Problem 2 - Advanced Engineering Mathematics

Question

The general solution of the linear system \(\mathbf{X}^{\prime}=\mathbf{A} \mathbf{X}\) 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 \(\mathbf{X}(0)=(1,1)\) $$ \mathbf{A}=\left(\begin{array}{rr} -1 & -2 \\ 3 & 4 \end{array}\right), \quad \mathbf{X}(t)=c_{1}\left(\begin{array}{r} 1 \\ -1 \end{array}\right) e^{t}+c_{2}\left(\begin{array}{r} -4 \\ 6 \end{array}\right) e^{2 t} $$

Step-by-Step Solution

Answer

The solution is unstable and diverges from (0,0). The graph shows rapid growth.

Step 1: Analyze the Eigenvalues

To determine the nature of the solution near (0, 0), we begin by finding the eigenvalues of the matrix \( \mathbf{A} = \begin{pmatrix} -1 & -2 \ 3 & 4 \end{pmatrix} \). The characteristic equation is \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). Calculating the determinant gives: \( \lambda^2 - (3)\lambda + 2 = 0 \). The roots of this quadratic are \( \lambda_1 = 1 \) and \( \lambda_2 = 2 \). These positive eigenvalues indicate that the solution exhibits unbounded growth (instability) near (0, 0).

Step 2: Nature of the Solution

The general solution \( \mathbf{X}(t) = c_1\begin{pmatrix} 1 \ -1 \end{pmatrix} e^t + c_2\begin{pmatrix} -4 \ 6 \end{pmatrix} e^{2t} \) shows two terms. The term with \( e^{2t} \), associated with the larger eigenvalue \( \lambda_2 = 2 \), grows faster and dominates as \( t \rightarrow \infty \). Solutions diverge exponentially from any equilibrium near (0, 0), confirming instability.

Step 3: Initial Condition Application

Apply the initial condition \( \mathbf{X}(0) = (1, 1) \) to find \( c_1 \) and \( c_2 \). Substituting \( t = 0 \), we get \( c_1\begin{pmatrix} 1 \ -1 \end{pmatrix} + c_2\begin{pmatrix} -4 \ 6 \end{pmatrix} = \begin{pmatrix} 1 \ 1 \end{pmatrix} \). Solving the system \( c_1 - 4c_2 = 1 \) and \( -c_1 + 6c_2 = 1 \), we find \( c_1 = 5 \) and \( c_2 = 1 \). Thus, the specific solution is \( \mathbf{X}(t) = 5\begin{pmatrix} 1 \ -1 \end{pmatrix} e^t + \begin{pmatrix} -4 \ 6 \end{pmatrix} e^{2t} \).

Step 4: Graph the Solution

Using a graphing utility or software, plot the trajectory of the solution \( \mathbf{X}(t) = 5\begin{pmatrix} 1 \ -1 \end{pmatrix} e^t + \begin{pmatrix} -4 \ 6 \end{pmatrix} e^{2t} \) starting at the point (1, 1). As the graph is constructed, observe the rapid divergence from the origin due to the dominance of the \( e^{2t} \) term, reflecting the exponential growth characteristic.

Key Concepts

Eigenvalues

Eigenvalues play a crucial role in understanding the behavior of linear differential equations. In this context, eigenvalues are derived from the coefficients in a system of differential equations represented by a matrix. The matrix \( \mathbf{A} \), given in our exercise as \( \begin{pmatrix} -1 & -2 \ 3 & 4 \end{pmatrix} \), serves as the starting line to find these eigenvalues.
To determine the eigenvalues, we use the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \), which translates here to a quadratic equation. Solving this, we find \( \lambda_1 = 1 \) and \( \lambda_2 = 2 \).
The meanings of these eigenvalues are essential:

• If both eigenvalues are positive, as in this exercise, the system is unstable.
• The solution grows away from the equilibrium point \((0,0)\).

This provides insight into the dynamics of the system described by the differential equation. Understanding eigenvalues helps predict whether solutions converge or diverge as time advances.

System Stability

System stability refers to how solutions of a differential equation behave over time. For the given system \( \mathbf{X}' = \mathbf{A} \mathbf{X} \), stability is primarily determined by the signs of the eigenvalues of matrix \( \mathbf{A} \).
In our case, both eigenvalues are positive \((\lambda_1 = 1, \lambda_2 = 2)\), indicating that the system is unstable. This means solutions will not remain close to the equilibrium point \((0,0)\).
In practical terms:

• A stable system would see trajectories that stay near, or return to, the equilibrium as time progresses.
• An unstable system, like ours, will see solutions moving further from equilibrium, indicating divergent behavior.

Graphing such a system, students will observe how solutions, influenced by the faster-growing exponential term, tend to drift away from the equilibrium point. This visual representation reinforces the concept of instability.

Exponential Growth

Exponential growth in differential equations refers to solutions that rise rapidly over time. In our exercise \( \mathbf{X}(t) = c_1 \begin{pmatrix} 1 \ -1 \end{pmatrix} e^t + c_2 \begin{pmatrix} -4 \ 6 \end{pmatrix} e^{2t} \), one exponential term \( e^{2t} \) is dominant.
This occurs because its associated eigenvalue \( \lambda_2 = 2 \) is larger than the other one. As \( t \rightarrow \infty \), the solution primarily behaves like the \( e^{2t} \) component.
Key points to remember:

• Large eigenvalues lead to faster growing terms in the solution.
• The larger the eigenvalue, the more rapidly the associated solution will grow.

Especially in systems of linear differential equations with positive eigenvalues, this exponential growth signifies that the solution trajectory will diverge from the equilibrium point over time. Teaching students to identify which exponential terms will dominate can help them understand the system's long-term behavior.