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|−α∗u+|u|p−2u,x∈Ω,t>0,u=∂u∂ν=0,x∈∂Ω,t>0,u(x,0)=u0(x),ut(x,0)=u1(x),x∈Ω, | (1.1) |
where
2≤p{<∞,n=1,2,3,4,≤2+4n−4,n≥5. | (1.2) |
The initial value
|x|−α∗u={∫Ω|x−y|−αu(y)dy,if α≠0,∫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
The potential term
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
CX→Y=supϕ∈X∖{0}‖ϕ‖Y‖ϕ‖X. | (1.3) |
● The norm of the Lebesgue space
● The inner product of the Hilbert space
● The norm of the Sobolev space
‖ϕ‖2H2=‖Δϕ‖2+‖ϕ‖2. |
Definition 1.1. Assume
u∈C([0,T];H20(Ω))∩C1([0,T];L2(Ω)) |
such that
∫Ωutϕdx+t∫0∫ΩΔuΔϕdxdτ+t∫0∫Ωuϕdxdτ=t∫0∫Ω(σ|x|−α∗u+|u|p−2u)ϕdxdτ+∫Ωu1ϕdx | (1.4) |
holds for any
The local-well posedness of solutions to problem (1.1) is the following theorem:
Theorem 1.2. Assume
u∈C([0,T];H20(Ω))∩C1([0,T];L2(Ω)). |
The solution
1.
2.
limt↑Tmax(‖u(t)‖H2+‖ut(t)‖)=∞, |
i.e., the solution blows up at a finite time
Furthermore, then energy
E(t)=E(0),0≤t<Tmax, | (1.5) |
where
E(t)=12‖ut‖2+J(u)(t),0≤t<Tmax, | (1.6) |
and
E(0)=E(t)|t=0=12‖u1‖2+J(u0). | (1.7) |
Moreover, we have
ddt(ut,u)=‖ut‖2−I(u). | (1.8) |
Here
J(ϕ)=12‖Δϕ‖2+12‖ϕ‖2−1p‖ϕ‖pp−σ2∫Ω∫Ωϕ(x)ϕ(y)|x−y|αdxdy, | (1.9) |
and
I(ϕ)=‖Δϕ‖2+‖ϕ‖2−‖ϕ‖pp−σ∫Ω∫Ωϕ(x)ϕ(y)|x−y|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
1.
2.
Then the weak solution got in Theorem 1.2 blows up in finite time, i.e.,
1. If
Tmax≤{2‖u0‖2(p−2)(u0,u1), if E(0)≤0 and (u0,u1)>0;4‖u0‖(p−2)√−2E(0), if E(0)<0 and (u0,u1)=0;¯T, if E(0)<0 and (u0,u1)<0, |
where
¯T=−4E(0)‖u0‖2+2(−(u0,u1)+√(u0,u1)2−2E(0)‖u0‖2)2−2E(0)(p−2)√(u0,u1)2−2E(0)‖u0‖2. |
2. If
Tmax≤{(pΛ+2)‖u0‖2(p−2)(u0,u1), if E(0)≤0 and (u0,u1)>0;2(pΛ+2)‖u0‖(p−2)√−2E(0), if E(0)<0 and (u0,u1)=0. |
Here,
σ∗=infϕ∈H20(Ω)∖{0}|‖Δϕ‖2+‖ϕ‖2∫Ω∫Ω|x−y|−αϕ(x)ϕ(y)dxdy| | (1.11) |
and
Λ={(|σ|−σ∗)R−α|Ω|p2‖u0‖2−p,if α≤0;κ(|σ|−σ∗)|Ω|(p−2)22p+p(2n−α)−2nn‖u0‖2−p,if 0<α<n, | (1.12) |
where
R=supx,y∈Ω|x−y|,κ=πα2Γ(n−α2)Γ(2n−α2){Γ(n2)Γ(n)}−n−αn. |
Remark 1. There are two remarks on the above theorem.
