arXiv:0805.3465v3 [math.AP] 24 Oct 2008
Global well-posedness of the critical Burgers equation in critical Besov spaces Changxing Miao1 and Gang Wu2 1
2
Institute of Applied Physics and Computational Mathematics, P.O. Box 8009, Beijing 100088, P.R. China. (miao
[email protected]) The Graduate School of China Academy of Engineering Physics, P.O. Box 2101, Beijing 100088, P.R. China. (
[email protected])
Abstract We make use of the method of modulus of continuity [12] and Fourier localization technique [1] to prove the global well-posedness of the critical Burgers 1 p equation ∂t u + u∂x u + Λu = 0 in critical Besov spaces B˙ p,1 (R) with p ∈ [1, ∞), √ where Λ = −△.
2000 Mathematics Subject Classification: 35K55, 35Q53 Key words and phrases: Burgers equation; Modulus of continuity; Fourier localization; Global well-posedness; Besov spaces
1
Introduction
We consider the Burgers equation with fractional dissipation in R, ( ∂t u + u∂x u + Λα u = 0 u(x, 0) = u0 (x),
(1.1)
where 0 ≤ α ≤ 2 and the operator Λα is defined by Fourier transform F(Λα u)(ξ) = |ξ|α Fu(ξ). The Burgers equation (1.1) with α = 0 and α = 2 has received an extensive amount of attention since the studies by Burgers in the 1940s. If α = 0, the equation is perhaps the most basic example of a PDE evolution leading to shocks; if α = 2, it provides an accessible model for studying the interaction between nonlinear and dissipative phenomena. Recently, in [12] for the periodic case authors give a complete study for general α ∈ [0, 2], see also [2, 9, 11, 14]. In particular, for α = 1, with help 1
of the method of modulus of continuity they proved the global well-posedness of the 1 equation in the critical Hilbert space H 2 (T1 ). In this paper, we study the following critical case, ( ∂t u + u∂x u + Λu = 0 u(x, 0) = u0 (x).
(1.2)
We use similar arguments as in [1]. Making use of Fourier localization technique and the method of modulus of continuity [12], we prove the global well-posedness of the 1 critical Burgers equation (1.2) in critical Besov spaces B˙ p (R) with p ∈ [1, ∞). p,1
1
p is the critical space under the scaling invariance. That It is well known that B˙ p,1 is, if u(x, t) is a solution of (1.2), then uλ (x, t) = u(λx, λt) is also a solution of the same equation and kuλ (·, t)k p1 ≈ ku(·, λt)k p1 .
B˙ p,1
B˙ p,1
Now we give out our main results. The first main result is the following: 1
p Theorem 1.1. Let u0 ∈ B˙ p,1 (R) with p ∈ [1, ∞), then the critical Burgers equation (1.2) has a unique global solution u such that 1
1
+1 p p ) ∩ L1loc (R+ ; B˙ p,1 ). u ∈ C(R+ ; B˙ p,1
Remark 1.1. Because of the restriction of the smooth index s stemming from the a priori estimate for the transport-diffusion equation (see Theorem 1.2), we can not get the result for the limit case p = ∞. Remark 1.2. The corresponding question for the quasi-geostrophic equation has been a focus of significant effort (see e.g. [1, 4, 5, 8, 13, 18, 19]) and the critical Q-G equation has been recently resolved in [13] for periodic case. Based on [13], Abidi-Hmidi in [1] and Dong-Du in [8] give the corresponding result for Cauchy problem of the critical Q-G equation in the framework of Besov space and Sobolev space, respectively. After the present paper is completed, Prof. J.Wu and H.Dong informed us that the authors in [9] gave the global well-posedness for the critical fractal Burgers equation 1 in inhomogeneous space H 2 by similar argument in [8]. In order to prove this theorem, we first prove the local well-posedness which is the major part of this paper. Next we make use of the modulus of continuity [12] to get the global well-posedness. We mention that the property allowing us to remove the periodicity is the spatial decay of the solution. The key of proving the local well-posedness is an optimal a priori estimate for the following transport-diffusion equation in RN : ( ∂t u + v · ∇u + νΛα u = f (T D)ν,α u(x, 0) = u0 (x), where v is a given vector field which needs not to be divergence free, u0 is the initial data, f is a given external force term, ν ≥ 0 is a constant, 0 ≤ α ≤ 2. Our second main result is the following: 2
Theorem 1.2. Let 1 ≤ ρ1 ≤ ρ ≤ ∞, 1 ≤ p ≤ p1 ≤ ∞ and 1 ≤ r ≤ ∞. Let s ∈ R satisfy the following N N s −N min , , if div v = 0 . or s > −1 − N min p1 p′ p1 p′
There exists a constant C > 0 depending only on N , α, s, p, p1 and r, such that for any smooth solution u of (T D)ν,α with ν ≥ 0, we have the following a priori estimate: 1 1 −1 (1.3) ν ρ kuk ρ s+ αρ ≤ CeCZ(T ) ku0 kB˙ s + ν ρ1 kf k ρ s−α+ ρα with Z(T ) :=
RT 0
e B˙ p,r L T
p,r
k∇v(t)k
N
p B˙ p11,∞ ∩L∞
e 1 B˙ p,r L T
1
d t.
Besides if u = v, then for all s > 0(s > −1 if div v = 0), the estimate (1.3) holds RT with Z(T ) = 0 k∇v(t)kL∞ d t.
Remark 1.3. When α = 2, the above a priori estimate has been proved by R.Danchin in [6]. In this paper, we extend Danchin’s results to the general case α ∈ [0, 2]. The proof ’s key is the use of Lagrangian coordinates transformation together with an important commutator estimate. The rest of this paper is arranged as follows: In Section 2, we recall some definitions and properties about homogeneous Besov spaces, and we will also list some useful lemmas. In Section 3, we prove Theorem 1.2. In Section 4, we prove the local well-posedness. In Section 5, we give the blow-up criterion. In Section 6, we complete the proof of the global well-posedness. Notation: Throughout the paper, C stands for a constant which may be different in each occurrence. We shall sometimes use the notation A . B instead of A ≤ CB and A ≈ B means that A . B and B . A.
