Problem 20 - Calculus for Biology and Medicine

Question

Show that \(\left[\begin{array}{l}0 \\ 0\end{array}\right]\) is an equilibrium of $$ \left[\begin{array}{l} x_{1}(t+1) \\ x_{2}(t+1) \end{array}\right]=\left[\begin{array}{rr} 0.1 & 0.4 \\ 0.1 & -0.2 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$ and determine its stability.

Step-by-Step Solution

Answer

The point \(\left[ \begin{array}{c} 0 \\ 0 \end{array} \right]\) is an equilibrium and it is stable with eigenvalues 0.1 and -0.2.

Step 1: Define Equilibrium Point

An equilibrium point is where the system does not change over time. For the matrix system given, \(\left[ \begin{array}{l} x_1(t+1) \ x_2(t+1) \end{array}\right] = \left[ \begin{array}{rr} 0.1 & 0.4 \ 0.1 & -0.2 \end{array}\right] \left[ \begin{array}{l} x_1(t) \ x_2(t) \end{array}\right]\), we set \(\left[ \begin{array}{l} x_1(t+1) \ x_2(t+1) \end{array}\right] = \left[ \begin{array}{l} x_1(t) \ x_2(t) \end{array}\right] =\left[ \begin{array}{l} 0 \ 0 \end{array}\right]\) to find that point. This gives the equation \(A\mathbf{x} = \mathbf{x}\).

Step 2: Check the Equilibrium Condition

Substitute \(\mathbf{x}=\left[ \begin{array}{l} 0 \ 0 \end{array}\right]\), in the equation \(\mathbf{x}(t+1) = A\mathbf{x}(t)\). We must verify that \(A \left[ \begin{array}{l} 0 \ 0 \end{array}\right] = \left[ \begin{array}{l} 0 \ 0 \end{array}\right]\). Doing so: \[\left[ \begin{array}{rr} 0.1 & 0.4 \ 0.1 & -0.2 \end{array}\right] \left[ \begin{array}{l} 0 \ 0 \end{array}\right] = \left[ \begin{array}{l} (0.1 \times 0) + (0.4 \times 0) \ (0.1 \times 0) + (-0.2 \times 0) \end{array}\right] = \left[ \begin{array}{l} 0 \ 0 \end{array}\right] \] This confirms that \(\left[ \begin{array}{l} 0 \ 0 \end{array}\right]\) is an equilibrium point.

Step 3: Analyze Stability Using Eigenvalues

Determine stability by finding the eigenvalues of matrix \(A\). The eigenvalues are found by solving \(\det(A - \lambda I) = 0\).Substitute \(A\) and \(\lambda I\) into the determinant equation: \[\det \left( \left[ \begin{array}{rr} 0.1 & 0.4 \ 0.1 & -0.2 \end{array}\right] - \lambda \left[ \begin{array}{rr} 1 & 0 \ 0 & 1 \end{array}\right] \right) = \det \left( \left[ \begin{array}{rr} 0.1-\lambda & 0.4 \ 0.1 & -0.2-\lambda \end{array}\right] \right)\]This evaluates to: \[(0.1-\lambda)(-0.2-\lambda) - (0.4 \times 0.1) = 0\]Solving gives the characteristic equation: \[\lambda^2 + 0.1\lambda - 0.02 = 0\]

Step 4: Solve For Eigenvalues

Solve the characteristic equation \(\lambda^2 + 0.1\lambda - 0.02 = 0\) by using the quadratic formula \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a=1,\ b=0.1,\ c=-0.02\). Calculate:\[\lambda = \frac{-0.1 \pm \sqrt{(0.1)^2 - 4 \times 1 \times (-0.02)}}{2 \times 1}\]\[\lambda = \frac{-0.1 \pm \sqrt{0.01 + 0.08}}{2}\]\[\lambda = \frac{-0.1 \pm \sqrt{0.09}}{2} = \frac{-0.1 \pm 0.3}{2}\]Thus, \(\lambda_1 = 0.1\) and \(\lambda_2 = -0.2\).

Step 5: Determine Stability from Eigenvalues

The stability of the equilibrium point \(\left[ \begin{array}{l} 0 \ 0 \end{array}\right]\) depends on whether all eigenvalues have an absolute value less than 1. The eigenvalues are \(\lambda_1 = 0.1\) and \(\lambda_2 = -0.2\), both of which have absolute values less than 1: \(|0.1| < 1\) and \(|-0.2| < 1\). Therefore, \(\left[ \begin{array}{l} 0 \ 0 \end{array}\right]\) is stable.

Key Concepts

Stability Analysis

Stability analysis is a vital part of understanding how equilibrium points behave over time in dynamic systems. In the given problem, we need to evaluate the stability of the equilibrium point \( \begin{bmatrix} 0 \ 0 \end{bmatrix} \)when subjected to a set of linear difference equations. In simple terms, stability analysis helps us understand if small perturbations around the equilibrium will diminish over time (stable) or grow (unstable).

The process typically involves determining whether all eigenvalues of the matrix associated with the system have magnitudes (absolute values) less than one. If they do, then the equilibrium point is stable, implying that any small disturbance from this point will eventually return to equilibrium. Conversely, if any eigenvalue exceeds one in absolute value, the point is unstable. Understanding these properties helps predict the future behavior of dynamic systems, which is crucial in many fields such as physics, engineering, and economics.

Eigenvalues

Eigenvalues play a critical role in determining the behavior of dynamic systems, particularly their stability. In linear algebra, eigenvalues are scalars associated with a square matrix, offering deep insights into the matrix's properties.

To find eigenvalues, we typically solve the characteristic equation derived from the matrix. The general form of this equation is \( \det(A - \lambda I) = 0 \), where:

• \( A \) is the matrix.
• \( \lambda \) is a scalar (the eigenvalue).
• \( I \) is the identity matrix of the same order as \( A \).

For the given matrix \( \begin{bmatrix} 0.1 & 0.4 \ 0.1 & -0.2 \end{bmatrix} \),we've calculated and solved the characteristic equation to determine the eigenvalues: \( \lambda_1 = 0.1 \) and \( \lambda_2 = -0.2 \).

The important takeaway is that eigenvalues reveal essential properties of a system, like stability and oscillatory behaviors. These values help us conclude that since both eigenvalues have absolute values less than one, the equilibrium point in our exercise is stable.

Linear Algebra

Linear algebra is a branch of mathematics centered around vector spaces and linear mappings between these spaces. It's fundamental to many areas of mathematics and applicable sciences, including the study of equilibrium points and stability in dynamic systems.

In this exercise, we use linear algebra to analyze a system of linear equations represented by a 2x2 matrix. The representation makes it easier to apply systematic methods to find equilibrium points and determine stability. With linear algebra, we can apply various matrix operations to find characteristics like eigenvalues, which help in understanding the dynamics of systems.

• Matrix multiplication is used to define the system dynamics.
• Setting the system's output equal to the input vector helps identify equilibrium points.
• Eigenvalue calculations determine stability.

Linear algebra provides the framework, enabling us to handle complex problems by breaking them down into manageable linear systems. This is especially useful for analyzing dynamic systems, facilitating clear understanding and modeling of real-world phenomena.