Question

Describe and graph trajectories of the given system. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} -4 & 3 \\ -2 & -11 \end{array}\right] \mathbf{y} $$

Short answer

Describe and graph the trajectories of the given system of linear differential equations represented by the matrix: $$ \mathbf{A} = \left[\begin{array}{rr} -4 & 3 \\ -2 & -11 \end{array}\right] $$ Solution: The trajectories of the system converge to the origin, as the eigenvalues are both negative. The general solution for the given system is: $$ \mathbf{y}(t) = c_1 e^{-5 t}\mathbf{v}_1 + c_2 e^{-6 t}\mathbf{v}_2 = c_1 e^{-5 t}\left[\begin{array}{r} -3 \\ 1 \end{array}\right] + c_2 e^{-6 t}\left[\begin{array}{r} -\frac{3}{2} \\ 1 \end{array}\right] $$ Where \(c_1\) and \(c_2\) are constants, and the eigenvectors are \(\mathbf{v}_1=\begin{bmatrix}-3\\1\end{bmatrix}\) and \(\mathbf{v}_2=\begin{bmatrix}-\frac{3}{2}\\1\end{bmatrix}\). The phase portrait consists of two lines corresponding to these eigenvectors, showing trajectories that approach the origin as time increases.

Step-by-step solution

Determine the eigenvalues

First, we need to find the eigenvalues of the given matrix: $$ \mathbf{A} = \left[\begin{array}{rr} -4 & 3 \\ -2 & -11 \end{array}\right] $$ To do this, we must solve the characteristic equation, given by the determinant of \((\mathbf{A} - \lambda \mathbf{I})\): $$ \operatorname{det}\left(\mathbf{A} - \lambda \mathbf{I}\right) = \left|\begin{array}{cc} -4 - \lambda & 3 \\ -2 & -11 - \lambda \end{array}\right| = (\lambda + 4)(\lambda + 11) - (-2)\times3=0 $$ This simplifies to: $$ \lambda^2 +15\lambda + 34 = 0 $$

Find the eigenvalues

Now we need to find the roots of the quadratic equation we obtained in Step 1: $$ \lambda^2 +15\lambda + 34 = 0 $$ Using the quadratic formula, we get: $$ \lambda = \frac{-15 \pm \sqrt{(-15)^2 - 4(1)(34)}}{2} = -5, -6 $$ So the eigenvalues of the matrix are \(\lambda_1 = -5\) and \(\lambda_2 = -6\).

Calculate the eigenvectors for each eigenvalue

Next, we have to find the eigenvectors corresponding to each eigenvalue. For \(\lambda_1 = -5\), we need to solve the following system of linear equations: $$ (\mathbf{A} - (-5)\mathbf{I})\mathbf{x} = \left[\begin{array}{rr} 1 & 3 \\ -2 & -6 \end{array}\right]\mathbf{x} = \mathbf{0} $$ The row reduced echelon form of this matrix is: $$ \left[\begin{array}{rr} 1 & 3 \\ 0 & 0 \end{array}\right] $$ We can see that the eigenvector \(\mathbf{v}_1\) for \(\lambda_1 = -5\) can be given as: $$ \mathbf{v}_1 = \left[\begin{array}{r} -3 \\ 1 \end{array}\right] $$ For \(\lambda_2 = -6\), we need to solve the following system: $$ (\mathbf{A} - (-6)\mathbf{I})\mathbf{x} = \left[\begin{array}{rr} 2 & 3 \\ -2 & -5 \end{array}\right]\mathbf{x} = \mathbf{0} $$ The row reduced echelon form of this matrix is: $$ \left[\begin{array}{rr} 1 & \frac{3}{2}\\ 0 & 0 \end{array}\right] $$ We can see that the eigenvector \(\mathbf{v}_2\) for \(\lambda_2 = -6\) can be given as: $$ \mathbf{v}_2 = \left[\begin{array}{r} -\frac{3}{2} \\ 1 \end{array}\right] $$

Formulate the general solution

Using the eigenvalues and eigenvectors we found in Steps 2 and 3, we can form the general solution for the given system: $$ \mathbf{y}(t) = c_1 e^{-5 t}\mathbf{v}_1 + c_2 e^{-6 t}\mathbf{v}_2 = c_1 e^{-5 t}\left[\begin{array}{r} -3 \\ 1 \end{array}\right] + c_2 e^{-6 t}\left[\begin{array}{r} -\frac{3}{2} \\ 1 \end{array}\right] $$ Where \(c_1\) and \(c_2\) are constants.

