Section 5.9. Almost linear system Systems

Objective:

1. Definition of an almost linear system

2. Definition of Jacobian matrix

3. Find the linearization of an almost linear system near it critical points and classify their stability.

In this section, we try to understand the behavior of equilibrium solutions or the stability of the critical points of an autonomous system. The key point is using linearization to approximate the behavior. Recall in calculus I, [latex]y=f(x)[/latex], [latex]y[/latex] is s function of [latex]x[/latex]. We can find the linearization of [latex]f(x)[/latex] near [latex]x=a[/latex]: [latex]L(x)=f(a)+f'(a)(x-a)[/latex]. We use [latex]L(x)[/latex] to approximate the value of [latex]f(x)[/latex] near [latex]x=a[/latex]. 

 

Definition: A differential equation with the form below is called an almost linear autonomous system of differential equation.  \begin{align*} \overrightarrow{x’} & =A\overrightarrow{x}+\overrightarrow{g(x_{1},x_{2})} \end{align*}  where [latex]g(x_{1},x_{2})[/latex] is small in term of  \[\lim_{(x_{1},x_{2})\rightarrow(0,0)}\frac{|\overrightarrow{g(x_{1},x_{2})}|}{\sqrt{x_{1}^{2}+x_{2}^{2}}}=0.\] Notice the definition can be extended to any critical points, [latex](a_{1},a_{2})[/latex]. We call this system is locally linear near [latex](a_{1},a_{2})[/latex]. 

 

 

Theorem: Given a nonlinear system of differential equations: \begin{align*} \frac{dx_{1}}{dt} & =F(x_{1},x_{2})\\ \frac{dx_{2}}{dt} & =G(x_{1},x_{2}). \end{align*} The system is locally linear in the neighborhood of a critical point [latex](a_{1},a_{2})[/latex] whenever the functions [latex]F(x_{1},x_{2})[/latex] and [latex]G(x_{1},x_{2})[/latex] have continuous partial derivatives up to order two.

 

We are now ready to apply the linearization to the critical point [latex](a_{1},a_{2})[/latex] of a locally linear system of differential equations: \begin{align*} \frac{dx_{1}}{dt} & =F(x_{1},x_{2})\\ \frac{dx_{2}}{dt} & =G(x_{1},x_{2}). \end{align*}

Since the system locally linear we can rewrite the system as  \begin{align*} \frac{dx_{1}}{dt} & =F(x_{1},x_{2})=F(a_{1},a_{2})+F_{x_{1}}(a_{1},a_{2})(x_{1}-a_{1})+F_{x_{2}}(a_{1},a_{2})(x_{2}-a_{2})+p(x_{1},x_{2})\\ \frac{dx_{2}}{dt} & =G(x_{1},x_{2})=G(a_{1},a_{2})+G_{x_{1}}(a_{1},a_{2})(x_{1}-a_{1})+G_{x_{2}}(a_{1},a_{2})(x_{2}-a_{2})+q(x_{1},x_{2}) \end{align*} where \[\lim_{(x_{1},x_{2})\rightarrow(a_{1},a_{2})}\frac{p(x_{1},x_{2})}{\sqrt{(x_{1}-a_{1})^{2}+(x_{2}-a_{2})^{2}}}=0.\] and  \[ \lim_{(x_{1},x_{2})\rightarrow(a_{1},a_{2})}\frac{q(x_{1},x_{2})}{\sqrt{(x_{1}-a_{1})^{2}+(x_{2}-a_{2})^{2}}}=0.\]

