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I'm solving the following Schroedinger equation in the domain $r>0$

$\psi''(r) + \left(E-\frac{a}{r^b}\right)\psi(r)=0 $,

where $0 < b < 2$ and $a, E$ are positive constants. Primarily I'm interested in the asymptotical power behavior of the solution as $r\to 0$. To be complete in the description of the problem, I fix my boundary conditions at $r\to +\infty$ as a plane wave ansatz.

I did a lot of DSolve with Mathematica and found out that $\psi(r)\to const$ as $r\to 0$. It gave me a hint for the power series solution, that one of the terms in the expansion

$\psi(r) = \sum\limits_i a_i r^{\alpha_i}$

for some (noninteger) $\alpha_i>0$ should cancel with $a/r^b$. However, even with this assumption there are a bunch of terms which do not cancel. Power expansion does not seem to work in this case as well as the WKB approximation (double checked numerically). What are the other known methods to find an approximate asymptotic behavior in this case? A good reference would be great too!

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Probably a stupid and unhelpful comment, but have you tried letting $r=1/s$ with $s \to \infty$? –  Zen Harper Feb 23 '11 at 8:01
    
Well, it's just changing $r^\alpha \to s^{-\alpha}$, it does not affect power counting... –  Peter Feb 23 '11 at 18:44
    
I am not an expert on differential equations, so ignore my comments if they sound stupid! Letting $s=1/r$ pushes the singularity out to infinity, but I'm not sure it's of much use. But also, if you let $\psi = \exp f$ then you change the second order linear equation into a first order nonlinear one. Maybe this is not simpler, but at least it's different! –  Zen Harper Feb 25 '11 at 1:58
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1 Answer

up vote 4 down vote accepted

As $r$ approaches zero, the coefficient of $\psi$ becomes dominated by the contribution of $-a/r^b$. This means that in the vicinity of zero your solution is dominated by solution of the following equation $$ \psi''-\frac{a}{r^b}\psi=0. $$ This can be demonstrated more rigorously (and, also, refined to the higher accuracy) by scaling into the vicinity of $r=0$. This equation possesses an explicit solution in terms of the modified Bessel functions: $$ \psi=C_1\sqrt{r}K_{\frac{1}{2-b}}\left(\frac{2\sqrt{a}}{b-2}r^{1-b/2}\right)+C_2\sqrt{r}I_{\frac{1}{2-b}}\left(\frac{2\sqrt{a}}{b-2}r^{1-b/2}\right). $$ The latest edition of NIST Handbook tells us that for $z\to 0$ $$ I_{\nu}(z)\sim (\frac{1}{2}z)^{\nu}/\Gamma(\nu+1) \quad\mbox{and}\quad K_{\nu}(z)\sim \frac{1}{2}\Gamma(\nu)(\frac{1}{2}z)^{-\nu} $$ so the actual constant at $r\to0$ depends only on $C_1$, unless $C_1$ vanishes. It seems pretty clear that there are no singularities here; typically you should expect that $$ \psi\sim \frac{C_1}{\Gamma(\frac{3-b}{2-b})}\left(\frac{\sqrt{a}}{b-2}\right)^{\frac{1}{2-b}} \quad\mbox{for}\quad r\to 0. $$ Specific values of $C_1$ and $C_2$ are less trivial to obtain, but they can be found by matching this limiting solution to another solution that is valid further away from $r=0$.

The method of matched asymptotic expansions is often used to solve this type of problems asymptotically, see e.g. Kevorkian & Cole (1996) "Multiple Scale and Singular Perturbation Methods", Springer. In this method you would need to construct two solutions: "inner" solution, valid as $r\to 0$, (it is the solution derived above) and "outer" solution, valid as $r\to\infty$ (which you took to be a plane wave). Some ingenuity may be needed to ensure that the ranges of validity of these two solutions overlap, so that they can actually be matched.

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Thanks! I was just working with Mathematica, apparently it didn't know about that solution. So yeah, it's a good point to check with thextbooks from time to time. –  Peter Mar 2 '11 at 19:08
    
I knew this equation from Section 8.4 of Gradshteyn and Ryzhik's "Table of Integrals, Series and Products" (this is where they collected various properties of Bessel functions). Maple 7 was also albe to integrate it. However, more generally, you should treat yourself to a copy of Polyanin and Zaitsev's "Handbook of Exact Solutions for Ordinary Differential Equations" which, I suspect, still beats any current computer algebra system hands down. –  Aleksey Pichugin Mar 2 '11 at 19:23
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