Convergence to equilibrium of solutions to a nonautonomous semilinear viscoelastic equation with finite or infinite memory

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Convergence to equilibrium of solutions to a nonautonomous semilinear viscoelastic equation with finite or infinite memory Hassan Yassine Lebanese University, Faculty of Sciences, Department of Mathematics, Baalbek-Zahle, Lebanon Received 1 February 2017; revised 17 May 2017

Abstract In this paper we consider the nonautonomous semilinear viscoelastic equation τ utt − u +

k(s)u(t − s)ds + f (x, u) = g, τ ∈ {t, ∞}, 0

in R+ × , with Dirichlet boundary conditions and finite (τ = t) or infinite (τ = ∞) memory. Here  is a bounded domain in Rn with smooth boundary and the nonlinearity f :  × R+ → R is analytic in the second variable, uniformly with respect to the first one. For this equation, we derive an appropriate Lyapunov function and we use the Łojasiewicz–Simon inequality to show that the dissipation given by the memory term is strong enough to prove the convergence to a steady state for any global bounded solution. In addition, we discuss the rate of convergence to equilibrium which is polynomial or exponential, depending on the Łojasiewicz exponent and the decay of the time-dependent right-hand side g. © 2017 Elsevier Inc. All rights reserved. MSC: 28C15; 46E05; 28C99 Keywords: Evolutionary integral equation; Semilinear; Stabilization; Łojasiewicz–Simon inequality

E-mail address: [email protected]. http://dx.doi.org/10.1016/j.jde.2017.08.015 0022-0396/© 2017 Elsevier Inc. All rights reserved.

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1. Introduction and assumptions The aim of this paper is the study of the convergence and decay rate to a steady state of global bounded solutions of the following nonautonomous semilinear viscoelastic equation with finite memory ⎧ t utt − u + 0 k(s)u(t − s)ds + f (x, u) = g in R+ × , ⎪ ⎪ ⎪ ⎨ u = 0 on R+ × , ⎪ u(0) = u0 in , ⎪ ⎪ ⎩ ut (0) = u1 in ,

(1)

and with infinite memory ⎧ ∞ utt − u + 0 k(s)u(t − s)ds + f (x, u) = g in R+ × , ⎪ ⎪ ⎪ ⎨ u = 0 on R+ × , ⎪ u(−t) = u0 (t) for t ≥ 0, ⎪ ⎪ ⎩ ut (0) = u1 in ,

(2)

where  ⊆ Rn (n ≥ 1) is a bounded open set with smooth boundary  and the functions u0 , u1 :  → R and u0 : R ×  → R are given initial data. The relaxation function k, the nonlinearity f and the forcing term g will be specified later. These problems model some phenomena in  viscoelasticity. Here, we understand k(s)u(t − s)ds, f (x, u), and g(t) to be the viscoelastic term, the source term, and the exterior forcing term, respectively; we refer the reader to [12,15, 31] for a discussion on how these models arise. See also [16–18] and the references therein for more details concerning the physical phenomena which are modelled by differential equations with memory. In the present work we study the asymptotic behaviour of global bounded weak solutions for problems (1) and (2) as t → ∞. The proof is based on the construction of an appropriate new Lyapunov functional, compactness properties, and on the Łojasiewicz–Simon inequality. In particular, when the kernel k(t) decays exponentially, we show that, if g tends to 0 sufficiently fast at infinity, any bounded solution of (1) and (2) having a relatively compact range in the energy space converges to a single steady state. We also give a condition implying existence of bounded global solutions. Finally, we show that the decay rate to equilibrium is either exponential or polynomial. The asymptotic behaviour as time goes to infinity of global solutions of problems (1) and (2) has been studied by many mathematicians, and related results concerning existence, convergence to equilibrium, and blow-up of solutions have been recently established. For the autonomous linear case (i.e. f = g = 0), a number of papers have appeared covering a large class of kernels k, which guarantee stability and show the relation between the decay rate of k and the asymptotic behaviour of solutions of the considered problem. Examples of such results can be found, for instance, in [4,13,16,19,26–29] and the references therein. An autonomous semilinear initial value problem related to Eq. (1) is considered in [8], namely t utt − u + 0

k(s)u(t − s)ds + |u|ρ u = 0 in R+ × ,

(3)

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3

with Dirichlet boundary conditions. Following the method introduced by Komornik and Zuazua [25], the article [8] uses a Lyapunov type technique for some perturbed energy to obtain the convergence results. This work extended the results of [9], in which equation (3) was considered with additional linear damping. For more results in the case of finite memory we refer to AlabauBoussouira et al. [3], where general decay estimates results are obtained by an approach based on integral inequalities and multiplier techniques (see also [7] and the references therein). For equation (3) in the case of infinite memory we refer to [11,14]. Note that, for all models cited above, except the model considered in [7], the autonomous case, i.e. g = 0, has been considered. Motivated by the above research, we consider in the present work the case when g = 0 and the nonlinearity f is analytic. We may refer to the work of M.A. Jendoubi [23,24] (see also, A. Haraux and M.A. Jendoubi [22]), where convergence to equilibrium as t → ∞ was obtained by the well known Łojasiewicz–Simon approach for bounded solutions of the wave equation with linear dissipation and analytic nonlinearity. Concerning integro-differential equations with analytic nonlinearity, there are only a few papers that derive results on convergence to equilibrium. In [1,2] a convergence result for a phase-field model with memory was proved. The article [10] contains a convergence result for an abstract second order equation where the dissipation is both frictional and with memory. In the finite dimensional case, a convergence result was proved in [32,35] for ordinary differential equations of order less than 1, and of order between 1 and 2 in time. The crucial step for such problems is to find a suitable Lyapunov functional to investigate the asymptotic behaviour of global bounded solutions. We point out that, in general, the construction of a Lyapunov functional for problems with memory is a highly nontrivial task. In fact, in both equations, the dissipation given by the basic energy estimates (11) and (37) (see below) is not strong enough to prove the convergence with the Łojasiewicz–Simon approach. To overcome the difficulty due to the weaker dissipation, it is necessary to introduce suitable perturbations which may vary from problem to problem to cancel out some undesired terms and produce some new dissipation. Another technical difficulty consists in the nonautonomous case. The forcing term g introduces non-standard difficulties and plays a crucial role on the type of decay rate to equilibrium of global bounded solutions. Here, referring to [34], the given convergence proof is direct and naturally generalizes the autonomous case. Let us explain in detail our assumptions on the nonlinearity f , the kernel k and the forcing term g. The nonlinearity f = f (x, u) :  × R → R is assumed to be a C 2 function satisfying the following assumptions: (F1) f is analytic in the second variable, uniformly with respect to x ∈  and u in bounded subsets of R. (F2) One has f (·, 0) ∈ L∞ (), and there exist constants ρ0 ≥ 0 and α > 0, (N − 2)α < 2 such that

|

∂f (x, u)| ≤ ρ0 (1 + |u|α ) for every u ∈ R, x ∈ . ∂u

In addition, we consider here the nonautonomous case; we assume that, for some δ > 0, the function g ∈ L1 (R+ ; L2 ()) ∩ L2 (R+ ; L2 ()) satisfies the polynomial decay condition

