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Preprint submitted on 2 Aug 2018
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ON LOWER BOUNDS AT BLOW UP OF SCALE INVARIANT NORMS FOR THE NAVIER-STOKES
EQUATIONS
Jean-Yves Chemin, Isabelle Gallagher, Ping Zhang
To cite this version:
Jean-Yves Chemin, Isabelle Gallagher, Ping Zhang. ON LOWER BOUNDS AT BLOW UP OF
SCALE INVARIANT NORMS FOR THE NAVIER-STOKES EQUATIONS. 2018. �hal-01849318�
ON LOWER BOUNDS AT BLOW UP OF SCALE INVARIANT NORMS FOR THE NAVIER-STOKES EQUATIONS
JEAN-YVES CHEMIN, ISABELLE GALLAGHER, AND PING ZHANG
Abstract. In this work we investigate the problem of preventing the incompressible 3D Navier-Stokes from developing singularities with the control of one component of the velocity field only in
L∞norm in times with values in a scaling invariant space. We introduce a space
”almost” invariant under the action of the scaling such that if one component measured in this space remains small enough, then there is no blow up.
Keywords: Incompressible Navier-Stokes Equations, Blow-up criteria, Anisotropic Littlewood-Paley Theory
AMS Subject Classification (2000): 35Q30, 76D03 1. Introduction
The purpose of this paper is the investigation of the possible behaviour of a solution of the incompressible Navier-Stokes equation in R
3near the (possible) blow up time. Let us recall the form of the incompressible Navier-Stokes equation
(N S)
∂
tv + div(v ⊗ v) − ∆v + ∇p = 0 , div v = 0 and v|
t=0= v
0,
where v = (v
1, v
2, v
3) stands for the velocity of the fluid and p for the pressure.
It is well known that the system has two main properties related to its physical origin:
• the scaling invariance which the fact that if (t, x) is a solution on [0, T ] then for any pos- itive real number λ, the vector field u
λ(t, x) def = λu(λ
2t, λx) is a solution on [0, λ
−2T ];
• the dissipation of energy which writes
(1) 1
2 kv(t)k
2L2+ Z
t0
k∇v(t
0)k
2L2dt
0≤ 1
2 kv
0k
2L2.
The first type of results which describe the behaviour of a (regular) solution just before the blow up are those which are a consequence of existence theorem for initial data in spaces more regular than the scaling. The seminal text [?] of J. Leray already pointed out in 1934 that the life span T
?(u
0) of the regular solution associated with an initial data in the Sobolev space H
1( R
3) is greater than ck∇u
0k
−4L2; then applying this result with u(t) as an initial data gives immediatly that, if T
?(u
0) is finite, then
(2) k∇u(t)k
4L2≥ c
T
?(u
0) − t which implies that
Z
T?(u0) 0k∇u(t)k
4L2dt = ∞.
More generally, it is very classical result that for any γ in ]0, 1[ we have (3) T
?(u
0) ≥ c
γku
0k
−1 γ
H˙12+2γ
which leads to ku(t)k
˙H12+2γ
≥ c
γ(T
?(u
0) − t)
γ·
Let us notice that the formula is scaling invariant abd comes from the resolution of (N S) with a fixed point argument follwing the Kato method. Morever, E. Poulon proved in [] that
1
a regular initila data exists which blows up at finite time then an initial data u
0exists in the unit sphere of ˙ H
12+γsuch that
T
?(u
0) = inf
T
?(u
0) , u
0∈ H ˙
12+γ, ku
0k
˙H12+γ
= 1 .
Assertion (3) can be generalized to the norm associated with the greatest space which is translation invariant, continuously included the space of tempered distribution S
0( R
3) and the norm has the same scaling as ˙ H
12+2γ. As pointed out by Y. Meyer in Lemma 9 of [?], this space is the Besov space ˙ B
∞,∞−1+2γwhich can be defined as the space of distributions such that
(4) kuk
B˙−1+2γ∞,∞
def = sup
t>0
t
12−2γke
t∆uk
L∞.
