Parabolic cylinder function

Remarks

The function $D_\nu$ solves Weber's equation, and it is the solution with $D_\nu(z) \sim z^\nu e^{-z^2/4}$ as $z \to +\infty$. The functions $D_\nu(z)$ and $D_\nu(-z)$ are solutions too, and they are linearly independent unless $\nu$ is a nonnegative integer. Another common notation is $U(a, z) = D_{-a - \frac12}(z)$. Example: for $\nu = 0$, the first coefficient is $\sqrt{\pi}/\Gamma(\frac12) = 1$ and the second is $0$, since $1/\Gamma(0) = 0$; so $D_0(z) = e^{-z^2/4}$. Check: $y = e^{-z^2/4}$ gives $y'' = \left( \frac{z^2}{4} - \frac12 \right) y$, so $y'' + \left( \frac12 - \frac{z^2}{4} \right) y = 0$. More generally, for $\nu = n \in \N$, one has $D_n(z) = 2^{-n/2} e^{-z^2/4} H_n(z / \sqrt{2})$ with $H_n$ the Hermite polynomials. Non-example: $y = e^{z^2/4}$ solves Weber's equation for $\nu = -1$, since $y'' = \left( \frac12 + \frac{z^2}{4} \right) y$, but it is not $D_{-1}$: it grows as $z \to +\infty$ instead of decaying.