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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

$r'=r(1-r)$

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..

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    $\begingroup$ 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
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    $\begingroup$ 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
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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).

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  • $\begingroup$ 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
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    $\begingroup$ @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
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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.

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