1. Firstly, by Lemma 2.2, if
σ∗≥{Rα(CH20→L1)−2,α∈(−∞,0],1κ(CH20→L2n2n−α)−2,α∈(0,n), |
where
2. Secondly, for
Theorem 1.3 is above the case
d=inf{J(ϕ):ϕ∈N}, | (1.13) |
where
N={ϕ∈H20(Ω)∖{0}:I(ϕ)=0}. | (1.14) |
Here
Remark 2. If we assume
d≥p−22p((1−|σ|σ∗)C2H20→Lp)pp−2, |
where
Theorem 1.4. (Global existence for
Theorem 1.5. (Blow-up for
1.
2.
where
Tmax≤{2‖u0‖2(p−2)(u0,u1), if E(0)≤d and (u0,u1)>0;4‖u0‖(p−2)√2(d−E(0)), if E(0)<d and (u0,u1)=0;ˆT, if E(0)<d and (u0,u1)<0, |
where
ˆT=4(d−E(0))‖u0‖2+2(−(u0,u1)+√(u0,u1)2+(2(d−E(0)))‖u0‖2)2(2(d−E(0)))(p−2)√(u0,u1)2+(2(d−E(0)))‖u0‖2. |
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
1q+n−θn+1r=2. |
Let
|∫Rn∫Rnu(x)v(y)|x−y|n−θdxdy|≤κ‖u‖q‖v‖r. |
If
q=r=2nn+θ, |
then
C=κ:=πn−θ2Γ(θ2)Γ(n+θ2){Γ(n2)Γ(n)}−θn. | (2.15) |
If
|∫Ω∫Ωu(x)v(y)|x−y|n−θdxdy|≤C‖u‖q‖v‖r. | (2.16) |
Lemma 2.2. Let
σ∗≥{Rα(CH20→L1)−2,α∈(−∞,0],1κ(CH20→L2n2n−α)−2,α∈(0,n), |
where
R=supx,y∈Ω|x−y|<∞. |
Proof.
|∫Ω∫Ω|x−y|−αϕ(x)ϕ(y)dxdy|≤(∫Ω|ϕ(x)|dx)2R−α≤C2H20→L1R−α(‖Δϕ‖2+‖ϕ‖2). |
|∫Ω∫Ω|x−y|−αϕ(x)ϕ(y)dxdy|≤κ‖ϕ‖22n2n−α≤κC2H20→L2n2n−α(‖Δϕ‖2+‖ϕ‖2). |
Lemma 2.3. [29,30] Suppose
F′′(t)F(t)−(1+r)(F′(t))2≥0, | (2.17) |
where
T≤F(0)rF′(0)<+∞ | (2.18) |
and
Lemma 2.4. Assume
d=inf{supλ≥0J(λϕ):ϕ∈H20(Ω)∖{0}}, | (2.19) |
and
d≥p−22p((1−|σ|σ∗)C2H20→Lp)pp−2, | (2.20) |
where
Proof. Firstly, we prove (2.19). For any
ˆλ=(‖Δϕ‖2+‖ϕ‖2−σ∫Ω∫Ω|x−y|−αϕ(x)ϕ(y)dxdy‖ϕ‖pp)1p−2, | (2.21) |
such that
supλ≥0J(λϕ)=J(ˆλϕ),ˆλϕ∈N. | (2.22) |
Then,
inf{supλ≥0J(λϕ):ϕ∈H20(Ω)∖{0}}=inf{J(ˆλϕ):ϕ∈H20(Ω)∖{0}}≥inf{J(ϕ):ϕ∈N}. |
On the other hand, for any
inf{J(ϕ):ϕ∈N}=inf{supλ≥0J(λϕ):ϕ∈N}≥inf{supλ≥0J(λϕ):ϕ∈H20(Ω)∖{0}}. |
Then (2.19) follows from the above two inequalities.
