Problem 34 - Calculus for Biology and Medicine

Question

Find all nonnegative equilibria of $$ \begin{array}{l} x_{1}(t+1)=x_{2}(t) \\ x_{2}(t+1)=\frac{1}{2} x_{1}(t)+\frac{1}{3} x_{2}(t)-x_{2}^{2}(t) \end{array} $$ and analyze their stability.

Step-by-Step Solution

Answer

There are two nonnegative equilibria: (0, 0) and (1/3, 1/3), both of which are unstable.

Step 1: Understand Equilibrium Condition

An equilibrium of the system occurs when the variables do not change over time: \(x_1(t+1) = x_1(t)\) and \(x_2(t+1) = x_2(t)\). We substitute these into the given system equations to determine equilibrium points.

Step 2: Set Up Equilibrium Equations

From the first equation, equating \(x_1(t+1) = x_1(t)\) gives us \(x_1 = x_2\). From the second equation, equating \(x_2(t+1) = x_2(t)\) gives us \(x_2 = \frac{1}{2}x_1 + \frac{1}{3}x_2 - x_2^2\).

Step 3: Solve for Equilibrium Points

Substitute \(x_1 = x_2\) into the second equilibrium equation: \(x_2 = \frac{1}{2}x_2 + \frac{1}{3}x_2 - x_2^2\). Simplify to find \(x_2 = \frac{5}{6}x_2 - x_2^2\). Rearrange to: \(0 = \frac{1}{3}x_2 - x_2^2\). Factor to get \(x_2(x_2 - \frac{1}{3}) = 0\). Thus, \(x_2 = 0\) or \(x_2 = \frac{1}{3}\).

Step 4: Determine Corresponding \(x_1\) Values

Since \(x_1 = x_2\), for \(x_2 = 0\), \(x_1 = 0\). For \(x_2 = \frac{1}{3}\), \(x_1 = \frac{1}{3}\). Thus, the equilibria points are \((0, 0)\) and \((\frac{1}{3}, \frac{1}{3})\).

Step 5: Stability Analysis Using Jacobian

Construct the Jacobian matrix at any point \((x_1, x_2)\): \[ J = \begin{bmatrix} 0 & 1 \ \frac{1}{2} & \frac{1}{3} - 2x_2 \end{bmatrix} \]. Evaluate at both equilibrium points: \((0, 0)\) and \((\frac{1}{3}, \frac{1}{3})\).

Step 6: Evaluate Jacobian at \((0, 0)\)

Compute \(J(0, 0) = \begin{bmatrix} 0 & 1 \ \frac{1}{2} & \frac{1}{3} \end{bmatrix} \). The characteristic equation is \( \lambda^2 - \frac{1}{3}\lambda - \frac{1}{2} = 0 \). Calculate to find the eigenvalues, both of which have non-zero real parts indicating instability.

Step 7: Evaluate Jacobian at \((\frac{1}{3}, \frac{1}{3})\)

Compute \(J(\frac{1}{3}, \frac{1}{3}) = \begin{bmatrix} 0 & 1 \ \frac{1}{2} & -\frac{1}{3} \end{bmatrix} \). The characteristic equation is \( \lambda^2 + \frac{1}{3}\lambda - \frac{1}{2} = 0 \). Calculate to find the eigenvalues, both of which have non-zero real parts indicating instability.

Key Concepts

Stability Analysis

Stability analysis is a crucial step in understanding the behavior of dynamic systems. It tells us whether small disturbances around an equilibrium point will decay over time, leading the system back to the equilibrium point, or grow larger, causing the system to move away.
In our problem, finding stability involves examining two equilibrium points derived from the equations:

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

To analyze their stability, we can look at how the system responds to small changes around these points by using mathematical tools like the Jacobian matrix and eigenvalues. If disturbances decay, the equilibrium is stable; if they grow, the equilibrium is unstable. Here, both equilibrium points were found to be unstable, meaning that the system will not return to these points if perturbed.

Jacobian Matrix

The Jacobian matrix is an essential component in the stability analysis of dynamical systems. It provides a linear approximation of the system's behavior near an equilibrium point. In simple terms, the Jacobian helps us understand how slight changes in a system's state variables affect the overall system dynamics.
For a system of equations, the Jacobian matrix consists of partial derivatives of the system's functions. For our analyzed system, the Jacobian matrix \(J\) at a generic point \((x_1, x_2)\) is:\[J = \begin{bmatrix} 0 & 1 \ \frac{1}{2} & \frac{1}{3} - 2x_2 \end{bmatrix}\]Evaluating the Jacobian at equilibrium points gives insight into the system's local behavior:

• \(J(0, 0) = \begin{bmatrix} 0 & 1 \ \frac{1}{2} & \frac{1}{3} \end{bmatrix}\)
• \(J(\frac{1}{3}, \frac{1}{3}) = \begin{bmatrix} 0 & 1 \ \frac{1}{2} & -\frac{1}{3} \end{bmatrix}\)

By exploring these matrices, we gain insight into which equilibria might be stable or unstable based on the resulting eigenvalues.

Eigenvalues

Eigenvalues are a fundamental concept in understanding the stability of equilibrium points in a dynamical system. They are derived from the Jacobian matrix and tell us how solutions to the system behave in the vicinity of an equilibrium point.
For our dynamic system, the eigenvalues of the Jacobian matrix at an equilibrium help determine stability:

• If all eigenvalues have negative real parts, the equilibrium is stable.
• If any eigenvalue has a positive real part, the equilibrium is unstable.

In the exercise, computing the eigenvalues involves inserting the Jacobian into the characteristic equation and solving for \(\lambda\). For the point \((0, 0)\), the characteristic equation is:\[ \lambda^2 - \frac{1}{3}\lambda - \frac{1}{2} = 0 \]For the point \((\frac{1}{3}, \frac{1}{3})\), it is:\[ \lambda^2 + \frac{1}{3}\lambda - \frac{1}{2} = 0 \]In both cases, the eigenvalues indicate that the equilibrium points are unstable, as none have all negative real parts.