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[2] | Yi Cheng, Ying Chu . A class of fourth-order hyperbolic equations with strongly damped and nonlinear logarithmic terms. Electronic Research Archive, 2021, 29(6): 3867-3887. doi: 10.3934/era.2021066 |
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[5] | Shuting Chang, Yaojun Ye . Upper and lower bounds for the blow-up time of a fourth-order parabolic equation with exponential nonlinearity. Electronic Research Archive, 2024, 32(11): 6225-6234. doi: 10.3934/era.2024289 |
[6] | Abdelhadi Safsaf, Suleman Alfalqi, Ahmed Bchatnia, Abderrahmane Beniani . Blow-up dynamics in nonlinear coupled wave equations with fractional damping and external source. Electronic Research Archive, 2024, 32(10): 5738-5751. doi: 10.3934/era.2024265 |
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[8] | Mohammad M. Al-Gharabli, Adel M. Al-Mahdi . Existence and stability results of a plate equation with nonlinear damping and source term. Electronic Research Archive, 2022, 30(11): 4038-4065. doi: 10.3934/era.2022205 |
[9] | Huafei Di, Yadong Shang, Jiali Yu . Existence and uniform decay estimates for the fourth order wave equation with nonlinear boundary damping and interior source. Electronic Research Archive, 2020, 28(1): 221-261. doi: 10.3934/era.2020015 |
[10] | Yuchen Zhu . Blow-up of solutions for a time fractional biharmonic equation with exponentional nonlinear memory. Electronic Research Archive, 2024, 32(11): 5988-6007. doi: 10.3934/era.2024278 |
In this paper, we deal with the following plate equation with nonlinear damping and a logarithmic source term
{utt+Δ2u+|ut|m−2ut=|u|p−2ulog|u|k,(x,t)∈Ω×R+,u=∂u∂ν=0,(x,t)∈∂Ω×R+,u(x,0)=u0(x),ut(x,0)=u1(x),x∈Ω, | (1) |
where
2<p<2(n−2)n−4ifn≥5;2<p<+∞ifn≤4. | (2) |
The logarithmic nonlinearity is of much interest in many branches of physics such as nuclear physics, optics and geophysics (see [5,6,15] and references therein). It has also been applied in quantum field theory, where this kind of nonlinearity appears naturally in cosmological inflation and in super symmetric field theories [4,13].
Let us review somework with logarithmic term which is closely related to the problem (1). Birula and Mycielski[6,7] studied the following problem
{utt−uxx+u−εulog|u|2=0,(x,t)∈[a,b]×(0,T),u(a,t)=u(b,t)=0,t∈(0,T),u(x,0)=u0(x),ut(x,0)=u1(x),x∈[a,b], | (3) |
which is a relativistic version of logarithmic quantum mechanics and can also be obtained by taking the limit
utt−Δu=ulog|u|k, | (4) |
in
utt−Δu+u+ut+|u|2u=ulog|u| | (5) |
to study the dynamics of Q-ball in theoretical physics. A numerical research was given in that work, while, there was no theoretical analysis for this problem. For the initial boundary value problem of(5), Han [17] obtained the global existence of weak solution in
utt−Δu+u+(g∗Δu)(t)+h(ut)ut+|u|2u=ulog|u|k. |
Recently, Al-Gharabli and Messaoudi [1] considered the following plate equation with logarithmic source term
{utt+Δ2u+u+h(ut)=ulog|u|k,(x,t)∈Ω×R+,u=∂u∂ν=0,(x,t)∈∂Ω×R+,u(x,0)=u0(x),ut(x,0)=u1(x),x∈Ω, | (6) |
where
utt+Δ2u+u−∫t0g(t−s)Δ2u(s)ds=ulog|u|k,(x,t)∈Ω×R+, | (7) |
they established the existence of solutions and proved an explicit and general decay rate result. However, there is no information on the finite or infinite blow up results in these researches [1,2,3].
At the same time, there are many results concerning the existence and nonexistence on evolution equation with polynomial source term. For example, for plate equation with polynomial source term
{utt+Δ2u+|ut|m−2ut=|u|p−2u,(x,t)∈Ω×R+,u=∂u∂ν=0,(x,t)∈∂Ω×R+,u(x,0)=u0(x),ut(x,0)=u1(x),x∈Ω, |
established an existence result and showed that the solution continues to exist globally if
To the best of our knowledge, there are few results on the evolution equation with the nonlinear logarithmic source term
utt−Δu+μ1ut(x,t)+μ2ut(x,t−τ)=|u|p−2ulog|u|k, | (8) |
obtained the local existence by using the semigroup theory and proved a finite time blow-up result when the initial energy is negative. Of course, these results also hold for the equation (8) without delay term (i.e.
