Discontinuous Galerkin methods for fractional diffusion problems

Transcription

1 Discontinuous Galerkin methods for fractional diffusion problems Bill McLean Kassem Mustapha School of Maths and Stats, University of NSW KFUPM, Dhahran Leipzig, 7 October, 2010

2 Outline Sub-diffusion Equation Various Numerical Methods Discontinuous Galerkin Methods

3 Part I Sub-diffusion Equation

4 Classical diffusion Density of particles u = u(x, t) satisfies u t (K u) = f (x, t) for x Ω and t > 0, for a diffusivity K > 0 and source density f.

5 Classical diffusion Density of particles u = u(x, t) satisfies u t (K u) = f (x, t) for x Ω and t > 0, for a diffusivity K > 0 and source density f. Initial condition u(x, 0) = u 0 (x) for x Ω.

6 Classical diffusion Density of particles u = u(x, t) satisfies u t (K u) = f (x, t) for x Ω and t > 0, for a diffusivity K > 0 and source density f. Initial condition u(x, 0) = u 0 (x) for x Ω. Zero-flux boundary condition u/ n = 0 for x Ω.

7 Classical diffusion Density of particles u = u(x, t) satisfies u t (K u) = f (x, t) for x Ω and t > 0, for a diffusivity K > 0 and source density f. Initial condition u(x, 0) = u 0 (x) for x Ω. Zero-flux boundary condition u/ n = 0 for x Ω. Absorbing boundary condition u = 0 for x Ω.

8 Anomalous sub-diffusion Density satisfies u t (ω ν K u) t = f (x, t) for x Ω and t > 0, where 0 < ν < 1 and ω ν (t) = tν 1 Γ(ν).

9 Anomalous sub-diffusion Density satisfies u t (ω ν K u) t = f (x, t) for x Ω and t > 0, where 0 < ν < 1 and ω ν (t) = tν 1 Γ(ν). Riemann Liouville fractional integral of order ν: ω ν v(t) = t 0 (t s) ν 1 v(s) ds. Γ(ν) Thus, (ω ν v) t = t 1 ν v.

10 Interpretation of ν Fractional diffusion equation arises from a CTRW model. The mean-square displacement of a particle is proportional to t ν.

11 Interpretation of ν Fractional diffusion equation arises from a CTRW model. The mean-square displacement of a particle is proportional to t ν. Classical diffusion coincides with the limiting case ν 1:

12 Interpretation of ν Fractional diffusion equation arises from a CTRW model. The mean-square displacement of a particle is proportional to t ν. Classical diffusion coincides with the limiting case ν 1: ω ν (t) = t ν 1 /Γ(ν) 1,

13 Interpretation of ν Fractional diffusion equation arises from a CTRW model. The mean-square displacement of a particle is proportional to t ν. Classical diffusion coincides with the limiting case ν 1: ω ν (t) = t ν 1 /Γ(ν) 1, (ω ν K u)(t) t 0 K u(x, s) ds,

14 Interpretation of ν Fractional diffusion equation arises from a CTRW model. The mean-square displacement of a particle is proportional to t ν. Classical diffusion coincides with the limiting case ν 1: ω ν (t) = t ν 1 /Γ(ν) 1, (ω ν K u)(t) t 0 K u(x, s) ds, (ω ν K u) t K u.

15 Hilbert space setting Write Au = (K u) and B ν v(t) = (ω ν v) t.

16 Hilbert space setting Write Au = (K u) and B ν v(t) = (ω ν v) t. Inner product and norm in L 2 (Ω): u, v = uv dx and u = u, u. Ω

17 Hilbert space setting Write Au = (K u) and B ν v(t) = (ω ν v) t. Inner product and norm in L 2 (Ω): u, v = uv dx and u = u, u. Bilinear form on H 1 (Ω): A(u, v) = Au, v = Ω Ω K u v dx.

18 Hilbert space setting Write Au = (K u) and B ν v(t) = (ω ν v) t. Inner product and norm in L 2 (Ω): u, v = uv dx and u = u, u. Bilinear form on H 1 (Ω): A(u, v) = Au, v = Ω Ω K u v dx. Weak formulation: u t, v + A(B ν u, v) = f, v for all v H 1 (Ω).

19 Separation of variables Complete orthonormal eigensystem: Aφ m = λ m φ m for m = 0, 1, 2,... φ m, φ n = δ mn, 0 λ 0 λ 1 λ 2, λ m as m.

20 Separation of variables Complete orthonormal eigensystem: Aφ m = λ m φ m for m = 0, 1, 2,... φ m, φ n = δ mn, 0 λ 0 λ 1 λ 2, λ m as m. Fourier modes u m (t) = u(t), φ m satisfy scalar IVP: u m + λ m B ν u m = f m (t), u m (0) = u 0m, where f m (t) = f (t), φ m, u 0m = u 0, φ m.

