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Attractors for a nonautonomous reaction-diffusion equation with delay


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Introduction and statement of the problem

We consider the following nonautonomous functional reaction-diffusion equation

tu(t,x)Δu(t,x)=f(u(t,x))+b(t,ut)(x)+g(t,x)in(τ,)×Ω,u=0on(τ,)×Ω,u(τ,x)=u0(x),τRandxΩ,u(τ+θ,x)=φ(θ,x),θ[r,0]andxΩ,$$\begin{array}{} \displaystyle \left\{ \begin{array}{l} \frac{\partial}{\partial t} u(t,x) - \Delta u(t,x) = f(u(t,x)) + b(t,u_t)(x) + g(t,x)\; \mbox{in}\; (\tau, \infty) \times \Omega\,,\\ u= 0 \;\mbox{on}\; (\tau, \infty) \times \partial{\Omega}\,,\\ u(\tau,x)=u^0(x),\; \tau \in \mathbb{R}\; \mbox{and} \;x \in \Omega \,,\\ u(\tau+\theta,x)= \varphi(\theta,x),\; \theta \in [-r,0] \;\mbox{and}\; x \in \Omega\,, \end{array} \right. \end{array}$$

where Ω ⊂ ℝ is a bounded domain with smooth boundary Ω, τ ∈ ℝ, u0L2(Ω) is the initial condition in τ and φL2([–r,0];L2(Ω)) is also the initial condition in [τr,τ],r > 0 is the length of the delay effect. For the rest we assume following assumptions conditions:

H1) Concerning the nonlinearity, we assume that fC1(ℝ, ℝ) there exist positive constants c, μ0, μ1, k and p > 2 N2pp2$\begin{array}{} \frac{2p}{p-2} \end{array}$ such that

cμ0|u|pf(u)ucμ1|u|puR,$$\begin{array}{} \displaystyle -c - \mu_0 \vert u \vert^p \leq f(u)u\leq c - \mu_1 \vert u \vert^p \; \forall u \in \mathbb{R}, \end{array}$$

f(u)f(v)(uv)k(uv)2u,vR.$$\begin{array}{} \displaystyle \left(f(u)-f(v)\right)(u-v) \leq k (u-v)^2 \; \forall u, v \in \mathbb{R}. \end{array} $$

Let us denote by

F(u):=0uf(s)ds.$$\begin{array}{} \displaystyle F(u) := \int_{0}^{u} f(s) ds\,. \end{array}$$

From (1.2), there exist positive constants l,cμ0,μ1$\begin{array}{} \mu'_0, \mu'_1 \end{array}$ such that

|f(u)|l|u|p1+1uR,$$\begin{array}{} \displaystyle \vert f(u) \vert \leq l\left(\vert u \vert^{p-1} + 1\right)\; \forall u\in \mathbb{R}, \end{array}$$

cμ0|u|pF(u)cμ1|u|puR.$$\begin{array}{} \displaystyle -c' - \mu'_0 \vert u \vert^p \leq F(u)\leq c' - \mu'_1 \vert u \vert^p \; \forall u \in \mathbb{R}. \end{array}$$

H2) The operator b : ℝ × L2([–r,0];L2(Ω)) → L2(Ω) is a time-dependent external force with delay, such that

For all ϕL2([–r,0]; L2(Ω)) the function ℝ∋ tb(t,ϕ) ∈ L2(Ω) is measurable;

b(t,0) = 0 for all t ∈ ℝ;

Lb > 0 s.t ∀ t ∈ ℝ and ∀ ϕ1, ϕ2L2([–r,0];L2(Ω));

b(t,ϕ1)b(t,ϕ2)Lbϕ1ϕ2L2([r,0];L2(Ω));$$\begin{array}{} \displaystyle \Vert b(t,\phi_1)-b(t,\phi_2)\Vert \leq L_b \Vert \phi_1-\phi_2 \Vert_{L^2([-r,0];L^2(\Omega))}\,; \end{array}$$

Cb > 0 s.t ∀ tτ, and ∀ u, vL2([τ-r, t]; L2(Ω));

τtb(s,us)b(s,vs)2dsCbτrtu(s)v(s)2ds.$$\begin{array}{} \displaystyle \int_{\tau}^{t}\Vert b(s,u_s)-b(s,v_s)\Vert^2 ds \leq C_b \int_{\tau-r}^{t}\Vert u(s)-v(s) \Vert^2 ds\,. \end{array}$$

Remark 1

From (I)-(III), for T > τ and uL2([τr,T];L2(Ω)) the function ℝ∋ tb(t,ϕ) ∈ L2(Ω) is measurable and belongs to L((τ,T);L2(Ω)).

H3) The function gLloc2$\begin{array}{} L^2_{loc} \end{array}$ (ℝ; L2(Ω)) is an another nondelayed time-dependent external force.

For more details on differential equations with delay, we refer the reader to J. Wu [9] and J.K. Hale [5]. The purpose of this paper is to discuss the existence of pullback 𝒟-attractor in L2(Ω)× L2([–r,0];L2(Ω)) by using a priori estimates of solutions to the problem (1.1).

This work is motivated by the work of T. Caraballo and J. Real. [1], where they proved the existence of pullback attractors for the following 2D-Navier Stokes model with delays:

utνΔu+i=12uiuxi=fp+g(t,ut)in(τ,)×Ω,divu=0in(τ,)×Ω,u=0on(τ,)×Ω,u(τ,x)=u0(x),xΩ,u(t,x)=ϕ(tτ,x),t(τh,τ)andxΩ,$$\begin{array}{} \displaystyle \left\{ \begin{array}{l} \frac{\partial u}{\partial t} - \nu\Delta u + \sum_{i=1}^{2} u_i \frac{\partial u}{\partial x_i}= f - \nabla p + g(t,u_t)\; \mbox{in}\; (\tau, \infty) \times \Omega\,,\\ div\; u= 0 \;\mbox{in}\; (\tau, \infty) \times \Omega\,,\\ u= 0 \;\mbox{on}\; (\tau, \infty) \times \partial\,{\Omega}\,,\\ u(\tau,x)=u_0(x), \;x \in \Omega \,,\\ u(t,x)= \phi(t-\tau,x),\; t \in (\tau-h,\tau) \;\mbox{and}\; x \in \Omega\,, \end{array} \right. \end{array}$$

where ν > 0 is the kinematic viscosity, u is the velocity field of the fluid, p the pressure, τ ∈ ℝ the initial time, u0 the initial velocity field, f a nondelayed external force field, g another external force with delay and ϕ the initial condition in (–h,0), where h is a fixed positive number.

On the other hand, the problem (1.1) without critical nonlinearity was treated by J. Li and J. Huang in [6], where they proved the existence of uniform attractor for the following non-autonomous parabolic equation with delays:

u(t,x)t+Au(t,x)+bu(t,x)=F(ut)(x)+g(t,x)xinΩ,u(τ,x)=u0(x),u(τ+θ,x)=ϕ(θ,x),θ(r,0).$$\begin{array}{} \displaystyle \left\{ \begin{array}{l} \frac{\partial u(t,x)}{\partial t} + A u(t,x) + b u(t,x) = F(u_t)(x) + g(t,x)\; x\;\mbox{in}\; \Omega\,,\\ u(\tau,x)=u_0(x),\; u(\tau+\theta,x)= \phi(\theta,x),\; \theta \in (-r,0) \,. \end{array} \right. \end{array}$$

Here Ω is a bounded domain in ℝn0 with smooth boundary, b ≥ 0, A is a densely-defined self-adjoint positive linear operator with domain D(A)⊂ L2(Ω) and with compact resolvent, F is the nonlinear term which is locally Lipschitz continuous for the initial condition, g is an external force.

In [3], J.Garcia-Luengo and P.Marin-Rubio treated the following reaction-diffusion equation with non-autonomous force in H–1 and delays under measurability conditions on the driving delay term:

utΔu=f(u)+g(t,ut)+k(t)in(τ,)×Ω,u=0on(τ,)×Ω,u(τ+s,x)=ϕ(s,x),s[r,0]andxΩ,$$\begin{array}{} \displaystyle \left\{ \begin{array}{l} \frac{\partial u}{\partial t} - \Delta u = f(u) + g(t,u_t) + k(t)\; \mbox{in}\; (\tau, \infty) \times \Omega\,,\\ u= 0 \;\mbox{on}\; (\tau, \infty) \times \partial\,{\Omega}\,,\\ u(\tau+s,x)= \phi(s,x),\; s \in [-r,0] \;\mbox{and}\; x \in \Omega\,, \end{array} \right. \end{array}$$

where τ ∈ ℝ, fC(ℝ) the nonlinear term with critical exponent, g is an external force with delay, kLloc2$\begin{array}{} L^{2}_{loc} \end{array}$ (ℝ;H–1(Ω)) a time-dependent force, ϕ the initial condition and h the lenght of the delay effect. In this work, the authors checked the existence of pullback 𝒟-attractor in C([–h,0]; L2(Ω)).

This paper is organized as follows. In section 2, we will prove the existence of weak solutions to the problem (1.1) by using the Faedo-Galerkin approximations, as well as the uniqueness and the continuous dependence of solution with respect to initial conditions. In section 3, we recall some definitions and abstract results on pullback 𝒟-attractor. Then we can prove the existence of pullback 𝒟-attractor for the nonautonomous problem with delay.

Existence and uniqueness of solution

First we give the concept of the solution.

Definition 1

A weak solution of(1.1)is a function uL2([τr,T];L2(Ω)) such that for all T > τ we have

uL2((τ,T);H01(Ω))Lp((τ,T);Lp(Ω))C([τ,T];L2(Ω))$$\begin{array}{} \displaystyle u \in L^2((\tau,T);H^1_0(\Omega)) \cap L^p((\tau,T);L^p(\Omega)) \cap C([\tau,T]; L^2(\Omega)) \end{array}$$

and

utL2([τ,T];L2(Ω)),$$\begin{array}{} \displaystyle \frac{\partial u}{\partial t} \in L^2([\tau, T]; L^2(\Omega))\,, \end{array}$$

with u(t) = φ(tτ), for t ∈ [τr,τ], and it satisfies

τTu,v+τTΩuv=τTΩf(u)v+τTb(t,ut),v+τTΩgv+u0,v(τ),$$\begin{array}{} \displaystyle \int^{T}_{\tau} - \langle u,v'\rangle + \int_{\tau}^{T}\int_{\Omega} \nabla u \nabla v = \int_{\tau}^{T}\int_{\Omega} f(u)v + \int_{\tau}^{T} \langle b(t,u_t), v\rangle \nonumber\\ \displaystyle \qquad \qquad\qquad\qquad\qquad\qquad~+ \int_{\tau}^{T}\int_{\Omega} gv + \langle u^0,v(\tau)\rangle \,, \end{array}$$

for all test functions vL2([τ, T]; H01$\begin{array}{} \displaystyle H^{1}_{0} \end{array}$ (Ω)) and v′∈ L2([τ, T]; H–1(Ω)) such that v(T) = 0.

Theorem 1

Assume that gLlog2$\begin{array}{} \displaystyle L^{2}_{log} \end{array}$ (ℝ;L2(Ω)), b and f satisfy (I)-(IV) and(1.2)-(1.5)respectively and ifλ1 > 1 + Cb/2, Then for all T > τ and all (u0,φ) in L2(Ω)× L2([–r,0];L2(Ω)), there exists a unique weak solution u to the problem(1.1).

Proof

Let us consider {ek}k ≥ 1, the complete basis of H01$\begin{array}{} \displaystyle H^{1}_{0} \end{array}$ (Ω) which is given by the orthonormal eigenfunctions of Δ in L2(Ω). We consider

um(t)=k=1mγk,m(t)ek,m=1,2,$$\begin{array}{} \displaystyle u^m(t) = \sum_{k=1}^{m} \gamma_{k,m}(t) e_k\,, \; m = 1, 2, \ldots \end{array}$$

which is the approximate solutions of Faedo-Galerkin of order m, that is

dumdt,ek+Δum,ek=f(um),ek+b(t,utm),ek+g,ekum(τ),ek=Pmu0,ek=u0,eki.e.Pmum(τ)u0inL2(Ω)um(τ+θ),ek=Pmφ(θ),ek=φ(θ),ekθ(r,0)$$\begin{array}{} \displaystyle \left\{ \begin{array}{l} \langle \frac{du^m}{dt},e_k\rangle + \langle \Delta u^m , e_k\rangle = \langle f(u^m), e_k\rangle+\langle b(t,u_{t}^{m}), e_k\rangle + \langle g,e_k\rangle \\ \langle u^m(\tau),e_k\rangle= \langle P_m u^0 ,e_k\rangle=\langle u^0,e_k\rangle \; \mbox{i.e.}\; P_mu^{m}(\tau) \to u^{0}\; \mbox{in}\; L^2(\Omega) \\ \langle u^{m}(\tau + \theta),e_k\rangle=\langle P_m \varphi(\theta),e_k\rangle=\langle \varphi(\theta),e_k\rangle \; \forall \theta \in (-r,0) \end{array}\right. \end{array}$$

for all k = 1 … m. Where γk,m(t) = 〈 um(t), ek 〉 denote the Fourier coefficients; such that γm,kC1((τ, T); ℝ) ∩ L2((τr, T), ℝ), γk,m$\begin{array}{} \gamma'_{k,m} \end{array}$ (t) is absolutely continuous, and Pmu(t) = k=1m$\begin{array}{} \sum^{m}_{k=1} \end{array}$u,ekek is the orthogonal projection of L2(Ω) and H01$\begin{array}{} \displaystyle H^{1}_{0} \end{array}$ (Ω) in Vm = span{e1, …, em}.

It is well-known that the above finite-dimensional delayed system is well-posed (e.g. cf. [2]), at least locally. We will provide a priori estimates for the Faedo-Galerkin approximate solutions.

For all m ∈ ℕand all T > τ, the sequence {um} is bounded in

L((τ,T);L2(Ω))L2((τ,T);H01(Ω))Lp((τ,T);Lp(Ω)).$$\begin{array}{} \displaystyle L^{\infty} ((\tau, T);L^2(\Omega))\cap L^{2} ((\tau, T);H^1_0(\Omega)) \cap L^{p} ((\tau, T);L^p(\Omega))\,. \end{array}$$

Multiplying (1.1) by um and integrating over Ω, we obtain

12ddtum(t)2+um(t)2=Ωf(um)um+Ωb(t,utm)um+Ωgum.$$\begin{array}{} \displaystyle \frac{1}{2} \frac{d}{dt} \Vert u^m (t) \Vert^2 + \Vert \nabla u^m (t) \Vert^2 = \int_{\Omega} f(u^m)u^m + \int_{\Omega} b(t,u_t^m)u^m + \int_{\Omega} gu^m \,. \end{array}$$

Using the hypothesis (1.2) and the Young inequality, we get

12ddtum(t)2+um(t)2c|Ω|μ1um(t)p+12b(t,utm)2+12um(t)2+12g(t)2+12um(t)2.$$\begin{array}{} \displaystyle \quad\frac{1}{2} \frac{d}{dt} \Vert u^m (t) \Vert^2 + \Vert \nabla u^m (t) \Vert^2 \\ \displaystyle \leq c\vert \Omega \vert - \mu_1\Vert u^m(t) \Vert^p + \frac{1}{2}\Vert b(t, u_t^m) \Vert^2 + \frac{1}{2} \Vert u^m(t) \Vert^2 + \frac{1}{2}\Vert g(t)\Vert^2 + \frac{1}{2}\Vert u^m(t) \Vert^2\,. \end{array}$$

So, one has

ddtum(t)2+2um(t)2+2μ1um(t)p2c|Ω|+b(t,utm)2+g(t)2+um(t)2.$$\begin{array}{} \displaystyle \quad\frac{d}{dt} \Vert u^m (t) \Vert^2 + 2\Vert \nabla u^m (t) \Vert^2 + 2\mu_1\Vert u^m(t) \Vert^p\\ \leq 2c\vert \Omega \vert + \Vert b(t, u_t^m) \Vert^2 + \Vert g(t)\Vert^2 + \Vert u^m(t) \Vert^2\,. \end{array}$$

After integrating this last estimate over [τ,t], τtT, we use (II) and (IV), so we get

um(t)2+2τtum(s)2ds+2μ1τtum(s)pds2c|Ω|(tτ)+um(τ)2+Cbτrtum(s)2ds+τtg(s)2ds+τtum(s)2ds,2c|Ω|(tτ)+um(τ)2+Cbτrτum(s)2ds+Cbτtum(s)2ds+τtg(s)2ds+τtum(s)2ds.$$\begin{array}{} \displaystyle \quad\Vert u^m (t) \Vert^2 + 2\int_{\tau}^{t} \Vert \nabla u^m (s) \Vert^2 ds + 2\mu_1 \int_{\tau}^{t} \Vert u^m(s) \Vert^p ds \\ \displaystyle \leq 2c \vert \Omega \vert (t-\tau) + \Vert u^m (\tau) \Vert^2 + C_b \int_{\tau -r}^{t} \Vert u^m(s) \Vert^2 ds\\ \displaystyle + \int_{\tau}^{t} \Vert g(s)\Vert^2 ds + \int_{\tau}^{t} \Vert u^m(s) \Vert^2 ds \,,\\ \displaystyle\leq 2c\vert \Omega \vert (t-\tau) + \Vert u^m (\tau) \Vert^2 + C_b\int_{\tau -r}^{\tau} \Vert u^m(s) \Vert^2 ds + C_b\int_{\tau}^{t} \Vert u^m(s) \Vert^2 ds \\ \displaystyle + \int_{\tau}^{t} \Vert g(s)\Vert^2 ds + \int_{\tau}^{t} \Vert u^m(s) \Vert^2 ds \,.\end{array}$$

