Problem 12 - Biocalculus Calculus for the Life Sciences

Question

Find all equilibria. Then find the linearization near each equilibrium. $$\begin{array}{l}{\frac{d x_{1}}{d t}=\ln x_{1}-x_{2}} \\ {\frac{d x_{2}}{d t}=x_{1}\left(1-x_{1}-x_{2}\right)}\end{array}$$

Step-by-Step Solution

Answer

Equilibrium is at (1, 0). Jacobian matrix is \( \begin{bmatrix} 1 & -1 \\ -1 & -1 \end{bmatrix} \).

Step 1: Set Derivatives to Zero

Equilibrium points occur when the derivative of the system is zero, meaning neither variable changes over time. Set the first equation, \( \frac{d x_1}{dt} = \ln x_1 - x_2 \), equal to zero, resulting in \( \ln x_1 = x_2 \). Set the second equation, \( \frac{d x_2}{dt} = x_1(1 - x_1 - x_2) \), equal to zero, resulting in \( x_1(1 - x_1 - x_2) = 0 \).

Step 2: Solve for Equilibrium Points

From the second equation, \( x_1(1 - x_1 - x_2) = 0 \), we have two cases: Case 1: \( x_1 = 0 \) leading to \( x_2 = \ln 0 \), which is undefined, so it provides no valid equilibrium in this context. Case 2: \( 1 - x_1 - x_2 = 0 \) implies \( x_2 = 1 - x_1 \). Substitute \( x_2 = \ln x_1 \) into \( x_2 = 1 - x_1 \) to obtain \( \ln x_1 = 1 - x_1 \). Solve this equation to find the equilibrium point \( x_1 = 1 \), which gives \( x_2 = 0 \) as the equilibrium point.

Step 3: Linearize Near Each Equilibrium

To linearize near the equilibrium \((x_1, x_2) = (1, 0)\), compute the Jacobian matrix of the system. The partial derivatives of \( \frac{d x_1}{dt} \) are \( \frac{\partial}{\partial x_1}(\ln x_1 - x_2) = \frac{1}{x_1} \) and \( \frac{\partial}{\partial x_2}(\ln x_1 - x_2) = -1 \). The partial derivatives of \( \frac{d x_2}{dt} \) are \( \frac{\partial}{\partial x_1}(x_1(1-x_1-x_2)) = 1 - 2x_1 - x_2 \) and \( \frac{\partial}{\partial x_2}(x_1(1-x_1-x_2)) = -x_1 \). Evaluate these at \((x_1, x_2) = (1, 0)\) to get the Jacobian matrix: \[ J = \begin{bmatrix} 1 & -1 \ -1 & -1 \end{bmatrix} \]

Key Concepts

Equilibrium Points

In a dynamical system, equilibrium points are states where the system doesn't change over time. These are important because they often represent stable configurations or long-term behaviors of the system. To find an equilibrium point, we set the derivatives of the system's equations to zero. In our exercise, this involved solving the equations:

• \( \frac{d x_1}{dt} = \ln x_1 - x_2 = 0 \)
• \( \frac{d x_2}{dt} = x_1(1 - x_1 - x_2) = 0 \)

By setting these to zero, we determined that one possible equilibrium occurs when \( x_1 = 1 \) and \( x_2 = 0 \). This corresponds to a state where all variables are constant, indicating no change over time.

Linearization

Once equilibrium points are found, the next step in analyzing them is linearization. This involves approximating the system near the equilibrium by using a linear model. The linear model can reveal the stability properties of the equilibrium. For the given system, we substitute small perturbations around the equilibrium point into the system equations and simplify. This approximation gives a linear representation that is easier to work with compared to the nonlinear original system. Here, we performed linearization around the equilibrium \((1, 0)\), which helps in understanding how small changes may evolve around this point.

Jacobian Matrix

The Jacobian matrix is a crucial tool for linearization in multivariable calculus. It consists of all first-order partial derivatives of the system. Essentially, it’s a matrix that represents the best linear approximation of a multivariable function around a certain point.To calculate the Jacobian matrix for our system, we need the partial derivatives of our differential equations:

• From \( \frac{d x_1}{dt} = \ln x_1 - x_2 \):
• \( \frac{\partial}{\partial x_1}(\ln x_1 - x_2) = \frac{1}{x_1} \)
• \( \frac{\partial}{\partial x_2}(\ln x_1 - x_2) = -1 \)
• From \( \frac{d x_2}{dt} = x_1(1-x_1-x_2) \):
• \( \frac{\partial}{\partial x_1}(x_1(1 - x_1 - x_2)) = 1 - 2x_1 - x_2 \)
• \( \frac{\partial}{\partial x_2}(x_1(1 - x_1 - x_2)) = -x_1 \)

Evaluating these at \((x_1, x_2) = (1, 0)\), we get the Jacobian matrix: \[ J = \begin{bmatrix} 1 & -1 \ -1 & -1 \end{bmatrix} \]This matrix helps in understanding the local dynamics near the equilibrium point.