Graph the trajectories

Finally, we can plot the trajectories of the system based on the general solution. The eigenvalues are both negative, so we expect the trajectories to converge to the equilibrium point \((0, 0)\). We can see that there are two eigenvectors, so the phase portrait will have two lines corresponding to these eigenvectors. The plot will show trajectories that approach the origin as time increases. It's important to note that for any initial condition, the trajectory will be a linear combination of the eigenvectors. In summary, the plot will reveal the trajectories of the system are converging to the origin along the lines formed by the eigenvectors. The general solution gives us insights into how the system evolves over time, and the plot helps to visualize this information.

Key concepts

Eigenvalues and Eigenvectors

Understanding eigenvalues and eigenvectors is crucial for solving systems of differential equations. Eigenvalues, denoted by \(\lambda\), are scalars that satisfy the characteristic equation \(\det(\mathbf{A} - \lambda \mathbf{I}) = 0\), where \(\mathbf{A}\) represents the coefficient matrix of the system, and \(\mathbf{I}\) is the identity matrix. Eigenvectors are non-zero vectors that, when multiplied by matrix \(\mathbf{A}\), scale by the corresponding eigenvalue without changing their direction.

Eigenvalues and eigenvectors play a significant role in predicting the behavior of the system's trajectories. For instance, negative eigenvalues suggest that the system's trajectories will converge to the equilibrium point, typically the origin, as time goes on, indicating a stable system. The eigenvectors, on the other hand, provide the direction along which the system's trajectories converge to or diverge from the equilibrium.

Phase Portrait

A phase portrait is a geometric representation of the trajectories of a dynamical system in the phase plane. Each point in the phase plane represents the state of the system at a given time, and the curves drawn are the trajectories that show how these states evolve over time.

In our given system, the phase portrait consists of trajectories that are influenced by the eigenvectors and their corresponding eigenvalues. Negative eigenvalues indicate that all trajectories will move towards the equilibrium state as time progresses. By plotting these trajectories, we can visualize the stability of the system and the long-term behavior of the states.

General Solution of Differential Equations

The general solution of a system of linear differential equations can be expressed as a linear combination of terms that involve eigenvalues and their corresponding eigenvectors. Specifically, for a two-dimensional system like the one in our exercise, the general solution has the form \(\mathbf{y}(t) = c_1 e^{\lambda_1 t}\mathbf{v}_1 + c_2 e^{\lambda_2 t}\mathbf{v}_2\), where \(c_1\) and \(c_2\) are constants that depend on initial conditions of the system. By plugging in different initial conditions, we obtain specific solutions that are part of the phase portrait.

This general solution gives us a comprehensive overview of the possible behaviors of the system over time. Given the eigenvalues are both negative in our example, the exponential terms will decay to zero as time increases, which aligns with the expected convergence towards the origin observed in the phase portrait.

Characteristic Equation

The characteristic equation of a matrix is a polynomial equation derived from the matrix's eigenvalues. It is obtained by setting the determinant of \(\mathbf{A} - \lambda \mathbf{I}\) equal to zero. For a 2x2 matrix, the characteristic equation generally takes the form of a quadratic equation \(\lambda^2 + b\lambda + c = 0\), where \(b\) and \(c\) are coefficients that depend on the elements of the matrix \(\mathbf{A}\).

The roots of the characteristic equation are the eigenvalues of the matrix, and they are fundamentally important in determining the behavior of the linear system. In our case, the characteristic equation \(\lambda^2 +15\lambda + 34 = 0\) results in two distinct negative eigenvalues, which tells us immediately about the system's nature of converging towards the equilibrium.

Linear Algebra

Linear algebra is the branch of mathematics concerning linear equations, linear functions, and their representations through matrices and vector spaces. It's the key foundation for understanding systems of differential equations. In the context of our exercise, linear algebraic concepts like matrices, eigenvectors, and eigenvalues are essential tools used to solve the system of equations and analyze the stability and dynamics of the system.

Eigenvectors are found by solving linear systems produced after substituting each eigenvalue into \(\mathbf{A} - \lambda \mathbf{I})\mathbf{x} = \mathbf{0}\). The row reduced echelon form helps solve these systems efficiently. The applicability of linear algebra in this context underscores its importance for students aiming to understand and solve real-world problems modeled by differential equations.