21 Laplace transformation Notation: ˆv(z) = L{v(t)} = 0 e zt v(t) dt.

22 Laplace transformation Notation: ˆv(z) = L{v(t)} = We transform 0 e zt v(t) dt. u m + λ m B ν u m = f m (t), u m (0) = u 0m, to obtain zû m (z) u 0m + λ m z 1 ν û m (z) = ˆf m (z).

23 Laplace transformation Notation: ˆv(z) = L{v(t)} = We transform 0 e zt v(t) dt. u m + λ m B ν u m = f m (t), u m (0) = u 0m, to obtain zû m (z) u 0m + λ m z 1 ν û m (z) = ˆf m (z). Thus, û m (z) = u 0m + ˆf m (z) z + λ m z 1 ν.

24 Explicit solution Mittag Leffler function satisfies { } E ν ( λt ν ) = L 1 1 z + λz 1 ν so { } 1 e λt = L 1, z + λ t u m (t) = E ν ( λ m t ν ( )u 0m + E ν λm (t s) ν) f m (s) ds. 0

25 Explicit solution Mittag Leffler function satisfies { } E ν ( λt ν ) = L 1 1 z + λz 1 ν so { } 1 e λt = L 1, z + λ t u m (t) = E ν ( λ m t ν ( )u 0m + E ν λm (t s) ν) f m (s) ds. 0 Hence, where t u(t) = E(t)u 0 + E(t s)f (s) ds 0 E(t)v = E ν ( λ m t ν ) v, φ m φ m. m=0

26 Positivity Plancherel Theorem implies 0 B ν v(t)v(t) dt = sin( 1 2 πν) π 0 y 1 ν ˆv(iy) 2 dy 0.

27 Positivity Plancherel Theorem implies 0 B ν v(t)v(t) dt = sin( 1 2 πν) π 0 y 1 ν ˆv(iy) 2 dy 0. So T 0 A(B ν u, u) dt = T λ m B ν u m (t)u m (t) dt 0. m=0 0

28 Stability via energy argument Take v = u in the weak formulation: u, u + A(B ν u, u) = f, u.

29 Stability via energy argument Take v = u in the weak formulation: u, u + A(B ν u, u) = f, u. Since u, u = 1 d 2 dt u 2, integrating from t = 0 to t = T gives T 1 2 u(t ) u 0 2 f u dt. 0

30 Stability via energy argument Take v = u in the weak formulation: u, u + A(B ν u, u) = f, u. Since u, u = 1 d 2 dt u 2, integrating from t = 0 to t = T gives T 1 2 u(t ) u 0 2 f u dt. 0 Existence of a unique mild solution follows, with t u(t) u f (s) ds. 0

31 Regularity for homogeneous problem If f (t) 0 then u(t) = E(t)u 0 = E ν ( λ m t ν ) u 0, φ m φ m. m=0

32 Regularity for homogeneous problem If f (t) 0 then u(t) = E(t)u 0 = E ν ( λ m t ν ) u 0, φ m φ m. m=0 The mth mode of u 0 is damped by the factor E ν ( λ m t ν ) λ 1 m t ν /Γ(1 ν),

33 Regularity for homogeneous problem If f (t) 0 then u(t) = E(t)u 0 = E ν ( λ m t ν ) u 0, φ m φ m. m=0 The mth mode of u 0 is damped by the factor implying the smoothing property E ν ( λ m t ν ) λ 1 m t ν /Γ(1 ν), t q A µ u (q) (t) Ct µν u 0, 0 µ 1, q = 0, 1, 2,....

34 Regularity for homogeneous problem If f (t) 0 then u(t) = E(t)u 0 = E ν ( λ m t ν ) u 0, φ m φ m. m=0 The mth mode of u 0 is damped by the factor implying the smoothing property E ν ( λ m t ν ) λ 1 m t ν /Γ(1 ν), t q A µ u (q) (t) Ct µν u 0, 0 µ 1, q = 0, 1, 2,.... For classical diffusion, E 1 ( λ m t) = e λmt leads to stronger smoothing: t q A µ u (q) (t) Ct µ u 0, 0 µ <.

35 Regularity for inhomogenous problem If u 0 = 0 then u(t) = t 0 E(t s)f (s) ds

36 Regularity for inhomogenous problem If u 0 = 0 then u(t) = and we can show t q+1 t q Au (q) (t) Ct ν 0 E(t s)f (s) ds t j=0 0 s j f (j) (s) ds.