Motivated by the above mentioned papers, our purpose in this research is to investigative the existence, energy decay and finite time blow-up of the solution to the initial boundary value problem (1). We note here that (i) the term
The rest of this article is organized as follows: Section 2 is concerned with some notation and some properties of the potential well. In Sect. 3, we present the existence and uniqueness of local solutions to (1) by using the contraction mapping principle. In Sect. 4, we prove the global existence and energy decay results. The proof of global existence result is based on the potential well theory and the continuous principle; while for energy decay result, the proof is based on the Nakao's inequality and some techniques on logarithmic nonlinearity. In Sect. 5, we prove the the finite time blow-up when the initial energy is negative. In Sect. 6, we establish the finite time blow-up result for problem (1) with
We give some material needed in the proof of our results. We use the standard Lebesgue space
Firstly, we introduce the Sobolev's embedding inequality : assume that
‖u‖p≤Cp‖Δu‖2,for u∈H20(Ω) | (9) |
where
Suppose (2) holds, we define
α∗:={2nn−4−p if n≥5,+∞ if n≤4 |
for any
Definition 2.1. A function
u(0)=u0,ut(0)=u1 |
and
⟨utt,v⟩+(Δu,Δv)+∫Ω|ut|m−2utvdx=∫Ω(|u|p−2ulog|u|k)vdx | (10) |
for any
Now, we introduce the energy functional
J(u)=J(u(t))=J(t)=12‖Δu‖2−1p∫Ω|u|plog|u|kdx+kp2‖u‖pp, | (11) |
and
I(u)=I(u(t))=I(t)=‖Δu‖2−∫Ω|u|plog|u|kdx. | (12) |
From the definitions (11) and (12), we have
J(u)=1pI(u)+(12−1p)‖Δu‖2+kp2‖u‖pp. | (13) |
The following lemmas play an important role in the studying the properties of the potential well.
Lemma 2.2. Let
(ⅰ):
(ⅱ): there exists a unique
(ⅲ):
Proof. We know
g(λ)=J(λu)==12λ2‖Δu‖2−kpλp∫Ω|u|plog|u|dx−kpλplogλ‖u‖pp+kp2λp‖u‖pp. |
It is obvious that
g′(λ)=λ(‖Δu‖2−kλp−2∫Ω|u|plog|u|dx−kλp−2logλ‖u‖pp) | (14) |
and
g′′(λ)=‖Δu‖2−k(p−1)λp−2∫Ω|u|plog|u|dx−k(p−1)λp−2logλ‖u‖pp−kλp−2‖u‖pp. |
From (14) and
g′(λ)|λ=λ∗=0, |
then we obtain
‖Δu‖2=kλ∗p−2∫Ωu2log|u|dx+kλ∗p−2logλ∗‖u‖pp. |
Substituting the above equation into
g′′(λ∗)=−k(p−2)λ∗p−2∫Ω|u|plog|u|dx−k(p−2)λ∗p−2logλ∗‖u‖pp−kλ∗p−2‖u‖pp=−(p−2)‖Δu‖2−kλ∗p−2‖u‖pp<0. |
From these and
From (12) and (14), we have
I(λu)=λddλJ(λu)=λg′(λ){>0,0<λ<λ,=0,λ=λ∗,<0,λ∗<λ<+∞. |
Then, we could define the potential well depth of the functional
d=inf{supλ≥0J(λu)|u∈H20(Ω)∖{0}}. | (15) |
We also define the well-known Nehari manifold
N={u|u∈H20(Ω)∖{0},I(u)=0}. |
As in [29,34], that the mountain pass level
d=infu∈NJ(u). |
It is easy to see that
Lemma 2.3. Assume that
r(α):=(αkCp+α∗)1p+α−2. |
Then, for any
(ⅰ) : if
(ⅱ) : if
Proof. Since
I(u)=‖∇u‖22−k∫Ω|u|plog|u|dx>‖∇u‖22−kα‖u‖p+αp+α≥‖∇u‖22−kCp+α∗α‖∇u‖p+α2=kCp+α∗α‖∇u‖22(rp+α−2(α)−‖∇u‖p+α−22). | (16) |
Obviously, the results can be obtained from the above inequality (16).
