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Non-linear ODE, plane-polar coordinates.

  1. Mar 13, 2012 #1
    1. The problem statement, all variables and given/known data

    I have:

    [itex]\dot{x}=4x+y-x(x^2+y^2)[/itex]
    [itex]\dot{y}=4y-x-y(x^2+y^2)[/itex]

    And I need to find [itex]\dot{r}[/itex] and [itex]\dot{\theta}[/itex]

    2. The attempt at a solution

    I got as far as:

    [itex]\dot{x}=r(\text{sin}(\theta)-\text{cos}(\theta)(r^2-4))[/itex]
    [itex]\dot{y}=r(-\text{sin}(\theta)(r^2-4)-\text{cos}(\theta))[/itex]

    How do I go from here to [itex]\dot{r}[/itex] and [itex]\dot{\theta}[/itex] ?
     
    Last edited: Mar 13, 2012
  2. jcsd
  3. Mar 13, 2012 #2

    LCKurtz

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    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.
     
  4. Mar 13, 2012 #3
    Ok, so I have:

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

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

    Is there a "non-tedious" way of solving this?
     
  5. Mar 13, 2012 #4

    HallsofIvy

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    I can't speak for what you would consider tedious but the standard method would be to multiply the first equation by [itex]cos(\theta)[/itex], the second equation by [itex]sin(\theta)[/itex], and add the two equations. Then turn around and multiply the first equation by [itex]sin(\theta)[/itex], the second equation by [itex]cos(\theta)[/itex], then subtract.
     
  6. Mar 13, 2012 #5
    Thank you, sir.
     
  7. Mar 13, 2012 #6
    Have I got this second part right?

    You mean, I presume, [itex]\dot{r}[/itex], 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?
     
    Last edited by a moderator: Mar 13, 2012
  8. Mar 13, 2012 #7
    [itex]\sin\theta\dot{x}-cos\theta\dot{y}:[/itex]

    [itex]\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)][/itex]

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

    [itex]-r\dot{\theta}=r[/itex]

    [itex]\dot{\theta}=\frac{r}{-r}=-1[/itex]
     
    Last edited: Mar 13, 2012
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