Services & ResourcesWolfram Forums
 MathGroup Archive
 2005 January February March April May June July August September October November December

Re: Solving a weakly singular integral equation - Take 2.

• To: mathgroup at smc.vnet.net
• Subject: [mg54594] Re: Solving a weakly singular integral equation - Take 2.
• From: Paul Abbott <paul at physics.uwa.edu.au>
• Date: Wed, 23 Feb 2005 03:13:10 -0500 (EST)
• Organization: The University of Western Australia
• Sender: owner-wri-mathgroup at wolfram.com

In article <cvc8c1$qt5$1 at smc.vnet.net>,
Zaeem Burq <Z.Burq at ms.unimelb.edu.au> wrote:

> I am still trying to solve a Linear Volterra type-2 integral equation by
> using successive approximations method. I am mainly interested in the
> behaviour of the solution near zero.
>
> The unknown function is f[t]. Define
>
> p[x_]:= Exp[- 0.5 * x^2]/Sqrt[2 Pi] (* Gaussian probability density
> function*)

0.5 is _not_ the same as 1/2. Use 1/2 instead.

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

> c[t_]:= Sqrt[2.1 t Log[Log[1/t]] ]

Since 2.1 is a parameter, something like

c[a_:2.1][t_] = Sqrt[a t Log[Log[1/t]]]

would be better.

Where does this function come from? For positive a and t, c only makes
sense for 0 < t < 1/E.

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

1.5 is _not_ the same as 3/2. Use 3/2 instead.

G[a_:2.1][s_,t_] = (c[a][t] - c[a][s]) p[(c[a][t] - c[a][s])/Sqrt[t-s]]/
(t-s)^(3/2)

> 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

and again ...

Note that

Simplify[(c[a][t] p[c[a][t]/Sqrt[t]])/t^(3/2), t > 0]

has a 1/t singularity multiplied by complicated logarithmic terms. Are
you sure that this is the correct form for the inhomogenous term? It is
the same as G[s,t] with the c[s] terms set to zero -- but c[s] is zero
only at s = 1/E.

> As you can see, the kernel has a $(t-s)^1.5$ in the denominator, i.e.,
> singularities along the diagonal (s=t). These singularities are
> appropriately killed by the function p,

I still have no idea what you mean by this. How is the singularity
killed by p? As far as I can see it is not!

> but Mathematica seems to have trouble with them:
>
> Since c is differentiable (except at zero), the factor c[t] - c[s] in the
> numerator of the kernel is O(t-s) when t-s is small and t > 0. This means
> that the denominator (t-s)^1.5 is only as bad as Sqrt[t-s]. This combined
> with the exponential function, means that the kernel is actually at-least
> Holder continuous of order -0.5.

You can see this directly via series expansion of G[a][s, t]. Entering

Simplify[G[a][s, t] + O[t, s]^(1/2), {a > 0, t > s > 0}] //
Normal // InputForm

yields

(a (1 + Log[s] Log[-Log[s]]))/
(2 Sqrt[2 Pi] Sqrt[t-s] Log[s] Sqrt[a s Log[-Log[s]]])

> Even when t=0, the exponential function kills the pole at zero.

As far as I can see, this is not correct.

> It is known from other sources, that this integral
> equation has a unique solution.

For exactly this equation? What is the reference for this solution? What
form does it take?

Unless the above issues are resolved, solving the integral equation via
successive approximations is doomed to fail.

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: Re: Re: Bug in 5.1??
• Next by Date: Re: Re: Re: Bug in 5.1??
• Previous by thread: Solving a weakly singular integral equation - Take 2.
• Next by thread: Re: Solving a weakly singular integral equation - Take 2.