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Chapter 10: Problem 14

L. F. Richardson proposed the following model to describe the spread of war fever. If \(y=f(t)\) is the percentage of the population advocating war at time \(t\), the rate of change of \(f(t)\) at any time is proportional to the product of the percentage of the population advocating war and the percentage not advocating war. Set up a differential equation that is satisfied by \(y=f(t)\), and sketch a solution. (Source: Econometrics.)

Short Answer

Expert verified
\( \frac{dy}{dt} = k y (1 - y) \). Sketch: Solutions tend towards \( y = 0 \) or \( y = 1 \).

Step by step solution

01

Identify the Variables

Let \( y = f(t) \) represent the percentage of the population advocating war at time \( t \).
02

Understand the Model

According to L. F. Richardson's model, the rate of change of \( f(t) \) is proportional to the product of the percentage of the population advocating war (\( y = f(t) \)) and the percentage not advocating war (\( 1 - y \)).
03

Formulate the Differential Equation

Translate the proportional relationship into a differential equation. Since the rate of change of \( f(t) \) is proportional to \( y(1 - y) \), we can write: \[ \frac{dy}{dt} = k y (1 - y) \] where \( k \) is the constant of proportionality.
04

Sketch the Solution

To sketch the solution, analyze the behavior of the differential equation. The equilibrium points occur when \( \frac{dy}{dt} = 0 \). This happens when \( y = 0 \) or \( y = 1 \). The solutions approach these equilibrium points depending on the initial value of \( y \).
05

Plot the Equilibria

Draw a horizontal line at \( y = 0 \) and another at \( y = 1 \). These lines represent the stable equilibria. If the initial percentage advocating war (\( y \)) is between 0 and 1, it will tend toward one of these equilibria over time.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

War Fever Model
The war fever model, proposed by L. F. Richardson, is a fascinating look into how opinions on war spread within a population. The model uses differential equations to represent the rate at which war fever, or the advocacy for war, spreads over time.

If you imagine a population at any point in time, a certain percentage advocates for war, while the rest are against it. This model focuses on how these percentages change. By understanding this, we get insights into how collective opinions can evolve and how a single viewpoint can potentially dominate over time.
L. F. Richardson
L. F. Richardson was a British physicist and psychologist who contributed significantly to the field of mathematical modeling in social sciences.

His work included the study of conflicts and how they spread, much like diseases. This approach was quite innovative for his time, as it applied mathematical techniques to understand social phenomena.

By proposing the war fever model, Richardson aimed to describe how war advocacy could grow within a population, highlighting the interplay between those for and against war. This model helps us quantify and predict changes in public opinion on a large scale.
Proportionality Constant
In Richardson's war fever model, a key element is the proportionality constant, denoted by the letter \(k\). This constant determines how strongly the rate of change in war advocacy is related to its influencing factors.

In the differential equation \(\frac{dy}{dt} = k y (1 - y)\), \(k\) affects how quickly the population shifts between advocacy and opposition to war.

If \(k\) is large, the opinion shifts rapidly. If \(k\) is small, the change is more gradual. Understanding \(k\) is crucial because it reveals how sensitive the population is to changes in war fever.
Equilibrium Points
Equilibrium points in the war fever model are where the population stabilizes in terms of war advocacy. These points occur when there is no change in the percentage advocating for war, meaning \(\frac{dy}{dt} = 0\).

For this model, the equilibrium points are at \(y = 0\) and \(y = 1\). At \(y = 0\), no one advocates for war. At \(y = 1\), everyone does.

The behavior of the model shows that the population will tend toward these equilibrium points over time. If the initial percentage of advocates is high, the population will likely move toward complete advocacy. Conversely, if it's low, the population will move towards no advocacy.
Differential Equation Solution Sketch
To sketch the solution for the differential equation \(\frac{dy}{dt} = k y (1 - y)\), consider the equilibrium points and the initial value of \(y\).

First, draw horizontal lines at \(y = 0\) and \(y = 1\). These are your equilibrium points. Then, analyze the initial position of \(y\).

For values of \(y\) slightly above 0, \(y\) will increase towards 1 indicating growth in war advocacy. For values of \(y\) slightly below 1, \(y\) will decrease towards 0 indicating a decline.

This sketch gives a visual representation of how war fever evolves, helping to simplify the understanding of the model's dynamics.

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Most popular questions from this chapter

Some homeowner's insurance policies include automatic inflation coverage based on the U.S. Commerce Department's construction cost index (CCI). Each year, the property insurance coverage is increased by an amount based on the change in the CCI. Let \(f(t)\) be the CCI at time \(t\) years since January 1,1990 , and let \(f(0)=100\). Suppose that the construction cost index is rising at a rate proportional to the CCI and the index was 115 on January 1, 1992. Construct and solve a differential equation satisfied by \(f(t)\). Then, determine when the CCI will reach 200 .

The differential equation \(y^{\prime}=2 t y+e^{t^{2}}, y(0)=5\), has solution \(y=(t+5) e^{t^{2}} .\) In the following table, fill in the second row with the values obtained from the use of a numerical method and the third row with the actual values calculated from the solution. What is the greatest difference between corresponding values in the second and third rows? $$ \begin{array}{c|c|c|c|c|c|c|c|c|c|c|c} t_{i} & 0 & .2 & .4 & .6 & .8 & 1 & 1.2 & 1.4 & 1.6 & 1.8 & 2 \\ \hline y_{i} & 5 & & & & & & & & & & \\ \hline y & 5 & 5.412 & & & & & & & & & 382.2 \\ \hline \end{array} $$

Use Euler's method with \(n=4\) to approximate the solution \(f(t)\) to \(y^{\prime}=2 t-y+1, y(0)=5\) for \(0 \leq t \leq 2\) Estimate \(f(2)\).

A certain drug is administered intravenously to a patient at the continuous rate of \(r\) milligrams per hour. The patient's body removes the drug from the bloodstream at a rate proportional to the amount of the drug in the blood, with constant of proportionality \(k=.5\) (a) Write a differential equation that is satisfied by the amount \(f(t)\) of the drug in the blood at time \(t\) (in hours). (b) Find \(f(t)\) assuming that \(f(0)=0\). (Give your answer in terms of \(r\).) (c) In a therapeutic 2-hour infusion, the amount of drug in the body should reach 1 milligram within 1 hour of administration and stay above this level for another hour. However, to avoid toxicity, the amount of drug in the body should not exceed 2 milligrams at any time. Plot the graph of \(f(t)\) on the interval \(1 \leq t \leq 2\), as \(r\) varies between 1 and 2 by increments of . \(1 .\) That is, plot \(f(t)\) for \(r=1,1.1,1.2,1.3, \ldots, 2 .\) By looking at the graphs, pick the values of \(r\) that yield a therapeutic and nontoxic 2 -hour infusion.

Solve the given equation using an integrating factor. Take \(t>0\). $$ y^{\prime}=e^{-t}(y+1) $$
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