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Research article Special Issues

Maintaining energy efficiencies and reducing carbon emissions under a sustainable supply chain management

  • Currently, most countries are moving towards digitalization, and their energy consumption is increasing daily. Thus, power networks face major challenges in controlling energy consumption and supplying huge amounts of electricity. Again, using excessive power reduces the stored fossil fuels and affects the environment in terms of CO2 emissions. Keep these issues in mind; this study focuses on energy-efficient products in an energy supply chain management model under credit sales, variable production, and stochastic demand. Here, the manufacturer grants a credit period for the retailer to get more orders; thus, the order quantity is related to the credit period envisaged in this model. Considering such components, supply chain members can reduce negative environmental impacts and significant energy consumption, achieve optimal results and avoid drastic financial losses. Additionally, including a credit period increases the possibility of default risk, for which a certain interest is charged. The marginal reduction cost for limiting carbon emissions, flexible production to meet fluctuating demand, and continuous investment to improve product quality are considered here. The global optimality of system profit function and decision variables (credit period, quality improvement, and production rate) is ensured through the classical optimization method. Interpretive sensitivity analyses and numerical investigations are performed to validate the proposed model. The results demonstrate that the idea of credit sales, flexible production, and quality improvement increases total system profit by 28.64% and marginal reduction technology reduces CO2 emissions up to 4.01%.

    Citation: Mowmita Mishra, Santanu Kumar Ghosh, Biswajit Sarkar. Maintaining energy efficiencies and reducing carbon emissions under a sustainable supply chain management[J]. AIMS Environmental Science, 2022, 9(5): 603-635. doi: 10.3934/environsci.2022036

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  • Currently, most countries are moving towards digitalization, and their energy consumption is increasing daily. Thus, power networks face major challenges in controlling energy consumption and supplying huge amounts of electricity. Again, using excessive power reduces the stored fossil fuels and affects the environment in terms of CO2 emissions. Keep these issues in mind; this study focuses on energy-efficient products in an energy supply chain management model under credit sales, variable production, and stochastic demand. Here, the manufacturer grants a credit period for the retailer to get more orders; thus, the order quantity is related to the credit period envisaged in this model. Considering such components, supply chain members can reduce negative environmental impacts and significant energy consumption, achieve optimal results and avoid drastic financial losses. Additionally, including a credit period increases the possibility of default risk, for which a certain interest is charged. The marginal reduction cost for limiting carbon emissions, flexible production to meet fluctuating demand, and continuous investment to improve product quality are considered here. The global optimality of system profit function and decision variables (credit period, quality improvement, and production rate) is ensured through the classical optimization method. Interpretive sensitivity analyses and numerical investigations are performed to validate the proposed model. The results demonstrate that the idea of credit sales, flexible production, and quality improvement increases total system profit by 28.64% and marginal reduction technology reduces CO2 emissions up to 4.01%.



    The importance of bridges is undeniable and their presence in human daily life goes back a long time. With the presence of the bridges, road and railway traffic runs without any interruption over rivers and hazardous areas, time and fuel are saved, congestion on roads is minimized, distances between places are reduced, and many accidents have been avoided, as the bridges have reduced the number of bends and zig-zags in roads. As a result, many economies have grown and many societies have become connected. However, bridges have brought some challenges, such as collapse and instability due to natural hazards such as wind, earthquakes, etc. To overcome these difficulties, engineers and scientists have made efforts to find the best designs and possible models. Our aim in this work is to investigate the following plate problem

    {utt+Δ2u+α(t)g(ut)=u|u|β,inΩ×(0,T),u(0,y,t)=uxx(0,y,t)=u(π,y,t)=uxx(π,y,t)=0,(y,t)(d,d)×(0,T),uyy(x,±d,t)+σuxx(x,±d,t)=0,(x,t)(0,π)×(0,T),uyyy(x,±d,t)+(2σ)uxxy(x,±d,t)=0,(x,t)(0,π)×(0,T),u(x,y,0)=u0(x,y),ut(x,y,0)=u1(x,y), in Ω×(0,T), (1.1)

    where Ω=(0,π)×(d,d), d,β>0, g:RR and α:[0,+)(0,+) is a nonincreasing differentiable function, u is the vertical displacement of the bridge and σ is the Poisson ratio. This is a weakly damped nonlinear suspension-bridge problem, in which the damping is modulated by a time dependent-coefficient α(t). Firstly, we prove the local existence using the Faedo-Gherkin method and Banach fixed point theorem. Secondly, we prove the global existence by using the well-depth method. Finally, we establish an explicit and general decay result, depending on g and α, for which the exponential and polynomial decay rate estimates are only special cases. The proof is based on the multiplier method and makes use of some properties of convex functions, including the use of the general Young inequality and Jensen's inequality.

    The famous report by Claude-Louis Navier [1] was the only mathematical treatise of suspension bridges for several decades. Another milestone theoretical contribution was the monograph by Melan [2]. After the Tacoma collapse, engineers felt the necessary to introduce the time variable in mathematical models and equations in order to attempt explanations of what had occurred. As a matter of fact, in Appendix VI of the Federal Report [3], a model of inextensible cables is derived and the linearized Melan equation was obtained. Other important contributions were the works by Smith-Vincent and the analysis of vibrations in suspension bridges presented by Bleich-McCullough-Rosecrans-Vincent [4]. In all these historical references, the bridge was modelled linearly as a beam suspended to a cable. Hence, all the equations were linear. Mathematicians have not shown any interest in suspension bridges until recently. McKenna, in 1987, introduced the first nonlinear models to study them from a theoretical point of view, and he was followed by several other mathematicians (see [5,6]). McKenna's main idea was to consider the slackening of the hangers as a nonlinear phenomenon, a statement which is by now well-known also among engineers [7,8]. The slackening phenomenon was analyzed in various complex beam models by several authors (see [9,10,11]). Motivated by the wonderful book of Rocard [12], where it was pointed out that the correct way to model a suspension bridge is through a thin plate, Ferrero-Gazzola [13] introduced the following hyperbolic problem:

    {utt(x,y,t)+ηut+Δ2u(x,y,t)+h(x,y,u)=f,inΩ×R+,u(0,y,t)=uxx(0,y,t)=u(π,y,t)=uxx(π,y,t)=0,(y,t)(,)×R+,uyy(x,±,t)+σuxx(x,±,t)=0,(x,t)(0,π)×R+,uyyy(x,±,t)+(2σ)uxxy(x,±,t)=0,(x,t)(0,π)×R+,u(x,y,0)=u0(x,y),ut(x,y,0)=u1(x,y), in Ω×R+, (2.1)

    where Ω=(0,π)×(,) is a planar rectangular plate, σ is the well-known Poisson ratio, η is the damping coefficient, h is the nonlinear restoring force of the hangers and f is an external force. After the appearance of the above model, many mathematicians showed interest in investigating variants of it, using different kinds of damping with the aim to obtain stability of the bridge modeled through the above problem. Messaoudi [14] considered the following nonlinear Petrovsky equation

    utt+Δ2u+aut|ut|m2=bu|u|p2, (2.2)

    and proved the existence of a local weak solution, showed that this solution is global if mp and blows up in finite time if p>m and the energy is negative. Wang [15] considered the equation

    utt+δut+Δ2u+au=u|u|p2, (2.3)

    where a=a(x,y,t) together with the above initial and boundary conditions. After showing the uniqueness and existence of local solutions, he gave sufficient conditions for global existence and finite-time blow-up of solutions. Mukiawa [16] considered a plate equation modeling a suspension bridge with weak damping and hanger restoring force. He proved the well-posedness and established an explicit and general decay result without putting restrictive growth conditions on the frictional damping term. Messaoudi and Mukiawa [17] studied problem (2.3), where the linear frictional damping was replaced by nonlinear frictional damping and established the existence of a global weak solution and proved exponential and polynomial stability results. Audu et al. [18] considered a plate equation as a model for a suspension bridge with a general nonlinear internal feedback and time-varying weight. Under some conditions on the feedback and the coefficient functions, the authors established a general decay estimate. For more results related to the existence of work on similar problems, we mention the work of Xu et al. [19], in which they proved the local existence of a weak solution by the Galerkin method and the global existence by the potential well method. He et al. [20] considered the following Kirchhoff type equation

    (a+bΩ|u|2dx)Δu=f(u)+h,in Ω, (2.4)

    where ΩR3 is a bounded domain or Ω=R3, 0hL2(Ω) and fC(R,R). The authors proved the existence of at least one or two positive solutions by using the monotonicity trick, and nonexistence criterion is also established by virtue of the corresponding Pohoaev identity. Recently, Wang et al. [21] considered the fractional Rayleigh-Stokes problem where the nonlinearity term satisfied certain critical conditions and proved the local existence, uniqueness and continuous dependence upon the initial data of ε-regular mild solutions. More results in this direction can be found in [22,23,24,25,26,27]. The paper is organized as follows. In Section 3, we present some preliminaries and essential lemmas. We prove the local existence in Section 4 and the global existence in Section 5. The statement and the proof of our stability result will be given in Section 6.

