_{1}

In this paper, we are interested in the local existence for the Boussinesq equations with the slip boundary conditions. Energy method and Gerlakin approach are employed in this paper to get the main result.

The Boussinesq equations are as follows:

{ u t + ( u ⋅ ∇ ) u + ∇ p = γ Δ u + θ f , θ t + ( u ⋅ ∇ ) θ = ε Δ θ , d i v u = 0 , ( x , t ) ∈ Ω × ( 0 , t ) (1.1)

where Ω is a bounded smooth domain of R 3 , u = ( u 1 , u 2 , u 3 ) and θ represent density and temperature, p is the pressure function, f is the external force, γ ≥ 0 , ε ≥ 0 represent viscous coefficient and thermal conductivity coefficient.

In this paper, the initial data is given by

( u , θ ) | t = 0 = ( u 0 , θ 0 ) (1.2)

and the boundary condition is

u ⋅ n | ∂ Ω = 0 , ∂ θ ∂ n | ∂ Ω = α θ , ( S u ⋅ n ) τ | ∂ Ω = ( B u ) τ (1.3)

Boussinesq equations are the classical model of fluid mechanics. There are a lot of important applications in marine ecology and weather forecasting. There are a lot of related conclusions about 2-D Boussinesq equations. Ye Zhuan [

However, for the initial questions and the boundary questions, the stations are more complicated and challenging. In fact, in fluid mechanics, two boundary conditions are considered mainly. One is the Diriclet boundary condition: u = 0 , x ∈ ∂ Ω , and another is the famous boundary condition proposed by Navier:

u ⋅ n = 0 , ( S u ⋅ n ) τ = − α u τ , x ∈ ∂ Ω (1.4)

where n is the unit outward normal on ∂ Ω , u τ is the tangential part of u , S u denotes the deformation tensor: S u = 1 2 ( ∇ u + ∇ u t ) .

Now, what we need to do is the study for the local existence of the problem (1.1) - (1.3). Usually, the followed existences are considered. First, the smooth of the existence is so small [

{ ρ t + d i v ( ρ u ) = 0 , ( ρ u j ) t + d i v ( ρ u j u ) + P ( ρ ) x j = μ Δ u j + λ d i v u x j + ρ f j , (1.5)

where d i v ( ⋅ ) denotes divergence.

Satisfying the bounded conditions:

{ u ⋅ υ = 0 , μ ∏ ∇ u υ = ∏ B u , x ∈ Ω , t > 0 (1.6)

where υ ( x ) is the unit outward normal on ∂ Ω , ∏ ( x ) is the projection onto the tangent plant to ∂ Ω at x, then the existence of the solutions can be given. T. Hmidi [

In this paper, we consider the Boussinesq equations. Because the temperature and the fluid are coupled together, the study becomes more difficult. The key we solve the problem is how to deal with the temperature.

First, we define a function

W = { w ∈ ( H 1 ( Ω ) ) 3 : w ( x ) ⋅ n ( x ) = 0 , x ∈ ∂ Ω } (1.7)

Definition 1.1. For a fixed time T , ( u , θ ) is called the solutions for (1.1) - (1.3) on [ 0 , T ] if the following holds:

1) ( u , θ ) ∈ C ( [ 0 , T ] ; H 1 ( Ω ) ) ∩ L 2 ( [ 0 , T ] ; W 2 , q ( Ω ) ) ,

( u t , θ t ) ∈ C ( [ 0 , T ] ; L 2 ( Ω ) ) ∩ L 2 ( [ 0 , T ] ; H 1 ( Ω ) ) .

2) For any times t 1 , t 2 ∈ [ 0 , T ] ,

∫ t 1 t 2 ∫ Ω ( u t + ( u ⋅ ∇ ) u + ∇ p ) φ d x d t = ∫ t 1 t 2 ∫ Ω ( γ Δ u + θ f ) φ d x d t ,

∫ t 1 t 2 ∫ Ω ( θ t + ( u ⋅ ∇ ) u ) ϕ d x d t = ∫ t 1 t 2 ∫ Ω ( ε Δ θ ) ϕ d x d t ,

for φ , ϕ Lipschitz Ω × [ 0 , T ] with ϕ , φ ∈ W .

Remark: the system parameters Ω and B will be assumed to satisfy the following conditions:

1) Ω is a bounded open set in R 3 with a C 3 boundary.

2) B is a C 1 in a neighborhood of ∂ Ω .

Theorem 1.2. Assume the hypotheses hold, let q ∈ ( 3 , 6 ] be fixed constant and the ( u 0 , θ 0 ) , p and f satisfy the following conditions:

( p , f ) ∈ C ( [ 0 , T ] ; L 2 ( Ω ) ) ∩ L 2 ( [ 0 , T ] ; L q ( Ω ) ) ;

( p t , f t ) ∈ L 2 ( [ 0 , T ] ; L 2 ( Ω ) ) ;

( S u ⋅ n ) τ | ∂ Ω = ( B u ) τ .

