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Determine the unique non constant periodic solution of the two dimensional system

$$ \dot{x}=x-y-x\left(x^2+y^2\right)\\ \dot{y}=x+y-y\left(x^2+y^2\right) $$

and find its characteristic multipliers.

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closed as off-topic by Willie Wong, астон вілла олоф мэллбэрг, Jack's wasted life, suomynonA, user223391 Nov 20 '16 at 3:35

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  • 2
    $\begingroup$ (1) First edit your question to make sure your parentheses are paired. (2) How does the function $f(x,y) = x^2 + y^2$ evolve? $\endgroup$ – Willie Wong Nov 18 '16 at 21:38
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Here's a good hint: try converting the system to polar coordinates: We know that $$\begin{align}r^2 = x^2 + y^2 &\implies 2r\dot r = 2x\dot x + 2y\dot y \\&\implies \dot r = {x\dot x + y\dot y \over r}\end{align}$$ and similarly we have that $$\dot \theta = {x\dot y - y\dot x \over r^2}.$$

So, let us determine what $\dot r$ is: $$\begin{align}\dot r & = {x\dot x + y\dot y \over r^2}\\ &= {x\left(x-y-x\left(x^2+y^2\right)\right) + y\left(x+y-y\left(x^2+y^2\right)\right) \over r^2}\\ &= {-r^2(r^2-1) \over r} \\&=-r(r^2-1),\end{align}$$ where the third line occurs because we let $x=r\cos\theta$, and $y=r\sin\theta$. Now, obviously $\dot r = 0$ when $r=0,\pm1$.

Notice that the origin is a source (and is the only critical point of the system) because if we take $$\begin{align}f(x,y) : = \begin{bmatrix}x-y-x\left(x^2+y^2\right)\\ x+y-y\left(x^2+y^2\right)\end{bmatrix} &\implies f(x,y) = 0 \implies (x,y) = (0,0) \\&\implies Df(x,y) = \begin{bmatrix}-3x^2-y^2+1 & -2xy-1\\1-2xy & -x^2-3y^2+1\end{bmatrix}\\&\implies Df(0,0) = \begin{bmatrix}1 & -1\\1 & 1\end{bmatrix}\\&\implies \text{eigenvalues of $1\pm i$}\\ &\implies (0,0) \text{ is a source since $1$ is positive.}\end{align}$$ Now, since we've established that $(0,0)$ is a source, we will ignore $r=0$, and we will ignore $r=-1$ since in polar it is the same as $r=1$. Now, consider $r=2$: $$\dot r(2) = -2(4-1) < 0.$$ Since $\dot r$ is negative, that means the flow of the system is getting pulled inward, but since the origin is a source, flow is coming outward. By the Poincaré-Bendixson Theorem, we have that there is a periodic orbit at $r=1$, and it is a circle with radius $1$ (we should also verify this by showing $\dot \theta$ is constant which I will not do here). So the periodic orbit can be determined by the function, say $$\phi_0(t) = \pmatrix{\cos(t)\\\sin(t)}.$$

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    $\begingroup$ Note that you can write periodic solution this way not because it's lying on unit circle, but because it's also a trajectory of $\dot{x} = y, \dot{y} = -x$ (term $x^2+y^2-1$ in rhs of original system vanishes over the unit circle), which can be written this way. $\endgroup$ – Evgeny Nov 19 '16 at 9:32

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