Problem 5 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

For the symbiosis system, $$ \begin{aligned} x_{n+1}-x_{n} &=\frac{5}{98} \cdot x_{n}\left(1+0.4 y_{n}-x_{n}\right) \\ y_{n+1}-y_{n} &=\frac{7}{120} \cdot y_{n}\left(1+\frac{5}{7} x_{n}-y_{n}\right) \end{aligned} $$ (1.76,2.0) is an equilibrium point. Choose the initial value, \(\left(x_{0}, y_{0}\right)=(1.76,2.1)\) and compute \(\left(x_{1}, y_{1}\right),\) \(\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right) .\)

Step-by-Step Solution

Answer

(x_1, y_1) = (1.8531, 2.13144), (x_2, y_2) = (1.94714, 2.15823), (x_3, y_3) = (2.04802, 2.18871).

Step 1: Find difference for x

Calculate the change in \( x \) from \( x_0 \) to \( x_1 \). Substitute in the given equation:\[ x_{1} - 1.76 = \frac{5}{98} \times 1.76 \times (1 + 0.4 \times 2.1 - 1.76) \]Simplify: \( 1 + 0.4 \times 2.1 - 1.76 = 1.04 \).Thus,\[ x_{1} - 1.76 = \frac{5}{98} \times 1.76 \times 1.04 \]Calculate:\( x_{1} - 1.76 \approx 0.0931 \).Therefore, \( x_{1} = 1.76 + 0.0931 = 1.8531 \).

Step 2: Compute y_1

Calculate the change in \( y \) from \( y_0 \) to \( y_1 \). Substitute in the given equation:\[ y_{1} - 2.1 = \frac{7}{120} \times 2.1 \times \left(1 + \frac{5}{7} \times 1.76 - 2.1\right) \]Simplify: \( 1 + \frac{5}{7} \times 1.76 - 2.1 = 0.2571 \).Thus,\[ y_{1} - 2.1 = \frac{7}{120} \times 2.1 \times 0.2571 \]Calculate:\( y_{1} - 2.1 \approx 0.03144 \).Therefore, \( y_{1} = 2.1 + 0.03144 = 2.13144 \).

Step 3: Calculate x_2

Now calculate \( x_{2} \) using \( x_{1} = 1.8531 \) and \( y_{1} = 2.13144 \):\[ x_{2} - 1.8531 = \frac{5}{98} \times 1.8531 \times (1 + 0.4 \times 2.13144 - 1.8531) \]Simplify and compute:\( 1 + 0.4 \times 2.13144 - 1.8531 = 0.99948 \).Thus,\[ x_{2} - 1.8531 = \frac{5}{98} \times 1.8531 \times 0.99948 \]Calculate:\( x_{2} - 1.8531 \approx 0.09404 \).Therefore, \( x_{2} = 1.8531 + 0.09404 = 1.94714 \).

Step 4: Compute y_2

Calculate \( y_{2} \) using \( x_{1} \) and previously found \( y_{1} \):\[ y_{2} - 2.13144 = \frac{7}{120} \times 2.13144 \times \left(1 + \frac{5}{7} \times 1.8531 - 2.13144\right) \]Simplify and compute:\( 1 + \frac{5}{7} \times 1.8531 - 2.13144 = 0.20944 \).Thus,\[ y_{2} - 2.13144 = \frac{7}{120} \times 2.13144 \times 0.20944 \]Calculate:\( y_{2} - 2.13144 \approx 0.02679 \).Therefore, \( y_{2} = 2.13144 + 0.02679 = 2.15823 \).

Step 5: Find x_3

Calculate \( x_{3} \) using \( x_{2} = 1.94714 \) and \( y_{2} = 2.15823 \):\[ x_{3} - 1.94714 = \frac{5}{98} \times 1.94714 \times (1 + 0.4 \times 2.15823 - 1.94714) \]Simplify and compute:\( 1 + 0.4 \times 2.15823 - 1.94714 = 1.01419 \).Thus,\[ x_{3} - 1.94714 = \frac{5}{98} \times 1.94714 \times 1.01419 \]Calculate:\( x_{3} - 1.94714 \approx 0.10088 \).Therefore, \( x_{3} = 1.94714 + 0.10088 = 2.04802 \).

