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I'm working towards the solution to a problem about involving the Landau-Zener transition, but I'm finding some difficulties. I need to estimate $ \,_1 F_1\left(\frac{\mathrm i}{4\epsilon},\frac12;\frac{\mathrm i t^2}{\epsilon}\right)$ as $\epsilon\rightarrow 0^+$, but I am not able to find any books or articles on this. So my first question is: how can we get the asymptotic series of this function near $\epsilon=0$?

Moreover, I have a further question: does the solution of the following Cauchy problem $$ \begin{cases} \epsilon^2x''(t)- 2 \mathrm i \epsilon h(t)x'(t)+x(t)=0,\\ x(0)=1,\\ x'(0)=0 \end{cases} $$ admit an universal asymptotic expansion as $\epsilon\rightarrow 0^+$ when $h(t)$ is an arbitrary function?

Edit: I tried to derive $\cos\frac t\epsilon$, but it leads to a divergent remainder. I plotted the function to reveal the difference. 1F1

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    $\begingroup$ Using Mehler-Heine asymptotics, $$\,_1 F_1\left(\frac{\mathrm i}{4\epsilon},\frac12;\frac{\mathrm i t^2}{\epsilon}\right)\sim\,_0 F_1\left(-,\frac12;\frac{-\mathrm t^2}{4\epsilon^2}\right)\sim\cos(t/\epsilon)\,\,\mathrm{as}\,\, \epsilon\rightarrow 0^+$$ $\endgroup$
    – Jorge Zuniga
    Commented Aug 30 at 5:03
  • $\begingroup$ @JorgeZuniga Thank you! Is there any article explaining how it is derived? $\endgroup$
    – Gnaphalium
    Commented Sep 2 at 3:28
  • $\begingroup$ Mehler Heine asymptotics is associated to hypergeometric polynomials but it can be generalized to hypergeometric functions as well. Pochammers' $(a)_k\sim a^k$ as $a\rightarrow\infty$. This asymptotically translates parameters to argument. The price you pay is that just the dominant or leading term is retained. For most cases this is enough. Here the argument is $\frac{i}{4\epsilon}\cdot\frac{it^2}{\epsilon}=\frac{- t^2}{4\epsilon^2}$ and$\,_0F_1(-,b,-z^2/4)=\Gamma(b)(z/2)^{1-b}J_{b-1}(z)$. Put $b=1/2$ and you are done $\endgroup$
    – Jorge Zuniga
    Commented Sep 2 at 13:51
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    $\begingroup$ Check this out publications.mfo.de/bitstream/handle/mfo/1069/… $\endgroup$
    – Jorge Zuniga
    Commented Sep 2 at 13:56
  • $\begingroup$ @JorgeZuniga, if you decide to post your comments as an answer, I'll upvote it $\endgroup$
    – Daniele Tampieri
    Commented Sep 2 at 16:08

1 Answer 1

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I had put it as a comment, but this answers the first question. For Pochhammers $(a)_n\sim\,a^n \,\,\mathrm{ as} \,\,a\rightarrow\infty$. Replacing this in the Taylor series summands of the hypergeometric $\,_1F_1$ function, the asymptotic variable is moved from the parameter list into the argument, a classical case that has been exhaustively studied. See for example DLMF. On applying this, the asymptotic series is lost after the first term, however the leading or dominant term of the asymptotic expansion is retained which is enough for many situations. Mehler Heine Asymptotics is usually associated to hypergeometric orthogonal polynomials but it works on generalized hypergeometric functions as well. See this Bracciali & Moreno-Balcazar report. For this particular case $$\,_1 F_1\left(\frac{\mathrm i}{4\epsilon},\frac12;\frac{\mathrm i t^2}{\epsilon}\right)\sim\,_0 F_1\left(-,\frac12;-\frac{t^2}{4\,\epsilon^2}\right)\,\,\mathrm{as}\,\,\epsilon\rightarrow\,0^+$$ but in terms of Bessel $J$ $$\,_0 F_1\left(-,b;-\frac{z^2}{4}\right)=\Gamma(b)\,\left(\frac z2\right)^{-(b-1)}\,J_{b-1}(z) $$ and $$\,_0 F_1\left(-,\frac 12;-\frac{z^2}{4}\right)=\cos(z)$$ Finally,$$\,_1 F_1\left(\frac{\mathrm i}{4\epsilon},\frac12;\frac{\mathrm i t^2}{\epsilon}\right)\sim\cos\left(\frac{t}{\epsilon}\right)\,\,\mathrm{as}\,\,\epsilon\rightarrow\,0^+$$

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  • $\begingroup$ I've edited the question. I cannot judge whether it's my misunderstanding or mistake. $\endgroup$
    – Gnaphalium
    Commented Sep 3 at 4:01
  • $\begingroup$ Of course, this gives a divergent but bounded remainder. $\endgroup$
    – Jorge Zuniga
    Commented Sep 3 at 13:24

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