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sup (1 + t)

1+δ

t∈R+

∞

g(s) 22 ds < ∞,

(G1)

t

or the exponential decay condition ∞ sup e

g(s) 22 ds < ∞. δt

t∈R+

(G2)

t

Concerning the kernel k, we assume that (K1) k : R+ → R+ is a C 1 function such that: ∞ 1−

k(s)ds = μ > 0. 0

(K2) There exists a positive constant a0 such that k  (t) ≤ −a0 k(t), t ≥ 0. The remaining part of this paper is organized as follows: In Section 2 we give some preliminaries and we state our main results. The convergence result and the estimate on the convergence rate of solutions to problem (1)/(2) are proved in Section 3. Section 4 is devoted to proving boundedness of any global solution to problem (1)/(2), under a supplementary condition on the nonlinearity f . 2. Preliminaries and main results Throughout we assume the following: • The inner products (respectively the norms) on the spaces H01 (), H −1 () and L2 () are denoted by (·, ·)H 1 () , (·, ·)∗ and (·, ·)2 (respectively, by · H 1 () , · ∗ and · 2 ). The norm 0 0 on Lp () is denoted by · p . • For any Hilbert space H, we let L2k (R+ ; H) be the weighted space with respect to the measure k(s)ds, L2k (R+ ; H) = {u : R+

∞ → H,

k(s) u(s) 2H ds < ∞}, 0

endowed with the inner product ∞ (φ, ψ)L2 (R+ ;H) =

k(s)(φ(s), ψ(s))H ds.

k

0

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• We denote by C (sometimes Ci ) a generic positive constant which may vary from line to line, which may depend on g, f , k, and the measure of , but which can be chosen independently of t ∈ R+ . Following the approach of Dafermos [13], we consider η = η(t, s, ·), the relative history of u, defined as η(t, s, ·) = u(t, ·) − u(t − s, ·). Hence, the Eq. (2) turns into the system ∞ ⎧ utt − μu − 0 k(s)η(t, s)ds + f (x, u) = g in R+ × , ⎪ ⎪ ⎪ + ⎪ ⎪ ⎨ ut = ηt + ηs in R × , u = 0 on R+ × , ⎪ ⎪ ⎪ u(0) = u0 , ut (0) = u1 in , ⎪ ⎪ ⎩ η(t, 0) = 0, η(0, ·) = η0 , where u0 = u0 (0) and η0 (s) = u0 (0) − u0 (s). We recall that, ∗ a weak solution of Eq. (1) is a function u : R+ → H01 (), u ∈ C(R+ ; H01 ()) ∩ C 1 (R+ ; L2 ()) such that for any v ∈ H01 () we have d (ut , v)2 + (∇u, ∇v)2 − dt

t k(t − s)(∇u(s), ∇v)2 ds + (f (u), v)2 = (g, v)2 . 0

∗ A weak solution of Eq. (2) is a function u : R+ → H01 (), u ∈ C(R+ ; H01 ()) ∩ C 1 (R+ ; L2 ()) η ∈ C(R+ ; L2k (R+ ; H01 ())), such that for any v ∈ H01 () we have d (ut , v)2 + μ(∇u, ∇v)2 + dt

∞ k(s)(∇η(t, s), ∇v)2 ds + (f (u), v)2 = (g, v)2 . 0

For more regular data, one should expect more regular solutions. In particular, we call ∗ a strong solution of Eq. (1) any function u ∈ C(R+ ; H 2 ()) ∩ C 1 (R+ ; H01 ()) ∩ C 2 (R+ ; L2 ()), satisfying the equation (1) for all t ∈ R+ .

(4)

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∗ A strong solution of Eq. (2) is any function u ∈ C(R+ ; H 2 ()) ∩ C 1 (R+ ; H01 ()) ∩ C 2 (R+ ; L2 ()), η, ηs ∈ C(R+ ; L2k (R+ ; H01 ())), satisfying the equation (2) for all t ∈ R+ . Existence, uniqueness, and regularity results can be found in [7,14,30]. The proof of our convergence result is based on the Łojasiewicz–Simon inequality for the energy functional Eμ : H01 () −→ R given by Eμ (u) =

μ 2



 |∇u|2 dx +



F (x, u)dx, 

u where F (x, u) = 0 f (x, s) ds (x ∈ , u ∈ R). By the regularity and growth condition on f , the function Eμ is twice continuously Fréchet differentiable [33]. If ∇Eμ (u) ∈ H −1 () and ∇ 2 Eμ (u) ∈ L(H01 (), H −1 ()) denote the first and second derivative at a point u ∈ H01 (), respectively, then for all φ, ψ ∈ H01 ()  (∇Eμ (u), ψ)H −1 (),H 1 () = μ

 ∇u∇ψ dx +

0



f (x, u)ψ dx 

= (−μu + f (x, u), ψ)H −1 (),H 1 () , 0

(5)

and  (∇ Eμ (u)φ, ψ)H −1 (),H 1 () = μ

 ∇φ∇ψ dx +

2

0



∂f (x, u)φψ dx. ∂u



The proof of the following proposition can be found in [22]. Proposition 1. Under assumptions (F1) and (F2) on the function f the energy functional Eμ ∈ C 2 (H01 ()) satisfies the Łojasiewicz–Simon inequality near every equilibrium point φ ∈ H01 (), that is, for every φ ∈ S, S = {φ ∈ H 2 () ∩ H01 () : − μφ + f (x, φ) = 0}, there exist βφ > 0, σφ > 0 and 0 < θφ ≤

1 2

such that

|Eμ (φ) − Eμ (ψ)|1−θφ ≤ βφ − μψ + f (x, ψ) ∗ for all ψ ∈ H01 () such that φ − ψ H 1 () < σφ . The number θφ is called the Łojasiewicz 0 exponent of Eμ at φ.

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In this paper, we prove convergence to equilibrium of any solution having relatively compact range in the energy space. The following assumption implies the boundedness of any global solution for problems (1)/(2): (F3) There exist λ < μλ1 and C ≥ 0 such that for every u ∈ R and every x ∈ , F (x, u) ≥ −λ

u2 − C, 2

where λ1 > 0 is the best constant in the following Poincaré inequality   2 |∇u| ≥ λ1 |u|2 (u ∈ H01 ()). 



