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
Best Answer
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.