# Global Attractor of Thermoelastic Coupled Beam Equations with Structural Damping

Authors: Peirong Shi; Danxia Wang; Weiyi Chen
Journal: Mathematical Problems in Engineering
 (2017)

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## Abstract

In this paper, we study the existence of a global attractor for a class ofn-dimension thermoelastic coupled beam equations with structural damping utt+Δ2u+Δ2ut-[σ(∫Ω‍(∇u)2dx)+ϕ(∫Ω‍∇u∇utdx)]Δu+f1(u)+g(ut)+νΔθ=q(x),inΩ×R+, and θt-Δθ+f2(θ)-νΔut=0. Here Ω is a bounded domain of RN, and σ(·) and ϕ(·) are both continuous nonnegative nonlinear real functions and q is a static load. The source terms f1(u) and f2(θ) and nonlinear external damping g(ut) are essentially |u|ρu, |θ|ϱθ, and |ut|rut respectively.


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## Body

## 1. Introduction

This problem is based on the equation(1)utt+uxxxx-α+β∫0Lux2dxuxx=0,which was proposed by Woinowsky-Krieger [1] as a model for vibrating beam with hinged ends.Without thermal effects, Ball [2] studied the initial-boundary value problem of more general beam equation (2)utt+uxxxx-M∫0Lux2dxuxx=0subjected to homogeneous boundary condition. Ma and Narciso [3] proved the existence of global solutions and the existence of a global attractor for the Kirchhoff-type beam equation (3)utt+Δ2u-M∫Ω∇u2dxΔu+fu+gut=hx,without structural damping, subjected to the conditions (4)u=∂u∂v=0on∂Ω×R+.In fact, the plate equations without thermal effects were studied by several authors; we quote, for instance, [4–8].In the following we also make some comments about previous works for the long-time dynamics of thermoelastic coupled beam system with thermal effects.Giorgi et al. [9] studied a class of one-dimensional thermoelastic coupled beam equations (5)utt+Δ2u-β+∇uL20,l2Δu-Δutt+fu+Δθ=f,θt-Δθ-Δut=gand gave the existence and uniqueness of global weak solution and the existence of global attractor under Dirichlet boundary conditions. Barbosa and Ma [10] studied the long-time behavior for a class of two-dimension thermoelastic coupled beam equation (6)utt+Δ2u-M∫Ω∇u2dxΔu-Δutt+fu+νΔθ=hx,θt-ωΔθ-1-ω∫0tkt-sΔθds-νΔut=0subjected to the conditions (7)u=Δu=0,θ=0.In addition, we also refer the reader to [11–15] and the references therein.A mathematical problem is the nonlinearn-dimension thermoelastic coupled beam equations with structural damping which arise from the model of the nonlinear vibration beam with Fourier thermal conduction law: (8)utt+Δ2u+Δ2ut-σ∫Ω∇u2dx+ϕ∫Ω∇u∇utdxΔu+f1u+gut+νΔθ=qx,inΩ×R+,(9)θt-Δθ+f2θ-νΔut=0with the initial conditions (10)ux,0=u0x,utx,0=u1x,θx,0=θ0xand the boundary conditions(11)u∂Ω=∂u∂v∂Ω=0,θ∂Ω=0.To the our best knowledge, the existence of global attractor for thermoelastic coupled beam equations was not considered in the presence of nonlinear structure damping. Here the unknown function u(x,t) is the elevation of the surface of beam; u0(x) and u1(x) are the given initial value functions; the subscript t denotes derivative with respect to t and the assumptions on nonlinear functions σ(·), ϕ(·), f1(·), f2(·), g(·), and the external force function q(x) will be specified later.Our fundamental assumptions onσ(·), ϕ(·), f1(·), f2(·), g(·), and q(x) are given as follows.Assumption 1.
We assume thatσ(·)∈C1(R) satisfying (12)σzz≥σ~z≥0,∀z≥0,where σ~(z)=∫0z‍σ(s)ds. This condition is promptly satisfied if σ(·) is nondecreasing with σ(0)=0.Assumption 2.
We also assume thatϕ(·)∈C1(R) satisfying ϕ(0)=0 and ϕ(·) is nondecreasing and (13)ϕss≥0,∀s∈R+.Assumption 3.
The functionf1(·):R→R is of class C1(R) and satisfies f(0)=0, and there exist constants k and ρ≥0 such that (14)f1u-f1v≤k11+uρ+vρu-v,∀u,v∈R,(15)-a0≤f^1u≤12f1uu+a1,where f^1(z)=∫0z‍f1(s)ds.Assumption 4.
The functionf2(·):R→R is of class C1(R) and satisfies f2(0)=0, and there exist constants k2,k3 and ϱ≥0 such that (16)f2θ-f2θ~≤k21+θϱ+θ~ϱθ-θ~,∀θ,θ~∈R,(17)f2θ-f2θ~θ-θ~≥k3θ-θ~ϱ+2,∀θ,θ~∈R.Assumption 5.
The functiong(·):R→R is of class C1(R) and satisfies g(0)=0, and there exist constants k4,k5 and r≥0 such that(18)gu-gvu-v≥k4u-vr+2,∀u,v∈R,(19)gu-gv≤k51+ur+vru-v,∀u,v∈R.Assumption 6.
q
(
x
)
∈
L
2
(
Ω
).Under the above assumptions, we prove the existence of global solutions and the existence of a global attractor of extensible beam equation system (8)–(11). And the paper is organized as follows. In Section 2, we introduce some Sobolev spaces. In Section 3, we discuss the existence and uniqueness of global strong solution and weak solution. In Sections 4 and 5, we establish the result of the existence of a global attractor.


