Problem 32 - 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. $$ A=\left[\begin{array}{rr} -5 & -2 \\ 6 & 3 \end{array}\right] $$

Step-by-Step Solution

Answer

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

Step 1: Determine the Eigenvalues of Matrix A

First, we need to find the eigenvalues of matrix \( A \). The eigenvalues \( \lambda \) can be found by solving the characteristic equation \( \text{det}(A - \lambda I) = 0 \), where \( I \) is the identity matrix. The characteristic equation is: \[\begin{vmatrix}-5 - \lambda & -2 \6 & 3 - \lambda\end{vmatrix} = 0\]Calculate the determinant to find the characteristic polynomial: \((-5 - \lambda)(3 - \lambda) - (-2)(6) = \lambda^2 + 2\lambda - 3 = 0\). Solve this to find \( \lambda_{1} = -3 \) and \( \lambda_{2} = 1 \).

Step 2: Classify the Equilibrium Point

With the eigenvalues known, we can classify the stability of the equilibrium point \((0,0)\). The eigenvalues are \( \lambda_{1} = -3 \) and \( \lambda_{2} = 1 \). Since one eigenvalue is negative (\( \lambda_{1} = -3 \)) and the other is positive (\( \lambda_{2} = 1 \)), the system contains both attracting and repelling components. This indicates a saddle point.

Step 3: Conclude the Stability Analysis

Since the equilibrium point \((0,0)\) has one negative eigenvalue and one positive eigenvalue, the system diverges in one direction and converges in another. Therefore, the equilibrium point is classified as a saddle point, which means it is unstable.

Key Concepts

Differential Equations

Differential equations are fundamental in understanding how systems evolve over time. They express the rate at which one variable changes with respect to another.
For example, when analyzing the change in a system like temperature or population, differential equations can model these dynamically. This helps predict future states of the system.

• In our exercise, we consider a simple linear differential equation of the form \(\frac{d \mathbf{x}}{d t} = A \mathbf{x}(t)\).
• This equation involves the derivative of \(\mathbf{x}\), representing how the state of the system \(\mathbf{x}(t)\) changes over time.

The matrix \(A\) plays a crucial role here. It's the constant that influences how the change happens. By analyzing this matrix, we can gain insights into the behavior of the system over time.

Eigenvalues

Eigenvalues are vital in the stability analysis of differential equations. They are scalar values associated with a matrix that provide insight into system behavior.
To find these, we solve the characteristic equation \(\text{det}(A - \lambda I) = 0\). This equation arises from requiring nontrivial solutions for \(A \mathbf{x} = \lambda \mathbf{x}\), where \(I\) is the identity matrix.

• For our matrix \(A\), we find eigenvalues \(\lambda_1 = -3\) and \(\lambda_2 = 1\).
• These values tell us about expansion and contraction directions of the system.

The sign of these eigenvalues is critical. A negative eigenvalue indicates a direction of contraction (pulling in), while a positive eigenvalue signals expansion (pushing out). This is foundational for determining system stability.

Equilibrium Point Classification

Classifying equilibrium points helps determine the stability of a system at rest. An equilibrium point, such as \((0,0)\) here, is where the system's derivative equals zero, meaning there's no change at that point if at rest.
After finding the eigenvalues, we use them to classify the equilibrium:

• - If both eigenvalues are negative, we have a stable equilibrium or a sink, as all paths lead to rest.
• - If both are positive, the equilibrium is unstable or a source, since all paths move away from the point.
• - With one negative and one positive eigenvalue, we face a saddle point like in our solution.

A saddle point indicates partial stability. Paths approach in some directions but diverge in others, leading to instability in the system.

Matrix Algebra

Matrix algebra is crucial for solving systems of linear equations and understanding dynamics in systems.
A matrix is a set of numbers arranged in rows and columns, and they are used to represent linear transformations. In our exercise, the matrix \(A\) represents the dynamics of the system.

• We compute the determinant and solve characteristic equations using matrix algebra.
• This aids in finding eigenvalues and eigenvectors, which are essential for analyzing system stability.

Matrix operations like addition, multiplication, and finding inverses allow us to manipulate data conveniently to solve complex problems. In stability analysis, they help model and predict system behaviors over time.