Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2013, Article ID 372726, 11 pages http://dx.doi.org/10.1155/2013/372726 Research Article Hyperbolic Relaxation of a Fourth Order Evolution Equation Renato Colucci and Gerardo R. Chacón Departamento de MatemaΜticas, Pontificia Universidad Javeriana, Cra. 7 No. 43-82, BogotaΜ, Colombia Correspondence should be addressed to Renato Colucci; renatocolucci@hotmail.com Received 26 November 2012; Revised 30 January 2013; Accepted 3 February 2013 Academic Editor: Juan J. Nieto Copyright © 2013 R. Colucci and G. R. ChacoΜn. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We propose a hyperbolic relaxation of a fourth order evolution equation, with an inertial term ππ’π‘π‘ , where π ∈ (0, 1]. We prove the existence of several absorbing sets having different regularities and the existence of a global attractor that is bounded in π»4 (πΌ) × {π»2 (πΌ) ∩ π»01 (πΌ)}. 1. Introduction Leting πΌ ⊂ R be an open interval, with |πΌ| ≤ 1, we consider the following initial-boundary value problem for π’ : πΌ × R+ → R: ππ’π‘π‘ + π’π‘ = − π2 π’π₯π₯π₯π₯ 1 + πσΈ σΈ (π’π₯ ) π’π₯π₯ , 2 (π₯, π‘) ∈ πΌ × R+ , π’ (π₯, 0) = π’0 (π₯) , in πΌ, π’π‘ (π₯, 0) = π’1 (π₯) , in πΌ, π’ = π’π₯π₯ = 0, (1) in ππΌ, π‘ ≥ 0, where the function π(π) = (π2 − 1)2 is the so-called doublewell potential, 0 < π βͺ 1 and π ∈ (0, 1] are positive parameters. Problem (1), with π = 0, was proposed in [1] where the global dynamics was studied. In particular, the dynamical behavior of the solutions for small values of the parameter π was studied by means of numerical experiments. The existence of three well-differentiated time scales with peculiar dynamical behavior was showen. In the first time scale of order π(π2 ) there is the formation of microstructure (see [2]) in the region where the gradient of the initial datum falls in the nonconvex region of π; this phenomenon produces a drastic reduction of the energy of the initial datum. In the second time scale of order π(1) the equation exhibits a heat equation-like behavior in the convex regions while slow motion in the nonconvex ones. In the last time scale of order π(π−2 ) the equation shows a finite-dimensional behavior: the solution is approximately the union of consecutive segments and the dynamic is slow. In [3], the third time scale was studied; the authors proved the existence of a global attractor Aπ ⊂ πΏ2 (πΌ) (see [4]) that is bounded in π»2 (πΌ). The time for which the solutions enter the absorbing set Bπ is of order π(π−2 ) and it is consistent with the estimates found in [1]. Moreover the authors proved the existence of an exponential attractor Iπ with finite fractal dimension of order π(π−10 ). In [5] the authors proved the existence of an inertial manifold Mπ (see [6]) whose dimension is of order π(π−19 ), and by the πdimensional volume elements methods (see [7]) an estimate of the dimension of the global attractor of order π(π−1 ) was found. This estimate is also consistent with the numerical experiments developed in [1]; in fact it was found that the wave length of the microstructure is of order π(π−1 ). In the last years the viscous and no viscous hyperbolic relaxation of the Cahn-Hilliard equation has been extensively investigated. The model was proposed in [8] while in [9] the existence of a family of exponential attractors was proved. The viscous and nonviscous perturbation has been studied in [10] where the existence of a family of global attractors that are upper semicontinuous with respect to the vanishing of perturbations parameters was proved. These results have been extended in 2 and 3 dimensions; see for example [11, 12] and the references therein. Due to the similarity of problem (1) to the Cahn-Hilliard equation we consider it interesting to study the hyperbolic 2 Abstract and Applied Analysis relaxation of the fourth order evolution equation proposed in [1]. In particular if V is the solution of the Cahn-Hilliard equation Vπ‘ + Δ [π2 ΔV − πσΈ (V)] = 0, (2) π₯ ∈ {0, 1} , (3) then π₯ π’ (π₯) = ∫ V (π ) ππ , (4) 0 is the solution of (1), with π = 0, with the corresponding boundary conditions: π’ = π’π₯π₯ = 0, Μ (π’, V) = πβVβ2 + 1 π2 σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅©2 + π½πβ¨V, π’β©, πΈ 2 σ΅© σ΅© (12) where π½ ∈ [0, π2 ]. with Neumann boundary conditions: π π V= ΔV = 0, ππ ππ Given functions π’ ∈ π»2 (πΌ) ∩ π»01 (πΌ) and V ∈ πΏ2 (πΌ) we define Μ on Hπ,π by the function πΈ π₯ ∈ {0, 1} . (5) In the present work we put the problem in the correct mathematical framework and prove the existence of a global attractor Aπ,π while we have left the proof of the existence of exponential attractors for a forthcoming paper. In Sections 2 and 3 we define the solution semigroup in the appropriate spaces and present some important energy estimates. In Section 4 we prove the existence of several absorbing sets with different regularities while in the last section we prove the existence of the global attractor. 