Research article

Decay result in a problem of a nonlinearly damped wave equation with variable exponent

  • Received: 19 August 2021 Accepted: 19 November 2021 Published: 24 November 2021
  • MSC : 35B37, 35L55, 74D05, 93D15, 93D20

  • In this work we study a wave equation with a nonlinear time dependent frictional damping of variable exponent type. The existence and uniqueness results are established using Fadeo-Galerkin approximation method. We also exploit the Komornik lemma to prove the uniform stability result for the energy associated to the solution of the problem under consideration.

    Citation: Mohammad Kafini, Jamilu Hashim Hassan, Mohammad M. Al-Gharabli. Decay result in a problem of a nonlinearly damped wave equation with variable exponent[J]. AIMS Mathematics, 2022, 7(2): 3067-3082. doi: 10.3934/math.2022170

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  • In this work we study a wave equation with a nonlinear time dependent frictional damping of variable exponent type. The existence and uniqueness results are established using Fadeo-Galerkin approximation method. We also exploit the Komornik lemma to prove the uniform stability result for the energy associated to the solution of the problem under consideration.



    In this work we are concerned with the decay rate of the following problem with nonlinear damping of variable exponent

    {utt(x,t)Δu(x,t)+α(t)[ut(x,t)+ut(x,t)|ut|m(x)2(x,t)]=0,inΩ×(0,T),u=0,onΩ×(0,T),u(x,0)=u0(x),ut(x,0)=u1(x),inΩ, (1.1)

    where T>0 and Ω is a bounded domain of Rn(n1). The functions u0, u1 are initial data and the variable exponent m()C(¯Ω) is a given functions satisfying

    1<m1m(x)m2<2, (1.2)

    where

    m1:=infxΩm(x),   m2:=supxΩm(x),   2={2nn2,ifn3,,ifn<3,

    and also satisfies the log-Hölder continuity condition:

    |m(x)m(y)|Alog|xy|, (1.3)

    for x,yΩ, with |xy|<δ, A>0 and 0<δ<1. The function α:[0,)(0,) is a bounded nonincreasing C1function and

    α0>0  such that  α(t)α0, t0. (1.4)

    Problems with variable exponents appear as a direct consequence of the advancement of science and technology. Many physical and engineering models require more sophisticated mathematical functional spaces to be studied and well understood. For example, in fluid dynamics, the electrorheological fluids (smart fluids) have the property that the viscosity changes when exposed to an electrical field. More examples are found in studying models of the image processing and filtration processes through a porous media. The Lebesgue and Sobolev spaces with variable exponents proved to be efficient tools to study such problems. More details on applications of these problems can be found in ([1,2,3]).

    A lot of papers in the literature dealt with stabilization of wave equations with different types of nonlinearities such as linear, polynomial and logarithmic. For instance, the following problem was studied by Nakao [4].

    uttΔu+|ut|m2ut+|u|p2u=0,  in  Ω×(0,),

    where m,p>2 and Ω Rn (n1) is a bounded domain. He showed that, with Dirichlet-boundary conditions, the problem has a unique global weak solution if 2p2(n1)/(n2), n3 and a global unique strong solution if p>2(n1)/(n2), n3. In both cases, he proved that the energy of the solution decays algebraically if m>2 and decays exponentially if m=2. Benaissa and Messaoudi [5] considered

    uttΔu+a(1+|ut|m2)ut=|u|p2u, in   Ω×(0,),

    where m,p>2 and showed, for small initial data in an appropriate function space, that the problem has a global weak solution which decays exponentially even if m>2. We also mention here the work of Mustafa and Messaoudi [6], where they considered

    uttΔu+α(t)g(ut)=0,  in  Ω×(0,),

    and established an explicit and general decay rate result, without imposing any restrictive growth assumption on the frictional damping term.

