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

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+# 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.