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Dysbiosis, small intestinal bacterial overgrowth and biofilms in autism and chronic illness

  • Received: 28 February 2018 Accepted: 22 May 2018 Published: 07 June 2018
  • Recent evidences highlight that alteration of gut microbiota homeostasis could trigger several human pathologies, among them autism spectrum disorders (ASD). This short hypothesis article summarizes the recent literature and offers a novel, complementary and biomedical drugs/natural agents-combined therapy for treating gastrointestinal issues and microbial biofilms in ASD and chronic illness.

    Citation: Anju Usman Singh. Dysbiosis, small intestinal bacterial overgrowth and biofilms in autism and chronic illness[J]. AIMS Molecular Science, 2018, 5(2): 160-165. doi: 10.3934/molsci.2018.2.160

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  • Recent evidences highlight that alteration of gut microbiota homeostasis could trigger several human pathologies, among them autism spectrum disorders (ASD). This short hypothesis article summarizes the recent literature and offers a novel, complementary and biomedical drugs/natural agents-combined therapy for treating gastrointestinal issues and microbial biofilms in ASD and chronic illness.


    We consider the initial boundary value problem of the following porous elastic system with nonlinear or linear weak damping terms and nonlinear source terms

    $ {uttμuxxbϕx+g1(ut)=f1(u,ϕ),          x(0,L), t[0,T),ϕttδϕxx+bux+ξϕ+g2(ϕt)=f2(u,ϕ), x(0,L), t[0,T),u(x,0)=u0(x),  ut(x,0)=u1(x),             x(0,L),ϕ(x,0)=ϕ0(x),  ϕt(x,0)=ϕ1(x),            x(0,L),u(0,t)=u(L,t)=ϕ(0,t)=ϕ(L,t)=0,    t[0,T), $ (1.1)

    where $ u(x, t) $ and $ \phi(x, t) $ are the displacement of the solid elastic material and the volume fraction, respectively, $ \mu, \ b, \ \delta $ and $ \xi $ are coefficients with physical meaning satisfying

    $ \mu > 0, \ b\neq0, \ \delta > 0, \ \xi > 0 \ \text{and}\ \mu\xi > b^2, $

    $ u_0, \ u_1, \phi_0 $ and $ \phi_1 $ are given initial data, and the assumptions of weak damping terms $ g_1, \ g_2 $ and nonlinear source terms $ f_1, \ f_2 $ will be given in Section 2 by Assumption 2.1 and Assumption 2.2, respectively.

    In the physical view, elastic solid with voids is an important extension of the classical elasticity theory. It allows the processing of porous solids in which the matrix material is elastic and the interstices are void of material (see [8,20] and references therein). Porous media reflects the properties of many materials in the real world, including rocks, soil, wood, ceramics, pressed powder, bones, natural gas hydrates and so on. Due to the diversity of porous media and its special physical properties, such models were widely applied in the past few decades in the petroleum industry, engineering, etc (see [1,12,13,16,17,19]).

    As mathematical efforts, Goodman and Cowin [2,8] established the continuum theory and the variational principle of granular materials. Then Nunziato and Cowin [3,18] developed the linear and nonlinear theories of porous elastic materials. In recent years, the study of the porous elastic system also attracted a lot of attention [5,6,7,21,22]. We particularly mention that Freitas et.al. in [5] studied the problem (1.1) and proved the global existence and finite time blowup of solutions. Especially, they built up the continuous dependence on initial data of the local solution in the following version

    $ ˆE(t)eC0tˆE(0), C0>0, $ (1.2)

    which can also be extended to the global solution with the same form. By denoting $ z = (u, \phi) $ and $ \tilde{z} = (\tilde{u}, \tilde{\phi}) $ the global solutions to problem (1.1) corresponding to the initial data $ z_{0} $, $ z_{1} $ and $ \tilde{z}_{0} $, $ \tilde{z}_{1} $, respectively, $ \widehat{E}(0) $ is the distance of two sets of different initial data

    $ z_0, \tilde{z}_0 \in V: = {H_{0}^1(0, L)}\times {H_{0}^1(0, L)}, $

    and

    $ z_{1}, \tilde{z}_{1} \in {L^2(0, L)}\times {L^2(0, L)}, $

    that is

    $ ˆE(0):=12z1˜z122+12z0˜z02V, $

    and $ \widehat{E}(t) $ is the distance of solutions induced by these two sets of different initial data

    $ ˆE(t):=12zt˜zt22+12z˜z2V. $

    The growth estimate (1.2) indicates that the growth of the distance of solutions $ \widehat{E}(t) $ is bounded by an exponential growth bound with time $ t $. In other words, as the time $ t $ goes to infinity, the distance of solutions $ \widehat{E}(t) $ of the system is bounded by a very large bound, by which it is hard to explain the solutions $ z $ and $ \tilde{z} $ of such a dissipative system with the initial data $ z_0, z_1 $ and $ \tilde{z}_0, \tilde{z}_1 $, respectively, as both of them are expected to decay to zero as the time $ t $ goes to infinity. Hence, the estimate on the growth of the distance of solutions $ \widehat{E}(t) $ is proposed to be improved to reflect the decay properties with time $ t $ to be consistent with the dissipative behavior of the system. To achieve this, the efforts in the present paper are illustrated by two new continuous dependence results on the initial data for the global-in-time solution. Especially, it is found that the system with the linear damping term behaves differently from that with the nonlinear damping term. Hence in the present paper, we adopt two different estimate strategies to deal with the problem and derive two different conclusions:

    $ ({\rm{i}}) $ For the linear damping case, i.e., $ g_1(u_t) $ and $ g_2(\phi_t) $ take the linear form and satisfy Assumption 2.1, we have

    $ ˆE(t)C1(ˆE(0)+C2(ˆE(0))a2)ρeC3t, $ (1.3)

    where the positive constants $ C_1, C_2, C_3, a, \rho $ are independent of initial data.

    $ ({\rm{ii}}) $ For the nonlinear damping case, i.e., $ g_1(u_t) $ and $ g_2(\phi_t) $ take the nonlinear form and satisfy Assumption 2.1, we have

    $ ˆE(t)C5(ˆE(0)+C6(ˆE(0))b02)κeC7t, $ (1.4)

    where $ 0 < \kappa < 1 $, and the positive constants $ C_5, C_6, C_7, b_0 $ are dependent of initial data.

    By observing (1.3) and (1.4), we find that these two continuous dependence results can reasonably reflect the decay property of the dissipative system (1.1). The difference between (1.3) and (1.4) is that the parameters in (1.3) do not depend on the initial data, while the parameters in (1.4) depend on the initial data. Hence although (1.3) and (1.4) are in the similar form, we present and prove them separately.

    Additionally, to develop the finite time blowup of the solution to problem (1.1) at the arbitrary positive initial energy level derived in [22], we estimate the lower bound of the blowup time in the present paper for the nonlinear weak damping case by noticing that the linear weak damping case was discussed in [22]. For more relative works on the blowup of solutions to the hyperbolic equations at high initial energy, please refer to [10,11,14,15,25]. We can also refer to [9,23,24] for the works about the blowup of solutions to parabolic equations.

    The rest of the present paper is organized as follows. In Section 2, we give some notations, assumptions about damping terms and source terms, and functionals and manifolds for the potential well theory. In Section 3, we deal with the continuous dependence on initial data of the global solution for the linear weak damping case. In Section 4, we establish the continuous dependence on initial data of the global solution for the nonlinear weak damping case. In Section 5, we estimate the lower bound of blowup time at the arbitrarily positive initial energy level for the nonlinear weak damping case.

