# Asymptotic boundary conditions for differential equation

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!

• 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. – bbgodfrey Sep 7 at 14:20
• 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. – xzczd Sep 8 at 4:02
• @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. – Lele Sep 8 at 9:06

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}]


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}]


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


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}]


This is already very close to the asymptotic result:

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


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]


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]


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]


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:

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