Problem 8 - Advanced Engineering Mathematics

Question

The general solution of the linear system \(\mathbf{X}^{\prime}=\mathbf{A X}\) is given. (a) In each case discuss the nature of the solution in a neighborhood of \((0,0)\). (b) With the aid of a graphing utility plot the solution that satisfies \(\mathbf{X}(0)=(1,1)\). $$ \begin{aligned} &\mathbf{A}=\left(\begin{array}{ll} -1 & 5 \\ -1 & 1 \end{array}\right) \\ &\mathbf{X}(t)=c_{1}\left(\begin{array}{c} 5 \cos 2 t \\ \cos 2 t-2 \sin 2 t \end{array}\right)+c_{2}\left(\begin{array}{c} 5 \sin 2 t \\ 2 \cos 2 t+\sin 2 t \end{array}\right) \end{aligned} $$

Step-by-Step Solution

Answer

The solution exhibits oscillatory behavior around the origin due to the periodic sine and cosine terms.

Step 1: Determine Eigenvalues and Eigenvectors

For the matrix \( \mathbf{A} = \begin{pmatrix} -1 & 5 \ -1 & 1 \end{pmatrix} \), we need to find its eigenvalues by solving the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). Calculating the determinant gives: \( \det\begin{pmatrix} -1 - \lambda & 5 \ -1 & 1 - \lambda \end{pmatrix} = (\lambda + 1)(\lambda - 1) - 5 = \lambda^2 - 1 - 5 = \lambda^2 - 6 \). Solving \( \lambda^2 - 6 = 0 \), we find the eigenvalues \( \lambda = \pm \sqrt{6} \). Corresponding eigenvectors can be found using \( (\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = \mathbf{0} \).

Step 2: Analyze Nature of Solutions Near the Origin

Given the general solution: \[ \mathbf{X}(t) = c_1 \begin{pmatrix} 5 \cos 2t \ \cos 2t - 2 \sin 2t \end{pmatrix} + c_2 \begin{pmatrix} 5 \sin 2t \ 2 \cos 2t + \sin 2t \end{pmatrix} \], the presence of terms \( \cos 2t \) and \( \sin 2t \) implies that the solutions rotate around the origin. This solution suggests a center or spiral, as the cosine and sine functions repeat periodically without growing or decaying.

Step 3: Apply Initial Condition

Substitute the initial condition \( \mathbf{X}(0) = (1,1) \) into the general solution to find constants \( c_1 \) and \( c_2 \): \[ \begin{pmatrix} 1 \ 1 \end{pmatrix} = c_1 \begin{pmatrix} 5 \ 1 \end{pmatrix} + c_2 \begin{pmatrix} 0 \ 2 \end{pmatrix} \]. From this system, solve for \( c_1 \) and \( c_2 \) by comparison of components: \( 5c_1 = 1 \) giving \( c_1 = \frac{1}{5} \), and \( c_1 + 2c_2 = 1 \) giving \( c_2 = \frac{2}{5} \).

Step 4: Construct Specific Solution

With \( c_1 = \frac{1}{5} \) and \( c_2 = \frac{2}{5} \), the specific solution becomes: \[ \mathbf{X}(t) = \frac{1}{5} \begin{pmatrix} 5 \cos 2t \ \cos 2t - 2 \sin 2t \end{pmatrix} + \frac{2}{5} \begin{pmatrix} 5 \sin 2t \ 2 \cos 2t + \sin 2t \end{pmatrix} \]. Simplify this to express \( \mathbf{X}(t) \) fully in terms of \( t \): \( \mathbf{X}(t) = \begin{pmatrix} \cos 2t + 2 \sin 2t \ \frac{1}{5}(\cos 2t - 2 \sin 2t) + \frac{2}{5}(2 \cos 2t + \sin 2t) \end{pmatrix} \).

Step 5: Visualize the Solution

On a graphing utility, plot the solution \( \mathbf{X}(t) = \begin{pmatrix} \cos 2t + 2 \sin 2t \ \frac{3}{5} \cos 2t + \frac{4}{5} \sin 2t \end{pmatrix} \). This is a parameterized curve in the plane, which exhibits oscillatory behavior as it revolves around the origin over time. Analyze the nature of this path to confirm the behavior discussed in the previous steps.

Key Concepts

Eigenstates

Eigenstates are fundamental concepts in linear algebra and quantum mechanics. They are the state vectors associated with distinct eigenvalues of a linear transformation represented by a matrix. In the context of differential equations, eigenstates help us understand how solutions evolve over time.
To find eigenstates, we begin with the characteristic equation of the matrix \( \mathbf{A} \), \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \), which gives the eigenvalues. Each eigenvalue corresponds to a distinct eigenstate, or eigenvector, that satisfies the equation \( (\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = \mathbf{0} \).

• In our exercise, the eigenvalues determined are \( \lambda = \pm \sqrt{6} \), signaling the nature of the system's solutions.
• The eigenvectors corresponding to these values provide directions in the solution space along which the system behaves in a predictable manner, either decaying, growing, or oscillating.

General Solution

The general solution of a linear differential system combines all possible behaviors characterized by the eigenvalues and eigenvectors. It describes how any initial condition evolves over time.
For our linear system, the general solution is given as: \[ \mathbf{X}(t) = c_1 \begin{pmatrix} 5 \cos 2t \ \cos 2t - 2 \sin 2t \end{pmatrix} + c_2 \begin{pmatrix} 5 \sin 2t \ 2 \cos 2t + \sin 2t \end{pmatrix} \] Here, the constants \( c_1 \) and \( c_2 \) are determined by initial conditions.

• This expression shows how the system behaves as time \( t \) progresses, indicating combinations of rotations and potential translations in the solution space.
• Each component in the sum corresponds to an eigenstate-derived behavior.

Oscillatory Behavior

Oscillatory behavior in systems of differential equations is characterized by solutions that revolve around a point, usually the origin, without converging or diverging. The presence of sine and cosine functions in the general solution is indicative of such behavior.

This is particularly evident in the trigonometric functions \( \cos 2t \) and \( \sin 2t \) in our solution's expression. These terms lead to a repetitive motion, suggesting that the solution pivots around the origin.

• Instead of fixed or diverging paths, these functions ensure that the trajectory of the solution is cyclical.
• The coefficients in front of the trigonometric functions influence the amplitude and frequency of the oscillations, yet the origin remains the center of motion.

Initial Value Problem

An initial value problem requires solving a differential equation subject to an initial condition at a specific point. This helps to find the particular solution that passes through given initial values.
In our example, we have the initial condition \( \mathbf{X}(0) = (1,1) \). By substituting \( t = 0 \) in the general solution and using this condition, we extract the specific constants \( c_1 \) and \( c_2 \) that mold the general solution to meet these initial requirements.

• The solution starts by setting up the expression \( \begin{pmatrix} 1 \ 1 \end{pmatrix} = \frac{1}{5} \begin{pmatrix} 5 \ 1 \end{pmatrix} + \frac{2}{5} \begin{pmatrix} 0 \ 2 \end{pmatrix} \), determining that \( c_1 = \frac{1}{5} \) and \( c_2 = \frac{2}{5} \).
• These constants specify the unique path the system follows, driven by its initial position in the plane.