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Synchronization for a class of complex-valued memristor-based competitive neural networks(CMCNNs) with different time scales

  • In this paper, the synchronization problem of complex-valued memristive competitive neural networks(CMCNNs) with different time scales is investigated. Based on differential inclusions and inequality techniques, some novel sufficient conditions are derived to ensure synchronization of the drive-response systems by designing a proper controller. Finally, a numerical example is provided to illustrate the usefulness and feasibility of our results.

    Citation: Yong Zhao, Shanshan Ren. Synchronization for a class of complex-valued memristor-based competitive neural networks(CMCNNs) with different time scales[J]. Electronic Research Archive, 2021, 29(5): 3323-3340. doi: 10.3934/era.2021041

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  • In this paper, the synchronization problem of complex-valued memristive competitive neural networks(CMCNNs) with different time scales is investigated. Based on differential inclusions and inequality techniques, some novel sufficient conditions are derived to ensure synchronization of the drive-response systems by designing a proper controller. Finally, a numerical example is provided to illustrate the usefulness and feasibility of our results.



    The equation:

    {tu+xf(u)β22xu+δ3xu+κu+γ2|u|u=0,0<t<T,xR,u(0,x)=u0(x),xR, (1.1)

    was originally derived in [14,17] with f(u)=au2 focusing on microbubbles coated by viscoelastic shells. These structures are crucial in ultrasound diagnosis using contrast agents, and the dynamics of individual coated bubbles are explored, taking into account nonlinear competition and dissipation factors such as dispersion, thermal effects, and drag force.

    The coefficients β2, δ, κ, and γ2 are related to the dissipation, the dispersion, the thermal conduction dissipation, and to the drag force, repsctively.

    If κ=γ=0, we obtain the Kudryashov-Sinelshchikov [18] Korteweg-de Vries-Burgers [3,20] equation

    tu+axu2β22xu+δ3xu=0, (1.2)

    that models pressure waves in liquids with gas bubbles, taking into account heat transfer and viscosity. The mathematical results on Eq (1.2) are the following:

    ● analysis of exact solutions in [13],

    ● existence of the traveling waves in [2],

    ● well-posedness and asymptotic behavior in [7,11].

    If β=0, we derive the Korteweg-de Vries equation:

    tu+axu2+δ3xu=0, (1.3)

    which describes surface waves of small amplitude and long wavelength in shallow water. Here, u(t,x) represents the wave height above a flat bottom, x corresponds to the distance in the propagation direction, and t denotes the elapsed time. In [4,6,10,12,15,16], the completele integrability of Eq (1.3) and the existence of solitary wave solutions are proved.

    Through the manuscript, we will assume

    ● on the coefficients

    β,δ,κ,γR,β,δ,γ0; (1.4)

    ● on the flux f, one of the following conditions:

    f(u)=au2+bu3, (1.5)
    fC1(R),|f(u)|C0(1+|u|),uR, (1.6)

    for some positive constant C0;

    ● on the initial value

    u0H1(R). (1.7)

    The main result of this paper is the following theorem.

    Theorem 1.1. Assume Eqs (1.5)–(1.7). For fixed T>0, there exists a unique distributional solution u of Eq (1.1), such that

    uL(0,T;H1(R))L4(0,T;W1,4(R))L6(0,T;W1,6(R))2xuL2((0,T)×R). (1.8)

    Moreover, if u1 and u2 are solutions to Eq (1.1) corresponding to the initial conditions u1,0 and u2,0, respectively, it holds that:

    u1(t,)u2(t,)L2(R)eC(T)tu1,0u2,0L2(R), (1.9)

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

    Observe that Theorem 1.1 gives the well-posedness of (1.1), without conditions on the constants. Moreover, the proof of Theorem 1.1 is based on the Aubin-Lions Lemma [5,21]. The analysis of Eq (1.1) is more delicate than the one of Eq (1.2) due to the presence of the nonlinear sources and the very general assumptions on the coefficients.

    The structure of the paper is outlined as follows. Section 2 is dedicated to establishing several a priori estimates for a vanishing viscosity approximation of Eq (1.1). These estimates are crucial for proving our main result, which is presented in Section 3.

