I am currently stuck at solving an ODE of the form \begin{equation} 0=\psi''(x)+(\varepsilon-(\alpha x^2-\beta)^2)\psi(x) \end{equation} where $\alpha,\beta$ and $\varepsilon$ are real parameters. Anticipating $\psi\sim\mathrm{e}^{-\alpha x^3/3}$ for large $x$, I made the Ansatz $\psi(x) = f(x)\mathrm{e}^{-\alpha x^3/3}$ and obtain the ODE \begin{equation} 0=f''(x)-2\alpha x^2 f'(x)+(\varepsilon-\beta^2+2\alpha\beta x^2-2\alpha x)f(x). \end{equation} Actually, I am looking for a normalized solution $\sim\mathrm{e}^{-\alpha|x|^3/3}$, where extra care needs to be taken around $x=0$. (I am a physics student and not so much concerned about mathematical rigor right now.) But any solution would be a first step of course. I tried a variety of things to solve the second equation:

  • A powerseries approach $f(x)=\sum_{n=0}^\infty a_n x^n$. I obtain\begin{equation}a_{n+2}=\frac{(\varepsilon-\beta^2)a_n-2\alpha(n+1)a_{n-1}+2\alpha\beta a_{n-2}}{(n+2)(n+1)}\end{equation} for $n\geq 2$, which does not look very promising.
  • I also checked A. Polyanins and V.F. Zaitsevs Handbook of Exact Solutions of Ordinary Differential Equations but failed to transform the above ODEs into one given in the book. Wolfram Mathematica was not helpful either.
  • I transformed the equation into an system of first order linear ODEs with $g=f'$. However such an approach is only helpful for constant coefficients, right?

I am thankful for any ideas or hints. A particular solution would be very helpful of course, as it would allow me to reduce the order of the ODE and find all solutions. Note that, while $\alpha$ and $\beta$ are given, I expect constraints on $\varepsilon$ for (normalized) solutions to exist, i.e. I would also appreciate a solution for special values of $\varepsilon$, e.g. $\varepsilon=\beta^2$.

Sorry for any grammatical errors. Thank you for your help!

  • $\begingroup$ Is there a reason why you expect a closed form solution to exist? Also, I can't see why the power series approach you took 'does not look very promising'. Whats wrong with it? $\endgroup$
    – mattos
    Dec 3, 2019 at 12:27
  • $\begingroup$ No, there might be no closed form solution, but I thought the equation looks somewhat standard. Well, the power series approach might be helpful, but I thought the recursion relation is too complicated... Do you think I should try to investigate it further? $\endgroup$
    – ciabatta
    Dec 3, 2019 at 12:30
  • 1
    $\begingroup$ Well, in my opinion, that isn't a complicated recursion. Unless someone comes up with a clever transformation or finds the result in a book somewhere, I would say the power series is your best choice. You could also compute the solution numerically and see if you can glean any information from it. +1 for the good question by the way. $\endgroup$
    – mattos
    Dec 3, 2019 at 12:40
  • 1
    $\begingroup$ If you consider $ε$ as small against the other term, then the WKB approximation gives basis solutions $(αx^2−β)^{-\frac12}\exp(\pm(\frac13αx^3−βx))$ for large $x$, which in the main is $\exp(\pm\frac13αx^3)/x$. $\endgroup$ Dec 3, 2019 at 12:50
  • $\begingroup$ Okay, thank you both for the quick reply. I will check both suggestions. $\endgroup$
    – ciabatta
    Dec 3, 2019 at 13:01

1 Answer 1



$\psi''(x)+(\varepsilon-(\alpha x^2-\beta)^2)\psi(x)=0$

$\psi''(x)-(\alpha^2x^4-2\alpha\beta x^2+\beta^2-\varepsilon)\psi(x)=0$

Let $\psi(x)=e^{nx^3}u(x)$ ,

Then $\psi'(x)=e^{nx^3}u'(x)+3nx^2e^{nx^3}u(x)$


$\therefore e^{nx^3}u''(x)+6nx^2e^{nx^3}u'(x)+(9n^2x^4+6nx)e^{nx^3}u(x)-(\alpha^2x^4-2\alpha\beta x^2+\beta^2-\varepsilon)e^{nx^3}u(x)=0$

$u''(x)+6nx^2u'(x)+((9n^2-\alpha^2)x^4+2\alpha\beta x^2+6nx+\varepsilon-\beta^2)u(x)=0$

Choose $9n^2-\alpha^2=0$ , i.e. $n=\dfrac{\alpha}{3}$ , the ODE becomes

$u''(x)+2\alpha x^2u'(x)+(2\alpha\beta x^2+2\alpha x+\varepsilon-\beta^2)u(x)=0$

Let $u(x)=e^{kx}v(x)$ ,

Then $u'(x)=e^{kx}v'(x)+ke^{kx}v(x)$


$\therefore e^{kx}v''(x)+2ke^{kx}v'(x)+k^2e^{kx}v(x)+2\alpha x^2(e^{kx}v'(x)+ke^{kx}v(x))+(2\alpha\beta x^2+2\alpha x+\varepsilon-\beta^2)e^{kx}v(x)=0$

$v''(x)+(2\alpha x^2+2k)v'(x)+(2\alpha(k+\beta)x^2+2\alpha x+k^2+\varepsilon-\beta^2)v(x)=0$

Choose $k=-\beta$ , the ODE becomes

$v''(x)+2(\alpha x^2-\beta)v'(x)+(2\alpha x+\varepsilon)v(x)=0$

Which relates to Heun's Triconfluent Equation.


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