Volterra integral equations

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Anonymous

Volterra integral equations 2016-01-23 22:50:28 Post No. 7805285

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Volterra integral equations 2016-01-23 22:50:28 Post No. 7805285

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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

>>

[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.

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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

[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]?

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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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