    As we mentioned earlier, modern technology and engineering required the use of variable exponents nonlinearities and the Lebesgue and Sobolev spaces with variable exponents as well. In this regard, we mention the work of Ghegal et al. [7] where, in a bounded domain, the following equation is considered

    uttΔu+|ut|m()2ut=|u|p()2u,  in   Ω×(0,).

    Under suitable conditions on the initial data and the variable exponents, the authors used stable-set method to prove a global existence result. Then, by applying an integral inequality due to Komornik, they obtained the stability result. More results can be found in ([8,9,10]).

    Hyperbolic problems involving variable-exponent nonlinearities with delay are also considered. For instance, Kafini and Messaoudi [11] studied the problem

    uttΔu+μ1|ut|m(x)2ut+μ2|ut|m(x)2(tτ)ut(tτ)=bu|u|p(x)2.

    For b>0, they established a global nonexistence result under suitable conditions on μ1,μ2,m(),p() and the initial data. While, for b=0, they obtained a decay result which is of either polynomial or exponential type depending on the nature of m().

    Recently, Messaoudi in [12] considered the problem

    uttdiv(|u|r()2u)Δut+|ut|m()2ut=0,    Ω×(0,T),

    and established several decay results depending on the nature of variable exponents r() and m(). See [13,14,15,16,17], for more results on the local existence and blow up for some problems with variable exponent nonlinearities.

    Fractional derivatives have been also influenced by variable orders. One can see variable-order fractional differential equations: mathematical foundations, physical models, numerical methods and applications as in [18]. Analyzing a variable-order time-fractional wave equation, which models, e.g., the vibration of a membrane in a viscoelastic environment examined in [19]. See also[20,21,22] for more details.

    In our work, we aim to study the nonlinear wave Eq (1.1) with nonlinear feedback having a variable exponent m(x) and a time-dependent coefficient α(t). We establish a decay result of an exponential and polynomial type under specific conditions on both m() and α(t) and the initial data. This paper consists of three sections in addition to the introduction. In Section 2, we recall the basic definitions of the variable exponent Lebesgue spaces Lp()(Ω), the Sobolev spaces W1,p()(Ω), as well as some of their properties. Section 3 is devoted to the existence and uniqueness of a weak global solution. In the last section, we show the decay result.

    In this section, we present some materials needed for the statement and the proof of our results. In what follows, we give definitions and properties related to Lebesgue and Sobolev spaces with variable exponents, see [23,24] for more details.

    Let Ω be a domain of Rn with n2 and p:Ω[1,] be a measurable function. The Lebesgue space Lp()(Ω) with a variable exponent p() is defined by

    Lp()(Ω)={v:ΩR; measurable such that ϱp()(λv)<+, for some λ>0},

    where

    ϱp()(v):=Ω|v(x)|p(x)dx.

    The Luxembourg-type norm is given by

    vp():=inf{λ>0:Ω|v(x)λ|p(x)dx1}.

    The space Lp()(Ω), equipped with the above norm, is a Banach space.

    Lemma 2.1. (Hölder's inequality) Let p,q,s1 be measurable functions defined on Ω suchthat

    1s(y)=1p(y)+1q(y),  for a.e.yΩ.

    If fLp()(Ω) and gLq()(Ω), then fgLs()(Ω) and

    fgs()2fp()gq().

    Lemma 2.2. If p:Ω[1,) is a measurable function and1p1p(x)p2<, then

    min{vp1p(),vp2p()}ϱp()(v)max{vp1p(),vp2p()},

    fora.e. xΩ and for any vLp()(Ω).

    Lemma 2.3. [12] If p:Ω[1,) is a measurable function and 1p1p(x)p2<, then

    Ω|v(x)|p(x)dxvp1p1+vp2p2,     vLp()(Ω).

    The variable-exponent Sobolev space W1,p()(Ω) is defined as

    W1,p()(Ω)={vLp()(Ω) such that v exists and |v|Lp()(Ω)}.

