Problem 17 - Calculus for Biology and Medicine

Question

Show that 0 0 is an equilibrium of $$ \left[\begin{array}{l} x_{1}(t+1) \\ x_{2}(t+1) \end{array}\right]=\left[\begin{array}{ll} -0.7 & 0 \\ -0.3 & 0.2 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$ and determine its stability.

Step-by-Step Solution

Answer

The point \((0, 0)\) is a stable equilibrium.

Step 1: Define the system

We are given a linear discrete-time system described by the equation: \[ \begin{bmatrix} x_{1}(t+1) \ x_{2}(t+1) \end{bmatrix} = \begin{bmatrix} -0.7 & 0 \ -0.3 & 0.2 \end{bmatrix} \begin{bmatrix} x_{1}(t) \ x_{2}(t) \end{bmatrix}. \]To show that the point \((0, 0)\) is an equilibrium point, we need to verify that when \((x_1(t), x_2(t)) = (0, 0)\), the resulting \((x_1(t+1), x_2(t+1))\) remains \((0, 0)\). This preserves the equilibrium.

Step 2: Verify the equilibrium point

Substitute \((x_1(t), x_2(t)) = (0, 0)\) into the system:\[\begin{bmatrix} x_{1}(t+1) \ x_{2}(t+1) \end{bmatrix} = \begin{bmatrix} -0.7 & 0 \ -0.3 & 0.2 \end{bmatrix} \begin{bmatrix} 0 \ 0 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}.\]This proves that \((0, 0)\) is an equilibrium point, as the system output remains \((0, 0)\) when starting from \((0, 0)\).

Step 3: Find eigenvalues for stability analysis

The stability of the equilibrium point can be determined by the eigenvalues of the matrix \(A = \begin{bmatrix} -0.7 & 0 \ -0.3 & 0.2 \end{bmatrix}\). We need to find the eigenvalues by solving the characteristic equation \(\det(A - \lambda I) = 0\).

Step 4: Calculate the characteristic equation

Subtract \(\lambda I\) from \(A\) which gives:\[A - \lambda I = \begin{bmatrix} -0.7 & 0 \ -0.3 & 0.2 \end{bmatrix} - \begin{bmatrix} \lambda & 0 \ 0 & \lambda \end{bmatrix} = \begin{bmatrix} -0.7 - \lambda & 0 \ -0.3 & 0.2 - \lambda \end{bmatrix}.\]Therefore, the characteristic equation is determined by:\[\det(A - \lambda I) = (-0.7 - \lambda)(0.2 - \lambda) - (0)(-0.3) = (-0.7 - \lambda)(0.2 - \lambda).\]

Step 5: Solve for eigenvalues

Solving the equation \((-0.7 - \lambda)(0.2 - \lambda) = 0\) gives us two solutions: \[\lambda_1 = -0.7 \, \text{and} \, \lambda_2 = 0.2.\]

Step 6: Determine stability

For stability, the modulus of each eigenvalue must be less than 1. We have: \[|\lambda_1| = |-0.7| = 0.7 < 1\] and \[|\lambda_2| = |0.2| = 0.2 < 1.\] Since both eigenvalues have moduli less than 1, the equilibrium point \((0, 0)\) is stable.

Key Concepts

Stability Analysis

In stability analysis, we aim to understand how a system behaves over time, particularly if it stays close to an equilibrium point. An equilibrium point is where the system does not change, meaning all derivatives or updates equal zero in that state. For the given discrete-time linear system, we found that

• At the point i.e., the system remains unchanged, highlighting a state of balance or equilibrium.

To determine stability, we analyze the system's response to small disturbances. A stable system will return to equilibrium after a disturbance, and this stability is often assessed using the system's eigenvalues.

Eigenvalues

Eigenvalues play a central role in determining the stability of a system. In any linear system, they indicate how the system will behave over time. For our discrete-time system matrix

• By solving , we determine the eigenvalues .

These eigenvalues inform us about the system's response:

• If the modulus of all eigenvalues is less than 1, the system is stable.
• If any eigenvalue has a modulus greater than or equal to 1, the system is unstable.
• In our case, both eigenvalues are less than 1 in magnitude, indicating the equilibrium is stable.

Understanding eigenvalues is crucial because they reveal how the system returns to equilibrium and the speed of convergence or divergence.

Linear Systems

Linear systems are characterized by linear equations that describe the relationship between inputs and outputs. Our system uses a linear transformation of state variables represented by matrices. The equation

• shows how the state at the next time step ( ) is a linear combination of the current state.

Such systems have predictable and analyzable behavior because linear equations provide simple solutions. The neat matrix form allows methods like eigenvalue analysis to quickly provide insights. Linear systems are often easier to study, making them a cornerstone in control theory and system dynamics.

Discrete-Time Systems

Discrete-time systems analyze changes at specific intervals. Designed for systems that evolve at distinct time steps, as opposed to continuously. Examples include digital signals or sample data in data analysis. In our case:

• represents the time-based transformation using a matrix, which updates state variables at each discrete step by multiplying with the current state.
• Allows easy computational modeling and simulation for predicting future states based on current observations.

Understanding discrete-time systems is vital in digital control systems, as they process data in digital format, aligning with technological advancements in computational efficiency.