2
Preliminaries
Let us first recall the Littlewood-Paley Theory. Let χ and ϕ be acouple of smooth N |ξ| ≤ 4 , radial functions valued in [0,1] such that χ is supported in the ball ξ ∈ R 3 ϕ is supported in the shell ξ ∈ RN 43 ≤ |ξ| ≤ 83 and X χ(ξ) + ϕ(2−q ξ) = 1, ∀ξ ∈ RN ; X
q∈N
ϕ(2−q ξ) = 1,
q∈Z
∀ξ ∈ RN \{0}.
Denoting ϕq (ξ) = ϕ(2−q ξ) and hq = F −1 ϕq , we define the homogeneous dyadic blocks as Z −q ˙ hq (y)u(x − y) d y, ∀q ∈ Z. ∆q u := ϕ(2 D)u = RN
3
We can also define the following low-frequency cut-off: X ˙ j u. ∆ S˙ q u := j≤q−1
Definition 2.1. Let Sh′ be the space of temperate distributions u such that lim S˙ q u = 0,
q→−∞
The formal equality u=
X
S ′.
in
˙ qu ∆
q∈Z
holds in Sh′ and is called the homogeneous Littlewood-Paley decomposition. It has nice properties of quasi-orthogonality: ˙ q′ ∆ ˙ qu ≡ 0 ∆
if
|q ′ − q| ≥ 2
˙ q v) ≡ 0 ˙ q′ (S˙ q−1 u∆ and ∆
|q ′ − q| ≥ 5.
if
(2.1)
Let us now define the homogeneous Besov spaces: Definition 2.2. For s ∈ R, (p, r) ∈ [1, ∞]2 and u ∈ Sh′ , we set kukB˙ s := p,r
X q∈Z
and kukB˙ s
p,∞
˙ q ukrLp 2qsr k∆
1 r
if
r 0, there exists a positive constant C = C(N, s) such that kuvkB˙ s ≤ C(kukL∞ kvkB˙ s + kvkL∞ kukB˙ s ). p,r
p,r
4
p,r
In our next study we require two kinds of coupled space-time Besov spaces. The first one is defined by the following manner: for T > 0 and ρ ∈ [1, ∞], we denote by s the set of all tempered distribution u satisfying LρT B˙ p,r kukLρ B˙ s
p,r
T
X 1
r qsr ˙ r
:= 2 k∆q ukLp
q∈Z
LρT
< ∞.
s which is the set of all tempered distribution u e ρ B˙ p,r The second mixed space is L T satisfying 1 X r qsr ˙ r < ∞. 2 k∆q ukLρ Lp kukLe ρ B˙ s := T
p,r
T
q∈Z
Let us remark that, by virtue of the Minkowski inequality, we have kukLe ρ B˙ s ≤ kukLρ B˙ s
if
ρ ≤ r,
kukLρ B˙ s ≤ kukLe ρ B˙ s
if
ρ ≥ r.
T
p,r
T
p,r
and T
p,r
T
p,r
Now we give some useful lemmas. Lemma 2.1. (cf. [10, 17]) Let φ be a smooth function supported in the shell ξ ∈ RN R1 ≤ |ξ| ≤ R2 , 0 < R1 < R2 . There exist two positive constants κ and C depending only on φ such that for all 1 ≤ p ≤ ∞, τ ≥ 0 and λ > 0, we have α
α
kφ(λ−1 D)e−τ Λ ukLp ≤ Ce−κτ λ kφ(λ−1 D)ukLp .
Lemma 2.2. (cf. [6]) Let v be a smooth vector field. Let ψt be the solution to Z t v(τ, ψτ (x)) d τ. ψt (x) = x + 0
Then for all t ∈ R+ , the flow ψt is a C 1 diffeomorphism over RN and one has k∇ψt±1 kL∞ ≤ eV (t) ,
k∇ψt±1 − IdkL∞ ≤ eV (t) − 1, Z t k∇2 ψt±1 kL∞ ≤ eV (t) k∇2 v(τ )kL∞ eV (τ ) d τ, 0
where V (t) =
Rt 0
k∇v(τ )kL∞ d τ .
Lemma 2.3. (cf. [3]) Let v be a given vector field belonging to L1loc (R+ ; Lip). For ˙ q u and denote by ψq the flow of the regularized vector field S˙ q−1 v. q ∈ Z we set uq := ∆ α ˙ Then for u ∈ Bp,∞ with α ∈ [0, 2) and p ∈ [1, ∞] we have α
kΛα (uq ◦ ψq ) − (Λα uq ) ◦ ψq kLp ≤ CeCV (t) V 1− 2 (t)2qα kuq kLp , where V (t) =
Rt 0
k∇v(τ )kL∞ d τ and C = C(α, p) > 0 is a constant.
