Section 5.8. Autonomous Systems

Objective:

1. Definition of Autonomous Systems

2. Finding the critical points of an Autonomous Systems

Recall that we have [latex]y'=y^{2}-2y[/latex], a differential equation that is not linear and we call it autonomous differential equation. We find the equilibrium solution by setting [latex]y^{2}-2y=0[/latex], hence [latex]y=0[/latex] and [latex]y=2[/latex] 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{align*} \frac{dx_{1}}{dt} & =F(x_{1},x_{2})\\ \frac{dx_{2}}{dt} & =G(x_{1},x_{2}). \end{align*} Notice [latex]F(x_{1},x_{2})[/latex] and [latex]G(x_{1},x_{2})[/latex] are functions of [latex]x_{1}[/latex] and [latex]x_{2}[/latex] only, i.e. the variable [latex]t[/latex] is not involved in both functions. If the degree of [latex]F(x_{1},x_{2})[/latex] and [latex]G(x_{1},x_{2})[/latex] in the variables [latex]x_{1}[/latex] and [latex]x_{2}[/latex] are all linear, then the system could be rewritten as [latex]\overrightarrow{x'}=A\overrightarrow{x}[/latex] which we know how to find the solution set. In this section, our focus is on the case that [latex]F(x_{1},x_{2})[/latex] and [latex]G(x_{1},x_{2})[/latex] are not linear functions. 

(b) [latex](a_{1},a_{2})[/latex] is called a critical point of the autonomous differential equations if [latex]F(a_{1},a_{2})=0=G(a_{1},a_{2})[/latex]. 

Example 1: Find critical point of the given system of differential equations.  \begin{align*} \frac{dx_{1}}{dt} & =(x_{1}-x_{2})(1-x_{1}-x_{2})\\ \frac{dx_{2}}{dt} & =x_{1}(2+x_{2}). \end{align*}

Exercise 1: Find critical point of the given system of differential equations.  \begin{align*} \frac{dx_{1}}{dt} & =(x_{1}-2x_{2})(x_{1}-x_{2})\\ \frac{dx_{2}}{dt} & =(x_{1}-2)(1-x_{2}). \end{align*}

Example 2: Find critical point of the given system of differential equations.  \begin{align*} \frac{dx_{1}}{dt} & =2x_{1}-x_{1}^{2}-x_{1}x_{2}\\ \frac{dx_{2}}{dt} & =3x_{2}-2x_{2}^{2}-3x_{1}x_{2}. \end{align*}

Exercise 2: Find critical point of the given system of differential equations. \begin{align*} \frac{dx_{1}}{dt} & =3x_{2}+x_{2}^{2}-x_{1}x_{2}\\ \frac{dx_{2}}{dt} & =-x_{1}+x_{1}^{2}-3x_{1}x_{2}. \end{align*}

Example 3: Find critical point of the given system of differential equations. \begin{align*} \frac{dx_{1}}{dt} & =x_{2}(1+x_{1}-x_{2})\\ \frac{dx_{2}}{dt} & =-x_{1}+x_{2}-4x_{1}x_{2}. \end{align*}

Exercise 3: Find critical point of the given system of differential equations.  \begin{align*} \frac{dx_{1}}{dt} & =2x_{2}-x_{1}-x_{1}x_{2}\\ \frac{dx_{2}}{dt} & =x_{1}(3x_{1}-x_{2}+1). \end{align*}

Example 4: Find critical point of the given system of differential equations.  \begin{align*} \frac{dx_{1}}{dt} & =(x_{2}+1)(x_{1}-x_{2})\\ \frac{dx_{2}}{dt} & =(x_{1}-4x_{1}^{2})x_{2}. \end{align*}

Exercise 4: Find critical point of the given system of differential equations. \begin{align*} \frac{dx_{1}}{dt} & =x_{1}(x_{2}-3x_{2}^{2})\\ \frac{dx_{2}}{dt} & =(x_{1}+2)(3x_{1}-x_{2}). \end{align*}

Group Work: 

Find critical point of the given system of differential equations. 

1. \begin{align*} \frac{dx_{1}}{dt} & =(x_{1}+3)(2+x_{2}-x_{2}^{2})\\ \frac{dx_{2}}{dt} & =(x_{1}-2)(x_{1}-2x_{2}). \end{align*}

2.  \begin{align*} \frac{dx_{1}}{dt} & =x_{1}(x_{1}-3x_{2}-1)\\ \frac{dx_{2}}{dt} & =(x_{1}+2)(x_{2}-2). \end{align*}

3. \begin{align*} \frac{dx_{1}}{dt} & =x_{1}(x_{1}-x_{2}+2)\\ \frac{dx_{2}}{dt} & =x_{1}+3x_{1}x_{2}-x_{2}. \end{align*}

4.  \begin{align*} \frac{dx_{1}}{dt} & =x_{1}(x_{2}-x_{1})\\ \frac{dx_{2}}{dt} & =(x_{1}+2)(3x_{1}-x_{2}-2). \end{align*}