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Asymptotic behavior for the Navier–Stokes equations in 2D exterior domains
Journal of Functional Analysis 240 (2006) 508–529 www.elsevier.com/locate/jfa
Asymptotic behavior for the Navier–Stokes equations in 2D exterior domains ✩ Hyeong-Ohk Bae a,∗ , Bum Ja Jin b a Department of Mathematics, Ajou University, Suwon 443-749, Republic of Korea b Korea Science Academy, Busanjin-Gu, Busan-Si, Republic of Korea
Received 23 December 2005; accepted 18 April 2006 Available online 16 June 2006 Communicated by H. Brezis
Abstract We show that the Lp spatial–temporal decay rates of solutions of incompressible flow in an 2D exterior domain. When a domain has a boundary, pressure term makes an obstacle since we do not have enough information on the pressure term near the boundary. To overcome the difficulty, we adopt the ideas in He, Xin [C. He, Z. Xin, Weighted estimates for nonstationary Navier–Stokes equations in exterior domain, Methods Appl. Anal. 7 (3) (2000) 443–458], and our previous results [H.-O. Bae, B.J. Jin, Asymptotic behavior of Stokes solutions in 2D exterior domains, J. Math. Fluid Mech., in press; H.-O. Bae, B.J. Jin, Temporal and spatial decay rates of Navier–Stokes solutions in exterior domains, submitted for publication]. For the spatial decay rate estimate, we first extend temporal decay rate result of the Navier–Stokes solutions for general Lp space when the initial velocity is in L2σ ∩ Lr , 1 < r q < ∞ (1 < r < q = ∞). © 2006 Elsevier Inc. All rights reserved. Keywords: 2D Navier–Stokes; Temporal–spatial decay rate; Exterior domain
1. Introduction Let u and p be the velocity and pressure, respectively, of the incompressible fluid in an exterior domain. Let Ω be the exterior of a simply connected set B in R2 which contains the origin and is contained in a bounded ball, for example, the unit ball. Furthermore, we assume that the boundary ✩
This work was supported by Korea Research Foundation Grant (KRF-2003-015-C00052).
* Corresponding author. Fax: +82 31 216 3027.
E-mail addresses: [email protected] (H.-O. Bae), [email protected] (B.J. Jin). 0022-1236/$ – see front matter © 2006 Elsevier Inc. All rights reserved. doi:10.1016/j.jfa.2006.04.029
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509
∂B is smooth enough to get the unique existence of the solution. We consider the Navier–Stokes equations in Ω: ∂ ∂t u + (u · ∇)u − u + ∇p
=0
∇ ·u=0
in (x, t) ∈ Ω × (0, ∞)
(1.1)
with no slip boundary condition u(x, t) = 0 for x ∈ ∂Ω and initial data u(x, 0) = u0 (x). Temporal decay estimates have been considered for various domains by Kato [21], Fujigaki and Miyakawa [16], Miyakawa and Schonbek [25], Miyakawa [24], Wiegner [29], Schonbek [26], Galdi and Maremonti [18], Kozono and Ogawa [22], Borchers and Miyakawa [9,10], Bae and Choe [4], Bae [2] and spatial decay estimates have been considered by He [19], He and Xin [20], Takahashi [28], Farwig and Sohr [15], Brandolese [12], Bae and Jin [5,6,8], Bae [1,3], etc. When the whole space R3 has been considered, we have uniform upper and lower bounds in 2 L space with spatial weight in [6,8]. While a fluid region with boundary has been considered, the situation is no longer simple. Since we do not have pressure representation in arbitrary domains, it is difficult to make use of the energy method. In [20], He and Xin have considered an exterior domain problem, by removing pressure term in a solution representation, they have obtained decay rate results for the strong solution, that is, for small data. They obtained the spatial decay of a strong solution: 3(1− 1 ) 1 + |x|2 2 7 p u ∈ L∞ 0, ∞; Lp (Ω)
for 7 < p ∞.
Very recently, we [8] also obtained the decays for weak and strong solutions for 3D exterior domains. More precisely, there is a suitably weak solution u of (1.1) satisfying the asymptotic property: for any sufficiently small δ > 0 there is a positive constant cδ depending on δ and independent of t such that for 1 < r < 65 , |x|u(t)
5
L2
3
cδ (1 + t) 4 − 2r +δ
for any t > 0. For strong solutions, for p 2 and for small δ > 0, 2 |x| u(t)
Lp
1− 32 ( 1r − p1 )+δ
cδ (1 + t)
.
However, the above results for exterior domains are all for the 3D Navier–Stokes equations. The temporal decay rates for the 2D Navier–Stokes is known by Kozono and Ogawa [22] and Dan and Shibata [14]: −1+1 uq = o t 2 q
for q 2,
1 u∞ = o t − 2 ,
if u0 ∈ L2σ . Borchers and Miyakawa [11] obtained the temporal L2 estimate of the Navier–Stokes solutions in 2D exterior domains: 1
1
u2 ct − r + 2 ,
1 < r 2,
if u0 ∈ Lr ∩ L2σ . It was difficult for us to understand some parts of their proof. We provide a different proof for more general estimates, 1 < q ∞, in Section 3 independently, which is
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stated in Theorem 1.1. Combining the series of papers of Dan and Shibata [13,14] for the 2D Stokes equations, we [7] estimated the temporal–spatial decay rates for the 2D Stokes solutions for the exterior domains: uq ct
− 1r + q1
,
1 < r q < ∞, or 1 < r < q = ∞,
if u0 ∈ Lr ∩ L2σ . In this paper, we intend to derive spatial–temporal decay estimates of solutions for the Navier– Stokes equations in the 2D exterior domains. For this purpose we modify the ideas in Bae, Jin [8] and He, Xin [20] done for 3D cases, and we can avoid the computations involving with the pressure term. For our main temporal–spatial estimates, we first obtain the temporal estimates as mentioned above. We state our main theorems below, of which proofs are the main objectives of the subsequent sections. Theorem 1.1. Let 1 < r q < ∞ or 1 < r < q = ∞. Assume u0 ∈ L2σ ∩ Lr . Let u be the solution of Navier–Stokes equations of two-dimensional exterior domain Ω with the initial velocity u0 . Then we have u(t)
Lq
−1+1 =O t r q
for 1 < r q < ∞ or 1 < r < q = ∞. Moreover, we have ∇u(t)
Lq
−1+1−1 =O t r q 2
for 1 < r q 2. Theorem 1.2. Let Ω be a two-dimensional exterior domain. Let u0 ∈ Lr (Ω) ∩ L2 (Ω) with 2r 2p 1 < r p+2 < 2 p < ∞, and ∇ · u0 = 0. We also assume |x|u0 ∈ L 2−r . Let 0 < α 1. Then there is a strong solution u of the Navier–Stokes equation (1.1) satisfying the following asymptotic property: for any small δ > 0 1 + |x| α u(t)
Lp (Ω)
− 1 + 1 + α +δ =O t 2 p 2
for any large t. In Section 2 we state some well-known facts and lemmas for our theorems, and in Section 3 we provide temporal decays. Section 4 is the proof of Theorem 1.2. 2. Preliminaries p
We use the usual notation that Lσ means the set of measurable functions in Lp with divergence free. Let PΩ be the Leray projection from L2 (Ω) → L2σ (Ω), and let −AΩ = PΩ be the generator of semigroup e−AΩ t . Let us recall the following well-known decay estimates of the Stokes semigroup in a two-dimensional exterior domain (refer to [14,27]).
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511
Proposition 2.1. Let f ∈ Lrσ . Then it holds that −A t e Ω f
Lq
ct
Lq
ct
−( 1r − q1 )
fr ,
t > 0,
(2.1)
for 1 < r q < ∞, 1 < r < q = ∞; −A t ∇e Ω f
−( 1r − q1 )− 12
fr ,
t > 0,
(2.2)
for 1 < r q 2. Suppose u is a solution of Navier–Stokes equations of two-dimensional exterior domain when the initial velocity u0 ∈ L2σ . Dan and Shibata [13] showed the following (see also Kozono, Ogawa [22]). Proposition 2.2. u(t)
Lq
∇u(t)
−( 1 − 1 ) for 2 q < ∞, =o t 2 q
L2
1 = o t− 2
(2.3)
as t goes to infinity. Moreover, u(t)
L∞
1 = o t− 2 .
Lemma 2.3. Let a < 1, b > 0, d < 1. If b + d < 1, then t
(t − s)−a (s + 1)−b s −d ds ct 1−a−d (1 + t)−b ;
0
if b + d = 1, then t
(t − s)−a (s + 1)−b s −d ds ct −a ln (t + 1);
0
if b + d > 1, then t
(t − s)−a (s + 1)−b s −d ds ct −a .
0
Proof. Let I =
t
t/2 (t
− s)−a (s + 1)−b s −d ds, II =
−b −d
I c(1 + t)
t
t
t/2
t/2 0
(t − s)−a (s + 1)−b s −d ds. Then
(t − s)−a ds = c(1 + t)−b t −d t 1−a = c(1 + t)−b t 1−a−d ,
(2.4)
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II ct
−a
t/2 (s + 1)−b s −d ds. 0
If t 2, then t/2 t/2 1 −b −d −b −d (s + 1) s ds = (s + 1) s ds + (s + 1)−b s −d ds 0
1
0
t/2 c
−b−d
(s + 1)
1 ds + c
1
⎧ 1−b−d ⎨ c(t + 1) c ln (t + 1) ⎩ c
s −d ds
0
if b + d < 1, if b + d = 1, if b + d > 1.
If t 2, then t/2 1 −b −d (s + 1) s ds c (s + 1)−b s −d ds c. 0
0
Hence we have ⎧ −a 1−b−d ⎨ t (t + 1) −a II c t ln (t + 1) ⎩ t −a
if b + d < 1, if b + d = 1, if b + d > 1.
Hence we have t 0
(t − s)−a (s + 1)−b s −d ds ⎧ 1−a−d (t + 1)−b + t −a (t + 1)1−b−d ⎪ ⎨t c t 1−a−d (t + 1)−b + t −a ln (t + 1) ⎪ ⎩ 1−a−d t (t + 1)−b + t −a
Lemma 2.4. Let α <
1 2
if b + d < 1, if b + d = 1, if b + d > 1.
2
and > 0 be given constants. Let us consider an integral inequality
X(t) ct
−α
+ t
−1/2
t 0
1
s − 2 X(s) ds.
