In this paper, we study the global existence and uniqueness of strong solutions for the Baer-Nunziato two-phase flow model in a bounded domain with a no - slip boundary. The global existence and uniqueness of strong solutions are obtained when the initial value is near the equilibrium state in <i>H</i><sup>2</sup> (Ω) . Furthermore, the exponential convergence rates of the pressure and velocity are also proved by delicate energy methods.
In this paper, we are interested in a version of one velocity Baer-Nunziato model with dissipation for the mixture of two compressible fluids in a smooth bounded domain Ω ⊂ ℝ 3 . The system is as follows:
{ α t + ( u ⋅ ∇ ) α = 0, 0 ≤ α ≤ 1, ρ t + div ( ρ u ) = 0, z t + div ( z u ) = 0, ( ( ρ + z ) u ) t + div ( ( ρ + z ) u ⊗ u ) + ∇ P ( f ( α ) ρ , g ( α ) z ) = div S , (1.1)
Here, t ≥ 0 and x = ( x 1 , x 2 , x 3 ) ∈ Ω . The variables z, u = ( u 1 , u 2 , u 3 ) and α denote the density of the fluid, the velocity field of the fluid and the volume fraction, respectively. ρ is the density of the particles in the mixture. Where f , g : ( 0,1 ) ↦ [ 0, ∞ ) are given functions, and satisfying
f ( α ) : = α 1 / γ + , g ( α ) : = ( 1 − α ) 1 / γ − , γ ± > 0 ; (1.2)
P = P ( α , ρ , z ) is pressure satisfying
P = α ρ γ + + ( 1 − α ) z γ − (1.3)
And
S = μ ( ∇ u + ∇ u T ) + λ ( div ( u ) ) I (1.4)
( I is the identity tensor) is the viscous stress tensor. The constant viscosity coefficients satisfy standard physical assumptions
μ > 0 , 3 λ + 2 μ ≥ 0. (1.5)
In fact, the system (1.1) is derived by Antonin Novotry in [
{ ( α ± ) t + v I ⋅ ∇ α ± = 0, ( α ± ρ ± ) t + div ( α ± ρ ± u ± ) = 0, ( α ± ρ ± u ± ) t + div ( α ± ρ ± u ± ⊗ u ± ) + ∇ ( α ± P ± ( ρ ± ) ) − P I ∇ ( α ± ) = α ± μ ± ( Δ u ± ) + α ± ( μ ± + λ ± ) ∇ div u ± , 0 ≤ α ± ≤ 1, α + + α − = 1, (1.6)
in the above ( α ± , α ± ρ ± ≥ 0, u ± ∈ ℝ 3 ) —concentrations, densities, velocities of the ± species—are unknown functions, n P ± are two (different) given functions defined on [ 0, ∞ ) and P I , v I are conveniently chosen quantities—they represent pressure and velocity at the interface. In the multifluid modeling, there are many possibilities about how the quantities v I , P I could be chosen, and there is no consensus about this choice.
As [
μ ± : = μ , λ ± : = λ , v I = u ± : = u
α P ± ( s ) = F ± ( f ± ( α ) s ) for all α ∈ ( 0,1 ) , s ∈ [ 0, ∞ )
with some functions F ± defined on [ 0, ∞ ) and functions f ± defined on ( 0,1 ) . The two velocity Baer-Nunziato system reduces to the one velocity Baer-Nunziato system.
System (1.1) corresponds to the barotropic and viscous version of the five-equation model of two-phase flows derived by Allaire, Clerc and Kokh in [
The results of weak solutions to multi-fluid models are in the mathematical literature in a short supply. It is convenient to quote [
P ( ρ , z ) = − b 1 ( ρ , z ) + b 1 ( ρ , z ) 2 + b 2 ( z ) , (1.7)
where b 1 and b 2 are linear functions with respect to each variable. And the study of two-phase flow models in a bounded domain is becoming increasingly popular [
To overcome the difficulties arising from the non-dissipation on ρ , z , we first rewrite system (1.1) in a more suitable form. The crucial idea is that instead of the variables ( α , ρ , z , u ) , we study the variables ( α , c , P , u ) . Let
c : = α ρ γ + − ( 1 − α ) z γ − , m : = ρ + z . (1.8)
By a direct calculation, from (1.3) and (1.8), we have
α ρ γ + = P + c 2 , ( 1 − α ) z γ − = P − c 2 , (1.9)
then the system (1.1) clearly can be written in terms of the variables ( α , c , P , u ) , that is
{ α t + ( u ⋅ ∇ ) α = 0, 0 ≤ α ≤ 1 c t + u ⋅ ∇ c + B 1 div u = 0, P t + u ⋅ ∇ P + B 2 div u = 0, m u t + ( m u ⋅ ∇ ) u + ∇ P = μ Δ u + ( λ + μ ) ∇ div u , (1.10)
with the initial and boundary conditions
{ ( α , c , P , u ) ( x , 0 ) = ( α 0 , c 0 , P 0 , u 0 ) ( x ) , x = ( x 1 , x 2 , x 3 ) ∈ Ω , u ( x , t ) | ∂ Ω = 0 , t ≥ 0 , 1 | Ω | ∫ Ω P 0 d x = P ¯ 0 , (1.11)
where B 1 = γ + − γ − 2 P + γ + + γ − 2 c , B 2 = γ + + γ − 2 P + γ + − γ − 2 c and P ¯ 0 is a positive constant.
Now, we are in a position to state our main results:
Theorem 1.1. Let P ¯ 0 > 0 , c ¯ and α ¯ are three constants, assume the initial boundary value ( α 0 , c 0 , P 0 − P ¯ 0 , u 0 ) ∈ H 2 ( Ω ) satisfies the compatibility conditions, i.e.
