# Using eigenvalues to determine the stability/behaviour of the system

first time I've been on physics.se but have used the math and cs before...

Anyway, here's my question:

If we have a damped pendulum described by the equation $$y'' + ay' + b = 0 , a>0$$ Using the conversion

$x_0=y$ and $x_1=y'$ we can convert this into a set of first-order ODEs as follows: $$x_0'=x_1$$ $$x_1'=-ax_1+b$$ or in matrix form : $$\mathbf{x'}=\begin{bmatrix}0 & 1\\0 & -a\end{bmatrix}\mathbf{x}+\begin{bmatrix}0 \\ b\end{bmatrix}$$

Now, the exercise where I'm doing this now says we can use the eigenvalues to determine the stability of the system the eigenvalues of the system.

The eigenvalues are $\lambda = 0$ and $\lambda = -a$...

I know that an eigenvalue of 0 means a neutral equilibrium such that a small change in the variable means it remains at the changed variable and that a negative eigenvalue means a small change will return to its original value, but I'm struggling to see how this relates to the physical situation of the damped oscillator, and what it means about the behaviour of the system..

Any thoughts would be appreciated

-
See the article on Lyapunov Stability on Wikipedia en.wikipedia.org/wiki/Lyapunov_stability, your system has an input, so you can focus on that part. –  recipriversexclusion Oct 15 '12 at 17:06
That doesn't look like the equation for a pendulum. Where's the restoring force? –  user1631 Oct 15 '12 at 17:40

Let me expand on Art's observation a bit.

In the equation, $b$ should be replaced by $by$, it's the restoring force of user1631, too. There is no "vector term" in the matrix equation anymore: the equation or set of equations is linear. The matrix is $$\pmatrix{0&1\\ b & -a}$$ The eigenvalues are solutions to $$(\lambda-0)(\lambda+a) + b = 0$$ i.e. $$\lambda = \frac{ -a\pm \sqrt{a^2 - 4b } }{2}$$ where we have somewhat permuted letters relatively to the usual formula fo the quadratic equation. For $a^2-4b\gt 0$, the eigenvalues are real and negative, implying damped motion: the oscillator will never swing to the opposite sign. For $a^2-4b\lt 0$, the eigenvalues are complex, complex conjugate to each other, and having a negative real part. It means that the motion is still damped by it's also oscillating. At any rate, due to the damping, the behavior is stable for all cases.

-
ah alright... the question obviously has a typo in my book... I really couldn't make sense of the situation but this makes perfect sense now .. and thanks for the explanation after, answers my original question too –  Ronald Oct 15 '12 at 19:39
hey, I was wondering if you could help with one other thing... that is whether there is a way to tell which eigenvalue corresponds to which variable.. not necessarily for a damped pendulum, but for any system in general. Like, for example, if we get a positive real and negative real eigenvalue. –  Ronald Oct 16 '12 at 16:23
Dear Ronald, not sure whether I understand but if I do, eigenvalues generally correspond to the behavior of all degrees of freedom, or their mixture. It's not one-for-one. If you're asking what a complex eigenvalue means, the imaginary part means that the solution is oscillating (in this case). Complex eigenvalues of real matrices/ equations are always paired to eigenvalues that are complex conjugates to each other, one for cos and one for sin, expressing oscillations. The real part determines the damping, exponential increase or decrease. –  Luboš Motl Oct 18 '12 at 4:46