J Geom Anal https://doi.org/10.1007/s12220-018-0048-9
N–S Systems via Q–Q−1 Spaces Jie Xiao1 · Junjie Zhang2
Received: 24 April 2018 © Mathematica Josephina, Inc. 2018
Abstract Under (α, p, n − 1) ∈ (−∞, 1) × (2, ∞) × N, this paper uses Qα (Rn ) −1 n n n n n and Q−1 α (R ) := div Qα (R ) (covering BMO(R ) andBMO (R )) where f ∈ | f (x)| t 2 f (x)|2 sup Qα (Rn ) ⇔ Rn 1+|x| n+1 dx < ∞ and I ×(0,(I )) |∇e coordinate cubes ⎧ I ⎪ 1 sn as α ∈ (−∞, 0); ⎪ 2 ⎨ 2 ωα (It ) dxdt < ∞ with (0, 1] s → ωα (s) = s n ln es as α = 0; t n−1 ⎪ ⎪ ⎩ n−2α s as α ∈ (0, 1), to demonstrate that the incompressible Navier–Stokes system ; u − (u · ∇)u + ∇p = ∂t u and div u = 0 in R1+n + has a unique mild soluu(0, x) = a(x) as x ∈ Rn tion under a −1 n n being sufficiently small; however, its steady state Qα (R ) u − (u · ∇)u + ∇p = 0 and div u = 0 in Rn ; has only zero solution u(x) → 0 as ∞ ← x ∈ Rn n p(n−2) under u ∈ BMO−1 (Rn ) ∩ L p, 2 (Rn ) .
JX was supported by NSERC of Canada (#202979); JZ was supported by the Fundamental Research Funds for the Central Universities of China (#2016YJS154) and NSERC of Canada (#202979).
B
Jie Xiao
[email protected] Junjie Zhang
[email protected]
1
Department of Mathematics and Statistics, Memorial University, St. John’s, NL A1C 5S7, Canada
2
Department of Mathematics, Beijing Jiaotong University, Beijing 100044, China
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J. Xiao, J. Zhang
Keywords N–S systems · Q–Q−1 spaces · Mild–null solutions Mathematics Subject Classification 30H25 · 35Q30 · 42B37 · 46E35
1 Statement of Theorem 1.3 This paper presents an avenue to understanding the role that the scaling-invariant spaces n≥2 ) Q1>α>−∞ (Rn≥2 ) and Q−1 1>α>−∞ (R := div Q1>α>−∞ (Rn≥2 ), . . . , Q1>α>−∞ (Rn≥2 )
n copies
play in the regularity theory of the n-dimensional incompressible Navier–Stokes (N– S) system and its steady state via considering the well-posedness of a mild solution for small initial data and the zero nature of a smooth solution with the vanishing boundary value at infinity. 1.1 Definitions of Q–Q−1 Spaces and N–S Systems To begin with, let us recall a variant of the original definition of Q spaces of several real variables introduced in [8, Lemma 2.2]—for (α, n) ∈ (−∞, ∞) × N let Q α (Rn ) be the space of all measurable functions with | f | Q α (Rn ) = sup
I ∈F
2α−n (I )
dy | f (x + y) − f (x)| dx n+2α |y| 2
|y|<(I )
I
1 2
< ∞,
where F is the family of all cubes in Rn with edges parallel to the coordinate axes and (I ) stands for the edge length of I ∈ F . Of significant interest is that α → Q α (Rn ) is decreasing. Moreover, if n ≥ 2 then (cf. [8, Theorem 2.3], [33, Theorem 4.1] and [32, Theorem 3.1]): ⎧ BMO(Rn ) ⎪ ⎪ ⎪ ⎨New subspace of BMO(Rn ) Q α (Rn ) = α ⎪ (−)− 2 L 2,2α (Rn ) between E(Rn ) and BMO(Rn ) ⎪ ⎪ ⎩ {constants}
123
as as as as
α α α α
∈ (−∞, 0); = 0; ∈ (0, 1); ∈ [1, ∞),
N–S Systems via Q–Q−1 Spaces
where • There are two constants c1,α , c2,α such that
α
(−) 2 f (x) = c1,α
Rn
= c2,α
Rn
α f (x) − f (y) dy and (−)− 2 f (x) |x − y|n+α f (y) dy |y − x|n−α
hold for any sufficiently smooth function f on Rn . • Under ( p, λ) ∈ [1, ∞) × (−∞, ∞) the symbol L p,λ (Rn ) represents the ( p, λ)Campanato space of all measurable functions f with (cf. [12]) f L p,λ (Rn ) =
sup
(r0 ,x0 )∈(0,∞)×Rn
λ−n r0
B(x0 ,r0 )
f (y) − −
B(x0 ,r0 )
p f (x) dx dy
1
p
<∞
for which −B(x0 ,r0 ) is the integral mean of f over the Euclidean ball B(x0 , r0 ) centred at x0 with radius r0 . • E(Rn ) is the conformal energy space of all functions f with their gradients ∇ f being n-integrable on Rn , and BMO(Rn ) = Q −n/2 (Rn ) (cf. [17]) is the John-Nirenberg space (cf. [10,17]) of all measurable functions f on Rn with an equivalent (BMO, 1 ≤ p < ∞)-norm f BMO(Rn ), p := sup
I ∈F
I
| f (x) − f I | p dx n (I )
1
p
< ∞ via f I := I
f (x)dx n . (I )
Furthermore, a combination of [5, Theorem 3.3], [31, Definition 1.3] and the forthcoming Propositions 2.1 and 2.2 motivates Qα (Rn ) based on a Carleson-type measure: Definition 1.1 Let (α, n) ∈ (−∞, ∞) × N. (i) Qα (Rn ) is the class of all measurable functions f on Rn obeying ⎧ | f (x)| ⎪ dx < ∞; ⎪ ⎨ Rn 1+|x|n+1 1 2 ωα (It ) 2 t f (x)|2 ⎪ f ⎪ dxdt < ∞, Qα (Rn ) := sup n−1 ⎩ I ×(0,(I )) |∇e t I ∈F
where • ∇ and are the gradient and Laplace operators in the space variable, respectively. ⎧ • ⎪s n as α ∈ (−∞, 0); ⎪ ⎨ 2 e n (0, 1] s → ωα (s) = s ln s as α = 0; ⎪ ⎪ ⎩ n−2α s as α ∈ (0, 1).
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•
et
(ii) The symbol
2
n
f (x) = (4π t 2 )− 2
Rn
e
− |x−y| 2 4t
2
f (y)dy.
