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.


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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Willie Wong, астон вілла олоф мэллбэрг, Jack's wasted life, suomynonA, Community
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 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

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 $$\phi_0(t) = \pmatrix{\cos(t)\\\sin(t)}.$$

  • 1
    $\begingroup$ 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

Not the answer you're looking for? Browse other questions tagged or ask your own question.