Problem 9 - Calculus for Biology and Medicine

Question

Find all equilibria of each system of differential equations and determine the stability of each equilibrium. \(\frac{d x_{1}}{d t}=4 x_{1}\left(1-x_{1}\right)-2 x_{1} x_{2}\) \(\frac{d x_{2}}{d t}=x_{2}\left(2-x_{2}\right)-x_{2}\)

Step-by-Step Solution

Answer

Equilibria: \((0,0), (1,0), (0,1), \left(\frac{1}{2}, 1\right)\). Stability: \((0,0)\) unstable, \((1,0)\) saddle, \((0,1)\) saddle, \(\left(\frac{1}{2}, 1\right)\) neutrally stable.

Step 1: Find Equilibria

To find the equilibria, set the right-hand side of each differential equation to zero. For equation (1), \( \frac{dx_1}{dt} = 4x_1(1-x_1) - 2x_1x_2 = 0 \). Factor and solve: \( x_1[4(1-x_1) - 2x_2] = 0 \). This gives: \( x_1 = 0 \) or \( 4 - 4x_1 - 2x_2 = 0 \). For equation (2), \( \frac{dx_2}{dt} = x_2(2-x_2) - x_2 = 0 \). Factor and solve: \( x_2[2-x_2-1] = 0 \), giving solutions: \( x_2 = 0 \) or \( x_2 = 1 \). The potential equilibria are found by substituting these values into each other.

Step 2: Solve Equilibrium Equations

Substitute the equilibrium conditions \( x_1 = 0 \) and \( x_2 = 0 \), as well as combinations such as \( x_1 = 0, x_2 = 1 \), \( x_1 = 1, x_2 = 0 \), etc. Another condition from equation (1) gives \( x_2 = 2 - 2x_1 \). Solve these alongside \( x_2 = 0 \) or \( x_2 = 1 \) to find equilibrium points. The solution gives equilibria as \( (x_1, x_2) = (0,0), (1,0), (0,1), \text{ and } \left(\frac{1}{2}, 1 \right) \).

Step 3: Assess Stability Around Each Equilibrium

Linearize the system near each equilibrium point using the Jacobian matrix. The Jacobian \( J \) is computed as: \[ J = \begin{bmatrix} \frac{\partial f}{\partial x_1} & \frac{\partial f}{\partial x_2} \ \frac{\partial g}{\partial x_1} & \frac{\partial g}{\partial x_2} \end{bmatrix} \]. Calculate the partial derivatives for each component function: \( \frac{\partial f}{\partial x_1} = 4 - 8x_1 - 2x_2 \), \( \frac{\partial f}{\partial x_2} = -2x_1 \), \( \frac{\partial g}{\partial x_1} = 0 \), \( \frac{\partial g}{\partial x_2} = 1 - 2x_2 \). Evaluate \( J \) at each equilibrium point to determine its matrix form.

Step 4: Compute Eigenvalues for Stability Analysis

Evaluate the Jacobian matrix at each equilibrium point and calculate the eigenvalues. For \( (0,0) \), the Jacobian is \( \begin{bmatrix} 4 & 0 \ 0 & 1 \end{bmatrix} \), with eigenvalues 4 and 1, indicating an unstable node. For \( (1,0) \), the Jacobian is \( \begin{bmatrix} -4 & -2 \ 0 & 1 \end{bmatrix} \), with eigenvalues -4 and 1, indicating a saddle point. For \( (0,1) \), the Jacobian is \( \begin{bmatrix} 2 & 0 \ 0 & -1 \end{bmatrix} \), with eigenvalues 2 and -1, indicating a saddle point. For \( \left(\frac{1}{2}, 1 \right) \), Jacobian \( \begin{bmatrix} 0 & -1 \ 0 & -1 \end{bmatrix} \), gives eigenvalues 0 and -1, indicating neutral stability.

Key Concepts

Equilibria

Equilibria in a system of differential equations are the points where the system remains constant over time. For each differential equation in our system, we must set the rate of change to zero and solve for the variable values that satisfy this condition. This process involves setting \( \frac{dx_1}{dt} \) and \( \frac{dx_2}{dt} \) to zero, leading us to solve two algebraic equations.

• For the first equation, \( x_1 = 0 \) or \( 4 - 4x_1 - 2x_2 = 0 \)
• For the second equation, \( x_2 = 0 \) or \( x_2 = 1 \)

By substituting these into each other, we find the combinations of \( x_1 \) and \( x_2 \) that lead to the system equilibrating. Thus, our equilibria are points where the system's dynamic behavior ceases.

Stability Analysis

Stability analysis helps us understand the behavior of a system near its equilibria. By analyzing stability, we determine whether small perturbations around the equilibrium points die out, stay constant, or grow over time.

• If perturbations die out, the equilibrium is deemed stable.
• If perturbations grow, it's considered unstable.
• If they stay constant, the stability could be neutral.

In this exercise, stability is determined by examining eigenvalues of the Jacobian matrix at each equilibrium point. This leads us to characterize equilibrium as nodal, saddle, or neutral.
Understand that different stability types indicate different dynamical behaviors of the solution trajectories around those points.

Jacobian Matrix

The Jacobian matrix is a tool used to linearize a system of differential equations around an equilibrium point. It's a square matrix consisting of the first-order partial derivatives of the system functions. The Jacobian provides local approximations of the system's behavior.
For our system, the Jacobian is calculated as \[ J = \begin{bmatrix} \frac{\partial f}{\partial x_1} & \frac{\partial f}{\partial x_2} \ \frac{\partial g}{\partial x_1} & \frac{\partial g}{\partial x_2} \end{bmatrix} \].
This matrix is evaluated at each equilibrium, helping us determine how small deviations affect the system. The Jacobian plays a critical role in determining the nature and stability of each equilibrium through its eigenvalues.

Eigenvalues

Eigenvalues of a matrix provide deep insights into the transformation it represents. In the stability analysis of equilibria, they reveal crucial information about stability.

• A positive real part of an eigenvalue signifies that perturbations grow, indicating instability.
• A negative real part shows that perturbations shrink, implying stability.
• Eigenvalues with zero real part require additional analysis for stability confirmation.

For each equilibrium point, the Jacobian matrix was used to calculate these eigenvalues, which allowed us to distinguish the stability types—unstable node, saddle point, or neutral stability—based on their values.
Understanding eigenvalues is vital for predicting the behavior of dynamical systems near equilibrium points.