# Fourier Series and Sturm-Liouville Eigenvalue Problems

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1 Fourier Series and Sturm-Liouville Eigenvalue Problems 2009

2 Outline Functions Fourier Series Representation Half-range Expansion Convergence of Fourier Series Parseval s Theorem and Mean Square Error Complex Form of Fourier Series Inner Products Orthogonal Functions Self-adjoint Operators Sturm-Liouville Eigenvalue Probelms

3 Periodic Functions Definition (Periodic Function) A 2L-periodic function f : R R {± } is a function such that there exists a constant L > 0 such that f(x) = f(x + 2L), x R. (1) Here 2L is called the fundamental period or just period. For example, f(x) = sin(x) is a 2π-periodic funtion with period 2π, since sin(x + 2π) = sin(x).

4 Even and Odd Functions Definition (Even and Odd Functions) A function f is even if and only if f( x) = f(x), x. A function f is odd if and only if f( x) = f(x), x. For example sin x is an odd function since sin( x) = sin x. cos x is an even function since cos( x) = cos x. Note that even function is symmetric about the y-axis. On the other hand, odd function is symmetric about the origin.

5 Examples of Even and Odd Functions

6 Piecewise Continuous Functions Definition (Piecewise Continuous Functions) A function f is said to be piecewise continuous on the interval [a, b] if 1. f(a+) and f(b ) exist, and 2. f is defined and continous on (a, b) except at a finite number of points in (a, b) where the left and right limits exits

7 Piecewise Smooth Functions Definition (Piecewise Smooth Functions) A function f, defined on the interval [a, b], is said to be piecewise smooth if f and f are piecewise continous on [a, b]. Thus f is piecewise smooth if 1. f is piecewise continous on [a, b], 2. f exists and is continous in (a, b) except possibly at finitely many points c where the one-sided limits lim x c f (x) and lim x c + f (x) exist. Furthermore, lim x a + f (x) and lim x b f (x) exist.

8 Examples of Piecewise Functions Figure: A piecewise smooth function. Figure: Another piecewise smooth function.

9 Some useful integrations If f is even. Then, for any a R a f(x) dx = 2 a a 0 If f is odd. Then, for any a R a a f(x) dx = 0. f(x) dx. If f is piecewise continuous and 2L-periodic. Then, for any a R 2L f(x) dx = 0 a a+2l f(x) dx.

10 Orthogonal Functions Definition (Orthogonal Functions) Two functions f and g are said to be orthogonal in the interval [a, b] if b a f(x)g(x) dx = 0. (2) We will come back to orthogonal functions again later.

11 Orthogonal Properties of Trigonometric Functions The orthogonal properties of sine and cosine functions are summarised as follow: π π π π π π where δ mn is the Kronecker s delta. cos mx sin nx dx = 0, (3) cos mx cos nx dx = πδ mn, (4) sin mx sin nx dx = πδ mn (5)

12 Kronecker s Delta Definition (Kronecker s Delta) δ mn = { 1, m = n 0, m n. (6)

13 Fourier Series Theorem (Fourier Series Representation) Suppose f is a 2L-periodic piecewise smooth function, then Fourier series of f is given by f(x) = a (a n cos nωx + b n sin nωx) (7) n=1 and the Fourier series converges to f(x) if f is continuous at x and to 1 [f(x+) + f(x )] otherwise. 2 Here ω = 2π is called the fundamental frequency, while the 2L amplitudes a 0, a n, and b n are called Fourier coefficients of f and they are given by the Euler formula.

14 Euler Formula Definition (Euler Formula) The Fourier coefficients for a 2L-periodic function f are given by a 0 = 1 L a n = 1 L b n = 1 L L L L f(x) dx, L f(x) cos nωx dx = 1 L L for n = 1, 2,.... L f(x) sin nωx dx = 1 L L L L L f(x) cos nπx L f(x) sin nπx L dx, dx.

