# Global Existence of Solutions for the Viscoelastic Kirchhoff Equation with Logarithmic Source Terms

Authors: Nadia Mezouar; Salah Mahmoud Boulaaras; Ali Allahem
Journal: Complexity
 (2020)

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## Abstract

In this paper, a nonlinear viscoelastic Kirchhoff equation in a bounded domain with a time-varying delay term and logarithmic nonlinearity in the weakly nonlinear internal feedback is considered, where the global and local existence of solutions in suitable Sobolev spaces by means of the energy method combined with Faedo-Galerkin procedure is proved with respect to the condition of the weight of the delay term in the feedback and the weight of the term without delay and the speed of delay. Furthermore, a general stability estimate using some properties of convex functions is given. These results extend and improve many results in the literature.


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## Body

## 1. Introduction

### 1.1. Model

In this paper, we consider the global existence and decay properties of solutions for the initial boundary value problem of the following viscoelastic nondegenerate Kirchhoff equation of the form:(1)utlutt−M∇u2Δu−Δutt+∫0tht−sΔusds+μ1g1utx,t+μ2g2utx,t−τt=νulnu,inΩ×0,+∞,ux,t=0,on∂Ω×0,+∞,ux,0=u0x,utx,0=u1x,inΩ,utx,t−τ0=f0x,t−τ0,inΩ×0,τ0,where Ω is a bounded domain in ℝn, n∈ℕ∗, with a smooth boundary ∂Ω, l>0, ν, μ1, and μ2 are positive real numbers, h is a positive function which decays exponentially, τt>0 is a time-varying delay, g1 and g2 are two functions, and the initial data u0,u1,f0 are in a suitable function space. Mr=a+brγ is a C1-function for r≥0, with a,b>0 and γ≥1.In the absence of delay term (i.e.,μ2=0), Han and Wang in [1] considered the following nonlinear viscoelastic equation with damping:(2)utlutt−Δu−Δutt+∫0tht−sΔusds+utx,t=0,inΩ×0,+∞.Time delay is often present in applications and practical problems. In recent years, the control of PDEs with time delay effects has become an active area of research (see, for example, [2–4]). For example, in [5], it has been proven that a small delay in a boundary control could turn a well-behaved hyperbolic system into a wild one, thus showing that delay can be a source of instability.Wu [6] treated problem (1) for a constant time delay τ and g1x=g2x=x. He proved the local existence result using the Faedo-Galerkin method and established the decay result employing suitable Lyapunov functionals under appropriate conditions on μ1 and μ2 and on the kernel h.Benaissa et al. [7] considered the case of constant time delay τ, with l=0 and Mr=1. They proved the global existence and uniform decay for the following problem:(3)utt−Δu+∫0tht−sΔusds+μ1g1utx,t+μ2g2utx,t−τ=0,inΩ×0,+∞.The same problem (3) was also treated by Kirane and Said-Houari [8] for g1x=g2x=x and a homogeneous right hand side with τ, a constant time delay. Daewook [9] considered a viscoelastic Kirchhoff equation, with a time-varying delay and a nonlinear source term, given as(4)utt−Mx,t,∇u2Δu+∫0tht−sdivax∇usds+umu+μ1utx,t+μ2utx,t−τt=0,inΩ×0,+∞,This equation describes axially moving viscoelastic materials. Using the smallness condition with respect to Kirchhoff coefficient and the relaxation function and by assuming0≤m≤2/n−2 if n>2 or 0≤m if n≤2, he obtained the uniform decay rate of the Kirchhoff-type energy.In [10], the authors studied homogeneous problem (1) without the viscoelastic term, with l=0 and Mr=1. In addition, μ1g1 and μ2g2 are multiplied by a positive nonincreasing function σ of C1R+ satisfying ∫0+∞σsds=+∞ and σ′t≤cσt. They proved the global existence, and using a multiplier method with some properties of convex functions to get decay rate of the energy (when t goes to infinity) depends on the function σ and on the function H which represents the growth at the origin of g1.Apart from the aforesaid attention given to polynomial nonlinear terms, logarithmic nonlinearity has also received a great deal of interest from both physicists and mathematicians. This type of nonlinearity was introduced in the nonrelativistic wave equations describing spinning particles moving in an external electromagnetic field and also in the relativistic wave equation for spinless particles [11]. Moreover, the logarithmic nonlinearity appears in several branches of physics such as inflationary cosmology [12], nuclear physics [13], optics [14], and geophysics [15]. With all this specific underlying meaning in physics, the global-in-time well-posedness of solution to the problem of evolution equation with such logarithmic-type nonlinearity captures lots of attention. Birula and Mycielski [16, 17] studied the following problem:(5)utt−uxx+u−εulnu2=0,ina,b×0,T,ua,t=ub,t=0,0,T,ux,0=u0x,utx,0=u1x,ina,b,which is a relativistic version of logarithmic quantum mechanics and can also be obtained by taking the limit p goes to 1 for the p-adic string equation [18, 19]. In [20], Cazenave and Haraux considered(6)utt−Δu=ulnuk,inℝ3,and they established the existence and uniqueness of the solution for the Cauchy problem. Gorka [21] used some compactness arguments and obtained the global existence of weak solutions, for all(7)u0,u1∈H01Ω×L2a,b,to initial boundary value problem (5) in the one-dimensional case. Bartkowski and Górka [22] proved the existence of classical solutions and investigated the weak solutions for the corresponding one-dimensional Cauchy problem for equation (6). Hiramatsu et al. [23] introduced the following equation:(8)utt−Δu+u+ut+u2u=ulnu,to study the dynamics of Q-ball in theoretical physics and presented a numerical study. However, there was no theoretical analysis for the problem. In [24], Han proved the global existence of weak solutions, for all(9)u0,u1∈H01Ω×L2Ω,to initial boundary value problem (8) in ℝ3.In the present paper, we investigate the stabilization of a dynamic model describing a string with a rigid surface and an interior somehow permissive to slight deformations. This leads to a varying material densityutl and a Kirchhoff term M∇u2 that depends on ∇u2. We prove the existence of global solutions in suitable Sobolev spaces by combining the energy method with the Fadeo-Galerkin procedure. We also establish an explicit and general decay result using a perturbed energy method with some techniques due to Mustafa and Messaoudi [25], as well as some properties of convex functions. These convexity arguments were introduced and developed by Lasiecka et al. [26–28] and used, with appropriate modifications, by Liu and Zuazua [29], Alabau-Boussouira [30], and others.The paper is organized as follows: In Section2, we give some hypotheses and state our main result. Then, in Section 3, we prove the global existence of weak solutions. Furthermore, in Section 4, the uniform decay of the energy is derived.


### 1.2. Formulation of the Results

We denote by (., .) the inner product inL2Ω and the corresponding norm by .2. Now, we introduce, as in [31], the new variable:(10)zx,ρ,t=utx,t−ρτt,x∈Ω,ρ∈0,1,t>0.Then, we have(11)τtztx,ρ,t+1−ρτ′tzρx,ρ,t=0,inΩ×0,1×0,+∞.Therefore, problem (1) is equivalent to(12)utlutt−M∇u2Δu−Δutt+∫0tht−sΔusds+μ1g1utx,t+μ2g2zx,1,t=νulnuinΩ×0,+∞,τtztx,ρ,t+1−ρτ′tzρx,ρ,t=0,inΩ×0,1×0,+∞,ux,t=0,on∂Ω×0,∞,zx,0,t=utx,t,onΩ×0,∞,ux,0=u0x,utx,0=u1x,inΩ,zx,ρ,0=f0x,−ρτ0,inΩ×0,1.To state and prove our result, we need some assumptions.(A1) Assume thatl satisfies(13)0<l≤2n−2,ifn>2,0<l<∞,ifn≤2.(A2) The relaxation functionh:ℝ+⟶ℝ+ is a bounded C1 function, such that(14)a−∫0∞hsds=k>0,h0>0,and suppose that there exists a positive constantζ satisfying(15)h′t≤−ζht.(A3)g1:ℝ⟶ℝ is a nondecreasing function of class C1 and H:ℝ+⟶ℝ+ is convex, increasing and of class C1ℝ+∩C2]0,+∞[, satisfying(16)H0=0andHis linear,on0,εorH′0=0andH″>0,on0,εsuch thatc1s≤g1s≤c2s,ifs≥εs2+g1s2≤H−1sg1s,ifs≤ε,whereε,c1,andc2 are positive constants. g2:ℝ⟶ℝ is an odd nondecreasing function of class C1ℝ such that there exist c3,α1,andα2>0,(17)g2′s≤c3,(18)α1sg2s≤Gs≤α2sg1s,Gs=∫0sg2rdr.(A4)τ is a function in W2,∞0,T, T>0, such that(19)0<τ0≤τt≤τ1,∀t>0,τ′t≤d<1,∀t>0,whereτ0 and τ1 are positive numbers.(A5) We also assume that(20)μ2<α11−dα21−α1dμ1.(A6)(21)1<ν<2πke3.We define the energy associated to the solution of system (12) by(22)Et=1l+2utl+2l+2+b2γ+1∇u2γ+1+12a−∫0thsds∇u2+12∇ut2−ν2∫Ωu2lnudx+ν4u2+12ho∇ut+ξτt∫Ω∫01Gzx,ρ,tdρdx,whereξ is a positive constant such that(23)μ21−α1α11−d<ξ<μ1−α2μ2α2,(24)hovt=∫0tht−sv.,t−v.,s2ds.Theorem 1.
(Global Existence). Letu0,u1,f0∈H2Ω∩H01Ω×H01Ω×H01Ω,H10,1 satisfy the compatibility condition:(25)f0.,0=u1.
Assume that (A1)–(A6) hold under smallness condition on the initial datau0,u1. Then, problem (1) admits a weak solution:(26)u∈L∞0,∞;H2Ω∩H01Ω,ut∈L∞0,∞;H01Ω,utt∈L∞0,∞;L2ΩTheorem 2.
(Uniform Decay Rates of Energy). Assume that (A1)–(A6) hold and ifE0 is positive and bounded, then for every t0>0, there exist positive constants w1,w2,w3, and ε0 such that the solution energy of (1) satisfies(27)Et≤w3H1−1w1t+w2,∀t≥t0,where(28)H1t=∫t11H2sds,H2t=tH′ε0t.Here,H1 is strictly decreasing and convex on 0,1 with limt⟶0H1t=+∞.


## 1.1. Model

In this paper, we consider the global existence and decay properties of solutions for the initial boundary value problem of the following viscoelastic nondegenerate Kirchhoff equation of the form:(1)utlutt−M∇u2Δu−Δutt+∫0tht−sΔusds+μ1g1utx,t+μ2g2utx,t−τt=νulnu,inΩ×0,+∞,ux,t=0,on∂Ω×0,+∞,ux,0=u0x,utx,0=u1x,inΩ,utx,t−τ0=f0x,t−τ0,inΩ×0,τ0,where Ω is a bounded domain in ℝn, n∈ℕ∗, with a smooth boundary ∂Ω, l>0, ν, μ1, and μ2 are positive real numbers, h is a positive function which decays exponentially, τt>0 is a time-varying delay, g1 and g2 are two functions, and the initial data u0,u1,f0 are in a suitable function space. Mr=a+brγ is a C1-function for r≥0, with a,b>0 and γ≥1.In the absence of delay term (i.e.,μ2=0), Han and Wang in [1] considered the following nonlinear viscoelastic equation with damping:(2)utlutt−Δu−Δutt+∫0tht−sΔusds+utx,t=0,inΩ×0,+∞.Time delay is often present in applications and practical problems. In recent years, the control of PDEs with time delay effects has become an active area of research (see, for example, [2–4]). For example, in [5], it has been proven that a small delay in a boundary control could turn a well-behaved hyperbolic system into a wild one, thus showing that delay can be a source of instability.Wu [6] treated problem (1) for a constant time delay τ and g1x=g2x=x. He proved the local existence result using the Faedo-Galerkin method and established the decay result employing suitable Lyapunov functionals under appropriate conditions on μ1 and μ2 and on the kernel h.Benaissa et al. [7] considered the case of constant time delay τ, with l=0 and Mr=1. They proved the global existence and uniform decay for the following problem:(3)utt−Δu+∫0tht−sΔusds+μ1g1utx,t+μ2g2utx,t−τ=0,inΩ×0,+∞.The same problem (3) was also treated by Kirane and Said-Houari [8] for g1x=g2x=x and a homogeneous right hand side with τ, a constant time delay. Daewook [9] considered a viscoelastic Kirchhoff equation, with a time-varying delay and a nonlinear source term, given as(4)utt−Mx,t,∇u2Δu+∫0tht−sdivax∇usds+umu+μ1utx,t+μ2utx,t−τt=0,inΩ×0,+∞,This equation describes axially moving viscoelastic materials. Using the smallness condition with respect to Kirchhoff coefficient and the relaxation function and by assuming0≤m≤2/n−2 if n>2 or 0≤m if n≤2, he obtained the uniform decay rate of the Kirchhoff-type energy.In [10], the authors studied homogeneous problem (1) without the viscoelastic term, with l=0 and Mr=1. In addition, μ1g1 and μ2g2 are multiplied by a positive nonincreasing function σ of C1R+ satisfying ∫0+∞σsds=+∞ and σ′t≤cσt. They proved the global existence, and using a multiplier method with some properties of convex functions to get decay rate of the energy (when t goes to infinity) depends on the function σ and on the function H which represents the growth at the origin of g1.Apart from the aforesaid attention given to polynomial nonlinear terms, logarithmic nonlinearity has also received a great deal of interest from both physicists and mathematicians. This type of nonlinearity was introduced in the nonrelativistic wave equations describing spinning particles moving in an external electromagnetic field and also in the relativistic wave equation for spinless particles [11]. Moreover, the logarithmic nonlinearity appears in several branches of physics such as inflationary cosmology [12], nuclear physics [13], optics [14], and geophysics [15]. With all this specific underlying meaning in physics, the global-in-time well-posedness of solution to the problem of evolution equation with such logarithmic-type nonlinearity captures lots of attention. Birula and Mycielski [16, 17] studied the following problem:(5)utt−uxx+u−εulnu2=0,ina,b×0,T,ua,t=ub,t=0,0,T,ux,0=u0x,utx,0=u1x,ina,b,which is a relativistic version of logarithmic quantum mechanics and can also be obtained by taking the limit p goes to 1 for the p-adic string equation [18, 19]. In [20], Cazenave and Haraux considered(6)utt−Δu=ulnuk,inℝ3,and they established the existence and uniqueness of the solution for the Cauchy problem. Gorka [21] used some compactness arguments and obtained the global existence of weak solutions, for all(7)u0,u1∈H01Ω×L2a,b,to initial boundary value problem (5) in the one-dimensional case. Bartkowski and Górka [22] proved the existence of classical solutions and investigated the weak solutions for the corresponding one-dimensional Cauchy problem for equation (6). Hiramatsu et al. [23] introduced the following equation:(8)utt−Δu+u+ut+u2u=ulnu,to study the dynamics of Q-ball in theoretical physics and presented a numerical study. However, there was no theoretical analysis for the problem. In [24], Han proved the global existence of weak solutions, for all(9)u0,u1∈H01Ω×L2Ω,to initial boundary value problem (8) in ℝ3.In the present paper, we investigate the stabilization of a dynamic model describing a string with a rigid surface and an interior somehow permissive to slight deformations. This leads to a varying material densityutl and a Kirchhoff term M∇u2 that depends on ∇u2. We prove the existence of global solutions in suitable Sobolev spaces by combining the energy method with the Fadeo-Galerkin procedure. We also establish an explicit and general decay result using a perturbed energy method with some techniques due to Mustafa and Messaoudi [25], as well as some properties of convex functions. These convexity arguments were introduced and developed by Lasiecka et al. [26–28] and used, with appropriate modifications, by Liu and Zuazua [29], Alabau-Boussouira [30], and others.The paper is organized as follows: In Section2, we give some hypotheses and state our main result. Then, in Section 3, we prove the global existence of weak solutions. Furthermore, in Section 4, the uniform decay of the energy is derived.