is finite. The generalization of the bound given by (3) to this norms can be done (see for instance Theorem1.3 of [?]) we recall here
Theorem 1.1. For any γ in the interval ]0, 1/2[, a constant c
γexists such that for any regular initial data u
0, its life span T
?(u
0) satisfies
(5) T
?(u
0) ≥ c
γku
0k
−1 γ
B˙∞,∞−1+2γ
which leads to ku(t)k
B˙−1+2γ∞,∞
≥ c
γ(T
?(u
0) − t)
γ·
This result as an analog for a global regularity uder the smallness condition which is the Koch and Tataru theorem (see [?] ) which claims that an initial data which have a small norm in the space BM O
−1( R
3) generates a global unique solution (which turns out to be as regular as the initial data). The space BM O
−1(R
3) is a very slightly bigger space than ˙ B
∞,2−1(R
3) defined by
(6) kuk
2˙B∞,2−1
def =
Z
∞0
ke
t∆uk
L∞dt < ∞
and very slightly smaller than the space ˙ B
∞,∞−1. Let us notice that classical space ˙ H
12of L
3( R
3) are continuous embedded in BM O
−1( R
3).
Let us point out that the proof of all these results do not use the special structure of (N S) and in particular all the above results are true for any systems of the type
(GN S) ∂
tu − ∆u + X
i,j
A
i,j(D)(u
iu
j)
where A
i,j(D) are smooth homogenenous Fourier multipliers of order 1. The problem investi- gated here is to improve the description of the behavior of the solution near a possible blow up using the special structure of the non linear term of the Navier-Stokes equation.
One major achievement in this field is the work [?] by L. Escauriaza, G. Ser¨ egin and V.
Sverak which proves that
(7) T
?(u
0) < ∞ = ⇒ lim sup
t→T?(u0)
ku(t)k
˙H12
= ∞.
A different context consists in formulating a condition which involves only one component of the velocity field. The first result in that direction is obtained in a pioneer work by J.
Neustupa and P. Penel (see [?]) but the norm involved was not scaling invariant. A lot of works (see [?, 4, ?, ?, ?, ?, ?, ?, ?]) established conditions of the type
Z
T?0
kv
3(t, ·)k
pLqdt = ∞ or Z
T?0
k∂
jv
3(t, ·)k
pLqdt = ∞ with relations on p and q which do not make these quantities scaling invariant.
2
The first result in that direction using scaling invariant condition has been proved by the first and the third author in [6]. It claims that for any regular intial data with derivative in L
32(R
3), then for any unit vector σ of R
3, we have
(8) T
?< ∞ = ⇒
Z
T?0
kv(t) · σk
pH˙12+ 2p
dt = ∞ .
for any p in the interval ]4, 6[. It has been generalized by Z. Zhang and the first and the third author for any p greater than 4 in [8]. This is the analog of the integral condition of (2) for only a component.
The motivation of the issue raised in this paper is what happens for the above criteria for p = ∞? in other term, it is possible to extend L. Escauriaza, G. Ser¨ egin and V. Sverak criteria (7) only for on component. This question sems to ambitious for the time being. Indeed, following the work [?] by G. Koch, F. Planchon and the second author
1one way to understand L. Escauriaza, G. Ser¨ egin and V. Sverak in the following : assume that a solution exists such that the ˙ H
12norm remains bounded near the blow up time. The first step consists in proving that the solution tends weakly to 0 when t tends to the blow up time. The second step consists in proving a backforward uniqueness result which implies that the solution is 0 which of course contradicts the fact that its blows up at finite time.
The first step relies in particular on the fact that the Navier-Stokes system (N S) is globally wellposed for small data in ˙ H
12. In our context, the equivalent is that if kv
0· σk
˙H12
is small enough for some unit vector σ of R
3, then there is a global regular solution. Such a result, assuming it is true, seems out of reach for the time being.
The result we prove in this paper is that if there is a blow up, it is not possible that a component of the velocity field tends to 0 to fast. More precisely, we are going to prove the following theorem.
Theorem 1.2. A positive constant c
0exists so that for any initial data v
0in H
1( R
3) with associate solution v of (N S) blowing up at a finite time T
?, for any unit vector σ of R
3, there holds
∀t < T
?, sup
t0∈[t,T?[
kv(t
0) · σk
˙H12
≥ c
0log
−12e + kv(t)k
4L2T
?− t
·
The other result we prove here in that if we reinforce slightly the ˙ H
12norm remains (almost) scaling invariant.
