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Research article Special Issues

Advancing document-level event extraction: Integration across texts and reciprocal feedback

  • Received: 17 August 2023 Revised: 16 October 2023 Accepted: 22 October 2023 Published: 03 November 2023
  • The primary objective of document-level event extraction is to extract relevant event information from lengthy texts. However, many existing methods for document-level event extraction fail to fully incorporate the contextual information that spans across sentences. To overcome this limitation, the present study proposes a document-level event extraction model called Integration Across Texts and Reciprocal Feedback (IATRF). The proposed model constructs a heterogeneous graph and employs a graph convolutional network to enhance the connection between document and entity information. This approach facilitates the acquisition of semantic information enriched with document-level context. Additionally, a Transformer classifier is introduced to transform multiple event types into a multi-label classification task. To tackle the challenge of event argument recognition, this paper introduces the Reciprocal Feedback Argument Extraction strategy. Experimental results conducted on both our COSM dataset and the publicly available ChFinAnn dataset demonstrate that the proposed model outperforms previous methods in terms of F1 value, thus confirming its effectiveness. The IATRF model effectively solves the problems of long-distance document context-aware representation and cross-sentence argument dispersion.

    Citation: Min Zuo, Jiaqi Li, Di Wu, Yingjun Wang, Wei Dong, Jianlei Kong, Kang Hu. Advancing document-level event extraction: Integration across texts and reciprocal feedback[J]. Mathematical Biosciences and Engineering, 2023, 20(11): 20050-20072. doi: 10.3934/mbe.2023888

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  • The primary objective of document-level event extraction is to extract relevant event information from lengthy texts. However, many existing methods for document-level event extraction fail to fully incorporate the contextual information that spans across sentences. To overcome this limitation, the present study proposes a document-level event extraction model called Integration Across Texts and Reciprocal Feedback (IATRF). The proposed model constructs a heterogeneous graph and employs a graph convolutional network to enhance the connection between document and entity information. This approach facilitates the acquisition of semantic information enriched with document-level context. Additionally, a Transformer classifier is introduced to transform multiple event types into a multi-label classification task. To tackle the challenge of event argument recognition, this paper introduces the Reciprocal Feedback Argument Extraction strategy. Experimental results conducted on both our COSM dataset and the publicly available ChFinAnn dataset demonstrate that the proposed model outperforms previous methods in terms of F1 value, thus confirming its effectiveness. The IATRF model effectively solves the problems of long-distance document context-aware representation and cross-sentence argument dispersion.



    In this paper, we consider the following plate equation with Hardy-Hénon potential and polynomial nonlinearity:

    {utt+Δ2u+u=σ|x|αu+|u|p2u,xΩ,t>0,u=uν=0,xΩ,t>0,u(x,0)=u0(x),ut(x,0)=u1(x),xΩ, (1.1)

    where ΩRn is a bounded domain with smooth boundary Ω, ν is the unit outer normal to the boundary Ω at x, α(,n) and σR are constants, and

    2p{<,n=1,2,3,4,2+4n4,n5. (1.2)

    The initial value (u0,u1)H20(Ω)×L2(Ω). Here,

    |x|αu={Ω|xy|αu(y)dy,if α0,Ωu(y)dy,if α=0.

    Plate equations have been investigated for many years due to their importance in some physical areas such as vibration and elasticity theories of solid mechanics. For instance, in the case when σ is identically zero, equation (1.1) becomes an equation with polynomial nonlinearity which arises in aeroelasticity modeling (see, for example, [14,15]), and the problem with (or without) damping, memory, time-delay etc. were studied extensively (see [17,21,25,26,27,33,34,37,40,55,44] and references therein for the topics on well-posedness, global existence, finite time blow-up, global attractor etc.).

    The potential term |x|αu is known in the literature as Hardy potential if α>0, while if α<0 it is known as Hénon potential. This type of potential is important in analyzing many aspects of physical phenomena with singular poles (at origin). For example, the Efimov states (the circumstances that the two-particle attraction is so weak that any two bosons can not form a pair, but the three bosons can be stable bound states): see e.g., [16]; effects on dipole-bound anions in polar molecules: see e.g., [4,7,28]; capture of matter by black holes (via near-horizon limits): see e.g., [9,19]; the motions of cold neutral atoms interacting with thin charged wires (falling in the singularity or scattering): see e.g., [5,12]; the renormalization group of limit cycle in nonrelativistic quantum mechanics: see e.g., [6,8]; and so on. The are a lot of studies of evolution equations with this type of potentials, see, for example, [1,2,3,45,32,50] for parabolic equation, [11,49,52,39] for wave equations, and [24,22,23,42,47,48,53] for Schrödinger equation. However, as far as we know, there seems little studies of fourth-order plate equation with Hardy-Hénon potentials.

    Motivated by the previous studies, in this paper, we will consider a fourth-order plate equation with Hardy-Hénon potential and polynomial nonlinearity, i.e, problem (1.1). We mainly concern with the well-posedness, and the conditions on global existence and finite time blow-up. To state the main results of this paper, we first introduce some notations used in this paper:

    ● Let (X,X) and (Y,Y) be two Banach space such that XY continuously. Then we denotes by CXY the optimal constant of the embedding, i.e.,

    CXY=supϕX{0}ϕYϕX. (1.3)

    ● The norm of the Lebesgue space Lp(Ω), 1p is denoted by p. Especially, we denote 2 by for simplicity.

    ● The inner product of the Hilbert space L2(Ω) is denoted by (,).

    ● The norm of the Sobolev space H20(Ω) is denoted by H2 and

    ϕ2H2=Δϕ2+ϕ2.

    Definition 1.1. Assume α(,n) and σR. Let T>0, u0H20(Ω), and u1L2(Ω). By a weak solution to problem (1.1), we mean a function

    uC([0,T];H20(Ω))C1([0,T];L2(Ω))

    such that u(0)=u0, ut(0)=u1, and

    Ωutϕdx+t0ΩΔuΔϕdxdτ+t0Ωuϕdxdτ=t0Ω(σ|x|αu+|u|p2u)ϕdxdτ+Ωu1ϕdx (1.4)

    holds for any ϕH20(Ω) and 0<tT.

    The local-well posedness of solutions to problem (1.1) is the following theorem:

    Theorem 1.2. Assume p satisfies (1.2), α(,n) and σR. Let (u0,u1)H20(Ω)×L2(Ω). Then there exists a positive constant T depending only on u0H2+u1 such that problem (1.1) admits a weak solution

    uC([0,T];H20(Ω))C1([0,T];L2(Ω)).

    The solution u can be extended to a maximal weak solution in [0,Tmax) such that either

    1. Tmax=, i.e., the problem admits a global weak solution; or

    2. Tmax<, and

    limtTmax(u(t)H2+ut(t))=,

    i.e., the solution blows up at a finite time Tmax.

    Furthermore, then energy E(t) is conservative, i.e.,

    E(t)=E(0),0t<Tmax, (1.5)

    where

    E(t)=12ut2+J(u)(t),0t<Tmax, (1.6)

    and

    E(0)=E(t)|t=0=12u12+J(u0). (1.7)

    Moreover, we have (ut,u)C1[0,Tmax), and

    ddt(ut,u)=ut2I(u). (1.8)

    Here J:H20(Ω)R is a functional defined by

    J(ϕ)=12Δϕ2+12ϕ21pϕppσ2ΩΩϕ(x)ϕ(y)|xy|αdxdy, (1.9)

    and I:H20(Ω)R is a functional defined by

    I(ϕ)=Δϕ2+ϕ2ϕppσΩΩϕ(x)ϕ(y)|xy|αdxdy. (1.10)

    Based on Theorem 1.2, we study the conditions on global existence and finite time blow-up. The first result is about the case that the initial energy is non-positive.

    Theorem 1.3. Assume p satisfies (1.2), α(,n) and σR. Let (u0,u1)H20(Ω)×L2(Ω) satisfy

    1. E(0)<0; or

    2. E(0)=0 and (u0,u1)>0.

    Then the weak solution got in Theorem 1.2 blows up in finite time, i.e., Tmax<. Moreover, Tmax satisfies the following estimates:

    1. If |σ|σ, then

    Tmax{2u02(p2)(u0,u1), if E(0)0 and (u0,u1)>0;4u0(p2)2E(0), if E(0)<0 and (u0,u1)=0;¯T, if E(0)<0 and (u0,u1)<0,

    where

    ¯T=4E(0)u02+2((u0,u1)+(u0,u1)22E(0)u02)22E(0)(p2)(u0,u1)22E(0)u02.

    2. If |σ|>σ and (u0,u1)0, then

    Tmax{(pΛ+2)u02(p2)(u0,u1), if E(0)0 and (u0,u1)>0;2(pΛ+2)u0(p2)2E(0), if E(0)<0 and (u0,u1)=0.

    Here,

    σ=infϕH20(Ω){0}|Δϕ2+ϕ2ΩΩ|xy|αϕ(x)ϕ(y)dxdy| (1.11)

    and

    Λ={(|σ|σ)Rα|Ω|p2u02p,if α0;κ(|σ|σ)|Ω|(p2)22p+p(2nα)2nnu02p,if 0<α<n, (1.12)

    where

    R=supx,yΩ|xy|,κ=πα2Γ(nα2)Γ(2nα2){Γ(n2)Γ(n)}nαn.

    Remark 1. There are two remarks on the above theorem.

    1. Firstly, by Lemma 2.2, if α(,n), σ is well-defined and

    σ{Rα(CH20L1)2,α(,0],1κ(CH20L2n2nα)2,α(0,n),

    where CH20L1 is the optimal constant of the embedding H20(Ω)L1(Ω), CH20L2n2nα is the optimal constant of the embedding H20(Ω)L2n2nα(Ω). So Theorem 1.3 makes sense.

    2. Secondly, for |σ|>σ and (u0,u1)<0. Due to technique reasons, we only show the solution will blow up in finite time, but the upper bound of the blow-up time Tmax is not given. We left the study of this problem as an open question.

