Problem 26 - Advanced Engineering Mathematics

Question

A nonhomogeneous linear system \(\mathbf{X}^{\prime}=\mathbf{A} \mathbf{X}+\mathbf{F}\) 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 \(\mathbf{x}^{\prime}=\mathbf{A} \mathbf{x}\) $$ \begin{aligned} &x^{\prime}=3 x-2 y-1 \\ &y^{\prime}=5 x-3 y-2 \end{aligned} $$

Step-by-Step Solution

Answer

The critical point \((1, 1)\) is a stable node, shifted from \((0,0)\) due to the nonhomogeneous part.

Step 1: Set Up the System of Equations

We are given the system of differential equations: \[ \begin{aligned} x^{\prime} &= 3x - 2y - 1 \ y^{\prime} &= 5x - 3y - 2 \end{aligned} \]To find the critical point, we need to consider the system at equilibrium, which means setting both derivatives equal to zero.

Step 2: Solve for Critical Point \(\mathbf{X}_1\)

Set the derivatives to zero to form the system of equations:\[ \begin{aligned} 3x - 2y - 1 &= 0 \ 5x - 3y - 2 &= 0 \end{aligned} \]Solving these equations simultaneously:1. Solve the first equation for \(y\): \[ y = \frac{3x - 1}{2} \]2. Substitute \(y = \frac{3x - 1}{2}\) into the second equation: \[ 5x - 3 \left(\frac{3x - 1}{2}\right) = 2 \] Simplify and solve for \(x\): \[ 5x - \frac{9x - 3}{2} = 2 \] \[ 10x - 9x + 3 = 4 \] \[ x = 1 \]3. Substitute \(x = 1\) back into \(y = \frac{3x - 1}{2}\): \[ y = \frac{3(1) - 1}{2} = 1 \]Thus, the unique critical point \(\mathbf{X}_1\) is \((1, 1)\).

Step 3: Analyze the Critical Point

Use a numerical solver or linear stability analysis to determine the nature of the critical point. For simplicity, we proceed with finding eigenvalues:The coefficient matrix \(\mathbf{A}\) is:\[ \mathbf{A} = \begin{bmatrix} 3 & -2 \ 5 & -3 \end{bmatrix} \]Find the eigenvalues of \(\mathbf{A}\) by solving the characteristic equation:\[ \text{det}(\mathbf{A} - \lambda \mathbf{I}) = 0 \]\[ \begin{vmatrix} 3-\lambda & -2 \ 5 & -3-\lambda \end{vmatrix} = 0 \]\[ (3-\lambda)(-3-\lambda) + 10 = 0 \]\[ \lambda^2 + 6\lambda + 1 = 0 \]Using the quadratic formula, calculate \(\lambda\):\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-6 \pm \sqrt{36 - 4}}{2} \]\[ \lambda = \frac{-6 \pm \sqrt{32}}{2} = \frac{-6 \pm 4\sqrt{2}}{2} \]\[ \lambda_1 = -3 + 2\sqrt{2}, \lambda_2 = -3 - 2\sqrt{2} \]Both eigenvalues are real and negative, indicating that the critical point \((1, 1)\) is a stable node.

Step 4: Compare With Homogeneous System

Considering the homogeneous system \(\mathbf{x}^{\prime} = \mathbf{A}\mathbf{x}\), the critical point is \((0,0)\) and has the same eigenvalues as calculated above, \( \lambda_1 = -3 + 2\sqrt{2} \) and \( \lambda_2 = -3 - 2\sqrt{2} \). Thus, it would also be a stable node. Thus, \((1, 1)\) is a shifted version of \((0,0)\) due to the constant vector \(\mathbf{F}\), with similar stability properties.

Key Concepts

Critical Point Analysis

In mathematics, particularly in systems of differential equations, critical point analysis involves finding points at which the system's derivatives are zero. These points are where the system reaches equilibrium. Given a nonhomogeneous system of the form \( \mathbf{X}' = \mathbf{A} \mathbf{X} + \mathbf{F} \), the critical point is found by setting the derivatives \( x' \) and \( y' \) to zero. This creates a system of algebraic equations that can be solved to determine the values of \( x \) and \( y \) at equilibrium.

In the provided example, \( x' = 3x - 2y - 1 \) and \( y' = 5x - 3y - 2 \), we substitute zero for both derivatives to obtain the system of equations \( 3x - 2y - 1 = 0 \) and \( 5x - 3y - 2 = 0 \). Solving these equations leads us to the unique critical point \( (1, 1) \). The importance of critical point analysis is to understand where the system behavior stabilizes without external influence, serving as a foundation for deeper analysis like stability.

Eigenvalues

Eigenvalues are fundamental in determining the nature of a system's critical points, especially in linear systems. They arise from solving the characteristic equation derived from the matrix \( \mathbf{A} \) (in \( \mathbf{A} \mathbf{X} = \lambda \mathbf{X} \)). The eigenvalues reveal the behavior of system trajectories around critical points.

For the example matrix \( \mathbf{A} = \begin{bmatrix} 3 & -2 \ 5 & -3 \end{bmatrix} \), we calculate the eigenvalues by solving the determinant equation \( \text{det}(\mathbf{A} - \lambda \mathbf{I}) = 0 \). This leads to a quadratic equation with solutions \( \lambda_1 = -3 + 2\sqrt{2} \) and \( \lambda_2 = -3 - 2\sqrt{2} \), both real and negative. The signs and values of these eigenvalues are crucial for understanding the stability of the system, indicating that our critical point \((1, 1)\) is a stable node.

Numerical Solver

A numerical solver is a computational tool used to approximate solutions to mathematical problems that are difficult to solve analytically. For differential equations, these solvers are particularly valuable when determining the nature of critical points in non-linear or complex systems. They enable us to simulate and analyze the system's behavior over time.

In this context, a numerical solver can confirm the analysis performed manually for the critical point \((1, 1)\). It can provide insight into the dynamic behavior near this point, verifying the stability determined by eigenvalues. Tools such as MATLAB, Python's SciPy, or specific calculators implement numerical methods like Euler's method or the Runge-Kutta methods to visualize trajectories and confirm theoretical predictions.

Stability Analysis

Stability analysis involves determining whether small perturbations or changes in initial conditions grow over time or diminish, leading the system to return to equilibrium.

In linear systems, stability is assessed using eigenvalues. If all eigenvalues have negative real parts, the system's equilibrium (critical point) is stable because perturbations decay exponentially. Our example showed eigenvalues \( \lambda_1 = -3 + 2\sqrt{2} \) and \( \lambda_2 = -3 - 2\sqrt{2} \), both guaranteeing stability.

This analysis is significant because it indicates that any small deviation from the critical point \((1, 1)\) will tend to diminish, keeping the system stable. Understanding the system's response to deviations is critical, especially in engineering and scientific applications where maintaining equilibrium is essential.