Vanishing heat conductivity limit for the 2D Cahn-Hilliard-Boussinesq system

  • Zaihong Jiang1Email author and

    Affiliated with

    • Jishan Fan2

      Affiliated with

      Boundary Value Problems20112011:54

      DOI: 10.1186/1687-2770-2011-54

      Received: 18 October 2011

      Accepted: 22 December 2011

      Published: 22 December 2011

      Abstract

      This article studies the vanishing heat conductivity limit for the 2D Cahn-Hilliard-boussinesq system in a bounded domain with non-slip boundary condition. The result has been proved globally in time.

      2010 MSC: 35Q30; 76D03; 76D05; 76D07.

      Keywords

      Cahn-Hilliard-Boussinesq inviscid limit non-slip boundary condition

      1 Introduction

      Let Ω ⊆ ℝ2 be a bounded, simply connected domain with smooth boundary ∂Ω, and n is the unit outward normal vector to ∂Ω. We consider the following Cahn-Hilliard-Boussinesq system in Ω × (0, ∞) [1]:
      t u + ( u ) u + π - Δ u = μ ϕ + θ e 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ1_HTML.gif
      (1.1)
      div u = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ2_HTML.gif
      (1.2)
      t θ + u θ = ε Δ θ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ3_HTML.gif
      (1.3)
      t ϕ + u ϕ = Δ μ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ4_HTML.gif
      (1.4)
      - Δ ϕ + f ( ϕ ) = μ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ5_HTML.gif
      (1.5)
      u = 0 , θ = 0 , ϕ n = μ n = 0 o n Ω × ( 0 , ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ6_HTML.gif
      (1.6)
      ( u , θ , ϕ ) ( x , 0 ) = ( u 0 , θ 0 , ϕ 0 ) ( x ) , x Ω , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ7_HTML.gif
      (1.7)

      where u, π, θ and ϕ denote unknown velocity field, pressure scalar, temperature of the fluid and the order parameter, respectively. ε > 0 is the heat conductivity coefficient and e2 : = (0, 1) t . μ is a chemical potential and f ( ϕ ) : = 1 4 ( ϕ 2 - 1 ) 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_IEq1_HTML.gif is the double well potential.

      When ϕ = 0, (1.1), (1.2) and (1.3) is the well-known Boussinesq system. In [2] Zhou and Fan proved a regularity criterion ω = ˙ c u r l u L 1 ( 0 , T ; , 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_IEq2_HTML.gif for the 3D Boussinesq system with partial viscosity. Later, in [3] Zhou and Fan studied the Cauchy problem of certain Boussinesq-α equations in n dimensions with n = 2 or 3. We establish regularity for the solution under u L 1 ( 0 , T ; , 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_IEq3_HTML.gif. Here , 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_IEq4_HTML.gif denotes the homogeneous Besov space. Chae [4] studied the vanishing viscosity limit ε → 0 when Ω = ℝ2. The aim of this article is to prove a similar result. We will prove that

      Theorem 1.1. Let ( u 0 , θ 0 ) H 0 1 H 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_IEq5_HTML.gif, ϕ0H4, div u0 = 0 in Ω and ϕ 0 n = μ 0 n = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_IEq6_HTML.gif on∂Ω. Then, there exists a positive constant C independent of ε such that
      u ε L ( 0 , T ; H 2 ) C , θ ε L ( 0 , T ; H 2 ) C , ϕ ε L ( 0 , T ; H 4 ) C , t ( u ϵ , θ ε , ϕ ϵ ) L 2 ( 0 , T ; L 2 ) C , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ8_HTML.gif
      (1.8)
      for any T > 0, which implies
      ( u ε , θ ε , ϕ ε ) ( u , θ , ϕ ) s t r o n g l y i n L 2 ( 0 , T ; H 1 ) w h e n ε 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ9_HTML.gif
      (1.9)

      Here, (u, θ, ϕ) is the solution of the problem (1.1)-(1.7) with ε = 0.

      2 Proof of Theorem 1.1

      Since (1.9) follows easily from (1.8) by the Aubin-Lions compactness principle, we only need to prove the a priori estimates (1.8). From now on, we will drop the subscript ε and throughout this section C will be a constant independent of ε.

