Differentiation and discrete translation of real functions Consider the problem of finding differentiable functions $y:\mathbb R\to\mathbb R$ with the property that:
$$\begin{equation}y'(x)=y(x+\xi)\end{equation}$$
for some $\xi\in\mathbb R$. Note that when $\xi=0$, $y(x)=C\exp x$ and when $\xi=-{\pi\over 2}$,$y(x)=C\cos x$. This is a suprisingly "varied" solution set.
My question is simple: what can be said about solutions to the problem for general $\xi$?
 A: Since this is a delay differential equation, our intuition should be that, if we choose to set $y$ to some function in $[0,\xi]$, we can extrapolate from there to solve the whole thing. In particular, suppose we knew the values in $[0,\xi]$, then we could find the values for $[\xi,2\xi]$ by setting $y(x)=y'(x-\xi)$, which is already known for any $x\in [\xi,2\xi]$. We could also work backwards to obtain the values in $[-\xi,0]$ by integrating
$$y(x)=y(0)-\int_x^0 y'(z) dz=y(0)-\int_x^0y(z+\xi)dz$$
which, is of course, known whenever $x$ is in $[-\xi,0]$. We could, of course, repeat the process, tacking further intervals of the form $[c\xi,(c+1)\xi]$ onto either side of the domain.
However, this won't work for every choice of $y$ in the domain $[0,\xi]$. We need $y$ to be infinitely differentiable in some open interval around that, otherwise we can't take derivatives of it forever. Note that, from the equation $y'(x)=y(x+\xi)$ we have that, more generally, $y^{(n+1)}(x)=y^{(n)}(x+\xi)$ by differentiating $n$ times, which, in particular, tells us we have to choose $y$ such that $y^{(n+1)}(0)=y^{(n)}(\xi)$. However, though this may limit the number of analytic solutions, we can create as many smooth solutions as we want. For instance, setting $\xi=1$ and defining the starting interval as something like
$$y(x)=e^{-\frac{1}{x(x-1)}}$$
where $f(0)=f(1)=0$ can be expanded into valid solution across all of $\mathbb{R}$, since the function has that $y^{(n)}(0)=y^{(n)}(1)=0$.
