Convergence analysis of a Petrov-Galerkin method for fractional wave problems with nonsmooth data
aa r X i v : . [ m a t h . NA ] A p r Convergence analysis of a Petrov-Galerkinmethod for fractional wave problems withnonsmooth data ∗ Hao Luo † , Binjie Li ‡ , Xiaoping Xie § School of Mathematics, Sichuan University, Chengdu 610064, China
Abstract
This paper analyzes the convergence of a Petrov-Galerkin method fortime fractional wave problems with nonsmooth data. Well-posedness andregularity of the weak solution to the time fractional wave problem arefirstly established. Then an optimal convergence analysis with nonsmoothdata is derived. Moreover, several numerical experiments are presentedto validate the theoretical results.
Keywords: fractional wave problem, regularity, Petrov-Galerkin, convergenceanalysis, nonsmooth data.
Let
T > ⊂ R d ( d = 1 , ,
3) be a convex d -polytope.This paper considers the following time fractional wave problem: D α ( u − u − tu ) − ∆ u = f in Ω × (0 , T ), u = 0 on ∂ Ω × (0 , T ), u (0) = u in Ω, u t (0) = u in Ω, (1)where 1 < α <
2, D α is a Riemann-Liouville fractional differential operator oforder α , and u , u and f are given data.In recent years, the time fractional wave problem (1) has attracted muchattention. It has been applied to model the anomalous process which may occurin anomalous transport or diffusion in heterogeneous media [31]. In addition, thesolution to the time fractional wave problem governs the propagation of stresswaves in viscoelastic media [13, 14]. For more details related to the applicationsof problem (1), we refer the reader to [3, 16].Let us first summarize some regularity results of the fractional wave problem.In [11], Bazhlekova considered the Duhamel-type representation of the solution ∗ This work was supported in part by National Natural Science Foundation of China(11771312). † Email: [email protected] ‡ Corresponding author. Email: [email protected] § Email: [email protected]
1o the fractional wave equation by the Mittag-Leffler function; however, theauthor did not investigate the regularity of the solution. Later on, in [12],Bazhlekova obtained the maximal L p -regularity estimate k u k L p (0 ,T ; L q (Ω)) + k D α u k L p (0 ,T ; L q (Ω)) + k ∆ u k L p (0 ,T ; L q (Ω)) C k f k L p (0 ,T ; L q (Ω)) , where 1 < p, q < ∞ . Sakamoto et al. [19] introduced a weak solution to thefractional wave equation by means of the eigenfunction expansions. They es-tablished the well-posedness of the weak solution and derived several regularityestimates in the continuous vector-valued spaces.Then, let us review the numerical treatments for the fractional wave equa-tion. In [20], two kinds of finite difference methods for the computation of frac-tional derivatives were presented: the first method, called L -type scheme, usesthe Lagrange interpolation technique; the second one, called G -type method, isbased on the Gr¨unwald-Letnikov definition. Sun et al. [33] developed a Crank-Nicolson scheme by the L O ( τ − α ) for C solutions. Jin et al. [4] analyzed the G O ( τ ) and O ( τ ), respectively. Inour previous work [6], a time-spectral method for fractional wave problems wasdesigned, which possesses exponential decay in temporal discretization, underthe condition that the solution is smooth enough. Recently, to conquer thesingularity in time variable, Li et al. [7] presented a space-time finite elementmethod for problem (1), and proved that high-order temporal accuracy canstill be achieved if appropriate graded temporal grids are adopted. Under someconditions, problem (1) is equivalent to an integro-differential model, and thereare many works on the numerical methods for this model; see [9, 10, 18] andthe references therein. To our knowledge, except for [4], no work available isdevoted to the numerical analysis for problem (1) with nonsmooth data.This motivates us to consider the numerical analysis for problem (1) withlow regularity data. In this paper, we first introduce a weak solution of problem(1) by the variational approach and establish the regularity results of the weaksolution in the case u = u = 0. Then by means of the famous transpositionmethod [17], the weak solution and its regularity of problem (1) are also con-sidered with more general data. Finally, under the condition that u = u = 0,for a Petrov-Galerkin method we obtain the following error estimates: • if f ∈ L (0 , T ; L (Ω)), then k ( u − U ) ′ k H ( α − / (0 ,T ; L (Ω)) + k u − U k C ([0 ,T ]; ˙ H (Ω)) C (cid:0) τ ( α − / + η ( α, τ, h ) (cid:1) k f k L (0 ,T ; L (Ω)) , (2)where η ( α, τ, h ) := ( h − /α if 1 < α / ,τ − / h if 3 / < α < • if f ∈ H − α (0 , T ; L (Ω)), then k u − U k C ([0 ,T ]; ˙ H (Ω)) C (cid:0) τ (3 − α ) / + η ( α, τ, h ) (cid:1) k f k H − α (0 ,T ; L (Ω)) , (3) (cid:13)(cid:13) u ′ − U ′ (cid:13)(cid:13) H ( α − / (0 ,T ; L (Ω)) C (cid:0) τ (3 − α ) / + η ( α, τ, h ) (cid:1) k f k H − α (0 ,T ; L (Ω)) , (4) η ( α, τ, h ) := h if 1 < α < / , (cid:0) | log h | (cid:1) h if α = 3 / ,h /α − + τ / − α h if 3 / < α < , and η ( α, τ, h ) := ( h /α − if 1 < α / ,τ / − α h if 3 / < α < . We note that, if 1 < α / u and (3) is optimal and nearly optimal with respect to theregularity of u for 1 < α < / α = 3 /
2, respectively. This is verifiedby our numerical experiments. If 3 / < α <
2, then all the estimates (2), (3)and (4) are optimal with respect to the regularity of u provided that h Cτ α/ .However, numerical results also indicate the optimal accuracy with respect tothe regularity without this requirement.The remainder of this paper consists of five sections. Firstly, some con-ventions and Sobolev spaces are introduced in Section 2. Secondly, severalfundamental properties of the fractional calculus operators are summarized inSection 3. Thirdly, the well-posedness and regularity of the weak solution toproblem (1) are rigorously established in Section 4.1. Fourthly, the convergenceof a Petrov-Galerkin method is derived in Section 5. Finally, in Section 6 nu-merical experiments are presented to verify the theoretical results and Section 7provides some concluding remarks. First of all, let us introduce some conventions: for a Lebesgue measurable set ω of R l ( l = 1 , , , H γ ( ω ) ( γ ∈ R ) and H β ( ω ) ( β >
0) denote two standardSobolev spaces [21, Chapter 34] and the symbol h p, q i ω means R ω pq whenever pq ∈ L ( ω ); for a Banach space X , X ∗ is the dual space of X and h· , ·i X meansthe duality pairing between X ∗ and X ; if X and Y are two Banach spaces, then[ X, Y ] θ, is the interpolation space constructed by the famous K -method [21,Chapter 22]; the symbol C × denotes a generic positive constant depending onlyon its subscript(s) × , and its value may differ at each occurrence.Next, we form some Hilbert spaces on the eigenvectors of − ∆ and presentsome basic properties of these spaces. It is well known that there exists anorthonormal basis { φ n : n ∈ N } of L (Ω) such that ( − ∆ φ n = λ n φ n in Ω ,φ n = 0 on ∂ Ω , where { λ n : n ∈ N } is a positive non-decreasing sequence and λ n → ∞ as n → ∞ . For any γ ∈ R , define˙ H γ (Ω) := ( ∞ X n =0 c n φ n : ∞ X n =0 λ γn c n < ∞ ) ∞ X n =0 c n φ n , ∞ X n =0 d n φ n ! ˙ H γ (Ω) := ∞ X n =0 λ γn c n d n , for all P ∞ n =0 c n φ n , P ∞ n =0 d n φ n ∈ ˙ H γ (Ω). Denote by k·k ˙ H γ (Ω) the induced normwith respect to this inner product. We see that ˙ H γ (Ω) is a separable Hilbertspace with an orthonormal basis { λ − γ/ n φ n : n ∈ N } and the space ˙ H − γ (Ω) isthe dual space of ˙ H γ (Ω) in the following sense * ∞ X n =0 c n φ n , ∞ X n =0 d n φ n + ˙ H γ (Ω) := ∞ X n =0 c n d n , for all P ∞ n =0 c n φ n ∈ ˙ H − γ (Ω) and P ∞ n =0 d n φ n ∈ ˙ H γ (Ω). Furthermore, it is clearthat ˙ H (Ω) = L (Ω) and ˙ H (Ω) coincides with H (Ω) with equivalent norms.Hence, for 0 < γ <
1, by the theory of interpolation spaces [21], ˙ H γ (Ω) coin-cides with H γ (Ω) = [ L (Ω) , H (Ω)] γ, with equivalent norms. As [30, Corollary9.1.23] implies k v k H (Ω) C Ω k v k ˙ H (Ω) ∀ v ∈ ˙ H (Ω) , the space ˙ H γ (Ω) is continuously embedded into H (Ω) ∩ H γ (Ω) if 1 < γ < −∞ < a < b < ∞ . Now we introducesome Sobolev spaces as follows. For any m ∈ N , define H m ( a, b ) := { v ∈ H m ( a, b ) : v ( k ) ( b ) = 0 , k < m, k ∈ N } , H m ( a, b ) := { v ∈ H m ( a, b ) : v ( k ) ( a ) = 0 , k < m, k ∈ N } , where v ( k ) is the k -th weak derivative of v , and endow those two spaces withthe following norms k v k H m ( a,b ) := (cid:13)(cid:13) v ( m ) (cid:13)(cid:13) L ( a,b ) , ∀ v ∈ H m ( a, b ) , k v k H m ( a,b ) := (cid:13)(cid:13) v ( m ) (cid:13)(cid:13) L ( a,b ) , ∀ v ∈ H m ( a, b ) , respectively. For k − < γ < k, k ∈ N > , define H γ ( a, b ) := [ H k − ( a, b ) , H k ( a, b )] γ − k +1 , , H γ ( a, b ) := [ H k − ( a, b ) , H k ( a, b )] γ − k +1 , . By [17, Chapter 1], we have the following standard results: if 0 < γ < /
2, then H γ ( a, b ), H γ ( a, b ) and H γ ( a, b ) are equivalent; if m + 1 / < γ < m + 1 , m ∈ N ,then H γ ( a, b ) = { v ∈ H m ( a, b ) : v ( m ) ( b ) = 0 , ( b − t ) m − γ v ( m ) ∈ L ( a, b ) } , H γ ( a, b ) = { v ∈ H m ( a, b ) : v ( m ) ( a ) = 0 , ( t − a ) m − γ v ( m ) ∈ L ( a, b ) } , with equivalent norms; if m γ m + 1 / , m ∈ N , then H γ ( a, b ) = { v ∈ H m ( a, b ) : ( b − t ) m − γ v ( m ) ∈ L ( a, b ) } , H γ ( a, b ) = { v ∈ H m ( a, b ) : ( t − a ) m − γ v ( m ) ∈ L ( a, b ) } ,
4n the sense of equivalent norms. For γ >
0, the spaces H γ ( a, b ) and H γ ( a, b )can be defined equivalently as the domains of fractional power of second or-der differential operators (see [23, 25, 32]). For γ >
0, denote by H − γ ( a, b )and H − γ ( a, b ) the dual spaces of H γ ( a, b ) and H γ ( a, b ), respectively. Since H γ ( a, b ) and H γ ( a, b ) are reflexive, they are the dual spaces of H − γ ( a, b ) and H − γ ( a, b ), respectively. Moreover, by [1, Theorems 1.18 and 1.23] and Theo-rems 12.2-12.6 of [17, Chapter 1], we readily conclude the following lemma. Lemma 2.1. If < θ < and β, γ ∈ R then (cid:2) H β ( a, b ) , H γ ( a, b ) (cid:3) θ, = H (1 − θ ) β + θγ ( a, b ) , (cid:2) H β ( a, b ) , H γ ( a, b ) (cid:3) θ, = H (1 − θ ) β + θγ ( a, b ) , (5) with equivalent norms. Remark 2.1.