Lemma 2.4. Assume the notations in Lemma 2.2 hold, we have
0<r∗:=supα∈(0,α∗)=(αkCp+α∗)1p+α−2≤r∗:=supα∈(0,α∗)(αkBp+α)1p+α−2|Ω|αp(p+α−2)<+∞, |
where
Proof. It is obvious that
γ(α)=(αkBp+α)1p+α−2|Ω|αp(p+α−2),α∈(0,+∞). |
For any
‖u‖p≤|Ω|αp(p+α)‖u‖p+α. |
Then, noticing
C∗=supu∈H20∖{0}‖u‖p+α‖Δu‖2≥|Ω|−αp(p+α)supu∈H20∖{0}‖u‖p‖Δu‖2≥|Ω|−αP(p+α)B, |
which implies
(αkCp+α∗)1p+α−2≤(αkBp+α)1p+α−2|Ω|αp(p+α−2), |
that is
Now, we will prove
Case a. If
r∗=supα∈(0,2nn−4−p)γ(α)≤maxα∈[0,2nn−4−p]γ(α)<+∞ |
Case b. If
h(α):=log[γ(α)]=1p+α−2[logα−logk−(p+α)logB]+αp(p+α−2)log|Ω|. |
Hence
h′(α)=p2+pα−2p+pαlogk−pαlogα+2pαlogB+pαlog|Ω|−2αlog|Ω|pα(p+α−2)2. |
For simplicity, we set
g(α):=p2+pα−2p+pαlogk−pαlogα+2pαlogB+pαlog|Ω|−2αlog|Ω|, |
then
g′(α)=p+plogk−plogα−p+2plogB+plog|Ω|−2log|Ω|=plogkB2|Ω|1−2pα, |
which yields that the function
On the one hand, due to
limα→0+g(α)=p2−2p>0 |
which implies that
While on the other hand, we can deduce that
limα→+∞g(α)=limα→+∞(p2−2p+pα[1+log(kB2|Ω|1−2p)]−logα)=−∞, |
which together with
Noting the relation between
r∗=supα∈(0,+∞)σ(α)=eh(α∗)<+∞. |
Making using of the Lemmas 2.2 and 2.3, we obtain the following corollary.
Corollary 1. Assume that
(ⅰ): if
(ⅱ): if
for any
Lemma 2.5. Assume that
Proof. (ⅰ) For the case
(ⅱ) For the case
J(u)=(12−1p)‖Δu‖22+kp2‖u‖pp≥(p−22p)r2∗>0. |
We define energy for the problem (1), which obeys the following energy equality of the weak solution
E(t)+∫t0‖uτ‖mmdτ=E(0), for all t∈[0,T) | (17) |
where
E(t)=12‖ut‖2+12‖Δu‖2−1p∫Ω|u|plog|u|kdx+kp2‖u‖pp, |
E(0)=12‖u1‖2+12‖Δu0‖2−1p∫Ω|u0|plog|u0|kdx+kp2‖u0‖pp. |
It is obvious that
E(t)=12‖ut‖2+J(u). |
Taking
ddtE(t)=−‖ut‖mm. | (18) |
Now, we define the subsets of
W={u∈H10(Ω)|J(u)<d,I(u)>0},V={u∈H10(Ω)|J(u)<d,I(u)<0}, | (19) |
where
In order to establish the global existence and blow-up results of solution, we have to prove the following invariance sets of
Lemma 2.6. If
(ⅰ):
(ⅱ):
Proof. It follows from the definition of weak solution and (17) that
12‖ut‖2+J(u)≤12‖u1‖2+J(u0)<d, for any t∈[0,T). | (20) |
(ⅰ) Arguing by contradiction, we assume that there exists a number
It follows from (20) that (a) is impossible. If (b) holds, then by the definition of
(ⅱ) The proof is similar to the proof of (ⅰ). We omit it.
In this section, we are concerned with the local existence and uniqueness for the solution of the problem (1). The idea comes from [14,28,38], where the source term is polynomial. First, we give a technical lemma given in [22] which plays an important role in the uniqueness of the solution.
Lemma 3.1. ([22]) For every
j(s)=|s|p−2log|s|,p>2 |
satisfies
|j(s)|≤A+|s|p−2+ε. |
Theorem 3.2. Suppose that
Proof. For every
H:=C([0,T];H20(Ω))∩C1([0,T];L2(Ω)) |
