Problem 31 - Calculus for Biology and Medicine

Question

To derive the model for the growth or decline of the population of cooperators interacting in a snowdrift game, we modeled the proportion of cooperators using a model. $$ \frac{d x}{d t}=k n x(1-x)(-(b-c / 2) x+(b-c)) $$ where \(b>0\) represents the benefit of interaction if one player is a cooperator and \(c>0\) is the cost of cooperation. In Section 8.3.4 we analyzed Equation \((8.68)\) if \(b>c / 2 .\) Determine the equilibria and what their stability is if \(b=c / 2\).

Step-by-Step Solution

Answer

Equilibria are at \(x=0\) and \(x=1\), with \(x=0\) unstable and \(x=1\) stable.

Step 1: Understanding the Equation

We are given the differential equation that models the change in the proportion of cooperators in a snowdrift game. The equation is: \[ \frac{d x}{d t}=k n x(1-x)(-(b-c / 2) x+(b-c)) \] where \(b\) is the benefit and \(c\) is the cost. We want to analyze the equilibria and stability when \(b = \frac{c}{2}\).

Step 2: Substituting Condition b = c/2

Plug in \(b = \frac{c}{2}\) into the equation: \[ \frac{d x}{d t} = k n x(1-x)(-(\frac{c}{2} - \frac{c}{2}) x + (\frac{c}{2} - c)) \]. Simplifying, we get: \[ \frac{d x}{d t} = k n x(1-x)(0 \cdot x - \frac{c}{2}) \]. Therefore, \[ \frac{d x}{d t} = -\frac{k n c}{2} x(1-x) \].

Step 3: Finding Equilibria

The equilibria occur where \(\frac{d x}{d t} = 0\). From our simplified equation, set \(-\frac{k n c}{2} x(1-x) = 0\). This equation is zero if either \(x = 0\) or \(x = 1\). Thus, the potential equilibria are \(x = 0\) and \(x = 1\).

Step 4: Stability Analysis of Equilibria

To determine stability, examine the sign of the derivative around each equilibrium: \[ \frac{d(\frac{d x}{d t})}{dx} = -\frac{k n c}{2}((1-x) - x) = -\frac{k n c}{2}(1-2x) \]. Evaluate at each equilibrium: - At \(x = 0\), \(\frac{d}{dx} = -\frac{k n c}{2} > 0\), indicating instability.- At \(x = 1\), \(\frac{d}{dx} = \frac{k n c}{2} < 0\), indicating stability.

Key Concepts

Differential Equations

A differential equation is a mathematical equation that relates some function with its derivatives. In the context of population dynamics, it helps us understand how a population changes over time. Here, we are studying the proportion of cooperators in a snowdrift game using a specific differential equation. The form of the equation provided is a first-order ordinary differential equation. It tells us how the cooperator proportion, denoted by \(x\), changes as time progresses (\(t\)).The specific equation given is: \[\frac{d x}{d t} = k n x(1-x)(-(b-c / 2) x+(b-c))\]This part of the equation, \(\frac{d x}{d t}\), denotes the rate of change of the proportion of cooperators. The terms inside the parentheses, \(-(b-c / 2) x+(b-c)\), are responsible for capturing the interactions among players, where \(b\) is the benefit and \(c\) is the cost of cooperation.To analyze the equation, we need to substitute specific values for \(b\) and \(c\) and simplify the expression to know more about the dynamics of this population game.

Stability Analysis

Once we've identified the equilibria of a system, stability analysis helps us determine whether these points are stable or unstable. In simpler terms, it tells us if small changes in \(x\) will return to that equilibrium over time or diverge away.The stability depends on the derivative of the rate of change at the equilibrium point. In our solution, this is given as:\[\frac{d(\frac{d x}{d t})}{dx} = -\frac{k n c}{2}(1-2x)\]For stability analysis:

• If the derivative at an equilibrium point is positive, the small deviations from that point will grow, indicating instability.
• Conversely, if the derivative is negative, deviations decrease, pulling the state back to equilibrium, which indicates stability.

Using this, we found:- At \(x = 0\), the derivative is positive, so this point is unstable.- At \(x = 1\), the derivative is negative, which makes this point stable.

Equilibrium Points

Equilibrium points in the context of differential equations occur where the rate of change is zero, meaning the system experiences no net change at these points. For our population model, equilibria occur when:\[\frac{d x}{d t} = 0\]Plugging in the condition \(b = \frac{c}{2}\), we simplified the equation, leading us to the form:\[\frac{d x}{d t} = -\frac{k n c}{2} x(1-x)\]By setting the rate of change to zero, we solve for the points where either \(x = 0\) or \(x = 1\). These are the equilibrium points of our system.

• \(x = 0\): Here, no cooperators are present in the population. This inactivity represents an equilibrium where no one cooperates.
• \(x = 1\): The entire population consists of cooperators, representing complete cooperation.

Understanding these conditions provides insight into how the population's cooperative behavior can remain unchanged under certain dynamics.