Equating Coefficients of Cos and Sin

Question

I've got a nonlinear system \begin{align} x'&=\frac{1}{2}x-y-\frac{1}{2}(x^3+y^2x)\\ y'&=x+\frac{1}{2}y-\frac{1}{2}(y^3+x^2y) \end{align}

I am to analyse the system when the system is changed to polar coordinates. The writer of the book states that he computes the following \begin{align} r'\cos\theta-r(\sin\theta)\theta'&=x'=\frac{1}{2}(r-r^3)\cos\theta-r\sin\theta\\ r'\sin\theta+r(\cos\theta)\theta'&=y'=\frac{1}{2}(r-r^3)\sin\theta+r\cos\theta \end{align}

And by equating the coefficients of $\cos\theta,\sin\theta$ he gets the following system \begin{align} r'&=r(1-r^2)/2\\ \theta'&=1 \end{align}

My issues is then:

I'm not sure where he gets the following identities \begin{align} r'\cos\theta-r(\sin\theta)\theta'&=x'\\ r'\sin\theta+r(\cos\theta)\theta'&=y' \end{align}

And I'm not that familiar with the equating coefficients method, so I've read a bit about it, and I can't seem to compute the presented system.

If I were to make a phase portrait of the system, would the axis be $r'$ and $\theta'$?

Source: [c. 8, p. 162], Differential Equations, Dynamical Systems & An Introduction to Chaos (2nd Ed) by Morris W. Hirsch, Stephen Smale and Robert L. Devaney.

Answer 1

$$\begin{cases} x=r\cos(\theta) \\ y=r\sin(\theta) \end{cases}\qquad \begin{cases} x'=r'\cos(\theta)-r\sin(\theta)\theta' \\ y'=r'\sin(\theta)+r\cos(\theta)\theta' \end{cases} $$ $$\begin{cases} x'=r'\cos(\theta)-r\sin(\theta)\theta'=\frac12r\cos(\theta)-r\sin(\theta)-\frac12r^3\cos(\theta) \\ y'=r'\sin(\theta)+r\cos(\theta)\theta'=r\cos(\theta)+\frac12 r\sin(\theta)-\frac12r^3\sin(\theta) \end{cases} $$

$x'\cos(\theta)+y'\sin(\theta)=r'=\left(\frac12r\cos(\theta)-r\sin(\theta)-\frac12r^3\cos(\theta) \right)\cos(\theta)+\left(r\cos(\theta)+\frac12 r\sin(\theta)-\frac12r^3\sin(\theta) \right)\sin(\theta)$

and after simplification : $$r'=\frac12r-\frac12r^3$$

$-x'\sin(\theta)+y'\cos(\theta)=r\theta' =-\left(\frac12r\cos(\theta)-r\sin(\theta)-\frac12r^3\cos(\theta) \right)\sin(\theta)+\left(r\cos(\theta)+\frac12 r\sin(\theta)-\frac12r^3\sin(\theta) \right)\cos(\theta)$

and after simplification :

$$r\theta'=r\qquad;\qquad \theta'=1$$

The equations in the book are correct.

Answer 2

He differentiates $x=r\cos\theta$ with the product rule.
Then multiply the third equation by $\cos\theta$, plus multiply the fourth equation by $\sin\theta$. That eliminates the $\theta^\prime$ terms.

Answer 3

Or one can set $z=x+iy$, then $$ z'=(\tfrac12+i)z-\tfrac12z|z|^2=\tfrac12(1+2i-|z|^2)z $$ so that for radius and angle one gets \begin{alignat}1 \frac{d}{dt}|z|^2&=2Re(\bar zz')&&=(1-|z|^2)|z|^2\\ \frac{d}{dt}\arg(z)&=Im(\frac{z'}{z})&&=1 \end{alignat}

Answer 4

While the method in the book is correct, it is often easier to derive the polar coordinates version using \begin{align} rr'&=xx'+yy'\\ r^2\theta'&=xy'-yx' \end{align} which is obtained by differentiating $$ r^2=x^2+y^2,\quad\tan\theta=\frac{y}{x}. $$ Indeed, the $r'$-equation is clear, and differentiating $\tan\theta=y/x$ gives $$ (\tan\theta)'=\theta'\sec^2\theta=\theta'(1+\tan^2\theta)=\theta'\frac{x^2+y^2}{x^2}=r^2\theta' $$ in the LHS and $$ \left(\frac{y}{x}\right)'=\frac{xy'-yx'}{x^2} $$ in the RHS.