Notes on Whittaker & Watson, Chapter XVI, part 3

Whittaker & Watson, A Course of Modern Analysis. Chapter XVI, The Confluent Hypergeometric Function.

16.5 The parabolic cylinder functions. Weber’s equation.

The equation $$ \frac{d^2w}{dz^2}+\left(2k-\frac14z^2\right)w=0 $$ is satisfied by $w=z^{-\frac12}W_{k,-\frac14}(\frac12z^2)$​, and the parabolic cylinder function $$ D_n(z)=2^{\frac12n+\frac14}z^{-\frac12}W_{\frac12n+\frac14,-\frac14}\left(\frac12z^2\right) $$ satisfies Weber’s equation $$ \frac{d^2D_n(z)}{dz^2}+\left(n+\frac12-\frac14z^2\right)D_n(z)=0. $$ Expressed in $M_{k,m}$, we have $$ D_n(z)=\frac{\Gamma(\frac12)2^{\frac12n+\frac14}z^{-\frac12}}{\Gamma(\frac12-\frac12n)}M_{\frac12n+\frac14,-\frac14}\left(\frac12z^2\right) + \frac{\Gamma(-\frac12)2^{\frac12n+\frac14}z^{-\frac12}}{\Gamma(-\frac12n)}M_{\frac12n+\frac14,\frac14}\left(\frac12z^2\right) $$ which is one-valued throughout $z\in\mathbb{C}$.

We have the asymptotic expansion of $D_n(z)$ for $\arg(z)<\frac34\pi$​ $$ D_n(z)\sim e^{-\frac14z^2}z^n\left(1-\frac{n(n-1)}{2z^2}+\frac{n(n-1)(n-2)(n-3)}{2\cdot4z^4}-\cdots\right). $$ 16.51 The second solution of Weber’s equation.

Weber’s equation is invariant under the transform $n\to-n-1,z\to\pm iz$. Thus, $D_{-n-1}(ze^{\frac12\pi i})$​​ is the second solution with $-\frac34\pi<\arg(z)<\frac14\pi$​.

16.511 The relation between the functions $D_n(z), D_{-n-1}(\pm iz)$ $$ D_n(z)=\frac{\Gamma(n+1)}{\sqrt{2\pi}}\left(e^{\frac12n\pi i}D_{-n-1}(iz)+e^{-\frac12n\pi i}D_{-n-1}(-iz)\right). $$ 16.52 The general asymptotic expansion of $D_n(z)$.

For $\frac14\pi<\arg(z)<\frac54\pi$​, $$ \begin{align*}D_n(z)\sim & \ e^{-\frac14z^2}z^n\left(1-\frac{n(n-1)}{2z^2}+\frac{n(n-1)(n-2)(n-3)}{2\cdot4z^4}-\cdots\right)\\ & -\frac{\sqrt{2\pi}}{\Gamma(-n)}e^{n\pi i}e^{\frac14z^2}z^{-n-1}\left(1+\frac{n(n+1)}{2z^2}+\frac{n(n+1)(n+2)(n+3)}{2\cdot4z^4}+\cdots\right), \end{align*} $$ which is obtained from the identity $D_n(z)=e^{n\pi i}D_{n}(-z)+\frac{\sqrt{2\pi}}{\Gamma(-n)}e^{(n+1)\pi i}D_{-n-1}(-iz)$​​​.

Similarily, for $-\frac54\pi<\arg(z)<-\frac14\pi$​, $$ \begin{align*}D_n(z)\sim & \ e^{-\frac14z^2}z^n\left(1-\frac{n(n-1)}{2z^2}+\frac{n(n-1)(n-2)(n-3)}{2\cdot4z^4}-\cdots\right)\\ & -\frac{\sqrt{2\pi}}{\Gamma(-n)}e^{-n\pi i}e^{\frac14z^2}z^{-n-1}\left(1+\frac{n(n+1)}{2z^2}+\frac{n(n+1)(n+2)(n+3)}{2\cdot4z^4}+\cdots\right), \end{align*} $$ 16.6 A contour integral for $D_n(z)$​​​ $$ D_n(z) = -\frac{\Gamma(n+1)}{2\pi i}e^{-\frac14z^2}\int_\infty^{(0+)}e^{-zt-\frac12t^2}(-t)^{-n-1}dt. $$ [There seems to be a typo in the text. The integral should satisfy the equation for $e^{\frac14z^2}D_n(z)$ instead of $e^{-\frac14z^2}D_n(z)$​​.]

16.61 Recurrence formulae for $D_n(z)$.

By differentiation under the integral sign with respect to $t$ in the integral representation of $D_n$, we have $$ D_{n+1}(z)-zD_n(z)+nD_{n-1}(z)=0. $$ By differentiation with respect to $z$, we have $$ D_n'(z)+\frac12zD_n(z)-nD_{n-1}(z)=0. $$ 16.7 Properties of $D_n(z)$ when $n$ is an integer.

With the change of variables $t=v-z$, we have $$ D_n(z)=\frac{(-1)^{n}n!e^{\frac14z^2}}{2\pi i}\int^{(z+)}\frac{e^{-\frac12v^2}}{(z-v)^{n+1}}dv = (-1)^{n}e^{\frac14z^2}\frac{d^n}{dz^n}e^{-\frac12z^2}. $$ Orthogonality $$ \int_{-\infty}^{\infty}D_m(x)D_n(x)=(2\pi)^\frac12n!\delta_{mn} $$ can be obtained by integrating the differential equation $D''_mD_n-D_mD''_n+(m-n)D_mD_n=0$​, and the induction $$ \begin{align*}(n+1)\int_{-\infty}^{\infty}D_n^2(x)dx &= \int_{-\infty}^{\infty}D_n(x)\left[\frac12xD_{n+1}(x)+D'_{n+1}(x)\right]dx & \text{recurrence}\\ & = D_n(x)D_{n+1}(x)\Big|_{-\infty}^{\infty}+\int_{-\infty}^{\infty}D_{n+1}(x)\left[\frac12xD_{n}(x)-D'_{n}(x)\right]dx \\ & = \int_{-\infty}^{\infty}D_{n+1}(x)D_{n+1}(x)dx. & \text{recurrence} \end{align*} $$