Evolution Equations governed by Lipschitz Continuous Non-autonomous Forms
aa r X i v : . [ m a t h . A P ] N ov Evolution Equations governed by LipschitzContinuous Non-autonomous Forms ∗ Ahmed Sani and Hafida LaasriApril 10, 2018
Abstract
We prove L -maximal regularity of linear non-autonomous evolution-ary Cauchy problem˙ u ( t ) + A ( t ) u ( t ) = f ( t ) for a.e. t ∈ [0 , T ] , u (0) = u , where the operator A ( t ) arises from a time depending sesquilinear form a ( t, ., . ) on a Hilbert space H with constant domain V. We prove themaximal regularity in H when these forms are time Lipschitz continuous.We proceed by approximating the problem using the frozen coefficientmethod developed in [9], [10] and [13]. As a consequence, we obtain aninvariance criterion for convex and closed sets of H. Key words:
Sesquilinear forms, non-autonomous evolution equations, maximalregularity, convex sets.
MSC:
In this paper we study non-autonomous evolutionary linear Cauchy-problems˙ u ( t ) + A ( t ) u ( t ) = f ( t ) , u (0) = u , (1.1)where the operators A ( t ) , t ∈ [0 , T ] , arise from sesquilinear forms on Hilbertspaces. More precisely, throughout this work H and V are two separable Hilbertspaces. The scalar products and the corresponding norms on H and V will bedenoted by ( . | . ), ( . | . ) V , k . k and k . k V , respectively. We assume that V ֒ → d H ;i.e., V is densely embedded into H and k u k ≤ c H k u k V ( u ∈ V ) (1.2)for some constant c H > . Let V ′ denote the antidual of V. The duality between V ′ and V is denoted by ∗ Work partly supported by DFG (JA 735/8-1) ., . i . As usual, we identify H with H ′ . It follows that
V ֒ → H ∼ = H ′ ֒ → V ′ andso V is identified with a subspace of V ′ . These embeddings are continuous and k f k V ′ ≤ c H k f k ( f ∈ V ′ ) (1.3)with the same constant c H as in (1.2) (see e.g., [7]).For a non-autonomous form a : [0 , T ] × V × V → C such that a ( t, ., . ) is sesquilinear for all t ∈ [0 , T ], a ( ., u, v ) is measurable for all u, v ∈ V, | a ( t, u, v ) | ≤ M k u k V k v k V ( t ∈ [0 , T ] , u, v ∈ V )and Re a ( t, u, u ) + ω k u k ≥ α k u k V ( t ∈ [0 , T ] , u ∈ V )for some α > , M ≥ ω ∈ R , for each t ∈ [0 , T ] we can associate a uniqueoperator A ( t ) ∈ Ł( V, V ′ ) such that a ( t, u, v ) = hA ( t ) u, v i for all u, v ∈ V. It isa known fact that −A ( t ) with domain V generates a holomorphic semigroup( T t ( s )) s ≥ on V ′ . Observe that kA ( t ) u k V ′ M k u k V for all u ∈ V and all t ∈ [0 , T ]. It is worth to mention that the mapping t
7→ A ( t ) is stronglymeasurable by the Dunford-Pettis Theorem [2] since the spaces are assumed tobe separable and t
7→ A ( t ) is weakly measurable. Thus t
7→ A ( t ) u is Bochnerintegrable on [0 , T ] with values in V ′ for all u ∈ V. The following well known maximal regularity result is due to J. L. Lions.
Theorem 1.1.
