Compact almost automorphic solutions for semilinear parabolic evolution equations
aa r X i v : . [ m a t h . A P ] A p r Compact almost automorphic solutions forsemilinear parabolic evolution equations
Brahim Es-sebbar Khalil Ezzinbi Kamal Khalil , ∗ April 9, 2020 Cadi Ayyad University, Faculty of Sciences and Technology Gueliz, Marrakech, Morocco. Department of Mathematics, Faculty of Sciences Semlalia, Cadi Ayyad University, Marrakesh B.P.2390-40000, Morocco.
Abstract.
In this paper, using the subvariant functional method due to Favard [14], we provethe existence of a unique compact almost automorphic solution for a class of semilinear evolutionequations in Banach spaces provided the existence of at least one bounded solution on the right halfline. More specifically, we improve the assumptions in [9], we show that the almost automorphyof the coefficients in a weaker sense (Stepanov almost automorphy of order 1 ≤ p < ∞ ) is enoughto obtain solutions that are almost automorphic in a strong sense (Bochner almost automorphy).For that purpose we distinguish two cases, p = 1 and p >
1. The main difficulty in this work,is to prove the existence of at least one solution with relatively compact range while the forcingterm is not necessarily bounded. Moreover, we propose to study a large class of reaction-diffusionproblems with unbounded forcing terms.
Keywords.
Almost automorphic solutions; C -semigroup; Nonlinear evolution equations;Reaction-diffusion equations; Subvariant functional method; Stepanov almost automorphic func-tions. The concept of almost automorphy introduced by Bochner [7] is not restricted just to continuousfunctions. One can generalize that notion to measurable functions with some suitable conditionsof integrability, namely, Stepanov almost automorphic functions, see [5, 7, 13]. That is a Stepanovalmost automorphic function is neither continuous nor bounded necessarily.Now, consider the following semilinear evolution equation in a Banach space X : x ′ ( t ) = Ax ( t ) + f ( t, x ( t )) for t ∈ R , (1.1)where ( A, D ( A )) is generator of a C -semigroup ( T ( t )) t ≥ on X . The function f : R × X −→ X islocally integrable of order 1 ≤ p < ∞ in t and continuous in x .The study of existence of almost periodic and almost automorphic solutions to equation (1.1)in infinite dimensional Banach spaces was deeply investigated in the last decades, we refer to[4, 5, 9, 10, 18, 20, 26, 27, 28]. Recently, in [4], the authors studied equation (1.1), in the paraboliccontext, that is when ( A, D ( A )) generates an analytic semigroup ( T ( t )) t ≥ on a Banach space X which has an exponential dichotomy on R and f is Stepanov almost periodic of order 1 ≤ p < ∞ and Lipschitzian with respect to x . They proved the existence and uniqueness of almost periodicsolutions for equation (1.1). In [10], the authors proved the existence and uniqueness of almostautomorphic solutions for equation (1.1) in the case where ( A, D ( A )) generates an exponentiallystable C -semigroup ( T ( t )) t ≥ on a Banach space X and f is S p -almost automorphic in t (of order1 < p < ∞ ) and Lipschitzian with respect to x . Moreover, in [9], the authors proved the existenceof compact almost automorphic solutions for equation (1.1) provided that ( A, D ( A )) generates acompact C -semigroup and f is almost automorphic in the classical sense uniformly with respectto x . The results are obtained using the subvariant functional method introduced in [15]. ∗ Corresponding author: [email protected]
1n this paper, using the subvariant functional method, we give sufficient conditions insuring theexistence and uniqueness of compact almost automorphic solutions to equation (1.1). We assumethat the C -semigroup ( T ( t )) t ≥ is compact and f is just Stepanov almost automorphic of order1 ≤ p < ∞ . Actually, we begin by introducing a very useful characterization of uniformly Stepanovalmost automorphic functions (Lemma 3.1). That is a uniformly S p -almost automorphic functionis pointwise S p -almost automorphic and uniformly continuous on compact sets of X in a weaksense. Then we prove that a mild solution x of equation (1.1) given by: x ( t ) = T ( t − s ) x ( s ) + Z ts T ( t − σ ) f ( σ, x ( σ )) dσ for t ≥ s, t, s ∈ R with relatively compact range is uniformly continuous, see Theorem 3.1. Technically, we distinguishtwo cases, 1 < p < ∞ and p = 1. Therefore, using Ascoli-Arzela Theorem and an extractiondiagonal argument, we construct a mild solution x on R such that its range is relatively compact,provided that equation (1.1) has at least a mild solution x on the right half line with relativelycompact range satisfying the following inclusion: { x ( t ) : t ∈ R } ⊂ { x ( t ) : t ≥ t } , see Lemma 3.5. Note that, the assumption of existence of mild solutions on the right half line withrelatively compact range of equation (1.1) is strongly needed in our work. Consequently, sufficientconditions insuring that assumption are improved in Lemmas 3.6, 3.8 and Proposition 3.1. Here,the situation is quite difficult since f is not necessarily bounded. Indeed, we consider two cases,1 < p < ∞ and p = 1. For 1 < p < ∞ , in Lemma 3.6, we prove that equation (1.1) has a mildsolution on R + with relatively compact range provided it has at least a bounded mild solutionon R + . In general, this result does not hold for p = 1 without any additional assumptions on f , see Lemma 3.8. Furthermore, when f satisfies hypothesis (H5) below, the result also holdsin general, see Proposition 3.1. After proving that, we give our main results, we show that, ifequation (1.1) has at least a bounded mild solution x on the right half line, then it has a uniquecompact almost automorphic solution provided the existence of a unique mild solution of equation(1.1) which minimizes a given subvariant functional. Here, we differentiate two cases, 1 < p < ∞ ,see Theorem 3.2 and the case p = 1, see Theorem 3.3.Furthermore, to illustrate our theoretical results, we study a class of reaction diffusion problemsof the form: ∂∂t v ( t, ξ ) = n X k =1 ∂ ∂ξ i v ( t, ξ ) + g ( v ( t, ξ )) + h ( t, ξ ) for t ∈ R and ξ ∈ Ω ,v ( t, ξ ) = 0 for t ∈ R , ξ ∈ ∂ Ω , (1.2)in a bounded open domain Ω ⊂ R n , n ≥ ∂ Ω and h : R × Ω → R isnot necessarily bounded. To satisfy our assumptions, we introduce the main fact that, equation(1.2) has a unique bounded global solution on R + , see Theorem 4.2. Our result is of independentinterest and it generalizes many works in the literature, see [17, 23]. Therefore, we give our mainTheorem of the existence of a unique compact almost automorphic solution to equation (1.2), seeTheorem 4.3.The work is organized as follows: Section 2 is devoted to preliminaries for Stepanov almostautomorphic functions and the subvariant minimizing functional. In Section 3, we give our mainresults of this work. We begin by the characterization of uniformly Stepanov almost automorphicfunctions. After that, we prove under the weak assumption of Stepanov almost automorphy on f that every mild solution of (1.1) with relatively compact range is uniformly continuous. In Section3.1 we give sufficient conditions to the existence of solutions with relatively compact ranges toequation (1.1) on the right half line. Section 3.2 is devoted to main Theorems 3.2 and 3.3, weprove the existence of compact almost automorphic solutions through a minimizing some subvariantfunctional provided the existence of a bounded mild solution on the right half line. Finally, Section4 concerns the application (1.2). Throughout this work, ( X, k · k ) is a Banach space. BC ( R , X ) equipped with the supremum norm,the Banach space of bounded continuous functions f from R into X . For 1 ≤ p < ∞ , q denotes2 .1 Almost automorphic functions its conjugate exponent defined by 1 p + 1 q = 1 if p = 1 and q = ∞ if p = 1. By L ploc ( R , X ) (resp. L p ( R , X )), we designate the space (resp. the Banach space) of all equivalence classes of measurablefunctions f from R into X such that k f ( · ) k p is locally integrable (resp. integrable). Definition 2.1 (H. Bohr) [8] A continuous function f : R −→ X is said to be almost periodic iffor every ε > , there exists l ε >
0, such that for every a ∈ R , there exists τ ∈ [ a, a + l ε ] satisfying: k f ( t + τ ) − f ( t ) k < ε for all t ∈ R . The space of all such functions is denoted by AP ( R , X ) . Definition 2.2 (S. Bochner) [7] A continuous function f : R −→ X is called almost automor-phic if for every sequence ( σ n ) n ≥ of real numbers, there exist a subsequence ( s n ) n ≥ ⊂ ( σ n ) n ≥ and g : R −→ X such that, for each t ∈ R g ( t ) =: lim n f ( t + s n ) and f ( t ) = lim n g ( t − s n ) . If the above limits hold uniformly on compact subsets of R , then f is called compact almostautomorphic.In the sequel, AA ( R , X ) (resp. KAA ( R , X )) denotes the space of almost automorphic (resp.compact almost automorphic) X -valued functions. Proposition 2.1 [12] Let f ∈ AA ( R , X ). Then, f is compact almost automorphic if and only ifit is uniformly continuous. Remark 2.1 (i)
By the pointwise convergence, the function g in Definition 2.2 is only measurableand bounded but not necessarily continuous. If one of the two convergences in Definition 2.2 isuniform on R , then f is almost periodic. For more details about this topic we refer the reader tothe book [21]. (ii) An almost automorphic function may not be uniformly continuous. In fact, the real function f ( t ) = sin (cid:18)
12 + cos( t ) + cos( √ t ) (cid:19) for t ∈ R , belong to AA ( R , R ), but is not uniformly continuous.Then, we have the following inclusions: AP ( R , X ) ⊂ KAA ( R , X ) ⊂ AA ( R , X ) ⊂ BC ( R , X ) . (2.1) Definition 2.3 [13]
Let 1 ≤ p < ∞ . A function f ∈ L ploc ( R , X ) is said to be bounded in the senseof Stepanov if sup t ∈ R Z [ t,t +1] k f ( s ) k p ds ! p = sup t ∈ R Z [0 , k f ( t + s ) k p ds ! p < ∞ . The space of all such functions is denoted by BS p ( R , X ) and is provided with the following norm: k f k BS p := sup t ∈ R Z [ t,t +1] k f ( s ) k p ds ! p = sup t ∈ R k f ( t + · ) k L p ([0 , ,X ) . Then, the following inclusions hold: BC ( R , X ) ⊂ BS p ( R , X ) ⊂ L ploc ( R , X ) . (2.2)3 .2 A subvariant minimizing functional Definition 2.4 (Bochner transform) [11] Let f ∈ L ploc ( R , X ) for 1 ≤ p < ∞ . The Bochnertransform of f is the function f b : R −→ L p ([0 , , X ) defined for all t ∈ R by( f b ( t ))( s ) = f ( t + s ) for s ∈ [0 , . Now, we give the definition of almost automorphy in the sense of Stepanov.