Proposition 2. Under the hypothesis (F3), let u be a global weak solution of (1)/(2). Then (u(t), ut (t)) is bounded in H01 () × L2 (). Our first main result, which establishes the asymptotic behaviour of global solutions to the problem (1)/(2), reads as follows. Theorem 3. Suppose that f satisfies (F1), (F2), that the kernel k satisfies (K1), (K2) and that g satisfies either (G1) or (G2). Let u be a global weak solution of equation (1)/(2) and assume also that: (T1) (u, ut ) is bounded in H01 () × L2 () (T2) {u(t) : t ≥ 0} is relatively compact in H01 (). Then there exists φ ∈ S such that

ut (t) 2 + u(t) − φ H 1 () −→ 0 as t → ∞. 0

(6)

The rate of convergence is stated in the following theorem. For its proof we use the estimates obtained in the proof of Theorem 3 for the Lyapunov function constructed there and the differential inequality given below (Lemma 8). In particular, we show that the Łojasiewicz exponent θ in the Łojasiewicz–Simon inequality and the type of the growth condition on g together determine the decay rate of the solution u to the steady state φ. Theorem 4. Under the assumptions of Theorem 3, let θ = θφ be the Łojasiewicz exponent of Eμ at φ where φ is given by (6). Then, the following assertions hold: (i) If θ ∈ (0, 12 ) and g satisfies the polynomial decay (G1), then there exists constants C, ξ > 0 such that for all t ≥ 0 we have

u(t) − φ 2 ≤ C(1 + t)−ξ , where

 ξ=

θ inf{ 1−2θ , 2δ } if g = 0, θ 1−2θ

if g = 0.

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1 (ii) If θ = and g satisfies the exponential growth (G2), then there exist constants C, κ > 0 2 such that

u(t) − φ 2 ≤ Ce−κt , t ≥ 0. 3. Proof of convergence results In this section we prove first Theorems 3 and 4 for the model (2). Then, building on this proof and by the construction of the corresponding Lyapunov energy, we will show that Theorems 3 and 4 are also valid for the model (1). 3.1. The case of infinite memory To prove our main theorems, we need to introduce some notation and to prove some intermediate results. Let us recall that the ω-limit set of a continuous function u : R+ → H01 () is defined by ω(u) = {φ ∈ H01 () : ∃ tn → +∞ such that lim u(tn ) − φ H 1 () = 0}. n→∞

0

From well-known results on dynamical systems [21], it follows that if u is a continuous function having relatively compact range, then its ω-limit set is a non-empty, compact, and connected subset of H01 (). Accordingly, we review in the following lemma some information on the structure of this set and we prove some auxiliary results. Lemma 5. Let u : R+ → H01 () be a weak solution of equation (1) and assume that the assumptions of Theorem 3 hold. Then: ∞ (i) ut ∈ L2 (R+ ; L2 ()), ∇Eμ (u) ∈ L2 (R+ ; H −1 ()) and 0 k(s) ∇η(t, s) 22 ds ∈ L1 (R+ ). (ii) The function Eμ is constant on ω(u), and Eμ (φ) = lim Eμ (u(t)) = E∞ = const < ∞ for all φ ∈ ω(u). t→∞

(iii) lim ut (t) 2 = lim t→∞

∞

t→∞ 0

k(s) ∇η(t, s) 22 ds = 0.

(iv) The ω-limit set of u is a subset of the set of stationary solutions. (v) There exists a uniform Łojasiewicz exponent θ ∈]0, 12 ], β > 0 and T > 0 such that for all t ≥ T |Eμ (u(t)) − E∞ |1−θ ≤ β ∇Eμ (u(t)) ∗ .

(7)

Proof of Lemma 5. We start by deriving suitable energy estimates. We proceed in several steps. For a strong solution, all calculations below are justified. By a simple density argument, they remain valid for any weak solution. We start by

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The basic energy estimate. Let G0 : R+ → R be the function defined by 1 G0 (t) = ut 22 + Eμ (u(t)) + 2

∞ (g(s), ut (s))2 ds. t

By multiplying Eq. (4) by ut , integrating over  and using integration by parts, one can easily find that d G0 (t) = − dt

∞ k(s)(∇η(t, s), ∇ut (t))2 ds.

(8)

0

Noting that ut = ηt + ηs , η(t, 0) = lim k(t) = 0, an integration by parts yields t→∞

∞ −

∞ k(s)(∇η(t, s), ∇ut (t))2 ds = −

0

k(s)(∇η(t, s), ∇ηs (t, s) + ∇ηt (t, s))2 ds 0

∞

1 ∞ 1 d 2 k(s) ∇η(t, s) 2 ds + k  (s) ∇η(t, s) 22 ds. =− 2 dt 2 0

(9)

0

So, let G : R+ → R be the function defined by 1 G(t) = G0 (t) + 2

∞ k(s) ∇η(t, s) 22 ds 0

1 1 = ut 22 + Eμ (u(t)) + 2 2

∞ k(s) ∇η(t, s) 22 ds

(10)

0

∞ + (g(s), ut (s))2 ds. t

Combining (8)–(10) and using the assumption (K2), we obtain d 1 G(t) ≤ dt 2

∞

k  (s) ∇η(t, s) 22 ds

0

≤

1 4

∞

k  (s) ∇η(t, s) 22 ds −

0

a0 4

∞ k(s) ∇η(t, s) 22 ds ≤ 0.

(11)

0

Hence the energy function G is decreasing and, since it is also bounded from below, the limit lim G(t) = inf G(t) = G∞ exists.

t→∞

t≥0

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Modifying the basic energy to control ut 2 . Let C0 = λ11 and ε > 0 be a real positive number which will be fixed in the sequel. Lemma 6. Let ∞ I (t) = −(ut (t),

1 k(s)η(t, s)ds)2 + 2

∞

g(t) 22 , t ∈ R+ . t

0

Then, for all t ≥ 0, we have d ε k(0)C0 (1 − μ) I (t) ≤ −

ut (t) 22 + ∇Eμ (u(t)) 2∗ − dt 2 8 2(1 − μ) + (1 − μ)(1 +

2 C0 + ) ε 2

∞

k  (s) ∇η(t, s) 22 ds

0

∞ k(s) ∇η(t, s) 22 ds.

(12)

0

Proof of Lemma 6. Multiplying Eq. (4) by

∞ 0

k(s)η(t, s)ds, we obtain

∞

∞ k(s)η(t, s)ds)2 + (∇Eμ (u),

(utt , 0

0

∞ −(

∞

∞ k(s)η(t, s)ds)2 = (g(t),

k(s)η(t, s)ds, 0

k(s)η(t, s)ds)2

0

k(s)η(t, s)ds)2 .

(13)

0

Using the fact that ηt (t, s) = −ηs (t, s) + ut (t), we find ∞ (utt ,

d k(s)η(t, s)ds)2 = (ut , dt

0

d = (ut , dt

∞

∞ k(s)η(t, s)ds)2 − (ut ,

k(s)ηt (t, s)ds)2

0

0

∞

∞ k(s)η(t, s)ds)2 + (ut ,

0

k(s)ηs (t, s)ds)2 0

− (1 − μ) ut (t) 22 . Integrating by parts with respect to s in the infinite memory integral, and using the fact that lim k(s) = 0 and η(t, 0) = 0, we obtain: s→∞

∞ (utt , 0

d k(s)η(t, s)ds)2 = ut , dt

∞ k(s)η(t, s)ds 0

− (1 − μ) ut (t) 22 .

∞

2

− (ut , 0

k  (s)η(t, s)ds)2

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Then, combining this equality and (13), one obtains: d − (ut , dt

∞

∞ k(s)η(t, s)ds)2 = −(1 − μ) ut (t) 22

− (ut ,

0

k  (s)η(t, s)ds)2

0

∞ + (∇Eμ (u),

∞ k(s)η(t, s)ds)2 − (g(t),

0

0

∞ −(

k(s)η(t, s)ds)2

∞ k(s)η(t, s)ds,

0

k(s)η(t, s)ds)2 .

