Using polar co-ordinates to solve the dynamical system: $$x^\prime = x-y-x^{3}$$
$$y^\prime = x+y-y^{3}$$ I would say that I mostly understand this question and need to manipulate the equations using the following known conversions: $x = r$cos$(\theta)$, $y=r$sin$(\theta)$, $x^{2} + y^{2} = r^{2}$ and $rr^\prime = xx^\prime + yy^\prime $ and $\theta' = \dfrac{x y' - y x'}{r^2}$
I have made a a start with the following attempt but have become stuck at this stage: \begin{align} rr^\prime &= xx^\prime + yy^\prime \\ &=x(x-y-x^{3}) + y(x+y-y^{3}) \\ &=x^{2} -xy - x^{4} + yx + y^{2} - y^{4} \\ &=(x^{2} + y^{2}) - (x^{4} + y^{4}) \\ &=(r^{2}) - (x^{4} + y^{4}) \end{align} It is at this point that I am stuck as I struggle to see how I can simplify the last line further in terms of r, even though I am fairly certain this is the course of action I need to take. Any help/advice would be appreciated.
You could use the first two relations you mentioned: $ x = r \cos\theta $ and $ y = r \sin\theta $. The term $ x^4 + y^4 $ then becomes $ r^4( \cos^4\theta + \sin^4\theta ) $, and you end up with $ r' = r - r^3 (\cos^4\theta + \sin^4\theta) $. You can then do a similar development for $\theta' = \frac{xy' - yx'}{r^2} $ to obtain a system in the form \begin{align} r' &= f(r, \theta)\\ \theta' &= g(r, \theta) \end{align}