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 <r<1) , \ int_0^infty f_0(q) J_0 (qr) , mathrm{d} q &= 0 ,quadqquadqquadquad (r>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 <r <1 $, 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