Research article

Large time behavior of strong solution to the magnetohydrodynamics system with temperature-dependent viscosity, heat-conductivity, and resistivity

  • Received: 26 October 2024 Revised: 30 December 2024 Accepted: 20 January 2025 Published: 21 February 2025
  • In this paper, we investigate an initial boundary value problem of a planar magnetohydrodynamics system with temperature-dependent viscosity, heat conductivity, and resistivity. When all of the relative coefficients mentioned above are power functions of temperature, the existence and uniqueness of a global-in-time non-vacuum strong solutions are proved under some special assumptions. At the same time, we obtain the nonlinear exponential stability of the solution. In fact, the initial data could be large if the power of viscosity is small enough.

    Citation: Dandan Song, Xiaokui Zhao. Large time behavior of strong solution to the magnetohydrodynamics system with temperature-dependent viscosity, heat-conductivity, and resistivity[J]. Electronic Research Archive, 2025, 33(2): 938-972. doi: 10.3934/era.2025043

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  • In this paper, we investigate an initial boundary value problem of a planar magnetohydrodynamics system with temperature-dependent viscosity, heat conductivity, and resistivity. When all of the relative coefficients mentioned above are power functions of temperature, the existence and uniqueness of a global-in-time non-vacuum strong solutions are proved under some special assumptions. At the same time, we obtain the nonlinear exponential stability of the solution. In fact, the initial data could be large if the power of viscosity is small enough.



    The governing equations of a planar magnetohydrodynamics (MHD) compressible flow can be written in Lagrange variable form as:

    vt=ux, (1.1)
    ut+(P+12|b|2)x=(μuxv)x, (1.2)
    wtbx=(λwxv)x, (1.3)
    (vb)twx=(νbxv)x, (1.4)
    (e+u2+|w|2+v|b|22)t+(u(P+12|b|2)wb)x=(κθxv+μuuxv+λwwxv+νbbxv)x. (1.5)

    Here t>0 represents the time, and xΩ=(0,1) denotes the Lagrange mass coordinate. The unknown functions v(x,t)>0, u(x,t), w=(w1(x,t),w2(x,t)), b=(b1(x,t),b2(x,t)), θ(x,t)>0, e, and P are the specific volume of the gas, longitudinal velocity, transverse velocity, transverse magnetic field, absolute temperature, internal energy, and pressure, respectively. μ and λ are the viscosity of the flow, ν is the resistivity, and κ is the heat conductivity.

    In this paper, we consider the MHD flow of a perfect gas. Thus, P and e satisfy:

    P=Rθvande=Cvθ+Const. (1.6)

    Here R>0 denotes the specific gas constant, and Cv>0 stands for the heat capacity at constant volume. It is assumed that μ, λ, ν, and κ satisfy

    μ=˜μθα,λ=˜λθα,ν=˜νθα,andκ=˜κθβ, (1.7)

    which contain the positive constants ˜μ>0, ˜λ>0, ˜ν>0, ˜κ>0, α>0, and β>0.

    The systems (1.1)–(1.7) are supplemented with initial conditions

    (v,u,w,b,θ)(x,0)=(v0,u0,w0,b0,θ0), (1.8)

    and boundary ones

    (u,w,b,θx)|Ω=0. (1.9)

    Obviously, the initial data (1.8) should be compatible with the boundary conditions (1.9).

    When w=b=0, the Eqs (1.1)–(1.5) are converted into the compressible Navier–Stokes equation, which can be derived from Boltzmann's equation, assuming that the space and time scales are larger than all inherent scale–lengths, such as the Debye length or the gyro-radii of the charged particles [1,2,3,4,5]. Also, one can deduce from the Chapman–Enskog expansion for the first level of approximation in kinetic theory that the viscosity μ and λ may depend on the temperature or the density (see Chapman and Cowling [6]). Experimental results [7] show that the transport coefficients μ and κ vary according to gas temperature and density at very high temperatures and densities.

    The central point of magnetohydrodynamics theory is that conductive fluids can support magnetic fields. Li and Shang [8] proved the existence and uniqueness of the global-in-time classical solution to the initial-boundary value problem when the viscosity, resistivity, and heat conductivity depend on the specific volume v and the temperature θ. In that paper, the coefficients are assumed to be proportional to h(v)θα, where h(v) is a non-degenerate and smooth function satisfying some natural conditions, and the absolute value of the exponent α is sufficiently small. It's worth noting that Li and Shang considered the planar compressible magnetohydrodynamic system for the domain [0,1]×R2. Besides, Huang et al. [9] proved the large-time behavior of strong solutions to equations of compressible planar magnetohydrodynamic flow with the heat conductivity is the power function of temperature. Similar results can be observed in various other reports [10,11,12,13,14,15,16].

    Recently, Sun et al. [17] verified the existence and uniqueness of a global-in-time non-vacuum strong solution to a one-dimensional compressible Navier–Stokes system for a viscous and heat-conducting ideal polytropic gas. It was assumed that the viscosity μ and heat conductivity κ depend on temperature θ with μ(θ)=θα and κ(θ)=θβ for sufficiently small α>0 and arbitrary β0.

    Before presenting our main results, we need to provide some explanations of the symbols first. Throughout this paper, the positive general constant C will be different in different lines. For 1p, and integer k0, we adopt the simplified notations for the standard Sobolev space as follows:

    Lp=Lp(Ω),Wk,p=Wk,p(Ω),Hk=Wk,2(Ω).

    Without loss of generality, we assume that ˜λ=˜μ=˜ν=˜κ=R=cv=1, and

    10v0dx=1,10(θ0+u20+|w0|2+v0|b0|22)dx=1. (1.10)

    Inspired above, we have the following main results.

    Theorem 1.1. For given positive constants M0>0 and V0>0. Assume that

    (v0,u0,w0,b0,θ0)H2M0,infx[0,1]{v0,θ0}V0>0.

    Then there exist ϵ0>0, C0>0 and C1>0 which depend only on β, M0, V0, such that the initial boundary value problem (1.1)–(1.9) with 0αε0 (see (3.5)) admits a unique global-in-time strong solution (v,u,w,b,θ)(x,t) on [0,1]×[0,+) satisfying

    C10v(x,t)C0,C1θ(x,t)C11, (1.11)

    and

    {(v,u,w,b,θ)C([0,+);H2),vxL2(0,+;H1),(ux,wx,bx,θx)L2(0,+;H2).

    Furthermore, for any t0, it holds that

    (v1,u,w,b,θ1)H1Ceη0t,

    where C,η0>0 are some positive constants.