We will give more details about how to derive (5) . As ( H ( a, b )) ∗ and H ( a, b ) are continuously embedded in H − ( a, b ) and H ( a, b ) , respec-tively, by Theorem 12.3 of [17, Chapter 1] we have that L ( a, b ) is continu-ously embedded in [ H − ( a, b ) , H ( a, b )] / , . Conversely, since H − ( a, b ) and H ( a, b ) are continuously embedded in H − ( a, b ) and H ( a, b ) , respectively,by Theorem 12.4 of [17, Chapter 1] we have that [ H − ( a, b ) , H ( a, b )] / , is continuously embedded in L ( a, b ) . Therefore, [ H − ( a, b ) , H ( a, b )] / , = L ( a, b ) with equivalent norms. Then by [1, Theorem 1.23] we obtain that, forany < θ < , H θ ( a, b ) = [ L ( a, b ) , H ( a, b )] θ, = (cid:2) [ H − ( a, b ) , H ( a, b )] / , , H ( a, b ) (cid:3) θ, = [ H − ( a, b ) , H ( a, b )] (1+ θ ) / , , with equivalent norms. The other cases are derived similarly. Finally, let us introduce some vector-valued spaces. Let X be a separableHilbert space with an orthonormal basis { e n : n ∈ N } . For any γ ∈ R , define H γ ( a, b ; X ) := ( ∞ X n =0 c n e n : ∞ X n =0 k c n k H γ ( a,b ) < ∞ ) , and endow this space with the norm (cid:13)(cid:13)(cid:13)(cid:13) ∞ X n =0 c n e n (cid:13)(cid:13)(cid:13)(cid:13) H γ ( a,b ; X ) := ∞ X n =0 k c n k H γ ( a,b ) ! / . The space H γ ( a, b ; X ) can be defined analogously. It is evident that both H γ ( a, b ; X ) and H γ ( a, b ; X ) are reflexive. In addition, the space H − γ ( a, b ; X )is the dual space of H γ ( a, b ; X ) in the sense that * ∞ X n =0 c n e n , ∞ X n =0 d n e n + H γ ( a,b ; X ) := ∞ X n =0 h c n , d n i H γ ( a,b ) , for all P ∞ n =0 c n e n ∈ H − γ ( a, b ; X ) and P ∞ n =0 d n e n ∈ H γ ( a, b ; X ), and the space H − γ ( a, b ; X ) is the dual space of H γ ( a, b ; X ) in the sense that * ∞ X n =0 c n e n , ∞ X n =0 d n e n + H γ ( a,b ; X ) := ∞ X n =0 h c n , d n i H γ ( a,b ) , P ∞ n =0 c n e n ∈ H − γ ( a, b ; X ) and P ∞ n =0 d n e n ∈ H γ ( a, b ; X ). Moreover,we use C ([ a, b ]; X ) to denote the continuous X -valued space. Lemma 2.2.
Assume that s, r, β, γ ∈ R and < θ < . If v ∈ H β (0 ,
1; ˙ H r (Ω)) ∩ H γ (0 ,
1; ˙ H s (Ω)) , then k v k H (1 − θ ) β + θγ (0 ,
1; ˙ H (1 − θ ) r + θs (Ω)) C β,γ,θ k v k − θ H β (0 ,
1; ˙ H r (Ω))) k v k θ H γ (0 ,
1; ˙ H s (Ω)) . (6) Proof.
By definition, there exists a unique decomposition v = P ∞ n =0 v n φ n , suchthat k v k H β (0 ,
1; ˙ H r (Ω)) = ∞ X n =0 λ rn k v n k H β (0 , , k v k H γ (0 ,
1; ˙ H s (Ω)) = ∞ X n =0 λ sn k v n k H γ (0 , . Therefore, by [1, Corollary 1.7] and Lemma 2.1 we have k v k H (1 − θ ) β + θγ (0 ,
1; ˙ H (1 − θ ) r + θs (Ω)) = ∞ X n =0 λ (1 − θ ) r + θsn k v n k H (1 − θ ) β + θγ (0 , C β,γ,θ ∞ X n =0 λ (1 − θ ) r + θsn k v n k H β (0 , , H γ (0 , θ, C β,γ,θ ∞ X n =0 λ (1 − θ ) r + θsn k v n k − θ ) H β (0 , k v n k θ H γ (0 , = C β,γ,θ ∞ X n =0 (cid:0) λ rn k v n k H β (0 , (cid:1) − θ (cid:0) λ sn k v n k H γ (0 , (cid:1) θ C β,γ,θ k v k − θ ) H β (0 ,
1; ˙ H r (Ω)) k v k θ H γ (0 ,
1; ˙ H s (Ω)) , which implies (6). (cid:4) In this section, we firstly summarize several fundamental properties of fractionalcalculus operators, then we generalize the fractional integral operator and provesome useful results. Assume that −∞ < a < b < ∞ and X is a separableHilbert space. Definition 3.1.
For γ > , define (cid:0) D − γa + v (cid:1) ( t ) := 1Γ( γ ) Z ta ( t − s ) γ − v ( s ) d s, t ∈ ( a, b ) , (cid:0) D − γb − v (cid:1) ( t ) := 1Γ( γ ) Z bt ( s − t ) γ − v ( s ) d s, t ∈ ( a, b ) , or all v ∈ L ( a, b ; X ) , where Γ( · ) is the Gamma function and L ( a, b ; X ) de-notes the X-valued Bochner integrable space. In addition, let D a + and D b − bethe identity operator on L ( a, b ; X ) . For j − < γ j with j ∈ N > , define D γa + v := D j D γ − ja + v, D γb − v := ( − D) j D γ − jb − v, for all v ∈ L ( a, b ; X ) , where D is the first-order differential operator in thedistribution sense. Lemma 3.1 ([27]) . Let v ∈ L ( a, b ) . If γ, β > , then D − γa + D − βa + v = D − γ − βa + v, D − γb − D − βb − v = D − γ − βb − v. If γ > β > , then D γa + D − βa + v = D γ − βa + v, D γb − D − βb − v = D γ − βb − v. Lemma 3.2 ([3]) . Assume that γ > . If w, v ∈ L ( a, b ) , then (cid:10) D − γa + w, v (cid:11) ( a,b ) = (cid:10) w, D − γb − v (cid:11) ( a,b ) . If v ∈ L ( a, b ) , then (cid:13)(cid:13) D − γa + v (cid:13)(cid:13) L ( a,b ) ( b − a ) γ Γ( γ + 1) k v k L ( a,b ) , (cid:13)(cid:13) D − γb − v (cid:13)(cid:13) L ( a,b ) ( b − a ) γ Γ( γ + 1) k v k L ( a,b ) . Lemma 3.3. If γ, β > , then (cid:13)(cid:13) D − γa + v (cid:13)(cid:13) H β + γ ( a,b ) C β,γ k v k H β ( a,b ) ∀ v ∈ H β ( a, b ) , (7) (cid:13)(cid:13) D − γb − v (cid:13)(cid:13) H β + γ ( a,b ) C β,γ k v k H β ( a,b ) ∀ v ∈ H β ( a, b ) . (8) Proof.
As the proof of (8) is analogous to that of (7) and the case γ, β ∈ N istrivial, we only prove (7) for the case that γ / ∈ N or β / ∈ N .We first use the standard scaling argument to prove the case β = 0 and0 < γ <
1. By definition we have H γ ( a, b ) = (cid:2) H ( a, b ) , H ( a, b ) (cid:3) γ, , and this space is endowed with the following norm k w k H γ ( a,b ) = (cid:18)Z ∞ (cid:0) t − γ K ( t, w ) (cid:1) d tt (cid:19) / ∀ w ∈ H γ ( a, b ) , where K ( t, w ) := inf w = w + w w ∈ H ( a,b ) , w ∈ H ( a,b ) k w k H ( a,b ) + t k w k H ( a,b ) , < t < ∞ , for all w ∈ H γ ( a, b ). For any v ∈ H ( a, b ), define b v ( s ) := v (cid:0) a + ( b − a ) s (cid:1) , < s < , K ( t, D − γa + v ) = ( b − a ) / γ K (cid:0) t/ ( b − a ) , D − γ b v (cid:1) , < t < ∞ . Since using [7, Lemma A.4] gives (cid:13)(cid:13) D − γ w (cid:13)(cid:13) H γ (0 , C γ k w k H (0 , ∀ w ∈ H (0 , , it follows that (cid:13)(cid:13) D − γa + v (cid:13)(cid:13) H γ ( a,b ) = ( b − a ) / (cid:13)(cid:13) D − γ b v (cid:13)(cid:13) H γ (0 , C γ ( b − a ) / k b v k H (0 , = C γ k v k H ( a,b ) . This proves (7) for β = 0 and 0 < γ < β ∈ N and m < γ < m + 1 , m ∈ N . Since v ∈ H β ( a, b ), by Lemma 3.1, it is evident thatD m + βa + D − γa + v = D m + β − γa + D − βa + D βa + v = D m − γa + D βa + v. Therefore, a direct calculation yields K ( t, D − γa + v ) = inf D − γa + v = w + w w ∈ H m + β ( a,b ) ,w ∈ H m + β +1 ( a,b ) k w k H m + β ( a,b ) + t k w k H m + β +1 ( a,b ) = inf D m − γa + D βa + v = v + v v ∈ H ( a,b ) , v ∈ H ( a,b ) k v k H ( a,b ) + t k v k H ( a,b ) = K ( t, D m − γa + D βa + v ) , for all 0 < t < ∞ , which implies that (cid:13)(cid:13) D − γa + v (cid:13)(cid:13) H β + γ ( a,b ) = (cid:13)(cid:13)(cid:13) D m − γa + D βa + v (cid:13)(cid:13)(cid:13) H γ − m ( a,b ) . Consequently, by the previous case, we have (cid:13)(cid:13) D − γa + v (cid:13)(cid:13) H β + γ ( a,b ) C β,γ (cid:13)(cid:13)(cid:13) D βa + v (cid:13)(cid:13)(cid:13) H ( a,b ) = C β,γ k v k H β ( a,b ) . This proves (7) for the case β ∈ N and m < γ < m + 1 , m ∈ N .Finally it remains to consider the case γ > n < β < n + 1 , n ∈ N .Since we have proved that (cid:13)(cid:13) D − γa + w (cid:13)(cid:13) H γ + n ( a,b ) C β,γ k w k H n ( a,b ) ∀ w ∈ H n ( a, b ) , (cid:13)(cid:13) D − γa + w (cid:13)(cid:13) H γ + n +1 ( a,b ) C β,γ k w k H n +1 ( a,b ) ∀ w ∈ H n +1 ( a, b ) , applying the theory of interpolation spaces [21, Lemma 22.3] gives (cid:13)(cid:13) D − γa + v (cid:13)(cid:13) H β + γ ( a,b ) C β,γ k v k H β ( a,b ) , for any v ∈ H β ( a, b ). This completes the proof of this lemma. (cid:4) emark 3.1. In [32, Theorem 2.1], Lemma 3.3 has been proved for β = 0 and γ . Lemma 3.4.
Let v ∈ L ( a, b ) and β > γ > . If D γa + v ∈ H β − γ ( a, b ) , then k v k H β ( a,b ) C β,γ (cid:13)(cid:13) D γa + v (cid:13)(cid:13) H β − γ ( a,b ) . (9) If D γb − v ∈ H β − γ ( a, b ) , then k v k H β ( a,b ) C β,γ (cid:13)(cid:13) D γb − v (cid:13)(cid:13) H β − γ ( a,b ) . (10) Proof.
Let us first prove (9). Suppose that k < γ k + 1 , k ∈ N . By definition,D γa + v = D k +1 D γ − k − a + v, then applying D − k − a + on both sides of the above equation and using integral byparts yield that(D γ − k − a + v )( t ) = (D − k − a + D γa + v )( t ) + k X i =0 c i ( t − a ) i Γ( i + 1) , a < t < b, (11)where c i ∈ R . Moreover, since by Lemma 3.1D γa + D k +1 − γa + D γ − k − a + v = D γa + v, D γa + D k +1 − γa + D − k − a + D γa + v = D γa + D − γa + D γa + v = D γa + v, applying D γa + D k +1 − γa + on both sides of (11) implies k X i =0 c i ( t − a ) i Γ( i − k ) = 0 , a < t < b. Therefore, it follows that c i = 0 for 0 i k , which, together with (11), givesD γ − k − a + v = D − k − a + D γa + v. By Lemma 3.1, applying D k +1 − γa + on both sides of the above equation yields that v = D − γa + D γa + v . Hence, by Lemma 3.3, k v k H β ( a,b ) = (cid:13)(cid:13) D − γa + D γa + v (cid:13)(cid:13) H β ( a,b ) C β,γ (cid:13)(cid:13) D γa + v (cid:13)(cid:13) H β − γ ( a,b ) , which proves (9). As (10) can be proved similarly, this completes the proof. (cid:4) Lemma 3.5. If γ > , then D − γa + D γa + v = v ∀ v ∈ H γ ( a, b ) , (12)D − γb − D γb − v = v ∀ v ∈ H γ ( a, b ) . (13) Proof.
