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Chapter 13: Problem 6

For the Lotka-Volterra system (13.18) show that the trajectories in the phase plane are given by \(f(x, y)=\) constant as in (13.20). In the first quadrant \(x \geq 0, y \geq 0\), the intersections of a line \(y=\) constant with a trajectory are given by \(-c \ln x+d x=\) constant. Hence show that there are 0,1 or 2 such intersections, so that the equilibrium point \((c / d, a / b)\) fore this system is a true centre (i.e. it cannot be a spiral point). Using the substitution \(x=\mathrm{e}^{p}, y=\mathrm{e}^{q}\), show that the system takes on the Hamiltonian canonical form (13.22).

Short Answer

Expert verified
Based on the step-by-step solution: 1. The given Lotka-Volterra system is represented by the set of equations: \[\begin{align} \frac{dx}{dt} &= x(a - by) \\ \frac{dy}{dt} &= -y(c - dx) \end{align}\] 2. The trajectories in the phase plane for the Lotka-Volterra system are given by the function \(f(x, y) = \text{constant}\). This has been shown by differentiating \(f(x, y)\) with respect to \(t\) and showing that the result is zero. 3. There can be either 0, 1, or 2 intersections of a line \(y = k = \text{constant}\) with a trajectory on the phase plane. 4. The equilibrium point for the system is a true center. 5. After substituting \(x = e^p, y = e^q\), the system takes on the Hamiltonian canonical form with the Hamiltonian function \(H(p, q) = -da + c \ln a - bb + a \ln b\).

Step by step solution

01

Write down the Lotka-Volterra system (13.18) and the function (13.20)

The Lotka-Volterra system (13.18) is given by: \begin{align} \frac{dx}{dt} &= x(a - by) \\ \frac{dy}{dt} &= -y(c - dx) \end{align} And the function (13.20) is: \begin{equation} f(x, y) = dx - c \ln x + by - a \ln y \end{equation}
02

Show that the trajectories in the phase plane are given by the function \(f(x, y) = \text{constant}\)

To show that the trajectories are given by the function \(f(x, y) = \text{constant}\), we need to differentiate \(f(x, y)\) with respect to \(t\) and show that the result is zero. \begin{align} \frac{d}{dt}f(x, y) &= \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} \\ &= (-c/x)(x(a-by)) + (b)(-y(c-dx)) \end{align} Using the given system, substitute \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\): \begin{align} \frac{d}{dt}f(x, y) &= -c(a-by) + b(-y)(c-dx) \\ &= -ac + bcy + bcy -bdx \\ &= 0 \end{align} Thus, \(f(x, y) = \text{constant}\).
03

Prove that there are 0, 1, or 2 intersections of a line \(y\) = constant with a trajectory

Consider a line \(y = k\), where \(k\) is a constant. At the intersection points, the following equation holds: \begin{equation} -c \ln x + dx = f(x, k) = \text{constant} \end{equation} This equation represents a transcendental equation. The number of intersections depends on the relative positions of the line \(y = k\) and the equilibrium point \((c/d, a/b)\). There can be following scenarios: - No intersection: line \(y = k\) lies below the equilibrium point. - One intersection: line \(y = k\) passes through the equilibrium point. - Two intersections: line \(y = k\) lies above the equilibrium point. Therefore, there are 0, 1, or 2 intersections.
04

Show that the equilibrium point is a true center

Since the equation in Step 3 has no more than two intersections, the equilibrium point \((c/d, a/b)\) cannot be a spiral point. It can only be a center or a saddle point. However, the saddle point has more than two intersections, so the only possibility left is a true center.
05

Use substitution \(x = e^p, y = e^q\) and transform the system into Hamiltonian canonical form (13.22)

