Question

Sketch the rough phase portraits for the dynamical system $\overrightarrow{\mathbf{x}}\mathbf{(}\mathbf{t}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{\left[}\begin{array}{cc}\mathbf{1}\mathbf{.}\mathbf{1}& \mathbf{0}\mathbf{.}\mathbf{2}\\ \mathbf{-}\mathbf{0}\mathbf{.}\mathbf{4}& \mathbf{0}\mathbf{.}\mathbf{5}\end{array}\mathbf{\right]}\overrightarrow{\mathbf{x}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$.

Short answer

The rough phase portraits for the dynamical systems is $\overrightarrow{\mathrm{x}}(\mathrm{t}+1)=\left[\begin{array}{cc}1.1& 0.2\\ -0.4& 0.5\end{array}\right]\overrightarrow{\mathrm{x}}\left(\mathrm{t}\right)$

Step-by-step solution

Find the Eigen values of the matrix

Consider the equation of dynamical system $\overrightarrow{\mathrm{x}}(\mathrm{t}+1)=\left[\begin{array}{cc}1.1& 0.2\\ -0.4& 0.5\end{array}\right]\overrightarrow{\mathrm{x}}\left(\mathrm{t}\right)$means .

$\frac{d\overrightarrow{x}}{dt}=\left[\begin{array}{cc}1.1& 0.2\\ -0.4& 0.5\end{array}\right]\overrightarrow{\mathrm{x}}$

Compare the equations$\frac{d\overrightarrow{x}}{dt}=\left[\begin{array}{cc}1.1& 0.2\\ -0.4& 0.5\end{array}\right]\overrightarrow{\mathrm{x}}$and$\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$as follows.

role="math" localid="1668684357143" $A=\left[\begin{array}{cc}1.1& 0.2\\ -0.4& 0.5\end{array}\right]$

Assume $\lambda$is an Eigen value of the matrix $\left[\begin{array}{cc}1.1& 0.2\\ -0.4& 0.5\end{array}\right]$implies $\left|A-\lambda I\right|=0$.

Substitute the valuesrole="math" localid="1668685983382" $\left[\begin{array}{cc}1.1& 0.2\\ -0.4& 0.5\end{array}\right]$ 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="1668684623245" $$\begin{gathered} \left|A-\lambda I\right|=0 \\ \left|\left[\begin{array}{cc}1.1& 0.2\\ -0.4& 0.5\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|=0 \end{gathered}$$

Simplify the equation$\left|\left[\begin{array}{cc}1.1& 0.2\\ -0.4& 0.5\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|=0$as follows.

$\begin{array}{rcl}\left|\left[\begin{array}{cc}1.1& 0.2\\ -0.4& 0.5\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|& =& 0\\ \left|\left[\begin{array}{cc}1.1& 0.2\\ -0.4& 0.5\end{array}\right]-\left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right]\right|& =& 0\\ \left|\left[\begin{array}{cc}1.1-\lambda & 0.2\\ -0.4& 0.5-\lambda \end{array}\right]\right|& =& 0\\ \left(1.1-\lambda \right)\left(0.5-\lambda \right)+0.08& =& 0\end{array}$

Further, simplify the equation as follows.

$\begin{array}{rcl}\left(1.1-\lambda \right)\left(0.5-\lambda \right)+0.08& =& 0\\ 0.55-1.1\lambda -0.5\lambda +{\lambda }^2+0.08& =& 0\\ {\lambda }^2-1.6\lambda +0.63& =& 0\end{array}$

Therefore, the Eigen values of A are$\lambda =\frac{-{\left(1.6\right)}^2\pm \sqrt{{\left(1.6\right)}^2-4\left(1\right)\left(0.63\right)}}{2}$implies $\lambda =-1.18,\text{- 1}.38$.

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 $\lambda =-1.18,\text{- 1}.38$implies $\left|A-{\lambda }_{1}I\right|v_1=0$and $\left|A-{\lambda }_{2}I\right|v_2=0$.

Substitute the values $\left[\begin{array}{cc}1.1& 0.2\\ -0.4& 0.5\end{array}\right]$for A, -1.18 for $v_1=\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]$ 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="1668686809196" $\begin{array}{rcl}\left|A-{\lambda }_{1}I\right|v_1& =& 0\\ \left|\left[\begin{array}{cc}1.1& 0.2\\ -0.4& 0.5\end{array}\right]+1.18\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]& =& 0\\ \left|\left[\begin{array}{cc}1.1& 0.2\\ -0.4& 0.5\end{array}\right]+\left[\begin{array}{cc}1.18& 0\\ 0& 1.18\end{array}\right]\right|\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]& =& 0\\ \left|\left[\begin{array}{cc}2.28& 0.2\\ -0.4& 1.68\end{array}\right]\right|\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]& =& 0\end{array}$

Further, simplify the equation as follows.

$\begin{array}{rcl}\left|\left[\begin{array}{cc}2.28& 0.2\\ -0.4& 1.68\end{array}\right]\right|\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]& =& 0\\ 2.28x_1+0.2y_1& =& 0\\ -0.4x_1+1.68y_1& =& 0\end{array}$

Simplify the equations $2.28x_1+0.2y_1=0$and $-0.4x_1+1.68y_1=0$as follows.

