NON-AUTONOMOUS REACTION-DIFFUSION EQUATIONS WITH VARIABLE EXPONENTS AND LARGE DIFFUSION

. In this work we prove continuity of solutions with respect to initial conditions and a couple of parameters and we prove upper semicontinuity of a family of pullback attractors for the problem τs , under homogeneous Neumann boundary conditions, u τs ∈ H := L 2 (Ω) , Ω ⊂ R n ( n ≥ 1) is a smooth bounded domain, B : H → H is a globally Lipschitz map with Lipschitz constant L ≥ 0, D s ∈ [1 , ∞ ), C ( · ) ∈ L ∞ ([ τ,T ]; R + ) is bounded from above and below and is monotonically nonincreasing in time, p s ( · ) ∈ C (¯Ω), p − s := min x ∈ ¯Ω p s ( x ) ≥ p, p + s := max x ∈ ¯Ω p s ( x ) ≤ a, for all s ∈ N , when p s ( · ) → p in L ∞ (Ω) and D s → ∞ as s → ∞ , with a,p > 2

(1) Non-autonomous evolution equations with spatially variable exponents and timedependent operator had been considered by Kloeden-Simsen [13]. They considered the diffusion coefficient depending on time and monotonically nonincreasing in time in order to guarantee existence and uniqueness of a strong global solution by using an abstract result of Yotsutani [20]. Here, we also will consider a time-dependent operator, but we will consider the time-dependent term in the perturbation part and we will consider the diffusion coefficient uniform in time, i.e., the diffusion coefficient will be a constant which does not depend on time.
We will prove continuity of the flows and upper semicontinuity of the family of pullback attractors {U s } s∈N as s goes to infinity for the problem (1) with respect to the couple of parameters (D s , p s ), where p s is the variable exponent and D s is the diffusion coefficient.
The autonomous version with C(·) ≡ 1 was considered in [18] where the authors proved the upper semicontinuity of the global attractors. The main different point in this work is that here we investigate the pullback asymptotic dynamics of problem (1) with respect to the variable exponents p s and the diffusion coefficients D s when the couple of parameters changes at the same time. The main novelty appears in the construction of a bounded global solution for the evolution process.
The paper is organized as follows. In Section 2 we present properties on the operator and guarantee the existence of global solutions. In Section 3 we obtain uniform estimates for solutions of (1). Existence of the pullback attractor is then obtained in Section 4. In Section 5 we prove that the solutions {u s } of the PDE (1) go to the solution u of the limit problem which will be an ODE, and, after that, we obtain in Section 6 the upper semicontinuity of the family of pullback attractors for the problem (1). Finally, in Section 7 we will present some examples with numerical simulation for the limit problem.
2. The properties of the operator and existence of solution. In this section we will not vary s, so, for simplicity, we will just write p s (·) = p(·) and D s = D. We present the operator, prove its properties and we assure existence of a unique global solution for the problem (1).
Let Ω ⊂ R n , n ≥ 1, be a bounded smooth domain, H := L 2 (Ω) and X := W 1,p(·) (Ω) with p − > 2. Then X ⊂ H ⊂ X * with continuous and dense embeddings. We refer the reader to [8,9,10,11] to see properties of the Lebesgue and Sobolev spaces with variable exponents. In particular, with for u ∈ L p(·) (Ω). The generalized Sobolev space is defined as From [11] we know that W 1,p(·) (Ω) is a Banach Space with the norm Consider the operator A(t) defined in X such that for each u ∈ X associate the following element of its dual space X * , A(t)u : X → R given by We will show that the operator A(t) : X → X * is monotone, hemicontinuous and coercive. For this purpose we need first to prove some estimates on the operator and the following two lemmas will be very useful.
Proof. Let u, v ∈ X. For each x ∈ Ω fixed, we have for all j ≥ j o . As lim j→∞ min{1, α} Lemma 2.6. For each t ≥ 0, the operator A(t) : X → X * is hemicontinuous.
