Problem 10 - Biocalculus Calculus for the Life Sciences

Question

Find all equilibria. Then find the linearization near each equilibrium. $$ \begin{array}{l} \frac{d x_{1}}{d t}=x_{1}-2 x_{1}^{2}-6 x_{1} x_{2} \\ \frac{d x_{2}}{d t}=2 x_{2}-8 x_{2}^{2}-2 x_{1} x_{2} \end{array} $$

Step-by-Step Solution

Answer

Equilibria are \((0,0), (0, \frac{1}{4}), (1, 0)\); linearizations are evaluated using Jacobian matrices at these points.

Step 1: Set the system to zero for equilibria

To find the equilibria, set each equation to zero: \( x_1 - 2x_1^2 - 6x_1x_2 = 0 \) and \( 2x_2 - 8x_2^2 - 2x_1x_2 = 0 \).

Step 2: Solve for potential equilibria in each equation

Factor both equations. For the first equation, factor as \( x_1(1 - 2x_1 - 6x_2) = 0 \). For the second equation, factor as \( 2x_2(1 - 4x_2 - x_1) = 0 \).

Step 3: Solve the factored system

This yields potential solutions: \( x_1 = 0 \) or \( 1 - 2x_1 - 6x_2 = 0 \) for the first equation, and \( x_2 = 0 \) or \( 1 - 4x_2 - x_1 = 0 \) for the second equation.

Step 4: Find combinations of solutions

By solving combinations of \( x_1 = 0, \ x_2 = 0, \ 1 - 2x_1 - 6x_2 = 0, \) and \( 1 - 4x_2 - x_1 = 0 \), equilibria points are: \((0,0), (0, \frac{1}{4}), (1, 0)\).

Step 5: Linearize the system

Calculate the Jacobian matrix: \( J = \begin{bmatrix} 1 - 4x_1 - 6x_2 & -6x_1 \ -2x_2 & 2 - 16x_2 - 2x_1 \end{bmatrix} \).

Step 6: Evaluate the Jacobian at each equilibrium

For each equilibrium, substitute the respective \((x_1, x_2)\) values into the Jacobian to find linearizations:- At \((0,0)\), \( J = \begin{bmatrix} 1 & 0 \ 0 & 2 \end{bmatrix} \).- At \((0, \frac{1}{4})\), \( J = \begin{bmatrix} -\frac{1}{2} & 0 \ -\frac{1}{2} & 0 \end{bmatrix} \).- At \((1,0)\), \( J = \begin{bmatrix} -3 & -6 \ 0 & 0 \end{bmatrix} \).

Step 7: Analyze stability using eigenvalues

For \((0,0), (0, \frac{1}{4})\) and \((1,0)\), calculate eigenvalues of their respective Jacobians. The signs of these eigenvalues indicate stability characteristics of each equilibrium.

Key Concepts

Linearization

Linearization is a fundamental technique in analyzing complex dynamical systems. It involves approximating a non-linear function with a linear function near a specified point, usually an equilibrium. This approximation simplifies the analysis as linear systems are easier to solve. In practice, this is done by finding the system's derivatives and constructing the Jacobian matrix.

In the given exercise, linearization is applied at each equilibrium point:

• We first identify the equilibria by solving the equations when all derivatives equal zero.
• Then, we calculate the partial derivatives of each equation with respect to each variable, assembling these into the Jacobian matrix.

By substituting the equilibrium values into the Jacobian, you obtain a linear approximation of the system dynamics about that point. This approximation helps to infer how the system behaves near an equilibrium.

Jacobian Matrix

The Jacobian matrix is a cornerstone tool in differential equations and dynamical systems. It consists of all first-order partial derivatives of a vector-valued function. For our exercise, the Jacobian matrix is derived from the two equations describing how the variables change over time. Each element of the Jacobian is a derivative that defines how sensitive a particular equation is to a small change in one of the state variables.

In this exercise, the Jacobian matrix \[J = \begin{bmatrix} 1 - 4x_1 - 6x_2 & -6x_1 \ -2x_2 & 2 - 16x_2 - 2x_1 \end{bmatrix}\] encodes the local dynamics near any point \((x_1, x_2)\) on the state space. Each entry in the matrix provides insight on how changes in one variable affect another. By calculating the Jacobian at each equilibrium, you glean critical information used in assessing how stable an equilibrium point is.

Stability Analysis

Stability analysis explores the behavior of a dynamical system near its equilibria. After computing the Jacobian, the next step is to determine the system's stability by evaluating the Jacobian's eigenvalues at each equilibrium point.

The nature of the eigenvalues provides insights:

• If all eigenvalues have negative real parts, the equilibrium is stable (attracts nearby points).
• If any eigenvalue has a positive real part, the equilibrium is unstable (repels nearby points).
• If eigenvalues are complex conjugates with positive real parts, the system may exhibit oscillatory instability.

In the exercise, after calculating the Jacobians at points \((0,0), (0, \frac{1}{4}),\) and \((1,0)\), we derive its eigenvalues to ascertain stability. This systematic approach is essential in understanding how systems behave under perturbations near equilibrium conditions.