    This space is a Banach space with respect to the norm

    vW1,p()(Ω)=vp()+vp().

    Suppose p() satisfies (1.3). Then the space W1,p()0(Ω) is defined to be the closure of C0(Ω) in W1,p()(Ω). The definition of the space W1,p()0(Ω) is usually different from the constant exponent case. However, under condition (1.3) both definitions coincide. The dual space of W1,p()0(Ω) is W1,p()0(Ω) defined in the same way as in the classical Sobolev spaces, where

    1p()+1p()=1.

    Lemma 2.4. (Poincaré's inequality) Let Ω be a bounded domain of Rn and p() satisfies (1.2) and (1.3), then

    vp()Cvp(),for allvW1,p()0(Ω),

    where C is a positive constant depends on p() and Ω. In particular, the space W1,p()0(Ω) has an equivalent norm given by

    vW1,p()0(Ω)=vp().

    Lemma 2.5. If p:¯Ω[1,) is continuous and 

    2p1p(x)p22nn2, n3,

    then the embedding H1(Ω)Lp()(Ω) is continuous and compact.

    Lemma 2.6. [27] Let E:R+R+ be a nonincreasing function and ϕ:R+R be an increasing C1 -function satisfying

    ϕ(0)=0 and ϕ(t)+  as   t+.

    Assume further, that there exist q0, A>0 such that

    SEq+1(t)ϕ(t)dtAE(S),  S>0.

    Then, t  0,

    E(t)CE(0)(1+ϕ(t))1/q,ifq>0,E(t)CE(0)eωϕ(t),ifq=0,

    where C and ω are positive constants independent of the initial energy E(0).

    Definition 2.7. Given the initial data (u0,u1)H10(Ω)×L2(Ω), a function u defined on Ω×(0,T) is called a weak solution of problem (1.1) if

    uL((0,T);H10(Ω)),utL((0,T);L2(Ω))Lm()(Ω×(0,T))

    and it verifies the variational equation

    utt,w+(u,w)+α(t)[(ut,w)+(|ut|m(x)2ut,w)]=0,  wC0(Ω).

    We introduce the energy functional associated to problem (1.1) as

    E(t):=12||ut||22+12||u||22,   t0. (2.1)

    Lemma 2.8. Let u be the solution of (1.1). Then, 

    E(t)=α(t)Ω(|ut|2+|ut|m(x))dx0, t0. (2.2)

    Proof. Multiplying Eq (1.1) by ut and integrating over Ω, the result follows.

    Remark 2.9. In the sequel, we use C to denote a generic constant which may differ from one place to another.

    The following theorem states our existence and uniqueness results, which are the main focus of this section.

    Theorem 3.1. Assume that the variable exponent m() satisfies conditions (1.2) and (1.3). Then, for any initial data u0H10(Ω),u1L2(Ω), problem (1.1)admits a unique global weak solution.

    Proof. To prove the existence of a weak solution to (1.1), we make use of the Galerkin approximation method. For that reason we assume {vj}j1 is an orthogonal basis for H10(Ω) and orthonormal in L2(Ω). We find a solution of the form

    uk(x,t)=kj=1ajk(t)vj(x),   ajk(t)=uk(t),vj,

    to the approximate problem

    (uktt,vj)+(uk,vj)+α(t)[(ukt,vj)+(|ukt|m()2ukt,vj)]=0, (3.1)

    where

    uk(x,0)=uk0(x)=kj=1(uk0,vj)vju0  strongly in H10(Ω),ukt(x,0)=uk1(x)=kj=1(uk1,vj)vju1  strongly in L2(Ω). (3.2)

    This system, by the standard ODE theory has a unique solution guaranteed on [0,tk), 0<tkT. Next, we need to show that this solution can be extended to the maximal interval [0,T), k1 and for any T>0.