endowed with the norm
‖u(t)‖H=(maxt∈[0,T](‖Δu(t)‖22+‖ut(t)‖22))12. |
For every given
{vtt+Δ2v+|vt|m−2vt=|u|p−2ulog|u|k,(x,t)∈Ω×R+,v=∂v∂ν=0,(x,t)∈∂Ω×R+,v(x,0)=u0(x),vt(x,0)=u1(x),x∈Ω. | (21) |
We shall prove that the problem (21) admits a unique solution
Let
u0h=h∑i=1(∫ΩΔu0Δwi)wi and u1h=h∑i=1(∫Ωu1wi)wi |
such that
vh(t)=h∑i=1γih(t)ωi, | (22) |
solves the following problem
{∫Ω(v′′h+Δ2vh+|v′h|m−2v′h−|u|p−2ulog|u|k)ηdx=0,vk(0)=u0h,v′h(0)=u1h. | (23) |
For
{γ′′ih(t)+λiγih(t)+ci|γ′ih(t)|m−2γ′ih(t)=ψi(t),γih(0)=∫Ωu0ωi,γ′ih(0)=∫Ωu1ωi, |
where
ci=‖ωi‖mm,ψi(t)=∫Ω|u(t)p−2u(t)log|u|kωidx∈C[0,T]. |
Then the above problem admits a unique local solution
Taking
‖v′h(t)‖2+‖Δvh(t)‖2+2∫t0‖v′h(τ)‖mmdτ=‖v1h‖2+‖Δv0h‖2+2∫t0∫Ω|u|p−2ulog|u|kv′h | (24) |
for every
2∫t0∫Ω|u|p−2ulog|u|kv′h |
≤2∫t0∫Ω||u|p−1log|u|k||v′h|≤∫t0∫Ω(C||u|p−1log|u||mm−1+∫t0‖v′h‖mm). | (25) |
In order to estimate (25), we focus on the logarithmic term. Here we denote
∫Ω‖u|p−1log|u||mm−1dx=∫Ω1‖u|p−1log|u||mm−1dx+∫Ω2‖u|p−1log|u||mm−1dx. |
By a simple calculation, we obtain
infs∈(0,1)sp−1logs=−1e(p−1), |
which implies
∫Ω1‖u|p−1log|u||mm−1dx≤[e(p−1)]−mm−1|Ω|. |
Let
ρ=2nn−4⋅m−1m−p+1>0forn≥5; any positive ρ,n≤4. |
By the Sobolev embedding from
∫Ω2‖u|p−1log|u||mm−1dx≤ρ−m−1m∫Ω2(|u|p−1+ρ)m−1mdx≤ρ−m−1m∫Ω2|u|2nn−4dx≤ρ−m−1m∫Ω|u|2nn−4dx=ρ−m−1m‖u‖2nn−42nn−4≤C‖u‖2nn−4H20≤C. |
The proof of the case
2∫t0∫Ω|u|p−2ulog|u|kv′h≤CT+∫t0‖v′h‖mm. | (26) |
Substituting this inequality into (24), we obtain
‖v′h(t)‖2+‖Δvh(t)‖2+∫t0‖v′h(τ)‖mmdτ≤C, | (27) |
where
vh(t) is bounded in L∞([0,T],H20(Ω)),v′h(t) is bounded in Lm[(0,T],Lm(Ω))∩L∞([0,T],L2(Ω)),[2mm]vh′′(t) is bounded in L2([0,T],H−2(Ω)). | (28) |
Hence, up to a subsequence, we could pass to the limit in (23) and obtain a weak solution
To prove the uniqueness, arguing by contradiction: if
‖wt−vt‖2+‖Δw−Δv‖2+2∫t0∫Ω(|wτ|m−2wτ−|vτ|m−2vτ)(wτ−vτ)=0. | (29) |
It follows from the following element inequality
(|φ|m−2φ−|ψ|m−2ψ)(φ−ψ)≥C|φ−ψ|m for m≥2, |
that (29) can make to be
‖wt−vt‖2+‖v−w‖2H20+C∫T0‖wτ−vτ‖mm≤0. |
Therefore, we have
Now, we are in the position to prove Theorem 3.1. For
R2:=2(‖u1‖2+‖Δu0‖2), |
and
BRT:={u∈H|u(0,x)=u0(x),ut(0,x)=u1(x),‖u‖H≤R} |
for every
Claim.
In fact, assume that
‖vt(t)‖2+‖Δv(t)‖2≤‖u1‖2+‖Δu0‖2+CR2nn−4T≤R22+CR2nn−4T |
for
Next we show that
⟨vtt,η⟩+(Δv,Δη)+∫Ω(|v1t|m−2v1t−|v2t|m−2v2t)ηdx=∫Ω(|w1|p−2w1log|w1|k−|w2|p−2w2log|w2|k)ηdx, | (30) |
for any
Taking
∫Ω(|v1t|m−2v1t−|v2t|m−2v2t)(v1t−v2t)dx≥0, |
and integrating both sides of (30) over
‖vt‖2+‖Δv‖2≤2k‖|w1|p−2w1log|w1|−|w2|p−2w2log|w2|‖‖vt‖, | (31) |
We need estimating the logarithmic term in (31) by using Lemma 3.1. By the similar argument as [22], we give the sketch of the proof.