n n n Q−1 (R ) := div Q (R ), . . . , Q (R ) α α α
n copies
n is the divergence space of Qα (Rn ). In other words, f ∈ Q−1 α (R ) if and only if n there are f 1 , . . . , f n ∈ Qα (R ) such that
f (x) = div f 1 (x), . . . , f n (x) =
n
∂x j f j (x1 , . . . , xn )
j=1
with f Q−1 n := inf α (R )
n
f j Qα (Rn )
j=1
for which the infimum is taken over all representations f = div( f 1 , . . . , f n ). Consequently, α → Qα (Rn ) is decreasing, in particular, if n − 1 ∈ N then ⎧ n ⎪ ∀ α ∈ (−∞, 0); ⎨BMO(R ) n Qα (R ) = New space between C I S(Rn ) and BMO(Rn ) ∀ α ∈ [0, 1); ⎪ ⎩ {constants} ∀ α ∈ [1, ∞), and
⎧ −1 n ⎪ ∀ α ∈ (−∞, 0); ⎨BMO (R ) −1 n Qα (R ) = New space between L 2,n−2 (Rn ) and BMO−1 (Rn ) ∀ α ∈ [0, 1); ⎪ ⎩ {0} ∀ α ∈ [1, ∞),
where C I S(Rn ) and L p,n−2 (Rn ) are the conformally invariant Sobolev and the Morrey p [1, ∞) p-space of all C 1 - and L loc -functions on Rn with ⎧ 1 ⎪ 2 ⎪ 2 ⎪ |∇ f (x)| ⎪ ⎪ f C I S(Rn ) := sup <∞ n−2 dx ⎪ I ⎪ (I ) ⎪ I ∈F ⎪ ⎪ ⎨and 1 ⎪ p ⎪ p ⎪ | f (x)| ⎪ f p,n−2 n := sup ⎪ <∞ n−2 dx ⎪ L (R ) I ⎪ (I ) ⎪ I ∈F ⎪ ⎪ ⎩ respectively,
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N–S Systems via Q–Q−1 Spaces
as well as BMO(Rn ) = Q− n2 (Rn ) and BMO−1 (Rn ) = Q−1 (Rn ). −n 2
n≥2 ) is of imporAccordingly, an exploration of Q1>α>−∞ (Rn≥2 ) and Q−1 1>α>−∞ (R tant interest; see e.g. [5,6,16,20,30,34,35]. In fact, we find that these two spaces are naturally applicable in the study of the evolutionary incompressible N–S system and its steady state as seen below. n Definition 1.2 Let n − 1 ∈ N, R1+n + = (0, ∞) × R and p be a pressure. Then,
(i) the n-dimensional incompressible N–S system is ⎧ 1+n ⎪ ⎨u − (u · ∇)u + ∇p = ∂t u in R+ ; div u = 0 in R1+n + ; ⎪ ⎩ u(0, x) = a(x) as x ∈ Rn .
(1.1)
(ii) the steady state of the n-dimensional incompressible N–S system is ⎧ n ⎪ ⎨u − (u · ∇)u + ∇p = 0 in R ; div u = 0 in Rn ; ⎪ ⎩ u(x) → 0 as ∞ ← x ∈
(1.2) Rn .
1.2 Principal Theorem and Its Remark Stemming from [19,26,27,31,32] (coupled with [11,22,24]), this paper shows the following result. Theorem 1.3 Let (α, p, n − 1) ∈ (−∞, 1) × (2, ∞) × N and p be a pressure. (i) There is 0 > 0 such that if the incompressible initial data a satisfies a = (a1 , . . . , an )
n Q−1 α (R )
n :=
n
a j Q−1 n ≤ 0 α (R )
j=1
then (1.1) has a unique mild solution
t
u = (u 1 , . . . , u n ) = et a −
e(t−s) P∇ · (u ⊗ u)ds,
0
where ⎧ t e a = et a1 , . . . , et an ; ⎪ ⎪ ⎪ ⎨ P = {P } jk j,k=1,...,n = {δ jk + R j Rk } j,k=1,...,n ; ⎪ Kronecker symbol; ⎪δ jk = the√ ⎪ −1 −1 √ ⎩ = ∂x j − = the Riesz transform, R j = ∂ j −
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with u X n (R1+n ) := α
+
n
u j X α (R1+n ) < ∞ +
j=1
as well as u j X α (R1+n ) := +
1
r 2 |u j (r, x)|
sup (r,x)∈R1+n +
+
r2
sup (r,x)∈R1+n +
0
|u j (t, y)| ωα 2
|y−x|
√t dydt r
n
t2
1 2
.
(ii) Any smooth solution u = (u 1 , . . . , u n ) of (1.2) is zero under u 1 , . . . , u n ∈ BMO−1 (Rn ) ∩ L p,
p(n−2) 2
(Rn ).
Remark 1.4 In order to put Theorem 1.3 into historical perspective, we make two comments: (i) Thanks to (cf. [8, Theorem 2.3 and Remark 2.11] and Proposition 2.1) BMO(Rn ) = Qα∈(−∞,0) (Rn ) ⊃ Q0 (Rn ) ⊃ Qα∈(0,1) (Rn ), Theorem 1.3(i) under α = 0 is known; see [19,31,32]. However, Theorem 1.3(i) under α = 0 reveals a brand-new phenomenon which can be clearly seen from Proposition 2.1, and so there not only exists a real difference amongst α ∈ (−∞, 0), α = 0 and α ∈ (0, 1); in fact, X α∈(0,1) (R1+n + ) is more regular 1+n ) which is in turn more regular than X than X α=0 (R1+n α∈(−∞,0) (R+ ), but also + the known cases α = 0 of Theorem 1.3(i) cannot be employed to deduce the remaining one α = 0 of Theorem 1.3(i). Furthermore, note that Theorem 1.3(i) can be extended to the conformal Morrey space L 2,n−2 (Rn ) (cf. [32, Theorem 3.3] and [18,23]) and even the conformal Lebesgue space L n (Rn ) ⊆ L 2,n−2 (Rn ) (cf. [7] for n = 3), and yet Theorem 1.3(i) can be pushed a little bit further—in −1 (Rn ) denotes accordance with [15, Theorem 1.1] and [9, Theorem 1.1], if B˙ ∞,∞ the scaling-invariance Besov space of all tempered distributions f on Rn with (cf. [3]) f B˙ ∞,∞ −1 (Rn ) :=
sup
√ t t|e f (y)| < ∞,
(t,y)∈R1+n +
then for a smooth solution u = (u 1 , . . . , u n ) of (1.1) and the first possible blow-up time t0 there exists a constant c0 > 0 depending on n ≥ 3 such that the estimation sup
( j,t)∈{1,...,n}×(t0 − 0 ,t0 )
123
u j (t, ·) B˙ ∞,∞ −1 (Rn ) ≤ c0 for some 0 > 0
N–S Systems via Q–Q−1 Spaces
ensures that t0 is not a blow-up time and u can be continued past t0 and so to (0, ∞); however, this is not against the ill-posedness result in [2]. (ii) As a limiting situation of Theorem 1.3(i), Theorem 1.3(ii) is new and its non-trivial argument has such a consequence of special interest that the classical Liouvilletype problem for the stationary incompressible N–S system is solvable under BMO(R2 ) ∩ BMO−1 (R2 )-constraint: if 2 u = (u 1 , u 2 ) ∈ BMO(R2 ) ∩ BMO−1 (R2 ) is a smooth solution of u − (u · ∇)u + ∇p = 0 in R2 ; div u = 0 in R2 , then u is a constant. Meanwhile, upon observing 2n
L n−2 (Rn ) ⊆ L
2n n−2 ,n
(Rn )
we find that if 2n
u j ∈ L n−2 (Rn ) ∩ BMO−1 (Rn ) in Theorem 1.3(i) then u ≡ 0 and hence this slightly strong hypothesis is natural and actually comes from both the kinetic energy structure and the scalinginvariance nature of (1.2). If W˙ 1,2 (Rn ) is the kinetic energy space—the closure of all C0∞ (Rn )-functions (infinitely differentiable functions with compact support in Rn ) f with respect to f W˙ 1,2 (Rn ) =
