Fourier Series. Department of Mathematical and Statistical Sciences University of Alberta

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1 1 Lecture Notes on Partial Differential Equations Chapter IV Fourier Series Ilyasse Aksikas Department of Mathematical and Statistical Sciences University of Alberta

2 DEFINITIONS 2 Before discussing how to solve linear partial differential equations by the method of separation of variables, it is important to summarize some of the important results on Fourier series. Joseph Fourier developed this type of series in his famous treatise on heat flow in the early 18s. 1 Definitions Definition 1.1. A function f is said to be piecewise continuous on [a, b] if there exists finitely many points a = x 1 < x 2 <... < x n = b such that f is continuous on [x i, x i+1 ] and lim x x + i f(x), lim x x i+1 f(x) exist for i = 1, 2,...,n 1.

3 DEFINITIONS 3 Fig. 1 Piecewise continuous function. Definition 1.2. A function f is said to be periodic, with a period p if f(x + p) = f(x) for all x. Example f(x) = sin(x) is a periodic function with period p = 2π. 2. f(x) = tan(x) is a periodic function with a period p = π.

4 DEFINITIONS 4 Fig. 2 Periodic function. Definition 1.3. Two functions f and g are said to be orthogonal with respect to the weight function q on [a, b] if b a f(x)g(x)q(x)dx =

5 DEFINITIONS 5 For a given function q, it is often possible to find an infinite sequence of functions φ 1, φ 2,... such that b a φ n (x)φ m (x)q(x)dx = if m n The sequence {φ n } n 1 is called an orthogonal system. The norm of φ n is defined as follows b φ n = φ 2 n(x)q(x)dx Moreover if φ n = 1, {φ n } n 1 is called an orthonormal system. Example 1.2. a 1. {sin(nx)} n 1 is an orthogonal system on the interval [, π] with respect to q(x) = 1. Indeed, sin(nx) sin(mx)dx = 1 2 [cos((m n)x) cos((m + n)x)] dx

6 FOURIER SERIES 6 For n m [ 1 = 2(m n) sin((m n)x) 1 2(m + n) 2 2. { π sin(nx)} n 1 is an orthonormal system. ] π sin((m + n)x) =. 2 Fourier Series Let f be a piecewise continuous function defined on [, π]. Then one can express f as a linear combination of sine and cosine functions as follows : f(x) 1 2 a + a k cos(kx) + b k sin(kx) where a, a k and b k are constants to be determined.

7 FOURIER SERIES 7 In order to find a, one has ( ) π 1 f(x)dx = 2 a + a k cos(kx) + b k sin(kx) dx = 1 2 a 2π + (a k cos(kx)dx + b k ) sin(kx)dx = a = 1 π f(x)dx On the other hand, to find a n, f(x) cos(nx)dx = ( 1 2 a cos(nx) + ) a k cos(kx) cos(nx) + b k sin(kx) cos(nx) dx

8 FOURIER SERIES 8 Similarly = + a n = a n = 1 π b n = 1 π cos(nx) cos(nx)dx + = a n π f(x) cos(nx)dx, n = 1, 2,... f(x) sin(nx)dx, n = 1, 2,... a n and b n are called Fourier coefficients. Example 2.1. Let us consider the function f(x) = { 1 x < 1 x π By using the formula of a, a n and b n, we find that a = 1 π f(x)dx = 1 π ( 1)dx + 1 π (1)dx =.

9 FOURIER SERIES 9 a n = 1 π = 1 π f(x) cos(nx)dx = 1 cos(nx)dx + 1 π π ( [ 1 n sin(nx)] + [ 1 ) n sin(nx)]π =. cos(nx)dx b n = 1 π = 1 π f(x) sin(nx)dx = 1 sin(nx)dx+ 1 π π ( [ 1 n cos(nx)] + [ 1 ) n cos(nx)]π sin(nx)dx = 1 ([cos() cos( nπ)] + [cos() cos(nπ)]) nπ = 2 2 (1 cos(nπ)) = nπ nπ (1 ( 1)n ). Thus the Fourier series of f is given by 2 f(x) nπ (1 ( 1)n )sin(nx). n=1

10 FOURIER SINE + COSINE SERIES 1 Theorem 2.1. If f is a periodic function with period 2π and f and f are piecewise continuous on [, π], then the fourier series 1 2 a + a k cos(kx) + b k sin(kx) is convergent. The sum of the Fourier series is equal to f(x) at all numbers x where f is continuous. At the numbers x where f is discontinuous, the sum of the Fourier series is the average of the right and the left limits, that is 1 2 (f(x+ ) + f(x )). 3 Fourier Sine + Cosine Series In this section we show that the series of sines only (and the series of cosines only) are special cases of a Fourier series.