37 Regularity for inhomogenous problem If u 0 = 0 then u(t) = and we can show t q+1 t q Au (q) (t) Ct ν 0 E(t s)f (s) ds t j=0 0 s j f (j) (s) ds. Analysis of DG methods uses estimates of the form t ν+q 1 Au (q) (t) Ct σ 1 with σ > 0.

38 Regularity for inhomogenous problem If u 0 = 0 then u(t) = and we can show t q+1 t q Au (q) (t) Ct ν 0 E(t s)f (s) ds t j=0 0 s j f (j) (s) ds. Analysis of DG methods uses estimates of the form t ν+q 1 Au (q) (t) Ct σ 1 with σ > 0. For instance, σ = ν/2 valid if A 1/2 u 0 < and t j f (j) (t) Ct ν/2 1, 0 j q + 1.

39 Part II Various Numerical Methods

40 Implicit finite difference scheme Langlands and Henry, J. Comput. Phys. 205: , Consider u t (ω ν u xx ) t = 0 for 0 x L and 0 t T, with u(x, 0) = u 0 (x), u x (0, t) = 0 = u x (L, t).

41 Implicit finite difference scheme Langlands and Henry, J. Comput. Phys. 205: , Consider u t (ω ν u xx ) t = 0 for 0 x L and 0 t T, with u(x, 0) = u 0 (x), u x (0, t) = 0 = u x (L, t). For step sizes x = L/P and t = T /N, seek U n p u(x p, t n ) for (x p, t n ) = (p x, n t), where 0 p P and 0 n N.

42 Approximations A formula of Oldham and Spanier gives ( (ω ν v) t tν 1 νv 0 p n Γ(1 + ν) n 1 ν + (Vp V j p j 1 ) [ (n j+1) ν (n j) ν]) at (x, t) = (x p, t n ). j=1

43 Approximations A formula of Oldham and Spanier gives ( (ω ν v) t tν 1 νv 0 p n Γ(1 + ν) n 1 ν + (Vp V j p j 1 ) [ (n j+1) ν (n j) ν]) at (x, t) = (x p, t n ). Combine with j=1 u t Un p Up n 1 t u x Un p+1 Un p 1, 2 x u xx Un p+1 2Un p + U n p 1 x 2,

44 Final scheme Write then w j = (j + 1) ν j ν and δ 2 U n p = Un p+1 2Un p + U n p 1 x 2, U n p U n 1 p t ( = tν 1 νn ν 1 δ 2 U 0 Γ(1 + ν) p+ n j=1 ) w n j (δ 2 Up δ j 2 Up j 1 ) for 1 n N and 0 p P,

45 Final scheme Write then w j = (j + 1) ν j ν and δ 2 U n p = Un p+1 2Un p + U n p 1 x 2, U n p U n 1 p t ( = tν 1 νn ν 1 δ 2 U 0 Γ(1 + ν) p+ n j=1 ) w n j (δ 2 Up δ j 2 Up j 1 ) for 1 n N and 0 p P, with U n 1 = U n 1 and U n P+1 = Un P 1.

46 Final scheme Write then with w j = (j + 1) ν j ν and δ 2 U n p = Un p+1 2Un p + U n p 1 x 2, U n p U n 1 p t ( = tν 1 νn ν 1 δ 2 U 0 Γ(1 + ν) p+ n j=1 ) w n j (δ 2 Up δ j 2 Up j 1 ) for 1 n N and 0 p P, U n 1 = U n 1 and U n P+1 = Un P 1. Scheme is implicit: put a = w 0 t ν /Γ(1 + ν) then U n p aδ 2 U n p = U n 1 p + (terms in δ 2 U 0 p,..., δ 2 U n 1 p ).

47 Convergence behaviour Scheme is unconditionally stable, and computational experiments indicate that U n p = u(x p, t n ) + O( t ν + x 2 ).

48 Convergence behaviour Scheme is unconditionally stable, and computational experiments indicate that U n p = u(x p, t n ) + O( t ν + x 2 ). As ν 1 we have w j = (j + 1) ν j ν 1 so t ν 1 ( νn ν 1 δ 2 Up 0 + Γ(1 + ν) n j=1 ) w n j (δ 2 Up j δ 2 Up j 1 ) δ 2 Up n, and we recover the implicit Euler scheme for classical diffusion: U n p U n 1 p t = δ 2 U n p.

49 Convolution quadrature Lubich, Numer. Math. 52: , Seek weights φ j such that (f g)(t n ) n φ n j g(t j ). j=0

50 Convolution quadrature Lubich, Numer. Math. 52: , Seek weights φ j such that (f g)(t n ) n φ n j g(t j ). j=0 Laplace inversion formula gives t 1 (f g)(t) = e z(t s)ˆf (z) dz g(s) ds 0 2πi Γ }{{} = 1 ˆf (z) 2πi Γ f (t s) t 0 e z(t s) g(s) ds dz.

