Introduction

Consider the following 2nd order system (Mass-Damper-Spring):

$\ddot{x} + \dot{x} = 3x + x^2 = 0 \Longleftrightarrow \ddot{x} + c \dot{x} + f(x) = 0$

$c$ is the dampening force and $f(x)$ is the non-linear spring.

The system can be placed into the $\dot{\underline{x}} = \underline{f}(\underline{x})$ form:

$\begin{cases} \dot{x}_1 = x_2 \\ \dot{x}_2 = -3x_1 - x_1^2 - x_2 \end{cases}$

Analysis of the system: Manual

Analysis of the system $\Rightarrow$ Phase Plane $(x_1, x_2)$:

Step 1: Find the Equillibrium Points:

$\begin{cases} f_1(x_1, x_2) = 0 \\ f_2(x_2, x_2) = 0 \end{cases} \Rightarrow \begin{cases} x_2 && = 0 \\ -3x_1 - x_1^2 - x_2 && = 0 \, \Rightarrow \, -3x_1 - x_1^2 = 0 \end{cases}$

$\begin{cases} x_2 = 0 \\ x_1(-3 - x_1) = 0 \end{cases} \Rightarrow$ 2 equillibrium points $\Rightarrow \begin{cases} \underline{x}_1 = [0, 0]^T \\ \underline{x}_2 = [-3, 0]^T \end{cases}$

Step 2: Linearize the System Around the Equillibrium Points:

$\displaystyle\dot{\underline{x}} = \underline{f}(\underline{x}) \Rightarrow \approx \underline{f}(\underline{x}^*) + \left . \frac{\partial \underline{f}}{\partial \underline{x}} \right |_{\underline{x}^*} (\underline{x} - \underline{x}^*)$

$\displaystyle A = \left . \frac{\partial\underline{f}}{\partial\underline{x}} \right |_{\underline{x}_{1, \, 2}^*} = \begin{bmatrix} 0 && 1 \\ -3 - 2x_1 && -1 \end{bmatrix}_{\underline{x}_1^*, \, \underline{x}_2^*}$

For $\underline{x} = \underline{x}_1$

We have the following system:

$\dot{\underline{x}} = A_1 \underline{x} \Rightarrow A_1 = \begin{bmatrix} 0 && 1 \\ -3 && -1 \end{bmatrix}$

Look for the behavior of the system near $\underline{x}_1 = \underline{0}$. e-value analysis:

$\det (A_1 - \lambda I) = \begin{vmatrix} -\lambda && 1 \\ -3 && -1 - \lambda \end{vmatrix} = \lambda \big(\lambda + 1\big) + 3 = 0$

$\Rightarrow \lambda^2 + \lambda + 3 = 0 \Rightarrow \lambda_{12} = -\frac{1}{2} \left( 1 \pm \sqrt{1 - 12}\right) = -\frac{1}{2} \left( 1 \pm \sqrt{11} j\right)$

For $\underline{x} = \underline{x}_2$

We have the following system:

$\dot{\underline{x}} = A_2 \underline{x} \Rightarrow A_2 = \begin{bmatrix} 0 && 1 \\ 3 && -1 \end{bmatrix}$

Look for the behavior of the system near $\underline{x}_2 = \underline{0}$. e-value analysis:

$\left|(A - \lambda I ) \right| = \begin{vmatrix} -\lambda && 1 \\ 3 && -1 - \lambda \end{vmatrix} = 0$

$\Rightarrow \lambda \left(\lambda + 1 \right) - 3 = 0 \Rightarrow \, \lambda^2 + \lambda - 3 = 0$

$\lambda = -\frac{1}{2} \left ( 1 \pm \sqrt{1 + 12} \right) = - \frac{1}{2} \pm \frac{1}{2} \sqrt{13}$

Note that both solutions of $\lambda$ are REAL! Therefore:

$\begin{matrix} \lambda_1 = -\frac{1}{2} \left( 1 - \sqrt{13} \right ) > 0 \\ \lambda_2 = -\frac{1}{2} \left( 1 + \sqrt{13} \right ) < 0 \end{matrix} \Rightarrow \text{ Unstable } \Rightarrow \underline{\text{Saddle Point}}$

Compute the e-vectors $\underline{v}_1,\, \underline{v}_2$ to determine the direction of convergence

Step 3: Put Everything Together:

!!!!! TODO: Insert Graphic !!!!!

Analysis of the System: Python

Given that:

$\begin{cases} \dot{x}_1 = f_1(x_1, x_2) = x_2 \\ \dot{x}_2 = f_2(x_2, x_2) = -3x_1 - x_1^2 - x_2 \end{cases}$

We can define two variables to these functions:

We can solve the system of equations using the solve function.

We get the same solutions that we found previously.

To linearize the system, we need to take the derivatives of each function with respect to $x_1$ and $x_2$. We'll compile this into a matrix A_1 like so:

Now we let $\dot{\underline{x}} = A_1 \, \underline{x}$. Therefore A_1 would be:

Next we find the determinate of $A_1 - \lambda I$

Analysis of the System: Numpy Meshgrid