Problem 8 - Calculus for Biology and Medicine

Question

Find all equilibria of each system of differential equations and use the analytical approach to determine the stability of each equilibrium. $$ \begin{array}{l} \frac{d x_{1}}{d t}=-x_{1}+3 x_{1}\left(1-x_{1}-x_{2}\right) \\ \frac{d x_{2}}{d t}=-x_{2}+5 x_{2}\left(1-x_{1}-x_{2}\right) \end{array} $$

Step-by-Step Solution

Answer

Equilibria points are (0,0), (\( \frac{2}{3}, 0 \)), and (0, \( \frac{4}{5} \)); all are unstable.

Step 1: Understand the System

We are given a system of differential equations: \( \frac{d x_{1}}{d t} = -x_{1} + 3x_{1}(1-x_{1}-x_{2}) \) and \( \frac{d x_{2}}{d t} = -x_{2} + 5x_{2}(1-x_{1}-x_{2}) \). Our task is to find equilibrium points where both derivatives are zero.

Step 2: Set the System to Zero

To find equilibria, set both equations to zero: \( -x_{1} + 3x_{1}(1-x_{1}-x_{2}) = 0 \) and \( -x_{2} + 5x_{2}(1-x_{1}-x_{2}) = 0 \). These will give us conditions for \( x_1 \) and \( x_2 \) in equilibrium.

Step 3: Solve First Equation for Equilibria

Factor the first equation: \( x_{1}(-1 + 3(1-x_{1}-x_{2})) = 0 \). The solutions are \( x_1 = 0 \) or \( -1 + 3 - 3x_{1} - 3x_{2} = 0 \). Simplifying gives \( 2 = 3x_1 + 3x_2 \) leading to \( x_2 = \frac{2}{3} - x_1 \).

Step 4: Solve Second Equation for Equilibria

Similarly, factor the second equation: \( x_{2}(-1 + 5 - 5x_{1} - 5x_{2}) = 0 \). The solutions are \( x_2 = 0 \) or \( 4 = 5x_1 + 5x_2 \), simplifying to \( x_2 = \frac{4}{5} - x_1 \).

Step 5: Combine Solutions to Find Equilibria Points

Equate \( \frac{2}{3} - x_1 \) and \( \frac{4}{5} - x_1 \) to find common points. Upon solving, we find no overlap in the region where both terms are real, indicating only the edge cases need checking: (0,0), (\( \frac{2}{3}, 0 \)), and (0, \( \frac{4}{5} \)).

Step 6: Assess Stability via Jacobian

Calculate the Jacobian matrix: \[ J = \begin{bmatrix} \frac{\partial \dot{x_1}}{\partial x_1} & \frac{\partial \dot{x_1}}{\partial x_2} \ \frac{\partial \dot{x_2}}{\partial x_1} & \frac{\partial \dot{x_2}}{\partial x_2} \end{bmatrix} \]. This results in \[ J = \begin{bmatrix} \frac{d }{d x_1}(-x_1+3x_1(1-x_1-x_2)) & \frac{d }{d x_2}(-x_1+3x_1(1-x_1-x_2)) \ \frac{d }{d x_1}(-x_2+5x_2(1-x_1-x_2)) & \frac{d }{d x_2}(-x_2+5x_2(1-x_1-x_2)) \end{bmatrix} \].

Step 7: Evaluate Jacobian at Equilibria

Substitute each equilibrium point into the Jacobian to determine stability: (0,0) is unstable, (\( \frac{2}{3}, 0 \)) has eigenvalues indicating instability, and (0, \( \frac{4}{5} \)) indicates instability.

Key Concepts

Equilibrium Points

Equilibrium points are where a system comes to rest, meaning no net change in the system's components. To find these, we identify the points where both system derivatives are zero. For a differential equation system, these points lie where rates of change reach zero for all involved variables. In this problem, for the system \( \frac{d x_1}{d t} = -x_1 + 3x_1(1-x_1-x_2) \) and \( \frac{d x_2}{d t} = -x_2 + 5x_2(1-x_1-x_2) \), equilibrium means solving \( -x_1 + 3x_1(1-x_1-x_2) = 0 \) and \( -x_2 + 5x_2(1-x_1-x_2) = 0 \). By setting these equations to zero, we focus on the values that bring about a state of unchanging behavior. These points represent critical moments where the system won't drift, allowing us to understand potential stable states or transitions.

Stability Analysis

Stability analysis helps us determine whether these equilibrium points are stable, meaning small disturbances won't lead to significant changes in the system over time. An unstable point suggests any slight perturbation will tend to grow, driving the system away from equilibrium. - We determine stability by calculating the eigenvalues of the Jacobian matrix at each equilibrium point. - If all eigenvalues have negative real parts, the system is stable at that point. - Having eigenvalues with positive real parts signals instability. In the problem's solution, after evaluating the Jacobian at the relevant points, (0,0) and others showed instability. This means that the system at these points, when disturbed slightly, will not return to equilibrium but rather diverge or move away.

Jacobian Matrix

The Jacobian matrix plays a crucial role in assessing stability by capturing how small changes in variables impact the system's behavior. Each element in the Jacobian represents a partial derivative of the system's equations, essentially measuring the sensitivity of each equation to each variable. - For a system \( \frac{d x_1}{d t} \) and \( \frac{d x_2}{d t} \), the Jacobian matrix \( J \) is given by: \[ J = \begin{bmatrix} \frac{\partial \dot{x_1}}{\partial x_1} & \frac{\partial \dot{x_1}}{\partial x_2} \ \frac{\partial \dot{x_2}}{\partial x_1} & \frac{\partial \dot{x_2}}{\partial x_2} \end{bmatrix} \] - Evaluating this matrix at each equilibrium point determines stability. - Eigenvalues of the Jacobian provide insight: stable equilibrium points yield all negative eigenvalues. The matrix's structure reveals the interdependencies of the system's variables, highlighting critical balance points.