Convergence analysis of a discontinuous Galerkin method for a sub-diffusion equation

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1 Numer Algor (29 52:69 88 DOI 1.17/s ORIGINAL PAPER Convergence analysis of a discontinuous Galerkin method for a sub-diffusion equation William McLean Kassem Mustapha Received: 16 September 28 / Accepted: 2 December 28 / Published online: 19 December 28 Springer Science + Business Media, LLC 28 Abstract We employ a piecewise-constant, discontinuous Galerkin method for the time discretization of a sub-diffusion equation. Denoting the maximum time step by k, we prove an a priori error bound of order k under realistic assumptions on the regularity of the solution. We also show that a spatial discretization using continuous, piecewise-linear finite elements leads to an additional error term of order h 2 max(1, log k 1. Some simple numerical examples illustrate this convergence behaviour in practice. Keywords Non-uniform time steps Memory term Finite elements 1 Introduction We study the initial-boundary value problem t u + t α Au = f for < t < T, withu = u at t =, (1 in the case 1 <α<, where the fractional power of t is interpreted in the Riemann Liouville sense and typically A = 2 on a bounded domain. We thank the University of New South Wales for financial support provided by a Faculty Research Grant. W. McLean (B School of Mathematics and Statistics, The University of New South Wales, Sydney 252, Australia w.mclean@unsw.edu.au K. Mustapha Department of Mathematics and Statistics, King Fahd University of Petroleum and Minerals, Dhahran, 31261, Saudi Arabia kassem@kfupm.edu.sa

2 7 Numer Algor (29 52:69 88 (Section 2 sets out precise technical assumptions. The problem (1 provides a macroscopic, continuum model of sub-diffusion, with u giving the density of the diffusing particles that have mean-square displacement proportional to t 1+α ; see the survey paper by Metzler and Klafter [5]. In the limiting case α = we have Brownian motion, and u obeys the classical heat equation. However, for 1 <α< the fractional-order derivative makes the evolution equation nonlocal in time. In a previous work [9], we studied discontinuous Galerkin methods for the time discretization of (1 inthecase <α<1, proving optimal error bounds for piecewise-constant and piecewise-linear trial functions. The convergence analysis for the present case 1 <α< is more difficult and we treat only piecewise-constants. Thus, our scheme is essentially a modified implicit Euler method. By proceeding much as in [9], we prove sub-optimal convergence of order k 1+α in Section 3, wherek denotes the maximum time-step, assuming suitable control of Au. Subsequently, in Section 4 we show an optimal convergence rate of order k, but with Au appearing in the error bound. Since Au (t and Au (t are typically singular at t =, to achieve the convergence rates cited above we must in practice employ non-uniform time steps. Langlands and Henry [2] have also studied a scheme of implicit Euler type for (1, but with a different treatment of the fractional derivative and employing only uniform time steps. They provided a partial error analysis and presented numerical experiments indicating O(k 1+α convergence when α = 1/2. Their method also incorporated the usual second-order central finite difference approximation of Au = u xx, giving an additional error term of order h 2 for a uniform spatial step-size h. Cuesta et al. [1] studied formally second-order accurate, convolution quadrature schemes for (1. They worked with the integrated form of the evolution equation, in our notation u + t 1 α u = u + t 1 f, and proved O(k 2 convergence if <α<1but only O(k 2+α convergence if 1 <α<. Schädle et al. [3, 1] have developed fast algorithms for evaluating convolution quadrature sums and for reducing the memory requirements of such methods. Another kind of scheme, [4, 6, 7], involving Laplace transformation combined with quadrature along a contour in the complex plane, provides spectral accuracy for the time discretization, but appears to offer little scope for handling nonlinear versions of (1. The formulation of our numerical method in Section 2 allows for a spatial discretization by a Galerkin method using a trial space in D(A 1/2, the domain of A 1/2. For instance, our approximate solution U(x, t may be continuous and piecewise-linear in x but discontinuous and piecewise-constant in t. We find in Theorem 1 that the method is unconditionally stable even when we choose a different trial space for each time step combined with arbitrarily-spaced time levels. This robustness is a significant advantage of the discontinuous Galerkin method and allows great flexibility in the choice of mesh. The error bounds in Sections 3 and 4 deal with the time discretization only, since we assume that the trial space for the spatial discretization is the whole of D(A 1/2.Section5 investigates the additional error incurred when

