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Math Help - Linear stability analysis

  1. #1
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    Linear stability analysis

    For the system
     \dot{x} = -y + ax(x^{2}+y^{2})
     \dot{y} = x + ay(x^{2}+y^{2})

    linearise the system around the fixed point and comment on the stability properties for  a < 0, a = 0, a > 0.

    The system can be written in polar coordinates as
     \dot{r} = ar^{3}
     \dot{\theta} = 1
    By solving these equations, comment on the stability properties of the fixed point for  a < 0, a = 0, a > 0.

    I've found the fixed point to be (0,0). I found that the fixed point is a centre, and so stable but not asymptotically stable for all a - but this seems to simple when the question specifically mentions  a < 0, a = 0, a > 0. .

    I've converted the equations to polar coords and solved to get  r = \sqrt{\frac{r_{0}}{1-2ar_{0}t}} but don't know where to go from here to comment on stability.

    Any help would be really appreciated
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  2. #2
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    Re: Linear stability analysis

    Messing around with this it becomes clear that

    i) if a=0, then (0,0) is a center

    ii) if a>0, the trajectory is an asymptotically unstable spiral about (0,0)

    iii) if a<0, the trajectory is an asymptotically stable spiral about (0,0)

    I'm not seeing from the linearization why this is so yet. My first cut had $a$ dropping out of the system matrix at (0,0).

    I'll continue to look at this.
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  3. #3
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    Re: Linear stability analysis

    Quote Originally Posted by sophiew3 View Post
    For the system
     \dot{x} = -y + ax(x^{2}+y^{2})
     \dot{y} = x + ay(x^{2}+y^{2})

    linearise the system around the fixed point and comment on the stability properties for  a < 0, a = 0, a > 0.

    The system can be written in polar coordinates as
     \dot{r} = ar^{3}
     \dot{\theta} = 1
    By solving these equations, comment on the stability properties of the fixed point for  a < 0, a = 0, a > 0.

    I've found the fixed point to be (0,0). I found that the fixed point is a centre, and so stable but not asymptotically stable for all a - but this seems to simple when the question specifically mentions  a < 0, a = 0, a > 0. .

    I've converted the equations to polar coords and solved to get  r = \sqrt{\frac{r_{0}}{1-2ar_{0}t}} but don't know where to go from here to comment on stability.

    Any help would be really appreciated
    You've basically completed the problem.

    Just look at your solution for $r(t)$ as $t\to \infty$ under different values of $a$ and with an initial condition slightly away from the critical point, i.e. it's been perturbed.
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  4. #4
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    Re: Linear stability analysis

    I think you've missed the point of the question. If you linearize about the critical point, it predicts a center. The eigenvalues are \pm i. Introducing polar coordinates the stability is easily determined and depends on the sign of a (whether positive, negative or zero).

    So what does this say? That the dynamics of the actual system cannot be predicted from the linear system. Why? The critical point is non-hyperbolic, i.e. the eigenvalues of the linear system have zero real part. The Hartman–Grobman states that the behaviour of a dynamical system near a hyperbolic equilibrium point is qualitatively the same as the behaviour of its linearization near this equilibrium point. (Note: the real part of the eigenvalues have nonzero real part).
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