(14)

0

Next, note the following estimates: ∞ (ut , 0

∞

1−μ 1 2 k (s)η(t, s)ds)2 ≤ −k  (s) −k  (s) η(t, s) 2 ds)22

ut (t) 2 + ( 2 2(1 − μ) 

0

≤

1−μ k(0)C0

ut (t) 22 − 2 2(1 − μ)

∞

k  (s) ∇η(t, s) 22 ds,

(15)

0

∞ k(s)η(t, s)ds)2 ≤ ∇Eμ (u) ∗

(∇Eμ (u),

∞

0

k(s) ∇η(t, s) 2 ds 0

∞ ε 2 2 ≤ ∇Eμ (u) ∗ + ( k(s) ∇η(t, s) 2 ds)2 8 ε 0

∞

ε 2 2 ≤ ∇Eμ (u) ∗ + ( k(s) k(s) ∇η(t, s) 2 ds)2 8 ε 0

ε 2(1 − μ) ≤ ∇Eμ (u) 2∗ + 8 ε

∞ k(s) ∇η(t, s) 22 ds,

(16)

0

∞ (g(t),

∞

1 1 2 k(s)η(t, s)ds)2 ≤ g(t) 2 + ( k(s) k(s) η(t, s) 2 ds)2 2 2

0

0

1 C0 (1 − μ) ≤ g(t) 22 + 2 2

∞ k(s) ∇η(t, s) 22 ds 0

1 d =− 2 dt

∞

g(s) 22 ds t

C0 (1 − μ) + 2

∞ k(s) ∇η(t, s) 22 ds, 0

(17)

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∞ ∞ ∞ −( k(s)η(t, s)ds, k(s)η(t, s)ds)2 ≤ ( k(s) ∇η(t, s) 2 ds)2 0

0

0

∞ ≤ (1 − μ)

k(s) ∇η(t, s) 22 ds.

(18)

0

Inserting estimates (15)–(18) into the equality (14), we obtain (12).

2

Modifying the basic energy to control ∇Eμ (u) ∗ . Lemma 7. Define J (t) = (−μu + f (u), ut )∗ + C0

∞

g(s) 22 ds t

= (∇Eμ (u), ut )∗ + C0

∞

g(s) 22 ds, t ∈ R+ ,

t

where C0 is the best constant in the inequality u 2∗ ≤ C0 u 22 , u ∈ L2 (). Then there exists a positive constant C1 such that, for all t ≥ 0, d 1 J (t) ≤ − ∇Eμ (u) 2∗ + C1 ut 22 + (1 − μ) dt 2

∞ k(s) ∇η(t, s) 22 ds.

(19)

0

Proof of Lemma 7. By differentiating J and using Eq. (4), the Cauchy–Schwarz inequality, Young’s inequality, and (K1) we obtain d J (t) = (∇Eμ (u), utt )∗ + (∇ 2 Eμ (u)ut , ut )∗ − C0 g 22 dt ∞ = (∇Eμ (u), g − ∇Eμ (u) + k(s)η(t, s)ds)∗ + (∇ 2 Eμ (u)ut , ut )∗ − C0 g 22 0

1 ≤ − ∇Eμ (u) 2∗ + g 2∗ +

2

∞

k(s)η(t, s)ds 2∗ + (∇ 2 Eμ (u)ut , ut )∗ − C0 g 22

0

1 ≤ − ∇Eμ (u) 2∗ + (1 − μ) 2

∞ k(s) ∇η(t, s) 22 ds + (∇ 2 Eμ (u)ut , ut )∗ .

(20)

0

In order to estimate the term (∇ 2 E(u)ut , ut )∗ let K = (−)−1 : H −1 () → H01 () be the duality mapping. We equip H −1 () with the inner product:

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13

(g1 , g2 )∗ = (Kg1 , Kg2 )H 1 () , g1 , g2 ∈ H −1 (). 0

Note that K ◦ ∇ 2 E(v) ∈ L(H01 ()). Moreover, by the definition of E and (−), for all u, v ∈ H01 (), we have K ◦ ∇ 2 E(u)v = v + (−)−1

 ∂f ∂u

(x, u)v .

From this, the growth assumption on f and the Sobolev embedding theorem, it is not difficult to deduce that the operator K ◦ ∇ 2 E(v) extends to a bounded linear operator on L2 () for every v ∈ H01 (), and K ◦ ∇ 2 E : H01 () → L(L2 ()) maps bounded sets into bounded sets. In addition, for all u ∈ L2 (), v ∈ H −1 (), we have (u, v)∗ = (Ku, Kv)H 1 () = (−(Ku), Kv)H −1 (),H 1 () 0

0

= (u, Kv)H −1 (),H 1 () = (u, Kv)2 . 0

Then: (∇ 2 E(u)ut , ut )∗ = (K ◦ ∇ 2 E(u)ut , ut )2 ≤ K ◦ ∇ 2 E(u) L(L2 ()) ut 22 ≤ C1 ut 22 . The claim follows from this inequality and (20).

(21)

2

Constructing a suitable new Lyapunov function. Set H0 (t) = G(t) + ε2 I (t) + ε 3 J (t) 1 1 = ut 22 + Eμ (u(t)) + 2 2

∞ k(s) ∇η(t, s) 22 ds

∞ + (g(s), ut (s))2 ds t

0

∞ − ε 2 (ut (t),

k(s)η(t, s)ds)2 + (ε 3 C0 +

ε2 ) 2

∞

g(s) 22 ds t

0

+ ε (∇Eμ (u), ut )∗ 3

Then, by combining (11), (12), and (19), we have for all t ≥ 0, d a0 H0 (t) ≤ − dt 4

∞ k(s) ∇η(t, s) 22 ds 0



+ ε2 −

k(0)C0 1 + ( − ε2 ) 4 2(1 − μ)

ε (1 − μ)

ut (t) 22 + ∇Eμ (u(t)) 2∗ + 2 8

∞ 0

k  (s) ∇η(t, s) 22 ds

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14

2 C0 (1 − μ)(1 + + ) ε 2

∞ k(s) ∇η(t, s) 22 ds



0

+ε

3



1 − ∇Eμ (u) 2∗ + C1 ut 22 + (1 − μ) 2

∞

k(s) ∇η(t, s) 22 ds .

0

Thus, if we choose ε > 0 small enough, we see that there exists a constant C2 > 0 such that for every t ≥ 0, d H0 (t) ≤ −C2 ut 22 + ∇Eμ (u) 2∗ + dt

∞ k(s) ∇η(t, s) 22 ds 0

∞

(−k  )(s) ∇η(t, s) 22 ds

+



0

∞



≤ −C2 ut 22 + ∇Eμ (u) 2∗

+

k(s) ∇η(t, s) 22 ds .

