# Global Attractor for Coupled Beam Equations with Nonhomogeneous Boundaries Conditions

Authors: Yuying Li; Haifeng Zhang
Journal: Mathematical Problems in Engineering
 (2018)

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## Abstract

We study the longtime behavior for thermoelastic coupled beam equations with strong damping and past history thermal memoryutt+uxxxx+uxxxxt+θxx-M∫0lux2dxuxx=hx, in 0,l×R+, and θt-ιθxx-1-ι∫0∞ksθxxt-sds-uxxt+pθ=qx, with 1/2≤ι<1, where we consider the nonlinear boundary conditions u0,t=ux0,t=uxxl,t=0, uxxxl,t+uxxxt(l,t)-M∫0lux2dxux(l,t)=ful,t+gutl,t,θ0,t=θxl,t=0 and certain initial conditions. To this problem little is known with nonlinear boundary conditions. This system arises from a model of the nonlinear thermoelastic coupled vibration beam with a nonlinear damping acting on x=l.


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## Body

## 1. Introduction

In this paper, we are concerned with the longtime dynamics for the 1-dimensional nonlinear thermoelastic coupled beam equations with strong damping and past history thermal memory(1)utt+uxxxx+uxxxxt+θxx-M∫0lux2dxuxx=hx,in0,l×R+,(2)θt-ιθxx-1-ι∫0∞ksθxxt-sds-uxxt+pθ=qx,with12≤ι<1.on which we consider the nonlinear boundary conditions (3)u0,t=ux0,t=uxxl,t=0,(4)uxxxl,t+uxxxtl,t-M∫0lux2dxuxl,t=ful,t+gutl,t,(5)θ0,t=θxl,t=0for every t>0 and the initial conditions (6)ux,0=u0x,utx,0=u1x,θx,0=θ0xfor every x∈0,l. This problem arises from a model of the nonlinear thermoelastic coupled vibration beam with a nonlinear damping acting on the x=l end, which simultaneously considers the medium damping, the viscous effect, the nonlinear constitutive relation, and thermoelasticity based on a theory of non-Fourier heat flux.Here the unknown functionsux,t and θx,t are the elevation of the surface of beam and vertical component of the temperature gradient, respectively, u0(x), u1(x) and θ0(x) are the given initial value functions, the subscripts t and x denote the derivatives with respect to t and x, respectively, M· is the nonlinearity of the material and continuous nonnegative nonlinear real function, f(u(l)) and p(θ) and g(ut(l)) are essentially ulρu(l) and θϱθ and utlrut(l), respectively, with ρ,ϱ,r>0, k:R+↦R is memory kernel and assumed to be a positive bounded convex function vanishing at infinity, and h(x) is the lateral load distribution, q(x) is the external heat supply. The detailed assumptions on nonlinear functions σ·, f·, p·, and g· and the external force functions h(x) and q(x) will be specified later.It is well known that the infinite dimensional dynamical systems determined by the elastic beam are different because of the difference of boundary conditions. Berti et al. [1] and Fastovska T. [2] and Giorgi et al. [3] and Wu [4] all considered the longtime dynamics of one-dimensional thermoelastic coupled beam equations under null boundary conditions. Barbose et al. [5] studied the longtime behavior for a class of two-dimensional thermoelastic coupled beam equations (7)utt+△2u-M∫Ω∇u2dx△u-△utt+fu+ν△θ=hxθt-ι△θ-1-ι∫0∞kt-s△θds-ν△ut=0subjected to the hinged conditions (8)u=△u=0,θ=0,x∈Γ.Dell’Oro et al. [6] dealt with the long-term properties of the thermoelastic nonlinear string-beam system related to the well-known Lazer-McKenna suspension bridge model (9)utt+uxxxx-β+uxL20,12uxx+u-v++θxx=fvtt-vxx-u-v++φx=gθt-θxx-utxx=hφt-φxx+vtx=0We also refer the reader to [7–11] and the references therein for thermoelastic coupled beam equations with null boundary conditions. However, abovementioned longtime dynamics for the thermoelastic coupled beam equations are all subjected to null boundary conditions.Under nonlinear boundary conditions, we refer the reader to the following works. One of the first studies in this direction was done by Pazoto and Perla Menzala [12], where stabilization of a thermoelastic extensible beam was considered. Motivated by the result, Ma proved the existence of global solutions and the existence of a global attractor for the Kirchhoff-type beam equation (10)utt+uxxxx-M∫0Lux2dxuxx=qx,under the nonlinear boundary conditions (11)u0,t=ux0,t=uxxL,t=0,uxxxL,t-M∫0Lux2dxuxL,t=fuL,t+gutL,t,in [13] and [14], respectively. In addition, we also found the works [15].In this paper, we use the method of the asymptotically compact property of the solution semigroupSt to prove the existence of a global attractor for the system (1)-(6).


