Idea Transcript
Fourier Series Mathematicians of the eighteenth century, including Daniel Bernoulli and Leonard Euler, expressed the problem of the vibratory motion of a stretched string through partial differential equations that had no solutions in terms of ‘‘elementary functions.’’ Their resolution of this difficulty was to introduce infinite series of sine and cosine functions that satisfied the equations. In the early nineteenth century, Joseph Fourier, while studying the problem of heat flow, developed a cohesive theory of such series. Consequently, they were named after him. Fourier series and Fourier integrals are investigated in this and the next chapter. As you explore the ideas, notice the similarities and differences with the chapters on infinite series and improper integrals.
PERIODIC FUNCTIONS A function f ðxÞ is said to have a period T or to be periodic with period T if for all x, f ðx þ TÞ ¼ f ðxÞ, where T is a positive constant. The least value of T > 0 is called the least period or simply the period of f ðxÞ. EXAMPLE 1. The function sin x has periods 2!; 4!; 6!; . . . ; since sin ðx þ 2!Þ; sin ðx þ 4!Þ; sin ðx þ 6!Þ; . . . all equal sin x. However, 2! is the least period or the period of sin x. EXAMPLE 2. The period of sin nx or cos nx, where n is a positive integer, is 2!=n. EXAMPLE 3. The period of tan x is !. EXAMPLE 4. A constant has any positive number as period.
f (x)
x
x
x
(a)
Period
f (x)
Period
f (x)
Period
Other examples of periodic functions are shown in the graphs of Figures 13-1(a), (b), and (c) below.
(b)
(c)
Fig. 13-1
336 Copyright 2002, 1963 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
CHAP. 13]
337
FOURIER SERIES
FOURIER SERIES Let f ðxÞ be defined in the interval ð#L; LÞ and outside of this interval by f ðx þ 2LÞ ¼ f ðxÞ, i.e., f ðxÞ is 2L-periodic. It is through this avenue that a new function on an infinite set of real numbers is created from the image on ð#L; LÞ. The Fourier series or Fourier expansion corresponding to f ðxÞ is given by 1 ! a0 X n!x n!x" þ bn sin þ an cos ð1Þ L L 2 n¼1 where the Fourier coefficients an and bn are 8 ð 1 L n!x > > > dx ¼ f ðxÞ cos a < n L L #L ð > 1 L n!x > > : bn ¼ dx f ðxÞ sin L #L L
n ¼ 0; 1; 2; . . .
ð2Þ
ORTHOGONALITY CONDITIONS FOR THE SINE AND COSINE FUNCTIONS Notice that the Fourier coefficients are integrals. These are obtained by starting with the series, (1), and employing the following properties called orthogonality conditions: (a)
ðL
cos
m!x n!x cos dx ¼ 0 if m 6¼ n and L if m ¼ n L L
sin
m!x n!x sin dx ¼ 0 if m 6¼ n and L if m ¼ n L L
sin
m!x n!x cos dx ¼ 0. Where m and n can assume any positive integer values. L L
#L
(b)
ðL
#L
(c)
ðL
#L
(3)
An explanation for calling these orthogonality conditions is given on Page 342. Their application in determining the Fourier coefficients is illustrated in the following pair of examples and then demonstrated in detail in Problem 13.4. EXAMPLE 1. To determine the Fourier coefficient a0 , integrate both sides of the Fourier series (1), i.e., ðL ðL ðL X 1 n a0 n!x n!xo f ðxÞ dx ¼ an cos dx þ dx þ bn sin L L #L #L 2 #L n¼1 Now
ðL
a0 dx ¼ a0 L; #L 2
ðL
#l
sin
n!x dx ¼ 0; L
ðL
#L
cos
n!x 1 dx ¼ 0, therefore, a0 ¼ L L
ðL
#L
f ðxÞ dx
!x EXAMPLE 2. To determine a1 , multiply both ð L sides of (1) by cos L and then integrate. Using the orthogonality 1 !x f ðxÞ cos dx. Now see Problem 13.4. conditions (3)a and (3)c , we obtain a1 ¼ L #L L
If L ¼ !, the series (1) and the coefficients (2) or (3) are particularly simple. case has the period 2!.
