Problem 29 - Biocalculus Calculus for the Life Sciences

Question

$$\begin{array}{c}{29-31 \text { Consumer resource models often have the following }} \\ {\text { general form }} \\ {R^{\prime}=f(R)-g(R, C) \quad C^{\prime}=\varepsilon g(R, C)-h(C)} \\ {\text { where } R \text { is the number of individuals of the resource and } C \text { is }} \\ {\text { the number of consumers. The function } f(R) \text { gives the rate of }} \\\ {\text { replenishment of the resource, } g(R, C) \text { describes the rate of }}\end{array}$$ $$ \begin{array}{l}{\text { replenishment of the resource, } g(R, C) \text { describes the rate of }} \\ {\text { consumption of the resource, and } h(C) \text { is the rate of loss of the }} \\ {\text { consumer. The constant } \varepsilon, \text { where } 0<\varepsilon<1, \text { is the conversion }} \\\ {\text { efficiency of resources into consumers. Find all equilibria of the }} \\\ {\text { following examples and determine their stability properties. }}\end{array}$$ $$\begin{array}{l}{\text { A chemostat is an experimental consumer-resource }} \\\ {\text { system. If the resource is not self-reproducing, then it can }} \\\ {\text { be modeled by choosing } f(R)=\theta, g(R, C)=b R C, \text { and }} \\ {h(C)=\mu C . \text { Suppose } \theta=2, b=1, \varepsilon=1, \text { and } \mu=1}\end{array}$$

Step-by-Step Solution

Answer

Equilibria: (2,0) (unstable), (1,2) (stable).

Step 1: Understand the Model Equations

Given the model equations: \( R' = f(R) - g(R, C) \) and \( C' = \varepsilon g(R, C) - h(C) \), We have specific functions to substitute: \( f(R) = \theta \), \( g(R, C) = bRC \), and \( h(C) = \mu C \). With \( \theta = 2 \), \( b = 1 \), \( \varepsilon = 1 \), and \( \mu = 1 \), the equations become: \[ R' = 2 - RC \] \[ C' = RC - C \].

Step 2: Find Equilibria

To find equilibria, set \( R' = 0 \) and \( C' = 0 \):1. From \( R' = 2 - RC = 0 \), we get \( 2 = RC \).2. From \( C' = RC - C = 0 \), we get \( RC = C \), assuming \( C eq 0 \), which results in \( R = 1 \).Substituting \( R = 1 \) in \( 2 = RC \) gives \( 2 = 1\cdot C \), so \( C = 2 \).Equilibria are: \( (R, C) = (2, 0) \) and \( (1, 2) \).

Step 3: Determine Stability using Jacobian Matrix

Construct the Jacobian matrix for the system:\[ J = \begin{bmatrix} -C & -R \ C & R - 1 \end{bmatrix} \]Evaluate \( J \) at each equilibrium:

Step 3.1: Evaluate Jacobian at (2, 0)

The Jacobian at \( (R, C) = (2, 0) \) is:\[ J = \begin{bmatrix} 0 & -2 \ 0 & 1 \end{bmatrix} \]The eigenvalues of this matrix are \( \lambda_1 = 1 \) and \( \lambda_2 = 0 \). Since one eigenvalue is positive, the equilibrium at \( (2, 0) \) is unstable.

Step 3.2: Evaluate Jacobian at (1, 2)

The Jacobian at \( (R, C) = (1, 2) \) is:\[ J = \begin{bmatrix} -2 & -1 \ 2 & 0 \end{bmatrix} \]The characteristic equation is given by: \[ \lambda^2 + 2\lambda + 2 = 0 \]Solving this quadratic using the quadratic formula, \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), gives \( \lambda = -1 \pm i \).The eigenvalues are complex with negative real parts, indicating that the equilibrium at \( (1, 2) \) is stable (spiral sink).

Key Concepts

Equilibrium Analysis

In consumer-resource models, equilibrium analysis involves finding the conditions where the system reaches a state of balance. This occurs when the change rates of both consumers and resources are zero. To determine equilibria, we set the derivatives of the resource and consumer equations to zero. For the given problem, the equations are:

• \( R' = f(R) - g(R, C) \)
• \( C' = \varepsilon g(R, C) - h(C) \)

Substituting the specific functions and constants: \( f(R) = 2 \), \( g(R, C) = RC \), and \( h(C) = C \), the equations become \( R' = 2 - RC \) and \( C' = RC - C \). Setting \( R' = 0 \) and \( C' = 0 \) helps us find the equilibrium points. Solving these gives us the equilibria points \((R, C) = (2, 0)\) and \((1, 2)\). These points indicate where the growth of resources matches their consumption by consumers and where consumer numbers stabilize relative to the available resources.

Jacobian Matrix

The Jacobian matrix is a crucial tool in analyzing systems of differential equations, such as those found in consumer-resource models. It provides a linear approximation of the system by representing it as a matrix of partial derivatives. For our system, the Jacobian matrix \( J \) is:\[J = \begin{bmatrix} \frac{\partial R'}{\partial R} & \frac{\partial R'}{\partial C} \ \frac{\partial C'}{\partial R} & \frac{\partial C'}{\partial C} \end{bmatrix} = \begin{bmatrix} -C & -R \ C & R - 1 \end{bmatrix}\]Evaluating this matrix at each equilibrium point helps us understand the local behavior of the system around those points. The nature of the eigenvalues derived from this matrix determines stability, with real and positive eigenvalues suggesting instability, and complex eigenvalues frequently indicating stability under certain conditions.

Stability of Equilibria

The stability of equilibria in consumer-resource models reveals whether small perturbations will decay or grow, impacting the system's long-term behavior. To assess this, we analyze the eigenvalues of the Jacobian matrix evaluated at each equilibrium point. For instance, at equilibrium \((2, 0)\), the Jacobian matrix yields eigenvalues \( \lambda_1 = 1 \) and \( \lambda_2 = 0 \). A positive eigenvalue indicates instability, suggesting that the system is repelled away from this equilibrium point.

Conversely, for equilibrium \((1, 2)\), the eigenvalues \( \lambda = -1 \pm i \) are complex with negative real parts. Such eigenvalues denote a stable equilibrium known as a spiral sink. This means any disturbances around this point will decay over time, and the system will return to equilibrium, characterized by oscillatory behavior. Understanding these stability properties is vital for predicting the dynamics of ecosystems, ensuring sustainable management practices.