Then there is a small T ∗ > 0 and a unique strong solution ( u , θ ) to the initial boundary value problems (1.2), (1.3) such that:

( u , θ ) ∈ C ( [ 0 , T * ] ; H 1 ( Ω ) ) ∩ L 2 ( [ 0 , T * ] ; W 2 , q ( Ω ) ) ;

( u t , θ t ) ∈ C ( [ 0 , T * ] ; L 2 ( Ω ) ) ∩ L 2 ( [ 0 , T * ] ; H 1 ( Ω ) ) .

In this part, we introduce the regularity of Laplace and the − Δ . First we assume the condition (1.2) holds, Ω and B satisfy the above hypotheses. Considering the following problem: find u : Ω ¯ → R 3 such that:

{ L u + β u = g , x ∈ Ω u ⋅ n = 0 , ( S ( u ) ⋅ n ) τ = ( B u ) τ , x ∈ ∂ Ω (2.1)

here L u = − γ Δ u , fixed β ∈ R , given g : Ω → R . Then the corresponding weak form is achieved.

Use ω ∈ W to multiply by the above differential Equation (2.1), we can get:

A β ( u , ω ) = ∫ Ω g ⋅ ω d x (2.2)

where A β : W × W → R is the bilinear form:

A β ( u , ω ) = γ ( − ∫ ∂ Ω A u ⋅ ω d s + ∫ Ω S u : S ω d x ) (2.3)

Obviously, A β is continuous on W × W if B is bounded, and we can use trace theorem to show that A β is coercive if β is enough large depending on γ , Ω , B . In this case, there is a bounded operator S β : L 2 → W satisfying

A β ( S β g , ω ) = ∫ Ω g ⋅ ω d x (2.4)

for all ω ∈ W . Furthermore, because the embedding W → L 2 is compact, S β is a compact operator from L 2 to L 2 and the symmetry condition guarantees S β is self-adjoint.

The following lemmas are taken from document [

Lemma 2.1. Assume that Ω ⊂ R 3 is abounded open set with a C 3 boundary and that B ∈ L ∞ ( ∂ Ω ) is a symmetric matrix. For β enough large, there is a compact self-adjoint operator S β : L 2 ( Ω ) → L 2 ( Ω ) , whose range is contained in W and for which (2.2) holds for g ∈ L 2 ( Ω ) and ω ∈ W . At the same time, there is an orthogonal basis { ω k } k for L 2 whose elements are in W ∩ C ∞ ( Ω ) , and which are eigenfunctions of S β

S β ω k = γ k − 1 ω k

where γ k → ∞ .

Lemma 2.2. Assume that m ≥ 0 , Ω ⊂ R 3 is abounded open set with a C m + 1 boundary and that B is a C m + 1 ( ∂ Ω ) mapping from a neighborhood of ∂ Ω into the set of 3 × 3 matrices, then there exists a constant C m = C ( γ , Ω , B ) such that if u is a solution of (2.1) in the sense of (2.2) where β and g ∈ H m ( Ω ) , then u ∈ H m + 2 ( Ω ) and

‖ u ‖ H m + 2 ( Ω ) ≤ C m ( ‖ u ‖ L 2 ( Ω ) + ‖ g ‖ H m ( Ω ) ) (2.5)

For the operator − Δ , consider the following problem: find θ : Ω ¯ → R 3 such that

{ − Δ θ + δ θ = g ¯ , x ∈ Ω ∂ θ ∂ n | ∂ Ω = α θ (2.6)

where δ ∈ R is fixed, g ¯ : Ω → R is given.

Use ϕ ∈ H 1 ( Ω ) to multiply by the above differential equation, and integration by parts, then we get

B δ ( θ , ϕ ) = ∫ Ω g ¯ ⋅ ϕ d x (2.7)

where B δ : H 1 × H 1 → R is the bilinear form:

B δ ( θ , ϕ ) = ∫ Ω ∇ θ ⋅ ∇ φ + δ θ φ d x (2.8)

Similar to the Lame operator, there exist a bounded operator Γ δ : L 2 ( Ω ) → H 1 meeting

B δ ( Γ δ g , ϕ ) = ∫ Ω g ¯ ⋅ ϕ d x (2.9)

for all φ ∈ H 1 and for δ large enough. Parallel to Lemmas 2.1 - 2.2, we can get the following lemmas.

Lemma 2.3. Assume that Ω ⊂ R 3 is abounded open set with a C 3 boundary. For γ large enough, there exist a compact self-adjoint operator Γ γ : L 2 ( Ω ) → L 2 ( Ω ) , whose range is contained in H 1 and for which (2.9) holds for g ¯ ∈ L 2 ( Ω ) and ϕ ∈ H 1 . At the same time, there is an orthogonal basis { ϕ k } k for L 2 whose elements are in H 1 ∩ C ∞ ( Ω ) and which are eigenfunctions of Γ γ , Γ γ ϕ k = γ k − 1 ¯ ω k , where γ k ¯ → ∞ .