    We denote the $ L^{2} $-inner product by

    $ (u, v): = \int_{0}^{L}{uv}{\rm d}x, $

    and the norm of $ L^{p}(0, L) $ by

    $ \|u\|_{p}: = \left( \int_{0}^{L}|u|^{p}{\rm d}x \right)^{\frac{1}{p}}. $

    As we are dealing with the system of two equations, for $ z = (u, \phi) $ and $ \hat{z} = (\hat{u}, \hat{\phi}) $, we introduce

    $ (z, \hat{z}): = (u, \hat{u})+(\phi, \hat{\phi}) $

    and

    $ zp:=(upp+ϕpp)1p. $ (2.1)

    Further, we consider the Hilbert space

    $ V = {H_{0}^1(0, L)}\times {H_{0}^1(0, L)} $

    with inner products given by

    $ (z,ˆz)V:=L0(μuxˆux+δϕxˆϕx+ξϕˆϕ+b(uxˆϕ+ϕˆux))dx $ (2.2)

    for $ z = (u, \phi) $, $ \hat{z} = (\hat{u}, \hat{\phi}) $, where $ \mu $, $ \delta $, $ \xi $, $ b $ are the coefficients of the terms in the equations in problem (1.1). Therefore, we have

    $ z2V:=L0(μu2x+δϕ2x+ξϕ2+2buxϕ)dx. $ (2.3)

    The norm $ \|z\|_{V} $ is equivalent to the corresponding usual norm on $ V $, i.e., $ {H_{0}^1(0, L)}\times {H_{0}^1(0, L)} $, introduced in [20]. For $ 1 < q < +\infty $, we define

    $ Rq:=supzV{0}zqqzqV, $ (2.4)

    which means

    $ zqqRqzqV $ (2.5)

    for $ z\in V $. Here, due to $ H^{1}_{0}(0, L)\hookrightarrow L^{q}(0, L) $ for $ 1 < q < +\infty $, we see $ 0 < R_{q} < +\infty $. And we denote

    $ \mathcal{F}(z): = (f_{1}(u, \phi), f_{2}(u, \phi)) $

    and

    $ \mathcal{G}(z_{t}): = (g_{1}(u_{t}), g_{2}(\phi_{t})), $

    where $ f_{j}(u, \phi) $, $ j = 1, 2 $, are the source terms, and $ g_{1}(u_{t}) $ and $ g_{2}(\phi_{t}) $ are the damping terms in the equations in problem (1.1).

    We give the following assumptions about damping terms, i.e., $ g_{1}(u_{t}) $ and $ g_{2}(\phi_{t}) $, and source terms, i.e., $ f_{j}(u, \phi) $, $ j = 1, 2 $, in the equations in problem (1.1).

    Assumption 2.1. (Damping terms) Let $ g_{1}, g_{2}: \mathbb{R}\rightarrow \mathbb{R} $ be continuous, monotone increasing functions with $ g_{1}(0) = g_{2}(0) = 0 $. In addition, there exist positive constants $ \alpha > 0 $ and $ \beta > 0 $ such that

    ${\rm{(i)}}$ for $ |s|\geq1 $

    $ α|s|m+1g1(s)sβ|s|m+1, m1; $ (2.6)

    and

    $ α|s|r+1g2(s)sβ|s|r+1, r1; $ (2.7)

    ${\rm{(ii)}}$ for $ |s| < 1 $

    $ α|s|ˆm|g1(s)|β|s|ˆm, ˆm1; $ (2.8)

    and

    $ α|s|ˆr|g2(s)|β|s|ˆr, ˆr1. $ (2.9)

    Assumption 2.2. (Source terms) For the functions $ f_{j}\in C^{1}(\mathbb{R}^{2}) $, $ j = 1, 2 $, there exists a positive constant $ C $ such that

    $ |fj(η)|C(|η1|p1+|η2|p1+1), p>1. $ (2.10)

    where $ \eta = (\eta_{1}, \eta_{2})\in\mathbb{R}^{2} $, $ f_{j}(\eta) = f_{j}(\eta_{1}, \eta_{2}) $, $ j = 1, 2 $, and

    $ fj:=(fjη1,fjη2). $

    There exists a nonnegative function $ F\in C^{2}(\mathbb{R}^{2}) $ satisfying

    $ F=F $ (2.11)

    and

    $ F(λη)=λp+1F(η) $ (2.12)

    for all $ \lambda > 0 $, where $ F(\eta) = F(\eta_{1}, \eta_{2}) $ and

    $ F:=(Fη1,Fη2). $ (2.13)

    According to [5], Assumption 2.2 implies that there exists a constant $ M > 0 $ such that

    $ F(z)M(|u|p+1+|ϕ|p+1). $ (2.14)

    Next, we recall some functionals and manifolds for the potential well theory. We recall the potential energy functional

    $ J(z):=12z2VL0F(z)dx $ (2.15)

    and the Nehari functional

    $ I(z):=z2V(p+1)L0F(z)dx. $

    The energy functional is defined as

    $ E(z(t),zt(t)):=12zt22+J(z). $ (2.16)

    And the Nehari manifold is defined as

    $ N:={zV{0}| I(z)=0}. $

    Then we can define the depth of the potential well

    $ d:=infzNJ(z). $

    By above, we introduce the stable manifold

    $ W:={zV| J(z)<d, I(z)>0}{0}. $

    Next, since we need to apply the decay rate of the energy in investigating continuous dependence on the initial data of the solution, we recall the following notations used in the investigation of the decay rate of the energy in [5]

    $ ˆd:=sups[0,+)M(s)=M(s0)=p12(p+1)((p+1)MRp+1)2p1, $ (2.17)

    where

    $ M(s):=12s2MRp+1sp+1, $ (2.18)

    and $ \mathcal{M}(s) $ attains the maximum value at

    $ s0:=((p+1)MRp+1)1p1. $ (2.19)

    Here, Proposition 2.11 in [5] shows the fact $ \hat{d}\leq d $.

    In this section, we consider the model equations in (1.1) with the linear weak damping terms, i.e., $ r = m = \hat{r} = \hat{m} = 1 $. First, we need the following decay result of the energy.

    Lemma 3.1. (Decay of the energy) Let Assumption 2.1 and Assumption 2.2 hold with $ r = m = \hat{r} = \hat{m} = 1 $. For any $ 0 < \sigma < 1 $, if $ \mathcal{E}(z_{0}, z_{1}) < \sigma \hat{d} $ and $ z_{0}\in W $, then one has

    $ E(z(t),zt(t))<K0eλ0t $ (3.1)

    for $ t > 0 $, where $ \lambda_{0} $ and $ K_{0} $ will be defined in the proof.

    Proof. We define

    $ H(t):=E(z(t),zt(t))+ε(z,zt), $

    where $ \varepsilon > 0 $. Here, according to Cauchy-Schwartz inequality, Young inequality, and (2.5), we have

    $ H(t)E(z(t),zt(t))+εz2zt2E(z(t),zt(t))+ε2z22+ε2zt22E(z(t),zt(t))+ε2R2z2V+ε2zt22E(z(t),zt(t))+εmax{R2,1}(12z2V+12zt22) $ (3.2)

    and

    $ H(t)E(z(t),zt(t))εz2zt2E(z(t),zt(t))εmax{R2,1}(12z2V+12zt22). $ (3.3)

    According to Theorem 2.12(ⅳ) in [5], we know

    $ 12z2V+12zt22p+1p1E(z(t),zt(t)), $ (3.4)

    which means that (3.2) and (3.3) turn to

    $ H(t)E(z(t),zt(t))+εmax{R2,1}(p+1)p1E(z(t),zt(t)) $ (3.5)

    and

    $ H(t)E(z(t),zt(t))εmax{R2,1}(p+1)p1E(z(t),zt(t)). $ (3.6)

    According to (3.5) and (3.6), we know

    $ α1E(z(t),zt(t))H(t)α2E(z(t),zt(t)), $ (3.7)

    where

    $ \alpha_{1}: = 1-\frac{\varepsilon\max\{R_{2}, 1\}(p+1)}{p-1} $

    and

    $ \alpha_{2}: = 1+\frac{\varepsilon\max\{R_{2}, 1\}(p+1)}{p-1}. $

    We calculate the derivative of the auxiliary functional $ H(t) $ with respect to time $ t $ as