Secondly, we prove (2.20), by (1.11), we have
d=p−22pinf{ˆλp‖ϕ‖pp:ϕ∈H20(Ω)∖{0}}=p−22p(infϕ∈H20(Ω)∖{0}‖Δϕ‖2+‖ϕ‖2−σ∫Ω∫Ω|x−y|−αϕ(x)ϕ(y)dxdy‖ϕ‖2p)pp−2≥p−22p(infϕ∈H20(Ω)∖{0}‖Δϕ‖2+‖ϕ‖2−|σ||∫Ω∫Ω|x−y|−αϕ(x)ϕ(y)dxdy|‖ϕ‖2p)pp−2≥p−22p(infϕ∈H20(Ω)∖{0}(1−|σ|σ∗)‖Δϕ‖2+‖ϕ‖2‖ϕ‖2p)pp−2=p−22p((1−σσ∗)C2H20→Lp)pp−2. | (2.23) |
Lemma 2.5. Assume
W={ϕ∈H20(Ω):I(ϕ)>0,J(ϕ)<d}∪{0}. | (2.24) |
V={u∈H20(Ω):I(ϕ)<0,J(ϕ)<d}, | (2.25) |
where
‖ϕ‖pp≤2pdp−2,∀ϕ∈W | (2.26) |
and
‖Δϕ‖2+‖ϕ‖2≥((1−|σ|σ∗)(CH20→Lp)−p)2p−2,∀ϕ∈V, | (2.27) |
where
W1={ϕ∈H20(Ω):J(ϕ)<d,‖Δϕ‖2+‖ϕ‖2<2pdp−2+σCϕ}, | (2.28) |
and
V1={ϕ∈H20(Ω):J(ϕ)<d,‖Δϕ‖2+‖ϕ‖2>2pdp−2+σCϕ}. | (2.29) |
Here,
Cϕ=∫Ω∫Ωϕ(x)ϕ(y)|x−y|αdxdy. |
Proof. Step 1. We prove (2.26). By the definition of
2J(ϕ)−I(ϕ)=p−2p‖ϕ‖pp,∀ϕ∈H20(Ω). | (2.30) |
For any
Step 2. We prove (2.27). For any
(CH20→Lp)p(‖Δϕ‖2+‖ϕ‖2)p2≥‖ϕ‖pp>‖Δϕ‖2+‖ϕ‖2−σ∫Ω∫Ωϕ(x)ϕ(y)|x−y|αdxdy≥(1−|σ|σ∗)(‖Δϕ‖2+‖ϕ‖2), | (2.31) |
which implies (2.27).
Step 3. We show that
Firstly, for any
‖Δϕ‖2+‖ϕ‖2>2pdp−2+σ∫Ω∫Ωϕ(x)ϕ(y)|x−y|αdxdy,12(‖Δϕ‖2+‖ϕ‖2)<d+1p‖ϕ‖pp+σ2∫Ω∫Ωϕ(x)ϕ(y)|x−y|αdxdy, |
which implies
‖ϕ‖pp>2pdp−2. |
Thus, by the definition of
I(ϕ)=‖Δϕ‖2+‖ϕ‖2−‖ϕ‖pp−σ∫Ω∫Ωϕ(x)ϕ(y)|x−y|αdxdy≤2d−p−2p‖ϕ‖pp<0, |
which, together with
Secondly, we prove
‖Δϕ‖2+‖ϕ‖2−σ∫Ω∫Ωϕ(x)ϕ(y)|x−y|αdxdy≥(1−|σ|σ∗)((1−|σ|σ∗)(CH20→Lp)−p)2p−2>0. |
Then, in view of
‖ϕ‖pp>‖Δϕ‖2+‖ϕ‖2−σ∫Ω∫Ωϕ(x)ϕ(y)|x−y|αdxdy, |
we get
‖ϕ‖2pp−2p>(‖Δϕ‖2+‖ϕ‖2−σ∫Ω∫Ωϕ(x)ϕ(y)|x−y|αdxdy)2p−2⇔‖ϕ‖2pp−2p>(‖Δϕ‖2+‖ϕ‖2−σ∫Ω∫Ωϕ(x)ϕ(y)|x−y|αdxdy)pp−2−1⇔‖Δϕ‖2+‖ϕ‖2−σ∫Ω∫Ωϕ(x)ϕ(y)|x−y|αdxdy >(‖Δϕ‖2+‖ϕ‖2−σ∫Ω∫Ωϕ(x)ϕ(y)|x−y|αdxdy‖ϕ‖2p)pp−2, |
which, together with the second line of (2.23), implies
‖Δϕ‖2+‖ϕ‖2−σ∫Ω∫Ωϕ(x)ϕ(y)|x−y|αdxdy>2pdp−2. |
Since
Step 4. We show that
Firstly we prove
‖Δϕ‖2+‖ϕ‖2−σ∫Ω∫Ωϕ(x)ϕ(y)|x−y|αdxdy>‖ϕ‖pp,12(‖Δϕ‖2+‖ϕ‖2)<d+1p‖u‖pp+12σ∫Ω∫Ωϕ(x)ϕ(y)|x−y|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
Secondly we prove
\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
Lemma 2.6. Assume
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
1. If there exists a
2. If there exists a
where
Proof. Firstly, we proof the first part by contradiction argument. Actually, if the conclusion is incorrect, by using
\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)),
\begin{equation} J(u)(t_1)\leq E(t_1) = E(t_0) < d. \end{equation} | (2.35) |
Then by
Secondly, we proof the second part. In 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
Due to
\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
If
\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
\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
\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
\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
\begin{equation*} \|\Phi\|_H = \sqrt{(\Phi, \Phi)_H}, \; \; \; \Phi\in H. \end{equation*} |
Suppose
\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
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
1.