    Theorem 1.3 is above the case E(0)0. In order to derive some results for the case E(0)>0, we use the potential well method (see, for example, [10,20,35,36,41,43,51]). Let

    d=inf{J(ϕ):ϕN}, (1.13)

    where J is the functional defined by (1.9), and N is the Nehari manifold defined as

    N={ϕH20(Ω){0}:I(ϕ)=0}. (1.14)

    Here I is the functional defined by (1.10).

    Remark 2. If we assume p satisfies (1.2), α(,n), and σ(σ,σ), where σ is the positive constant defined by (1.11), by Lemma 2.4, d is a positive constant and

    dp22p((1|σ|σ)C2H20Lp)pp2,

    where CH20Lp is the optimal constant of the embedding H20(Ω)Lp(Ω).

    Theorem 1.4. (Global existence for 0<E(0)d). Assume p satisfies (1.2), α(,n), and σ(σ,σ), where σ is the positive constant defined by (1.11). Let u be the weak solution got in Theorem 1.2 with (u0,u1)H20(Ω)×L2(Ω) satisfying I(u0)>0 and 0<E(0)d, where E(0),I,d are defined in (1.7), (1.10), (1.13) respectively. Then u exists globally, i.e., Tmax=.

    Theorem 1.5. (Blow-up for 0<E(0)d) Assume p satisfies (1.2), α(,n), and σ(σ,σ), where σ is the positive constant defined by (1.11). Let u be the weak solution got in Theorem 1.2 with (u0,u1)H20(Ω)×L2(Ω) satisfying

    1. E(0)<d,I(u0)<0; or

    2. E(0)=d,I(u0)<0,(u0,u1)>0,

    where E(0),I,d are defined in (1.7), (1.10), (1.13) respectively. Then u blows up in finite time, i.e., Tmax<. Moreover, Tmax satisfies the following estimates:

    Tmax{2u02(p2)(u0,u1), if E(0)d and (u0,u1)>0;4u0(p2)2(dE(0)), if E(0)<d and (u0,u1)=0;ˆT, if E(0)<d and (u0,u1)<0,

    where

    ˆT=4(dE(0))u02+2((u0,u1)+(u0,u1)2+(2(dE(0)))u02)2(2(dE(0)))(p2)(u0,u1)2+(2(dE(0)))u02.

    The organization of the rest of this paper is as follows: In Section 2, we give some preliminaries, which will be used in this paper; In Section 3, we study the well-posed of solutions by semigroup theory and prove Theorem 1.2; In Section 4, we study the conditions on global existence and finite time blow-up and prove Theorems 1.3, 1.4 and 1.5.

    The following well-known Hardy-Littlewood-Sobolev inequality can be found in [31]:

    Lemma 2.1. Let q,r>1 and 0<θ<n with

    1q+nθn+1r=2.

    Let uLq(Rn) and vLr(Rn). Then there exists a sharp positive constant C depending only on n,α and q such that

    |RnRnu(x)v(y)|xy|nθdxdy|κuqvr.

    If

    q=r=2nn+θ,

    then

    C=κ:=πnθ2Γ(θ2)Γ(n+θ2){Γ(n2)Γ(n)}θn. (2.15)

    If uLq(Ω) and vLr(Ω) with q and r satisfying the assumptions, by valuing them 0 in RnΩ, it also holds

    |ΩΩu(x)v(y)|xy|nθdxdy|Cuqvr. (2.16)

    Lemma 2.2. Let α(,n) and σ be the constant defined in (1.11). Then σ is well-defined and

    σ{Rα(CH20L1)2,α(,0],1κ(CH20L2n2nα)2,α(0,n),

    where CH20L1 is the optimal constant of the embedding H20(Ω)L1(Ω), CH20L2n2nα is the optimal constant of the embedding H20(Ω)L2n2nα(Ω), κ is the constant given in (2.15) with θ=nα, and

    R=supx,yΩ|xy|<.

    Proof.  If α(,0]_, we have,

    |ΩΩ|xy|αϕ(x)ϕ(y)dxdy|(Ω|ϕ(x)|dx)2RαC2H20L1Rα(Δϕ2+ϕ2).

     If α(0,n)_, by Hardy-Littlewood-Sobolev inequality (see (2.16) of Lemma 2.1) with q=r=2n2nα and H20(Ω)L2n2nα(Ω) with constant CH20L2n2nα,

    |ΩΩ|xy|αϕ(x)ϕ(y)dxdy|κϕ22n2nακC2H20L2n2nα(Δϕ2+ϕ2).

    Lemma 2.3. [29,30] Suppose F(t)C2[0,T) is a nonnegative function satisfying

    F(t)F(t)(1+r)(F(t))20, (2.17)

    where 0<T+ and r is a positive constant. If F(0)>0 and F(0)>0, then

    TF(0)rF(0)<+ (2.18)

    and F(t)+ as tT.

    Lemma 2.4. Assume p satisfies (1.2), α(,n), and σ(σ,σ), where σ is the positive constant defined by (1.11). Let d be the constant defined in (1.13). Then we have

    d=inf{supλ0J(λϕ):ϕH20(Ω){0}}, (2.19)

    and

    dp22p((1|σ|σ)C2H20Lp)pp2, (2.20)

    where CH20Lp is the optimal constant of the embedding H20(Ω)Lp(Ω).

    Proof. Firstly, we prove (2.19). For any ϕH20(Ω){0}, since σ[0,σ), by means of a simple calculation, we find there exists a unique positive constant ˆλ defined by

    ˆλ=(Δϕ2+ϕ2σΩΩ|xy|αϕ(x)ϕ(y)dxdyϕpp)1p2, (2.21)

    such that

    supλ0J(λϕ)=J(ˆλϕ),ˆλϕN. (2.22)

    Then,

    inf{supλ0J(λϕ):ϕH20(Ω){0}}=inf{J(ˆλϕ):ϕH20(Ω){0}}inf{J(ϕ):ϕN}.

    On the other hand, for any ϕN, we have ˆλ=1. Then

    inf{J(ϕ):ϕN}=inf{supλ0J(λϕ):ϕN}inf{supλ0J(λϕ):ϕH20(Ω){0}}.

    Then (2.19) follows from the above two inequalities.

    Secondly, we prove (2.20), by (1.11), we have

    d=p22pinf{ˆλpϕpp:ϕH20(Ω){0}}=p22p(infϕH20(Ω){0}Δϕ2+ϕ2σΩΩ|xy|αϕ(x)ϕ(y)dxdyϕ2p)pp2p22p(infϕH20(Ω){0}Δϕ2+ϕ2|σ||ΩΩ|xy|αϕ(x)ϕ(y)dxdy|ϕ2p)pp2p22p(infϕH20(Ω){0}(1|σ|σ)Δϕ2+ϕ2ϕ2p)pp2=p22p((1σσ)C2H20Lp)pp2. (2.23)

    Lemma 2.5. Assume p satisfies (1.2), α(,n), and σ(σ,σ), where σ is the positive constant defined by (1.11). Let

    W={ϕH20(Ω):I(ϕ)>0,J(ϕ)<d}{0}. (2.24)
    V={uH20(Ω):I(ϕ)<0,J(ϕ)<d}, (2.25)

    where d is the positive constant defined in (1.13). Then

    ϕpp2pdp2,ϕW (2.26)

    and

    Δϕ2+ϕ2((1|σ|σ)(CH20Lp)p)2p2,ϕV, (2.27)

    where CH20Lp is the optimal constant of the embedding H20(Ω)Lp(Ω). Moreover, W=W1, and V1=V, where

    W1={ϕH20(Ω):J(ϕ)<d,Δϕ2+ϕ2<2pdp2+σCϕ}, (2.28)

    and

    V1={ϕH20(Ω):J(ϕ)<d,Δϕ2+ϕ2>2pdp2+σCϕ}. (2.29)

    Here,

    Cϕ=ΩΩϕ(x)ϕ(y)|xy|αdxdy.

    Proof. Step 1. We prove (2.26). By the definition of J and I (see (1.9) and (1.10)), we have

    2J(ϕ)I(ϕ)=p2pϕpp,ϕH20(Ω). (2.30)

    For any ϕW, since I(ϕ)>0 and J(ϕ)<d, (2.26) follows from the above inequality.

    Step 2. We prove (2.27). For any ϕV, by the definition of I in (1.10), |σ|<σ, it follows from (1.11) and I(ϕ)<0 that

    (CH20Lp)p(Δϕ2+ϕ2)p2ϕpp>Δϕ2+ϕ2σΩΩϕ(x)ϕ(y)|xy|αdxdy(1|σ|σ)(Δϕ2+ϕ2), (2.31)

    which implies (2.27).

    Step 3. We show that V1=V.

    Firstly, for any ϕV1, by (1.9) and the definition of V1, we get

    Δϕ2+ϕ2>2pdp2+σΩΩϕ(x)ϕ(y)|xy|αdxdy,12(Δϕ2+ϕ2)<d+1pϕpp+σ2ΩΩϕ(x)ϕ(y)|xy|αdxdy,

    which implies

    ϕpp>2pdp2.

    Thus, by the definition of I in (1.10) and the above two inequalities,

    I(ϕ)=Δϕ2+ϕ2ϕppσΩΩϕ(x)ϕ(y)|xy|αdxdy2dp2pϕpp<0,

    which, together with J(ϕ)<d, implies ϕV, then V1V.

    Secondly, we prove VV1. For any ϕV, since |σ|<σ, it follow from (2.27) and (2.31) that

    Δϕ2+ϕ2σΩΩϕ(x)ϕ(y)|xy|αdxdy(1|σ|σ)((1|σ|σ)(CH20Lp)p)2p2>0.