11 FOURIER SINE + COSINE SERIES Fourier Sine Series Definition 3.1. An odd function is a function with the property f( x) = f(x). Example f(x) = x f(x) = sin(4x). Remark 3.1. The integral of an odd function over a symmetric interval is zero. Let us calculate the Fourier coefficients of an odd function : b n = 1 π a n = 1 π a = 1 π f(x) sin(nx)dx = 2 π f(x)dx = f(x) cos(nx)dx = f(x) sin(nx)dx

12 FOURIER SINE + COSINE SERIES 12 Since a n =, all the cosine functions will not appear in the Fourier series of an odd function. The Fourier series of an odd function is an infinite series of odd functions (sines) : f(x) b n sin(nx) n=1 Now assume that the function f is only given for x π and not necessarily odd. In this case f can be extended as an odd function. This extension is called the odd extension of f. Moreover the Fourier series of the odd extension of f only involves sines : the odd extension of f(x) b n sin(nx), x π. n=1 However, we are only interested in what happens [, π]. In this interval f is identical to its odd extension :

13 FOURIER SINE + COSINE SERIES 13 where f(x) b n sin(nx), x π, n=1 b n = 2 π f(x) sin(nx)dx 3.2 Fourier Cosine Series Definition 3.2. An even function is a function with the property f( x) = f(x). The sine coefficients of a Fourier series will be zero for an even function, b n = 1 π f(x) sin(nx)dx =. The Fourier series of an even function is an infinite series of even

14 FOURIER SINE + COSINE SERIES 14 functions (cosines) : f(x) a 2 + n=1 a n cos(nx). The coefficients of the cosines may be evaluated using information about f(x) only on [, π], since a n = 1 π a = 1 π f(x)dx = 2 π f(x) cos(nx)dx = 2 π f(x)dx f(x) cos(nx)dx If the function f is only given for x π and not necessarily even, then f can be extended as an even function, which called the even extension of f. Moreover the Fourier series of the even extension of f only involves cosines :

15 FOURIER SINE + COSINE SERIES 15 the even extension of f(x) a n cos(nx), x π. n= In the interval [, π], the function f is identical to its even extension : where a n = 2 π f(x) a n cos(nx), x π, n=1 a n = 2 f(x) sin(nx)dx π Example 3.2. Let us consider the function f(x) = 1 on [, π]. The Fourier cosine series has coefficients a = 2 π 1dx = 2 cos(nx)dx =

16 DIFFERENTIATION AND INTEGRATION OF FOURIER SERIES 16 Then f(x) a 2 + n=1 a n cos(nx) = 1. 4 Differentiation and Integration of Fourier Series Let us consider a continuous function on the interval [, π], with Fourier series f(x) a 2 + then f has a Fourier series f (x) n=1 a n cos(nx) + b n sin(nx), na n sin(nx) + nb n cos(nx). n=1 That is f (x) can be written as a Fourier series

17 DIFFERENTIATION AND INTEGRATION OF FOURIER SERIES 17 where f (x) ã n cos(nx) + b n sin(nx), n=1 ã n = b n n = n π bn = a n n = n π f(x) cos(nx)dx f(x) sin(nx)dx. For any function f on the interval [, π], its integral has Fourier series (assume that a = ) x F(x) = f( x)d x ã 2 + ã n cos(nx) + b n sin(nx) where ã n = 1 n b n, n=1 bn = 1 n a n, and ã = 1 π f(x)dx.

