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If the eigenvalues and eigenvectors of a planar, autonomous, linear system are complex, then there are no straight line solutions. Use \([\mathrm{v}, \mathrm{e}]=\mathrm{eig}(\mathrm{A})\) to demonstrate that the eigenvalues and eigenvectors of the system \(x^{\prime}=2.9 x+2.0 y, y^{\prime}=-5.0 x-3.1 y\) are complex, then use pplane6 to show that solution trajectories spiral in the phase plane.

Short Answer

Expert verified
The eigenvalues are complex, indicating spiral trajectories in the phase plane.

Step by step solution

01

Formulate the System in Matrix Form

The given system of equations is \( x' = 2.9x + 2.0y \) and \( y' = -5.0x - 3.1y \). We can represent this system in matrix form as \[ \begin{pmatrix} x' \ y' \end{pmatrix} = \begin{pmatrix} 2.9 & 2.0 \ -5.0 & -3.1 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} \].
02

Calculate the Eigenvalues

The eigenvalues of the matrix \( A = \begin{pmatrix} 2.9 & 2.0 \ -5.0 & -3.1 \end{pmatrix} \) are determined by solving the characteristic equation \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix. Calculate the determinant of \( \begin{pmatrix} 2.9 - \lambda & 2.0 \ -5.0 & -3.1 - \lambda \end{pmatrix} \).
03

Solve the Characteristic Equation

The characteristic equation becomes: \((2.9 - \lambda)(-3.1 - \lambda) - (2.0)(-5.0) = 0\). Expanding this we get: \[\lambda^2 - (2.9 - 3.1)\lambda + (2.9)(-3.1) + 10 = 0\]. Simplifying, we have \[\lambda^2 + 0.2\lambda + 0.98 = 0\].
04

Find the Eigenvalues from the Quadratic Formula

Solve \( \lambda^2 + 0.2\lambda + 0.98 = 0 \) using the quadratic formula \[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \], where \( a = 1, b = 0.2, c = 0.98\). Calculate \( b^2 - 4ac = 0.04 - 3.92 = -3.88 \), which is negative, indicating the eigenvalues are complex.
05

Interpret Eigenvalues and Determine Nature of Solutions

Since the eigenvalues found are complex, this confirms that any trajectories will not follow straight lines, as complex eigenvalues result in trajectories that spiral in the phase plane.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Systems
In mathematics, a linear system is a collection of linear equations that share a common set of variables. Such systems can be represented in matrix form, which allows for a clearer visualization and simplicity in calculations. This is especially useful when dealing with multiple linear equations simultaneously.

A linear system can usually be expressed using matrices, where each equation in the system corresponds to a row in the matrix. For example, the system given in the exercise involves two equations represented in matrix form as \( \begin{pmatrix} x' \ y' \end{pmatrix} = A \begin{pmatrix} x \ y \end{pmatrix} \), with a specific matrix \( A \).

One key trait of linear systems is that they overlap with other mathematical topics, like linear algebra and differential equations. These systems often require solving for variables or understanding the behavior of the system's solutions. Understanding the composition of linear systems helps lay the foundation for more advanced topics in mathematics.
Autonomous Systems
An autonomous system, in the context of differential equations, refers to a system where the rules governing the behavior of the system do not explicitly depend on time. This means that the system's behavior is solely determined by the current state, making calculations simpler in some respects.

Autonomous systems are vital in studying dynamic systems such as those that model physical phenomena. Because they don't change over time unless the state changes, they provide a reliable model of behavior under constant conditions.

For example, in the given exercise, the system \( x' = 2.9x + 2.0y \) and \( y' = -5.0x - 3.1y \) is autonomous because the equations lack a time variable. Exploring these kinds of systems can reveal behaviors like equilibrium points and consistent patterns over time.

Understanding autonomous systems is crucial for analyzing trajectories and predicting system behavior in mathematical modeling.
Differential Equations
A differential equation involves equations that describe how one variable changes with respect to another, typically involving derivatives. These equations are powerful tools in modeling situations where change is fundamental, such as physics, engineering, and economics.

In the context of the exercise, the differential equations \( x' = 2.9x + 2.0y \) and \( y' = -5.0x - 3.1y \) are involved in determining how the variables \( x \) and \( y \) change with respect to time. Solving these equations provides insights on how variables evolve and interact over time.

