Problem 37 - 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. \(A=\left[\begin{array}{rr}-3 & -1 \\ 1 & -6\end{array}\right]\)

Step-by-Step Solution

Answer

The equilibrium at (0,0) is stable and is a sink.

Step 1: Find the Eigenvalues of Matrix A

To determine the stability of the equilibrium, we first need to find the eigenvalues of matrix \( A \). The eigenvalues are the solutions of the characteristic equation given by \( \text{det}(A - \lambda I) = 0 \), where \( I \) is the identity matrix. Calculate the determinant:\[\text{det}\left(\begin{bmatrix} -3-\lambda & -1 \ 1 & -6-\lambda \end{bmatrix}\right) = (-3-\lambda)(-6-\lambda) - (-1)(1)\]This simplifies to:\[\lambda^2 + 9\lambda + 17 = 0\]

Step 2: Solve for Eigenvalues

Using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), solve the quadratic equation \( \lambda^2 + 9\lambda + 17 = 0 \).Plugging in the values \( a = 1, b = 9, c = 17 \):\[\lambda = \frac{-9 \pm \sqrt{81 - 68}}{2} = \frac{-9 \pm \sqrt{13}}{2}\]This gives the eigenvalues \( \lambda_1 = \frac{-9 + \sqrt{13}}{2} \) and \( \lambda_2 = \frac{-9 - \sqrt{13}}{2} \).

Step 3: Analyze the Signs of the Eigenvalues

Determine the nature of the equilibrium by analyzing the signs of \( \lambda_1 \) and \( \lambda_2 \).Since both eigenvalues \( \lambda_1 \) and \( \lambda_2 \) are real and their product (the constant term of the quadratic, 17 in this case) is positive, both eigenvalues must have the same sign. Here both \( \lambda_1 \approx -1.697 \) and \( \lambda_2 \approx -7.303 \) are negative.

Step 4: Determine Stability and Type of Equilibrium

Since both eigenvalues are negative, the equilibrium point at \((0,0)\) is stable and classified as a sink. In this case, trajectories in the phase space approach the equilibrium point as time goes to infinity.

Key Concepts

Eigenvalues

To understand how systems behave, especially those described by differential equations, it's essential to grasp the concept of eigenvalues. Eigenvalues are special numbers that are drawn from a matrix, and they have important implications for stability analysis. When dealing with a 2x2 matrix, the eigenvalues are determined by solving the characteristic equation. This is often represented by the equation \( \text{det}(A - \lambda I) = 0 \), where \( A \) is your matrix and \( I \) is the identity matrix. For example, when you have the matrix \( A = \begin{bmatrix} -3 & -1 \ 1 & -6 \end{bmatrix} \), you substitute this into the characteristic equation. You then solve for \( \lambda \), which gives you the eigenvalues. In our scenario, this ends up with a handy quadratic equation: \( \lambda^2 + 9\lambda + 17 = 0 \). By solving this, you get two real eigenvalues: \( \lambda_1 = \frac{-9 + \sqrt{13}}{2} \) and \( \lambda_2 = \frac{-9 - \sqrt{13}}{2} \). Understanding these helps to decipher whether points will move towards or away from an equilibrium in a system.

Stability Analysis

Stability analysis is a tool used to determine whether a system will return to equilibrium after being perturbed, or whether it will diverge. When examining linear systems like ours, eigenvalues play a crucial role. If both eigenvalues of the matrix have negative real parts, the equilibrium is stable, meaning that the system will return to its equilibrium state over time. Conversely, if any eigenvalue has a positive real part, the system is unstable and moves away from equilibrium. Indeed, the sign of the eigenvalues tells you how trajectories behave. For example, in our exercise, the eigenvalues are approximately \( \lambda_1 \approx -1.697 \) and \( \lambda_2 \approx -7.303 \). Both are negative, indicating that the system is stable. Whenever the eigenvalues are negative, the system's trajectories head back toward the equilibrium point, as seen in stable sink behavior.

Equilibrium Classification

When classifying equilibrium points, the nature of the eigenvalues once again provides insight into the behavior of the system. Typically, equilibrium points can be classified as sinks, sources, or saddle points based on the signs of the eigenvalues.

• **Sink:** If all eigenvalues have negative real parts, the system is stable and all trajectories lead into the equilibrium point. This is what we describe as a sink. Our exercise is a classic example, with both eigenvalues being negative, indicating that the point \((0,0)\) acts as a sink.
• **Source:** When all eigenvalues have positive real parts, the system is unstable, and trajectories will repel from the equilibrium. This is known as a source.
• **Saddle Point:** Mixed eigenvalue signs indicate neither full attraction nor repulsion; the system can be part stable and part unstable, leading to a saddle point where some directions lead into the equilibrium and others lead away.

Understanding the classification helps pinpoint how systems behave over time, facilitating easier predictions and long-term behavior analysis.