Electronic Journal of Differential Equations, Vol. 2016 (2016), No. 73, pp. 1–11.
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 EulerBoussinesq equation. In particular, it is shown that if
Z T∗
Z T∗
kΛ1−α θ(t)kLr
√
dt < ∞ or
kΛ1−α θkḂ 0
sup
dt < ∞,
∞,∞
r log r
r≥2
0
0
then the local solution can be continued to the global one. This is an improvement of classical Lipschitz-type blow-up criterion (k∇θkL1 L∞ ) in terms of the
t
temperature θ.
1. Introduction
The 2D incompressible generalized Boussinesq equations with the fractional
Laplacian dissipation is of the type
∂t v + (v · ∇)v + νΛβ v + ∇π = θe2 ,
(x, t) ∈ R2 × (0, ∞),
∂t θ + (v · ∇)θ + κΛα θ = 0,
∇ · v = 0,
v(x, 0) = v 0 ,
(1.1)
θ(x, 0) = θ0 ,
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,
Z
α f (ξ) = |ξ|α fb(ξ),
Λd
fb(ξ) =
e−ix·ξ f (x) dx.
R2
The study of the standard 2D incompressible Boussinesq system (with ν > 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].
2010 Mathematics 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.
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C. QIAN
EJDE-2016/73
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 criterion k∇θkL1t L∞
in the framework of L2 . The purposes of this article is to establish some blow-up
criteria better than k∇θkL1t L∞ .
In view of [17], one can establish the local well-posedness results for the system
(1.1) with 0 < α ≤ 2, κ > 0 and
s
v 0 ∈ Bp,1
(R2 )
where s ≥ 1 +
follows
2
p
with
div v 0 = 0,
s−α
θ0 ∈ Bp,1
(R2 ),
(1.2)
with p ∈]1, ∞[. Here, we define the function space of solutions as
s−α
s
s
XTs,p := C([0, T ]; Bp,1
(R2 )) × C([0, T ]; Bp,1
(R2 )) ∩ L1 ([0, T ]; Bp,1
(R2 ) ,
(1.3)
s−α
Bp,1
(R2 )
s
where
and Bp,1
(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,p
T ∗ be the local unique solution of
(1.1) with initial data satisfying (1.2), where T ∗ is the maximal existence time. If
Z T∗
kΛ1−α θ(t)kLr
√
dt < ∞,
(1.4)
sup
r log r
r≥2
0
or
Z ∗
T
0
kΛ1−α θkḂ 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 functions χ, ϕ ∈ S(Rd ) supported in B = {ξ ∈
Rd : |ξ| ≤ 4/3} and C = {ξ ∈ Rd : 3/4 ≤ |ξ| ≤ 8/3} respectively, such that
X
ϕ(2−j ξ) = 1, ∀ξ ∈ Rd \{0},
j∈Z
χ(ξ) +
X
ϕ(2−j ξ) = 1, ∀ξ ∈ Rd .
j≥0
0
For u ∈ S , we set
∆−1 u = χ(D)u ∀q ∈ N, ∆q u = ϕ(2−q D)u ∀q ∈ Z,
∆˙ q u = ϕ(2−q D)u.
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2D BOUSSINESQ EQUATIONS
3
The following low-frequency cut-off will be also used:
X
X
Sq u =
∆j u and Ṡq u =
∆˙ j u
−1≤j≤q−1
j≤q−1
We now recall the definitions of Besov spaces through the dyadic decomposition.
s
Let s ∈ R, p, q ∈ [1, ∞], the inhomogeneous Besov space Bp,q
(Rd ) is the set of
tempered distribution u such that
s
kukBp,q
:= 2js k∆j ukLp `q < ∞.
To define the homogeneous Besov spaces, we first denote by S 0 /P the space of
s
tempered distributions modulo polynomials. Thus we define the space Ḃp,q
(Rd ) as
0
the set of distribution u ∈ S /P such that
kuk s := 2js k∆˙ j ukLp q < ∞.