## 1.2. Formulation of the Results

We denote by (., .) the inner product inL2Ω and the corresponding norm by .2. Now, we introduce, as in [31], the new variable:(10)zx,ρ,t=utx,t−ρτt,x∈Ω,ρ∈0,1,t>0.Then, we have(11)τtztx,ρ,t+1−ρτ′tzρx,ρ,t=0,inΩ×0,1×0,+∞.Therefore, problem (1) is equivalent to(12)utlutt−M∇u2Δu−Δutt+∫0tht−sΔusds+μ1g1utx,t+μ2g2zx,1,t=νulnuinΩ×0,+∞,τtztx,ρ,t+1−ρτ′tzρx,ρ,t=0,inΩ×0,1×0,+∞,ux,t=0,on∂Ω×0,∞,zx,0,t=utx,t,onΩ×0,∞,ux,0=u0x,utx,0=u1x,inΩ,zx,ρ,0=f0x,−ρτ0,inΩ×0,1.To state and prove our result, we need some assumptions.(A1) Assume thatl satisfies(13)0<l≤2n−2,ifn>2,0<l<∞,ifn≤2.(A2) The relaxation functionh:ℝ+⟶ℝ+ is a bounded C1 function, such that(14)a−∫0∞hsds=k>0,h0>0,and suppose that there exists a positive constantζ satisfying(15)h′t≤−ζht.(A3)g1:ℝ⟶ℝ is a nondecreasing function of class C1 and H:ℝ+⟶ℝ+ is convex, increasing and of class C1ℝ+∩C2]0,+∞[, satisfying(16)H0=0andHis linear,on0,εorH′0=0andH″>0,on0,εsuch thatc1s≤g1s≤c2s,ifs≥εs2+g1s2≤H−1sg1s,ifs≤ε,whereε,c1,andc2 are positive constants. g2:ℝ⟶ℝ is an odd nondecreasing function of class C1ℝ such that there exist c3,α1,andα2>0,(17)g2′s≤c3,(18)α1sg2s≤Gs≤α2sg1s,Gs=∫0sg2rdr.(A4)τ is a function in W2,∞0,T, T>0, such that(19)0<τ0≤τt≤τ1,∀t>0,τ′t≤d<1,∀t>0,whereτ0 and τ1 are positive numbers.(A5) We also assume that(20)μ2<α11−dα21−α1dμ1.(A6)(21)1<ν<2πke3.We define the energy associated to the solution of system (12) by(22)Et=1l+2utl+2l+2+b2γ+1∇u2γ+1+12a−∫0thsds∇u2+12∇ut2−ν2∫Ωu2lnudx+ν4u2+12ho∇ut+ξτt∫Ω∫01Gzx,ρ,tdρdx,whereξ is a positive constant such that(23)μ21−α1α11−d<ξ<μ1−α2μ2α2,(24)hovt=∫0tht−sv.,t−v.,s2ds.Theorem 1.
(Global Existence). Letu0,u1,f0∈H2Ω∩H01Ω×H01Ω×H01Ω,H10,1 satisfy the compatibility condition:(25)f0.,0=u1.
Assume that (A1)–(A6) hold under smallness condition on the initial datau0,u1. Then, problem (1) admits a weak solution:(26)u∈L∞0,∞;H2Ω∩H01Ω,ut∈L∞0,∞;H01Ω,utt∈L∞0,∞;L2ΩTheorem 2.
(Uniform Decay Rates of Energy). Assume that (A1)–(A6) hold and ifE0 is positive and bounded, then for every t0>0, there exist positive constants w1,w2,w3, and ε0 such that the solution energy of (1) satisfies(27)Et≤w3H1−1w1t+w2,∀t≥t0,where(28)H1t=∫t11H2sds,H2t=tH′ε0t.Here,H1 is strictly decreasing and convex on 0,1 with limt⟶0H1t=+∞.


## 2. Preliminaries

Lemma 1.
(Sobolev–Poincaré’s Inequality). Letq be a number with(29)2≤q≤+∞n=1,2or2≤q≤2nn−2n≥3.
Then, there exists a constantCs=CsΩ,q such that(30)uq≤Cs∇u,foru∈H01Ω.Lemma 2.
(see [32, 33]) (Logarithmic Sobolev Inequality). Let u be any function in H01Ω and σ>0 be any number. Then,(31)∫Ωu2lnudx≤12u2lnu2+σ22π∇u2−1+lnσu2.Lemma 3.
(see [20]) (Logarithmic Gronwall Inequality). Let C>0 and φ∈L10,T;ℝ+ and assume that the function w:0,T⟶1,∞ satisfies(32)wt≤C1+∫0tφswslnwsds,∀t∈0,T.
Then,(33)wt≤CexpC∫0tφsds,∀t∈0,T.Lemma 4.
Letϵ0∈0,1. Then, there exists dϵ0>0 such that(34)slns≤s2+dϵ0s1−ϵ0,∀s>0.Proof.
Letrs=sϵ0lns−s. Notice that r is continuous on 0,∞, and its limit at 0+ is 0+ and its limit at ∞ is −∞. Then, r has a maximum dε0 on 0,∞, so the proof is complete.
The following lemma states an important property of the convolution operator.Lemma 5.
(see [34]). For h,φ∈C10,+∞,ℝ, we have(35)∫Ωh∗φφtdx=−12htφt2+12h′oφt−12ddthoφt−∫0thsdsφ2.Remark 1.
Let us denote byΦ∗ the conjugate function of the differentiable convex function Φ, i.e.,(36)Φ∗s=supt∈ℝ+st−Φt.
Then,Φ∗ is the Legendre transform of Φ, which is given by (see Arnold [35], p. 61-62)(37)Φ∗s=sΦ′−1s−ΦΦ′−1s,ifs∈0,Φ′r,and Φ∗ satisfies the generalized Young inequality:(38)AB≤Φ∗A+ΦB,ifA∈0,Φ′rB∈0,r.Lemma 6.
Letu,z be a solution of problem (12). Then, the energy functional defined by (22) satisfies(39)E′t≤−λ∫Ωutg1utdx−β∫Ωzx,1,tg2zx,1,tdx−12ht∇ut2+12h′o∇ut≤0,where λ=μ1−ξα2−μ2α2 and β=ξ1−dα1−μ21−α1.Proof.
Multiplying the first equation in (12) by ut, integrating over Ω, and using integration by parts, we get(40)ddt1l+2utl+2l+2+b2γ+1∇u2γ+1+12a∇u2+12∇ut2−ν2∫Ωu2lnudx+ν4u2+∫Ω∫0tht−s∇us∇uttdsdx+μ1∫Ωutx,tg1utx,tdx+μ2∫Ωutx,tg2zx,1,tdx=0.
Consequently, by applying Lemma5, equation (40) becomes(41)ddt1l+2utl+2l+2+b2γ+1∇u2γ+1+12a−∫0thsds∇u2+12∇ut2−ν2∫Ωu2lnudx+ν4u2+12ho∇ut+12ht∇ut2−12h′o∇ut+μ1∫Ωutx,tg1utx,tdx+μ2∫Ωutx,tg2zx,1,tdx=0.
Multiplying the second equation in (12) by ξg2z and integrating the result over Ω×0,1, we obtain(42)ξτt∫Ω∫01ztx,ρ,tg2zx,ρ,tdρdx=−ξ∫Ω∫011−ρτ′t∂∂ρGzx,ρ,tdρdx.
Consequently,(43)ddtξτt∫Ω∫01Gzx,ρ,tdρdx=ξτ′t∫Ω∫01Gzx,ρ,tdρdx−ξ∫Ω∫011−ρτ′t∂∂ρGzx,ρ,tdρdx=−ξ∫Ω∫01∂∂ρ1−ρτ′tGzx,ρ,tdρdx=−ξ1−τ′t∫ΩGzx,1,tdx+ξ∫ΩGutx,tdx.
Combining (41) and (43), we obtain(44)E′t=−ξ1−τ′t∫ΩGzx,1,tdx+ξ∫ΩGutx,tdx−12ht∇ut2+12h′o∇ut−μ1∫Ωutx,tg1utx,tdx−μ2∫Ωutx,tg2zx,1,tdx.
From (18) and A4, we get(45)E′t≤−μ1−ξα2∫Ωutx,tg1utx,tdx−ξ1−dα1∫Ωzx,1,tg2zx,1,tdx−μ2∫Ωutx,tg2zx,1,tdx−12ht∇ut2+12h′o∇ut.
Using (18) and Remark 1, we obtain(46)G∗s=sg2−1s−Gg2−1s,∀s≥0.
Hence,(47)G∗g2zx,1,t=zx,1,tg2x,1,t−Gzx,1,t≤1−α1zx,1,tg2zx,1,t.
Using (18) and (38) with A=g2zx,1,t and B=utx,t, we have from (45) that(48)E′t≤−μ1−ξα2∫Ωutx,tg1utx,tdx−ξ1−dα1∫Ωzx,1,tg2zx,1,tdx+μ2∫ΩGutx,t+G∗g2zx,1,tdx−12ht∇ut2+12h′o∇ut≤−μ1−ξα2−μ2α2∫Ωutx,tg1utx,tdx−ξ1−dα1−μ21−α1∫Ωzx,1,tg2zx,1,tdx−12ht∇ut2+12h′o∇ut.
This completes the proof.


## 3. Proof of Theorem1

### 3.1. Local Existence

Throughout this section, we assumeu0∈H2Ω∩H01Ω,u1∈H01Ω and f0∈H01Ω,H10,1. We will use the Faedo-Galerkin method to prove the existence of a solution to problem (1). Let T>0 be fixed and let wk,k∈ℕ, be a basis of H2Ω∩H01Ω, and let Vk be the space generated by wk. Now, we define, for 1≤j≤k, the sequence ϕjx,ρ as follows:(49)ϕjx,0=wj.Then, we may extendϕjx,0 by ϕjx,ρ over L2Ω×0,1 such that ϕjj forms a basis of L2Ω,H10,1 and denote Zk as the space generated by ϕk. We construct approximate solutions uk,zk, k=1,2,3,…, in the form(50)ukt=∑j=1kcjktwj,zkt=∑j=1kdjktϕj,where cjk and djkj=1,2,…,k are determined by the following ordinary differential equations:(51)utkluttk,wj+M∇ukt2∇uk,∇wj+∇uttk,∇wj−∫0tht−s∇uks,∇wjds+μ1g1utk,wj+μ2g2zk.,1,wj=ν∫Ωwjuklnukdx,1≤j≤k,zkx,0,t=utkx,t,(52)uk0=u0k=∑j=1ku0,wjwj⟶u0,inH2Ω∩H01Ωask⟶+∞,(53)utk0=u1k=∑j=1ku1,wjwj⟶u1,inH01Ωask⟶+∞,(54)τtztk+1−ρτρ′k,ϕj=0,1≤j≤k,(55)zkρ,0=z0k=∑j=1kf0,ϕjϕj⟶f0,inH01Ω,H10,1ask⟶+∞.Noting thatl/2l+1+1/2l+1+1/2=1, from the generalized Hölder inequality, we obtain(56)utkluttk,wj=∫Ωutkluttkwjdx≤∫Ωutk2l+1dxl/2l+1uttk2l+1wj2.SinceA1 holds, according to Sobolev, embedding the nonlinear term utkluttk,wj in (51) makes sense.The standard theory of ODE guarantees that systems (51)–(55) have a unique solution in 0,tk, with 0<tk<T, by Zorn lemma since the nonlinear terms in (51) are locally Lipschitz continuous. Note that ukt is of class C2. In the next step, we obtain a priori estimate for the solution of systems (51)–(55), so that it can be extended to 0,T and that the local solution is uniformly bounded independently of k and t.


