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I know that the general form solution to the Hermite differential equation $$ y''-2xy'+2\lambda y=0$$ is $$y(x)=a_1 M(-\frac{\lambda}{2},\frac{1}{2},x^2)+a_2 H(\lambda,x),$$ where $M(\cdot,\cdot,\cdot)$ is a confluent hypergeometric function of the first kind, and $H(\cdot,\cdot)$ is a Hermite polynomial.

For a general value of $\lambda$ (negative and non-integer real valued), is there a special solution to the Hermite differential equation such that it's first order derivative goes to zero for $x\rightarrow-\infty$? In other words, is there a parametric characterization of $a_2$ as a function of $a_1$ and $\lambda$ such that $y'(x)\rightarrow 0$ as $x\rightarrow-\infty$?

From mathematica numerical calculations it seems that such a special case exists. However, I was not able to characterize it explicitly. I would appreciate any help on this. Thank you!

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$$y(x)=a_1 M(-\frac{\lambda}{2},\frac{1}{2},x^2)+a_2 H(\lambda,x)\Rightarrow$$ $$y'(x)=-2a_1\lambda x M(1-\tfrac{1}{2}{\lambda},\tfrac{3}{2},x^2)+2 {a_2} {\lambda} H({\lambda}-1,x)$$ The asymptotics for $x\rightarrow-\infty$ and $\lambda$ negative non-integer is $$y'(x)\rightarrow\frac{2 e^{x^2}(-x)^{-\lambda} }{\sqrt{\pi }\, \Gamma \left(-\lambda/2\right)}\bigl(a_2 \Gamma \left(-\lambda/2\right) \Gamma (\lambda+1)\sin \pi \lambda -\pi a_1\bigr).$$ So this vanishes if $$a_2 \Gamma \left(-\lambda/2\right) \Gamma (\lambda+1)\sin \pi \lambda =\pi a_1.$$

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  • $\begingroup$ Thank you Carlo. Why does the $a_1$ term diverge as $e^{\frac{x^2}{2}}$? From en.wikipedia.org/wiki/… it seems that it should diverge as $e^{x^2}$? Also why does the $a_2$ term diverge as $e^{x^2}$? I couldn't find any pointers for this. I really appreciate your help!! $\endgroup$ – Jackie Lu Mar 25 '19 at 18:19
  • $\begingroup$ OK, I see I was using a different "M" function, the one from en.wikipedia.org/wiki/Whittaker_function; let me try and correct. $\endgroup$ – Carlo Beenakker Mar 25 '19 at 21:05
  • $\begingroup$ Thank you. Can you kindly point to me the reference of the asymptotics for H(,)? I couldn't find anything for negative non-integer $\lambda$. Would really appreciate your help. $\endgroup$ – Jackie Lu Apr 3 '19 at 20:12
  • $\begingroup$ this is the expansion I used: wolframalpha.com/input/?i=HermiteH%5B+lambda-1,+-x%5D $\endgroup$ – Carlo Beenakker Apr 3 '19 at 20:19

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