Global regularity of solutions of 2D Boussinesq equations with fractional diffusion

Nonlinear Analysis 72 (2010) 677–681

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Global regularity of solutions of 2D Boussinesq equations with fractional diffusion Xiaojing Xu School of Mathematical Sciences, Beijing Normal University, Laboratory of Mathematics and Complex Systems, Ministry of Education, Beijing 100875, People’s Republic of China

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Article history: Received 24 June 2009 Accepted 6 July 2009 MSC: 76B03 76D03 76D05

abstract The goal of this work is to study the Boussinesq equations for an incompressible fluid in R2 , with diffusion modeled by fractional Laplacian. The existence, the uniqueness and the regularity of solution has been proved. © 2009 Elsevier Ltd. All rights reserved.

Keywords: Fractional diffusion Regularity Global existence Boussinesq equations

1. Introduction We consider the Cauchy problem of 2D fractal diffusion Boussinesq equations for an incompressible fluid flows in R2

 ∂u α    ∂ t + (u · ∇)u + ν(−∆) u + ∇ P = θ e2 ,    ∂θ + (u · ∇)θ + κ(−∆)β θ = 0,  ∂t     div u = 0, u(x, 0) = u0 (x), θ (x, 0) = θ0 (x),

(x, t ) ∈ R2 × (0, ∞), (1.1)

where u(x, t ) = (u1 , u2 ) is the velocity vector field, P = P (x, t ) is the scalar pressure, θ (x, t ) is the scalar temperature, ν ≥ 0 is the viscosity, κ ≥ 0 is the thermal diffusivity, e = (0, 1), α, β ∈ (0, 1), and (−∆)α is the pseudodifferential operator defined via the Fourier transform \ (− ∆)α v(ξ ) = |ξ |2αb v(ξ ). In the following, for simplicity of notation, we denote

Λ = (−∆)1/2 . The Boussinesq system plays an important role in atmospheric sciences (see e.g. [1]), as well as a model in many geophysical applications (see e.g. [2]). For this reason, this system is studied systematically by scientists from different domains. We contribute to this theory by mathematical studies of these equations with the fractional Laplacian.

E-mail address: [email protected]. 0362-546X/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2009.07.008

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X. Xu / Nonlinear Analysis 72 (2010) 677–681

The global-in-time regularity of solutions to (1.1) with α = β = 1 and ν, κ > 0 is well-known (see e.g. [3]). However, the regularity or singularity questions of the case of (1.1) with ν = κ = 0 is a rather challenging problem in mathematical fluid mechanics (see e.g. [4–9]). Recently, the global regularity of solutions to (1.1) with partial viscosity terms has been studied by many authors (see e.g. [10–15]), mainly for α = β = 1 either ν = 0, κ > 0 or ν > 0, κ = 0. In this paper, we consider another case when partial viscosity is divided into two parts, namely, α, β ∈ (0, 1), ν, κ > 0, which enter the velocity and thermal equation separately. Similar modifications of MHD equations were considered in [16,17]. Our main result is stated as follows. Theorem 1.1. Let ν > 0, κ > 0 be fixed, α ∈ [ 12 , 1), β ∈ (0, 21 ], α + β = 1, and div u0 = 0. Let m > 2 be an integer, and (u0 , θ0 ) ∈ H m (R2 ). Then, there exists a unique solution (u, θ ) to the Cauchy problem (1.1) such that

θ ∈ C ([0, ∞); H m (R2 )) ∩ L2 (0, ∞; H m+β (R2 )), and u ∈ C ([0, ∞); H m (R2 )) ∩ L2 (0, ∞; H m+α (R2 )). Remark 1.1. For simplicity of the exposition, we formulate and prove Theorem 1.1 in the critical case α + β = 1, only. One can easily verify that, by arguments from this work, we can obtain an analogous result for 1 ≤ α + β ≤ 2.  2. Preliminary lemmas First of all, let us give a positive inequality which is important for establishing the energy inequalities in framework of Lp . Lemma 2.1 (Positive Inequality). Let 0 ≤ α ≤ 2. For every p > 1, we have

