Nonlinear Analysis 103 (2014) 87–97

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Nonlinear Analysis

journal homepage: www.elsevier.com/locate/na

On the exponential type explosion ofNavier–Stokes equationsJamel Benameur ∗

Department of Mathematics, College of Sciences, King Saud University, Riyadh 11451, Saudi Arabia

a r t i c l e i n f o

Article history:Received 2 May 2013Accepted 18 March 2014Communicated by Enzo Mitidieri

MSC:35-XX35Qxx35Q3035D35

Keywords:Incompressible fluidsNavier–Stokes equationsRegularity of generalized solutionsSobolev spacesBlow-up criterion

a b s t r a c t

The classical results on the explosion of the maximal solution of incompressibleNavier–Stokes equations are of type c(T ∗

− t)−σ0 for some σ0 > 0. Inspired by the worksBenameur and Selmi (2012) [15], Chemin (2004) [16], we use the Sobolev–Gevrey spaces toget better explosion results, precisely if ea|D|

1/σu0

∈ Hs(R3), then |ea|D|1/σ

u(t)|Hs is at leastof the order (T ∗

− t)−σ1 exp(c(T ∗− t)−σ2 ) for some σ1 > 0 and σ2 > 0. Fourier analysis

and standard techniques are used.© 2014 Elsevier Ltd. All rights reserved.

1. Introduction

The incompressible Navier–Stokes system in Cartesian coordinates is given by:∂tu − ν1u + (u · ∇)u = −∇p, in R+× R3,

div u = 0 in R+× R3,

u(0) = u0 in R3,

(NS)

where ν > 0 is the viscosity of the fluid, u = u(t, x) = (u1, u2, u3) and p = p(t, x) denote respectively, the unknownvelocity and the unknown pressure of the fluid at the point (t, x) ∈ R+

× R3, (u · ∇u) := u1∂1u + u2∂2u + u3∂3u, andu0

= (u01(x), u

02(x), u

03(x)) is a given initial velocity. If u0 is quite regular, the pressure p is determined. The system (NS)

has the scaling property: If u(t, x) is a solution of the initial data u0(x), then for any λ > 0, λu(λ2t, λx) is a solution of(NS) with the initial data λu0(λx). Our problem is the type of the blow-up criterion of the solution if the maximal timeT ∗ is finite. Precisely, the question posed by K. Ammari [1]: Is the type of explosion due to the chosen space or to thenon-linear part of the Navier–Stokes equations? In the literature, there are several authors who have studied the problemof explosion of a non-global solution of (NS) system, and all the obtained results do not exceed C(T ∗

− t)−σ for some σ ≥ 0,precisely: if u is a maximal solution of (NS) in C([0, T ∗), B) with T ∗ < ∞, then C(T ∗

− t)−σ≤ ∥u(t)∥B. The classical

result due to J. Leray [2,3]: if u is a non regular solution of the system (NS) and T ∗ is the first time where u is not regular,

∗ Tel.: +966 533087448; fax: +966 14676512.E-mail addresses: jamelbenamer@yahoo.fr, jbenameur@ksu.edu.sa.

http://dx.doi.org/10.1016/j.na.2014.03.0110362-546X/© 2014 Elsevier Ltd. All rights reserved.

88 J. Benameur / Nonlinear Analysis 103 (2014) 87–97

then cν3/4(T ∗− t)−1/4

≤ ∥u(t)∥H1 . Also, if u0∈ Hs(R3), with s > 5/2, T. Kato [4] showed that there is an existence

and uniqueness for a local solution of (NS) in C([0, T ];Hs(R3)) ∩ C1([0, T ];Hs−1(R3)). For more information, J. Beale,T. Kato, A. Majda [5], and T. Kato, G. Ponce [6] proved the following blow-up result: let u ∈ C([0, T ∗);Hs(R3)) be a maximalsolution of (NS) given in [4] with T ∗ is finite, then ∥∇ × u∥L1([0,T∗),L∞(R3)) = +∞. To prove this result, the authors usedthe following estimate ∂t∥u(t)∥2

Hs ≤ C∥∇u(t)∥L∞∥u(t)∥2Hs , which gives c(T ∗

− t)−1≤ ∥u(t)∥Hs . To improve this result,

some authors treat the dependence of the growth of ∥u(t)∥Hs with respect to the index of regularity. In [7], the authorproved that, if u ∈ C([0, T ∗);Hk(R3))with k > 5/2 an integer, is the maximal solution of Euler or Navier–Stokes equations,then C(T ∗

− t)−2k/5≤ ∥u(t)∥Hk . In [8], I showed that the quality of the explosion depends on the index of regularity.

