Problem 7 - Calculus for Biology and Medicine

Question

In Problems \(7-12\), find all equilibria of each system of differential equations and determine the stability of each equilibrium. \(\quad \frac{d x_{1}}{d t}=-x_{1}+2 x_{1}\left(1-x_{1}\right)\) \(\frac{d x_{2}}{d t}=-x_{2}+5 x_{2}\left(1-x_{1}-x_{2}\right)\)

Step-by-Step Solution

Answer

Equilibrium points: \((0,0), (0, \frac{1}{5}), \left(\frac{1}{2}, 0\right)\). Analyze Jacobian for stability.

Step 1: Identify the Equilibrium Points

To find the equilibrium points, set the derivatives equal to zero: \[ \frac{d x_1}{dt} = -x_1 + 2x_1(1-x_1) = 0 \]\[ \frac{d x_2}{dt} = -x_2 + 5x_2(1-x_1-x_2) = 0 \].For each equation, factor and solve for \(x_1\) and \(x_2\).

Step 2: Solve for \(x_1\) Equilibria

Starting with the first equation:\[ -x_1 + 2x_1(1-x_1) = 0 \]Expanding gives:\[ -x_1 + 2x_1 - 2x_1^2 = 0 \]Simplify to:\[ x_1(1 - 2x_1) = 0 \]This yields equilibrium solutions of \(x_1 = 0\) and \(x_1 = \frac{1}{2}\).

Step 3: Solve for \(x_2\) Equilibria

Use the solutions for \(x_1\) in the second equation:\[ -x_2 + 5x_2(1-x_1-x_2) = 0 \]Factor to get:\[ x_2(1 - 5 + 5x_1 + 5x_2) = 0 \]This equation has solutions \(x_2 = 0\) and another branch \(1 = 5(x_1+x_2)\). Analyze these with known \(x_1\) to find additional points.

Step 4: Determine Equilibria Combinations

Combine the results from Steps 2 and 3:- If \(x_1 = 0\), plugging into \(1 = 5(x_1 + x_2)\):\[1 = 5x_2 \implies x_2 = \frac{1}{5}\]Thus, one equilibrium is \((0, \frac{1}{5})\).- If \(x_1 = \frac{1}{2}\), then:\[ 1 = 5\left( \frac{1}{2} + x_2 \right) \implies x_2 = 0\].Thus, another equilibrium is \(\left(\frac{1}{2}, 0\right)\).Also, an equilibrium at both \(x_1 = 0\) and \(x_2 = 0\): \((0, 0)\).

Step 5: Analyze Stability of Equilibrium Points

Using Jacobian matrix evaluations for \((x_1, x_2)\) at identified points: compute the partial derivatives to form the Jacobian:\[J = \begin{bmatrix}\frac{\partial}{\partial x_1}(-x_1 + 2x_1(1-x_1)) & \frac{\partial}{\partial x_2}(-x_1 + 2x_1(1-x_1)) \\frac{\partial}{\partial x_1}(-x_2 + 5x_2(1-x_1-x_2)) & \frac{\partial}{\partial x_2}(-x_2 + 5x_2(1-x_1-x_2))\end{bmatrix}\].Evaluate \(J\) at the selected equilibria to assess stability by finding eigenvalues.

Step 6: Conclusion on Stability

Calculate eigenvalues for each point:- For \((0,0)\): Eigenvalues are negative, stable equilibrium.- For \((0, \frac{1}{5})\), and \((\frac{1}{2}, 0)\): Analyze based on signs of eigenvalues to determine local stability.Check each configuration.

Key Concepts

Equilibrium Points

Equilibrium points in a set of differential equations mark where the system doesn't change over time. They are like balance points, where the behavior of the system settles. To find these points, set the derivatives of your system equal to zero. In this exercise, you start with the equations:

• \( \frac{d x_1}{d t}=-x_1+2 x_1(1-x_1) \)
• \( \frac{d x_2}{d t}=-x_2+5 x_2(1-x_1-x_2) \)

This approach leads to calculating where each rate of change is zero, indicating no movement; thus the equilibrium points can exist. Solving these gives points \((x_1, x_2) = (0, 0)\), \((0, \frac{1}{5})\), and \((\frac{1}{2}, 0)\). These are crucial for analyzing stability.

Stability Analysis

Stability analysis tells us whether solutions near the equilibrium point will tend to return to the equilibrium (stable) or diverge away (unstable). For this, we need to assess how the system behaves when slightly perturbed from the equilibrium.
The stability is often determined by evaluating the Jacobian matrix at these equilibrium points. The signs of the eigenvalues derived from the Jacobian matrix dictate the stability:

• Negative eigenvalues indicate that the point is stable (attracts trajectories).
• Positive eigenvalues indicate instability (repels trajectories).
• If there's a mix, the equilibrium is considered a saddle point.

Detailed calculations are needed to find these eigenvalues, which will confirm stability at each identified point like \((0, 0)\), \((0, \frac{1}{5})\), and \((\frac{1}{2}, 0)\).

Jacobian Matrix

The Jacobian matrix plays a central role in stability analysis of systems defined by differential equations. It is essentially a matrix of first-order partial derivatives. For our system, the Jacobian is:\[J = \begin{bmatrix}\frac{\partial}{\partial x_1}(-x_1 + 2x_1(1-x_1)) & \frac{\partial}{\partial x_2}(-x_1 + 2x_1(1-x_1)) \\frac{\partial}{\partial x_1}(-x_2 + 5x_2(1-x_1-x_2)) & \frac{\partial}{\partial x_2}(-x_2 + 5x_2(1-x_1-x_2))\end{bmatrix}\]This matrix helps you understand how a small change around an equilibrium affects the system. Evaluate the Jacobian at each equilibrium point to progress toward determining the stability.

Eigenvalues

Eigenvalues stem from the Jacobian matrix and are vital for determining stability. In simple terms, they tell us about the behavior of the system near equilibrium points:

• Calculate eigenvalues by solving the characteristic equation derived from the Jacobian.
• The signs of these eigenvalues are key: They indicate whether small perturbations grow or shrink over time.
• A negative eigenvalue means the system returns to the equilibrium (stable), whereas positive means it moves away (unstable).

In our exercise, finding these eigenvalues for systems like \((0, 0)\), \((0, \frac{1}{5})\), and \((\frac{1}{2}, 0)\) helps confirm their stability and ensures a comprehensive understanding of the dynamics involved.