Problem 24 - 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) Find all equilibria and classify them, by linearizing the system near each equilibrium. (c) Draw the directions of the vector field on the zero isoclines, and in the regions between the zero isoclines.

Step-by-Step Solution

Answer

Equilibria are: \((0,0)\), \((0,10)\), \((5,0)\), \((4,2)\). Classify using eigenvalues at each equilibrium.

Step 1: Find the Zero Isoclines

**Zero Isoclines** are the lines where the derivative of a variable is zero. For \( \frac{dx_1}{dt} = 0 \), solve \( x_1(10 - x_1 - 2x_2) = 0 \). This gives the zero isoclines: \( x_1 = 0 \) or \( 10 = x_1 + 2x_2 \). For \( \frac{dx_2}{dt} = 0 \), solve \( x_2(10 - 2x_1 - x_2) = 0 \). This gives the zero isoclines: \( x_2 = 0 \) or \( 10 = 2x_1 + x_2 \).

Step 2: Graph the Zero Isoclines

Plot the equations from Step 1 on the \((x_1, x_2)\) plane. The line \(x_1 = 0\) is the y-axis, the line \(x_2 = 0\) is the x-axis, \(10 = x_1 + 2x_2\) is a line with intercepts \((10,0)\) on the \(x_1\)-axis and \((0,5)\) on the \(x_2\)-axis, and \(10 = 2x_1 + x_2\) is a line with intercepts \((5,0)\) on the \(x_1\)-axis and \((0,10)\) on the \(x_2\)-axis.

Step 3: Find Equilibria

Equilibria occur where both derivatives are simultaneously zero. Solve the system obtained from setting both expressions to zero:1. \( x_1(10 - x_1 - 2x_2) = 0 \)2. \( x_2(10 - 2x_1 - x_2) = 0 \)Substitute \( x_1 = 0 \), then from the second equation, either \( x_2 = 0 \) or \( 10 = x_2 \). Therefore, equilibria at \((0,0)\) and \((0,10)\). Similarly, substitute \( x_2 = 0 \) and solve. You find an equilibrium at \((5,0)\).Last, solve \( 10 = x_1 + 2x_2 \) and \( 10 = 2x_1 + x_2 \) simultaneously, leading to equilibrium at \((4,2)\).

Step 4: Linearize Near Equilibria

To classify equilibria, linearize the system using the Jacobian matrix \( J \), which involves taking partial derivatives of the system with respect to \(x_1\) and \(x_2\). The Jacobian is:\[ J = \begin{bmatrix}10 - 2x_1 - 2x_2 & -2x_1 \-2x_2 & 10 - 2x_1 - x_2 \\end{bmatrix} \]Evaluate \( J \) at each equilibrium:- At \((0,0)\), \((0,10)\), \((5,0)\), and \((4,2)\). Determine eigenvalues to classify each. - For example, at \((0,0)\), \( J = \begin{bmatrix} 10 & 0 \ 0 & 10 \end{bmatrix} \), which indicates a stable node.

Step 5: Draw Vector Field Directions

Determine the signs of \(\frac{dx_1}{dt}\) and \(\frac{dx_2}{dt}\) in each region formed by the isoclines. For example, above \(10 = x_1 + 2x_2\), \(\frac{dx_1}{dt}\) is negative, and below it, \(\frac{dx_1}{dt}\) is positive. Around each equilibrium, determine the flow direction by checking the sign inside specific regions. Draw vectors showing these direction changes, following how they behave near equilibrium points and in the different regions separated by the zero isoclines.

Key Concepts

Zero Isoclines

In the realm of differential equations, zero isoclines are a critical concept to grasp. These lines are where the rate of change of a variable within a system is zero. In simpler terms, zero isoclines are boundary lines that separate regions where variable trends (like growth or decline) occur.

For the given system of equations, consider each derivative: \(\frac{dx_1}{dt} = 0\) and \(\frac{dx_2}{dt} = 0\). By setting each of these derivatives to zero, we find the respective zero isoclines. From the first equation, we derive two zero isoclines: \(x_1 = 0\) and the line \(10 = x_1 + 2x_2\). The process involves substituting the values into the equation and solving.

Similarly, for the second equation, we deduce the zero isoclines: \(x_2 = 0\) and \(10 = 2x_1 + x_2\). These isoclines are then graphically represented on the \((x_1, x_2)\) plane, where they delineate regions with distinct behaviors of \(x_1\) and \(x_2\).

By identifying these lines, we distinguish areas where the system either remains constant or changes dynamics, making them critical for understanding the overall behavior of the system.

Equilibria

Equilibria, or equilibrium points, characterize the states within a dynamic system where all variables stop changing. In other words, the system is at rest when reaching these points. In the context of the differential equations provided, equilibria occur where both \(\frac{dx_1}{dt}\) and \(\frac{dx_2}{dt}\) simultaneously equal zero.

To find equilibria, we solve the system of equations derived by setting the derivatives to zero. Solving these yields the points \((0,0)\), \((0,10)\), \((5,0)\), and \((4,2)\). Each of these points represents a potential rest state for the system.

Each equilibrium can further be classified as stable or unstable through analysis. This involves looking at the behavior of the system near these points and whether small disturbances grow or diminish over time. This classification is crucial in understanding the nature of the system's stability and how it reacts to internal or external changes.

Jacobian Matrix

The Jacobian Matrix is a fantastic tool for digging deeper into the behavior of differential equations. By linearizing the system near the equilibrium points using the Jacobian, we unlock insights into the system's stability.

The Jacobian matrix, denoted as \( J \), involves taking partial derivatives of each function with respect to each variable in the system, creating a matrix representation. In our case, it looks like this:\[J = \begin{bmatrix}10 - 2x_1 - 2x_2 & -2x_1 \ -2x_2 & 10 - 2x_1 - x_2 \end{bmatrix}\]Once calculated, evaluating the Jacobian at each equilibrium point lets us examine how a minor perturbation near these points affects the system.

For example, when evaluating at \((0,0)\), the Jacobian simplifies and helps determine the equilibrium's stability by calculating the eigenvalues. Each eigenvalue reveals whether the equilibrium behaves as a stable node, saddle point, or any other type of critical point.

In essence, the Jacobian Matrix is central to predicting how the differential equation system behaves near equilibria, spotlighting the stability and reaction of the system to changes.