Problem 18 - Biocalculus Calculus for the Life Sciences

Question

A Jacobian matrix and two equlibria are given. Determine if each is locally stable, unstable, or if the analysis is inconclusive. $$J=\left[ \begin{array}{cc}{-\frac{1}{1+x_{2}}} & {\frac{x_{1}}{\left(1+x_{2}\right)^{2}}} \\\ {-1+\frac{x_{2}}{\left(1+x_{1}\right)^{2}}} & {-\frac{1}{1+x_{1}}}\end{array}\right]$$ $$\begin{array}{l}{\text { (i) } \hat{x}_{1}=-2, \hat{x}_{2}=-2} \\ {\text { (ii) } \hat{x}_{1}=\frac{1}{2}, \hat{x}_{2}=-\frac{3}{4}}\end{array}$$

Step-by-Step Solution

Answer

Equilibrium (i) is inconclusive; Equilibrium (ii) is locally unstable.

Step 1: Substitute Equilibrium Points into Jacobian

For each equilibrium point, substitute the given \( x_1 \) and \( x_2 \) into the Jacobian matrix. For (i) \((\hat{x}_1, \hat{x}_2) = (-2, -2)\): \[J = \begin{bmatrix}-\frac{1}{1 + (-2)} & \frac{-2}{(1 + (-2))^2} \-1 + \frac{-2}{(1 + (-2))^2} & -\frac{1}{1 + (-2)}\end{bmatrix} = \begin{bmatrix}1 & -2 \ 4 & 1\end{bmatrix}\]For (ii) \((\hat{x}_1, \hat{x}_2) = \left(\frac{1}{2}, -\frac{3}{4} \right)\): \[J = \begin{bmatrix}-\frac{1}{1 + (-3/4)} & \frac{1/2}{(1 + (-3/4))^2} \-1 + \frac{-3/4}{(1 + 1/2)^2} & -\frac{1}{1 + 1/2}\end{bmatrix} = \begin{bmatrix}-4 & 8 \-1.5 & -\frac{2}{3}\end{bmatrix}\]

Step 2: Calculate Eigenvalues for Equilibrium (i)

Determine the eigenvalues of the matrix \(\begin{bmatrix}1 & -2 \ 4 & 1\end{bmatrix}\). Calculate the determinant of \( (J - \lambda I) \): \[det(J - \lambda I) = \begin{vmatrix}1-\lambda & -2 \ 4 & 1-\lambda\end{vmatrix} = (1-\lambda)^2 - (-2)(4) = \lambda^2 - 2\lambda + 9\]Solve \( \lambda^2 - 2\lambda + 9 = 0 \). Use the quadratic formula:\[\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{4 - 36}}{2} = 1 \pm i\sqrt{5}\]

Step 3: Determine Stability for Equilibrium (i)

Since the eigenvalues \(1 \pm i\sqrt{5}\) have non-zero imaginary parts, the equilibrium at \((\hat{x}_1, \hat{x}_2) = (-2, -2)\) is a center or a focus, and thus the analysis is inconclusive.

Step 4: Calculate Eigenvalues for Equilibrium (ii)

Determine the eigenvalues of the matrix \(\begin{bmatrix}-4 & 8 \-1.5 & -\frac{2}{3}\end{bmatrix}\). Calculate the determinant of \( (J - \lambda I) \): \[det(J - \lambda I) = \begin{vmatrix}-4-\lambda & 8 \ -1.5 & -\frac{2}{3}-\lambda\end{vmatrix} = (-4-\lambda)(-\frac{2}{3}-\lambda) - (8)(-1.5)\]\[= (\lambda + 4)(\lambda + \frac{2}{3}) - 12 = \lambda^2 + \frac{2}{3}\lambda + 4\lambda + \frac{8}{3} - 12\]\[= \lambda^2 + \frac{14}{3}\lambda - \frac{28}{3}\]Solve \( \lambda^2 + \frac{14}{3}\lambda - \frac{28}{3} = 0 \) using the quadratic formula:\[\lambda = \frac{-\frac{14}{3} \pm \sqrt{(\frac{14}{3})^2 + \frac{112}{3}}}{2} = \frac{-\frac{14}{3} \pm \sqrt{\frac{784}{9} + \frac{1008}{9}}}{2} = \frac{-\frac{14}{3} \pm \frac{188}{9}}{2}\]

Step 5: Determine Stability for Equilibrium (ii)

The calculations yield real and distinct eigenvalues. As one of them is negative, the equilibrium point is a saddle point, making it locally unstable.

Key Concepts

Local Stability Analysis

Analyzing local stability helps us understand the behavior of a system near its equilibrium points. This involves using the Jacobian matrix, which represents the system's linearized dynamics near these points. By examining how small disturbances evolve in the vicinity of an equilibrium, we can tell if the system will return to equilibrium or deviate further away.
In this context, local stability analysis revolves around two key factors: the Jacobian matrix of the system and its eigenvalues. For a given set of equilibrium points, we substitute these values into the Jacobian to form a specific matrix.
Then, we calculate the eigenvalues, which are critical in determining the nature of the equilibrium. If all eigenvalues have negative real parts, the system is locally stable at that point as any small disturbance will decay. Conversely, if any eigenvalue has a positive real part, the system will diverge, indicating instability. When the eigenvalues are purely imaginary, the analysis might be inconclusive, signaling a need for further investigation.

Eigenvalues

Eigenvalues are central to local stability analysis. They help determine how a system behaves near an equilibrium point. When we compute eigenvalues from the Jacobian matrix, we are essentially looking for the roots of the characteristic equation.

The eigenvalues can tell us:

• Stability: If every eigenvalue has a negative real part, the equilibrium is stable.
• Instability: If any eigenvalue has a positive real part, the equilibrium is unstable.
• Inconclusive: Purely imaginary eigenvalues suggest more complex dynamics like periodic solutions.

This is why calculating eigenvalues doesn't just provide numerical insights but also offers qualitative behavior of dynamical systems around equilibrium points.
For example, complex eigenvalues with non-zero imaginary parts, like in equilibrium (i), hint at oscillatory behavior, making the analysis inconclusive. Whereas for equilibrium (ii), the presence of real eigenvalues with different signs confirms local instability.

Equilibrium Points

Equilibrium points are where the system stays constant, meaning there is no net change in its state. These points are found by solving the system's differential equations when set to zero.
Once determined, their nature is revealed through local stability analysis. The equilibrium can be stable, unstable, or the analysis can be inconclusive, depending on the eigenvalues derived from the Jacobian matrix at these points.

• Stable points mean the system will return to equilibrium after a disturbance.
• Unstable points mean the system will move away upon disturbances.
• Inconclusive points require further analysis to understand the system's behavior.

For our exercise, at equilibrium (i) \((-2, -2)\), the analysis doesn’t fully determine the status because of complex eigenvalues. However, for equilibrium (ii) \(\left(\frac{1}{2}, -\frac{3}{4}\right)\), the system is certainly unstable due to the presence of real eigenvalues with differing signs, indicating a saddle point reached by trajectories only via specific paths.