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Just a quick question! I'm trying to solve a Volterra equation using the Laplace transform.

$$\ u(t) = t - \int_{0}^{t} (t-s) u(s) ds $$

where kernel $\ k(t,s) = t-s $. I've the solution for this question but what happens if the kernel is simplified to $\ k(s) = s$ such that: $$\ u(t) = t - \int_{0}^{t} s u(s) ds $$

In this case, $\ f(t)=t $ but $\ k(t) = 0$. Then the Laplace transform of k would be $\ L[k]=0 $? It doesn't feel right. What do you think?

Another question: I saw another Volterra question here: solution of a volterra equation but why did he differentiate the equation instead of using the Laplace transform method?

Thanks!!

hardmath
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beginner
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  • You seem to have substituted $t=0$ only in one place, but you must replace all occurrences of $t$ in the first equation, which gives $u(0)=0$ as expected. – hardmath Oct 23 '17 at 18:07
  • @hardmath, Hi there. Sorry, I edited the question. I meant to say that the kernel is simplified to $ s$ and no longer $ (t - s)$. – beginner Oct 23 '17 at 19:01
  • The notation is still difficult to follow, in part because you omitted "the Volterra equation" per se, so that the mentions of $f(t)$ and $k(t)$ are not grounded. – hardmath Oct 23 '17 at 19:31
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    @hardmath oh, I'm sorry. So, how do I solve the 2nd equation? – beginner Oct 23 '17 at 20:39

1 Answers1

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This is the OP's "second equation" that is to be solved (see Comments):

$$ \ u(t) = t - \int_{0}^{t} s\; u(s)\; ds $$

A typical approach to solving such integral equations is by fixed point iteration.

That is, suppose we define $u_0(t) \equiv 0$. Then for successive indexes we compute:

$$ u_{k+1}(t) = t - \int_0^t s\; u_k(s)\; ds $$

As a result we get these first few iterations:

$$ u_1(t) = t - 0 = t $$

$$ u_2(t) = t - \frac{t^3}{3} $$

$$ u_3(t) = t - \frac{t^3}{3} + \frac{t^5}{15} $$

$$ u_4(t) = t - \frac{t^3}{3} + \frac{t^5}{15} - \frac{t^7}{105} $$

Clearly (by induction) the limit of sequence $u_k$ is the power series:

$$ u(t) = \sum_{n=0}^{\infty} (-1)^n \frac{t^{2n+1}}{(2n+1)!!} $$

I don't know right off what the "closed form" of this power series might be, but the Answers to the previous Question Double factorial series suggest that it is related to an exponential integral.

hardmath
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