2.
Here
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
\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
1.
2.
\begin{equation*} \lim\limits_{t\uparrow {T_{\max}}}\|U(t)\|_H = \infty, \end{equation*} |
i.e., the solution blows up at a finite time
Furthermore, if
By introduction
\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
\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
Lemma 3.3. Let
Proof. Let
\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
\begin{equation*} \|\Phi\|_H = (\Phi, \Phi)_H = \|\phi_1\|_{H^2}+\|\varphi_2\|. \end{equation*} |
Let
\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
\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
\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
\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
Next, we show (3.45) admits a mild solution.
Lemma 3.4. Assume
\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
1.
2.
\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
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
\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
Next we show
Case 1.
\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.
\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
\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.
\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
\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.
\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.
\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
Next we prove (3.50). Suppose firstly
\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
In general case
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
We denote the inner produce of the Hilbert space
\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
\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
\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
\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
\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
Step 2. Proof of
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
\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,
Step 3. Proof of the equality (1.5). The energy identity (1.5) follows from (3.50) directly. In fact by using
\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
\begin{equation*} E(t) = E(0)\leq0, \; \; \; 0\leq t < {T_{\max}}. \end{equation*} |
By the definitions of
\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
\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:
Case 1.
\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
\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.
\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.
\begin{equation*} \beta = -2E(0)\hbox{ and }\gamma = \frac{\|u_0\|}{\sqrt{ \beta}}, \end{equation*} |
then
\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.
\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
\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.
\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) |
\begin{equation*} \frac{1}{\frac{2n}{2n- \alpha}}+\frac{ \alpha}{n}+\frac{1}{\frac{2n}{2n- \alpha}} = 2 \end{equation*} |
and
\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
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.
\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
\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
\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
\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) |
\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*} |
\begin{equation*} \beta = -2E(0)\hbox{ and }\gamma = \frac{\|u_0\|}{\sqrt{ \beta}}, \end{equation*} |
then
\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.
By (4.60),
\frac{d}{dt}(u_t, u)\geq -I(u)(t)\geq-2E(0) > 0. |
Integrating this inequality from
\begin{equation*} (u_t, u)\geq (u_0, u_1)-2E(0)t. \end{equation*} |
Let
Proof of Theorem 1.4. Let
Step 1.
\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
\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
\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
Step 2.
\begin{equation*} J(u)(t)\leq d, \; \; \; t\in[0, {T_{\max}})\hbox{ and } J(u_0) = d. \end{equation*} |
If
● there exists a constant
In fact, if the claim is not true, there must exists a constant
Proof of Theorem 1.5. Let
If
Since
\begin{equation} I(u)\leq pE(0)-\frac p2\|u_t\|^2-pd, \; \; \; 0\leq t < {T_{\max}}. \end{equation} | (4.82) |
Let
\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.
\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.
\begin{equation*} \beta = 2(d-E(0)), \; \; \; \gamma = \frac{\|u_0\|}{\sqrt{ \beta}}, \end{equation*} |
then
\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.
\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
\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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