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indicate how they can be employed to solve initial value problems with periodic forcing terms. Find the formal solution of the initial value problem $$ y^{\prime \prime}+\omega^{2} y=f(t), \quad y(0)=1, \quad y^{\prime}(0)=0 $$ where \(f\) is periodic with period 2 and $$ f(t)=\left\\{\begin{aligned} 1-t, & 0 \leq t<1 \\\\-1+t, & 1 \leq t<2 \end{aligned}\right. $$ See Problem 8.

Short Answer

Expert verified
Answer: The primary approach used to solve the initial value problem with a periodic forcing term in this exercise is Floquet theory and Fourier series.

Step by step solution

01

Compute the Fourier series of the forcing term

First, we need to find the Fourier series of f(t) to analyze the given periodic forcing term. The formula for the Fourier series of f(t) is given by: $$ f(t) = \frac{a_{0}}{2} + \sum_{n=1}^{\infty}a_{n}\cos(\frac{n\pi t}{1})+\sum_{n=1}^{\infty}b_{n}\sin(\frac{n\pi t}{1}) $$ We need to compute the coefficients a鈧, a鈧, and b鈧. a鈧 = (1/2) 鈭個虏 f(t) dt a鈧 = 鈭個虏 f(t) cos(n蟺t) dt, for n 鈮 1 b鈧 = 鈭個虏 f(t) sin(n蟺t) dt, for n 鈮 1 To compute the coefficients, it is essential to divide the integration period into two parts 0 鈮 t < 1 and 1 鈮 t < 2, since f(t) has different values in these intervals: a鈧 = (1/2) ((鈭個鹿 (1-t) dt) + (鈭孤 (-1+t) dt) a鈧 = 鈭個鹿 (1-t) cos(n蟺t) dt + 鈭孤 (-1+t) cos(n蟺t) dt b鈧 = 鈭個鹿 (1-t) sin(n蟺t) dt + 鈭孤 (-1+t) sin(n 蟺t) dt Now, evaluate these integrals to determine the values of a鈧, a鈧, and b鈧.
02

Use Floquet Theory to find undetermined coefficients

Once the Fourier series of the forcing term is found, we can write the general solution for the homogeneous part of the equation: $$ y_{h}(t) = c_{1}\cos(\omega t) + c_{2}\sin(\omega t) $$ Then, we can write the particular solution y鈧(t) of the inhomogeneous equation using Floquet theory: $$ y_{p}(t) = \frac{a_{0}}{2\omega^{2}} + \sum_{n=1}^{\infty}\left(\frac{a_{n}\cos(\frac{n\pi t}{1})}{\omega^{2}-\frac{n^{2}\pi^{2}}{1^2}}+\frac{b_{n}\sin(\frac{n\pi t}{1})}{\omega^{2}-\frac{n^{2}\pi^{2}}{1^2}}\right) $$ Now, add the homogeneous solution and the particular solution to find the general solution for the given differential equation: $$ y(t) = y_{h}(t) + y_{p}(t) $$
03

Apply initial conditions

Once we have found the general solution using the homogeneous and particular solutions, it is time to apply the given initial conditions to find the exact formal solution: y(0) = 1 y'(0) = 0 Using the equation of y(t), we can determine the corresponding values of the constants c鈧 and c鈧.
04

Final formal solution

With the determined values of the coefficients c鈧 and c鈧, we can now write the final formal solution: $$ y(t) = c_{1}\cos(\omega t) + c_{2}\sin(\omega t) + \frac{a_{0}}{2\omega^{2}} + \sum_{n=1}^{\infty}\left(\frac{a_{n}\cos(\frac{n\pi t}{1})}{\omega^{2}-\frac{n^{2}\pi^{2}}{1^2}}+\frac{b_{n}\sin(\frac{n\pi t}{1})}{\omega^{2}-\frac{n^{2}\pi^{2}}{1^2}}\right) $$ This is the formal solution for the given initial value problem with a periodic forcing term.

