Spectral Method for Unbounded Domains

Transcription

1 Spectral Method for Unbounded Domains 1. Laguerre Polynomials/Functions 2. Hermite Polynomials/Functions 3. Approximation by Laguerre and Hermite Polynomials/Functions 4. Spectral Methods Using Laguerre and Hermite Functions 5. Mapped Spectral Methods and Rational Approximations May 8,

2 1 Laguerre Polynomials/Functions 1.1 (Generalized) Laguerre Polynomials Laguerre polynomials, denoted by L n (x), are orthogonal w.r.t. ω(x)=e x, on the half liner + (0,+ ), i.e. 6 + L n (x)l m (x)e x dx=δ mn. (1) 0 The Generalized Laguerre polynomials (GLPs), denoted by L (α) n (x),α> 1 are orthongal w.r.t. ω α (x)=x α e x onr +, i.e. + L n (x)l m (x)ω α (x)dx=γ (α) n δ mn, (2) where 0 γ (α) n = Γ(n+α+1). (3) n! 2

3 Basic properties of GLPs 1. Three-term recurrence formula (n+1)l (α) n+1 (x)=(2n+α+1 x)l (α) n (x) (n+α)l (α) n 1 (x), L 0 (α) (x)=1, L 1 (α) (x)= x+α Leading coefficient: k (α) n = ( 1)n, n! value at 0: L (α) n (0)= Γ(n+α+1) n!γ(α+1) 3

4 3. Sturm Liouville equation (λ n =n!) ) x α e x x (x α+1 e x x L (α) n (x) +λ n L (α) n (x)=0, (4) 4. x 2 x L (α) n (x)+(α+1 x) x L (α) n (x)+λ n L (α) n (x)=0. (5) { } (α) x L n orthogonal w.r.t. ω α x L n (α) (x) x L m (α) (x)ω α+1 (x)dx=λ n γ n (α) δ mn. (6) 4

5 5. Rodrigues s formula Explicit expression 6. Derivatives L n (α) (x)= x α e x n! L n (α) (x)= k=0 d n dx n{xn+α e x }. (7) n ( 1) k k! ( ) n+a x n k k n 1 x L (α) n (x) = L (α+1) n 1 (x)= L (α) k (x), k=0 L (α) n (x) = x L (α) n (x) x L (α) n+1 (x), x x L (α) n (x) = nl (a) n (x) (n+α)l (α) n 1 (x), (8) 5

6 7. Asymptotic properties of the GLPs. For large x, L n (α) (x) is dominated by leading term ( 1) n x n n! Theorem of Szegö (1975) 1 [ L (α) n (x)= nα 2e x cos ( 2α+1 2 nx πx α π ) + O(1) nx ] (9) for all x [c/n,b]. upper bounds L (α) n (x) ( L (α) n (0)e x/2, 2 L (α) n (0) ) ifα 0, e x/2, ifα<0. (10) 6

7 7

8 1.2 Generalized Laguerre Functions (GLFs) Definition: Lˆn (α) e x/2 L (α) (x)6 n (x), x R +, α> 1. GLFs are orthogonal w.r.t weight function ωˆα=x α, i.e. + Lˆn (α) (x)lˆn (α) (x)ωˆα(x)dx=γ (α) n δ nm. (11) Introducing then 0 ˆx= x (12) x L (α) n (x)=e x/2 ˆxLˆn (α) (x). (13) 8

9 Basic properties of GLFs Three-term recurrence relation (n+1)lˆn+1 (α) (x)=(2n+α+1 x)lˆn (α) (x) (n+α)lˆn 1 (α) (x), Decay property Lˆn Lˆn Lˆ0 (α) (x)=e x/2, Lˆ1 (α) (x)=(α+1 x)e x/2. (α) (x) (α) (x) 0, x + (14) 1, x [0,+ ) (15) Sturm Liouville equation ) x α e x/2 x (x α+1 e x/2 ˆxLˆn (α) (x) +nlˆn (α) (x)=0. (16) 9

10 Orthogonality + ˆxLˆn (α) (x) ˆxLˆm (α) (x)ωˆα+1 (x)dx=λ n γ (α) n δ mn. (17) 0 Derivative relation n 1 ˆxLˆn (α) (x) = Lˆn 1 (α+1) (x)= Lˆk (α) (x), k=0 Lˆn (α) (x) = ˆxLˆn (α) (x) ˆxLˆn+1 (α) (x), x ˆxL (α) n (x) = nlˆn (a) (x) (n+α)lˆn 1 (α) (x), (18) 10