Definition 1.1. Let E > 0 be given, let us define H ˙
1 2
log,E
the space of distributions a in the homogeneous space H ˙
12( R
3) such that
kak
2H˙
12 log,E
def
= Z
R3
|ξ| log
2E|ξ| + e
| b a(ξ)|
2dξ < ∞ .
Our theorem is the following.
Theorem 1.3. Let E > 0 be given. A positive constant c
0exists such that if v the solution of (N S) associated with an initial data v
0belongs to H
1( R
3), and the maximal time of existence T
?(v
0) is finite, then
∀σ ∈ S
2, lim sup
t→T?(v0)
kv(t) · σk
H˙
1 2 log,E
≥ c
0.
Take care of the constant E
Let us make some comments about the result and in particular about the k · klog, E norm.
It is possible to bound the life span by this norm is the following way.
1
See also [?] for a more elementary approach
3Theorem 1.4. Let E > 0 be given. Let v
0be an initial data in H
1 2
log,E
. Then the maximal time of existence T
?of the solution v to (N S) in the space C[0, T
?[; H
12) satisfies
(9) T
?≥ cE
2exp −Ckv
0k
H˙
1 2 log,E
def= T (E).
Morever, we have, if T
?is finite
kv(t)k
H˙
12 log,E
≥ c log E
2T
?− t
·
Proof. It is wellknown that a criteria having regular solution of (N S) up a time T is that (10) I (T ) def =
Z
T0
Z
R3
e
−t|ξ|2|ξ|
3| b v
0(ξ)|
2dξdt = ε with ε << 1.
For some parameter λ which will be choosen later on, let write that
I (T ) ≤ 1
log
2 √λT
+ e
Z
T0
Z
√ T|ξ|≥λ
e
−t|ξ|2|ξ|
3log
2(|ξ|E + e)| b v
0(ξ)|
2dξdt
+ Z
T0
Z
√T|ξ|≤λ
e
−t|ξ|2|ξ|
3| v b
0(ξ)|
2dξdt
≤
kv
0k
2H˙
1 2 log,E
log
2 √λT
+ e + λ
Z
R3
Z
∞0
e
−t|ξ|21
t
12|ξ| |ξ|
2dt
|ξ| | b v
0(ξ)|
2dξdt
≤
kv
0k
2H˙
1 2 log,E
log
2√λ
T
+ e + λkv
0k
2H˙12
.
Choosing
λ = ε
2kv
0k
2H˙12
and T = ε
2 Ekv
0k
−2H˙12
exp −Ckv
0k
H˙
1 2 log,E
.
ensures Condition (13) once observed that as kv
0k
2H˙
1 2 log,E
is greater than kv
0k
2H˙12
(and thus large) it is not relevant in the above formula defining T . 2
Explain the reason of this shifting The structure of the paper is the following:
in the first section, we reduce the proof of Theorem 1.2 and Theorem 1.3 to the proof of two lemmas related of the estimation of expression of the type
Z
R3
∂
iv
j(x)∂
3v
j(x)∂
iv
j(x)dx
where i is in {1, 2}. These expressions show up when we do L
2energy estimates for ∇
hv. Here we face the difficulty that we cannot control ∂
3v in any sense using only k∇vk
L2and k∇∂
3vk
L2. Before going on, let us introduce some notation that will be used in all that follows. By a . b, we mean that there is a uniform constant C, which may be different on different lines, such that a ≤ Cb. We denote by (a|b)
L2the L
2( R
3) inner product of a and b. L
pT(L
qh(L
rv)) stands for the space L
p([0, T ]; L
q( R
xh; L
r( R
x3))) with x
h= (x
1, x
2), and ∇
h= (∂
x1, ∂
x2),
4
∆
h= ∂
x21+ ∂
x22. Finally, we always denote c
k,`k,`∈Z2
(resp. (c
j)
j∈Z) to be a generic element of `
2( Z
2) (resp.`
2( Z )) so that X
k,`∈Z2
c
2k,`= 1 (resp. X
j∈Z
c
2j= 1) .