Given f ∈ L (0 , T ; V ′ ) and u ∈ H, there is a unique solution u ∈ MR ( V, V ′ ) := L (0 , T ; V ) ∩ H (0 , T ; V ′ ) of ˙ u ( t ) + A ( t ) u ( t ) = f ( t ) , u (0) = u . (1.4)Note that M R ( V, V ′ ) ֒ → d C ([0 , T ]; H ) (see [19, p.106]), so the condition u (0) = u in (1.4) makes sense and the solution is continuous on [0 , T ] withvalues in H. The proof of Theorem 1.1 can be given by an application of Lions’ Representa-tion Theorem [14] (see also [19, p. 112] and [22, Chapter 3]) or by Galerkin’smethod [8, XVIII Chapter 3, p. 620]. We refer also to an alternative proofgiven by Tanabe [21, Section 5.5].In Section 3, we give an other proof by using the approach of frozen coefficientdeveloped in [9], [10] and [13], from which we derive the criterion for invarianceof convex closed sets established by [3] and also the recent result given by [4]for Lipschitz continuous forms.Let Λ := (0 = λ < λ < ... < λ n +1 = T ) be a subdivision of [0 , T ] . Weapproximate (1.1) by (1.5), obtained when the generators A ( t ) are frozen onthe interval [ λ k , λ k +1 [ . More precisely, let A Λ : [0 , T ] → L ( V, V ′ ) be given by A Λ ( t ) := (cid:26) A k for λ k ≤ t < λ k +1 , A n for t = T, with A k x := 1 λ k +1 − λ k Z λ k +1 λ k A ( r ) u d r ( u ∈ V, k = 0 , , ..., n ) . t
7→ A ( t ) is, as mentioned above, strongly Bochner-integrable.We show (see Theorem 3.2) that for all u ∈ H and f ∈ L (0 , T ; V ′ ) the non-autonomous problem˙ u Λ ( t ) + A Λ ( t ) u Λ ( t ) = f ( t ) , u Λ (0) = u (1.5)has a unique solution u Λ ∈ M R ( V, V ′ ) which converges in M R ( V, V ′ ) as | Λ | → u := lim | Λ |→ u Λ solves uniquely (1 . . Let C be a closed convex subset of the Hilbert space H and let P : H → C bethe orthogonal projection onto C . As a consequence of Theorem 3.2 we obtain:If u ∈ C , P ( V ) ⊂ V and Re a ( t, P v, v − P v ) ≥ t ∈ [0 , T ] and for all v ∈ V, then u ( t ) ∈ C for all t ∈ [0 , T ] , where u is the solution of (1.4) with f = 0 . In the autonomous case condition(1.6) is also necessary for the invariance of C , see [17]. More recently, for f = 0the invariance of C under the solution of (1.4) was proved by Arendt, Dier andOuhabaz [3] provided thatRe a ( t, P v, v − P v ) ≥ h f ( t ) , v − P v i (1.7)for almost every t ∈ [0 , T ] and for all v ∈ V. Theorem 1.1 establishes L -maximal regularity of the Cauchy problem (1.4)in V ′ assuming only that t a ( t, u, v ) is measurable for all u, v ∈ V . However,in applications to boundary valued problems, only the part A ( t ) of A ( t ) in H does realize the boundary conditions in question. Thus one is interested in L -maximal regularity in H : Problem 1.2. If f ∈ L (0 , T ; H ) and u ∈ V , does the solution of (1.5) belongto MR ( V, H ) := L (0 , T ; V ) ∩ H (0 , T ; H )?This Problem 1.2 is asked (for u = 0) by Lions [14, p. 68] and is, to ourknowledge, still open. Note that if a (or equivalently A ) is a step function theanswer to Problem 1.2 is affirmative. In fact, for u ∈ V and f ∈ L (0 , T ; H ) thesolution u Λ of (1.5) belongs to MR ( V, H ) ∩ C ([0 , T ]; V ) (see Section 3). Thus,Problem 1.2 can be reformulated as follows: Problem 1.3. If f ∈ L (0 , T ; H ) and u ∈ V , does the solution of (1.5) convergein MR ( V, H ) as | Λ | → t ) on a ( t, ., . ) . For symmetric forms, Lionsproved L -maximal regularity in H for u = 0 ( respectively for u ∈ D ( A (0)))provided a ( ., u, v ) ∈ C [0 , T ] (respectively a ( ., u, v ) ∈ C [0 , T ] ) for all u, v ∈ V, [14, p. 68 and p. 94)]. Moreover, a combination of [14, Theorem 1.1, p. 129] and[14, Theorem 5.1, p. 138] shows that if a ( ., u, v ) ∈ C [0 , T ] for all u, v ∈ V , then(1.5) has L -maximal regularity in H. Bardos [6] gave also a positive answerto Problem (1.2) under the assumptions that the domains of both A ( t ) / and A ( t ) ∗ / coincide with V and that A ( . ) / is continuously differentiable with3alues in Ł( V, V ′ ). We mention also a result of Ouhabaz and Spina [15] andOuhabaz and Haak [11]. They proved L -maximal regularity for (possibly non-symmetric) forms such that a ( ., u, v ) ∈ C α [0 , T ] for all u, v ∈ V and some α > .The result in [15] concerns the case u = 0 and the one in [11] concerns the case u in the real-interpolation space ( H, D ( A (0))) / , . In Section 4 , we are concerned with a recent result obtained in [4]. Assumethat the sesquilinear form a can be written as a ( t, u, v ) = a ( t, u, v ) + a ( t, u, v )where a is symmetric, bounded (i.e a ( t, u, v ) M k u kk v k , M ≥
0) andcoercive as above and piecewise Lipschitz-continuous on [0 , T ] with Lipschitzconstant L , whereas a : [0 , T ] × V × H → C satisfies | a ( t, u, v ) | ≤ M k u k V k v k H and a ( ., u, v ) is measurable for all u ∈ V , v ∈ H . Furthermore, let B : [0 , T ] → Ł( H ) be strongly measurable such that k B ( t ) k Ł( H ) ≤ β for all t ∈ [0 , T ] and0 < β ≤ ( B ( t ) g | g ) H for g ∈ H , k g k H = 1, t ∈ [0 , T ] . Then, the following resultis proved in [4, Corollary 4.3] :
Theorem 1.4.