5
Lemma 2.4. Let σ ∈ R and 1 ≤ p ≤ p1 ≤ ∞, p2 := (1/p − 1/p1 )−1 . Let Rq := ˙ q u − [∆ ˙ q , v · ∇]u. There exists a constant C = C(N, σ) such that (S˙ q−1 v − v) · ∇∆ X ′ qσ ˙ q′ ukLp kS˙ q′ −1 ∇vkL∞ 2q σ k∆ 2 kRq kLp ≤ C |q ′ −q|≤4
+
X
q ′ ≥q−3
+
′ ˙ q ukLp ˙ q′ ∇vkL∞ 2qσ k∆ 2q−q k∆
X
(q−q ′′ )(σ−1− pN ) q ′ pN
2
1
2
1
|q ′ −q|≤4 q ′′ ≤q ′ −2
X
+
′′
(q−q ′ ) σ+N min( p1 , p1′ )
2
1
q ′ ≥q−3 |q ′′ −q ′ |≤1 q ′ pN
×2
˙ q′ ∇vkLp1 2q σ k∆ ˙ q′′ ukLp k∆
q−q ′
2
1
q′′ σ ˙ ˙ ˙ k∆q′ ∇vkLp1 + k∆q′ div vkLp1 2 k∆q′′ ukLp ,
and the third term in the right-hand side may be replaced by X ′ ˙ q′ ∇vkLp1 kS˙ q′ −1 ∂j ukLp2 . C 2q (σ−1) k∆ |q ′ −q|≤4
Besides if u = v, the following estimate holds true: X ′ ˙ q′ ukLp kS˙ q′ −1 ∇ukL∞ 2q σ k∆ 2qσ kRq kLp ≤ C |q ′ −q|≤4
+
X
q ′ ≥q−3
+
′ ˙ q ukLp ˙ q′ ∇ukL∞ 2qσ k∆ 2q−q k∆
X
q ′ ≥q−3 |q ′′ −q ′ |≤1 q−q ′
× 2
′
2(q−q )σ q′′ σ ˙ ˙ ˙ k∆q′ ∇ukL∞ + k∆q′ div ukL∞ 2 k∆q′′ ukLp .
R.Dancin in [7] gave the proof for the nonhomogeneous case. For the convenience of the reader, we will give the proof for the homogeneous case in the appendix.
3
Proof of Theorem 1.2
Proof of Theorem 1.2. Here we only prove the case α ∈ [0, 2) (for the case α = 2, see [6]). ˙ q u and fq := ∆ ˙ q f . Applying ∆ ˙ q to (T D)ν,α yields Let uq := ∆ ∂t uq + S˙ q−1 v · ∇uq + νΛα uq = fq + Rq ˙ q , v · ∇]u. with Rq := (S˙ q−1 v − v) · ∇uq − [∆ 6
Let ψq be the flow of the regularized vector field S˙ q−1 v. Denote u ¯q := uq ◦ ψq , ¯ ¯ q := Rq ◦ ψq . Then we have fq := fq ◦ ψq and R ¯ q + νGq ∂t u ¯q + νΛα u ¯q = f¯q + R
(3.1)
with Gq := Λα (uq ◦ ψq ) − (Λα uq ) ◦ ψq . ˙ j to (3.1) and using Lemma 2.1, we get Applying ∆ jα ˙ j u0,q kLp + ˙ ju k∆ ¯q (t)kLp . e−κνt2 k∆
Z
t
jα
e−κν(t−τ )2
0 ¯ ˙ ˙ ¯ ˙ j Gq kLp d τ. × k∆j fq kLp + k∆j Rq kLp + νk∆
(3.2)
Now from Lemma 2.3 we have
α
˙ j Gq (t)kLp ≤ CeCV (t) V 1− 2 (t)2qα kuq kLp . k∆
(3.3)
According to Bernstein lemma and Lemma 2.2, we can get ˙ j f¯q (t)kLp . 2−j k∇∆ ˙ j f¯q kLp k∆
. 2−j k(∇fq ) ◦ ψq kLp k∇ψq kL∞
(3.4)
1
. 2−j k∇fq kLp kJψq−1 kLp ∞ k∇ψq kL∞
. eCV (t) 2q−j kfq kLp . Arguing similarly as in deriving (3.4), we obtain
˙ jR ¯ q (t)kLp . eCV (t) 2q−j kRq kLp . k∆ According to Lemma 2.4, we get ˙ jR ¯ q (t)kLp . eCV (t) 2q−j cq (t)2−qs Z ′ (t)ku(t)k ˙ s , k∆ B
(3.5)
p,r
with kcq (t)kℓr = 1. Plugging (3.3), (3.4) and (3.5) into (3.2), taking the Lρ norm over [0, t] and mul1 q(s+ α ) ρ , we obtain tiplying both sides by ν ρ 2 1
α
˙ ju ¯q kLρt Lp . 2 ν ρ 2q(s+ ρ ) k∆
(q−j)α ρ
− ρ1′
+ν
1
1 ρ
1
jα ˙ j u0,q kLp (1 − e−κνρt2 ) ρ 2qs k∆
(q−j)(1+ α + ρα′ ) CV (t) q(s− ρα′ ) ρ 1 1
q(s+ α ) ρ
+ν 2
e
2
2
) (q−j)(1+ α ρ
+2
(q−j)α CV (t)
e
Z
0
t
2
V
1− α 2
t
(t)kuq kLρt Lp
cq (τ )Z ′ (τ )eCV (τ ) ku(τ )kB˙ s d τ. p,r
Let M0 ∈ Z to be fixed hereafter. Decomposing X ˙ ju ∆ ¯q ◦ ψq−1 , uq = S˙ q−M0 u ¯q ◦ ψq−1 + j≥q−M0
7
kfq kLρ1 Lp
(3.6)
we have for all t ∈ [0, T ], kuq kLρt Lp ≤ eCV (t) kS˙ q−M0 u ¯q kLρt Lp +
X
j≥q−M0
˙ ju k∆ ¯q kLρt Lp .