#### 3.1.1. The First Estimate

Since the sequencesu0k,u1k, and z0k converge and from Lemma 6, we can find a positive constant C1 independent of k such that(57)Ekt−Ek0≤−λ∫0t∫Ωutkg1utkdxds−β∫0t∫Ωzkx,1,sg2zkx,1,sdxds−12∫0ths∇uks2ds+12∫0th′o∇uksds≤−λ∫0t∫Ωutkg1utkdxds−β∫0t∫Ωzkx,1,sg2zkx,1,sdxds.Ash is a positive nonincreasing function, we get(58)Ekt+λ∫0t∫Ωutkg1utkdxds+β∫0t∫Ωzkx,1,sg2zkx,1,sdxds≤Ek0≤C1,where(59)Ekt=1l+2utkl+2l+2+b2γ+1∇uk2γ+1+12a−∫0thsds∇uk2+12∇utk2+12ho∇ukt−ν2∫Ωuk2lnukdx+ν4uk2+ξτt∫Ω∫01Gzkx,ρ,tdρdx.By applying the Logarithmic Sobolev inequality, (58) yields(60)utkl+2l+2+k−νσ22π∇uk2+∇utk2+ν2+ν1+lnσuk2+h∘∇ukt+∫Ω∫01Gzkx,ρ,tdρdx+∫0t∫Ωutkg1utkdxds+∫0t∫Ωzkx,1,sg2zkx,1,sdxds≤C2+uk2lnuk2,where C2 is a positive constant depending only on u0H01,u1H01,l,γ,ξ,τ1,λ, and β.By choosing(61)e−3/2<σ<2πkν,we obtain k−νσ2/2π>0 and ν/2+ν1+lnσ>0.This selection is possible thanks toA6. So we get(62)utkl+2l+2+∇uk2+∇utk2+uk2+h∘∇ukt+∫Ω∫01Gzkx,ρ,tdρdx+∫0t∫Ωutkg1utkdxds+∫0t∫Ωzkx,1,sg2zkx,1,sdxds≤c1+uk2lnuk2.Let us note that(63)ukt=uk0+∫0tusksds.Then, by using Cauchy Schwarz’s inequality, we get(64)uk2≤uk02+2∫0tusksds2≤uk02+2T∫0tusks2ds.Hence, (62) gives(65)uk2≤C1+∫0tuk2lnuk2ds,where C=max2Tc,2uk02. Applying the Logarithmic Gronwall inequality to (65), we obtain(66)uk2≤CeCT.Hence, from (58), we obtain the first estimate:(67)utkl+2l+2+∇uk2+∇utk2+uk2+h°∇ukt+∫Ω∫01Gzkx,ρ,tdρdx+∫0t∫Ωutkg1utkdxds+∫0t∫Ωzkx,1,sg2zkx,1,sdxds≤c1+CeCTlnCeCT=A1,The estimate implies that the solutionuk,zk exists in 0,T and it yields(68)ukis bounded inLloc∞0,∞,H1Ω,(69)utkis bounded inLloc∞0,∞,H01Ω,(70)Gzkx,ρ,tis bounded inLloc∞0,∞,L1Ω×0,1,(71)utktg1utktis bounded inL1Ω×0,T,(72)zkx,1,tg2zkx,1,tis bounded inL1Ω×0,T.


#### 3.1.2. The Second Estimate

Replacingwj by −Δwj in (51), multiplying by ctjk, and summing over j from 1 to k, it follows that(73)∫Ωutktluttkt−Δutkdx+∫ΩM∇uk2ΔukΔutkdx+∫ΩΔuttkΔutkdx−∫0tht−s∫ΩΔukΔutkdxds−μ1∫ΩΔutkg1utkdx−μ2∫ΩΔutkg2zkx,1,tdx=−ν∫ΩΔutkuklnukdx.Noting thatM∇uk2≥a and by using Lemma 5, we obtain(74)−∫ΩutktluttktΔutkdx+12ddta−∫0thsdsΔuk2+Δutk2+hoΔuk+12htΔuk2−12h′oΔuk+μ1∫Ω∇utk2g1′utkdx+μ2∫Ω∇utk∇zkx,1,tg2′zkx,1,tdx≤−ν∫ΩΔutkuklnukdx.By using the Green formula, we have(75)−∫ΩutktluttkΔutkdx=ddt∫Ωutktl∇utk2dx−l+1∫Ωutktl∇uttk∇utkdx.Consequently, equation (74) yields(76)12ddt2∫Ωutktl∇utk2dx+a−∫0thsdsΔuk2+Δutk2+hoΔuk−l+1∫Ωutktl∇uttk∇utkdx+12htΔuk2−12h′oΔuk+μ1∫Ω∇utk2g1′utkdx+μ2∫Ω∇utk∇zkx,1,tg2′zkx,1,tdx≤−ν∫ΩΔutkuklnukdx.To estimate the term on the right-hand side of (76), we apply Lemma 4 with ϵ0=1/2 and use repeatedly Young’s, Cauchy-Schwartz’s, and the embedding inequalities as follows:(77)ν∫ΩΔutkuklnukdx≤ν∫ΩΔutkuk2+dϵ0ukdx≤νη∫ΩΔutk2dx+14η∫Ωuk2+dϵ0uk2dx≤νη∫ΩΔutk2dx+c4η∫Ωuk4dx+∫Ωukdx≤νηΔutk2+c4η∇uk4+uk,η>0.Combining (76) and (77) to have(78)12ddt2∫Ωutktl∇utk2dx+a−∫0thsdsΔuk2+Δutk2+hoΔuk−l+1∫Ωutktl∇uttk∇utkdx+12htΔuk2−12h′oΔuk+μ1∫Ω∇utk2g1′utkdx+μ2∫Ω∇utk∇zkx,1,tg2′zkx,1,tdx≤νηΔutk2+c4η∇uk4+uk.Replacingϕj by −Δϕj in (54), multiplying by djk, and summing over j from 1 to k, it follows that(79)τt1−τ′tρ∫Ω∇ztk∇zkdx+∫Ω∇zρk∇zkdx=0.Then, we get(80)12ddtτt1−τ′tρ∇zk2−τt1−τ′tρ′∇zk2+12ddρ∇zk2=0.We integrate over0,1, and we find(81)12ddt∫01τt1−τ′tρ∇zkx,ρ,t2dρ+12∇zkx,1,t2=12∫01τt1−τ′tρ′∇zkx,ρ,t2dρ+12∇utkt2.Combining (78) and (81) and using (A2), we get(82)12ddt2∫Ωutktl∇utk2dx+a−∫0thsdsΔuk2+Δutk2+hoΔuk+∫01τt1−τ′tρ∇zkx,ρ,t2dρ+μ1∫Ω∇utk2g1′utkdx+12∇zkx,1,t2≤l+1∫Ωutktl∇uttk∇utkdx−μ2∫Ω∇utk∇zkx,1,tg2′zkx,1,tdx+12∫01τt1−τ′tρ′∇zkx,ρ,t2dρ+12∇utkt2+νηΔutk2+c4η∇uk4+uk.From the first estimate (67) and Young’s inequality, we get(83)∫Ωutkl∇uttkt∇utkdx≤A1l/l+2+1/2∇uttk2≤η∇uttk2+A12l/l+2+14η.Using (17) and Chaucy-Schwarz’s inequality, we obtain(84)∫Ω∇utk∇zkx,1,tg2′zkx,1,tdx≤c32ε2∫Ω∇utk2dx+ε2∫Ω∇zkx,1,t2dx.Taking into account (83) and (84) into (82) yields(85)12ddt2∫Ωutktl∇utk2dx+a−∫0thsdsΔuk2+Δutk2+hoΔuk+∫01τt1−τ′tρ∇zkx,ρ,t2dρ+μ1∫Ω∇utk2g1′utkdx+c∇zkx,1,t2≤l+1η∇uttk2+Cη′A1+12∫01τt1−τ′tρ′∇zkx,ρ,t2dρ+c′∇utk2+νηΔutk2+c4ηθ4+uk.Multiplying (51) by cttjk and summing over j from 1 to k, it follows that(86)∫Ωutkluttk2dx+∫ΩM∇uk2∇uk∇uttkdx+∇uttk2−∫0tht−s∫Ω∇uks∇uttktdxds+μ1∫Ωuttkg1utkdx+μ2∫Ωuttkg2zkx,1,tdx=ν∫Ωuttkuklnukdx.Then,(87)∫Ωutkluttk2dx+∇uttk2=−∫ΩM∇uk2∇uk∇uttkdx+∫0tht−s∫Ω∇uks∇uttktdxds−μ1∫Ωuttkg1utkdx−μ2∫Ωuttkg2zkx,1,tdx+ν∫Ωuttkuklnukdx.Differentiating (54) with respect to t, we get(88)τt1−ρτ′t′ztk+τt1−ρτ′tzttk+ztρk,ϕj=0.Multiplying bydtjk and summing over j from 1 to k, it follows that(89)τt1−ρτ′t′ztk2+12τt1−ρτ′tddtztk2+12ddρztk2=0.Then, we have(90)12τt1−ρτ′t′ztk2+12ddtτt1−ρτ′tztk2+12ddρztk2=0.Integrating over0,1 with respect to ρ, we obtain(91)12∫01τt1−ρτ′t′ztk2dρ+12ddt∫01τt1−ρτ′tztk2dρ+12ztkx,1,t2−12uttkx,t2=0.Summing (87) and (91), we get(92)∫Ωutkluttk2dx+∇uttk2+12ddt∫01τt1−ρτ′tztk2dρ+12ztkx,1,t2=12uttkx,t2−12∫01τt1−ρτ′t′ztk2dρ−∫ΩM∇uk2∇uk∇uttkdx+∫0tht−s∫Ω∇uks∇uttktdxds+ν∫Ωuttkuklnukdx−μ1∫Ωuttkg1utkdx−μ2∫Ωuttkg2zkx,1,tdx.By Cauchy-Schwarz’s, Sobolev’s, and Young’s inequalities, the right hand side of (92) can be estimated as follows:(93)∫ΩM∇uk2∇uk∇uttkdx≤a+b∇uk2γ∫Ω∇uk∇uttkdx≤a+bE02γ/2γ+1∫Ω∇uk∇uttkdx≤η∇uttk2+m024η∇uk2,(94)∫0tht−s∫Ω∇uks∇uttktdxds≤η∇uttk2+14η∫Ω∫0tht−s∇uksds2dx≤η∇uttk2+14η∫0thsds∫Ω∫0tht−s∇uks2dsdx≤η∇uttk2+14ηa−k∫Ω∫0tht−s∇uks2dsdx≤η∇uttk2+14ηa−k∫0tht−s∇uks2ds≤η∇uttk2+14ηa−kh0∫0t∇uks2ds,and from (16),(95)∫Ωuttkg1utkdx≤12∫Ωuttk2dx+12∫Ωg1utkdx≤12∫Ωuttk2dx+12∫utk≥εg1utk2dx+12∫utk≤εg1utk2dx≤12∫Ωuttk2dx+12∫utk≥εutkg1utkdx+12∫ΩH−1utkg1utkdx.Using Lemma6, Jensen’s inequality, and the concavity of H−1, we obtain(96)∫Ωuttkg1utkdx≤12∫Ωuttk2dx+12∫utk≥εutkg1utkdx+cH−1∫Ωutkg1utkdx+k∫Ωuttkuklnukdx≤12∫Ωuttk2dx+12∫utk≥εutkg1utkdx+c′H∗1+c′′∫Ωutkg1utkdx≤12uttk2dx+c′H∗1+c′′∫Ωutkg1utkdx≤Cs∇uttk2+c′H∗1+c−E′.From (17) (that is, g2s≤cs,∀s∈ℝ), we get(97)∫Ωuttkg2zkx,1,tdx≤12∫Ωuttk2dx+12∫Ωg2zkx,1,t2dx≤12uttk2+c3∫Ωzkx,1,tg2zkx,1,tdx≤Cs∇uttk2+c′−E′.Similar to (77), we get(98)ν∫Ωuttkuklnukdx≤cη∇uttk2+c4η∇uk4+uk.Substituting (93)–(98) into (92) yields(99)∫Ωutkluttk2dx+∇uttk2+12ddt∫01τt1−ρτ′tztk2dρ+12ztkx,1,t2≤2η+3Cs∇uttk2++m024η∇uk2+14ηa−kh0∫0t∇uks2ds+c−E′−12∫01τt1−ρτ′t′ztk2dρ+cη∇uttk2+c4η∇uk4+uk.Combining (85) and (99), we get(100)12ddt2∫Ωutktl∇utk2dx+a−∫0thsdsΔuk2+Δutk2+hoΔuk+∫01τt1−τ′tρ∇zkx,ρ,t2dρ+∫01τt1−ρτ′tztk2dρ+cEt+∫Ωutkluttk2dx+12ztkx,1,t2+1−l+3η−3Cs∇uttk2+μ1∫Ω∇utk2g1′utkdx+c∇zkx,1,t2≤Cη′A1+12∫01τt1−ρτ′t′∇zkx,ρ,t2dρ+c′∇utk2+m024η∇uk2+cη∇uttk2+c44η∇uk4+uk+νηΔutk2.Then, from (67) and by integration over 0,t, (100) yields(101)a−∫0thsdsΔuk2+Δutk2+hoΔuk+∫01τt1−τ′tρ∇zkx,ρ,t2dρ+∫01τt1−ρτ′tztk2dρ+cEt+2∫Ωutktl∇utk2dx+2∫0t∫Ωutkluttk2dxds+c∫0tztkx,1,t2ds+21−l+3−cη−3Cs∫0t∇uttk2ds+c∗∫0t∇zkx,1,t2ds≤Cη′A1+c′+m022η+12ηa−kh0TT+E0+A2+νη∫0tΔutk2ds+c∗′′∫0t∫01τt1−τ′tρ′∇zkx,ρ,t2dρds+c′′∫0t∫01τt1−ρτ′t′ztk2dρds.For a suitableη, we get(102)Δuk2+Δutk2+hoΔuk+∫01τt1−τ′tρ∇zkx,ρ,t2dρ+∫01τt1−ρτ′tztk2dρ+∫0t∇uttk2ds≤Cδ′A1+A1′T+A2′+c∗′′∫0t∫01τt1−ρτ′t′∇zkx,ρ,t2dρds+c′′∫01τt1−ρτ′t′ztk2dρ+c∫0tΔutk2ds.Using Gronwall lemma, we obtain(103)Δuk2+Δutk2+hoΔuk+∫01τt1−τ′tρ∇zkx,ρ,t2dρ+∫01τt1−ρτ′tztk2dρ+∫0t∇uttk2ds≤C3.We observe from the estimates (67) and (103) that there exists a subsequence um of uk and functions u,z,χ,andψ such that(104)um⇀uweakly star inL∞0,T,H2Ω∩H01Ω,(105)utm⇀utweakly star inL∞0,T,H02Ω,(106)g1utm⇀χweakly star inL2Ω×0,T,(107)uttm⇀uttweakly star inL20,T,H01Ω,(108)zm⇀zweakly star inL∞0,T,H01Ω,L20,1,(109)ztm⇀ztweakly star inL∞0,T,L2Ω×0,1,(110)g2zmx,1,t⇀ψweakly star inL2Ω×0,T.Now, we will prove thatu is the solution of (1). First, we will treat the nonlinear terms.(1)
Termutklutk: from the first estimate (67) and Lemma 1, we deduce(111)utklutkL20,T,L2Ω=∫0Tutk2l+12l+1dt≤Cs2l+1∫0T∇utk2l+1dt≤Cs2l+1A12l+1T.On the other hand, from Aubin-Lions theorem (see Lions [36]), we deduce that there exists a subsequence of um, still denoted by um such that(112)utm⟶utstrongly inL20,T,L2Ω,which implies that(113)utm⟶utalmost everywhere inA.Hence,(114)utmlutm⟶utlutalmost everywhere inA,where A=Ω×0,T. Thus, using (117), (114), and Lions Lemma, we derive(115)utmlutm⇀utlutweakly inL20,T,L2Ω,(116)zm⟶zstrongly inL20,T,L2Ω,which implies zm⟶z almost everywhere in A.(2) Termuklnuk: using (103), we have uk being bounded in L∞0,T,H02Ω which implies the boundedness of uk in L2A. Similarly, utk is bounded in L2A. Then, from Aubin-Lions theorem, we find a subsequence such that(117)um⟶ustrongly inL2A,which implies(118)um⟶ualmost everywhere inA.Since the maps⟶νslns is continuous, we have the following convergence:(119)νumlnum⟶νulnualmost every where inA.Using the embedding ofH01Ω in L∞Ω, it is clear that νumlnum is bounded in L∞Ω×0,T. Next, taking into account the Lebesgue bounded convergence theorem Ωis bounded, we get(120)νumlnum⟶νulnustrongly inL20,T;L2Ω.Lemma 7.
For eachT>0,g1ut,g2zx,1,t∈L1A, and g1u′L1A,g2zx,1,tL1A≤K, where K is a constant independent of t.Proof.
By (A2) and (118), we have(121)g1utmx,t⟶g1utx,talmost everywhere inA,0≤utkx,tg1utmx,t⟶utx,tg1utx,talmost everywhere inA.
Hence, by (71) and Fatou’s Lemma, we have(122)∫0T∫Ωutx,tg1utx,tdxdt≤K1,forT>0.
By using Cauchy-Schwarz’s inequality, (96), and (122), we have(123)∫0T∫Ωg1utx,tdxdt≤cA1/2∫0T∫Ωutx,tg1utx,tdxdt1/2≤cA1/2K11/2≡K.Lemma 8.
We haveg1utk⇀g1utweak inL2Ω×0,T and g2zk⇀g2zweak inL2Ω×0,T.Proof.
LetE⊂Ω×0,T and set(124)E1=x,t∈E,g1utkx,t≤1E,E2=EE1,where E is the measure of E. If Mr=infs,s∈ℝandgs≥r,(125)∫Eg1utkdxdt≤cE+M1E−1∫E2utkg1utkdxdt.
By applying (71), we deduce that supk∫Eg1utkdxdt⟶0 as E⟶0. From Vitali’s convergence theorem, we deduce that(126)g1utk⟶g1utinL1Ω×0,T.
Hence,(127)g1utk⇀g1utweak inL2Ω×0,T.
Similarly, we have(128)g2zk⇀g2zweak inL2Ω×0,T.Remark 2.
By using (103) and from (104) and (105) combined with the Aubin-Lions compactness lemma, we deduce(129)∫0TM∇ukt2Δukt,wθtdt⟶∫0TM∇ut2Δut,wθtdt,ask⟶∞.
By multiplying (51) and (54) by θt∈D0,T and by integrating over 0,T, it follows that(130)∫0Tutktluttkt,wjθtdt+∫0TM∇ukt2∇ukt,∇wjθtdt+∫0T∇uttk,∇wjθtdt−∫0T∫0tht−s∇uks,∇wjθtdsdt+μ1∫0Tg1utk,wjθtdt+μ2∫0Tg2zk.,1,wjθtdt=ν∫0Tukslnuks,wjθtdt,∫0T∫01∫Ωτtztk+1−ρτ′tzρkϕjθtdxdρdt=0,for all j=1,…,k.
The convergence of (104)–(110), (115), and (124)–(129) is sufficient to pass to the limit in (130) in order to obtain(131)∫0T∫Ωutlutt−M∇ut2Δu−Δutt+∫0tht−sΔusds+μ1g1ut+μ2g2z.,1wθtdsdt=ν∫0Tuslnus,wθtdt,∫0T∫01∫Ωτtzt+1−ρτ′tzρϕjθtdxdρdt=0.
Then, problem (1) has a weak solution on 0,T, T>0.


