Airy function

Remarks

The two functions are fixed by their initial values $\operatorname{Ai}(0) = \frac{1}{3^{2/3}\,\Gamma(2/3)}$, $\operatorname{Ai}'(0) = -\frac{1}{3^{1/3}\,\Gamma(1/3)}$, $\operatorname{Bi}(0) = \frac{1}{3^{1/6}\,\Gamma(2/3)}$, $\operatorname{Bi}'(0) = \frac{3^{1/6}}{\Gamma(1/3)}$, where $\Gamma$ is the gamma function. Their Wronskian is $\operatorname{Ai}(z) \operatorname{Bi}'(z) - \operatorname{Ai}'(z) \operatorname{Bi}(z) = \frac{1}{\pi}$, so they are independent. Example: $\operatorname{Ai}$ is the solution that tends to $0$ as $x \to +\infty$, where $\operatorname{Ai}(x) \sim \frac{e^{-\zeta}}{2\sqrt{\pi}\, x^{1/4}}$ with $\zeta = \frac{2}{3} x^{3/2}$; for $x > 0$ it equals $\frac{1}{\pi} \sqrt{\frac{x}{3}}\, K_{1/3}(\zeta)$ with the modified Bessel function $K_{1/3}$. A non-example: $\operatorname{Ai} + \operatorname{Bi}$ solves the equation but is not $\operatorname{Ai}$, since it grows like $\operatorname{Bi}$ as $x \to +\infty$. For $x \to -\infty$, $\operatorname{Ai}(x) \sim \frac{1}{\sqrt{\pi}\, \abs{x}^{1/4}} \sin\left( \frac{2}{3} \abs{x}^{3/2} + \frac{\pi}{4} \right)$.