From 4600f54a45ccde53bb6ca4e170c829b97187b838 Mon Sep 17 00:00:00 2001
From: Dane Sabo <dane.sabo@pitt.edu>
Date: Mon, 9 Sep 2024 14:54:16 -0400
Subject: [PATCH] vault backup: 2024-09-09 14:54:16

---
 .../2024-09-09 Frameworks and Review.md       | 41 +++++++++++++++++--
 1 file changed, 38 insertions(+), 3 deletions(-)

diff --git a/300s School/301. ME 2016 - Nonlinear Dynamical Systems 1/2024-09-09 Frameworks and Review.md b/300s School/301. ME 2016 - Nonlinear Dynamical Systems 1/2024-09-09 Frameworks and Review.md
index 08960100..99d7b902 100644
--- a/300s School/301. ME 2016 - Nonlinear Dynamical Systems 1/2024-09-09 Frameworks and Review.md	
+++ b/300s School/301. ME 2016 - Nonlinear Dynamical Systems 1/2024-09-09 Frameworks and Review.md	
@@ -84,7 +84,42 @@ $$\frac{dV}{dt} <0 \rightarrow V \text{continuously decreasing in time until } \
 - Stable fixed poitn: $V(x^\star)$ is (a local) minimum of $V(x)$
 - Unstable fixed point: $V(x^\star)$ is (a local) maximum of $V(x)$. 
 # A review of Linear Differential Equations
-$$\dot x = \frac{dx}{dt} = A x + F$$
+## Foundation
+$$\dot{\vec{x}} = \frac{dx}{dt} = \bf{A} \vec{x} + \vec{F}$$
 Homogeneous (no input/forcing):
-$$\dot x = A x$$
-$$x(t) = C e^{At}$$
+$$\dot{\vec{x}} = \bf{A} \vec{x}$$
+$$\vec{x}(t) = C e^{\bf{A}t}$$
+This e to the matrix A is kinda gross. We represent this as follows:
+$$ e^{\bf{A}} = \sum_{k=0}^\inf \frac{\bf{A}^k}{k!}$$so then:
+$$ e^{\bf{A}t} = \sum_{k=0}^\inf \frac{\bf{A}^k t^k}{k!}$$
+But since $\frac{d}{dt}e^{\bf{A}t} = \bf{A}e^{\bf{A}t}$
+Some proof that xdot = Ax
+
+## Diagonalization
+Suppose:
+$$ \dot{\vec{x}} = \bf{A}\vec{x} $$
+with A diagonal. Then
+$$ \dot{x}_n = a_{nn} x_n$$
+This is nice. So we should try and diagonalize things...
+- Find $\bf{P}$ such that $\bf P^{-1} \bf A \bf P = \bf D$
+Then we have new coordinates:
+- $\vec{x} = \bf{P} \vec y$
+- $\vec{\dot{x}} = \bf{P} \vec{\dot{y}}$
+- then $\dot{\vec{y}} = \bf{P^{-1}AP}\vec{y}$
+Choose P to be column wise eigenvectors of A, with diagonal matrix D having the eigenvalues of A. 
+This isn't always possible.
+## Fundamental Matrix
+$$ e^{\bf{A}t} = \Psi(t) = \text{Fundamental matrix of } \dot{\vec{x}} = \bf{A} \vec{x}$$
+How do we calculate this thing?
+- series calculation of $A^m$
+- Smarter computer program
+- Laplace Transform
+Consider: $\dot{\vec{x}} = \bf{A}\vec{x} + \vec{F}$
+$$ \vec{x}(t) = \bf{\Psi} \vec{c} \leftarrow \text{autonomous}$$
+or the total solution:
+$$ \vec{x}(t) = \bf{\Psi} \vec{c} + \int_{t_0}^t \bf{\Psi}^{-1}(\bf{I}) \bf{F}(I)dI + \bf{\Psi}(t_0)\vec{c}$$
+$$ \vec{x}(t_0) = \bf{\Psi}(t_0)\vec{c}$$
+Then the solution is 
+$$\vec{x}(t) = e^{\bf{A}t}\vec{c} + e^{\bf{A}t} \int_{t_0}^t e^{-\bf{A}I} F(T) \delta T$$
+Then using the Laplace transform:
+$$e^{\bf{A}t} = \mathcal{L}^{-1} \{ (sI-\bf{A})^-1 \} $$