Definition 2.5 [13] Let 1 ≤ p < ∞ . A function f ∈ L ploc ( R , X ) is said to be almost automorphicin the sense of Stepanov (or S p -almost automorphic), if for every sequence ( σ n ) n ≥ of real numbers,there exists a subsequence ( s n ) n ≥ ⊂ ( σ n ) n ≥ and g ∈ L ploc ( R , X ), such that, for each t ∈ R lim n (cid:18)Z t +1 t k f ( s + s n ) − g ( s ) k p ds (cid:19) p = 0 and lim n (cid:18)Z t +1 t k g ( s − s n ) − f ( s ) k p ds (cid:19) p = 0 . The space of all such functions is denoted by
AAS p ( R , X ) . Theorem 2.1 [13] The following are equivalent: (i) f is S p -almost automorphic. (ii) For every sequence ( σ n ) n ≥ of real numbers, there exists a subsequence ( s n ) n ≥ ⊂ ( σ n ) n ≥ for each t ∈ R lim n,m (cid:18)Z t +1 t k f ( τ + s n − s m ) − f ( τ ) k p dτ (cid:19) p = 0 . Remark 2.2 [4] (i)
Every almost automorphic function is S p -almost automorphic for 1 ≤ p < ∞ . (ii) For all 1 ≤ p ≤ p < ∞ , if f is S p -almost automorphic, then f is S p -almost automorphic. (iii) The Bochner transform of an X -valued function is a L p ([0 , , X )-valued function. Moreover,a function f is S p -almost automorphic if and only if f b is (Bochner) almost automorphic. (iv) A function ϕ ( t, s ) for t ∈ R , s ∈ [0 ,
1] is the Bochner transform of a function f (i.e., ∃ f : R −→ X such that ( f b ( t ))( s ) = ϕ ( t, s ) , t ∈ R , s ∈ [0 , ϕ ( t + τ, s − τ ) = ϕ ( t, s )for all t ∈ R , s ∈ [0 ,
1] and τ ∈ [ s − , s ] . Definition 2.6
Let 1 ≤ p < ∞ . A function f : R × X −→ Y such that f ( · , x ) ∈ L ploc ( R , Y )for each x ∈ X is said to be almost automorphic in t uniformly with respect to x in X if foreach compact set K in X , for every sequence ( σ n ) n ≥ of real numbers, there exists a subsequence( s n ) n ≥ ⊂ ( σ n ) n ≥ and g ( · , x ) ∈ L ploc ( R , Y ) for each x ∈ X , such thatlim n sup x ∈ K (cid:18)Z t +1 t k f ( s + s n , x ) − g ( s, x ) k p ds (cid:19) p = 0 , lim n sup x ∈ K (cid:18)Z t +1 t k g ( s − s n , x ) − f ( s, x ) k p ds (cid:19) p = 0(2.3)are well-defined for each t ∈ R . The space of all such functions is denoted by
AAS p U ( R × X, Y ) . The results in this section was introduced by [9] as a generalisation of the works by Fink, see [15].Let K be a compact subset of X . By F K , we denote the set of mild solutions x on R of equation(1.1) with relatively compact range and by C K ( R , X ) the set C K ( R , X ) = { x ∈ C ( R , X ) : x ( t ) ∈ K for all t ∈ R } . Note that F K ⊂ C K ( R , X ). Definition 2.7
A functional λ K : C K ( R , X ) → R is called a subvariant functional associated tothe compact set K , if λ K satisfies the following conditions: (i) λ K is invariant by translation: λ K ( x τ ) = λ K ( x ) for each τ ∈ R , where x τ ( . ) = x ( τ + . ). (ii) λ K is lower semicontinuous for the topology of compact convergence: if lim n → + ∞ x n = y uniformlyon each compact subset of R , then λ K ( y ) ≤ lim inf n → + ∞ λ K ( x n ).4 efinition 2.8 A function x ∗ : R −→ X is called a minimal K -valued solution of equation (1.1)if the following hold x ∗ ∈ F K and λ K ( x ∗ ) = inf x ∈F K λ K ( x ) (2.4)An example of subvariant functional is given by: λ K ( x ) = sup t ∈ R Φ( x ( t )) where Φ ∈ C ( K, R ) . In particular, we have λ K ( x ) = sup t ∈ R k x ( t ) k . (1.1) In this section, we prove the existence of compact almost automorphic solutions to the semilinearevolution equation (1.1).By a mild solution of equation (1.1), we mean a continuous function x : R −→ X satisfying thefollowing variation of constants formula: x ( t ) = T ( t − σ ) x ( σ ) + Z tσ T ( t − s ) f ( s, x ( s )) ds for all t ≥ σ. (3.1)Now, we give the following hypotheses: (H1) ( A, D ( A )) is the infinitesimal generator of a C -semigroup ( T ( t )) t ≥ on X . (H2) ( T ( t )) t ≥ is compact, i.e., T ( t ) is a compact operator for each t > (H3) For all
R >
0, the function sup k x k≤ R k f ( · , x ) k ∈ L ploc ( R , R ) such thatsup t ∈ R Z t +1 t sup k x k≤ R k f ( s, x ) k p ds ! p < + ∞ . (H4) The function f is given by f ( t, x ) := ϕ ( t ) g ( x ) + ψ ( t ), where ϕ ∈ AAS ( R , R ) ∩ L ∞ ( R , R ), ψ ∈ AAS ( R , X ) and g ∈ C ( X, X ) transform bounded subsets of X into bounded. Remark 3.1
Hypothesis (H4) implies (H3) . Definition 3.1
Let p ∈ [1 , ∞ ) and f : R × X −→ X be such that f ( · , x ) ∈ BS p ( R , X ) for each x ∈ X . Then, f is said to be S p -uniformly continuous with respect to the second argument oneach compact K ⊂ X if, for all K ⊂ X compact for all ε > δ K,ε such that for all x , x ∈ K , we have k x − x k ≤ δ K,ε = ⇒ (cid:18)Z t +1 t k f ( s, x ) − f ( s, x ) k p ds (cid:19) p ≤ ε for all t ∈ R . (3.2) Lemma 3.1
Let p ∈ [1 , ∞ ) and f : R × X −→ X be such that f ( · , x ) ∈ L ploc ( R , X ) for each x ∈ X .Then, f ∈ AAS p U ( R × X, X ) if and only if the following hold: (i)
For each x ∈ X , f ( · , x ) ∈ AAS p ( R , X ) . (ii) f is S p -uniformly continuous with respect to the second argument on each compact K ⊂ X . Proof.
Let f ∈ AAS p U ( R × X, X ) and f b : R × X −→ L p ([0 , , X ) be the Bochner transformassociated to f . It follows in view of [6, Proposition 5.5], that (i) is clearly satisfied and that: foreach compact K ⊂ X, for all ε > δ K,ε such that for all x , x ∈ K one has k x − x k ≤ δ K,ε = ⇒ k f b ( t, x ) − f b ( t, x ) k p ≤ ε for all t ∈ R . Since k f b ( t, x ) − f b ( t, x ) k p = Z [0 , k ( f b ( t, x ))( s ) − ( f b ( t, x ))( s ) k p ds ! p = (cid:18)Z t +1 t k f ( s, x ) − f ( s, x ) k p ds (cid:19) p for all t ∈ R .
