Question

In each of Problems 23 and 24 ; (a) Find the eigenvalues of the given system. (b) Choose an initial point (other than the origin) and draw the corresponding trajectory in the \(x_{1} x_{2}\) -plane. Also draw the trajectories in the \(x_{1} x_{1}-\) and \(x_{2} x_{3}-\) planes. (c) For the initial point in part (b) draw the corresponding trajectory in \(x_{1} x_{2} x_{3}\) -space. $$ \mathbf{x}^{\prime}=\left(\begin{array}{ccc}{-\frac{1}{4}} & {1} & {0} \\\ {-1} & {-\frac{1}{4}} & {0} \\ {0} & {0} & {-\frac{1}{4}}\end{array}\right) \mathbf{x} $$

Short answer

**Short Answer:** Eigenvalues of the given matrix are \(\lambda = -\frac{1}{4}\) with an algebraic multiplicity of 3. The trajectory of the system in the \(x_1x_2\)-plane, \(x_1x_1-\) and \(x_2x_3-\) planes, and in the \(x_1x_2x_3\)-space can be computed using numerical methods or software starting from the initial point (1, 1, 0).

Step-by-step solution

Calculate the eigenvalues

To find the eigenvalues of the given matrix, we first need to compute its characteristic polynomial. This is done by finding the determinant of the matrix subtracted by the identity matrix multiplied by the eigenvalue (\(\lambda\)). The matrix we have is given as: $$ A = \left(\begin{array}{ccc}{-\frac{1}{4}} & {1} & {0} \\\ {-1} & {-\frac{1}{4}} & {0} \\\ {0} & {0} & {-\frac{1}{4}}\end{array}\right) $$ So, for the characteristic polynomial, we have: $$ \det(A - \lambda I) = \left|\begin{array}{ccc}{-\frac{1}{4} - \lambda} & {1} & {0} \\\ {-1} & {-\frac{1}{4} - \lambda} & {0} \\\ {0} & {0} & {-\frac{1}{4} - \lambda}\end{array}\right| $$ Calculating the determinant, we get: $$ (-\frac{1}{4} - \lambda) \left((-\frac{1}{4} - \lambda) (-\frac{1}{4}- \lambda)\right) - 1(-1)(0) $$ $$ (-\frac{1}{4} - \lambda)^3 $$ Setting this equal to zero and finding the roots of the polynomial, we get \(\lambda = -\frac{1}{4}\). Since all the roots are the same, this eigenvalue has algebraic multiplicity 3.

Choose an initial point and draw the corresponding trajectory in the \(x_1x_2\)-plane

Let us choose an initial point in the \(x_1x_2\)-plane, say \((x_1, x_2) = (1, 1)\) and \(x_3 = 0\). Then, we can write the given system of differential equations as: $$ \begin{aligned} x_1' &= -\frac{1}{4}x_1 + x_2 \\ x_2' &= -x_1 -\frac{1}{4}x_2 \\ x_3' &= -\frac{1}{4}x_3 \end{aligned} $$ We can use numerical methods or software to solve these equations (starting from the initial point) and plot the trajectory of the system in the \(x_1x_2\)-plane.

Draw the trajectories in the \(x_1x_1-\) and \(x_2x_3-\) planes

We now want to find the trajectories in the \(x_1x_1-\) and \(x_2x_3-\) planes. To do this, we can use the same set of equations and set \(x_2 = 0\) for the \(x_1x_1-\)plane equation, and \(x_1 = 0\) for the \(x_2x_3-\)plane equation. We can then use numerical methods or software to solve these equations and plot the points in the required plane.

Draw the trajectory in the \(x_1x_2x_3\)-space

For this step, we use the initial point chosen in part (b), which is \((x_1, x_2, x_3) = (1, 1, 0)\), and solve the given set of equations with this initial point. We can then plot the trajectory of the system in the \(x_1x_2x_3\)-space using numerical methods or visualization software.

Key concepts

Characteristic Polynomial

The characteristic polynomial is a fundamental concept in linear algebra, particularly when analyzing linear systems of differential equations like in our exercise. It is derived from the matrix that represents the system and encapsulates vital information about its behavior, especially the eigenvalues.

To find the characteristic polynomial of a matrix, we calculate the determinant of the matrix after subtracting the product of an eigenvalue and the identity matrix, which is denoted as \(A - \lambda I\). The characteristic polynomial for the given matrix \( A \) in our exercise is found by setting up:
\[\det(A - \lambda I) = \left|\begin{array}{ccc}{-\frac{1}{4} - \lambda} & {1} & {0} \ {-1} & {-\frac{1}{4} - \lambda} & {0} \ {0} & {0} & {-\frac{1}{4} - \lambda}\end{array}\right|\]
which simplifies to \( (-\frac{1}{4} - \lambda)^3 \), indicating all eigenvalues are identical, \( \lambda = -\frac{1}{4} \). Knowing these eigenvalues is crucial as they dictate the behavior of solutions to the differential equations system.

Differential Equations

Differential equations describe how a particular quantity changes over time and are widely used in physics, engineering, economics, biology, and many other fields. In our problem, we are given a system of linear differential equations that describe the evolution of state variables \( x_1, x_2, \text{and} x_3 \).

To solve the system, we often start by finding the eigenvalues, which then inform us about the solution's behavior. If an eigenvalue is negative, the corresponding system's component decays over time; if it's positive, it grows. The system in our exercise is:
\[\begin{aligned} x_1' &= -\frac{1}{4}x_1 + x_2 \ x_2' &= -x_1 -\frac{1}{4}x_2 \ x_3' &= -\frac{1}{4}x_3\end{aligned}\]
The solutions of the system can be complex and depend on the initial conditions, as we see when we select the initial point (1, 1, 0) and note the interplay between \( x_1' \) and \( x_2' \), which affects the trajectory in phase space.

Trajectory in Phase Space

A trajectory in phase space offers a complete description of the system's state over time. It shows us where the system starts and how it evolves according to the differential equations governing its behavior. In the context of our problem, trajectories in different planes (\(x_1x_2\)-plane, \(x_1x_1\)-plane, and \(x_2x_3\)-plane) highlight various aspects of the system's dynamics.

For instance, plotting the trajectory in the \(x_1x_2\)-plane after setting \(x_3 = 0\) and using our initial point (1, 1) helps visualize how the system evolves in a two-dimensional space, ignoring the \(x_3\) component. In contrast, for the full \(x_1x_2x_3\)-space, we consider all three variables, providing a three-dimensional view of the system's state over time. While the exercise simplifies by using zero values for certain variables in some planes, it’s vital to remember that the complete dynamics are captured only in the full three-dimensional phase space, which requires solving the entire system of differential equations with the given initial conditions.