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

**Re: Re: Re: Bug in 5.1??**

**Re: Re: Re: Bug in 5.1??**

**Solving a weakly singular integral equation - Take 2.**

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