finding solution to function$f^{n}(x)=f(x+k)

Question

according to question Finding solutions to $f'(x) = f(x + k)$

i ask generalization of this question

i am trying to find non-trivial functions $f \colon \mathbb R \to \mathbb R$ that $f^{n}(x) = f(x+k)$ with $k \in \mathbb R$,and $f^{n}$ is n'th derivate $f$

For $k<1$, I've found functions $f(x)= a^x$ that $a>1$,of course for large $n$ and some $a$ ,this is hold for $k\ge 1$

However, for $k>-1$,and $n$ be even I can only find a solution $f(x) = a^{-x}$, that $a>1$ .of course

for large $n$ and some $a$ ,this is hold for $k\le {-1}$

is there any other solution for values of $k$ and $n$?

Answer 1

I will give you (almost) the same answer than for the case $n=1$. Let $a=\alpha+\beta i$, $\alpha,\beta\in\mathbb{R}$ be any complex solution of $a^n=e^{ka}$. A solution is given by $$a=-\frac{n}{k}W(-\frac{k}{n})$$ where $W$ is the Lambert or product logarithm function., but in general there are infinitely many solutions. Then $e^{\alpha x}\cos(\beta x)$ and $e^{\alpha x}\sin(\beta x)$ solve the equation. Are these the only solutions? No.

Given any $C^\infty$ function $\phi$ with compact support in $[0,k]$, it can be extended to $\mathbb{R}$ in such a way that it verifies the equation. I assume now that $k>0$. Then define $f$ on $[k,2k]$ as $\phi^{(n)}(x-k)$, on $[2k,3k]$ as $\phi^{(2n)}(x-k)$, and so on. I leave to you the details of how to extend the solution to $(-\infty,0]$.

Since the equation is linear, any linear combination of the solutions will also be a solution.

Answer 2

Also note that if $D$ is the derivative $f\mapsto f\\ ^'$ and $S_\tau$ is the translation $f\mapsto f(\cdot + \tau)$ you want $f\in\ker (D-S_\tau)^n$ with $\tau:=k/n,$ that consists in solving $n$ times the the inhomogeneous equation $f\\ ^ ' (x)-f(x+\tau)=h(x).$ Solve in $f:=f_{i+1}$ putting recursively $h:=f_i,$ starting from $f_0:=0.$