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

[fluidistic]

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.

 

[Vargo]

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

 

[fluidistic]

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.