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Stability of 2 critical points in a system of DE's

  1. Jun 17, 2012 #1

    fluidistic

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    Gold Member

    1. The problem statement, all variables and given/known data
    Basically I found the following system of DE's:
    [itex]\frac{dx}{dt}=y[/itex]
    [itex]\frac{dy}{dt}=-\frac{g}{l} \sin x - \frac{cy}{ml}[/itex]. (Damped pendulum)
    I'm asked to analize the stability of the critical points [itex]x=0[/itex], [itex]y=0[/itex] and [itex]x=\pi[/itex], [itex]y=0[/itex].
    Using intuition the first point is asymptotically stable while the second point is unstable.


    2. Relevant equations
    I tried to put the system under matrix form and then check out the eigenvalues of a matrix but I have some problems.


    3. The attempt at a solution
    [itex]\begin {bmatrix} \frac{dx}{dt} \\ \frac{dy}{dt} \end{bmatrix} =A \begin {bmatrix} x \\ y \end{bmatrix}[/itex] where [itex]A= \begin {bmatrix} a& b \\ c & d \end{bmatrix}[/itex].
    I tried to determine the elements of the matrix A but was unable to perform this without making the assumption that [itex]\sin x \approx x[/itex] (for small x or x close to pi I guess?). Making that assumption I reached [itex]A\approx \begin {bmatrix} 0 &1 \\ -\frac{g}{l} & -\frac{c}{ml} \end{bmatrix}[/itex]. I found the eigenvalues to be [itex]\lambda _1 =- \frac{c}{2ml} + \sqrt {\frac{c^2}{ml^2} - \frac{4g}{l}}[/itex] and [itex]\lambda _2 =- \frac{c}{2ml} - \sqrt {\frac{c^2}{ml^2} - \frac{4g}{l}}[/itex].
    So that I have a bunch of possible cases which doesn't look good at all to me. Furthermore I never used to information that the critical points are [itex](0,0)[/itex] and [itex](\pi , 0 )[/itex] yet.
    Can someone make some comments so far? Thanks in advance.
     
  2. jcsd
  3. Jun 17, 2012 #2
    Hello,

    First of all, suppose you have a system in the general form X'=F(X), and let X_0 be a particular point. Then

    $$ (X-X_0)' = F(X) ≈ dF_{X_0}(X-X_0) + F(X_0).$$

    Being a critical point just means that F(X_0)=0, so you have a homogeneous linear term on the right side of your linear approximation.

    $$ (X-X_0)' = A (X-X_0),$$
    where [itex] A = dF_{X_0} [/itex].

    You have analyzed A at X_0 = (0,0), now you just have to calculate the linearization at the other point and look at its eigenvalues.

    See http://en.wikipedia.org/wiki/Stability_theory#Stability_of_fixed_points
     
  4. Jun 18, 2012 #3

    fluidistic

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    Gold Member

    Hmm sorry I have no idea on what I've done. How do you know I "analyzed A at X_0 = (0,0)"? Where did I use the critical point (0,0)?
    In another source I've read that [itex]\begin {bmatrix} x' \\ y' \end{bmatrix} =[/itex][itex] \begin {bmatrix} \frac{ \partial f (x_0, y_0 )}{\partial x } & \frac{ \partial f (x_0, y_0 )}{\partial y } \\ \frac{ \partial g (x_0, y_0 )}{\partial x } &\frac{ \partial g (x_0, y_0 )}{\partial y } \end{bmatrix}[/itex][itex] \begin {bmatrix} x \\ y \end{bmatrix}[/itex] where [itex]f(x,y)=y[/itex] and [itex]g(x,y)=-\frac{g}{l} \sin x - \frac{cy}{ml}[/itex]. Following this information I reach, for the critical point (0,0), [itex]A=\begin {bmatrix} 0 &1 \\ -\frac{g}{l} & 0 \end{bmatrix}[/itex] giving me the eigenvalues [itex]\pm i \sqrt {\frac{g}{l}}[/itex]. According to https://controls.engin.umich.edu/wiki/index.php/EigenvalueStability, this means that (0,0) behaves like an undamped oscillator because my 2 eigenvalues are complex with 0 as real part. This result makes no sense, I'm supposed to find an asymptotically critical point for (0,0)...

    Edit: Nevermind, I made a mistake for A, I fall over the result in my first post again. Ok now I understand better. Thanks.
     
    Last edited: Jun 18, 2012
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