DIRICHLET CONDITIONS Suppose that (1) (2)
f ðxÞ is defined except possibly at a finite number of points in ð#L; LÞ f ðxÞ is periodic outside ð#L; LÞ with period 2L
The function in this
338
FOURIER SERIES
(3)
[CHAP. 13
f ðxÞ and f 0 ðxÞ are piecewise continuous in ð#L; LÞ.
Then the series (1) with Fourier coefficients converges to ðaÞ ðbÞ
f ðxÞ if x is a point of continuity f ðx þ 0Þ þ f ðx # 0Þ if x is a point of discontinuity 2
Here f ðx þ 0Þ and f ðx # 0Þ are the right- and left-hand limits of f ðxÞ at x and represent lim f ðx þ !Þ and !!0þ lim f ðx # !Þ, respectively. For a proof see Problems 13.18 through 13.23.
!!0þ
The conditions (1), (2), and (3) imposed on f ðxÞ are sufficient but not necessary, and are generally satisfied in practice. There are at present no known necessary and sufficient conditions for convergence of Fourier series. It is of interest that continuity of f ðxÞ does not alone ensure convergence of a Fourier series.
ODD AND EVEN FUNCTIONS A function f ðxÞ is called odd if f ð#xÞ ¼ #f ðxÞ. Thus, x3 ; x5 # 3x3 þ 2x; sin x; tan 3x are odd functions. A function f ðxÞ is called even if f ð#xÞ ¼ f ðxÞ. Thus, x4 ; 2x6 # 4x2 þ 5; cos x; ex þ e#x are even functions. The functions portrayed graphically in Figures 13-1(a) and 13-1ðbÞ are odd and even respectively, but that of Fig. 13-1(c) is neither odd nor even. In the Fourier series corresponding to an odd function, only sine terms can be present. In the Fourier series corresponding to an even function, only cosine terms (and possibly a constant which we shall consider a cosine term) can be present.
HALF RANGE FOURIER SINE OR COSINE SERIES A half range Fourier sine or cosine series is a series in which only sine terms or only cosine terms are present, respectively. When a half range series corresponding to a given function is desired, the function is generally defined in the interval ð0; LÞ [which is half of the interval ð#L; LÞ, thus accounting for the name half range] and then the function is specified as odd or even, so that it is clearly defined in the other half of the interval, namely, ð#L; 0Þ. In such case, we have 8 ð 2 L n"x > > > a dx for half range sine series ¼ 0; b ¼ f ðxÞ sin n < n L 0 L ð4Þ ð > 2 L n"x > > dx for half range cosine series f ðxÞ cos : bn ¼ 0; an ¼ L 0 L PARSEVAL’S IDENTITY If an and bn are the Fourier coefficients corresponding to f ðxÞ and if f ðxÞ satisfies the Dirichlet conditions. Then (See Problem 13.13.)
1 L
ðL
#L
f f ðxÞg2 dx ¼
1 a20 X þ ða2 þ b2n Þ 2 n¼1 n
(5)
CHAP. 13]
FOURIER SERIES
339
DIFFERENTIATION AND INTEGRATION OF FOURIER SERIES Differentiation and integration of Fourier series can be justified by using the theorems on Pages 271 and 272, which hold for series in general. It must be emphasized, however, that those theorems provide sufficient conditions and are not necessary. The following theorem for integration is especially useful. Theorem.
The Fourier series corresponding to ð f ðxÞ may be integrated term by term from a to x, and the x
resulting series will converge uniformly to
a
f ðxÞ dx provided that f ðxÞ is piecewise continuous in
#L @ x @ L and both a and x are in this interval.