## 2. Basic Spaces

Our analysis is based on the following Sobolev spaces. Let(20)U=θ∈H2Ω∩L2Ω;θ∂Ω=0.Then for regular solutions we consider the phase space (21)H1=H4Ω∩H02Ω×H4Ω∩H02Ω×U.In the case of weak solutions we consider the phase space (22)H0=H02Ω×L2Ω×L2Ω.In H0 we adopt the norm defined by (23)u,v,θH02=Δu2+v2+θ2.


## 3. The Existence of Global Solutions

Firstly, using the classical Galerkin method, we can establish the existence and uniqueness of regular solution to problem (8)–(11). We state it as follows.Theorem 7.
Under assumptions(H1)–(H6), for any initial data (u0,u1,θ0)∈H1, then problem (8)–(11) has a unique regular solution (u,θ) with (24)u∈L∞R+,H4Ω∩H02Ω,ut∈L∞R+,H4Ω∩H02Ω,utt∈L∞R+,L2Ω,θ∈L∞R+,U,θt∈L∞R+,L2Ω.Proof.
Let us consider the variational problem associated with (8)–(11): find (u(t),θ(t))∈H4(Ω)∩H02(Ω)×U such that (25)∫Ωuttωdx+∫ΩΔuΔωdx+∫ΩΔutΔωdx+σ∇u2∫Ω∇u∇ωdx+ϕ∫Ω∇ut∇ωdx∫Ω∇ut∇ωdx+∫Ωf1uωdx+∫Ωgutωdx+ν∫ΩΔθωdx=∫Ωqxωdx,∫Ωθtϖdx-∫ΩΔθϖdx+∫Ωf2θϖdx-ν∫ΩΔutϖdx=0for all ω∈H02(Ω) and ϖ∈U. This is done with the Galerkin approximation method which is standard. Here we denote the approximate solution by (um(t),θm(t)). We can get the theorem by proving the existence of approximation solution, the estimate of approximation solution, convergence, and uniqueness. In the following we give the estimates of approximation solution and the proof of uniqueness of solution.
Estimate 1. In the first approximate equation and the second approximate equation of (25), respectively putting ω=utm(t) and ϖ=θm(t) and making a computation of addition and considering σ^(z)=∫0z‍σ(s)ds and f^1(z)=∫0z‍f1(s)ds, by using Schwarz inequality, and then integrating from 0 to t<tm, we see that(26)utm2+Δum2+σ^∇um2+∫Ωf^1umdx+θm2+∫0tΔutm2dt+2∫0t∇θm2dt+2∫0tϕ∫Ω∇u∇utdx∫Ω∇u∇utdxds+2∫0t∫Ωgutmutmdxds+2∫0t∫Ωf2θmθmdxds≤1λ1∫0tqx2ds+utm02+Δum02+σ^∇um02+∫Ωf^1um0dx+θm02independent of m and t, where λ1 is the same as λ1 of (47). Taking into account assumptions (13), (17), and (18) of ϕ(·), f2(·), and g(·) and Assumption 6, we see that there exists M1>0 depending only on T such that (27)utm2+Δum2+σ^∇um2+∫Ωf^1umdx+θm2≤M1for all t∈[0,T] and for all m∈N.
Estimate 2. In the first approximate equation and the second approximate equation of (25), respectively, putting ω=Δ2utm(t) and ϖ=Δθm(t) and making a computation of addition by using Schwarz inequality and Young inequality and considering the assumptions of σ(·),ϕ(·),f1(·),g(·), f2(·), and q(·), we see that there exists M2>0 depending only on T such that (28)Δutm2+Δ2um2+∇θm2≤M2for all t∈[0,T] and for all m∈N.
Estimate 3. In the first approximate equation and the second approximate equation of (25), respectively integrating by parts with ω=uttm(0) and ϖ=θtm(0) with t=0 and using Schwarz inequality and Young inequality, we see that there exists M3,M4>0 depending only on T such that (29)uttm02≤M3,θtm02≤M4for all t∈[0,T] and for all m∈N.
Estimate 4. Let us fix t,ξ>0 such that ξ<T-t. Respectively taking the difference of the first approximate equation and the second approximate equation of (25) with t=t+ξ and t=t and respectively replacing ω by utm(t+ξ)-utm(t) and ϖ by θm(t+ξ)-θm(t), we can find constants M5,M6>0, depending only on T, such that (30)uttm2+Δutm2≤M5,θtm2≤M6,∀m∈N,∀t∈0,T.
Estimate 5. Taking the scalar product in H with ω=Δθm for the second approximate equation of (25), after a computation we can find a constant M7>0, depending only on T such that (31)Δθm2≤M7,∀m∈N,∀t∈0,T.With the estimates 1-2 and 4-5, we can get the necessary compactness in order to pass approximate equation of (25) to the limit. Then it is a matter of routine to conclude the existence of global solutions in [0,T].