Lemma 2.4. Assume that m ≥ 0 , Ω ⊂ R 3 is abounded open set with a C m + 1 boundary and that B is a C m + 1 ( ∂ Ω ) mapping from a neighborhood of ∂ Ω into the set of 3 × 3 matrices, then there exists a constant C m ¯ = C ¯ ( Ω ) such that if θ is a solution of (2.6) in the sense of (2.7) where g ¯ ∈ H m ( Ω ) , then θ ∈ H m + 2 ( Ω ) and

‖ θ ‖ H m + 2 ( Ω ) ≤ C m ¯ ( ‖ θ ‖ L 2 ( Ω ) + ‖ g ¯ ‖ H m ( Ω ) ) (2.10)

In this part, we need the following prior estimates to prove the local existence of the solution. Assume that the following inequalities hold:

‖ ( u 0 , θ 0 ) H 2 ‖ + 1 ≤ C 0 (3.1)

sup 0 ≤ t ≤ T ∗ ( ‖ u ( t ) ‖ H 2 + ‖ u t ( t ) ‖ L 2 ) + ∫ 0 T ∗ ( ‖ ∇ u t ‖ L 2 2 + ‖ ∇ 2 u ‖ L q 2 ) d t + 1 ≤ C 1 (3.2)

where 1 ≤ C 0 ≤ C 1 , 0 < T ∗ ≤ T .

Remark: C is a constant if be not added.

Lemma 3.1.1. ‖ θ ‖ L 2 2 + ∫ 0 t ‖ ∇ θ ‖ L 2 2 d s ≤ C

Proof: Multiplying the second equation of (1.1) by θ and integrating over Ω , one has

1 2 d d t ∫ Ω θ 2 d x = ∫ Ω ε Δ θ ⋅ θ − ( u ⋅ ∇ ) θ ⋅ θ d x (3.1.1)

∫ Ω ε Δ θ ⋅ θ d x ≤ C ( ∫ ∂ Ω ∂ θ ∂ n θ d s − ∫ Ω ( ∇ θ ) 2 d x ) = C ( ∫ ∂ Ω α θ 2 d s − ∫ Ω ( ∇ θ ) 2 d x ) ≤ C ∫ Ω ( θ 2 + | θ | ⋅ | ∇ θ | ) d x + C ∫ Ω ( ∇ θ ) 2 d x ≤ C ( ‖ θ ‖ L 2 + ‖ ∇ θ ‖ L 2 + C δ ‖ θ ‖ L 2 + δ ‖ ∇ θ ‖ L 2 ) (3.1.2)

∫ Ω ( u ⋅ ∇ ) θ ⋅ θ d x = ∫ Ω u ⋅ ∇ θ ⋅ θ d x ≤ ‖ u ‖ L 3 ‖ θ ‖ H 1 ‖ ∇ θ ‖ L 2 ≤ ‖ u ‖ L 2 1 2 ‖ u ‖ H 1 1 2 ( ‖ θ ‖ L 2 + ‖ ∇ θ ‖ L 2 ) ‖ ∇ θ ‖ L 2 ≤ C ( ‖ θ ‖ L 2 ‖ ∇ θ ‖ L 2 + ‖ ∇ θ ‖ L 2 2 ) ≤ C ( δ ‖ ∇ θ ‖ L 2 2 + C δ ‖ θ ‖ L 2 2 + ‖ ∇ θ ‖ L 2 2 ) (3.1.3)

Substituting (3.1.2) - (3.1.3) into (3.1.1), letting δ small enough and using Gronwall’s inequality:

‖ θ ‖ L 2 2 + ∫ 0 t ‖ ∇ θ ‖ L 2 2 d s ≤ C

Lemma 3.1.2. ‖ ∇ θ ‖ L 2 2 + ∫ 0 t ‖ θ t ‖ L 2 2 d s ≤ C

Proof: Multiplying the second equation of (1.1) by θ t and integrating over Ω , one has

∫ Ω θ t 2 d x = ∫ Ω ε Δ θ ⋅ θ t − ( u ⋅ ∇ ) θ ⋅ θ t d x (3.1.4)

∫ Ω ε Δ θ ⋅ θ t d x = ∫ ∂ Ω ε ∂ θ ∂ n θ t d s − ∫ Ω ε ∇ θ ⋅ ∇ θ t d x = 1 2 d d t ∫ ∂ Ω ε α θ 2 d s − 1 2 d d t ∫ Ω ε ( ∇ θ ) 2 d x ≤ C d d t ∫ Ω ( | θ | 2 + | θ | ⋅ | ∇ θ | ) + ( ∇ θ ) 2 d x ≤ C d d t ( C δ ‖ ∇ θ ‖ L 2 2 + δ ‖ θ ‖ L 2 2 + ‖ ∇ θ ‖ L 2 2 ) (3.1.5)

∫ Ω ( u ⋅ ∇ ) θ ⋅ θ t d x = ∫ Ω u ⋅ ∇ θ ⋅ θ t d x ≤ ‖ u ‖ L ∞ ‖ θ t ‖ L 2 ‖ ∇ θ ‖ L 2 ≤ C ( δ ‖ θ t ‖ L 2 2 + C δ ‖ ∇ θ ‖ L 2 2 ) (3.1.6)

Substituting (3.1.5) - (3.1.6) into (3.1.4), letting δ small enough and using Gronwall’s inequality,