2. Preliminaries Μ ⋅) induces a Proposition 1. For all π½ ∈ [0, π2 ] the function πΈ(⋅, norm equivalent to the norm on Hπ,π . Proof. By Schwartz inequality (11) and using the fact that π½ ≤ π2 < 1, π ∈ (0, 1], we get 1 1 σ΅© σ΅©2 π½π β¨V, π’β© ≤ π½π βVββπ’β ≤ π2 βVβ2 + π4 σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© 2 2 π π 2 σ΅© σ΅©2 ≤ βVβ2 + σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© = β(π’, V)β2Hπ,π . 2 2 (13) Μ V) we From the previous inequality and by definition of πΈ(π’, get Μ (π’, V) ≤ 3β(π’, V)β2 . πΈ Hπ,π (14) By a different application of Schwartz inequality and from (11) we get π½π β¨V, π’β© ≤ π½π βVβ βπ’β ≤ π2 βVβ2 + π½2 βπ’β2 4 π4 π 2 σ΅© σ΅©2 ≤ πβVβ + βπ’β2 ≤ πβVβ2 + σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© . 4 4 (15) 2 We begin this section by defining the following Hilbert spaces that will be helpful for our analysis: 2 Hπ,π = [π» (πΌ) ∩ π»01 2 (πΌ)] × πΏ (πΌ) , 2 (6) Vπ,π = π· (π΄) × π» (πΌ) , (7) π· (π΄) = {π’ ∈ π»4 (πΌ) : π’ = π’π₯π₯ = 0 in ππΌ} (8) where is the domain of the differential operator π΄ = π4 /ππ₯4 . The above spaces are equipped with the norms β(π’, V)β2Hπ,π = β(π’, V)β2Vπ,π = π2 σ΅©σ΅© σ΅©σ΅©2 π 2 σ΅©π’ σ΅© + βVβ , 2 σ΅© π₯π₯ σ΅© 2 π2 σ΅©σ΅© σ΅©σ΅©2 σ΅©π’ σ΅© + 2 σ΅© xπ₯π₯π₯ σ΅© π σ΅©σ΅© σ΅©σ΅©2 σ΅©V σ΅© , 2 σ΅© π₯π₯ σ΅© (9) (10) where β ⋅ β represents the πΏ2 norm. We will denote by β¨⋅, ⋅β© the inner product in πΏ2 (πΌ). We recall that for all π’ ∈ π»2 (πΌ)∩π»01 (πΌ) with |πΌ| ≤ 1, we have σ΅© σ΅© σ΅© σ΅© βπ’β ≤ σ΅©σ΅©σ΅©π’π₯ σ΅©σ΅©σ΅© ≤ σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© . (11) Throughout the paper we will use two norms that are equivalent to (9) and (10) in order to simplify the computation. Combining (13) and (15) we get Μ (π’, V) ≥ max { π βVβ2 , 1 π2 σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅©2 } , πΈ 2 4 σ΅© σ΅© (16) and as a consequence Μ (π’, V) ≥ 1 β(π’, V)β2 . πΈ Hπ,π 3 (17) The proof of the following theorem follows from classical applications of the Faedo-Galerkin method. We will only show a Lipschitz estimate that will be needed for further computations. Theorem 2. For every (π’0 , π’1 ) ∈ Hπ,π there exists a unique solution π’(π‘) for the initial value problem (1) such that π’ ∈ πΆπ (R+ ; π»2 (πΌ) ∩ π»01 (πΌ)) ∩ πΆπ1 (R+ ; πΏ2 (πΌ)) . (18) If, moreover, (π’0 , π’1 ) ∈ Vπ,π , then: π’ ∈ πΆπ (R+ ; π· (π΄)) ∩ πΆπ1 (R+ ; π»2 (πΌ) ∩ π»01 (πΌ)) ∩ πΆπ2 (R+ ; πΏ2 (πΌ)) . (19) Abstract and Applied Analysis 3 Proposition 3. For any constant π ≥ 0 there exists a positive constant πΎ = πΎ(π ) such that, for any initial data π’1 (0), π’2 (0) with βπ’π (0)βπ»π,π ≤ π , π = 1, 2 one has σ΅©σ΅© σ΅© (πΎ2 /π2 )π‘ σ΅© σ΅©σ΅©π’1 (0) − π’2 (0)σ΅©σ΅©σ΅© , σ΅©σ΅©ππ,π (π‘)π’1 (0) − ππ,π (π‘)π’2 (0)σ΅©σ΅©σ΅© σ΅©π»π,π σ΅© σ΅© σ΅©π»π,π ≤ π (20) 3. A Priori Estimates In this section we provide useful a priori estimates of energy type. Equation (1) admits a Liapunov functional of the form: σ΅© σ΅©2 1 σ΅© σ΅©2 1 πΈ (π’, π’π‘ ) = πσ΅©σ΅©σ΅©π’π‘ σ΅©σ΅©σ΅© + π2 σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© + ∫ π (π’π₯ ) ππ₯, 2 2 πΌ that is not increasing along the solutions; in fact if we multiply (1) by π’π‘ and integrate over πΌ we obtain where ππ,π (π‘) is the solution semigroup of the problem (1). Proof. Let π’1 , π’2 , two solutions of (1) with initial data π’1 (0) and π’2 (0). Let, π€ = π’1 − π’2 then we write ππ€π‘π‘ + π€π‘ + π2 π€π₯π₯π₯π₯ = 1 π {πσΈ (π’1π₯ ) − πσΈ (π’2π₯ )} . 2 ππ₯ (21) π σ΅© σ΅©2 πΈ (π’, π’π‘ ) = −σ΅©σ΅©σ΅©π’π‘ σ΅©σ΅©σ΅© . ππ‘ π‘ π π σ΅©σ΅© σ΅©σ΅©2 π2 π σ΅©σ΅© σ΅©2 σ΅© σ΅©2 σ΅©π€ σ΅© + σ΅©π€ σ΅©σ΅© + σ΅©σ΅©π€ σ΅©σ΅© 2 ππ‘ σ΅© π‘ σ΅© 2 ππ‘ σ΅© π₯π₯ σ΅© σ΅© π‘ σ΅© (28) Then for any fixed initial data (π’0 , π’1 ) ∈ Hπ,π we have that the corresponding solution satisfies π σ΅© σ΅©2 π2 σ΅© σ΅©2 σ΅© σ΅©2 sup { σ΅©σ΅©σ΅©π’π‘ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© } = supσ΅©σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Hπ,π 2 2 π‘≥0 π‘≥0 = −2 ∫ π€π₯π₯ π€π‘ ππ₯ πΌ 2 + 6 ∫ {π’1π₯ π€π₯π₯ + π’2π₯x π€π₯ [π’1π₯ + π’2π₯ ]} π€π‘ ππ₯ (29) ≤ πΈ (π’0 , π’1 ) . πΌ σ΅© σ΅©2 σ΅© σ΅© σ΅©σ΅© σ΅© σ΅©σ΅© σ΅© ≤ 2 σ΅©σ΅©σ΅©π€π₯π₯ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π€π‘ σ΅©σ΅©σ΅© + 6σ΅©σ΅©σ΅©π’1π₯ σ΅©σ΅©σ΅©∞ σ΅©σ΅©σ΅©π€π₯π₯ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π€π‘ σ΅©σ΅©σ΅© Moreover if we integrate over R+ we get the integral control 1/2 πΌ ∞ πΌ σ΅© σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅© σ΅©2 ≤ 2πΎ σ΅©σ΅©σ΅©π€π₯π₯ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π€π‘ σ΅©σ΅©σ΅© ≤ πΎ2 σ΅©σ΅©σ΅©π€π₯π₯ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π€π‘ σ΅©σ΅©σ΅© , (31) By summing (27) and (31) we get where we have used the following inequality (see [13] and inequality (11)): (23) The constant πΎ can be explicitly computed using the estimates of the absorbing set presented in the next section. From the last inequality we get (24) and then using Gronwall’s lemma we get σ΅©σ΅© σ΅© (πΎ2 /π2 )π‘ σ΅© σ΅©σ΅©(π€(0), π€π‘ (0))σ΅©σ΅©σ΅© . σ΅©σ΅©(π€, π€π‘ )σ΅©σ΅©σ΅©Hπ,π ≤ π σ΅© σ΅©Hπ,π π½ π σ΅© σ΅©2 βπ’β2 + π½π2 σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© 2 ππ‘ σ΅© σ΅©4 σ΅© σ΅©2 + 2π½σ΅©σ΅©σ΅©π’π₯ σ΅©σ΅©σ΅©4 = 2π½σ΅©σ΅©σ΅©π’π₯ σ΅©σ΅©σ΅© . π½π ∫ π’π‘π‘ π’ππ₯ + π 2 σ΅©σ΅© σ΅©σ΅© σ΅© } σ΅©π€ σ΅©σ΅© σ΅©σ΅©π€ σ΅©σ΅© π2 σ΅© π₯π₯ σ΅© σ΅© π‘ σ΅© π σ΅©σ΅© 2πΎ2 σ΅© σ΅©2 σ΅©2 σ΅©σ΅©(π€, π€π‘ )σ΅©σ΅©σ΅©Hπ,π ≤ 2 σ΅©σ΅©σ΅©(π€, π€π‘ )σ΅©σ΅©σ΅©Hπ,π , ππ‘ π (30) We consider an important estimate that will be useful later. Let π½ > 0 be a parameter to be determined later; if we multiply (1) by π½π’ we obtain σ΅© σ΅©2 σ΅© σ΅©2 1/2 σ΅© σ΅©σ΅© σ΅© ×(σ΅©σ΅©σ΅©π’1π₯π₯ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π’2π₯π₯ σ΅©σ΅©σ΅© ) } σ΅©σ΅©σ΅©π€π₯π₯ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π€π‘ σ΅©σ΅©σ΅© σ΅©σ΅© σ΅©σ΅©2 σ΅©2 2 σ΅© σ΅©2 2σ΅© σ΅©σ΅©π’π₯ σ΅©σ΅©∞ ≤ πΆ σ΅©σ΅©σ΅©π’π₯ σ΅©σ΅©σ΅©π»1 (πΌ) ≤ 2πΆ σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© . σ΅© σ΅©2 ∫ σ΅©σ΅©σ΅©π’π‘ σ΅©σ΅©σ΅© ππ‘ ≤ πΈ (π’0 , π’1 ) . 0 (22) σ΅© σ΅©2 σ΅© σ΅© ≤ 2 {1 + 18πΆ2 σ΅©σ΅©σ΅©π’1π₯π₯ σ΅©σ΅©σ΅© + 12 σ΅©σ΅©σ΅©π’2π₯π₯ σ΅©σ΅©σ΅© ≤ 2 {1 + (36πΆ2 + 24√2) (27) Moreover, integrating the previous inequality on (0, π‘) with respect to time we get σ΅© σ΅©2 ∫ σ΅©σ΅©σ΅©π’π‘ σ΅©σ΅©σ΅© ππ + πΈ (π’, π’π‘ ) = πΈ (π’0 , π’1 ) . 0 We multiply the above equation by π€π‘ in πΏ2 (πΌ); then σ΅© σ΅© 2 2 + 12 σ΅©σ΅©σ΅©π’2π₯π₯ σ΅©σ΅©σ΅© (∫ π€π₯2 π€π‘2 [π’1π₯ + π’2π₯ ] ππ₯) (26) (25) π π σ΅© σ΅©2 πΈ + π½π β¨π’π‘ , π’β© + (1 − π½π) σ΅©σ΅©σ΅©π’π‘ σ΅©σ΅©σ΅© ππ‘ ππ‘ π½ π σ΅© σ΅©2 σ΅© σ΅©4 βπ’β2 + π½π2 σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© + 2π½σ΅©σ΅©σ΅©π’π₯ σ΅©σ΅©σ΅©4 2 ππ‘ σ΅© σ΅©2 = 2π½σ΅©σ΅©σ΅©π’π₯ σ΅©σ΅©σ΅© . + (32) Using the expression of the energy we can rewrite the previous in the following way: π π σ΅© σ΅©2 πΈ + π½π β¨π’π‘ , π’β© + (1 − 3π½π) σ΅©σ΅©σ΅©π’π‘ σ΅©σ΅©σ΅© ππ‘ ππ‘ σ΅©σ΅© σ΅©σ΅©4 σ΅©π’π₯ σ΅© |πΌ| + π½β¨π’π‘ , π’β© + 2π½ (πΈ + σ΅© σ΅©4 − ) = 0. 2 2 (33) We will estimate some of the terms of the previous inequality in the following lemma. 4 Abstract and Applied Analysis Lemma 4. Fix π ≤ 1 then for all π½ ∈ (0, 1/3) one has 4. Absorbing Sets σ΅© σ΅©2 σ΅© σ΅©4 2π½2 πβ¨π’π‘ , π’β© ≤ (1 − 3π½π) σ΅©σ΅©σ΅©π’π‘ σ΅©σ΅©σ΅© + π½σ΅©σ΅©σ΅©π’π₯ σ΅©σ΅©σ΅©4 + π½ |πΌ| + π½β¨uπ‘ , π’β©. (34) Proof. By Holder’s inequality we get σ΅© σ΅©2 σ΅© σ΅©2 σ΅© σ΅©4 2σ΅©σ΅©σ΅©π’π₯ σ΅©σ΅©σ΅© ≤ 2σ΅©σ΅©σ΅©π’π₯ σ΅©σ΅©σ΅©4 |πΌ|1/2 ≤ σ΅©σ΅©σ΅©π’π₯ σ΅©σ΅©σ΅©4 + |πΌ| , (35) then we use PoincareΜ’s inequality and the fact that |πΌ| ≤ 1 to conclude that σ΅©σ΅© σ΅©σ΅©4 σ΅©π’π₯ σ΅© |πΌ| (36) βπ’β2 ≤ σ΅© σ΅©4 + . 2 2 Therefore, for a positive constant π1 to be determined later, we have that π σ΅© σ΅©2 1 σ΅© σ΅© β¨π’π‘ , π’β© ≤ σ΅©σ΅©σ΅©π’π‘ σ΅©σ΅©σ΅© βπ’β ≤ 1 σ΅©σ΅©σ΅©π’π‘ σ΅©σ΅©σ΅© + βπ’β2 2 2π1 σ΅© σ΅©4 π1 σ΅©σ΅© σ΅©σ΅©2 1 σ΅©σ΅©σ΅©π’π₯ σ΅©σ΅©σ΅©4 |πΌ| ≤ σ΅©σ΅©π’π‘ σ΅©σ΅© + ( + ). 2 2π1 2 2 (37) 1 σ΅© σ΅©2 σ΅© σ΅©2 1 σ΅© σ΅©2 |πΌ| ∫ π (π’π₯ ) ππ₯ ≤ σ΅©σ΅©σ΅©π’π₯ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π₯ σ΅©σ΅©σ΅©πΏ∞ (πΌ) − σ΅©σ΅©σ΅©π’π₯ σ΅©σ΅©σ΅© + 2 πΌ 2 2 ≤ 1 σ΅©σ΅© σ΅©σ΅©2 σ΅©σ΅© σ΅©σ΅©2 |πΌ| σ΅©π’ σ΅© σ΅©π’ σ΅© + 2 σ΅© π₯ σ΅© σ΅© π₯π₯ σ΅© 2 (44) 1 σ΅©σ΅© σ΅©σ΅©4 |πΌ| σ΅©π’ σ΅© + . 2 σ΅© π₯π₯ σ΅© 2 We will use this inequality several times in what follows. ≤ 4.1. Absorbing Set in Hπ,π Proposition 5. For all π 0 > 1 the set Consequently, by choosing the parameters π1 = 1/12 and π½ < 1/3 we have that (2π½2 π − π½) β¨π’π‘ , π’β© σ΅¨ σ΅¨ π σ΅© σ΅©2 ≤ σ΅¨σ΅¨σ΅¨σ΅¨2π½2 π − π½σ΅¨σ΅¨σ΅¨σ΅¨ 1 σ΅©σ΅©σ΅©π’π‘ σ΅©σ΅©σ΅© 2 σ΅¨σ΅¨ 2 σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨2π½2 π − π½σ΅¨σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨ σ΅©σ΅© σ΅©σ΅©4 σ΅¨σ΅¨σ΅¨2π½ π − π½σ΅¨σ΅¨σ΅¨ +σ΅¨ |πΌ| σ΅©σ΅©π’π₯ σ΅©σ΅©4 + 4π1 4π1 In this section we will show the existence of several absorbing sets for the solution semigroup of (1) in the space Hπ,π . Also, assuming further regularity of the initial data, the existence of a more regular absorbing sets is also shown. First, notice the following estimate of the nonlinear term of the energy functional defined in (26). Using (23) we get π΅Φ := {(π’, π’π‘ ) ∈ Hπ,π : Φ (π’, π’π‘ ) ≤ π 0 } (45) is bounded, absorbing, and positively invariant for the semigroup ππ,π (π‘) in Hπ,π . Proof. The set is positively invariant, in fact if (π’0 , π’1 ) ∈ Hπ,π with Φ(π’0 , π’1 ) ≤ π 0 we have, from (42), that (38) σ΅© σ΅©4 σ΅© σ΅©2 ≤ (1 − 3π½π) σ΅©σ΅©σ΅©π’π‘ σ΅©σ΅©σ΅© + π½σ΅©σ΅©σ΅©π’π₯ σ΅©σ΅©σ΅©4 + π½ |πΌ| . Φ (π’, π’π‘ ) ≤ [π 0 − 1] π−2π½π‘ + 1 ≤ π 0 , ∀π‘ > 0. (46) The boundness of π΅Φ follows directly from (42). Now suppose that (π’0 , π’1 ) are such that Φ(π’0 , π’1 ) ≤ π , with π > π 0 then again from (42) we get Φ (π’, π’π‘ ) ≤ π 0 , ∀π‘ ≥ πΦ , (47) where Now, using the previous lemma and (33) we conclude that π π πΈ + π½π β¨π’π‘ , π’β© + 2π½2 π β¨π’π‘ , π’β© + 2π½πΈ ≤ 2π½ |πΌ| . ππ‘ ππ‘ (39) πΦ := 1 π −1 log { }. 2π½ π 0 − 1 (48) Then π΅Φ is absorbing for the semigroup ππ,π (π‘). Proposition 6. For all π 1 > 1 one has that the set If we set Φ (π’, π’π‘ ) = Φ (π‘) = πΈ (π’, π’π‘ ) + π½πβ¨π’π‘ , π’β©, (40) Μ (π’, π’π‘ ) ≤ π 1 } π΅πΈΜ := {(π’, π’π‘ ) ∈ Hπ,π : πΈ (49) is absorbing for the semigroup ππ,π (π‘). then from (39) we get π Φ (π‘) + 2π½Φ (π‘) ≤ 2π½. ππ‘ (41) Now, integrating we have that −2π½π‘ Φ (π‘) ≤ [Φ (π’0 , π’1 ) − 1] π + 1. (42) The previous inequality will be used in the next section to show the existence of absorbing sets for the problem (1). We remark that the following inequality holds for the Μ and Φ: functions πΈ, πΈ, Μ (π‘) + 1 ∫ π (π’π₯ ) ππ₯ ≥ πΈ Μ ≥ 0. Φ (π‘) = πΈ 2 πΌ (43) Proof. We have from (42) and (43) that Μ (π’, π’π‘ ) ≤ [Φ (π’0 , π’1 ) − 1] π−2π½π‘ + 1. πΈ (50) Μ 0 , π’1 ) ≤ π , with π > π 1 . Then, Now, take (π’0 , π’1 ) such that πΈ(π’ by (42) and (44) we get that Μ (π‘) ≤ Φ (π‘) ≤ πΈ Μ (π‘) + 1 σ΅©σ΅©σ΅©π’π₯ σ΅©σ΅©σ΅©4 + |πΌ| πΈ 2 σ΅© σ΅©4 2 Μ (π‘) + 1 σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅©4 + |πΌ| ≤πΈ 2σ΅© σ΅© 2 Μ 2 + |πΌ| . Μ (π‘) + 4 πΈ(π‘) ≤πΈ 2 π 2 (51) Abstract and Applied Analysis 5 Proposition 8. There exists π 3 > 0 such that the closed ball Therefore, Μ (π’, π’π‘ ) ≤ {πΈ Μ 0 , π’1 )2 + |πΌ| − 1} π−2π½π‘ + 1 Μ (π’0 , π’1 ) + 4 πΈ(π’ πΈ π2 2 4 |πΌ| ≤ {π + 2 π 2 + − 1} π−2π½π‘ + 1. π 2 (52) Consequently, we have that Μ (π’, π’π‘ ) ≤ π 1 , πΈ ∀π‘ ≥ ππΈΜ , (53) ππΈΜ = 2 π + (4/π ) π + (|πΌ| /2) − 1 1 }. log { 2π½ π 1 − 1 (54) Now using the equivalence of the norm on Hπ,π and (52) we get σ΅©σ΅© σ΅©2 σ΅© σ΅©2 σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Hπ,π ≤ 3 {3σ΅©σ΅©σ΅©(π’0 , π’1 )σ΅©σ΅©σ΅©Hπ,π + Proof. If we multiply (1) by π’π₯π₯π‘ + π’π₯π₯ and integrate over πΌ we obtain 1 σ΅©σ΅© σ΅©σ΅©2 σ΅©π’ σ΅© + πβ¨π’π‘π₯ , π’π₯ β©} 2 σ΅© π₯σ΅© σ΅© σ΅©2 σ΅© σ΅©2 + (1 − π) σ΅©σ΅©σ΅©π’π‘π₯ σ΅©σ΅©σ΅© + π2 σ΅©σ΅©σ΅©π’π₯π₯π₯ σ΅©σ΅©σ΅© 1 1 = − β¨[πσΈ (π’π₯ )]π₯ , π’π₯π₯π‘ β© − β¨[πσΈ (π’π₯ )]π₯ , π’π₯π₯ β© . 2 2 (63) Let us denote the differential term of the previous inequality as 36 σ΅©σ΅© |πΌ| σ΅©4 + − 1} π−2π½π‘ + 3. σ΅©(π’ , π’ )σ΅©σ΅© π2 σ΅© 0 1 σ΅©Hπ,π 2 (55) Ψ0 (π‘) = Let π 2 > √3. If (π’0 , π’1 ) ∈ Hπ,π is such that σ΅© σ΅©σ΅© σ΅©σ΅©(π’0 , π’1 )σ΅©σ΅©σ΅©Hπ,π ≤ π , then we have that σ΅©σ΅© σ΅© σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Hπ,π ≤ π 2 , ∀π‘ ≥ ππ», (57) 3 {3π 2 + (36/π2 ) π 4 + |πΌ| /2 − 1} 1 ππ» = . log 2π½ π 22 − 3 (58) π σ΅©σ΅© σ΅©σ΅©2 π2 σ΅©σ΅© σ΅©2 σ΅©π’ σ΅© + σ΅©σ΅©π’π₯π₯π₯ σ΅©σ΅©σ΅© + π β¨π’π‘π₯ , π’π₯ β© 2 σ΅© π‘π₯ σ΅© 2 1 σ΅© σ΅©2 + σ΅©σ΅©σ΅©π’π₯ σ΅©σ΅©σ΅© . 2 (56) where 1 1σ΅© σ΅© σ΅© σ΅©2 − β¨[πσΈ (π’π₯ )]π₯ , π’π₯π₯ β© ≤ σ΅©σ΅©σ΅©σ΅©πσΈ σΈ (π’π₯ )σ΅©σ΅©σ΅©σ΅©∞ σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© 2 2 σ΅© σ΅©2 σ΅© σ΅©2 ≤ 2 (3σ΅©σ΅©σ΅©π’π₯ σ΅©σ΅©σ΅©∞ + 1) σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© σ΅© σ΅©2 σ΅© σ΅©2 ≤ 2 (3σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© + 1) σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© Proposition 7. For all π 2 > √3 one has that the ball ≤ σ΅© σ΅© π΅Hπ,π := {(π’, π’π‘ ) ∈ Hπ,π : σ΅©σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Hπ,π ≤ π 2 } (59) is an absorbing set for the semigroup ππ,π (π‘) in Hπ,π . 2 4 2 6π 2 π ( + 1) := πΆ1 . π2 2 π2 Ψ1 = 1 2 ππ₯; ∫ πσΈ σΈ (π’π₯ ) π’π₯π₯ 4 πΌ (66) then we can rewrite the the second term of r.h.s of (63) (60) equipped with the norm π σ΅© σ΅©2 π 2 σ΅© σ΅©σ΅© σ΅©2 σ΅©2 σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Uπ,π = σ΅©σ΅©σ΅©π’π‘π₯ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π’π₯π₯π₯ σ΅©σ΅©σ΅© . 