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Cauchy problem of Schrödinger-Improved Boussinesq Systems on the torus
J. Math. Anal. Appl. 415 (2014) 217–239
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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Cauchy problem of Schrödinger-Improved Boussinesq Systems on the torus Sijia Zhong 1 Department of Mathematics, Southeast University, Nanjing 210096, China
a r t i c l e
i n f o
Article history: Received 15 July 2013 Available online 27 January 2014 Submitted by T. Witelski Keywords: Schrödinger-Improved Boussinesq System Local well-posedness Global well-posedness Noise
a b s t r a c t In this paper, we will study the local well-posedness of Schrödinger-Improved Boussinesq System with additive noise in Td , d 1, and we will also study the global well-posedness of dimension 1 case with the initial data (u0 , v1 , v2 ) ∈ L2 × L2 × L2 almost surely, gaining some exponential growth of L2 norm of v. © 2014 Elsevier Inc. All rights reserved.
1. Introduction Our aim in the present paper is to study the Cauchy problem for a stochastic Schrödinger-Improved Boussinesq System with an additive noise in the Schrödinger equation on the torus, which has spatial correlations “as rough” as our techniques allow, i.e. ⎧ ∂2B ⎪ ⎪ , u + Δu = vu + φ i∂ ⎪ t ⎨ ∂t∂x (1.1) ∂t2 v − Δv − Δ∂t2 v = Δ|u|2 , ⎪ ⎪ ⎪ ⎩ u(0, x) = u0 (x), v(0, x) = v1 (x), ∂t v(0, x) = v2 (x), where u and v are random process defined for (t, x) ∈ R+ × Td . B is a two parameter Brownian motion on R+ × Td , that is, a zero mean Gaussian process whose correlation function is given by E B(t, x)B(s, y) = (t ∧ s)(x ∧ y) 2
∂ B for t, s 0, x, y ∈ Td . φ is a bounded linear operator on L2 defined by a kernel c(x, y), then φ ∂t∂x is a space–time noise with correlation function
1
This work is supported by JSPS Postdoctoral Fellowship for Foreign Researchers and NSFC 11001049, KJ2010411.
0022-247X/$ – see front matter © 2014 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jmaa.2014.01.065
S. Zhong / J. Math. Anal. Appl. 415 (2014) 217–239
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∂2B ∂2B (t, x)φ (s, y) E φ ∂t∂x ∂t∂x
= c(x, y)δt−s .
The case φ = Id, i.e. c(x, y) = δx−y , corresponds to the space time white noise, which requires the regularity of u0 to be s < − d2 . Unfortunately, we can only treat the case s 0 for d = 1 and s > d2 − 1 for d 2 from the crucial bilinear analysis. So we assume that φ is a Hilbert–Schmidt operator from L2 (Td ) into H s (Td ). We will denote by L0,s 2 the space of such operators, endowed by its natural norm φ2L0,s = 2
ns φen 2 2 , L
n∈Z
here the notation · will be defined in Section 2. 2 d in·x Now, denote by W (t) = ∂B } n∈Z βn en a cylindrical wiener process on L (T ), where {en } = {e ∂x = 2 is a complete orthonormal system in L and {βn } is a sequence of mutually independent Brownian motions in a fixed probability space (Ω, F, p) associated with a filtration {Ft }t0 . The process φW is then a φφ∗ -Wiener process, i.e. Gaussian process with law (N (0, tφφ∗ )t0 ), and the Itô form of Schrödinger part of the system (1.1) is written as i du + Δu dt = vu dt + φ dW.
(1.2)
On the other hand, for the Boussinesq equation, denote v± = (i∂t ± 1)v, then it is reduced to
√ √ i∂t v± ∓ v± = w( −Δ )2 |u|2 − −Δ −2 v, v± (0, x) = iv2 ± v1 ,
where w(k) =
√ k . 1+k2
(1.3)
We will consider the mild form of (1.2) and (1.3), i.e.
⎧ t t ⎪ v+ − v− ⎪ itΔ i(t−s)Δ ⎪ ⎪ u(s) ds − i ei(t−s)Δ φ dW (s), u(t) = e u0 − i e ⎪ ⎪ 2 ⎨ 0
0
t t ⎪ ⎪ √ √ ⎪ v+ − v− ⎪ ∓it ∓i(t−s) 2 2 ⎪ (s) ds. w( −Δ ) |u| (s) ds + i e∓i(t−s) −Δ −2 ⎪ ⎩ v± = e v±0 − i e 2 0
(1.4)
0
And our main result of this paper is: Theorem 1.1. Suppose s k2 0, and |s − k| < 1 for d = 1, s > k2 + d4 − 12 > d2 − 1 and |s − k| < 1 for d 2. Assume φ ∈ L0,s 2 for s defined above. Let (u0 (w, x), v±0 (w, x)) be F0 -measurable, with (u0 (w, x), v±0 (w, x)) ∈ H s (Td )×H k (Td ) for almost surely ω ∈ Ω, then there is a stopping time Tω > 0 and a unique process (u, v± ) solution of (1.4) in s,b k,b C [0, Tω ]; H s × H k ∩ X[0,T × X±,[0,T ω] ω] almost surely. s,b k,b Here X[0,T , X±,[0,T are the Bourgain spaces, whose definitions will be given in the next section. ω] ω]
Remark 1.2. By the same argument as in the paper of de Bouard, Debussche, and Tsutsumi [10], Φ(t, x) = t i(t−s)Δ e φ dW (s) can not be in X s,b for b 12 , so we will choose b < 12 . However in the proof, we can see 0 that even in the deterministic case, for s, k < 0, the key bilinear estimates will fail.
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By the idea of [6], there is the following global result, Theorem 1.3. For s = k = 0, d = 1, problem (1.4) is globally well-posed almost surely. Now, we mention some results on the deterministic problems of the Schrödinger-Improved Boussinesq and the relevant systems. Those suggest what happens in the case of the stochastic perturbation. ⎧ ⎪ ⎨ i∂t u + Δu = vu, ∂t2 v − Δv = Δ|u|2 , ⎪ ⎩ u(0, x) = u0 (x), v(0, x) = v1 (x),
∂t v(0, x) = v2 (x),
(u0 , v1 , v2 ) ∈ H s × H k × H k−1 ,
is called the Zakharov system, which was introduced in [18] to describe the long wave Langmuir turbulence in a plasma. There are plenty of papers studying the local and global well posedness of this system. For example: [12,2,3] and [6] for Rd , in which case the lowest allowed regularity for (u0 , v1 , v2 ) is (s, k) = (0, − 12 ) d−4 for d = 1, 2 and ( d−3 2 +, 2 +) for d 3. Particularly, in [6], Colliander, Holmer, and Tzirakis proved the global existence of the solution for (s, k) = (0, − 12 ), d = 1 by the conservation of L2 norm of u. For the periodic case, Takaoka in [17] gained the lowest allowed regularity (s, k) = (0, − 12 ) of dimension 1 for the nonresonant case and ( 12 , 0) for the resonant case. Recently, Kishimoto [15] studied the higher dimensional d−2 torus, and got the result that for d = 2, (s, k) = ( 12 , 0) and d 3, ( d−1 2 +, 2 +). Kishimoto also studied the global well posedness of the problem in 2d case by the improved I-method and got the result of (s, k) = ( 23 +, 0). In [1], Akahori studied the Schrödinger-Improved Boussinesq System in Rd , which has a less singular nonlinearity ⎧ ⎪ ⎨ i∂t u + Δu = vu, ∂t2 v − Δv − Δ∂t2 v = Δ|u|2 , ⎪ ⎩ u(0, x) = u0 (x), v(0, x) = v1 (x),
(1.5) ∂t v(0, x) = v2 (x),
(u0 , v1 , v2 ) ∈ H s × H k × H k .
He proved that for −
1 < s < 1, 4
−
1 < k < 1, 2
d 3,
system (1.5) is locally well-posed, and for s = k = 0,
d 2,
it is globally well-posed. (In [16], Ozawa and Tsutaya studied the local well-posedness of s = k = 0, d 3.) In this paper, we will not only consider the periodic case, but also with a stochastic perturbation. The idea for proving Theorem 1.1 is classical. First, we will find some bilinear estimates in X s,b for b < 12 , then by the contraction mapping principle, we show that almost surely there is a unique solution to (1.4). The idea for proving Theorem 1.3 is inspired by Colliander, Holmer, and Tzirakis [6] for Zakharov system. However, different from the deterministic case, the L2 -norm of u(t) does not conserved any more. So we use some idea from Kim [14] and so on. First, we define some stopping time δN (w) (ω ∈ Ω) by comparing the size of u(t)L2 + v+ (t)L2 + v− (t)L2 with N . Then by the same calculation of de Bouard and Debussche [8], we can see that the expected value of u(t)2L2 is controlled by T . With the help of these, combining the idea in [6], we also show the expected value of v+ (t)L2 + v− (t)L2 also only depends on T , but not on N , which shows the limitation of the shopping time δN goes to infinite.
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Remark 1.4. Similar to the proof here, we can also get the global well-posedness of Schrödinger–Boussinesq system with an additive noise (please refer to Farah and Pastor [11] for the deterministic case). However, for the nonresonant Zakharov system, it seems that the case (s, k) = (0, − 12 ) depends heavily on choosing b = 12 . The paper will be organized as follows: in Section 2, we will give some notations and lemmas which will be used later. In Section 3, the key bilinear estimates will be proved, and finally, in Section 4, we will finish the proof of the theorems. 2. Notations and lemmas A B means there is some constant C > 0 such that A B, while A ∼ B means both A B and B A. fˆ means the Fourier transformation of f with respect tot or x variable, while f˜ means Fourier transformation of f with respect to both t and x variables. x = 1 + |x|2 . k,b Bourgain spaces X s,b and X± are defined as spaces equipped with the norm b f X s,b = ns τ + n2 f˜(τ, n) L2 l2
τ n
and f X k,b = nk τ ± 1b f˜(τ, n) L2 l2 . ±
τ n
s,b We also define spaces XIs,b and X±,I restricted to time interval I with the norms
f X s,b = inf gX s,b I
g|I =f
and f X s,b = inf gX s,b . ±,I
g|I =f
±
From now on, we omit I when there is no confusion. Assume ψ is a smooth cut off function, which is defined by
1 |t| 12 , ψ(t) = 0 |t| > 1. Denote ψδ (t) = ψ( δt ). Lemma 2.1. (See [6,12].) Suppose 0 < δ 1. (1) S(t)f C(R
t ,H
s)
Cf H s ,
(2.1)
here S(t) = eitΔ or e∓it . (2) For 0 b 12 , ψδ (t)S(t)f δ 12 −b f H s , A k,b here A = X s,b or X± .
(2.2)
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(3) For 0 b <
1 2
221
k,−b and A = X s,−b or X± ,
t ψδ (t) S(t − s)F (s) ds 0
Cδ 2 −b F A . 1
(2.3)
C([0,δ],H s )
(4) For 0 b < 12 , 0 b, b + b 1, t ψδ (t) F (s) ds 0
Cδ 1−b−b F H −b ,
(2.4)
t
Htb
and t ψδ (t) S(t − s)F (s) ds 0
Cδ 1−b−b F A(−b ) ,
(2.5)
A(b)
k,c here A(c) = X s,c or X± . 1 (5) For 0 b b < 2 ,
f A(b) Cδ b −b f A(b ) ,
(2.6)
s,c s,c here A(c) = X[0,δ] or X±,[0,δ] .
Lemma 2.2. (See [12,13].) (1) Let α, β > 0, such that α + β > 1, there exists ∞
c dτ , τ α τ − aβ aγ
(2.7)
1 < ∞, nα a + bnn1 + n2 β
(2.8)
−∞
here γ = min{α, β, α + β − 1}. (2) For α, β > 0, α + 2β > 1, then
n∈Z
for β > 12 ,
n∈Z
1 < ∞, a + bnn1 + n2 β
(2.9)
and for α, β > 0, α + β > 1,
n∈Z
1 < ∞. nα a + nβ
(2.10)
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Lemma 2.3. (See Chapter V of [4] and [5].) (1) uL4 (T1+1 ) u
3
X 0, 8
.
(2.11)
(2) Suppose IN √−Δ2N (u) = u and IL√−Δ2L (v) = v, then s1 uvL2 (T1+d ) min{N, L} uX 0,b vX 0,b , with s1 >
d 2
− 1, b = 14 (1 +
d 2
(2.12)
− s1 )+ < 12 , and d 2.
Lemma 2.4. (See Proposition 2.1 of [9].) For s, b ∈ R with b < 12 , and assume that φ ∈ L0,s 2 , then Φ(t, x) = t i(t−s)Δ 2 s,b e φ dW (s) is in L (Ω, X ) with the smooth cut off function ψ defined as above, 0 E ψΦ2X s,b Cφ2L0,s ,
(2.13)
2
here C depends on b and ψ. Remark 2.5. (1) The authors in [9] consider the KdV equation, but Proposition 2.1 is not dependent on the KdV operator. (2) If we replace ψ(t) with ψδ (t), then by the calculation of [9], we have E ψδ Φ2X s,b Cδ 1−2b φ2L0,s ,
for 0 < δ < 1,
(2.14)
for δ 1.
(2.15)
2
and E ψδ Φ2X s,b Cδ 4 φ2L0,s , 2
The idea for proving the following lemma is exactly the same as Proposition 3.2 of [8], and so we omit it here (we should use Itô formula and some sequence of approximation first). Lemma 2.6. For φ ∈ L0,s 2 , and a stopping time T > 0 a.s. then for t < T , the solution (u, v± ) to the system (1.4) with the initial data (u0 , v±0 ), we have
u(t) 2 2 = u0 2 2 − 2 L L
t u(s, x)φen dx dβn (s) + tφ2L0,s .
