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Consider the following set of equations: $$ \begin{array}{l} y = f(x) \\ \frac{dy}{dx} = f(y) \end{array}$$ These can be written as finding some differentiable function $f(x)$ such that $$ f^{\prime} = f(f(x)) $$ For example, say $y(0) = 1$. Then $\left. \frac{dy}{dx} \right|_{x=0}$ is determined by the value of $y(1)$. The derivative at the $x=0$ had better be negative, otherwise by the time the function gets to 1, the value will be too great and will contradict the alleged value of hte derivative at $x=0$.

Many years ago I tried various techniques to find a solution (other than the trivial $f(x) = 0$) to this equation. It has properties akin to a delay equation, but the delay is variable and strongly depends on the solution itself. I tried expanding as a series; that fails spectacularly. I tried eigenvalue tricks, without any notable progress. Fourier analysis looks good, until you contemplate things like $\sin( \sin( \ldots \sin(x)) \ldots)$ that emerge and that made me give up. I still think if there are solutions there will be periodic solutions.l

I suspect solving this problem is hard, but perhaps somebody can prove that no non-trivial solution exists.

Edit after seeing the good complex-valued solution provided by JJaquelin:

Can anybody find a *real differentiable $f(x)$ other than the trivial $f(x)=0$ that satisfies the conditions, or prove that no such function exists.

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  • $\begingroup$ Maybe mathoverflow is a better place for this question. $\endgroup$ – Moishe Kohan Jun 4 '14 at 3:39
  • $\begingroup$ @studiosus: I'm not so sure about that. I've asked a similar question there, but it got moved to this site. $\endgroup$ – Lucian Jun 4 '14 at 9:14
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This is not the a complete answer. Only two solutions are shown below. Furthermore, I plugged them into the initial equation and it's agree.

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

Search of solutions on the form $f(x)=a x^b$

$$a b x^{b-1} = a(a x^b)^b = a^{b+1} x^{b^2} \rightarrow \begin{cases}b^2 = b-1\\ a^{b+1} = a b \rightarrow a = b^{1/b}\end{cases}$$

$$b=\frac{1}{2}\pm i\frac{\sqrt{3}}{2}; \frac{1}{b}=\frac{1}{2}\mp i\frac{\sqrt{3}}{2}; a=\left(\frac{1}{2}\pm i\frac{\sqrt{3}}{2}\right)^{\frac{1}{2}\mp i\frac{\sqrt{3}}{2}}$$

First solution: $f(x)=\left(\frac{1}{2}+ i\frac{\sqrt{3}}{2}\right)^{\frac{1}{2}- i\frac{\sqrt{3}}{2}}x^{\frac{1}{2}+ i\frac{\sqrt{3}}{2}}$

Second solution: $f(x)=\left(\frac{1}{2}- i\frac{\sqrt{3}}{2}\right)^{\frac{1}{2}+ i\frac{\sqrt{3}}{2}}x^{\frac{1}{2}- i\frac{\sqrt{3}}{2}}$

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  • $\begingroup$ Yes, this presents solutions to the problem as posed. I had been looking for real-valued, differentiable f(x), and there I suspect there are no non-trivial solutions but cannot prove it. $\endgroup$ – Mark Fischler Jun 4 '14 at 15:57
  • $\begingroup$ In particular, the real part of your first answer does not by itself solve the equation (I plotted it using alpha). $\endgroup$ – Mark Fischler Jun 4 '14 at 16:27
  • $\begingroup$ Of course, the real part is not solution of the equation. It is clear that the two solutions are complex solutions. I was not looking especially for real solution. $\endgroup$ – JJacquelin Jun 4 '14 at 17:55
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It looks like the contraction mapping theorem can be applied here to guarantee a local solution. Consider $$Af(x)=b+\int_0^x f(f(t))dt,\qquad x\in[-a,a]$$ defined on $C^1[-a,a]$, endowed with norm $\|f\|=\|f\|_\infty+\|f'\|_\infty$ (or any other equivalent norm). Then it's not hard to show that if $B$ is the ball $B=\{f\in C^1[-a,a]: \|f\|\le a\}$, then for small enough $a$ and small enough $b$ (depending on $a$) $A:B\to B$, and the operator $A^2$ is a contraction on $B$. This guarantees the existence and the uniqueness of a solution in $B$, satisfying the initial condition $f(0)=b$.

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  • $\begingroup$ What is $b$ here? $\endgroup$ – Cameron Williams Apr 29 '16 at 20:10
  • $\begingroup$ b represents the initial value: f(0)=b. $\endgroup$ – Carlo Morpurgo Apr 29 '16 at 21:13
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Just playing around a bit with JJacquelin's answer.

Regarding $f(x) =\left(\frac{1}{2}+ i\frac{\sqrt{3}}{2}\right)^{\frac{1}{2}- i\frac{\sqrt{3}}{2}}x^{\frac{1}{2}+ i\frac{\sqrt{3}}{2}} $, since $\frac{1}{2}+ i\frac{\sqrt{3}}{2} =e^{i\pi/3} $ and $\frac{1}{2}- i\frac{\sqrt{3}}{2} =e^{-i\pi/3} $, this becomes

$\begin{array}\\ f(x) &=\left(e^{i\pi/3}\right)^{e^{-i\pi/3}}x^{e^{i\pi/3}}\\ &=e^{i(\pi/3)e^{-i\pi/3}}x^{e^{i\pi/3}}\\ &=e^{i(\pi/3)(\cos(\pi/3)-i\sin(\pi/3)}x^{\cos(\pi/3)+i\sin(\pi/3)}\\ &=e^{(\pi/3)(i\cos(\pi/3)+\sin(\pi/3)}x^{\cos(\pi/3)+i\sin(\pi/3)}\\ &=e^{(\pi/3)\sin(\pi/3)}e^{(\pi/3)(i\cos(\pi/3)}x^{\cos(\pi/3)+i\sin(\pi/3)}\\ &=e^{(\pi/3)\sin(\pi/3)}e^{(\pi/3)(i\cos(\pi/3)}x^{\cos(\pi/3)}x^{i\sin(\pi/3)}\\ &=e^{(\pi/3)\sin(\pi/3)}x^{\cos(\pi/3)}e^{(\pi/3)(i\cos(\pi/3)}x^{i\sin(\pi/3)}\\ &=e^{(\pi/3)\sin(\pi/3)}x^{\cos(\pi/3)}e^{(\pi/3)(i\cos(\pi/3)}e^{i\ln(x)\sin(\pi/3)}\\ &=e^{(\pi/3)\sin(\pi/3)}x^{\cos(\pi/3)}e^{i(\ln(x)\sin(\pi/3)+\cos(\pi/3))}\\ &=e^{(\pi/3)\sin(\pi/3)}x^{\cos(\pi/3)}(\cos(\ln(x)\sin(\pi/3)+\cos(\pi/3))+i\sin(\ln(x)\sin(\pi/3)+\cos(\pi/3)))\\ \end{array} $

Don't know if this is any use, but, hey.

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