(2.5)
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513
We assume that
lim t
−
t
t→0+
1
s − 2 X(s) ds = 0.
0
Then, there is 0 so that if 0 , then X(t) ct −α for some c independent of t. Proof. Set J (t) =
t 0
1
s − 2 X(s) ds, then J (t) satisfies the inequality 1 d J ct −α t − 2 + t −1 J (t). dt
This can be rewritten by 1 d − t J (t) ct −α− 2 − . dt If we integrate over (0, t) we have
t
−
t J (t) c
1
s −α− 2 − ds.
0
Here, we used the hypothesis limt→0+ t − J (t) = 0. Suppose that α + < 12 . By Lemma 2.3, t
1
1
s −α− 2 − ds ct −α+ 2 − ,
0
therefore, we have 1
J (t) ct −α+ 2 . Replace (2.5) with (2.6) to conclude that X(t) ct −α .
(2.6) 2
Lemma 2.5. Suppose X(t) satisfies the following inequality: for all t > 0
X(t) ct
−β
t + 0
1
1
(t − s)− 2 s − 2 X(s) ds.
(2.7)
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Then X(t) ct
−β
+ ct
− 12
t
1
τ − 2 X(s) ds,
0
where c is independent of and t. Proof. Let I =
t
t/2 (t
1
1
− s)− 2 s − 2 X(s) ds, II =
I ct
− 12
t
− 12
(t − s)
t/2 0
1
1
(t − s)− 2 s − 2 X(s) ds. Then we observe
II ct
X(s) ds,
−1/2
t/2 1 s − 2 X(s) ds. 0
t/2
If we iterate (2.7) to I , we have t
1
t2I c
1
(t − s)− 2 X(s) ds
t/2
t
− 12
(t − s)
c
s
−β
+
− 12 − 12
(s − τ )
τ
X(τ ) dτ ds
0
t/2
t
− 12 −β
(t − s)
=c
s
s
t s ds + c
1
1
1
(t − s)− 2 (s − τ )− 2 τ − 2 X(τ ) dτ ds
t/2 0
t/2
= I1 + I2 .
(2.8) 1
Direct computation shows that I1 ct −β+ 2 , and if we apply Fubini’s theorem to I2 we have t/2 t I2 = c
1
1
1
(t − s)− 2 (s − τ )− 2 τ − 2 X(τ ) ds dτ
0 t/2
t t + c
1
1
1
(t − s)− 2 (s − τ )− 2 τ − 2 X(τ ) ds dτ
t/2 τ
t t/2 − 12 − 12 − 12 = c τ X(τ ) (t − s) (s − τ ) ds dτ 0
t/2
t + c t/2
τ
− 12
t
− 12
(t − s)
X(τ ) τ
− 12
(s − τ )
ds dτ
H.-O. Bae, B.J. Jin / Journal of Functional Analysis 240 (2006) 508–529
t c
515
1
τ − 2 X(τ ) dτ.
0
This follows computations such that for 0 < τ < t t
− 12
(t − s)
− 12
(s − τ )
t−τ ds = c
τ
and that for 0 < τ <
1
1
z− 2 (t − τ − z)− 2 dz = c,
0 t 2
t
− 12
(t − s)
− 12
(s − τ )
t ds c
1
1
(t − s)− 2 (s − τ )− 2 ds = c,
τ
t/2
for some constant c independent of t owing to Lemma 2.3. Hence, we conclude that 1
I ct −β + ct − 2
t
1
τ − 2 X(τ ) dτ.
0
Combining (2.7) and the estimates of I and II, we complete the proof.
2
3. Temporal decays In this section we obtain the temporal decay rates for the Navier–Stokes solutions. We assume that u is a solution of (1.1) satisfying Proposition 2.2. Lemma 3.1. −A (t−s) e Ω PΩ (u · ∇)u (s)
1
c(t − s)− 2 u ⊗ uLq ,
Lq
t >s>0
(3.1)
for 2 q < ∞. ∞ (Ω). We observe that for Proof. Let φ ∈ C0,σ
1 q
=1−
1 q
e−AΩ (t−s) PΩ (u · ∇)u (s), φ = u ⊗ u, ∇e−AΩ (t−s) φ 1 cu ⊗ uLq ∇e−AΩ (t−s) φ L cu ⊗ uLq (t − s)− 2 φLq
q
by (2.2). Hence we have −A (t−s) e Ω PΩ (u · ∇)u (s)
1
Lq
c(t − s)− 2 u ⊗ uLq .
For our main theorem in this section we need the following lemma.
2
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Lemma 3.2. For all t > 0, u(t)
Lq
ct
− 12 + q1
,
u(t)
Lq
ct
− 12 + q1
,
if u0 ∈ L2σ for q 2, and
if u0 ∈ Lq ∩ L2σ for 1 < q < 2. Proof. It is already well known that u(t) ∈ Lq for 2 q ∞ if u0 ∈ L2σ . Moreover, u(t)Lq − 12 + q1
for 2 q ∞. Hence we have only to show that u(t) ∈ Lq for 1 < q < 2, if u0 ∈ L2σ ∩ Lq . By Duhamel principle, u can be represented by
ct
u(t) = e
−AΩ t
t u0 +
e−AΩ (t−s) PΩ (u · ∇)u (s) ds = I + II.
0
By (2.1), I and II are estimated by t I q cu0 Lq ,
IIq c
(u · ∇)u
Lq
ds.
0
And by Hölder inequality we observe that (u · ∇)u
u
− 12 + 2−q 2q
−1+ q1
Lq
2q 2−q
∇uL2 ,
and by (2.3), we have u
2q 2−q
cs
= cs
,
∇u(t)
Hence, t IIq c
s
− 32 + q1
ds = ct
− 12 + q1
.
0
By the estimates of I and II, we conclude that u(t) for 1 < q < 2, if u0 ∈ Lq ∩ L2σ .
2
Lq
ct
− 12 + q1
for all t > 0,
1
L2
s− 2 .
(3.2)
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517
Remark 3.3. We note that u0 ∈ L2σ ∩ Lr , 1 < r q < 2, implies u0 ∈ Lq by interpolation. We note that < min{ q1 , 12 }, then
lim t
−
t
1 s − 2 u(s)Lq ds = 0,
t→0+
0
since by Proposition 2.2 t s
u(s)Lq ds c
− 12
t
0
1
s− 2 s
− 12 + q1
1
ds ct q ,
q 2,
0
and t s
u(s)Lq ds c
− 12
0
t
− 12 + q1
1
s − 2 (1 + s)
1
− 12 + q1
ds ct 2 (1 + t)
,
1 < q < 2.
0
Theorem 3.4. Let 1 < r q < ∞ or 1 < r < q = ∞. Assume u0 ∈ L2σ ∩ Lr . Let u be the solution of Navier–Stokes equations of two-dimensional exterior domain Ω with the initial velocity u0 , so that (2.3) and (2.4) holds. Moreover, we assume that for a given small > 0 1
uL∞ t − 2
for all t.
(3.3)
Then we have u(t)
Lq
−1+1 =O t r q
(3.4)
for 1 < r q < ∞ or 1 < r < q = ∞. Moreover, we have 1 1 1 ∇u(t) = O t − r + q − 2 L q
(3.5)
for 1 < r q 2. Remark 3.5. As a mater of fact, the hypothesis (3.3) does not lose a generality, and can be removed. According to (2.4) there is time T () so that u(t)L∞ t −1/2 for t T (). We consider v(x, t) = u(x, t + T ()). By Lemma 3.2 it is shown that u(t) ∈ Lp ∩ L2σ for each t > 0, if u0 ∈ Lp ∩ L2σ , 1 < p < ∞. Hence we have v(x, 0) = u(x, T ()) ∈ Lr ∩ Lq ∩ L2σ if u0 ∈ Lr ∩ Lq ∩ L2σ . And v satisfies all the hypothesis of Theorem 3.4. Therefore v satisfies (3.4) and (3.5). This again implies that the estimate (3.4) and (3.5) for u. This idea is stated in [11]. Proof of Theorem 3.4. By Duhamel principle, u can be represented by u = I + II, where I and II are defined in (3.2). By (2.1) we have that I q ct
−( 1r − q1 )
(3.6)
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for 1 < r q < ∞ or 1 < r < q = ∞, if u0 ∈ Lr . By (2.2) we also observe that ∇I Lq ct
− 1r + q1 − 12
for 1 < r q 2, if u0 ∈ Lr . Step 1. At this step we will prove the inequality (3.4) for 1 < r q < ∞,
2q q+2
< r. (Note that
2q 2+q
2.) By the hypothesis, we have that 1 u ⊗ uLq cu(s)L u(s)L cs − 2 u(s)L . ∞
q
q
If we apply (3.1) to II we obtain t IILq c
1 1 (t − s)− 2 s − 2 u(s)L ds.
(3.7)
q
0
Combining (3.6) and (3.7), we have u(t)
Lq
ct
−( 1r − q1 )
t + c
1 1 (t − s)− 2 s − 2 u(s)L ds q
(3.8)
0
for 1 < r q < ∞. Let X(t) = u(t)Lq , then (3.8) is rewritten by X(t) ct
−( 1r − q1 )
t + c
1
1
(t − s)− 2 s − 2 X(s) ds.
(3.9)
0
By Lemma 2.5, we have X(t) ct
−( 1r − q1 )
1
+ ct − 2
t
1
s − 2 X(s) ds.
(3.10)
0
Recall Remark 3.3 that if < min{ q1 , 12 }, 1 < q ∞, we have
lim t
−
t
t→0+
1
s − 2 X(s) ds = 0.
0
Hence applying Lemma 2.4 we conclude that there is 0 so that for 0 we have X(t) ct
−( 1r − q1 )
(3.11)
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for
1 r
−
1 q
< 12 , i.e., for
2q q+2
< r q < ∞.
Step 2. At this step we will prove the inequality (3.4) for 1 < r Take p ∈ (1, r). By (2.1), we have −A (t−s) e Ω PΩ (u · ∇)u
Let m =
2p 2−p .
519
Lq
2q q+2 ,
r q < ∞.
PΩ (u · ∇)uLp −1+1 −1+1 c(t − s) p q (u · ∇)uLp c(t − s) p q u − p1 + q1
c(t − s)
Since m > r > p =
2m 2+m ,
2p 2−p
∇uL2 .
by the result of Step 1 we have 1
1
uLm ct − r + m
if u0 ∈ Lr ∩ L2σ .