∂ t l u ( 0 ) | ∂ Ω = 0 , l = 0 , 1 ,
where ∂ t u ( x ,0 ) = μ Δ u 0 + ( λ + μ ) ∇ div u 0 m 0 − m 0 ( u 0 ⋅ ∇ ) u 0 − ∇ P 0 is the lth derivative at t = 0 of any solution of the system (1.10) - (1.11), as calculated from
(1.10) to yield an expression in terms of α 0 , c 0 , P 0 , u 0 . Then there exists a constant ε 0 such that if
‖ ( α 0 − α ¯ , c 0 − c ¯ , P 0 − P ¯ 0 , u 0 ) ‖ 2 ≤ ε 0 (1.12)
then the initial boundary value problem (1.10) - (1.11) admits a unique solution ( α , c , P , u ) globally in the time with P > 0 , which satisfies
α − α ¯ , P − P ¯ , c − c ¯ ∈ C 0 ( [ 0, ∞ ) ; H 2 ( Ω ) ) ∩ C 1 ( [ 0, ∞ ) ; H 1 ( Ω ) ) , u ∈ C 0 ( [ 0, ∞ ) ; H 2 ( Ω ) ∩ H 0 1 ( Ω ) ) ∩ C 1 ( [ 0, ∞ ) ; L 2 ( Ω ) ) , (1.13)
where
P ¯ ( t ) = 1 | Ω | ∫ Ω P ( x , t ) d x . (1.14)
Moreover, there exist two positive constant C 0 , η 0 such that for any t > 0 , it holds that
‖ ( P − P ¯ , u ) ( t ) ‖ 2 2 + ∫ 0 t ‖ ( P − P ¯ ) ( τ ) ‖ 2 2 + ‖ ( u ) ( τ ) ‖ 3 2 d τ ≤ C 0 ‖ ( P 0 − P ¯ 0 , u 0 ) ‖ 2 2 , (1.15)
‖ ( P − P ¯ , u ) ( t ) ‖ 2 2 + ‖ ∂ t ( P − P ¯ , u ) ( t ) ‖ L 2 2 ≤ C 0 ‖ ( P 0 − P ¯ 0 , u 0 ) ( t ) ‖ 2 2 exp { − η 0 t } , (1.16)
‖ ( α − α ¯ ) ( t ) ‖ 2 ≤ C 0 ‖ α 0 ‖ 2 exp { C 0 ‖ ( P 0 − P ¯ 0 , u 0 ) ‖ 2 } . (1.17)
‖ ( c − c ¯ ) ( t ) ‖ 2 ≤ C 0 exp { C 0 ‖ ( P 0 − P ¯ 0 , u 0 ) ‖ 2 } ( ‖ u 0 ‖ 2 + ‖ ( P 0 − P ¯ 0 , u 0 ) ‖ 2 2 ) . (1.18)
Finally, lim t → ∞ P ¯ ( t ) exists and let lim t → ∞ P ¯ ( t ) = P ˜ , the following convergence rate holds
| P ˜ − P ¯ ( t ) | ≤ C 0 ‖ ( α 0 , c 0 , P 0 − P ¯ , u 0 ) ‖ 2 2 exp { − η 0 t } . (1.19)
Throughout this paper, C denotes the generic positive constant depending only on the initial data and physical coefficients but independent of time t. Moreover, the norms in Sobolev spaces H m ( Ω ) and W m , p ( Ω ) are denoted respectively by ‖ ⋅ ‖ m and ‖ ⋅ ‖ m , p , for m ≥ 0 and p ≥ 1 . Particularly, for m = 0 , we will simply use ‖ ⋅ ‖ L 2 and ‖ ⋅ ‖ L p . As usual, 〈 ⋅ , ⋅ 〉 denotes the inner-product in L 2 ( Ω ) . Finally, ∇ = ( ∂ 1 , ∂ 2 , ∂ 3 ) , ∂ i = ∂ x i ( i = 1 , 2 , 3 ) and for any integer l ≥ 0 , ∇ l f denotes all derivatives of order l of the function f.
The rest of this paper is organized in the following way. In the next section, we show some useful inequalities. In Section 3, we obtained some a priori estimates and hence the global existence by the energy estimate method. As a by-product, we get the time decay estimates of the solutions.
In this section, we first introduce some Sobolev’s inequalities that will be used frequently in later articles (cf. [
Lemma 2.1. [
( i ) ‖ f ‖ L ∞ ≤ C ‖ f ‖ 2 , ( ii ) ‖ f ‖ L p ≤ C ‖ f ‖ 1 , 2 ≤ p ≤ 6, (2.1)
for some contants C > 0 depending only on Ω .
Due to the slip boundary condition, the classical energy estimates can’t be applied directly to spatial derivatives. In order to get the estimates on the tangential derivatives of the solutions ( P , u ) , we introduce the following lemmas on the stationary Stokes equations, c.f. [
Lemma 2.2. [
{ − μ Δ u + ∇ P = g , div u = f , u | ∂ Ω = 0 , (2.2)
where f ∈ H k + 1 ( Ω ) , g ∈ H k ( k ≥ 0 ) . Then the above problem has a solution ( P , u ) ∈ H k + 1 × ( H k + 2 ∩ H 0 1 ) which is unique modulo a constant of integration for P. Moreover, this solution satisfies:
‖ u ‖ k + 2 2 + ‖ ∇ P ‖ k 2 ≤ C ( ‖ f ‖ k + 1 2 + ‖ g ‖ k 2 ) . (2.3)
In this section, we will prove the global existence and large-time behavior of the solution with the small initial data. Firstly, we start with the local existence and uniqueness of the strong solution to the initial boundary value problem (1.10) - (1.11).
Proposition 3.1. (Local existence) Let ( α 0 , c 0 , P 0 , u 0 ) ∈ H 2 ( Ω ) such that
inf x ∈ Ω { P 0 ( x ) } > 0 , ∂ t l u 0 | ∂ Ω = 0 , l = 0 , 1.
Then there exists a positive constant T and C, such that the initial value problem (1.10) - (1.11) has a unique solution ( α , c , P , u ) ∈ C ( [ 0, T ] ; H 2 ( Ω ) ) satisfying
inf t ∈ [ 0, T ] , x ∈ Ω ¯ { P ( t , x ) } > 0, c t , P t ∈ C ( [ 0, T ] ; H 1 ( Ω ) ) , u t ∈ C ( [ 0, T ] ; L 2 ( Ω ) ) , u , α ∈ L 2 ( [ 0, T ] ; H 2 ( Ω ) )
Furthermore, the following estimates hold,
‖ α ( t ) ‖ 2 + ‖ c ( t ) ‖ 2 + ‖ P ( t ) ‖ 2 + ‖ u ( t ) ‖ 2 ≤ C ( ‖ α 0 ‖ 2 + ‖ c 0 ‖ 2 + ‖ P 0 ‖ 2 + ‖ u 0 ‖ 2 ) .
Next, we will establish some a priori estimates of the solution ( α , c , P , u ) . We first make the a priori assumption that
‖ ( α − α ¯ , c − c ¯ , P − P ¯ , u ) ( t ) ‖ 2 + | P ¯ ( t ) − P ¯ 0 | ≤ ε , for any t ≥ 0, (3.1)
where ε > 0 is sufficiently small. By the Sobolev inequality, we have
1 2 P ¯ 0 ≤ P ( t ) ≤ 2 P ¯ 0 , 1 C ≤ m ( t ) ≤ C , for any t ≥ 0. (3.2)
This will be often used in the rest of paper.
In order to deduce a prior estimate, We first use the energy estimation method to estimate the lower derivative of ( P , u ) .
