ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu
BLOW-UP CRITERION FOR THE 2D EULER-BOUSSINESQ SYSTEM IN TERMS OF TEMPERATURE
CHENYIN QIAN
Abstract. In this article, we study the blow-up slutions for the 2D Euler- Boussinesq equation. In particular, it is shown that if
Z T∗
0
sup
r≥2
kΛ1−αθ(t)kLr
√rlogr dt <∞ or Z T∗
0
kΛ1−αθkB˙0∞,∞dt <∞,
then the local solution can be continued to the global one. This is an improve- ment of classical Lipschitz-type blow-up criterion (k∇θkL1
tL∞) in terms of the temperatureθ.
1. Introduction
The 2D incompressible generalized Boussinesq equations with the fractional Laplacian dissipation is of the type
∂tv+ (v· ∇)v+νΛβv+∇π=θe2, (x, t)∈R2×(0,∞),
∂tθ+ (v· ∇)θ+κΛαθ= 0,
∇ ·v= 0,
v(x,0) =v0, θ(x,0) =θ0,
(1.1)
where v = v(x, t) = (v1(x, t), v2(x, t)), π = π(x, t) and θ = θ(x, t) stand for, respectively, the velocity vector field, the pressure and the temperature. Here, the constants ν ≥ 0 and κ ≥ 0 denote viscosity coefficient and thermal diffusivity coefficient respectively, and e2= (0,1). Λ =√
−∆ is the Zygmund operator, and Λα is defined by the Fourier transform,
Λdαf(ξ) =|ξ|αfb(ξ), fb(ξ) = Z
R2
e−ix·ξf(x) dx.
The study of the standard 2D incompressible Boussinesq system (withν >0, κ >0 and α= β = 2 in (1.1)) can be traced to 1980s (see [5]). Later, there are many works considering the global well-posedness problem for the standard 2D Boussinesq system without the viscosity (ν = 0, κ > 0) or without the thermal diffusivity (ν >0,κ= 0), see [1, 7, 14, 15, 19].
2010Mathematics Subject Classification. 35Q35, 35B35, 35B65, 76D05.
Key words and phrases. 2D Boussinesq equation; blow-up criterion; Besov space;
transport equation.
c
2016 Texas State University.
Submitted December 18, 20915. Published March 15, 2016.
1
Recently, the 2D incompressible Boussinesq system with fractional Laplacian generalizations have attracted considerable attention. For example, Hmidi and Zerguine obtained the global well-posedness of Euler-Boussinesq system (1.1) with 1< α≤2 in [17]. Hmidi, Keraani and Rousset obtained the global well-posedness for the Euler-Boussinesq system with critical dissipation (namely, α= 1) in [16].
There are many other related results to the Boussinesq equations system (1.1), we refer the reader to [2, 3, 6, 9, 11, 18, 12, 13, 21]. From above mentioned result, we see that it is difficult for us to get the global well-posedness for the system (1.1) with super-critical dissipation (α+β < 1). Therefore, people may turn to the blow-up criteria in terms of velocity (v) and temperature (θ). As for the blow-up criterion for Boussinesq equations in terms of θ, we refer readers to [8] (see also [7]), in which the authors Chae and Nam obtained the blow-up criterionk∇θkL1
tL∞
in the framework of L2. The purposes of this article is to establish some blow-up criteria better thank∇θkL1
tL∞.
In view of [17], one can establish the local well-posedness results for the system (1.1) with 0< α≤2, κ >0 and
v0∈Bsp,1(R2) with divv0= 0, θ0∈Bp,1s−α(R2), (1.2) wheres≥1 +2p withp∈]1,∞[. Here, we define the function space of solutions as follows
XTs,p:=C([0, T];Bsp,1(R2))× C([0, T];Bp,1s−α(R2))∩L1([0, T];Bp,1s (R2)
, (1.3) whereBs−αp,1 (R2) andBp,1s (R2) are Besov spaces (see Section 2). Now, we give the main results of this article.
Theorem 1.1. Let 1/2 < α≤1 and (v, θ)∈χs,pT∗ be the local unique solution of (1.1)with initial data satisfying (1.2), whereT∗ is the maximal existence time. If
Z T∗
0
sup
r≥2
kΛ1−αθ(t)kLr
√rlogr dt <∞, (1.4)
or
Z T∗
0
kΛ1−αθkB˙∞,∞0 dt <∞, (1.5) then the solution (v, θ)can be continued beyond T∗.
