Problem 22 - Calculus for Biology and Medicine

Question

For make vector field plots of each of the differential equations. Find any equilibria of each differential equation and use your vector field plot to classify whether each equilibrium is stable or unstable. $$ \frac{d N}{d t}=N^{3} e^{-N} $$

Step-by-Step Solution

Answer

The equilibrium is at \( N = 0 \) and it is unstable.

Step 1: Investigate Equilibria

Equilibrium points occur where the differential equation is equal to zero. For the given differential equation: \( \frac{dN}{dt} = N^3 e^{-N} \), set it to zero: \( N^3 e^{-N} = 0 \).Since \( e^{-N} \) is never zero, \( N^3 = 0 \) gives us the equilibrium point at \( N = 0 \).

Step 2: Analyze Stability

To determine the stability of the equilibrium point, let's consider the sign of \( \frac{dN}{dt} \) around \( N = 0 \).- For \( N < 0 \), \( N^3 < 0 \) making \( \frac{dN}{dt} < 0 \), so \( N \) decreases further away from the equilibrium.- For \( N > 0 \), \( N^3 > 0 \) making \( \frac{dN}{dt} > 0 \), so \( N \) increases further away from the equilibrium.Thus, \( N = 0 \) is an unstable equilibrium.

Step 3: Plot the Vector Field

Create the vector field plot for the differential equation by plotting vectors at various points \( N \). The direction of each vector is determined by \( \frac{dN}{dt} = N^3 e^{-N} \), which gives insight into how \( N \) changes over time:- At \( N = 0 \), slope is \(0\).- At \( N > 0 \), vectors point upward as \( N^3 e^{-N} > 0 \).- At extreme \( N \), vectors tend to level out due to the decay of \( e^{-N} \).This plot visibly confirms that \( N = 0 \) is unstable since the trajectories move away from it.

Key Concepts

Differential Equations

Differential equations describe how a change in one variable corresponds to a change in another. They are an essential part of mathematical modeling of physical systems. In simple terms, a differential equation links a function with its derivatives, showing how the function changes. For example, the stopping force on a car is proportional to its speed, which is modeled as a differential equation.
In the case of the given differential equation \( \frac{dN}{dt} = N^3 e^{-N} \), it tells us how \( N \) changes over time \( t \). The equation comprises a continuous function \( N \), modified by the exponential decay \( e^{-N} \). This decay affects how quickly \( N \) changes, especially as \( N \) becomes very large.
Differential equations like this one are useful because they help us predict future developments of \( N \) based on its current value. They find applications in numerous fields, including physics, biology, and economics.

Equilibrium Points

Equilibrium points are crucial in understanding the behavior of differential equations. They represent states where the system described by the differential equation experiences no change over time. In essence, at equilibrium, the rate of change is zero.
To find the equilibrium point for \( \frac{dN}{dt} = N^3 e^{-N} \), we set the equation to zero since equilibrium happens when there is no rate of change: \( N^3 e^{-N} = 0 \).

• \( e^{-N} \) is always positive, therefore it never reaches zero.
• This leaves \( N^3 = 0 \), which solves to give us \( N = 0 \) as the equilibrium point.

At this point, \( N \) does not change, as confirmed by the zero rate of change. Equilibria are fundamental since they help us identify critical points where stability analysis can be conducted.

Stability Analysis

Stability analysis delves into how equilibrium points behave in the face of small disturbances. Once equilibrium points are determined, understanding their stability helps predict whether a system will return to these points or drift away.
For the differential equation \( \frac{dN}{dt} = N^3 e^{-N} \), stability is analyzed by looking at how \( N \) changes around the equilibrium point \( N = 0 \).

• If \( N < 0 \): \( N^3 < 0 \) leads to \( \frac{dN}{dt} < 0 \). This means \( N \) moves further from zero. Thus, disturbances in the negative direction move \( N \) away from equilibrium.
• If \( N > 0 \): \( N^3 > 0 \) leads to \( \frac{dN}{dt} > 0 \). Here, \( N \) also moves further from zero, so disturbances in the positive direction move \( N \) away.

With this behavior, \( N = 0 \) is determined to be an unstable equilibrium because any tiny push in either direction will cause \( N \) to shift away, rather than return to equilibrium. Confirming this through a vector field plot solidifies our understanding by visualizing trajectories leading away from \( N = 0 \).