Making use of mean value theorem, we have, for
||w1|p−2w1log|w1|−|w2|p−2w2log|w2||=k|1+(p−1)log|θw1+(1−θ)w2|||θw1+(1−θ)w2|p−2|w1−w2|. |
Then, it follows from Lemma 3.1 that
||w1|p−2w1log|w1|−|w2|p−2w2log|w2||≤k|θw1+(1−θ)w2|p−2|w1−w2|+k(p−1)A|w1−w2|+k(p−1)|θw1+(1−θ)w2|p−2+ε|w1−w2|≤k(|w1|+|w2|)p−2|w1−w2|+k(p−1)A|w1−w2|+k(p−1)(|w1|+|w2|)p−2+ε|w1−w2|. |
Since
∫Ω[(|w1|+|w2|)p−2|w1−w2|]2dx≤C(∫Ω(|w1|+|w2|)2(p−1)dx)(p−2)/(p−1)×(∫Ω|w1−w2|2(p−1)dx)1/(p−1)≤C[‖w1‖2(p−1)L2(p−1)+‖w2‖2(p−1)L2(p−1)](p−2)/(p−1)‖w1−w2‖2L2(p−1)≤C[‖w1‖2(p−1)H20(Ω)+‖w2‖2(p−1)H20(Ω)](p−2)/(p−1)‖w1−w2‖2H20(Ω)≤CR2(p−2)‖w1−w2‖2H20(Ω). |
By the similar argument, we have
∫Ω[(|w1|+|w2|)p−2+ε|w1−w2|]2dx≤C(∫Ω(|w1|+|w2|)2(p−2+ε)(p−1)/(p−2)dx)(p−2)/(p−1) |
×(∫Ω|w1−w2|2(p−1)dx)1/(p−1)≤(∫Ω(|w1|+|w2|)2(p−1)+2ε(p−1)/(p−2)dx)(p−2)/(p−1)‖w1−w2‖2L2(p−1). |
Using (2), we can choose sufficiently small
ˉp=2(p−1)+2ε(p−1)p−2≤2nn−4, |
which yields that
∫Ω[(|w1|+|w2|)p−2+ε|w1−w2|]2dx≤C[‖w1‖ˉpLˉp(Ω)+‖w2‖ˉpLˉp(Ω)](p−2)/(p−1)‖w1−w2‖2L2(p−1)≤CRˉp(p−2)/(p−1)‖w1−w2‖2H20(Ω). |
Noticing
‖|w1|p−2w1log|w1|−|w2|p−2w2log|w2|‖≤C(Rp−2+1+Rˉp(p−2)/2(p−1))‖w1−w2‖H20(Ω). |
Thus, it follows from (31) that
‖Φ(w1)−Φ(w2)‖H=‖v1−v2‖H≤C(Rp−2+1+Rˉp(p−2)/2(p−1))T‖w1−w2‖H. | (32) |
We choose
In this section, we consider the global existence and energy decay of the solution for problem (1). First, we introduce the following lemmas which play an important role in studying the decay estimate of global solution for the problem (1).
Lemma 4.1. [33] Let
ϕ(t)1+r≤ω0(ϕ(t)−ϕ(t+1))on[0,T], |
where
(ⅰ): if
ϕ(t)≤(ϕ(0)−r+ω−10r[t−1]+)−1ron[0,T]; |
(ⅱ): if
ϕ(t)≤ϕ(0)e−ω1[t−1]+on[0,T], |
where
Now, we establish the global existence and energy decay results.
Theorem 4.2. Let
E(t)≤Ke−κt,ifm=2; |
and
E(t)≤(E(0)−m−22+(m−2)τ2[t−1]+)−2m−2,ifm>2, |
where
Proof. Step 1.. Global existence. It suffices to show that
d>E(0)≥E(t)=12‖ut‖2+J(u)=12‖ut‖2+1pI(u)+(12−1p)‖Δu‖2+kp2‖u‖pp>12‖ut‖2+p−22p‖Δu‖2, | (33) |
which yields that
‖ut‖2+‖Δu‖2≤2pp−2d<+∞. |
The above inequality and the continuation principle imply the global existence, i.e.
