/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 Consider two interacting populat... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider two interacting populations, \(x\) and \(y,\) that are mutually symbiotic: the presence of \(x\) enhances the growth of \(y\) and the presence of \(y\) enhances the growth of \(x\). A dynamic relation between \(x\) and \(y\) may take the form $$ \begin{aligned} x^{\prime}(t) &=r_{x} \times x(t) \times(1-a x(t)+b y(t)) \\ y^{\prime}(t) &=r_{y} \times y(t) \times(1+c x(t)-d y(t)) \end{aligned} $$ Describe the roles of the parameters \(a, b, c,\) and \(d\) in Equations \(18.63 .\)

Short Answer

Expert verified
Parameters \(a\) and \(d\) control intraspecies competition, while \(b\) and \(c\) enhance interspecies symbiosis effects.

Step by step solution

01

Understand the equations

The equations describe the rate of change of two interacting populations, \(x\) and \(y\). The first part of each equation, \(r_x \times x(t)\) and \(r_y \times y(t)\), represents the natural growth rate of each population. The products describe how these rates are modified by the presence of both populations through the factors inside the parentheses.
02

Analyze the role of parameter 'a'

In the first equation, \(a\) is a parameter that affects the self-regulation of population \(x\). It is negative, indicating that an increase in \(x(t)\) has a dampening effect on its own growth due to intraspecies competition for resources.
03

Analyze the role of parameter 'b'

The parameter \(b\) in the first equation represents the positive impact of population \(y\) on the growth of population \(x\). It suggests that as \( y(t) \) increases, it enhances the growth of \(x\), emphasizing a symbiotic relationship where \(y\) supports \(x\).
04

Analyze the role of parameter 'c'

In the second equation, \(c\) is a parameter that indicates the beneficial impact of population \(x\) on the growth of population \(y\). The presence of \(x(t)\), thus increases the growth rate of \(y(t)\), contributing to the symbiotic relationship between \(x\) and \(y\).
05

Analyze the role of parameter 'd'

The parameter \(d\) in the second equation represents internal competition within the \(y\) population. It works similarly to \(a\) in the first equation and suggests that as \(y(t)\) increases, it has a self-regulating effect that decreases its growth rate.

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Key Concepts

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

Differential Equations
Differential equations are mathematical tools used to model how things change over time. In the context of symbiotic population dynamics, these equations help us understand the interaction between two populations. Here, we have two equations governing the rates of change for populations \(x\) and \(y\). These rates are influenced by several factors: natural growth rates and interactions between the populations.
The differential equations given in the problem are:\[\begin{aligned} x'(t) &= r_{x} \cdot x(t) \cdot (1 - ax(t) + by(t)) \ y'(t) &= r_{y} \cdot y(t) \cdot (1 + cx(t) - dy(t)) \end{aligned}\]
  • \(r_x \) and \(r_y \) represent the intrinsic growth rates of populations \(x\) and \(y\) without any limiting factors.
  • The terms \(-ax(t)\) and \(-dy(t)\) introduce negative feedback which models intraspecies competition.
  • The terms \(+by(t)\) and \(+cx(t)\) provide positive feedback representing mutualistic interactions between populations.
This mathematical modeling technique provides an insightful way to predict how populations might change and interact over time, considering various ecological factors.
Intraspecies Competition
Intraspecies competition is an essential concept in ecology, describing how members of the same species vie for the same resources. This can affect the growth rate of the population. In the differential equations presented, this concept is shown through the parameters \(a\) and \(d\).

The presence of \(-ax(t)\) in the model indicates that as the population of \(x\) grows, individuals compete more intensely for limited resources, such as food and space. This increased competition leads to a self-regulating effect where the growth rate slows down.

Similarly, the term \(-dy(t)\) depicts the internal competition in population \(y\). As \(y\) becomes more crowded, the pressure on resources increases, lessening the growth of \(y\).
  • \(a\): Self-regulation parameter for population \(x\)
  • \(d\): Self-regulation parameter for population \(y\)
Understanding intraspecies competition helps in predicting how natural constraints can limit population size, preventing indefinite growth.
Mutualism in Biology
Mutualism is a type of symbiotic relationship where both species involved benefit from the interaction. In the equations given, this beneficial relationship is illustrated by the parameters \(b\) and \(c\). These parameters signify how each population aids the other's growth.

The parameter \(b\) signifies the influence of population \(y\) on population \(x\). The positive term \(+by(t)\) indicates that as \(y\) grows, it provides benefits to \(x\), boosting its growth rate.

Conversely, parameter \(c\) reflects the converse impact: \(+cx(t)\) shows that as \(x\) increases, it fosters an environment where \(y\) can grow more robustly.
  • \(b\): Mutualism factor contributing growth from \(y\) to \(x\)
  • \(c\): Mutualism factor contributing growth from \(x\) to \(y\)
Mutualism facilitates the coexistence and thriving of both populations, highlighting interdependencies in ecosystems that promote biodiversity stability.

