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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.

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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—making 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—important 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.

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Most popular questions from this chapter

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

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

The equation of motion of an undamped pendulum is \(d^{2} \theta / d t^{2}+\omega^{2} \sin \theta=0,\) where \(\omega^{2}=g / L .\) Let \(x=\theta, y=d \theta / d t\) to obtain the system of equations $$ d x / d t=y, \quad d y / d t=-\omega^{2} \sin x $$ (a) Show that the critical points are \((\pm n \pi, 0), n=0,1,2, \ldots,\) and that the system is almost lincar in the neighborhood of cach critical point. (b) Show that the critical point \((0,0)\) is a (stable) center of the corresponding linear system. Using Theorem 9.3.2 what can be said about the nonlinear system? The situation is similar at the critical points \((\pm 2 n \pi, 0), n=1,2,3, \ldots\) What is the physical interpretation of these critical points? (c) Show that the critical point \((\pi, 0)\) is an (unstable) saddle point of the corresponding linear system. What conclusion can you draw about the nonlinear system? The situation is similar at the critical points \([\pm(2 n-1) \pi, 0], n=1,2,3, \ldots\) What is the physical interpretation of these critical points? (d) Choose a value for \(\omega^{2}\) and plot a few trajectories of the nonlinear system in the neighborhood of the origin. Can you now draw any further conclusion about the nature of the critical point at \((0,0)\) for the nonlinear system? (e) Using the value of \(\omega^{2}\) from part (d) draw a phase portrait for the pendulum. Compare your plot with Figure 9.3 .5 for the damped pendulum.

Consider the ellipsoid $$ V(x, y, z)=r x^{2}+\sigma y^{2}+\sigma(z-2 r)^{2}=c>0 $$ (a) Calculate \(d V / d t\) along trajectories of the Lorenz equations \((1) .\) (b) Determine a sufficient condition on \(c\) so that every trajectory crossing \(V(x, y, z)=c\) is directed inward. (c) Evaluate the condition found in part (b) for the case \(\sigma=10, b=8 / 3, r=28\)

We will prove part of Theorem 9.3 .2 : If the critical point \((0,0)\) of the almost linear system $$ d x / d t=a_{11} x+a_{12} y+F_{1}(x, y), \quad d y / d t=a_{21} x+a_{22} y+G_{1}(x, y) $$ is an asymptotically stable critical point of the corresponding linear system $$ d x / d t=a_{11} x+a_{12} y, \quad d y / d t=a_{21} x+a_{22} y $$ then it is an asymptotically stable critical point of the almost linear system (i). Problem 12 deals with the corresponding result for instability. Consider the linear system (ii). (a) Since \((0,0)\) is an asymptotically stable critical point, show that \(a_{11}+a_{22}<0\) and \(\left.a_{11} a_{22}-a_{12} a_{21}>0 . \text { (See Problem } 21 \text { of Section } 9.1 .\right)\) (b) Construct a Liapunov function \(V(x, y)=A x^{2}+B x y+C y^{2}\) such that \(V\) is positive definite and \(\hat{V}\) is negative definite. One way to ensure that \(\hat{V}\) is negative definite is to choose \(A, B,\) and \(C\) so that \(\hat{V}(x, y)=-x^{2}-y^{2} .\) Show that this leads to the result $$ \begin{array}{l}{A=-\frac{a_{21}^{2}+a_{22}^{2}+\left(a_{11} a_{22}-a_{12} a_{21}\right)}{2 \Delta}, \quad B=\frac{a_{12} a_{22}+a_{11} a_{21}}{\Delta}} \\\ {C=-\frac{a_{11}^{2}+a_{12}^{2}+\left(a_{11} a_{22}-a_{12} a_{21}\right)}{2 \Delta}}\end{array} $$ where \(\Delta=\left(a_{11}+a_{22}\right)\left(a_{11} a_{22}-a_{12} a_{21}\right)\) (c) Using the result of part (a) show that \(A>0\) and then show (several steps of algebra are required) that $$ 4 A C-B^{2}=\frac{\left(a_{11}^{2}+a_{12}^{2}+a_{21}^{2}+a_{22}^{2}\right)\left(a_{11} a_{22}-a_{12} a_{21}\right)+2\left(a_{11} a_{22}-a_{12} a_{21}\right)^{2}}{\Delta^{2}}>0 $$ Thus by Theorem 9.6.4, \(V\) is positive definite.
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