Problem 6 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Draw the nullclines and some direction arrows and analyze the equilibria of the following symbiosis models. $$ \begin{aligned} \text { a. } \quad & x^{\prime}(t)=0.2 \times x(t) \times(1-0.5 x(t)+0.4 y(t)) \\\ & y^{\prime}(t)=0.1 \times y(t) \times(1+0.4 x(t)-0.5 y(t)) \\ \text { b. } \quad & x^{\prime}(t)=0.2 \times x(t) \times(1-0.8 x(t)+0.4 y(t)) \\\ & y^{\prime}(t)=0.1 \times y(t) \times(1+0.4 x(t)-0.5 y(t)) \\ \text { c. } \quad & x^{\prime}(t)=0.2 \times x(t) \times(1-0.5 x(t)+0.4 y(t)) \\\ & y^{\prime}(t)=0.1 \times y(t) \times(1+0.4 x(t)-0.2 y(t)) \\ \text { d. } x^{\prime}(t) &=0.2 \times x(t) \times(1-5 x(t)+10 y(t)) \\ & y^{\prime}(t)=0.1 \times y(t) \times(1+2 x(t)-5 y(t)) \\ \text { e. } \quad & x^{\prime}(t)=0.2 \times x(t) \times(1-1.1 x(t)+y(t)) \\ & y^{\prime}(t)=0.1 \times y(t) \times(1+x(t)-y(t)) \\ \text { f. } & x^{\prime}(t)=0.2 \times x(t) \times(1-0.9 x(t)+y(t)) \\ & y^{\prime}(t)=0.1 \times y(t) \times(1+x(t)-y(t)) \end{aligned} $$

Step-by-Step Solution

Answer

Identify nullclines, solve for equilibrium points, and analyze stability using direction arrows and Jacobian matrix.

Step 1: Nullclines of the Symbiosis Model

To find the nullclines, we set the derivatives \(x'(t)\) and \(y'(t)\) to zero and solve the resulting equations. For part a, the nullcline equations are:\[x'(t) = 0 : 0.2 \cdot x(t) \cdot (1 - 0.5x(t) + 0.4y(t)) = 0 \y'(t) = 0 : 0.1 \cdot y(t) \cdot (1 + 0.4x(t) - 0.5y(t)) = 0\]This gives us the conditions \( x(t) = 0 \), \( y(t) = 0 \), and equations \(1-0.5x(t)+0.4y(t)=0\) and \(1+0.4x(t)-0.5y(t)=0\) which describe the curves on the \(xy\)-plane where the system is at rest. Repeat similar calculations for the other cases (b to f).

Step 2: Solve for Nullcline Equations

Solve the equations obtained from Step 1. For part a,1. **From \(x'(t) = 0\):** - Case 1: \(x(t) = 0\) - Case 2: \(y(t) = \frac{1+0.5x(t)}{0.4}\).2. **From \(y'(t) = 0\):** - Case 1: \(y(t) = 0\) - Case 2: \(y(t) = \frac{1+0.4x(t)}{0.5}\).Repeat for each model b to f.

Step 3: Equilibrium Points Analysis

Solve simultaneous equations formed by nullclines to find equilibrium points. For part a, solve:\[1 - 0.5x + 0.4y = 0 \1 + 0.4x - 0.5y = 0\]Solving these equations gives equilibrium points \((x^*, y^*)\). Repeat for cases b to f, finding the specific equilibrium points for each system.

Step 4: Direction Fields and Arrows

To understand the dynamics near equilibrium points, sketch a direction field by plugging several (\(x,y\)) pairs into the original differential equations and observe the direction of \((x'(t), y'(t))\). For instance, where \(x'(t)\) is positive and \(y'(t)\) is negative, draw arrows indicating movement in the xy-plane in that direction. Apply this analysis in vicinity of the equilibrium points to determine stability.

Step 5: Stability Analysis of Equilibria

Examine the signs of \(x'(t)\) and \(y'(t)\) around equilibrium points to determine their stability. If both derivatives are negative, the point is stable/attractive; if both can switch signs under small perturbations, the point is unstable. For Part a, evaluate the Jacobian matrix at equilibrium points to determine the type (stable node, saddle point, etc.). Repeat the process for b to f using each model's specific parameters in the system.

Key Concepts

Nullclines

Nullclines are a foundational concept when analyzing systems of differential equations, especially in biological models like the ones analyzed in this exercise.
Nullclines help us understand where changes in the system pause or stabilize momentarily. These are curves in the phase plane where the time derivative of either variable, say, \(x\) or \(y\), is zero, meaning that variable does not change at those points.

Finding nullclines involves setting equations \(x'(t)\) and \(y'(t)\) to zero and solving for the conditions under which these derivatives vanish.
For example, in the exercise, the nullcline equations for parts like (a) are derived by setting the differential equations of \(x'(t) = 0.2 \times x(t) \times (1 - 0.5x(t) + 0.4y(t)) = 0\) and \(y'(t) = 0.1 \times y(t) \times (1 + 0.4x(t) - 0.5y(t)) = 0\) to zero.

By finding the nullclines in this manner, we can outline particular conditions in which the system does not alter in time—this provides critical insight into the natural steady states of dynamic biological interactions, guiding further examination toward equilibrium points and their stability.

Equilibrium Points

Equilibrium points are specific points where both variables in our system remain constant over time, as all derivatives are zero at these points.
To find these points, we solve the system of equations formed by intersecting nullclines.

In simpler terms, equilibrium points come about due to the intersection of nullclines, meaning these points satisfy all the equations of the nullclines simultaneously. For instance, in our example, we solve the equations \(1 - 0.5x + 0.4y = 0\) and \(1 + 0.4x - 0.5y = 0\) to find the exact coordinates \((x^*, y^*)\) indicating equilibrium.
Each point tells us where the organisms' interactions reach a state of balance, essential for understanding how populations might maintain or alter their sizes over time without external influences.

Equilibrium points can dictate the fate of interacting species in symbiosis models, offering insights into natural coexistence or potential extinction scenarios, based on system parameters and initial conditions.

Stability Analysis

Stability analysis determines whether equilibrium points identified remain steady under small disturbances.
We assess this by examining the Jacobian matrix or observing the signs of the derivatives around these equilibrium points.

Stable equilibrium points attract nearby starting points in the phase space, pulling trajectories towards them over time. This means both \(x'(t)\) and \(y'(t)\) remain subtle in changes and tend toward zero.
Unstable equilibria exhibit the opposite behavior, where minor variations can grow, potentially leading the system to diverge or approach other dynamics. In practical scenarios, unstable points may represent unsustainable population conditions or volatile environments.

To systematically assess stability, calculate the Jacobian matrix at equilibrium points and evaluate its eigenvalues. The sign of these eigenvalues gives hints about the stability classification:

• If all eigenvalues are negative, the equilibrium is stable (a local attractor).
• If any eigenvalue is positive, the equilibrium can be unstable.
• Mixed signs could indicate saddle points, showing partial stability.

This analysis tells us whether the populations in a model system, like our symbiosis models, coexist sustainably or disappear after minor changes, enriching our understanding of complex ecological interactions.