11 1.3 Laguerre Gauss Type Quadarature Theorem 1. Let {x α j,ω α j } N j=0 be the set of LG or LGR nodes and weights. Laguerre Gauss (LG): {x α j } N j=0 are zeros of L (α) N+1 (x), ω α Γ(N +α+1) j = (N +1)! = Γ(N +α+1) (N +α+1)(n +1)! 1 L (α) N (x α j ) x L (α) N+1 (x α j ) x j α [ L N (α) (x j α ) ], 0 j N. 2 (19) Laguerre Gauss Radau (LGR): x α 0 =0 and {x α j } N j=1 are zeros of x L (α) N+1 (x); ω α 0 = (α+1)γ2 (α+1)n!, ω α Γ(N +α+2) j = = Γ(N +α+1) 1 [ N!(N +α+1) x L (α) N (x α j ) Γ(N +α+1) 1 [ ] N!(N +α+1) L (α) N (x α 2, 1 j<n. j ) ] 2 (20) with δ=1,0 for LG and LGR quadrature, respectively, we have 0 + N p(x)x α e x dx= j=0 p(x j α )ω j α, p P 2N+δ, (21) 11

12 Remark 2. Denote {x j,ω j } N j=0 type nodes and weigths LG case: {x j } N j=0 ω j = the usual (α=0) Laguerre Gauss zeros of L N+1 (x), x j (N +1)L 2, 0 j N. (22) N (x j ) LGR case: x 0 =0, {x j } N j=1 zeros of x L N+1 (x), ω j = 1 (N +1)L 2, 0 j N. (23) N (x j ) We find from (9) that weights are exponentially small for large x j. L N (x) e x/2 (Nx) 1/4 ω j e x x j j N 12

13 Theorem 3. Let {x α j,ω α j } N j=0 and weights. Define be the LG or LGR quadrature nodes and ωˆjα =e x j α ω j α, 0 j N, (24) PˆN 6 { φ:φ=e x/2 ψ, ψ P N }. (25) Then we have the modified quadrature formula 0 + N p(x)x α dx= j=0 p(x j α )ωˆjα, p Pˆ2N+δ, (26) where δ=1,0 for the modified Laguerre Gauss rule and the modified Laguerre Gauss Radau rule, respectively. Remark 4. For α = 0, ωˆj = ωˆj=o ( ) x j /N. 1 (N +1) [ LˆN(x j ) ] 2, 0 j N 13

14 Computation of Nodes and Weights Eigenvalue method to calculate the nodes {x α N j } j=0 A N+1 = a 0 b 1 b 1 a 1 b 2 b N 1 a N 1 b N b N a N, a j =2j+α+1, 0 j N; b j =j(j+α), 1 j N. First calculate weights {ωˆjα } N j=0, then compute {ω α j } N j=0 by ω j α =e x j α ωˆjα, 0 j N. Nodes distribution: min j {x α j } N 1, max j {x α α j } 4N, min j x j+1 x α j N 1 14

15 15

16 1.4 Interpolation and Discrete Laguerre Transform {x α j,ωˆjα } N j=0 u,v N,ωˆα= j=0 modified LG or LGR nodes and weights, define N u(x α j )v(x α j )ωˆjα, u N,ωˆα= u,u N,ωˆα. The exactness of the Gauss quadrature implies p,q N,ωˆα=(p,q) ωˆα, p q Pˆ2N+δ, (27) where δ=1,0 for the modified LG and LGR quadrature, respectively. Define interpolation Î N (α) :C[0,+ ) PˆN as ( ) Î (α) N u (x)= N n=0 ũ α n Lˆn (α) (x). sucht that ( ) Î (α) N u (x α j )=u(x α j ), 0 j N. 16

17 Given the physical values {u(x α j )} N j=0, the coefficients {ũ α N n } n=0 be determined by the forward discrete transform: ũ n α = 1 γ n (α) N j=0 u(x α j )Lˆn (α) (x α j )ωˆjα, 0 n N. can Backward discrete transform calculates {u(x α j )} N j=0 using {ũ α n } n=0 : N u(x j α )= n=0 ũ α n Lˆn (α) (x α j ), 0 j N. The above definitions extends to usual Laguerre-Gauss transform directly (by removing ˆ ). 17

18 1.5 Differentiation Differentiation in Physical Space Differentiation in Laguerre polynomials space P N : {h j } Lagrange polynomials assoc. with LGR points {x α j } j=0 u P N, u (m) =(u (m) (x α 0 ),,u (m) (x α N )) T u6, u (0), then N. u (m) =D m u, m 1, (28) where D=(d kj ) k,j=0,,n, d kj =h j (x k ). Differentiation in Laguerre function space PˆN: ĥ j (x)=e x/2 /e x α j /2 h j (x). (29) then ĥ j (x α k )=δ kj, 0 k,j N; PˆN= span { ĥ j :0 j N }. Moreover ĥ j PˆN and dˆkj ĥ α 6 j (xk )=e x k α /2 /e x α j /2 d kj 1 2 δ kj, 0 j,k N. (30) 18

19 Differentiation in spectral space u(x)= N û n L (α) n (x), û n = 1 (α) n=0 γ n 0 + u(x)l n (a) (x)ω α (x)dx, N u (x)= n=1 N û n x L n (α) (x)= n=0 û n (1) L n (α) (x) P N 1, û N (1) =0. The coefficients { (1) û } N 1 N n can be calculated from {û n=0 n } n=1 { (1) û n 1 û N (1) =0. =û n (1) û n, n=n,n 1,,1, Similarly, for the Laguerre function case (in space PˆN) vˆn (1) =vˆn+1 (1) 1 2 (vˆn+vˆn+1 ), n=n 1,,0, vˆn (1) = 1 vˆn. 2 by (31) (32) 19