5t follows that (3.2) holds and then (ii) is achieved.Conversely, let f : R × X −→ X be a function such that f ( · , x ) ∈ L ploc ( R , X ) for each x ∈ X. Assume that f satisfy (i)-(ii) . Let us fix a compact subset K in X and ε > . Since K is compact,it follows that there exists a finite subset { x , ..., x n } ⊂ K ( n ∈ N ∗ ) such that K ⊆ n [ i =1 B ( x i , δ K,ε ) . For x ∈ K there exist i = 1 , ..., n satisfying k x − x i k ≤ δ K,ε . Let ( σ n ) n ≥ be a sequence of realnumbers and its subsequence ( s n ) n ≥ such that (cid:18)Z t +1 t k f ( s + s l − s k , x ) − f ( s, x ) k p ds (cid:19) p ≤ (cid:18)Z t +1 t k f ( s + s l − s k , x ) − f ( s + s l − s k , x i ) k p ds (cid:19) p + (cid:18)Z t +1 t k f ( s + s l − s k , x i ) − f ( s, x i ) k p ds (cid:19) p + (cid:18)Z t +1 t k f ( s, x i ) − f ( s, x ) k p ds (cid:19) p . (3.3)Using (i) , it holds that f ( · , x i ) ∈ AAS p ( R , X ). Hence, for k, l large enough, for each t ∈ R (cid:18)Z t +1 t k f ( s + s l − s k , x i ) − f ( s, x i ) k p ds (cid:19) p ≤ ε . (3.4)Otherwise, since k x − x i k ≤ δ K,δ and by (ii) we claim that, for each k, l ∈ N (cid:18)Z t +1 t k f ( s + s l − s k , x ) − f ( s + s l − s k , x i ) k p ds (cid:19) p ≤ ε t ∈ R (3.5)and (cid:18)Z t +1 t k f ( s, x ) − f ( s, x i ) k p ds (cid:19) p ≤ ε t ∈ R . (3.6)Consequently, if we replace (3.4), (3.5) and (3.6) in (3.3) then, for k, l large enough we havesup x ∈ K (cid:18)Z t +1 t k f ( s + s l − s k , x ) − f ( s, x ) k p ds (cid:19) p ≤ ε. for each t ∈ R . (cid:4) Remark 3.2 (a)
In view of Lemma 3.1. If f is the function defined by (H4) , then f ∈ AAS U ( R × X, X ). (b) Using Lemma 3.1, we can prove easily that the function g in (2.3) is S p -uniformly continuouswith respect to the second argument on each compact K ⊂ X . Lemma 3.2
Let f ∈ AAS p U ( R × X, X ) for p ∈ [1 , ∞ ). Then, for each compact K in X , we have (i) k p = sup t ∈ R sup x ∈ K (cid:16)R t +1 t k f ( s, x ) k p ds (cid:17) p < + ∞ . (ii) l p = sup t ∈ R sup x ∈ K (cid:16)R t +1 t k g ( s, x ) k p ds (cid:17) p < + ∞ , where g is the function defined in (2.3). Proof. (i)
Let K ⊂ X be compact and let δ > { x , ..., x n } ⊂ K ( n ∈ N ∗ )such that K ⊆ n [ i =1 B ( x i , δ ) . Therefore, for each x ∈ K there exist i = 1 , ..., n satisfying k x − x i k ≤ δ .Then, for all x ∈ K and ε >
0, we have (cid:18)Z t +1 t k f ( s, x ) k p ds (cid:19) p ≤ (cid:18)Z t +1 t k f ( s, x ) − f ( s, x i ) k p ds (cid:19) p + n X i =1 (cid:18)Z t +1 t k f ( s, x i ) k p ds (cid:19) p for all t ∈ R . Hence, by Lemma 3.1, we claim thatsup t ∈ R sup x ∈ K (cid:18)Z t +1 t k f ( s, x ) k p ds (cid:19) p ≤ ε + n X i =1 sup t ∈ R (cid:18)Z t +1 t k f ( s, x i ) k p ds (cid:19) p < ∞ . ii) Let K ⊂ X be compact and let ( σ n ) n ≥ be a sequence. Since f ∈ AAS p U ( R × X, X ), it followsthat there exists a subsequence ( s n ) n ≥ ⊂ ( σ n ) n ≥ such that (2.3) holds. Hence, for all n ∈ N (cid:18)Z t +1 t k g ( s, x ) k p ds (cid:19) p ≤ sup x ∈ K (cid:18)Z t +1 t k f ( s + s n , x ) − g ( s, x ) k p ds (cid:19) p + sup x ∈ K (cid:18)Z t +1 t k f ( s + s n , x ) k p ds (cid:19) p ≤ sup x ∈ K (cid:18)Z t +1 t k f ( s + s n , x ) − g ( s, x ) k p ds (cid:19) p + k p for all x ∈ K, t ∈ R . Therefore, for n large enough, we havesup x ∈ K (cid:18)Z t +1 t k g ( s, x ) k p ds (cid:19) p ≤ k p for all t ∈ R . This proves the result. (cid:4)
Lemma 3.3 [3, Theorem 4.1.2] Let Y be a normed space and ( T i ) i ∈ T be any family of boundedlinear operators on Y such that sup i ∈ I | T i | < ∞ . If D is a dense subset of Y , and if for each y ∈ DT i y → T y as i → ∞ , for some bounded linear operator T . Then for every compact set K ⊂ Y sup y ∈ K k T i y − T y k → i → ∞ . The following Lemma is needed in the proof of Theorem 3.1.
Lemma 3.4
Let ( x n ) n be a sequence in a Banach space X and x ∈ X . If every subsequence( x ′ n ) n ⊂ ( x n ) n has a subsequence ( x ′′ n ) n ⊂ ( x ′ n ) n ⊂ ( x n ) n that converges to x , then the wholesequence ( x n ) n converges to x . Proof.
By contradiction. (cid:4)
Theorem 3.1
Let f ∈ AAS p U ( R × X, X ) for p ∈ [1 , ∞ ). Assume that (H1) holds. Then, a mildsolution of equation (1.1) with relatively compact range is uniformly continuous. Proof.