## 2. Basic Assumptions and Basic Lemmas

Our fundamental assumptions onM·, f·, p·, g·, μ(s) and h(x), and q(x) are given as follows.We assume thatM·∈C1(R) satisfying (12)Mzz≥M^z≥0,∀z≥0where M^z=∫0zMsds. This condition is promptly satisfied if M· is nondecreasing with M0=0.The functionf·:R→R is of class C1(R) and satisfies f0=0, and there exist constants k and ρ>0 such that (13)fu-fv≤k11+uρ+vρu-v,∀u,v∈R,(14)-a0≤f^u≤12fuu+a1,where f^z=∫0zfsds.The functionp·:R→R is of class C1(R) and satisfies p(0)=0, and there exist constants k2 and ϱ>0 such that (15)pθ-pθ~θ-θ~≥k2θ-θ~ϱ+2,∀θ,θ~∈R,The function g·:R→R is of class C1(R) and satisfies g(0)=0, and there exist constants k3,k4 and r>ϱ such that (16)gu-gvu-v≥k3u-vr+2,∀u,v∈R,and(17)gu-gv≤k41+ur+vru-v,∀u,v∈R,The assumptions on k(s) are as follows: k is vanishing at ∞; moreover (18)-1-ιk′s=μswhere μ∈C1(R+)∩L1(R+) and there exists a constant δ1>0, ∀s∈R+ such that (19)μ′s≤-δ1μs,s≥0.Finally, we assume that hx,qx∈L20,l.To prove the main result, we need the following Lemmas.Lemma 1 (see [16]).
Letφ(t) be a nonnegative continuous function defined on 0,T, 1<T≤∞, which satisfies supt≤s≤t+1φs1+η≤M0(φ(t)-φ(t+1))+M1, 0≤t≤T-1, where M0,M1,η are positive constants. Then we have (20)φt≤M0-1ηt-1++sup0≤s≤1⁡φs-η-1/η+M11/η+1,0≤t≤TLemma 2 (see [17]).
Assume that, for any bounded positive invariant setB⊂H and for any ε>0, there exists T=T(ε,B) such that d(S(T)x,S(T)y)≤ε+ϖT(x,y),∀x,y∈B, where ϖT:H×H→R satisfies, for any sequence zn⊂B, liminfm→∞liminfn→∞ϖTzn,zm=0. Then S(t) is asymptotically smooth.
Under our hypotheses onM·, f·, p·, g·, k(s), h(x), and q(x), we can derive the existence and uniqueness of global solutions and the existence of a global attractor for 1-dimensional nonlinear thermoelastic coupled equations systems (1)-(6) step by step.


## 3. Basic Space and Global Solutions

First we proceed as in Barbose et al. [5], Giorgi [18], and Dafermos [19] and define a new variable χ=χt(x,s) by (21)χ=χtx,s=∫0sθx,t-τdτ,t,s∈0,∞×R+From the definition of χ, for all t≥0, we have χt(x,0)=0 in 0,l,t∈R+ and χ0(s)=χ0(s) in 0,l, s∈R+, where χ0(s)=∫0sθ0τdτ, s∈R+. Differentiate (21) with respect to t and s on both sides, respectively, and make the sum to get (22)χt+χs=θx,tin0,l×R+×R+So (23)χxxt+χxxs=θxxx,tin0,l×R+×R+Thus (24)χxxs=θxxt-sTherefore thermal memory can be rewritten to be (25)-∫0∞ksθxxt-sds=-∫0∞ksdχxx=-ksχxx0∞+∫0∞k′sχxxdsThen, from the assumption (19) of kernel k(s), problems (1)-(6) are transformed into the new system (26)utt+uxxxx+uxxxxt+θxx-M∫0lux2dxuxx=hx(27)θt-ιθxx-∫0∞μsχxxtsds-uxxt+pθ=qx(28)χtts=θt-χsts,in0,l×R+×R+with the initial conditions (29)ux,0=u0x,utx,0=u1x,θx,0=θ0xin0,l(30)χ0x,s=χ0x,sin0,l×R+and nonhomogeneous boundary conditions(31)u0,t=ux0,t=uxxl,t=0,uxxxl,t+uxxxtl,t-M∫0lux2dxuxl,t=ful,t+gutl,t,θ0,t=θxl,t=0,χt0,s=χxtl,s=0.Our analysis is based on the following Sobolev spaces. Let(32)L20,l,U=u∈H10,l;u0=0V=u∈H20,l;u0=ux0=0W=u∈H40,l∩V;uxxl=0X=θ∈H10,l,θ0=0Y=θ∈H20,l;θ0=θxl=0and with respect to the new variable χ, we define the weighted space (33)Lμ2R+,X=χ:R+⟶X∣∫0∞μsχX2dswhich is a Hilbert space with innerproduct and the norm defined by(34)χ,χ~=χ,χ~Lμ2R+,X=∫0∞μs∫0lχxχ~xdxds,χμ,X2=∫0∞μsχX2dsMotivated by the boundary condition (4), we assume, for regular solutions, that data u0 and u1 satisfy the compatibility condition (35)uxxx0l+uxxx1l-Mux02ux0l=fu0l+gu1lThen for regular solutions we consider the phase space (36)H1=u0,u1,θ,χ∈W×W×Y×Lμ2R+,Ysatisfing35.In the case of weak solutions we consider the phase space (37)H0=V×L20,l×L20,l×Lμ2R+,XIn H0 we adopt the norm defined by(38)u,v,θ,χH02=uxx2+v2+θ2+χμ,X2.Using the classical Galerkin method, we can establish the existence and uniqueness of regular solution and weak solution to problems (26)-(31) as in the work of Ma et al. [14] and Cavalcanti et al. [20].Theorem 3.
