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.


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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.