Problem 14 - 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. $$\begin{array}{l}{J=\left[ \begin{array}{cc}{-1+2 x_{1}} & {0} \\ {0} & {-\frac{1}{3}+2 x_{2}}\end{array}\right]} \\ {\text { (i) } \hat{x}_{1}=-1, \hat{x}_{2}=0} \\ {\text { (ii) } \hat{x}_{1}=2, \hat{x}_{2}=\frac{1}{3}}\end{array}$$

Step-by-Step Solution

Answer

Equilibrium (i) is locally stable; equilibrium (ii) is unstable.

Step 1: Identify the Jacobian Matrix at Each Equilibrium

The given Jacobian matrix is \[ J = \begin{bmatrix} -1 + 2x_1 & 0 \ 0 & -\frac{1}{3} + 2x_2 \end{bmatrix} \]. We will evaluate this matrix at each equilibrium point \((\hat{x}_1, \hat{x}_2)\).

Step 2: Evaluate the Jacobian at Equilibrium (i)

For equilibrium point (i) \( (\hat{x}_1 = -1, \hat{x}_2 = 0) \), substitute into the Jacobian: \[ J_i = \begin{bmatrix} -1 + 2(-1) & 0 \ 0 & -\frac{1}{3} + 2(0) \end{bmatrix} = \begin{bmatrix} -3 & 0 \ 0 & -\frac{1}{3} \end{bmatrix} \].

Step 3: Determine Stability at Equilibrium (i)

The eigenvalues of \( J_i \) are on the main diagonal: \(-3\) and \(-\frac{1}{3}\). Since both eigenvalues are negative, equilibrium (i) is locally stable.

Step 4: Evaluate the Jacobian at Equilibrium (ii)

For equilibrium point (ii) \( (\hat{x}_1 = 2, \hat{x}_2 = \frac{1}{3}) \), substitute into the Jacobian: \[ J_{ii} = \begin{bmatrix} -1 + 2(2) & 0 \ 0 & -\frac{1}{3} + 2\left(\frac{1}{3}\right) \end{bmatrix} = \begin{bmatrix} 3 & 0 \ 0 & 0 \end{bmatrix} \].

Step 5: Determine Stability at Equilibrium (ii)

The eigenvalues of \( J_{ii} \) are \(3\) and \(0\). Since one eigenvalue is positive, equilibrium (ii) is unstable.

Key Concepts

Equilibrium Points

Equilibrium points in systems of equations are specific points where the system doesn't change. Think of these points like places of rest, where movement stops. When dealing with differential equations, the system is in balance at these points.
To find equilibrium points, you need the system's derivatives to equal zero at these specific spots. This is like finding the zero-energy state of the system.

• At equilibrium, the system stays put if undisturbed.
• In mathematics and physical systems, equilibrium points are crucial in understanding dynamics.

For the given exercise, we had two equilibrium points: (i) \((\hat{x}_1 = -1, \hat{x}_2 = 0)\) and (ii) \((\hat{x}_1 = 2, \hat{x}_2 = \frac{1}{3})\). These points were where we evaluated the Jacobian matrix, which is a tool to study how the system behaves nearby.

Local Stability

Local stability is about the behavior of the system in the immediate vicinity of an equilibrium point. Imagine a marble sitting perfectly on a flat surface. If you slightly nudge it, will it return to its position or roll away? That's the essence of local stability.

• If the equilibrium is locally stable, small disturbances will die out over time.
• If the equilibrium is unstable, even small disturbances grow, taking the system away.

In our exercise, the stability was determined based on the signs of the eigenvalues of the Jacobian matrix. At equilibrium (i), with eigenvalues \(-3\) and \(-\frac{1}{3}\), the negative signs indicated stability. On the other hand, equilibrium (ii) had eigenvalues \(3\) and \(0\), meaning instability due to the presence of a positive eigenvalue.

Eigenvalues

Eigenvalues are numbers that give us insight into a matrix's properties. They're key to understanding how things evolve over time in dynamic systems. In the context of stability analysis, eigenvalues of a Jacobian matrix at an equilibrium point inform us about its local stability.
To find eigenvalues, solve the characteristic equation \(\det(J - \lambda I) = 0\), where \(\lambda\)s are eigenvalues. This analysis is like checking the system's response to minor disturbances.

• Negative eigenvalues imply that the disturbances shrink, indicating local stability.
• Positive eigenvalues suggest disturbances grow, leading to instability.
• An eigenvalue of zero typically indicates a need for deeper analysis.

In the exercise, the eigenvalues were found directly from the main diagonal of the Jacobian matrices at given equilibrium points. This simplified the process as the matrices were diagonal, making the eigenvalues straightforward to identify.