Let u ∈ V , f ∈ L (0 , T ; H ) . Then there exists a unique M R ( V, H ) satisfying ˙ u ( t ) + B ( t ) A ( t ) u ( t ) = f ( t ) a.e. u (0) = u . Moreover k u k MR ( V,H ) ≤ C h k u k V + k f k L (0 ,T ; H ) i , (1.8) where the constant C depends only on β , β , M , M , α, T, L and γ . In the special case where B = I and a = a (or equivalently a = 0) we proofthat Problem 1.3 has a positive answer.We emphasize that our result on approximation may be applied to concretelinear evolution equations. For example, to evolution equation governed byelliptic operator in nondivergence form on a domain Ω with time dependingcoefficients ˙ u ( t ) − X i,j ∂ i a ij ( t, . ) ∂ j u ( t ) = f ( t ) u (0) = u ∈ H (Ω) . with an appropriate Lipschitz continuity property on the coefficients with re-spect to t and boundary conditions such as Neumann or non-autonomous Robinboundary conditions. Acknowledgment
The authors are most grateful to Wolfgang Arendt and Omar El-Mennaoui forfruitful discussions on maximal regularity and invariance criterion for the non-autonomous linear Cauchy problem.
Consider a continuous and H - elliptic sesquilinear form a : V × V → C . Thismeans, respectively | a ( u, v ) | ≤ M k u k V k v k V for some M ≥ u, v ∈ V, (2.1)4e a ( u ) + ω k u k ≥ α k u k V for some α > , ω ∈ R and all u ∈ V. (2.2)Here and in the following we shortly write a ( u ) for a ( u, u ) . The operator
A ∈ Ł( V, V ′ ) associated with a on V ′ is defined by hA u, v i = a ( u, v ) ( u, v ∈ V ) . Seen as an unbounded operator on V ′ with domain D ( A ) = V, the operator −A generates a holomorphic C − semigroup T on V ′ . The semigroup is bounded ona sector if ω = 0, in which case A is an isomorphism. Denote by A the part of A on H ; i.e., D ( A ) := { u ∈ V : A u ∈ H } Au = A u. It is a known fact that − A generates a holomorphic C -semigroup T on H and T = T | H is the restriction of the semigroup generated by −A to H. Then A isthe operator induced by a on H. We refer to [12],[16] and [21, Chap. 2].
Remark 2.1.
The sesquilinear form a satisfies condition (2.2) if and only if theform a ω given by a ω ( u, v ) := a ( u, v ) + ω ( u | v )is coercive. Moreover, if T ω (respectively A ω ) denotes the semigroup (respec-tively the operator) associated with a ω , then T ω ( t ) = e − ωt T ( t ) and A ω = ω + A for all t ≥ . Then it is possible to choose, without loss of generality, a coercive(i.e., ω = 0 . )The following maximal regularity results are well known: If u ∈ H, f ∈ L ( a, b ; V ′ ) then the function u ( t ) = T ( t ) u + Z ta T ( t − r ) f ( r )d r belongs to L ( a, b ; V ) ∩ H ( a, b ; V ′ ) and is the unique solution of the autonomousinitial value problem˙ u ( t ) + A u ( t ) = f ( t ) , t.a.e on [ a, b ] ⊂ [0 , T ] , u ( a ) = u . (2.3)Recall that the maximal regularity space MR ( a, b ; V, V ′ ) := L ( a, b ; V ) ∩ H ( a, b ; V ′ ) (2.4)is continuously embedded in C ([ a, b ] , H ) and if u ∈ MR ( a, b ; V, V ′ ) then thefunction k u ( . ) k is absolutely continuous on [ a, b ] and ddt k u ( . ) k = 2 Re h ˙ u, u i (2.5)see e.g., [19, Chapter III, Proposition 1.2] or [21, Lemma 5.5.1]. For [ a, b ] = [0 , T ]we shortly denote M R ( a, b ; V, V ′ ) by M R ( V, V ′ ) in (2.4).Furthermore, if ( f, u ) ∈ L ( a, b ; H ) × V then the solution u of (2.3) belongs tothe maximal regularity space M R ( a, b ; D ( A ) , H ) := L ( a, b ; D ( A )) ∩ H ( a, b ; H ) (2.6)5hich is equipped with the norm k . k MR given for all u ∈ M R ( a, b ; D ( A ) , H ) by k u k MR := Z ba k u ( t ) k dt + Z ba k ˙ u ( t ) k dt + Z ba k Au ( t ) k dt. (2.7)The maximal regularity space M R ( a, b ; D ( A ) , H ) is continuously embedded into C ([ a, b ]; V ) , [8, Exemple 1, page 577]. If the form a is symmetric, then foreach u ∈ M R ( a, b ; D ( A ) , H ) , the function a ( u ( . )) belongs to W , ( a, b ) and thefollowing product formula holds ddt a ( u ( t )) = 2( Au ( t ) | ˙ u ( t )) for a.e. t ∈ [ a, b ] , (2.8)for the proof we refer to [5, Lemma 3.1].The following lemma gives a locally uniform estimate for the solution of theautonomous problem. This estimate will play an important role in the study ofthe convergence in Theorem 5.1. Lemma 2.2. [5, Theorem 3.1] Let a be a continuous and H -elliptic sesquilinearform. Assume the form a is symmetric. Let f ∈ L ( a, b ; H ) and u ∈ V. Let u ∈ M R ( a, b ; D ( A ) , H ) be such that ˙ u ( t ) + Au ( t ) = f ( t ) , t.a.e on [ a, b ] ⊂ [0 , T ] , u ( a ) = u . (2.9) Then there exists a constant c > such that sup s ∈ [ a,b ] k u ( s ) k V ≤ c h k u ( a ) k V + k f k L ( a,b ; H ) i (2.10) where c = c ( M, α, ω, T ) > is independent of f, u and [ a, b ] ⊂ [0 , T ] . For the sake of completeness, we include here a simpler proof in the nonrestrictive case ω = 0 . Proof.
We use the same technique as in the proof of [5, Theorem 3.1]. Forsimplicity and according to Remark 2.1 we may assume without loss of generalitythat ω = 0 in (2.2). For almost every t ∈ [ a, b ]( ˙ u ( t ) | ˙ u ( t )) + ( Au ( t ) | ˙ u ( t )) = ( f ( t ) | ˙ u ( t )) . The rule formula (2.8) and the Cauchy-Schwartz inequality together with theYoung inequality applied to the term on the right-hand side of the above equalityimply that, for almost every t ∈ [ a, b ]12 k ˙ u ( t ) k + 12 ddt a ( u ( t )) ≤ k f ( t ) k . Integrating this inequality on [ a, t ] , it follows that Z ta k ˙ u ( s ) k ds + a ( u ( t )) ≤ a ( u ( a )) + Z ta k f ( s ) k ds. Z ta k ˙ u ( s ) k ds + α k u ( t ) k V ≤ M k u ( a ) k V + k f k L ( a,b ; H ) (2.11)for almost every t ∈ [0 , T ] . It follows thatsup t ∈ [ a,b ] k u ( t ) k V ≤ α (cid:16) M k u ( a ) k V + k f k L ( a,b ; H ) (cid:17) (2.12)which gives the desired estimate. Remark 2.3.
Lemma 2.2 says that the constant c in (2.12) depends onlyon M, α, ω and T, but it does not depend on the subinterval [ a, b ] or on otherproperties of a . V ′ Let
H, V be the Hilbert spaces explained in the previous sections. Let
T > a : [0 , T ] × V × V → C be a non-autonomous form , i.e., a ( t, ., . ) is sesquilinear for all t ∈ [0 , T ], a ( ., u, v )is measurable for all u, v ∈ V, | a ( t, u, v ) | ≤ M k u k V k v k V ( t ∈ [0 , T ] , u, v ∈ V ) (3.1)and Re a ( t, u, u ) + ω k u k ≥ α k u k V ( t ∈ [0 , T ] , u ∈ V ) (3.2)for some α > , M ≥ ω ∈ R . We recall that, for all t ∈ [0 , T ] we denote by A ( t ) ∈ Ł( V, V ′ ) the operatorassociated with the form a ( t, ., . ) in V ′ and by T t the analytic C -semigroupgenerated by −A ( t ) on V ′ . Consider the non-autonomous Cauchy problem˙ u ( t ) + A ( t ) u ( t ) = f ( t ) , for a.e t ∈ [0 , T ] , u (0) = u . (3.3)In this section, we are interested in the well-posedness of (3.3) in V ′ with L -maximal regularity. The case where a is independent on t is described in theprevious section.The