Precisely, let u ∈ C([0, T ∗),Hs(R3)), (s > 5/2), be a maximal solution of (NS), given by [4]. Suppose that T ∗ < ∞, thenC(T ∗

− t)−s/3≤ ∥u(t)∥Hs , in which we can take s > 3/2. In [9], for the periodic case, I proved that, if the maximal solution

u ∈ C([0, T ∗);Hs(T3)), s > 5/2, of Navier–Stokes equations, then C(T ∗− t)−2s/5

≤ ∥u(t)∥Hs , and other results in Hs′ fors′ ∈ [1, s]. So, the idea is to find a functional space Bα,s(R3) which has the form

Bα,s(R3) =

f ∈ L2(R3);

R3

(1 + |ξ |2)sα(|ξ |)

2|f (ξ)|2dξ < ∞

,

with s > 3/2 and the function α is defined and increasing from [0, ∞) to (0, ∞). The choice of s > 3/2 is not optimal forthe existence and uniqueness, we can take s > 1/2. But this condition is necessary to estimate ∥u(t)∥L1 for finding a newblow up result. This space is equipped with the norm

∥f ∥Bα,s = ∥α(|D|)f ∥Hs

and the associated inner product

⟨f /g⟩Bα,s = ⟨α(|D|)f /α(|D|)g⟩Hs .

For the existence and uniqueness of the solution, a sufficient condition is the following

α(|ξ |) ≤ Cα(|ξ − η|)α(|η|), ∀ξ, η ∈ R3. (1)

We can suppose that C = 1, simply to change α by Cα. If we take β = ln(α), the condition (1) becomes

β(|ξ |) ≤ β(|ξ − η|) + β(|η|), ∀ξ, η ∈ R3.

A sufficient condition is

β(|ξ |) ≤ a|ξ |, ∀ξ ∈ R3

where a > 0.Therefore, we consider a function α which satisfies

α(|ξ |) ≤ ea|ξ |, ∀ξ ∈ R3.

Then, we choose the following family

α(y) = eayr, ∀y ∈ [0, ∞),

with a > 0 and r ∈ (0, 1]. To address the main issue, we will use the Sobolev–Gevrey space Hsa,σ (R3), a ≥ 0 and s ∈ R,

σ =1r ∈ [1, ∞), with the norm

∥f ∥Hsa,σ = ∥ea|D|

1/σf ∥Hs .

Such Sobolev–Gevrey spaces have been used in other contexts, e.g., the recent works [10–12].Now we are ready to state the main result.

Theorem 1.1. Let a, s, σ ∈ R such that a > 0, s > 3/2 and σ > 1. Let u0∈Hs

a,σ (R3)3

such that div u0= 0. Then, there

is a unique time T ∗∈ (0, ∞] and a unique solution u ∈ C([0, T ∗),Hs

a,σ (R3)) of Navier–Stokes equations (NS) such thatu ∈ C([0, T ∗

],Hsa,σ (R3)). Moreover, if T ∗ < ∞, then

C1(T ∗− t)−s/3 exp

aC2(T ∗

− t)−13σ

≤ ∥u(t)∥Hs

a,σ , ∀t ∈ [0, T ∗), (2)

where C1 = C1(s, u0, σ ) > 0 and C2 = C2(s, u0, σ ) > 0.

In this short paragraph, we explain our choice and its novelty. In the literature of the study of Navier–Stokes equations,there are different types of spaces that are used. The first type is the homogeneous functional space, that is to say∥λf (λ ·)∥E = λµ

∥f ∥E, ∀λ > 0; For example: Hs(R3), Lp(R3), and Cr(R3). There are also spaces that are almosthomogeneous (homogeneous for only λ = 2k, k ∈ Z) and satisfy, for all λ > 0, ∥λf (λ ·)∥E ≃ λµ

∥f ∥E , that is to sayc1λµ

∥f ∥E ≤ ∥λf (λ ·)∥E ≤ c2λµ∥f ∥E . The homogeneous Besov spaces give a typical example for this kind. In the same frame,

J. Benameur / Nonlinear Analysis 103 (2014) 87–97 89

there are homogeneous quantities used; for example: ∥u0∥1/2L2 ∥∇u0

∥1/2L2 , ∥u0

∥1/2L2 ∥u0

∥L∞ . A second type of space used is anon-homogeneous space, but only which is finite intersections of homogeneous spaces B = ∩

Nk=1 Bk. If f ∈ B, then its norm

is ∥f ∥B =N

k=1 ∥f ∥Bk ; and for all λ > 0, ∥λf (λ ·)∥B ≃N

k=1 λµk∥f ∥Bk ; for example, Hs= Hs

∩ L2 and Cr= L∞

∩ Cr . In allthese cases, we have for all λ > 0

f ∈ B ⇐⇒ λf (λ ·) ∈ B. (3)Then, for any result on the solution u(t, x) of (NS) on the interval [0, T ∗), we have the same result for the solutionλu(λ2t, λx)on the interval [0, T∗

λ2). Unlike the above types, there is another kind of space where property (3) becomes not satisfied. Our

space Bα,s(R3) is of this type and we have an alternative property for (3). For all λ > 0, we can write

∥λf (λ ·)∥2Bα,s

= λ−1

η

(1 + λ2|η|

2)sα(λ|η|)

2|f (η)|2dη

then α(z) = α( zλ) and

f ∈ Bα,s(R3) ⇐⇒ λf (λ ·) ∈ Bα( ·

λ ),s(R3),

which replaces (3). Precisely, if α(z) = ez and f (x) = F −1(e−32 |ξ |), we have f ∈ Bα,s(R3) and λf (λ ·) ∈ Bα,s(R3) for λ ≥ 3/2.