### 3.2. Global Existence

To state and prove our global existence, we introduce the following functionals:(132)It=a−∫0thsds∇u2+∇ut2+ho∇ut−3ν∫Ωu2lnudx,(133)Jt=12a−∫0thsds∇u2+12∇ut2+12ho∇ut−ν2∫Ωu2lnudx+ν4u2=13a−∫0thsds∇u2+∇ut2+ho∇ut+ν4u2+16It.We note that(134)Et=1l+2utl+2l+2+b2γ+1∇u2γ+1+Jt+ξτt∫Ω∫01Gzx,ρ,tdρdx.Lemma 9.
The following inequalities hold:(135)−d0Ωcs3∇u3/2≤∫Ωu2lnudx≤cs3∇u3,∀u∈H01Ω,where d0=sup0<s<1slns=2/e, Ω is the Lesbegue measure of Ω, and cs is the smallest embedding constant:(136)u3≤cs∇u,∀u∈H01Ω.Proof.
LetΩ1=x∈Ω:u≤1 and Ω2=x∈Ω:ut>1. By using (136), we have(137)∫Ωu2lnudx=∫Ω1u2lnudx+∫Ω2u2lnudx≤∫Ω2u2lnudx≤∫Ω2u3dx≤∫Ω2u3dx≤cs3∇u3.
On the other hand, using Hölder’s inequality and (136), we get(138)−∫Ωu2lnudx=−∫Ω1u2lnudx−∫Ω2u2lnudx≤−∫Ω1u2lnudx=∫Ω1u2lnudx≤d0∫Ω1u3/2dx≤d0Ω∫Ω1u3dx1/2≤d0Ωcs3∇u3/2.
Hence, (135) is obtained.Lemma 10.
Assume that (A1)–(A6) hold. Letu0,u1∈H2Ω∩H01Ω×H01Ω such that(139)I0>0,27νcs3E0k1/2≤k.
Then,(140)It>0,∀t∈0,T.Proof.
SinceI0>0 and I is continuous on 0,T, there exists t∗∈0,T such that It>0, for all t∈0,t∗. Let us denote by t0 the largest real number in 0,T such that I>0 on 0,t0. We assume by contradiction that t0≠T, so we have It0=0 and from (133), we have(141)∇ut2≤3kJt≤3kEt≤3kE0,∀t∈0,t0.
The last inequality is obtained from Lemma6. If ∇ut02=0, then (132) and (135) give(142)0=It0=a−∫0t0hsds∇ut02+∇utt02+ho∇ut0−3ν∫Ωu2t0lnut0dx=ho∇ut0=∫0t0hs∇us2ds.
Consequently, ifh>0 on 0,t0, we get(143)∇us2=0,∀s∈0,t0.
Then,(144)It=0,∀s∈0,t0,which is not true since I>0 on 0,t0. If there exists t∗ in 0,t0 such that ht∗=0, then let t1∈0,t0 be the smallest real number such that ht1=0. Because h0>0 and h is positive, nonincreasing, and continuous on ℝ+, then t1>0 and h=0 on t1,+∞. Therefore, from (142), we deduce that(145)0=∫0t0hs∇us2ds=∫0t1hs∇us2ds.Then,(146)∇us2=0,∀s∈0,t1.
As given above, we get a contradiction with the fact thatI>0 on 0,t0. Then, we conclude that ∇ut02>0. From (132), we have(147)It0≥k∇ut02−3ν∫Ωut02lnut0dx.
By using (135) and (141), we have(148)It0≥k−3ν3kE01/2∇ut02.
By recalling (139), we arrive at It0>0, which contradicts the assumption that It0=0. Hence, t0=T and then I>0 on 0,T.
This completes the proof of Theorem1.


## 3.1. Local Existence

Throughout this section, we assumeu0∈H2Ω∩H01Ω,u1∈H01Ω and f0∈H01Ω,H10,1. We will use the Faedo-Galerkin method to prove the existence of a solution to problem (1). Let T>0 be fixed and let wk,k∈ℕ, be a basis of H2Ω∩H01Ω, and let Vk be the space generated by wk. Now, we define, for 1≤j≤k, the sequence ϕjx,ρ as follows:(49)ϕjx,0=wj.Then, we may extendϕjx,0 by ϕjx,ρ over L2Ω×0,1 such that ϕjj forms a basis of L2Ω,H10,1 and denote Zk as the space generated by ϕk. We construct approximate solutions uk,zk, k=1,2,3,…, in the form(50)ukt=∑j=1kcjktwj,zkt=∑j=1kdjktϕj,where cjk and djkj=1,2,…,k are determined by the following ordinary differential equations:(51)utkluttk,wj+M∇ukt2∇uk,∇wj+∇uttk,∇wj−∫0tht−s∇uks,∇wjds+μ1g1utk,wj+μ2g2zk.,1,wj=ν∫Ωwjuklnukdx,1≤j≤k,zkx,0,t=utkx,t,(52)uk0=u0k=∑j=1ku0,wjwj⟶u0,inH2Ω∩H01Ωask⟶+∞,(53)utk0=u1k=∑j=1ku1,wjwj⟶u1,inH01Ωask⟶+∞,(54)τtztk+1−ρτρ′k,ϕj=0,1≤j≤k,(55)zkρ,0=z0k=∑j=1kf0,ϕjϕj⟶f0,inH01Ω,H10,1ask⟶+∞.Noting thatl/2l+1+1/2l+1+1/2=1, from the generalized Hölder inequality, we obtain(56)utkluttk,wj=∫Ωutkluttkwjdx≤∫Ωutk2l+1dxl/2l+1uttk2l+1wj2.SinceA1 holds, according to Sobolev, embedding the nonlinear term utkluttk,wj in (51) makes sense.The standard theory of ODE guarantees that systems (51)–(55) have a unique solution in 0,tk, with 0<tk<T, by Zorn lemma since the nonlinear terms in (51) are locally Lipschitz continuous. Note that ukt is of class C2. In the next step, we obtain a priori estimate for the solution of systems (51)–(55), so that it can be extended to 0,T and that the local solution is uniformly bounded independently of k and t.


### 3.1.1. The First Estimate

Since the sequencesu0k,u1k, and z0k converge and from Lemma 6, we can find a positive constant C1 independent of k such that(57)Ekt−Ek0≤−λ∫0t∫Ωutkg1utkdxds−β∫0t∫Ωzkx,1,sg2zkx,1,sdxds−12∫0ths∇uks2ds+12∫0th′o∇uksds≤−λ∫0t∫Ωutkg1utkdxds−β∫0t∫Ωzkx,1,sg2zkx,1,sdxds.Ash is a positive nonincreasing function, we get(58)Ekt+λ∫0t∫Ωutkg1utkdxds+β∫0t∫Ωzkx,1,sg2zkx,1,sdxds≤Ek0≤C1,where(59)Ekt=1l+2utkl+2l+2+b2γ+1∇uk2γ+1+12a−∫0thsds∇uk2+12∇utk2+12ho∇ukt−ν2∫Ωuk2lnukdx+ν4uk2+ξτt∫Ω∫01Gzkx,ρ,tdρdx.By applying the Logarithmic Sobolev inequality, (58) yields(60)utkl+2l+2+k−νσ22π∇uk2+∇utk2+ν2+ν1+lnσuk2+h∘∇ukt+∫Ω∫01Gzkx,ρ,tdρdx+∫0t∫Ωutkg1utkdxds+∫0t∫Ωzkx,1,sg2zkx,1,sdxds≤C2+uk2lnuk2,where C2 is a positive constant depending only on u0H01,u1H01,l,γ,ξ,τ1,λ, and β.By choosing(61)e−3/2<σ<2πkν,we obtain k−νσ2/2π>0 and ν/2+ν1+lnσ>0.This selection is possible thanks toA6. So we get(62)utkl+2l+2+∇uk2+∇utk2+uk2+h∘∇ukt+∫Ω∫01Gzkx,ρ,tdρdx+∫0t∫Ωutkg1utkdxds+∫0t∫Ωzkx,1,sg2zkx,1,sdxds≤c1+uk2lnuk2.Let us note that(63)ukt=uk0+∫0tusksds.Then, by using Cauchy Schwarz’s inequality, we get(64)uk2≤uk02+2∫0tusksds2≤uk02+2T∫0tusks2ds.Hence, (62) gives(65)uk2≤C1+∫0tuk2lnuk2ds,where C=max2Tc,2uk02. Applying the Logarithmic Gronwall inequality to (65), we obtain(66)uk2≤CeCT.Hence, from (58), we obtain the first estimate:(67)utkl+2l+2+∇uk2+∇utk2+uk2+h°∇ukt+∫Ω∫01Gzkx,ρ,tdρdx+∫0t∫Ωutkg1utkdxds+∫0t∫Ωzkx,1,sg2zkx,1,sdxds≤c1+CeCTlnCeCT=A1,The estimate implies that the solutionuk,zk exists in 0,T and it yields(68)ukis bounded inLloc∞0,∞,H1Ω,(69)utkis bounded inLloc∞0,∞,H01Ω,(70)Gzkx,ρ,tis bounded inLloc∞0,∞,L1Ω×0,1,(71)utktg1utktis bounded inL1Ω×0,T,(72)zkx,1,tg2zkx,1,tis bounded inL1Ω×0,T.