Using the substitution \(x = e^p, y = e^q\), we transform the system: \begin{align} \frac{dx}{dt} &= \frac{d(e^p)}{dt} = e^p \frac{dp}{dt} \\ \frac{dy}{dt} &= \frac{d(e^q)}{dt} = e^q \frac{dq}{dt} \end{align} Plug the substituted variables into the Lotka-Volterra system: \begin{align} e^p \frac{dp}{dt} &= e^p(a - be^q) \\ e^q \frac{dq}{dt} &= -e^q(c - de^p) \end{align} Divide by \(e^p\) and \(e^q\), respectively: \begin{align} \frac{dp}{dt} &= a - be^q \\ \frac{dq}{dt} &= -(c - de^p) \end{align} This system now takes the Hamiltonian canonical form: \begin{align} \frac{dp}{dt} &= -\frac{\partial H}{\partial q} \\ \frac{dq}{dt} &= \frac{\partial H}{\partial p} \end{align} Where \(H(p, q) = -da + c \ln a - bb + a \ln b\) is the Hamiltonian function.

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Key Concepts

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

Phase Plane Trajectories
A key concept in analyzing systems like the Lotka-Volterra model is the idea of phase plane trajectories. These are paths that represent the state of the system (comprised of populations in this biological model) over time. By graphing these trajectories, we can visually observe how the system evolves.

In the case of the Lotka-Volterra system, to understand what these trajectories look like, we examine the function derived from the system's differential equations. Here, we consider a particular function, f(x, y), that remains constant along any trajectory of the system. Mathematically, it is demonstrated that the derivative of this function with respect to time is zero along a path, confirming that it is indeed constant for the phase plane trajectories.

The importance of these trajectories lies in their capacity to describe the behavior of predator and prey populations without solving the differential equations at each time step. Instead, one can analyze the shape and location of these paths to predict long-term behaviors such as stability or oscillations in populations.
Equilibrium Point
An equilibrium point is essentially a state where the system settles, and once reached, no further change occurs if undisturbed. In dynamics, it's akin to a ball coming to rest in a trough – at this point, the forces balance out, and the ball stops rolling.

In the Lotka-Volterra model, we find an equilibrium point by setting the rate of change of both populations to zero. For the given system, the equilibrium is at \((c/d, a/b)\). The behavior around this point is significant: it can either be stable (attract nearby points), unstable (repel them), or neutral (neither attract nor repel, but make them circulate around).

The exercise explores the intersections of a horizontal line with the system's trajectories. Depending on the location of this line relative to the equilibrium point, we may have none, one, or two intersections. This helps us determine that the point in question is a true center — it is not a spiral since spirals would have an infinite number of such intersections. Recognizing an equilibrium point as a true center is powerful: it informs us that the system's populations will oscillate without ever settling down or spiraling out of control.
Hamiltonian Canonical Form
The Hamiltonian canonical form is a representation used in physics and mathematics to simplify complex systems of differential equations by describing the evolution of a system over time in terms of energy conservation. A Hamiltonian system has pairs of conjugate variables and follows a specific set of rules governed by the Hamiltonian function, an analogue to the total energy of the system.

The Lotka-Volterra system can be transformed into this form by a change of variables, switching from the populations (x, y) to a new set of variables (p, q), which are related to the former through an exponential function. This transformation simplifies the original equations into a different pair, which resemble the canonical set of Hamilton's equations. This not only offers a fresh perspective on the system but also opens up a rich toolkit of analytical methods from the realm of Hamiltonian mechanics.

Being able to cast the Lotka-Volterra system into Hamiltonian form is more than a mathematical trick — it links population dynamics to energy conservation concepts and allows for exploitation of the symmetries in the system, which can lead to further insights into the system's qualitative behavior.

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Most popular questions from this chapter

A simple model for the dynamics of malaria due to Ross (1911) and Macdonald (1952) is $$ \begin{aligned} &\dot{x}=\left(\frac{a b M}{N}\right) y(1-x)-r x \\ &\dot{y}=a x(1-y)-\mu y \end{aligned} $$ where: \(x, y\) are the infected proportions of the human host, female mosquito populations, \(N, M\) are the numerical sizes of the human, female mosquito populations, \(a\) is the biting rate by a single mosquito, \(b\) is the proportion of infected bites that result in infection, \(r, \mu\) are per capita rates of recovery, mortality for humans, mosquitoes, respectively. Show that the disease can maintain itself within these populations or must die out according as $$ R=\frac{M}{N} \frac{a^{2} b}{\mu r}>1 \text { or }<1. $$

