[PDF] [PDF] 104 Fourier Cosine and Sine Series - Berkeley Math

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10.4 Fourier Cosine and Sine Series

To solve a partial dierential equation, typically we represent a function by a trigonometric series consisting

of only sine functions or only cosine functions.

Recall that the Fourier series for an odd function dened on [L;L] consists entirely of sine terms. Thus

we might achieve f(x) =1X n=1a nsinnxL (1)

by articially extending the functionf(x);0< x < Lto the interval (L;L) in such a way that the extended

function is odd. That is, f o(x) =f(x);0< x < L; f(x);L < x <0; and extendingfo(x) to allxusing 2L-periodicity.fo(x) is an extension off(x) becausefo(x) =f(x) on

(0;L). This extension is called theodd2L-periodic extensionoff(x). The resulting Fourier series expansion

is called a half-range expansion forf(x) because it represents the functionf(x) on (0;L). Similarly, theeven2L-periodic extensionoff(x) as the function f e(x) =f(x);0< x < L; f(x);L < x <0; withfe(x+ 2L) =fe(x). To illustrate the various extensions, let's consider the functionf(x) =x;0< x < . If we extendf(x) to the interval (;) using-periodicity, then the extensionfis given by e f(x) =x;0< x < x+; < x <0; with ef(x+ 2) =ef(x). In the previous quiz we saw that the Fourier series foref(x) is e f(x)2 1X k=11k sin2kx;

which consists of both odd functions (the sine terms) and even functions (the constant term), because the

-periodic extensionef(x) is neither an even nor an odd function. The odd 2-periodic extension off(x) is

justfo(x) =x; < x < , which has the Fourier series expansion f o(x)21X n=1(1)n+1n sinnx:(2) Becausefo(x) =f(x) on the interval (0;), the expansion in (2) is a half-range expansion forf(x). The even 2-periodic extension off(x) is the functionfe(x) =jxj; < x < , which has the Fourier series expansion f e(x) =2 4 1 X k=11(2k1)2cos(2k1)x(3) (see Example 2 inx10.3 lecture notes). The preceding three extensions, the-periodic functionef(x), the odd 2-periodic functionfo(x), and

the even 2-periodic functionfe(x), are natural extensions off(x). The Fourier series expansions forfo(x)

andfe(x), given in (2) and (3) equalf(x) on the interval (0;). This motivates the following denitions.

1 Denition.Letf(x)be piecewise continuous on the interval[0;L]. The Fourier cosine series off(x)on [0;L]is a 02 +1X n=1a ncosnxL ;(4) where a n=2L Z L 0 f(x)cosnxL dx; n= 0;1;::::(5)

The Fourier sine series off(x)on[0;L]is

1 X n=1b nsinnxL ;(6) where b n=2L Z L 0 f(x)sinnxL dx; n= 1;2;::::(7)

The trigonometric series in (4) is the Fourier series forfe(x), the even 2L-periodic extension off(x). The

trigonometric series in (6) is the Fourier series forfo(x), the odd 2L-periodic extension off(x). These are

calledhalf-range expansionsforf(x). Example 1.Determine (a) the-periodic extensionef, (b) the odd2-periodic extensionfo, and (c) the even2-periodic extensionfe, forf(x) =x;0< x < . Example 2.Compute the Fourier sine series forf(x) =x;0< x < . Example 3.Compute the Fourier cosine series forf(x) =ex;0< x <1.

A mathematical model for source-less the heat

ow in a uniform wire whose ends are kept at constant

temperature 0 is the following initial value problem, whereu(x;t) is the temperature in the wire at location

xand timet: @u@t (x;t) =@2u@x

2(x;t);0< x < L;t >0 (8)

u(0;t) =u(L;t) = 0; t >0 (9) u(x;0) =f(x);0< x < L:(10) Using the method of separation of variables, we may derive the following solution: u(x;t) =1X n=1c ne(n=L)2tsinnxL :(11)

Example 4.Find the solution to the heat problem

@u@t = 5@2u@x

2;0< x < ;t >0

u(0;t) =u(;t) = 0; t >0 u(x;0) =x(x);0< x < : 2quotesdbs_dbs17.pdfusesText_23

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