Under the assumptions onM·, f·, p·, g·, k(s), h(x), and q(x), for any initial data (u0,u1,θ0,χ0)∈H1, there exists a unique regular solution (u,θ,χ) of problems (26)-(31) such that(39)u∈Lloc∞R+;V∩C00,∞;V∩C10,∞;L20,l,utt∈L∞R+;L20,lθ,θt∈L∞R+,L20,l,χ,χt∈L∞R+,Lμ2R+,X.In addition, if the initial data (u0,u1,θ0,χ0)∈H0, there exists a unique weak solution (u,θ,χ) of problems (26)-(31) such that (40)u,θ,χ∈CR+,H0which depends continuously on initial data with respect to the norm of H0. What is more, in both cases, is that (41)ut2+uxx2+M^ux2+θ2+χtμ2≤Cwhere C is a constant and C denotes different constant in different expression in this paper.
The existence of weak solutions follows from density arguments as shown in [14, 21].
Theorem3 implies that problems (26)-(31) define a nonlinear C0−semigroup S(t) on H0. Moreover, the operator S(t) defined in H0 meets the usual semigroup properties S(t+τ)=S(t)S(τ), S(0)=I, ∀t,τ∈R.


## 4. Global Attractor

The main result of this paper reads as follows.Theorem 4.
Assume the hypotheses of Theorem3; then the corresponding semigroup S(t) of problem (26)-(31) 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 (42)Et=12ut2+uxx2+M^ux2+θ2+f^ul-∫0lhudxAlso, since u(0)=ux(0)=0, the following inequalities hold: (43)ul2≤∫0ldx∫0lux2dx=lux2and (44)uxl2≤∫0ldx∫0luxx2dx=luxx2Let us fix an arbitrary bounded set B⊂H0 and consider the solutions of problems (26)-(31) given by ut,utt,θt=Stu0,u1,θ0 with u0,u1,θ0∈B. Our analysis is based on the modified energy function (45)E~t=Et+a0+1λ1h2≥0,withf^1u≥-a0where λ1>0 is the first eigenvalue of the operator ∂xx in H2(0,l). It is easy to see that E~(t) dominates ut,utt,θtH02 and uxxs2≤4E~(s).
By multiplying (26) by u and integrating over 0,l(46)uxx2=ut2-ddtut,u-∫0luxxtuxxdx+uxlθl-∫0lθuxxdx-M∫0lux2dxux2-fulul-gutlul+∫0lhudxand inserting it into E~(t), then integrating it from t1 to t2, and considering (12) and (14), we obtain that (47)∫t1t2E~tds≤∫t1t2ut2ds-12∫0lutt2ut2dx-∫0lutt1ut1dx-∫t1t212∫0luxxtuxxdxds+∫t1t212uxlθlds-12∫t1t2∫0lθuxxdxds-12∫t1t2gutlulds+a0+a1+1λ1q12-12∫t1t2∫0lhudxds+12∫t1t2θ2dswhere t1,t2∈t,t+1.
Now let us begin to estimate the right hand side of (47) to use the above Lemma 1 of Nakao.
