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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 |
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Nothing is new under the Moon... http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=321705 |
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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 $m=2$. |
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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 $b$. |
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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. |
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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 $f(a)=a$. |
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