Ḃp,q
`
We point out that if s > 0 then we have
s
(Rd )
Bp,q
s
(Rd ) ∩ Lp (Rd ) and
= Ḃp,q
s
kukBp,q
≈ kukḂ s + kukLp .
p,q
In our next study we require two kinds of coupled space-time Besov spaces. The
s
first is defined in the following manner: for T > 0 and q ≥ 1, we denote by LrT Ḃp,q
the set of all tempered distribution u satisfying
kukLr Ḃ s := k 2qs k∆˙ q ukLp q kLrT < ∞.
p,q
T
`
s
e r Ḃp,q
which is the set of tempered distribution u
The second mixed space is L
T
satisfying
kukLer Ḃ s := 2qs k∆˙ q ukLrT Lp q < ∞.
T
p,q
`
3. Proof of main results
In this section we use Φk to denote function of the form
Φk (t) = C0 exp(. . . exp(C0 t) . . . ),
| {z }
k times
where C0 depends 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
Z t
Φk (τ )dτ ≤ Φk (t) and exp
Φk (τ )dτ ≤ Φk+1 (t).
0
0
Firstly, we introduce a pseudo-differential operator Rα defined by Rα := Λ−α ∂1 =
Λ1−α R, 0 < α < 1, where R := ∂Λ1 is the usual Riesz transform. For (1.1), the
vorticity equation is
∂t ω + v · ∇ω = ∂1 θ,
(3.1)
and the acting of Rα on the temperature equation, we have
∂t Rα θ + v · ∇Rα θ + κΛα Rα θ = −[Rα , v · ∇]θ.
(3.2)
Denote z = ω + Rα θ. Thus we obtain
∂t z + v · ∇z = −[Rα , v · ∇]θ.
(3.3)
Firstly, we give the following crucial Lemmas which are useful for us to proof the
main results.
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Lemma 3.1 ([4]). Assume that v is divergence-free and that f satisfies the transport
equation on Rd ,
∂t f + v · ∇f = g,
(3.4)
f |t=0 = f0 .
There exists a constant C, depending only on d, such that for all 1 ≤ p, r ≤ ∞ and
t ∈ R+ , we have
Z t
0
k∇v(τ )kL∞ dτ .
+ kgkLe1 (B 0 ) 1 +
(3.5)
kf kLe∞ (B 0 ) ≤ C kf0 kBp,r
t
p,r
t
p,r
0
Lemma 3.2 ([17]). Let v be a solution of the incompressible Euler system on R2
∂t v + v · ∇v + ∇π = f,
v(x, 0) = v 0 ,
(3.6)
div v = 0.
Then for s > −1, (p, r) ∈ (1, ∞) × [1, ∞] we have
Z t
s
s dτ ,
kvkLe∞ (B s ) ≤ CeCV (t) kv 0 kBp,r
+
e−CV (τ ) kf (τ )kBp,r
p,r
t
with V (t) :=
Rt
0
(3.7)
0
k∇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
s
s+1−α + kθkLp
k[Rα , v · ∇]θkBp,r
≤ Ck∇vkLp kθkB∞,r
,
(3.8)
(ii) for every (r, %) ∈ [1, ∞] × (1, ∞) and > 0, there exists a constant C > 0
such that
0
+1−α + kθkL%
k[Rα , v · ∇]θkB∞,r
≤ C(kωkL∞ + kωkL% ) kθkB∞,r
.
(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 and v be a smooth divergence-free vector-field of
R2 . Let θ be a smooth solution of
∂t θ + v · ∇θ + κΛα θ = f,
θ(x, 0) = θ0 .
Then (i) for every ρ ∈ [1, ∞], p ∈ [1, ∞], s > −1 one has (see [17])
Z t
1
s (1 + t ρ ) +
s )
Ts (τ )dτ + kf kL1t (Bp,1
kθk eρ s+ αρ ≤ CeCV (t) kθ0 kBp,1
Lt (Bp,1 )
(3.10)
(3.11)
0
with
Z
t
k∇v(τ )kL∞ dτ,
V (t) :=
0
s 1
Ts (t) := k∇θ(t)kL∞ kv(t)kBp,1
[1,∞) (s).
