Problem 35 - Calculus for Biology and Medicine

Question

We consider differential equations of the form \(\frac{d \mathbf{x}}{d t}=A \mathbf{x}(t)\) where $$ A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] $$ The eigenvalues of A will be real, distinct, and nonzero. Analyze the stability of the equilibrium \((0,0)\), and classify the equilib\mathrm{\\{} r i u m ~ a c c o r d i n g ~ t o ~ w h e t h e r ~ i t ~ i s ~ a ~ s i n k , ~ a ~ s o u r c e , ~ o r ~ a ~ s a d d l e ~ point. \(A=\left[\begin{array}{rr}1 & 3 \\ 1 & -1\end{array}\right]\)

Step-by-Step Solution

Answer

The equilibrium \((0,0)\) is a saddle point as one eigenvalue is positive and the other is negative.

Step 1: Find the Eigenvalues of A

To find the eigenvalues of matrix \( A \), we solve the characteristic equation: \( \det(A - \lambda I) = 0 \). For matrix \( A = \begin{bmatrix} 1 & 3 \ 1 & -1 \end{bmatrix} \), the equation becomes: \[ \det\left(\begin{bmatrix} 1-\lambda & 3 \ 1 & -1-\lambda \end{bmatrix}\right) = (1-\lambda)(-1-\lambda) - 3 \ = \lambda^2 - 5 + 3 \]. Solving \( \lambda^2 + \lambda - 4 = 0 \) gives the eigenvalues.

Step 2: Solve the Characteristic Equation

The characteristic equation is \( \lambda^2 + \lambda - 4 = 0 \). Using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 1, c = -4 \), we get: \[ \lambda = \frac{-1 \pm \sqrt{1 + 16}}{2} = \frac{-1 \pm \sqrt{17}}{2} \]. The eigenvalues are \( \lambda_1 = \frac{-1 + \sqrt{17}}{2} \) and \( \lambda_2 = \frac{-1 - \sqrt{17}}{2} \).

Step 3: Analyze Stability and Type of Equilibrium

Since the eigenvalues \( \lambda_1 = \frac{-1 + \sqrt{17}}{2} \) and \( \lambda_2 = \frac{-1 - \sqrt{17}}{2} \) are real and distinct, we analyze their signs. \( \lambda_1 \) is positive, indicating instability, and \( \lambda_2 \) is negative, indicating stability. Since one eigenvalue is positive and the other is negative, the equilibrium point \((0,0)\) is a saddle point.

Key Concepts

Eigenvalues

Eigenvalues are essential in understanding the dynamics of differential equations. In our problem, we need to determine the eigenvalues of a 2x2 matrix \( A \), given by \[ A = \begin{bmatrix} 1 & 3 \ 1 & -1 \end{bmatrix} \]. Eigenvalues are special numbers associated with a matrix, and they are found by solving the characteristic equation derived from the matrix. They help describe the behavior of a system represented by the matrix.
Generally, eigenvalues are found by solving the determinant equation:

• \( \det(A - \lambda I) = 0 \)

Where \( \lambda \) represents the eigenvalues, and \( I \) is the identity matrix. In this exercise, solving the characteristic equation gives us two eigenvalues, \( \lambda_1 = \frac{-1 + \sqrt{17}}{2} \) and \( \lambda_2 = \frac{-1 - \sqrt{17}}{2} \), which are crucial for further analysis of system behavior.

Stability Analysis

Stability analysis in the context of differential equations helps us understand how the system behaves over time. For a 2x2 matrix like ours, the signs of the eigenvalues are key indicators of the system's stability at the equilibrium point (often the origin, such as (0,0) in our case).
Stability is determined by the eigenvalues:

• If all eigenvalues have negative real parts, the system is stable and will settle to equilibrium over time.
• If at least one eigenvalue has a positive real part, the system is unstable, meaning perturbations will grow, moving the system away from equilibrium.

For our matrix with eigenvalues \( \lambda_1 = \frac{-1 + \sqrt{17}}{2} \) and \( \lambda_2 = \frac{-1 - \sqrt{17}}{2} \), \( \lambda_1 \) is positive, indicating some instability in the system. However, \( \lambda_2 \) is negative, indicating some stability. This mix leads to an interesting classification at the equilibrium, something we refer to as a saddle point.

Saddle Point

The term "saddle point" in the context of differential equations refers to a type of equilibrium point. It is characterized by having both stable and unstable directions. In simpler terms, a system with a saddle point has one path that leads towards stability, as in returning to equilibrium, and another path that moves away from stability.
A mathematical system like the one in our example may fluctuate in behavior depending on initial conditions. Here, the equilibrium point (often marked as (0,0) ) is a saddle point, because one eigenvalue is positive, causing instability, and the other is negative, causing stability.

• This combination makes it non-attractive overall, as the stable path influences only movement towards equilibrium over some trajectories, while other paths will diverge away.
• Such mixed behavior is said to be neither a sink (collision of trajectories) nor a source (dispersion of trajectories).

Characteristic Equation

The characteristic equation is central in finding eigenvalues for a matrix, which are used to analyze linear systems. It often appears in the form \( \det(A - \lambda I) = 0 \), where \( A \) is the original matrix, \( \lambda \) are the eigenvalues, and \( I \) is the identity matrix.
For the matrix given in our problem: \[ A = \begin{bmatrix} 1 & 3 \ 1 & -1 \end{bmatrix} \], constructing the characteristic equation involves:

• Subtracting \( \lambda \) times the identity matrix from \( A \).
• Calculating the determinant of the resulting matrix.

This results in \( \lambda^2 + \lambda - 4 = 0 \) by calculating and simplifying the determinant. Solving this quadratic gives us the eigenvalues. The characteristic equation, thus, serves as a gateway to understanding the system's overall behavior through its eigenvalues.