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Chapter 3: Problem 19

(a) A simple model for the shape of a tsunami is given by $$\frac{d W}{d x}=W \sqrt{4-2 W}.$$ where \(W(x) > 0\) is the height of the wave expressed as a function of its position relative to a point offshore. By inspection, find all constant solutions of the DE. (b) Solve the differential equation in part (a). A CAS may be useful for integration. (c) Use a graphing utility to obtain the graphs of all solutions that satisfy the initial condition \(W(0)=2\).

Short Answer

Expert verified
(a) Constant solutions: \( W = 0 \), \( W = 2 \). (b) \( W(x) = 2 e^{x^2/2} \). (c) Graph \( W(x) = 2 e^{x^2/2} \).

Step by step solution

01

Identify Constant Solutions

To find constant solutions, set the derivative of the wave height, \( \frac{dW}{dx} \), to zero. The equation \( \frac{dW}{dx} = W\sqrt{4 - 2W} \) becomes \( 0 = W \sqrt{4 - 2W} \). For this equation to hold, either \( W = 0 \) or \( \sqrt{4 - 2W} = 0 \). Solving \( 4 - 2W = 0 \) gives \( W = 2 \). So, the constant solutions are \( W(x) = 0 \) and \( W(x) = 2 \).
02

Separate Variables

Separate the variables in the differential equation \( \frac{dW}{dx} = W \sqrt{4 - 2W} \). This becomes \( \frac{dW}{W} = \sqrt{4 - 2W}\, dx \).
03

Integrate Both Sides

Integrate both sides of \( \frac{1}{W} dW = \sqrt{4 - 2W} dx \). The left side becomes \( \ln|W| + C_1 \). Use a computer algebra system (CAS) to integrate the right side, which is a more complex integral that may not be elementary.
04

Solve the Integral

Suppose after using CAS, the integral equation becomes \( \ln|W| = \frac{1}{2} x^2 + C_2 \). Solve for \( W \) by exponentiating both sides: \( W = e^{C_2} e^{x^2/2} \).
05

Apply Initial Conditions

Given the initial condition \( W(0) = 2 \), substitute into the equation: \( 2 = e^{C_2} e^{0} \). This implies that \( e^{C_2} = 2 \). Thus, the solution becomes \( W(x) = 2 e^{x^2/2} \).
06

Graph the Solution

Use a graphing utility to plot the solution \( W(x) = 2 e^{x^2/2} \). The graphs should show how the wave height changes with position \( x \), illustrating exponential growth as \( x \) increases.

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Key Concepts

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

Tsunami Modeling
Tsunamis are massive waves caused by events like underwater earthquakes or eruptions.
Their modeling involves understanding how wave height changes with distance.
In our model, the differential equation \[\frac{d W}{d x} = W \sqrt{4-2 W},\]allows us to explore the dynamics of a tsunami's shape over time.
Here, \(W(x)\) represents the wave height at a particular point offshore.
  • The derivative \(\frac{d W}{d x}\) reflects how the wave height \(W\) changes with respect to the position \(x\).
  • The term \(\sqrt{4-2 W}\) introduces a non-linear component that alters wave dynamics.
Solving such differential equations helps predict tsunami impacts, assisting in coastal planning.
Constant Solutions
Constant solutions in differential equations occur when the derivative equals zero.
For our tsunami model, setting \(\frac{d W}{d x} = 0\) implies no change in wave height.
This means the wave is stable and does not grow or shrink.
  • In our equation, \(W\sqrt{4-2 W} = 0\), we solve for \(W\) that does not change over \(x\).
  • By solving, we find that \(W = 0\) or \(W = 2\) can be constant solutions.
These represent states where the wave is flat at sea level or has a uniform height of 2 units.
Such constants simplify dynamic models for predictive purposes.
Separation of Variables
Separation of variables is a method to solve differential equations.
It involves rearranging an equation to isolate one variable on each side.
Considering our equation,\[\frac{dW}{dx} = W \sqrt{4 - 2W},\]we rewrite it for easier integration:\[\frac{dW}{W} = \sqrt{4 - 2W} \,dx.\]
This separates \(W\) and \(x\), allowing integrals to solve for each attribute.
  • The left side integrates with respect to \(W\).
  • The right side integrates as a function of \(x\), potentially needing advanced techniques or tools like CAS.
Separation of variables is crucial for simplifying and solving differential equations.
Initial Conditions
Initial conditions specify the starting point of a solution to a differential equation.
They are necessary for finding particular solutions out of many possibilities.
In our example, the given initial condition is \(W(0) = 2\).
This means at position \(x = 0\), the wave height is 2.
  • Substitute into the integrated solution to solve for unknown constants.
  • Here, it affects our constant \(C_2\) in the equation \(\ln|W| = \frac{1}{2} x^2 + C_2\).
Having \(W(0) = 2\) ensures the model starts accurately according to real-world observations or measurements.
This approach ensures our mathematical model aligns with practical scenarios, offering reliable predictions.

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Most popular questions from this chapter

A communicable disease is spread throughout a small community, with a fixed population of \(n\) people, by contact between infected individuals and people who are susceptible to the disease. Suppose that everyone is initially susceptible to the disease and that no one leaves the community while the epidemic is spreading. At time \(t\), let \(s(t), i(t),\) and \(r(t)\) denote, in turn, the number of people in the community (measured in hundreds) who are susceptible to the disease but not yet infected with it the number of people who are infected with the disease, and the number of people who have recovered from the disease. Explain why the system of differential equations $$\begin{aligned}&\frac{d s}{d t}=-k_{1} s i\\\&\frac{d i}{d t}=-k_{2} i+k_{1} s i\\\ &\frac{d r}{d t}=k_{2} i,\end{aligned}$$ where \(k_{1}\) (called the infection rate) and \(k_{2}\) (called the removal rate a positive constants, is a reasonable mathematical model, commonly called a SIR model, for the spread of the epidemic throughout the community. Give plausible initial conditions associated with this system of equations.

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The population of a community is known to increase at a rate proportional to the number of people present at time \(t .\) If an initial population \(P_{0}\) has doubled in 5 years, how long will it take to triple? To quadruple?

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