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Chapter 6: Problem 6

The next three exercises deal with competition models of increasing complexity. We assume \(N_{1}, N_{2} \geq 0\) in all cases. Finally, suppose that both species have finite carrying capacities: $$ \begin{aligned} &\dot{N}_{1}=r_{1} N_{1}\left(1-N_{1} / K_{1}\right)-b_{1} N_{1} N_{2} \\ &\dot{N}_{2}=r_{2} N_{2}\left(1-N_{2} / K_{2}\right)-b_{2} N_{1} N_{2} \end{aligned} $$ a) Nondimensionalize the model. How many dimensionless groups are needed? b) Show that there are four qualitatively different phase portraits, as far as longterm behavior is concemed. c) Find conditions under which the two species can stably coexist. Explain the biological meaning of these conditions. (Hint: The carrying capacities reflect the competition within a species, whereas the \(b\) 's reflect the competition between species.)

Short Answer

Expert verified
In summary, the competition model with two species having finite carrying capacities can be nondimensionalized into two dimensionless groups, \(\alpha\) and \(\beta\). Through a phase portrait analysis, we find four qualitatively different long-term behaviors, one of which allows for stable coexistence of both species. For the two species to coexist stably, the competition between species needs to be stronger than the competition within species 1 and the growth rate of species 2 must exceed the combined effect of competition and carrying capacity of species 2.

Step by step solution

01

Nondimensionalize the model

To nondimensionalize the model, we introduce the following dimensionless variables: \(\bar{N}_{1}=N_{1} / K_{1}, \bar{N}_{2}=N_{2} / K_{2}, \bar{t}=\left(r_{1} / K_{1}\right) t\). Then, we have: \[ \begin{aligned} \frac{d\bar{N}_{1}}{d\bar{t}} &= \frac{K_{1}}{r_{1}} \dot{N}_{1} \\ \frac{d\bar{N}_{2}}{d\bar{t}} &= \frac{K_{2}}{r_{1}} \dot{N}_{2} \end{aligned} \] Substituting the given equations for \(\dot{N}_{1}\) and \(\dot{N}_{2}\) into these, we get: \[ \begin{aligned} \frac{d\bar{N}_{1}}{d\bar{t}} &= \bar{N}_{1} (1-\bar{N}_{1}) - \frac{b_{1}K_{1}}{r_{1}} \bar{N}_{1} \bar{N}_{2} \\ \frac{d\bar{N}_{2}}{d\bar{t}} &= \frac{r_{2}}{r_{1}} \bar{N}_{2} (1 - \bar{N}_{2}) - \frac{b_{2}K_{1}}{r_{1}} \bar{N}_{1} \bar{N}_{2} \end{aligned} \] Thus, we have two dimensionless groups: \(\alpha = \frac{b_{1}K_{1}}{r_{1}}\) and \(\beta = \frac{r_{2} K_{1}}{r_{1} K_{2}}\).
02

Analyzing the phase portraits

The long-term behavior of the model is determined by analyzing the nullclines: points where the rate of change for each species is zero (i.e., \(d\bar{N}_{1}/d\bar{t} = 0\) and \(d\bar{N}_{2}/d\bar{t} = 0\)). The nullclines equations are: \[ \begin{aligned} 1-\bar{N}_{1} - \alpha \bar{N}_{1}\bar{N}_{2} &= 0 \\ 1-\bar{N}_{1} - \frac{\beta}{\alpha} \bar{N}_{1}\bar{N}_{2} &= 0 \end{aligned} \] By analyzing the nullclines, we can observe that there are four qualitatively different phase portraits determined by the relative values of \(\alpha\) and \(\beta\): 1. \(\alpha > 1\) and \(\beta > \alpha\): Both species coexist with nonzero populations. 2. \(\alpha > 1\) and \(\beta < \alpha\): Species 1 wins and species 2 goes extinct. 3. \(\alpha < 1\) and \(\beta > \alpha\): Species 2 wins and species 1 goes extinct. 4. \(\alpha < 1\) and \(\beta < \alpha\): Both species go extinct.
03

Finding conditions for stable coexistence

To find the conditions under which the two species can stably coexist, we should be in the case where \(\alpha > 1\) and \(\beta > \alpha\). This translates into the following inequalities: \[ \begin{aligned} \frac{b_{1}K_{1}}{r_{1}} &> 1 \\ \frac{r_{2}K_{1}}{r_{1}K_{2}} &> \frac{b_{1}K_{1}}{r_{1}} \end{aligned} \] This implies that: \[ \begin{aligned} b_{1} &> r_{1} / K_{1} \\ r_{2} &> b_{1} K_{2} \end{aligned} \] The first inequality indicates that the competition between species should be stronger than the competition within species 1, while the second inequality shows that the growth rate of species 2 should be larger than the product of the competition term of species 1 and the carrying capacity of species 2. The biological meaning behind these conditions is that, in order for the two species to coexist stably, the interspecific competition should be more intense than the intraspecific competition for species 1 and the growth rate of species 2 must be larger than the consequence of that competition.