Z Rn

(Λα w)|w|p−2 w dx ≥ C (p)

Z

 Rn

α

p

Λ 2 |w| 2

2

dx

for all w ∈ Lp (Rn ) such that Λα w ∈ Lp (Rn ), where C (p) =

(2.2) 4(p−1) . p2

Inequality (2.2) is well-known in the theory of sub-Markovian operators and its statement and the proof is given e.g. in [18, Theorem 2.1 combined with the Beurling–Deny condition (1.7)]. Observe that if α = 2, integrating by parts we obtain (2.2) with the equality. In the following, we state a blow-up criterion for smooth solutions of (1.1), which plays the crucial role in the proof of our main result. Theorem 2.1 (Blow-up Criterion). Let α, β ∈ [0, 2], ν ≥ 0, κ ≥ 0. Suppose (u0 , θ0 ) ∈ H m (R2 ) with m > 2 being an integer. Then, there exists a unique local classical solution (u, θ ) ∈ C ([0, T ); H m (R2 )) of problem (1.1) for some T = T (ku0 kH m (R2 ) , kθ0 kH m (R2 ) ). Moreover, the solution remains in H m (R2 ) up to a time T1 > T , namely (u, θ ) ∈ C ([0, T1 ]; H m (R2 )) if and only if T

Z

k∇θ (τ )kL∞ dτ < ∞.

(2.3)

0

Theorem 2.1, in the inviscid case ν = 0 and κ ≥ 0, was proved in [4]. The arguments from [4] works with minor changes also for problem (1.1) with the fractional diffusion, due to inequality (2.2). By this reason, we skip details of the proof of Theorem 2.1. Hence, in order to prove Theorem 1.1, it suffices to show that (2.3) holds true for the smooth solutions (u, θ ) to the Cauchy problem (1.1).

3. Proof of theorem In this section, we first show some a priori estimates for a smooth solution (u, θ ) ∈ C ([0, T ); H m (R2 )) with m > 2 to (1.1), then prove that (2.3) is valid. For simplicity, let ν = κ = 1.

X. Xu / Nonlinear Analysis 72 (2010) 677–681

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3.1. Preliminary estimates

• Estimate of kθkL∞ (0,∞;Lp (R2 )) . Let p ≥ 2. Multiplying the second equation in (1.1) by |θ |p−2 θ and integrating over R2 , we deduce that 1 d p dt

kθ (t )kpLp +

t

Z

(−∆)β θ |θ |p−2 θ dx = 0,

0

where we have used the divergence free condition. This identity together with Lemma 2.1, allows us to get

kθ (t )kpLp + C (p)

t

Z 0

p

kΛβ |θ | 2 (τ )k2L2 dτ ≤ kθ0 kpLp .

(3.4)

In particular, when p = 2, we have

kθ (t )k2L2 +

t

Z 0

kΛβ θ (τ )k2L2 dτ ≤ kθ0 k2L2 .

(3.5)

• Estimate of kukL∞ (0,∞;L2 (R2 )) . Multiplying the first equation of (1.1) by u, and integrating it over R2 , we have 1 d 2 dt

ku(t )k + 2 L2

Z

α

R2

u (−∆) u dx =

Z

θ e2 |u| dx − 2

R2

Z

(u · ∇)|u| dx − 2

R2

Z R2

∇ P u dx.

This identity together with inequality (2.2) and the divergence free condition, yield that 1 d 2 dt

ku(t )k2L2 +

Z R2

|Λα u|2 dx =

Z R2

θ e2 u dx ≤ kθ (t )kL2 ku(t )kL2 .

By (3.5) and the Hölder inequality, we deduce that

ku(t )k2L2 + 4

t

Z 0

kΛα u(τ )k2L2 dτ ≤ 4kθ0 k2L2 T 2 + 2ku0 k2L2 .