    Then, in view of I(ϕ)<0, i.e.,

    ϕpp>Δϕ2+ϕ2σΩΩϕ(x)ϕ(y)|xy|αdxdy,

    we get

    ϕ2pp2p>(Δϕ2+ϕ2σΩΩϕ(x)ϕ(y)|xy|αdxdy)2p2ϕ2pp2p>(Δϕ2+ϕ2σΩΩϕ(x)ϕ(y)|xy|αdxdy)pp21Δϕ2+ϕ2σΩΩϕ(x)ϕ(y)|xy|αdxdy         >(Δϕ2+ϕ2σΩΩϕ(x)ϕ(y)|xy|αdxdyϕ2p)pp2,

    which, together with the second line of (2.23), implies

    Δϕ2+ϕ2σΩΩϕ(x)ϕ(y)|xy|αdxdy>2pdp2.

    Since J(ϕ)<d, the above inequality infers that ϕV1, and then VV1.

    Step 4. We show that W=W1.

    Firstly we prove WW1. In fact, for any ϕW, if ϕ=0, it is obvious that ϕW1; if ϕ0, in view of the definition of W, we get

    Δϕ2+ϕ2σΩΩϕ(x)ϕ(y)|xy|αdxdy>ϕpp,12(Δϕ2+ϕ2)<d+1pupp+12σΩΩϕ(x)ϕ(y)|xy|αdxdy.

    The above two inequalities imply

    \begin{equation*} \Delta\phi\|^2+\|\phi\|^2 < \frac{2pd}{p-2}+\sigma \int\limits_\Omega \int\limits_ \Omega\frac{\phi(x)\phi(y)}{|x-y|^ \alpha}dxdy, \end{equation*}

    which, together with J(u)<d , implies \phi\in \mathcal{W}_1 , then \mathcal{W}\subset \mathcal{W}_1 .

    Secondly we prove \mathcal{W}_1\subset\mathcal{W} by contradiction argument. If there exists \phi\in \mathcal{W}_1\setminus \mathcal{W} . Then we have

    \begin{align} &J(\phi) < d\; \; \; \; \; \hbox{(by (2.28))}, \end{align} (2.32)
    \begin{align} &\|\Delta\phi\|^2+\|\phi\|^2 < \frac{2pd}{p-2}+\sigma \int\limits_\Omega \int\limits_ \Omega\frac{\phi(x)\phi(y)}{|x-y|^ \alpha}dxdy\; \; \; \; \; \hbox{(by (2.28))}, \end{align} (2.33)
    \begin{align} &I(\phi)\leq 0, \; \; \phi\neq0\; \; \; \; \; \hbox{(by (2.24) and (2.32))}. \end{align} (2.34)

    If I(\phi)<0 , the by (2.32), we get \phi\in \mathcal{V} (see (2.25)), and then \phi\in \mathcal{V}_1 (since \mathcal{V} = \mathcal{V}_1 has been proved in Step 3), which, together with (2.29), contradicts (2.33); If I(\phi) = 0 , then by \phi\neq0 , we get \phi\in \mathcal{N} (see (1.14)), and then J(\phi)\geq d (see (1.13)), which contradicts (2.32). So, \phi\in \mathcal{W} , and then \mathcal{W}_1\subset \mathcal{W} .

    Lemma 2.6. Assume p satisfies (1.2), \alpha\in(-\infty, n) , and \sigma\in(-\sigma^*, \sigma^*) , where \sigma^* is the positive constant defined by (1.11). Let

    u\in C \left([0, {T_{\max}});H_0^2( \Omega) \right)\cap C^1 \left([0, {T_{\max}});L^2( \Omega) \right)

    be the maximal weak solution to (1.1) with initial value (u_0, u_1)\in H_0^2( \Omega)\times L^2( \Omega) got in Theorem 1.2.

    1. If there exists a t_0\in [0, T_{\max}) such that E(t_0)<d , then u(t)\in \mathcal{W} for t\in[t_0, {T_{\max}}) provided that u(t_0) \in \mathcal{W} ;

    2. If there exists a t_0\in [0, T_{\max}) such that either E(t_0)<d or E(t_0) = d and (u_t, u)_{t = t_0}\geq 0 , then u(t)\in \mathcal{V} for t\in[t_0, {T_{\max}}) provided that u(t_0) \in \mathcal{V} ,

    where E(t) is the energy functional defined in (1.6), \mathcal{W} and \mathcal{V} is is the sets defined in (2.24) and (2.25) respectively, d constant defined in (1.13).

    Proof. Firstly, we proof the first part by contradiction argument. Actually, if the conclusion is incorrect, by using u\in C \left([0, {T_{\max}});H_0^2( \Omega) \right)\cap C^1 \left([0, {T_{\max}});L^2( \Omega) \right) and u(t_0)\in \mathcal{W} , there must exist a t_1\in(t_0, {T_{\max}}) such that

    \begin{equation*} u(t)\in \mathcal{W}, \; \; \; t\in[t_0, t_1);\hbox{ and }u(t_1)\in \partial \mathcal{W}. \end{equation*}

    Since the energy is conservative (see (1.5)), E(t) = \frac{1}{2}\left\| u_t\right\|^{2}+J(u)(t) (see (1.6)), and E(0)<d , we get

    \begin{equation} J(u)(t_1)\leq E(t_1) = E(t_0) < d. \end{equation} (2.35)

    Then by u(t_1)\in \partial \mathcal{W} and the definition of \mathcal{W} (see (2.24)), it follows I(u)(t_1) = 0 and u(t_1)\neq0 . So u(t_1)\in \mathcal{N} (see the definition of \mathcal{N} in (1.14)), then by the definition of d (see (1.13)), it follows J(u)(t_1)\geq d , which contradicts (2.35).

    Secondly, we proof the second part. In the case E\left(t_{0}\right)<d , in view of (2.27), the proof is similar to the first part. We only prove the case

    \begin{equation*} E\left(t_{0}\right) = d \text { and }(u_t, u)_{t = t_0}\geq 0 \end{equation*}

    in detail. Arguing by contradiction, if the conclusion is incorrect, by u\left(t_{0}\right) \in \mathcal{V} and u \in C\left(\left[0, T_{\max }\right) ; H_{0}^{2}\right), we obtain that there must exist a t_{1} \in\left(t_{0}, T_{\max }\right) such that u(\cdot, t) \in \mathcal{V} , t \in\left[t_{0}, t_{1}\right) and u(t_1)\in \partial \mathcal{V} , i.e. (see (2.25)),

    (i) : J\left(u\right)\left(t_{1}\right)<d, I\left(u\right)\left(t_{1}\right) = 0 ; or

    (ii): J\left(u\right)\left(t_{1}\right) = d, I\left(u\right)\left(t_{1}\right) \leq 0 .

    Due to u(t) \in \mathcal{V} for any t \in\left[t_{0}, t_{1}\right) and u \in C\left(\left[0, T_{\max }\right) ; H_0 ^{2}\right) , by (2.27), we get

    \begin{align} \|\Delta u(t_1)\|^2+\|u(t_1)\|^2&\geq \left( \left(1-\frac{|\sigma|}{\sigma^*} \right) \left( C_{H_0^2 \rightarrow L^p} \right)^{-p} \right)^{\frac{2}{p-2}}. \end{align} (2.36)

    If (i) is true, by using I\left(u\left(t_{1}\right)\right) = 0 and (2.36), we have u\left(t_{1}\right) \in \mathcal{N} (see (1.14)), which implies J\left(u\right)\left(t_{1}\right) \geq d (see (1.13)), a contradiction.

    If (ii) is true, by (1.5) and E(t) = \frac{1}{2}\left\|u_t\right\|^{2}+J(u)(t) (see (1.6)), we get

    \begin{equation} d = E(t_0) = \frac{1}{2}\left\|u_{t}\left(t_{1}\right)\right\|^{2}+J\left(u\right)\left(t_{1}\right). \end{equation} (2.37)

    Combining (2.37) and J\left(u\right)\left(t_{1}\right) = d, we have

    \begin{equation} \left\|u_{t}\left( t_{1}\right)\right\|^{2} = 0. \end{equation} (2.38)

    Utilizing Cauchy-Schwartz's inequality, we obtain that

    \begin{equation} (u_t, u)|_{t = t_1} \leq\left\|u_{t}\left(t_{1}\right)\right\| \left\|u\left(t_{1}\right)\right\| = 0. \end{equation} (2.39)

    Integrating (1.8) over [0, t] , we obtain

    \begin{equation} (u_t, u)+ \int\limits_0^tI(u)(\tau)d\tau- \int\limits_0^t\|u_{\tau}\|^2d\tau = (u_1, u_0), \; \; \; 0\leq t < {T_{\max}}. \end{equation} (2.40)

    By (2.40), we get

    \begin{equation*} \begin{split} &(u_t, u)|_{t = t_0}+ \int\limits_0^{t_0}I(u)(\tau)d\tau- \int\limits_0^{t_0}\|u_{\tau}\|^2d\tau\\ = &(u_t, u)|_{t = t_1}+ \int\limits_0^{t_1}I(u)(\tau)d\tau- \int\limits_0^{t_1}\|u_{\tau}\|^2d\tau. \end{split} \end{equation*}

    Then,

    \begin{equation*} (u_t, u)|_{t = t_0}-(u_t, u)|_{t = t_1} = \int\limits_{t_0}^{t_1}I(u)(\tau)d\tau- \int\limits_{t_0}^{t_1}\|u_{\tau}\|^2d\tau. \end{equation*}

    Since I(u)(t)<0 (by using u(t)\in \mathcal{V} for t\in [t_0, t_1) ) and (u_t, u)|_{t = t_0}\geq0 , we get from the above equality that

    \begin{equation*} (u_t, u)|_{t = t_1} = (u_t, u)|_{t = t_0}- \int\limits_{t_0}^{t_1}I(u)(\tau)d\tau+ \int\limits_{t_0}^{t_1}\|u_{\tau}\|^2d\tau > 0, \end{equation*}

    which contradicts (2.39).

    In this section, we study local well-posedness of solutions to (1.1) by semigroup theory. To this end, first, we introduce some fundamental theory on semigroup theory.