3 Numer Algor (29 52: the trial space is a proper subspace of D(A 1/2. For simplicity, we focus on using continuous piecewise linears with a fixed spatial mesh, showing that the error associated with the spatial discretization is of order h 2 l(k, where l(k = max(1, log k 1. (2 Although the scheme treated here is only first-order accurate with respect to t, our analysis provides a foundation for studying the convergence of higherorder discontinuous Galerkin methods for (1when 1 <α<. 2 Notation and assumptions To formulate our initial-boundary value problem (1 in an abstract setting, let H be a real Hilbert space with inner product, and norm. We suppose that A is a self-adjoint linear operator with dense domain D(A H, thata possesses a complete orthonormal eigensystem {φ m } m=1, Aφ m = λ m φ m and φ m,φ m = δ m,m for m, m 1, and that λ 1 λ 2 λ 3. The associated bilinear form, A(u,v= Au, v = λ m u,φ m φ m, v for u, v D(A 1/2, (3 m=1 is thus positive semi-definite. For instance, when H = L 2 (Ω for a bounded, Lipschitz domain Ω R d and A = 2 with homogeneous Dirichlet or Neumann boundary conditions, we have A(u, v = Ω u vdx. Let ω α (t = tα 1 for < t < and <α<, Γ(α and denote the Riemann Liouville fractional integration operator of order α by B α v(t = t ω α (t sv(s ds for < t <. We extend this definition to obtain a fractional derivative by setting B α v(t = (B 1+α v (t for 1 <α<. Since the Laplace transform of ω α (t is z α, one finds that, for suitably regular v, with lim B αv(t = v (t and lim B α v(t = α 1 α +1 B v(t lim α B α v(t = v(t. t v(s ds,

4 72 Numer Algor (29 52:69 88 Furthermore, by applying the Parseval Plancherel theorem we may show v(tb α v(t dt = 1 y α cos(πα ˆv(iy 2 dy for 1 <α<1, 2π (4 where ˆv denotes the Laplace transform of v. Assuming 1 <α< and setting t α = B α in (1, our problem is to find u :[, T] H satisfying u + B α Au = f (t for < t < T, with u( = u, (5 ( for a given initial datum u H and inhomogeneous term f L 1 (, T; H. By writing v m (t = v(t, φ m, we see from (3 and(4 that T A ( B α v(t, v(t T dt = λ m v m (tb α v m (t dt. (6 m=1 Using this positivity property, one may show via an energy argument that the abstract initial-value problem (5 admits a unique mild solution [7, 8], and that this solution is stable in the sense that u(t u +2 t f (s ds for t T. (7 To discretize in time, we introduce grid points = t < t 1 < t 2 < < t N = T, define the half-open interval I n = (t n 1, t n ] with length k n = t n t n 1 for 1 n N, and denote the maximum step-size by k = max 1 n N k n. In practice, our chief interest is in a smoothly-graded mesh of the form t n = (n/n γ T for n N, withγ 1. (8 Such meshes compensate for the singular behaviour of the solution of (5 as t. Typically, we have positive constants M and σ such that t 1+α Au (t +t 2+α Au (t Mt σ 1 for < t T. (9 For instance, if f and u D(A r for some r >, then(9 holds with σ = (1 + αr; see[7, Theorem 2.1] and [1, equation (8.2]. For each time interval I n we select a subspace S n D(A 1/2 and let W denote the resulting piecewise-constant trial space. Thus, U W has the form U(t = U n for t I n, whereu n S n. In particular, U(t n = U n and we write U+ n = U(t+ n = lim t t n U(t and [U] n = U+ n U n,withu( = U.SinceU is constant on I n+1, we have U+ n = U n+1 and so [U] n = U n+1 U n. For any continuous test function v :[t n 1, t n ] D(A 1/2, the solution of the initial value problem (5 satisfies [ u (t, v(t + A ( B α u(t, v(t ] dt = f (t, v(t dt, (1 I n I n