### 3.1.2. The Second Estimate

Replacingwj by −Δwj in (51), multiplying by ctjk, and summing over j from 1 to k, it follows that(73)∫Ωutktluttkt−Δutkdx+∫ΩM∇uk2ΔukΔutkdx+∫ΩΔuttkΔutkdx−∫0tht−s∫ΩΔukΔutkdxds−μ1∫ΩΔutkg1utkdx−μ2∫ΩΔutkg2zkx,1,tdx=−ν∫ΩΔutkuklnukdx.Noting thatM∇uk2≥a and by using Lemma 5, we obtain(74)−∫ΩutktluttktΔutkdx+12ddta−∫0thsdsΔuk2+Δutk2+hoΔuk+12htΔuk2−12h′oΔuk+μ1∫Ω∇utk2g1′utkdx+μ2∫Ω∇utk∇zkx,1,tg2′zkx,1,tdx≤−ν∫ΩΔutkuklnukdx.By using the Green formula, we have(75)−∫ΩutktluttkΔutkdx=ddt∫Ωutktl∇utk2dx−l+1∫Ωutktl∇uttk∇utkdx.Consequently, equation (74) yields(76)12ddt2∫Ωutktl∇utk2dx+a−∫0thsdsΔuk2+Δutk2+hoΔuk−l+1∫Ωutktl∇uttk∇utkdx+12htΔuk2−12h′oΔuk+μ1∫Ω∇utk2g1′utkdx+μ2∫Ω∇utk∇zkx,1,tg2′zkx,1,tdx≤−ν∫ΩΔutkuklnukdx.To estimate the term on the right-hand side of (76), we apply Lemma 4 with ϵ0=1/2 and use repeatedly Young’s, Cauchy-Schwartz’s, and the embedding inequalities as follows:(77)ν∫ΩΔutkuklnukdx≤ν∫ΩΔutkuk2+dϵ0ukdx≤νη∫ΩΔutk2dx+14η∫Ωuk2+dϵ0uk2dx≤νη∫ΩΔutk2dx+c4η∫Ωuk4dx+∫Ωukdx≤νηΔutk2+c4η∇uk4+uk,η>0.Combining (76) and (77) to have(78)12ddt2∫Ωutktl∇utk2dx+a−∫0thsdsΔuk2+Δutk2+hoΔuk−l+1∫Ωutktl∇uttk∇utkdx+12htΔuk2−12h′oΔuk+μ1∫Ω∇utk2g1′utkdx+μ2∫Ω∇utk∇zkx,1,tg2′zkx,1,tdx≤νηΔutk2+c4η∇uk4+uk.Replacingϕj by −Δϕj in (54), multiplying by djk, and summing over j from 1 to k, it follows that(79)τt1−τ′tρ∫Ω∇ztk∇zkdx+∫Ω∇zρk∇zkdx=0.Then, we get(80)12ddtτt1−τ′tρ∇zk2−τt1−τ′tρ′∇zk2+12ddρ∇zk2=0.We integrate over0,1, and we find(81)12ddt∫01τt1−τ′tρ∇zkx,ρ,t2dρ+12∇zkx,1,t2=12∫01τt1−τ′tρ′∇zkx,ρ,t2dρ+12∇utkt2.Combining (78) and (81) and using (A2), we get(82)12ddt2∫Ωutktl∇utk2dx+a−∫0thsdsΔuk2+Δutk2+hoΔuk+∫01τt1−τ′tρ∇zkx,ρ,t2dρ+μ1∫Ω∇utk2g1′utkdx+12∇zkx,1,t2≤l+1∫Ωutktl∇uttk∇utkdx−μ2∫Ω∇utk∇zkx,1,tg2′zkx,1,tdx+12∫01τt1−τ′tρ′∇zkx,ρ,t2dρ+12∇utkt2+νηΔutk2+c4η∇uk4+uk.From the first estimate (67) and Young’s inequality, we get(83)∫Ωutkl∇uttkt∇utkdx≤A1l/l+2+1/2∇uttk2≤η∇uttk2+A12l/l+2+14η.Using (17) and Chaucy-Schwarz’s inequality, we obtain(84)∫Ω∇utk∇zkx,1,tg2′zkx,1,tdx≤c32ε2∫Ω∇utk2dx+ε2∫Ω∇zkx,1,t2dx.Taking into account (83) and (84) into (82) yields(85)12ddt2∫Ωutktl∇utk2dx+a−∫0thsdsΔuk2+Δutk2+hoΔuk+∫01τt1−τ′tρ∇zkx,ρ,t2dρ+μ1∫Ω∇utk2g1′utkdx+c∇zkx,1,t2≤l+1η∇uttk2+Cη′A1+12∫01τt1−τ′tρ′∇zkx,ρ,t2dρ+c′∇utk2+νηΔutk2+c4ηθ4+uk.Multiplying (51) by cttjk and summing over j from 1 to k, it follows that(86)∫Ωutkluttk2dx+∫ΩM∇uk2∇uk∇uttkdx+∇uttk2−∫0tht−s∫Ω∇uks∇uttktdxds+μ1∫Ωuttkg1utkdx+μ2∫Ωuttkg2zkx,1,tdx=ν∫Ωuttkuklnukdx.Then,(87)∫Ωutkluttk2dx+∇uttk2=−∫ΩM∇uk2∇uk∇uttkdx+∫0tht−s∫Ω∇uks∇uttktdxds−μ1∫Ωuttkg1utkdx−μ2∫Ωuttkg2zkx,1,tdx+ν∫Ωuttkuklnukdx.Differentiating (54) with respect to t, we get(88)τt1−ρτ′t′ztk+τt1−ρτ′tzttk+ztρk,ϕj=0.Multiplying bydtjk and summing over j from 1 to k, it follows that(89)τt1−ρτ′t′ztk2+12τt1−ρτ′tddtztk2+12ddρztk2=0.Then, we have(90)12τt1−ρτ′t′ztk2+12ddtτt1−ρτ′tztk2+12ddρztk2=0.Integrating over0,1 with respect to ρ, we obtain(91)12∫01τt1−ρτ′t′ztk2dρ+12ddt∫01τt1−ρτ′tztk2dρ+12ztkx,1,t2−12uttkx,t2=0.Summing (87) and (91), we get(92)∫Ωutkluttk2dx+∇uttk2+12ddt∫01τt1−ρτ′tztk2dρ+12ztkx,1,t2=12uttkx,t2−12∫01τt1−ρτ′t′ztk2dρ−∫ΩM∇uk2∇uk∇uttkdx+∫0tht−s∫Ω∇uks∇uttktdxds+ν∫Ωuttkuklnukdx−μ1∫Ωuttkg1utkdx−μ2∫Ωuttkg2zkx,1,tdx.By Cauchy-Schwarz’s, Sobolev’s, and Young’s inequalities, the right hand side of (92) can be estimated as follows:(93)∫ΩM∇uk2∇uk∇uttkdx≤a+b∇uk2γ∫Ω∇uk∇uttkdx≤a+bE02γ/2γ+1∫Ω∇uk∇uttkdx≤η∇uttk2+m024η∇uk2,(94)∫0tht−s∫Ω∇uks∇uttktdxds≤η∇uttk2+14η∫Ω∫0tht−s∇uksds2dx≤η∇uttk2+14η∫0thsds∫Ω∫0tht−s∇uks2dsdx≤η∇uttk2+14ηa−k∫Ω∫0tht−s∇uks2dsdx≤η∇uttk2+14ηa−k∫0tht−s∇uks2ds≤η∇uttk2+14ηa−kh0∫0t∇uks2ds,and from (16),(95)∫Ωuttkg1utkdx≤12∫Ωuttk2dx+12∫Ωg1utkdx≤12∫Ωuttk2dx+12∫utk≥εg1utk2dx+12∫utk≤εg1utk2dx≤12∫Ωuttk2dx+12∫utk≥εutkg1utkdx+12∫ΩH−1utkg1utkdx.Using Lemma6, Jensen’s inequality, and the concavity of H−1, we obtain(96)∫Ωuttkg1utkdx≤12∫Ωuttk2dx+12∫utk≥εutkg1utkdx+cH−1∫Ωutkg1utkdx+k∫Ωuttkuklnukdx≤12∫Ωuttk2dx+12∫utk≥εutkg1utkdx+c′H∗1+c′′∫Ωutkg1utkdx≤12uttk2dx+c′H∗1+c′′∫Ωutkg1utkdx≤Cs∇uttk2+c′H∗1+c−E′.From (17) (that is, g2s≤cs,∀s∈ℝ), we get(97)∫Ωuttkg2zkx,1,tdx≤12∫Ωuttk2dx+12∫Ωg2zkx,1,t2dx≤12uttk2+c3∫Ωzkx,1,tg2zkx,1,tdx≤Cs∇uttk2+c′−E′.Similar to (77), we get(98)ν∫Ωuttkuklnukdx≤cη∇uttk2+c4η∇uk4+uk.Substituting (93)–(98) into (92) yields(99)∫Ωutkluttk2dx+∇uttk2+12ddt∫01τt1−ρτ′tztk2dρ+12ztkx,1,t2≤2η+3Cs∇uttk2++m024η∇uk2+14ηa−kh0∫0t∇uks2ds+c−E′−12∫01τt1−ρτ′t′ztk2dρ+cη∇uttk2+c4η∇uk4+uk.Combining (85) and (99), we get(100)12ddt2∫Ωutktl∇utk2dx+a−∫0thsdsΔuk2+Δutk2+hoΔuk+∫01τt1−τ′tρ∇zkx,ρ,t2dρ+∫01τt1−ρτ′tztk2dρ+cEt+∫Ωutkluttk2dx+12ztkx,1,t2+1−l+3η−3Cs∇uttk2+μ1∫Ω∇utk2g1′utkdx+c∇zkx,1,t2≤Cη′A1+12∫01τt1−ρτ′t′∇zkx,ρ,t2dρ+c′∇utk2+m024η∇uk2+cη∇uttk2+c44η∇uk4+uk+νηΔutk2.Then, from (67) and by integration over 0,t, (100) yields(101)a−∫0thsdsΔuk2+Δutk2+hoΔuk+∫01τt1−τ′tρ∇zkx,ρ,t2dρ+∫01τt1−ρτ′tztk2dρ+cEt+2∫Ωutktl∇utk2dx+2∫0t∫Ωutkluttk2dxds+c∫0tztkx,1,t2ds+21−l+3−cη−3Cs∫0t∇uttk2ds+c∗∫0t∇zkx,1,t2ds≤Cη′A1+c′+m022η+12ηa−kh0TT+E0+A2+νη∫0tΔutk2ds+c∗′′∫0t∫01τt1−τ′tρ′∇zkx,ρ,t2dρds+c′′∫0t∫01τt1−ρτ′t′ztk2dρds.For a suitableη, we get(102)Δuk2+Δutk2+hoΔuk+∫01τt1−τ′tρ∇zkx,ρ,t2dρ+∫01τt1−ρτ′tztk2dρ+∫0t∇uttk2ds≤Cδ′A1+A1′T+A2′+c∗′′∫0t∫01τt1−ρτ′t′∇zkx,ρ,t2dρds+c′′∫01τt1−ρτ′t′ztk2dρ+c∫0tΔutk2ds.Using Gronwall lemma, we obtain(103)Δuk2+Δutk2+hoΔuk+∫01τt1−τ′tρ∇zkx,ρ,t2dρ+∫01τt1−ρτ′tztk2dρ+∫0t∇uttk2ds≤C3.We observe from the estimates (67) and (103) that there exists a subsequence um of uk and functions u,z,χ,andψ such that(104)um⇀uweakly star inL∞0,T,H2Ω∩H01Ω,(105)utm⇀utweakly star inL∞0,T,H02Ω,(106)g1utm⇀χweakly star inL2Ω×0,T,(107)uttm⇀uttweakly star inL20,T,H01Ω,(108)zm⇀zweakly star inL∞0,T,H01Ω,L20,1,(109)ztm⇀ztweakly star inL∞0,T,L2Ω×0,1,(110)g2zmx,1,t⇀ψweakly star inL2Ω×0,T.Now, we will prove thatu is the solution of (1). First, we will treat the nonlinear terms.(1)
Termutklutk: from the first estimate (67) and Lemma 1, we deduce(111)utklutkL20,T,L2Ω=∫0Tutk2l+12l+1dt≤Cs2l+1∫0T∇utk2l+1dt≤Cs2l+1A12l+1T.On the other hand, from Aubin-Lions theorem (see Lions [36]), we deduce that there exists a subsequence of um, still denoted by um such that(112)utm⟶utstrongly inL20,T,L2Ω,which implies that(113)utm⟶utalmost everywhere inA.Hence,(114)utmlutm⟶utlutalmost everywhere inA,where A=Ω×0,T. Thus, using (117), (114), and Lions Lemma, we derive(115)utmlutm⇀utlutweakly inL20,T,L2Ω,(116)zm⟶zstrongly inL20,T,L2Ω,which implies zm⟶z almost everywhere in A.(2) Termuklnuk: using (103), we have uk being bounded in L∞0,T,H02Ω which implies the boundedness of uk in L2A. Similarly, utk is bounded in L2A. Then, from Aubin-Lions theorem, we find a subsequence such that(117)um⟶ustrongly inL2A,which implies(118)um⟶ualmost everywhere inA.Since the maps⟶νslns is continuous, we have the following convergence:(119)νumlnum⟶νulnualmost every where inA.Using the embedding ofH01Ω in L∞Ω, it is clear that νumlnum is bounded in L∞Ω×0,T. Next, taking into account the Lebesgue bounded convergence theorem Ωis bounded, we get(120)νumlnum⟶νulnustrongly inL20,T;L2Ω.Lemma 7.
For eachT>0,g1ut,g2zx,1,t∈L1A, and g1u′L1A,g2zx,1,tL1A≤K, where K is a constant independent of t.Proof.
By (A2) and (118), we have(121)g1utmx,t⟶g1utx,talmost everywhere inA,0≤utkx,tg1utmx,t⟶utx,tg1utx,talmost everywhere inA.
Hence, by (71) and Fatou’s Lemma, we have(122)∫0T∫Ωutx,tg1utx,tdxdt≤K1,forT>0.
By using Cauchy-Schwarz’s inequality, (96), and (122), we have(123)∫0T∫Ωg1utx,tdxdt≤cA1/2∫0T∫Ωutx,tg1utx,tdxdt1/2≤cA1/2K11/2≡K.Lemma 8.
We haveg1utk⇀g1utweak inL2Ω×0,T and g2zk⇀g2zweak inL2Ω×0,T.Proof.
LetE⊂Ω×0,T and set(124)E1=x,t∈E,g1utkx,t≤1E,E2=EE1,where E is the measure of E. If Mr=infs,s∈ℝandgs≥r,(125)∫Eg1utkdxdt≤cE+M1E−1∫E2utkg1utkdxdt.
By applying (71), we deduce that supk∫Eg1utkdxdt⟶0 as E⟶0. From Vitali’s convergence theorem, we deduce that(126)g1utk⟶g1utinL1Ω×0,T.
Hence,(127)g1utk⇀g1utweak inL2Ω×0,T.
Similarly, we have(128)g2zk⇀g2zweak inL2Ω×0,T.Remark 2.
By using (103) and from (104) and (105) combined with the Aubin-Lions compactness lemma, we deduce(129)∫0TM∇ukt2Δukt,wθtdt⟶∫0TM∇ut2Δut,wθtdt,ask⟶∞.
By multiplying (51) and (54) by θt∈D0,T and by integrating over 0,T, it follows that(130)∫0Tutktluttkt,wjθtdt+∫0TM∇ukt2∇ukt,∇wjθtdt+∫0T∇uttk,∇wjθtdt−∫0T∫0tht−s∇uks,∇wjθtdsdt+μ1∫0Tg1utk,wjθtdt+μ2∫0Tg2zk.,1,wjθtdt=ν∫0Tukslnuks,wjθtdt,∫0T∫01∫Ωτtztk+1−ρτ′tzρkϕjθtdxdρdt=0,for all j=1,…,k.
The convergence of (104)–(110), (115), and (124)–(129) is sufficient to pass to the limit in (130) in order to obtain(131)∫0T∫Ωutlutt−M∇ut2Δu−Δutt+∫0tht−sΔusds+μ1g1ut+μ2g2z.,1wθtdsdt=ν∫0Tuslnus,wθtdt,∫0T∫01∫Ωτtzt+1−ρτ′tzρϕjθtdxdρdt=0.
Then, problem (1) has a weak solution on 0,T, T>0.


