Problem 17 - Calculus for Biology and Medicine

Question

Interactions between different subpopulations need not be competitive. In fact, different subpopulations may share resources, and the presence of many subpopulations may provide a pool of genetic diversity that helps the population of organisms to react to changing conditions. We will model cooperation between subpopulations by again assuming that the extinction rate depends on \(p\), but now \(m(p)=a-b p\), where a and \(b\) are both positive constants. So \(m(p)\) decreases as \(p\) increases. Ourmodel for the number of subpopulations then becomes: $$ \frac{d p}{d t}=c p(1-p)-(a-b p) p $$ We will analyze this model in Problems 17 and \(18 .\) (a) Find the equilibrium values of \(p\) (your answer will depend on the constant \(c\) ). You may assume \(c>1\). (b) What is the condition on \(c\) for \(p\) to have a nontrivial equilibrium (i.e., an equilibrium in which \(\hat{p} \in(0,1])\) ? (c) Show that if your condition from (b) is met, then the nontrivial equilibrium is also stable.Assume that the number of subpopulations obeys (8.61) with \(a=2, b=1\), and \(c\) some unknown (positive) constant.

Step-by-Step Solution

Answer

Equilibrium points are \( p = 0 \) and \( p = \frac{a-c}{b+c} \).

Step 1: Derive the Equilibrium Points

The equilibrium values of \( p \) occur when \( \frac{d p}{d t} = 0 \). Set the equation to zero:\[c p(1-p)-(a-b p) p = 0\]Simplify this to get:\[p ( cp - c - a + bp ) = 0\]This gives us two equilibrium points by setting each factor to zero:1. \( p = 0 \)2. Combining terms, we get \( cp - a + bp - c = 0 \), or \( p = \frac{a-c}{b+c} \).

Key Concepts

Equilibrium Analysis

The concept of equilibrium in a biological model refers to a steady state where the population does not change over time. For the model given, equilibrium occurs when the rate of change of the population density, \( \frac{d p}{d t} \), equals zero. This means that the rate at which subpopulations increase and decrease is balanced.

In our model, this equilibrium condition unfolds when the term \[ c p(1-p) - (a-b p) p = 0 \].
By setting this expression to zero, we deduce that the population has reached a point where it no longer grows or shrinks. Solving this equation leads us to possible equilibrium points. In this case, the mathematical arrangement gives:

• \( p = 0 \): a situation where no individuals exist within the subpopulation.
• \( p = \frac{a-c}{b+c} \): a potential non-zero equilibrium point dependent on the constants \(a\), \(b\), and \(c\).

Finding and understanding these equilibrium points is crucial as it tells us under what conditions a population will remain constant over time.

Nontrivial Equilibrium Conditions

A nontrivial equilibrium is one where the population density, \( \hat{p} \), remains within the interval \((0,1] \). This means that there is a stable population that neither goes extinct nor grows indefinitely. In the context of our model, we need to determine the specific conditions under which such a nontrivial equilibrium exists.

To ensure the balance point \( \frac{a-c}{b+c} \) is within the specified range, we solve the inequality:
\( 0 < \frac{a-c}{b+c} \leq 1 \)

• From \( \frac{a-c}{b+c} > 0 \), we find that \( c < a \). This ensures that the extinction rate \(a\) is greater than the rate affected by cooperation \(c\), allowing the population to persist.
• From \( \frac{a-c}{b+c} \leq 1 \), it follows trivially since all parameters are positive and \(c< a\) satisfies the condition.

This analysis confirms that the presence of a nontrivial equilibrium essentially depends on the specific relationship between the parameters of our model.

Stability of Equilibrium Points

Once we've identified the equilibrium points, we must determine their stability. A stable equilibrium means that if the population is perturbed, it will return to this equilibrium, while an unstable one will not.

Stability analysis typically involves examining the derivative of the function around the equilibrium point. We assess the stability of \( \hat{p} = \frac{a-c}{b+c} \) by computing the derivative of \( \frac{d p}{d t} \) with respect to \( p \) and evaluating this derivative at the equilibrium point.

If the derivative is negative, the equilibrium is considered stable. In our model, the differential expression:

• The derivative around the nontrivial point \( \hat{p} \) results in a negative value when \( c < a \), confirming the stability of the equilibrium point under the given conditions.

This means the community of subpopulations does not only achieve balance but withstands small changes in population density, underlining the cooperative dynamics among them.