5 Numer Algor (29 52: whereas for the discontinuous Galerkin solution U W we require that U n 1 +, [ Xn U (t, X(t + A ( B α U(t, X(t ] dt I n = U n 1, X+ n 1 + f (t, X(t dt, for X W and 1 n N. (11 I n Given a suitable U u, the equations (11 determine U(t for t T. In fact, since U (t = on I n and U+ n 1 = U n, we may rewrite (11 as a modified backward-euler scheme, U n U n 1,χ + A(B n α k U,χ= f n,χ for all χ S n, (12 n where f n = k 1 n f (t dt is the average value of f over I n, and likewise I n B n α v = 1 k n I n B α v(t dt = 1 k n [ (B 1+α v(t n (B 1+α v(t n 1 ]. Defining β nn = I n ω 1+α (t n s ds = ω 2+α (k n = k 1+α n /Γ (2 + α and [ β nj = ω1+α (t n 1 s ω 1+α (t n s ] ds for 1 j n 1, I j and noting that these weights are positive because ω 1+α (t is positive and strictly decreasing for < t <, we have in effect a nonlocal finite difference approximation for the fractional derivative, B α v(t n B n α V = k 1 n ( β nn V n β nj V j if v V W. (13 Notice that the scheme (12 is implicit: to find U n we must solve the variational problem U n,χ + β nn A(U n,χ= U n 1 + k n f n,χ β nj A(U j,χ for all χ S n. n 1 + (14 We remark that since ω 1 (t 1, ifα then β nn k n and β nj for j < n, so our method reduces to the classical implicit Euler scheme for a parabolic PDE, but with the average value f n in place of the pointwise value f (t n. 3 Error from the time discretization We prove stability for our numerical scheme by adapting a proof from [8, Lemma 2.2].

6 74 Numer Algor (29 52:69 88 Theorem 1 Given U H and f L 1 ( (, T; H, the discontinuous Galerkin method (11 has a unique solution U W.Furthermore,U n D(A for n 1, and U n U +2 N k n f n for n N. n=1 Proof Putting χ = φ m in (14, and writing Um n = U n,φ m and f m n = f n,φ m, we find that When n = 1, m=1 (1 + β nn λ m U n m = U n 1 m m=1 n 1 + k f n m n + β nj λ m Um j. ( AU 1 2 = (λ m Um 1 λ 2 m ( 2 = U m 1 + β 11 λ +k f 1 m 1 2 U + k 1 f 1 2 <. m (β 11 2 Thus, U 1 D(A, and by induction on n we find that U n D(A for 1 n N. To prove the stability estimate, we note that U n U n 1, 2U n = U n U n 1, U n + U n 1 + U n U n 1, U n U n 1 = U n 2 U n [U] n 1 2, so taking χ = 2k n U n in (12 gives U n 2 U n [U] n k n A(B n α U, U n = 2k n f n, U n. (15 Summing from n = 1 to N, and noting that, by (6, we see that N k n A(B n α U, U n = T n=1 A ( B α U(t, U(t dt, N 1 N U N 2 U 2 + [U] n 2 2 k n f n, U n. n= n=1 Choosing n such that U n =max 1 n N U n, we have n ( U n 2 U k n f n U n U n U +2 n=1 and the estimate follows at once. N n=1 k n f n,

7 Numer Algor (29 52: To bound the error in U, we define the piecewise-constant interpolant Πu(t = u(t n for t I n and 1 n N, with Πu( = u(, and put θ = U Πu and η = Πu u, so that U u = θ + η. We may estimate η from the integral representation η(t = tn t u (s ds for t I n, (16 and in the proof of the next result we use Theorem 1 to estimate θ in terms of η. Theorem 2 If S n = D(A 1/2 for 1 n N, then for the discontinuous Galerkin method (11 we have the error bound U n u(t n U 4 u + (t t j 1 1+α Au (t dt. Γ(2 + α I j Proof By letting v(t = χ S n in (1, we see that the solution of the continuous problem satisfies u(t n u(t n 1, χ + k n A (B n α u,χ = k n f n,χ, (17 whereas by (12, U n U n 1,χ + k n A (B n α U,χ = k n f n,χ. (18 It follows that θ n = U n u(t n S n = D(A 1/2 satisfies θ n θ n 1,χ + k n A (B n α θ,χ = k n A (B n α η, χ = k n B n α Aη, χ for all χ S n. Thus, θ is the discontinuous Galerkin solution for the problem with initial datum θ = U u and inhomogeneous term B α Aη, and so by Theorem 1, U n u(t n = θ n U u +2 k j B α j Aη for 1 n N. (19 Since k n B n α Aη = B 1+α Aη(t n B 1+α Aη(t n 1, we define { ω 1+α (t n s >, t n 1 < s < t n, Λ n (s = ω 1+α (t n s ω 1+α (t n 1 s <, < s < t n 1, and δ n (t = t t j 1 Λ n (s ds for t I j, (2

