Problem 27 - Calculus for Biology and Medicine

Question

Show that the equilibrium \(\left[\begin{array}{l}0 \\ 0\end{array}\right]\) of $$ \begin{array}{l} x_{1}(t+1)=\frac{x_{2}(t)}{4\left(1+x_{1}^{2}(t)\right)} \\ x_{2}(t+1)=\frac{2 x_{1}(t)}{1+x_{2}^{2}(t)} \end{array} $$ is locally stable.

Step-by-Step Solution

Answer

The equilibrium point \([0,0]\) is not locally stable based on the linearization analysis.

Step 1: Linearization of the System

We need to linearize the system around the equilibrium point \([0, 0]\). The given dynamical system is:\[ x_1(t+1) = \frac{x_2(t)}{4(1+x_1^2(t))} \]\[ x_2(t+1) = \frac{2x_1(t)}{1+x_2^2(t)} \]To linearize, we calculate the Jacobian matrix of the system at the equilibrium.

Step 2: Compute Partial Derivatives

The Jacobian matrix \(J\) is given by the partial derivatives of the system. Calculate:- \(\frac{\partial}{\partial x_1} \left( \frac{x_2}{4(1+x_1^2)} \right) \), \(\frac{\partial}{\partial x_2} \left( \frac{x_2}{4(1+x_1^2)} \right) \) for the first equation.- \(\frac{\partial}{\partial x_1} \left( \frac{2x_1}{1+x_2^2} \right) \), \(\frac{\partial}{\partial x_2} \left( \frac{2x_1}{1+x_2^2} \right) \) for the second equation.

Step 3: Evaluate the Jacobian at Equilibrium Point

Insert \( x_1 = 0 \) and \( x_2 = 0 \) into the partial derivatives to find the Jacobian matrix at \([0,0]\).The derivatives evaluate to:\[J = \begin{bmatrix}0 & \frac{1}{4} \2 & 0\end{bmatrix}\]

Step 4: Determine Eigenvalues of the Jacobian

Calculate the eigenvalues of the Jacobian matrix \( J \).The characteristic equation is obtained by solving:\[ \text{det}(J - \lambda I) = 0 \]where \( I \) is the identity matrix. Substitute and solve:\[ (\lambda)(\lambda) - (2)(\frac{1}{4}) = 0 \]\[ \lambda^2 - \frac{1}{2} = 0 \]\[ \lambda = \pm \sqrt{\frac{1}{2}} \]

Step 5: Analyze Stability

The eigenvalues are \( \lambda = \pm \sqrt{\frac{1}{2}} \). Since both eigenvalues are real and at least one is positive, the linearized system indicates that the equilibrium is not stable.However, due to the exercise asking about local stability, this implies we need to review the system's nature considering perturbations that affect stability on a nonlinear level.

Key Concepts

Equilibrium Point

An equilibrium point in a dynamical system is a state where the system remains unchanged if it starts there. In other words, it's a point where all motion ceases, and the system stays at rest unless disturbed by an external force. The importance of identifying equilibrium points in stability analysis is crucial. They act as "resting positions" for the system.

• To find an equilibrium point, set the system equations to zero.
• In our case, the equilibrium point is \([0, 0]\) from the system of nonlinear equations.
• Equilibrium points help to predict how the system behaves in the long run.

At \([0, 0]\), both equations balance perfectly without any change, hence they are equilibrium points.

Linearization

Linearization is a technique used to approximate a nonlinear system near its equilibrium point with a linear one. This approach simplifies the analysis because linear systems are easier to handle mathematically. By examining the behavior of this simpler system, we can infer the local dynamics of the original nonlinear system.

• We start by determining the Jacobian matrix of the system at the equilibrium point.
• Linearization involves finding partial derivatives, forming the Jacobian matrix and evaluating it at \([0, 0]\).
• This method provides a local view, meaning it is valid only near the equilibrium point.

Overall, linearization helps visualize how small deviations from equilibrium evolve, providing insights into stability.

Jacobian Matrix

The Jacobian matrix is a matrix of all first-order partial derivatives of a vector-valued function. It plays a significant role in determining the stability of the equilibrium point.

• It provides information on how small changes in the variables \(x_1\) and \(x_2\) affect changes in the system outputs.
• For the given system, the Jacobian matrix helps in understanding the linear approximation of the system at \[ [0,0] \].
• In the solution, the Jacobian is calculated as:\[J = \begin{bmatrix}0 & \frac{1}{4} \ 2 & 0\end{bmatrix}\]

Calculating the Jacobian matrix requires knowledge of partial derivatives and is essential for further stability assessment through eigenvalues.

Eigenvalues

Eigenvalues derived from the Jacobian matrix assessment help determine stability. They are calculated from the characteristic equation, which is formed using the Jacobian matrix.

• Find the determinant of \( J - \lambda I \) and set it to zero.
• Solving for \( \lambda \) gives the eigenvalues which indicate the system's behavior near the equilibrium.
• For our exercise, the eigenvalues are \( \lambda = \pm \sqrt{\frac{1}{2}} \).

Eigenvalues tell us how small perturbations at equilibrium grow out over time. If any eigenvalue is positive, it indicates instability locally, like in our example.

Dynamical System

A dynamical system is a system in which a fixed rule describes the time dependence of a point in a geometrical space. Essentially, it is a model used to describe the complex systems that evolve over time according to specific rules.

• These systems can be continuous or discrete, linear or nonlinear.
• The example provided is a discrete, nonlinear dynamical system represented by iterative equations.
• Understanding these systems helps in predicting the long-term behavior of various scientific phenomena.

In essence, analyzing a dynamical system allows us to understand how the state of a system changes over time and helps in characterizing stability around its equilibrium points.