20 2 Hermite Polynomials/Functions 2.1 Hermite polynomials 1. H m onr6 (x) defined (,+ ), weight ω(x)=e x2 H m (x)h n (x)ω(x)dx=γ n δ mn, γ n = π 2 n n!. (33) 2. Three-term recurrence relation H n+1 (x)=2xh n (x) 2nH n 1 (x), n 1, (34) Leading coefficient k n =2 n. H 0 (x)=1, H 1 (x)=2x. 3. Connection with Laguerre polynomials H 2n (x) = ( 1) n 2 2n n!l n ( 1/2) (x 2 ), H 2n+1 (x) = ( 1) n 2 2n+1 n!xl n (1/2) (x 2 ). (35) 20

21 4. Values at 0: H 2n (0)=( 2) n (2n 1)!!, H 2n+1 (0)=0. (36) 5. Sturm Liouville equation: e x2 ( e x 2 H n (x) ) +λ n H n (x)=0, λ n =2n, (37) 6. Rodrigues formula explicit expression H n (x) 2xH n (x)+λ n H n (x)=0. (38) H n (x)=( 1) n x2 dn { } e e x 2, (39) dx n [n/2] H n (x)= k=0 ( 1) k n! k!(n 2k)! (2x)n 2k. (40) 21

22 7. Upper bound H n (x) <c2 n/2 n! e x 2 /2, c (41) 8. Derivatives H n (x)=λ n H n 1 (x), n 1. (42) H n (x)=2xh n (x) H n+1 (x), n 0. (43) 9. Asymptotics Γ(n/2+1) e x2 /2 H n (x)=cos n! x + 3 ( sin 2n+1 6 2n+1 ( 2n+1 x nπ 2 x nπ 2 ) ) +O(n 1 ). (44) 22

23 2.2 Hermite functions Ĥ n (x)= 1 e x2 /2 H π 1/4 n (x), n 0, x R (45) 2 n n! Orthogonality R Ĥ n (x)ĥ m (x)dx=δ mn. (46) 23

24 Three-term recurrence 2 n Ĥ n+1 (x)=x Ĥ n (x) Ĥ n 1 (x),n 1 n+1 n+1 Ĥ 0 =π 1/4 e x2 /2, Ĥ 1 (x)= 2 π 1/4 xe x2 /2. (47) Differentiation equation Ĥ n (x)+(2n+1 x 2 )Ĥ n (x)=0. (48) 24

25 Derivatives Ĥ n (x) = = 2nĤn 1 (x) xĥ n (x) n Ĥ 2 n 1 (x) n+1 Ĥ 2 n+1 (x) (49) leads to Ĥ m (x)ĥn(x)dx= R n(n 1) 2, m=n 2, n+ 1 2, m=n, (n+1)(n+2) 2, m=n+2, 0, otherwise. (50) 25

26 Asymptotic (Ĥ n (x) n 1/4, n ) Ĥ n (x)=π 1/4 2 n n! [n/2] k=0 ( 1) k ( x n 2k 4 k e ) x2 /2. (51) k!(n 2k)! 26

27 2.3 Hermite-Gauss Quadrature N Theorem 5. (Hermite-Gauss Quadrature) Let {x j } j=0 H N+1 (x), and let {ω j } N j=0 be given by be zeros of Then, + ω j = π 2 N N! (N +1)H 2, 0 j N. (52) N (x j ) N p(x)e x2 dx= j=0 p(x j )ω j, p P 2N+1. (53) 27

28 N Theorem 6. (Modified Hermite-Gauss Quadrature) Let {x j,ω j } j=0 be the Hermite-Gauss quadrature nodes and weights, Define Then ωˆj=e x j 2 ω j = 1 (N +1)Ĥ 2, 0 j N. (54) N (x j ) where + N p(x)q(x)dx= j=0 p(x j )q(x j )ωˆj, p q Pˆ2N+1, (55) PˆM 6 { φ:φ=e x2 /2 ψ, ψ P M }. (56) 28

29 Compuation of Nodes and Weights Zeros {x j } N j=0 A N+1 = of H n+1 (x) are eigenvalues of a 0 b 1 b 1 a 1 b 2 b N 1 a N 1 b N b N a N, (57) a j =0, 0 j N; b j =j/2, 1 j N. Computing {ω j } N j=0 using (52) is not stable. Instead using ωˆj are given by (54). Node distribution: max j x j 2N, ω j =e x j 2 ωˆj, minj x j x j 1 1/ N 29

30 30

31 2.4 Interpolation and Discrete Hermite Transforms Using Hermite polynomials in P N : I h N v P N intepolates v C(R) at Hermite-Gauss pts {x j } N j=0 : N I N h v= n=0 ṽ n H n (x); (I N h v)(x j )=v(x j ), 0 j N. Forward discrete transform: ṽ n = 1 N γ n j=0 Backward discrete transform v(x j )H n (x j )ω j, 0 n N (58) N v(x j )= n=0 ṽ n H n (x j ), 0 j N. (59) 31