By the fact that λ1u2 ≤ ‖ ∇ u2, one has

um(t)2+2τtum(s)2ds+2μ1τtum(s)pds2c|Ω|(tτ)+um(τ)2+Cbτrτum(s)2ds+Cbλ11τtum(s)2ds+τtg(s)2ds+λ11τtum(s)2ds.$$\begin{array}{} \displaystyle \quad\Vert u^m (t) \Vert^2 + 2\int_{\tau}^{t} \Vert \nabla u^m (s) \Vert^2 ds + 2\mu_1\int_{\tau}^{t} \Vert u^m(s) \Vert^p ds\\ \displaystyle \leq 2c\vert \Omega \vert (t-\tau) + \Vert u^m (\tau) \Vert^2 + C_b\int_{\tau -r}^{\tau} \Vert u^m(s) \Vert^2 ds + C_b\lambda_1^{-1}\int_{\tau}^{t} \Vert \nabla u^m(s) \Vert^2 ds \\ \displaystyle + \int_{\tau}^{t} \Vert g(s)\Vert^2 ds + \lambda_1^{-1} \int_{\tau}^{t} \Vert \nabla u^m(s) \Vert^2 ds\,. \end{array}$$

Then, we find

um(t)2+(2Cbλ11λ11)τtum(s)2ds+2μ1τtum(s)pds2c|Ω|(tτ)+um(τ)2+Cbτrτum(s)2ds+τtg(s)2ds,2c|Ω|(Tτ)+um(τ)2+Cbτrτum(s)2ds+τtg(s)2ds.$$\begin{array}{} \displaystyle \quad\Vert u^m (t) \Vert^2 + (2-C_b\lambda_1^{-1}-\lambda_1^{-1})\int_{\tau}^{t} \Vert \nabla u^m (s) \Vert^2 ds + 2\mu_1\int_{\tau}^{t} \Vert u^m(s) \Vert^p ds \\ \displaystyle \leq 2c\vert \Omega \vert (t-\tau) +\Vert u^m (\tau) \Vert^2 + C_b\int_{\tau -r}^{\tau} \Vert u^m(s) \Vert^2 ds + \int_{\tau}^{t} \Vert g(s)\Vert^2 ds \,,\\ \displaystyle \leq 2c\vert \Omega \vert (T-\tau) +\Vert u^m (\tau) \Vert^2 + C_b\int_{\tau -r}^{\tau} \Vert u^m(s) \Vert^2 ds + \int_{\tau}^{t} \Vert g(s)\Vert^2 ds \,. \end{array}$$

Since gLlog2$\begin{array}{} L^{2}_{log} \end{array}$ (ℝ,L2(Ω)) and for λ1 > 1+ Cb/2, we deduce by this last estimate that for all T > τ, the sequence

{um} is bounded inL((τ,T);L2(Ω))L2((τ,T);H01(Ω))Lp((τ,T);Lp(Ω)).$$\begin{array}{} \displaystyle \{u^m\}\;\mbox{ is bounded in}\; L^{\infty} ((\tau, T);L^2(\Omega))\cap L^{2} ((\tau, T);H^1_0(\Omega)) \cap L^{p} ((\tau, T);L^p(\Omega))\,. \end{array}$$

Also, the estimate (2.1) implies that the local solution can extended to the interval [τ, T].

{f(um)}isboundedinLq((τ,T);Lq(Ω)).$$\begin{array}{} \displaystyle \{f(u^m)\} \; {is\, bounded \,in}\; L^{q}((\tau, T);L^{q}(\Omega))\,. \end{array}$$

Using (1.4), we have

f(um(t)Lq(Ω)q=Ω|f(um(t,x))|qdx,lqΩ|um(t,x)|p1+1qdx.$$\begin{array}{} \displaystyle \Vert f(u^m(t) \Vert_{L^q(\Omega)}^q = \int_{\Omega}\vert f(u^m(t,x)) \vert^q dx\,,\\ \displaystyle \qquad \qquad\qquad~ \leq l^q\int_{\Omega}\left(\vert u^m(t,x) \vert^{p-1} + 1\right)^q dx\,. \end{array}$$

By the convexity of the power and the fact that p = q(p–1), one has

f(um(t))Lq(Ω)q2q1lqΩ|um(t,x)|q(p1)dx+2q1lq|Ω|,2q1lqum(t)Lp(Ω)q(p1)+2q1lq|Ω|,2q1lqum(t)Lp(Ω)p+2q1lq|Ω|.$$\begin{array}{} \displaystyle \Vert f(u^m(t)) \Vert_{L^q(\Omega)}^q \leq 2^{q-1}l^q\int_{\Omega}\vert u^m(t,x) \vert^{q(p-1)} dx + 2^{q-1}l^q \vert \Omega \vert \,,\\ \displaystyle \qquad\qquad\qquad\quad\leq 2^{q-1}l^q \Vert u^m(t) \Vert_{L^p(\Omega)}^{q(p-1)} + 2^{q-1}l^q \vert \Omega \vert \,,\\ \displaystyle \qquad\qquad\qquad\quad\leq 2^{q-1}l^q \Vert u^m(t) \Vert_{L^p(\Omega)}^p + 2^{q-1}l^q \vert \Omega \vert\,. \end{array}$$

Integrating this last estimate over [τ,t], τtT, one obtains

τtf(um(s))Lq(Ω)qds2q1lqτtum(s)Lp(Ω)pds+2q1lq|Ω|(tτ)$$\begin{array}{} \displaystyle \int_{\tau}^{t}\Vert f(u^m(s)) \Vert_{L^q(\Omega)}^q ds \leq 2^{q-1}l^q \int_{\tau}^{t} \Vert u^m(s) \Vert_{L^p(\Omega)}^p ds + 2^{q-1}l^q \vert \Omega \vert(t-\tau) \end{array}$$

From (2.1) we deduce that the term τtum(s)Lp(Ω)p$\begin{array}{} \int_{\tau}^{t} \Vert u^m(s) \Vert_{L^p(\Omega)}^p \end{array}$ds is bounded, so by this last estimate we conclude that {f(um)} is bounded in Lq((τ, T);Lq(Ω)), for all T > τ.

tum$\begin{array}{} \left\{\frac{\partial }{\partial t}u^m\right\} \end{array}$is bounded in L2((τ,T);L2(Ω)).

Now, multiplying (1.1) by umt$\begin{array}{} \frac{\partial u^m}{\partial t} \end{array}$ and integrating over Ω, one has

ddtum(t)2+12ddtum(t)2=Ωf(um)umt+Ωb(t,utm)umt+Ωgumt.$$\begin{array}{} \displaystyle \quad-\left\Vert\frac{d}{d t} u^m(t) \right \Vert^2 + \frac{1}{2} \frac{d}{dt} \Vert \nabla u^m(t) \Vert^2 \nonumber\\ \displaystyle = \int_{\Omega} f(u^m)\frac{\partial u^m}{\partial t}+ \int_{\Omega} b(t,u^m_t)\frac{\partial u^m}{\partial t} + \int_{\Omega} g\frac{\partial u^m}{\partial t}\,. \end{array}$$

On the other hand, we have

ddtF(u)=dFduut,=f(u)ut.$$\begin{array}{} \displaystyle \frac{d}{d t} F(u) = \frac{dF}{d u} \,\frac{\partial u}{\partial t}\,,\\ \displaystyle \qquad\qquad= f(u) \, \frac{\partial u}{\partial t} \,. \end{array}$$

So

ddtΩF(u)=Ωf(u)ut.$$\begin{array}{} \displaystyle \frac{d}{d t}\int_{\Omega} F(u) = \int_{\Omega} f(u)\frac{\partial u}{\partial t} \,. \end{array}$$

Using this last equality in (2.4), we find

ddtum(t)2+12ddtum(t)2=ddtΩF(um)+Ωb(t,utm)umt+Ωgumt.$$\begin{array}{} \displaystyle \left\Vert\frac{d}{d t} u^m(t) \right \Vert^2 + \frac{1}{2} \frac{d}{dt} \Vert \nabla u^m(t) \Vert^2 = \frac{d}{dt} \int_{\Omega} F(u^m)+ \int_{\Omega} b(t,u^m_t)\frac{\partial u^m}{\partial t} + \int_{\Omega} g\frac{\partial u^m}{\partial t}\,. \end{array}$$

From (1.5) and Cauchy inequality, we have

ddtum(t)2+12ddtum(t)2,ddtΩ(cμ1|u(t,x)|p)dx+ε12b(t,utm)2+12ε1ddtum(t)2+ε22g(t)2+12ε2ddtum(t)2.$$\begin{array}{} \displaystyle \quad\left\Vert\frac{d}{d t} u^m(t) \right \Vert^2 + \frac{1}{2} \frac{d}{dt} \Vert \nabla u^m(t) \Vert^2 \,,\\ \displaystyle \leq \frac{d}{dt} \int_{\Omega} (c' - \mu'_1 \vert u(t,x) \vert^p) dx + \frac{\varepsilon_1}{2}\Vert b(t,u^m_t) \Vert^2 + \frac{1}{2 \varepsilon_1}\left\Vert \frac{d }{d t}u^m(t)\right\Vert^2 \\ \displaystyle + \frac{\varepsilon_2}{2}\Vert g(t) \Vert^2 + \frac{1}{2 \varepsilon_2}\left \Vert \frac{d}{d t} u^m(t) \right \Vert^2 \,. \end{array}$$

After simplification, one obtains

21ε11ε2ddtum(t)2+ddtum(t)2+2μ1um(t)Lp(Ω)pε1b(t,utm)2+ε2g(t)2.$$\begin{array}{} \displaystyle \quad\left (2-\frac{1}{ \varepsilon_1}-\frac{1}{ \varepsilon_2}\right)\left\Vert\frac{d}{d t} u^m(t) \right\Vert^2 + \frac{d}{dt} \left(\Vert \nabla u^m(t) \Vert^2 + 2 \mu'_1 \Vert u^m(t) \Vert^p_{L^p(\Omega)}\right) \\ \leq {\varepsilon_1}\Vert b(t,u^m_t) \Vert^2 + {\varepsilon_2}\Vert g(t) \Vert^2 \,. \end{array}$$

We can choose ε1 = ε2 = 2 to get

ddtum(t)2+ddtum(t)2+2μ1um(t)Lp(Ω)p2b(t,utm)2+2g(t)2.$$\begin{array}{} \displaystyle \left\Vert\frac{d}{d t} u^m(t) \right\Vert^2 + \frac{d}{dt} \left(\Vert \nabla u^m(t) \Vert^2 + 2 \mu'_1 \Vert u^m(t) \Vert^p_{L^p(\Omega)}\right) \leq 2\Vert b(t,u^m_t) \Vert^2 + 2\Vert g(t) \Vert^2 \,. \end{array} $$

Integrating this last estimate over [τ,t] and using (II) and (IV), one has

τtddsum(s)2ds+um(t)2+2μ1um(t)Lp(Ω)pum(τ)2+2μ1um(τ)Lp(Ω)p+2Cbτrtum(s)2ds+2τtg(s)2ds,um(τ)2+2μ1um(τ)Lp(Ω)p+2Cbτrτum(s)2ds+2Cbτtum(s)2ds+2τtg(s)2ds$$\begin{array}{} \displaystyle \quad \int_{\tau}^{t}\left\Vert\frac{d}{d s} u^m(s) \right\Vert^2 ds +\Vert \nabla u^m(t) \Vert^2 + 2 \mu'_1 \Vert u^m(t) \Vert^p_{L^p(\Omega)} \\ \displaystyle \leq \Vert \nabla u^m(\tau) \Vert^2 + 2 \mu'_1 \Vert u^m(\tau) \Vert^p_{L^p(\Omega)} + 2C_b\int_{\tau -r}^{t}\Vert u^m(s) \Vert^2 ds+ 2 \int_{\tau}^{t} \Vert g(s) \Vert^2 ds \,,\\ \displaystyle \leq \Vert \nabla u^m(\tau) \Vert^2 + 2 \mu'_1 \Vert u^m(\tau) \Vert^p_{L^p(\Omega)} + 2C_b\int_{\tau -r}^{\tau}\Vert u^m(s) \Vert^2 ds \\ \displaystyle + 2C_b\int_{\tau}^{t}\Vert u^m(s) \Vert^2 ds+ 2 \int_{\tau}^{t} \Vert g(s) \Vert^2 ds \end{array}$$

Since λ1u2 ≤ ‖ ∇u2, one has

τtddsum(s)2ds+um(t)2+2μ1um(t)Lp(Ω)pum(τ)2+2μ1um(τ)Lp(Ω)p+2Cbτrτum(s)2ds+2Cbλ11τtum(s)2ds+2τtg(s)2ds$$\begin{array}{} \displaystyle \quad\int_{\tau}^{t}\left\Vert\frac{d}{d s} u^m(s) \right\Vert^2 ds +\Vert \nabla u^m(t) \Vert^2 + 2 \mu'_1 \Vert u^m(t) \Vert^p_{L^p(\Omega)} \\ \displaystyle \leq \Vert \nabla u^m(\tau) \Vert^2 + 2 \mu'_1 \Vert u^m(\tau) \Vert^p_{L^p(\Omega)} + 2C_b\int_{\tau -r}^{\tau}\Vert u^m(s) \Vert^2 ds \\ \displaystyle + 2C_b \lambda_1^{-1}\int_{\tau}^{t}\Vert \nabla u^m(s) \Vert^2 ds + 2 \int_{\tau}^{t} \Vert g(s) \Vert^2 ds \end{array}$$

From (2.1), we have τtum(s)2ds$\begin{array}{} \int_{\tau}^{t}\Vert \nabla u^m(s) \Vert^2 ds \end{array}$

is bounded and since gLloc2$\begin{array}{} L^2_{loc} \end{array}$ (ℝ;L2(Ω)), this last estimate gives that

tumis bounded inL2((τ,T);L2(Ω)),$$\begin{array}{} \displaystyle \left\{\frac{\partial }{\partial t}u^m\right\} \; \mbox{is bounded in}\; L^2((\tau,T);L^2(\Omega)) \,, \end{array}$$

for all T > τ.

From the claims (1), (2) and (3), the hypothesis (IV) and the remark (1)1, we can extract a subsequence (relabelled the same) such that

umuweakly* inL((τ,T);L2(Ω)),umuweakly inL2((τ,T);H01(Ω)),umuweakly inLp((τ,T);Lp(Ω)),umtutstrongly inL2((τ,T);L2(Ω)),f(um)σweakly inLq((τ,T);Lq(Ω)),b(.,u.m)b(.,u.)strongly inL2((τ,T);L2(Ω)).$$\begin{array}{} \displaystyle u^m \rightharpoonup u \;\mbox{weakly* in}\; L^{\infty}((\tau,T);L^2(\Omega)), \\ \displaystyle u^m \rightharpoonup u \;\mbox{weakly in} \; L^{2}((\tau,T);H^{1}_0(\Omega)),\\ \displaystyle u^m \rightharpoonup u \;\mbox{weakly in} \; L^{p}((\tau,T);L^p(\Omega)),\\ \displaystyle \frac{\partial u^m}{\partial t} \rightarrow \frac{\partial u}{\partial t} \;\mbox{strongly in}\; L^{2}((\tau,T);L^2(\Omega)),\\ \displaystyle f(u^m) \rightharpoonup \sigma' \;\mbox{weakly in}\; L^{q}((\tau,T);L^q(\Omega)),\\ \displaystyle b(.,u^m_.) \rightarrow b(.,u_.) \;\mbox{strongly in}\; L^{2}((\tau,T);L^2(\Omega))\,. \end{array}$$

By the Aubin-Lions lemma of compactness, we conclude that umu strongly in L2((τ,T);L2(Ω)). Thus umu a.e [τ,T]× Ω.

Since f is continuous, we deduce that f(um) → f(u) a.e [τ,T]× Ω. So from (2.3) and (lemma 1.3 in [7], p.12) we can identify σ′ with f(u).

To prove that u(τ) = u0, we put vC1((τ,T); H01$\begin{array}{} \displaystyle H^{1}_{0} \end{array}$ (Ω)) such that v(T) = 0 and we note from (1.1) that

τTu,v+τTΩuv=τTΩf(u)v+τTb(t,ut),v+τTΩgv+u(τ),v(τ).$$\begin{array}{} \displaystyle \int^{T}_{\tau} - \langle u,v'\rangle + \int_{\tau}^{T}\int_{\Omega} \nabla u \nabla v = \int_{\tau}^{T}\int_{\Omega} f(u)v + \int_{\tau}^{T} \langle b(t,u_t), v\rangle \\ \displaystyle \qquad \qquad\qquad\qquad\qquad\qquad~+ \int_{\tau}^{T}\int_{\Omega} gv + \langle u(\tau),v(\tau)\rangle \,. \end{array}$$

In a similar way, from the Faedo-Galerkin approximations, we have

τTum,vτTΩumv=τTΩf(um)v+τTb(t,utm),v+τTΩgv+um(τ),v(τ).$$\begin{array}{} \displaystyle \int^{T}_{\tau} - \langle u^m,v'\rangle \int_{\tau}^{T}\int_{\Omega} \nabla u^m \nabla v = \int_{\tau}^{T}\int_{\Omega} f(u^m)v + \int_{\tau}^{T} \langle b(t,u^m_t), v\rangle \\ \displaystyle \qquad\qquad\qquad\qquad\qquad\qquad\quad + \int_{\tau}^{T}\int_{\Omega} gv + \langle u^m(\tau),v(\tau)\rangle \,. \end{array}$$

Using the fact that um(τ) → u0 in L2(Ω) and (2.6) to find

τTu,v+τTΩuv=τTΩf(u)v+τTb(t,ut),v+τTΩgv+u0,v(τ).$$\begin{array}{} \displaystyle \int^{T}_{\tau} - \langle u,v'\rangle + \int_{\tau}^{T}\int_{\Omega} \nabla u \nabla v = \int_{\tau}^{T}\int_{\Omega} f(u)v + \int_{\tau}^{T} \langle b(t,u_t), v\rangle \\ \displaystyle \qquad\qquad\qquad\qquad\qquad\qquad~+ \int_{\tau}^{T}\int_{\Omega} gv + \langle u^0,v(\tau)\rangle \,. \end{array}$$

Since v(τ) is given arbitrarily, comparing (2.7) and (2.9) we deduce that u(τ) = u0.

