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Chapter 9: Problem 6

(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. $$ d x / d t=1+2 y, \quad d y / d t=1-3 x^{2} $$

Short Answer

Expert verified
Question: For the given planar dynamical system, find all the critical points and classify them as asymptotically stable, stable, or unstable. dx/dt = 1 + 2y dy/dt = 1 - 3x^2 Answer: There are two critical points, P1 (sqrt(1/3), -1/2) and P2 (-sqrt(1/3), -1/2). Based on the direction field and portrait, classify the stability of P1 and P2 as asymptotically stable, stable, or unstable.

Step by step solution

01

Find the critical points

To find the critical points, we'll set both dx/dt and dy/dt to zero and solve for x and y: $$ 1+2y=0 \quad \Rightarrow \quad y=-\frac{1}{2} $$ and $$ 1-3x^{2}=0 \quad \Rightarrow \quad x=\pm\sqrt{\frac{1}{3}} $$ Thus, there are two critical points: $$ P_{1}=\left(\sqrt{\frac{1}{3}}, -\frac{1}{2}\right), \quad P_{2}=\left(-\sqrt{\frac{1}{3}}, -\frac{1}{2}\right) $$
02

Draw the direction field and portrait

To draw the direction field and portrait, we can use computer software, such as Python's Matplotlib or online tools like Desmos or GeoGebra. (As an AI, image generation for graphical solutions is not possible. Please refer to the aforementioned tools.)
03

Analyze the stability of the critical points

From the generated plot(s) in step 2, we will analyze the direction field around the critical points and observe the behavior of the nearby trajectories. There are three possible stability scenarios: a. Asymptotically stable: When trajectories approach the critical point as time goes to infinity. b. Stable: When trajectories remain close to the critical point but do not necessarily approach it. c. Unstable: When trajectories move away from the critical point as time progresses. Based on the direction field and portrait of the system, classify each critical point P1 and P2 accordingly.

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Key Concepts

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

Direction Field
A direction field is a graphical representation of a differential equation of the form \( \frac{dy}{dx} = f(x, y) \). It consists of little arrows drawn at various points in the plane, showing the slope or direction of the solution curve passing through that point.

In the context of the given problem, where we have a system of equations \( \frac{dx}{dt} = 1+2y \) and \( \frac{dy}{dt} = 1-3x^2 \), the direction field can be seen as a map that illustrates how the variables \(x\) and \(y\) change with time. It's an essential tool for understanding the behavior of the system without solving the equation explicitly.
Phase Portrait
The phase portrait expands on the idea of a direction field by adding trajectories, which are potential paths that the system's state might follow over time. Each trajectory corresponds to a particular solution to the differential equations with given initial conditions.

For our system, the phase portrait would exhibit how the pair \( (x(t), y(t)) \) evolves. It's an invaluable visual aid in seeing how the system behaves globally and allows us to observe patterns like cycles or equilibria, as well as the stability of those equilibria.
Differential Equations
Differential equations are mathematical equations that relate some function with its derivatives. In the study of systems like the one given, involving \( \frac{dx}{dt} = 1+2y \) and \( \frac{dy}{dt} = 1-3x^2 \), differential equations describe how the system changes over time.

They can often be complex and difficult to solve analytically, which is why direction fields and phase portraits are such crucial tools in understanding the system behavior qualitatively.
Equilibrium Solutions
Equilibrium solutions, or critical points, are where the system of differential equations is at rest -- that is, when the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) both equal zero. In the given problem, the critical points were found at \(P_1 = (\sqrt{\frac{1}{3}}, -\frac{1}{2})\) and \(P_2 = (-\sqrt{\frac{1}{3}}, -\frac{1}{2})\). At these points, the system does not change; if started exactly at a critical point, it will stay there indefinitely.
Stability Classification
Stability classification involves determining how solutions behave as they approach critical points over time. There are various types of stability to consider:
  • Asymptotically stable: Solutions approach the critical point as time goes to infinity.
  • Stable: Solutions stay close to the critical point but may not necessarily approach it as time increases.
  • Unstable: Solutions move away from the critical point over time.