Proof. We have to show that A(t)(u + λv) A(t)u as λ → 0 for all u, v ∈ X. As p − > 1, X is a reflexive Banach Space (see [9,11]), so we just have to show that for all φ ∈ X. Let u, v, φ ∈ X and λ ∈ (−1, 1). Define Thus, Note that with g(·) ∈ L 1 (Ω). We obtain by the Dominated Convergence Theorem that Thus, as a consequence of Lemmas 2.4, 2.5 and 2.6 we have that the realization operator A H (t) : is a maximal monotone operator in H. We will show that A H (t) is the subdifferential of the following convex, proper and lower semicontinuous map ϕ t p(·) : H → R given by Lemma 2.7. The map ϕ t p(·) is convex and proper.
Proof. Let u ∈ X. Then, u ∈ L p(·) (Ω) and ∇u ∈ L p(·) (Ω). So, Therefore ϕ t p(·) is proper. Since the application γ p is convex for γ > 0 given, u, v ∈ X and 0 ≤ λ ≤ 1 we have Lemma 2.8. The map ϕ t p(·) is lower semicontinuous. Proof. Let (u n ) be a sequence such that u n → u in H. We have to show that On the other hand, if lim inf n→∞ ϕ p(·) (u n ) = a < +∞ then there is a subsequence Then, Similarly we have that ρ(∇u nj ) ≤ p + δ. (7) Using (6), (7) and the Lemma 2.1 we obtain Therefore, ||u nj || X is a bounded sequence in the reflexive Banach space X. So, (u nj ) has a subsequence (which we still denote by (u nj )) such that u nj v in X for some v ∈ X. As H * ⊂ X * we have u nj v in H and by the uniqueness of the weak limit u = v ∈ X.
We observe that the function ψ t : X → X * , defined by ψ t (w) := ϕ t p(·) (w) for any w ∈ X, is Gateaux differentiable. So, it follows from Example 1, p. 54 in [2], that u ∈ D(∂ψ t ) and ∂ψ t (u) consists of a single element, namely the Gateaux differential of ψ t at u, i.e., ∇ u ψ t (u) = ∂ψ t (u). As u nj u in X and ∅ = ∂ψ t (u) ∈ X * we obtain Let us consider now the subdifferential ∂ϕ t p(·) which is the restriction of ∂ψ t to the set K t := {u ∈ X : A(t)u ∈ H}.
and ∂ϕ t p(·) are both maximal monotone operators in H, so it is enough to show that for any u ∈ H, for all ξ ∈ X. If ξ ∈ H − X, then ϕ t p(·) (ξ) = ∞ and consequently (8) holds.
Observe that if we would consider the nonlinear absorption term with a different variable exponent from the ones in the diffusion term, we could not define a maximal monotone operator of the subdifferential type.
2.1. Existence of solution. We will show the existence of a unique global solution by using [20]. Consider the problem Assumption A. Let T > τ be fixed. A.1 : There is a set τ / ∈ Z ⊂ [τ, T ] of zero measure such that ϕ t is a lower semicontinuous proper convex function from H into (−∞, ∞] with a non-empty effective domain for each t ∈ [τ, T ] − Z; A.2 : For any positive integer r there exist a constant K r > 0, an absolutely continuous function g r : [τ, T ] → R with g r ∈ L β (τ, T ) and a function of bounded variation and where α is some fixed constant 0 ≤ α ≤ 1 and Theorem 2.10. [20] Suppose that Assumption (A) is satisfied. Then, for each f ∈ L 2 (τ, T ; H) and u 0 ∈ D(ϕ τ ) the equation (9) has a unique strong solution u on Using the monotonicity of the operator and the Gronwall Lemma we obtain the following Lemma 2.11. If f, g ∈ L 2 (τ, T ; H) and u, v are the solutions of the equations The proof of the following theorem is analogous to the proof of Theorem 1 in [16]. Let us consider now the problem with our specific operator Proof. Taking Z as the empty set, ϕ t p(·) is lower semicontinuous proper convex function for each t ∈ [τ, T ]. Consider r a positive integer, K r := r and α := 1 2 .