8 76 Numer Algor (29 52:69 88 so that, using (16 and reversing the order of integration, tn k n B n α Aη = Λ n (saη(s ds = Λ n (s I j s = tn t j Au (t dt ds δ n (tau (t dt. (21 Figure 1 illustrates the behaviour of the function δ n, and we see that k j B α j Aη j i=1 I i δ j (t Au (t dt = If t I i then δ i (t > but δ j (t < for j > i,so t ( δ j (t =δ i (t δ j (t = Λ i (s j=i = t t i 1 t j=i+1 ( ω 1+α (t i s t i 1 i=1 j=i+1 I i δ j (t Au (t dt. j=i Λ j (s ds ( ω1+α (t j s ω 1+α (t j 1 s ds j=i+1 ( = 2ω1+α (t i s ω 1+α (t n s ds t i 1 2ω 2+α (k i 2ω 2+α (t i t 2ω 2+α (t t i 1, (22 where the final step relies on the fact that < 1 + α<1. We see at once that if Au (t is bounded then the sum appearing in the error bound of Theorem 2 is O(k 1+α. The next theorem shows that this convergence rate holds also when Au (t is singular at t = provided we use a Fig. 1 The function defined in (

9 Numer Algor (29 52: suitable mesh grading. More precisely, we suppose that the time steps satisfy, for some γ 1, k n Ck min ( 1, tn 1 1/γ and t n Ct n 1 for 2 n N, (23 with ck γ k 1 Ck γ. (24 For example, these bounds are satisfied in the case of our standard example (8. Theorem 3 Let S n = D(A 1/2 for 1 n N, assume that u has the regularity property (9 with <σ <1 + α, and put γ = (1 + α/σ > 1. If (23 and (24 hold, then k γσ, for 1 γ<γ, U n u(t n U u +CM k log(t n /t 1, for γ = γ, tn σ (1+α/γ k 1+α, for γ>γ. Proof We estimate n I j (t t j 1 1+α Au (t dt from the error bound of Theorem 2. For the first subinterval, (24 implies that k1 t 1+α Au (t dt M I 1 t σ 1 dt = Mσ 1 k σ 1 CMkγσ, (25 and for the subsequent intervals (23 gives (t t j 1 1+α Au (t dt M I j k 1+α j t σ 1 (1+α dt I j tn CMk 1+α t σ (1+α/γ 1 dt. The error bound now follows from tn k σ (1+α/γ t σ (1+α/γ 1 1 Ck γσ (1+α, γ < γ, dt C log(t n /t 1, γ = γ t 1, tn σ (1+α/γ, γ > γ. t 1 (26 4 Improved convergence rate The O(k 1+α error bound from Theorem 3 is suboptimal, and we prove below that the error is in fact O(k assuming u has the regularity properties (9. For δ j defined by (2, we put Δ j (t = t t 1 δ j (s ds for t 1 t t j and j 2.

10 78 Numer Algor (29 52:69 88 Our proof depends on the following properties of these functions. D1: Δ j (t Δ j (t j 1 for t 1 t t j 1. D2: Δ j (t j Δ j (t Δ j (t j 1 for t j 1 t t j. D3: Δ j (t j <. The first two properties are clear because Δ j (t 1 = and Δ j (t = δ j(t > for t 1 t t j 1, whereas Δ j (t = δ j(t < for t j 1 t t j. The third property requires some additional assumptions on the mesh. Lemma 1 A sufficient condition for D3 is that k j k j 1 and k i k i 1 for 3 i j 1. t j t i t j 1 t i 1 Proof Let Ω ji = (t i sω 1+α (t j s ds = I i We have Δ 2 (t 1 = and, for j 3, i=2 ki t j 1 j 1 Δ j (t j 1 = δ j (t dt = t 1 I i sω 1+α (t j t i + s ds for 2 i j n. t t i 1 Λ j (s ds dt j 1 = (t i s [ ω 1+α (t j 1 s ω 1+α (t j s ] j 1 ds = (Ω j 1,i Ω ji, I i i=2 i=2 (27 with so Δ j (t j 1 Δ j (t j = δ j (t dt = ω 1+α (t j s ds dt I j I j t j 1 = (t j sω 1+α (t j s ds = Ω jj for j 2, (28 I j t j 1 Δ j (t j = Ω jj (Ω j 1,i Ω ji = Ω j2 + i=2 i=3 j (Ω ji Ω j 1,i 1. Since Ω jj =(1+αω 3+α (k j, the assumption k j k j 1 ensures Ω jj Ω j 1, j 1. For 3 i j 1, the substitution s = (t j t i y gives ( x Ω ji = (t j t i 2+α ki F where F(x = yω 1+α (1 + y dy. t j t i

