Problem 23 - Advanced Engineering Mathematics

Question

A nonhomogeneous linear system \(\mathbf{X}^{\prime}=\mathbf{A X}+\mathbf{F}\) is given. (a) In each case determine the unique critical point \(\mathbf{X}_{1}\). (b) Use a numerical solver to determine the nature of the critical point in part (a). (c) Investigate the relationship between \(\mathbf{X}_{1}\) and the critical point \((0,0)\) of the homogeneous linear system \(\mathbf{X}^{\prime}=\mathbf{A X}\). $$ \begin{aligned} &x^{\prime}=2 x+3 y-6 \\ &y^{\prime}=-x-2 y+5 \end{aligned} $$

Step-by-Step Solution

Answer

The critical point is \((-3, 4)\) and is a saddle point, like \((0,0)\).

Step 1: Write the System in Matrix Form

Let's express the given system of differential equations in matrix form. The system is:\(\begin{aligned} x^{\prime} &= 2x + 3y - 6, \ y^{\prime} &= -x - 2y + 5 \end{aligned}\)This can be written as\(\mathbf{X}^{\prime} = \begin{bmatrix} x^{\prime} \ y^{\prime} \end{bmatrix} = \begin{bmatrix} 2 & 3 \ -1 & -2 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} + \begin{bmatrix} -6 \ 5 \end{bmatrix}\)where \(\mathbf{A} = \begin{bmatrix} 2 & 3 \ -1 & -2 \end{bmatrix}\) and \(\mathbf{F} = \begin{bmatrix} -6 \ 5 \end{bmatrix}\).

Step 2: Find the Critical Point \(\mathbf{X}_{1}\)

The critical point \(\mathbf{X}_{1} = (x_1, y_1)\) occurs where \(\mathbf{X}^{\prime}=0\).Solving \(\begin{bmatrix} 2 & 3 \ -1 & -2 \end{bmatrix} \begin{bmatrix} x_1 \ y_1 \end{bmatrix} + \begin{bmatrix} -6 \ 5 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\), we get:1. \(2x_1 + 3y_1 - 6 = 0\)2. \(-x_1 - 2y_1 + 5 = 0\)Solving the first equation for \(x_1\): \(x_1 = 3 - \frac{3}{2}y_1\). Substitute \(x_1\) in the second equation:\(-3 + \frac{3}{2}y_1 - 2y_1 + 5 = 0\),\(-3 + \frac{3}{2}y_1 - \frac{4}{2}y_1 + 5 = 0\),\(\frac{-1}{2}y_1 + 2 = 0\).So, \(y_1 = 4\).Substitute \(y_1 = 4\) back to find \(x_1\): \(x_1 = 3 - \frac{3}{2}(4) = -3\).Thus, the critical point \(\mathbf{X}_{1} = (-3, 4)\).

Step 3: Determine the Nature of the Critical Point

To determine the nature of the critical point, calculate the eigenvalues of the matrix \(\mathbf{A}\).The characteristic polynomial is given by \(\det(\mathbf{A} - \lambda \mathbf{I}) = 0\):\(\begin{vmatrix} 2-\lambda & 3 \ -1 & -2-\lambda \end{vmatrix} = (2-\lambda)(-2-\lambda) + 3\).Solving this for \(\lambda\) gives us the eigenvalues.\(\lambda^2 - 4 = 0\) \implies \(\lambda = \pm 2\).Since the eigenvalues are real and distinct (one positive and one negative), the critical point \((-3, 4)\) is a saddle point.

Step 4: Investigate the Relationship with \((0,0)\)

The homogeneous system \(\mathbf{X}^{\prime} = \mathbf{AX}\) has a critical point at \((0,0)\). Its nature, given the matrix \(\mathbf{A}\), is also a saddle point since the eigenvalues remain the same.The critical points \((-3, 4)\) and \((0,0)\) have the same nature (saddle points). The addition of \(\mathbf{F}\) shifts the critical point from the origin to \((-3, 4)\), but does not alter the nature of the critical point.

Key Concepts

Critical Points in Nonhomogeneous Systems

Critical points are special points on a graph where the derivatives of function components zero out. In the context of a nonhomogeneous linear system, a critical point occurs when the derivatives \(\mathbf{X}'\) disappear, resulting in a state of equilibrium. Here's how to identify a critical point in a nonhomogeneous system, given by \(\mathbf{X}'=\mathbf{AX}+\mathbf{F}\).

• Solve the equation \(\mathbf{AX}+\mathbf{F}=0\) to find \(\mathbf{X}\), which represents the critical point, denoted as \(\mathbf{X}_1 = (x_1, y_1)\).
• In this case, the critical point was solved to be \(-3, 4\).

Critical points are crucial for analyzing the system's behavior over time, as they help determine stability and predict long-term behavior.

Eigenvalues and their role in System Dynamics

Eigenvalues are numbers that provide key insights into the dynamics of a system. They derive from the characteristic equation produced by a matrix (in this case, matrix \(\mathbf{A}\)), which characterizes the linear system.

• To find them, solve the equation \(\det(\mathbf{A} - \lambda \mathbf{I}) = 0\) where \(\lambda\) represents the eigenvalues.
• For the matrix \(\mathbf{A}\) in our example, the eigenvalues were found to be \(+2\) and \(-2\).

These values indicate how the system evolves over time from any given point, especially in response to initial fluctuations or disturbances. Real eigenvalues with differing signs, like ours, point to a system with a saddle point character.

Understanding Saddle Points

A saddle point in the context of linear systems is a critical point that is neither entirely stable nor unstable. It occurs when eigenvalues include both positive and negative values, suggesting different directions of stability.

• Saddle points indicate that some trajectories approach the critical point, while others diverge away from it.
• Because one eigenvalue is positive and the other negative in our system, \(-3, 4\) is classified as a saddle point.

This interesting behavior implies that while part of the system may stabilize, other parts may shift away, leading to potential instability depending on initial conditions.

The Concept of Homogeneous Systems

In contrast to nonhomogeneous systems, homogeneous systems assume the absence of constant forcing terms, represented solely by \(\mathbf{X}'=\mathbf{AX}\). They are simpler to analyze because they have their critical point naturally at the origin \(\mathbf{X} = (0, 0)\).

• Such systems rely only on the eigenvectors and eigenvalues from \(\mathbf{A}\) to describe behavior.
• In our exercise, the homogeneous system’s critical point \(\mathbf{X}=(0,0)\) was also identified as a saddle point due to the same eigenvalues.

Homogeneous systems help us understand the intrinsic dynamics uncoupled from external influences, serving as great benchmarks for more complex nonhomogeneous analyses.