# Polar coordinates differential equation

I have the following ODE:

$$\dot x=-y(x^2+y^2), \dot y=x(x^2+y^2)$$

I want to sketch the phase portrait (manually) and I want to find the flow $\phi_t$, the orbit $O(x_0)$ and the limit set $\omega(x_0)$

I start by taking polar coordinates and change the system to $\dot r=-r^3\sin\theta, \dot\theta=r^3\cos\theta$

The phase portrait then looks like the one a stable centre, right?

How can I continue to find the flow of the function, i.e the solution of the differential equation?

We are given:

$$\dot x=-y(x^2+y^2), \dot y=x(x^2+y^2)$$

Recall, when we are doing polar coordinates, we have

$$x = r \cos \theta, y = r \sin \theta, x^2+y^2 = r^2$$

When we differentiate this, we have:

$$2 x x' + 2 y y' = 2 r r'$$

This gives us:

$$rr' = x(-y(x^2+y^2)) + y(x(x^2+y^2)) = 0 \rightarrow r' = 0$$

To find the angle, we take:

$$\dfrac{r \sin \theta}{r \cos \theta} = \tan \theta = \dfrac{y}{x}$$

Using the quotient rule gives us:

$$\theta' = \dfrac{x y' - y x'}{r^2} = \dfrac{x^2(x^2+y^2)+y^2(x^2+y^2)}{r^2}= \dfrac{r^4}{r^2} = r^2$$

Thus in polar coordinates, the original system becomes

• $r' = 0$
• $\theta' = r^2$

Can you now solve this system and draw the phase portrait?

The phase portrait should look like: • Thanks for your answwer. As solution I get $r(t)=r_0$ which is simply some constant, therefore for $\theta(t)=tr_0^2+\theta_0$. The phase portrait is then just on point in the centre and some circles around this centre, right? – Ulrich Otto Oct 27 '13 at 21:04
• Do you also have an idea about the orbit and the limit set? – Ulrich Otto Oct 28 '13 at 0:40
• Thanks again, I will try to find a way to identify the orbit and the limit. My last question relats to stability: What about stability of the solution $x(t)$ with $x(0)=x_0$ with repsect to small perturbations of the initial value $x_0$? – Ulrich Otto Oct 28 '13 at 15:57
• What do you differentiate to get $2xx'+2yy'=2rr'$? – The Coding Wombat Feb 18 '19 at 12:15
• What is the derivative of $$x^2 + y^2 = r^2$$ – Moo Feb 19 '19 at 13:06