Problem 64 - Calculus for Biology and Medicine

Question

Draw the vector field plot of the differential equation. Then, using the given initial conditions, sketch the solutions (i.e., draw a graph showing the dependent variable as a function of the independent variable). \(\frac{d N}{d t}=(N-1)(N+1)(N-4)\) (a) \(N(0)=0\), (b) \(N(0)=2\), (c) \(N(0)=6\), (d) \(N(0)=-2\).

Step-by-Step Solution

Answer

Equilibrium points are at \(N = 1, -1, 4\). Solutions for initial conditions move as per stability analysis.

Step 1: Understand the Differential Equation

The given differential equation is \(\frac{d N}{d t}=(N-1)(N+1)(N-4)\). This equation describes how the variable \(N\) changes over time \(t\). The right-hand side can be expanded or analyzed to find equilibrium points, where \(\frac{d N}{d t} = 0\).

Step 2: Find Equilibrium Points

Equilibrium points occur when \(\frac{d N}{d t} = 0\). For \((N-1)(N+1)(N-4) = 0\), the equilibrium points are found by setting each factor equal to zero: \(N = 1\), \(N = -1\), and \(N = 4\). At these values, \(N\) is constant over time.

Step 3: Analyze Stability at Equilibrium Points

Investigate the stability of each equilibrium point by observing the sign of \(\frac{d N}{d t}\) around the points. For example, near \(N=1\), \(\frac{d N}{d t} < 0\) for \(N < 1\) and \(\frac{d N}{d t} > 0\) for \(N > 1\), making \(N=1\) an unstable equilibrium. Repeat this analysis for \(N = -1\) and \(N = 4\).

Step 4: Sketch the Vector Field

For each region defined by the equilibrium points \(N = -1\), \(N = 1\), and \(N = 4\), draw small arrows indicating the direction in which \(N\) moves. For \(N < -1\), \(-1 < N < 1\), \(1 < N < 4\) and \(N > 4\), draw the arrows indicating whether \(\frac{d N}{d t}\) is positive (arrow pointing right) or negative (arrow pointing left).

Step 5: Sketch the Solutions for Initial Conditions

For each given initial condition, sketch how \(N\) evolves over time starting from that specific initial value: - (a) \(N(0) = 0\): Starts between -1 and 1, \(N\) moves towards -1. - (b) \(N(0) = 2\): Starts between 1 and 4, \(N\) moves towards 4. - (c) \(N(0) = 6\): Starts above 4, \(N\) moves towards infinity. - (d) \(N(0) = -2\): Starts less than -1, \(N\) moves further negative.

Key Concepts

Vector Field

In the study of differential equations, a vector field provides a visual representation of how a variable, such as \( N \), shifts over time relative to another variable, like \( t \). Imagine small arrows that indicate the direction and speed of \( N \) at any given point. This visual approach paints a clearer picture of how solutions to the differential equation evolve.
For example, consider the differential equation \( \frac{d N}{d t} = (N-1)(N+1)(N-4) \). To draw the vector field for this equation, observe separate regions divided by equilibrium points. Before plotting, determine where the derivative is positive or negative across these regions.
When \( \frac{d N}{d t} > 0 \), draw arrows pointing to the right (indicating increasing \( N \)), and when \( \frac{d N}{d t} < 0 \), draw arrows pointing to the left (indicating decreasing \( N \)). This method helps predict how \( N \) behaves depending on its initial state.

Equilibrium Points

Equilibrium points in a differential equation represent the values where the change in the variable over time is zero. In other words, these are the points where \( N \) remains constant without shifting up or down along the graph. In our case, the differential equation \( \frac{d N}{d t} = (N-1)(N+1)(N-4) \) hints at such points.
Find them by setting the derivative equal to zero:

• \( N = 1 \)
• \( N = -1 \)
• \( N = 4 \)

Each value represents an equilibrium point. At these points, \( N \) does not change, acting as a tipping or balancing point in the system, akin to a seesaw. They can be likened to a pause in the evolution of a system's behavior. Understanding such points is crucial for predicting system dynamics and eventual outcomes.

Stability Analysis

Stability analysis involves determining whether small disturbances near an equilibrium point will diminish over time or amplify, causing the system to move away from that point. To understand this, consider the nature of the differential equation \( \frac{d N}{d t} = (N-1)(N+1)(N-4) \) near its equilibrium points.
Evaluate the signs of the changes in \( N \) around each equilibrium point to deduce stability:

• At \( N = -1 \), if moving away from -1 in either direction decreases the function's value (\( \frac{d N}{d t} > 0 \)), it indicates stability.
• At \( N = 1 \), if a minor shift leads \( N \) in both directions to move away from that point (\( \frac{d N}{d t} > 0 \) or \( < 0 \)), it is unstable.
• At \( N = 4 \), if moving away leads to \( N \) increasing but coming back to the point is challenging (\( \frac{d N}{d t} < 0 \)), it indicates instability.

Stability analysis provides insight into how external influences impact the steady state of a system and whether eventual return to equilibrium is feasible.

Initial Conditions

Initial conditions specify the starting point for a differential equation's solution. These act like a starting line marker and influence how the solution's trajectory or path develops over time. In practical terms, it's like a race starting at different locations.
Take, for example, our differential equation: with specified conditions given by

• \( N(0) = 0 \)
• \( N(0) = 2 \)
• \( N(0) = 6 \)
• \( N(0) = -2 \)

Each of these starts dictates different behaviors in \( N \).
Initially,

• For \( N(0) = 0 \), \( N \) tends towards -1.
• From \( N(0) = 2 \), \( N \) trends toward 4.
• Starting with \( N(0) = 6 \), \( N \) goes off to infinity.
• And \( N(0) = -2 \) leads \( N \) further negative.

These initial conditions underscoring distinct outcomes show how initial scenarios define trajectories and shed light on dynamic systems.