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Research article

On the initial-boundary value problem for a Kuramoto-Sinelshchikov type equation

  • Received: 28 July 2020 Accepted: 23 August 2020 Published: 08 September 2020
  • The Kuramoto-Sinelshchikov equation describes the evolution of a phase turbulence in reaction-diffusion systems or the evolution of the plane flame propagation, taking in account the combined influence of diffusion and thermal conduction of the gas on the stability of a plane flame front. In this paper, we prove the well-posedness of the classical solutions for the initial-boundary value problem for this equation, under appropriate boundary conditions.

    Citation: Giuseppe Maria Coclite, Lorenzo di Ruvo. On the initial-boundary value problem for a Kuramoto-Sinelshchikov type equation[J]. Mathematics in Engineering, 2021, 3(4): 1-43. doi: 10.3934/mine.2021036

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  • The Kuramoto-Sinelshchikov equation describes the evolution of a phase turbulence in reaction-diffusion systems or the evolution of the plane flame propagation, taking in account the combined influence of diffusion and thermal conduction of the gas on the stability of a plane flame front. In this paper, we prove the well-posedness of the classical solutions for the initial-boundary value problem for this equation, under appropriate boundary conditions.



    In this paper, we investigate the well-posedness of the classical solutions for the equation:

    tu+κxu2+qxu3+ν2xu+δ3xu+β24xu=0, (1.1)

    with κ,q,ν,δ,βR and β0.

    We are interested in the initial-boundary value problem for this equation. More precisely we consider the following boundary conditions

    {u(t,0)=g(t),t>0,xu(t,0)=h(t),t>0,g,hW1,(0,),g(0)=u0(0),q=0, (1.2)
    {u(t,0)=g(t),t>0,2xu(t,0)=0,t>0,gW1,(0,),g(0)=u0(0),q=0, (1.3)
    {xu(t,0)=0,t>0,3xu(t,0)=0,t>0,q=a2,δ=0, (1.4)
    {2xu(t,0)=0,t>0,3xu(t,0)=0,t>0,q=a20, (1.5)
    {u(t,0)=0,t>0,xu(t,0)=0,t>0. (1.6)

    Moreover, we augment (1.1) with the following initial datum:

    u(0,x)=u0(x),x>0, (1.7)

    on which assume

    u0H2(0,). (1.8)

    Assuming q=0, (1.1) reads

    tu+κxu2+ν2xu+δ3xu+β24xu=0, (1.9)

    (1.9) arises in interesting physical situations, for example as a model for long waves on a viscous fluid owing down an inclined plane [50]. and to derive drift waves in a plasma [20]. (1.9) was derived also independently by Kuramoto [30,31,32] as a model for phase turbulence in reaction-diffusion systems and by Sivashinsky [47] as a model for plane flame propagation, describing the combined influence of diffusion and thermal conduction of the gas on the stability of a plane flame front.

    Equation (1.9) also describes incipient instabilities in a variety of physical and chemical systems [7,23,33]. Moreover, (1.9), which is also known as the Benney-Lin equation [4,41], was derived by Kuramoto in the study of phase turbulence in the Belousov-Zhabotinsky reaction [38].

    The dynamical properties and the existence of exact solutions for (1.9) have been investigated in [21,26,28,43,44,51]. In [3,6,22], the control problem for (1.9) with periodic boundary conditions, and on a bounded interval are studied, respectively. In [8], the problem of global exponential stabilization of (1.9) with periodic boundary conditions is analyzed. In [24], it is proposed a generalization of optimal control theory for (1.9), while in [42] the problem of global boundary control of (1.9) is considered. In [45], the existence of solitonic solutions for (1.9) is proven. In [5,18,48], the well-posedness of the Cauchy problem for (1.9) is proven, using the energy space technique, a priori estimates together with an application of the Cauchy-Kovalevskaya and the fixed point method, respectively. In particular, in [18], the well-posedness of the Cauchy problem for (1.1) is proven. In [12], following [9,10,37,46] it is proven that, when ν,δ,β2 go to zero, the solution of (1.9) converges to the unique entropy one of the Burgers equation. Finally, the initial-boundary value problem for (1.9), under the conditions (1.2) is analyzed in [39,40], in a quarter plane and in a bounded domain, respectively, using the energy space technique, under appropriate assumptions on κ,ν,δ,β.

    Taking q=ν=β=0 in (1.1), we have the Korteweg-de Vries equation [27]

    tu+κxu2+δ3xu=0, (1.10)

    that has a very wide range of applications, such as magnetic fluid waves, ion sound waves, and longitudinal astigmatic waves.

    From a mathematical point of view, in [16,19,25], the Cauchy problem for (1.10) is studied, while in [29], the author reviewed the travelling wave solutions for (1.10). Moreover, in [14,37,46], the convergence of the solution of (1.10) to the unique entropy one of the Burgers equation is proven.

    Taking κ=r=δ=μ=β=0, (1.1) becomes

    tu+qxu3+δ3xu=0, (1.11)

    which is known as the modified Korteweg-de Vries equation.

    [1,2,15,34,35,36] show that (1.11) is a non-slowly-varying envelope approximation model that describes the physics of few-cycle-pulse optical solitons. In [19,25], the Cauchy problem for (1.11) is studied, while, in [13,46], the convergence of the solution of (1.11) to the unique entropy solution of the following scalar conservation law

    tu+qxu3=0. (1.12)

    The main result of this paper is the following theorem.

    Theorem 1.1. Fix T>0. The initial value problems (1.1)-(1.2)-(1.7), (1.1)-(1.3)-(1.7), (1.1)-(1.4)-(1.7), (1.1)-(1.5)-(1.7), (1.1)-(1.6)-(1.7), admit an unique solution

    uH1((0,T)×(0,))L(0,T;H2(0,)). (1.13)

    Moreover, if u1 and u2 are two solutions of the same initial-boundary value problem for (1.1), we have

    u1(t,)u2(t,)L2(0,)eC(T)tu1,0u2,0L2(0,), (1.14)

    for some suitable C(T)>0, and every tT.

    Compared to [39], Theorem 1.1 gives the well-posedness of the initial-boundary value problem (1.1)-(1.2)-(1.7), without any additional assumption on the constants. The proof of Theorem 1.1 relies on deriving suitable a priori estimates together with an application of the Cauchy-Kovalevskaya Theorem [49].

    The paper is organized as follows. In Sections 2, 3, 4, 5, 6, we prove Theorem 1.1 for (1.1)-(1.2)-(1.7), (1.1)-(1.3)-(1.7), (1.1)-(1.4)-(1.7), (1.1)-(1.5)-(1.7), (1.1)-(1.6)-(1.7), respectively.

    In this section, we prove Theorem 1.1 for (1.1)-(1.2)-(1.7).

    Let us prove some a priori estimates on u, denoting with C0 the constants which depend only on the data, and with C(T), the constants which depend also on T.

    Following [11,17], we introduce the auxiliary variable:

    v(t,x)=u(t,x)g(t)ex[g(t)+h(t)]xex. (2.1)

    Observe that

    tv(t,x)=tu(t,x)g(t)ex[g(t)+h(t)]xex,xv(t,x)=xu(t,x)h(t)ex+[g(t)+h(t)]xex,2xv(t,x)=2xu(t,x)+2h(t)ex+g(t)ex[g(t)+h(t)]xex,3xv(t,x)=3xu(t,x)3h(t)ex2g(t)ex+[g(t)+h(t)]xex,4xv(t,x)=4xu(t,x)+4h(t)ex+3g(t)ex[g(t)+h(t)]xex. (2.2)

    In particular, thanks to (1.1)-(1.2)-(1.7), (2.1) and (2.2),

    v(t,0)=u(t,0)g(t)=0,xv(t,0)=xu(t,0)h(t)=0. (2.3)

    Moreover, thanks to (1.2) and (2.2), we have that

    v02L2(0,)u02L2(0,). (2.4)

    Again by (1.1)-(1.2)-(1.7), we have the following equation for v:

    tv+2κvxv+ν2xv+δ3xv+β24xv=g(t)ex+[g(t)+h(t)]xex2κh(t)exv+2κ[g(t)+h(t)]xexv2κg(t)exxv2κg(t)h(t)e2x+2κg(t)[g(t)+h(t)]xe2x2κ[g(t)+h(t)]xexxv2κh(t)[g(t)+h(t)]xe2x+2κ[g(t)+h(t)]2x2e2x+2h(t)ex+g(t)ex[g(t)+h(t)]xex3h(t)ex2g(t)ex+[g(t)+h(t)]xex+4h(t)ex+3g(t)ex[g(t)+h(t)]xex. (2.5)

    Lemma 2.1. Fix T>0. There exists a constant C(T)>0, such that

    v(t,)2L2(0,)+β2eC0t2t0eC0s2xv(s,)2L2(0,)dsC(T), (2.6)

    for every 0tT. In particular, we have that

    t0xv(s,)2L2(0,)dsC(T), (2.7)

    for every 0tT.