    Suppose that H is a Hilbert space with inner product (\cdot, \cdot)_H and norm

    \begin{equation*} \|\Phi\|_H = \sqrt{(\Phi, \Phi)_H}, \; \; \; \Phi\in H. \end{equation*}

    Suppose F is a nonlinear operator from H into H . F is said to satisfy the local Lipschitz condition if for any positive constant M>0 , there is a positive constant L_M depending only on M such that when U, V\in H , \|U\|_H\leq M and \|V\|_H\leq M ,

    \begin{equation} \|F(U)-F(V)\|_H\leq M\|U-V\|_H. \end{equation} (3.41)

    Consider the following abstract semilinear evolution equation

    \begin{equation} \left\{\begin{split} & U_t+AU = F(U), \; \; \; t > 0, \\ &U(0) = U_0, \end{split} \right. \end{equation} (3.42)

    where A:D(A)\rightarrow H is a densely defined linear operator on H , i.e., A is linear and D(A) is dense in H , where D(A) = \{\Phi\in H: A\Phi\in H\} .

    First, we introduce the Lumer-Phillips theorem (see, for example, [38,Theorem 1.2.3] and [54,Lemma 2.2.3]):

    Lemma 3.1. The necessary and sufficient conditions for A generating a contraction C_0 -semigroup \{e^{tA}\}_{t\geq0} on H are

    1. (A\Phi, \Phi)_H\leq0 for all \Phi\in D(A) , and

    2. R(I-A) = H .

    Here R(I+A) = \{\Phi+A\Phi:\Phi\in D(A)\} is the range of the operator.

    Next, we introduce the local well-posedness results for (3.42), which can be found in [54,Theorems 2.5.4 and 2.5.5]:

    Lemma 3.2. Suppose that A generates a contraction C_0 -semigroup \{e^{tA}\}_{t\geq0} on H , and F is a nonlinear operator from H into H satisfying the local Lipschitz condition. Then for any U_0\in H , there is a positive constant T depending only on \|U_0\|_H such that problem (3.42) admits a unique local mild solution U(t) , i.e., U\in C([0, T], H) and satisfies

    \begin{equation} U(t) = e^{tA}U_0+ \int\limits_0^te^{(t-\tau)A}F(U(\tau))d\tau, \ \ t\in[0, T]. \end{equation} (3.43)

    The solution U can be extended to a maximal mild solution in [0, {T_{\max}}) such that either

    1. {T_{\max}} = \infty , i.e., the problem admits a global mild solution; or

    2. {T_{\max}}<\infty , and

    \begin{equation*} \lim\limits_{t\uparrow {T_{\max}}}\|U(t)\|_H = \infty, \end{equation*}

    i.e., the solution blows up at a finite time {T_{\max}} .

    Furthermore, if u_0\in D(A) , then u\in C \left([0, {T_{\max}});D(A) \right)\cap C^1 \left([0, {T_{\max}});H \right) is classical solution.

    By introduction U = (u, v): = (u, u_t) , U_0 = (u_0, u_1) , and

    \begin{equation} \begin{split} A& = \left(\begin{array}{cc} 0 & \quad I\\ -\Delta^2-I & \quad 0\\ \end{array}\right), \\F(U)& = \left( \begin{array}{c} 0 \\ \sigma\left|x\right|^{-\alpha}* u+\left|u\right|^{p-2}u \\ \end{array} \right), \end{split} \end{equation} (3.44)

    where I is the identity operator, (1.1) can be equivalently written as the following system

    \begin{equation} \left\{\begin{array}{ll} {U_t = AU+F(U)} \quad & {x \in \Omega, t > 0}, \\ {U = \frac{\partial U}{\partial \nu} = 0, } & {x \in \partial \Omega, t > 0} \\ {U(x, 0) = U_{0}(x), } \; \; \; & {x \in \Omega}. \end{array}\right. \end{equation} (3.45)

    In the next lemma, we show A generates a contraction semigroup \{e^{tA}\}_{t\geq0} on H_0^2( \Omega)\times L^2( \Omega) .

    Lemma 3.3. Let A be the operator defined in (3.44). Then A generates a contraction semigroup \{e^{tA}\}_{t\geq0} on H_0^2( \Omega)\times L^2( \Omega) .

    Proof. Let H: = H_0 ^2( \Omega)\times L^2( \Omega) , then H is a Hilbert space with inner produce (\cdot, \cdot)_H defined as

    \begin{equation} (\Phi, \Psi)_H: = \int\limits_{ \Omega} \left(\Delta\varphi_1\Delta\psi_1+\varphi_1\psi_1+\varphi_2\psi_2 \right) dx, \end{equation} (3.46)

    where \Psi = (\varphi_1, \varphi_2), \; \Psi = (\psi_1, \psi_2)\in H . Then

    \begin{equation*} \|\Phi\|_H = (\Phi, \Phi)_H = \|\phi_1\|_{H^2}+\|\varphi_2\|. \end{equation*}

    Let A be the linear operator defined in (3.44), then

    \begin{equation*} A:D(A) = (H^4( \Omega) \cap H_0 ^2( \Omega))\times H_0 ^2( \Omega)\subset H\rightarrow H. \end{equation*}

    Next we show A generates a C_0 -semigroup on H by using Lemma 3.1. It is obvious D(A) is dense in H , and for any \Phi = (\varphi_1, \varphi_2)\in D(A) , we have

    \begin{equation} \begin{split} (A\Phi, \Phi)_H = & \left( \left(\varphi_2, -\Delta^2\varphi_1-\varphi_1 \right), (\varphi_1, \varphi_2) \right)_H\\ = & \int\limits_{ \Omega} \left(\Delta\varphi_2\Delta\varphi_1+\varphi_2\varphi_1+ \left(-\Delta^2\varphi_1-\varphi_1 \right)\varphi_2 \right) dx\\ = &0. \end{split} \end{equation} (3.47)

    Next we show R(I-A) = H . Fixed any f = (f_1, f_2)\in H , since f_1+f_2\in L^2( \Omega) , by standard theory of elliptic equation, the following problem

    \begin{equation} \left\{\begin{split} &\Delta^2 u+2u = f_1+f_2, \; \; \; &&x\in \Omega, \\ &u = \frac{ \partial u}{ \partial \nu} = 0, &&x\in \partial \Omega \end{split} \right. \end{equation} (3.48)

    admits a unique solution u\in H^4( \Omega)\cap H_0^2( \Omega) . Let v = u-f_1\in H_0^2( \Omega) . Then U = (u, v) satisfies

    \begin{align*} (I-A)U = &\left(\begin{array}{cc} I & \quad -I\\ \Delta^2+I & \quad I\\ \end{array}\right)\left( \begin{array}{c} u \\ v \\ \end{array} \right)\\ = &\left( \begin{array}{c} u-v \\ \Delta^2 u+u+v \\ \end{array} \right)\\ = &\left( \begin{array}{c} f_1 \\ \Delta^2 u+2u-f_1 \\ \end{array} \right)\\ = &\left( \begin{array}{c} f_1 \\ f_2 \\ \end{array} \right) = f, \end{align*}

    which implies R(I-A) = H . Then by Lemma 3.1, A generates a contraction C_0 -semigroup on H .

    Next, we show (3.45) admits a mild solution.

    Lemma 3.4. Assume \alpha\in(-\infty, n) and \sigma\geq0 . Let H be the Hilbert space defined in Lemma 3.3, and U_0 = (u_0, u_1)\in H = H_0^2( \Omega)\times L^2( \Omega) . Then there exists a positive constant T depending only on \|U_0\|_H = \|u_0\|_{H^2}+\|u_1\| such that problem (3.45) admits a unique mild solution U(t) , i.e., U = (u, u_t)\in C \left([0, T]; H \right) and satisfies

    \begin{equation} U(t) = e^{tA}U_0+ \int\limits_0^te^{(t-\tau)A}F(U(\tau))d\tau, \; \; \; 0\leq t\leq T. \end{equation} (3.49)

    The solution U can be extended to a maximal weak solution in [0, {T_{\max}}) such that either

    1. {T_{\max}} = \infty , i.e., the problem admits a global mild solution; or

    2. {T_{\max}}<\infty , and

    \begin{equation*} \lim\limits_{t\uparrow {T_{\max}}}\|U(t)\|_H = \lim\limits_{t\uparrow {T_{\max}}} \left(\|u(t)\|_{H^2}+\|u_t(t)\| \right) = \infty, \end{equation*}

    i.e., the solution blows up at a finite time {T_{\max}} .

    Furthermore, it holds

    \begin{equation} \|U(t)\|_H = \|U_0\|_H+2 \int\limits_0^t(F(U(\tau)), U(\tau))_Hd\tau. \end{equation} (3.50)

    Proof. Let F be the nonlinear function defined in (3.44). In view of Lemmas 3.2 and 3.3, to prove local existence, uniqueness, and extension of mild solutions, we only need to show F:H\rightarrow H satisfying the local Lipschitz condition.

    \underline{\text{First,we}\ \text{show}\ F(H)\subset H}. For any U = (u, v)\in H , by (3.44), to prove F(U)\in H , we only need to prove,

    \begin{equation} \sigma\left|x\right|^{-\alpha}* u+\left|u\right|^{p-2}u\in L^2( \Omega), \; \; \; \forall u\in H_0^2( \Omega). \end{equation} (3.51)

    Since H_0^2( \Omega)\hookrightarrow L^{2(p-1)}( \Omega) (see (1.2)), it is obvious |u|^{p-2}u\in L^2( \Omega) .

    Next we show \left|x\right|^{-\alpha}* u\in L^2( \Omega) . According the range of \alpha , we divide the proof into two cases: \alpha\leq 0 and \alpha\in(0, n) .

