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Chapter 5: Problem 21

In this problem we examine a continuous plant-herbivore model. We shall define \(q\) as the chemical state of the plant. Low values of \(q\) mean that the plant is toxic; higher values mean that the herbivores derive some nutritious value from it. Consider a situation in which plant quality is enhanced when the vegetation is subjected to a low to moderate level of herbivory, and declines when herbivory is extensive. Assume that herbivores whose density is \(I\) are small insects (such as scale bugs) that attach themselves to one plant for long periods of time. Further assume that their growth rate depends on the quality of the vegetation they consume. Typical equations that have been suggested for such a system are $$ \begin{aligned} \frac{d q}{d t} &=K_{1}-K_{2} q I\left(t-I_{0}\right) \\ \frac{d l}{d t} &=K_{3} I\left(1-\frac{K_{4} I}{q}\right) \end{aligned} $$ (a) Explain the equations, and suggest possible meanings for \(K_{1}, K_{2}, I_{0}, K_{3}\), and \(K_{4}\). (b) Show that the equations can be written in the following dimensionless form: $$ \begin{aligned} &\frac{d q}{d t}=1-K q I(I-1), \\ &\frac{d I}{d t}=\alpha I\left(1-\frac{I}{q}\right) . \end{aligned} $$ (c) Find qualitative solutions using phase-plane methods. Is there a steady state? What are its stability properties? (d) Interpret your solutions in part (c).

Short Answer

Expert verified
The equations describe plant quality decreasing with high herbivory and herbivore growth depending on plant quality. Stability depends on specific parameter values.

Step by step solution

01

Explain the equations

Examine the equations provided. The first equation describes the change in chemical state of the plant, \(\frac{dq}{dt} = K_1 - K_2 q I(t - I_0)\), where \(q\) is the plant quality, \(I\) is the herbivore density, and \(I_0\) is a threshold density. The second equation describes the change in herbivore density, \(\frac{dl}{dt} = K_3 I\left(1 - \frac{K_4 I}{q}\right)\), where \(K_3\) is the growth rate constant and \(K_4\) relates the herbivore density to plant quality.
02

Determine the meanings of constants

\(K_1\) could represent the natural replenishment or enhancement rate of plant quality. \(K_2\) might describe the reduction impact of herbivores on plant quality. \(I_0\) is the herbivore density threshold above which plant quality decreases. \(K_3\) represents the growth rate constant of herbivores. \(K_4\) is a scaling factor relating herbivore density to the plant's quality.
03

Rewrite equations in dimensionless form

Introduce dimensionless variables \(q'\) and \(I'\) where \(q = q' q_0\) and \(I = I' I_0\). Choose appropriate scaling factors \(q_0\) and \(I_0\) such that \(K_1/K_2 I_0 q_0 = 1\) and \(K_3/K_4 = \alpha\). Rewrite the equations: \(\frac{dq'}{dt'} = 1 - K q' I'(I' - 1)\) and \(\frac{dI'}{dt'} = \alpha I'\left(1 - \frac{I'}{q'}\right)\). Here, the constants \(K\) and \(\alpha\) are dimensionless.
04

Analyze phase-plane for qualitative solutions

Plot the nullclines for \(\frac{dq'}{dt'} = 0\) and \(\frac{dI'}{dt'} = 0\). The nullcline \(\frac{dq'}{dt'} = 0\) yields \(1 - K q' I'(I' - 1) = 0\), while \(\frac{dI'}{dt'} = 0\) gives \(\alpha I'\left(1 - \frac{I'}{q'}\right) = 0\). Identify intersections to locate steady states.
05

Determine stability of steady state

Analyze the Jacobian matrix at the steady state. If the real parts of the eigenvalues of the Jacobian are negative, the steady state is stable. Otherwise, it is unstable.
06

Interpret the results

The model shows that low to moderate herbivory can improve plant quality, but extensive herbivory deteriorates it. Analyze the impacts on biodiversity and ecosystem stability.