Uniqueness. Let (u,θ),(v,θ~) be two solutions of (8)–(11) with the same initial data. Then writing p=u-v,ϑ=θ-θ~ and taking the difference (25) with u=u,θ=θ and u=v,θ=θ~ and respectively replacing ω,ϖ by pt,ϑ and then making a computation of addition, we have (32)12ddtpt2+Δp2+ϑ2+Δpt2+Δϑ2+J1+J2+∫Ωf1u-f1vptdx+∫Ωgut-gvtptdx+∫Ωf2θ-f2θ~ϑtdx=0,where J1=∫Ω‍[σ(∫Ω‍(∇u)2dx)Δu-σ(∫Ω‍(∇v)2dx)Δv]ptdx and J2=∫Ω‍[ϕ(∫Ω‍∇u∇utdx)Δu-ϕ(∫Ω‍∇v∇vtdx)Δv]ptdx. Using Mean Value Theorem and the Young inequalities combined with the estimates 1-2 and 4-5, we deduce that for some constant M8>0, (33)ddtpt2+Δp2+ϑ2≤M8pt2+Δp2+ϑ2,∀t∈0,T.Then from Gronwall’s Lemma we see that u=v,θ=θ~. The proof of Theorem 7 is completed.Theorem 8.
Under the assumptions of Theorem7, if the initial data (u0,u1,θ0)∈H0, there exists a unique weak solution of problem (8)–(11) which depends continuously on initial data with respect to the norm of H0.Proof.
By using density arguments, we can obtain the existence of a weak solution inH0.
Let us consider{u0,u1,θ0}∈H1¯. Since H1 is dense in H0, then there exists {uμ0,uμ1,θμ0}⊂H1, such that (34)uμ0⟶u0inH02Ω;uμ1⟶u1inL2Ω;θμ0⟶θ0inL2Ω.We observe that for each μ∈N, there exists (uμ,θμ), smooth solution of the initial-boundary value problem (8)–(11) which satisfies(35)uttμ+Δ2uμ+Δ2utμ-σ∫Ω∇uμ2dx+ϕ∫Ω∇uμ∇utμdxΔuμ+f1uμ+gutμ+νΔθμ=qx,θtμ-Δθμ+f2θμ-νΔutμ=0.Respectively multiplying the first equation in (35) by uμ and multiplying the second equation in (35) by θμ and integrating over Ω and taking the sum and then considering the arguments used in the estimate of the existence of solution, we obtain (36)utμ2+Δuμ2+θμ2≤C0,where C0 is a positive constant independent of μ∈N.
DefiningZμ,σ=uμ-uσ,Z~μ,σ=θμ-θ~μ:μ,σ∈N, following the steps already used in the uniqueness of regular solution for (8)–(11), and considering the convergence given in (34), we deduce that there exists (u,θ) such that (37)uμ⟶ustronglyinC0,T;H02Ω,utμ⟶utstronglyinC0,T;L2Ω,θμ⟶θstronglyinC0,T;L2Ω.From the above convergence, we can pass to the limit using standard arguments in order to obtain (38)uttμ+Δ2uμ+Δ2utμ-σ∫Ω∇uμ2dx+ϕ∫Ω∇uμ∇utμdxΔuμ+f1uμ+gutμ+νΔθμ=qx,θtμ-Δθμ+f2θμ-νΔutμ=0.Theorem 8 is proved.Remark 9.
In both cases(39)ut2+Δu2+uρ+2ρ+2+θ2≤C,where C is a constant depending on the initial data in different expression.In addition, in this paper,C denotes different constant in different expression.Remark 10.
Theorem8 implies that problem (8)–(11) defines a nonlinear C0-semigroup S(t) on H0. Indeed, let us set S(t)(u0,u1,θ0)=(u(t),ut(t),θ(t)), where u is the unique solution corresponding to initial data (u0,u1,θ0)∈H0. Moreover, the operator S(t) defined in H0 meets the usual semigroup properties (40)St,τ=StSτ,∀t,τ∈R,S0=I.To prove the main result, we need the following Lemma 11 of Nakao and Lemma 12Lemma 11 (see [16]).
Letφ(t) be a nonnegative continuous function defined on [0,T),1<T≤∞, which satisfies (41)supt≤s≤t+1φs1+η≤M0φt-φt+1+M1,0≤t≤T-1,where M0,M1,η are positive constants. Then we have (42)φt≤M0-1ηt-1++sup0≤s≤1φs-η-1/η+M11/η+1,0≤t≤T.Lemma 12 (see [17]).
Assume that for any bounded positive invariant setB⊂H and for any ε>0, there exists T=T(ε,B) such that(43)dSTx,STy≤ε+ϖTx,y,∀x,y∈B,where ϖT:H×H→R satisfies for any sequence {zn}⊂B(44)liminfm→∞liminfn→∞ϖTzn,zm=0.Then S(t) is asymptotically smooth.


