In the era of big data, interval-valued data is quite common in real life and can be used to describe the uncertainty of variables. In this paper, we introduced random effects panel interval-valued data models based on the center and range method and constructed a Bayesian method for the models, including estimation and prediction. Some simulation studies indicate that the proposed Bayesian method performs well. Finally, our proposed panel interval-valued data Bayesian models were applied in forecasting of the Air Quality Index, and the experimental evaluation of actual data sets shows the advantages and the performance of our proposed models.
Citation: Dengke Xu, Linlin Shen, Yuanyang Tangzhu, Shiqi Ke. Bayesian analysis of random effects panel interval-valued data models[J]. Electronic Research Archive, 2025, 33(5): 3210-3224. doi: 10.3934/era.2025141
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In the era of big data, interval-valued data is quite common in real life and can be used to describe the uncertainty of variables. In this paper, we introduced random effects panel interval-valued data models based on the center and range method and constructed a Bayesian method for the models, including estimation and prediction. Some simulation studies indicate that the proposed Bayesian method performs well. Finally, our proposed panel interval-valued data Bayesian models were applied in forecasting of the Air Quality Index, and the experimental evaluation of actual data sets shows the advantages and the performance of our proposed models.
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
\begin{equation*} {T_{\max}}\leq\left\{ \begin{array}{ll} \frac{(p\Lambda+2)\|u_0\|^2}{(p-2)(u_0, u_1)}, \; \; \; & \ if\ E(0)\leq 0 \ and\ (u_0, u_1) > 0; \\ \frac{2(p\Lambda+2)\|u_0\|}{(p-2)\sqrt{-2E(0)}}, \; \; \; & \ if\ E(0) < 0 \ and\ (u_0, u_1) = 0. \end{array} \right. \end{equation*} |
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
Δϕ‖2+‖ϕ‖2<2pdp−2+σ∫Ω∫Ωϕ(x)ϕ(y)|x−y|αdxdy, |
which, together with
Secondly we prove
J(ϕ)<d(by (2.28)), | (2.32) |
‖Δϕ‖2+‖ϕ‖2<2pdp−2+σ∫Ω∫Ωϕ(x)ϕ(y)|x−y|αdxdy(by (2.28)), | (2.33) |
I(ϕ)≤0,ϕ≠0(by (2.24) and (2.32)). | (2.34) |
If
Lemma 2.6. Assume
u∈C([0,Tmax);H20(Ω))∩C1([0,Tmax);L2(Ω)) |
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
u(t)∈W,t∈[t0,t1); and u(t1)∈∂W. |
Since the energy is conservative (see (1.5)),
J(u)(t1)≤E(t1)=E(t0)<d. | (2.35) |
Then by
Secondly, we proof the second part. In the case
E(t0)=d and (ut,u)t=t0≥0 |
in detail. Arguing by contradiction, if the conclusion is incorrect, by
Due to
‖Δu(t1)‖2+‖u(t1)‖2≥((1−|σ|σ∗)(CH20→Lp)−p)2p−2. | (2.36) |
If
If
d=E(t0)=12‖ut(t1)‖2+J(u)(t1). | (2.37) |
Combining (2.37) and
‖ut(t1)‖2=0. | (2.38) |
Utilizing Cauchy-Schwartz's inequality, we obtain that
(ut,u)|t=t1≤‖ut(t1)‖‖u(t1)‖=0. | (2.39) |
Integrating (1.8) over
(ut,u)+t∫0I(u)(τ)dτ−t∫0‖uτ‖2dτ=(u1,u0),0≤t<Tmax. | (2.40) |
By (2.40), we get
(ut,u)|t=t0+t0∫0I(u)(τ)dτ−t0∫0‖uτ‖2dτ=(ut,u)|t=t1+t1∫0I(u)(τ)dτ−t1∫0‖uτ‖2dτ. |
Then,
(ut,u)|t=t0−(ut,u)|t=t1=t1∫t0I(u)(τ)dτ−t1∫t0‖uτ‖2dτ. |
Since
(ut,u)|t=t1=(ut,u)|t=t0−t1∫t0I(u)(τ)dτ+t1∫t0‖uτ‖2dτ>0, |
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
‖Φ‖H=√(Φ,Φ)H,Φ∈H. |
Suppose
‖F(U)−F(V)‖H≤M‖U−V‖H. | (3.41) |
Consider the following abstract semilinear evolution equation
{Ut+AU=F(U),t>0,U(0)=U0, | (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
U(t)=etAU0+t∫0e(t−τ)AF(U(τ))dτ, t∈[0,T]. | (3.43) |
The solution
1.