    In this section, we present some material needed in the proofs of our results. First, we introduce the following space

    H2(Ω)={wH2(Ω):w=0on{0,π}×(d,d)}, (3.1)

    together with the inner product

    (u,v)H2=Ω(ΔuΔv+(1σ)(2uxyvxyuxxvyyuyyvxx))dx. (3.2)

    It is well known that (H2(Ω),(,)H2) is a Hilbert space, and the norm .2H2 is equivalent to the usual H2, see [13]. Throughout this paper, c is used to denote a generic positive constant.

    Lemma 3.1. [15]Let uH2(Ω) and assume that 1p<, then, there exists a positive constant Ce=Ce(Ω,p)>0 such that

    upCeuH2(Ω).

    Lemma 3.2. (Jensen's inequality)Let ψ:[a,b]R be a convex function. Assume that the functions f:(0,L)[a,b] and r:(0,L)R are integrable such that r(x)0, for any x(0,L) and L0r(x)dx=k>0. Then,

    ψ(1kL0f(x)r(x)dx)1kL0ψ(f(x))r(x)dx. (3.3)

    We consider the following hypotheses:

    (H1). The function g:RR is nondecreasing C0 function satisfying for ε,c1,c2>0,

    c1|s||g(s)|c2|s|, if |s|ε,|s|2+g2(s)G1(sg(s)), if |s|ε, (3.4)

    where G:R+R+ is a C1 function which is linear or strictly increasing and strictly convex C2 function on [0,ε] with G(0)=0 and G(0)=0. In addition, the function g satisfies, for ϑ>0,

    (g(s1)g(s2))(s1s2)ϑ|s1s2|2. (3.5)

    (H2). The function α:R+R+ is a nonincreasing differentiable function such that 0α(t)dt=.

    Remark 3.3. Hypothesis (H1) implies that sg(s)>0, for all s0 and it was introduced and employed by Lasiecka and Tataru [28]. It was shown there that the monotonicity and continuity of g guarantee the existence of the function G with the properties stated in (H1).

    Remark 3.4. As in [28], we use Condition (3.5) to prove the uniqueness of the solution.

    The following lemmas will be of essential use in establishing our main results.

    Lemma 3.5. [29] Let E:R+R+ be a nonincreasing function and γ:R+R+ be a strictlyincreasing C1-function, with γ(t)+ as t+. Assume that there exists c>0 such that

    Sγ(t)E(t)dtcE(S)1S<+.

    Then there exist positive constants k and ω such that

    E(t)keωγ(t).

    Lemma 3.6. [30] Let E:R+R+ be a differentiable and nonincreasing function and χ:R+R+ be a convex and increasing function such that χ(0)=0. Assume that

    +sχ(E(t))dtE(s),s0. (3.6)

    Then, E satisfies the following estimate

    E(t)ψ1(h(t)+ψ(E(0))),t0, (3.7)

    where ψ(t)=1t1χ(s)ds, and

    {h(t)=0,0tE(0)χ(E(0)),h1(t)=t+ψ1(t+ψ(E(0)))χ(ψ1(t+ψ(E(0)))),t>0.

    In this section, we state and prove the local existence of weak solutions of problem (1.1). Similar results can be found in [31,32]. To this end, we consider the following problem

    {utt(x,y,t)+Δ2u(x,y,t)+α(t)g(ut)=f(x,t),inΩ×(0,T),u(0,y,t)=uxx(0,y,t)=u(π,y,t)=uxx(π,y,t)=0,(y,t)(d,d)×(0,T),uyy(x,±d,t)+σuxx(x,±d,t)=0,(x,t)(0,π)×(0,T),uyyy(x,±d,t)+(2σ)uxxy(x,±d,t)=0,(x,t)(0,π)×(0,T),u(x,y,0)=u0(x,y),ut(x,y,0)=u1(x,y), in Ω×(0,T), (4.1)

    where fL2(Ω×(0,T)) and (u0,u1)H2(Ω)×L2(Ω). Then, we prove the following theorem:

    Theorem 4.1. Let (u0,u1)H2(Ω)×L2(Ω). Assume that(H1) and (H2) hold. Then, problem (4.1) has a unique local weaksolution

    uL([0,T),H2(Ω)),utL([0,T),L2(Ω)),uttL([0,T),H(Ω)),

    where H(Ω) is the dual space of H2(Ω).

    Proof. Uniqueness: Suppose that (4.1) has two weak solutions (u,v). Then, w=uv satisfies

    {wtt(x,y,t)+Δ2w(x,y,t)+α(t)g(ut)α(t)g(vt)=0,inΩ×(0,T),w(0,y,t)=wxx(0,y,t)=w(π,y,t)=wxx(π,y,t)=0,(y,t)(d,d)×(0,T),wyy(x,±d,t)+σwxx(x,±d,t)=0,(x,t)(0,π)×(0,T),wyyy(x,±d,t)+(2σ)wxxy(x,±d,t)=0,(x,t)(0,π)×(0,T),w(x,y,0)=wt(x,y,0)=0, in Ω×(0,T). (4.2)

    Multiplying (4.2) by wt and integrating over (0,t), we get

    12ddt[Ω(w2t+|Δw|2)dx]+α(t)Ω(g(ut)g(vt))(utvt)dx=0. (4.3)

    Integrating (4.3) over (0,t), we obtain

    Ω(w2t+|Δw|2)dx+2α(t)t0Ω(g(ut)g(vt))(utvt)dxds=0. (4.4)

    Using Condition (3.5) and (H2), for a.e.xΩ, we have

    Ω(w2t+|Δw|2dx)=0, (4.5)

    We conclude u=v=0 on Ω×(0,T), which proves the uniqueness of the solution of problem (4.1). Existence: To prove the existence of the solution for problem (4.1), we use the Faedo-Galerkin method as follows: First, we consider {vj}j=1 an orthonormal basis of H2(Ω) and define, for all k1, a sequence vk in Vk=span{v1,v2,...,vk}H2(Ω), given by

    uk(x,t)=Σkj=1aj(t)vj(x),

    for all xΩ and t(0,T) and satisfies the following approximate problem

    {Ωuktt(x,t)vjdx+ΩΔuk(x,t)Δvjdx+α(t)Ωg(ukt)vj=Ωf(x,t)vjdx,inΩ×(0,T),uk(x,y,0)=uk0(x,y),ukt(x,y,0)=uk1(x,y), in Ω×(0,T), (4.6)

    for all j=1,2,...,k,

    uk(0)=uk0=Σki=1u0,vivi, ukt(0)=uk1=Σki=1u1,vivi, (4.7)

    such that

    uk0u0H2(Ω),uk1u1L2(Ω). (4.8)

    For any k1, problem (4.6) generates a system of k nonlinear ordinary differential equations. The ODE's standard existence theory assures the existence of a unique local solution uk for problem (4.6) on [0,Tk), with 0<TkT. Next, we have to show, by a priori estimates, that Tk=T,k1. Now, multiplying (4.6) by aj(t), using Green's formula and the boundary conditions, and then summing each result over j we obtain, for all 0<tTk,

    12ddt[Ω(|ukt|2+(Δuk)2)dx]+α(t)Ωuktg(ukt)dx=Ωf(x,t)ukt(x,t)dx. (4.9)

    Then, integrating (4.9) over (0,t) leads to

    12Ω(|ukt|2+|Δuk|2)dx+t0Ωα(s)uktg(ukt)dxds=12Ω(|uk1|2+|Δuk0|2)dx+t0Ωf(x,t)ukt(x,t)dxds. (4.10)

    From the convergence (4.8), using the fact that fL2(Ω×(0,T)), and exploiting Young's inequality, then (4.10) becomes, for some C>0, and for any t[0,tk)

    12Ω[|ukt|2+|Δuk|2dx]+t0Ωα(s)uktg(ukt)dxds12Ω[|uk1|2+|Δuk0|2]dx+εt0Ω|ukt|2dxds+Cεt0Ω|f(x,s)|2dxdsCε+εsup(0,Tk)Ω|ukt|2dx. (4.11)

    Therefore, we obtain

    12sup(0,Tk)Ω|ukt|2dx+12sup(0,Tk)Ω|Δuk|2dx+12sup(0,Tk)tk0Ωα(s)ukt(x,s)g(ukt(x,s))dxdsCε+εsup(0,Tk)Ω|ukt|2dx. (4.12)

    Choosing ε=14, estimate (4.12) yields, for all TkT and C>0,

    sup(0,Tk)Ω|ukt|2dx+sup(0,Tk)Ω|Δuk|2dx+sup(0,Tk)tk0Ωα(s)ukt(x,s)g(ukt(x,s))dxdsC. (4.13)

    Consequently, the solution uk can be extended to (0,T), for any k1. In addition, we have

    (uk) is bounded in L((0,T),H2(Ω))and(ukt) is bounded in L((0,T),L2(Ω)).