    Proof. Let 0tT. We begin by observing that, thanks to (2.3),

    4κ0v2xvdx=0,2δ0v3xvdx=2δ0xv2xv=0,2β0v4xvdx=2β20xv3xvdx=2β22xv(t,)2L2(0,). (2.8)

    Therefore, by (2.8), multiplying (2.5) by 2v, an integration on (0,) gives

    ddtv(t,)2L2(0,)+2β22xv(t,)2L2(0,)=2ν0v2xvdx+2g(t)0exvdx+2[g(t)+h(t)]0xexvdx+4κ[g(t)+h(t)]0xexv2dx4κg(t)Rexvxvdx4κ[g(t)+h(t)]0xexvdx4κ[g(t)+h(t)]0xexvxvdx4κh(t)[g(t)+h(t)]0xe2xvdx+4κ[g(t)+h(t)]20x2e2xvdx+4h(t)0exvdx+2g(t)0exvdx2[g(t)+h(t)]0xexvdx6h(t)0exvdx4g(t)0exvdx+2[g(t)+h(t)]0xexvdx+8h(t)0exvdx+6g(t)0exvdx2[g(t)+h(t)]0xexvdx. (2.9)

    Observe that, for each x(0,),

    ex1,xexe,0e2xdx=120x2e4xdx=132,0x2e2xdx=14,0x4e4xdx=3128. (2.10)

    Due (1.2), (2.10) and the Young inequality,

    2|ν|0|v||2xv|dx=20|νvβ||β2xv|dxν2β2v(t,)2L2(0,)+β22xu(t,)2L2(0,),2|g(t)|0ex|v|dx2C00ex|v|dxC0+C0v(t,)2L2(0,),2|g(t)+h(t)|0xex|v|dx2C00xex|v|dxC0+C0v(t,)2L2(0,),4|κ||g(t)+h(t)|0xexv2dxC00xexv2dxC0v(t,)2L2(0,),4|κ||g(t)|0ex|v||xv|dx2C00ex|v||xv|dx2C00|v||xv|dxC0v(t,)2L2(0,)+C0xv(t,)2L2(0,),4|κ||g(t)+h(t)|0xex|v|dx2C00xex|v|dxC0+C0v(t,)2L2(0,),4κ|g(t)+h(t)|0xex|v||xv|dx2C00xex|v||xv|dxC0v(t,)2L2(0,)+C0xv(t,)2L2(0,)4|κ||h(t)||g(t)+h(t)|0xe2x|v|dx2C00xe2x|v|dxC0+C0v(t,)2L2(0,),4|κ|[g(t)+h(t)]20x2e2x|v|dx2C00x2e2x|v|dxC0+C0v(t,)2L2(0,),4|h(t)|0ex|v|dx2C00ex|v|dxC0+C0v(t,)2L2(0,),2|g(t)|0ex|v|dx2C00ex|v|dxC0+C0v(t,)2L2(0,),2|g(t)+h(t)|0xex|v|dx2C00xex|v|dxC0+C0v(t,)2L2(0,),6|h(t)|0ex|v|dx2C00ex|v|dxC0+C0v(t,)2L2(0,),4|g(t)|0ex|v|dx2C00ex|v|dxC0+C0v(t,)2L2(0,),2|g(t)+h(t)]|0xex|v|dx2C00xex|v|dxC0+C0v(t,)2L2(0,),8|h(t)|0ex|v|dx2C0Rex|v|dxC0+C0v(t,)2L2(0,),6|g(t)|0ex|v|dx2C00ex|v|dxC0+C0v(t,)2L2(0,),2|g(t)+h(t)|0xex|v|dx2C00xex|v|dxC0+C0v(t,)2L2(0,).

    It follows from (2.9) that

    ddtv(t,)2L2(0,)+β22xv(t,)2L2(0,)C0v(t,)2L2(0,)+C0xv(t,)2L2(0,)+C0. (2.11)

    Thanks to (2.3),

    C0xv(t,)2L2(0,)=C00xvxvdx=C00v2xvdx.

    Therefore, by the Young inequality,

    C0xv(t,)2L2(0,)0|C0vβ||β2xv|dxC0v(t,)2L2(0,)+β222xv(t,)2L2(0,). (2.12)

    Consequently, by (2.11),

    ddtv(t,)2L2(0,)+β222xv(t,)2L2(0,)C0v(t,)2L2(0,)+C0.

    By the Gronwall Lemma and (2.4), we have

    v(t,)2L2(0,)+β2eC0t2t0eC0s2xv(s,)2L2(0,)dsC0eC0t+C0tC(T),

    which gives (2.6).

    Finally, we prove (2.7). By (2.6), and (2.12),

    C0xv(t,)2L2(0,)C(T)+β222xv(t,)2L2(0,).

    Integrating on (0,t), by (2.6), we have

    C0t0xv(s,)2L2(0,)dsC(T)t+β22t02xv(s,)2L2(0,)dsC(T),

    which gives (2.7).

    Lemma 2.2. Fix T>0. There exists a constant C(T)>0, such that

    u(t,)L2(0,)C(T), (2.13)
    t0xu(s,)2L2(0,)dsC(T), (2.14)
    t02xu(s,)2L2(0,)dsC(T), (2.15)

    for every 0tT.

    Proof. Let 0tT. We begin by proving (2.13). Observe that, by (2.1),

    u2(t,x)=[v(t,x)+g(t)ex+[g(t)+h(t)]xex]=v2(t,x)+g2(t)e2x+[g(t)+h(t)]x2e2x+2g(t)exv(t,x)+2[g(t)+h(t)]xexv+2g(t)[g(t)+h(t)]xe2x.

    Consequently, an integration on (0,) gives

    u(t,)2L2(0,)=v(t,)2L2(0,)+g2(t)0e2xdx+[g(t)+h(t)]20x2e2xdx+2g(t)0exvdx+2[g(t)+h(t)]0xexvdx+2g(t)[g(t)+h(t)]x0xe2xdx. (2.16)

    Observe that

    0xe2x=14. (2.17)

    Due to (1.2), (2.10), (2.17) and the Young inequality,

    g2(t)0e2xdxC00e2xdxC0,[g(t)+h(t)]20x2e2xdxC00x2e2xdxC0,2|g(t)|0ex|v|dx2C00ex|v|dxC0+C0v(t,)2L2(0,)2|g(t)+h(t)|0xex|v|dx2C00xex|v|dxC0+C0v(t,)2L2(0,)2|g(t)||g(t)+h(t)|]x0xe2xdxC0x0xe2xdxC0.

    Therefore, by (2.6) and (2.16),

    u(t,)2L2(0,)C0v(t,)2L2(0,)+C0C(T),

    that is (2.13).

    We prove (2.14). By (2.5), we have that

    (xu(t,x))2=[xv(t,x)+h(t)ex[g(t)+h(t)]xex]2=(xv(t,x))2+h2(t)e2x+[g(t)+h(t)]2x2e2x+2h(t)exxv(t,x)2[g(t)+h(t)]xexxv(t,x)2h(t)[g(t)+h(t)]xe2x.

    Therefore, an integration on (0,) gives

    xu(t,)2L2(0,)=xv(t,)2L2(0,)+h2(t)0e2xdx+[g(t)+h(t)]20x2e2xdx+2h(t)0exxvdx2[g(t)+h(t)]0xexxvdx2h(t)[g(t)+h(t)]0xe2xdx. (2.18)

    Due to (1.2), (2.10), (2.17) and the Young inequality,

    h2(t)0e2xdxC00e2xdxC0,[g(t)+h(t)]20x2e2xdxC00x2e2xdxC0,2|h(t)|0ex|xv|dx2C00ex|xv|dxC0+C0xv(t,)2L2(0,),2|g(t)+h(t)|0xex|xv|dx2C00xex|xv|dxC0+C0xv(t,)2L2(0,),2|h(t)||g(t)+h(t)|0xe2xdxC00xe2xdxC0.

    It follows from (2.18) that

    xu(t,)2L2(0,)C0xv(t,)2L2(0,)+C0.

    Integrating on (0,t), by (2.7), we have that

    t0xu(s,)2L2(0,)dsC0t0xv(s,)2L2(0,)ds+C0tC(T),

    which gives (2.14).

    Finally, we prove (2.15). By (2.2), we have that

    (2xu(t,x))2=[2xv(t,x)[2h(t)+g(t)]ex+[g(t)+h(t)]xex]2=(2xv(t,x))2+[2h(t)+g(t)]2e2x+[g(t)+h(t)]2x2e2x2[2h(t)+g(t)]ex2xv+2[g(t)+h(t)]xex2xv2[2h(t)+g(t)][g(t)+h(t)]xe2x.

    An integration on (0,) gives

    2xu(t,)2L2(0,)=2xv(t,)2L2(0,)+[2h(t)+g(t)]20e2xdx+[g(t)+h(t)]20x2e2xdx2[2h(t)+g(t)]0ex2xvdx+2[g(t)+h(t)]0xex2xvdx2[2h(t)+g(t)][g(t)+h(t)]0xe2xdx. (2.19)

    Due to (1.2), (2.10), (2.17) and the Young inequality,

    [2h(t)+g(t)]20e2xdxC00e2xdxC0,[g(t)+h(t)]20x2e2xdxC00x2e2xdxC0,2|2h(t)+g(t)|0ex|2xv|dx2C00ex|2xv|dxC0+C02xv(t,)2L2(0,),2|g(t)+h(t)|0xex|2xv|dx2C00xex|2xv|dxC0+C02xv(t,)2L2(0,),2|2h(t)+g(t)||g(t)+h(t)|0xe2xdxC00xe2xdxC0.