Let k ∈ N > satisfy that k − < γ k . For any v ∈ H γ ( a, b ), sinceLemma 3.3 implies D γ − ka + v ∈ H k ( a, b ), a straightforward computation yieldsthat D − γa + D γa + v = D − γa + D k D γ − ka + v = D k D − γa + D γ − ka + v = D k D − ka + v = v, which proves (12). An analogous argument proves (13) and thus concludes theproof of this lemma. (cid:4) emma 3.6. If β > γ > , then C β,γ k v k H β ( a,b ) (cid:13)(cid:13) D γa + v (cid:13)(cid:13) H β − γ ( a,b ) C β,γ k v k H β ( a,b ) ∀ v ∈ H β ( a, b ) , (14) C β,γ k v k H β ( a,b ) (cid:13)(cid:13) D γb − v (cid:13)(cid:13) H β − γ ( a,b ) C β,γ k v k H β ( a,b ) ∀ v ∈ H β ( a, b ) . (15) Proof.
Since the proof of (15) is similar to that of (14), we only prove (14). Ifwe can prove (cid:13)(cid:13) D γa + v (cid:13)(cid:13) H β − γ ( a,b ) C β,γ k v k H β ( a,b ) (16)for all v ∈ H β ( a, b ), then, by Lemma 3.4, k v k H β ( a,b ) C β,γ (cid:13)(cid:13) D γa + v (cid:13)(cid:13) H β − γ ( a,b ) . Hence it suffices to prove (16). By Lemmas 3.1 and 3.5,D γa + v = D γa + D − βa + D βa + v = D γ − βa + D βa + v, and using Lemma 3.3 gives (cid:13)(cid:13) D γa + v (cid:13)(cid:13) H β − γ ( a,b ) = (cid:13)(cid:13)(cid:13) D γ − βa + D βa + v (cid:13)(cid:13)(cid:13) H β − γ ( a,b ) C β,γ (cid:13)(cid:13)(cid:13) D βa + v (cid:13)(cid:13)(cid:13) H ( a,b ) . Let k ∈ N > satisfy that k − < β k . Invoking Lemma 3.3 again implies that (cid:13)(cid:13)(cid:13) D βa + v (cid:13)(cid:13)(cid:13) H ( a,b ) = (cid:13)(cid:13) D k D β − ka + v (cid:13)(cid:13) H ( a,b ) = (cid:13)(cid:13) D β − ka + v (cid:13)(cid:13) H k ( a,b ) C β k v k H β ( a,b ) , which, together with the previous inequality, proves (16). This finishes the proofof this lemma. (cid:4) Remark 3.2.
In [32, Theorem 2.1], an alternative proof of (14) has been givenfor β and γ = β . Lemma 3.7. If β, γ > , then k v k H β ( a,b ) C β,γ (cid:13)(cid:13) D − γa + v (cid:13)(cid:13) H β + γ ( a,b ) ∀ v ∈ H β ( a, b ) , k v k H β ( a,b ) C β,γ (cid:13)(cid:13) D − γb − v (cid:13)(cid:13) H β + γ ( a,b ) ∀ v ∈ H β ( a, b ) . Proof.
Combining Lemmas 3.5 and 3.6, we obtain k v k H β ( a,b ) = (cid:13)(cid:13) D γa + D − γa + v (cid:13)(cid:13) H β ( a,b ) C β,γ (cid:13)(cid:13) D − γa + v (cid:13)(cid:13) H β + γ ( a,b ) ∀ v ∈ H β ( a, b ) , k v k H β ( a,b ) = (cid:13)(cid:13) D γb − D − γb − v (cid:13)(cid:13) H β ( a,b ) C β,γ (cid:13)(cid:13) D − γb − v (cid:13)(cid:13) H β + γ ( a,b ) ∀ v ∈ H β ( a, b ) . This concludes the proof. (cid:4)
Lemma 3.8 ([5, 29]) . If − / < γ < / and v ∈ H max { ,γ } ( a, b ) , then cos( γπ ) (cid:13)(cid:13) D γa + v (cid:13)(cid:13) L ( a,b ) (cid:10) D γa + v, D γb − v (cid:11) ( a,b ) sec( γπ ) (cid:13)(cid:13) D γa + v (cid:13)(cid:13) L ( a,b ) , cos( γπ ) (cid:13)(cid:13) D γb − v (cid:13)(cid:13) L ( a,b ) (cid:10) D γa + v, D γb − v (cid:11) ( a,b ) sec( γπ ) (cid:13)(cid:13) D γb − v (cid:13)(cid:13) L ( a,b ) . Moreover, if v, w ∈ H γ ( a, b ) with < γ < / , then (cid:10) D γa + v, w (cid:11) H γ ( a,b ) = (cid:10) D γa + v, D γb − w (cid:11) ( a,b ) = (cid:10) D γb − w, v (cid:11) H γ ( a,b ) . emma 3.9. If β > and γ < / , then k v k C [0 , C β,γ k v k (1 / − γ ) / ( β − γ ) H β (0 , k v k ( β − / / ( β − γ ) H γ (0 , , (17) for all v ∈ H β (0 , .Proof. Let s := max { , γ } . Since v ∈ H β (0 , v ( t ) = (cid:10) v ′ , v (cid:11) (0 ,t ) = (cid:10) v ′ , D − st − D st − v (cid:11) (0 ,t ) = (cid:10) D − s v ′ , D st − v (cid:11) (0 ,t ) = (cid:10) D − s v, D st − v (cid:11) (0 ,t ) (cid:13)(cid:13) D − s v (cid:13)(cid:13) L (0 ,t ) k D st − v k L (0 ,t ) C s (cid:13)(cid:13) D − s v (cid:13)(cid:13) L (0 ,t ) k D s v k L (0 ,t ) C s (cid:13)(cid:13) D − s v (cid:13)(cid:13) L (0 , k D s v k L (0 , C s k v k H − s (0 , k v k H s (0 , , for all 0 t
1. It follows that k v k C [0 , C s k v k / H s (0 , k v k / H − s (0 , . Additionally, by Lemma 2.1 and [1, Corollary 1.7] we have k v k H s (0 , C β,γ k v k ( s − γ ) / ( β − γ ) H β (0 , k v k ( β − s ) / ( β − γ ) H γ (0 , , k v k H − s (0 , C β,γ k v k (1 − γ − s ) / ( β − γ ) H β (0 , k v k ( β + s − / ( β − γ ) H γ (0 , . Consequently, combining the above three estimates proves (17). (cid:4)
Lemma 3.10. If v ∈ H β (0 , with β > / , then k v k C [0 , C β ǫ k v k − ǫ H / (0 , k v k ǫ H β (0 , , (18) for all < ǫ / max { , β } .Proof. For any w ∈ L (0 , w to ( −∞ ,
0) by zero and denote thisextension by e w . Let n ∈ N satisfy that n − < β n . Following the proofof [17, Theorem 8.1], we define an extension operator E : L (0 , → L ( R ) bythat, for any w ∈ L (0 , Ew )( t ) := e w ( t ) , if t < , n +1 X j =1 γ j e w ( j + 1 − jt ) , if t > , where γ j is defined by n +1 X j =1 ( − j ) k γ j = 1 , k n. Since a straightforward computation gives k Ew k H k ( R ) C k k w k H k (0 , ∀ w ∈ H k (0 , , k n, k Ev k H (1 − ǫ ) / ǫβ ( R ) C β k v k [ H / (0 , , H β (0 , ǫ, . In addition, [21, (23.11)] implies that (cid:16) Z R (cid:0) | ξ | (cid:1) (1 − ǫ ) / ǫβ | ( F Ev )( ξ ) | d ξ (cid:17) / C β k Ev k H (1 − ǫ ) / ǫβ ( R ) , where F is the Fourier transform operator. Moreover, using [22, (1.2.4) and(1.2.5)] yields k Ev k L ∞ ( R ) C β √ ǫ (cid:16) Z R (cid:0) ξ (cid:1) (1 − ǫ ) / ǫβ | ( F Ev )( ξ ) | d ξ (cid:17) / . Therefore it follows that k v k C [0 , = k Ev k L ∞ ( R ) C β √ ǫ k v k [ H / (0 , , H β (0 , ǫ, . Since borrowing the proof of [1, Corollary 1.7] gives k v k [ H / (0 , , H β (0 , ǫ, √ ǫ k v k − ǫ H / (0 , k v k ǫ H β (0 , , we finally obtain (18) by the above two inequalities. This concludes the proofof this lemma. (cid:4) Now let us generalize the fractional integral operator as follows. Recall thatin this section X denotes a separable Hilbert space. Assume that β, γ > v ∈ H − β ( a, b ; X ). If 0 < γ β , then define D − γa + v ∈ H γ − β ( a, b ; X ) by that (cid:10) D − γa + v, w (cid:11) H β − γ ( a,b ; X ) := (cid:10) v, D − γb − w (cid:11) H β ( a,b ; X ) , (19)for all w ∈ H β − γ ( a, b ; X ); if γ > β , then define D − γa + v ∈ H γ − β ( a, b ; X ) by thatD − γa + v := D β − γa + D − βa + v (20)By Lemmas 3.3, 3.6 and 3.7, the generalized left-sided fractional integral oper-ator D − γa + : H − β ( a, b ; X ) → H γ − β ( a, b ; X )is well-defined for all β, γ >
0. Symmetrically, we can generalize the right-sided fractional integral operator as follows. Assume that β, γ > v ∈ H − β ( a, b ; X ). Define D − γb − v ∈ H γ − β ( a, b ; X ) by that (cid:10) D − γb − v, w (cid:11) H β − γ ( a,b ; X ) := (cid:10) v, D − γa + w (cid:11) H β ( a,b ; X ) , for all w ∈ H β − γ ( a, b ; X ). Lemma 3.11. If β, γ > , then C β,γ k v k H − β ( a,b ) (cid:13)(cid:13) D − γa + v (cid:13)(cid:13) H γ − β ( a,b ) C β,γ k v k H − β ( a,b ) ∀ v ∈ H − β ( a, b ) , (21) C β,γ k v k H − β ( a,b ) (cid:13)(cid:13) D − γb − v (cid:13)(cid:13) H γ − β ( a,b ) C β,γ k v k H − β ( a,b ) ∀ v ∈ H − β ( a, b ) . (22) roof. Since the proofs of (21) and (22) are similar, we only give the proof ofthe former.Let us first prove that (cid:13)(cid:13) D − γa + v (cid:13)(cid:13) H γ − β ( a,b ) C β,γ k v k H − β ( a,b ) ∀ v ∈ H − β ( a, b ) . (23)If γ β , then by Lemma 3.3 and definition (19), (cid:10) D − γa + v, w (cid:11) H β − γ ( a,b ) = (cid:10) v, D − γb − w (cid:11) H β ( a,b ) k v k H − β ( a,b ) (cid:13)(cid:13) D − γb − w (cid:13)(cid:13) H β ( a,b ) C β,γ k v k H − β ( a,b ) k w k H β − γ ( a,b ) , for all w ∈ H β − γ ( a, b ), which proves (23) for γ β . If γ > β , then by definition(20) and Lemma 3.3 and the previous case, we have (cid:13)(cid:13) D − γa + v (cid:13)(cid:13) H γ − β ( a,b ) = (cid:13)(cid:13) D β − γa + D − βa + v (cid:13)(cid:13) H γ − β ( a,b ) C β,γ (cid:13)(cid:13) D − βa + v (cid:13)(cid:13) L ( a,b ) C β,γ k v k H − β ( a,b ) . This proves (23) for the case γ > β .Then it remains to prove that k v k H − β ( a,b ) C β,γ (cid:13)(cid:13) D − γa + v (cid:13)(cid:13) H γ − β ( a,b ) ∀ v ∈ H − β ( a, b ) . (24)If γ β , then by definition (19) and Lemmas 3.5 and 3.6 h v, w i H β ( a,b ) = (cid:10) v, D − γb − D γb − w (cid:11) H β ( a,b ) = (cid:10) D − γa + v, D γb − w (cid:11) H β − γ ( a,b ) (cid:13)(cid:13) D − γa + v (cid:13)(cid:13) H γ − β ( a,b ) (cid:13)(cid:13) D − γb − w (cid:13)(cid:13) H β − γ ( a,b ) C β,γ (cid:13)(cid:13) D − γa + v (cid:13)(cid:13) H γ − β ( a,b ) k w k H β ( a,b ) , for all w ∈ H β ( a, b ), so that we have k v k H − β ( a,b ) C β,γ (cid:13)(cid:13) D − γa + v (cid:13)(cid:13) H γ − β ( a,b ) . If γ > β , then by definition (20) and Lemmas 3.5 and 3.6, an evident calculationgives h v, w i H β ( a,b ) = (cid:10) v, D − βb − D βb − w (cid:11) H β ( a,b ) = (cid:10) D − βa + v, D βb − w (cid:11) ( a,b ) = (cid:10) D γ − βa + D − γa + v, D βb − w (cid:11) ( a,b ) (cid:13)(cid:13) D γ − βa + D − γa + v (cid:13)(cid:13) L ( a,b ) (cid:13)(cid:13) D βb − w (cid:13)(cid:13) L ( a,b ) C β,γ (cid:13)(cid:13) D − γa + v (cid:13)(cid:13) H γ − β ( a,b ) k w k H β ( a,b ) , for all w ∈ H β ( a, b ), which implies that k v k H − β ( a,b ) C β,γ (cid:13)(cid:13) D − γa + v (cid:13)(cid:13) H γ − β ( a,b ) . This proves (24) and thus completes the proof of this lemma. (cid:4) emma 3.12. If v ∈ H γ ( a, b ) with < γ < / , then (cid:12)(cid:12)(cid:10) D γa + v, D − γa + v (cid:11) H γ ( a,b ) (cid:12)(cid:12) C γ k v k H γ ( a,b ) k v k H − γ ( a,b ) . (25) Proof.
