Finding a particular solution for $\frac{\mathrm{d}^2R}{\mathrm{d}r^2}+\frac1r\frac{\mathrm{d}R}{\mathrm{d}r}+\alpha^2R=J_0(\alpha r)$

Question

Motivation

I have the following non-homogeneous Bessel differential equation $$\frac{\mathrm{d}^2R}{\mathrm{d}r^2}+\frac1r\frac{\mathrm{d}R}{\mathrm{d}r}+\alpha^2R=J_0(\alpha r)$$

I want to find the general solution for this ODE. I know that the general solution can be written as

$$R=R_h+R_p$$

where $R_h$ is the basis for the homogeneous ODE and is known to be

$$R_h=C_1 \, J_0(\alpha r) + C_2 \, Y_0(\alpha r).$$

Next, for finding the particular solution $R_p$ one can use the method of variation of parameters to obtain a particular solution using the homogeneous ones. However, this will lead to some messy solution with hard integrals! As the MAPLE or WOLFRAM both use this technique their result is not valuable for me. I know that

$$R_p=\frac1{2 \alpha^2}\left[\alpha rJ_1(\alpha r)\right] \tag{*}$$

and it can be verified by putting it into the ODE. In fact, I saw this in some published paper.

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Question

Is there an elegant method to obtain such a particular solution mentioned in $(*)$?

Answer 1

Start with the equation $$ \frac{d^2}{dr^2}J_{0}(\alpha r)+\frac{1}{r}\frac{d}{dr}J_{0}(\alpha r)+\alpha^2J_{0}(\alpha r)=0. $$ $J_0$ has an everywhere convergent power series. So you can differentiate with respect to $\alpha$, and interchange orders of differentiation: $$ \frac{d^2}{dr^2}(rJ_{0}'(\alpha r))+\frac{1}{r}\frac{d}{dr}(rJ_0'(\alpha r))+\alpha^2(rJ_0'(\alpha r))+2\alpha J_{0}(\alpha r)=0. $$ Therefore $R(r)=-\frac{r}{2\alpha}J_{0}'(\alpha r)=-\frac{1}{2\alpha^2}(\alpha r)J_{0}'(\alpha r)$ is a solution of $$ R''(r)+\frac{1}{r}R'(r)+\alpha^2R(r)=J_0(\alpha r). $$

Answer 2

Using variation of the constant method. Looking for the solution in the form $$R_p = C(r)J_0(\alpha r),$$ then $$\dfrac d{dr}R_p = \dfrac d{dr}C(r)J_0(\alpha r) + C(r)\dfrac d{dr}J_0(\alpha r),$$ $$\dfrac{d^2}{dr^2}R_p = \dfrac{d^2}{dr^2}C(r)J_0(\alpha r) + 2\dfrac d{dr}C(r)\dfrac d{dr}J_0(\alpha r) + C(r)\dfrac{d^2}{dr^2}J_0(\alpha r).$$ Substitution to non-homogenius equation gives: $$J_0(\alpha r)\dfrac{d^2}{dr^2}C(r) + \left(2\dfrac{d}{dr}J_0(\alpha r)+\dfrac1rJ_0(\alpha r)\right)\dfrac{d}{dr}C(r) + \left(\dfrac{d^2}{dr^2}J_0(\alpha r)+\dfrac1r\dfrac d{dr}J_0(\alpha r)+\alpha^2J_0(\alpha r)\right)C(r) = J_0(\alpha r),$$ $$J_0(\alpha r)\dfrac{d^2}{dr^2}C(r) + \left(\dfrac1rJ_0(\alpha r) + 2\dfrac d{dr}J_0(\alpha r)\right)\dfrac{d}{dr}C(r) = J_0(\alpha r),$$ Do substitution: $$D(r)=\dfrac{d}{dr}C(r),$$ then $$J_0(\alpha r)\dfrac{d}{dr}D(r) + \dfrac1rJ_0(\alpha r)D(r) + 2\dfrac d{dr}J_0(\alpha r)D(r) = J_0(\alpha r),$$ $$J_0^2(\alpha r)D(r) + 2rJ_0(\alpha r)\dfrac d{dr}J_0(\alpha r)D(r) + rJ_0^2(\alpha r)\dfrac{d}{dr}D(r) = rJ_0^2(\alpha r),$$ $$\dfrac{d}{dr}\left(rJ_0^2(\alpha r)D(r)\right) = rJ_0^2(\alpha r)$$
Use the formula $\int zJ_0^2(z)dz = \dfrac{z^2}2\left(J_0^2(z)+J_1^2(z)\right):$ $$rJ_0^2(\alpha r)D(r) = \dfrac{r^2}{2}\left(J_0^2(\alpha r)+J_1^2(\alpha r)\right),$$ $$D(r) = \dfrac{r}{2}\dfrac{J_1^2(\alpha r)+J_1^2(\alpha r)}{J_0^2(\alpha r)},$$ $$\dfrac d{dr}C(r) = \dfrac1{2\alpha}\dfrac{d}{dr}\left(\dfrac{rJ_1(\alpha r)}{J_0(\alpha r)}\right),$$ where $$\dfrac{dJ_0(\alpha r)}{dr}=-\alpha J_1(\alpha r),\quad \dfrac{dJ_1(\alpha r)}{dr}=\alpha\left(J_0(\alpha r)-\dfrac1{\alpha r}J_1(\alpha r)\right)$$

$$C(r) = \dfrac r{2\alpha}\dfrac{J_1(\alpha r)}{J_0(\alpha r)},$$ $$\boxed{R_p = \dfrac r{2\alpha}J_1(\alpha r)}$$

And the same way for $Y_0(\alpha r).$