Are functions of the form $f(x)=A\sin(x)+B\cos(x)$ the only ones satisfying $f'(x)=f(x+\pi/2)$?

Question

Suppose $f : \mathbb{R} \rightarrow \mathbb{R}$ is a circular function with fundamental period $2\pi$, or else that $f$ is the zero function. In other words, assume that $f$ is of the form $$f(x) = A \sin (x) + B \cos (x).$$

Then $f$ necessarily satisfies

$$f'(x) = f(x+\pi/2). \tag{$*$}$$

Question. Do there exist analytic functions not of this form satisfying $(*)$?

Note that the question is whether there exist analytic functions not of the form $x \mapsto A\sin(x)+B\cos(x)$ that nevertheless satisfy $(*)$. That is, there's no assumption of periodicity in the question.

Answer 1

If you apply the starred result four times, you arrive at $$ f''''(x) = f(x + 2\pi) = f(x), $$ so we can solve $f''''(x) = f(x)$ as a start to finding all possible solutions to the original equation (and some spurious ones as well).

I haven't had my morning caffeine yet, so I can't write down all solutions of that off the top of my head. Maybe linear combinations of $e^{bx}$, where $b$ is a fourth root of unity? I think that's right.

Clearly two of those (for $b = i$) give you sines and cosines. Ah...and the other two are $e^x$ and $e^{-x}$, which aren't periodic.

So a function satisfying (*) is a linear combination of $\sin(x), \cos(x)$, $\exp(x)$, and $\exp(-x)$. If it's periodic, then the coefficients of the last two are zero, and it must be a combination of sine and cosine, and we're done.

Answer 2

Edit: In earlier versions of the question it seemed to me the OP was assuming that $f$ has period $2\pi$. The words "That is" in the question seems to take for granted that a linear combination of $\sin$ and $\cos$ is the only $2\pi$-periodic solution. This doesn't seem quite so evident to me; I give a proof below. I doubt that there are more aperiodic analytic solutions, no proof yet.

Assuming first that $f$ does have period $2\pi$: Say $$f(t)\sim\sum c_ne^{int}.$$

The condition $f'(t)=f(t+\pi/2)$ implies that $f$ is infinitely differentiable, so there's no problem with convergence, termwise differentiation, etc. So $f'(t)=f(t+\pi/2)$ says exactly $$inc_n=e^{in\pi/2}c_n.$$If you look at the absolute value of both sides you see $$c_n=0\quad(|n|\ne1).$$So $$f(t)=c_1e^{it}+c_{-1}e^{-it}=A\sin(t)+B\cos(t).$$