(2.16)
2
n∈Z 0 d T
Moreover, by Martingale inequality,
E
2 sup u(t) L2 2u0 2L2 + CT,
t∈[0,T ]
here C depends on φL0,s . 2
(2.17)
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3. Bilinear estimate Proposition 3.1. Suppose d = 1, s k2 0, and |s − k| < 1, or d 2, s > Then, there is some b < 12 , such that
k 2
+ d4 − 12 >
d 2
− 1, and |s − k| < 1.
uvX s,−b uX s,b vX k,b
(3.1)
2 |u| k,−b u2 s,b . X X
(3.2)
±
and ±
Proof. Take + for example and assume by density that u, v ∈ C0∞ (R × Td ). [Proof of (3.1)] By duality, we just need to prove J = uvw dx dt uX s,b vX k,b wX −s,b , +
R×Td
here w ∈ X −s,b . It also equals to prove
A= n0 =n1 +n2 τ =τ +τ 0 1 2
n0 s f (τ1 , n1 )g(τ2 , n2 )h(τ0 , n0 ) dτ1 dτ2 n1 s n2 k τ0 + n20 b τ1 + n21 b τ2 + 1b
f L2τ ln2 gL2τ ln2 hL2τ ln2 ,
(3.3)
with f, g, h ∈ L2τ ln2 . d = 1. Case 1. |n1 | ∼ |n0 |. τ0 + n20 + τ1 + n21 + |τ2 + 1| n20 − n21 − 1 = (n0 − n1 )(n0 + n1 ) − 1 = n2 (n0 + n1 ) − 1, and denote M = max{|τ0 + n20 |, |τ1 + n21 |, |τ2 + 1|}. Subcase 1.1. |n2 | 1, i.e. |n0 − n1 | 1. By Schwarz inequality,
n0 s f (τ1 , n1 )g(τ2 , n2 )h(τ0 , n0 ) A= dτ1 dτ2 n1 s n2 k τ0 + n20 b τ1 + n21 b τ2 + 1b n =n +i |i|1
0
1
τ0 =τ1 +τ2
|i|1 n0 =n1 +i
|f (τ1 , n1 )g(τ2 , i)h(τ0 , n1 + i)| dτ1 dτ2 τ0 + n20 b τ1 + n21 b τ2 + 1b
f (n1 )
g(i) h(n + i) sup 1 L2 L2 L2 τ
|i|1 n1
τ
τ
τ0
f (n1 ) 2 g(i) 2 h(n1 + i) 2 sup L L L τ
|i|1 n1
f (n1 )
|i|1 n1
g(i) |i|1
with k 0 and b = 12 −.
L2τ
L2τ
τ
τ
τ0
1 dτ1 τ0 + n20 2b τ1 + n21 2b τ2 + 12b
1 2 2b τ0 + n0 τ0 + 1 + n21 4b−1
12
12
g(i) 2 h(n1 + i) 2 L L τ
τ
f L2τ ln2 hL2τ ln2 f L2τ ln2 gL2τ ln2 hL2τ ln2 ,
(3.4)
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Subcase 1.2. n0 + n1 = 0, so n2 = 2n0 . Similarly to the above case, A
n0 τ =τ +τ 0 1 2
f (τ1 , n1 )g(τ2 , n2 )h(τ0 , n0 ) dτ1 dτ2 n0 k τ0 + n20 b τ1 + n21 b τ2 + 1b
f (−n0 ) 2 g(2n0 ) 2 h(n0 ) 2 L L L τ
n0
τ
τ
f ln∞ L2τ gL2τ ln2 hL2τ ln2 f L2τ ln2 gL2τ ln2 hL2τ ln2 ,
(3.5)
with k 0. Subcase 1.3. |n2 | 2 and |n0 + n1 | 1. In this case |n0 | |n1 |, and 1 ∼ |n2 ||n0 + n1 | = |n2 ||2n0 − n2 |. M n2 (n0 + n1 ) − 1 = |n2 |n0 + n1 − n2 a. M = |τ0 + n20 |. By Schwarz inequality, A hL2τ ln2
n0 s f (τ1 , n1 )g(τ2 , n2 ) dτ 1 2 2 s k b b b n1 n2 τ0 + n0 τ1 + n1 τ2 + 1 L2 l 2
n1
τ n
12 2 1 2 f (τ1 , n1 )g(τ2 , n2 ) dτ1 C(τ0 , n0 ) hL2τ ln2
2 L2τ ln
n1
1
sup C(τ0 , n0 ) 2 f L2τ ln2 gL2τ ln2 hL2τ ln2 ,
(3.6)
τ0 ,n0
where C(τ0 , n0 ) =
n1 2s n2 2k τ0
n2
n0 2s dτ2 . + n20 2b τ1 + n21 2b τ2 + 12b
Then by Lemma 2.2, C(τ0 , n0 )
n2
n2
1 dτ2 n2 2k+2b 2n0 − n2 2b τ1 + n21 2b τ2 + 12b
1 1, n2 2k+2b 2n0 − n2 2b τ0 + (n0 − n2 )2 + 14b−1
with k > −1 and b = 12 − > 14 . b. M = |τ1 + n21 |. Similar to case a, it can be reduced to C(τ1 , n1 ) =
n2
n0 2s dτ2 , n1 2s n2 2k τ0 + n20 2b τ1 + n21 2b τ2 + 12b
then we can treat it in the same way as case a. c. M = |τ2 + 1|. In this case C(τ2 , n2 ) =
n1
n0 2s dτ1 n1 2s n2 2k τ0 + n20 2b τ1 + n21 2b τ2 + 12b
(3.7)
S. Zhong / J. Math. Anal. Appl. 415 (2014) 217–239
n1
1 dτ1 n2 2k+2b n0 + n1 2b τ0 + n20 2b τ1 + n21 2b
n2 + 2n1
n1
225
2b τ
1 1, 2 4b−1 2 + n2 − 2n2 n1
(3.8)
with b = 12 − > 13 . Case 2. |n1 | |n0 | so |n2 | ∼ |n0 |. In this case M |n0 |2 . Subcase 2.1. M = |τ0 + n20 |. C(τ0 , n0 )
n1
n1
1 dτ1 n1 2s n2 2k−2s+4b τ1 + n21 2b τ2 + 12b
1 1, n1 2k+4b τ0 + 1 + n21 4b−1
(3.9)
with b = 12 − and k + 1 > s. Subcase 2.2. M = |τ1 + n21 |. As in Subcase 2.1, it can be reduced to estimating C(τ1 , n1 ) =
n0 2s dτ2 + n20 2b τ1 + n21 2b τ2 + 12b
n2
n1 2s n2 2k τ0
n2
1 dτ2 n2 2k−2s+4b τ0 + n20 2b τ2 + 12b
n2
1 1, n2 2k−2s+4b n20 + τ1 − 14b−1
(3.10)
with s 0 and k + 1 > s. Subcase 2.3. M = |τ2 + 1|. As before, we just need to estimate C(τ2 , n2 ) =
n1
n1
n0 2s dτ1 n1 2s n2 2k τ0 + n20 2b τ1 + n21 2b τ2 + 12b n1
2k+4b τ
n1
n1
2k+4b τ
1 2 2b 2 2b dτ1 0 + n0 τ1 + n1
1 1 = 2 2 4b−1 2k+2b τ + n2 + 2n n 4b−1 n 2 + n0 − n1 1 2 1 2 2 n 1
1,
(3.11)
with k > − 12 . Case 3. |n2 | ∼ |n1 | |n0 |. and in this case M |n1 |2 . The way to deal with this case is similar to Case 2. In fact, it will be better. Conclusively, for d = 1, when s, k 0 and k + 1 > s, (3.1) holds. d 2. Denote s0 = d2 − 1+, and take b = 12 −. Case 1. |n1 | |n0 |. Let N = (N0 , N1 , N2 ) be dyadic integer, i.e. Nj = 2mj , mj = 0, 1, 2, . . . , j = 0, 1, 2. So J
N
where
I(N ),
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226
I(N ) =
u
N1 N2
v
w
N0
dx dt,
R×Td
with uN1 means the support of u ˆ(n1 ) is on N1 |n1 | 2N1 , so are v N2 and wN0 . Subcase 1.1. N2 N0 ∼ N1 . ˜ ), I(N ) ∼ I(N
(3.12)
(k)
with the support M = {(n0 , n1 , n2 ) ∈ Z3d : Ni sup{|ni |} 2Ni , i = 0, 1, 2, k = 1, . . . , d}. Cover M with disjoint interiors Qα = n ∈ Zd : sup n(k) − α(k) N2 , here α ∈ Zd , and project uN1 and wN0 onto each Qα , then by Lemma 2.3 (2)
1 N2 uN wαN0 dx dt α v
˜ ) I(N R×Td
α
N0 1 uN α wα α
t,x
+
δ b N2s0 −k δ
N v 2 2 L
N 1 wα 0 0,b v N2 0,0 N2s0 uN α X 0,b X X
α
b
L2t,x
N2s0 −k
δ b N2s0 −k
N0 N1 N0 N1 N0 N1
s
N uα 1
X s,b
N wα 0
X −s,b
N v 2 k,b X +
α
s
N 2 uα 1
12
X s,b
s
α
N u 1
N 2 wα 0 −s,b X
12
N v 2 k,b X +
α
X s,b
N w 0
X −s,b
N v 2 k,b , X
(3.13)
+
here δ is the length of time interval. Thus for N0 ∼ N1 i.e. there is some c > 0 such that N1 = 2j N0 , |j| c, and k > s0
N
I(N ) ∼
˜ ) I(N
N
s N0 uN1 s,b wN0 −s,b v N2 k,b δ X X X+ N1 N s N0 uN1 s,b wN0 −s,b v k,b δb X+ X X N1 N0 ,N1
j u2 N0 s,b wN0 −s,b v k,b δb 2js X X X b
N2s0 −k
+
−cjc
δ
b
−cjc
δ
b
−cjc
N0
12
12 2j N 2 N 2 0 0 u w vX k,b 2 X s,b X −s,b js
+
N0
2
js
N0
uX s,b wX −s,b vX k,b δ b uX s,b wX −s,b vX k,b . +
+
(3.14)
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Subcase 1.2. N2 ∼ N0 N1 . The first step is the same as in (3.11) except that there is no need to do the decomposition onto Qα . Therefore
N
s N0 uN1 s,b wN0 −s,b v N2 k,b X X X+ N1 N s
N s0 −k N0 b u 1 s,b wN0 −s,b v N2 k,b δ N2 X X X+ N1 N
N0 s uN1 s,b wN0 −s,b v k,b . δb X+ X X N1
I(N ) δ b
N0s0 N2−k
(3.15)
N0 ,N1
Since N0 N1 , there is some c > 0 such that N1 = 2j N0 , with j −c, and
N0 s uN1 s,b wN0 −s,b X X N1
N0 ,N1
=
j−c
=
2−js
j u2 N0
X
N 0 s,b w
X −s,b
N0
2−js
2j N 2 u 0
X
12
N 2 w 0 s,b
X −s,b
N0
j−c
12
N0
2−js uX s,b wX −s,b = uX s,b wX −s,b ,
(3.16)
j−c
with s > 0. Conclusively, for s > 0 and k > s1 >
d 2
− 1,
J uX s,b wX −s,b vX k,b . +
Case 2. |n0 | |n1 |. So |n0 | ∼ |n2 | and M |n0 |2 , here the definition of M is the same as the one in dimension 1 case. Subcase 2.1. M = |τ0 + n20 |. The same as in Case d = 1, 1
A sup C(τ0 , n0 ) 2 f L2τ ln2 gL2τ ln2 hL2τ ln2 ,
(3.17)
τ0 ,n0
with C(τ0 , n0 ) = with b = 12 −, s < k + 1 and k >
n1
n0 2s dτ1 n1 2s n2 2k τ0 + n20 2b τ1 + n21 2b τ2 + 12b
n1
n1 2s n2 2k−2s+4b τ1
1
1 2 4b−1 0 + 1 + n1
n1
n1
n1
1 1, n1 2k+4b
d 2
− 1.
2k+4b τ
+ n21 2b τ2 + 12b
dτ1
S. Zhong / J. Math. Anal. Appl. 415 (2014) 217–239
228
Subcase 2.2. M = |τ1 + n21 |. A
f (τ1 , n1 )g(τ2 , n2 )h(τ0 , n0 ) dτ1 dτ2 n1 s n2 k−s+2b τ0 + n20 b τ2 + 1b
n0 =n1 +n2
n0 =n1 +n2
n0 =n1 +n2
1 f (n1 ) 2 g(n2 ) 2 h(n0 ) 2 sup Lτ Lτ Lτ n1 k+2b τ1
1 τ0 + n20 2b τ2 + 12b
dτ2
12 1 1 f (n1 ) 2 g(n2 ) 2 h(n0 ) 2 sup Lτ Lτ Lτ n1 k+2b τ1 + n20 − 14b−1 τ1
1 f (n1 ) 2 g(n2 ) 2 h(n0 ) 2 k+2b L L Lτ τ τ n1 n0 =n1 +n2
1 f (n1 )L2 g(n2 )L2 l2 hL2τ ln2 τ τ n n1 k+2b n1 1 f L2τ ln2 gL2τ ln2 hL2τ ln2 n1 k+2b 2 f L2τ ln2 gL2τ ln2 hL2τ ln2 , l
12
(3.18)
n
with k > d2 − 1 and s < k + 1. Subcase 2.3. M = |τ2 + 1|. We can reduced the proof to estimating C(τ2 , n2 ) =
n1
1 dτ1 n1 2s n0 2k−2s τ0 + n20 2b τ1 + n21 2b τ2 + 12b
n1
1 dτ1 , n1 2k+4b τ0 + n20 2b τ1 + n21 2b
(3.19)
which can be dealt with exactly similarly to Subcase 2.1. Consequently, for 0 < s < k + 1 and k > d2 − 1, A f L2τ ln2 gL2τ ln2 hL2τ ln2 . Thus (3.1) is proved for d 2. [Proof of (3.2)] The same as (3.1), we reduce the proof to estimating
B=
n0 =n1 +n2τ =τ +τ 0 1 2
n0 k f (τ1 , n1 )g(τ2 , n2 )h(τ0 , n0 ) dτ1 dτ2 n1 s n2 s τ0 + 1b τ1 + n21 b τ2 − n22 b
f L2τ ln2 gL2τ ln2 hL2τ ln2 ,
(3.20)
for f , g, h ∈ L2τ ln2 . d = 1. Case 1. |n0 | 1, so |n1 | ∼ |n2 |, then B=
|n0 |1 n2 =n0 −n1τ0 =τ1 +τ2
|n0 |1 n2 =n0 −n1τ0 =τ1 +τ2 1
n0 k f (τ1 , n1 )g(τ2 , n2 )h(τ0 , n0 ) dτ1 dτ2 n1 s n2 s τ0 + 1b τ1 + n21 b τ2 − n22 b |f (τ1 , n1 )g(τ2 , n2 )h(τ0 , n0 )| dτ1 dτ2 n1 2s τ0 + 1b τ1 + n21 b τ2 − n22 b
sup C(τ1 , n1 ) 2 f L2τ ln2 gL2τ ln2 hL2τ ln2 τ1 ,n1
(3.21)
S. Zhong / J. Math. Anal. Appl. 415 (2014) 217–239
229
with
C(τ1 , n1 ) =
|n0 |1
1 dτ0 n1 4s τ0 + 12b τ1 + n21 2b τ2 − n22 2b
|n0 |1
n1
4s τ
1
+
n21 2b τ1
1 1, + 1 − (n0 − n1 )2 4b−1
for s 0 and b = 12 − > 14 . Case 2. n1 − n2 = 0, i.e. n1 = n2 and n0 = 2n1 , the same as before, 1
B sup C(τ0 , n0 ) 2 f L2τ ln2 gL2τ ln2 hL2τ ln2 ,
(3.22)
τ0 ,n0
and
C(τ0 , n0 ) =
n0 =2n1 ,n2 =n1 ,n1
n1 2s n2 2s τ0
n0 2k dτ1 + 12b τ1 + n21 2b τ2 − n22 2b
1 1, n0 4s−2k τ0 + 12b τ0 4b−1
with s k2 and b = 12 − > 14 . Case 3. |n0 | 2 and |n1 − n2 | 1. Then |τ0 + 1| + τ1 + n21 + τ2 − n22
1 n21 − n22 − 1 = n0 (n1 − n2 ) − 1 = |n0 |n1 − n2 − |n0 ||n1 − n2 |, n0
and denote M = max{|τ0 + 1|, |τ1 + n21 |, |τ2 − n22 |}. By symmetry argument, we suppose |n1 | |n2 |, so |n0 | |n1 |. Subcase 3.1. |n1 | |n2 |, so |n0 | ∼ |n1 | and M |n0 |2 ∼ |n1 |2 . a. M = |τ0 + 1|. Then 1
B sup C(τ0 , n0 ) 2 f L2τ ln2 gL2τ ln2 hL2τ ln2 ,
(3.23)
τ0 ,n0
and
1 dτ2 2s−2k n 2s τ + 12b τ + n2 2b τ − n2 2b n 1 2 0 1 2 1 2 n2 1 1 dτ2 n1 2s−2k+4b n2 2s τ1 + n21 2b τ2 − n22 2b
1 1, 4s−2k+4b τ + n2 − 2n n 4b−1 n 2 0 0 2 0 n
C(τ0 , n0 ) =
2
1 with 2s − 2k + 4b 0, 4s − 2k + 4b + 4b − 1 > 1, b = 12 −, i.e. s > max{k − 1, k−1 2 }, and b = 2 −. 2 b. M = |τ1 + n1 |. Similarly, it is reduced to estimate
C(τ1 , n1 ) =
n2
n0 2k dτ2 n1 2s n2 2s τ0 + 12b τ1 + n21 2b τ2 − n22 2b
(3.24)
S. Zhong / J. Math. Anal. Appl. 415 (2014) 217–239
230
1
n2
n1 2s−2k+4b n2 2s
n2
1 dτ2 τ0 + 12b τ2 − n22 2b
1 1, n2 4s−2k+4b τ1 + 1 + n22 4b−1
with 2s − 2k + 4b 0 and 4s − 2k + 4b + 2(4b − 1) > 1, i.e. s > max{k − 1, c. M = |τ2 − n22 |, then
(3.25)
k− 32 2
}, and b = 12 −.
