Problem 15 - Calculus for Biology and Medicine

Question

Assume that $$ \begin{array}{l} \frac{d x_{1}}{d t}=x_{1}\left(10-2 x_{1}-x_{2}\right) \\ \frac{d x_{2}}{d t}=x_{2}\left(10-x_{1}-2 x_{2}\right) \end{array} $$ (a) Graph the zero isoclines. (b) Show that \(\left(\frac{10}{3}, \frac{10}{3}\right)\) is an equilibrium, and use the analytical approach to determine its stability.

Step-by-Step Solution

Answer

Zero isoclines are linear and cross at equilibrium \( \left( \frac{10}{3}, \frac{10}{3} \right) \), which is stable.

Step 1: Understanding Zero Isoclines

Zero isoclines are the curves where the derivatives are zero; they help us understand the behavior of a system. For \( \frac{d x_1}{d t} = 0 \), we set \( x_1(10 - 2x_1 - x_2) = 0 \), leading to either \( x_1 = 0 \) or \( 10 - 2x_1 - x_2 = 0 \). For \( \frac{d x_2}{d t} = 0 \), we set \( x_2(10 - x_1 - 2x_2) = 0 \), leading to either \( x_2 = 0 \) or \( 10 - x_1 - 2x_2 = 0 \).

Step 2: Plotting the Zero Isoclines

We graph the zero isoclines derived: For the first equation, we have the lines \( x_1 = 0 \) and \( x_2 = 10 - 2x_1 \). For the second, \( x_2 = 0 \) and \( x_1 = 10 - 2x_2 \). Plotting these lines on a plane provides the analysis framework.

Step 3: Finding Equilibrium Points

To find equilibria, set both derivatives to zero simultaneously. Solve \( 10 - 2x_1 - x_2 = 0 \) and \( 10 - x_1 - 2x_2 = 0 \) together. Adding these equations gives \( 20 = 3x_1 + 3x_2 \), leading directly to \( x_1 = \frac{10}{3} \) and \( x_2 = \frac{10}{3} \).

Step 4: Stability Analysis

Evaluate the Jacobian matrix of the system at the equilibrium point \( \left( \frac{10}{3}, \frac{10}{3} \right) \). The partial derivatives are \( A = 10 - 4x_1 - x_2 \) and \( B = -x_1 \). At equilibrium, calculate: \[ A = 10 - 4 \left( \frac{10}{3} \right) - \frac{10}{3}, B = - \frac{10}{3}, C = - \frac{10}{3}, D = 10 - \frac{10}{3} - 4 \left( \frac{10}{3} \right) \].

Step 5: Determining Stability

Calculate the eigenvalues of the Jacobian matrix using \( |J - \lambda I| = 0 \). Determine the sign of eigenvalues: if real parts are negative, the stationary point is stable. Calculations show real negative eigenvalues, indicating local stability.

Key Concepts

Zero Isoclines in Differential Equations

Zero isoclines are a fundamental concept in differential equations used to identify where a dynamic system experiences no change in certain state variables. This means that at these points, the derivative, or rate of change, of the variable equals zero. They serve as helpful tools in visualizing the behavior and possible equilibrium configurations of systems.

Consider the given set of differential equations:

• \( \frac{dx_1}{dt} = x_1(10 - 2x_1 - x_2) \)
• \( \frac{dx_2}{dt} = x_2(10 - x_1 - 2x_2) \)

To find the zero isoclines, we set the expressions to zero. This gives us conditions: either the variable itself is zero, or the remainder of the expression equals zero. For our equations, these lead to:

• For \( x_1 \), the zero isoclines are defined by the lines \( x_1 = 0 \) or \( x_2 = 10 - 2x_1 \).
• For \( x_2 \), the zero isoclines are defined by the lines \( x_2 = 0 \) or \( x_1 = 10 - 2x_2 \).

Plotting these lines on a coordinate plane helps us gain insight into the dynamics of the system, offering a framework to understand potential steady states or changes in behavior.

Equilibrium Points in Systems of Equations

Equilibrium points are crucial for understanding and analyzing dynamical systems. At these points, the system doesn't change over time, meaning each variable's derivative is zero. These points represent the steady-state solutions for the system.

In the system given by the differential equations, to find the equilibrium points, we solve the equations when both derivatives are zero:

• \( 10 - 2x_1 - x_2 = 0 \)
• \( 10 - x_1 - 2x_2 = 0 \)

By solving these equations simultaneously, we discover that one equilibrium point is at \( (\frac{10}{3}, \frac{10}{3}) \). This means, at this specific point, the system reaches a steady state where no further changes occur to either variable. Equilibrium points like these are essential to evaluate the overall behavior of a system, as they highlight the conditions under which the system remains unchanged.

Stability Analysis of Equilibrium Points

Stability analysis of equilibrium points helps to predict how a system will react to small disturbances. In other words, it determines whether a system, when slightly perturbed from an equilibrium point, will return to the equilibrium or deviate further away.

To perform the stability analysis, we use the Jacobian matrix, which comprises the first-order partial derivatives of the system of equations around the equilibrium point. For point \( (\frac{10}{3}, \frac{10}{3}) \), the Jacobian matrix elements are determined:

• \( A = 10 - 4x_1 - x_2 \)
• \( B = -x_1 \)
• \( C = -x_2 \)
• \( D = 10 - x_1 - 4x_2 \)

By evaluating these derivatives at the equilibrium point and calculating the eigenvalues of the resultant matrix, we find the nature of the equilibrium. If the real parts of the eigenvalues are negative, the equilibrium point is stable, meaning it will return to stability if disturbed. In this case, the calculated negative eigenvalues indicate that \( (\frac{10}{3}, \frac{10}{3}) \) is a locally stable point in this system. Understanding stability is essential for predicting the long-term behavior of dynamical systems.