## 4. The Existence of Absorbing Set

The main result of an absorbing set reads as follows.Theorem 13.
Assume the hypotheses of Theorem8; then the corresponding semigroup S(t) of problem (8)–(11) has an absorbing set B in H0.Proof.
Now we show that semigroupS(t) has as absorbing set B in H0. Firstly, we can calculate the total energy functional(45)Et=12ut2+Δu2+σ^∇u2+θ2+∫Ωf^1udx-∫Ωqutdx.Let us fix an arbitrary bounded set B⊂H0 and consider the solutions of problem (8)–(11) given by (u(t),ut(t),θ(t))=S(t)(u0,u1,θ0) with (u0,u1,θ0)∈B. Our analysis is based on the modified energy function (46)E~t=Et+a0Ω+1λ1q2,where λ1>0 is the first eigenvalue of the operator Δ in H02(Ω); that is, λ1 satisfies (47)u2≤1λ1Δu2,∀u∈H02Ω.It is easy to see thatE~(t) dominates (u(t),ut(t),θ(t))H02 and E~(t)≥1/4Δut2. By multiplying (8) by u and integrating over Ω, we have (48)Δu2=-σ∫Ω∇u2dx+ϕ∫Ω∇u∇utdx∇u2-∫Ωf1uudx+ut2-ddtut,u-∫Ωgutudx+∫Ωqutdx-∫ΩΔuΔutdx+υ∫Ω∇θ∇udx.Inserting (48) into E~(t), we obtain(49)E~t=ut2+12σ^∫Ω∇u2dx-σ∫Ω∇u2dx∇u2-12ϕ∫Ω∇u∇utdx∇u2+12θ2-12∫ΩΔuΔutdx+∫Ωf^1u-12f1uudx-12ddt∫Ωutudx-12∫Ωgutudx+ν2∫Ω∇θ∇udx+a0Ω+1λ1q2-12∫Ωqutdx.Considering (12) and (15) and integrating from t1 to t2 for (49), we obtain that (50)∫t1t2E~tds≤∫t1t2ut2ds-12∫t1t2ϕ∫Ω∇u∇utdx∇u2ds+12∫t1t2θ2ds-12∫t1t2∫ΩΔuΔutdxds-12∫Ωutt2ut2dx-∫Ωutt1ut1dx-12∫t1t2∫Ωgutudxds+ν2∫t1t2∫Ω∇θ∇udsdx+a0+a1Ω+1λ1q2+12∫t1t2∫Ωqutdxds,where t1,t2∈[t,t+1].Now let us begin to estimate the right hand side of (50) to use the above Lemma 11 of Nakao.First, by multiplying (8) by ut and multiplying (9) by θ and integrating over Ω and then taking the sum, we have (51)ddtEt+Δut2+∇θ2=-∫Ωgututdx-ϕ∫Ω∇u∇utdx∫Ω∇u∇utdx-∫Ωf2θθdx.Then integrating from t to t+1, we get (52)Et+1-Et+∫tt+1Δut2ds+∫tt+1∇θ2ds=-∫tt+1∫Ωgututdxds-∫tt+1ϕ∫Ω∇u∇utdx∫Ω∇u∇utdxds-∫tt+1∫Ωf2θθdxds.Taking into account assumptions (13), (17), and (18) of ϕ(·), f2(·), and g(·), we have(53)Et≥Et+1.Then we define an auxiliary function I2(t) by putting (54)I2t=Et-Et+1≥0.Thus it is obvious that (55)∫tt+1Δut2≤I2t,(56)∫tt+1∇θ2≤I2t.Noting that r/r+2+2/r+2=1 and using twice Holder inequalities and considering assumption (18) of g(·), we have (57)∫tt+1ut2ds≤Ωr/r+2∫tt+1∫Ωutr+2dxds2/r+2≤Ωr/r+21k4∫tt+1∫Ωgututdxds2/r+2≤Ωr/r+21k4It4/r+2.Using the Mean Value Theorem with ϕ(0)=0 and considering the estimate of (39) and then using Young inequality combined with (55), we have (58)12∫t1t2ϕ∫Ω∇u∇utdx∇u2ds=12∫t1t2ϕ′ξ1∫Ω∇u∇utdx∇u2ds≤12∫t1t2CΔutΔuds≤C4η∫t1t2Δut2ds+η4∫t1t2Δu2ds≤C4η∫t1t2Δut2ds+ηsupt≤s≤t+1E~s≤C4ηI2t+ηsupt≤s≤t+1E~s,where ξ1 is among 0 and ∫Ω‍∇u∇utdx.Sinceθ2≤1/λ2∇θ2, ∀θ∈H01, from (56) we obtain (59)12∫t1t2θ2ds≤12λ2I2t,where λ2 is the first eigenvalue of the operator ∇ in H01(Ω).Using Young inequality, we