2. The life span of (N S) computed with log -type norms The goal of this section is to present the proof of Theorem 1.4.
Proof of Theorem 1.4. We simply prove an a priori estimate for a regular solution v of (N S) on [0, T
?[ and skip the classical regularization process. Let us define
(11) T def = sup n
T < T
?/ ∀ t ≤ T , kv(t)k
H˙
1 2 log,E
≤ 2kv
0k
H˙
1 2 log,E
o ,
and also v
],Λdef = F
−11
{E|ξ|≥Λ}b v(ξ)
and v
[,Λdef = v − v
],Λfor some positive constant Λ to be chosen later on. Then we observe from Definition 1.1 that
(12)
kv
],Λk
2H˙12
≤ Z
{E|ξ|≥Λ}
1
log
2E|ξ| + e |ξ| log
2E|ξ| + e
| v(ξ)| b
2dξ
≤ 1 log
2Λ
Z
{E|ξ|≥Λ}
|ξ| log
2E|ξ| + e
| b v(ξ)|
2dξ
≤ 1 log
2Λ kvk
2H˙
1 2 log,E
.
Let us proceed to the energy type estimate for (N S) in the k · k
H˙
1 2 log,E
norm. This gives
(13) 1
2 d
dt kv(t)k
2H˙
1 2 log,E
+ k∇v(t)k
2H˙
1 2 log,E
= −(v · ∇v|v)
H˙
12 log,E
.
We claim that
(14)
(v · ∇v|v)
H˙
1 2 log,E
≤ C min n
kvk
˙H12
k∇vk
2H˙
12 log,E
, kvk
L∞kvk
H˙
1 2 log,E
k∇vk
H˙1 2 log,E
o .
Assuming (14) to be true, then using Bernstein’s inequality (see [1, Lemma 2.1], or Lemma A.1 for an anisotropic version used later in this paper), we infer from (11) and (12) that, for any t less than T ,
(v · ∇v|v)
H˙
1 2 log,E
≤ C kv
],Λk
˙H12
k∇vk
2H˙
12 log,E
+ kv
[,Λk
L∞kvk
H˙
1 2 log,E
k∇vk
H˙1 2 log,E
≤ Ckvk
H˙
12 log,E
(log Λ)
−1+ 1 4
k∇vk
2H˙
1 2 log,E
+ Ckv
[,Λk
2L∞kvk
2H˙
1 2 log,E
≤ Ckv
0k
H˙
1 2 log,E
(log Λ)
−1+ 1 4
k∇vk
2H˙
12 log,E
+ C
0Λ
3E
3kvk
2L2kvk
2H˙
12 log,E
.
Now let us choose
Λ = exp C
1kv
0k
H˙
1 2 log,E
for some large enough constant C
1. We deduce from the conservation of energy and from (13) that for any t less than T ,
d
dt kv(t)k
2H˙
12 log,E
+ k∇v(t)k
2H˙
12 log,E
≤ C
0E
3exp 3C
1kv
0k
H˙
1 2 log,E
kv
0k
2L2kv(t)k
2H˙
12 log,E
.
5
Then Gronwall’s Lemma implies that for any t less than T , kv(t)k
2H˙
12 log,E
≤ kv
0k
2H˙
12 log,E
exp
C
0tkv
0k
2L2E
3exp 3C
1kv
0k
H˙
1 2 log,E
.
This ensures (9).
Now let us prove (??). In view of (9), we have d
dE T (E) = T(E) 3
E − C d dE kv
0k
H˙
1 2 log,E
.
And we observe from Definition 1.1 that d
dE kv
0k
H˙
12 log,E
= kv
0k
−1H˙
1 2 log,E
Z
R3
|ξ|
2log(E|ξ| + e)
E|ξ| + e | b v
0(ξ)|
2dξ
≤ 1
E kv
0k
−1H˙
1 2 log,E
Z
R3
|ξ| log(E|ξ| + e)| b v
0(ξ)|
2dξ
≤ kv
0k
˙H12
E ,
since kv(t)k
˙H12
≤ kv(t)k
H˙
1 2 log,E
, so this gives rise to d
dt T (E) ≥ 3 − Ckv
0k
˙H12
T(E) E · Then under the assumption that kv
0k
˙H12
≤
C2we obtain d
dt T(E) ≥ T (E) E , so that T (E ) → ∞ as E → ∞.
