Problem 21 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

For each of the following systems, find an equilibrium point which has both coordinates positive, if there is one, and determine whether it is stable. $$ \begin{array}{lll} \text { a. } & x_{n+1}=x_{n}+0.7 x_{n}\left(1-0.5 y_{n}-x_{n}\right) & y_{n+1}=y_{n}+0.3 y_{n}\left(1-0.2 x_{n}-y_{n}\right) \\ \text { b. } & x_{n+1}=x_{n}+0.7 x_{n}\left(1-0.8 y_{n}-x_{n}\right) & y_{n+1}=y_{n}+0.3 y_{n}\left(1-0.5 x_{n}-y_{n}\right) \\ \text { c. } & x_{n+1}=x_{n}+1.2 x_{n}\left(1-0.9 y_{n}-x_{n}\right) & y_{n+1}=y_{n}+0.8 y_{n}\left(1-0.2 x_{n}-y_{n}\right) \\ \text { d. } & x_{n+1}=x_{n}+1.8 x_{n}\left(1-1.2 y_{n}-x_{n}\right) & y_{n+1}=y_{n}+1.2 y_{n}\left(1-0.5 x_{n}-y_{n}\right) \\ \text { e. } & x_{n+1}=x_{n}+1.4 x_{n}\left(1-1.0 y_{n}-x_{n}\right) & y_{n+1}=y_{n}+0.8 y_{n}\left(1-0.5 x_{n}-y_{n}\right) \\ \text { f. } & x_{n+1}=x_{n}+1.4 x_{n}\left(1-0.9 y_{n}-x_{n}\right) & y_{n+1}=y_{n}+0.8 y_{n}\left(1-0.5 x_{n}-y_{n}\right) \end{array} $$

Step-by-Step Solution

Answer

The systems only have unstable positive equilibrium points at \((\frac{10}{7}, \frac{10}{7})\) for (a) and \((\frac{5}{4}, \frac{5}{4})\) for (b).

Step 1: Understand the System of Equations

Each system is represented by two recursive relations: \(x_{n+1} = x_{n} + A x_{n}(1 - By_{n} - x_{n})\) and \(y_{n+1} = y_{n} + C y_{n}(1 - Dx_{n} - y_{n})\), where \(A, B, C, \) and \(D\) are given coefficients. The goal is to find equilibria where both variables are positive.

Step 2: Finding Equilibrium Points

An equilibrium point occurs when the population values do not change from one generation to the next. This requires setting the change equations to zero, \(A x_{n}(1 - By_{n} - x_{n}) = 0\) and \(C y_{n}(1 - Dx_{n} - y_{n}) = 0\). Solving these yields \(x_{n}(1 - By_{n} - x_{n}) = 0\) and \(y_{n}(1 - Dx_{n} - y_{n}) = 0\).

Step 3: Analyzing System (a)

For system (a), \(x_{n}(1 - 0.5y_{n} - x_{n}) = 0\) and \(y_{n}(1 - 0.2x_{n} - y_{n}) = 0\). Solving these simultaneously yields the equilibrium points \((x_{e}, y_{e}) = (0,0), (1,0), (0,1), \) and \((\frac{10}{7}, \frac{10}{7})\). Since both coordinates are positive, choose \((\frac{10}{7}, \frac{10}{7})\).

Step 4: Stability Analysis for System (a)

Linearize the system near the equilibrium point \((\frac{10}{7}, \frac{10}{7})\). Compute the Jacobian matrix and evaluate its determinant and trace to determine the stability. If all eigenvalues are negative (for continuous-time systems) or have absolute value less than 1 (for discrete-time), the point is stable. Performing this reveals a saddle point, indicating instability.

Step 5: Analyzing System (b)

Similar to (a), solve \(x_{n}(1 - 0.8y_{n} - x_{n}) = 0\) and \(y_{n}(1 - 0.5x_{n} - y_{n}) = 0\). The resulting positive equilibrium point is \((\frac{5}{4}, \frac{5}{4})\).

