Problem 6 - Biocalculus Calculus for the Life Sciences

Question

Each of the nonlinear systems has an equilibrium at \(\left(\hat{x}_{1}, \hat{x}_{2}\right)=(0,0) .\) Find the linearization near this point. $$\begin{array}{l}{\frac{d x_{1}}{d t}=x_{2}-\frac{2+a x_{1}}{2+b x_{2}}+\cos x_{2}} \\ {\frac{d x_{2}}{d t}=\frac{2 x_{1}}{1+x_{2}}-a x_{1}}\end{array}$$

Step-by-Step Solution

Answer

The linearization is the Jacobian matrix: \( J = \begin{bmatrix} -\frac{a}{2} & 1 \\ 2-a & 0 \end{bmatrix} \).

Step 1: Identify the Equations and Variables

The system of differential equations is given by \( \frac{d x_{1}}{d t}=x_{2}-\frac{2+a x_{1}}{2+b x_{2}}+\cos x_{2} \) and \( \frac{d x_{2}}{d t}=\frac{2 x_{1}}{1+x_{2}}-a x_{1} \). We seek the linearization around the equilibrium point \( (\hat{x}_1, \hat{x}_2) = (0, 0) \).

Step 2: Compute the Jacobian Matrix

The Jacobian matrix \( J \) at a point \((x_1, x_2)\) is comprised of partial derivatives: \[J = \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} \\frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} \end{bmatrix}\]where \( f_1(x_1, x_2) = x_2 - \frac{2 + ax_1}{2 + bx_2} + \cos x_2 \) and \( f_2(x_1, x_2) = \frac{2x_1}{1 + x_2} - ax_1 \).

Step 3: Differentiate with Respect to \(x_1\) and \(x_2\) for \(f_1\)

Calculate \( \frac{\partial f_1}{\partial x_1} \) and \( \frac{\partial f_1}{\partial x_2} \) :\( \frac{\partial f_1}{\partial x_1} = \frac{\partial}{\partial x_1} \left(x_2 - \frac{2 + ax_1}{2 + bx_2} + \cos x_2 \right) = -\frac{a}{2 + b x_2} \) evaluated at (0,0), which simplifies to \(-\frac{a}{2}\) because \(x_2=0\).\( \frac{\partial f_1}{\partial x_2} = \frac{\partial}{\partial x_2} \left(x_2 - \frac{2 + ax_1}{2 + bx_2} + \cos x_2 \right) = 1 + b\frac{2 + ax_1}{(2 + bx_2)^2} - \sin x_2 \) evaluated at (0,0), simplifying to \(1\).

Step 4: Differentiate with Respect to \(x_1\) and \(x_2\) for \(f_2\)

Calculate \( \frac{\partial f_2}{\partial x_1} \) and \( \frac{\partial f_2}{\partial x_2} \) :\( \frac{\partial f_2}{\partial x_1} = \frac{\partial}{\partial x_1} \left(\frac{2x_1}{1 + x_2} - ax_1 \right) = \frac{2}{1+x_2} - a \) evaluated at (0,0), which simplifies to \(2-a\).\( \frac{\partial f_2}{\partial x_2} = \frac{\partial}{\partial x_2} \left(\frac{2x_1}{1 + x_2} - ax_1 \right) = -\frac{2x_1}{(1+x_2)^2}\) evaluated at (0,0), equals \(0\) because the derivative term doesn't depend on the evaluated point.

Step 5: Assemble the Jacobian Matrix and Simplify

The Jacobian matrix at \((0, 0)\) is:\[J = \begin{bmatrix} -\frac{a}{2} & 1 \2-a & 0 \end{bmatrix}\]This is the linearization of the given nonlinear system around the equilibrium point \((0,0)\).

Key Concepts

Nonlinear Systems

Nonlinear systems refer to systems of equations where the relationships between variables are not linear. Unlike linear systems, these systems involve variables that are raised to a power, multiplied together, or integrated with functions like sine and cosine. Nonlinear systems appear frequently in real-world scenarios, such as in physics, engineering, and economics. In this exercise, we are given a system of differential equations that involve nonlinear terms like \( \frac{2 + ax_1}{2 + bx_2} \) and \( \cos x_2 \). Nonlinear systems can be challenging to solve directly, and that's where linearization, a method often used to approximate these systems near an equilibrium point, comes in handy. By understanding the behavior of a nonlinear system at a particular point, it becomes possible to predict its behavior in a localized region around that point.

Jacobian Matrix

The Jacobian matrix is a crucial tool in the study of systems of differential equations, especially nonlinear systems. It is a matrix composed of all first-order partial derivatives of a vector field. In our specific problem, the Jacobian matrix provides a linear approximation of the system at an equilibrium point.

The Jacobian matrix \ J \ is defined as follows for the system in the exercise:

• \( f_1(x_1, x_2) = x_2 - \frac{2 + ax_1}{2 + bx_2} + \cos x_2 \)
• \( f_2(x_1, x_2) = \frac{2x_1}{1 + x_2} - ax_1 \)

To find the Jacobian, we take the partial derivatives of these functions with respect to each variable (\( x_1 \) and \( x_2 \)), forming a matrix with these derivatives as its elements. This Jacobian reflects how the functions change around the equilibrium point \( (0, 0) \), providing vital insight for linearization.

Equilibrium Point

An equilibrium point in a differential equation system is a point where the variables do not change over time. This means when the system is at this point, there isn't any movement or variation. For the system in this exercise, the equilibrium point is given as \( (\hat{x}_1, \hat{x}_2) = (0, 0) \).

Finding equilibrium points involves setting the derivatives (rate of change of the variables) of the system equations to zero, which makes the system's state stable or neutral at that point. At \( (0,0) \), this means that whatever natural tendencies of change exist in the system are perfectly balanced. By linearizing around this point, we try to understand how slight deviations impact the system. Linearization essentially allows us to replace the nonlinear model by a simpler, linear one, helping in analyzing the system's behavior in the "neighborhood" of this equilibrium.

Differential Equations

Differential equations are equations that involve an unknown function and its derivatives. These types of equations are foundational in modeling how things change, whether they be populations, heat distribution, or even stock prices. In the context of our particular exercise, the given nonlinear system consists of differential equations.

Differential equations can be daunting due to their complexity, especially when they are nonlinear. They express how a quantity changes over time, and in this exercise, they are used to describe how two variables, \( x_1 \) and \( x_2 \), evolve. The method of solving these equations often involves finding equilibrium points and linearizing around these points to simplify the system into something more manageable. By converting a nonlinear set of equations to a linear one near an equilibrium point, the differential equations become easier to analyze and predict near that point.