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Most popular questions from this chapter

Anderson and May \(^{4}\) give the following model of immune effector cells (helper and cytotoxic T-cells), \(E,\) that limit viral population, \(V\), growth in a human body. $$ \begin{array}{l} d E / d t=\Lambda-\mu E+\epsilon V E \\ d V / d t=r V-\sigma V E \end{array} $$ a. \(\Lambda\) is intrinsic production rate of effector cells from bone marrow. Give similar meaning to each of the other four terms on the RHS of Equations 18.64 . b. Find the equilibrium effector cell population, \(\hat{E},\) in the absence of virus. c. Suppose an inoculum \(V_{0}\) of virus is introduced into the body with \(E=\hat{E}\). Find conditions on \(r\), \(\sigma,\) and \(\hat{E}\) in order that the viral population will increase. d. The Jacobian matrix at any \((E, V)\) is $$ J(E, V)=\left[\begin{array}{rc} -\mu+\epsilon V & \epsilon E \\ -\sigma V & r-\sigma V \end{array}\right] $$ Show that $$ J(\hat{E}, 0)=\left[\begin{array}{rr} -\mu & \sigma \frac{\Lambda}{\mu} \\ 0 & r-\sigma \frac{\Lambda}{\mu} \end{array}\right] $$ e. The characteristic values of the upper diagonal matrix \(J(\hat{E}, 0)\) are the diagonal entries, \(-\mu\) and \(r-\sigma \frac{\Lambda}{\mu} .\) What happens to a small introduction of virus into a healthy individual if both characteristic values are negative? f. In order that the viral population to expand it is necessary that \(r-\sigma \frac{\Lambda}{\mu}>0 .\) What is the role of \(r\) in the model? g. If that condition is met and the viral population increases, show that there will be an equilibrium state, $$ E^{*}=r / \sigma, \quad V^{*}=\frac{\mu r-\Lambda \sigma}{\epsilon r} $$ It is clear that in order for \(V^{*}\) to be positive, we must have (again) $$ \mu r-\Lambda \sigma>0 \quad \text { so that } \quad r>\frac{\Lambda \sigma}{\mu} $$ h. Anderson and May report this system to be asymptotically stable, at \(\left(E^{*}, V^{*}\right),\) but only weakly so, meaning that it is subject to wide oscillations. Show that $$ J\left(E^{*}, V^{*}\right)=\left[\begin{array}{cc} -\frac{\Lambda \sigma}{r} & \epsilon \frac{r}{\sigma} \\ -\sigma \frac{\mu r-\Lambda \sigma}{\epsilon r} & 0 \end{array}\right] $$ i. The characteristic equation of a \(2 \times 2\) matrix \(M\) is is \(s^{2}-\operatorname{trace}(M) s+\operatorname{det}(M)=0\). Show that the characteristic equation of \(J\left(E^{*}, V^{*}\right)\) is $$ s^{2}+\frac{\Lambda \sigma}{r} s+\mu r-\Lambda \sigma=0 $$ j. The roots of the characteristic equation 18.65 are $$ \frac{-\frac{\Lambda \sigma}{r} \pm \sqrt{\left(\frac{\Lambda \sigma}{r}\right)^{2}-4(\mu r-\Lambda \sigma)}}{2} $$ Argue that the real part is negative so that the system is stable. Note: Two cases: $$ \left(\frac{\Lambda \sigma}{r}\right)^{2}-4(\mu r-\Lambda \sigma)>0 \quad \text { and } \quad\left(\frac{\Lambda \sigma}{r}\right)^{2}-4(\mu r-\Lambda \sigma)<0 $$ k. Use \(\Lambda=1, \mu=0.5, \epsilon=0.02, r=0.25,\) and \(\sigma=0.01\) and compute \(E^{*}, V^{*},\) and the stability at \(\left(E^{*}, V^{*}\right)\) l. Let \(E_{0}=2, V_{0}=1, \Lambda=1, \mu=0.5, \epsilon=0.02, r=0.25,\) and \(\sigma=0.01 .\) Approximate the solutions to Equations 18.64 using the trapezoid rule. Observe that the rise in viral load precedes the increase in effector cells.