get(60)12∫t1t2∫ΩΔuΔutds≤14η∫t1t2Δut2ds+η4∫t1t2Δu2ds≤14ηIt2+ηsupt≤s≤t+1E~s.Since (57), in view of the Mean Value Theorem for integral, there exist number t1∈[t,t+1/4] and number t2∈[t+3/4,t+1] such that (61)utt12≤Ωr/r+24k4It4/r+2,utt22≤Ωr/r+24k4It4/r+2.Thus from Schwarz inequality combined with (47) and (61), we have(62)12∫Ωutt2ut2dx-∫Ωutt1ut1dx≤12λ1utt2Δut2+utt1Δut1≤2λ1k4Ωr/2r+2It2/r+2supt≤s≤t+1Δu≤2/λ1k4Ωr/2r+22ηIt4/r+2+ηsupt≤s≤t+1E~s.Considering assumption (19) of g(·) and using Young inequality and Holder inequality with r+1/r+2+1/r+2=1, then from (47) and (57), we have (63)12∫t1t2∫Ωgutudxds≤k52∫t1t2∫Ω1+utrutudxds≤k524ηλ1∫tt+1∫Ωuts2dxds+ηλ14∫tt+1∫Ωus2dxds+k52∫t1t2utr+1udxds≤k524ηλ1∫tt+1∫Ωuts2dxds+ηλ14∫tt+1∫Ωus2dxds+k52∫tt+1∫Ωutr+2dxr+1/r+2∫Ωur+2dx1/r+2ds≤k524ηλ1∫tt+1∫Ωuts2dxds+ηλ14∫tt+1∫Ωus2dxds+k52∫tt+1∫Ωutr+2dxr+1/r+2r+2/r+1dsr+1/r+2∫tt+1∫Ωur+2dx1/r+2r+2ds1/r+2=k524ηλ1∫tt+1∫Ωuts2dxds+ηλ14∫tt+1∫Ωus2dxds+k52∫tt+1∫Ωutr+2dxdsr+1/r+2∫tt+1∫Ωur+2dxds1/r+2≤k524ηλ1Ωr/r+21k4It4/r+2+ηsupt≤s≤t+1E~s+k5μ2k4It2r+1/r+2supt≤s≤t+1Δu≤k524ηλ1Ωr/r+21k4It4/r+2+ηsupt≤s≤t+1E~s+k52μ24ηk42It4r+1/r+2+ηsupt≤s≤t+1E~s.Also by Young inequality, we have (64)ν2∫t1t2∫Ω∇θ∇udxds≤ν2∫t1t2∇θ∇uds≤∫t1t2ν24ηλ3∇θ2ds+∫t1t2ηλ34∇u2ds≤ν24ηλ3I2t+ηsupt≤s≤t+1E~s,where λ3>0 is the first eigenvalue of the operator ∇ in H01(Ω); that is, λ3 satisfies (65)∇u2≤1λ3Δu2,∀u∈H01Ω.Finally using Young inequality again, we get that (66)12∫t1t2∫Ωqutdxds≤14ηλ1q2+ηsupt≤s≤t+1E~s.Inserting (57)–(60) and (62), (63), (64), and (66) into (50), we obtain (67)∫t1t2E~sds≤Ωr/r+21k4+2/λ1k4Ωr/2r+22η+k524ηλ1Ωr/r+21k4It4/r+2+C4η+12λ2+14η+ν24ηλ3It2+k52μ24ηk42It4r+1/r+2+6ηsupt≤s≤t+1E~s+a0+a1Ω+1λ1+14ηλ1q2.For the left hand side of (67), we use the Mean Value Theorem; then there exists number τ∈[t1,t2] such that (68)∫t1t2E~sds≥12E~t+1=12E~t-It2.So we conclude that (69)E~t≤It2+2∫t1t2E~sds.Inserting (67) into (69), we obtain that (70)E~t≤2Ωr/r+21k4+2/λ1k4Ωr/2r+22η+k528ηλ1Ωr/r+21k4It4/r+2+1+C4η+12λ2+14η+ν24ηλ3It2+k52μ24ηk42It4r+1/r+2+12ηsupt≤s≤t+1E~s+a0+a1Ω+21λ1+14ηλ1q2.Letting 0<η<1/12 and noting that I(t)2r/r+2 and I(t)4r/r+2 are bounded with estimate (39), then from (70), we get (71)E~t≤21-12ηIt4/r+2Ωr/r+21k4+2/λ1k4Ωr/2r+22η+k524ηλ1Ωr/r+21k4+1+C4η+12λ2+14η+ν24ηλ3C+k52μ24ηk42C+11-12ηa0+a1Ω+21-12η1λ1+14ηλ1q2,where C is a constant which depends on B.Set(72)C1=21-12ηΩr/r+21k4+2/λ1k4Ωr/2r+22η+k524ηλ1Ωr/r+21k4+1+C4η+12λ2+14η+ν24ηλ3C+k52μ24ηk42C;then (70) can be rewritten as(73)E~t1+r/2≤C1E~t-E~t+1+21-12ηa0+a1Ω+21-12η1λ1+14ηλ1q21+r/2.Using Nakao’s Lemma 11, we conclude that(74)E~t≤C1-1r2t-1++E~0-r/2-2/r+11-12ηa0+a1Ω+21-12η1λ1+14ηλ1q2.As r→∞, the first term of the right side of (74) goes to zero; thus, with E~(t), we conclude (75)B=u,v,θ∈H0∣Δu2+v2+θ2≤81-12ηa0+a1Ω+81-12η1λ1+14ηλ1q2is an absorbing set for S(t) in H0.