    Remark 1.1. From the view of physics, the resistance is a function of temperature (e.g., [18]). This implies that our results is physical. Dou et al.'s 2021 study[19], published in Scientific Reports, delves into a variety of issues, including Enhanced Oil Recovery, where the technology can potentially improve the extraction of residual oil from oil fields. The equations presented in our study, especially those related to magnetic force distribution and the relationship between magnetic force and displacement, are pivotal for understanding and optimizing these applications. They assist in predicting the behavior of magnetic foams under various conditions, which is crucial for designing effective systems in the aforementioned fields.

    In this paper, we try to use the framework of Li and Liang [20] to prove the global wellposedness of the solution. It should be emphasized that the key step is to derive the time–independent positive lower and upper bounds of specific volume and temperature. The foremost obstacles lie in the strong non-linearities caused by the temperature-dependent viscosity, resistivity, and heat-conductivity from Eq (1.7). Fortunately, these nonlinear terms are involved with μx, λx, νx, μt, λt, or νt which can be controlled by the smallness of α. With the help of upper and lower bounds of the specific volume, we can then estimate the higher-order derivatives of the solutions, and the upper and lower bounds of the temperature.

    The rest of this paper is organized as follows. Section 2 is devoted to a discussion of a number of a priori estimates independent of time, which are required to extend the local solution to the time global. Based on the estimates given in Section 2, the main results of Theorem 1.1 are established in Section 3.

    For constants N,m1,m2, and T, define

    X(0,T;m1,m2,N):={(v,u,θ,w,b):(v1,u,w,b,θ1)C([0,T];H2),vxL2(0,T;H1),(ux,θx,wx,bx)L2(0,T;H2),vtC([0,T];H1),(ut,θt,wt,bt)L2(0,T;H1),v(x,t)m1,θ(x,t)m2,E(0,T)N2,(x,t)[0,1]×[0,T]},

    where

    E(0,T):=sup0tT(u,w,b,vx,θx)2H1+T0θt2L2dt.

    The main purpose of this section is to derive certain t-dependent a priori estimates for the variables (v,u,θ,w,b) in the function space X(0,T;m1,m2,N), relevant to the initial boundary value problem (1.1)–(1.9) for T>0 and 0<mi1(i=1,2), 2N<+. It follows from Sobolev's inequality that

    m1v(x,t)2N,m2θ(x,t)2N,for(x,t)[0,1]×[0,T].

    Firstly, let us derive the time-independent lower and upper bounds of v.

    Lemma 2.1. Assume that the conditions listed in Theorem 1.1 hold; then there exists a constant 0<ϵ11 depending only on β, M0, and V0, such that if

    mα22,Nα2,αH(m1,m2,N)ϵ1, (2.1)

    where

    H(m1,m2,N):=(m11+m12+N+1)8.

    Then for (x,t)[0,1]×[0,T],

    C10v(x,t)C0. (2.2)

    Here (and in what follow), C0,Ci(i=1,2,,10), and C denote some generic positive constants depending only on β, (v0,u0,w0,b0,θ0)H2, infx(0,1)v0(x), and infx(0,1)θ0(x).

    Proof. The proof is divided into four steps.

    Step 1. (Basic energy estimate)

    According to (1.1), (1.5), (1.9), and (1.10), for t>0, one has

    10v(x,t)dx=1,10(θ+u2+|w|2+v|b|22)(x,t)dx=1. (2.3)

    In light of (2.1), it is deduced that

    θα+θαL([0,1]×[0,T])mα2+(2N)α2+4=6. (2.4)

    Simplifying (1.5) gives

    θt+θvux=(θβθxv)x+θα(u2x+|wx|2+|bx|2)v. (2.5)

    Then, multiplying (1.1)–(1.4), and (2.5) by (1v1), u, w, b, and (1θ1), respectively, and integrating them over [0,1]×[0,T], together with (2.4) gives

    sup0tT10(u2+|w|2+v|b|22+(vlnv)+(θlnθ))dx+T0W(s)dsE0, (2.6)

    where

    W(t)=10(θαθ2xvθ2+θα(u2x+|wx|2+|bx|2)vθ)(x,t)dx, (2.7)

    and E0 is the initial total entropy defined by

    E0=10(u20+|w0|2+v|b0|22+(v0lnv0)+(θ0lnθ0))dx.

    Step 2. (Representation formula for v)

    First, (1.2) can be written as

    ut+(P+12|b|2)x=μ(lnv)xt+μxuxv,

    that is,

    (uμ)t+g+(μ1(P+12|b|2))x=(lnv)xt, (2.8)

    where

    g=(μ1)tu(μ1)x(P+12|b|2)μxuxμv.

    Integrating (2.8) over [0,t]×[x1(t),x], it follows

    xx1(t)(uμu0μ0)dξ+10xx1(t)gdξds+t0(P+12|b|2μP+12|b|2μ(x1))ds=lnv(x,t)lnv(x1(t),t)[lnv0(x)lnv(x1(t),0)],

    where x1(t)[0,1] is determined by the following steps and μ0=μ(θ0). Moreover, for ease of notation, define

    F=uxvμ1(P+12|b|2)x0g(ζ)dζ,φ=t0F(x,s)ds+x0u0μ0dζ.

    Based on the above definitions that

    φx=uμ,φt=F. (2.9)

    It is easy to show that

    t0[μ1(P+12|b|2)(x1(t),s)+x1(t)0g(ξ,s)dξ]ds=t0(uxμF)(x1(t),s)ds=t0[(lnv)tF](x1(t),s)ds=lnv(x1(t),t)lnv(x1(t),0)t0F(x1(t),s)ds. (2.10)

    With the help of (1.1) and (2.9) that

    (vφ)t(uφ)x=vφtuφx=vFu2μ=uxvμ(P+12|b|2)vx0g(ξ)dξu2μ. (2.11)

    Integrating (2.11) over [0,t]×[0,1], it follows

    10vφdx=10v0x0u0μ0dξdxt010[vμ(P+12|b|2)+vx0g(ξ)dξ+u2μ]dxds. (2.12)

    Hence, according to the mean value theorem, there exists x1(t)[0,1] such that φ(x1(t),t)=10vφdx. By the definition of φ, (2.9) and (2.12), one obtains

    t0F(x1(t),s)ds=φ(x1(t),t)x1(t)0u0μ0dξ=10v0x0u0μ0dξdxt010[vμ(P+12|b|2)+vx0g(ξ)dξ+u2μ]dxdsx1(t)0u0μ0dξ. (2.13)

