Problem 4 - 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. $$\begin{array}{l}{\frac{d x_{1}}{d t}=5\left(1+\cos x_{1}\right)+a x_{1}-b x_{2}-10} \\ {\frac{d x_{2}}{d t}=3 x_{2}+b x_{1} x_{2}}\end{array}$$

Step-by-Step Solution

Answer

Jacobian matrix at \((0, 0)\) is \(J = \begin{bmatrix} a & -b \\ 0 & 3 \end{bmatrix}\).

Step 1: Evaluate Partial Derivatives

To find the linearization near the equilibrium point \((0, 0)\), we need to calculate the partial derivatives of each function with respect to each variable at the point \((0, 0)\). The first function is \(f_1(x_1, x_2) = 5(1 + \cos x_1) + ax_1 - bx_2 - 10\) and the second function is \(f_2(x_1, x_2) = 3x_2 + bx_1x_2\). We will calculate \(\frac{\partial f_1}{\partial x_1}\), \(\frac{\partial f_1}{\partial x_2}\), \(\frac{\partial f_2}{\partial x_1}\), and \(\frac{\partial f_2}{\partial x_2}\).

Step 2: Calculate Partial Derivative of \(f_1\)

\[\frac{\partial f_1}{\partial x_1} = \frac{\partial}{\partial x_1}\left(5(1 + \cos x_1) + ax_1 - bx_2 - 10\right) = -5\sin x_1 + a.\] Evaluating at \((0, 0)\), we get \(-5\sin(0) + a = a\).

Step 3: Calculate Partial Derivative of \(f_1\) with respect to \(x_2\)

\[\frac{\partial f_1}{\partial x_2} = \frac{\partial}{\partial x_2}\left(5(1 + \cos x_1) + ax_1 - bx_2 - 10\right) = -b.\] Evaluating at \((0, 0)\) gives \(-b\).

Step 4: Calculate Partial Derivative of \(f_2\)

\[\frac{\partial f_2}{\partial x_1} = \frac{\partial}{\partial x_1}\left(3x_2 + bx_1x_2\right) = bx_2.\] Evaluating at \((0, 0)\) results in \(0\).

Step 5: Calculate Partial Derivative of \(f_2\) with respect to \(x_2\)

\[\frac{\partial f_2}{\partial x_2} = \frac{\partial}{\partial x_2}\left(3x_2 + bx_1x_2\right) = 3 + bx_1.\] Evaluating at \((0, 0)\), we obtain \(3\).

Step 6: Construct the Jacobian Matrix

The Jacobian matrix at \((0, 0)\) is formed using the partial derivatives calculated: \[J = \begin{bmatrix} a & -b \ 0 & 3 \end{bmatrix}.\]

Step 7: State the Linearization

The linearization of the system at \((0, 0)\) is given by the system \[\frac{d}{dt} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} a & -b \ 0 & 3 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix}.\]

Key Concepts

Linearization

When dealing with nonlinear systems, sometimes they can be complex and difficult to analyze directly. Luckily, linearization offers a way to simplify these systems near a point of interest, often an equilibrium point. It involves approximating the system by a linear one, which is much easier to work with. Think of it as "straightening" the curve of a nonlinear equation near a given point to better understand its behavior in that region.
To linearize a nonlinear system, you need to know how to compute partial derivatives and construct a Jacobian matrix. First, compute the partial derivatives of each function with respect to each variable, evaluated at the equilibrium point. Once you have these, they are organized into a matrix known as the Jacobian matrix. This matrix is crucial as it contains all the information about how the system behaves linearly around the given point.
This powerful method forms the backbone of analyzing nonlinear systems in numerous fields, including physics and engineering.

Equilibrium Points

Equilibrium points in a system are where the system remains steady if left undisturbed. It's where nothing changes over time. For a dynamical system described by a set of differential equations, these points are solutions where all derivatives are zero.
To identify equilibrium points, you set the system's equations to zero and solve for the variables. Consider them like resting points of the system, where it can potentially sit indefinitely without external influences.
Once found, these equilibrium points serve as anchors from which we can analyze the behavior of the system both locally and globally. They are key in simplifying complex systems and play a vital role in understanding stability and dynamics.

Jacobian Matrix

The Jacobian matrix is a fundamental tool in math and engineering that describes how a system transforms near a particular point. It's a matrix consisting of the first partial derivatives of a vector-valued function.
Essentially, the Jacobian matrix provides a snapshot of how small changes in input variables of a function will affect the output. In the context of linearization, it gives the coefficients of the linear approximation of the system:

• The entries are calculated by taking partial derivatives of each function with respect to each variable.
• It is evaluated at the equilibrium point to see how the system behaves near that point.

This matrix isn't just practical for linear approximation. It also plays a crucial role in determining the stability of an equilibrium point and how the system behaves when nudged slightly from this point. Understanding and calculating the Jacobian matrix provides deep insight into how nonlinear systems operate near equilibrium.