‖ ∇ θ ‖ L 2 2 + ∫ 0 t ‖ θ t ‖ L 2 2 d s ≤ C

Lemma 3.1.3. ‖ θ t ‖ L 2 2 + ∫ 0 t ‖ ∇ θ t ‖ L 2 2 d s ≤ C

Proof: Differentiating the second equation of (1.1) with respect to t , multiplying the second equation of (1.1) by θ t and integrating over Ω , one has

1 2 d d t ∫ Ω θ t 2 d x = ∫ Ω ε Δ θ t ⋅ θ t − u t ⋅ ∇ θ ⋅ θ t − u ⋅ ∇ θ t ⋅ θ t d x (3.1.7)

∫ Ω ε Δ θ t ⋅ θ t d x = ∫ ∂ Ω ε ∂ θ t ∂ n θ t d s − ∫ Ω ε ( ∇ θ t ) 2 d x ≤ C ( ∫ ∂ Ω θ t 2 d s − ∫ Ω ( ∇ θ t ) 2 d x ) ≤ C ( ∫ Ω θ t 2 + ∇ θ t ⋅ θ t d x + ‖ ∇ θ t ‖ L 2 2 ) ≤ C ( ‖ θ t ‖ L 2 2 + ‖ ∇ θ t ‖ L 2 ‖ θ t ‖ L 2 + ‖ ∇ θ t ‖ L 2 2 ) ≤ C ( ‖ θ t ‖ L 2 2 + C δ ‖ ∇ θ t ‖ L 2 2 + δ ‖ θ t ‖ L 2 2 + ‖ ∇ θ t ‖ L 2 2 ) (3.1.8)

∫ Ω u t ⋅ ∇ θ ⋅ θ t d x ≤ ‖ u t ‖ L 3 ‖ ∇ θ ‖ L 2 ‖ θ t ‖ H 1 ≤ C ‖ u t ‖ L 2 1 2 ‖ θ ‖ L 6 1 2 ‖ θ t ‖ H 1 ≤ C ( C δ ‖ θ t ‖ L 2 2 + δ ‖ u t ‖ L 2 ‖ θ ‖ H 1 + C κ ‖ ∇ θ t ‖ L 2 2 + κ ‖ u t ‖ L 2 ‖ θ ‖ H 1 ) (3.1.9)

∫ Ω u ⋅ ∇ θ t ⋅ θ t d x ≤ ‖ u ‖ L ∞ ‖ θ t ‖ L 2 ‖ ∇ θ t ‖ L 2 ≤ C δ ‖ θ t ‖ L 2 2 + δ ‖ ∇ θ t ‖ L 2 2 (3.1.10)

Substituting (3.1.8) - (3.1.10) into (3.1.7), letting δ and κ small enough and using Gronwall’s inequality,

‖ θ t ‖ L 2 2 + ∫ 0 t ‖ ∇ θ t ‖ L 2 2 d s ≤ C

Lemma 3.2.1. ‖ u ‖ L 2 2 + ∫ 0 t ‖ ∇ u ‖ L 2 2 d s ≤ C

Proof: Multiplying the first equation of (1.1) by u and integrating over Ω , one has

1 2 d d t ∫ Ω u 2 d x = ∫ Ω γ Δ u ⋅ u + u 2 ⋅ ∇ u − ∇ p ⋅ u + θ f ⋅ u d x (3.2.1)

∫ Ω γ Δ u ⋅ u d x = ∫ Ω 2 γ d i v ( S u ) ⋅ u d x = 2 γ ( ∫ ∂ Ω S u ⋅ n ⋅ u d s − ∫ Ω S u ⋅ ∇ u d x ) ≤ C ( ∫ ∂ Ω B u ⋅ u d s + ∫ Ω S u ⋅ ∇ u d x ) ≤ C ( ‖ u ‖ L 2 2 + ‖ ∇ u ‖ L 2 2 ) (3.2.2)

∫ Ω ( u ⋅ ∇ ) u ⋅ u d x = ∫ Ω u 2 ⋅ ∇ u d x ≤ ‖ u ‖ L 3 ‖ u ‖ H 1 ‖ ∇ u ‖ L 2 ≤ ‖ u ‖ L 2 1 2 ‖ u ‖ H 1 1 2 ( ‖ u ‖ L 2 + ‖ ∇ u ‖ L 2 ) ‖ ∇ u ‖ L 2 ≤ C ( ‖ u ‖ L 2 ‖ ∇ u ‖ L 2 + ‖ ∇ u ‖ L 2 2 ) ≤ C ( δ ‖ ∇ u ‖ L 2 2 + C δ ‖ u ‖ L 2 2 + ‖ ∇ u ‖ L 2 2 ) (3.2.3)

∫ Ω ∇ p ⋅ u d x = ‖ ∇ p ‖ L 2 ‖ u ‖ L 2 ≤ C δ ‖ u ‖ L 2 2 + δ ‖ ∇ p ‖ L 2 2 (3.2.4)

∫ Ω θ f ⋅ u d x = ‖ θ ‖ L 2 ‖ f ‖ L ∞ ‖ u ‖ L 2 ≤ C ( C δ ‖ u ‖ L 2 2 + δ ‖ θ ‖ L 2 2 ) (3.2.5)

Substituting (3.2.2) - (3.2.5) into (3.2.1), letting δ small enough and using Gronwall’s inequality,