To prove that uC([τ,T];L2(Ω)), we put wm = umu then we have

twmΔwm=f(um)f(u)+b(t,utm)b(t,ut).$$\begin{array}{} \displaystyle \frac{\partial}{\partial t} w^m - \Delta w^m = f(u^m)-f(u) + b(t,u^m_t) - b(t,u_t) \,. \end{array}$$

Multiplying this equation by wm and integrating over Ω, we obtain

ddtwm(t)2+2wm(t)2=2Ωf(um)f(u)wm+2Ω(b(t,utm)b(t,ut))(umu).$$\begin{array}{} \displaystyle \frac{d}{dt} \Vert w^m (t) \Vert^2 + 2\Vert \nabla w^m (t) \Vert^2 = 2 \int_{\Omega}\left(f(u^m)-f(u)\right)w^m \\ \displaystyle \qquad\qquad\qquad\qquad\qquad\quad\,\,\,\,\,\,+ 2\int_{\Omega}(b(t,u_t^m)-b(t,u_t))(u^m-u) \,. \end{array}$$

By (1.3), (I) and (1.6), we get

ddtwm(t)2+2wm(t)22kwm(t)2+2LbwtmL2([r,0];L2(Ω))2.$$\begin{array}{} \displaystyle \frac{d}{dt} \Vert w^m (t) \Vert^2 + 2\Vert \nabla w^m (t) \Vert^2 \leq 2 k \Vert w^m(t) \Vert^2 + 2 L_b \Vert w^m_t \Vert^2_{L^2([-r,0];L^2(\Omega))} \,. \end{array}$$

Integrating over [τ,t], we get

wm(t)2wm(τ)2+2τtwm(s)2ds2kτtwm(s)2+2Lbτtr0wm(s+θ)2dθds,2kτtwm(s)2+2Lbr0τrtwm(s)2dsdθ,2kτtwm(s)2+2Lbrτrτwm(s)2ds+2Lbrτtwm(s)2ds.$$\begin{array}{} \displaystyle \quad\Vert w^m (t) \Vert^2 - \Vert w^m (\tau) \Vert^2 + 2\int_{\tau}^{t} \Vert \nabla w^m (s) \Vert^2 ds\\ \displaystyle \leq 2k \int_{\tau}^{t}\Vert w^m(s) \Vert^2 + 2 L_b \int_{\tau}^{t}\int_{-r}^{0}\Vert w^m(s+\theta) \Vert^2 d\,\theta ds \,,\\ \displaystyle \leq 2k \int_{\tau}^{t}\Vert w^m(s) \Vert^2 + 2 L_b \int_{-r}^{0} \int_{\tau-r}^{t}\Vert w^m(s) \Vert^2 ds d\theta \,,\\ \displaystyle \leq 2k \int_{\tau}^{t}\Vert w^m(s) \Vert^2 + 2 L_br \int_{\tau-r}^{\tau}\Vert w^m(s) \Vert^2 ds + 2 L_br \int_{\tau}^{t}\Vert w^m(s) \Vert^2 ds\,. \end{array} $$

Therefore by by this last estimate, we can deduce that

wm(t)2wm(τ)2+2Lbrτrτwm(s)2ds+(2k+2Lbr)τtwm(s)2ds.$$\begin{array}{} \displaystyle \Vert w^m (t) \Vert^2 \leq \Vert w^m (\tau) \Vert^2 + 2 L_br \int_{\tau-r}^{\tau}\Vert w^m(s) \Vert^2 ds + (2k+2 L_br) \int_{\tau}^{t}\Vert w^m(s) \Vert^2 ds\,. \end{array}$$

Applying the Gronwall lemma to this estimate, we obtain

wm(t)2wm(τ)2+2Lbrτrτwm(s)2dse(2k+2Lbr)(tτ).$$\begin{array}{} \displaystyle \Vert w^m (t) \Vert^2 \leq \left(\Vert w^m (\tau) \Vert^2 + 2 L_br \int_{\tau-r}^{\tau}\Vert w^m(s) \Vert^2 ds\right)e^{(2k+2 L_br) (t-\tau)}\,. \end{array}$$

Since um(τ)→ u0 and um(τ + θ)→ φ(θ), the estimate (2.10) shows that umu uniformly in C([τ,T];L2(Ω)).

Finally, we prove the uniqueness and continuous dependence of the solution. Let u1; u2 be two solutions of problem (1.1) with the initial conditions u0,1, u0,2 and φ1, φ2. Denoting that w = u1u2 and repeating the argument as in the proof of (2.10), we find

w(t)2w(τ)2+2Lbrτrτw(s)2dse(2k+2Lbr)(tτ).$$\begin{array}{} \displaystyle \Vert w (t) \Vert^2 \leq \left(\Vert w (\tau) \Vert^2 + 2 L_br \int_{\tau-r}^{\tau}\Vert w(s) \Vert^2 ds\right)e^{(2k+2 L_br) (t-\tau)}\,. \end{array}$$

and this completes the proof of the theorem. ◼

Existence of pullback D-attractors
Preliminaries of pullback D-attractors

First, we give some basic definitions and an abstract result on the existence of pullback attractors, which we need to obtain our results (we refer the reader to [2,3,4,8]). Let (X,d) be a complete metric space, 𝒫(X) be the class of nonempty subsets of X, and suppose 𝒟 is a nonempty class of parameterized sets = {D(t) : t ∈ ℝ}⊂ 𝒫(X).

Definition 2

A two parameter family of mappings U(t,τ) : XX tτ, τ ∈ ℝ, is called to be a process if

S(τ,τ)x = {x},∀ τ ∈ ℝ, xY;

S(t,s)S(s,τ)x = S(t,τ)x, ∀ tsτ, τ ∈ ℝ, xX.

Definition 3

A family of bounded sets B̂ = {B(t) : t ∈ ℝ}∈ 𝒟 is called pullback 𝒟-absorbing for the processS(t,τ)} if for any t ∈ ℝ and for any D̂ ∈ 𝒟, there exists τ0(t,) ≤ t such that

S(t,τ)D(τ)B(t)forallττ0(t,D^).$$\begin{array}{} \displaystyle S(t,\tau)D(\tau) \subset B(t)\quad \it{for\, all }\; \tau \leq \tau_0(t,\widehat{D}) \,. \end{array}$$

Definition 4

The process S(t,τ) is said to be pullback 𝒟-asymptotically compact if for all t ∈ ℝ, all D̂ ∈ 𝒟, any sequence τ n → -∞, and any sequence xnD(τn), the sequence {S(t,τn)xn} is relatively compact in X.

Definition 5

A family  = {A(t) : t ∈ ℝ}⊂ 𝒫(X) is said to be a pullback 7 𝒟-attractor forS(t,τ)} if

A(t) is compact for all t ∈ ℝ;

 is invariant; i.e., S(t,τ)A(τ) = A(t), for all tτ;

 is pullback 𝒟-attracting; i.e.,

limτdist(S(t,τ)D(τ),A(t))=0,$$\begin{array}{} \displaystyle \lim_{\tau \to -\infty} dist(S(t,\tau)D(\tau), A(t))=0\,, \end{array}$$

for all D̂ ∈ 𝒟 and all t ∈ ℝ;

If {C(t) : t ∈ ℝ} is another family of closed attracting sets then A(t) ⊂ C(t), for all t ∈ ℝ.

Theorem 2

Let us suppose that the process {S(t,τ)} is pullback 𝒟-asymptotically compact, and B̂ = {B(t) : t ∈ ℝ}∈ 𝒟 is a family of pullback 𝒟-absorbing sets for {S(t,τ)}. Then there exists a pullback 𝒟-attractor {A(t) : t ∈ ℝ} such that

A(t)=stτsS(t,τ)B(τ)¯.$$\begin{array}{} \displaystyle A(t) = \bigcap_{s\leq t}\overline{\bigcup_{\tau \leq s}S(t,\tau) B(\tau)}\,. \end{array}$$

Construction of the associated process

Now, we will apply the above results in the phase space H: = L2(Ω) × L2([–r,0]; L2(Ω)), which is a Hilbert space with the norm

(u0,φ)H2=u02+r0φ(θ)2dθ,$$\begin{array}{} \displaystyle \Vert (u^0,\varphi) \Vert^2_H = \Vert \nabla u^0 \Vert^2 + \int_{-r}^{0}\Vert \varphi (\theta) \Vert^2 d\theta \,, \end{array}$$

with (u0,φ)∈ H. To this aim, We consider gLloc2$\begin{array}{} L^{2}_{loc} \end{array}$ (ℝ;L2(Ω)), b: ℝ × L2([–r,0];L2(Ω)) → L2(Ω) with the hypotheses (I)-(IV) and fC1(ℝ;ℝ) verifying (1.2)-(1.5). Then the family of mappings

S(t,τ):HH(u0,φ)S(t,τ)(u0,φ)=(u(t),ut),$$\begin{array}{} \displaystyle S(t,\tau) : H\rightarrow H \\ \displaystyle \qquad\quad~~(u^0,\varphi) \longmapsto S(t,\tau)(u^0,\varphi) = (u(t),u_t) \,, \end{array} $$

with tτ, τ ∈ ℝ and u is the weak solution to (1.1), defines a process.

On the other hand, we construct the family of mappings

U(t,τ):HC([r,0];L2(Ω))(u0,φ)U(t,τ)(u0,φ)=ut,tτ+r,$$\begin{array}{} \displaystyle U(t,\tau) : H\rightarrow C([-r,0];L^2(\Omega)) \\ \displaystyle \qquad\qquad (u^0,\varphi) \longmapsto U(t,\tau)(u^0,\varphi) = u_t \,,\; \forall t \geq \tau + r\,, \end{array}$$

which we will use in our analysis. Of course, it is sensible to expect that the both operators should be related. Let us consider the linear mapping

j:C([r,0];L2(Ω))l2(Ω)×C([r,0];L2(Ω))φj(φ)=(φ(0),φ).$$\begin{array}{} \displaystyle j : C([-r,0]; L^2(\Omega))\rightarrow l^2(\Omega) \times C([-r,0]; L^2(\Omega)) \\ \displaystyle ~~~~~\varphi \longmapsto j(\varphi) = (\varphi(0),\varphi) \,. \end{array}$$

This map is obviously continuous from C([–r,0]; L2(Ω)) into H. We note that for all (u0,φ) ∈ H provided that tτ + r, so we write

S(t,τ)(u0,φ)=j(U(t,τ)(u0,φ)),(u0,φ)H,tτ+r.$$\begin{array}{} \displaystyle S(t,\tau)(u^0,\varphi) = j(U(t,\tau)(u^0,\varphi))\,,\; \forall (u^0,\varphi) \in H\,,\; \forall t \geq \tau + r\,. \end{array}$$

To check the continuity of the process, we need the following lemma.

Lemma 1

Let (u0,φ), (v0,ϕ)∈ H be two couples of initial conditions for the problem(1.1)and u, v be the corresponding solutions to(1.1). Then there exists a positive constant ν: = 2(12+k+Cb2λ1)>0,$\begin{array}{} 2(\frac{1}{2}+k+\frac{C_b}{2}-\lambda_1)>0\,, \end{array}$such that

u(t)v(t)2u0v02+Cbφϕ2eν(tτ),tτ.$$\begin{array}{} \displaystyle \Vert u(t)-v(t) \Vert^2 \leq \left(\Vert u^0-v^0 \Vert^2 + C_b \Vert \varphi - \phi \Vert^2\right) e^{\nu (t-\tau)}\,,\; \forall t\geq \tau\,. \end{array}$$

It also holds

utvtC([r,0];L2(Ω))2u0v02+Cbφϕ2eν(trτ),tτ+r.$$\begin{array}{} \displaystyle \Vert u_t-v_t \Vert_{C([-r,0];L^2(\Omega))}^2 \leq \left(\Vert u^0-v^0 \Vert^2 + C_b \Vert \varphi - \phi \Vert^2\right) e^{\nu (t-r-\tau)}\,,\; \forall t\geq \tau + r\,. \end{array}$$

Proof

From (1.1), one has

t(uv)Δ(uv)=f(u)f(v)+b(t,ut)b(t,vt).$$\begin{array}{} \displaystyle \frac{\partial}{\partial t} (u-v) - \Delta (u-v) = f(u) -f(v) + b(t,u_t) - b(t,v_t)\,. \end{array}$$

We put w = uv, we obtain

wtΔw=f(u)f(v)+b(t,ut)b(t,vt).$$\begin{array}{} \displaystyle \frac{\partial w}{\partial t} - \Delta w = f(u) -f(v) + b(t,u_t) - b(t,v_t)\,. \end{array}$$

Multiplying this equation by w and integrating it over Ω, one gets

12ddtw(t)2+w(t)2=Ωf(u)f(v)w+Ωb(t,ut)b(t,vt)w.$$\begin{array}{} \displaystyle \frac{1}{2} \frac{d}{dt} \Vert w(t) \Vert^2 + \Vert \nabla w(t) \Vert^2 = \int_{\Omega} \left( f(u)-f(v)\right)w + \int_{\Omega} \left( b(t,u_t) - b(t,v_t)\right)w\,. \end{array}$$

Using (1.3) and Cauchy-Schwarz inequality, one has

12ddtw(t)2+w(t)2kw(t)2+b(t,ut)b(t,vt)w(t).$$\begin{array}{} \displaystyle \frac{1}{2} \frac{d}{dt} \Vert w(t) \Vert^2 + \Vert \nabla w(t) \Vert^2 \leq k\Vert w(t) \Vert^2 + \Vert b(t,u_t) - b(t,v_t) \Vert \Vert w(t) \Vert\,. \end{array}$$

Since λ1w(t) ‖2 ≤ ‖ ∇ w(t) ‖2 and by the Young inequality, one finds

ddtw(t)2+2λ1w(t)2ddtw(t)2+2w(t)2,2kw(t)2+b(t,ut)b(t,vt)2+w(t)2.$$\begin{array}{} \displaystyle \frac{d}{dt} \Vert w(t) \Vert^2 + 2 \lambda_1\Vert w(t) \Vert^2 \leq \frac{d}{dt} \Vert w(t) \Vert^2 + 2 \Vert \nabla w(t) \Vert^2\,,\\ \displaystyle \qquad\qquad\qquad\qquad\qquad\quad\leq 2k\Vert w(t) \Vert^2 + \Vert b(t,u_t) - b(t,v_t) \Vert^2 + \Vert w(t) \Vert^2 \,. \end{array}$$

Therefore, one has

ddtw(t)2212+kλ1w(t)2+b(t,ut)b(t,vt)2.$$\begin{array}{} \displaystyle \frac{d}{dt} \Vert w(t) \Vert^2 \leq 2\left(\frac{1}{2}+k -\lambda_1\right) \Vert w(t) \Vert^2 + \Vert b(t,u_t) - b(t,v_t) \Vert^2\,. \end{array}$$

Integrating this last estimate from τ to t and using (1.7), one obtains

w(t)2w(τ)2+212+kλ1τtw(s)2ds+τtb(s,us)b(s,vs)2ds,w(τ)2+212+kλ1τtw(s)2ds+Cbτrtw(s)2ds,w(τ)2+212+kλ1τtw(s)2ds+Cbτrτw(s)2ds+Cbτtw(s)2ds,w(τ)2+Cbτrτw(s)2ds+212+k+Cb2λ1τtw(s)2ds.$$\begin{array}{} \displaystyle \Vert w(t) \Vert^2 \leq \Vert w(\tau) \Vert^2 + 2 \left(\frac{1}{2}+k -\lambda_1\right) \int_{\tau}^{t} \Vert w(s) \Vert^2 ds \\ \displaystyle \qquad\quad~+ \int_{\tau}^{t} \Vert b(s,u_s) - b(s,v_s) \Vert^2 ds\,,\\ \displaystyle \qquad\quad~\leq \Vert w(\tau) \Vert^2 + 2 \left(\frac{1}{2}+k -\lambda_1\right) \int_{\tau}^{t} \Vert w(s) \Vert^2 ds + C_b\int_{\tau -r}^{t} \Vert w(s) \Vert^2 ds\,,\\ \displaystyle \qquad\quad~\leq \Vert w(\tau) \Vert^2 + 2 \left(\frac{1}{2}+k -\lambda_1\right) \int_{\tau}^{t} \Vert w(s) \Vert^2 ds + C_b\int_{\tau -r}^{\tau} \Vert w(s) \Vert^2 ds \\ \displaystyle \qquad\quad~+ C_b \int_{\tau }^{t} \Vert w(s) \Vert^2 ds\,,\\ \displaystyle \qquad\quad~\leq \Vert w(\tau) \Vert^2 + C_b\int_{\tau -r}^{\tau} \Vert w(s) \Vert^2 ds + 2 \left(\frac{1}{2}+k + \frac{C_b}{2}-\lambda_1\right) \int_{\tau}^{t} \Vert w(s) \Vert^2 ds\,. \end{array}$$

By the Gronwall lemma, for all tτ, one deduces

w(t)2w(τ)2+Cbτrτw(s)2dseν(tτ),u0v02+CbφϕL2([r,0];L2(Ω))2eν(tτ),$$\begin{array}{} \displaystyle \Vert w(t) \Vert^2 \leq \left(\Vert w(\tau) \Vert^2 + C_b \int_{\tau -r}^{\tau} \Vert w(s) \Vert^2 ds\right) e^{\nu (t-\tau)}\,,\\ \displaystyle \qquad\quad~ \leq \left(\Vert u^0 -v^0 \Vert^2 + C_b \Vert \varphi - \phi \Vert_{L^2([-r,0];L^2(\Omega))}^2 \right) e^{\nu (t-\tau)}\,, \end{array}$$

and by this last estimate, we proved (3.3). Now, assume that tτ + r, so t + θτ for all θ ∈ [–r,0] and one has

w(t+θ)2u0v02+CbφϕL2([r,0];L2(Ω))2eν(t+θτ),u0v02+CbφϕL2([r,0];L2(Ω))2eν(trτ).$$\begin{array}{} \displaystyle \Vert w(t+\theta) \Vert^2 \leq \left(\Vert u^0 -v^0 \Vert^2 + C_b \Vert \varphi - \phi \Vert_{L^2([-r,0];L^2(\Omega))}^2 \right) e^{\nu (t+\theta-\tau)}\,,\\ \displaystyle \qquad\qquad~~~~\leq \left(\Vert u^0 -v^0 \Vert^2 + C_b \Vert \varphi - \phi \Vert_{L^2([-r,0];L^2(\Omega))}^2 \right) e^{\nu (t-r-\tau)}\,. \end{array}$$

Hence, we conclude

wtC([r,0];L2(Ω))u0v02+CbφϕL2([r,0];L2(Ω))2eν(trτ).$$\begin{array}{} \displaystyle \Vert w_t \Vert_{C([-r,0];L^2(\Omega))} \leq \left(\Vert u^0 -v^0 \Vert^2 + C_b \Vert \varphi - \phi \Vert_{L^2([-r,0];L^2(\Omega))}^2 \right) e^{\nu (t-r-\tau)}\,. \end{array}$$

By this last estimate we finished the proof of this lemma. ◼

Theorem 3

Under the previous assumptions, the mapping S(.,.) defined in(3.1), is a continuous process for all τt.