B sup C(τ2 , n2 )f L2τ ln2 gL2τ ln2 hL2τ ln2 ,
(3.26)
τ2 ,n2
and
n0 2k dτ1 n1 2s n2 2s τ0 + 12b τ1 + n21 2b τ2 − n22 2b n1
1 1 dτ1 2s−2k+4b n 2s 2b τ + n2 2b n τ + 1 1 2 0 1 1 n
C(τ2 , n2 ) =
1
n1
1 n1
2s−2k+4b τ
2
+ 1 − n21 4b−1
1
(3.27)
with s 0, and 2s − 2k + 4b + 2(4b − 1) > 1, i.e. s 0, s > k − 32 and b = 12 −. Subcase 3.2. |n1 | ∼ |n2 |, so |n0 | |n1 |, and M |n0 ||n1 − n2 |. a. M = |τ0 + 1|, the proof is the same as in Subcase 3.1 a,
C(τ0 , n0 ) =
n2
n2
n1 2s n2 2s τ0
n0 2k dτ2 + 12b τ1 + n21 2b τ2 − n22 2b
n0 2k−2b . n2 4s n1 − n2 2b τ0 + n20 − 2n0 n2 4b−1
(3.28)
If 0 k < 12 , then there is b = 12 − such that k < b, and (3.28)
n2
1 n2 4s 2n2
− n0
2b τ
0
+ n20 − 2n0 n2 4b−1
1,
(3.29)
with s 0 and b = 12 −. If k 12 > b, then (3.28)
n2
n2
with 4s − 2k + 4b > 1, i.e. s >
k− 12 2
n2 4s−2k+2b 2n2
1 − n0 2b τ0 + n20 − 2n0 n2 4b−1
1 1, n2 4s−2k+2b 2n2 − n0 2b and b = 12 −.
(3.30)
S. Zhong / J. Math. Anal. Appl. 415 (2014) 217–239
231
b. M = |τ1 + n21 |, so
C(τ1 , n1 )
n0 2k dτ2 n2 4s+2b n1 − n2 2b τ0 + 12b τ2 + n22 2b
n2
n2
1 1, n2 4s−2k+2b n2 − n1 2b τ1 + 1 + n22 4b−1
(3.31)
k− 1
with s > 2 2 and b = 12 −. c. M = |τ2 − n22 |, by the symmetry of n1 and n2 , this case is exactly the same as b. Consequently, for d = 1, s k2 0, s > k − 1 and b = 12 −, (3.2) exists. d 2. By the same symbol as before, it equals to prove J=
¯N2 wN0 dx dt u2X s,b wX −k,b , u N1 u
(3.32)
+
N −k,b with N = (N0 , N1 , N2 ), Ni = 2mj , mj = 0, 1, 2, . . . , j = 0, 1, 2, w ∈ X+ and uN1 means the support of u ˆ(n1 ) is on N1 |n1 | < 2N1 , so are uN2 and wN0 . Without losing of generality, suppose N1 N2 . Case 1. N1 ∼ N2 N0 . By Lemma 2.3 (2), we have
J
uN1 uN2 2 wN0 2 L L t,x
N
d
N12
−1+ N1 u
N
d
N12
−1−2s+
t,x
X 0,b
N u 2 0,b wN0 0,0 X X +
N0k δ b uN1 X s,b uN2 X s,b wN0 X −k,b u2X s,b wX −k,b , +
+
(3.33)
N
with d2 − 1 − 2s + k < 0 i.e. s > k2 + d4 − 12 and b = 12 −. Case 2. N1 N2 , so N1 ∼ N0 . (3.32) equals to prove B=
n0 =n1 +n2 τ =τ +τ 0 1 2
n0 k f ((τ1 , n1 )g(τ2 , n2 )h(τ0 , n0 )) dτ1 dτ2 n1 s n2 s τ0 + 1b τ1 + n21 b τ2 − n22 b
f L2τ ln2 gL2τ ln2 hL2τ ln2 ,
(3.34)
with f , g, h ∈ L2τ ln2 . Since |τ0 + 1| + τ1 + n21 + τ2 − n22 n21 − n22 − 1 = n0 (n1 − n2 ) − 1 1 |n0 |2 ∼ |n1 |2 , = |n0 |n1 − n2 − n0 by N1 ∼ N0 N2 . Put M = max{|τ0 + 1|, |τ1 + n21 |, |τ2 − n22 |}. Subcase 2.1. M = |τ0 + 1|. 1
B sup C(τ0 , n0 ) 2 f L2τ ln2 gL2τ ln2 hL2τ ln2 , τ0 ,n0
and
S. Zhong / J. Math. Anal. Appl. 415 (2014) 217–239
232
C(τ0 , n0 ) =
n2
n0 2k dτ2 n1 2s n2 2s τ0 + 12b τ1 + n21 2b τ2 − n22 2b
n2
1 dτ2 n1 2s−2k+4b n2 2s τ1 + n21 2b τ2 − n22 2b
n2
n2
n2
4s−2k+4b τ
1 2 4b−1 0 + n0 − 2n0 n2
1 1, n2 4s−2k+4b
(3.35)
with 2s − 2k + 4b 0 and 4s − 2k + 4b > d, i.e. s > k − 1, s > Subcase 2.2. M = |τ1 + n21 |.
k 2
+
d 4
− 12 , and b = 12 −.
1
B sup C(τ1 , n1 ) 2 f L2τ ln2 gL2τ ln2 hL2τ ln2 , τ1 ,n1
and C(τ1 , n1 ) =
n2
n0 2k dτ2 n1 2s n2 2s τ0 + 12b τ1 + n21 2b τ2 − n22 2b
n2
1 dτ2 n1 2s−2k+4b n2 2s τ1 + n21 2b τ2 − n22 2b
n2
1 n2
4s−2k+4b τ
1
+ 1 + n22 4b−1
1,
(3.36)
with s > k − 1, s > k2 + d4 − 12 , and b = 12 −. Subcase 2.3. M = |τ2 − n22 |. B
n0 =n1 +n2τ =τ +τ 0 1 2
1 dτ1 dτ2 n1 s−k+2b n2 s τ0 + 1b τ1 + n21 b
1
n0 =n1 +n2
n2 2s−k+2b
1
n0 =n1 +n2
n2 2s−k+2b
1 2s−k+2b
f (n1 ) f (n1 ) f (n1 )
g(n2 ) 2 h(n0 ) 2 sup L2 L L τ
τ
τ
τ2
g(n2 ) 2 h(n0 ) 2 sup L2 L L τ
τ
τ
1 dτ1 τ1 + τ2 + 12b τ1 + n21 2b
τ2
1 τ2 + 1 − n21 4b−1
12
12
g(n2 ) 2 h(n0 ) 2 L L
τ τ n2
1 f (n0 − n2 )L2 g(n2 )L2 l2 hL2τ ln2 τ τ n 2s−k+2b n2
n0 =n1 +n2
L2τ
n2
1 f L2 l2 gL2 l2 hL2 l2 f L2 l2 gL2 l2 hL2 l2 , τ n τ n τ n τ n τ n τ n n2 2s−k+2b 2 ln
with s − k + 2b 0 and 2s − k + 2b > d, i.e. s > k − 1, s > k2 + d4 − Consequently, for s > max{k − 1, k2 + d4 − 12 }, (3.2) holds. 2
1 2
and b = 12 −.
(3.37)
S. Zhong / J. Math. Anal. Appl. 415 (2014) 217–239
233
Remark 3.2. It is obvious that the requirement s, k 0 is necessary. In fact, for d = 1, Subcase 1.2 would be a good example to prove that k 0 is optimal for (3.1). Correspondingly, if we take n1 = −n2 = N and |τ1 + n21 |, |τ2 − n22 | 1, then we can show s 0 is necessary for (3.2). Then for d 2, since n = (n(1) , n ), with n ∈ Zd−1 , we can choose some similar example. 4. Proof of Theorems 1.1 and 1.3 4.1. Proof of Theorem 1.1 concluded k,b k,b Proof. Let B = {(u, v+ , v− ) ∈ X s,b × X+ × X− : uX s,b M δ 2 −b (u0 H s + φL0,2 ), v+ X k,b + 2 1
+
v− X k,b M δ 2 −b (v+0 H k + v−0 H k )}, 1
−
t itΔ
B(u, v± ) = e
u0 − iψδ
e
i(t−s)Δ v+
− v− u(s) ds − iψδ 2
0
t ei(t−s)Δ φ dW (s), 0
and B± (u, v± ) = e∓it v±0 − iψδ
t
√ √ v+ − v− (s) ds. e∓i(t−s) w( −Δ )2 |u|2 (s) ds + iψδ e∓i(t−s) −Δ−2 2
0
0
t
Then for (u, v+ , v− ) ∈ B, 0 < δ < 1, by Lemmas 2.1, 2.4, Bienaymé–Chebyshev inequality and Proposition 3.1, we have B(u, v± ) s,b Cδ 12 −b u0 H s + δ 1−2b C v+ uX s,−b + v− uX s,−b + δ 12 −b Cφ 0,s L2 X 1 1 δ 2 −b u0 H s + δ 1−2b v+ X s,b + v− X s,−b uX s,b + δ 2 −b φL0,s , 2 B+ (u, v± )
k,b X+
Cδ
1 2 −b
+ B− (u, v± ) X k,b −
v+0 H k + v−0 H k + δ
t √ 2 v − v + − ∓i(t−s) −2 |u| X k,−b + ψδ e (s) ds −Δ ± 2
1−2b
0
δ 2 −b v±0 H k + δ 1−2b u2X s,b 1
(4.1)
−
+
t v − v + − (s) ds + ψδ e∓i(t−s) 2 0
,
k,b X±
(4.2)
k,b X±
B(u1 , v±1 ) − B(u2 , v±2 ) s,b X 1−2b (v+1 + v−1 )u1 − (v+2 + v−2 )u2 X s,−b δ δ 1−2b v+1 X k,b + v−1 X k,b u1 − u2 X s,b + v+1 − v+2 X k,b + v−1 − v−2 X k,b u2 X s,b , +
−
+
−
(4.3) and B+ (u1 , v±1 ) − B+ (u2 , v±2 ) k,b + B− (u1 , v±1 ) − B− (u2 , v±2 ) k,b X+ X− t 1−2b i∓(t−s) δ u1 X s,b + u2 X s,b u1 − u2 X s,b + ψδ e (v±1 − v±2 )(s) ds 0
for almost surely ω ∈ Ω.
, k,b X±
(4.4)
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234
On the other hand, by Lemma 2.1, t ∓i(t−s) v+ (s) ds ψδ e 0
δ 1−b v+ X k,0 = δ 1−b v+ L2t Hxk = δ 1−b v+ X k,0 δv+ X k,b , ±
k,b X±
t
and the same calculation is valid for ψδ Therefore, taking δ ∼ min
v+0 H k
0
2
−3b
(4.5)
+
e∓i(t−s) v− (s) dsX k,b . ±
31
1 + v−0 H k
+
,
v+0 H k + v−0 H k u0 2H s + φ2L0,s
31 2
−3b
,
u0 H s
2
1 + φL0,s 2
31 2
−3b
,
with b < 12 , (B(u, v± ), B± (u, v± )) are contraction map on B. So there is a unique pathwise solution (u, v± ) k,b . to (1.4) with the initial data (u0 , v±0 ) ∈ H s × H k in X s,b × X± Then, by Lemma 2.1 and Theorem 6.10 of [7], there exists 1 uH s u0 H s + Cδ 2 −b v+ uX s,−b + Cv− uX s,−b + φL0,s 2 1 u0 H s + δ 2 −b v+ X k,b + v− X k,b uX s,b + φL0,s 2 + − 3 −3b u0 H s + δ 2 v+0 H k + v−0 H k u0 H s + φL0,s . 2
(4.6)
And 1 1 v± H k v±0 H k + Cδ 2 −b |u|2 X k,−b + Cδ 2 +b v+ X k,b + v− X k,b ±
v±0 H k v±0 H k
+
+δ u2X s,b + δ v+0 H k + v−0 H k 3 + δ 2 −3b u0 2H s + δ v+0 H k + v−0 H k .
−
1 2 −b
Hence there is a unique pathwise solution (u, v+ , v− ) on H s × H k × H k .