2 2 (65) Let us define 4.2. Absorbing Set in Uπ,π . Now suppose that the initial data has some additional regularity. Then we can prove the existence of more regular absorbing sets. Let us define the following space: Uπ,π = {π»3 (πΌ) ∩ π»2 (πΌ) ∩ π»01 (πΌ)} × π»1 (πΌ) , (64) We will estimate the right hand side of (63). The first term can be estimated as follows: Thus, we have proved the following. Then we have the following. (62) is a bounded absorbing set for ππ,π (π‘) in Uπ,π . π π σ΅©σ΅© σ΅©σ΅©2 π2 σ΅©σ΅© σ΅©2 { σ΅©π’ σ΅© + σ΅©σ΅©π’π₯π₯π₯ σ΅©σ΅©σ΅© + ππ‘ 2 σ΅© π‘π₯ σ΅© 2 where 4 σ΅© σ΅© B3 = {(π’, π’π‘ ) ∈ Uπ,π : σ΅©σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Uπ,π ≤ π 3 } , (61) 1 π 1 2 − β¨[πσΈ (π’π₯ )]π₯ , π’π₯π₯π‘ β© = − Ψ1 + ∫ πσΈ σΈ σΈ (π’π₯ ) π’π₯π‘ π’π₯π₯ , 2 ππ‘ 4 πΌ (67) and estimate 6 Abstract and Applied Analysis 1 2 ∫ πσΈ σΈ σΈ (π’π₯ ) π’π₯π‘ π’π₯π₯ 4 πΌ Then from (72) we get 1 π Μ Ψ + Ψ ≤ πΆ + π2 := πΆ, ππ‘ 2 σ΅© σ΅© ≤ 6σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅©∞ ∫ π’π₯ π’π₯π‘ π’π₯π₯ ππ₯ πΌ σ΅© σ΅© σ΅© σ΅©σ΅© σ΅© ≤ 6σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅©∞ σ΅©σ΅©σ΅©π’π₯π‘ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π₯ π’π₯π₯ σ΅©σ΅©σ΅© and by Gronwall’s lemma we obtain σ΅© σ΅© σ΅© σ΅© σ΅© σ΅©2 ≤ 6σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅©∞ σ΅©σ΅©σ΅©π’π₯π‘ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© σ΅©1/2 σ΅© σ΅© σ΅© σ΅©5/2 σ΅© ≤ 6σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π₯π₯π₯ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π₯π‘ σ΅©σ΅©σ΅© 5/2 √2 ≤ 6( π 2 ) π (73) Μ π−(1/2)π‘ + 2πΆ, Μ Ψ (π‘) ≤ (Ψ (0) − 2πΆ) (68) σ΅©σ΅© σ΅©1/2 σ΅© σ΅© σ΅©σ΅©π’π₯π₯π₯ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π₯π‘ σ΅©σ΅©σ΅© σ΅©σ΅© σ΅©2 σ΅© σ΅©2 Μ π−(1/2)π‘ σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Uπ,π ≤ 2 (σ΅©σ΅©σ΅©(π’0 , π’1 )σ΅©σ΅©σ΅©Uπ,π + π2 − 2πΆ) (74) Μ + 2π1 . + 4πΆ 5 ≤ π σ΅©σ΅© σ΅©σ΅©2 18 √2 σ΅© σ΅© σ΅©π’ σ΅© + ( π 2 ) σ΅©σ΅©σ΅©π’π₯π₯π₯ σ΅©σ΅©σ΅© 2 σ΅© π₯π‘ σ΅© π π Then if (π’0 , π’1 ) ∈ Uπ,π such that 5 2 σ΅©σ΅© σ΅© σ΅©σ΅©(π’0 , π’1 )σ΅©σ΅©σ΅©Uπ,π ≤ π , π σ΅© σ΅©2 1 18 √2 π2 σ΅© σ΅©2 ≤ σ΅©σ΅©σ΅©π’π₯π‘ σ΅©σ΅©σ΅© + [ ( π 2 ) ] + σ΅©σ΅©σ΅©π’π₯π₯π₯ σ΅©σ΅©σ΅© 2 2 ππ π 2 := we have that π σ΅©σ΅© σ΅©σ΅©2 π2 σ΅©σ΅© σ΅©2 σ΅©σ΅©π’π₯π‘ σ΅©σ΅© + σ΅©σ΅©π’π₯π₯π₯ σ΅©σ΅©σ΅© + πΆ2 . 2 2 Then if π ∈ (0, 1/2) and setting Ψ := Ψ0 + Ψ1 , πΆ := πΆ1 + πΆ2 we have π Ψ+ ππ‘ π σ΅©σ΅© σ΅©σ΅©2 π2 σ΅©σ΅© σ΅©2 σ΅©π’ σ΅© + σ΅©σ΅©π’π₯π₯π₯ σ΅©σ΅©σ΅© ≤ πΆ. 2 σ΅© π₯π‘ σ΅© 2 σ΅©σ΅© σ΅© σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Uπ,π ≤ π 3 , ππ := 2 log ( 1 σ΅© σ΅©2 σ΅© σ΅©2 Ψ ≥ σ΅©σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Uπ,π + σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© 2 σ΅© σ΅© σ΅© σ΅© σ΅© σ΅©2 − π σ΅©σ΅©σ΅©π’π‘π₯ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π₯ σ΅©σ΅©σ΅© − σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© ∀π‘ ≥ ππ, (76) where (69) To conclude the proof we note that π σ΅© σ΅©2 σ΅© σ΅©2 ≥ σ΅©σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Uπ,π − σ΅©σ΅©σ΅©π’π‘π₯ σ΅©σ΅©σ΅© − 4 (75) Μ 2π 2 + 2π2 − 4πΆ ), 2 Μ π 3 − 4πΆ − 2π1 (77) and this concludes the proof. 3 σ΅©σ΅© σ΅©σ΅©2 σ΅©π’ σ΅© 2 σ΅© π₯π₯ σ΅© (70) 1σ΅© 3 σ΅©2 ≥ σ΅©σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Uπ,π − 2 π 22 2 π Proposition 9. There exists π 4 > 0 such that the closed ball 1σ΅© σ΅©2 := σ΅©σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Uπ,π − π1 , 2 σ΅© σ΅© B4 = {(π’, π’π‘ ) ∈ Vπ,π : σ΅©σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Vπ,π ≤ π 4 } and that π σ΅© σ΅©2 σ΅© σ΅©2 σ΅© σ΅©2 Ψ ≤ σ΅©σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Uπ,π + σ΅©σ΅©σ΅©π’π‘π₯ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© 2 σ΅©σ΅© σ΅©σ΅©2 σ΅©σ΅© σ΅©σ΅©2 + σ΅©σ΅©π’π₯π₯ σ΅©σ΅© (3σ΅©σ΅©π’π₯ σ΅©σ΅©∞ + 1) 2 6 σ΅© σ΅©2 ≤ 2σ΅©σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Uπ,π + 2 π 22 ( 2 π 22 + 2) π π σ΅©2 σ΅© := 2σ΅©σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Uπ,π + π2 . 4.3. Absorbing Set in Vπ,π . We consider the space Vπ,π as defined in (7), equipped with the norm defined in (10). By considering more regular initial conditions we can prove the following result. is a bounded absorbing set for the semigroup ππ,π (π‘) in Vπ,π . Proof. Multiply (1) by π’π₯π₯π₯π₯π‘ + π’π₯π₯π₯π₯ ; then we obtain (71) 2 π π σ΅©σ΅© σ΅©2 π σ΅© σ΅©2 { σ΅©σ΅©π’π‘π₯π₯ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π’π₯π₯π₯π₯ σ΅©σ΅©σ΅© ππ‘ 2 2 1σ΅© σ΅©2 +π β¨π’π‘π₯π₯ , π’π₯π₯ β© + σ΅©σ΅©σ΅©π’π₯π₯π₯ σ΅©σ΅©σ΅© } 2 σ΅© σ΅© σ΅© σ΅©2 + (1 − π) σ΅©σ΅©σ΅©π’π₯π₯π‘ σ΅©σ΅©σ΅© + π2 σ΅©σ΅©σ΅©π’π₯π₯π₯π₯ σ΅©σ΅©σ΅© From the previous inequalities we get 1 σ΅©σ΅© σ΅©2 σ΅© σ΅©2 σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Uπ,π − π1 ≤ Ψ ≤ 2σ΅©σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Uπ,π + π2 . 2 (78) (72) = 1 β¨[πσΈ (π’π₯ )]π₯ , π’π₯π₯π₯π₯π‘ + π’π₯π₯π₯π₯ β© . 