    It follows from (2.19) that

    2xu(t,)2L2(0,)2xv(t,)2L2(0,)+C0.

    Integrating on (0,t), by (2.6), we have that

    t02xu(s,)2L2(0,)dst02xv(s,)2L2(0,)ds+C0tC(T),

    which gives (2.15).

    Lemma 2.3. Fix T>0. There exists a constant C(T)>0, such that

    t0xu(s,)2L(0,)dsC(T), (2.20)
    t0u(s,)xu(s,)2L2(0,)dsC(T), (2.21)

    for every 0tT.

    Proof. Let 0tT. We begin by proving (2.20). Thanks to (1.2) and the Young inequality,

    (xu(t,x))2=2x0xu2xudy+h2(t)20|xu||2xu|dx+C0xu(t,)2L2(0,)+2xu(t,)2L2(0,)+C0.

    Hence,

    xu(t,)2L(0,)xu(t,)2L2(0,)+2xu(t,)2L2(0,)+C0 (2.22)

    Integrating on (0,t), by (2.14) and (2.15), we have that

    t0xu(s,)2L(0,)dst0xu(s,)2L2(0,)ds+t02xu(s,)2L2(0,)ds+C0tC(T),

    which gives (2.20).

    Finally, we prove (2.21). Thanks to (2.13), we have that

    u(t,)xu(s,)2L2(0,)ds=0u2(xu)2dxxu(t,)2L(0,)u(t,)2L2(0,)C(T)xu(t,)2L(0,).

    (2.21) follows from (2.20) and an integration on (0,t).

    Lemma 2.4. Fix T>0. There exists a constant C(T)>0, such that

    2xu(t,)2L2(0,)+δ2+24β4xu(t,)2L2(0,) (2.23)
    +β22t04xu(s,)2L2(0,)ds+δ2+246β203xu(s,)2L2(0,)dsC(T),t0(2xu(s,0))2dsC(T),t0(3xu(s,0))2dsC(T), (2.24)
    xuL((0,T)×(0,))C(T), (2.25)
    uL((0,T)×(0,))C(T), (2.26)

    for every 0tT.

    Proof. Let 0tT. Consider an positive constant A, which will be specified later. Multiplying (1.1)-(1.2)-(1.7) by

    24xu2A2xu,

    we have that

    (24xu2A2xu)tu+2κ(24xu2A2xu)uxu+ν(24xu2A2xu)2xu+δ(24xu2A2xu)3xu+β2(24xu2A2xu)4xu=0. (2.27)

    Observe that, thanks to (1.1)-(1.2)-(1.7),

    0(24xu2A2xu)tudx=23xu(t,0)tu(t,0)203xutxudx+2Axu(t,0)tu(t,0)+Addtxu(t,)2L2(0,)=2g(t)3xu(t,0)+2h(t)2xu(t,0)+2Ah(t)g(t)+ddt(2xu(t,)2L2(0,)+Axu(t,)2L2(0,)),2Aβ202xu4xudx=2Aβ22xu(t,0)3xu(t,0)+2Aβ23xu(t,)2L2(0,). (2.28)

    Therefore, thanks to (2.28), an integration of (2.27) on (0,) gives

    ddt(2xu(t,)2L2(0,)+Axu(t,)2L2(0,))+2β24xu(t,)2L2(0,)+2Aβ23xu(t,)2L2(0,)=4κ0uxu4xudx+4Aκ0uxu2xudx2ν04xu2xudx2Aν2xu(t,)2L2(0,)2δ04xu3xudx+2Aδ02xu3xudx+2g(t)3xu(t,0)2h(t)2xu(t,0)2Ah(t)g(t)2Aβ22xu(t,0)3xu(t,0). (2.29)

    Due to the Young inequality,

    4|κ|0|uxu||4xu|dx=20|2κuxuβD1||βD14xu|dx4κ2β2D1u(t,)xu(t,)2L2(0,)+D1β24xu(t,)2L2(0,),4A|κ|0|uxu||2xu|dx=20|2Aκuxu||2xu|dx4A2κ2u(t,)xu(t,)2L2(0,)+2xu(t,)2L2(0,),2|ν|0|4xu||2xu|dx=20|βD14xu||ν2xuβD1|dxD1β24xu(t,)2L2(0,)+ν2β2D12xu(t,)2L2(0,),2|δ|0|4xu||3xu|dx=20|βD14xu||δ3xuβD1|dxD1β24xu(t,)2L2(0,)+δ2β2D13xu(t,)2L2(0,),2A|δ|0|2xu||3xu|dx=A0|δ2xuβ||β3xu|dxAδ2β22xu(t,)2L2(0,)+Aβ23xu(t,)2L2(0,),

    where D1 is a positive constant, which will be specified later. It follows from (2.29) that

    ddt(2xu(t,)2L2(0,)+Axu(t,)2L2(0,))+β2(23D1)4xu(t,)2L2(0,)+(Aβ2δ2β2D1)3xu(t,)2L2(0,)4κ2(1β2D1+A2)u(t,)xu(t,)2L2(0,)+(1+ν2β2D1+2A|ν|+Aδ2β2)2xu(t,)2L2(0,)+2|g(t)||3xu(t,0)|+2|h(t)||2xu(t,0)|+2A|h(t)||g(t)|+2Aβ2|2xu(t,0)||3xu(t,0)|.

    Choosing D1=3, we have that

    ddt(2xu(t,)2L2(0,)+Axu(t,)2L2(0,))+β24xu(t,)2L2(0,)+(Aβ2δ23β2)3xu(t,)2L2(0,)4κ2(13β2+A2)u(t,)xu(t,)2L2(0,)+(1+ν23β2+2A|ν|+Aδ2β2)2xu(t,)2L2(0,)+2|g(t)||3xu(t,0)|+2|h(t)||2xu(t,0)|+2A|h(t)||g(t)|+2Aβ2|2xu(t,0)||3xu(t,0)|. (2.30)

    Due to (1.2) and the Young inequality,

    2|g(t)||3xu(t,0)|(g(t))2+(3xu(t,0))2C0+(3xu(t,0))2,2|h(t)||2xu(t,0)|(h(t))2+(2xu(t,0))2C0+(2xu(t,0))2,2A|h(t)||g(t)|AC0,2Aβ2|2xu(t,0)||3xu(t,0)|A2β4(2xu(t,0))2+(3xu(t,0))2.

    Consequently, by (2.30),

    ddt(2xu(t,)2L2(0,)+Axu(t,)2L2(0,))+β24xu(t,)2L2(0,)+(Aβ2δ23β2)3xu(t,)2L2(0,)4κ2(13β2+A2)u(t,)xu(t,)2L2(0,)+(1+ν23β2+2A|ν|+Aδ2β2)2xu(t,)2L2(0,)+2(3xu(t,0))2+(1+A2β4)(2xu(t,0))2+C0(1+A). (2.31)

    Thanks to the Young inequality,

    2(3xu(t,0))2=403xu4xudx0|43xuβ||β4xu|dx8β23xu(t,)2L2(0,)+β224xu(t,)2L2(0,),(1+A2β4)(2xu(t,0))2=2(1+A2β4)02xu3xudx0|2(1+A2β4)2xuβA||βA3xu|dx2(1+A2β4)22xu(t,)2L2(0,)+Aβ223xu(t,)2L2(0,). (2.32)

    It follows from (2.31) that

    ddt(2xu(t,)2L2(0,)+Axu(t,)2L2(0,))+β224xu(t,)2L2(0,)+(Aβ22δ2+243β2)3xu(t,)2L2(0,)4κ2(13β2+A2)u(t,)xu(t,)2L2(0,)+(1+ν23β2+2A|ν|+Aδ2β2+2(1+A2β4)2)2xu(t,)2L2(0,)+C0(1+A).

    Choosing

    A=δ2+24β4, (2.33)

    we have that

    ddt(2xu(t,)2L2(0,)+δ2+24β4xu(t,)2L2(0,))+β224xu(t,)2L2(0,)+δ2+246β23xu(t,)2L2(0,)C0u(t,)xu(t,)2L2(0,)+C02xu(t,)2L2(0,)+C0.

    Integrating on (0,t), by (1.8), (2.15) and (2.21), we get

    2xu(t,)2L2(0,)+δ2+24β4xu(t,)2L2(0,)+β22t04xu(s,)2L2(0,)ds+δ2+246β203xu(s,)2L2(0,)dsC0+C0t0u(s,)xu(s,)2L2(0,)ds+C0t02xu(s,)2L2(0,)ds+C0tC(T),

    which gives (2.23).

    (2.24) follows from (2.15), (2.23), (2.33) and an integration of (2.32) on (0,t), while (2.22) and (2.23) give (2.25).

    Finally, we prove (2.26). Due to (1.2), (2.13), (2.23) and the Hölder inequality,

    u2(t,x)=2x0uxudy+g2(t)20|u||xu|dx+C0u(t,)L2(0,)xu(t,)L2(0,)+C0C(T).

    Hence,

    u2L((0,T)×(0,))C(T),

    which gives (2.26).

