Is it possible to solve the eigenvalue problem of $$y''(x) - 2\gamma\, y'(x) + [\lambda^2 + \gamma^2 - (\frac{x^2}{2}+\alpha)^2 + x]\, y(x)=0$$ where $\lambda$ is the eigenvalue and $\alpha,\gamma$ are parameters. The boundary condition is $y(\pm\infty)=0$.

Or instead of the eigenvalue problem, can I just solve it with freely running $\lambda$? Then probably I can tackle the eigenproblem by imposing the boundary condition. Hopefully, it is related to or can be transformed into some classical form with special function?


2 Answers 2


$y''(x)-2\gamma y'(x)+\left(\lambda^2+\gamma^2-\left(\dfrac{x^2}{2}+\alpha\right)^2+x\right)y(x)=0$

$y''(x)-2\gamma y'(x)-\left(\dfrac{x^4}{4}+\alpha x^2-x+\alpha^2-\lambda^2-\gamma^2\right)y(x)=0$

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

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


$\therefore e^{nx^3}u''(x)+6nx^2e^{nx^3}u'(x)+(9n^2x^4+6nx)e^{nx^3}u(x)-2\gamma(e^{nx^3}u'(x)+3nx^2e^{nx^3}u(x))-\left(\dfrac{x^4}{4}+\alpha x^2-x+\alpha^2-\lambda^2-\gamma^2\right)e^{nx^3}u(x)=0$

$u''(x)+(6nx^2-2\gamma)u'(x)+\left(\dfrac{(36n^2-1)x^4}{4}-(6\gamma n+\alpha)x^2+(6n+1)x-\alpha^2+\lambda^2+\gamma^2\right)u(x)=0$

Choose $36n^2-1=0$ , i.e. $n=-\dfrac{1}{6}$ , the ODE becomes


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

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


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

$v''(x)-(x^2+2\gamma-2k)v'(x)-((\alpha-\gamma+k)x^2+\alpha^2-\lambda^2-\gamma^2-k^2+2\gamma k)v(x)=0$

Choose $k=\gamma-\alpha$ , the ODE becomes


Which relates to Heun's Triconfluent Equation.

Alternatively, choose $n=\dfrac{1}{6}$ , the ODE becomes


Choose another $k=\alpha+\gamma$ , the ODE simplify to


Which relates to Heun's Triconfluent Equation.

  • 1
    $\begingroup$ Thank you! I see it relates to Heun's functions. But could you show how it relates to the triconfluent one? Looks somewhat different? BTW, is $y(x)=e^{-x(\frac{1}{6}x^2+\alpha-\gamma)}u(x)$ a better choice or not? $\endgroup$
    – xiaohuamao
    Commented Oct 6, 2020 at 5:50
  • 1
    $\begingroup$ Thanks for the update! May I ask a question? To write down a complete general solution, is it necessary to also add the other one using a slightly different transform like $y(x)=e^{x(\frac{1}{6}x^2+\alpha+\gamma)}u(x)$ (just like this answer)? I think for 2nd-order ODEs this is usually the case and the two linearly independent solutions are known (e.g., Kummer and Tricomi for confluent hypergeometric equation). But for this one, the choice of the 2nd transform seems not quite unique and the linear independence is not clear either? $\endgroup$
    – xiaohuamao
    Commented Oct 6, 2020 at 18:45
  • $\begingroup$ How to make the coefficient of $v'$ match the form in the linked equation? By an imaginary shift in $x$? $\endgroup$
    – xiaohuamao
    Commented Oct 6, 2020 at 18:47

Starting with $$ y''(x) - 2\gamma\, y'(x) + [\lambda^2 + \gamma^2 - (\frac{x^2}{2}+\alpha)^2 + x]\, y(x)=0, $$ let $y = e^{\gamma t} f$. Then the above is reduced to potential form $$ (e^{\gamma t}f''+2\gamma e^{\gamma t}f'+\gamma^2e^{\gamma t}f) -2\gamma(e^{\gamma t}f'+\gamma e^{\gamma t}f) +[\lambda^2 + \gamma^2 - (\frac{x^2}{2}+\alpha)^2 + x]e^{\gamma t}f=0 \\ f''+(\lambda^2-(\frac{x^2}{2}+\alpha)^2+x)f=0. $$ This may be written as an eigenfunction problem in standard potential form: $$ -f''+\left((\frac{x^2}{2}+\alpha)^2-x\right)f=\lambda^2 f. $$


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