2.
limt↑Tmax‖U(t)‖H=∞, |
i.e., the solution blows up at a finite time
Furthermore, if
By introduction
A=(0I−Δ2−I0),F(U)=(0σ|x|−α∗u+|u|p−2u), | (3.44) |
where
{Ut=AU+F(U)x∈Ω,t>0,U=∂U∂ν=0,x∈∂Ω,t>0U(x,0)=U0(x),x∈Ω. | (3.45) |
In the next lemma, we show
Lemma 3.3. Let
Proof. Let
(Φ,Ψ)H:=∫Ω(Δφ1Δψ1+φ1ψ1+φ2ψ2)dx, | (3.46) |
where
‖Φ‖H=(Φ,Φ)H=‖ϕ1‖H2+‖φ2‖. |
Let
A:D(A)=(H4(Ω)∩H20(Ω))×H20(Ω)⊂H→H. |
Next we show
(AΦ,Φ)H=((φ2,−Δ2φ1−φ1),(φ1,φ2))H=∫Ω(Δφ2Δφ1+φ2φ1+(−Δ2φ1−φ1)φ2)dx=0. | (3.47) |
Next we show
{Δ2u+2u=f1+f2,x∈Ω,u=∂u∂ν=0,x∈∂Ω | (3.48) |
admits a unique solution
(I−A)U=(I−IΔ2+II)(uv)=(u−vΔ2u+u+v)=(f1Δ2u+2u−f1)=(f1f2)=f, |
which implies
Next, we show (3.45) admits a mild solution.
Lemma 3.4. Assume
U(t)=etAU0+t∫0e(t−τ)AF(U(τ))dτ,0≤t≤T. | (3.49) |
The solution
1.
2.
limt↑Tmax‖U(t)‖H=limt↑Tmax(‖u(t)‖H2+‖ut(t)‖)=∞, |
i.e., the solution blows up at a finite time
Furthermore, it holds
‖U(t)‖H=‖U0‖H+2t∫0(F(U(τ)),U(τ))Hdτ. | (3.50) |
Proof. Let
σ|x|−α∗u+|u|p−2u∈L2(Ω),∀u∈H20(Ω). | (3.51) |
Since
Next we show
Case 1.
R=supx,y∈Ω|x−y|<∞. |
Then, by Hölder's inequality,
‖|x|−α∗u‖2=∫Ω(∫Ω|x−y|−αu(y)dy)2dx≤R−2α|Ω|(∫Ωu(y)dy)2≤R−2α|Ω|2‖u‖2<∞. | (3.52) |
Case 2.
12n2n−α+αn+12n2n−α=2, |
by using (2.16) with
∫Ω(|x|−α∗u)(x)ϕ(x)dx=∫Ω∫Ωu(y)ϕ(x)|x−y|αdxdy≤κ‖u‖2n2n−α‖ϕ‖2n2n−α≤κC2Ω‖u‖‖ϕ‖. |
Then we get
‖|x|−α∗u‖=supϕ∈L2(Ω),‖ϕ‖=1∫Ω(|x|−α∗u)(x)ϕ(x)dx≤κC2Ω‖u‖<∞. | (3.53) |
So (3.51) is true.
‖U1‖H=‖u1‖H2+‖v1‖≤M,‖U2‖H=‖u2‖H2+‖v2‖≤M, | (3.54) |
where
χ={R−α|Ω|,α≤0κC2Ω,0<α<n. |
Then, by (3.52) and (3.53),
‖F(U1)−F(U2)‖H≤|σ|‖|x|−α∗(u1−u2)‖+‖|u1|p−2u1−|u2|p−2u2‖≤|σ|χ‖u1−u2‖+(p−2)‖1∫0|θu1+(1−θ)u2|p−2dθ(u1−u2)‖. |
Case 1.
‖1∫0|θu1+(1−θ)u2|p−2dθ(u1−u2)‖≤‖1∫0|θu1+(1−θ)u2|p−2dθ‖∞‖u1−u2‖≤(‖u1‖∞+‖u2‖∞)p−2‖u1−u2‖≤(2CH20→L∞M)p−2‖u1−u2‖. |
Case 2.
