Problem 43 - 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 complex conjugates. Analyze the stability of the equilibrium \((0,0)\), and classify the equilibrium according to whether it is a stable spiral, an unstable spiral, or a center. $$ A=\left[\begin{array}{rr} 2 & -1 \\ 3 & 0 \end{array}\right] $$

Step-by-Step Solution

Answer

The equilibrium (0,0) is an unstable spiral.

Step 1: Identify the problem

We need to determine the stability and type of equilibrium at the origin (0,0) for a given differential equation \( \frac{d \mathbf{x}}{d t}=A \mathbf{x}(t) \) with a specific matrix \( A \).

Step 2: Write matrix A

We are given the matrix \( A = \left[ \begin{array}{rr} 2 & -1 \ 3 & 0 \end{array} \right] \).

Step 3: Calculate the characteristic equation

The characteristic equation of matrix \( A \) is determined by \( \det(A - \lambda I) = 0 \). For our matrix, it is \( \left| \begin{array}{rr} 2 - \lambda & -1 \ 3 & -\lambda \end{array} \right| = 0 \).

Step 4: Solve the determinant

Calculate the determinant: \( (2 - \lambda)(-\lambda) - (-1)(3) = \lambda^2 - 2\lambda + 3 = 0 \).

Step 5: Find the eigenvalues

Solve the quadratic equation \( \lambda^2 - 2\lambda + 3 = 0 \) using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), which gives \( \lambda = \frac{2 \pm \sqrt{-8}}{2} = 1 \pm i \sqrt{2} \).

Step 6: Identify the type of eigenvalues

The eigenvalues \( 1 \pm i \sqrt{2} \) have positive real parts and non-zero imaginary parts, indicating that they are complex conjugates with positive real parts.

Step 7: Classify the equilibrium point

Eigenvalues with positive real parts indicate that the system has an unstable spiral at the equilibrium point (0,0).

Key Concepts

Eigenvalues

Eigenvalues are fundamental concepts in the study of differential equations and matrices. They help us understand the behavior of systems described by these equations. In the context of a 2x2 matrix like in our problem, eigenvalues act as roots of the characteristic polynomial obtained from the matrix.

• To find eigenvalues, we solve the characteristic equation which is derived from the determinant of the matrix but with an identity matrix, multiplied by a scalar, subtracted from it: \( \det(A - \lambda I) = 0 \).
• In our example, the matrix \( A = \left[ \begin{array}{rr} 2 & -1 \ 3 & 0 \end{array} \right] \) results in a characteristic equation: \( \lambda^2 - 2\lambda + 3 = 0 \).
• Solving this quadratic equation using the quadratic formula, \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we find two complex eigenvalues: \( 1 \pm i \sqrt{2} \).

These complex eigenvalues indicate oscillatory behavior due to their imaginary parts, and their stability depends on the real parts, as discussed in the equilibrium stability section.

Equilibrium Stability

Equilibrium stability is crucial when analyzing the behavior of systems governed by differential equations. It tells us whether small perturbations or deviations from an equilibrium point lead to system stability or instability.

• For linear systems like the one in the exercise, stability is determined by the real parts of the eigenvalues.
• If the real parts are positive, as with eigenvalues \( 1 \pm i \sqrt{2} \) in our example, the equilibrium point \((0,0)\) is unstable. This scenario is known as an unstable spiral.
• Conversely, if the real parts were negative, the equilibrium would be stable, attracting trajectories towards it. If the real parts were zero, we would have a center, indicating stability but with no tendency to spiral inwards or outwards.

The stability classification, which is heavily influenced by the sign of the real parts of the eigenvalues, helps predict long-term behavior of dynamical systems near equilibrium.

Characteristic Equation

The characteristic equation plays a vital role in determining the eigenvalues of a matrix. Constructing this equation involves forming the determinant of the matrix \( A \) minus \( \lambda \) times the identity matrix \( I \).

• For the matrix in our example, \( A = \left[ \begin{array}{rr} 2 & -1 \ 3 & 0 \end{array} \right] \), the characteristic equation is \( \det(A - \lambda I) = 0 \).
• The calculation begins with creating the matrix \( A - \lambda I = \left[ \begin{array}{rr} 2 - \lambda & -1 \ 3 & -\lambda \end{array} \right] \).
• Next, solve the determinant of this matrix, yielding the polynomial \( \lambda^2 - 2\lambda + 3 = 0 \).

This quadratic equation encapsulates the dynamics and characteristics of the system, offering insight into the nature of equilibrium through its roots, the eigenvalues. Understanding and solving the characteristic equation is key to exploring how different perturbations can affect system stability.