That is:

∥λf (λ ·)∥2Bα,s

= λ−1

η

(1 + λ2|η|

2)se2λ−

32

|η|dη = ∞, ∀λ ≥ 3/2.

We finish by explaining how the scaling property does not work here, let u0∈ Bα,s(R3), and u the unique solution

of (NS) such that u ∈ C([0, T ∗∗),Hs(R3)) and u ∈ C([0, T ∗), Bα,s(R3)) satisfying u ∈ C([0, T ∗∗],Hs(R3)) and u ∈

C([0, T ∗], Bα,s(R3)). As Bα,s(R3) ↩→ Hs(R3), then T ∗

≤ T ∗∗. The key point of this paper is to prove T ∗∗= T ∗, which is

equivalent to prove that, if ∥u(t)∥Bα,s is not bounded, then ∥u(t)∥Hs is also not bounded. If we change u0(x) by λu0(λx), thenthe corresponding solution uλ = λu(λ2t, λ2x) satisfies

uλ ∈ C([0, T ∗∗/λ2),Hs(R3))

uλ ∈ C([0, T ∗/λ2), Bα( ·

λ ),s(R3)).

Therefore, the problem is when α(z) is stronger than all polynomials (for example α = eaz). The default of this space puts adifficulty on getting a general result for explosion in C([0, T ∗), Bα,s(R3)).

To solve this problem, we choose α of type eaz1/σ

, with a > 0 and σ > 1. With this choice, we give a blow up criterionof the solution of (NS) at least of the order c1(T ∗

− t)−γ1ec2(T∗−t)−γ2 . The proof of this result is based on an energy estimate

given by:

∥u(t)∥2L2(R3)

+ 2ν t

0∥∇u(τ )∥2

L2(R3)dτ ≤ ∥u0

∥2L2(R3)

, (4)

and a result of explosion onu(t, ξ) in L1(R3) (see (21)). To explain this choice, let h ∈ H(C) be an entire function such thath(z) =

∞

n=0 anzn, with a0 > 0 and an ≥ 0 for all n ∈ N. We assume that there exist C > 0, σ > 1 and θ ∈ (0, 1) such that

h((z1 + z2)1/σ ) ≤ C[h(z1/σ1 )]θh(z1/σ2 ), ∀0 ≤ z1 ≤ z2. (5)

Taking α(z) = h(z1/σ ), we get (ν/2)(T ∗− t)−1

≤ ∥[α(|ξ |)]θu(t)∥L1 . By induction, we get (ν/2)(T ∗− t)−1

≤ ∥[α(|ξ |)]θnu(t)∥L1 . The Dominated Convergence Theorem gives the fundamental inequality (ν/2)(T ∗

− t)−1≤ ∥u(t)∥L1 . Combining

this inequality with ∥u(t)∥L1 ≤ Cδ∥u0∥1− 3

2δL2 ∥u(t)∥

32δHδ for δ ≥ δ0, with a uniform estimate of Cδ , we obtain:

C1(T ∗− t)−

2s3σ bk

C2

(T ∗ − t)1/3σ

k≤

ξ

bk(|ξ |1/σ )k|ξ |

2s|u(t, ξ)|2dξ, bk =

kp=0

apak−p; C1, C2 > 0,

which implies C1(T ∗− t)−

s3σ h(C2(T ∗

− t)−13σ ) ≤ ∥u(t)∥Bα,s . To be precise and work on non-empty sets, we treat the case

α(z) = f (z1/σ ) with f (z) = ez, σ > 1 and θ = 1/σ , where it is easy to verify that the property (5) is satisfied. But forσ = 1, (5) is not satisfied. Precisely, if u ∈ C([0, T ∗),Hs

a,1(R3)) is the maximal solution of (NS) with T ∗ < ∞. Is (2) holds

for σ = 1? Inspired by the proof of Theorem 1.1, it is enough to prove that if u(t) is not bounded in Hsa,1(R

3), then ∥u(t)∥L1

is not bounded, which is equivalent to prove that if ∥u(t)∥L1 is bounded then u(t) is bounded in Hsa,1(R

3).This problem is similar to that treat by Foias and Temam in [13], but their analysis does not suffice here. Begin by

recalling this result (see [13]). If u0∈ H1(T3) and assuming that we have a global solution, then ∥ea|D|u(t)∥2

H1 ≤ 2 + 2M21

where M1 = supt≥0 ∥u(t)∥H1 and a = C(1 + M21 )

−1. If we assume that u is non regular and T ∗ is the maximal time ofregularity, and we want to use the same method then eat |D|u(t) ∈ H1(T3) for all t ∈ [t0, T ∗) for some t0 ∈ (0, T ∗) withat ∼ C(1+ supz∈[0,t] ∥u(z)∥2

H1)−1 which tends to zero as t tends to T ∗. This method does not give a uniform parameter a for

all t ∈ [t0, T ∗). However, this method does not work in our case, since the parameter a is given at the beginning.