    Lemma 2.5. Fix T>0. There exists a constant C(T)>0, such that

    t0tu(s,)2L2(0,)dsC(T), (2.34)

    for every 0tT.

    Proof. Let 0tT. Multiplying (1.1)-(1.2)-(1.7) by 2tu, an integration on (0,) gives

    2tu(t,)2L2(0,)=4κ0uxutudx2ν02xutudx2δ03xutudx2β204xutudx. (2.35)

    Due to (2.23) and the Young inequality,

    4|κ|0uxutudx=20|2κuxuD2||D2tu|dx4κ2D2u(t,)xu(t,)2L2(0,)+D2tu(t,)2L2(0,),2|ν|0|2xu||tu|dx=20|ν2xuD2||D2tu|dxν2D22xu(t,)2L2(0,)+D2tu(t,)2L2(0,)C(T)D2+D2tu(t,)2L2(0,),2|δ|0|3xu||tu|dx=20|δ3xuD2||D2tu|dxδ2D23xu(t,)2L2(0,)+D2tu(t,)2L2(0,),2β20|4xu||tu|dx=20|β24xuD2||D2tu|dxβ4D24xu(t,)2L2(0,)+D2tu(t,)2L2(0,).

    Consequently, by (2.35),

    2(12D2)tu(t,)2L2(0,)C(T)D2+4κ2D2u(t,)xu(t,)2L2(0,)+δ2D23xu(t,)2L2(0,)+β4D24xu(t,)2L2(0,).

    Choosing D2=14, we have that

    tu(t,)2L2(0,)C(T)+16κ2u(t,)xu(t,)2L2(0,)+4δ23xu(t,)2L2(0,)+4β44xu(t,)2L2(0,).

    Integrating on (0,t), by (2.21) and (2.23), we get

    t0tu(s,)2L2(0,)dsC(T)t+16κ2t0u(s,)xu(t,)2L2(0,)ds+4δ2t03xu(s,)2L2(0,)ds+4β4t04xu(s,)2L2(0,)dsC(T),

    which gives (2.34).

    Now, we prove Theorem 1.1.

    Proof. Fix T>0. Thanks to Lemmas 2.2, 2.4, 2.5 and the Cauchy-Kovalevskaya Theorem [49], we have that u is solution of (1.1)-(1.2)-(1.7) and (1.13) holds.

    We prove (1.14). Let u1 and u2 be two solutions of (1.1)-(1.2)-(1.7), which verify (1.13), that is

    {tu1+2κu1xu1+ν2xu1+δ3xu1+β24xu1=0,t>0,x>0,u1(t,0)=g(t),t>0,xu1(t,0)=h(t),t>0,u1(0,x)=u1,0(x),x>0,{tu2+2κu2xu2+ν2xu2+δ3xu2+β24xu2=0,t>0,x>0,u2(t,0)=g(t),t>0,xu2(t,0)=h(t),t>0,u2(0,x)=u2,0(x),x>0.

    Then, the function

    ω=u1u2 (2.36)

    solves the following initial-boundary value problem:

    {tω+2κ(u1xu1u2xu2)+ν2xω+δ3xω+β24xω=0,t>0,x>0,ω(t,0)=0,t>0,xω(t,0)=0,t>0,ω(0,x)=u1,0(x)u2,0(x),x>0. (2.37)

    Observe that, thanks to (2.36),

    u1xu1u2xu2=u1xu1u1xu2+u1xu2u2xu2=u1xω+xu2ω.

    Therefore, (2.37) is equivalent to the following equation:

    tω+2κu1xω+2κxu2ω+ν2xω+δ3xω+β24xω=0. (2.38)

    Moreover, since u1,u2L(0,T;H2(0,)), we have that

    xu1L((0,T)×(0,)),xu2L((0,T)×(0,))C(T). (2.39)

    Observe again that, thanks to (2.37),

    4κ0u1ωxωdx=2κ0xu1ω2dx,2δ0ω3xωdx=2δ0xω2xωdx=0,2β20ω4xωdx=2β20xω3xωdx=2β22xω(t,)2L2(0,). (2.40)

    Therefore, multiplying (2.38) by 2ω, thanks to (2.40), an integration on (0,) gives

    ddtxω(t,)2L2(0,)+2β22xω(t,)2L2(0,)=2κ0(xu12xu2)ω2dx2ν0ω2xωdx. (2.41)

    Due to (2.39) and the Young inequality,

    2|κ|0|xu12xu2|ω2dxC(T)ω(t,)2L2(0,),2|ν|0|ω||2xω|dx=20|νωβ||β2xω|dxν2β2ω(t,)2L2(0,)+β22xω(t,)2L2(0,).

    Therefore, by (2.41),

    ddtxω(t,)2L2(0,)+β22xω(t,)2L2(0,)C(T)ω(t,)2L2(0,).

    The Gronwall Lemma and (2.37) gives

    xω(t,)2L2(0,)+β2eC(T)tt0eC(T)s2xω(s,)2L2(0,)dseC(T)tω02L2(0,). (2.42)

    (1.14) follows from (2.36) and (2.42).

    In this section, we prove Theorem 1.1 for (1.1)-(1.3)-(1.7).

    Let us prove some a priori estimates on u.

    Following [11,17], we consider the following function:

    v(t,x)=u(t,x)g(t)ex. (3.1)

    Observe that

    tv(t,x)=tu(t,x)g(t)ex,xv(t,x)=xu(t,x)+g(t)ex,2xv(t,x)=2xu(t,x)g(t)ex,3xv(t,x)=3xu(t,x)+g(t)ex,4xv(t,x)=4xug(t)ex. (3.2)

    By (1.1)-(1.3)-(1.7) and (3.2),

    v(t,0)=u(t,0)g(t)=0,2xv(t,0)=g(t), (3.3)

    while, by (1.3) and (3.2), we have (2.4).

    Again by (1.1)-(1.3)-(1.7) and (3.2), we have the following equation for v.

    tv+2κvxv+ν2xv+δ3xv+β24xv=g(t)ex+2κg(t)exv2κg(t)exxv+2κg2(t)e2xνg(t)exδg(t)exβ2g(t)ex. (3.4)

    We prove the following result.

    Lemma 3.1. Fix T>0. There exists a constant C(T)>0, such that

    v(t,)2L2(0,)+β2eC0t42t0eC0s2xv(s,)2L2(0,)dsC(T), (3.5)

    for every 0tT. In particular, we have (2.7). Moreover,

    t0(xv(s,0))2dsC(T), (3.6)

    for every 0tT.

    Proof. Let 0tT. We begin by observing that, thanks to (3.3),

    20vtvdx=ddtv(t,)2L2(0,),2δ0v3xv=2δ0xv2xvdx,2β20v4xvdx=2β20xv3xvdx=2β2xv(t,0)2xv(t,0)+2β22xv(t,)2L2(0,)=2β2g(t)xv(t,0)+2β22xv(t,)2L2(0,). (3.7)

    Consequently, multiplying (3.4) by 2v, thanks to (3.7), an integration of (3.7) on (0,) gives

    ddtv(t,)2L2(0,)+2β22xv(t,)2L2(0,)=2g(t)0exvdx+4κg(t)0exv2dx4κg(t)0exvxvdx+4κg2(t)0e2xvdx2νg(t)0exvdx2δg(t)0exvdxβ2g(t)0exvdx2ν0v2xvdx+2δ0xv2xvdx2β2g(t)xv(t,0). (3.8)

    Since

    0e4xdx=14, (3.9)

    thanks to (1.3), (2.10), (3.9) and the Young inequality,

    2|g(t)|0ex|v|dx2C00ex|v|dxC0+C0v(t,)2L2(0,),4|κ||g(t)|0exv2dxC00exv2dxC0v(t,)2L2(0,),4|κ||g(t)|0ex|v||xv|dx2C00ex|v||xv|dx2C00|v||xv|dxC0v(t,)2L2(0,)+C0xv(t,)2L2(0,),4|κ|g2(t)0e2x|v|dx2C00e2x|v|dxC0+C0v(t,)2L2(0,),2|ν+δ+β||g(t)|0ex|v|dx2C00ex|v|dxC0+C0v(t,)2L2(0,),2|ν|0v|2xv|dx=20|νvβ||β2xv|dxν2β2v(t,)2L2(0,)+β22xv(t,)2L2(0,),2|δ|0|xv||2xv|dx=0|2δxvβ||β2xv|dx2δ2β2xv(t,)2L2(0,)+β222xv(t,)2L2(0,),2β2|g(t)||xv(t,0)|2C0|xv(t,0)|C0+C0(xv(t,0))2.

    It follows from (3.8) that

    ddtv(t,)2L2(0,)+β222xv(t,)2L2(0,)C0+C0v(t,)2L2(0,)+C0xv(t,)2L2(0,)+C0(xv(t,0))2. (3.10)

    Observe that, by the Young inequality,

    C0(xv(t,0))2=2C00xv2xvdx2C00|xv||2xv|dx20|3C0xvβ||β32xv|dxC0xv(t,)2L2(0,)+β232xv(t,)2L2(0,). (3.11)

    Consequently, by (3.10),

    ddtv(t,)2L2(0,)+β262xv(t,)2L2(0,)C0+C0v(t,)2L2(0,)+C0xv(t,)2L2(0,). (3.12)

    Observe that, thanks to (3.3),

    C0xv(t,)2L2(0,)=C00xvxvdx=C00vxvdx.