Step 2.. We claim that there exists constant
I(u)≥θ‖Δu‖2. | (34) |
In fact, it follows from
d≤J(λ0u(t))=1pI(λ0u)+(12−1p)‖Δ(λ0u)‖2+kp2‖λ0u‖pp=p−22pλ20‖Δu‖2+kp2λp0‖u‖pp=λp0(p−22pλ2−p0‖Δu‖2+kp2‖u‖pp)≤λp0(p−22p‖Δu‖2+kp2‖u‖pp)<λp0E(0), |
which implies that
λ0>(dE(0))1p>1. | (35) |
It follows from (12) that
0=I(λ0u)=‖Δ(λ0u)‖2−∫Ω|λ0u|plog|λ0u|kdx=λ20‖Δu‖2−λp0k∫Ω|u|plog|u|dx−(λp0klogλ0)‖u‖pp=λp0I(u)−λp0‖Δu‖2+λ20‖Δu‖2−(λp0klogλ0)‖u‖pp=λp0I(u)−(λp0−λ20)‖Δu‖2−(λp0klogλ0)‖u‖pp. |
Combining this equality with (35), we have
λp0I(u)=(λp0−λ20)‖Δu‖2+(λp0klogλ0)‖u‖pp≥(λp0−λ20)‖Δu‖2, |
which implies that
I(u)≥(1−λ2−p0)‖Δu‖2. |
Hence, the inequality (34) holds with
Step 3.. Energy decay. By integrating (18) over
E(t)−E(t+1)≡D(t)m, | (36) |
where
D(t)m=∫t+1t‖uτ‖mmdτ. | (37) |
In view of (37) and the embedding
∫t+1t∫Ω|ut|2dxdt≤c(Ω)D(t)2. | (38) |
Thus, from (38), there exist
‖ut(ti)‖22≤4c(Ω)D(t)2,i=1,2. | (39) |
On the other hand, multiplying
∫t2t1I(u)dt=∫t2t1‖ut‖2dt+(ut(t1),u(t1))−(ut(t2),u(t2))−∫t2t1∫Ω|ut|m−2utudxdt. | (40) |
It follows from (33) that
|∫t2t1∫Ω|um−2tutudxdt|≤∫t2t1‖u‖m‖ut‖m−1mdt≤C∫t2t1‖Δu‖‖ut‖m−1mdt≤C(2pp−2)12supt1≤s≤t2E(s)12∫t2t1‖ut‖m−1mdt≤C(2pp−2)12supt1≤s≤t2E(s)12D(t)m−1. | (41) |
By using (33) and (39), we also have
‖ut(ti)‖2‖u(ti)‖2≤C1D(t)supt1≤s≤t2E(s)12,i=1,2. | (42) |
Combining (38), (41) with (42), we have from (40) that
∫t2t1I(u)dt≤c(Ω)D(t)2+2C1D(t)supt1≤s≤t2E(s)12+C(2pp−2)12D(t)m−1supt1≤s≤t2E(s)12. | (43) |
Moreover, using (33) and (34), it is easy to see that
‖u‖pp≤Cpp‖Δu‖p≤Cpp(2pp−2E(0))p−221θI(u). |
Thus, we deduce that
E(t)≤12‖ut‖2+C2I(u), | (44) |
where
∫t2t1E(t)dt≤12∫t2t1‖ut‖22dt+C2∫t2t1I(u)dt. | (45) |
By integrating (18) over
E(t)=E(t2)+∫t2t‖ut‖mmds |
Since
E(t2)≤2∫t2t1E(t)dt. |
Then, in view of (36), we have
E(t)=E(t+1)+D(t)m≤E(t2)+D(t)m≤2∫t2t1E(t)dt+D(t)m. |
Thus, combining (38) with (45), we get that
E(t)≤(c(Ω)+2c(Ω)C2)D(t)2+D(t)m2+C2[2C1D(t)+C(2p(p−2))12D(t)m−1]supt1≤s≤t2E(s)12≤(c(Ω)+2c(Ω)C2)D(t)2+D(t)m+2C2[2C1D(t)+C(2p(p−2))12D(t)m−1]E(t)12. |
Hence, it follows from Young's inequality that
E(t)≤C3[D(t)2+D(t)m+D(t)2(m−1)] | (46) |
holds with some positive constant
E(t)≤C3[1+D(t)m−2+D(t)2m−4]D(t)2≤C3[1+E(0)m−2m+E(0)2m−4m]D(t)2, |
which implies that
E(t)m2≤(C4(E(0)))m2D(t)m=(C4(E(0)))m2(E(t)−E(t+1)), |
where
limE(0)→0C4(E(0))=C3 |
Hence, the energy decay estimates hold with
K=E(0)eκ,κ=log3C33C3−1andτ=(C4(E(0)))−m2. | (47) |
In this section, we will establish that the solution of problem (1) blows up in finite time provided
Lemma 5.1. Assume that (2) holds. Then there exists a positive constant
(∫Ω|u|plog|u|kdx)s/p≤C[∫Ω|u|plog|u|kdx+‖Δu‖22], |
for any
Lemma 5.2. Assume that (2) holds. Then there exists a positive constant
‖u‖pp≤C[∫Ω|u|plog|u|kdx+‖Δu‖2], |
for any
Lemma 5.3. Assume that (2) holds. Then there exists a positive constant
‖u‖sp≤C[‖u‖pp+‖∇u‖22], |
for any
The proof of lemma 5.1-5.3 is similar to the proof in [22], we omit the details.