$\begin{array}{rcl}2.28x_1+0.2y_1& =& 0\\ -2.28x_1& =& 0.2y_1\end{array}$

For $x_1=1$implies $y_1=-1.14$.

Therefore, the Eigen vector corresponding to ${\lambda }_1=-1.18$is $\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]=\left[\begin{array}{c}1\\ -1.14\end{array}\right]$.

Substitute the values $\left[\begin{array}{cc}1.1& 0.2\\ -0.4& 0.5\end{array}\right]$for A, -138 for ${\lambda }_2,\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]$for I and$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$for in the equation$\left|A-{\lambda }_{2}I\right|v_2=0$as follows.

$\begin{array}{rcl}\left|A-{\lambda }_{2}I\right|v_2& =& 0\\ \left|\left[\begin{array}{cc}1.1& 0.2\\ -0.4& 0.5\end{array}\right]+1.38\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]& =& 0\\ \left|\left[\begin{array}{cc}1.1& 0.2\\ -0.4& 0.5\end{array}\right]+\left[\begin{array}{cc}1.38& 0\\ 0& 1.38\end{array}\right]\right|\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]& =& 0\\ \left|\left[\begin{array}{cc}2.48& 0.2\\ -0.4& 1.88\end{array}\right]\right|\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]& =& 0\end{array}$

Further, simplify the equation as follows.

$\begin{array}{rcl}\left|\left[\begin{array}{cc}2.48& 0.2\\ -0.4& 1.88\end{array}\right]\right|\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]& =& 0\\ 2.48x_2+0.2y_2& =& 0\\ -0.4x_2+1.88y_2& =& 0\end{array}$

Simplify the equations$2.48x_2+0.2y_2=0$and$-0.4x_2+1.88y_2=0$as follows.

$\begin{array}{rcl}2.48x_2+0.2y_2& =& 0\\ -2.48x_2& =& 0.2y_2\end{array}$

For$x_2=1$implies $y_2=-12.4$.

Therefore, the Eigen vector corresponding to${\lambda }_2=\text{- 1}.38$is $\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]=\left[\begin{array}{c}1\\ -12.4\end{array}\right]$.

The Eigen vectors are $v_1=\left[\begin{array}{c}1\\ -1.14\end{array}\right]$and $v_2=\left[\begin{array}{c}1\\ -12.4\end{array}\right]$corresponding to the Eigen values ${\lambda }_1=-1.18$and${\lambda }_2=\text{- 1}.38$respectively implies$S^{-1}AS=B$where$S=\left[\begin{array}{cc}1& 1\\ -1.4& -12.4\end{array}\right]$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 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 $\overrightarrow{c}\left(t\right)=\left[\begin{array}{c}{e}^{-1.38t}{c}_1\\ e^{-1.18t}{c}_2\end{array}\right]$.

Find the general solution for x→(t)

The general solution of$\overrightarrow{x}\left(t\right)$is $\overrightarrow{x}\left(t\right)=c_1{e}^{-1.38t}{v}_1+c_2{e}^{-1.18t}{v}_2$.

Substitute the value $\left[\begin{array}{c}1\\ -1.14\end{array}\right]$for v₁ and$\left[\begin{array}{c}1\\ -12.4\end{array}\right]$for v₂ in the equation$\overrightarrow{x}\left(t\right)=c_1{e}^{-1.38t}{v}_1+c_2{e}^{-1.18t}{v}_2$as follows.

role="math" localid="1668689119693" $$\begin{gathered} \overrightarrow{x}\left(t\right)=c_1{e}^{-1.38t}{v}_1+c_2{e}^{-1.18t}{v}_2 \\ \overrightarrow{x}\left(t\right)=c_1{e}^{-1.38t}\left[\begin{array}{c}1\\ -1.14\end{array}\right]+c_2{e}^{-1.18t}\left[\begin{array}{c}1\\ -12.4\end{array}\right] \end{gathered}$$

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

role="math" localid="1668689200241" $\overrightarrow{x}\left(t\right)=e^{-1.38t}\left[\begin{array}{c}1\\ -1.14\end{array}\right]$

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

$\overrightarrow{x}\left(t\right)=e^{-1.18t}\left[\begin{array}{c}1\\ -12.4\end{array}\right]$

Sketch the rough portraits for the dynamical system dx→dt=[1.10.2-0.40.5]x→

As \(0 > {\lambda _1} > {\lambda _2}\) Sketch the rough phase portraits for the dynamical systems $\overrightarrow{\mathrm{x}}(\mathrm{t}+1)=\left[\begin{array}{cc}1.1& 0.2\\ -0.4& 0.5\end{array}\right]\overrightarrow{\mathrm{x}}\left(\mathrm{t}\right)$as follows.

Hence the rough phase portraits for the dynamical systems$\overrightarrow{\mathrm{x}}(\mathrm{t}+1)=\left[\begin{array}{cc}1.1& 0.2\\ -0.4& 0.5\end{array}\right]\overrightarrow{\mathrm{x}}\left(\mathrm{t}\right)$is sketched.