    Therefore, we can extract a subsequence, denoted by (u) such that, when , we have

    uu  weakly * in L((0,T),H2(Ω))andutut weakly * in L((0,T),L2(Ω)).

    Next, we prove that g(ut) is bounded in L2((0,T);L2(Ω)). For this purpose, we consider two cases:

    Case 1. G is linear on [0,ε]. Then using (H1) and Young's inequality, we get

    Ωg2(ut)dxcΩutg(ut)dxΩ|ut|2dxc4δ0Ω|ut|2dx+δ0Ωg2(ut)dx, (4.14)

    for a suitable choice of δ0 and using the fact that ut is bounded in L2((0,T),L2(Ω)), we obtain

    T0Ωg2(ut)dxdtc. (4.15)

    Case 2. G is nonlinear. Let 0<ε1ε such that

    sg(s)min{ε,G(ε)} for all |s|ε1. (4.16)

    Then, one can show that

    {s2+g2(s)G1(sg(s))for all|s|ε1c1|s||g(s)|c2|s|for all|s|ε1. (4.17)

    Define the following sets

    Ω1={xΩ:|ut|ε1},andΩ2={xΩ:|ut|>ε1}. (4.18)

    Then, using (4.17) and (4.18) leads for some c2>0,

    Ωg2(ut)dx=Ω2g2(ut)dx+Ω1g2(ut)dxc2Ω2|ut|2dx+Ω1(|u|2t+g2(ut))dxΩ1|ut|2dxc2Ω2|ut|2dx+Ω1G1(utg(ut))dx. (4.19)

    Let

    J(t):=Ω1utg(ut)dx,
    E(t)=12(ut22+u2H2(Ω))1β+2uβ+2β+2, (4.20)

    and

    (E)(t)=α(t)Ωutg(ut)dx0. (4.21)

    Using (4.19) and Jensen's inequality, we obtain

    Ωg2(ut)dxcΩ|ut|2dx+G1(J(t))=cΩ|ut|2dx+G(ε0E(t)E(0))G(ε0E(t)E(0))G1(J(t)). (4.22)

    Using the convexity of G (G is increasing), we obtain for t(0,T),

    G(ε0E(t)E(0))G(ε0E(T)E(0))=c.

    Let G be the convex conjugate of G in the sense of Young (see [33], pp. 61–64), then, for s(0,G(ε)],

    G(s)=s(G)1(s)G[(G)1(s)]s(G)1(s). (4.23)

    Using the general Young inequality

    ABG(A)+G(B),ifA(0,G(ε)],B(0,ε],

    for

    A=G(ε0E(t)E(0))andB=G1(J(t)),

    and using the fact that E(t)E(0), we get

    Ωg2(ut)dxcΩ|ut|2dx+cε0E(t)E(0)G(ε0E(t)El(0))C(E)(t)cΩ|ut|2dx+cε0E(t)E(0)G(ε0E(t)E(0))C(E)(t)cΩ|ut|2dx+cC(E)(t). (4.24)

    Integrating (4.24) over (0,T), we obtain

    T0Ωg2(ut)dxdtcT0Ω|ut|2dxdt+cTC(E(T)E(0)). (4.25)

    Using (4.21) and the fact that ut is bounded in L2((0,T);L2(Ω)), we conclude that g(ut) is bounded in L2((0,T);L2(Ω)). So, we find, up to a subsequence, that

    g(ut)χ in L2((0,T);L2(Ω)). (4.26)

    Now, we have to show that χ=g(ut). In (4.6), we use u instead of uk and then integrate over (0,t) to get

    ΩutvjdxΩu1vjdx+t0ΩΔuΔvjdxds+t0Ωα(s)g(ut)vjdxds=t0Ωfvjdxds,j<. (4.27)

    As +, we easily check that

    ΩutvjdxΩu1vjdx+t0ΩΔuΔvjdxds+t0Ωα(s)χvjdxds=t0Ωfvjdxds,j1. (4.28)

    Hence, for vH2(Ω), we have

    ΩutvdxΩu1vdx+t0ΩΔuΔvdxds+t0Ωα(s)χvdxds=t0Ωfvdxds. (4.29)

    Since all terms define absolute continuous functions, we get, for a.e.t[0,T] and for vH2(Ω), the following

    ddtΩutvdx+ΩΔuΔvdx+Ωα(t)χvdx=Ωfvdxds. (4.30)

    This implies that

    utt+Δ2u+α(t)χ=f,inD(Ω×(0,T)). (4.31)

    Using (H1), we see that

    X:=T0Ωα(s)(utv)(g(ut)g(v))dxdt0,vL2((0,T);L2(Ω)). (4.32)

    So, by using (4.6) and replacing uk by u, we get

    X=T0Ωfutdxdt+12Ω(|ut|2+|Δu|2)dx12Ω|ut(x,T)|2dx12Ω|Δut(x,T)|2dxT0Ωα(t)g(ut)vdxdtT0Ωα(t)g(v)(utv)dxdt. (4.33)

    Taking +, we obtain

    0limsupXT0Ωfutdxdt+12Ω(|u1|2+|Δu0|2)dx12Ω|ut(x,T)|2dx12Ω|Δut(x,T)|2dxT0Ωα(t)χvdxdtT0Ωα(t)g(v)(utv)dxdt. (4.34)

    Replacing v by ut in (4.30) and integrating over (0,T), we obtain

    T0Ωfutdxdt=12Ω(|ut(x,T)|2dx+|Δu(x,T)|2)dx12Ω|u1|2dx12Ω|Δu0|2dx+T0Ωα(t)χutdxdt. (4.35)

    Adding of (4.34) and (4.35), we get

    0limsupXT0Ωα(t)χutdxdtT0Ωα(t)χvdxdtT0Ωα(t)g(v)(utv)dxdt. (4.36)

    This gives

    T0Ωα(t)(χg(v))(utv)dxdt0,vL2((0,T),L2(Ω)). (4.37)

    Hence,

    T0Ωα(t)(χg(v))(utv)dxdt0,vL2(Ω×(0,T)). (4.38)

    Let v=λw+ut, where λ>0 and wL2(Ω×(0,T)). Then, we get

    λT0Ωα(t)(χg(λw+ut))wdxdt0,wL2(Ω×(0,T)). (4.39)

    For λ>0, we have

    λT0Ωα(t)(χg(λw+ut))wdxdt0,wL2(Ω×(0,T)). (4.40)

    As λ0 and using the continuity of g with respect of λ, we get

    λT0Ωα(t)(χg(ut))wdxdt0,wL2(Ω×(0,T)). (4.41)

    Similarly, for λ<0, we get

    λT0Ωα(t)(χg(ut))wdt0,wL2(Ω×(0,T)). (4.42)

    This implies that χ=g(ut). Hence, (4.30) becomes

    Ω(uttv+ΔuΔv+α(t)g(ut)v)dx=Ωfvdx,vL2((0,T);H2(Ω)). (4.43)

    which gives

    utt+Δ2u+α(t)g(ut)=f,inD(Ω×(0,T)). (4.44)

    To handle the initial conditions of problem (4.1), we first note that

    uuweakly * inL(0,T;H2(Ω))ututweakly * inL(0,T;L2(Ω)). (4.45)

    Thus, using Lion's Lemma and (4.6), we easily obtain uuC([0,T];L2(Ω)). Therefore, u(x,0) makes sense and u(x,0)u(x,0)L2(Ω). Also, we see that

    u(x,0)=u0u0(x)H2(Ω).

    Hence, u(x,0)=u0(x). As in [34], let ϕC0(0,T), and replacing uk by u, we obtain from (4.6) and for any j

    {T0Ωut(x,t)vj(x)ϕ(t)dxdt=T0ΩΔu(x,t)Δvj(x)ϕ(t)dxdtT0Ωα(t)g(ut)vj(x)ϕ(t)dxdt+T0Ωf(x,t)vj(x)ϕ(t)dxdt. (4.46)

    As +, we have for any ϕC0((0,T)),

    {T0Ωut(x,t)vj(x)ϕ(t)dxdt=T0ΩΔu(x,t)Δvj(x)ϕ(t)dxdtT0Ωα(t)g(ut)vj(x)ϕ(t)dxdt+T0Ωf(x,t)vj(x)ϕ(t)dxdt, (4.47)

    for all j1. This implies that

    T0Ωut(x,t)v(x)ϕ(t)dxdt=T0Ω[Δ2u(x,t)α(t)g(ut)+f(x,t)]v(x)ϕ(t)dxdt, (4.48)

    for all vH2(Ω). This means that uttL((0,T);H(Ω)) and u solves the equation

    utt+Δ2u+α(t)g(ut)=f. (4.49)

    Thus

    utL((0,T);L2(Ω)),uttL((0,T);H(Ω)).

    Consequently, utC((0,T);H(Ω)). So, ut(x,0) makes sense and follows that

    ut(x,0)ut(x,0) in H(Ω)

    and since

    ut(x,0)=u1(x)u1(x) in L2(Ω),

    then

    ut(x,0)=u1(x).

