Chapter 9. Fourier Series A. the Fourier Sine Series
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Vector Spaces in Physics 8/6/2015 Chapter 9. Fourier Series A. The Fourier Sine Series The general solution. In Chapter 8 we found solutions to the wave equation for a string fixed at both ends, of length L, and with wave velocity v, yn x, t A n sin k n x cos n t k n n L 1 n 2L , n = 1, 2, 3, . (9-1) n v n n L v fn n 2L We are now going to proceed to describe an arbitrary motion of the string with both ends fixed in terms of a linear superposition of normal modes: nx general solution, string , (9-2) y x, t Ann sin cos t n1 L fixed at both ends nv where and the coefficients An are to be determined. Here are some things to be noted: n L (1) We are assuming for the moment that this equation is true - that is, in the limit where we include an infinite number of terms, that the infinite series is an arbitrarily good approximation to the solution y x, t . (2) Each term in the series is separately a solution to the wave equation satisfying the boundary conditions, and so the series sum itself is such a solution. (3) This series is sort of like an expansion of a vector in terms of a set of basis vectors. In this picture the coefficients An are the coordinates of the function . (4) We still have to specify initial conditions and find a method to ensure that they are satisfied. This is the next order of business. Initial conditions. The solution above satisfies the boundary conditions appropriate to the problem, y0, t y L , t 0. But no initial conditions have been proposed. A complete set of initial conditions would consist of specifying the displacement of the string at some initial time, say t = 0, and the velocity of the string at the same time. These conditions might look as follows: 9 - 1 Vector Spaces in Physics 8/6/2015 y x,0 f x y . (9-3) gx() t t0 It might be noticed by the astute observer that if we take the partial derivative of equation (9-2) with respect to time and set t = 0, it vanishes identically. (Each term would have a factor of sinnt , which vanishes at t = 0.) This means that we have already built into this solution the condition that the string is not moving at t = 0; or, in terms of the initial conditions just stated, gx 0 . (9-4). This is a limitation; if we wanted equation (9-2) to be completely general, we would have to add another set of terms multiplied by factors of cosnt . This makes things quite a bit more complicated and does not add very much to the understanding of Fourier series; we will just live with the limitation to wave motions where the string is stationary at t = 0. This leaves us with the condition on the displacement at t = 0, which takes on the form nx y x,0 f x An sin . (9-5). n1 L The series nx f x An sin Fourier sine series. (9-6). n1 L is a very famous equation in mathematics, representing the statement that any function of x defined on the interval [0,L] and vanishing at the ends of the interval can be represented as a linear superposition of the sine waves vanishing at the ends of the interval. We will now spend some time seeing how this works. Orthogonality. nx If the functions sin are to play the role of basis vectors in this process, it would be nice if L they were orthogonal. To define orthogonality we need to define an inner product. For functions defined on the interval [0,L], a useful definition of an inner product is the following: L u, v u x v x dx inner product (9-7) x0 Orthogonality of two functions ux and vx means that uv,0 . For the functions , if this inner product is evaluated, the result is n x m xL n x m x L sin ,sin sin sin dx (9-8) nm LLLLx0 2 9 - 2 Vector Spaces in Physics 8/6/2015 where nm is the Kronecker delta symbol. (The proof of this fact is left to the problems.) So, the functions almost constitute an orthonormal basis for the space of functions we are 2 considering. They could be properly normalized by multiplying each one by a factor of . This L is not usually done, just to keep the equations simpler. Now we need a way to find the coefficients An ,n = 1, 2, 3, .... If we remember the method for determining the coordinates of vectors, it is very easy: 2 mx Am f x,sin LL inversion of Fourier series (9-9) 2 L mx f xsin dx LLx0 This important relation can be verified as follows: nx f x An sin n1 L 22LLm x n x m x f xsin dx A sin sin dx n LLLLLxx00n1 2 L n x m x Asin sin dx (9-10) n LLLn1 x0 2 L An nm L n1 2 Am Here we have of course used the orthogonality relation, equation 9-8. Thus for any given initial condition fx we can calculate the coefficients An, and use equation 9-2 to calculate the position nx at any time, to anysin desired accuracy. L Completeness. nx The functions sin are not only orthogonal. They are a “complete” representation of L functions of x over the interval 0 ≤ x ≤ L, meaning that such functions can be represented arbitrarily well by a linear combination of these sine functions. A somewhat more precise statement of this completeness condition is that given by Dirichlet: For any function which is piecewise continuous over the interval [0,L], the Fourier series nx An sin (9-11) n1 L converges to f(x) at every point. At points of discontinuity, the series converges to the average of f(x-) and f(x+). 9 - 3 Vector Spaces in Physics 8/6/2015 Figure9-1. Initial conditions for the triangle wave, initially displaced by an amount h at the center. So we have found one of the most important expansions of a function as a series of orthogonal functions: The Fourier Sine Series 0 < x < L nx f x An sin (9-12) n1 L with 2 L mx A f x sin dx m LLx0 Example. Calculate the Fourier coefficients of a Fourier sine series for the symmetrical triangle wave, shown in figure 9-1. Solution. The function f(x) can be written 9 - 4 Vector Spaces in Physics 8/6/2015 2hL xx,0 L 2 fx (9-13) 2hL L x, x L L 2 and then the coefficients An can be evaluated: 2 L mx A f xsin dx m LLx0 2LL/2 2h m x 2 2 h m x xsin dx L x sin dx (9-14) LLLLLLx0 x L / 2 4hLLL/2 m x 4 h m x 4 h m x xsin dx x sin dx sin dx 22 LLx0LLLL x L / 2 x L / 2 There are various ways to do tiresome integrals like these. My preferred method is to change to mx dimensionless variables, then do the integrals by parts. So I will change to the variable u , L giving 2 44h Lm/2 m h L m A usin u du u sin u du sin udu m 2 L mu0 u m / 2 L m u m / 2 mm/2 44hhm/2 m m uucos uu cos cos udu sin udu cos u 22u0 m / 2 u m / 2 m u0 u m / 2 m 4h m m m / 2 m 4hm 22 2 cos m cos m sin u sinum cos cos m 22 u0 um /2 m 2 8hm sin m22 2 (9-15) This finally simplified rather well. We could leave the result as it is, but another change shows the m pattern to the values of the coefficients better. We observe that sin vanishes for m even; and 2 for m odd, m m1 sin 1 2 (m odd) (9-16) 2 Thus the result is 0,m even m1 Am 8h (9-17) 1 2 ,m odd m22 The corresponding series for f(x) is 9 - 5 Vector Spaces in Physics 8/6/2015 8h x 8 h 3 x 8 h 5 x fx sin sin sin . . .. 2LLL9 2 25 2 (9-18) x35 x x h .81057 sin .09006 sin .03242 sin . . .. LLL The contribution from each of the first five terms is plotted in figure 9-2. Note that it is possible just from comparing the function to be represented with each normal mode to determine which ones will have a zero coefficient. For the case of the triangle wave, the n = 2 and n = 4 sine waves are anti-symmetric about the center of the interval (x = 0.5), and the triangle wave is symmetric about this point. This makes it pretty clear that none of the even-n modes are going to contribute. It is also fairly clear that for the odd-n modes, the contributions from integrating over the intervals [0,0.5] and [0.5,1] will be equal. So, retrospectively, one could have just done the first half of the integral (the easier half), for odd n, multiplying by 2 to account for the other half of the integral. 9 - 6 Vector Spaces in Physics 8/6/2015 nx Figure 9-2.