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

Question

I'm trying to find non-trivial functions $f \colon \mathbb R \to \mathbb R $ that $f'(x) = f(x+k)$ with $k \in \mathbb R$.

For $k \le 0$, I've found functions based on $f(x)= e^x$, such as $f(x) = e^{x \frac{W(k)}{k}}$, where $W(k)$ is the Lambert W function.

However, for $k>0$, I can only find a solution if $k=2\pi n+\frac{\pi}{2} $ with $n\in\mathbb N$. The solution is $f(x) = \sin x$. Are there solutions for other values of $k$?

I was hoping that for $k>0$, a function whose graphic is similar to $f(x)=-e^x$ could exist, but it seems it doesn't exist. Or does it?

The method I used is basically trial and error. What other method could I have used?

Thanks.

Answer 1

This is a kind (a very simple one indeed) of a differential-delay equation. The theory is well-established, and fully described in the book of J. Hale and S. Lunel, Introduction to Functional Differential Equations, Springer-Verlag (1993).

Answer 2

Let $a=\alpha+\beta i$, $\alpha,\beta\in\mathbb{R}$, be a complex number such that $a=e^{ak}$. In terms of $\alpha$ and $\beta$ this means that $\alpha=e^{\alpha k}\cos(\beta k)$ and $\beta=e^{\alpha k}\sin(\beta k)$. Then $f(x)=Ae^{\alpha x}\cos(\beta x)+Be^{\alpha x}\sin(\beta x)$ is a solution for any choice of $A,B\in \mathbb{R}$. Are there other solutions? Yes.

In the following I assume $k>0$. Let $\phi$ be a $C^\infty$ function with compact support in $[0,k]$. Let $n$ be a non negative integer. Define $g$ on $[nk,(n+1)k]$ by $g(x)=\phi^{(n)}(x-nk)$. Then $g$ verifies $g'(x)=g(x+k)$ on $[0,+\infty)$. We need to extend $g$ to $(-\infty,0]$. For this define indectively $g$ on $[-(n+1)k,-nk]$ by $g(x)=-\int_{x+k}^{-(n-1)k}g(s)ds$.

The general solution is of the form $f+g$ for some choice of $\alpha,\beta$ and $\phi$.

Answer 3

i could not Edit my answer so i have posted new answer

1)let $f(x)=a^{x}$ in which $a>1$ so

$f'(x)=a^{x}\log a$ and $f(x+k)=a^{x+k}$ then $k=\log \log a/\log a$ this is hold for $k<1$

2)let $f(x)=g(x)a^{-x}$ in which, $1<a<1.76$, and $g(x)=-1$, if, $x>1$, $g(x)=1$

if $0<x<1$ ,otherwise $g(x)=0$, so for $0<x<1$,

$f'(x)=-a^{-x}\log a$ and $f(x+k)=-a^{-x-k}$ then $k=-\log \log a/\log a$ this is hold for $k\ge 1$

Answer 4

Any periodic function which contains a scaled or translated version of its own derivative, for example sine or cosine , or any finite or infinite sum of multiple periodic functions which also yields a periodic function which is its own derivative, can be expressed in the format which you are asking for. (Assuming that you are considering the sine solution you listed as a non-trivial solution)

For example, for $k=1$, you can transform the domain and range of the function $y=sin(x)$ to $y=2\pi sin(2\pi x -\pi/2)$, or for arbitrary $k$, $$y=k sin(\frac{2\pi x}{k} - \pi/2)$$

You can create a similar function for cosine by adding a different phase shift to the domain. So for arbitary $k$, you could use sine or cosine with the domain scaled and translated and the range scaled as necessary to get a solution of the format you'd like.

$$ y = c g(\frac{a x}{k}+b) $$

If you are looking for a non-periodic trivial solution, then it's a different story and answer, delayed differential equations, as pointed out by Denis Serre above.

Answer 5

DEs are my weak subject so I probably should not comment on this, but if I tried to solve it I would look for a solution of the type:

$$A e^{\beta x} \sin(ax+b) + B e^{\beta x} \cos(ax+b) $$

My reasoning being: $\sin (ax+b)$ looks like the natural choice but you get an extra constant so you need to introduce an exponential to kill it...