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Consider the following autonomous system in $\mathbb{R}^2$:

$$\dot{x}_1 = -x_2 + (1 - x_1^2 - x_2^2)$$ $$\dot{x}_2 = x_1 + (x_1^2 + x_2^2 - 1)(1 - x_1 x_2)$$

First, for every $t_0 \in \mathbb{R}$ and $\psi \in \mathbb{R}$, I want to explicitly find a solution of the system, with the initial value problem $x(t_0) = \pmatrix{cos\psi \\ sin\psi}$. As a tip, I'm supposed to do this by "sharply looking".

Next, I want to prove that every inital value problem $x(0) = x_0$ that satisfies $||x_0||_2 < 1$ has a unique, on all of $\mathbb{R}$ defined solution, that stays within the open circle $\{x \in \mathbb{R}^2: ||x||_2 < 1\}$.

I must admit that I sat here for several minutes and tried to "look sharply", but so far without success. What am I supposed to see that I fail to see? Are there any "tricks" or formulas that help simplify the original system or so?

For the second part, it sounds like one could apply the Picard-Lindelöf theorem, but I'm not so sure on how one could do that. So far, I've only seen applications of the Picard-Lindelöf theorem where we conclude the existence of a unique solution on an interval $[a, b] \subseteq \mathbb{R}$, and not on the entirety of $\mathbb{R}$.

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You are supposed to see that on the unit circle you have $$ 1-x_1^2+x_2^2=0 $$ and that the vector field is tangential to the unit circle. Uniqueness provides the separation of the in- and outside of the unit circle.

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  • $\begingroup$ Thanks, I totally missed that the equation for the unit circle was part of the system. That sounds reasonable. So the initial value problem could be interpreted that we "start" at a specific point on the unit circle? I still have trouble figuring out how the solution $x(t)$ would look like for the initial value problem. $\endgroup$ – moran Nov 6 '15 at 20:20
  • $\begingroup$ The solution of course is $x(t)=(\cos (t+ψ),\sin (t+ψ))$. But that only works on the unit circle. $\endgroup$ – LutzL Nov 6 '15 at 21:52

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