Problem 21
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. $$ \frac{d N}{d t}=N \ln (2 / N), N>0 $$
Step-by-Step Solution
Verified Answer
The equilibrium point is \(N = 2\). It is stable, as shown by the vector field plot.
1Step 1: Finding Equilibria
To find the equilibrium points of the differential equation \(\frac{dN}{dt}=N\ln(2/N)\), we set \(\frac{dN}{dt} = 0\). This simplifies to \(N\ln(2/N) = 0\). Since \(N > 0\), this implies \(\ln(2/N) = 0\), which gives \(N = 2\). Thus, the equilibrium point is \(N = 2\).
2Step 2: Analyzing Stability
To determine the stability of the equilibrium point \(N = 2\), we consider small perturbations around \(N = 2\). For \(N < 2\), \(\ln(2/N) > 0\), hence \(\frac{dN}{dt} > 0\), and \(N\) increases towards 2. For \(N > 2\), \(\ln(2/N) < 0\), hence \(\frac{dN}{dt} < 0\), and \(N\) decreases towards 2. This behavior indicates that the equilibrium point at \(N = 2\) is stable.
3Step 3: Sketching the Vector Field
To sketch the vector field, we plot \(\frac{dN}{dt} = N\ln(2/N)\) on the \(N\) axis. For values of \(N < 2\), \(\frac{dN}{dt} > 0\) and the arrows point towards increasing \(N\). For \(N > 2\), \(\frac{dN}{dt} < 0\) and the arrows point towards decreasing \(N\). At the equilibrium point \(N = 2\), the arrows from both sides point towards it, reaffirming its stability.
4Step 4: Interpret the Graphical Solution
From the vector field plot, the equilibrium point \(N = 2\) is clearly visible as the state where the vectors converge, indicating stability. The arrows indicate the direction of \(N\) over time, supporting our analytical result that \(N\) will settle at 2 for any starting value \(N > 0\).
Key Concepts
Equilibrium PointsStability AnalysisVector Field Plots
Equilibrium Points
In the context of differential equations, equilibrium points are vital as they represent states where the system does not change. For the given equation \( \frac{dN}{dt} = N \ln(2/N) \), equilibrium points are determined by setting the equation to zero, \( \frac{dN}{dt} = 0 \). This is because, at equilibrium, the rate of change is zero, implying that the system is steady.
From this setup, it simplifies to \( N \ln(2/N) = 0 \). Given \( N > 0 \), the solution involves finding when the natural logarithm \( \ln(2/N) = 0 \). Solving this yields the equilibrium point \( N = 2 \). This is the value where the system balances itself.
Finding equilibrium points is an essential first step in understanding the long-term behavior of a differential system because it indicates where the solutions will eventually settle, provided certain conditions.
From this setup, it simplifies to \( N \ln(2/N) = 0 \). Given \( N > 0 \), the solution involves finding when the natural logarithm \( \ln(2/N) = 0 \). Solving this yields the equilibrium point \( N = 2 \). This is the value where the system balances itself.
Finding equilibrium points is an essential first step in understanding the long-term behavior of a differential system because it indicates where the solutions will eventually settle, provided certain conditions.
Stability Analysis
Stability analysis is crucial to understand how small changes near equilibrium points affect a system. In our problem, once we identified the equilibrium point as \( N = 2 \), we need to figure out if it's stable or unstable.
To assess stability, examine the behavior of \( N \) when it's slightly less or more than 2:
Stability in this context implies that the system returns to equilibrium after small disturbances, akin to a ball settling back into a bowl.
To assess stability, examine the behavior of \( N \) when it's slightly less or more than 2:
- For \( N < 2 \), \( \ln(2/N) > 0 \), implying \( \frac{dN}{dt} > 0 \), leading \( N \) to increase towards 2.
- For \( N > 2 \), \( \ln(2/N) < 0 \), meaning \( \frac{dN}{dt} < 0 \), making \( N \) decrease towards 2.
Stability in this context implies that the system returns to equilibrium after small disturbances, akin to a ball settling back into a bowl.
Vector Field Plots
Vector field plots visually illustrate how solutions to differential equations evolve over time. For our equation \( \frac{dN}{dt} = N \ln(2/N) \), vector fields help us understand the system dynamics intuitively.
By plotting \( \frac{dN}{dt} \) against \( N \), we observe:
Vector field plots are powerful tools for predicting the behavior of differential equations, offering a visual means to verify stability and directions of change.
By plotting \( \frac{dN}{dt} \) against \( N \), we observe:
- Arrows pointing towards increasing \( N \) for \( N < 2 \), confirming positive growth rate.
- Arrows towards decreasing \( N \) for \( N > 2 \), confirming a negative growth rate.
- Convergence of arrows at \( N = 2 \), emphasizing stability as both sides draw towards it.
Vector field plots are powerful tools for predicting the behavior of differential equations, offering a visual means to verify stability and directions of change.
Other exercises in this chapter
Problem 19
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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
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A reversible chemical reaction between chemicals \(A\) and \(B\) produces a product C: \(A+B \rightleftharpoons\). We modeled this reaction in Section 8.3.3 usi
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For each of the Problems 13-24 you should determine whether the problem needs to be solved using separation of variables or integrating factors (some of the pro
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