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

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

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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

1 Answer 1

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.

Lastly, I've noted that you haven't included neither the domain nor initial/boundary conditions in any of your questions. This very problematic, as the I/B conds. are part of the differential operator, and strongly determine the solution.

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