    $ H(t)=ddtE(z(t),zt(t))+εzt22+ε(ztt,z). $ (3.8)

    In (3.8), we have

    $ ddtE(z(t),zt(t))=12ddtzt22+12ddtz2V+L0ddtF(z)dx=12ddt(ut22+ϕt22)+12L0ddt(μu2x+δϕ2x+ξϕ2+2buxϕ)dx+L0ddtF(z)dx=L0(ututt+ϕtϕtt)dx+L0(μuxuxt+δϕxϕxt+ξϕϕt+buxtϕ+buxϕt)dx+L0F(z)ztdx. $ (3.9)

    Here, the notation $ \nabla F $ is defined by (2.13). Thus, according to (2.11), we know $ \nabla F(z) = \mathcal{F}(z) $, which means (3.9) turns to

    $ ddtE(z(t),zt(t))=L0(ututt+ϕtϕtt)dx+L0(μuxuxt+δϕxϕxt+ξϕϕt+buxtϕ+buxϕt)dx+L0F(z)ztdx=L0(ututt+ϕtϕtt)dx+L0(μuxuxt+δϕxϕxt+ξϕϕt+buxtϕ+buxϕt)dx+L0(f1(u,ϕ)ut+f2(u,ϕ)ϕt)dx=L0(ututt+ϕtϕtt)dx+L0(μuxuxt+δϕxϕxt+ξϕϕt+buxϕt)dxL0butϕxdx+L0(f1(u,ϕ)ut+f2(u,ϕ)ϕt)dx=L0(ututt+μuxuxtbutϕxf1(u,ϕ)ut)dx+L0(ϕtϕtt+δϕxϕxt+buxϕt+ξϕϕtf2(u,ϕ)ϕt)dx. $ (3.10)

    Testing the both sides of the first equation in (1.1) by $ u_{t} $ and integrating both sides over $ [0, L] $, we have

    $ L0(ututt+μuxuxtbutϕxf1(u,ϕ)ut)dx=L0g1(ut)utdx. $ (3.11)

    And testing the both sides of the second equation in (1.1) by $ \phi_{t} $ and integrating both sides over $ [0, L] $, we have

    $ L0(ϕtϕtt+δϕxϕxt+buxϕt+ξϕϕtf2(u,ϕ)ϕt)dx=L0g2(ϕt)ϕtdx. $ (3.12)

    By substituting (3.11) and (3.12) into (3.10), we have

    $ ddtE(z(t),zt(t))=L0g1(ut)utdxL0g2(ϕt)ϕtdx. $ (3.13)

    Next, we use Assumption 2.1 to deal with (3.13). In Assumption 2.1, for $ |s|\geq1 $, according to (2.6) with $ m = 1 $ and (2.7) with $ r = 1 $, we know that

    $ α|s|2gj(s)sβ|s|2, j=1,2. $ (3.14)

    Then taking the absolute value of (3.14) gives

    $ α|s||gj(s)|β|s|, j=1,2. $ (3.15)

    For $ |s| < 1 $, according to (2.8) with $ \hat{m} = 1 $ and (2.9) with $ \hat{r} = 1 $, we know that (3.15) also holds. Meanwhile, since $ g_{1}(0) = g_{2}(0) = 0 $ and $ g_{j}(s) $, $ j = 1, 2 $, are assumed to be the increasing functions, for $ j = 1, 2 $, we know $ g_{j}(s) > 0 $ for $ s > 0 $ and $ g_{j}(s) < 0 $ for $ s < 0 $, which gives $ g_{j}(s)s\geq0 $, $ j = 1, 2 $, for $ s\in \mathbb{R} $. Thus, we have

    $ L0g1(ut)utdx+L0g2(ϕt)ϕtdx=L0|g1(ut)ut|dx+L0|g2(ϕt)ϕt|dxαut22+αϕt22=αzt22, $

    which makes (3.13) turn to

    $ ddtE(z(t),zt(t))αzt22. $ (3.16)

    We deal with the term $ \varepsilon\left(z_{tt}, z\right) $ in (3.8). Testing the both sides of the first equation in problem (1.1) by $ u $ and integrating both sides over $ [0, L] $, we have

    $ (utt,u)=μux22b(ux,ϕ)(g1(ut),u)+(f1(u,ϕ),u). $ (3.17)

    And testing the both sides of the second equation in problem (1.1) by $ \phi $ and integrating both sides over $ [0, L] $, we have

    $ (ϕtt,ϕ)=δϕx22b(ux,ϕ)ξϕ22(g2(ϕt),ϕ)+(f2(u,ϕ),ϕ). $ (3.18)

    By (3.17) plus (3.18), we have

    $ (ztt,z)=L0(μu2x+δϕ2x+ξϕ2+2buxϕ)dx(g1(ut),u)(g2(ϕt),ϕ)+(f1(u,ϕ),u)+(f2(u,ϕ),ϕ)=ϕ2V(G(zt),z)+(F(z),z)ϕ2V+|(G(zt),z)|+(F(z),z). $ (3.19)

    According to (3.16) and (3.19), we know that (3.8) turns to

    $ H(t)αzt22+εzt22εz2V+ε|(G(zt),z)|+ε(F(z),z). $ (3.20)

    Next, we deal with the term $ \varepsilon\left|\left(\mathcal{G}(z_{t}), z\right)\right| $ in (3.20). By using (3.15) and Hölder inequality, we know

    $ |(G(zt),z)|=|(g1(ut),u)+(g2(ϕt),ϕ)||(g1(ut),u)|+|(g2(ϕt),ϕ)|L0|g1(ut)||u|dx+L0|g2(ϕt)||ϕ|dxβL0|ut||u|dx+βL0|ϕt||ϕ|dxβut2u2+βϕt2ϕ22βzt2z2. $ (3.21)

    Then, We deal with $ \varepsilon\left(\mathcal{F}(z), z\right) $ in (3.20). Here, we first need to give

    $ F(z)z=(p+1)F(z). $ (3.22)

    For all $ \lambda > 0 $, taking the derivative of both sides of (2.12) with respect to $ \lambda $, we know

    $ ddλF(λz)=F(λz)z=ddλλp+1F(z)=(p+1)λpF(z), $ (3.23)

    where $ \nabla F $ is defined by (2.13). By taking $ \lambda = 1 $ in (3.23) and using (2.11), we obtain (3.22). According to (3.22) and (2.14), we have

    $ (F(z),z)=L0F(z)zdx=(p+1)L0F(z)dx(p+1)Mzp+1p+1. $ (3.24)

    By using (2.5), (3.24) turns to

    $ (F(z),z)(p+1)MRp+1zp+1V=(p+1)MRp+1zp1Vz2V. $ (3.25)

    Then, we estimate the term $ \|z\|^{p-1}_{V} $ in (3.25). According to Theorem 2.12 (ⅱ) in [5], we know $ z(t)\in W $ for $ t > 0 $. By using $ I(z(t)) > 0 $, i.e., $ z(t)\in W $, we have

    $ (p+1)L0F(z(t))dx<z(t)2V, $

    which means

    $ J(z(t))=12z(t)2VL0F(z(t))dx>12z(t)2V1p+1z(t)2V=p12(p+1)z(t)2V. $ (3.26)

    Meanwhile, according to (3.16), i.e.,

    $ \frac{{\rm d}}{{\rm d}t}\mathcal{E}(z(t), z_{t}(t))\leq0, $

    we have $ \mathcal{E}(z(t), z_{t}(t))\leq\mathcal{E}(z_{0}, z_{1}) $. Thus, we know

