Problem 24 - 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 \(\mathbf{X}^{\prime}=\mathbf{A X}\). $$ \begin{aligned} &x^{\prime}=-5 x+9 y+13 \\ &y^{\prime}=-x-11 y-23 \end{aligned} $$

Step-by-Step Solution

Answer

The critical point is \((-1, -2)\), a stable node, analogous to the homogeneous system's origin.

Step 1: Identify the System of Equations

We have a nonhomogeneous linear system:\[\begin{align*}\frac{dx}{dt} &= -5x + 9y + 13, \ \frac{dy}{dt} &= -x - 11y - 23.\end{align*}\] The matrix \( \mathbf{A} \) for the homogeneous system is \( \begin{pmatrix} -5 & 9 \ -1 & -11 \end{pmatrix} \) and \( \mathbf{F} \) is \( \begin{pmatrix} 13 \ -23 \end{pmatrix} \).

Step 2: Determine the Critical Point \( \mathbf{X}_1 \)

The critical point \( \mathbf{X}_1 \) satisfies \( \mathbf{A} \mathbf{X}_1 + \mathbf{F} = \mathbf{0} \). Hence, solve \( \begin{pmatrix} -5 & 9 \ -1 & -11 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} + \begin{pmatrix} 13 \ -23 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \).

Step 3: Solve the System for \( \mathbf{X}_1 \)

To find \( x \) and \( y \), solve the equation system:\[\begin{align*}-5x + 9y + 13 &= 0, \-x - 11y - 23 &= 0.\end{align*}\] Solve the second equation for \( x \): \( x = -11y - 23 \). Substitute this expression in the first equation.

Step 4: Substitute and Solve the Resulting Equation

Substitute \( x = -11y - 23 \) into \( -5x + 9y + 13 = 0 \) to get:\[-5(-11y - 23) + 9y + 13 = 0.\] Simplify and solve for \( y \): \( 55y + 115 + 9y + 13 = 0 \Rightarrow 64y = -128 \Rightarrow y = -2 \).

Step 5: Find \( x \) Using \( y = -2 \)

Use \( y = -2 \) in \( x = -11y - 23 \) to find \( x \):\[ x = -11(-2) - 23 = 22 - 23 = -1. \] Thus, the critical point \( \mathbf{X}_1 = (-1, -2) \).

Step 6: Analyze the Critical Point Using Eigenvalues

To analyze the nature of the critical point, find the eigenvalues of \( \mathbf{A} = \begin{pmatrix} -5 & 9 \ -1 & -11 \end{pmatrix} \) by solving \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). This determinant equals \( \lambda^2 + 16\lambda + 64 = 0 \), which simplifies to \( (\lambda + 8)^2 = 0 \). So, the eigenvalues are \( \lambda = -8 \) (repeated).

Step 7: Determine Stability from Eigenvalues

Since the eigenvalues are negative real numbers, the critical point \( \mathbf{X}_1 = (-1, -2) \) is a stable node. It indicates that trajectories are attracted toward this point as time progresses.

Step 8: Compare \( \mathbf{X}_1 \) with Homogeneous Critical Point

For the homogeneous system \( \mathbf{X}' = \mathbf{A} \mathbf{X} \), the critical point is \( (0,0) \). The nonhomogeneous critical point \( \mathbf{X}_1 = (-1, -2) \) reflects the translation effect due to \( \mathbf{F} \). The nature and stability are similar to the homogeneous system's critical point but shifted in the plane.

Key Concepts

Critical Points

In the study of differential systems, critical points are essential to understand. A critical point, or equilibrium point, is a point in the phase space where the system's derivatives equal zero. This means for a point \(\mathbf{X}_1\), the system does not change over time. In our nonhomogeneous system, we find \(\mathbf{X}_1\) such that \( \mathbf{A} \mathbf{X}_1 + \mathbf{F} = \mathbf{0} \). Solving for \(x\) and \(y\) leads to finding these points where the system is at rest. Identifying these is crucial because they help determine the behavior of solutions around these points, giving insights into the system's dynamics.

Eigenvalues

Eigenvalues play a pivotal role in analyzing linear systems. They are parameters derived from the coefficient matrix \( \mathbf{A} \) that indicate the nature of the system's behavior near critical points. To find eigenvalues, solve the characteristic equation \(\det(\mathbf{A} - \lambda \mathbf{I}) = 0\). For our matrix, \( \mathbf{A} = \begin{pmatrix} -5 & 9 \ -1 & -11 \end{pmatrix} \), this led to the eigenvalues \( \lambda = -8 \) repeated. Knowing the eigenvalues allows us to predict if trajectories in the phase plane are attracted to or repelled from the critical point over time, offering a window into the system's temporal behavior.

Stability Analysis

Stability analysis examines how solutions behave as time progresses, particularly around critical points. In our scenario, with eigenvalues being \(\lambda = -8\), both negative, it suggests stability. Negative real eigenvalues mean that the critical point is a stable node, where trajectories move inward as time increases. For complex systems, knowing how to evaluate stability through eigenvalues provides a clearer picture of system resilience and can predict whether small disturbances will grow or fade away. This understanding lends itself to numerous applications, from engineering to biology, wherever system dynamics are crucial.

Homogeneous Systems

Homogeneous systems are forms of differential equations without the forcing function, \(\mathbf{F}\). Mathematically, it's expressed as \( \mathbf{X}' = \mathbf{A} \mathbf{X} \). These systems have a significant role due to their simplicity and their role as a baseline when comparing with nonhomogeneous ones. The critical point for this system is typically the origin, \((0,0)\). Comparing the nonhomogeneous critical point \(\mathbf{X}_1 = (-1, -2)\) to the origin highlights the effects of \(\mathbf{F}\) as it shifts the system’s equilibrium. Understanding both homogeneous and nonhomogeneous systems provides a fuller picture of the dynamic behavior and potential transformations induced by external factors.