Problem 24 - Calculus for Biology and Medicine

Question

24\. Two Hot and Cold Lovers Imagine that Romeo and Juliet both behave in the same way Romeo does in Example \(4 ;\) that is, their affections are modeled by a system of equations $$ \begin{array}{l} \frac{d J}{d t}=-b R \\ \frac{d R}{d t}=-b J \end{array} $$ where \(b\) is a positive constant. (a) By writing the system (11.55) as a matrix equation classify the equilibrium \((0,0)\); i.e., is it a stable node, spiral, saddle point, and so on? (b) Find the eigenvector directions for the equilibrium. (c) Based on your answers to (a) and (b), what is the fate of Romeo and Juliet's relationship as \(t \rightarrow \infty\) ?

Step-by-Step Solution

Answer

The equilibrium is a saddle point; the relationship is unstable and diverging as time goes to infinity.

Step 1: Write the system as a matrix equation

Rewrite the given equations \(\frac{dJ}{dt}=-bR\) and \(\frac{dR}{dt}=-bJ\) in matrix form. This gives us a matrix \(A\) such that \(\frac{d}{dt} \begin{bmatrix} J \ R \end{bmatrix} = A\begin{bmatrix} J \ R \end{bmatrix}\). Thus, \(A = \begin{bmatrix} 0 & -b \ -b & 0 \end{bmatrix}\).

Step 2: Determine the eigenvalues

For a matrix \(A\), the eigenvalues \(\lambda\) are found by solving the characteristic equation \(\det(A - \lambda I) = 0\). For \(A = \begin{bmatrix} 0 & -b \ -b & 0 \end{bmatrix}\), \(\det\begin{pmatrix} -\lambda & -b \ -b & -\lambda \end{pmatrix} = \lambda^2 - b^2 = 0\). Thus, \(\lambda = \pm b\).

Step 3: Classify the equilibrium point

The eigenvalues \(\lambda = \pm b\) are real and distinct, with opposite signs. This implies that the equilibrium point \((0,0)\) is a saddle point.

Step 4: Find the eigenvectors

For \(\lambda = b\), solve \(\begin{bmatrix} 0 & -b \ -b & 0 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = b \begin{bmatrix} x \ y \end{bmatrix}\) which gives \(x = -y\). The eigenvector is \(\begin{bmatrix} 1 \ -1 \end{bmatrix}\). For \(\lambda = -b\), solve \(\begin{bmatrix} 0 & -b \ -b & 0 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = -b \begin{bmatrix} x \ y \end{bmatrix}\) which gives \(x = y\). The eigenvector is \(\begin{bmatrix} 1 \ 1 \end{bmatrix}\).

Step 5: Interpret the fate of the relationship

Since the equilibrium point is a saddle point and the eigenvector directions are \(\begin{bmatrix} 1 \ -1 \end{bmatrix}\) and \(\begin{bmatrix} 1 \ 1 \end{bmatrix}\), the system is unstable. As time \(t\to \infty\), Romeo and Juliet's affection will not converge to a stable state. Their relationship is likely oscillatory and diverging.

Key Concepts

Eigenvalues

Understanding eigenvalues is fundamental when analyzing systems of linear differential equations. In this context, eigenvalues help determine the behavior of the solutions near an equilibrium point. An equilibrium point in a system of equations is a point where the system remains unchanged over time without external influences. To find the eigenvalues of a matrix, we solve the characteristic equation.
This involves finding the roots of the determinant equation \( \det(A - \lambda I) = 0 \). In our exercise, \( A = \begin{bmatrix} 0 & -b \ -b & 0 \end{bmatrix} \) yields the characteristic equation \( \lambda^2 - b^2 = 0 \). Solving this gives eigenvalues \( \lambda = \pm b \).
The presence of real and distinct eigenvalues \( \lambda = b \) and \( \lambda = -b \), one positive and one negative, indicates a saddle point, suggesting instability in the equilibrium.

Eigenvectors

Once eigenvalues are calculated, finding eigenvectors is the next step. Eigenvectors are non-zero vectors that change by only a scalar factor when the linear transformation described by the matrix is applied. In simpler terms, they point in a direction that is stretched or compressed by the transformation.
For each eigenvalue \( \lambda \), we solve the equation \((A - \lambda I)v = 0\) to find the corresponding eigenvector \( v \). For our matrix, with eigenvalues \( \lambda = b \) and \( \lambda = -b \), the respective eigenvectors are determined as \( \begin{bmatrix} 1 \ -1 \end{bmatrix} \) and \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \).
These vectors indicate the directions along which the system's behavior diverges or converges, further supporting the classification of the equilibrium as a saddle point.

Saddle Point

A saddle point occurs at the equilibrium of a system of differential equations when the eigenvalues are real and have opposite signs. This indicates that in some directions, solutions may diverge, while in others, they converge. The behavior is akin to a saddle on a horse, which is stable in one direction but unstable in the perpendicular direction.
For our scenario, the matrix \( A \) yielded eigenvalues \( \lambda = \pm b \). The positive and negative eigenvalues mean the trajectories of the system will move away from the equilibrium point. Hence, the equilibrium point \((0,0)\) is classified as a saddle point, confirming instability in the system.
This instability is significant because it implies sensitivity to initial conditions; small perturbations can result in large deviations in the trajectory over time.

Stability Analysis

Stability analysis involves assessing whether small perturbations in a system will die out or grow over time. It's a crucial concept for understanding the long-term behavior of systems modeled by differential equations. Stability near an equilibrium point is often determined using eigenvalues of the system's matrix.
For the system in question, we consider the eigenvalues \( \lambda = \pm b \). Since they have opposite signs, the system is unstable, indicating the equilibrium is not a point of attraction but rather a point from which trajectories diverge.

• A positive eigenvalue suggests the direction of instability, demonstrating growth away from equilibrium.
• A negative eigenvalue might typically suggest stability, but the presence of both indicates a saddle point.

This instability implies that the affection dynamics between Romeo and Juliet are unlikely to reach a steady state, thus their relationship is predicted to oscillate and ultimately diverge over time.