MathGroup Archive 2005

[Date Index] [Thread Index] [Author Index]

Search the Archive

Solving a weakly singular integral equation - Take 2.

Dear all,

using successive approximations method. I am mainly interested in the
behaviour of the solution near zero.

The unknown function is f[t]. Define



kernel *)

The integral equation is:


appropriately killed by the function p, but Mathematica seems to have
trouble with them:

Holder continuous of order -0.5. Even when t=0, the exponential function
kills the pole at zero. It is known from other sources, that this integral
equation has a unique solution.
I wrote down the following routine based on a paper in the Mathematica
Journal: n is the number of iterations of the approximation process:

\!\(Clear[c, approxsoln, K]\n
  n = 3\n
  \(p[x_] := \[ExponentialE]\^\(\(-x\^2\)/2\)\/\@\(2  \[Pi]\);\)\n
  \(c[t_] := Sqrt[2.1\ t\ Log[Log[1/t]]];\)\n
  \(approxsoln[x_] = 0;\)\n
        t_] := \(\(\ \)\(\((c[t] - c[s])\)\ p[\(c[t] - c[s]\)\/\@\(t - 
\)\/\((t - s)\)\^1.5`;\)\n
  \(For[j = 1, 
      j <= \ n, \(j++\), \[IndentingNewLine]values = 
        Table[{t, \(\(\ \)\(c[t] p[c[t]\/\@t]\)\)\/\((t)\)\^1.5`\  - 
              NIntegrate[G[s, t]*approxsoln[s], {s, 0, t}]}, {t, 
            0.0000000001,  .01,  .01\/10}]; \n
      approxsoln[t_] = 
        InterpolatingPolynomial[values, t]; \[IndentingNewLine]Print[
  Plot[approxsoln[x], {x, 0,  .01}]\)


1. I have had to truncated the integral away from zero - i.e., s runs from
0 to t, but t runs from 0.0000000001 to 0.01. It should run from 0 to 0.1.
This is not entirely satisfactory, as I am mainly interested in behaviour
near zero.

2. As the code stands, Mathematica tells me that the numerical integrals
converge too slowly.

I'd appreciate any help.

Best, Zaeem.


Zaeem Burq
PhD Stochastic Processes,
Dept. of Mathematics and Statistics, 

Room 201, Richard Berry Building
University of Melbourne,
Parkville, VIC 3052.

ph: 8344 4248.

  • Next by Date: Re: Plotting symbolic representation of numerica function
  • Previous by thread: Re: Printing numbers