Are there any techniques for solving a differential equation of the form $f ' (x) = f( f( x ) )$? 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.
 A: 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 $b$.
A: 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.
A: 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 $f(a)=a$. 
A: Nothing is new under the Moon... 
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=321705
A: 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 $m=2$.
