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(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. $$ d x / d t=y(2-x-y), \quad d y / d t=-x-y-2 x y $$

Short Answer

Expert verified
Answer: The process of classifying the critical points of a system of first-order, autonomous differential equations involves finding critical points by setting both dx/dt and dy/dt equal to zero, drawing a direction field and portrait of the system using computational tools, and finally, analyzing the behavior of the solutions around each critical point in the plots to determine whether each point is asymptotically stable, stable, or unstable, and classify its type.

Step by step solution

01

Find the critical points

To find the critical points, set both dx/dt and dy/dt equal to zero and solve for x and y. $$ y(2-x-y) = 0,\quad -x-y-2xy = 0 $$ There are three critical points: (0, 0), (2, 0), and (1, 1).
02

Draw direction field and portrait of the system

This step requires using a computer to draw the direction field and portrait of the system using the given differential equations. Students can use tools like MATLAB, Python, or online tools to plot the direction field and portrait of the system. However, since creating the computer-generated direction field and portrait isn't possible in this format, we will move on to the next step.
03

Classify the critical points

In order to classify the critical points, students need to analyze the plot(s) made in step 2, which is not possible to display here. However, a general approach to classifying the critical points is provided here: 1. In the direction field, look for the behavior of the solutions around each critical point. 2. If the solutions around the critical point seem to converge towards it, then it is asymptotically stable. 3. If the solutions around the critical point neither converge nor diverge, the point is considered stable. 4. If the solutions around the critical point seem to move away from it, then it is unstable. 5. Depending on the direction field's behavior, the classification type of the critical point can be determined (e.g., node, saddle, spiral). After students plot the direction field using a computer, they need to follow these guidelines to classify the behavior of each critical point found in step 1.

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Key Concepts

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

Critical Points
Critical points in differential equations are the values of the variables where the derivatives are zero. They effectively represent the points where the change halts, meaning they're positions of rest or equilibrium within the system. These points are found by setting the differential equations to zero and solving for the variables involved. In our example, we have two equations: \( y(2-x-y) = 0 \) and \( -x-y-2xy = 0 \). To find the critical points, we solve these equations simultaneously. By solving these, we arrive at three critical points:
  • \((0, 0)\)
  • \((2, 0)\)
  • \((1, 1)\)
These critical points indicate where motions described by the system cease momentarily.
Equilibrium Solutions
An equilibrium solution is a special kind of solution to a differential equation where the solution remains constant over time. Simply put, it is a state in which the system's variables do not change as time progresses. This occurs because at these points, the rates of change for the system are zero, maintaining a balance or steady state.For the given system, the critical points that we've found—\((0, 0), (2, 0), \) and \((1, 1)\)—serve as equilibrium solutions. They are the values of \(x\) and \(y\) where the system remains static. Identifying and understanding these equilibrium solutions helps us predict the long-term behavior of the system under analysis.
Direction Field
A direction field, also known as a slope field, is a visual representation of a differential equation. It helps us understand how solutions behave without actually solving the differential equation explicitly. Direction fields are composed of small line segments or vectors that indicate the slope of the solution curves at various points in the plane.By using computational tools like MATLAB or Python, we can plot these fields by evaluating the differential equations at various grid points in the \((x, y)\) plane. While we can't display such plots here, the primary aim is:
  • To visualize and confirm the trajectories of solutions close to critical points.
  • To perceive the general direction the system will follow over time.
  • To assist in classifying the nature of critical points such as nodes, spirals, or saddles.
Direction fields are an essential tool in analyzing differential equations by offering insights that balance both algebraic and geometric perspectives.
Asymptotic Stability
Asymptotic stability is a property of an equilibrium solution in which, over time, solutions that begin near this equilibrium tend to move closer and closer to it. This indicates that any slight disturbance to the system will dissipate over time, with the system behavior converging to a specific stable state.To determine asymptotic stability for our critical points—\((0, 0), (2, 0), \) and \((1, 1)\)—we use the direction field plotted through computational methods. Here are the common classifications:
  • An asymptotically stable equilibrium point is where the trajectories converge towards it, illustrating passive response to small disturbances.
  • A stable but not asymptotically stable point is where trajectories stay near but don't necessarily approach the point.
  • An unstable point is where trajectories diverge away, leading the system to a different state.
This classification helps us understand whether or not a system will naturally return to equilibrium after a disturbance or if it will move away, displaying its long-term predictability or volatility.

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Most popular questions from this chapter

Consider the autonomous system $$ d x / d t=x, \quad d y / d t=-2 y+x^{3} $$ (a) Show that the critical point \((0,0)\) is a saddle point. (b) Sketch the trajectories for the corresponding linear system and show that the trajectory for which \(x \rightarrow 0, y \rightarrow 0\) as \(t \rightarrow \infty\) is given by \(x=0\). (c) Determine the trajectories for the nonlinear system for \(x \neq 0\) by integrating the equation for \(d y / d x\). Show that the trajectory corresponding to \(x=0\) for the linear system is unaltered, but that the one corresponding to \(y=0\) is \(y=x^{3} / 5 .\) Sketch several of the trajectories for the nonlinear system.

Two species of fish that compete with each other for food, but do not prey on each other, are bluegill and redear. Suppose that a pond is stocked with bluegill and redear, and let \(x\) and \(y\) be the populations of bluegill and redear, respectively, at time \(t\). Suppose further that the competition is modeled by the equations $$\frac{d x}{d t}=x\left(\epsilon_{1}-\sigma_{1} x-\alpha_{1} y\right), \frac{d y}{d t}=y\left(\epsilon_{2}-\sigma_{2} y-\alpha_{2} x\right)$$ a. If \(\epsilon_{2} / \alpha_{2}>\epsilon_{1} / \sigma_{1}\) and \(\epsilon_{2} / \sigma_{2}>\epsilon_{1} / \alpha_{1},\) show that the only equilibrium populations in the pond are no fish, no redear, or no bluegill. What will happen for large \(t ?\) b. If \(\epsilon_{1} / \sigma_{1}>\epsilon_{2} / \alpha_{2}\) and \(\epsilon_{1} / \alpha_{1}>\epsilon_{2} / \sigma_{2}\), show that the only equilibrium populations in the pond are no fish, no redear, or no bluegill. What will happen for large \(t ?\)

Prove that for the system $$ d x / d t=F(x, y), \quad d y / d t=G(x, y) $$ there is at most one trajectory passing through a given point \(\left(x_{0}, y_{0}\right)\) Hint: Let \(C_{0}\) be the trajectory generated by the solution \(x=\phi_{0}(t), y=\psi_{0}(t),\) with \(\phi_{0}\left(l_{0}\right)=\) \(x_{0}, \psi_{0}\left(t_{0}\right)=y_{0},\) and let \(C_{1}\) be trajectory generated by the solution \(x=\phi_{1}(t), y=\psi_{1}(t)\) with \(\phi_{1}\left(t_{1}\right)=x_{0}, \psi_{1}\left(t_{1}\right)=y_{0}\). Use the fact that the system is autonomous and also the existence and uniqueness theorem to show that \(C_{0}\) and \(C_{1}\) are the same.

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

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?

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