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Chapter 9: Problem 6

(a) Find the eigenvalues and eigenvectors. (b) Classify the critical point \((0,0)\) as to type and determine whether it is stable, asymptotically stable, or unstable. (c) Sketch several trajectories in the phase plane and also sketch some typical graphs of \(x_{1}\) versus \(t .\) (d) Use a computer to plot accurately the curves requested in part (c). \(\frac{d \mathbf{x}}{d t}=\left(\begin{array}{ll}{2} & {-5} \\ {1} & {-2}\end{array}\right) \mathbf{x}\)

Short Answer

Expert verified
Short Answer: The given system of ordinary differential equations (ODEs) has a critical point at (0,0) that is a saddle point, making it unstable. The general solution is given by \(\mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2\), and the behavior of \(x_1(t) = c_1 e^{t} + c_2 e^{-3t}\) depends on the initial conditions represented by the constants \(c_1\) and \(c_2\). Sketching trajectories in the phase plane and graphs of \(x_1\) versus \(t\) can provide insights into the system's behavior. For more accurate plots, use software like MATLAB, Wolfram Mathematica, or Python with appropriate libraries.

Step by step solution

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01

Find the eigenvalues

To find the eigenvalues, we need to compute the characteristic equation of the given matrix: \(\det(A - \lambda I)=\left|\begin{array}{cc}{2-\lambda} & {-5} \\\ {1} & {-2-\lambda}\end{array}\right| = 0\) Expanding, we get: \((2-\lambda)(-2-\lambda)+(-5)(1) = 0\) \(\lambda^2-4\lambda-3=-3\lambda^2+12\lambda+15=0\) Now, solve the quadratic equation for the eigenvalues: \(\lambda_1 = 1\) and \(\lambda_2 = -3\)
02

Find the eigenvectors

For each eigenvalue, we will find the corresponding eigenvector. For \(\lambda_1 = 1\): \((A - \lambda_1 I) \mathbf{v}_1 = (A - I) \mathbf{v}_1 = \left(\begin{array}{cc}{1} & {-5} \\\ {1} & {-3}\end{array}\right)\mathbf{v}_1 = 0\) The eigenvector \(\mathbf{v}_1\) satisfies: \(\left(\begin{array}{l}{1}\\{1}\end{array}\right)k = \mathbf{v}_1\) For \(\lambda_2 = -3\): \((A - \lambda_2 I) \mathbf{v}_2 = (A + 3I) \mathbf{v}_2 = \left(\begin{array}{cc}{5} & {-5} \\\ {1} & {1}\end{array}\right)\mathbf{v}_2 = 0\) The eigenvector \(\mathbf{v}_2\) satisfies: \(\left(\begin{array}{l}{1}\\{-1}\end{array}\right)k = \mathbf{v}_2\)
03

Analyze the critical point \((0,0)\)

The eigenvalues and eigenvectors provide information about the behavior of the critical point \((0,0)\). Both eigenvalues are real, so the phase plane solutions will be either nodes or saddles. In this case, since one eigenvalue is positive \((\lambda_1 = 1)\) and the other is negative \((\lambda_2 = -3)\), the critical point is a saddle point, making it unstable and not asymptotically stable.
04

Sketch trajectories in the phase plane

The eigenvectors corresponding to the eigenvalues provide us with the directions of the trajectories in the phase plane. The eigenvector associated with the positive eigenvalue \(\lambda_1\) provides the unstable direction, while the eigenvector associated with the negative eigenvalue \(\lambda_2\) gives the stable direction. Using the eigenvectors calculated in step 2, we can sketch the trajectories in the phase plane, with each trajectory starting and ending along the eigenvectors' directions.
05

Sketch \(x_1\) versus \(t\) graphs

Based on the eigenvalues, the general solution of the linear system can be expressed as: \(\mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2\) Focusing on \(x_1(t)\): \(x_1(t) = c_1 e^{t} + c_2 e^{-3t}\) Depending on the initial conditions (i.e., the constants \(c_1\) and \(c_2\)), \(x_1(t)\) can show different behaviors. For positive \(c_1\) and \(c_2\), the curve will start at a positive value and exponentially approach zero as \(t\) goes to infinity. For negative \(c_1\) and positive \(c_2\), the curve will start at a negative value and exponentially approach zero as \(t\) goes to infinity. Different curves can be sketched based on various initial conditions.
06

Use a computer for accurate plots

To get accurate plots for both phase plane trajectories and \(x_1\) vs \(t\) graphs, you can use software like MATLAB, Wolfram Mathematica, or Python with libraries like NumPy, SciPy, and Matplotlib. These tools will let you analyze the system more accurately and visualize the results better.

Key Concepts

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

Eigenvalues and Eigenvectors
Understanding the concepts of eigenvalues and eigenvectors is crucial for analyzing linear transformations represented by matrices. In the context of dynamical systems, eigenvalues reveal important information about the system's behavior near critical points.