15 Collorary If f is even and 2L-periodic, then the Fourier series representation is f(x) = a a n cos nωx. n=1 If f is odd and 2L-periodic, then the Fourier series representation is f(x) = b n sin nωx. n=1 Here, a 0, a n, and b n are given by the Euler formula.

16 Examples of Fourier Series (Odd function, digital impulses) Find the Fourier representation of the periodic function f(x) with period 2π, where { 1, π < x < 0, f(x) = 1, 0 < x < π. [Answer:] f(x) = n odd 4 sin nx nπ = k=1 4 sin(2k 1)x (2k 1)π.

17 Graphs for Digital Pulse Train and its Fourier Series

18 Examples of Fourier Series (Even function) Find the Fourier series for f(x) = x if 1 < x < 1 and f(x + 2) = f(x). [Answer:] f(x) = 1 2 n=1 4 cos(2n 1)x (2n 1) 2 π 2.

19 Graphs for f(x) = x, f(x + 2) = f(x)

20 Examples of Fourier Series Find the Fourier series of the 2-periodic function f(x) = x 3 + π if 1 < x < 1. [Answer:] f(x) = π + n=1 ( 1) n 12 2n2 π 2 n 3 π 3 sin nπx.

21 Graphs for f(x) = x 3 + π, f(x + 2) = f(x)

22 Full-range Extensions Consider a function f that is only defined in the interval [0, p). We could always extension the function outside the range to produce a new function. Of course, we have infinite many ways to extend the function, but here we will focus only on three specific extensions. Definition (Full-range Periodic Extension) The full-range periodic extension g of a function f defined in [0, p) is a p-periodic function given by g(x) = f(x) if 0 x < p, g(x) = g(x + p) if x < 0, g(x) = g(x p) if x p.

23 Half-range Periodic Extensions Definition (Half-range Even Periodic Extension) The half-range even periodic extension f e of a function f defined in [0, p) is a 2p-periodic even function given by { f(x) 0 x < p, f e (x) = f( x) p x < 0.

24 Half-range Periodic Extensions Definition (Half-range Odd Periodic Extension) The half-range odd periodic extension f o of a function f defined in [0, p) is a 2p-periodic odd function given by { f(x) 0 x < p, f e (x) = f( x) p x < 0.

25 Examples Periodic Extensions

26 Full-range Fourier Series for f defined on [0, p) Theorem If f(x) is a piecewise smooth function defined on an interval [0, p), then f has a full-range Fourier series expansion f(x) = a (a n cos nωx + b n sin nωx), 0 x < p, (8) n=1 where ω = 2π and the Fourier coefficients p a 0 = 1 p f(x) dx, a n = 1 p f(x) cos nωx dx, and p/2 0 p/2 0 b n = 1 p f(x) sin nωx dx. p/2 0

27 Fourier Cosine Series for f defined on [0, p) Theorem If f(x) is a piecewise smooth function defined on an interval [0, p), then f has a half-range Fourier cosine series expansion f(x) = a a n cos nωx, 0 x < p, (9) n=1 where ω = π p and the Fourier coefficients a 0 = 2 p and a n = 2 p p 0 f(x) cos nωx dx. p 0 f(x) dx,

28 Fourier Since Series for f defined on [0, p) Theorem If f(x) is a piecewise smooth function defined on an interval [0, p), then f has a half-range Fourier sine series expansion where ω = π p b n = 2 p p 0 f(x) = b n sin nωx, 0 x < p, (10) n=1 and the Fourier coefficients f(x) sin nωx dx.

29 Examples Consider a signal f(t) = t measured from an experiment over the duration given by 0 <= t < 4. Sketch the full-range periodic extension of f(t). Find the corresponding Fourier expansion of f(t). Sketch the half-range even extension of f(t) and find the corresponding Fourier cosine expansion of f(t). Sketch the half-range odd extension of f(t) and find the corresponding Fourier sine expansion of f(t).