    $ p12(p+1)z(t)2VJ(z(t))E(z(t),zt(t))E(z0,z1), $ (3.27)

    i.e.,

    $ z(t)p1V(2(p+1)p1E(z0,z1))p12, $

    for $ t > 0 $, which implies that (3.25) turns to

    $ (F(z),z)(p+1)MRp+1(2(p+1)p1E(z0,z1))p12z2V. $ (3.28)

    Due to $ \mathcal{E}(z_{0}, z_{1}) < \sigma \hat{d} $ being assumed, we know that (3.28) turns to

    $ (F(z),z)σp12z2V, $ (3.29)

    where $ \hat{d} $ is defined by (2.17). Substituting (3.21) and (3.29) into (3.20), we have

    $ H(t)αzt22+εzt22+εσp12z2Vεz2V+2εβzt2z2. $ (3.30)

    By using Young inequality for $ \delta_{0} > 0 $ and inequality (2.5) for $ q = 2 $, we know that (3.30) turns to

    $ H(t)αzt22+εzt22+εσp12z2Vεz2V+εβδ0zt22+εβδ0R2z2V=(αεεβδ0)zt22ε(1σp12βδ0R2)z2V. $ (3.31)

    In (3.31), we choose $ \delta_{0} > 0 $ to make $ 1-\sigma^{\frac{p-1}{2}}-\beta\delta_{0} R_{2} > 0 $ hold, where $ 1-\sigma^{\frac{p-1}{2}} > 0 $ due to $ \sigma\in(0, 1) $. Then, we select $ \varepsilon > 0 $ such that $ \alpha-\varepsilon-\frac{\varepsilon\beta}{\delta_{0}} > 0 $ and

    $ \alpha_{1} = 1-\frac{\varepsilon\max\{R_{2}, 1\}(p+1)}{p-1} > 0. $

    To deal with (3.31), we first have

    $ (αεεβδ0)zt22+ε(1σp12βδ0R2)z2V=2(αεεβδ0)12zt22+2ε(1σp12βδ0R2)12z2Vmin{2(αεεβδ0),2ε(1σp12βδ0R2)}(12zt22+12z2V). $ (3.32)

    According to Theorem 2.12 (ⅳ) in [5], (3.32) turns to

    $ (αεεβδ0)zt22+ε(1σp12βδ0R2)z2Vmin{2(αεεβδ0),2ε(1σp12βδ0R2)}E(z(t),zt(t)). $ (3.33)

    Due to (3.7), i.e., $ H(t)\leq \alpha_{2}\mathcal{E}(z(t), z_{t}(t)) $, (3.33) turns to

    $ (αεεβδ0)zt22+ε(1σp12βδ0R2)z2Vmin{2(αεεβδ0),2ε(1σp12βδ0R2)}α2H(t). $ (3.34)

    Thus, we know that (3.31) implies

    $ H(t)λ0H(t), $ (3.35)

    where

    $ λ0:=min{2(αεεβδ0),2ε(1σp12βδ0R2)}α2. $ (3.36)

    By using Gronwall's inequality, (3.35) gives

    $ H(t)eλ0tH(0). $ (3.37)

    According to (3.7), (3.37), and the assumptions $ \mathcal{E}(z_{0}, z_{1}) < \sigma\hat{d} $ and $ 0 < \sigma < 1 $, we have

    $ E(z(t),zt(t))α2E(z0,z1)α1eλ0t<α2σˆdα1eλ0t<K0eλ0t, $ (3.38)

    where

    $ K0:=α2ˆdα1. $ (3.39)

    Theorem 3.2. (Continuous dependence on initial data for linear weak damping case) Let Assumption 2.1 and Assumption 2.2 hold with $ r = m = \hat{r} = \hat{m} = 1 $. For any $ 0 < \sigma < 1 $, suppose $ \mathcal{E}(z_{0}, z_{1}) < \sigma \hat{d} $, $ z_{0}\in W $, $ \mathcal{E}(\tilde{z}_{0}, \tilde{z}_{1}) < \sigma \hat{d} $ and $ \tilde{z}_{0}\in W $. Let $ z = (u, \phi) $ and $ \tilde{z} = (\tilde{u}, \tilde{\phi}) $ be the global solutions to problem (1.1) with the initial data $ z_{0} $, $ z_{1} $, and $ \tilde{z}_{0} $, $ \tilde{z}_{1} $, respectively. Then one has

    $ ˆE(t)C1(ˆE(0)+C2(ˆE(0))a2)ρeC3t, $ (3.40)

    where

    $ C1:=(1+C4eC4λ0(p1)λ0(p1))ρ(4(p+1)K0p1)1ρ,C2:=2a2Nλ1,C3:=λ0(1ρ),C4:=43CR124R124(p1)(2(p+1)K0p1)p1,0<a<min{2λ0ˉMC+λ0,1},0<ρ<1,$ (3.41)

    $ \lambda_{0} $ and $ K_{0} $ are defined by (3.36) and (3.39), respectively, $ R_{4(p-1)} $ is the best embedding constant defined in (2.4) taking $ q = 4(p-1) $,

    $ λ1:=λ0(2a)aˉMC2, $
    $ N:=21aC(2K0)2a2+23aCR122(2(p+1)K0p1)12(2K0)1a2, $

    and

    $ ˉM:=max{23252R124(2R4(p1)(2(p+1)σˆdp1)2(p1)+L)12,1}. $ (3.42)

    Proof. We denote $ w: = z-\tilde{z} $. According to the proof of Theorem 2.5 in [5], we notice that

    $ ˆE(t)ˆE(0)+t0L0(F(z(τ))F(˜z(τ)))wt(τ)dxdτ $ (3.43)

    holds by Assumption 2.1 and Assumption 2.2. In the following, we shall finish this proof by considering the following two steps. In Step $ \rm{I} $, we shall derive a similar estimate of the growth of $ \widehat{E}(t) $ to (135) in [5]. As we build this estimate for the global solution instead of the local solution treated in [5], we have to rebuild all the necessary estimates based on the conditions for global existence theory.

    Step Ⅰ: Global estimate of $ \widehat{E}(t) $ for global solution.

    We estimate the term $ \int_{0}^{t}\int_{0}^{L}\left(\mathcal{F}(z(\tau))-\mathcal{F}(\tilde{z}(\tau))\right) w_{t}(\tau){\rm d}x{\rm d}\tau $ in (3.43) as follows

    $ L0(F(z(t))F(˜z(t)))wtdx=L0(f1(z)f1(˜z))(ut˜ut)dx+L0(f2(z)f2(˜z))(ϕt˜ϕt)dxL0|f1(z)f1(˜z)||ut˜ut|dx+L0|f2(z)f2(˜z)||ϕt˜ϕt|dx. $ (3.44)

    Here, according to the proof of Lemma 3.2 in [5], we notice that (2.10) in Assumption 2.2 gives

    $ |fj(z)fj(˜z)|C|z˜z|(|u|p1+|˜u|p1+|ϕ|p1+|˜ϕ|p1+1), j=1,2, $ (3.45)

    which means (3.44) turns to

    $ L0(F(z(t))F(˜z(t)))wtdxL0C|z˜z|(|u|p1+|˜u|p1+|ϕ|p1+|˜ϕ|p1+1)|ut˜ut|dx:=A1+L0C|z˜z|(|u|p1+|˜u|p1+|ϕ|p1+|˜ϕ|p1+1)|ϕt˜ϕt|dx:=A2. $ (3.46)

    Next, we deal with $ A_{1} $ and $ A_{2} $ separately. For $ A_{1} $, by Hölder inequality and Young inequality, we have

    $ A1C(L0|z˜z|2(|u|p1+|˜u|p1+|ϕ|p1+|˜ϕ|p1+1)2dx)12(L0|ut˜ut|2dx)12C2L0|z˜z|2(|u|p1+|˜u|p1+|ϕ|p1+|˜ϕ|p1+1)2dx+C2L0|ut˜ut|2dx. $ (3.47)