51 Auxiliary ODE Put so that y(t; z) = t 0 e z(t s) g(s) ds (f g)(t) = 1 ˆf (z)y(t; z) dz. 2πi Γ

52 Auxiliary ODE Put so that y(t; z) = t 0 e z(t s) g(s) ds (f g)(t) = 1 ˆf (z)y(t; z) dz. 2πi Γ Since y = zy + g and y(0) = 0, we have y(t n ; z) Y n (z) where Y n Y n 1 t = zy n + g(t n ), Y 0 = 0,

53 Auxiliary ODE Put so that y(t; z) = t 0 e z(t s) g(s) ds (f g)(t) = 1 ˆf (z)y(t; z) dz. 2πi Γ Since y = zy + g and y(0) = 0, we have y(t n ; z) Y n (z) where so Y n Y n 1 t = zy n + g(t n ), Y 0 = 0, (f g)(t n ) 1 ˆf (z)y n dz. 2πi Γ

54 Generating functions Write then δ(ζ) t Ỹ (ζ) = Y n ζ n, n=0 Ỹ (ζ) = zỹ (ζ) + g(ζ), δ(ζ) = 1 ζ,

55 Generating functions Write then so δ(ζ) t Ỹ (ζ) = Y n ζ n, n=0 Ỹ (ζ) = zỹ (ζ) + g(ζ), δ(ζ) = 1 ζ, Ỹ (ζ) = ( δ(ζ) t 1 z ) 1 g(ζ),

56 Generating functions Write then so δ(ζ) t Ỹ (ζ) = Y n ζ n, n=0 Ỹ (ζ) = zỹ (ζ) + g(ζ), δ(ζ) = 1 ζ, Ỹ (ζ) = ( δ(ζ) t 1 z ) 1 g(ζ), and by Cauchy s theorem, f g(ζ) 1 ˆf (z)ỹ (ζ) dz = ˆf ( δ(ζ) t 1) g(ζ). 2πi Γ

57 Definition of weights Compute φ 0, φ 1,... so that ˆf ( δ(ζ)/ t ) = φ n ζ n = φ(ζ) n=0

58 Definition of weights Compute φ 0, φ 1,... so that ˆf ( δ(ζ)/ t ) = φ n ζ n = φ(ζ) n=0 then f g(ζ) φ(ζ) g(ζ)

59 Definition of weights Compute φ 0, φ 1,... so that ˆf ( δ(ζ)/ t ) = φ n ζ n = φ(ζ) n=0 then f g(ζ) φ(ζ) g(ζ) and so (f g)(t n ) n φ n j g(t j ). j=0

60 Explicit finite difference scheme Yuste and Acedo, SIAM J. Numer. Anal. 42: , Consider u t K(ω ν u xx ) t = 0 and u( L, t) = 0 = u(l, t).

61 Explicit finite difference scheme Yuste and Acedo, SIAM J. Numer. Anal. 42: , Consider u t K(ω ν u xx ) t = 0 and u( L, t) = 0 = u(l, t). Since L{(ω ν v) t } = z 1 ν ˆv(z) we have n (ω ν v) t t ν 1 ω n j Vp j at (x, t) = (x p, t n ), j=0 where ω j ζ j = (1 ζ) 1 ν. j=0

62 Forward difference in time Finite difference approximations u t Un+1 p Up n t lead to the scheme U n+1 p and U n p t u xx δ 2 U n p = Un p+1 2Un p + U n p 1 x 2 n = K t ν 1 ω n j δ 2 Up. j j=0

63 Forward difference in time Finite difference approximations u t Un+1 p Up n t lead to the scheme U n+1 p and U n p t u xx δ 2 U n p = Un p+1 2Un p + U n p 1 x 2 n = K t ν 1 ω n j δ 2 Up. j j=0 If ν 1 then ω j 0 for j 1 and we recover the explicit Euler scheme for classical diffusion: U n+1 p U n p t = Kδ 2 U n p.