1 |∇ f (x)| dx 2
Rn
2
,
then (cf. [33, Theorem 4.1] and [32, Theorem 3.1]) ⎧ 1,2 2 2 ⎪ ⎪W˙ (R ) ⊂ BMO(R ); ⎨ L n (Rn ) ⊆ L 2,n−2 (Rn ) ⊂ BMO−1 (Rn ); ⎪ ⎪ ⎩ f 2n f W˙ 1,2 (Rn ) ∀ (n − 2, f ) ∈ N × C0∞ (Rn ), n L n−2 (R )
and consequently, Theorem 1.3(ii) implies that any smooth solution u = (u 1 , . . . , u n ) of (1.2) with u 1 , . . . , u n ∈ BMO−1 (Rn ) ∩ W˙ 1,2 (Rn ) or 2n
u 1 , . . . , u n ∈ L 2,n−2 (Rn ) ∩ L n−2 (Rn )
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J. Xiao, J. Zhang
must be zero. This assertion is worth being compared with the existing twofold results: on the one hand, if u j ∈ W˙ 1,2 (Rn ) then u = (u 1 , . . . , u n ) must be zero under n = 3 (cf. [11]) or under n = 3 and one more condition u j ∈ BMO−1 (R3 ) (cf. [26, Theorem 1.1] which includes u j ∈ L 6 (R3 )); on the other hand, since R3
−1 9 |u(x)| 2 ln 2 + |u(x)|−1 dx ≤ (ln 2)−1
9
R3
|u(x)| 2 dx
[4, Theorem 1.1] (saying that if the left side of the last inequality is finite then u of (1.2) under n = 3 must be zero) improves the well-known result (stating that if the right side of the last inequality is finite then u of (1.2) under n = 3 must be zero) in [11], and if u j ∈ L 2,2 (R3 ) then the smooth solution u = (u 1 , u 2 , u 3 ) of (1.2) (without assuming lim|x|→∞ u(x) = 0) in R3 must be zero (cf. [27, Theorem 1.1]). Since −1 (Rn ), BMO−1 (Rn ) ⊂ B˙ ∞,∞
it is our conjecture that if −1 u 1 , . . . , u n ∈ B˙ ∞,∞ (Rn ) ∩ L 2< p<∞,
p(n−2) 2
(Rn )
then any smooth solution u = (u 1 , . . . , u n ) of (1.2) must be zero—this is due to the fact that (cf. [29, Theorem 2 and its remark]) √ r |∇er f (x)| < ∞
sup (r,x)∈R1+n +
amounts to (under 0 < β < 1)
r2
sup (r,x)∈R1+n +
0
|y−x|
t ∇e f (y)2
t r2
n +β 2
dydt n
t2
1 2
< ∞.
The rest of the paper is organized as follows. The second section presents the basic n characterization of Q1>α>−∞ (Rn ) and Q−1 1>α>−∞ (R ) whose the case α = 0 is of particular importance due to Propositions 2.1 and 2.2. The third section is used to prove Theorem 1.3(i). The fourth section is designed to validate Theorem 1.3(ii). Notation 1.5 Nevertheless, the coming-up-next symbolic conventions are necessary for us to proceed: • Z stands for all integers and ⎧ n Z = Z × · · · × Z; ⎪ ⎪ ⎨ n times
⎪ [0, 1]n = [0, 1] × · · · × [0, 1] . ⎪ ⎩
n times
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N–S Systems via Q–Q−1 Spaces
For r > 0 and I ∈ F , r I is the cube with the same centre as I and side-length r (I ), while I + x denotes I translated by x ∈ Rn . Similarly, a Euclidean ball in Rn with centre x and radius r will be denoted by B = B(x, r ), and its volume, dilates and translates by |B|, ρ B, B + y, respectively. • The characteristic function of a set E will be denoted by 1 E . • U V means U ≤ C V for a positive constant C; U ≈ V means both U V and V U . n 2 Qα (Rn ) and Q−1 α (R ) under (α, n − 1) ∈ (−∞, 1) × N n This section documents the fundamental structure of Qα (Rn ) and Q−1 α (R ) based on 1+n the Carleson-type measures over R+ whose location at α = 0 needs special attention 2 thanks to ω0 (s) = s n ln es .
2.1 Qα (Rn ) under (α, n − 1) ∈ (−∞, 1) × N Proposition 2.1 Let (α, n − 1) ∈ (−∞, 1) × N and f be a measurable function with Rn
| f (x)|(1 + |x|1+n )−1 dx < ∞.
Then f Qα=0 (Rn ) ≈ | f | Q α=0 (Rn ) but | f | Q0 (Rn ) f Q0 (Rn ) [ f ]Q0 (Rn ) , where [ f ]Q0 (Rn ) := sup
I ∈F
|y|<(I )
I
| f (x + y) − f (x)|2 dx
1 2 dy e(I ) 3 ln . n |y| (I )|y|
Proof Since Proposition 2.1 under α = 0 is known (cf. [5,8,31,32]), it remains to check the case α = 0. Firstly, let us prove | f | Q0 (Rn ) f Q0 (Rn ) . On the one hand, let F(t, x) = et
2
f (x)
|x − y|2 exp − f (y)dy 4t 2 Rn |z|2 exp − f (x + t z) dz. 4 Rn n
= (4π t 2 )− 2 n
= (4π )− 2
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Then ⎧ F(0, x) = f (x); ⎪ ⎪ ⎨ 2 n ∂x j F(t, x) = Rn ∂ y j (4π t 2 )− 2 exp − |y| f (x + y) − f (x) dy; 2 4t ⎪ ⎪ |y|2 ⎩∂ 2 −n min t −1−n , |y|−1−n . y j (4π t ) 2 exp − 4t 2
(2.1)
Note that the triangle inequality gives ⎧ | f (x + y)− f (x)| ≤A1 + A2 ; ⎪ ⎪ ⎨ y F(0, x); A1 := F |y| 2 ,x + 2 − ⎪ ⎪ ⎩A := F |y| , x + y − y − F(0, x + y). 2 2 2 so it suffices to show −n Bi := (I )
|Ai | dx 2
|y|<(I )
I
dy f 2Q0 (Rn ) ∀ I ∈ F and i = 1, 2. |y|n
Due to the symmetry of A1 and A2 about x + 2y , it is enough to estimate B1 . To do so, upon letting ∇s,z = (∂s , ∂z 1 , . . . , ∂z n ) and noticing A1
1 d t|y| ty dt = F ,x + dt 2 2 0 1 |y| y dt, ≤ , ty · ∇s,z F(s, z) (s,z)= t|y| 2 2 2 ,x+ 2 0
we use the Minkowski inequality, the representation of F(t, x) via f (x + t z) and the change of variables s = 2−1 |y|t to deduce
1 |A1 | dx 2
I
2
21 |y| y 2 dt ∇s,z F(s, z) , ≤ dx ty · (s,z)= t|y| 2 2 2 ,x+ 2 0 I 1 ≤ ∂s F(s, z)(s,z)= t|y| ,x+ t y I
0
1
2
2
1 2 |y| 2 + ∇z F(s, z)(s,z)= t|y| ,x+ t y dt dx 2 2 2
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N–S Systems via Q–Q−1 Spaces 1
≤
I
0
∂s F(s, z)(s,z)= t|y| ,x+ t y 2
2
2 + ∇z F(s, z)(s,z)= t|y| ,x+ t y dx 2
2
|y| dt 2
1 2 2 y dx ∇ F s, x + s ds. |y| I
|y| 2
2
1
0
This, along with |y| ≤ (I ), gives −n B1 (I )
|y|<(I )
(I ) s
|y| 2
0
2
1 |∇ F(t, x)|2 dx
4I
2
dt
2
dy |y|n
−n ds (I ) s 0 0 2 1 s ds ∇ F(t(I ), ·) L 2 (4I ) dt ≈ (I )2−n . s 0 0
∇ F(t, ·) L 2 (4I ) dt
Note that (cf. [21] or [1, Lemma 11])
1 s
0
2 |φ(t)| dt
1
μ(s) ds
0
s |φ(t)|2 ν(t) dt ⇔ sup
0
s∈(0,1)
0
ν(t)
dt −1 < ∞.
