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 equilibrium according to whether it is a sink, a source, or a saddle point. $$ A=\left[\begin{array}{rr} 0 & -2 \\ -1 & 3 \end{array}\right] $$

Step-by-Step Solution

Answer

The equilibrium \((0,0)\) is a source and unstable.

Step 1: Identify the System and Equilibrium

The system in question is a linear system of equations given by \( \frac{d \mathbf{x}}{d t} = A \mathbf{x}(t) \), where \( A = \left[\begin{array}{rr} 0 & -2 \ -1 & 3 \end{array}\right] \). The equilibrium point under consideration is \((0,0)\).

Step 2: Calculate the Eigenvalues

To analyze the stability, we first need to find the eigenvalues of matrix \( A \). The eigenvalues are found by solving the characteristic equation: \( \text{det}(A - \lambda I) = 0 \), where \( A - \lambda I = \left[\begin{array}{rr} 0-\lambda & -2 \ -1 & 3-\lambda \end{array}\right] \).

Step 3: Solve the Characteristic Equation

Calculate the determinant for \( A - \lambda I \):\[ \text{det}(A - \lambda I) = (0-\lambda)(3-\lambda) - (-2)(-1) = \lambda^2 - 3\lambda + 2 = 0 \]Solve this quadratic equation for eigenvalues:\( \lambda = \frac{3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} \), which simplifies to \( \lambda = \frac{3 \pm 1}{2} \). Thus, \( \lambda_1 = 2 \) and \( \lambda_2 = 1 \).

Step 4: Analyze Stability from Eigenvalues

The eigenvalues \( \lambda_1 = 2 \) and \( \lambda_2 = 1 \) are both positive and real. In a dynamical system, positive real eigenvalues indicate that the trajectories are moving away from the equilibrium point \((0,0)\). Thus, \((0,0)\) is an unstable point.

Step 5: Classify the Equilibrium

Since both eigenvalues are positive, which means that trajectories diverge from the equilibrium, we classify this type of equilibrium as a **source**. In dynamical systems, a source indicates that trajectories are repelled from the neighborhood of the equilibrium point.

Key Concepts

Stability Analysis of Differential Equations

When analyzing a differential equation system like \( \frac{d \mathbf{x}}{dt} = A \mathbf{x} \), determining the stability of equilibrium points is crucial. Stability analysis helps us understand how solutions behave over time, specifically near equilibrium points. In this context, the equilibrium point is \((0,0)\). This analysis involves examining if solutions are drawn towards this point (stable) or repelled (unstable). The method requires understanding the behavior of eigenvalues derived from the matrix \(A\).The stability is assessed by the signs of the eigenvalues:

• Stable Equilibrium: If all eigenvalues have negative real parts. Trajectories converge to the equilibrium.
• Unstable Equilibrium: If any eigenvalue has a positive real part. Trajectories move away from the equilibrium.
• Semi-Stable: If there are zero real parts, indicating more analysis is needed.

In our case, both eigenvalues are positive, which means the system is unstable, and the trajectories are repelled from the origin.

Understanding Eigenvalues

Eigenvalues are just numbers that give us important information about a matrix in a differential equations system. They are found by solving the characteristic equation of the matrix \(A\). This equation is derived from subtracting \(\lambda I\) (where \(I\) is the identity matrix) from \(A\) and setting its determinant to zero.For matrix \(A\), the characteristic equation was \( \lambda^2 - 3\lambda + 2 = 0 \). Solving it provided two eigenvalues: \( \lambda_1 = 2 \) and \( \lambda_2 = 1 \).Why are eigenvalues so important?

• Indicate Stability: They help in determining if an equilibrium is stable or unstable.
• Inform Dynamics: Positive real parts suggest that solutions expand or move away from equilibrium, while negative ones suggest convergence.
• Classification: The combination of eigenvalues also helps categorize the type of equilibrium being observed.

In our exercise, the positive eigenvalues tell us that the (0,0) equilibrium point is unstable and behaves like a source.

Equilibrium Classification: Sink, Source, and Saddle Point

Classifying equilibrium points in a dynamical system is instrumental in predicting long-term behavior.The terms "sink," "source," and "saddle" describe the nature of equilibrium points based on eigenvalues' properties.

• Sink: All eigenvalues have negative real parts. Trajectories move towards the equilibrium point. It is stable.
• Source: All eigenvalues have positive real parts. Trajectories diverge from the equilibrium point. It is unstable. In our example, since both \( \lambda_1 = 2 \) and \( \lambda_2 = 1 \) are positive, the system is classified as a source.
• Saddle Point: Mixed-sign eigenvalues. Trajectories can both approach and recede from the equilibrium. It is unstable.

Determining these classifications aids significantly in understanding system behavior near equilibrium points. In the exercise provided, the equilibrium at \((0,0)\) is a source, indicating that the system's trajectories will move away from this point over time, emphasizing its unstable nature.