2. Notation and preliminaries
We begin this section with dyadic decomposition. Let S be the Schwartz class of rapidly decreasing functions. Let functionsχ, ϕ∈ S(Rd) supported inB={ξ∈ Rd:|ξ| ≤4/3} andC={ξ∈Rd : 3/4≤ |ξ| ≤8/3} respectively, such that
X
j∈Z
ϕ(2−jξ) = 1,∀ξ∈Rd\{0},
χ(ξ) +X
j≥0
ϕ(2−jξ) = 1,∀ξ∈Rd.
Foru∈ S0, we set
∆−1u=χ(D)u ∀q∈N, ∆qu=ϕ(2−qD)u ∀q∈Z,
∆˙qu=ϕ(2−qD)u.
The following low-frequency cut-off will be also used:
Squ= X
−1≤j≤q−1
∆ju and S˙qu= X
j≤q−1
∆˙ju
We now recall the definitions of Besov spaces through the dyadic decomposition.
Let s ∈ R, p, q ∈ [1,∞], the inhomogeneous Besov space Bp,qs (Rd) is the set of tempered distributionusuch that
kukBsp,q := 2jsk∆jukLp
`q <∞.
To define the homogeneous Besov spaces, we first denote by S0/P the space of tempered distributions modulo polynomials. Thus we define the space ˙Bp,qs (Rd) as the set of distributionu∈ S0/P such that
kukB˙sp,q := 2jsk∆˙jukLp
`q <∞.
We point out that ifs >0 then we haveBsp,q(Rd) = ˙Bsp,q(Rd)∩Lp(Rd) and kukBsp,q ≈ kukB˙sp,q+kukLp.
In our next study we require two kinds of coupled space-time Besov spaces. The first is defined in the following manner: forT >0 andq≥1, we denote byLrTB˙sp,q the set of all tempered distribution u satisfying
kukLr
TB˙sp,q:=k
2qsk∆˙qukLp
`q
kLrT <∞.
The second mixed space is LerTB˙p,qs which is the set of tempered distribution u satisfying
kukLerTB˙p,qs :=
2qsk∆˙qukLrTLp
`q
<∞.
3. Proof of main results In this section we use Φk to denote function of the form
Φk(t) =C0exp(. . .exp
| {z }
ktimes
(C0t). . .),
whereC0depends on the involved norms of the initial data and its value may vary from line to line up to some absolute constants. We will make an intensive use (without mentioning it) of the following trivial facts
Z t
0
Φk(τ)dτ ≤Φk(t) and expZ t 0
Φk(τ)dτ
≤Φk+1(t).
Firstly, we introduce a pseudo-differential operatorRαdefined byRα:= Λ−α∂1= Λ1−αR,0 < α < 1, where R := ∂Λ1 is the usual Riesz transform. For (1.1), the vorticity equation is
∂tω+v· ∇ω=∂1θ, (3.1)
and the acting ofRα on the temperature equation, we have
∂tRαθ+v· ∇Rαθ+κΛαRαθ=−[Rα, v· ∇]θ. (3.2) Denotez=ω+Rαθ. Thus we obtain
∂tz+v· ∇z=−[Rα, v· ∇]θ. (3.3) Firstly, we give the following crucial Lemmas which are useful for us to proof the main results.
Lemma 3.1([4]). Assume thatvis divergence-free and thatfsatisfies the transport equation onRd,
∂tf +v· ∇f =g,
f|t=0=f0. (3.4)
There exists a constantC, depending only ond, such that for all 1≤p, r≤ ∞and t∈R+, we have
kfk
Le∞t (Bp,r0 )≤C kf0kB0 p,r+kgk
Le1t(Bp,r0 )
1 +
Z t
0
k∇v(τ)kL∞dτ
. (3.5)
Lemma 3.2 ([17]). Let v be a solution of the incompressible Euler system onR2
∂tv+v· ∇v+∇π=f, v(x,0) =v0,
divv= 0.
(3.6) Then for s >−1,(p, r)∈(1,∞)×[1,∞] we have
kvkLe∞t (Bsp,r)≤CeCV(t)
kv0kBsp,r+ Z t
0
e−CV(τ)kf(τ)kBsp,rdτ
, (3.7) withV(t) :=Rt
0k∇v(τ)kL∞dτ.