## 3.1.1. The First Estimate

Since the sequencesu0k,u1k, and z0k converge and from Lemma 6, we can find a positive constant C1 independent of k such that(57)Ekt−Ek0≤−λ∫0t∫Ωutkg1utkdxds−β∫0t∫Ωzkx,1,sg2zkx,1,sdxds−12∫0ths∇uks2ds+12∫0th′o∇uksds≤−λ∫0t∫Ωutkg1utkdxds−β∫0t∫Ωzkx,1,sg2zkx,1,sdxds.Ash is a positive nonincreasing function, we get(58)Ekt+λ∫0t∫Ωutkg1utkdxds+β∫0t∫Ωzkx,1,sg2zkx,1,sdxds≤Ek0≤C1,where(59)Ekt=1l+2utkl+2l+2+b2γ+1∇uk2γ+1+12a−∫0thsds∇uk2+12∇utk2+12ho∇ukt−ν2∫Ωuk2lnukdx+ν4uk2+ξτt∫Ω∫01Gzkx,ρ,tdρdx.By applying the Logarithmic Sobolev inequality, (58) yields(60)utkl+2l+2+k−νσ22π∇uk2+∇utk2+ν2+ν1+lnσuk2+h∘∇ukt+∫Ω∫01Gzkx,ρ,tdρdx+∫0t∫Ωutkg1utkdxds+∫0t∫Ωzkx,1,sg2zkx,1,sdxds≤C2+uk2lnuk2,where C2 is a positive constant depending only on u0H01,u1H01,l,γ,ξ,τ1,λ, and β.By choosing(61)e−3/2<σ<2πkν,we obtain k−νσ2/2π>0 and ν/2+ν1+lnσ>0.This selection is possible thanks toA6. So we get(62)utkl+2l+2+∇uk2+∇utk2+uk2+h∘∇ukt+∫Ω∫01Gzkx,ρ,tdρdx+∫0t∫Ωutkg1utkdxds+∫0t∫Ωzkx,1,sg2zkx,1,sdxds≤c1+uk2lnuk2.Let us note that(63)ukt=uk0+∫0tusksds.Then, by using Cauchy Schwarz’s inequality, we get(64)uk2≤uk02+2∫0tusksds2≤uk02+2T∫0tusks2ds.Hence, (62) gives(65)uk2≤C1+∫0tuk2lnuk2ds,where C=max2Tc,2uk02. Applying the Logarithmic Gronwall inequality to (65), we obtain(66)uk2≤CeCT.Hence, from (58), we obtain the first estimate:(67)utkl+2l+2+∇uk2+∇utk2+uk2+h°∇ukt+∫Ω∫01Gzkx,ρ,tdρdx+∫0t∫Ωutkg1utkdxds+∫0t∫Ωzkx,1,sg2zkx,1,sdxds≤c1+CeCTlnCeCT=A1,The estimate implies that the solutionuk,zk exists in 0,T and it yields(68)ukis bounded inLloc∞0,∞,H1Ω,(69)utkis bounded inLloc∞0,∞,H01Ω,(70)Gzkx,ρ,tis bounded inLloc∞0,∞,L1Ω×0,1,(71)utktg1utktis bounded inL1Ω×0,T,(72)zkx,1,tg2zkx,1,tis bounded inL1Ω×0,T.


## 3.1.2. The Second Estimate

Replacingwj by −Δwj in (51), multiplying by ctjk, and summing over j from 1 to k, it follows that(73)∫Ωutktluttkt−Δutkdx+∫ΩM∇uk2ΔukΔutkdx+∫ΩΔuttkΔutkdx−∫0tht−s∫ΩΔukΔutkdxds−μ1∫ΩΔutkg1utkdx−μ2∫ΩΔutkg2zkx,1,tdx=−ν∫ΩΔutkuklnukdx.Noting thatM∇uk2≥a and by using Lemma 5, we obtain(74)−∫ΩutktluttktΔutkdx+12ddta−∫0thsdsΔuk2+Δutk2+hoΔuk+12htΔuk2−12h′oΔuk+μ1∫Ω∇utk2g1′utkdx+μ2∫Ω∇utk∇zkx,1,tg2′zkx,1,tdx≤−ν∫ΩΔutkuklnukdx.By using the Green formula, we have(75)−∫ΩutktluttkΔutkdx=ddt∫Ωutktl∇utk2dx−l+1∫Ωutktl∇uttk∇utkdx.Consequently, equation (74) yields(76)12ddt2∫Ωutktl∇utk2dx+a−∫0thsdsΔuk2+Δutk2+hoΔuk−l+1∫Ωutktl∇uttk∇utkdx+12htΔuk2−12h′oΔuk+μ1∫Ω∇utk2g1′utkdx+μ2∫Ω∇utk∇zkx,1,tg2′zkx,1,tdx≤−ν∫ΩΔutkuklnukdx.To estimate the term on the right-hand side of (76), we apply Lemma 4 with ϵ0=1/2 and use repeatedly Young’s, Cauchy-Schwartz’s, and the embedding inequalities as follows:(77)ν∫ΩΔutkuklnukdx≤ν∫ΩΔutkuk2+dϵ0ukdx≤νη∫ΩΔutk2dx+14η∫Ωuk2+dϵ0uk2dx≤νη∫ΩΔutk2dx+c4η∫Ωuk4dx+∫Ωukdx≤νηΔutk2+c4η∇uk4+uk,η>0.Combining (76) and (77) to have(78)12ddt2∫Ωutktl∇utk2dx+a−∫0thsdsΔuk2+Δutk2+hoΔuk−l+1∫Ωutktl∇uttk∇utkdx+12htΔuk2−12h′oΔuk+μ1∫Ω∇utk2g1′utkdx+μ2∫Ω∇utk∇zkx,1,tg2′zkx,1,tdx≤νηΔutk2+c4η∇uk4+uk.Replacingϕj by −Δϕj in (54), multiplying by djk, and summing over j from 1 to k, it follows that(79)τt1−τ′tρ∫Ω∇ztk∇zkdx+∫Ω∇zρk∇zkdx=0.Then, we get(80)12ddtτt1−τ′tρ∇zk2−τt1−τ′tρ′∇zk2+12ddρ∇zk2=0.We integrate over0,1, and we find(81)12ddt∫01τt1−τ′tρ∇zkx,ρ,t2dρ+12∇zkx,1,t2=12∫01τt1−τ′tρ′∇zkx,ρ,t2dρ+12∇utkt2.Combining (78) and (81) and using (A2), we get(82)12ddt2∫Ωutktl∇utk2dx+a−∫0thsdsΔuk2+Δutk2+hoΔuk+∫01τt1−τ′tρ∇zkx,ρ,t2dρ+μ1∫Ω∇utk2g1′utkdx+12∇zkx,1,t2≤l+1∫Ωutktl∇uttk∇utkdx−μ2∫Ω∇utk∇zkx,1,tg2′zkx,1,tdx+12∫01τt1−τ′tρ′∇zkx,ρ,t2dρ+12∇utkt2+νηΔutk2+c4η∇uk4+uk.From the first estimate (67) and Young’s inequality, we get(83)∫Ωutkl∇uttkt∇utkdx≤A1l/l+2+1/2∇uttk2≤η∇uttk2+A12l/l+2+14η.Using (17) and Chaucy-Schwarz’s inequality, we obtain(84)∫Ω∇utk∇zkx,1,tg2′zkx,1,tdx≤c32ε2∫Ω∇utk2dx+ε2∫Ω∇zkx,1,t2dx.Taking into account (83) and (84) into (82) yields(85)12ddt2∫Ωutktl∇utk2dx+a−∫0thsdsΔuk2+Δutk2+hoΔuk+∫01τt1−τ′tρ∇zkx,ρ,t2dρ+μ1∫Ω∇utk2g1′utkdx+c∇zkx,1,t2≤l+1η∇uttk2+Cη′A1+12∫01τt1−τ′tρ′∇zkx,ρ,t2dρ+c′∇utk2+νηΔutk2+c4ηθ4+uk.Multiplying (51) by cttjk and summing over j from 1 to k, it follows that(86)∫Ωutkluttk2dx+∫ΩM∇uk2∇uk∇uttkdx+∇uttk2−∫0tht−s∫Ω∇uks∇uttktdxds+μ1∫Ωuttkg1utkdx+μ2∫Ωuttkg2zkx,1,tdx=ν∫Ωuttkuklnukdx.Then,(87)∫Ωutkluttk2dx+∇uttk2=−∫ΩM∇uk2∇uk∇uttkdx+∫0tht−s∫Ω∇uks∇uttktdxds−μ1∫Ωuttkg1utkdx−μ2∫Ωuttkg2zkx,1,tdx+ν∫Ωuttkuklnukdx.Differentiating (54) with respect to t, we get(88)τt1−ρτ′t′ztk+τt1−ρτ′tzttk+ztρk,ϕj=0.Multiplying bydtjk and summing over j from 1 to k, it follows that(89)τt1−ρτ′t′ztk2+12τt1−ρτ′tddtztk2+12ddρztk2=0.Then, we have(90)12τt1−ρτ′t′ztk2+12ddtτt1−ρτ′tztk2+12ddρztk2=0.Integrating over0,1 with respect to ρ, we obtain(91)12∫01τt1−ρτ′t′ztk2dρ+12ddt∫01τt1−ρτ′tztk2dρ+12ztkx,1,t2−12uttkx,t2=0.Summing (87) and (91), we get(92)∫Ωutkluttk2dx+∇uttk2+12ddt∫01τt1−ρτ′tztk2dρ+12ztkx,1,t2=12uttkx,t2−12∫01τt1−ρτ′t′ztk2dρ−∫ΩM∇uk2∇uk∇uttkdx+∫0tht−s∫Ω∇uks∇uttktdxds+ν∫Ωuttkuklnukdx−μ1∫Ωuttkg1utkdx−μ2∫Ωuttkg2zkx,1,tdx.By Cauchy-Schwarz’s, Sobolev’s, and Young’s inequalities, the right hand side of (92) can be estimated as follows:(93)∫ΩM∇uk2∇uk∇uttkdx≤a+b∇uk2γ∫Ω∇uk∇uttkdx≤a+bE02γ/2γ+1∫Ω∇uk∇uttkdx≤η∇uttk2+m024η∇uk2,(94)∫0tht−s∫Ω∇uks∇uttktdxds≤η∇uttk2+14η∫Ω∫0tht−s∇uksds2dx≤η∇uttk2+14η∫0thsds∫Ω∫0tht−s∇uks2dsdx≤η∇uttk2+14ηa−k∫Ω∫0tht−s∇uks2dsdx≤η∇uttk2+14ηa−k∫0tht−s∇uks2ds≤η∇uttk2+14ηa−kh0∫0t∇uks2ds,and from (16),(95)∫Ωuttkg1utkdx≤12∫Ωuttk2dx+12∫Ωg1utkdx≤12∫Ωuttk2dx+12∫utk≥εg1utk2dx+12∫utk≤εg1utk2dx≤12∫Ωuttk2dx+12∫utk≥εutkg1utkdx+12∫ΩH−1utkg1utkdx.Using Lemma6, Jensen’s inequality, and the concavity of H−1, we obtain(96)∫Ωuttkg1utkdx≤12∫Ωuttk2dx+12∫utk≥εutkg1utkdx+cH−1∫Ωutkg1utkdx+k∫Ωuttkuklnukdx≤12∫Ωuttk2dx+12∫utk≥εutkg1utkdx+c′H∗1+c′′∫Ωutkg1utkdx≤12uttk2dx+c′H∗1+c′′∫Ωutkg1utkdx≤Cs∇uttk2+c′H∗1+c−E′.From (17) (that is, g2s≤cs,∀s∈ℝ), we get(97)∫Ωuttkg2zkx,1,tdx≤12∫Ωuttk2dx+12∫Ωg2zkx,1,t2dx≤12uttk2+c3∫Ωzkx,1,tg2zkx,1,tdx≤Cs∇uttk2+c′−E′.Similar to (77), we get(98)ν∫Ωuttkuklnukdx≤cη∇uttk2+c4η∇uk4+uk.Substituting (93)–(98) into (92) yields(99)∫Ωutkluttk2dx+∇uttk2+12ddt∫01τt1−ρτ′tztk2dρ+12ztkx,1,t2≤2η+3Cs∇uttk2++m024η∇uk2+14ηa−kh0∫0t∇uks2ds+c−E′−12∫01τt1−ρτ′t′ztk2dρ+cη∇uttk2+c4η∇uk4+uk.Combining (85) and (99), we get(100)12ddt2∫Ωutktl∇utk2dx+a−∫0thsdsΔuk2+Δutk2+hoΔuk+∫01τt1−τ′tρ∇zkx,ρ,t2dρ+∫01τt1−ρτ′tztk2dρ+cEt+∫Ωutkluttk2dx+12ztkx,1,t2+1−l+3η−3Cs∇uttk2+μ1∫Ω∇utk2g1′utkdx+c∇zkx,1,t2≤Cη′A1+12∫01τt1−ρτ′t′∇zkx,ρ,t2dρ+c′∇utk2+m024η∇uk2+cη∇uttk2+c44η∇uk4+uk+νηΔutk2.Then, from (67) and by integration over 0,t, (100) yields(101)a−∫0thsdsΔuk2+Δutk2+hoΔuk+∫01τt1−τ′tρ∇zkx,ρ,t2dρ+∫01τt1−ρτ′tztk2dρ+cEt+2∫Ωutktl∇utk2dx+2∫0t∫Ωutkluttk2dxds+c∫0tztkx,1,t2ds+21−l+3−cη−3Cs∫0t∇uttk2ds+c∗∫0t∇zkx,1,t2ds≤Cη′A1+c′+m022η+12ηa−kh0TT+E0+A2+νη∫0tΔutk2ds+c∗′′∫0t∫01τt1−τ′tρ′∇zkx,ρ,t2dρds+c′′∫0t∫01τt1−ρτ′t′ztk2dρds.For a suitableη, we get(102)Δuk2+Δutk2+hoΔuk+∫01τt1−τ′tρ∇zkx,ρ,t2dρ+∫01τt1−ρτ′tztk2dρ+∫0t∇uttk2ds≤Cδ′A1+A1′T+A2′+c∗′′∫0t∫01τt1−ρτ′t′∇zkx,ρ,t2dρds+c′′∫01τt1−ρτ′t′ztk2dρ+c∫0tΔutk2ds.Using Gronwall lemma, we obtain(103)Δuk2+Δutk2+hoΔuk+∫01τt1−τ′tρ∇zkx,ρ,t2dρ+∫01τt1−ρτ′tztk2dρ+∫0t∇uttk2ds≤C3.We observe from the estimates (67) and (103) that there exists a subsequence um of uk and functions u,z,χ,andψ such that(104)um⇀uweakly star inL∞0,T,H2Ω∩H01Ω,(105)utm⇀utweakly star inL∞0,T,H02Ω,(106)g1utm⇀χweakly star inL2Ω×0,T,(107)uttm⇀uttweakly star inL20,T,H01Ω,(108)zm⇀zweakly star inL∞0,T,H01Ω,L20,1,(109)ztm⇀ztweakly star inL∞0,T,L2Ω×0,1,(110)g2zmx,1,t⇀ψweakly star inL2Ω×0,T.Now, we will prove thatu is the solution of (1). First, we will treat the nonlinear terms.(1)
Termutklutk: from the first estimate (67) and Lemma 1, we deduce(111)utklutkL20,T,L2Ω=∫0Tutk2l+12l+1dt≤Cs2l+1∫0T∇utk2l+1dt≤Cs2l+1A12l+1T.On the other hand, from Aubin-Lions theorem (see Lions [36]), we deduce that there exists a subsequence of um, still denoted by um such that(112)utm⟶utstrongly inL20,T,L2Ω,which implies that(113)utm⟶utalmost everywhere inA.Hence,(114)utmlutm⟶utlutalmost everywhere inA,where A=Ω×0,T. Thus, using (117), (114), and Lions Lemma, we derive(115)utmlutm⇀utlutweakly inL20,T,L2Ω,(116)zm⟶zstrongly inL20,T,L2Ω,which implies zm⟶z almost everywhere in A.(2) Termuklnuk: using (103), we have uk being bounded in L∞0,T,H02Ω which implies the boundedness of uk in L2A. Similarly, utk is bounded in L2A. Then, from Aubin-Lions theorem, we find a subsequence such that(117)um⟶ustrongly inL2A,which implies(118)um⟶ualmost everywhere inA.Since the maps⟶νslns is continuous, we have the following convergence:(119)νumlnum⟶νulnualmost every where inA.Using the embedding ofH01Ω in L∞Ω, it is clear that νumlnum is bounded in L∞Ω×0,T. Next, taking into account the Lebesgue bounded convergence theorem Ωis bounded, we get(120)νumlnum⟶νulnustrongly inL20,T;L2Ω.Lemma 7.
For eachT>0,g1ut,g2zx,1,t∈L1A, and g1u′L1A,g2zx,1,tL1A≤K, where K is a constant independent of t.Proof.
By (A2) and (118), we have(121)g1utmx,t⟶g1utx,talmost everywhere inA,0≤utkx,tg1utmx,t⟶utx,tg1utx,talmost everywhere inA.
Hence, by (71) and Fatou’s Lemma, we have(122)∫0T∫Ωutx,tg1utx,tdxdt≤K1,forT>0.
By using Cauchy-Schwarz’s inequality, (96), and (122), we have(123)∫0T∫Ωg1utx,tdxdt≤cA1/2∫0T∫Ωutx,tg1utx,tdxdt1/2≤cA1/2K11/2≡K.Lemma 8.
We haveg1utk⇀g1utweak inL2Ω×0,T and g2zk⇀g2zweak inL2Ω×0,T.Proof.
LetE⊂Ω×0,T and set(124)E1=x,t∈E,g1utkx,t≤1E,E2=EE1,where E is the measure of E. If Mr=infs,s∈ℝandgs≥r,(125)∫Eg1utkdxdt≤cE+M1E−1∫E2utkg1utkdxdt.
By applying (71), we deduce that supk∫Eg1utkdxdt⟶0 as E⟶0. From Vitali’s convergence theorem, we deduce that(126)g1utk⟶g1utinL1Ω×0,T.
Hence,(127)g1utk⇀g1utweak inL2Ω×0,T.
Similarly, we have(128)g2zk⇀g2zweak inL2Ω×0,T.Remark 2.
By using (103) and from (104) and (105) combined with the Aubin-Lions compactness lemma, we deduce(129)∫0TM∇ukt2Δukt,wθtdt⟶∫0TM∇ut2Δut,wθtdt,ask⟶∞.
By multiplying (51) and (54) by θt∈D0,T and by integrating over 0,T, it follows that(130)∫0Tutktluttkt,wjθtdt+∫0TM∇ukt2∇ukt,∇wjθtdt+∫0T∇uttk,∇wjθtdt−∫0T∫0tht−s∇uks,∇wjθtdsdt+μ1∫0Tg1utk,wjθtdt+μ2∫0Tg2zk.,1,wjθtdt=ν∫0Tukslnuks,wjθtdt,∫0T∫01∫Ωτtztk+1−ρτ′tzρkϕjθtdxdρdt=0,for all j=1,…,k.
The convergence of (104)–(110), (115), and (124)–(129) is sufficient to pass to the limit in (130) in order to obtain(131)∫0T∫Ωutlutt−M∇ut2Δu−Δutt+∫0tht−sΔusds+μ1g1ut+μ2g2z.,1wθtdsdt=ν∫0Tuslnus,wθtdt,∫0T∫01∫Ωτtzt+1−ρτ′tzρϕjθtdxdρdt=0.
Then, problem (1) has a weak solution on 0,T, T>0.