    To establish existence, we utilize a vanishing viscosity approximation of equation (1.1), as discussed in [19]. Let 0<ε<1 be a small parameter, and denote by uεC([0,T)×R) the unique classical solution to the following problem [1,9]:

    {tuε+xf(uε)β22xuε+δ3xuε+κu+γ2|u|u=ε4xuε,0<t<T,xR,uε(0,x)=uε,0(x),xR, (2.1)

    where uε,0 is a C approximation of u0, such that

    uε,0H1(R)u0H1(R). (2.2)

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

    We begin by proving the following lemma:

    Lemma 2.1. Let T>0 be fixed. There exists a constant C(T)>0, which does not depend on ε, such that

    uε(t,)2L2(R)+2γ2e|κ|tt0Re|κ|su2ε|uε|dsdx+2β2e|κ|tt0e|κ|sxuε(s,)2L2(R)ds+2εe|κ|tt0e|κ|s2xuε(s,)2L2(R)C(T), (2.3)

    for every 0tT.

    Proof. For 0tT. Multiplying equations (2.1) by 2uε, and integrating over R yields

    ddtuε(t,)2L2(R)=2Ruεtuεdx=2Ruεf(uε)xuεdx=0+2β2Ruε2xuεdx2δRuε3xuεdxκuε(t,)2L2(R)2γ2R|uε|u2εdx2εRuε4xuεdx=2β2xuε(t,)2L2(R)+2δRxuε2xuεdxκuε(t,)2L2(R)2γ2R|uε|u2εdx+2εRxuε3xuεdx=2β2xuε(t,)2L2(R)κuε(t,)2L2(R)2γ2R|uε|u2εdx2ε2xuε(t,)2L2(R).

    Thus, it follows that

    ddtuε(t,)2L2(R)+2β2xuε(t,)2L2(R)+2γ2R|uε|u2εdx+2ε2xuε(t,)2L2(R)=κuε(t,)2L2(R)|κ|uε(t,)2L2(R).

    Therefore, applying the Gronwall's lemma and using Eq (2.2), we obtain

    uε(t,)2L2(R)+2β2e|κ|tt0e|κ|sxuε(s,)2L2(R)ds+2γ2e|κ|tt0Re|κ|t|uε|u2εdsdx+2ε2xuε(t,)2L2(R)+2εe|κ|tt0e|κ|s2xuε(s,)2L2(R)dsC0e|κ|tC(T),

    which gives Eq (2.3).

    Lemma 2.2. Fix T>0 and assume (1.5). There exists a constant C(T)>0, independent of ε, such that

    uεL((0,T)×R)C(T), (2.4)
    xuε(t,)2L2(R)+β2t02xuε(s,)2L2(R)ds (2.5)
    +2εt03xuε(s,)2L2(R)dsC(T),t0xuε(s,)4L4(R)dsC(T), (2.6)

    holds for every 0tT.

    Proof. Let 0tT. Consider A,B as two real constants, which will be specified later. Thanks to Eq (1.5), multiplying Eq (2.1) by

    22xuε+Au2ε+Bu3ε,

    we have that

    (22xuε+Au2ε+Bu3ε)tuε+2a(22xuε+Au2ε+Bu3ε)uεxuε+3b(22xuε+Au2ε+Bu3ε)u2εxuεβ2(22xuε+Au2ε+Bu3ε)2xuε+δ(22xuε+Au2ε+Bu3ε)3xuε+κ(22xuε+Au2ε+Bu3ε)uε+γ2(22xuε+Au2ε+Bu3ε)|uε|uε=ε(22xuε+Au2ε+Bu3ε)4xuε. (2.7)