    Case 1. \alpha\leq 0 . Since \Omega is bound, we have

    \begin{equation*} R = \sup\limits_{x, y\in \Omega}|x-y| < \infty. \end{equation*}

    Then, by Hölder's inequality,

    \begin{equation} \begin{split} \left\||x|^{- \alpha}*u\right\|^2& = \int\limits_ \Omega \left( \int\limits_ \Omega|x-y|^{- \alpha}u(y)dy \right)^2dx\\ &\leq R^{-2 \alpha}| \Omega| \left( \int\limits_ \Omega u(y)dy \right)^2\\ &\leq R^{-2 \alpha}| \Omega|^2\|u\|^2 < \infty. \end{split} \end{equation} (3.52)

    Case 2. 0< \alpha<n . Since \alpha<n , it follows \frac{2n}{2n- \alpha}<2 . For any \phi\in L^2( \Omega) , there exists a positive C_ \Omega depending only on \Omega such that \|\phi\|_{\frac{2n}{2n- \alpha}}\leq C_ \Omega\|\phi\| . Since

    \begin{equation*} \frac{1}{\frac{2n}{2n- \alpha}}+\frac{ \alpha}{n}+\frac{1}{\frac{2n}{2n- \alpha}} = 2, \end{equation*}

    by using (2.16) with q = r = \frac{2n}{2n- \alpha} , \theta = n- \alpha , we get

    \begin{align*} \int\limits_{ \Omega} \left(|x|^{- \alpha}*u \right)(x)\phi(x)dx& = \int\limits_ \Omega \int\limits_ \Omega\frac{u(y)\phi(x)}{|x-y|^ \alpha}dxdy\\ &\leq\kappa\|u\|_{\frac{2n}{2n- \alpha}}\|\phi\|_{\frac{2n}{2n- \alpha}}\\ &\leq \kappa C_ \Omega^2\|u\|\|\phi\|. \end{align*}

    Then we get

    \begin{equation} \begin{split} \left\||x|^{- \alpha}*u\right\|& = \sup\limits_{\phi\in L^2( \Omega), \|\phi\| = 1} \int\limits_{ \Omega} \left(|x|^{- \alpha}*u \right)(x)\phi(x)dx\\ &\leq \kappa C_ \Omega^2\|u\| < \infty. \end{split} \end{equation} (3.53)

    So (3.51) is true.

    \underline{\text{Next}\ \text{we}\ \text{show}\ F\ \text{is}\ \text{locally}\ \text{Lipschitz}\ \text{continuous}}. Let U_1 = (u_1, v_1)\in H and U_2 = (u_2, v_2)\in H be such that

    \begin{equation} \|U_1\|_H = \|u_1\|_{H^2}+\|v_1\|\leq M, \; \; \|U_2\|_H = \|u_2\|_{H^2}+\|v_2\|\leq M, \end{equation} (3.54)

    where M is a positive constant. Let

    \begin{equation*} \chi = \left\{ \begin{array}{ll} R^{- \alpha}| \Omega|, \; \; \; & \alpha\leq0 \\ \kappa C_ \Omega^2, & 0 < \alpha < n. \end{array} \right. \end{equation*}

    Then, by (3.52) and (3.53),

    \begin{align*} \|F(U_1)-F(U_2)\|_H\leq&|\sigma|\left\||x|^{- \alpha}*(u_1-u_2)\right\|+\left\||u_1|^{p-2}u_1-|u_2|^{p-2}u_2\right\|\\ \leq&|\sigma|\chi\|u_1-u_2\|\\ &+(p-2)\left\| \int\limits_0^1| \theta u_1+(1- \theta) u_2|^{p-2}d \theta(u_1-u_2)\right\|. \end{align*}

    Case 1. n = 1, 2, 3, 4 . Since H_0^2( \Omega)\hookrightarrow L^\infty( \Omega) with optimal constant C_{H_0^2 \rightarrow L^\infty} , in view of (3.54), we have

    \begin{align*} \left\| \int\limits_0^1| \theta u_1+(1- \theta) u_2|^{p-2}d \theta(u_1-u_2)\right\|\leq &\left\| \int\limits_0^1| \theta u_1+(1- \theta) u_2|^{p-2}d \theta\right\|_\infty\|u_1-u_2\|\\ \leq & \left(\|u_1\|_\infty+\|u_2\|_\infty \right)^{p-2}\|u_1-u_2\|\\ \leq & \left(2C_{H_0^2 \rightarrow L^\infty} M \right)^{p-2}\|u_1-u_2\|. \end{align*}

    Case 2. n\geq 5 . Since H_0^2( \Omega)\hookrightarrow L^\frac{2n}{n-4} ( \Omega) with optimal constant C_{H_0^2 \rightarrow L^\frac{2n}{n-4}} and H_0^2( \Omega)\hookrightarrow L^\frac{n(p-2)}{2}( \Omega) with optimal constant C_{H_0^2 \rightarrow L^\frac{n(p-2)}{2}} , in view of (3.54), we have

    \begin{align*} &\left\| \int\limits_0^1| \theta u_1+(1- \theta) u_2|^{p-2}d \theta(u_1-u_2)\right\|^2\\ \leq & \left( \int\limits_ \Omega \left( \int\limits_0^1| \theta u_1+(1- \theta) u_2|^{p-2}d \theta \right)^\frac{n}{2}dx \right)^{\frac{n}{4}}\|u_1-u_2\|_{\frac{2n}{n-4}}^2\\ \leq & 2^{\frac{np}{2}} \left( C_{H_0^2 \rightarrow L^\frac{2n}{n-4}} \right)^2 \left( \int\limits_ \Omega \left(|u_1|^{\frac{n(p-2)}{2}}+|u_2|^{\frac{n(p-2)}{2}} \right) dx \right)^{\frac{n}{4}}\|u_1-u_2\|_{H^2}^2\\ \leq & 2^{\frac{np}{2}+\frac{n}{4}} \left( C_{H_0^2 \rightarrow L^\frac{2n}{n-4}} \right)^2 \left( C_{H_0^2 \rightarrow L^\frac{n(p-2)}{2}}M \right)^{\frac{n^2(p-2)}{8}}\|u_1-u_2\|_{H^2}^2 \end{align*}

    In view of the above three inequalities, we get F is locally Lipschitz continuous. Then the local existence and extension of mild solutions follows.

    Next we prove (3.50). Suppose firstly U_0\in D(A) = (H^4( \Omega) \cap H_0 ^2( \Omega))\times H_0 ^2( \Omega) , then by Lemma 3.2, U\in C \left([0, {T_{\max}});D(A) \right)\cap C^1 \left([0, {T_{\max}});H \right) is a classical solution. Then it follows from (3.45) and (3.47) that

    \begin{align*} \frac{1}{2}\frac{d}{dt}\|U(t)\|_H^2& = (U, U_t)_H\\ & = (U, A(U))+(U, F(U)) = (U, F(U)), \; \; \; 0\leq t < {T_{\max}}. \end{align*}

    For fixed t\in[0, {T_{\max}}) , integrating the above equality over [0, t] , we get (3.50).

    In general case U_0\in H , since D(A) is dense in H , we approximate U_0 by a sequence \{U_{n0}\}_{n = 1}^\infty , and then we pass to the limit to obtain (3.50).

    Proof of Theorem 1.2. Step 1. Existence of maximal weak solution. By Lemma 3.4 and Definition 1.1, to show the existence of maximal weak solution, we only need to prove the mild solution U = (u, u_t) got in Lemma 3.4 satisfies (1.4).

    We denote the inner produce of the Hilbert space L^2( \Omega)\times L^2( \Omega) by \left(\left(\cdot, \cdot \right)\right) , i.e.,

    \begin{equation*} \left(\left( U, V \right)\right) = \int\limits_ \Omega(u_1v_1+u_2v_2)dx, \; \; \; \forall\; U = (u_1, u_2), \; V = (v_1, v_2)\in L^2( \Omega)\times L^2( \Omega). \end{equation*}

    Since C_0^\infty( \Omega) is dense in H_0^2( \Omega) , by density arguments, we only need to prove (1.4) with \phi\in C_0^\infty( \Omega) . Let U(t) = (u, u_t)\in C \left([0, {T_{\max}}); H_0^2( \Omega)\times L^2( \Omega) \right) be the mild solution of (3.45) got in Lemma 3.4 and \Phi = (0, \phi) . For fixed t\in[0, {T_{\max}}) , by using (3.49), we get

    \begin{equation*} \left(\left( U, \Phi \right)\right) = \left(\left( e^{tA}U_0, \Phi \right)\right)+ \left(\left( \int\limits_0^t e^{(t-\tau)A}F(U(\tau)), \Phi \right)\right). \end{equation*}

    We differentiate to obtain

    \begin{equation} \begin{split} \frac{d}{dt} \left(\left( U, \Phi \right)\right) = \frac{d}{dt} \left(\left( e^{tA}U_0, \Phi \right)\right)+\frac{d}{dt} \left(\left( \int\limits_0^t e^{(t-\tau)A}F(U(\tau)), \Phi \right)\right). \end{split} \end{equation} (3.55)

    Now, using the standard properties of the semigroup (see for example, [54,Chapter 2]), we obtain

    \begin{equation} \begin{split} \frac{d}{dt} \left(\left( e^{tA}U_0, \Phi \right)\right) = & \left(\left( e^{tA}U_0, A^*\Phi \right)\right)+ \left(\left( e^{tA}U_0, \Phi_t \right)\right) \end{split} \end{equation} (3.56)

    where

    \begin{equation*} A^* = \left(\begin{array}{cc} 0 & \quad -\Delta^2-I\\ I & \quad\; \; 0\\ \end{array}\right) \end{equation*}

    is the adjoint operator of A ; and

    \begin{equation} \begin{split} \frac{d}{dt} \left(\left( \int\limits_0^t e^{(t-\tau)A}F(U(\tau)), \Phi \right)\right)&\\ = \left(\left( F(U(t)), \Phi \right)\right)&+ \left(\left( \int\limits_0^t e^{(t-\tau)A}F(U(\tau)), A^*\Phi \right)\right)\\ &+ \left(\left( \int\limits_0^t e^{(t-\tau)A}F(U(\tau)), \Phi_t \right)\right). \end{split} \end{equation} (3.57)

    Then it follows from (3.55)-(3.57) and (3.49) that

    \begin{equation} \frac{d}{dt} \left(\left( U, \Phi \right)\right) = \left(\left( F(U(t)), \Phi \right)\right)+ \left(\left( U, A^*\Phi \right)\right)+ \left(\left( U, \Phi_t \right)\right). \end{equation} (3.58)