11 Numer Algor (29 52: The first of the stated conditions ensures t j t i = (t j 1 t i 1 + (k j k i t j 1 t i 1, and the second ensures F ( k i /(t j t i F ( k i 1 /(t j 1 t i 1 because F is a monotone increasing function. Therefore, Ω ji Ω j 1,i 1 for 3 i j and property D3 holds. We now verify that the sufficient condition of Lemma 1 holds for our standard mesh. Lemma 2 D3 is satisfied for the graded mesh (8. Proof Recall that t i = (i/n γ T. The first inequality of Lemma 1 follows from j γ ( j 1 γ = γ = γ j j 1 j 1 j 2 t γ 1 dt γ and to prove the second we note that i [i γ (i 1 γ ][( j 1 γ (i 1 γ ]=γ 2 so i γ (i 1 γ j γ i γ j j 1 (t 1 γ 1 dt t γ 1 dt = ( j 1 γ ( j 2 γ, = γ 2 i s γ 1 ds i 1 j 1 i 1 = γ 2 i 1 i 2 = γ 2 i 1 i 2 γ 2 i 1 i 2 i 1 j i j i j i j 1 i 1 t γ 1 dt s γ 1 t γ 1 dt ds (s + 1 γ 1 (t 1 γ 1 dt ds (st + t s 1 γ 1 dt ds (st γ 1 dt ds =[(i 1 γ (i 2 γ ][ j γ i γ ], (i 1γ (i 2 γ ( j 1 γ (i 1 γ for 3 i j 1. Theorem 4 If S n = D(A 1/2 for 1 n N and if the mesh is of the form (8, then U n u(t n U 4 u + t 1+α Au (t dt Γ(2 + α I 1 + 2t1+α n Γ(2 + α k n Au 6 (t n + k j t 1+α j Au (t dt. Γ(2 + α I j

12 8 Numer Algor (29 52:69 88 Proof Since Δ j (t 1 =, integration by parts gives t j δ j (tau (t dt = δ j (tau (t dt Δ j (t j Au (t j + I 1 t 1 I 1 t j Δ j (tau (t dt for j 2, and hence (21 shows that k j B α j Aη δ j (t Au (t dt + Δ j (t j Au (t j + j i=2 I i Δ j (t Au (t dt. Recalling (22 and interchanging the order of the double sum, we see that k j B α I j Aη 2 ω 2+α (t Au (t dt + Δ j (t j Au (t j 1 + Put Δ j (t Au (t dt. (29 i=2 I i j=i G j = j i=2 I i (t i tω 1+α (t j t dt and observe from (27and(28that Δ j (t j = G j 1 G j for j 2, with G j ( t max k j i ω 1+α (t j t dt k j ω 2+α (t j. (3 2 i j t 1 Property D3 implies Δ j (t j Au (t j =(G j G j 1 Au (t j = G j Au (t j Au (t j+1 + Au (t j+1 G j 1 Au (t j G j Au (t j+1 Au (t j +G j Au (t j+1 G j 1 Au (t j G j Au (t j+1 G j 1 Au (t j +G j Au I (t dt, j

