Volterra integral equations

[math]u(x) = f(x) + \int_a^x K(x,s)x(s)ds[/math]

I tried to somehow use the fact that it can be rewritten as the Fredholm integral equation in a triangle area:
[math]\hat{K}(x,s)=\begin{cases}
K(x,s) & a\leq s\leq x\\
0 & x\leq s\leq b
\end{cases}[/math]

but so far no luck.

>>7805285
Assume separability. For the zeroing condition to hold, the s-dependent functions in the expansion of the kernel need to also depend on x (or how else do you make them zero?). This clearly contradicts separability.

>>7805447
But can't I zero out the x-dependant function in those cases? Let's assume
[math]K(x,s ) = \sum_{i=1}^NA_i(x)B_i(s )[/math]
and then
[math]\forall x<s A_i(x)=0[/math]?

>>7805314
I don't know what this means. You have K(x,s) in there... is that a typo? How is x an argument of K and also a function of s and also the argument of u and also a limit of integration?

>>7805489
But then the x functions will depend on s? Still not separable.

>>7805628
Hmm, yeah, that makes sense. Thank you.
>>7805604
Oh yeah, there's a typo, let me fix this:
[math]u(x) = f(x) + \int_a^x K(x,s)u(s)ds, x \in [a,b][/math]