First, multiplying (26) by ut and integrating over 0,l, multiplying (27) by θ and integrating over 0,l, and then taking the sum and integrating from t to t+1, we get (48)∫tt+1uxxt22dτ+ι∫tt+1θx2dτ+∫tt+1∫0lpθθdxdτ+∫tt+1gutlutlds=Et-Et+1+∫tt+1∫0lqθdxdτ+∫tt+1uxtlθldτUsing Schwarz inequality and Young inequality with ϱ+1/ϱ+2+1/ϱ+2=1 and imbedding inequality, we get (49)∫tt+1∫0lqθdxdτ≤∫tt+1qθdτ≤ε-1/ϱ+1lϱ/2+14qϱ+2/ϱ+1+εlϱ/2+1∫tt+1θϱ+2dτ≤ε-1/ϱ+1l(ϱ/2+1)4qϱ+2/ϱ+1+ε∫tt+1θϱ+2ϱ+2dτConsidering (44) and using Young inequality, we have (50)∫tt+1uxtlθldτ≤∫tt+1uxxtθxdτ≤∫tt+113ιuxxt2+3ι4θx2dτWith the assumption (15) of f·, we have (51)12∫tt+1∫0lfθθdxdτ≥k22∫tt+1∫0lθϱ+2dxdτ=k22∫tt+1θϱ+2ϱ+2dτAs ε≤k2/2 and 1/2≤ι<1, combined with (50), from (48) we obtain (52)13∫tt+1uxxt2dτ+ι4∫tt+1θx2dτ+12∫tt+1∫0lpθθdxdτ+∫tt+1gutlutlds≤Et-Et+1+ε-1/ϱ+1lϱ/2+14qϱ+2/ϱ+1Taking into account assumption (15) of p· and assumption (16) of g·, we have (53)Et+ε-1/ϱ+1lϱ/2+14qϱ+2/ϱ+1≥Et+1,and then we define an auxiliary function I2(t) by putting (54)I2t=Et+ε-1/ϱ+1l(ϱ/2+1)4qϱ+2/ϱ+1-Et+1≥0.And we can get(55)∫tt+1uxxt22dτ≤3I2t,∫tt+1θx2dτ≤4ιI2t∫tt+1∫0lpθθdxdτ≤2I2t,∫tt+1gutlutldxdτ≤I2tUsing twice holder inequality with r/r+2+2/r+2=1 and considering the assumption (16) of g·, we obtain that (56)∫tt+1ut2ds=∫tt+1∫0lut2dxds≤∫tt+1∫0l1r+2/rdxr/r+2∫0lutr+2dxr+2/2ds=lr/r+2∫tt+1∫0lutr+2dxr+2/2ds≤lr/r+2∫tt+11r+2/rr/r+2∫tt+1∫0lutr+2dxds2/r+2≤lr/r+2∫tt+1∫0lutr+2dxds2/r+2≤lr/r+21k3∫tt+1∫0lgututdxds2/r+2≤lr/r+21k3It4/r+2Since (56), 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 (57)utt12≤lr/r+24k3It4/r+2andutt22≤lr/r+24k3It4/r+2Thus from Schwarz inequality combined with (57), noting that uxx(s)2≤4E~(s), we have (58)12∫0lutt2ut2dx-∫0lutt1ut1dx≤12λ1utt2uxxt2+utt1uxxt1≤1λ1lr/2r+22k3It2/r+2supt≤s≤t+1⁡uxx≤1λ1ηlr/r+24k3It4/r+2+ηsupt≤s≤t+1⁡E~sUsing Schwarz inequality and Young inequality combined with the first inequality of (55), we get (59)12∫t1t2∫0luxxuxxtdxds≤34ηI2t+ηsupt≤s≤t+1⁡E~sConsidering (44) and using Young inequality combined with the second inequality of (55), we have (60)∫t1t212uxlθlds≤∫t1t2uxxθxds≤∫t1t21ηθx2+η4uxx2ds≤4ηιI2t+ηsupt≤s≤t+1E~sUsing twice Holder inequality with ϱ/ϱ+2+2/ϱ+2=1, combined with the assumption (15) on p(θ) and the fourth inequality of (55), we have (61)12∫tt+1θ2ds≤12∫tt+1∫0l1ϱ+2/ϱdxϱ/ϱ+2∫0lθϱ+2dxϱ+2/2ds≤12lϱ/ϱ+2∫tt+11ϱ+2/ϱdsϱ/ϱ+2∫tt+1∫0lθϱ+2dxds2/ϱ+2≤12lϱ/ϱ+2∫tt+11k2∫0lpθθdxds2/ϱ+2≤12lϱ/ϱ+22k22/ϱ+2It4/ϱ+2Thus using Young inequality and considering (61), we have (62)-12∫t1t2∫0lθuxxdxds≤∫t1t214ηθ2ds+∫t1t2η4uxx2ds≤14η2k22/ϱ+2lϱ/ϱ+2It4/ϱ+2+ηsupt≤s≤t+1⁡E~sConsidering the assumptions (17) of g· combined with (43), we obtain (63)12∫t1t2gutlulds≤k42∫t1t2lutluxds+k42∫t1t2utlr+1luxdsWith(64)ux2=uxx,t-ux0,t2=∫0xuxxdx2≤∫0xdx∫0xuxx2dx≤x∫0xuxx2ddxwe