Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Consider the linear system $$ d x / d t=a_{11} x+a_{12} y, \quad d y / d t=a_{21} x+a_{22} y $$ where \(a_{11}, \ldots, a_{22}\) are real constants. Let \(p=a_{11}+a_{22}, q=a_{11} a_{22}-a_{12} a_{21},\) and \(\Delta=\) \(p^{2}-4 q\). Show that the critical point \((0,0)\) is a (a) Node if \(q>0\) and \(\Delta \geq 0\) (b) Saddle point if \(q<0\); (c) Spiral point if \(p \neq 0\) and \(\Delta<0\); (d) Center if \(p=0\) and \(q>0\). Hint: These conclusions can be obtained by studying the eigenvalues \(r_{1}\) and \(r_{2}\). It may also be helpful to establish, and then to use, the relations \(r_{1} r_{2}=q\) and \(r_{1}+r_{2}=p\).

Short Answer

Expert verified
Answer: The behavior of the critical point (0,0) can be classified as a node, saddle point, spiral point, or center, depending on the values of p, q, and Δ. To determine the specific behavior, find the eigenvalues r1 and r2 of the system's matrix and analyze the critical point according to the given conditions.

Step by step solution

01

Write down the linear system's matrix

First, let's write down the matrix of the linear system: $$ A=\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} $$
02

Solve the characteristic equation

Next, we need to find the eigenvalues \(r_1\) and \(r_2\) of the matrix \(A\). To do this, we solve the characteristic equation: $$ \det(A - rI) = 0 $$ where \(I\) is the identity matrix. In this case, the characteristic equation is $$ \begin{vmatrix} a_{11}-r & a_{12} \\ a_{21} & a_{22}-r \end{vmatrix} = (a_{11}-r)(a_{22}-r) - a_{12}a_{21} = r^2 - pr + q = 0 $$ Using the given relations \(r_1r_2 = q\) and \(r_1+r_2=p\), we will find the behavior of the critical point \((0,0)\).
03

Analyze the critical point using eigenvalues

Now, let's analyze the critical point according to the conditions given in the problem statement: (a) Node if \(q>0\) and \(\Delta \geq 0\) If \(q>0\), then \(r_1r_2 > 0\). This implies that either both eigenvalues are positive or both eigenvalues are negative. Moreover, if \(\Delta \geq 0\), the eigenvalues are real. Therefore, the critical point \((0,0)\) is a node because the solutions either move away from or towards the critical point in every direction. (b) Saddle point if \(q<0\) If \(q<0\), then \(r_1r_2 < 0\). This means that one eigenvalue is positive and one is negative, making the critical point a saddle point because the solutions move away from the critical point in one direction and towards it in another direction. (c) Spiral point if \(p \neq 0\) and \(\Delta<0\) If \(p\neq 0\) and \(\Delta < 0\), the eigenvalues are complex conjugates with non-zero real parts. This makes the critical point a spiral point because the solutions move away or towards the critical point in a spiral pattern. (d) Center if \(p=0\) and \(q>0\) If \(p=0\) and \(q>0\), the eigenvalues are purely imaginary (\(r_1 = ai\) and \(r_2 = -ai\) for some real number \(a\)). This results in the critical point being a center because the solutions move around it along closed curves.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Eigenvalues and Eigenvectors
In the study of differential equations, eigenvalues and eigenvectors play a crucial role in understanding the dynamics of linear systems. An eigenvalue, denoted as \(r\), is a scalar that results from the transformation of a vector by a square matrix, and the corresponding eigenvector is a non-zero vector that changes at most by its scalar factor during this transformation.

For the linear system given in the exercise, the eigenvalues are found by solving the characteristic equation \( r^2 - pr + q = 0 \) for the matrix that represents the system. The roots of this equation, \(r_1\) and \(r_2\), help determine the behavior of the system near the critical point \( (0,0) \). The relationships \(r_1r_2 = q\) and \(r_1+r_2=p\) further assist us in characterizing the stability and nature of the critical point, based on the aforementioned conditions.
Stability of Critical Points
The stability of a critical point refers to the system's behavior in the vicinity of that point. When analyzing a linear system of differential equations, the stability can be determined using the eigenvalues of the system's coefficient matrix.