    By the similar process, we can deal with $ A_{2} $ as

    $ A2C2L0|z˜z|2(|u|p1+|˜u|p1+|ϕ|p1+|˜ϕ|p1+1)2dx+C2L0|ϕt˜ϕt|2dx. $ (3.48)

    According to (3.47), (3.48) and Hölder inequality, we know that (3.46) turns to

    $ L0(F(z(t))F(˜z(t)))wtdxCL0|z˜z|2(|u|p1+|˜u|p1+|ϕ|p1+|˜ϕ|p1+1)2dx+C2wt22C(L0|z˜z|4dx)12(L0(|u|p1+|˜u|p1+|ϕ|p1+|˜ϕ|p1+1)4dx)12+C2wt22. $ (3.49)

    In (3.49), by noticing $ z = (u, \phi) $, $ \tilde{z} = (\tilde{u}, \tilde{\phi}) $, we see that

    $ (L0|z˜z|4dx)12=(L0((|u˜u|2+|ϕ˜ϕ|2)12)4dx)12=(L0(|u˜u|4+|ϕ˜ϕ|4+2|u˜u|2|ϕ˜ϕ|2)dx)12=(L0(|u˜u|4+|ϕ˜ϕ|4)dx+L02|u˜u|2|ϕ˜ϕ|2dx)12. $ (3.50)

    By using Hölder inequality and Young inequality, we know (3.50) turns to

    $ (L0|z˜z|4dx)12(L0(|u˜u|4+|ϕ˜ϕ|4)dx+2u˜u24ϕ˜ϕ24)12(2u˜u44+2ϕ˜ϕ44)12=212z˜z24. $ (3.51)

    Next, we deal with $ \int_{0}^{L}(|u|^{p-1}+|\tilde{u}|^{p-1}+|\phi|^{p-1}+|\tilde{\phi}|^{p-1}+1)^{4}{\rm d}x $ in (3.49). For $ k_{1} $, $ k_{2} $, $ k_{3} $, $ k_{4} $, $ k_{5}\geq0 $, we have

    $ (k1+k2+k3+k4+k5)4(5max{k1,k2,k3,k4,k5})4=54max{k41,k42,k43,k44,k45}54(k41+k42+k43+k44+k45). $

    From above observation, we have

    $ L0(|u|p1+|˜u|p1+|ϕ|p1+|˜ϕ|p1+1)4dx54L0(|u|4(p1)+|˜u|4(p1)+|ϕ|4(p1)+|˜ϕ|4(p1)+1)dx. $ (3.52)

    According to (3.51) and (3.52), (3.49) turns to

    $ L0(F(z)F(˜z))wtdxC212z˜z24(54L0(|u|4(p1)+|˜u|4(p1)+|ϕ|4(p1)+|˜ϕ|4(p1)+1)dx)12+C2wt22=C21252z˜z24(L0(|u|4(p1)+|ϕ|4(p1))dx+L0(|˜u|4(p1)+|˜φ|4(p1))dx+L)12+C2wt22=C21252z˜z24(z4(p1)4(p1)+˜z4(p1)4(p1)+L)12+C2wt22. $ (3.53)

    By using (2.5), we know that (3.53) turns to

    $ L0(F(z)F(˜z))wtdxC21252R124z˜z2V(R4(p1)z4(p1)V+R4(p1)˜z4(p1)V+L)12+C2wt22. $ (3.54)

    According to (3.27) and the assumptions $ \mathcal{E}(z_{0}, z_{1}) < \sigma \hat{d} $ and $ \mathcal{E}(\tilde{z}_{0}, \tilde{z}_{1}) < \sigma \hat{d} $, we have

    $ z2V<2(p+1)σˆdp1 $ (3.55)

    and

    $ ˜z2V<2(p+1)σˆdp1. $ (3.56)

    Substituting (3.55) and (3.56) into (3.54), we have

    $ L0(F(z)F(˜z))wtdxC21252R124(2R4(p1)(2(p+1)σˆdp1)2(p1)+L)12w2V+C2wt22ˉMC(12wt22+12w2V)=ˉMCˆE(t). $ (3.57)

    Due to (3.57), we know

    $ t0L0(F(z(τ))F(˜z(τ)))wtdxdτˉMCt0ˆE(τ)dτ. $ (3.58)

    Substituting (3.58) into (3.43), we have

    $ ˆE(t)ˆE(0)+ˉMCt0ˆE(τ)dτ. $ (3.59)

    Then, by a variation of Gronwall's inequality (see Appendix), we have

    $ ˆE(t)ˆE(0)eˉMCt. $ (3.60)

    As the growth estimate (3.60) we derived in Step $ \rm{I} $ does not reflect the decay of the solution, we shall deal with the decay terms and the non-decay terms separately in Step $ \rm{II} $ to upgrade the results obtained in Step $ \rm{I} $, i.e., (3.60), by giving an improved estimate to reflect the dissipative property of the system (1.1).

    Step Ⅱ: Decay estimate of $ \widehat{E}(t) $.

    According to (3.46), we have

    $ L0(F(z(t))F(˜z(t)))wtdxL0C|z˜z|(|u|p1+|˜u|p1+|ϕ|p1+|˜ϕ|p1)|ut˜ut|dx:=A3+L0C|z˜z|(|u|p1+|˜u|p1+|ϕ|p1+|˜ϕ|p1)|ϕt˜ϕt|dx:=A4+L0C|z˜z||ut˜ut|dx+L0C|z˜z||ϕt˜ϕt|dx. $ (3.61)

    By the similar process dealing with $ A_{1} $ and $ A_{2} $, we can treat $ A_{3} $ and $ A_{4} $ as

    $ A3C2L0|z˜z|2(|u|p1+|˜u|p1+|ϕ|p1+|˜ϕ|p1)2dx+C2L0|ut˜ut|2dx $ (3.62)

    and

    $ A4C2L0|z˜z|2(|u|p1+|˜u|p1+|ϕ|p1+|˜ϕ|p1)2dx+C2L0|ϕt˜ϕt|2dx. $ (3.63)

    By the similar process of obtaining (3.54), i.e.,

    $ A1+A2C21252R124z˜z2V(R4(p1)z4(p1)V+R4(p1)˜z4(p1)V+L)12+C2wt22, $

    we can use (3.62) and (3.63) to give

    $ A3+A4C21242R124R124(p1)z˜z2V(z4(p1)V+˜z4(p1)V)12+C2wt22. $ (3.64)

    Due to (3.26), (2.16) and Lemma $ 3.1 $, we know

    $ 12zt22+p12(p+1)z2VE(z(t),zt(t))<K0eλ0t $ (3.65)

    and

    $ 12˜zt22+p12(p+1)˜z2VE(˜z(t),˜zt(t))<K0eλ0t. $ (3.66)

    According to (3.65) and (3.66), we have

    $ z4(p1)V<(2(p+1)p1K0)2(p1)e2λ0(p1)t $

    and

    $ ˜z4(p1)V<(2(p+1)p1K0)2(p1)e2λ0(p1)t, $

    which mean

    $ (z4(p1)V+˜z4(p1)V)12<212(2(p+1)p1K0)p1eλ0(p1)t. $ (3.67)

    By substituting (3.67) into (3.64), we obtain

    $ A3+A4C4eλ0(p1)t12z˜z2V+C2zt˜zt22C4eλ0(p1)t(12z˜z2V+12zt˜zt22)+C2zt˜zt22. $ (3.68)

