Problem 25
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 \(\mathbf{x}^{\prime}=\mathbf{A} \mathbf{x}\) $$ \begin{aligned} &x^{\prime}=0.1 x-0.2 y+0.35 \\ &y^{\prime}=0.1 x+0.1 y-0.25 \end{aligned} $$
Step-by-Step Solution
Verified Answer
(a) Critical point: (0.5, 2). (b) Saddle point. (c) Both are saddle points, but (0.5, 2) is shifted.
1Step 1: Identify the system equations
The given linear nonhomogeneous system can be written asthe following equations: 1. \( x' = 0.1x - 0.2y + 0.35 \)2. \( y' = 0.1x + 0.1y - 0.25 \).Here, the matrix \( \mathbf{A} \) is \( \begin{pmatrix} 0.1 & -0.2 \ 0.1 & 0.1 \end{pmatrix} \) and the vector \( \mathbf{F} \) is \( \begin{pmatrix} 0.35 \ -0.25 \end{pmatrix} \).
2Step 2: Find the critical point \( \mathbf{X}_1 \)
The critical point \( \mathbf{X}_1 \) is found by setting the derivative equal to 0, i.e., solving the system of equations given by:- \( 0 = 0.1x - 0.2y + 0.35 \)- \( 0 = 0.1x + 0.1y - 0.25 \).Solving these equations simultaneously, we get:- From equation 2: \( 0.1x + 0.1y = 0.25 \) gives \( x + y = 2.5 \).- Substitute \( x = 2.5 - y \) into equation 1: \( 0 = 0.1(2.5 - y) - 0.2y + 0.35 \). Simplifying: \( 0 = 0.25 - 0.1y - 0.2y + 0.35 \) \( 0 = 0.6 - 0.3y \) gives \( y = 2 \).- Substitute \( y = 2 \) back into \( x = 2.5 - y \): \( x = 2.5 - 2 = 0.5 \).Thus, \( \mathbf{X}_1 = \begin{pmatrix} 0.5 \ 2 \end{pmatrix} \).
3Step 3: Determine the nature of the critical point \( \mathbf{X}_1 \)
To determine the nature of the critical point \( \mathbf{X}_1 \), we examine the eigenvalues of matrix \( \mathbf{A} \).- The characteristic equation of \( \mathbf{A} \) is \( \lambda^2 - 0.2\lambda - 0.03 \).- Solving gives eigenvalues \( \lambda_1 \approx 0.25 \) and \( \lambda_2 \approx -0.05 \).- Since the eigenvalues have opposite signs, \( \mathbf{X}_1 \) is a saddle point.
4Step 4: Compare with the homogeneous system
The homogeneous system \( x' = 0.1x - 0.2y \), \( y' = 0.1x + 0.1y \) has its critical point at the origin (0,0).- Both systems have the same matrix \( \mathbf{A} \), so the nature of the critical points (saddle points) is the same.- However, \( \mathbf{X}_1 = \begin{pmatrix} 0.5 \ 2 \end{pmatrix} \) is shifted due to the nonhomogeneous term \( \mathbf{F} \).
Key Concepts
Critical PointsNumerical SolversHomogeneous Linear SystemsSaddle PointsEigenvalues
Critical Points
In the context of a linear system, critical points are where the system's state doesn't change, so the derivatives are zero. To find them, we solve the system of equations with the derivatives set to zero. For example, given the equations \( x' = 0.1x - 0.2y + 0.35 \) and \( y' = 0.1x + 0.1y - 0.25 \), setting the derivatives \( x' \) and \( y' \) to zero allows us to find the critical point \( \mathbf{X}_1 \), which was calculated as \( \begin{pmatrix} 0.5 \ 2 \end{pmatrix} \). Some properties of critical points include:
- They indicate system stability or instability.
- Understanding them helps predict long-term system behavior.
Numerical Solvers
Numerical solvers are computational tools that help us determine the nature of critical points and other system behaviors when analytical solutions are complex or impossible. In our system, after finding the critical point \( \mathbf{X}_1 \), a numerical solver can simulate the system to approximate how it behaves around this point.Utilizing numerical solvers comes with benefits:
- They handle complex, non-linear systems that are difficult to solve by hand.
- They provide visual insights through graphical outputs and simulations.
Homogeneous Linear Systems
Homogeneous linear systems are a particular type of system characterized by having all constant or non-variable terms equal to zero, as shown in the system: \[\begin{aligned}&x^{\prime}=0.1 x-0.2 y \&y^{\prime}=0.1 x+0.1 y\end{aligned}\]This system serves as a foundational topic since it helps explain behavior without external forces or shifts.Key features include:
- Critical points are usually at the origin \((0,0)\).
- The behavior of solutions is driven entirely by the matrix \(\mathbf{A}\).
Saddle Points
Saddle points are special types of critical points where the system's stability is neutral – neither fully stable nor unstable. In our linear system, by calculating the eigenvalues and finding opposite signs, we determined that \( \mathbf{X}_1 \) is a saddle point. This means:
- One eigenvalue suggests a direction of stability.
- The other eigenvalue indicates instability in another direction.
Eigenvalues
Eigenvalues are key to understanding system stability, as they are derived from the matrix \( \mathbf{A} \) in the linear system. By examining eigenvalues, we can identify the nature of critical points. For example, the characteristic equation of our matrix resulted in eigenvalues \( \lambda_1 \approx 0.25 \) and \( \lambda_2 \approx -0.05 \).Understanding eigenvalues allows us to:
- Predict if the system will converge or diverge over time.
- Determine if critical points are stable, unstable, or saddle points.
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