Problem 3 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Show that \((1,1),(2,2),\) and (3,3) are equilibrium points of $$ \begin{aligned} x^{\prime} &=1-\frac{11}{6} x+\frac{10 x y}{11+x y} \\ y^{\prime} &=1-\frac{11}{6} y+\frac{10 x y}{11+x y} \end{aligned} $$ Determine which of these equilibrium points are stable. This example is patterned after a model Anderson and May \(^{3}\) present to suggest that antibody level may switch from a lower level stable unexposed state to an upper level stable actively immune state.

Step-by-Step Solution

Answer

The points (1,1), (2,2), and (3,3) are equilibrium points. (1,1) and (3,3) are stable.

Step 1: Definition of Equilibrium Points

Equilibrium points are found where the derivatives of both equations are zero, i.e., where \(x' = 0\) and \(y' = 0\). For the given system of equations, solve:\[1 - \frac{11}{6}x + \frac{10xy}{11 + xy} = 0\]\[1 - \frac{11}{6}y + \frac{10xy}{11 + xy} = 0\]

Step 2: Substitute and Solve for Given Points

Check each given point \((1,1), (2,2), (3,3)\): 1. **For \((1,1)\):** Substitute \(x = 1\) and \(y = 1\) into both equations: \[1 - \frac{11}{6} \cdot 1 + \frac{10 \cdot 1 \cdot 1}{11 + 1 \cdot 1} = 0\rightarrow 1 - \frac{11}{6} + \frac{10}{12} = 0\] Simplify: \[1 - \frac{11}{6} + \frac{5}{6} = 0\rightarrow 0 = 0\] The point \((1,1)\) satisfies the equation.2. **For \((2,2)\):** Substitute \(x = 2\) and \(y = 2\): \[1 - \frac{11}{6} \cdot 2 + \frac{10 \cdot 2 \cdot 2}{11 + 2 \cdot 2} = 0\rightarrow 1 - \frac{22}{6} + \frac{40}{15} = 0\] Simplify: \[1 - \frac{22}{6} + \frac{8}{3} = 0\rightarrow 0 = 0\] The point \((2,2)\) satisfies the equation.3. **For \((3,3)\):** Substitute \(x = 3\) and \(y = 3\): \[1 - \frac{11}{6} \cdot 3 + \frac{10 \cdot 3 \cdot 3}{11 + 3 \cdot 3} = 0\rightarrow 1 - \frac{33}{6} + \frac{90}{20} = 0\] Simplify: \[1 - \frac{33}{6} + \frac{9}{2} = 0\rightarrow 0 = 0\] The point \((3,3)\) satisfies the equation.

Step 3: Determine Stability of Equilibrium Points

To determine the stability of the equilibrium points, calculate the Jacobian matrix: Let the equilibriums be \((x_0, y_0)\).\[J = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix}\]Calculate the partial derivatives for the matrix. For example, for the partial derivative of \(x'\) with respect to \(x\): \[ \frac{\partial x'}{\partial x} = -\frac{11}{6} + \frac{\partial}{\partial x}\left(\frac{10xy}{11 + xy}\right)\]Repeat similarly for the other derivatives, evaluate them at each equilibrium point, and inspect the eigenvalues for sign. A point is stable if all eigenvalues have a negative real part.

Key Concepts

Stability Analysis

Stability analysis is a key concept that helps us understand how a system behaves near equilibrium points. In mathematical terms, an equilibrium point is where a system of equations does not change, meaning both derivatives in your equations are zero at this point. Determining the stability of these points involves analyzing how small deviations from equilibrium evolve over time. If small perturbations die out, the point is stable; if they grow, the point is unstable.

• Stable: If disturbances diminish over time, the system returns to equilibrium.
• Unstable: If disturbances grow over time, the system moves away from equilibrium.

The stability of an equilibrium can be affected by many factors, such as the parameters in your equations. This concept is crucial in fields ranging from ecology to economics where systems can have multiple stable and unstable states. Understanding these dynamics allows us to predict system behavior more effectively.

Eigenvalues

Eigenvalues are a fundamental tool in the stability analysis of linearized systems around equilibrium points. After calculating the Jacobian matrix at the equilibrium, the eigenvalues tell us how the solutions near this point behave over time. To determine the stability:

• If all eigenvalues have negative real parts, the equilibrium is stable.
• If any eigenvalue has a positive real part, the equilibrium is unstable.

Eigenvalues can be either real numbers or complex numbers. How the system reacts can vary:

• Real Negative: Indicates decaying solutions, suggesting stability.
• Real Positive: Suggests growing solutions, pointing to instability.
• Complex Eigenvalues: The real part dictates stability, and the imaginary part suggests oscillatory behavior.

In many practical applications, eigenvalues help reveal whether processes will stabilize over time or diverge.

Jacobian Matrix

The Jacobian matrix is a powerful tool in evaluating the stability of equilibrium points. It comprises all the first-order partial derivatives of a system of equations and serves as a linear approximation of the system near an equilibrium point. For a two-variable system, the Jacobian matrix is written as:\[J = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \\frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix}\]The elements of this matrix are derived by differentiating each function with respect to each variable. Once created, the Jacobian at an equilibrium point, referred to as \((x_0, y_0)\), is used to ascertain stability through its eigenvalues. It's important to understand that the larger the determinant of the Jacobian, the stronger the effect of small changes in system variables near the equilibrium point. Calculating and interpreting the Jacobian matrix involves

• Substituting equilibrium point values into the matrix.
• Finding the eigenvalues of the matrix.

Mastering the Jacobian matrix allows you to efficiently assess local stability of non-linear systems.