# ODE system, plane-polar coordinates

#### Jason

##### New member
I have:

$\dot{x}=4x+y-x(x^2+y^2)$
$\dot{y}=4y-x-y(x^2+y^2)$

And I need to find $\dot{r}$ and $\dot{\theta}$

I got as far as:

$\dot{x}=r(\text{sin}(\theta)-\text{cos}(\theta)(r^2-4))$
$\dot{y}=r(-\text{sin}(\theta)(r^2-4)-\text{cos}(\theta))$

How do I go from here to $\dot{r}$ and $\dot{\theta}$ ?

#### Ackbach

##### Indicium Physicus
Staff member
Assuming you're using the usual $x=r\cos(\theta),\;y=r\sin(\theta)$, then the product and chain rules give you
\begin{align*}
\dot{x}&=\dot{r}\cos(\theta)-r\sin(\theta)\, \dot{\theta}\\
\dot{y}&=\dot{r}\sin(\theta)+r\cos(\theta)\, \dot{\theta}.
\end{align*}
Plug all of these into your DE's. Can you continue from here?

#### Jason

##### New member
Do I, for example, set:

$\dot{r}\cos(\theta)-r\sin(\theta)\, \dot{\theta}=r(\text{sin}(\theta)-\text{cos}(\theta)(r^2-4))$

and solve from there?

#### Ackbach

##### Indicium Physicus
Staff member
Do I, for example, set:

$\dot{r}\cos(\theta)-r\sin(\theta)\, \dot{\theta}=r(\text{sin}(\theta)-\text{cos}(\theta)(r^2-4))$

and solve from there?
[EDIT] You are correct. Do the same for the other equation. What do you notice about how $\dot{r}$ and $\dot{\theta}$ appear in those two equations?

• Jason

#### Jason

##### New member
Not sure yet, just messing around with the equations at the moment (any tips are always appreciated).

#### Jester

##### Well-known member
MHB Math Helper
You can also use

$r \dot{r} = x \dot{x} + y \dot{y}$

$r^2 \dot{\theta} = x \dot{y} - y \dot{x}$

simplify and use $x^2+y^2 = r^2$ where possible.

#### Ackbach

##### Indicium Physicus
Staff member
Not sure yet, just messing around with the equations at the moment (any tips are always appreciated).
You can either use Jester's trick, or you can see that $\dot{r}$ and $\dot{\theta}$ appear linearly in the two equations. What does that suggest to you?

#### Jason

##### New member
I ended up just solving the two equations:

$\dot{r}=r(4-r^2)$ and $\dot{\theta}=-1$

------------------------------

I found three non-equilibrium solutions:

$r=2$ and $\theta=-1$

When $r<2$, $\dot{r}>0$, so $r$ increases and solutions move out toward the $r=2$ circle.

When $r>2$, $\dot{r}<0$, so $r$ decreases and solutions move in toward the $r=2$ circle.

I assume this means the cycle is stable? The only problem I have is how do I determine if the trajectories are clockwise or anti-clockwise?

Last edited:

#### Ackbach

##### Indicium Physicus
Staff member
I ended up just solving the two equations:

$\dot{r}=r(4-r^2)$ and $\dot{\theta}=-1$

------------------------------

I found three non-equilibrium solutions:

$r=2$ and $\theta=-1$

When $r<2$, $\dot{r}>0$, so $r$ increases and solutions move out toward the $r=2$ circle.

When $r>2$, $\dot{r}<0$, so $r$ decreases and solutions move in toward the $r=2$ circle.

I assume this means the cycle is stable? The only problem I have is how do I determine if the trajectories are clockwise or anti-clockwise?
You can solve this DE exactly. The $\theta$ DE is straight-forward integration, and the $r$ equation is separable.

#### Jester

##### Well-known member
MHB Math Helper
I ended up just solving the two equations:

$\dot{r}=r(4-r^2)$ and $\dot{\theta}=-1$

------------------------------

I found three non-equilibrium solutions:

$r=2$ and $\theta=-1$

When $r<2$, $\dot{r}>0$, so $r$ increases and solutions move out toward the $r=2$ circle.

When $r>2$, $\dot{r}<0$, so $r$ decreases and solutions move in toward the $r=2$ circle.

I assume this means the cycle is stable? The only problem I have is how do I determine if the trajectories are clockwise or anti-clockwise?
Since $\dot{\theta} = -1$, then $\theta$ is decreasing meaning clockwise.

• Jason