−1
1 s μ(t) dt
So, upon choosing ⎧ ⎪ μ(t) = et ; ⎪ ⎨ 2 ; ν(t) = et ln et ⎪ ⎪ ⎩ φ(t) = ∇ F(t(I ), ·) L 2 (4I ) , we obtain that the last supremum is finite, thereby getting
1 s
0
0
2 ∇ F(t(I ), ·) L 2 (4I ) dt
ds s
1 0
e 2 ∇ F(t(I ), ·) 2L 2 (4I ) t ln dt t
and so
e 2 ∇ F(t(I ), ·) 2L 2 (4I ) t ln dt t 0 (I ) t dt ≈ ∇ F(t, ·) 2L 2 (4I ) ω0 n−1 (I ) t 0
B1 (I )
2−n
1
f 2Q0 (Rn ) .
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Secondly, let us check f Q0 (Rn ) [ f ]Q0 (Rn ) . It is enough to validate
(I )
|∂x j F(t, x)| ω0 2
I
0
sup
J ∈F
J
J
t (I )
dxdt t n−1
| f (x) − f (y)|2 e(J ) 3 dxdy. n ln |x − y| (J )|x − y|
According to (2.1) and the first half of the argument for [8, Lemma 3.1], we have ∂x j F(t, ·) 2L 2 (Rn )
t
−2n−2
+
t |ζ |=1
0
∞
t
|ζ |=1
2
f (· + r ζ ) − f L 2 (Rn ) dζ r
f (· + r ζ ) − f L 2 (Rn ) dζ r
−2
n−1
dr
2 dr
.
On the one hand, via choosing ⎧ 2 ⎪ ⎨μ(s) = s −2n−1 ln e(I ) ; s (0, (I )] s → 3 ⎪ ⎩ν(s) = s 1−2n ln e(I ) , s and calculating
(I )
μ(s) ds
t
t
0
−1 ν(s) ds
t
−2n
e(I ) 3 e(I ) −3 2n ln t ln ≈ 1, t t
we apply the above-mentioned Hardy inequality to get
t
2 e(I ) 2 f (· + r ζ ) − f L 2 (Rn ) dζ r n−1 dr t ln dt t |ζ |=1 0 0 (I ) e(I ) 3 dr f (· + r ζ ) − f 2L 2 (Rn ) dζ ln r r |ζ |=1 0 e(I ) 3 dy 2 | f (x + y) − f (x)| dx ln . |y| |y|n |y|<(I ) Rn (I )
t −2n−2
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N–S Systems via Q–Q−1 Spaces
On the other hand, via taking ⎧ 2 ⎪ ⎨μ(s) = s ln e(I ) ; s 3 ⎪ ⎩ν(s) = s 3 ln e(I ) , s and estimating t μ(s) ds 0
∞
ν(s)
t
−1
ds
e(I ) −2 e(I ) 2 t ln t ln ≈ 1, t t
we achieve the following form of the Hardy inequality (cf. [21] or [1, Lemma 11]) 2 (I ) ∞ ∞ |ψ(t)| dt μ(s) ds |ψ(t)|2 ν(t) dt, 0
s
0
thereby getting (I ) ∞
2 e(I ) 2 f (· + r ζ ) − f L 2 (Rn ) dζ r −2 dr t ln dt t |ζ |=1 0 t ∞ e(I ) 3 dr f (· + r ζ ) − f 2L 2 (Rn ) dζ ln r r |ζ |=1 0 e(I ) 3 dy 2 | f (x + y) − f (x)| dx ln . |y| |y|n Rn Rn
Accordingly, we obtain (I ) 0
I
Rn
Rn
t (I )
dxdt t n−1 | f (x + y) − f (x)|2 e(I ) 3 n ln |y| dxdy. (I )|y|
|∂x j F(t, x)|2 ω0
Now, for J ∈ F with (J ) ≥ 3(I ) and the same centre as I ’s let ϕ be a function with ⎧ ⎪ 0 ≤ ϕ ≤ 1; ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ϕ = 1 on (2/3)J ; supp(ϕ) ⊂ (3/4)J ; ⎪ −1 ⎪ ⎪ |ϕ(x) − ϕ(y)| (J ) |x − y|; ⎪ ⎪ ⎪ ⎩ f = f + ( f − f )ϕ + ( f − f )(1 − ϕ). J J J Following the second half of the argument for [8, Lemma 3.1] and noticing the basic estimate f BMO(Rn ) [ f ]Q0 (Rn ) f Q0<α<1 (Rn ) ,
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J. Xiao, J. Zhang
we can readily find
(I )
0
t (I )
|∂x j F(t, x)| ω0 2
I
sup
dxdt t n−1
| f (x + y) − f (x)|2 e(J ) 3 ln dxdy, n |y| (J )|y|
|y|<(J )
J ∈F
J
as desired. n 2.2 Q−1 α (R ) under (α, n − 1) ∈ (−∞, 1) × N
Proposition 2.1 leads to the following assertion. Proposition 2.2 Let (α, n − 1) ∈ (−∞, 1) × N and f be a tempered distribution on n Rn . Then f ∈ Q−1 α (R ) if and only if f Q−1 := n α (R ),∗
r2
sup (r,x)∈R1+n +
B(x,r )
0
|e
t
1 √ t dxdt 2 f (y)| ωα < ∞. n r t2 (2.2) 2
Proof Since Proposition 2.2 under α = 0 is known (cf. [19,22,31,32]), it suffices to handle the case α = 0. n On the one hand, suppose that f ∈ Q−1 0 (R ). From Definition 1.1(ii) it follows n that there exist f 1 , . . . , f n ∈ Q0 (R ) such that f (x) =
n
∂x j f j (x).
