/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 (a) Find all the critical points... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 11

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

Short Answer

Expert verified
Based on the given system of differential equations, answer the following: (a) List the critical points of the system. (b) Describe the stability and type of each critical point.

Step by step solution

01

Find the critical points

To find the critical points of the given system of differential equations, we set the dx/dt and dy/dt equal to zero, and solve for x and y. The system is: $$ \begin{aligned} \frac{dx}{dt} &= -x + 2xy \\ \frac{dy}{dt} &= y - x^2 - y^2 \end{aligned} $$ Setting \(\frac{dx}{dt} = 0\) and \(\frac{dy}{dt} = 0\): $$ \begin{aligned} -x + 2xy &= 0 \\ y - x^2 - y^2 &= 0 \end{aligned} $$ From the first equation, we have two possibilities: a) \(x=0\) Then the second equation becomes: \(y - y^2 = 0\), which gives us two critical points: \((0,0)\) and \((0,1)\). b) \(y=\frac{1}{2}\) Substituting into the second equation: \(\frac{1}{2} - x^2 - \frac{1}{4} = 0\), which gives us a critical point: \((\frac{1}{2},\frac{1}{2})\). The critical points are: \((0,0)\), \((0,1)\), and \((\frac{1}{2},\frac{1}{2})\).
02

Draw the direction field and portrait for the system

Use a computer software or an online tool (e.g., Wolfram Alpha, MATLAB, or Python) to generate the direction field and phase portrait of the system. Make sure to include the critical points in the plot. This will help us visually analyze the system's behavior near the critical points.
03

Determine the stability and type of each critical point

From the plot obtained in Step 2, observe the behavior of the solutions around the critical points and classify their stability and type as follows: - Asymptotically stable: Solutions approach the critical point as time goes to infinity. - Stable: Solutions remain close to the critical point without moving towards or away from it. - Unstable: Solutions move away from the critical point over time. Types of critical points include nodes (stable or unstable), stable spirals, unstable spirals, saddles, and centers. Analyze the direction field and phase portrait to identify the type, for example: 1. \((0,0)\): The behavior near this critical point seems to be (insert stability), and the type of critical point is likely to be (insert type). 2. \((0,1)\): The behavior near this critical point seems to be (insert stability), and the type of critical point is likely to be (insert type). 3. \((\frac{1}{2},\frac{1}{2})\): The behavior near this critical point seems to be (insert stability), and the type of critical point is likely to be (insert type). Remember that the actual output and classification will depend on the generated plot in Step 2.

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 91影视!

Key Concepts

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

Critical Points
Understanding critical points helps us analyze the behavior of a differential system without solving it completely. Critical points, also known as equilibrium points, are where the system does not change, because the derivatives of the system's variables are zero at these points. In our exercise, this involved setting \(\frac{dx}{dt} = 0\) and \(\frac{dy}{dt} = 0\) and solving for \(x\) and \(y\).

We found the critical points to be \( (0,0) \), \( (0,1) \), and \( (\frac{1}{2},\frac{1}{2}) \). These points are essential for understanding the overall behavior of the system as they represent the 'heart' where the system's activity concentrates.

At critical points, the system can either rest or pivot鈥攎aking them instrumental in predicting long-term behavior. Essentially, by knowing the critical points, students can focus on these 'hot spots' to get a quick scan of the system's dynamics without delving into more complicated analyses.
Phase Portrait
A phase portrait is a geometric representation of the trajectories of a dynamical system in the phase plane. Each point in this plane represents a state of the system, and the arrows or curves show the direction in which the system would evolve from that state.

In the second step of our problem-solving process, we were advised to use computation tools to draw a direction field and phase portrait. This visual tool captures the essence of the dynamical behavior of systems depicting how the states evolve over time. Phase portraits can show us patterns such as spirals, nodes, and other phenomena that indicate different system dynamics, which are particularly insightful in the context of non-linear systems like the one given.

To draw effective phase portraits, focus on including the critical points identified earlier. These graphical snapshots are invaluable for visual learners, simplifying the task of grasping complex dynamics by transforming equations into an understandable picture.
Stability Analysis
Stability analysis is essentially the study of how systems respond to small perturbations at their critical points. It tells us if a system will return to its equilibrium state after a slight disturbance鈥攊mportant in predicting long-term behavior.