    Substituting (2.13) into (2.10) gives

    t0[μ1(P+12|b|2)(x1(t),s)+x1(t)0g(ξ,s)dξ]ds=lnv(x1(t),t)lnv(x1(t),0)10v0x0u0μ0dξdx+x1(t)0u0μ0dξ+t010[vμ(P+12|b|2)+vx0g(ξ)dξ+u2μ]dxds. (2.14)

    Moreover, substituting (2.14) into (2.9),

    t0μ1(P+12|b|2)ds+t0x0gdξdst010[vμ(P+12|b|2)+vx0g(ξ)dξ+u2μ]dxds+xx1(t)(uμu0μ0)dξ+10v0x0u0μ0dξdxx1(t)0u0μ0dξ=lnv(x,t)lnv(x,0). (2.15)

    It follows from (2.15) that

    v=B1AD, (2.16)

    where

    A:=exp{t0[μ1(P+12|b|2)+x0gdζ]ds};B:=exp{t010[vμ(P+12|b|2)+vx0g(ζ)dζ+u2μ]dxds};D:=v0exp{xx1(t)(uμu0μ0)dζ+10v0x0u0μ0dζdxx1(t)0u0μ0dζ}.

    From (2.16), one has

    vD1B=A. (2.17)

    Furthermore, we define

    J:=1μ(P+12|b|2)+x0dζ.

    Then, multiplying (2.17) by J, we have

    vD1BJ=tA.

    Since A(0)=1, integrating the above equality over (0,t) with respect to time, one obtains

    v=DB1+t0B(s)B(t)D(t)D(s)v[1μ(P+12|b|2)+x0gdξ]ds. (2.18)

    Step 3. (Lower bound for v)

    Applying Jensen's inequality to the convex function θlnθ leads to

    10θdxln10θdx10(θlnθ)dx,

    which, together with (2.3) and (2.6), leads to

    ˉθ(t)=10θ(x,t)dx[α1,1], (2.19)

    where 0<α1<α2 are two roots of

    xlnx=e0.

    The expression of D and

    xx1(t)uμdξ10θαudxm2α2+10u2dxC,

    imply that

    C1DC. (2.20)

    Next, we will estimate B. It follows from (2.3) and (2.4) that

    10(vμ(P+12|b|2)+u2μ)dx=10θα(θ+12v|b|2+u2)dxmα210(θ+12v|b|2+u2)dx4, (2.21)

    and

    10(vμ(P+12|b|2)+u2μ)dx10θα(θ+12v|b|2+u2)dx(2N)α10θdxα14. (2.22)

    On the other hand, by the expression of g, there exists a sufficiently small ϵ>0 such that

    10x0g(ξ)dξ=10vx0αθα1(θtu+θx(P+12|b|2)θαθxuxv)dξdxαmα1210v10(θtu+m11θθx+12θx|b|2mα2m11θxux)dξdxαmα12(θt2L2+u2L2+m11θxL2θL2+θxL2uxL2)ϵt. (2.23)

    Putting (2.21)–(2.23) into the expression of B, we will find that there exist C2,C3, such that

    eC2tB(t)eC3t.

    That means

    eC2(ts)B(s)B(t)eC3(ts).

    Thus, for 0t<t0, one has

    vDB1Cεt0eC2(ts)dsCeCt0Cε(1eC2t0).

    For large enough t>t0, it follows

    infxΩv(x,t)Ct0B(s)B(t)θ1αdsCε(1eC3t). (2.24)

    Therefore, one needs the estimates of θ and B(s)B(t). By the mean value theorem and (2.3), there exists x2(t)[0,1], such that C1θ(x2(t),t)C. Based on Cauchy-Schwarz's inequality, one has

    |[ln(θ+1)]β2+1[ln(θ(x2(t),t)+1)]β2+1|=|xx2(ln(θ+1))βθxv(θ+1)v(ζ)dζ|C(10(ln(θ+1))βθ2xvθ2dx)12(10vdx)12C(10(ln(θ+1))βθ2xvθ2dx)12,

    which implies

    θCCW(t).

    From (2.18), (2.20), (2.21), and (2.23), one has

    t0B(s)B(t)θ1αdst0B(s)B(t)(110θβθ2xvθ2dx)dsCeCtC{t20B(s)B(t)10θβθ2xvθ2dxds+tt2B(s)B(t)10θβθ2xvθ2dxds}CeCtCeC2tCtt210θβθ2xvθ2dxdsC. (2.25)

    For the large enough time T0, when t>T0, plugging (2.25) into (2.24) gives

    infxΩv(x,t)C.

    Step 4. (Upper bound for v)

    According to Holder's inequality, for 0<β1, one has

    |θ12(x,t)θ12(x2(t),t)|10θ12θxdxv12(10θβθ2xvθ2dx)12(10θ1βdx)12v12(10θβθ2xvθ2dx)12. (2.26)

    That means

    θ(x,t)C+v10θβθ2xvθ2dx. (2.27)

    For 1<β<, one has

    |θβ2(x,t)θβ2(x2(t),t)|10θβ21θxθdx(10θβθ2xvθ2dx)12(10θ1βdx)12(10θβθ2xvθ2dx)12,

    which means

    θ(x,t)C+10θβθ2xvθ2dx. (2.28)

    Then the standard calculations give

    maxx[0,1]|b|2(x,t)C10|bbx|dxC10θα|bx|2vθdx+C10vθ1α|b|2dxC10θα|bx|2vθdx+C. (2.29)

    It follows from the expression of v and (2.26)–(2.29) that

    vCeCt+Ct0eC(ts)((1+v)10θβθ2xvθ2dx+10θα|bx|2vθdx)ds.

    By using Grönwall's inequality, one has

    vC.

    Up to now, the proof of Lemma 2.1 has been finished.