Let x be a solution of equation (1.1) such that K = { x ( t ) : t ∈ R } ⊂ X is compact. Wedistinguish two cases: The case p ∈ (1 , ∞ ) . Since ( T ( t )) t ≥ is a C -semigroup on X , it follows in view of Lemma 3.3,that lim t → sup x ∈ K k T ( t ) x − x k = 0 . (3.7)Furthermore, the C -semigroup ( T ( t )) t ≥ on X satisfies k T ( t ) k ≤ M e ωt for t ≥ M ≥ ω ∈ R . Let q be such that 1 p + 1 q = 1. Using H¨older inequality, we have for all t, s ∈ R , k x ( t ) − x ( s ) k ≤ k T ( t − s ) x ( s ) − x ( s ) k + Z ts k T ( t − σ ) f ( σ, x ( σ )) k dσ ≤ sup x ∈ K k T ( t − s ) x − x k + M (cid:18)Z ts e qω ( t − σ ) dσ (cid:19) q (cid:18)Z ts k f ( σ, x ( σ )) k p dσ (cid:19) p ≤ sup x ∈ K k T ( t − s ) x − x k + k p M ( t − s + 3) p (cid:18)Z t − s e qωσ dσ (cid:19) q , t ≥ s. s and t , we claim that k x ( t ) − x ( s ) k ≤ sup x ∈ K k T ( | t − s | ) x − x k + k p M ′ ( | t − s | +3) p Z | t − s | e qωσ dσ ! q . From Z | t − s | e qωσ dσ ! q → | t − s |→ k x ( t ) − x ( s ) k→ | t − s |→ . Consequently, x is uniformly continuous. The case p = 1 . To show that x is uniformly continuous, we take two real sequences ( t n ) n and( s n ) n such that | t n − s n | → n → ∞ and we prove that y n = x ( t n ) − x ( s n ) → n → ∞ . Infact, consider the sequence ( h n ) n defined by h n = t n − s n . Then, we have lim n →∞ h n = 0. Assumewithout loss of generality that 0 ≤ h n ≤ n ∈ N . Thus, we have x ( t n ) = x ( s n + h n ) = T ( h n ) x ( s n ) + Z s n + h n s n T ( s n + h n − s ) f ( s, x ( s )) ds = T ( h n ) x ( s n ) + Z h n T ( h n − s ) f ( s + s n , x ( s + s n )) ds. Let M ≥ ω ∈ R such that k T ( t ) k ≤ M e ωt for all t ≥
0. Then k x ( s n + h n ) − x ( s n ) k ≤ k T ( h n ) x ( s n ) − x ( s n ) k + Z h n k T ( h n − s ) f ( s + s n , x ( s + s n )) k ds ≤ sup y ∈ K k T ( h n ) y − y k + M Z h n e ω ( h n − s ) k f ( s + s n , x ( s + s n )) k ds ≤ sup y ∈ K k T ( h n ) y − y k + M e | ω | h n Z h n k f ( s + s n , x ( s + s n )) k ds. (3.8)The semigroup ( T ( t )) t ≥ is strongly continuous, then for each y ∈ K , T ( h n ) y → y as, n → ∞ .This implies that sup n k T ( h n ) y k < ∞ for each y ∈ X and thus by the Banach-Steinhaus’s Theoremsup n k T ( h n ) k < ∞ . It follows from Lemma 3.3 thatsup y ∈ K k T ( h n ) y − y k → n → ∞ . (3.9)Let y ′ n = x ( s ′ n + h ′ n ) − x ( s ′ n ) be a subsequence of y n . From the S -almost automorphy of f , thereexist a subsequence ( s ′′ n ) n ⊂ ( s ′ n ) n and there exist g : R × X → X with g ( · , x ) ∈ L loc ( R , X ) foreach x ∈ X such that for every t ∈ R , we havesup x ∈ K Z t +1 t k f ( s + s ′′ n , x ) − g ( s, x ) k ds → n → ∞ . (3.10)Let ( h ′′ n ) n be the corresponding subsequence of ( h ′ n ) n . We can assume that 0 ≤ h n ≤ n ∈ N . Then, we have Z h ′′ n k f ( s + s ′′ n , x ( s + s ′′ n )) k ds ≤ Z h ′′ n k f ( s + s ′′ n , x ( s + s ′′ n )) − g ( s, x ( s + s ′′ n )) k ds + Z h ′′ n k g ( s, x ( s + s ′′ n )) k ds ≤ sup x ∈ K Z h ′′ n k f ( s + s ′′ n , x ) − g ( s, x ) k ds + Z h ′′ n k g ( s, x ( s + s ′′ n )) k ds ≤ sup x ∈ K Z k f ( s + s ′′ n , x ) − g ( s, x ) k ds + Z h ′′ n k g ( s, x ( s + s ′′ n )) k ds. Using (3.10) we claim that lim n →∞ sup x ∈ K Z k f ( s + s ′′ n , x ) − g ( s, x ) k ds = 0.On the other hand, let δ > x , ..., x m ∈ K , m ∈ N ∗ , such that , for each8 ∈ K there exists i = 1 , ..., m satisfies k x − x i k ≤ δ . Let ε >
0, then, for all s ∈ R for all n ∈ N ,there exists i ( s, n ) = 1 , ..., m with k x ( s + s ′′ n ) − x i ( s,n ) k ≤ δ . Hence, by Remark 3.2, we obtain Z h ′′ n k g ( s, x ( s + s ′′ n )) k ds ≤ Z k g ( s, x ( s + s ′′ n )) − g ( s, x i ( s,n ) ) k ds + m X i =1 Z h ′′ n k g ( s, x i ) k ds ≤ ε + m X i =1 Z h ′′ n k g ( s, x i ) k ds for all ε > . By Lemma 3.2, we claim that, for each i = 1 , ..., m , we have Z h ′′ n k g ( s, x i ) k ds → n → ∞ . Then, Z h ′′ n k g ( s, x ( s + s ′′ n )) k ds → n → ∞ . Therefore, Z h ′′ n k f ( s + s ′′ n , x ( s + s ′′ n )) k ds → n → ∞ . (3.11)Consequently, we conclude from (3.8), (3.9) and (3.11) that k y ′′ n k = k x ( t ′′ n ) − x ( s ′′ n ) k = k x ( s ′′ n + h ′′ n ) − x ( s ′′ n ) k → n → ∞ , Finally, using Lemma 3.4, the whole sequence y n = x ( t n ) − x ( s n ) converges to 0. We concludethat the mild solution x is uniformly continuous. (cid:4) Lemma 3.5
Let f ∈ AAS p ( R × X, X ) for p ∈ [1 , ∞ ). Assume that (H1)-(H2) hold. If equation(1.1) has at least a mild solution x : [ t , + ∞ ) −→ X with relatively compact range. Then, thereexists a mild solution x on R of equation (1.1) such that { x ( t ) : t ∈ R } ⊂ { x ( t ) : t ≥ t } . (3.12) Proof.
Let K := { x ( t ) : t ≥ t } ⊂ X . By assumption K is compact. Moreover, the mild solution x is given by: x ( t ) = T ( t − s ) x ( s ) + Z ts T ( t − σ ) f ( σ, x ( σ )) dσ for t ≥ s ≥ t . Let ( t ′ n ) n be a sequence of real numbers such thatlim n → + ∞ t ′ n = + ∞ . Since f ∈ AAS p U ( R × X, X ), it follows that there exist a subsequence ( t n ) n ⊂ ( t ′ n ) n and g : R × K → X such that for each t ∈ R , we havelim n sup x ∈ K (cid:18)Z t +1 t k f ( s + s n , x ) − g ( s, x ) k p ds (cid:19) p = 0 . (3.13)For n ∈ N sufficiently large, the function t x ( · + t n ) ∈ K is defined on R . Let u n ( t ) := x ( t + t n )for all t ∈ R and the set U n = { u n : n ∈ N } . Then, U n ⊂ C ( R , X ) and ( u n ) n satisfies for each t ∈ R u n ( t ) ∈ K . Therefore, for each t ∈ R , the subset U n ( t ) = { u n ( t ) : n ∈ N } is relativelycompact in X . By Theorem 3.1 and using Arzela-Ascoli’s Theorem and an extraction diagonalargument applied to the sequence ( u n ) n , we claim that there exist x ∗ ∈ C ( R , X ) and a subsequenceof ( t n ) n such that x ( t + t n ) → x ∗ ( t ) as n → + ∞ , (3.14)uniformly on each compact subset of R . Hence, for n ∈ N sufficiently large, we have x ( t + t n ) = T ( t − s ) x ( s + t n ) + Z ts T ( t − σ ) f ( σ + t n , x ( σ + t n )) dσ, t ≥ s. (3.15)9et t, s ∈ R , t ≥ s . Then, using (3.14), we claim thatlim n → + ∞ T ( t − s ) x ( s + t n ) = T ( t − s ) x ∗ ( s ) . (3.16)Now, we prove thatlim n → + ∞ Z ts T ( t − σ ) f ( σ + t n , x ( σ + t n )) dσ = Z ts T ( t − σ ) g ( σ, x ∗ ( σ )) dσ. (3.17)In fact, we have (cid:18)Z t +1 t k f ( σ + t n , x ( σ + t n )) − g ( σ, x ∗ ( σ )) k p dσ (cid:19) p ≤ (cid:18)Z t +1 t k f ( σ + t n , x ( σ + t n )) − g ( σ, x ( σ + t n )) k p dσ (cid:19) p + (cid:18)Z t +1 t k g ( σ, x ( σ + t n )) − g ( σ, x ∗ ( σ )) k p dσ (cid:19) p ≤ sup x ∈ K (cid:18)Z t +1 t k f ( σ + t n , x ) − g ( σ, x ) k p dσ (cid:19) p | {z } (I) + (cid:18)Z t +1 t k g ( σ, x ( σ + t n )) − g ( σ, x ∗ ( σ )) k p dσ (cid:19) p | {z } (II) , t ∈ R . Using (3.13), we obtain that (I) → n → ∞ . On the other hand, by Remark 3.2, it followsthat g is S p -uniformly continuous. Therefore, we have (II) → n → ∞ . Consequently, for each t ∈ R K p ( t, n ) := (cid:18)Z t +1 t k f ( σ + t n , x ( σ + t n )) − g ( σ, x ∗ ( σ )) k p dσ (cid:19) p → n → ∞ . (3.18)Thus, for p ∈ [1 , ∞ ), we have k Z ts T ( t − σ ) f ( σ + t n , x ( σ + t n )) dσ − Z ts T ( t − σ ) g ( σ, x ∗ ( σ )) dσ k≤ Z ts k T ( t − σ ) [ f ( σ + t n , x ( σ + t n )) − g ( σ, x ∗ ( σ ))] k dσ ≤ M [ t ]+1 X m =[ s ] e | ω | ( t − m ) Z m +1 m k f ( σ + t n , x ( σ + t n )) − g ( σ, x ∗ ( σ )) k dσ ≤ M [ t ]+1 X m =[ s ] e | ω | ( t − m ) (cid:18)Z m +1 m k f ( σ + t n , x ( σ + t n )) − g ( σ, x ∗ ( σ )) k p dσ (cid:19) p = M [ t ]+1 X m =[ s ] e | ω | ( t − m ) k p ( m, n )Therefore, using the limit in (3.18) we obtain (3.17). Consequentely, by (3.16) and (3.17) we deducethat x ∗ ( t ) = T ( t − s ) x ∗ ( s ) + Z ts T ( t − σ ) g ( σ, x ∗ ( σ )) dσ for t ≥ s. Hence, x ∗ is a mild solution with a relatively compact range of the following equation x ′∗ ( t ) = Ax ∗ ( t ) + g ( t, x ∗ ( t )) , t ∈ R . (3.19)By taking the sequence ( − t n ) n instead of ( t n ) n , we apply the same construction to the mild solution x ∗ of equation (3.19). Then, we obtain the existence of a mild solution x with relatively compactrange of equation (1.1). Consequently, x is a mild solution satisfying (3.12). (cid:4) .1 Solutions with relatively compact ranges In this section, we investigate the relative compactness of the range of bounded solutions to equation(1.1) on the right half-line. We distinguish, two cases, p ∈ (1 , ∞ ) and p = 1.To conclude our results, we use the Kuratowski measure of noncompactness α of bounded subsets B in X defined by: α ( B ) := inf { ε > B has a finite cover of balls with diameter < ε } . (3.20)The measure α satisfies the following properties: (a) α ( B ) = 0 ⇔ B is relatively compact. (b) α ( B + B ) ≤ α ( B ) + α ( B ) for all B and B bounded in X . (c) α ( B (0 , ε ) = 2 ε for all ε > . For more details, we refer to [19].