case where a is a step function is also easy to describe. In fact, let Λ =(0 = λ < λ < ... < λ n +1 = T ) be a subdivision of [0 , T ] . Let a k : V × V → C for k = 0 , , ..., n a finite family of continuous and H -elliptic forms. The associated operators aredenoted by A k ∈ Ł( V, V ′ ) . Let T k denote the C − semigroup generated by −A k on V ′ for all k = 0 , ...n. The function a Λ : [0 , T ] × V × V → C (3.4)7efined by a Λ ( t ; u, v ) := a k ( u, v ) for λ k ≤ t < λ k +1 and a Λ ( T ; u, v ) := a n ( u, v ) , is strongly measurable on [0 , T ]. Let A Λ : [0 , T ] → Ł( V, V ′ )be given by A Λ ( t ) := A k for λ k ≤ t < λ k +1 , k = 0 , , ..., n, and A Λ ( T ) := A n . For each subinterval [ a, b ] ⊂ [0 , T ] such that λ m − ≤ a < λ m < ... < λ l − ≤ b < λ l we define the operators P Λ ( a, b ) ∈ L ( V ′ ) by P Λ ( a, b ) := T l − ( b − λ l − ) T l − ( λ l − − λ l − ) ... T m ( λ m +1 − λ m ) T m − ( λ m − a ) , (3.5)and for λ l − ≤ a ≤ b < λ l by P Λ ( a, b ) := T l − ( b − a ) . (3.6)It is easy to see, that for all u ∈ H and f ∈ L ( a, b, V ′ ) the function u Λ ( t ) = P Λ ( a, t ) u + Z ta P Λ ( r, t ) f ( r )d r (3.7)belongs to MR ( a, b ; V, V ′ ) and is the unique solution of the initial value problem˙ u Λ ( t ) + A Λ ( t ) u Λ ( t ) = f ( t ) , for a.e t ∈ [ a, b ] ⊂ [0 , T ] , u Λ ( a ) = u . The product given by (3.5)-(3.6) and also the existence of a limit of this productas | Λ | converges to 0 uniformly on [ a, b ] ⊂ [0 , T ] , was studied in [9],[13] and [10].This leads to a theory of integral product, comparable to that of the classicalRiemann integral. The notion of product integral has been introduced by V.Volterra at the end of 19th century. We refer to A. Slavík [20] and the referencestherein for a discussion on the work of Volterra and for more details on productintegration theory.Consider now the general case where a : [0 , T ] × V × V → C is a non-autonomous form and let A ( t ) ∈ Ł( V, V ′ ) be the associated operator with a ( t, ., . ) on V ′ . We want to approximate a and A by step functions. Let Λ := (0 = λ < λ < ... < λ n +1 = T ) be a subdivision of [0 , T ] and a Λ : [0 , T ] × V × V → C and A Λ : [0 , T ] → L ( V, V ′ ) be as above where A k are associated with thesesquilinear forms a k ( u, v ) := 1 λ k +1 − λ k Z λ k +1 λ k a ( r ; u, v )d r for u, v ∈ V, k = 0 , , ..., n. (3.8)Note that a k satisfies (3.1) and (3.2), k = 0 , , ...n , we then have for all u ∈ V A k u := 1 λ k +1 − λ k Z λ k +1 λ k A ( r ) u d r. (3.9)Let u ∈ H and f ∈ L (0 , T ; V ′ ) and let u Λ ∈ MR ( V, V ′ ) denote the uniquesolution of˙ u Λ ( t ) + A Λ ( t ) u Λ ( t ) = f ( t ) , for a.e t ∈ [0 , T ] , u Λ (0) = u (3.10)8ecall that u Λ is given explicitly by (3.5)-(3.7).For simplicity and according to Remark 2.1, we may assume without loss ofgenerality that ω = 0 in (3.2). In fact, let u Λ ∈ M R ( V, V ′ ) and v Λ ( t ) := e − wt u Λ ( t ) . Then u Λ satisfies (3.10) if and only if v Λ satisfies˙ v Λ ( t ) + ( ω + A Λ ( t )) v Λ ( t ) = e − wt f ( t ) t − a . e . on [0 , T ] , v Λ (0) = u (3.11)In the sequel, ω = 0 will be our assumption. Lemma 3.1.
Let u ∈ H and f ∈ L (0 , T ; V ′ ) . Let u Λ ∈ MR ( V, V ′ ) be thesolution of (3.10). Then there exists a constant c > independent of f, u and Λ such that Z t k u Λ ( s ) k V ds ≤ c h Z t k f ( s ) k V ′ ds + k u k i , (3.12) for a.e t ∈ [0 , T ] . Proof.
