Problem 28 - Calculus for Biology and Medicine

Question

Show that the equilibrium \(\left[\begin{array}{l}0 \\ 0\end{array}\right]\) of $$ \begin{array}{l} x_{1}(t+1)=\frac{3 x_{2}(t)}{1+x_{1}^{2}(t)} \\ x_{2}(t+1)=\frac{2 x_{1}(t)}{1+x_{2}^{2}(t)} \end{array} $$ is unstable.

Step-by-Step Solution

Answer

The equilibrium is unstable due to purely imaginary eigenvalues, suggesting oscillatory behavior without damping.

Step 1: Define the equilibrium point

An equilibrium point is where the system does not change. For the given system of equations, the equilibrium point is \(\begin{pmatrix} 0 \ 0 \end{pmatrix}\). At this point, both equations are satisfied for \(x_1(t+1) = x_1(t)\) and \(x_2(t+1) = x_2(t)\).

Step 2: Linearize the system

To analyze the stability, we need to linearize the system near the equilibrium point. Compute the Jacobian matrix of the system at \(\begin{pmatrix} 0 \ 0 \end{pmatrix}\). The Jacobian matrix is:\[J = \begin{bmatrix} \frac{\partial x_1(t+1)}{\partial x_1(t)} & \frac{\partial x_1(t+1)}{\partial x_2(t)} \ \frac{\partial x_2(t+1)}{\partial x_1(t)} & \frac{\partial x_2(t+1)}{\partial x_2(t)} \end{bmatrix}\]

Step 3: Compute partial derivatives

Calculate the partial derivatives:- \(\frac{\partial x_1(t+1)}{\partial x_1(t)} = 0\) since there is no \(x_1(t)\) in \(x_1(t+1)\).- \(\frac{\partial x_1(t+1)}{\partial x_2(t)} = \frac{3}{1+(0)^2} = 3\).- \(\frac{\partial x_2(t+1)}{\partial x_1(t)} = \frac{2}{1+(0)^2} = 2\).- \(\frac{\partial x_2(t+1)}{\partial x_2(t)} = 0\) since there is no \(x_2(t)\) in \(x_2(t+1)\).

Step 4: Construct the Jacobian matrix

Using the computed partial derivatives, we construct the Jacobian matrix:\[J = \begin{bmatrix} 0 & 3 \ 2 & 0 \end{bmatrix}\]

Step 5: Determine eigenvalues of the Jacobian

To determine the stability, calculate the eigenvalues of \(J\). The eigenvalues \(\lambda\) can be found from the characteristic equation:\[\det(J - \lambda I) = 0\]This gives:\[\begin{vmatrix} -\lambda & 3 \ 2 & -\lambda \end{vmatrix} = \lambda^2 - 6 = 0\]So, \(\lambda^2 = -6\), implying \(\lambda = \pm i\sqrt{6}\).

Step 6: Analyze eigenvalues for stability

The eigenvalues \(\lambda = \pm i\sqrt{6}\) are purely imaginary, indicating that the equilibrium point \(\begin{pmatrix} 0 \ 0 \end{pmatrix}\) is not asymptotically stable. Since the real parts of the eigenvalues are zero, this system might exhibit oscillatory behavior and is therefore unstable.

Key Concepts

Equilibrium Point

An equilibrium point in a dynamic system is a crucial concept as it represents the state where the system does not evolve over time. In other words, at this point, there are no changes in the system's state variables as time progresses.

For the given system of difference equations:

• \( x_{1}(t+1)=\frac{3 x_{2}(t)}{1+x_{1}^{2}(t)} \)
• \( x_{2}(t+1)=\frac{2 x_{1}(t)}{1+x_{2}^{2}(t)} \)

The equilibrium point is \([0, 0]\). This means that when both \(x_1\) and \(x_2\) are zero, the future values of \(x_1\) and \(x_2\) will also remain zero, satisfying the condition \(x_1(t+1) = x_1(t)\) and \(x_2(t+1) = x_2(t)\). Finding equilibrium points involves solving where both updates of the system equal the current state, which is typically the backdrop for further stability analysis.

Jacobian Matrix

The Jacobian matrix is a powerful tool used in mathematics to study the behavior near an equilibrium point, especially in non-linear systems. It is essentially a matrix of all first-order partial derivatives of a vector-valued function.

In the context of this system, the Jacobian matrix will help in linearizing the system around the equilibrium point \([0, 0]\). Matrix linearization provides a simplified way to understand how small changes near an equilibrium point affect the system's behavior.

For our exercise, to derive the Jacobian matrix, we perform the following:

• Calculate \( \frac{\partial x_1(t+1)}{\partial x_1(t)} \)and \(\frac{\partial x_1(t+1)}{\partial x_2(t)}\)
• Calculate \(\frac{\partial x_2(t+1)}{\partial x_1(t)} \) and \( \frac{\partial x_2(t+1)}{\partial x_2(t)}\)

After calculating, the Jacobian for this system at \([0, 0]\) becomes \(J = \begin{bmatrix} 0 & 3 \ 2 & 0 \end{bmatrix}\). Such linearization is crucial for studying stability because it approximates the non-linear dynamics with linear ones around the equilibrium.

Eigenvalues

Eigenvalues are fundamental to understanding the stability of a system when studying the Jacobian matrix. They give insight into how perturbations near equilibrium points evolve over time.

To assess stability using eigenvalues:

• First, solve the characteristic equation \( \det(J - \lambda I) = 0 \) where \( I \) is the identity matrix.
• For this exercise, it results in \( \lambda^2 = -6 \) giving eigenvalues \( \lambda = \pm i\sqrt{6} \).

These eigenvalues are purely imaginary, \( \pm i\sqrt{6} \). In terms of stability, an eigenvalue with a real part that is zero suggests a neutral stability. As the system may not return to equilibrium or show asymptotic behavior, but rather exhibit oscillations. Thus, the equilibrium point \([0, 0]\) is considered unstable. This behavior is typical in systems that can sustain harmonic oscillations.