SPECTRAL METHODS: ORTHOGONAL POLYNOMIALS

Transcription

1 SPECTRAL METHODS: ORTHOGONAL POLYNOMIALS 31 October, 2007

2 1 INTRODUCTION 2 ORTHOGONAL POLYNOMIALS Properties of Orthogonal Polynomials 3 GAUSS INTEGRATION Gauss- Radau Integration Gauss -Lobatto Integration 4 JACOBI POLYNOMIALS Legendre Polynomials Chebychev Polynomials 5 EXAMPLE

3 Preview of Spectral Methods What are Spectral Methods? The main components for their formulation are Trial functions Test functions The three types of Spectral Schemes are;

4 Preview of Spectral Methods What are Spectral Methods? The main components for their formulation are Trial functions Test functions The three types of Spectral Schemes are; Galerkin

5 Preview of Spectral Methods What are Spectral Methods? The main components for their formulation are Trial functions Test functions The three types of Spectral Schemes are; Galerkin Collocation

6 Preview of Spectral Methods What are Spectral Methods? The main components for their formulation are Trial functions Test functions The three types of Spectral Schemes are; Galerkin Collocation Tau

7 What choice for the trial function ω(x)? Periodic Problem :ω(x) Trigonometric Polynomials Non-Periodic Problem :ω(x) Orthogonal Polynomials

8 Orthogonal Polynomials Sturm -Liouville problems (SLP)

9 Orthogonal Polynomials Sturm -Liouville problems (SLP) A Sturm - Liouville problem is an eigenvalue problem of the form (pu ) + qu = λωu in the interval ( 1, 1) with boundary condition for u where p, q and ω are given p C 1 ( 1, 1), q is bounded, ω is the weight function

10 Orthogonal Polynomials Sturm -Liouville problems (SLP) A Sturm - Liouville problem is an eigenvalue problem of the form (pu ) + qu = λωu in the interval ( 1, 1) with boundary condition for u where p, q and ω are given p C 1 ( 1, 1), q is bounded, ω is the weight function How is spectral accuracy guaranteed?

11 Properties of Orthogonal Polynomials Given ( 1, 1) and weight ω(x) > 0 on ( 1, 1) and ω L 1 ( 1, 1). The weighted Sobolev L 2 ω( 1, 1) is defined by L 2 ω ( 1, 1) = {p : 1 1 The inner product of L 2 ( 1, 1)is given by ω (p, g) ω = 1 1 } p 2 (x)ω(x)dx < + p(x)g(x)ω(x)dx and the norm p L 2 ω = (p, p) 1/2 ω

12 Properties of Orthogonal Polynomials Given ( 1, 1) and weight ω(x) > 0 on ( 1, 1) and ω L 1 ( 1, 1). The weighted Sobolev L 2 ω( 1, 1) is defined by L 2 ω ( 1, 1) = {p : 1 1 The inner product of L 2 ( 1, 1)is given by ω (p, g) ω = 1 1 } p 2 (x)ω(x)dx < + p(x)g(x)ω(x)dx and the norm p L 2 ω = (p, p) 1/2 ω

13 Properties of Orthogonal Polynomials A system of algebraic polynomials {p k } k=0,1... with degree k is said to be orthogonal in L 2 ω( 1, 1) if (p k, p m ) ω = 0, m k ie 1 1 p k (x)p m (x)ω(x)dx = 0 whenever m k

14 Properties of Orthogonal Polynomials A system of algebraic polynomials {p k } k=0,1... with degree k is said to be orthogonal in L 2 ω( 1, 1) if (p k, p m ) ω = 0, m k ie 1 1 p k (x)p m (x)ω(x)dx = 0 whenever m k A series of a function u L 2 ( 1, 1) can be represented in terms ω of the system {p k } by where Su = û k p k k=0 û k = 1 1 p k 2 u(x)p k (x)ω(x)dx ω 1 is the polynomial transform of u

15 Properties of Orthogonal Polynomials A system of algebraic polynomials {p k } k=0,1... with degree k is said to be orthogonal in L 2 ω( 1, 1) if (p k, p m ) ω = 0, m k ie 1 1 p k (x)p m (x)ω(x)dx = 0 whenever m k A series of a function u L 2 ( 1, 1) can be represented in terms ω of the system {p k } by where Su = û k p k k=0 û k = 1 1 p k 2 u(x)p k (x)ω(x)dx ω 1 is the polynomial transform of u

16 Existence and Uniqueness of Orthogonal Polynomials Lemma If a sequence of polynomials {p k } k=0 is orthogonal then the polynomial p N+1 (x) is orthogonal to any polynomial q of degree N or less

