Please tell me where I made a mistake? Or maybe I used the wrong method to solve it?
Link to my attempted solution: https://ru.overleaf.com/read/xcxsthdjpnmx#17beab
For the successive approximation method, I did the calculations in Wolfram Mathematica. For the second method - manually.
$\sigma_{\phi End}(\rho)=\sigma_{\phi Begin}(rho)+\int_{\rho}^{bEnd}\sigma_{\phi Begin}(s)\cdot\dfrac{s\cdot(a^2-\kappa\cdot\rho^2)}{\rho^2\cdot(a^2+\kappa\cdot s^2)}\,ds= \\ \\ =\sigma_{\phi Begin}(rho)-\int_{bEnd}^{\rho}\sigma_{\phi Begin}(s)\cdot\dfrac{s\cdot(a^2-\kappa\cdot\rho^2)}{\rho^2\cdot(a^2+\kappa\cdot s^2)}\,ds \\ $
Since this integral equation has a variable lower limit and a non-zero term, this is a Volterra equation of the 2nd kind.
Solution by the method of successive approximations:
General formula:
$K_{n+1}(\rho,s)=\int_{bEnd}^{\rho}K_{n}(\rho,t)\cdot K_{n}(t,s)\,ds \\ K1(\rho,s)=\dfrac{s}{a^2+\kappa\cdot s^2}; \\ K2(\rho,s)=\dfrac{s\cdot\ln\left(\dfrac{a^2+\kappa\cdot\rho^2}{a^2+\kappa\cdot s^2}\right)}{2\cdot\kappa\cdot(a^2+\kappa\cdot s^2)}; \\ \\ K3(\rho,s)=\dfrac{s\cdot\ln^3\left(\dfrac{a^2+\kappa\cdot\rho^2}{a^2+\kappa\cdot s^2}\right)}{48\cdot\kappa^3\cdot(a^2+\kappa\cdot s^2)}; \\ \\ \\ K4(\rho,s)=\dfrac{s\cdot\ln^7\left(\dfrac{a^2+\kappa\cdot\rho^2}{a^2+\kappa\cdot s^2}\right)}{645120\cdot\kappa^7\cdot(a^2+\kappa\cdot s^2)}; \\ \\ \\ K5(\rho,s)=\dfrac{s\cdot\ln^{15}\left(\dfrac{a^2+\kappa\cdot\rho^2}{a^2+\kappa\cdot s^2}\right)}{42849873690624000\cdot\kappa^{15}\cdot(a^2+\kappa\cdot s^2)};$
I didn’t see a general pattern in the results, so I tried to solve it using another method:
$\sigma_{\phi End}(\rho)=\sigma_{\phi Begin}(\rho)+\int_{\rho}^{bEnd}\sigma_{\phi Begin}(s)\cdot\dfrac{s\cdot(a^2-\kappa\cdot\rho^2)}{\rho^2\cdot(a^2+\kappa\cdot s^2)}\,ds \\ \dfrac{\rho^2}{a^2-\kappa\cdot\rho^2}\cdot\sigma_{\phi End}(\rho)=\dfrac{\rho^2}{a^2-\kappa\cdot\rho^2}\cdot\sigma_{\phi Begin}(\rho)+\int_{\rho}^{bEnd}\sigma_{\phi Begin}(s)\cdot\dfrac{s}{a^2+\kappa\cdot s^2}\,ds \\ \left(\dfrac{\rho^2}{a^2-\kappa\cdot\rho^2}\cdot\sigma_{\phi}(\rho)\right)^{'}_{\rho}=...=\dfrac{2\cdot a^2\cdot\rho\cdot\sigma_{\phi}(\rho)+\rho^2\cdot\sigma_{\phi}(\rho)^{'}_{\rho}\cdot(a^2-\kappa\cdot\rho^2)}{(a^2-\kappa\cdot\rho^2)^2} \\ \int_{\rho}^{bEnd}\sigma_{\phi Begin}(s)\cdot\dfrac{s}{a^2+\kappa\cdot s^2}\,ds=-\sigma_{\phi Begin}(\rho)\cdot\dfrac{\rho}{a^2+\kappa\cdot\rho^2} \\ \dfrac{2\cdot a^2\cdot\rho\cdot\sigma_{\phi End}(\rho)+\rho^2\cdot\sigma_{\phi End}(\rho)^{'}_{\rho}\cdot(a^2-\kappa\cdot\rho^2)}{(a^2-\kappa\cdot\rho^2)^2}= \\ =\dfrac{2\cdot a^2\cdot\rho\cdot\sigma_{\phi Begin}(\rho)+\rho^2\cdot\sigma_{\phi Begin}(\rho)^{'}_{\rho}\cdot(a^2-\kappa\cdot\rho^2)}{(a^2-\kappa\cdot\rho^2)^2}-\sigma_{\phi Begin}(\rho)\cdot\dfrac{\rho}{a^2+\kappa\cdot\rho^2}\,(*) \\ \dfrac{2\cdot a^2\cdot\rho\cdot\sigma_{\phi Begin}(\rho)}{(a^2-\kappa\cdot\rho^2)^2}-\sigma_{\phi Begin}(\rho)\cdot\dfrac{\rho}{a^2+\kappa\cdot\rho^2}=\dfrac{2\cdot a^2\cdot\rho\cdot\sigma_{\phi Begin}(\rho)}{(a^2-\kappa\cdot\rho^2)^2}\cdot(a^2+ \\ +\kappa\cdot\rho^2)-\sigma_{\phi Begin}(\rho)\cdot\dfrac{\rho}{a^2+\kappa\cdot\rho^2}\cdot(a^2-\kappa\cdot\rho^2)^2= \\ =\dfrac{2\cdot a^4\cdot\rho\cdot\sigma_{\phi B}(\rho)+2\cdot\kappa\cdot a^2\cdot\rho^3\cdot\sigma_{\phi B}(\rho)-\sigma_{\phi B}(\rho)\cdot a^4\cdot\rho+2\cdot\sigma_{\phi B}(\rho)\cdot\kappa\cdot a^2\cdot\rho^3-\sigma_{\phi B}(\rho)\cdot\kappa\cdot\rho^5}{(a^2+\kappa\cdot\rho^2)\cdot(a^2-\kappa\cdot\rho^2)^2}= \\ =\dfrac{a^4\cdot\rho\cdot\sigma_{\phi B}(\rho)+4\cdot\kappa\cdot a^2\cdot\rho^3\cdot\sigma_{\phi B}(\rho)-\sigma_{\phi B}(\rho)\cdot\kappa\cdot\rho^5}{(a^2+\kappa\cdot\rho^2)\cdot(a^2-\kappa\cdot\rho^2)^2}$
Returning to (*) there is also a rather unpleasant expression. Please tell me where I made a mistake in the calculations? Or tell me what method can be used to solve this equation?