(4.7)
2
4.2. Proof of Theorem 1.3 The idea is similar to [6], although the mass uL2 here is not conserved. Proof. Since the constant C before the last term of (4.1) depends on ω ∈ Ω, to make it clearly, we consider the following system instead.
t itΔ
u(t) = e
u0 − iψδ (t)
ei(t−s)Δ
1 v+ − v− u (s) ds − iψL δ b− 2 ψδ ΦX 0,b ψN ΦL2 ψδ (t)Φ(t), 2
(4.8)
0
and v± (t) = e∓it v±0 − iψδ (t)
t
√ √ v+ − v− (s) ds, e∓i(t−s) w( −Δ )2 |u|2 (s) − −Δ−2 2
(4.9)
0
where ψ, ψδ , ψL and ψN are the truncation function defined in Section 2. Let L, N ∈ N and 0 < δ < 1.
S. Zhong / J. Math. Anal. Appl. 415 (2014) 217–239
235
On the other hand, for s = k = 0, the estimate in Proposition 3.1 can be improved. Since by Lemma 2.3 (1), for b = 38 , uvw dx dt uL4t,x vL2t,x wL4t,x u
3
X 0, 8
vX 0,0 w ±
3
3
X 0, 8
δ 8 u
3
X 0, 8
v
T×T 0, 3
3
0, 3
Therefore, let B = {(u, v+ , v− ) ∈ X 0, 8 × X+ 8 × X− 8 : u v−
0, 3
X− 8
0, 3
X± 8
w
3
X 0, 8
.
1
Cδ 8 (u0 L2 + L), v+
3
X 0, 8
0, 3 8
X+
+
1 8
Cδ (v+0 L2 + v−0 L2 )}, t itΔ
B(u, v± ) = e
u0 − iψδ
ei(t−s)Δ
1 v+ − v− u(s) ds − iψL δ − 8 ψδ Φ 0, 38 ψN ΦL2 ψδ (t)Φ(t), X 2
0
and ∓it
B± (u, v± ) = e
t v±0 − iψδ
e
∓i(t−s)
√ −2 v+ − v− w( −Δ ) |u| (s) ds − Δ (s) ds. 2 √
2
2
0
Now, for ω ∈ Ω, we define a stopping time δL (ω) by 1 δL (ω) = inf δ > 0: δ − 8 ψδ Φ
3
X 0, 8
>L .
(4.10)
Then, for (u, v± ) ∈ B, 0 < δ < δL , we have B(u, v± )
3
X 0, 8
1 1 1 Cδ 8 u0 L2 + δ 4 C v+ u 0,− 38 + v− u 0,− 38 + Cδ 8 L X X 1 5 1 δ 8 u0 L2 + δ 8 v+ 0, 38 + v− 0, 38 u 0, 38 + δ 8 L, X+
B+ (u, v± )
0, 3 8
X+
+ B− (u, v± )
0, 3
X− 8
1 1 Cδ 8 v+0 L2 + v−0 L2 + δ 4 |u|2
0,− 3 X± 8
1 5 δ 8 v+0 L2 + v−0 L2 + δ 8 u2
3
X 0, 8
B(u1 , v±1 ) − B(u2 , v±2 )
3 X 0, 8
(4.11)
X
X−
+ δ v+
+ δ v+
0, 3 X+ 8
0, 3 X+ 8
+ v−
+ v−
0, 3 X− 8
0, 3 X− 8
,
(4.12)
1 δ 4 (v+1 + v−1 )u1 − (v+2 + v−2 )u2 0,− 38 X 5 8 δ v+1 0, 38 + v−1 0, 58 u1 − u2 0, 38 X+
5 + δ 8 v+1 − v+2
X
X−
0, 3 X+ 8
+ v−1 − v−2
0, 3 X− 8
u2
3
,
(4.13)
.
(4.14)
X 0, 8
and B+ (u1 , v±1 ) − B+ (u2 , v±2 ) 5 δ 8 u1
0, 3 X+ 8
3
X 0, 8
+ u2
3
X 0, 8
+ B− (u1 , v±1 ) − B− (u2 , v±2 )
u1 − u2
3
X 0, 8
+ δ v+1 − v+2
0, 3
X− 8
0, 3 8
X+
+ v−1 − v−2
0, 3
X− 8
S. Zhong / J. Math. Anal. Appl. 415 (2014) 217–239
236
Hence we should choose δ satisfying ⎧ 3 ⎪ δ 4 v+0 L2 + v−0 L2 1, ⎪ ⎪ ⎪ ⎪ ⎨ δ 78 u 2 + L2 δ 18 v 2 + v 2 , 0 L +0 L −0 L 3 ⎪ ⎪ δ 4 u0 L2 + L 1, ⎪ ⎪ ⎪ ⎩ δ 1, 0, 3
3
(4.15)
0, 3
so that there is a unique solution (u(t), v+ (t), v− (t)) ∈ X 0, 8 × X+ 8 × X− 8 to system (4.8)–(4.9) with the initial data (u0 , v±0 ). Then, by Lemma 2.1 again, we have 1 uL2 u0 L2 + Cδ 8 v+ u 0,− 38 + Cv− u 0,− 38 + N X X 1 2 u0 L2 + δ v+ 0, 38 + v− 0, 38 u 0, 38 + N X+
u0 L2
X
X−
3 + δ 4 v+0 L2 + v−0 L2 u0 L2 + L + N.
(4.16)
And 1 n± L2 v±0 L2 + Cδ 8 |u|2
0,− 3 8
X±
7 + Cδ 8 v+
0, 3 8
X+
+ v−
0, 3
X− 8
1 v±0 L2 + δ 2 u2 0, 3 + δ v+0 L2 + v−0 L2 8 X 2 3 4 v±0 L2 + δ u0 L2 + L + δ v+0 L2 + v−0 L2 .
(4.17)
Hence (u, v± ) is a solution to (4.8)–(4.9) in L2 × L2 for t ∈ [0, δ ∧ δL ]. On the other hand, from Remark 2.5 (2) and Bienaymé–Chebyshev inequality, we can see that
1 P δL (ω) 2
Cφ2 0,s 1 1 L2 8 P ∃δ s.t. δ , ψδ Φ 0, 38 > δ L . 2 X 2 L
(4.18)
Denote ΩL = {ω: δL (ω) 12 }. So for ω ∈ ΩLc , δ ∧ δL = δ. Then, for ω ∈ ΩLc and N above, define another stopping time δN by δN (ω) = inf δ > 0, Φ(t) L2 > N, t ∈ [0, δ] . And for any M 1, define ⎧ 0 if u0 L2 + v±0 L2 M, ⎪ ⎪ ⎪ ⎪ ⎨ inf{t > 0: uLMN (t) 2 + v±LMN (t) 2 M } if u0 2 + v±0 2 < M, L L L L δM (ω) = ⎪ 2 ∞ if u + v ⎪ 0 L ±0 L2 < M, ⎪ ⎪ ⎩ and the set {t > 0: uLMN (t)L2 + v±LMN (t)L2 M } is empty,
(4.19)
where (uLMN (t), v±LM N (t)) is the solution to (4.8)–(4.9) with the initial data (u0 , v±0 ), for all t ∈ [0, δM (ω)∧ δN (ω)]. Let δM N = min{δM , δN }.
S. Zhong / J. Math. Anal. Appl. 415 (2014) 217–239
237
Then, by the pathwise uniqueness
uL1M 1N 1 (t), v±L1M 1N 1 (t) ≡ uL2M 2N 2 (t), v±L2M 2N 2 (t) ,
on [0, δM 1N 1 ] ∩ [0, δM 2N 2 ].
Thus δM N is nondecreasing in M . Set δ∞ (ω) = lim
lim
lim δM N (ω),
L→∞ N →∞ M →∞
and for any t ∈ [0, δ∞ ), let (u(t), v± (t)) = (uLMN (t), v±LM N (t)), with t ∈ [0, δM N ]. Therefore, (u(t), v± (t)) is the solution to (1.4) with the initial data (u0 , v±0 ), for all t ∈ [0, δ∞ (ω)). Claim 4.1. Let ω ∈ ΩLc and L, N ∈ N be fixed. For each T 0, sup t∈[0,δM N ∧T ]
2 2 v± (t) 2 2CT (L +supt∈[0,δM N ∧T ] u(t) L2 ) . L
(4.20)
Proof. If for all t ∈ [0, δM N ∧ T ] v± (t)
L2
C
u(t)
sup
L2
t∈[0,δM N ∧T ]
+L ,
(4.21)
then (4.20) is obvious. Otherwise, there is some t0 ∈ [0, δM N ∧ T ], such that v± (t)
sup
L2
t∈[0,δM N ∧t0 )
u(t)
sup t∈[0,δM N ∧T ]
L2
+ L,
and v± (t0 )
L2
∼
sup t∈[0,δM N ∧T ]
u(t) 2 + L. L
Choose this t0 to be the starting point. Then, by the local well-posed result, there is some δ < for t ∈ [t0 , t0 + δ], there is a unique pathwise solution to (4.8)–(4.9) in X u
3
X 0, 8
1 Cδ 8 u(t0 ) L2 + L , v+
0, 3 X+ 8
+ v−
0, 3 X− 8
0, 38
0, 38
0, 38
× X+ × X−
1 2
such that
satisfying,
1 Cδ 8 v+ (t0 ) L2 + v− (t0 ) L2 .
(4.22)
Here, δ should satisfy (4.22) with u(t0 )L2 + L v+ (t0 )L2 + v− (t0 )L2 , which requires that δ ∼ 4 (v+ (t0 )L2 + v− (t0 )L2 )− 3 . By (4.17), for t ∈ [t0 , t0 + δ] there exists, v+ (t) 2 + v− (t) 2 v+ (t0 ) 2 + v− (t0 ) 2 L L L L v+ (t0 ) L2 + v− (t0 ) L2 v+ (t0 ) L2 + v− (t0 ) L2 v+ (t0 ) 2 + v− (t0 ) 2 L
L
2 3 + δ 4 u(t0 ) L2 + L + δ v+ (t0 ) L2 + v− (t0 ) L2 2 3 1 + δ 4 u(t0 ) L2 + L + δ 4 2 1 + Cδ 4 1 + u(t0 ) L2 + L 1 u(t) 2 2 . + Cδ 4 L2 + sup (4.23) L t∈[0,δM N ∧T ]
S. Zhong / J. Math. Anal. Appl. 415 (2014) 217–239
238
1
Hence the increment of L2 norm of (v+ , v− ) from t0 to t0 + δ is Cδ 4 (L2 + supt∈[0,δM N ∧T ] u(t)2L2 ), and it takes m steps for the left hand side of (4.23) to double, where m= So the total time is δm =
v+ (t0 )L2 + v− (t0 )L2 . Cδ (L2 + supt∈[0,δM N ∧T ] u(t)2L2 ) 1 4
δ( v+ (t0 ) L2 + v− (t0 ) L2 ) 1
Cδ 4 (L2 +supt∈[0,δ
M N ∧T ]
u 2L2 )
∼
1 L2 +supt∈[0,δ
of (v+ (t0 )L2 + v− (t0 )L2 ). Therefore, before we reach time T , we need to carry out
T δm
M N ∧T ]
u(t) 2L2
, which is independent T δm 2 ∧T { u(t) L2 })
iterations. Since 2
supt∈[δM N ∧T ] {u(t)2L2 }), the left hand side of (4.23) is smaller than 2T (L +supt∈δM N initial one. Hence 2 2 v+ (t) 2 + v− (t) 2 C v+ (t0 ) 2 + v− (t0 ) 2 2T (L +supt∈[0,δM N ∧T ] u(t) L2 ) L L L L 2 2 C u(t0 ) L2 + L 2T (L +supt∈[0,δM N ∧T ] u(t) L2 ) 2 2 u(t) 2 + L 2T (L +supt∈[0,δM N ∧T ] u(t) L2 ) C sup L
T (L2 + times the
t∈[0,δM N ∧T ]
2CT (L
2
+supt∈[0,δ
M N ∧T ]
u(t) 2L2 )
,
(4.24)
for t ∈ [0, δM N ∧ T ]. Therefore the claim is proved. 2 Since by Lemma 2.6, for T 0 fixed, we have u(t) 2 2 2u0 2 2 + T C(1 + T ). E sup L L
(4.25)
t∈[0,δM N ∧T ]
Then from (4.20), we have sup
t∈[0,δM N ∧T ]
ln v± (t) L2 CT L2 +
sup t∈[0,δM N ∧T ]
u(t) 2 2 , L
(4.26)
so there exists E
sup t∈[0,δM N ∧T ]
ln v± (t) L2 CT L2 + u0 2L2 + T C L2 + T 2 .
(4.27)
We can see that the right hand side of (4.25) and (4.27) only depends on T 0 and L ∈ N, but is independent of M , hence, with the help of Bienaymé–Tchebysheff’s inequality and Theorem 6.10 of [7], we have v+ (t) 2 + v− (t) 2 ln M u(t) 2 2 M 2 + P P (δM N T ) P sup ln sup L L L t∈[0,δM N ∧T ]
·P
T 1 2
t∈[0,δM N ∧T ]
ΩL type sets + P 2
2
sup t∈[0,δM N ∧T ]
Φ(t) 2 2 > N 2 · P T ΩL type sets 1 L 2
C(1 + T ) C(L + T ) CT C(T ) CT + + 2 2 M ln M L N 2 L2
(4.28)
for all M ∈ N. So we obtain lim P (δM N T ) C(T )
M →∞
1 1 → 0, N 2 L2
as N → ∞ and L → ∞.
Therefore δ∞ = ∞ for almost all ω, which shows the global well-posedness.