30 Convergence of Fourier Series In the Fourier Series Representation Theorem, we were saying that for every 2L-periodic piecewise smooth function f, we could construct a partial sum s N (x) = a N 2 + (a n cos nωx + b n sin nωx). n=1 And, when N, the partial sum s N (x) converges to f(x), if f(x) is continuous for all x; 1 [f(x+) + f(x )], at the discontinuous points, or jumps. 2

31 Figure: s N (x) converges to f(x), except at the jumps. Figure: s N (x) converges uniformly on the interval [-1, 1]

32 Pointwise Convergence Definition (Pointwise Convergence) A sequence of functions {s n } is said to converge pointwise to the function f on the set E, if the sequence of numbers {s n (x)} converges to the number f(x), for each x in E. Or, if x E, s n (x) f(x), then {s n } converge pointwise to f.

33 Uniform Convergence Definition (Uniform Convergence) We say that s n converges to f uniformly on a set E, and we write s n f uniformly on E if, given ɛ > 0, we can find a positive integer N such that for all n N s n (x) f(x) < ɛ, x E. Definition (Uniform Convergence Series) A series s(x) = k=0 u k(x) is said to converge uniformly to f(x) on a set E if the sequence of partial sums s n (x) = n k=0 u k(x) converges uniformly to f(x).

34 Note that if a sequence of partial sums s n converges uniformly to f, then s n is also pointwise convergence. However, the converse is not always true. In order to determine is a s n is uniformly convergence, we use the WeierstraßM-test.

35 WeierstraßM-Test Theorem (WeierstraßM-Test) Let {u k } k=0 be a sequence of real- or complex-valued functions on E. If there exists a sequence {M k } k=0 of nonnegative real numbers such that the following two conditions hold: u k (x) M k, x E, and Then k=0 M k <. u k (x) converges uniformly on E. k=0

36 Gibbs Phenomena Here is an example of non-uniform convergence. The peaks remain same height but the width of the peaks changes. N=10 N=30 N= sqr(x) S10(x) 1.5 sqr(x) S30(x) 1.5 sqr(x) S50(x) N=10 N=30 N= err(x) 0.8 err(x) 0.8 err(x)

37 Mean Square Error Since s N converges to f only when N, for most practical purposes, we need N to be large but finite. Thus, we are approximating f with s N, and it is important for us to keep track of the error of the approximation. Definition (Mean Square Error) The mean square error of the partial sum s N relative to f is E N = 1 2L = 1 2L L L L [f(x) s N (x)] 2 dx L[f(x)] 2 dx 1 4 a N n=1 ( ) a 2 n + b 2 n.

38 Mean Square Approximation Theorem (Mean Square Approximation) Suppose that f is square integrable, i.e. L L f(x) 2 dx on [ L, L]. Then s N, the Nth partial sum of the Fourier series of f, approximates f in the mean square sense with an error E N that decreases to zero as N. lim E N = 1 L N 2L L[f(x)] 2 dx 1 4 a N n=1 ( ) a 2 n + b 2 n = 0.

39 Bessel s Inequality and Parseval s Identity Since E N > 0, from the definition of E N, we get Definition (Bessel s Inequality) 1 4 a N n=1 (a 2 n + b 2 n) 1 2L L L [f(x)] 2 dx. A stronger result is when taking the limit N Definition (Parseval s Identity) 1 4 a n=1 (a 2 n + b 2 n) = 1 2L L L [f(x)] 2 dx.

40 Multiplication Theorem A generalisation of the Parseval s identity is the multiplication theorem. Theorem (Multiplication (Inner Product) Theorem) If f and g are two 2L-periodic piecewise smooth functions 1 L f(x)g(x) dx = 2L L n= c n d n where c n and d n are the Fourier coefficients for the complex Fourier series of f and g respectively.

