# Volterra integral equation

I have to find an approximate numerical solution for the equation

$$F(x) - \lambda \int\limits_1^{x} \text{d}s \;s^2 F(s) Z(x-s) = G(x)$$ $$Z(s) = (\psi''(1-2\ h\ i\ s)- 0.5 \psi''(1-2\ h\ i\ s))$$ $$G(s) = (\psi'(1-2\ h\ i\ s)- 0.5 \psi'(1-2\ h\ i\ s))$$ \ Where $\psi$ indicates the polygamma function. Since both the kernel and $G$ are bounded on the interval $[0,T]$, there should be an unique solution to this equation. A way to find it is to consider the series defined by: $$F(x) = \sum\limits_{n=0}^\infty \lambda^n F_n(x)$$ $$F_0 (x) = G(x)$$ $$F_n (x) =\int\limits_0^{x-1} \text{d}t \; (x-s)^2 Z(s) F_{n-1}(x-s)$$

Z[x_,y_] := PolyGamma[2, 1 - 2 I (x-y)] - 0.5 PolyGamma[2, 1 - I(x-y)]

G[x_] :=
PolyGamma[1, 1 - 2 I x] - 0.5 PolyGamma[1, 1 - I x]

l1 = Table[{x, NIntegrate[s^2 G[s] Z[x,s], {s, 1, x }]}, {x,
1, 10000, 100}] // Gives the numerical values of the first function in the series.

f1 = Interpolation[l1, InterpolationOrder -> 2] // Interpolates the first  function.

l2 = Table[{x, NIntegrate[(s)^2 f1[s] Z[2 I (x-s)], {s, 1, x }]}, {x,
1, 10000, 100}] // Gives the numerical values of the second term in the series.

f2 = Interpolation[l2, InterpolationOrder -> 2] // Interpolates the second  function.


Etcetera.

Now, if i want to improve this to get higher precision, or perhaps if you have ideas to get really to the solution $F(x)$. Or maybe to write an unique code that does all the steps up to order $N$, as i don't know how to do that efficiently.. Thanks

• Writing the equations in Mathematica code would be a start :) – Öskå Nov 23 '14 at 15:27
• What is his? Please specify – chris Nov 23 '14 at 16:08
• I'm not sure what to make of the "polygamy function". Maybe I'll ask my wives.. – Daniel Lichtblau Nov 23 '14 at 17:12
• If this is a question about Mathematica, I would suggest a web search for 'volterra mathematica stackexchange" but without the quotes. There are several links to related questions on this forum and some have code that might be adaptable to your specific case (hard to tell since, as @Oska noted, you provided no code to start with). – Daniel Lichtblau Nov 23 '14 at 17:17
• ditto a search for "Fredholm equations" – Mike Honeychurch Nov 23 '14 at 22:35