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I am trying to solve the following equation ($\ell\in \mathbb{N}$) $$f''(x)+\left(1-\dfrac{\ell(\ell+1)}{x^2}+\frac{2}{x}\right)f(x)=0,$$ with these two asymptotic boundary conditions (b.c.) on the function and its derivative: $$\lim_{x\rightarrow0}f'(x)=(\ell+1)[f(x)/x],$$ $$\lim_{x\rightarrow\infty}f(x)=\frac{1}{2i}\left[e^{ix}-e^{-i(x-\ell\pi)}\right].$$

The equation DiffEq = f''[x] + (1 - l(l+1)/x^2 + 2/x) f[x] == 0 can be analytically solved with DSolve, but I do not understand how to impose these two "analytical" b.c. to get rid of the two integration constants. I mean, they are not merely "numerical" b.c. and therefore I cannot simply do something like DSolve[{DiffEq,f[a]==b,f[c]==d},f[x],x], with a,b,c,d being some numbers.

Does anybody have an idea?

Thanks a lot!

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    $\begingroup$ Try applying Asymptotic with Assumptions -> x > 0 && l >= 0 && l ∈ Integers to the analytical solution at x ->0 and x -> Infinity. Note, however, that I tried doing so and ran into numerous complications. $\endgroup$ – bbgodfrey Sep 7 at 14:20
  • $\begingroup$ Are you only interested in symbolic solution? Also, can you add some background info? I know this is an equation appeared in separation of variable of Laplacian equation in spherical coordinates, but never see these b.c.s before. $\endgroup$ – xzczd Sep 8 at 4:02
  • $\begingroup$ @xzczd yes I am interested in the symbolic solution at first, in a second moment in the numerical one, though. The b.c.s are nothing but the usual one, stated in a different manner: at the origin, you want the reduced radial wavefunction to be zero, at infinity you want it to oscillate, as in the free-particle problem. These have been taken from a paper I am studying (arxiv.org/pdf/1901.10030.pdf), but I have also simplified it a little bit... there you can find more details, but in essence, it does not change. $\endgroup$ – Lele Sep 8 at 9:06
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DSolve[y''[x] + (1 - (l (l + 1))/x^2 + 2/x) y[x] == 0, y[x], x]
(*{{y[x] -> 
C[1] WhittakerM[-I, 1/2 (1 + 2 l), 2 I x] + 
C[2] WhittakerW[-I, 1/2 (1 + 2 l), 2 I x]}}*)

without any restrict for $l$ or $x$ or $y$.

This general solution offers the integration factors for matching the first requirement:

Table[Limit[
    D[FunctionExpand[WhittakerW[-I, 1/2 (1 + 2 l), 2 I x]], x], 
    x -> 0], {l, 1, 10, 1}]

output

Table[Limit[D[FunctionExpand[WhittakerM[-I, 1/2 (1 + 2 l), 2 I x]], x], 
  x -> 0], {l, 1, 10, 1}]

(*{0, 0, 0, 0, 0, 0, 0, 0, 0, 0}*)

Table[Limit[D[FunctionExpand[WhittakerM[-I, 1/2 (1 + 2 l),2 I x]], x], 
  x -> 0], {l, -10, 0, 1}]

(*{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1}*)

Table[Limit[(l + 1) FunctionExpand[
    WhittakerM[-I, 1/2 (1 + 2 l),2 I x]/x], x -> 0], {l, 1, 10, 1}]

(*{0, 0, 0, 0, 0, 0, 0, 0, 0, 0}*)

Table[Limit[(l + 1) FunctionExpand[
    WhittakerM[-I, 1/2 (1 + 2 l),2 I x]/x], x -> 0], {l, -10, 0, 1}]

(*{ComplexInfinity, ComplexInfinity, ComplexInfinity, ComplexInfinity, \
ComplexInfinity, ComplexInfinity, ComplexInfinity, ComplexInfinity, \
ComplexInfinity, Indeterminate, 1}*)

$$ \lim_{x\rightarrow 0} f'\left(x\right) =(\ell +1)\left(\frac{f (x)}{x}\right) $$

has not the power to give more information than that l<=0 give no contribution at x==0 to the solution. This condition can be satisfied!