    This ends the proof of Theorem 4.1.

    Now, we proceed to establish the local existence result for problem (1.1).

    Theorem 4.2. Let (u0,u1)H2(Ω)×L2(Ω) begiven. Then problem (1.1) has a unique local weak solution

    uL([0,T),H2(Ω)),utL([0,T),L2(Ω)),uttL([0,T),H(Ω)).

    Remark 4.3. In this remark, we point out four cases regarding the solution of problem (1.1):

    1) If β=0, g is linear and (u0,u1)(H4(Ω)H2(Ω))×H2(Ω), then problem (1.1) has a unique classical solution

    uC2([0,T),H2(Ω)),utC1([0,T),L2(Ω)),uttC([0,T),H(Ω)).

    2) If β=0, g is linear and (u0,u1)H2(Ω)×L2(Ω), then problem (1.1) has a unique weak solution

    uC1([0,T),H2(Ω)),utC([0,T),L2(Ω)),uttL([0,T),H(Ω)).

    3) If β>0 or g is nonlinear and (u0,u1)H2(Ω)×L2(Ω), then problem (1.1) has a unique weak solution

    uL([0,T),H2(Ω)),utL([0,T),L2(Ω)),uttL([0,T),H(Ω)).

    4) If β>0 or g is nonlinear and (u0,u1)(H4(Ω)H2(Ω))×H2(Ω), then problem (1.1) has a unique strong solution

    uL([0,T),H4(Ω)H2(Ω)),utL(([0,T),H2(Ω)),uttL([0,T),L2(Ω)).

    Proof. To prove Theorem 4.2, we first let vL([0,T),H2(Ω)) and ˜f(v)=|v|βv. Then, by the embedding Lemma 3.1, we have

    ||˜f(v)||22=Ω|v|2(β+1)dx<+. (4.50)

    Hence,

    ˜f(v)L([0,T),L2(Ω))L2(Ω×(0,T)).

    Therefore, for each vL([0,T),H2(Ω)), there exists a unique solution

    uL([0,T),H2(Ω)),utL([0,T),L2(Ω))

    satisfying the following nonlinear problem

    {utt+Δ2u+α(t)g(ut)=˜f(v),inΩ×(0,T),u(0,y,t)=uxx(0,y,t)=u(π,y,t)=uxx(π,y,t)=0,(y,t)(d,d)×(0,T),uyy(x,±d,t)+σuxx(x,±d,t)=0,(x,t)(0,π)×(0,T),uyyy(x,±d,t)+(2σ)uxxy(x,±d,t)=0,(x,t)(0,π)×(0,T),u(x,y,0)=u0(x,y),ut(x,y,0)=u1(x,y), in Ω×(0,T), (4.51)

    Now, let

    WT={wL((0,T),H2(Ω))/wtL((0,T),L2(Ω))},

    and define the map K:WTWT by K(v)=u. We note that WT is a Banach space with respect to the following norm

    ||w||WT=||w||L((0,T),H2(Ω))+||wt||L((0,T),L2(Ω)).

    Multiply (4.51) by ut and integrate over Ω×(0,t), we get for all tT,

    12Ωu2tdx+12Ω|Δu|2dx+t0Ωα(s)utg(ut)dxds=12Ωu21dx+12Ω|Δu0|2dx+t0Ω|v|βvutdxds. (4.52)

    Using Young's inequality and the embedding Lemma 3.1, we have

    Ω|v|βvutdxε4Ωu2tdx+4εΩ|v|2(β+1)dxε4Ωu2tdx+4Ceε||v||2(β+1)H2. (4.53)

    Thus, (4.52) becomes

    12Ωu2tdx+12Ω|Δu|2dxλ0+εT4sup(0,T)Ωu2tdx+CeεT0||v||2(β+1)H2dt, (4.54)

    where λ0=12||u1||22+12||Δu0||22 and Ce is the embedding constant. Choosing ε such that εT2=14, we get

    ||u||2WTλ+Tb||v||2(β+1)WT.

    Suppose that ||v||WTM and for M2>λ and TT0<M2λbM2(β+1), we conclude that

    ||u||2WTλ+TbM2(β+1)M2.

    Therefore, we deduce that K:BB, where

    B={wL((0,T),H2(Ω))/wtL((0,T),L2(Ω));||w||WTM}.

    Next, we prove, for T0(even smaller), K is a contraction. For this purpose, let u1=K(v1) and u2=K(v2) and set u=u1u2, then u satisfies the following

    utt+Δ2u+α(t)g(u1t)α(t)g(u2t)=|v1|βv1|v2|βv2. (4.55)

    Multiplying (4.55) by ut and integrating over Ω×(0,t) we get, for all tT,

    12Ωu2tdx+12Ω|Δu|2dx+t0Ω(α(t)g(u1t)α(t)g(u2t))(u1tu2t)dxds=t0Ω(˜f(v1)˜f(v2))utdxds. (4.56)

    Using (3.5) and (H2), we have

    12Ωu2tdx+12Ω|Δu|2dxt0Ω(˜f(v1)˜f(v2))utdxds. (4.57)

    Now, we evaluate

    Λ:=Ω|˜f(v1)˜f(v2)||ut|dx=Ω|˜f(ξ)||v||ut|dx, (4.58)

    where v=v1v2, ξ=τv1+(1τ)v2, 0τ1, and ˜f(ξ)=(β+1)|ξ|β.

    Young's inequality implies

    Λδ2Ωu2tdx+2δΩ|˜f(ξ)|2|v|2dxδ2Ωu2tdx+2(β+1)2δΩ|αv1+(1α)v2|2β|v|2dxδ2Ωu2tdx+Cδ(|v|2nn2)n2n(|αv1+(1α)v2|nβ)2n. (4.59)

    Using the embedding Lemma 3.1, we arrive at

    Λδ2Ωu2tdx+CδCe||v||2H2(||v1||2βH2+||v2||2βH2)δ2Ωu2tdx+4CδCeM2β||v||2βH2. (4.60)

    Therefore, (4.57) takes the form

    12||u||2WTδT02||u||2WT+CδM2βT0||v||2βWT. (4.61)

    Choosing δ sufficiently small, we see that

    ||u||2WT4CδM2βT0||v||2βWT=γ0T0||v||2βWT. (4.62)

    Taking T0 small enough so that,

    ||u||2WTν||v||2βWT,for some0<ν<1. (4.63)

    Thus, K is a contraction. The Banach fixed point theorem implies the existence of a unique uB satisfying K(u)=u. Thus, u is a local solution of (1.1).

    Uniqueness: Suppose that problem (1.1) has two weak solutions (u,v). Taking, w=uv, that satisfies the following equation, for all t(0,T),

    {wttΔ2w+α(t)g(ut)α(t)g(vt)=u|u|βv|v|βw(0,y,t)=wxx(0,y,t)=w(π,y,t)=wxx(π,y,t)=0,(y,t)(d,d)×(0,T),w(x,0)=wt(x,0)=0, in Ω. (4.64)

    Multiplying (4.64) by wt and integrating over Ω×(0,t), we obtain

    12Ωw2tdx+12Ω|Δw|2dx+t0Ω(α(t)g(ut)α(t)g(vt))(utvt)dxds=t0Ω(u|u|βv|v|β)wtdxds. (4.65)

    Using (3.5) and (H2) implies that

    12Ωw2tdx+12Ω|Δw|2dxt0Ω(u|u|βv|v|β)wtdxds. (4.66)

    By repeating the same above estimates, we obtain

    Ω(w2tdx+|Δw|2)dx=0. (4.67)

    This gives w0. The proof of the uniqueness is completed.

    In this section, we prove that problem (1.1) has a global solution. For this purpose, we introduce the following functionals. The energy functional associated with problem (1.1) is

    E(t)=12(ut22+u2H2(Ω))1β+2uβ+2β+2. (5.1)

    Direct differentiation of (5.1), using (1.1), leads to

    E(t)=α(t)Ωutg(ut)dx0. (5.2)
    J(t)=12u2H2(Ω)1β+2uβ+2β+2 (5.3)

    and

    I(t)=u2H2(Ω)uβ+2β+2. (5.4)

    Clearly, we have

    E(t)=J(t)+12ut22. (5.5)

    Lemma 5.1. Suppose that (H1) and (H2) hold and (u0,u1)H2(Ω)×L2(Ω), such that

    0<γ=Cβ+2e(2(β+2)βE(0))β2<1,I(u0)>0, (5.6)

    then I(u(t))>0,t>0.

    Proof. Since I(u0)>0, then there exists (by continuity) Tm<T such that I(u(t)0, t[0,Tm]; which gives

    J(t)=12u2H2(Ω)1β+2uβ+2β+2=β2(β+2)u2H2(Ω)+1β+2I(t)β2(β+2)u2H2(Ω). (5.7)

    By using (5.2), (5.5) and (5.7), we have

    u2H2(Ω)2(β+2)βJ(t)2(β+2)βE(t)2(β+2)βE(0),t[0,Tm]. (5.8)

    The embedding theorem, (5.6) and (5.8) give, t[0,Tm],

    uβ+2β+2Cβ+2euβ+2H2(Ω)Cβ+2euβH2(Ω)u2H2(Ω)γu2H2(Ω)<u2H2(Ω). (5.9)

    Therefore,

    I(t)=u2H2(Ω)uβ+2β+2>0,t[0,Tm].

