# Special Differential Equation (continued-2)

May you help me out in solving inhomogeneous differential equation looking like[this is radial part of Schrodinger equation]:

$$R^{\prime\prime}+\frac{1}{r}R^{\prime}-\frac{a^{2}}{r^{2}}R+b^{2}R-\frac{c^{2}}{r}\delta(r-\rho)R-d^{2}R=F\exp(ikr)$$

-
I added LaTeX formatting; please check whether it is what you meant. –  Johnny Westerling Oct 29 '12 at 15:19
Is $\delta(r-\rho)$ a dirac delta, or a step function or what? –  ja72 Oct 29 '12 at 15:41
It's a dirac delta function. –  nagendra Oct 29 '12 at 15:56
Where does $r$ lie? –  Pragabhava Oct 31 '12 at 4:00
r goes like 0<r<infinity. –  nagendra Oct 31 '12 at 13:38

First of all, let's write the equation in a cleaner way: $$R'' + \frac{1}{r}R' +\left(b^2 - d^2 - \tfrac{a^2}{r^2}\right)R = F e^{ikr} + \tfrac{c^2}{r} \delta(r-\rho) R \quad \Longrightarrow$$ $$\frac{d}{dr}\left\{r\frac{d R}{dr}\right\} + r \left(b^2 - d^2 - \tfrac{a^2}{r^2}\right)R = r F e^{ikr} + c^2 \delta(r-\rho)R$$ Now, supposing $F$ is a constant, integrating both sides of the equality \begin{multline} \int_{\rho-\epsilon}^{\rho+\epsilon} \frac{d}{dr}\left\{r\frac{d R}{dr}\right\} dr + \int_{\rho-\epsilon}^{\rho+\epsilon} r \left(b^2 - d^2 - \tfrac{a^2}{r^2}\right)R = \\ \int_{\rho-\epsilon}^{\rho+\epsilon}\left\{r F e^{ikr} + c^2 \delta(r-\rho) R\right\}dr. \end{multline} Supposing $R$ continous a.e., the second term on the LHS and the exponential term in the RHS will vanish as $\epsilon \to 0$, leaving $$\int_{\rho-\epsilon}^{\rho+\epsilon} \frac{d}{dr}\left\{r\frac{d R}{dr}\right\} dr= c^2 R(\rho),$$ and then you have the jump comdition $$\lim_{r \to \rho^+} \left(r\frac{d R}{dr}\right) - \lim_{r \to \rho^-} \left(r\frac{d R}{dr}\right) = c^2 R(\rho).$$
Now (where is $r$?), for $0 \le r < \rho$ $$R'' + \frac{1}{r}R' + \left(b^2 - d^2 - \tfrac{a^2}{r^2}\right)R = F e^{ikr},$$ and the same equation applies for $\rho < r < \infty$.
Note: I'm only going to solve the case $|d| \le |b|$, and $\textbf{F = 0}$. The inhomogeneous case can be solved the same way using variation of parameters (the case $|b| < |d|$ is analogous).
In this case, assuming that $r \in \mathbb{R}^+$, then $$R(r) = \begin{cases}k_1 J_a\big(\sqrt{b^2 - d^2}r\big), & 0\le r < \rho \\ \\ k_2 J_{a}\big(\sqrt{b^2 - d^2}r\big) + k_3 J_{-a}\big(\sqrt{b^2 - d^2}r\big), & \rho < r < \infty\end{cases}$$ where $J_{\pm a}$ are the Bessel functions of order $\pm a$. To determine the values of $k_1$, $k_2$ and $k_3$ you have, for example \begin{align} R_0&=k_1 J_a(0)\\ 0 &= k_2 J_a\big(\sqrt{b^2 - d^2}\rho\big) + k_3 J_{-a}\big(\sqrt{b^2 - d^2}\rho\big) - k_1 J_a\big(\sqrt{b^2 - d^2}\rho\big) \end{align} and \begin{multline} \frac{k_1}{\rho} J_a\big(\sqrt{b^2 - d^2}\rho\big) = \\ k_2 J_a'\big(\sqrt{b^2 - d^2}\rho\big) + k_3 J_{-a}'\big(\sqrt{b^2 - d^2}\rho\big) - k_1 J_a'\big(\sqrt{b^2 - d^2}\rho\big), \end{multline} where the first equation is the boundary condition, the second is the continuity condition, and the third is the jump condition.