The asymptotic behaviour can be treated with AsymptoticDSolveValue:

lsol1=AsymptoticDSolveValue[{y''[x] + (1 - (l (l + 1))/x^2 + 2/x) y[x] == 
   0}, y[x], {x, 0, 1}]

output

AsymptoticDSolveValue[{y''[x] + (1 - (l (l + 1))/x^2 + 2/x) y[x] == 
   0}, y[x], {x, 0, 3}]

output

is again independent of values for x and l. But the contributions from the integration constant c2 have to be discarded.

Table[(lsol1 /. C[2] -> 0)/x // FullSimplify, {l, 1, 10, 1}]

output

This is already very close to the asymptotic result:

Table[((l + 1) lsol1 /. C[2] -> 0)/x // FullSimplify, {l, 1, 10, 1}]

output

Table[(((l + 1) lsol1 /. C[2] -> 0)/x // FullSimplify) /. x -> 0, {l, 
  1, 10, 1}]

(*{0, 0, 0, 0, 0, 0, 0, 0, 0, 0}*)

This matches already the result in the closed-form.

$$ \lim_{x\rightarrow \infty }f(x) =\frac{1}{2 i}\left(e^{i x}-e^{-(i (x-\pi \ell ))}\right). $$

Can be solved in Mathematica with

FunctionExpand[WhittakerM[-I, 1/2 (1 + 2 l), 2 I x]]

From the discussion near x==0 the parameter c2==0 solves the problem. But it got clear that this solution diverges strong to infinity.

So for the solution for very large x C2 has to be zero!

To get a first impression l==1:

ReImPlot[{Re@
   FunctionExpand[WhittakerM[-I, 1/2 (1 + 2 1), 2 I x]], (3/(
    2 I) (E^(I x) - E^(-I ( x - 1 π))))}, {x, 0, 35}, 
 PlotRange -> Full]

ReImPlot of two functions under consideration

Examples that sustain the hypothesis are

ReImPlot[{FunctionExpand[
   WhittakerM[-I, 1/2 (1 + 2 2), 2 I x]], (30/(
    2 I) (E^(I x) - E^(-I ( x - 2 π))))}, {x, 1335, 1355}, 
 PlotRange -> Full]

Plot for large values for l==2

ReImPlot[{FunctionExpand[
   WhittakerM[-I, 1/2 (1 + 2 3), 2 I x]], (-400/(
    2 I) (E^(I x) - E^(-I ( x - 3 \[Pi]))))}, {x, 1335, 1355}, 
 PlotRange -> Full]

Plot for large values for l==3

This has close relevance to be true.

The is the Wolfram function repository https://functions.wolfram.com/ that kind of help:

symptotic series expansions (6 formulas)

From that this

asymptotic approximation

seems to the the most interesting:

$$ M_{\nu ,\mu }(z)\propto \Gamma (2 \mu +1) \left(\frac{e^{z/2} \left(O\left(\frac{1}{z}\right)+1\right) z^{-\nu }}{\Gamma \left(\mu -\nu +\frac{1}{2}\right)}+\frac{e^{-\frac{z}{2}} \left(O\left(\frac{1}{z}\right)+1\right) \left(z^{\mu +\frac{1}{2}} (-z)^{-\mu +\nu -\frac{1}{2}}\right)}{\Gamma \left(\mu +\nu +\frac{1}{2}\right)}\right)/;(\left| z\right| \to \infty) $$

With the parameters from Your question:

solution match from Wolfram functions repository

This has the two summands with the exponential function, but not the phase at first glance.

Thanks for Your interest. Hope this provides best advances.

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    $\begingroup$ Your conclusion, then, is that there is no solution for the problem as stated? $\endgroup$ – bbgodfrey Sep 7 at 15:51
  • $\begingroup$ Wow, thank you for your deep answer! I am not understanding, though, what is your conclusion regarding the solution at infinity, especially for what regards finding the integration constant C[1]. $\endgroup$ – Lele Sep 8 at 9:08

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