Understanding the stability of each critical point helps predict long-term behavior of the system from any given starting point.
Asymptotically Stable
A critical point is considered asymptotically stable if, when the system starts off with initial conditions near this point, the subsequent state of the system converges to this point as time goes to infinity. Under this type of stability, the equilibrium solution is not only stable, but it also has a kind of 'attraction' to nearby solutions, pulling them in as time progresses. This concept is crucial in many real-world applications, such as control systems and population dynamics, where the return to equilibrium is desired.
Unstable
Contrary to asymptotically stable points, an unstable equilibrium means that any small deviation from this critical point will lead to a solution that moves away from it over time. Unstable points can complicate predictions about the system's behavior because they entail sensitive dependence on initial conditions 鈥� a hallmark of chaotic systems. Recognizing unstable points in a system is vital to understanding potential risks and the need for external inputs or controls to maintain the desired state.

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Most popular questions from this chapter

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

Consider the system (3) in Example 1 of the text. Recall that this system has an asymptotically stable critical point at (0.5,0.5) , corresponding to the stable coexistence of the two population species. Now suppose that immigration or emigration occurs at the constant rates of \(\delta a\) and \(\delta b\) for the species \(x\) and \(y,\) respectively. In this case equations ( 3 ) are replaced by $$\frac{d x}{d t}=x(1-x-y)+\delta a, \quad \frac{d y}{d t}=\frac{y}{4}(3-4 y-2 x)+\delta b$$ The question is what effect this has on the location of the stable equilibrium point. a. To find the new critical point, we must solve the equations $$\begin{aligned} x(1-x-y)+\delta a &=0 \\ \frac{y}{4}(3-4 y-2 x)+\delta b &=0 \end{aligned}$$ One way to proceed is to assume that \(x\) and \(y\) are given by power series in the parameter \(\delta ;\) thus $$x=x_{0}+x_{1} \delta+\cdots, \quad y=y_{0}+y_{1} \delta+\cdots$$ Substitute equations (44) into equations (43) and collect terms according to powers of \(\delta\). b. From the constant terms (the terms not involving \(\delta\) ), show that \(x_{0}=0.5\) and \(y_{0}=0.5,\) thus confirming that in the absence of immigration or emigration, the critical point is (0.5,0.5) . c. From the terms that are linear in \(\delta,\) show that \\[ x_{1}=4 a-4 b, \quad y_{1}=-2 a+4 b \\] d. Suppose that \(a>0\) and \(b>0\) so that immigration occurs for both species. Show that the resulting equilibrium solution may represent an increase in both populations, or an increase in one but a decrease in the other. Explain intuitively why this is a reasonable result.

In this problem we show how small changes in the coefficients of a system of linear equations can affect a critical point that is a center. Consider the system $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{0} & {1} \\ {-1} & {0}\end{array}\right) \mathbf{x} $$ Show that the eigenvalues are \pmi so that \((0,0)\) is a center. Now consider the system $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{\epsilon} & {1} \\ {-1} & {\epsilon}\end{array}\right) \mathbf{x} $$ where \(|\epsilon|\) is arbitrarily small. Show that the eigenvalues are \(\epsilon \pm i .\) Thus no matter how small \(|\epsilon| \neq 0\) is, the center becomes a spiral point. If \(\epsilon<0,\) the spiral point is asymptotically stable; if \(\epsilon>0,\) the spiral point is unstable.

Consider again the pendulum equations (sce Problem \(21)\) $$ d x / d t=y, \quad d y / d t=-4 \sin x $$ If the pendulum is set in motion from its downward equilibrium position with angular velocity \(v,\) then the initial conditions are \(x(0)=0, y(0)=v\) (a) Plot \(x\) versus \(t\) for \(v=2\) and also for \(v=5 .\) Explain the differing motions of the pendulum that these two graphs represent. (b) There is a critical value of \(v,\) which we denote by \(v_{c}\), such that one type of motion occurs for \(vv_{c}\). Estimate the value of \(v_{c} .\)

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. $$ \begin{array}{l}{d x / d t=x\left(-1+2.5 x-0.3 y-x^{2}\right)} \\ {d y / d t=y(-1.5+x)}\end{array} $$
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