17 Existence and Uniqueness of Orthogonal Polynomials Theorem For any positive function ω(x) L 1 ( 1, 1), a unique set of Monic orthogonal polynomials {p k }, which can be constructed as follows and where and p 0 = 1, p 1 = x α 1 with α 1 = ω(x)xdx/ ω(x)dx 1 p k+1 (x) = (x α k+1 )p k (x) β k+1 p k 1 (x) k 1 β k+1 = α k+1 = xω(x)p 2 (x)dx/ ω(x)p 2 (x)dx k k 1 1 xω(x)p k (x)p k 1 (x)dx/ ω(x)p 2 (x)dx k 1 1

18 The Spectral Representation of Function Orthogonal projections on the space of the polynomials of degree N, N P N u = û k p k k=1 The completeness of {p k } = u P N u 0 as N u L 2 ( 1, 1) ω

19 How to compute the integral 1 1 u(x)p k (x)ω(x)dx = (u, p k ) ω

20 How to compute the integral 1 1 u(x)p k (x)ω(x)dx = (u, p k ) ω By Gauss Integration. Let x 0 < x 1 <... x N be the roots of (N + 1) th orthogonal polynomials p N+1 (x) and let ω 0... ω N be the solution of the linear system N 1 (x j ) k ω j = x k ω(x)dx; 0 k N. j=0 1 Then ω j > 0 for j = 0, 1... N > and 1 N x k p(x)ω(x)dx = (x j ) k ω j p p 2N+1 1 j=0

21 Gauss- Radau Integration The Way Out Is As Follows Gauss Radau Integration Consider the polynomial q(x) = p N+1 (x) + ap N (x) Let 1 = x 0 < x 1 < x N be the (N + 1) roots of the polynomial and ω 0,... ω N be the solution of the linear system N (x j ) k ω j = j=0 1 1 x k ω(x)dx. Then 1 1 x k p(x)ω(x)dx = N (x j ) k ω j p p 2N j=0

22 Gauss -Lobatto Integration The Way Out Is As Follows Gauss Lobatto Integration Consider the polynomial q(x) = P N+1 (x) + ap N (x) + bp N 1 (x) let 1 = x 0 < x 1 < x N = 1 be the roots of the polynomial and ω 0,... ω N be the solution of the linear system N (x j ) k ω j = j=0 1 1 x k ω(x)dx Then 1 1 x k p(x)ω(x)dx = N (x j ) k ω j p P 2N 1 j=0

23 Gauss -Lobatto Integration Gauss- Lobatto The Gauss-Lobatto points for the Jacobi Polynomials corresponding to the weight ω(x) = (1 x) α (1 + x) α for N = 8 and 1/2 α 1/2 Figure: The Gauss-Lobatto points for N = 8 Jacobi polynomials with the weight function ω(x) = (1 x) α (1 + x) β

24 Gauss -Lobatto Integration The Interpolating Polynomial The interpolating polynomial associated with { } N x j j=0 I N u is defined as a polynomial of degree less than or equal to N, such that I N u(x j ) = u(x j ) j = 0, 1... N Hence I N u = N ũ k p k k=0 The discrete polynomial coefficients of u and its inverse relationship is ũ k = 1 γ k N j=0 u(x j )p k (x j )ω j k = 0,... N

25 Gauss -Lobatto Integration The Interpolating Polynomial The discrete polynomial coefficient ũ k can also be expressed in terms of continuous coefficients û k as ũ k = û k + 1 (p γ l, p k ) N û l k = 0... N k l>n = I N u = p N u + R N u where R N u = { } N 1 (p γ l, p k ) N û l p k k k=0 l>n is known as the aliasing error (as a result of the interpolation)

26 Jacobi Polynomials The Jacobi Polynomials is denoted by J α,β n (x) with ω(x) = (1 x) α (1 + x) β for α, β > 1 on ( 1, 1).

27 Jacobi Polynomials The Jacobi Polynomials is denoted by J α,β n (x) with ω(x) = (1 x) α (1 + x) β for α, β > 1 on ( 1, 1). Also normalized by where Γ(x) is a gamma function. J α,β Γ(n + α + 1) n (1) = n!γ(β + 1)

28 Jacobi Polynomials The Jacobi Polynomials is denoted by J α,β n (x) with ω(x) = (1 x) α (1 + x) β for α, β > 1 on ( 1, 1). Also normalized by J α,β Γ(n + α + 1) n (1) = n!γ(β + 1) where Γ(x) is a gamma function. Satisfies the orthogonality condition. 1 1 J α,β n (x)j α,β m (x)(1 x) α (1 + x) β ) dx = 0 n m

