Problem 34 - 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.$$ A=\left[\begin{array}{rr} -2 & 4 \\ 2 & -5 \end{array}\right] $$

Step-by-Step Solution

Answer

The equilibrium point \((0,0)\) is a sink and is stable, as both eigenvalues are negative.

Step 1: Determine the Eigenvalues

To analyze stability, we first find the eigenvalues of the matrix \( A \). The eigenvalues can be determined by solving the characteristic equation \( \det(A - \lambda I) = 0 \). For this particular matrix \( A = \begin{bmatrix} -2 & 4 \ 2 & -5 \end{bmatrix} \), calculate the determinant: \[ \det \left( \begin{bmatrix} -2 & 4 \ 2 & -5 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \right) = \det \begin{bmatrix} -2 - \lambda & 4 \ 2 & -5 - \lambda \end{bmatrix} \] \[ = (-2 - \lambda)(-5 - \lambda) - (4 \times 2) \]\[ = (\lambda + 2)(\lambda + 5) - 8 \] \[ = \lambda^2 + 7\lambda + 10 - 8 \]\[ = \lambda^2 + 7\lambda + 2 = 0 \].

Step 2: Solve the Characteristic Equation

The characteristic polynomial from Step 1 is \( \lambda^2 + 7\lambda + 2 = 0 \). Use the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve for \( \lambda \), where \( a = 1, b = 7, c = 2 \). \[ \lambda = \frac{-7 \pm \sqrt{7^2 - 4 \times 1 \times 2}}{2 \times 1} \] \[ = \frac{-7 \pm \sqrt{49 - 8}}{2} \] \[ = \frac{-7 \pm \sqrt{41}}{2} \].The solutions are two distinct real eigenvalues: \( \lambda_1 = \frac{-7 + \sqrt{41}}{2} \) and \( \lambda_2 = \frac{-7 - \sqrt{41}}{2} \).

Step 3: Classify the Equilibrium Point

The eigenvalues \( \lambda_1 \) and \( \lambda_2 \) are both real and distinct. We can classify the equilibrium point by observing the signs of these eigenvalues:- If both eigenvalues are negative, then \((0,0)\) is a sink (stable).- If both eigenvalues are positive, then \((0,0)\) is a source (unstable).- If they have opposite signs, then \((0,0)\) is a saddle point (unstable).Compute the values:- \( \lambda_1 = \frac{-7 + \sqrt{41}}{2} \approx -0.7 \),- \( \lambda_2 = \frac{-7 - \sqrt{41}}{2} \approx -6.3 \).Both eigenvalues are negative, indicating that \((0,0)\) is a 2sink" and thus stable.

Key Concepts

Eigenvalues

Eigenvalues are fundamental in analyzing systems, especially when dealing with matrices in differential equations. They provide crucial insights into the behavior of a system over time. To find the eigenvalues of a matrix like \( A \), you solve the characteristic equation \( \det(A - \lambda I) = 0 \). Here, \( \lambda \) represents the eigenvalues, and \( I \) is the identity matrix. In the given exercise, the characteristic polynomial derived is \( \lambda^2 + 7\lambda + 2 = 0 \). By solving this quadratic equation, the eigenvalues \( \lambda_1 \) and \( \lambda_2 \) are found. These values help determine how the system evolves. Distinct real eigenvalues indicate a clear behavior in the system's dynamics.

Matrix Stability

Matrix stability refers to whether the solutions to a differential system tend toward a point (stable) or away from it (unstable) as time progresses. The stability of an equilibrium point, such as \((0,0)\) in this system, is primarily determined by the eigenvalues of the system matrix.

• If all eigenvalues have negative real parts, the system is stable, meaning solutions converge to the equilibrium.
• If any eigenvalue has a positive real part, the system is unstable, and solutions move away from the equilibrium.
• If eigenvalues are zero or purely imaginary, further analysis is needed, as conclusions about stability are not straightforward.

In this exercise, both eigenvalues are negative \( (\lambda_1 \approx -0.7 \) and \( \lambda_2 \approx -6.3) \), confirming that the system is stable and the solutions will eventually converge to the equilibrium point \((0,0)\).

Equilibrium Classification

In the context of differential equations, equilibrium classification is about determining the nature of equilibrium points based on the eigenvalues derived from the system’s matrix. The classification helps understand how the system behaves when slightly disturbed from equilibrium.

• A **sink** is where all eigenvalues are negative; the equilibrium is attracting, and solutions near this point will return to it, indicating stability.
• A **source** has positive eigenvalues; the equilibrium is repelling, leading to instability as solutions move away.
• A **saddle point** involves eigenvalues with different signs; this results in a directionally unstable equilibrium, with solutions approaching from specific directions but departing from others.

In the problem, both eigenvalues are negative, making \((0,0)\) a sink, showing that solutions gravitate towards this equilibrium point, confirming system stability.