On the existence and uniqueness of weak solutions to time-fractional elliptic equations with time-dependent variable coefficients
aa r X i v : . [ m a t h . A P ] J a n On the existence and uniqueness of weaksolutions to time-fractional elliptic equations withtime dependent variable coefficients
H.T. Tuan ∗ January 29, 2021
Abstract
This paper is devoted to discussing the existence and uniqueness ofweak solutions to time-fractional elliptic equations having time-dependentvariable coefficients. To obtain the main result, our strategy is to com-bine the Galerkin method, a basic inequality for the fractional derivativeof convex Lyapunov candidate functions, the Yoshida approximation se-quence and the weak compactness argument.
Keywords and phases : Time fractional derivatives, Existence and uniqueness,Weak solution, Elliptic equations, Variable coefficients
Diffusion equations with fractional-order derivatives in time (which is calledas time-fractional diffusion equations) have been introduced in Physics by Nig-matullin [N86] to describe super slow diffusion process in a porous medium withthe structure type of fractal geometry (Koch’s tree). Then, by the probabilisticpoint of view, in the paper [MK2000], Metzler and Klafter have pointed out thata time-fractional diffusion equation generates a non-Markovian diffusion processwith a long memory. After that, Roman and Alemany [RA94] have consideredcontinuous-time random walks on fractals and observed that the average prob-ability density of random walks on fractals obeys a diffusion equation with afractional time derivative asymptotically. Another context where such systemsappear is the modelling of evolution processes in materials with memory, seee.g., [P93, C99]. ∗ [email protected], Institute of Mathematics, Vietnam Academy of Science and Tech-nology, 18 Hoang Quoc Viet, 10307 Ha Noi, Viet Nam N the set of natural numbers and by R the setof real numbers. For any natural d ∈ N , let R d be the d -dimensional Euclideanspace. For a open subset Ω of R d , let C ∞ c (Ω) denote the space of infinitelydifferentiable functions f : Ω → R with the compact support in Ω, L p (Ω),2 ∈ N , be the set of measurable functions f : Ω → R such that Z Ω | f ( x ) | p dx < ∞ ,H (Ω) be the Sobolev space containing all locally integrable functions f : Ω → R such that f and its weak derivatives belong to L (Ω), H (Ω) be the closure of C ∞ c (Ω) in H (Ω), and H − (Ω) be the dual space of H (Ω). Fix T >
0, denoteby W ([0 , T ]; R ) the space of functions f : [0 , T ]; R such that f and its weakderivative belong to the space L ([0 , T ]; R ) and W ([0 , T ]; H − (Ω)) by the spaceof functions f : [0 , T ] → H − (Ω) such that f and the weak derivative belong tothe space L ([0 , T ]; H − (Ω)). We briefly recall an abstract framework of fractional calculus.Let α ∈ (0 , , T ] ⊂ R and x : [0 , T ] → R be a measurable function suchthat R T | x ( τ ) | dτ < ∞ . The right-handed Riemann–Liouville integral operatorof order α is defined by( I α x )( t ) := 1Γ( α ) Z t ( t − τ ) α − x ( τ ) dτ, where Γ( · ) is the Gamma function. The left-handed Riemann–Liouville integraloperator of order α is defined by( I αT − x )( t ) := 1Γ( α ) Z Tt ( τ − t ) α − x ( τ ) dτ. We have the following result on the relation between the right-handed and left-handed Riemann–Liouville integral operators.
Theorem 2.1 (The Hardy–Littlewood form of fractional integration by parts) . If p > q >
1, 0 < α < p + q − ≤ α , and f ∈ L p ([0 , T ]; R ) , g ∈ L q ([0 , T ]; R ) , then Z T ( I α f ( t )) g ( t ) dt = Z T f ( t ) I αT − g ( t ) dt. Proof.
See [LY37]. 3he right-handed
Riemann–Liouville fractional derivative RL D αa + x of x on [0 , T ]is defined by RL D α x ( t ) := ( DI − α x )( t ) for almost t ∈ [0 , T ] , where D = ddt is the usual derivative. The left-handed Riemann–Liouville frac-tional derivative RL D αT − x of x on [0 , T ] is defined by RL D αT − x ( t ) := − ( DI − αT − x )( t ) for almost t ∈ [0 , T ] . Theorem 2.2.