COMPLEX NOTATION FOR FOURIER SERIES Using Euler’s identities, ei! ¼ cos ! þ i sin !; e#i! ¼ cos ! # i sin ! ð6Þ pffiffiffiffiffiffi where i ¼ #1 (see Problem 11.48, Chapter 11, Page 295), the Fourier series for f ðxÞ can be written as f ðxÞ ¼
1 X
1 2L
ð7Þ
f ðxÞe#in"x=L dx
ð8Þ
n¼#1
where cn ¼
cn ein"x=L
ðL
#L
In writing the equality (7), we are supposing that the Dirichlet conditions are satisfied and further that f ðxÞ is continuous at x. If f ðxÞ is discontinuous at x, the left side of (7) should be replaced by ðf ðx þ 0Þ þ f ðx # 0Þ : 2 BOUNDARY-VALUE PROBLEMS Boundary-value problems seek to determine solutions of partial differential equations satisfying certain prescribed conditions called boundary conditions. Some of these problems can be solved by use of Fourier series (see Problem 13.24). EXAMPLE.
The classical problem of a vibrating string may be idealized in the following way. See Fig. 13-2.
Suppose a string is tautly stretched between points ð0; 0Þ and ðL; 0Þ. Suppose the tension, F, is the same at every point of the string. The string is made to vibrate in the xy plane by pulling it to the parabolic position gðxÞ ¼ mðLx # x2 Þ and releasing it. (m is a numerically small positive constant.) Its equation will be of the form y ¼ f ðx; tÞ. The problem of establishing this equation is idealized by (a) assuming that the constant tension, F, is so large as compared to the weight wL of the string that the gravitational force can be neglected, (b) the displacement at any point of the string is so small that the length of the string may be taken as L for any of its positions, and (c) the vibrations are purely transverse. w @2 y The force acting on a segment PQ is !x 2 ; g @t x < x1 < x þ !x; g & 32 ft per sec:2 . If # and $ are the angles that F makes with the horizontal, then the vertical Fig. 13-2
340
FOURIER SERIES
[CHAP. 13
difference in tensions is Fðsin ! " sin "Þ. This is the force producing the acceleration that accounts for ( ) " the vibratory motion. tan ! tan " @y ðx þ !x; tÞ" Now Ffsin ! " sin "g ¼ F pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi " pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi & Fftan ! " tan "g ¼ F 2 # 2 @x 1 þ tan " 1 þ tan ! @y ðx; tÞ , where the squared terms in the denominator are neglected because the vibrations are small. @x Next, equate the two forms of the force, i.e., " # @y @y w @2 y ðx þ !x; tÞ " ðx; tÞ ¼ !x 2 F @x @x g @t rffiffiffiffiffi Fg , the resulting equation is divide by !x, and then let !x ! 0. After letting ! ¼ w 2 @2 y 2@ y ¼ ! @t2 @x2 This homogeneous second partial derivative equation is the classical equation for the vibrating string. Associated boundary conditions are yð0; tÞ ¼ 0; yðL; tÞ ¼ 0; t > 0 The initial conditions are yðx; 0Þ ¼ mðLx " x2 Þ;
@y ðx; 0Þ ¼ 0; 0 < x < L @t
The method of solution is to separate variables, i.e., assume yðx; tÞ ¼ GðxÞHðtÞ Then upon substituting GðxÞ H 00 ðtÞ ¼ !2 G 00 ðxÞ HðtÞ Separating variables yields G 00 H 00 ¼ k; ¼ !2 k; where k is an arbitrary constant G H Since the solution must be periodic, trial solutions are pffiffiffiffiffiffi pffiffiffiffiffiffi GðxÞ ¼ c1 sin "k x þ c2 cos "k x; < 0 pffiffiffiffiffiffi pffiffiffiffiffiffi HðtÞ ¼ c3 sin ! "k t þ c4 cos ! "k t Therefore
y ¼ GH ¼ ½c1 sin
pffiffiffiffiffiffi pffiffiffiffiffiffi pffiffiffiffiffiffi pffiffiffiffiffiffi "k x þ c2 cos "k x(½c3 sin ! "k t þ c4 cos ! "k t(
The initial condition y ¼ 0 at x ¼ 0 for all t leads to the evaluation c2 ¼ 0. Thus pffiffiffiffiffiffi pffiffiffiffiffiffi pffiffiffiffiffiffi y ¼ ½c1 sin "k x(½c3 sin ! "k t þ c4 cos ! "k t(
pffiffiffiffiffiffi pffiffiffiffiffiffi Now the boundary condition y ¼ 0 at x ¼ L, thus 0 ¼ ½c1 sin "k L(½c3 sin ! "k t þ pffiffiffiimpose ffiffiffi c4 cos ! "k t(: c1 6¼ 0 as that would imply y ¼ 0 and a trivial solution. The next simplest solution results from the h pffiffiffiffiffiffi n# n# ih n# n# i t þ c4 cos ! t and the first factor is zero when choice "k ¼ , since y ¼ c1 sin x c3 sin ! L L l L x ¼ L.
