If $f(x)=\int_{x-1}^x f(s)ds$, is $f$ constant? Periodic?

Solution 1:

Here is a typical argument using characteristic equation:

Let $\alpha \in \mathbb{C}\setminus\{0\}$ solve the equation $\alpha = 1-e^{-\alpha}$. One can indeed prove that such solution exists. One such solution is numerically given by $-2.08884 + 7.46149 i$. Then

$$ \int_{x-1}^{x} e^{\alpha t} \, \mathrm{d}t = \frac{1 - e^{-\alpha}}{\alpha} e^{\alpha x} = e^{\alpha x}, $$

hence $f(x) = e^{\alpha x}$ is one (complex-valued) solution of the equation

$$ f(x) = \int_{x-1}^{x} f(t) \, \mathrm{d}t \tag{*}$$

If one is interested in real-valued solutions only, then one can consider both the real part and the imaginary part of $e^{\alpha x}$. In particular, this tells that there exists an analytic solution of $\text{(*)}$ which is neither constant nor having real-period.

Addendum. We prove the following claim:

Claim. There exists a non-zero solution of $\alpha = 1 - e^{-\alpha}$ in $\mathbb{C}$.

Proof. We first note that $\varphi(x) = x(1-\log x)$ satisfies $\varphi(0^+) = 0$ and $\varphi(1) = 1$. Next, let $k$ be a positive integer. Then

1. There exists $y \in (2k\pi, (2k+\frac{1}{2})\pi)$ such that $ \varphi(\sin(y)/y) = \cos (y) $, by the intermediate-value theorem.

2. Set $x = \log(\sin(y)/y)$.

We claim that $ \alpha = x + iy $ solves the equation. Indeed, it is clear that $ e^{-x}\sin y = y $ holds. Moreover,

$$ (1-x)e^x = \varphi(\sin(y)/y) = \cos(y), $$

and so, $1 - x = e^{-x}\cos(y)$. Combining altogether,

$$ 1 - \alpha = 1 - x - iy = e^{-x}\cos(y) - ie^{-x}\sin(y) = e^{-x-iy} = e^{-\alpha}. $$

Therefore the claim follows. ////

(A careful inespection shows that this construction produces all the solutions of $\alpha = 1 - e^{-\alpha}$ in the upper half-plane.)

Solution 2:

Define $f(x)=3x^2-4x+1$ for $x\in(0,1)$ so the differential equation is true for $x=1$.
For $x\in(1,2)$, solve the differential equation $$\frac{df}{dx}=f(x)-(3(x-1)^2-4(x-1)+1)$$ Iterate the procedure, for $x\in (2,3)$ and so on.