Problem 15 - Calculus for Biology and Medicine

Question

In Problems 13 and 14 we assumed that the per colony extinction rate was proportional to \(p .\) This means that the per colony extinction rate goes to 0 for small \(p .\) This may not be realisticsubpopulations may still go extinct even if they are not competing among themselves. One way to model this is to say that the per colony extinction rate is a function \(m(p)\) of \(p\). In Problems 15 and 16 we will assume that \(m(p)=a+b p\) for some constants \(a, b>0 .\) That is, the extinction rate increases with \(p\) because of competition between subpopulations, but \(m(p)\) does not vanish as \(p \rightarrow 0\). Then our model for proportion of occupied sites must be modified to: $$ \frac{d p}{d t}=c p(1-p)-(a+b p) p $$ where \(c, a, b\) are all positive constants. Assuming that the subpopulations obey the differential Equation (8.60) and the coefficients are \(a=1, b=2\), but \(c\) is allowed to take any value: (a) Find the equilibrium values of \(p\) (your answer will depend on the unknown coefficient \(c\) ). (b) What are the conditions on \(c\) for \(p\) to have a nontrivial equilibrium, that is, an equilibrium in which \(p \in(0,1] ?\) (c) Show that if your condition from (b) is met, then the nontrivial equilibrium is also stable.

Step-by-Step Solution

Answer

Equilibrium values are \(p = 0\) and \(p = \frac{c-1}{c+2}\). For nontrivial equilibrium, \(c > 1\). This nontrivial equilibrium is stable if \(c > 1\).

Step 1: Set the Differential Equation to Zero

To find the equilibrium values of \( p \), we need to set the differential equation to zero:\[\frac{dp}{dt} = cp(1-p) - (a + bp)p = 0\]Plug in the coefficients \( a = 1 \) and \( b = 2 \):\[cp(1-p) - (1 + 2p)p = 0\]

Step 2: Simplify the Equation

Simplify the equation by distributing and combining like terms:\[cp - cp^2 - p - 2p^2 = 0\]Rearrange terms:\[(c - 1)p - (c + 2)p^2 = 0\]

Step 3: Factor the Equation

Factor out the common term \( p \):\[p((c - 1) - (c + 2)p) = 0\]This gives the solutions:\[p = 0 \quad \text{or} \quad p = \frac{c-1}{c+2}\]

Step 4: Determine Conditions for Nontrivial Equilibrium

For a nontrivial equilibrium, \( p \in (0,1] \). Thus, the solution \( p = \frac{c-1}{c+2} \) must satisfy:\[0 < \frac{c-1}{c+2} \leq 1\]The inequality \( \frac{c-1}{c+2} > 0 \) is satisfied when \( c > 1 \). The condition that \( \frac{c-1}{c+2} \leq 1 \) is always true for \( c > 1 \). Therefore, the condition for a nontrivial equilibrium is \( c > 1 \).

Step 5: Check Stability of Nontrivial Equilibrium

The stability of the equilibrium can be determined by examining the derivative of the right-hand side of the differential equation, evaluated at the equilibrium point.The derivative \( \frac{d}{dp} \left[ cp(1-p) - (1 + 2p)p \right] \) simplifies to:\[\frac{d}{dp} = c - 2cp - 1 - 4p\]Evaluate this derivative at \( p = \frac{c-1}{c+2} \):\[c - 2c\left(\frac{c-1}{c+2}\right) - 1 - 4\left(\frac{c-1}{c+2}\right)\]This value must be negative for stability. Simplifying gives:\[-\frac{(c-1)(c+3)}{c+2} < 0\]Since \( c > 1 \), this expression is indeed negative, indicating stability.

Key Concepts

Equilibrium Analysis

In the study of differential equations, equilibrium analysis is a critical tool. It allows us to find the states at which a system remains unchanged over time. For our population dynamics model, this means finding the values of the variable, in this case, proportion of occupied sites \( p \), that keeps the time derivative \( \frac{dp}{dt} \) equal to zero.

• First, we set the differential equation: \[ \frac{dp}{dt} = cp(1-p) - (a + bp)p = 0 \]
• This equation represents the change in proportion over time.

By analyzing the equilibrium at \( p = 0 \) and \( p = \frac{c-1}{c+2} \), we identify two potential equilibrium points. The first corresponds to a completely unoccupied state, while the second requires further exploration for nontrivial conditions.

Stability Analysis

Stability analysis helps us determine whether an equilibrium point will resist small disturbances over time or lead the system back to equilibrium after disturbances. This is crucial in understanding the long-term behavior of our population dynamics model. We use the first derivative test here:

• We take the derivative of the differential equation: \[ \frac{d}{dp} \left[ cp(1-p) - (1 + 2p)p \right] \]
• Evaluating this at \( p = \frac{c-1}{c+2} \) helps us determine stability.

For stability, this derivative must be negative, ensuring any small perturbation in \( p \) decay back to equilibrium. Our calculations showed stability when \( c > 1 \), meaning for appropriate conditions, the population dynamics will revert to equilibrium.

Mathematical Modeling

Mathematical modeling is an essential aspect that underpins the structured analysis of this exercise. It allows real-world phenomena, like population dynamics, to be represented using equations. These models help in predicting how populations might change over time under various conditions by:

• Incorporating known data and variables, like extinction rates \( m(p) \) and population interactions.
• Allowing exploration of different scenarios by adjusting constants \( a, b, c \).

This model uses constants \( a = 1 \), \( b = 2 \), and variable \( c \) to explore how competition between subpopulations and inherent extinction rates affect occupied proportions \( p \). Through this symbolic representation, insights are gained about the effects and stability of various population levels.

Population Dynamics

In this context, population dynamics refer to the ways populations change in size and composition over time. Differential equations are especially useful to model and predict these changes by considering factors like birth rates, death rates, and competitive interactions among subpopulations.
The equation \( \frac{d p}{d t} = cp(1-p) - (a+b p)p \) represents underlying biological processes:

• The term \( cp(1-p) \) models growth potential with a carrying capacity constraint.
• The term \( - (a+b p)p \) takes into account both constant and competition-driven extinction.

Together, these terms capture the complex mechanisms driving changes in occupied site proportions, providing a simplified view of how interactions and environmental constraints impact population survival and sustainability.