    Therefore, by the Young inequality,

    C0xv(t,)2L2(0,)20|7C0v2β||β2xv7|dxC0v(t,)2L2(0,)+β272xv(t,)2L2(0,). (3.13)

    It follows from (3.12) that

    ddtv(t,)2L2(0,)+β2422xv(t,)2L2(0,)C0+C0v(t,)2L2(0,). (3.14)

    By (2.4) and the Gronwall Lemma, we get

    v(t,)2L2(0,)+β2eC0t42t0eC0s2xv(s,)2L2(0,)dsC0eC0t+C0eC0tt0eC0sdsC(T),

    which gives (3.5).

    We prove (2.7). By (3.5) and (3.13), we have that

    C0xv(t,)2L2(0,)C(T)+β272xv(t,)2L2(0,).

    Integrating on (0,t), by (3.5), we have (2.7).

    Finally, (3.6) follows from (2.7), (3.5), (3.11) and an integration on (0,t).

    Lemma 3.2. Fix T>0. There exists a constant C(T)>0, such that

    t0(xu(s,0))2dsC(T), (3.15)

    for every 0tT. In particular, (2.13), (2.14), (2.15), (2.20), (2.21) hold.

    Proof. Let 0tT. We prove (3.15). We begin by observing that, thanks (3.2),

    xu(t,0)=xv(t,0)g(t).

    Therefore, by (1.3) and the Young inequality,

    (xu(t,0))2=(xv(t,0))2+g2(t)2g(t)xv(t,0)(xv(t,0))2+g2(t)+2|g(t)||xv(t,0)|2(xv(t,0))2+2g2(t)2(xv(t,0))2+C0,

    Integrating on (0,t), by (3.6), we get

    t0(xu(s,0))2ds2t0(xv(s,0))2ds+C0tC(T),

    which gives (3.15).

    Arguing as in Lemma 2.2, we have (2.13), (2.14) and (2.15).

    We prove (2.20). Thanks to the Young inequality,

    (xu(t,x))2=2x0|xu||2xu|dx+2(xu(t,0))220|xu||2xu|dx+2(xu(t,0))2xu(t,)2L2(0,)+2xu(t,)2L2(0,)+2(xu(t,0))2.

    Therefore,

    xu(t,)2L(0,)xu(t,)2L2(0,)+2xu(t,)2L2(0,)+2(xu(t,0))2. (3.16)

    Integrating on (0,t), by (2.14), (2.15) and (3.15), we have (2.20).

    Finally, arguing as in Lemma 2.3, we have (2.20).

    Lemma 3.3. Fix T>0. There exists a constant C(T)>0, such that

    xu(t,)2L2(0,)+β2t03xu(s,)2L2(0,)dsC(T), (3.17)

    for every 0tT. In particular, (2.26) holds.

    Proof. Let 0tT. We begin by observing that

    20u2xudx=2xu(t,0)tu(t,0)+ddtxu(t,)2L2(0,),2δ02xu3xudx=0,2β202xu4xudx=2β23xu(t,)2L2(0,). (3.18)

    Since, thanks (1.1)-(1.3)-(1.7), tu(t,0)=g(t), multiplying (1.1)-(1.3)-(1.7) by 22xu, thanks to (3.18), an integration on (0,) gives

    ddtxu(t,)2L2(0,)+2β23xu(t,)2L2(0,)=2g(t)xu(t,0)+4κ0uxu2xudx+2ν2xu(t,)2L2(0,). (3.19)

    Due to (1.3) and the Young inequality,

    2|g(t)||xu(t,0)|2C0|xu(t,0)|C0+C0(xu(t,0))24|κ|0|uxu||2xu|dx2κ2u(t,)xu(t,)2L2(0,)+2xu(t,)2L2(0,).

    Consequently, by (3.19),

    ddtxu(t,)2L2(0,)+2β23xu(t,)2L2(0,)C0+C0(xu(t,0))2+2κ2u(t,)xu(t,)2L2(0,)+(1+|ν|)2xu(t,)2L2(0,).

    Integrating on (0,t), by (1.8), (2.15) and (2.21), we get

    xu(t,)2L2(0,)+2β2t03xu(s,)2L2(0,)dsC0+C0t+C0t0(xu(s,0))2ds+2κ2t0u(s,)xu(s,)2L2(0,)ds+(1+|ν|)t02xu(s,)2L2(0,)dsC(T),

    which gives (3.17).

    Finally, arguing as in Lemma 2.4, we have (2.26).

    Lemma 3.4. Fix T>0. There exists a constant C(T)>0, such that

    2xu(t,)2L2(0,)+β2t04xu(s,)2L2(0,)dsC(T), (3.20)

    for ever 0tT. In particular, we have (2.25) and

    t0(3xu(s,0))2dsC(T), (3.21)

    for every 0tT. Moreover, (2.34) holds.

    Proof. Let 0tT. We begin by observing that, thanks to (1.1)-(1.3)-(1.7),

    204xutudx=23xu(t,0)tu(t,0)203xutxudx=23xu(t,0)tu(t,0)+ddt2xu(t,)2L2(0,),2ν02xu4xudx=2ν3xu(t,)2L2(0,). (3.22)

    Since, thanks to (1.1)-(1.3)-(1.7), tu(t,0)=g(t), multiplying (1.1)-(1.3)-(1.7) by 24xu, thanks (3.22), an integration on (0,) gives

    ddt2xu(t,)2L2(0,)+2β24xu(t,)2L2(0,)=2g(t)3xu(t,0)4κ0uxu4xudx+2ν3xu(t,)2L2(0,)2δ03xu4xudx. (3.23)

    Due to (1.3) and the Young inequality,

    2|g(t)||3xu(t,0)|2C0|3xu(t,0)|C0+C0(3xu(t,0))2,4|κ|0|uxu||4xu|dx=20|2κuxuβ||β4xu|dx4κ2β2u(t,)xu(t,)2L2(0,)+β24xu(t,)2L2(0,)2|δ|0|3xu||4xu|dx=0|2δ3xuβ||β4xu|dx2δ2β23xu(t,)2L2(0,)+β224xu(t,)2L2(0,).

    It follows from (3.23) that

    ddt2xu(t,)2L2(0,)+β224xu(t,)2L2(0,)C0+C0(3xu(t,0))2+4κ2β2u(t,)xu(t,)2L2(0,)+(2|ν|+2δ2β2)3xu(t,)2L2(0,). (3.24)

    Thanks to the Young inequality,

    C0(3xu(t,0))2=2C003xu4xudx20|3C03xuβ||β4xu3|dxC03xu(t,)2L2(0,)+β234xu(t,)2L2(0,). (3.25)

    Consequently, by (3.24),

    ddt2xu(t,)2L2(0,)+β264xu(t,)2L2(0,)C0+4κ2β2u(t,)xu(t,)2L2(0,)+C03xu(t,)2L2(0,).

    Integrating on (0,t), by (1.8), (2.21) and (3.17), we get

    2xu(t,)2L2(0,)+β26t04xu(s,)2L2(0,)C0+C0t+4κ2β2t0u(s,)xu(s,)2L2(0,)ds+C0t03xu(s,)2L2(0,)dsC(T),

    which gives (3.20).

    We prove (2.25). Thanks to the Hölder inequality,

    (xu(t,0))2=20xu2xudx20|xu||2xu|dx2xu(t,)L2(0,)2xu(t,)L2(0,). (3.26)

    Therefore, by (3.17) and (3.20),

    (xu(t,0))2C(T). (3.27)

    By (3.16), (3.17) and (3.20), we obtain that

    xu2L((0,T)×(0,))C(T),

    which gives (2.25).

    Finally, (3.21) follows from (3.17), (3.20), (3.25) and an integration on (0,t), while arguing as in Lemma 2.5, we have (2.34).

    Arguing as in Section 2, we have Theorem 1.1.

    In this section, we prove Theorem 1.1 for (1.1)-(1.4)-(1.7).

    Let us prove some a priori estimates on u.

    We begin by proving the following lemma.

    Lemma 4.1. Fix T>0. There exists a constant C(T)>0, such that

    u(t,)2L(0,)+β2eC0tt0eC0s2xu(s,)2L2(0,)ds+a2eC0tt0eC0su4(s,0)dsC(T). (4.1)

    for every 0tT. In particular, we have (2.14), (2.20), (2.21) and

    t0u2(s,0)dsC(T), (4.2)

    for every 0tT.

    Proof. Let 0tT. Multiplying (1.1)-(1.4)-(1.7) by 2v, thanks to (1.1)-(1.4)-(1.7), an integration on (0,) give

    ddtu(t,)2L2(0,)=20utudx=4κ0u2xudx+6a20u3xudx2ν0u2xudx2β20u4xudx=4κ3u3(t,0)3a22u4(t,0)2ν0u2xudx+2β20xu3xudx=4κ3u3(t,0)3a22u4(t,0)2ν0u2xudx2β22xu(t,)2L2(0,).