    Lemma 2.2. Assume that the conditions listed in Lemma 2.1 hold; then for any p>0, there exists some positive constant C(p) such that

    sup0tT10θ1pdx+T010(θβθ2xθp+1+u2x+|wx|2+|bx|2θp)dxdtC(p). (2.30)

    Proof. From (2.6), we see that (2.30) holds for p=1. Then we assume p>0 and p1. Multiplying (2.5) by θp and integrating by parts, one can arrive at

    1p1(10θ1pdx)t+p10θβθ2xvθp+1dx+10θα(u2x+|wx|2+|bx|2)vθpdx=10(θ1p1)uxvdx+10uxvdxC(p)10|θ121|(θ12p+1)|ux|dx+10vtvdxC(p)maxx[0,1]|θ121|10(θ12p+1)|ux|dx+10(lnv)tdxC(p)maxx[0,1]|θ121|(10(θ12p)|ux|dx+10|ux|dx)+(10lnvdx)tC(p)maxx[0,1]|θ121|(10vθ1pθαdx)12(10θαu2xvθpdx)12+C(p)maxx[0,1]|θ121|(10u2xvθdx)12(10vθdx)12+(10lnvdx)t1210θαu2xvθpdx+C(p)maxx[0,1]|θ121|(10vθ1pθαdx+1)+(10lnvdx)t. (2.31)

    Moreover, it follows from (2.3) and (2.19) that

    α110θdx10(θ+ηθα(u2+|w|2+v|b|2)2)dx1.

    For any real number q, it follows from (2.2) and (2.4) that

    |1ˉθq|=|10ddη(10(θ+ηθα(u2+|w|2+v|b|2)2)dx)qdη|=|10q(10θ+ηθα(u2+|w|2+v|b|2)2dx)q1dη10θα(u2+|w|2+v|b|2)2dx|C(q)maxx[0,1](|u|+|w|+|b|)(10(u2+|w|2+v|b|2)dx)12C(q)10(|ux|+|wx|+|bx|)dxC(q)(10θα(u2x+|wx|2+|bx|2)vθdx)12(10vθ1αdx)C(q)W12(t). (2.32)

    After that, for β(0,1), it follows from (2.27), (2.28), and (2.19) that

    maxx[0,1]|θ121|maxx[0,1]|θ12ˉθ12|+maxx[0,1]|ˉθ121|C10θ12|θx|dx+CW12(t)C(10θβθ2xvθ2dx)12(10vθ1βdx)12+CW12(t)CW12(t), (2.33)

    when β1, one has

    maxx[0,1]|θ121|maxx[0,1]|θ12ˉθ12|+maxx[0,1]|ˉθ121|maxx[0,1]|θβ2ˉθβ2|+CW12(t)C10θβ21|θx|dx+CW12(t)C(10θβθ2xvθ2dx)12(10vθ1βdx)12+CW12(t)CW12(t). (2.34)

    Therefore, for β>0, it follows from (2.33), (2.34), and (2.6) that

    T0maxx[0,1](θ121)2dtC. (2.35)

    Finally, we see that for p[0,1], one has

    10θ1pdx10θdx+1C.

    And for β0, it follows from (2.33), (2.34), and (2.6) that

    sup0t<10|lnv|dxC. (2.36)

    As a result, according to (2.33), (2.34), (2.7), (2.35), and Grönwall's inequality, we derive (2.30) from (2.31), which finishes the proof of Lemma 2.2.

    Lemma 2.3. Assume that the conditions listed in Lemma 2.1 hold; then for all T>0,

    sup0tT10v2xdx+T010(v2x(1+θ)+u2x+|wx|2+|bx|2)dxdtC4. (2.37)

    Proof. First, integrating (2.5) over [0,1]×[0,t], by (2.36), one has

    T010θα(u2x+|wx|2+|bx|2)vdxdt=10θdx10θ0dx+T010θ1vuxdxdt+10lnvdx10lnv0dx12T010θαu2xvdxdt+CT010(θ1)2vθαdxdt+C12T010θαu2xvdxdt+CT0maxx[0,1]|θ121|210(θ12+1)2dxdt+C12T010θαu2xvdxdt+CT0maxx[0,1]|θ121|2dt+C12T010θαu2xvdxdt+CT0W(t)dt+CC,

    thus it follows from (2.2) and (2.4) that

    T010(u2x+|wx|2+|bx|2)dxdtC. (2.38)

    Next, since

    (θαvxv)t=θα(vxv)t+αθα1θtvxv=θα(vtv)x+αθα1θtvxv=(θαvtv)xαθα1θxvtv+αθα1θtvxv=(θαvtv)x+αθα1v(vxθtθxvt),

    the momentum equation (1.2) can be rewritten as

    (uθαvxv)t=(θv+12|b|2)xαθα1v(vxθtθxvt). (2.39)

    Multiplying (2.39) by (uθαvxv) and integrating it over [0,1]×[0,t] yields that for any t[0,T],

    1210(uθαvxv)2dx1210(uθαvxv)2(x,0)dx=T010(θvxv2θxvbbx)(uθαvxv)dxdtT010αθα1v(vxθtθxvt)(uθαvxv)dxdt=T010(θα+1v2xv3)dxdt+T010θuvxv2dxdtT010θxv(uθαvxv)dxdtT010bbx(uθαvxv)dxdtT010αθα1v(vxθtθxvt)(uθαvxv)dxdt=T010(θα+1v2xv3)dxdt+4i=1Ii. (2.40)

    Each Ii(i=1,2,3,4) can be estimated as follows. First, based on Cauchy's inequality, we have

    |I1|=|T010θuvxv2dxdt|18T010θα+1v2xv3dxdt+CT010θ1αu2vdxdt18T010θα+1v2xv3dxdt+C, (2.41)

    where it has been used

    T010θ1αu2vdxdtCT010θu2dxdtCT0maxx[0,1]|u|210θdxdtCT0maxx[0,1]|u|2dtCT010u2xdxdtC.

    Next, by using (2.4), (2.6), and (2.30) with p=β, it follows

    |I2|=|T010θxv(uθαvxv)dxdt|18T010θα+1v2xv3dxdt+CT010θ1αu2vdxdt+CT010θαθ2xvθdxdt18T010θα+1v2xv3dxdt+C. (2.42)

    Combining (2.42) with Cauchy's inequality leads to

    |I3|=|T010bbx(uθαvxv)dxdt|=T010(|bx|2+|b|2(uθαvxv)2)dxdtCT0W(t)10(uθαvxv)2dxdt+C. (2.43)

    Rewriting (2.5) as

    θt=θβθxvxv2+βθβ1θ2xv+θβθxxv+θα(u2x+|wx|2+|bx|2)vθvux.

    We set

    Y=(uvθαvx)(vxθtθxvt)=θαv2xθtuvvxθtθαθxuxvx+uvθxux=uvθxux+(θαθt+uθβθxv)v2x(βθβ1uθ2x+θβuθxx+θαu(u2x+|wx|+|bx|2)θuux+θαθxux)vx=uvθxux+R1v2x+R2vx, (2.44)

    where

    R1:=θαθt+uθβθxv,

    and

    R2:=(βθβ1uθ2x+θβuθxx+θαu(u2x+|wx|+|bx|2)θuux+θαθxux).