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Key Concepts

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

Phase-Plane Analysis
Phase-plane analysis is a powerful graphical method for studying the behavior of systems of two-dimensional differential equations. It involves plotting the trajectories of the system's state variables against each other, rather than against time.
In the context of the continuous plant-herbivore model, the state variables are plant quality, denoted as \( q \), and herbivore density, denoted as \( I \). By plotting these variables against each other, we can visualize how the system evolves over time.
The key components of phase-plane analysis include:
  • Trajectories: Paths that show how the system evolves from various initial conditions.
  • Nullclines: Curves where each state variable's rate of change is zero. For our model, we have \( \frac{dq}{dt} = 0 \) and \( \frac{dI}{dt} = 0 \) nullclines.
  • Fixed Points: Intersections of nullclines, indicating steady states where the system doesn't change.
Analyzing these components helps to understand the system dynamics and predict long-term behavior.
Steady State Stability
Steady state stability in a dynamical system refers to whether a system tends to return to a steady state after a small disturbance. To analyze the stability, we use the Jacobian matrix at the steady state.

In our continuous plant-herbivore model, the steady state is found by setting the rate of change equations to zero and solving for \( q \) and \( I \). Once found, we evaluate the Jacobian matrix at this point.

The Jacobian matrix for our dimensionless equations is:
\[ \begin{bmatrix} \frac{\partial (1 - K q I (I-1))}{\partial q} & \frac{\partial (1 - K q I (I-1))}{\partial I} \ \frac{\partial (\alpha I(1 - \frac{I}{q}))}{\partial q} & \frac{\partial (\alpha I(1 - \frac{I}{q})}{\partial I} \ \end{bmatrix} \]

By calculating the eigenvalues of the Jacobian, we can determine stability:
  • Negative Real Parts: The steady state is stable.
  • Positive Real Parts: The steady state is unstable.
Thus, understanding the stability properties helps to predict whether the system will remain in equilibrium under small disturbances.
Dimensionless Equations
Dimensionless equations simplify the analysis by reducing the number of parameters. This is achieved by introducing scaled variables to remove units.

In our plant-herbivore model:
\[ \frac{dq}{dt} = K_1 - K_2 q I (t - I_0) \ \frac{dI}{dt} = K_3 I (1 - \frac{K_4 I}{q}) \]

Transforming these to dimensionless form involves:
  • Introducing dimensionless variables \( q' \) and \( I' \) such that \( q = q_0 q' \) and \( I = I_0 I' \).
  • Choosing scaling factors so that parameters combine into fewer dimensionless groups.

Results in:
\[ \frac{dq'}{dt'} = 1 - K q' I'(I' - 1) \ \frac{dI'}{dt'} = \alpha I' (1 - \frac{I'}{q'}) \]
This transformation simplifies the analysis and helps generalize the behavior across different parameter values.
Herbivore-Plant Interaction
The interaction between herbivores and plants is central to the model. It involves how plant quality and herbivore density affect each other.

Key points in the model:
  • Plant Quality (\( q \)): Improves at low herbivory levels but deteriorates when herbivory is extensive.
  • Herbivore Density (\( I \)): Depends on plant quality for growth.

The equations capturing these interactions are:
\[ \frac{dq}{dt} = K_1 - K_2 q I (t - I_0) \ \frac{dI}{dt} = K_3 I (1 - \frac{K_4 I}{q}) \]

These equations imply:
  • Moderate herbivory can benefit plant quality by stimulating growth or chemical changes.
  • Excessive herbivory harms plant quality, which then reduces herbivore growth.

Understanding these interactions helps predict how ecosystems respond to varying herbivore pressures, affecting biodiversity and stability.

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Most popular questions from this chapter

The following equations were given by J. S. Griffith \((1971, \mathrm{pp} .118-122)\), as a model for the interactions of messenger RNA \(M\) and protein \(E\) : $$ \dot{M}=\frac{a K E^{m}}{1+K E^{m}}-b M, \quad \dot{E}=c M-d E . $$ (See problem 25 in Chapter 7 for an interpretation.) (a) Show that by changing units one can rewrite these in terms of dimensionless variables, as follows $$ \dot{M}=\frac{E^{m}}{1+E^{m}}-\alpha M, \quad \dot{E}=M-\beta E . $$ Find \(\alpha\) and \(\beta\) in terms of the original parameters. (b) Show that one steady state is \(E=M=0\) and that others satisfy \(E^{m-1}=\alpha \beta\left(1+E^{m}\right)\). For \(m=1\) show that this steady state exists only if \(\alpha \beta \leq 1\). (c) Case 1. Show that for \(m=1\) and \(\alpha \beta>1\), the only steady state \(E=M=0\) is stable. Draw a phase-plane diagram of the system. (d) Case 2. Show that for \(m=2\), at steady state $$ E=\frac{1 \pm\left(1+4 \alpha^{2} \beta^{2}\right)^{1 / 2}}{2 \alpha \beta} $$ Conclude that there are two solutions if \(2 \alpha \beta<1\), one if \(2 \alpha \beta=1\), and none if \(2 \alpha \beta>1\). (e) Case 2 continued. For \(m=2\) and \(2 \alpha \beta<1\), show that there are two stable steady states (one of which is at \(E=M=0\) ) and one saddle point. Draw a phase-plane diagram of this system.