13 Numer Algor (29 52: so, since G 1 =, Δ j (t j Au (t j =(G n G n 1 Au (t n + Δ j (t j Au (t j G n Au (t n + G j Au I (t dt. j To estimate the double sum in (29, write Δ j (t = g j 1 (t g j (t where i 1 g j (t = (t l zω 1+α (t j z dz I l l=2 t + (t zω 1+α (t j z dz for t I i and 2 i j. t i 1 If t I i,then [ Δ j (t = g j 1 (t g j (t ] = g i (t g n (t g i (t i = G i j=i+1 j=i+1 and, using properties D2 and D3, Δ i (t max ( Δ i (t i 1, Δ i (t i = max ( g i 1 (t i 1 g i (t i 1, G i G i 1 = max ( G i 1 g i (t i 1, G i G i 1 max(gi 1, G i = G i, so Δ j (t Au (t dt 2G i Au I (t dt. i i=2 I i j=i i=2 The desired error bound now follows by (19and(3. We easily deduce the O(k convergence rate given an appropriate mesh grading. Theorem 5 Let S n = D(A 1/2 for 1 n N and suppose that u has the regularity property (9 with <σ <1.Ifweusethegradedmesh(8,then k γσ, 1 γ<1/σ, U n u(t n U u +CM k log(t n /t 1, γ = 1/σ, tn σ 1/γ k, γ > 1/σ. Proof Consider the terms in the error bound of Theorem 4. We have already dealt with the contribution from the integral over I 1 ;see(25. Using (9 and (23 we find that tn 1+α k n Au (t n CMk n tn σ 1 CMktn σ 1/γ,

14 82 Numer Algor (29 52:69 88 and, cf. (26, k j t 1+α j Au (t dt C I j 5 Error from the space discretization tn k j t 1 j t 2+α Au (t dt CMk t σ 1/γ 1 dt. I j t 1 Suppose now that S n is a finite-dimensional subspace of D(A 1/2, and introduce the Ritz projector R n : D(A 1/2 S n associated with the (strictly positive-definite bilinear form A(u, v + u, v. Thus, A(R n v,χ+ R n v,χ = A(v,χ+ v, χ for all χ S n, and we put θ(t = U n R n u(t n and ξ(t = R n u(t n u(t for t I n, so that U u = θ + ξ with θ W. Recall that η = Πu u. Lemma 3 If S 1 S 2 S 2 S n then θ n U R u +2 ( ξ j ξ j 1 +k j B α j (Aη Πξ. (31 Proof Let W(t = R n u(t n for t I n, or in other words W = u + ξ. By(17, W n W n 1,χ +k n A (B n α W,χ =k n f n,χ + ξ n ξ n 1,χ + k n A (B n α ξ,χ, and since θ = U W we see from (18 that for χ S n, θ n θ n 1,χ + k n A (B n α θ,χ = ξ n ξ n 1,χ k n A (B n α ξ,χ. If j n, thenχ S n S j so for s I j, A ( ξ(s, χ = A ( R j u(t j u(s, χ = A ( u(t j u(s, χ + u(t j R j u(t j, χ = A ( η(s, χ Πξ(s, χ = Aη(s Πξ(s, χ and therefore A (B n α ξ,χ = B n α (Aη Πξ,χ, which shows that θ is the discontinuous Galerkin solution of the problem with initial datum θ = U R u and source term equal to k 1 n (ξ n ξ n 1 B α (Aη Πξ for t I n.

15 Numer Algor (29 52: Hence, the desired estimate follows at once from the stability result of Theorem 1. We have already estimated the term n k j B j α Aη in our analysis of the time discretization, and recalling (13 we may write The triangle inequality gives j 1 k j B α j Πξ = β jjξ j β ji ξ i. i=1 k j B α j Πξ j β ji ξ i = i=1 ξ i i=1 β ji, j=i and using (22 we obtain so β ji = j=i I i ( Λ i (s j=i+1 Λ j (s ds ds 2ω 2+α (k i = 2k1+α i Γ(2 + α, k j B j α Πξ 2 Γ(2 + α k 1+α j ξ j. (32 However, we would need k j not k 1+α j on the right-hand side of (32 to establish optimal convergence for the spatial discretization. This difficulty is related to the one we faced in Section 3 for the pure time discretization, and the solution follows similar lines to the method of Section 4. Lemma 4 k j B α j Πξ tn 1+α ( ξ n +2 Γ(2 + α ξ j+1 ξ j. Proof Let B ji = i l=1 β jl for 1 i j 1 so that We have j 1 β j1 = B j1 and β ji = B ji B j,i 1 for 2 i j 1. β ji ξ i = B j1 ξ 1 + (B ji B j,i 1 ξ i = B j, j 1 ξ j B ji (ξ i+1 ξ i for j 2, i=1 j 1 i=2 j 2 i=1