The formula Z T f ( t ) RL D α g ( t ) dt = Z T g ( t ) RL D αT − f ( t ) dt is valid for 0 < α < f ∈ I αT − ( L p ), g ∈ I α ( L q ) and p + q − ≤ α . Proof.
See [SKM93, Corollary 2, pp. 46].The right-handed
Caputo fractional derivative of x on [0 , T ] is defined by C D α x ( t ) = RL D αa + ( x ( t ) − x (0)) for almost t ∈ [0 , T ]and the left-handed Caputo fractional derivative of x on [0 , T ] is defined by C D αT − x ( t ) = RL D αT − ( x ( t ) − x ( T )) for almost t ∈ [0 , T ] . We have a sufficient condition for the existence of fractional derivative.
Theorem 2.3.
Let f ∈ AC ([0 , T ]; R ), α ∈ (0 , RL D α f and RL D αT − f exist almost everywhere. Moreover, RL D α f, RL D αT − f ∈ L r ([0 , T ]; R ), 1 ≤ r ≤ α , and RL D α f ( t ) = 1Γ(1 − α ) h f (0) t α + Z t f ′ ( τ )( t − τ ) α dτ i , RL D αT − f ( t ) = 1Γ(1 − α ) h f ( T )( T − t ) α − Z Tt f ′ ( τ )( τ − t ) α dτ i . Proof.
See [SKM93, Lemma 2.2, pp. 35–36].
Definition 2.4.
Let u ∈ L ([0 , T ]; H (Ω)). We define the weak Riemann–Liouville fractional derivative of the order α of u , RL D α u ( t ), as below Z T ϕ ( t ) RL D α u ( t ) dt = Z T RL D αT − ϕ ( t ) u ( t ) dt for all ϕ ∈ C ∞ c ((0 , T ); R ). 4 Weak solutions to time-fractional elliptic equa-tions
Let Ω be a bounded domain in R d with the boundary ∂ Ω ∈ C , T > α ∈ (0 , T = (0 , T ] × Ω. We consider the equation of the order α∂ α u ( t, x ) ∂t α − d X i,j =1 ∂ x i ( a ij ( t, x ) ∂ x j u ( t, x )) + d X j =1 b j ( t, x ) ∂ x j u ( t, x ) + c ( t, x ) u ( t, x ) = f ( t )(1)for ( t, x ) ∈ Ω T , where ∂ α u ( t, · ) ∂t α is the weak Riemann–Liouville fractional deriva-tive of the order α of u with respect to the time variable t and(a1) a ij , b j , c ∈ L ∞ (Ω T ; R ) for all 1 ≤ i, j ≤ d ;(a2) a ij = a ji for all 1 ≤ i, j ≤ d ;(a3) there exists θ > P di,j =1 a ij ( t, x ) ξ i ξ j ≥ θ k ξ k for a.e. t ∈ (0 , T ), x ∈ Ω and for all ξ ∈ R d ;(a4) f ∈ L ∞ ([0 , T ]; H − (Ω)).Assume that u ( t, x ) = 0 on [0 , T ] × ∂ Ω . (2)Denote a ( u, v ; t ) := Z Ω X ≤ i,j ≤ d a ij ∂ x j u∂ x i v + X ≤ j ≤ d b j ∂ x j uv + cuv for all u, v ∈ H (Ω) and u ( t ) := [ u ( t )]( x ) for all t ∈ [0 , T ], x ∈ Ω. Definition 3.1 (Weak solution) . A function u : [0 , T ] → H (Ω) is a weaksolution to the problem (1) with the condition (2) if(i) u ∈ L ([0 , T ]; H (Ω)) and RL D α u ∈ L ([0 , T ]; H − (Ω));(ii) for all v ∈ H (Ω) h RL D α u ( t ) , v i H − × H + a ( u ( t ) , v ; t ) = h f ( t ) , v i H − × H for a.e. t ∈ (0 , T ), where h· , ·i H − × H is the duality pairing between H − (Ω) and H (Ω). 