64 Convergence behaviour Von-Neumann stability condition: K tν x ν.

65 Convergence behaviour Von-Neumann stability condition: K tν x ν. Formally, error is O( t + x 2 ), but in practice....

66 Implicit convolution quadrature scheme Cuesta, Lubich and Palencia, Math. Comp., 75: , Work with integrated equation where F (t) = t 0 f (s) ds. u + ω ν Au = u 0 + F (t),

67 Implicit convolution quadrature scheme Cuesta, Lubich and Palencia, Math. Comp., 75: , Work with integrated equation where F (t) = t 0 f (s) ds. Compute weights φ n satisfying n=0 u + ω ν Au = u 0 + F (t), ( ) φ n ζ n ( ) δ(ζ) ν = ˆω ν δ(ζ)/ t =, t so that ω ν v n φ n j V j. j=0

68 Convergence behaviour Rearrange equation as (I + ω ν A)(u u 0 ) = ω 1+ν (t)au 0 + F (t), and discretize: n U n u 0 + φ n j A(U j u 0 ) = ω 1+ν (t n )Au 0 + F (t n ). j=0

69 Convergence behaviour Rearrange equation as and discretize: U n u 0 + (I + ω ν A)(u u 0 ) = ω 1+ν (t)au 0 + F (t), n φ n j A(U j u 0 ) = ω 1+ν (t n )Au 0 + F (t n ). j=0 Error analysis shows U n u(t n ) C t 1+ν A ρ u 0 t 1+ν(1 ρ) n + C t 2 ( A σ f (0) t 1 νσ n for 0 ρ 2 and 0 σ 1. + f (0) + tn 0 ) f (s) ds

70 Laplace transforms and quadrature McLean and Thomée, J. Integral Equations Appl., 22: 57 94, We consider u t + (ω ν Au) t = f (t).

71 Laplace transforms and quadrature McLean and Thomée, J. Integral Equations Appl., 22: 57 94, We consider u t + (ω ν Au) t = f (t). Take Laplace transforms to obtain zû(z) u 0 + z 1 ν Aû(z) = ˆf (z),

72 Laplace transforms and quadrature McLean and Thomée, J. Integral Equations Appl., 22: 57 94, We consider u t + (ω ν Au) t = f (t). Take Laplace transforms to obtain zû(z) u 0 + z 1 ν Aû(z) = ˆf (z), or (zi + z 1 ν A)û(z) = u 0 + ˆf (z).

73 Laplace transforms and quadrature McLean and Thomée, J. Integral Equations Appl., 22: 57 94, We consider u t + (ω ν Au) t = f (t). Take Laplace transforms to obtain or zû(z) u 0 + z 1 ν Aû(z) = ˆf (z), (zi + z 1 ν A)û(z) = u 0 + ˆf (z). Thus, we can find û(z) by solving an elliptic boundary value problem with complex coefficients.

74 Resolvent estimate Spectrum of A lies in [0, ) so for each ϕ > 0, (λi A) 1 C ϕ λ for arg λ ϕ.

75 Resolvent estimate Spectrum of A lies in [0, ) so for each ϕ > 0, (λi A) 1 C ϕ λ for arg λ ϕ. Thus, (zi + A) 1 C ϕ z for arg z π ϕ.

76 Resolvent estimate Spectrum of A lies in [0, ) so for each ϕ > 0, (λi A) 1 C ϕ λ for arg λ ϕ. Thus, We have so (zi + A) 1 C ϕ z for arg z π ϕ. (zi + z 1 ν A)û(z) = u 0 + ˆf (z), (z ν I + A)û(z) = z ν 1 [u 0 + ˆf (z)],

77 Resolvent estimate Spectrum of A lies in [0, ) so for each ϕ > 0, (λi A) 1 C ϕ λ for arg λ ϕ. Thus, We have so (zi + A) 1 C ϕ z which is solvable with for arg z π ϕ. (zi + z 1 ν A)û(z) = u 0 + ˆf (z), (z ν I + A)û(z) = z ν 1 [u 0 + ˆf (z)], û(z) C ϕ u 0 + ˆf (z) z for arg z π ϕ. ν

78 Integral representation Laplace inversion formula: u(t) = 1 2πi Γ e zt û(z) dz.

79 Integral representation Laplace inversion formula: u(t) = 1 2πi Γ e zt û(z) dz. Deform contour Γ to the curve with parametric representation z(ξ) = a + b ( 1 sin(δ iξ) ), < ξ <, which is the left branch of a hyperbola with asymptotes y = ±(x a b) cot δ.

80 Integral representation Laplace inversion formula: u(t) = 1 2πi Γ e zt û(z) dz. Deform contour Γ to the curve with parametric representation z(ξ) = a + b ( 1 sin(δ iξ) ), < ξ <, which is the left branch of a hyperbola with asymptotes y = ±(x a b) cot δ. Must assume ˆf (z) is analytic to the right of Γ.

81 Quadrature We have u(t) = 1 e z(ξ)t û ( z(ξ) ) z (ξ) dξ. 2πi

82 Quadrature We have u(t) = 1 e z(ξ)t û ( z(ξ) ) z (ξ) dξ. 2πi Integrand exhibits a double-exponential decay as ξ, since e z(ξ)t = e at e bt(1 sin δ cosh ξ).