Define g r : [τ, T ] → R with g r (t) := t + r, and h r (t) := r. We have that g r is an absolutely continuous function g r = 1 ∈ L 2 (τ ; T ) and h r is a bounded variation function. For all t ∈ [τ, T ], w ∈ D(ϕ t p(·) ) = X with w ≤ r and s ∈ [t, T ]. Consider the elementw := w ∈ X = D(ϕ s p(·) ). We will check thatw satisfies (10) and (11). Note that . Therefore, we obtain from Theorem 2.10 the existence of a global solution to the problem (12).
Theorem 2.14. The problem (1) has a unique global strong solution.
3. Uniform estimates. From now on we denote X s := W 1,ps(·) (Ω) and Y := W 1,p (Ω). It is a known result that X s ⊂ H with continuous and dense embeddings (see [8,17]). Moreover, it is easy to see that u s H ≤ 4(|Ω| + 1) 2 u s Xs , for all u s ∈ X s and for all s ∈ N.
Lemma 3.1. Let u s be a solution of (1) with u s (τ ) = u τ s ∈ H. Given T 1 > 0, there exists a positive number r 0 = r 0 (T 1 ) such that u s (t) H ≤ r 0 , for each t ≥ T 1 + τ and s ∈ N. Furthermore, given a bounded set B ⊂ H, there existsD 1 > 0 such that u s (t) H ≤D 1 , for all t ≥ τ and s ∈ N such that u τ s ∈ B.
and the first part of the lemma is proved. The second part of the lemma follows from the Gronwall-Bellman Lemma.
Lemma 3.2. Let u s be a solution of (1). Given T 2 > 0, there exists a positive constant r 1 > 0, independent of s, such that for every t ≥ T 2 + τ and s ∈ N.
As a consequence of Lemma 3.2, we have Corollary 2. a) Let u s be a solution of problem (1). Given T 2 > 0 there exists a positive constant r 2 , independent of s, such that for all t ≥ T 2 + τ and s ∈ N. b) There exist bounded sets B 1 in Y and B s 1 in X s such that , for each t ∈ R, Proof. a) By Lemma 3.2 there exists r 1 > 0 such that Thus, for all t ≥ T 2 + τ and s ∈ N and the result follows with r 2 := 2(|Ω| + 1)r 1 . b) It follows from Lemma 3.2 and the previous item a). c) By b) there exists a bounded set B 1 in Y such that A s ⊂ B 1 for all s ∈ N. Since Y ⊂ H with continuous and compact embedding, the result is proved.

5.
The limit problem and convergence properties. Our objective in this section is to prove that the limit problem of problem (1) as D s increases to infinity and p s (·) → p > 2 in L ∞ (Ω) as s → ∞ is described by a non-autonomous ordinary differential equation. Firstly we observe that the gradient of the solutions u s of problem (1) converge in norm to zero as s → ∞, which allows us to guess the limit problem du dt (t) + C(t)|u(t)| p−2 u(t) =B(u(t)), t > τ, withB := B |R if we identify R with the constant functions which are in H, since Ω is a bounded set. The proof of the next result is analogous as in [18] and we will not present much details in the proof here since the non-autonomous term C(t) did not presented difficulties for this result. Now we present our main result. The novelty here is that it is necessary to construct a backwards-bounded global solution for the evolution process whereas in the autonomous case it was necessary to construct a bounded complete trajectory for a semigroup. Proof. Let τ ∈ R be fixed. Consider an arbitrary sequence {w s } s∈N with w s ∈ A s (τ ) for each s ∈ N. In order to obtain lim s→+∞ dist(A s (τ ), A ∞ (τ )) = 0 it is sufficient to show that {w s } s∈N has a convergent subsequence whose limit belongs to A ∞ (τ ) (see Lemma 3.2 in [4]). We know from Corollary 2 (c) that Our goal is to show that w ∈ A ∞ (τ ). By the characterization given in Theorem 1.17 in [4] it is sufficient to show that there exists a backwards-bounded global solution ξ(·) with w = ξ(τ ). Let {S sj (t, τ )} t≥τ and {S ∞ (t, τ )} t≥τ be the evolution process associated with (1) and (19), respectively.