41 Complex Form of Fourier Series Theorem (Complex Form of Fourier Series) Let f be a 2L-periodic piecewise smooth function. The complex form of the Fourier series of f is n= c n e inωx, where the frequency ω = 2π/2L and the Fourier coefficients c n = 1 L 2L L f(x)e inωx dx, n = 0, ±1, ±2,.... For all x, the complex Fourier series converges to f(x) if f is continuous at x, and to 1 [f(x+) + f(x )] otherwise. 2

42 Relations of Complex and Real Fourier Coefficients c n = c n; c 0 = 1 2 a 0 c n = 1 2 (a n ib n ); c n = 1 2 (a n + ib n ). a 0 = 2c 0 ; a n = c n + c n ; b n = i(c n c n ).

43 Complex Form of Parseval s Identity Theorem (Complex Form of Parseval s Identity) Suppose f is a square integrable 2L-periodic piecewise smooth function on [ L, L]. Then 1 2L L L [f(x)] 2 dx = n= c n c n = n= where c n is the complex Fourier coefficients of f. c n 2

44 Example Find the complex Fourier series for the 2π-periodic function f(x) = e x defined in ( π, π).

45 Frequency Spectra The distribution of the magnitude of complex Fourier coefficients c n in frequency domain is called the amplitude spectrum of f. The distribution of p 0 = c 0 2 and p n = c n 2 in frequency domain is called the power spectrum of f.

46 Inner Products Definition (Inner Products) Let ψ and φ be (possibly complex) functions of x on the interval (a, b). Then the inner product of ψ and φ is ψ φ = b a ψ (x)φ(x) dx. Note that the notation of inner product in some books is (φ, ψ) = b a ψ (x)φ(x) dx. Please take note on the order of φ and ψ in the brackets.

47 Norm Definition (Norm) Let f be (possibly complex) function of x on the interval (a, b). Then the norm of f is ψ = ( b 1/2 f f = f(x) dx) 2. a

48 Orthogonal Functions Definition (Orthogonal Functions) The functions f and g are called orthogonal on the interval (a, b) if their inner product is zero, f g = b a f (x)g(x) dx = 0. Definition (Orthogonal Set of Functions) A set of functions {F 1, F 2, F 3,... } defined on the interval (a, b) is called an orthogonal set if F n 0 for all n; and F m F n = 0, for m n.

49 Normalisation and Orthonormal Set Definition (Normalisation) A normalised function f n for a function F n, with F n 0, is defined as f n (x) = F n(x) F n. Definition (Orthonormal Set of Functions) A set of functions {f 1, f 2, f 3,... } defined on the interval (a, b) is called an orthonormal set if f n = 1 for all n; and f m f n = 0, for m n; or simply, f m f n = δ mn, where δ mn is the Kronecker s delta.

50 Generalized Fourier Series Theorem (Generalized Fourier Series) If {f 1, f 2, f 3,... } is a complete set of orthogonal functions on (a, b) and if f can be represented as a linear combination of f n, then the generalised Fourier series of f is given by f(x) = where a n = fn f f n 2 a n f n (x) = n=1 n=1 f n f f n 2 f n(x), is the generalised Fourier coefficient.

51 Parseval s Identity Theorem (Generalized Parseval s Identity) If {f 1, f 2, f 3,... } is a complete set of orthogonal functions on (a, b) and let f be such that f is finite. Then b a f(x) 2 dx = n=1 f n f 2 f n 2.

52 Orthogonality with respect to a Weight, w(x) (Inner product) f g = b a (Orthogonality) f m f n = f (x)g(x)w(x) dx b a (Generalised Fourier Series) f n f f(x) = f n f n(x) 2 n=1 f m(x)f n (x)w(x) dx = f m 2 δ mn (Generalised Parseval s Identity) b f(x) 2 f n f 2 w(x) dx = f n 2 a n=1

53 Adjoint and Self-adjoint Operators Definition (Adjoints of Differential Operators) Suppose u and v are (possible complex) functions of x in (a, b) and let L be a linear differential operator. Then, the formal adjoint M of L is another operator such that for all u and v f L[g] = M[f] g. Definition (Self-adjoint Operators) Suppose M is the formal adjoint operator for a linear operator L in space S. If M = L, then the operator L is said to be formally self-adjoint or formally Hermitian.

