Non-linear ODE, plane-polar coordinates.

[spitz]

Homework Statement

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}

2. The attempt at a solution

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} ?

 

[LCKurtz]

Homework Statement

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}

2. The attempt at a solution

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} ?

For the left sides of these last two equations you have $x = r\cos\theta,\ y = r\sin\theta$ to work with. Remembering that everything is a function of $t$, you can caculate $\dot x$ and $\dot y$ by differentiating those, using the chain rule.

 

[spitz]

Ok, so I have:

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

\dot{r}\sin(\theta)+r\cos(\theta)\dot\theta=r(-\sin(\theta)(r^2-4)-\cos(\theta))

Is there a "non-tedious" way of solving this?

 

[HallsofIvy]

I can't speak for what you would consider tedious but the standard method would be to multiply the first equation by cos(\theta), the second equation by sin(\theta), and add the two equations. Then turn around and multiply the first equation by sin(\theta), the second equation by cos(\theta), then subtract.

 

[spitz]

Thank you, sir.

 

[spitz]

Have I got this second part right?

Then turn around and multiply the first equation by sin(\theta), the second equation by cos(\theta), then subtract.

-r\sin^2(\theta)\dot{\theta}-r\cos^2(\theta)\dot{\theta}=r(sin^2(\theta)+cos^2(\theta))

-r(sin^2(\theta)+cos^2(\theta))\dot{\theta}=r

\dot{\theta}=-1

You mean, I presume, \dot{r}, not r, on the left of both of those two equations. Now, how did you get "-1" on the right of the last equation? It looks to me like you have r on the right of both first two equations. Isn't r- r= 0?

 

[spitz]

\sin\theta\dot{x}-cos\theta\dot{y}:

\dot{r}\sin\theta\cos\theta-r\sin^2\theta\dot{\theta}-\dot{r}\sin\theta\cos\theta-r\cos^2\theta\dot{\theta}=r[\sin^2\theta+\cos^2\theta+(sin\theta\cos\theta-\sin\theta\cos\theta)(4-r^3)]

-r(\sin^2 \theta+\cos^2 \theta)\dot{\theta}=r(\sin^2 \theta+\cos^2 \theta)

-r\dot{\theta}=r

\dot{\theta}=\frac{r}{-r}=-1