Lemma 3.1. Under the conditions of Theorem 1.1 and (3.1), there exists a positive constant C such that for any t ≥ 0 , it holds
1 2 d d t ∫ Ω m | u | 2 + ( P − P ¯ ) 2 B ¯ 2 d x + μ ∫ Ω | ∇ u | 2 d x + ( μ + λ ) ∫ Ω | div u | 2 d x ≤ C ε ( ‖ ∇ P ‖ L 2 2 + ‖ ∇ u ‖ L 2 2 ) . (3.3)
Proof.Multiplying (1.10)4 by u, and integrating on Ω , using integration by parts, we get
1 2 d d t ∫ Ω m | u | 2 d x + μ ∫ Ω | ∇ u | 2 d x + ( μ + λ ) ∫ Ω | div u | 2 d x + ∫ Ω u ⋅ ∇ P d x = 0. (3.4)
In order to get the estimate of P, we shall deduce the equation of P ¯ . By integrating (1.10)3 over Ω gives
∫ Ω P t d x + ∫ Ω u ⋅ ∇ P d x + ∫ Ω B 2 div u d x = ∫ Ω P t d x + ∫ Ω u ⋅ ∇ P d x − ∫ Ω u ⋅ ∇ B 2 d x = 0. (3.5)
Therefore,
∫ Ω P t d x = ∫ Ω u ⋅ ∇ ( B 2 − P ) d x . (3.6)
Then, we have
P ¯ t ( t ) = 1 | Ω | ∫ Ω P t ( x , t ) d x = 1 | Ω | ∫ Ω u ⋅ ∇ ( B 2 − P ) d x . (3.7)
Combining the above equality, (3.1) and (3.2), we obtain
| P ¯ t | ≤ C ‖ u ‖ L 2 ( ‖ ∇ B 2 ‖ L 2 + ‖ ∇ P ‖ L 2 ) ≤ C ‖ u ‖ L 2 ( ‖ ∇ c ‖ L 2 + ‖ ∇ P ‖ L 2 ) . (3.8)
Now, we rewrite Equation (1.10)3 in the linear form as the following
( P − P ¯ ) t B ¯ 2 + div u + P ¯ t + ( B 2 − B ¯ 2 ) div u + u ⋅ ∇ P B ¯ 2 = 0. (3.9)
Multiplying the above equality by P − P ¯ and integrating over Ω gives
1 2 d d t ∫ Ω ( P − P ¯ ) 2 B ¯ 2 d x + ∫ Ω div u ( P − P ¯ ) d x = − 1 2 d d t ∫ Ω ( P − P ¯ ) 2 ( B ¯ 2 ) P ¯ P ¯ t B ¯ 2 2 d x + ∫ Ω P ¯ t + ( B 2 − B ¯ 2 ) div u + u ⋅ ∇ P B ¯ 2 ( P − P ¯ ) d x , (3.10)
Adding (3.4) to (3.10), we find
1 2 d d t ∫ Ω m | u | 2 + ( P − P ¯ ) 2 B ¯ 2 d x + μ ∫ Ω | ∇ u | 2 d x + ( μ + λ ) ∫ Ω | div u | 2 d x = − 1 2 d d t ∫ Ω ( P − P ¯ ) 2 ( B ¯ 2 ) P ¯ P ¯ t B ¯ 2 2 d x + ∫ Ω P ¯ t + ( B 2 − B ¯ 2 ) div u + u ⋅ ∇ P B ¯ 2 ( P − P ¯ ) d x . (3.11)
By using (3.1), (3.2), (3.8), Lemma 2.1, Hölder’s inequality and Poincaré’s inequality, the right terms of the above equation can be estimated as follows:
| ∫ Ω ( P − P ¯ ) 2 ( B ¯ 2 ) P ¯ P ¯ t B ¯ 2 2 d x | ≤ | ( B ¯ 2 ) P ¯ P ¯ t B ¯ 2 | ‖ P − P ¯ ‖ L 2 2 ≤ C ‖ u ‖ L 2 ( ‖ ∇ c ‖ L 2 + ‖ ∇ P ‖ L 2 ) ‖ P − P ¯ ‖ L 2 2 ≤ C ε ‖ ∇ P ‖ L 2 2 . (3.12)
| ∫ Ω P ¯ t + u ⋅ ∇ P B ¯ 2 ( P − P ¯ ) d x | ≤ | C B ¯ 2 | ( | P ¯ t | ‖ P − P ¯ ‖ L 2 + ‖ u ‖ L 3 ‖ ∇ P ‖ L 2 ‖ P − P ¯ ‖ L 6 ) ≤ C ( ‖ u ‖ L 2 ( ‖ ∇ c ‖ L 2 + ‖ ∇ P ‖ L 2 ) ‖ P − P ¯ ‖ L 2 + ‖ ∇ u ‖ L 2 ‖ ∇ P ‖ L 2 2 ) ≤ C ε ( ‖ ∇ u ‖ L 2 2 + ‖ ∇ P ‖ L 2 2 ) . (3.13)
| ∫ Ω ( B 2 − B ¯ 2 ) div u B ¯ 2 ( P − P ¯ ) d x | ≤ ∫ Ω | B 2 − B ¯ 2 B ¯ 2 | div u ( P − P ¯ ) d x ≤ C ‖ B 2 − B ¯ 2 B ¯ 2 ‖ L 3 ‖ div u ‖ L 2 ‖ P − P ¯ ‖ L 6 ≤ C ε ‖ div u ‖ L 2 ‖ ∇ P ‖ L 2 ≤ C ε ( ‖ ∇ u ‖ L 2 2 + ‖ ∇ P ‖ L 2 2 ) . (3.14)
Plugging (3.12)-(3.14) into (3.11) yields (3.3). The proof of Lemmas 3.1 is completed.
Next, we give the energy estimate of the time derivative for ( P , u ) .
