Volterra equation for a Bessel type IVP that appears in inverse scattering

Question

I have the following i.v.p. (Colton Kress-Inverse acoustic and electromagnetic scattering theory, Springer) $$y''(r)+(k^2n(r)-\frac{l(l+1)}{r^2})y(r)=0$$ $$y(0)=0, y'(0)=1$$ using the Liouville transformation: $z(ξ)=y(r)n(r)^{1/4},\ ξ=\int_0^{r}n(t)dt$, the problem is transformed to $$z''(ξ)+(k^2-\frac{l(l+1)}{ξ^2}-g(ξ))z(ξ)=0$$ $$z(0)=0,\ z'(0)=\frac{1}{n(0)^{1/4}}$$ for a suitable function g.

I want to write the solution as a Volterra integral equation. In the case that l=0, it is easy to find that $$z(ξ)=\frac{sin(kξ)}{kn(0)^{1/4}}+1/k\int_0^{ξ}sin(k(ξ-t))g(t)z(t)dt$$ If $l>0$, I would expect solutions in the form $\sqrt{ξ} J_{l+1/2}(ξ)$ but I can't find the way.

Answer 1

Using Maple I am obtaining

$$z \left( \xi \right) =C\sqrt {\xi}{{\rm J_{L+1/2}}\left(\,\xi\,k\right)} +1/2\,\sqrt {\xi}\pi \, \left( {{\rm J_{L+1/2}}\left(\,\xi\,k\right)} \int _{0}^{\xi}\!\sqrt {\eta}{{\rm Y_{L+1/2,}}\left(\,\eta\,k\right)}g \left( \eta \right) z \left( \eta \right) {d\eta}- {{\rm Y_{L+1/2,}}\left(\,\xi\,k\right)}\int _{0}^{\xi}\!\sqrt {\eta} {{\rm J_{L+1/2,}}\left(\,\eta\,k\right)}g \left( \eta \right) z \left( \eta \right) {d\eta} \right) $$

Do you agree?

Answer 2

$y''(r)+\left(k^2n(r)-\dfrac{l(l+1)}{r^2}\right)y(r)=0$

$\int_0^ry''(t)~dt+\int_0^r\left(k^2n(t)-\dfrac{l(l+1)}{t^2}\right)y(t)~dt=0$

$[y'(t)]_0^r+\int_0^r\left(k^2n(t)-\dfrac{l(l+1)}{t^2}\right)y(t)~dt=0$

$y'(r)-y'(0)+\int_0^r\left(k^2n(t)-\dfrac{l(l+1)}{t^2}\right)y(t)~dt=0$

$y'(r)+\int_0^r\left(k^2n(t)-\dfrac{l(l+1)}{t^2}\right)y(t)~dt-1=0$

$\int_0^ry'(t)~dt+\int_0^r\int_0^r\left(k^2n(t)-\dfrac{l(l+1)}{t^2}\right)y(t)~dt~dt=\int_0^rdt$

$[y(t)]_0^r+\int_0^r\int_0^r\left(k^2n(t)-\dfrac{l(l+1)}{t^2}\right)y(t)~dt~dt=[t]_0^r$

$y(r)-y(0)+\int_0^r\int_0^r\left(k^2n(t)-\dfrac{l(l+1)}{t^2}\right)y(t)~dt~dt=r$

$y(r)+\int_0^r\int_0^r\left(k^2n(t)-\dfrac{l(l+1)}{t^2}\right)y(t)~dt~dt=r$