Lemma 3.3. Let α∈(0,1),v be a smooth divergence-free vector field.
(i) for every(s, p, r)∈(−1, α)×[2,∞]×[1,∞], there exists a constant C >0 such that
k[Rα, v· ∇]θkBsp,r ≤Ck∇vkLp
kθkBs+1−α
∞,r +kθkLp
, (3.8)
(ii) for every (r, %)∈[1,∞]×(1,∞) and >0, there exists a constantC > 0 such that
k[Rα, v· ∇]θkB0∞,r ≤C(kωkL∞+kωkL%)
kθkB+1−α
∞,r +kθkL%
. (3.9) Proof. Part (i) can be found in [21, Proposition 3.3]. For the second part, we can imitate the proof of [16, Theorem 3.3(2)] to get (ii). We omit the detail here.
Lemma 3.4. Let α∈[0,2],κ >0 andv be a smooth divergence-free vector-field of R2. Letθ be a smooth solution of
∂tθ+v· ∇θ+κΛαθ=f,
θ(x,0) =θ0. (3.10)
Then (i) for every ρ∈[1,∞],p∈[1,∞],s >−1 one has (see [17]) kθk
Leρt(Bs+
α ρ
p,1 )≤CeCV(t)
kθ0kBp,1s (1 +t1ρ) + Z t
0
Ts(τ)dτ+kfkL1 t(Bsp,1)
(3.11) with
V(t) :=
Z t
0
k∇v(τ)kL∞dτ, Ts(t) :=k∇θ(t)kL∞kv(t)kBsp,11[1,∞)(s).
(ii) for everyp∈[1,∞], we have theLp estimates (see[10]) kθkLp≤ kθ0kLp+
Z t
0
kf(τ)kLpdτ. (3.12)
(iii) for every p∈[1,∞](see [16]) kθkLe∞t B0p,1 ≤C kθ0kB0
p,1+kfkL1
tB0p,1
1 +
Z t
0
k∇v(τ)kL∞dτ
. (3.13)
(iv) for p∈(1,∞), ρ∈ [1,∞] and f = 0, there exists a constant C such that (see [17])
sup
q∈N
2qαρk∆qθkLρ
tLp≤Ckθ0kLp+Ckθ0kL∞kωkL1
tLp. (3.14) Proof of Theorem 1.1. By using the above Lemmas and the equation (3.3), we first estimatek∇v(t)kL∞. We do this in several steps.
Step 1: Estimation ofkω(t)kLa, for some max{2, p}< a <∞. From (3.3), for any t < T∗, by Lemma 3.4 (ii) we have
kz(t)kLa ≤ kz(0)kLa+ Z t
0
k[Rα, v· ∇]θ(τ)kLadτ. (3.15) Note that (see [10, p 516])Rt
0ΛαRαθ|Rαθ|a−2Rαθdτ ≥0, by (3.2) we have kRαθ(t)kLa ≤ kRαθ0kLa+
Z t
0
k[Rα, v· ∇]θ(τ)kLadτ. (3.16) Since v0 ∈ Bp,1s , s ≥ 1 + 2/p, we see that ω0 ∈ Lp ∩L∞, and for θ0 ∈ Bs−αp,1 with 0 < α ≤ 1, we have θ0 ∈ Lp∩L∞ note that p < a < ∞, the embedding Bp,1s−α ,→Bs−α−
2 p+a2
a,1 ,→B1−α+a,1 a2 ,→B1−αa,1 , we have Rαθ0 ∈La. From (3.15) and (3.16) we obtain
kωkLa≤ kz(t)kLa+kRαθ(t)kLa
≤C kω0kLp∩L∞+kθ0kBs−α p,1
+C Z t
0
k[Rα, v· ∇]θ(τ)kLadτ. (3.17) Using the classical embeddingBa,10 ,→La, and Lemma 3.3 (i), we have
k[Rα, v· ∇]θ(t)kLa ≤Ck[Rα, v· ∇]θ(t)kB0 a,1
≤Ck∇v(t)kLa kθ(t)kB1−α
∞,1 +kθ(t)kLa
. (3.18)
Fromkθ(t)kLa≤ kθ0kLa,∀t≥0, it follows that kω(t)kLa ≤C
kω0kLp∩L∞+kθ0kBs−α
p,1
+C Z t
0
kω(τ)kLa
kθ(τ)kB1−α
∞,1 +kθ0kLa
dτ.