Lemma 3.2. Under the conditions of Theorem 1.1 and (3.1), there exists a positive constant C such that for any t ≥ 0 it holds
1 2 d d t ∫ Ω m | u t | 2 + ( P t − P ¯ t ) 2 B ¯ 2 d x + μ ∫ Ω | ∇ u t | 2 d x + ( μ + λ ) ∫ Ω | div u t | 2 d x ≤ C ε ( ‖ ∇ u ‖ 1 2 + ‖ ∇ u t ‖ L 2 2 ) . (3.15)
Proof.Differentiating (1.10)4 and (3.9) with respect to t, then multiplying the result by u t and ( P − P ¯ ) t respectively, then summing up and integrating on Ω , we get
1 2 d d t ∫ Ω m | u t | 2 + ( P t − P ¯ t ) 2 B ¯ 2 d x + μ ∫ Ω | ∇ u t | 2 d x + ( μ + λ ) ∫ Ω | div u t | 2 d x = − 1 2 ∫ Ω ( m t | u t | 2 − ( P t − P ¯ t ) 2 ( B ¯ 2 ) P ¯ P ¯ t ( B ¯ 2 ) 2 ) d x − ∫ Ω ( m u ⋅ ∇ u ) t u t d x − ∫ Ω ( P ¯ t + ( B 2 − B ¯ 2 ) div u + u ⋅ ∇ P B ¯ 2 ) t ( P t − P ¯ t ) d x (3.16)
We use the boundary conditions u t | ∂ Ω = 0 . For the first term on the right hand side of the above equality, we have from (3.1), (3.2) and Lemma 2.1, Hölder’s inequality and Poincaré’s inequality that
| ∫ Ω m t | u t | 2 + ( P t − P ¯ t ) 2 ( B ¯ 2 ) P ¯ P ¯ t ( B ¯ 2 ) 2 | d x = | ∫ Ω div ( m u ) | u t | 2 + ( P t − P ¯ t ) 2 ( B ¯ 2 ) P ¯ P ¯ t ( B ¯ 2 ) 2 | d x = | ∫ Ω 2 m u ∇ u t u t − ( P t − P ¯ t ) 2 ( B ¯ 2 ) P ¯ P ¯ t ( B ¯ 2 ) 2 | d x ≤ C ‖ m ‖ L ∞ ‖ u ‖ L 3 ‖ ∇ u t ‖ L 2 ‖ u t ‖ L 6 + | ( B ¯ 2 ) P ¯ P ¯ t ( B ¯ 2 ) 2 | ( ‖ P t ‖ L 2 2 + ‖ P ¯ t ‖ L 2 2 ) ≤ C ε ‖ ∇ u ‖ L 2 ‖ ∇ u t ‖ L 2 2 + C ε ‖ ∇ u ‖ 1 2 ≤ C ε ( ‖ ∇ u ‖ 1 2 + ‖ ∇ u t ‖ L 2 2 ) (3.17)
By (3.1) and (1.10)2 yields
‖ P t ‖ L 2 ≤ C ( ‖ u ⋅ ∇ P ‖ L 2 + ‖ ∇ u ‖ L 2 ) ≤ C ( ‖ u ‖ L ∞ ‖ ∇ P ‖ L 2 + ‖ ∇ u ‖ L 2 ) ≤ C ‖ ∇ u ‖ 1 . (3.18)
Similarly, for the second term, we have
| ∫ Ω ( m u ⋅ ∇ u ) t u t d x | = | ∫ Ω ( m t u ⋅ ∇ u u t + m u t ∇ u u t + m u ⋅ ∇ u t u t ) d x | = | ∫ Ω ( − ( ∇ m u + m div u ) u ⋅ ∇ u u t + m u t ∇ u u t + m u ⋅ ∇ u t u t ) d x | ≤ C ( ‖ ∇ m ‖ L 3 ‖ u ‖ L ∞ 2 ‖ ∇ u ‖ L 2 ‖ u t ‖ L 6 + ‖ m ‖ L ∞ ‖ div u ‖ L 2 ‖ u ‖ L ∞ ‖ ∇ u ‖ L 3 ‖ u t ‖ L 6 + ‖ m ‖ L ∞ ‖ ∇ u ‖ L 3 ‖ u t ‖ L 3 2 + ‖ m ‖ L ∞ ‖ u ‖ L 3 ‖ ∇ u t ‖ L 2 ‖ u t ‖ L 6 ) ≤ C ε ( ‖ ∇ u ‖ L 2 2 + ‖ ∇ u t ‖ L 2 2 ) , (3.19)
where is using
‖ ∇ m ‖ L 3 ≤ ‖ ∇ m ‖ 1 ≤ C ( ‖ ∇ P ‖ 1 + ‖ ∇ c ‖ 1 ) . (3.20)
Then, combing (3.1), (1.10)2, (3.7) and (3.18), we can obtain
| P ¯ t t | ≤ 1 | Ω | ∫ Ω ( u t ⋅ ∇ ( B 2 − P ) + u ⋅ ∇ ( B 2 − P ) t ) d x ≤ C [ ‖ u t ‖ L 2 ( ‖ ∇ P ‖ L 2 + ‖ ∇ c ‖ L 2 ) + ‖ ∇ u ‖ L 2 ( ‖ P t ‖ L 2 + ‖ c t ‖ L 2 ) ] ≤ C ε ( ‖ u t ‖ L 2 + ‖ ∇ u ‖ L 2 + ‖ c t ‖ L 2 ) ≤ C ε ( ‖ u t ‖ L 2 + ‖ ∇ u ‖ L 2 ) , (3.21)
Finally, for the last term, we have
| ∫ Ω ( P ¯ t + ( B 2 − B ¯ 2 ) div u + u ⋅ ∇ P B ¯ 2 ) t ( P t − P ¯ t ) d x | ≤ | ∫ Ω { P ¯ t t + [ ( B 2 ) P P t + ( B 2 ) c c t − ( B ¯ 2 ) P ¯ P ¯ t ] div u + ( B 2 − B ¯ 2 ) div u t } ( P t − P ¯ t ) B ¯ 2 d x | + | ∫ Ω ( u t ⋅ ∇ P + u ⋅ ∇ P t ) ( P t − P ¯ t ) B ¯ 2 d x | + | ∫ Ω { P ¯ t + ( B 2 − B ¯ 2 ) div u + u ⋅ ∇ P } ( B ¯ 2 ) P ¯ P ¯ t ( P t − P ¯ t ) ( B ¯ 2 ) 2 d x | ≤ C ε ( ‖ ∇ u ‖ 1 2 + ‖ ∇ u t ‖ L 2 2 ) . (3.22)
Substituting (3.17), (3.19), (3.22) into (3.16) gives (3.15).
