Problem 16 - Calculus for Biology and Medicine

Question

Assume that $$ \begin{array}{l} \frac{d x_{1}}{d t}=x_{1}\left(10-x_{1}-2 x_{2}\right) \\ \frac{d x_{2}}{d t}=x_{2}\left(10-2 x_{1}-x_{2}\right) \end{array} $$ (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

The zero isoclines are graphed based on the conditions \( x_1=0 \), \( x_2=0 \), \( 10-x_1-2x_2=0 \), and \( 10-2x_1-x_2=0 \). The point \( \left(\frac{10}{3}, \frac{10}{3}\right) \) is an equilibrium and is stable since the eigenvalues of the evaluated Jacobian are negative.

Step 1: Identify Zero Isoclines for the System

Zero isoclines occur when the time derivative is zero for each equation. For \( \frac{d x_1}{d t} = 0 \), this gives \( x_{1} \left( 10 - x_{1} - 2 x_{2} \right) = 0 \). Therefore, the zero isoclines for \( \frac{d x_1}{d t} = 0 \) are \( x_1 = 0 \) and \( 10 - x_1 - 2 x_2 = 0 \). For \( \frac{d x_2}{d t} = 0 \), the zero isoclines are \( x_2 = 0 \) and \( 10 - 2 x_1 - x_2 = 0 \).

Step 2: Graph the Zero Isoclines

Plot the zero isoclines on a graph. For the line \( 10 - x_1 - 2x_2 = 0 \), rearranging gives \( x_1 = 10 - 2x_2 \). This line intersects the \( x_1 \)-axis at 10 and has a slope of -2. For \( 10 - 2x_1 - x_2 = 0 \), rearranging gives \( x_2 = 10 - 2x_1 \), with an intersection at point \( (5, 0) \) and slope of -0.5. Also include the lines \( x_1 = 0 \) and \( x_2 = 0 \).

Step 3: Find Equilibrium Points

Equilibrium points satisfy both \( \frac{d x_1}{d t} = 0 \) and \( \frac{d x_2}{d t} = 0 \). Solving the two equations \( 10 - x_1 - 2x_2 = 0 \) and \( 10 - 2x_1 - x_2 = 0 \) simultaneously gives \( x_1 = \frac{10}{3} \) and \( x_2 = \frac{10}{3} \). This confirms \( \left( \frac{10}{3}, \frac{10}{3} \right) \) as an equilibrium point.

Step 4: Derive the Jacobian Matrix

The Jacobian matrix is used to study stability. The system of equations is given by \( f(x_1, x_2) = x_1(10 - x_1 - 2x_2) \) and \( g(x_1, x_2) = x_2(10 - 2x_1 - x_2) \). The Jacobian matrix \( J \) is \[ J = \begin{bmatrix} \frac{\partial f}{\partial x_1} & \frac{\partial f}{\partial x_2} \ \frac{\partial g}{\partial x_1} & \frac{\partial g}{\partial x_2} \end{bmatrix} = \begin{bmatrix} 10 - 2x_1 - 2x_2 & -2x_1 \ -2x_2 & 10 - 2x_1 - 2x_2 \end{bmatrix}. \]

Step 5: Evaluate Jacobian at Equilibrium

Substitute \( x_1 = \frac{10}{3} \) and \( x_2 = \frac{10}{3} \) into the Jacobian matrix. This results in the evaluated Jacobian: \[ J = \begin{bmatrix} -\frac{10}{3} & -\frac{20}{3} \ -\frac{20}{3} & -\frac{10}{3} \end{bmatrix}. \]

Step 6: Determine Stability via Eigenvalues

Compute the eigenvalues of the Jacobian matrix by solving the characteristic equation \( \det(J - \lambda I) = 0 \). With \( I \) as the identity matrix, the equation is \[ \left(-\frac{10}{3} - \lambda\right)^2 - \left(-\frac{20}{3}\right)^2 = 0. \] Solving this gives eigenvalues \( \lambda_1 = -10 \) and \( \lambda_2 = -\frac{10}{3} \), which are both negative, indicating that the equilibrium point is stable.

Key Concepts

Equilibrium Points

In a system of differential equations, equilibrium points—also known as steady states—are crucial because they represent conditions where the system doesn't change over time. Specifically, these are the points where the derivatives are zero.

To find equilibrium points, we set the right-hand side of the differential equation to zero and solve the resulting algebraic system. Using the exercise's equations:

• For \( \frac{d x_1}{d t} \), solve \( x_1(10 - x_1 - 2x_2) = 0 \).
• For \( \frac{d x_2}{d t} \), solve \( x_2(10 - 2x_1 - x_2) = 0 \).

By solving these, we find that the equilibrium occurs at \( \left( \frac{10}{3}, \frac{10}{3} \right) \). At this point, neither \( x_1 \) nor \( x_2 \) changes, meaning it's a steady state configuration of the system.

Jacobian Matrix

The Jacobian Matrix is a powerful tool in analyzing the stability of equilibrium points in a system of non-linear equations. It contains partial derivatives that show how each variable in the system influences each other.

For a system with variables \( x_1 \) and \( x_2 \), the Jacobian matrix is expressed as:\[ J = \begin{bmatrix} \frac{\partial f}{\partial x_1} & \frac{\partial f}{\partial x_2} \ \frac{\partial g}{\partial x_1} & \frac{\partial g}{\partial x_2} \end{bmatrix} \]Where \( f \) and \( g \) are your system's equations.

• The top left, \( \frac{\partial f}{\partial x_1} \), tells how \( x_1 \) affects its rate of change in the first equation.
• \( \frac{\partial f}{\partial x_2} \) shows how \( x_2 \) impacts the first equation.
• Similarly, the terms in the second row reflect influences in the second equation.

In our exercise, by plugging in the equilibrium points into the Jacobian matrix, we assess the system's behavior near these steady states. This aids in determining if perturbations or small changes increase (instability) or decrease (stability) over time.

Zero Isoclines

Zero isoclines are lines in the phase plane where the rate of change of one of the variables is zero. These lines help divide the phase plane into regions where the directions of the trajectories can vary significantly.

For the equations given, we find zero isoclines by setting the derivatives to zero:

• For \( \frac{d x_1}{d t} = 0 \): Consider \( x_1(10 - x_1 - 2x_2) = 0 \). Here, zero isoclines are at \( x_1=0 \) and \( 10 - x_1 - 2x_2 = 0 \).
• For \( \frac{d x_2}{d t} = 0 \): We use \( x_2(10 - 2x_1 - x_2) = 0 \). This results in zero isoclines at \( x_2=0 \) and \( 10 - 2x_1 - x_2 = 0 \).

Graphically, plotting zero isoclines provides a roadmap for understanding how the trajectories evolve in different regions. At intersections, we might find equilibrium points, offering further insights into the system's behavior.