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ODE system, plane-polar coordinates

Jason

New member
Feb 5, 2012
28
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
Jan 26, 2012
4,184
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
Feb 5, 2012
28
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
Jan 26, 2012
4,184
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

New member
Feb 5, 2012
28
Not sure yet, just messing around with the equations at the moment (any tips are always appreciated).
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
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
Jan 26, 2012
4,184
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
Feb 5, 2012
28
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
Jan 26, 2012
4,184
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
Jan 26, 2012
183
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.