Chapter 5.2, Problem 1E

Interpretation Introduction

Interpretation:

To find characteristic polynomial ${\text{λ}}^{\text{2}}\text{- 5λ + 6 = 0}$ for the system of linear equations $\dot{\text{x}}\text{ = 4x - y}$ and $\dot{\text{y}}\text{ = 2x + y}$, using $\dot{\text{x}}\text{ = Ax}$. Also find eigenvalues and eigenvectors of matrix A.

Solve the given system of linear equations and write the general solution.

Classify the fixed points at the origin.

Solve the system subject to the primary condition ${\text{(x}}_{\text{0}}{\text{,y}}_{\text{0}}\text{) = (3,4)}$.

Concept Introduction:

Equations for two dimensional linear system are $\dot{\text{x}}\text{ = ax + by}$ and $\dot{\text{y}}\text{ = cx + dy}$.

The above linear system expressed in the form $\dot{\text{x}}\text{ = Ax}$.

The standard characteristics polynomials is

${\text{λ}}^{\text{2}}\text{- τλ + Δ = 0}$, where $\text{τ}$ is trace ofmatrix A, $\text{λ}$ is corresponding eigenvalue, and $\text{Δ}$ is the determinant of matrix A.

Answer to Problem 1E

Solution:

The characteristic polynomial for the linear system of equation is ${\text{λ}}^{\text{2}}\text{- 5λ + 6 = 0}$. Eigenvalues and eigenvectors of matrix A are ${\text{λ}}_{\text{1}}\text{= 3}$, ${\text{λ}}_2\text{= 2}$, $\overrightarrow{v_1}=\left(\begin{array}{l}\text{1}\\ \text{1}\end{array}\right)$ and $\overrightarrow{v_2}=\left(\begin{array}{l}\text{1}\\ \text{2}\end{array}\right)$.

Overall solution of the system is $\text{X(t) =}\left(\begin{array}{l}\text{x}\\ \text{y}\end{array}\right){\text{= c}}_1{e}^{{\lambda }_{1}t}\overrightarrow{{\text{v}}_1}+{\text{c}}_2{e}^{{\lambda }_{2}t}\overrightarrow{{\text{v}}_2}={\text{c}}_1{e}^{3t}\left(\begin{array}{l}\text{1}\\ \text{1}\end{array}\right)+{\text{c}}_2{e}^{2t}\left(\begin{array}{l}\text{1}\\ 2\end{array}\right)$.

The nature of fixed points for the given linear equations is unstable and growing.

The solution of the system for initial condition ${\text{(x}}_{\text{0}}{\text{,y}}_{\text{0}}\text{) = (3,4)}$ is $\begin{array}{l}\text{x(t)}=2e^{3t}+e^{2t}\\ \text{y(t)}=2e^{3t}+2e^{2t}\end{array}$.

Explanation of Solution

The linear system equations are

$\dot{\text{x}}\text{ = 4x - y}$

And,

$\dot{\text{y}}\text{ = 2x + y}$.

The above linear equations can be written in matrix form as

$\dot{\text{x}}\text{ = Ax}$

$\begin{array}{l}\dot{\text{x}}=\left(\begin{array}{l}\dot{x}\\ \dot{y}\end{array}\right)\\ \text{A= }\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\\ \text{x=}\left(\begin{array}{l}x\\ y\end{array}\right)\end{array}$

$\left(\begin{array}{c}\dot{\text{x}}\\ \dot{\text{y}}\end{array}\right)\text{ = }\left(\begin{array}{cc}\text{a}& \text{b}\\ \text{c}& \text{d}\end{array}\right)\left(\begin{array}{c}\text{x}\\ \text{y}\end{array}\right)$

$\left(\begin{array}{c}\dot{\text{x}}\\ \dot{\text{y}}\end{array}\right)\text{ = }\left(\begin{array}{cc}4& -1\\ 2& 1\end{array}\right)\left(\begin{array}{c}\text{x}\\ \text{y}\end{array}\right)$

The standard characteristic polynomial is

${\text{λ}}^{\text{2}}\text{- τλ + Δ = 0}$

Here, $\text{λ}$ is the corresponding eigenvalue, $\text{τ}$ is trace of the matrix $\left(\text{a + d}\right)$, and $\Delta$ is $\left(\text{ad - bc}\right)$- determinant of the matrix A.