Lemma 5.4. Assume that (2) and
‖u‖mm≤C[(∫Ω|u|plog|u|kdx)mp+‖Δu‖2mp], |
for any
Proof. Noting
Now we are in the position to state and prove the blow up result for
Theorem 5.5. Suppose that the conditions in Lemma 5.4 hold. Then the solution to the problem (1) blows up in finite time provided that
Proof. We denote
E(t)≤E(0)<0,H′(t)=−E′(t)=‖ut‖mm. |
and
0<H(0)≤H(t)≤1p∫Ω|u|plog|u|kdx. | (48) |
We define
L(t)=H1−β(t)+ε∫Ωuutdx,t≥0, |
where
2(p−m)(m−1)p2<β<p−m2(m−1)p<1. | (49) |
By taking a derivation of
L′(t)=(1−β)H−β(t)H′(t)+ε‖ut‖2−ε‖Δu‖2−ε∫Ω|ut|m−2utudx+ε∫Ω|u|plog|u|kdx. |
Adding and subtracting
L′(t)=(1−β)H−β(t)H′(t)+εp(1−a)+22‖ut‖2+εp(1−a)−22‖Δu‖2+εp(1−a)H(t)−ε∫Ω|ut|m−2utudx+εa∫Ω|u|plog|u|kdx+ε(1−a)kp‖u‖pp. | (50) |
In view of Young's inequality, we have
∫Ω|ut|m−2utudx≤δmm‖u‖mm+m−1mδ−m/(m−1)‖ut‖mm |
for any
L′(t)≥[(1−β)H−β(t)−m−1mεδ−m/(m−1)]‖ut‖mm−εδmm‖u‖mm+εp(1−a)+22‖ut‖2+εp(1−a)−22‖Δu‖2+εp(1−a)H(t)+εa∫Ω|u|plog|u|kdx+ε(1−a)kp‖u‖pp. | (51) |
Since the integral is taken over the
L′(t)≥[(1−β)−m−1mεM]H−β(t)‖ut‖mm−εM1−mmHβ(m−1)‖u‖mm+εp(1−a)+22‖ut‖2+εp(1−a)−22‖Δu‖2+εp(1−a)H(t)+εa∫Ω|u|plog|u|kdx+ε(1−a)kp‖u‖pp. | (52) |
Making using of (48), Lemma 5.4 and Young's inequality, we find
Hβ(m−1)‖u‖mm≤(∫Ω|u|plog|u|kdx)β(m−1)‖u‖mm≤C[(∫Ω|u|plog|u|kdx)β(m−1)+mp+(∫Ω|u|plog|u|kdx)β(m−1)‖Δu‖2mp]≤C[(∫Ω|u|plog|u|kdx)β(m−1)+mp+(∫Ω|u|plog|u|kdx)β(m−1)⋅pp−m+‖Δu‖2]. |
Hence, it follows from Lemma 5.1 that
2<β(m−1)p+m≤pand2<β(m−1)p2p−m≤p. |
Thus, Lemma 5.1 implies
Hβ(m−1)‖u‖mm≤C(∫Ω|u|plog|u|kdx+‖Δu‖2). | (53) |
Combining (52) and (53), we have
L′(t)≥[(1−β)−m−1mεM]H−β(t)‖ut‖mm+εp(1−a)+22‖ut‖2 |
+ε[p(1−a)−22−M1−mmC]‖Δu‖2+ε[a−M1−mmC]∫Ω|u|plog|u|kdx+εp(1−a)H(t)+ε(1−a)kp‖u‖pp. | (54) |
Now, we choose
p(1−a)−22>0 |
and
p(1−a)−22−M1−mmC>0anda−M1−mmC>0. |
Once
(1−β)−m−1mεM>0andL(0)=H1−β(0)+ε∫Ωu0u1dx>0. |
Thus, for some constant
L′(t)≥γ[H(t)+‖ut‖2+‖Δu‖2+‖u‖pp+∫Ω|u|plog|u|kdx]. | (55) |
Consequently we have
L(t)≥L(0),for allt>0. |
On the other hand, using Lemma 5.3, by the same method as in [32], we can deduce
L11−β(t)≤C[H(t)+‖ut‖2+‖Δu‖2+‖u‖pp],t≥0. | (56) |
Combining (55) and (56), we obtain
L′(t)≥λL11−β(t),t≥0. | (57) |
where
Lβ/(1−β)(t)≥1L−β/(1−β)(0)−λtβ/(1−β). |
which implies that
T≤T∗=1−βλβLβ/(1−β)(0). |
This completes the proof of Theorem 5.1.
In this section, we consider the problem (1) with the linear damping term, i.e.
Lemma 6.1. [14] Let
Lemma 6.2. Suppose that
∫Ωu0u1dx≥0. |
Let
Proof. Let
⟨utt,u⟩=ddt(ut,u)−‖ut‖2fora.e.t≥0, |
Moreover, by testing the equation with
⟨utt,u⟩+‖Δu‖2+(ut,u)=∫Ω|u|plog|u|kdx, |
which implies
ddt((ut,u)+12‖u‖2)=‖ut‖2−I(u). |
Hence, if
G′(t)+G(t)=2‖ut‖2−2I(u(t))>0fora.e.t∈[0,T). |
Therefore, it follows from Lemma 6.1 with
Lemma 6.3. Let
‖u1‖2−2(u1,u0)+ΛE(0)<0, | (58) |
where
Proof. If this was not the case, by the continuity of
(u1,u0)≤‖u1‖‖u0‖≤12(‖u1‖2+‖u0‖2). | (59) |
By Lemma 6.2, (58) and (59), we deduce that
F(t)=‖u(t)‖2>‖u0‖2≥2(u1,u0)−‖u1‖2>ΛE(0)fort∈(0,t0), | (60) |
which implies
F(t0)=‖u(t0)‖2>ΛE(0) | (61) |
by the continuity of
E(0)≥E(t0)=1pI(u(t0))+(12−1p)‖Δu(t0)‖2+kp2‖u(t0)‖pp≥p−22p‖Δu(t0)‖2 |
that is
‖Δu(t0)‖2≤2pp−2E(0). |
Hence, we have
F(t0)=‖u(t0)‖2≤B0‖Δu(t0)‖2≤2B0pp−2E(0)=ΛE(0), |
which is a contradiction with (61). The proof is complete.