    Therefore, we have that

    ddtu(t,)2L2(0,)+2β22xu(t,)2L2(0,)+3a22u4(t,0)=4κ3u3(t,0)2ν0u2xudx. (4.3)

    Due to the Young inequality,

    4|κ|3|u(t,0)|3=2(2|κu(t,0)|3D3)(D3u2)4κ29D3u2(t,0)+D3u4(t,0)=2(2κ29D3D3)(D3u2(t,0))+D3u4(t,0)4κ481D3+2D3u4(t,0),2|ν|0|u||2xu|dx20|νuβ||β2xu|dxν2β2u(t,)2L2(0,)+β22xu(t,)2L2(0,),

    where D3 is a positive constant, which will be specified later. Consequently, by (4.3),

    ddtu(t,)2L2(0,)+β22xu(t,)2L2(0,)+(3a222D3)u4(t,0)4κ481D3+ν2β2u(t,)2L2(0,).

    Choosing D3=a22, we have that

    ddtu(t,)2L2(0,)+2β22xu(t,)2L2(0,)+a2u4(t,0)C0+C0u(t,)2L2(0,).

    By the Gronwall Lemma and (1.8), we get

    u(t,)2L2(0,)+2β2eC0tt0eC0s2xu(s,)2L2(0,)ds+a2eC0tt0eC0su4(s,0)dsC0eC0t+C0eC0tt0eC0sdsC(T),

    which gives (4.1).

    We prove (2.14). Thanks to (1.1)-(1.4)-(1.7), (4.1) and the Young inequality,

    xu(t,)2L2(0,)=0xuxudx=0u2xudx0|u||2xu|dx12u(t,)2L2(0,)+122xu(t,)2L2(0,)C(T)+122xu(t,)2L2(0,).

    (2.14) follows from (4.1) and an integration on (0,t).

    We prove (2.20). By (3.16) and (1.1)-(1.4)-(1.7), we have that

    xu(t,)2L(0,)xu(t,)2L2(0,)+2xu(t,)2L2(0,) (4.4)

    Therefore, an integration on (0,t), (2.14) and (4.1) gives (2.20).

    Arguing as in Lemma 2.3, we have (2.21).

    Finally, we prove (4.2). We begin by observing that, by the Young inequality,

    u2(t,0)12+12u4(t,0).

    (4.2) follows from (4.1) and an integration on (0,t).

    Lemma 4.2. Fix T>0. There exists a constant C(T)>0, such that (2.26) holds. In particular, we have that

    xu(t,)2L2(0,)+2β2t03xu(s,)2L2(0,)dsC(T), (4.5)
    t0(2xu(s,0))2dsC(T), (4.6)

    for every 0tT.

    Proof. Let 0tT. Multiplying (1.1)-(1.4)-(1.7) by 22xu, thanks to (1.1)-(1.4)-(1.7), an integration on (0,) gives

    ddtxu(t,)2L2(0,)=202xutudx=4κ0uxu4xudx6a20u2xu2xudx+2ν2xu(t,)2L2(0,)+2β202xu4xudx=4κ0uxu4xudx6a20u2xu2xudx+2ν2xu(t,)2L2(0,)2β23xu(t,)2L2(0,).

    Therefore, we have that

    ddtxu(t,)2L2(0,)+2β23xu(t,)2L2(0,)=4κ0uxu2xudx6a20u2xu2xudx+2ν2xu(t,)2L2(0,). (4.7)

    Due to the Young inequality,

    4|κ|0|uxu||4xu|dx2κ2u(t,)xu(t,)2L2(0,)+22xu(t,)2L2(0,),6a20|u2xu||2xu|dx3a40u4(xu)2dx+32xu(t,)2L2(0,)3a2u2L((0,T)×(0,))u(t,)xu(t,)2L2(0,)+32xu(t,)2L2(0,).

    It follows from (4.7) that

    ddtxu(t,)2L2(0,)+2β23xu(t,)2L2(0,)(2κ2+3a2u2L((0,T)×(0,)))u(t,)xu(t,)2L2(0,)+(2|ν|+4)2xu(t,)2L2(0,).

    Integrating on (0,t), by (1.8), (2.21) and (4.1), we have that

    xu(t,)2L2(0,)+2β2t03xu(s,)2L2(0,)dsC0+(2κ2+3a2u2L((0,T)×(0,)))t0u(s,)xu(s,)2L2(0,)ds+(2|ν|+4)t02xu(s,)2L2(0,)dsC(T)(1+u2L((0,T)×(0,))). (4.8)

    We prove (2.26). Thanks to (4.1), (4.8) and the Hölder inequality,

    u2(t,x)=2x0uxudy20uxudx20|u||xu|dx2u(t,)L2(0,))xu(t,)L2(0,)C(T)(1+u2L((0,T)×(0,))).

    Therefore,

    u4L((0,T)×(0,))C(T)u2L((0,T)×(0,))C(T)0, (4.9)

    which gives (2.26).

    (4.5) follows from (2.26) and (4.8).

    Finally, we prove (4.6). Thanks to the Young inequality,

    (2xu(t,0))2=202xu3xudx20|2xu||3xu|dx2xu(t,)2L2(0,)+3xu(t,)2L2(0,).

    (4.1), (4.5) and an integration on (0,t) give (4.6).

    Lemma 4.3. Fix T>0. There exists a constant C(T)>0, such that

    2xu(t,)2L2(0,)+β2t04xu(s,)2L2(0,)dsC(T), (4.10)

    for every 0tT. In particular, (2.25) holds. Moreover, we have (2.34).

    Proof. Let 0tT. Since, by (1.1)-(1.4)-(1.7), 2txu(t,0)=0, multiplying (1.1)-(1.4)-(1.7) by 24xu, thanks to (1.1)-(1.4)-(1.7), an integration on (0,t) gives

    ddt2xu(t,)2L2(0,)=204xutudx=4κ0uxu4xudx+6a20u2xu4xudx2ν02xu4xudx2β24xu(t,)2L2(0,)=4κ0uxu4xudx+6a20u2xu4xudx+2ν3xu(t,)2L2(0,)2β24xu(t,)2L2(0,).

    Therefore, we have

    ddt2xu(t,)2L2(0,)+2β24xu(t,)2L2(0,)=4κ0uxu4xudx+6a20u2xu4xudx+2ν3xu(t,)2L2(0,). (4.11)

    Due to (2.26), (4.5) and the Young inequality,

    4|κ|0|u||xu||4xu|dx4|κ|uL((0,T)×(0,))0|xu||4xu|dxC(T)0|xu||4xu|dx=0|C(T)xuβ||β4xu|dxC(T)xu(t,)2L2(0,)+β224xu(t,)2L2(0,)C(T)+β224xu(t,)2L2(0,),6a20u2|xu||4xu|dx6a2u2L((0,T)×(0,))0|xu||4xu|dxC(T)0|xu||4xu|dx=0|C(T)xuβ||β4xu|dxC(T)xu(t,)2L2(0,)+β224xu(t,)2L2(0,)C(T)+β224xu(t,)2L2(0,).

    It follows from (4.11) that

    ddt2xu(t,)2L2(0,)+β24xu(t,)2L2(0,)C(T)+2|ν|3xu(t,)2L2(0,).

    Integrating on (0,t), by (1.8) and (4.5), we get

    2xu(t,)2L2(0,)+β2t04xu(s,)2L2(0,)dsC0C(T)t+2|ν|t03xu(s,)2L2(0,)dsC(T),

    which gives (4.10).

    (2.25) follows from (4.4), (4.5) and (4.10).

    Finally, we prove (2.34). Multiplying (1.1)-(1.4)-(1.7) by 2tu, an integration on (0,) give

    2tu(t,)2L2(0,)=4κ0uxutudx+6a20u2xutudx2ν02xutudx2β204xutudx. (4.12)

    Due to (2.26), (4.5), (4.10) and the Young inequality,

    4|κ|0|u||xu||tu|dx4|κ|uL((0,T)×(0,))0|xu||tu|dx2C(T)0|xu||tu|dx=20|C(T)xuD4||D4tu|dxC(T)D4xu(t,)2L2(0,)+D4tu(t,)2L2(0,)C(T)D4+D4tu(t,)2L2(0,),6a20u2|xu||tu|dx6a2u2L((0,T)×(0,))0|xu||tu|dx2C(T)0|xu||tu|dx=20|C(T)xuD4||D4tu|dxC(T)D4xu(t,)2L2(0,)+D4tu(t,)2L2(0,)C(T)D4+D4tu(t,)2L2(0,),2|ν|0|2xu||tu|dx=20|ν2xuD4||D4tu|dxν2D42xu(t,)2L2(0,)+D4tu(t,)2L2(0,)C(T)D4+D4tu(t,)2L2(0,),2β20|4xu||tu|dx=20|β24xuD2||D4tu|dxβ4D44xu(t,)2L2(0,)+D4tu(t,)2L2(0,).

    Therefore, by (4.12),

    2(12D4)tu(t,)2L2(0,)C(T)D4+β4D44xu(t,)2L2(0,).

    Choosing D4=14, we have

    tu(t,)2L2(0,)C(T)+4β44xu(t,)2L2(0,).

    An integration on (0,t) and (4.10) give (2.34).

    Arguing as in Section 2, we have Theorem 1.1.

    In this section, we prove Theorem 1.1 for (1.1)-(1.5)-(1.7).

    Let us prove some a priori estimates on u.