83 Quadrature We have u(t) = 1 e z(ξ)t û ( z(ξ) ) z (ξ) dξ. 2πi Integrand exhibits a double-exponential decay as ξ, since e z(ξ)t = e at e bt(1 sin δ cosh ξ). Choose a quadrature step k > 0, then u(t) U N (t) = k 2πi N j= N e z j tû(z j )z j, where z j = z(ξ j ), z j = z (ξ j ), ξ j = jk.

84 Exponential convergence Given N and a time scale T, can choose a, b, δ so that, U N (t) u(t) Ce µn( u 0 + max ˆf (z) ) for T t 2T, where µ > 0 and the max is over z in a certain set including Γ.

85 Exponential convergence Given N and a time scale T, can choose a, b, δ so that, U N (t) u(t) Ce µn( u 0 + max ˆf (z) ) for T t 2T, where µ > 0 and the max is over z in a certain set including Γ. We have to solve 2N + 1 elliptic problems to find the û(z j ).

86 Exponential convergence Given N and a time scale T, can choose a, b, δ so that, U N (t) u(t) Ce µn( u 0 + max ˆf (z) ) for T t 2T, where µ > 0 and the max is over z in a certain set including Γ. We have to solve 2N + 1 elliptic problems to find the û(z j ). The elliptic problems can be solved in parallel.

87 Exponential convergence Given N and a time scale T, can choose a, b, δ so that, U N (t) u(t) Ce µn( u 0 + max ˆf (z) ) for T t 2T, where µ > 0 and the max is over z in a certain set including Γ. We have to solve 2N + 1 elliptic problems to find the û(z j ). The elliptic problems can be solved in parallel. A variation on the method avoids using ˆf (z) and achieves accuracy O(e µ N ). The elliptic problems take the form t ) (z ν I + A)w(z, t) = z (e ν 1 zt u 0 + e z(t s) f (s) ds. 0

88 Part III Discontinuous Galerkin Methods

89 Trial space Nonuniform grid: 0 = t 0 < t 1 < t 2 < < t N = T with k n = t n t n 1 and k = max 1 n N k n.

90 Trial space Nonuniform grid: 0 = t 0 < t 1 < t 2 < < t N = T with k n = t n t n 1 and k = max 1 n N k n. For each half-open interval I n = (t n 1, t n ] we choose a subspace S n D(A 1/2 ) H 1 (Ω).

91 Trial space Nonuniform grid: 0 = t 0 < t 1 < t 2 < < t N = T with k n = t n t n 1 and k = max 1 n N k n. For each half-open interval I n = (t n 1, t n ] we choose a subspace S n D(A 1/2 ) H 1 (Ω). Let W r denote the space of piecewise polynomial functions V of order r: V (t) = r a p t p 1 for t I n, with a p S n. p=1

92 Trial space Nonuniform grid: 0 = t 0 < t 1 < t 2 < < t N = T with k n = t n t n 1 and k = max 1 n N k n. For each half-open interval I n = (t n 1, t n ] we choose a subspace S n D(A 1/2 ) H 1 (Ω). Let W r denote the space of piecewise polynomial functions V of order r: V (t) = r a p t p 1 for t I n, with a p S n. p=1 Since V may be discontinuous at t n, write V n = V (t n ) = V (t n ), V n + = V (t + n ), [V ] n = V n + V n.

93 Fractional derivative of a discontinuous function Integration by parts shows that for t I n, n 1 B ν V (t) = ω ν (t)v+ 0 + ω ν (t t j )[V ] j + continuous terms. j=1 Thus, B ν V (t) is left-continuous at t = t n but behaves like (t t n 1 ) ν 1 as t t + n 1.

94 Weak formulation Exact solution satisfies [ u, v + A(B ν u, v) ] dt = f, v dt I n I n for any continuous test function v : [t n 1, t n ] D(A 1/2 ).

95 Weak formulation Exact solution satisfies [ u, v + A(B ν u, v) ] dt = f, v dt I n I n for any continuous test function v : [t n 1, t n ] D(A 1/2 ). Discontinuous Galerkin (DG) solution U W r defined by U n 1 +, V n I n [ U, V + A(B ν U, V ) ] dt for every test function V W r, with U 0 u 0. = U n 1, V n I n f, V dt

96 Piecewise-constant case (implicit Euler) When r = 1, we have U(t) = U n for t I n, and the DG method reduces to U n, χ + k n A( B n νu, χ) = U n 1, χ + k n f n, χ for all χ S n, where B ν n = 1 k n I n B ν U dt and f n = 1 k n I n f dt.

