Problem 54 - 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. \(A=\left[\begin{array}{rr}0 & -3 \\ 2 & 2\end{array}\right]\)

Step-by-Step Solution

Answer

The equilibrium \((0,0)\) is an unstable spiral because the real parts of the eigenvalues are positive.

Step 1: Write the System of Equations

We start by writing the given differential equation in its natural form. For a matrix \( A = \begin{bmatrix} 0 & -3 \ 2 & 2 \end{bmatrix} \), the system of differential equations is: \( \frac{d \mathbf{x}}{d t} = A \mathbf{x}(t) \). Expressing \( \mathbf{x}(t) \) as \( \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix} \), we have two equations: \( \begin{align*} \frac{dx_1}{dt} &= -3x_2 \ \frac{dx_2}{dt} &= 2x_1 + 2x_2 \end{align*} \).

Step 2: Find the Eigenvalues of A

To classify the equilibrium, we need to find the eigenvalues of matrix \( A \). The eigenvalues are found by solving \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix. Therefore, we have: \( \det \begin{bmatrix} -\lambda & -3 \ 2 & 2 - \lambda \end{bmatrix} = (-\lambda)(2 - \lambda) - (-3)(2) = 0 \). Simplifying gives the characteristic equation: \( \lambda^2 - 2\lambda + 6 = 0 \).

Step 3: Solve the Characteristic Equation

Solve the quadratic equation \( \lambda^2 - 2\lambda + 6 = 0 \) using the quadratic formula: \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = -2 \), and \( c = 6 \). This gives \( \lambda = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(6)}}{2(1)} = \frac{2 \pm \sqrt{4 - 24}}{2} = \frac{2 \pm \sqrt{-20}}{2} \). So, \( \lambda = 1 \pm i\sqrt{5} \).

Step 4: Analyze the Stability of the Equilibrium

The eigenvalues \( 1 \pm i\sqrt{5} \) are complex with a positive real part \( 1 \). This indicates that the system exhibits exponential growth with rotational behavior, which means that the equilibrium point \( (0,0) \) is an unstable spiral.

Key Concepts

Eigenvalues

When analyzing linear differential equations, eigenvalues play a crucial role in understanding how the system behaves over time. In our problem, we are given a matrix \( A = \begin{bmatrix} 0 & -3 \ 2 & 2 \end{bmatrix} \). To find the eigenvalues, we solve the characteristic equation \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix.

The characteristic equation is derived by calculating the determinant of \( A - \lambda I \), which yields \( \lambda^2 - 2\lambda + 6 = 0 \). Solving this quadratic equation using the quadratic formula gives the eigenvalues \( \lambda = 1 \pm i\sqrt{5} \). These eigenvalues are complex, indicating certain rotational behaviors in the system's response.

Understanding these eigenvalues helps us interpret the system's future behavior, especially around points of equilibrium. In this case, the real part of the eigenvalues is positive, leading us to crucial insights in the subsequent stability analysis.

Stability Analysis

Stability analysis is vital for understanding how solutions to differential equations behave as time progresses. When examining our system, the eigenvalues \( 1 \pm i\sqrt{5} \) offer key information about its stability.

• The real part of the eigenvalues is positive, which signifies that the solutions grow exponentially over time.
• The imaginary part implies the presence of oscillations or rotational motion around the equilibrium point.

Together, these facts tell us that our system does not return to equilibrium after small disturbances. Instead, due to the positive real part, any small deviation will grow, spiraling outward as time advances. This growth is a hallmark of instability in the system. Therefore, the equilibrium point \((0, 0)\) is unstable, specifically forming what is known as an unstable spiral in the context of linear differential equations.

Equilibrium Classification

For differential equations, classifying the type of equilibrium is essential for predicting how systems react under perturbations. In our analysis, the classification hinges on the nature of the eigenvalues derived earlier.

An equilibrium point such as \((0,0)\) is examined based on the real and imaginary parts of its eigenvalues:

• If the real part is positive, as seen here with \( \lambda = 1 \pm i\sqrt{5} \), it means any small displacement from the equilibrium point will result in growth away from that equilibrium.
• The presence of an imaginary part signifies spiraling motion, causing trajectories in the phase space to rotate along their path.

Given these conditions, we conclude that \((0, 0)\) is classified as an unstable spiral. Such a classification is crucial because it gives us insight into both quantitative and qualitative behaviors of solutions, underlining the fact that minor changes can lead to significant divergence over time.