    Since U = (u, u_t) and \Phi = (0, \phi) , we have

    \begin{align*} \left(\left( U, \Phi \right)\right) = & \int\limits_ \Omega u_t\phi dx, \\ \left(\left( F(U(t)), \Phi \right)\right) = & \int\limits_ \Omega \left(\sigma|x|^{- \alpha}*u+|u|^{p-2}u \right)\phi dx, \\ \left(\left( U, A^*\Phi \right)\right) = & \left(\left( (u, u_t), (-\Delta^2\phi-\phi, 0) \right)\right) = - \int\limits_ \Omega \left( u\Delta^2\phi+u\phi \right) dx, \\ \left(\left( U, \Phi_t \right)\right) = & \int\limits_ \Omega u_t\phi_t dx. \end{align*}

    Then it follows from (3.58) that

    \begin{align*} \frac{d}{dt} \int\limits_ \Omega u_t\phi dx+ \int\limits_ \Omega \left( u\Delta^2\phi+u\phi \right) dx = \int\limits_ \Omega \left(\sigma|x|^{- \alpha}*u+|u|^{p-2}u \right)\phi dx+ \int\limits_ \Omega u_t\phi_t dx. \end{align*}

    Since u\in C \left([0, {T_{\max}}); H_0^2( \Omega) \right) , integrating by parts, we get

    \begin{equation} \begin{split} &\frac{d}{dt} \int\limits_ \Omega u_t\phi dx+ \int\limits_ \Omega \left( \Delta u\Delta\phi+u\phi \right) dx\\ = & \int\limits_ \Omega \left(\sigma|x|^{- \alpha}*u+|u|^{p-2}u \right)\phi dx+ \int\limits_ \Omega u_t\phi_t dx. \end{split} \end{equation} (3.59)

    Note \phi_t = 0 , integrating in time over [0, t] for any t\in(0, {T_{\max}}) , we obtain (1.4).

    Step 2. Proof of (u_t, u) in C^1[0, {T_{\max}}) and the equality (1.8). Since

    u(t)\in C \left([0, {T_{\max}});H_0^2( \Omega) \right)\hbox{ and }u_t\in C \left([0, {T_{\max}});L^2( \Omega) \right),

    by taking \phi = u(t) in (3.59), we get

    \begin{equation*} \begin{split} \frac{d}{dt}(u_t, u)& = \|u_t\|^2- \left(\overbrace{\|\Delta u\|^2+\|u\|^2-\|u\|_p^p-\sigma \int\limits_ \Omega \int\limits_ \Omega\frac{u(x, t)u(y, t)}{|x-y|^ \alpha}dxdy}^{ = I(u)} \right)\\ &\in C[0, {T_{\max}}), \end{split} \end{equation*}

    i.e, (u_t, u) in C^1[0, {T_{\max}}) and (1.8) holds.

    Step 3. Proof of the equality (1.5). The energy identity (1.5) follows from (3.50) directly. In fact by using U = (u, u_t) , (3.44), and (3.46), we have

    \begin{align*} \|U\|_H& = \|\Delta u\|^2+\|u\|^2+\|u_t\|^2, \\ (F(U), U)_H& = \frac{d}{dt} \left(\frac{\sigma}{2} \int\limits_{ \Omega} \int\limits_{ \Omega}\frac{u(x, t)u(y, t)}{|x-y|^ \alpha}dxdy+\frac{1}{p}\|u\|_p^p \right). \end{align*}

    Then by (3.50), we get (1.5).

    Proof of Theorem 1.3. Let u\in C \left([0, {T_{\max}});H_0^2( \Omega) \right)\cap C^1 \left([0, {T_{\max}});L^2( \Omega) \right) be the maximal weak solution got in Theorem 1.2. By E(0)\leq0 and (1.5), it holds,

    \begin{equation*} E(t) = E(0)\leq0, \; \; \; 0\leq t < {T_{\max}}. \end{equation*}

    By the definitions of J and I (see (1.9) and (1.10)), we get

    \begin{align*} I(u)& = 2J(u)-\frac{p-2}{p}\|u\|_p^p, \\ I(u)& = pJ(u)-\frac{p-2}{2} \left(\|\Delta u\|^2+\|u\|^2-\sigma\int_ \Omega\int_ \Omega\frac{u(x, t)u(y, t)}{|x-y|^ \alpha}dxdy \right). \end{align*}

    By (1.5) and (1.6), it follows J(u) = E(0)-\frac{1}{2}\|u_t\|^2 . Then, by the above two inequalities, we get

    \begin{equation} I(u) = 2E(0)-\|u_t\|^2-\frac{p-2}{p}\|u\|_p^p \end{equation} (4.60)

    and

    \begin{equation} \begin{split} I(u) = &pE(0)-\frac{p}{2}\|u_t\|^2\\ &-\frac{p-2}{2} \left(\|\Delta u\|^2+\|u\|^2-\sigma\int_ \Omega\int_ \Omega\frac{u(x, t)u(y, t)}{|x-y|^ \alpha}dxdy \right). \end{split} \end{equation} (4.61)

    By (1.11), we get

    \begin{equation} \left|\sigma \int_ \Omega\int_ \Omega\frac{u(x, t)u(y, t)}{|x-y|^ \alpha}dxdy \right|\leq\|\Delta u\|^2+\|u\|^2\hbox{ if }|\sigma|\leq\sigma^*. \end{equation} (4.62)

    In the following we divide the proof into two cases: |\sigma|\leq\sigma^* and |\sigma|>\sigma^* .

    Case 1. |\sigma|\leq\sigma^* . It follows from (4.61) and (4.62) that

    \begin{equation} I(u)\leq p E(0)-\frac{p}{2}\|u_t\|^2. \end{equation} (4.63)

    Let

    \begin{equation} h(t) = \left\| u(t)\right\|^2+\beta (t+\gamma)^2, \; \; \; 0\leq t < {T_{\max}}, \end{equation} (4.64)

    where \beta\geq0 and \gamma\geq0 are two constants to be determined later. Then by using (1.8) and (4.63), we have

    \begin{align} h'(t)& = 2(u_t, u)+2 \beta(t+\gamma), \end{align} (4.65)
    \begin{align} h''(t)& = 2\|u_t\|^2-2I(u)(t)+2 \beta\geq-2pE(0)+(p+2)\|u_t\|^2+2 \beta. \end{align} (4.66)

    By Cauchy-Schwartz's inequality,

    \begin{equation} \begin{split} (h'(t))^2&\leq4 \left(\|u_t\|\|u\|+ \beta(t+\gamma) \right)^2\\ & = 4 \left(\left\| u\right\|^2 \left\|u_t\right\|^2+\beta ^2 (t+\gamma)^2 +2\beta (t+\gamma)\left\| u\right\| \left\| u_t\right\| \right) \\ &\leq 4 \left(\left\| u\right\|^2 \left\|u_t\right\|^2+\beta ^2 (t+\gamma)^2 + \beta (t+\gamma)^2\left\| u_t\right\|^2+\beta \left\| u\right\|^2 \right) \\ &\leq 4 \left(\left(\left\| u\right\|^2+\beta (t+\gamma)^2 \right)\left(\left\|u_t\right\|^2+\beta \right) \right)\\ & = 4h(t)\left(\left\|u_t\right\|^2+\beta \right). \end{split} \end{equation} (4.67)

    Then by (4.66) and (4.67), it follows

    \begin{equation} \begin{split} h''(t)h(t)- \left(1+\frac{p-2}{4} \right)(h'(t))^2&\geq p(-2E(0)- \beta)h(t)\\&\geq 0\hbox{ for }0\leq \beta\leq-2E(0). \end{split} \end{equation} (4.68)

    Subcase 1. E(0)\leq 0 and (u_0, u_1)>0 . We take \beta = \gamma = 0 , then h(0) = \|u_0\|^2>0 and h'(0) = 2(u_0, u_1)>0 . Then, it follows from Lemma 2.3 that h blows up at a finite time \hat T , \hat T\geq {T_{\max}} (by Theorem 1.2), and

    \begin{equation*} \hat T\leq\frac{h(0)}{\frac{p-2}{4}h'(0)} = \frac{2\|u_0\|^2}{(p-2)(u_0, u_1)}. \end{equation*}

    Subcase 2. E(0)<0 and (u_0, u_1) = 0 . We take

    \begin{equation*} \beta = -2E(0)\hbox{ and }\gamma = \frac{\|u_0\|}{\sqrt{ \beta}}, \end{equation*}

    then h(0) = \|u_0\|^2+ \beta\gamma^2 = 2\|u_0\|^2>0 , h'(0) = 2 \beta\gamma = 2\sqrt{-2E(0)}\|u_0\|>0 . Then, it follows from Lemma 2.3 that h blows up at a finite time \hat T , \hat T\geq {T_{\max}} (by Theorem 1.2), and

    \begin{equation*} \hat T\leq\frac{h(0)}{\frac{p-2}{4}h'(0)} = \frac{4\|u_0\|}{(p-2)\sqrt{-2E(0)}}. \end{equation*}

    Subcase 3. E(0)<0 and (u_0, u_1)<0 . We take

    \begin{equation*} \beta = -2E(0), \; \; \; \gamma = \frac{-(u_0, u_1)+\sqrt{(u_0, u_1)^2+ \beta\|u_0\|^2}}{ \beta}, \end{equation*}

    then,

    \begin{align*} h(0)& = \|u_0\|^2+ \beta\gamma^2 = \|u_0\|^2+\left(-(u_0, u_1)+\sqrt{(u_0, u_1)^2+ \beta\|u_0\|^2}\right)^2/ \beta > 0, \\ h'(0)& = 2(u_0, u_1)+2 \beta\gamma = 2\sqrt{(u_0, u_1)^2+ \beta\|u_0\|^2)} > 0. \end{align*}

    Then, it follows from Lemma 2.3 that h blows up at a finite time \hat T , \hat T\geq {T_{\max}} (by Theorem 1.2), and

    \begin{equation*} \begin{split} \hat T&\leq\frac{h(0)}{\frac{p-2}{4}h'(0)} = \frac{2\left(\|u_0\|^2+\frac{\left(-(u_0, u_1)+\sqrt{(u_0, u_1)^2+ \beta\|u_0\|^2)}\right)^2}{ \beta}\right)}{(p-2)\sqrt{(u_0, u_1)^2+ \beta\|u_0\|^2)}}\\ & = \frac{2 \beta\left\|u_0\right\|^2+2\left(-(u_0, u_1)+\sqrt{(u_0, u_1)^2+ \beta\|u_0\|^2}\right)^2}{ \beta(p-2)\sqrt{(u_0, u_1)^2+ \beta\left\|u_0\right\|^2}}\\ & = \frac{-4E(0)\left\|u_0\right\|^2+2\left(-(u_0, u_1)+\sqrt{(u_0, u_1)^2-2E(0)\|u_0\|^2}\right)^2}{-2E(0)(p-2)\sqrt{(u_0, u_1)^2-2E(0)\left\|u_0\right\|^2}} \end{split} \end{equation*}

    Case 2. |\sigma|>\sigma^* . Firstly we estimate \int_ \Omega\int_ \Omega|x-y|^{- \alpha}u(x, t)u(y, t)dxdy .