    Then from (2.44), one has

    |I4|=|T010αθα1v2(uvθxux+R1v2x+R2vx)dxdt||T010αθα1uθxuxvdxdt|+|T010αθα1v2R1v2xdxdt|+|T010αθα1v2R2vxdxdt|:=3i=1Ji. (2.45)

    Each Ji(i=1,2,3) can be estimated as follows. First, by means of Cauchy's inequality

    J1T0αθ1+αβ2(10θβθ2xvθ2dx+10θαu2xvθdx)dtC. (2.46)

    According to the definition of R1, one has

    J2=|T010(αθα1v2xv2uθβθxv+αθα1v2xv2θαθt)dxdt|=|T010αθα+β1uv2xθxv3dxdt|+|T010αθ2α1v2xθtv2dxdt|T0αθα+β12vxuv(10θα+1v2xv3dx+10θβθ2xvθ2dx)dt+T0αθ32(α1)vxv(10θα+1v2xv3dx+10θ2tdx)dt18T010θα+1v2xv3dxdt+CT010θβθ2xvθ2dxdt+CαT010θ2tdxdt18T010θα+1v2xv3dxdt+C. (2.47)

    It follows from the definition of R2 that

    R22C(θ2αθ2xu2x+u2β2θ2β2θ4x+θ2βu2θ2xx+u2θ2αu4x+u2θ2α|wx|4+u2θ2α|bx|4+u2xθ2α+2),

    then, one has

    J3=|T010αθα1v2R2vxdxdt|18T010θα+1v2xv3dxdt+CT010α2θα3R22vdxdt18T010θα+1v2xv3dxdt+C. (2.48)

    Inserting (2.46)–(2.48) into (2.45) gives

    |I4|14T010θα+1v2xv3dxdt+C. (2.49)

    Putting (2.41)–(2.43), and (2.49) into (2.40), combining Grönwall's inequality gives

    10(uvxv)2dx+T010θα+1v2xv3dxdtC.

    Note that

    10(uvxv)2dx=10(u22uvxv+v2xv2)2dx=10u2dx+10v2xv2dx210uvxvdx,

    that means

    10u2dx+10v2xv2dx+T010θα+1v2xv3dxdtC+210uvxvdx1210v2xv2dx+C10u2dx+CC.

    On the other hand, it follows from (2.33) and (2.34) that

    10v2xdx=10v2x(1θ)dx+10θv2xdxCmaxx[0,1]|θ121|210v2xdx+10θv2xdxCW(t)10v2xdx+10θv2xdx.

    Together with (2.38), the proof of Lemma 2.3 has been completed.

    Lemma 2.4. Assume that the conditions listed in Lemma 2.1 hold; then for all T>0, one has

    sup0tT10(|bx|2+|wx|2)dx+T010(|bt|2+|wt|2+|bxx|2+|wxx|2)dxdtC5. (2.50)

    Proof. First, rewrite (1.3) as

    wt=θαwxxvθαwxvxv2+αθα1θxwxv+bx. (2.51)

    Multiplying (2.51) by wxx and integrating over [0,1]×[0,T], one obtains:

    1210|wx|2dx+T010θα|wxx|2vdxdt=T010θαvxwxwxxv2dxdt+T010αθα1wxwxxvdxdt+T010bxwxxdxdt=3i=1Ii. (2.52)

    Each Ii(i=1,2,3) is estimated as follows. From Cauchy's inequality and (2.4), it shows

    I1=T010θαvxwxwxxv2dxdt18T010θα|wxx|2vdxdt+CT010θα|wx|2v2xv3dxdt18T010θα|wxx|2vdxdt+CT0maxx[0,1]|wx|210θαv2xv3dxdt18T010θα|wxx|2vdxdt+CT0maxx[0,1]|wx|2dt14T010θα|wxx|2vdxdt+C, (2.53)

    where it has been used

    T0maxx[0,1]|wx|2dtT010|wx||wxx|dxdt+CT010|wx|2dxdt18T010θα|wxx|2vdxdt+CT010|wx|2dxdt18T010θα|wxx|2vdxdt+C. (2.54)

    Next, from the a priori assumption, one has

    I2=T010αθα1θxwxwxxvdxdt18T010θα|wxx|2vdxdt+CT010α2θα2θ2x|wx|2vdxdt18T010θα|wxx|2vdxdt+CT0maxx[0,1]|wx|2|α|210θα2θ2xvdxdt18T010θα|wxx|2vdxdt+CT0maxx[0,1]|wx|2|α|2mα22θ2x2L2dt18T010θα|wxx|2vdxdt+CT0maxx[0,1]|wx|2|α|2H(m1,m2,N)dt18T010θα|wxx|2vdxdt+CT0maxx[0,1]|wx|2dt14T010θα|wxx|2vdxdt+C. (2.55)

    Furthermore, according to (2.37), one has

    I3=T010bxwxxdxdt18T010θα|wxx|2vdxdt+CT010|bx|2θαvdxdt18T010θα|wxx|2vdxdt+C. (2.56)

    Substituting (2.53), (2.55), and (2.56) into (2.52), one has

    10|wx|2dx+T010|wxx|2dxdtC. (2.57)

    Combining (2.51) with (2.54) gives

    T010|wt|2dxdtT010(θ2αw2xxv2+θ2α|wx|2v2xv4+α2θ2α2θ2x|wx|2v2+|bx|2)dxdtCT0maxx[0,1]|wx|2dt+CC. (2.58)

    Next, rewrite (1.4) as

    bt=θαbxxv2θαbxvxv3+αθα1θxbxv2uxbv+wxv. (2.59)

    Multiplying (2.59) by bxx and integrating the result over [0,1]×[0,T] yields

    1210|bx|2dx+T010θα|bxx|2v2dxdt=T010αθα1θxbxbxxv2dxdt+T010θαvxbxbxxv3dxdt+T010uxbbxxvdxdtT010wxbxxvdxdt:=4i=1Ji. (2.60)

    Each Ji(i=1,2,3) is estimated as follows. From Cauchy's inequality and (2.4), one has

