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.


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


1 Answer 1


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.

  • $\begingroup$ Thanks Paul a lot for your formidable answer. How do you mean by standard methods? The ones based on expanding the unknown solution into a polynomial function and solving a linear system for the unknown coefficients? A reference or article would be helpful $\endgroup$
    – keynes
    Commented Nov 17, 2019 at 11:09
  • $\begingroup$ Combining your great approach to this one one can nicely obtain the solution of the problem. Thanks for helping! $\endgroup$
    – keynes
    Commented Nov 17, 2019 at 13:27
  • 1
    $\begingroup$ Happy it helped! $\endgroup$
    – Paul Enta
    Commented Nov 17, 2019 at 13:29
  • $\begingroup$ 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 $r > 1$. $\endgroup$
    – keynes
    Commented Nov 18, 2019 at 10:50
  • $\begingroup$ Sorry about that! I'll delete this answer soon. (Also, thank you for the generous bounty you wanted to give). $\endgroup$
    – Paul Enta
    Commented Nov 18, 2019 at 17:53

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