5 .1 Galerkin approximation solution Let { e j } ∞ j =1 be smooth functions which constitutes an orthonormal basic of L (Ω) and a basis of H (Ω) such that(i) − ∆ e k = λ k e k , k ∈ N ;(ii) Z Ω e j e j = ( , i = j , i = j and Z Ω De i De j = ( λ i , i = j , i = j For the existence of these functions, see [E98, p. 335]. Fix N ∈ N . Let E N = span { e , . . . , e N } and P N be the project map from L (Ω) to E N definedby P N u = P Ni =1 c i e i with u ∈ L (Ω) has the form u = P ∞ i =1 c i e i in which c i = R Ω ue i dx , i ∈ N . In this section, using Galerkin method, we will constructapproximation solutions to the problem (1)–(2) in E N .Let the function u N : [0 , T ] → E N having the form u N ( t ) = N X i =1 c i ( t ) e i , t ∈ [0 , T ] , (3)here c i ( · ), 1 ≤ i ≤ N , is continuous and has the Riemann–Liouville fractionalderivative of the order α on [0 , T ]. Assume that( RL D α u N , v ) L + a ( u N ( t ) , v ; t ) = h f ( t ) , v i H − × H (4)for a.e. t ∈ (0 , T ) and u N (0) = 0 . (5)We obtain the following result. Proposition 3.2.
For any N ∈ N , there exists a unique solution to the problem(4)–(5) having the form (3). Proof.
Consider the function u N : [0 , T ] → E N having the form u N ( t ) = N X i =1 c i ( t ) e i , t ∈ [0 , T ] , where c i ( · ), 1 ≤ i ≤ N , is continuous and has the Riemann–Liouville fractionalderivative of the order α on [0 , T ]. To u N ( · ) is a solution to (4)–(5) then c i ( · ),6 ≤ i ≤ N , have to satisfy the following condition( N X i =1 RL D α c i ( t ) e i , e j ) L + a ( N X i =1 c i ( t ) e i , e j ; t ) (6)= RL D α c j ( t ) + N X i =1 c i ( t ) a ( e i , e j ; t ) = h f ( t ) , e j i H − × H = f j ( t )for 1 ≤ j ≤ N and almost every t ∈ (0 , T ]. Moreover, c j (0) = 0 , ≤ j ≤ N. (7)Put −→ c ( t ) = ( c ( t ) , . . . , c N ( t )) T , A ( t ) = ( a ( e i , e j ; t )) ≤ i,j ≤ N , and −→ f ( t ) = ( f ( t ) , . . . , f N ( t )) T . Then the system (6)–(7) is rewritten in the form RL D α −→ c ( t ) + A ( t ) −→ c ( t ) = C D α −→ c ( t ) + A ( t ) −→ c ( t ) = −→ f ( t ) , t ∈ (0 , T ] , (8) −→ c (0) = 0 . (9)Hence, the system (8)–(9) has a solution on [0 , T ] if and only if the followingintegral equation has a continuous solution −→ c ( t ) = − α ) Z t ( t − τ ) α − A ( τ ) −→ c ( τ ) dτ + 1Γ( α ) Z t ( t − τ ) α − −→ f ( τ ) dτ, t ∈ [0 , T ] . (10)On the space C ([0 , T ]; R N ), we establish an operator as T ϕ ( t ) = − α ) Z t ( t − τ ) α − A ( τ ) ϕ ( τ ) dτ + 1Γ( α ) Z t ( t − τ ) α − −→ f ( τ ) dτ, t ∈ [0 , T ] . For any γ >
0, define a norm k · k γ on C ([0 , T ]; R N ) by k ϕ k γ := max t ∈ [0 ,T ] k ϕ ( t ) k exp ( γt ) . It is obvious that ( C ([0 , T ]; R N ) , k · k γ ) is a Banach space. On the other hand,for any ϕ, ˜ ϕ ∈ C ([0 , T ]; R N ) and t ∈ [0 , T ], we have kT ϕ ( t ) − T ˜ ϕ ( t ) k exp ( γt ) ≤ ess sup t ∈ [0 ,T ] k A ( t ) k Γ( α ) Z t ( t − τ ) α − exp ( − γ ( t − τ )) k ϕ ( τ ) − ˜ ϕ ( τ ) k exp ( γτ ) dτ ≤ ess sup t ∈ [0 ,T ] k A ( t ) k γ α Γ( α ) Z γt u α − exp ( − u ) du k ϕ − ˜ ϕ k γ ≤ ess sup t ∈ [0 ,T ] k A ( t ) k γ α k ϕ − ˜ ϕ k γ . Hence, for γ > T is contractive in ( C ([0 , T ]; R N ) , k·k γ ) and has a unique fixed point which is also the solution to the system (10).The proof is complete. 7e now give some estimate of the Galerkin approximation solution. Proposition 3.3.
