 # Interraption:To show that the system \displaystyle\dot{{r}} Maiclubk 2020-11-03 Answered

Interraption: To show that the system $\stackrel{˙}{r}=r\left(1-{r}^{2}\right),\stackrel{˙}{\theta }=1$ is equivalent to $\stackrel{˙}{x}=x-y-x\left({x}^{2}+{y}^{2}\right),\stackrel{˙}{y}=x+y-y\left({x}^{2}+{y}^{2}\right)$ for polar to Cartesian coordinates.
A limit cycle is a closed trajectory. Isolated means that neighboring trajectories are not closed.
A limit cycle is said to be unstable or half stable, if all neighboring trajectories approach the lemin cycle.
These systems oscillate even in the absence of external periodic force.

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The coordinate system can be either the Cartesian system or the polar system. In case of the polar system, a point is determined by a distance and angle from a reference point while as in Cartesian coordinate system each point is determined by a pair of numerical coordinates.
It is given that $x=r\mathrm{cos}\theta \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y=r\mathrm{sin}\theta$
Square and add the above equations.
${x}^{2}+{y}^{2}={r}^{2}{\mathrm{cos}}^{2}\theta +{r}^{2}{\mathrm{sin}}^{2}\theta$
${x}^{2}+{y}^{2}={r}^{2}\left({\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta \right)$
Substitute $\left({\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta \right)=1$ so that
${x}^{2}+{y}^{2}={r}^{2}$
Now,
$x=r\mathrm{cos}\theta$
Differentiate.
$\stackrel{˙}{x}=\stackrel{˙}{r}\theta -r\theta \mathrm{sin}\theta$
Substitute into the above equation.
$\stackrel{˙}{x}=\left(\frac{x\left(r\left(1-{r}^{2}\right)\right)}{r}\right)-y$
$\stackrel{˙}{x}=x\left(1-{r}^{2}\right)-y$
Substitute ${r}^{2}={x}^{2}+{y}^{2}$
$\stackrel{˙}{x}=x\left(1-{x}^{2}-{y}^{2}\right)-y$
$\stackrel{˙}{x}=x-y-x\left({x}^{2}+{y}^{2}\right)$
Thus, it is shown that $\stackrel{˙}{x}=x-y-x\left({x}^{2}+{y}^{2}\right)$
Now,
$y=\mathrm{sin}\theta$
Differentiate.
$\stackrel{˙}{y}=\stackrel{˙}{r}\mathrm{sin}\theta +r\theta \mathrm{cos}\theta$
Substitute $\stackrel{˙}{r}=r\left(1-{r}^{2}\right),\mathrm{sin}\theta =\frac{y}{r},\stackrel{˙}{\theta }=1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}x=r\mathrm{cos}\theta$ into the above equation.
$\stackrel{˙}{y}=\left(\frac{y\left(r\left(1-{r}^{2}\right)\right)}{r}\right)+x$
$\stackrel{˙}{y}=y\left(1-{r}^{2}\right)+x$
But ${r}^{2}={x}^{2}+{y}^{2}$
$\stackrel{˙}{y}=x+y-y\left({x}^{2}+{y}^{2}\right)$
Therefore, it is shown that $\stackrel{˙}{y}=x+y-y\left({x}^{2}+{y}^{2}\right)$
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