- #1
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I'm interested in studying the following equation, solving for [itex]\phi(s)[/itex] given [itex] y(s)[/itex]:
[tex]\int_0^s \frac{2r}{\sqrt{s^2-r^2}} \phi(r)dr=y(s)[/tex]
or in more standard form:
[tex]\int_0^s K(s,r)\phi(r)=y(s)[/tex]
This is how I think it should be approached:
The kernel,[itex]K(s,r)[/itex] is singular at [itex]s=r[/itex]. Thus, the first step is to transform it to a finite kernel via composition with [itex] \sqrt{\xi-r}[/itex]. Once this is done, then the resulting equation can be further transformed to a Volterra equation of the second kind which then can be solved by Picard's process of successive approximations. I realize effecting the integrations likely becomes intractable but I'd still like to determine the solution format.
Can anyone tell me if this is the correct approach to follow? I'll spend time with this approach and report here my progress.
Thanks,
Salty
[tex]\int_0^s \frac{2r}{\sqrt{s^2-r^2}} \phi(r)dr=y(s)[/tex]
or in more standard form:
[tex]\int_0^s K(s,r)\phi(r)=y(s)[/tex]
This is how I think it should be approached:
The kernel,[itex]K(s,r)[/itex] is singular at [itex]s=r[/itex]. Thus, the first step is to transform it to a finite kernel via composition with [itex] \sqrt{\xi-r}[/itex]. Once this is done, then the resulting equation can be further transformed to a Volterra equation of the second kind which then can be solved by Picard's process of successive approximations. I realize effecting the integrations likely becomes intractable but I'd still like to determine the solution format.
Can anyone tell me if this is the correct approach to follow? I'll spend time with this approach and report here my progress.
Thanks,
Salty