Next, we will localize ∂ Ω when estimate the boundary solutions. Refer to [
Lemma 3.3. Under the conditions of Theorem 1.1 and (3.1), there exists a positive constant C such that for any t ≥ 0 , it holds
1 2 d d t ∫ Ω m | ∇ u χ 0 | 2 + | ∇ P χ 0 | 2 B ¯ 2 d x + μ ∫ Ω | ∇ 2 u χ 0 | 2 d x + ( μ + λ ) ∫ Ω | ∇ div u χ 0 | 2 d x ≤ C ε ( ‖ ∇ u ‖ 1 2 + ‖ ∇ P ‖ 1 2 + ‖ ∇ u t ‖ L 2 2 ) + C ‖ ∇ u ‖ L 2 ( ‖ ∇ 2 u ‖ L 2 + ‖ ∇ P ‖ L 2 ) . (3.23)
1 2 d d t ∫ Ω m | ∇ 2 u χ 0 | 2 + | ∇ 2 P χ 0 | 2 B ¯ 2 d x + μ ∫ Ω | ∇ 3 u χ 0 | 2 d x + ( μ + λ ) ∫ Ω | ∇ 2 div u χ 0 | 2 d x ≤ C ε ( ‖ ∇ u ‖ 2 2 + ‖ ∇ P ‖ 1 2 + ‖ ∇ u t ‖ L 2 2 ) + C ‖ ∇ 2 u ‖ L 2 ( ‖ ∇ 3 u ‖ L 2 + ‖ ∇ 2 P ‖ L 2 ) . (3.24)
Proof.Differentiating (1.10)4 and (3.9) with respect to x i , multiplying the resulting equations by u x i χ 0 2 , P x i χ 0 2 respectively, then summing up and integrating on Ω , we have
1 2 d d t ∫ Ω m | u x i χ 0 | 2 + | P x i χ 0 | 2 B ¯ 2 d x + μ ∫ Ω | ∇ u x i χ 0 | 2 d x + ( μ + λ ) ∫ Ω | div u x i χ 0 | 2 d x = 1 2 ∫ Ω m t | u x i χ 0 | 2 − | P x i χ 0 | 2 ( B ¯ 2 ) P ¯ P ¯ t B ¯ 2 2 d x − ∫ Ω [ m x i u t + ( m u ⋅ ∇ u ) x i ] u x i χ 0 2 d x − ∫ Ω ( P ¯ t + ( B 2 − B ¯ 2 ) div u + u ⋅ ∇ P B ¯ 2 ) x i P x i χ 0 2 d x − μ ∫ Ω u x i ∇ u x i ∇ χ 0 2 d x − ( μ + λ ) ∫ Ω div u x i u x i ∇ χ 0 2 d x + ∫ Ω P x i u x i ∇ χ 0 2 d x ≤ C ε ( ‖ ∇ u ‖ 1 2 + ‖ ∇ P ‖ 1 2 + ‖ ∇ u t ‖ L 2 2 ) + C ‖ ∇ u ‖ L 2 ( ‖ ∇ 2 u ‖ L 2 + ‖ ∇ P ‖ L 2 ) . (3.25)
which gives (3.23). Repeating the above procedure again for 2nd order spatial derivatives we get the following
1 2 d d t ∫ Ω m | u x i x j χ 0 | 2 + | P x i x j χ 0 | 2 B ¯ 2 d x + μ ∫ Ω | ∇ u x i x j χ 0 | 2 d x + ( μ + λ ) ∫ Ω | div u x i x j χ 0 | 2 d x = 1 2 ∫ Ω m t | u x i x j χ 0 | 2 − | P x i x j χ 0 | 2 ( B ¯ 2 ) P ¯ P ¯ t B ¯ 2 2 d x − ∫ Ω [ m x i x j u t + m x i u t x j + m x j u t x i + ( m u ⋅ ∇ u ) x i x j ] u x i x j χ 0 2 d x − ∫ Ω ( P ¯ t + ( B 2 − B ¯ 2 ) div u + u ⋅ ∇ P B ¯ 2 ) x i x j P x i x j χ 0 2 d x − μ ∫ Ω u x i x j ∇ u x i x j ∇ χ 0 2 d x − ( μ + λ ) ∫ Ω div u x i x j u x i x j ∇ χ 0 2 d x + ∫ Ω P x i x j u x i x j ∇ χ 0 2 d x ≤ C ε ( ‖ ∇ u ‖ 1 2 + ‖ ∇ P ‖ 1 2 + ‖ ∇ u t ‖ L 2 2 ) + C ‖ ∇ u ‖ L 2 ( ‖ ∇ 2 u ‖ L 2 + ‖ ∇ P ‖ L 2 ) . (3.26)
which implies (3.24). So, the Lemma 3.3 can be finished.
Refer to [
1) The surface O j ∩ ∂ Ω is the image of a smooth vector function z j ( y 1 , y 2 ) = ( z 1 j , z 2 j , z 3 j ) ( y 1 , y 2 ) (e.g. take the local geodesic polar coordinate), satifying
| z y 1 j | = 1 , z y 1 j ⋅ z y 2 j = 0 and | z y 2 j | ≥ δ > 0 , (3.27)
where δ is a positive constant independent of 1 ≤ j ≤ N .
2) For any x = ( x 1 , x 2 , x 3 ) ∈ O j is represented by
x i = Ψ i ( y ) = y 3 η i j ( z j ( y 1 , y 2 ) + z i j ( y 1 , y 2 ) ) , for i = 1 , 2 , 3 (3.28)
where η j ( y 1 , y 2 ) = ( η 1 j , η 2 j , η 3 j ) ( z j ( y 1 , y 2 ) ) represents the internal unit normal vector at the point z j ( y 1 , y 2 ) of the surface ∂ Ω .
We omit the subscript j in what follows for the simplicity of presentation. For k = 1 , 2 , we define the unit vectors
e 1 = z y 1 and e 2 = z y 2 | z y 2 | ,
Then Frenet-Serret’s formal gives that there exist smooth functions ( α 1 , β 1 , γ 1 , α 2 , β 2 , γ 2 ) of ( y 1 , y 2 ) satisfying
∂ ∂ y 1 ( e 1 e 2 η ) i = ( 0 − γ 1 − α 1 γ 1 0 − β 1 α 1 β 1 0 ) ( e 1 e 2 η ) i ,
∂ ∂ y 1 ( e 1 e 2 η ) i = ( 0 − γ 2 − α 2 γ 2 0 − β 2 α 2 β 2 0 ) ( e 1 e 2 η ) i ,
where e m i denote the i-th component of e m . An elementary calculation shows that the Jacobian J of the transform (3.28) is
J = Ψ y 1 × Ψ y 2 ⋅ η = | z y 2 | + ( α 1 | z y 2 | + β 2 ) y 3 + ( α 1 β 2 − α 2 β 1 ) y 3 2 (3.29)
By (3.29), we have the transform (3.28) is regular by choosing y 3 so small that J ≥ δ 2 for some positive δ . Therefore, the inverse function of Ψ ( y ) : = ( Ψ 1 , Ψ 2 , Ψ 3 ) ( y ) exists, and we use y = Ψ − 1 ( x ) denote it. Using a straightforward calculation, ( y 1 , y 2 , y 3 ) x i ( x ) can be expressed by
{ ∂ x i y 1 = 1 J ( Ψ y 2 × Ψ y 3 ) i = 1 J ( A e i 1 + B e i 2 ) = : r 1 i , ∂ x i y 2 = 1 J ( Ψ y 3 × Ψ y 1 ) i = 1 J ( C e i 1 + D e i 2 ) = : r 2 i , ∂ x i y 3 = 1 J ( Ψ y 1 × Ψ y 2 ) i = η i = : r 1 i , (3.30)
where A = | z y 2 | + β 2 y 3 , B = − y 3 α 2 , C = − β 1 y 3 , D = 1 + α 1 y 3 , and J = A D − B C ≥ δ 2 . It’s easy to find out, (3.30) gives
∑ i = 1 3 a 3 i 2 = | n | 2 = 1 , r 1 i r 3 i = r 2 i r 3 i = 0 , J 2 = ( A C + B D ) 2 − ( A 2 + B 2 ) ( C 2 + D 2 ) (3.31)
and
∂ x i = r k i ∂ y k (3.32)
where we have used the Einstein convention of summing over repeated indices.