There exists a positive constant C , depending on Ω, T andthe coefficients of the equation (1), such that for all N ∈ N it holds that k u N k L ([0 ,T ]; L (Ω)) + k u N k L ([0 ,T ]; H (Ω)) + k RL D α u N k L ([0 ,T ]; H − (Ω)) ≤ C k f k L ∞ ([0 ,T ]; H − (Ω) . Proof.
First, using [T18, Theorem 2], we have h RL D α u N ( t ) , u N ( t ) i H − × H = h C D α u N ( t ) , u N ( t ) i H − × H = (cid:16) N X i =1 C D α c i ( t ) e i , N X i =1 c i ( t ) e i (cid:17) L = N X i =1 c i ( t ) C D α c i ( t ) ≥ N X i =1 C D α ( c i ( t )) = 12 C D α N X i =1 ( c i ( t )) = 12 C D α k u N ( t ) k L (Ω) . On the other hand, by [E98, Theorem 3, p. 300], there exist β > ν ≥ β k u N ( t ) k H ≤ a ( u N ( t ) , u N ( t ); t ) + ν k u N ( t ) k L for almost every t ∈ (0 , T ) . Moreover, from the assumption of f and the Cauchy inequality h f ( t ) , u N ( t ) i H − × H ≤ k f ( t ) k H − (Ω) k u N ( t ) k H (Ω) ≤ β k f ( t ) k H − (Ω) + β k u N ( t ) k H (Ω) . Thus, for almost every t ∈ (0 , T ), C D α k u N ( t ) k L (Ω) ≤ ν k u N ( t ) k L (Ω) + 12 β k f ( t ) k H − (Ω) . Put v ( t ) := k u N ( t ) k L , h ( t ) := β k f ( t ) k H − and use the comparison principlefor solutions to fractional differential equation and the variation of constantsformula for solutions to the equations (see [CT17, Lemma 3.1]), we obtain theestimate v ( t ) ≤ Z t ( t − τ ) α − E α,α (2 ν ( t − τ ) α ) h ( τ ) dτ for a.a. t ∈ [0 , T ] . k u N ( t ) k L (Ω) ≤ T α E α,α +1 (2 νT α ) k f k L ∞ (0 ,T ; H − (Ω)) β for a.a. t ∈ [0 , T ] . (11)Next, by the similar arguments as above, we see that C D α k u N ( t ) k L (Ω) + β k u N ( t ) k H (Ω) ≤ β k f ( t ) k H − (Ω) + 2 ν k u N ( t ) k L (Ω) . Furthermore, C D α k u N ( t ) k L (Ω) = RL D α k u N ( t ) k L (Ω) = ddt I − α ( k u N ( t ) k L (Ω) ) ,I − α ( k u N ( t ) k L (Ω) ) | T + β k u N k L (0 ,T ; H (Ω)) ≤ T k f k L ∞ (0 ,T ; H − (Ω)) β + 2 ν Z T k u N ( t ) k L (Ω) dt, Thus, β k u N k L ([0 ,T ]; H (Ω)) ≤ T k f k L ∞ ([0 ,T ]; H − (Ω)) β + 2 ν Z T k u N ( t ) k L (Ω) dt. This combines with (11) implies that there is a constant C > k u N k L ([0 ,T ]; H (Ω)) ≤ C k f k L ∞ ([0 ,T ]; H − (Ω)) . (12)Finally, fix any v ∈ H (Ω) with k v k H ≤
1, and write v = v + v , where v ∈ span { e , . . . , e N } and ( e i , v ) = 0, 1 ≤ i ≤ N . Using the estimate concerningthe bilinear operator a ( · , · ) and the facts that( RL D α u N ( t ) , v ) L = h RL D α u N ( t ) , v i H − × H , ( RL D α u N ( t ) , v ) L + a ( u N ( t ) , v ; t ) = h f ( t ) , v i H − × H and k v k H (Ω) ≤ k v k H (Ω) ≤
