A special function solution of a fourth-order ODE

Question

I want to consider the solutions of the following fourth-order ODE: $$ f^{(4)}(t)+a tf^{(1)}(t)+b f(t)=0, \tag{$\ast$}$$ where $a,b$ are complex parameters. It turns out that with a Fourier transformation, we get a first-order equation like for the Airy equation $$ f^{(2)}(t)- tf(t)=0. $$ This is due to the multiplicative factor $t$ which becomes $-id/d\tau$ on the Fourier side, whereas the Fourier transforms of $f^{(4)}(t), f^{(1)}(t)$ are $(i\tau)^4\hat f(\tau), i\tau\hat f(\tau)$. Eventually, we find a first-order differential equation on $\hat f$ with $0$ as a regular singular point, so that we can solve $(\ast)$ explicitly.

My question. The special functions solutions of $(\ast)$ are essentially the inverse Fourier transform (say in the tempered distribution sense) of $\tau^2 e^{i\tau^4}$. Do they have a name? Are they studied systematically somewhere?

Answer 1

you ask whether the inverse Fourier transform of $\tau^2 e^{i\tau^4}$ is some named special function; as indicated by Johannes Trost, it's a hypergeometric function,

$$\int_{-\infty}^\infty \tau^2 e^{i\tau^4}e^{-i\omega\tau}\,d\tau=-2e^{i\pi/8} \frac{d^2}{d\omega^2}{\cal F}(\omega),$$

$${\cal F}(\omega)= \Gamma \left(\frac{5}{4}\right) \, _0F_2\left(\frac{1}{2},\frac{3}{4};\frac{i \omega^4}{2^{8}}\right)-\frac{i\omega^2}{8} \Gamma \left(\frac{3}{4}\right) \, _0F_2\left(\frac{5}{4},\frac{3}{2};\frac{i \omega^4}{2^{8}}\right)$$

I don't think it will get any simpler than this...