32 Using Hermite functions in PˆN Î h N u PˆN intepolates u C(R) at Hermite-Gauss pts {x j } N j=0 : N Î N h u= n=0 ũ n Ĥ n (x); ( ÎN h u ) (x j )=u(x j ), 0 j N. Forward discrete transform: ũ n = 1 N γ n j=0 Backward discrete transform u(x j )Ĥ n (x j )ωˆj, 0 n N (60) N u(x j )= n=0 ũ n Ĥ n (x j ), 0 j N. (61) 32

33 2.5 Differentiation Differentiation in Physical Space Hermite polynomials case (space P N ): u (m) =D m u, m 1, (62) where the entries of D are given by H N (x k ) 1, k j, d kj = H N (x j ) x k x j x k, k=j. Hermite functions case (space PˆN): dˆkj= x k δ kj + e x k 2 /2 Ĥ N (x k ) 1 e d, k j, kj= x 2 Ĥ N (x j ) x k x j j/2 0, k=j. (63) (64) 33

34 Differentiation in Frequency Space For u P N, u P N 1 : N u(x)= n=0 N û n H n (x), u (x)= n=1 N û n H n (x)= n=0 û n (1) H n (x), û N (1) =0; û n (1) =2(n+1)û n+1, n=n 1,N 2,,0. (65) For v PˆN, v PˆN+1: v(x)= N vˆnĥ n (x), v (x)= N N+1 vˆnĥ n (x)= vˆn (1) Ĥ n (x). n=0 n=0 n=0 vˆn (1) = n+1 vˆn+1 n vˆn 1, n=n +1,N,,0. (66) 2 2 with vˆ 1 =vˆn+1 =vˆn+2 =0. 34

35 3 Approximation Estimates 3.1 Inverse Inequalities For Laguerre approximation, ω α =x α e x and ωˆα=x α. Theorem 7. For α> 1 and any φ P N, x m φ ωα+m N m/2 φ ωα, m 0. (67) Corollary 8. For α> 1 and any ψ PˆN, ˆxm ψ ωˆα+m Nm/2 ψ ωˆα, m 0. (68) Theorem 9. For α 0 and any φ P N, x m φ ωα N m φ ωα, m 0. (69) 35

36 Corollary 10. For α 0 and any ψ PˆN, ˆxm ψ ωˆα N m ψ ωˆα, m 0. (70) For the Hermite case, only one weight ω(x)=e x2. We have Theorem 11. For any φ P N, x φ ω N φ ω, moreover, let ˆ= x +x. Then for any ψ PˆN, ˆxψ N ψ. 36

37 3.2 Orthogonal Projections 2 Laguerre polynomial L ωα projection: Π N,α :L 2 ωα (R + ) P N we have Define (Π N,α u u,v N ) ωα =0, v N P N (71) Π N,α u= N n=0 û n (α) L n (α) (x), û n (α) = 1 γ n (α) ( u,l n (α) ) ω α. B m α (R + {u: k 2 )6 x u L ωα+k (R + ), 0 k m}, (72) u Bα m=( m k=0 k 2 x u ωα+k ) 1/2, u Bα m= m x u ωα+m. Theorem 12. Let α> 1. If u B m α (R + ) and 0 m N +1, then l (N m+1)! x (Π N,α u u) ωα+l m (N l+1)! x u ωα+m. (73) 37

38 Laguerre function case: ωˆα=x α For u L 2 ωˆa (R + ), we have ue x/2 L 2 ωα (R + ). Define Clearly Define ΠˆN,αu=e x/2 Π N,α ( ue x/2 ). PˆN (74) ( ΠˆN,α u u,v N ) ωˆα=0, v N PˆN. m { Bˆα (R+ u: ˆx k )6 2 u L ωˆα+k (R + ), 0 k m }, (75) ( u m Bˆα = m k=0 Theorem 13. ˆxk u 2 ωˆα+k ) 1/2, u Bˆ m α = ˆxm u ωˆα+m. m Let α> 1. If u Bˆα (R+ ) and 0 m N +1, then ˆxl (Π N,α u u) ωˆα+l (N m+1)! (N l+1)! ˆxm u ωˆα+m. (76) 38

39 Remark 14. Remarks on Laguerre approximation: 1. The convergence rate is about N (l m)/2, is only half of the classical Jacobi approximation. This is a direct consequence of the linear growth of the eigenvalues in the Sturm Liouville problem. 2. B m α (R + ) includes functions that do not decay at infinity. A fast convergence rate in ωα+l norm does not mean that the error would decay rapidly for large x. m 3. u Bˆα (R+ ) requires u decays at infinity. So theorem (13) doesn t apply to functions like sin(x). On the other hand, for u(x)=(1+x) h and u(x)=sin(kx)/(1+x) h, we have for both ˆxm u ωˆα+m < if m<2h α 1, so we have u ΠˆN,α u ωˆa N (2h α 1)/2. (77) 39