CHAP. 13]
341
FOURIER SERIES
@y ðx; 0Þ ¼ 0, 0 < x < L can be considered. @t @y h n! ih n! n! n! n! i ¼ c1 sin x c3 " cos " t $ c4 " sin " t @t L L L L L
With this equation in place the boundary condition
At t ¼ 0
Since c1 6¼ 0 and sin
h n! i n! 0 ¼ c1 sin x c3 " L L n! x is not identically zero, it follows that c3 ¼ 0 and that L h n! ih n! n! i x c4 " cos " t y ¼ c1 sin L L L
The remaining initial condition is
yðx; 0Þ ¼ mðLx $ x2 Þ; 0 < x < L When it is imposed mðLx $ x2 Þ ¼ c1 c4 "
n! n! sin x L L
However, this relation cannot be satisfied for all x on the interval ð0; LÞ. Thus, the preceding extensive analysis of the problem of the vibrating string has led us to an inadequate form n! n! n! y ¼ c1 c4 " sin x cos " t L L L and an initial condition that is not satisfied. At this point the power of Fourier series is employed. In particular, a theorem of differential equations states that any finite sum of a particular solution also is a solution. Generalize this to infinite sum and consider y¼
1 X
bn sin
n¼1
n! n! x cos " t L L
with the initial condition expressed through a half range sine series, i.e., 1 X
bn sin
n¼1
n! x ¼ mðLx $ x2 Þ; L
t¼0
According to the formula of Page 338 for coefficient of a half range sine series ðL L n!x bn ¼ ðLx $ x2 Þ sin dx 2m L 0 That is L b ¼ 2m n
ðL
Lx sin
0
n!x dx $ L
ðL
x2 sin
0
n!x dx L
Application of integration by parts to the second integral yields ðL ðL L n!x L3 L n!x bn ¼ L x sin dx þ cos 2x dx cos n! þ 2m L L n! 0 0 n! When integration by parts is applied to the two integrals of this expression and a little algebra is employed the result is bn ¼
4L2 ð1 $ cos n!Þ ðn!Þ3
342
FOURIER SERIES
[CHAP. 13
Therefore, y¼
1 X
bn sin
n¼1
n! n! x cos " t L L
with the coefficients bn defined above.
ORTHOGONAL FUNCTIONS Two vectors A and B are called orthogonal (perpendicular) if A " B ¼ 0 or A1 B1 þ A2 B2 þ A3 B3 ¼ 0, where A ¼ A1 i þ A2 j þ A3 k and B ¼ B1 i þ B2 j þ B3 k. Although not geometrically or physically evident, these ideas can be generalized to include vectors with more than three components. In particular, we can think of a function, say, AðxÞ, as being a vector with an infinity of components (i.e., an infinite dimensional vector), the value of each component being specified by substituting a particular value of x in some interval ða; bÞ. It is natural in such case to define two functions, AðxÞ and BðxÞ, as orthogonal in ða; bÞ if ðb a
AðxÞ BðxÞ dx ¼ 0
ð9Þ
A vector A is called a unit vector or normalized vector if its magnitude is unity, i.e., if A " A ¼ A2 ¼ 1. Extending the concept, we say that the function AðxÞ is normal or normalized in ða; bÞ if ðb a
fAðxÞg2 dx ¼ 1
ð10Þ
From the above it is clear that we can consider a set of functions f#k ðxÞg; k ¼ 1; 2; 3; . . . ; having the properties ðb a
ðb a
#m ðxÞ#n ðxÞ dx ¼ 0
f#m ðxÞg2 dx ¼ 1
m 6¼ n
m ¼ 1; 2; 3; . . .
ð11Þ ð12Þ
In such case, each member of the set is orthogonal to every other member of the set and is also normalized. We call such a set of functions an orthonormal set. The equations (11) and (12) can be summarized by writing ðb a
#m ðxÞ#n ðxÞ dx ¼ $mn
ð13Þ
where $mn , called Kronecker’s symbol, is defined as 0 if m 6¼ n and 1 if m ¼ n. Just as any vector r in three dimensions can be expanded in a set of mutually orthogonal unit vectors i; j; k in the form r ¼ c1 i þ c2 j þ c3 k, so we consider the possibility of expanding a function f ðxÞ in a set of orthonormal functions, i.e., f ðxÞ ¼
1 X n¼1
cn #n ðxÞ
a@x@b
ð14Þ
As we have seen, Fourier series are constructed from orthogonal functions. Generalizations of Fourier series are of great interest and utility both from theoretical and applied viewpoints.
CHAP. 13]
343
FOURIER SERIES
Solved Problems FOURIER SERIES 13.1. Graph each of the following functions. ! 3 0 #5 < x < 0 > 3 0 < x < 5 > > : 3=2 x¼5
then the series will converge to f ðxÞ for #5 @ x @ 5.
13.6. Expand f ðxÞ ¼ x2 ; 0 < x < 2! in a Fourier series if (a) the period is 2!, (b) the period is not specified. (a) The graph of f ðxÞ with period 2! is shown in Fig. 13-7 below.
CHAP. 13]
347
FOURIER SERIES
f (x)
4p 2
_ 6p
_ 4p
_ 2p
x
O
2p
4p
6p
Fig. 13-7
Period ¼ 2L ¼ 2! and L ¼ !. Choosing c ¼ 0, we have ð ð 1 cþ2L n!x 1 2! 2 f ðxÞ cos x cos nx dx an ¼ dx ¼ L c L ! 0 $ " # " #%& " # 1 sin nx % cos nx % sin nx &&2! 4 2 ðx Þ þ2 ¼ % ð2xÞ & ¼ n2 ; ! n n2 n3 0 If n ¼ 0; a0 ¼
1 !
ð 2! 0
x2 dx ¼
8!2 : 3
ð n!x 1 2! 2 dx ¼ x sin nx dx L ! 0 c $ " # " #%&2! ' cos nx( 1 sin nx cos nx && %4! ¼ ¼ þ ð2Þ % ð2xÞ % 2 ðx2 Þ % & 3 ! n n n n 0
bn ¼
1 L
ð cþ2L
n 6¼ 0
f ðxÞ sin
# 1 " 4!2 X 4 4! sin nx : þ cos nx % Then f ðxÞ ¼ x ¼ n 3 n2 n¼1 2
This is valid for 0 < x < 2!. At x ¼ 0 and x ¼ 2! the series converges to 2!2 .
(b) If the period is not specified, the Fourier series cannot be determined uniquely in general.
13.7. Using the results of Problem 13.6, prove that
1 1 1 !2 . þ þ þ & & & ¼ 6 12 22 32
At x ¼ 0 the Fourier series of Problem 13.6 reduces to
1 4!2 X 4 þ . 2 3 n n¼1
By the Dirichlet conditions, the series converges at x ¼ 0 to 12 ð0 þ 4!2 Þ ¼ 2!2 . Then
1 1 X 4!2 X 4 1 !2 þ ¼ 2!2 , and so ¼ . 2 2 3 6 n n n¼1 n¼1
ODD AND EVEN FUNCTIONS, HALF RANGE FOURIER SERIES 13.8. Classify each of the following functions according as they are even, odd, or neither even nor odd. $ 2 0 0.
13.45. (a) Solve
(b) Give a possible physical interpretation of the problem and solution. Ans:
2
2
ðaÞ Uðx; tÞ ¼ 3e"2! t sin !x " 2e"50! t sin 5!x.
@U @2 U ¼ 2 subject to the conditions Uð0; tÞ ¼ 0; Uð6; tÞ ¼ 0; Uðx; 0Þ ¼ @t @x physically. % 1 $ X 1 " cosðm!=3Þ "m2 !2 t=36 m!x e 2 sin Ans: Uðx; tÞ ¼ m! 6 m¼1
13.46. Solve
#
1 0