Since by Lemma 3.8, (cid:13)(cid:13) D γb − D − γa + v (cid:13)(cid:13) L ( a,b ) C γ (cid:13)(cid:13) D γa + D − γa + v (cid:13)(cid:13) L ( a,b ) = C γ (cid:13)(cid:13) D − γa + v (cid:13)(cid:13) L ( a,b ) , it follows that (cid:12)(cid:12)(cid:10) D γa + v, D − γa + v (cid:11) H γ ( a,b ) (cid:12)(cid:12) = (cid:12)(cid:12)(cid:10) D γa + v, D γb − D − γa + v (cid:11) ( a,b ) (cid:12)(cid:12) (cid:13)(cid:13) D γa + v (cid:13)(cid:13) L ( a,b ) (cid:13)(cid:13) D γb − D − γa + v (cid:13)(cid:13) L ( a,b ) C γ (cid:13)(cid:13) D γa + v (cid:13)(cid:13) L ( a,b ) (cid:13)(cid:13) D − γa + v (cid:13)(cid:13) L ( a,b ) C γ k v k H γ ( a,b ) k v k H − γ ( a,b ) , by Lemmas 3.6 and 3.11. This completes the proof. (cid:4) Lemma 3.13.
Assume that γ < β + 1 / . If v ∈ H β ( a, b ) , then D γa + v ∈ H β − γ ( a, b ) and (cid:10) D γa + v, w (cid:11) H γ − β ( a,b ) = D D βa + v, D γ − βb − w E ( a,b ) (26) for all w ∈ H γ − β ( a, b ) . If v ∈ H β ( a, b ) , then D γb − v ∈ H β − γ ( a, b ) and (cid:10) D γb − v, w (cid:11) H γ − β ( a,b ) = D D βb − v, D γ − βb − w E ( a,b ) (27) for all w ∈ H γ − β ( a, b ) .Proof. As the proof of (27) is similar to that of (26), we only prove (26).Firstly, if 0 < γ β , then by Lemma 3.6, it is obvious that D γa + v ∈ H β − γ ( a, b ), and (26) holds indeed by by Lemma 3.5 and definition (19).Next we consider the case γ β > γ . By Lemmas 3.3 and 3.11, wehave D γa + v ∈ H β − γ ( a, b ). If β >
0, then by Lemmas 3.1 and 3.5, it is evidentthat D γ − βa + D βa + v = D γa + D − βa + D βa + v = D γa + v. If β <
0, then by definition (20), D D γ − βb − w, D βa + v E ( a,b ) = D w, D γ − βa + D βa + v E H β − γ ( a,b ) = (cid:10) w, D γa + v (cid:11) H β − γ ( a,b ) = (cid:10) D γa + v, w (cid:11) H γ − β ( a,b ) , for all w ∈ H γ − β ( a, b ), which proves (26) for γ β > γ .Then let us consider the case γ γ − / < β < γ . By Lemma 3.11,we have D γa + v ∈ H β − γ ( a, b ), and using Lemmas 3.1 and 3.5 and definition (19)gives (cid:10) D γa + v, w (cid:11) H γ − β ( a,b ) = (cid:10) v, D γb − w (cid:11) H − β ( a,b ) = D v, D γb − D β − γb − D γ − βb − w E H − β ( a,b ) = D v, D βb − D γ − βb − w E H − β ( a,b ) = D D βa + v, D γ − βb − w E ( a,b ) , w ∈ H γ − β ( a, b ). This proves (26) for γ γ − / < β < γ .Finally it remains to consider the case γ > γ − / < β < γ . Let k ∈ N satisfy k − < γ k . If β >
0, then by Lemmas 3.2 and 3.6, a directmanipulation implies that (cid:10) D γa + v, φ (cid:11) = D D k D γ − ka + v, φ E = ( − k D D γ − ka + v, D k φ E ( a,b ) = ( − k D v, D γ − kb − D k φ E ( a,b ) = (cid:10) v, D γb − φ (cid:11) ( a,b ) = D v, D βb − D γ − βb − φ E ( a,b ) = D D βa + v, D γ − βb − φ E ( a,b ) C β,γ k v k H β ( a,b ) k φ k H γ − β ( a,b ) , for all φ ∈ C ∞ ( a, b ). If β <
0, then 0 < γ < /
2. By Lemmas 3.6 and 3.11 anddefinition (20), for all φ ∈ C ∞ ( a, b ), a similar deduction gives (cid:10) D γa + v, φ (cid:11) = D D D γ − a + v, φ E = − D D γ − a + v, D φ E ( a,b ) = − D D γ − β − a + D βa + v, D φ E ( a,b ) = − D D βa + v, D γ − β − b − D φ E ( a,b ) = D D βa + v, D γ − βb − φ E ( a,b ) C β,γ k v k H β ( a,b ) k φ k H γ − β ( a,b ) . Above, we use h· , ·i to denote the dual pair between the dual space of C ∞ ( a, b )and C ∞ ( a, b ). Since 0 < γ − β < /
2, it is clear that C ∞ ( a, b ) is dense in H γ − β ( a, b ). Consequently, we conclude that D γa + v ∈ H β − γ ( a, b ) and (26)holds indeed. This concludes the proof of this lemma. (cid:4) Lemma 3.14. If max { β, β + γ } < s + 1 / , then D γa + D βa + v = D β + γa + v ∀ v ∈ H s ( a, b ) , (28)D γb − D βb − v = D β + γb − v ∀ v ∈ H s ( a, b ) . (29) Proof.
Let us first prove thatD s − βa + D βa + v = D sa + v ∀ v ∈ H s ( a, b ) . (30)By Lemmas 3.6 and 3.11, it is evident that D sa + v ∈ L ( a, b ). In addition,applying Lemma 3.13 yields that D D s − βa + D βa + v, φ E ( a,b ) = D D βa + v, D s − βb − φ E H β − s ( a,b ) = D D sa + v, D β − sb − D s − βb − φ E ( a,b ) = (cid:10) D sa + v, φ (cid:11) ( a,b ) , for any φ ∈ C ∞ ( a, b ), which proves (30).Then we turn to the proof of (28). Since using Lemma 3.13 implies thatD βa + v ∈ H s − β ( a, b ) and both D γa + D βa + v and D β + γa + v belong to H s − γ − β ( a, b ),by (30), we have D D γa + D βa + v, w E H β + γ − s ( a,b ) = D D s − βa + D βa + v, D β + γ − sb − w E ( a,b ) = D D sa + v, D β + γ − sb − w E ( a,b ) = D D β + γa + v, w E H β + γ − s ( a,b ) , w ∈ H β + γ − s ( a, b ), which proves (28). As (29) can be proved analogously,we finish the proof of this lemma. (cid:4) Define c W := H α/ (0 , T ; L (Ω)) ∩ H α/ (0 , T ; ˙ H (Ω)) ,W := H α/ (0 , T ; L (Ω)) ∩ H − α/ (0 , T ; ˙ H (Ω)) , and endow these two spaces with the norms k·k c W := (cid:16) k·k H α/ (0 ,T ; L (Ω)) + k·k H α/ (0 ,T ; ˙ H (Ω)) (cid:17) / , k·k W := (cid:16) k·k H α/ (0 ,T ; L (Ω)) + k·k H − α/ (0 ,T ; ˙ H (Ω)) (cid:17) / , respectively. Assuming thatD α/ ( u + tu ) ∈ W ∗ and f ∈ c W ∗ , we call u ∈ W a weak solution to problem (1) if (cid:10) D α/ u, v (cid:11) H α/ (0 ,T ; L (Ω)) + (cid:10) ∇ D − α/ u, ∇ D − α/ T − v (cid:11) Ω × (0 ,T ) = (cid:10) f, D − α/ T − v (cid:11) c W + (cid:10) D α/ ( u + tu ) , v (cid:11) W (31)for all v ∈ W . Remark 4.1.
Notice that H α/ (0 , T ; L (Ω)) = H α/ (0 , T ; L (Ω)) , H − α/ (0 , T ; ˙ H (Ω)) = H − α/ (0 , T ; ˙ H (Ω)) , with equivalent norms. Hence, by Lemmas 3.3 and 3.8, it is easy to verify thateach term in (31) makes sense. By Lemmas 3.3, 3.6 and 3.8 and the well-known Lax-Milgram theorem, aroutine argument yields that the above weak solution is well-defined.
Theorem 4.1.