    We begin by proving the following lemma.

    Lemma 5.1. Fix T>0. There exists a constant C(T)>0, such that

    u(t,)2L(0,)+β2eC0t6t0eC0s2xu(s,)2L2(0,)ds+a2eC0t2t0eC0su4(s,0)dsC(T). (5.1)

    for every 0tT. In particular, we have (2.14), (2.20), (2.21) and (4.2).

    Proof. Let 0tT. We begin by observing that, thanks to (1.1)-(1.5)-(1.7),

    4κ0u2xudx=4κ3u3(t,0),6a20u2xudx=3a22u4(t,0),2δx0u3xu=2δ0xu2xudx,2β20u4xudx=2β20u3xudx=2β22xu(t,)2L2(0,). (5.2)

    Therefore, multiplying (1.1)-(1.5)-(1.7) by 2u, thanks to (5.2), an integration on (0,) gives

    ddtu(t,)2L2(0,)+2β22xu(t,)2L2(0,)+3a22u4(t,0)=4κ3u2(t,0)2ν0u2xudx+2δ0xu2xudx. (5.3)

    Due to the Young inequality,

    4|κ|3u2(t,0)8κ29a2u2(t,0)+a22u4(t,0),2|ν|0|u||2xu|dx=0|2νuβ||β2xu|dx2ν2β2u(t,)2L2(0,)+β222xu(t,)2L2(0,),2|δ|0|xu||2xu|dx=0|2δxuβ||β4xu|dx2δ2β2xu(t,)2L2(0,)+β222xu(t,)2L2(0,).

    It follows from (5.3) that

    ddtu(t,)2L2(0,)+β22xu(t,)2L2(0,)+a2u4(t,0)8κ29a2u2(t,0)+2ν2β2u(t,)2L2(0,)+2δ2β2xu(t,)2L2(0,)C0u2(t,0)+C0u(t,)2L2(0,)+C0xu(t,)2L2(0,). (5.4)

    Observe that, by the Hölder inequality,

    xu(t,)2L2(0,)=0xuxudx=u(t,0)xu(t,0)0u2xudx|u(t,0)||xu(t,0)|+0|u||2xu|dx|u(t,0)||xu(t,0)|+u(t,)L2(0,)2xu(t,)L2(0,). (5.5)

    Moreover, by the Young inequality,

    |u(t,0)||xu(t,0)|12D5u2(t,0)+D52(xu(t,0))2, (5.6)

    where D5 is a positive constant, which will be specified later. Observe that, by the Young inequality,

    D52(xu(t,0))2=D50xu2xudx0|xu||D52xu|dx12xu(t,)2L2(0,)+D252xu(t,)2L2(0,). (5.7)

    It follows from (5.5), (5.6) and (5.7) that

    12xu(t,)2L2(0,)12D5u2(t,0)+D252xu(t,)2L2(0,)+u(t,)L2(0,)2xu(t,)L2(0,),

    that is

    xu(t,)2L2(0,)1D5u2(t,0)+2D252xu(t,)2L2(0,)+2u(t,)L2(0,)2xu(t,)L2(0,). (5.8)

    Due to the Young inequality,

    2u(t,)L2(0,)2xu(t,)L2(0,)1D6u(t,)2L2(0,)+D62xu(t,)2L2(0,),

    where D6 is a positive constant, which will be specified later. It follows from (5.8) that

    xu(t,)2L2(0,)1D5u2(t,0)+2D252xu(t,)2L2(0,)+1D6u(t,)2L2(0,)+D62xu(t,)2L2(0,). (5.9)

    Consequently, by (5.4) and (5.9),

    ddtu(t,)2L2(0,)+(β22D25D6)2xu(t,)2L2(0,)+a2u4(t,0)(C0+1D5)u2(t,0)+(C0+1D6)u(t,)2L2(0,).

    Choosing

    D5=|β|2,D6=13, (5.10)

    we have that

    ddtu(t,)2L2(0,)+β262xu(t,)2L2(0,)+a2u4(t,0)C0u2(t,0)+C0u(t,)2L2(0,). (5.11)

    Due to the Young inequality,

    C0u2(t,0)C0+a22u4(t,0).

    It follows from (5.11) that

    ddtu(t,)2L2(0,)+β262xu(t,)2L2(0,)+a22u4(t,0)C0+C0u(t,)2L2(0,).

    By the Gronwall Lemma and (1.8), we get

    u(t,)2L2(0,)+β2eC0t6t0eC0s2xu(s,)2L2(0,)ds+a2eC0t2t0eC0su4(s,0)dsC0+C0eC0tt0eC0sdsC(T),

    which gives (5.1).

    Arguing as in Lemma 4.1, we have (4.2), while (4.2), (5.1), (5.9), (5.10) and an integration on (0,t) give (2.14).

    We prove (2.20). By (3.16), (5.7) and (5.10), we have that

    xu(t,)2L(0,)C0xu(t,)2L2(0,)+C02xu(t,)2L2(0,).

    (2.14), (5.1) and an integration on (0,t) give (2.20).

    Arguing as in Lemma 2.3, we have (2.21), while (3.15) follows from (2.14), (5.1), (5.7), (5.10) and an integration on (0,t).

    Lemma 5.2. Fix T>0. There exists a constant C(T)>0, such that (2.26) holds. In particular,

    2xu(t,)2L2(0,)+β22t04xu(s,)2L2(0,)dsC(T), (5.12)

    for every 0tT. Moreover, we have (2.25),

    xu(t,)L2(0,)C(T), (5.13)
    t03xu(s,)2L2(0,)dsC(T), (5.14)

    for every 0tT.

    Proof. Let 0tT. We begin by observing that, thanks (1.1)-(1.5)-(1.7),

    204xutudx=203xutxudx=ddt2xu(t,)2L2(0,),2δ03xu4xudx=0. (5.15)

    Therefore, multiplying (1.1)-(1.5)-(1.7) by 24xu, by (5.15) an integration on (0, gives

    ddt2xu(t,)2L2(0,)+2β24xu(t,)2L2(0,)=4κ0uxu4xudx+6a20u2xu4xudx2ν02xu4xudx. (5.16)

    Due to the Young inequality,

    4|κ|0|uxu||4xu|dx=0|4κuxuβ||β4xu|dx8κ2β2u(t,)xu(t,)2L2(0,)+β224xu(t,)2L2(0,),6a20|u2xu||4xu|dx=0|6a2u2xuβ||β4xu|dx18a4β20u4(xu)2dx+β224xu(t,)2L2(0,)18a4β2u2L((0,)×(0infty))u(t,)xu(t,)2L2(0,)+β224xu(t,)2L2(0,),2|ν|0|2xu||4xu|dx=0|2ν2xuβ||β4xu|dx2ν2β22xu(t,)2L2(0,)+β224xu(t,)2L2(0,).

    It follows from (5.16) that

    ddt2xu(t,)2L2(0,)+β224xu(t,)2L2(0,)C0(1+u2L((0,)×(0,)))u(t,)xu(t,)2L2(0,)+C02xu(t,)2L2(0,).

    Integrating on (0,), by (1.8), (2.21) and (5.1), we have that

    2xu(t,)2L2(0,)+β22t04xu(s,)2L2(0,)dsC0+C0(1+u2L((0,)×(0,)))t0u(s,)xu(s,)2L2(0,)ds+C0t02xu(s,)2L2(0,)dsC(T)(1+u2L((0,)×(0,))). (5.17)

    We prove that

    xu(t,)2L2(0,)C(T)(1+u2L((0,)×(0,))). (5.18)

    We begin by observing that, by (5.1), (5.9) and (5.10),

    xu(t,)2L2(0,)C0u2(t,0)+C(T)(1+2xu(t,)2L2(0,)). (5.19)

    Due to (5.1) and the Young inequality,

    C0u2(t,0)=2C00uxudx2C00|u||xu|dxC0u(t,)L2(0,)+12xu(t,)2L2(0,)C(T)+12xu(t,)2L2(0,).

    Therefore, by (5.18),

    xu(t,)2L2(0,)C(T)(1+2xu(t,)2L2(0,)). (5.20)

    (5.17) and (5.19) give (5.20).

    We prove (2.26). Thanks to (5.1), (5.19) and the Hölder inequality, we have (4.9), which gives (2.26).

    (5.12) and (5.13) follows from (2.26), (5.17) and (5.18), respectively, while (3.16), (5.7), (5.10) (5.12) and (5.13) gives (2.25).

    Finally, we prove (5.14). We begin by observing that, thanks to (1.1)-(1.5)-(1.7),

    3xu(t,)2L2(0,)=03xu3xudx=02xu3xudx.

    Therefore, by (5.12) and the Young inequality,

    3xu(t,)2L2(0,)0|2xu||3xu|dx122xu(t,)2L2(0,)+124xu(t,)2L2(0,)C(T)++124xu(t,)2L2(0,).

    (5.14) follows from (5.12) and an integration on (0,t).

    Lemma 5.3. Fix T>0. There exists a constant C(T)>0, such that (2.34) holds.

    Proof. Let 0tT. Arguing as in Lemma 4.3, we have that

    tu(t,)2L2(0,)C(T)+4β44xu(t,)2L2(0,)+2|δ|0|3xu|tu|dx. (5.21)

    Due to the Young inequality,

    2|δ|0|3xu|tu|dx2δ23xu(t,)2L2(0,)+12tu(t,)2L2(0,).