get (65)ux2=∫0lux2dx≤∫0lx∫0xuxx2dxdx≤∫0lxdx∫0luxx2dx≤l2uxx2Using Young inequality combined with (43) and (65) and the first inequality of (55), we have (66)k42∫t1t2lutluxds≤k4l3/22∫t1t2utluxxds≤1ηk4l3/222∫t1t2utl2ds+η4∫t1t2uxx2ds≤1ηk4l3/222l∫t1t2uxt2ds+η4supt≤s≤t+1⁡uxx≤1ηk4l3/222l3∫t1t2uxxt2ds+ηsupt≤s≤t+1⁡E~s≤3ηk4l3/222l3I2t+ηsupt≤s≤t+1⁡E~sand using twice Holder inequality with r+1/r+2+1/r+2=1 combined with the embedding theorem and Young inequality, (67)k42∫t1t2utlr+1luxds≤k4l2∫t1t2utr+2ldsr+1/r+2∫t1t2uxr+2ds1/r+2≤1k3r+1/r+2k4l2I2r+1/r+2t∫t1t2∫0l1r+2/rdxr/r+2∫0luxr+2dxds1/r+2=1k3r+1/r+2k4llr/r+222I2r+1/r+2t∫t1t2∫0luxr+2dxds1/r+2=1k3r+1/r+2k4llr/r+222I2r+1/r+2t∫t1t2uxr+2r+2ds1/r+2≤1k3r+1/r+2k4llr/r+222I2r+1/r+2tsupt≤s≤t+1⁡uxsr+2≤1k3r+1/r+2k4llr/r+222I2r+1/r+2tsupt≤s≤t+1⁡uxx≤1η1k3r+1/r+2k4llr/r+2222I4r+1/r+2t+η4supt≤s≤t+1⁡uxx2≤1η1k3r+1/r+2k4llr/r+2222I4r+1/r+2t+ηsupt≤s≤t+1⁡Es~Thus we can obtain (68)12∫t1t2gutlulds≤3ηk4l3/222l3I2t+1η1k3r+1/r+2k4llr/r+2222I4r+1/r+2t+2ηsupt≤s≤t+1⁡E~sFinally using Young inequality, we get that (69)12∫t1t2∫0lhutdxds≤14ηλ1h2+ηsupt≤s≤t+1⁡E~sInserting (56), (58)-(62), and(68)-(69) into (47), we obtain (70)∫t1t2E~sds≤lr/r+21k3+1λ1ηlr/r+24k3It4/r+2+3η4+4ηι+3ηk4l3/222l3I2t+12lϱ/ϱ+22k22/ϱ+2+14η2k22/ϱ+2lϱ/ϱ+2I4/ϱ+2t+1η1k3r+1/r+2k4llr/r+2222I4r+1/r+2t+7ηsupt≤s≤t+1⁡E~s+a0+a1+1λ1+14ηλ1h2For the left hand side of (70), we use the Mean Value Theorem; then there exists number τ∈t1,t2 such that (71)∫t1t2E~sds≥12E~t+1=12E~t-It2+ε-1/ϱ+1lϱ/2+18qϱ+2/ϱ+1So we conclude that (72)E~t≤It2+2∫t1t2E~sds-ε-1/ϱ+1l(ϱ/2+1)8qϱ+2/ϱ+1Inserting (70) into (72), we obtain that (73)E~t≤2lr/r+21k3+1λ1ηlr/r+24k3It4/r+2+1+23η4+4ηι+3ηk4l3/222l3I2t+212lϱ/ϱ+22k22/ϱ+2+14η2k22/ϱ+2lϱ/ϱ+2I4/ϱ+2t+2η1k3r+1/r+2k4llr/r+2222I4r+1/r+2t+14ηsupt≤s≤t+1⁡E~s+2a0+a1+1λ1+214ηλ1h2+ε-1/ϱ+1l(ϱ/2+1)8qϱ+2/ϱ+1Letting 0<η<1/14, considering the boundedness of It2r/r+2, It4(r-ϱ)/(r+2)(ϱ+2), and It4r/r+2 and then setting C1=1/1-14η{2(lr/r+21/k3 + 1/λ1ηlr/r+24/k3) + {[1+(3η/4+4/ηι + 1/ηk4l3/2/22l3)] + 2(1/2lϱ/ϱ+2·2/k22/ϱ+2 + 1/4η2/k22/ϱ+2lϱ/ϱ+2) + 2/η·1/k3r+1/r+2k4llr/r+22/22}C}, from (73) we get (74)E~t≤C1It4/r+2+21-14ηa0+a1+11-14η21λ1+14ηλ1h2+ε-1/ϱ+1lϱ/2+18qϱ+2/ϱ+1,and then (74) can be rewritten as (75)E~t1+r/2≤C1E~t-E~t+1+21-14ηa0+a1+11-14η21λ1+14ηλ1h2+ε-1/ϱ+1lϱ/2+18qϱ+2/ϱ+11+ϱ/2Using Nakao’s Lemma 1, we conclude that (76)E~t≤C1-1r2t-1++E~0-r/2-2/r+21-14ηa0+a1+11-14η2λ1+24ηλ1h2+ε-1/ϱ+1lϱ/2+18qϱ+2/ϱ+1As t→∞, the first term of the right side of (76) goes to zero; thus with E~t≥1/4uxx2+vx2+θ2+χμ2, we conclude that(77)B=u,v,θ,χ∈H0∣uxx2+vx2+θ2+χμ2≤81-14ηa0+a1+41-14η2λ1+24ηλ1h2+ε-1/ϱ+1lϱ/2+18qϱ+2/ϱ+1is an absorbing set for S(t) in H0. Theorem 4 is proved.The main result of a global attractor reads as follows.Theorem 5.
