# Solving a dual integral equation involving a zeroth-order Bessel function

Consider the following dual integral equations

\begin{align} \int_0^\infty q^3 f_0(q) J_0 (qr) \, \mathrm{d} q &= g(r) \qquad\qquad\quad (01) \, , \end{align} where $$g(r) = \frac{9}{4\pi} \frac{16h^6-72h^4 r^2 + 18h^2 r^4 +r^6}{(h^2+r^2)^{11/2}} \, .$$

We search a solution of the integral form $$$$f_0 (q) = \int_0^1 \lambda(t) \sin (qt) \, \mathrm{d} t \, , \label{integralWithHeaviside_sin}$$$$ which clearly satisfies the integral equation for $$r>1$$ by making use of the relation $$$$\int_0^\infty J_0 (qr) \sin(qt) \, \mathrm{d} q = \frac{H(t-r)}{(t^2-r^2)^{1/2}} \, ,$$$$ where $$H(\cdot)$$ denotes Heaviside function.

For $$0, it follows from three successive integration by parts that $$$$\label{longEqBending} \begin{split} \int_0^\infty & J_0(qr) \, \mathrm{d} q \int_0^1 q^3 \lambda(t)\sin(qt) \, \mathrm{d} t \\ &= \int_0^\infty J_0(qr) \, \mathrm{d} q \bigg( \left(\lambda''(1)-q^2 \lambda(1)\right) \cos(q) + q \lambda'(1) \sin(q) -\lambda''(0) + q^2 \lambda(0) \\ &\qquad- \int_0^1 \lambda'''(t) \cos(qt) \, \mathrm{d} t \bigg) \, . \end{split}$$$$

For the integral on the right-hand side of the latter equation to be convergent, we require that $$\lambda(0)=\lambda(1) =\lambda'(1)=0$$. Thus, the latter equation becomes $$$$\label{secondTermBending} \int_0^\infty J_0(qr) \, \mathrm{d} q \int_0^1 q^3 \lambda(t)\sin(qt) = - \frac{\lambda''(0)}{r} - \int_0^r \frac{\lambda'''(t) \, \mathrm{d} t}{(r^2-t^2)^{1/2}} \, ,$$$$ after using identity $$$$\int_0^\infty J_0 (qr) \cos(qt) \, \mathrm{d} q = \frac{H(r-t)}{(r^2-t^2)^{1/2}} \, . \label{integralWithHeaviside_cos}$$$$

Thus, the integral equation can be simplified as $$$$\frac{\lambda''(0)}{r} + \int_0^r \frac{\lambda'''(t) \, \mathrm{d} t}{(r^2-t^2)^{1/2}} = -g(r) \, ,$$$$

By multiplying both members of the latter equation by $$r/(s^2-r^2)^{1/2}$$ and integrating with respect to $$r$$ from 0 to $$s$$, the resulting equation reads $$$$\lambda''(s) = - \frac{24 h^3}{\pi^2 } \frac{s(3h^2 - 5s^2)}{(s^2+h^2)^5} \, .$$$$

The latter equation can now be easily solved. But the problem is that we have required that $$\lambda(0)=\lambda(1)=0$$ but also $$\lambda'(1)=0$$. As the final equation is a second order ODE, only 2 boundary conditions are in principal required. I would be grateful if someone here could be of help and clarify how this could be explained.

Thank you!

• The first thing I would do is to find the function $\lambda$ and then check if the conditions are satisfied (or better yet, that it makes your intermediate expression convergent, which is the whole reason you introduced them). – Yuriy S Nov 29 '18 at 13:59
• @YuriyS Apparently the expression of λ as calculated from the last math step with λ(0)=λ(1)=0 does not satisfy the condition λ′(1)=0. Accordingly, one intermediate step would in principle NOT lead to a convergence of the overall integral. Maybe the first derivative should just be by construction discontinuous? Any inputs are highly desirable – Daddy Nov 29 '18 at 16:43