j=1
Via the Minkowski inequality we get n ∂x j f j f Q−1 (Rn ),∗ = 0
j=1
n Q−1 0 (R ),∗
n
f j Q0 (Rn ) = f Q−1 (Rn ) . 0
j=1
On the other hand, suppose that (2.2) is valid for α = 0. Then f Q−1 (Rn ),∗ < ∞. 0 For simplicity set ∂ j = ∂x j and f j,k (x) = ∂ j ∂k (−)−1 f (x) = ∂x j ∂xk (−)−1 f (x) ∀ j, k = 1, . . . , n. n As long as f j,k Q−1 (Rn ),∗ < ∞ is proved, we will achieve f ∈ Q−1 0 (R ) thanks to 0
f j,k Q−1 (Rn ),∗ < ∞ ⇒ f k = −∂k (−)−1 f ∈ Q0 (Rn ), 0
and there exists the representation f =
123
n
k=1 ∂k f k
via
N–S Systems via Q–Q−1 Spaces
⎛ ⎞ n n n √ √ √ ⎝ ∂k f k ⎠ (ξ ) = − −1ξk fˆk (ξ ) = −1ξk −1ξk |ξ |−2 fˆ(ξ ) = fˆ(ξ ). k=1
k=1
k=1
Now, it remains to validate f j,k Q−1 (Rn ),∗ < ∞. In doing so, let us take a function 0 φ ∈ C0∞ (Rn ) such that ⎧ supp(φ) ⊂ B(0, 1) = {x ∈ Rn : |x| < 1}; ⎪ ⎪ ⎪ ⎨ φ(x)dx = 1; Rn ⎪ φr (x) = r −n φ( rx ); ⎪ ⎪ ⎩ gr (t, x) = φr ∗ ∂ j ∂k (−)−1 et f (x). Then et f j,k (x) = ∂ j ∂k (−)−1 et f (x) = fr (t, x) + gr (t, x). 1 (Rn ) stands for the predual of B −1 (Rn ), then ˙ ∞,∞ If B˙ 1,1
gr (t, ·) L ∞ (Rn ) φ B˙ 1
1,1 (R
n)
∂ j ∂k (−)−1 et f B˙ ∞,∞ −1 (Rn )
r −1 f B˙ ∞,∞ −1 (Rn ) r −1 f BMO−1 (Rn ) r −1 f Q−1 (Rn ),∗ , 0
and hence √ t dydt |gr (t, x)| ω0 n r t2 0 B(x,r ) n 2 2 r2 2 t er dydt 2 −2 ln f Q−1 (Rn ),∗ r n 2 er t 0 t2 0 B(x,r ) r 2 2 2 er 2 −2 ln f Q−1 (Rn ),∗r dt t 0 0
r2
2
f 2Q−1 (Rn ),∗ . 0
Next, we estimate fr . Upon taking ϕ ∈ C0∞ (Rn ) with ⎧ n ⎪ ⎪ϕ(y) = 1 ∀ y ∈ B(0, 10) = {x ∈ R : |x| < 10}; ⎪ ⎨ϕ (y) = ϕ y−x ; r,x r ⎪ fr (y) = Fr,x (t, y) + G r,x (t, y); ⎪ ⎪ ⎩ G r,x (y) = ∂ j ∂k (−)−1 ϕr,x et f (y) − φr ∗ ∂ j ∂k (−)−1 ϕr,x et f (y),
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J. Xiao, J. Zhang
and using the Plancherel formula for the space variable, we obtain √ t dt n r t2 0 √ r2 ∧ t dt 2 −1 t ∂ j ∂k (−) ϕr,x e f 2 n ω0 n L (R ) r t2 0 r 2 2 √t dt −2 t y j yk |y| (ϕr,x e f )(y) dy ω0 n n r t2 0 R √ r2 2 t dt ϕr,x et f 2 n ω0 n (R ) L r t2 0 √ r2 t dt ϕr,x et f 2 2 n ω0 n . L (R ) r t2 0
r2
∂ j ∂k (−)−1 ϕr,x et f 2L 2 (Rn ) ω0
Similarly, there holds √ r2 t dt φr ∗ ∂ j ∂k (−)−1 ϕr,x et f 2L 2 (Rn ) ω0 n r t2 0 √ r2 t dt ϕr,x et f 2 2 n ω0 n . L (R ) r t2 0 Thus we can obtain the following inequality for G r,x : √ √ r2 r2 t dt t dt t 2 ϕ G r,x (t, ·) 2L 2 (Rn ) ω0 e f ω r,x n n . L 2 (Rn ) 0 r r 2 t t2 0 0 Meanwhile, in order to bound Fr,x , upon noticing |Fr,x (t, y)|2 dy r n+1 |et f (w)|2 |x − w|−(n+1) dw B(x,r )
|w−x|≥10r
(as shown in [22, p.161]), we get √ r 2 t dt 2 |Fr,x (t, y)| dy ω0 n r t2 0 B(x,r ) 2
√ dt t r n+1 |x − w|−(n+1) |et f (w)|2 ω0 dw n r t2 |w−x|≥10r 0 2 √ ∞ r t dt n+1 −(n+1) t 2 |x − w| |e f (w)| ω0 r dw n l l+1 r t2 0 l=1 10 r ≤|w−x|<10 r 2 √ ∞ r t dt −l(n+1) t 2 10 |e f (w)| ω0 dw n r t2 |w−x|<10l+1 r 0 r
l=1
f 2Q−1 (Rn ),∗ . 0
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N–S Systems via Q–Q−1 Spaces
The estimates for G r,x and Fr,x give 0
r2
B(x,r )
| fr (t, y)|2 ω0
√ t dt 2 . n dy f −1 n Q0 (R ),∗ r 2 t
The foregoing argument actually reveals f j,k Q−1 (Rn ),∗ < ∞.
0
3 Proof of Theorem 1.3(i) This section is devoted to demonstrating the first part of Theorem 1.3. 3.1 Quadratic Estimate for an Integral Operator Lemma 3.1 Let (α, n − 1) ∈ (−∞, 1) × N, ϑ(t, x) be a function on R1+n + and ⎧ ⎨ϑ(t, x) = 0t e(t−s) ϑ(s, x)ds; √ r2 t dydt sup n . ⎩[ϑ]α = 0 B(x,r ) |ϑ(t, y)|ωα r 2 t
(r,x)∈(0,1)×Rn
Then, (i) the global estimate is
∞ 0
√ dt ϑ(t, ·) 2L 2 (Rn ) ωα ( t) n t2
∞ 0
√ dt ϑ(t, ·) 2L 2 (Rn ) ωα ( t) n . t2
(ii) the local estimate is 2 1 t √ √ dt −et ϑ(s, ·)ds 2 n ωα ( t) n2 t 0 0 L (R ) 1 √ dt [ϑ]α ϑ(t, ·) L 1 (Rn ) ωα ( t) n . t2 0 Proof The case α = 0 of Lemma 3.1 is known (cf. [19,22,31,32]), so it remains to check the case α = 0. (i) Without loss of generality, we may define ϑ = 0 = ϑ for t ∈ (−∞, 0). Upon setting |x|2 t − n2 e (0, x) = (4π t) exp − 4t
and defining ω(t, x) =
et (0, x) as 0 as
t > 0; t ≤ 0,
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J. Xiao, J. Zhang
we get ϑ(t, x) =
Rn
Rn
ω(t − s, x − y)ϑ(s, y) dyds,
and hence becomes a convolution operator over R1+n . Since ω(t, ·)(ζ ) =
Rn
ω(t, x)e−2π
√
−1x·ζ
dx = −(2π )2 |ζ |2 e−(2π )
2 t|ζ |2
∀ t > 0,
we have ·)(ζ ) = ϑ(t,
=
R1+n R1+n
√ −2π −1x·ζ
ω(t − s, x − y)ϑ(s, y)e dx dyds √ −2π −1(u+y)·ζ ϑ(s, y) ω(t − s, u)e du dyds Rn
Rn
t
√ 2 2 ϑ(s, y)e−2π −1y·ζ −(2π )2 |ζ |2 e−(2π ) (t−s)|ζ | dyds 0 Rn t 2 2 = −(2π )2 ·)(ζ )ds. |ζ |2 e−(2π ) (t−s)|ζ | ϑ(s,
=
0
The above equality, together with the Fubini theorem and the Plancherel formula, implies
∞
√ dt ϑ(t, ·) 2L 2 (Rn ) ω0 ( t) n t 2 0 ∞ √ ·)(x)|2 dx ω0 ( t) dtn = |ϑ(t, n t2 R 0 2 ∞ t 2 √ dt |ζ | |ϑ(s, ·)(ζ )|ds dζ ω0 ( t) n 2 2 t2 0 0 e(2π ) (t−s)|ζ | Rn ∞ 2 t √ dt |ζ |2 | ϑ(s, ·)(ζ )|ds ω0 ( t) n dζ 2 (t−s)|ζ |2 (2π ) t2 Rn 0 0 e 2 ∞ ∞ √ dt |ζ |2 1{0≤s≤t} |ϑ(s, ·)(ζ )|ds ω0 ( t) n dζ. 2 2 e(2π ) (t−s)|ζ | t2 0 0 Rn
An application of the Schur lemma deduces that if ⎧ √ 1 2 ⎪ ⎨ J (s, ζ ) = |ϑ(s, ·)(ζ )| ω0 ( n s) ; s2 √ 1 √ 1 ⎪ ⎩ K (s, t) = 1{0≤s≤t} |ζ |2 ω0 ( n s) − 2 ω0 (n t) 2 , (2π )2 (t−s)|ζ |2 e
123
s2
t2
N–S Systems via Q–Q−1 Spaces
then
∞
0
≈
0
0
≈
0
∞
1{0≤s≤t} |ζ |2 |ϑ(s, ·)(ζ )|
0
∞ ∞ ∞ ∞
2 2 e(2π ) (t−s)|ζ |
2 ds
2
K (s, t)J (s, ζ )ds
√ dt ω0 ( t) n t2
dt
0
(J (t, ζ ))2 dt √ dt |ϑ(t, ·)(ζ )|2 ω0 ( t) n . t2
Accordingly, we can use the Plancherel formula again to derive ∞ ∞ √ dt √ dt 2 ϑ(t, ·) L 2 (Rn ) ω0 ( t) n ϑ(t, ·) 2L 2 (Rn ) ω0 ( t) n t2 t2 0 0 as desired. (ii) In the sequel, ·, · stands for the inner product acting on L 2 (Rn ). So 2 t 1 √ √ dt −et ϑ(s, ·)ds 2 n ω0 ( t) n2 t 0 0 L (R ) 2 1 t √ √ dt −et = ϑ(s, y)ds dy ω0 ( t) n t2 Rn 0 0 1 t t √ √ √ dt = −et ϑ(s, ·), −et ϑ(h, ·)dsdh ω0 ( t) n t2 0 0 0 1 √ dt ϑ(s, ·), =2 (−)e2t ϑ(h, ·)ω0 ( t) n dsdh t2 s 0
0
Thus we obtain 2 1 t √ −et ϑ(s, ·)ds 2 0
0
L
sup
(r,x)∈(0,1)×Rn
r −n
√ dt ω0 ( t) n [ϑ]0 t2 (Rn )
r2 0
0
1
B(x,r )
|ϑ(s, y)| dyds.