(3.19)
According to Gronwall’s inequality we obtain
kωkLa ≤CeCteCkθkL1tB1−α∞,1. (3.20) Next we estimate kθkL1
tB∞,11−α, letN ∈ N, by the Littlewood-Paley decomposition, by condition (1.4) we see that
kθkL1 tB1−α∞,1
≤ kSNθkL1
tB1−α∞,1 +X
q≥N
2q(1−α)k∆qθkL1 tL∞
≤Ctkθ0kL∞+C Z t
0
X
0≤q<N
k∆qΛ1−αθ(τ)kL∞dτ +X
q≥N
2q(1−α)k∆qθkL1 tL∞
≤Ctkθ0kL∞+C Z t
0
X
0≤q<N
2q2bk∆qΛ1−αθ(τ)kLbdτ+X
q≥N
2q(1−α)k∆qθkL1 tL∞
≤C X
0≤q<N
2q2bp blogb
Z t
0
sup
r≥2
kΛ1−αθ(τ)kLr
√rlogr dτ +Ctkθ0kL∞+CX
q≥N
2q(1−α+a2)k∆qθkL1 tLa
≤Ctkθ0kL∞+C2N2bp
blogb+CX
q≥N
2q(1−α+a2)k∆qθkL1 tLa.
Since α >1/2, we choose a large enough such that 1−2α+ 4/a <0, and using Lemma 3.4 (iv), we have
X
q≥N
2q(1−α+2a)k∆qθkL1
tLa
≤ X
q≥N
2q(1−2α+a2)
kθ0kLa+kθ0kL∞
Z t
0
kω(τ)kLadτ
≤Ckθ0kLa+ 2N(1−2α+a2)kθ0kL∞ Z t
0
kω(τ)kLadτ.
(3.21)
Therefore,
kθkL1
tB∞,11−α ≤C(t+ 1)kθ0kL∞∩La+C2N2bb12+ + 2N(1−2α+a2)kθ0kL∞
Z t
0
kω(τ)kLadτ,
for any 0< <1/2. Settingb=N and selectingN as follows N =hlog e+Rt
0kω(τ)kLadτ (2α−1−2/a) log 2
i + 2.
Then kθkL1
tB1−α∞,1 ≤C(t+ 1)kθ0kL∞∩La+Ch log
e+ Z t
0
kω(τ)kLadτi12+
. (3.22) Combining (3.20) with (3.22), we have
kθkL1
tB∞,11−α≤C(t+ 1)kθ0kL∞∩La+Ch log
e+ Z t
0
kω(τ)kLadτi12+
≤C(t+ 1)kθ0kL∞∩La+Ch log
e+tCeCteCkθkL1tB1−α∞,1i12+
≤C(t2+t+ 1) kθ0kL∞∩La+ 1
+Ckθk12+
L1tB1−α∞,1,
(3.23)
it follows that
kθkL1
tB∞,11−α≤C(1 +t+t2).
By (3.20), we obtain
kω(t)kLa≤Φ1(t), (3.24)
where Φk(t), k= 1,2,3, . . . is same the as as in subsection 3.3.
Step 2: Estimation ofkω(t)kL∞. By using the maximum principle for the transport equation (3.3) we have
kz(t)kL∞≤ kz(0)kL∞+ Z t
0
k[Rα, v· ∇]θ(τ)kL∞dτ, (3.25) By using Lemma 3.4 (ii) withp=∞, we see that (3.2) follows that
kRαθ(t)kL∞ ≤ kRαθ0kL∞+ Z t
0
k[Rα, v· ∇]θ(τ)kL∞dτ. (3.26) Forθ0∈Bp,1s−α with 0< α≤1, s≥1 + 2/p, we have Bp,1s−α,→Bs−α−
2 p
∞,1 ,→B∞,11−α, we haveRαθ0∈L∞. From (3.25) and (3.26) we obtain
kωkL∞ ≤ kz(t)kL∞+kRαθ(t)kL∞
≤ kω0kL∞+kθ0kBs−α
p,1 + 2 Z t
0
k[Rα, v· ∇]θ(τ)kL∞dτ. (3.27) Using the classical embeddingB∞,10 ,→L∞, and Lemma 3.3 (i), we have
k[Rα, v· ∇]θkL∞≤Ck[Rα, v· ∇]θkB0
∞,1
≤C(kωkL∞+kωkLa)
kθkB+1−α
∞,1 +kθkLa
. (3.28)
Combining (3.27) and (3.28) we have kω(t)kL∞
≤C+C Z t
0
(kω(τ)kL∞+kω(τ)kLa)
kθ(τ)kB+1−α
∞,1 +kθ(τ)kLa
dτ
≤C+kωkL∞t La
kθkL1
tB+1−α∞,1 +tkθ0kLa
+C Z t
0
kω(τ)kL∞
kθ(τ)kB+1−α
∞,1 +kθ0kLa
dτ.