    Replace vj by ukt in (3.1) to get

    ddt[ukt22+uk22]+2α(t)[ukt22+Ω|ukt|m()dx]=0,

    and integrate over (0,t) for t(0,tk) to arrive at

    ukt22+uk22+2t0α(s)[ukt(s)22+Ω|ukt|m()(s)dx]ds=uk122+uk022C,  k1. (3.3)

    Hence, the solution can be extended to [0,T), for any given T>0.

    Using (1.4), we arrive at

    ukt22+uk22+2α0t0[ukt(s)22+Ω|ukt|m()(s)dx]dsC,

    where we can conclude that

    uk is bounded in L((0,T);H10(Ω))
    ukt is bounded in L((0,T);L2(Ω))
    ukt is bounded in Lm()(Ω×(0,T)).

    Therefore, we can extract subsequences, still denoted by uk and ukt, such that

    uku weakly star in L((0,T);H10(Ω))
    uktut weakly star in L((0,T);L2(Ω)).

    As ukt is bounded in Lm()(Ω×(0,T)), then |ukt|m()2ukt is bounded in Lm()m()1(Ω×(0,T)). Hence,

    |ukt|m()2uktψ    weakly in    Lm()m()1(Ω×(0,T)).

    To show that ψ=|ut|m()2ut, we integrate (3.1) over (0,t) to get, j=1,...,k,

    ΩuktvjdxΩuk1vjdx+t0Ωukvjdx+t0α(s)Ω(ukt(s)+|ukt|m()2ukt(s))vjdxds=0.

    Now, letting k+ and differentiating the latter result with respect to t gives

    ddtΩutvdx+Ωuvdx+α(t)Ω(ut+ψ)vdx=0,   vH10(Ω). (3.4)

    Hence,

    uttΔu+α(t)(ut+ψ)=0,    in  D(Ω×(0,T)).

    If we define

    χk=2T0α(t)Ω(|ukt|m()2ukt|v|m()2v)(uktv)dxdt, vLm()((0,T);H10(Ω)),

    and

    A(v)=|v|m()2v,

    then we have

    χk=2T0α(t)Ω(A(ukt)A(v))(uktv)dxdt0, vLm()((0,T);H10(Ω)).

    Using Eq (3.3), we get

    χk=uk122+uk022Ω(|ukt(T)|2+|uk(T)|2)dx2T0α(t)Ω|ukt|2dxdt2T0α(t)ΩA(ukt)vdxdt2T0α(t)ΩA(v)(uktv)dxdt.

    As k+,

    0limsupkχku122+u022Ω(|ut(T)|2+|u(T)|2)dx2T0α(t)Ω|ut|2dxdt2T0α(t)Ωψvdxdt2T0α(t)ΩA(v)(utv)dxdt. (3.5)

    Integration of (3.4) over (0,T) after replacing v by ut give

    Ω|ut(T)|2dx+Ω|u(T)|2dxu122u022+2T0α(t)Ω(|ut|2+ψut)dxdt=0. (3.6)

    Adding (3.5) and (3.6) give

    0limsupkχk2T0α(t)Ωψutdxdt2T0α(t)Ωψvdxdt2T0α(t)ΩA(v)(utv)dxdt=2T0α(t)Ω(ψA(v))(utv)dxdt,   vLm()((0,T);H10(Ω)).

    Thus, by the density of H10(Ω) in Lm()(Ω) we have

    T0Ω(ψA(v))(utv)dxdt0,   vLm()(Ω×(0,T)).

    If we let v=λw+ut for wLm()(Ω×(0,T)) then

    T0Ω(ψA(λw+ut))wdxdt0,   wLm()(Ω×(0,T)).

    As 0<λ0, we have,

    T0Ω(ψA(ut))wdxdt0,   wLm()(Ω×(0,T)).

    Similarly, if 0>λ0, we have,

    T0Ω(ψA(ut))wdxdt0,   wLm()(Ω×(0,T)).

    This implies that ψ=A(ut)=|ut|m()2ut.

