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

Question

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 $q<n$. 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.

Answer 1

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 $$ g(x) \sim x^{\beta} \exp\left(-\frac{x^\alpha}{\alpha}\right), \qquad \text{as } x \to +\infty. $$

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[1] C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer-Verlag, 1999.