(ii) for every p ∈ [1, ∞], we have the Lp estimates (see [10])
Z t
0
kθkLp ≤ kθ kLp +
kf (τ )kLp dτ.
0
(3.12)
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2D BOUSSINESQ EQUATIONS
5
(iii) for every p ∈ [1, ∞] (see [16])
0
0
kθkLe∞ B 0 ≤ C kθ0 kBp,1
+ kf kL1t Bp,1
t
t
Z
1+
p,1
k∇v(τ )kL∞ dτ .
(3.13)
0
(iv) for p ∈ (1, ∞), ρ ∈ [1, ∞] and f = 0, there exists a constant C such that
(see [17])
α
(3.14)
sup 2q ρ k∆q θkLρt Lp ≤ Ckθ0 kLp + Ckθ0 kL∞ kωkL1t Lp .
q∈N
Proof of Theorem 1.1. By using the above Lemmas and the equation (3.3), we first
estimate k∇v(t)kL∞ . We do this in several steps.
Step 1: Estimation of kω(t)kLa , for some max{2, p} < a < ∞. From (3.3), for any
t < T ∗ , by Lemma 3.4 (ii) we have
Z t
kz(t)kLa ≤ kz(0)kLa +
k[Rα , v · ∇]θ(τ )kLa dτ.
(3.15)
0
Note that (see [10, p 516])
kRα θ(t)kLa
Rt
α
Λ Rα θ|Rα θ|a−2 Rα θdτ ≥ 0, by (3.2) we have
Z t
0
a
≤ kRα θ kL +
k[Rα , v · ∇]θ(τ )kLa dτ.
(3.16)
0
0
s−α
Since v ∈
s ≥ 1 + 2/p, we see that ω 0 ∈ Lp ∩ L∞ , and for θ0 ∈ Bp,1
0
p
∞
with 0 < α ≤ 1, we have θ ∈ L ∩ L note that p < a < ∞, the embedding
0
s
Bp,1
,
2
s−α− 2 + a
s−α
Bp,1
,→ Ba,1 p
(3.16) we obtain
2
1−α+ a
,→ Ba,1
1−α
,→ Ba,1
, we have Rα θ0 ∈ La . From (3.15) and
kωkLa ≤ kz(t)kLa + kRα θ(t)kLa
≤ C kω 0 kLp ∩L∞ + kθ0 kB s−α + C
p,1
Using the classical embedding
0
Ba,1
t
Z
k[Rα , v · ∇]θ(τ )kLa dτ.
(3.17)
0
a
,→ L , and Lemma 3.3 (i), we have
0
k[Rα , v · ∇]θ(t)kLa ≤ Ck[Rα , v · ∇]θ(t)kBa,1
≤ Ck∇v(t)kLa kθ(t)kB 1−α + kθ(t)kLa .
(3.18)
∞,1
From kθ(t)kLa ≤ kθ0 kLa , ∀t ≥ 0, it follows that
kω(t)kLa ≤ C kω 0 kLp ∩L∞ + kθ0 kB s−α
p,1
Z t
+C
kω(τ )kLa kθ(τ )kB 1−α + kθ0 kLa dτ.
(3.19)
∞,1
0
According to Gronwall’s inequality we obtain
kωkLa ≤ CeCt e
CkθkL1 B 1−α
t
∞,1
.
(3.20)
Next we estimate kθkL1 B 1−α , let N ∈ N, by the Littlewood-Paley decomposition,
t ∞,1
by condition (1.4) we see that
kθkL1 B 1−α
t
∞,1
≤ kSN θkL1 B 1−α +
t
X
∞,1
q≥N
2q(1−α) k∆q θkL1t L∞
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C. QIAN
≤ Ctkθ0 kL∞ + C
t
Z
X
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k∆q Λ1−α θ(τ )kL∞ dτ +
0 0≤q<N
≤ Ctkθ0 kL∞ + C
t
Z
X
q≥N
2
X
X
2q b k∆q Λ1−α θ(τ )kLb dτ +
0 0≤q<N
≤C
X
q 2b
2
p
t
kΛ1−α θ(τ )kLr
√
sup
dτ
r log r
r≥2
b log b
0
2
X
+ Ctkθ0 kL∞ + C
2q(1−α) k∆q θkL1t L∞
q≥N
Z
0≤q<N
2q(1−α) k∆q θkL1t L∞
2q(1−α+ a ) k∆q θkL1t La
q≥N
2
≤ Ctkθ0 kL∞ + C2N b
p
b log b + C
X
2
2q(1−α+ a ) k∆q θkL1t La .