      First, by the maximum principle, it follows from (1.2), (1.3), and (1.6) that
      θ L ( 0 , T ; L ) θ 0 L C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ10_HTML.gif
      (2.1)
      Testing (1.3) by θ, using (1.2) and (1.6), we see that
      1 2 d d t θ 2 d x + ε θ 2 d x = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equa_HTML.gif
      whence
      ε θ L 2 ( 0 , T ; H 1 ) C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ11_HTML.gif
      (2.2)
      Testing (1.1) and (1.4) by u and μ, respectively, using (1.2), (1.6), (2.1), and summing up the result, we find that
      d d t 1 2 u 2 + 1 2 ϕ 2 + f ( ϕ ) d x + u 2 + μ 2 d x = θ e 2 u d x θ L 2 u L 2 C u L 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equb_HTML.gif
      which gives
      ϕ L ( 0 , T ; H 1 ) C , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ12_HTML.gif
      (2.3)
      u L ( 0 , T ; L 2 ) + u L 2 ( 0 , T ; H 1 ) C , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ13_HTML.gif
      (2.4)
      μ L 2 ( 0 , T ; L 2 ) C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ14_HTML.gif
      (2.5)
      Testing (1.4) by ϕ, using (1.2), (1.5) and (1.6), we infer that
      1 2 d d t ϕ 2 d x + Δ ϕ 2 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 1 2 Δ ϕ 2 d x + 1 2 ϕ 2 d x , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equc_HTML.gif
      which leads to
      ϕ L 2 ( 0 , T ; H 2 ) C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ15_HTML.gif
      (2.6)
      We will use the following Gagliardo-Nirenberg inequality:
      ϕ L 2 C ϕ L 6 ϕ H 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ16_HTML.gif
      (2.7)
      It follows from (2.6), (2.7), (2.5), (2.3) and (1.5) 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 ( 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 6 2 ϕ H 2 2 d t C + C ϕ L ( 0 , T ; H 1 ) 2 0 T ϕ H 2 2 d t C , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ17_HTML.gif
      (2.8)
      which yields
      ϕ L 2 ( 0 , T ; H 3 ) C , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ18_HTML.gif
      (2.9)
      ϕ L 4 ( 0 , T ; L ) C , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ19_HTML.gif
      (2.10)
      ϕ L 2 ( 0 , T ; L ) C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ20_HTML.gif
      (2.11)
      Testing (1.4) by Δ2ϕ, using (1.5), (2.4), (2.3), (2.10) and (2.11), we derive
      1 2 d d t Δ ϕ 2 d x + Δ 2 ϕ 2 d x = - u ϕ Δ 2 ϕ d x + Δ ( ϕ 3 - ϕ ) Δ 2 ϕ d x u L 2 ϕ L Δ 2 ϕ L 2 + Δ ( ϕ 3 - ϕ ) L 2 Δ 2 ϕ L 2 C ϕ L Δ 2 ϕ L 2 + C ( ϕ L 2 Δ ϕ L 2 + ϕ L ϕ L ϕ L 2 + Δ ϕ L 2 ) Δ 2 ϕ L 2 C ϕ L Δ 2 ϕ L 2 + C ( ϕ L 2 Δ ϕ L 2 + ϕ H 2 ϕ L + Δ ϕ L 2 ) Δ 2 ϕ L 2 1 2 Δ 2 ϕ L 2 2 + C ϕ L 2 + C ϕ L 4 Δ ϕ L 2 2 + C ϕ L 2 ϕ H 2 2 + C Δ ϕ L 2 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equd_HTML.gif
      which implies
      ϕ L ( 0 , T ; H 2 ) + ϕ L 2 ( 0 , T ; H 4 ) C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ21_HTML.gif
      (2.12)
      Testing (1.1) by -Δu + ∇π, using (1.2), (1.6), (2.12), (2.1) and (2.4), we reach
      1 2 d d t u 2 d x + ( - Δ u + π ) 2 d x = ( μ ϕ + θ e 2 - u u ) ( - Δ u + π ) d x ( μ L 2 ϕ L + θ L 2 + u L 4 u L 4 ) - Δ u + π L 2 C ( ϕ L + 1 + u L 2 1 2 u L 2 1 2 u L 2 1 2 Δ u L 2 1 2 ) - Δ u + π L 2 C ϕ L 2 + C + C u L 2 4 + 1 2 - Δ u + π L 2 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Eque_HTML.gif
      which yields
      u L ( 0 , T ; H 1 ) + u L 2 ( 0 , T ; H 2 ) C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ22_HTML.gif
      (2.13)
      Here, we have used the Gagliardo-Nirenberg inequalities:
      u L 4 2 C u L 2 u L 2 , u L 4 2 C u L 2 u H 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equf_HTML.gif
      and the H2-theory of the Stokes system:
      u H 2 + π H 1 C - Δ u + π L 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ23_HTML.gif
      (2.14)
      Similarly to (2.13), we have
      t u L 2 ( 0 , T ; L 2 ) C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ24_HTML.gif
      (2.15)
      (1.1), (1.2), (1.6) and (1.7) can be rewritten as
      t u - Δ u + π = g : = μ ϕ + θ e 2 - u u , i n Ω × ( 0 , ) , u = 0 , o n Ω × ( 0 , ) , u ( x , 0)  =  u 0 ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equg_HTML.gif
      Using (2.12), (2.1), (2.13), and the regularity theory of Stokes system, we have
      t u L 2 ( 0 , T ; L p ) + u L 2 ( 0 , T ; W 2 , p ) C g L 2 ( 0 , T ; L p ) C μ L 2 ( 0 , T ; L ) ϕ L ( 0 , T ; L p ) + C θ L ( 0 , T ; L ) + C u L ( 0 , T ; L 2 p ) u L 2 ( 0 , T ; L 2 p ) C , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ25_HTML.gif
      (2.16)

      for any 2 < p < ∞.

