Temporal-spatial decays for the Stokes flows in half spaces

In this paper, we prove new estimates on temporal-spatial decays in L1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$L^1$$\end{document} for solutions to the Stokes equations in the half spaces. Our approach is based on the Ukai’s solution formula of the Stokes equations and heat kernel estimates.


Introduction
In this paper, we consider the Stokes equation where • n ≥ 2, and R n + = {x = (x , x n ) : x ∈ R n−1 , x n > 0} is the upper-half space of R n ; • u = (u 1 (x, t), u 2 (x, t), . . . , u n (x, t)) and p = p(x, t) are unknown velocity vector and the pressure, respectively; • the initial data u 0 = u 0 (x) is assumed to satisfy a compatibility condition: ∇ · u 0 = 0 in R n + and the normal component of u 0 equals to zero on ∂R n + . The estimates of the solution to the Stokes equation play an important role in the study of the temporal and the spatial decays for weak solutions to the Navier-Stokes equation. See [6,10,12,13,15] and the references therein. Denote by A the Stokes operator −P Δ in R n + , where P is the projection: L r (R n + ) → L r σ (R n + ), 1 < r < ∞. Then the solution to (1) can be written as u(t, x) = e −tA u 0 . In the whole space R n , the flow e −tA behaves similarly to the heat Laplacian semigroup e tΔ ; hence, for all 1 ≤ p ≤ q ≤ ∞. However, this is not the case in the half-space R n + . Precisely, in the half-space case, we have for all 1 < p ≤ q ≤ ∞ or 1 ≤ p < q ≤ ∞. See [10]. It is worth noticing that unlike the whole space case, (2) does not hold true for p = q = 1. In fact, it was proved in [7] that there exists a function u 0 ∈ L 1 (R n + ) such that e −tA u 0 / ∈ L 1 (R n + ). In order to deal with L 1 (R n + ) estimates, we need certain extra condition on the initial data u 0 . We first recall the main result in [1,Theorem 1.1]. Theorem A. Assume that the initial date u 0 satisfies ∇ · u = 0 in R n + and R n−1 u 0 (y , y n )dy = 0 a.e. y n > 0.
Then, there is a constant C independent of u 0 such that e −tA u 0 H 1 (R n + ) ≤ Ct −1 R n + y n |y ||u 0 (y)|dy (4) and e −tA u 0 H 1 (R n + ) ≤ Ct −1/2 R n + |y ||u 0 (y)|dy, where H 1 (R n + ) = inf F H 1 (R n ) : F ∈ H 1 (R n ), F | R n + = f with H 1 (R n ) being the Hardy space on R n . (See (19) for the precise definition of the Hardy space).
Then in [12] a stronger condition than (3) is considered. A vector a = (a 1 , a 2 , . . . , a n ) on R n + , n ≥ 2 satisfies the tangential parity condition if x j and even in x k ; and a 1 ( Then it was proved that Theorem B. [12] Assume that the initial date u 0 = (u 1 0 , u 2 0 , . . . , u n 0 ) satisfies ∇·u = 0 in R n + and u n 0 | ∂R n + = 0. Suppose further that u 0 satisfies the tangential parity condition. Then, for any 0 < α ≤ 1 and t > 0 It is important to note that the method in [1] used to obtain the estimates (4) and (5) can not be applied to derive the estimate (6). This is a reason why in [12], the tangential parity condition is assumed to obtain the estimate (6). We would like to emphasize that the tangential parity condition is stronger than condition (3). It is worth noticing that such estimates (4) and (5) are not useful to study the long-time behaviour of solutions to the corresponding Navier-Stokes equation under the tangential parity condition. See for example [12].This show the motivation of the estimate (6).
The main aim of this paper is to prove the following general estimate.
Theorem 1.1. Assume that the initial date u 0 satisfies ∇ · u 0 = 0 in R n + and the condition (3). Then for any t > 0, and 0 < α ≤ 1 and 0 ≤ β ≤ 1, Some comments are in order: (a) Since the tangential parity condition is stronger than (3), Theorem 1.1 is still true under the tangential parity condition. Therefore it is also interesting to study the L 1 decay estimates for the solutions to the Navier-Stokes equations. However, we do not pursue this problem in this paper. Such a result should be elsewhere. (b) It is easy to see that the estimate (7) turns out to be the estimates (4) and (5) when we choose (α, β) = (1, 1) and (α, β) = (1, 0). Moreover, using the inequality |y | ≤ |y| and y n ≤ |y|, the estimate (7) deduces, for α ∈ (0, 2] which extends the range of α ∈ (0, 1] in (6) under the weaker assumption (3). We would like to point out that the approaches in [1,12] can not be applicable to derive the estimate (7). More importantly, although the decay estimates in [1,12] hold true, there are gaps in their proofs. For example (please see the definitions of Λ and E(t) in Sect. 3), both papers used the fact that eE(t)u 0 ∈ H 1 (R n ), but this statement was not proved and it is not clear if this is true. See [1, p.17] and [12, p.1542]. Moreover, one of the key ingredients is proving the following ∂ j Λ −1 eE(t)u 0 ∈ H 1 (R n ). However, what was proved in [1,12] . See the proof of [1, Lemma 3.3] and [12, Lemma 2.2]. Therefore, our approach fills in the gaps of the proofs of Theorem A and Theorem B in [1,12].
The organization of the paper is as follows. In Sect. 2, we prove some kernel estimates which play an essential role in proving temporal-spatial decay estimates for the Stokes equation. The proof of Theorem 1.1 will be given in Sect. 3.
Throughout the paper, we always use C and c to denote positive constants that are independent of the main parameters involved but whose values may differ from line to line. We will write A B if there is a universal constant C so that A ≤ CB and A ∼ B if A B and B A.

