# Real analysis: solve a dual integral equation that involves a zero-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 (0 one) ,, 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 look for a solution of the integral form.
$$begin {equation} f_0 (q) = int_0 ^ 1 lambda (t) sin (qt) , mathrm {d} t , label {integralWithHeaviside_sin} end {equation}$$
that clearly satisfies the integral equation for $$r> 1$$ making use of the relationship
$$begin {equation} int_0 ^ infty J_0 (qr) sin (qt) , mathrm {d} q = frac {H (tr)} {(t ^ 2-r ^ 2) ^ {1/2}} , , end {equation}$$
where $$H ( cdot)$$ Denotes the Heaviside function.

by $$0 , follows three successive integrations by parts that
$$begin {equation} 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 & # 39; & # 39; (1) -q ^ 2 lambda (1) right) cos (q) + q lambda & # 39; (1) without (q) – lambda & # 39; & # 39; (0) + q ^ 2 lambda (0) – int_0 ^ 1 lambda & # 39; & # 39; & # 39; (t) cos (qt) , mathrm {d} t bigg) ,. end {split} end {equation}$$

For the integral on the right side of the last equation to be convergent, we require that $$lambda (0) = lambda (1) = lambda & # 39; (1) = 0$$.
Thus, the last equation becomes
$$begin {equation} label {secondTermBending} int_0 ^ infty J_0 (qr) , mathrm {d} q int_0 ^ 1 q ^ 3 lambda (t) sin (qt) = – frac { lambda & # 39; & # 39; (0)} {r} – int_0 ^ r frac { lambda & # 39; & # 39; & # 39; (t) , mathrm {d} t} {(r ^ 2-t ^ 2) ^ {1/2}} ,, end {equation}$$
after using the identity
$$begin {equation} int_0 ^ infty J_0 (qr) cos (qt) , mathrm {d} q = frac {H (rt)} {(r ^ 2-t ^ 2) ^ {1/2}} , . label {integralWithHeaviside_cos} end {equation}$$

Therefore, the integral equation can be simplified as
$$begin {equation} frac { lambda & # 39; & # 39; (0)} {r} + int_0 ^ r frac { lambda & # 39; & # 39; & # 39; (t) , mathrm {d} t} {(r ^ 2-t ^ 2) ^ {1/2}} = -g (r) ,, end {equation}$$

By multiplying both members of the last equation by $$r / (s ^ 2-r ^ 2) ^ {1/2}$$ and integrating with respect to $$r$$ from 0 to $$s$$, the resulting equation reads
$$begin {equation} & # 39; & # 39; (s) = – frac {24 h ^ 3} { pi ^ 2} frac {s (3h ^ 2 – 5s ^ 2)} {(s ^ 2 + h ^ 2) ^ 5} ,. end {equation}$$

The last equation can now be easily solved.
But the problem is that we have required that $$lambda (0) = lambda (1) = 0$$ but also $$lambda & # 39; (1) = 0$$.
Since the final equation is second order EDO, only 2 boundary conditions are required.
I would be glad if someone here could clarify how this could be explained.

Thank you