97 Piecewise-constant case (implicit Euler) When r = 1, we have U(t) = U n for t I n, and the DG method reduces to U n, χ + k n A( B n νu, χ) = U n 1, χ + k n f n, χ for all χ S n, where B ν n = 1 k n I n B ν U dt and f n = 1 k n I n f dt. In other words, U n U n 1, χ + A( B νu, n χ) = f n, χ. k n

98 Discrete fractional derivative We find ( B νv n = kn 1 β nn V n n 1 ) β nj V j where β nn = ω ν (t n s) ds = ω 1+ν (k n ) = kn ν /Γ(1 + ν) > 0, I n [ β nj = ων (t n 1 s) ω ν (t n s) ] ds > 0, 1 j n 1. I j j=1

99 Discrete fractional derivative We find ( B νv n = kn 1 β nn V n n 1 ) β nj V j where β nn = ω ν (t n s) ds = ω 1+ν (k n ) = kn ν /Γ(1 + ν) > 0, I n [ β nj = ων (t n 1 s) ω ν (t n s) ] ds > 0, 1 j n 1. I j Thus, n 1 U n, χ + β nn A(U n, χ) = U n 1 + k n f n, χ + β nj A(U j, χ). j=1 j=1

100 Discrete fractional derivative We find ( B νv n = kn 1 β nn V n n 1 ) β nj V j where β nn = ω ν (t n s) ds = ω 1+ν (k n ) = kn ν /Γ(1 + ν) > 0, I n [ β nj = ων (t n 1 s) ω ν (t n s) ] ds > 0, 1 j n 1. I j Thus, n 1 U n, χ + β nn A(U n, χ) = U n 1 + k n f n, χ + β nj A(U j, χ). If ν 1 then β nn k n and β nj 0. j=1 j=1

101 Stability via energy argument Taking V = U and using U, U = 1 2 (d/dt) U 2 we find ( 1 2 U j 2 + U j 1 + 2) + A(B ν U, U) dt I j = U j 1, U j I j f, U dt.

102 Stability via energy argument Taking V = U and using U, U = 1 2 (d/dt) U 2 we find ( 1 2 U j 2 + U j 1 + 2) + A(B ν U, U) dt I j Summing over j and using positivity we find j=1 = U j 1, U j I j f, U dt. n 1 ( U n 2 + U [U] j 2 2 U 0, U+ 0 + tn 0 ) f, U dt,

103 Stability via energy argument Taking V = U and using U, U = 1 2 (d/dt) U 2 we find ( 1 2 U j 2 + U j 1 + 2) + A(B ν U, U) dt I j Summing over j and using positivity we find j=1 = U j 1, U j I j f, U dt. n 1 ( U n 2 + U [U] j 2 2 U 0, U+ 0 + tn 0 ) f, U dt, and then tn ) U n C r ( U 0 + f dt. 0

104 Projection operator Define Πu on I j to be the unique polynomial of order r satisfying Πu(t j ) = u(t j ) and I j [u(t) Πu(t)]t p 1 dt = 0 for p = 1, 2,..., r 1.

105 Projection operator Define Πu on I j to be the unique polynomial of order r satisfying Πu(t j ) = u(t j ) and I j [u(t) Πu(t)]t p 1 dt = 0 for p = 1, 2,..., r 1. If r = 2 then Πu(t) = u(t n ) + u(t n) ū n (t t n ) for t I n, k n /2 where ū n = kn 1 I n u(t) dt.

106 Error from the time discretization Suppose S n = L 2 (Ω), so there is no space discretization.

107 Error from the time discretization Suppose S n = L 2 (Ω), so there is no space discretization. Write v J = sup t J v(t) and J n = (0, t n ].

108 Error from the time discretization Suppose S n = L 2 (Ω), so there is no space discretization. Write v J = sup t J v(t) and J n = (0, t n ]. Optimal approximation: u Πu In C r k q 1 n I n u (q) (t) dt for q = 1, 2,..., r.

109 Error from the time discretization Suppose S n = L 2 (Ω), so there is no space discretization. Write v J = sup t J v(t) and J n = (0, t n ]. Optimal approximation: u Πu In C r k q 1 n Split the DG error into I n u (q) (t) dt for q = 1, 2,..., r. U u = θ + η where θ = U Πu W r and η = Πu u.