(3.7)
By Lemma A.1 in [6], we have 1
kS˙ q−M0 u ¯q kLp . kJψq−1 kLp ∞ (2−q k∇Jψq−1 kL∞ kJψq kL∞ + 2−M0 k∇ψq−1 kL∞ )kuq kLp . This together with Lemma 2.2 and Bernstein lemma leads to kS˙ q−M0 u ¯q kLρt Lp . eCV (t) (eCV (t) − 1 + 2−M0 )kuq kLρt Lp . ˙ j u0,q = 0 for |j − q| > 1, from (3.6) we get As ∆ X 1 q(s+ α ) ˙ ρ k∆ ¯q kLρt Lp νρ2 ju
(3.8)
j≥q−M0
1
qα
.(1 − e−κνρt2 ) ρ 2qs ku0,q kLp + ν 1 ρ
M0 α CV (t)
+ν 2
e
+ 2M0 (1+α)
Z
t 0
V
1− α 2
− ρ1′
) q(s+ α ρ
(t)2
1
q(s− ρα′ )
2M0 (1+α) eCV (t) 2
1
kfq kLρ1 Lp t
(3.9)
kuq kLρt Lp
cq (τ )Z ′ (τ )eCV (τ ) ku(τ )kB˙ s d τ. p,r
Plugging (3.8) and (3.9) into (3.7) yields that 1
q(s+ α ) ρ
νρ2
qα
1
kuq kLρt Lp ≤ C(1 − e−κνρt2 ) ρ 2qs ku0,q kLp −1 q(s− ρα′ ) ′ 1 kfq k ρ1 p + CeCV (t) ν ρ1 2M0 (1+α) 2 Lt L 1 α α q(s+ ρ ) + 2−M0 + 2M0 α V 1− 2 (t) ν ρ 2 kuq kLρt Lp Z t + 2M0 (1+α) cq (τ )Z ′ (τ )ku(τ )kB˙ s d τ . p,r
0
Choose M0 to be the unique integer such that 2C2−M0 ∈ ( 18 , 14 ] and T1 to be the largest real number such that 2−M0 α 2 2−α T1 ≤ T and CV (T1 ) ≤ C0 with C0 = min ln 2, . α 8C 2 Thus for t ∈ [0, T1 ], there exists a constant C1 such that 1 1 q(s+ α ) −κνρt2qα ρ qs ρ ku k ρ p ≤ C ) 2 ku0,q kLp νρ2 1 (1 − e q Lt L −
1 ′
q(s−
α ′
)
ρ1 kfq kLρ1 Lp + ν ρ1 2 t Z t cq (τ )Z ′ (τ )ku(τ )kB˙ s d τ . + p,r
0
Taking ℓr norm yields 1 ρ
ν kuk
s+ α e ρ B˙ p,r ρ L t
− 1′ ≤ C1 ku0 kB˙ s + ν ρ1 kf k p,r
s− α′ ρ ρ1 ˙ e Lt Bp,r 1
8
+
Z
t 0
Z ′ (τ )ku(τ )kB˙ s d τ . (3.10) p,r
Splitting [0, T ] into m subintervals like as [0, T1 ], [T1 , T2 ] and so on, such that C
Z
Tk+1
k∇v(t)kL∞ d t ≈ C0 .
Tk
Arguing similarly as in deriving (3.10), we get for all t ∈ [Tk , Tk+1 ], 1
ν ρ kuk
s+ α ρ eρ L B˙ [Tk ,t] p,r
− 1′ ≤ C1 ku(Tk )kB˙ s + ν ρ1 kf k p,r
Z
+
t
Tk
Z ′ (τ )ku(τ )kB˙ s
p,r
dτ .
e ρ1 L [T
k
s− α′
ρ 1 B˙ ,t] p,r
By a standard induction argument, it can be shown that 1 ρ
ν kuk
≤
s+ α e ρ B˙ p,r ρ L t
C1k+1
ku0 kB˙ s + ν
− ρ1′
1
p,r
kf k
s− α′ e ρ1 B˙ p,r ρ1 L t
+
Z
t 0
Z ′ (τ )ku(τ )kB˙ s d τ . p,r
Since the number of such subintervals is m ≈ CV (T )C0−1 , one can readily conclude that up to a change of C, 1
ν ρ kuk
s+ α e ρ B˙ p,r ρ L T
− 1′ ≤ CeCV (T ) ku0 kB˙ s + ν ρ1 kf k p,r
+
Z
T
0
′
s− α′
e ρ1 B˙ p,r ρ1 L T
(3.11)
Z (t)ku(t)kB˙ s d t . p,r
Of course, the above inequality is valid for all ρ ∈ [ρ1 , ∞]. Choosing first ρ = ∞ in (3.11) and applying Gronwall lemma leads to 1 −1 kukLe ∞ B˙ s ≤ CeCZ(T ) ku0 kB˙ s + ν ρ1 kf k ρ s−α+ ρα . (3.12) T
e 1 B˙ p,r L T
p,r
p,r
1
Now plugging (3.12) into (3.11) yields the desired estimate for general ρ.
4
Local well-posedness
In this section, we prove the following result: 1
p Proposition 4.1. Let u0 ∈ B˙ p,1 (R) with p ∈ [1, ∞), then there exists T > 0 such that the equation (1.2) has a unique solution u such that 1
1
e ∞ B˙ p ∩ L1 B˙ p +1 . u∈L T p,1 T p,1 1
e ∞ B˙ p +β . Besides for all β ∈ R+ , we have tβ u ∈ L p,1 T
Proof. We prove this proposition by making use of an iterative method. Step 1: approximation solution. 9
Let u0 := e−tΛ u0 (x) and let un+1 be the solution of the linear equation ( ∂t un+1 + un ∂x un+1 + Λun+1 = 0 un+1 (x, 0) = u0 (x). 1
+1
p Obviously u0 ∈ L1 (R+ ; B˙ p,1 ), thus according to Theorem 1.2, we have ∀n ∈ N, 1
1
+1
e ∞ (R+ ; B˙ p ) ∩ L1 (R+ ; B˙ p ). un ∈ L p,1 p,1 Step 2: uniform bounds. Now we intend to obtain uniform bounds, with respect to the parameter n, for some T > 0 independent of n. By making use of Lemma 2.4 and similar arguments as in the proof of Theorem 1.2, for all T > 0 such that Z T kun (τ )k p1 +1 d τ ≤ CC0 , B˙ p,1
0
we have kun+1 k
1+1 e 2 B˙ p 2 L t p,1
+ kun+1 k
1 p +1 L1t B˙ p,1
≤C
X q∈Z
q
q 1 ˙ q u0 kLp (1 − e−κt2 ) 2 2 p k∆
+ Ckun k
1+1 2
e 2 B˙ p L t p,1
kun+1 k
1+1 2
e 2 B˙ p L t p,1
.