For the Galton board of Fig. \(13.25\) we may arrange things so that each piece of lead shot has an equal chance of rebounding just to the left or to the right at each direct encounter with a scattering pin at each level. Show that the probabilities of each piece of shot passing between the pins along a particular row \(n\) are then given by \(\left(\begin{array}{c}n \\\ r\end{array}\right)\left(\frac{1}{2}\right)^{n}\) where the binomial coefficient \(\left(\begin{array}{l}n \\ r\end{array}\right)=n ! /[(n-r) ! r !]\) and \(r=0,1, \ldots, n\). Use the result \(\left(\begin{array}{c}n+1 \\\ r+1\end{array}\right)=\left(\begin{array}{c}n \\\ r\end{array}\right)+\left(\begin{array}{c}n \\ r+1\end{array}\right)\) to generate the probability distribution for row \(n=\) 16. (For large numbers of pieces of shot and large \(n\) the distribution of shot in the collection compartments approximates the standard normal error curve \(y=k \exp \left(-x^{2} / 2 s^{2}\right)\) where \(k, s\) are constants.)

The relativistic equivalent of the simple harmonic oscillator equation for a spring with constant \(k\) and a rest mass \(m_{0}\) attached is $$ \frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{m_{0} y}{\sqrt{1-y^{2} / c^{2}}}\right)+k x=0 \quad \text { with } \quad \dot{x}=y $$ where \(c\) is the speed of light. Show that the phase trajectories are given by $$ m_{0} c^{2} / \sqrt{1-y^{2} / c^{2}}+\frac{1}{2} k x^{2}=\text { constant } $$ and sketch the phase portrait for this system.

Show that Rayleigh's equation in the form (13.28) has a single critical point at \((0,0)\) and that this is always unstable. Making use of substitution \(\dot{x}=v / \sqrt{3}\) show that \(v\) satisfies the Van der Pol equation (13.29).

Consider the Lorenz system \((13.33)\). (a) Show that the origin \(P_{1}(0,0,0)\) is a critical point and that its stability depends on eigenvalues \(\lambda\) satisfying the cubic $$ (\lambda+\beta)\left[\lambda^{2}+(\sigma+1) \lambda+\sigma(1-\rho)\right]=0 $$ Hence show that \(P_{1}\) is asymptotically stable only when \(0<\rho<1\) (b) Show that there are two further critical points $$ P_{2}, P_{3} \equiv[\pm \sqrt{\beta(\rho-1)}, \pm \sqrt{\beta(\rho-1)},(\rho-1)] $$ when \(\rho>1\), and that their stability depends on eigenvalues \(\lambda\) satisfying the cubic $$ \lambda^{3}+\lambda^{2}(\sigma+\beta+1)+\lambda \beta(\sigma+\rho)+2 \sigma \beta(\rho-1)=0 $$ (c) Show that when \(\rho=1\) the roots of the cubic in (b) are \(0,-\beta,-(1+\sigma)\) and that in order for the roots to have the form \(-\mu, \pm \mathrm{i} \nu\) (with \(\mu, \nu\) real) we must have $$ \rho=\rho_{\text {crit }}=\frac{\sigma(\sigma+\beta+3)}{(\sigma-\beta-1)}>0 $$ (d) By considering how the roots of the cubic in (b) change continuously with \(\rho\) (with \(\left(\sigma, \beta\right.\) kept constant), show that \(P_{2}, P_{3}\) are asymptotically stable for \(1<\rho<\rho_{\text {crit }}\) and unstable for \(\rho>\rho_{\text {crit }}\). (e) Show that if \(\bar{z}=z-\rho-\sigma\), then $$ \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} t}\left(x^{2}+y^{2}+\bar{z}^{2}\right)=-\sigma x^{2}-y^{2}-\beta\left[\bar{z}+\frac{1}{2}(\rho+\sigma)\right]^{2}+\frac{1}{4} \beta(\rho+\sigma)^{2} $$ so that \(\left(x^{2}+y^{2}+\bar{z}^{2}\right)^{1 / 2}\) decreases for all states outside any sphere which contains a particular ellipsoid (implying the existence of an attractor).
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