Therefore, in each O j , (1.10)3 - (1.10)4 can be rewritten in the local coordinates ( y 1 , y 2 , y 3 ) as follows:
E p : = d P d t + B ¯ 2 J [ ( A e 1 + B e 2 ) ⋅ u y 1 + ( C e 1 + D e 2 ) ⋅ u y 2 + J η ⋅ u y 3 ] = g ,
E u : = m u t − μ J 2 [ ( A 2 + B 2 ) u y 1 y 1 + 2 ( A C + B D ) u y 1 y 2 + ( C 2 + D 2 ) u y 2 y 2 + J 2 u y 3 y 3 ] + one order terms of u + 1 J ( A e 1 + B e 2 ) [ μ + λ B ¯ 2 d P d t + P ] y 1 + 1 J ( C e 1 + D e 2 ) [ μ + λ B ¯ 2 d P d t + P ] y 2 + η [ μ + λ B ¯ 2 d P d t + P ] y 3 = h ,
where
d d t = ∂ t + u ⋅ ∇ ( denotes the material derivative ) , g = − ( B 2 − B ¯ 2 ) div u , h = m u ⋅ ∇ u + μ + λ B ¯ 2 ∇ g . J 2 = ( A C + B D ) 2 − ( A 2 + B 2 ) ( C 2 + D 2 )
Let us denote the tangential derivatives by ∂ = ( ∂ y 1 , ∂ y 2 ) and χ j be arbitrary but fixed-function in C 0 ∞ ( O j ) . Obviously, x j ∂ k u = 0 on ∂ Ω j − 1 , where 0 ≤ k ≤ 2 and Ω j − 1 ( y ) : = { y | y = Ψ − 1 ( x ) , x ∈ Ω j = O j ∩ Ω } . Estimating the tangential derivatives in the similar way as the above lemma, we have
Lemma 3.4. Under the conditions of Theorem 1.1 and (3.1), there exists a positive constant C such that for any t ≥ 0,1 ≤ j ≤ N , it holds
d d t ∫ Ω j − 1 m | ∂ u χ j | 2 + | ∂ P χ j | 2 B ¯ 2 d y + ∫ Ω j − 1 | ∂ ∇ u χ j | 2 d y + ∫ Ω j − 1 | ∂ d P d t χ j | 2 d y ≤ C ε ( ‖ ∇ u ‖ 1 2 + ‖ ∇ u t ‖ L 2 2 + ‖ ∇ P ‖ 1 2 ) + C ‖ ∇ u ‖ L 2 ( ‖ ∇ u ‖ 1 + ‖ ∇ P ‖ L 2 ) , (3.33)
d d t ∫ Ω j − 1 m | ∂ 2 u χ j | 2 + | ∂ 2 P χ j | 2 B ¯ 2 d y + ∫ Ω j − 1 | ∂ 2 ∇ u χ j | 2 d y + ∫ Ω j − 1 | ∂ 2 d P d t χ j | 2 d y ≤ C ε ( ‖ ∇ u ‖ 2 2 + ‖ ∇ u t ‖ L 2 2 + ‖ ∇ P ‖ 1 2 ) + C ‖ ∇ 2 u ‖ L 2 ( ‖ ∇ u ‖ 2 + ‖ ∇ 2 P ‖ L 2 ) . (3.34)
Next, we begin to deduce the estimates of derivatives in the normal directions.
Lemma 3.5. Under the conditions of Theorem 1.1 and (3.1), there exists a positive constant C such that for any t ≥ 0 , k + l = 1 , k , l ≥ 0 , 1 ≤ j ≤ N , it holds
d d t ∫ Ω j − 1 | P y 3 χ j | 2 d y + ∫ Ω j − 1 | ( d P d t ) y 3 χ j | 2 d y ≤ C [ ‖ ∇ u ‖ L 2 2 + ‖ u t ‖ L 2 2 + ε ( ‖ ∇ P ‖ 1 2 + ‖ ∇ u ‖ 1 2 ) + ∫ Ω j − 1 | ∂ ∇ u χ j | 2 d y ] , (3.35)
d d t ∫ Ω j − 1 | ∂ k ∂ y 3 l + 1 P χ j | 2 d y + ∫ Ω j − 1 | ∂ k ∂ y 3 l + 1 ( d P d t ) y 3 χ j | 2 d y ≤ C [ ‖ ∇ u ‖ 1 2 + ‖ u t ‖ 1 2 + ε ( ‖ ∇ P ‖ 1 2 + ‖ ∇ 2 u ‖ 1 2 ) + ∫ Ω j − 1 | ∂ k + 1 ∂ y 3 l ∇ u χ j | 2 d y ] , (3.36)
Proof.First, using ∂ y 3 ( E P − g ) = 0 and η ( E u − h ) = 0 , that is the following forms:
( d P d t ) y 3 + B ¯ 2 J [ ( A e 1 + B e 2 ) ⋅ u y 1 y 3 + ( C e 1 + D e 2 ) ⋅ u y 2 y 3 + J η ⋅ u y 3 y 3 ] + one order terms of u = g y 3 , (3.37)
η m u t − μ J 2 [ ( A 2 + B 2 ) η u y 1 y 1 + 2 ( A C + B D ) η u y 1 y 2 + ( C 2 + D 2 ) η u y 2 y 2 + J 2 η u y 3 y 3 ] + one order terms of u + [ μ + λ B ¯ 2 d P d t + P ] y 3 = η h , (3.38)
In order to eliminate u y 3 y 3 in equation (3.37), we use (3.37) × μ B ¯ 2 + (3.38) yields:
2 μ + λ B ¯ 2 ( d P d t ) y 3 + P y 3 = − μ J 2 [ ( A 2 + B 2 ) η u y 1 y 1 + 2 ( A C + B D ) η u y 1 y 2 + ( C 2 + D 2 ) η u y 2 y 2 ] − η m u t − μ J [ ( A e 1 + B e 2 ) ⋅ u y 1 y 3 + ( C e 1 + D e 2 ) ⋅ u y 2 y 3 + J η ⋅ u y 3 y 3 ] + one order terms of u + η h + μ B ¯ 2 g y 3 = Φ . (3.39)
Multiply that by χ j 2 ( d P d t ) y 3 and integrating on Ω j − 1 , we obtain
1 2 d d t ∫ Ω j − 1 | P y 3 χ j | 2 d y + 2 μ + λ B ¯ 2 ∫ Ω j − 1 | ( d P d t ) y 3 χ j | 2 d y = ∫ Ω j − 1 − ( u ⋅ ∇ P ) y 3 P y 3 χ j 2 + ( d P d t ) y 3 Φ χ j 2 d y . = : K 1 + K 2 . (3.40)
Estimate each term at the right end of the above equation,
| K 1 | ≤ | ∫ Ω j − 1 u y 3 ⋅ ∇ P P y 3 χ j 2 d y | + 1 2 | ∫ Ω j − 1 ( P y 3 ) 2 div ( u χ j 2 ) d y | ≤ C ‖ ∇ u ‖ 1 ‖ ∇ P ‖ 1 2 ≤ C ε ‖ ∇ P ‖ 1 2 , (3.41)
and
| K 2 | ≤ 2 μ + λ 2 B ¯ 2 ∫ Ω j − 1 | ( d P d t ) y 3 χ j | 2 d y + C ∫ Ω j − 1 | Φ χ j | 2 d y ≤ 2 μ + λ 2 B ¯ 2 ∫ Ω j − 1 | ( d P d t ) y 3 χ j | 2 d y + C ∫ Ω j − 1 | ∂ ∇ u χ j | 2 d y + C ( ‖ ∇ u ‖ L 2 2 + ‖ u t ‖ L 2 2 + ε ‖ ∇ u ‖ 1 2 ) . (3.42)
Substituting (3.41) and (3.42) into (3.40) get
d d t ∫ Ω j − 1 | P y 3 χ j | 2 d y + 2 μ + λ B ¯ 2 ∫ Ω j − 1 | ( d P d t ) χ j | 2 d y ≤ C [ ‖ ∇ u ‖ L 2 2 + ‖ u t ‖ L 2 2 + ε ( ‖ ∇ u ‖ 1 2 + ‖ ∇ P ‖ 1 2 ) + ∫ Ω j − 1 | ∂ ∇ u χ j | 2 d y ] , (3.43)
which implies (3.35).