By Lebesgue theorem, there exist T > 0 and an absolute constant ε0 > 0 such that X q 1 q ˙ q u0 kLp ≤ ε0 (1 − e−κT 2 ) 2 2 p k∆ (4.1) q∈Z
and kun+1 k
1+1 2
e 2 B˙ p L T p,1
+ kun+1 k
1 +1
p L1T B˙ p,1
≤ 2ε0 .
(4.2) 1
p On the other hand, by Theorem 1.2 and the Sobolev embedding B˙ p,1 ֒→ L∞ , we have
C
ku
n+1
k
1
e ∞ B˙ p L p,1 T
≤ Ce
RT 0
kun (τ )k
1 p +1 B˙ p,1
dτ
ku0 k
1
p B˙ p,1
≤ Cku0 k
1
p B˙ p,1
.
Combining the above results, we have proved that the sequence (un )n∈N is uniformly 1 1 e∞ B˙ p ∩ L1 B˙ p +1 . bounded in L T
p,1
T
p,1
Step 3: strong convergence.
1
e∞ B˙ p . We first prove that (un )n∈N is a Cauchy sequence in L p,1 T 10
Let (n, m) ∈ N2 , n > m and un,m := un − um . One easily verifies that ( ∂t un+1,m+1 + un ∂x un+1,m+1 + Λun+1,m+1 = −un,m ∂x um+1 un+1,m+1 (x, 0) = 0. According to Theorem 1.2, we have n+1,m+1
ku
Z
Ckun k
k
1 e ∞ B˙ p L p,1 T
1 +1 ˙ p L1 B T p,1
≤ Ce
T
0
kun,m ∂x um+1 (τ )k
1
p B˙ p,1
d τ.
(4.3)
1
p By Proposition 2.1 and the embedding B˙ p,1 ֒→ L∞ , we have
kun,m ∂x um+1 k
1
p B˙ p,1
. kun,m k
1
p B˙ p,1
kum+1 k
1 +1
p B˙ p,1
.
Substituting this into (4.3) yields ku
n+1,m+1
k
1 e ∞ B˙ p L p,1 T
≤ Cku
n,m
Ckun k
k
1 e ∞ B˙ p L p,1 T
1 +1 ˙ p L1 B T p,1
e
Z
T
kum+1 (τ )k
1 +1
p B˙ p,1
0
d τ.
By (4.1), we can choose ε0 small enough such that kun+1,m+1 k
1
e ∞ B˙ p L p,1 T
≤ ǫkun,m k
1
e ∞ B˙ p L p,1 T
with ǫ < 1. Now we can get by induction kun+1,m+1 k
1
e ∞ B˙ p L p,1 T
≤ ǫm+1 kun,0 k
1
e ∞ B˙ p L p,1 T
≤ Cǫm+1 ku0 k
1
p B˙ p,1
.
1
e ∞ B˙ p . Thus there exists u ∈ This implies that (un )n∈N is a Cauchy sequence in L p,1 T 1
1
e ∞ B˙ p such that un converges strongly to u in L e ∞ B˙ p . Fatou lemma and (4.2) ensure L p,1 p,1 T T 1
+1 p . Thus by passing to the limit into the approximation equation, we that u ∈ L1T B˙ p,1 1
1
e∞ B˙ p ∩ L1 B˙ p +1 . can get a solution to (1.2) in L p,1 T p,1 T Step 4: uniqueness.
Let u1 and u2 be two solutions of the equation (1.2) with the same initial data 1 1 e ∞ B˙ p ∩ L1 B˙ p +1 . Let u1,2 := u1 − u2 , then we have and belonging to the space L p,1 T T p,1 ( ∂t u1,2 + u1 ∂x u1,2 + Λu1,2 = −u1,2 ∂x u2 u1,2 (x, 0) = 0. By similar arguments as in Step 3, we have ku1,2 k
1 e ∞ B˙ p L t p,1
Cku1 k
≤ Ce
1 +1 ˙ p L1 t Bp,1
Z
t 0
ku1,2 k
Gronwall’s inequality ensures that u1 = u2 , ∀t ∈ [0, T ]. 11
1
˙p e∞ L τ Bp,1
ku2 (τ )k
1 +1
p B˙ p,1
d τ.
Step 5: smoothing effect. We will prove that for all β ∈ R+ , we have C(β+1)kuk β
kt uk
1 +β
e ∞ B˙ p L p,1 T
1 +1 ˙ p L1 B T p,1
≤ Cβ e
kuk
1
e ∞ B˙ p L p,1 T
.
(4.4)
It is obvious that (
∂t (tβ u) + u∂x (tβ u) + Λ(tβ u) = βtβ−1 u (tβ u)(x, 0) = 0.
When β = 1, by Theorem 1.2, we have Ckuk
ktu(t)k
1 +1
e ∞ B˙ p L p,1 T
1 +1 ˙ p L1 B T p,1
≤ Ce
kuk
1
e ∞ B˙ p L p,1 T
.
Suppose (4.4) is true for n, we will prove it for n + 1. Applying Theorem 1.2 to the equation of tn+1 u yields that Ckuk
kt
n+1
u(t)k
1 +n+1
e ∞ B˙ p L p,1 T
1 +1 ˙ p L1 B T p,1
≤ C(n + 1)e
ktn uk
1 +n
e ∞ B˙ p L p,1 T
C(n+2)kuk
1 +1 ˙ p L1 B T p,1
≤ Cn e
kuk
1
e ∞ B˙ p L p,1 T
.
For general β ∈ R+ , obviously [β] ≤ β ≤ [β] + 1. Thus by the following interpolation ktβ uk
1 +β
e ∞ B˙ p L p,1 T
[β]+1−β
. kt[β] uk
1 +[β]
e ∞ B˙ p L p,1 T
β−[β]
kt[β]+1 uk
1 +[β]+1
e ∞ B˙ p L p,1 T
,
we can get the estimate for general β ∈ R+ .