    Observe that

    R(22xuε+Au2ε+Bu3ε)tuεdx=ddt(xuε(t,)2L2(R)+A3Ru3εdx+B4Ru4εdx),2aR(22xuε+Au2ε+Bu3ε)uεxuεdx=4aRuεxuε2xuεdx,3bR(22xuε+Au2ε+Bu3ε)u2εxuεdx=6bRu2εxuε2xuεdx,β2R(22xuε+Au2ε+Bu3ε)2xuεdx=2β22xuε(t,)2L2(R)+2Aβ2Ruε(xuε)2dx+3Bβ2Ru2ε(xuε)2dx,δR(22xuε+Au2ε+Bu3ε)3xuεdx=2AδRuεxuε2xuεdx3BδRu2εxuε2xuεdx,κR(22xuε+Au2ε+Bu3ε)uεdx=2κxuε(t,)2L2(R)+AκRu3εdx+BκRu4εdx,γ2R(22xuε+Au2ε+Bu3ε)|uε|uεdx=2γ2R|uε|uε2xuεdx+Aγ2R|u|u3εdx+Bγ2R|uε|u4dx,εR(22xuε+Au2ε+Bu3ε)4xuεdx=2ε3xuε(t,)2L2(R)+2AεRuεxuε3xuεdx+3BεRu2εxuε3xuεdx=2ε3xuε(t,)2L2(R)AεR(xuε)3dx6BεRuε(xuε)22xuεdx3Bεuε(t,)2xuε(t,)2L2(R)=2ε3xuε(t,)2L2(R)AεR(xuε)3dx+2BεR(xuε)4dx3Bεuε(t,)2xuε(t,)2L2(R).

    Therefore, an integration on R gives

    ddt(xuε(t,)2L2(R)+A3Ru3εdx+B4Ru4εdx)+β22xuε(t,)2L2(R)+2ε3xuε(t,)2L2(R)=(4a+Aδ)Ruεxuε2xuεdx3(2b+Bδ)Ru2εxuε2xuεdx2Aβ2Ruε(xuε)2dx3Bβ2Ru2ε(xuε)2dxκxuε(t,)2L2(R)Aκ3Ru3εdxBκ4Ru4εdx+2γ2R|uε|uε2xuεdxAγ2R|uε|u3εdxBγ2R|uε|u4εdxAεR(xuε)3dx+2BεR(xuε)4dx3Bεuε(t,)2xuε(t,)2L2(R).

    Taking

    (A,B)=(4aδ,2bδ),

    we get

    ddt(xuε(t,)2L2(R)4a3δRu3εdxbδRu4εdx)+2β22xuε(t,)2L2(R)+2ε3xuε(t,)2L2(R)=8aβ2δRuε(xuε)2dx+6bβ2δRu2ε(xuε)2dxκxuε(t,)2L2(R)+4aκ3δRu3εdx+bκ2Ru4εdx+2γ2R|uε|uε2xuεdx+4aγ2δR|uε|u3εdx+2bγ2δR|uε|u4εdx+4aεδR(xuε)3dx4bεδR(xuε)4dx+6bεδuε(t,)2xuε(t,)2L2(R). (2.8)

    Since 0<ε<1, due to the Young inequality and (2.3),

    8aβ2δR|uε|(xuε)2dx4Ru2ε(xuε)2dx+4a2β4δ2xuε(t,)2L2(R)4uε2L((0,T)×R)xuε(t,)2L2(R)+4a2β4δ2xuε(t,)2L2(R)C0(1+uε2L((0,T)×R))xuε(t,)2L2(R),|6bβ2δ|Ru2ε(xuε)2dx|6bβ2δ|uε2L((0,T)×R)xuε(t,)2L2(R),|4aκ3δ|R|uε|3dx|4aκ3δ|uεL((0,T)×R)uε(t,)2L2(R)C(T)uεL((0,T)×R),|bκ2|Ru4εdx|bκ2|uε2L((0,T)×R)uε(t,)2L2(R)C(T)uε2L((0,T)×R),2γ2R|uε|uε2xuεdx2R|γ2|uε|uεβ||β2xuε|dxγ4β2Ruε4dx+β22xuε(t,)2L2(R)γ4β2uε2L((0.T)×R)uε(t,)2L2(R)+β22xuε(t,)2L2(R)C(T)uε2L((0,T)×R)+β22xuε(t,)2L2(R),|4aγ2δ|R|uε||uε|3dx=|4aγ2δ|Ru4εdx|4aγ2δ|uε2L((0,T)×R)uε(t,)2L2(R)C(T)uε2L((0,T)×R),|2bγ2δ|R|uε|uε4dx|2bγ2δ|uε3L((0,T)×R)uε(t,)2L2(R)C(T)uε3L((0,T)×R),|4aεδ|R|xuε|3dx|4aεδ|xuε(t,)2L2(R)+|4aεδ|R(xuε)4dx|4aδ|xuε(t,)2L2(R)+|4aεδ|R(xuε)4dx.

