Problem 39 - 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. \(A=\left[\begin{array}{ll}0 & -2 \\ 1 & -3\end{array}\right]\)

Step-by-Step Solution

Answer

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

Step 1: Find the Characteristic Equation

First, find the characteristic equation for the matrix \(A\). The characteristic equation is given by \(\text{det}(A - \lambda I) = 0\), where \(I\) is the identity matrix. Substitute \(A = \begin{bmatrix} 0 & -2 \ 1 & -3 \end{bmatrix}\) , and \(\lambda I = \begin{bmatrix} \lambda & 0 \ 0 & \lambda \end{bmatrix}\). Compute \(\det(A - \lambda I) = \begin{vmatrix} -\lambda & -2 \ 1 & -3-\lambda \end{vmatrix} = 0\).

Step 2: Solve the Determinant

Calculate the determinant: \(\det(A - \lambda I) = (-\lambda)(-3-\lambda) - (-2)(1) = \lambda(3+\lambda) + 2 = \lambda^2 + 3\lambda + 2 \)Set the determinant equal to zero: \(\lambda^2 + 3\lambda + 2 = 0 \)

Step 3: Calculate the Eigenvalues

Solve the quadratic equation \(\lambda^2 + 3\lambda + 2 = 0\) by factoring: \((\lambda + 1)(\lambda + 2) = 0\)This gives the eigenvalues \(\lambda_1 = -1\) and \(\lambda_2 = -2\).

Step 4: Analyze Stability Using Eigenvalues

The equilibrium point \((0,0)\) is stable if all eigenvalues are negative. Here, both eigenvalues \(-1\) and \(-2\) are negative.

Step 5: Classify the Equilibrium

Since both eigenvalues are negative, \((0,0)\) is a sink. This means trajectories are drawn into \((0,0)\) as time progresses, indicating stability.

Key Concepts

Eigenvalues

In the context of linear differential equations, eigenvalues are fundamental because they determine the behavior of solutions over time. When dealing with the system of differential equations given by \( \frac{d \mathbf{x}}{dt} = A \mathbf{x} \), where \( A \) is a matrix, the eigenvalues give insight into the "directions" along which the system evolves.
To find the eigenvalues of a matrix \( A \), we solve the characteristic equation, \( \text{det}(A - \lambda I) = 0 \). Here, \( I \) is the identity matrix, and \( \lambda \) represents potential eigenvalues.
In the exercise, after substituting the given matrix \( A \) and computing the determinant, the characteristic equation becomes \( \lambda^2 + 3\lambda + 2 = 0 \). Solving this quadratic equation through factoring, we find the eigenvalues \( \lambda_1 = -1 \) and \( \lambda_2 = -2 \).
Eigenvalues like these, which are both real and negative, imply specific dynamic behaviors, guiding us towards understanding system stability.

Stability Analysis

Stability analysis is a crucial process to determine how solutions to a differential equation behave as time progresses. For an equilibrium point, such as \((0,0)\), the nature of stability is primarily determined by the eigenvalues of the matrix \( A \).
If an equilibrium point is stable, small perturbations will die out, and the system will return to that point over time. To determine stability, we consider the signs of eigenvalues:

• If all eigenvalues have negative real parts, the equilibrium is stable (often called asymptotically stable). Any small departure from the equilibrium decays over time.
• If any eigenvalue has a positive real part, the equilibrium is unstable, meaning perturbations grow, and the system diverges from the equilibrium point.

In the provided solution, both eigenvalues are \( -1 \) and \( -2 \), which are negative. This indicates that the equilibrium \((0,0)\) is stable, as any deviation from this point diminishes over time, causing the trajectories of the differential equation to be pulled towards the equilibrium.

Equilibrium Classification

Once we've established stability, classifying the type of equilibrium provides further insight into the system's dynamics. The classification depends on the characteristics of the eigenvalues:

• A sink occurs when all eigenvalues are real and negative. This is the case in our example, where trajectories are attracted towards the equilibrium, making the point stable.
• A source is present if all eigenvalues are real and positive, leading to trajectories moving away from the equilibrium point, signaling instability.
• A saddle point arises if there are eigenvalues with mixed signs or complex numbers, suggesting that the system will move away from the equilibrium along certain directions while possibly approaching it in others.

In this exercise, both eigenvalues are negative, classifying the equilibrium point \((0,0)\) as a sink. Therefore, we can say confidently that over time, the behaviors described by the differential equation will result in states converging towards this equilibrium, indicating a predictable and stable system.