$\text{τ = a + d}$

$\text{τ = 4 + 1}$

$\text{τ = 5}$

$\text{Δ = ad - bc}$

$\text{Δ = }\left(\text{4}\right)\left(\text{1}\right)\text{ - }\left(\text{-1}\right)\left(\text{2}\right)$

$\Delta \text{ = 6}$

Substitute $\text{5}$ for $\text{τ}$, $\text{6}$ for $\Delta$ in the above standard characteristic polynomial, and solve.

${\text{λ}}^{\text{2}}\text{- 5λ + 6 = 0}$

This is the necessarycharacteristic polynomial.

To find the eigenvalues of the above characteristic polynomial, find its roots.

$\begin{array}{l}\text{λ=}\frac{\text{-b ± }\sqrt{{\text{b}}^{\text{2}}\text{-4ac}}}{\text{2a}}\\ \text{b = -5,a = 1,c =6}\\ \text{λ=}\frac{\text{5 ± }\sqrt{{\text{5}}^{\text{2}}\text{- (4×1×6)}}}{\text{2×1}}\\ \text{λ=}\frac{\text{5 ± }\sqrt{\text{25-24}}}{\text{2}}\\ \text{λ= }\frac{\text{5 ± 1}}{\text{2}}\\ {\text{λ}}_{\text{1}}{\text{=3 and λ}}_{\text{2}}\text{=2}\end{array}$

Using characteristic equation, find eigenvectors for the given eigenvalues.

$\begin{array}{l}\text{Ax = λx}\\ \text{x =}\left(\begin{array}{l}{\text{v}}_{\text{1}}\\ {\text{v}}_{\text{2}}\end{array}\right)\\ \left(\begin{array}{cc}\text{4}& \text{-1}\\ \text{2}& \text{1}\end{array}\right)\left(\begin{array}{l}{\text{v}}_{\text{1}}\\ {\text{v}}_{\text{2}}\end{array}\right)\text{= 3}\left(\begin{array}{l}{\text{v}}_{\text{1}}\\ {\text{v}}_{\text{2}}\end{array}\right)\\ {\text{4v}}_{\text{1}}{\text{- v}}_{\text{2}}{\text{=3v}}_{\text{1}}\\ {\text{v}}_{\text{1}}{\text{-v}}_{\text{2}}\text{=0}\\ {\text{v}}_{\text{1}}{\text{= v}}_{\text{2}}\\ {\text{2v}}_{\text{1}}{\text{+v}}_{\text{2}}{\text{=3v}}_{\text{2}}\\ {\text{v}}_{\text{1}}{\text{-v}}_{\text{2}}\text{=0}\\ {\text{v}}_{\text{1}}{\text{= v}}_{\text{2}}\\ \text{so first vector for λ = 3 is ,}\\ \overrightarrow{v_1}=\left(\begin{array}{l}{\text{v}}_{\text{1}}\\ {\text{v}}_{\text{2}}\end{array}\right)\text{= }\left(\begin{array}{l}\text{1}\\ \text{1}\end{array}\right)\\ \text{for λ=3}\\ \left(\begin{array}{cc}\text{1}& \text{-1}\\ \text{2}& \text{-2}\end{array}\right)\left(\begin{array}{l}{\text{v}}_{\text{1}}\\ {\text{v}}_{\text{2}}\end{array}\right)\text{= 0}\\ \text{for }\lambda \text{=2}\\ \left(\begin{array}{cc}\text{4}& \text{-1}\\ \text{2}& \text{1}\end{array}\right)\left(\begin{array}{l}{\text{v}}_{\text{1}}\\ {\text{v}}_{\text{2}}\end{array}\right)\text{= 2}\left(\begin{array}{l}{\text{v}}_{\text{1}}\\ {\text{v}}_{\text{2}}\end{array}\right)\\ 4{\text{v}}_{\text{1}}-{\text{v}}_{\text{2}}=2{\text{v}}_{\text{1}}\\ {\text{v}}_{\text{2}}=2{\text{v}}_{\text{1}}\\ {\text{when v}}_{\text{1}}=1\\ {\text{v}}_{\text{2}}=2\\ \overrightarrow{v_2}=\left(\begin{array}{l}{\text{v}}_{\text{1}}\\ {\text{v}}_{\text{2}}\end{array}\right)\text{= }\left(\begin{array}{l}\text{1}\\ \text{2}\end{array}\right)\end{array}$

General solution of the system is written as

$\text{X(t) =}\left(\begin{array}{l}\text{x}\\ \text{y}\end{array}\right){\text{= c}}_1{e}^{{\lambda }_{1}t}\overrightarrow{{\text{v}}_1}+{\text{c}}_2{e}^{{\lambda }_{2}t}\overrightarrow{{\text{v}}_2}={\text{c}}_1{e}^{3t}\left(\begin{array}{l}\text{1}\\ \text{1}\end{array}\right)+{\text{c}}_2{e}^{2t}\left(\begin{array}{l}\text{1}\\ 2\end{array}\right)$

Here, ${\text{c}}_1$ and ${\text{c}}_2$ are constants.

