Problem 63 - Calculus for Biology and Medicine

Question

We consider differential equations of the form \(\frac{d x}{d t}=A x(t)\) where \(A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]\) Analyze the stability of the equilibrium \(\mathbf{( 0 , 0 ) , \text { and classify the }}\) equilibrium. \(A=\left[\begin{array}{rr}-2 & 3 \\ 1 & -4\end{array}\right]\)

Step-by-Step Solution

Answer

The equilibrium \((0, 0)\) is a stable node.

Step 1: Define the System

We begin with the system of differential equations given by \(\frac{dx}{dt} = Ax(t)\), where \(A\) is the coefficient matrix \(\begin{pmatrix}-2 & 3 \ 1 & -4\end{pmatrix}\). The equilibrium point for this system is \((0, 0)\).

Step 2: Find the Eigenvalues of the Matrix A

The stability of the equilibrium at the origin can be determined by finding the eigenvalues of \(A\). We find the eigenvalues by solving the characteristic equation \(\text{det}(A - \lambda I) = 0\), where \(I\) is the identity matrix. The characteristic equation for \(A\) is \((-2 - \lambda)(-4 - \lambda) - (3 \times 1) = 0\).

Step 3: Solve the Characteristic Equation

Expand the determinant to obtain \(\lambda^2 + 6\lambda + 5 = 0\). This is a quadratic equation, which we solve using the quadratic formula \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a=1\), \(b=6\), and \(c=5\).

Step 4: Calculate the Eigenvalues

Using the quadratic formula, we find: \(\lambda = \frac{-6 \pm \sqrt{6^2 - 4 \times 1 \times 5}}{2}\), leading to \(\lambda = \frac{-6 \pm \sqrt{16}}{2}\). Simplifying, we have \(\lambda_1 = -1\) and \(\lambda_2 = -5\).

Step 5: Analyze Stability

Since both eigenvalues \(\lambda_1 = -1\) and \(\lambda_2 = -5\) have negative real parts, the equilibrium point \((0, 0)\) is stable. The fact that both eigenvalues are real and negative indicates that the equilibrium is a stable node.

Key Concepts

Stability Analysis

When examining differential equations, one crucial aspect is understanding how the system behaves over time. This is determined through stability analysis. In our example, the differential equation is expressed as \( \frac{dx}{dt} = Ax(t) \), where \( A \) is a given matrix.In simple terms, stability analysis tells us if small deviations from an equilibrium point will die out or grow. If they die out, the system returns to an equilibrium, making it stable. Conversely, if deviations grow, the equilibrium is unstable. Stability depends heavily on the eigenvalues of the matrix \( A \), which we will explore next.

By finding out where the equilibrium point lies and understanding the nature of the solutions around it, we can classify the system dynamics. In our exercise, the equilibrium is at \( (0, 0) \). This analysis helps determine whether the system will naturally calm down to a rest state or diverge.

Eigenvalues

The eigenvalues of matrix \( A \) play a critical role in assessing the system's stability. To find these eigenvalues, you first set up the characteristic equation \( \text{det}(A - \lambda I) = 0 \), where \( I \) is the identity matrix corresponding to \( A \).Next, you solve this equation to find \( \lambda \), which represents the eigenvalues. Here's a helpful reminder:- When the real parts of all eigenvalues are negative, the equilibrium is stable.- If any eigenvalue has a positive real part, the equilibrium is unstable.- Complex eigenvalues with negative real parts also indicate stability but typically lead to oscillatory behavior.

In our scenario, solving the characteristic equation for \( A \) results in two eigenvalues: \( \lambda_1 = -1 \) and \( \lambda_2 = -5 \). Since both have negative real parts, the equilibrium point \((0, 0)\) is stable. These real negative eigenvalues suggest that any disturbance from the equilibrium will decrease exponentially over time, leading the system back to the equilibrium position.

Equilibrium Points

Equilibrium points, sometimes called fixed points, are solutions to the differential equation where the system does not change over time. For our particular example, the equilibrium point \((0, 0)\) is where the functions \( x(t) \) do not vary when \( t \) changes.To figure out these points, set the \( \frac{dx}{dt} \) equation to zero and solve for \( x(t) \). The matrix \( A \) influences the character and stability of these points. Understanding - Where these points lie,- How system trajectories behave around them,- And if they are stable all come together in stability analysis.

The stability of equilibrium points offers insights into how solutions evolve. In simple terms, if you nudge the system slightly, does it come back to balance, or does it spiral out of control? In our problem, since both eigenvalues are negative, any small deviation will diminish over time, classifying the equilibrium as a stable node.