Heat kernel estimates
In this section, we prove a number of kernel estimates which play an important role in the proof of the main result.
for all y ∈ R d and t > 0. As a consequence, It is obvious that It remains to estimate the sum in (8). To this end, we write This proves the first inequality.
The second inequality follows directly from the first one and the following inequality . This completes our proof.
Then there exists C such that (i) for every y ∈ R d and t > 0, Temporal-spatial decays for the Stokes flows Page 5 of 22 23 Proof. (i) The proof of this item is simple and we omit the details. ( For the term E 1 , we have This completes our proof.
Hence, it suffices to show that To do this, we consider these two terms separately. Estimate of F 11 We have NoDEA Temporal-spatial decays for the Stokes flows Page 7 of 22 23 Using e u − 1 u α for α, u ∈ [0, 1], Owing to the following inequality where in the last inequality we used Lemma 2.1.
For the term G 2 , we have Similarly to the estimate of G 1 , we have For the term G 21 , we write For the term H 2 , note that |x − y| ∼ |x| as |x| > 2|y|. Hence, by Lemma 2.1, Collecting the estimates of H 1 , H 2 , G 21 , G 22 , we come up with G 2 |y| α t α/2−γ/2 . Taking this and the estimate of G 1 into account, we have F 11 |y| α t α/2−γ/2 . Estimate ofF 12 We have We take care of M 2 first. Since x, y < 0 in this situation, we have This, along with the following inequality e u − 1 u α for α, u ∈ [0, 1], yields At this stage, using the argument in the estimations of H 1 and H 2 , we arrive at This, along with the estimate of M 2 , implies This completes our proof. Proof. We write As a consequence, dx, By making use of the following inequalities It remains to estimate E 2 . To do this, using the following inequality we have where in the last inequality we used Lemma 2.1.
for every s, t > 0, x ∈ R and y > 0, and for some c > 0.
Proof. In order to prove the lemma. We will employ the following simple inequality whose proof will be omitted: let a > 0. Then there exists c > 0 such that for any s, t > 0 and x, y ∈ R, we have We consider two cases: t < |y| 2 and t ≥ |y| 2 . Case 1: t < |y| 2 In this situation, we have for some c > 0, where we used (9) in the second inequality.
For E 2 , we write Using the inequality e u − 1 u α for α ∈ [0, 1] and u 1, we have for some c > 0, where we used e − (y+z) 2 2c 0 t (since y, z ≥ 0) in the second inequality and (9) in the last inequality. For the term E 1 , we notice that yz > 4t ≥ 4y 2 , and hence z > 4y. In this situation, Therefore, for some c > 0, where we used the following estimate 2c t yz t α y α t α/2 in the third inequality and used (9) in the last inequality.
This completes our proof.
Using the fact that for some c > 0, .
For the second term, we note that a(t+s) .
Hence, for s + t > |y| 2 , we consider two cases.

Case 2:
If |x| ≥ 2|y|, then we have |x − y| ∼ |y|. If |x||y| ≥ (s + t) 2 , then we have we have Otherwise, if |x||y| ≥ (s + t) 2 , then we write where in the second inequality we used the following inequality |1 − e x | |x| α for all x with |x| 1. Therefore, This, along with Lemma 2.2, implies as desired. This completes our proof.
Proof. The proof of this Lemma is similar to that of Lemma 2.7 and we would like to leave the details for the interested reader.

Temporal-spatial decay estimates for the Stokes equations
This section is devoted to prove the temporal-spatial decay estimates for the Stokes equations in Theorem 1.1. We first make some conventions of notations. For x ∈ R n , we denote its tangential components (x 1 , . . . , x n−1 ) by x ∈ R n−1 , so that x = (x , x n ). We set ∂ j = ∂/∂x j , ∇ = (∂ 1 , . . . , ∂ n−1 ), Δ = n−1 j=1 ∂ 2 j , and Δ n = ∂ 2 n . The Riesz transforms R j (j = 1, . . . , n), S j (j = 1, . . . , n − 1), and the operator Λ are defined by We set R = (R 1 , . . . , R n−1 ), S = (S 1 , . . . , S n−1 ), and define where r is the restriction operator from R n to R n + , and e is the zero extension operator from R n + to R n defined by ef = f for x n ≥ 0, 0 for x n = 0. We also define the operators E(t) and F (t) by where G t is the Gauss kernel H t (x) = (4πt) −n/2 e −|x| 2 /4t . Note that E(t) and F (t) are the semigroups generated by the Dirichlet Laplacian and Neumann Laplacian, respectively. We recall the following result in [19] regarding the solution formula of the Stokes equation (1).
It is easy to see that the solution u defined in Theorem 3.1 is given as a restriction rū ofū = (ū 1 , . . . ,ū n−1 ,ū n ) := (ū ,ū n ) of the form In order to prove Theorem 1.1 we need the following representations for u in [1,12]. (14) and (15). Then we havē

Proposition 3.2. Letū is given by
where E(t) is an operator defined by for all x ∈ R n and f ∈ L 1 (R n + ). We now recall the definition of the Hardy space H 1 (R n ). See [18]. A function f ∈ L 1 (R n ) is said to be in Hardy space where H t is the heat kernel of the Laplacian defined by 4s .
In the sequel, we denote and Then we have In order to prove Theorem 1.1 we need the following auxiliary estimates.
Proof. Proof of (21) Denote by a * the odd extension of a from R n + to R n , i.e.
Then we have E(t)a(x) = e tΔ a * (x), which implies that for s > 0, e sΔ E(t)a(x) = e (s+t)Δ a * (x). − H 1,s+t (x n + y n )]a(y , y n )dy n dy dx.