Problem 11 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

Equilibria: (0,0) (unstable) and (2,-2) (stable).

Step 1: Find Equilibrium System

To find the equilibria of the system, set the right-hand side of each differential equation to zero: 1. \(x_{1} x_{2} - 2x_{2} = 0\)2. \(x_{1} + x_{2} = 0\)

Step 2: Solve the First Equation for Equilibrium Points

From \(x_{1} x_{2} - 2x_{2} = 0\), factor the expression:\[x_{2}(x_{1} - 2) = 0\]Thus, \(x_{2} = 0\) or \(x_{1} = 2\).

Step 3: Solve the Equilibrium System Jointly

Combine the solutions of the first equation with the constraint from the second equation, \(x_{1} + x_{2} = 0\):- If \(x_{2} = 0\), then from \(x_{1}+x_{2}=0\), we have \(x_{1}=0\). So, one equilibrium is \((x_{1}, x_{2}) = (0, 0)\).- If \(x_{1} = 2\), then \(x_{2} = -2\), giving the second equilibrium, \((x_{1}, x_{2}) = (2, -2)\).

Step 4: Linearize the System Around Equilibria

To determine stability, we calculate the Jacobian matrix:\[\begin{bmatrix}\frac{\partial}{\partial x_1}(x_1 x_2 - 2x_2) & \frac{\partial}{\partial x_2}(x_1 x_2 - 2x_2) \frac{\partial}{\partial x_1}(x_1 + x_2) & \frac{\partial}{\partial x_2}(x_1 + x_2) \end{bmatrix} = \begin{bmatrix}x_2 & x_1 - 2 \1 & 1\end{bmatrix}\]

Step 5: Evaluate Jacobian at Equilibria

Calculate the Jacobian at each equilibrium:- At \((0,0)\): \[J = \begin{bmatrix}0 & -2 \1 & 1\end{bmatrix}\]- At \((2, -2)\): \[J = \begin{bmatrix}-2 & 0 \1 & 1\end{bmatrix}\]

Step 6: Determine Stability using Eigenvalues

Find the eigenvalues of each Jacobian matrix.- For \(J(0,0)\): * Characteristic equation: \(\lambda^2 - \lambda + 2 = 0\) * Eigenvalues: \(\lambda = \frac{1 \pm i\sqrt{7}}{2}\) (complex with positive real parts, unstable)- For \(J(2,-2)\): * Characteristic equation: \(\lambda^2 + \lambda + 2 = 0\) * Eigenvalues: \(\lambda = \frac{-1 \pm i\sqrt{7}}{2}\) (complex with negative real parts, stable)

Key Concepts

Equilibria

When analyzing a system of differential equations, one of the fundamental steps is finding its equilibria. Equilibria are points where the system doesn't change over time. It's like finding places in a landscape where the ground is perfectly flat—no rolling or sliding occurs. In mathematical terms, equilibria happen when the derivative (or rate of change) of each parameter is equal to zero.

For a system of equations, we set each equation to zero and solve for the variables involved. In the original problem, we had:

• Equation 1: \(x_1 x_2 - 2x_2 = 0\)
• Equation 2: \(x_1 + x_2 = 0\)

By solving these equations, we determine the points in the system that do not evolve over time. For this system, the equilibria are found at the points \((0, 0)\) and \((2, -2)\). Each of these represents a stable condition of the system—if the system starts at one of these points, it remains there as time progresses.

Stability

Stability in a system of differential equations indicates how the system behaves when it is slightly disturbed from an equilibrium point. We naturally want to know whether such disturbances will decay, leaving the system unchanged, or grow over time, pulling the system away from equilibrium.

To assess stability, we examine the eigenvalues of the Jacobian matrix at each equilibrium. The sign of the real part of these eigenvalues tells us about the stability:

• If the real parts are negative, the equilibrium is stable—disturbances shrink over time.
• If they are positive, the equilibrium is unstable—disturbances grow.
• If half of the real parts are positive and half are negative, the equilibrium is a saddle point.

In the provided solution, the stability at \((0, 0)\) was found to be unstable due to eigenvalues having positive real parts, while \((2, -2)\) is stable due to negative real parts. Think of stability as a mountain or valley: down a slope is stable, up a slope is unstable.

Jacobian Matrix

The Jacobian matrix is a crucial tool in understanding the behavior of systems of differential equations. It works like a key that unlocks the dynamics of the system around equilibrium points. It provides a linear approximation of the system near these points, allowing us to study how small perturbations change or stabilize over time.

In our system, the Jacobian is given by:\[J = \begin{bmatrix}\frac{\partial}{\partial x_1}(x_1 x_2 - 2x_2) & \frac{\partial}{\partial x_2}(x_1 x_2 - 2x_2) \\frac{\partial}{\partial x_1}(x_1 + x_2) & \frac{\partial}{\partial x_2}(x_1 + x_2)\end{bmatrix}\]Which simplifies to:\[J = \begin{bmatrix}x_2 & x_1 - 2 \1 & 1\end{bmatrix}\]By evaluating this matrix at each equilibrium, we extract crucial characteristics about the system's stability and behavior. It's akin to mapping a terrain—where the map gives insights into the paths of least resistance or challenge.