MHB ODE system, plane-polar coordinates

[Jason4]

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]

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?

 

[Jason4]

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]

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?

 

[Jason4]

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

 

[Mathwebster]

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]

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?

 

[Jason4]

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?

 

[Ackbach]

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.

 

[Mathwebster]

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.