Problem (31) admits a unique solution u ∈ W and k u k W C α (cid:16) k f k c W ∗ + (cid:13)(cid:13) D α/ ( u + tu ) (cid:13)(cid:13) W ∗ (cid:17) . Remark 4.2. If u ∈ L (Ω) and u ∈ ˙ H − (Ω) , then a simple calculation yieldsthat D α/ ( u + tu ) ∈ W ∗ . Therefore the weak solution is well-defined by (31) for u ∈ L (Ω) , u ∈ ˙ H − (Ω) and f ∈ c W ∗ . u = u = 0. Let us first define y ∈ H α/ (0 , T ) by that (cid:10) D α/ y, z (cid:11) H α/ (0 ,T ) + λ (cid:10) D − α/ y, z (cid:11) (0 ,T ) = (cid:10) D − α/ g, z (cid:11) H α/ (0 ,T ) (32)for all z ∈ H α/ (0 , T ), where g ∈ H − α/ (0 , T ) and λ is a positive constant.Similar to problem (31), problem (32) admits a unique solution y ∈ H α/ (0 , T )and k y k H α/ (0 ,T ) + λ / k y k H − α/ (0 ,T ) C α k g k H − α/ (0 ,T ) . Lemma 4.1. If γ > − α/ and g ∈ H γ (0 , T ) , then D α + γ y + λ D γ y = D γ g, (33) k y k H α + γ (0 ,T ) + λ k y k H γ (0 ,T ) C α,γ k g k H γ (0 ,T ) . (34) Moreover, if − α γ < / then λ γ − / /α k y k C [0 ,T ] C α,γ,T k g k H γ (0 ,T ) , (35) and if γ = 1 / then λ − ǫ/ k y k C [0 ,T ] C α,T ǫ k g k H / (0 ,T ) , (36) for all < ǫ / (2 α + 1) .Proof. Firstly, let us consider the following problem: find w ∈ H α/ (0 , T ) suchthat (cid:10) D α/ w, z (cid:11) H α/ (0 ,T ) + λ (cid:10) D − α/ w, z (cid:11) (0 ,T ) = (cid:10) D γ g, z (cid:11) (0 ,T ) (37)for all z ∈ H α/ (0 , T ). Since using Lemma 3.6 yields (cid:13)(cid:13) D γ g (cid:13)(cid:13) L (0 ,T ) C γ k g k H γ (0 ,T ) , (38)by Lemmas 3.3, 3.6 and 3.8 and the Lax-Milgram theorem, we claim that prob-lem (37) admits a unique solution w ∈ H α/ (0 , T ). In addition, inserting z = w into (37) gives k w k H α/ (0 ,T ) + λ k w k H − α/ (0 ,T ) C α (cid:13)(cid:13) D γ g (cid:13)(cid:13) L (0 ,T ) k w k L (0 ,T ) . (39)Since Lemma 2.1 and [1, Corollary 1.7] imply k w k L (0 ,T ) C α k w k / H α/ (0 ,T ) k w k / H − α/ (0 ,T ) , a simple calculation gives, by (39), that λ / k w k H α/ (0 ,T ) + λ / k w k H − α/ (0 ,T ) C α (cid:13)(cid:13) D γ g (cid:13)(cid:13) L (0 ,T ) . (40)Observe that by (37)D α/ w + λ D − α/ w = D γ g in L (0 , T ) , (41)17nd then multiplying D α/ w on both sides of the above equality and integratingon (0 , T ) yields that (cid:13)(cid:13) D α/ w (cid:13)(cid:13) L (0 ,T ) = (cid:10) D γ g, D α/ w (cid:11) (0 ,T ) − λ (cid:10) D α/ w, D − α/ w (cid:11) (0 ,T ) (cid:13)(cid:13) D γ g (cid:13)(cid:13) L (0 ,T ) (cid:13)(cid:13) D α/ w (cid:13)(cid:13) L (0 ,T ) + C α λ k w k H − α/ (0 ,T ) k w k H α/ (0 ,T ) , by Lemma 3.12. Thus using (40) and Young’s inequality with ǫ yields (cid:13)(cid:13) D α/ w (cid:13)(cid:13) L (0 ,T ) C α (cid:13)(cid:13) D γ g (cid:13)(cid:13) L (0 ,T ) , and it follows from (41) that λ (cid:13)(cid:13) D − α/ w (cid:13)(cid:13) L (0 ,T ) = (cid:13)(cid:13) D α/ w − D γ g (cid:13)(cid:13) L (0 ,T ) C α (cid:13)(cid:13) D γ g (cid:13)(cid:13) L (0 ,T ) . Hence, by Lemmas 3.4 and 3.11, combining (38) and the above two estimatesyields (cid:13)(cid:13) w (cid:13)(cid:13) H α/ (0 ,T ) + λ k w k H − α/ (0 ,T ) C α,γ k g k H γ (0 ,T ) . (42)Secondly, let us prove (33). By the facts w ∈ H α/ (0 , T ) and Lemma 3.14,applying D − α/ − γ on both sides of (41) yields (cid:0) D α/ + λ D − α/ (cid:1) D − α/ − γ w = D − α/ g. Therefore, y := D − α/ − γ w is the solution to problem (32), and using Lem-mas 3.3 and 3.11 and (42) proves (34). Hence by the fact y ∈ H α + γ (0 , T ), therelation w = D α/ γ y , Lemma 3.14 and (41), we obtain thatD α + γ y + λ D γ y = D γ g, which proves (33).Finally, it remains to prove (35) and (36). If 1 − α γ < /
2, then byLemma 3.9 and (34), λ γ − / /α k y k C [0 ,T ] C α,γ,T k y k (1 / − γ ) /α H α + γ (0 ,T ) (cid:0) λ k y k H γ (0 ,T ) (cid:1) γ − / /α C α,γ,T (cid:0) k y k H α + γ (0 ,T ) + λ k y k H γ (0 ,T ) (cid:1) C α,γ,T k g k H γ (0 ,T ) , which proves (35). If γ = 1 /
2, then using Lemma 3.10 and (34) gives λ − ǫ/ k y k C [0 ,T ] C α,T ǫ (cid:0) λ k y k H / (0 ,T ) (cid:1) − ǫ/ k y k ǫ/ H α +1 / (0 ,T ) C α,T ǫ k g k H / (0 ,T ) , for all 0 < ǫ / (2 α + 1). This proves (36) and thus completes the proof ofthis lemma. (cid:4) emma 4.2. If u = u = 0 and f ∈ H − α/ (0 , T ; L (Ω)) ∪ H − α/ (0 , T ; ˙ H − (Ω)) , then the solution to problem (31) is given by that u ( t ) = ∞ X n =0 y n ( t ) φ n , t T, (43) where y n satisfies (cid:10) D α/ y n , z (cid:11) H α/ (0 ,T ) + λ n (cid:10) D − α/ y n , z (cid:11) (0 ,T ) = D D − α/ (cid:10) f, φ n (cid:11) ˙ H (Ω) , z E (0 ,T ) (44) for all z ∈ H α/ (0 , T ) .Proof. Let us first consider the case that f ∈ H − α/ (0 , T ; L (Ω)). Similar toproblem (32), problem (44) admits a unique solution y n ∈ H α/ (0 , T ), and k y n k H α/ (0 ,T ) + λ n k y n k H − α/ (0 ,T ) C α (cid:13)(cid:13) h f, φ n i ˙ H (Ω) (cid:13)(cid:13) H − α/ (0 ,T ) . Since ∞ X n =0 (cid:13)(cid:13) h f, φ n i ˙ H (Ω) (cid:13)(cid:13) H − α/ (0 ,T ) = ∞ X n =0 (cid:13)(cid:13) h f, φ n i L (Ω) (cid:13)(cid:13) H − α/ (0 ,T ) = k f k H − α/ (0 ,T ; L (Ω)) , it follows that ∞ X n =0 (cid:16) k y n k H α/ (0 ,T ) + λ n k y n k H − α/ (0 ,T ) (cid:17) C α k f k H − α/ (0 ,T ; L (Ω)) . Therefore, u defined by (43) belongs to W and satisfies that k u k W C α k f k H − α/ (0 ,T ; L (Ω)) . Next, let us verify that u is the solution to problem (31). For any v ∈ W ,there exists a unique decomposition v = P ∞ n =0 v n φ n , and ∞ X n =0 (cid:16) λ n k v n k H − α/ (0 ,T ) + k v n k H α/ (0 ,T ) (cid:17) = k v k W . It is evident that (cid:10) D α/ u, v (cid:11) H α/ (0 ,T ; L (Ω)) = ∞ X n =0 (cid:10) D α/ y n , v n (cid:11) H α/ (0 ,T ) , (cid:10) ∇ D − α/ u, ∇ D − α/ T − v (cid:11) Ω × (0 ,T ) = ∞ X n =0 λ n (cid:10) D − α/ y n , v n (cid:11) (0 ,T ) . Since f ∈ H − α/ (0 , T ; L (Ω)) ⊂ c W ∗ , we also have (cid:10) f, D − α/ T − v (cid:11) c W = ∞ X n =0 D D − α/ (cid:10) f, φ n (cid:11) ˙ H (Ω) , v n E (0 ,T ) . u given by (43) fulfills(31) for all v ∈ W , and therefore it is indeed the solution to problem (31). Sincethe proof of the case that f ∈ H − α/ (0 , T ; ˙ H − (Ω)) is similar. This completesthe proof of the lemma. (cid:4) Combining Lemmas 4.1 and 4.2, we readily conclude the following regularityresults for the weak solution u to problem (1). Theorem 4.2.
Assume that β > , γ > − α/ or β > − , γ > − α/ . If u = u = 0 and f ∈ H γ (0 , T ; ˙ H β (Ω)) , then D α + γ u − ∆ D γ u = D γ f in L (0 , T ; ˙ H β (Ω)) , (45) and k u k H α + γ (0 ,T ; ˙ H β (Ω)) + k u k H γ (0 ,T ; ˙ H β (Ω)) C α,γ k f k H γ (0 ,T ; ˙ H β (Ω)) . Moreover, if − α γ < / then k u k C ([0 ,T ]; ˙ H β +(2 γ − /α (Ω)) C α,γ,T k f k H γ (0 ,T ; ˙ H β (Ω)) , and if γ = 1 / then k u k C ([0 ,T ]; ˙ H β − ǫ (Ω)) C α,T ǫ k f k H / (0 ,T ; ˙ H β (Ω)) , for all < ǫ / (2 α + 1) . Remark 4.3.
By Lemmas 4.1 and 4.2, if − α γ < / , then the seriesin (43) converges to u in ˙ H β +(2 γ − /α (Ω) uniformly on [0 , T ] , so that u ∈ C ([0 , T ]; ˙ H β +(2 γ − /α (Ω)) . Similarly, if γ = 1 / , then u ∈ C ([0 , T ]; ˙ H β − ǫ (Ω)) for all < ǫ / (2 α + 1) . Remark 4.4.
We observe that [12, Theorem 4.21] has already contained theregularity estimate k u k L p (0 ,T ; L q (Ω)) + (cid:13)(cid:13) D α u (cid:13)(cid:13) L p (0 ,T ; L q (Ω)) + k ∆ u k L p (0 ,T ; L q (Ω)) C α,p,q k f k L p (0 ,T ; L q (Ω)) , where < p, q < ∞ . In the case p = q = 2 , this result implies k u k H α (0 ,T ; L (Ω)) + k u k L (0 ,T ; ˙ H (Ω)) C α k f k L (0 ,T ; L (Ω)) . Remark 4.5.
Let us introduce a simple example to demonstrate that a smoothsource term f with f (0) = 0 can not guarantee a smooth solution. For any β > , define the Mittag-Leffler function E α,β ( z ) by E α,β ( z ) := ∞ X n =0 z n Γ( nα + β ) , z ∈ C , and this function admits a growth estimate that [16] | E α,β ( − t ) | C α,β t , t > . (46)20 or any t > , let w ( t ) := t α / Γ( α + 1) ,y ( t ) := − λ − E α, ( − λt α ) + λ − , where λ is a positive constant, then a straightforward calculation yields D α y + λy = 1 , t > . If < α < / , then by (46) we obtain λ − /α k ( y − w ) ′′ k L (0 ,T ) = Z T λ − /α t α − E α, α − ( − λt α ) d t C α Z T λ − /α t α − (1 + λt α ) − d t C α (cid:18) Z λ − /α λ − /α t α − d t + Z ∞ λ − /α λ − /α t α − d t (cid:19) C α . If / α < , then an analogous deduction gives λ − /α k ( y − w ) ′′′ k L (0 ,T ) C α . Now, assume that u = u = 0 . If < α < / and f ( t ) = v ∈ ˙ H /α − (Ω) , t T , then by the techniques used in the proof of Theorem 4.2 we obtain k u − t α / Γ( α + 1) v k H (0 ,T ; L (Ω)) C α k v k ˙ H /α − (Ω) . Similarly, if / α < and f ( t ) = v ∈ ˙ H /α − (Ω) , t T , then k u − t α / Γ( α + 1) v k H (0 ,T ; L (Ω)) C α k v k ˙ H /α − (Ω) . Consequently, the temporal regularity of u is essentially determined by t α v , andits temporal regularity can not exceed H α +1 / . Analogously to Theorem 4.2, we have the following theorem.
Theorem 4.3.
Assume that β > , γ > − α/ or β > − , γ > − α/ . If q ∈ H γ (0 , T ; ˙ H β (Ω)) , then there exists a unique w ∈ H α + γ (0 , T ; ˙ H β (Ω)) ∩ H γ (0 , T ; ˙ H β (Ω)) such that D α + γT − w − ∆ D γT − w = D γT − q in L (0 , T ; ˙ H β (Ω)) , and k w k H α + γ (0 ,T ; ˙ H β (Ω)) + k w k H γ (0 ,T ; ˙ H β (Ω)) C α,γ k q k H γ (0 ,T ; ˙ H β (Ω)) . .2 Transposition method In this subsection, we use the transposition method [17] to investigate the reg-ularity of problem (1) with more general data. Let G := H α (0 , T ; L (Ω)) ∩ L (0 , T ; ˙ H (Ω)) , and endow this space with the norm k·k G := max n k·k H α (0 ,T ; L (Ω)) , k·k L (0 ,T ; ˙ H (Ω)) o . In addition, define G := (cid:8)(cid:0) D α − T − v (cid:1) (0) : v ∈ G (cid:9) , G := (cid:8)(cid:0) D α − T − v (cid:1) (0) : v ∈ G (cid:9) , and we equip them respectively with the norms k v k G := inf v ∈ G (D α − T − v )(0)= v k v k G , k v k G := inf v ∈ G (D α − T − v )(0)= v k v k G . Assuming that f ∈ G ∗ , u ∈ G ∗ and u ∈ G ∗ , we call u ∈ L (0 , T ; L (Ω)) aweak solution to problem (1) if (cid:10) u, D αT − v − ∆ v (cid:11) Ω × (0 ,T ) = h f, v i G + (cid:10) u , (cid:0) D α − T − v (cid:1) (0) (cid:11) G + (cid:10) u , (cid:0) D α − T − v (cid:1) (0) (cid:11) G (47)for all v ∈ G . Theorem 4.4.