Assume the hypotheses of Theorem3; then the corresponding semigroup S(t) of problems (26)-(31) is asymptotically smooth in space H0.Proof.
We are going to apply Lemmas1 and 2 to prove the asymptotic smoothness. Given initial data (u0,u1,θ0,χ0) and (v0,v1,θ~0,χ~0)∈B in a bounded invariant set B⊂H0, let (u,θ,χ), (v,θ~,χ~) be the corresponding weak solutions of problems (27)–(31). Then the differences w=u-v, ϑ=θ-θ~, Π=χ-χ~ are the weak solutions of (78)wtt+wxxxx+wxxxxt+ϑxx-Mux2wxx=ΔMvxx,ϑt-ιϑxx-∫0∞μsΠxxtsds-wxxt+Δp=0,Πtt=-Πst+ϑ,w0=u0-v0,wt0=u1-v1,ϑ0=θ0-θ~0,Π0=χ0-χ~0,w0=wx0=wxxl=0,ϑ0=ϑxl=0,Π0=Πxl=0,wxxxl+wxxxtl-Mux2wxl,t-ΔMvxl,t=Δf+Δgwhere (79)ΔM=Mux2-Mvx2,Δp=pθ-pθ~,Δf=ful,t-fvl,t,Δg=gutl,t-gvtl,tLet us define(80)Ewt=wt2+wxx2+Mux2wx2+ϑ2+Πμ2.We can assume formally that w is sufficiently regular by density. Then, multiplying the first equation in (78) by wt and integrating over 0,l, multiplying the second equation in (78) by ϑ and integrating over 0,l and taking the inner with Π for the third equation in (78) in space Lμ2(R+,X), and then taking the sum, we get (81)12ddtEwt+wxxt2+ιϑx2+∫0∞μs∫0lΠxsΠxdxds+∫0lΔpϑdx+Δgwtl=-M′ux2wx2∫0luxxutdx-ΔMvxlwtl-wxtlϑl+∫0lΔMvxxwtdx-Δfwtl
Let us estimate the right hand side of (81).
Considering the continuity ofM′· and the estimates (41) (82)-M′ux2wx2∫0luxxutdx≤Cwx2.Applying the mean value theorem, then considering (44) and the estimates (41), we have (83)-ΔMvxlwtl≤M′ξ1wxux+vxvxlwtl≤Cwxwtl≤Cwxr+2/r+1+k34wtlr+2where ξ1 is among ux2 and vx2. Also considering (43) and (44), then using Young inequality, we have (84)wxtlϑldτ≤wxxtϑx≤wxxtϑx≤13ιwxxt2+3ι4ϑx2Also applying the mean value theorem combined with the estimates (41), we get (85)ΔM∫0lvxxwtdx≤Cwx.Considering that the assumption (13) of f· combined with the estimates (41) and (43), then Young inequality (r+1/r+2+1/r+2=1) (86)-Δfwtl≤k11+luxρ+lvxρlwxwtl≤3k1Cwxwtl≤Cwxr+2/r+1+k34wtlr+2On the other hand, combining with the assumption (19) on μ(s)(87)∫0∞μs∫0lΠxsΠxdxds=-12∫0∞μ′sΠx2ds≥δ12Πμ2Considering the assumption (15) on p·(88)∫0lΔpϑdx≥k2ϑϱ+2ϱ+2,Also by view of the assumption (16) on g·, we have (89)Δgwtl≥k3utl-vtlr+2=k3wtlr+2,Thus by inserting (82)-(89) into (81), with 1/2≤ι<1, we get that (90)12ddtEwt+13wxxt2+ι4ϑx2+δ12Πμ2+k32wtlr+2+k2ϑϱ+2ϱ+2≤Cwx+wx2+wxr+2/r+1Then integrating from t to t+1 and defining an auxiliary function F2(t), we get (91)13∫tt+1wxxt2ds+ι4∫tt+1ϑx2ds+∫tt+1δ12Πμ2ds+k32∫tt+1wtlr+2ds+k2∫tt+1ϑϱ+2ϱ+2ds≤12Ewt-Ewt+1+C∫tt+1wx+wx2+wxr+2/r+1ds=Ft2Thus we can get (92)13∫tt+1wxxt2ds≤Ft2,ι4∫tt+1ϑx2ds≤Ft2,∫tt+1δ12Πμ2ds≤Ft2,k32∫tt+1wtlr+2ds≤F2t,(93)k2∫tt+1ϑϱ+2ϱ+2ds≤Ft2Then by multiplying first equation in (78) by w and integrating over 0,l, then integrating from t1 to t2, we get that (94)∫t1t2wxx2+σux2wx2ds=-∫0lwtt2wt2dx+∫0lwtt1wt1dx+∫t1t2wt2dt-∫t1t2Δfwlds-∫t1t2Δgwlds-∫t1t2ΔMvxlwlds+∫t1t2ΔM∫0lvxxwdxds+wxlϑl-∫t1t2∫0lϑwxxdxdsNow let us estimate the right hand side of (94). Firstly, from the first inequality of (92), we infer that (95)∫tt+1wt2ds≤∫tt+11λ1wxxt2ds≤3λ1F2tThus we combine with the first inequality of (92), and there exists t1∈t,t+1/4 and t2∈t+3/4,t+1 such that (96)wtt12≤6λ1F2tandwtt22≤6λ1F2t,then we can deduce that (97)-∫0lwtt2wt2dx+∫0lwtt1wt1dx≤26λ1Ftsupt≤σ≤t+1⁡wxx≤48λ12Ft2+18supt≤σ≤t+1⁡Ewσ.The assumption (13) on f· combined with (43) and the estimates (41) implies that (98)∫t1t2J3wldxds≤∫t1t2k11+uρ+vρw2ldxds≤C∫t1t2wx2ds,By the assumption (17) on g· combined with (43) and the estimates (41), and using the Holder inequality and Young inequality, we have (99)-∫t1t2Δgwlds≤k4∫t1t21+utlr+vtlrwtlwlds≤k4∫t1t21+luxtr+lvxtrlwxtlwxds≤C∫t1t2wxtwxds≤8λ2C2∫t1t2wxt2ds+λ232∫t1t2wx2ds≤8λ2C2∫t1t2wxt2ds+132∫t1t2wxx2ds≤8λ2C2∫t1t2wxxt2ds+132supt≤σ≤t+1⁡Ewσ≤CFt2+132supt≤σ≤t+1⁡Ewσwhere λ2>0 is the first eigenvalue of the operator ∂x in H1(0,l). Applying the Mean value theorem and using Schwarz inequality and Young inequality, we have (100)-∫t1t2ΔMvxlwlds≤∫t1t2σ′ξ2wxux+vxvxlwtlds≤∫t1t2Cwxwtlds≤∫t1t2Cwxr+2/r+1ds+∫t1t2k34wtlr+2dswhere ξ2 is among ux2 and vx2. Also apply the Mean value theorem combined with the estimates (41) to get (101)∫t1t2J1∫0lvxxwdxds≤C∫t1t2wx2ds,Using Young inequality,we have (102)∫t1t2-wxlϑlds≤∫t1t2wxxϑxds≤∫t1t28ηϑx2+η32wxx2ds≤32ηιF2t+η8supt≤s≤t+1⁡E~sFinally, using twice Holder inequality with ϱ/ϱ+2+2/ϱ+2=1, combined with the fifth inequality of (92), we have (103)∫tt+1ϑ2ds≤∫tt+1∫0l1ϱ+2/ϱdxϱ/ϱ+2∫0lϑϱ+2dxϱ+2/2ds≤lϱ/ϱ+2∫tt+11ϱ+2/ϱdsϱ/ϱ+2∫tt+1∫0lϑϱ+2dxds2/ϱ+2≤lϱ/ϱ+21k22/ϱ+2Ft4/ϱ+2,thus by Schwarz inequality and Young inequality, we get (104)∫t1t2∫0lϑwxxdxds≤8lϱ/ϱ+21k22/ϱ+2Ft4/ϱ+2+132supt≤σ≤t+1⁡EwσBy inserting (95)and (97)-(104) into (94), we obtain that (105)12∫t1t2wxx2+σux2wx2ds≤C∫tt+1wx2ds+92lϱ/ϱ+21k22/ϱ+2Ft4/ϱ+2+123λ1+48λ12+C+32ηιF2t+532supt≤σ≤t+1⁡EwσThen from the definition of Ew(t) combined with the third