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

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SUMMARY

The discussion centers on analyzing the stability of critical points in a damped pendulum system represented by the differential equations dx/dt = y and dy/dt = -g/l sin(x) - (c/m)y. The critical points under examination are (0,0) and (π,0). The analysis involves constructing the Jacobian matrix A and calculating its eigenvalues. The eigenvalues for the point (0,0) are found to be complex, indicating that this point behaves like an undamped oscillator, contrary to the expectation of asymptotic stability.

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Homework Statement


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

Homework Equations


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

The Attempt at a Solution


\begin {bmatrix} \frac{dx}{dt} \\ \frac{dy}{dt} \end{bmatrix} =A \begin {bmatrix} x \\ y \end{bmatrix} where A= \begin {bmatrix} a& b \\ c & d \end{bmatrix}.
I tried to determine the elements of the matrix A but was unable to perform this without making the assumption that \sin x \approx x (for small x or x close to pi I guess?). Making that assumption I reached A\approx \begin {bmatrix} 0 &1 \\ -\frac{g}{l} & -\frac{c}{ml} \end{bmatrix}. I found the eigenvalues to be \lambda _1 =- \frac{c}{2ml} + \sqrt {\frac{c^2}{ml^2} - \frac{4g}{l}} and \lambda _2 =- \frac{c}{2ml} - \sqrt {\frac{c^2}{ml^2} - \frac{4g}{l}}.
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 (0,0) and (\pi , 0 ) yet.
Can someone make some comments so far? Thanks in advance.
 
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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 A = dF_{X_0}.

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
 
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 \begin {bmatrix} x' \\ y' \end{bmatrix} =\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}\begin {bmatrix} x \\ y \end{bmatrix} where f(x,y)=y and g(x,y)=-\frac{g}{l} \sin x - \frac{cy}{ml}. Following this information I reach, for the critical point (0,0), A=\begin {bmatrix} 0 &1 \\ -\frac{g}{l} & 0 \end{bmatrix} giving me the eigenvalues \pm i \sqrt {\frac{g}{l}}. 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:

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