• Estimate of kω(t )kL∞ (0,∞;L2 (R2 )) .

Taking the operation curl on both sides of the first equation in (1.1) and denoting ω = curl u = ∂x1 u2 − ∂x2 u1 , we get

ωt + (−∆)α ω + (u · ∇)ω = −θx1 .

(3.6)

Multiplying the above equality by ω, integrating over R , and using the divergence free condition for u, we find, after integration by parts and by the Cauchy inequality 2

1 d 2 dt

kω(t )k2L2 + kΛα ωk2L2 =

Z

1 2

R2

Z =− R2

≤

1 2

(u · ∇)|ω|2 dx −

Z R2

θx1 ωdx

θx1 ωdx 1

kΛα ωk2L2 + kΛβ θ k2L2 . 2

Here, in the last inequality, we have used the Parseval theorem and the relation α + β = 1. Thus, we have d dt

kω(t )k2L2 + kΛα ωk2L2 ≤ kΛβ θk2L2 .

By virtue of estimate (3.5), we deduce that

kω(t )k + 2 L2

t

Z

α

kΛ ω(τ )k 0

2 d L2

τ ≤ kω k + 2 0 L2

t

Z 0

kΛβ θ (τ )k2L2 dτ ≤ C (kω0 kL2 , kθ0 kL2 , T ).

(3.7)

3.2. Estimate of kΛα DωkL∞ (0,∞;L2 (R2 )) We first compute the derivative D = (∂x1 , ∂x2 ) of both sides of (3.6), and then take L2 inner product with Dω. After integration by parts, using the Parseval inequality and the interpolation inequality (recall that α + β = 1), we obtain

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X. Xu / Nonlinear Analysis 72 (2010) 677–681

1 d 2 dt

Z

kDω(t )k2L2 + kΛα Dωk2L2 = −

2

ZR =−

R2

≤ k∇ uk

Z

[D(u · ∇)ω]Dω dx −

ZR

[(Du · ∇)ω]Dω dx − 2

L 1−α

kDωkL2 kDωk

2

L 1−α

Dθx1 Dω dx

R2

1

1

+ kΛα Dωk2L2 + kΛβ Dθ k2L2

2

Lα

2

3α−1

≤ k∇ uk

Dθx1 Dω dx

2

2

1−α

1

1

2

2

kDωkL2 α kΛα DωkL2α + kΛα Dωk2L2 + kΛβ Dθ k2L2

2α

3

1

L 1−α

4

2

≤ C k∇ uk 3α−2 1 kDωk2L2 + kΛα Dωk2L2 + kΛβ Dθ k2L2 ,

(3.8)

where we have used the assumptions α ≥ 12 and div u = 0. Next, computing the derivative ∇ ⊥ = (−∂x2 , ∂x1 ) of the second equation from (1.1), we easily show that

∇ ⊥ θt + ∇ ⊥ [(u · ∇)θ] + (−∆)β ∇ ⊥ θ = 0.

(3.9)

We multiply the above equality by ∇ θ , and integrate it over R . Similar arguments as those in (3.8) lead to ⊥

1 d 2 dt

2

k∇ ⊥ θ (t )k2L2 + kΛβ ∇ ⊥ θ k2L2 ≤ −

Z R2

= k∇ uk ≤ k∇ uk 1

≤

2

(u · ∇)∇ ⊥ θ ∇ ⊥ θ dx − k∇ ⊥ θ kL2 k∇ ⊥ θk

2

L 1−α

R2

∇ ⊥ θ · ∇ u ∇ ⊥ θ dx

2

Lα

k∇ ⊥ θ kL2 kΛβ ∇ ⊥ θ kL2

2

L 1−α

k∇ uk2

Z

1

2

L 1−α

k∇ ⊥ θk2L2 + kΛβ ∇ ⊥ θ k2L2 .

(3.10)

2

Now, combining (3.8) with (3.10), one can show the function X (t ) = kDω(t )kL2 + k∇ ⊥ θ (t )kL2 satisfies the inequality d dt

2α

X (t ) ≤ C (k∇ uk 3α−2 1 + k∇ uk2

2

L 1−α

L 1−α

)X (t ).