(3.29)
We claim that
kθkL1
tB∞,1+1−α ≤Φ1(t), (3.30) for some suitable >0. In fact that, from step 1 and the Lemma 3.4 (iv) we have
kθkLe1tBαa,∞ ≤Φ1(t), and then for everyσ < α, we have
kθkL1
tBa,1σ ≤ kθk
Le1tBαa,∞ ≤Φ1(t).
We choose thatσ=α−1/a, and we keep in mind thatasatisfies 1−2α+ 4/a <0, therefore 2α−1−4/a >0, we choose satisfies 0 < <2α−1−3/a, then we have+ 1−α < α−3/a=σ−2/a(here we note that the selected parameterais make sureα >3/a, and we haveσ−2/a >0). Therefore, we have the embedding Ba,1α−1/a = Bσa,1 ,→ Bσ−2/a∞,1 ,→ B+1−α∞,1 , and we finally obtain (3.30). Thus, by (3.30) and step 1, one has
kω(t)kL∞ ≤Φ1(t) +C Z t
0
kω(τ)kL∞
kθ(τ)kB+1−α
∞,1 +kθ0kLa
dτ. (3.31)
Again using (3.30) and Gronwall’s inequality, we obtain kω(t)kL∞ ≤Φ2(t).
Step 3: Estimation ofk∇v(t)kL∞. By using Lemma 3.1 and Lemma 3.4(iii), from (3.2) and (3.3) we have
kz(t)kB0
∞,1+kRαθ(t)kB0
∞,1 ≤
C+k[Rα, v· ∇]θkL1 tB0∞,1
(1+k∇vkL1
tL∞), (3.32) Thanks to Lemma 3.3, and by step 1, step 2, we have
k[Rα, v· ∇]θkL1
tB0∞,1
≤C Z t
0
(kω(τ)kL∞+kω(τ)kLa)
kθ(τ)kB+1−α
∞,1 +kθ(τ)kLa
dτ ≤Φ2(t). (3.33) Therefore, we have
kω(t)kB0
∞,1 ≤ kz(t)kB0
∞,1+kRαθ(t)kB0
∞,1 ≤Φ2(t)
1 +k∇vkL1 tL∞
. On the other hand, we have
k∇v(t)kL∞≤ k∇∆−1v(t)kL∞+X
q∈N
k∆q∇v(t)kL∞
≤ kω(t)kLa+kω(t)kB0
∞,1
≤Φ1(t) +kω(t)kB0
∞,1.
(3.34)
Using (3.34), we have kω(t)kB0
∞,1≤Φ2(t) 1 +
Z t
0
kω(τ)kB0
∞,1dτ .
From Gronwall’s inequality we obtain kω(t)kB0
∞,1 ≤ Φ3(t). Going back to (3.34), we obtain
k∇vkL∞≤Φ3(t).
Next, we give the estimate ofkv(t)kBp,1s ,kθ(t)kBs−α
p,1 . Sinces≥1 + 2/p, we have Bp,1s−α,→B∞,11−α. By Lemma 3.4(i), we have
kθ(t)k
Le∞t (B∞,11−α)+ Z t
0
kθ(τ)kB1
∞,1dτ ≤CeCV(t)(1 +t)kθ0kB1−α
∞,1
≤CeCV(t)(1 +t)kθ0kBs−α p,1 .
(3.35) On the other hand, by Lemma 3.2 we have
kv(t)k
Le∞t (Bs−αp,1 )≤CeCV(t)
kv0kBsp,1+ Z t
0
kθ(τ)kBs−α
p,1 dτ .
