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A regularity criterion for the Cahn-Hilliard-Boussinesq system with zero viscosity

Abstract

This paper studies a coupled Cahn-Hilliard-Boussinesq system with zero viscosity. We prove a regularity criterion in terms of vorticity in the homogeneous Besov space B ˙ , 0 .

MSC:35Q30, 76D03, 76D05, 76D07.

1 Introduction

In this paper, we study the following Cahn-Hilliard-Boussinesq system with zero viscosity [1]:

t u+(u)u+π=μϕ+θ e 3 ,
(1.1)
divu=0,
(1.2)
t θ+uθ=Δθ,
(1.3)
t ϕ+uϕ=Δμ,
(1.4)
Δϕ+ f (ϕ)=μ,
(1.5)
(u,θ,ϕ)(x,0)=( u 0 , θ 0 , ϕ 0 )(x),x R 3 ,
(1.6)

with u the fluid velocity field, θ the temperature, ϕ the order parameter, and π the pressure, are the unknowns. e 3 := ( 0 , 0 , 1 ) t . μ is the chemical potential. f(ϕ):= 1 4 ( ϕ 2 1 ) 2 is the double well potential.

When θ=ϕ0, (1.1) and (1.2) are the well-known Euler system; Kozono, Ogawa and Taniuchi [2] proved the following regularity criterion:

ω:=curlu L 1 (0,T; B ˙ , 0 ).
(1.7)

Here B ˙ , 0 denotes the homogeneous Besov space.

When ϕ=0, (1.1), (1.2), and (1.3) are the well-known Boussinesq system with zero viscosity; Fan and Zhou [3] also showed the regularity criterion (1.7).

When u=0, (1.4) and (1.5) are the well-known Cahn-Hilliard system.

It is easy to show that the problem (1.1)-(1.6) has a unique local smooth solution, thus we omit the details here. However, the global regularity is still open. The aim of this paper is to study the blow up criterion. We will prove the following.

Theorem 1.1 Let T>0 and u 0 H 3 , θ 0 H 2 , ϕ 0 H 3 with div u 0 =0 in R 3 . Suppose that (u,θ,ϕ) is a local smooth solution to the problem (1.1)-(1.6). Then (u,θ,ϕ) is smooth up to time T provided that (1.7) is satisfied.

We will use the following logarithmic Sobolev inequality [2]:

u L C ( 1 + curl u B ˙ , 0 log ( e + Λ 3 u L 2 ) ) ,
(1.8)

and the bilinear product and commutator estimates due to Kato and Ponce [4]:

Λ s ( f g ) L p C ( Λ s f L p 1 g L q 1 + f L p 2 Λ s g L q 2 ) ,
(1.9)
Λ s ( f g ) f Λ s g L p C ( f L p 1 Λ s 1 g L q 1 + Λ s f L p 2 g L q 2 ) ,
(1.10)

with s>0, Λ:= ( Δ ) 1 / 2 and 1 p = 1 p 1 + 1 q 1 = 1 p 2 + 1 q 2 .

2 Proof of Theorem 1.1

First, it follows from (1.2), (1.3), and the maximum principle that

θ L ( 0 , T ; L ) C.
(2.1)

Testing (1.3) by θ, using (1.2), we see that

1 2 d d t θ 2 dx+ | θ | 2 dx=0,

whence

θ L ( 0 , T ; L 2 ) + θ L 2 ( 0 , T ; H 1 ) C.
(2.2)

Testing (1.4) by ϕ, using (1.2) and (1.5), we find that

1 2 d d t ϕ 2 d x + | Δ ϕ | 2 d x = f ( ϕ ) Δ ϕ d x = ( ϕ 3 ϕ ) Δ ϕ d x = 3 ϕ 2 | ϕ | 2 d x ϕ Δ ϕ d x ϕ Δ ϕ d x ϕ L 2 Δ ϕ L 2 1 2 Δ ϕ L 2 2 + 1 2 ϕ L 2 2 ,

which gives

ϕ L ( 0 , T ; L 2 ) + ϕ L 2 ( 0 , T ; H 2 ) C.
(2.3)

Testing (1.1) and (1.4) by u and μ, respectively, using (1.2), (1.5), and (2.2), summing up the result, we deduce that

d d t 1 2 | ϕ | 2 + f ( ϕ ) + 1 2 u 2 d x + | μ | 2 d x = θ e 3 u d x θ L 2 u L 2 C u L 2 ,

which gives

ϕ L ( 0 , T ; H 1 ) C,
(2.4)
u L ( 0 , T ; L 2 ) C,
(2.5)
μ L 2 ( 0 , T ; L 2 ) C.
(2.6)

In the following calculations, we will use the following Gagliardo-Nirenberg inequality:

ϕ L 2 C ϕ L 2 Δ ϕ L 2 .
(2.7)

It follows from (2.6), (1.5), (2.3), (2.4), and (2.7) that

0 T | Δ ϕ | 2 d x d t = 0 T | ( f ( ϕ ) μ ) | 2 d x d t C 0 T | μ | 2 d x d t + C 0 T | f ( ϕ ) | 2 d x d t C + C 0 T | ( ϕ 3 ϕ ) | 2 d x d t C + C 0 T ϕ 4 | ϕ | 2 d x d t C + C ϕ L ( 0 , T ; L 2 ) 2 0 T ϕ L 4 d t C + C 0 T ϕ L 4 d t C + C 0 T ϕ L 2 2 Δ ϕ L 2 2 d t C + C sup t ϕ ( t ) L 2 2 0 T Δ ϕ L 2 2 d t C ,
(2.8)

which yields

ϕ L 2 ( 0 , T ; L ) C.
(2.9)