2 (79) Abstract and Applied Analysis 7 We call π0 (π‘) the differential term of the previous inequality and we estimate the r.h.s.: 1 β¨[πσΈ (π’π₯ )]π₯ , π’π₯π₯π₯π₯ β© 2 4 1 1 2 2 = − β¨πσΈ σΈ σΈ (π’π₯ ) π’π₯π₯ , π’π₯π₯π₯ β© − β¨[πσΈ σΈ (π’π₯ )]π₯π₯ , π’π₯π₯π₯ β© 2 2 2 2 = −12 ∫ π’π₯ π’π₯π₯ π’π₯π₯π₯ ππ₯ − 2 ∫ (3π’π₯2 − 1) π’π₯π₯π₯ ππ₯ πΌ 12 2 π + 1) := π΄ 1 . π2 2 := π2 σ΅©σ΅© σ΅©2 σ΅©σ΅©π’π₯π₯π₯π₯ σ΅©σ΅©σ΅© + π΄ 2 , 2 1 2 ]π₯ , π’π₯π₯π‘ β© β¨[πσΈ σΈ σΈ (π’π₯ ) π’π₯π₯ 2 = σ΅© σ΅©2 σ΅© σ΅© σ΅©2 σ΅© σ΅© σ΅© ≤ 2σ΅©σ΅©σ΅©π’π₯π₯π₯ σ΅©σ΅©σ΅© + 12σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅©∞ σ΅©σ΅©σ΅©π’π₯ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π₯π₯π₯ σ΅©σ΅©σ΅© σ΅© σ΅©2 σ΅© σ΅©2 ≤ 2σ΅©σ΅©σ΅©π’π₯π₯π₯ σ΅©σ΅©σ΅© (6σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© + 1) ≤ π2 π 32 ( σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅©σ΅©π’π₯π₯π₯π₯ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π’π₯π‘ σ΅©σ΅©σ΅© π 2 π 3/2 1 2 π2 σ΅© σ΅©2 ≤ 2 (12 ⋅ 23/4 2 7/23 ) + π 32 + σ΅©σ΅©σ΅©π’π₯π₯π₯π₯ σ΅©σ΅©σ΅© 2π π 2 π 1 = − β¨[πσΈ (π’π₯ )]π₯π₯ , π’π₯π₯π₯ β© 2 πΌ ≤ (12 ⋅ 2 3/2 2 3/4 π 2 π 3 ) 2 π7/2 1 3 + πσΈ σΈ σΈ (π’π₯ ) 2π’π₯π₯ π’π₯π₯π₯ , π’π₯π₯π‘ β© β¨ππV (π’π₯ ) π’π₯π₯ 2 3 = 12 ∫ π’π₯π₯ π’π₯π₯π‘ ππ₯ + 24 ∫ π’π₯ π’π₯π₯ π’π₯π₯π₯ π’π₯π₯π‘ ππ₯ πΌ πΌ σ΅© σ΅©2 σ΅© σ΅© ≤ 12 (σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅©∞ σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© +2σ΅©σ΅©σ΅©π’π₯ σ΅©σ΅©σ΅©∞ σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅©∞ σ΅©σ΅©σ΅©π’π₯π₯π₯ σ΅©σ΅©σ΅©) σ΅©σ΅©σ΅©π’π₯π₯π‘ σ΅©σ΅©σ΅© (80) σ΅© σ΅©2 σ΅© σ΅© ≤ 12 (σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π₯π₯π₯ σ΅©σ΅©σ΅© Let us define 1 2 π1 (π‘) := ∫ πσΈ σΈ (π’π₯ ) π’π₯π₯π₯ ππ₯; 4 πΌ σ΅© σ΅©3/2 σ΅© σ΅©3/2 σ΅© σ΅© +2σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π₯π₯π₯ σ΅©σ΅©σ΅© ) σ΅©σ΅©σ΅©π’π₯π₯π‘ σ΅©σ΅©σ΅© (81) ≤ then σ΅© σ΅©3/2 σ΅© σ΅©3/2 2 +2σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π₯π₯π₯ σ΅©σ΅©σ΅© ) 1 β¨[πσΈ (π’π₯ )]π₯ , π’π₯π₯π₯π₯π‘ β© 2 2√2π 22 π 3 πσ΅© 2π 2 π 3 3/2 σ΅©2 ≤ σ΅©σ΅©σ΅©π’π₯π₯π‘ σ΅©σ΅©σ΅© + 72[ + 2( ) ] 2 π3 π2 1 = − β¨[πσΈ (π’π₯ )]π₯π₯ , π’π₯π₯π₯π‘ β© 2 1 2 = − β¨πσΈ σΈ σΈ (π’π₯ ) π’π₯π₯ , π’π₯π₯π₯π‘ β© 2 1 − β¨πσΈ σΈ (π’π₯ ) π’π₯π₯π₯ , π’π₯π₯π₯π‘ β© 2 =− + 1 π 2 ]π₯ , π’π₯π₯π‘ β© π1 + β¨[πσΈ σΈ σΈ (π’π₯ ) π’π₯π₯ ππ‘ 2 1 2 ππ₯. ∫ πσΈ σΈ σΈ (π’π₯ ) π’π₯π‘ π’π₯π₯π₯ 4 πΌ We estimate the last two terms of the previous equality: 1 2 ππ₯ ∫ πσΈ σΈ σΈ (π’π₯ ) π’π₯π‘ π’π₯π₯π₯ 4 πΌ π σ΅©σ΅© σ΅©2 σ΅© σ΅© σ΅©2 σ΅© σ΅©π’ σ΅©σ΅© + 72 (σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π₯π₯π₯ σ΅©σ΅©σ΅© 2 σ΅© π₯π₯π‘ σ΅© := π σ΅©σ΅© σ΅©2 σ΅©π’ σ΅©σ΅© + π΄ 3 . 2 σ΅© π₯π₯π‘ σ΅© (83) (82) Then if we set π(π‘) = π0 (π‘) + π1 (π‘) we get π σ΅©2 σ΅© π (π‘) + σ΅©σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Vπ,π ≤ π΄ 1 + π΄ 2 + π΄ 3 := π΄. ππ‘ Moreover we have 2 = 6 ∫ π’π₯ π’π₯π‘ π’π₯π₯π₯ ππ₯ πΌ σ΅© σ΅© σ΅© σ΅© σ΅© σ΅©σ΅© σ΅© ≤ 6σ΅©σ΅©σ΅©π’π₯π₯π₯ σ΅©σ΅©σ΅©∞ σ΅©σ΅©σ΅©π’π₯ σ΅©σ΅©σ΅©∞ σ΅©σ΅©σ΅©π’π₯π‘ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π₯π₯π₯ σ΅©σ΅©σ΅© σ΅© σ΅©3/2 σ΅© σ΅©1/2 σ΅© σ΅© σ΅© σ΅©2 ≤ 6σ΅©σ΅©σ΅©π’π₯π₯π₯ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π₯π₯π₯π₯ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π₯π‘ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© ≤ (12 ⋅ 23/4 π 22 π 33/2 σ΅©σ΅© σ΅©1/2 σ΅© σ΅© ) σ΅©σ΅©π’π₯π₯π₯π₯ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π₯π‘ σ΅©σ΅©σ΅© 7/2 π 2 πσ΅© σ΅© σ΅©2 σ΅©2 π (π‘) ≥ σ΅©σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Vπ,π − σ΅©σ΅©σ΅©π’π₯π₯π‘ σ΅©σ΅©σ΅© − 4 1σ΅© 1 σ΅©2 ≥ σ΅©σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Vπ,π − π 22 2 2 1σ΅© σ΅©2 := σ΅©σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Vπ,π − π1 , 2 1 σ΅©σ΅© σ΅©σ΅©2 σ΅©π’ σ΅© 2 σ΅© π₯π₯ σ΅© (84) 8 Abstract and Applied Analysis σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅© π (π‘) ≤ σ΅©σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Vπ,π + π σ΅©σ΅©σ΅©π’π₯π₯π‘ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© such that (πΆ1) 1 σ΅© σ΅©2 1 σ΅© σ΅© σ΅© σ΅©3 + σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©σ΅©πσΈ σΈ (π’π₯ )σ΅©σ΅©σ΅©σ΅©∞ σ΅©σ΅©σ΅©π’π₯π₯π₯ σ΅©σ΅©σ΅© 2 4 σ΅©σ΅© σ΅©σ΅©2 σ΅© σ΅©2 ≤ 2σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©Vπ,π + σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅© (94) σ΅© σ΅© lim { sup σ΅©σ΅©σ΅©σ΅©ππ,π (π‘) π§σ΅©σ΅©σ΅©σ΅©H } = 0. π‘ → ∞ π§∈H π,π (95) π,π (πΆ2) π,π σ΅©2 σ΅© σ΅©2 σ΅© + (3σ΅©σ΅©σ΅©π’π₯ σ΅©σ΅©σ΅©∞ + 1) σ΅©σ΅©σ΅©π’π₯π₯π₯ σ΅©σ΅©σ΅© σ΅© σ΅©2 ≤ 2σ΅©σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Vπ,π σ΅© σ΅© sup sup σ΅©σ΅©σ΅©σ΅©πΏ π,π (π‘)π§σ΅©σ΅©σ΅©σ΅©V < ∞, π,π π‘≥0 π§∈H Let B ⊂ Hπ,π be a fixed bounded set and let (π’0 , π’1 ) ∈ B ⊂ Hπ,π . We will define the decomposition of ππ,π as follows: 2 6 2 + 2 π 22 + ( 2 π 22 + 1) 2 π 32 π π π ππ,π (π‘) (π’0 , π’1 ) = (π’, π’π‘ ) , σ΅© σ΅©2 := 2σ΅©σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Vπ,π + π2 . πΏ π,π (π‘) (π’0 , π’1 ) = (β, βπ‘ ) , (85) ππ,π (π‘) (π’0 , π’1 ) = (V, Vπ‘ ) , Then putting all together we get 1 σ΅©σ΅© σ΅©2 σ΅© σ΅©2 σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Vπ,π − π1 ≤ π (π‘) ≤ 2σ΅©σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Vπ,π + π2 . 