5
Blow-up criterion
In this section, we prove the following blow-up criterion: 1
e ∞ B˙ p ∩ Proposition 5.1. Let T ∗ be the maximum local existence time of u in L p,1 T 1
+1
p . If T ∗ < ∞, then L1T B˙ p,1
Z Proof. Suppose
R T∗
∀t ∈ [0, T ∗ ),
0
T∗ 0
k∂x u(t)kL∞ d t = ∞.
k∂x u(t)kL∞ d t be finite, then by Theorem 1.2, we have
ku(t)k
1 p B˙ p,1
≤ MT ∗ := CeC
12
R T∗ 0
k∂x u(t)kL∞ d t
ku0 k
1
p B˙ p,1
< ∞.
(5.1)
Let Te > 0 such that
X e q 1 (1 − e−κT 2 ) 2 MT ∗ ≤ ε0 ,
(5.2)
q∈Z
where ε0 is the absolute constant emerged in the proof of Proposition 4.1. Now (5.1) and (5.2) imply that X e q 1 q ˙ ∀t ∈ [0, T ∗ ), (1 − e−κT 2 ) 2 2 p k∆ q u(t)kLp ≤ ε0 . q∈Z
This together with the local existence theory ensures that, there exists a solution u e(t) on [0, Te) to (1.2) with the initial datum u(T ∗ − Te/2). By uniqueness, u e(t) = ∗ ∗ e e u(t + T − T /2) on [0, T /2) so that u e extends the solution u beyond T .
6
Global well-posedness
In this section, making use of the method of modulus of continuity [12], with help of similar arguments as in [1], we give the proof of the global well-posedness. Let T ∗ be the maximal existence time of the solution u to (1.2) in the space 1 1 e ∞ ([0, T ∗ ); B˙ p ) ∩ L1 ([0, T ∗ ); B˙ p +1 ). From Proposition 4.1, there exists T0 > 0 L p,1 p,1 loc such that ∀t ∈ [0, T0 ], tk∂x u(t)kL∞ ≤ Cku0 k p1 . B˙ p,1
Let λ be a positive real number that will be fixed later and T1 ∈ (0, T0 ). We define the set I := {T ∈ [T1 , T ∗ ); ∀t ∈ [T1 , T ], ∀x 6= y ∈ R, |u(x, t) − u(y, t)| < ωλ (|x − y|)}, where ω : R+ −→ R+ is strictly increasing, concave, ω(0) = 0, ω ′ (0) < +∞, limξ→0+ ω ′′ (ξ) = −∞ and ωλ (|x − y|) = ω(λ|x − y|). The function ω is a modulus of continuity chosen as in [12]. We first prove that T1 belongs to I under suitable conditions over λ. Let C0 be a large positive number such that 2ku0 kL∞ < ω(C0 ) < 3ku0 kL∞ . Since ω is strictly increasing, then by maximum principle we have λ|x − y| ≥ C0 ⇒ |u(x, T1 ) − u(y, T1 )| ≤ 2ku0 kL∞ < ωλ (|x − y|). On the other hand we have from Mean Value Theorem |u(x, T1 ) − u(y, T1 )| ≤ |x − y|k∂x u(T1 )kL∞ . Let 0 < δ0 < C0 . Then by the concavity of ω we have λ|x − y| ≤ δ0 ⇒ ωλ (|x − y|) ≥ 13
ω(δ0 ) λ|x − y|. δ0
(6.1)
If we choose λ so that λ>
δ0 k∂x u(T1 )kL∞ , ω(δ0 )
then we get 0 < λ|x − y| ≤ δ0 ⇒ |u(x, T1 ) − u(y, T1 )| < ωλ (|x − y|). Let us now consider the case δ0 ≤ λ|x − y| ≤ C0 . By Mean Value Theorem and the increasing property of ω, we can get |u(x, T1 ) − u(y, T1 )| ≤
C0 k∂x u(T1 )kL∞ λ
Choosing λ such that λ>
and
ω(δ0 ) ≤ ωλ (|x − y|).
C0 k∂x u(T1 )kL∞ , ω(δ0 )
thus we get δ0 ≤ λ|x − y| ≤ C0 ⇒ |u(x, T1 ) − u(y, T1 )| < ωλ (|x − y|). All the preceding conditions over λ can be obtained if we take λ=
ω −1 (3ku0 kL∞ ) k∂x u(T1 )kL∞ . 2ku0 kL∞
(6.2)
From the construction, the set I is an interval of the form [T1 , T∗ ). We have three possibilities which will be discussed separately. Case 1: The first possibility is T∗ = T ∗ . In this case we necessarily have T ∗ = ∞ because the Lipschitz norm of u does not blow up. Case 2: The second possibility is T∗ ∈ I and we will show that is not possible. Let C0 satisfy (6.1), then for all t ∈ [T1 , T ∗ ), we have λ|x − y| ≥ C0 ⇒ |u(x, t) − u(y, t)| < ωλ (|x − y|). 1
p Since ∂x u(t) belongs to C((0, T ∗ ); B˙ p,1 ), then for ε > 0 there exist η0 , R > 0 such that ∀t ∈ [T∗ , T∗ + η0 ],
k∂x u(t)kL∞ ≤ k∂x u(T∗ )kL∞ +
ε 2
ε c and k∂x u(T∗ )kL∞ (B(0,R) ) ≤ , 2
where B(0,R) is the ball of radius R and with center the origin. Hence for λ|x−y| ≤ C0 and x or y ∈ B c C0 , we have for ∀t ∈ [T∗ , T∗ + η0 ] (0,R+
λ
)
c |u(x, t) − u(y, t)| ≤ |x − y|k∂x u(t)kL∞ (B(0,R) ) ≤ ε|x − y|.
