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Volterra integral equations
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Why can't the Kernel in the Volterra integral equation ever be separable? My textbook says it's obvious due to K(x,s)=0 at s>x, but I don't see how that helps to prove it.
u(x) = f(x) + \int_a^x K(x,s)x(s)ds
>>
$u(x) = f(x) + \int_a^x K(x,s)x(s)ds$
>>
I tried to somehow use the fact that it can be rewritten as the Fredholm integral equation in a triangle area:
$\hat{K}(x,s)=\begin{cases} K(x,s) & a\leq s\leq x\\ 0 & x\leq s\leq b \end{cases}$

but so far no luck.
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>>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.
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>>7805447
But can't I zero out the x-dependant function in those cases? Let's assume
$K(x,s ) = \sum_{i=1}^NA_i(x)B_i(s )$
and then
$\forall x<s A_i(x)=0$?
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>>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?
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>>7805489
But then the x functions will depend on s? Still not separable.
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>>7805628
Hmm, yeah, that makes sense. Thank you.
>>7805604
Oh yeah, there's a typo, let me fix this:
$u(x) = f(x) + \int_a^x K(x,s)u(s)ds, x \in [a,b]$