Find a non-trivial solution to the problem $f:\;\mathbb{R}\to \mathbb{R},\;f^\prime(x) = f(x-1)$.

Question

I'm trying to figure out if there are non-trivial solutions to the problem $f:\;\mathbb{R}\to \mathbb{R},\;f^\prime(x) = f(x-1)$.

So far I've come to the conclusions that $f^{(n)}(x) = f(-n)$

With the Taylor series around $x = 0$ being $f(x) = f(0) + f(-1)x + f(-2)x^2/2! + \ldots$

From the Taylor series, we know that if $f(n) = 0$ for every $n$, then there is no non-trivial solution, because the series would give $f(x) \equiv 0$. So the solution is not something like $f(x) = g(x)\cos(\pi x)$ for example. So maybe there's an arithmetic series $a_n = f(-n) \neq 0$ that we can figure out?

How to proceed from here? Thank you.

Answer 1

One can look for solutions in the form $f(x) = e^{bx}$. Then to have $f^{'}(x) = f(x-1)$ we would have to have $be^{bx} = e^{b(x-1)}$ so we must have $b=e^{-b}$. The last equation has a solution since at the function $g(b) = b-e^{-b}$ changes sign between $0$ and $1$.