Problem 40 - 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. \(A=\left[\begin{array}{ll}1 & -2 \\ 1 & -3\end{array}\right]\)

Step-by-Step Solution

Answer

The equilibrium \((0,0)\) is a saddle point, indicating instability.

Step 1: Analyze the given matrix

The matrix \( A \) given in the problem is: \[ A = \begin{bmatrix} 1 & -2 \ 1 & -3 \end{bmatrix} \] Our task is to find the equilibrium behavior of the differential equation \( \frac{d \mathbf{x}}{dt} = A \mathbf{x}(t) \).

Step 2: Calculate the eigenvalues of A

To determine the eigenvalues, solve the characteristic equation \( \det(A - \lambda I) = 0 \). Compute: \[ \det \begin{bmatrix} 1-\lambda & -2 \ 1 & -3-\lambda \end{bmatrix} = (1-\lambda)(-3-\lambda) - (-2)(1) = 0 \] Expanding this gives: \[ (1-\lambda)(-3-\lambda) + 2 = \lambda^2 + 2\lambda - 3 + 2 = \lambda^2 + 2\lambda - 1 = 0 \].

Step 3: Solve the characteristic equation

Solve the quadratic equation \( \lambda^2 + 2\lambda - 1 = 0 \) using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = 2 \), and \( c = -1 \): \[ \lambda = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} \] Simplifying, the eigenvalues are \( \lambda_1 = -1 + \sqrt{2} \) and \( \lambda_2 = -1 - \sqrt{2} \).

Step 4: Analyze the stability and classify the equilibrium point

Since \( \lambda_1 = -1 + \sqrt{2} \) and \( \lambda_2 = -1 - \sqrt{2} \), the eigenvalues are real, distinct, and nonzero. Evaluating further, \( -1 + \sqrt{2} \) has a positive value and \( -1 - \sqrt{2} \) has a negative value. When one eigenvalue is positive and the other negative, the equilibrium point \((0,0)\) is a saddle point, indicating instability.

Key Concepts

Eigenvalues

To fully understand the behavior of a system described by a matrix differential equation, finding its eigenvalues is vital. It's a bit like uncovering the hidden characteristics of the matrix. In this context, the eigenvalues help us understand how systems evolve over time.

The eigenvalues are determined by setting up and solving the characteristic equation: \ \( \det(A - \lambda I) = 0 \). This equation calculates where the matrix, when shifted by an amount \( \lambda \) times the identity matrix \( I \, \, \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \), becomes singular, meaning it loses its invertibility. Solving this for our specific matrix A reveals two eigenvalues, \( \lambda_1 = -1 + \sqrt{2} \) and \( \lambda_2 = -1 - \sqrt{2} \).

These eigenvalues are crucial because they tell us about the system's potential trajectories and asymptotic behavior. Since they are both real and distinct, we know that the related differential equation solutions will not oscillate, as they might with complex eigenvalues.

Equilibrium Analysis

Equilibrium analysis is a core concept when examining differential equations, making it essential to understand. An equilibrium point is where the system doesn't change over time, where the derivative \( \frac{d \mathbf{x}}{dt} \) equals zero.

For the system to be at equilibrium, the equation \( A \mathbf{x}(t) = 0 \) must hold true since \( \frac{d \mathbf{x}}{dt} = A \mathbf{x}(t) \). In simpler terms, when the product of the matrix A and the state vector \( \mathbf{x}(t) \) renders zero, the system is said to be in equilibrium. The point \( (0,0) \) is often a natural equilibrium point for linear systems as shown in this problem.

Understanding equilibrium points is critical as they represent the steady states. In real-life systems, this could translate to scenarios like a pendulum at rest or water levels in interconnected tanks. The manner in which trajectories near these points behave informs us of the system stability.

Stability and Classification

Determining the stability of equilibrium points in a system is key to understanding how the system behaves under small disturbances. This exercise involves classifying the equilibrium point based on its eigenvalues.

In our case, the eigenvalues are \( -1 + \sqrt{2} \) and \( -1 - \sqrt{2} \). Notably, one eigenvalue is positive while the other is negative:

• \( \lambda_1 = -1 + \sqrt{2} > 0 \)
• \( \lambda_2 = -1 - \sqrt{2} < 0 \)

This tells us that the equilibrium point \( (0,0) \) is a saddle point. In the world of differential equations, a saddle point indicates instability. Systems at these points can respond dramatically to small perturbations, diverging away rather than returning to equilibrium.

Knowing whether a point is a sink, source, or saddle can help predict the behavior of complex systems immensely. A sink draws trajectories in, showing a stable system. A source pushes them out, suggesting instability. Here, the saddle tells us the system might be wildly unstable in some directions yet stable in others.