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Chapter 9: Problem 25

In this problem we show how small changes in the coefficients of a system of linear equations can affect a critical point that is a center. Consider the system $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{0} & {1} \\ {-1} & {0}\end{array}\right) \mathbf{x} $$ Show that the eigenvalues are \pmi so that \((0,0)\) is a center. Now consider the system $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{\epsilon} & {1} \\ {-1} & {\epsilon}\end{array}\right) \mathbf{x} $$ where \(|\epsilon|\) is arbitrarily small. Show that the eigenvalues are \(\epsilon \pm i .\) Thus no matter how small \(|\epsilon| \neq 0\) is, the center becomes a spiral point. If \(\epsilon<0,\) the spiral point is asymptotically stable; if \(\epsilon>0,\) the spiral point is unstable.

Short Answer

Expert verified
#Answer# For the first system with matrix A, the critical point (0,0) is a center as the eigenvalues are purely imaginary: \(\lambda = \pm i\). For the second system with matrix B, the critical point becomes a spiral point with eigenvalues \(\lambda = \epsilon \pm i\) where \(|\epsilon| \neq 0\). If \(\epsilon<0\), the spiral point is asymptotically stable; if \(\epsilon>0\), the spiral point is unstable. Small changes in the coefficients of a system of linear equations can change a critical point that is a center into a spiral point.

Step by step solution

01

Eigenvalues for the first system

For the first system with the matrix A: $$ A = \left(\begin{array}{rr}{0} & {1} \\\ {-1} & {0}\end{array}\right) $$ We need to find the eigenvalues by solving the characteristic equation: $$ \det(A - \lambda I) = 0 $$ Where \(\lambda\) are the eigenvalues, and I is the identity matrix.
02

Characteristic equation for the first system

The characteristic equation for the first system is: $$ \left|\begin{array}{cc}{0 - \lambda} & {1} \\\ {-1} & {0 - \lambda}\end{array}\right| = (\lambda)(-\lambda) - (-1)(1) = \lambda^2 + 1 = 0 $$
03

Eigenvalues of the first system

Finding the solutions for the characteristic equation, we get the eigenvalues: \(\lambda^2 + 1 = 0 \implies \lambda = \pm i\) Since the eigenvalues are purely imaginary, the critical point (0,0) is a center.
04

Eigenvalues for the second system

Now, we will consider the second system with matrix B: $$ B = \left(\begin{array}{rr}{\epsilon} & {1} \\\ {-1} & {\epsilon}\end{array}\right) $$ Again, we need to find the eigenvalues by solving the characteristic equation: $$ \det(B - \lambda I) = 0 $$
05

Characteristic equation for the second system

The characteristic equation for the second system is: $$ \left|\begin{array}{cc}{\epsilon-\lambda} & {1} \\\ {-1} & {\epsilon-\lambda}\end{array}\right| = (\epsilon-\lambda)^2 - (-1)(1) = \lambda^2 -2\epsilon \lambda+\epsilon^2 + 1 = 0 $$
06

Eigenvalues of the second system

Finding the solutions for the characteristic equation, we get the eigenvalues by solving the quadratic equation: \(\lambda = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a} = \frac{2\epsilon \pm \sqrt{(-2\epsilon)^{2}-4(1)(\epsilon^2 + 1)}}{2(1)}\) \(\lambda = \epsilon \pm i\) No matter how small \(|\epsilon| \neq 0\), the center becomes a spiral point. If \(\epsilon<0\), the spiral point is asymptotically stable; if \(\epsilon>0\), the spiral point is unstable. Therefore, we have shown that small changes in the coefficients of a system of linear equations can change a critical point that is a center into a spiral point.

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Key Concepts

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

Linear Differential Equations
Linear differential equations are mathematical expressions that model the behavior of dynamic systems using linear combinations of unknown variables and their derivatives. They serve as foundational blocks for understanding more complex dynamics in various scientific and engineering disciplines. An important characteristic of linear systems is that their solutions can be expressed as a superposition of other solutions,
making the mathematical analysis more tractable. Understanding linear differential equations involves identifying critical points, which are the solutions where the system is at equilibrium, and determining their stability. This requires the use of eigenvalues, a concept that lies at the heart of the study of system dynamics.
Center and Spiral Point Stability
In the context of linear differential equations, a critical point can exhibit different types of stability. A 'center' is a type of critical point around which the system's trajectories circulate without converging or diverging from the point itself. This behavior is characteristic when the eigenvalues of the system's matrix are purely imaginary numbers, implying the motion is periodic. On the contrary, a 'spiral point' refers to a critical point where trajectories exhibit a spiral motion, with convergence or divergence from the point depending on the stability of the system. The system stability is determined by the real part of the eigenvalues: a negative real part indicates asymptotic stability (converging spirals), while a positive real part signifies instability (diverging spirals).
Studying the stability of these points is crucial for predicting the long-term behavior of dynamical systems.
Characteristic Equation
The characteristic equation is a fundamental tool in the analysis of linear differential equations, specifically when determining the eigenvalues of a system. It is derived from the system's matrix by subtracting a scalar multiple of the identity matrix and setting the determinant to zero. The solutions to this polynomial equation are the eigenvalues that inform about the system's stability and behavior.
For a 2x2 matrix, the characteristic equation is usually a quadratic equation of the form \(\text{det}\(A - \lambda I\) = \lambda^2 - \text{trace}\(A\) \lambda + \text{det}\(A\) = 0\), where \lambda\ represents the eigenvalues. The coefficients and discriminant of this equation carry significant importance as they determine whether the eigenvalues are real, purely imaginary, or complex, each implying different system behaviors such as nodes, centers, or spirals.

