Section 5.8. Autonomous Systems

Objective:

1. Definition of Autonomous Systems

2. Finding the critical points of an Autonomous Systems

Recall that we have $y^{\prime }=y^2-2y$, a differential equation that is not linear and we call it autonomous differential equation. We find the equilibrium solution by setting $y^2-2y=0$, hence $y=0$ and $y=2$ are called critical points. We draw the vector field to decide the solution is a stable or unstable solution. We are doing exactly the same process for a system of first order differential equations. 

Definition: (a) A differential equation with the form below is called an autonomous system of differential equation.  $$\begin{array}{rl}\frac{d{x}_1}{dt}& =F(x_1,x_2)\\ \frac{d{x}_2}{dt}& =G(x_1,x_2).\end{array}$$ Notice $F(x_1,x_2)$ and $G(x_1,x_2)$ are functions of $x_1$ and $x_2$ only, i.e. the variable $t$ is not involved in both functions. If the degree of $F(x_1,x_2)$ and $G(x_1,x_2)$ in the variables $x_1$ and $x_2$ are all linear, then the system could be rewritten as $\overrightarrow{x^{\prime }}=A\overrightarrow{x}$ which we know how to find the solution set. In this section, our focus is on the case that $F(x_1,x_2)$ and $G(x_1,x_2)$ are not linear functions. 

(b) $(a_1,a_2)$ is called a critical point of the autonomous differential equations if $F(a_1,a_2)=0=G(a_1,a_2)$. 

Example 1: Find critical point of the given system of differential equations.  $$\begin{array}{rl}\frac{d{x}_1}{dt}& =(x_1-x_2)(1-x_1-x_2)\\ \frac{d{x}_2}{dt}& =x_1(2+x_2).\end{array}$$

Exercise 1: Find critical point of the given system of differential equations.  $$\begin{array}{rl}\frac{d{x}_1}{dt}& =(x_1-2x_2)(x_1-x_2)\\ \frac{d{x}_2}{dt}& =(x_1-2)(1-x_2).\end{array}$$

Example 2: Find critical point of the given system of differential equations.  $$\begin{array}{rl}\frac{d{x}_1}{dt}& =2x_1-x_1^2-x_1{x}_2\\ \frac{d{x}_2}{dt}& =3x_2-2x_2^2-3x_1{x}_2.\end{array}$$

Exercise 2: Find critical point of the given system of differential equations. $$\begin{array}{rl}\frac{d{x}_1}{dt}& =3x_2+x_2^2-x_1{x}_2\\ \frac{d{x}_2}{dt}& =-x_1+x_1^2-3x_1{x}_2.\end{array}$$

Example 3: Find critical point of the given system of differential equations. $$\begin{array}{rl}\frac{d{x}_1}{dt}& =x_2(1+x_1-x_2)\\ \frac{d{x}_2}{dt}& =-x_1+x_2-4x_1{x}_2.\end{array}$$

Exercise 3: Find critical point of the given system of differential equations.  $$\begin{array}{rl}\frac{d{x}_1}{dt}& =2x_2-x_1-x_1{x}_2\\ \frac{d{x}_2}{dt}& =x_1(3x_1-x_2+1).\end{array}$$

Example 4: Find critical point of the given system of differential equations.  $$\begin{array}{rl}\frac{d{x}_1}{dt}& =(x_2+1)(x_1-x_2)\\ \frac{d{x}_2}{dt}& =(x_1-4x_1^2)x_2.\end{array}$$

Exercise 4: Find critical point of the given system of differential equations. $$\begin{array}{rl}\frac{d{x}_1}{dt}& =x_1(x_2-3x_2^2)\\ \frac{d{x}_2}{dt}& =(x_1+2)(3x_1-x_2).\end{array}$$

Group Work: 

Find critical point of the given system of differential equations. 

1. $$\begin{array}{rl}\frac{d{x}_1}{dt}& =(x_1+3)(2+x_2-x_2^2)\\ \frac{d{x}_2}{dt}& =(x_1-2)(x_1-2x_2).\end{array}$$

2.  $$\begin{array}{rl}\frac{d{x}_1}{dt}& =x_1(x_1-3x_2-1)\\ \frac{d{x}_2}{dt}& =(x_1+2)(x_2-2).\end{array}$$

3. $$\begin{array}{rl}\frac{d{x}_1}{dt}& =x_1(x_1-x_2+2)\\ \frac{d{x}_2}{dt}& =x_1+3x_1{x}_2-x_2.\end{array}$$

4.  $$\begin{array}{rl}\frac{d{x}_1}{dt}& =x_1(x_2-x_1)\\ \frac{d{x}_2}{dt}& =(x_1+2)(3x_1-x_2-2).\end{array}$$