Problem 37

Question

Denote the size of a population by $N(t)$, and assume that $N(t)$ satisfies $$ \frac{d N}{d t}=N e^{-a N}-N^{2} $$ where $a$ is a positive constant. (a) Show that the nontrivial equilibrium $N^{*}$ satisfies $$ e^{-a N^{*}}=N^{*} $$ (b) Assume now that the nontrivial equilibrium $N^{*}$ is a function of the parameter $a$. Use implicit differentiation to show that $N^{*}$ is a decreasing function of $a$.

Step-by-Step Solution

Verified

Answer

(a) $N^*$ satisfies $e^{-a N^*} = N^*$. (b) $N^*$ decreases with increasing $a$.

1Step 1: Analyze the Equilibrium Condition

To find the equilibrium points, we need to set $\frac{dN}{dt} = 0$. This simplifies the differential equation to: \[ N e^{-aN} - N^2 = 0 \] Factor out $N$: \[ N(e^{-aN} - N) = 0 \] The equilibrium points are where this expression equals zero, which gives us two cases: 1. $ N = 0 $ (trivial equilibrium) 2. $ e^{-aN} = N $ (nontrivial equilibrium) Thus, we have shown that the nontrivial equilibrium $N^*$ satisfies $e^{-a N^*} = N^*$.

2Step 2: Differentiate the Equilibrium Equation Implicitly

We have the equation for equilibrium: \[ e^{-a N^*} = N^* \] Differentiate both sides with respect to $a$, using implicit differentiation. Note that $N^*$ is a function of $a$. The derivative of the left side is: \[ \frac{d}{da}[e^{-a N^*}] = (-N^*e^{-a N^*}) - a(e^{-a N^*})\frac{dN^*}{da} \] The derivative of the right side is: \[ \frac{d}{da}[N^*] = \frac{dN^*}{da} \] Equating the two expressions gives us the derivative relation.

3Step 3: Solve for $\frac{dN^*}{da}$

From Step 2, we have: \[ -N^* e^{-aN^*} - a e^{-aN^*} \frac{dN^*}{da} = \frac{dN^*}{da} \] Rearrange to solve for $\frac{dN^*}{da}$: \[ -a e^{-aN^*} \frac{dN^*}{da} - \frac{dN^*}{da} = N^* e^{-aN^*} \] Factor out $\frac{dN^*}{da}$: \[ \frac{dN^*}{da}(-a e^{-aN^*} - 1) = N^* e^{-aN^*} \] Thus,\[ \frac{dN^*}{da} = \frac{N^* e^{-aN^*}}{-a e^{-aN^*} - 1} \]

4Step 4: Analyze the Sign of $\frac{dN^*}{da}$

Since $N^* e^{-aN^*} = N^*$, the term $N^* e^{-aN^*}$ is positive given that $N^* > 0$. The denominator $-a e^{-aN^*} - 1$ simplifies to $ -1 - a N^* < 0$ because $a N^* > 0$ and $ -a e^{-aN^*} = -a N^*$ due to the equilibrium condition. Therefore, the numerator is positive and the denominator is negative, making $\frac{dN^*}{da}$ negative. Thus, $N^*$ is a decreasing function of $a$.

Key Concepts

Differential EquationsEquilibrium AnalysisImplicit Differentiation

Differential Equations

Differential equations are fundamental tools in modeling and understanding population dynamics. They express how a quantity such as population size changes over time. In our exercise, the rate of change of the population size $N(t)$ is given by the differential equation: \[ \frac{dN}{dt} = N e^{-aN} - N^2 \] Differential equations like this one are handy because they help us predict how population size evolves. Each term in the equation represents a different factor influencing the growth or decline of the population.

The term $N e^{-aN}$ signifies the growth influenced by a factor that diminishes with larger population size due to the presence of $e^{-aN}$, which is an exponential decay factor.
On the flip side, $N^2$ represents the decline due to intra-population competition, which grows with the square of the size of the population.

Thus, a balance between these terms is necessary to find where the population neither grows nor shrinks, which brings us to the idea of equilibrium points.

Equilibrium Analysis

An important aspect when dealing with population modeling is finding the equilibrium points. These points occur when the rate of change $\frac{dN}{dt}$ is zero, meaning the population size stays constant over time. To find these, we set the differential equation to zero and solve: \[ N(e^{-aN} - N) = 0 \] From this factorized form, we gather two possible scenarios for equilibria:

$N = 0$, known as the trivial equilibrium, where no population exists.
$e^{-aN} = N$, known as the nontrivial equilibrium, for a stable and non-zero population size $N^*$.

By understanding these equilibrium points, especially the nontrivial one, we can grasp how various factors interplay to stabilize or destabilize a population over time.

Implicit Differentiation

When dealing with variables that depend on parameters, implicit differentiation is a useful technique. For the given problem, $N^*$ (the nontrivial equilibrium) is influenced by the parameter $a$. We use implicit differentiation on the equilibrium equation $e^{-aN^*} = N^*$: First, differentiate both sides with respect to $a$:

The left side, $e^{-aN^*}$, becomes $-N^* e^{-aN^*} - a e^{-aN^*} \frac{dN^*}{da}$ using the chain rule.
The right side, $N^*$, becomes $\frac{dN^*}{da}$ since it's a direct derivative.

Equating both derivatives gives: \[ -N^* e^{-aN^*} - a e^{-aN^*} \frac{dN^*}{da} = \frac{dN^*}{da} \] This equation indicates that $\frac{dN^*}{da}$ can be derived, showing how the equilibrium changes with the parameter $a$. Solving this carefully helps us understand that $N^*$ decreases as $a$ increases. This reveals how sensitive a stable population is to changes in environmental parameters.

Problem 37

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