    It follows from Eq (2.8) that

    ddt(xuε(t,)2L2(R)4a3δRu3εdxbδRu4εdx)+β22xuε(t,)2L2(R)+2ε3xuε(t,)2L2(R)C0(1+uε2L((0,T)×R))xuε(t,)2L2(R)+C(T)uεL((0,T)×R)+C(T)uε2L((0,T)×R)+C(T)uε3L((0,T)×R)+C0εR(xuε)4dx+C0εuε(t,)2xuε(t,)2L2(R)+C0xuε(t,)2L2(R). (2.9)

    [8, Lemma 2.3] says that

    R(xuε)4dx9Ru2ε(2xuε)2dx9uε2L((0,T)×R)2xuε(t,)2L2(R). (2.10)

    Moreover, we have that

    uε(t,)2xuε(t,)2L2(R)=Ru2ε(2xuε)2dxuε2L((0,T)×R)2xuε(t,)2L2(R). (2.11)

    Consequentially, by Eqs (2.9)–(2.11), we have that

    ddt(xuε(t,)2L2(R)4a3δRu3εdxbδRu4εdx)+β22xuε(t,)2L2(R)+2ε3xuε(t,)2L2(R)C0(1+uε2L((0,T)×R))xuε(t,)2L2(R)+C(T)uεL((0,T)×R)+C(T)uε2L((0,T)×R)+C(T)uε3L((0,T)×R)+C0εuε2L((0,T)×R)2xuε(t,)2L2(R)+C0xuε(t,)2L2(R).

    An integration on (0,t) and Eqs (2.2) and (2.3) give

    xuε(t,)2L2(R)4a3δRu3εdxbδRu4εdx+β2t02xuε(s,)2L2(R)ds+2εt03xuε(s,)2L2(R)dsC0(1+uε2L((0,T)×R))t0xuε(s,)2L2(R)ds+C(T)uεL((0,T)×R)t+C(T)uε2L((0,T)×R)t+C(T)uε3L((0,T)×R)t+C0εuε2L((0,T)×R)t02xuε(s,)2L2(R)ds+C0t0xuε(s,)2L2(R)dsC(T)(1+uεL((0,T)×R)+uε2L((0,T)×R)+uε3L((0,T)×R)).

    Therefore, by Eq (2.3),

    xuε(t,)2L2(R)+β2t02xuε(s,)2L2(R)ds+2εt03xuε(s,)2L2(R)dsC(T)(1+uεL((0,T)×R)+uε2L((0,T)×R)+uε3L((0,T)×R))+4a3δRu3εdx+bδRu4εdxC(T)(1+uεL((0,T)×R)+uε2L((0,T)×R)+uε3L((0,T)×R))+|4a3δ|R|uε|3dx+|bδ|Ru4εdxC(T)(1+uεL((0,T)×R)+uε2L((0,T)×R)+uε3L((0,T)×R))+|4a3δ|uεL((0,T)×R)uε(t,)2L2(R)+|bδ|uε2L((0,T)×R)uε(t,)2L2(R)C(T)(1+uεL((0,T)×R)+uε2L((0,T)×R)+uε3L((0,T)×R)). (2.12)

    We prove Eq (2.4). Thanks to the Hölder inequality,

    u2ε(t,x)=2xuεxuεdx2R|uε||xuε|dx2uε(t,)L2(R)xuε(t,)L2(R).

    Hence, we have that

    uε(t,)4L(R)4uε(t,)2L2(R)xuε(t,)2L2(R). (2.13)

    Thanks to Eqs (2.3) and (2.12), we have that

    uε4L((0,T)×R)C(T)(1+uεL((0,T)×R)+uε2L((0,T)×R)+uε3L((0,T)×R)). (2.14)

    Due to the Young inequality,

    C(T)uε3L((0,T)×R)12uε4L((0,T)×R)+C(T)uε2L((0,T)×R),C(T)uεL((0,T)×R)C(T)uε2L((0,T)×R)+C(T).

    By Eq (2.14), we have that

    12uε4L((0,T)×R)C(T)uε2L((0,T)×R)C(T)0,

    which gives Eq (2.4).

    Equation (2.5) follows from Eqs (2.4) and (2.12).