2σ΅© where β and V are solutions of the following problems: (86) πβπ‘π‘ + βπ‘ + π2 βπ₯π₯π₯π₯ − πΌβπ₯π₯ From the previous inequality we get π π 1 Μ π (π‘) + π (π‘) ≤ π΄ + 2 := π΄ ππ‘ 2 2 1 = πσΈ σΈ (βπ₯ ) βπ₯π₯ − πΌπ’π₯π₯ , 2 (87) Μ Μ π−(1/2)π‘ + 2π΄. π (π‘) ≤ [π (0) − 2π΄] βπ‘ (π₯, 0) = 0, (88) β = βπ₯π₯ = 0, Μ + 2π1 . + 4π΄ in ππΌ, πVπ‘π‘ + Vπ‘ + π2 Vπ₯π₯π₯π₯ − πΌVπ₯π₯ Using again (86) we get σ΅©2 σ΅© σ΅©2 σ΅©σ΅© Μ π−(1/2)π‘ σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Vπ,π ≤ 2 [2σ΅©σ΅©σ΅©(π’0 , π’1 )σ΅©σ΅©σ΅©Vπ,π + π2 − 2π΄] (97) β (π₯, 0) = 0, and by Gronwall’s lemma 1 1 = πσΈ σΈ (π’π₯ ) π’π₯π₯ − πσΈ σΈ (βπ₯ ) βπ₯π₯ , 2 2 (89) (98) V (0) = π’0 , Then if (π’0 , π’1 ) ∈ Vπ,π such that σ΅© σ΅©σ΅© σ΅©σ΅©(π’0 , π’1 )σ΅©σ΅©σ΅©Vπ,π ≤ π , then there exists π 4 > 0: σ΅©σ΅© σ΅© σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Vπ,π ≤ π 4 , (96) Vπ‘ (0) = π’1 , V = Vπ₯π₯ = 0, (90) in ππΌ, where ∀π‘ ≥ ππ, (91) where Μ 2 (2π 2 + π2 − 2π΄) ], ππ := 2 log [ Μ − 2π1 π 2 − 4π΄ (92) 4 6√2 π > 6√6. π 2 (99) Before showing that the semigroups πΏ π,π (π‘) and ππ,π (π‘) satisfy the conditions (94) and (95), respectively, we consider the following lemma (see [9]) that will be useful for the sequel. Lemma 10. Let π : R+ → R be an absolutely continuous function which fulfills, for some ] > 0 and almost every π‘ ≥ 0, the differential inequality and this concludes the proof. 5. Global Attractor In this section we will show the existence of a global attractor for the semigroup ππ,π (π‘) in Hπ,π . Since we have already proved the existence of the absorbing set in Hπ,π , then it is sufficient (see, e.g., [4] or [14] for general results or [15] for a recent application on a weakly damped wave equation) to prove that, for any fixed bounded set B ⊂ Hπ,π , the solution semigroup ππ,π (π‘) admits the decomposition: ππ,π (π‘) = πΏ π,π (π‘) + ππ,π (π‘) , πΌ≥ (93) π π (π‘) + 2]π (π‘) ≤ π (π‘) π (π‘) , ππ‘ (100) where π is a positive function satisfying π‘ ∫ π (π¦) ππ¦ ≤ ]π‘ + π, 0 ∀π‘ ≥ 0. (101) π‘ ≥ 0. (102) Then π (π‘) ≤ ππ (0) π−]π‘ , Abstract and Applied Analysis 9 5.1. The Semigroup πΏ π,π (π‘). Due to the similarity of problem (97) to (1), Proposition 9 still holds in this setting. We will omit the proof. Consequently we get that condition (94) holds by noting that (β(0), βπ‘ (0)) = (0, 0). Moreover by (97), multiplying the equation by βπ‘ and integrating over πΌ, we get that there exists a constant π > 0 such that where ππ₯ (π₯) is between βπ₯ (π₯) and π’π₯ (π₯) and satisfies βππ₯ β∞ ≤ βVπ₯ β∞ . Moreover we have β¨π (π’π₯ ) − π (βπ₯ ) − πσΈ (βπ₯ ) Vπ₯ , 1β© = β¨πσΈ (βπ₯ + πΜπ₯ ) − πσΈ (βπ₯ ) , Vπ₯ β© = β¨πσΈ σΈ (π π₯ ) πΜπ₯ , Vπ₯ β© π π σ΅©σ΅© σ΅©σ΅©2 π2 σ΅©σ΅© σ΅©σ΅©2 { σ΅©β σ΅© + σ΅©σ΅©βπ₯π₯ σ΅©σ΅© ππ‘ 2 σ΅© π‘ σ΅© 2 1 πΌ σ΅© σ΅©2 + ∫ π (βπ₯ ) ππ₯ + σ΅©σ΅©σ΅©βπ₯ σ΅©σ΅©σ΅© + πΌβ¨π’π₯π₯ , ββ©} 2 πΌ 2 σ΅© σ΅©2 σ΅© σ΅© + σ΅©σ΅©σ΅©βπ‘ σ΅©σ΅©σ΅© = πΌβ¨π’π‘ , βπ₯π₯ β© ≤ π σ΅©σ΅©σ΅©π’π‘ σ΅©σ΅©σ΅© . = 12 ∫ π π₯2 πΜπ₯ Vπ₯ ππ₯ − 4 ∫ ππ₯ Vπ₯ ππ₯ π‘ π‘ 1/2 ≤ π΄ + π΅π‘ (104) . Consequently, for any πΎ > 0 there exists a constant πΆπΎ ≥ 0 such that π‘ σ΅© σ΅©2 ∫ σ΅©σ΅©σ΅©βπ‘ σ΅©σ΅©σ΅© ππ ≤ πΎ π‘ + πΆπΎ . 0 πΌ (103) Then integrating the above equation on (0, π‘) and using (30) we get that there exist constants π΄ > 0 and π΅ > 0 such that σ΅© σ΅©2 σ΅© σ΅© ∫ σ΅©σ΅©σ΅©βπ‘ σ΅©σ΅©σ΅© ππ ≤ π΄ + π ∫ σ΅©σ΅©σ΅©π’π‘ σ΅©σ΅©σ΅© ππ 0 0 (105) 5.2. The Semigroup ππ,π (π‘). Let π½ > 0 be a parameter to be determined later and let us multiply, in πΏ2 (πΌ), (98) by 2Vπ‘ +π½V. Then we get where πΜπ₯ (π₯) is between 0 and Vπ₯ (π₯), π π₯ (π₯) is between 0 and πΜπ₯ (π₯) and βπ π₯ β∞ ≤ βπΜπ₯ β∞ ≤ βVπ₯ β∞ . Moreover we have used that Vπ₯2 ≥ πΜπ₯ Vπ₯ ≥ 0. We denote as F the differential term of (106) and prove that it induces a norm that is equivalent to that of Hπ,π . Then from (99) and (107) we have that σ΅© σ΅©2 F (π‘) ≥ πσ΅©σ΅©σ΅©Vπ‘ σ΅©σ΅©σ΅© − σ΅© σ΅©2 σ΅© σ΅©2 + (2 − ππ½) σ΅©σ΅©σ΅©Vπ‘ σ΅©σ΅©σ΅© + π2 π½σ΅©σ΅©σ΅©Vπ₯π₯ σ΅©σ΅©σ΅© − 3 σ΅© σ΅©2 σ΅© σ΅©2 σ΅© σ΅©2 ≤ πσ΅©σ΅©σ΅©Vπ‘ σ΅©σ΅©σ΅© + π½σ΅©σ΅©σ΅©Vπ₯π₯ σ΅©σ΅©σ΅© + π2 σ΅©σ΅©σ΅©Vπ₯π₯ σ΅©σ΅©σ΅© 2 σ΅© σ΅©2 σ΅© σ΅©2 σ΅© σ΅©2 + πΌσ΅©σ΅©σ΅©Vπ₯π₯ σ΅©σ΅©σ΅© + 4 (3σ΅©σ΅©σ΅©π’π₯ σ΅©σ΅©σ΅©∞ + 1) σ΅©σ΅©σ΅©Vπ₯π₯ σ΅©σ΅©σ΅© (106) 12√2π 2 3 σ΅© σ΅©2 σ΅© σ΅©2 ≤ πσ΅©σ΅©σ΅©Vπ‘ σ΅©σ΅©σ΅© + (π½ + π2 + πΌ + + 4) σ΅©σ΅©σ΅©Vπ₯π₯ σ΅©σ΅©σ΅© 2 π = β¨βπ‘π₯ , πσΈ (π’π₯ ) − πσΈ (βπ₯ ) − πσΈ σΈ (βπ₯ ) Vπ₯ β© . 