Is there any techniques for solving a differential equation including iterated function? f ' (x) = f( f( x ) ) - MathOverflow

Is there any techniques for solving a differential equation including iterated function? f ' (x) = f( f( x ) )

2012-10-30T10:19:11Z

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

Answer by Brendan McKay for Is there any techniques for solving a differential equation including iterated function? f ' (x) = f( f( x ) )

2012-10-30T11:16:39Z

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 $b$.

Answer by Alex Gavrilov for Is there any techniques for solving a differential equation including iterated function? f ' (x) = f( f( x ) )

2012-10-30T12:37:45Z

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 $f(a)=a$.

Answer by fedja for Is there any techniques for solving a differential equation including iterated function? f ' (x) = f( f( x ) )

2012-11-01T02:36:50Z

Nothing is new under the Moon...

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

Answer by Anixx for Is there any techniques for solving a differential equation including iterated function? f ' (x) = f( f( x ) )

2012-11-01T03:06:48Z

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 $m=2$.

Answer by Anixx for Is there any techniques for solving a differential equation including iterated function? f ' (x) = f( f( x ) )

2012-11-01T17:19:43Z

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.