    J1=T010αθα1θxbxbxxv2dxdt18T010θα|bxx|2v2dxdt+CT010α2θα2θ2x|bx|2v2dxdt18T010θα|bxx|2v2dxdt+CT0maxx[0,1]|bx|2|α|210θα2θ2xv2dxdt18T010θα|bxx|2v2dxdt+CT0maxx[0,1]|bx|2|α|2mα22θ2x2L2dt18T010θα|wxx|2v2dxdt+CT0maxx[0,1]|bx|2|α|2H(m1,m2,N)dt18T010θα|bxx|2v2dxdt+CT0maxx[0,1]|bx|2dt18T010θα|bxx|2v2dxdt+C, (2.61)

    where it has been used

    T0maxx[0,1]|bx|2dtT010|bx||bxx|dxdt+CT010|bx|2dxdt18T010θα|bxx|2v2dxdt+CT010|bx|2dxdt18T010θα|bxx|2v2dxdt+C. (2.62)

    It follows from (2.2), (2.4), (2.37), and (2.62) that

    J2=T010θαvxbxbxxv3dxdt18T010θα|bxx|2v2dxdt+CT0maxx[0,1]|bx|210θαv2xv4dxdt18T010θα|bxx|2v2dxdt+CT0maxx[0,1]|bx|2dt14T010θα|bxx|2v2dxdt+C. (2.63)

    From (2.2)–(2.4), one has

    J3=T010uxbbxxvdxdt18T010θα|bxx|2v2dxdt+max(x,t)[0,1]×[0,T]|b|2CT010u2xdxdt18T010θα|bxx|2v2dxdt+max(x,t)[0,1]×[0,T]|b|218T010θα|bxx|2v2dxdt+Csup0<t<T10|b||bx|dx18T010θα|bxx|2v2dxdt+18sup0<t<T|bx|2+C. (2.64)

    According to (2.37), it follows

    J4=T010wxbxxvdxdt18T010θα|bxx|2v2dxdt+CT010|wx|2dxdt18T010θα|bxx|2v2dxdt+C. (2.65)

    Inserting (2.61), (2.63)–(2.65) into (2.60), which implies

    sup0<t<T10|bx|2dx+T010|bxx|2dxdtC. (2.66)

    From (2.59), one obtains

    T010|bt|2dxdtCT010(θ2α|bxx|2v4+α2θ2α2θ2x|bx|2v4+θ2α|bx|2v2xv6+u2x|b|2v2+|wx|2v2)dxdtC. (2.67)

    Therefore, it follows from (2.57), (2.58), (2.66), and (2.67) that (2.50) is correct. Then the Lemma 2.4 has been proved.

    Lemma 2.5. Assume that the conditions listed in Lemma 2.1 hold; then for all T>0,

    T010θ2xdxdtC6.

    Proof. For the case of β>1, setting p=β1 in (2.30) will give

    T010θ2xdxdtC. (2.68)

    For 0<β1, multiplying (2.5) by θ1β2 and integrating by parts, it gives

    24β(10θ2β2dx)t+2β210θβ2θ2xvdx=10θ2β2uxvdx+10θα+1β2(u2x+|wx|2+|bx|2)vdx=10(ˉθ2β2θ2β2)uxvdx+10(1ˉθ2β2)uxvdx10uxvdx+10θα+1β2(u2x+|wx|2+|bx|2)vdx:=4i=1Ii. (2.69)

    Each Ii(i=1,2,3,4) can be estimated as follows. First, by (2.7), one has

    I1=10(ˉθ2β2θ2β2)uxvdx=10|ˉθ1β4θ1β4|(ˉθ1β4+θ1β4)|ux|dxCmaxx[0,1]|ˉθ1β4θ1β4|(10(θ2β2+1)dx)12(10u2xdx)1218maxx[0,1]|ˉθ1β4θ1β4|+C10(θ2β2+1)dx10u2xdx18(10θβ4|θx|dx)2+C10(θ2β2+1)dx10u2xdx1810θβ2θ2xvdx+C10θβθ2xvθ2dx+C10(θ2β2+1)dx10u2xdx1810θβ2θ2xvdx+CW(t)+C10θ2β2dx10u2xdx+C10u2xdx. (2.70)

    According to (2.32), one obtains

    I2=10(1ˉθ2β2)uxvdxmaxx[0,1]|1ˉθ2β2|10uxdxmaxx[0,1]|1ˉθ2β2|2+C10u2xdxCW(t)+C10u2xdx. (2.71)

    It follows

    I3=10uxvdxC10u2xdxC. (2.72)

    Next, it follows from (2.4) that

    I4=10θα+1β2(u2x+|wx|2+|bx|2)vdxCmaxx[0,1](|θ1β2ˉθ1β2|+1)10(u2x+|wx|2+|bx|2)dx1810θβ2|θx|dx10(u2x+|wx|2+|bx|2)dx+C10(u2x+|wx|2+|bx|2)dx1810θβ2θ2xvdx+C10θβθ2xvθ2dx+C10(u2x+|wx|2+|bx|2)dx+C(10u2xdx)2+C(10|wx|2dx)2+C(10|bx|2dx)21810θβ2θ2xvdx+CW(t)+C10(u2x+|wx|2+|bx|2)dx+C(10u2xdx)2+C(10|wx|2dx)2+C(10|bx|2dx)2. (2.73)

    Substituting (2.70)–(2.73) into (2.69), integrating on [0,T], and combining (2.37) and Grönwall's inequality, one has when 0<β1,

    T010θ2xdxdtC. (2.74)

    Then from (2.68) and (2.74) the proof of Lemma 2.5 has ended.

    Lemma 2.6. Assume that the conditions listed in Lemma 2.1 hold; then for all T>0, one has

    sup0tT10u2xdx+T010u2xxdxdtC7. (2.75)

    Proof. Rewrite (1.2) as

    ut=θαuxxvθαuxvxv2+αθα1θxuxvθxv+θvxv2bbx. (2.76)

    Multiplying (2.76) by uxx and integrating the result over [0,1]×[0,T], it shows

    1210u2xdx+T010θαu2xxvdxdtT010θαvxuxuxxv2dxdtT010αθα1θxuxuxxvdxdt+T010θxuxxvdxdtT010θvxuxxv2dxdt+T010bbxuxxdxdt=5i=1Ji. (2.77)

    Each Ji can be estimated as follows. First, according to (2.37), one has

    J1=T010θαvxuxuxxv2dxdt18T010θαu2xxvdxdt+CT0maxx[0,1]|ux|210v2xdxdt18T010θαu2xxvdxdt+CT0maxx[0,1]|ux|2dt14T010θαu2xxvdxdt+C, (2.78)

    where it has been used

    T0maxx[0,1]|ux|2dtT010|ux||uxx|dxdt+CT010u2xdxdt18T010θαu2xxvdxdt+CT010u2xdxdt18T010θαu2xxvdxdt+C. (2.79)