90 J. Benameur / Nonlinear Analysis 103 (2014) 87–97

The remainder of this paper is organized as follows: In Section 2,we give some notations andwe showpreliminary resultswhich will be very useful for this paper. Section 3 is devoted to a detailed study of the system (NS) in the space Bα,s(R3). InSection 4, we prove the main result of this paper where we used standard Fourier techniques.

2. Notations and preliminary results

2.1. Notations

In this short section, we collect some notations and definitions that will be used later.

• The Fourier transformation is normalized as

F (f )(ξ) =∧

f (ξ) =

R3

exp(−ix · ξ)f (x)dx, ξ = (ξ1, ξ2, ξ3) ∈ R3.

• The inverse Fourier formula is

F −1(g)(x) = (2π)−3

R3exp(iξ .x)f (ξ)dξ, x = (x1, x2, x3) ∈ R3.

• For s ∈ R,Hs(R3) denotes the usual non-homogeneous Sobolev space on R3 and ⟨., .⟩Hs(R3) denotes the usual scalarproduct on Hs(R3).

• For s ∈ R, Hs(R3) denotes the usual homogeneous Sobolev space on R3 and ⟨., .⟩Hs(R3) denotes the usual scalar producton Hs(R3).

• If f = (f1, f2, f3) and g = (g1, g2, g3) are two vector fields, we set

f ⊗ g := (g1f , g2f , g3f ),

and

div(f ⊗ g) := (div(g1f ), div(g2f ), div(g3f )).

2.2. Preliminary results

Lemma 2.1. For δ > 3/2, we have

∥f ∥L1(R3) ≤ Cδ∥f ∥1− 3

2δL2 ∥f ∥

32δHδ(R3)

, (6)

with Cδ = 2

π3

( 2δ

3 − 1)3/(4δ) + 2δ

3 − 1−1+ 3

4δ.

Moreover, for all δ0 > 3/2, there is M(δ0) > 0 such that

Cδ ≤ M(δ0), ∀δ ≥ δ0. (7)

Proof. For λ > 0, we have

∥f ∥L1(R3) =

|ξ |<λ

1 × |f (ξ)|dξ +

|ξ |>λ

1|ξ |δ

× |ξ |δ|f (ξ)|dξ .

Cauchy–Schwarz inequality yields

∥f ∥L1(R3) ≤

|ξ |<λ

dξ1/2

∥f ∥L2 +

|ξ |>λ

1|ξ |2δ

dξ1/2

∥f ∥Hδ

≤ 2

π

3λ3/2

∥f ∥L2 + 2

π

31

√2δ − 3

λ−

δ− 3

2

∥f ∥Hδ

≤ 2

π

3ϕ(λ)

where ϕ(λ) := Aλ3/2+ Bλ−

δ− 3

2

, and with A = ∥f ∥L2 , and B =

1√2δ−3

∥f ∥Hδ .Differentiating ϕ, we get

ϕ′(λ) :=32Aλ1/2

−

δ −

32

Bλ−

δ− 1

2

= λ1/2

32A −

δ −

32

Bλ−δ

.

J. Benameur / Nonlinear Analysis 103 (2014) 87–97 91

Then

ϕ′(λ) = 0 ⇒ λ = λ0 =

2δ3

− 1B/A

1/δ

.

As

ϕ(λ0) = A

2δ3

− 1B/A

3/(2δ)

+ B

2δ3

− 1B/A

−1+ 32δ

=

2δ3

− 13/(2δ)

+

2δ3

− 1−1+ 3

2δA1− 3

2δ B32δ

=

2δ3

− 13/(2δ)

+

2δ3

− 1−1+ 3

2δ

(2δ − 3)−34δ ∥f ∥

1− 32δ

L2∥f ∥

32δHδ

=

2δ3

− 13/(4δ)

+

2δ3

− 1−1+ 3

4δ2−

34δ ∥f ∥

1− 32δ

L2 ∥f ∥32δHδ

≤

2δ3

− 13/(4δ)

+

2δ3

− 1−1+ 3

4δ

∥f ∥1− 3

2δL2

∥f ∥32δHδ .

Therefore,

∥f ∥L1(R3) ≤ 2

π

3

2δ3

− 13/(4δ)

+

2δ3

− 1−1+ 3

4δ

∥f ∥1− 3

2δL2

∥f ∥32δHδ ,

and (6) is proved.

To prove (7), put g(δ) = 2

π3

( 2δ

3 − 1)3/(4δ) + ( 2δ3 − 1)−1+ 3

4δ

. Clearly g is continuous on (3/2, ∞) and limδ→∞ g(δ) =

2

π3 . Then, for every δ0 > 3/2, g is bounded on [δ0, ∞).In the following lemma we characterize the algebra cases of the space Bα,s(R3).