‖ u ‖ L 2 2 + ∫ 0 t ‖ ∇ u ‖ L 2 2 d s ≤ C

Lemma 3.2.2. ‖ ∇ u ‖ L 2 2 + ∫ 0 t ‖ u t ‖ L 2 2 d s ≤ C

Proof: Multiplying the first equation of (1.1) by u t and integrating over Ω , one has

∫ Ω u t 2 d x = ∫ Ω γ Δ u ⋅ u t + u ⋅ ∇ u ⋅ u t − ∇ p ⋅ u t + θ f ⋅ u t d x (3.2.6)

∫ Ω γ Δ u ⋅ u t d x = ∫ Ω 2 γ d i v ( S u ) ⋅ u t d x = 2 γ ( ∫ ∂ Ω S u ⋅ n ⋅ u t d s − ∫ Ω S u ⋅ ∇ u t d x ) ≤ C ( ∫ ∂ Ω B u ⋅ u t d s + ∫ Ω S u ⋅ ∇ u t d x ) ≤ C d d t ( ‖ u ‖ L 2 2 + ‖ ∇ u ‖ L 2 2 ) (3.2.7)

∫ Ω ( u ⋅ ∇ ) u ⋅ u t d x = ∫ Ω u ⋅ ∇ u ⋅ u t d x ≤ ‖ u ‖ L ∞ ‖ u t ‖ L 2 ‖ ∇ u ‖ L 2 ≤ C ( C δ ‖ u t ‖ L 2 2 + δ ‖ ∇ u ‖ L 2 2 ) (3.2.8)

∫ Ω ∇ p ⋅ u t d x = ‖ ∇ p ‖ L 2 ‖ u t ‖ L 2 ≤ C δ ‖ u t ‖ L 2 2 + δ ‖ ∇ p ‖ L 2 2 (3.2.9)

∫ Ω θ f ⋅ u t d x = ‖ θ ‖ L 2 ‖ f ‖ L ∞ ‖ u t ‖ L 2 ≤ C ( C δ ‖ u t ‖ L 2 2 + δ ‖ θ ‖ L 2 2 ) (3.2.10)

Substituting (3.2.7) - (3.2.10) into (3.2.6), letting δ small enough and using Gronwall’s inequality,

‖ ∇ u ‖ L 2 2 + ∫ 0 t ‖ u t ‖ L 2 2 d s ≤ C

Lemma 3.2.3. ‖ u t ‖ L 2 2 + ∫ 0 t ‖ ∇ u t ‖ L 2 2 d s ≤ C

Proof: Differentiating the first equation of (1.1) with respect to t , multiplying the first equation of (1.1) by u t and integrating over Ω , one has

∫ Ω u t t u t d x = ∫ Ω γ Δ u t ⋅ u t + u t ⋅ ∇ u ⋅ u t + u ⋅ ∇ u t ⋅ u t − ∇ p t ⋅ u t + θ t f ⋅ u t + θ f t ⋅ u t d x (3.2.11)

∫ Ω γ Δ u t ⋅ u t d x = ∫ Ω 2 γ d i v ( S u t ) ⋅ u t d x = 2 γ ( ∫ ∂ Ω S u t ⋅ n ⋅ u t d s − ∫ Ω S u t ⋅ ∇ u t d x ) ≤ C ( ∫ ∂ Ω B u t ⋅ u t d s + ∫ Ω S u t ⋅ ∇ u t d x ) ≤ C ( ‖ u t ‖ L 2 2 + ‖ ∇ u t ‖ L 2 2 ) (3.2.12)

∫ Ω u t ⋅ ∇ u ⋅ u t d x ≤ ‖ u t ‖ L 3 ‖ ∇ u ‖ L 2 ‖ u t ‖ H 1 ≤ C ‖ u t ‖ L 2 1 2 ‖ u t ‖ L 6 1 2 ‖ u t ‖ H 1 ≤ C ( C δ ‖ u t ‖ L 2 2 + δ ‖ u t ‖ L 2 ‖ u t ‖ H 1 + C κ ‖ ∇ u t ‖ L 2 2 + κ ‖ u t ‖ L 2 ‖ u t ‖ H 1 ) (3.2.13)

∫ Ω u ⋅ ∇ u t ⋅ u t d x ≤ ‖ u ‖ L ∞ ‖ u t ‖ L 2 ‖ ∇ u t ‖ L 2 ≤ C δ ‖ u t ‖ L 2 2 + δ ‖ ∇ u t ‖ L 2 2 (3.2.14)

∫ Ω ∇ p t ⋅ u t d x = ‖ ∇ p t ‖ L 2 ‖ u t ‖ L 2 ≤ C δ ‖ u t ‖ L 2 2 + δ ‖ ∇ p t ‖ L 2 2 (3.2.15)

∫ Ω θ t f ⋅ u t d x = ‖ θ t ‖ L 2 ‖ f ‖ L ∞ ‖ u t ‖ L 2 ≤ C ( C δ ‖ u t ‖ L 2 2 + δ ‖ θ t ‖ L 2 2 ) (3.2.16)

