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I have an equation which is similar to Hermite's differential equation $y''(x)+xy'(x) +\lambda y(x)=0$ with decaying boundary conditions at infinity $y\rightarrow 0$ as $x\rightarrow\pm\infty$. I believe the eigenvalues take the form $\lambda_n =-n$ where $n=0,1,2,...$ from the power series solution.

The eigenfunction which satisfies the boundary condition is $\phi_n(x) = A_n e^{-y^2/2}H_{\lambda_n-1}\left(\frac{y}{\sqrt 2}\right)$. All of which are Hermite polynomials for negative integers. The recurrence relation $2 n H_{n-1}(x) = 2x H_n(x)-H_{n+1}(x)$ is no helpful since it gives indeterminate values.

But I have seen them that these polynomial of non-negative orders are expressed in terms of error function, but I could not find any reference to this. Is there any way to derive them in terms of error function? It would be helpful if someone clarifies about the orthogonality condition for negative integers. Is it the same as Hermite's polynomials of positive integers?

Betelgeuse
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2 Answers2

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For $H(-n,z)$ there is the rather complicated expression with the error function and $H(n,z)$ from http://functions.wolfram.com/07.01.03.0014.01 . It seems simpler to use Kummer’s confluent hypergeometric function, see http://functions.wolfram.com/07.01.02.0001.01

$$H(\nu, z) = 2^{\nu} \sqrt{\pi}\Big(\frac{1}{\Gamma((1 - \nu)/2)} {_1F_1}(-\nu/2, 1/2, z^2) -\frac{2z}{\Gamma(-\nu/2)}{_1F_1}((1 - \nu)/2, 3/2, z^2))\Big)$$

I dont know about orthogonality for negative indices.

gammatester
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  • For the relation between usual Hermite polynomials and the error function mentionned here, just plug $$ e^{-x^2}= \frac{d}{dx} \int_0^x e^{-t^2}\, dt = \frac{\sqrt{\pi}}{2} \mathop{erf}(x)$$ in the usual definition of a Hermite polynomial.

  • More generally, one could define $H_{\lambda},\ \lambda\in \mathbb{R}$ as one of the two linearly independent solutions of $$ y''(x) - 2 x\, y(x) + \lambda\, y(x) =0 \tag{H}\label{H}$$ as is suggested here or kind of, here (Caveat: I'm not certain this definition coincides with what one finds elsewhere...). Choosing a solution with only non-trivial odd powers in its series expansion is coherent with the fact that $e^{-x^2}$ only has non-trivial even powers and that a primitive of it, will only have odd powers.

The question that brought me to this answer is in fact the relation between your equation $y''(x) + x\, y(x) + \lambda\, y(x) =0 $ and (\ref{H}). A simple change of variable $x \longleftrightarrow -x$ does not help but Wolfram alpha suggests to factorize out a $e^{-\frac{x^2}{2}}$ factor: one indeed checks that if $y(x) =e^{-\frac{x^2}{2}}\, u(x)$ satisfies the equation with $+x\, y'(x)$ then $u$ satisfies: $$ u''(x)-x\, u'(x) + (\lambda-1)\, u(x) =0 $$

Let me also add a link to a similar question.

Noix07
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