# Definition of elementary and special functions

This is a (perhaps) naive question, but one that I have been thinking about lately. Is it a true statement that all functions (elementary or special) can be defined as the solution to a particular differential equation? That is, any function $f(x)$ can be defined by a solution to

$\quad F(x,y,y',..,y^{(n)}) = 0$

with (possibly) appropriate boundary conditions. For example, $e^{x}$ can be defined as the solution to:

$y' - y = 0 \quad$ where $\quad y(0)=1$.

A series solution gives you exactly the power series of the exponential function and we use this as its definition. I think the same is true for the trigonometric functions and Bessel functions (and others). Are there any special functions that cannot be defined through the solution of a differential equation?

Thanks.

• Any differentiable function $f$ satisfies the differential equation $f' = g$ where $g = f'$, so some clarification is needed to get anything nontrivial: perhaps the differential equation should be of the form $F(x, y, ..., y^{(n)}) = 0$ where $F$ is a rational function. – Robert Israel Apr 6 '16 at 15:49
• Pretty sure the Riemann zeta function doesn't satisfy any normal kind of differential equation? Someone correct me if I'm wrong – ClassicStyle Apr 6 '16 at 16:03
• Thanks @RobertIsrael Have changed my question accordingly and hope that makes more sense. A quick search seems to suggest the Riemann Zeta function is a counterexample here - at least for an algebraic differential equation – AloneAndConfused Apr 6 '16 at 16:11
• It may satisfy some strange DE but it is known that the zeta function and gamma function both do not satisfy algebraic differential equations... – ClassicStyle Apr 6 '16 at 16:12

The function $f(x)=\begin{cases}-1 \text{ if }x<0 \\ 1 \text{ if } x\geq 0 \end{cases}$ can not be the solution of a differential equation, not even if we consider the weak derivative