If both eigenvalues are real and have the same sign, the critical point is a node; it's stable if both eigenvalues are negative (attracting node), and unstable if they are positive (repelling node). In the case where the eigenvalues are real but with opposite signs, the critical point becomes a saddle point, which is inherently unstable as trajectories approach the point along one eigendirection but repel along another. Conversely, if the eigenvalues are purely imaginary, we encounter a center, around which trajectories circulate in a neutral stability, neither approaching nor repelling from the critical point. Lastly, when the eigenvalues are complex with non-zero real parts, a spiral point emerges, indicating either a stable or unstable spiral, depending on the direction of the spiral, which is dictated by the sign of the real part of the eigenvalues.
Phase Plane Analysis
Phase plane analysis is a graphical method to study the behavior of solutions to a system of two first-order linear differential equations. This analysis involves plotting the trajectories of solutions in a plane, which is known as the phase plane, with one variable on the x-axis and the other on the y-axis.

In the context of the given exercise, the dynamics of the system near the critical point \( (0,0) \) can be visualized on the phase plane. The nature of the critical point, influenced by the eigenvalues and eigenvectors, defines the shape and stability of trajectories on this plane. For example, a node will appear as trajectories that stem from or converge to a single point, a saddle point will have trajectories that bend away in one direction while coming close in another, a center will show circular paths around the critical point, and a spiral point will have trajectories that wrap around the point in a spiral manner.

The visual examination of phase plane plots offers an intuitive understanding of system dynamics that can complement the algebraic insights provided by the eigenvalues. Moreover, phase plane analysis is not just restricted to linear systems and can also be applied to study the qualitative behavior of non-linear systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The equation of motion of a spring-mass system with damping (see Section 3.8) is $$ m \frac{d^{2} u}{d t^{2}}+c \frac{d u}{d t}+k u=0 $$ where \(m, c,\) and \(k\) are positive. Write this second order equation as a system of two first order equations for \(x=u, y=d u / d t .\) Show that \(x=0, y=0\) is a critical point, and analyze the nature and stability of the critical point as a function of the parameters \(m, c,\) and \(k .\) A similar analysis can be applied to the electric circuit equation (see Section 3.8) $$L \frac{d^{2} I}{d t^{2}}+R \frac{d I}{d t}+\frac{1}{C} I=0.$$

A generalization of the damped pendulum equation discussed in the text, or a damped spring-mass system, is the Liénard equation $$ \frac{d^{2} x}{d t^{2}}+c(x) \frac{d x}{d t}+g(x)=0 $$ If \(c(x)\) is a constant and \(g(x)=k x,\) then this equation has the form of the linear pen- \(\text { dulum equation [replace }\sin \theta \text { with } \theta \text { in Eq. ( } 12) \text { of Section } 9.2]\); otherwise the damping force \(c(x) d x / d t\) and restoring force \(g(x)\) are nonlinear. Assume that \(c\) is continuously differentiable, \(g\) is twice continuously differentiable, and \(g(0)=0 .\) (a) Write the Lienard equation as a system of two first equations by introducing the variable \(y=d x / d t\). (b) Show that \((0,0)\) is a and \(g\), then the critical point is asymptotically stable, (c) Show that if \(c(0) \geq 0\) and \(g^{\prime}(0)>0\), then the critical point is asymptotically stable, and that if \(c(0)<0\) or \(g^{\prime}(0)<0\), then the critical point is asymptotically stable, and that if \(c \text { ( } 0)<0\) or \(g^{\prime}(0)<0\), then the critical point is unstable of \(x=0\). Hint: Use Taylor series to approximate \(c\) and \(g\) in the neighborhood of \(x=0\)

Given that \(x=\phi(t), y=\psi(t)\) is a solution of the autonomous system $$ d x / d t=F(x, y), \quad d y / d t=G(x, y) $$ for \(\alpha

(a) Find an equation of the form \(H(x, y)=c\) satisfied by the trajectories. (b) Plot several level curves of the function \(H\). These are trajectories of the given system. Indicate the direction of motion on each trajectory. $$ d x / d t=2 y, \quad d y / d t=8 x $$

(a) Find all the critical points (equilibrium solutions). (b) Use a computer to draw a direction field and portrait for the system. (c) From the plot(s) in part (b) determine whether each critical point is asymptotically stable, stable, or unstable, and classify it as to type. $$ d x / d t=y(2-x-y), \quad d y / d t=-x-y-2 x y $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free