    Secondly, combining (2.30) with p=β+1α, one obtains

    J2=T010αθα1θxuxuxxvdxdt18T010θαu2xxvdxdt+CT0maxx[0,1]|ux|2|α|2mα2210θ2xdxdt18T010θαu2xxvdxdt+CT0maxx[0,1]|ux|2dt1410θαu2xxvdx+C. (2.80)

    Next, from (2.2) and (2.4), one has

    J3=T010θxuxxvdxdt18T010θαu2xxvdxdt+CT010θ2xdxdt18T010θαu2xxvdxdt+C. (2.81)

    Furthermore, from (2.37), one has

    J4=T010θvxuxxv2dxdt18T010θαu2xxvdxdt+CT010θ2v2xdx18T010θαu2xxvdxdt+CT0(maxx[0,1](θˉθ)2+1)10v2xdxdt18T010θαu2xxvdxdt+CT0maxx[0,1](θˉθ)2dt+CT010v2xdxdt18T010θαu2xxvdx+CT010θ2xdxdt18T010θαu2xxvdx+C, (2.82)

    where we have used

    T0maxx[0,1]|θˉθ|2dtC.

    In fact, for 0<β<2,

    T0maxx[0,1]|θˉθ|2dtCT010|θx|2dxdtC,

    and for β2,

    T0maxx[0,1]|θˉθ|2dtCT0maxx[0,1]|θβ2ˉθβ2|2dtCT010θβ2|θx|2dxdtC(T010θβθ2xvθ2dxdt)12(T010vdxdt)12C.

    Finally, it follows from (2.62) that

    J5=T010bbxuxxdxdt18T010θαu2xxvdxdt+CT0maxx[0,1]|bx|210v|b|2dxdt18T010θαu2xxvdxdt+CT0maxx[0,1]|bx|2dt1810θαu2xxvdx+C. (2.83)

    Substituting (2.78) and (2.80)–(2.83) into (2.77) gives

    10u2xdx+T010u2xxdxdtC. (2.84)

    On the other hand, from (1.2), one has

    |ut|2C(u2xx+v2xu2x+α2θ2α2θ2xu2x+θ2x+θ2v2x+|b|2|bx|2),

    and

    T010u2tdxdtC.

    Combining this with (2.84), the proof of Lemma 2.6 has been finished.

    Lemma 2.7. Assume that the conditions listed in Lemma 2.1 hold; then for all T>0,

    C1θ(x,t)C11, (2.85)
    sup0tT10θ2xdx+T010(θ2t+θ2xx)dxdtC8. (2.86)

    Proof. Multiplying (2.5) by θ gives

    12ddt10θ2dx+10θβθ2xvdx=10(1θ2)uxvdx(10lnvdx)t+10θα+1(u2x+|wx|2+|bx|2)vdxCmaxx[0,1](|1θ|2+u2x+|wx|2+|bx|2)(10lnvdx)t.

    It follows from (2.32), (2.74), and (2.6) that

    T0maxx[0,1](θ1)2dtCT0maxx[0,1](θˉθ)2dt+CT0maxx[0,1](ˉθ1)2dtCT010θ|θx|dxdt+CT0V(t)dtT0θ2xdxdt+CT010θ2dxdt+CT0V(t)dtCT010θ2dxdt+C.

    By combining this with (2.3), (2.37), and Grönwall's inequality, one has

    T010θβθ2xdxdtC. (2.87)

    Next, multiplying (2.5) by θβθt and integrating it over (0,1), by (2.4), one has

    12ddt10(θβθx)2vdx+10θβθ2tdx=1210(θβθx)2v2uxdx10θβ+1θtuxvdx+10θα+βθt(u2x+|wx|2+|bx|2)vdx1210θβθ2tdx+Cmaxx[0,1]|ux|θβ210θ3β2θ2xv2dx+C10θβ+2u2xdx+10θβ(u4x+|wx|4+|bx|4)dx=1210θβθ2tdx+3i=1Ii. (2.88)

    Moreover, each Ii(i=1,2,3) can be estimated as follows. First,

    I1=Cmaxx[0,1]|ux|θβ210θ3β2θ2xv2dxCmaxx[0,1]u2xθβ+C(10θ32βθ2xdx)2Cmaxx[0,1]u2xmaxx[0,1](1+θ2β+2)+C10θβθ2xdx10θ2βθ2xdx. (2.89)

    Second,

    I2C10(1+θ2β+2)u2xdxCmaxx[0,1]u2xmaxx[0,1](1+θ2β+2). (2.90)

    Finally,

    I3Cmaxx[0,1](u4x+|wx|4+|bx|4)maxx[0,1](1+θ2β+2).

    Substituting (2.88)–(2.90) into (2.87) gives

    12ddt10(θβθx)2vdx+10θβθ2tdxCmaxx[0,1](u2x+u4x+|wx|4+|bx|4)maxx[0,1](1+θ2β+2)+C10θβθ2xdx10θ2βθ2xdx.

    Direct calculations yield

    maxx[0,1]|θβ+1ˉθβ+1|C+C10θ2βθ2xdx, (2.91)

    and

    maxx[0,1](1+θ2β+2)maxx[0,1](1+θβ+1)2C(10θβ|θx|dx)2C10θ2βθ2xdx.

    From this and (2.91), and integrating over [0,T], together with Grönwall's inequality, one has

    sup0tT10θ2βθ2xdx+T010θβθ2tdxdtC. (2.92)

    Combining with (2.91), one has

    max(x,t)[0,1]×[0,T]θ(x,t)C. (2.93)

    On the one hand, (2.93) gives

    T010(θβ+1ˉθβ+1)2dxdtCT010θ2βθ2xdxdtC. (2.94)

    Together with (2.4), (2.37), (2.91), and (2.92), one has

    T0|ddt10(θβ+1ˉθβ+1)2dx|dtCT010(θβ+1ˉθβ+1)2dxdt+CT010(θ2βθ2t+ˉθ2t)dxdtCT010u2xdxdt+CC. (2.95)

    Combining (2.37), (2.93), and (2.94) leads to

    limt+10(θβ+1ˉθβ+1)2dx=0.

    Then combining (2.91) gives

    maxx[0,1](θβ+1ˉθβ+1)410(θβ+1ˉθβ+1)2dx10θ2βθ2xdx0,ast+. (2.96)

    Therefore, it follows from (2.19) and (2.95) that there exists some T0, such that

    θ(x,t)γ12,

    for all (x,t)[0,1]×[T0,+).