Sketch the phase-plane behavior of the following systems of linear equations and classify the stability characteristic of the steady state at \((0,0)\) : (a) \(\frac{d x}{d t}=-2 y\), $$ \frac{d y}{d t}=x \text {. } $$ (b) \(\frac{d x}{d t}=3 x+2 y\), $$ \frac{d y}{d t}=4 x+y . $$ (c) \(\begin{aligned} \frac{d x}{d t} &=2 x+y, \\ \frac{d y}{d t} &=x+2 y . \end{aligned}\) (d) $$ \begin{aligned} &\frac{d x}{d t}=5 x+8 y \\ &\frac{d y}{d t}=-3 x-5 y \end{aligned} $$ (e) $$ \begin{aligned} &\frac{d x}{d t}=-4-2 y \\ &\frac{d y}{d t}=3 x-y \end{aligned} $$ (f) \(\begin{aligned} \frac{d x}{d t} &=x-4 y \\ \frac{d y}{d t} &=x+y \end{aligned}\)

For the following first-order ordinary differential equations, sketch solution curves \(y(t)\) by first plotting the tangent vectors specified by the differential equations: (a) \(\frac{d y}{d t}=y^{2}\). (d) \(\frac{d y}{d t}=y e^{(r-1)}\) (b) \(\frac{d y}{d t}=1-\frac{y}{1+y}\). (e) \(\frac{d y}{d t}=\sin y \cos y\). (c) \(\frac{d y}{d t}=y(y-2)\).

Write a system of linear first-order ODEs whose solutions have the following qualitative behaviors: (a) \((0,0)\) is a stable node with eigenvalues \(\lambda_{1}=-1\) and \(\lambda_{2}=-2\). (b) \((0,0)\) is a saddle point with eigenvalues \(\lambda_{1}=-1\) and \(\lambda_{2}=3\). (c) \((0,0)\) is a center with eigenvalues \(\lambda=\pm 2 i\). (d) \((0,0)\) is an unstable node with eigenvalues \(\lambda_{1}=2\) and \(\lambda_{2}=3\). Hint: Use the fact that \(\lambda_{1}\) and \(\lambda_{2}\) are eigenvalues of a matrix \(A\), then $$ \begin{aligned} &\lambda_{1}+\lambda_{2}=\operatorname{Tr} \mathrm{A}=a_{11}+a_{22} \\ &\lambda_{1} \lambda_{2}=\operatorname{det} \mathrm{A}=a_{11} a_{22}-a_{12} a_{21} . \end{aligned} $$ Note that there will be many possible choices for each of the above.

Sketch the nullelines in the xy phase plane, identify steady states, and draw directions of arrows on the nullclines for the following systems of first- order equations: (a) \(\frac{d x}{d t}=y^{2}-x^{2}\), $$ \frac{d y}{d t}=x-1 \text {. } $$ (b) \(\frac{d x}{d t}=x\left(y^{2}-y\right)\), $$ \frac{d y}{d r}=x-y \text {. } $$ (c) \(\frac{d x}{d t}=x^{2}+y\), $$ \frac{d y}{d t}=-y \text {. } $$ (d) \(\frac{d x}{d t}=-x y\), \(\frac{d y}{d t}=(1+x)(1-y)\). (e) \(\frac{d x}{d t}=x^{2}-y\), $$ \frac{d y}{d t}=y^{2}-x \text {. } $$ (f) \(\frac{d x}{d t}=\frac{-x y}{1+x}+x\), \(\frac{d y}{d t}=\frac{x y}{1+x}-y .\) (g) \(\frac{d x}{d t}=x y(1-x)+C\), \(\frac{d y}{d t}=y\left(1-\frac{y}{x}\right) .\)
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