To conclude the proof of the theorem we use again the fact that kv(t)k
˙H12
≤ kv(t)k
H˙
1 2 log,E
, so we deduce from (13) and (14) that
d
dt kv(t)k
2H˙
1 2 log,E
+ k∇v(t)k
2H˙
1 2 log,E
≤ Ckv(t)k
H˙
12 log,E
k∇v(t)k
2H˙
1 2 log,E
,
which implies that for t ≤ T d
dt kv(t)k
2H˙
1 2 log,E
+ 1 − 2Ckv
0k
H˙
12 log,E
k∇v(t)k
2H˙
1 2 log,E
≤ 0 . Thus in particular if kv
0k
H˙
1 2 log,E
≤ c
0≤
4C1, we achieve kv(t)k
H˙
1 2 log,E
≤ kv
0k
H˙
1 2 log,E
for t ≤ T .
This in turn shows that T = T
?= ∞. The theorem is proved, up to the proof of (14).
Proof of (14). We observe from Definition 1.1 that
(15) a ∈ H ˙
1 2
log,E
= ⇒ k∆
jak
L2. c
j2
−j2log
−1E2
j+ e kak
H˙1 2 log,E
.
Applying Bony’s decomposition (A.2) to v · ∇v gives v · ∇v =
3
X
k=1
T
vk∂
kv + T
∂kvv
k+ R(v
k, ∂
kv)
.
6
Considering the support of the Fourier transform of the terms in T
vk∂
kv and due to the fact that kS
jvk
L∞. c
j2
jkvk
˙H12
, we have k∆
jT
vk∂
kvk
L2. X
|j0−j|≤4
kS
j0−1vk
L∞k∆
j0∇vk
L2. X
|j0−j|≤4
c
j02
j0
2
log
−1E2
j0+ e kvk
˙H12
k∇vk
H˙
1 2 log,E
. c
j2
2jlog
−1E2
j+ e kvk
˙H12
k∇vk
H˙
1 2 log,E
.
Similarly noting that kS
jvk
L∞. kvk
L∞, we find k∆
jT
∂kvv
kk
L2. X
|j0−j|≤4
c
j02
j0
2
log
−1E2
j0+ e
kvk
L∞kvk
H˙
1 2 log,E
. c
j2
j2log
−1E2
j+ e
kvk
L∞kvk
H˙
1 2 log,E
.
Along the same lines, we have
k∆
jT
∂kvv
kk
L2. X
|j0−j|≤4
kS
j0−1∇vk
L∞k∆
j0vk
L2.
Then due to kS
j∇vk
L∞. c
j2
2jkvk
˙H12
and kS
j∇vk
L∞. c
j2
jkvk
L∞, by applying Bernstein’s inequality, we obtain
k∆
jT
∂kvv
kk
L2. c
j2
j2log
−1E2
j+ e min
kvk
˙H12
k∇vk
H˙
12 log,E
, kvk
L∞kvk
H˙
12 log,E
.
Finally, we get, by applying Bernstein’s inequality again, that k∆
jR(v
k, ∂
kv)k
L2. 2
3j2X
j0≥j−3
k∆
j0vk
L2k ∆ e
j0∇vk
L2. 2
3j2X
j0≥j−3
c
j02
−j0log
−1E2
j0+ e kvk
˙H12
k∇vk
H˙
1 2 log,E
. 2
3j2log
−1E2
j+ e X
j0≥j−N0
c
j02
−j0kvk
˙H12
k∇vk
H˙
1 2 log,E
. c
j2
j2log
−1E2
j+ e kvk
˙H12
k∇vk
H˙
1 2 log,E
.
On the other hand due to div v = 0, one has P
k
R(v
k, ∂
kv) = div R(v, v) so k∆
jX
k
R(v
k, ∂
kv)k
L2. 2
jX
j0≥j−N0
k∆
j0vk
L∞k ∆ e
j0vk
L2. 2
jX
j0≥j−N0
c
j02
−j0
2
log
−1E2
j0+ e
kvk
L∞kvk
H˙
12 log,E
. c
j2
2jlog
−1E2
j+ e
kvk
L∞kvk
H˙
1 2 log,E
.
As a result, it turns out that
(16) k∆
j(v · ∇v)k
L2. c
j2
j2log
−1E2
j+ e min
kvk
˙H12
k∇vk
H˙
12 log,E
, kvk
L∞kvk
H˙
12 log,E
.