Lemma 2.2. Let s > 3/2. Let α : [0, ∞) → (0, ∞) be increasing function and satisfying condition (1). For every f , g ∈

Bα,s(R3), we have fg ∈ Bα,s(R3), and

∥fg∥Bα,s ≤ C(s)∥F (α(|D|)f )∥L1∥g∥Bα,s + ∥F (α(|D|)g)∥L1∥f ∥Bα,s

. (8)

∥fg∥Bα,s ≤ C(s)∥f ∥Bα,s∥g∥Bα,s . (9)

Proof. • Proof of (8): We have

∥fg∥2Bα,s

=

ξ

(1 + |ξ |2)sα(|ξ |)

2|fg(ξ)|2dξ

=

ξ

(1 + |ξ |2)sα(|ξ |)

2|(f ∗g)(ξ)|2dξ

≤

ξ

(1 + |ξ |2)sα(|ξ |)

2|f | ∗ |g|(ξ)

2dξ

≤

ξ

η

(1 + |ξ |2)s/2α(|ξ |)|f (ξ − η)| |g(η)|dη

2dξ

≤

ξ

|η|<|ξ−η|

(1 + |ξ |2)s/2α(|ξ |)|f (ξ − η)||g(η)|dη

+

|η|>|ξ−η|

(1 + |ξ |2)s/2α(|ξ |)|f (ξ − η)||g(η)|dη

2dξ .

Combining (1) and the elementary inequality

(1 + |ξ |2)s/2 ≤ 2s(1 + (max(|ξ − η|, |η|))2)s/2,

92 J. Benameur / Nonlinear Analysis 103 (2014) 87–97

we get

∥fg∥2Bα,s

≤ 22s

ξ

I1(ξ) + I2(ξ)

2dξ

≤ 22s+2

ξ

I1(ξ)2 + I2(ξ)2dξ,

where

I1(ξ) =

η

(1 + |ξ − η|2)s/2α(|ξ − η|)|f (ξ − η)|α(|η|)|g(η)|dη

I2(ξ) =

η

α(|ξ − η|)|f (ξ − η)| · (1 + |η|2)s/2α(|η|)|g(η)|dη.

By define the following functions:

F1(ξ) := (1 + |ξ |2)s/2α(|ξ |)|f (ξ)|

F2(ξ) := α(|ξ |)|f (ξ)|

G1(ξ) := (1 + |ξ |2)s/2|α(|ξ |)|g(ξ)|

G2(ξ) := α(|ξ |)|g(ξ)|,

we obtain

∥fg∥2Bα,s

≤ 22s+2∥F1 ∗ G2 |||

2L2 +|||F2 ∗ G1∥

2L2

.

Young inequality yields

∥fg∥2Bα,s

≤ 22s+2∥F1∥2

L2∥G2∥2L1 + ∥F2∥2

L1∥G1∥2L2

.

Then, we can deduce inequality (8).• Proof of (9): By (8) it suffices to estimate ∥G1∥

2L1 and ∥F2∥2

L1 .Cauchy–Schwarz inequality gives

∥G1∥2L1 ≤

ξ

1(1 + |ξ |2)s

dξ∥G1∥2Hs

≤ 4π

∞

0

ρ2

(1 + ρ2)sdρ∥G1∥

2Hs

≤ 4π 1

0ρ2dρ +

∞

1ρ2−2sdρ

∥G1∥

2Hs

≤ 4π13

+1

2s − 3

∥G1∥

2Hs .

The same approach is followed for ∥F2∥2L1. Then, the desired result is proved, and the proof of Lemma 2.2 is finished.

3. Study of (NS) in the space Bα,s(R3)

Theorem 3.1. Let s > 3/2. Let α : [0, ∞) → (0, ∞) be an increasing function and satisfying condition (1). Let u0∈

Bα,s(R3)3

such that div u0= 0. Then, there is a unique time T ∗

∈ (0, ∞] and a unique solution u ∈ C([0, T ∗), Bα,s(R3))

of Navier–Stokes equations (NS) such that u ∈ C([0, T ∗], Bα,s(R3)). Moreover, if T ∗ is finite, then

lim supt↗T∗

∥u(t)∥Bα,s = ∞. (10)

Proof. The integral form of Navier–Stokes equations is

u = eνt∆u0+ B(u, u),

where

B(u, v) = −

t

0eν(t−τ)∆P

div (u ⊗ v)

J. Benameur / Nonlinear Analysis 103 (2014) 87–97 93

and

F (Pf )(ξ) =f (ξ) −(f (ξ) · ξ)

|ξ |2ξ .

To estimate B(u, v) in Bα,s(R3), we have

∥B(u, v)∥Bα,s(t) ≤

t

0∥eν(t−τ)∆P

div (u ⊗ v)

∥Bα,s(τ )dτ

≤ ν−1/2 t

0(t − τ)−1/2

∥u ⊗ v∥Bα,s(τ )dτ ,

with

∥u ⊗ v∥2Bα,s

(τ ) =

ξ

(1 + |ξ |2)sα(|ξ |)

2|F (u ⊗ v)(τ , ξ)|2dξ

≤

ξ

(1 + |ξ |2)sα(|ξ |)

2η

|u(τ , ξ − η)| · |v(τ , η)|dη2

dξ

≤

ξ

(1 + |ξ |2)sα(|ξ |)

η

|u(τ , ξ − η)| · |v(τ , η)|dη2

dξ .