√ dt ϑ(t, ·) L 1 (Rn ) ω0 ( t) n . t2
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J. Xiao, J. Zhang
3.2 Well-Posedness for Incompressible N–S System It should be noticed that the case α = 0 of Theorem 1.3(i) is known (cf. [19,31,32]). So it suffices to demonstrate the case α = 0. Here it is perhaps appropriate to point out that our logarithm-oriented approach is a non-trivial enhance of the idea introduced by [19] and developed in [31,32] as well as in [22]. Because a combination of [22, Lemma 16.1] and Proposition 2.2 yields √ t t e a j L ∞ (Rn ) sup
t
x∈Rn
0
|e B(x,t)
s
dyds a j (y)| tn 2
21
a j Q−1 (Rn ) ∀ t > 0, 0,∗
according to Picard’s contraction principle, it remains to verify the bilinear operator
t
(u, v)(t, x) =
e(t−s) P∇ · (u ⊗ v)(x) ds
0 1+n 1+n n n is bounded from X 0n (R1+n + ) × X 0 (R+ ) to X 0 (R+ ). This will proceed in two steps. ∞ Step 1 One establishes the L -bound: 1 t 2 |(u, v)(t, x)| u X n (R1+n ) v X n (R1+n ) . sup (3.1) 0
(t,x)∈R1+n +
This follows from two considerations. First, if e(t−s) P∇ · (u ⊗ v) L ∞ (Rn )
t 2
+
+
0
≤ s < t then
u L ∞ (Rn ) v L ∞ (Rn ) √ t −s 1
(t − s)− 2 s −1 u X n (R1+n ) v X n (R1+n ) , +
0
0
+
and hence t t 1 |e(t−s) P∇ · (u ⊗ v)(x)|ds t −1 (t − s)− 2 ds u X n (R1+n ) v X n (R1+n ) t 2
t 2
0
+
1
t − 2 u X n (R1+n ) v X n (R1+n ) . 0
+
0
+
Second, if 0 < s < 2t , then |e(t−s) P∇ · (u ⊗ v)(x)| |u(s, y)||v(s, y)| # n+1 dy Rn t 2 + |x − y| −(n+1) √ t(|k| + 1) k∈Zn
123
|u(s, √ x−y∈ t(k+[0,1]n )
y)||v(s, y)|dy,
0
+
N–S Systems via Q–Q−1 Spaces
and hence (via the Cauchy–Schwarz inequality) t 2 |e(t−s) P∇ · (u ⊗ v)(x)|ds 0
t 2
0
√
t(|k| + 1)
−(n+1)
k∈Zn
|u(s, √ x−y∈ t(k+[0,1]n )
y)||v(s, y)|dyds
1
t − 2 u X n (R1+n ) v X n (R1+n ) . 0
+
+
0
The above two treatments are placed together to derive (3.1): t t 2 1 |(u, v)| + |e(t−s) P∇ · (u ⊗ v)(x)|ds t − 2 u X n (R1+n ) v X n (R1+n ) . t 2
0
0
+
0
+
Step 2 One establishes the L 2 -bound: √ ⎞ 21 ⎛ r2 ω0 r s sup ⎝ |(u, v)(s, y)|2 dyds ⎠ n 2 1+n s 0 B(x,r ) (r,x)∈R+ u X n (R1+n ) v X n (R1+n ) . +
0
0
+
(3.2)
To do so, define 1r,x (y) = 1|y−x|<10r (y), i.e. the characteristic function on the ball {y ∈ Rn : |y − x| < 10r }. We divide the bilinear operator (u, v) into three parts: (u, v) = 1 (u, v) − 2 (u, v) − 3 (u, v), where ⎧ s (s−h) ⎪ P∇ · ((1 − 1r,x )u ⊗ v)dh; ⎨1 (u, v) = √ 0 e s √ −1 2 (u, v) = ( −) P∇ · 0 e(s−h) (( −)−1 (I − eh )(1r,x )u ⊗ v)dh; ⎪ √ √ ⎩ s 3 (u, v) = ( −)−1 P∇ · −es 0 (1r,x )u ⊗ v)dh , and I stands for the identity operator. In what follows, we shall estimate |i (u, v)| for i = 1, 2, 3, respectively. Estimate for 1 (u, v): When 0 < s < r 2 and |y − x| < r , the Cauchy–Schwarz inequality produces |1 (u, v)|
s
|z−x|≥10r
0
0
r2
|z−x|≥10r
|u(h, z)||v(h, z)| √ n+1 dzdh s − h + |y − z| |u(h, z)||v(h, z)| dzdh |x − z|n+1
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J. Xiao, J. Zhang
r2
|z−x|≥10r
0
r2
×
|z−x|≥10r
0
r
−1
|u(h, z)|2 dzdh |x − z|n+1
1
|v(h, z)|2 dzdh |x − z|n+1
2
1 2
u X n (R1+n ) v X n (R1+n ) . +
0
0
+
So √ t dydt D1 := |1 (u, v)| ω0 n r t2 0 B(x,r ) 2 √ r t dydt −2 r ω0 u 2X n (R1+n ) v 2X n (R1+n ) n + + r 0 0 t2 0 B(x,r ) 2 r2 er ln √ ≈ r −2 dt u 2X n (R1+n ) v 2X n (R1+n ) + + 0 0 t 0
r2
2
u 2X n (R1+n ) v 2X n (R1+n ) . 0
+
0
+
Estimate for 2 (u, v): Employing L 2 -boundedness of the Riesz transform and Lemma 3.1(i) yields √ t dydt D2 := |2 (u, v)| ω0 n r t2 0 B(x,r ) √ r2 t dt = 2 (u, v) 2L 2 (B) ω0 n r t2 0 r 2 s √ 2 (s−h) −1 h e ( −) (I − e )(1r,x )u ⊗ v dh
r2
0
2
0
r2
0
√ 2 ( −)−1 (I − es )(1r,x )u ⊗ v 2 L
√ t dt ω0 n r t2 L 2 (Rn )
√ s ds ω0 n . n (R ) r s2
Since sup s −1 (1 − e−s ) < ∞, 2
s∈(0,∞)
√ √ the operator ( −)−1 (I − es ) is bounded on L 2 (Rn ) with its operator norm s. Upon setting (1r,x )u ⊗ v = (1r,x (y))u(s, y) ⊗ v(s, y) =: M(s, y),
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N–S Systems via Q–Q−1 Spaces
we have
r2
D2
s M(s, 0
y) 2L 2 (Rn ) ω0
√ s ds n D21 D22 r s2
where ⎧ ⎨D21 := sM(s, y) L ∞ ((0,r 2 )×Rn ) ; √ 2 ⎩D22 := 0r Rn |M(s, y)|ω0 r s
dyds n
s2
.