7
Let us now return to the proof of (14). Observing that (v · ∇v|v)
H˙
12 log,E
∼ X
j∈Z
2
jlog
2E2
j+ e
∆
j(v · ∇v)|∆
jv
L2
,
from which, using also (16), we deduce that (v · ∇v|v)
H˙
1 2 log,E
≤ X
j∈Z
2
jlog
2E2
j+ e
k∆
j(v · ∇v)k
L2k∆
jvk
L2. X
j∈Z
c
2jmin kvk
˙H12
k∇vk
H˙
1 2 log,E
, kvk
L∞kvk
H˙
1 2 log,E
k∇vk
H˙
1 2 log,E
. min kvk
˙H12
k∇vk
H˙
1 2 log,E
, kvk
L∞kvk
H˙
1 2 log,E
k∇vk
H˙
1 2 log,E
.
This completes the proof of (14), hence of the theorem. 2
3. Anisotropic lower bounds at blow up
The goal of this section is to present the proof of Theorems 1.2 and 1.3. We choose σ = (0, 0, 1) in what follows. Let us define
(17) I(v
h, v
3) def =
2
X
i,m=1
Z
R3
∂
iv
3∂
3v
m(x)∂
iv
m(x)dx .
Let us state the two main lemmas leading to the theorem.
2Lemma 3.1. For any positive constant E
0, we have that
(18) I(v
h, v
3) . kv
3k
˙H12
log
12k∇v
3k
2L2E
0kv
3k
4H˙12
+ e
k∇
hvk
2˙H1
+ k∂
3v
hk
2L2E
0k∇v
3k
2L2.
Lemma 3.2. Let E
0∼ kv
0k
2L2, we have that
(19) I(v
h, v
3) . kv
3k
H˙
1 2 log
k∇
hv
hk
2˙H1
+ kv
3k
˙H12
k∂
3v
hk
2L2E
02·
We admit these lemmas for the time being.
Proof of Theorem 1.2. As in [4], we perform L
2energy estimate in the momentum equation of (N S) with −∆
hv. This can be interpreted as a ˙ H
1energy estimate for the horizontal
2
Here
E0, a scaling parameter which will be chosen later, is homogeneous to a kinetic energy.
8
variables. Indeed we have
(20)
1 2
d
dt k∇
hvk
2L2+ k∇
hvk
2˙H1
=
3
X
j=1
E
j(v) with
E
1(v) def = −
2
X
i=1
∂
iv
h· ∇
hv
h∂
iv
hL2
,
E
2(v) def = −
2
X
i=1
∂
iv
h· ∇
hv
3∂
iv
3L2
,
E
3(v) def = −
2
X
i=1
(∂
iv
3∂
3v
h∂
iv
h)
L2and E
4(v) def = −
2
X
i=1
(∂
iv
3∂
3v
3∂
iv
3)
L2. Let div
hv
hdef = ∂
1v
1+ ∂
2v
2. A direct computation shows that
E
1(v) = − Z
R3
div
hv
hX
2i,j=1
(∂
iv
j)
2+ ∂
1v
2∂
2v
1− ∂
1v
1∂
2v
2dx ,
which together with div v = 0 ensure that E
1(v) =
Z
R3
∂
3v
3X
2i,j=1
(∂
iv
j)
2+ ∂
1v
2∂
2v
1− ∂
1v
1∂
2v
2dx .
Then it follows from the laws of product in Besov spaces (see [1]) that (21)
|E
1(v)| . k∂
3v
3k
B˙−
1 2,∞2
k(∇
hv
h)
2k
B˙
1 2 2,1
. kv
3k
˙H12
k∇
hv
hk
2˙H1
. Similarly, we have
(22)
|E
2(v)| . k∇
hv
3k
B˙−
1 2,∞2
k∇
hv
h· ∇
hv
3k
B˙
1 2 2,1
. kv
3k
˙H12
k∇
hv
hk
H˙1k∇
hv
3k
H˙1, and
(23)
|E
4(v)| . k∂
3v
3k
B˙−
1 2,∞2
k(∇
hv
3)
2k
B˙
1 2 2,1
. kv
3k
H˙12
k∇
hv
3k
2˙H1
.
It remains to handle the estimate of E
3(v). It follows from Lemma 3.1 that (24) E
3(v) . kv
3k
˙H12
log
12k∇
hvk
2L2E
0kv
3k
4H˙12
+ e
k∇
hvk
2˙H1
+ k∂
3v
hk
2L2E
0k∇v
3k
2L2.