Using property (1), we get

∥u ⊗ v∥2Bα,s

(τ ) ≤ c

ξ

(1 + |ξ |2)s

η

α(|ξ − η|)|u(τ , ξ − η)| · α(|η|)|v(τ , η)|dη2

dξ

≤ c∥U .V∥Hs

where

U = F −1(α(|ξ |)|u(τ , ξ)|) and V = F −1(α(|ξ |)|v(τ , ξ)|).

As s > 3/2, we get

∥u ⊗ v∥Bα,s ≤ c∥U∥Hs∥V∥Hs = c∥u∥Bα,s∥v∥Bα,s .

Therefore,

∥B(u, v)∥Bα,s(t) ≤ cν−1/2T 1/2∥u∥L∞([0,T ],Bα)∥v∥L∞([0,T ],Bα,s). (11)

Now, recall the following result of the Fixed Point Theorem (see [14]).

Lemma 3.1. Let X be an abstract Banach space with norm ∥ ∥ and let B : X × X −→ X be a bilinear operator, such that for anyx1, x2 ∈ X,

∥B(x1, x2)∥ ≤ c0∥x1∥ ∥x2∥.

Then, for any y ∈ X such that

4c0∥y∥ < 1

the equation

x = y + B(x, x)

has a solution x in X. In particular the solution satisfies

∥x∥ ≤ 2∥y∥

and it is the only one such that

∥x∥ <12c0

.

Combining inequality (11) and Lemma 3.1, and choosing a small time T , we guarantee the existence and uniqueness of asolution in the space C([0, T ], Bα,s(R3)).

Now, we want to prove (10) by contradiction. For this, let u be a maximal solution of (NS) with T ∗ < ∞. If we assumethat ∥u(t)∥Bα,s is bounded on [0, T ∗), then u(t) is a Cauchy sequence in Bα,s(R3) at the time T ∗. Then, u(t) converges tou∗

∈ Bα,s(R3). So, if the system of Navier–Stokes is starting by u∗, we get a unique solution extends u, which is absurd.

94 J. Benameur / Nonlinear Analysis 103 (2014) 87–97

Theorem 3.2. Let s > 3/2. Let α : [0, ∞) → (0, ∞) be an increasing function satisfying condition (1). Let u0∈Bα,s(R3)

3such that div u0

= 0 and u ∈ C([0, T ∗), Bα,s(R3)) the maximal solution of Navier–Stokes equations (NS) given by Theorem 3.1.If T ∗ < ∞, then

(ν/2)1/2√T ∗ − t

≤ ∥Fα(|D|)u

(t)∥L1 , ∀t ∈ [0, T ∗), (12)

ν/2T ∗ − t

≤ cs∥u(t)∥Bα,s , ∀t ∈ [0, T ∗). (13)

Proof. Clearly inequality (13) is derived from (12) by applying Cauchy–Schwarz inequality.

Proof of (12). This proof is done in two steps.

Step 1: Taking the scalar product in Bα,s(R3), we obtain

12∂t∥u(t)∥2

Bα,s+ ν∥∇u(t)∥2

Bα,s≤ |⟨α(|D|)(u(t) · ∇u(t))/α(|D|)u(t)⟩Hs |

≤ |⟨α(|D|)div(u ⊗ u)(t)/α(|D|)u(t)⟩Hs |

≤ |⟨α(|D|)(u ⊗ u)(t)/∇α(|D|)u(t)⟩Hs |

≤ ∥α(|D|)(u ⊗ u)(t)∥Hs(R3)∥∇u(t)∥Bα,s .

Using (8) we obtain

12∂t∥u(t)∥2

Bα,s+ ν∥∇u(t)∥2

Bα,s≤ 2s+2

∥F (α(|D|)u(t))∥L1∥u(t)∥Bα,s∥∇u(t)∥Bα,s .

The elementary inequality xy ≤x22 +

y2

2 gives

12∂t∥u(t)∥2

Bα,s+ ν∥∇u(t)∥2

Bα,s≤ 22s+2ν−1

∥F (α(|D|)u(t))∥2L1∥u(t)∥

2Bα,s

+ν

2∥∇u(t)∥2

Bα,s

and

∂t∥u(t)∥2Bα,s

+ ν∥∇u(t)∥2Bα,s

≤ 22s+3ν−1∥F (α(|D|)u(t))∥2

L1∥u(t)∥2Bα,s

.

Gronwall lemma implies that, for every 0 ≤ t ≤ T < T ∗

∥u(T )∥2Bα,s

≤ ∥u(t)∥2Bα,s

e22s+3ν−1 T

t ∥F (α(|ξ |)u(τ ))∥2L1

dτ.

The fact that lim supt↗T∗ ∥u(t)∥Bα,s = ∞ implies T∗

t∥F (α(|D|)u(τ ))∥2

L1dτ = ∞, ∀0 ≤ t < T ∗. (14)

Step 2: Returning to the Navier–Stokes system and taking the Fourier transform in the first equation, we get

12∂t |u(t, ξ)|2 + ν|ξ |

2|u(t, ξ)|2 + Re

F (u · ∇u)(t, ξ) ·u(t, −ξ)

= 0.

