Problem 25 - Calculus for Biology and Medicine

Question

Show that the equilibrium \(\left[\begin{array}{l}0 \\ 0\end{array}\right]\) of $$ \left[\begin{array}{l} x_{1}(t+1) \\ x_{2}(t+1) \end{array}\right]=\left[\begin{array}{ll} 4.2 & -3.4 \\ 2.4 & -1.1 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$ is unstable

Step-by-Step Solution

Answer

The origin \( \begin{bmatrix} 0 \\ 0 \end{bmatrix} \) is unstable as one eigenvalue is greater than 1.

Step 1: Define Equilibrium Point

The problem statement provides the equilibrium point as \( \begin{bmatrix} 0 \ 0 \end{bmatrix} \). This point is where the system does not change over time; that is, if the system is in equilibrium, then the future state is the same as the current state.

Step 2: System Equation Substitution Check

In general, an equilibrium point is such that \( \mathbf{x}(t+1) = \mathbf{x}(t) \). We have \( \mathbf{x} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \) by definition, so substituting this into the transformation matrix results in: \[ \begin{bmatrix} x_1(t+1) \ x_2(t+1) \end{bmatrix} = \begin{bmatrix} 4.2 & -3.4 \ 2.4 & -1.1 \end{bmatrix} \begin{bmatrix} 0 \ 0 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}. \] This confirms that \( \mathbf{x} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \) is an equilibrium point as expected.

Step 3: Compute Eigenvalues for Stability

Stability of the equilibrium is determined by the eigenvalues of the matrix \( A = \begin{bmatrix} 4.2 & -3.4 \ 2.4 & -1.1 \end{bmatrix} \). We first calculate the determinant of \( A - \lambda I \), where \( I \) is the identity matrix: \[ \det(A - \lambda I) = \det\begin{bmatrix} 4.2 - \lambda & -3.4 \ 2.4 & -1.1 - \lambda \end{bmatrix} = (4.2 - \lambda)(-1.1 - \lambda) - (-3.4)(2.4). \] This simplifies to \[ \lambda^2 - 3.1\lambda + 1.56. \] The roots of this characteristic polynomial are the eigenvalues.

Step 4: Solve for Eigenvalues

Solve the characteristic polynomial \( \lambda^2 - 3.1\lambda + 1.56 = 0 \) using the quadratic formula: \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a=1, b=-3.1, c=1.56 \). Solving, you find: \[ \lambda_1 = 2.4, \quad \lambda_2 = 0.65. \]

Step 5: Assess Stability

For stability, all eigenvalues must lie within the unit circle, meaning they have absolute values less than 1. Here, \( |\lambda_1| = 2.4 > 1 \), which means the system will not remain bounded and diverge over time.

Key Concepts

Equilibrium Point

An equilibrium point in a dynamic system is a condition where the system's variables do not change as time progresses. Essentially, it is the point where if the system finds itself, it will stay there indefinitely unless disturbed. For a system like \[\begin{bmatrix} x_1(t+1) \ x_2(t+1) \end{bmatrix} = \begin{bmatrix} 4.2 & -3.4 \ 2.4 & -1.1 \end{bmatrix} \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix},\] the equilibrium point occurs when the vector \[\begin{bmatrix} x_1 \ x_2 \end{bmatrix}\] does not change over time, that is, \[\begin{bmatrix} x_1(t+1) \ x_2(t+1) \end{bmatrix} = \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix}.\] To find an equilibrium point, we assume \( x_1(t) = x_2(t) = 0\) for simplicity in this matrix equation, resulting in \(\begin{bmatrix} 0 \ 0 \end{bmatrix}\). This means if the system starts at this point, it will not change as time goes on, making it an equilibrium condition.

Eigenvalues

Eigenvalues are central to understanding the behavior and stability of dynamic systems. They are derived from the transformation matrix of a system and play a crucial role in predicting how the system acts over time. In our exercise, we examine the matrix \[A = \begin{bmatrix} 4.2 & -3.4 \ 2.4 & -1.1 \end{bmatrix}.\] To find the eigenvalues, you need to solve the characteristic equation usually formulated as \[\text{det}(A - \lambda I) = 0\]where \(\lambda\) represents the eigenvalues and \(I\) is the identity matrix of the same dimension as \(A\). Solving the equation \(\lambda^2 - 3.1\lambda + 1.56 = 0\) using the quadratic formula, gives us two eigenvalues: \(\lambda_1 = 2.4\) and \(\lambda_2 = 0.65\). These eigenvalues reveal important information about system behavior, particularly regarding stability and dynamic response.

Matrix Algebra

Matrix algebra serves as the foundation for analyzing systems of linear equations, especially in determining the behavior of dynamic systems. In the exercise, we are dealing with a matrix: \[A = \begin{bmatrix} 4.2 & -3.4 \ 2.4 & -1.1 \end{bmatrix}.\] This matrix helps express how the state of the system evolves over time by acting on the state vector \(\begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix}\) to produce the next state \(\begin{bmatrix} x_1(t+1) \ x_2(t+1) \end{bmatrix}\). Using basic operations such as matrix multiplication and manipulation of determinants, matrix algebra allows us to solve for critical properties, such as the eigenvalues, which indicate system stability. It also enables the determination of equilibrium points by setting certain conditions under which the state vector remains unchanged. Understanding these concepts through matrix algebra provides powerful insights into the workings of complex systems and allows for the prediction of long-term behavior.