Step 6: Calculate y_3

Calculate \( y_{3} \) using \( x_{2} = 1.94714 \) and \( y_{2} = 2.15823 \):\[ y_{3} - 2.15823 = \frac{7}{120} \times 2.15823 \times \left(1 + \frac{5}{7} \times 1.94714 - 2.15823\right) \]Simplify and compute:\( 1 + \frac{5}{7} \times 1.94714 - 2.15823 = 0.22863 \).Thus,\[ y_{3} - 2.15823 = \frac{7}{120} \times 2.15823 \times 0.22863 \]Calculate:\( y_{3} - 2.15823 \approx 0.03048 \).Therefore, \( y_{3} = 2.15823 + 0.03048 = 2.18871 \).

Key Concepts

Symbiosis Systems

In the study of mathematical biology, symbiosis systems describe the interaction between species that live closely together. Such systems can be represented using mathematical models that show how different species affect one another. For instance, in the provided exercise, we are dealing with a two-species symbiosis model where the interaction between species is represented by difference equations. Each equation describes how one species' population at the next step is affected by both its current population and the current population of the other species.

These interactions might be beneficial like mutualism or antagonistic like parasitism, depending on how the species influence each other's growth. In our system:

• The presence of one species contributes positively or negatively to the other's growth rate.
• Equilibrium points, like the suggested \((1.76, 2.0)\), represent populations where the system stabilizes, meaning populations remain constant if outsiders do not disrupt them.
• Such equilibrium states are crucial for understanding sustainable populations over time.

This model, using mathematical equations, helps illustrate the potential outcomes from different initial populations and allows predictions on how these populations may evolve or stabilize over time.

Difference Equations

Difference equations are a core part of modeling dynamic systems that evolve over discrete time steps, like the symbiosis system in the exercise. Unlike differential equations, which apply to continuous time models, difference equations apply when changes occur at specific intervals, like generations in population studies.

In our symbiosis problem:

• The equation for each species calculates the change in its population from one step to the next.
• They incorporate factors such as the previous population and the influence of the population of other species.

The beauty of difference equations lies in their simplicity and effectiveness in predicting future states of a system given an initial state. The formula for change in population involves constants like \(\frac{5}{98}\) or \(\frac{7}{120}\), representing growth rates, and these constants are crucial to understanding how populations interact over time.

By solving these equations step by step:

• Students can calculate future populations, predicting the next few points, such as \((x_1, y_1), (x_2, y_2),\) and so on.
• This iterative approach helps visualize the trajectory of population changes.

Understanding each variable's contribution to the equation allows for deeper insights into how symbiosis affects population dynamics.

Iterative Calculations

Iterative calculations are essential for solving difference equations in symbiosis systems. They involve starting with an initial state and and using repeated application of the same mathematical process to calculate future states. Each iteration builds upon previously calculated values, mimicking how real systems evolve over time.

In our exercise, the iterations were performed as follows:

• Given initial values \((x_0, y_0) = (1.76, 2.1)\), we calculate subsequent populations using the provided difference equations.
• Each iterative step involves calculating the next state (e.g., \(x_1, y_1\)) based on the current state (e.g., \(x_0, y_0\)).
• After solving for \((x_{n+1}, y_{n+1})\), this becomes the new current state, and the process repeats.
• With each step, precision in calculation is crucial, as small errors can significantly impact the final outcome.

Iterations are particularly illustrative of how systems gradually shift toward equilibrium or grow/decline based on their initial parameters and interactions between species. This incremental process allows for clear visualization of the system's evolution and highlights how initial choices and errors can propagate through a sequence of outcomes.