‖1∫0|θu1+(1−θ)u2|p−2dθ(u1−u2)‖2≤(∫Ω(1∫0|θu1+(1−θ)u2|p−2dθ)n2dx)n4‖u1−u2‖22nn−4≤2np2(CH20→L2nn−4)2(∫Ω(|u1|n(p−2)2+|u2|n(p−2)2)dx)n4‖u1−u2‖2H2≤2np2+n4(CH20→L2nn−4)2(CH20→Ln(p−2)2M)n2(p−2)8‖u1−u2‖2H2 |
In view of the above three inequalities, we get
Next we prove (3.50). Suppose firstly
12ddt‖U(t)‖2H=(U,Ut)H=(U,A(U))+(U,F(U))=(U,F(U)),0≤t<Tmax. |
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
((U,V))=∫Ω(u1v1+u2v2)dx,∀U=(u1,u2),V=(v1,v2)∈L2(Ω)×L2(Ω). |
Since
((U,Φ))=((etAU0,Φ))+((t∫0e(t−τ)AF(U(τ)),Φ)). |
We differentiate to obtain
ddt((U,Φ))=ddt((etAU0,Φ))+ddt((t∫0e(t−τ)AF(U(τ)),Φ)). | (3.55) |
Now, using the standard properties of the semigroup (see for example, [54,Chapter 2]), we obtain
ddt((etAU0,Φ))=((etAU0,A∗Φ))+((etAU0,Φt)) | (3.56) |
where
A∗=(0−Δ2−II0) |
is the adjoint operator of
ddt((t∫0e(t−τ)AF(U(τ)),Φ))=((F(U(t)),Φ))+((t∫0e(t−τ)AF(U(τ)),A∗Φ))+((t∫0e(t−τ)AF(U(τ)),Φt)). | (3.57) |
Then it follows from (3.55)-(3.57) and (3.49) that
ddt((U,Φ))=((F(U(t)),Φ))+((U,A∗Φ))+((U,Φt)). | (3.58) |
Since
((U,Φ))=∫Ωutϕdx,((F(U(t)),Φ))=∫Ω(σ|x|−α∗u+|u|p−2u)ϕdx,((U,A∗Φ))=(((u,ut),(−Δ2ϕ−ϕ,0)))=−∫Ω(uΔ2ϕ+uϕ)dx,((U,Φt))=∫Ωutϕtdx. |
Then it follows from (3.58) that
ddt∫Ωutϕdx+∫Ω(uΔ2ϕ+uϕ)dx=∫Ω(σ|x|−α∗u+|u|p−2u)ϕdx+∫Ωutϕtdx. |
Since
ddt∫Ωutϕdx+∫Ω(ΔuΔϕ+uϕ)dx=∫Ω(σ|x|−α∗u+|u|p−2u)ϕdx+∫Ωutϕtdx. | (3.59) |
Note
Step 2. Proof of
u(t)∈C([0,Tmax);H20(Ω)) and ut∈C([0,Tmax);L2(Ω)), |
by taking
ddt(ut,u)=‖ut‖2−(=I(u)⏞‖Δu‖2+‖u‖2−‖u‖pp−σ∫Ω∫Ωu(x,t)u(y,t)|x−y|αdxdy)∈C[0,Tmax), |
i.e,
Step 3. Proof of the equality (1.5). The energy identity (1.5) follows from (3.50) directly. In fact by using
‖U‖H=‖Δu‖2+‖u‖2+‖ut‖2,(F(U),U)H=ddt(σ2∫Ω∫Ωu(x,t)u(y,t)|x−y|αdxdy+1p‖u‖pp). |
Then by (3.50), we get (1.5).
Proof of Theorem 1.3. Let
E(t)=E(0)≤0,0≤t<Tmax. |
By the definitions of
I(u)=2J(u)−p−2p‖u‖pp,I(u)=pJ(u)−p−22(‖Δu‖2+‖u‖2−σ∫Ω∫Ωu(x,t)u(y,t)|x−y|αdxdy). |
By (1.5) and (1.6), it follows
I(u)=2E(0)−‖ut‖2−p−2p‖u‖pp | (4.60) |
and
I(u)=pE(0)−p2‖ut‖2−p−22(‖Δu‖2+‖u‖2−σ∫Ω∫Ωu(x,t)u(y,t)|x−y|αdxdy). | (4.61) |
By (1.11), we get
|σ∫Ω∫Ωu(x,t)u(y,t)|x−y|αdxdy|≤‖Δu‖2+‖u‖2 if |σ|≤σ∗. | (4.62) |
In the following we divide the proof into two cases:
Case 1.
