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Chapter 5: Problem 24

In modeling the effect of spruce budworm on forest, Ludwig et al. (1978) defined the following set of variables for the condition of the forest: \(S(t)=\) total surface area of trees \(E(t)=\) energy reserve of trees. They considered the following set of equations for these variables in the presence of a constant budworm population \(B\) : $$ \begin{aligned} &\frac{d S}{d t}=r_{S} S\left(1-\frac{S}{K_{S}} \frac{K_{E}}{E}\right), \\ &\frac{d E}{d t}=r_{E} E\left(1-\frac{E}{K_{E}}\right)-P \frac{B}{S} . \end{aligned} $$ The factors \(r, K\), and \(P\) are to be considered constant. *(a) Interpret possible meanings of these equations. (b) Sketch nullclines and determine how many steady states exist. (c) Draw a phase-plane portrait of the system. Show that the outcomes differ qualitatively depending on whether \(B\) is small or large. *(d) Interpret what this might imply biologically. [Note: You may wish to consult Ludwig et al. (1978) or to return to parts (a) and (d) after reading Chapter 6.]

Short Answer

Expert verified
Small budworm populations lead to stable forests; large populations cause forest instability.

Step by step solution

01

Interpret Equation 1

Analyze the first equation: \(\frac{d S}{d t}=r_{S} S\left(1-\frac{S}{K_{S}} \frac{K_{E}}{E}\right)\).This represents the rate of change of the total surface area of trees:\(\frac{dS}{dt}\). The term \(r_{S}S\) suggests that without any limiting factors, the surface area grows exponentially.The factor \(\left(1-\frac{S}{K_{S}} \frac{K_{E}}{E}\right)\) is a limiting factor that reduces the growth rate as \(S\) increases relative to \(K_S\) and \(E\) decreases relative to \(K_E\).
02

Interpret Equation 2

Analyze the second equation: \(\frac{d E}{d t}=r_{E} E\left(1-\frac{E}{K_{E}}\right)-P \frac{B}{S} \). This represents the rate of change of the energy reserve of trees: \(\frac{dE}{dt}\). The term \(r_EE\left(1-\frac{E}{K_E}\right)\) suggests exponential growth limited by the energy carrying capacity \(K_E\). The term \(P \frac{B}{S}\) represents the energy loss due to the constant budworm population \(B\), affecting the trees' energy reserve inversely relative to the surface area \(S\).
03

Sketch Nullclines

Nullclines are where the rate of change of variables equals zero. For \(\frac{dS}{dt}=0\):\[r_{S} S\left(1-\frac{S}{K_{S}} \frac{K_{E}}{E}\right) = 0\Rightarrow S = 0 \text { or } E = \frac{K_E K_S}{S}\]. For \(\frac{dE}{dt}=0\):\[r_{E} E\left(1-\frac{E}{K_{E}}\right)-P \frac{B}{S} = 0\Rightarrow E = 0 \text{ or } E = K_E \left(1-\frac{PB}{r_E S}\right)\].
04

Determine Steady States

Steady states occur at the intersection points of the nullclines. Solve the equations:\[ S = 0 \text { and } E = K_E \frac{K_S}{S}\]\[ E = 0 \text { and } E = K_E \left(1 - \frac{P B}{r_E S} \right)\].This reveals multiple intersections, indicating more than one steady state.
05

Draw Phase-Plane Portrait

Create a phase-plane diagram with axes \(S\) and \(E\). Plot the nullclines and steady states. Determine vector field trajectories. Nullcline intersections yield steady states. Analyze how trajectories change with different \(B\) values.
06

Analyze Small vs. Large B

For small \(B\), the energy loss term \(P\frac{B}{S}\) is minimal, leading to stable forestry indicated by trajectories converging to steady states with larger \(S\) and \(E\). For large \(B\), this term significantly reduces \(E\), leading to instability and possibly an unsustainable forestry system.
07

Biological Interpretation

These findings imply that smaller budworm populations allow forests to sustain or regenerate. High budworm populations cause excessive energy loss, thus affecting forest health and stability adversely.

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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 describe how quantities change over time. In this exercise, we have two such equations modeling the condition of forests affected by spruce budworm.

The first equation, \(\frac{dS}{dt} = r_S S \bigg(1 - \frac{S}{K_S} \frac{K_E}{E}\bigg)\), portrays the rate at which the total surface area of trees, \(S(t)\), changes. The term \(r_S S\) shows that, without limitations, the tree surface area would grow exponentially.