    On the other hand, for p>2, multiplying (2.5) by 1θp, one has

    1p1(10(1θ)p1dx)t+10μu2xvθpdx10uxvθp1dx1210μu2xvθpdx+12101μvθp2dx.

    That means

    1θp2Lp1ddt1θLp1C1θp2Lp2C1θp2Lp1, (2.97)

    where the positive constant C independent of p and T. (2.97) gives

    sup0<t<Tθ1Lp1C(T+1).

    Letting p+, there exists a positive constant C1γ12 such that

    θ(x,t)C1,

    for all (x,t)[0,1]×[0,T0]. Combining this, (2.96), and (2.92) yields that for all (x,t)[0,1]×[0,+),

    C1θC11. (2.98)

    Together with (2.91), one has

    sup0tT10θ2xdx+T010θ2tdxdtC. (2.99)

    Finally, it follows from (2.5) that

    θβθxxv=θt+θvuxβθβ1θ2xv+θβvxθxv2θα(u2x+|wx|2+|bx|2)v,

    from this and (2.37), (2.97), (2.98) yields

    (2.100)

    Combining with (2.98)–(2.100), the proof of Lemma 2.7 has been finished.

    Lemma 2.8. Assume that the conditions listed in Lemma 2.1 hold; then for all , one has

    (2.101)

    Proof. First, differentiating (1.2) with respect to shows

    Multiplying the above equation by , one obtains after integration by parts,

    (2.102)

    where in the last inequality it has been used

    Next, differentiating (1.3) with respect to shows

    Multiplying the above equation by , one also gets after integration by parts,

    (2.103)

    Similarly, differentiating (1.4) with respect to shows

    Multiplying the above by , one gets after integration by parts that

    (2.104)

    At the end, differentiating (2.5) with respect to shows

    Multiplying the above by and integrating by parts, one obtains

    (2.105)

    According to (2.86), one has

    Combining (2.102)–(2.105) and Grönwall's inequality, we deduce

    (2.106)

    Finally, we rewrite (1.2) as

    It follows from (2.106), (2.85), (2.86) and (2.37) that

    (2.107)

    Similarly, rewriting (2.5) as

    Using (2.5), (2.106), (2.85), (2.86), (2.37), and (2.107), one has

    (2.108)

    Next, rewriting (1.3) as

    Then it follows that

    (2.109)

    At the end, rewriting (1.4) as

    thus, one has

    (2.110)

    Combining (2.106)–(2.110), the proof of Lemma 2.8 has been proved.

    Lemma 2.9. Assume that the conditions listed in Lemma (2.1) hold; then for all , one has

    (2.111)

    Proof. First, differentiating (1.2) with respect to gives

    (2.112)

    Multiplying (2.112) by , and integrating it over , we arrive at

    Together with (2.37), (2.75), (2.85), (2.86), (2.101), and Grönwall's inequality, one has

    That means

    (2.113)

    Furthermore, (2.112) can be written as

    Together with (2.37), (2.75), (2.85), (2.86), (2.101), and (2.113), one has

    (2.114)

    Next, differentiating (2.5) with respect to , it shows

    Thus, one has

    (2.115)

    Similarly, differentiating (1.4) with respect to , one has

    It implies that

    (2.116)

    Finally, differentiating (1.3) with respect to gives

    Then, one has

    (2.117)

    Combining with (2.113)–(2.117), we obtain (2.111). The Lemma 2.9 has been proved.

    With all the a priori estimates in Section 2 at hand, we will complete the proof of Theorem 1.1. For this purpose, it will be shown that the existence and uniqueness of local solutions to the initial-boundary value problem (1.1)–(1.9), which can be obtained by using the Banach theorem and the contractivity of the operator defined by the linearization of the problem on a small time interval.

    Lemma 3.1. Letting the (1.10) holds, then there exists , depending only on , and , such that the initial boundary value problem (1.1)–(1.9) has a unique solution .

    Proof of Theorem 1.1 First, using Lemma 3.1, the problem (1.1)–(1.9) has a unique solution , where .

    For the positive constants with being small enough such that

    (3.1)

    where is chosen in Lemma 2.1, one deduces from Lemmas 2.1–2.9 with that the solution satisfies

    (3.2)

    and

    (3.3)

    where is chosen in Section 2, and . It follows from Lemmas 2.8 and 2.9 that . If one takes as the initial data and applies Lemma 3.1 again, the local solution can be extended to the time interval with . Moreover, one obtains

    and

    Combining with (3.2), (3.3) implies

    and

    (3.4)

    Taking , where is chosen in (3.1) and is chosen to be such that

    where is chosen in Lemma 2.1. Then one can employ Lemmas 2.1–2.9 with to infer the local solution satisfies (3.2) and (3.3).

    Thus, choosing

    (3.5)

    and repeating the above procedure, one can then extend the solution step by step to a global one provided that . Furthermore, one derives the initial boundary value problem (1.1)–(1.9) has a unique global solution satisfying (3.2) and (3.3). Moreover, .

    The large-time behavior (1.11) follows from Lemmas 2.3–2.9 by using a standard argument (see Reference [21]).

    First, similar to (2.6), multiplying (1.1) by , (1.2) by , (1.3) by , (1.4) by , (2.5) by and adding them altogether, integrating the resultant equality over , one has after using (2.2) and (2.85) that

    where (and in what follows) and denote some generic positive constants depending only on and .

    By means of (2.87), (2.2), (2.85), (2.86), (2.101) and Sobolev's inequality, one obtains

    (3.6)

    Next, multiplying (2.39) by and integrating the resultant equality over , using (2.2), (2.85), (2.50), (2.75), and Poincare's inequality yields that

    (3.7)

    By the virtue of (2.50), (2.75), and (2.54), one obtains that

    (3.8)

    Furthermore, adding (3.5) multiplied by , (2.7) multiplied by , and (3.4) multiplied by , which satisfy

    From (3.6) and choosing suitably small, it follows

    where

    By using Cauchy–Schwarz's inequality, one obtains

    which along with Poincare's inequality, yields that

    Finally, using Poincare's inequality and (3.7) implies that

    where it has been used the conservation of energy implied by (1.5), (2.30), and the following fact:

    By means of (3.6) and (3.8), one obtains that

    Thus, the proof of Theorem 1.1 has been completed.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The authors are grateful to the anonymous referees for their helpful comments and valuable suggestions, which have improved the presentation of the manuscript. This work was supported by the Jiangxi Provincial Natural Science Foundation (Grant Nos. 20232BAB211001, 20242BAB25006).

    The authors declare there are no conflicts of interest.



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