2
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Acknowledgments The author would like to thank Professor Yoshio Tsutsumi for suggesting this interesting problem and a lot of discussion. And it is a pleasure to thank Department of Mathematics of Kyoto University for its hospitality. References [1] T. Akahori, Well-posedness for the Schrödinger-improved Boussinesq system and related bilinear estimates, Funkcial. Ekvac. 50 (3) (2007) 469–489. [2] I. Bejenaru, S. Herr, Convolutions of singular measures and applications to the Zakharov system, J. Funct. Anal. 261 (2) (2011) 478–506. [3] I. Bejenaru, S. Herr, J. Holmer, D. Tataru, On the 2D Zakharov system with L2 Schröinger data, Nonlinearity 22 (2009) 1063–1089. [4] J. Bourgain, Global Solutions of Nonlinear Schrödinger Equation, Amer. Math. Soc. Colloq. Publ., vol. 46, American Mathematical Society, Providence, RI, ISBN 0-8218-1919-4, 1999, viii+182 pp. [5] N. Burq, P. Gérard, N. Tzvetkov, Bilinear eigenfunction estimates and the nonlinear Schrödinger equation on surfaces, Invent. Math. 159 (1) (2005) 187–223. [6] J. Colliander, J. Holmer, N. Tzirakis, Low regularity global well-posedness for the Zakharov and Klein–Gordon–Schrödinger systems, Trans. Amer. Math. Soc. 360 (9) (2008) 4619–4638. [7] G. Da Prato, J. Zabczyk, Stochastic Equations in Infinite Dimensions, Encyclopedia Math. Appl., Cambridge University Press, Cambridge, England, 1992. [8] A. de Bouard, A. Debussche, The stochastic nonlinear Schrödinger equation in H 1 , Stoch. Anal. Appl. 21 (1) (2003) 97–126. [9] A. de Bouard, A. Debussche, Y. Tsutsumi, White noise driven Korteweg–de Vries equation, J. Funct. Anal. 169 (2) (1999) 532–558. [10] A. de Bouard, A. Debussche, Y. Tsutsumi, Periodic solutions of the Korteweg–de Vries equation driven by white noise, SIAM J. Math. Anal. 36 (3) (2004/2005) 815–855. [11] L.G. Farah, A. Pastor, On the periodic Schrödinger–Boussinesq system, J. Math. Anal. Appl. 368 (1) (2010) 330–349. [12] J. Ginibre, Y. Tsutsumi, G. Velo, On the Cauchy problem for the Zakharov system, J. Funct. Anal. 151 (1997) 384–436. [13] C.E. Kenig, G. Ponce, L. Vega, A bilinear estimate with applications to the KdV equation, J. Amer. Math. Soc. 9 (2) (1996) 573–603. [14] J.U. Kim, Invariant measures for a stochastic nonlinear Schrödinger equation, Indiana Univ. Math. J. 55 (2) (2006) 687–717. [15] N. Kishimoto, Local well-posedness for the Zakharov system on higher dimensional torus, preprint. [16] T. Ozawa, K. Tsutaya, On the Cauchy problems for Schrödinger-improved Boussinesq equations, Adv. Stud. Pure Math. 47 (2007) 291–301. [17] H. Takaoka, Well-posedness for the Zakharov system with the periodic boundary condition, Differential Integral Equations 12 (1999) 789–810. [18] V.E. Zakharov, Collapse of Langmuir waves, Sov. Phys. JETP 35 (1972) 908–914.
Contents lists available at ScienceDirect
Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Cauchy problem of Schrödinger-Improved Boussinesq Systems on the torus Sijia Zhong 1 Department of Mathematics, Southeast University, Nanjing 210096, China
a r t i c l e
i n f o
Article history: Received 15 July 2013 Available online 27 January 2014 Submitted by T. Witelski Keywords: Schrödinger-Improved Boussinesq System Local well-posedness Global well-posedness Noise
a b s t r a c t In this paper, we will study the local well-posedness of Schrödinger-Improved Boussinesq System with additive noise in Td , d 1, and we will also study the global well-posedness of dimension 1 case with the initial data (u0 , v1 , v2 ) ∈ L2 × L2 × L2 almost surely, gaining some exponential growth of L2 norm of v. © 2014 Elsevier Inc. All rights reserved.
1. Introduction Our aim in the present paper is to study the Cauchy problem for a stochastic Schrödinger-Improved Boussinesq System with an additive noise in the Schrödinger equation on the torus, which has spatial correlations “as rough” as our techniques allow, i.e. ⎧ ∂2B ⎪ ⎪ , u + Δu = vu + φ i∂ ⎪ t ⎨ ∂t∂x (1.1) ∂t2 v − Δv − Δ∂t2 v = Δ|u|2 , ⎪ ⎪ ⎪ ⎩ u(0, x) = u0 (x), v(0, x) = v1 (x), ∂t v(0, x) = v2 (x), where u and v are random process defined for (t, x) ∈ R+ × Td . B is a two parameter Brownian motion on R+ × Td , that is, a zero mean Gaussian process whose correlation function is given by E B(t, x)B(s, y) = (t ∧ s)(x ∧ y) 2
∂ B for t, s 0, x, y ∈ Td . φ is a bounded linear operator on L2 defined by a kernel c(x, y), then φ ∂t∂x is a space–time noise with correlation function
1
This work is supported by JSPS Postdoctoral Fellowship for Foreign Researchers and NSFC 11001049, KJ2010411.
0022-247X/$ – see front matter © 2014 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jmaa.2014.01.065
S. Zhong / J. Math. Anal. Appl. 415 (2014) 217–239
218
∂2B ∂2B (t, x)φ (s, y) E φ ∂t∂x ∂t∂x
= c(x, y)δt−s .
The case φ = Id, i.e. c(x, y) = δx−y , corresponds to the space time white noise, which requires the regularity of u0 to be s < − d2 . Unfortunately, we can only treat the case s 0 for d = 1 and s > d2 − 1 for d 2 from the crucial bilinear analysis. So we assume that φ is a Hilbert–Schmidt operator from L2 (Td ) into H s (Td ). We will denote by L0,s 2 the space of such operators, endowed by its natural norm φ2L0,s = 2
ns φen 2 2 , L
n∈Z
here the notation · will be defined in Section 2. 2 d in·x Now, denote by W (t) = ∂B } n∈Z βn en a cylindrical wiener process on L (T ), where {en } = {e ∂x = 2 is a complete orthonormal system in L and {βn } is a sequence of mutually independent Brownian motions in a fixed probability space (Ω, F, p) associated with a filtration {Ft }t0 . The process φW is then a φφ∗ -Wiener process, i.e. Gaussian process with law (N (0, tφφ∗ )t0 ), and the Itô form of Schrödinger part of the system (1.1) is written as i du + Δu dt = vu dt + φ dW.
(1.2)
On the other hand, for the Boussinesq equation, denote v± = (i∂t ± 1)v, then it is reduced to
√ √ i∂t v± ∓ v± = w( −Δ )2 |u|2 − −Δ −2 v, v± (0, x) = iv2 ± v1 ,
where w(k) =
√ k . 1+k2
(1.3)
We will consider the mild form of (1.2) and (1.3), i.e.
⎧ t t ⎪ v+ − v− ⎪ itΔ i(t−s)Δ ⎪ ⎪ u(s) ds − i ei(t−s)Δ φ dW (s), u(t) = e u0 − i e ⎪ ⎪ 2 ⎨ 0
0
t t ⎪ ⎪ √ √ ⎪ v+ − v− ⎪ ∓it ∓i(t−s) 2 2 ⎪ (s) ds. w( −Δ ) |u| (s) ds + i e∓i(t−s) −Δ −2 ⎪ ⎩ v± = e v±0 − i e 2 0
(1.4)
0
And our main result of this paper is: Theorem 1.1. Suppose s k2 0, and |s − k| < 1 for d = 1, s > k2 + d4 − 12 > d2 − 1 and |s − k| < 1 for d 2. Assume φ ∈ L0,s 2 for s defined above. Let (u0 (w, x), v±0 (w, x)) be F0 -measurable, with (u0 (w, x), v±0 (w, x)) ∈ H s (Td )×H k (Td ) for almost surely ω ∈ Ω, then there is a stopping time Tω > 0 and a unique process (u, v± ) solution of (1.4) in s,b k,b C [0, Tω ]; H s × H k ∩ X[0,T × X±,[0,T ω] ω] almost surely. s,b k,b Here X[0,T , X±,[0,T are the Bourgain spaces, whose definitions will be given in the next section. ω] ω]
Remark 1.2. By the same argument as in the paper of de Bouard, Debussche, and Tsutsumi [10], Φ(t, x) = t i(t−s)Δ e φ dW (s) can not be in X s,b for b 12 , so we will choose b < 12 . However in the proof, we can see 0 that even in the deterministic case, for s, k < 0, the key bilinear estimates will fail.
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By the idea of [6], there is the following global result, Theorem 1.3. For s = k = 0, d = 1, problem (1.4) is globally well-posed almost surely. Now, we mention some results on the deterministic problems of the Schrödinger-Improved Boussinesq and the relevant systems. Those suggest what happens in the case of the stochastic perturbation. ⎧ ⎪ ⎨ i∂t u + Δu = vu, ∂t2 v − Δv = Δ|u|2 , ⎪ ⎩ u(0, x) = u0 (x), v(0, x) = v1 (x),
∂t v(0, x) = v2 (x),
(u0 , v1 , v2 ) ∈ H s × H k × H k−1 ,
is called the Zakharov system, which was introduced in [18] to describe the long wave Langmuir turbulence in a plasma. There are plenty of papers studying the local and global well posedness of this system. For example: [12,2,3] and [6] for Rd , in which case the lowest allowed regularity for (u0 , v1 , v2 ) is (s, k) = (0, − 12 ) d−4 for d = 1, 2 and ( d−3 2 +, 2 +) for d 3. Particularly, in [6], Colliander, Holmer, and Tzirakis proved the global existence of the solution for (s, k) = (0, − 12 ), d = 1 by the conservation of L2 norm of u. For the periodic case, Takaoka in [17] gained the lowest allowed regularity (s, k) = (0, − 12 ) of dimension 1 for the nonresonant case and ( 12 , 0) for the resonant case. Recently, Kishimoto [15] studied the higher dimensional d−2 torus, and got the result that for d = 2, (s, k) = ( 12 , 0) and d 3, ( d−1 2 +, 2 +). Kishimoto also studied the global well posedness of the problem in 2d case by the improved I-method and got the result of (s, k) = ( 23 +, 0). In [1], Akahori studied the Schrödinger-Improved Boussinesq System in Rd , which has a less singular nonlinearity ⎧ ⎪ ⎨ i∂t u + Δu = vu, ∂t2 v − Δv − Δ∂t2 v = Δ|u|2 , ⎪ ⎩ u(0, x) = u0 (x), v(0, x) = v1 (x),
(1.5) ∂t v(0, x) = v2 (x),
(u0 , v1 , v2 ) ∈ H s × H k × H k .
He proved that for −
1 < s < 1, 4
−
1 < k < 1, 2
d 3,
system (1.5) is locally well-posed, and for s = k = 0,
d 2,
it is globally well-posed. (In [16], Ozawa and Tsutaya studied the local well-posedness of s = k = 0, d 3.) In this paper, we will not only consider the periodic case, but also with a stochastic perturbation. The idea for proving Theorem 1.1 is classical. First, we will find some bilinear estimates in X s,b for b < 12 , then by the contraction mapping principle, we show that almost surely there is a unique solution to (1.4). The idea for proving Theorem 1.3 is inspired by Colliander, Holmer, and Tzirakis [6] for Zakharov system. However, different from the deterministic case, the L2 -norm of u(t) does not conserved any more. So we use some idea from Kim [14] and so on. First, we define some stopping time δN (w) (ω ∈ Ω) by comparing the size of u(t)L2 + v+ (t)L2 + v− (t)L2 with N . Then by the same calculation of de Bouard and Debussche [8], we can see that the expected value of u(t)2L2 is controlled by T . With the help of these, combining the idea in [6], we also show the expected value of v+ (t)L2 + v− (t)L2 also only depends on T , but not on N , which shows the limitation of the shopping time δN goes to infinite.
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Remark 1.4. Similar to the proof here, we can also get the global well-posedness of Schrödinger–Boussinesq system with an additive noise (please refer to Farah and Pastor [11] for the deterministic case). However, for the nonresonant Zakharov system, it seems that the case (s, k) = (0, − 12 ) depends heavily on choosing b = 12 . The paper will be organized as follows: in Section 2, we will give some notations and lemmas which will be used later. In Section 3, the key bilinear estimates will be proved, and finally, in Section 4, we will finish the proof of the theorems. 2. Notations and lemmas A B means there is some constant C > 0 such that A B, while A ∼ B means both A B and B A. fˆ means the Fourier transformation of f with respect tot or x variable, while f˜ means Fourier transformation of f with respect to both t and x variables. x = 1 + |x|2 . k,b Bourgain spaces X s,b and X± are defined as spaces equipped with the norm b f X s,b = ns τ + n2 f˜(τ, n) L2 l2
τ n
and f X k,b = nk τ ± 1b f˜(τ, n) L2 l2 . ±
τ n
s,b We also define spaces XIs,b and X±,I restricted to time interval I with the norms
f X s,b = inf gX s,b I
g|I =f
and f X s,b = inf gX s,b . ±,I
g|I =f
±
From now on, we omit I when there is no confusion. Assume ψ is a smooth cut off function, which is defined by
1 |t| 12 , ψ(t) = 0 |t| > 1. Denote ψδ (t) = ψ( δt ). Lemma 2.1. (See [6,12].) Suppose 0 < δ 1. (1) S(t)f C(R
t ,H
s)
Cf H s ,
(2.1)
here S(t) = eitΔ or e∓it . (2) For 0 b 12 , ψδ (t)S(t)f δ 12 −b f H s , A k,b here A = X s,b or X± .
(2.2)
S. Zhong / J. Math. Anal. Appl. 415 (2014) 217–239
(3) For 0 b <
1 2
221
k,−b and A = X s,−b or X± ,
t ψδ (t) S(t − s)F (s) ds 0
Cδ 2 −b F A . 1
(2.3)
C([0,δ],H s )
(4) For 0 b < 12 , 0 b, b + b 1, t ψδ (t) F (s) ds 0
Cδ 1−b−b F H −b ,
(2.4)
t
Htb
and t ψδ (t) S(t − s)F (s) ds 0
Cδ 1−b−b F A(−b ) ,
(2.5)
A(b)
k,c here A(c) = X s,c or X± . 1 (5) For 0 b b < 2 ,
f A(b) Cδ b −b f A(b ) ,
(2.6)
s,c s,c here A(c) = X[0,δ] or X±,[0,δ] .
Lemma 2.2. (See [12,13].) (1) Let α, β > 0, such that α + β > 1, there exists ∞
c dτ , τ α τ − aβ aγ
(2.7)
1 < ∞, nα a + bnn1 + n2 β
(2.8)
−∞
here γ = min{α, β, α + β − 1}. (2) For α, β > 0, α + 2β > 1, then
n∈Z
for β > 12 ,
n∈Z
1 < ∞, a + bnn1 + n2 β
(2.9)
and for α, β > 0, α + β > 1,
n∈Z
1 < ∞. nα a + nβ
(2.10)
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Lemma 2.3. (See Chapter V of [4] and [5].) (1) uL4 (T1+1 ) u
3
X 0, 8
.
(2.11)
(2) Suppose IN √−Δ2N (u) = u and IL√−Δ2L (v) = v, then s1 uvL2 (T1+d ) min{N, L} uX 0,b vX 0,b , with s1 >
d 2
− 1, b = 14 (1 +
d 2
(2.12)
− s1 )+ < 12 , and d 2.
Lemma 2.4. (See Proposition 2.1 of [9].) For s, b ∈ R with b < 12 , and assume that φ ∈ L0,s 2 , then Φ(t, x) = t i(t−s)Δ 2 s,b e φ dW (s) is in L (Ω, X ) with the smooth cut off function ψ defined as above, 0 E ψΦ2X s,b Cφ2L0,s ,
(2.13)
2
here C depends on b and ψ. Remark 2.5. (1) The authors in [9] consider the KdV equation, but Proposition 2.1 is not dependent on the KdV operator. (2) If we replace ψ(t) with ψδ (t), then by the calculation of [9], we have E ψδ Φ2X s,b Cδ 1−2b φ2L0,s ,
for 0 < δ < 1,
(2.14)
for δ 1.