Therefore, Gronwall’s inequality and the embedding inequality and estimate (3.7) yield that X (t ) ≤ CX (0) exp

t

Z

2α

(k∇ u(τ )k 3α−2 1 + k∇ u(τ )k2

≤ CX (0) exp

t

0

( ≤ CX (0) exp T

L 1−α

L 1−α

0

Z

2

) dτ



2α

(kΛα ω(τ )kL32α−1 + kΛα ω(τ )k2L2 )dτ

4α−2 3α−1

t

Z

α

kΛ ω(τ )k 0

2 d L2

τ

2α  3α− 1



t

Z

) α

kΛ ω(τ )k

+ 0

2 d L2

τ

≤ C (T , ku0 kH 2 , kθ0 kH 1 ). Finally, by virtue of estimate (3.8), we deduce

kDω(t )k2L2 +

t

Z 0

kΛα Dω(τ )k2L2 dτ ≤ C (T , ku0 kH 2 , kθ0 kH 1 ).

(3.11)

3.3. Estimate of k∇θ kL∞ (0,∞;L∞ (R2 )) Multiplying (3.9) by |∇ ⊥ θ|p−2 ∇ ⊥ θ , and integrating it over R2 , we have 1 d p dt

k∇ ⊥ θ (t )kpLp + C (p)k∇ ⊥ θ kp

Z

2p L 1−α

≤ R2

∇ ⊥ θ · ∇ u|∇ ⊥ θ|p−2 ∇ ⊥ θ dx

≤ k∇ ukL∞ k∇ ⊥ θ kpLp . Using Gronwall’s inequality and the obvious identity k∇ Q kLp = k∇ ⊥ Q kLp , we easily show that

k∇θ (t )kLp ≤ C k∇θ0 kLp exp

t

Z 0

 k∇ u(τ )kL∞ dτ .

(3.12)

X. Xu / Nonlinear Analysis 72 (2010) 677–681

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This inequality together with the Gagliardo–Nirenberg inequality, allows us to obtain t

Z

k∇ u(τ )kL∞ dτ ≤ C

t

Z 0

0

≤ CT

α

1

kω(τ )kL12+α kΛ1+α ω(τ )kL12+α dτ

2α+1 α+1

t

Z

kω(τ )k 0

2 d L2

τ

 1+α2α

t

Z +C 0

kΛ1+α ω(τ )k2L2 dτ

≤ C (T , ku0 kH 2 , kθ0 kH 1 ). Using Sobolev embedding

kf (x)kLp (R2 ) ≤ C kf (x)kH s (R2 ) ,

s > 1,

where C is independent of p ∈ [2, ∞], then we have

k∇θ (t )kLp ≤ C kθ0 kH m exp

t

Z

k∇ u(τ )kL∞ dτ



≤ C for all t ∈ [0, T ],

0

where C is independent of p. Passing to the limit p → ∞ in above inequality, we obtain

k∇θ (t )kL∞ ≤ C (T , ku0 kH 2 , kθ0 kH m ),

∀t ∈ [0, T ].

This implies that condition (2.3) holds true, and according to Theorem 2.1, we obtain a unique solution of (1.1) such that (u, θ ) ∈ C ([0, ∞); H m (R2 )). By (3.10), (3.11) and the iteration process, we construct u ∈ L2 (0, ∞; H m+α (R2 )) and θ ∈ L2 (0, ∞; H m+β (R2 )), and we complete the proof of Theorem 1.1. Acknowledgments This paper was partially written while the author enjoyed the hospitality and support of the Laboratoire d’Analyse et de Mathématiques Appliquées, Université de Marne-la-Vallée. The author would like to thank Marco Cannone for his invitation and useful discussion, and the referee for his useful comments and suggestions. This paper is partially supported by the NSF of China ([10601009]; [10701014]). References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18]

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