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Stability of dynamical system

  1. Dec 15, 2012 #1
    Suppose we have a dynamical system [itex]x_{t+1} = Ax_{t}[/itex] where [itex]A[/itex] is matrix, [itex]x[/itex] is vector. We suppose that $x$ always grow as time goes on.

    If we treat equilibrium as the whole time evolution(path) of [itex]x[/itex] given [itex]x_0 = a[/itex] and no disturbance to the value of [itex]x[/itex] - that is $x$ follows from the initial condition, how would we be able to define stability of the system? What would be the equation?
     
  2. jcsd
  3. Dec 15, 2012 #2

    pasmith

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    Homework Helper

    I think you mean that [itex]\|x_t\|[/itex] grows.

    What do you mean by stability? Do you mean that a small change in the initial condition tends to 0 as [itex]t \to \infty[/itex]? The formal expression of that is that the solution starting at [itex]x_0[/itex] is stable if and only if there exists some [itex]\epsilon > 0[/itex] such that for any solution [itex]y_t[/itex] starting at [itex]y_0[/itex], if [itex]\|y_0 - x_0\| < \epsilon[/itex] then [itex]\|y_t - x_t\| \to 0[/itex] as [itex]t \to \infty[/itex].

    If so your system is unstable, since for [itex]\|x_t\|[/itex] to grow there must exist an eigenvalue [itex]\lambda[/itex] of A such that [itex]|\lambda| > 1[/itex]. This eigenvalue has a corresponding eigenvector [itex]v[/itex], and we can assume that [itex]\|v\| = 1[/itex]. Let [itex]\epsilon > 0[/itex], and [itex]y_0 = x_0 + \frac12\epsilon v[/itex]. Then [itex]\|y_0 - x_0\| = \frac12 \epsilon < \epsilon[/itex], and
    [tex]
    \|y_t - x_t\| = \left\|\frac12 \epsilon A^t v \right\| = \frac12 \epsilon |\lambda^t|
    [/tex]
    which tends to infinity as [itex]t \to \infty[/itex] for any strictly positive [itex]\epsilon[/itex].

    Alternatively you could define stability to mean that the solution starting at [itex]x_0[/itex] is stable if and only if, for all finite [itex]t[/itex] and all [itex]\epsilon > 0[/itex], there exists [itex]\delta > 0[/itex] such that for any solution [itex]y_t[/itex] starting at [itex]y_0[/itex], if [itex]\|x_0 - y_0\| < \delta[/itex] then [itex]\|x_t - y_t\| < \epsilon[/itex]. That's equivalent to requiring that [itex]x_t - y_t[/itex] is a continuous function of [itex]x_0 - y_0[/itex], which in this case it is.
     
    Last edited: Dec 15, 2012
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