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Dane Sabo 2024-09-09 14:29:31 -04:00
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.obsidian/workspace.json vendored
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@ -48,7 +48,7 @@ How is this useful to us engineers?
We are going to see systems that are nonlinear, and they can give us ideas about where things could blow up. In our second example, we have generally a pretty safe area below x = 2. Anywhere below there, we know we're going to end up at x = -2, but above x =2, all hell breaks loose.
This is what we care about. We want to know where in our nonlinear system domains things can become dangerous.
## How do we numerically get a time domain response?
# How do we numerically get a time domain response?
Numericaly:
$$ \dot x = f(x) $$
$$\frac{dx}{dt} = \lim_{\Delta t \rightarrow 0} \frac{f(x(t+\Delta t))-f(x(x))}{\Delta t} $$
@ -60,4 +60,31 @@ This is the tangent (or the secant while $\Delta t =/ 0$)
> There are better methods!
> ode45() <- Variable Step Runge-Kutta
We're going to use a lot of odeint in SciPy
We're going to use a lot of odeint in SciPy
*Insert code here*.
# Geometric Analyses in Higher Dimensions
We want to do geometric analysis with n-dimensions, not just one.
## Linear stability for a nonlinear system
1. Find the fixed equilibrium points
$$\dot x = 0 = f(x)$$
This equation is solved by points $x^\star$
Define: $$ \epsilon(t) = x(t) - x^\star \rightarrow x(t) = x^\star + \epsilon(t)$$
$$\frac{dx}{dt} = f(x)$$
$$\frac{d\epsilon}{dt} = f(x^\star +\epsilon)$$
$$= f(x^\star) + \epsilon f'(x^\star) + \Theta(\epsilon^2)\rightarrow \epsilon f'(x^\star) + \Theta(\epsilon^2)$$
$$\frac{d\epsilon}{dt} = \epsilon f'(x^\star) + \Theta(\epsilon^2) \rightarrow \epsilon f'(x^\star)$$
- If $\epsilon f'(x^\star) > 0 \rightarrow \text{Unstable}$
- If $\epsilon f'(x^\star) < 0 \rightarrow \text{Stable}$
- If $\epsilon f'(x^\star) = 0 \rightarrow \text{Inconclusive}$
## Energy Motivation
$$\dot{x} = f(x) = -\frac{dV}{dt} \text{V = potential}$$
$$\frac{dV}{dx} = -\frac{dx}{dt}$$
$$\frac{dV}{dt} <0 \rightarrow V \text{continuously decreasing in time until } \frac{dV}{dt} = 0$$
- Stable fixed poitn: $V(x^\star)$ is (a local) minimum of $V(x)$
- Unstable fixed point: $V(x^\star)$ is (a local) maximum of $V(x)$.
# A review of Linear Differential Equations
$$\dot x = \frac{dx}{dt} = A x + F$$
Homogeneous (no input/forcing):
$$\dot x = A x$$
$$x(t) = C e^{At}$$