∫ Ω θ f t ⋅ u t d x = ‖ θ ‖ L 2 ‖ f t ‖ L ∞ ‖ u t ‖ L 2 ≤ C ( C δ ‖ u t ‖ L 2 2 + δ ‖ θ ‖ L 2 2 ) (3.2.17)

Substituting (3.2.12)-(3.2.17) into (3.2.11), letting δ and κ small enough and using Gronwall’s inequality,

‖ u t ‖ L 2 2 + ∫ 0 t ‖ ∇ u t ‖ L 2 2 d s ≤ C

Lemma 3.3.1. ‖ ∇ θ ‖ H 1 ≤ C

Proof: According to the second equation of (1.1), we have − Δ θ = − θ t ε − ( u ⋅ ∇ ) θ ε , then use the elliptic regularity:

‖ θ ‖ H 2 ≤ C ( ‖ − θ t − ( u ⋅ ∇ ) θ ‖ L 2 + ‖ θ ‖ L 2 ) ≤ C ( ‖ θ t ‖ L 2 + ‖ ∇ θ ⋅ u ‖ L 2 + ‖ θ ‖ L 2 )

‖ ∇ θ ‖ H 1 ≤ ‖ θ ‖ H 2 ≤ C ( ‖ θ t ‖ L 2 + ‖ ∇ θ ⋅ u ‖ L 2 + ‖ θ ‖ L 2 ) ≤ C ( ‖ θ t ‖ L 2 + ‖ u ‖ L ∞ ‖ ∇ θ ‖ L 2 + ‖ θ ‖ L 2 ) ≤ C

Lemma 3.3.2. ∫ 0 t ‖ ∇ θ ‖ W 1 , q 2 d s ≤ C

Proof: According to the second equation of (1.1), we have − Δ θ = − θ t ε − ( u ⋅ ∇ ) θ ε , then use the elliptic regularity:

‖ θ ‖ W 2 , q ≤ C ( ‖ − θ t − ( u ⋅ ∇ ) θ ‖ L q + ‖ θ ‖ L q ) ≤ C ( ‖ θ t ‖ L q + ‖ ∇ θ ⋅ u ‖ L q + ‖ θ ‖ L q )

Both sides of the above inequality multiply by itself, then

‖ θ ‖ W 2 , q 2 ≤ C ( ‖ θ t ‖ L q 2 + ‖ ∇ θ ⋅ u ‖ L q 2 + ‖ θ ‖ L q 2 )

‖ ∇ θ ‖ W 1 , q 2 ≤ ‖ θ ‖ W 2 , q 2 ≤ C ( ‖ θ t ‖ L q 2 + ‖ ∇ θ ⋅ u ‖ L q 2 + ‖ θ ‖ L q 2 ) ≤ C ( ‖ θ t ‖ H 1 2 + ‖ u ‖ L ∞ 2 ‖ ∇ θ ‖ H 1 2 + ‖ θ ‖ H 1 2 )

And integrating over ( 0 , t ) , we have

∫ 0 t ‖ ∇ θ ‖ W 1 , q 2 d s ≤ ∫ 0 t C ( ‖ θ t ‖ H 1 2 + ‖ u ‖ L ∞ 2 ‖ ∇ θ ‖ H 1 2 + ‖ θ ‖ H 1 2 ) d s ≤ C

Lemma 3.3.3. ‖ ∇ u ‖ H 1 ≤ C

Proof: According to the first equation of (1.1), we have − Δ u = − u t γ − ( u ⋅ ∇ ) u γ − ∇ p γ + θ f γ , then use the elliptic regularity:

‖ u ‖ H 2 ≤ C ( ‖ − u t − ( u ⋅ ∇ ) u − ∇ p + θ f ‖ L 2 + ‖ u ‖ L 2 ) ≤ C ( ‖ u t ‖ L 2 + ‖ ( u ⋅ ∇ ) u ‖ L 2 + ‖ ∇ p ‖ L 2 + ‖ θ f ‖ L 2 + ‖ u ‖ L 2 )

‖ ∇ u ‖ | | H 1 ≤ ‖ u ‖ H 2 ≤ C ( ‖ u t ‖ L 2 + ‖ ( u ⋅ ∇ ) u ‖ L 2 + ‖ ∇ p ‖ L 2 + ‖ θ f ‖ L 2 + ‖ u ‖ L 2 ) ≤ C ( ‖ u t ‖ L 2 + ‖ u ‖ L 3 ‖ ∇ u ‖ H 1 + ‖ ∇ p ‖ L 2 + || f || L ∞ ‖ θ ‖ L 2 + ‖ u ‖ L 2 ) ≤ C ( ‖ u t ‖ L 2 + ‖ u ‖ L 3 1 2 ‖ u ‖ H 1 1 2 ‖ ∇ u ‖ H 1 + ‖ ∇ p ‖ L 2 + ‖ f ‖ L ∞ ‖ θ ‖ L 2 + ‖ u ‖ L 2 ) ≤ C

Lemma 3.3.4. ∫ 0 t ‖ ∇ u ‖ W 1 , q 2 d s ≤ C

Proof: According to the first equation of (1.1), we have − Δ u = − u t γ − ( u ⋅ ∇ ) u γ − ∇ p γ + θ f γ , then use the elliptic regularity:

‖ u ‖ W 2 , q ≤ C ( ‖ − u t − ( u ⋅ ∇ ) u − ∇ p + θ f ‖ L q + ‖ u ‖ L q ) ≤ C ( ‖ u t ‖ L q + ‖ ( u ⋅ ∇ ) u ‖ L q + ‖ ∇ p ‖ L q + ‖ θ f ‖ L q + ‖ u ‖ L q )

Both sides of the above inequality multiply by itself, then

‖ u ‖ W 2 , q 2 ≤ C ( ‖ u t ‖ L q 2 + ‖ ( u ⋅ ∇ ) u ‖ L q 2 + ‖ ∇ p ‖ L q 2 + ‖ θ f ‖ L q 2 + ‖ u ‖ L q 2 )

‖ ∇ u ‖ W 1 , q 2 ≤ ‖ u ‖ W 2 , q 2 ≤ C ( ‖ u t ‖ L q 2 + ‖ ( u ⋅ ∇ ) u ‖ L q 2 + ‖ ∇ p ‖ L q 2 + ‖ θ f ‖ L q 2 + ‖ u ‖ L q 2 ) ≤ C ( ‖ u t ‖ H 1 2 + ‖ u ‖ L ∞ 2 ‖ ∇ u ‖ H 1 2 + ‖ ∇ p ‖ L q 2 + ‖ f ‖ L ∞ 2 ‖ θ ‖ H 1 2 + ‖ u ‖ H 1 2 )

And integrating over ( 0 , t ) , we have ∫ 0 t ‖ ∇ u ‖ W 1 , q 2 d s ≤ C .

First, we consider the following linearized system:

{ u t + ( v ⋅ ∇ ) u + ∇ p = γ Δ u + θ f , θ t + ( u ⋅ ∇ ) θ = ε Δ θ , d i v u = 0 , ( x , t ) ∈ Ω × ( 0 , t ) (4.1)

Lemma 4.1. Let Ω be a bounded domain in R 3 with smooth boundary, when 3 < q ≤ 6 , we have ( u 0 , θ 0 ) ∈ H 1 ( Ω ) . Assume that

v ∈ L ∞ ( [ 0 , T ] ; H 2 ( Ω ) ) ∩ L 2 ( [ 0 , T ] ; W 2 , q ( Ω ) ) ,

v t ∈ L ∞ ( [ 0 , T ] ; L 2 ) ∩ L 2 ( [ 0 , T ] ; H 1 )

with the boundary conditions:

v ⋅ n | ∂ Ω = 0 , ( S v ⋅ n ) τ | ∂ Ω = ( B v ) τ

Then there is a unique strong solution ( u , θ ) meeting (1.1) - (1.3) such tat

( u , θ ) ∈ C ( [ 0 , T ] ; H 2 ( Ω ) ) ∩ L 2 ( [ 0 , T ] ; W 2 , q ( Ω ) ) ;

( u t , θ t ) ∈ C ( [ 0 , T ] ; L 2 ( Ω ) ) ∩ L 2 ( [ 0 , T ] ; H 1 ( Ω ) ) (4.2)

Proof: It follows from Theorem 4 in chapter 5.9 [

Next, Gerlakin approach is applied to prove the local existence of the solution of Equation (4.1).

Assume that { ϕ l } 1 m and { ω l } 1 m respectively representing the eigenvectors of the operator − Δ and the operator L are smooth functions.

V m = s p a n { ϕ 1 , ϕ 2 , ⋯ , ϕ m } , W m = s p a n { ω 1 , ω 2 , ⋯ , ω m }

for a positive constant m fixed, let

θ m ( t ) = ∑ l = 1 m a m l ( t ) ϕ l , u m ( t ) = ∑ l = 1 m b m l ( t ) ω l (4.3)

We hope that the coefficients a m l ( t ) , b m l ( t ) satisfy:

{ ∫ Ω u t m ω m + v m ∇ u m u t m ω m + ∇ p m u t m ω m d x = ∫ Ω γ Δ u m u t m ω m + θ m f m u t m ω m d x ∫ Ω θ t m ϕ m + u m ∇ θ m ϕ m d x = ∫ Ω ε Δ θ m ϕ m d x (4.4)

Thus we seek functions ( u m , θ m ) that satisfy the “projection” (4.4) of problem (4.1) onto the finite dimensional subspace spanned by ( { ω l } 1 m , { ϕ l } 1 m ) .

It follows from Theorem 1 in chapter 7 [

Similar to the prior estimates in part 3, we have

sup 0 ≤ t ≤ T ∗ ( ‖ u m ( t ) ‖ H 2 + ‖ u t m ( t ) ‖ L 2 ) + ∫ 0 T ∗ ( ‖ ∇ u t m ‖ L 2 2 + ‖ ∇ 2 u m ‖ L q 2 ) d t + 1 ≤ C 1

where C has no connection with m , then we have

( u m , θ m ) → ( u , θ ) in L ∞ ( 0 , T ; H 1 )

( u m , θ m ) → weakly ( u , θ ) in L 2 ( 0 , T ; W 2 , q )

( u m , θ m ) → weakly * ( u , θ ) in L ∞ ( 0 , T ; H 2 )

( ∂ t u m , ∂ t θ m ) → weak ( u t , θ t ) in L 2 ( 0 , T ; H 1 )

It follows from Theorem 3 in chapter 7 [

The proof of Lemma 4.1 is completed.