Since [latex](a_{1},a_{2})[/latex] is a critical point, i.e. [latex]F(a_{1},a_{2})=0[/latex] and [latex]G(a_{1},a_{2})=0,[/latex] we further rewrite the system as \begin{align*}\frac{d}{dt}\left[\begin{array}{c} x_{1}-a_{1}\\ x_{2}-a_{2} \end{array}\right] & =\left[\begin{array}{cc} F_{x_{1}}(a_{1},a_{2}) & F_{x_{2}}(a_{1},a_{2})\\ G_{x_{1}}(a_{1},a_{2}) & G_{x_{2}}(a_{1},a_{2}) \end{array}\right]\left[\begin{array}{c} x_{1}-a_{1}\\ x_{2}-a_{2} \end{array}\right]+\left[\begin{array}{c} p(x_{1},x_{2})\\ q(x_{1},x_{2}) \end{array}\right] \end{align*} using the fact that [latex]d(x_{1}-a_{1})/dt=dx_{1}/dt[/latex] and [latex]d(x_{2}-a_{2})/dt=dx_{2}/dt[/latex]. \begin{align*} \frac{d}{dt}\left[\begin{array}{c} x_{1}-a_{1}\\ x_{2}-a_{2} \end{array}\right] & =\left[\begin{array}{cc} F_{x_{1}}(a_{1},a_{2}) & F_{x_{2}}(a_{1},a_{2})\\ G_{x_{1}}(a_{1},a_{2}) & G_{x_{2}}(a_{1},a_{2}) \end{array}\right]\left[\begin{array}{c} x_{1}-a_{1}\\ x_{2}-a_{2} \end{array}\right] \end{align*} is called the linearization of the system \begin{align*} \frac{dx_{1}}{dt} & =F(x_{1},x_{2})\\ \frac{dx_{2}}{dt} & =G(x_{1},x_{2}) \end{align*}  near [latex](a_{1},a_{2})[/latex]. The matrix \[ \left[\begin{array}{cc} F_{x_{1}}(x_{1},x_{2}) & F_{x_{2}}(x_{1},x_{2})\\ G_{x_{1}}(x_{1},x_{2}) & G_{x_{2}}(x_{1},x_{2}) \end{array}\right] \] is called the Jacobian matrix of the functions [latex]F(x_{1},x_{2})[/latex] and [latex]G(x_{1},x_{2})[/latex]. We assume \[ \text{det}\left[\begin{array}{cc} F_{x_{1}}(a_{1},a_{2}) & F_{x_{2}}(a_{1},a_{2})\\ G_{x_{1}}(a_{1},a_{2}) & G_{x_{2}}(a_{1},a_{2}) \end{array}\right] \neq 0 \] and use the linear system to approximate the critical point behavior. 

 

 

Example 1: Discuss the type and stability of the critical points by examining the corresponding linear system. \begin{align*} \frac{dx_{1}}{dt} & =x_{1}+x_{2}^{2}\\ \frac{dx_{2}}{dt} & =x_{1}+x_{2}. \end{align*}

 

 

 

Exercise 1: Discuss the type and stability of the critical points by examining the corresponding linear system. \begin{align*}  \frac{dx_{1}}{dt} & =x_{1}-2x_{2}\\ \frac{dx_{2}}{dt} & =x_{1}^{2}+x_{2}. \end{align*}

 

 

 

Example 2: Discuss the type and stability of the critical points by examining the corresponding linear system. \begin{align*} \frac{dx_{1}}{dt} & =x_{1}-x_{1}^{2}-x_{1}x_{2}\\ \frac{dx_{2}}{dt} & =3x_{2}-x_{1}x_{2}-2x_{2}^{2}. \end{align*}

 

 

 

Exercise 2: Discuss the type and stability of the critical points by examining the corresponding linear system. \begin{align*} \frac{dx_{1}}{dt} & =x_{1}-x_{1}x_{2}-2x_{2}\\ \frac{dx_{2}}{dt} & =x_{1}^{2}+x_{1}x_{2}+2x_{1}.  \end{align*}

 

 

 

Example 3: Discuss the type and stability of the critical points by examining the corresponding linear system. \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 3: Discuss the type and stability of the critical points by examining the corresponding linear system. \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*}

 

 

Group Work: 

Discuss the type and stability of the critical points by examining

the corresponding linear system.

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

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