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Most popular questions from this chapter

Each of Problems I through 6 can be interpreted as describing the interaction of two species with populations \(x\) and \(y .\) In each of these problems carry out the following steps. $$ \begin{array}{l}{\text { (a) Draw a direction field and describe how solutions seem to behave. }} \\ {\text { (b) Find the critical points. }} \\ {\text { (c) For each critical point find the corresponding linear system. Find the eigenvalues and }} \\ {\text { eigenvectors of the linear system; classify each critical point as to type, and determine }} \\ {\text { whether it is asymptotically stable, stable, or unstable. }}\end{array} $$ $$ \begin{array}{l}{\text { (d) Sketch the trajectories in the neighborhood of each critical point. }} \\ {\text { (c) Compute and plot enough trajectories of the given system to show clearly the behavior of }} \\ {\text { the solutions. }} \\ {\text { (f) Determine the limiting behavior of } x \text { and } y \text { as } t \rightarrow \infty \text { and interpret the results in terms of }} \\ {\text { the populations of the two species. }}\end{array} $$ $$ \begin{array}{l}{d x / d t=x(1.5-0.5 x-y)} \\ {d y / d t=y(0.75-y-0.125 x)}\end{array} $$

For certain \(r\) intervals, or windows, the Lorenz equations exhibit a period- doubling property similar to that of the logistic difference equation discussed in Section \(2.9 .\) Careful calculations may reveal this phenomenon. (a) One period-doubling window contains the value \(r=100 .\) Let \(r=100\) and plot the trajectory starting at \((5,5,5)\) or some other initial point of your choice. Does the solution appear to be periodic? What is the period? (b) Repeat the calculation in part (a) for slightly smaller values of \(r .\) When \(r \cong 99.98\), you may be able to observe that the period of the solution doubles. Try to observe this result by performing calculations with nearby values of \(r\). (c) As \(r\) decreases further, the period of the solution doubles repeatedly. The next period doubling occurs at about \(r=99.629 .\) Try to observe this by plotting trajectories for nearby values of \(r .\)

(a) Determine all critical points of the given system of equations. (b) Find the corresponding linear system near each critical point. (c) Find the eigenalues of each linear system. What conclusions can you then draw about the nonlinear system? (d) Draw a phase portrait of the nonlinear system to confirm your conclusions, or to extend them in those cases where the linear system does not provide definite information about the nonlinear system. $$ d x / d t=2 x+y+x y^{3}, \quad d y / d t=x-2 y-x y $$

Consider the system \(\mathbf{x}^{\prime}=\mathbf{A} \mathbf{x},\) and suppose that \(\mathbf{A}\) has one zero eigenvalue. (a) Show that \(\mathbf{x}=\mathbf{0}\) is a critical point, and that, in addition, every point on a certain straight line through the origin is also a critical point. (b) Let \(r_{1}=0\) and \(r_{2} \neq 0,\) and let \(\boldsymbol{\xi}^{(1)}\) and \(\boldsymbol{\xi}^{(2)}\) be corresponding eigenvectors. Show that the trajectories are as indicated in Figure \(9.1 .8 .\) What is the direction of motion on the trajectories?

(a) By solving the equation for \(d y / d x,\) show that the equation of the trajectories of the undamped pendulum of Problem 19 can be written as $$ \frac{1}{2} y^{2}+\omega^{2}(1-\cos x)=c $$ where \(c\) is a constant of integration. (b) Multiply Eq. (i) by \(m L L^{2} .\) Then express the result in terms of \(\theta\) to obtain $$ \frac{1}{2} m L^{2}\left(\frac{d \theta}{d t}\right)^{2}+m g L(1-\cos \theta)=E $$ where \(E=m L^{2} c .\) (c) Show that the first term in Eq. (ii) is the kinetic energy of the pendulum and that the second term is the potential energy due to gravity. Thus the total energy \(E\) of the pendulum is constant along any trajectory; its value is determined by the initial conditions.
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