Obsidian/300s School/ME 2016 - Nonlinear Dynamical Systems 1/2024-09-23 Temporary Title.md

title	allDay	startTime	endTime	date	completed
Temporary Title	false	12:30	14:30	2024-09-23	null

Motivation for Nonlinear Dynamics

1. Started with an example of a simple spring-mass. This is a linear system by default
2. Then we talked about an undamped pendulum. Linear for only small angle assumption... EOM:

\ddot{\theta} = -\frac{g}{l} \sin(\theta)

Small angle assumption makes this just like \ddot x+ \omega^2 x = 0 To look at stability, we could look at the poles. But for us we're going to do things in the time domain and look at first order systems. $$ \left[ \matrix{ \ddot x \ \dot x} \right] = \left[ \matrix{ 0 & -\omega^2 \ 1 & 0 } \right] \left[ \matrix{ \dot x \ x } \right] $$ Recall from last time:

• \Delta = \det(A)
• \tau = \text{tr}(A) When we look at these for this matrix, we see we should have a center. Our general solution should be: x(t) = A \sin(\omega t + \phi) y(t) = \dot x(t) = A \cos(\omega t + \phi) Recall the trig identity \sin(x)^2 + \cos(x)^2 = 1 Then:

x^2 + \frac{y^2}{\omega^2} = A^2

and for the \omega = \sqrt{\frac{k}{m}} case

\frac{k}{2} x^2 + \frac{m}{2} \dot x^2 = \frac{k A^2}{2}

This is recognizably the sum of energy for a system!

Damped Oscillation (Linear)

m \ddot x + c \dot x + k x = 0 \ddot x + \beta \dot x + \omega^2 x = 0, \beta>0, \omega>0

Where \dot x = y and \dot y = -\beta y - \omega^2 x $$ \left[ \matrix{ \dot x \ \dot y} \right] = \left[ \matrix{ 0 & 1 \ -\omega^2 & - \beta } \right] \left[ \matrix{ y \ x } \right] $$ Now we look at our parabola chart to see where we are: \tau^2 - 4 \Delta = \beta^2 - 4 \omega^2 Stable Spiral (Underdamped)

• $z < 0 \rightarrow \tau^2 - 4 \Delta<0 \rightarrow \tau^2 < 4 \Delta$$ Degenerate Node (Stable) ((Critically Damped))
• $z = 0 \rightarrow \tau^2 = 4 \Delta$$ Stable Node (Overdamped)
• $z > 0 \rightarrow \tau^2 - 4 \Delta>0 \rightarrow \tau^2 > 4 \Delta$$

These things relate directly to the eigenvectors and eigenvectors of the system.

Eigenvalues and Eigenvectors

z = \beta^2 - 4 \omega^2 Solving for lambda (\det[A - I\lambda]) will lead us towards the same expression for z. This is what's under the square root.