We now present the main blow-up result for the weak solution of problem (1) with
Theorem 6.4. Assume the conditions of Lemma 6.3 hold. Then the weak solution
Proof. It follows from Lemma 6.3 that
η(t)=‖u‖2+∫t0‖u(τ)‖2dτ+(T0−t)‖u0‖2. |
Notice
η(t)≥ϱforallt∈[0,T0]. | (62) |
Moreover,
η′(t)=2(ut,u)+‖u‖2−‖u0‖2=2(ut,u)+2∫t0(uτ,u)dτ, | (63) |
hence, we have
η′′(t)=2‖ut‖2+2⟨utt,u⟩+2(ut,u)=2(‖ut‖2−‖Δu‖2+∫Ω|u|plog|u|kdx)=2‖ut‖2−2I(u(t)). | (64) |
It follows from (63) that
(η′(t))2=4((ut,u)2+2(ut,u)∫t0(uτ,u)dτ+(∫t0(uτ,u)dτ)2). |
By the Cauchy-Schwarz inequality, we obtain
‖ut‖2‖u‖2≥(ut,u)2 |
∫t0‖u‖2dτ∫t0‖uτ‖2dτ≥(∫t0(uτ,u)dτ)2 |
and
2(ut,u)∫t0(uτ,u)dτ≤2‖ut‖‖u‖(∫t0‖uτ‖2dτ)12(∫t0‖u‖2dτ)12≤‖ut‖2∫t0‖u‖2dτ+‖u‖2∫t0‖uτ‖2dτ. |
Combining the above inequalities, we have
η′(t)2≤4(‖ut‖2‖u‖2+‖ut‖2∫t0‖u‖2dτ+‖u‖2∫t0‖uτ‖2dτ+∫t0‖uτ‖2dτ∫t0‖u‖2dτ)=4(‖u‖2+∫t0‖u‖2dτ)(‖ut‖2+∫t0‖uτ‖2dτ)≤4η(t)(‖ut‖2+∫t0‖uτ‖2dτ). | (65) |
Hence, it follows from (64) and (65) that
η′′(t)η(t)−p+24η′(t)2≥η(t)(η′′(t)−(p+2)(‖ut‖2+∫t0‖uτ‖2dτ))=η(t)(−p‖ut‖2−2‖Δu‖2+2∫Ω|u|plog|u|kdx−(p+2)∫t0‖uτ‖2dτ). | (66) |
Now, we define
ξ(t)=−p‖ut‖2−2‖Δu‖2+2∫Ω|u|plog|u|kdx−(p+2)∫t0‖uτ‖2dτ. |
Noticing
ξ(t)=(p−2)‖Δu‖2−2pE(t)−(p+2)∫t0‖uτ‖2dτ+2kp‖u‖pp=(p−2)‖Δu‖2−2pE(0)+(p−2)∫t0‖uτ‖2dτ+2kp‖u‖pp≥(p−2)‖Δu‖2−2pE(0). |
From (60) and Lemma 6.2, we deduce that
2pE(0)<p−2B0‖u0‖2<p−2B0‖u‖2<(p−2)‖Δu‖2. |
which yields that
η′′(t)−p+24η′(t)2≥ϱς,t∈[0,T0], |
which implies that
(η−p−24(t))′′≤−p−24ϱς(η(t))−p+64<0. |
Hence, it follows that there exists a
limt→T∗η−p−24(t)=0, |
that is
limt→T∗η(t)=+∞. |
In turn, this implies that
limt→T∗‖Δu(t)‖2=+∞. | (67) |
In fact, if
limt→T∗∫t0‖u(τ)‖2dτ=+∞ |
so that (67) is also satisfied. Hence (67) is a contraction with
The author would like to thank Professor Hua Chen for the careful reading of this paper and for the valuable suggestions to improve the presentation of the paper. This project is supported by NSFC (No. 11801145), Key Scientific Research Foundation of the Higher Education Institutions of Henan Province, China (Grant No.19A110004 and the Fund of Young Backbone Teacher in Henan Province (NO. 2018GGJS068, 21420048).
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