Since u Λ ∈ M R ( V, V ′ ) , it follows from (2.5) ddt k u Λ ( t ) k = 2 Re h ˙ u Λ ( t ) , u Λ ( t ) i = 2 Re h f ( t ) − A Λ ( t ) u Λ ( t ) , u Λ ( t ) i = − a Λ ( t, u Λ ( t ) , u Λ ( t )) + 2 Re h f ( t ) , u Λ ( t ) i for almost every t ∈ [0 , T ] . Integrating this equality on (0 , t ) , by coercivity ofthe form a and the Cauchy-Schwartz inequality we obtain k u Λ ( t ) k + 2 α Z t k u Λ ( s ) k V ds ≤ Z t k f ( s ) k V ′ k u Λ ( s ) k V ds + k u k . Inequality (3.12) follows from this estimate and the standard inequality ab ≤
12 ( a ε + εb ) ( ε > , a, b ∈ R ) . Let | Λ | := max j =0 , ,...,n ( λ j +1 − λ j ) denote the mesh of the subdivision Λ of [0 , T ] . The main result of this section is the following
Theorem 3.2.
Let f ∈ L (0 , T ; V ′ ) and u ∈ H. Then the solution u Λ of(3.10) converges weakly in M R ( V, V ′ ) as | Λ | −→ and u := lim | Λ |→ u Λ is theunique solution of (1.4).Proof. To prove that lim u Λ exists as | Λ | −→ , it suffices, by the compactnessof bounded sets of L (0 , T, V ) , to show that it exists u ∈ M R ( V, V ′ ) such thatevery convergent subsequence of u Λ converges to u . We then begin with theuniqueness. Uniqueness:
Let u ∈ MR ( V, V ′ ) be a solution of (1.4) with f = 0and u (0) = 0 . Then ddt k u ( t ) k = 2Re h ˙ u ( t ) , u ( t ) i = − hA ( t ) u ( t ) , u ( t ) i = − a ( t, u ( t ) , u ( t )) . ddt k u ( t ) k ≤ − α k u ( t ) k V and since u (0) = 0 , it follows that u ( t ) = 0 for a.e. t ∈ [0 , T ] . Existence:
Let u ∈ H and f ∈ L (0 , T ; V ′ ) . Let u Λ ∈ MR ( V, V ′ ) be thesolution of (3.10). Since u Λ is bounded in L (0 , T ; V ) by Lemma 3.1, we canassume (after passing to a subsequence) that u Λ ⇀ u in L (0 , T ; V ) as | Λ | goes to 0. Let now g ∈ L (0 , T ; V ) . We have A ∗ Λ g −→ A ∗ g in L (0 , T ; V ′ ) [10,Lemma 2.3 and Lemma 3.1]. Since Z T hA Λ ( s ) u Λ ( s ) , g ( s ) i ds = Z T h u Λ ( s ) , A ∗ Λ ( s ) g ( s ) i ds, it follows that Z T hA Λ ( s ) u Λ ( s ) , g ( s ) i ds → Z T hA ( s ) u ( s ) , g ( s ) i ds or, in other words, A Λ u Λ ⇀ A u in L (0 , T ; V ′ ) and so ˙ u Λ converges weakly in L (0 , T ; V ′ ) by (3.10).Thus, letting | Λ | → u ( t ) + A ( t ) u ( t ) = f ( t ) t − a . e . on [0 , T ] , Since MR ( V, V ′ ) ֒ → C ([0 , T ]; H ) , we have also that u Λ ⇀ u in C ([0 , T ]; H ) andin particular u Λ (0) ⇀ u (0) in H, so that u satisfies (1.4). This completes theproof. We use the same notations as in the previous sections. We consider a non-autonomous form a : [0 , T ] × V × V → C . Let A ( t ) ∈ Ł( V, V ′ ) be the associateoperator. In this section we give a other proof of a known invariance criterionfor the non-autonomous homogeneous Cauchy-problem˙ u ( t ) + A ( t ) u ( t ) = 0 t -a.e. on [0 , T ] , u (0) = u . (4.1)Let C be a closed convex subset of the Hilbert space H and let P : H → C be the orthogonal projection onto C ; i.e. for x ∈ H, P x is the unique element x C in C such that Re( x − x C | y − x C ) ≤ y ∈ C . Recall, that the closed convex set C is invariant for the Cauchy problem (4 . u ∈ C the solution u of (4 . u ( t ) ∈ C for all t ∈ [0 , T ] . Recently, Arendt et al. [3] proved that the C is invariant for the inhomogenous Cauchy problem (1 .
4) provided that
P V ⊂ V and Re a ( t, P v, v − P v ) ≥ Re h f ( t ) , v − P v i for all v ∈ V and for a.e t ∈ [0 , T ] . As consequence of our approach, we obtain easily Theorem 2.2 in [3] for thehomogeneous Cauchy problem from Theorem 3.2.10 heorem 4.1.
Let a be a non-autonomous form on V. Let C be a closed convexsubset of H. Then the convex set C is invariant for the Cauchy problem (4 . provided that P V ⊂ V and Re a ( t, P v, v − P v ) ≥ for all v ∈ V and a.e. t ∈ [0 , T ] . Proof.
