Problem 48 - 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. \(A=\left[\begin{array}{ll}2 & -4 \\ 2 & -3\end{array}\right]\)

Step-by-Step Solution

Answer

The equilibrium \((0,0)\) is a stable spiral, as it has complex eigenvalues with negative real parts.

Step 1: Calculate the Characteristic Polynomial

Given the matrix\[A = \begin{pmatrix} 2 & -4 \ 2 & -3 \end{pmatrix}\]We need to find the characteristic polynomial. The characteristic polynomial is \(\det(A - \lambda I)\), where \(\lambda\) is a scalar and \(I\) is the identity matrix. Calculate:\[A - \lambda I = \begin{pmatrix} 2 - \lambda & -4 \ 2 & -3-\lambda \end{pmatrix}\]Compute the determinant:\[\begin{vmatrix} 2 - \lambda & -4 \ 2 & -3-\lambda \end{vmatrix} = (2-\lambda)(-3-\lambda) - (2)(-4)\]Simplifying, we get:\[\lambda^2 + \lambda + 14\] as the characteristic polynomial.

Step 2: Determine the Eigenvalues

Solve the characteristic equation \(\lambda^2 + \lambda + 14 = 0\). To find the eigenvalues, apply the quadratic formula:\[\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \(a = 1\), \(b = 1\), \(c = 14\). Thus,\[\lambda = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 14}}{2}\]\[\lambda = \frac{-1 \pm \sqrt{-55}}{2}\]Calculating, we find:\[\lambda = \frac{-1 \pm i\sqrt{55}}{2}\]The eigenvalues are complex conjugates with a real part of \(-\frac{1}{2}\).

Step 3: Determine Stability from Eigenvalues

Analyzing the eigenvalues \(\lambda = \frac{-1 \pm i\sqrt{55}}{2}\), notice the real part is \(-\frac{1}{2}\), which is negative. When the real part of complex eigenvalues is negative, the equilibrium point \((0,0)\) is stable.Therefore, since the eigenvalues are complex with negative real parts, the equilibrium point is a stable spiral.

Step 4: Classify the Equilibrium

Based on the eigenvalue analysis, where the eigenvalues have a negative real part, the equilibrium at \((0,0)\) is classified as a **stable spiral**. This means that trajectories spiral into the origin, becoming more tightly wound as time progresses.

Key Concepts

Eigenvalues

In the study of differential equations, eigenvalues play a crucial role in determining the dynamics of a system. They are key to understanding the behavior of solutions to the differential equation system, especially around equilibrium points. In this exercise, we have a matrix \[ A = \begin{pmatrix} 2 & -4 \ 2 & -3 \end{pmatrix} \]When determining eigenvalues, the characteristic equation derived from this matrix helps us identify these crucial numbers. In our case, the eigenvalues turned out to be complex conjugates: \[ \lambda = \frac{-1 \pm i\sqrt{55}}{2} \]### Quick Checklist:- **Identify the characteristic equation**: Start with \(\det(A - \lambda I)\).- **Solve for \(\lambda\)**: Use the quadratic formula when needed.- **Interpret the eigenvalues**: Here, they are complex numbers with negative real parts, indicating specific stability characteristics.

Stability Analysis

Stability analysis in differential equations is a method used to predict how perturbations or changes affect the system around an equilibrium point. Understanding the stability of a system guides us to predict the system's behavior over time.

For systems with complex conjugate eigenvalues like \[ \lambda = \frac{-1 \pm i\sqrt{55}}{2} \], attention is paid to the real part of the eigenvalues, which is crucial in determining stability:- **Real part**: If it’s negative, as in this example, the equilibrium is stable, leading us to a stable spiral classification.- **Behavioral insight**: When the real part is negative, any initial movement towards or away from the equilibrium will slow down and reverse direction over time, spiraling in a stable manner.

Equilibrium Points

Equilibrium points are solutions to differential equations where the system doesn't change over time. For the equation provided, \[ \frac{d \mathbf{x}}{dt} = A \mathbf{x}(t) \], these points are crucial, as they can determine the long-term behavior of the system. In our problem, \( (0,0) \) is the equilibrium point for matrix \[ A = \begin{pmatrix} 2 & -4 \ 2 & -3 \end{pmatrix} \].**Why it's important:**- **Rest state**: The system remains unchanged if started exactly at an equilibrium.- **Close encounters**: An analysis of what happens when the system starts near these points reveals stability or instability. In our case, the equilibrium point is a stable spiral because of the nature of our eigenvalues.- **Predictive power**: Knowing the equilibrium helps in predicting future behavior of the system.

Characteristic Polynomial

To find eigenvalues, one must first establish the characteristic polynomial from the matrix representing the system. This polynomial, usually expressed from \( \det(A - \lambda I) \), contains the eigenvalues as its roots.

In the given example, the characteristic polynomial was calculated as \( \lambda^2 + \lambda + 14 = 0 \). This equation gives insight into the general form of the solution without solving the system itself. **Key Steps to Remember:**- **Setup**: Use \( A - \lambda I \) to derive the matrix that feeds into the determinant.- **Compute determinant**: This often leads to a quadratic expression. - **Understand roots**: The roots of this polynomial are the eigenvalues that dictate system behavior.