Problem 11 - 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. $$ \begin{array}{l} \frac{d x_{1}}{d t}=x_{1}-x_{2} \\ \frac{d x_{2}}{d t}=x_{1} x_{2}-x_{2} \end{array} $$

Step-by-Step Solution

Answer

The equilibria are \( (0,0) \) and \( (1,1) \), both are unstable.

Step 1: Find Equilibria

To find the equilibria of the system, we set the derivatives to zero: \( \frac{d x_{1}}{d t}=0 \) and \( \frac{d x_{2}}{d t}=0 \). This gives us the equations: \( x_{1} - x_{2} = 0 \) and \( x_{1}x_{2} - x_{2} = 0 \). Solving the first equation results in \( x_{1} = x_{2} \). Substituting this into the second equation gives \( x_{2}x_{2} - x_{2} = 0 \) or \( x_{2}(x_{2} - 1) = 0 \), leading to \( x_{2} = 0 \) or \( x_{2} = 1 \). Thus, the equilibria are \( (x_1, x_2) = (0, 0) \) and \( (x_1, x_2) = (1, 1) \).

Step 2: Linearize the System

To analyze the stability, we linearize the system around the equilibria by finding the Jacobian matrix: \( J = \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} \ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} \end{bmatrix} \). For our system, \( f_1 = x_1 - x_2 \) and \( f_2 = x_1 x_2 - x_2 \). Calculating, we find: \( J = \begin{bmatrix} 1 & -1 \ x_2 & x_1 - 1 \end{bmatrix} \).

Step 3: Evaluate Jacobian at Equilibria

Substitute the equilibrium points into the Jacobian. For \( (0, 0) \), \( J = \begin{bmatrix} 1 & -1 \ 0 & -1 \end{bmatrix} \). For \( (1, 1) \), \( J = \begin{bmatrix} 1 & -1 \ 1 & 0 \end{bmatrix} \).

Step 4: Determine Stability using Eigenvalues

Compute the eigenvalues of the Jacobian matrix for each equilibrium to determine stability. For \( (0, 0) \), the characteristic equation is \( \text{det}(J - \lambda I) = \begin{vmatrix} 1 - \lambda & -1 \ 0 & -1 - \lambda \end{vmatrix} = (1-\lambda)(-1-\lambda) = 0 \), yielding eigenvalues \( \lambda_1 = 1 \) and \( \lambda_2 = -1 \). Because there's a positive eigenvalue, \( (0, 0) \) is unstable. For \( (1, 1) \), the characteristic equation is \( \begin{vmatrix} 1 - \lambda & -1 \ 1 & - \lambda \end{vmatrix} = (1-\lambda)(-\lambda) + 1 = \lambda^2 - \lambda + 1 = 0 \). Solving gives eigenvalues \( \lambda_{1,2} = \frac{1 \pm \sqrt{-3}}{2} = \frac{1}{2} \pm i\frac{\sqrt{3}}{2} \), which are complex with a positive real part. Thus, \( (1, 1) \) is also unstable.

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve an unknown function and its derivatives. They are used to describe various phenomena, such as motion, heat, electricity, and more.
The given system of differential equations in the problem describes the rates of change of two variables, \( x_1 \) and \( x_2 \), over time \( t \):

• \( \frac{d x_{1}}{d t} = x_{1} - x_{2} \)
• \( \frac{d x_{2}}{d t} = x_{1} x_{2} - x_{2} \)

Each equation in the system allows us to understand how the variable it describes changes in response to the other variables. This type of system often requires stability analysis to understand how solutions behave over time.

Equilibria

In the context of differential equations, an equilibrium point is a set of values for the variables where their rates of change are zero. This means once the system reaches these values, it will stay there unless disturbed.
To find equilibria in our system, we set the derivatives equal to zero and solve the resulting set of equations:

• \( x_{1} - x_{2} = 0 \)
• \( x_{1} x_{2} - x_{2} = 0 \)

Solving these, we find the solutions to be the points \( (x_1, x_2) = (0, 0) \) and \( (1, 1) \). These points are significant because they are potential endpoints of the system's behavior over time.

Jacobian Matrix

The Jacobian matrix is a tool used in the study of differential equations to understand how small changes in the system's variables affect the system as a whole. It is particularly useful in linearizing non-linear systems around equilibria. For our system, the Jacobian matrix is given by:

• \[J = \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} \\frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} \end{bmatrix}\]

where \( f_1 = x_{1} - x_{2} \) and \( f_2 = x_{1} x_{2} - x_{2} \). Calculating these partial derivatives, we find the Jacobian for this system to be:

• \[J = \begin{bmatrix} 1 & -1 \x_2 & x_1 - 1 \end{bmatrix}\]

By evaluating the Jacobian at each equilibrium point, we gain essential insights into the system's local behavior around these points.

Eigenvalues

Eigenvalues are scalar values that provide key insights into the behavior of a linear system around an equilibrium point. They help determine the stability of an equilibrium by analyzing the sign of their real parts. If any eigenvalue of a system has a positive real part, the system is unstable near the equilibrium point.
For the first equilibrium point \( (0, 0) \), the Jacobian matrix yields eigenvalues \( \lambda_1 = 1 \) and \( \lambda_2 = -1 \), indicating instability because one of the eigenvalues is positive.
For the second point \( (1, 1) \), the eigenvalues \( \lambda_{1,2} = \frac{1}{2} \pm i\frac{\sqrt{3}}{2} \) are complex with positive real parts, also indicating instability.
Understanding the eigenvalues helps us predict how small perturbations at the equilibria will grow or decay, giving us crucial stability information.