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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.

j.c.
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frigen
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    I think this is more suitable for math.stackexchange.com – Beni Bogosel Oct 30 '12 at 10:49
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    Beni, why do you say that? – Vidit Nanda Oct 30 '12 at 11:57
  • @Vel Nias I think math analysis questions are not welcome here. – Anixx Nov 01 '12 at 11:03
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    @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. – fedja Nov 01 '12 at 12:33
  • Is that a delay differential equation? – Zsbán Ambrus Nov 01 '12 at 14:03
  • Zsbán, a delay differential equation would be something like $f'(x) = f(f(x-a))$. The right side of the equation above does not exhibit dependence on the backward/forward trajectory and hence there is no non-zero "delay" – Vidit Nanda Nov 02 '12 at 18:00
  • What a nice trick for parties lol – Joseph Victor Nov 02 '12 at 18:46
  • @ViditNanda is what you wrote really a delay differential equation, as it appears in applications? Wouldn't that rather be something like $f'(x)=g(f(x-a))$ where $g$ is given? – Michael Bächtold Feb 17 '21 at 16:18

5 Answers5

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Nothing is new under the Moon...

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

fedja
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  • wow! Thank you very much! Did you solve it by yourself? You're amazing! – frigen Nov 01 '12 at 03:17
  • I see no solution following the link. – Anixx Nov 01 '12 at 03:27
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    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. – fedja Nov 01 '12 at 03:32
  • @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." – Anixx Nov 01 '12 at 03:34
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    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 :)). – fedja Nov 01 '12 at 03:46
  • @fedja for the formulas look below. I do not know whether a formula exists if to apply your limitations (requiring the solution to be real) – Anixx Nov 01 '12 at 03:54
  • Unfortunately, to make sense of complex solutions, one has first to figure out where the function is defined. The composition is tricky because your complex powers take real values sometimes and I do not see why the branch used for $f$ agrees at those points with the branch used for $f(f)$. The fact that these formal solutions fail to be differentiable at $0$ quite miserably adds to the reasons I call them "pseudo-solutions". They can, perhaps, be made some rigorous sense of but that will require more than just using the power differentiation rule... – fedja Nov 01 '12 at 04:09
  • See my other answer regarding real solution. – Anixx Nov 01 '12 at 18:06
  • @fedja There is a 404 error in the link. Does it happen for you? – Tyma Gaidash Nov 28 '22 at 12:54
  • @TymaGaidash Yes. Apparently they deleted some old posts :-( – fedja Dec 02 '22 at 19:46
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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 $m=2$.

Anixx
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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 $b$.

Brendan McKay
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    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... – fedja Nov 01 '12 at 02:41
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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.

Anixx
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    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 $-20$. – fedja Nov 02 '12 at 02:50
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    @fedja yes, the proof in the linked paper requires p<1 or a Siegel number. I wiil add this to the answer. – Anixx Nov 02 '12 at 10:25
  • @Anixx Could you please explain the formulas for $d_n$ in more detail? For example, when I try to obtain $d_1$, I get $d_1 = d_0 \times B_{0,0}(,\cdot,) = d_0\times1=d_0=p \ne p^2$. – colt_browning Jul 09 '17 at 12:19
  • @colt_browning sorry I posted this answer 5 years ago, I think this formula is given in the linked paper. – Anixx Jul 09 '17 at 12:42
  • @Anixx I see, but that paper doesn't even mention Bell polynomials, so I had a hope that you remember this. If you don't, never mind. – colt_browning Jul 09 '17 at 19:57
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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 $f(a)=a$.

Alex Gavrilov
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