A friend told me to solve the following differential equation:


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.

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    $\begingroup$ The constant function f(x) = 0 works.... $\endgroup$ Sep 2, 2014 at 12:02
  • $\begingroup$ 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. $\endgroup$ Sep 2, 2014 at 12:32
  • $\begingroup$ 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. $\endgroup$
    – Bumblebee
    Sep 2, 2014 at 12:35
  • $\begingroup$ @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 $f(\alpha) = f^{\prime}(\alpha) = f^{\prime \prime}(\alpha) = \cdots = 0$. $\endgroup$ Sep 3, 2014 at 14:18
  • 1
    $\begingroup$ @davincisghost Yes, after your original comment I was of course looking for nontrivial holomorphic solutions. In a way $\infty$ is a solution of $2k=k-1$ :) $\endgroup$ Sep 4, 2014 at 6:42

1 Answer 1


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


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.


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