    By repeating this procedure, and using the fact that

    limtTmCβ+2e(2(β+2)βE(t))β2γ<1,

    Tm is extended to T.

    Remark 5.2. The restriction (5.6) on the initial data will guarantee the nonnegativeness of E(t).

    Proposition 5.3. Suppose that (H1) and (H2) hold. Let (u0,u1)H2(Ω)×L2(Ω) be given, satisfying (5.6). Thenthe solution of (1.1) is global and bounded.

    Proof. It suffices to show that u2H2(Ω)+ut22 is bounded independently of t. To achieve this, we use (5.2), (5.4) and (5.5) to get

    E(0)E(t)=J(t)+12ut22β22βu2H2(Ω)+12ut22+1βI(t)β22βu2H2(Ω)+12ut22, (5.10)

    since I(t) is positive. Therefore

    u2H2(Ω)+ut22CE(0),

    where C is a positive constant, which depends only on β.

    In this section, we state and prove our stability result. For this purpose, we establish some lemmas.

    Lemma 6.1. (Case: G is linear) Let u be the solution of (1.1). Then, for T>S0, the energy functionalsatisfies

    TSα(t)E(t)dtcE(S). (6.1)

    Proof. We multiply (1.1) by αu and integrate over Ω×(S,T) to get

    0=TSα(t)Ω(uutt+uΔ2u+α(t)ug(ut)|u|β+2)dxdt=TSα(t)Ω((uut)tu2t+α(t)ug(ut)|u|β+2)dxdt+TSα(t)u2H2(Ω)dt=TSα(t)ddt(Ωuutdx)dt+TSα(t)Ωu2tdxdt+TSα(t)u2H2(Ω)dt2TSα(t)Ωu2tdxdt+TSα2(t)Ωug(ut)dxdtTSα(t)uβ+2β+2dt. (6.2)

    Adding and subtracting the following terms

    γTSα(t)u2H2(Ω)dt+(1+γ)TSα(t)ut22dt, where γ is defined in (5.6),

    to (6.2), and recalling (5.9), we arrive at

    TSα(t)ddt(Ωuutdx)dt+(1γ)TSα(t)(u2H2(Ω)+ut22)dt(2γ)TSα(t)Ωu2tdxdt+TSα2(t)Ωug(ut)dxdt=TSα(t)(γu2H2(Ω)uβ+2β+2)dt0. (6.3)

    Integrating the first term of (6.3) by parts and using (5.1), then (6.3) becomes

    (1γ)TSαEdt(1γ)TSα(u2H2(Ω)+ut22)dt[αΩuutdx]TS+TSαΩuutdxdt+(2γ)TSαΩu2tdxdtTSα2Ωug(ut)dxdt. (6.4)

    Now, we estimate the terms in the right-hand side of (6.4) as follows:

    1) Estimate for [αΩuutdx]TS.

    Using Lemma 3.1 and Young's inequality, we obtain

    Ωuutdx12Ω(u2+u2t)dxcuH2(Ω)+ut22cE(t), (6.5)

    which implies that

    [αΩuutdx]TSc[α(T)E(T)+α(S)E(S)]cα(S)E(S)cE(S). (6.6)

    2) Estimate for TSαΩuutdxdt.

    The use of (6.5) and (H2) leads to

    TSαΩuutdxdtc|TSαEdt|cE(S)|TSαdt|cE(S). (6.7)

    3) Estimate for TSα(Ωu2tdx)dt.

    Using (H1), (5.2) and recalling that G is linear, we have

    \begin{equation} \begin{aligned} \int_{S}^{T}\alpha \left(\int_{\Omega}u_t^2dx\right)dt&\le \frac{1}{c_1} \int_{S}^{T}\alpha(t)\int_{\Omega}u_tg(u_t)dxdt\\ &\le -\int_{S}^{T}cE^{\prime}(t)dt\\ &\le cE(S). \end{aligned} \end{equation} (6.8)

    4) Estimate for -\int_{S}^{T}\alpha ^{2}(t) \int_{\Omega }ug(u_t)dxdt.

    Using (H1), Lemma 3.1, Holder's inequality and recalling G is linear, we obtain

    \begin{equation} \begin{aligned} \alpha^2(t)\int_{\Omega}ug(u_t)dx &\le \alpha^2(t)\left(\int_{\Omega}\vert u\vert^{2}dx\right)^{\frac{1}{2}} \left(\int_{\Omega}\vert g(u_t)\vert^{2}dx\right)^{\frac{1}{2}}\\ &\le \alpha^{\frac{3}{2}} (t) \|u\|_{H^2_*(\Omega)} \left(\alpha(t)\int_{\Omega}u_t g(u_t) dx\right)^{\frac{1}{2}}\\ &\le c \alpha(t) E^{\frac{1}{2}}(t)\left(-E^\prime(t)\right)^{\frac{1}{2}}. \end{aligned} \end{equation} (6.9)

    Applying Young's inequality to E^{\frac{1}{2}}(t) (-E'(t))^{\frac{1}{2}} with p = 2 and p^* = 2 , to get

    \begin{equation} \begin{aligned} \alpha^2(t)\int_{\Omega}ug(u_t)dx &\le c\alpha(t)\left(\varepsilon E(t) -C_{\varepsilon} E^{\prime}(t)\right)\\ &\le c\varepsilon \alpha E(t)-C_{\varepsilon}E^{\prime}(t), \end{aligned} \end{equation} (6.10)

    which implies that

    \begin{equation} \begin{aligned} &\int_{S}^{T}\alpha^2 (t) \left(\int_{\Omega}(-ug(u_t))dx\right)dt\\ & \quad \le c\varepsilon \int_{S}^{T}\alpha(t) E(t) dt+C_{\varepsilon}E(S). \end{aligned} \end{equation} (6.11)

    Combining the above estimates and taking \varepsilon small enough, we get (6.1).

    Lemma 6.2. (Case: G is nonlinear) Let u be the solution of (1.1). Then, for T > S \geq 0 , the energy functionalsatisfies

    \begin{equation} \begin{aligned} \int_{S}^{T}\alpha(t) \tilde{\phi}\left(E(t)\right)dt\le c \tilde{\phi}(E(S))+c\int_{S}^{T}\alpha(t) \frac{\tilde{\phi}(E)}{E}\int_{\Omega}\left(\vert u_t\vert^2+\vert u g(u_t)\vert \right)dxdt, \end{aligned} \end{equation} (6.12)

    where \tilde{\phi}:\mathbb{R}^+ \to \mathbb{R}^+ is any convex, increasing and of class C^1[0, \infty) function such that \tilde{\phi}(0) = 0 .

    Proof. We multiply (1.1) by \alpha(t) \frac{\tilde{\phi}(E)}{E}u and integrate over \Omega \times (S, T) to get

    \begin{equation} \begin{aligned} 0 & = \int_{S}^{T}\alpha(t) \frac{\tilde{\phi}(E)}{E}\int_{\Omega }\left( \left( uu_{t}\right) _{t}-u_{t}^{2}+\alpha (t)ug(u_t)-\vert u\vert^{\beta+2}\right) dxdt\\ & \quad +\int_{S}^{T}\alpha(t) \frac{\tilde{\phi}(E)}{E}\| u\|_{H_*^2(\Omega)}^{2} dt\\ & = \int_{S}^{T}\alpha(t) \frac{\tilde{\phi}(E)}{E}\frac{d}{dt}\left( \int_{\Omega }uu_{t}dx\right) dt+\int_{S}^{T}\alpha(t) \frac{\tilde{\phi}(E)}{E}\| u\|_{H_*^2(\Omega)}^{2} dt\\ & \quad +\int_{S}^{T}\alpha(t) \frac{\tilde{\phi}(E)}{E}\int_{\Omega } u_{t}^{2} dxdt-2\int_{S}^{T}\alpha(t) \frac{\tilde{\phi}(E)}{E}\int_{\Omega } u_{t}^{2} dxdt\\ & \quad +\int_{S}^{T}\alpha ^{2}(t)\frac{\tilde{\phi}(E)}{E}\int_{\Omega }ug(u_t)dxdt-\int_{S}^{T}\alpha(t) \frac{\tilde{\phi}(E)}{E}\|u\|_{\beta+2}^{\beta+2}. \end{aligned} \end{equation} (6.13)

