# Higher Airy functions: an exponentially decreasing solution of $f^{(n)}(x) = (-1)^n \, x^a \, f(x)$

Consider the differential equation $$f^{(n)}(x) = (-1)^n\, x^a \, f(x),$$ where $$a > 0$$ is a real number.

My numerical experiments suggest that this equation has a unique solution on $$\mathbb{R}_+$$ satisfying $$f(x) \sim_{x \to \infty} x^{-\beta} e^{-x^\alpha/\alpha},$$ where $$\alpha = 1+\frac{a}{n}$$, $$\beta = \frac{a}{2n}$$.

Is this known?

The solution would be a generalization of the Airy function, which is the unique bounded solution of $$f''(x) = x f(x)$$ (that is, $$n=2$$, $$a=1$$). The case $$a=1$$ is known: analogously to the Airy function, one constructs the solution using the integral $$f(x) = \int_{\mathbb{R}} \exp\left[i \left(\frac{t^{n+1}}{n+1} + tx \right)\right] \, dt.$$ With some analysis one checks that this function satisfies the differential equation and has the right asymptotic as $$x \to \infty$$.

If $$n=2$$ and $$a$$ is an integer, the required solution can be constructed using a much more complicated integral found in "A generalization of the Airy integral" by Gundersen and Steinbart.

If $$a$$ is an integer, the existence of such a solution probably follows from "The possible orders of solutions for linear differential equations with polynomial coefficients" by Gundersen, Steinbart, and Wang (but I still have to check). However, the statement seems to be just as true for any real $$a$$.

Here is a slightly more general conjecture. For any complex number $$c \not=0$$ the differential equation $$f^{(n)}(x) = c^n \, x^a \, f(x)$$ has a solution satisfying $$f(x) \sim_{x \to \infty} x^\beta e^{c \, x^\alpha/\alpha},$$ for $$\alpha = 1+\frac{a}n$$ and $$\beta = \frac{ac}{2n}$$.

It is easy to write a basis of solutions of this differential equation using (sort of) hypergeometric functions. The first solution is $$f_0(x) = 1 + \frac{c^n \, x^{a+n}}{(a+n)(a+n-1) \dots (a+1)} + \frac{c^{2n} \, x^{2(a+n)}}{(a+n)(a+n-1) \dots (a+1) \; \cdot \; (2a+2n)(2a+2n-1)\dots(2a+n+1)} + \dots.$$ It is straightforward to check that the series converges for any positive $$x$$ and that $$f_0$$ is, indeed, a solution of the equation. Similarly one can construct a solution $$f_q(x) = x^q + \dots$$ for any integer $$q. However all these solutions seem to have the same asymptotic expansions as $$x \to \infty$$ (not just the leading term, but the whole asymptotic expansion!), and one needs a clever linear combination of them to find solutions that are exponentially smaller.

For the differential equation $$f^{(n)}(x) = c^n x^a f(x)$$ with $$c \neq 0$$ a complex number and $$a > 0$$ a real number, the possible behaviors at infinity are known, see Example 5 Section 3.4 in [1]. We get the following possible behaviors at infinity $$f(x) \sim \gamma \, x^{\beta} \exp\left(\omega \, c \, \frac{x^\alpha}{\alpha}\right), \qquad \text{as } x \to +\infty,$$ with $$\alpha = 1 + \frac{a}{n}$$, $$\beta = a \frac{1-n}{2n}$$, $$\omega^n = 1$$ a $$n$$-th root of the unity, and $$\gamma$$ a constant. From this, if $$\min\{\Re(c \, \omega) \mid \omega^n = 1\}$$ is reach for an unique $$\omega_c$$ then there exists an unique solution $$g$$ such that $$g(x) \sim x^{\beta} \exp\left(\omega_c \, c \, \frac{x^\alpha}{\alpha}\right), \qquad \text{as } x \to +\infty.$$
If we return to your first example with $$c = -1$$, we see that $$\min\{\Re(-\omega) \mid \omega^n = 1\}$$ is uniquely reach for $$\omega = 1$$ therefore there exits an unique solution $$g$$ such that $$g(x) \sim x^{\beta} \exp\left(-\frac{x^\alpha}{\alpha}\right), \qquad \text{as } x \to +\infty.$$