54 Example: Adjoint for L[φ] p(x)dφ/dx Suppose L[φ] p(x)dφ/dx, then u L[v] = b [ u p d a dx = M[u] v ] v dx = [u pv] ba b a [ ] d v dx (p u) dx In the last step, we set the boundary term to zero. The the adjoint for the operator L consists of Formal adjoint M[φ] d dx (p φ); and Boundary conditions [u pv] b a = 0. Furhermore, if p(x) is a pure imaginary constant, then M = L. i.e. L is self-adjoint.

55 Adjoint for 2 nd Order Linear Differential Operators Suppose L[φ] a 0 (x)φ + a 1 (x)φ + a 2 (x)φ. Then, u L[v] = [u a 0 v + u a 1 v v(a 0 u ) ] b a + b a v [(a 0 u ) (a 1 u ) + a 2 u ] dx = M[u] v with appropriate choice of boundary conditions. Note that M[φ] a 0 φ + (2a 0 a 1 )φ + (a 2 a 1 + a 0)φ. M can be made self-adjoint if a 1 = a 0. The self-adjoint operator S is S[φ] = d ( a 0 (x) dφ ) + a 2 (x)φ. dx dx The neccessary boundary condition is [a 0 u v a 0 v(u ) ] b a = 0.

56 Eigenvalue Problems & Sturm-Liouville Equation Definition (Eigenvalue Problem) The eigenvalue problem associated to a differential operator L is the equation Ly + λy = 0, where λ is called the eigenvalue, and y is called the eigenfunction. It is possible to find a weight factor w(x) > 0 for L such that S[y] w(x)l[y] is self-adjoint. The resulting eigenvalue equation is called the Sturm-Liouville Equation Definition (Sturm-Liouville Equation) [S + λw(x)] y = d dx a < x < b. ( a 0 (x) dy dx ) + a 2 (x)y + λw(x)y = 0 for

57 Regular Sturm-Liouville Problems Definition (Regular Sturm-Liouville Problem) A regular SL problem is a boundary value problem on a closed finite interval [a, b] of the form ( d a 0 (x) dy ) + a 2 (x)y + λw(x)y = 0, a < x < b, dx dx satisfying regularity conditions and boundary conditions c 1 y(a) + c 2 y (a) = 0, d 1 y(b) + d 2 y (b) = 0, where at least one of c 1 and c 2 and at least one of d 1 and d 2 are non-zero.

58 Singular Sturm-Liouville Problems Definition (Regularity Conditions) The regularity conditions of a regular SL problem are a 0 (x), a 0(x), a 2 (x) and w(x) are continuous in [a, b]; a 0 (x) > 0 and w(x) > 0. Definition (Singular Sturm-Liouville Problem) A singular SL problem is a boundary value problem consists of Sturm-Liouville equation, but either fails the regularity conditions; or infinite boundary conditions; or one or more of the coefficients become singular.

59 Solutions of SL Problems A trivial (not useful) solution to the SL problem is y = 0. Other non-trivial solutions would be the eigenfunctions y m, and for each of these eigenfunctions there is a corresponding eigenvalue λ m. There are infinite many of these eigenfunctions, and the set of eigenfunctions {y 1, y 2,..., y m,... } forms a complete orthogonal set of functions that span the infinite dimensional Hilbert space. Any function f in the Hilbert space can be expressed as a linear combination of the eigenfunctions, f(x) = a n y n (x). n=1

60 Eigenvalues of Sturm-Liouville Problems Theorem (Sturm-Liouville Problem) The eigenvalues and eigenfunctions of a SL problem has the properties of All eigenvalues are real and compose a countably infinite collections satisfying λ 1 < λ 2 < λ 3 <... where λ j as j. To each eigenvalue λ j there corresponds only to one independent eigenfunction y j (x). The eigenfunctions y j (x), j = 1, 2,..., compose a complete orthogonal set with appropriate to the weight functions w(x) in doubly-integrable functions space L 2 (a, b).