For D21 , we have D21 ≤
1 2
sup s u(s, ·) L ∞ (Rn )
s∈(0,∞)
1 2
sup s v(s, ·) L ∞ (Rn )
s∈(0,∞)
u X n (R1+n ) v X n (R1+n ) . 0
+
0
+
For D22 , we can use the Cauchy–Schwarz inequality to estimate D22 u X n (R1+n ) v X n (R1+n ) . +
0
+
0
Consequently, we get D2 u 2X n (R1+n ) v 2X n (R1+n ) . +
0
0
+
Estimate for 3 (u, v): Similarly for 2 (u, v), we utilize L 2 -boundedness of the Riesz transform and Lemma 3.1(ii) to derive √ t dydt n r t2 0 B(x,r ) 2 √ r2 t √ t dt t −e M(s, ·)ds n 2 ω0 r t2 0 0 L 2 1 t √ dt −et r4 M(r 2 s, ·) ds 2 ω0 (t) n2 t 0 0 L 2 2 u X n (R1+n ) v X n (R1+n ) .
D3 :=
r2
|3 (u, v)|2 ω0
0
+
0
+
Finally, putting the estimates for D1 , D2 , D3 together implies the desired (3.2):
r2 0
√ t dydt |(u, v)(t, y)| ω0 D1 + D2 + D3 n r t2 B(x,r ) 2
u 2X n (R1+n ) v 2X n (R1+n ) . 0
+
0
+
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J. Xiao, J. Zhang
4 Proof of Theorem 1.3(ii) This section is devoted to demonstrating the second part of Theorem 1.3. 4.1 Criterion for a Multiplicative Inequality Lemma 4.1 Let (γ , η, n − 1) ∈ (0, ∞) × (0, 1) × N and Ξ be a tempered distribution on Rn . Then the following three statements are equivalent: (i) The quadratic Trudinger-type inequality:
2γ 2 1+γ | f (x)| Ξ (x) dx ∇ f L1+γ f ∀ f ∈ C0∞ (Rn ). 2 (Rn ) L 2 (Rn ) n 2
R
(ii) The quadratic Nash-type inequality:
Rn
| f (x)|2 Ξ (x) dx ∇ f 2L 2 (Rn ) + −γ f 2L 2 (Rn ) ∀ ( , f ) ∈ (0, ∞) × C0∞ (Rn ).
(iii) Ξ = div(∇−1 Ξ ) = ∇ · ∇−1 Ξ with ⎧ γ −1 +η √ ⎪ γ +1 n) ⎪ ⎪ Lip − (R as γ ∈ (0, 1); ⎪ η ⎨ −1 −1 n as γ = 1; ∂x1 ( Ξ ), . . . , ∂xn ( Ξ ) ∈ BMO(R ) ⎪ √ γ −1 ⎪ ⎪ ⎪ ⎩ − γ +1 Q γ −1 (Rn ) as γ ∈ (1, ∞). γ +1
Proof This is a by-product of [25, Theorem V], [31, Theorem 1.2(i)], Proposition 2.1 and the basic equivalence (cf. [28]) f ∈ Lip 1−γ (Rn ) ⇔ sup 1+γ
x,y∈Rn
| f (x) − f (y)| 1−γ
|x − y| 1+γ
<∞⇔ f ∈
√ γ −1 +η γ +1 Lipη (Rn ) . −
Here, it is perhaps appropriate to mention that the divergence space of a suitable −1 (Rn ); see also Remark 1.4. formulation of limη→0 Lipη (Rn ) may be regarded as B˙ ∞,∞ 4.2 Zero Solution to Steady State of Incompressible N–S System The forthcoming proof of Theorem 1.3(ii) actually improves the argument for [26, Theorem 1.1] (coupled implicitly with Lemma 4.1). Suppose that u = (u 1 , . . . , u n ) is a smooth solution of (1.2) satisfying u 1 , . . . , u n ∈ BMO−1 (Rn ) ∩ L p,
123
p(n−2) 2
(Rn ) under 2 < p < ∞.
N–S Systems via Q–Q−1 Spaces
n Step 1 One establishes the Caccioppoli-type estimate for u ∈ BMO−1 (Rn ) : ∇u L 2 (B(x, r )) u − u B(x,r ) L p (B(x,r )) 2 ∀ (r, x, p) ∈ (0, ∞) × Rn × (2, ∞), n r n−2 p 2 r 2 (4.1) where u B(x,r ) still denotes the mean value of u over B(x, r ). In order to prove (4.1), for 2−1r ≤ t < s ≤ r we take a non-negative cut-off function χ ∈ C0∞ (B(x, r )) obeying ⎧ ⎪ ⎨χ = 1 χ =0 ⎪ ⎩ |∇χ | (s − t)−1
in B(x, t); in Rn \ B(x, s); in Rn .
For simplicity, upon setting u˘ = u − u B(x,r ) = u 1 − (u 1 ) B(x,r ) , . . . , u n − (u n ) B(x,r ) , we gain that for a given 2 < p < ∞ there is a function Φ (cf. [26, 2.1]) obeying: ⎧ ⎪ n ⎪ ⎨Φ|R \B(x,r ) = 0; divΦ = ∇χ · u˘ = div(χ u); ˘ ⎪ ⎪ ⎩ ∇Φ p ∇χ · u ˘ L p B(x,s) (s − t)−1 u ˘ L p (B(x,s)) . L B(x,s) Now, we utilize χ u˘ − Φ to test (1.2), integrate by parts over B(x, s), and find χ (y)|∇u(y)|2 dy =: E1 + E2 + E3 + E4 , B(x,s)
where ⎧ ⎪ E1 ⎪ ⎪ ⎪ ⎨E 2 ⎪ E 3 ⎪ ⎪ ⎪ ⎩E 4
= − B(x,s) ∇u(y) : ∇χ (y) ⊗ u(y) ˇ dy; = B(x,s) ∇Φ(y) : ∇u(y) dy; ˇ dy; = − B(x,s) u(y) · ∇u(y) · χ (y)u(y) = B(x,s) u(y) · ∇u(y) · Φ(y) dy.