Proof

The proof of this theorem is as the proof of Theorem 9 in [1]. The uniqueness of the solutions implies that S(.,.) is a process. For the continuity of S(.,.), we use the previous lemma. We consider (u0,φ), (v0,ϕ)∈ H and u, v are their corresponding solutions. Firstly, if we take tτ + r, it follows from (3.4)

utvtL2([r,0];L2(Ω))2=r0u(t+θ)v(t+θ)2dθ,r0sups[r,0]u(t+s)v(t+s)2dθ,ru0v02+Cbφϕ2eν(trτ).$$\begin{array}{} \displaystyle \Vert u_t -v_t \Vert^2_{L^2([-r,0];L^2(\Omega))} = \int_{-r}^{0} \Vert u(t+\theta) - v(t+\theta) \Vert^2 d\theta\,,\\ \displaystyle \qquad\qquad\qquad\qquad\quad~\leq \int_{-r}^{0} \sup_{s\in [-r,0]}\Vert u(t+s) - v(t+s) \Vert^2 d\theta\,,\\ \displaystyle \qquad\qquad\qquad\qquad\quad~\leq r \left(\Vert u^0-v^0 \Vert^2 + C_b \Vert \varphi - \phi \Vert^2\right) e^{\nu (t-r-\tau)}\,. \end{array}$$

Now, for t ∈[τ,τ + r], we deduce

utvtL2([r,0];L2(Ω))2=r0u(t+θ)v(t+θ)2dθ,ru0v02+(Cbr+1)φϕ2eν(trτ).$$\begin{array}{} \displaystyle \Vert u_t -v_t \Vert^2_{L^2([-r,0];L^2(\Omega))} = \int_{-r}^{0} \Vert u(t+\theta) - v(t+\theta) \Vert^2 d\theta\,,\\ \displaystyle \qquad\qquad\qquad\qquad\quad~\leq \left(r\Vert u^0-v^0 \Vert^2 + (C_br+1) \Vert \varphi - \phi \Vert^2\right) e^{\nu (t-r-\tau)}\,. \end{array}$$

So, for all tτ, we have

utvtL2([r,0];L2(Ω))2ru0v02+(Cbr+1)φϕ2eν(trτ).$$\begin{array}{} \displaystyle \Vert u_t -v_t \Vert^2_{L^2([-r,0];L^2(\Omega))} \leq \left(r\Vert u^0-v^0 \Vert^2 + (C_br+1) \Vert \varphi - \phi \Vert^2\right) e^{\nu (t-r-\tau)}\,. \end{array}$$

Hence, by this last estimate and (3.3) we deduce the continuity of S(t,τ).◼

Existence of pullback D-absorbing set in C([–r,0]; L2(Ω)) and H

Firstly, we need to the following lemma, it relates the absorption properties for the mappings with those of process S in the fact that, proving those for U yields to similar properties for S.

Lemma 2

Assume that the family of bounded sets {B(t): t ∈ ℝ} in the space C([–r,0]; L2(Ω)) is pullback 𝒟-absorbing for the mapping U(.,.). Then the family of bounded sets {j(B(t)): t ∈ ℝ} in L2(Ω) × C([–r,0]; L2(Ω)) is pullback 𝒟-absorbing for the process S(.,.).

Proof

Let {D(t): t ∈ ℝ} be a family bounded sets in H, so there exists T > r such that

U(t,τ)D(τ)B(t),tτT.$$\begin{array}{} \displaystyle U(t,\tau)D(\tau) \subset B(t)\,,\; \forall t-\tau \geq T\,. \end{array}$$

On the other hand, we have

S(t,τ)(u0,φ)=j(U(t,τ)(u0,φ)),$$\begin{array}{} \displaystyle S(t,\tau)(u^0,\varphi) = j(U(t,\tau)(u^0,\varphi))\,, \end{array}$$

it follows that

S(t,τ)(u0,φ)=j(U(t,τ)(u0,φ))j(B(t)),tτT.$$\begin{array}{} \displaystyle S(t,\tau)(u^0,\varphi) = j(U(t,\tau)(u^0,\varphi)) \subset j(B(t))\,,\; \forall t-\tau \geq T\,. \end{array}$$

Remark 2

Noticing that the word absorbing used in this papier should be interpreted in a generalized sense, since U is not a process.

Now, we need the following estimations.

Lemma 3

Assume thatgLloc2(R;L2(Ω)),$\begin{array}{} \displaystyle g\in L^2_{loc}(\mathbb{R};L^2(\Omega))\,, \end{array}$there exists a small enough α < 2λ1 – 2–Cbsuch that

teαtg(s)2ds<,$$\begin{array}{} \displaystyle \int_{-\infty}^{t} e^{\alpha t} \Vert g(s) \Vert^2 ds \lt \infty \,, \end{array}$$

the functionfsatisfies(1.2)-(1.5)andbfulfills conditions (I)-(IV) and

τteσsb(s,us)b(s,vs)2dsCbτrteσsu(s)v(s)2ds.$$\begin{array}{} \displaystyle \int_{\tau}^{t} e^{\sigma s} \Vert b(s,u_s) - b(s,v_s) \Vert^2 ds \leq C_b \int_{\tau -r}^{t} e^{\sigma s} \Vert u(s)-v(s) \Vert^2 ds\,. \end{array}$$

Then we have

u(t)2eα(tτ)u(τ)2+Cbeα(tτ)τrτu(s)2ds+2c|Ω|α11eα(tτ)+eαtteαsg(s)2ds,$$\begin{array}{} \displaystyle \Vert u(t) \Vert^2 \leq e^{-\alpha(t- \tau)}\Vert u(\tau) \Vert^2 + C_b e^{-\alpha (t-\tau)}\int_{\tau-r}^{\tau} \Vert u(s)\Vert^2 ds \nonumber\\ \displaystyle \qquad\quad\,\,+\,2 c\vert \Omega \vert \alpha^{-1} \left(1- e^{-\alpha (t-\tau)}\right) + e^{-\alpha t}\int_{-\infty}^{t} e^{\alpha s} \Vert g(s) \Vert^2 ds \,, \end{array}$$

and

ηeαtτteαsu(s)2ds+2μ1eαtτteαsu(s)Lp(Ω)pdseα(tτ)u(τ)2+Cbeα(tτ)τrτu(s)2ds+2c|Ω|α11eα(tτ)+eαtteαsg(s)2ds,$$\begin{array}{} \displaystyle \quad\eta e^{-\alpha t}\int_{\tau}^{t} e^{\alpha s}\Vert u(s) \Vert^2 ds + 2\mu_1e^{-\alpha t} \int_{\tau}^{t} e^{\alpha s}\Vert u(s) \Vert^p_{L^p(\Omega)} ds \nonumber\\ \displaystyle \leq e^{-\alpha(t- \tau)}\Vert u(\tau) \Vert^2 + C_b e^{-\alpha (t-\tau)}\int_{\tau-r}^{\tau} \Vert u(s)\Vert^2 ds + 2 c\vert \Omega \vert \alpha^{-1} \left(1- e^{-\alpha (t-\tau)}\right) \nonumber\\ \displaystyle +\, e^{-\alpha t}\int_{-\infty}^{t} e^{\alpha s} \Vert g(s) \Vert^2 ds \,, \end{array}$$

where η : = 2λ1 – 2 – αCb.

Proof

Multiplying (1.1) by u and integrating over Ω, one has

12ddtu(t)2+u(t)2=Ωf(u)u+Ωb(t,ut)u+Ωgu.$$\begin{array}{} \displaystyle \frac{1}{2} \frac{d}{dt} \Vert u(t) \Vert^2 + \Vert \nabla u(t) \Vert^2 = \int_{\Omega} f(u) u + \int_{\Omega} b(t,u_t) u + \int_{\Omega} g u\,. \end{array}$$

By (1.2), Cauchy-Shwarz and Young inequalities, we obtain

12ddtu(t)2+u(t)2+μ1u(t)Lp(Ω)pc|Ω|+12b(t,ut)2+12g(t)2+u(t)2.$$\begin{array}{} \displaystyle \frac{1}{2} \frac{d}{dt} \Vert u(t) \Vert^2 + \Vert \nabla u(t) \Vert^2 + \mu_1 \Vert u(t) \Vert^p_{L^p(\Omega)} \leq c\vert \Omega \vert + \frac{1}{2} \Vert b(t,u_t)\Vert^2 + \frac{1}{2}\Vert g(t) \Vert^2 + \Vert u(t) \Vert^2\,. \end{array}$$

Since λ1u2 ≤ ∥∇u2 and after calculation, one has

ddtu(t)2+2(λ11)u(t)2+2μ1u(t)Lp(Ω)p2c|Ω|+b(t,ut)2+g(t)2.$$\begin{array}{} \displaystyle \frac{d}{dt} \Vert u(t) \Vert^2 + 2(\lambda_1 - 1)\Vert u(t) \Vert^2 + 2\mu_1 \Vert u(t) \Vert^p_{L^p(\Omega)} \leq 2 c\vert \Omega \vert + \Vert b(t,u_t)\Vert^2 + \Vert g(t) \Vert^2 \,. \end{array}$$

Now, we multiply this last estimate by eαt such that 0 < α < 2λ1 –2 – Cb, so one gets

eαtddtu(t)2+2(λ11)eαtu(t)2+2μ1eαtu(t)Lp(Ω)p2c|Ω|eαt+eαtb(t,ut)2+eαtg(t)2.$$\begin{array}{} \quad \displaystyle e^{\alpha t} \frac{d}{dt} \Vert u(t) \Vert^2 + 2(\lambda_1 - 1)e^{\alpha t}\Vert u(t) \Vert^2 + 2\mu_1 e^{\alpha t}\Vert u(t) \Vert^p_{L^p(\Omega)} \\ \leq 2 c\vert \Omega \vert e^{\alpha t} + e^{\alpha t} \Vert b(t,u_t)\Vert^2 + e^{\alpha t} \Vert g(t) \Vert^2 \,. \end{array}$$

On the other hand, we have

ddteαtu(t)2=αeαtu(t)2+eαtddtu(t)2$$\begin{array}{} \displaystyle \frac{d}{dt}\left(e^{\alpha t}\Vert u(t) \Vert^2\right) = \alpha e^{\alpha t} \Vert u(t) \Vert^2 + e^{\alpha t} \frac{d}{dt} \Vert u(t) \Vert^2 \end{array}$$

We substitute (3.9) in this equality, we find

ddteαtu(t)2αeαtu(t)22(λ11)eαtu(t)22μ1eαtu(t)Lp(Ω)p+2c|Ω|eαt+eαtb(t,ut)2+eαtg(t)2.$$\begin{array}{} \displaystyle \frac{d}{dt}\left(e^{\alpha t}\Vert u(t) \Vert^2\right) \leq \alpha e^{\alpha t} \Vert u(t) \Vert^2 -2 (\lambda_1 - 1)e^{\alpha t}\Vert u(t) \Vert^2 - 2\mu_1 e^{\alpha t}\Vert u(t) \Vert^p_{L^p(\Omega)} \\ \displaystyle \qquad\qquad\qquad\quad\,+ 2 c\vert \Omega \vert e^{\alpha t} + e^{\alpha t} \Vert b(t,u_t)\Vert^2 + e^{\alpha t} \Vert g(t) \Vert^2 \,. \end{array}$$

Integrating this last estimate over [τ, t], one obtains

eαtu(t)2eατu(τ)2+2c|Ω|α1(eαteατ)+(α+22λ1)τteαsu(s)2ds2μ1τteαsu(s)Lp(Ω)pds+τteαsb(s,us)2ds+τteαsg(s)2ds.$$\begin{array}{} \displaystyle e^{\alpha t}\Vert u(t) \Vert^2 \leq e^{\alpha \tau}\Vert u(\tau) \Vert^2 + 2 c\vert \Omega \vert \alpha^{-1} (e^{\alpha t}- e^{\alpha \tau}) \\ \displaystyle \qquad\qquad\,\,\,\,+\, (\alpha +2-2 \lambda_1) \int_{\tau}^{t} e^{\alpha s}\Vert u(s) \Vert^2 ds - 2\mu_1 \int_{\tau}^{t} e^{\alpha s}\Vert u(s) \Vert^p_{L^p(\Omega)} ds\\ \displaystyle \qquad\qquad\,\,\,\,+ \int_{\tau}^{t} e^{\alpha s} \Vert b(s,u_s)\Vert^2 ds + \int_{\tau}^{t} e^{\alpha s} \Vert g(s) \Vert^2 ds \,. \end{array}$$

Using (3.6) and (II), one has

eαtu(t)2eατu(τ)2+2c|Ω|α1(eαteατ)+(α+22λ1)τteαsu(s)2ds2μ1τteαsu(s)Lp(Ω)pds+Cbτrteαsu(s)2ds+τteαsg(s)2ds.$$\begin{array}{} \displaystyle e^{\alpha t}\Vert u(t) \Vert^2 \leq e^{\alpha \tau}\Vert u(\tau) \Vert^2 + 2 c\vert \Omega \vert \alpha^{-1} (e^{\alpha t}- e^{\alpha \tau}) \nonumber\\ \displaystyle \qquad\qquad\,\,\,\,+\, (\alpha +2-2 \lambda_1) \int_{\tau}^{t} e^{\alpha s}\Vert u(s) \Vert^2 ds - 2\mu_1 \int_{\tau}^{t} e^{\alpha s}\Vert u(s) \Vert^p_{L^p(\Omega)} ds \nonumber\\ \displaystyle \qquad\qquad\,\,\,\,+ \,C_b\int_{\tau-r}^{t} e^{\alpha s} \Vert u(s)\Vert^2 ds + \int_{\tau}^{t} e^{\alpha s} \Vert g(s) \Vert^2 ds \,. \end{array}$$

On the other hand, we have

τrteαsu(s)2ds=τrτeαsu(s)2ds+τteαsu(s)2ds,eαττrτu(s)2ds+τteαsu(s)2ds.$$\begin{array}{} \displaystyle \int_{\tau-r}^{t} e^{\alpha s} \Vert u(s)\Vert^2 ds = \int_{\tau-r}^{\tau} e^{\alpha s} \Vert u(s)\Vert^2 ds + \int_{\tau}^{t} e^{\alpha s} \Vert u(s)\Vert^2 ds \,,\nonumber\\ \displaystyle \qquad\qquad\qquad\qquad\leq e^{\alpha \tau}\int_{\tau-r}^{\tau} \Vert u(s)\Vert^2 ds + \int_{\tau}^{t} e^{\alpha s} \Vert u(s)\Vert^2 ds \,. \end{array}$$

So by (3.10) and (3.11), one finds

eαtu(t)2eατu(τ)2+2c|Ω|α1(eαteατ)+(α+22λ1+Cb)τteαsu(s)2ds2μ1τteαsu(s)Lp(Ω)pds+Cbeαττrτu(s)2ds+τteαsg(s)2ds.$$\begin{array}{} \displaystyle e^{\alpha t}\Vert u(t) \Vert^2 \leq e^{\alpha \tau}\Vert u(\tau) \Vert^2 + 2 c\vert \Omega \vert \alpha^{-1} (e^{\alpha t}- e^{\alpha \tau}) \\ \displaystyle \qquad\qquad\,\,\,\,+\, (\alpha +2-2 \lambda_1 + C_b) \int_{\tau}^{t} e^{\alpha s}\Vert u(s) \Vert^2 ds - 2\mu_1 \int_{\tau}^{t} e^{\alpha s}\Vert u(s) \Vert^p_{L^p(\Omega)} ds \\ \displaystyle \qquad\qquad\,\,\,\,+\, C_be^{\alpha \tau}\int_{\tau-r}^{\tau} \Vert u(s)\Vert^2 ds + \int_{\tau}^{t} e^{\alpha s} \Vert g(s) \Vert^2 ds \,. \end{array}$$

Hence, by (3.5) we obtain

u(t)2+(2λ1α2Cb)eαtτteαsu(s)2ds+2μ1eαtτteαsu(s)Lp(Ω)pdseα(tτ)u(τ)2+2c|Ω|α11eα(tτ)+Cbeα(tτ)τrτu(s)2ds+eαtteαsg(s)2ds.$$\begin{array}{} \displaystyle \,\,\,\,\Vert u(t) \Vert^2 + (2 \lambda_1 - \alpha - 2 - C_b) e^{-\alpha t}\int_{\tau}^{t} e^{\alpha s}\Vert u(s) \Vert^2 ds \\ \displaystyle + 2\mu_1e^{-\alpha t} \int_{\tau}^{t} e^{\alpha s}\Vert u(s) \Vert^p_{L^p(\Omega)} ds \\ \displaystyle \leq e^{-\alpha(t- \tau)}\Vert u(\tau) \Vert^2 + 2 c\vert \Omega \vert \alpha^{-1} \left(1- e^{-\alpha (t-\tau)}\right) + C_b e^{-\alpha (t-\tau)}\int_{\tau-r}^{\tau} \Vert u(s)\Vert^2 ds \\ \displaystyle + e^{-\alpha t}\int_{-\infty}^{t} e^{\alpha s} \Vert g(s) \Vert^2 ds \,. \end{array}$$