    To handle the initial conditions, we use Lions' Lemma [25], to obtain, up to a subsequence, that

    uku in C([0,T];L2(Ω)).

    Therefore, uk(,0) makes sense and uk(,0)u(,0) in L2(Ω). Also, by density we have

    uk(,0)=uk0u0  in  H10(Ω),

    hence u(,0)=u0.

    For the other condition, as in [26], we obtain from (3.1) and for any jk and ϕC0(0,T),

    T0Ωuktvj(x)ϕ(t)dxdt=T0Ωukvj(x)ϕ(t)dxdt+T0α(t)Ω(ukt+|ukt|m()2ukt)vj(x)ϕ(t)dxdt.

    As k+, we obtain that, for all vH10(Ω),

    T0Ωutv(x)ϕ(t)dxdt=T0Δuα(t)(ut+|ut|m()2ut),v(x)ϕ(t)dt.

    This implies that

    uttLm()m()1([0,T);H1(Ω)),

    and u solves the equation

    uttΔu+α(t)(ut+|ut|m()2ut)=0.

    Therefore,

    utC([0,T);H1(Ω)),

    where ukt(,0) makes sense and ukt(,0)ut(,0) in H1(Ω). But we have

    ukt(,0)=uk1u1  in  L2(Ω).

    So ut(,0)=u1.

    To prove the uniqueness, we assume u and v are two solutions of (3.1). Then w=uv satisfies the following problem

    {wttΔw+α(t)(wt+|ut|m()2ut|vt|m()2vt)=0inΩ×(0,T),w=0,onΩ×(0,T),w(x,0)=w0(x),wt(x,0)=w1(x),inΩ.

    Multiply the equation by wt and integrate over Ω, to obtain

    12ddt[Ω(|wt|2+|w|2)dx]+α(t)Ω[|wt|2+(|ut|m()2ut|vt|m()2vt)(utvt)]dx=0.

    Integration over (0,t), to get

    Ω(|wt|2+|w|2)dx+2t0α(t)Ω[|wt|2+(|ut|m()2ut|vt|m()2vt)(utvt)]dxdt=0.

    Using the fact that

    (|a|m()2a|b|m()2b)(ab)0, a,bR  and a.e xΩ,

    we obtain

    Ω(|wt|2+|w|2)dx=0.

    This implies that w=C=0, since w=0 on Ω. Hence, the uniqueness. This completes the proof of Theorem 3.1.

    Theorem 4.1. Let (u0,u1)H10(Ω)×L2(Ω) be given. Assume that 0α(τ)dτ= and m()C(¯Ω) that satisfies 

    2m1m(x)m2<2.

     Then, the solution energy (2.1) satisfies, for two positive constants k1,k2,

    E(t)k1ek2t0α(s)ds,t0. (4.1)

    Proof. Multiply (1.1) by αuEq(t) and integrate over Ω×(s,T), 0<s<T, to obtain

    TsαEq(t)Ω(uuttuΔu+α(uut+uut|ut|m(x)2))dxdt=0,

    which gives

    TsαEq(t)Ω(ddt(uut)u2t+|u|2+α(uut+uut|ut|m(x)2))dxdt=0, (4.2)

    for q0 to be specified later.

    Recalling the fact that Ω(|u|2+u2t)dx=2E(t) and using the relation

    ddt(αEq(t)Ωuutdx)=αEq(t)Ωuutdx+qαEq1(t)E(t)Ωuutdx+αEq(t)ddtΩuutdx,

    equation (4.2) becomes

    2TsαEq+1(t)dt=Tsddt(αEq(t)Ωuutdx)Tsα2Eq(t)Ωuutdxdt+qTsαEq1(t)E(t)Ωuutdxdt+TsαEq(t)Ωuutdxdt+2TsαEq(t)Ωu2tdxdtTsα2Eq(t)Ωuut(x,t)|ut|m(x)2dxdt. (4.3)

