Chapter 7.2, Problem 15E

Interpretation Introduction

Interpretation:

For the system equation $\dot{\text{x}}\text{ = x}\left(\text{2 - x - y}\right)$ and ${\dot{\text{y}}\text{ = y(4x - x}}^{\text{2}}\text{- 3)}$, with no closed orbits, find the fixed points and classify them. Also sketch the phase portrait.

Concept Introduction:

• Determine the Jacobian matrix for system equation and fixed points.

• Identify the stability using Eigen values.

• Sketch the phase portrait.

Answer to Problem 15E

Solution:

1. The fixed points are $\left(\text{x,y}\right)\text{ = }\left(\text{0,0}\right)\left(\text{1,1}\right)\left(\text{2,0}\right)\text{and}\left(\text{3,-1}\right)$ and $\left(\text{3,-1}\right)$. The point $\left(\text{x,y}\right)\text{ = }\left(\text{0,0}\right)$ is a saddle point, $\left(\text{x,y}\right)\text{ = }\left(\text{1,1}\right)$ is a stable spiral, the point $\left(\text{x,y}\right)\text{ = }\left(\text{2,0}\right)$ is a saddle point and $\left(\text{x,y}\right)\text{ = }\left(\text{3,-1}\right)$ is a stable spiral.

2. The phase portrait of the given system equation is shown.

Explanation of Solution

In the linear system the determination of the matrix A is defined by $\text{Δ}$. If det (A) is positive so the Eigen values are real and have opposite signs and the fixed point is a saddle point.

Here, $\text{x}$ is the position coordinate. Position is the function of time hence

$\dot{\text{x}}\text{= }\frac{\text{dx}}{\text{dt}}$

1. Consider the given system equations

   $\dot{\text{x}}\text{ = x}\left(\text{2 - x - y}\right)$

   ${\dot{\text{y}}\text{ = y(4x - x}}^{\text{2}}\text{- 3)}$

   which has no closed orbits in the region. Now we will determine the fixed points of the given linear system by using the Jacobian matrix A and it’s determinant.

   The Jacobian matrix form of the linear system is,

   $\text{A = }\left(\begin{array}{cc}\frac{\text{d}\dot{\text{x}}}{\text{dx}}& \frac{\text{d}\dot{\text{x}}}{\text{dy}}\\ \frac{\text{d}\dot{\text{y}}}{\text{dx}}& \frac{\text{d}\dot{\text{y}}}{\text{dy}}\end{array}\right)$

   Now using $\dot{\text{x}}$ and $\dot{\text{y}}$ equations, the values are,

   $\text{A = }\left(\begin{array}{cc}\frac{\text{d}\dot{\text{x}}}{\text{dx}}& \frac{\text{d}\dot{\text{x}}}{\text{dy}}\\ \frac{\text{d}\dot{\text{y}}}{\text{dx}}& \frac{\text{d}\dot{\text{y}}}{\text{dy}}\end{array}\right)$

   $\text{A = }\left(\begin{array}{cc}\frac{\text{d}\left(\text{x}\left(\text{2 - x - y}\right)\right)}{\text{dx}}& \frac{\text{d}\left(\text{x}\left(\text{2 - x - y}\right)\right)}{\text{dy}}\\ \frac{\text{d}\left({\text{y(4x - x}}^{\text{2}}\text{- 3)}\right)}{\text{dx}}& \frac{\text{d}\left({\text{y(4x - x}}^{\text{2}}\text{- 3)}\right)}{\text{dy}}\end{array}\right)$

   $\text{A = }\left(\begin{array}{cc}\left(\text{2 - 2x - y}\right)& \text{-x}\\ \text{y}\left(\text{4 - 2x}\right)& \left({\text{4x - x}}^{\text{2}}\text{- 3}\right)\end{array}\right)$

   So the matrix form of the given linear system is,

   $\text{A = }\left(\begin{array}{cc}\text{2 - 2x - y}& \text{-x}\\ \text{y}\left(\text{4 - 2x}\right)& {\text{4x - x}}^{\text{2}}\text{- 3}\end{array}\right)$

   Let the first derivative of the linear system be equal to zero, so the fixed values are determined as; Determinant of matrix A and it’s determinant

   $\frac{\text{d}\dot{\text{x}}}{\text{dx}}\text{ = 0, }\frac{\text{d}\dot{\text{x}}}{\text{dy}}\text{ = 0}$

   So,

   $\text{2 - 2x - y = 0}$

   $\text{-x = 0}$

   Similarly, $\frac{\text{d}\dot{\text{y}}}{\text{dx}}\text{ = 0, }\frac{\text{d}\dot{\text{y}}}{\text{dy}}\text{ = 0}$

