Problem 13 - Calculus for Biology and Medicine

Question

Assume that \(a>0\). Find all point equilibria of $$ \begin{array}{l} \frac{d x_{1}}{d t}=x_{2}\left(x_{1}-a\right) \\ \frac{d x_{2}}{d t}=x_{2}^{2}-x_{1} \end{array} $$ and characterize their stability.

Step-by-Step Solution

Answer

The equilibria are (0, 0) (neutrally stable), (a, \sqrt{a}) (unstable), and (a, -\sqrt{a}) (stable).

Step 1: Determine the Equilibrium Points

To find the equilibrium points, set the derivatives equal to zero: \[\frac{dx_1}{dt} = x_2(x_1 - a) = 0 \] and \[\frac{dx_2}{dt} = x_2^2 - x_1 = 0.\] From the first equation, we have either \(x_2 = 0\) or \(x_1 = a\). Substitute these into the second equation to find corresponding \(x_1\) values.

Step 2: Calculate Equilibrium for x_2 = 0

If \(x_2 = 0\), then the second equation becomes: \[0^2 - x_1 = 0 \] Solving for \(x_1\), we get \(x_1 = 0\). Thus, one equilibrium point is \((x_1, x_2) = (0, 0)\).

Step 3: Calculate Equilibrium for x_1 = a

If \(x_1 = a\), substitute into the second equation: \[x_2^2 - a = 0 \] Thus, \(x_2^2 = a\) and \(x_2 = \pm\sqrt{a}\). Therefore, the equilibrium points here are \((a, \sqrt{a})\) and \((a, -\sqrt{a})\).

Step 4: Stability Analysis at (0,0)

The linearization of the system around \((0, 0)\) involves computing the Jacobian matrix: \[J = \begin{bmatrix}\frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} \\frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} \end{bmatrix} = \begin{bmatrix}0 & x_1 - a \-1 & 2x_2 \end{bmatrix}\] At \((0, 0)\), the Jacobian \(J = \begin{bmatrix}0 & -a \-1 & 0\end{bmatrix}\). The eigenvalues satisfy \(\lambda^2 + a = 0\), giving \(\lambda = \pm i\sqrt{a}\). The eigenvalues are purely imaginary, indicating a center and thus neutrally stable.

Step 5: Stability Analysis at (a, \sqrt{a})

For \((a, \sqrt{a})\), substitute into the Jacobian: \[J = \begin{bmatrix}0 & 0 \-1 & 2\sqrt{a} \end{bmatrix}\] The eigenvalues satisfy \(\lambda (\lambda - 2\sqrt{a}) = 0\), giving \(\lambda_1 = 0\) and \(\lambda_2 = 2\sqrt{a}\). As \(\lambda_2 > 0\), \((a, \sqrt{a})\) is an unstable node.

Step 6: Stability Analysis at (a, -\sqrt{a})

For \((a, -\sqrt{a})\), the Jacobian is: \[J = \begin{bmatrix}0 & 0 \-1 & -2\sqrt{a} \end{bmatrix}\] The characteristic equation is \(\lambda (\lambda + 2\sqrt{a}) = 0\), giving \(\lambda_1 = 0\) and \(\lambda_2 = -2\sqrt{a}\). As \(\lambda_2 < 0\), \((a, -\sqrt{a})\) is a stable node.

Key Concepts

Stability Analysis

A stability analysis helps determine the behavior of a system near its equilibrium points. In dynamical systems, each equilibrium point can either attract nearby trajectories (stable), repel trajectories (unstable), or exhibit neutral behavior. This characterization requires examining small perturbations around the equilibrium points.

For this particular system, stability around an equilibrium point is assessed using linearization, which involves examining the behavior of the system's linear approximation. The key tool for this purpose is the Jacobian matrix, which we will discuss further. However, understanding stability requires looking at the eigenvalues of this Jacobian matrix at each equilibrium.

In the current exercise, we find three equilibrium points:

• (0, 0)
• (a, \(\sqrt{a}\))
• (a, \, \(-\sqrt{a}\))

Through the stability analysis:- A point is neutrally stable if the real parts of all eigenvalues are zero.- An equilibrium is unstable if at least one eigenvalue has a positive real part. - It is stable if all eigenvalues have strictly negative real parts.

This approach lets us conclude the nature of each equilibrium point:

• (0, 0) is neutrally stable.
• (a, \(\sqrt{a}\)) is unstable.
• (a, \, \(-\sqrt{a}\)) is stable.

Jacobian Matrix

The Jacobian matrix is central to linearizing a dynamic system around equilibrium points. It provides information about how the system behaves when it's slightly perturbed from an equilibrium. Essentially, it contains all the first-order partial derivatives of the system's functions.

For a dynamical system described by equations, the Jacobian matrix \( J \) is structured with elements derived from these equations:

• The partial derivatives of the first function with respect to each variable form the first row.
• The partial derivatives of the second function with respect to each variable form the second row, and so on for systems with more variables.

In our example, the dynamical system:\[\frac{d x_{1}}{d t}=x_{2}(x_{1}-a) \\frac{d x_{2}}{d t}=x_{2}^{2}-x_{1}\]has a Jacobian matrix given by:\[J = \begin{bmatrix}\frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} \\frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2}\end{bmatrix} = \begin{bmatrix}x_2 & x_1 - a \-1 & 2x_2\end{bmatrix}\]Evaluating this Jacobian at specific equilibrium points provides values that we'll use in subsequent stability analyses through eigenvalues.

Eigenvalues

Eigenvalues are vital in stability analysis as they predict how a system responds when disturbed away from equilibrium. They are extracted from the Jacobian matrix associated with the linearized system at any equilibrium point.

To find eigenvalues, you need to solve the characteristic equation derived from the Jacobian:\[\text{det}(J - \lambda I) = 0\]where \( \lambda \) denotes the eigenvalue and \( I \) is the identity matrix.

In practice:

• An eigenvalue with a positive real part indicates instability (the system moves away from equilibrium).
• A negative eigenvalue suggests stability (the system returns to equilibrium).
• A purely imaginary eigenvalue points to neutral stability or oscillatory behavior.

For example, analyzing the Jacobian matrix at

• (0, 0) results in purely imaginary eigenvalues \( \lambda = \pm i\sqrt{a} \), indicating neutral stability.
• At (a, \( \sqrt{a} \)), one eigenvalue is zero and another positive \(2\sqrt{a}\), leading to instability.
• At (a, \(-\sqrt{a}\)), with eigenvalues one zero, another negative \(-2\sqrt{a}\), establishes stability.

Understanding eigenvalues helps predict the long-term behavior of systems and their reaction to disturbances, which is critical in fields like engineering and physics.