(22)

0

Hence the energy function H0 is decreasing and, since it is also bounded from below, the limit lim H0 (t) = inf H0 (t) = H∞ exists.

t→∞

t≥0

From this and the inequality (22) we obtain (i). Let φ ∈ ω(u). Then there exists an unbounded increasing sequence (tn ) in R+ such that u(tn ) → φ in H01 (). Since ut ∈ L2 (R+ , L2 ()), we have tn +s

u(tn + s) = u(tn ) +

ut (ρ) dρ → φ in L2 () for every s ∈ [0, 1]. tn

This, together with the relative compactness of the trajectory in H01(), implies that: u(tn + s) → φ in H01 () for every s ∈ [0, 1]. Hence, by the continuity of Eμ , Eμ (u(tn + s)) → Eμ (φ) in R for every s ∈ [0, 1]. Consequently, using the dominated convergence theorem, 1 Eμ (φ) = lim

Eμ (u(tn + s))ds.

n→∞ 0

Therefore, by integrating G(tn + ·) over (0, 1), we obtain

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15

1 Eμ (φ) = lim

G(tn + s)ds = G∞ .

n→∞ 0

Here we have used (i), the integrability of g and the fact that ut is bounded in L2 (). Since φ was chosen arbitrarily in ω(u), this implies that Eμ is constant on ω(u). Moreover, since u has compact range in H01 (), we obtain lim Eμ (u(t)) = G∞ = E∞ .

t→∞

Property (iii) is an immediate consequence of (ii) and (10). In order to prove (iv), let φ ∈ ω(u) and choose tn → ∞ such that u(tn ) → φ in H01 (). We have already seen that this implies u(tn + s) → φ in H01 () for every s ∈ [0, 1]. Hence ∇Eμ (u(tn + s)) → ∇Eμ (φ) in H −1 () for every s ∈ [0, 1].

(23)

Next, note the following estimate ∞

k(s)η(t, s)ds 2∗

≤

∞

0

k(s) ∇η(t, s) 2 ds

2

0

∞ ≤ k L1 (R+ )

k(s) ∇η(t, s) 22 ds.

(24)

0

By using (23), (24), the dominated convergence theorem, (4), (i) and (iii) we infer (in H −1 ()) that: 1 ∇Eμ (φ) =

1 ∇Eμ (φ)dτ = lim

∇Eμ (u(tn + τ ))dτ

n→∞

0

0

1 = lim

n→∞ 0



∞ − utt +

k(s)η(t, s)ds + g (tn + τ )dτ

0

1 ∞

 g + k(s)η(t, s)ds (tn + τ )dτ = lim ut (tn ) − ut (tn + 1) + n→∞

0

0

= 0. By Proposition 1, for all φ ∈ ω(u) there exist constants σφ > 0, βφ > 0 and 0 < θφ ≤ |Eμ (ψ) − Eμ (φ)|1−θφ ≤ βφ ∇Eμ (ψ) ∗ for every ψ ∈ Bσφ (φ) = {ψ ∈ H01 () : φ − ψ H 1 () ≤ σφ }. 0

1 2

such that (25)

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16

By continuity of Eμ , we can choose σφ > 0 small enough such that |Eμ (ψ) − Eμ (φ)| ≤ 1 for every ψ ∈ Bσφ (φ).

(26)

In addition, since ω(u) is compact, there exists a finite family of open balls B(φi , σφi ) covering ω(u), where φi ∈ ω(u) and σφi > 0 is such that (25) and (26) hold. Since lim dist(u(t), ω(u)) t→∞  = 0, there exists T > 0 such that u(t) ∈ B(φi , σφi ) for all t ≥ T . Choosing β = sup βφi , θ = inf θφi we obtain (v). 2 Now, we are in a position to prove Theorem 3. Proof of Theorem 3. Let H : R+ → R be the function given by H (t) = H0 (t) − E∞ . By (22), d H (t) ≤ −C2 ut 22 + ∇Eμ (u) 2∗ + dt

∞

k(s) ∇η(t, s) 22 ds .

(27)

0

Then H (t) is decreasing. In addition lim H (t) = 0. Then H (t) is positive. Now, we consider t→∞

two possibilities: If the function g satisfies the polynomial growth (G1), then, for θ as in Lemma 5, let θ0 ∈ (0, θ ] δ . Note that (7) is satisfied with θ replaced be such that (1 + δ)(1 − θ0 ) > 1, that is θ0 < 1+δ by θ0 . Then, by applying the Cauchy–Schwarz inequality and Young’s inequality, we obtain 1−θ0  θ 2(1−θ0 ) H (t)1−θ0 ≤ C ut 2 + |Eμ (u) − E∞ |(1−θ0 ) + ∇Eμ (u) ∗ + ut 2 0

+

∞

k(s) ∇η(t, s) 22

ds

1 2(1−θ0 ) 2

∞

(1−θ0 ) +

g(s) 22 ds t

0

∞

(1−θ0 )  + (g(s), ut (s))2 ds .

(28)

t



(1−θ0 ) ∞ Next, we will control the term t (g(s), ut (s))2 ds by the other terms of the right-hand side of this inequality. Using the Cauchy–Schwarz inequality and Young’s inequality, ∞ ∞ ∞ 1 2 2 (g(s), ut (s))2 ds ≤

g(s) 2 ds + C2 ut (s) 22 ds, C2 t

t

t

where C2 is given by (27). In addition, H (t) positive and decreasing to 0. Then, by (27),

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17

∞ 1

ut (s) 22 ≤ H (t), t ∈ R+ . C2 t

Hence, ∞ ∞ 1 2 (g(s), ut (s))2 ds ≤

g(s) 22 ds + H (t). C2 t

t

Using this inequality, the definition of H , the Cauchy–Schwarz inequality and Young’s inequality, we obtain ∞ 1 1  1−θ θ (g(s), ut (s))2 ds ≤ C ut 22 + |Eμ (u) − E∞ |+ ∇Eμ (u) ∗ 0 + ut 20 t

∞ +

k(s) ∇η(t, s) 22

∞  ds + g(s) 22 ds . t

0

It follows from this inequality and Young’s inequality that

(1−θ0 )  ∞ 2(1−θ0 ) (g(s), ut (s))2 ds ≤ C ut 2 + |Eμ (u) − E∞ |(1−θ0 ) + ∇Eμ (u) ∗ t 1−θ0 θ0

+ ut 2

+

∞

k(s) ∇η(t, s) 22 ds

1 2(1−θ0 ) 2

0

+



∞

g(s) 22 ds

(1−θ0 ) 

,

t

which together with (28) implies that 1−θ0  θ 2(1−θ0 ) H (t)1−θ0 ≤ C ut 2 + |Eμ (u) − E∞ |(1−θ0 ) + ∇Eμ (u) ∗ + ut 2 0

+

∞ 0

k(s) ∇η(t, s) 22

ds

1 2(1−θ0 ) 2

∞

(1−θ0 )  +

g(s) 22 ds . t

By Lemma 5 (iii) and (v), choosing T > 0 sufficiently large so as to ensure that both ∞

ut (t) 2 ≤ 1, 0 k(s) ∇η(t, s) 22 ds ≤ 1, and inequality (7) hold for all t ≥ T , and taking into 0 account that 1−θ θ0 ≥ 1, and 2(1 − θ0 ) ≥ 1 it follows that for every t ≥ T we have

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18

1−θ0

H (t)

