Finding the particular solution to $ \phi''(r) + \frac{1}{r}\phi'(r)+C_1\phi(r) = C_2 J_0(\alpha r/a) $

Question

My question is similar to this one, but it's slightly different:

$$ \phi''(r) + \frac{1}{r}\phi'(r)+C_1\phi(r) = C_2 J_0(\alpha r/a) $$

where $C_1$, $C_2$ and $a$ are constants. For the particular solution, I tried a solution similar to the form that was outlined in that question, namely:

$$ \phi_p (r) = C r J_1 (\alpha r/a) $$

but after plugging it in I couldn't get $0=0$, though I admit I am not very familiar with Bessel functions and their identities. So my question is what would be the particular solution to try, as well as the homogeneous solution (if it's different from the referenced question)?

Answer 1

I think that the particular solution is rather $$\phi_p (r) = K J_0\left(\frac{\alpha r}{a}\right)$$ If you replace in the equation and use the relation between Bessel functions of different orders, you should get $$J_0\left(\frac{r \alpha }{a}\right) \frac{a^2 ({C_2}-{C_1} K)+\alpha ^2 K}{a^2}=0\implies K=\frac{a^2 {C_2}}{a^2 {C_1}-\alpha ^2} $$

Answer 2

Claude's fine solution of course fails when $J_0(\alpha r/a)$ is already a solution of the homogeneous differential equation. That is, when $C_1 = \alpha^2/a^2$. But (unlike trigonometric cases) you cannot simply multiply by $r$ to get your new candidates for undetermined coefficients.

In that case, variation of parameters still produces a solution. Maple gets

$$ \phi_p(r) = \frac{-\pi r^2 C_2}{4}\,{{ J}_{1}\left({\frac {\alpha\, r}{a}}\right)} \left( {{ J}_{0}\left({\frac {\alpha\,r }{a}}\right)}{{ Y}_{1}\left({\frac {\alpha\,r}{a}}\right)}-{{ J} _{1}\left({\frac {\alpha\,r}{a}}\right)}{{ Y}_{0}\left({\frac { \alpha\,r}{a}}\right)} \right) $$