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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. $$ \begin{array}{l}{d x / d t=x(1-0.5 y)} \\ {d y / d t=y(-0.25+0.5 x)}\end{array} $$

Short Answer

Expert verified
#Short answer# The populations of the two species will oscillate around the equilibrium point (0.5, 2) as time goes to infinity. This indicates that the species will coexist in an oscillatory manner, with fluctuations in their populations over time.

Step by step solution

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01

Draw a direction field and describe how solutions seem to behave

First, we notice that the given system of differential equations are: $$ \frac{dx}{dt} = x(1 - 0.5y), \\ \frac{dy}{dt} = y(-0.25 + 0.5x). $$ To draw a direction field, we plug in values for \(x\) and \(y\) to determine the behavior of the trajectories in different regions of the \(xy\)-plane. For example, if \(x>0, y>0\) then both \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) will be positive, implying that trajectories move upwards and to the right. We can do this for other regions of the \(xy\)-plane and prepare a direction field accordingly.
02

Find the critical points

A critical point \((x_0, y_0)\) is when both \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) are zero. Equating the given equations to zero and solving for \(x\) and \(y\), we find that the critical points are: \((x_0, y_0) = (0, 0)\) and \((x_1, y_1) = (0.5, 2)\).
03

Determine eigenvalues, eigenvectors, classify critical points and their stability

At each critical point, we can linearize the system of differential equations by computing the matrix \(J\) of partial derivatives: $$ J = \begin{pmatrix} \frac{\partial (1 - 0.5 y)}{\partial x} & \frac{\partial (1 - 0.5 y)}{\partial y} \\ \frac{\partial (-0.25 + 0.5 x)}{\partial x} & \frac{\partial (-0.25 + 0.5 x)}{\partial y} \end{pmatrix} = \begin{pmatrix} 1-0.5y & -0.5x \\ 0.5y & -0.25+0.5x \end{pmatrix}. $$ Now, we evaluate the Jacobian matrix \(J\) at each critical point and find the eigenvalues and eigenvectors. At \((x_0, y_0) = (0, 0)\): $$ J = \begin{pmatrix} 1 & 0 \\ 0 & -0.25 \end{pmatrix}. $$ Eigenvalues are \(\lambda_1 = 1\) and \(\lambda_2 = -0.25\). Since the eigenvalues have different signs, the critical point \((0, 0)\) is a saddle point, which is unstable. At \((x_1, y_1) = (0.5, 2)\): $$ J = \begin{pmatrix} 0 & -0.25 \\ 1 & 0 \end{pmatrix}. $$ The characteristic equation for this matrix is: $$ \lambda^2 + \frac{1}{4} = 0. $$ This gives the complex eigenvalues \(\lambda_3 = i/2\) and \(\lambda_4 = -i/2\). A pair of purely imaginary eigenvalues indicates that the critical point \((0.5, 2)\) is a center, and the stability cannot be determined from the eigenvalues.
04

Sketch trajectories in the neighborhood of critical points

Near the saddle point \((0, 0)\), trajectories move away from the point, indicating the population will not stabilize at (0, 0). Near the center \((0.5, 2)\), the trajectories will appear as closed curves, implying the oscillatory behavior of the populations of the two species around this point.
05

Draw a phase portrait for the system

Using the information about the trajectories and the critical points, we can now draw a phase portrait for the system. The phase portrait will show the direction of the trajectories in each region of the \(xy\)-plane and the behavior near the critical points mentioned in Step 4.
06

Determine the limiting behavior of \(x\) and \(y\) as \(t \rightarrow \infty\) and interpret the results

As \(t \rightarrow \infty\), the system tends to the critical point \((0.5, 2)\), which is a center. This indicates that the two populations will oscillate around the equilibrium point \((0.5, 2)\). Therefore, both species will coexist in an oscillatory manner, with fluctuations in their populations over time.

