Consider the following dual integral equations (which arise in some elasticity situations) $$ \int_0^\infty q^3 f(q) J_0 (qr) \, \mathrm{d}q = g(r)\, , \quad\quad (0<r<R) \, , \\ \int_0^\infty f(q) J_0 (qr) \, \mathrm{d}q = 0 \, , \quad\quad (r>R) \, , $$ wherein $g(r)$ is a known function.
For its solution, I put $$ f (q) = \int_0^R \chi(t) \sin (qt) \, \mathrm{d} t \, , $$ which clearly satisfies the equation for $r>R$ owing to the identity $$ \int_0^\infty J_0 (qr) \sin(qt) \, \mathrm{d} q = \frac{H(t-r)}{(t^2-r^2)^{1/2}} \, , $$ with $H$ being the Heaviside step function. Substituting this solution form into the equation for $0<r<R$ and integrating successively by parts, we obtain $$ \int_0^R \chi(t) \, \mathrm{d} t \int_0^\infty q^3 J_0 (qr) \sin (qt) \, \mathrm{d} q = \\\int_0^\infty J_0(qr) \, \mathrm{d} q \bigg( \left(\chi''(R)-q^2 \chi(R)\right) \cos(qR) + q \chi'(R) \sin(qR) \\-\chi''(0) + q^2 \chi(0) - \int_0^R \chi'''(t) \cos(qt) \, \mathrm{d} t \bigg) \, . $$
By making use of $$ \int_0^\infty J_0 (qr) \cos(qt) \, \mathrm{d} q = \frac{H(r-t)}{(r^2-t^2)^{1/2}} \, , $$ an under some assumptions, we can write $$ -\int_0^r \frac{\chi'''(t) \, \mathrm{d}t}{(r^2-t^2)^{1/2}} = g(r) \, , \quad\quad (0<r<R) \, . $$
Multiplying both members by $r/(s^2-r^2)^{1/2}$, integrating with respect to $r$ between 0 and $s$ and interchanging the integration order we obtain the following $$ -\frac{\pi}{2} \chi''(s) = \int_0^s \frac{r g(r) \, \mathrm{d}r}{(s^2-r^2)^{1/2}}=: h(s) \, , \quad\quad (0<s<R) \, , $$ upon using the result $$ \int_t^s \frac{r\, \mathrm{d} r}{(s^2-r^2)^{1/2} (r^2-t^2)^{1/2}} = \frac{\pi}{2} \, . $$
Finally, by integrating the equation giving $\chi''(s)$ twice we obtain the solution up to 2 unknown integration constants. My question is what are the assumptions that should be made for the determination of the unknown integration constants, notably in after the integration by part?
Your help and hints are very appreciated.
Thanks
Federiko