Exercise 18.3 .6 Interpret the output of the MATLAB program. $$ \begin{array}{l} \mathrm{A}=[0.770 .1 ; 0.0680 .9] \\ \mathrm{T} 1=[\exp (\mathrm{A}) \operatorname{expm}(\mathrm{A})] \\ \mathrm{T} 2=[\log (\mathrm{A}) \operatorname{logm}(\mathrm{A})] \\ \mathrm{T} 3=[\exp (\log (\mathrm{A})) \operatorname{expm}(\operatorname{logm}(\mathrm{A}))] \\ \mathrm{T} 4=\left[\log \left(\mathrm{A}^{-2}\right) 2 * \log (\mathrm{a})\right] \end{array} $$ \(\mathrm{T} 5=\left[\operatorname{logm}\left(\mathrm{A}^{-2}\right) 2 * \operatorname{logm}\left(\mathrm{A}^{-2}\right)\right]\) \(\mathrm{T} 6=[\exp (2 * \mathrm{~A}) \exp (\mathrm{A}) * \exp (\mathrm{A})]\) \(\mathrm{T} 7=[\operatorname{expm}(2 * \mathrm{~A}) \operatorname{expm}(\mathrm{A}) * \operatorname{expm}(\mathrm{A})]\) $$ \text { Output } $$ $$ \begin{array}{rrrrr} \mathrm{T} 1= & 2.1598 & 1.1052 & 2.1674 & 0.2309 \\ & 1.0704 & 2.4596 & 0.1570 & 2.4676 \\ \mathrm{~T} 2= & -0.2614 & -2.3026 & -0.2666 & 0.1204 \\ & -2.6882 & -0.1054 & 0.0819 & -0.1100 \\ \mathrm{~T} 3= & 0.7700 & 0.1000 & 0.7700 & 0.1000 \\ & 0.0680 & 0.9000 & 0.0680 & 0.9000 \\ \mathrm{~T} 4= & -0.5113 & -1.7898 & -0.5227 & -4.6052 \\ & -2.1754 & -0.2024 & -5.3765 & -0.2107 \\ \mathrm{~T} 5= & -0.5331 & 0.2408 & -0.5331 & 0.2408 \\ & 0.1637 & -0.2201 & 0.1637 & -0.2201 \\ \mathrm{~T} 6= & 4.6646 & 1.2214 & 5.8475 & 5.1052 \\ & 1.1457 & 6.0496 & 4.9444 & 7.2326 \\ \mathrm{~T} 7= & 4.7341 & 1.0703 & 4.7341 & 1.0703 \\ & 0.7278 & 6.1254 & 0.7278 & 6.1254 \end{array} $$

Find the local linear approximation to the system $$ \begin{array}{l} x^{\prime}=x-x^{2}-x y \\ y^{\prime}=y-0.5 x y-2 y^{2} \end{array} $$ a. At the equilibrium point (0,0) . b. At the equilibrium point (0,0.5) . c. At the equilibrium point (1,0) . d. At the equilibrium point \((2 / 3,1 / 3)\). For each of the local linear approximations, determine whether it is stable.

Draw the nullclines and some direction arrows and analyze the equilibria of the following competition models. $$ \begin{array}{ll} \text { a. } & x^{\prime}(t)=0.2 \times x(t) \times(1-0.2 x(t)-0.4 y(t)) \\ & y^{\prime}(t)=0.1 \times y(t) \times(1-0.4 x(t)-0.5 y(t)) \\ \text { b. } & x^{\prime}(t)=0.2 \times x(t) \times(1-0.2 x(t)-0.8 y(t)) \\ & y^{\prime}(t)=0.1 \times y(t) \times(1-0.4 x(t)-0.5 y(t)) \\ \text { c. } & x^{\prime}(t)=0.2 \times x(t) \times(1-0.6 x(t)-0.4 y(t)) \\ & y^{\prime}(t)=0.1 \times y(t) \times(1-0.4 x(t)-0.5 y(t)) \\ \text { d. } & x^{\prime}(t)=0.2 \times x(t) \times(1-0.4 x(t)-0.4 y(t)) \\ & y^{\prime}(t)=0.1 \times y(t) \times(1-0.3 x(t)-0.5 y(t)) \end{array} $$

Draw the nullclines and some direction arrows and analyze the equilibria of the following symbiosis models. $$ \begin{aligned} \text { a. } \quad & x^{\prime}(t)=0.2 \times x(t) \times(1-0.5 x(t)+0.4 y(t)) \\\ & y^{\prime}(t)=0.1 \times y(t) \times(1+0.4 x(t)-0.5 y(t)) \\ \text { b. } \quad & x^{\prime}(t)=0.2 \times x(t) \times(1-0.8 x(t)+0.4 y(t)) \\\ & y^{\prime}(t)=0.1 \times y(t) \times(1+0.4 x(t)-0.5 y(t)) \\ \text { c. } \quad & x^{\prime}(t)=0.2 \times x(t) \times(1-0.5 x(t)+0.4 y(t)) \\\ & y^{\prime}(t)=0.1 \times y(t) \times(1+0.4 x(t)-0.2 y(t)) \\ \text { d. } x^{\prime}(t) &=0.2 \times x(t) \times(1-5 x(t)+10 y(t)) \\ & y^{\prime}(t)=0.1 \times y(t) \times(1+2 x(t)-5 y(t)) \\ \text { e. } \quad & x^{\prime}(t)=0.2 \times x(t) \times(1-1.1 x(t)+y(t)) \\ & y^{\prime}(t)=0.1 \times y(t) \times(1+x(t)-y(t)) \\ \text { f. } & x^{\prime}(t)=0.2 \times x(t) \times(1-0.9 x(t)+y(t)) \\ & y^{\prime}(t)=0.1 \times y(t) \times(1+x(t)-y(t)) \end{aligned} $$

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