    Finally, we prove Eq (2.6). We begin by observing that, from Eqs (2.4) and (2.10), we have

    xuε(t,)4L4(R)C(T)2xuε(t,)2L2(R).

    An integration on (0,t) and Eqs (2.5) give Eq (2.6).

    Lemma 2.3. Fix T>0 and assume (1.6). There exists a constant C(T)>0, independent of ε, such that Eq (2.4) holds. Moreover, we have Eqs (2.5) and (2.6).

    Proof. Let 0tT. Multiplying Eq (2.1) by 22xuε, an integration on R gives

    ddtxuε(t,)2L2(R)=2R2xuεtuεdx=2Rf(uε)xuε2xuεdx2β22xuε(t,)2L2(R)2δR2xuε3xuεdx2κRuε2xuεdx2γ2R|uε|uε2xuεdx+2εR2xuε4xuεdx=2Rf(uε)xuε2xuεdx2β22xuε(t,)2L2(R)+2κxuε(t,)2L2(R)+2γ2R|uε|uε2xuεdx2ε3xuε(t,)2L2(R).

    Therefore, we have that

    ddtxuε(t,)2L2(R)+2β22xuε(t,)2L2(R)+2ε3xuε(t,)2L2(R)=2Rf(uε)xuε2xuεdx+2κxuε(t,)2L2(R)+2γ2R|uε|uε2xuεdx. (2.15)

    Due Eqs (1.6) and (2.3) and the Young inequality,

    2R|f(uε)||xuε||2xuε|dxC0R|xuε2xuε|dx+C0R|uεxuε||2xuε|dx=2R|C03xuε2β||β2xuε3|dx+2R|C03uεxuε2β||3β2xuε|dxC0xuε(t,)2L2(R)+C0Ru2ε(xuε)2dx+2β232xuε(t,)2L2(R)C0xuε(t,)2L2(R)+C0uε2L((0,T)×R)xuε(t,)2L2(R)+2β232xuε(t,)2L2(R)C0(1+uε2L((0,T)×R))xuε(t,)2L2(R)+2β232xuε(t,)2L2(R),2γ2R|uε|uε2xuεdx2γ2Ru2ε|2xuε|dx=2R|3γ2u2εβ||β2xuε3|dx3γ4β2Ru4εdx+β232xuε(t,)2L2(R)3γ4β2uε2L((0,T)×R)uε(t,)2L2(R)+β232xuε(t,)2L2(R)C(T)uε2L((0,T)×R)+β232xuε(t,)2L2(R).

    It follows from Eq (2.15) that

    ddtxuε(t,)2L2(R)+β22xuε(t,)2L2(R)+2ε3xuε(t,)2L2(R)C0(1+uε2L((0,T)×R))xuε(t,)2L2(R)+C(T)uε2L((0,T)×R).

    Integrating on (0,t), by Eq (2.3), we have that

    xuε(t,)2L2(R)+β2t02xuε(s,)2L2(R)ds+2εt03xuε(s,)2L2(R)C0+C0(1+uε2L((0,T)×R))t0xuε(s,)2L2(R)ds+C(T)uε2L((0,T)×R)tC(T)(1+uε2L((0,T)×R)). (2.16)

    Thanks to Eqs (2.3), (2.13), and (2.16), we have that

    uε4L((0,T)×R)C(T)(1+uε2L((0,T)×R)).

    Therefore,

    uε4L((0,T)×R)C(T)uε2L((0,T)×R)C(T)0,

    which gives (2.4).

    Equation (2.5) follows from (2.4) and (2.16), while, arguing as in Lemma 2.2, we have Eq (2.6).

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

    t0xuε(s,)6L6(R)dsC(T), (2.17)

    for every 0tT.

    Proof. Let 0tT. We begin by observing that,

    R(xuε)6dxxuε(t,)4L(R)xuε(t,)2L2(R). (2.18)

    Thanks to the Hölder inequality,

    (xuε(t,x))2=2xxuε2xuεdy2R|xuε||2xuε|dx2xuε(t,)L2(R)2xuε(t,)2L2(R).

    Hence,

    u(t,)4L(R)4xuε(t,)2L2(R)2xuε(t,)2L2(R).

    It follows from Eq (2.18) that

    R(xuε)6dx4xuε(t,)4L2(R)2xuε(t,)2L2(R).

