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Differential equation, limit cycles, stability, phase plane. By 11/12

  1. Dec 10, 2012 #1
    1. The problem statement, all variables and given/known data

    The trajectory of an arrow in space obeys the following system of equations:

    [itex]\dot{x}[/itex] = y+[itex](x^2+y^2-3)^2 (x^3-x+xy^2)[/itex]

    [itex]\dot{y}[/itex] = y+[itex](x^2+y^2-3)^2 (y^3-y+x^2y)[/itex]

    1. Questions

    a) Derive an ODE for the radial coordiante r(t) = [itex]\sqrt[]{x^2(t)+y^2(t)}[/itex]

    b) Show that the system admits two limit cycles and classify their stability

    c) The target is standing at the origin (0,0) of the phase plane. Let Po=(x(0),y(0))[itex]\neq[/itex] (0,0) denote the initial position of the arrow at the time t = 0. Determine and plot the region in the (x,y) plane containing all the points Po such that the arrow will hit the target at some instant of time t > 0


    3. The attempt at a solution

    I derived ODE in radial coordiante and got:

    [itex]\dot{r}[/itex]=[itex](r^2-3)^2(r^2-1)r[/itex]

    So we have 5 points to check the stability: [itex]0, 1,-1,\sqrt{3},-\sqrt{3} [/itex]

    Than I calculate the derivative of [itex]\dot{r}[/itex] and evaluate in those 5 points: I get five zeros so they are undecided.

    Then I daw the plot of [itex]\dot{r}[/itex] and if the [itex]\dot{r}[/itex]>0 it does mean that arrow on the phase line is to the right and if [itex]\dot{r}[/itex]<0 the arrow is to the left.

    So The phase line looks like this: <- (-\sqrt{3}) <- (-1) -> (0) <- (1) -> (\sqrt{3}) ->

    Is that correct? What should I do next?
     
  2. jcsd
  3. Dec 10, 2012 #2

    pasmith

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    Homework Helper

    Are you sure these are correct? Because they tell me that
    [tex]r\dot r = x\dot x + y\dot y[/tex]
    will not be independent of [itex]\theta[/itex], because the linear terms give me
    [tex]xy + y^2 = r^2(\cos\theta\sin\theta + \sin^2\theta)[/tex].

    Assuming that you are correct that
    [tex]\dot r = r(r^2 - 3)^2(r^2 - 1) = f(r)[/tex]
    then since by convention [itex]r \geq 0[/itex], there is a fixed point at the origin and two orbits, at [itex]r = \sqrt 3[/itex] and [itex]r = 1[/itex].

    To determine the stability, you look at the sign of [itex]f'(r)[/itex]. If it's negative then the point is stable, and if it's positive then the point is unstable. If it's zero then it's inconclusive and you have to look at higher derivatives.

    Alternatively, in one dimension r is increasing if [itex]\dot r > 0[/itex] and decreasing if [itex]\dot r < 0[/itex]. By inspection, [itex]\dot r < 0[/itex] if [itex]0 < r < 1[/itex] and [itex]\dot r > 0[/itex] when [itex]1 < r < \sqrt 3[/itex] and [itex]r > \sqrt 3[/itex]. So the origin is stable, the orbit at 1 is unstable, and the orbit at [itex]\sqrt 3[/itex] is stable from below and unstable from above.

    For part (c) you need to think about what trajectories eventually end up at the origin.
     
  4. Dec 10, 2012 #3
    There is a typo: shoul be -x instead of one of y

    [itex]\dot{x}[/itex] = y+[itex](x^2+y^2-3)^2 (x^3-x+xy^2)[/itex]

    [itex]\dot{y}[/itex] = -x+[itex](x^2+y^2-3)^2 (y^3-y+x^2y)[/itex]
     
  5. Dec 10, 2012 #4

    pasmith

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    Homework Helper

    Then your result for [itex]\dot r[/itex] is correct.
     
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