MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to solve this differential equation but I have no idea how.

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

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

P.S. I don't think chain rule is useful for this

share|cite|improve this question
I think this is more suitable for – Beni Bogosel Oct 30 '12 at 10:49
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 1 '12 at 11:03
@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 1 '12 at 12:33
Is that a delay differential equation? – Zsbán Ambrus Nov 1 '12 at 14:03
up vote 15 down vote accepted

Nothing is new under the Moon...

share|cite|improve this answer
wow! Thank you very much! Did you solve it by yourself? You're amazing! – frigen Nov 1 '12 at 3:17
I see no solution following the link. – Anixx Nov 1 '12 at 3:27
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 1 '12 at 3:32
@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 1 '12 at 3:34
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 1 '12 at 3:46

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


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 $m=2$.

share|cite|improve this answer
See also my answer regarding real solutions below. – Anixx Nov 2 '12 at 10:36

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 $b$.

share|cite|improve this answer
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 1 '12 at 2:41

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 $f(a)=a$.

share|cite|improve this answer
Yes. +1 And I will give the expansion in another answer. – Anixx Nov 1 '12 at 11:04

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:

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


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.

share|cite|improve this answer
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 2 '12 at 2:50
@fedja yes, the proof in the linked paper requires p<1 or a Siegel number. I wiil add this to the answer. – Anixx Nov 2 '12 at 10:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.