∞ 1 ≤ C ut 2 + ∇Eμ (u) ∗ + ( k(s) ∇η(t, s) 22 ds 2

(29)

0

+



∞ (1−θ0 )

g(s) 22 ds . t

If there exists T0 ≥ T such that H (T0 ) = 0, then H (t) = 0 for all t ≥ T0 . By the inequality (27) we obtain ut (t, .) 2 = 0 for all t ≥ T0 . In this case, u is a stationary solution and, in particular, a convergent solution. We may therefore suppose in the following that H (t) > 0 for all t ≥ T , therefore we can differentiate H (t)θ0 and, using (29), (27), (G1), we have for all t ≥ T −

d d H (t)θ0 = −θ0 H (t)θ0 −1 H (t) dt dt ∞  C ut 22 + ∇Eμ (u) 2∗ + 0 k(s) ∇η(t, s) 22 ds ≥  ∞ 1 ( ut 2 + ∇Eμ (u) ∗ + 0 k(s) ∇η(t, s) 22 ds 2 + (1 + t)−(1+δ)(1−θ0 )   ≥ C ut 2 + ∇Eμ (u) ∗ +

∞ k(s) ∇η(t, s) 22 ds

1 2

(30)

0 −(1+δ)(1−θ0 )

− C(1 + t)

.

d H (t)θ0 + C(1 + t)−(1+δ)(1−θ0 ) is intedt grable on [T , +∞), we obtain that ut 2 is integrable on [T , +∞), which implies that lim u(t, ·) From the above inequality and the fact that the term −

t→∞

exists in L2 (). By the relative compactness of the range of u in H01 (), lim u(t, ·) exists t→∞

in H01 (). This is the claim. Finally, when the growth condition in g is exponential, we replace, in the inequality (30), the term (1 + t)−(1+δ)(1−θ0 ) by the term e−δt (1−θ0 ) which is integrable too, and then the same conclusion holds. This completes the proof of Theorem 3. 2 Now, we shall prove the exponential or polynomial decay of solutions to Eq. (2), depending on the Łojasiewicz exponent and the type of the decay condition on g. The following lemma is used in the proof of the polynomial convergence rate. Its proof can be found in [6]. 1,1 Lemma 8. Let ζ ∈ Wloc (R+ , R+ ). We suppose that there exist constants K1 > 0, K2 ≥ 0, k > 1 and λ > 0 such that for almost every t ≥ 0 we have

d ζ (t) + K1 ζ (t)k ≤ K2 (1 + t)−λ . dt

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19

Then there exists a positive constant m such that ζ (t) ≤ m(1 + t)−ν , where ν = inf{

1 λ , }. k−1 k

Proof of Theorem 4. We proceed in two steps. Step 1 (Polynomial decay). First, we note that the inequalities (29) and (30) are satisfied when θ0 is replaced by the initial exponent θ given by Lemma 5 (v). By using (29) together with (G1) and Young’s inequality, we obtain for every t ∈ [T , ∞[ H (t)2(1−θ) ≤ C ut 22 + ∇Eμ (u) 2∗ ∞ +

(31)

k(s) ∇η(t, s) 22 ds + (1 + t)−2(1+δ)(1−θ) .

0

From this inequality and (27), we obtain the following differential inequality for every t ≥ T C3

d H (t) + H (t)2(1−θ) ≤ C(1 + t)−2(1+δ)(1−θ) . dt

(32)

Then we may apply Lemma 8 in order to obtain H (t) ≤ C(1 + t)−γ ,

(33)

1 where γ = inf{ 1−2θ , 1 + δ}. Using again (27), we have

−

d H (t) ≥ C ut (t) 22 . dt

Integrating this inequality over [t, 2t] (t ≥ T ) and using (33) and the fact that H (t) ≥ 0, we obtain 2t

ut (s) 22 ds ≤ C(1 + t)−γ .

t

Note that for every t ∈ R+ , 2t t

2t 1

ut (s) 2 ds ≤ t ( ut (s) 22 ds) 2 . 1 2

t

It follows that 2t

ut (s) 2 ds ≤ C(1 + t) t

1−γ 2

for every t ≥ T .

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20

Therefore we obtain for every t ≥ T k+1 ∞ ∞ 2 t ∞   1−γ 1−γ

ut (s) 2 ds ≤

ut (s) 2 ds ≤ C (2k t) 2 ≤ C(1 + t) 2 .

k=0

t

k=0

2k t

Then, for all t ≥ T ∞

u(t) − φ 2 ≤ C ut (s) 2 ds ≤ C(1 + t)−ξ , where ξ = inf{ t

Step 2 (Exponential decay). Suppose that θ = tial growth (G2). Then (32) becomes

1 2

θ δ , }. 1 − 2θ 2

and that g satisfies the exponen-

d H (t) ≤ −C4 H (t) + C5 e−δt , dt where C4 = Now, let

1 C3

and C3 can be chosen large enough to ensure that C4 < δ.

K(t) = H (t) − C5 e

−C4 t

t

e−(δ−C4 )s ds.

0

Then d d K(t) = H (t) − C5 e−δt + C4 C5 e−C4 t dt dt

t

e−(δ−C4 )s ds

0

≤ −C4 K(t). This yields K(t) ≤ e−C4 t , and therefore H (t) ≤ e

−C4 t

t



1 + C5

e−(δ−C4 )s ds ≤ Ce−C4 t .

(34)

0

On the other hand, from the inequality (30) (when g satisfies the exponential decay and θ0 = θ = 12 ) we have for every t ≥ T −

1 δt d H (t) 2 + Ce− 2 ≥ C ut (t) 2 . dt

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21

Integrating this inequality over the interval [t, ∞) (t ≥ T ), we obtain ∞ 1 δt

u(t) − φ 2 ≤ ut (s) 2 ds ≤ CH (t) 2 + Ce− 2 . t

This inequality together with the inequality (34) implies the claim.

2

3.2. The case of finite memory In this section we shall prove Theorems 3 and 4 for equation (1). We let t k ◦ v(t) =

k(t − s) v(t) − v(s) 22 ds. 0

The following lemma represents a version of Lemma 5 for equation (1). Lemma 9. Let u : R+ → H01 () be a weak solution of equation (1) and assume that the assumptions of Theorem 3 hold. Then: (i) ut ∈ L2 (R+ ; L2 ()), ∇Eμ (u) ∈ L2 (R+ ; H −1 ()) and k ◦ ∇u(t) ∈ L1 (R+ ). (ii) The function Eμ is constant on ω(u), and lim Eμ (u(t)) = Eμ (φ) = E∞ = const < ∞, for all φ ∈ ω(u).

t→∞

(iii) lim ut (t) 2 = lim k ◦ ∇u(t) = 0. t→∞

t→∞

(iv) The ω-limit set of u is a subset of the set of stationary solutions. (v) There exists a uniform Łojasiewicz exponent θ ∈]0, 12 ], β > 0 and T > 0 such that for all t ≥ T |Eμ (u(t)) − E∞ |1−θ ≤ β ∇Eμ (u) ∗ . Proof of Lemma 9. The proof of this lemma is analogous to that of Lemma 5; the only difference concerns the energy estimate, so we outline this part. Let U0 : R+ → R be the function defined by 1 U0 (t) = ut 22 + E(u(t)) + 2

∞ (g(s), ut (s))2 ds, t

where E is the energy functional defined on H01 () by 1 E(u) = 2



 |∇u| dx + 2



1 = Eμ (u) + 2

F (x, u)dx 

∞ k(s)ds ∇u(t) 22 . 0

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22

The function U0 is differentiable and d U0 (t) = dt

t k(t − s)(∇u(s), ∇ut (t))2 ds.