Let u ∈ C and let u Λ ∈ MR ( V, V ′ ) be the solution of (4 . . The function u Λ is given explicitly by (3.5)-(3.6). From Theorem 2.1 in [17] (or Theorem 2.2in [16]), it follows easily that u Λ ( t ) ∈ C if and only if P V ⊂ V andRe a k ( P v, v − P v ) ≥ v ∈ V and k = 0 , , ..., n. Recall that a k is given by (3.8). The inequality above holds if and only ifRe a ( t, P v, v − P v ) ≥ t ∈ [0 , T ] . Let now u be the solution of (4 . . ByTheorem 3.2 we have u Λ ⇀ u in MR ( V, V ′ ) ֒ → d C ([0 , τ ] , H ) . The claim followsfrom the fact that the weak closure of the convex set C is equal to its normclosure. Theorem 4.2.
Assume that the non-autonomous form a is symmetric and ac-cretive. The convex set C is invariant for the homogeneous Cauchy problem (4 . provided that P V ⊂ V and a ( t, P v, P v ) ≤ a ( t, v, v ) for a.e. t ∈ [0 , T ] . Proof.
Let u Λ ∈ MR ( V, V ′ ) be the solution of (4 . . By Theorem 2.2 in [16], wehave u Λ ( t ) ∈ C if and only if P V ⊂ V and a k ( P v, P v ) ≥ a k ( v, v ) for all v ∈ V and k = 0 , , ..., n. This inequality holds if and only if Re a ( t, P v, P v ) ≥ a ( t, v, v ) for a.e. t ∈ [0 , T ]and for all v ∈ V. The claim follows from the fact that t u Λ converge weakly in C ([0 , τ ] , H ) to the solution of (4 . . H Recall that
V, H denote two separable Hilbert spaces and a : [0 , T ] × V × V → C is a non-autonomous form introduced in the previous section. We adopt herethe notations of Sections 3. We consider the Hilbert space M R ( V, H ) := L (0 , T ; V ) ∩ H (0 , T ; H )with norm k u k MR ( V,H ) := k u k L (0 ,T ; V ) + k u k H (0 ,T ; H ) . Let Λ be a subdivision of [0 , T ] and let f ∈ L (0 , T ; H ) and u ∈ V. The solution u Λ of (3.10) belongs to MR ( V, H ) and u Λ ∈ C ([0 , T ] , V ) . In fact, let A k be givenby (3.9) and let A k be the part of A k in H. Then it is not difficult to see that u Λ | [ λk,λk +1[ ∈ M R ( λ k , λ k +1 ; D ( A k ) , H ) , k = 0 , , , ..., n. (5.1)Note, that on each interval [ λ k , λ k +1 [ the solution u Λ coincides with the solutionof the autonomous Cauchy problem˙ u k ( t ) + A k u k ( t ) = f ( t ) t − a . e . o n ( λ k , λ k +1 ) , u k ( λ k ) = u k − ( λ k ) ∈ V which belongs to M R ( λ k , λ k +1 ; D ( A k ) , H ) , see Section 2.11e assume in addition that a is symmetric ; i.e., a ( t, u, v ) = a ( t, v, u ) ( t ∈ [0 , T ] , u, v ∈ V ) , (5.2)and Lipschitz continuous i.e., there exists a positive constant L such that | a ( t, u, v ) − a ( s, u, v ) | ≤ L | t − s |k u k V k v k V ( t, s ∈ [0 , T ] , u, v ∈ V ) (5.3)For simplicity, we assume in the following that the subdivision Λ of [0 , T ] isuniform, i.e., λ i +1 − λ i = λ j +1 − λ j for all ( i, j ) ∈ { , , , ..., n } . Theorem 5.1 below, shows that the solution u Λ of (3.10) converges weaklyin M R ( V, H ) and so the limit u, which is the solution of (1.4), belongs to themaximal regularity space M R ( V, H ) . This gives an other proof of Theorem 5.1in [4] with a symmetric and B = Id.
Theorem 5.1.
