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Consider the following ODE eigenproblem of $y(x)$ \begin{equation} y'' + [\varepsilon + b^2 x - (a + \frac{b^2}{2}x^2)^2 ] y=0 \end{equation} with eigenvalue $\varepsilon$, real constants $a,b$. The boundary condition is $y(\pm\infty)=0$. Numerically, this turns out to have well-behaved eigensolutions.

My question is how to see the typical length scale of the eigensolution $y(x)$, i.e., how it asymptotically decays. For instance, if $y(x)\sim e^{-x^2/c^2}$, $c$ is the length scale I mean.


This ODE can also be shown to have the following general solution \begin{equation} y(x)= \sum_{s=\pm} C_s\, e^{-arx_s - \frac{x_s^3}{2}} \mathscr{H}_\mathrm{T}(\alpha,\beta_s,\gamma,x_s) \end{equation} with integration constants $C_\pm$, $r=(\frac{3}{b^2})^{\frac{1}{3}}$, $\alpha=r^2\varepsilon,\beta_\pm=\pm3,\gamma=2ra,x_\pm=\pm x/r$ and $\mathscr{H}_\mathrm{T}$ the triconfluent Heun's function. However, its asymptotics is not solely determined by the exponential factor, because $\mathscr{H}_\mathrm{T}$ is not truncated to be a finite polynomial for these $\beta$'s, although overall $y(x)$ decays well. So it's not clear to me whether this general solution helps the above question.

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The quick-and-dirty way to guess the asymptotic behavior is to substitute in a WKB ansatz $y = e^{S(x)}$ and keep only the leading order terms, which here would be the highest power of $x$ and the highest power of $S'(x)$, namely $(S')^2 - (x^2 b^2/2)^2 = \text{l.o.t}$. The solution is $S(x) \sim \pm x^3 b^2/6$ as $|x| \to \infty$. This is exactly the leading asymptotic for $S(x)$ captured by your general solution.

One can go through a longer algorithmic procedure (for any singular ODE with meromorphic coefficients, of which yours is a special case) to get very precise information about the full asymptotic expansion. I give a reference and go through an example calculation for a different equation in this answer.

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  • $\begingroup$ Thank you. Maybe I missed something basic. Using the general solution form with the boundary condition, I did numerically obtain quickly decayed solutions $y(x)$ as a whole. However, as I mentioned, the H_T does not truncate and diverges well beyond the exponential suppression, i.e., each of $s=\pm$ part in the solution diverges at both $x=\pm\infty$ (numerically I can also see this). Only the sum gives the finite result. This seems different from the cubic exponential asymptotics. Or I missunderstood? $\endgroup$
    – xiaohuamao
    May 26 at 12:35
  • $\begingroup$ @xiaohuamao The logic that I outlined does not depend on any representation exact solutions and deduces the asymptotics directly from the equation. I don't really know anything about H_T or how it is used to get the exact solution, so I can't really comment on your numerical experiments. If you want to know the asymptotics of H_T, instead of $y(x)$, you should probably make that the focus of your question. $\endgroup$ May 26 at 13:06
  • $\begingroup$ Thank you very much. In other words, if I substitute $y(x)=\exp{[-arx_s - \frac{x_s^3}{2}]}v(x)$, the equation for $v(x)$ by all means diverges faster than the $\exp[\pm x^3b^2/6]$. What I said is not just numerics, it is an exact result from the polynomial truncation condition of H_T (in the link in my question). I somewhat feel this suggests something else than the simple asymptotics. But I'm not familiar with this field and probably being stupid. So I just would like to confirm with experts: you don't find any contradiction between the asymptotics and what I described? $\endgroup$
    – xiaohuamao
    May 26 at 13:27
  • $\begingroup$ @xiaohuamao You say that "the equation for $v(x)$ by all means diverges faster than the $\exp[\pm x^3 b^2/6]$." I really don't see why you are so convinced of this. The reference you linked says nothing of the kind, unless I'm missing something very basic too. $\endgroup$ May 26 at 13:42
  • $\begingroup$ Yes, the link only says it does not truncate in this case. Since the H_T equation has an irregular singularity at $\infty$, diverging faster than that exponential is possible, I suppose. Then, by numerics in Mathematica, each part does diverge that way for all the eigenvalues I tried. And once I sum the two parts, everything looks perfect from all that I expect. Therefore I feel this is the case. So if we assume this is true, do you find any contradiction or not? $\endgroup$
    – xiaohuamao
    May 26 at 13:52

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