Problem 34 - 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. \(A=\left[\begin{array}{rr}2 & 4 \\ 2 & -2\end{array}\right]\)

Step-by-Step Solution

Answer

The equilibrium at \((0,0)\) is a saddle point and it is unstable.

Step 1: Find the Eigenvalues of A

To determine the eigenvalues of matrix \( A \), we use the characteristic equation \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix. Calculating \( A - \lambda I \) gives us:\[ A - \lambda I = \begin{bmatrix} 2 - \lambda & 4 \ 2 & -2 - \lambda \end{bmatrix} \]The determinant is then:\[ \det(A - \lambda I) = (2 - \lambda)(-2 - \lambda) - 8 = \lambda^2 - 6 = 0 \]Solving \( \lambda^2 - 6 = 0 \) yields the eigenvalues \( \lambda_1 = \sqrt{6} \) and \( \lambda_2 = -\sqrt{6} \).

Step 2: Analyze Stability and Classification

The eigenvalues \( \lambda_1 = \sqrt{6} \) and \( \lambda_2 = -\sqrt{6} \) are real and distinct. Since \( \lambda_1 > 0 \) and \( \lambda_2 < 0 \), there is one positive eigenvalue and one negative eigenvalue. This implies that as \( t \to \infty \), one component of the solution will go to infinity while another will shrink to zero. Therefore, the equilibrium point \((0,0)\) is a saddle point. This implies that the equilibrium point is unstable.

Key Concepts

Understanding Eigenvalues

Eigenvalues are a crucial concept in analyzing systems of differential equations. Simply put, they are numbers associated with a square matrix that reveal important properties about the system's behavior.
For the matrix given in the exercise:

• We start by finding a special equation called the characteristic equation: \( \ ext{det}(A - \lambda I) = 0 \).
• In our example, solving this equation allows us to determine the eigenvalues of the matrix \( A \).
• These eigenvalues are \( \lambda_1 = \sqrt{6} \) and \( \lambda_2 = -\sqrt{6} \).

Why are eigenvalues so important? They help us predict the behavior of solutions over time and determine if a system is stable or not. It's fascinating that such small numbers can tell us so much about the dynamics of a system!

Stability Analysis Around Equilibrium Points

Stability analysis is a vital step in understanding how systems behave near equilibrium points. The goal is to determine whether these points attract or repel nearby solutions over time.
Our equilibrium point is \((0,0)\) in the example. For stability:

• We look at the signs of the eigenvalues.
• Here, \( \lambda_1 = \sqrt{6} \) is positive, and \( \lambda_2 = -\sqrt{6} \) is negative.
• A positive eigenvalue indicates an unstable direction as it suggests solutions will grow over time.
• A negative eigenvalue suggests solutions will shrink, indicating a stable direction.

This mix of positive and negative eigenvalues means our equilibrium is unstable due to the presence of the positive eigenvalue. Understanding stability helps us predict where solutions will move if the system is slightly perturbed.

Equilibrium Classification: Sink, Source, Saddle

Once we've examined eigenvalues and stability, the next step is classifying equilibrium points. We can categorize them as a sink, source, or saddle point based on the eigenvalue signs.
The classification criteria are:

• If both eigenvalues are negative, the equilibrium is a **sink** and is stable (solutions tend to converge to this point).
• If both eigenvalues are positive, the equilibrium is a **source** and is unstable (solutions tend to diverge).
• If one eigenvalue is positive and the other is negative, the equilibrium is a **saddle point**, which is unstable (solutions diverge along one direction and converge along another).

For our matrix:- We find that the point \((0,0)\) is classified as a saddle point because it has one positive and one negative eigenvalue.
This classification tells us that trajectories will have both convergence and divergence depending on the initial conditions, making the system quite complex.