16 84 Numer Algor (29 52:69 88 and thus k j B α j Πξ =β 11 ξ 1 + j 1 β jjξ j β ji ξ i i=1 β 11 ξ 1 + β jj B j, j 1 ξ j + Put ω j = ω 2+α (t j and observe that j 2 B ji ξ i+1 ξ i. j=3 i=1 (33 β jj B j, j 1 = t j ω 1+α (t j t dt t j 1 ω 1+α (t j 1 t dt = ω j ω j 1 for j 2, with β 11 = ω 1 = ω 1 ω,so β 11 ξ 1 + β jj B j, j 1 ξ j = (ω j ω j 1 ξ j ( = ω j ξ j ξ j+1 + ξ j+1 ω j 1 ξ j ( ω j ξ j+1 ξ j + ω j ξ j+1 ω j 1 ξ j = ω n 1 ξ n + ω j ξ j+1 ξ j ω n ( ξ n + ξ j+1 ξ j. Turning to the double sum in (33, we reverse the order of summation to obtain j 2 n 2 B ji ξ i+1 ξ i = ξ i+1 ξ i j=3 i=1 i=1 j=i+2 and the result follows from the estimate ti [ B ji = ω1+α (t j 1 t ω 1+α (t j t ] dt j=i+2 j=i+2 B ji, = ti [ ω1+α (t i+1 t ω 1+α (t n t ] dt ω i+1 ω n.

17 Numer Algor (29 52: We now focus on the concrete case when H = L 2 (Ω for a bounded, convex, polyhedral domain Ω, anda is a strongly-elliptic, second-order, self-adjoint partial differential operator with spectrum in [,. Put Ḣ r (Ω = D(A r/2 ={u L 2 (Ω : A r/2 u < } for r 2. In the case of homogeneous Dirichlet boundary conditions, Ḣ r (Ω ={v H r (Ω : v = on Ω } for r = 1, 2, whereas for homogeneous Neumann boundary conditions we have Ḣ 1 (Ω = H 1 (Ω and Ḣ 2 (Ω ={v H 2 (Ω : n v = on Ω }. We triangulate Ω and define the usual continuous, piecewise-linear finite element space S h Ḣ 1 (Ω, whereh denotes the maximum diameter of the elements. Let the mesh be quasi-uniform so that the Ritz projector R h has the approximation property v R h v Ch 2 v H 2 (Ω for v Ḣ 2 (Ω. (34 For simplicity, we take S n = S h for 1 n N; thus, we employ a fixed trial space for all time levels. Recall the notation l(k = max(1, log k 1 from (2. Theorem 6 Assume, in addition to the regularity property (9 with <σ <1, that u Ḣ2 (Ω M and u(t Ḣ2 (Ω + t u (t Ḣ2 (Ω M for < t T. (35 If S n = S h for 1 n N,andifweusethegradedmesh(8,then k γσ, 1 γ <1/σ, U n u(t n U R h u +CMh 2 l(t n /t 1 +CM k log(t n /t 1, γ =1/σ, tn σ 1/γ k, γ >1/σ. Proof Let ω n = ω 2+α (t n = tn 1+α /Γ (2 + α. Comparing the error bound of Lemma 3 with the one obtained in (19 for the case S n = D(A 1/2,andthen applying Lemma 4, we see that it suffices to estimate ( 2ω n ξ j ξ j 1 +k j B α j Πξ (2 + 4ω n ξ j ξ j 1 +2ω n ξ n. Since the Ritz projector R n = R h is independent of n, ξ j ξ j 1 = (R h I ( u(t j u(t j 1 = (R h Iu (t dt for j 2, I j and the approximation property (34 combined with the additional regularity assumption (35 gives tn ξ j ξ j 1 Ch 2 u (t Ḣ2 (Ω dt CMh2 log(t n /t 1. t 1