q≥N
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
2
2q(1−α+ a ) k∆q θkL1t La
q≥N
≤
X
2
q(1−2α+ a
)
2
0
kθ k
La
0
+ kθ k
Z
t
kω(τ )kLa dτ
L∞
(3.21)
0
q≥N
2
≤ Ckθ0 kLa + 2N (1−2α+ a ) kθ0 kL∞
Z
t
kω(τ )kLa dτ.
0
Therefore,
2
1
kθkL1 B 1−α ≤ C(t + 1)kθ0 kL∞ ∩La + C2N b b 2 +
t ∞,1
Z t
2
+ 2N (1−2α+ a ) kθ0 kL∞
kω(τ )kLa dτ,
0
for any 0 < < 1/2. Setting b = N and selecting N as follows
h log e + R t kω(τ )k a dτ i
L
0
+ 2.
N=
(2α − 1 − 2/a) log 2
Then
Z t
h
i 21 +
kθkL1 B 1−α ≤ C(t + 1)kθ0 kL∞ ∩La + C log e +
kω(τ )kLa dτ
.
t
∞,1
(3.22)
0
Combining (3.20) with (3.22), we have
0
kθkL1 B 1−α ≤ C(t + 1)kθ k
t
∞,1
Z t
i 21 +
+ C log e +
kω(τ )kLa dτ
h
L∞ ∩La
0
h
Ckθk
1
1−α
Lt B∞,1
≤ C(t + 1)kθ0 kL∞ ∩La + C log e + tCeCt e
1
+
≤ C(t2 + t + 1) kθ0 kL∞ ∩La + 1 + CkθkL2 1 B 1−α ,
t
i 21 +
(3.23)
∞,1
it follows that
kθkL1 B 1−α ≤ C(1 + t + t2 ).
t
∞,1
By (3.20), we obtain
kω(t)kLa ≤ Φ1 (t),
(3.24)
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2D BOUSSINESQ EQUATIONS
7
where Φk (t), k = 1, 2, 3, . . . is same the as as in subsection 3.3.
Step 2: Estimation of kω(t)kL∞ . By using the maximum principle for the transport
equation (3.3) we have
Z t
kz(t)kL∞ ≤ kz(0)kL∞ +
k[Rα , v · ∇]θ(τ )kL∞ dτ,
(3.25)
0
By using Lemma 3.4 (ii) with p = ∞, we see that (3.2) follows that
Z t
0
∞
∞
kRα θ(t)kL ≤ kRα θ kL +
k[Rα , v · ∇]θ(τ )kL∞ dτ.
(3.26)
0
2
s−α− p
s−α
s−α
For θ0 ∈ Bp,1
with 0 < α ≤ 1, s ≥ 1 + 2/p, we have Bp,1
,→ B∞,1
0
∞
we have Rα θ ∈ L . From (3.25) and (3.26) we obtain
kωkL∞ ≤ kz(t)kL∞ + kRα θ(t)kL∞
Z t
0
0
≤ kω kL∞ + kθ kB s−α + 2
k[Rα , v · ∇]θ(τ )kL∞ dτ.
p,1
Using the classical embedding
0
B∞,1
1−α
,→ B∞,1
,
(3.27)
0
∞
,→ L , and Lemma 3.3 (i), we have
0
k[Rα , v · ∇]θkL∞ ≤ Ck[Rα , v · ∇]θkB∞,1
≤ C(kωkL∞ + kωkLa ) kθkB +1−α + kθkLa .
