This is a worked out example in my book, but I am having a little trouble understanding it:

Consider the system of equations:

$$x'=y+x(1-x^2-y^2)$$ $$y'=-x+y(1-x^2-y^2)$$

The orbits and limit sets of this example can be easily determined by using polar coordinates. (My question: what is the motivation for that thinking? What should clue me in to thinking that I should use polar coordinates?)

The polar coordinate satisfies $r^2=x^2+y^2$ so by differentiating with respect to t and using the differential equations we get:

$r\cdot r'=x\cdot x'+y\cdot y'$ (I am unclear about how the book even got this first equation from $r^2=x^2+y^2$)

$=x\cdot y+x^2(1-r^2)-x\cdot y+y^2(1-r^2)$ Substitute in $x'$ and $y'$ and then multiple out and replace with $r$, I get this step

$=r^2(1-r^2)$ cancel terms, I get this step too


similarly, the angle variable $\theta$ satisfies $\tan\theta=\frac yx$, so the derivative with respect to $t$ yields $\sec^2(\theta)\theta'=x^{-2}[x^2+xy(1-r^2)-y^2-xy(1-r^2)]=-\frac{r^2}{x^2}$ so $\theta=1$

Thus the solution goes clockwise around the origin at unit angular speed.

I don't understand the $\theta$ step at all or how they reached the conclusion of clockwise around the origin with unit angular speed..

But then it just jumps to saying "the origin is a fixed point, so α(0)=ω(0)={0} but I have no idea how they reached this conclusion..

  • 1
    $\begingroup$ Regarding the first question: Anything dependent on $x^2+y^2$ (such as $1-x^2-y^2 = 1 - (x^2+y^2)$) lends itself to conversion into polar (or spherical or cylindrical) coordinates. $\endgroup$ – AlexR Nov 10 '14 at 16:41
  • 2
    $\begingroup$ Regarding the second one: Differentiate $r^2 = x^2+y^2$ to obtain $2rr' = 2xx'+2yy'$. $\endgroup$ – AlexR Nov 10 '14 at 16:44
  • $\begingroup$ @AlexR oh I get it now, thanks~ $\endgroup$ – Math Major Nov 10 '14 at 16:45
  • $\begingroup$ If you get an answer on this site wich solves your question, you should accept it to signal that you no longer need help with the question. $\endgroup$ – AlexR Nov 10 '14 at 16:53
  • $\begingroup$ @AlexR it won't let me accept it yet...it says I have to wait a few minutes... $\endgroup$ – Math Major Nov 10 '14 at 16:53

Firsly, anything dependent on $x^2+y^2$ (such as $1-x^2-y^2 = 1-(x^2+y^2)$) lends itself to conversion into polar (or spherical or cylindrical) coordinates. We obtain $$\begin{align*} r^2 & = x^2+y^2 \\ \Rightarrow \frac d{dt} r^2 & = \frac d{dt} x^2 + \frac d{dt} y^2 \\ \Rightarrow 2rr' & = 2xx' + 2yy' \\ \Rightarrow rr' & = xx'+yy' \end{align*}$$ This trick of differentiating both sides has also been applied to $$\begin{align*} \tan \theta & = \frac yx \\ \Rightarrow \sec^2 \theta \cdot \theta' & = \frac{x'y - y'x}{x^2} \end{align*}$$ The last equation was then manipulated to arrive at $\theta' = 1$ (error in your typeset I guess).

  • $\begingroup$ Thanks! But I am still wondering where did the tanθ come from? And how did they arrive at the conclusion of counter clockwise? $\endgroup$ – Math Major Nov 10 '14 at 16:52
  • 1
    $\begingroup$ @MathMajor Counter-clockwise is the usual polar coordinate transform: $$(\theta, r) \mapsto (r \cos\theta, r \sin\theta)$$ This also gives $$\frac yx = \frac{r\sin \theta}{r\cos\theta} = \tan\theta$$ $\endgroup$ – AlexR Nov 10 '14 at 16:54

When you see an $x^2 + y^2$ in a problem that is probably going to be tractable, one good first thing to try is to see how the problem looks transformed to polar coordinates. (Even in a problem that comes about in real life, where you don't know the solution will be possible to obtain, this is a good first shot.)

When $r^2 = x^2 + y^2$ it is valid to take the derivative of each side with respect to $t$. For example, $${d \over dt} (r^2) = 2r {dr \over dt} = 2rr' $$ The book got the equation you present by doing this on both sides of the equation, and dividing by 2.

I see another answer and that you are unclear on how $\tan \theta$ comes in. That is easy: In polar coordinates, $y/x = \tan \theta$. Draw yourself a right triangle with one point at the origin, the right angle on the X axis and the third point at $(x,y)$ and ask yourself what $\tan \theta$ would be in that picture.

By the way, the solution that approaches $r=1$ is in fact a stable orbit, since if $r = 1+ \epsilon$, $r' = -\epsilon r$ so the solution will approach $r=1$ in an exponential fachion.

Then you have seen how to get from that to $$ r' = r(1-r) $$ The first thing to notice about that equation is that considering $r$ alone (without $\theta$) it will have fixed points wherever $r' = 0$, that is, at $r=0$ and at $r=1$. But then $r=0$ is a fixed point of the whole system since at $r=0$, $\theta$ is irrelevant.

However, that fixed point is an unstable fixed point, since near the origin with $r(t=0) = \epsilon$, the solution behaves like $r = \epsilon e^t$, exploding away from the origin.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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