Dynamical System transformation

Question

How can the system $$\frac{dx}{dt}=-y+\epsilon x(x^2+y^2)$$$$\frac{dy}{ dt}=x+\epsilon y(x^2+y^2)$$ be transformed into $$\frac{dr}{dt}=\epsilon r^3$$ $$\frac{d\theta}{dt}=1$$ via polar coordinates?

I sub in $x=r\cos(\theta)$ and $y=r\sin(\theta)$ into both of the original equations, but there are $\frac{dr}{dt}$ and $\frac{d\theta}{dt}$ terms in both equations then. And adding/subtracting the equations doesn't seem to produce anything 'nice'.

Answer 1

Well, to obtain this form you can just differentiate $r^2(t) = x^2(t) + y^2(t)$ and $\theta(t) = \arctan{\frac{y(t)}{x(t)}}$. Let's check:

$$ \frac{dr^2}{dt} = 2 r \frac{dr}{dt} $$ but also $$ \frac{dr^2}{dt} = \frac{d}{dt}(x^2(t) + y^2(t)) = 2 x \dot{x} + 2 y \dot{y} = 2 \epsilon (x^2+y^2)^2 = 2 \epsilon r^4 $$ From this follows that $2 r \frac{dr}{dt} = 2 \epsilon r^4 $ and $\frac{dr}{dt} = \epsilon r^3$.

Similar with $\dot{\theta}(t)$:

$$ \frac{d}{dt} \left (\arctan{\frac{y}{x}} \right) = \frac{\dot{y}x - y \dot{x}}{x^2+y^2} \equiv 1 $$

Answer 2

The given equations are

$\dfrac{dx}{dt} = - y + \epsilon x (x^2 + y^2), \tag{0A}$

$\dfrac{dx}{dt} = x + \epsilon y (x^2 + y^2); \tag{0B}$

we have

$r^2 = x^2 + y^2; \tag{1}$

thus,

$2r\dfrac{dr}{dt} = 2x\dfrac{dx}{dt} + 2y \dfrac{dy}{dt}, \tag{2}$

whence

$r \dfrac{dr}{dt} = x\dfrac{dx}{dt} + y \dfrac{dy}{dt}; \tag{3}$

using (0A)-(0B) in (3) yields

$r \dfrac{dr}{dt} = x(-y + \epsilon x(x^2 + y^2)) + y(x + \epsilon y(x^2 + y^2))$ $= -xy + \epsilon x^2 (x^2 + y^2) + xy + \epsilon y^2 (x^2 + y^2) = \epsilon (x^2 + y^2)(x^2 + y^2)$ $= \epsilon r^2 r^2 = \epsilon r^4; \tag{4}$

assuming $r \ne 0$ it follows that

$\dfrac{dr}{dt} = \epsilon r^3. \tag{5}$

We further have

$x = r\cos \theta, \tag{6}$

$y = r\sin \theta; \tag{7}$

thus

$\dfrac{dx}{dt} = \dfrac{dr}{dt} \cos \theta - r\dfrac{d\theta}{dt} \sin \theta; \tag{8}$

$\dfrac{dy}{dt} = \dfrac{dr}{dt} \sin \theta + r\dfrac{d\theta}{dt} \cos \theta; \tag{9}$

using (0A), (1), (5), (6) and (7) in (8),

$-r \sin \theta + \epsilon r^3 \cos \theta = \dfrac{dr}{dt} \cos \theta - r\dfrac{d\theta}{dt} \sin \theta = \epsilon r^3 \cos \theta - r\dfrac{d\theta}{dt} \sin \theta, \tag{10}$

whence

$-r\sin \theta = - r\dfrac{d\theta}{dt} \sin \theta, \tag{11}$

and whence

$\sin \theta \ne 0 \Rightarrow \dfrac{d\theta}{dt} = 1; \tag{12}$

additionally, using (0B), (1), (5), (6), (7) in (9):

$r \cos \theta + \epsilon r^3 \sin \theta = \dfrac{dr}{dt} \sin \theta + r\dfrac{d\theta}{dt} \cos \theta = \epsilon r^3 \sin \theta + r\dfrac{d\theta}{dt} \cos \theta, \tag{13}$

or

$r\cos \theta = r\dfrac{d\theta}{dt} \cos \theta ; \tag{14}$

finally,

$\cos \theta \ne 0 \Rightarrow \dfrac{d \theta}{dt} = 1. \tag{15}$

Since $\cos \theta$, $\sin \theta$ cannot both vanish, we have shown that

$\dfrac{d \theta}{dt} = 1 \tag{16}$

in all cases. Together with (5), this completes the transformation to polars. Note that this derivation covers the cases $x = 0$, $y = 0$, in which the transformations

$\tan \theta = \dfrac{y}{x}; \;\; \cot \theta = \dfrac{x}{y} \tag{17}$

may become problematic.