## 5. The Existence of a Global Attractor

The main result of a global attractor reads as follows.Theorem 14.
Assume the hypotheses of Theorem8; then the corresponding semigroup S(t) of problem (8)–(11) is asymptotic compactness.Proof.
We are going to apply Lemmas11 and 12 to prove the asymptotic smooth. Given initial data (u0,u1,θ0) and (v0,v1,θ~0) in a bounded invariant set B⊂H0, let (u,θ),(v,θ~) be the corresponding weak solutions of problem (8)–(11). Then the differences w=u-v,ϑ=θ-θ~ are the weak solutions of (76)wtt+Δ2w+Δ2wt+νΔϑ+Δg=σ∇u2Δu-σ∇v2Δv+Δϕ-Δf1,ϑt-Δϑ+Δf2-νΔwt=0,w=∂w∂v=0,ϑ=0,w0=u0-v0,wt0=u1-v1,ϑ0=θ0-θ~0,where (77)Δϕ=ϕ∫Ω∇u∇utdxΔu-ϕ∫Ω∇v∇vtdxΔv,Δg=gut-gvt,Δf1=f1u-f1v,Δf2=f2θ-f2θ~.Let us define (78)Ewt=wt2+Δw2+σ∇u2∇w2+ϑ2.As before, by density, we can assume formally that w is sufficiently regular. Then, multiplying the first equation in (76) by wt and integrating over Ω and multiplying the second equation in (76) by ϑ and integrating over Ω and then taking the sum, we get (79)12ddtEwt+Δwt2+∇ϑ2+∫ΩΔgwtdx+∫ΩΔf2ϑdx=-σ′∇u2∇w2∫ΩΔuutdx+Δσ∫ΩΔvwtdx+∫ΩΔϕwtdx-∫ΩΔf1wtdx,where(80)Δσ=σ∇u2-σ∇v2.Let us estimate the right hand side of (79).
Considering the continuity ofσ′(·) and estimate (39)(81)-σ′∇u2∇w2≤C∇w2,Applying the Mean Value Theorem combined with estimate (39), by Young inequality, we get (82)Δσ∫ΩΔvwtdx≤C∇wr+2/r+1+k44wtr+2r+2.Also use the Mean Value Theorem combined with estimate (39) and Young inequality to get(83)∫ΩΔϕwtdx=∫Ωϕ∫Ω∇u∇utdxΔwwtdx-∫Ωϕ∫Ω∇u∇utdx-ϕ∫Ω∇v∇vtdxΔvwtdx=∫Ωϕ′ξ2∫Ω∇u∇utdxΔwwtdx-∫Ωϕ′ξ3∫Ω∇w∇wtdxΔvwtdx≤CwΔwt+CwΔwtwt≤14Δwt2+C2w2+14Δwt2+C2w2wt2≤14Δwt2+C2∇w2+C2w2wtr+22≤14Δwt2+C2∇w2+C4k3w2r+2/r+k34wtr+2r+2,where ξ2 is among 0 and ∫Ω‍∇u∇utdx, and ξ3 is among ∫Ω‍∇u∇utdx and ∫Ω‍∇v∇vtdx.
By the Holder inequality, Minkowski inequality combined with the estimate of (39), and Young inequality, we obtain (84)∫ΩΔf1wtdx≤C∇wr+2/r+1+k44wtr+2r+2.On the other hand, considering assumptions (17) and (18) of f2(·) and g(·), (85)∫ΩΔf2ϑdx≥k3ϑϱ+2ϱ+2,∫ΩΔgwtdx≥k4wtr+2r+2.Thus by inserting (81)–(85) into (79), we get that (86)12ddtEwt+14Δwt2+∇ϑ2+k44wtr+2r+2+k3ϑϱ+2ϱ+2≤C∇w2+∇wr+2/r+1+∇w2r+2/r.Then integrating from t to t+1 and defining an auxiliary function F2(t), we get (87)14∫tt+1Δwt2ds+∫tt+1∇ϑ2ds+k44∫tt+1wtr+2r+2ds+k3∫tt+1ϑϱ+2ϱ+2ds≤Ewt-Ewt+1+C∫tt+1∇w2+∇wr+2/r+1+∇w22r+2/rds=Ft2.It is obvious that(88)Ewt+1≥Ett,(89)14∫tt+1Δwt2ds≤Ft2,∫tt+1∇ϑ2ds≤F2t,(90)k44∫tt+1wtr+2r+2ds≤Ft2,k3∫tt+1ϑϱ+2ϱ+2ds≤Ft2.Then by multiplying first equation in (76) by w and integrating over Ω again, we obtain that (91)Δw2+σ∇u2∇w2=-ddt∫Ωwtwdx+wt2-∫ΩΔ2wtwdx+Δσ∫ΩΔvwdx-∫ΩΔf1wdx-∫ΩΔgwdx+∫Ωϕ∫Ω∇u∇utdx-ϕ∫Ω∇v∇vtdxΔvwdx-ϕ∫Ω∇u∇utdx∇w2-ν∫ΩϑΔwdx.Integrating from t1 to t2, we get (92)∫t1t2Δw2+σ∇u2∇w2ds=∫Ωwtt2wt2dx-∫Ωwtt1wt1dx+∫t1t2wt2dt-∫t1t2∫ΩΔwtΔwdxds+∫t1t2Δσ∫ΩΔvwdxds-∫t1t2∫ΩΔf1wdxds-∫t1t2∫ΩΔgwdxds+∫t1t2∫Ωϕ∫Ω∇u∇utdx-ϕ∫Ω∇v∇vtdxΔvwdxds-∫t1t2ϕ∫Ω∇u∇utdx∇w2ds+ν∫t1t2∫Ω∇ϑ∇wdxds.Now let us estimate the right hand side of (92). Firstly, from the first inequality of (90), by holder inequality we infer that (93)∫tt+1wt2ds=∫tt+1∫Ωwt2dxds≤Ωr/r+2∫tt+1∫Ωwt2r+2/2dxds2/r+2≤CFt4/r+2;thus there exists t1∈[t,t+1/4] and t2∈[t+3/4,t+1] such that (94)wtt12≤CF4/r+2t,wtt22≤CF4/r+2t;then we can deduce that(95)∫Ωwtt2wt2dx-∫Ωwtt1wt1dx≤CFt4/r+2+18supt≤σ≤t+1Ewσ.Use Schwarz inequality combined with the estimate of (39) and Holder inequality to obtain (96)∫t1t2∫ΩΔwtΔwdxds≤C∫t1t2Δwtds≤C∫t1t21ds1/2∫t1t2Δwt2ds1/2≤CFt.Apply the Mean Value Theorem combined with estimate (39) to get (97)∫t1t2Δσ∫ΩΔvwdxds≤C∫t1t2∇w2ds.Assumption (14) of f(·) and the estimate of (39) imply that(98)∫t1t2∫ΩΔf1wdxds≤C∫t1t2∇w2ds.Also from assumption (19) of g(·) and the estimate of (39) combined with (94), we have (99)∫t1t2∫ΩΔgwdxds≤C∫t1t2wtΔwds≤CFt4/r+2+18supt≤σ≤t+1Ewσ.Using the Mean Value Theorem and considering the assumption of ϕ(·) and the estimate of (39), we have (100)∫t1t2∫Ωϕ∫Ω∇u∇utdx-ϕ∫Ω∇v∇vtdxΔvwdxds=∫t1t2ϕ′ξ4∫Ω∇w∇wtdx∫ΩΔvwdxds≤C∫t1t2wwtds≤C∫t1t2∇w2ds+C∫t1t2wt2ds,(101)∫t1t2ϕ∫Ω∇u∇utdx∇w2ds≤C∫t1t2∇w2ds,where ξ4 is among ∫Ω‍∇u∇utdx and ∫Ω‍∇v∇vtdx.
Finally, use Young inequality to get(102)ν∫t1t2∫Ω∇ϑ∇wdxds≤ν22∫t1t2∇ϑ2+12∫t1t2∇w2ds.By inserting (93) and (95)–(102) into (92), we obtain that (103)∫t1t2Δw2+σ∇u2∇w2ds≤3C∫t1t2∇w2ds+2C∫t1t2wt2ds+C∫t1t2∇ϑ2ds+2CFt4/r+2+14supt≤σ≤t+1Ewσ+CFt.Considering (89) and (93), from (103), we have (104)∫t1t2Δw2+σ∇u2∇w2ds≤3C∫tt+1∇wds+CFt+CF2t+4CFt4/r+2+14supt≤σ≤t+1Ewσ.Using Holder inequality with 