29 Jacobi Polynomials The Jacobi Polynomials is denoted by J α,β n (x) with ω(x) = (1 x) α (1 + x) β for α, β > 1 on ( 1, 1). Also normalized by J α,β Γ(n + α + 1) n (1) = n!γ(β + 1) where Γ(x) is a gamma function. Satisfies the orthogonality condition 1 1 J α,β n (x)j α,β m (x)(1 x) α (1 + x) β ) dx = 0 n m. satisfies the singular Sturm-Liouville Problem (1 x) α (1 + x) β d dx { (1 x) α+1 (1 + x) β+1} d dx Jα,β n (x) +n(n + 1α + β)j α,β n (x) = 0

30 Legendre Polynomials Legendre polynomials Denoted by L k (x), k = 0, 1... are eigenfunctions of SLP with. Normalized by ((1 x 2 )L k (x)) + k(k + 1)L k (x) = 0 p(x) = 1 x 2, q(x) = 0, ω(x) = 1 L k (x) = 1 [k/2] ( )( ) k 2k 2l 2 k ( 1) l x k 2l l k l=0

31 Legendre Polynomials Properties of Legendre polynomials L k (x) 1, 1 x 1, L k (±1) = (±1) k, L (x) 1 (k + 1), 1 x 1, k 2 1 ( L 2 (x) dx = k k 2) 1 The expansion of any u L 2 ω( 1, 1) u(x) = in terms of the L k s is û k L k (x), û k = (k ) u(x)l k (x)dx k=0 1

32 Legendre Polynomials Discrete Legendre Series The explicit formulas for the quadrature points and weights are Legendre Gauss(LG) Legendre Gauss -Radau(LGR) Legendre Gauss -Lobatto(LGL)

33 Legendre Polynomials Discrete Legendre Series LG, LGR, LGL ω j = x j (j = 0... N) zeros of L N+1 2 (1 x 2 j )[L N+1 (x j)] 2 j = 0... N

34 Legendre Polynomials Discrete Legendre Series LG, LGR, LGL ω j = x j (j = 0... N) zeros of L N+1 2 (1 x 2 j )[L N+1 (x j)] 2 j = 0... N ω 0 = x j (j = 0... N), zeros of L N + L N+1 2 (N + 1) 2, ω 1 1 x j = j (N + 1) 2, j = 1,..., N. [L N+1 (x j )] 2

35 Legendre Polynomials Discrete Legendre Series LG, LGR, LGL ω j = x j (j = 0... N) zeros of L N+1 2 (1 x 2 j )[L N+1 (x j)] 2 j = 0... N ω 0 = x j (j = 0... N), zeros of L N + L N+1 2 (N + 1) 2, ω 1 1 x j = j (N + 1) 2, j = 1,..., N. [L N+1 (x j )] 2 x 0 = 1, x N = 1, x j (j = 0... N 1), zeros of L N 2 1 ω j = (N + 1) [L N (x j )] 2 for all j = 0... N

36 Legendre Polynomials Discrete Legendre Series Continue The normalization factor is given by γ k = (k ) 1 for k < N γ N = { (N ) for Gauss and Gauss-Radau formulas, 2/N for the Gauss-Lobatto formula.

37 Legendre Polynomials Differentiation of Legendre Polynomials That is if u = k=0 ûk L k then u can be represented as where u = k=0 û (1) k = (2k + 1) k=1 û (1) L k k p=k+1 p+k,odd û p k 0 The recursion relation is given by [ û(1) u k 1 (x) = 2k 1 û ] (1) k+1 L k 2k + 3 (x)

38 Chebychev Polynomials Chebyshev Polynomials The Chebyshev Polynomial of the first kind is denoted by T k (x), k = 0, 1... are the eigenfunctions of SLP ( 1 x 2 T k(x)) + k 2 1 x 2 T k(x) = 0 with p(x) = (1 x 2 ) 1 2, q(x) = 0 and ω(x) = (1 x 2 ) 1 2 The chebyshev polynomial can be expressed in a power series as T k (x) = k [k/2] k (k l 1) ( 1) 2 l!(k 2l)! )2xk 2l l=0

39 Chebychev Polynomials Chebyshev Polynomials The trigonometric relation cos (k + 1)θ + cos (k 1)θ = 2 cos θ cos kθ gives the recursion relation with T 0 (x) 1 and T 1 (x) x T k+1 (x) = 2xT k T k 1 (x)

40 Chebychev Polynomials Properties of Chebyshev Polynomials where T k (x) 1, 1 x 1, T k (±1) = (±1) k, T k(x) (k 2 ), 1 x 1, 1 1 T k(±) = (±) k+1 k 2, T 2 dx π k(x) = c k 1 x 2, c k = { 2, k = 0 1, k 1. The Chebyshev expansion of a function u L 2 w ( 1, 1) is u(x) = û k L k (x), k=0 π 1 û k = c k u(x)t k (x)dx 2 1

