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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)?

Shawn Zamperini
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2 Answers2

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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 $$J_0\left(\frac{r \alpha }{a}\right) \frac{a^2 ({C_2}-{C_1} K)+\alpha ^2 K}{a^2}=0\implies K=\frac{a^2 {C_2}}{a^2 {C_1}-\alpha ^2} $$

Claude Leibovici
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  • This is the solution! Much thanks. But this isn't to say @GEdgar's response is incorrect, only that in my case $C_1 \neq \alpha^2/a^2$, so Claude's solution applies here (fortunately too, because the math is easier!). – Shawn Zamperini Sep 04 '18 at 12:10
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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

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

GEdgar
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  • I'd like to make sure I understand your solution. $C_1$ and $C_2$ are known constants, i.e. a grouping of variables that I figured was easier to just group together (so $C_1$ does not equal $\alpha^2/a^2$ in this case). But if the solution to the homogeneous equation is, as referenced in the other question, $\phi = A J_0(\alpha r/a) + B Y_0(\alpha r/a)$, then I guess your point still stands. Does that sound correct? – Shawn Zamperini Sep 02 '18 at 16:45