Thus, for η : = 2 λ1α – 2 – Cb > 0, by this last estimate we get

u(t)2eα(tτ)u(τ)2+Cbeα(tτ)τrτu(s)2ds+2c|Ω|α11eα(tτ)+eαtteαsg(s)2ds,$$\begin{array}{} \displaystyle \Vert u(t) \Vert^2 \leq e^{-\alpha(t- \tau)}\Vert u(\tau) \Vert^2 + C_b e^{-\alpha (t-\tau)}\int_{\tau-r}^{\tau} \Vert u(s)\Vert^2 ds \\ \displaystyle \qquad\quad\,\,+ 2 c\vert \Omega \vert \alpha^{-1} \left(1- e^{-\alpha (t-\tau)}\right) + e^{-\alpha t}\int_{-\infty}^{t} e^{\alpha s} \Vert g(s) \Vert^2 ds \,, \end{array}$$

and

ηeαtτteαsu(s)2ds+2μ1eαtτteαsu(s)Lp(Ω)pdseα(tτ)u(τ)2+Cbeα(tτ)τrτu(s)2ds+2c|Ω|α11eα(tτ)+eαtteαsg(s)2ds,$$\begin{array}{} \displaystyle \quad\eta e^{-\alpha t}\int_{\tau}^{t} e^{\alpha s}\Vert u(s) \Vert^2 ds + 2\mu_1e^{-\alpha t} \int_{\tau}^{t} e^{\alpha s}\Vert u(s) \Vert^p_{L^p(\Omega)} ds \\ \displaystyle \leq e^{-\alpha(t- \tau)}\Vert u(\tau) \Vert^2 + C_b e^{-\alpha (t-\tau)}\int_{\tau-r}^{\tau} \Vert u(s)\Vert^2 ds + 2 c\vert \Omega \vert \alpha^{-1} \left(1- e^{-\alpha (t-\tau)}\right) \\ \displaystyle+ e^{-\alpha t}\int_{-\infty}^{t} e^{\alpha s} \Vert g(s) \Vert^2 ds \,, \end{array}$$

for all tτ. So by these two estimations the proof of the lemma is finished. ◼

Proposition 1

Under the assumptions inlemma (3). Then the family {B1(t) : t ∈ ℝ} given by

B1(t)=B¯C([r,0];L2(Ω))(0,R1(t)),$$\begin{array}{} \displaystyle B_1(t) = \overline{B}_{C([-r,0]; L^2(\Omega))}(0, R_1(t))\,, \end{array}$$

with

R12(t)=eαr2c|Ω|α1+eαtteαtg(s)2ds,tR;$$\begin{array}{} \displaystyle R_1^2(t)= e^{\alpha r} \left(2 c \vert \Omega \vert \alpha^{-1} + e^{-\alpha t} \int_{-\infty}^{t} e^{\alpha t} \Vert g(s) \Vert^2 ds \right) \,,\; \forall t \in \mathbb{R}\,; \end{array}$$

is pullback 𝓓-absorbing for the mappingU(t, τ). Moreover, the family {B0(t) : t ∈ ℝ} given by

B0(t)=B¯L2(Ω))(0,R1(t))×B¯L2([r,0];L2(Ω))0,rR1(t)H,tR,$$\begin{array}{} \displaystyle B_0(t) = \overline{B}_{L^2(\Omega))}(0, R_1(t)) \times \overline{B}_{L^2([-r,0]; L^2(\Omega))}\left(0, \sqrt{r}R_1(t)\right) \subset H\,,\; \forall t\in \mathbb{R}\,, \end{array}$$

is pullback 𝓓-absorbing for the process S defined by(3.1).

Proof

The first part may be proved as follows.

By definition, we have

U(t,τ)(u0,φ)C([r,0];L2(Ω))2=sups[r,0]u(t+s)2.$$\begin{array}{} \displaystyle \Vert U(t,\tau)(u^0,\varphi) \Vert_{C([-r,0];L^2(\Omega))}^2 = \sup_{s\in[-r,0]} \Vert u(t+s) \Vert^2 \,. \end{array}$$

From (3.7), if we take tτ + r, so t + θτ. Then one has

u(t+θ)2eα(t+θτ)u(τ)2+Cbeα(t+θτ)φL2([r,0];L2(Ω))2+2c|Ω|α11eα(t+θτ)+eα(t+θ)t+θeαsg(s)2ds,$$\begin{array}{} \displaystyle \Vert u(t+\theta) \Vert^2 \leq e^{-\alpha(t+\theta- \tau)}\Vert u(\tau) \Vert^2 + C_b e^{-\alpha (t+\theta-\tau)} \Vert \varphi\Vert_{L^2([-r,0];L^2(\Omega))}^2 \\ \displaystyle \qquad\qquad\quad+ 2 c\vert \Omega \vert \alpha^{-1} \left(1- e^{-\alpha (t+\theta-\tau)}\right) + e^{-\alpha (t+\theta)}\int_{-\infty}^{t+\theta} e^{\alpha s} \Vert g(s) \Vert^2 ds \,, \end{array}$$

which implies that

sups[r,0]u(t+s)2eα(trτ)u(τ)2+Cbeα(trτ)φL2([r,0];L2(Ω))2+2c|Ω|α1eαreαreα(tτ)+eα(tr)teαsg(s)2ds,$$\begin{array}{} \displaystyle \sup_{s\in [-r,0]}\Vert u(t+s) \Vert^2 \leq e^{-\alpha(t-r- \tau)}\Vert u(\tau) \Vert^2 + C_b e^{-\alpha (t-r-\tau)}\Vert \varphi\Vert_{L^2([-r,0];L^2(\Omega))}^2 \nonumber\\ \displaystyle \qquad\qquad\qquad\qquad+ \,2 c\vert \Omega \vert \alpha^{-1} e^{\alpha r}\left(e^{-\alpha r}- e^{-\alpha (t-\tau)}\right) \nonumber\\ \displaystyle \qquad\qquad\qquad\qquad+\, e^{-\alpha (t-r)}\int_{-\infty}^{t} e^{\alpha s} \Vert g(s) \Vert^2 ds \,, \end{array}$$

On the one hand, we have

φL2([r,0];L2(Ω))2=r0φ(θ)2dθ,r0sups[r,0]φ(s)2dθ,rφC([r,0];L2(Ω))2.$$\begin{array}{} \displaystyle \Vert \varphi\Vert_{L^2([-r,0];L^2(\Omega))}^2 = \int_{-r}^{0} \Vert \varphi(\theta)\Vert^2 d\theta \,,\nonumber\\ \displaystyle \qquad\qquad\qquad\quad\,\,\leq \int_{-r}^{0} \sup_{s\in [-r,0]}\Vert \varphi(s)\Vert^2 d\theta\,,\nonumber\\ \displaystyle \qquad\qquad\qquad\quad\,\,\leq r \Vert \varphi \Vert^2_{C([-r,0];L^2(\Omega))}\,. \end{array}$$

Therefore by (3.12), (3.13) and the fact that u(τ) = φ (0), we obtain

sups[r,0]u(t+s)2eα(trτ)φ(0)2+Cbreα(trτ)φC([r,0];L2(Ω))2+2c|Ω|α1eαreαreα(tτ)+eα(tr)teαsg(s)2ds,(1+Cbr)eα(trτ)φC([r,0];L2(Ω))2+2c|Ω|α1eαr+eα(tr)teαsg(s)2ds.$$\begin{array}{} \displaystyle \quad\sup_{s\in [-r,0]}\Vert u(t+s) \Vert^2 \leq e^{-\alpha(t-r- \tau)}\Vert \varphi(0) \Vert^2 + C_b r e^{-\alpha (t-r-\tau)}\Vert \varphi\Vert_{C([-r,0];L^2(\Omega))}^2 \\ \displaystyle +\, 2 c\vert \Omega \vert \alpha^{-1} e^{\alpha r}\left(e^{-\alpha r}- e^{-\alpha (t-\tau)}\right) + e^{-\alpha (t-r)}\int_{-\infty}^{t} e^{\alpha s} \Vert g(s) \Vert^2 ds \,,\\ \displaystyle \leq (1+C_b r) e^{-\alpha (t-r-\tau)} \Vert \varphi\Vert_{C([-r,0];L^2(\Omega))}^2 + 2 c\vert \Omega \vert \alpha^{-1} e^{\alpha r} \\ \displaystyle + \,e^{-\alpha (t-r)}\int_{-\infty}^{t} e^{\alpha s} \Vert g(s) \Vert^2 ds \,. \end{array}$$

Then, we find

U(t,τ)(u0,φ)C([r,0];L2(Ω))2=sups[r,0]u(t+s)2,(1+Cbr)eα(trτ)φC([r,0];L2(Ω))2+eαr2c|Ω|α1+eαtteαsg(s)2ds,$$\begin{array}{} \displaystyle \quad\Vert U(t,\tau)(u^0,\varphi) \Vert_{C([-r,0];L^2(\Omega))}^2 = \sup_{s\in [-r,0]}\Vert u(t+s) \Vert^2 \,,\nonumber\\ \leq (1+C_b r) e^{-\alpha (t-r-\tau)} \Vert \varphi\Vert_{C([-r,0];L^2(\Omega))}^2 \nonumber\\ +\, e^{\alpha r} \left(2 c\vert \Omega \vert \alpha^{-1} + e^{-\alpha t}\int_{-\infty}^{t} e^{\alpha s} \Vert g(s) \Vert^2 ds\right) \,, \end{array}$$

for all (u0, φ) ∈ H and all tτ + r.

Let 𝓡 be the set of all functions ρ : ℝ ⟶ (0, + ∞) such that

limteαtρ2(t)=0.$$\begin{array}{} \displaystyle \lim_{t\rightarrow -\infty} e^{\alpha t} \rho^{2}(t) =0\,. \end{array}$$

By 𝓓 we denote the class of all families = {D(t) : t ∈ ℝ} ⊂ 𝓟(C([–r, 0];L2(Ω))) such that D(t) ⊂ BC([–r, 0];L2(Ω))(0,ρ(t)), for some ρ ∈ 𝓡, where we denote by BC([–r, 0];L2(Ω))(0, ρ(t)) the closed ball in C([–r, 0];L2(Ω)) centered at 0 with radius ρ(t). Let

R12(t)=eαr2c|Ω|α1+eαtteαsg(s)2ds.$$\begin{array}{} \displaystyle R_1^2(t)= e^{\alpha r} \left(2 c\vert \Omega \vert \alpha^{-1} + e^{-\alpha t}\int_{-\infty}^{t} e^{\alpha s} \Vert g(s) \Vert^2 ds\right)\,. \end{array}$$

Thus, for all ∈ 𝓓 and all t ∈ ℝ, by (3.14) there exists τ0 (, t) ≤ t such that

U(t,τ)(u0,φ)C([r,0];L2(Ω))2R12(t),$$\begin{array}{} \displaystyle \Vert U(t,\tau)(u^0,\varphi) \Vert_{C([-r,0];L^2(\Omega))}^2 \leq R_1^2(t)\,, \end{array}$$

for all ττ0(, t); i.e., B1(t) = BC([–r, 0]; L2(Ω))(0, R1(t)) is pullback 𝓓-absorbing for the mapping U(t, τ).

Concerning the second part, we observe that {j(B(t)), t ∈ ℝ} is a family of pullback 𝓓-absorbing sets for the process S. On the other hand, since

φL2([r,0];L2(Ω))2rφC([r,0];L2(Ω))2,$$\begin{array}{} \displaystyle \Vert \varphi\Vert_{L^2([-r,0];L^2(\Omega))}^2 \leq r \Vert \varphi\Vert_{C([-r,0];L^2(\Omega))}^2 \,, \end{array}$$

and

j(B(t))=(φ(0),φ):φB¯C([r,0];L2(Ω))(0,R1(t)),$$\begin{array}{} \displaystyle j(B(t)) = \left\{(\varphi(0),\varphi)\,:\; \varphi \in \overline{B}_{C([-r,0]; L^2(\Omega))}(0, R_1(t))\right\}\,, \end{array}$$

we deduce that

j(B(t))B¯L2(Ω))(0,R1(t))×B¯L2([r,0];L2(Ω))0,rR1(t)=B0(t),$$\begin{array}{} \displaystyle j(B(t)) \subset \overline{B}_{L^2(\Omega))}(0, R_1(t)) \times \overline{B}_{L^2([-r,0]; L^2(\Omega))}\left(0, \sqrt{r}R_1(t)\right) = B_0(t)\,, \end{array}$$

which implies that the family {B0(t): t ∈ ℝ} is pullback 𝓓-absorbing sets for the process S. ◼

Existence of pullback D-absorbing set in C([r,0];H01(Ω))$\begin{array}{} \displaystyle C([-r,0]; H^1_0(\Omega)) \end{array}$
Proposition 2

Suppose that conditions oflemma (3)are satisfied, if there exists a sufficiently smallαsuch that

α<α<min2λ11λ1,2μ1.$$\begin{array}{} \displaystyle \alpha \lt \alpha^* \lt \min \left\{ 2 \frac{\lambda_1 -1}{\lambda_1}\,,\, 2\mu_1 \right\}\,. \end{array}$$

Then the family {B2(t) : t ∈ ℝ} given by

B2(t)=B¯C([r,0];H01(Ω))(0,R2(t)),$$\begin{array}{} \displaystyle B_2(t) = \overline{B}_{C([-r,0]; H^1_0(\Omega))}(0, R_2(t))\,, \end{array}$$

where

R22(t)=2c|Ω|α1eαr+2Cbα1η1eαr+2Cbη1eα(tr)teαsg(s)2ds+2eα(tr)teαsg(s)2ds,tR,$$\begin{array}{} \displaystyle R_2^2(t)= 2c \vert \Omega \vert \left({\alpha^*}^{-1} e^{\alpha^* r }+ 2C_b \alpha^{-1} \eta^{-1} e^{\alpha r}\right) \\ \displaystyle \qquad\,\,\,\,+\, 2C_b \eta^{-1} e^{-\alpha (t-r)}\int_{-\infty}^{t} e^{\alpha s} \Vert g(s) \Vert^2 ds + 2 e^{-\alpha^* (t-r)} \int_{-\infty}^{t} e^{\alpha^* s}\Vert g(s) \Vert^2 ds \,,\; \forall t\in \mathbb{R}\,, \end{array}$$

is pullback 𝓓-absorbing for the mapping U(t, τ).

Proof

Multipying (1.1) by u+ut$\begin{array}{} \displaystyle u+\frac{\partial u}{\partial t} \end{array}$ and integrating over Ω, we obtain

ddtu(t)2+12ddtu(t)2+u(t)2=Ωf(u)u+ut+Ωb(t,ut)u+ut+Ωgu+ut.$$\begin{array}{} \displaystyle \quad \left\Vert \frac{d}{dt} u(t) \right\Vert^2 + \frac{1}{2} \frac{d}{dt} \left(\Vert u(t) \Vert^2 + \Vert \nabla u(t) \Vert^2\right) \\ \displaystyle = \int_{\Omega} f(u)\left( u+\frac{\partial u}{\partial t} \right) + \int_{\Omega} b(t,u_t) \left( u+\frac{\partial u}{\partial t} \right) + \int_{\Omega} g\left( u+\frac{\partial u}{\partial t} \right)\,. \end{array}$$

Using (1.2), (1.5), (2.5) and Young inequality, one finds

2ddtu(t)2+ddtu(t)2+u(t)2+2μ1u(t)Lp(Ω)p+2u(t)2+2μ1u(t)Lp(Ω)p2c|Ω|+2b(t,ut)2+2g(t)2+2ddtu(t)2+2u(t)2.$$\begin{array}{} \displaystyle \,\,\,\,\,2\left\Vert \frac{d}{dt} u(t) \right\Vert^2 + \frac{d}{dt} \left(\Vert u(t) \Vert^2 + \Vert \nabla u(t) \Vert^2 + 2\mu'_1 \Vert u(t) \Vert_{L^p(\Omega)}^p \right) \\ \displaystyle + 2 \Vert \nabla u(t) \Vert^2 + 2\mu_1 \Vert u(t) \Vert_{L^p(\Omega)}^p \\ \displaystyle \leq 2c \vert\Omega \vert + 2\Vert b(t,u_t) \Vert^2 + 2\Vert g(t) \Vert^2 + 2\left\Vert \frac{d}{dt} u(t) \right\Vert^2 + 2\Vert u(t) \Vert^2\,. \end{array}$$

By the fact that λ1u2 ≤ ∥∇u2, after simplification one has

ddtu(t)2+u(t)2+2μ1u(t)Lp(Ω)p+2(1λ11)u(t)2+2μ1u(t)Lp(Ω)p2c|Ω|+2b(t,ut)2+2g(t)2.$$\begin{array}{} \displaystyle \quad\frac{d}{dt} \left(\Vert u(t) \Vert^2 + \Vert \nabla u(t) \Vert^2 + 2\mu'_1 \Vert u(t) \Vert_{L^p(\Omega)}^p \right) \\ \displaystyle + 2(1-\lambda_1^{-1}) \Vert \nabla u(t) \Vert^2 + 2\mu_1 \Vert u(t) \Vert_{L^p(\Omega)}^p \\ \displaystyle \leq 2c \vert\Omega \vert + 2\Vert b(t,u_t) \Vert^2 + 2\Vert g(t) \Vert^2 \,. \end{array}$$

Since α in lemma (3) is small enough, we can choose a positive constant α sufficiently small with α < α* < min2λ11λ1,2μ1,$\begin{array}{} \displaystyle \min \left\{ 2 \frac{\lambda_1 -1}{\lambda_1}\,,\, 2\mu_1 \right\}\,, \end{array}$ such that

2(1λ11)u(t)2α(u(t)2+u(t)2).$$\begin{array}{} \displaystyle 2(1-\lambda_1^{-1}) \Vert \nabla u(t) \Vert^2 \geq \alpha^* (\Vert u(t) \Vert^2 + \Vert \nabla u(t) \Vert^2)\,. \end{array}$$

