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