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Key Concepts

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

Nondimensionalization
Nondimensionalization might sound complex, but it's just a mathematical technique that helps simplify equations. In ecological models, it brings clarity by scaling variables and parameters to make them dimensionless. Imagine dealing with animals; rather than counting every individual, you might refer to the ratio of individuals to their environment's capacity, like using percentages in day-to-day life.
Nondimensionalizing can highlight key parameters and interactions in an ecosystem. For our exercise, nondimensionalization reduces our variables through scaling based on carrying capacity and time. We replaced populations \(N_1\) and \(N_2\) by \(\bar{N}_{1} = N_{1}/K_{1}\) and \(\bar{N}_{2} = N_{2}/K_{2}\), respectively. Time also transforms, becoming \(\bar{t} = \left( r_{1}/K_{1} \right) t \).
This process reveals dimensionless groups like \(\alpha\) and \(\beta\), derived from factors like competition strength and growth rate. By making sense of the model without specific units, you can see patterns more clearly.
Carrying Capacity
The carrying capacity is an important concept in ecology. It's the maximum population size that the environment can sustainably support. Think of it like a venue's maximum occupancy when hosting an event 鈥� everything works well until capacity is exceeded.
Carrying capacity, denoted as \(K_1\) and \(K_2\) for species 1 and 2 respectively, guides how species grow. When a population reaches its carrying capacity, its growth tends to zero to avoid overusing resources. In our exercise, the carrying capacity impacts the equations for both species: reducing reproduction rates as population approaches it.
This concept helps clarify the structure and limits within biological populations. Standing at its core, carrying capacity helps us understand resource limitations and pressures within ecosystems.
Interspecific Competition
Interspecific competition refers to when species compete for the same resources, like two animals vying for the same fruit tree in the jungle. In our model, such competition impacts growth rates due to shared resource usage.
The equations incorporate this competition through terms like \(b_1 N_1 N_2 \) and \(b_2 N_1 N_2 \), symbolizing the interactive effects between species 1 and 2. If competition is fierce, it may limit one or both populations unless they find a balance.
Understanding interspecific competition offers insights into ecological dynamics. It implies that if one species overpowers the other, it might lead toward extinction of the weaker species unless conditions change, promoting stable coexistence.
Phase Portraits
Phase portraits are graphical tools used to visualize how systems evolve over time. By examining phase portraits, we can represent possible trajectories of populations in a model.
In our exercise, the phase portraits show four distinct scenarios based on parameters \(\alpha\) and \(\beta\), reflecting different possible outcomes for species.
  • Both may flourish together.
  • One species might dominate, causing the other's extinction.
  • They might both face extinction if competition is too intense.
Analyzing these scenarios helps us predict long-term behaviors from varying parameter interactions. Understanding phase portraits thus assists scientists in foreseeing ecological scenarios and planning management strategies to support biodiversity.

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Most popular questions from this chapter

Consider the system \(\ddot{x}=x^{3}-x\). a) Find all the equilibrium points and classify them. b) Find a conserved quantity. c) Sketch the phase portait.

(Reversible system on a cylinder) While studying chaotic streamlines inside a drop immersed in a steady Stokes flow, Stone ct al. (I991) encountered the system $$ \dot{x}=\frac{\sqrt{2}}{4} x(x-1) \sin \phi, \quad \dot{\phi}=\frac{1}{2}\left[\beta-\frac{1}{\sqrt{2}} \cos \phi-\frac{1}{8 \sqrt{2}} x \cos \phi\right] $$ where \(0 \leq x \leq 1\) and \(-\pi \leq \phi<\pi\) Since the system is \(2 \pi\)-periodic in \(\phi\), it may be considered as a vector ficld on a cylinder. (See Section \(6.7\) for another vector field on a cylinder.) The \(x\)-axis runs along the cylinder, and the \(\phi\)-axis wraps around it. Note that the cylindrical phase space is finite, with edges given by the circles \(x=0\) and \(x=1\) a) Show that the system is reversible. b) Verify that for \(\frac{9}{8 \sqrt{2}}>\beta>\frac{1}{2}\), the system has three fixed points on the cylinder. one of which is a saddle. Show that this saddle is connected to itself by a homoclinic orbit that winds around the waist of the cylinder. Using reversibility, prove that there is a band of closed orbits sandwiched between the circle \(x=0\). and the homoclinic orbit. Sketch the phase portrait on the cylinder, and check. your results by numerical integration. c) Show that as \(\beta \rightarrow \frac{1}{\sqrt{2}}\) from above, the saddle point moves toward the circle \(x=0\), and the homoclinic orbit tightens like a noose. Show that all the closed orbits disappear when \(\beta=\frac{1}{\sqrt{2}}\). d) For \(0<\beta<\frac{1}{\sqrt{2}}\), show that there are two saddle points on the edge \(x=0\). Plot the phase portrait on the cylinder.

Investigate the stability of the equilibrium points of the system \(\vec{x}=(x-a)\left(x^{2}-a\right)\) for all real values of the parameter \(a .\) (Hints: It might help to graph the right-hand side. An alternative is to rewrite the equation as \(\ddot{x}=-V^{\prime}(x)\) for a suitable potential energy function \(V\) and then use your intuition about particles moving in potentials.)

For cach of the following systems, find the fixed points. Then sketch the nullclines, the vector field, and a plausible phase portrait. $$ \dot{x}=x(x-y), \dot{y}=y(2 x-y) $$

(Nonlinear terms can change a star into a spiral) Here's another example that shows that borderline fixed points are sensitive to nonlinear terms. Consider the system in polar coordinates given by \(\dot{r}=-r, \dot{\theta}=1 / \ln r\) a) Find \(r(t)\) and \(\theta(t)\) explicitly, given an initial condition \(\left(r_{0}, \theta_{0}\right)\). b) Show that \(r(t) \rightarrow 0\) and \(|\theta(t)| \rightarrow \infty\) as \(t \rightarrow \infty\). Therefore the origin is a stable spiral for the nonlinear system. c) Write the system in \(x, y\) coordinates. d) Show that the linearized system about the origin is \(\dot{x}=-x, \dot{y}=-y\). Thus the origin is a stable star for the linearized system.
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