should also choose PlotPoints to ensure that ContourPlot evaluates the potential at the point on the edge of the disk.

Now extract the information we need from this, and form the data required for the curves in the other quadrants.

Finally, we can Show all this, together with the lines omitted from contplot.

Whatever method one uses to solve the integral equation of this example, the computations become more critical as decreases. Cooke [5] takes his computations down to , but records a value for obtained by Fox and Blake using the methods of Fox and Goodwin [3]; for this value of they find a scaled capacitance of 9.233. Using our approach, we need more iterations, but otherwise there are no problems.

Since this value is within 0.01 percent of that given by the asymptotic formula, valid for , obtained by Chew and Kong [7], namely

there is little practical need to carry out the computations with the integral equation for smaller values of . (See Chew and Kong's paper and the references therein for an account of the history of this formula; it is the product of over a century of effort by a large number of mathematicians.) However, Mathematica will do this if required. It is obviously sensible to reorganize the computations presented in this rather discursive fashion into a more efficient program, but the main changes required are in the interpolation. For nonsmall , the graph of is close to a parabola and InterpolatingPolynomial is appropriate, but as decreases the gradients become large near the ends of the interval and interpolation by neither polynomials nor splines is appropriate; Fit with suitable basis functions is more effective. Since this modification will be employed later in a different example, we will not illustrate it here.

It should be noted that the familiar "rule of thumb" criteria for stopping a sequence of iterations (e.g., continue iterating until the difference between two consecutive iterates is less than some predetermined small value) become more and more inadequate as decreases, because the operator involved approaches the identity as . This feature is independent of the means we use to represent our solution: the effective application of our method in dealing with nontrivial problems depends on some judicious experimentation with the method of interpolation employed and the number of iterations used.

Fabrikant [8] tabulates many values for the scaled capacitance, down to , using a different method based on integral equations that produces upper and lower bounds. However, an asymptotic approach yields results of much greater accuracy for such values of . Fabrikant also quotes an error of approximately 0.01 percent for , but his error estimates inevitably increase as decreases, while the asymptotic formula becomes more accurate for smaller values of .



     
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