## 3.2. Global Existence

To state and prove our global existence, we introduce the following functionals:(132)It=a−∫0thsds∇u2+∇ut2+ho∇ut−3ν∫Ωu2lnudx,(133)Jt=12a−∫0thsds∇u2+12∇ut2+12ho∇ut−ν2∫Ωu2lnudx+ν4u2=13a−∫0thsds∇u2+∇ut2+ho∇ut+ν4u2+16It.We note that(134)Et=1l+2utl+2l+2+b2γ+1∇u2γ+1+Jt+ξτt∫Ω∫01Gzx,ρ,tdρdx.Lemma 9.
The following inequalities hold:(135)−d0Ωcs3∇u3/2≤∫Ωu2lnudx≤cs3∇u3,∀u∈H01Ω,where d0=sup0<s<1slns=2/e, Ω is the Lesbegue measure of Ω, and cs is the smallest embedding constant:(136)u3≤cs∇u,∀u∈H01Ω.Proof.
LetΩ1=x∈Ω:u≤1 and Ω2=x∈Ω:ut>1. By using (136), we have(137)∫Ωu2lnudx=∫Ω1u2lnudx+∫Ω2u2lnudx≤∫Ω2u2lnudx≤∫Ω2u3dx≤∫Ω2u3dx≤cs3∇u3.
On the other hand, using Hölder’s inequality and (136), we get(138)−∫Ωu2lnudx=−∫Ω1u2lnudx−∫Ω2u2lnudx≤−∫Ω1u2lnudx=∫Ω1u2lnudx≤d0∫Ω1u3/2dx≤d0Ω∫Ω1u3dx1/2≤d0Ωcs3∇u3/2.
Hence, (135) is obtained.Lemma 10.
Assume that (A1)–(A6) hold. Letu0,u1∈H2Ω∩H01Ω×H01Ω such that(139)I0>0,27νcs3E0k1/2≤k.
Then,(140)It>0,∀t∈0,T.Proof.
SinceI0>0 and I is continuous on 0,T, there exists t∗∈0,T such that It>0, for all t∈0,t∗. Let us denote by t0 the largest real number in 0,T such that I>0 on 0,t0. We assume by contradiction that t0≠T, so we have It0=0 and from (133), we have(141)∇ut2≤3kJt≤3kEt≤3kE0,∀t∈0,t0.
The last inequality is obtained from Lemma6. If ∇ut02=0, then (132) and (135) give(142)0=It0=a−∫0t0hsds∇ut02+∇utt02+ho∇ut0−3ν∫Ωu2t0lnut0dx=ho∇ut0=∫0t0hs∇us2ds.
Consequently, ifh>0 on 0,t0, we get(143)∇us2=0,∀s∈0,t0.
Then,(144)It=0,∀s∈0,t0,which is not true since I>0 on 0,t0. If there exists t∗ in 0,t0 such that ht∗=0, then let t1∈0,t0 be the smallest real number such that ht1=0. Because h0>0 and h is positive, nonincreasing, and continuous on ℝ+, then t1>0 and h=0 on t1,+∞. Therefore, from (142), we deduce that(145)0=∫0t0hs∇us2ds=∫0t1hs∇us2ds.Then,(146)∇us2=0,∀s∈0,t1.
As given above, we get a contradiction with the fact thatI>0 on 0,t0. Then, we conclude that ∇ut02>0. From (132), we have(147)It0≥k∇ut02−3ν∫Ωut02lnut0dx.
By using (135) and (141), we have(148)It0≥k−3ν3kE01/2∇ut02.
By recalling (139), we arrive at It0>0, which contradicts the assumption that It0=0. Hence, t0=T and then I>0 on 0,T.
This completes the proof of Theorem1.