1/ϱ+2+ϱ+1/ϱ+2=1, (105)∫t1t2ϑ2ds≤C∫t1t2ϑϱ+2ds≤C∫t1t21ϱ+2/ϱ+2dsϱ+1/ϱ+2∫t1t2ϑϱ+2ϱ+2ds≤C∫t1t2ϑϱ+2ϱ+2ds≤CF2t.Then from the definition of Ew(t) and (93), (104), and (105), we obtain that (106)∫t1t2Ewsds≤5CFt4/r+2+2CF2t+CFt+14supt≤σ≤t+1Ewσ+3C∫tt+1∇wds.For (106), by using Mean Value theorem, there exists t∗∈[t1,t2] such that(107)Ewt∗≤10CFt4/r+2+2CFt+4CF2t+12supt≤σ≤t+1Ewσ+6C∫tt+1∇wds.From (87), we see that (108)Ewt≤Ewt+1+F2t.Let Ew=supt≤σ≤t+1Ew(σ) with τ∈[t,t+1]; then integrate (86) over [t,τ] and over [t∗,t+1] to have (109)supt≤σ≤t+1Ewσ≤Ewτ≤Ewt+1+F2t+C∫tt+1∇w2+∇w2r+2/rds≤Ewt∗+F2t+C∫tt+1∇w2+∇w2r+2/rds.Inserting (107) into (109), we obtain (110)supt≤σ≤t+1Ewσ≤10CFt4/r+2+2CFt+4CF2t+12supt≤σ≤t+1Ewσ+6C∫tt+1∇wds+F2t+C∫tt+1∇w2+∇w2r+2/rds.Therefore from the boundary of 1+F(t)(1-4/r+2)+F(t)2-4/r+2, we have (111)supt≤σ≤t+1Ewσ≤CFt4/r+2+C∫tt+1∇w+∇w2+∇w2r+2/rds.Therefore (112)supt≤σ≤t+1Ewσ1+r/2≤CEwt-Ewt+1+Csup0≤σ≤T∫σσ+1∇w+∇w2+∇w2r+2/rds.From Nakao’s Lemma 11, there exists CB>0 and CT>0 such that (113)Ewt≤CBt-1+-2/r+CTsup0≤σ≤T∫σσ+1∇w+∇w2+∇w2r+2/rds2/r+2,0≤t≤T.From the definition of Ew(t), we have (114)w,wt,ϑH0≤CBt-1+-2/r+CTsup0≤σ≤T∫σσ+1∇w+∇w2+∇w2r+2/rds2/r+2.Given ε>0, we choose T large such that(115)CBt-1+-2/r≤ε,and define ϖT:H0×H0→R as (116)ϖTu0,u1,θ0,v0,v1,θ~0=CTsup∫σσ+1∇w+∇w2+∇w2r+2/rds2/r+2.Then from (114)–(116), we get (117)STu0,u1,θ0-STv0,v1,θ~0H0≤ε+ϖTu0,u1,θ0,v0,v1,θ~0for all (u0,u1,θ0),(v0,v1,θ~0)∈B.
Let(un0,un1,θn0) be a given sequence of initial data in B. Then the corresponding sequence (un,utn,θn) of solutions of the problem (8)–(11) is uniformly bounded in H0, because B is bounded and positively invariant. So {un} is bounded in C([0,∞),H02(Ω))∩C1([0,∞),L2(Ω)). Since H02(Ω)↪H01(Ω) compactly, there exists a subsequence unk which converges strongly in C([0,T],H01(Ω)). Therefore (118)limk→∞liml→∞∫0T∇unks-∇unls+∇unks-∇unls2+∇unks-∇unls2r+2/rds=0,limk→∞liml→∞ϖTunk0,unk1,θnk0,unl0,unl1,θnl0=0.So S(t) is asymptotically smooth in H0. That is, Lemma 12 holds. Thus Theorem 14 is proved.In view of Theorems13 and 14, we have the following.Theorem 15.
The corresponding semigroupS(t) of problem (8)–(11) has a compact global attractor in the phase space H0.



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*Source: 2908964-2017-03-22.xml*