    The first term in the right side of (4.3) is estimated, using Poincaré's inequality, (2.2) and the fact that

    Ωuutdx12Ω(|u|2+u2t)dxCΩ(|u|2+u2t)dxCE(t),

    to have

    |Tsddt(αEq(t)Ωuutdx)dt|C[α(s)Eq+1(s)+α(T)Eq+1(T)]Cα(0)Eq(0)E(s)CE(s). (4.4)

    Using Young's inequality, the second term leads to

    |Tsα2Eq(t)Ωuutdxdt|TsEq(t)[δCα(t)Ω|u|2dx+C4δα(t)Ωu2tdx]dtδCTsαEq+1(t)dtC4δTsEq(t)E(t)dt,  δ>0.

    Taking δ=1/2C, we get

    |Tsα2Eq(t)Ωuutdxdt|12TsαEq+1(t)dt+CE(s). (4.5)

    Similar to the first term, we have

    |qTsαEq1(t)E(t)Ωuutdxdt|CTsEq(t)E(t)dtCEq+1(s)CE(s). (4.6)

    The fourth term:

    |Tsα(t)Eq(t)Ωuutdxdt|CTs|α(t)|Eq+1(t)dtCEq+1(s)Ts|α(t)|dtCEq+1(s)α(s)CE(s). (4.7)

    The fifth term:

    2TsαEq(t)Ωu2tdxdt2TsEq(t)E(t)dtCEq+1(s)CE(s). (4.8)

    The last term in the right-hand side of (4.3) is handled by using Young's inequality with

    a(x)=m(x)m(x)1 and a(x)=m(x).

    So, for a.e. xΩ, ε>0, and

    cε(x)=ε1m(x)(m(x))m(x)(m(x)1)m(x)1,

    we have

    |Tsα2Eq(t)Ωuut|ut|m(x)2dxdt|CTsαEq(t)[εΩ|u(t)|m(x)dx+Ωcε(x)|ut(t)|m(x)dx]dtCTsαEq(t)[ε(Ω|u(t)|m1dx+Ω|u(t)|m2dx)+Ωcε(x)|ut(t)|m(x)dx]dtCTsαEq(t)[ε(||u(t)||m12+||u(t)||m22)+Ωcε(x)|ut(t)|m(x)dx]dtCTsαEq(t)[ε(Em12(t)+Em22(t))+Ωcε(x)|ut(t)|m(x)dx]dtεCEm121(0)TsαEq+1(t)dt+CTsαEq(t)Ωcε(x)|ut(t)|m(x)dxdt. (4.9)

    If we fix ε=1/2CEm121(0), noting that cε(x) is bounded since m(x) is bounded, then (4.9) becomes

    |Tsα2Eq(t)Ωuut|ut|m(x)2dxdt|12TsαEq+1(t)dtcTsEq(t)E(t)dt12TsαEq+1(t)dt+CE(s). (4.10)

    Combining (4.3)(4.10) and taking T we arrive at

    sαEq+1(t)dtCE(s).

    Therefore, (4.1) is established by the virtue of Lemma 2.6 for q=0 and ϕ(t)=t0α(s)ds.

    Example 1. If we take α(t)=1 and m(x)=2 then we have ϕ(t)=t and hence, for two positive constants k1,k2,

    E(t)k1ek2t, t0.

    The next theorem handles the case: 1<m1<2.

    Theorem 4.2. (Polynomial Decay) Let (u0,u1)H10(Ω)×L2(Ω) be given. Assume that 0α(τ)dτ= and m()C(¯Ω) and satisfies (1.2). Assume further that m1<2. Then, the solution energy (1.1) satisfies, for some positive constant K,

    E(t)K(1+t0α(τ)dτ)1m12m1 t0. (4.11)

    Proof. We follow the same steps in the proof of the previous theorem. But we have to re-estimate the last term in (4.3). For this purpose, we define

    Ω1={xΩ | m(x)<2}  and  Ω2={xΩ | m(x)2}.

