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I am trying to solve the following differential equation

$$f ' (x) = f( f( x ) ),$$

but I have no idea how. I don't think the chain rule is useful for this.

Although I don't think this differential equation is solvable, I'd like to know if there is any interesting approach to solve a differential equation of this kind, or, at least, a non-trivial solution of the equation.

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    $\begingroup$ I think this is more suitable for math.stackexchange.com $\endgroup$
    – Beni Bogosel
    Commented Oct 30, 2012 at 10:49
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    $\begingroup$ Beni, why do you say that? $\endgroup$
    – Vidit Nanda
    Commented Oct 30, 2012 at 11:57
  • $\begingroup$ @Vel Nias I think math analysis questions are not welcome here. $\endgroup$
    – Anixx
    Commented Nov 1, 2012 at 11:03
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    $\begingroup$ @Anixx. For the best of my knowledge, this claim is plain wrong. However, I agree that the remarks like "this is homework" or "ask that on MSE" that are not supported by any evidence that the person making them can solve the problem himself can be somewhat irritating... As to Beni's recommendation itself, MSE is not a bad site per se but it is just DROWNED in "homeworks" nowadays. MO and AoPS are much better choices for something nontrivial IMHO. $\endgroup$
    – fedja
    Commented Nov 1, 2012 at 12:33
  • $\begingroup$ Is that a delay differential equation? $\endgroup$
    – Zsbán Ambrus
    Commented Nov 1, 2012 at 14:03

5 Answers 5

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Nothing is new under the Moon...

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=321705

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  • $\begingroup$ wow! Thank you very much! Did you solve it by yourself? You're amazing! $\endgroup$
    – frigen
    Commented Nov 1, 2012 at 3:17
  • $\begingroup$ I see no solution following the link. $\endgroup$
    – Anixx
    Commented Nov 1, 2012 at 3:27
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    $\begingroup$ Meaning you haven't scrolled down or you failed to understand what is written there? In the latter case you are welcome to ask questions. $\endgroup$
    – fedja
    Commented Nov 1, 2012 at 3:32
  • $\begingroup$ @fedja I only see the supposed proofs of existence. Regarding the solution, you yourself wrote "I have no hope for an explicit elementary formula for it." $\endgroup$
    – Anixx
    Commented Nov 1, 2012 at 3:34
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    $\begingroup$ Existence and uniqueness of a one-parameter family of solutions, to be exact ;). Do you believe one can come with a formula? We can, probably, give it a shot and try to prove that the functions in question are not elementary but that is quite another story (and, most likely, quite a non-trivial one given that there exist formal elementary pseudo-solutions like the ones you mentioned). If you are interested in such a project, I can think of what might be the right approach here (but a bit later :)). $\endgroup$
    – fedja
    Commented Nov 1, 2012 at 3:46
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There are two closed form solutions:

$$\displaystyle f_1(x) = e^{\frac{\pi}{3} (-1)^{1/6}} x^{\frac{1}{2}+\frac{i \sqrt{3}}{2}}$$ $$\displaystyle f_2(x) = e^{\frac{\pi}{3} (-1)^{11/6}} x^{\frac{1}{2}+\frac{i \sqrt{3}}{2}}$$

The solution technique can be found in this paper.

For a general case, solution of the equation

$$f'(z)=f^{[m]}(z)$$

has the form

$$f(z)=\beta z^\gamma$$

where $\beta$ and $\gamma$ should be obtained from the system

$$\gamma^m=\gamma-1$$ $$\beta^{\gamma^{m-1}+...+\gamma}=\gamma$$

In your case $m=2$.

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  • $\begingroup$ See also my answer regarding real solutions below. $\endgroup$
    – Anixx
    Commented Nov 2, 2012 at 10:36
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I don't know, but one answer is $f(x)=ax^c$ where $a=\frac12(\sqrt {3}+i){ e^{\frac16\pi\sqrt {3}}}$ and $c=\frac12+\frac12i\sqrt{3}$. Another is obtained by taking the complex conjugate of both $a$ and $b$.

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    $\begingroup$ The only unclear thing is where this function is defined. Note that complex powers of real numbers are complex and complex powers of complex numbers are branching like crazy... $\endgroup$
    – fedja
    Commented Nov 1, 2012 at 2:41
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And regarding real solutions to the question, Alex Gavrilov is completely correct. A Taylor expansion at fixed point $p$ gives us the real solution. Existence of this solution is proven in the paper which I already referenced from my another answer.

$$f(z)=\sum_{n=0}^\infty \frac{d_n (z-p)^n}{n!}$$

where $d_n$ is defined as follows:

$$d_0=p$$ $$d_{n+1}=\sum _{k=0}^n d_k \operatorname{B}_{n,k}(d_1,...,d_{n-k+1})$$

where $B_{n,k}$ are the Bell polynomials

This gives the following starting coefficients:

$$d_1=p^2$$ $$d_2=p^3+p^4$$ $$d_3=p^4 + 4 p^5 + p^6 + p^7$$ $$d_4=p^5 + 11 p^6 + 11 p^7 + 8 p^8 + 4 p^9 + p^{10} + p^{11}$$

etc.

The fixed point $p$ here serves as a parameter, which determines the family of solutions. According the linked theorem, the expansion should converge in the neighborhood of $p$ for $0 < |p| < 1 $ or $p$ being a Siegel number.

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    $\begingroup$ Anixx, this becomes boring. Yeah, if $p$ is small enough, this has a chance to work (though I wonder how you prove that there are no other solutions). However, for large $p$, you have a polynomial with the leading term $p^N$ with $N\approx k^2$ for the $k$'th coefficient. The miraculous cancellations can shave only something like $8^{N}$ off it for a typical $p$ (Remez). So, you are left with $(p/8)^{ck^2}$ which eats up the factorial and the geometrical progression for breakfast and happily flies to infinity by the lunchtime if $p$ is like $-20$. $\endgroup$
    – fedja
    Commented Nov 2, 2012 at 2:50
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    $\begingroup$ @fedja yes, the proof in the linked paper requires p<1 or a Siegel number. I wiil add this to the answer. $\endgroup$
    – Anixx
    Commented Nov 2, 2012 at 10:25
  • $\begingroup$ @Anixx Could you please explain the formulas for $d_n$ in more detail? For example, when I try to obtain $d_1$, I get $d_1 = d_0 \times B_{0,0}(\,\cdot\,) = d_0\times1=d_0=p \ne p^2$. $\endgroup$
    – colt_browning
    Commented Jul 9, 2017 at 12:19
  • $\begingroup$ @colt_browning sorry I posted this answer 5 years ago, I think this formula is given in the linked paper. $\endgroup$
    – Anixx
    Commented Jul 9, 2017 at 12:42
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    $\begingroup$ In fact, the recursion seems to work if the sum over $k$ goes from $1$ rather than $0$, the initial conditions are $d_0=d_1=p$, and then the following starting coefficients are as you say but with index shifted by $1$: $d_2=p^2$, etc. $\endgroup$
    – colt_browning
    Commented Sep 22 at 22:35
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For what I know, the standard method is the Taylor series expansion at a fixed point, i.e. at a point $x=a$ such that $f(a)=a$.

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  • $\begingroup$ Yes. +1 And I will give the expansion in another answer. $\endgroup$
    – Anixx
    Commented Nov 1, 2012 at 11:04

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