The case p ∈ (1 , ∞ ) .Lemma 3.6 Assume that (H1)-(H3) hold. If equation (1.1) has at least a bounded mild solution x : [ t , + ∞ ) −→ X . Then, its range { x ( t ) : t ≥ t } is relatively compact in X . Proof. (i)
For 0 < ε <
1, define x ( t ) = T ( ε ) x ( t − ε ) + Z tt − ε T ( t − σ ) f ( σ, x ( σ )) dσ for t > t + 1 . Let t > t + 1. Then, we have T ( ε ) x ( t − ε ) ∈ T ( ε ) B (0 , M ) (3.21)where M := sup t ≥ t k x ( t ) k < + ∞ and B (0 , M ) is the closed ball of center 0 and radius M .Using H¨older inequality, we obtain k Z tt − ε T ( t − σ ) f ( σ, x ( σ )) dσ k ≤ M e | ω | ε ε q (cid:18)Z tt − ε k f ( σ, x ( σ )) k p dσ (cid:19) p ≤ M e | ω | ε ε q ˜ k p := δ ( ε ) , where ˜ k p = sup t ∈ R Z tt − sup x ∈ B (0 ,M ) k f ( σ, x ) k p dσ ! p .Hence, Z tt − ε T ( t − σ ) f ( σ, x ( σ )) dσ ∈ B (0 , δ ( ε )) . Consequently, { x ( t ) : t ≥ t } ⊂ { x ( t ) : t ∈ [ t , t + 1] } ∪ T ( ε ) B (0 , M ) ∪ B (0 , δ ( ε )) . Thus, by the continuity of x , the set { x ( t ) : t ∈ [ t , t + 1] } is compact. Hence α ( { x ( t ) : t ∈ [ t , t + 1] } ) = 0 . Therefore, from hypothesis (H2) , we claim that α ( { x ( t ) : t ≥ t } ) ≤ δ ( ε ) for all ε > . Since δ ( ε ) → ε →
0. It follows that α ( { x ( t ) : t ≥ t } ) = 0 . (cid:4) The case p = 1 . In this case, the following Lemmas are needed. 11 .1 Solutions with relatively compact ranges
Lemma 3.7
Let ρ ∈ AAS ( R , X ). Then, for all 0 < ε <
1, the following function: x ερ ( t ) := Z tt − ε T ( t − s ) ρ ( s ) ds for t ≥ t , t ∈ R (3.22)has a relatively compact range on [ t , ∞ ) . Proof.
Let 0 < ε < t ′ n ) n ⊂ R be a sequence. Since ρ ∈ AAS ( R , X ) it follows that, thereexist a subsequence ( t n ) n ⊂ ( t ′ n ) n and g ∈ L loc ( R , X ) such that for each t ∈ R Z t +1 t k ρ ( t n + s ) − g ( s ) k ds → n → ∞ . (3.23)Let y ερ := Z ε T ( σ ) g ( − σ ) dσ ∈ X . Then, k x ερ ( t n ) − y ερ k = k Z t n − εt n T ( t n − s ) ρ ( s ) ds − Z ε T ( σ ) g ( − σ ) dσ k = k Z ε T ( σ ) [ ρ ( t n − σ ) − g ( − σ )] dσ k≤ M e | ω | Z k ρ ( t n − σ ) − g ( − σ ) k dσ. Using (3.23), we obtain that Z k ρ ( t n − σ ) − g ( − σ ) k ds → n → ∞ , independently of t and ε . Therefore x ερ ( t n ) → y ερ as n → ∞ . Consequently, x ερ has a relatively compact range on [ t , ∞ ) . (cid:4) Lemma 3.8
Let f : R × X −→ X satisfy (H4) with φ, ψ ∈ AAS ( R , X ). Assume that (H1)-(H2) hold. If equation (1.1) has at least a bounded mild solution x : [ t , + ∞ ) −→ X , then x has a relatively compact range. Proof.
Let 0 < ε <
1. Define x ( t ) = T ( ε ) x ( t − ε ) + Z tt − ε T ( t − σ ) f ( σ, x ( σ )) dσ for t > t + 1 . Let t > t + 1. Then, we have T ( ε ) x ( t − ε ) ∈ T ( ε ) B (0 , M ) (3.24)where M := sup t ≥ t k x ( t ) k < + ∞ and B (0 , M ) is the closed ball of center 0 and radius M .It suffices to prove that the measure of noncompactness α of Z tt − ε T ( t − σ ) f ( σ, x ( σ )) dσ equalszero. In fact, by assumption on f , we obtain that Z tt − ε T ( t − σ ) f ( σ, x ( σ )) dσ = Z tt − ε T ( t − σ ) φ ( σ ) g ( x ( σ )) dσ + Z tt − ε T ( t − σ ) ψ ( σ ) dσ k Z tt − ε T ( t − σ ) φ ( σ ) g ( x ( σ )) dσ k ≤ M e | ω | ε | φ | ∞ sup x ∈ B (0 ,M ) k g ( x ) k ε := γ ( ε ) . Hence, Z tt − ε T ( t − σ ) f ( σ, x ( σ )) dσ ∈ B (0 , γ ( ε )) + { x εψ ( t ) : t > t + 1 } . By Lemma 3.7 and (H2) , it follows that α ( { x ( t ) : t ≥ t } ) ≤ γ ( ε ) . Since γ ( ε ) → ε →
0. We claim that α ( { x ( t ) : t ≥ t } ) = 0 . .2 Almost automorphic solutions minimizing a subvariant functional (cid:4) In general when f does not satisfy (H4) , Lemma 3.8 cannot hold. However, under the additionalassumption (H5) below, we have the following main result which is of independent interest. (H5) For all bounded subset B ⊂ X , the set Γ := { f ( t, x ) : ( t, x ) ∈ R × X } is relativelycompact in X . Proposition 3.1
Let f : R × X −→ X such that f ( · , x ) ∈ BS ( R , X ) for each x ∈ X . Assumethat (H1) - (H2) and (H5) are satisfied. If equation (1.1) has at least a bounded mild solution x : [ t , + ∞ ) −→ X , then, its range is relatively compact in X . Proof.
Let 0 < ε <
1. The mild solution x satisfies the following: x ( t ) = T ( ε ) x ( t − ε ) + Z tt − ε T ( t − σ ) f ( σ, x ( σ )) dσ for t ≥ t . Let x ε ( t ) := Z tt − ε T ( t − σ ) f ( σ, x ( σ )) dσ. As in the proof of Lemma 3.8, it remins to prove that,the set { x ε ( t ) : t > t + 1 } is relatively compact in X . In fact, let ( t ′ n ) n ⊂ R and for each t ∈ R , we define the sequence( f ( t ′ n − t, x ( t ′ n − t ))) n . Using (H5) , there exist as subsequence ( t n ) n ⊂ ( t ′ n ) n and g : R −→ X measurable such that for each t ∈ R f ( t n − t, x ( t n − t )) → g ( t ) as n → ∞ . (3.25)Given y ε := Z ε T ( σ ) g ( σ ) dσ. Then k x ε ( t n ) − y ε k = k Z t n − εt n T ( t n − s ) f ( s, x ( s )) ds − Z ε T ( σ ) g ( σ ) dσ k≤ M e | ω | ε Z ε k f ( t n − σ, x ( t n − σ )) − g ( σ ) k dσ. Hence, using (3.25), it follows in view of dominated convergence Theorem that Z ε k f ( t n − σ, x ( t n − σ )) − g ( σ ) k dσ → n → ∞ , which implies that k x ε ( t n ) − y ε k → n → ∞ . Therefore, α ( { x ε ( t ) : t > t + 1 } ) . Consequently, using hypothesis (H2) , we obtain that α ( { x ( t ) : t ≥ t } ) = 0 . (cid:4) In the case p ∈ (1 , ∞ ), we have the following main Theorem. Theorem 3.2
Let f ∈ AAS p U ( R × X, X ) and K be a compact subset of X . Assume that (H1)-(H3) are satisfied. Moreover, if equation (1.1) has at least a mild solution x defined and boundedon [ t , + ∞ ). Then, the following hold: (i) If λ K is a subvariant functional associated to the compact set K , then equation (1.1) hasat least a minimal K -valued solution x . (ii) If equation(1.1) has a unique minimal K -valued solution x , then x is compact almostautomorphic. 13 .2 Almost automorphic solutions minimizing a subvariant functional Proof.
