Problem 8 - Biocalculus Calculus for the Life Sciences

Question

Find all equilibria. Then find the linearization near each equilibrium. $$\begin{array}{l}{\frac{d x_{1}}{d t}=x_{2}-5 x_{1} x_{2}} \\ {\frac{d x_{2}}{d t}=2 x_{1}-6 x_{1} x_{2}}\end{array}$$

Step-by-Step Solution

Answer

Equilibrium points: (0, 0), (1/5, 0), (0, 1/3), (1/5, 1/3). Linearized systems given by evaluating the Jacobian at these points.

Step 1: Identify Equilibrium Points

To find the equilibria, set the right-hand sides of the given differential equations to zero. Start with the equations: \( \frac{d x_1}{d t} = x_2 - 5x_1 x_2 = 0 \) and \( \frac{d x_2}{d t} = 2x_1 - 6x_1 x_2 = 0 \). Solve for \(x_1\) and \(x_2\):1. From the first equation: \(x_2(1 - 5x_1) = 0\), giving \(x_2 = 0\) or \(x_1 = \frac{1}{5}\).2. From the second equation: \(x_1(2 - 6x_2) = 0\), giving \(x_1 = 0\) or \(x_2 = \frac{1}{3}\). The combinations of these solutions provide equilibrium points: \((0, 0), (\frac{1}{5}, 0), (0, \frac{1}{3}), (\frac{1}{5}, \frac{1}{3})\).

Step 2: Calculate Jacobian Matrix

To linearize near each equilibrium, compute the Jacobian matrix of the system. The Jacobian matrix \(J\) is given by the partial derivatives of each function with respective variables:\[J = \begin{bmatrix}\frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} \\frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2}\end{bmatrix}\]where \(f_1 = x_2 - 5x_1x_2\) and \(f_2 = 2x_1 - 6x_1x_2\).Calculate the partial derivatives: 1. \(\frac{\partial f_1}{\partial x_1} = -5x_2\), \(\frac{\partial f_1}{\partial x_2} = 1 - 5x_1\).2. \(\frac{\partial f_2}{\partial x_1} = 2 - 6x_2\), \(\frac{\partial f_2}{\partial x_2} = -6x_1\).

Step 3: Linearize at Equilibrium Points

Substitute each equilibrium point into the Jacobian matrix to obtain the linearized system:1. At \((0, 0)\): \[ J = \begin{bmatrix} 0 & 1 \ 2 & 0 \end{bmatrix} \] 2. At \((\frac{1}{5}, 0)\): \[ J = \begin{bmatrix} 0 & 0 \ 2 & -\frac{6}{5} \end{bmatrix} \] 3. At \((0, \frac{1}{3})\): \[ J = \begin{bmatrix} -\frac{5}{3} & \frac{2}{3} \ 0 & 0 \end{bmatrix} \] 4. At \((\frac{1}{5}, \frac{1}{3})\): \[ J = \begin{bmatrix} -1 & 0 \ 0 & -\frac{1}{5} \end{bmatrix} \]

Key Concepts

Differential Equations

Differential equations are fundamental tools in modeling a wide array of systems where changes occur over time or space. Whether describing populations, economies, or physical processes, differential equations help capture the relationship between changing variables. In this exercise, we deal with a system of differential equations that model how two variables, \(x_1\) and \(x_2\), evolve over time:

• \( \frac{d x_1}{d t} = x_2 - 5 x_1 x_2 \)
• \( \frac{d x_2}{d t} = 2 x_1 - 6 x_1 x_2 \)

To find equilibrium points, we solve these equations by setting the derivatives to zero, which implies no change over time, meaning the system is at rest. Here, the solutions provide equilibrium points:

• \((0, 0)\)
• \((\frac{1}{5}, 0)\)
• \((0, \frac{1}{3})\)
• \((\frac{1}{5}, \frac{1}{3})\)

At these points, the system either maintains its state or changes in a predictable manner.

Linearization

Linearization is a powerful technique when dealing with nonlinear differential equations, like those in our exercise. The idea is to approximate the system near an equilibrium point using linear equations, making it easier to analyze. This is because linear equations are simpler to handle than their nonlinear counterparts. By examining how changes in the variables affect the system at equilibrium, linearization helps us understand the local behavior of the system.
To linearize a system, we use the Jacobian matrix, evaluating it at each equilibrium point. This transformation converts complex interactions into a simpler, approximated version, giving insight into stability, direction, and speed of state changes near equilibrium.
In essence, linearization provides a simplified snapshot of a dynamic system in the neighborhood of a stable point, making analysis and predictions intuitive and manageable.

Jacobian Matrix

The Jacobian matrix is a crucial component in the study of dynamical systems, allowing us to understand and predict system behavior near equilibrium points. This matrix consists of partial derivatives, which measure how each variable influences the rate of change of the others. For the given system, the Jacobian matrix \( J \) is:

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

Where:

• \( f_1 = x_2 - 5x_1x_2 \)
• \( f_2 = 2x_1 - 6x_1x_2 \)

Calculating its entries involves finding these derivatives:

• \( \frac{\partial f_1}{\partial x_1} = -5x_2 \)
• \( \frac{\partial f_1}{\partial x_2} = 1 - 5x_1 \)
• \( \frac{\partial f_2}{\partial x_1} = 2 - 6x_2 \)
• \( \frac{\partial f_2}{\partial x_2} = -6x_1 \)

When substituting equilibrium points into this matrix, we obtain crucial insights into the stability and interactions of the system at those points. This matrix forms the foundation for linearization, offering a mathematically precise picture of local dynamics.