# Solving numerically a system of two dual integral equations for the unknowns $f_1(t)$ and $f_2(t)$

The two unknown functions $$f_1(t)$$ and $$f_2(t)$$ are the solutions of the following system of dual integral equations defined by the double integrals \begin{align} \int_0^\infty \mathrm{d}\lambda \, \lambda^{-\frac{1}{2}} J_{1}(\lambda r) \int_0^1 \mathrm{d}t \Big( \left( 1 + e^{-\lambda}(\lambda-1) \right) J_{\frac{3}{2}} (\lambda t) f_1(t) -e^{-\lambda} \lambda J_{\frac{1}{2}} (\lambda t) f_2(t) \Big) = -\frac{16r}{\left(1+4r^2\right)^{\frac{3}{2}}} \, , \\ \int_0^\infty \mathrm{d}\lambda \, \lambda^{-\frac{1}{2}} J_{0}(\lambda r) \int_0^1 \mathrm{d}t \Big( -e^{-\lambda} \lambda J_{\frac{3}{2}} (\lambda t) f_1(t) +\left( 1 + e^{-\lambda}(\lambda+1) \right) J_{\frac{1}{2}} (\lambda t) f_2(t) \Big) = \frac{16\left(1+2r^2\right)}{\left(1+4r^2\right)^{\frac{3}{2}}} \, , \end{align} where $$r \in [0,1]$$, and $$J_\alpha$$ denotes the $$\alpha$$th order Bessel function of the first kind. i have spend a couple days trying to figure out how this can be solved analytically, eventually by reducing the integrals to classical Abel integral equation, but unfortunately without success.

Probably the problem can better be dealt with using numerical evaluations.

Does anyone here have perhaps an idea on how to proceed with those equations? Your help is highly appreciated.

EDIT:

Applying inverse Hankel transform does not hold here because $$r \in [0,1]$$. This would be possible if the integral equation is defined for $$r \in [0,\infty)$$. The answer below gives the solution in this case.

Thank you

Not an answer: the following proposes a solution when the integral equations are defined for $$r\in[0,\infty).$$
These equations can be decoupled by remarking that the outer integrals on $$\lambda$$ can be put in the form of a Hankel transform defined as $$\begin{equation} \mathcal{H}_\nu\left[F(\lambda);\lambda\to r\right]=\int_0^\infty F(\lambda)J_\nu(\lambda r)\sqrt{\lambda r}\,d\lambda \end{equation}$$ This transform is self-inverting if $$\nu>-1/2$$. The system can be written as \begin{align} r^{-1/2}\mathcal{H}_1\left[\lambda^{-1}\int_0^1 \mathrm{d}t \Big( \left( 1 + e^{-\lambda}(\lambda-1) \right) J_{\frac{3}{2}} (\lambda t) f_1(t) -e^{-\lambda} \lambda J_{\frac{1}{2}} (\lambda t) f_2(t) \Big);\lambda\to r\right] = -\frac{16r}{\left(1+4r^2\right)^{\frac{3}{2}}} \\ r^{-1/2}\mathcal{H}_0\left[\lambda^{-1} \int_0^1 \mathrm{d}t \Big( -e^{-\lambda} \lambda J_{\frac{3}{2}} (\lambda t) f_1(t) +\left( 1 + e^{-\lambda}(\lambda+1) \right) J_{\frac{1}{2}} (\lambda t) f_2(t) \Big);\lambda\to r\right] = \frac{16\left(1+2r^2\right)}{\left(1+4r^2\right)^{\frac{3}{2}}} \end{align} Using tabulated Hankel transforms from Ederlyi TII 8.5.18 and 8.5.19 p.24, we find \begin{align} \mathcal{H}_1\left[\frac{-16r^{3/2}}{\left(1+4r^2\right)^{\frac{3}{2}}};r\to\lambda\right]&=-2\lambda^{1/2}e^{-\lambda/2}\\ \mathcal{H}_0\left[\frac{16r^{1/2}\left(1+2r^2\right)}{\left(1+4r^2\right)^{\frac{3}{2}}};r\to\lambda\right]&=\mathcal{H}_0\left[\frac{8r^{1/2}}{\left(1+4r^2\right)^{\frac{3}{2}}};r\to\lambda\right] +\mathcal{H}_0\left[\frac{8r^{1/2}}{\left(1+4r^2\right)^{\frac{1}{2}}};r\to\lambda\right]\\ &=16\lambda^{1/2}e^{-\lambda/2}+8\lambda^{-1/2}e^{-\lambda/2} \end{align} Hankel inversion of the above system gives thus \begin{align} \left( 1 + e^{-\lambda}(\lambda-1) \right)\int_0^1 J_{\frac{3}{2}} (\lambda t) f_1(t)\,dt -e^{-\lambda} \lambda\int_0^1 J_{\frac{1}{2}} (\lambda t) f_2(t) \,dt &= -2\lambda^{3/2}e^{-\lambda/2} \\ -e^{-\lambda} \lambda \int_0^1 J_{\frac{3}{2}} (\lambda t) f_1(t)\,dt +\left( 1 + e^{-\lambda}(\lambda+1) \right)\int_0^1 J_{\frac{1}{2}} (\lambda t) f_2(t)\,dt &= 16\lambda^{3/2}e^{-\lambda/2}+8\lambda^{1/2}e^{-\lambda/2} \end{align} Defining $$\begin{equation} F_1\left( \lambda \right)=\int_0^1 J_{\frac{3}{2}} (\lambda t) f_1(t)\,dt\;;\;F_2\left( \lambda \right)=\int_0^1 J_{\frac{1}{2}} (\lambda t) f_2(t)\,dt \end{equation}$$ the system can be solved to find explicit expressions for $$F_{1,2}(\lambda)$$, from which $$f_{1,2}(r)$$ can be numerically evaluated using standard methods for integral equations.
• We actually overlooked an important detail here. The variable $r \in [0,1]$ thus employing inverse Hankel transform here is not possible... These equations do not apply if $r > 1$. Nov 18, 2019 at 10:50