Hence II is estimated by t IILq c
1 − p1 + q1 − 1r + m − 12
(t − s)
s
s
ds = ct
− 1r + q1
.
0
Step 3. At this step we will prove the inequality (3.4) for 1 < r < q = ∞. 2r Take p ∈ (1, 2) satisfying 2+r < p < r. By (2.1), we have −A (t−s) e Ω PΩ (u · ∇)u Let m =
2p 2−p .
Since m > r > p =
2m 2+m ,
L∞
− p1
c(t − s)
u
2p 2−p
∇uL2 .
by the result of Step 1 we have 1
1
uLm ct − r + m
if u0 ∈ Lr ∩ L2σ .
Hence II is estimated by t IIL∞ c
1 − p1 − 1r + m − 12
(t − s)
s
s
1
ds = ct − r .
0
Step 4. At this step we will prove the inequality (3.5) for 1 < r q 2. (Note that Take 1 < p < r. By (2.2) we have that −A (t−s) ∇e Ω PΩ (u · ∇)u
Lq
PΩ (u · ∇)uLp .
− p1 + q1 − 12
c(t − s)
We note that PΩ (u · ∇)u
Lp
c(u · ∇)uLp cu
2p
L 2−p
∇uL2 .
2q q+2
1.)
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Let m = 2p/(2 − p). Note that 2 < m < ∞. By (3.11) of Step 1, we observe that u(s)Lm 1 1 cs − r + m if u0 ∈ L2σ ∩ Lr . Hence ∇II is estimated by t ∇IILq c
1 − p1 + q1 − 12 − 1r + 2−p 2p − 2
(t − s)
s
s
ds ct
− 1r + q1 − 12
.
2
0
4. Spatial–temporal estimates: proof of Theorem 1.2 In this section, modifying idea in He, Xin [20] and Bae, Jin [8], and using the temporal estimates in the previous section, we will derive the decay rates for solutions: u|x|
Lp
− 1 + 1 + 1 +δ =O t r p 2
2p for any small δ > 0, 1 < r p+2 < 2 < p < ∞. Let Ω be the exterior of a simply connected set 3 B in R which contains the origin and is contained in a bounded ball, for example, the unit ball. We further assume that the boundary ∂B is smooth. We use the notation · p = · Lp (Ω) , for 2p short. From now on we assume that r, p with 1 < r p+2 < 2 < p < ∞ are given numbers. 1 Let N be the fundamental function of −, that is, N = N (x − y) = 2π log |x − y|. Define φ be a smooth function which vanishes on the boundary ∂Ω rapidly enough and set
v(x) =
N (x − y) φ(y)(∇ × u)(y) dy.
R3
By definition of v we have −v = φ∇ × u. Remark 4.1. In general, the identity
N (x − y)∂yi f (y) dy = ∂xi
Ω
N (x − y)f (y) dy Ω
does not hold except f = 0 on the boundary ∂Ω. Therefore, we choose φ which can give reasoning of integrations by parts to have following identity for non-vanishing f on ∂Ω:
N (x − y)∂yi (φf )(y) dy = ∂xi Ω
N (x − y)(φf )(y) dy. Ω
In this section, we define φ(x) = |x|χ(|x|), where χ is a nonnegative cut-off function with χ ∈ C0∞ [0, ∞), χ(s) = 0 for s 1, χ(s) = 1 for s 2. Observe that φ = O(|x|) as |x| → ∞, and that |∇φ| c, ∇ 2 φ = O(1/|x|) as |x| → ∞.
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521
Remark 4.2. Observe that ∇ × v = φu + ∇ × N ∗ (∇φ) × u − ∇ N ∗ (u · ∇φ) . Let 2 s < ∞. We have ∇ × vs cφus + c∇ N ∗ |∇φ||u| s . By Sobolev inequality 2 ∇ N ∗ |∇φ||u| ∇ 2 N ∗ |∇φ||u| s and by Calderon–Zygmund inequality 2 ∇ N ∗ |∇φ||u|
2s s+2
c|∇φ|u
2s s+2
2s s+2
,
.
Recall ∇φ = O(1). Hence we have ∇ × vs cφus + cu
2s 2+s
.
We consider the fundamental solution for the nonstationary Stokes equation, written by Vti (x) = V i (x, t) = Γt (x)ei + ∇
∂ (N ∗ Γt )(x) ∂xi
and Q(x, t) = −δ(t)∇N, where Γt (x) = Γ (x, t) = (4πt)−1 e−|x| /4t (refer to [23]). Set θt (x) = θ (x, t) = (N ∗ Γt )(x), and ωti (x) = ωi (x, t) = θ (x, t)ei , i = 1, 2, 3, where ei is the standard unit vector of which the ith term is 1. As a matter of fact, we consider the 2-dimensional vectors as 3-dimensional vectors of which the third component is zero. Then, by Biot–Savart’s law, we have 2
∇ × ∇ × ωi = −ωi + ∇ div ωi = V i , hence, we have the identity ∇y × φ(y)∇y × ωi (x − y, t − τ ) = φ(y)V i (x − y, t − τ ) + R1i (x, y, t − τ ), where R1i (x, y, t − τ ) = ∇φ(y) × ∇y × ωi (x − y, t − τ ). It is easy to check that R1i = 0 on ∂Ω. We apply ∇y × [φ(y)∇y × ωi (x − y, t − τ )] as a test function to (1.1), and integrate over Ω × (0, t − ), and then we have
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t−
∂u − u + (u · ∇)u · ∇y × φ(y)∇y × ωi (x − y, t − τ ) dy dτ ∂t
0 Ω
t− =−
∇p(y) · ∇y × φ(y)∇y × ωi (x − y, t − τ ) dy dτ = 0.
0 Ω
Let R2i (x, y, t − τ ) = −2 ∇φ(y) · ∇ V i (x − y, t − τ ) − φ(y)V i (x − y, t − τ ). Taking integration by parts and observing (−∂/∂τ − y )V i (x − y, t − τ ) = 0, the above identity becomes t−
∂ u(y, τ ) · R2i (x, y, t − τ ) − + y R1i (x, y, t − τ ) dy dτ ∂τ
0 Ω
+
u(y, t − ) · ∇y × φ(y)∇y × ωi (x − y, ) dy
Ω
−
u0 (y) · ∇y × φ(y)∇y × ωi (x − y, t) dy
Ω
t− =−
(u · ∇)u(y, τ ) · φ(y)V i (x − y, t − τ ) + R1i (x, y, t − τ ) dy dτ.
0 Ω
We observe that lim
→0
u(y, t − ) · ∇y × φ(y)∇y × ωi (x − y, ) dy
Ω
= −∇x × N ∗ φ(y)∇y × u · ei ≡ −(∇x × v) · ei = −(∇x × v)i and
u0 (y) · ∇y × φ(y)∇y × ωi (x − y, t) dy = − ∇x × N ∗ φ(y)∇y × u0 · ei ∗ Γt = −(∇x × v0 )i ∗ Γt ,
where v0 = N ∗ [φ(y)∇y × u0 ], and v = N ∗ [φ(y)∇y × u]. Since ∇ × φ(y)(∇ × u)(y) = ∇ × ∇ × (φu) − ∇ × (∇φ) × u = −(φu) + ∇ (u · ∇)φ − ∇ × (∇φ) × u , we have that
(4.1)
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523
(∇x × v)i = ui φ + ∂xi N ∗ (u · ∇)φ − ei · ∇ × N ∗ (∇φ × u) , (∇x × v0 )i = ui0 φ + ∂xi N ∗ (u0 · ∇)φ − ei · ∇ × N ∗ (∇φ × u0 ) .
(4.2)
Then we get the following integral representation on ∇ × v from (4.1): t (∇x × v)i =
(u · ∇)u(y, τ ) · φ(y)V i (x − y, t − τ ) dy dτ 0
t (u · ∇)u(y, τ ) · R1i (x, y, t − τ ) dy dτ
+ 0
t + (∇x × v0 )i ∗ Γt −
u(y, τ ) · (∂τ + y )R1i (x, y, t − τ ) dy dτ 0
t
u(y, τ ) · R2i (x, y, t − τ ) dy dτ
+ 0
= I1i
+ I2i + · · · + I5i .
With the help of (4.2) and Sobolev’s inequality, one has uφp ∇ × vp + ∇N ∗ (u∇φ)p ∇ × vp + ∇ 2 N ∗ (u∇φ) ∇ × vp + ct
− 1r + p1 + 12
2p p+2
∇ × vp + cu
2p p+2
(4.3)
.
Therefore, it is enough to estimate ∇ × vp to get our temporal–spatial decay rates. Before going on we remind that ∇u(t) c(t + 1)− 1r + 12 t − 12 , 2
u(t) c(t + 1)− 1r + 12 , 2 1
if u0 ∈ L2σ , 1
1
since u(t)2 c, ∇u(t)2 ct − 2 , and if u0 ∈ L2σ ∩ Lr , then u(t)2 ct − r + 2 and 1 1 1 ∇u(t)2 ct − r + 2 − 2 . From straightforward calculations we have that for 1 < s < ∞ 2+β ∂ θt−τ s c∂ β Γt−τ s , and for 1 s ∞ β |β| ∂ Γt−τ c(t − τ )− 2 −1+ 1s . s Therefore, by the help of the generalized Minkowski’s, Young’s convolution and Hölder’s inequalities, we obtain the estimates for I1 as follows:
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t
φ(u · ∇)u
I1 p
t 2p p+2
Γt−τ 2 dτ c
0
1
φup ∇u2 (t − τ )− 2 dτ
0
t
1
1
1
1
uφp (1 + τ )− r + 2 τ − 2 (t − τ )− 2 dτ
0
since
1 p
+1=
2+p 2p
+
t I1 p c
1 2
and V i 2 cΓ 2 . By (4.3), I1 is bounded by
1 1 1 1 −1+ 1 +1 ∇ × vp + τ r p 2 (1 + τ )− r + 2 τ − 2 (t − τ )− 2 dτ
0
t c
1
1
1
1
∇ × vp τ − 2 (1 + τ )− r + 2 (t − τ )− 2 dτ
0
t +c
1
1
(1 + τ )− r + 2 τ
− 1r + p1
1
(t − τ )− 2 dτ
0
t c
− 1r + 12 − 12
∇ × vp (1 + τ )
τ
− 12
(t − τ )
t dτ + c
0
− 1r + p1
1
(t − τ )− 2 dτ
0
t c
τ
1
1
1
1
1
∇ × vp (1 + τ )− r + 2 τ − 2 (t − τ )− 2 dτ + ct 2
− 1r + p1
(4.4)
.