3 σ΅© σ΅©2 π 2 σ΅© σ΅©2 := πσ΅©σ΅©σ΅©Vπ‘ σ΅©σ΅©σ΅© + π΄ σ΅©σ΅©σ΅©Vπ₯π₯ σ΅©σ΅©σ΅© 2 2 2 σ΅© σ΅© ≤ πσ΅©σ΅©σ΅©(V, Vπ‘ )σ΅©σ΅©σ΅©Hπ,π , We estimate the term involving π and its derivatives: π½ β¨πσΈ (π’π₯ ) − πσΈ (βπ₯ ) , Vπ₯ β© 2 (110) π½ = − β¨πσΈ σΈ (ππ₯ ) Vπ₯ , Vπ₯ β© 2 = 2π½ ∫ (1 − 3ππ₯2 ) Vπ₯2 ππ₯ πΌ σ΅© σ΅© σ΅© σ΅©2 ≥ −6π½σ΅©σ΅©σ΅©Vπ₯ σ΅©σ΅©σ΅©∞ σ΅©σ΅©σ΅©Vπ₯ σ΅©σ΅©σ΅© 6π½√2 σ΅©σ΅© σ΅©σ΅©2 ≥− π 2 σ΅©σ΅©Vπ₯ σ΅©σ΅© , π (109) σ΅© σ΅©2 π½ + π2 σ΅©σ΅©σ΅©Vπ₯π₯ σ΅©σ΅©σ΅© + βVβ2 2 σ΅© σ΅© σ΅©2 σ΅©σ΅© σ΅©σ΅©2 σ΅©σ΅©σ΅© σΈ σΈ + πΌσ΅©σ΅©Vπ₯ σ΅©σ΅© + σ΅©σ΅©π (ππ₯ )σ΅©σ΅©σ΅©σ΅©∞ σ΅©σ΅©σ΅©Vπ₯ σ΅©σ΅©σ΅© π½ β¨πσΈ (π’π₯ ) − πσΈ (βπ₯ ) , Vπ₯ β© 2 − π σ΅©σ΅© σ΅©σ΅©2 π½ 2 σ΅©V σ΅© − βVβ 2 σ΅© π‘σ΅© 2 σ΅© σ΅©2 σ΅© σ΅©2 6π½√2 σ΅©σ΅© σ΅©σ΅©2 + π2 σ΅©σ΅©σ΅©Vπ₯π₯ σ΅©σ΅©σ΅© + πΌσ΅©σ΅©σ΅©Vπ₯ σ΅©σ΅©σ΅© − π 2 σ΅©σ΅©Vπ₯ σ΅©σ΅© π π σ΅© σ΅©2 σ΅© σ΅©2 σ΅© σ΅©2 ≥ σ΅©σ΅©σ΅©Vπ‘ σ΅©σ΅©σ΅© + π2 σ΅©σ΅©σ΅©Vπ₯π₯ σ΅©σ΅©σ΅© ≥ σ΅©σ΅©σ΅©(V, Vπ‘ )σ΅©σ΅©σ΅©Hπ,π . 2 Moreover we have σ΅© σ΅©2 π σ΅© σ΅©2 π½ F (π‘) ≤ πσ΅©σ΅©σ΅©Vπ‘ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©Vπ‘ σ΅©σ΅©σ΅© + βVβ2 2 2 π½ 2 σ΅© σ΅©2 βVβ + πΌσ΅©σ΅©σ΅©Vπ₯ σ΅©σ΅©σ΅© 2 +β¨π (π’π₯ ) − π (βπ₯ ) − πσΈ (βπ₯ ) Vπ₯ , 1β©} πΌ σ΅© σ΅©2 ≥ −4σ΅©σ΅©σ΅©Vπ₯ σ΅©σ΅©σ΅© , π σ΅©σ΅© σ΅©σ΅©2 σ΅© σ΅©2 {πσ΅©V σ΅© + ππ½β¨Vπ‘ , Vβ© + π2 σ΅©σ΅©σ΅©Vπ₯π₯ σ΅©σ΅©σ΅© ππ‘ σ΅© π‘ σ΅© + (108) where we have used the same argument as (107) and (108) and where (107) π΄= 12√2π 2 2 (π½ + π2 + πΌ + + 4) , π2 π π = max {3, π΄} . (111) Then we have σ΅©σ΅© σ΅©2 σ΅© σ΅©2 σ΅©σ΅©(V, Vπ‘ )σ΅©σ΅©σ΅©Hπ,π ≤ F (π‘) ≤ πσ΅©σ΅©σ΅©(V, Vπ‘ )σ΅©σ΅©σ΅©Hπ,π . (112) 10 Abstract and Applied Analysis We estimate the right hand side of (106): Then by Lemma 10 and from (105) we conclude that there exists a positive constant πΎ such that β¨βπ‘π₯ , πσΈ (π’π₯ ) − πσΈ (βπ₯ ) − πσΈ σΈ (βπ₯ ) Vπ₯ β© σΈ σΈ σΈ σΈ σΈ = β¨βπ‘ , π (βπ₯ ) βπ₯π₯ Vπ₯ + (π (βπ₯ ) − π (π’π₯ )) π’π₯π₯ β© = β¨βπ‘ , 24βπ₯ βπ₯π₯ Vπ₯ + 12 (βπ₯2 − π’π₯2 ) π’π₯π₯ β© = β¨βπ‘ , 12βπ₯ Vπ₯ (2βπ₯π₯ − π’π₯π₯ ) − 12Vπ₯ π’π₯ π’π₯π₯ β© 1/2 Acknowledgment πΌ σ΅© σ΅© 2 4 + 12 σ΅©σ΅©σ΅©βπ‘ σ΅©σ΅©σ΅© (∫ π’π₯π₯ Vπ₯ ) 1/2 The authors would like to thank the editor and the referees for their helpful suggestions and remarks. This work was partially supported by projects PPTA 4476 and PPTA 5571 of Pontificia Universidad Javeriana, BogotaΜ. πΌ σ΅© σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© ≤ 24 σ΅©σ΅©σ΅©βπ‘ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©Vπ₯ σ΅©σ΅©σ΅©∞ σ΅©σ΅©σ΅©βπ₯ σ΅©σ΅©σ΅©∞ σ΅©σ΅©σ΅©Vπ₯π₯ σ΅©σ΅©σ΅© σ΅© σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© + 12 σ΅©σ΅©σ΅©βπ‘ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©Vπ₯ σ΅©σ΅©σ΅©∞ σ΅©σ΅©σ΅©π’π₯π₯ σ΅©σ΅©σ΅©∞ σ΅©σ΅©σ΅©Vπ₯ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅©2 σ΅© σ΅© σ΅© σ΅© ≤ 12 σ΅©σ΅©σ΅©βπ‘ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©Vπ₯π₯ σ΅©σ΅©σ΅© (2 σ΅©σ΅©σ΅©βπ₯π₯ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π’π₯π₯π₯ σ΅©σ΅©σ΅©) σ΅© σ΅© σ΅© σ΅©2 ≤ π σ΅©σ΅©σ΅©βπ‘ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©Vπ₯π₯ σ΅©σ΅©σ΅© References 2π2 σ΅©σ΅© σ΅©σ΅©2 π½π2 σ΅©σ΅© σ΅©σ΅©2 } σ΅©V σ΅© , σ΅©β σ΅© + π½π2 σ΅© π‘ σ΅© 2 σ΅© π₯π₯ σ΅© (113) where π= (118) Theorem 11. The semigroup ππ,π (π‘) possesses a connected global attractor Aπ,π ⊂ Hπ,π that is bounded in Vπ,π . = β¨βπ‘ , 12Vπ₯ [−2Vπ₯π₯ βπ₯ − Vπ₯ π’π₯π₯ ]β© ≤{ and consequently from (112) we conclude σ΅©2 σ΅© σ΅©2 −(π½/2π)π‘ σ΅©σ΅© . σ΅©σ΅©(π’, π’π‘ )σ΅©σ΅©σ΅©Hπ,π ≤ ππΎσ΅©σ΅©σ΅©(π’0 , π’1 )σ΅©σ΅©σ΅©Hπ,π π (117) This shows that the semigroup ππ,π (π‘) satisfies condition (95) and hence we have proven the following. = β¨βπ‘ , 12Vπ₯ [βπ₯ (βπ₯π₯ − Vπ₯π₯ ) − π’π₯ π’π₯π₯ ]β© σ΅© σ΅© 2 2 βπ₯ ) ≤ 24 σ΅©σ΅©σ΅©βπ‘ σ΅©σ΅©σ΅© (∫ Vπ₯2 Vπ₯π₯ F (π‘) ≤ πΎF (0) π−(π½/2π)π‘ , σΈ σΈ 12√2 (2π 2 + π 3 ) . π (114) Then it remains to estimate the following term: σ΅© σ΅©2 σ΅© σ΅©2 (2 − ππ½) σ΅©σ΅©σ΅©Vπ‘ σ΅©σ΅©σ΅© + π½π2 σ΅©σ΅©σ΅©Vπ₯π₯ σ΅©σ΅©σ΅© σ΅© σ΅©2 π½ + πΌπ½σ΅©σ΅©σ΅©Vπ₯ σ΅©σ΅©σ΅© − β¨πσΈ (π’π₯ ) − πσΈ (βπ₯ ) , Vπ₯ β© 2 π σ΅©σ΅© σ΅©σ΅©2 σ΅© σ΅©2 ≥ π½ σ΅©σ΅©Vπ‘ σ΅©σ΅© + π½π2 σ΅©σ΅©σ΅©Vπ₯π₯ σ΅©σ΅©σ΅© 2 (115) √ 2 σ΅© σ΅©2 σ΅© σ΅©2 + πΌπ½σ΅©σ΅©σ΅©Vπ₯ σ΅©σ΅©σ΅© − 6π½ π 2 σ΅©σ΅©σ΅©Vπ₯ σ΅©σ΅©σ΅© π π σ΅© σ΅©2 σ΅© σ΅©2 ≥ π½ σ΅©σ΅©σ΅©Vπ‘ σ΅©σ΅©σ΅© + π½π2 σ΅©σ΅©σ΅©Vπ₯π₯ σ΅©σ΅©σ΅© . 2 Then, using the above estimates, we can rewrite (106) in the following way: π½ π2 σ΅©σ΅© σ΅©σ΅©2 σ΅©σ΅© σ΅©σ΅©2 π F (π‘) + F (π‘) ≤ σ΅©β σ΅© σ΅©V σ΅© ππ‘ π 2π½π2 σ΅© π‘ σ΅© σ΅© π₯π₯ σ΅© ≤ π2 σ΅©σ΅© σ΅©σ΅©2 σ΅©σ΅© σ΅©2 σ΅©σ΅©βπ‘ σ΅©σ΅© σ΅©σ΅©(V, Vπ‘ )σ΅©σ΅©σ΅©Hπ,π 4 π½π ≤ π2 σ΅©σ΅© σ΅©σ΅©2 σ΅©β σ΅© F (π‘) . π½π4 σ΅© π‘ σ΅© (116) [1] G. 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