On the other hand we have from the concavity of ω λ|x − y| ≤ C0 ⇒
ω(C0 ) λ|x − y| ≤ ωλ (|x − y|). C0 14
Thus if we take ε sufficiently small such that ε
0 such that ∀t ∈ [T∗ , T∗ + η1 ] k∂x u(t)kL∞ (B
C ) (0,R+ 0 ) λ
0 such that for all t ∈ [T∗ , T∗ + η2 ] ∀x, y ∈ B(0,R+ C0 ) , δ1 ≤ λ|x − y|; |u(x, t) − u(y, t)| < ωλ (|x − y|). λ
Taking η = min(η0 , η1 , η2 ), we obtain that T∗ + η ∈ I which contradicts the fact that T∗ is maximal. Case 3: The last possibility is that T∗ does not belong to I. By the continuity in time of u, there exist x 6= y such that u(x, T∗ ) − u(y, T∗ ) = ωλ (ξ), 15
with
ξ = |x − y|.
We will show that this scenario can not occur and more precisely: f ′ (T∗ ) < 0 where f (t) := u(x, t) − u(y, t). This is impossible since f (t) ≤ f (T∗ ), ∀t ∈ [0, T∗ ]. The proof is the same as [12] and for the convenience of the reader we sketch out the proof. From the regularity of the solution we see that the equation can be defined in the classical manner and f ′ (T∗ ) = u(y, T∗ )∂x u(y, T∗ ) − u(x, T∗ )∂x u(x, T∗ ) + Λu(y, T∗ ) − Λu(x, T∗ ). From [12] we have u(y, T∗ )∂x u(y, T∗ ) − u(x, T∗ )∂x u(x, T∗ ) ≤ ωλ (ξ)ωλ′ (ξ). Again from [12] Z ξ 1 2 ωλ (ξ + 2η) + ωλ (ξ − 2η) − 2ωλ (ξ) dη Λu(y, T∗ ) − Λu(x, T∗ ) ≤ π 0 η2 Z 1 ∞ ωλ (2η + ξ) − ωλ (2η − ξ) − 2ωλ (ξ) + dη π ξ η2 2
≤ λJ(λξ), where 1 J(ξ) = π
Z
ξ 2
ω(ξ + 2η) + ω(ξ − 2η) − 2ω(ξ) dη η2 0 Z 1 ∞ ω(2η + ξ) − ω(2η − ξ) − 2ω(ξ) + d η. π ξ η2 2
Thus we get f ′ (T∗ ) ≤ λ(ωω ′ + J)(λξ). Now, we choose the same function as [12] ( ω(ξ) =
ξ√ , 1+4π ξ0 ξ
Cξ0 log ξ,
if 0 ≤ ξ ≤ ξ0 ; if ξ ≥ ξ0 ,
here ξ0 is sufficiently large number and Cξ0 is chosen to provide continuity of ω. It is shown in [12], ∀ξ 6= 0, ω(ξ)ω ′ (ξ) + J(ξ) < 0.
Thus we can get that f ′ (T∗ ) < 0.
Combining the above discussion, we conclude that T ∗ = ∞ and ∀t ∈ [T1 , ∞),
k∂x ukL∞ ≤ λω ′ (0) = λ.
The value of λ is given by (6.2).
16
7
Appendix --
Commutator Estimate
In this appendix, we give the proof of Lemma 2.4. By Bony’s decomposition, we have ˙ q u − [∆ ˙ q , v · ∇]u Rq = (S˙ q−1 v − v) · ∇∆ ˙ q ]∂j u + T˙ ˙ v j − ∆ ˙ q T˙∂ u v j = [T˙vj , ∆ j ∂j ∆q u j j ˙ ,∆ ˙ q u) − ∂j ∆ ˙ q R(v ˙ , u) + ∂j R(v ˙ q R(div ˙ ˙ ˙ q u) + ∆ v, u) − R(div v, ∆
(7.1)
˙ qu + (S˙ q−1 v − v) · ∇∆
=: Rq1 + Rq2 + Rq3 + Rq4 + Rq5 + Rq6 . Above, the summation convention over repeated indices has been used. The notation T˙ stands for homogeneous Bonys paraproduct which is defined by X ˙ q′ g, T˙f g := S˙ q′ −1 f ∆ q ′ ∈Z
and R˙ stands for the remainder operator defined by X ˙ ˙ q′ +1 g). ˙ q′ −1 g + ∆ ˙ q′ g + ∆ ˙ q′ f (∆ R(f, g) := ∆ q ′ ∈Z
Note that
˙ q′ vkLa , ˙ q′ ∇vkLa ≈ 2q k∆ k∆ ′
∀a ∈ [1, ∞], q ′ ∈ Z.