(2.15)
2
and E ψδ Φ2X s,b Cδ 4 φ2L0,s , 2
The idea for proving the following lemma is exactly the same as Proposition 3.2 of [8], and so we omit it here (we should use Itô formula and some sequence of approximation first). Lemma 2.6. For φ ∈ L0,s 2 , and a stopping time T > 0 a.s. then for t < T , the solution (u, v± ) to the system (1.4) with the initial data (u0 , v±0 ), we have
u(t) 2 2 = u0 2 2 − 2 L L
t u(s, x)φen dx dβn (s) + tφ2L0,s .
(2.16)
2
n∈Z 0 d T
Moreover, by Martingale inequality,
E
2 sup u(t) L2 2u0 2L2 + CT,
t∈[0,T ]
here C depends on φL0,s . 2
(2.17)
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3. Bilinear estimate Proposition 3.1. Suppose d = 1, s k2 0, and |s − k| < 1, or d 2, s > Then, there is some b < 12 , such that
k 2
+ d4 − 12 >
d 2
− 1, and |s − k| < 1.
uvX s,−b uX s,b vX k,b
(3.1)
2 |u| k,−b u2 s,b . X X
(3.2)
±
and ±
Proof. Take + for example and assume by density that u, v ∈ C0∞ (R × Td ). [Proof of (3.1)] By duality, we just need to prove J = uvw dx dt uX s,b vX k,b wX −s,b , +
R×Td
here w ∈ X −s,b . It also equals to prove
A= n0 =n1 +n2 τ =τ +τ 0 1 2
n0 s f (τ1 , n1 )g(τ2 , n2 )h(τ0 , n0 ) dτ1 dτ2 n1 s n2 k τ0 + n20 b τ1 + n21 b τ2 + 1b
f L2τ ln2 gL2τ ln2 hL2τ ln2 ,
(3.3)
with f, g, h ∈ L2τ ln2 . d = 1. Case 1. |n1 | ∼ |n0 |. τ0 + n20 + τ1 + n21 + |τ2 + 1| n20 − n21 − 1 = (n0 − n1 )(n0 + n1 ) − 1 = n2 (n0 + n1 ) − 1, and denote M = max{|τ0 + n20 |, |τ1 + n21 |, |τ2 + 1|}. Subcase 1.1. |n2 | 1, i.e. |n0 − n1 | 1. By Schwarz inequality,
n0 s f (τ1 , n1 )g(τ2 , n2 )h(τ0 , n0 ) A= dτ1 dτ2 n1 s n2 k τ0 + n20 b τ1 + n21 b τ2 + 1b n =n +i |i|1
0
1
τ0 =τ1 +τ2
|i|1 n0 =n1 +i
|f (τ1 , n1 )g(τ2 , i)h(τ0 , n1 + i)| dτ1 dτ2 τ0 + n20 b τ1 + n21 b τ2 + 1b
f (n1 )
g(i) h(n + i) sup 1 L2 L2 L2 τ
|i|1 n1
τ
τ
τ0
f (n1 ) 2 g(i) 2 h(n1 + i) 2 sup L L L τ
|i|1 n1
f (n1 )
|i|1 n1
g(i) |i|1
with k 0 and b = 12 −.
L2τ
L2τ
τ
τ
τ0
1 dτ1 τ0 + n20 2b τ1 + n21 2b τ2 + 12b
1 2 2b τ0 + n0 τ0 + 1 + n21 4b−1
12
12
g(i) 2 h(n1 + i) 2 L L τ
τ
f L2τ ln2 hL2τ ln2 f L2τ ln2 gL2τ ln2 hL2τ ln2 ,
(3.4)
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224
Subcase 1.2. n0 + n1 = 0, so n2 = 2n0 . Similarly to the above case, A
n0 τ =τ +τ 0 1 2
f (τ1 , n1 )g(τ2 , n2 )h(τ0 , n0 ) dτ1 dτ2 n0 k τ0 + n20 b τ1 + n21 b τ2 + 1b
f (−n0 ) 2 g(2n0 ) 2 h(n0 ) 2 L L L τ
n0
τ
τ
f ln∞ L2τ gL2τ ln2 hL2τ ln2 f L2τ ln2 gL2τ ln2 hL2τ ln2 ,
(3.5)
with k 0. Subcase 1.3. |n2 | 2 and |n0 + n1 | 1. In this case |n0 | |n1 |, and 1 ∼ |n2 ||n0 + n1 | = |n2 ||2n0 − n2 |. M n2 (n0 + n1 ) − 1 = |n2 |n0 + n1 − n2 a. M = |τ0 + n20 |. By Schwarz inequality, A hL2τ ln2
n0 s f (τ1 , n1 )g(τ2 , n2 ) dτ 1 2 2 s k b b b n1 n2 τ0 + n0 τ1 + n1 τ2 + 1 L2 l 2
n1
τ n
12 2 1 2 f (τ1 , n1 )g(τ2 , n2 ) dτ1 C(τ0 , n0 ) hL2τ ln2
2 L2τ ln
n1
1
sup C(τ0 , n0 ) 2 f L2τ ln2 gL2τ ln2 hL2τ ln2 ,
(3.6)
τ0 ,n0
where C(τ0 , n0 ) =
n1 2s n2 2k τ0
n2
n0 2s dτ2 . + n20 2b τ1 + n21 2b τ2 + 12b
Then by Lemma 2.2, C(τ0 , n0 )
n2
n2
1 dτ2 n2 2k+2b 2n0 − n2 2b τ1 + n21 2b τ2 + 12b
1 1, n2 2k+2b 2n0 − n2 2b τ0 + (n0 − n2 )2 + 14b−1
with k > −1 and b = 12 − > 14 . b. M = |τ1 + n21 |. Similar to case a, it can be reduced to C(τ1 , n1 ) =
n2
n0 2s dτ2 , n1 2s n2 2k τ0 + n20 2b τ1 + n21 2b τ2 + 12b
then we can treat it in the same way as case a. c. M = |τ2 + 1|. In this case C(τ2 , n2 ) =
n1
n0 2s dτ1 n1 2s n2 2k τ0 + n20 2b τ1 + n21 2b τ2 + 12b
(3.7)
S. Zhong / J. Math. Anal. Appl. 415 (2014) 217–239
n1
1 dτ1 n2 2k+2b n0 + n1 2b τ0 + n20 2b τ1 + n21 2b
n2 + 2n1
n1
225
2b τ
1 1, 2 4b−1 2 + n2 − 2n2 n1
(3.8)
with b = 12 − > 13 . Case 2. |n1 | |n0 | so |n2 | ∼ |n0 |. In this case M |n0 |2 . Subcase 2.1. M = |τ0 + n20 |. C(τ0 , n0 )
n1
n1
1 dτ1 n1 2s n2 2k−2s+4b τ1 + n21 2b τ2 + 12b
1 1, n1 2k+4b τ0 + 1 + n21 4b−1
(3.9)
with b = 12 − and k + 1 > s. Subcase 2.2. M = |τ1 + n21 |. As in Subcase 2.1, it can be reduced to estimating C(τ1 , n1 ) =
n0 2s dτ2 + n20 2b τ1 + n21 2b τ2 + 12b
n2
n1 2s n2 2k τ0
n2
1 dτ2 n2 2k−2s+4b τ0 + n20 2b τ2 + 12b
n2
1 1, n2 2k−2s+4b n20 + τ1 − 14b−1
(3.10)
with s 0 and k + 1 > s. Subcase 2.3. M = |τ2 + 1|. As before, we just need to estimate C(τ2 , n2 ) =
n1
n1
n0 2s dτ1 n1 2s n2 2k τ0 + n20 2b τ1 + n21 2b τ2 + 12b n1
2k+4b τ
n1
n1
2k+4b τ
1 2 2b 2 2b dτ1 0 + n0 τ1 + n1
1 1 = 2 2 4b−1 2k+2b τ + n2 + 2n n 4b−1 n 2 + n0 − n1 1 2 1 2 2 n 1
1,
(3.11)
with k > − 12 . Case 3. |n2 | ∼ |n1 | |n0 |. and in this case M |n1 |2 . The way to deal with this case is similar to Case 2. In fact, it will be better. Conclusively, for d = 1, when s, k 0 and k + 1 > s, (3.1) holds. d 2. Denote s0 = d2 − 1+, and take b = 12 −. Case 1. |n1 | |n0 |. Let N = (N0 , N1 , N2 ) be dyadic integer, i.e. Nj = 2mj , mj = 0, 1, 2, . . . , j = 0, 1, 2. So J
N
where
I(N ),
S. Zhong / J. Math. Anal. Appl. 415 (2014) 217–239
226
I(N ) =
u
N1 N2
v
w
N0
dx dt,
R×Td
with uN1 means the support of u ˆ(n1 ) is on N1 |n1 | 2N1 , so are v N2 and wN0 . Subcase 1.1. N2 N0 ∼ N1 . ˜ ), I(N ) ∼ I(N
(3.12)
(k)
with the support M = {(n0 , n1 , n2 ) ∈ Z3d : Ni sup{|ni |} 2Ni , i = 0, 1, 2, k = 1, . . . , d}. Cover M with disjoint interiors Qα = n ∈ Zd : sup n(k) − α(k) N2 , here α ∈ Zd , and project uN1 and wN0 onto each Qα , then by Lemma 2.3 (2)
1 N2 uN wαN0 dx dt α v
˜ ) I(N R×Td
α
N0 1 uN α wα α
t,x
+
δ b N2s0 −k δ
N v 2 2 L
N 1 wα 0 0,b v N2 0,0 N2s0 uN α X 0,b X X
α
b
L2t,x
N2s0 −k
δ b N2s0 −k
N0 N1 N0 N1 N0 N1
s
N uα 1
X s,b
N wα 0
X −s,b
N v 2 k,b X +
α
s
N 2 uα 1
12
X s,b
s
α
N u 1
N 2 wα 0 −s,b X
12
N v 2 k,b X +
α
X s,b
N w 0
X −s,b
N v 2 k,b , X
(3.13)
+
here δ is the length of time interval. Thus for N0 ∼ N1 i.e. there is some c > 0 such that N1 = 2j N0 , |j| c, and k > s0
N
I(N ) ∼
˜ ) I(N
N
s N0 uN1 s,b wN0 −s,b v N2 k,b δ X X X+ N1 N s N0 uN1 s,b wN0 −s,b v k,b δb X+ X X N1 N0 ,N1
j u2 N0 s,b wN0 −s,b v k,b δb 2js X X X b
N2s0 −k
+
−cjc
δ
b
−cjc
δ
b
−cjc
N0
12
12 2j N 2 N 2 0 0 u w vX k,b 2 X s,b X −s,b js
+
N0
2
js
N0
uX s,b wX −s,b vX k,b δ b uX s,b wX −s,b vX k,b . +
+
(3.14)
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227
Subcase 1.2. N2 ∼ N0 N1 . The first step is the same as in (3.11) except that there is no need to do the decomposition onto Qα . Therefore
N
s N0 uN1 s,b wN0 −s,b v N2 k,b X X X+ N1 N s
N s0 −k N0 b u 1 s,b wN0 −s,b v N2 k,b δ N2 X X X+ N1 N
N0 s uN1 s,b wN0 −s,b v k,b . δb X+ X X N1
I(N ) δ b
N0s0 N2−k
(3.15)
N0 ,N1
Since N0 N1 , there is some c > 0 such that N1 = 2j N0 , with j −c, and
N0 s uN1 s,b wN0 −s,b X X N1
N0 ,N1
=
j−c
=
2−js
j u2 N0
X
N 0 s,b w
X −s,b
N0
2−js
2j N 2 u 0
X
12
N 2 w 0 s,b
X −s,b
N0
j−c
12
N0
2−js uX s,b wX −s,b = uX s,b wX −s,b ,
(3.16)
j−c
with s > 0. Conclusively, for s > 0 and k > s1 >
d 2
− 1,
J uX s,b wX −s,b vX k,b . +
Case 2. |n0 | |n1 |. So |n0 | ∼ |n2 | and M |n0 |2 , here the definition of M is the same as the one in dimension 1 case. Subcase 2.1. M = |τ0 + n20 |. The same as in Case d = 1, 1
A sup C(τ0 , n0 ) 2 f L2τ ln2 gL2τ ln2 hL2τ ln2 ,
(3.17)
τ0 ,n0
with C(τ0 , n0 ) = with b = 12 −, s < k + 1 and k >
n1
n0 2s dτ1 n1 2s n2 2k τ0 + n20 2b τ1 + n21 2b τ2 + 12b
n1
n1 2s n2 2k−2s+4b τ1
1
1 2 4b−1 0 + 1 + n1
n1
n1
n1
1 1, n1 2k+4b
d 2
− 1.
