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 equilibrium according to whether it is a sink, a source, or a saddle point.3 $$ A=\left[\begin{array}{rr} -3 & -1 \\ 1 & -6 \end{array}\right] $$

Step-by-Step Solution

Answer

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

Step 1: Define the Problem

We have a differential equation \( \frac{d \mathbf{x}}{d t} = A \mathbf{x}(t) \) with matrix \( A = \left[ \begin{array}{rr} -3 & -1 \ 1 & -6 \end{array} \right] \). Our goal is to analyze the stability of the equilibrium at \( (0,0) \) and classify it as a sink, source, or saddle point.

Step 2: Calculate the Characteristic Equation

To analyze the stability of the equilibrium point, we first need to calculate the eigenvalues of matrix \( A \). We start by finding the characteristic equation, which is given by \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix. Thus, \( A - \lambda I = \left[ \begin{array}{rr} -3 - \lambda & -1 \ 1 & -6 - \lambda \end{array} \right] \).

Step 3: Solve the Determinant Equation

Calculate the determinant: \( \det(A - \lambda I) = (-3 - \lambda)(-6 - \lambda) - (-1)(1) = (\lambda + 3)(\lambda + 6) - 1 \). This simplifies to \( \lambda^2 + 9\lambda + 18 - 1 = \lambda^2 + 9\lambda + 17 = 0 \).

Step 4: Find the Eigenvalues

Solve the quadratic equation \( \lambda^2 + 9\lambda + 17 = 0 \) using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 9, c = 17 \). Calculate \( b^2 - 4ac = 81 - 68 = 13 \). Thus, \( \lambda = \frac{-9 \pm \sqrt{13}}{2} \), giving two distinct, real eigenvalues.

Step 5: Analyze the Equilibrium Type

The eigenvalues are \( \lambda_1 = \frac{-9 + \sqrt{13}}{2} \) and \( \lambda_2 = \frac{-9 - \sqrt{13}}{2} \). Since both eigenvalues are negative, the origin \( (0,0) \) is asymptotically stable, and the equilibrium is a sink.

Key Concepts

Eigenvalues

Eigenvalues are key to understanding the behavior of a differential equation's solutions, especially when dealing with systems described by matrices. When calculating eigenvalues, you are essentially looking for the special numbers, called the eigenvalues, that represent the growth rates of the vector directions in a system. In our example, we use a matrix \( A \), and calculate the eigenvalues by solving the characteristic equation, \( \det(A - \lambda I) = 0 \). This method involves:

• Setting up the matrix \( A - \lambda I \), where \( I \) is the identity matrix
• Finding the determinant of the resulting matrix
• Solving the resulting polynomial equation for \( \lambda \)

The values of \( \lambda \) you find are the eigenvalues. In this exercise, the eigenvalues help us determine the system's long-term behavior without solving the system explicitly. The eigenvalues can tell us whether solutions grow, decay, or change direction, making them critical in stability analysis and classification of equilibrium points.
In our specific case, the eigenvalues were calculated as \( \lambda_1 = \frac{-9 + \sqrt{13}}{2} \) and \( \lambda_2 = \frac{-9 - \sqrt{13}}{2} \), both of which are real and negative.

Stability of Equilibrium

The stability of equilibrium points in a system of differential equations often hinges on the eigenvalues of the system's matrix. Stability indicates whether, over time, the system's solutions will gravitate towards an equilibrium point, move away, or perhaps neither. To determine stability, remember:

• If all eigenvalues of the matrix have negative real parts, the equilibrium point is asymptotically stable—solutions move towards the equilibrium as time tends to infinity, often referred to as a "sink".
• If any eigenvalue has a positive real part, the point is unstable—solutions tend to move away from the equilibrium, called a "source".
• If eigenvalues have both positive and negative real parts, the point is a saddle point, indicating instability as some solutions move towards and others move away from the equilibrium.

In the given exercise, both eigenvalues are real and negative, indicating the system is asymptotically stable, making the equilibrium point \((0,0)\) a sink. This means that any perturbations in the system will decrease over time, eventually bringing the state back to the equilibrium point.

Linear Algebra

Linear algebra provides the foundational tools to tackle systems represented by matrices, as is typical within differential equations. By leveraging linear algebra concepts like matrices, determinants, and eigenvectors, we can simplify complex systems and analyze their behavior efficiently. The essential steps usually involve:

• Understanding how transformations are represented by matrices.
• Calculating determinants, which inform about invertibility and eigenvalues.
• Finding eigenvectors, which represent directions preserved by the transformation.
• Using these concepts to understand how the state of the system evolves over time.

For instance, in our exercise, linear algebra allows us to compose the differential equation as a matrix problem, check its stability, and understand the long-term evolution without necessarily solving the differential equation directly. It aids us in predicting dynamics by understanding the influence matrices exert on vector spaces, thereby making it easier to classify equilibrium points and their nature based on their respective eigenvalues.