Question

Sketch the rough phase portraits for the dynamical system $\overrightarrow{\mathbf{x}}\left(t+1\right)\mathbf{=}\left[\begin{array}{cc}0.9& 0.2\\ 0.2& 1.2\end{array}\right]\overrightarrow{\mathbf{x}}\left(t\right)$.

Short answer

The rough phase portraits for the dynamical systems $\overrightarrow{\mathrm{x}}(\mathrm{t}+1)=\left[\begin{array}{cc}0.9& 0.2\\ 0.2& 1.2\end{array}\right]\overrightarrow{\mathrm{x}}\left(\mathrm{t}\right)$is:

Step-by-step solution

Find the Eigen values of the matrix

Consider the equation of dynamical system $\overrightarrow{\mathrm{x}}(\mathrm{t}+1)=\left[\begin{array}{cc}0.9& 0.2\\ 0.2& 1.2\end{array}\right]\overrightarrow{\mathrm{x}}\left(\mathrm{t}\right)$means .

role="math" localid="1668682841033" $\frac{d\overrightarrow{x}}{dt}=\left[\begin{array}{cc}0.9& 0.2\\ 0.2& 1.2\end{array}\right]\overrightarrow{\mathrm{x}}$

Compare the equations role="math" localid="1668669101929" $\frac{d\overrightarrow{x}}{dt}=\left[\begin{array}{cc}0.9& 0.2\\ 0.2& 1.2\end{array}\right]\overrightarrow{\mathrm{x}}$and $\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$as follows.

$A=\left[\begin{array}{cc}0.9& 0.2\\ 0.2& 1.2\end{array}\right]$

Assume$\lambda$is an Eigen value of the matrix role="math" localid="1668669206898" $\left[\begin{array}{cc}0.9& 0.2\\ 0.2& 1.2\end{array}\right]$implies $\left|A-\lambda I\right|=0$.

Substitute the values role="math" localid="1668670930937" $\left[\begin{array}{cc}0.9& 0.2\\ 0.2& 1.2\end{array}\right]$ for A and $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ for I in the equation $\left|A-\lambda I\right|=0$as follows.

role="math" localid="1668670293862" $$\begin{gathered} \left|A-\lambda I\right|=0 \\ \left|\left[\begin{array}{cc}0.9& 0.2\\ 0.2& 1.2\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|=0 \end{gathered}$$

Simplify the equation$\left|\left[\begin{array}{cc}0.9& 0.2\\ 0.2& 1.2\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|=0$as follows.

$\begin{array}{rcl}\left|\left[\begin{array}{cc}0.9& 0.2\\ 0.2& 1.2\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|& =& 0\\ \left|\left[\begin{array}{cc}0.9& 0.2\\ 0.2& 1.2\end{array}\right]-\left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right]\right|& =& 0\\ \left|\left[\begin{array}{cc}0.9-\lambda & 0.2\\ 0.2& 1.2-\lambda \end{array}\right]\right|& =& 0\\ \left(0.9-\lambda \right)\left(1.2-\lambda \right)-0.04& =& 0\end{array}$

Further, simplify the equation as follows.

$\begin{array}{rcl}\left(0.9-\lambda \right)\left(1.2-\lambda \right)-0.04& =& 0\\ 1.08-0.9\lambda -1.2\lambda +{\lambda }^2-0.04& =& 0\\ {\lambda }^2-2.1\lambda -1.04& =& 0\end{array}$

Therefore, the Eigen values of A are$\lambda =\frac{-{\left(2.1\right)}^2\pm \sqrt{{\left(2.1\right)}^2-4\left(1\right)\left(1.04\right)}}{2}$implies $\lambda =-1.9,\text{- 2}.5$.

Find the Eigen vector corresponding to the Eigen values

Assume $v_1=\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]$and $v_2=\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]$are Eigen vector corresponding to $\lambda =-1.9,-25$implies $\left|A-{\lambda }_{1}I\right|v_1=0$and $\left|A-{\lambda }_{2}I\right|v_2=0$.

Substitute the values role="math" localid="1668671936150" $\left[\begin{array}{cc}0.9& 0.2\\ 0.2& 1.2\end{array}\right]$ for A, -1.9 for ${\lambda }_1,\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]$ for v₁ and $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ for I in the equation $\left|A-{\lambda }_{1}I\right|v_1=0$as follows.

role="math" localid="1668672511880" $\begin{array}{rcl}\left|A-{\lambda }_{1}I\right|v{}_1& =& 0\\ \left|\left[\begin{array}{cc}0.9& 0.2\\ 0.2& 1.2\end{array}\right]-1.9\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]& =& 0\\ \left|\left[\begin{array}{cc}0.9& 0.2\\ 0.2& 1.2\end{array}\right]-\left[\begin{array}{cc}1.9& 0\\ 0& 1.9\end{array}\right]\right|\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]& =& 0\\ \left|\left[\begin{array}{cc}2.8& 0.2\\ 0.2& 3.1\end{array}\right]\right|\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]& =& 0\end{array}$

Further, simplify the equation as follows.

$\begin{array}{rcl}\left|\left[\begin{array}{cc}2.8& 0.2\\ 0.2& 3.1\end{array}\right]\right|\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]& =& 0\\ 2.8x_1+0.2y_1& =& 0\\ 0.2x_1+3.1y_1& =& 0\end{array}$

Simplify the equations $2.8x_1+0.2y_1=0$and $0.2x_1+3.1y_1=0$as follows.

$\begin{array}{rcl}2.8x_1+0.2y_1& =& 0\\ -2.8x_1& =& 0.2y_1\end{array}$

For $x_1=1$implies $y_1=-1.4$.

Therefore, the Eigen vector corresponding to ${\lambda }_1=-1.9$is $\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]=\left[\begin{array}{c}1\\ -1.4\end{array}\right]$.