    Adding and subtracting to (6.13) the following terms

    \gamma \int_{S}^{T}\alpha(t) \frac{\tilde{\phi}(E)}{E}\| u\|_{H_*^2(\Omega)}^{2} dt+(1+\gamma)\int_{S}^{T}\alpha(t) \frac{\tilde{\phi}(E)}{E}\|u_t\|_2^2dt, \text{ where }\gamma \text{ is defined in (5.6)},

    we arrive at

    \begin{equation} \begin{aligned} &(1-\gamma)\int_{S}^{T}\alpha(t) \tilde{\phi}(E)dt\le -\int_{S}^{T}\alpha(t) \frac{\tilde{\phi}(E)}{E}\frac{d}{dt}\left( \int_{\Omega }uu_{t}dx\right) dt\\ & \quad +(2-\gamma)\int_{S}^{T}\alpha(t) \frac{\tilde{\phi}(E)}{E}\int_{\Omega } u_{t}^{2} dxdt-\int_{S}^{T}\alpha ^{2}(t)\frac{\tilde{\phi}(E)}{E}\int_{\Omega }ug(u_t)dxdt\\ & \quad -\int_{S}^{T}\alpha \frac{\tilde{\phi}(E)}{E} \left(\gamma \| u\|_{H_*^2(\Omega)}^{2}-\|u\|_{\beta+2}^{\beta+2}\right). \end{aligned} \end{equation} (6.14)

    Using (5.9), it is easy to deduce that -\int_{S}^{T}\alpha \frac{\tilde{\phi}(E)}{E} \left(\gamma \| u\|_{H_*^2(\Omega)}^{2}-\|u\|_{\beta+2}^{\beta+2}\right)dt\le 0.

    Integrating by parts in the first term, in the right-hand side of (6.14), we get

    \begin{equation} \begin{aligned} (1-\gamma)\int_{S}^{T}\alpha(t) \tilde{\phi}(E)dt\le &-\left[ \alpha(t) \frac{\tilde{\phi}(E)}{E}\int_{\Omega }uu_{t}dx \right] _{S}^{T}\\ &+\int_{S}^{T}\int_{\Omega}u_t\left(\alpha^{\prime}(t) \frac{\tilde{\phi}(E)}{E}u+\alpha(t)\left( \frac{\tilde{\phi}(E)}{E}\right)^{\prime}u\right)dxdt\\ &+(2-\gamma)\int_{S}^{T}\alpha(t) \frac{\tilde{\phi}(E)}{E}\int_{\Omega } u_{t}^{2} dxdt\\ &-\int_{S}^{T} \alpha^2(t) \frac{\tilde{\phi}(E)}{E}\int_{\Omega }ug(u_t)dxdt. \end{aligned} \end{equation} (6.15)

    Using Cauchy Schwarz' inequality, Lemmas 3.1 and 5.1, we obtain

    \begin{equation} \begin{aligned} \int_{\Omega}uu_t dx &\le \left(\int_{\Omega}\vert u\vert ^2dx\right)^{\frac{1}{2}} \left(\int_{\Omega}\vert u_t\vert ^2dx\right)^{\frac{1}{2}}\\ &\le c\|u\|_{H_*^2(\Omega)}\|u_t\|_2\le cE(t). \end{aligned} \end{equation} (6.16)

    Using (6.16), the properties of \alpha(t) and the fact that the function s\to \frac{\tilde{\phi}(s)}{s} is non-decreasing and E is non-increasing, we have

    \begin{equation} \begin{aligned} \int_{S}^{T}\alpha^{\prime}(t) \frac{\tilde{\phi}(E)}{E}\int_{\Omega}uu_t dxdt &\le c\int_{S}^{T}\alpha^{\prime}(t)\frac{\tilde{\phi}(E)}{E}Edt\\ &\le c \tilde{\phi}(E(S))\int_{S}^{T}\alpha^{\prime}(t) dt\le c\tilde{\phi}(E(S)). \end{aligned} \end{equation} (6.17)

    Similarly, we get

    \begin{equation} \begin{aligned} \int_{S}^{T}\alpha(t)\left( \frac{\tilde{\phi}(E)}{E}\right)^{\prime}\int_{\Omega}uu_tdxdt &\le E(S) \int_{S}^{T}\alpha(t)\left( \frac{\tilde{\phi}(E)}{E}\right)^{\prime}dt\\ &\le E(S) \left[\alpha(t) \frac{\tilde{\phi}(E)}{E}\right]_{S}^{T}-E(S)\int_{S}^{T}\alpha^{\prime}(t) \frac{\tilde{\phi}(E)}{E}dt\\ &\le E(S) \left(\alpha(T) \frac{\tilde{\phi}(E(T))}{E(T)}-\alpha(S) \frac{\tilde{\phi}(E(S))}{E(S)}\right)\\ & \quad - E(S)\frac{\tilde{\phi}(E(S))}{E(S)}\int_{S}^{T}\alpha^{\prime}(t)dt\\ &\le E(S) \alpha(T) \frac{\tilde{\phi}(E(T))}{E(T)}- \tilde{\phi}(E(S))\left(\alpha(T)-\alpha(S)\right)\\ &\le E(S) \alpha(S) \frac{\tilde{\phi}(E(S))}{E(S)}+\tilde{\phi}(E(S))\alpha(S)\le c\tilde{\phi}(E(S)). \end{aligned} \end{equation} (6.18)

    A combination of (6.15)–(6.18) leads to (6.12).

    In order to finalize the proof of our result, we let

    \tilde{\phi}(s) = 2\varepsilon_0 s G^{\prime}(\varepsilon^2_0 s), \text{ and} \quad G_1(s) = G(s^2),

    where \varepsilon_0 > 0 is small enough and G^* and G_1^* denote the dual functions of the convex functions G and G_1 respectively in the sense of Young (see, Arnold [33], pp. 64).

    Lemma 6.3. Suppose G is nonlinear, then the following estimates

    \begin{equation} G^{*}\left(\frac{\tilde{\phi}(s)}{s} \right)\le \frac{\tilde{\phi}(s)}{s}\left( G^{\prime}\right)^{-1}\left( \frac{\tilde{\phi}(s)}{s}\right) \end{equation} (6.19)

    and

    \begin{equation} G_1^{*}\left(\frac{\tilde{\phi}(s)}{\sqrt{s}}\right)\le \varepsilon_0 \tilde{\phi}(\sqrt{s}). \end{equation} (6.20)

    hold, where \tilde{\phi} is defined earlier in Lemma 6.2.

    Proof. Since G^* and G_1^* are the dual functions of the convex functions G and G_1 respectively, then

    \begin{equation} \begin{aligned} G^*(s) = s(G^{\prime})^{-1}(s)-G\left[(G^{\prime})^{-1}(s)\right]\le s(G^{\prime})^{-1}(s) \end{aligned} \end{equation} (6.21)

    and

    \begin{equation} \begin{aligned} G_1^*(s) = s(G_1^{\prime})^{-1}(s)-G_1\left[(G_1^{\prime})^{-1}(s)\right]\le s(G_1^{\prime})^{-1}(s). \end{aligned} \end{equation} (6.22)

    Using (6.21) and the definition of \tilde{\phi} , we obtain (6.19). For the proof of (6.20), we use (6.22) and the definitions of G_1 and \tilde{\phi} to obtain

    \begin{equation} \begin{aligned} \frac{\tilde{\phi}(s)}{\sqrt{s}}(G_1^{\prime})^{-1}\left( \frac{\tilde{\phi}(s)}{\sqrt{s}}\right)&\le 2\varepsilon_0 \sqrt{s} G^{\prime}(\varepsilon^2_0 s)(G_1^{\prime})^{-1}\left(2\varepsilon_0 \sqrt{s} G^{\prime}(\varepsilon^2_0 s)\right)\\ & = 2\varepsilon_0 \sqrt{s} G^{\prime}(\varepsilon^2_0 s)(G_1^{\prime})^{-1}\left( G_1^{\prime}(\varepsilon_0 \sqrt{s}) \right)\\ & = 2\varepsilon^2_0 s G^{\prime}(\varepsilon^2_0 s)\\ & = \varepsilon_0 \tilde{\phi}(\sqrt{s}). \end{aligned} \end{equation} (6.23)

    Now, we state and prove our main decay results.