Eigenvalues are special numbers associated with a square matrix that characterize the factor by which eigenvectors are scaled during a linear transformation. To find eigenvalues for a matrix, we solve the characteristic equation, which is obtained by setting the determinant of the matrix minus \(\lambda I\) to zero. Here, \(\lambda\) represents the eigenvalue and \(I\) represents the identity matrix.

Eigenvectors, on the other hand, are non-zero vectors that only change by a scalar factor when the matrix is applied to them. This scalar factor is the eigenvalue corresponding to that eigenvector. The eigenvector points in a direction that is preserved under the linear transformation described by the matrix. In the exercise provided, after finding eigenvalues, eigenvectors were calculated by substituting each eigenvalue back into the equation \(A - \lambda I)\mathbf{v} = 0\) and solving for the vector \(\mathbf{v}\).

For dynamical systems, the eigenvalues indicate the rate of separation or convergence of trajectories in the system, and the eigenvectors indicate the direction along which these trajectories occur.
Phase Plane Analysis
Phase plane analysis is a graphical method to study two-dimensional autonomous systems. The 'phase plane' is a visual representation of all possible states of a system, where each state is a point in two-dimensional space. Trajectories in the phase plane show how the states evolve over time given different initial conditions.

In the phase plane, critical points are states where the system does not change; that is, the derivative or rate of change \(\frac{d\mathbf{x}}{dt}\) is zero. When we find the eigenvalues and eigenvectors of the system, they help us to draw the trajectories. Stable and unstable manifolds are constructed along the directions given by the eigenvectors. These manifolds are paths that trajectories tend to follow as time progresses.

In the given exercise, trajectories that start near the critical point \(0,0\) will move away along the direction of the eigenvector associated with the positive eigenvalue, and move toward the critical point along the direction of the eigenvector associated with the negative eigenvalue. This behavior is typical of a saddle point, as mentioned in the solution.
Stability of Critical Points
The stability of critical points in a dynamical system is determined by the nature of the eigenvalues derived from the system's linearization around these points. When all eigenvalues have negative real parts, the critical point is stable, meaning that trajectories nearby will approach the critical point as time goes to infinity. This is referred to as an asymptotically stable point.

If any of the eigenvalues have positive real parts, the critical point is unstable, as trajectories tend to move away from the point as time progresses. A critical point is classified as asymptotically stable, stable (neutrally stable), or unstable based on the sign of the real parts of its eigenvalues.

For the critical point \(0,0\) in the provided example, because we have both a positive and negative eigenvalue, the point is classified as unstable (specifically, a saddle point), and trajectories will move away from it. It's important to note that even a single positive eigenvalue can render a system unstable, regardless of the other eigenvalues.

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Most popular questions from this chapter

Verify that \((0,0)\) is a critical point, show that the system is almost linear, and discuss the type and stability of the critical point \((0,0)\) by examining the corresponding linear system. $$ d x / d t=x-y^{2}, \quad d y / d t=x-2 y+x^{2} $$

If \(x=r \cos \theta, y=r \sin \theta,\) show that \(y(d x / d t)-x(d y / d t)=-r^{2}(d \theta / d t)\)

In this problem we indicate how to show that the trajectories are ellipses when the eigen- values are pure imaginary. Consider the system $$ \left(\begin{array}{l}{x} \\\ {y}\end{array}\right)^{\prime}=\left(\begin{array}{ll}{a_{11}} & {a_{12}} \\\ {a_{21}} & {a_{22}}\end{array}\right)\left(\begin{array}{l}{x} \\\ {y}\end{array}\right) $$ (a) Show that the eigenvalues of the coefficient matrix are pure imaginary if and only if $$ a_{11}+a_{22}=0, \quad a_{11} a_{22}-a_{12} a_{21}>0 $$ (b) The trajectories of the system (i) can be found by converting Eqs. (i) into the single equation $$ \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{a_{21} x+a_{22} y}{a_{11} x+a_{12} y} $$ Use the first of Eqs. (ii) to show that Eq. (iii) is exact. (c) By integrating Eq. (iii) show that $$ a_{21} x^{2}+2 a_{22} x y-a_{12} y^{2}=k $$ where \(k\) is a constant. Use Eqs. (ii) to conclude that the graph of Eq. (iv) is always an ellipse. Hint: What is the discriminant of the quadratic form in Eq. (iv)?

Can be interpreted as describing the interaction of two species with population densities \(x\) and \(y .\) In each of these problems carry out the following steps. (a) Draw a direction field and describe how solutions seem to behave. (b) Find the critical points. (c) For each critical point find the corresponding linear system. Find the eigenvalues and eigenvectors of the linear system; classify each critical point as to type, and determine whether it is asymptotically stable, or unstable. (d) Sketch the trajectories in the neighborhood of each critical point. (e) Draw a phase portrait for the system. (f) Determine the limiting behavior of \(x\) and \(y\) as \(t \rightarrow \infty\) and interpret the results in terms of the populations of the two species. $$ \begin{array}{l}{d x / d t=x(1.5-0.5 y)} \\ {d y / d t=y(-0.5+x)}\end{array} $$

(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. $$ d x / d t=2 x+y+x y^{3}, \quad d y / d t=x-2 y-x y $$
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