Problem (47) admits a unique solution u , and k u k L (0 ,T ; L (Ω)) C α (cid:16) k f k G ∗ + k u k G ∗ + k u k G ∗ (cid:17) . (48) Proof.
For any v ∈ L (0 , T ; L (Ω)), Theorem 4.3 implies that there exists aunique w ∈ G such that D αT − w − ∆ w = v and k w k G C α k v k L (0 ,T ; L (Ω)) . Moreover, for any w ∈ G , Lemma 3.6 implies that (cid:13)(cid:13) D αT − w (cid:13)(cid:13) L (0 ,T ; L (Ω)) C α k w k H α (0 ,T ; L (Ω)) . Therefore, applying the Babuˇska-Lax-Milgram theorem [15] yields that thereexists a unique u ∈ L (0 , T ; L (Ω)) such that (47) and (48) hold. This completesthe proof of this theorem. (cid:4) Remark 4.6. If u ∈ ˙ H − / (Ω) and u ∈ ˙ H − / (Ω) , then an evident computa-tion implies that u ∈ G ∗ and u ∈ G ∗ . Hence, the weak solution is well-definedby (47) for u ∈ ˙ H − / (Ω) , u ∈ ˙ H − / (Ω) and f ∈ G ∗ . Moreover, if f ∈ c W ∗ and D α/ ( u + tu ) ∈ W ∗ , then the weak solutions defined by (31) and (47) areessentially the same. A Petrov-Galerkin Method
Given J ∈ N > , let 0 = t < t < · · · < t J = T be a partition of [0 , T ]. Set τ j := t j − t j − and I j := ( t j − , t j ) for each 1 j J , and define τ := max j J τ j .Let K h be a shape-regular triangulation of Ω consisting of d -simplexes, and weuse h to denote the maximum diameter of the elements in K h . Define S h := (cid:8) v h ∈ H (Ω) : v h | K ∈ P ( K ) , ∀ K ∈ K h (cid:9) , c W τ := { b w τ ∈ L (0 , T ) : b w τ | I j ∈ P ( I j ) , ∀ j J } ,W τ := { w τ ∈ H (0 , T ) : w τ | I j ∈ P ( I j ) , ∀ j J } . Above and throughout, P k ( O )( k = 0 ,
1) denotes the set of polynomials definedon O with degree k , where O is either an interval or an element of K h .Moreover, define c W τ ⊗ S h := span { b w τ v h : b w τ ∈ c W τ , v h ∈ S h } ,W τ ⊗ S h := span { w τ v h : w τ ∈ W τ , v h ∈ S h } . Assuming that u , u ∈ S ∗ h and f ∈ ( c W τ ⊗ S h ) ∗ , we define an approximation U ∈ W τ ⊗ S h to problem (1) by that U (0) = u ,h and (cid:10) D α − U ′ , V (cid:11) Ω × (0 ,T ) + h∇ U, ∇ V i Ω × (0 ,T ) = h f, V i c W τ ⊗ S h + (cid:10) u , (cid:10) D α t, V (cid:11) (0 ,T ) (cid:11) S h (49)for all V ∈ c W τ ⊗ S h , where u ,h is the Ritz projection of u onto S h if u ∈ H (Ω)and u ,h is the L (Ω)-orthogonal projection of u onto S h if u ∈ S ∗ h \ H (Ω).Here the Ritz projection R h : H (Ω) → S h is defined by that h∇ ( v − R h v ) , ∇ v h i Ω = 0 ∀ v h ∈ S h , for all v ∈ H (Ω).Similar to [6, Theorem 4.1], we have the following stability result. Theorem 5.1.
Problem (49) admits a unique solution U . Moreover, if u ∈ ˙ H (Ω) , u ∈ L (Ω) and f ∈ H (1 − α ) / (0 , T ; L (Ω)) , then k U ′ k H ( α − / (0 ,T ; L (Ω)) + k U k C ([0 ,T ]; ˙ H (Ω)) C α,T (cid:0) k u k ˙ H (Ω) + k u k L (Ω) + k f k H (1 − α ) / (0 ,T ; L (Ω)) (cid:1) . Remark 5.1. By (31) , it is natural to develop another numerical algorithm forproblem (1) as follows: given u , u ∈ S ∗ h and f ∈ c W ∗ , seek U ∈ c W τ ⊗ S h suchthat (cid:10) D α/ U, V (cid:11) Ω × (0 ,T ) + (cid:10) ∇ D − α/ U, ∇ V (cid:11) Ω × (0 ,T ) = (cid:10) f, D − α/ T − V (cid:11) c W + (cid:10) u , h D α/ , V i (0 ,T ) (cid:11) S h + (cid:10) u , h D α/ t, V i (0 ,T ) (cid:11) S h (50) for all V ∈ c W τ ⊗ S h . Analogous to Theorem 4.1, if D α/ ( u + tu ) ∈ W ∗ , then k U k H α/ (0 ,T ; L (Ω)) + k U k H − α/ (0 ,T ; ˙ H (Ω)) C α (cid:0) k f k c W ∗ + (cid:13)(cid:13) D α/ ( u + tu ) (cid:13)(cid:13) W ∗ (cid:1) . ence, by Theorem 5.1 this algorithm is more robust than algorithm (49) ; how-ever, the computational cost of this algorithm is larger than that of the latter.To the author’s knowledge, this algorithm has never been proposed before. Wewill pay more attention to this algorithm in future works. This subsection considers the convergence analysis of the Petrov-Galerkin methodwith u = u = 0. Let σ := max i,j J { τ i /τ j } and ρ := max j J max i J | i − j | { τ i /τ j } . In what follows, a . b means that there exists a positive constant C , dependingonly on α, ρ, T, Ω and the shape-regular parameter of K h , such that a Cb . Inaddition, a ∼ b means that a . b . a . The main results are the following twotheorems. Theorem 5.2. If f ∈ H − α (0 , T ; L (Ω)) , then k u − U k C ([0 ,T ]; ˙ H (Ω)) . (cid:0) τ (3 − α ) / + ε ( α, τ, h ) (cid:1) k f k H − α (0 ,T ; L (Ω)) , (51) k u ′ − U ′ k H ( α − / (0 ,T ; L (Ω)) . (cid:0) τ (3 − α ) / + ε ( α, τ, h ) (cid:1) k f k H − α (0 ,T ; L (Ω)) , (52) where ε ( α, τ, h ) := h i f < α < / , (cid:0) | log h | (cid:1) h i f α = 3 / ,h /α − + C σ τ / − α h i f / < α < , and ε ( α, τ, h ) := ( h /α − i f < α / ,C σ τ / − α h i f / < α < . (53) Theorem 5.3. If f ∈ L (0 , T ; L (Ω)) , then k ( u − U ) ′ k H ( α − / (0 ,T ; L (Ω)) + k u − U k C ([0 ,T ]; ˙ H (Ω)) . (cid:0) τ ( α − / + ε ( α, τ, h ) (cid:1) k f k L (0 ,T ; L (Ω)) , (54) where ε ( α, τ, h ) := ( h − /α i f < α / ,C σ τ − / h i f / < α < . (55) Remark 5.2. If < α / , then by Lemma 2.2 and Theorem 4.2, estimates (52) and (54) are optimal with respect to the regularity of u ; moreover, (51) isoptimal and nearly optimal with respect to the regularity of u for < α < / and α = 3 / , respectively. If / < α < , then all the estimates (51) , (52) and (54) are optimal with respect to the regularity of u provided that the temporalgrid is quasi-uniform and h Cτ α/ for some positive constant C . However,numerical results indicate that the requirement h Cτ α/ is unnecessary.
24o prove the above two theorems, we need several interpolation operatorsas follows. Define W τ := { w τ ∈ H (0 , T ) : w τ | I j ∈ P ( I j ) , ∀ j J } . For 0 j J , set ω j := ( t , t ) , j = 0 , ( t J − , t J ) , j = J, ( t j − , t j +1 ) , j < J, and let Π j be the L -orthogonal projection operator onto P ( ω j ). We introducethe Cl´ement interpolation operator Q τ : L (0 , T ) → W τ by that, for all v ∈ L (0 , T ), ( Q τ v )( t j ) = (Π j v )( t j ) , j J. For each 1 j J , define c W τ,j := (cid:8) b w τ ∈ L (0 , T ) : b w τ | I τ ∈ P ( I τ ) , ∀ I τ ∈ T j (cid:9) , where T j := ( (cid:8) ω i : 0 i < j/ (cid:9) ∪ (cid:8) I i : j < i J (cid:9) , j is odd , (cid:8) ω i − : 1 i j/ (cid:9) ∪ (cid:8) I i : j < i J (cid:9) , j is even . Let P τ,j be the L -orthogonal projection operator onto c W τ,j , and define a familyof modified Cl´ement interpolation operators Q τ,j : L (0 , T ) → W τ by that, forany v ∈ L (0 , T ), ( h v − Q τ,j v, i ω i = 0 if 1 i < j and | i − j | is odd,( Q τ,j v )( p ) = ( Q τ v )( p ) if p is a node of the partition T j .By definition, it is evident that Q τ, = Q τ .For the above interpolant operators and the Ritz projection operator R h , wehave the following standard results [2, 24, 28], which will be used implicitly inour proofs. If v ∈ H γ (0 , T ) with 1 / < γ
2, then k ( I − Q τ ) v k C [0 ,T ] C ρ,γ,T τ γ − / k v k H γ (0 ,T ) ;if v ∈ H γ (0 , T ) with 0 γ
2, then k ( I − Q τ ) v k L (0 ,T ) C ρ,γ τ γ k v k H γ (0 ,T ) . If v ∈ H γ (0 , T ) with 0 γ
1, then k ( I − P τ,j ) v k L (0 ,tj ) C γ τ γ k v k H γ (0 ,t j ) , for all 1 j J . If v ∈ ˙ H r (Ω) with 1 r
2, then k ( I − R h ) v k L (Ω) + h k ( I − R h ) v k ˙ H (Ω) . h r k v k ˙ H r (Ω) . Except for those well-known results, we also need to establish some nonstandarderror estimates of the interpolation operator Q τ,j .25 emma 5.1. If v ∈ L (0 , T ) and z ∈ H γ (0 , T ) with γ , then h ( I − Q τ,j ) v, z i (0 ,t j ) C γ τ γ k ( I − Q τ,j ) v k L (0 ,t j ) k z k H γ (0 ,t j ) (56) for all j J .Proof. If j is even then, by the definitions of P τ,j and Q τ,j , h ( I − Q τ,j ) v, z i (0 ,t j ) = h ( I − Q τ,j ) v, ( I − P τ,j ) z i (0 ,t j ) k ( I − Q τ,j ) v k L (0 ,t j ) k ( I − P τ,j ) z k L (0 ,t j ) C γ τ γ k ( I − Q τ,j ) v k L (0 ,t j ) k z k H γ (0 ,t j ) , which proves (56). If j is odd then, also by the definitions of P τ,j and Q τ,j , h ( I − Q τ,j ) v, z i (0 ,t j ) = h ( I − Q τ,j ) v, z i (0 ,t ) + h ( I − Q τ,j ) v, ( I − P τ,j ) z i ( t ,t j ) k ( I − Q τ,j ) v k L (0 ,t j ) (cid:0) k z k L (0 ,t ) + k ( I − P τ,j ) z k L ( t ,t j ) (cid:1) k ( I − Q τ,j ) v k L (0 ,t j ) (cid:0) k z k L (0 ,t ) + C γ τ γ k z k H γ (0 ,t j ) (cid:1) . (57)Since Lemma 3.5 implies that z = D − γ D γ z , by Lemmas 3.2 and 3.6 we have k z k L (0 ,t ) = (cid:13)(cid:13) D − γ D γ z (cid:13)(cid:13) L (0 ,t ) C γ t γ (cid:13)(cid:13) D γ z (cid:13)(cid:13) L (0 ,t ) C γ t γ k z k H γ (0 ,t ) . Therefore, combining the above inequality and (57) proves (56) in