## Related questions with answers

*Either by hand or using a computer software package, sketch the phase plane diagrams for the given system. Discuss the stability of each limit cycle and critical point. Here $x=r \cos \theta$ and $y=r \sin \theta$.*

$\frac{d x}{d t}=y+x \sin (1 / r)$,

$\frac{d y}{d t}=-x+y \sin (1 / r)$

Solution

VerifiedMultiplying the first equation by $x$, the second one by $y$ and adding, we get

$\begin{align*} x\frac{dx}{dt}+y\frac{dy}{dt}&=x\big(y+x\sin(1/r)\big)+y\big(-x+y\sin(1/r)\big)\\ &=xy+x^2\sin(1/r)-xy+y^2\sin(1/r)\\ &\qquad\text{Simplify}\\ &=x^2\sin(1/r)+y^2\sin(1/r)\\ &\qquad\text{Take out common factor}\\ &=(x^2+y^2)\sin(1/r)\\ &\qquad\text{Use the fact that $x^2+y^2=r^2$}\\ &=r^2\sin(1/r) \end{align*}$

Using the identity

$x\frac{dx}{dt}+y\frac{dy}{dt}=r\frac{dr}{dt},$

we have

$r\frac{dr}{dt}=r^2\sin(1/r).$

Simplify

$\frac{dr}{dt}=r\sin(1/r).$

We have that $dr/dt=0$ when $\sin(1/r)=0\implies 1/r=n\pi\implies r=1/(n\pi),~n=1,2,\ldots$. Thus, the origin ($r=0$) is not an isolated critical point. Note that $dr/dt>0$ ($\sin(1/r)>0$) for $2n\pi<1/r<(2n+1)\pi$ or $1\big/[(2n+1)\pi]<r<1\big/(2n\pi)$, and $dr/dt<0$ for $1\big/(2n\pi)<r<1\big/[(2n-1)\pi]$.

Consequently, the trajectories spiral into the limit cycles $r=1/(2n\pi)$ (hence, stable) and away from the limit cycles $r=1\big/[(2n+1)\pi]$ (hence, unstable). {\bf The origin is a stable critical point}. To realize this, notice that for every $\epsilon>0$ there exists a positive integer $n_\epsilon$ such that $1/(2n_\epsilon\pi)<\epsilon$, so we can simply choose $\delta_\epsilon=1/(2n_\epsilon\pi)$. Then, the fact that $r=1/(2n_\epsilon\pi)$ is a stable limit cycle guarantees that any trajectory that begins within $\delta_\epsilon$ of the origin remains within $\epsilon$ of the origin.

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