Question

Describe and graph trajectories of the given system. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} -1 & -10 \\ -5 & 4 \end{array}\right] \mathbf{y} $$

Short answer

Question: Describe and graph the trajectories for the system of linear differential equations given in matrix form: $$\frac{d\mathbf{y}}{dt}=A\mathbf{y}$$, where $$A = \left[\begin{array}{rr} -1 & -10 \\ -5 & 4 \end{array}\right].$$ Answer: The eigenvalues of the matrix A are -6 and 7. The corresponding eigenvectors are [2, 1] and [-5, -3]. The general solution of the system is given by: $$\mathbf{y}(t) = c_1 \mathrm{e}^{-6t} \begin{bmatrix}2 \\ 1\end{bmatrix} + c_2 \mathrm{e}^{7t} \begin{bmatrix}-5 \\ -3\end{bmatrix},$$ where \(c_1\) and \(c_2\) are arbitrary constants. The trajectories can be plotted using a numerical method with the phase portrait showing the eigenvectors' directions. Near eigenvector [2, 1], trajectories tend to flow away from the origin, while near eigenvector [-5, -3], trajectories tend to flow towards the origin.

Step-by-step solution

Find the eigenvalues and eigenvectors of the given system matrix

First, we'll find the eigenvalues and eigenvectors for the given matrix: $$ A = \left[\begin{array}{rr} -1 & -10 \\ -5 & 4 \end{array}\right] $$ To find the eigenvalues, we'll solve the characteristic equation of the system, det\((A-\lambda I) = 0\): $$ \mathrm{det} \left(\left[\begin{array}{rr} -1-\lambda & -10 \\ -5 & 4-\lambda \end{array}\right] \right) = (\lambda+1)(\lambda-4) + 50 = \lambda^2 -3\lambda - 46 = (\lambda+6)(\lambda-7) $$ The eigenvalues are \(\lambda_1 = -6\) and \(\lambda_2 = 7\). Next, we'll find the eigenvectors for each eigenvalue: For \(\lambda_1 = -6\): $$ (A+6I)\mathbf{v_1} = \left[\begin{array}{rr} 5 & -10 \\ -5 & 10 \end{array}\right] \begin{bmatrix} v_{11}\\ v_{12} \end{bmatrix} = 0 $$ Let \(v_{11} = 2\), then \(v_{12} = 1\). So, \(\mathbf{v_1} = \begin{bmatrix}2 \\ 1\end{bmatrix}\). For \(\lambda_2 = 7\): $$ (A-7I)\mathbf{v_2} = \left[\begin{array}{rr} -8 & -10 \\ -5 & -3 \end{array}\right] \begin{bmatrix} v_{21}\\ v_{22} \end{bmatrix} = 0 $$ Let \(v_{21} = -5\), then \(v_{22} = -3\). So, \(\mathbf{v_2} = \begin{bmatrix}-5 \\ -3\end{bmatrix}\).

Construct the general solution

Using the eigenvalues (i.e., \(\lambda_1 = -6\), \(\lambda_2 = 7\)) and eigenvectors (i.e., \(\mathbf{v_1} = \begin{bmatrix}2 \\ 1\end{bmatrix}\), \(\mathbf{v_2} = \begin{bmatrix}-5 \\ -3\end{bmatrix}\)), we'll construct the general solution to the given system of linear differential equations: $$ \mathbf{y}(t) = c_1 \mathrm{e}^{\lambda_1 t} \mathbf{v_1} + c_2 \mathrm{e}^{\lambda_2 t} \mathbf{v_2} = c_1 \mathrm{e}^{-6t} \begin{bmatrix}2 \\ 1\end{bmatrix} + c_2 \mathrm{e}^{7t} \begin{bmatrix}-5 \\ -3\end{bmatrix} $$ Note that \(c_1\) and \(c_2\) are arbitrary constants.

Plot the trajectories

To graph the trajectories of the given system, we'll need to use a numerical method (like Euler's method) to plot the phase portrait. We'll also plot the eigenvectors as they lie on the trajectories. The eigenvectors show the direction in which the trajectories move. The principal trajectories are those along with the eigenvectors, which when extended would be: $$ \mathbf{y}(t) = k_1 \begin{bmatrix}2 \\ 1\end{bmatrix} $$ $$ \mathbf{y}(t) = k_2 \begin{bmatrix}-5 \\ -3\end{bmatrix} $$ Where \(k_1\) and \(k_2\) are arbitrary constants. Using a mathematical software package or plotting tool, plot the phase portrait of the given system, along with the eigenvectors. The trajectories extend in the directions of the eigenvectors. For the provided example, the trajectories near \(\mathbf{v_1}\) tend to flow away from the origin due to the negative eigenvalue, and near \(\mathbf{v_2}\) will flow towards the origin due to the positive eigenvalue.

Key concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors play a crucial role in understanding systems of linear differential equations. An eigenvalue is a scalar that indicates how much the eigenvector is scaled during a linear transformation. To find eigenvalues, we look at the matrix representing our system and solve the characteristic equation, which is derived by setting the determinant of the matrix minus a scalar times the identity matrix to zero.

Once we have the eigenvalues, we find the corresponding eigenvectors. These are non-zero vectors that change by only a scalar factor when that linear transformation is applied. For the system in the exercise, the eigenvalues are found to be \(\lambda_1 = -6\) and \(\lambda_2 = 7\). The associated eigenvectors, which are vectors that point in the direction of the trajectories of our system, are determined through further matrix calculations. They give us a direct visualization of the system's behavior at different points and are essential for graphing the system's trajectories.

Phase Portrait

A phase portrait is a geometric representation of the trajectories of a dynamical system in the phase plane. Each trajectory represents the state of the system at a given time, with different initial conditions producing different trajectories. When plotting a phase portrait for a system of linear differential equations, the eigenvectors provide the direction and the eigenvalues dictate the nature of the trajectories.

Eigenvectors corresponding to negative eigenvalues will see their trajectories move away from the origin and eigenvectors with positive eigenvalues will have trajectories that move toward the origin. The phase portrait provides a visual summary of the system's behavior over time. By looking at the phase portrait of the system given in the exercise, we can predict the stability and type of equilibrium of the system based on the nature of the trajectories and their relation to eigenvectors.

Characteristic Equation

The characteristic equation of a matrix is a fundamental tool for determining the eigenvalues of the system. It is formed by taking the determinant of the matrix after subtracting a scalar multiple of the identity matrix. We put this determinant equal to zero: \( \det(A - \lambda I) = 0 \).

This resultant polynomial equation in terms of \( \lambda \) is what we call the characteristic equation, and solving for \( \lambda \) gives us the system's eigenvalues. In the provided exercise, solving the characteristic equation \( \lambda^2 - 3\lambda - 46 = 0 \) resulted in two distinct eigenvalues, indicating distinct behaviors along each associated eigenvector in the system's phase portrait.

Numerical Methods for Differential Equations

In practice, many systems of differential equations cannot be solved analytically, and numerical methods are employed to obtain approximate solutions. Methods such as Euler's method, Runge-Kutta, and others allow for the approximation of solutions by discretizing the equations and evaluating them at sequential points.

These methods can be used to plot the trajectories of a system's phase portrait by iteratively calculating the state of the system at a series of time steps. In the exercise, a numerical method would be necessary to graph the trajectories as analytical solutions are not readily available for all initial conditions. By implementing numerical techniques, students can visualize the dynamic behaviors of the system and understand how the state evolves over time.