    Substituting (3.68) into (3.61) and using Hölder inequality and (2.5), we have

    $ L0(F(z(t))F(˜z(t)))wtdxC4eλ0(p1)tˆE(t)+C2zt˜zt22+L0C|z˜z||ut˜ut|dx+L0C|z˜z||ϕt˜ϕt|dxC4eλ0(p1)tˆE(t)+C2zt˜zt22+Cz˜z2ut˜ut2+Cz˜z2ϕt˜ϕt2C4eλ0(p1)tˆE(t)+C2zt˜zt22+2Cz˜z2zt˜zt2C4eλ0(p1)tˆE(t)+C2zt˜zt22+2CR122z˜zVzt˜zt2. $ (3.69)

    According to (3.60), we know

    $ zt˜zt2(2ˆE(0)eˉMCt)12, $

    i.e.,

    $ zt˜zta2(2ˆE(0))a2eaˉMCt2 $ (3.70)

    for $ 0 < a < 1 $. Meanwhile, combining (3.65) and (3.66), we also have

    $ zt˜zt2zt2+˜zt22(2K0)12eλ02t, $ (3.71)

    i.e.,

    $ zt˜zt1a221a(2K0)1a2eλ0(1a)2t, $ (3.72)

    and

    $ z˜zVzV+˜zV2(2(p+1)K0p1)12eλ02t. $ (3.73)

    Combining (3.70), (3.71) and (3.72), we have

    $ zt˜zt22=zt˜zt2zt˜zta2zt˜zt1a222a(2K0)2a2(2ˆE(0))a2eλ0(2a)aˉMC2t. $ (3.74)

    We choose $ 0 < a < \min\left\{\frac{2\lambda_{0}}{\bar{M}C+\lambda_{0}}, 1\right\} $ such that

    $ λ0(2a)aˉMC>0 $ (3.75)

    in (3.74). Meanwhile, according to (3.70), (3.72) and (3.73), we notice that

    $ z˜zVzt˜zt2=z˜zVzt˜zta2zt˜zt1a222a(2(p+1)K0p1)12(2K0)1a2(2ˆE(0))a2eλ0(2a)aˉMC2t. $ (3.76)

    Due to (3.74) and (3.76), we see that (3.69) turns to

    $ L0(F(z(t))F(˜z(t)))wtdxC4eλ0(p1)tˆE(t)+21aC(2K0)2a2(2ˆE(0))a2eλ0(2a)aˉMC2t+23aCR122(2(p+1)K0p1)12(2K0)1a2(2ˆE(0))a2eλ0(2a)aˉMC2t. $ (3.77)

    By substituting (3.77) into (3.43), we obtain

    $ ˆE(t)ˆE(0)+C4t0eλ0(p1)τˆE(τ)dτ+(2ˆE(0))a2D, $ (3.78)

    where

    $ D:=Nt0eλ1τdτ=Nλ1Nλ1eλ1t. $ (3.79)

    Here, according to (3.79), we notice that $ D\leq\frac{N}{\lambda_{1}} $, which means that (3.78) turns to

    $ ˆE(t)ˆE(0)+(2ˆE(0))a2Nλ1+C4t0eλ0(p1)τˆE(τ)dτ, $ (3.80)

    i.e.,

    $ eλ0(p1)tˆE(t)eλ0(p1)t(ˆE(0)+(2ˆE(0))a2Nλ1)+C4eλ0(p1)tt0eλ0(p1)τˆE(τ)dτ. $ (3.81)

    We define

    $ F(t):=t0eλ0(p1)τˆE(τ)dτ. $ (3.82)

    Thus, we can rewrite (3.81) as

    $ F(t)eλ0(p1)t(ˆE(0)+(2ˆE(0))a2Nλ1)+C4eλ0(p1)tF(t). $ (3.83)

    By applying Gronwall's inequality, (3.83) gives

    $ F(t)(ˆE(0)+(2ˆE(0))a2Nλ1)eC4t0eλ0(p1)τdτt0eλ0(p1)τdτ=(ˆE(0)+(2ˆE(0))a2Nλ1)eC4λ0(p1)(1eλ0(p1)t)1eλ0(p1)tλ0(p1)(ˆE(0)+(2ˆE(0))a2Nλ1)eC4λ0(p1)λ0(p1), $

    which means (3.80) turns to

    $ ˆE(t)(ˆE(0)+(2ˆE(0))a2Nλ1)(1+C4eC4λ0(p1)λ0(p1)). $ (3.84)

    For $ 0 < \rho < 1 $, according to (3.84), we have

    $ ˆE(t)=ˆE(t)ρˆE(t)1ρ(ˆE(0)+(2ˆE(0))a2Nλ1)ρ(1+C4eC4λ0(p1)λ0(p1))ρˆE(t)1ρ. $ (3.85)

    Here, by using Young inequality, we know

    $ ˆE(t)1ρ=(12zt˜zt22+12z˜z2V)1ρ(12(zt2+˜zt2)2+12(zV+˜zV)2)1ρ=(12zt22+zt2˜zt2+12˜zt22+12z2V+zV˜zV  +12˜z2V)1ρ(zt22+˜zt22+z2V+˜z2V)1ρ $ (3.86)

    According to (3.65) and (3.66), we know

    $ p12(p+1)(zt22+z2V)<K0eλ0t $ (3.87)

    and

    $ p12(p+1)(˜zt22+˜z2V)<K0eλ0t. $ (3.88)

    By substituting (3.87) and (3.88) into (3.86), we have

    $ ˆE(t)1ρ(4(p+1)K0p1)1ρeλ0(1ρ)t, $

    which means that (3.85) turns to (3.40).

    In this section, we consider the continuous dependence of the global solution on the initial data for the nonlinear weak damping case of the model equations in problem (1.1) by supposing that $ m\geq1 $, $ r\geq1 $, and $ \hat{m} = \hat{r} = 1 $ in Assumption 2.1, which means that the weak damping terms $ g_j(s) $, $ j = 1, 2 $, take the nonlinear form for $ |s|\geq1 $ and linear form for $ |s| < 1 $. These conditions are applied to improve the estimate (1.2) and reflect the decay property of (1.1), which was clearly clarified in Corollary 2.14 in [5], that is, the condition $ \hat{m} = \hat{r} = 1 $ is necessary to obtain the exponential decay of the energy, which helps to get the exponential decay, and the absence of such linear condition can only lead to the polynomial decay of the energy. Hence although we discuss the nonlinear weak damping case here, we still need to assume that the terms $ g_j(s) $, $ j = 1, 2 $, take the linear form for $ |s| < 1 $.

    Theorem 4.1. (Continuous dependence on initial data for nonlinear weak damping case) Let Assumption 2.1 and Assumption 2.2 hold with $ \hat{r} = \hat{m} = 1 $, $ \mathcal{E}(z_{0}, z_{1}) < \mathcal{M}(s_0-\nu) $, $ \mathcal{E}(\tilde{z}_{0}, \tilde{z}_{1}) < \mathcal{M}(s_0-\nu) $, $ \|z_{0}\|_{V}\leq s_{0}-\nu $, and $ \|\tilde{z}_{0}\|_{V}\leq s_{0}-\nu $ for some $ \nu > 0 $. Let $ z = (u, \phi) $ and $ \tilde{z} = (\tilde{u}, \tilde{\phi}) $ are the global solutions to the problem (1.1) with the initial data $ z_{0} $, $ z_{1} $, and $ \tilde{z}_{0} $, $ \tilde{z}_{1} $, respectively, where $ \mathcal{M} $ and $ s_{0} $ are defined in (2.18) and (2.19), respectively. Then one has

    $ ˆE(t)C5(ˆE(0)+C6(ˆE(0))b02)κeC7t, $ (4.1)

    where

    $ 0<κ<1,C5:=(1+C8TeC8Tθ0(p1)θ0(p1))κ(4(p+1)eθ+˜θˆdp1)1κ,C6:=2b02N1λ2,C7:=θ0(1κ)T,C8:=43R124R124(p1)C(2(p+1)ˆdp1eθ+˜θ)p1,θ0:=θ+˜θ|θ˜θ|2=min{θ,˜θ}, $ (4.2)

    and $ \theta > 0 $, $ \tilde{\theta} > 0 $, and $ T > 0 $ satisfy

    $ E(z(t),zt(t))eθE(z0,z1)eθTt $ (4.3)

    and

    $ E(˜z(t),˜zt(t))e˜θE(˜z0,˜z1)e˜θTt, $ (4.4)
    $ b0:=θ0θ0+ˉMCT, $ (4.5)