1, we obtain |h RL D α u N ( t ) , v i H − × H | ≤ | a ( u N ( t ) , v ; t ) | + |h f ( t ) , v i H − × H |≤ C k u N ( t ) k H (Ω) k v k H (Ω) + k f ( t ) k H − (Ω) k v k H (Ω) . Hence, k RL D α u N ( t ) k H − (Ω) ≤ C k u N ( t ) k H (Ω) + k f ( t ) k H − (Ω) which together with (13) and the estimate (12) implies k RL D α u N k L ([0 ,T ]; H − (Ω)) ≤ C k f k L ∞ ([0 ,T ]; H − (Ω)) . (13)From (11), (12) and (13), the proof is complete.9 .2 Existence and uniqueness of weak solutions We now in a position to state the main result of the paper.
Theorem 3.4.
Consider the problem (1)–(2). Suppose that assumptions (a1)–(a4) hold. Then, this problem has a unique weak solution.
Proof.
First, we prove the system (1)–(2) has at least one weak solution. From(12) and (13), by the Banach–Aloaglu theorem, there exist a sequence { n k } ∞ k =1 such that u n k ⇀ u in L ([0 , T ]; H (Ω)) , (14) RL D α u n k ⇀ v in L ([0 , T ]; H − (Ω)) . (15)Let ϕ ∈ C ∞ c ((0 , T ); R ) and ψ ∈ H (Ω) be arbitrary. Then, Z T h v ( t ) , φ ( t ) ψ i H − × H dt = lim k →∞ Z T h RL D α u n k ( t ) , φ ( t ) ψ i H − × H dt = lim k →∞ Z T φ ( t ) h RL D α u n k ( t ) , ψ i H − × H dt = lim k →∞ Z T RL D αT − φ ( t ) h u n k ( t ) , ψ i H − × H dt = Z T RL D αT − φ ( t ) h u ( t ) , ψ i H − × H dt = Z T h RL D α u ( t ) , ϕ ( t ) ψ i H − × H dt = Z T ϕ ( t ) h RL D α u ( t ) , ψ i H − × H dt, which implies v ( t ) = RL D α u ( t ) . (16)Fix N, M ∈ N and N > M . For any ϕ ∈ C ∞ ([0 , T ]; R ) and w ∈ E M , we seethat Z T h RL D α u N ( t ) , ϕ ( t ) w i H − × H dt = Z T ϕ ( t ) h RL D α u N ( t ) , w i H − × H → Z T ϕ ( t ) h RL D α u ( t ) , w i H − × H , Z T a ( u N ( t ) , ϕ ( t ) w ; t ) dt = Z T ϕ ( t ) a ( u N ( t ) , w ; t ) dt → Z T ϕ ( t ) a ( u ( t ) , w ; t ) dt, Z T h f ( t ) , ϕ ( t ) w i H − × H dt = Z T ϕ ( t ) h f ( t ) , w i H − × H dt. Thus, h RL D α u ( t ) , w i H − × H + a ( u ( t ) , w ; t ) = h f ( t ) , w i H − × H (17)for all w ∈ E M . Moreover, S M ∈ N E M is dense in H (Ω) which implies that (17)is also true for all w ∈ H (Ω) and thus the problem (1)–(2) has a weak solution.Next, we give a proof for the uniqueness of the weak solution to (1)–(2). Supposethat u , u are two weak solution to this system. Define u := u − u . Then, u satisfies h ( k ∗ u ) ′ ( t ) , v i H − × H + a ( u ( t ) , v ; t ) = 0 , v ∈ H (Ω) , a.a. t ∈ (0 , T ) . Let v = u ( t ), then for almost every t ∈ (0 , T ) h RL D α u ( t ) , u ( t ) i H − × H + a ( u ( t ) , u ( t ); t ) = h ( k ∗ u ) ′ ( t ) , u ( t ) i H − × H (18)+ a ( u ( t ) , u ( t ); t ) = 0 , where k ( t ) = − α ) t α for t >