By the same way, using ∂ k ∂ y 3 l to (3.38), multiplying the resulting equations by χ j 2 ∂ k ∂ y 3 l + 1 ( d P d t ) , then when k + l = 1 we have
d d t ∫ Ω j − 1 | ∂ k ∂ y 3 l + 1 P χ j | 2 d y + 2 μ + λ B ¯ 2 ∫ Ω j − 1 | ∂ k ∂ y 3 l + 1 ( d P d t ) y 3 χ j | 2 d y ≤ C [ ‖ ∇ u ‖ 1 2 + ‖ u t ‖ 1 2 + ε ( ‖ ∇ 2 u ‖ 1 2 + ‖ ∇ P ‖ 1 2 ) + ∫ Ω j − 1 | ∂ k + 1 ∂ y 3 l ∇ u χ j | 2 d y ] . (3.44)
which implies (3.36). The proof of Lemma3.5 is completed.
Finally, we use Lemma 2.2 to deduce the estimates on the tangential derivatives of ( P , u ) .
Lemma 3.6. Under the conditions of Theorem 1.1 and (3.1), there exists a positive constant C such that for any t ≥ 0 , it holds
‖ ∇ 2 u ‖ L 2 2 + ‖ ∇ P ‖ L 2 2 ≤ C ( ‖ d P d t ‖ 1 2 + ‖ u t ‖ L 2 2 + ‖ ∇ 2 u ‖ L 2 2 ‖ ∇ u ‖ 1 2 ) , (3.45)
∫ Ω j − 1 | ∂ ∇ 2 u χ j | 2 d y + ∫ Ω j − 1 | ∂ ∇ P χ j | 2 d y ≤ C ( ‖ ∇ u ‖ 1 2 + ‖ u t ‖ 1 2 + ‖ ∇ P ‖ L 2 2 + ‖ ∇ 3 u ‖ L 2 2 ‖ ∇ u ‖ 1 2 + ‖ ∇ P ‖ L 2 + ‖ ∇ d P d t ‖ L 2 ) + C ∫ Ω j − 1 | ∂ ∇ d P d t χ j | 2 d y . (3.46)
Proof.We rewrite the perturbed equations as the Stokes problem:
{ div u = − 1 B 2 d P d t , − μ Δ u + ∇ P = ( λ + μ ) ∇ div u − ( m u t + m u ⋅ ∇ u ) , u | ∂ Ω = 0, (3.47)
where applying Lemma 2.2 to (3.47), one can easily get (3.45).
Next we prove (3.46). To do this, by applying χ j ∂ to equation (3.47)2, we have
{ div ( χ j ∂ u ) = − χ j ∂ ( 1 B 2 d P d t ) + ∇ χ j ∂ u , − μ Δ ( χ j ∂ u ) + ∇ ( χ j ∂ P ) = − 2 μ ∇ χ j ∇ ( ∂ u ) − Δ χ j ∂ u + ∇ χ j ∂ P − ( λ + μ ) χ j ∇ ∂ ( 1 B 2 d P d t ) − χ j ∂ ( m u t + m u ⋅ ∇ u ) , χ j ∂ u | ∂ Ω j − 1 = 0, (3.48)
Using the Lemma 2.2 to (3.48) gives (3.46). The proof of Lemma 3.6 is completed.
Now, let’s start proving Theorem 1.1. We will do it by four steps.
Step 1: We first estimate the lower order derivatives for ( P , u ) . Suppose D be a fixed but large positive constant. Let D2 × ((3.3) + (3.15)) + D × ((3.23) + (3.33)) + (3.35), there exists a function H 1 ( P , u ) which is equivalent to ‖ u ‖ 2 + ‖ P − P ¯ ‖ 2 + ‖ u t ‖ 2 + ‖ P t − P ¯ t ‖ 2 + ‖ ∇ P ‖ 2 and satisfies
d d t { H 1 + ∫ Ω m | ∇ u χ 0 | 2 d x + ∑ j = 1 N ∫ Ω j − 1 m | ∂ u χ j | 2 d y } + D ( ‖ ∇ u ‖ L 2 2 + ‖ ∇ u t ‖ L 2 2 ) + ‖ ∇ d P d t ‖ L 2 2 + ∫ Ω | ∇ 2 u χ 0 | 2 d x + ∑ j = 1 N ∫ Ω j − 1 | ∂ ∇ u χ j | 2 d y ≤ 1 D 1 / 3 ( ‖ ∇ 2 u ‖ L 2 2 + ‖ ∇ P ‖ L 2 2 ) + C ε ‖ ∇ 2 P ‖ L 2 2 . (3.49)
Substituting equation (3.45) into the above equation and using d P d t = − B 2 div u , we obtain
d d t { H 1 + ∫ Ω m | ∇ u χ 0 | 2 d x + ∑ j = 1 N ∫ Ω j − 1 m | ∂ u χ j | 2 d y } + ‖ ∇ u ‖ 1 2 + ‖ ∇ P ‖ L 2 2 + ‖ ∇ u t ‖ L 2 2 + ‖ ∇ d P d t ‖ L 2 2 ≤ C ε ‖ ∇ 2 P ‖ L 2 2 . (3.50)
where D is enough large, ε is arbitrarily small.
