Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2005
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2005

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

Search the Archive

Re: Solving a weakly singular integral equation

  • To: mathgroup at smc.vnet.net
  • Subject: [mg53827] Re: Solving a weakly singular integral equation
  • From: Paul Abbott <paul at physics.uwa.edu.au>
  • Date: Sat, 29 Jan 2005 06:02:35 -0500 (EST)
  • Organization: The University of Western Australia
  • References: <ctahsj$b14$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

In article <ctahsj$b14$1 at smc.vnet.net>,
 Zaeem Burq <Z.Burq at ms.unimelb.edu.au> wrote:

> I am trying to solve a non-linear Volterra type-2 integral equation by
> using successive approximations method. 

As far as I can see, your integral equation is not nonlinear (it is 
certainly linear in f).

> I am mainly interested in the behaviour of the solution near zero.

I believe that the solution is singular there so successive (numerical) 
approximation is unlikely to help.
 
> The unknown function is f[t]. Define
> 
> p[x_]:= Exp[- 0.5 * x^2]/(2 Pi) (* Gaussian probability density function*)

This is not a pdf (it does not integrate to 1). It would be better to 
define this as

  p[x_] = Exp[-x^2/2]/Sqrt[2 Pi]

where exact quantities replace approximate ones.

> c[t]:= Sqrt[t] + 2t/9

As you write later, this is

 c[t_] = Sqrt[t] + 2t/9;

Note that it is usually more efficient to use = instead of := in 
function definitions when the right-hand side is explicit.

> G[s_,t_]:= (c[t] - c[s]) p[(c[t]-c[s])/Sqrt[t-s]] / (t-s)^1.5  (* The 
> kernel *)

Write this as

 G[s_,t_] = (c[t] - c[s]) p[(c[t]-c[s])/Sqrt[t-s]] / (t-s)^(3/2);

Note, however, that this expression is real only for t >= s.
 
> The integral equation is:
> 
> f[t] = c[t]p[c[t]/Sqrt[t]] / t^1.5 - \int_{0}^{t} G[s,t] f[s] ds

In Mathematica notation, you are trying to solve

 f[t]==c[t] p[c[t]/Sqrt[t]]/t^(3/2) - Integrate[G[s, t] f[s], {s, 0, t}]

> As you can see, the perturbation function has an apparant singularity at 
> 0, and the kernel has singularities along the diagonal (s=t). All the 
> singularities are appropriately killed by the function p, but Mathematica 
> seems to have trouble with them. 

Series expansion of the inhomogenous term about t = 0, i.e.,

  (c[t] p[c[t]/Sqrt[t]])/t^(3/2) + O[t]

shows that it goes like

  1/(Sqrt[2 E Pi] t)

as t -> 0. Hence the singularity is not eliminated by the function p. 

Morevover, I think it is possible to show that there is _no_ solution to 
this integral equation: first note that, using the change of variables
s -> x t, the integral part reads

 t Integrate[G[x t, t] f[x t], {x, 0, 1}]

Now, if f[t] ~ t^q as t -> 0, then series expansion of the integrand 
with respect to t leads to computing the following integral:

  Integrate[(E^((Sqrt[x] - 1)/(2(Sqrt[x] + 1)))*
   (1 - Sqrt[x]) x^q)/(1 - x)^(3/2), {x, 0, 1}]

which diverges if q <= -1. But, to agree with the expansion of the 
inhomogenous term, we require a term with q = -1.

On the other hand, a slight modification of the inhomogenous term, say 
changing t^(3/2) -> t, leads to

 f[t] == c[t] p[c[t]/Sqrt[t]]/t - Integrate[G[s, t] f[s], {s, 0, t}]

an equation that can be easily solved by appropriate (using fractional 
powers) series expansion of f.

For some relevant references on integral equations see

   http://mathworld.wolfram.com/IntegralEquation.html

Note that Numerical Recipes is online at

  http://www.library.cornell.edu/nr/bookcpdf.html

The following sections are relevant:

  18.2 Volterra Equations 794
  18.3 Integral Equations with Singular Kernels 797

The method that you were trying to apply to this problem looks like that 
outlined in the paper by Stan Richardson on successive approximation 
solutions to integral equations, which appeared in The Mathematica 
Journal 9(2): 460-482.

Cheers,
Paul

-- 
Paul Abbott                                   Phone: +61 8 6488 2734
School of Physics, M013                         Fax: +61 8 6488 1014
The University of Western Australia      (CRICOS Provider No 00126G)         
35 Stirling Highway
Crawley WA 6009                      mailto:paul at physics.uwa.edu.au 
AUSTRALIA                            http://physics.uwa.edu.au/~paul


  • Prev by Date: Re: Algebra of Einstein velocity addition
  • Next by Date: Re: Nonlinear Curve Fitting
  • Previous by thread: Re: Solving a weakly singular integral equation
  • Next by thread: A question about pattern