(35)

0

A direct calculations yields t

t  1 d k(t − s)(∇u(s), ∇ut (t))2 ds = ( k(s)ds) ∇u(t) 22 − k ◦ ∇u(t) 2 dt

0

0

1 1 + k  ◦ ∇u(t) − k(t) ∇u(t) 22 . 2 2

(36)

Then, let U : R+ → R be the function defined by t 1 1 U (t) = U0 (t) − ( k(s)ds) ∇u(t) 22 + k ◦ ∇u(t) 2 2 0

1 1 = ut 22 + Eμ (u(t)) + k ◦ ∇u(t) 2 2 ∞ ∞ 1 2 k(s)ds ∇u(t) 2 + (g(s), ut (s))2 ds. + 2 t

t

By combining (35), (36) and the assumption (K2), we infer that d 1 U (t) ≤ k  ◦ ∇u(t) − dt 2 1  ≤ k ◦ ∇u(t) − 4

1 k(t) ∇u(t) 22 2 a0 1 k ◦ ∇u(t) − k(t) ∇u(t) 22 ≤ 0. 4 2

(37)

Hence the energy function U is decreasing and, since it is also bounded from below, the limit lim U (t) = inf U (t) = U∞ exists.

t→∞

t≥0

Modifying the basic energy to control ut 2 . Let ε > 0 be a real positive number which will be fixed in the sequel. Lemma 10. Let t I1 (t) = −(ut (t), 0

Then, for every t ∈ R+ ,

1 k(t − s)(u(t) − u(s))ds)2 + 2

∞

g(s) 22 ds, t ∈ R+ . t

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23

t d ε ε (1 − μ)k(t)

∇u(t) 22 I1 (t) ≤ −( k(s)ds − ) ut (t) 22 + ∇Eμ (u(t)) 2∗ + dt 2 8 2a0 0

+ (1 − μ)(

C0 + 3 2 C0 (k(t) − k(0))  + )k ◦ ∇u + k ◦ ∇u. 2 ε 2ε

(38)

Proof of Lemma 10. By differentiating I1 and using (1), we easily see that t t d 2 I1 (t) = −( k(s)ds) ut 2 − (ut (t), k  (t − s)(u(t) − u(s))ds)2 dt 0

0

t − (utt ,

1 k(t − s)(u(t) − u(s))ds)2 − g(t) 22 2

0

t = −(

t k(s)ds) ut 22

− (ut (t),

0

k  (t − s)(u(t) − u(s))ds)2

(39)

0

t

t k(t − s)u(s)ds − g,

+ (−u + f (u) + 0

k(t − s)(u(t) − u(s))ds)2 0

1 − g(t) 22 . 2 We have t −(ut (t),

1 k (t − s)(u(t) − u(s))ds)2 ≤ 2ε 

0

t

2 −k  (t − s) −k  (t − s) (u(t) − u(s)) 2 ds

0

ε + ut 22 2 C0 (k(t) − k(0)) ≤ 2ε

t

k  (t − s) (∇u(t) − ∇u(s)) 22 ds

0

ε + ut 22 2 C0 (k(t) − k(0))  ε ≤ k ◦ ∇u + ut 22 . 2ε 2 In addition, observe that t

t k(t − s)u(s)ds =

0

t k(t − s)(u(s) − u(t))ds +

0

k(s)u(t)ds 0

(40)

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24

t =

∞ k(t − s)(u(s) − u(t))ds + u(t)

0

k(s)ds

(41)

0

∞ −u(t)

k(s)ds. t

It follows that t (−u + f (u) +

t k(t − s)u(s)ds − g,

0

0

∞ = (∇Eμ (u) − u(t)

t k(s)ds − g,

t

k(t − s)(u(t) − u(s))ds)2

(42)

0

t −(

k(t − s)(u(t) − u(s))ds)2

t k(t − s)(u(t) − u(s))ds,

0

k(t − s)(u(t) − u(s))ds)2 . 0

The terms on the right hand side of (42) can be controlled in the following way: t k(t − s)(u(t) − u(s))ds)2

(∇Eμ (u), 0

t ≤ ∇Eμ (u) ∗

k(t − s) (∇u(t) − ∇u(s)) 2 ds 0

ε 2 ≤ ∇Eμ (u) 2∗ + 8 ε

t

2 k(t − s) k(t − s) (∇u(t) − ∇u(s)) 2 ds 0

ε 2 ≤ ∇Eμ (u) 2∗ + ( 8 ε

t

t k(t − s) (∇u(t) − ∇u(s)) 22 ds

k(s)ds) 0

0

2(1 − μ) ε )k ◦ ∇u, ≤ ∇Eμ (u) 2∗ + ( 8 ε ∞

t

t

1 ≤ u(t) 2

k(t − s)(u(t) − u(s))ds)2

k(s)ds,

(−u(t)

0

∞ k(s)ds 2∗ t

1 +

2

t k(t − s)(∇u(t) − ∇u(s))ds 22 0

(43)

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1 ≤ 2

∞

25

2 1−μ k(s)ds ∇u(t) 22 + ( )k ◦ ∇u 2

t

∞ 1−μ 1−μ ≤ ( k(s)ds ∇u(t) 22 + ( )k ◦ ∇u 2 2 t

≤ t (g(t),

1−μ (1 − μ)k(t)

∇u(t) 22 + ( )k ◦ ∇u, 2a0 2

1 1 k(t − s)(u(t) − u(s))ds)2 ≤ g(t) 22 +

2 2

0

(44) t k(t − s)(u(t) − u(s))ds 22 0

1 C0 (1 − μ) ≤ g(t) 22 + ( )k ◦ ∇u, 2 2

(45)

t t − ( k(t − s)(u(t) − u(s))ds, k(t − s)(u(t) − u(s))ds)2 0

0

t =

k(t − s)(∇u(t) − ∇u(s))ds 22 0

≤ (1 − μ)k ◦ ∇u. By combining (39)–(46) we obtain (38).

(46) 2

Modifying the basic energy to control ∇Eμ (u) ∗ . Lemma 11. Let J1 (t) = (−μu + f (u), ut )∗ + C0

∞

g(s) 22 ds t

= (∇Eμ (u), ut )∗ + C0

∞

g(s) 22 ds t ∈ R+ .

t

Then, for every t ∈ R+ , we have d 1 (1 − μ)k(t)

∇u 22 , J1 (t) ≤ − ∇Eμ (u) 2∗ + C1 ut 22 + (1 − μ)k ◦ ∇u + dt 4 a0 where C1 is given by (19).