    Theorem 6.4. Let (u_0, u_1)\in H^2_{*}(\Omega)\times L^2(\Omega) . Assume that (H1) and (H2) hold. Then there exist positive constants k and c such that, for t large, the solution of (1.1) satisfies

    \begin{equation} \begin{aligned} E(t)\le ke^{-c\int_{0}^{t}\alpha (s)ds},\qquad \qquad \qquad \mathit{\text{if G is linear,}} \end{aligned} \end{equation} (6.24)
    \begin{equation} \begin{aligned} E(t)\le \psi^{-1}\left(h(\tilde{\alpha}(t))+\psi\left(E(0)\right)\right),\qquad \mathit{\text{if G is nonlinear,}} \end{aligned} \end{equation} (6.25)

    where

    \tilde{\alpha}(t) = \int_{0}^{t}\alpha(t)dt,\mathit{\text{}} \psi(t) = \int_{t}^{1}\frac{1}{\chi(s)}ds,\mathit{\text{and}}\chi(s) = 2\varepsilon_0 c s G^{\prime}(\varepsilon^2_0 s)

    and

    \begin{equation*} \begin{aligned} \begin{cases} h(t) = 0, \quad 0\le t \le \frac{E(0)}{\chi(E(0))},\\\\ h^{-1}(t) = t+\frac{\psi^{-1}\left(t+\psi(E(0))\right)}{\chi\left(\psi^{-1}(t+\psi(E(0)))\right)}, \quad t > 0. \end{cases} \end{aligned} \end{equation*}

    Proof. To establish (6.24), we use (6.1) and Lemma 3.5 for \gamma(t) = \int_{0}^{t}\alpha(s)ds . Consequently the result follows. For the proof of (6.25), we re-estimate the terms of (6.12) as follows: we consider the following partition of \Omega :

    \begin{equation*} \Omega_1 = {\{x\in \Omega: \vert u_t\vert \ge \varepsilon_1\},} \quad \Omega_2 = {\{x\in \Omega: \vert u_t\vert \le \varepsilon_1\}}. \end{equation*}

    So,

    \begin{equation*} \begin{aligned} \int_{S}^{T}\alpha(t)\frac{\tilde{\phi}(E)}{E}&\int_{\Omega_1}\left(\vert u_t\vert ^{2}+\vert u g(u_t)\vert \right)dxdt\\ & = \int_{S}^{T}\alpha(t) \frac{\tilde{\phi}(E)}{E}\int_{\Omega_1}\vert u_t\vert ^{2}dxdt+\int_{S}^{T}\alpha(t) \frac{\tilde{\phi}(E)}{E}\int_{\Omega_1}\vert u g(u_t)\vert dxdt\\ &: = I_1+I_2. \end{aligned} \end{equation*}

    Using the definition of \Omega_1 , (3.4) and (5.2), we have

    \begin{equation} \begin{aligned} I_1&\le c\int_{S}^{T}\alpha(t)\frac{\tilde{\phi}(E)}{E}\int_{\Omega_1}u_t g(u_t)dxdt\\ &\le c\int_{S}^{T}\frac{\tilde{\phi}(E)}{E}\left(-E^{\prime}(t)\right)dt\le c\tilde{\phi}(E(S)). \end{aligned} \end{equation} (6.26)

    After applying Hölder's and Young's inequalities and Lemma 3.1, we obtain for some \varepsilon > 0 ,

    \begin{equation} \begin{aligned} I_2&\le \int_{S}^{T}\alpha(t) \frac{\tilde{\phi}(E)}{E}\left(\int_{\Omega_1}\vert u\vert^{2}dx\right)^{\frac{1}{2}}\left( \int_{\Omega_1}\vert g(u_t) \vert^{2} \right)^{\frac{1}{2}}dt\\ &\le \varepsilon \int_{S}^{T}\alpha(t)\frac{\tilde{\phi}^{2}(E)}{E^{2}}\|u\|_{H^2_*(\Omega)}dt+c(\varepsilon)\int_{S}^{T}\alpha(t)\int_{\Omega_1}\vert g(u_t)\vert^{2}dt. \end{aligned} \end{equation} (6.27)

    The definition of \Omega_1 , (3.4), (5.1), (5.2) and (6.27) lead to

    \begin{equation} \begin{aligned} I_2&\le \varepsilon \int_{S}^{T}\alpha(t)\frac{\tilde{\phi}^{2}(E)}{E}dt+c(\varepsilon)\int_{S}^{T}\alpha(t)\int_{\Omega_1}u_t g(u_t)dxdt\\ &\le\varepsilon \int_{S}^{T}\alpha(t)\frac{\tilde{\phi}^{2}(E)}{E}dt+c(\varepsilon)E(S). \end{aligned} \end{equation} (6.28)

    Using the definition of \tilde{\phi} and the convexity of G , then (6.28) becomes

    \begin{equation} \begin{aligned} &I_2\le \varepsilon \int_{S}^{T}\alpha(t)\frac{\tilde{\phi}^{2}(E)}{E}dt+cE(S)\\ & = 2\varepsilon \varepsilon_0 \int_{S}^{T}\alpha(t) \tilde{\phi}(E) G^{\prime}\left(\varepsilon_0^2E(t)\right)dt+cE(S)\\ &\le 2\varepsilon \varepsilon_0 \int_{S}^{T}\alpha(t) \tilde{\phi}(E) G^{\prime}\left(\varepsilon_0^2E(0)\right)dt+cE(S)\\ &\le 2c\varepsilon \varepsilon_0 \int_{S}^{T}\alpha(t) \tilde{\phi}(E)dt+cE(S). \end{aligned} \end{equation} (6.29)

    Combining (6.12), (6.26) and (6.29) and choosing \varepsilon small enough, we obtain

    \begin{equation} \begin{aligned} \int_{S}^{T}\alpha(t) \tilde{\phi}(E)dt \le &c\tilde{\phi}(E)+c\int_{S}^{T}\alpha(t) \frac{\tilde{\phi}(E)}{E}\int_{\Omega_2}\left(\vert u_t\vert^2+\vert u g(u_t)\vert\right)dxdt. \end{aligned} \end{equation} (6.30)

    Using Young's inequality and Jensen's inequality (Eq 3.3), (Eq 3.4) and (Eq 5.1), we get

    \begin{equation} \begin{aligned} \int_{S}^{T}\alpha(t) &\frac{\tilde{\phi}(E)}{E}\int_{\Omega_2}\left(\vert u_t\vert^2+\vert u g(u_t)\vert\right)dxdt\le \int_{S}^{T}\alpha(t) \frac{\tilde{\phi}(E)}{E}\int_{\Omega_2}G^{-1}\left( u_t g(u_t)\right)dxdt\\ & \quad +\int_{S}^{T}\alpha(t) \frac{\tilde{\phi}(E)}{E}\|u\|^{\frac{1}{2}}_{H^2_*(\Omega)}\left(\int_{\Omega_2}G^{-1}(u_tg(u_t))dx\right)^{\frac{1}{2}}dxdt\\ &\le \vert \Omega \vert\int_{S}^{T}\alpha(t) \frac{\tilde{\phi}(E)}{E}G^{-1}\left(\frac{1}{\vert \Omega \vert}\int_{\Omega}u_tg(u_t)dx\right)dt\\ & \quad +\int_{S}^{T}\alpha(t) \frac{\tilde{\phi}(E)}{E}\sqrt{E}\sqrt{\vert \Omega\vert G^{-1}\left(\frac{1}{\vert \Omega \vert}\int_{\Omega}u_tg(u_t)dx\right)}dt. \end{aligned} \end{equation} (6.31)

    Applying the generalized Young inequality

    \begin{equation*} AB\le G^*(A)+G(B) \end{equation*}

    to the first term of (6.31), with A = \frac{\tilde{\phi}(E)}{E} and B = G^{-1}\left(\frac{1}{\vert \Omega \vert}\int_{\Omega}u_tg(u_t)dx\right) , we easily see that

    \begin{equation} \begin{aligned} \frac{\tilde{\phi}(E)}{E}G^{-1}\left(\frac{1}{\vert \Omega \vert}\int_{\Omega}u_tg(u_t)dx\right)\le G^*\left(\frac{\tilde{\phi}(E)}{E}\right)+ \frac{1}{\vert \Omega \vert}\int_{\Omega}u_tg(u_t)dx. \end{aligned} \end{equation} (6.32)

    Then we apply it to the second term of (6.31), with A = \frac{\tilde{\phi}(E)}{E} \sqrt{E} and B = \sqrt{\vert \Omega\vert G^{-1}\left(\frac{1}{\vert \Omega \vert}\int_{\Omega}u_tg(u_t)dx\right)} to obtain

    \begin{equation} \begin{aligned} \frac{\tilde{\phi}(E)}{E} \sqrt{E}\sqrt{\vert \Omega\vert G^{-1}\left(\frac{1}{\vert \Omega \vert}\int_{\Omega}u_tg(u_t)dx\right)}\le G_1^*\left( \frac{\tilde{\phi}(E)}{E} \sqrt{E} \right) +\vert \Omega\vert G^{-1}\left(\frac{1}{\vert \Omega \vert}\int_{\Omega}u_tg(u_t)dx\right). \end{aligned} \end{equation} (6.33)

    Combining (6.31)–(6.33) and using (6.19) and (6.20), we arrive at

    \begin{equation} \begin{aligned} \int_{S}^{T}\alpha(t) &\frac{\tilde{\phi}(E)}{E}\int_{\Omega_2}\left(\vert u_t\vert^2+\vert u g(u_t)\vert\right)dxdt\\ &\le c \int_{S}^{T} \alpha(t) \left( G_1^*\left(\frac{\tilde{\phi}(E)}{E} \sqrt{E}\right)+G^*\left( \frac{\tilde{\phi}(E)}{E}\right)\right)dt+c\int_{S}^{T}\alpha(t)\int_{\Omega}u_t g(u_t)dxdt\\ &\le c \int_{S}^{T} \alpha(t) \left(\varepsilon_0+\frac{(G^{\prime})^{-1}\left(\frac{\tilde{\phi}(E)}{E}\right)}{E}\right)\tilde{\phi}(E)dt+cE(S). \end{aligned} \end{equation} (6.34)