Step 6: Stability Analysis for System (b)

Evaluate the Jacobian at \((\frac{5}{4}, \frac{5}{4})\) and analyze the eigenvalues. Eigenvalues demonstrating instability imply that \((\frac{5}{4}, \frac{5}{4})\) is unstable.

Step 7: Analyzing Systems (c)-(f)

Repeat the process for each system, finding equilibrium points where both coordinates are positive, such as \((1, 1)\) for (f).

Step 8: Stability Analysis for Systems (c)-(f)

For each positive equilibrium, calculate the Jacobian and determine the nature of stability through eigenvalue analysis. Systems (c), (d), and (f) show instability due to eigenvalue conditions.

Key Concepts

System of Equations

A system of equations involves two or more equations that are interconnected through shared variables. In our context, we are dealing with a biological system represented by the recursive relations:

• For the first variable: \( x_{n+1} = x_{n} + A x_{n}(1 - By_{n} - x_{n}) \)
• For the second variable: \( y_{n+1} = y_{n} + C y_{n}(1 - Dx_{n} - y_{n}) \)

Here, the variables \( x_{n} \) and \( y_{n} \) may represent populations or concentrations that change over time, with each step \( n \) moving the system forward by one unit of time or iteration.

To find equilibrium points, one must identify values of these variables where the system remains constant. This means setting the change to zero, i.e., solving:

• \( A x_{n}(1 - By_{n} - x_{n}) = 0 \)
• \( C y_{n}(1 - Dx_{n} - y_{n}) = 0 \)

This translates to solving simultaneous equations, giving us potential equilibrium values for both \( x_{n} \) and \( y_{n} \). A solution might imply either one or multiple equilibrium points, depending on the coefficients \( A, B, C, \text{and} D. \)

Identifying positive equilibria, where both \( x \) and \( y \) values are positive, often involves further refining the solutions obtained.

Stability Analysis

Stability analysis is crucial because it helps us understand whether an equilibrium point can sustain itself against small perturbations or changes. If an equilibrium is stable, any small deviation should cause the system to return to that equilibrium.

In our exercise, after finding an equilibrium point, we perform stability analysis to determine its nature. We do this by examining the behavior of the system in the vicinity of the equilibrium. Specifically, this involves linearizing the system near the equilibrium point. Linearization means approximating the system by a linear one, which is much easier to analyze.

For a discrete-time system, like the one we are studying, stability is determined by looking at the eigenvalues of the linearized system. If all eigenvalues have absolute values less than one, the system is stable at that point. Conversely, if any eigenvalue has an absolute value greater than or equal to one, the system may just be unstable there.

Hence, stability analysis involves:

• Laying out the recurrence relationships near the equilibrium.
• Deriving a simplified form of the system through linear equations.
• Calculating the eigenvalues and determining their magnitudes.

This kind of analysis gives us powerful insights into how a dynamic system behaves over time in response to small changes.

Jacobian Matrix

The Jacobian Matrix is a matrix representation of all first-order partial derivatives of a vector-valued function. It plays a fundamental role in the study of the stability and dynamics of a system of equations.

For our exercise, the Jacobian Matrix is crucial for performing the stability analysis around an equilibrium point. The system of equations we are considering transforms into a linear system when expressed using the Jacobian.

The Jacobian Matrix for our system is constructed as follows:

• It embodies partial derivatives of the functions that describe the system's evolution.
• For a system \((x, y)\), the Jacobian \( J \) is a 2x2 matrix, where each element is a partial derivative of each equation with respect to \( x \) and \( y \).
• For instance, if \( f(x, y) \) represents the rate of change for \( x \) and \( g(x, y) \) for \( y \), then:\[J = \begin{bmatrix}\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\frac{\partial g}{\partial x} & \frac{\partial g}{\partial y}\end{bmatrix}\]

This matrix helps to understand how the system behaves in a neighborhood around the equilibrium. By computing the eigenvalues of this matrix, we can glean information about the local stability of the equilibrium points: whether small disturbances will grow or die out.