41 Chebychev Polynomials Discrete Chebyshev Series The explicit formulas for the quadrature points and weights are Chebyshev Gauss(CG) Chebyshev Gauss -Radau(CGR) Chebyshev Gauss -Lobatto(CGL)

42 Chebychev Polynomials Discrete Chebyshev Series Chebyshev Gauss(CG), Chebyshev Gauss -Radau(CGR), Chebyshev Gauss -Lobatto(CGL) x j = cos (2j + 1)π 2N + 2, ω j = π N + 1, j = 0,..., N

43 Chebychev Polynomials Discrete Chebyshev Series Chebyshev Gauss(CG), Chebyshev Gauss -Radau(CGR), Chebyshev Gauss -Lobatto(CGL) x j = cos (2j + 1)π 2N + 2, ω j = π N + 1, j = 0,..., N x j = cos 2πj 2N + 1, ω j = { π 2N+1, j = 0, π 2N+2, j = 0,..., N

44 Chebychev Polynomials Discrete Chebyshev Series Chebyshev Gauss(CG), Chebyshev Gauss -Radau(CGR), Chebyshev Gauss -Lobatto(CGL) x j = cos (2j + 1)π 2N + 2, ω j = π N + 1, j = 0,..., N x j = cos 2πj 2N + 1, ω j = x j = cos jπ 2N, ω j = { π 2N+1, j = 0, π 2N+2, j = 0,..., N { π 2N, j = 0, N, π N, j = 1,... N 1

45 Chebychev Polynomials The Chebyshev transform space This is given by where c k = C k = 2 cos πjk Nc j c k N { 2, j = 0, N, 1, j = 1,... N 1 The inverse transform is represented by (C 1 ) jk = cos πjk N Both transforms can be evaluated by the Fast Fourier Transform

46 Chebychev Polynomials The normalization factors γ k is given by γ k = π 2 c k for k < N γ N = { π 2 for Gauss and Gauss-Radau formulas, π for the Gauss-Lobatto formula The aliasing error for the Chebyshev Gauss-Lobatto points is given by ũ k = û k + j=2mn±k j>n û j

47 Chebychev Polynomials Differentiation of Chebyshev polynomials The derivative of a function u expanded in Chebyshev polynomial is given by u = û (1) T k k where û (1) k = 2 c k k=0 p=k+1 p+k odd pû p k 0 The above expression is a consequence of the relation and finally we obtain 2T k (x) = 1 k + 1 T k+1 (x) 1 k 1 T k 1 (x) 2kû k = c k 1 û (1) k 1 û(1) k+1

48 Chebychev Polynomials Differentiation of Chebyshev Polynomials Continue The recursion relation is given by c k û (1) = û (1) + 2(k + 1)û(1), 0 k N 1 k+2 k+1 The generalization relation is given by c k û (q) = û (q) + 2(k + 1)û(q 1), k 0 k+2 k 1 The coefficients of the second derivative are û (2) k = 1 c k p=k+2 p+k even p(p 2 k 2 )û p, k 0

49 Chebychev Polynomials A simple Differential equation with boundary conditions Let consider the 1 D second-order linear PDE d 2 u dx 2 4du dx + 4u = ex + C, x [ 1, 1] with the Dirichlet boundary conditions u( 1) = 0 and u(1) = 0 and where C is a constant: C = 4e/(1 + e 2 ). The exact solution of the system is u(x) = e x sinh 1 sinh 2 e2x + C 4

50 Chebychev Polynomials Solving by Chebyshev spectral mehthod Look for a numerical solution by the first Chebyshev polynomials: T 0 (x), T 1 (x), T 2 (x), T 3 (x) and T 4 (x), N = 4. Expand the source u(x) = e x + C onto the Chebyshev Polynomials p 4 u(x) = 4 ũ n T n (x) n=0 and with and I 4 u(x) = 4 û n T n (x) n=0 2 1 dx ũ n = T π(1 + δ 0n ) n (x)u(x) 1 1 x 2 û n = 2 π(1 + δ 0n ) 4 w i T n (x i )u(x i ) n=0

51 Chebychev Polynomials Solving by Chebyshev spectral mehthod where the x i s being the 5 Gauss -Lobatto quadrature points for the weight w = (1 x 2 ) 1/2 : x i = { cos(iπ/4), 0 i 4} = { 1, 1 1 2, 0, 2 } The continuous coefficient is obtained as follows û N ũ N û N -ũ N

52 Chebychev Polynomials The source and its Chebyshev interpolant

53 Chebychev Polynomials Interpolation error and the aliasing error Figure: N = 4(5 Chebyshev polynomials)

54 Chebychev Polynomials Thank you