Key Concepts

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

Direction Field
The direction field is a visual tool that provides insight into the behavior of solutions to a system of differential equations without actually solving the equations. It consists of tiny vectors (or arrows) at various points in the variables' plane, indicating the direction in which the solution moves at that point. The direction and magnitude of the vectors are determined by the differential equations. For example, in our population dynamics problem, by evaluating and plotting the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) at various points, we can visualize how the population densities are expected to change over time. This approach reveals whether the populations are increasing, decreasing, or remaining constant at each point in the plane. The direction field is especially helpful for identifying general trends and patterns in the system's behavior.
Critical Points
Critical points, also known as equilibrium points, are vital in analyzing differential equations as they represent states where the system does not change, hence no movement occurs. In population dynamics, these points indicate stable conditions where the species' populations remain constant. To find the critical points, we set the time derivatives \( \frac{dx}{dt} = 0 \) and \( \frac{dy}{dt} = 0 \) and solve the system of equations. The solution yields points in the plane where the vectors in the direction field are zero-length, indicating no movement. In our exercise, the critical points are identified as \( (x_0, y_0) = (0, 0) \) and \( (x_1, y_1) = (0.5, 2) \) which represent two different potential long-term behaviors of the populations.
Eigenvalues and Eigenvectors
Eigenvalues and eigenvectors are powerful mathematical concepts used in the analysis of linear systems which arise from linearizing nonlinear systems at critical points. They help in classifying the type and determining the stability of these critical points. An eigenvalue gives you information about the potential growth, decay, or oscillation in magnitude of solutions nearby the critical points, while eigenvectors indicate the direction in which these behaviors occur. For stable systems, the real parts of all eigenvalues are negative, causing solutions to decay towards the critical point over time, suggesting a return to equilibrium after a disturbance. Our exercise demonstrates the process of finding eigenvalues and eigenvectors for the Jacobian matrix at each critical point and thus classifying them.
Phase Portrait
A phase portrait is an advanced graph that encompasses a set of trajectories representing possible states of a dynamical system. Each trajectory shows a possible path that system could follow over time in the state-space. It is a comprehensive picture which combines the information from the direction field and the behavior near critical points. By examining a phase portrait, one can deduce the stability of an ecosystem and predict how it may respond to different starting conditions. In our population model, the phase portrait reveals that trajectories near the saddle point move away, indicating instability, whereas near the center, trajectories form closed loops, suggesting a stable, oscillating behavior.
Asymptotic Stability
Asymptotic stability is a fundamental concept concerning the long-term behavior of dynamical systems. A critical point is considered asymptotically stable if solutions starting near the point tend to it as time goes to infinity. This indicates a strong form of stability where not only does the system return to equilibrium after perturbations, but it also remains there. Conversely, an unstable critical point will lead to solutions that diverge away. In our case study, we look at the eigenvalues of the Jacobian matrix to determine stability. If all eigenvalues have negative real parts, the critical point is asymptotically stable, whereas if any eigenvalue has a positive real part, it is considered unstable. Purely imaginary eigenvalues, however, require further investigation as they alone cannot determine stability.

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Most popular questions from this chapter

(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. $$ d x / d t=-(x-y)(1-x-y), \quad d y / d t=x(2+y) $$

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

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.

Each of Problems I through 6 can be interpreted as describing the interaction of two species with populations \(x\) and \(y .\) In each of these problems carry out the following steps. $$ \begin{array}{l}{\text { (a) Draw a direction field and describe how solutions seem to behave. }} \\ {\text { (b) Find the critical points. }} \\ {\text { (c) For each critical point find the corresponding linear system. Find the eigenvalues and }} \\ {\text { eigenvectors of the linear system; classify each critical point as to type, and determine }} \\ {\text { whether it is asymptotically stable, stable, or unstable. }}\end{array} $$ $$ \begin{array}{l}{\text { (d) Sketch the trajectories in the neighborhood of each critical point. }} \\ {\text { (c) Compute and plot enough trajectories of the given system to show clearly the behavior of }} \\ {\text { the solutions. }} \\ {\text { (f) Determine the limiting behavior of } x \text { and } y \text { as } t \rightarrow \infty \text { and interpret the results in terms of }} \\ {\text { the populations of the two species. }}\end{array} $$ $$ \begin{array}{l}{d x / d t=x(1.5-0.5 x-y)} \\ {d y / d t=y(0.75-y-0.125 x)}\end{array} $$

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

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