The fixed points at the origin can never be stable if both eigenvalues are positive and real. Hence the nature of fixed points for the given linear equations is unstable and growing. This means the stream lines are moving out from the center of the origin. Also, $\text{ τ = 5 and Δ=6 }$ yield unstable node.

Initial condition is ${\text{(x}}_{\text{0}}{\text{,y}}_{\text{0}}\text{) = (3,4)}$.

$\text{X(t) =}\left(\begin{array}{l}\text{x(t)}\\ \text{y(t)}\end{array}\right){\text{= c}}_1{e}^{{\lambda }_{1}t}\overrightarrow{{\text{v}}_1}+{\text{c}}_2{e}^{{\lambda }_{2}t}\overrightarrow{{\text{v}}_2}={\text{c}}_1{e}^{3t}\left(\begin{array}{l}\text{1}\\ \text{1}\end{array}\right)+{\text{c}}_2{e}^{2t}\left(\begin{array}{l}\text{1}\\ 2\end{array}\right)$

Substitute $\text{t = 0}$ in the obtained equation.

$\begin{array}{l}\text{X(0) =}\left(\begin{array}{l}\text{x(0)}\\ \text{y(0)}\end{array}\right){\text{= c}}_1{e}^{{\lambda }_{1}0}\overrightarrow{{\text{v}}_1}+{\text{c}}_2{e}^{{\lambda }_{2}0}\overrightarrow{{\text{v}}_2}={\text{c}}_1\left(\begin{array}{l}\text{1}\\ \text{1}\end{array}\right)+{\text{c}}_2\left(\begin{array}{l}\text{1}\\ 2\end{array}\right)\\ \left(\begin{array}{l}\text{x(0)}\\ \text{y(0)}\end{array}\right)={\text{c}}_1\left(\begin{array}{l}\text{1}\\ \text{1}\end{array}\right)+{\text{c}}_2\left(\begin{array}{l}\text{1}\\ 2\end{array}\right)\\ \left(\begin{array}{l}\text{3}\\ \text{4}\end{array}\right)={\text{c}}_1\left(\begin{array}{l}\text{1}\\ \text{1}\end{array}\right)+{\text{c}}_2\left(\begin{array}{l}\text{1}\\ 2\end{array}\right)\\ {\text{c}}_1+{\text{c}}_2=3\\ {\text{c}}_1+2{\text{c}}_2=4\end{array}$

Solving for ${\text{c}}_1$ and ${\text{c}}_2$,

${\text{c}}_1=2,{\text{c}}_2=1$.

Substituting this values in general solution of time t,

$\begin{array}{l}\left(\begin{array}{l}\text{x(t)}\\ \text{y(t)}\end{array}\right)\text{= }2e^{{\lambda }_{1}t}\overrightarrow{{\text{v}}_1}+1e^{{\lambda }_{2}t}=2e^{3t}\left(\begin{array}{l}\text{1}\\ \text{1}\end{array}\right)+e^{2t}\left(\begin{array}{l}\text{1}\\ 2\end{array}\right)\\ \left(\begin{array}{l}\text{x(t)}\\ \text{y(t)}\end{array}\right)\text{= }2e^{{\lambda }_{1}t}\overrightarrow{{\text{v}}_1}+1e^{{\lambda }_{2}t}=e^{3t}\left(\begin{array}{l}\text{2}\\ 2\end{array}\right)+e^{2t}\left(\begin{array}{l}\text{1}\\ 2\end{array}\right)\\ \text{x(t)}=2e^{3t}+e^{2t}\\ \text{y(t)}=2e^{3t}+2e^{2t}\end{array}$

This $\text{x(t)}$ and $\text{y(t)}$ is the solution for the given initial conditions.

Conclusion

The characteristic equation, eigenvalues, and eigenvectors are found for the given system of linear equations. Also general solution and solution of the given initial condition is found.