Applying Λ 3 to (1.1), testing by Λ 3 u, using (1.2), (1.9), (1.10), and noting that

Δϕϕ= j j ( j ϕϕ) 1 2 | ϕ | 2 ,

we derive

1 2 d d t | Λ 3 u | 2 d x = ( Λ 3 ( u u ) u Λ 3 u ) Λ 3 u d x j Λ 3 j ( j ϕ ϕ ) Λ 3 u d x + Λ 3 θ e 3 Λ 3 u d x C u L Λ 3 u L 2 2 + C ϕ L Λ 5 ϕ L 2 Λ 3 u L 2 + C Λ 3 θ L 2 Λ 3 u L 2 C u L Λ 3 u L 2 2 + C ϕ L 2 Λ 3 u L 2 2 + C Λ 3 u L 2 2 + ϵ Λ 5 ϕ L 2 2 + ϵ Λ 3 θ L 2 2
(2.10)

for any 0<ϵ<1.

Taking Δ to (1.3), testing by Δθ, using (1.2), (1.10), and (2.1), we obtain

1 2 d d t | Δ θ | 2 d x + | Δ θ | 2 d x = ( Δ ( u θ ) u Δ θ ) Δ θ d x C u L Δ θ L 2 2 + C θ L 4 Δ u L 4 Δ θ L 2 C u L Δ θ L 2 2 + C θ L 1 / 2 Δ θ L 2 1 / 2 u L 1 / 2 Δ u L 2 1 / 2 Δ θ L 2 C u L Δ θ L 2 2 + C ( u L Δ θ L 2 + Δ u L 2 ) Δ θ L 2 C u L Δ θ L 2 2 + C Δ θ L 2 2 + C Δ u L 2 2 .
(2.11)

Here we have used the Gagliardo-Nirenberg inequalities:

θ L 4 2 C θ L Δ θ L 2 , Δ u L 4 2 C u L Δ u L 2 .

Taking Δ to (1.4), testing by Δϕ, using (1.5), (1.2), and (1.10), we have

1 2 d d t | Δ ϕ | 2 d x + | Λ 5 ϕ | 2 d x = i ( Δ ( u i i ϕ ) u i Δ i ϕ ) Δ ϕ d x + Δ Δ f ( ϕ ) Δ ϕ d x C u L Δ ϕ L 2 2 + C ϕ L Λ 3 u L 2 Δ ϕ L 2 + Δ f ( ϕ ) Δ 2 ϕ d x .
(2.12)

Using (2.4), we get

Δ f ( ϕ ) Δ 2 ϕ d x C ( | Δ ϕ | + | ϕ 2 | | Δ ϕ | + | ϕ | | ϕ | | 2 ϕ | + | ϕ | 3 ) | Δ 2 ϕ | d x C ( Δ ϕ L 2 + ϕ L 2 Δ ϕ L 2 + ϕ L ϕ L 2 2 ϕ L + ϕ L 2 ϕ L 2 ) Λ 5 ϕ L 2 C Δ ϕ L 2 2 + C ϕ L 4 Δ ϕ L 2 2 + C ϕ L 2 ϕ L Λ 5 ϕ L 2 + C ϕ L 2 Λ 5 ϕ L 2 + 1 4 Λ 5 ϕ L 2 2 .
(2.13)

Now we use the following Gagliardo-Nirenberg inequalities:

2 ϕ L C Δ ϕ L 2 3 / 4 Λ 5 ϕ L 2 1 / 4 , ϕ L 2 C Δ ϕ L 2 Λ 3 ϕ L 2 .

We obtain

Δ f ( ϕ ) Δ 2 ϕ d x C Δ ϕ L 2 2 + C ϕ L 4 Δ ϕ L 2 2 + C ϕ L 8 / 3 Δ ϕ L 2 2 + C Δ ϕ L 2 2 Λ 3 ϕ L 2 2 + 1 2 Λ 5 ϕ L 2 .
(2.14)

Combining (2.10), (2.11), (2.12), and (2.14), taking ϵ small enough, using (1.8), (2.5), Gronwall’s inequality and

ϕ L 4 ( 0 , T ; L ) C,

we conclude that

u L ( 0 , T ; H 3 ) C , θ L ( 0 , T ; H 2 ) + θ L 2 ( 0 , T ; H 3 ) C , ϕ L ( 0 , T ; H 3 ) + ϕ L 2 ( 0 , T ; H 5 ) C .

This completes the proof.

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Acknowledgements

This project is funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, under Grant No. 27-130-35-HiCi. The authors, therefore, acknowledge technical and financial support of KAU.

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Correspondence to Yong Zhou.

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Ma, C., Alzahrani, F.S., Hayat, T. et al. A regularity criterion for the Cahn-Hilliard-Boussinesq system with zero viscosity. J Inequal Appl 2014, 287 (2014). https://doi.org/10.1186/1029-242X-2014-287

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Keywords

  • regularity criterion
  • Cahn-Hilliard-Boussinesq
  • zero viscosity