Step 2: In this step, we will estimate the higher order derivatives for ( P , u ) . Let l = 0 in (3.36), by D × (2(3.24) + (3.34)) + (3.36), we get
d d t { D ∫ Ω m | ∇ 2 u χ 0 | + | ∇ 2 P χ 0 | 2 B ¯ 2 d x + D ∑ j = 1 N ∫ Ω j − 1 m | ∂ 2 u χ j | 2 + | ∂ 2 P χ j | 2 B ¯ 2 d y + ∫ Ω j − 1 | ∂ ∂ P y 3 χ j | 2 d y } + ∫ Ω | ∇ 3 u χ 0 | 2 d x + ∫ Ω | ∇ 2 d P d t χ 0 | 2 d x + ∑ j = 1 N ∫ Ω j − 1 | ∂ 2 ∇ u χ j | 2 + | ∂ ∇ d P d t χ j | 2 d y ≤ C D ( ‖ ∇ u ‖ 1 2 + ‖ ∇ u t ‖ L 2 2 ) + C D ε ( ‖ ∇ 2 u ‖ 1 2 + ‖ ∇ P ‖ 1 2 ) + C D ‖ ∇ 2 u ‖ L 2 ( ‖ ∇ u ‖ 2 + ‖ ∇ 2 P ‖ L 2 ) (3.51)
Then by taking l = 1 in (3.36) and substituting (3.46) into (3.36), we have
d d t ∫ Ω j − 1 | ∂ y 3 2 P χ j | 2 d y + ∫ Ω j − 1 | ∂ y 3 2 ( d P d t ) χ j | 2 d y ≤ C [ ‖ ∇ u ‖ 1 2 + ‖ u t ‖ 1 2 + ‖ ∇ P ‖ L 2 ( ∥ ∇ P ∥ L 2 + ‖ ∇ d P d t ‖ L 2 ) + δ ‖ ∇ P ‖ L 2 2 + ‖ ∇ 2 u ‖ L 2 2 ] (3.52)
By the same way, D × (3.51) + (3.52), there exists H 2 ( P ) which is equivalent to ‖ ∇ 2 P ‖ L 2 2 , such that
d d t { D 2 ∫ Ω m | ∇ 2 u χ 0 | d x + D 2 ∑ j = 1 N ∫ Ω j − 1 m | ∂ u χ j | 2 d y + H 2 } + ∫ Ω | ∇ 3 u χ 0 | 2 d x + ∑ j = 1 N ∫ Ω j − 1 | ∂ 2 ∇ u χ j | 2 + ∫ Ω | ∇ 2 d P d t | 2 d x ≤ C D 2 ( ‖ ∇ u ‖ 1 2 + ‖ ∇ u t ‖ L 2 2 + ‖ ∇ P ‖ L 2 2 ) + C D 2 δ ( ‖ ∇ P ‖ 1 2 + ‖ ∇ 2 u ‖ 1 2 ) + C D 2 ‖ ∇ 2 u ‖ L 2 ( ‖ ∇ u ‖ 2 + ‖ ∇ 2 P ‖ L 2 ) . (3.53)
Applying Lemma 2.2 to (3.47), we obtain
‖ ∇ 3 u ‖ L 2 2 + ‖ ∇ 2 P ‖ L 2 2 ≤ C ( ‖ ∇ u ‖ 1 2 + ‖ u t ‖ 1 2 ‖ ∇ P ‖ L 2 2 + ‖ ∇ d P d t ‖ 1 2 + ‖ ∇ u ‖ 1 2 ‖ ∇ 3 u ‖ L 2 2 ) . (3.54)
Step 3: Establish the energy inequality of Gronwall-type. An application of the L p -estimate of elliptic system to (1.10)4 gives
‖ ∇ 2 u ‖ L 2 2 ≤ C ( ‖ u t ‖ L 2 2 + ‖ ∇ P ‖ L 2 + ‖ ∇ u ‖ L 2 2 ) (3.55)
Consider D4 × (3.50) + D × (3.53) + (3.54), there exists a function H 3 ( P , u ) which is equivalent to ‖ P − P ¯ ‖ 2 2 + ‖ u ‖ 2 2 + ‖ P t − P ¯ t ‖ L 2 2 + ‖ u t ‖ L 2 2 such that
d H 3 d t + C H 3 + C ‖ ∇ 3 u ‖ L 2 2 ≤ 0, (3.56)
where we have used the Poincaré’s inequality ‖ P t − P ¯ t ‖ L 2 ≤ ‖ ∇ P ( t ) ‖ . Integrating the above inequality over [0,t] get (1.15). Using Gronwall’s inequality to (3.56), it is clear that there exist two positive constant C 1 and η 1 such that
H 3 ( P , u ) ≤ C 1 H 3 ( P 0 , u 0 ) e − η 1 t , (3.57)
which together with (1.10)3 yield (1.16).
Step 4: Finally, we prove (1.17), (1.18) and (1.19), and we can conclude the energy estimates on α , c as following:
d d t ‖ α − α ¯ ‖ 2 2 ≤ C ‖ u ‖ 2 2 ‖ α − α ¯ ‖ 2 2 ,
d d t ‖ c − c ¯ ‖ 2 2 ≤ C ‖ u ‖ 2 ‖ c − c ¯ ‖ 2 2 + ‖ B 1 ‖ 2 2 ,
By Gronwall’s inequality, we get
‖ α − α ¯ ‖ 2 2 ≤ C ‖ α 0 − α ¯ ‖ 2 exp { C ∫ 0 t ‖ u ( τ ) ‖ 2 d τ }
‖ c − c ¯ ‖ 2 2 ≤ C exp { C 1 ∫ 0 t ‖ u ( τ ) ‖ 2 d τ } ( ‖ u 0 ‖ 2 + ∫ 0 t ‖ B 1 ‖ 2 2 d τ ) ,
By simple calculation, implies (1.17) and (1.18). To prove (1.19), we first show that l i m t → ∞ P ¯ ( t ) exists. In fact, for any arbitrary positive constant ε , there exists a positive constant T = max { 1, ln η 0 ε C 0 − η 0 } such that for any t 2 > t 1 > T , it holds that
| P ¯ ( t 2 ) − P ¯ ( t 1 ) | = | ∫ t 1 t 2 P ¯ τ d τ | ≤ C 0 ∫ t 1 t 2 e − η 0 τ d τ ≤ C 0 η 0 e − η 0 t 1 < ε ,
which implies that l i m t → ∞ P ¯ ( t ) exists. Now, setting P ˜ = lim t → ∞ P ¯ ( t ) , and combining (3.8), we obtain
‖ P ˜ − P ¯ ( t ) ‖ = | ∫ t ∞ P ¯ τ d τ | ≤ C ∫ t ∞ ‖ u ( τ ) ‖ ( ‖ ∇ P ( τ ) ‖ + ‖ ∇ u ( τ ) ‖ ) d τ ,
which together with (1.16) implies (1.19).
Finally, we have finished the proof of Theorem 1.1.
The author declares no conflicts of interest regarding the publication of this paper.
Kou, W.H. (2022) Global Existence of Solutions for Baer-Nunziato Two-Phase Flow Model in a Bounded Domain. Open Journal of Applied Sciences, 12, 631-649. https://doi.org/10.4236/ojapps.2022.124043