2k+4b τ
+ n21 2b τ2 + 12b
dτ1
S. Zhong / J. Math. Anal. Appl. 415 (2014) 217–239
228
Subcase 2.2. M = |τ1 + n21 |. A
f (τ1 , n1 )g(τ2 , n2 )h(τ0 , n0 ) dτ1 dτ2 n1 s n2 k−s+2b τ0 + n20 b τ2 + 1b
n0 =n1 +n2
n0 =n1 +n2
n0 =n1 +n2
1 f (n1 ) 2 g(n2 ) 2 h(n0 ) 2 sup Lτ Lτ Lτ n1 k+2b τ1
1 τ0 + n20 2b τ2 + 12b
dτ2
12 1 1 f (n1 ) 2 g(n2 ) 2 h(n0 ) 2 sup Lτ Lτ Lτ n1 k+2b τ1 + n20 − 14b−1 τ1
1 f (n1 ) 2 g(n2 ) 2 h(n0 ) 2 k+2b L L Lτ τ τ n1 n0 =n1 +n2
1 f (n1 )L2 g(n2 )L2 l2 hL2τ ln2 τ τ n n1 k+2b n1 1 f L2τ ln2 gL2τ ln2 hL2τ ln2 n1 k+2b 2 f L2τ ln2 gL2τ ln2 hL2τ ln2 , l
12
(3.18)
n
with k > d2 − 1 and s < k + 1. Subcase 2.3. M = |τ2 + 1|. We can reduced the proof to estimating C(τ2 , n2 ) =
n1
1 dτ1 n1 2s n0 2k−2s τ0 + n20 2b τ1 + n21 2b τ2 + 12b
n1
1 dτ1 , n1 2k+4b τ0 + n20 2b τ1 + n21 2b
(3.19)
which can be dealt with exactly similarly to Subcase 2.1. Consequently, for 0 < s < k + 1 and k > d2 − 1, A f L2τ ln2 gL2τ ln2 hL2τ ln2 . Thus (3.1) is proved for d 2. [Proof of (3.2)] The same as (3.1), we reduce the proof to estimating
B=
n0 =n1 +n2τ =τ +τ 0 1 2
n0 k f (τ1 , n1 )g(τ2 , n2 )h(τ0 , n0 ) dτ1 dτ2 n1 s n2 s τ0 + 1b τ1 + n21 b τ2 − n22 b
f L2τ ln2 gL2τ ln2 hL2τ ln2 ,
(3.20)
for f , g, h ∈ L2τ ln2 . d = 1. Case 1. |n0 | 1, so |n1 | ∼ |n2 |, then B=
|n0 |1 n2 =n0 −n1τ0 =τ1 +τ2
|n0 |1 n2 =n0 −n1τ0 =τ1 +τ2 1
n0 k f (τ1 , n1 )g(τ2 , n2 )h(τ0 , n0 ) dτ1 dτ2 n1 s n2 s τ0 + 1b τ1 + n21 b τ2 − n22 b |f (τ1 , n1 )g(τ2 , n2 )h(τ0 , n0 )| dτ1 dτ2 n1 2s τ0 + 1b τ1 + n21 b τ2 − n22 b
sup C(τ1 , n1 ) 2 f L2τ ln2 gL2τ ln2 hL2τ ln2 τ1 ,n1
(3.21)
S. Zhong / J. Math. Anal. Appl. 415 (2014) 217–239
229
with
C(τ1 , n1 ) =
|n0 |1
1 dτ0 n1 4s τ0 + 12b τ1 + n21 2b τ2 − n22 2b
|n0 |1
n1
4s τ
1
+
n21 2b τ1
1 1, + 1 − (n0 − n1 )2 4b−1
for s 0 and b = 12 − > 14 . Case 2. n1 − n2 = 0, i.e. n1 = n2 and n0 = 2n1 , the same as before, 1
B sup C(τ0 , n0 ) 2 f L2τ ln2 gL2τ ln2 hL2τ ln2 ,
(3.22)
τ0 ,n0
and
C(τ0 , n0 ) =
n0 =2n1 ,n2 =n1 ,n1
n1 2s n2 2s τ0
n0 2k dτ1 + 12b τ1 + n21 2b τ2 − n22 2b
1 1, n0 4s−2k τ0 + 12b τ0 4b−1
with s k2 and b = 12 − > 14 . Case 3. |n0 | 2 and |n1 − n2 | 1. Then |τ0 + 1| + τ1 + n21 + τ2 − n22
1 n21 − n22 − 1 = n0 (n1 − n2 ) − 1 = |n0 |n1 − n2 − |n0 ||n1 − n2 |, n0
and denote M = max{|τ0 + 1|, |τ1 + n21 |, |τ2 − n22 |}. By symmetry argument, we suppose |n1 | |n2 |, so |n0 | |n1 |. Subcase 3.1. |n1 | |n2 |, so |n0 | ∼ |n1 | and M |n0 |2 ∼ |n1 |2 . a. M = |τ0 + 1|. Then 1
B sup C(τ0 , n0 ) 2 f L2τ ln2 gL2τ ln2 hL2τ ln2 ,
(3.23)
τ0 ,n0
and
1 dτ2 2s−2k n 2s τ + 12b τ + n2 2b τ − n2 2b n 1 2 0 1 2 1 2 n2 1 1 dτ2 n1 2s−2k+4b n2 2s τ1 + n21 2b τ2 − n22 2b
1 1, 4s−2k+4b τ + n2 − 2n n 4b−1 n 2 0 0 2 0 n
C(τ0 , n0 ) =
2
1 with 2s − 2k + 4b 0, 4s − 2k + 4b + 4b − 1 > 1, b = 12 −, i.e. s > max{k − 1, k−1 2 }, and b = 2 −. 2 b. M = |τ1 + n1 |. Similarly, it is reduced to estimate
C(τ1 , n1 ) =
n2
n0 2k dτ2 n1 2s n2 2s τ0 + 12b τ1 + n21 2b τ2 − n22 2b
(3.24)
S. Zhong / J. Math. Anal. Appl. 415 (2014) 217–239
230
1
n2
n1 2s−2k+4b n2 2s
n2
1 dτ2 τ0 + 12b τ2 − n22 2b
1 1, n2 4s−2k+4b τ1 + 1 + n22 4b−1
with 2s − 2k + 4b 0 and 4s − 2k + 4b + 2(4b − 1) > 1, i.e. s > max{k − 1, c. M = |τ2 − n22 |, then
(3.25)
k− 32 2
}, and b = 12 −.
B sup C(τ2 , n2 )f L2τ ln2 gL2τ ln2 hL2τ ln2 ,
(3.26)
τ2 ,n2
and
n0 2k dτ1 n1 2s n2 2s τ0 + 12b τ1 + n21 2b τ2 − n22 2b n1
1 1 dτ1 2s−2k+4b n 2s 2b τ + n2 2b n τ + 1 1 2 0 1 1 n
C(τ2 , n2 ) =
1
n1
1 n1
2s−2k+4b τ
2
+ 1 − n21 4b−1
1
(3.27)
with s 0, and 2s − 2k + 4b + 2(4b − 1) > 1, i.e. s 0, s > k − 32 and b = 12 −. Subcase 3.2. |n1 | ∼ |n2 |, so |n0 | |n1 |, and M |n0 ||n1 − n2 |. a. M = |τ0 + 1|, the proof is the same as in Subcase 3.1 a,
C(τ0 , n0 ) =
n2
n2
n1 2s n2 2s τ0
n0 2k dτ2 + 12b τ1 + n21 2b τ2 − n22 2b
n0 2k−2b . n2 4s n1 − n2 2b τ0 + n20 − 2n0 n2 4b−1
(3.28)
If 0 k < 12 , then there is b = 12 − such that k < b, and (3.28)
n2
1 n2 4s 2n2
− n0
2b τ
0
+ n20 − 2n0 n2 4b−1
1,
(3.29)
with s 0 and b = 12 −. If k 12 > b, then (3.28)
n2
n2
with 4s − 2k + 4b > 1, i.e. s >
k− 12 2
n2 4s−2k+2b 2n2
1 − n0 2b τ0 + n20 − 2n0 n2 4b−1
1 1, n2 4s−2k+2b 2n2 − n0 2b and b = 12 −.
(3.30)
S. Zhong / J. Math. Anal. Appl. 415 (2014) 217–239
231
b. M = |τ1 + n21 |, so
C(τ1 , n1 )
n0 2k dτ2 n2 4s+2b n1 − n2 2b τ0 + 12b τ2 + n22 2b
n2
n2
1 1, n2 4s−2k+2b n2 − n1 2b τ1 + 1 + n22 4b−1
(3.31)
k− 1
with s > 2 2 and b = 12 −. c. M = |τ2 − n22 |, by the symmetry of n1 and n2 , this case is exactly the same as b. Consequently, for d = 1, s k2 0, s > k − 1 and b = 12 −, (3.2) exists. d 2. By the same symbol as before, it equals to prove J=
¯N2 wN0 dx dt u2X s,b wX −k,b , u N1 u
(3.32)
+
N −k,b with N = (N0 , N1 , N2 ), Ni = 2mj , mj = 0, 1, 2, . . . , j = 0, 1, 2, w ∈ X+ and uN1 means the support of u ˆ(n1 ) is on N1 |n1 | < 2N1 , so are uN2 and wN0 . Without losing of generality, suppose N1 N2 . Case 1. N1 ∼ N2 N0 . By Lemma 2.3 (2), we have
J
uN1 uN2 2 wN0 2 L L t,x
N
d
N12
−1+ N1 u
N
d
N12
−1−2s+
t,x
X 0,b
N u 2 0,b wN0 0,0 X X +
N0k δ b uN1 X s,b uN2 X s,b wN0 X −k,b u2X s,b wX −k,b , +
+
(3.33)
N
with d2 − 1 − 2s + k < 0 i.e. s > k2 + d4 − 12 and b = 12 −. Case 2. N1 N2 , so N1 ∼ N0 . (3.32) equals to prove B=
n0 =n1 +n2 τ =τ +τ 0 1 2
n0 k f ((τ1 , n1 )g(τ2 , n2 )h(τ0 , n0 )) dτ1 dτ2 n1 s n2 s τ0 + 1b τ1 + n21 b τ2 − n22 b
f L2τ ln2 gL2τ ln2 hL2τ ln2 ,
(3.34)
with f , g, h ∈ L2τ ln2 . Since |τ0 + 1| + τ1 + n21 + τ2 − n22 n21 − n22 − 1 = n0 (n1 − n2 ) − 1 1 |n0 |2 ∼ |n1 |2 , = |n0 |n1 − n2 − n0 by N1 ∼ N0 N2 . Put M = max{|τ0 + 1|, |τ1 + n21 |, |τ2 − n22 |}. Subcase 2.1. M = |τ0 + 1|. 1
B sup C(τ0 , n0 ) 2 f L2τ ln2 gL2τ ln2 hL2τ ln2 , τ0 ,n0
and
S. Zhong / J. Math. Anal. Appl. 415 (2014) 217–239
232
C(τ0 , n0 ) =
n2
n0 2k dτ2 n1 2s n2 2s τ0 + 12b τ1 + n21 2b τ2 − n22 2b
n2
1 dτ2 n1 2s−2k+4b n2 2s τ1 + n21 2b τ2 − n22 2b
n2
n2
n2
4s−2k+4b τ
1 2 4b−1 0 + n0 − 2n0 n2
1 1, n2 4s−2k+4b
(3.35)
with 2s − 2k + 4b 0 and 4s − 2k + 4b > d, i.e. s > k − 1, s > Subcase 2.2. M = |τ1 + n21 |.
k 2
+
d 4
− 12 , and b = 12 −.
1
B sup C(τ1 , n1 ) 2 f L2τ ln2 gL2τ ln2 hL2τ ln2 , τ1 ,n1
and C(τ1 , n1 ) =
n2
n0 2k dτ2 n1 2s n2 2s τ0 + 12b τ1 + n21 2b τ2 − n22 2b
n2
1 dτ2 n1 2s−2k+4b n2 2s τ1 + n21 2b τ2 − n22 2b
n2
1 n2
4s−2k+4b τ
1
+ 1 + n22 4b−1
1,
(3.36)
with s > k − 1, s > k2 + d4 − 12 , and b = 12 −. Subcase 2.3. M = |τ2 − n22 |. B
n0 =n1 +n2τ =τ +τ 0 1 2
1 dτ1 dτ2 n1 s−k+2b n2 s τ0 + 1b τ1 + n21 b
1
n0 =n1 +n2
n2 2s−k+2b
1
n0 =n1 +n2
n2 2s−k+2b
1 2s−k+2b
f (n1 ) f (n1 ) f (n1 )
g(n2 ) 2 h(n0 ) 2 sup L2 L L τ
τ
τ
τ2
g(n2 ) 2 h(n0 ) 2 sup L2 L L τ
τ
τ
1 dτ1 τ1 + τ2 + 12b τ1 + n21 2b
τ2
1 τ2 + 1 − n21 4b−1
12
12
g(n2 ) 2 h(n0 ) 2 L L
τ τ n2
1 f (n0 − n2 )L2 g(n2 )L2 l2 hL2τ ln2 τ τ n 2s−k+2b n2
n0 =n1 +n2
L2τ
n2
1 f L2 l2 gL2 l2 hL2 l2 f L2 l2 gL2 l2 hL2 l2 , τ n τ n τ n τ n τ n τ n n2 2s−k+2b 2 ln
with s − k + 2b 0 and 2s − k + 2b > d, i.e. s > k − 1, s > k2 + d4 − Consequently, for s > max{k − 1, k2 + d4 − 12 }, (3.2) holds. 2
1 2
and b = 12 −.
(3.37)
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Remark 3.2. It is obvious that the requirement s, k 0 is necessary. In fact, for d = 1, Subcase 1.2 would be a good example to prove that k 0 is optimal for (3.1). Correspondingly, if we take n1 = −n2 = N and |τ1 + n21 |, |τ2 − n22 | 1, then we can show s 0 is necessary for (3.2). Then for d 2, since n = (n(1) , n ), with n ∈ Zd−1 , we can choose some similar example. 4. Proof of Theorems 1.1 and 1.3 4.1. Proof of Theorem 1.1 concluded k,b k,b Proof. Let B = {(u, v+ , v− ) ∈ X s,b × X+ × X− : uX s,b M δ 2 −b (u0 H s + φL0,2 ), v+ X k,b + 2 1
+
v− X k,b M δ 2 −b (v+0 H k + v−0 H k )}, 1
−
t itΔ
B(u, v± ) = e
u0 − iψδ
e
i(t−s)Δ v+
− v− u(s) ds − iψδ 2
0
t ei(t−s)Δ φ dW (s), 0
and B± (u, v± ) = e∓it v±0 − iψδ
t
√ √ v+ − v− (s) ds. e∓i(t−s) w( −Δ )2 |u|2 (s) ds + iψδ e∓i(t−s) −Δ−2 2
0
0
t
Then for (u, v+ , v− ) ∈ B, 0 < δ < 1, by Lemmas 2.1, 2.4, Bienaymé–Chebyshev inequality and Proposition 3.1, we have B(u, v± ) s,b Cδ 12 −b u0 H s + δ 1−2b C v+ uX s,−b + v− uX s,−b + δ 12 −b Cφ 0,s L2 X 1 1 δ 2 −b u0 H s + δ 1−2b v+ X s,b + v− X s,−b uX s,b + δ 2 −b φL0,s , 2 B+ (u, v± )
k,b X+
Cδ
1 2 −b
+ B− (u, v± ) X k,b −
v+0 H k + v−0 H k + δ
t √ 2 v − v + − ∓i(t−s) −2 |u| X k,−b + ψδ e (s) ds −Δ ± 2
1−2b
0
δ 2 −b v±0 H k + δ 1−2b u2X s,b 1
(4.1)
−
+
t v − v + − (s) ds + ψδ e∓i(t−s) 2 0
,
k,b X±
(4.2)
k,b X±
B(u1 , v±1 ) − B(u2 , v±2 ) s,b X 1−2b (v+1 + v−1 )u1 − (v+2 + v−2 )u2 X s,−b δ δ 1−2b v+1 X k,b + v−1 X k,b u1 − u2 X s,b + v+1 − v+2 X k,b + v−1 − v−2 X k,b u2 X s,b , +
−
+
−
(4.3) and B+ (u1 , v±1 ) − B+ (u2 , v±2 ) k,b + B− (u1 , v±1 ) − B− (u2 , v±2 ) k,b X+ X− t 1−2b i∓(t−s) δ u1 X s,b + u2 X s,b u1 − u2 X s,b + ψδ e (v±1 − v±2 )(s) ds 0
for almost surely ω ∈ Ω.