0
Since R1i = ∇φ(y) × ∇y × ωi (x − y, t − τ ), we also have t I2 p c
(u · ∇u)∇φ ∇ωi dτ 1 p
0
t c
u2 ∇u2 ∇ 2 ωi
2p p+2
dτ
0
t c
1
1
1
− 12 + p1
(1 + τ )− r + 2 τ − 2 (t − τ )
ct
− 1r + p1 + 12
(4.5)
.
0
Applying Young’s convolution inequality to I3 , we have that (∇ × v0 ) ∗ Γt c∇ × v0 p
2r 2−r
Γt s c∇ × v0
2r 2−r
t
− 1r + 12 + p1
,
H.-O. Bae, B.J. Jin / Journal of Functional Analysis 240 (2006) 508–529
where
1 s
=
1 p
−
1 r
525 2r
+ 32 . By Remark 4.2 we note that if u0 ∈ Lr and φu0 ∈ L 2−r , then ∇ × v0 ∈
2r
2r
L 2−r . Hence, if u0 ∈ Lr and φu0 ∈ L 2−r , then I3 p ct
− 1r + 12 + p1
(4.6)
.
Since ∂t Γ + Γ = 0, we have that ∂t ωi + ωi = N ∗ (∂t Γ + Γ ) = 0. Hence (∂τ + y )R1i = ∇φ(y) × ∇y × ∂t ωi + ωi + 2∇∂k φ × ∂k ∇y × ωi + ∇φ × ∇y × ωi = 2∂k (∇φ) × ∂k ∇y × ωi + ∇φ × ∇y × ωi . Therefore, I4 is bounded by t I4 p c
(u × ∇∂k φ) ∗ ∂k ∇ × ωi + (u × ∇φ) ∗ ∇ × ωi dτ p p
0
t c
u|x|−1
2p p+2
2 i ∇ ω + |x|−2 u ∇ωi dτ 2 1 p
2p p+2
2 i ∇ ω + |x|−2 u ∇ 2 ωi 2 1
2p p+2
1 −1+ 1 (t − τ )− 2 + |x|−2 u1 (t − τ ) 2 p dτ.
0
t c
u|x|−1
2p p+2
dτ
0
t c
u|x|−1
0
If we use the weighted inequality (5.3) in Galdi [17, p. 61], we have u|x|−1 Let q > 1. Let
1 q
2p p+2
∇u
2p p+2
ct
1 − 1r + p+2 2p − 2
= ct
− 1r + p1
.
= 1 − q1 .
1 1 −2 1 1 |x| u uq |x|−2 cq uq cq t − r + q = cq t − r +1− q . 1 q
Here we used the fact |x|−2 ∈ Lq (Ω) for q > 1. Therefore, we have t I4 p
−1+ 1 1 − 1 +1− q1 −1+ 1 cτ r p (t − τ )− 2 + cq τ r (t − τ ) 2 p dτ
0
ct
− 1r + p1 + 12
+ cq t
− 1r + p1 + 32 − q1
(4.7)
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H.-O. Bae, B.J. Jin / Journal of Functional Analysis 240 (2006) 508–529
for any q > 1. Let δ = 1 − q1 , then δ > 0 and δ goes to zero as q goes to 1. Hence the above estimate can be rewritten by I4 p ct
− 1r + p1 + 12
+ cδ t
− 1r + p1 + 12 +δ
.
Now we consider the estimate of I5 . Observe that t I5 p
(u∂j φ) ∗ ∂j V i + (uφ) ∗ V i dτ = P1 + P2 . p p
0
For P2 we can estimate in a similar way to the first term of (4.7). For P1 , observe that (u∂j φ) ∗ ∂j V i cu∂j φ2 ∇V i p
2p p+2
cu2 ∇Γ
2p p+2
,
and t P1
1
−1+ p1
1
τ − r + 2 (t − τ )
dτ ct
− 1r + p1 + 12
.
0
Therefore, we have I5 2 ct
− 1r + 12 + p1
(4.8)
.
Finally, from (4.4)–(4.8) we have t ∇ × vp
1
1
1
1
(1 + τ )− r + 2 τ − 2 (t − τ )− 2 ∇ × vp dτ
0
+ ct
− 1r + 12 + p1
+ cδ t
− 1r + 12 + p1 +δ
.
Now we define Bp (t) by Bp (t) = ∇ × v(t)p . Then Bp (t) satisfies t Bp (t)
1
1
1
1
(1 + τ )− r + 2 τ − 2 (t − τ )− 2 Bp (τ ) dτ
0
+ ct
− 1r + 12 + p1
+ cδ t
− 1r + 12 + p1 +δ
.
By the same reasoning as in the proof of Lemma 2.5, inequality (4.9) can be reduced to
(4.9)
H.-O. Bae, B.J. Jin / Journal of Functional Analysis 240 (2006) 508–529
Bp (t) c1 t
− 12
t
1
1
527
1
(1 + τ )− r + 2 τ − 2 Bp (τ ) dτ
0
+ ct
− 1r + 12 + p1
+ cδ t
− 1r + 12 + p1 +δ
(4.10)
for some constant c1 > 0. t 1 1 1 Set Yp (t) = 0 (1 + τ )− r + 2 τ − 2 Bp (τ ) dτ . Then Yp (t) satisfies 1 1 1 1 d −1+ 1 Yp c1 (1 + t)− r + 2 τ −1 Yp (t) + ct r p (1 + t)− r + 2 dt
+ cδ t
− 1r + p1 +δ
1
1
(1 + t)− r + 2 .
Let be small positive number so that 1 −
1 r
+
1 p
1
(4.11)
− > 0. Take large time T so that 1
c1 (1 + T )− r + 2 . Then (4.11) reduced to 1 1 1 1 d −1+ 1 − 1 + 1 +δ Yp t −1 Yp (t) + ct r p (1 + t)− r + 2 + cδ t r p (1 + t)− r + 2 dt
t −1 Yp (t) + ct
− 1r + p1
for t > T (here, we have considered δ < 1r − 12 ). Now by the same reasoning as in the proof of Lemma 2.4, it holds that for t > T () d − − 1 + 1 − t Yp (t) ct r p dt
for t > T .
Now integrate over [T , t] to obtain
t
−
Yp (t) T− Yp (T ) +
t τ
− 1r + p1 −
dτ
T
1− 1 + 1 − 1− 1 + 1 − T− Yp (T ) + c t r p − T r p ,
t > T .
Hence Yp (t) is bounded by 1− 1 + 1 1− 1 + 1 − Yp (t) T− Yp (T )t + c t r p − T r p t , Note that t < t
− 1r + p1 +1
since < 1 −
1 r
+ p1 . Hence we have
Yp (t) ct
1− 1r + p1
,
t > T .
t > T .
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H.-O. Bae, B.J. Jin / Journal of Functional Analysis 240 (2006) 508–529
By (4.10) the above estimate of Yp (t) implies that 1
Bp (t) c1 t − 2 Yp (t) + ct
− 1r + 12 + p1
+ cδ t
− 1r + 12 + p1 +δ
cδ t
− 1r + 12 + p1 +δ
,
t > T .
This implies that 1 1 1 ∇ × v(t) cδ t − r + 2 + p +δ , p
t > T .
Therefore we conclude that 1 1 1 ∇ × v(t) = O t − r + 2 + p +δ p for any small δ > 0. Combining this estimate and (4.3), we conclude that 1 1 1 |x|u(t) = O t − r + 2 + p +δ p for any δ > 0, which completes the proof. References [1] H.-O. Bae, Analyticity and asymptotics for the Stokes solutions in weighted space, J. Math. Anal. Appl. 269 (2002) 149–171. [2] H.-O. Bae, Temporal decays in L1 and L∞ for the Stokes flow, J. Differential Equations 222 (2006) 1–20. [3] H.-O. Bae, Temporal and spatial decays for the Stokes flow, submitted for publication. [4] H.-O. Bae, H.J. Choe, Decay rate for the incompressible flows in half spaces, Math. Z. 238 (2001) 799–816. [5] H.-O. Bae, B.J. Jin, Upper and lower bounds of temporal and spatial decays for the Navier–Stokes equations, J. Differential Equations 209 (2005) 365–391. [6] H.-O. Bae, B.J. Jin, Temporal and spatial decays for the Navier–Stokes equations, Proc. Roy. Soc. Edinburgh Sect. A 135 (2005) 461–477. [7] H.-O. Bae, B.J. Jin, Asymptotic behavior of Stokes solutions in 2D exterior domains, J. Math. Fluid Mech., in press. [8] H.-O. Bae, B.J. Jin, Temporal and spatial decay rates of Navier–Stokes solutions in exterior domains, submitted for publication. [9] W. Borchers, T. Miyakawa, Algebraic L2 decay for Navier–Stokes flows in exterior domains, Acta Math. 165 (1990) 189–227. [10] W. Borchers, T. Miyakawa, Algebraic L2 decay for Navier–Stokes flows in exterior domains, II, Hiroshima Math. J. 21 (1991) 621–640. [11] W. Borchers, T. Miyakawa, L2 -Decay for Navier–Stokes flows in an unbounded domains, with application to exterior stationary flows, Arch. Ration. Mech. Anal. 118 (1992) 273–295. [12] L. Brandolese, Space–time decay of Navier–Stokes flows invariant under rotations, Math. Ann. 329 (4) (2004) 685–706. [13] W. Dan, Y. Shibata, On the Lq –Lr estimates of the Stokes semigroup in a 2-dimensional exterior domain, J. Math. Soc. Japan 51 (1) (1999) 181–207. [14] W. Dan, Y. Shibata, Remark on the Lq –Lr estimate of the Stokes semigroup in a 2-dimensional exterior domain, Pacific J. Math. 189 (2) (1999) 223–239. [15] R. Farwig, H. Sohr, Global estimates in weighted spaces of weak solutions of the Navier–Stokes equations in exterior domains, Arch. Math. 67 (1996) 319–330. [16] Y. Fujigaki, T. Miyakawa, Asymptotic profiles of nonstationary incompressible Navier–Stokes flows in the whole space, SIAM J. Math. Anal. 33 (3) (2001) 523–544. [17] G. Galdi, An Introduction to the Mathematical Theory of the Navier–Stokes Equations, vol. 1, Linearized Steady Problems, Springer Tracts Nat. Philos., vol. 38, Springer, Berlin, 1994. [18] G.P. Galdi, P. Maremonti, Monotonic decreasing and asymptotic behaviour of the kinetic energy for weak solutions of the Navier–Stokes equations in exterior domains, Arch. Ration. Mech. Anal. 94 (1986) 253–266.
