Dynamical Systems, Part 2


filed under : #dynamicalSystems #math #math324

This is a supplement for my Math 324 class for the summer of 2018.

Last time, we talked about fixed points for general systems $\dot{x} = v(x)$. In this post we'll focus on linear systems $\dot{x} = Ax$, where $A$ is a $n \times n$ matrix.

Linear Systems Review

For the linear system $\dot{x} = Ax$, a solution is given by $\phi_t(x) = e^{tA}x$, where

$$ e^{tA} = [1 + tA + \frac{t^2}{2!}A^2 + \frac{t^3}{3!}A^3 + \dots]. $$
When $A$ has $n$ linearly independent eigenvectors $K_1, \dots, K_n$, then the general solution takes the form
$$ X = C_1e^{\lambda_1 t}K_1 + \cdots + C_ne^{\lambda_n t}K_n, $$
where $\lambda_i$ is the associated eigenvalue for $K_i$.

For complex eigenvalues $\lambda_j = \alpha_j + i\beta_j$ and $\bar{\lambda}_j = \alpha_j - i\beta_j$, we can write the solution as

$$ \begin{align*} x_j + x_{j+1} & = e^{t\alpha_j}(K_R\cos{t\beta} - K_I\sin{t\beta})\\ & + e^{t\alpha_j}(K_R\sin{t\beta} + K_I\cos{t\beta}).\end{align*} $$
Here the associated eigenvectors for $\lambda_j$ and $\bar{\lambda}_j$ are $K_R \pm iK_I$. respectfully.

Suppose $x_1(t), \dots, x_n(t)$ are $n$ linearly independent solutions to $\dot{x} = Ax$, then the matrix $X(t) = [x_1(t), \dots, x_n(t)]$ is called the The fundamental solution matrix. We then have

$$ e^{At} = X(t)X(0)^{-1}. $$

Let's do an example. Let's try to find $e^{At}$ for

$$ \begin{pmatrix}1 & -1 \\\ 2 & -2 \end{pmatrix}. $$
We first see that the characteristic equation is
$$ det(A - \lambda I) = \lambda(\lambda + 1). $$
Therefore the eigenvalues are $\lambda_1 = 0$ and $\lambda_2 = -1$. The associated eigenvectors are
$$ K_1 = \begin{pmatrix}1 \\ 1\end{pmatrix} \quad \quad K_2 = \begin{pmatrix}1 \\ 2 \end{pmatrix}. $$
Thus, the fundamental matrix is given by
$$ X(t) = \begin{pmatrix}1 & e^{-t} \\ 1 & 2e^{-t} \end{pmatrix}. $$
And $X(0)^{-1}$ is calculated as
$$ X(0)^{-1} = \begin{pmatrix}1 & 1 \\ 1 & 2 \end{pmatrix}^{-1} = \begin{pmatrix}2 & -1 \\ -1 & 1\end{pmatrix}. $$
Finally, we get
$$ e^{At} = X(t)X(0)^{-1} = \begin{pmatrix}2-e^{-t} & -1 + e^{-t} \\ 2 - 2e^{-t} & -1 + e^{-t}\end{pmatrix}. $$

Exercise 1. Solve $\dot{x} = Ax$ and find $e^{At}$ for
$$ A = \begin{pmatrix} 4 & 3 \\ -4 & 4 \end{pmatrix}. $$

Classification of Fixed Points of Linear System

For a linear system $\dot{x} = Ax$ with $\det(A) \ne 0$, there's only one fixed point: $x = 0$. For $2$ dimensional systems, let's break it down based on eigenvalues:

Exercise 2. Determine the type of each fixed point for the following linear system:
$$ \begin{array}{l}\dot{x} = y \\ \dot{y} = -x\end{array} $$

Nonlinear System

For the nonlinear system $\dot{x} = v{x}$, we can still approximate what the fixed points are going to look like. Suppose $\bar{x}$ is a fixed point so that $v(\bar{x}) = 0$. Consider $A$ to be the Jacobian of $v$ at $\bar{x}$, then we can study the behavior of $v$ around $\bar{x}$ by studying the linear system $\dot{x} = Ax$.

A Pendulum Example

We can model a pendulum with the differential equation $\ddot{\theta} + k\sin \theta = 0$, where $k > 0$. Take $x_1 = \theta$ and $x_2 = \dot{\theta}$. Then we have the following nonlinear system:

$$ \begin{array}{l}\dot{x_1} = x_2 \\ \dot{x_2} = -k\sin{x_1}\end{array} $$

The Jacobian of this system is given by

$$ J(x) = \begin{pmatrix}0 & 1 \\ -k\cos{x_1} & 0 \end{pmatrix}. $$
We only have to deal with two fixed points: $(0,0)$ and $(0, \pi)$. The Jacobian at these points are,
$$ A_1 = \begin{pmatrix}0 & 1 \\ -k & 0 \end{pmatrix} \mbox{ and } A_2 = \begin{pmatrix}0 & 1 \\ k & 0 \end{pmatrix}. $$

Therefore at $(0,0)$, we have a stable center as the eigenvalues of $A_2$ are $\pm i\sqrt{k}$, and $(0, \pi)$ is unstable as the eigenvalues are $\pm \sqrt{k}$.