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I want to write this ODE system in polar coordinates (r,$\theta$).

$$\dot x =x-y-x^3 $$

$$\dot y = x+y-y^3$$

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closed as off-topic by Mark Fantini, Ivo Terek, user98602, user147263, Przemysław Scherwentke Jan 5 '15 at 2:45

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  • $\begingroup$ What's meaning of $\dot x$? $\endgroup$ – Paul Jan 5 '15 at 0:53
  • $\begingroup$ It's a notation for x'(t) $\endgroup$ – Mary Jan 5 '15 at 0:54
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We use:

$$x = r \cos \theta, y = r \sin \theta, r^2 = x^2 + y^2, \theta = \tan^{-1}\left(\frac{y}{x}\right)$$

From:

$$x^2 + y^2 = r^2 \implies x x' + y y' = r r'$$

We have:

$$\begin{align} x x' + y y' &= x(x-y-x^3) + y(x+y-y^3) \\ &= r \cos \theta(r \cos \theta-r \sin \theta - (r \cos \theta)^3) + r \sin \theta(r \cos \theta + r \sin \theta -(r \sin \theta)^3) \\&= r r'\end{align}$$

We also have:

$$\theta = \tan^{-1}\left(\frac{y}{x}\right) \implies \theta' = \dfrac{xy'-yx'}{r^2}$$

Make the appropriate substitutions, do some trig simplifications and cleanup to derive $r'$ and $\theta'$.

Update

You should end up with:

$$r' = -\frac{1}{4} r \left(r^2 \cos (4 \theta)+3 r^2-4\right)$$

$$\theta' = -\left(\left(r^2-1\right) \cos (2 \theta)+\sin (2 \theta)\right)$$

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  • $\begingroup$ yes, no problem! Thank you so much for your help @Amzoti . $\endgroup$ – Mary Jan 5 '15 at 1:22
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So $r$ and $\theta$ are function of $t$, let $x=r\cos\theta$ and $y=r\sin \theta$, then you will have

$$r'\cos \theta-r\sin\theta \times (\theta)'=r\cos\theta-r\sin \theta-(r\cos\theta)^3;$$

$$r'\sin \theta+r\cos\theta \times (\theta)'=r\cos\theta+r\sin \theta-(r\sin\theta)^3$$

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  • $\begingroup$ Exactly, but I want to write it in terms of r'= (...) and $\theta ' = (...)$. $\endgroup$ – Mary Jan 5 '15 at 1:03
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    $\begingroup$ It is not difficult for you to complete it, i believe. $\endgroup$ – Paul Jan 5 '15 at 1:09
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    $\begingroup$ Yes, I will. However, I suggest you that you should think by yourself for half an hour at least.. $\endgroup$ – Paul Jan 5 '15 at 1:12
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    $\begingroup$ I've just noticed you're missing $\theta '$ . It's also a function of t. $\endgroup$ – Mary Jan 5 '15 at 1:12
  • $\begingroup$ Yes, I'm sorry. $\endgroup$ – Paul Jan 5 '15 at 1:14

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