    $ \bar{M} $ is defined in (3.42),

    $ N1:=(8eθ+˜θˆd)2b02C2+2CR122(8eθ+˜θˆd)1b02(8(p+1)p1eθ+˜θˆd)12, $

    and

    $ λ2:=θ0(2b0)2Tb0ˉMC2=θ0(2b0)b0ˉMCT2T. $

    Proof. Due to Corollary 2.14 in [5], for any $ T > 0 $, we know that there exist $ \theta $ and $ \tilde{\theta} $ to make (4.3) and (4.4) hold, where $ \theta $ is dependent on $ \mathcal{E}(z_{0}, z_{1}) $ and $ T $, and $ \tilde{\theta} $ is dependent on $ \mathcal{E}(\tilde{z}_{0}, \tilde{z}_{1}) $ and $ T $. According to Proposition 2.11 in [5], the assumptions $ \mathcal{E}(z_{0}, z_{1}) < \mathcal{M}(s_0-\nu) $, $ \|z_{0}\|_{V}\leq s_{0}-\nu $, and $ \mathcal{E}(\tilde{z}_{0}, \tilde{z}_{1}) < \mathcal{M}(s_0-\nu) $, $ \|\tilde{z}_{0}\|_{V}\leq s_{0}-\nu $ give $ z_{0}\in W $ and $ \tilde{z}_{0}\in W $, respectively. Here $ \mathcal{M}(s_0-\nu) < \hat{d} $ can be observed according to (2.17). Thus, we know (3.26) also holds. According to these facts and (2.16), we have

    $ 12zt22+p12(p+1)z2V<E(z(t),zt(t))eθE(z0,z1)eθTt<eθˆdeθTt, $ (4.6)

    and

    $ 12˜zt22+p12(p+1)˜z2V<E(˜z(t),˜zt(t))e˜θE(˜z0,˜z1)e˜θTt<e˜θˆde˜θTt. $ (4.7)

    Due to (4.2), we know that (4.6) and (4.7) turn to

    $ 12zt22+p12(p+1)z2V<eθˆdeθTt<eθ+˜θˆdeθ0Tt $ (4.8)

    and

    $ 12˜zt22+p12(p+1)˜z2V<e˜θˆde˜θTt<eθ+˜θˆdeθ0Tt, $ (4.9)

    respectively. According to (4.8) and (4.9), we have

    $ z4(p1)V<(2(p+1)ˆdp1eθ+˜θ)2(p1)e2θ0(p1)Tt $

    and

    $ ˜z4(p1)V<(2(p+1)ˆdp1eθ+˜θ)2(p1)e2θ0(p1)Tt, $

    which mean

    $ (z4(p1)V+˜z4(p1)V)12<(2(p+1)ˆdp1eθ+˜θ)p1212eθ0(p1)Tt. $ (4.10)

    Next, we need to use the estimate (3.64) to continue this proof. More precisely, by substituting (4.10) into (3.64), we obtain

    $ A3+A4C8eθ0(p1)Tt12z˜z2V+C2wt22C8eθ0(p1)Tt(12z˜z2V+12zt˜zt22)+C2wt22. $ (4.11)

    By substituting (4.11) into (3.61) and the similar process of obtaining (3.69), we have

    $ L0(F(z(t))F(˜z(t)))wtdxC8eθ0(p1)TtˆE(t)+C2wt22+2Cz˜z2zt˜zt2C8eθ0(p1)TtˆE(t)+C2wt22+2CR122z˜zVzt˜zt2. $ (4.12)

    According to (3.60), we know

    $ zt˜zt2(2ˆE(0)eˉMCt)12, $

    i.e.,

    $ zt˜ztb02(2ˆE(0))b02eb0ˉMCt2, $ (4.13)

    where $ b_0 $ is defined by (4.5). Meanwhile, combining (4.8) and (4.9), we know

    $ zt˜zt2zt2+˜zt2(8eθ+˜θˆd)12eθ02Tt, $ (4.14)

    i.e.,

    $ zt˜zt1b02(8eθ+˜θˆd)1b02eθ0(1b0)2Tt. $ (4.15)

    and

    $ z˜zVzV+˜zV(8(p+1)p1eθ+˜θˆd)12eθ02Tt, $ (4.16)

    where $ b_0 > 0 $ and $ 1-b_0 > 0 $ are ensured by (4.5). According to (4.13), (4.14) and (4.15), we have

    $ zt˜zt22=zt˜zt2zt˜ztb02zt˜zt1b02(2ˆE(0))b02(8eθ+˜θˆd)2b02e(θ0(2b0)2Tb0ˉMC2)t. $ (4.17)

    According to (4.5), we have

    $ 0<b0<2θ0θ0+ˉMCT, $ (4.18)

    i.e.,

    $ θ0(2b0)2Tb0ˉMC2>0 $

    in (4.17). Meanwhile, according to (4.13), (4.15) and (4.16), we notice that

    $ z˜zVzt˜zt2=z˜zVzt˜ztb02zt˜zt1b02(2ˆE(0))b02(8eθ+˜θˆd)1b02(8(p+1)p1eθ+˜θˆd)12  e(θ0(2b0)2Tb0ˉMC2)t. $ (4.19)

    Due to (4.17) and (4.19), we know that (4.12) turns to

    $ L0(F(z(t))F(˜z(t)))wtdxC8eθ0(p1)TtˆE(t)+(2ˆE(0))b02(8eθ+˜θˆd)2b02Ce(θ0(2b0)2Tb0ˉMC2)t2+2CR122(2ˆE(0))b02(8eθ+˜θˆd)1b02(8(p+1)p1eθ+˜θˆd)12e(θ0(2b0)2Tb0ˉMC2)t. $ (4.20)

    By substituting (4.20) into (3.43), we obtain

    $ ˆE(t)ˆE(0)+C8t0eθ0(p1)TτˆE(τ)dτ+(2ˆE(0))b02D1, $ (4.21)

    where

    $ D1:=N1t0eλ2τdτ=N1λ2N1λ2eλ2t. $ (4.22)

    Here, according to (4.22), we notice that $ D_{1}\leq\frac{N_{1}}{\lambda_{2}} $, which means that (4.21) turns to

    $ ˆE(t)ˆE(0)+(2ˆE(0))b02N1λ2+C8t0eθ0(p1)TτˆE(τ)dτ, $ (4.23)

    i.e.,

    $ eθ0(p1)TtˆE(t)eθ0(p1)Tt(ˆE(0)+(2ˆE(0))b02N1λ2)+C8eθ0(p1)Ttt0eθ0(p1)TτˆE(τ)dτ. $ (4.24)

    By similar process of obtaining (3.84), we have

    $ ˆE(t)(ˆE(0)+(2ˆE(0))b02N1λ2)(1+C8TeC8Tθ0(p1)θ0(p1)). $ (4.25)

    For $ 0 < \kappa < 1 $, according to (4.25), we know

    $ ˆE(t)=ˆE(t)κˆE(t)1κ(ˆE(0)+(2ˆE(0))b02N1λ2)κ(1+C8TeC8Tθ0(p1)θ0(p1))κˆE(t)1κ. $ (4.26)

    By the similar process of obtaining (3.86), we have

    $ ˆE(t)1κ(zt22+˜zt22+z2V+˜z2V)1κ. $ (4.27)

    According to (4.8) and (4.9), we know

    $ p12(p+1)(zt22+z2V)<eθ+˜θˆdeθ0Tt $ (4.28)

    and

    $ p12(p+1)(˜zt22+˜z2V)<eθ+˜θˆdeθ0Tt. $ (4.29)

    By substituting (4.28) and (4.29) into (4.27), we have

    $ ˆE(t)1κ(4(p+1)eθ+˜θˆdp1)1κeθ0(1κ)Tt, $

    which means that (4.26) turns to (4.1).