But, for any ε > 0

12∂t |u(t, ξ)|2 =

12∂t

|u(t, ξ)|2 + ε

=

|u(t, ξ)|2 + ε∂t

|u(t, ξ)|2 + ε,

and

∂t

|u(t, ξ)|2 + ε + ν|ξ |2 |u(t, ξ)|2

|u(t, ξ)|2 + ε+

F (u · ∇u)(t, ξ) ·u(t, −ξ)|u(t, ξ)|2 + ε

= 0.

Then

∂t

|u(t, ξ)|2 + ε + ν|ξ |2 |u(t, ξ)|2

|u(t, ξ)|2 + ε≤ |F (u · ∇u)(t, ξ)|.

J. Benameur / Nonlinear Analysis 103 (2014) 87–97 95

Integrating on [t, T ] ⊂ [0, T ∗), we obtain|u(T , ξ)|2 + ε + ν|ξ |

2 T

t

|u(τ , ξ)|2|u(τ , ξ)|2 + ε

dτ ≤

|u(t, ξ)|2 + ε +

T

t|F (u · ∇u)(τ , ξ)|dτ .

Taking ε → 0, we have

|u(T , ξ)| + ν|ξ |2 T

t|u(τ , ξ)|dτ ≤ |u(t, ξ)| +

T

t|F (u · ∇u)(τ , ξ)|dτ

≤ |u(t, ξ)| +

T

t|ξ∥F (u∇u)(τ , ξ)|(τ , ξ)|dτ

≤ |u(t, ξ)| +

T

t|u| ∗ξ |∇u|(τ , ξ)|(τ , ξ)|dτ

≤ |u(t, ξ)| +

T

t

η

|u(τ , ξ − η)∥∇u|(τ , η)|dηdτ .

Multiplying the last equation by α(|ξ |) and using (1), we obtain

α(|ξ |)|u(T , ξ)| + ν|ξ |2 T

tα(|ξ |)|u(τ , ξ)|dτ

≤ α(|ξ |)|u(t, ξ)| +

T

t

η

α(|ξ − η|)|u(τ , ξ − η)|α(|η|)|u|(τ , η)|dηdτ .

Integrating over ξ ∈ R3 and using Young inequality, we get

∥F (α(|D|)u)(T )∥L1 + ν

T

t∥F (α(|D|)1u)∥L1 ≤ ∥F (α(|D|)u)(t)∥L1 +

T

t∥F (α(|D|)u)∥L1∥F (α(|D|)∇u)∥L1 .

Cauchy–Schwarz inequality yields

∥F (α(|D|)u)(T )∥L1 + ν

T

t∥F (α(|D|)1u)∥L1 ≤ ∥F (α(|D|)u)(t)∥L1 +

T

t∥F (α(|D|)u)∥3/2

L1∥F (α(|D|)∇u)∥1/2

L1.

Inequality xy ≤x22 +

y2

2 gives

∥F (α(|D|)u)(T )∥L1 +ν

2

T

t∥F (α(|D|)1u)∥L1 ≤ ∥F (α(|D|)u)(t)∥L1 +

1ν

T

t∥F (α(|D|)u)∥3

L1 .

By Gronwall lemma, we obtain, for 0 ≤ t ≤ T < T ∗

∥F (α(|D|)u)(T )∥L1 ≤ ∥F (α(|D|)u)(t)∥L1 expν−1

T

t∥F (α(|D|)u(τ ))∥2

L1dτ

or

∥F (α(|D|)u)(T )∥2L1 exp

−2ν−1

T

t∥F (α(|D|)u(τ ))∥2

L1dτ

≤ ∥F (α(|D|)u)(t)∥2L1 .

Integrating over [t0, T ] ⊂ [0, T ∗), we get

1 − exp−2ν−1

T

t0∥F (α(|D|)u(τ ))∥2

L1dτ

≤ 2ν−1∥F (α(|D|)u)(t0)∥2

L1(T − t0).

Inequality (14) gives, if T → T ∗,

1 ≤ 2ν−1∥F (α(|D|)u(t0))∥2

L1(T∗− t0), ∀t0 ∈ [0, T ∗),

then (12) is proved and the proof of Theorem 3.2 is finished.

4. Blow up criterion in Sobolev–Gevrey spaces

In this section we prove Theorem 1.1.• Starting by proving the technical result: For all r ∈ (0, 1]

(x + y)r ≤ yr + rxr , ∀0 ≤ x ≤ y. (15)

96 J. Benameur / Nonlinear Analysis 103 (2014) 87–97

Taylor formula applying to the function

f (t) = (1 + t)r

on the interval [0, z] ⊂ [0, 1], gives

(1 + z)r = 1 + rz +r(r − 1)(1 + c)r−2

2z2,

for some c ∈ [0, z]. As r, z ∈ (0, 1), we get

(1 + z)r ≤ 1 + rz ≤ 1 + rzr .

By taking z =xy , we can deduce inequality (15).