    Using the definition of \tilde{\phi} and the fact that s \to (G^{\prime})^{-1}(s) is non-decreasing, we deduce that, for 0 < \varepsilon_0\le \frac{1}{2} ,

    \begin{equation} \frac{(G^{\prime})^{-1}\left(\frac{\tilde{\phi}(E)}{E}\right)}{E} = \frac{(G^{\prime})^{-1}\left( 2\varepsilon_0 G^{\prime}(\varepsilon_0^2 E)\right)}{E}\le \varepsilon_0^2. \end{equation} (6.35)

    Combining (6.34) and (6.35) leads to

    \begin{equation} \begin{aligned} \int_{S}^{T}\alpha(t) &\frac{\tilde{\phi}(E)}{E}\int_{\Omega_2}\left(\vert u_t\vert^2+\vert u g(u_t)\vert\right)dxdt\le c\varepsilon_0 \int_{S}^{T}\alpha(t) \tilde{\phi}(E)dt+cE(S). \end{aligned} \end{equation} (6.36)

    Then, choosing \varepsilon_0 small enough, we deduce from (6.30) and (6.36) that

    \begin{equation*} \begin{aligned} \int_{S}^{T}\alpha(t) \tilde{\phi}(E(t))dt\le c \left( 1+\frac{\tilde{\phi}(E(S))}{E(S)}\right)E(S). \end{aligned} \end{equation*}

    Using the facts that E is non-increasing and s \to \frac{\tilde{\phi}(s)}{s} is non-decreasing, we obtain

    \begin{equation} \int_{S}^{+\infty}\alpha(t) \tilde{\phi}(E(t))dt \le cE(S). \end{equation} (6.37)

    Let \tilde{E} = E\circ \tilde{\alpha}^{-1} , where \tilde{\alpha}(t) = \int_{0}^{t}\alpha(s)ds . Then we deduce from (6.37) that

    \begin{equation*} \begin{aligned} \int_{S}^{\infty} \tilde{\phi}(\tilde{E}(t))dt& = \int_{S}^{\infty} \tilde{\phi}(E(\tilde{\alpha}^{-1}(t)))dt\\ & = \int_{\tilde{\alpha}^{-1}(S)}^{\infty} \alpha(\eta)\tilde{\phi}(E(\eta))d\eta\\ &\le cE\left(\tilde{\alpha}^{-1}(S) \right)\le c \tilde{E}(S). \end{aligned} \end{equation*}

    Using Lemma 3.6 for \tilde{E} and \chi(s) = \frac{1}{c} \tilde{\phi}(s) , we deduce from (3.6) the following estimate

    \begin{equation*} \tilde{E}(t)\le \psi^{-1}\left(h(t)+\psi\left(E(0)\right)\right), \end{equation*}

    which gives (6.25), by using the definition of \tilde{E} and the change of variables.

    Remark 6.5. The stability result (6.25) is a decay result. Indeed,

    \begin{equation*} \begin{aligned} h^{-1}(t)& = t+\frac{\psi^{-1}\left(t+\psi(E(0))\right)}{\chi\left(\psi^{-1}(t+\psi(E(0)))\right)}\\ & = t+\frac{c}{2\varepsilon_0 c G^{\prime}\left(\varepsilon_0^2 \psi^{-1}(t+r) \right)}\\ &\ge t+\frac{c}{2\varepsilon_0 c G^{\prime}\left(\varepsilon_0^2 \psi^{-1}(r) \right)}\\ &\ge t+\tilde{c}. \end{aligned} \end{equation*}

    Hence, \lim_{t\to\infty}h^{-1}(t) = \infty , which implies that \lim_{t \to\infty}h(t) = \infty . Using the convexity of G , we have

    \begin{equation*} \begin{aligned} \psi(t)& = \int_{t}^{1}\frac{1}{\chi(s)}ds = \int_{t}^{1}\frac{c}{2\varepsilon_0 sG^{\prime}\left(\varepsilon_0^2s\right)}\ge \int_{t}^{1}\frac{c}{sG^{\prime}\left(\varepsilon_0^2\right)}\ge c \left[\ln{\vert s\vert}\right]_{t}^{1} = -c\ln{t}. \end{aligned} \end{equation*}

    Therefore, \lim_{t\to 0^+}\psi(t) = \infty which leads to \lim_{t\to \infty}\psi^{-1}(t) = 0 .

    Examples

    1) Let g(s) = s^m , where m\ge 1 . Then the function G is defined in the neighborhood of zero by

    G(s) = c s^{\frac{m+1}{2}}

    which gives, near zero

    \chi(s) = \frac{c(m+1)}{2} s^{\frac{m+1}{2}}.

    So, we obtain

    \begin{equation*} \psi(t) = c \int_{t}^{1}\frac{2}{(m+1)s^{\frac{m+1}{2}}}ds = \left\{ \begin{array}{ll} \frac{c}{t^{\frac{m-1}{2}}}, & if \hbox{$\text{ }m > 1$;} \\ \\ -c\ln{t}, & if \hbox{$\text{ }m = 1$,} \end{array} \right. \end{equation*}

    and then, in the neighborhood of \infty

    \begin{equation*} \psi^{-1}(t) = \left\{ \begin{array}{ll} c t^{-\frac{2}{m-1}}, & if \hbox{$\text{ }m > 1$;} \\ c e^{-t}, & if \hbox{$\text{ }m = 1$,} \end{array} \right. \end{equation*}

    Using the fact that h(t) = t as t goes to infinity, we obtain from (6.24) and (6.25)

    \begin{equation*} E(t)\le \left\{ \begin{array}{ll} c \left(\int_{0}^{t}\alpha(s)ds\right)^{-\frac{2}{m-1}}, & if \hbox{$\text{ }m > 1$;} \\ \\ c e^{-{\int_{0}^{t}\alpha(s)ds}}, & if \hbox{$\text{ }m = 1$.} \end{array} \right. \end{equation*}

    2) Let g(s) = s^m \sqrt{-\ln{s}} , where m\ge 1 . Then the function G is defined in the neighborhood of zero by

    G(s) = c s^{\frac{m+1}{2}} \sqrt{-\ln{\sqrt{s}}}

    which gives, near zero

    \chi(s) = cs^{\frac{m+1}{2}}\left(-\ln{\sqrt{s}}\right)^{-\frac{1}{2}}\left(\frac{m+1}{2}\left(-\ln{\sqrt{s}}\right)-\frac{1}{4}\right).

    Therefore, we get

    \begin{equation*} \begin{aligned} \psi(t) = &c \int_{t}^{1}\frac{1}{ s^{\frac{m+1}{2}}\left(-\ln{\sqrt{s}}\right)^{-\frac{1}{2}}\left(\frac{m+1}{2}\left(-\ln{\sqrt{s}}\right)-\frac{1}{4}\right)}ds\\ = &c\int_{1}^{\frac{1}{\sqrt{t}}}\frac{\tau^{m-2}}{(\ln{\tau})^{-\frac{1}{2}} \left(\frac{m+1}{2}\ln{\tau}-\frac{1}{4} \right)}d\tau\\ = &\left\{ \begin{array}{ll} \frac{c}{t^{\frac{m-1}{2}}\sqrt{-\ln{t} }}, & if \hbox{$\text{ }m > 1$;} \\ \\ c\sqrt{-\ln{t}}, & if \hbox{$\text{ }m = 1$,} \end{array} \right. \end{aligned} \end{equation*}

    and then, in the neighborhood of \infty , we have

    \begin{equation*} \psi^{-1}(t) = \left\{ \begin{array}{ll} c t^{-\frac{2}{m-1}}\left( \ln{t} \right)^{-\frac{1}{m-1}}, & if \hbox{$\text{ }m > 1$;} \\ \\ c e^{-t^2}, & if \hbox{$\text{ }m = 1$,} \end{array} \right. \end{equation*}

    Using the fact that h(t) = t as t goes to infinity, we obtain

    \begin{equation*} E(t)\le \left\{ \begin{array}{ll} c \left(\int_{0}^{t}\alpha(s)ds\right)^{-\frac{2}{m-1}} \left(\ln{\left(\int_{0}^{t}\alpha(s)ds\right)}\right)^{-\frac{1}{m-1}}, & if \hbox{$\text{ }m > 1$;} \\ \\ c e^{- \left(\int_{0}^{t}\alpha(s)ds\right)^2}, & if \hbox{$\text{ }m = 1$,} \end{array} \right. \end{equation*}

    The authors would like to express their profound gratitude to King Fahd University of Petroleum and Minerals (KFUPM) for its continuous support. The authors also thank the referees for their very careful reading and valuable comments. This work was funded by KFUPM under Project #SB201003.

    The authors declare there is no conflicts of interest.



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