0. Motivated by Rico Zacher [VZ08, pp. 291–292], we will approximate the operator ddt ( k ∗ u ) by the sequences { ddt ( k n ∗ u ) } n ,where k n ( t ) := ns ( t ) = nE α ( − nt α ) (note that by using Laplace transform, wesee that s ( · ) is the unique solution to the equation s ( t ) + n ( l ∗ s )( t ) = 1, t > l ( t ) = t α − Γ( α ) , t >
0, see [KTT20, p. 3]). We rewrite the equation (18) asbelow h ( k n ∗ u ) ′ ( t ) , u ( t ) i H − × H + a ( u ( t ) , u ( t ); t ) = h n ( t ) , a.a. t ∈ (0 , T ) (19)with h n ( t ) := h ( k n ∗ u ) ′ ( t ) − ( k ∗ u ) ′ ( t ) , u ( t ) i H − × H , a.a. t ∈ (0 , T ) . By virtue of [Z09, Lemma 2.1],12 ddt ( k n ∗ k u ( · ) k L (Ω) )( t ) ≤ ( ddt ( k n ∗ u )( t ) , u ( t )) L (Ω) , a.a. t ∈ (0 , T ) . On the other hand, there exists ν ≥ a ( u ( t ) , u ( t ); t ) ≥ − ν k u ( t ) k L (Ω) . Hence, from (19), we have ddt ( k n ∗ k u ( · ) k L (Ω) )( t ) ≤ ν k u ( t ) k L (Ω) + 2 h n ( t ) , a.a. t ∈ (0 , T ) . This together with the positivity of l implies that l ∗ ddt ( k n ∗ k u ( · ) k L (Ω) ) ≤ ν ( l ∗ k u ( · ) k L (Ω) )( t ) + 2 l ∗ h n ( t ) , a.a. t ∈ (0 , T ) .
11e will show that lim n →∞ h n = 0 in L ([0 , T ]; R ) . (20)From the facts that u ∈ L ([0 , T ]; H − (Ω)), RL D α u = ddt ( k ∗ u ) ∈ L ([0 , T ]; H − (Ω)),[VZ08, Proposition 2.1, p. 293] and [VZ08, Example 2.1, p. 294], we have k ∗ k u ( · ) k H − (Ω) ∈ W ([0 , T ]; R ) . This implies that u ∈ D ( B ) , k u ( · ) k H − (Ω) ∈ D ( B ) , where B ( u ) = ddt k ∗ u, D ( B ) = { u ∈ L (0 , T ) : k ∗ u ∈ W ([0 , T ]; R ) } , and B ( u ) = ddt k ∗ u, D ( B ) = { u ∈ L ([0 , T ]; H − (Ω)) : k ∗ u ∈ W ([0 , T ]; H − (Ω)) } . Hence, by using [VZ08, Estimate (18), p. 292], we obtainlim n →∞ Z T k ddt [( k − k n ) ∗ u ]( t ) k H − (Ω) dt = lim n →∞ k ddt [( k − k n ) ∗ u ]( · ) k L ([0 ,T ]; H − (Ω)) = 0 . (21)By (21) and the Holder inequality, the following estimates holdlim n →∞ Z T | h n ( t ) | dt = lim n →∞ Z T |h ( k n ∗ u ) ′ ( t ) − ( k ∗ u ) ′ ( t ) , u ( t ) i H − × H | dt ≤ lim n →∞ Z T k ddt [( k − k n ) ∗ u ]( t ) k H − (Ω) k u ( t ) k H (Ω) ≤ lim n →∞ k ddt [( k − k n ) ∗ u ]( · ) k L ([0 ,T ]; H − (Ω)) k u k L ([0 ,T ]; H (Ω)) = 0 , which shows that lim n →∞ h n = 0 in L ([0 , T ]). Notice that( l ∗ k )( t ) = 1Γ( α )Γ(1 − α ) Z t s α − ( t − s ) − α ds = 1Γ( α )Γ(1 − α ) Z u − α (1 − u ) α − du = 1Γ( α )Γ(1 − α ) B ( α, − α )= 1 , ∀ t > ,l ∗ ddt ( k n ∗k u ( · ) k L (Ω) ) = ddt ( k n ∗ l ∗k u ( · ) k L (Ω) ) → ddt ( k ∗ l ∗k u ( · ) k L (Ω) ) = k u ( · ) k L (Ω) L ([0 , T ]; R ) as n → ∞ (see [VZ08, Estimate (19), p. 292]) and l ∗ h n → L ([0 , T ]; R ) as n → ∞ , we obtain k u ( t ) k L (Ω) ≤ ν ( l ∗ k u ( · ) k L (Ω) )( t ) , a.a. t ∈ (0 , T ) . (22)From (22), using a