Question

The matrices of the systems in exercises are singular. Describe and graph the trajectories of nonconstant solutions of the given systems. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 3 & -1 \\ -3 & 1 \end{array}\right] \mathbf{y} $$

Short answer

Answer: The trajectories approach the origin in the direction of the eigenvector (1, 3) and move away from the origin in the direction of the eigenvector (1, -1).

Step-by-step solution

Analyze the matrix

First, we must analyze the given matrix: $$ A = \left[\begin{array}{rr} 3 & -1 \\\ -3 & 1 \end{array}\right] $$ We can notice that the determinant of this matrix is 0 (since \(3*1 - (-1)*(-3) = 0\)), which means it is singular.

Rewrite the system as a set of differential equations

Next, we rewrite the system of first-order linear differential equations: $$ \begin{cases} \frac{dy_1}{dt} = 3y_1 - y_2, \\ \frac{dy_2}{dt} = -3y_1 + y_2. \end{cases} $$

Solve the differential equations

There are various methods to solve this system of differential equations. In this case, we can use the eigenvalue and eigenvector methods. Start by finding the eigenvalues and eigenvectors of the matrix A: $$ \det(A-\lambda I) = \left|\begin{array}{cc} 3-\lambda & -1 \\\ -3 & 1-\lambda \end{array}\right| = (3-\lambda)(1-\lambda) - (-1)*(-3) = \lambda^2 - 4\lambda $$ The eigenvalues are \(\lambda_1 = 0\) and \(\lambda_2 = 4\). Now, we find the eigenvectors associated with each eigenvalue: For \(\lambda_1 = 0\), we have \((A - 0I) \mathbf{v}_1 = 0\): $$ \left[\begin{array}{cc} 3 & -1\\ -3 & 1 \end{array}\right] \mathbf{v}_1=0 $$ We obtain the eigenvector \(\mathbf{v}_1 = \begin{bmatrix} 1 \\ 3 \end{bmatrix}\) For \(\lambda_2 = 4\), we have \((A - 4I) \mathbf{v}_2 = 0\): $$ \left[\begin{array}{cc} -1 & -1\\ -3 & -3 \end{array}\right] \mathbf{v}_2 =0 $$ We obtain the eigenvector \(\mathbf{v}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\)

Write the general solution

With the eigenvalues and eigenvectors found in Step 3, the general solution of the system of differential equations is: $$ \mathbf{y}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2 $$ Substituting the eigenvalues and eigenvectors into the general solution, we have: $$ \mathbf{y}(t) = c_1 e^{0t} \begin{bmatrix} 1 \\ 3 \end{bmatrix} + c_2 e^{4t} \begin{bmatrix} 1 \\ -1 \end{bmatrix} $$

Describe the trajectories

The trajectories of the general solution found in Step 4 can be analyzed in the \(y_1,y_2\) plane. For the first eigenvector \(\mathbf{v}_1\), it corresponds to the null eigenvalue case (as it is a singular matrix). In this case, all trajectories emanate from or converge to the origin along the direction of \(\mathbf{v}_1\). For the second eigenvector \(\mathbf{v}_2\), it corresponds to a positive eigenvalue \(\lambda_2 = 4\). This means that trajectories will grow exponentially in the direction of \(\mathbf{v}_2\) when moving away from the origin. Consequently, the trajectories of nonconstant solutions will move away from the origin in the direction of \(\mathbf{v}_2\) and approach the origin in the direction of \(\mathbf{v}_1\).

Graph the trajectories

To graph the trajectories, we need to plot the eigenvectors on the \(y_1,y_2\) plane. The eigenvector \(\mathbf{v}_1=\begin{bmatrix} 1 \\ 3 \end{bmatrix}\) determines the direction from which trajectories approach the origin, while the eigenvector \(\mathbf{v}_2=\begin{bmatrix} 1 \\ -1 \end{bmatrix}\) determines the direction of the growing exponential trajectories. As a result, the trajectories form a set of curves moving away from the origin in the direction of \(\mathbf{v}_2\) (1, -1) and approaching the origin in the direction of \(\mathbf{v}_1\) (1, 3).

Key concepts

Singular Matrix

A singular matrix is a square matrix that does not have an inverse. In more technical terms, a matrix is singular if its determinant is zero. This concept is crucial when dealing with systems of linear equations and differential equations because it implies that the system may not have a unique solution. For example, in the given exercise, the matrix \( A \) defined by \[ A = \left[\begin{array}{rr} 3 & -1 \ -3 & 1 \end{array}\right] \] has a determinant of zero, which signifies that it's a singular matrix.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental concepts in linear algebra, associated with matrices. An eigenvalue is a special scalar that, when multiplied by an associated eigenvector, leaves the vector unchanged after it has been transformed by a matrix. Mathematically, for a given matrix \( A \), if \( \lambda \) is an eigenvalue, then \( A\mathbf{v} = \lambda \mathbf{v} \), where \( \mathbf{v} \) is the corresponding eigenvector. They play a vital role in the stability analysis of systems of differential equations. In the above system, the eigenvalues \( \lambda_1 = 0 \) and \( \lambda_2 = 4 \) have profound implications on the behavior of the system's solutions over time.

System of Differential Equations

A system of differential equations comprises several equations that relate the rates of change of variables to each other. For the exercises provided, the system is a pair of first-order linear differential equations. These equations describe how two variables, \( y_1 \) and \( y_2 \) in our case, change with respect to an independent variable, often time. To solve these systems, one may employ several methods, including matrix methods involving eigenvalues and eigenvectors. Solving these systems helps to predict the trajectory of each variable over time and understand their dynamic interactions.

Phase Plane Analysis

Phase plane analysis is a visual method to analyze and understand the behavior of systems of first-order differential equations in two variables. By plotting the variables against each other, one can observe the trajectories or paths that solutions to the system take over time. These plots help identify the stability of equilibrium points and visualize the overall behavior of the system such as cycles, spirals, or lines, and in the case of the given exercise, trajectories that approach or move away from the origin based on the eigenvectors. The provided solution shows that the system has two types of trajectories corresponding to the null and positive eigenvalues obtained from the matrix.