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Key Concepts

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

Initial Value Problem
In mathematics and physics, an *Initial Value Problem* (IVP) is a differential equation coupled with a set of initial conditions. It aims to find a function that not only satisfies the differential equation but also adheres to these initial conditions at a specific point. Consider the equation we have: \[ y'' + \omega^2 y = f(t), \quad y(0) = 1, \quad y'(0) = 0 \]This equation defines our IVP, where we have specified initial conditions at time \( t = 0 \). These initial conditions provide a starting point to solve the differential equation. By integrating these conditions, we can find a unique solution whose behavior is fully determined by the initial state. Understanding IVP is critical as it serves as the foundation for modeling dynamic systems in natural and applied sciences.
Periodic Forcing
*Periodic Forcing* refers to a situation where an external force that acts on a system varies periodically with time. In our problem, the periodic forcing term is given by \( f(t) \), a piecewise function with a period of 2. It is defined as:
  • \( f(t) = 1-t \) for \( 0 \leq t < 1 \)
  • \( f(t) = -1+t \) for \( 1 \leq t < 2 \)
This type of term is crucial in modeling real-world systems like suspension springs that experience rhythmic external disturbances. To tackle such a forcing term mathematically, we use techniques like Fourier series to express it as a sum of sines and cosines. This transformation decomposes the forcing function into comprehensible parts, enabling us to find particular solutions efficiently for the differential equation.
Homogeneous Solution
The *Homogeneous Solution* of a differential equation is a solution that satisfies the associated homogeneous equation, meaning an equation with zero on the right-hand side. For our problem: \[ y'' + \omega^2 y = 0 \]The solution to this equation is:\[ y_{h}(t) = c_{1} \cos(\omega t) + c_{2} \sin(\omega t) \]In finding a general solution to the original inhomogeneous problem, the homogeneous solution represents the part of the solution that accounts for the system's "natural" behavior, without any external forces acting on it. This is crucial as it forms the basis to which we add the particular solution 鈥 which accounts for the periodic forcing 鈥 to address the full problem and ensure it aligns with initial conditions.
Floquet Theory
*Floquet Theory* is a powerful tool used to analyze differential equations with periodic coefficients. Although more complex than solving static problems, Floquet Theory provides a method for addressing periodic solutions in linear systems. When used in the context of our differential equation, it allows us to express the particular solution of the inhomogeneous equation as a Fourier series expansion: \[ y_{p}(t) = \frac{a_{0}}{2 \omega^2} + \sum_{n=1}^{\infty}\left(\frac{a_{n}\cos\left(\frac{n \pi t}{1}\right)}{\omega^2-\frac{n^2 \pi^2}{1^2}}+\frac{b_{n}\sin\left(\frac{n \pi t}{1}\right)}{\omega^2-\frac{n^2 \pi^2}{1^2}}\right) \]By applying Floquet Theory, we can essentially "break down" the problem using the periodic properties of the given system while allowing the precise periodic response to match the periodic forcing term. This theory simplifies and organizes the process of finding solutions that address complex, periodic disturbances.

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Most popular questions from this chapter

(a) Sketch the graph of the given function for three periods. (b) Find the Fourier series for the given function. $$ f(x)=\left\\{\begin{array}{lr}{x+1,} & {-1 \leq x < 0,} \\ {1-x,} & {0 \leq x < 1 ;}\end{array} \quad f(x+2)=f(x)\right. $$

Consider a rod of length 30 for which \(\alpha^{2}=1 .\) Suppose the initial temperature distribution is given by \(u(x, 0)=x(60-x) / 30\) and that the boundary conditions are \(u(0, t)=30\) and \(u(30, t)=0\) (a) Find the temperature in the rod as a function of position and time. (b) Plot \(u\) versus \(x\) for several values of \(t\). Also plot \(u\) versus \(t\) for several values of \(x\). (c) Plot \(u\) versus \(t\) for \(x=12\). Observe that \(u\) initially decreases, then increases for a while, and finally decreases to approach its steady-state value. Explain physically why this behavior occurs at this point.

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In each of Problems 15 through 22 find the required Fourier series for the given function and sketch the graph of the function to which the series converges over three periods. $$ \begin{array}{l}{f(x)=\left\\{\begin{array}{ll}{1,} & {0

Suppose that we wish to calculate values of the function \(g,\) where $$ g(x)=\sum_{n=1}^{\infty} \frac{(2 n-1)}{1+(2 n-1)^{2}} \sin (2 n-1) \pi x $$ It is possible to show that this series converges, albeit rather slowly. However, observe that for large \(n\) the terms in the series (i) are approximately equal to \([\sin (2 n-1) \pi x] /(2 n-1)\) and that the latter terms are similar to those in the example in the text, Eq. (6). (a) Show that $$ \sum_{n=1}^{\infty}[\sin (2 n-1) \pi x] /(2 n-1)=(\pi / 2)\left[f(x)-\frac{1}{2}\right] $$ where \(f\) is the square wave in the example with \(L=1\) (b) Subtract Eq. (ii) from Eq. (i) and show that $$ g(x)=\frac{\pi}{2}\left[f(x)-\frac{1}{2}\right]-\sum_{n=1}^{\infty} \frac{\sin (2 n-1) \pi x}{(2 n-1)\left[1+(2 n-1)^{2}\right]} $$ The series (iii) converges much faster than the series (i) and thus provides a better way to calculate values of \(g(x) .\)

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