How do you solve $f'(x) = f(f(x))$?

A friend told me to solve the following differential equation:

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

I have no idea how to solve this! This doesn't seem to be an ordinary differential equation and I can't even solve this numerically!

I think my friend is trolling me.

• The constant function f(x) = 0 works.... Sep 2, 2014 at 12:02
• Assume $f$ holomorphic and $f(0)=0$, say $0$ is a root of order $k$. Then $f(f(x))$ has a root of order $2k$ and $f'$ has a root of order $k-1$. Oops. Sep 2, 2014 at 12:32
• suppose $f(x)=x^n$ can satisfy this equation. Then $n-1=n^2.$ There are two complex solutions for this equation. I don't no they can be acceptable as an answers. But nothing seems like wrong. Sep 2, 2014 at 12:35
• @HagenvonEitzen The constant function $f(x) = 0$ cannot have a 'root of order $k$'. It has a zero at each $\alpha \in \mathbb{C}$, but none of its zeros is of finite order since $f(\alpha) = f^{\prime}(\alpha) = f^{\prime \prime}(\alpha) = \cdots = 0$. Sep 3, 2014 at 14:18
• @davincisghost Yes, after your original comment I was of course looking for nontrivial holomorphic solutions. In a way $\infty$ is a solution of $2k=k-1$ :) Sep 4, 2014 at 6:42

This question has been asked on MathOverflow, this answer is migrated from there: -

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 $$m=2$$.

Note: I merely brought this to your attention, all credit should go to Anixx on MO.