Second order linear ODE with polynomial coefficients

Question

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!

Answer 1

Hint:

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

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

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

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

$\psi''(x)=e^{nx^3}u''(x)+3nx^2e^{nx^3}u'(x)+3nx^2e^{nx^3}u'(x)+(9n^2x^4+6nx)e^{nx^3}u(x)=e^{nx^3}u''(x)+6nx^2e^{nx^3}u'(x)+(9n^2x^4+6nx)e^{nx^3}u(x)$

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

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

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

$u''(x)=e^{kx}v''(x)+ke^{kx}v'(x)+ke^{kx}v'(x)+k^2e^{kx}v(x)=e^{kx}v''(x)+2ke^{kx}v'(x)+k^2e^{kx}v(x)$

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

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

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.