    Therefore, by Eq (2.5),

    R(xuε)6dxC(T)2xuε(t,)2L2(R).

    An integration on (0,t) and Eq (2.5) gives (2.17).

    This section is devoted to the proof of Theorem 1.1.

    We begin by proving the following result.

    Lemma 3.1. Fix T>0. Then,

    the family {uε}ε>0 is compact in L2loc((0,T)×R). (3.1)

    Consequently, there exist a subsequence {uεk}kN and uL2loc((0,T)×R) such that

    uεku in L2loc((0,T)×R) and a.e. in (0,T)×R. (3.2)

    Moreover, u is a solution of Eq (1.1), satisfying Eq (1.8).

    Proof. We begin by proving Eq (3.1). To prove Eq (3.1), we rely on the Aubin-Lions Lemma (see [5,21]). We recall that

    H1loc(R)↪↪L2loc(R)H1loc(R),

    where the first inclusion is compact and the second one is continuous. Owing to the Aubin-Lions Lemma [21], to prove Eq (3.1), it suffices to show that

    {uε}ε>0 is uniformly bounded in L2(0,T;H1loc(R)), (3.3)
    {tuε}ε>0 is uniformly bounded in L2(0,T;H1loc(R)). (3.4)

    We prove Eq (3.3). Thanks to Lemmas 2.1–2.3,

    uε(t,)2H1(R)=uε(t,)2L2(R)+xuε(t,)2L2(R)C(T).

    Therefore,

    {uε}ε>0 is uniformly bounded in L(0,T;H1(R)),

    which gives Eq (3.3).

    We prove Eq (3.4). Observe that, by Eq (2.1),

    tuε=x(G(uε))f(uε)xuεκuεγ2|uε|uε,

    where

    G(uε)=β2xuεδ2xuεε3xuε. (3.5)

    Since 0<ε<1, thanks to Eq (2.5), we have that

    β2xuε2L2((0,T)×R),δ22xuε2L2((0,T)×R)C(T),ε23xuε2L2((0,T)×R)C(T). (3.6)

    Therefore, by Eqs (3.5) and (3.6), we have that

    {x(G(uε))}ε>0 is bounded in L2(0,T;H1(R)). (3.7)

    We claim that

    T0R(f(uε))2(xuε)2dtdxC(T). (3.8)

    Thanks to Eqs (2.4) and (2.5),

    T0R(f(uε))2(xuε)2dtdxf2L(C(T),C(T))T0xuε(t,)2L2(R)dtC(T).

    Moreover, thanks to Eq (2.3),

    |κ|T0R(uε)2dxC(T). (3.9)

    We have that

    γ2T0R(|uε|uε)2dsdxC(T). (3.10)

    In fact, thanks to Eqs (2.3) and (2.4),

    γ2T0R(|uε|uε)2dsdxγ2uε2L((0,T)×R)T0R(uε)2dsdxC(T)T0R(uε)2dsdxC(T).

    Therefore, Eq (3.4) follows from Eqs (3.7)–(3.10).

    Thanks to the Aubin-Lions Lemma, Eqs (3.1) and (3.2) hold.

    Consequently, arguing as in [5, Theorem 1.1], u is solution of Eq (1.1) and, thanks to Lemmas 2.1–2.3 and Eqs (2.4), (1.8) holds.

    Proof of Theorem 1.1. Lemma 3.1 gives the existence of a solution of Eq (1.1).

    We prove Eq (1.9). Let u1 and u2 be two solutions of Eq (1.1), which verify Eq (1.8), that is,

    {tui+xf(ui)β22xui+δ3xui+κui+γ2|ui|ui=0,0<t<T,xR,ui(0,x)=ui,0(x),xR,i=1,2.

