Consider the 2nd-order linear ODE $x f^{''}(x) + x (\beta - 2 \alpha x) \kappa / \sigma f^{'}(x) - 1 / \sigma \left[ 2 \alpha \kappa - \lambda^2 (\beta - 2 \alpha x)^2 \right] f(x) = 0$, where $\sigma>0$, $\beta>0$, $\alpha<0$ and $\kappa>0$. The ODE has a non-essential singularity at $x=0$, and may be regarded as an eigenvalue equation for $\lambda$. The ODE may be cast in the self-adjoint form, $\left[ e^{a x + b x^2} f_{\lambda}^{'}(x) \right]^{'} - 2 \alpha \kappa /\sigma e^{a x + b x^2} f_{\lambda}(x) + \lambda^2 (\beta - 2 \alpha x)^2 /x e^{a x + b x^2} f_{\lambda}(x) = 0 $, where $a = \kappa \beta / \sigma$ and $b = - \alpha \kappa / \sigma $. If one requires the solution be regular at the origin, then

  1. Is it possible to ensure the normalisation (for the eigenfunctions), $\int_0^{\infty} dx \left[ (\beta - 2 \alpha x)^2 / x \right] e^{a x + b x^2} f_{\lambda_1}(x) f_{\lambda_2}(x) = \delta(\lambda_1 - \lambda_2)$, where $\delta(x)$ is the Dirac delta function?

  2. What is the asymptotic behaviour for $f_{\lambda}(x)$? (I expect the set of eigenfunctions to be complete; I suppose that $f_{\lambda}(x) / \sqrt{\rho(x)}$ should behave like sine waves for large $x$ (??), where $\rho(x)$ is the weight function, and thus expect $f_{\lambda}(x) \sim \left[ x / (\beta - 2 \alpha x)^2 \right]^{1/2} e^{-(a x + b x^2)/2} $. Is this correct?)

  3. How may one relate an eigenfunction satisfying the above normalisation (if the latter is possible at all) with the Frobenius series solution at the origin. For example, the regular solution $f^{0}_{\lambda}(x)$ at the origin goes as $x^c$, $c>0$. One may define the series solution to be $f^{0}_{\lambda}(x) = x^c g(\lambda, x)$, where $g(\lambda, 0) = 1$. How does it relate to the eigenfunction satisfying the delta-function normalisation? i.e. What is $N(\lambda)$ in $f_{\lambda}(x) = N(\lambda) f^{0}_{\lambda}(x)$? Does $N(\lambda)$ depend on $\lambda$ at all?

Thanks a lot.

P.S.: I'm sorry. The claims above are almost all wrong. In the following I shall write down some of my observations. Would someone please comment (on whether the following arguments are valid or not.)

  1. Expanding the solution in a Frobenius series $f_{\lambda}(x) = x^c \sum_{i=0}^{\infty} a_i x^i $ gives the indicial eq. $a_0 c (-1 + c) = 0$, which implies $c=0$ or $c=1$. The physical situation dictates that one should choose the $c=1$ solution.
  2. Now we are left with one solution for each $\lambda$ (i.e. the one which is regular at the origin). The problem is whether $\lambda$ should be continuous or discrete. To answer this we have to turn to the asymptotics of the ODE for large $x$.
  3. If one performs the substitution $f_{\lambda}(x) \rightarrow e^{-(a x+b x^2)/2} g_{\lambda}(x)$, one obtains \begin{eqnarray} g^{''}(x) &=& \left\{ \left[ \frac{k_0^2}{4 \sigma^2} (- 2 \alpha x + \eta)^2 +\frac{k_0 \alpha}{\sigma} \right] - \frac{1}{\sigma} \frac{(- 2 \alpha x + \eta)^2}{x}\lambda^2 \right\} g(x) \\ &&=\left[ q(x) + p(x) \lambda^2 \right] g(x) . \end{eqnarray} Note that for a given $\lambda$, for large enough $x$, $q(x) + p(x) \lambda^2 > 0$. I seem to find a theorem which says in this case the function ceases to oscillate. If this is the case, then it means the function either decays or explodes for large $x$.
  4. Indeed, the only special term seems to be the $1/x$-term in $p(x)$. For large $x$, try dropping this term. Then $q(x) + p(x) \lambda^2$ is just a quadratic polynomial of $x$, and the equation is nothing but the parabolic cylinder differential equation. It seems that one solution decays as $e^{-b/2 x^2}$ while the other blows up (as $e^{b/2 x^2}$?).
  5. Summing up, the solution that is regular at the origin may blow up at infinity for arbitrary $\lambda$, and it decays only for some (discrete) values of $\lambda$. In the completeness relation for the eigenfuntions, one should sum over discrete values of $\lambda$ rather than enforcing a continuous summation.
  6. May one compare the above system with the Hermite diff. eq. or Laguerre diff. eq.?

1 Answer 1


Your equation has a regular singular point at 0 and an irregular singular point of rank 2 at infinity. This means it can be transformed to the biconfluent Heun equation. There is some literature on this equation. The book by Ronveaux is a place to start looking.

  • $\begingroup$ Thanks for pointing that out. I'll certainly take a look at the references to see what I'll find. $\endgroup$
    – user51524
    Jun 4, 2014 at 10:17

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