Let K := { x ( t ) : t ≥ t } , by Lemma 3.6, the set K is compact in X . (i) Define δ = inf x ∈F K λ K ( x ). Using Lemma 3.5, we claim that the set F K is nonempty.Hence, there exists a sequence ( x n ) n ⊂ F K , such thatlim n → + ∞ λ K ( x n ) = δ. (3.26)Given U n := { x n : n ∈ N } ⊂ C ( R , X ), by definition of F K , the set U n ( t ) := { x n ( t ) , n ∈ N } ⊂ K ,for each t ∈ R .Since f ∈ AAS p U ( R × X, X ), it follows by Theorem 3.1, that tha family U n is equicontinuous.Hence, in view of Arzela-Ascoli Theorem, we claim that U n is a relatively compact subset of C ( R , X )endowed with the topology of compact convergence. Therefore, there exists a subsequence of ( x n ) n (steel denoted by ( x n ) n ) such that x n ( t ) → x ∗ ( t ) as n → + ∞ , (3.27)uniformly on each compact subset of R . Obviously, x ∗ ∈ C ( R , X ) with range including in K (i.e., x ∗ ∈ C K R , X )).Since, for every n ∈ N , x n is a mild solution on R of equation (1.1), we claim (using the same proofas in Lemma 3.5) that x ∗ ( t ) = T ( t − s ) x ∗ ( s ) + Z ts T ( t − σ ) f ( σ, x ∗ ( σ )) dσ for t ≥ s. Thus x ∗ is a mild solution on R of equation (1.1). This implies that x ∗ ∈ F K and that δ ≤ λ K ( x ∗ ) . (3.28)By (3.27) and Definition 2.7, we obtain that λ K ( x ∗ ) ≤ lim inf n → + ∞ λ K ( x n ) . (3.29)From (3.27), (3.28) and (3.29), we deduce that λ K ( x ∗ ) = δ. (3.30)Consequently, x ∗ is a minimal K -valued solution: λ K ( x ∗ ) = inf x ∈F K λ K ( x ) . (3.31)This proves the existence of the minimal K -valued solution of equation (1.1). (ii) Let x ∗ be the unique minimal solution with relatively compact range of equation (1.1). Weshow that x ∗ is compact almost automorphic. In fact, let ( t ′ n ) n be a sequence of real numbers.Since f ∈ AAS p U ( R × X, X ) it follows that there exist a subsequence ( t n ) n ⊂ ( t ′ n ) n and a function g : R × X −→ X such that (2.3) holds.Define the set U n = { x n : n ∈ N } ⊂ C ( R , X ) where for each t ∈ R , x n ( t ) := x ∗ ( t + t n ).By assumption, for each t ∈ R , x n ( t ) ∈ K . Hence, for each t ∈ R U n ( t ) ⊂ K. Moreover, by Theorem 3.1, it holds in view of Arzela-Ascoli Theorem in C ( R , X ) endowed withtopology of compact convergence that, there exist y ∗ ∈ C ( R , X ) and a subsequence ( t n ) such that x n ( t ) → y ∗ ( t ) as n → ∞ , uniformly on compact subsets of R . Therefore, T ( t − s ) x n ( s ) → T ( t − s ) y ∗ ( s ) as n → ∞ , uniformly on compact subsets of R . On the other hand, since x ∗ is a mild solution of equation(1.1), we have x ∗ ( t + t n ) = T ( t − s ) x ∗ ( s + t n ) + Z ts T ( t − σ ) f ( σ + t n , x ∗ ( σ + t n )) dσ, t ≥ s. y ∗ ( t ) = T ( t − s ) y ∗ ( s ) + Z ts T ( t − σ ) f ( σ + t n , y ∗ ( σ )) dσ, t ≥ s. Consequently, y ∗ is a mild solution on R with relatively compact range in K of equation x ′ ( t ) = Ax ( t ) + g ( t, x ( t )) , t ∈ R . (3.32)By definition of the subvariant λ K , we obtain that λ K ( y ∗ ) ≤ λ K ( x ∗ ). Then, from (3.31) we deducethat λ K ( y ∗ ) ≤ inf x ∈F K λ K ( x ) . (3.33)Furthermore, since y ∗ is a mild solution with relatively compact range in K of equation (3.32), itfollows in view of (2.3) and by using the same reasoning as for x ∗ by taking the sequence ( − t n ) n instead of ( t n ) n , that there exists a mild solution z ∗ with relatively compact range in K of equation(1.1). Then, we have λ K ( z ∗ ) ≤ λ K ( y ∗ ) . Therefore, by (3.33), we obtain λ K ( z ∗ ) ≤ inf x ∈F K λ K ( x ) . Hence λ K ( z ∗ ) = inf x ∈F K λ K ( x ) . Therefore, z ∗ is a minimal solution with relatively compact range in K of equation (1.1). Byuniqueness we deduce that x ∗ = z ∗ . Consequently, x ∗ is compact almost automorphic. (cid:4) From Theorem 3.2, we deduce the following result.
Corollary 3.1
Let f ∈ AAS p U ( R × X, X ). Assume that (H1)-(H3) are satisfied. If equation(1.1) has a unique bounded mild solution x on R , then x is compact almost automorphic. Proof.
By assumptions, all hypotheses of Theorem 3.2 are satisfied and in particular, the set K := { x ( t ) : t ≥ } is compact. Hence, by taking λ K ≡
1, we conclude the result in view ofTheorem 3.2- (ii) . (cid:4) For p = 1, Theorem 3.2 and Corollary 3.1 become : Theorem 3.3
Let K be a compact subset of X . Assume that (H1) (H2) and (H4) are satisfied.Moreover, if equation (1.1) has at least a mild solution x defined and bounded on [ t , + ∞ ). Then,the following hold: (i) If λ K is a subvariant functional associated to the compact set K , then equation (1.1) has atleast a minimal K -valued solution x . (ii) If equation(1.1) has a unique minimal K -valued solution x , then x is compact almost auto-morphic. Proof.
We argue as the same as in the proof of Theorem 3.2, using Remarks 3.1, 3.2 and Lemma3.8 in place of Lemma 3.6. (cid:4)
Hence, we have the following Corollary.
Corollary 3.2
Let (H1) , (H2) and (H4) be satisfied. If equation (1.1) has a unique boundedmild solution x , then x is compact almost automorphic. Consider the following nonautonomous reaction-diffusion problem : ∂∂t v ( t, ξ ) = n X k =1 ∂ ∂ξ i v ( t, ξ ) + g ( v ( t, ξ )) + h ( t, ξ ) for t ∈ R and ξ ∈ Ω ,v ( t, ξ ) = 0 for t ∈ R , ξ ∈ ∂ Ω , (4.1)15here Ω ⊂ R n , n ≥ ∂ Ω, g : R → R and h : R × Ω → R .Given the Banach space X = C (cid:0) Ω (cid:1) := { x ∈ C (Ω , R ) : x | ∂ Ω = 0 } endowed with the uniform normtopology. Let h ( t, ξ ) = sin (cid:18)
12 + cost + cos √ t (cid:19) h ( ξ ) + a ( t ) for t ∈ R and ξ ∈ Ωwhere h ∈ X and a ( t ) = X n ≥ β n ( t ) such that for every n ≥ β n ( t ) = X i ∈ P n H ( n ( t − i )) , with P n = 3 n (2 Z + 1) and H ∈ C ∞ ( R , R ) with support in ( − , ) such that H ≥ , H (0) = 1 and Z − H ( s ) ds = 1 . Remark 4.1 [25] The function a ∈ C ∞ ( R , R ) but a / ∈ AA ( R , R ) since it is not bounded on R .However a ∈ AAS ( R , R ) . Define the operator A : D ( A ) ⊂ X → X by D ( A ) = ( x ∈ X ∩ H (Ω) : n X k =1 ∂ ∂ξ i x := ∆ x ∈ X ) ,Ax = ∆ x. It is well known that (
A, D ( A )) generates a compact C -semigroup ( T ( t )) t ≥ on X such that k T ( t ) k≤ M e − λ t for t ≥ , (4.2)where λ := min { λ : λ ∈ σ ( − A ) } >
0, see [17] for more details.Consequently, hypotheses (H1) and (H2) are satisfied. Moreover, we assume that g is locallyLipschitzian such that g (0) = 0 and lim sup | r |→ + ∞ g ( r ) r < λ . (4.3) Remark 4.2
Condition (4.3) implies: there exists ¯ M ≥ rg ( r ) ≤ Cr for | r |≥ ¯ M with C < λ . Consider G : X → X and H : R → X defined by G ( x )( ξ ) = g ( x ( ξ )) , x ∈ X, ξ ∈ Ωand H ( t )( ξ ) = h ( t, ξ ) , x ∈ X, ξ ∈ Ω . Hence, we denote by f : R × X → X the function defined by f ( t, x ) = G ( x ) + H ( t ) , t ∈ R , x ∈ X. (4.4)Then, equation (4.1) is equivalent to the following abstract evolution equation: x ′ ( t ) = Ax ( t ) + f ( t, x ( t )) for t ∈ R . (4.5)The corresponding initial value problem is the following: (cid:26) x ′ ( t ) = Ax ( t ) + f ( t, x ( t )) for t ≥ x (0) = x , (4.6)Now, under the above assumptions we prove the existence of a unique bounded global solution toequation (4.6) under the weak assumption that H (and hence f ) is not bounded on t , see Remark4.1. 16 heorem 4.1 For each x ∈ X there exist t ( x ) > x ( · , x )of equation (4.6) defined on [0 , t ( x )) such that t ( x ) = + ∞ or lim sup t → t ( x ) k x ( t, x ) k = + ∞ . (4.7) Proof.