(7.2)
Now let us estimate each term in (7.1). Bounds for 2qσ kRq1 kLp : ˙ q , we have By (2.1) and the definition of ∆ X ˙ q ]∂j ∆ ˙ q′ u Rq1 = [S˙ q′ −1 v j , ∆ |q ′ −q|≤4
=
X Z
N |q ′ −q|≤4 R
(7.3) ˙ q′ u(x − 2−q y) d y. h(y) S˙ q′ −1 v j (x) − S˙ q′ −1 v j (x − 2−q y) ∂j ∆
Applying Mean Value Theorem and Young’s inequality to (7.3) yields X ′ ˙ q′ ukLp . 2qσ kRq1 kLp ≤ C kS˙ q′ −1 ∇vkL∞ 2q σ k∆ |q ′ −q|≤4
Bounds for 2qσ kRq2 kLp : According to (2.1), we have Rq2 =
X
q ′ ≥q−3
˙ q u∆ ˙ q′ vj . S˙ q′ −1 ∂j ∆ 17
(7.4)
By (7.2), we can get 2qσ kRq2 kLp ≤ C ≤C
X
q ′ ≥q−3
X
q ′ ≥q−3
˙ q′ v j kL∞ kS˙ q′ −1 ∂j ∆ ˙ q ukLp 2qσ k∆ ′ ˙ q ukLp . ˙ q′ ∇vkL∞ 2qσ k∆ 2q−q k∆
(7.5)
Bounds for 2qσ kRq3 kLp : Again from (2.1), we have X X ˙ q′ v j ). ˙ q (∆ ˙ q′′ ∂j u∆ ˙ q (S˙ q′ −1 ∂j u∆ ˙ q′ vj ) = − ∆ ∆ Rq3 = − 1 p2
Therefore, denoting 2qσ kRq3 kLp ≤ C ≤C
(7.6)
|q ′ −q|≤4
|q ′ −q|≤4
q ′′ ≤q ′ −2
=
1 p
−
1 p1
and taking advantage of (7.2), we can obtain
X
˙ q′ v j kLp1 k∆ ˙ q′′ ∂j ukLp2 2qσ k∆
X
2
|q ′ −q|≤4 q ′′ ≤q ′ −2
(q−q ′′ )(σ−1− pN ) q ′ pN 1
2
|q ′ −q|≤4 q ′′ ≤q ′ −2
1
˙ q′ ∇vkLp1 2q σ k∆ ˙ q′′ ukLp . k∆ ′′
Note that, starting from the first equality of (7.6), one can alternately get X ˙ q′ v j kLp1 kS˙ q′ −1 ∂j ukLp2 2qσ kR3 kLp ≤ C 2qσ k∆
(7.7)
q
|q ′ −q|≤4
≤C
X
|q ′ −q|≤4
˙ q′ ∇vkLp1 kS˙ q′ −1 ∂j ukLp2 . 2q (σ−1) k∆ ′
(7.8)
Bounds for 2qσ kRq4 kLp : Rq4 =
X
|q ′ −q|≤2 |q ′′ −q ′ |≤1
˙ q′ vj ∆ ˙ q∆ ˙ q′′ u) − ∂j (∆
=: Rq4,1 + Rq4,2 .
X
˙ q (∆ ˙ q′ vj ∆ ˙ q′′ u) ∂j ∆
q ′ ≥q−3 |q ′′ −q ′ |≤1
By (7.2), we can get 2qσ kRq4,1 kLp ≤ C ≤C
X
˙ q′′ ukLp ˙ q′ ∇vkL∞ 2q′ σ k∆ k∆
X
2
|q ′ −q|≤2 |q ′′ −q ′ |≤1
q ′ pN
1
|q ′ −q|≤2 |q ′′ −q ′ |≤1
18
˙ q′ ∇vkLp1 2q′ σ k∆ ˙ q′′ ukLp . k∆
(7.9)
For Rq4,2 , we proceed differently according to the value of denote p13 := p1 + p11 and have 2qσ kRq4,2 kLp ≤ C ≤C ≤C
If
1 p
+
1 p1
X
) q( pN − N p
2q(1+σ) 2
q ′ ≥q−3 |q ′′ −q ′ |≤1
X
3
q pN
2q(1+σ) 2
q ′ ≥q−3 |q ′′ −q ′ |≤1
X
1
1 p
+
1 p1 .
If
1 p
+
1 p1
˙ q′′ ukLp3 ˙ q′ v∆ k∆
˙ q′ vkLp1 k∆ ˙ q′′ ukLp k∆
(q−q ′ )(1+σ+ pN ) q ′ pN
2
2
1
≤ 1, we
1
q ′ ≥q−3 |q ′′ −q ′ |≤1
(7.10)
˙ q′′ ukLp . ˙ q′ ∇vkLp1 2q′ σ k∆ k∆
> 1, taking p1 = p′ in the above computations yields
2qσ kRq4,2 kLp ≤ C ≤C ≤C
X
qN p′
˙ q′′ ukL1 ˙ q′ v∆ k∆
qN p′
˙ q′′ ukLp ˙ q′ vk p′ k∆ k∆ L
2q(1+σ) 2
q ′ ≥q−3 |q ′′ −q ′ |≤1
X
2q(1+σ) 2
q ′ ≥q−3 |q ′′ −q ′ |≤1
X
(q−q ′ )(1+σ+ N ) q ′ pN p′
2
2
q ′ ≥q−3 |q ′′ −q ′ |≤1
1
(7.11)
˙ q′ ∇vkLp1 2q′ σ k∆ ˙ q′′ ukLp . k∆
Putting (7.9), (7.10) and (7.11) together, we obtain qσ
2
kRq4 kLp
≤C
X
(q−q ′ ) 1+σ+N min( p1 + p1′ )
2
1
q ′ ≥q−3 |q ′′ −q ′ |≤1 q ′ pN 1
×2
(7.12)
˙ q′ ∇vkLp1 2q′ σ k∆ ˙ q′′ ukLp . k∆
Bounds for 2qσ kRq5 kLp : Similar computations yield qσ
2
kRq5 kLp
≤C
X
2
1
q ′ ≥q−3 |q ′′ −q ′ |≤1 q ′ pN 1
×2
˙ q′ div vkLp1 2q′ σ k∆ ˙ q′′ ukLp . k∆
Bounds for 2qσ kRq6 kLp : Rq6 = −
(q−q ′ ) σ+N min( p1 + p1′ )
X
q ′ ≥q−1
˙ q u, ˙ q ′ v · ∇∆ ∆
19
(7.13)
thus by Bernstein lemma, we have X ′ ˙ q ukLp . ˙ q′ ∇vkL∞ 2qσ k∆ 2qσ kRq6 kLp ≤ C 2q−q k∆
(7.14)
q ′ ≥q−1
Combining inequalities (7.4), (7.5), (7.7) or (7.8), (7.12), (7.13) and (7.14), we end up with the desired estimate for Rq . Straightforward modifications in the estimates for Rq3 , Rq4 and Rq5 leads to the desired estimate in the special case where u = v. Acknowledgments: The authors would like to thank Prof. P.Constantin for helpful comments and suggestions. The authors also thank Prof. H.Dong and J.Wu for kindly informing us the recent paper [9]. The authors were partly supported by the NSF of China (No.10725102).
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