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[19] C. He, Weighted energy inequalities for nonstationary Navier–Stokes equations, J. Differential Equations 148 (1998) 422–444. [20] C. He, Z. Xin, Weighted estimates for nonstationary Navier–Stokes equations in exterior domain, Methods Appl. Anal. 7 (3) (2000) 443–458. [21] T. Kato, Strong Lp solutions of the Navier–Stokes equations in Rn , with application to weak solutions, Math. Z. 192 (1984) 471–480. [22] H. Kozono, T. Ogawa, Two-dimensional Navier–Stokes flow in unbounded domains, Math. Ann. 297 (1993) 1–31. [23] O.A. Ladyzhenskaya, The Mathematical Theory of Viscous Incompressible Flow, Gordon & Breach, New York, 1969. [24] T. Miyakawa, On space–time decay properties of nonstationary incompressible Navier–Stokes flows in Rn , Funkcial. Ekvac. 43 (3) (2000) 541–557. [25] T. Miyakawa, M.E. Schonbek, On optimal decay rates for weak solutions to the Navier–Stokes equations in Rn , in: Proceedings of Partial Differential Equations and Applications, Olomouc, 1999, Math. Bohem. 126 (2) (2001) 443–455. [26] M.E. Schonbek, Large time behavior of solutions of the Navier–Stokes equations, Comm. Partial Differential Equations 11 (1986) 733–763. [27] Y. Shibata, On the decay estimate of the Stokes semigroup in a two-dimensional exterior domain, in: NSE-6 Conference, VSP/TEV, Vilnius, 1998, pp. 315–330. [28] S. Takahashi, A weighted equation approach to decay rate estimates for the Navier–Stokes equations, Nonlinear Anal. 37 (1999) 751–789. [29] M. Wiegner, Decay estimates for strong solutions of the Navier–Stokes equations in exterior domains, in: Navier– Stokes Equation and Related Nolinear Problems, Ferrara, 1999, Ann. Univ. Ferrara Sez. VII (N.S.) 46 (2000) 61–79.
Asymptotic behavior for the Navier–Stokes equations in 2D exterior domains ✩ Hyeong-Ohk Bae a,∗ , Bum Ja Jin b a Department of Mathematics, Ajou University, Suwon 443-749, Republic of Korea b Korea Science Academy, Busanjin-Gu, Busan-Si, Republic of Korea
Received 23 December 2005; accepted 18 April 2006 Available online 16 June 2006 Communicated by H. Brezis
Abstract We show that the Lp spatial–temporal decay rates of solutions of incompressible flow in an 2D exterior domain. When a domain has a boundary, pressure term makes an obstacle since we do not have enough information on the pressure term near the boundary. To overcome the difficulty, we adopt the ideas in He, Xin [C. He, Z. Xin, Weighted estimates for nonstationary Navier–Stokes equations in exterior domain, Methods Appl. Anal. 7 (3) (2000) 443–458], and our previous results [H.-O. Bae, B.J. Jin, Asymptotic behavior of Stokes solutions in 2D exterior domains, J. Math. Fluid Mech., in press; H.-O. Bae, B.J. Jin, Temporal and spatial decay rates of Navier–Stokes solutions in exterior domains, submitted for publication]. For the spatial decay rate estimate, we first extend temporal decay rate result of the Navier–Stokes solutions for general Lp space when the initial velocity is in L2σ ∩ Lr , 1 < r q < ∞ (1 < r < q = ∞). © 2006 Elsevier Inc. All rights reserved. Keywords: 2D Navier–Stokes; Temporal–spatial decay rate; Exterior domain
1. Introduction Let u and p be the velocity and pressure, respectively, of the incompressible fluid in an exterior domain. Let Ω be the exterior of a simply connected set B in R2 which contains the origin and is contained in a bounded ball, for example, the unit ball. Furthermore, we assume that the boundary ✩
This work was supported by Korea Research Foundation Grant (KRF-2003-015-C00052).
* Corresponding author. Fax: +82 31 216 3027.
E-mail addresses: [email protected] (H.-O. Bae), [email protected] (B.J. Jin). 0022-1236/$ – see front matter © 2006 Elsevier Inc. All rights reserved. doi:10.1016/j.jfa.2006.04.029
H.-O. Bae, B.J. Jin / Journal of Functional Analysis 240 (2006) 508–529
509
∂B is smooth enough to get the unique existence of the solution. We consider the Navier–Stokes equations in Ω: ∂ ∂t u + (u · ∇)u − u + ∇p
=0
∇ ·u=0
in (x, t) ∈ Ω × (0, ∞)
(1.1)
with no slip boundary condition u(x, t) = 0 for x ∈ ∂Ω and initial data u(x, 0) = u0 (x). Temporal decay estimates have been considered for various domains by Kato [21], Fujigaki and Miyakawa [16], Miyakawa and Schonbek [25], Miyakawa [24], Wiegner [29], Schonbek [26], Galdi and Maremonti [18], Kozono and Ogawa [22], Borchers and Miyakawa [9,10], Bae and Choe [4], Bae [2] and spatial decay estimates have been considered by He [19], He and Xin [20], Takahashi [28], Farwig and Sohr [15], Brandolese [12], Bae and Jin [5,6,8], Bae [1,3], etc. When the whole space R3 has been considered, we have uniform upper and lower bounds in 2 L space with spatial weight in [6,8]. While a fluid region with boundary has been considered, the situation is no longer simple. Since we do not have pressure representation in arbitrary domains, it is difficult to make use of the energy method. In [20], He and Xin have considered an exterior domain problem, by removing pressure term in a solution representation, they have obtained decay rate results for the strong solution, that is, for small data. They obtained the spatial decay of a strong solution: 3(1− 1 ) 1 + |x|2 2 7 p u ∈ L∞ 0, ∞; Lp (Ω)
for 7 < p ∞.
Very recently, we [8] also obtained the decays for weak and strong solutions for 3D exterior domains. More precisely, there is a suitably weak solution u of (1.1) satisfying the asymptotic property: for any sufficiently small δ > 0 there is a positive constant cδ depending on δ and independent of t such that for 1 < r < 65 , |x|u(t)
5
L2
3
cδ (1 + t) 4 − 2r +δ
for any t > 0. For strong solutions, for p 2 and for small δ > 0, 2 |x| u(t)
Lp
1− 32 ( 1r − p1 )+δ
cδ (1 + t)
.
However, the above results for exterior domains are all for the 3D Navier–Stokes equations. The temporal decay rates for the 2D Navier–Stokes is known by Kozono and Ogawa [22] and Dan and Shibata [14]: −1+1 uq = o t 2 q
for q 2,
1 u∞ = o t − 2 ,
if u0 ∈ L2σ . Borchers and Miyakawa [11] obtained the temporal L2 estimate of the Navier–Stokes solutions in 2D exterior domains: 1
1
u2 ct − r + 2 ,
1 < r 2,
if u0 ∈ Lr ∩ L2σ . It was difficult for us to understand some parts of their proof. We provide a different proof for more general estimates, 1 < q ∞, in Section 3 independently, which is
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H.-O. Bae, B.J. Jin / Journal of Functional Analysis 240 (2006) 508–529
stated in Theorem 1.1. Combining the series of papers of Dan and Shibata [13,14] for the 2D Stokes equations, we [7] estimated the temporal–spatial decay rates for the 2D Stokes solutions for the exterior domains: uq ct
− 1r + q1
,
1 < r q < ∞, or 1 < r < q = ∞,
if u0 ∈ Lr ∩ L2σ . In this paper, we intend to derive spatial–temporal decay estimates of solutions for the Navier– Stokes equations in the 2D exterior domains. For this purpose we modify the ideas in Bae, Jin [8] and He, Xin [20] done for 3D cases, and we can avoid the computations involving with the pressure term. For our main temporal–spatial estimates, we first obtain the temporal estimates as mentioned above. We state our main theorems below, of which proofs are the main objectives of the subsequent sections. Theorem 1.1. Let 1 < r q < ∞ or 1 < r < q = ∞. Assume u0 ∈ L2σ ∩ Lr . Let u be the solution of Navier–Stokes equations of two-dimensional exterior domain Ω with the initial velocity u0 . Then we have u(t)
Lq
−1+1 =O t r q
for 1 < r q < ∞ or 1 < r < q = ∞. Moreover, we have ∇u(t)
Lq
−1+1−1 =O t r q 2
for 1 < r q 2. Theorem 1.2. Let Ω be a two-dimensional exterior domain. Let u0 ∈ Lr (Ω) ∩ L2 (Ω) with 2r 2p 1 < r p+2 < 2 p < ∞, and ∇ · u0 = 0. We also assume |x|u0 ∈ L 2−r . Let 0 < α 1. Then there is a strong solution u of the Navier–Stokes equation (1.1) satisfying the following asymptotic property: for any small δ > 0 1 + |x| α u(t)
Lp (Ω)
− 1 + 1 + α +δ =O t 2 p 2
for any large t. In Section 2 we state some well-known facts and lemmas for our theorems, and in Section 3 we provide temporal decays. Section 4 is the proof of Theorem 1.2. 2. Preliminaries p
We use the usual notation that Lσ means the set of measurable functions in Lp with divergence free. Let PΩ be the Leray projection from L2 (Ω) → L2σ (Ω), and let −AΩ = PΩ be the generator of semigroup e−AΩ t . Let us recall the following well-known decay estimates of the Stokes semigroup in a two-dimensional exterior domain (refer to [14,27]).
H.-O. Bae, B.J. Jin / Journal of Functional Analysis 240 (2006) 508–529
511
Proposition 2.1. Let f ∈ Lrσ . Then it holds that −A t e Ω f
Lq
ct
Lq
ct
−( 1r − q1 )
fr ,
t > 0,
(2.1)
for 1 < r q < ∞, 1 < r < q = ∞; −A t ∇e Ω f
−( 1r − q1 )− 12
fr ,
t > 0,
(2.2)
for 1 < r q 2. Suppose u is a solution of Navier–Stokes equations of two-dimensional exterior domain when the initial velocity u0 ∈ L2σ . Dan and Shibata [13] showed the following (see also Kozono, Ogawa [22]). Proposition 2.2. u(t)
Lq
∇u(t)
−( 1 − 1 ) for 2 q < ∞, =o t 2 q
L2
1 = o t− 2
(2.3)
as t goes to infinity. Moreover, u(t)
L∞
1 = o t− 2 .