Consider τ j −∞. For simplicity, let us call {S j (t, τ )} t≥τ = {S sj (t, τ )} t≥τ , w 0 := w and w j := w sj the subsequence of {w s } such that w j → w 0 ∈ R as j → +∞. Without loss of generality, we can consider τ j < τ, ∀j ∈ N. By the invariance of the pullback attractors, Thus, w j = S j (τ, τ j )x j , for some x j ∈ A j (τ j ). From Theorem 5.3, Also, there exists x j,1 ∈ A j (τ j − 1) such that x j = S j (τ j , τ j − 1)x j,1 . Now consider the subsequence where K(H) := {K ⊂ H : K is a not vanish compact subset}. Then, up to a subsequence, we have From Lemma 6.2, we have w 1 ∈ R. By Theorem 5.3, Indeed, Again, by the invariance of the pullback attractors, there exists x j,2 ∈ A j (τ j − 2) such that Then, (up to a subsequence) we have as j → ∞. Indeed, Proceeding inductively we define
Let us show now that ξ is bounded (in particular backwards-bounded): First, note that for each t ∈ R, ξ(t) = S ∞ (t, τ − r)w r for some r ∈ {0, 1, 2, 3, . . .} and Using Lemma 6.2, we conclude that each term S ∞ (t, τ − r)w r is independent of x. Consequently, ξ(t) is a constant function on the the spatial variable x. Since for each t ∈ R and j ∈ N, and the last one is bounded (because is relatively compact) there is a constant C > 0 such that Thus, in particular, we have that ξ(·) is bounded in H. Then, there is a constant C > 0 such that So, we conclude that ξ : R → R is a bounded global solution with w = w 0 = ξ(τ ). 7. Some numerical simulation for the limit problem. In this section we made some numerical experiments with specific functions C(·) and B(·). These simulations show some information about the behavior of the solutions of the ordinary differential equation (19) which commands the pullback asymptotic dynamics of the family of partial differential equations (1) as s → +∞. By fixing the value of the constant p > 2 there is no great loss of information as we will see below (compare figures 4 and 5). We consider as examples the ordinary differential equation In our first example the theoretical hypotheses on C(·) will be satisfied with α = 0.1 and M = 1.1. Considering p = 4 and a monotonically nonincreasing function C(t) = 1.1, t < 0 e −t + 0.1, 0 ≤ t we have the following explicit solution x(t) = ± 10 √ 110+100K e −2 t , t < 0 ± 10 √ 10−100 e −2 t +200 e −t +100K e −2 t , 0 ≤ t where K is a constant to be determined by the initial conditions as well the sign ± to be used. These solutions are depicted in the next picture. For the existence of a strong solution of the ODE limit problem the coefficient C(·) does not need to be nonincreasing (see Remark 1). So, we also present below some examples with C(·) not monotone, only nonincreasing in some intervals. In all the next examples we have 0.1 ≤ C(t) ≤ 2.1.
Considering p = 4 and C(t) = e −t 2 + 0.1 we have the following source type solution, x(t) = ± 10 10 + 100 √ πERF (t − 1)e −2 t+1 + 100K e −2 t where is the error function. K is a constant to be determined by the initial conditions as well the sign ±. These solutions are depicted in the following figure.
With p = 4 and C(t) = cos(t) + 1.1 we have the following explicit solution, x(t) = ± 10 110 + 80 cos (t) + 40 sin (t) + 100K e −2 t where K is a constant to be determined by the initial conditions as well the sign ±. These solutions are depicted in the following figure.      Interpretation of the figures: In all the examples the solutions seems to approach to two orbits, more precisely, there are two orbits x * (·), y * (·) such that: (1) For each t ∈ R, x * (t), y * (t) ∈ A ∞ (t), x * (t) ≤ y * (t) and A ∞ (t) ⊆ [x * (t), y * (t)] := {r ∈ R; x * (t) ≤ r ≤ y * (t)}.