18 DIFFERENTIATION AND INTEGRATION OF FOURIER SERIES 18 Example 4.1. Let us consider the function The Fourier series is The Fourier series of is given by F(x) = f(x) = x { 1 x < 1 x π kodd f( x)d x = F(x) = π 2 + kodd 4 kπ sin(kx). { 1 x < 1 x π 4 k 2 π cos(kx).,

19 COMPLEX FOURIER SERIES 19 5 Complex Fourier Series Recall that e iθ = cos(θ) + isin(θ) then cos(θ) = eiθ + e iθ and sin(θ) = eiθ e iθ 2 2i Thus the Fourier series can be written using complex variables f(x) a 2 + = a 2 + = a 2 + a k cos(kx) + b k sin(kx) a k [ e ikx + e ikx 2 (a k ib k ) 2 ] [ e ikx e ikx ] + b k 2i e ikx + (a k + ib k ) e ikx 2

20 COMPLEX FOURIER SERIES 2 where = c + c k = (a k ib k ) 2 c k e ikx + c k e ikx c k = (a k + ib k ) 2 = 1 2π = 1 2π c = a 2 = 1 2π f(x)dx f(x)(cos(kx) isin(kx))dx = 1 2π In summary, the complex Fourier series is f(x) c k e ikx where c k = 1 2π f(x)(cos(kx)+isin(kx))dx = 1 2π f(x)e ikx dx. f(x)e iπx dx f(x)e iπx dx

21 COMPLEX FOURIER SERIES 21 Example 5.1. Let us consider the function f(x) = { 1 < x < 1 < x < π. The complex Fourier series coefficients are c k = 1 ( ) 1e ikx dx + 1e ikx dx 2π { ( ) 1 1 = 2π ik e ikx ] + 1 ik e ikx ] π, k 1 2π ( + π), k = { 1 ( = 2kiπ 1 e ikπ e ikπ + 1 ), k, k = { i ( = kπ (eikπ e ikπ ) ), k, k = = { i kπ (1 cos(kπ))), k, k =

22 FOURIER SERIES OVER ANY INTERVAL 22 = { 2i kπ, Then k odd, k even f(x) kodd 2i kπ eikx. 6 Fourier Series over any Interval In general, Fourier series, sine and cosine series may be defined not just over [, π] but over any interval [a, b]. Let us consider a function F(t) periodic with a period 2π and t [, π], then the Fourier series of F is given by F(t) a 2 + a k cos(kt) + b k sin(kt),

23 FOURIER SERIES OVER ANY INTERVAL 23 where a k = 1 π f(t) cos(kt)dt and b k = 1 π f(t) sin(kt)dt Let a and b be two constants. Define the new variable x as follows x = 1 1 (b + a) + (b a)t [a, b] 2 2π ( ) 2x b a t = π. b a Now let us define the function f by ( ) 1 1 f(x) = f (b + a) + (b a)t = F(t) 2 2π The Fourier series of the function f is f(x) a 2 + ( ) (2x b a) a k cos kπ +b k sin b a ( ) (2x b a) kπ b a

24 FOURIER SERIES OVER ANY INTERVAL 24 where and a k = 2 b a b k = 2 b a b a b a f(x) cos f(x) sin ( ) (2x b a) kπ dx b a ( ) (2x b a) kπ dx. b a In particular, if b = l, a = l and f is periodic with period 2l f(x) a 2 + ( ) ( ) kπx kπx a k cos + b k sin l l where a k = 1 l l l ( ) kπx f(x) cos dx and b k = 1 l l l l ( ) kπx f(x) sin dx l Example 6.1. Let us consider the function f defined as follows {, 2 x < f(x) = 2 x, < x 2.

25 FOURIER SERIES OVER ANY INTERVAL 25 By using the formula of a, a n and b n, we find that and a k = a = (2 x)dx, (2 x) cos kπx 2 dx, k = 1, 2,... b k = 1 (2 x) sin kπx dx, k = 1, 2, Evaluating these integrals gives a = 1 2, a k = 2 k 2 π 2 [1 ( 1)k ] and b k = 2 kπ where use has been made of the fact that cos(kπ) = ( 1) k and sin(kπ) =. Thus the Fourier series becomes f(x) = ( [1 ( 1) k ] π k 2 cos kπx π ) kπx sin. k 2

26 FOURIER SERIES OVER ANY INTERVAL 26 The sine series of a function f on [, l] is where f(x) a 2 + a k = 2 l l a k cos kπx l f(x) cos kπx l The cosine series of a function f on [, l] is where f(x) a k = 2 l l a k sin kπx l dx. f(x) sin kπx dx. l