40 H 1 -type projections. (Here we only consider the case α=0) Let ω(x)=e x. Denote H 1 0,ω (R + )={u H 1 ω (R + ):u(0)=0}, P 0 N ={φ P N :φ(0)=0}. Define Π 1,0 N :H 1 0,ω (R + ) P 0 N as (( 1,0 u ΠN u ) ),v N ω =0, v N P 0 N. (78) Theorem 15. If u H 1 0,ω (R + ) and x u B m 1 0 (R + ), then for 1 m N +1, 1,0 ΠN,α u u 1,ω (N m+1)! N! x m u ωm 1, (79) Proof: Let φ(x) = ΠN 1,0 u 0x (y) dy. Then u φ H 1 0,ω (R + ). (B.35b) 1,0 ΠN u u 1,ω φ u 1,ω c x (φ u) ω 40

41 Laguerre function case For u H 1 0 (R + ), we have ue x/2 H 1 0,ω (R + ). Define ΠˆN 1,0 u=e x/2 1,0 Π ( N ue x/2) PˆN 0. Theorem 16. For any u H 1 0 (R + ), we have ( (u ΠˆN 1,0 u ) ),v N + 1 4( 1,0 ) u ΠˆN u,v N =0, vn PˆN 0. (80) Let xˆ = x + 1. If u H (R + ) and ˆxu Bˆn m 1 (R + ), then 1,0 ΠˆN u u (N m+1)! 1 c ˆxm u N! ωˆm 1, (81) where c is a positive constant independent of m,n and u. 41

42 H 2 -type orthogonal projections Define H 2 0,ω (R + )={v H 2 ω (R + ):v(0)=v (0)=0}, X N =H 2 0,ω (R + ) P N, H 0 2 (R + )={v H 2 (R + ):v(0)=v (0)=0}, XˆN=H 0 2 (R + ) PˆN. Define Π 2,0 N :H 2 0,ω (R + ) X N, as (( v ΠN 2,0 u ),v N ) ω =0, v N X N. (82) Define ΠˆN 2,0 :H 2 0 (R + ) XˆN as ΠˆN 2,0 u=e x/2 2,0 Π ( N ue x/2). (83) 42

43 Theorem 17. If v H 2 0,ω (R + ) and 2 x v B m 2 0 (R + ) with 2 m N +1, then we have 2,0 ΠN v v 2,ω c (N m+1)! (N 1)! x m v ωm 2. (84) For u H 2 0 (R + ) and all u N XˆN, we have (( 2,0 u ΠˆN u ),u ) N + 1 (( 2,0 2 u ΠˆN u ) ) 1 (,u N + 2,0 ) 16 u ΠˆN u,u N =0. (85) Moreover, if u H 2 0 (R + ) and ˆx 2 u Bˆ0 m 2 (R + ) with 2 m N +1, then 2,0 ΠˆN u u (N m+1)! 2,ωˆ c ˆxm u (N 1)! ωˆm 2. (86) 43

44 Hermite projections (ω(x)=e x2 ) Π N :L ω 2 (R) P N : Clearly (u Π N u,v N ) ω =0, v N P N. (87) N Π N u(x)= n=0 û n H n (x), û n = 1 γ n (u,h n ) ω. Theorem 18. For any u H m ω (R) with 0 m N +1, l x (Π N u u) ω 2 (l m)/2 (N m+1)! m (N l+1)! x ω, 0 l m. (88) Reamrks: The L ω 2 (R)-orthogonal projection is simultaneously optimal in the H ω l (R)-norm with l 1. 44

45 Hermite function approximation For u L 2 (R), we have ue x2 /2 L ω 2 (R), define ( e x2 /2 Π ) ΠˆNu6 N ue x 2 /2 PˆN, (89) which satisfies ( ) ( ( u ΠˆNu,v N = ue x 2 /2 Π ) N ue x 2 /2,v N e ) x2 /2 ω =0, v N PˆN. Theorem 19. Let ˆx= x +x. For ˆxm u L 2 (R) with 0 m N+1, ( ˆxl ΠˆNu u ) 2 (l m)/2 (N m+1)! ˆxm u, 0 l m. (90) (N l+1)! Theorem 20. For any ˆxm u L 2 (R) with 2 m N +1, ( l x ΠˆNu u ) (N m+1)! ˆxm u, l=0,1,2. (91) 2 m (N l+1)! 45

46 3.3 Interpolations Theorem 21. (Laguerre-Gauss interpolation[guo et al 2006b]) Let α > 1. If u C(R + ) B m α (R + ) and x u B m 1 α (R + ) with 1 m N +1, then I (α) N u u ωα (N m+1)! N! ( ) m x u ωα+m 1+ lnn m x u ωα+m. m Theorem 22. Let α > 1. If u C(R + ) Bˆα (R+ ) and ˆxu Bˆα m 1 (R + ) with 1 m N +1 then Î (α) N u u (N m+1)! ( m ˆx u ωˆα N! ωˆα+m 1 + lnn m ) ˆx u ωˆα+m 46

