In general how do I find the autonomous system that corresponds to the flow $\phi_t$ and how to find the flow $\phi_t$ given the autonomous system.

Examples Part A: Given $$\phi_t(k_1,k_2)=(\frac{k_1}{1-k_1t},k_2 e^{-t}) $$ find the autonomous system which has this flow.

and Part B: Given $$x'=x^3 $$ find the flow $\phi_t$.

Specifically I am looking for the general strategy/properties of flows of non-linear systems that will help me solve problems of this type. For linear systems I understand the strategy but I do not know the steps for non-linear systems.

Notes Part A: Using the following relationship between the flow and the autonomous system $\dfrac{d}{dt}(\phi_t(x_0)) = f(\phi_t(x_0))$. We find the autonomous system that corresponds to the flow $\phi_t(k_1,k_2)=(\frac{k_1}{1-k_1t},k_2 e^{-t})$. \begin{align} \dfrac{d}{dt}\phi_t(k_1,k_2)&=(\frac{k_1^2}{(1-k_1t)^2},-k_2 e^{-t}) \\ f(\phi_t(k_1),\phi_t(k_2))&=(\phi_t^2(k_1),-\phi_t(k_2)) \end{align} So $f(x_1,x_2)=(x_1^2,-x_2)$ for $x_1,x_2 \in \mathbb{R}$

Part B: The solution to $x'=x^3$ is $x(t)=\pm \frac{1}{\sqrt{2(c-t)}}$ where $c$ is a constant of integration. With initial condition $x(0)=x_0$ we get $x(t)=\pm \frac{1}{\sqrt{\frac{1}{x_0^2}-2t}}$. Do I find $\phi_t(x_0)$ by solving $\phi_t'(x_0)=\pm \frac{1}{\sqrt{\frac{1}{\phi_t^2(x_0)}-2t}}$

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    $\begingroup$ I might misunderstood your question, but wouldn't you just take the derivative of $\phi_t$ and see what happens? By definition, $\phi_t(x_0)$ is the unique solution satisfying $\dfrac{d}{dt}(\phi_t(x_0)) = f(\phi_t(x_0)), \phi_0(x_0) = x_0$. $\endgroup$ – Chee Han Feb 25 '17 at 0:52
  • $\begingroup$ Use what @CheeHan said, and also recall what I said your other question about this subject - that the first flow can be split to form two independent flows. $\endgroup$ – Nadiels Feb 25 '17 at 0:55
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    $\begingroup$ Your $x$ component is the solution to $\dot x=x^2$, for $\dot x=x^3$ you would get a square root in the solution. Easy to see as you can transform the given solution as $\frac1{x}=\frac1{k_1}-t$ which loses the constant under differentiation. $\endgroup$ – LutzL Feb 25 '17 at 12:05
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    $\begingroup$ Yes, with more structure your task makes sense. The generic way to find the differential equation for $y(t)=F(t,k)$ is to solve this system for $k=G(t,y(t))$ and then differentiate to $0=G_t+G_y·y'$. $\endgroup$ – LutzL Feb 26 '17 at 12:03
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    $\begingroup$ In part B, you actually find $$x(t)=\frac{x_0}{\sqrt{1-2x_0^2t}}$$ hence your flow is defined by $$\phi_t(\xi)=\frac{\xi}{\sqrt{1-2\xi^2t}}$$ for every $(t,\xi)$ such that $2\xi^2t<1$. // Sorry to ask this but, are you sure to fully have in mind the definitions of the subject? $\endgroup$ – Did Feb 27 '17 at 16:16

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