   $\text{y}\left(\text{4 - 2x}\right)\text{ = 0}$

   $\text{x = 2}$

   $\text{y = 0}$

   And,

   ${\text{4x - x}}^{\text{2}}\text{- 3 = 0}$

   $\text{x = 1, 3}$

   Since the values are $\text{x = 1, 3}$ and $\text{x = 2}$ at $\text{y = 0}$ if the value $\text{y = 0}$ put in the linear system $\dot{\text{x}}\text{ = x}\left(\text{2 - x - y}\right)$ at $\dot{\text{x}}\text{ = 0}$ so the values are $\text{x = 0}$ and $\text{x = 2}$. Further at $\text{x = 1}$ put in $\dot{\text{x}}\text{ = x}\left(\text{2 - x - y}\right)$ at $\dot{\text{x}}\text{ = 0}$ so the value is $\text{y = 1}$ and at $\text{x = 3}$ the value is $\text{x = -1}$.

   Hence, the fixed points are $\left(\text{x,y}\right)\text{ = }\left(\text{0,0}\right)\left(\text{1,1}\right)\left(\text{2,0}\right)\text{and}\left(\text{3,-1}\right)$ and $\left(\text{3,-1}\right)$.

   Now the Eigen values at $\left(\text{x,y}\right)\text{ = }\left(\text{0,0}\right)$ are,

   ${\text{A}}_{\left(\text{0,0}\right)}\text{ = }\left(\begin{array}{cc}\text{2}& \text{0}\\ \text{0}& \text{-3}\end{array}\right)$

   $\left|\text{A - λI}\right|\text{ = 0}$

   $\left|\begin{array}{cc}\text{2 - λ }& \text{0}\\ \text{0}& \text{-3 - λ}\end{array}\right|\text{ = 0}$

   $\left(\text{-3 - λ}\right)\left(\text{2 - λ}\right)\text{= 0}$

   $\text{λ = 2, -3}$

   This Eigen value is a saddle point.

   Again the Eigen value at $\left(\text{x,y}\right)\text{ = }\left(\text{1,1}\right)$ are;

   ${\text{A}}_{\left(\text{1,1}\right)}\text{ = }\left(\begin{array}{cc}\text{-1}& \text{-1}\\ \text{2}& \text{0}\end{array}\right)$

   $\left|\text{A - λI}\right|\text{ = 0}$

   $\left|\begin{array}{cc}\text{-1 - λ }& -1\\ 2& \text{- λ}\end{array}\right|\text{ = 0}$

   $\text{λ(λ + 1) + 2 = 0}$

   ${\text{λ}}^2\text{ + λ + 2 = 0}$

   Take the roots of the quadratic equation,

   ${\text{λ}}_{\text{1,2}}\text{ = }\frac{-1\pm \sqrt{1-8}}{2}$

   ${\text{λ}}_{\text{1,2}}\text{ = }\frac{\text{-1 ± i}\sqrt{\text{7}}}{\text{2}}$

   This Eigen value is a stable spiral.

   And the Eigen value at $\left(\text{x,y}\right)\text{ = }\left(\text{2,0}\right)$ are;

   ${\text{A}}_{\left(\text{2,0}\right)}\text{ = }\left(\begin{array}{cc}\text{-2}& \text{-2}\\ \text{0}& \text{1}\end{array}\right)$

   $\left|\text{A - λI}\right|\text{ = 0}$

   $\left|\begin{array}{cc}\text{-2- }\lambda & \text{-2}\\ \text{0}& \text{1-λ}\end{array}\right|\text{ = 0}$

   $\left(\text{-2- }\lambda \right)\left(\text{1-λ}\right)\text{= 0}$

   $\text{λ = 1,-2}$

   This Eigen value is a saddle point.

   And the Eigen values at $\left(\text{x,y}\right)\text{ = }\left(\text{3,-1}\right)$ are;

   ${\text{A}}_{\left(\text{3,-1}\right)}\text{ = }\left(\begin{array}{cc}\text{-3}& \text{-3}\\ \text{2}& \text{0}\end{array}\right)$

   $\left|\text{A - λI}\right|\text{ = 0}$

   $\left|\begin{array}{cc}\text{-3- λ}& \text{-3}\\ \text{2}& \text{-λ}\end{array}\right|\text{ = 0}$

   $\text{λ}\left(\text{λ + 3}\right)\text{ + 6 = 0}$

   Taking the roots of the quadratic equation is,

   ${\text{λ}}^{\text{2}}\text{ + 3λ + 6 = 0}$

   ${\text{λ}}_{\text{1,2}}\text{ = }\frac{-3\pm \sqrt{9-24}}{2}$

   ${\text{λ}}_{\text{1,2}}\text{ = }\frac{\text{-3 ± i}\sqrt{\text{15}}}{\text{2}}$

   Therefore the Eigen value is a stable spiral.

2. Now sketch the phase portrait of the given linear system equation is as given below.