    \underline{\text{If}\ \alpha \le 0}, let R = \sup_{x, y\in \Omega}|x-y| , then by Hölder's inequality, we get

    \begin{equation} \begin{split} \left| \int\limits_ \Omega \int\limits_ \Omega\frac{u(x, t)u(y, t)}{|x-y|^ \alpha}dxdy\right|&\leq R^{- \alpha} \left( \int\limits_ \Omega|u| dx \right)^2\\ &\leq R^{- \alpha}| \Omega|^{\frac{2(p-1)}{p}}\|u\|_p^2. \end{split} \end{equation} (4.69)

    \underline{\text{If }0<\alpha <n}, since

    \begin{equation*} \frac{1}{\frac{2n}{2n- \alpha}}+\frac{ \alpha}{n}+\frac{1}{\frac{2n}{2n- \alpha}} = 2 \end{equation*}

    and \frac{2n}{2n- \alpha}<p , by using (2.16) with q = r = \frac{2n}{2n- \alpha} , \theta = n- \alpha and Hölder's inequality, we get

    \begin{equation} \begin{split} \left| \int\limits_ \Omega \int\limits_ \Omega\frac{u(x, t)u(y, t)}{|x-y|^ \alpha}dxdy\right|&\leq\kappa\|u\|_{\frac{2n}{2n- \alpha}}^2\\ &\leq\kappa| \Omega|^{\frac{p(2n- \alpha)-2n}{n}}\|u\|_p^2, \end{split} \end{equation} (4.70)

    where \kappa is the positive constant defined in (2.15).

    Let

    \begin{equation*} \Theta = \left\{ \begin{array}{ll} R^{- \alpha}| \Omega|^{\frac{2(p-1)}{p}}, \; \; \; & \hbox{if } \alpha\leq0, \\ \kappa| \Omega|^{\frac{p(2n- \alpha)-2n}{n}}, & \hbox{if }0 < \alpha < n. \end{array} \right. \end{equation*}

    By (4.69) and (4.70), we get

    \begin{equation} \left| \int\limits_ \Omega \int\limits_ \Omega\frac{u(x, t)u(y, t)}{|x-y|^ \alpha}dxdy\right|\leq\Theta\|u\|_p^2. \end{equation} (4.71)

    Subcase 1. E(0)<0 and (u_0, u_1)\geq0 ; or E(0) = 0 and (u_0, u_1)>0 . By (1.8), (4.60), and E(0)\leq 0 , we get

    \frac{d}{dt}(u_t, u)\geq0.

    Then

    \frac{d}{dt}\|u(t)\|^2 = (u_t, u)\geq (u_1, u_0)\geq0,

    and then \|u\|\geq\|u_0\| . Then it follows from the Hölder's inequality that

    \begin{equation*} \|u_0\|^2\leq\|u\|^2\leq| \Omega|^{\frac{p-2}{p}}\|u\|_p^2, \end{equation*}

    which, together with (4.71) implies

    \begin{equation} \left| \int\limits_ \Omega \int\limits_ \Omega\frac{u(x, t)u(y, t)}{|x-y|^ \alpha}dxdy\right|\leq\Theta\|u\|_p^{2-p}\|u\|_p^p\leq\Theta| \Omega|^{\frac{(p-2)^2}{2p}}\|u_0\|^{2-p}\|u\|_p^p. \end{equation} (4.72)

    In view of (1.11), (4.61) and (4.72), we obtain

    \begin{equation} \begin{split} I(u)\leq& pE(0)-\frac{p}{2}\|u_t\|^2\\ &-\frac{p-2}{2} \left(\|\Delta u\|^2+\|u\|^2-|\sigma|\left|\int_ \Omega\int_ \Omega\frac{u(x, t)u(y, t)}{|x-y|^ \alpha}dxdy\right| \right)\\ = &pE(0)-\frac{p}{2}\|u_t\|^2\\ &-\frac{p-2}{2} \left(\|\Delta u\|^2+\|u\|^2-\sigma^*\left|\int_ \Omega\int_ \Omega\frac{u(x, t)u(y, t)}{|x-y|^ \alpha}dxdy\right| \right)\\ &+\frac{p-2}{2}(|\sigma|-\sigma^*)\left|\int_ \Omega\int_ \Omega\frac{u(x, t)u(y, t)}{|x-y|^ \alpha}dxdy\right|\\ \leq& pE(0)-\frac{p}{2}\|u_t\|^2+\frac{p-2}{2}\Lambda\|u\|_p^p, \end{split} \end{equation} (4.73)

    where

    \begin{equation} \Lambda = (|\sigma|-\sigma^*)\Theta| \Omega|^{\frac{(p-2)^2}{2p}}\|u_0\|^{2-p}. \end{equation} (4.74)

    It follows (\frac{\Lambda}{2}\times(4.60)+\frac{1}{p}\times(4.73))\times\frac{2p}{p\Lambda+2} that

    \begin{equation} I(u)\leq\frac{2p(\Lambda+1)}{p\Lambda+2}E(0)-\frac{p(\Lambda+1)}{p\Lambda+2}\|u_t\|^2 \end{equation} (4.75)

    Let h be the function defined in (4.64). Then by using (1.8) and (4.75), we have

    \begin{equation} h'(t) = 2(u_t, u)+2 \beta(t+\gamma) \end{equation} (4.76)

    and

    \begin{equation} \begin{split} h''(t)& = 2\|u_t\|^2-2I(u)(t)+2 \beta\\ &\geq-\frac{4p(\Lambda+1)}{p\Lambda+2}E(0)+ \left(\frac{2p(\Lambda+1)}{p\Lambda+2}+2 \right)\|u_t\|^2+2 \beta. \end{split} \end{equation} (4.77)

    Then it follows from (4.67) that

    \begin{equation} \begin{split} &h''(t)h(t)- \left(1+\frac{p-2}{2(p\Lambda+2)} \right)(h'(t))^2\\ \geq&\frac{2p(\Lambda+1)}{p\Lambda+2} \left(-2E(0)- \beta \right) h(t)\geq 0\hbox{ for }0\leq \beta\leq-2E(0). \end{split} \end{equation} (4.78)

    \underline{\text{If}\ E(0)\le 0\ \text{and}\ \left( {{u}_{0}},{{u}_{1}} \right)>0} , we take \beta = \gamma = 0 , then h(0) = \|u_0\|^2>0 and h'(0) = 2(u_0, u_1)>0 . Then, it follows from Lemma 2.3 that h blows up at a finite time \hat T , \hat T\geq {T_{\max}} (by Theorem 1.2), and

    \begin{equation*} \hat T\leq\frac{h(0)}{\frac{p-2}{2(p\Lambda+2)}h'(0)} = \frac{(p\Lambda+2)\|u_0\|^2}{(p-2)(u_0, u_1)}. \end{equation*}

    \underline{\text{If }E(0)<0\ \text{and}\ \left( {{u}_{0}},{{u}_{1}} \right)=0}, we take

    \begin{equation*} \beta = -2E(0)\hbox{ and }\gamma = \frac{\|u_0\|}{\sqrt{ \beta}}, \end{equation*}

    then h(0) = \|u_0\|^2+ \beta\gamma^2 = 2\|u_0\|^2>0 , h'(0) = 2 \beta\gamma = 2\sqrt{-2E(0)}\|u_0\|>0 . Then, it follows from Lemma 2.3 that h blows up at a finite time \hat T , \hat T\geq {T_{\max}} (by Theorem 1.2), and

    \begin{equation*} \hat T\leq\frac{h(0)}{\frac{p-2}{2(p\Lambda+2)}h'(0)} = \frac{2(p\Lambda+2)\|u_0\|}{(p-2)\sqrt{-2E(0)}}. \end{equation*}

    Subcase 2. E(0)<0 and (u_0, u_1)<0 . We prove u blows up by contradiction argument. Assume u exists globally, i.e., {T_{\max}} = \infty .

    By (4.60), -I(u)(t)\geq-2E(0)>0 , t\in[0, \infty) . Then it follows (1.8),

    \frac{d}{dt}(u_t, u)\geq -I(u)(t)\geq-2E(0) > 0.

    Integrating this inequality from 0 to t we get

    \begin{equation*} (u_t, u)\geq (u_0, u_1)-2E(0)t. \end{equation*}

    Let t_0 = \frac{(u_0, u_1)}{2E(0)} , then (u(t_0), u_t(t_0))\geq0 . Moreover, E(t_0) = E(0)<0 . We come back to subcase 1. Then \|u\| will become infinite in finite time, which contradicts {T_{\max}} = \infty .