Lemma 2.3. Let a < 1, b > 0, d < 1. If b + d < 1, then t
(t − s)−a (s + 1)−b s −d ds ct 1−a−d (1 + t)−b ;
0
if b + d = 1, then t
(t − s)−a (s + 1)−b s −d ds ct −a ln (t + 1);
0
if b + d > 1, then t
(t − s)−a (s + 1)−b s −d ds ct −a .
0
Proof. Let I =
t
t/2 (t
− s)−a (s + 1)−b s −d ds, II =
−b −d
I c(1 + t)
t
t
t/2
t/2 0
(t − s)−a (s + 1)−b s −d ds. Then
(t − s)−a ds = c(1 + t)−b t −d t 1−a = c(1 + t)−b t 1−a−d ,
(2.4)
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II ct
−a
t/2 (s + 1)−b s −d ds. 0
If t 2, then t/2 t/2 1 −b −d −b −d (s + 1) s ds = (s + 1) s ds + (s + 1)−b s −d ds 0
1
0
t/2 c
−b−d
(s + 1)
1 ds + c
1
⎧ 1−b−d ⎨ c(t + 1) c ln (t + 1) ⎩ c
s −d ds
0
if b + d < 1, if b + d = 1, if b + d > 1.
If t 2, then t/2 1 −b −d (s + 1) s ds c (s + 1)−b s −d ds c. 0
0
Hence we have ⎧ −a 1−b−d ⎨ t (t + 1) −a II c t ln (t + 1) ⎩ t −a
if b + d < 1, if b + d = 1, if b + d > 1.
Hence we have t 0
(t − s)−a (s + 1)−b s −d ds ⎧ 1−a−d (t + 1)−b + t −a (t + 1)1−b−d ⎪ ⎨t c t 1−a−d (t + 1)−b + t −a ln (t + 1) ⎪ ⎩ 1−a−d t (t + 1)−b + t −a
Lemma 2.4. Let α <
1 2
if b + d < 1, if b + d = 1, if b + d > 1.
2
and > 0 be given constants. Let us consider an integral inequality
X(t) ct
−α
+ t
−1/2
t 0
1
s − 2 X(s) ds.
(2.5)
H.-O. Bae, B.J. Jin / Journal of Functional Analysis 240 (2006) 508–529
513
We assume that
lim t
−
t
t→0+
1
s − 2 X(s) ds = 0.
0
Then, there is 0 so that if 0 , then X(t) ct −α for some c independent of t. Proof. Set J (t) =
t 0
1
s − 2 X(s) ds, then J (t) satisfies the inequality 1 d J ct −α t − 2 + t −1 J (t). dt
This can be rewritten by 1 d − t J (t) ct −α− 2 − . dt If we integrate over (0, t) we have
t
−
t J (t) c
1
s −α− 2 − ds.
0
Here, we used the hypothesis limt→0+ t − J (t) = 0. Suppose that α + < 12 . By Lemma 2.3, t
1
1
s −α− 2 − ds ct −α+ 2 − ,
0
therefore, we have 1
J (t) ct −α+ 2 . Replace (2.5) with (2.6) to conclude that X(t) ct −α .
(2.6) 2
Lemma 2.5. Suppose X(t) satisfies the following inequality: for all t > 0
X(t) ct
−β
t + 0
1
1
(t − s)− 2 s − 2 X(s) ds.
(2.7)
514
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Then X(t) ct
−β
+ ct
− 12
t
1
τ − 2 X(s) ds,
0
where c is independent of and t. Proof. Let I =
t
t/2 (t
1
1
− s)− 2 s − 2 X(s) ds, II =
I ct
− 12
t
− 12
(t − s)
t/2 0
1
1
(t − s)− 2 s − 2 X(s) ds. Then we observe
II ct
X(s) ds,
−1/2
t/2 1 s − 2 X(s) ds. 0
t/2
If we iterate (2.7) to I , we have t
1
t2I c
1
(t − s)− 2 X(s) ds
t/2
t
− 12
(t − s)
c
s
−β
+
− 12 − 12
(s − τ )
τ
X(τ ) dτ ds
0
t/2
t
− 12 −β
(t − s)
=c
s
s
t s ds + c
1
1
1
(t − s)− 2 (s − τ )− 2 τ − 2 X(τ ) dτ ds
t/2 0
t/2
= I1 + I2 .
(2.8) 1
Direct computation shows that I1 ct −β+ 2 , and if we apply Fubini’s theorem to I2 we have t/2 t I2 = c
1
1
1
(t − s)− 2 (s − τ )− 2 τ − 2 X(τ ) ds dτ
0 t/2
t t + c
1
1
1
(t − s)− 2 (s − τ )− 2 τ − 2 X(τ ) ds dτ
t/2 τ
t t/2 − 12 − 12 − 12 = c τ X(τ ) (t − s) (s − τ ) ds dτ 0
t/2
t + c t/2
τ
− 12
t
− 12
(t − s)
X(τ ) τ
− 12
(s − τ )
ds dτ
H.-O. Bae, B.J. Jin / Journal of Functional Analysis 240 (2006) 508–529
t c
515
1
τ − 2 X(τ ) dτ.
0
This follows computations such that for 0 < τ < t t
− 12
(t − s)
− 12
(s − τ )
t−τ ds = c
τ
and that for 0 < τ <
1
1
z− 2 (t − τ − z)− 2 dz = c,
0 t 2
t
− 12
(t − s)
− 12
(s − τ )
t ds c
1
1
(t − s)− 2 (s − τ )− 2 ds = c,
τ
t/2
for some constant c independent of t owing to Lemma 2.3. Hence, we conclude that 1
I ct −β + ct − 2
t
1
τ − 2 X(τ ) dτ.
0
Combining (2.7) and the estimates of I and II, we complete the proof.
2
3. Temporal decays In this section we obtain the temporal decay rates for the Navier–Stokes solutions. We assume that u is a solution of (1.1) satisfying Proposition 2.2. Lemma 3.1. −A (t−s) e Ω PΩ (u · ∇)u (s)
1
c(t − s)− 2 u ⊗ uLq ,
Lq
t >s>0
(3.1)
for 2 q < ∞. ∞ (Ω). We observe that for Proof. Let φ ∈ C0,σ
1 q
=1−
1 q
e−AΩ (t−s) PΩ (u · ∇)u (s), φ = u ⊗ u, ∇e−AΩ (t−s) φ 1 cu ⊗ uLq ∇e−AΩ (t−s) φ L cu ⊗ uLq (t − s)− 2 φLq
q
by (2.2). Hence we have −A (t−s) e Ω PΩ (u · ∇)u (s)
1
Lq
c(t − s)− 2 u ⊗ uLq .
For our main theorem in this section we need the following lemma.
2
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Lemma 3.2. For all t > 0, u(t)
Lq
ct
− 12 + q1
,
u(t)
Lq
ct
− 12 + q1
,
if u0 ∈ L2σ for q 2, and
if u0 ∈ Lq ∩ L2σ for 1 < q < 2. Proof. It is already well known that u(t) ∈ Lq for 2 q ∞ if u0 ∈ L2σ . Moreover, u(t)Lq − 12 + q1
for 2 q ∞. Hence we have only to show that u(t) ∈ Lq for 1 < q < 2, if u0 ∈ L2σ ∩ Lq . By Duhamel principle, u can be represented by
ct
u(t) = e
−AΩ t
t u0 +
e−AΩ (t−s) PΩ (u · ∇)u (s) ds = I + II.
0
By (2.1), I and II are estimated by t I q cu0 Lq ,
IIq c
(u · ∇)u
Lq
ds.
0
And by Hölder inequality we observe that (u · ∇)u
u
− 12 + 2−q 2q
−1+ q1
Lq
2q 2−q
∇uL2 ,
and by (2.3), we have u
2q 2−q
cs
= cs
,
∇u(t)
Hence, t IIq c
s
− 32 + q1
ds = ct
− 12 + q1
.
0
By the estimates of I and II, we conclude that u(t) for 1 < q < 2, if u0 ∈ Lq ∩ L2σ .
2
Lq
ct
− 12 + q1
for all t > 0,
1
L2
s− 2 .
(3.2)
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517
Remark 3.3. We note that u0 ∈ L2σ ∩ Lr , 1 < r q < 2, implies u0 ∈ Lq by interpolation. We note that < min{ q1 , 12 }, then
lim t
−
t
1 s − 2 u(s)Lq ds = 0,
t→0+
0
since by Proposition 2.2 t s
u(s)Lq ds c
− 12
t
0
1
s− 2 s
− 12 + q1
1
ds ct q ,
q 2,
0
and t s
u(s)Lq ds c
− 12
0
t
− 12 + q1
1
s − 2 (1 + s)
1
− 12 + q1
ds ct 2 (1 + t)
,
1 < q < 2.
0
Theorem 3.4. Let 1 < r q < ∞ or 1 < r < q = ∞. Assume u0 ∈ L2σ ∩ Lr . Let u be the solution of Navier–Stokes equations of two-dimensional exterior domain Ω with the initial velocity u0 , so that (2.3) and (2.4) holds. Moreover, we assume that for a given small > 0 1
uL∞ t − 2
for all t.
(3.3)
Then we have u(t)
Lq
−1+1 =O t r q
(3.4)
for 1 < r q < ∞ or 1 < r < q = ∞. Moreover, we have 1 1 1 ∇u(t) = O t − r + q − 2 L q
(3.5)
for 1 < r q 2. Remark 3.5. As a mater of fact, the hypothesis (3.3) does not lose a generality, and can be removed. According to (2.4) there is time T () so that u(t)L∞ t −1/2 for t T (). We consider v(x, t) = u(x, t + T ()). By Lemma 3.2 it is shown that u(t) ∈ Lp ∩ L2σ for each t > 0, if u0 ∈ Lp ∩ L2σ , 1 < p < ∞. Hence we have v(x, 0) = u(x, T ()) ∈ Lr ∩ Lq ∩ L2σ if u0 ∈ Lr ∩ Lq ∩ L2σ . And v satisfies all the hypothesis of Theorem 3.4. Therefore v satisfies (3.4) and (3.5). This again implies that the estimate (3.4) and (3.5) for u. This idea is stated in [11]. Proof of Theorem 3.4. By Duhamel principle, u can be represented by u = I + II, where I and II are defined in (3.2). By (2.1) we have that I q ct
−( 1r − q1 )
(3.6)
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for 1 < r q < ∞ or 1 < r < q = ∞, if u0 ∈ Lr . By (2.2) we also observe that ∇I Lq ct
− 1r + q1 − 12
for 1 < r q 2, if u0 ∈ Lr . Step 1. At this step we will prove the inequality (3.4) for 1 < r q < ∞,
2q q+2
< r. (Note that
2q 2+q
2.) By the hypothesis, we have that 1 u ⊗ uLq cu(s)L u(s)L cs − 2 u(s)L . ∞
q
q
If we apply (3.1) to II we obtain t IILq c
1 1 (t − s)− 2 s − 2 u(s)L ds.