Therefore, by the embeddingB∞,10 ,→L∞we have Z t
0
k∇θ(τ)kL∞kv(τ)kBs−α p,1 dτ
≤ sup
0≤τ≤t
kv(τ)kBs−α
p,1
Z t
0
kθ(τ)kB1
∞,1dτ
≤CeCV(t)(1 +t)kθ0kBs−α
p,1
kv0kBp,1s + Z t
0
kθ(τ)kBs−α
p,1 dτ .
(3.36)
Applying (3.36) and (3.11), the estimate ofθ reads as follows kθ(t)k
Le∞t (Bp,1s−α)+ Z t
0
kθ(τ)kBs
p,1dτ
≤CeCV(t)
kθ0kBs−α
p,1 (1 +t) + Z t
0
k∇θ(τ)kL∞kv(τ)kBs−α p,1 dτ
≤CeCV(t)(1 +t)kθ0kBs−α
p,1
1 +kv0kBsp,1+ Z t
0
kθ(τ)kBs−α
p,1 dτ .
(3.37)
Combining the estimates ofv (applying Lemma 3.2 again) and (3.37), we have kv(t)kBsp,1+kθ(t)kBs−α
p,1 + Z t
0
kθ(τ)kBp,1s dτ
≤CeCV(t)(1 +t)kθ0kBs−α
p,1
1 +kv0kBsp,1+ Z t
0
kθ(τ)kBs−α
p,1 dτ
≤Φ4(t) 1 +
Z t
0
kθ(τ)kBs−α p,1 dτ
.
(3.38)
Therefore, by Gronwall’s inequality, we finally obtain kv(t)kBp,1s +kθ(t)kBs−α
p,1 + Z t
0
kθ(τ)kBsp,1dτ ≤Φ5(t). (3.39) This completes the first part (when assumption (1.4) holds) of this theorem.
The proof of the second part (when assumption (1.5) holds) of this theorem is quite similar to the one in the first part. The main difference is the estimates of kωkLa in step 1. We begin with (3.18), and by the embedding Ba,20 ,→ La and Lemma 3.3 (i), we have
k[Rα, v· ∇]θ(t)kLa≤Ck[Rα, v· ∇]θ(t)kB0 a,2
≤Ck∇v(t)kLa
kθ(t)kB1−α
∞,2 +kθ(t)kLa
. (3.40)
It follows that
kω(t)kLa ≤C
kω0kLp∩L∞+kθ0kBs−α p,1
+C Z t
0
kω(τ)kLa
kθ(τ)kB1−α
∞,2 +kθ0kLa
dτ.
(3.41)
According to Gronwall’s inequality we obtain
kωkLa ≤CeCteCkθkL1tB1−α∞,2. (3.42) The estimate of kθkL1
tB1−α∞,2 is as follows, let N ∈N, by the Littlewood-Paley de- composition, by condition (1.5) we see that
kθkL1 tB1−α∞,2
≤ kSNθkL1
tB1−α∞,2 +k(Id− SN)θkL1 tB1−α∞,1
≤ kSNθkL1
tB1−α∞,2 +X
q≥N
2q(1−α)k∆qθkL1
tL∞
≤Ctkθ0kL∞+C Z t
0
X
0≤q<N
k∆qΛ1−αθ(τ)k2L∞
1/2
dτ +X
q≥N
2q(1−α)k∆qθkL1 tL∞
≤C√ N
Z t
0
kΛ1−αθ(τ)kB˙0∞,∞dτ+Ctkθ0kL∞+C X
q≥N
2q(1−α+2a)k∆qθkL1 tLa
≤C√
N+Ctkθ0kL∞+CX
q≥N
2q(1−α+2a)k∆qθkL1 tLa. Then, as for (3.21), we can choose suitableN such that
kθkL1
tB1−α∞,2 ≤C(t+ 1)kθ0kL∞∩La+Ch log
e+ Z t
0
kω(τ)kLadτi1/2
. (3.43) Combining (3.42) with (3.43), we have
kθkL1
tB∞,21−α ≤Φ1(t).
For the rest, we can follow the same process as above, and complete the proof.
Acknowledgements. I would like to thank the referee for a careful reading of the work and many valuable comments. The research is supported by the Natural Science Foundation of Zhejiang Province (LQ16A010001) and NSFC 11501517.
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Chenyin Qian
Department of Mathematics, Zhejiang Normal University, Jinhua 321004, China E-mail address:[email protected]