Polar coordinates differential equation

Solution 1:

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: