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FOURIER SERIES
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DIFFERENTIATION AND INTEGRATION OF FOURIER SERIES Di erentiation and integration of Fourier series can be justi ed by using the theorems on Pages 271 and 272, which hold for series in general. It must be emphasized, however, that those theorems provide su cient conditions and are not necessary. The following theorem for integration is especially useful. Theorem. The Fourier series corresponding to f x may be integrated term by term from a to x, and the
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resulting series will converge uniformly to
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f x dx provided that f x is piecewise continuous in
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COMPLEX NOTATION FOR FOURIER SERIES Using Euler s identities, 6 p where i 1 (see Problem 11.48, 11, Page 295), the Fourier series for f x can be written as f x where cn 1 2L L
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ei cos  i sin ;
e i cos  i sin 
cn einx=L
f x e inx=L dx
In writing the equality (7), we are supposing that the Dirichlet conditions are satis ed and further that f x is continuous at x. If f x is discontinuous at x, the left side of (7) should be replaced by f x 0 f x 0 : 2 BOUNDARY-VALUE PROBLEMS Boundary-value problems seek to determine solutions of partial di erential 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 0; 0 and L; 0 . 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 g x m Lx x2 and releasing it. (m is a numerically small positive constant.) Its equation will be of the form y f x; t . 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 x; g % 32 ft per sec:2 . If and are the angles that F makes with the horizontal, then the vertical Fig. 13-2
FOURIER SERIES
[CHAP. 13
di erence in tensions is F sin sin . This is the force producing the acceleration that accounts for ( ) & the vibratory motion. tan tan @y x x; t Now Ffsin sin g F p p % Fftan tan g F 2 ' 2 @x 1 tan 1 tan @y x; t , 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 x x; t x; t x 2 F @x @x g @t r Fg , the resulting equation is divide by x, and then let x ! 0. After letting w 2 2 @ y @ y 2 2 @t2 @x This homogeneous second partial derivative equation is the classical equation for the vibrating string. Associated boundary conditions are y 0; t 0; y L; t 0; t > 0 The initial conditions are y x; 0 m Lx x2 ; @y x; 0 0; 0 < x < L @t
The method of solution is to separate variables, i.e., assume y x; t G x H t Then upon substituting G x H 00 t 2 G 00 x H t Separating variables yields G 00 H 00 k; 2 k; where k is an arbitrary constant G H Since the solution must be periodic, trial solutions are p p G x c1 sin k x c2 cos k x; < 0 p p H t c3 sin k t c4 cos k t Therefore y GH c1 sin p p p p k x c2 cos k x c3 sin k t c4 cos k t
The initial condition y 0 at x 0 for all t leads to the evaluation c2 0. Thus p p p y c1 sin k x c3 sin k t c4 cos k t
p p Now p impose the boundary condition y 0 at x L, thus 0 c1 sin k L c3 sin k t c4 cos k t : c1 6 0 as that would imply y 0 and a trivial solution. The next simplest solution results from the h p n n ih n n i t c4 cos t and the rst factor is zero when choice k , since y c1 sin x c3 sin L L l L x L.
CHAP. 13]
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