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Suppose we have the differential equation $$ -f''(x) + \Big( \frac{1}{4} x^2 - \nu - \frac{1}{2} \Big) \, f(x) = 0 \: , \quad \text{where } \nu \in \mathbb{R} \: . $$ The general solution to this equation is $f(x) = a \, D_\nu(x) + b \, D_\nu(-x)$, where $D_\nu$ is the parabolic cylinder function.

If $\nu$ is an non-negative integer, then $D_\nu(x) \propto \exp(-x^2/4)$ is nice and bounded. My question is: are there any $\nu$ other than the non-negative integers, such that $$ \lim_{x \to -\infty} D_\nu(x) = 0 $$ (ie. $D_\nu$ vanishes as $x$ goes to negative infinity)?

From trying various different $\nu$ numerically, it seems that the answer is no, however I wasn't able to prove it. The Digital Library of Mathematical Functions includes asymptotic expansions of the parabolic cylinder functions, I just don't know how to use them to prove this.

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  • $\begingroup$ For $\nu=0,1,2\ldots$ it does not become a polynomial but rather a polynomial times a decaying exponential. $\endgroup$
    – Gary
    Jul 8, 2021 at 14:22
  • $\begingroup$ @Gary Ayy, my bad! The part “polynomial (nice and bounded)” is nonsense by itself :) For $\nu \in \mathbb N$ the parabolic cylinder function becomes $$D_\nu(w) = 2^{\frac{\nu}{2}} \exp \big({-}\tfrac{w^2}{4} \big) \, H_\nu \big( \frac{w}{\sqrt{2}} \big)$$. Thanks for pointing this out, I'll edit the question! $\endgroup$
    – m93a
    Jul 8, 2021 at 14:29
  • $\begingroup$ Are you happy with the anwer I provided? $\endgroup$
    – Gary
    Jul 8, 2021 at 16:11

2 Answers 2

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By http://dlmf.nist.gov/12.9.E3, $$ D_\nu (xe^{ \pm \pi i} ) = e^{ \pm \pi \nu i} e^{ - \frac{1}{4}x^2 } x^\nu \left( {1 + \mathcal{O}\!\left( {\frac{1}{{x^2 }}} \right)} \right) + \frac{{\sqrt {2\pi } }}{{\Gamma ( - \nu )}}e^{\frac{1}{4}x^2 } x^{ - \nu - 1} \left( {1 + \mathcal{O}\!\left( {\frac{1}{{x^2 }}} \right)} \right) $$ as $x\to +\infty$. Since the parabolic cylinder function is a single valued entire function, taking the average of the two sides gives $$ D_\nu ( - x) = e^{ - \frac{1}{4}x^2 } x^\nu \left( {\cos (\pi \nu ) +\mathcal{O}\!\left( {\frac{1}{{x^2 }}} \right)} \right) + \frac{{\sqrt {2\pi } }}{{\Gamma ( - \nu )}}e^{\frac{1}{4}x^2 } x^{ - \nu - 1} \left( {1 + \mathcal{O}\!\left( {\frac{1}{{x^2 }}} \right)} \right), $$ as $x\to +\infty$. This shows that $\lim _{x \to - \infty } D_\nu (x) = 0$ if and only if $\nu$ is a non-negative integer.

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  • $\begingroup$ Thank you for the short and elegant proof! It's still unclear to me why do you take the average of the two estimates. Is it simply to make the result real? $\endgroup$
    – m93a
    Jul 8, 2021 at 16:23
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    $\begingroup$ Yes. This has to do with Stokes lines. If you like, you can use $$ D_\nu ( - x) = \cos (\pi \nu )D_\nu (x) + \frac{\pi }{{\Gamma ( - \nu )}}V\left( { - \nu - \tfrac{1}{2},x} \right) $$ (cf. dlmf.nist.gov/12.2.E15) together with dlmf.nist.gov/12.9.E1 and dlmf.nist.gov/12.9.E2 to get the same result without averaging. $\endgroup$
    – Gary
    Jul 8, 2021 at 16:31
  • $\begingroup$ Awesome, thanks! $\endgroup$
    – m93a
    Jul 8, 2021 at 16:39
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When $\nu$ is a positive even integer, you don't even need the asymptotic expansion - the Taylor series is enough.

One can write the parabolic cylinder function in terms of a confluent hypergeometric (for $\operatorname{Re}(z)>0$): $$D_{\nu}(z)=2^{\nu/2}e^{-z^2/4}~{}_1F_1\left(\frac{-\nu}{2};\frac{1}{2};\frac{z^2}{2}\right)$$

Wolfram Mathworld

Recalling that the series expansion for the generalized hypergeometric is $$_{p} F_{q}\left(\begin{matrix} a_{1} ,...,a_{p}\\ b_{1} ,...,b_{q} \end{matrix} ;z\right) =\sum _{k=0}^{\infty }\frac{\prod _{i=1}^{p}\frac{\Gamma ( a_{i} +k)}{\Gamma ( a_{i})}}{\prod _{j=1}^{q}\frac{\Gamma ( b_{j} +k)}{\Gamma ( b_{j})}}\frac{z^{k}}{k!}$$ We have that $$D_{\nu}(z)=2^{\nu/2}e^{-z^2/4}\sum_{k=0}^\infty \frac{\Gamma(-\nu/2+k)~\Gamma(1/2)}{\Gamma(1/2+k)\Gamma(-\nu/2)}~\frac{(z^2/2)^k}{k!}\tag{1}$$ Now, observe that $$\frac{\Gamma(-\nu/2+k)}{\Gamma(-\nu/2)}=\prod_{l=0}^{k-1}(-\nu/2+l)\tag{2}$$ So if $\nu$ is a positive even integer, then the series expansion (1) will eventually terminate, because if $k-1\geq -\nu/2$ then the product (2) will always contain at least one zero. But, a terminating series is just a polynomial! So the parabolic cylinder function will take the form $$D_{\text{positive even integer}}=\text{decaying exponential}~\cdot~\text{polynomial}$$ Since the exponential decay beats the polynomial growth, this function will vanish at large arguments.

Unfortunately the same approach doesn't work for the positive odd integers, but in this case you can use the asymptotic expansion in @Gary 's answer and make note of the poles of the Gamma function at the negative integers.

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    $\begingroup$ Does this answer the question? The OP knew that it vanishes at minus infiniy for positive integer values of $\nu$. Just look at dlmf.nist.gov/12.7.E2 $\endgroup$
    – Gary
    Jul 8, 2021 at 14:20

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