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In chapter 3 (example 4) of the book "Advanced Mathematical Methods for Scientists and Engineers", by Bender and Orszag, I want to get the approximate solution for $+\infty$ for the parabolic cylinder differential equation:

$$y'' + (\nu + 1/2 -x^2/4)y = 0.$$

First I get rid of the singularity at infinite using the following transformation:

$$y = \mathrm e^{S(x)}$$ yielding

$$S'' + S'^2 + \nu + 1/2 - x^2/4=0.$$

Then I assume the following approximations $S'' \ll S'^2$ and $\nu + 1/2 <\ll1/4 x^2$, which gives,

$$S(x) \approx \pm x^2/4$$ when $x \to \infty$.

This is the controlling factor of the general differential equation. To get the leading behavior I assume that

$$S(x) \approx \pm x^2/4 + C(x)$$ where $C(x) \ll \pm x^2/4 $.

I know that the answer to this problem is given by

$$y \approx C_1 x^{-(\nu+1)} \mathrm e^{x^2/4 }$$ and $$y \approx C_2 x^{\nu} \mathrm e^{-x^2/4 }.$$

However I don't know how to reach such result. The differential equation for $C(x)$ is given as $$(\pm 1/2 + C'') + (\pm x/2 + C')^2 + \nu + 1/2 - x^2/4 = 0,$$

Using the approximation for $C(x)$ I get that $C' \ll\pm x/2$ and $C'' \ll \pm 1/2$, hence, the differential equation would not depend on $C(x)$, which makes no sense.

I have tried to assume only one of these conditions at a time, however, even so, I do not get to the desired result.

Edison Santos
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1 Answers1

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By taking

$y=e^{S(x)}$

you have already pulled out the exponential behaviour in the solution.

As a result another exponential correction

$y=e^{S(x)}e^{C(x)}$

will not further elucidate $y$'s behaviour. So no surprise that the differential equation for $C$ becomes independent of it.

Try the following correction instead
$y=e^{S(x)}p(x)=e^{S(x)}e^{ln(p(x))}$

For $S(x)=k \frac{x^2}{4}$

  1. $k=1$ leads to $p''+x p'+(\nu+1)p=0$ which (ignoring $p''$ term) solves as $x^{-(\nu+1)}$

  2. $k=-1$ leads to $p''-x p'+\nu p=0$ which (ignoring $p''$ term) solves as $x^{(\nu)}$

lineage
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  • It is what I needed, thanks a lot.

    Just another question, I see your argument that we cannot use another exponential approximation for the function $C(x)$, however that is exactly how the book does for the modified Bessel function (it is example 2, page 92, if you are interested). In this case, such substitution works. Would you be able to explain why?

     – Edison Santos Dec 17 '19 at 14:46
  • can't seem to find the use of the second exponential. $w(x)$ is sub exponential(eqn 3.5.8a) – lineage Dec 17 '19 at 15:20