So, we can write

ddtγ1(t)+αγ1(t)2c|Ω|+2b(t,ut)2+2g(t)2,$$\begin{array}{} \displaystyle \frac{d}{dt}\gamma_1 (t) + \alpha^* \gamma_1 (t) \leq 2c \vert \Omega \vert + 2 \Vert b(t,u_t) \Vert^2 + 2 \Vert g(t) \Vert^2 \,, \end{array}$$

where

γ1(t)=u(t)2+u(t)2+2μ1u(t)Lp(Ω)p.$$\begin{array}{} \displaystyle \gamma_1 (t) = \Vert u(t) \Vert^2 + \Vert \nabla u(t) \Vert^2 + 2\mu'_1 \Vert u(t) \Vert_{L^p(\Omega)}^p \,. \end{array}$$

Multiplying (3.16) by eαt, one has

eαtddtγ1(t)+αeαtγ1(t)2c|Ω|eαt+2eαtb(t,ut)2+2eαtg(t)2.$$\begin{array}{} \displaystyle e^{\alpha^* t} \frac{d}{dt}\gamma_1 (t) + \alpha^* e^{\alpha^* t} \gamma_1 (t) \leq 2c \vert \Omega \vert e^{\alpha^* t} + 2 e^{\alpha^* t} \Vert b(t,u_t) \Vert^2+ 2 e^{\alpha^* t}\Vert g(t) \Vert^2 \,. \end{array}$$

On the other hand, we have

ddteαtγ1(t)=αeαtγ1(t)+eαtddtγ1(t)$$\begin{array}{} \displaystyle \frac{d}{dt}\left(e^{\alpha^* t} \gamma_1 (t)\right) = \alpha^* e^{\alpha^* t} \gamma_1 (t) + e^{\alpha^* t} \frac{d}{dt}\gamma_1 (t) \end{array}$$

Then, by (3.18) and (3.19), we obtain

ddteαtγ1(t)αeαtγ1(t)αeαtγ1(t)+2c|Ω|eαt+2eαtb(t,ut)2+2eαtg(t)2,2c|Ω|eαt+2eαtb(t,ut)2+2eαtg(t)2.$$\begin{array}{} \displaystyle \frac{d}{dt}\left(e^{\alpha^* t} \gamma_1 (t)\right) \leq \alpha^* e^{\alpha^* t} \gamma_1 (t) - \alpha^* e^{\alpha^* t} \gamma_1 (t) + 2c \vert \Omega \vert e^{\alpha^* t} \\ \displaystyle \qquad\qquad\qquad\,\,\,+\,2 e^{\alpha^* t} \Vert b(t,u_t) \Vert^2+ 2 e^{\alpha^* t}\Vert g(t) \Vert^2 \,,\\ \displaystyle \qquad\qquad\qquad\,\,\,\leq 2c \vert \Omega \vert e^{\alpha^* t} + 2 e^{\alpha^* t} \Vert b(t,u_t) \Vert^2 + 2 e^{\alpha^* t}\Vert g(t) \Vert^2 \,. \end{array}$$

Integrating this last one from τ to t, one gets

eαtγ1(t)eατγ1(τ)+2c|Ω|τteαsds+2τteαsb(s,us)2ds+2τteαsg(s)2ds,+eατγ1(τ)+2c|Ω|α1eαteατ+2τteαsb(s,us)2ds++2τteαsg(s)2ds.$$\begin{array}{} \displaystyle e^{\alpha^* t} \gamma_1 (t) \leq e^{\alpha^* \tau} \gamma_1 (\tau) + 2c \vert \Omega \vert \int_{\tau}^{t}e^{\alpha^* s} ds+ 2 \int_{\tau}^{t} e^{\alpha^* s} \Vert b(s,u_s) \Vert^2 ds\\ \displaystyle \qquad\qquad+\, 2 \int_{\tau}^{t} e^{\alpha^* s}\Vert g(s) \Vert^2 ds\,,\\ \displaystyle \qquad\qquad+\,\leq e^{\alpha^* \tau} \gamma_1 (\tau) + 2c \vert \Omega \vert {\alpha^*}^{-1} \left ( e^{\alpha^* t} - e^{\alpha^* \tau}\right) + 2 \int_{\tau}^{t} e^{\alpha^* s} \Vert b(s,u_s) \Vert^2 ds\\ \displaystyle \qquad\qquad+\,+\, 2 \int_{\tau}^{t} e^{\alpha^* s}\Vert g(s) \Vert^2 ds\,. \end{array}$$

From (3.5) and (3.6), one finds

eαtγ1(t)eατγ1(τ)+2c|Ω|α1eαteατ+2Cbτrteαsu(s)2ds+2teαsg(s)2ds,eατγ1(τ)+2c|Ω|α1eαteατ+2Cbeαττrτu(s)2ds+2Cbτteαsu(s)2ds+2teαsg(s)2ds.$$\begin{array}{} \displaystyle e^{\alpha^* t} \gamma_1 (t) \leq e^{\alpha^* \tau} \gamma_1 (\tau) + 2c \vert \Omega \vert {\alpha^*}^{-1} \left ( e^{\alpha^* t} - e^{\alpha^* \tau}\right) + 2C_b \int_{\tau-r}^{t} e^{\alpha^* s} \Vert u(s) \Vert^2 ds\\ \displaystyle \qquad\qquad+\, 2 \int_{-\infty}^{t} e^{\alpha^* s}\Vert g(s) \Vert^2 ds\,,\\ \displaystyle \qquad\qquad\leq e^{\alpha^* \tau} \gamma_1 (\tau) + 2c \vert \Omega \vert {\alpha^*}^{-1} \left ( e^{\alpha^* t} - e^{\alpha^* \tau}\right) + 2C_b e^{\alpha^* \tau}\int_{\tau-r}^{\tau} \Vert u(s) \Vert^2 ds\\ \displaystyle \qquad\qquad+\, 2C_b \int_{\tau}^{t} e^{\alpha^* s} \Vert u(s) \Vert^2 ds + 2 \int_{-\infty}^{t} e^{\alpha^* s}\Vert g(s) \Vert^2 ds\,. \end{array}$$

We multiply this estimate by eαt, we obtain

γ1(t)eα(tτ)γ1(τ)+2Cbeα(tτ)τrτu(s)2ds+2c|Ω|α11eα(tτ)+2Cbeαtτteαsu(s)2ds+2eαtteαsg(s)2ds.$$\begin{array}{} \displaystyle \gamma_1(t) \leq e^{-\alpha^* (t-\tau)} \gamma_1 (\tau) +2C_b e^{-\alpha^* (t-\tau)}\int_{\tau-r}^{\tau} \Vert u(s) \Vert^2 ds \nonumber\\ \displaystyle \qquad\,\,+\,2c \vert \Omega \vert {\alpha^*}^{-1} \left ( 1 - e^{-\alpha^* (t-\tau)}\right) + 2C_b e^{-\alpha^* t} \int_{\tau}^{t} e^{\alpha^* s} \Vert u(s) \Vert^2 ds \nonumber\\ \displaystyle \qquad\,\,+\,2 e^{-\alpha^* t} \int_{-\infty}^{t} e^{\alpha^* s}\Vert g(s) \Vert^2 ds\,. \end{array}$$

On the one hand, since H01(Ω)L2(Ω)andH01(Ω)Lp(Ω),$\begin{array}{} \displaystyle H^1_0(\Omega) \subset L^2(\Omega)\, {\rm and}\, H^1_0(\Omega) \subset L^p(\Omega)\,, \end{array}$we have

γ1(τ)=u(τ)2+u(τ)2+2μ1u(τ)Lp(Ω)p,(1+λ11)u(τ)2+2μ1u(τ)Lp(Ω)p,(1+λ11)u(τ)2+k1u(τ)p,k2(1+λ11)u(τ)p+k1u(τ)p,k3u(τ)p.$$\begin{array}{} \displaystyle \gamma_1 (\tau) = \Vert u(\tau) \Vert^2 + \Vert \nabla u(\tau) \Vert^2 + 2\mu'_1 \Vert u(\tau) \Vert_{L^p(\Omega)}^p \,,\nonumber\\ \displaystyle \qquad\,\,\,\leq (1+ \lambda_1^{-1}) \Vert \nabla u(\tau) \Vert^2 + 2\mu'_1 \Vert u(\tau) \Vert_{L^p(\Omega)}^p \,,\nonumber\\ \displaystyle \qquad\,\,\, \leq (1+ \lambda_1^{-1}) \Vert \nabla u(\tau) \Vert^2 + k_1 \Vert \nabla u(\tau) \Vert^p\,,\nonumber\\ \displaystyle \qquad\,\,\, \leq k_2 (1+ \lambda_1^{-1}) \Vert \nabla u(\tau) \Vert^p + k_1 \Vert \nabla u(\tau) \Vert^p \,,\nonumber\\ \displaystyle \qquad\,\,\,\leq k_3 \Vert \nabla u(\tau) \Vert^p\,. \end{array}$$

So, by (3.17), (3.20) and (3.21), one finds

u(t)2+u(t)2+2μ1u(t)Lp(Ω)pk3eα(tτ)u(τ)p+2Cbeα(tτ)τrτu(s)2ds+2c|Ω|α11eα(tτ)+2Cbeαtτteαsu(s)2ds+2eαtteαsg(s)2ds.$$\begin{array}{} \displaystyle \,\,\,\,\,\Vert u(t) \Vert^2 + \Vert \nabla u(t) \Vert^2 + 2\mu'_1 \Vert u(t) \Vert_{L^p(\Omega)}^p \leq k_3 e^{-\alpha^* (t-\tau)} \Vert \nabla u(\tau) \Vert^p \\ \displaystyle +2C_b e^{-\alpha^* (t-\tau)}\int_{\tau-r}^{\tau} \Vert u(s) \Vert^2 ds + 2c \vert \Omega \vert {\alpha^*}^{-1} \left ( 1 - e^{-\alpha^* (t-\tau)}\right) \\ \displaystyle + 2C_b e^{-\alpha^* t} \int_{\tau}^{t} e^{\alpha^* s} \Vert u(s) \Vert^2 ds + 2 e^{-\alpha^* t} \int_{-\infty}^{t} e^{\alpha^* s}\Vert g(s) \Vert^2 ds\,. \end{array}$$

From this last estimate and (3.8), we have

u(t)2k3eα(tτ)u(τ)p+2Cbeα(tτ)τrτu(s)2ds+2c|Ω|α11eα(tτ)+2eαtteαsg(s)2ds+2Cbeαtτteαsu(s)2ds,k3eα(tτ)u(τ)p+2Cbeα(tτ)τrτu(s)2ds+2c|Ω|α11eα(tτ)+2eαtteαsg(s)2ds+2Cbη1eα(tτ)u(τ)2+2Cb2η1eα(tτ)τrτu(s)2ds+4Cbc|Ω|α1η11eα(tτ)+2Cbη1eαtteαsg(s)2ds,k3eα(tτ)u(τ)p+2Cbη1eα(tτ)u(τ)2+2Cbeα(tτ)+Cbη1eα(tτ)τrτu(s)2ds+2c|Ω|α11eα(tτ)+4Cbc|Ω|α1η11eα(tτ)+2eαtteαsg(s)2ds+2Cbη1eαtteαsg(s)2ds.$$\begin{array}{} \displaystyle \Vert \nabla u(t) \Vert^2 \leq k_3 e^{-\alpha^* (t-\tau)} \Vert \nabla u(\tau) \Vert^p + 2C_b e^{-\alpha^* (t-\tau)}\int_{\tau-r}^{\tau} \Vert u(s) \Vert^2 ds \\ \displaystyle \qquad\qquad\,+\, 2c \vert \Omega \vert {\alpha^*}^{-1} \left ( 1 - e^{-\alpha^* (t-\tau)}\right) + 2 e^{-\alpha^* t} \int_{-\infty}^{t} e^{\alpha^* s}\Vert g(s) \Vert^2 ds \\ \displaystyle \qquad\qquad\, +\, 2C_b e^{-\alpha^* t} \int_{\tau}^{t} e^{\alpha^* s} \Vert u(s) \Vert^2 ds \,,\\ \displaystyle \qquad\qquad\, \leq k_3 e^{-\alpha^* (t-\tau)} \Vert \nabla u(\tau) \Vert^p + 2C_b e^{-\alpha^* (t-\tau)}\int_{\tau-r}^{\tau} \Vert u(s) \Vert^2 ds \\ \displaystyle \qquad\qquad\, + \,2c \vert \Omega \vert {\alpha^*}^{-1} \left ( 1 - e^{-\alpha^* (t-\tau)}\right) + 2 e^{-\alpha^* t} \int_{-\infty}^{t} e^{\alpha^* s}\Vert g(s) \Vert^2 ds \\ \displaystyle \qquad\qquad\, +\, 2C_b \eta^{-1} e^{-\alpha(t- \tau)}\Vert u(\tau) \Vert^2 + 2C^2_b \eta^{-1} e^{-\alpha (t-\tau)}\int_{\tau-r}^{\tau} \Vert u(s)\Vert^2 ds \\ \displaystyle \qquad\qquad\,+\, 4C_b c\vert \Omega \vert \alpha^{-1} \eta^{-1} \left(1- e^{-\alpha (t-\tau)}\right) + 2C_b \eta^{-1} e^{-\alpha t}\int_{-\infty}^{t} e^{\alpha s} \Vert g(s) \Vert^2 ds \,,\\ \displaystyle \qquad\qquad\,\leq k_3 e^{-\alpha^* (t-\tau)} \Vert \nabla u(\tau) \Vert^p + 2C_b \eta^{-1} e^{-\alpha(t- \tau)}\Vert u(\tau) \Vert^2\\ \displaystyle \qquad\qquad\,+\, 2C_b \left(e^{-\alpha^* (t-\tau)} + C_b \eta^{-1} e^{-\alpha (t-\tau)} \right) \int_{\tau-r}^{\tau} \Vert u(s)\Vert^2 ds \\ \displaystyle \qquad\qquad\,+\,2c \vert \Omega \vert {\alpha^*}^{-1} \left ( 1 - e^{-\alpha^* (t-\tau)}\right) + 4C_b c\vert \Omega \vert \alpha^{-1} \eta^{-1} \left(1- e^{-\alpha (t-\tau)}\right) \\ \displaystyle \qquad\qquad\,+\,2 e^{-\alpha^* t} \int_{-\infty}^{t} e^{\alpha^* s}\Vert g(s) \Vert^2 ds + 2C_b \eta^{-1} e^{-\alpha t}\int_{-\infty}^{t} e^{\alpha s} \Vert g(s) \Vert^2 ds \,. \end{array}$$

In the fact that

φL2([r,0];L2(Ω))2rφC([r,0];L2(Ω))2,$$\begin{array}{} \displaystyle \Vert \varphi \Vert_{L^2([-r,0]; L^2(\Omega))}^2 \leq r \Vert \varphi \Vert_{C([-r,0]; L^2(\Omega))}^2\,, \end{array}$$

one has

u(t)2k3eα(tτ)u(τ)p+2Cbη1eα(tτ)u(τ)2+2Cbreα(tτ)+Cbη1eα(tτ)φC([r,0];L2(Ω))2+2c|Ω|α11eα(tτ)+4Cbc|Ω|α1η11eα(tτ)+2eαtteαsg(s)2ds+2Cbη1eαtteαsg(s)2dsk3eα(tτ)u(τ)p+2Cbη1eα(tτ)u(τ)2+2Cbreα(tτ)+Cbη1eα(tτ)φC([r,0];L2(Ω))2+2c|Ω|α1+4Cbc|Ω|α1η1+2eαtteαsg(s)2ds+2Cbη1eαtteαsg(s)2ds.$$\begin{array}{} \quad\displaystyle \Vert \nabla u(t) \Vert^2 \leq k_3 e^{-\alpha^* (t-\tau)} \Vert \nabla u(\tau) \Vert^p + 2C_b \eta^{-1} e^{-\alpha(t- \tau)}\Vert u(\tau) \Vert^2 \nonumber\\ \displaystyle + 2C_br \left(e^{-\alpha^* (t-\tau)} + C_b \eta^{-1} e^{-\alpha (t-\tau)} \right) \Vert \varphi\Vert_{C([-r,0]; L^2(\Omega))}^2 \nonumber\\ \displaystyle + 2c \vert \Omega \vert {\alpha^*}^{-1} \left ( 1 - e^{-\alpha^* (t-\tau)}\right) + 4C_b c\vert \Omega \vert \alpha^{-1} \eta^{-1} \left(1- e^{-\alpha (t-\tau)}\right) \nonumber\\ \displaystyle + 2 e^{-\alpha^* t} \int_{-\infty}^{t} e^{\alpha^* s}\Vert g(s) \Vert^2 ds + 2C_b \eta^{-1} e^{-\alpha t}\int_{-\infty}^{t} e^{\alpha s} \Vert g(s) \Vert^2 ds \,\nonumber\\ \displaystyle \leq k_3 e^{-\alpha^* (t-\tau)} \Vert \nabla u(\tau) \Vert^p + 2C_b \eta^{-1} e^{-\alpha(t- \tau)}\Vert u(\tau) \Vert^2 \nonumber\\ + 2C_br \left(e^{-\alpha^* (t-\tau)} + C_b \eta^{-1} e^{-\alpha (t-\tau)} \right) \Vert \varphi\Vert_{C([-r,0]; L^2(\Omega))}^2 \nonumber\\ \displaystyle + 2c \vert \Omega \vert {\alpha^*}^{-1} + 4C_b c\vert \Omega \vert \alpha^{-1} \eta^{-1} \nonumber\\ \displaystyle + 2 e^{-\alpha^* t} \int_{-\infty}^{t} e^{\alpha^* s}\Vert g(s) \Vert^2 ds + 2C_b \eta^{-1} e^{-\alpha t}\int_{-\infty}^{t} e^{\alpha s} \Vert g(s) \Vert^2 ds\,. \end{array}$$