Differential equations can be categorized into linear and nonlinear types. Linear differential equations, like those in the exercise, have solutions that can often be superimposed. This property helps in analyzing and solving complex systems more effortlessly.

Mastering differential equations paves the way to understanding a vast range of scientific and engineering problems.
Eigenvalues and Eigenvectors
Eigenvalues and eigenvectors are fundamental concepts in linear algebra, often described in relation to matrices. They are essential for understanding the dynamics of systems represented in matrix form.

In essence, an eigenvalue is a scalar that indicates how much an eigenvector is stretched or compressed during a linear transformation. An eigenvector, on the other hand, is a direction along which this transformation occurs.

In the problem, we compute eigenvalues by solving the characteristic equation derived from the system matrix. The presence of complex eigenvalues, as found in the solution, implies that the system exhibits oscillatory behavior, such as spiraling trajectories in the phase plane.

Comprehending eigenvalues and eigenvectors is crucial as they offer insights into the stability and behavior of systems, including those modeled by differential equations. This understanding allows for predictions of system responses to various conditions, which is invaluable in engineering and physics.

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Most popular questions from this chapter

In Exercises \(21-30\), perform steps i) and ii) for the linear system $$ \mathbf{x}^{\prime}=\mathbf{A x}, $$ where \(\mathbf{A}\) is the given matrix. i) Find the type of the equilibrium point at the origin. Do this without the aid of any technology or pplane6. You may, of course, check your answer using pplane. ii) Use pplane6 to plot several solution curves - enough to fully illustrate the behavior of the system. You will find it convenient to use the "linear system" choice from the Gallery menu. If the eigenvalues are real, use the "Keyboard input" option to include straight line solutions starting at \(\pm 10\) times the eigenvectors. If the equilibrium point is a saddle point, compute and plot the separatrices. $$ \left[\begin{array}{rr} 6 & -4 \\ 18 & -6 \end{array}\right] $$

In Exercises \(1-4\), without the aid of technology, using only your algebra skills, sketch the nullclines and find the equilibrium point(s) of the assigned system. Indicate the flow of the the vector field along each nullcline, similar to that shown in Figure 13.1. Check your result with pplane6. If the Symbolic Toolbox is available, use the solve command to find the equilibrium point(s). $$ \begin{aligned} &x^{\prime}=x+2 y-4 \\ &y^{\prime}=2 x-y-2 \end{aligned} $$

If a system has one negative and one positive eigenvalue, then one straight line solution moves toward the origin and the other moves away. Consequently, general solutions (being linear combinations of straight line solutions) must do the same thing. Enter the system \(x^{\prime}=9 x-14 y, y^{\prime}=7 x-12 y\), in pplane6 and plot the straight line solutions. Plot several more solutions and note that they move toward the origin only to move away at the last moment. Select Solutions \(\rightarrow\) Find an equilibrium point, find the equilibrium point at the origin, then read its classification from the PPLANE6 Equilibrium point data window.

Consider the nonlinear system $$ \begin{aligned} &x^{\prime}=x(1-x)-x y, \\ &y^{\prime}=y(2-y)+2 x y . \end{aligned} $$ Show that \((-1 / 3,4 / 3)\) is an equilibrium point of the system. a) Without the use of technology, calculate the Jacobian of the system at the equilibrium point \((-1 / 3\), \(4 / 3)\). What is the equation of the linearization at this equilibrium point? Use [v,e]=eig( \(\mathrm{J})\) to find the eigenvalues and eigenvectors of this Jacobian. b) Enter the system in pplane6. Find the equilibrium point at \((-1 / 3,4 / 3)\). Does the data in the Equilibrium point data window agree with your findings in part (a)? Note: The eigenvalues of the Jacobian predict classification of the equilibrium point. In this case, the point \((-1 / 3,4 / 3)\) is a saddle because the eigenvalues are real and opposite in sign. c) Display the linearization. Does the equation of the linearization agree with your findings in part (a)?

In Exercises \(1-4\), without the aid of technology, using only your algebra skills, sketch the nullclines and find the equilibrium point(s) of the assigned system. Indicate the flow of the the vector field along each nullcline, similar to that shown in Figure 13.1. Check your result with pplane6. If the Symbolic Toolbox is available, use the solve command to find the equilibrium point(s). $$ \begin{aligned} &x^{\prime}=x+2 y \\ &y^{\prime}=2 x-y \end{aligned} $$

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