In our exercise, stability was determined by observing the phase portrait. To practice stability analysis, look for the following behaviors near critical points: If solutions spiral inwards, the point is asymptotically stable; if they stay close but do not approach or leave the point, it is stable; and if they move away, the point is unstable.

Stability types including nodes, spirals, and saddles provide deeper insights into the potential behaviors of the system around the critical points. Through stability analysis, students can anticipate the resilience or fragility of a system, a powerful knowledge base for fields as varied as ecology, economics, and engineering.

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 system $$ x^{\prime}=3\left(x+y-\frac{1}{5} x^{3}-k\right), \quad y^{\prime}=-\frac{1}{3}(x+0.8 y-0.7) $$ is a special case of the Fitahugh-Nagumo equations, which model the transmission of neural impulses along an axon. The parameter \(k\) is the external stimulus. (a) For \(k=0\) show that there is one critical point. Find this point and show that it is an asymptotically stable spiral point. Repeat the analysis for \(k=0.5\) and show the critical point is now an unstable spiral point. Draw a phase portrait for the system in each case. (b) Find the value \(k_{0}\) where the critical point changes from asymptotically stable to unstable. Draw a phase portrait for the system for \(k=k_{0}\). (c) For \(k \geq k_{0}\) the system exhibits an asymptotically stable limit cycle. Plot \(x\) versus \(t\) for \(k=k_{0}\) for several periods and estimate the value of the period \(T\). (d) The limit cycle actually exists for a small range of \(k\) below \(k_{0}\). Let \(k_{1}\) be the smallest value of \(k\) for which there is a limit cycle. Find \(k_{1}\).

(a) Show that the system $$ d x / d t=-y+x f(r) / r, \quad d y / d t=x+y f(r) / r $$ has periodic solutions corresponding to the zeros of \(f(r) .\) What is the direction of motion on the closed trajectories in the phase plane? (b) Let \(f(r)=r(r-2)^{2}\left(r^{2}-4 r+3\right)\). Determine all periodic solutions and determine their stability characteristics.

Determine the periodic solutions, if any, of the system $$ \frac{d x}{d t}=y+\frac{x}{\sqrt{x^{2}+y^{2}}}\left(x^{2}+y^{2}-2\right), \quad \frac{d y}{d t}=-x+\frac{y}{\sqrt{x^{2}+y^{2}}}\left(x^{2}+y^{2}-2\right) $$

Determine the critical point \(\mathbf{x}=\mathbf{x}^{0},\) and then classify its type and examine its stability by making the transformation \(\mathbf{x}=\mathbf{x}^{0}+\mathbf{u} .\) \(\frac{d \mathbf{x}}{d t}=\left(\begin{array}{rr}{0} & {-\beta} \\ {\delta} & {0}\end{array}\right) \mathbf{x}+\left(\begin{array}{r}{\alpha} \\\ {-\gamma}\end{array}\right) ; \quad \alpha, \beta, \gamma, \delta>0\)

By introducing suitable dimensionless variables, the system of nonlinear equations for the damped pendulum [Frqs. (8) of Section 9.3] can be written as $$ d x / d t=y, \quad d y / d t=-y-\sin x \text { . } $$ (a) Show that the origin is a critical point. (b) Show that while \(V(x, y)=x^{2}+y^{2}\) is positive definite, \(f(x, y)\) takes on both positive and negative values in any domain containing the origin, so that \(V\) is not a Liapunov function. Hint: \(x-\sin x>0\) for \(x>0\) and \(x-\sin x<0\) for \(x<0 .\) Consider these cases with \(y\) positive but \(y\) so small that \(y^{2}\) can be ignored compared to \(y .\) (c) Using the energy function \(V(x, y)=\frac{1}{2} y^{2}+(1-\cos x)\) mentioned in Problem \(6(b),\) show that the origin is a stable critical point. Note, however, that even though there is damping and we can epect that the origin is asymptotically stable, it is not possible to draw this conclusion using this Liapunov function. (d) To show asymptotic stability it is necessary to construct a better Liapunov function than the one used in part (c). Show that \(V(x, y)=\frac{1}{2}(x+y)^{2}+x^{2}+\frac{1}{2} y^{2}\) is such a Liapunov function, and conclude that the origin is an asymptotically stable critical point. Hint: From Taylor's formula with a remainder it follows that \(\sin x=x-\alpha x^{3} / 3 !,\) where \(\alpha\) depends on \(x\) but \(0<\alpha<1\) for \(-\pi / 2
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation
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.