Gronwall type inequality as in [H81, Lemma 7.1.1, p. 188],then k u ( t ) k L (Ω) = 0 a.e. in (0 , T ), that is, u = u . The proof is complete. Remark . The key point in the proof of the uniqueness of weak solution tothe problem (1)–(2) is to show that h n → L ([0 , T ]) as n → ∞ . Theapproach proposed in [Z09] can not apply directly to this situation because theoperator B = ddt ( k ∗ u ) with domain D ( B ) = { u ∈ L ([0 , T ]; H (Ω)) : ddt k ∗ u ∈ L ([0 , T ]; H − (Ω)) } is not m-accretive. Remark . In [B20E, Theorem 3.1], the author has proved the existence anduniqueness of a weak solution to a class of fractional diffusion equations withCaputo derivative. The right hand side of these equations has the same formas in the equation (1) in our paper. However, to obtain this result, they needadditional following assumptions • k ∂ t f ( t ) k H − (Ω) ≤ Ct − α ; • b j = 0 (1 ≤ j ≤ d ), ∂ t a i,j ∈ L ∞ (Ω T ) (1 ≤ i, j ≤ d ), and ∂ t c ∈ L ∞ (Ω T ),see [B20E, lemmas 3.3, 3.4]. In this paper, by another approach, we have studiedthe existence and uniqueness of a weak solution to the fractional-order equation(1) (with Riemann–Liouville fractional derivative) without the assumptions asmentioned above. Remark . An important step in proving the existence of the weak solution isto show that the function v in (15) is the fractional derivative RL D α u (in theweak sense) of the solution u . In this paper, we do that by using Definition 2.4.This definition is inspired by the integration by the parts formula for Riemann–Liouville fractional derivatives (Theorem 2.2). Unfortunately, this property isgenerally not true for Caputo fractional derivatives (except for the case wherethese two fractional derivatives coincide). Therefore, the approach as in thepresent paper cannot be applied to study the existence of weak solutions topartial differential equations with Caputo fractional derivatives in time. Acknowledgement
This research is supported by The International Center for Research and Post-graduate Training in Mathematics–Institute of Mathematics–Vietnam Academyof Science and Technology under the Grant ICRTM01-2020.09. A part of thispaper was completed while the author was a postdoc at the Vietnam Institute13or Advanced Study in Mathematics (VIASM). He would like to thank VIASMfor support and hospitality. The author also would like to thank Nguyen AnhTu, Ha Duc Thai, and the anonymous referee for the valuable time, construc-tive suggestions, and interesting comments that have helped him improve thequality and presentation of the paper.
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