• Now, for σ =1r ∈ [1, ∞), a > 0 and s ≥ 0, we consider the space

Hsa,σ (R3) := {f ∈ L2(R3), (1 + |ξ |

2)s/2ea|ξ |1/σf (ξ) ∈ L2(R3)}

equipped by the norm ∥f ∥Hsa,σ = ∥ea|D|

1/σf ∥Hs and its associated scalar product ⟨f /g⟩Hs

a,σ = ⟨ea|D|1/σ

f /ea|D|1/σ

g⟩Hs . The spacesHs

a,σ (R3) have the following properties:

(P1) Hsa,σ (R3) is a Hilbert space.

(P2) If 0 ≤ a1 < a2, then Hsa2,σ (R3) ↩→ Hs

a1,σ (R3).(P3) If s ∈ R, 0 < a and b > 0, then ∥

√−∆eb∆f ∥Hs

a,σ ≤1

√b∥f ∥Hs

a,σ .

Using inequality (15), we obtain

ea|ξ |1/σ

≤ eamax(|ξ−η|,|η|)1/σ eaσ min(|ξ−η|,|η|)1/σ (16)

ea|ξ |1/σ

≤ ea|ξ−η|1/σ

ea|η|1/σ

. (17)

Using these inequalities, we can deduce the following proposition which characterizes the algebra cases of the spaceHs

a,σ (R3).

Proposition 4.1. Let s > 3/2, there is a constant c(s) such that for every a > 0, σ ≥ 1, and for every f , g ∈ Hsa,σ (R3), we have

fg ∈ Hsa,σ (R3), and

∥fg∥Hsa,σ ≤ 2s+1

∥F

e

aσ |D|

1/σf

∥L1∥g∥Hsa,σ + ∥F

e

aσ |D|

1/σg

∥L1∥f ∥Hsa,σ

. (18)

∥fg∥Hsa,σ ≤ c(s)

∥f ∥Hs

aσ ,σ

∥g∥Hsa,σ + ∥g∥Hs

aσ ,σ

∥f ∥Hsa,σ

. (19)

In particular

∥fg∥Hsa,σ ≤ 2c(s)∥f ∥Hs

a,σ ∥g∥Hsa,σ . (20)

Proof. The proof of this proposition is similar to the proof of Lemma 2.2, and taking into account inequalities (16)–(17).

Proof of Theorem 1.1. Most of this proof is given by Theorems 3.1–3.2. Therefore, some details are omitted.

Step 1: The existence and uniqueness in C([0, T ],Hsa,σ (R3)) is given by Theorem 3.2.

Step 2: The maximal solution of Navier–Stokes equations becomes large if the maximum time is finite. If u is a maximalsolution in C([0, T ∗),Hs

a,σ (R3)) with T ∗ is finite, then

lim supt↗T∗

∥u(t)∥Hsa,σ = ∞.

Step 3: By copying the proof of Theorem 3.2 and taking into account inequality (8), we get T∗

t

F e

aσ |D|

1/σu(τ )

L1dτ = ∞, ∀t ∈ [0, T ∗)

andν/2

T ∗ − t≤

F e

aσ |D|

1/σu(t)

2L1

, ∀t ∈ [0, T ∗).

J. Benameur / Nonlinear Analysis 103 (2014) 87–97 97

Using ∥F (eaσ |D|

1/σu(t))∥L1 ≤ Cs∥u(t)∥Hs

aσ ,σ

and Hsa,σ (R3) ↩→ Hs

aσ ,σ

(R3), we ensure that u ∈ C([0, T ∗),Hsaσ ,σ

(R3))

and lim supt↗T∗ ∥u(t)∥Hsaσ ,σ

= ∞. Therefore,

ν/2T ∗ − t

≤

F e

aσ2 |D|

1/σu(t)

2L1

, ∀t ∈ [0, T ∗).

By induction, we can deduce that, for every n ∈ N

ν/2T ∗ − t

≤

F (ea

σn |D|1/σ

u(t))2L1

, ∀t ∈ [0, T ∗).

The Dominated Convergence Theorem gives, if n → ∞,

ν/2T ∗ − t

≤ ∥u(t)∥2L1 , ∀t ∈ [0, T ∗). (21)

Step 4: Combining inequalities (6)–(7) and (21), for δ0 = s and δ = s +k2σ , k ∈ Z+, we get

ν/2T ∗ − t

≤ M(s)2∥u0∥

2− 3s+ k

2σL2 ∥u(t)∥

3s+ k

2σ

Hs+ k2σ

, ∀t ∈ [0, T ∗).

Then, for t ∈ [0, T ∗), we have

C21

(T ∗ − t)2s/3

C2

(T ∗ − t)13σ

k≤ ∥u(t)∥2

Hs+ k2σ

, (k ≥ 0).

Then, for all k ≥ 0,

C21

(T ∗ − t)2s/3(2a)k

k!

C2

(T ∗ − t)13σ

k≤

ξ

(2a|ξ |1/σ )k

k!|ξ |

2s|u(t, ξ)|2dξ .

Summing the above inequalities over the set {k ≥ 0}, we get (2). Therefore, the proof of Theorem 1.1 is finished.

Acknowledgments

I am deeply grateful to the referee for his suggestions which helped me to improve this paper.I would like to express my sincere gratitude to Nader Masmoudi for his kind advice.This project was supported by King Saud University, Deanship of Scientific Research, College of Science Research Center.

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