inequality of (92), and (95) and (103), we obtain that (106)12∫t1t2Ewsds≤C∫tt+1wx2ds+5lϱ/ϱ+21k22/ϱ+2Ft4/ϱ+2+126λ1+48λ12+C+32ηι+1δ1F2t+532supt≤σ≤t+1⁡EwσFor (106), by using Mean Value theorem, there exists t∗∈t1,t2 such that (107)12t2-t1Ewt∗≤C∫tt+1wx2ds+5lϱ/ϱ+21k22/ϱ+2Ft4/ϱ+2+126λ1+48λ12+C+32ηι+1δ1F2t+532supt≤σ≤t+1⁡EwσConsidering that t1∈t,t+1/4 and t2∈t+3/4,t+1, we have (108)Ewt∗≤C∫tt+1wx2ds+20lϱ/ϱ+21k22/ϱ+2Ft4/ϱ+2+26λ1+48λ12+C+32ηι+1δ1F2t+58supt≤σ≤t+1⁡EwσFrom (91), we see that (109)Ewt≤Ewt+1+2F2tLet Ew(τ)=supt≤σ≤t+1Ew(σ) with τ∈t,t+1, then integrate (90) over t,τ and over t∗,t+1 to have (110)supt≤σ≤t+1⁡Ewσ=Ewτ≤Ewt+C∫tt+1wx+wx2+wxr+2/r+1ds≤Ewt+1+2F2t+C∫tt+1wx+wx2+wxr+2/r+1ds≤Ewt∗+2F2t+C∫tt+1wx+wx2+wxr+2/r+1dsInserting (108) into (110), we obtain (111)38supt≤σ≤t+1⁡Ewσ≤C∫tt+1wx+wx2+wxr+2/r+1ds+20lϱ/ϱ+21k22/ϱ+2Ft4/ϱ+2+CF2tTherefore from the boundary of Ft2ϱ/ϱ+2, we have (112)supt≤σ≤t+1⁡Ewσ≤CFt4/ϱ+2+Csup0≤σ≤T⁡∫σσ+1wx+wx2+wxr+2/r+1dsTherefore (113)supt≤σ≤t+1⁡Ewσ1+ϱ/2≤CEwt-Ewt+1+Csup0≤σ≤T∫σσ+1wx+wx2+wxr+2/r+1dsFrom Nakao’s Lemma 1, there exists CB>0 and CT>0 such that (114)Ewt≤CBt-1+-2/ϱ+CT(sup0≤σ≤T⁡∫σσ+1wx+wx2+wxr+2/r+12/ϱ+2,0≤t≤TFrom the definition of Ewt, we have (115)w,wt,ϑ,ΠH0≤CBt-1+-2/ϱ+CT(sup0≤σ≤T⁡∫σσ+1wx2+wxr+2/r+12/ϱ+2.Given ε>0, we choose T large such that (116)CBt-1+-2/ϱ≤εand define ϖT:H0×H0→R as (117)ϖTu0,u1,θ0,χ0,v0,v1,θ~0,χ~0=CTsup0≤σ≤T⁡∫σσ+1wx+wx2+wxr+2/r+1ds2/ϱ+2.Then from (115)-(117), we get (118)STu0,u1,θ0,χ0-STv0,v1,θ~0,χ~0H0≤ε+ϖTu0,u1,θ0,χ0,v0,v1,θ~0,χ~0for all (u0,u1,θ0,χ0),(v0,v1,θ~0,χ~0)∈B.
Let(un0,un1,θn0,χn0) be a given sequence of initial data in B. Then the corresponding sequence (un,utn,θn,χn) of solutions of the problems (26)-(31) is uniformly bounded in H0, because B is bounded and positively invariant. So un is bounded in C0,∞,V∩C10,∞,L20,l. Since V↪↪U compactly, there exists a subsequence unk which converges strongly in C0,T,U. Therefore (119)limk→∞⁡liml→∞⁡∫0Tuxnks-uxnls+uxnks-uxnls2+uxnks-uxnlsr+2/r+12/ϱ+2ds=0and (120)limk→∞⁡liml→∞⁡ϖTunk0,unk1,θnk0,χnk0,unl0,unl1,θnl0,χnk0=0.So S(t) is asymptotically smooth in H0. That is, Lemma 2 holds. Thus Theorem 5 is proved.Theorem 6.
The corresponding semigroupS(t) of problems (26)-(27) has a compact global attractor in the phase space H0.Proof.
In view of Theorems3–5, we directly get Theorem 6.



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*Source: 4816794-2018-08-30.xml*