47 Hermite Interpolation Theorem 23. ([Guo and Xu 2000])For u C(R) H ω m (R) with m 1, we have x l (I N h u u) ω N 1 6 +l m 2 x m u ω, 0 l m. (92) Theorem 24. Let ˆx = x+x. For u C(R) and ˆxm u L 2 (R) with fixed m 1, we have ( ˆxl Î h N u u ) 1 N 6 +l m 2 ˆxm u, 0 l m. (93) 47

48 4 Spectral Methods Using Laguerre and Hermite Functions 4.1 Laguerre-Galerkin Method u xx +γu=f, x R +,γ>0; u(0)=0, lim u(x)=0. (94) x + Weak formulation (for f (H 0 1 (R + )) ) { Findu H 0 1 (R + ) such that a(u,v)6 (u,v )+γ(u,v)=(f,v), v H 0 1 (R + ) (95) Laguerre spectral-galerkin approximation to (94) { Findu N PˆN 0 such that a(u N,v N )= ( Î N f,v N ), vn PˆN 0 (96) 48

49 Define φˆk(x)=(l k (x) L k+1 (x))e x/2 =Lˆk(x) Lˆk+1(x) (97) then PˆN 0 = span { φˆ0,φˆ1,,φˆn 1}. By setting u N = N 1 k=0 û k φˆk, u=(û 0,û 1,,û N 1 ) T ; f j = ( Î N f,φˆj), f =(f0,f 1,,f N 1 ) T ; s jk = ( ) φˆk,φˆj, S=(sjk ) 0 j,k N 1 ; m jk = ( φˆk,φˆj), M =(mjk ) 0 j,k N 1 we find M is a symmetric tridiagonal matrix and S=I 1 M, so (96) 4 reduces to ( ( 1) ) I+ γ 4 M u=f. (98) 49

50 Theorem 25. If u H 0 1 (R + ), ˆxu Bˆ0 m 1 (R + ), f C(R +) Bˆ0 k (R + ) and ˆxf Bˆ0 k 1 (R + ) with 1 k,m N +1, then we have u u N 1 + (N m+1)! N! (N k+1)! N! m ˆx u ωˆm 1 ( ˆxk f + lnn ˆxk ωˆk 1 f ωˆk). (99) Proof. Let e N =u N ΠˆN 1,0 u and ẽ N =u ΠˆN 1,0 u, then by a(u N u,v N )= ( Î N f f,v N ), vn PˆN 0 a(e N,v N )=a(ẽ N,v N )+ ( Î N f f,v N ), vn PˆN 0. Taking v N =e N, we find e N 1 c ( ẽ N 1 + ÎN f f ). 50

51 4.2 Hermite Galerkin Method u xx +γu=f, x R, γ>0; Weak formulation { Findu H 1 (R) such that ( x u, x v)+γ(u,v)=(f,v), v H 1 (R). lim u(x)=0. (100) x (101) Hermite Galerkin method { Findu N PˆN such that ( x u, x v)+γ(u,v)= ( Î N h f,v N ), vn PˆN. (102) 51

52 Theorem 26. Let γ > 0 and ˆx = x + x. If u H 1 (R) with ˆxm u L 2 (R), and f C(R) with ˆx k f L 2 (R) and fixed k,m 1, then u N u 1 N 1 m m 1 2 ˆx u +N 6 k 2 ˆxk f. (103) 4.3 Numerical Results Exact solutionu(x) x (0, ) x (, ) u 1 (x) e x sin(kx) e x2 sin(kx) u 2 (x) 1 1 (1+x) h (1+x 2 ) h u 3 (x) sin(kx) sin(kx) (1+x) h (1+x 2 ) h 52

53 Figure 1. (Laguerre) Exponetial decay solution u 1 (x) with k=4,γ=1. 53

54 Figure 2. (Laguerre) Algebraic decay solution u 2 (x): different h and N. 54

55 Figure 3. (Hermite) Exponetial decay u 1 (x), Algebraic deacy u 2 (x). 55

56 4.4 Scaling Factor Assume that u(x) ε forx>m. Scaling factor (x N is the largest L G-R point) β N =x N /M, Scaled equation β N 2 v yy +γv=g(y); v(0)=0, lim v(y)=0. (104) y + 56

57 Figure 4. Scaling effects. β N =

58 5 Mapped Spectral Methods 5.1 Mappings x=g(y;s), x Λ6 y ( 1,1), (0,+ ) or(,+ ) (105) with s>0 and dx dy =g (y;s)>0. g( 1;s)=0,g(1;s)=+, ifλ=(0,+ ) g(±1;s)=±, ifλ=(,+ ) (106) 58

59 1. Mapping between x (,+ ) and y ( 1,1) with s>0. Algebraic mapping: x= sy, y= 1 y 2 Logarithmic mapping: x. (107) x 2 +s 2 x= s 2 ln 1+y 1 y, y= tanh x s (108) Exponential mapping: x=sinh(sy), y= 1 s ln( ) x+ x 2 +1, (109) where y ( 1,1), x ( L s,l s ), and L s = sinh(s). 59

