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In the equation

$$R''(r)+\frac5rR'(r)+\frac4{r^2}LR(r)+2ER(r)-\frac2rUR(r)=0\tag1$$

$L$ and $U$ are constant matrices, and $E$ is a scalar. I'm looking for a solution, which decays to $0$ as $r\to\infty$. I've been able to find asymptotic solution looking like

$$R(r)\sim\exp\left(-\sqrt{-2E}r\right)\;\text{as }r\to\infty.\tag2$$

But I'd like to have at least one $U$-dependent term so as to know ratios of the components of $R$ at infinity. I tried to multiply by $r$ both sides of the equation and take $r\to\infty$, but this didn't work because the $R''$ and $2ER$ terms appeared to be dominant and went to infinity (these are the terms which helped me get $(2)$).

So how can I get the next term in the expansion of the solution at infinity?

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    i would put $R(r)=\exp(-\sqrt{-2E}r)+f(r,U)$ where $f(r,U)/\exp(...)\rightarrow 0$ as $r\rightarrow \infty$ and see if you can get something out of it – 2017-02-25
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    @tired the equation I get for $f$ then is almost as $(1)$, but with additional nonhomogeneous RHS, which is proportional to $\exp(-\sqrt{-2E}r)$. I'm not sure how $f(r,U)/\exp(...)\to0$ would help, since if I use it, I only get that the nonhomogeneity vanishes at infinity, which is obvious. – 2017-02-25
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    Do $L$ $U$ commute? If yes you can simultaneously diagonalize them and the equation will be solved in terms of hypergeometric functions. But it seems that you get this one by solving the Dirac equation, so they probably don't))) – 2017-02-27
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    @KirylPesotski no they don't commute. And no, this is not from Dirac equation, it's from a reduced Schrödinger equation. – 2017-02-28

1 Answers 1

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We have a nice fact about the terms with $L$ and $U$: one of the terms is much smaller than the other, and moreover, it's $L$-dependent term which goes to zero faster:

$$\frac4{r^2}L-\frac2rU\sim-\frac2rU\text{ as }r\to\infty.\tag{I}$$

This gives us the relation with $U$ but without $L$:

$$R''(r)+\frac5rR'(r)-\frac2rUR(r)\sim-2ER(r)\text{ as }r\to\infty.\tag{II}$$

Now we have only one matrix, $U$, and this allows us to switch to a basis where it's diagonal. If we denote

$$U=KWK^{-1},\tag{III}$$

where $W$ is a diagonal matrix with $u_i$ on the diagonal ($i=1,2,...,n$ for $n\times n$ matrix $U$), then, denoting

$$Q(r)=K^{-1}R(r),\tag{IV}$$

we can transform the relation $(\mathrm{II})$ to

$$Q''(r)+\frac5rQ'(r)-\frac2rWQ(r)\sim-2EQ(r)\text{ as }r\to\infty.\tag{V}$$

But since $W$ is diagonal, we can actually split this system into a set of independent relations for components $Q_i$ of $Q$:

$$Q_i''(r)+\frac5rQ_i'(r)-\frac2r u_i Q_i(r)\sim-2EQ_i(r)\text{ as }r\to\infty.\tag{VI}$$

These relations can be treated by the usual method of dominant balance as follows. Set

$$Q_i(r)=\exp(S_i(r)).\tag{VII}$$

We'll get

$$(S_i'(r))^2+S_i''(r)+\frac5r S_i'(r)\sim\frac2ru_i-2E.\tag{VIII}$$

Assuming $(S_i'(r))^2\gg S_i''(r)$ as $r\to\infty$, we drop $S_i''$ term, simplifying $(\mathrm{VIII})$ to

$$(S_i'(r))^2+\frac5r S_i'(r)\sim\frac2ru_i-2E.\tag{IX}$$

Solving quadratic equation for $S_i'(r)$ and integrating, we get

$$S_i(r)\sim-\frac52\ln r\pm\int\limits_1^r\sqrt{\frac{25}{4r^2}+\frac2ru_i-2E}\,\mathrm dr+C_i.\tag X$$

As we want the decaying solution, we choose $\pm\to-$. Thus we have a set of $Q_i$, each with its own arbitrary constant factor of $e^{C_i}$. Going back to original basis as given by $(\mathrm{IV})$, we finally get our $R(r)$, up to $n$ arbitrary constants, which should be defined from other considerations.