Then by inserting (21-24) into (20) gives rise to
(25) 1 2
d
dt k∇
hvk
2L2+ k∇
hvk
2˙H1
. kv
3k
˙H12
log
12k∇
hvk
2L2E
0kv
3k
4H˙12
+ e
k∇
hvk
2˙H1
+ k∂
3v
hk
2L2E
0k∇v
3k
2L2.
9
Now, a positive real number m being given, let us consider T < T e
?and T
?such that (26) sup
t∈[0,Te[
kv
3(t)k
˙H12
≤ m and T
?def = sup
t ∈ [0, T e [ / k∇
hvk
2L∞([0,t];L2)≤ 2k∇
hv
0k
2L2. Note that div v = 0, we have
(27)
k∇v
3(t)k
2L2=k∇
hv
3(t)k
2L2+ k∂
3v
3(t)k
2L2=k∇
hv
3(t)k
2L2+ k div
hv
h(t)k
2L2≤ 2k∇
hv
0k
2L2∀t ≤ T
?. Then for any t less than T
?, Inequality (25) writes
(28)
d
dt k∇
hvk
2L2+ 2k∇
hvk
2˙H1
≤ C
0m
log
12k∇
hv
0k
2L2E
0m
4+ e
k∇
hvk
2˙H1
+ k∂
3v
hk
2L2E
0k∇
hv
0k
2L2.
So that by time integration of (28) from [0, t] and using the L
2energy estimate on v, we infer that, for any t less than T
?,
k∇
hv(t)k
2L2+ 2 Z
t0
k∇
hv(t
0)k
2˙H1
dt
0≤ k∇
hv
0k
2L2+ C
0m log
12k∇
hv
0k
2L2E
0m
4+ e Z
t 0k∇
hv(t
0)k
2˙H1
dt
0+ C
0m k∇
hv
0k
2L2E
0Z
t0
k∂
3v
h(t
0)k
2L2dt
0≤ k∇
hv
0k
2L2+ C
0m log
12k∇
hv
0k
2L2E
0m
4+ e Z
t 0k∇
hv(t
0)k
2˙H1
dt
0+ C
0m kv
0k
2L2k∇
hv
0k
2L2E
0·
Choosing E
0= kv
0k
2L22C
0m in the above inequality implies that k∇
hv(t)k
2L2+
Z
t0
k∇
hv(t
0)k
2H˙1dt
0≤ 3
2 k∇
hv
0k
2L2+ C
1m log
12k∇
hv
0k
2L2kv
0k
2L2m
5+ e
Z
t0
k∇
hv(t
0)k
2˙H1
dt
0. Let us assume that
C
1m log
12k∇
hv
0k
2L2kv
0k
2L2m
5+ e
≤ 1 2 · This implies that m is smaller than 1/2C
1and thus that
(29) m log
12k∇
hv
0k
2L2kv
0k
2L2c + e
≤ 1 2C
2· Then we infer that, for any t less than T
?,
k∇
hv(t)k
2L2≤ 3
2 k∇
hv
0k
2L2.
This implies that T
?= T e and thus that T e is less than the blow-up time T
?. Moreover, thanks to (27) and Theorem 1.4 of [6], the solution v(t) is smooth for t ≤ T . e By contraposition
10
and (29), this implies that
(30) sup
t∈[0,T?[
kv
3(t)k
˙H12
≥ c
1log
−12k∇
hv
0k
2L2kv
0k
2L2c
0+ e
.
Let us translate this inequality in time. Let us define m(t) def = sup
t0∈[t,T?[
kv
3(t
0)k
˙H12
. Inequality (30) writes
k∇
hv(t)k
2L2kv(t)k
2L2≥ c
2exp c
3m
2(t)
·
Because of the energy estimate, we get, by integrating the above inequality over [t, T
∗[, that 1
2 kv(t)k
4L2≥ Z
T?t
k∇
hv(t
0)k
2L2dt
0kvk
2L∞ t (L2)≥ Z
T?t
k∇
hv(t
0)k
2L2kv(t
0)k
2L2dt
0≥ c
2Z
T?t
exp c
3m
2(t
0)
dt
0.
The function t 7−→ exp c
3m
2(t
0)
is a non decreasing function. Thus 1
2 kv(t)k
4L2≥ c
2exp c
3m
2(t)
(T
?− t).
