fof=f' (open question for me)

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blang

#1 h Dec 31, 2009, 8:46 AM • 3 Y

Y by Adventure10 and 2 other users

Hi !

An open question for me :

Find all $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f \circ f =f'$ .

Thanks !

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fedja

#2 h Dec 31, 2009, 7:52 PM • 6 Y

Y by pieater314159, Adventure10, Mango247, and 3 other users

Nice to have you back!

This is going to be long and technical, so I'm not even sure I'll be able to post the entire thing at once. OK, let's start at least

.

Step 1: $f\in C^1$ . Indeed, $f$ muct be at least continuous, whence the derivative is continuous due to the equation. Actually, you can get $C^\infty$ this way.

Step 2: $f$ cannot have a fixed point $a > 0$ . Indeed, then we would have $f(x)\ge a$ for all $x\ge a$ (you cannot cross the level $a$ down because every time you are at that level the derivative is $a > 0$ ), whence $f'(x)\ge a$ when $x\ge a$ , whence $f(x)\ge a + a(x - a)$ for $x\ge a$ , whence $f'(x) = f(f(x))\ge a + a^2(x - a)$ for $x > a$ , whence $f(x)\ge a + c(x - a)^2$ , so $f'(x)\ge a + c(f(x) - a)^2$ , but this inequality leads to a blow-up in finite time as $x$ increases.

Step 3: $f$ has a fixed point. Indeed, otherwise we have either $f(x) > x$ for all $x$ , or $f(x) < x$ for all $x$ . The first option will result in $f'(x) > x$ , in $f(x) > \frac 12 x^2$ , and, finally in $f'(x)\ge \frac 12 f(x)^2$ and a blow-up in finite time. The second option will result in $f'(x) < 0$ for all $x < 0$ , which is incompatible with going to $- \infty$ at $- \infty$ .

Step 4: Suppose that $f$ has $0$ as a fixed point. Then we have
$f(x) = \int_0^x f(f(t))\,dt.$
Take the largest interval $( - a,a)$ with the property that $|f\circ f|\le 1$ on $[ - a,a]$ (possibly the entire real line). Now, we have $|f(x)|\le |x|$ on $( - a,a)$ whence $f(x)\in( - a,a)$ for $x\in ( - a,a)$ and $|f'(x)| = |f(f(x))|\le |f(x)|$ on $( - a,a)$ . Now, Gronwall's lemma implies that $f\equiv 0$ on $( - a,a)$ , so if $a = + \infty$ , we are done and if $a < + \infty$ , then $a$ can be increased contrdicting the choice of $a$ .

Thus, we may assume that $f( - a) = - a$ for some $a > 0$ . We will show that for each $a > 0$ , there is a unique function $f$ satisfying the equation and this initial condition. Needless to say, I have no hope for an explicit elementary formula for it.

Step 5: Similarly to Step 2, we can show that $f(x)\le - a$ for $x\ge - a$ and $f(x)\ge - a$ for $x\le - a$ . Denoting $g(t) = - f( - a + t) - a$ and $h(t) = f( - a - t) + a$ , we get the system
$g'(t) = a - h(g(t)),\qquad h'(t) = a + g(h(t))$
with the initial data $g(0) = h(0) = 0$ to solve in non-negative smooth functions on $[0, + \infty)$ .

Step 6: $h$ is increasing and $h(t)\ge at$ . That is obvious.

Step 7: $g$ is increasing and tends to a finite positive limit as $t\to + \infty$ equal to the unique solution $L\le 1$ of the equation $h(L) = a$ . This is not so obvious but pretty standard. The key is the Lemma in the next step that should be applied to the solution $g$ and the constant solution $L$ to show that $g(t)\le L$ everywhere after which everything becomes clear.

Step 8:
Lemma: If $F\ge G$ are two smooth functions on and $f,g$ are two smooth solutions of $f' = F(f)$ , $g' = G(g)$ with $f(0)\ge g(0)$ , then $f\ge g$ on $[0, + \infty)$ .

Proof: Replace $F$ by $F_\delta = F + \delta$ with $\delta > 0$ and consider $f_\delta$ solving $f_\delta' = F_\delta(f_\delta), f_\delta(0) = f(0) + \delta$ . Then the graph of $f_\delta$ cannot cross or even touch that of $g$ because at the first crossing point we would have $f_\delta' - g'\ge\delta > 0$ . So, $f_\delta > g$ . But $f_\delta\to f$ as $\delta\to 0$ by the classical stability theorem.

Now we have come to the real meat but I'll have to interrupt my typing for at least half an hour. I'll continue as soon as I find free time.

This post has been edited 1 time. Last edited by fedja, Feb 15, 2010, 2:54 PM

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fedja

#3 h Dec 31, 2009, 10:56 PM • 3 Y

Y by pieater314159, Adventure10, and 1 other user

Uphh... Back to the proof. We have the system $g'=a-h(g),h'=a+g(h)$ . We want to solve it iteratively. Start with $g$ , find $h$ from the second equation, find new $g$ from the first equation for that $h$ , and so on. Everything will be fine if we are able to introduce some "distance" between possible functions $g$ , in which this single loop $g\mapsto h\mapsto g$ is contractive. So, here goes

Step 9: Our possible functions $g$ are $C^1$ , increasing, have positive derivative at $0$ and tend to some limit not greater than $1$ at $+\infty$ . This describes our space. For each pair $g,g_1$ of such functions, there exists the least $\lambda\ge 1$ such that $\lambda^{-1}g\le g_1\le \lambda g$ . This $\lambda$ (or, more precisely, $\lambda -1$ will measure how close $g$ and $g_1$ are. Note that it is not a distance but the fixed point theorem still holds in this strange space.

Step 10: Suppose that $g$ and $g_1$ are $\lambda$ -close, then
$h(\mu^{-1}t) \le h_1(t)\le h(\mu t)$ with $\mu-1=\frac{\lambda-1}{1+a}$ . Indeed, we know that $h$ solves $h'=a+g(h)$ , so $H(t)=H(\mu t)$ solves
$\begin{aligned} H'=\mu a+\mu g(H)&= a+ a(\mu-1)+\mu g(H)\ge a+ [a(\mu-1)+\mu] g(H) \\ &= a+\lambda g(H)\ge a+g_1(H)\,, \end{aligned}$
(we used that $0\le g\le 1$ ) so, by Lemma, $H\ge h_1$ . The other inequality is obtained by exchanging the roles of $g$ and $g_1$ .

Step 11: Now it remains to show that the above inequality for $h$ and $h_1$ implies that new $g$ and $g_1$ are $\mu$ -close. Again, consider $G=\mu g$ . It solves
$G'=\mu(a-h(\mu^{-1}G))_+\ge (a-h_1(G))_+$
and the lemma does the trick again (note that we can safely drop the values of $h,h_1$ exceeding $a$ in our differential equations for $g,g_1$ because we never get to that area.

The end!

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blang

#4 h Jan 1, 2010, 5:45 PM • 2 Y

Y by Adventure10, Mango247

Thanks, fedja !

I will study your solution.

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