Assume that a is symmetric and Lipschitz continuous. Let ( f, u ) ∈ L (0 , T ; H ) × V. Then u Λ , the solution of (3.10), converges weaklyin M R ( V, H ) as | Λ | −→ and u := lim | Λ |→ u Λ is the unique solution of (1.4).Moreover k u k MR ( V,H ) ≤ c h k u k V + k f k L (0 ,T ; H ) i , (5.4) where the constant c depends merely on α, c H , M and L .Proof. Let ( f, u ) ∈ L (0 , T ; H ) × V. Let u Λ ∈ M R ( V, H ) be the solution of(3.10). According to the proof of Theorem 3.2, it remains to prove that u Λ is bounded in M R ( V, H ) . We estimate first the derivative ˙ u Λ . Using (2.8) and(5.1) we obtain Z T k ˙ u Λ ( t ) k dt = Z T Re( −A Λ ( t ) u Λ ( t ) | ˙ u Λ ( t )) dt + Z T Re( f ( t ) | ˙ u Λ ( t )) dt = n − X k =0 Z λ k +1 λ k Re( −A Λ ( t ) u Λ ( t ) | ˙ u Λ ( t )) dt + Z T Re( f ( t ) | ˙ u Λ ( t )) dt = n − X k =0 Z λ k +1 λ k Re( −A k u Λ ( t ) | ˙ u Λ ( t )) dt + Z T Re( f ( t ) | ˙ u Λ ( t )) dt = − n − X k =0 Z λ k +1 λ k ddt a k ( u Λ ( t )) dt + Z T Re( f ( t ) | ˙ u Λ ( t )) dt For the first term on the right-hand side of the above equality − n − X k =0 Z λ k +1 λ k ddt a k ( u Λ ( t )) dt = − n − X k =0 (cid:16) a k ( u Λ ( λ k +1 )) − a k ( u Λ ( λ k )) (cid:17) = − n − X k =0 a k ( u Λ ( λ k +1 )) − n − X k = − a k +1 ( u Λ ( λ k +1 )) ! = − n − X k =0 (cid:16) a k ( u Λ ( λ k +1 )) − a k +1 ( u Λ ( λ k +1 )) (cid:17) − a n − ( u Λ ( λ n )) + a ( u Λ (0)) ≤ − n − X k =0 (cid:16) a k ( u Λ ( λ k +1 )) − a k +1 ( u Λ ( λ k +1 )) (cid:17) + M k u Λ (0) k V a we obtain | a k ( u Λ ( λ k +1 )) − a k +1 ( u Λ ( λ k +1 )) | ≤ L ( λ k +1 − λ k ) k u Λ ( λ k +1 )) k V (5.5)for every k = 0 , , ..., n − k = 0 , , , ..., n − t k ∈ [ λ k , λ k +1 [ be arbitrary. Then u Λ | [ tk,λk +1[ belongs to M R ( t k , λ k +1 ; D ( A k ) , H ) and k u Λ ( λ k +1 ) k V ≤ c h k u Λ ( t k ) k V + k f k L ( t k ,λ k +1 ; H ) i (5.6)where the constant c depends only on M, ω, α, c H and T (see Lemma 2.2).Inserting (5.6) into (5.5) we obtain then for every k = 0 , , ..., n − | a k ( u Λ ( λ k +1 )) − a k +1 ( u Λ ( λ k +1 )) |≤ c ( λ k +1 − λ k ) k u Λ ( t k ) k V + c ( λ k +1 − λ k ) k f k L (0 ,T ; H ) ≤ c Z λ k +1 λ k k u Λ ( s ) k V ds + c ( λ k +1 − λ k ) k f k L (0 ,T ; H ) , For the last inequality, t k is chosen such that( λ k +1 − λ k ) k u Λ ( t k ) k V = Z λ k +1 λ k k u Λ ( s ) k V ds using the mean value theorem and the fact that t k ∈ [ λ k , λ k +1 [ is arbitrary.Thus n − X k =0 | a k ( u Λ ( λ k +1 )) − a k +1 ( u Λ ( λ k +1 )) | ≤ c h k u Λ k L (0 ,T ; V ) + k f k L (0 ,T ; H ) i (5.7)for some c = c ( M, ω, α, c H , T, L ) (possibly different from the previous one). Itfollows Z T k ˙ u Λ ( t ) k dt ≤ c h k u Λ k L (0 ,T ; V ) + k f k L (0 ,T ; H ) i + Z T Re( f ( t ) | ˙ u Λ ( t )) dt + M k u k V Finally, from this inequality, the estimate (3.12) in Lemma 3.1, the Cauchy-Schwarz and the Young’s inequality applied the third term on right-hand side,it follows that there is a constant c = c ( M, ω, α, c H , T, L ) such that Z T k ˙ u Λ ( t ) k dt + Z T k u Λ ( t ) k V dt ≤ c (cid:16) k u k V + k f k L (0 ,T ; H ) (cid:17) This completes the proof.
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Hafida Laasri , Fachbereich C - Mathematik und Naturwissenschaften, Univer-sity of Wuppertal, Gaußstraße 20, 42097 Wuppertal, Germany, [email protected]
Ahmed Sani , Department of Mathematics, University Ibn Zohr, Faculty of Sci-ences, Agadir, Morocco, [email protected]@gmail.com