    Therefore, by (5.21), we have

    12tu(t,)2L2(0,)C(T)+4β44xu(t,)2L2(0,)+2δ23xu(t,)2L2(0,).

    (2.34) follows from (5.12), (5.14) and an integration on (0,t).

    Arguing as in Section 2, we have Theorem 1.1.

    In this section, we prove Theorem 1.1 for (1.1)-(1.6)-(1.7).

    Let us prove some a priori estimates on u.

    We begin by proving the following lemma.

    Lemma 6.1. Fix T>0. There exists a constant C(T)>0, such that

    u(t,)2L2(0,)+β2eC0t2t0eC0s2xu(s,)2L2(0,)dsC(T), (6.1)

    for every 0tT. In particular, we have (2.14), (2.20) and (2.21).

    Proof. Let 0tT. We begin by observing that, thanks to (1.1)-(1.6)-(1.7),

    4κ0u2xudx=0,6q0u3xudx=0,2ν0u2xudx=2νxu(t,)2L2(0,),2δ0u3xudx=2δ0xu2xudx,2β20u4xudx=2β20xu3xudx=2β22xu(t,)2L2(0,). (6.2)

    Therefore, multiplying (1.1)-(1.6)-(1.7) by 2v, thanks to (6.2), an integration on 0,) gives

    ddtu(t,)2L2(0,)+2β22xu(t,)2L2(0,)=2νxu(t,)2L2(0,)2δ0xu2xudx. (6.3)

    Due to the Young inequality,

    2|δ|0|xu||2xu|dx=20|δxuβ||β2xu|dxδ2β2xu(t,)2L2(0,)+β22xu(t,)2L2(0,).

    Consequently, by (6.3),

    ddtu(t,)2L2(0,)+β22xu(t,)2L2(0,)C0xu(t,)2L2(0,). (6.4)

    Observe that, by (5.5) and (1.1)-(1.6)-(1.7),

    C0xu(t,)2L2(0,)C0u(t,)L2(0,)2xu(t,)2L2(0,)

    Therefore, by the Young inequality,

    C0xu(t,)2L2(0,)C0u(t,)2L2(0,)+β222xu(t,)L2(0,). (6.5)

    It follows from (6.4) that

    ddtu(t,)2L2(0,)+β222xu(t,)2L2(0,)C0u(t,)2L2(0,).

    By the Gronwall Lemma and (1.8).

    u(t,)2L2(0,)+β2eC0t2t0eC0s2xu(s,)2L2(0,)dsC0eC0tC(T),

    which gives (6.1).

    (2.14) follows from (6.1) and an integration on (0,t), while, arguing as in Lemma 2.3, we have (2.20) and (2.21).

    Lemma 6.2. Fix T>0. There exists a constant C(T)>0, such that (2.26) holds. In particular,

    2xu(t,)2L2(0,)+4(4δ2+4)β4xu(t,)2L2(0,)+β22t04xu(s,)2L2(0,)ds+4δ2+2β2t03xu(s,)2L2(0,)dsC(T), (6.6)

    for every 0tT. Moreover, we have (2.24), (2.25) and (2.34).

    Proof. Let 0tT. Consider an positive constant B, which will be specified later. Observing that, since, thanks to (1.1)-(1.6)-(1.7), tu(t,0=txu(t,0), we have that

    204xutudx=203xutxudx=ddt2xu(t,)2L2(0,),2B02xutudx=Bddtxu(t,)2L2(0,),2Bβ202xu4xudx=2Bβ22xu(t,0)3xu(t,0)+2Bβ23xu(t,)2L2(0,). (6.7)

    Therefore, multiplying (1.1)-(1.6)-(1.7) by 24xu, thanks to (6.7), an integration on (0,) gives

    ddt(2xu(t,)2L2(0,)+Bxu(t,)2L2(0,))+2β24xu(t,)2L2(0,)+2Bβ23xu(t,)2L2(0,)=2Bβ22xu(t,0)3xu(t,0)4κ0uxu4xudx+4Bκ0uxu2xudx6q0u2xu4xudx+6Bq0u2xu2xudx2ν02xu4xudx+2Bν2xu(t,)2L2(0,)2δ03xu4xudx+2Bδ03xu2xudx. (6.8)

    Due to the Young inequality,

    2Bβ22xu(t,0)3xu(t,0)β4B2(2xu(t,0)2+(3xu(t,0))24|κ|0|uxu||4xu|dx=20|2κuxuβD7||βD74xu|dx4κ2β2D7u(t,)xu(t,)2L2(0,)+D7β24xu(t,)2L2(0,),4B|κ|0|uxu||2xu|dxu(t,)xu(t,)2L2(0,)+4B2κ22xu(t,)2L2(0,),6|q|0|u2xu||4xu|dx=20|3qu2xuβD7||βD74xu|dx9q2β2D70u4(xu)2dx+D7β24xu(t,)2L2(0,)9q2β2D7u2L((0,T)×(0,))u(t,)xu(t,)2L2(0,)+D7β24xu(t,)2L2(0,)6B|q|0|u2xu||2xu|dx0u4(xu)2dx+9q2B22xu(t,)2L2(0,)u2L((0,T)×(0,))u(t,)xu(t,)2L2(0,)+9q2B22xu(t,)2L2(0,)2|ν|0|2xu||4xu|dx=20|ν2xuβD7||βD74xu|dxν2β22xu(t,)2L2(0,)+β2D74xu(t,)2L2(0,),2|δ|0|3xu||4xu|dx=20|δ3xuβD7||βD74xu|dxδ2β2D73xu(t,)2L2(0,)+β2D74xu(t,)2L2(0,),2B|δ|0|3xu||2xu|dx=2B0|β3xu||δ2xuβ|dxBβ23xu(t,)2L2(0,)+Bδ2β22xu(t,)2L2(0,),

    where D7 is a positive constant, which will be specified later. It follows from (6.8) that

    ddt(2xu(t,)2L2(0,)+Bxu(t,)2L2(0,))+(24D7)β24xu(t,)2L2(0,)+(Bβ2δ2β2D7)3xu(t,)2L2(0,)C0B2(2xu(t,0)2+(3xu(t,0))2+C0(1+14+B2+B)2xu(t,)2L2(0,)+C0(1+1D7+(1+1D7)u2L((0,T)×(0,)))u(t,)xu(t,)2L2(0,).

    Choosing D7=14, we have that

    ddt(2xu(t,)2L2(0,)+Bxu(t,)2L2(0,))+β24xu(t,)2L2(0,)+(Bβ24δ2β2)3xu(t,)2L2(0,)B2(2xu(t,0)2+(3xu(t,0))2+C0(1+B2+B)2xu(t,)2L2(0,)+C0(1+u2L((0,T)×(0,)))u(t,)xu(t,)2L2(0,). (6.9)

    Observe that by the Young inequality,

    B2(2xu(t,0)2=B202xu3xudx0|2BB2xuβ||Bβ3xu|dxB3β22xu(t,)2L2(0,)+Bβ223xu(t,)2L2(0,),(3xu(t,0))2=203xu4xudx≤=0|23xuβ||β4xu|dx2β23xu(t,)2L2(0,)+β224xu(t,)2L2(0,).

    Consequently, by (6.9),

    ddt(2xu(t,)2L2(0,)+Bxu(t,)2L2(0,))+β224xu(t,)2L2(0,)+(Bβ224δ2+2β2)3xu(t,)2L2(0,)C0(1+u2L((0,T)×(0,)))u(t,)xu(t,)2L2(0,)+C0(1+B3+B2+B)2xu(t,)2L2(0,).

    Choosing

    B=4(4δ2+4)β4,

    we have that

    ddt(2xu(t,)2L2(0,)+4(4δ2+4)β4xu(t,)2L2(0,))+β224xu(t,)2L2(0,)+4δ2+2β23xu(t,)2L2(0,)C0(1+u2L((0,T)×(0,)))u(t,)xu(t,)2L2(0,)+C02xu(t,)2L2(0,).

    Integrating on (0,t), by (1.8), (2.21) and (6.1), we get

    2xu(t,)2L2(0,)+4(4δ2+4)β4xu(t,)2L2(0,)+β22t04xu(s,)2L2(0,)ds+4δ2+2β2t03xu(s,)2L2(0,)dsC0+C0(1+u2L((0,T)×(0,)))t0u(s,)xu(s,)2L2(0,)ds+C0t02xu(s,)2L2(0,)dsC(T)(1+u2L((0,T)×(0,))). (6.10)

    We prove (2.26). Thanks to (6.1), (6.10) and the Hölder inequality, we have (4.9), which gives (2.26).

    (6.6) follows from (2.26) and (6.10).

    Finally, arguing as in Lemma 2.4, we have (2.24) and (2.25), while arguing as in Lemma 5.3, (2.34) holds.

    Arguing as in Section 2, we have Theorem 1.1.

    The authors are members of the Gruppo Nazionale per l'Analisi Matematica, la Probabilità e le loro Applicazioni (GNAMPA) of the Istituto Nazionale di Alta Matematica (INdAM).

    The authors declare that there are no conflict of interest.



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