110 Error from the time discretization Suppose S n = L 2 (Ω), so there is no space discretization. Write v J = sup t J v(t) and J n = (0, t n ]. Optimal approximation: u Πu In C r k q 1 n Split the DG error into I n u (q) (t) dt for q = 1, 2,..., r. U u = θ + η where θ = U Πu W r and η = Πu u. Can show n 1 tn θ 2 J n + θ n 2 + [θ] j 2 C r B ν Aη, θ dt. j=1 0

111 Piecewise-constant case If r = 1 then tn 0 B ν Aη, θ dt = n B ν Aη dt, θ j = I j j=1 and k n B νaη n = tn 0 δ n (t)au (t) dt. n k j B νaη, j θ j j=1

112 Piecewise-constant case If r = 1 then tn 0 B ν Aη, θ dt = n B ν Aη dt, θ j = I j j=1 and k n B νaη n = tn 0 δ n (t)au (t) dt. n k j B νaη, j θ j j=1

113 Error analysis (r = 1) Easy to estimate n j=1 δ j(t) and hence show U n u(t n ) U 0 u Γ(1 + ν) n j=1 I j (t t j 1 ) ν Au (t) dt.

114 Error analysis (r = 1) Easy to estimate n j=1 δ j(t) and hence show U n u(t n ) U 0 u 0 + Tricky argument involving j (t) = t 0 4 Γ(1 + ν) n j=1 I j (t t j 1 ) ν Au (t) dt. δ j (s) ds for t 1 t t j, 2 j n, shows ( U n u(t n ) U 0 1 u t ν Au (t) dt Γ(1 + ν) I 1 n ) + 2k n tn ν Au (t n ) + 6 k j t I ν Au (t) dt. j j=2

115 Convergence behaviour (r = 1) Numer. Algorithms, 52: 69 88, Graded mesh: t n = (n/n) γ T for 0 n N, with γ 1.

116 Convergence behaviour (r = 1) Numer. Algorithms, 52: 69 88, Graded mesh: t n = (n/n) γ T for 0 n N, with γ 1. If t ν Au (t) Ct σ 1, then U n u(t n ) U 0 u 0 + Ck ν provided γ max(1, ν/σ).

117 Convergence behaviour (r = 1) Numer. Algorithms, 52: 69 88, Graded mesh: t n = (n/n) γ T for 0 n N, with γ 1. If t ν Au (t) Ct σ 1, then U n u(t n ) U 0 u 0 + Ck ν provided γ max(1, ν/σ). If also t ν+1 Au (t) Ct σ 1, then U n u(t n ) U 0 u 0 + Ck provided γ max(1, 1/σ).

118 Convergence behaviour (r = 1) Numer. Algorithms, 52: 69 88, Graded mesh: t n = (n/n) γ T for 0 n N, with γ 1. If t ν Au (t) Ct σ 1, then U n u(t n ) U 0 u 0 + Ck ν provided γ max(1, ν/σ). If also t ν+1 Au (t) Ct σ 1, then U n u(t n ) U 0 u 0 + Ck provided γ max(1, 1/σ). Spatial discretization with piecewise-linear finite elements leads to additional error of order h 2 log(t 1 /t n ).

119 Convergence behaviour for piecewise-linears (r = 2). Numer. Algorithms, to appear. Can show U u Jn U 0 u 0 + u Πu Jn + C t ν Au (t) dt I 1 + C n j=2 k 1+ν j I j Au (t) dt.

120 Convergence behaviour for piecewise-linears (r = 2). Numer. Algorithms, to appear. Can show U u Jn U 0 u 0 + u Πu Jn + C t ν Au (t) dt I 1 + C n j=2 k 1+ν j I j Au (t) dt. Gives error of order k 1+ν provided γ > (1 + ν)/σ.

121 Convergence behaviour for piecewise-linears (r = 2). Numer. Algorithms, to appear. Can show U u Jn U 0 u 0 + u Πu Jn + C t ν Au (t) dt I 1 + C n j=2 k 1+ν j I j Au (t) dt. Gives error of order k 1+ν provided γ > (1 + ν)/σ. In practice, observe O(k 2 ) provided γ > 2/σ.

122 Convergence behaviour for piecewise-linears (r = 2). Numer. Algorithms, to appear. Can show U u Jn U 0 u 0 + u Πu Jn + C t ν Au (t) dt I 1 + C n j=2 k 1+ν j I j Au (t) dt. Gives error of order k 1+ν provided γ > (1 + ν)/σ. In practice, observe O(k 2 ) provided γ > 2/σ. Spatial discretization with piecewise-linear finite elements again gives additional error of order h 2 log(t 1 /t n ).

123 Convergence behaviour for piecewise-linears (r = 2). Numer. Algorithms, to appear. Can show U u Jn U 0 u 0 + u Πu Jn + C t ν Au (t) dt I 1 + C n j=2 k 1+ν j I j Au (t) dt. Gives error of order k 1+ν provided γ > (1 + ν)/σ. In practice, observe O(k 2 ) provided γ > 2/σ. Spatial discretization with piecewise-linear finite elements again gives additional error of order h 2 log(t 1 /t n ). New paper proves error of order k ρ for ρ = min(2, ν).