However, growth is limited by the factor \(\bigg(1 - \frac{S}{K_S} \frac{K_E}{E}\bigg)\), which decreases growth as \(S\) increases relative to \(K_S\) and \(E\) decreases relative to \(K_E\).

The second equation, \(\frac{dE}{dt} = r_E E \bigg(1 - \frac{E}{K_E}\bigg) - P \frac{B}{S}\), reflects the rate of change of the trees' energy reserves, \(E(t)\). The term \(\text{r}_\text{E} \text{E} \bigg(1 - \frac{\text{E}}{\text{K}_\text{E}}\bigg)\) means the energy reserve grows exponentially but is capped by the energy carrying capacity, \(K_E\).

The last term, \(P \frac{B}{S}\), shows the energy loss due to the spruce budworm population \(B\), signifying that energy reserves are inversely related to the forest’s surface area.
nullclines
Nullclines are crucial for analyzing differential equations as they indicate where the rate of change of either \(S\) or \(E\) is zero.

For our system:
  • For \(\frac{dS}{dt} = 0\): \(r_S S \bigg(1 - \frac{S}{K_S} \frac{K_E}{E}\bigg) = 0\). Simplifying, this yields two nullclines: \(S = 0\) or \(E = \frac{K_E K_S}{S}\).

  • For \(\frac{dE}{dt} = 0\): \(r_E E \bigg(1 - \frac{E}{K_E}\bigg) - P \frac{B}{S} = 0\). This results in \(E = 0\) or \(E = K_E \bigg(1 - \frac{P B}{r_E S}\bigg)\).


Plotting these nullclines on the phase-plane, where the axes are \(S\) and \(E\), helps identify where the system might stabilize. Places where these nullclines intersect mark potential steady states for the system.
steady states
Steady states are the conditions where the system does not change over time. These states occur where the nullclines intersect.

To find steady states, we solve the nullcline equations together. For example:
  • Setting \(S = 0\ and \E = \frac{K_E K_S}{S}\) together can determine one set of intersections.

  • Combining \(E = 0\ and \E = K_E \bigg(1 - \frac{P B}{r_E S}\bigg)\) offers another set of intersections.


These intersections indicate multiple potential steady states, implying the forest has different conditions under which it can remain unchanged without external influences.
phase-plane analysis
Phase-plane analysis offers a visual way to study the dynamics of our differential equations. By plotting \(S\) and \(E\) and including the nullclines, we can visualize the system’s behavior.

Here’s how you can proceed:
  • Draw the \(S-E\) axes.

  • Plot the nullclines derived previously. These create curves that separate the plane into distinct regions.

  • Mark the intersection points of these nullclines. These point out the steady states.


Next, add vectors to show how \(S\) and \(E\) change in each region, forming a flow or trajectory. This step is critical because:
  • When \(B\) (constant budworm population) is small, trajectories generally move toward stable steady states meaning the forest remains healthy.

  • When \(B\) is large, trajectories might lead away from steady states, implying potential forest decline due to high-energy loss from budworm infestation.

This visual model demonstrates the varying forest outcomes based on the budworm population levels, guiding us to valid biological interpretations.

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Most popular questions from this chapter

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) \(\frac{d y}{d t}=y(y-2)\).

Curves in the plane. The locus of points for which \(x=y^{3}\) can be written in the form \((x(t), y(t))\) by choosing some parameter \(t\). For example, $$ x(t)=t, \quad y(t)=t^{3} . $$ Other choices are possible, for example, $$ x=s^{1 / 3}, \quad y=s \text {. } $$ These would depict the same curve but a different rate of motion along the curve. Then, using this parameterized form we can depict any position on the curve by the vector $$ \mathbf{x}(t)=\left(t, t^{3}\right) $$ and any tangent vector to the curve by the vector $$ \mathbf{v}(t)=\frac{d x}{d t}=\left(\frac{d x(t)}{d t}, \frac{d y(t)}{d t}\right)=\left(1,3 t^{2}\right) \text {. } $$ For example, at \(t=1, \mathbf{x}=(1,1)\) and \(\mathbf{v}=(1,3)\). (a) Using the parameterized form given here, sketch the curve, and compute the tangent vectors at points \((0,0),(2,8)\), and \((-1,-1)\). (b) Find a way of parameterizing the following curves, and determine the form of the tangent vector to the curve: (1) \(y=x(x-1)\). (4) \(x^{2}+y^{2}=1\). (2) \(y^{2}=\sin x\). (5) \(y=a x+b\). (3) \(x=1 / y\). (6) \(y=4 x^{2}\).

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.

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.

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).
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