If we take tτ + r i.e. t + θτ, it follows

u(t+θ)2k3eα(t+θτ)u(τ)p+2Cbη1eα(t+θτ)u(τ)2+2Cbreα(t+θτ)+Cbη1eα(t+θτ)φC([r,0];L2(Ω))2+2c|Ω|α11eα(t+θτ)+4Cbc|Ω|α1η11eα(t+θτ)+2eα(t+θ)t+θeαsg(s)2ds+2Cbη1eα(t+θ)t+θeαsg(s)2ds.$$\begin{array}{} \displaystyle \quad\Vert \nabla u(t+\theta) \Vert^2 \leq k_3 e^{-\alpha^* (t+\theta-\tau)} \Vert \nabla u(\tau) \Vert^p + 2C_b \eta^{-1} e^{-\alpha(t+\theta- \tau)}\Vert u(\tau) \Vert^2 \\ \displaystyle +\,2C_br \left(e^{-\alpha^* (t+\theta-\tau)} + C_b \eta^{-1} e^{-\alpha (t+\theta-\tau)} \right) \Vert \varphi\Vert_{C([-r,0]; L^2(\Omega))}^2 \\ \displaystyle +\,2c \vert \Omega \vert {\alpha^*}^{-1} \left ( 1 - e^{-\alpha^* (t+\theta-\tau)}\right)+ 4C_b c\vert \Omega \vert \alpha^{-1} \eta^{-1} \left(1- e^{-\alpha (t+\theta-\tau)}\right) \\ \displaystyle +\,2 e^{-\alpha^* (t+\theta)} \int_{-\infty}^{t+\theta} e^{\alpha^* s}\Vert g(s) \Vert^2 ds + 2C_b \eta^{-1} e^{-\alpha (t+\theta)}\int_{-\infty}^{t+\theta} e^{\alpha s} \Vert g(s) \Vert^2 ds \,. \end{array}$$

Hence,

U(t,τ)(u0,φ)C([r,0];H01(Ω))2=supθ[r,0]u(t+θ)2,k3eα(trτ)u(τ)p+2Cbη1eα(trτ)u(τ)2+2Cbreα(trτ)+Cbη1eα(trτ)φC([r,0];L2(Ω))2+2c|Ω|α11eα(trτ)+4Cbc|Ω|α1η11eα(trτ)+2eα(tr)teαsg(s)2ds+2Cbη1eα(tr)teαsg(s)2ds.$$\begin{array}{} \displaystyle \quad\Vert U(t,\tau)(u^0,\varphi) \Vert^2_{C([-r,0];H^1_0(\Omega))} = \sup_{\theta\in[-r,0]} \Vert \nabla u(t+\theta) \Vert^2\,,\\ \displaystyle \leq k_3 e^{-\alpha^* (t-r-\tau)} \Vert \nabla u(\tau) \Vert^p + 2C_b \eta^{-1} e^{-\alpha(t-r- \tau)}\Vert u(\tau) \Vert^2 \\ \displaystyle +\,2C_br \left(e^{-\alpha^* (t-r-\tau)} + C_b \eta^{-1} e^{-\alpha (t-r-\tau)} \right) \Vert \varphi\Vert_{C([-r,0]; L^2(\Omega))}^2 \\ \displaystyle +\,2c \vert \Omega \vert {\alpha^*}^{-1} \left ( 1 - e^{-\alpha^* (t-r-\tau)}\right)+ 4C_b c\vert \Omega \vert \alpha^{-1} \eta^{-1} \left(1- e^{-\alpha (t-r-\tau)}\right) \\ \displaystyle +\,2 e^{-\alpha^* (t-r)} \int_{-\infty}^{t} e^{\alpha^* s}\Vert g(s) \Vert^2 ds + 2C_b \eta^{-1} e^{-\alpha (t-r)}\int_{-\infty}^{t} e^{\alpha s} \Vert g(s) \Vert^2 ds \,. \end{array}$$

So, we obtain

U(t,τ)(u0,φ)C([r,0];H01(Ω))2k3eα(trτ)u(τ)p+2Cbη1eα(trτ)u(τ)2+2Cbreα(trτ)+Cbη1eα(trτ)φC([r,0];L2(Ω))2+2c|Ω|α1eαr+2Cbα1η1eαr+2Cbη1eα(tr)teαsg(s)2ds+2eα(tr)teαsg(s)2ds.$$\begin{array}{} \displaystyle \quad\Vert U(t,\tau)(u^0,\varphi) \Vert^2_{C([-r,0];H^1_0(\Omega))} \nonumber\\ \displaystyle \leq k_3 e^{-\alpha^* (t-r-\tau)} \Vert \nabla u(\tau) \Vert^p + 2C_b \eta^{-1} e^{-\alpha(t-r- \tau)}\Vert u(\tau) \Vert^2 \nonumber\\ \displaystyle +\,2C_br \left(e^{-\alpha^* (t-r-\tau)} + C_b \eta^{-1} e^{-\alpha (t-r-\tau)} \right) \Vert \varphi\Vert_{C([-r,0]; L^2(\Omega))}^2 \nonumber\\ \displaystyle +\, 2c \vert \Omega \vert \left({\alpha^*}^{-1} e^{\alpha^* r }+ 2C_b \alpha^{-1} \eta^{-1} e^{\alpha r}\right) \nonumber\\ \displaystyle + \,2C_b \eta^{-1} e^{-\alpha (t-r)}\int_{-\infty}^{t} e^{\alpha s} \Vert g(s) \Vert^2 ds \nonumber\\ \displaystyle + \,2 e^{-\alpha^* (t-r)} \int_{-\infty}^{t} e^{\alpha^* s}\Vert g(s) \Vert^2 ds \,. \end{array}$$

Similarly to the Lemma 3, let 𝓡 be the set of all functions ρ : ℝ ⟶ (0, + ∞) such that

limteαtρ2(t)=0,$$\begin{array}{} \displaystyle \lim_{t\rightarrow -\infty} e^{\alpha^* t} \rho^{2}(t) =0\,, \end{array}$$

by 𝓓 we denote the class of all families D^={D(t):tR}P(C([r,0];H01(Ω)))$\begin{array}{} \displaystyle \mathbf{\widehat{D}} = \{D(t) : t\in \mathbb{R} \} \subset \mathcal{P}(C([-r,0];H^1_0(\Omega))) \end{array}$ such that D(t) ⊂ B¯C([r,0];H01(Ω))(0,ρ(t)),$\begin{array}{} \displaystyle \mathbf{\overline{B}}_{C([-r,0];H^1_0(\Omega))}(0,\rho(t))\,, \end{array}$ for some ρ ∈ 𝓡, where we denote by B¯C([r,0];H01(Ω))(0,ρ(t))$\begin{array}{} \displaystyle \mathbf{\overline{B}}_{C([-r,0];H^1_0(\Omega))}(0,\rho(t)) \end{array}$ the closed ball in C([r,0];H01(Ω))$\begin{array}{} \displaystyle C([-r,0];H^1_0(\Omega)) \end{array}$ centered at 0 with radius ρ(t). Let

R22(t)=2c|Ω|α1eαr+2Cbα1η1eαr+2Cbη1eα(tr)teαsg(s)2ds+2eα(tr)teαsg(s)2ds.$$\begin{array}{} \displaystyle R_2^2(t)= 2c \vert \Omega \vert \left({\alpha^*}^{-1} e^{\alpha^* r }+ 2C_b \alpha^{-1} \eta^{-1} e^{\alpha r}\right) \\ \displaystyle \qquad\,\,\,\,+\,2C_b \eta^{-1} e^{-\alpha (t-r)}\int_{-\infty}^{t} e^{\alpha s} \Vert g(s) \Vert^2 ds + 2 e^{-\alpha^* (t-r)} \int_{-\infty}^{t} e^{\alpha^* s}\Vert g(s) \Vert^2 ds \,. \end{array}$$

Thus, for all ∈ 𝓓 and all t ∈ ℝ, by (3.23) there exists τ0 (, t) ≤ t such that

U(t,τ)(u0,φ)C([r,0];H01(Ω))2R22(t),$$\begin{array}{} \displaystyle \Vert U(t,\tau)(u^0,\varphi) \Vert_{C([-r,0];H^1_0(\Omega))}^2 \leq R_2^2(t)\,, \end{array}$$

for all ττ0(, t), this means that B2(t)=B¯C([r,0];H01(Ω))(0,R2(t))$\begin{array}{} \displaystyle B_2(t) = \overline{B}_{C([-r,0]; H^1_0(\Omega))}(0, R_2(t)) \end{array}$ is pullback 𝓓-absorbing for the mapping U(t, τ).

The proof of the proposition is completed. ◼

Existence of pullback D-attractor

To prove the existence of pullback 𝓓-attractor, we need to prove the following lemma.

Lemma 4

Assume that conditions oflemma (3)are satisfied. Then the process {S(t, τ)} corresponding to(1.1)is pullback 𝓓-asymptotically compact.

Proof

Let t ∈ ℝ, ∈ 𝓓, a sequences τnn→+∞ – ∞ and (u0, n, φnD(τn), be fixed. We have to check that the sequence

{S(t,τn)(u0,n,φn)}={(u(t,τn,(u0,n,φn)),ut(.,τn,(u0,n,φn)))},$$\begin{array}{} \displaystyle \{S(t,\tau_n)(u^{0,n},\varphi^n)\}= \{(u(t,\tau_n, (u^{0,n},\varphi^n)), u_t(.,\tau_n, (u^{0,n},\varphi^n)))\}\,, \end{array}$$

is relatively compact in H. In order to show this, we need to prove that the sequence

{U(t,τn)(u0,n,φn)}={ut(.,τn,(u0,n,φn))}$$\begin{array}{} \displaystyle \{U(t,\tau_n)(u^{0,n},\varphi^n)\} = \{u_t(.,\tau_n, (u^{0,n},\varphi^n))\} \end{array}$$

is relatively compact in C([–r, 0];L2(Ω)). To this end, we use the Ascoli-Arzela theorem. In other words, we check

the equicontinuity property for the sequence {ut(.,τn,(u0,n,φn))}:={utn(.)},i.e.ε>0,δ>0$\begin{array}{} \displaystyle \{u_t(.,\tau_n, (u^{0,n},\varphi^n))\}:= \{u_t^n(.)\}\,, \rm i.e.\,\, \forall \varepsilon \gt 0, \, \exists \delta \gt 0 \end{array}$ such that if |θ1θ2|δ,thenutn(θ1)utn(θ2)ε,forallθ1>θ2[r,0]$\begin{array}{} \displaystyle \vert \theta_1 -\theta_2 \vert \leq \delta \,,\, \rm then\, \Vert u^n_t(\theta_1) - u^n_t(\theta_2) \Vert \leq \varepsilon\,,\, for\, all\, \,\theta_1 \gt \theta_2 \in [-r,0]\, \end{array}$;

the uniform boundedness of {utn(θ)},$\begin{array}{} \displaystyle \{u^n_t(\theta)\}\,, \end{array}$ for all θ ∈ [–r, 0].

In order to prove (b), we consider un, u the corresponding solutions to (1.1), so by Lemma 1 we can deduce that {utn}$\begin{array}{} \displaystyle \{u^n_t\} \end{array}$ and {ut} are uniformly bounded in C([–r, 0]; L2(Ω)).

To prove (a), we proceed as follows:

utn(θ1)utn(θ2)=u(t+θ1)u(t+θ2),=t+θ2t+θ1u(s)ds,t+θ2t+θ1u(s)ds,t+θ2t+θ1(Δu(s)+f(u(s))+b(s,us)+g(s))ds.$$\begin{array}{} \displaystyle \Vert u^n_t(\theta_1) - u^n_t(\theta_2) \Vert= \Vert u(t+\theta_1) - u(t+\theta_2)\Vert \,,\nonumber\\ \displaystyle \qquad\qquad\qquad\quad\,\,\,\,\,= \left\Vert \int_{t+\theta_2}^{t+\theta_1} u'(s) ds \right\Vert\,,\nonumber\\ \displaystyle \qquad\qquad\qquad\quad\,\,\,\,\,\leq \int_{t+\theta_2}^{t+\theta_1} \Vert u'(s) \Vert ds \,,\nonumber\\ \displaystyle \qquad\qquad\qquad\quad\,\,\,\,\,\leq \int_{t+\theta_2}^{t+\theta_1} \bigg( \Vert \Delta u(s) \Vert + \Vert f(u(s))\Vert + \Vert b(s,u_s)\Vert \nonumber\\ \displaystyle \qquad\qquad\qquad\quad\,\,\,\,\,+\, \Vert g(s)\Vert\bigg) ds\,. \end{array}$$

Now, we estimate the terms on the right hand side of this inequality

From the Holder inequality, we have

t+θ2t+θ1Δu(s)dst+θ2t+θ1ds1/2t+θ2t+θ1Δu(s)2ds1/2,|θ1θ2|1/2t+θ2t+θ1Δu(s)2ds1/2.$$\begin{array}{} \displaystyle \int_{t+\theta_2}^{t+\theta_1} \Vert \Delta u(s) \Vert ds \leq \left(\int_{t+\theta_2}^{t+\theta_1} ds\right)^{1/2} \left(\int_{t+\theta_2}^{t+\theta_1} \Vert \Delta u(s) \Vert^2 ds\right)^{1/2}\,,\\ \displaystyle \qquad\qquad\qquad\quad\quad\leq \vert \theta_1 - \theta_2 \vert^{1/2} \left(\int_{t+\theta_2}^{t+\theta_1} \Vert \Delta u(s) \Vert^2 ds\right)^{1/2}\,. \end{array}$$

On the one hand, we have

Δu2λmu2.$$\begin{array}{} \displaystyle \Vert \Delta u \Vert^2 \leq \lambda_m \Vert \nabla u \Vert^2\,. \end{array}$$

So, using this inequality in (3.22) and integrating it over [t + θ2, t + θ1], one obtain

t+θ2t+θ1Δu(s)2dsλmt+θ2t+θ1u(s)2dsk3λmt+θ2t+θ1eα(sτ)u(τ)pds+2Cbη1λmt+θ2t+θ1eα(sτ)u(τ)2ds+2Cbrλmt+θ2t+θ1eα(sτ)+Cbη1eα(sτ)φC([r,0];L2(Ω))2ds+2c|Ω|λmt+θ2t+θ1α11eα(sτ)+Cbα1η11eα(sτ)ds+2λmt+θ2t+θ1eαsseαsg(s)2dsds+2Cbη1λmt+θ2t+θ1eαsseαsg(s)2dsds.$$\begin{array}{} \displaystyle \quad\int_{t+\theta_2}^{t+\theta_1} \Vert \Delta u(s) \Vert^2 ds \leq \lambda_m \int_{t+\theta_2}^{t+\theta_1} \Vert \nabla u(s) \Vert^2 ds \\ \displaystyle \leq k_3 \lambda_m \int_{t+\theta_2}^{t+\theta_1} e^{-\alpha^* (s-\tau)} \Vert \nabla u(\tau) \Vert^p ds + 2C_b \eta^{-1}\lambda_m \int_{t+\theta_2}^{t+\theta_1} e^{-\alpha(s- \tau)} \Vert u(\tau) \Vert^2 ds \\ \displaystyle +\,2C_br\lambda_m \int_{t+\theta_2}^{t+\theta_1} \left(e^{-\alpha^* (s-\tau)} + C_b \eta^{-1} e^{-\alpha (s-\tau)} \right) \Vert \varphi\Vert_{C([-r,0]; L^2(\Omega))}^2 ds \\ \displaystyle +\, 2c \vert \Omega \vert \lambda_m \int_{t+\theta_2}^{t+\theta_1}\left( {\alpha^*}^{-1} \left ( 1 - e^{-\alpha^* (s-\tau)}\right) + C_b \alpha^{-1} \eta^{-1} \left(1- e^{-\alpha (s-\tau)}\right)\right) ds \\ \displaystyle +\, 2 \lambda_m \int_{t+\theta_2}^{t+\theta_1} e^{-\alpha^* s} \int_{-\infty}^{s} e^{\alpha^* s'}\Vert g(s') \Vert^2 ds'ds \\ \displaystyle +\, 2C_b \eta^{-1} \lambda_m \int_{t+\theta_2}^{t+\theta_1} e^{-\alpha s}\int_{-\infty}^{s} e^{\alpha s'} \Vert g(s') \Vert^2 ds'ds \,. \end{array}$$

Therefore, one obtains

t+θ2t+θ1Δu(s)2dsk3λmu(τ)pα1eα(tτ)eαθ2eαθ1+2Cbη1λmu(τ)2α1eα(tτ)eαθ2eαθ1+2CbrλmφC([r,0];L2(Ω))2α1eα(tτ)eαθ2eαθ1+2Cb2rλmφC([r,0];L2(Ω))2η1α1eα(tτ)eαθ2eαθ1+2c|Ω|λmα11α1eα(tτ)eαθ2eαθ1+2c|Ω|Cbη1λmα11α1eα(tτ)eαθ2eαθ1+2λmα1eαθ2eαθ1eαtteαsg(s)2ds+2Cbη1λmα1eαθ2eαθ1eαtteαsg(s)2ds0whenθ1θ2.$$\begin{array}{} \displaystyle \quad\int_{t+\theta_2}^{t+\theta_1} \Vert \Delta u(s) \Vert^2 ds\\ \displaystyle \leq k_3 \lambda_m \Vert \nabla u(\tau) \Vert^p {\alpha^*}^{-1} e^{-\alpha^*(t- \tau)} \left(e^{-\alpha^* \theta_2} - e^{-\alpha^* \theta_1}\right)\\ \displaystyle +\, 2C_b \eta^{-1}\lambda_m \Vert u(\tau) \Vert^2 \alpha^{-1} e^{-\alpha (t- \tau)} \left(e^{-\alpha \theta_2} - e^{-\alpha \theta_1}\right)\\ \displaystyle +\, 2C_br\lambda_m \Vert \varphi\Vert_{C([-r,0]; L^2(\Omega))}^2 {\alpha^*}^{-1} e^{-\alpha^*(t- \tau)} \left(e^{-\alpha^* \theta_2} - e^{-\alpha^* \theta_1}\right)\\ \displaystyle +\,2C^2_br\lambda_m \Vert \varphi\Vert_{C([-r,0]; L^2(\Omega))}^2 \eta^{-1} \alpha^{-1} e^{-\alpha (t- \tau)} \left(e^{-\alpha \theta_2} - e^{-\alpha \theta_1}\right)\\ \displaystyle +\, 2c \vert \Omega \vert \lambda_m {\alpha^*}^{-1} \left ( 1 - {\alpha^*}^{-1} e^{-\alpha^*(t- \tau)} \left(e^{-\alpha^* \theta_2} - e^{-\alpha^* \theta_1}\right)\right)\\ \displaystyle +\,2c \vert \Omega \vert C_b \eta^{-1} \lambda_m \alpha^{-1} \left(1- {\alpha}^{-1} e^{-\alpha(t- \tau)} \left(e^{-\alpha \theta_2} - e^{-\alpha \theta_1}\right)\right)\\ \displaystyle +\, 2 \lambda_m {\alpha^*}^{-1} \left( e^{-\alpha^*\theta_2} - e^{-\alpha^*\theta_1} \right) e^{-\alpha^*t} \int_{-\infty}^{t} e^{\alpha^* s'}\Vert g(s') \Vert^2 ds'\\ \displaystyle + 2C_b \eta^{-1} \lambda_m \alpha^{-1} \left( e^{-\alpha \theta_2} - e^{-\alpha \theta_1} \right) e^{-\alpha t} \int_{-\infty}^{t} e^{\alpha s'}\Vert g(s') \Vert^2 ds'\\ \displaystyle \quad\to 0 \; \mbox{when} \; \theta_1 \to \theta_2\,. \end{array}$$