(47)

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26

Proof of Lemma 11. By differentiating J1 , using (1) and (41), we easily see that d J1 (t) = (∇Eμ (u), utt )∗ + (∇ 2 Eμ (u)ut , ut )∗ − C0 g(t) 22 dt t = (∇Eμ (u), g − ∇E(u) − k(t − s)u(s) ds) + (∇ 2 Eμ (u)ut , ut )∗ − C0 g(t) 22 0

t

∞

= (∇Eμ (u), g − ∇Eμ (u) −

k(t − s)(u(s) − u(t)) ds + u(t) t

0

+ (∇

2

k(s)ds)∗

Eμ (u)ut , ut )∗ − C0 g 22 .

Then, using the Cauchy–Schwarz inequality, Young’s inequality, (K1), (K2), and (21) we obtain d 1 J1 (t) ≤ − ∇Eμ (u) 2∗ + g 2∗ +

dt 4

t

∞ k(t

− s)(u(s) − u(t)) ds 2∗ + u(t) t

0

+ (∇

2

k(s)ds 2∗

Eμ (u)ut , ut )∗ − C0 g 22

1 (1 − μ)k(t) ≤ − ∇Eμ (u) 2∗ + C1 ut 22 + (1 − μ)k ◦ ∇u +

∇u 22 . 2 4 a0 Constructing a suitable new Lyapunov function. Let V0 (t) = U (t) + ε 2 I1 (t) + ε 3 J1 (t) 1 = ut 22 + Eμ (u(t)) + 2

∞ ∞ 1 (g(s), ut (s))2 ds + ( k(s)ds) ∇u(t) 22 2 t

t

1 + k ◦ ∇u(t) − ε 2 (ut (t), 2

t k(t − s)(u(t) − u(s))ds)2 0

+ ε (∇Eμ (u), ut )∗ + (ε 3

3

C0

ε2 + ) 2

∞

g(t) 22 . t

Then, for every t ∈ R+ , we have d 1 a0 1 V0 (t) ≤ k  ◦ ∇u(t) − k ◦ ∇u(t) − k(t) ∇u(t) 22 dt 4 4 2 t  ε ε (1 − μ)k(t) + ε 2 − ( k(s)ds − ) ut (t) 22 + ∇Eμ (u(t)) 2∗ +

∇u(t) 22 2 8 2a0 0

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27

C0 + 3 2 C0 (k(t) − k(0))  + )k ◦ ∇u + k ◦ ∇u 2 ε 2ε 1

(1 − μ)k(t) + ε 3 − ∇Eμ (u) 2∗ + C1 ut 22 + (1 − μ)k ◦ ∇u +

∇u(t) 22 . 4 a0 + (1 − μ)(

Choosing ε > 0 small enough, one sees that there exist T , C6 > 0 such that for all t ≥ T ,  d V0 (t) ≤ −C6 ut 22 + ∇Eμ (u) 2∗ − k  ◦ ∇u(t) + k ◦ ∇u(t) + k(t) ∇u(t) 22 dt  ≤ −C6 ut 22 + ∇Eμ (u) 2∗ + k ◦ ∇u(t) + k(t) ∇u(t) 22 .

(48)

Hence the energy function V0 is decreasing for large t (t ≥ T ) and, since it is also bounded from below, the limit lim V0 (t) = inf V0 (t) = V∞ exists.

t→∞

t≥0

Using the last inequality and (37), the proof of Lemma 9 can be completed in exactly the same way as that of Lemma 5, so we omit the details. 2 Proof of Theorem 3 for the equation (1). Let V (t) = V0 (t) − E∞ .

(49)

By (48) we have  d V (t) ≤ −C6 ut 22 + ∇Eμ (u) 2∗ + k ◦ ∇u(t) + k(t) ∇u(t) 22 , t ≥ T . dt On the other hand, using Lemma 9 and arguing as in the first case, we can show that, for T > 0 sufficiently large and for every t ≥ T , 1 V (t)1−θ0 ≤ C ut 2 + ∇Eμ (u) ∗ + (k ◦ ∇u(t)) 2 +

(50)

∞ (1−θ0 )

k(t) ∇u(t) 2 +

g(s) 22 ds . 

t

The rest of the proof of Theorem 3 can be done in exactly the same way as that for the first case and we do not go into detail. 2 Finally, the proof of the convergence rate for the second model is exactly the same as that for the first model and we omit it.

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4. Boundedness of global solutions In this section we give, under the hypothesis (F3), a proof of boundedness of global solutions of equation (2). The proof of the corresponding result for model (1) is analogous, so we omit it. Proof of Proposition 2. Let W0 : R+ → R be the function defined by 1 W0 (t) = ut 22 + Eμ (u(t)) − 2

t (g(s), ut (s))2 ds. 0

By multiplying Eq. (4) by ut , integrating over  and using integration by parts, one can easily find that d W0 (t) = − dt

∞ k(s)(∇η(t, s), ∇ut (t))2 ds.

(51)

0

Now, let W : R+ → R be the function defined by 1 W (t) = W0 (t) + 2

∞ k(s) ∇η(t, s) 22 ds 0

1 1 = ut 22 + Eμ (u(t)) + 2 2

∞ k(s) ∇η(t, s) 22 ds 0

t −

(g(s), ut (s))2 ds. 0

Then by combining (51) and (9), we obtain d 1 W (t) ≤ dt 2

∞

k  (s) ∇η(t, s) 22 ds ≤ 0.

0

Thus W is decreasing. In addition, from the condition (F2) we have  |F (x, u0 )|≤ C(1+ u0 α+2 1 ), H0



where C ≥ 0 is a constant depending only on the constants from condition (F2), the measure of  and the constant of the embedding H01 () → Lα+2 (). It follows from this inequality and the definition of W that there exists a constant C ≥ 0 such that 2 W (0) ≤ C (1+ u1 22 + ∇u0 22 + u0 α+2 1 + η0 L2 (R+ ;H 1 ()) ). H0

k

0

(52)

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29

On the other hand, using condition (F3), there exists C > 0 such that

∇u(t) 22 ≤ CEμ (u(t)) + C. Now, we have 1 W (t) ≥ ut 22 + Eμ (u(t)) − 2

t (g(s), ut (s))2 ds 0

t ≥ C( ∇u(t) 22 + ut (t) 22 ) −

(g(s), ut (s))2 ds + C. 0

Combining this inequality, (52), and the fact that W is decreasing, we obtain t ( ∇u(t) 22 + ut (t) 22 ) ≤ CW (0) + C

(g(s), ut (s))2 ds + C 0

t ≤C+C

g(s) 2 ut (s) 2 ds. 0

From this inequality and the Cauchy–Schwarz inequality we have

∇u(t) 22 ut (t) 22 + +1 2 2 t ≤ C + C g(s) 2 ut (s) 2 ds

∇u(t) 2 + ut (t) 2 ≤

0

t ≤ C7 + C8

g(s) 2 ( ut (s) 2 + ∇u(s) 2 )ds. 0

Using the Grönwall Lemma [5,20],

∇u(t) 2 + ut (t) 2 ≤ C7 eC8

t

0 g(s) 2 ds

≤ C.

2

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