61 Eigenfunction Expansions Theorem (Eigenfunction Expansions) If f L 2 (a, b) then eigenfunction expansion of f on {y 1, y 2,... } is where A n = y n f y n 2 = f(x) = b a A n y n, a < x < b, n=1 y n(x)f(x)w(x) dx/ b a y n (x) 2 w(x) dx.

62 Example SL Problem: Harmonic Equation An example of SL equation is the Harmonic Equation with boundary condition. ODE : y + λy = 0, 0 < x < L. Dirichlet Boundary condition: y(0) = y(l) = 0. Eigenvalues: λ n = kn 2 = n2 π 2, n = 1, 2,... ; L2 ( nπ ) Eigenfunctions: y n (x) = sin L x for n = 1, 2,... ; Eigenfunction expansion: f(x) = b n sin nπ L x. n=1

63 Example SL Problem: Harmonic Equation Again, the Harmonic Equation with another boundary condition. ODE : y + λy = 0, 0 < x < L. Neumann Boundary condition: y (0) = y (L) = 0. Eigenvalues: λ n = kn 2 = n2 π 2 L ; 2 ( nπ ) Eigenfunctions: y n (x) = cos L x for n = 0, 1, 2,... ; Eigenfunction expansion: f(x) = a a n cos nπ L x. n=1

64 Example SL Problem: Harmonic Equation The Harmonic Equation with periodic boundary condition. ODE: y + λy = 0, 0 < x < 2π. Periodic boundary condition: y(0) = y(2π). Eigenvalues: λ n = kn 2 = n 2 ; Eigenfunctions: y n (x) = e inx for n = 0, ±1, ±2,... ; Eigenfunction expansion: f(x) = c n e inx. n=

65 Example SL Problem: (Parametric) Bessel Equation Bessel Equation. x 2 y + xy + (λ 2 x 2 µ 2 )y = 0 or [x 2 y ] + (λ 2 x µ2 )y = 0 in the interval 0 < x < L. x Since a 0 (x) = x 2 is zero at x = 0, = singular SL problem. ODE: x 2 y + xy + (λ 2 x 2 µ 2 )y = 0; Boundary conditions: y(x) is bounded, and y(l) = 0; Eigenvalues: λ 2 = λ 2 n = αn, n = 1, 2,..., where α L n is the nth-root of J µ, i.e. J µ (α n ) = 0; Eigenfunctions: y n (x) = J µ (λ n x), n = 1, 2,... ; Weight: w(x) = x; Eigenfunction expansion: f(x) = n=1 a nj µ (λ n x).

66 Example SL Problem: Spherical Bessel Equation Bessel Equation. x 2 y + xy + (λ 2 x 2 n(n + 1))y = 0 in the interval 0 < x <. Since a 0 (x) = x 2 is zero at x = 0, = singular SL problem. ODE: x 2 y + xy + (λ 2 x 2 µ(µ + 1))y = 0; Boundary conditions: y(x) is bounded, and y(l) = 0; Eigenvalues: λ 2 = λ 2 n = αn L, n = 1, 2,..., where α n is the nth-root of j µ, i.e. j µ (α n ) = 0; Eigenfunctions: y n (x) = j µ (λ n x), n = 1, 2,... ; Weight: w(x) = x; Eigenfunction expansion: f(x) = n=1 a nj µ (λ n x).

67 Example SL Problem: Legendre Equation Legendre Equation. (1 x 2 )y 2xy + n(n + 1)y = 0 or [(1 x 2 )y ] + n(n + 1)y = 0, in the interval 1 < x < 1. a 0 (x) = (1 x 2 ) and a 0 (±1) = 0 = singular SL problem. ODE: (1 x 2 )y 2xy + λy = 0; Boundary conditions: y(x) is bounded at x = ±1; Eigenvalues: λ = λ n = n(n + 1), n = 0, 1, 2,... ; Eigenfunctions: y n (x) = P n (x), n = 0, 1, 2,... ; Eigenfunction expansion: f(x) = n=0 a np n (x).

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