(3.7)
q
0
Combining (3.6) and (3.7), we have u(t)
Lq
ct
−( 1r − q1 )
t + c
1 1 (t − s)− 2 s − 2 u(s)L ds q
(3.8)
0
for 1 < r q < ∞. Let X(t) = u(t)Lq , then (3.8) is rewritten by X(t) ct
−( 1r − q1 )
t + c
1
1
(t − s)− 2 s − 2 X(s) ds.
(3.9)
0
By Lemma 2.5, we have X(t) ct
−( 1r − q1 )
1
+ ct − 2
t
1
s − 2 X(s) ds.
(3.10)
0
Recall Remark 3.3 that if < min{ q1 , 12 }, 1 < q ∞, we have
lim t
−
t
t→0+
1
s − 2 X(s) ds = 0.
0
Hence applying Lemma 2.4 we conclude that there is 0 so that for 0 we have X(t) ct
−( 1r − q1 )
(3.11)
H.-O. Bae, B.J. Jin / Journal of Functional Analysis 240 (2006) 508–529
for
1 r
−
1 q
< 12 , i.e., for
2q q+2
< r q < ∞.
Step 2. At this step we will prove the inequality (3.4) for 1 < r Take p ∈ (1, r). By (2.1), we have −A (t−s) e Ω PΩ (u · ∇)u
Let m =
2p 2−p .
519
Lq
2q q+2 ,
r q < ∞.
PΩ (u · ∇)uLp −1+1 −1+1 c(t − s) p q (u · ∇)uLp c(t − s) p q u − p1 + q1
c(t − s)
Since m > r > p =
2m 2+m ,
2p 2−p
∇uL2 .
by the result of Step 1 we have 1
1
uLm ct − r + m
if u0 ∈ Lr ∩ L2σ .
Hence II is estimated by t IILq c
1 − p1 + q1 − 1r + m − 12
(t − s)
s
s
ds = ct
− 1r + q1
.
0
Step 3. At this step we will prove the inequality (3.4) for 1 < r < q = ∞. 2r Take p ∈ (1, 2) satisfying 2+r < p < r. By (2.1), we have −A (t−s) e Ω PΩ (u · ∇)u Let m =
2p 2−p .
Since m > r > p =
2m 2+m ,
L∞
− p1
c(t − s)
u
2p 2−p
∇uL2 .
by the result of Step 1 we have 1
1
uLm ct − r + m
if u0 ∈ Lr ∩ L2σ .
Hence II is estimated by t IIL∞ c
1 − p1 − 1r + m − 12
(t − s)
s
s
1
ds = ct − r .
0
Step 4. At this step we will prove the inequality (3.5) for 1 < r q 2. (Note that Take 1 < p < r. By (2.2) we have that −A (t−s) ∇e Ω PΩ (u · ∇)u
Lq
PΩ (u · ∇)uLp .
− p1 + q1 − 12
c(t − s)
We note that PΩ (u · ∇)u
Lp
c(u · ∇)uLp cu
2p
L 2−p
∇uL2 .
2q q+2
1.)
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Let m = 2p/(2 − p). Note that 2 < m < ∞. By (3.11) of Step 1, we observe that u(s)Lm 1 1 cs − r + m if u0 ∈ L2σ ∩ Lr . Hence ∇II is estimated by t ∇IILq c
1 − p1 + q1 − 12 − 1r + 2−p 2p − 2
(t − s)
s
s
ds ct
− 1r + q1 − 12
.
2
0
4. Spatial–temporal estimates: proof of Theorem 1.2 In this section, modifying idea in He, Xin [20] and Bae, Jin [8], and using the temporal estimates in the previous section, we will derive the decay rates for solutions: u|x|
Lp
− 1 + 1 + 1 +δ =O t r p 2
2p for any small δ > 0, 1 < r p+2 < 2 < p < ∞. Let Ω be the exterior of a simply connected set 3 B in R which contains the origin and is contained in a bounded ball, for example, the unit ball. We further assume that the boundary ∂B is smooth. We use the notation · p = · Lp (Ω) , for 2p short. From now on we assume that r, p with 1 < r p+2 < 2 < p < ∞ are given numbers. 1 Let N be the fundamental function of −, that is, N = N (x − y) = 2π log |x − y|. Define φ be a smooth function which vanishes on the boundary ∂Ω rapidly enough and set
v(x) =
N (x − y) φ(y)(∇ × u)(y) dy.
R3
By definition of v we have −v = φ∇ × u. Remark 4.1. In general, the identity
N (x − y)∂yi f (y) dy = ∂xi
Ω
N (x − y)f (y) dy Ω
does not hold except f = 0 on the boundary ∂Ω. Therefore, we choose φ which can give reasoning of integrations by parts to have following identity for non-vanishing f on ∂Ω:
N (x − y)∂yi (φf )(y) dy = ∂xi Ω
N (x − y)(φf )(y) dy. Ω
In this section, we define φ(x) = |x|χ(|x|), where χ is a nonnegative cut-off function with χ ∈ C0∞ [0, ∞), χ(s) = 0 for s 1, χ(s) = 1 for s 2. Observe that φ = O(|x|) as |x| → ∞, and that |∇φ| c, ∇ 2 φ = O(1/|x|) as |x| → ∞.
H.-O. Bae, B.J. Jin / Journal of Functional Analysis 240 (2006) 508–529
521
Remark 4.2. Observe that ∇ × v = φu + ∇ × N ∗ (∇φ) × u − ∇ N ∗ (u · ∇φ) . Let 2 s < ∞. We have ∇ × vs cφus + c∇ N ∗ |∇φ||u| s . By Sobolev inequality 2 ∇ N ∗ |∇φ||u| ∇ 2 N ∗ |∇φ||u| s and by Calderon–Zygmund inequality 2 ∇ N ∗ |∇φ||u|
2s s+2
c|∇φ|u
2s s+2
2s s+2
,
.
Recall ∇φ = O(1). Hence we have ∇ × vs cφus + cu
2s 2+s
.
We consider the fundamental solution for the nonstationary Stokes equation, written by Vti (x) = V i (x, t) = Γt (x)ei + ∇
∂ (N ∗ Γt )(x) ∂xi
and Q(x, t) = −δ(t)∇N, where Γt (x) = Γ (x, t) = (4πt)−1 e−|x| /4t (refer to [23]). Set θt (x) = θ (x, t) = (N ∗ Γt )(x), and ωti (x) = ωi (x, t) = θ (x, t)ei , i = 1, 2, 3, where ei is the standard unit vector of which the ith term is 1. As a matter of fact, we consider the 2-dimensional vectors as 3-dimensional vectors of which the third component is zero. Then, by Biot–Savart’s law, we have 2
∇ × ∇ × ωi = −ωi + ∇ div ωi = V i , hence, we have the identity ∇y × φ(y)∇y × ωi (x − y, t − τ ) = φ(y)V i (x − y, t − τ ) + R1i (x, y, t − τ ), where R1i (x, y, t − τ ) = ∇φ(y) × ∇y × ωi (x − y, t − τ ). It is easy to check that R1i = 0 on ∂Ω. We apply ∇y × [φ(y)∇y × ωi (x − y, t − τ )] as a test function to (1.1), and integrate over Ω × (0, t − ), and then we have
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H.-O. Bae, B.J. Jin / Journal of Functional Analysis 240 (2006) 508–529
t−
∂u − u + (u · ∇)u · ∇y × φ(y)∇y × ωi (x − y, t − τ ) dy dτ ∂t
0 Ω
t− =−
∇p(y) · ∇y × φ(y)∇y × ωi (x − y, t − τ ) dy dτ = 0.
0 Ω
Let R2i (x, y, t − τ ) = −2 ∇φ(y) · ∇ V i (x − y, t − τ ) − φ(y)V i (x − y, t − τ ). Taking integration by parts and observing (−∂/∂τ − y )V i (x − y, t − τ ) = 0, the above identity becomes t−
∂ u(y, τ ) · R2i (x, y, t − τ ) − + y R1i (x, y, t − τ ) dy dτ ∂τ
0 Ω
+
u(y, t − ) · ∇y × φ(y)∇y × ωi (x − y, ) dy
Ω
−
u0 (y) · ∇y × φ(y)∇y × ωi (x − y, t) dy
Ω
t− =−
(u · ∇)u(y, τ ) · φ(y)V i (x − y, t − τ ) + R1i (x, y, t − τ ) dy dτ.
0 Ω
We observe that lim
→0
u(y, t − ) · ∇y × φ(y)∇y × ωi (x − y, ) dy
Ω
= −∇x × N ∗ φ(y)∇y × u · ei ≡ −(∇x × v) · ei = −(∇x × v)i and
u0 (y) · ∇y × φ(y)∇y × ωi (x − y, t) dy = − ∇x × N ∗ φ(y)∇y × u0 · ei ∗ Γt = −(∇x × v0 )i ∗ Γt ,
where v0 = N ∗ [φ(y)∇y × u0 ], and v = N ∗ [φ(y)∇y × u]. Since ∇ × φ(y)(∇ × u)(y) = ∇ × ∇ × (φu) − ∇ × (∇φ) × u = −(φu) + ∇ (u · ∇)φ − ∇ × (∇φ) × u , we have that
(4.1)
H.-O. Bae, B.J. Jin / Journal of Functional Analysis 240 (2006) 508–529
523
(∇x × v)i = ui φ + ∂xi N ∗ (u · ∇)φ − ei · ∇ × N ∗ (∇φ × u) , (∇x × v0 )i = ui0 φ + ∂xi N ∗ (u0 · ∇)φ − ei · ∇ × N ∗ (∇φ × u0 ) .