    Thus,

    Ωuut|ut|m(x)2dx=Ω1uut|ut|m(x)2dx+Ω2uut|ut|m(x)2dx.

    Then we use Young's inequality and Poincaré's inequality, to get

    |Ω1uut|ut|m(x)2dx|δCΩ|u|2dx+14δΩ1|ut|2m(x)2dx. (4.12)

    In order to estimate the last term of (4.12), we define

    m3:=supxΩ1m(x)2.

    Then Hölder's inequality and the embedding give

    |αΩ1uut|ut|m(x)2dx|δCΩ|u|2dx+α4δ[Ω1|ut|2m12dx+Ω1|ut|2m32dx]δCΩ|u|2dx+Cα4δ[(Ω1|ut|2dx)m11+(Ω1|ut|2dx)m31]δCΩ|u|2dx+Cα4δ[(Ω|ut|2dx)m11+(Ω|ut|2dx)m31]δCΩ|u|2dx+Cα4δ[1+(Ω|ut|2dx)m3m1](Ω|ut|2dx)m11δCΩ|u|2dx+C4δ[1+(2E(0))m3m1](E(t))m11δCΩ|u|2dx+C4δ(E(t))m11. (4.13)

    Thus,

    |Tsα2Eq(t)Ω1uut|ut|m(x)2dxdt|δCTsαEq+1(t)dt+cδTsαEq(t)(E(t))m11dt. (4.14)

    Using Young's inequality, we obtain for any λ>0,

    Eq(t)(E(t))m11λ(E(t))q2m1+cλ(E(t)).

    If we let q+1=q2m1 hence q=2m1m11, then (4.14) implies that

    |Tsα2Eq(t)Ω1uut|ut|m(x)2dxdt|δCTsαEq+1(t)dt+λcδTsαEq+1(t)dt+cδcλTsα(E(t))dt.

    Then we choose δ=1/4C. After δ is fixed, we choose λ=1/4cδ to obtain

    |Tsα2Eq(t)Ω1uut|ut|m(x)2dxdt|12TsαEq+1(t)dt+CE(s). (4.15)

    Now over Ω2, we follow the same steps as in (4.9) to conclude that

    |Tsα2Eq(t)Ω2uut|ut|m(x)2dxdt|12TsαEq+1(t)dt+CE(s). (4.16)

    Combining (4.15) and (4.16), give

    |Tsα2Eq(t)Ωuut|ut|m(x)2dxdt|TsαEq+1(t)dt+CE(s). (4.17)

    Consequently, from (4.3)(4.8) and (4.17), we have

    TsαEq+1(t)dtCE(s).

    If we let T, then from Lemma 2.6 with ϕ(t)=t0α(τ)dτ and q=2m1m11>0, we arrive for some K>0,

    E(t)K(1+t0α(τ)dτ)1m12m1.

    This completes the proof.

    Example 2. If we take Ω=(0,1), α(t)=2+t1+t and m(x)=212+x, then we have ϕ(t)=t+ln(1+t), m1=3/2 and

    E(t)K(1+t+ln(1+t))1, t0,

    for a positive constant K.

    In this paper, we have shown that the time varying coefficient appears in the problem has a direct effect in well posednesss and the decay rates. In fact, we investigated the nonlinear wave Eq (1.1) with nonlinear feedback having a variable exponent m(x) and a time-dependent coefficient α(t). We established a decay result of an exponential and polynomial type under specific conditions on both m() and α(t) and the initial data.

    The authors would like to express their sincere thanks to King Fahd University of Petroleum and Minerals (KFUPM)/Interdisciplinary Research Center (IRC) for Construction and Building Materials for its support. This work has been funded by KFUPM under Project # SB191048.

    The authors declare that there is no conflict of interest regarding the publication of this paper.



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