60 2. Mapping between x (0,+ ) and y ( 1,1) with s>0: Algebraic mapping: Logarithmic mapping x=s 1+y x s, y= 1 y x+s. (110) x= s 2 ln3+y 1 y, y=1 2tanh ( x s ). (111) Exponential mapping ( x=sinh s 1+y ), y= 2 2 s ln( ) x+ x , (112) where x (0,L s ), L s = sinh(s). 60

61 5.2 Approximation by Mapped Jacobi Polynomials For the algebraic mapping, we define rational Jacobi polynomials: then Λ j α,β n,s J α,β (x)6 n (y) (113) j α,β n,s (x)j α,β n,s (x)ω α,β s (x)dx=γ α,β n δ mn, (114) where γ α,β n = ( J α,β α,β n,j ) n ω and a,β ω α,β s (x)=ω α,β (y) dy ( dx dx =ωα,β (y) dy ) 1>0 (115) Note that for mapping (107) and (110), we have respectively dx dy = s 1 y 2 3 =sω 3 2, 3 2, dx dy = 2s (1 y) 2 =2sω 2,0. 61

62 Figure 5. (Left) Graph of j 0,0 n,1 (x) with n=8, 12 using mapping (107). (Right) Graphs of j 0,0 8,s (x) with s=0.2, 0.6,1 using mapping (110). 62

63 Approximation V α,β N,s = span { j α,β n,s (x):n=0,1,,n }, s>0. (116) Consider the orthogonal projection π N,s α,β 2 :L α,β ωs (Λ) V α,β N,s : ( α,β ) πn,s u u,v N ω α,β=0, v N V α,β s N,s. (117) For a given mapping x=g(y,s), define a s (x)6 dx dy (>0), U s(y)6 u(x)=u(g(y,s)). (118) Then by defining D x a u6 du s, we have dx du s dy =a du s dx =D xu, d k U s dy k =D x k u. (119) 63

64 Define space m { } B α,β (Λ)= u:uis measurable m inλand u B α,β < (120) equipped with the norm and semi-norm ( u m B α,β = m k=0 D k 2 x u α+k,β+k ωs ) 1/2, m m = Dx u B α,β u ωs α+m,β+m. m Theorem 27. Let α,β> 1. If u B α,β (Λ), then for 0 m N+1, α,β πn,s u u ωs α,β and for 1 m N +1, ( α,β x πn,s u u ) α,β w s (N m+1)! (N +1)! (N m+1)! N! where ω s α,β (x)=ω α+1,β+1 g (y;s), y=h(x;s). (N +m) m 2 Dx m u ωs α+m,β+m, (121) (N +m) (1 m)/2 D x m u ωs α+m,β+m, (122 64

65 For algebraic mapping on (0, + ) 1. For u(x)= 1 (1+x) h, D x m u ωs α+m,β+m< if m<2h+α+1: α,β u πn,s u ωs α,β N (2h+α+1), (u(x)=(1+x) h ) (123) 2. For u(x)= sin(kx), D m (1+x) h x u ωs α+m,β+m< if m< 2h+α+1 : 3 α,β u πn,s u ( ) ωs α,β N (2h+α+1)/3, u(x)= sin(kx) (124) (1+x) h For algebraic mapping on (, + ) 1. Algebraic decay α,β u πn,s u ωs α,β N (2h+α+1), (u(x)=(1+x 2 ) h ) (125) 2. Algebraic decay with oscillation α,β u πn,s u ωs α,β N (2h+α+1)/2, (u(x)= sin(kx) ) (126) (1+x 2 ) h 65

66 Remark For u=(1+x) h and u=(1+x 2 ) h a. if h positive integer, expressed exactly by finite sum of mapped rational functions b. for other case, algebraic convergence rate but faster than approx. by Laguerre functions or Hermite functions. 2. For solutions with oscillation u(x) = sin(kx) (1+x) h, sin(kx) (1+x 2 ) h. The convergence rates are much slower than solution without oscillations. And also slower than Laguerre functions and Hermite functions methods.. 3. For solutions with exponetial decay at infinity, the convergence rate is faster than any algebraic rate. e c N 66

67 Mapped Interpolation Given Jacobi-Gauss points { ξ N,j α,β } N j=0 α,β, define ζ N,j,s g ( ξ α,β 6 N,j ; s ), and mapped Jacobi Gauss interpolation I α,β N,s :C(Λ) V N,s α,β as: I α,β N,s u V α,β N,s s.t. ( I α,β N,s u )( α,β ζ ) N,j,s =u ( a,β ) ζ N,j,s,j=0,1,,N. (127) m Theorem 29. Let α,β> 1. If u B α,β (Λ) with 1 m N+1, then ( α,β x IN,s u u ) α,β+n α,β IN,s u u α,β ω s ωs (N m+1)! N! (N +m) (1 m)/2 D x m u ωs α+m,β+m. (128) 67

68 Figure 6. (Left): Hermite Gauss points ( o ) vs. mapped Legendre Gauss points using the algebraic map (107) with s=1 ( ). min j x j+1 x j O(N 1 ) for mapped Legendre Gauss points. (Right) Laguerre Gauss Radau points ( o ) vs. mapped Legendre Gauss-Radau points using the algebraic map (110) with s=1 ( ). min j x j+1 x j O(N 2 ) for mapped Legendre Gauss Radau points. 68