    Then, the function

    ω(t,x)=u1(t,x)u2(t,x), (3.11)

    is the solution of the following Cauchy problem:

    {tω+x(f(u1)f(u2))β22xω+δ2xω+κω+γ2(|u1|u1|u2|u2)=0,0<t<T,xR,ω(0,x)=u1,0(x)u2,0(x),xR. (3.12)

    Fixed T>0, since u1,u2H1(R), for every 0tT, we have that

    u1L((0,T)×R),u2L((0,T)×R)C(T). (3.13)

    We define

    g=f(u1)f(u2)ω (3.14)

    and observe that, by Eq (3.13), we have that

    |g|fL(C(T),C(T))C(T). (3.15)

    Moreover, by Eq (3.11) we have that

    ||u1||u2|||u1u2|=|ω|. (3.16)

    Observe that thanks to Eq (3.11),

    |u1|u1|u2|u2=|u1|u1|u1|u2+|u1|u2|u2|u2=|u1|ω+u2(|u1||u2|). (3.17)

    Thanks to Eqs (3.14) and (3.17), Equation (3.12) is equivalent to the following one:

    tω+x(gω)β22xω+δ3xω+κω+γ2|u1|ω+γ2u2(|u1||u2|)=0. (3.18)

    Multiplying Eq (3.18) by 2ω, an integration on R gives

    dtdtω(t,)2L2(R)=2Rωtω=2Rωx(gω)dx+2β2Rω2xωdx2δRω3xωdx2κω(t,)2L2(R)2γ2R|u1|ω2dx2γ2Ru2(|u1||u2|)ωdx=2Rgωxωdx2β2xω(t,)2L2(R)+2δRxω2xωdx2κω(t,)2L2(R)2γ2R|u1|ω2dx2γ2Ru2(|u1||u2|)ωdx=2Rgωxωdx2β2xω(t,)2L2(R)2κω(t,)2L2(R)2γ2R|u1|ω2dx2γ2Ru2(|u1||u2|)ωdx.

    Therefore, we have that

    ω(t,)2L2(R)+2β2xω(t,)2L2(R)+2γ2R|u1|ω2dx=2Rgωxωdxκω(t,)2L2(R)2γ2Ru2(|u1||u2|)ωdx. (3.19)

    Due to Eqs (3.13), (3.15) and (3.16) and the Young inequality,

    2R|g||ω||xω|dx2C(T)R|ω||xω|dx=2R|C(T)ωβ||βxω|dxC(T)ω(t,)2L2(R)+β2xω(t,)2L2(R),2γ2R|u2||(|u1||u2|)||ω|dx2γ2u2L((0,T)×R)R|(|u1||u2|)||ω|dxC(T)ω(t,)2L2(R).

    It follows from Eq (3.19) that

    ω(t,)2L2(R)+β2xω(t,)2L2(R)+2γ2R|u1|ω2dxC(T)ω(t,)2L2(R).

    The Gronwall Lemma and Eq (3.12) give

    ω(t,)2L2(R)+β2eC(T)tt0eC(T)sxω(s,)2L2(R)ds+2γ2eC(T)tt0ReC(T)s|u1|ω2dsdxeC(T)tω02L2(R). (3.20)

    Equation (1.9) follows from Eqs (3.11) and (3.20).

    Giuseppe Maria Coclite and Lorenzo Di Ruvo equally contributed to the methodologies, typesetting, and the development of the paper.

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

    Giuseppe Maria Coclite is an editorial boardmember for [Networks and Heterogeneous Media] and was not involved inthe editorial review or the decision to publish this article.

    GMC is member of the Gruppo Nazionale per l'Analisi Matematica, la Probabilità e le loro Applicazioni (GNAMPA) of the Istituto Nazionale di Alta Matematica (INdAM). GMC has been partially supported by the Project funded under the National Recovery and Resilience Plan (NRRP), Mission 4 Component 2 Investment 1.4 -Call for tender No. 3138 of 16/12/2021 of Italian Ministry of University and Research funded by the European Union -NextGenerationEUoAward Number: CN000023, Concession Decree No. 1033 of 17/06/2022 adopted by the Italian Ministry of University and Research, CUP: D93C22000410001, Centro Nazionale per la Mobilità Sostenibile, the Italian Ministry of Education, University and Research under the Programme Department of Excellence Legge 232/2016 (Grant No. CUP - D93C23000100001), and the Research Project of National Relevance "Evolution problems involving interacting scales" granted by the Italian Ministry of Education, University and Research (MIUR Prin 2022, project code 2022M9BKBC, Grant No. CUP D53D23005880006). GMC expresses its gratitude to the HIAS - Hamburg Institute for Advanced Study for their warm hospitality.

    The authors declare there is no conflict of interest.



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