, k,b X±
(4.4)
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234
On the other hand, by Lemma 2.1, t ∓i(t−s) v+ (s) ds ψδ e 0
δ 1−b v+ X k,0 = δ 1−b v+ L2t Hxk = δ 1−b v+ X k,0 δv+ X k,b , ±
k,b X±
t
and the same calculation is valid for ψδ Therefore, taking δ ∼ min
v+0 H k
0
2
−3b
(4.5)
+
e∓i(t−s) v− (s) dsX k,b . ±
31
1 + v−0 H k
+
,
v+0 H k + v−0 H k u0 2H s + φ2L0,s
31 2
−3b
,
u0 H s
2
1 + φL0,s 2
31 2
−3b
,
with b < 12 , (B(u, v± ), B± (u, v± )) are contraction map on B. So there is a unique pathwise solution (u, v± ) k,b . to (1.4) with the initial data (u0 , v±0 ) ∈ H s × H k in X s,b × X± Then, by Lemma 2.1 and Theorem 6.10 of [7], there exists 1 uH s u0 H s + Cδ 2 −b v+ uX s,−b + Cv− uX s,−b + φL0,s 2 1 u0 H s + δ 2 −b v+ X k,b + v− X k,b uX s,b + φL0,s 2 + − 3 −3b u0 H s + δ 2 v+0 H k + v−0 H k u0 H s + φL0,s . 2
(4.6)
And 1 1 v± H k v±0 H k + Cδ 2 −b |u|2 X k,−b + Cδ 2 +b v+ X k,b + v− X k,b ±
v±0 H k v±0 H k
+
+δ u2X s,b + δ v+0 H k + v−0 H k 3 + δ 2 −3b u0 2H s + δ v+0 H k + v−0 H k .
−
1 2 −b
Hence there is a unique pathwise solution (u, v+ , v− ) on H s × H k × H k .
(4.7)
2
4.2. Proof of Theorem 1.3 The idea is similar to [6], although the mass uL2 here is not conserved. Proof. Since the constant C before the last term of (4.1) depends on ω ∈ Ω, to make it clearly, we consider the following system instead.
t itΔ
u(t) = e
u0 − iψδ (t)
ei(t−s)Δ
1 v+ − v− u (s) ds − iψL δ b− 2 ψδ ΦX 0,b ψN ΦL2 ψδ (t)Φ(t), 2
(4.8)
0
and v± (t) = e∓it v±0 − iψδ (t)
t
√ √ v+ − v− (s) ds, e∓i(t−s) w( −Δ )2 |u|2 (s) − −Δ−2 2
(4.9)
0
where ψ, ψδ , ψL and ψN are the truncation function defined in Section 2. Let L, N ∈ N and 0 < δ < 1.
S. Zhong / J. Math. Anal. Appl. 415 (2014) 217–239
235
On the other hand, for s = k = 0, the estimate in Proposition 3.1 can be improved. Since by Lemma 2.3 (1), for b = 38 , uvw dx dt uL4t,x vL2t,x wL4t,x u
3
X 0, 8
vX 0,0 w ±
3
3
X 0, 8
δ 8 u
3
X 0, 8
v
T×T 0, 3
3
0, 3
Therefore, let B = {(u, v+ , v− ) ∈ X 0, 8 × X+ 8 × X− 8 : u v−
0, 3
X− 8
0, 3
X± 8
w
3
X 0, 8
.
1
Cδ 8 (u0 L2 + L), v+
3
X 0, 8
0, 3 8
X+
+
1 8
Cδ (v+0 L2 + v−0 L2 )}, t itΔ
B(u, v± ) = e
u0 − iψδ
ei(t−s)Δ
1 v+ − v− u(s) ds − iψL δ − 8 ψδ Φ 0, 38 ψN ΦL2 ψδ (t)Φ(t), X 2
0
and ∓it
B± (u, v± ) = e
t v±0 − iψδ
e
∓i(t−s)
√ −2 v+ − v− w( −Δ ) |u| (s) ds − Δ (s) ds. 2 √
2
2
0
Now, for ω ∈ Ω, we define a stopping time δL (ω) by 1 δL (ω) = inf δ > 0: δ − 8 ψδ Φ
3
X 0, 8
>L .
(4.10)
Then, for (u, v± ) ∈ B, 0 < δ < δL , we have B(u, v± )
3
X 0, 8
1 1 1 Cδ 8 u0 L2 + δ 4 C v+ u 0,− 38 + v− u 0,− 38 + Cδ 8 L X X 1 5 1 δ 8 u0 L2 + δ 8 v+ 0, 38 + v− 0, 38 u 0, 38 + δ 8 L, X+
B+ (u, v± )
0, 3 8
X+
+ B− (u, v± )
0, 3
X− 8
1 1 Cδ 8 v+0 L2 + v−0 L2 + δ 4 |u|2
0,− 3 X± 8
1 5 δ 8 v+0 L2 + v−0 L2 + δ 8 u2
3
X 0, 8
B(u1 , v±1 ) − B(u2 , v±2 )
3 X 0, 8
(4.11)
X
X−
+ δ v+
+ δ v+
0, 3 X+ 8
0, 3 X+ 8
+ v−
+ v−
0, 3 X− 8
0, 3 X− 8
,
(4.12)
1 δ 4 (v+1 + v−1 )u1 − (v+2 + v−2 )u2 0,− 38 X 5 8 δ v+1 0, 38 + v−1 0, 58 u1 − u2 0, 38 X+
5 + δ 8 v+1 − v+2
X
X−
0, 3 X+ 8
+ v−1 − v−2
0, 3 X− 8
u2
3
,
(4.13)
.
(4.14)
X 0, 8
and B+ (u1 , v±1 ) − B+ (u2 , v±2 ) 5 δ 8 u1
0, 3 X+ 8
3
X 0, 8
+ u2
3
X 0, 8
+ B− (u1 , v±1 ) − B− (u2 , v±2 )
u1 − u2
3
X 0, 8
+ δ v+1 − v+2
0, 3
X− 8
0, 3 8
X+
+ v−1 − v−2
0, 3
X− 8
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Hence we should choose δ satisfying ⎧ 3 ⎪ δ 4 v+0 L2 + v−0 L2 1, ⎪ ⎪ ⎪ ⎪ ⎨ δ 78 u 2 + L2 δ 18 v 2 + v 2 , 0 L +0 L −0 L 3 ⎪ ⎪ δ 4 u0 L2 + L 1, ⎪ ⎪ ⎪ ⎩ δ 1, 0, 3
3
(4.15)
0, 3
so that there is a unique solution (u(t), v+ (t), v− (t)) ∈ X 0, 8 × X+ 8 × X− 8 to system (4.8)–(4.9) with the initial data (u0 , v±0 ). Then, by Lemma 2.1 again, we have 1 uL2 u0 L2 + Cδ 8 v+ u 0,− 38 + Cv− u 0,− 38 + N X X 1 2 u0 L2 + δ v+ 0, 38 + v− 0, 38 u 0, 38 + N X+
u0 L2
X
X−
3 + δ 4 v+0 L2 + v−0 L2 u0 L2 + L + N.
(4.16)
And 1 n± L2 v±0 L2 + Cδ 8 |u|2
0,− 3 8
X±
7 + Cδ 8 v+
0, 3 8
X+
+ v−
0, 3
X− 8
1 v±0 L2 + δ 2 u2 0, 3 + δ v+0 L2 + v−0 L2 8 X 2 3 4 v±0 L2 + δ u0 L2 + L + δ v+0 L2 + v−0 L2 .
(4.17)
Hence (u, v± ) is a solution to (4.8)–(4.9) in L2 × L2 for t ∈ [0, δ ∧ δL ]. On the other hand, from Remark 2.5 (2) and Bienaymé–Chebyshev inequality, we can see that
1 P δL (ω) 2
Cφ2 0,s 1 1 L2 8 P ∃δ s.t. δ , ψδ Φ 0, 38 > δ L . 2 X 2 L
(4.18)
Denote ΩL = {ω: δL (ω) 12 }. So for ω ∈ ΩLc , δ ∧ δL = δ. Then, for ω ∈ ΩLc and N above, define another stopping time δN by δN (ω) = inf δ > 0, Φ(t) L2 > N, t ∈ [0, δ] . And for any M 1, define ⎧ 0 if u0 L2 + v±0 L2 M, ⎪ ⎪ ⎪ ⎪ ⎨ inf{t > 0: uLMN (t) 2 + v±LMN (t) 2 M } if u0 2 + v±0 2 < M, L L L L δM (ω) = ⎪ 2 ∞ if u + v ⎪ 0 L ±0 L2 < M, ⎪ ⎪ ⎩ and the set {t > 0: uLMN (t)L2 + v±LMN (t)L2 M } is empty,
(4.19)
where (uLMN (t), v±LM N (t)) is the solution to (4.8)–(4.9) with the initial data (u0 , v±0 ), for all t ∈ [0, δM (ω)∧ δN (ω)]. Let δM N = min{δM , δN }.
S. Zhong / J. Math. Anal. Appl. 415 (2014) 217–239
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Then, by the pathwise uniqueness
uL1M 1N 1 (t), v±L1M 1N 1 (t) ≡ uL2M 2N 2 (t), v±L2M 2N 2 (t) ,
on [0, δM 1N 1 ] ∩ [0, δM 2N 2 ].
Thus δM N is nondecreasing in M . Set δ∞ (ω) = lim
lim
lim δM N (ω),
L→∞ N →∞ M →∞
and for any t ∈ [0, δ∞ ), let (u(t), v± (t)) = (uLMN (t), v±LM N (t)), with t ∈ [0, δM N ]. Therefore, (u(t), v± (t)) is the solution to (1.4) with the initial data (u0 , v±0 ), for all t ∈ [0, δ∞ (ω)). Claim 4.1. Let ω ∈ ΩLc and L, N ∈ N be fixed. For each T 0, sup t∈[0,δM N ∧T ]
2 2 v± (t) 2 2CT (L +supt∈[0,δM N ∧T ] u(t) L2 ) . L
(4.20)
Proof. If for all t ∈ [0, δM N ∧ T ] v± (t)
L2
C
u(t)
sup
L2
t∈[0,δM N ∧T ]
+L ,
(4.21)
then (4.20) is obvious. Otherwise, there is some t0 ∈ [0, δM N ∧ T ], such that v± (t)
sup
L2
t∈[0,δM N ∧t0 )
u(t)
sup t∈[0,δM N ∧T ]
L2
+ L,
and v± (t0 )
L2
∼
sup t∈[0,δM N ∧T ]
u(t) 2 + L. L
Choose this t0 to be the starting point. Then, by the local well-posed result, there is some δ < for t ∈ [t0 , t0 + δ], there is a unique pathwise solution to (4.8)–(4.9) in X u
3
X 0, 8
1 Cδ 8 u(t0 ) L2 + L , v+
0, 3 X+ 8
+ v−
0, 3 X− 8
0, 38
0, 38
0, 38
× X+ × X−
1 2
such that
satisfying,
1 Cδ 8 v+ (t0 ) L2 + v− (t0 ) L2 .
(4.22)
Here, δ should satisfy (4.22) with u(t0 )L2 + L v+ (t0 )L2 + v− (t0 )L2 , which requires that δ ∼ 4 (v+ (t0 )L2 + v− (t0 )L2 )− 3 . By (4.17), for t ∈ [t0 , t0 + δ] there exists, v+ (t) 2 + v− (t) 2 v+ (t0 ) 2 + v− (t0 ) 2 L L L L v+ (t0 ) L2 + v− (t0 ) L2 v+ (t0 ) L2 + v− (t0 ) L2 v+ (t0 ) 2 + v− (t0 ) 2 L
L
2 3 + δ 4 u(t0 ) L2 + L + δ v+ (t0 ) L2 + v− (t0 ) L2 2 3 1 + δ 4 u(t0 ) L2 + L + δ 4 2 1 + Cδ 4 1 + u(t0 ) L2 + L 1 u(t) 2 2 . + Cδ 4 L2 + sup (4.23) L t∈[0,δM N ∧T ]
S. Zhong / J. Math. Anal. Appl. 415 (2014) 217–239
238
1
Hence the increment of L2 norm of (v+ , v− ) from t0 to t0 + δ is Cδ 4 (L2 + supt∈[0,δM N ∧T ] u(t)2L2 ), and it takes m steps for the left hand side of (4.23) to double, where m= So the total time is δm =
v+ (t0 )L2 + v− (t0 )L2 . Cδ (L2 + supt∈[0,δM N ∧T ] u(t)2L2 ) 1 4
δ( v+ (t0 ) L2 + v− (t0 ) L2 ) 1
Cδ 4 (L2 +supt∈[0,δ
M N ∧T ]
u 2L2 )
∼
1 L2 +supt∈[0,δ
of (v+ (t0 )L2 + v− (t0 )L2 ). Therefore, before we reach time T , we need to carry out
T δm
M N ∧T ]
u(t) 2L2
, which is independent T δm 2 ∧T { u(t) L2 })
iterations. Since 2
supt∈[δM N ∧T ] {u(t)2L2 }), the left hand side of (4.23) is smaller than 2T (L +supt∈δM N initial one. Hence 2 2 v+ (t) 2 + v− (t) 2 C v+ (t0 ) 2 + v− (t0 ) 2 2T (L +supt∈[0,δM N ∧T ] u(t) L2 ) L L L L 2 2 C u(t0 ) L2 + L 2T (L +supt∈[0,δM N ∧T ] u(t) L2 ) 2 2 u(t) 2 + L 2T (L +supt∈[0,δM N ∧T ] u(t) L2 ) C sup L
T (L2 + times the
t∈[0,δM N ∧T ]
2CT (L
2
+supt∈[0,δ
M N ∧T ]
u(t) 2L2 )
,
(4.24)
for t ∈ [0, δM N ∧ T ]. Therefore the claim is proved. 2 Since by Lemma 2.6, for T 0 fixed, we have u(t) 2 2 2u0 2 2 + T C(1 + T ). E sup L L
(4.25)
t∈[0,δM N ∧T ]
Then from (4.20), we have sup
t∈[0,δM N ∧T ]
ln v± (t) L2 CT L2 +
sup t∈[0,δM N ∧T ]
u(t) 2 2 , L
(4.26)
so there exists E
sup t∈[0,δM N ∧T ]
ln v± (t) L2 CT L2 + u0 2L2 + T C L2 + T 2 .
(4.27)
We can see that the right hand side of (4.25) and (4.27) only depends on T 0 and L ∈ N, but is independent of M , hence, with the help of Bienaymé–Tchebysheff’s inequality and Theorem 6.10 of [7], we have v+ (t) 2 + v− (t) 2 ln M u(t) 2 2 M 2 + P P (δM N T ) P sup ln sup L L L t∈[0,δM N ∧T ]
·P
T 1 2
t∈[0,δM N ∧T ]
ΩL type sets + P 2
2
sup t∈[0,δM N ∧T ]
Φ(t) 2 2 > N 2 · P T ΩL type sets 1 L 2
C(1 + T ) C(L + T ) CT C(T ) CT + + 2 2 M ln M L N 2 L2
(4.28)
for all M ∈ N. So we obtain lim P (δM N T ) C(T )
M →∞
1 1 → 0, N 2 L2
as N → ∞ and L → ∞.
Therefore δ∞ = ∞ for almost all ω, which shows the global well-posedness.
2
S. Zhong / J. Math. Anal. Appl. 415 (2014) 217–239
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