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Consider the coupled system $$\dot{x}=-x+y, \,\,\, \dot{y}=-x-y.$$ Working with polar coordinates $r$ and $\theta$ such that $$x=r\cos\theta, \,\,\, y=r\sin\theta,$$ Obtain uncoupled differential equations for $r$ and $\theta$.

I did some research online and found out how to convert a differential equation to polar coordinates. I used the relationship $$r^2 = x^2+y^2$$ Then I differentiated this equation, and plugged in our $\dot{x}$ and $\dot{y}$ values given.

I got $\dot{r}=-r.$ Which I'm nearly certain is correct. Then to find $\dot{\theta}$ I used $\tan \theta = \dfrac{r\sin\theta}{r\cos\theta} =\dfrac{y}{x},$ differentiate it and I found $\dot{\theta} = -1$.

Solve the two differential equations from (i) to obtain $r(t)$ and $\theta(t)$

This seems ok, I just integrated my $\dot{r}$ and my $\dot{\theta}$ equations. I got $$r(t) = -\dfrac{r^2}{2}\,\,\, \text{and} \,\,\, \theta(t)= -\theta$$

Using the results from (ii) sketch the trajectory and identify the type and stability of the equilibrium point $q=(0,0)$.

This is the part that I'm writing the question about. I just want make sure I can understand this right. Our $\theta$ is negative does that mean it will be spiralling anti-clockwise. Also $r$ is negative so does that mean it will be decaying towards the equilibrium? If that's the case wouldn't we have a stable attractive node around the equilibrium?

Thanks very much.

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    $\begingroup$ Integration of $\dot r$, $\dot\theta$ should be with respect to $t$: $\dot r=-r$ $\Rightarrow$ $r(t)=e^{-t}r_0$ etc $\endgroup$ – A.Γ. May 7 '17 at 10:36
  • $\begingroup$ Ah thanks very much. Yeah I solved both separable differential equations. $\theta= -t$. Was my idea about the phase plane correct. I haven't actually solved any trajectories for polar coordinates. I know you can solve other systems with eigenvalues and eigenvectors and plot their trajectories. Not sure how it works in polar coordinates. $r(t)$ is also decaying. $\endgroup$ – Patrick Moloney May 7 '17 at 10:56
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    $\begingroup$ $\dot\theta<0$, hence, rotating clockwise (the negative angle direction), $\dot r<0$, hence, $r$ decreasing. The origin is the stable equilibrium, correct. $\endgroup$ – A.Γ. May 7 '17 at 11:03
  • $\begingroup$ Of course $\theta>0$ is taken as clockwise. I understand. I appreciate the support. $\endgroup$ – Patrick Moloney May 7 '17 at 11:06
  • $\begingroup$ Just to summarize what's been done here, the solution is $r=r_0e^{i\theta_0}e^{-(1+i)\theta}$. The first term is a scaling parameter, the second term is a rotation parameter, and the last term shows that this is logarithmic spiral with a flair coefficient of $-1$ (shrinking) and rotating clockwise. $\endgroup$ – Cye Waldman May 9 '17 at 20:48
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Your solutions of the differential equations are both wrong. The independent variable is $t$. $\dot r=-r$ has the well-known solution $r(t)=r_0·e^{-t}$ and $\dot θ=-1$ trivially $θ(t)=θ_0-t$.

Yes, you are right that the spirals are clock-wise. As the radius gets smaller in time, the image of the curve is an anti-clockwise spiral. This more intuitive interpretation can belongs to the time-inversed problem.

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