## 4. Uniform Decay of the Energy Proof of Theorem2

In this section, we study the solution’s asymptotic behavior of system (1).To prove our main result, we construct a Lyapunov functionalF equivalent to E. For this, we define some functionals which allow us to obtain the desired estimate.Lemma 11.
Letu,z be a solution of problem (12). Then, the functional(149)χt=τt∫Ω∫01e−2τtρGzx,ρ,tdρdxsatisfies the estimates(150)iχt≤1ξEt,iiχ′t≤−2τte−2τ1∫Ω∫01Gzx,ρ,tdρdx−α11−de−2τ1∫Ωzx,1,tg2zx,1,tdx+α2∫Ωutx,tg1utx,tdx.Proof.
(ii) Differentiating (149) with respect to t and using (16), (11), and A4, we get(151)ddtχt=τ′t∫Ω∫01e−2τtρGzx,ρ,tdρdx+τt∫Ω∫01e−2τtρ∂Gzx,ρ,t∂t−2τ′tρe−2τtρGzx,ρ,tdρdx=∫Ω∫01e−2τtρτ′tGzx,ρ,t+τt∂Gzx,ρ,t∂tdρdx−2∫Ω∫01τtτ′tρe−2τtρGzx,ρ,tdρdx=−∫Ω∫01e−2τtρ∂∂ρ1−ρτ′tGzx,ρ,tdρdx−2∫Ω∫01τtτ′tρe−2τtρGzx,ρ,tdρdx=−∫Ω∫01∂∂ρe−2τtρ1−τ′tρGzx,ρ,t+2τte−2τtρ1−τ′tρGzx,ρ,tdρdx−2τtτ′t∫Ω∫01ρe−2τtρGzx,ρ,tdρdx=∫ΩGutx,tdx−e−2τt1−τ′t∫ΩGzx,1,tdx−2τt∫Ω∫011−τ′tρ+τ′tρe−2τtρGzx,ρ,tdρdx≤−2χt+α2∫Ωutx,tg1utx,tdx−e−2τt1−dα1∫Ωzx,1,tg2zx,1,tdx.
Sincee−2τtρ is a decreasing function for ρ∈0,1 and τt∈τ0,τ1, we deduce(152)χt≥τt∫Ω∫01e−2τ1Gzx,ρ,tdρdx.
Thus, our proof is completed.Lemma 12.
Letu,z be a solution of problem (12). Then, the functional(153)ϕt=1l+1∫Ωutlutudx+∫Ω∇ut∇udxsatisfies the estimates(154)iϕt≤1l+2utl+2l+2+l+1−1l+2csl+22E0al/2+12∇u2+12∇ut2,iiϕ′t≤1l+1utl+2l+2−Ma∇u2∇u2+1+ηa−k∇u2+14ηho∇ut+∇ut2−μ1∫Ωux,tg1utx,tdx−μ2∫Ωux,tg2zx,1,tdx+ν∫Ωu2lnudx,where η>0 and cs is the Sobolev embedding constant.Proof.
(i)
From Young’s inequality, Sobolev embedding, and Lemma6, we deduce(155)ϕt≤1l+2utl+2l+2+l+1−1l+2ul+2l+2+12∇ut2+12∇u2≤1l+2utl+2l+2+l+1−1l+2csl+2∇ul+2+12∇ut2+12∇u2≤1l+2utl+2l+2+l+1−1l+2csl+22E0al/2+12∇u2+12∇ut2.(ii)
Differentiatingϕt with respect to t and using the first equation of (12), we get(156)ϕ′t=1l+1∫Ωutlut′udx+1l+1∫Ωutl+2dx+∫Ω∇utt∇udx+∫Ω∇ut∇utdx=∫Ωutluttudx+1l+1utl+2l+2−∫ΩΔuttudx+∇ut2=1l+1utl+2l+2+∫Ωutlutt−Δuttudx+∇ut2=1l+1utl+2l+2+∫ΩM∇u2Δu−∫0tht−sΔusds−μ1g1utx,t−μ2g2zx,1,t+νulnuudx+∇ut2=1l+1utl+2l+2−M∇u2∇u2+∫Ω∇ut∫0tht−s∇usdsdx−μ1∫Ωug1utx,tdx−μ2∫Ωug2zx,1,tdx+∇ut2+ν∫Ωu2lnudx.
By using Young’s inequality and Sobolev embedding, we can estimate the third term in the right side as follows:(157)∫Ω∇ut∫0tht−s∇usdsdx≤∫0tht−s∫Ω∇ut∇us−∇utdxds+∇ut2∫0tht−sds≤η∇ut2∫0thsds+14η∫0tht−s∇us−∇ut2ds+∇ut2∫0thsds≤1+ηa−k∇ut2+14ηho∇ut.
Thus, our proof is completed.Lemma 13.
Letu,z be a solution of problem (12). Then, the functional(158)ψt=∫ΩΔut−1l+1utlut∫0tht−sut−usdsdxsatisfies the estimates(159)iψt≤12∇ut2+12a−k+l+1−1l+2a−kl+2csl+24E0al/2ho∇ut+1l+2utl+2l+2,iiψ′t≤δa−k+b0M∇u2∇u2+2δa−k2∇u2+M04δ+2δ+14δ+μ1+μ2+νcs24δa−kho∇ut−h04δ1+cs2l+1h′o∇ut+δ+δa0l+1−∫0thsds∇ut2+μ1δg1utx,t2+μ2δg2zx,1,t2−1l+1∫0thsdsutl+2l+2,where(160)M0=a+b2E0aγ,a0=cs2l+12E0al,η>0,b0=ν2cs2maxΩa,cs2b,and cs is the Sobolev embedding constant.Proof.
(i)
We have(161)ψt=−∫Ω∇ut∫0tht−s∇ut−∇usdsdx−∫Ω1l+1utlut∫0tht−sut−usdsdx.We use Young’s and Hölder’s inequalities with the conjugate exponentsp=l+2/l+1 and q=l+2; the second term in the right hand side can be estimated as(162)−∫Ω1l+1utlut∫0tht−sut−usdsdx≤1l+2utl+2l+2+l+1−1l+2∫Ω∫0tht−sl+1/l+2ht−s1/l+2ut−usdsl+2dx≤1l+2utl+2l+2+l+1−1l+2∫0tht−sdsl+1∫Ω∫0tht−sut−usl+2dsdx≤1l+2utl+2l+2+l+1−1l+2a−kl+1csl+24E0al/2ho∇ut.We get the last inequality from (22) and Lemma 6. Similarly, we use Young’s and Hölder’s inequalities with p=q=2, we get(163)−∫Ω∇ut∫0tht−s∇ut−∇usdsdx≤12∇ut2+12∫Ω∫0tht−s∇ut−∇usds2dx≤12∇ut2+12a−kho∇ut.Combining (162) and (163), we deduce(164)ψt≤12∇ut2+12a−k+l+1−1l+2a−kl+2csl+24E0al/2ho∇ut+1l+2utl+2l+2.(ii)
We use the Leibniz formula and the first equation of (12), and we have(165)ψ′t=∫ΩΔutt−utlutt∫0tht−sut−usdsdx+∫ΩΔut−1l+1utlut∫0th′t−sut−us+ht−suttdsdx=∫ΩM∇u2∇ut∫0tht−s∇ut−∇usdsdx−∫Ω∫0tht−s∇usds∫0tht−s∇ut−∇usdsdx+μ1∫Ωg1utx,t∫0tht−sut−usdsdx+μ2∫Ωg2zx,1,t∫0tht−sut−usdsdx−ν∫Ωulnu∫0tht−sut−usdsdx−∫Ω∇ut∫0th′t−s∇ut−∇usdsdx−1l+1∫Ωutlut∫0th′t−sut−usdsdx−∇ut2∫0thsds−1l+1utl+2l+2∫0thsds=I1+I2+μ1I3+μ2I4+I5+I6+I7−∇ut2∫0thsds−1l+1utl+2l+2∫0thsds.
In what follows, we will estimateI1,…,I7. For I1, we use Hölder’s and Young’s inequalities with p=q=2, and we get(166)I1≤M∇u2∫Ω∇ut∫0thsds1/2∫0tht−s∇ut−∇us2ds1/2dx≤M∇u2δ∫Ω∇ut2∫0thsdsdx+14δ∫Ω∫0tht−s∇ut−∇us2dsdx≤M∇u2δ∇ut2∫0thsds+14δho∇ut≤δM∇u2∇ut2a−k+M04δho∇ut,where M0=a+b2E0/aγ obtained by recalling (22) and Lemma 6. Similarly,(167)I2≤δ∫Ω∫0tht−s∇usds2dx+14δ∫Ω∫0tht−s∇ut−∇usds2dx≤δ∫Ω∫0tht−s∇us−∇ut+∇utds2dx+14δ∫0thsdsho∇ut≤2δ∇ut2∫0thsds2dx+2δ+14δ∫0thsdsho∇ut≤2δ∇ut2a−k2+2δ+14δa−kho∇ut,(168)I3≤δg1utx,t2+cs24δa−kho∇ut,(169)I4≤δg2zx,1,t2+cs24δa−kho∇ut.
To estimateI5, we apply Lemma 4 with ϵ0=1/2 and use repeatedly Young’s, Cauchy-Schwartz’s, and the embedding inequalities, as follows:(170)I5≤ν∫Ωu2+dϵ0u∫0tht−sut−usdsdx≤ν14δ∫Ω∫0tht−sut−usds2dx+δ∫Ωu2+dϵ0u2dx≤cs2a−kν4δho∇ut+ν2δ∫Ωu4dx+∫Ωudx≤cs2a−kν4δho∇ut+ν2δ∇u4+Ωu≤cs2a−kν4δho∇ut+b0δM∇u2∇u2,where b0=ν/2cs2maxΩ/a,cs2/b. Also,(171)I6≤δ∫Ω∇ut2dx+14δ∫Ω∫0th′t−s∇ut−∇usds2dx≤δ∇ut2+14δ∫0th′t−sds∫Ω∫0th′t−s∇ut−∇us2dsdx.
Ash is a positive decreasing function, h′t−s=−h′t−s and then,(172)I6≤δ‖∇ut‖2+14δ∫0t−h′t−sds∫Ω∫0t−h′t−s∇ut−∇us2dsdx≤δ∇ut2−h04δh′o∇ut,(173)I7≤1l+1δ∫Ωutlut2dx+14δ∫Ω∫0th′t−sut−usds2dx≤1l+1δut2l+12l+1−h0cs24δh′o∇ut≤δcs2l+1l+1∇ut2l+1−h0cs24δl+1h′o∇ut≤δa0l+1∇ut2−h0cs24δl+1h′o∇ut,where a0=cs2l+12E0/al obtained by recalling (22) and Lemma 6. Combining (165) and (166)–(173), we finish the proof. Now, for M,ε1>0, we introduce the following functional:(174)Ft=MEt+ε1ϕt+ψt+χt.Lemma 14.
Letu,z be a solution of problem (12). Assume that (A1)–(A6) hold and(175)0<E0<minekπ4,k327ν2c∗6.
Then,Ft satisfies, along the solution and for some positive constants m,c>0, the following estimate:(176)F′t≤−mEt+cg1utx,t2+g2zx,1,t2+∫Ωux,tg1utx,tdx+∫Ωux,tg2zx,1,tdxand Ft∼Et.Proof.
By (ii) of Lemmas11–13 and Lemma 6 and by (A2), we deduce that for t≥t0>0:(177)F′t=ME′t+ε1ϕ′t+ψ′t+χ′t≤−Mλ−α2∫Ωutx,tg1utx,tdx−Mβ+α11−de−2τ1∫Ωzx,1,tg2zx,1,tdx−2τte−2τ1∫Ω∫01Gzx,ρ,tdρdx−ε1−δa−k+b0M∇u2∇u2−1l+1h0−ε1utl+2l+2−h0−δ1+a0l+1−ε1∇ut2−Mh12−ε11+ηa−k−2δa−k2∇ut2−ζM2−h04δ1+cs2l+1−ε14η+M04δ+2δ+14δ+μ1+μ2+νcs24δa−kho∇ut+μ1δg1utx,t2+μ2δg2zx,1,t2−ε1μ1∫Ωux,tg1utx,tdx−ε1μ2∫Ωux,tg2zx,1,tdx+ε1ν∫Ωu2lnudx,where(178)h0=∫0t0hsds,h1=minht,t≥t0>0.
We takeh0>ε1 and δ>0 sufficiently small such that(179)a4=h0−δ1+a0l+1−ε1>0,a2=ε1−δa−k+b0>0.
As long asε1 and δ are fixed, we choose M large enough such that(180)a1=Mλ−α2>0,a5=Mh12−ε11+ηa−k−2δa−k2>0,a6=ζM2−h04δ1+cs2l+1−ε14η+M04δ+2δ+14δ+μ1+μ2+νcs24δa−k>0.
Thus,(181)F′t≤−a31l+1utl+2l+2−a2M∇u2∇u2−a4∇ut2−a6ho∇ut−2τte−2τ1∫Ω∫01Gzx,ρ,tdρdx+cg1utx,t2+g2zx,1,t2+∫Ωux,tg1utx,tdx+∫Ωux,tg2zx,1,tdx+ε1ν∫Ωu2lnudx≤−m1Et+ε1−m12ν∫Ωu2lnudx+m1ν4u22+cg1utx,t2+g2zx,1,t2+∫Ωux,tg1utx,tdx+∫Ωux,tg2zx,1,tdx,where(182)m1=min2a2,2e−2τ1ξ,2a4,a3.
Using the Logarithmic Sobolev inequality, we get(183)F′t≤−m1Et−ε1−m12ν221+lnσ−lnu2u2+ε1−m12σ22π∇u2+m1ν4u2+cg1utx,t2+g2zx,1,t2+∫Ωux,tg1utx,tdx+∫Ωux,tg2zx,1,tdx.
From (61) and for(184)m1≤ε1≤m12ν+1,we have(185)m2=m12a−∫0thsds−ε1−m12σ22π≥m12k−ε1−m12kν>0m1ν4≤ε1−m12ν2.
This selection is possible thanks to (A6). So we get(186)F′t≤−mEt−ε1−m12ν21+2lnσ−lnu2u2F′t≤−mEt−ε1−m12ν21+2lnσ−lnu2u2+∫Ωux,tg1utx,tdx+∫Ωux,tg2zx,1,tdx,where m=minm1,m2. By recalling that E′≤0 and It>0 and using (133), (134), and (175), we obtain(187)lnu22≤ln4νJt≤ln4νEt≤ln4νE0≤lnekπν.
Takingσ satisfies(188)maxe−3/2,kπν≤σ≤2kπν.(So (61) is satisfied), and we guarantee(189)1+2lnσ−lnu2≥0,twhich completes the proof of (176). To prove Ft∼Et, we show that there exist two positive constants κ1 and κ2 such that(190)κ1Et≤Ft≤κ2Et.
From (i) of Lemmas11–13, (140), (133), and (134), we get κ>0 depending on ε1,a,l,cs,E0,k,andξ such that(191)ε1ϕt+ψt+χt≤κEt.
For a choice ofM large enough such that κ1=M−κ>0 and κ2=M+κ>0, we get our result. By the proof of Theorem 2 as given by Komornik [37], we consider the following partition of Ω:(192)Ω1=x∈Ω:ut≤ε,Ω2=x∈Ω:ut>ε.
We use Young’s inequality (withp=q=2), (22), and Lemma 6, and we have(193)∫Ωug1utdx+g1ut2≤δu2+14δ+1g1ut2≤δCs2∇u2+14δ+1∫Ω1H−1utg1utdx+c2∫Ω2utg1utdx≤2δCs2aEt+cδ∫Ω1H−1utg1utdx−CδE′t.
Similarly and by application of (17), we obtain(194)∫Ωug2zx,1,tdx+g2zx,1,t2≤δCs2∇u2+14δ+1c3∫Ωzx,1,tg2zx,1,tdx≤2δCs2aEt−Cδ′E′t.
Combining (193) and (194), (176) becomes(195)F′t≤−m−4δCsaEt−Cδ′′E′t+cδ∫Ω1H−1utg1utdx,where Cδ′′=Cδ′+Cδ. Now, for δ small enough such that d=m−4δCs/a>0, the function Lt=Ft+Cδ′′Et satisfies(196)L′t≤−dEt+c∫Ω1H−1utg1utdx,(197)Lt∼Et.Case 1.
H is linear on 0,ε; using (16) and Lemma 6, we deduce that(198)L′t≤−dEt−cE′t.
Thus,R=L+cE∼E satisfies(199)Rt≤R0e−c′t.
Hence,(200)Et≤CE0e−c′t.Case 2.
H is nonlinear on 0,ε; so we exploit Jensen’s inequality (see [2]) and the concavity of H−1 to obtain(201)H−11Ω1∫Ω1utg1utdx≥c∫Ω1H−1utg1utdx.
Then, (196) becomes(202)L′t≤−dEt+H−11Ω1∫Ω1utg1utdx.
Forε0<ε and w0>0, we define L0 by(203)L0t=H′ε0EtE0Lt+w0Et.
Then, we easily see that forb1,b2>0,(204)b1L0t≤Et≤b2L0t.
By recalling thatE′≤0,H′>0,andH′′>0 on 0,ε and using (202), we obtain(205)L0′t=ε0E′tE0H′′ε0EtE0Lt+H′ε0EtE0L′t+w0E′t≤−dEtH′ε0EtE0+cH′ε0EtE0H−11Ω1∫Ω1utg1utdx+w0E′t.
Using Remark1 with H∗, the convex conjugate of H in the sense of Young, we obtain(206)L0′t≤−dEtH′ε0EtE0+cH∗H′ε0EtE0L0′t≤−dEtH′ε0EtE0+cH∗H′ε0EtE0≤−dEtH′ε0EtE0+cε0EtE0H′ε0EtE0−w1E′t+w0E′t,where w1 is a positive constant depending of Ω1andα2 . By taking ε0 small enough such that dE0−cε0>0 and w0>w1, we obtain(207)L0′t≤−wEtE0H′ε0EtE0=−wH2EtE0,where H2t=tH′ε0t a positive increasing function on 0,1. By setting L1t=b1L0t/E0, we easily see that, by (204), we have(208)L1t∼Et.
Using (207), we arrive at(209)L1′t≤−w1H2L1t.
By recalling (28), we deduce(210)L1′t≤w11H1′L1t,which gives(211)H1L1t′≤w1.
A simple integration leads to(212)H1L1t≤w1t+H1L10.
Consequently,(213)L1t≤H1−1w1t+w2.
Using (208) and (213), we obtain (27). The proof is completed.



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