    Proof of Theorem 1.4. Let u\in C \left([0, {T_{\max}});H_0^2( \Omega) \right)\cap C^1 \left([0, {T_{\max}});L^2( \Omega) \right) be the weak solution got in Theorem 1.2 with (u_0, u_1)\in H_0^2( \Omega)\times L^2( \Omega) satisfying I(u_0)>0 and 0<E(0)\leq d .

    Step 1. E(0)<d ; or E(0) = d and \|u_1\|>0 . By (1.7), J(u_0)<d . Since, we also have I(u_0)>0 , we get u_0\in \mathcal{W} (see (2.24) for the definition of \mathcal{W} ). Then, by Lemma 2.6, u(t)\in \mathcal{W} for all t\in[0, {T_{\max}}) . Since \mathcal{W} = \mathcal{W}_1 (see Lemma 2.5), it follows from the definition of \mathcal{W}_1 (see (2.28)) that

    \begin{equation} \|\Delta u\|^2+\|u\|^2 < \frac{2pd}{p-2}+\sigma \int\limits_\Omega \int\limits_ \Omega\frac{u(x, t)u(y, t)}{|x-y|^ \alpha}dxdy, \; \; \; 0\leq t < {T_{\max}}. \end{equation} (4.79)

    By (1.11), it follows

    \begin{equation*} \left| \int\limits_\Omega \int\limits_ \Omega\frac{u(x, t)u(y, t)}{|x-y|^ \alpha}dxdy\right|\leq\frac{1}{\sigma^*} \left(\|\Delta u\|^2+\|u\|^2 \right), \end{equation*}

    which, together with (4.79) and |\sigma|<\sigma^* , implies

    \begin{equation} \|\Delta u\|^2+\|u\|^2\leq \left(1-\frac{|\sigma|}{\sigma^*} \right)^{-1}\frac{2pd}{p-2}, \; \; \; 0\leq t < {T_{\max}}. \end{equation} (4.80)

    Since E(t) = E(0) (see (1.5)), E(t) = \frac12\|u_t\|^2+J(u)(t) (see (1.6)), 2J(u)-I(u) = \frac{p-2}{p}\|u\|_p^p (see (2.30)), I(u)(t)>0 (since u(t)\in \mathcal{W} ) for all t\in[0, {T_{\max}}) , and E(0)\leq d , it holds

    \begin{equation} \|u_t\|^2 = 2E(0)-2J(u)(t)\leq 2d-I(u) < 2d, \; \; \; 0\leq t < {T_{\max}}. \end{equation} (4.81)

    By (4.80), (4.81), and Theorem 1.2, we get {T_{\max}} = \infty .

    Step 2. E(0) = d and \|u_1\| = 0 . Since E(t) = E(0) (see (1.5)), E(t) = \frac12\|u_t\|^2+J(u)(t) (see (1.6)) and E(0) = \frac{1}{2}\|u_1\|^2+J(u_0) (see (1.7)), we get

    \begin{equation*} J(u)(t)\leq d, \; \; \; t\in[0, {T_{\max}})\hbox{ and } J(u_0) = d. \end{equation*}

    If J(u(t))\equiv d for all t\in[0, {T_{\max}}) , then by (1.6), \|u_t\|\equiv0 for t\in[0, {T_{\max}}) , then u(t)\equiv u_0 , so by Theorem 1.2, {T_{\max}} = \infty ; If there exists t_1\in[0, {T_{\max}}) such that J(u)(t_1)<d , we claim

    ● there exists a constant \sigma>0 sufficient small, and a sequence \{t_n\}_{n = 1}^\infty such that \sigma\geq t_n\downarrow0 as n\uparrow\infty and J(u)(t_n)<d for n = 1, 2, \cdots .

    In fact, if the claim is not true, there must exists a constant \delta>0 such that J(u)(t) = d for t\in[0, \delta] . Then by analysis as the first case, u(t)\equiv u_0 for t\in[0, \delta] ; and then u(t)\equiv u_0 for t\in[\delta, 2\delta] , [2\delta, 3\delta] , etc. So u(t)\equiv u_0 for all t\in[0, {T_{\max}}) , then J(u)(t_1) = J(u_0) = d , a contradiction. So the claim is true, since I(u_0)>0 and I(u)(t)\in C[0, {T_{\max}}) , \lim_{n\uparrow\infty}I(u)(t_n) = I(u_0)>0 . So for n large enough, we have J(u)(t_n)<d and I(u)(t_n)>0 , we come back to step 1.

    Proof of Theorem 1.5. Let u\in C \left([0, {T_{\max}});H_0^2( \Omega) \right)\cap C^1 \left([0, {T_{\max}});L^2( \Omega) \right) be the weak solution got in Theorem 1.2 with (u_0, u_1)\in H_0^2( \Omega)\times L^2( \Omega) .

    If E(0)<d , by (1.7), we get J(u_0)<d ; If E(0) = d and (u_0, u_1)>0 , we must have \|u_1\|>0 , and by (1.7) again, we get J(u_0)<d . Now we prove \|u_1\|>0 by contradiction argument. In fact if \|u_1\| = 0 , by Cauchy-Schwartz's inequality, we have 0<(u_0, u_1)\leq\|u_0\|\|u_1\| = 0 , a contradiction.

    Since J(u_0)<d and I(u_0)<0 , we get u_0\in \mathcal{V} (see (2.25)). Then by Lemma 2.6, u(t)\in \mathcal{V} for t\in[0, {T_{\max}}) , where \mathcal{V} is the set defined in (2.25). Since \mathcal{V} = \mathcal{V}_1 (see Lemma 2.5), by (2.29) and (4.61),

    \begin{equation} I(u)\leq pE(0)-\frac p2\|u_t\|^2-pd, \; \; \; 0\leq t < {T_{\max}}. \end{equation} (4.82)

    Let h(t) be the function defined in (4.64), with \beta\geq0 and \gamma\geq0 to be determined later. Then by using (1.8) and (4.82), we have

    \begin{align} h'(t)& = 2(u_t, u)+2 \beta(t+\gamma), \end{align} (4.83)
    \begin{align} h''(t)& = 2\|u_t\|^2-2I(u)(t)+2 \beta\geq-2pE(0)+(p+2)\|u_t\|^2+2pd+2 \beta. \end{align} (4.84)

    Then it follows from (4.67) that

    \begin{equation} \begin{split} h''(t)h(t)- \left(1+\frac{p-2}{4} \right)(h'(t))^2&\geq p[2(d-E(0))- \beta]\\ &\geq0\hbox{ for }0\leq \beta\leq 2(d-E(0)). \end{split} \end{equation} (4.85)

    Case 1. E(0)\leq d and (u_0, u_1)>0 . We take \beta = \gamma = 0 , then h(0) = \|u_0\|^2>0 and h'(0) = 2(u_0, u_1)>0 . Then, it follows from Lemma 2.3 that h blows up at a finite time \hat T , \hat T\geq {T_{\max}} (by Theorem 1.2), and

    \begin{equation*} \hat T\leq\frac{h(0)}{\frac{p-2}{4}h'(0)} = \frac{2\|u_0\|^2}{(p-2)(u_0, u_1)}. \end{equation*}

    Case 2. E(0)<d and (u_0, u_1) = 0 . We take

    \begin{equation*} \beta = 2(d-E(0)), \; \; \; \gamma = \frac{\|u_0\|}{\sqrt{ \beta}}, \end{equation*}

    then h(0) = \|u_0\|^2+ \beta\gamma^2 = 2\|u_0\|^2>0 , h'(0) = 2 \beta\gamma = 2\sqrt{2(d-E(0))}\|u_0\|>0 . Then, it follows from Lemma 2.3 that h blows up at a finite time \hat T , \hat T\geq {T_{\max}} (by Theorem 1.2), and

    \begin{equation*} \hat T\leq\frac{h(0)}{\frac{p-2}{4}h'(0)} = \frac{4\|u_0\|}{(p-2)\sqrt{2(d-E(0))}}. \end{equation*}

    Case 3. E(0)<d and (u_0, u_1)<0 . We take

    \begin{equation*} \beta = 2(d-E(0)), \; \; \; \gamma = \frac{-(u_0, u_1)+\sqrt{(u_0, u_1)^2+ \beta\|u_0\|^2)}}{ \beta}, \end{equation*}

    then,

    \begin{align*} h(0)& = \|u_0\|^2+ \beta\gamma^2 = \|u_0\|^2+\left(-(u_0, u_1)+\sqrt{(u_0, u_1)^2+ \beta\|u_0\|^2)}\right)^2/ \beta > 0, \\ h'(0)& = 2(u_0, u_1)+2 \beta\gamma = 2\sqrt{(u_0, u_1)^2+ \beta\|u_0\|^2)} > 0. \end{align*}

    Then, it follows from Lemma 2.3 that h blows up at a finite time \hat T , \hat T\geq {T_{\max}} (by Theorem 1.2), and

    \begin{equation*} \begin{split} \hat T&\leq\frac{h(0)}{\frac{p-2}{4}h'(0)} = \frac{2\left(\|u_0\|^2+\frac{\left(-(u_0, u_1)+\sqrt{(u_0, u_1)^2+ \beta\|u_0\|^2)}\right)^2}{ \beta}\right)}{(p-2)\sqrt{(u_0, u_1)^2+ \beta\|u_0\|^2)}}\\ & = \frac{2 \beta\left\|u_0\right\|^2+2\left(-(u_0, u_1)+\sqrt{(u_0, u_1)^2+ \beta\|u_0\|^2}\right)^2}{ \beta(p-2)\sqrt{(u_0, u_1)^2+ \beta\left\|u_0\right\|^2}} \end{split} \end{equation*}
    \begin{equation*} \begin{split} & = \frac{4(d-E(0))\left\|u_0\right\|^2+2\left(-(u_0, u_1)+\sqrt{(u_0, u_1)^2+(2(d-E(0)))\|u_0\|^2}\right)^2}{(2(d-E(0)))(p-2)\sqrt{(u_0, u_1)^2+(2(d-E(0)))\left\|u_0\right\|^2}}. \end{split} \end{equation*}

    The authors would like to thank the referees for the comments and valuable suggestions.



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