Problem 23 - Calculus for Biology and Medicine

Question

Let $$ \begin{array}{l} \frac{d x_{1}}{d t}=x_{1}\left(2-x_{1}\right)-x_{1} x_{2} \\ \frac{d x_{2}}{d t}=x_{1} x_{2}-x_{2} \end{array} $$ (a) Graph the zero isoclines. (b) Show that \((1,1)\) is an equilibrium. Use the graphical approach to determine its stability. 24\. Let $$ \begin{array}{l} \frac{d x_{1}}{d t}=x_{1}\left(2-x_{1}^{2}\right)-x_{1} x_{2} \\ \frac{d x_{2}}{d t}=x_{1} x_{2}-x_{2} \end{array} $$ (a) Graph the zero isoclines. (b) Show that \((1,1)\) is an equilibrium. Use the graphical approach to determine its stability.

Step-by-Step Solution

Answer

For both systems, the equilibrium at \((1,1)\) is stable, as trajectories converge towards this point.

Step 1: Identify Zero Isoclines for the First System

For the system of equations \( \frac{d x_{1}}{d t} = x_{1}(2-x_{1}) - x_{1} x_{2} \) and \( \frac{d x_{2}}{d t} = x_{1} x_{2} - x_{2} \), set each differential equation to zero to find the zero isoclines.For \( \frac{d x_{1}}{d t} = 0 \):\[ x_{1}(2-x_{1}) - x_{1} x_{2} = 0 \] Simplify to find the isoclines: \( x_{1} = 0 \) or \( 2-x_{1} = x_{2} \).For \( \frac{d x_{2}}{d t} = 0 \):\[ x_{1} x_{2} - x_{2} = 0 \]\( x_{2} = 0 \) or \( x_{1} = 1 \).

Step 2: Identify the Equilibrium Point (First System)

Find where both zero isoclines intersect. This occurs where \( \frac{d x_{1}}{d t} = 0 \) and \( \frac{d x_{2}}{d t} = 0 \) simultaneously.From the intersection point of the equations: \( x_{1} = 1 \) and \( x_{2} = 1 \). So, \((1,1)\) is an equilibrium point.

Step 3: Determine Stability Using a Graphical Approach (First System)

On a phase plane, plot the zero isoclines: \( x_{1} = 0 \), \( 2 - x_{1} = x_{2} \), \( x_{1} = 1 \), and \( x_{2} = 0 \).Graphical analysis allows observing the direction of trajectories relative to zero isoclines. Analyze the motion, confirming that trajectories generally tend toward \((1,1)\), hence suggesting the stability.

Step 4: Identify Zero Isoclines for the Second System

For the system of equations \( \frac{d x_{1}}{d t} = x_{1}(2-x_{1}^{2}) - x_{1} x_{2} \) and \( \frac{d x_{2}}{d t} = x_{1} x_{2} - x_{2} \), again set each differential equation to zero.For \( \frac{d x_{1}}{d t} = 0 \):\[ x_{1}(2-x_{1}^{2}) - x_{1} x_{2} = 0 \] Simplify to find the isoclines: \( x_{1} = 0 \), \( 2 - x_{1}^{2} = x_{2} \).For \( \frac{d x_{2}}{d t} = 0 \):\[ x_{1} x_{2} - x_{2} = 0 \]\( x_{2} = 0 \) or \( x_{1} = 1 \).

Step 5: Identify the Equilibrium Point (Second System)

Find where both zero isoclines intersect: \( 2-x_{1}^{2} = 1 \) when \( x_{1} = 1 \) and \( x_{2} = 1 \). Therefore, the equilibrium point is also \((1,1)\).

Step 6: Determine Stability Using a Graphical Approach (Second System)

Plot the zero isoclines from Step 4. Analyze motion same as in Step 3. Visually determine trajectory behavior around \((1,1)\). As for both systems, paths converge to \((1,1)\), indicating stability at this point.

Key Concepts

Zero Isoclines

A zero isocline, in a system of differential equations, is a curve where the rate of change of a variable is zero. To understand zero isoclines, let's examine a given system of differential equations such as: \[ \begin{array}{l} \frac{d x_{1}}{d t} = x_{1}(2-x_{1}) - x_{1} x_{2} \ \frac{d x_{2}}{d t} = x_{1} x_{2} - x_{2} \end{array} \] To find the zero isoclines of this system:

• Set \( \frac{d x_{1}}{d t} = 0 \), resulting in the isoclines \( x_{1} = 0 \) or \( 2-x_{1} = x_{2} \).
• Set \( \frac{d x_{2}}{d t} = 0 \), leading to \( x_{2} = 0 \) or \( x_{1} = 1 \).

These equations represent lines or curves on the phase plane where one variable's rate of change is zero. As such, they provide critical insight into the dynamics of the system by highlighting regions where variables neither grow nor shrink.

Stability of Equilibrium

Equilibrium points in a differential system occur when all rates of change are zero simultaneously. In our example system, such a point can be located by finding where the zero isoclines intersect. The intersection of these isoclines is typically at points where the system balances out, often leading to an equilibrium.For the system above:

• The intersection of \( x_{1} = 1 \) and \( x_{2} = 1 \), derived from the zero isoclines, gives the equilibrium point \((1,1)\).

After identifying \((1,1)\) as an equilibrium, we need to determine its stability:- If nearby trajectories on the phase plane tend towards this point, it is stable.- If trajectories diverge away, it is unstable.Graphical analysis shows that in this system, paths generally converge to \((1,1)\), indicating a stable equilibrium where the system returns to balance after small disturbances.

Phase Plane Analysis

Phase plane analysis involves plotting trajectories of a system of differential equations on a coordinate plane. This plane, known as the phase plane, enables visualization of dynamic behavior and stability of the system’s equilibrium points.To conduct phase plane analysis for the system \( \frac{d x_{1}}{d t} = x_{1}(2-x_{1}) - x_{1} x_{2} \) and \( \frac{d x_{2}}{d t} = x_{1} x_{2} - x_{2} \):

• Plot the zero isoclines identified earlier: \( x_{1} = 0 \), \( 2 - x_{1} = x_{2} \), \( x_{1} = 1 \), and \( x_{2} = 0 \).
• Draw trajectories showing how solutions evolve over time to understand direction and behavior.

By analyzing these trajectories relative to the zero isoclines, we observe how the system behaves around the equilibrium point \((1,1)\). In this case, trajectory paths converge towards this point, reinforcing its stability and confirming the theoretical findings.