18 86 Numer Algor (29 52:69 88 Finally, ξ j = (R h Iu(t j CMh 2 for all j, and thus ξ 1 ξ + ξ n ξ + ξ 1 + ξ n CMh 2. 6 Numerical experiments We now apply the discontinuous Galerkin method (12 to some problems of the form (1. In each case the time interval is [, T] =[, 1] and we employ a time mesh of the form (8 for various choices of the mesh grading parameter γ 1. Since the source term f (t in each example is quite simple we computed the mean value f n analytically, but the stability estimate of Theorem 1 shows that it would suffice to use any O(k-approximation to f n. For instance, we could use f (t n f n if f/ t is bounded. 6.1 A purely time-dependent problem We consider du dt + d t ω α (t su(s ds = f (t for < t < T with u( = u. (36 dt Using the Mittag Leffler function E μ (x = p= xp /Ɣ(1 + pμ, we may write the exact solution as u(t = E α+1 ( t α+1 u + t E α+1 ( s α+1 f (t s ds. Choosing an initial datum u = 1 and a source term f (t = (α + 2t α+1, we find that u(t = E α+1 ( t α+1 + t Γ(α+ 3 ( 1 E α+1,2 ( t α+1. (37 Since the exact solution (37 behaves like t α+1 as t +, we see that the regularity condition (9 holds for σ 2 + 2α. Thus, by Theorem 5, if 1 <α< 1/2 so that γ = (2 + 2α 1 > 1, then we expect O(k γ(2+2α convergence if 1 γ<γ,ando(k convergence if γ>γ. The numerical results shown in Table 1 are consistent with these expectations, since for α =.7 we have 2 + 2α =.6 = 3/5 and γ = 5/3. Table 1 The maximum error U u for the purely time-dependent problem (36 with different mesh gradings, when α =.7 N γ = 1 γ = 2 γ = e e e e e e e e e e e e e e e e e e-4.96 Theory

19 Numer Algor (29 52: Table 2 The maximum error U u with different mesh gradings for a problem in one space dimension, when α =.6 N γ = 1 γ = 1.4 γ = e-2 7.e-3 7.6e e e e e e e e e e e e e e e e-4.99 Theory A problem in one space dimension Let Ω = (, 1 and Au = u xx, and assume that u = u(x, t satisfies homogeneous Dirichlet boundary conditions u(, t = = u(1, t for all t [, T] = [, 1]. We choose the initial datum u (x and source term f (x, t such that the exact solution is u(t, x = ( t α+1 Ɣ(α + 2 sin(π x. π 2 Once again, the regularity condition (9 holds for σ 2 + 2α. We apply the scheme of Theorem 6 for t n of the form (8 and with a uniform spatial mesh having N 1/2 subintervals, each of length h = 1/N 1/2 = k 1/2,so that h 2 = k and the convergence rate is determined by the time discretization (ignoring a possible logarithmic factor. For our discrete initial datum U we take the L 2 projection of u onto the continuous, piecewise-linear finite element space S h. The numerical results, shown in Table 2 for the case α =.6, are as expected, since 2 + 2α =.8 = 4/5 and γ = 5/4 = References 1. Cuesta, E., Lubich, C., Palencia, C.: Convolution quadrature time discretization of fractional diffusive-wave equations. Math. Comput. 75, (26 2. Langlands, T.A.M., Henry, B.I.: The accuracy and stability of an implicit solution method for the fractional diffusion equation. J. Comp. Phys. 25, (25 3. López-Fernández, M., Lubich, C., Schädle, A.: Adaptive, fast and oblivious convolution quadrature in evolution equations with memory. SIAM J. Sci. Comput. 3, (28 4. López-Fernandez, M., Palencia, C., Schädle, A.: A spectral order method for inverting sectorial Laplace transforms. SIAM J. Numer. Anal. 44, (26 5. Metzler, R., Klafter, J.: The random walk s guide to anomalous diffusion: a fractional dynamics approach. Phys. Rep. 339, 1 77 (2 6. McLean, W., Thomée, V.: Numerical solution via Laplace transforms of a fractional order evolution equation. J. Integral Equ. Appl. (28, in press 7. McLean, W., Thomée, V.: Maximum-norm error analysis of a numerical solution via Laplace transformation and quadrature of a fractional order evolution equation. IMA J. Numer. Anal. (28, in press

20 88 Numer Algor (29 52: McLean, W., Thomée, V., Wahlbin, L.B.: Discretization with variable time steps of an evolution equation with a positive-type memory term. J. Comput. Appl. Math. 69, ( Mustapha, K., McLean, W.: Discontinuous Galerkin method for an evolution equation with a memory term of positive type. Math. Comput. (28, in press 1. Schädle, A., López-Fernandez, M., Lubich, C.: Fast and oblivious convolution quadrature. SIAM J. Sci. Comput. 28, (26