(3.28)
∞,1
Combining (3.27) and (3.28) we have
kω(t)kL∞
Z
t
(kω(τ )kL∞ + kω(τ )kLa ) kθ(τ )kB +1−α + kθ(τ )kLa dτ
∞,1
0
a
≤ C + kωkL∞
kθkL1 B +1−α + tkθ0 kLa
t L
t ∞,1
Z t
+C
kω(τ )kL∞ kθ(τ )kB +1−α + kθ0 kLa dτ.
≤C +C
(3.29)
∞,1
0
We claim that
kθkL1 B +1−α ≤ Φ1 (t),
∞,1
t
(3.30)
for some suitable > 0. In fact that, from step 1 and the Lemma 3.4 (iv) we have
kθkLe1 B α
t
a,∞
≤ Φ1 (t),
and then for every σ < α, we have
σ
≤ kθkLe1 B α
kθkL1t Ba,1
t
a,∞
≤ Φ1 (t).
We choose that σ = α − 1/a, and we keep in mind that a satisfies 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 parameter a is
make sure α > 3/a, and we have σ − 2/a > 0). Therefore, we have the embedding
σ−2/a
α−1/a
+1−α
σ
, and we finally obtain (3.30). Thus, by
Ba,1
= Ba,1
,→ B∞,1 ,→ B∞,1
(3.30) and step 1, one has
Z t
kω(t)kL∞ ≤ Φ1 (t) + C
kω(τ )kL∞ kθ(τ )kB +1−α + kθ0 kLa dτ.
(3.31)
0
∞,1
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C. QIAN
EJDE-2016/73
Again using (3.30) and Gronwall’s inequality, we obtain
kω(t)kL∞ ≤ Φ2 (t).
Step 3: Estimation of k∇v(t)kL∞ . By using Lemma 3.1 and Lemma 3.4(iii), from
(3.2) and (3.3) we have
0
0
0
(1+k∇vkL1t L∞ ), (3.32)
kz(t)kB∞,1
+kRα θ(t)kB∞,1
≤ C + k[Rα , v · ∇]θkL1t B∞,1
Thanks to Lemma 3.3, and by step 1, step 2, we have
0
k[Rα , v · ∇]θkL1t B∞,1
Z t
(3.33)
(kω(τ )kL∞ + kω(τ )kLa ) kθ(τ )kB +1−α + kθ(τ )kLa dτ ≤ Φ2 (t).
≤C
∞,1
0
Therefore, we have
0
0
0
kω(t)kB∞,1
≤ kz(t)kB∞,1
+ kRα θ(t)kB∞,1
≤ Φ2 (t) 1 + k∇vkL1t L∞ .
On the other hand, we have
k∇v(t)kL∞ ≤ k∇∆−1 v(t)kL∞ +
X
k∆q ∇v(t)kL∞
q∈N
(3.34)
0
≤ kω(t)kLa + kω(t)kB∞,1
0
≤ Φ1 (t) + kω(t)kB∞,1
.
Using (3.34), we have
Z t
0
0
kω(t)kB∞,1
≤ Φ2 (t) 1 +
kω(τ )kB∞,1
dτ .
0
0
From Gronwall’s inequality we obtain kω(t)kB∞,1
≤ Φ3 (t). Going back to (3.34),
we obtain
k∇vkL∞ ≤ Φ3 (t).
s , kθ(t)k s−α . Since s ≥ 1 + 2/p, we have
Next, we give the estimate of kv(t)kBp,1
B
p,1
s−α
1−α
Bp,1
,→ B∞,1
. By Lemma 3.4(i), we have
Z t
1
kθ(τ )kB∞,1
dτ ≤ CeCV (t) (1 + t)kθ0 kB 1−α
kθ(t)kLe∞ (B 1−α ) +
∞,1
t
∞,1
0
≤ Ce
CV (t)
(3.35)
0
(1 + t)kθ kB s−α .
p,1
On the other hand, by Lemma 3.2 we have
Z t
s
kv(t)kLe∞ (B s−α ) ≤ CeCV (t) kv 0 kBp,1
+
kθ(τ )kB s−α dτ .