Fix R > t := log(2 M ( L R + λ ˜ M + λ ) λ ) λ − > M := e [ t ]+1 ([ t ] +2) k f ( · , k BS < ∞ .E = { u ∈ C ([0 , t ] , X ) : u (0) = x and k u ( t ) k ≤ R for all t ∈ [0 , t ] } ⊂ C ([0 , t ] , X ) . The set E equipped with the uniform norm topology is complete since it is closed in the Banachspace ( C ([0 , t ] , X ) , k · k ∞ ). Define the functional Φ : E −→ C ([0 , t ] , X ) byΦ( u )( t ) = T ( t ) x + Z t T ( t − s ) f ( s, u ( s )) ds for t ∈ [0 , t ] . Let u ∈ E , we show that Φ( u ) ∈ E . In fact, k Φ( u )( t ) k ≤ M e − λ t k x k + M Z t e − λ ( t − s ) k f ( s, u ( s )) k ds ≤ M e − λ t R + M Z t e − λ ( t − s ) [ k f ( s, u ( s )) − f ( s, k + k f ( s, k ] ds ≤ M e − λ t R + M e − λ t Z t e λ s L R k u ( s ) k ds + M e − λ t [ t ]+1 X k =0 Z k +1 k e λ s k f ( s, k ds ≤ e − λ t M (cid:18) L R λ + ˜ M (cid:19) R = R ≤ R for all t ∈ R . Then Φ( E ) ⊂ E . Now, let u, v ∈ E , then we have k Φ( u )( t ) − Φ( v )( t ) k ≤ M Z t e − λ ( t − s ) k f ( s, u ( s )) − f ( s, v ( s )) k ds ≤ e − λ t M L R λ ≤ e − λ t M (cid:18) L R λ + ˜ M + 1 (cid:19) k u − v k ∞ = k u − v k ∞ t ∈ R . Hence k Φ( u ) − Φ( v ) k ∞ ≤ k u − v k ∞ x to equation (4.6) defined on [0 , t ]. Let t ε > t be well chosen andconsider equation (4.6) for t ∈ [ t , t ε ] with initial data x t = x ( t ). Arguing as above, we canprove the existence of a unique mild solution x ε : [ t , t ε ] −→ X with x ε ( t ) = x t . Then, we define x = (cid:26) x in [0 , t ] ,x ε in [ t , t ε ] , By construction x : [0 , t ε ] −→ X is continuous. Therefore x is the unique solution of equation(4.6) in [0 , t ε ]. Hence, we adopt the same extension technique and we prove the existence ofa unique maximal solution x : [0 , t ( x )) −→ X where t ( x ) := sup { t > x ∈ C ([0 , t ] , X ) solution to (4.6) } . Now, we show formula (4.1). By contradiction, assume that t ( x ) < ∞ and there exists C ≥ k x ( t ) k ≤ C for all t ∈ [0 , t ( x )). We show that x is uniformlycontinuous on [0 , t ( x )). Let ( t n ) n , ( s n ) n ∈ [0 , t ( x )) be two sequences such that σ n = t n − s n → n → ∞ . Without loss of generality we assume that 0 ≤ σ n ≤ n ≥
0. Hence, we have k T ( t n ) x − T ( s n ) x k ≤ M k T ( σ n ) x − x k → n → ∞ (4.8)follows from the strong continuity of the semigroup ( T ( t )) t ≥ on X . Moreover, we have Z t n T ( t n − s ) f ( s, x ( s )) ds − Z s n T ( s n − s ) f ( s, x ( s )) ds = Z σ n T ( t n − s ) f ( s, x ( s )) ds − Z s n T ( s n − s ) [ f ( s, x ( s )) − f ( s − σ n , x ( s − σ n )] ds g it holds that f is also locally Lipschitzian on X and it iscontinuous on ∈ [0 , t ( x )] which implies thatsup ( t,x ) ∈ [0 ,t ( x )] × ¯ B (0 ,C ) k f ( t, x ) k = ˜ C < ∞ . Hence, k T ( t n − s ) f ( s, x ( s )) k ≤ M e λ t ( x ) ˜ C for all s ∈ [0 , t ( x )) . Then, by dominated convergence Theorem we claim that Z σ n T ( t n − s ) f ( s, x ( s )) ds → n → ∞ .Furthermore, k Z s n T ( s n − s ) [ f ( s, x ( s )) − f ( s − σ n , x ( s − σ n ))] ds k ≤ M λ Z t ( x )0 [ k f ( s, x ( s )) − f ( s, x ( s − σ n )) k + k f ( s, x ( s − σ n )) − f ( s − σ n , x ( s − σ n )) k ] ds ≤ M λ L C Z t ( x )0 k x ( s ) − x ( s − σ n ) k ds + M λ Z t ( x )0 k f ( s, x ( s − σ n )) − f ( s − σ n , x ( s − σ n )) k ds where M λ = M e λ t ( x ) . By the continuity of x , it follows using dominated convergence Theoremthat Z t ( x )0 k x ( s ) − x ( s − σ n ) k ds ; Z t ( x )0 k f ( s, x ( s − σ n )) − f ( s − σ n , x ( s − σ n )) k ds → n → ∞ . Thus k Z t n T ( t n − s ) f ( s, x ( s )) ds − Z s n T ( s n − s ) f ( s, x ( s )) ds k → n → ∞ . (4.9)Consequently, by (4.8) and (4.9) we claim that k x ( t n ) − x ( s n ) k → n → ∞ . Then, x is uniformly continuous on [0 , t ( x )) which implies that lim t → t ( x ) − x ( t ) = x t ( x ) ∈ X. There-fore, there exists η > y is solution to equation (4.6) in [ T ( x ) , T ( x ) + η ] with initialdata y ( T ( x )) = x t ( x ) . Hence z = (cid:26) x in [0 , t ( x )] ,y in [ t ( x ) , t ( x ) + η ] , defines a solution of equation (4.6) in [0 , t ( x ) + η ] which contradicts the fact that x is maximal.Consequently, formula (4.1) is satisfied. (cid:4) Lemma 4.1 [17, Theorem 8.3.7] Assume H = 0. Then, for each x ∈ X the following hold (i) Equation (4.6) has a unique global solution u x given by: u x ( t ) = T ( t ) x + Z t T ( t − s ) G ( x ( s )) ds, t ≥ . (ii) The solution u x is bounded on R + i.e., sup t ≥ k u x ( t ) k < ∞ . Theorem 4.2
Let p ∈ [1 , ∞ ). Then for each x ∈ X equation (4.6) has a unique global boundedmild solution x given by x ( t ) = T ( t ) x + Z t T ( t − s ) G ( x ( s )) ds + Z t T ( t − s ) H ( s ) ds, t ≥ . Proof.