Substitute the values $\left[\begin{array}{cc}0.9& 0.2\\ 0.2& 1.2\end{array}\right]$ for A, -2.5 for ${\lambda }_2,\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]$for v₂ and$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$for I in the equation$\left|A-{\lambda }_{2}I\right|v_2=0$as follows.

$\begin{array}{rcl}\left|A-{\lambda }_{2}I\right|v_2& =& 0\\ \left|\left[\begin{array}{cc}0.9& 0.2\\ 0.2& 1.2\end{array}\right]-1.6\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]& =& 0\\ \left|\left[\begin{array}{cc}0.9& 0.2\\ 0.2& 1.2\end{array}\right]+\left[\begin{array}{cc}-1.6& 0\\ 0& -1.6\end{array}\right]\right|\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]& =& 0\\ \left|\left[\begin{array}{cc}-0.7& 0.2\\ 0.2& -0.4\end{array}\right]\right|\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]& =& 0\end{array}$

Further, simplify the equation as follows.

$\begin{array}{rcl}\left|\left[\begin{array}{cc}-0.7& 0.2\\ 0.2& -0.4\end{array}\right]\right|\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]& =& 0\\ -0.7x_2+0.2y_2& =& 0\\ 0.2x_2-0.4y_2& =& 0\end{array}$

Simplify the equations $-0.7x_2+0.2y_2=0$and $0.2x_2-0.4y_2=0$as follows.

$\begin{array}{rcl}-0.7x_2+0.2y_2& =& 0\\ 0.7x_2& =& 0.2y_2\end{array}$

For$x_2=2$implies $y_2=7$.

Therefore, the Eigen vector corresponding to${\lambda }_2=-2.5$is $\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]=\left[\begin{array}{c}2\\ 7\end{array}\right]$.

The Eigen vectors are $v_1=\left[\begin{array}{c}1\\ -1.4\end{array}\right]$and $v_2=\left[\begin{array}{c}2\\ 7\end{array}\right]$corresponding to the Eigen values ${\lambda }_1=-1.9$and ${\lambda }_2=-2.5$respectively implies$S^{-1}AS=B$where$S=\left[\begin{array}{cc}1& 2\\ -1.4& 7\end{array}\right]$and $B=\left[\begin{array}{cc}-1.9& 0\\ 0& -2.5\end{array}\right]$.

Find the solution for dx→dt=[0.90.20.21.2]x→

Substitute the value SBS^-1 for A in the equation$\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$as follows.

$\begin{array}{rcl}\frac{d\overrightarrow{x}}{dt}& =& A\overrightarrow{x}\\ \frac{d\overrightarrow{x}}{dt}& =& SB{S}^{-1}\overrightarrow{x}\\ S^{-1}\frac{d}{dt}& =& B{S}^{-1}\overrightarrow{x}\\ \frac{d}{dt}\left(S^{-1}\overrightarrow{x}\right)& =& B\left(S^{-1}\overrightarrow{x}\right)\end{array}$

Assume$\overrightarrow{c}=S^{-1}\overrightarrow{x}$implies $\frac{d}{dt}\overrightarrow{c}=B\overrightarrow{c}$.

The solution of the equation$\frac{d}{dt}\overrightarrow{c}=B\overrightarrow{c}$is $\overrightarrow{c}\left(t\right)=\left[\begin{array}{c}{e}^{-2.5t}{c}_1\\ e^{-1.9t}{c}_2\end{array}\right]$.

Find the general solution for x→(t)

The general solution of$\overrightarrow{x}\left(t\right)$is $\overrightarrow{x}\left(t\right)=c_1{e}^{-2.5t}{v}_1+c_2{e}^{-1.9t}{v}_2$.

Substitute the value$\left[\begin{array}{c}1\\ -1.4\end{array}\right]$for v₁ and $\left[\begin{array}{c}2\\ 7\end{array}\right]$for v₂ in the equation $\overrightarrow{x}\left(t\right)=c_1{e}^{-2.5t}{v}_1+c_2{e}^{-1.9t}{v}_2$as follows.

$$\begin{gathered} \overrightarrow{x}\left(t\right)=c_1{e}^{-2.5t}{v}_1+c_2{e}^{-1.9t}{v}_2 \\ \overrightarrow{x}\left(t\right)=c_1{e}^{-2.5t}\left[\begin{array}{c}1\\ -1.4\end{array}\right]+c_2{e}^{-1.9t}\left[\begin{array}{c}2\\ 7\end{array}\right] \end{gathered}$$

For$c_1=1$and$c_2=0$the value of$\overrightarrow{x}\left(t\right)$is defined as follows.

$\overrightarrow{x}\left(t\right)=e^{-2.5t}\left[\begin{array}{c}1\\ -1.4\end{array}\right]$

For $c_1=0$and $c_2=1$the value of $\overrightarrow{x}\left(t\right)$is defined as follows.

$\overrightarrow{x}\left(t\right)=e^{-1.9t}\left[\begin{array}{c}2\\ 7\end{array}\right]$

Sketch the rough portraits for the dynamical system dx→dt=[0.90.20.21.2]x→

As \(0 > {\lambda _1} > {\lambda _2}\) Sketch the rough phase portraits for the dynamical systems $\overrightarrow{x}(t+1)=\left[\begin{array}{cc}0.9& 0.2\\ 0.2& 1.2\end{array}\right]\overrightarrow{\mathrm{x}}\left(t\right)$ as follows.

Hence the rough phase portraits for the dynamical systems $\overrightarrow{x}(t+1)=\left[\begin{array}{cc}0.9& 0.2\\ 0.2& 1.2\end{array}\right]\overrightarrow{\mathrm{x}}\left(t\right)$is sketched.