Hence, it follows that

t+θ2t+θ1Δu(s)ds|θ1θ2|1/2t+θ2t+θ1Δu(s)2ds1/20whenθ1θ2.$$\begin{array}{} \displaystyle \int_{t+\theta_2}^{t+\theta_1} \Vert \Delta u(s) \Vert ds \leq \vert \theta_1 - \theta_2 \vert^{1/2} \left(\int_{t+\theta_2}^{t+\theta_1} \Vert \Delta u(s) \Vert^2 ds\right)^{1/2}\\ \displaystyle \qquad\qquad\qquad\qquad\quad \to 0 \; \mbox{when} \; \theta_1 \to \theta_2\,. \end{array}$$

From the Holder inequality, we have

t+θ2t+θ1f(u(s))ds|θ1θ2|1/2t+θ2t+θ1f(u(s))2ds1/2.$$\begin{array}{} \displaystyle \int_{t+\theta_2}^{t+\theta_1} \Vert f(u(s))\Vert ds \leq \vert \theta_1 - \theta_2 \vert^{1/2} \cdot \left( \int_{t+\theta_2}^{t+\theta_1} \Vert f(u(s))\Vert^2 ds \right)^{1/2} \,. \end{array}$$

Using (1.4) and the convexity of the power, one gets

f(u(t))2=Ω|f(u(t,x))|2dx2l2u(t)2(p1)+2l2|Ω|.$$\begin{array}{} \displaystyle \Vert f(u(t))\Vert^2 = \int_{\Omega} \vert f(u(t,x))\vert^2 dx \,\\ \displaystyle \qquad\qquad\,\,\,\,\leq 2l^2 \Vert u(t) \Vert^{2(p-1)} + 2l^2 \vert \Omega \vert\,. \end{array}$$

Integrating this estimate over [t + θ2, t + θ1], one finds

t+θ2t+θ1f(u(s))2ds2l2t+θ2t+θ1u(s)2(p1)ds+2l2|Ω||θ1θ2|.$$\begin{array}{} \displaystyle \int_{t+\theta_2}^{t+\theta_1} \Vert f(u(s))\Vert^2 ds \leq 2l^2 \int_{t+\theta_2}^{t+\theta_1} \Vert u(s) \Vert^{2(p-1)} ds + 2l^2 \vert \Omega \vert \cdot \vert \theta_1 -\theta_2 \vert \,. \end{array}$$

Since λ1u2 ≤ ∥∇u2, we have

t+θ2t+θ1f(u(s))2ds2l2λ1(p1)t+θ2t+θ1u(s)2(p1)ds+2l2|Ω||θ1θ2|.$$\begin{array}{} \displaystyle \int_{t+\theta_2}^{t+\theta_1} \Vert f(u(s))\Vert^2 ds \leq 2l^2 \lambda_1^{(p-1)}\int_{t+\theta_2}^{t+\theta_1} \Vert \nabla u(s) \Vert^{2(p-1)} ds + 2l^2 \vert \Omega \vert \cdot \vert \theta_1 -\theta_2 \vert \,. \end{array}$$

From (3.22), one has

u(t)2(p1){k3eα(tτ)u(τ)p+2Cbη1eα(tτ)u(τ)2+2Cbreα(tτ)+Cbη1eα(tτ)φC([r,0];L2(Ω))2+2c|Ω|α1+4Cbc|Ω|α1η1+2eαtteαsg(s)2ds+2Cbη1eαtteαsg(s)2ds}(p1).$$\begin{array}{} \displaystyle \quad\Vert \nabla u(t) \Vert^{2(p-1)} \leq \bigg\{ k_3 e^{-\alpha^* (t-\tau)} \Vert \nabla u(\tau) \Vert^p + 2C_b \eta^{-1} e^{-\alpha(t- \tau)}\Vert u(\tau) \Vert^2 \\ \displaystyle +\, 2C_br \left(e^{-\alpha^* (t-\tau)} + C_b \eta^{-1} e^{-\alpha (t-\tau)} \right) \Vert \varphi\Vert_{C([-r,0]; L^2(\Omega))}^2 \\ \displaystyle +\, 2c \vert \Omega \vert {\alpha^*}^{-1} + 4C_b c\vert \Omega \vert \alpha^{-1} \eta^{-1} \\ \displaystyle +\, 2 e^{-\alpha^* t} \int_{-\infty}^{t} e^{\alpha^* s}\Vert g(s) \Vert^2 ds + 2C_b \eta^{-1} e^{-\alpha t}\int_{-\infty}^{t} e^{\alpha s} \Vert g(s) \Vert^2 ds \bigg\}^{(p-1)}\,. \end{array}$$

By applying the convexity of power three times, one gets

u(t)2(p1)22(p2)k3u(τ)p+2CbrφC([r,0];L2(Ω))2(p1)e(p1)α(tτ)+22(p2)2Cbη1u(τ)2+2Cb2rη1φC([r,0];L2(Ω))2(p1)e(p1)α(tτ)+22(p2)2c|Ω|α1+4Cbc|Ω|α1η1(p1)+23(p2)2(p1)teαsg(s)2ds(p1)e(p1)αt+23(p2)(2Cbη1)(p1)teαsg(s)2ds(p1)e(p1)αt.$$\begin{array}{} \displaystyle \quad\Vert \nabla u(t) \Vert^{2(p-1)} \\ \displaystyle \leq 2^{2(p-2)}\left(k_3 \Vert \nabla u(\tau) \Vert^p + 2C_br\Vert \varphi\Vert_{C([-r,0]; L^2(\Omega))}^2\right)^{(p-1)}e^{-(p-1)\alpha^* (t-\tau)} \\ \displaystyle +\, 2^{2(p-2)} \left(2C_b \eta^{-1} \Vert u(\tau) \Vert^2 + 2C^2_br \eta^{-1} \Vert \varphi\Vert_{C([-r,0]; L^2(\Omega))}^2 \right)^{(p-1)}e^{-(p-1)\alpha (t-\tau)} \\ \displaystyle +\, 2^{2(p-2)}\left(2c \vert \Omega \vert {\alpha^*}^{-1} + 4C_b c\vert \Omega \vert \alpha^{-1} \eta^{-1}\right)^{(p-1)}\\ \displaystyle + \,2^{3(p-2)} 2^{(p-1)} \left( \int_{-\infty}^{t} e^{\alpha^* s}\Vert g(s) \Vert^2 ds \right)^{(p-1)} e^{-(p-1)\alpha^* t}\\ \displaystyle +\, 2^{3(p-2)} (2C_b \eta^{-1})^{(p-1)} \left( \int_{-\infty}^{t} e^{\alpha s}\Vert g(s) \Vert^2 ds \right)^{(p-1)} e^{-(p-1)\alpha t}\,. \end{array}$$

Integrating it over [t + θ2, t + θ1], one obtains

t+θ2t+θ1u(s)2(p1)ds22(p2)k3u(τ)p+2CbrφC([r,0];L2(Ω))2(p1)t+θ2t+θ1e(p1)α(sτ)ds+22(p2)2Cbη1u(τ)2+2Cb2rη1φC([r,0];L2(Ω))2(p1)t+θ2t+θ1e(p1)α(sτ)ds+22(p2)2c|Ω|α1+4Cbc|Ω|α1η1(p1)|θ1θ2|+23(p2)2(p1)teαsg(s)2ds(p1)t+θ2t+θ1e(p1)αsds+23(p2)(2Cbη1)(p1)teαsg(s)2ds(p1)t+θ2t+θ1e(p1)αsds.$$\begin{array}{} \displaystyle \quad\int_{t+\theta_2}^{t+\theta_1}\Vert \nabla u(s) \Vert^{2(p-1)} ds\\ \displaystyle \leq 2^{2(p-2)}\left(k_3 \Vert \nabla u(\tau) \Vert^p + 2C_br\Vert \varphi\Vert_{C([-r,0]; L^2(\Omega))}^2\right)^{(p-1)} \int_{t+\theta_2}^{t+\theta_1} e^{-(p-1)\alpha^* (s-\tau)} ds \\ \displaystyle +\, 2^{2(p-2)} \left(2C_b \eta^{-1} \Vert u(\tau) \Vert^2 + 2C^2_br \eta^{-1} \Vert \varphi\Vert_{C([-r,0]; L^2(\Omega))}^2 \right)^{(p-1)}\int_{t+\theta_2}^{t+\theta_1} e^{-(p-1)\alpha (s-\tau)} ds \\ \displaystyle +\, 2^{2(p-2)}\left(2c \vert \Omega \vert {\alpha^*}^{-1} + 4C_b c\vert \Omega \vert \alpha^{-1} \eta^{-1}\right)^{(p-1)} \vert \theta_1- \theta_2 \vert \\ \displaystyle +\, 2^{3(p-2)} 2^{(p-1)} \left( \int_{-\infty}^{t} e^{\alpha^* s}\Vert g(s) \Vert^2 ds \right)^{(p-1)} \int_{t+\theta_2}^{t+\theta_1} e^{-(p-1)\alpha^* s} ds\\ \displaystyle +\, 2^{3(p-2)} (2C_b \eta^{-1})^{(p-1)} \left( \int_{-\infty}^{t} e^{\alpha s}\Vert g(s) \Vert^2 ds \right)^{(p-1)} \int_{t+\theta_2}^{t+\theta_1} e^{-(p-1)\alpha s} ds\,. \end{array}$$

Therefore, we get

t+θ2t+θ1u(s)2(p1)dsC1e(p1)α(tτ)e(p1)αθ2e(p1)αθ1+C2e(p1)α(tτ)e(p1)αθ2e(p1)αθ1+C3|θ1θ2|+C4e(p1)αte(p1)αθ2e(p1)αθ1+C5e(p1)αte(p1)αθ2e(p1)αθ10asθ1θ2.$$\begin{array}{} \displaystyle \int_{t+\theta_2}^{t+\theta_1}\Vert \nabla u(s) \Vert^{2(p-1)} ds \leq C'_1 e^{-(p-1)\alpha^*(t- \tau)} \left(e^{-(p-1)\alpha^* \theta_2} - e^{-(p-1)\alpha^* \theta_1}\right)\\ \displaystyle \qquad\qquad\qquad\qquad\qquad+\,\, C'_2 e^{-(p-1)\alpha(t- \tau)} \left(e^{-(p-1)\alpha \theta_2} - e^{-(p-1)\alpha \theta_1}\right)\\ \displaystyle \qquad\qquad\qquad\qquad\qquad+\,\, C'_3 \vert \theta_1- \theta_2 \vert + C'_4 e^{-(p-1)\alpha t} \left( e^{-(p-1)\alpha \theta_2} - e^{-(p-1)\alpha \theta_1}\right)\\ \displaystyle \qquad\qquad\qquad\qquad\qquad+\,\, C'_5 e^{-(p-1)\alpha^* t} \left( e^{-(p-1)\alpha^* \theta_2} - e^{-(p-1)\alpha^* \theta_1}\right)\\ \displaystyle \qquad\qquad\qquad\qquad\qquad\quad\to 0 \; \mbox{as}\; \theta_1 \to \theta_2\,. \end{array}$$

Hence by (3.27), (3.28) and this last estimate we deduce that

t+θ2t+θ1f(u(s))ds0asθ1θ2.$$\begin{array}{} \displaystyle \int_{t+\theta_2}^{t+\theta_1} \Vert f(u(s))\Vert ds \;\to 0 \; \mbox{as}\; \theta_1 \to \theta_2\,. \end{array}$$

Similarly, by the Holder inequality, we have

t+θ2t+θ1b(s,us)ds|θ1θ2|1/2t+θ2t+θ1b(s,us)2ds1/2.$$\begin{array}{} \displaystyle \int_{t+\theta_2}^{t+\theta_1} \Vert b(s,u_s)\Vert ds \leq \vert \theta_1 - \theta_2 \vert^{1/2} \cdot \left( \int_{t+\theta_2}^{t+\theta_1} \Vert b(s,u_s)\Vert^2 ds \right)^{1/2} \,. \end{array}$$

On the other hand, by (II), (1.7) and since λ1u2 ≤ ∥∇u2, one has

t+θ2t+θ1b(s,us)2dsCbt+θ2rt+θ1u(s)2dst+θ2rt+θ2u(s)2ds+t+θ2t+θ1u(s)2dsφL2([r,0];L2(Ω))2+λ11t+θ2t+θ1u(s)2ds.$$\begin{array}{} \displaystyle \int_{t+\theta_2}^{t+\theta_1} \Vert b(s,u_s)\Vert^2 ds \leq C_b \int_{t+\theta_2-r}^{t+\theta_1} \Vert u(s) \Vert^2 ds \nonumber\\ \displaystyle \qquad\qquad\qquad\qquad\,\,\,\,\leq \int_{t+\theta_2-r}^{t+\theta_2} \Vert u(s) \Vert^2 ds + \int_{t+\theta_2}^{t+\theta_1} \Vert u(s) \Vert^2 ds \nonumber\\ \displaystyle \qquad\qquad\qquad\qquad\,\,\,\,\leq \Vert \varphi \Vert^2_{L^2([-r,0]; L^2(\Omega))} + \lambda_1^{-1} \int_{t+\theta_2}^{t+\theta_1} \Vert \nabla u(s) \Vert^2 ds\,. \end{array}$$

By (3.26), it follows that

t+θ2t+θ1u(s)2ds0asθ1θ2.$$\begin{array}{} \displaystyle \int_{t+\theta_2}^{t+\theta_1} \Vert \nabla u(s) \Vert^2 ds \; \to 0 \; \mbox{as}\; \theta_1 \to \theta_2 \,. \end{array}$$

Then, (3.29), (3.30) and this last estimate, we deduce that

t+θ2t+θ1b(s,us)ds0whenθ1θ2.$$\begin{array}{} \displaystyle \int_{t+\theta_2}^{t+\theta_1} \Vert b(s,u_s)\Vert ds \; \to 0 \; \mbox{when}\; \theta_1 \to \theta_2 \,. \end{array}$$

Finally, we use the Holder inequality to obtain

t+θ2t+θ1g(s)ds|θ1θ2|1/2t+θ2t+θ1g(s)2ds1/2.$$\begin{array}{} \displaystyle \int_{t+\theta_2}^{t+\theta_1} \Vert g(s)\Vert ds \leq \vert \theta_1 - \theta_2 \vert^{1/2} \cdot \left(\int_{t+\theta_2}^{t+\theta_1} \Vert g(s)\Vert^2 ds\right)^{1/2} \,. \end{array}$$

Since gLloc2(R;L2(Ω)),$\begin{array}{} \displaystyle g\in L^2_{loc}(\mathbb{R}; L^2(\Omega))\,, \end{array}$ one gets

t+θ2t+θ1g(s)ds|θ1θ2|1/2gL2([t+θ2,t+θ1];L2(Ω))0whenθ1θ2.$$\begin{array}{} \displaystyle \int_{t+\theta_2}^{t+\theta_1} \Vert g(s)\Vert ds \leq \vert \theta_1 - \theta_2 \vert^{1/2} \cdot \Vert g \Vert_{L^2([t+\theta_2,t+\theta_1]; L^2(\Omega))}\\ \displaystyle \qquad\qquad\qquad\quad\,\,\,\,\,\, \to 0 \; \mbox{when}\; \theta_1 \to \theta_2 \,. \end{array}$$

Consequently, by 1), 2), 3), 4) and (3.25), we deduce that

u(t+θ1)u(t+θ2)0whenθ1θ2,$$\begin{array}{} \displaystyle \Vert u(t+\theta_1) - u(t+\theta_2)\Vert \; \to 0 \; \mbox{when}\; \theta_1 \to \theta_2 \,, \end{array}$$

and this ensures the equicontinuity property in C([–r, 0]; L2(Ω)); i.e. the sequence {U(t, τn)(u0, n, φn)} is relatively compact in C([–r, 0]; L2(Ω)).

Since we have S(t, τn)(u0, n, φn) = j(U(t, τn)(u0, n, φn)), so {S(t, τn)(u0, n, φn)} is relatively compact in the space L2(Ω) × C([–r, 0]; L2(Ω)) and by the continuous injection of L2(Ω) × C([–r, 0]; L2(Ω)) in H, we deduce that {S(t, τn)(u0, n, φn)} is relatively compact in H. The proof of this lemma is completed. ◼

By Proposition 1 and Lemma 4, we proved that the process S(t, τ) has a pullback 𝓓-absorbing set and it is pullback 𝓓-asymptotically compact, then by Theorem 2 we can deduce the following result.

Theorem 4

The process {S(t, τ)} corresponding to(1.1)has a pullback 𝓓-attractor = {A(t) : t ∈ ℝ} in H. Furetheremore, ÂL2(Ω) × C([–r, 0]; L2(Ω)).

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