It remains to estimate E1 , E2 , E3 , E4 , respectively. From Hölder’s inequality it follows that |∇u(y)||∇χ (y)||u(y)|dy ˇ |E1 | + |E2 | B(x,s)
(s − t)−1
1 |∇u(y)|2 dy B(x,s)
1
2
2 |u(y)| ˇ dy
2
B(x,s)
123
J. Xiao, J. Zhang
r 2p s−t
≈r
n( p−2) 2p
1
n( p−2)
|∇u(y)|2 dy
B(x,r )
B(x,s)
(s − t)
−1
1
2
p |u(y)| ˇ dy
p
∇u L 2 (B(x,s)) u ˇ L p (B(x,r )) .
In order to control E3 and E4 , we must use the assumption u 1 , . . . , u n ∈ BMO−1 (Rn ) which generates a skew symmetric matrix = (1k )nk=1 , . . . , (nk )nk=1 such that ⎧ n , . . . , div( )n ⎪ ⎪ u = (u , . . . , u ) = div = div( ) 1 n 1k nk ⎨ k=1 k=1 ; n u j = ∇ · ( j1 , . . . , jn ) = div( jk )k=1 ; ⎪ ⎪ ⎩ ∈ BMO(Rn ). jk Using integration by parts, Hölder’s inequality, p ∈ (2, ∞) and the skew symmetry of , we deduce (u(y) · ∇ u(y)) ˇ · χ (y)u(y) ˇ dy |E3 | = B(x,s) n ˇ = jk,k (y)uˇ i, j (y)uˇ i (y)χ (y) dy i, j,k=1 B(x,s) n ˇ = jk (y)uˇ i, j (y)uˇ i (y)χ,k (y) dy i, j,k=1 B(x,s) 1 1 2 2 2 2 ˇ (s − t)−1 |∇u(y)|2 dy |(y)| |u(y)| ˇ dy B(x,s) −1
(s − t) ≈r r
n( p−2) 2p n( p−2) 2p
∇u L 2 (B(x,s)) u ˇ L p (B(x,s))
(s − t)
−1
B(x,s)
2p
p−2 dy ˇ |(y)|
p−2 2p
B(x,s)
∇u L 2 B(x,s) u ˇ L p (B(x,r )) BMO(Rn ),2 p/( p−2)
ˇ L p (B(x,r )) BMO(Rn ) . (s − t)−1 ∇u L 2 B(x,s) u
In the above and below, f ,k (y) represents ∂ yk f (y) = ∂k f (y). Similarly, we have n ˇ |E4 | = jk,k (y)uˇ i, j (y)Φi (y) dy i, j,k=1 B(x,s)
123
N–S Systems via Q–Q−1 Spaces
n ˇ = jk (y)uˇ i, j (y)Φi,k (y) dy i, j,k=1 B(x,s) 1 1 2 2 2 2 2 ˇ |∇u(y)| dy |(y)| |∇Φ(y)| dy
B(x,s)
1 |∇u(y)|2 dy
2
B(x,s) −1
(s − t) r ≈r
B(x,s)
1 |∇Φ(y)| p dy
B(x,s)
p
B(x,s)
∇u L 2 (B(x,s)) u(y) ˇ L p (B(x,r ))
2p
p−2 dy ˇ |(y)|
2p
p−2 dy ˇ |(y)|
p−2 2p
p−2 2p
B(x,s)
n( p−2) 2p
(s − t)−1 ∇u L 2 (B(x,s)) u ˇ L p (B(x,r )) BMO(Rn ),2 p/( p−2)
n( p−2) 2p
ˇ L p (B(x,r )) BMO(Rn ) . (s − t)−1 ∇u L 2 B(x,s) u
Accordingly, we obtain |∇u(y)|2 dy B(x,t)
χ (y)|∇u(y)|2 dy B(x,s)
|E1 | + |E2 | + |E3 | + |E4 |
(s − t)−1 ∇u L 2 (B(x,s)) u ˇ L p (B(x,r )) 1 + BMO(Rn ),2 p/( p−2) n( p−2) ≈ r 2 p (s − t)−1 ∇u L 2 (B(x,s)) u ˇ L p (B(x,r )) 1 + BMO(Rn ) .
r
n( p−2) 2p
This, along with Young’s inequality ρτ ≤ 2−1 (ε−1 ρ 2 + ετ 2 ) ∀ (ρ, τ, ε) ∈ (0, ∞) × (0, ∞) × (0, ∞), produces a constant κ p > 0 depending only on p ∈ (2, ∞) such that ∇u 2 2 L
B(x,t)
≤ 2−2 ∇u 2 2 + κ pr L (B(x,s))
n( p−2) p
(s − t)−2 u ˇ 2L p (B(x,r )) .
Now, by iteration we get (4.1). p(n−2) Step 2 One utilizes u 1 , . . . , u n ∈ L 2< p<∞, 2 (Rn ) to verify u ≡ 0. p(n−2) If u 1 , . . . , u n ∈ L 2< p<∞, 2 (Rn ) then an application of (4.1) gives ∇u j L 2 (B(x, r )) u j 2
L 2< p<∞,
p(n−2) 2 (Rn )
< ∞,
123
J. Xiao, J. Zhang
whence u j W˙ 1,2 (Rn ) u j
L 2< p<∞,
p(n−2) 2 (Rn )
< ∞.
Upon utilizing the local Gagliardo–Nirenberg–Soblov inequality for uˇ j : uˇ j
qn
L n−q (B(x,r ))
∇u j L q (B(x,r )) ∀ (r, x) ∈ R1+n + and q ∈ [1, n)
and (4.1), we achieve the reverse Hölder inequality ∇u j L 2 (B(x, r )) −n 2 r n r 2
2
B(x,r )
|∇u j (y)|
pn p+n
p+n pn
dy
∀ (r, x) ∈ R1+n + .
(4.2)
Now, from the above-validated fact u j ∈ W˙ 1,2 (Rn ) it follows that pn
ϒ p := |∇u j | p+n ∈ L
2( p+n) pn
(Rn ).
Moreover, if M( f ) is the Hardy–Littlewood maximal function of a given function f on Rn : | f (y)| dy ∀ x ∈ Rn , M( f )(x) = sup r −n B(x,r )
r ∈(0,∞)
then the operator M is bounded on L q>1 (Rn ), and hence (4.2) derives 2( p+n) pn
M(ϒ p
2( p+n) )(x) M(ϒ p )(x) pn
under
2( p + n) > 1. pn
Consequently, 2( p+n) ∇u j 2L 2 (Rn ) = ϒ p pn
L 1 (Rn )
2( p+n) ≤ M ϒ p pn ϒ p
2( p+n) = ϒ p pn =
123
L 1 (Rn )
2( p+n) pn 2( p+n) L pn (Rn )
L 1 (Rn ) 2 ∇u j L 2 (Rn ) .
N–S Systems via Q–Q−1 Spaces
This last estimation shows that 2( p+n) pn M ϒ p
L 1 (Rn )
≈ ∇u j 2L 2 (Rn ) < ∞.
However, according to [14, Example 2.1.8] or [13] we have 2( p+n) pn
ϒp
2( p+n) ≡ 0 #⇒ M ϒ p pn (x) ≈ |x|−n ∀ |x| $ 1,
thereby reaching a contradiction ∞=
|x|$1
2( p+n) |x|−n dx M ϒ p pn
L 1 (Rn )
< ∞.
Consequently, ϒ p ≡ 0, and so u j is a constant. With the help of the above consideration, we use the boundary value condition lim u j (x) = 0
|x|→∞
to deduce that the constant u j is actually zero, and therefore u completely vanishes.
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