I(u)≤pE(0)−p2‖ut‖2. | (4.63) |
Let
h(t)=‖u(t)‖2+β(t+γ)2,0≤t<Tmax, | (4.64) |
where
h′(t)=2(ut,u)+2β(t+γ), | (4.65) |
h″(t)=2‖ut‖2−2I(u)(t)+2β≥−2pE(0)+(p+2)‖ut‖2+2β. | (4.66) |
By Cauchy-Schwartz's inequality,
(h′(t))2≤4(‖ut‖‖u‖+β(t+γ))2=4(‖u‖2‖ut‖2+β2(t+γ)2+2β(t+γ)‖u‖‖ut‖)≤4(‖u‖2‖ut‖2+β2(t+γ)2+β(t+γ)2‖ut‖2+β‖u‖2)≤4((‖u‖2+β(t+γ)2)(‖ut‖2+β))=4h(t)(‖ut‖2+β). | (4.67) |
Then by (4.66) and (4.67), it follows
h″(t)h(t)−(1+p−24)(h′(t))2≥p(−2E(0)−β)h(t)≥0 for 0≤β≤−2E(0). | (4.68) |
Subcase 1.
ˆT≤h(0)p−24h′(0)=2‖u0‖2(p−2)(u0,u1). |
Subcase 2.
β=−2E(0) and γ=‖u0‖√β, |
then
ˆT≤h(0)p−24h′(0)=4‖u0‖(p−2)√−2E(0). |
Subcase 3.
β=−2E(0),γ=−(u0,u1)+√(u0,u1)2+β‖u0‖2β, |
then,
h(0)=‖u0‖2+βγ2=‖u0‖2+(−(u0,u1)+√(u0,u1)2+β‖u0‖2)2/β>0,h′(0)=2(u0,u1)+2βγ=2√(u0,u1)2+β‖u0‖2)>0. |
Then, it follows from Lemma 2.3 that
ˆT≤h(0)p−24h′(0)=2(‖u0‖2+(−(u0,u1)+√(u0,u1)2+β‖u0‖2))2β)(p−2)√(u0,u1)2+β‖u0‖2)=2β‖u0‖2+2(−(u0,u1)+√(u0,u1)2+β‖u0‖2)2β(p−2)√(u0,u1)2+β‖u0‖2=−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 |
Case 2.
|∫Ω∫Ωu(x,t)u(y,t)|x−y|αdxdy|≤R−α(∫Ω|u|dx)2≤R−α|Ω|2(p−1)p‖u‖2p. | (4.69) |
12n2n−α+αn+12n2n−α=2 |
and
|∫Ω∫Ωu(x,t)u(y,t)|x−y|αdxdy|≤κ‖u‖22n2n−α≤κ|Ω|p(2n−α)−2nn‖u‖2p, | (4.70) |
where
Let
Θ={R−α|Ω|2(p−1)p,if α≤0,κ|Ω|p(2n−α)−2nn,if 0<α<n. |
By (4.69) and (4.70), we get
|∫Ω∫Ωu(x,t)u(y,t)|x−y|αdxdy|≤Θ‖u‖2p. | (4.71) |
Subcase 1.
