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I have a dynamical system on the plane given by $$\dot{x}=-y+x\left(1-\sqrt{x^2+y^2}\right)\\ \\ \dot{y}=y+x\left(1-\sqrt{x^2+y^2}\right)$$

I want to convert this into polar coordinates as it will be easier for the question I am attempting to solve (it gives a hint to convert this into polar coordinates).

Problem is I don't know how to do this I know all the relations for polar coordinates to relate it to Cartesian coordinates for example I know the bit in the bracket would be $(1-r)$ but I don't know what $\dot{x}$ would and how to convert it into polar coordinates.

Any help?

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closed as unclear what you're asking by Did, Leucippus, Daniel W. Farlow, choco_addicted, Charles Apr 23 '16 at 3:46

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  • $\begingroup$ Answer this question: if you know $\dot{x}$ and $\dot{y}$ and also know how $r$ and $\phi$ related to $x$ and $y$, can you write $\dot{r}$ and $\dot{\phi}$ ? $\endgroup$ – Evgeny Apr 12 '16 at 21:30
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    $\begingroup$ What exactly you don't understand? $\endgroup$ – Evgeny Apr 12 '16 at 21:38
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    $\begingroup$ As you know, $r^2 = x^2+y^2$. What would happen if you calculate time derivative of left and right side of this equality? $\endgroup$ – Evgeny Apr 12 '16 at 21:39
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    $\begingroup$ Yes, that's it. If you don't remember chain rule for derivatives, it's the right time to revise it, because it's used everywhere in differential equations. $\endgroup$ – Evgeny Apr 12 '16 at 21:58
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    $\begingroup$ Having just done this calculation I think the original system was missing a bracket. The solution is much nicer if $$ \dot{x}=\left(-y+x\right)\left(1-\sqrt{x^2+y^2}\right)\\ \\ \dot{y}=\left(y+x\right)\left(1-\sqrt{x^2+y^2}\right) $$ $\endgroup$ – Giskard Sep 18 '16 at 17:03
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I suggest that you try to understand how it works in general.

$1)$ From $x',y'$ to $r',\theta'$: $$ r' = \left(\sqrt{x^2 +y^2} \right)' = \frac{(x^2 +y^2)'}{2 \sqrt{x^2 +y^2}}=\frac{xx' +yy'}{r} $$ and $$ \theta' = \left(\arctan \frac{y}{x} \right)' = \frac{(y/x)'}{1+(y/x)^2} = \frac{y' x -x' y}{r^2}. $$

Now in the other direction.

$2)$ From $r',\theta'$ to $x',y'$: $$ x'= (r\cos\theta)'=r' \cos \theta -r \theta' \sin \theta $$ and $$ y'= (r\sin\theta)'= r' \sin \theta + r \theta' \cos \theta . $$

After understanding this you can proceed on your own.

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