    The finite time blowup at the positive initial energy level was established for the linear weak damping case and nonlinear weak damping case in [22], and for the linear weak damping case, the lower and upper bounds of the blowup time were also estimated there. Hence in this section, we shall estimate the lower bound of the blowup time at the positive initial energy level for the nonlinear weak damping case.

    Theorem 5.1. (Lower bound of blowup time for positive initial energy and nonlinear weak damping case) Let Assumption 2.1 and Assumption 2.2 hold, and $ \mathcal{E}(z_{0}, z_{1})\geq 0 $. Suppose $ z(x, t) $ is the solution to problem (1.1). If $ z(x, t) $ blows up at a finite time $ T_{0} $, then we have the estimate of blowup time

    $ T_{0} \geq \int_{G(0)}^\infty\frac{1}{C_{9}y^{p}+C_{10}y+C_{11}}{\rm d}y, $

    where

    $ C9:=(p+1)R2p22p2Mp,C10:=(p+1)M,C11:=(p+1)E(z0,z1)+(p+1)R2p22p2(E(z0,z1))p, $

    and

    $ G(0): = \|z_{0}\|^{p+1}_{p+1}. $

    Proof. Let $ z = (u, \phi) $ be a weak solution to problem (1.1). We suppose that such solution blows up at a finite time $ T_{0} $. Our goal is to obtain an estimate of the lower bound of $ T_{0} $.

    For $ t\in [0, T_{0}) $, we define

    $ G(t):=z(t)p+1p+1=u(t)p+1p+1+ϕ(t)p+1p+1, $ (5.1)

    then, by Hölder inequality and Young inequality, we have

    $ G(t)=(p+1)L0|u|p1uutdx+(p+1)L0|ϕ|p1ϕϕtdx(p+1)L0|u|p|ut|dx+(p+1)L0|ϕ|p|ϕt|dx(p+1)up2put2+(p+1)ϕp2pϕt2p+12(u2p2p+ut22+ϕ2p2p+ϕt22)=p+12(z2p2p+zt22). $ (5.2)

    Next task is to estimate the terms in the last line of (5.2). By (2.14) and (2.16), we obtain

    $ E(z(t),zt(t))=12zt22+12z2VL0F(z(t))dx12zt22+12z2VML0(|u|p+1+|ϕ|p+1)dx=12zt22+12z2VMzp+1p+1. $ (5.3)

    According to (3.16), we know

    $ E(z(t),zt(t))E(z0,z1), t[0,T0), $ (5.4)

    where $ \mathcal{E}(z_{0}, z_{1})\geq 0 $. We notice that (5.3) and (5.4) give

    $ zt22+z2V2E(z0,z1)+2MG(t), $ (5.5)

    which means

    $ z2V2E(z0,z1)+2MG(t), $ (5.6)

    and

    $ zt222E(z0,z1)+2MG(t). $ (5.7)

    Combining (2.5) and (5.6), we see

    $ z2p2pR2p(2E(z0,z1)+2MG(t))p. $ (5.8)

    By substituting (5.7) and (5.8) into (5.2), we have

    $ G(t)(p+1)R2p2(2E(z0,z1)+2MG(t))p+(p+1)(E(z0,z1)+MG(t)). $ (5.9)

    We consider the function $ h(x): = x^{p}, x > 0, p > 1 $. Since $ h^{''}(x) = p(p-1)x^{p-2} > 0 $, $ h(x) $ is a convex function. Thus it gives that

    $ h \left( \frac{\tilde{k}_{1}+\tilde{k}_{2}}{2} \right) \leq \frac{1}{2}h(\tilde{k}_{1})+\frac{1}{2}h(\tilde{k}_{2}), \ \ \tilde{k}_1, \tilde{k}_2\geq 0, $

    that is to say

    $ (\tilde{k}_{1}+\tilde{k}_{2})^{p}\leq 2^{p-1}(\tilde{k}_{1}^{p}+\tilde{k}_{2}^{p}). $

    Then, due to $ \mathcal{E}(z_{0}, z_{1})\geq0 $ and $ G(t)\geq0 $, we can get

    $ (2E(z0,z1)+2MG(t))p2p1((2E(z0,z1))p+(2MG(t))p), $ (5.10)

    which means that (5.9) turns to

    $ G(t)(p+1)R2p22p2Mp(G(t))p+(p+1)MG(t)+(p+1)E(z0,z1)+(p+1)R2p22p2(E(z0,z1))p, $

    i.e.,

    $ G(t)C9(G(t))p+C10G(t)+C111. $ (5.11)

    Recalling the assumption that the solution of problem (1.1) blows up in finite time $ T_{0} $, we have

    $ limtT0G(t)=limtT0z(t)p+1p+1=. $ (5.12)

    Then, integrating both sides of (5.11) on $ (0, T_{0}) $ and combining (5.12), we get

    $ \int_{G(0)}^\infty\frac{1}{C_{9}y^{p}+C_{10}y+C_{11}}{\rm d}y\leq T_{0}. $

    Thus, the proof of Theorem 5.1 is completed.

    In Sept Ⅰ of the proofs of Theorem $ 3.2 $, by the classical form of Gronwall's inequality (integral form) shown in Appendix B.2 of [4], we know that (3.59) gives

    $ ˆE(t)ˆE(0)(1+ˉMCteˉMCt). $ (6.1)

    In (6.1), the growth order of the distance of the solutions, i.e., $ \widehat{E}(t) $, is controlled by the product of an exponential function and a polynomial function, which is higher than that in (1.2) established for the local solution. In Sept Ⅱ of the proofs of Theorem $ 3.2 $, in order to build the growth estimate of $ \widehat{E}(t) $ in the same form as (1.2) for the global solution, i.e., (3.60), we need the following variation of Gronwall's inequality.

    Proposition 6.1. For a nonnegative, summable function $ \zeta(t) $ on $ [0, \bar{T}] $ with satisfying

    $ ζ(t)ˉC1t0ζ(τ)dτ+ˉC2 $ (6.2)

    for the constants $ \bar{C}_{1}, \bar{C}_{2}\geq0 $, one has

    $ ζ(t)ˉC2eˉC1t $ (6.3)

    for a.e. $ 0\leq t\leq \bar{T} $.

    Proof. We use the similar idea of proving the classical form of Gronwall's inequality shown by Appendix B in [4] to give the proofs. We first define the auxiliary function

    $ χ(t):=eˉC1tt0ζ(τ)dτ. $ (6.4)

    By direct calculation, we have

    $ χ(t)=eˉC1t(ζ(t)ˉC1t0ζ(τ)dτ). $ (6.5)

    Substituting (6.2) into (6.5), we have

    $ χ(t)eˉC1tˉC2, $

    which means

    $ t0χ(τ)dτt0eˉC1τˉC2dτ, $

    i.e.,

    $ χ(t)ˉC2ˉC1(1eˉC1t). $ (6.6)

    According to (6.4) and (6.6), we have

    $ t0ζ(τ)dτˉC2ˉC1(eˉC1t1). $ (6.7)

    Substituting (6.7) into (6.2), we obtain (6.3).

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

    Runzhang Xu was supported by the National Natural Science Foundation of China (12271122) and the Fundamental Research Funds for the Central Universities. Chao Yang was supported by the Ph.D. Student Research and Innovation Fund of the Fundamental Research Funds for the Central Universities (3072022GIP2403).

    The authors declare there is no conflict of interest.

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