# Asymptotic behavior of an ODE

Consider the following ODE eigenproblem of $$y(x)$$ $$$$y'' + [\varepsilon + b^2 x - (a + \frac{b^2}{2}x^2)^2 ] y=0$$$$ with eigenvalue $$\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 $$$$y(x)= \sum_{s=\pm} C_s\, e^{-arx_s - \frac{x_s^3}{2}} \mathscr{H}_\mathrm{T}(\alpha,\beta_s,\gamma,x_s)$$$$ with integration constants $$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.

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.
• 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? May 26 at 12:35
• @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. May 26 at 13:06
• 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? May 26 at 13:27
• @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. May 26 at 13:42
• 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? May 26 at 13:52