This writes
m(t) ≥ c log
−12e + kv(t)k
4L2T
?− t
and the theorem is proved provided that we prove Lemma 3.1. 2 Proof of Theorem 1.3. The proof follows exactly the same lines as the previous one, replacing the estimate of E
3(v) by Lemma 3.2 instead of Lemma 3.1. Estimate (25) becomes
(31) 1
2 d
dt k∇
hv(t)k
2L2+ k∇
hvk
2˙H1
≤ C
kv
3k
H˙
1 2 log
k∇
hvk
2˙H1
+ kv
3k
˙H12
k∂
3v
hk
2L2E
02so by time integration and thanks to the energy estimate we find that as long as t ≤ T
∗def
= n
T ∈]0, T
?[ / sup
t∈[0,T]
kv
3(t)k
H˙
1 2 log
≤ 1 2C
o ,
there holds
k∇
hv(t)k
2L2+ Z
t0
k∇
hv(t
0)k
2˙H1
dt
0≤ k∇
hv
0k
2L2+ 1 E
02Z
t0
k∂
3v
h(t
0)k
2L2dt
0≤ k∇
hv
0k
2L2+ kv
0k
2L2E
02, which together with (27) ensures that
sup
t∈[0,T∗]
k∇v
3(t)k
2L2≤ k∇
hv
0k
2L2+ kv
0k
2L2E
02·
Then Theorem 1.4 of [6] implies that the solution v(t) is smooth for t ≤ T
∗. Theorem 1.3
follows by contraposition. 2
11
Let us now present the proof of Lemma 3.1 and Lemma 3.2.
Proof of Lemma 3.1 . We first get, by applying Bony’s decomposition (A.2) to ∂
iv
3∂
iv
mwith i, m belnging to {1, 2} and then using Leibniz formula, that
(32)
∂
iv
3∂
iv
m= T
∂ivm∂
iv
3+ T
∂iv3∂
iv
m+ R(∂
iv
3, ∂
iv
m)
= ∂
iT
∂ivmv
3+ A(v
3, v
m) with A(v
3, v
m) = −T
∂2ivm
v
3+ T
∂iv3∂
iv
m+ R(∂
iv
3, ∂
iv
m) . Applying Lemma A.1 gives
kS
j(∂
iv
m)k
L∞. c
j2
j(
32−s2) k∇
hvk
H˙s2∀ s
2< 3 2 , and there holds
k∆
jT
∂ivmv
3k
L2. X
|j0−j|≤4
kS
j0−1(∂
iv
m)k
L∞k∆
j0v
3k
L2. X
|j0−j|≤4
c
j02
j(
32−s1−s2) k∇
hvk
H˙s2kv
3k
H˙s1. c
j2
j(
32−s1−s2) k∇
hvk
H˙s2kv
3k
H˙s1, that is
(33) kT
∂ivmv
3k
L2. k∇
hvk
H˙s2kv
3k
H˙s1with s
1+ s
2= 3
2 and s
2< 3 2 · Then by using integrating by parts, we obtain
(34)
Z
R3
∂
iT
∂ivmv
3∂
3v
mdx =
Z
R3
T
∂ivmv
3∂
i∂
3v
mdx
≤ kT
∂ivmv
3k
L2k∂
i∂
3v
mk
L2. kv
3k
˙H12
k∇
hvk
2˙H1
. Next we claim that
(35) kA(v
3, v
m)k
B˙s1+s2−11,2
. kv
3k
H˙s1k∂
iv
mk
H˙s2with s
1+ s
2> 1 and s
1, s
2∈]0, 1] . Indeed due to s
1+ s
2> 1, it is easy to observe from Lemma A.1 that
k∆
jR(∂
iv
3, ∂
iv
m)k
L1. X
j0≥j−3
k∆
j0∂
iv
3k
L2k ∆ e
j0∂
iv
mk
L2. X
j0≥j−3
c
j02
−j0(s1+s2−1)kv
3k
H˙s1k∂
iv
mk
H˙s2. c
j2
−j(s1+s2−1)kv
3k
H˙s1k∂
iv
mk
H˙s2. Similarly since s
2≤ 1, one has
k∆
jT
∂2ivm
v
3k
L1. X
|j0−j|≤4
kS
j0−1∂
i2v
mk
L2k∆
j0v
3k
L2. c
j2
−j(s1+s2−1)kv
3k
H˙s1k∂
iv
mk
H˙s2.
12