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

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.

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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).

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Tiny nitpick: it's usually backwards (delay differential equation). – J. M. Sep 24 2010 at 13:08
When there is only one shift (here, $k$), bacward and forward are only a matter of orientation of the real line, and thus are equivalent. The troubles come when there are two shifts $h$ and $k$ with opposite signs. There are very few works in this case ; I remember that S. Benzoni-Gavage had a paper on it. – Denis Serre Sep 24 2010 at 13:24
I mean your ordering of the words is backwards from what I usually see ("delay differential" instead of "differential delay"). ;) – J. M. Sep 24 2010 at 14:09
OK, so let me reverse everithing : forward and backward instead of backward and forward, and so on... Molière's play Le bourgeois gentilhomme' gives a funny list of permutation of a declaration of love. *Belle marquise, d'amour mourrir vos beaux yeux me font, D'amour mourrir, belle marquise, me font vos beux yeux, etc ... – Denis Serre Sep 24 2010 at 14:21
Thanks for the name ("delay differencial" or "differential delay"). I didn't know it (i had searched for it before if I knew the name :P) – Choli Sep 24 2010 at 14:38

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 $\phi$.

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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 $k\ge 1$

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 I'm going to delete your old answer. I have fixed the LaTeX in your new answer, but I still have no idea what it means. – S. Carnahan♦ Sep 26 2010 at 9:49 It appears to be a pair of Ansätze for the problem. However, it seems to be lacking a proof that given a particular $k$ one can actually find such an $a$. This would require rather more care than is being demonstrated here. – Yemon Choi Sep 26 2010 at 10:02 since $k\ge 1$ and $01$ then $g(x+k)=-1$,from this results we have $f(x+k)=-a^{-x-k}$,so $a^{-k}=log{a}$,hence $k=-loglog{a}/log{a}$ – Hashem sazegar Sep 26 2010 at 10:13 $a$ is larger than 1 and less than almost 1.76 so $k$ is almost >=1 – Hashem sazegar Sep 26 2010 at 10:21 Hashem: $k$ is given at the start, and $a$ must be found. You cannot say "if $a$ does this then $k$ will do this" - that is not a proof. – Yemon Choi Sep 26 2010 at 21:00
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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.

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 Yes, for me the sine is a non-trivial solution (trivial solution would be $f(x)=0$). I understand your point (transforming the sine so its period matches the delay ($k$) we want). However, I still don't get it. For example: $f(x) = a \sin (bx+c)$ Then $f'(x) = ab \cos (bx+c) = ab \sin (bx+c+\frac{\pi}{2})$ $f(x+k) = a \sin(bx+bk+c$ Therefore, if $f'(x)=f(x+k)$: $a=ab$ (1) and $bk-\frac{\pi}{2} = 2\pi n$ (2) From (1), either $a=0$ or $b=1$. We choose $b=1$ ($a=0 \to f(x)=0$). Then, in (2): $k-\frac{\pi}{2} = 2\pi n$, therefore the only solution appears when $k=2\pi n+\frac{\pi}{2}$. Am I wrong? – Choli Sep 24 2010 at 14:49 Choli, you are right. I will change my answer accordingly. – sleepless in beantown Sep 24 2010 at 18:23

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...

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