Problem 24 - Calculus for Biology and Medicine

Question

Show that the equilibrium \(\begin{array}{ll}0 & \text { of } \\ 0\end{array}\) $$ \left[\begin{array}{l} x_{1}(t+1) \\ x_{2}(t+1) \end{array}\right]=\left[\begin{array}{rr} 0.2 & 0.3 \\ -0.5 & -0.4 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$ is stable.

Step-by-Step Solution

Answer

The equilibrium is stable because both eigenvalues have magnitudes less than 1.

Step 1: Understand the System

We have a discrete-time dynamical system given by the matrix \( A = \begin{bmatrix} 0.2 & 0.3 \ -0.5 & -0.4 \end{bmatrix} \). We need to determine the stability of the equilibrium at the origin \( (x_1, x_2) = (0, 0) \). Stability requires the eigenvalues of matrix \( A \) to have magnitudes less than 1.

Step 2: Write the Characteristic Equation

To find the eigenvalues, we start by writing the characteristic equation. This is given by \( \text{det}(A - \lambda I) = 0 \), where \( I \) is the identity matrix. Substituting, we have:\[\text{det}\left(\begin{bmatrix} 0.2 - \lambda & 0.3 \ -0.5 & -0.4 - \lambda \end{bmatrix}\right) = 0.\]

Step 3: Compute the Determinant

Now compute the determinant:\[(0.2 - \lambda)(-0.4 - \lambda) - (0.3)(-0.5) = 0.\]Simplifying this, we have:\[ (-0.4 - \lambda)(0.2 - \lambda) + 0.15 = 0. \]

Step 4: Expand and Simplify

Expand the equation:\[\lambda^2 + 0.2\lambda + 0.4\lambda + 0.08 + 0.15 = 0.\]Then combine like terms:\[\lambda^2 + 0.6\lambda + 0.23 = 0.\]

Step 5: Solve for Eigenvalues

To find the eigenvalues, use the quadratic formula: \[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 1, b = 0.6, c = 0.23 \). Calculate:\[b^2 - 4ac = (0.6)^2 - 4 \cdot 1 \cdot 0.23 = 0.36 - 0.92 = -0.56.\]

Step 6: Determine Eigenvalues

Since \( b^2 - 4ac < 0 \), the eigenvalues are complex, implying:\[\lambda = \frac{-0.6 \pm \sqrt{-0.56}i}{2}.\]Calculate the magnitude of the eigenvalues:\[|\lambda| = \sqrt{\left(-0.3\right)^2 + \left(\frac{\sqrt{0.56}}{2}\right)^2} = \sqrt{0.09 + 0.14} = 0.5.\]

Step 7: Conclude Stability

Both eigenvalues have magnitudes equal to \( 0.5 \), which are less than 1. Therefore, the system is stable at the origin.

Key Concepts

Discrete-Time Dynamical Systems

In the realm of mathematics and systems theory, discrete-time dynamical systems are a fundamental concept. These systems describe how variables evolve over discrete time intervals. Instead of continuous change, as seen in differential equations, changes occur in steps. Consider a system that updates its state at every tick of a clock: that's the essence of discrete time.

Discrete-time dynamical systems are often defined using matrices to model linear transformations over time. For example, if we have two variables, these dynamics can be represented as:

• A state vector, often denoted as \(\begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix}\), which changes over discrete time t.
• A transformation matrix \(A\), such that the next state \(x(t+1)\) is derived from \(x(t)\) through matrix multiplication.

This approach helps analyze complex systems in various fields, such as economics, biology, and control systems. Understanding these principles enables us to predict and control system behaviors effectively.

Eigenvalues

The term eigenvalues, stemming from linear algebra, plays a vital role in understanding dynamical systems. Eigenvalues provide critical insights into the behavior of systems captured by matrices. Let's delve deeper.

For a given square matrix \(A\), an eigenvalue is a number that signifies how much the linear transformation stretches or compresses vectors when multiplied by the matrix. It is defined through the equation:

• \((A - \lambda I) = 0\)

Here, \(\lambda\) represents an eigenvalue of \(A\), and \(I\) is the identity matrix. Finding the eigenvalues of a system matrix can reveal valuable information, such as:

• If vectors grow, shrink, or rotate.
• Whether the equilibrium points are stable, neutral, or unstable.

Determining the eigenvalues is crucial for stability analysis, which is a key objective when examining discrete-time systems.

Characteristic Equation

To find eigenvalues of a matrix, we use the characteristic equation. This equation arises from the determinant condition \(\text{det}(A - \lambda I) = 0\). Let's break that down.

When you subtract \(\lambda I\) from a matrix \(A\), you essentially adjust the matrix in such a way that it reflects potential eigenvectors being stretched by \(\lambda\). Solving for \(\lambda\), where the determinant of the resulting matrix equals zero, gives us the characteristic equation and its roots (eigenvalues).

The significance of the characteristic equation includes:

• It's a polynomial equation where the degree corresponds to the size of the matrix. For a 2x2 matrix, it results in a quadratic equation.
• Solving the characteristic equation yields eigenvalues that determine system behavior over time.

Understanding how to derive and solve the characteristic equation is crucial in analyzing stability and system dynamics.

Equilibrium Stability

Equilibrium stability is a cornerstone concept in the study of dynamical systems. Equilibrium refers to a state where the system does not change over time. The stability of an equilibrium determines how a system responds to small disturbances.

An equilibrium point is said to be stable if, after a slight perturbation, the system 'returns' to the equilibrium point over time. In contrast, an unstable equilibrium will lead the system away after a disturbance.

Key factors that affect equilibrium stability include:

• The nature of eigenvalues. If all eigenvalues of a system matrix have magnitudes less than 1, the system is stable.
• If eigenvalues have a magnitude equal to or greater than 1, stability is not guaranteed, indicating potential instability.

Analyzing eigenvalues provides insight into whether perturbations will dissipate, indicating stability, or cause divergences. Such analysis helps in designing systems that are robust and reliable.