1
$\begingroup$

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 $n$?

$\endgroup$

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $f_0:=0.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.