Fix x ∈ X and let x be the unique maximal solution to (4.6) defined on [0 , t ( x )).Moreover, using Theorem 4.1 x satisfies formula (4.1). Note that a mild solution of equation (4.6)is given by x ( t ) = u x ( t ) + Z t T ( t − s ) H ( s ) ds, t ∈ [0 , t ( x )) . x is global. In fact, by Lemma 4.1- (i) , it holds that sup t ∈ [0 ,t ( x )) k u x ( t ) k < ∞ ,if it is not, t ( x ) < ∞ which contradicts the fact that u x is global. Hence k x ( t ) k ≤ k u x ( t ) k + Z t k T ( t − s ) H ( s ) k ds ≤ k u x ( t ) k + M Z t e − λ ( t − s ) k H ( s ) k ds ≤ k u x ( t ) k + M [ t ]+1 X k =0 Z k +1 k e − λ ( t − s ) k H ( s ) k ds ≤ sup t ∈ [0 ,t ( x )) k u x ( t ) k + M e λ e λ − k H k BS p , t ∈ [0 , t ( x )) . (4.10)Then sup t ∈ [0 ,t ( x )) k x ( t ) k < ∞ . Consequently, by (4.1), we claim that t ( x )) = ∞ . Thus x is global.To conclude, we show that x is bounded on R + . Using Lemma 4.1- (ii) we havesup t ≥ k u x ( t ) k < ∞ . Therefore, by (4.10) we can deduce that sup t ≥ k x ( t ) k < ∞ . (cid:4) In order to prove the existence of a compact almost automorphic solution for equation (4.1), weneed the following results.
Lemma 4.2
The function f satisfies Hypothesis (H4) with φ = 1 and ψ = H . That is f ∈ AAS U ( R × X, X ). Proof.
Let R ≥ B ⊂ X bounded by R . By (4.3), we have k G ( x ) k ≤ L R k x k + k G (0) k≤ L R .R for all x ∈ B. This proves the fact for G . To conclude, it suffices to prove that H ∈ AAS ( R , X ). In fact, let( t ′ n ) n ⊂ R be a sequence. Since a ∈ AAS ( R , R ) (see [25]) and b : t sin (cid:16) cost + cos √ t (cid:17) ∈ AAS ( R , R ), it follows that there exist ( t n ) n ⊂ ( t ′ n ) n and ˜ a, ˜ b ∈ L ∞ ( R , R ) such that, for each t ∈ R Z t +1 t | a ( s + t n ) − ˜ a ( s ) | ds → , Z t +1 t | b ( s + t n ) − ˜ b ( s ) | ds → n → ∞ . (4.11)and Z t +1 t | ˜ a ( s − t n ) − a ( s ) | ds → , Z t +1 t | ˜ b ( s − t n ) − b ( s ) | ds → n → ∞ . (4.12)Define ˜ H ( t )( ξ ) = ˜ b ( t ) h ( ξ ) + ˜ a ( t ) for t ∈ R and ξ ∈ Ω. Hence, by (4.11), we claim that for each t ∈ R Z t +1 t k H ( s + t n ) − ˜ H ( s ) k ds = k h k Z t +1 t | a ( s + t n ) − ˜ a ( s ) | ds + Z t +1 t | b ( s + t n ) − ˜ b ( s ) | ds → n → ∞ . By the same way, we prove that for each t ∈ R Z t +1 t k ˜ H ( s − t n ) − H ( s ) k ds → n → ∞ . By Remark (3.2)- (a) , we conclude that f ∈ AAS U ( R × X, X ).19
For the uniqueness we need the following additional assumption on g :The function r g ( r ) − r is nonincreasing on R . (4.13)Consider the function E : X → R defined by E ( x ) = 12 Z Ω | x ( ξ ) | dξ. (4.14) Lemma 4.3 [9] Let u and v be two mild solutions of equation (4.5). Then the following hold. (i) The function t E ( u ( t ) − v ( t )) is nonincreasing on R . (ii) If t E ( u ( t ) − v ( t )) is constant on R , then there exists w ∈ X such that u ( t ) − v ( t ) = w , t ∈ R , (4.15) G ( u ( t )) − G ( v ( t )) = − λ w , t ∈ R . (4.16)Moreover, for all θ ∈ [0 , θu + (1 − θ ) v is also a mild solution of equation (4.5). Theorem 4.3
Under the above assumptions. The following statements hold for equation (4.5): (i)
Equation (4.5) has a at least one mild solution x which is compact almost automorphic . If x is a mild solution which is only almost automorphic, then there exists w ∈ X such that x ( t ) = x ( t ) + w for t ∈ R . (4.17)Consequently x is compact almost automorphic. (ii) If the function r → g ( r ) − r is strictly decreasing on R , then the compact almost mild solutionis unique Proof.
We use Theorem 3.3 To prove the existence of a compact almost automorphic solution forequation (4.5). By Lemma 4.2 Hypotheses (H1) , (H2) and (H4) are satisfied. In addition, fromProposition 4.2, equation (4.6) has at least a mild solution x defined and bounded on [0 , + ∞ ).Let K := { x ( t ) : t ≥ } , by Lemma 3.8, K is compact in X . Let co ( K ) be the convex hull ofthe compact set K . Then co ( K ) is a compact convex subset of X . It is clear that K ⊂ co ( K ).Define the subvariant functional λ co ( K ) : F co ( K ) −→ [0 , ∞ ) by λ co ( K ) ( x ) = sup t ∈ R E ( x ( t ))associated to the compact set co ( K ). We prove that equation (4.5) has at most a minimal co ( K )-valued solution. Let u and v be two minimal co ( K )-valued solutions of equation (4.5). Defineinf x ∈F co ( K ) λ co ( K ) ( x ) := δ = sup t ∈ R E ( u ( t )) = sup t ∈ R E ( v ( t )) . (4.18) Case I.
Assume that E ( u ( t ) − v ( t ))) = c ∈ R + for all t ∈ R . By the convexity of co ( K ) andLemma 4.3, the set F co ( K ) is convex too. Then 12 u + 12 v ∈ F co ( K ) . Therefore δ ≤ sup t ∈ R E (cid:18) u ( t ) + 12 v ( t ) (cid:19) . (4.19)On the other hand, by parallelogram inequality, we have E (cid:18) u ( t ) + 12 v ( t ) (cid:19) + E (cid:18) u ( t ) − v ( t ) (cid:19) = 12 E ( u ( t )) + 12 E ( v ( t )) , Then sup t ∈ R E (cid:18) u ( t ) + 12 v ( t ) (cid:19) + 14 c ≤
12 sup t ∈ R E ( u ( t )) + 12 sup t ∈ R E ( v ( t )) . (4.20)Consequently, δ + 14 c ≤ δ + 12 δ c ≤
0. Hence, by definition of E , we obtain u = v . Case II : Let ( t n ) n be a sequence of real numbers such that lim n → + ∞ t n = −∞ . Since by Lemma 4.2, f ∈ AAS U ( R × X, X ). Then, by applying Theorem 3.2 two times for u and v respectively, weclaim that there exists a subsequence of ( t n ) n such that u ( t + t n ) → u ( t ) as n → + ∞ , (4.21) u ( t − t n ) → u ( t ) as n → + ∞ , (4.22) v ( t + t n ) → v ( t ) as n → + ∞ , (4.23) v ( t − t n ) → v ( t ) as n → + ∞ , (4.24)uniformly on each compact subset of R , where u and v are two minimal co ( K )-valued solutionsof equation (4.5). Using (4.21)-(4.24), we claim that for each t ∈ R , we havelim m → + ∞ lim n → + ∞ E ( u ( t + t n − t m ) − v ( t + t n − t m )) = E ( u ( t ) − v ( t )) . Moreover, since by Lemma 4.3, the function t → E ( u ( t ) − v ( t )) is nonincreasing on R and lim n → + ∞ t n = −∞ , it follow that for each t ∈ R lim n → + ∞ E ( u ( t + t n ) − v ( t + t n )) = sup τ ∈ R E ( u ( τ ) − v ( τ )) . Therefore, E ( u ( t ) − v ( t )) = sup τ ∈ R E ( u ( τ ) − v ( τ )) . (4.25)By (4.25) and Case I , we obtain that u = v . By definition of E , we deduce thatsup τ ∈ R E ( u ( τ ) − v ( τ )) = 0 , Consequently u = v and then equation (4.5) has at most one minimal co ( K )-valued solution. (i) Using Theorem 3.3, we claim that equation (4.5) has at least a compact almost automorphicsolution. Let x be a mild solution which is compact almost automorphic. If x is mild solutionwhich is almost automorphic, then by (16) and Lemma 4.3 respectively, the function t → E ( x ( t ) − x ( t )) is almost automorphic and nonincreasing. Therefore it is constant on R . Consequently, inview of Lemma 4.3 -(ii) , we obtain (4.17). (ii) If the function r → g ( r ) − r is strictly decreasing on R , then the uniqueness of the compactalmost automorphic solution results directly from (4.15) and (4.16). (cid:4) EFERENCES
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Brahim Es-Sebbar,
Cadi Ayyad University, Faculty of Sciences and Technology Gueliz, Marrakech,Morocco
E-mail address , B. Es-sebbar: [email protected]
Khalil Ezzinbi,
Department of Mathematics, Faculty of Sciences Semlalia, Cadi Ayyad University,Marrakesh B.P. 2390-40000, Morocco.
E-mail address , K. Ezzinbi: [email protected]
Kamal Khalil,
Department of Mathematics, Faculty of Sciences Semlalia, Cadi Ayyad University,Marrakesh B.P. 2390-40000, Morocco.
E-mail address , K. Khalil (Corresponding author): [email protected]
April 9, 2020