ddt(ut,u)≥0. |
Then
ddt‖u(t)‖2=(ut,u)≥(u1,u0)≥0, |
and then
‖u0‖2≤‖u‖2≤|Ω|p−2p‖u‖2p, |
which, together with (4.71) implies
|∫Ω∫Ωu(x,t)u(y,t)|x−y|αdxdy|≤Θ‖u‖2−pp‖u‖pp≤Θ|Ω|(p−2)22p‖u0‖2−p‖u‖pp. | (4.72) |
In view of (1.11), (4.61) and (4.72), we obtain
I(u)≤pE(0)−p2‖ut‖2−p−22(‖Δu‖2+‖u‖2−|σ||∫Ω∫Ωu(x,t)u(y,t)|x−y|αdxdy|)=pE(0)−p2‖ut‖2−p−22(‖Δu‖2+‖u‖2−σ∗|∫Ω∫Ωu(x,t)u(y,t)|x−y|αdxdy|)+p−22(|σ|−σ∗)|∫Ω∫Ωu(x,t)u(y,t)|x−y|αdxdy|≤pE(0)−p2‖ut‖2+p−22Λ‖u‖pp, | (4.73) |
where
Λ=(|σ|−σ∗)Θ|Ω|(p−2)22p‖u0‖2−p. | (4.74) |
It follows
I(u)≤2p(Λ+1)pΛ+2E(0)−p(Λ+1)pΛ+2‖ut‖2 | (4.75) |
Let
h′(t)=2(ut,u)+2β(t+γ) | (4.76) |
and
h″(t)=2‖ut‖2−2I(u)(t)+2β≥−4p(Λ+1)pΛ+2E(0)+(2p(Λ+1)pΛ+2+2)‖ut‖2+2β. | (4.77) |
Then it follows from (4.67) that
h″(t)h(t)−(1+p−22(pΛ+2))(h′(t))2≥2p(Λ+1)pΛ+2(−2E(0)−β)h(t)≥0 for 0≤β≤−2E(0). | (4.78) |
ˆT≤h(0)p−22(pΛ+2)h′(0)=(pΛ+2)‖u0‖2(p−2)(u0,u1). |
β=−2E(0) and γ=‖u0‖√β, |
then
ˆT≤h(0)p−22(pΛ+2)h′(0)=2(pΛ+2)‖u0‖(p−2)√−2E(0). |
Subcase 2.
By (4.60),
ddt(ut,u)≥−I(u)(t)≥−2E(0)>0. |
Integrating this inequality from
(ut,u)≥(u0,u1)−2E(0)t. |
Let
Proof of Theorem 1.4. Let
Step 1.
‖Δu‖2+‖u‖2<2pdp−2+σ∫Ω∫Ωu(x,t)u(y,t)|x−y|αdxdy,0≤t<Tmax. | (4.79) |
By (1.11), it follows
|∫Ω∫Ωu(x,t)u(y,t)|x−y|αdxdy|≤1σ∗(‖Δu‖2+‖u‖2), |
which, together with (4.79) and
‖Δu‖2+‖u‖2≤(1−|σ|σ∗)−12pdp−2,0≤t<Tmax. | (4.80) |
Since
‖ut‖2=2E(0)−2J(u)(t)≤2d−I(u)<2d,0≤t<Tmax. | (4.81) |
By (4.80), (4.81), and Theorem 1.2, we get
Step 2.
J(u)(t)≤d,t∈[0,Tmax) and J(u0)=d. |
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
I(u)≤pE(0)−p2‖ut‖2−pd,0≤t<Tmax. | (4.82) |
Let
h′(t)=2(ut,u)+2β(t+γ), | (4.83) |
h″(t)=2‖ut‖2−2I(u)(t)+2β≥−2pE(0)+(p+2)‖ut‖2+2pd+2β. | (4.84) |
Then it follows from (4.67) that
h″(t)h(t)−(1+p−24)(h′(t))2≥p[2(d−E(0))−β]≥0 for 0≤β≤2(d−E(0)). | (4.85) |
Case 1.
ˆT≤h(0)p−24h′(0)=2‖u0‖2(p−2)(u0,u1). |
Case 2.
β=2(d−E(0)),γ=‖u0‖√β, |
then
ˆT≤h(0)p−24h′(0)=4‖u0‖(p−2)√2(d−E(0)). |
Case 3.
β=2(d−E(0)),γ=−(u0,u1)+√(u0,u1)2+β‖u0‖2)β, |
then,
h(0)=‖u0‖2+βγ2=‖u0‖2+(−(u0,u1)+√(u0,u1)2+β‖u0‖2))2/β>0,h′(0)=2(u0,u1)+2βγ=2√(u0,u1)2+β‖u0‖2)>0. |
Then, it follows from Lemma 2.3 that
ˆT≤h(0)p−24h′(0)=2(‖u0‖2+(−(u0,u1)+√(u0,u1)2+β‖u0‖2))2β)(p−2)√(u0,u1)2+β‖u0‖2)=2β‖u0‖2+2(−(u0,u1)+√(u0,u1)2+β‖u0‖2)2β(p−2)√(u0,u1)2+β‖u0‖2 |
=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 authors would like to thank the referees for the comments and valuable suggestions.
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