69 5.3 Spectral Methods Using Mapped Jacobi Polynomials γu x (a(x) x u)=f, x Λ=(,+ ), γ>0 (129) For a given map x=g(y;s), the weighted weak form { 1 Findu B α,β (Λ) such that γ(u,v) ωs α,β+ ( ( α,β a(x) x u, )) 1 x vωs =(f,v) ωs α,β, v B α,β (Λ). (130) The corresponding mapped Jacobi Galerkin method is Findu N V α,β α,β N,s such that v N V N,s γ(u N,v N ) ωs α,β+ ( ( α,β a(x) x u N, x vn ω )) s = ( I α,β ) N,s f,v N ω s α,β. (131) 69

70 A second approach: Jacobi approximation for transformed problem: γu s (y) ( ) 1 a(g(y;s)) g (y;s) y g (y;s) yu s (y) =F s (y), (132) where U s (y)=u(g(y;s)) and F s (y)=f(g(y;s)). Let ωˆs α,β (y)=ω α,β (y)g (y;s). Jacobi-Gelerkin method for (132) is Findũ N P N s.t. ( ṽ N P N a(g(y;s)) γ(ũ N,ṽ N ) ω α,β+ ( α,β g (y;s) yũ N, )) y ṽn ωˆs = ( I α,β (133) N F s,ṽ N )ω α,β This approach is in general more difficult to analyze, but can be easily implemented using the standard Jacobi collocation method. 70

71 Error Estimates for a Model Problem γu(x) x 2 u(x)=f(x), x Λ=(0, ), γ>0; u(0)=0. (134) Weak formulation { Findu H 1 0,ω (Λ) such that a ω (u,v)=(f,v) ω, v H 1 0,ω (Λ), (135) where ω ω α,β 1 6 s, H 0,ω ={u H 1 ω (Λ):u(0)=0}, and a ω (u,v)=γ(u,v) ω +( x u, x (vω)), u,v H 1 0,ω (Λ). (136) DenoteX N = { u V α,β N,s :u(0)=0 }. Jacobi Galerkin approximation is { FinduN X N such that a ω (u N,v N )= ( I α,β ) N,s f,v N ω, v (137) N X n, 71

72 Lemma 30. Assume that d 1 = max x Λ ω 1 (x) x ω(x), d 2 = max x Λ ω 1 (x) x 2 ω(x) are finite. Then, for any u,v H ω 1 (Λ), a ω (u,v) (d 1 +1) u 1,ω v 1,ω +γ u ω v ω. If, in addition v 2 (x)ω (x) x=0 =0 and lim x v 2 ω (x) 0, then 2 a ω (v,v) v 1,ω +(γ d 2 /2) v 2 ω, v H 1 ω (Λ). (138) Theorem 31. Assume that the condition of Lemma 30 are satisfied and γ d 2 /2>0. Then the problem ( 137) admits a unique solution. We have u u N 1,ω inf u v N 1,ω + f α,β IN,s f v N X ω. (139) N 72

73 For mapped Legendre method (d 1 2, d 2 6) Corollary 32. Let u and u N be respectively the solution of ( 135) and ( 137) with (α,β)=(0,0) and the mapping ( 110) with s=1. Assume m that u B 0,0 (Λ) and f k B 0,0 (Λ), k,m 1 and γ>3, we have u u N 1,ω1 0,0 N 1 m D m x u m,m ω1 +N k D k x f ω1 k,k, (140) 73

74 For mapped Chebyshev method. We have d 1,d 2 =. but a ω (, ) is still continuous and coercive [cf. Guo et al. 2002]. Corollary 33. Let u and u N be the solution of (135) and (137) with (α,β)=( 1/2, 1/2) and the mapping ( 110) with s=1. Assume that m m u B 1/2, 1/2 (Λ) and f B 1/2, 1/2 (Λ) and that γ> 14, we have 17 u u N 1,ω1 1/2, 1/2 N 1 m D x m u ω1 m 1/2,m 1/2+N k D x k f ω1 k 1/2,k 1/2, (141) 74

75 Implementation and Numerical Results Take φ k (x)=j 0,0 0,0 s,k (x)+j s,k+1 then ω(x)= 2. (1+x) 2 with s=1 as basis of X N, (137) leads to a non-symmetric sparse system (M tridiagonal, S seven diagonal ) (γm +S)u=f (142) 75

76 Figure 7. (Left) u(x)=sin(2x)e x ; (Right) u(x)=1/(1+x) 5/2. 76

77 Figure 8. (Left) u(x)=sin(2x)/(1+x) 7/2 ; (Right) u(x)=sin(2x)e x2. 77

78 Figure 9. (Left)u(x)=1/(1+x 2 ) 5/2 ; (Right)u(x)=sin(2x)/(1+x 2 ) 7/2. 78

79 Conclusion 1. Two approaches to solve problem in unbounded domain a. Laguerre/Hermite method b. Mapped Jacobi method 2. The convergence rates of two approaches are compatible 3. Scaling parameter affects the convergence rates. 79