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.
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8$\begingroup$ The constant function f(x) = 0 works.... $\endgroup$– davincisghostSep 2, 2014 at 12:02
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$\begingroup$ 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. $\endgroup$– Hagen von EitzenSep 2, 2014 at 12:32
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$\begingroup$ 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. $\endgroup$– BumblebeeSep 2, 2014 at 12:35
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$\begingroup$ @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 $f(\alpha) = f^{\prime}(\alpha) = f^{\prime \prime}(\alpha) = \cdots = 0$. $\endgroup$– davincisghostSep 3, 2014 at 14:18
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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 $2k=k-1$ :) $\endgroup$– Hagen von EitzenSep 4, 2014 at 6:42
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1 Answer
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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.
answered Sep 2, 2014 at 12:23