Are there any nonlinear solutions to f(x+1) - f(x) = f'(x)? - MathOverflow

Are there any nonlinear solutions to f(x+1) - f(x) = f'(x)?

2010-01-14T06:54:48Z

(Asked by bcross at math.iuiui.edu on the Q&A board at JMM)

Are there any nonlinear solutions to $f(x+1) - f(x) = f'(x)$?

[Ed: please retag appropriately]

Answer by Mariano Suárez-Alvarez for Are there any nonlinear solutions to f(x+1) - f(x) = f'(x)?

2010-01-14T07:24:40Z

Theorem 1 in [Sugiyama, Shohei. On the existence and uniqueness theorems of difference-differential equations. Kōdai Math. Sem. Rep. 12 1960 179--190. MR0121552] (which you can probably get from here) gives an existence and uniqueness theorem which provides non-linear solutions on finite intervals.

Answer by Bjorn Poonen for Are there any nonlinear solutions to f(x+1) - f(x) = f'(x)?

2010-01-14T07:41:44Z

Yes, there exist nonlinear solutions.

Multiplying by $e^{x+1}$ and setting $g(x):=e^x f(x)$ transforms the question into finding a solution to $g(x+1)=eg'(x)$ not of the form $e^x(ax+b)$.

Start with any $C^\infty$ function on $\mathbb{R}$ whose Taylor series centered at $0$ and $1$ are identically $0$, but which is nonzero somewhere inside $(0,1)$. Restrict it to $[0,1]$. Let $g(x)$ on $[0,1]$ be this. Using $g(x+1):=eg'(x)$ for $x \in [0,1]$ extends $g(x)$ to a $C^\infty$ function $g(x)$ on $[0,2]$, which can then be extended to $[0,3]$, and so on. In the other direction, use $g(x) := \int_0^x e^{-1} g(t+1) dt$ to define $g(x)$ for $x \in [-1,0]$, and then for $x \in [-2,-1]$, and so on. These piece together to give a $C^\infty$ function $g(x)$ on all of $\mathbb{R}$. The corresponding $f(x)$ satisfies $f(0)=0$ and $f(1)=0$ but is not identically $0$, so it is not linear.

Answer by mnf for Are there any nonlinear solutions to f(x+1) - f(x) = f'(x)?

2010-01-14T14:43:41Z

This is an elaboration of Qiaochu Yuan's prior comment: there are complex solutions (in fact, infinitely many) to $e^t-1 = t$, and then $e^{tx}$ is a solution.