Next, the iteration method is used to prove the local existence of the solution of Boussinesq equations.

Construct approximate solutions of Boussinesq equations that meet the initial and boundary problems (1.2) - (1.3).

1) define u 0 = 0 ,

2) assume that k ≥ 1 , define v = u k − 1 ,

{ u t k + ( u k − 1 ⋅ ∇ ) u k + ∇ p k = γ Δ u k + θ k f k , θ t k + ( u k ⋅ ∇ ) θ k = ε Δ θ k , d i v u k = 0 , ( x , t ) ∈ Ω × ( 0 , t ) (5.1)

Initial conditions:

( u k , θ k ) | t = 0 = ( u 0 k , θ 0 k ) (5.2)

Boundary conditions:

u k ⋅ n | ∂ Ω = 0 , ∂ θ k ∂ n | ∂ Ω = α θ k , ( S u k ⋅ n ) τ | ∂ Ω = ( B u k ) τ (5.3)

According to Lemma 4.1, we can know that the problems (5.1) - (5.3) exist the local solutions ( u k , θ k ) . Furthermore, according to the prior estimates, we get

sup 0 ≤ t ≤ T ∗ ( ‖ u k ‖ H 2 + ‖ u t k ‖ L 2 ) + ∫ 0 T ∗ ( ‖ ∇ u t k ‖ L 2 2 + ‖ ∇ 2 u k ‖ L q 2 ) d t + 1 ≤ C 1 (5.4)

where C has no connection with k .

According to Aubin-Lions lemma, one has

( u k , θ k ) → ( u , θ ) in L ∞ ( 0 , T * ; H 1 )

( u , θ ) ∈ L ∞ ( [ 0 , T ∗ ] ; H 2 ( Ω ) ) ∩ L 2 ( [ 0 , T ∗ ] ; W 2 , q ( Ω ) )

( u t , θ t ) ∈ L ∞ ( [ 0 , T ∗ ] ; L 2 ( Ω ) ) ∩ L 2 ( [ 0 , T ∗ ] ; H 1 ( Ω ) )

Last, we show the continuity of u and θ over time.

‖ ( u t ) t ‖ W * = sup 0 ≤ t ≤ T * | 〈 ( u t ) t , ψ 〉 | ≤ sup 0 ≤ t ≤ T * ∫ Ω | γ Δ u t + u t ⋅ ∇ u + u ⋅ ∇ u t + ∇ p t + θ t f + θ f t | | ψ | d x ≤ sup 0 ≤ t ≤ T * C ( ( ‖ u t ‖ L 2 ‖ ∇ u ‖ L 3 + ‖ ∇ u t ‖ L 2 ‖ u ‖ L 3 + ‖ θ t ‖ L 2 ‖ f ‖ L 3 + ‖ θ ‖ L 3 ‖ f t ‖ L 2 ) ‖ ψ ‖ H 1 + ‖ ∇ p t ‖ L 2 ‖ ψ ‖ L 2 + ‖ ∇ u t ‖ L 2 ‖ ∇ ψ ‖ L 2 ) + sup 0 ≤ t ≤ T * ∫ ∂ Ω ∇ u t ψ d s

where W * denotes the dual space of W . It follows from Theorem 3 in chapter 5.9 [

‖ ( θ t ) t ‖ W * = sup ‖ φ ‖ H 1 = 1 | 〈 ( θ t ) t , φ 〉 | ≤ sup ‖ φ ‖ H 1 = 1 ∫ Ω | ε Δ θ t + u t ⋅ ∇ θ + u ⋅ ∇ θ t | | φ | d x ≤ sup ‖ φ ‖ H 1 = 1 C ( ( ‖ u t ‖ L 2 ‖ ∇ θ ‖ L 3 + ‖ ∇ θ t ‖ L 2 ‖ u ‖ L 3 ) ‖ φ ‖ H 1 + ‖ ∇ θ t ‖ L 2 ‖ ∇ φ ‖ L 2 ) + sup ‖ φ ‖ H 1 = 1 ∫ ∂ Ω ∇ θ t φ d s

where ( H 1 ( Ω ) ) * denotes the dual space of H 1 ( Ω ) . This can show ( θ t ) t ∈ L 2 ( 0 , T * ; ( H 1 ( Ω ) ) * ) . So we have θ t ∈ L 2 ( 0 , T * ; H 1 ( Ω ) ) . Similar to the proof of the continuity of u t , it is easy to know θ t ∈ C ( 0 , T * ; L 2 ) . Then according to the elliptic regularity, we have ∇ 2 θ ∈ C ( 0 , T * ; L 2 ) .

Above all, we complete the proof of the Theorem 1.2.

Lu, J. (2017) Local Existence for Boussinesq Equations with Slip Boundary Condition in a Bounded Domain. Journal of Applied Mathematics and Physics, 5, 1951-1963. https://doi.org/10.4236/jamp.2017.510165