the case that j is odd. This completes the proof of this lemma. (cid:4) Lemma 5.2. If v ∈ H γ (0 , T ) with γ , then k ( I − Q τ,j ) v k L (0 ,T ) C ρ,γ τ γ k v k H γ (0 ,T ) (58) for all j J .Proof. If j = 1 then (58) is standard, and so we assume that 2 j J . Bythe definition of Q τ,j , ( Q τ,j v ) (0) = ( Q τ v ) (0), ( Q τ,j v ) | ( t j ,T ) = ( Q τ v ) | ( t j ,T ) , and( Q τ,j v ) ( t i ) = ( Q τ v ) ( t i ) if 1 i j and j − i is even. If 1 i j − j − i is odd, then a straightforward calculation gives( Q τ,j v − Q τ v )( t i ) = 2 | ω i | Z ω i ( v − Q τ v )( t ) d t, (59)and hence k Q τ,j v − Q τ v k L ( ω i ) = r | ω i | | ( Q τ,j v − Q τ v )( t i ) | √ k ( I − Q τ ) v k L ( ω i ) , which implies k Q τ,j v − Q τ v k L (0 ,T ) √ k ( I − Q τ ) v k L (0 ,T ) . It follows that k ( I − Q τ,j ) v k L (0 ,T ) √ √ k ( I − Q τ ) v k L (0 ,T ) C ρ,γ τ γ k v k H γ (0 ,T ) . This proves (58) and thus concludes the proof. (cid:4) emma 5.3. Assume that < β < / . If v ∈ H γ (0 , T ) with β + 1 γ ,then (cid:13)(cid:13) ( v − Q τ,j v ) ′ (cid:13)(cid:13) H β (0 ,T ) C ρ,β,γ,T τ γ − − β k v k H γ (0 ,T ) , (60) for all j J .Proof. For simplicity, set g := ( v − Q τ,j v ) ′ . Following the proof of [8, Lemma4.1], we obtain k g k H β (0 ,T ) C β k g k H β (0 ,T ) C β,T (cid:0) I + I (cid:1) , where I := J X i =1 Z I i Z I i | g ( s ) − g ( t ) | | s − t | β d s d t, I := J X i =1 Z I i g ( t ) (cid:0) ( t i − t ) − β + ( t − t i − ) − β (cid:1) d t. As I can be estimated by that I J X i =1 τ γ − − β ) i Z I i Z I i | g ( s ) − g ( t ) | | s − t | γ − d s d t = J X i =1 τ γ − − β ) i Z I i Z I i | v ′ ( s ) − v ′ ( t ) | | s − t | γ − d s d t C γ,T τ γ − − β ) k v ′ k H γ − (0 ,T ) C γ,T τ γ − − β ) k v k H γ (0 ,T ) , it remains to prove I C ρ,β,γ,T τ γ − − β ) k v k H γ (0 ,T ) . (61)By [17, Theorem 11.2] and the fact that H γ − (0 ,
1) is continuously embeddedin H β (0 , Z | z ( t ) | (cid:0) t − β + (1 − t ) − β (cid:1) d t C β k z k H β (0 , C β,γ k z k H γ − (0 , C β,γ (cid:18) k z k L (0 , + Z Z | z ( s ) − z ( t ) | | s − t | γ − d s d t (cid:19) , where z ∈ H γ − (0 , I C β,γ J X i =1 (cid:18) τ − βi k g k L ( I i ) + τ γ − − β ) i Z I i Z I i | v ′ ( s ) − v ′ ( t ) | | s − t | γ − d s d t (cid:19) C β,γ,T (cid:18) J X i =1 τ − βi k g k L ( I i ) + τ γ − − β ) k v ′ k H γ − (0 ,T ) (cid:19) C β,γ,T (cid:18) J X i =1 τ − βi k g k L ( I i ) + τ γ − − β ) k v k H γ (0 ,T ) (cid:19) .
27o prove (61), it suffices to prove J X i =1 τ − βi k g k L ( I i ) C ρ,β,γ,T τ γ − − β ) k v k H γ (0 ,T ) . (62)By (59), we have k ( Q τ v − Q τ,j v ) ′ k L ( ω i ) C ρ | ω i | − k ( I − Q τ ) v k L ( ω i ) , for all 1 i j − j − i is odd. Hence it follows that J X i =1 τ − βi k ( Q τ v − Q τ,j v ) ′ k L ( I i ) C ρ J X i =1 τ − β − i k ( I − Q τ ) v k L ( I i ) , which yields the inequality J X i =1 τ − βi k g k L ( I i ) C ρ J X i =1 τ − β − i (cid:16) k ( I − Q τ ) v k L ( I i ) + τ i (cid:13)(cid:13) ( v − Q τ v ) ′ (cid:13)(cid:13) L ( I i ) (cid:17) . Therefore, by the standard estimates J X i =1 τ − βi k ( v − Q τ v ) ′ k L ( I i ) C ρ,β,γ,T τ γ − − β ) k v k H γ (0 ,T ) , J X i =1 τ − β − i k ( I − Q τ ) v k L ( I i ) C ρ,β,γ,T τ γ − − β ) k v k H γ (0 ,T ) , we obtain (62). This completes the proof of this lemma. (cid:4) Lemma 5.4. If w ∈ W τ , then k w k H β +1 (0 ,T ) C σ,β,γ,T τ γ − β − k w k H γ (0 ,T ) (63) for all β < / and γ β + 1 .Proof. If β = 0, then (63) is trivial for γ = 1 and standard for γ = 0, andhence applying [21, Lemma 22.3] yields (63) for 0 < γ <
1. It remains thereforeto prove (63) for 0 < β < / γ β + 1. To this end, we assume0 < β < /
2. For any 1 γ β + 1, following the proof of [8, Lemma 4.1], weobtain (cid:13)(cid:13) w ′ (cid:13)(cid:13) H β (0 ,T ) C β,T (cid:13)(cid:13) w ′ (cid:13)(cid:13) H β (0 ,T ) C β,T J X i =1 Z I i (cid:12)(cid:12) w ′ ( t ) (cid:12)(cid:12) (cid:16) ( t i − t ) − β + ( t − t i − ) − β (cid:17) d t C β,T J X i =1 τ γ − β − i Z I i (cid:12)(cid:12) w ′ ( t ) (cid:12)(cid:12) (cid:0) ( t i − t ) − γ + ( t − t i − ) − γ (cid:1) d t C σ,β,γ,T τ γ − β − (cid:13)(cid:13) w ′ (cid:13)(cid:13) H γ − (0 ,T ) . Hence, by the estimates k w k H β +1 (0 ,T ) C β k w ′ k H β (0 ,T ) and k w ′ k H γ − (0 ,T ) C γ k w k H γ (0 ,T ) , k w k H β +1 (0 ,T ) C β k w ′ k H β (0 ,T ) C σ,β,γ,T τ γ − β − k w ′ k H γ − (0 ,T ) C σ,β,γ,T τ γ − β − k w k H γ (0 ,T ) , namely (63) holds for 1 γ β + 1. In addition, for any 0 γ <
1, since wehave already proved k w k H (0 ,T ) C σ,γ,T τ γ − k w k H γ (0 ,T ) , k w k H β +1 (0 ,T ) C σ,β,T τ − β k w k H (0 ,T ) , it is clear that (63) holds. This completes the proof. (cid:4) Lemma 5.5. If f ∈ H − α (0 , T ; L (Ω)) , then k ( U − Q τ,j R h u ) ′ k H ( α − / (0 ,t j ; L (Ω)) + k ( U − Q τ,j R h u )( t j ) k ˙ H (Ω) . (cid:0) τ (3 − α ) / + ε ( α, τ, h ) (cid:1) k f k H − α (0 ,T ; L (Ω)) , (64) for all j J , where ε ( α, τ, h ) is defined by (53) .Proof. By (45) and (49), (cid:10) D α − ( u − U ) ′ , ξ ′ j (cid:11) Ω × (0 ,t j ) + (cid:10) ∇ ( u − U ) , ∇ ξ ′ j (cid:11) Ω × (0 ,t j ) = 0 , where ξ j = U − Q τ,j R h u . As ξ j (0) = 0, using integration by parts gives2 (cid:10) ∇ ξ j , ∇ ξ ′ j (cid:11) Ω × (0 ,t j ) = k ξ j ( t j ) k H (Ω) , and a simple calculation then yields (cid:10) D α − ξ ′ j , ξ ′ j (cid:11) Ω × (0 ,t j ) + 12 k ξ j ( t j ) k H (Ω) = (cid:10) ∇ ( u − Q τ,j R h u ) , ∇ ξ ′ j (cid:11) Ω × (0 ,t j ) + (cid:10) D α − ( u − Q τ,j R h u ) ′ , ξ ′ j (cid:11) Ω × (0 ,t j ) . It follows from Lemmas 3.6 and 3.8 that (cid:13)(cid:13) ξ ′ j (cid:13)(cid:13) H ( α − / (0 ,t j ; L (Ω)) + k ξ j ( t j ) k H (Ω) . E + E + E , where E := (cid:10) D α − ( u − Q τ,j u ) ′ , ξ ′ j (cid:11) Ω × (0 ,t j ) ,E := (cid:10) ∇ ( u − Q τ,j R h u ) , ∇ ξ ′ j (cid:11) Ω × (0 ,t j ) ,E := D D α − (cid:0) Q τ,j ( u − R h u ) (cid:1) ′ , ξ ′ j E Ω × (0 ,t j ) . Next, let us estimate E , E and E one by one. Since applying Lemma 5.3indicates k ( u − Q τ,j u ) ′ k H ( α − / (0 ,T ; L (Ω)) . τ (3 − α ) / k u k H (0 ,T ; L (Ω)) ,
29y Lemmas 3.6 and 3.8 and Theorem 4.2 we obtain E . (cid:13)(cid:13)(cid:13) D ( α − / ( u − Q τ,j u ) ′ (cid:13)(cid:13)(cid:13) L (0 ,t j ; L (Ω)) (cid:13)(cid:13)(cid:13) D ( α − / ξ ′ j (cid:13)(cid:13)(cid:13) L (0 ,t j ; L (Ω)) . k ( u − Q τ,j u ) ′ k H ( α − / (0 ,T ; L (Ω)) (cid:13)(cid:13) ξ ′ j (cid:13)(cid:13) H ( α − / (0 ,t j ; L (Ω)) . τ (3 − α ) / k f k H − α (0 ,T ; L (Ω)) (cid:13)(cid:13) ξ ′ j (cid:13)(cid:13) H ( α − / (0 ,t j ; L (Ω)) . By Lemma 5.1 and the definition of R h , E = (cid:10) ∇ ( u − Q τ,j u ) , ∇ ξ ′ j (cid:11) Ω × (0 ,t j ) = (cid:10) ( Q τ,j − I )∆ u, ξ ′ j (cid:11) Ω × (0 ,t j ) . τ ( α − / k ( I − Q τ,j ) u k L (0 ,T ; ˙ H (Ω)) (cid:13)(cid:13) ξ ′ j (cid:13)(cid:13) H ( α − / (0 ,t j ; L (Ω)) , so that using Theorem 4.2 and Lemma 5.2 gives E . τ (3 − α ) / k f k H − α (0 ,T ; L (Ω)) (cid:13)(cid:13) ξ ′ j (cid:13)(cid:13) H ( α − / (0 ,t j ; L (Ω)) . If 1 < α /
2, then applying Lemma 5.3 indicates (cid:13)(cid:13)(cid:0) Q τ,j ( u − R h u ) (cid:1) ′ (cid:13)(cid:13) H ( α − / (0 ,T ; L (Ω)) . k ( I − R h ) u k H ( α +1) / (0 ,T ; L (Ω)) . h /α − k u k H ( α +1) / (0 ,T ; ˙ H /α − (Ω)) , and if 3 / < α < (cid:13)(cid:13)(cid:0) Q τ,j ( u − R h u ) (cid:1) ′ (cid:13)(cid:13) H ( α − / (0 ,T ; L (Ω)) . C σ τ / − α (cid:13)(cid:13)(cid:0) Q τ,j ( u − R h u ) (cid:1) ′ (cid:13)(cid:13) H − α/ (0 ,T ; L (Ω)) . C σ τ / − α k ( I − R h ) u k H − α/ (0 ,T ; L (Ω)) . C σ τ / − α h k u k H − α/ (0 ,T ; ˙ H (Ω)) . Therefore, by Lemmas 2.2, 3.6 and 3.8 and Theorem 4.2 we get E . (cid:13)(cid:13)(cid:13) D ( α − / (cid:0) Q τ,j ( u − R h u ) (cid:1) ′ (cid:13)(cid:13)(cid:13) L (0 ,t j ; L (Ω)) (cid:13)(cid:13)(cid:13) D ( α − / ξ ′ j (cid:13)(cid:13)(cid:13) L (0 ,t j ; L (Ω)) . (cid:13)(cid:13)(cid:0) Q τ,j ( u − R h u ) (cid:1) ′ (cid:13)(cid:13) H ( α − / (0 ,T ; L (Ω)) (cid:13)(cid:13) ξ ′ j (cid:13)(cid:13) H ( α − / (0 ,t j ; L (Ω)) . ε ( α, τ, h ) k f k H − α (0 ,T ; L (Ω)) (cid:13)(cid:13) ξ ′ j (cid:13)(cid:13) H ( α − / (0 ,t j ; L (Ω)) . Finally, combining the estimates of E , E and E and the Young’s inequalitywith ǫ , we obtain that (cid:13)(cid:13) ξ ′ j (cid:13)(cid:13) H ( α − / (0 ,t j ; L (Ω)) + k ξ j ( t j ) k ˙ H (Ω) . (cid:0) τ (3 − α ) / + ε ( α, τ, h ) (cid:1) k f k H − α (0 ,T ; L (Ω)) , for all 1 j J . This proves (64) and thus concludes the proof. (cid:4) Remark 5.3.