t
p,1
p,1
0
∞
0
B∞,1
Therefore, by the embedding
,→ L we have
Z t
k∇θ(τ )kL∞ kv(τ )kB s−α dτ
p,1
0
Z t
1
≤ sup kv(τ )kB s−α
kθ(τ )kB∞,1
dτ
0≤τ ≤t
CV (t)
≤ Ce
p,1
(3.36)
0
0
0
Z
(1 + t)kθ kB s−α kv k
p,1
s
Bp,1
+
0
t
kθ(τ )kB s−α dτ .
p,1
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2D BOUSSINESQ EQUATIONS
9
Applying (3.36) and (3.11), the estimate of θ reads as follows
Z t
s dτ
kθ(τ )kBp,1
kθ(t)kLe∞ (B s−α ) +
t
p,1
0
Z t
≤ CeCV (t) kθ0 kB s−α (1 + t) +
k∇θ(τ )kL∞ kv(τ )kB s−α dτ
p,1
p,1
0
Z t
s
≤ CeCV (t) (1 + t)kθ0 kB s−α 1 + kv 0 kBp,1
+
kθ(τ )kB s−α dτ .
p,1
(3.37)
p,1
0
Combining the estimates of v (applying Lemma 3.2 again) and (3.37), we have
Z t
s
s dτ
kv(t)kBp,1 + kθ(t)kB s−α +
kθ(τ )kBp,1
p,1
0
Z t
CV (t)
0
0
s
(3.38)
≤ Ce
(1 + t)kθ kB s−α 1 + kv kBp,1
kθ(τ )kB s−α dτ
+
p,1
p,1
0
Z t
≤ Φ4 (t) 1 +
kθ(τ )kB s−α dτ .
p,1
0
Therefore, by Gronwall’s inequality, we finally obtain
Z t
s
s dτ ≤ Φ5 (t).
kv(t)kBp,1
+ kθ(t)kB s−α +
kθ(τ )kBp,1
p,1
(3.39)
0
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
0
kωkLa in step 1. We begin with (3.18), and by the embedding Ba,2
,→ La and
Lemma 3.3 (i), we have
0
k[Rα , v · ∇]θ(t)kLa ≤ Ck[Rα , v · ∇]θ(t)kBa,2
≤ Ck∇v(t)kLa kθ(t)kB 1−α + kθ(t)kLa .
(3.40)
∞,2
It follows that
kω(t)kLa ≤ C kω 0 kLp ∩L∞ + kθ0 kB s−α
p,1
Z t
+C
kω(τ )kLa kθ(τ )kB 1−α + kθ0 kLa dτ.
(3.41)
∞,2
0
According to Gronwall’s inequality we obtain
kωkLa ≤ CeCt e
CkθkL1 B 1−α
t
∞,2
.
(3.42)
The estimate of kθkL1 B 1−α is as follows, let N ∈ N, by the Littlewood-Paley det ∞,2
composition, by condition (1.5) we see that
kθkL1 B 1−α
t
∞,2
≤ kSN θkL1 B 1−α + k(Id − SN )θkL1 B 1−α
t ∞,2
t ∞,1
X
q(1−α)
≤ kSN θkL1 B 1−α +
2
k∆q θkL1t L∞
t
∞,2
q≥N
≤ Ctkθ0 kL∞ + C
Z t X
0
0≤q<N
k∆q Λ1−α θ(τ )k2L∞
1/2
dτ +
X
q≥N
2q(1−α) k∆q θkL1t L∞
10
C. QIAN
√ Z t 1−α
kΛ
θ(τ )kḂ 0
≤C N
0
√
∞,∞
≤ C N + Ctkθ0 kL∞ + C
EJDE-2016/73
dτ + Ctkθ0 kL∞ + C
X
2
2q(1−α+ a ) k∆q θkL1t La
q≥N
2
)
q(1−α+ a
X
2
k∆q θkL1t La .
q≥N
Then, as for (3.21), we can choose suitable N such that
Z t
i1/2
h
kω(τ )kLa dτ
.
kθkL1 B 1−α ≤ C(t + 1)kθ0 kL∞ ∩La + C log e +
t
∞,2
(3.43)
0
Combining (3.42) with (3.43), we have
kθkL1 B 1−α ≤ Φ1 (t).
t
∞,2
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: qcyjcsx@163.com