Problem 32 - Calculus for Biology and Medicine

Question

Show that \(\left[\begin{array}{c}0 \\ 0\end{array}\right]\) and \(\left[\begin{array}{r}-\pi \\ \pi\end{array}\right]\) are equilibria of $$ \begin{array}{l} x_{1}(t+1)=-x_{2}(t) \\ x_{2}(t+1)=\sin \left(x_{2}(t)\right)-x_{1}(t) \end{array} $$ and analyze their stability.

Step-by-Step Solution

Answer

Both vectors are equilibria; [0,0] is unstable, [-π,π] is generally unstable.

Step 1: Identify Equilibria

To identify equilibria, we need to set the system of equations to zero, since equilibria occur where the derivatives are zero. Therefore, find values of \( (x_1, x_2) \) for which \[-x_2 = x_1(t+1) = 0 \]and\[\sin(x_2) - x_1 = x_2(t+1) = 0 \]Analyzing the given vectors, substitute into these equations to verify if they are equilibria.

Step 2: Substitute \([0, 0]\) and Check

Substitute \( x_1 = 0 \) and \( x_2 = 0 \) into the equations:\[-0 = 0 \Rightarrow \text{True,} \]\[\sin(0) - 0 = 0 \Rightarrow \text{True.} \]Thus, \([0, 0]\) is an equilibrium point since both equations hold true.

Step 3: Substitute \([-\pi, \pi]\) and Check

Substitute \( x_1 = -\pi \) and \( x_2 = \pi \) into the equations:\[-\pi = -\pi \Rightarrow \text{True,} \]\[\sin(\pi) - (-\pi) = 0 \Rightarrow 0 + \pi = \pi \Rightarrow \text{True.} \]Thus, \([-\pi, \pi]\) is also an equilibrium point since both equations hold true.

Step 4: Analyze Stability of \([0, 0]\)

To determine stability, linearize the system around the equilibrium. This involves calculating the Jacobian matrix at the point:\[ J = \begin{bmatrix} \frac{\partial }{\partial x_1}(-x_2) & \frac{\partial }{\partial x_2}(-x_2) \ \frac{\partial }{\partial x_1}(\sin x_2 - x_1) & \frac{\partial }{\partial x_2}(\sin x_2 - x_1) \end{bmatrix} = \begin{bmatrix} 0 & -1 \ -1 & \cos x_2 \end{bmatrix} \]Evaluating at \([0, 0]\), the Jacobian matrix becomes:\[ J(0,0) = \begin{bmatrix} 0 & -1 \ -1 & 1 \end{bmatrix} \]Calculate the eigenvalues of this Jacobian to determine stability.

Step 5: Eigenvalue Calculation for \([0, 0]\)

Calculate eigenvalues from the determinant equation:\[ \text{det}(J - \lambda I) = \begin{vmatrix} -\lambda & -1 \ -1 & 1-\lambda \end{vmatrix} = \lambda^2 - \lambda - 1 = 0 \]The eigenvalues are found by solving:\[ \lambda = \frac{1 \pm \sqrt{5}}{2} \]Since one eigenvalue exceeds 0, \([0, 0]\) is a saddle point and hence unstable.

Step 6: Analyze Stability of \([-\pi, \pi]\)

Similarly, calculate the Jacobian at \([-\pi, \pi]\).Replace \( \cos x_2 \) in the Jacobian with \( \cos \pi = -1 \):\[ J(-\pi, \pi) = \begin{bmatrix} 0 & -1 \ -1 & -1 \end{bmatrix} \]Find eigenvalues from:\[ \text{det}(J - \lambda I) = \begin{vmatrix} -\lambda & -1 \ -1 & -1-\lambda \end{vmatrix} = \lambda^2 + \lambda = 0 \]Eigenvalues are: \( \lambda = 0 \) and \( \lambda = -1 \), indicating a non-isolated equilibrium point which makes it neutrally stable but generally unstable due to the zero eigenvalue.

Key Concepts

Stability Analysis

Stability analysis helps us determine whether small perturbations or changes in a system around its equilibria will die out or grow over time. When dealing with dynamic systems like the one in the exercise, the stability of equilibrium points is crucial to understanding how the system behaves in the long run.

The concept revolves around analyzing how the system reacts to disturbances. If small changes tend to disappear, the system is stable at that point. If they grow, the system is unstable.

Stability is often categorized into three types:

• Asymptotically Stable: Perturbations decay to zero over time.
• Unstable: Perturbations grow over time.
• Neutrally Stable: Perturbations neither decay nor grow significantly.

Stability analysis typically involves techniques like linearization and finding eigenvalues, which we will explore next.

Linearization

Linearization is a mathematical method used to approximate a nonlinear system by a linear one, around an equilibrium point. This simplifies the analysis by using a linear approximation to understand the behavior of the system near the equilibrium. This method is particularly efficient for small perturbations.

The process involves finding the first order Taylor expansion of the system equations around the equilibrium point. In simple terms, we determine the slope of the system's curve at the equilibrium, which can be represented by a matrix, often called the Jacobian matrix, as we will discuss next.

This approach allows us to use linear algebra techniques to analyze stability, making it a powerful tool in equilibrium analysis. While it might not capture behaviors far from the equilibrium, it provides accurate local insights.

Jacobian Matrix

The Jacobian matrix is a key component in the process of linearization. It is a matrix of all first-order partial derivatives of a vector-valued function. Essentially, it describes the rate of change of each output of a system with respect to each input.

For a system of equations, the Jacobian matrix provides a first-order approximation of how the system behaves near an equilibrium point, making it a significant tool in assessing stability. In the context of our exercise, the Jacobian matrix is given by:\[J = \begin{bmatrix} \frac{\partial }{\partial x_1}(-x_2) & \frac{\partial }{\partial x_2}(-x_2) \ \frac{\partial }{\partial x_1}(\sin x_2 - x_1) & \frac{\partial }{\partial x_2}(\sin x_2 - x_1) \end{bmatrix}\]Evaluating this at the equilibrium points, we obtain matrices that help understand how small perturbations can affect the system.

The next step is to analyze these matrices to find the eigenvalues, which will give us more insights into the stability characteristics of each equilibrium point.

Eigenvalues

Eigenvalues are a fundamental concept in the stability analysis of linear systems. They are scalars that give us significant insights into a system's dynamics at an equilibrium point, particularly when derived from the Jacobian matrix.

Calculating the eigenvalues of the Jacobian involves solving the characteristic equation:\[\text{det}(J - \lambda I) = 0\]where \( \lambda \) represents the eigenvalues and \( I \) is the identity matrix.

In stability analysis:

• If all eigenvalues have negative real parts, the equilibrium is asymptotically stable.
• If at least one eigenvalue has a positive real part, the equilibrium is unstable.
• If eigenvalues have zero real parts, it suggests neutral stability, but additional investigation is needed as behaviors can be unpredictable.

For our exercise, analyzing the eigenvalues at each equilibrium reveals the stability nature. These insights help predict the system's response to small disturbances around those equilibrium points.