Assume that f ∈ H − α (0 , T ; L (Ω)) with / < α < . ByLemma 2.2 and Theorem 4.2 we have u ∈ H ( α +1) / (0 , T ; ˙ H /α − (Ω)) ∩ H − α/ (0 , T ; ˙ H (Ω)) . herefore, ( R h u ) ′ = R h u ′ ∈ H ( α − / (0 , T ; ˙ H /α − (Ω)) may not make sensesince /α − < , but ( R h u ) ′ = R h u ′ ∈ H − α/ (0 , T ; ˙ H (Ω)) makes senseindeed. This is the reason why we use τ / − α k ( Q τ,j ( u − R h u )) ′ k H − α/ (0 ,T ; L (Ω)) to bound k ( Q τ,j ( u − R h u )) ′ k H ( α − / (0 ,T ; L (Ω)) , when estimating the term E in the proof of the above lemma. Analogously to Lemma 5.5, we have the following lemma.
Lemma 5.6. If f ∈ L (0 , T ; L (Ω)) , then k ( U − Q τ,j R h u ) ′ k H ( α − / (0 ,t j ; L (Ω)) + k ( U − Q τ,j R h u )( t j ) k ˙ H (Ω) . (cid:0) τ ( α − / + C σ τ − / h (cid:1) k f k L (0 ,T ; L (Ω)) , for all j J . Proof of Theorem 5.2.
By Lemma 2.2 and Theorem 4.2, we have k ( I − Q τ ) R h u k C ([0 ,T ]; ˙ H (Ω)) . τ (3 − α ) / k f k H − α (0 ,T ; L (Ω)) , k ( I − R h ) u k C ([0 ,T ]; ˙ H (Ω)) . h k f k H − α (0 ,T ; L (Ω)) if 1 < α < / ,h − ǫ ǫ k f k H / (0 ,T ; L (Ω)) if α = 3 / ,h /α − k f k H − α (0 ,T ; L (Ω)) if 3 / < α < , where 0 < ǫ /
2. Therefore, if α = 3 /
2, then letting ǫ = 1 / (2 + | log h | ) yields k ( I − R h ) u k C ([0 ,T ]; ˙ H (Ω)) . (cid:0) | log h | (cid:1) h k f k H / (0 ,T ; L (Ω)) . Since ( Q τ,j R h u )( t j ) = ( Q τ R h u )( t j ) for all 1 j J , we obtain k Q τ R h u − U k C ([0 ,T ]; ˙ H (Ω)) = max j J k ( Q τ R h u − U )( t j ) k ˙ H (Ω) = max j J k ( Q τ,j R h u − U )( t j ) k ˙ H (Ω) . (cid:0) τ (3 − α ) / + ε ( α, τ, h ) (cid:1) k f k H − α (0 ,T ; L (Ω)) , by Lemma 5.5, and hence, k R h u − U k C ([0 ,T ]; ˙ H (Ω)) k ( I − Q τ ) R h u k C ([0 ,T ]; ˙ H (Ω)) + k Q τ R h u − U k C ([0 ,T ]; ˙ H (Ω)) . (cid:0) τ (3 − α ) / + ε ( α, τ, h ) (cid:1) k f k H − α (0 ,T ; L (Ω)) . (65)Therefore, using the triangle inequality k u − U k C ([0 ,T ]; ˙ H (Ω)) k u − R h u k C ([0 ,T ]; ˙ H (Ω)) + k R h u − U k C ([0 ,T ]; ˙ H (Ω)) proves (51). Since the proof of Lemma 5.5 yields that (cid:13)(cid:13)(cid:0) u − Q τ,j R h u ) (cid:1) ′ (cid:13)(cid:13) H ( α − / (0 ,T ; L (Ω)) (cid:13)(cid:13) ( u − Q τ,j u ) ′ (cid:13)(cid:13) H ( α − / (0 ,T ; L (Ω)) + (cid:13)(cid:13)(cid:0) Q τ,j ( u − R h u ) (cid:1) ′ (cid:13)(cid:13) H ( α − / (0 ,T ; L (Ω)) . (cid:0) τ (3 − α ) / + ε ( α, τ, h ) (cid:1) k f k H − α (0 ,T ; L (Ω)) , (cid:4) Proof of Theorem 5.3.
In view of Lemma 5.6, the proof of the case 3 / < α < < α / k ( u − U ) ′ k H ( α − / (0 ,T ; L (Ω)) . k f k H (1 − α ) / (0 ,T ; L (Ω)) , by (52) and the fact L (0 , T ; L (Ω)) = (cid:2) H (1 − α ) / (0 , T ; L (Ω)) , H − α (0 , T ; L (Ω)) (cid:3) ( α − / (3 − α ) , , applying [21, Lemma 22.3] yields (cid:13)(cid:13) ( u − U ) ′ (cid:13)(cid:13) H ( α − / (0 ,T ; L (Ω)) . (cid:0) τ (3 − α ) / + h /α − (cid:1) ( α − / (3 − α ) k f k L (0 ,T ; L (Ω)) . Hence, from the inequality (cid:0) τ (3 − α ) / + h /α − (cid:1) ( α − / (3 − α ) < τ ( α − / + h − /α , it follows that k ( u − U ) ′ k H ( α − / (0 ,T ; L (Ω)) . (cid:0) τ ( α − / + h − /α (cid:1) k f k L (0 ,T ; L (Ω)) . (66)In addition, Theorems 4.2 and 5.1 imply k R h u − U k C ([0 ,T ]; ˙ H (Ω)) . k f k H (1 − α ) / (0 ,T ; L (Ω)) , and then, by this estimate and (65), proceeding as in the proof of (66) yields k R h u − U k C ([0 ,T ]; ˙ H (Ω)) . (cid:0) τ ( α − / + h − /α (cid:1) k f k L (0 ,T ; L (Ω)) . Therefore, since Theorem 4.2 gives k u − R h u k C ([0 ,T ]; ˙ H (Ω)) . h − /α k f k L (0 ,T ; L (Ω)) , we obtain k u − U k C ([0 ,T ]; ˙ H (Ω)) k u − R h u k C ([0 ,T ]; ˙ H (Ω)) + k R h u − U k C ([0 ,T ]; ˙ H (Ω)) . (cid:0) τ ( α − / + h − /α (cid:1) k f k L (0 ,T ; L (Ω)) . (67)Finally, combining (66) and (67) proves (54) and thus concludes the proofof this theorem. (cid:4) In this section, we present some numerical examples to validate Theorems 5.2and 5.3 in one dimensional case. We set Ω := (0 , T := 1, and use the uniformtemporal and spatial grids. Define E ( U ) := k ˜ u − U k C ([0 ,T ]; ˙ H (Ω)) , E ( U ) := (cid:13)(cid:13) D ( α − / (˜ u − U ) ′ (cid:13)(cid:13) L (0 ,T ; L (Ω)) , u is the numerical solution with τ = 2 − and h = 2 − . Here we observethat Lemma 3.6 implies k (˜ u − U ) ′ k H ( α − / (0 ,T ; L (Ω)) ∼ (cid:13)(cid:13) D ( α − / (˜ u − U ) ′ (cid:13)(cid:13) L (0 ,T ; L (Ω)) . It is easy to see that (49) yields a block triangular Toeplitz-like with tri-diagonalblock system, so that we can apply a fast direct O ( h − J (log J ) ) solver basedon the divide-and-conquer strategy [26] to solve this system efficiently. Addi-tionally, the calculation of E ( U ) involves only the matrix-vector multiplicationof a block triangular Toeplitz-like matrix, which can be completed within com-putational cost of O ( h − J log J ) by fast Fourier transform. Example 1.
This example adopts f ( x, t ) = t − . x − . , ( x, t ) ∈ Ω × (0 , T ) . The relationship between the spatial errors and the spatial step sizes are dis-played in Fig. 1. These numerical results indicate that E ( U ) ≈ O (cid:0) h − /α (cid:1) , E ( U ) ≈ O (cid:0) h − /α (cid:1) . The relationship between the errors and the temporal step sizes are plotted inFig. 2, which demonstrate that E ( U ) ≈ O (cid:0) τ ( α − / (cid:1) , E ( U ) ≈ O (cid:0) τ ( α − / (cid:1) . Therefore, if 1 < α / / < α <
2, numerical results also show the optimal accuracyof E ( U ) and E ( U ) with respect to the regularity, without the restriction that h Cτ α/ . -2 -1 h -1 E ( U ) α = 1 . α = 1 . α = 1 . -2 -1 h E ( U ) α = 1 . α = 1 . α = 1 . Figure 1:
Spatial errors of Example 1, τ = 2 − . -2 -1 τ E ( U ) α = 1 . α = 1 . α = 1 . -2 -1 τ -1 E ( U ) α = 1 . α = 1 . α = 1 . Figure 2:
Temporal errors of Example 1, h = 2 − . Example 2.
This example employs f ( x, t ) = t . − α x − . , ( x, t ) ∈ Ω × (0 , T ) , and the spatial errors and temporal errors are plotted in Figs. 3 and 4, respec-tively. These numerical results demonstrate that E ( U ) ≈ ( O (cid:0) τ (3 − α ) / + h (cid:1) if 1 < α / , O (cid:0) τ (3 − α ) / + h /α − (cid:1) if 3 / < α < , E ( U ) ≈ O (cid:0) τ (3 − α ) / + h /α − (cid:1) . Hence, if 1 < α /
2, then numerical results verify the theoretical predictionsof Theorem 5.2. But for 3 / < α <
2, numerical results also indicate that theconvergence rates of E ( U ) and E ( U ) are optimal with respect to the regularity,without the requirement h Cτ α/ . -3 -2 -1 h -3 -2 -1 E ( U ) α = 1 . α = 1 . α = 1 . α = 1 . α = 1 . -3 -2 -1 h -5 -4 -3 -2 -1 E ( U ) α = 1 . α = 1 . α = 1 . Figure 3:
Spatial errors of Example 2, τ = 2 − . -4 -3 τ -5 -4 -3 -2 -1 E ( U ) α = 1 . α = 1 . α = 1 . -3 -2 -1 τ -3 -2 -1 E ( U ) α = 1 . α = 1 . α = 1 . Figure 4:
Temporal errors of Example 2, h = 2 − . This paper concerns the convergence of a Petrov-Galerkin method for fractionalwave problems with nonsmooth data. The weak solution and its regularity arestudied by the variational approach. Optimal error estimate with respect tothe regularity of the solution under the norm C ([0 , T ]; ˙ H (Ω)) is derived if f ∈ L (0 , T ; L (Ω)) and 1 < α /
2, and numerical results validate this theoreticalresult. For 3 / < α <
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