      (2.16) gives
      u L 2 ( 0 , T ; L ) C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ26_HTML.gif
      (2.17)
      It follows from (1.3) and (1.6) that
      Δ θ = 0 o n Ω × ( 0 , ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ27_HTML.gif
      (2.18)
      Applying Δ to (1.3), testing by Δθ, using (1.2), (1.6), (2.16), (2.17) and (2.18), we obtain
      1 2 d d t Δ θ 2 d x + ε Δ θ 2 d x = - ( Δ ( u θ ) - u Δ θ ) Δ θ d x C ( Δ u L 4 θ L 4 + u L Δ θ L 2 ) Δ θ L 2 C ( Δ u L 4 + u L ) Δ θ L 2 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equh_HTML.gif
      which implies
      θ L ( 0 , T ; H 2 ) + ε θ L 2 ( 0 , T ; H 3 ) C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ28_HTML.gif
      (2.19)
      It follows from (1.3), (1.6), (2.19) and (2.13) that
      t θ L ( 0 , T ; L 2 ) C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ29_HTML.gif
      (2.20)
      Taking ∂ t to (1.4) and (1.5), testing by ∂ t ϕ, using (1.2), (1.6), (2.12), and (2.15), we have
      1 2 d d t t ϕ 2 d x + Δ t ϕ 2 d x = - t u ϕ t ϕ d x + Δ ( 3 ϕ 2 t ϕ - t ϕ ) t ϕ d x = - t u ϕ t ϕ d x + ( 3 ϕ 2 t ϕ - t ϕ ) Δ t ϕ d x t u L 2 ϕ L t ϕ L 2 + ( 3 ϕ L 2 + 1 ) t ϕ L 2 Δ t ϕ L 2 t u L 2 ϕ L t ϕ L 2 + 1 2 Δ t ϕ L 2 2 + C t ϕ L 2 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equi_HTML.gif
      which gives
      t ϕ L ( 0 , T ; L 2 ) + t ϕ L 2 ( 0 , T ; H 2 ) C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ30_HTML.gif
      (2.21)
      By the regularity theory of elliptic equation, it follows from (1.4), (1.5), (1.6), (2.21), (2.13) and (2.12) that
      ϕ L ( 0 , T ; H 4 ) C Δ ϕ L ( 0 , T ; H 2 ) C μ - f ( ϕ ) L ( 0 , T ; H 2 ) C μ L ( 0 , T ; H 2 ) + C f ( ϕ ) L ( 0 , T ; H 2 ) C Δ μ L ( 0 , T ; L 2 ) + C f ( ϕ ) L ( 0 , T ; H 2 ) C t ϕ + u ϕ L ( 0 , T ; L 2 ) + C f ( ϕ ) L ( 0 , T ; H 2 ) C t ϕ L ( 0 , T ; L 2 ) + C u L ( 0 , T ; L 4 ) ϕ L ( 0 , T ; L 4 ) + C f ( ϕ ) L ( 0 , T ; H 2 ) C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ31_HTML.gif
      (2.22)
      Taking ∂ t to (1.1), testing by ∂ t u, using (1.2), (1.6), (2.17), (2.22), (2.21) and (1.5), we conclude that
      1 2 d d t t u 2 d x + t u 2 d x = - t u u t u d x + ( t μ ϕ + μ t ϕ + t θ e 2 ) t u d x u L t u L 2 2 + ( t u L 2 ϕ L + μ L t ϕ L 2 + t θ L 2 ) t u L 2 u L t u L 2 2 + C ( Δ t ϕ L 2 + t ( ϕ 3 - ϕ ) L 2 + t ϕ L 2 + 1 ) t u L 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equj_HTML.gif
      which implies
      t u L ( 0 , T ; L 2 ) + t u L 2 ( 0 , T ; H 1 ) C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equ32_HTML.gif
      (2.23)
      Using (2.23), (2.22), (2.1), (2.13), (1.1), (1.2), (1.6) and the H2-theory of the Stokes system, we arrive at
      u L ( 0 , T ; H 2 ) C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-54/MediaObjects/13661_2011_Article_99_Equk_HTML.gif

      This completes the proof.

      Declarations

      Acknowledgements

      This study was supported by the NSFC (No. 11171154) and NSFC (Grant No. 11101376).

      Authors’ Affiliations

      (1)
      Department of Mathematics, Zhejiang Normal University
      (2)
      Department of Applied Mathematics, Nanjing Forestry University

      References

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      2. Fan Jishan, Zhou Yong: A note on regularity criterion for the 3D Boussinesq system with partial viscosity. Appl Math Lett 2009, 22: 802-805. 10.1016/j.aml.2008.06.041View ArticleMathSciNetMATH
      3. Zhou Yong, Fan Jishan: On the Cauchy problems for certain Boussinesq- α equations. Proc R Soc Edinburgh Sect A 2010, 140: 319-327. 10.1017/S0308210509000122View ArticleMathSciNetMATH
      4. Chae Dongho: Global regularity for the 2D Boussinesq equations with partial viscosity terms. Adv Math 2006, 203: 497-513. 10.1016/j.aim.2005.05.001View ArticleMathSciNetMATH

      Copyright

      © Jiang and Fan; licensee Springer. 2011

      This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://​creativecommons.​org/​licenses/​by/​2.​0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.