(4.2)
Then we get the following integral representation on ∇ × v from (4.1): t (∇x × v)i =
(u · ∇)u(y, τ ) · φ(y)V i (x − y, t − τ ) dy dτ 0
t (u · ∇)u(y, τ ) · R1i (x, y, t − τ ) dy dτ
+ 0
t + (∇x × v0 )i ∗ Γt −
u(y, τ ) · (∂τ + y )R1i (x, y, t − τ ) dy dτ 0
t
u(y, τ ) · R2i (x, y, t − τ ) dy dτ
+ 0
= I1i
+ I2i + · · · + I5i .
With the help of (4.2) and Sobolev’s inequality, one has uφp ∇ × vp + ∇N ∗ (u∇φ)p ∇ × vp + ∇ 2 N ∗ (u∇φ) ∇ × vp + ct
− 1r + p1 + 12
2p p+2
∇ × vp + cu
2p p+2
(4.3)
.
Therefore, it is enough to estimate ∇ × vp to get our temporal–spatial decay rates. Before going on we remind that ∇u(t) c(t + 1)− 1r + 12 t − 12 , 2
u(t) c(t + 1)− 1r + 12 , 2 1
if u0 ∈ L2σ , 1
1
since u(t)2 c, ∇u(t)2 ct − 2 , and if u0 ∈ L2σ ∩ Lr , then u(t)2 ct − r + 2 and 1 1 1 ∇u(t)2 ct − r + 2 − 2 . From straightforward calculations we have that for 1 < s < ∞ 2+β ∂ θt−τ s c∂ β Γt−τ s , and for 1 s ∞ β |β| ∂ Γt−τ c(t − τ )− 2 −1+ 1s . s Therefore, by the help of the generalized Minkowski’s, Young’s convolution and Hölder’s inequalities, we obtain the estimates for I1 as follows:
524
H.-O. Bae, B.J. Jin / Journal of Functional Analysis 240 (2006) 508–529
t
φ(u · ∇)u
I1 p
t 2p p+2
Γt−τ 2 dτ c
0
1
φup ∇u2 (t − τ )− 2 dτ
0
t
1
1
1
1
uφp (1 + τ )− r + 2 τ − 2 (t − τ )− 2 dτ
0
since
1 p
+1=
2+p 2p
+
t I1 p c
1 2
and V i 2 cΓ 2 . By (4.3), I1 is bounded by
1 1 1 1 −1+ 1 +1 ∇ × vp + τ r p 2 (1 + τ )− r + 2 τ − 2 (t − τ )− 2 dτ
0
t c
1
1
1
1
∇ × vp τ − 2 (1 + τ )− r + 2 (t − τ )− 2 dτ
0
t +c
1
1
(1 + τ )− r + 2 τ
− 1r + p1
1
(t − τ )− 2 dτ
0
t c
− 1r + 12 − 12
∇ × vp (1 + τ )
τ
− 12
(t − τ )
t dτ + c
0
− 1r + p1
1
(t − τ )− 2 dτ
0
t c
τ
1
1
1
1
1
∇ × vp (1 + τ )− r + 2 τ − 2 (t − τ )− 2 dτ + ct 2
− 1r + p1
(4.4)
.
0
Since R1i = ∇φ(y) × ∇y × ωi (x − y, t − τ ), we also have t I2 p c
(u · ∇u)∇φ ∇ωi dτ 1 p
0
t c
u2 ∇u2 ∇ 2 ωi
2p p+2
dτ
0
t c
1
1
1
− 12 + p1
(1 + τ )− r + 2 τ − 2 (t − τ )
ct
− 1r + p1 + 12
(4.5)
.
0
Applying Young’s convolution inequality to I3 , we have that (∇ × v0 ) ∗ Γt c∇ × v0 p
2r 2−r
Γt s c∇ × v0
2r 2−r
t
− 1r + 12 + p1
,
H.-O. Bae, B.J. Jin / Journal of Functional Analysis 240 (2006) 508–529
where
1 s
=
1 p
−
1 r
525 2r
+ 32 . By Remark 4.2 we note that if u0 ∈ Lr and φu0 ∈ L 2−r , then ∇ × v0 ∈
2r
2r
L 2−r . Hence, if u0 ∈ Lr and φu0 ∈ L 2−r , then I3 p ct
− 1r + 12 + p1
(4.6)
.
Since ∂t Γ + Γ = 0, we have that ∂t ωi + ωi = N ∗ (∂t Γ + Γ ) = 0. Hence (∂τ + y )R1i = ∇φ(y) × ∇y × ∂t ωi + ωi + 2∇∂k φ × ∂k ∇y × ωi + ∇φ × ∇y × ωi = 2∂k (∇φ) × ∂k ∇y × ωi + ∇φ × ∇y × ωi . Therefore, I4 is bounded by t I4 p c
(u × ∇∂k φ) ∗ ∂k ∇ × ωi + (u × ∇φ) ∗ ∇ × ωi dτ p p
0
t c
u|x|−1
2p p+2
2 i ∇ ω + |x|−2 u ∇ωi dτ 2 1 p
2p p+2
2 i ∇ ω + |x|−2 u ∇ 2 ωi 2 1
2p p+2
1 −1+ 1 (t − τ )− 2 + |x|−2 u1 (t − τ ) 2 p dτ.
0
t c
u|x|−1
2p p+2
dτ
0
t c
u|x|−1
0
If we use the weighted inequality (5.3) in Galdi [17, p. 61], we have u|x|−1 Let q > 1. Let
1 q
2p p+2
∇u
2p p+2
ct
1 − 1r + p+2 2p − 2
= ct
− 1r + p1
.
= 1 − q1 .
1 1 −2 1 1 |x| u uq |x|−2 cq uq cq t − r + q = cq t − r +1− q . 1 q
Here we used the fact |x|−2 ∈ Lq (Ω) for q > 1. Therefore, we have t I4 p
−1+ 1 1 − 1 +1− q1 −1+ 1 cτ r p (t − τ )− 2 + cq τ r (t − τ ) 2 p dτ
0
ct
− 1r + p1 + 12
+ cq t
− 1r + p1 + 32 − q1
(4.7)
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H.-O. Bae, B.J. Jin / Journal of Functional Analysis 240 (2006) 508–529
for any q > 1. Let δ = 1 − q1 , then δ > 0 and δ goes to zero as q goes to 1. Hence the above estimate can be rewritten by I4 p ct
− 1r + p1 + 12
+ cδ t
− 1r + p1 + 12 +δ
.
Now we consider the estimate of I5 . Observe that t I5 p
(u∂j φ) ∗ ∂j V i + (uφ) ∗ V i dτ = P1 + P2 . p p
0
For P2 we can estimate in a similar way to the first term of (4.7). For P1 , observe that (u∂j φ) ∗ ∂j V i cu∂j φ2 ∇V i p
2p p+2
cu2 ∇Γ
2p p+2
,
and t P1
1
−1+ p1
1
τ − r + 2 (t − τ )
dτ ct
− 1r + p1 + 12
.
0
Therefore, we have I5 2 ct
− 1r + 12 + p1
(4.8)
.
Finally, from (4.4)–(4.8) we have t ∇ × vp
1
1
1
1
(1 + τ )− r + 2 τ − 2 (t − τ )− 2 ∇ × vp dτ
0
+ ct
− 1r + 12 + p1
+ cδ t
− 1r + 12 + p1 +δ
.
Now we define Bp (t) by Bp (t) = ∇ × v(t)p . Then Bp (t) satisfies t Bp (t)
1
1
1
1
(1 + τ )− r + 2 τ − 2 (t − τ )− 2 Bp (τ ) dτ
0
+ ct
− 1r + 12 + p1
+ cδ t
− 1r + 12 + p1 +δ
.
By the same reasoning as in the proof of Lemma 2.5, inequality (4.9) can be reduced to
(4.9)
H.-O. Bae, B.J. Jin / Journal of Functional Analysis 240 (2006) 508–529
Bp (t) c1 t
− 12
t
1
1
527
1
(1 + τ )− r + 2 τ − 2 Bp (τ ) dτ
0
+ ct
− 1r + 12 + p1
+ cδ t
− 1r + 12 + p1 +δ
(4.10)
for some constant c1 > 0. t 1 1 1 Set Yp (t) = 0 (1 + τ )− r + 2 τ − 2 Bp (τ ) dτ . Then Yp (t) satisfies 1 1 1 1 d −1+ 1 Yp c1 (1 + t)− r + 2 τ −1 Yp (t) + ct r p (1 + t)− r + 2 dt
+ cδ t
− 1r + p1 +δ
1
1
(1 + t)− r + 2 .
Let be small positive number so that 1 −
1 r
+
1 p
1
(4.11)
− > 0. Take large time T so that 1
c1 (1 + T )− r + 2 . Then (4.11) reduced to 1 1 1 1 d −1+ 1 − 1 + 1 +δ Yp t −1 Yp (t) + ct r p (1 + t)− r + 2 + cδ t r p (1 + t)− r + 2 dt
t −1 Yp (t) + ct
− 1r + p1
for t > T (here, we have considered δ < 1r − 12 ). Now by the same reasoning as in the proof of Lemma 2.4, it holds that for t > T () d − − 1 + 1 − t Yp (t) ct r p dt
for t > T .
Now integrate over [T , t] to obtain
t
−
Yp (t) T− Yp (T ) +
t τ
− 1r + p1 −
dτ
T
1− 1 + 1 − 1− 1 + 1 − T− Yp (T ) + c t r p − T r p ,
t > T .
Hence Yp (t) is bounded by 1− 1 + 1 1− 1 + 1 − Yp (t) T− Yp (T )t + c t r p − T r p t , Note that t < t
− 1r + p1 +1
since < 1 −
1 r
+ p1 . Hence we have
Yp (t) ct
1− 1r + p1
,
t > T .
t > T .
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H.-O. Bae, B.J. Jin / Journal of Functional Analysis 240 (2006) 508–529
By (4.10) the above estimate of Yp (t) implies that 1
Bp (t) c1 t − 2 Yp (t) + ct
− 1r + 12 + p1
+ cδ t
− 1r + 12 + p1 +δ
cδ t
− 1r + 12 + p1 +δ
,
t > T .
This implies that 1 1 1 ∇ × v(t) cδ t − r + 2 + p +δ , p
t > T .
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