# Frequency-Uniform Decomposition, Function SpacesXp,qs, and Applications to Nonlinear Evolution Equations

Authors: Shaolei Ru; Jiecheng Chen
Journal: Journal of Function Spaces and Applications
 (2013)

---

## Abstract

By combining frequency-uniform decomposition withLp(ℓq), we introduce a new class of function spaces (denoted by Xp,qs). Moreover, we study the Cauchy problem for the generalized NLS equations and Ginzburg-Landau equations in Lr(0,T;Xp,1s).


---

## Body

## 1. Introduction and Notation

In this paper, we introduce a new space (denoted byXp,qs) by combining frequency-uniform decomposition with Lp(ℓq). In this new space, we discuss the semilinear estimates, Schwartz estimates, embedding properties between Xp,qs and Triebel-Lizorkin space Fp,qs, and so forth. Moreover, we study the well-posedness of NLS equations and Ginzburg-Landau equations in Lr(0,T;Xp,1s).In recent decades, the well-posedness of Schrödinger equations and Ginzburg-Landau equations is developed quickly. In this paper, we mainly study the local Cauchy problem of NLS equations and Ginzburg-Landau equations in this new spaceXp,qs. Many people have studied these problems in Bessel potential and Besov spaces, for example, see [1–9], and so forth. We know that Besov spaces and Triebel spaces are generated by combining Littlewood-Paley decomposition with ℓq(Lp) and Lp(ℓq) (one can refer to [10] for the properties of Besov and Triebel spaces.). Next, we give a brief introduction of these spaces (see [10, 11]).To define the Littlewood-Paley decomposition, it is convenient to consider a radial functionψ:ℝn→[0,1] such that
(1)ψ(ξ)={1,|ξ|≤1;smooth,1<|ξ|<2;0,|ξ|≥2,
and to define the sequence {φk}0∞ by
(2)φk(ξ)=φ(2-kξ)-φ(2-k+1ξ),k∈ℕ;φ0(ξ)=1-∑k=1∞φk(ξ)=ψ(ξ).
Since suppφ⊂{ξ:2-1≤|ξ|≤2}, we can easily see that suppφk⊂{ξ:2k-1≤|ξ|≤2k+1}, k∈ℕ, and suppφ0⊂{ξ:|ξ|≤2}. Define
(3)Δk=F-1φkF,k∈ℕ∪{0},
where F is the Fourier transform and F-1 is the inverse Fourier transform. {Δk}0∞ is said to be Littlewood-Paley decomposition operator. Formally, we see that
(4)∑k=0∞Δk=I,in𝒮′.
We now define the Besov space Bp,qs(ℝn) for -∞<s<∞ and 0<p,q≤∞ by
(5)Bp,qs(ℝn)={f∈S′(ℝn):∥f∥Bp,qs<∞},∥f∥Bp,qs:=(∑k=0∞2ksq∥Δkf∥pq)1/q,
and Triebel space Fp,qs(ℝn) for -∞<s<∞ and 0<p<∞, 0<q≤∞ by
(6)Fp,qs(ℝn)={f∈S′(ℝn):∥f∥Fp,qs<∞},∥f∥Fp,qs:=∥(∑k=0∞2ksq|Δkf|q)1/q∥p.Besov and Triebel spaces had been formulated during 1960s–1980s, which have been widely applied in recent years. Besov spacesBp,qs and Triebel spaces Fp,qs are introduced by combining Littlewood-Paly decomposition operator with the function spaces ℓq(Lp) and Lp(ℓq). Similar to Besov spaces and Triebel spaces, one can use frequency-uniform decomposition and ℓq(Lp), Lp(ℓq) to generate a new class of function spaces. Actually, the spaces introduced by frequency-uniform decomposition and ℓq(Lp) are modulation spaces. Many people have studied the well-posedness of evolution equations in these spaces (e.g., see [12–16], etc.). In this paper, we will consider the spaces generated by the frequency-uniform decomposition and Lp(ℓq). Firstly, we will recall the definition of the frequency-uniform decomposition and modulation spaces. In the 1930s, Wiener [17] first introduced the frequency-uniform decomposition. So, sometimes we call it Wiener decomposition of ℝn that is roughly denoted by
(7)□k~F-1χQkF,k∈ℤn,
where χE denotes the characteristic function on the set E. Since Qk is a translation of Q0, □k have the same localized structures in frequency space, which are said to be the frequency-uniform decomposition operators. Since χQk can not make differential operations, one needs to redefine it by smooth truncation function (see [12, 14, 18, 19]).Now we give an exact definition on modulation spaces. We first denote|ξ|∞:=maxi=1,...,n|ξi|. Let ρ∈S(ℝn):ℝn→[0,1] be a smooth radial bump function adapted to {ξ:|ξi|∞⩽1} with ρ(ξ)=1 if |ξ|∞⩽1/2 and ρ(ξ)=0 if |ξ|∞⩾1. Let ρk be a translation of ρ:
(8)ρk(ξ)=ρ(ξ-k),k∈ℤn.
Since ρk(ξ)=1 in the unit closed cube Qk:{ξ∈ℝn:|ξ-k|∞⩽1/2} and {Qk}k∈ℤn is a covering of ℝn, one has that for all ξ∈ℝn, ∑k∈ℤnρk(ξ)⩾1. Then, we write
(9)σk(ξ)=ρk(ξ)(∑k∈ℤn‍ρk(ξ))-1,k∈ℤn.
It is easy to see that
(***)|σk(ξ)|≥c,ξ∈Qk;suppσk⊂{ξ:|ξ-k|∞≤1};∑k∈ℤnσk(ξ)≡1.∀ξ∈ℝn;|Dασk(ξ)|≤C|α|,∀ξ∈ℝn,α∈(N∪{0})n.
Hence, the set
(10)Υn={{σk}k∈ℤn:{σk}k∈ℤnsatisfy(***)}
is nonvoid. Let {σk}k∈ℤn∈Υn be a function sequence. Denote
(11)□k:=F-1σkF,k∈ℤn,
which is said to be the frequency-uniform decomposition operator. For any k∈ℤn, we set |k|=|k1|+⋯+|kn|, 〈k〉=1+|k|. For any s∈ℝ, 0<p,q≤∞, we denote
(12)Mp,qs(ℝn)={(∑k∈ℤn‍〈k〉sq∥□kf∥Lpq)1/qf∈S′(ℝn):Mp,qs(ℝn)515=∥f∥Mp,qs=(∑k∈ℤn‍〈k〉sq∥□kf∥Lpq)1/q<∞}.Mp,qs:=Mp,qs(ℝn) is said to be a modulation space, which was first introduced by Feichtinger [19] in the case 1⩽p,q⩽∞ (Appendix Theorem E shows the properties on Mp,qs(ℝn)). For any 0⩽s<∞, 0<p,q⩽∞, we denote
(13)Ep,qs(ℝn)={(∑k∈ℤn‍2|k|sq∥kf∥Lpq)1/qf∈S′(ℝn):Ep,qs(ℝn)551=∥f∥Ep,qs=(∑k∈ℤn‍2|k|sq∥□kf∥Lpq)1/q<∞}.Ep,qs:=Ep,qs(ℝn) was first introduced by Baoxiang et al. [12].We now introduce a new class of function spaces (denoted byXp,qs) by combining the frequency-uniform decomposition with Lp(ℓq). For any 0<p<∞, 0<q⩽∞, 0⩽s<∞, we denote
(14)Xp,qs(ℝn):={∥(∑k∈ℤn‍2|k|sq∥kf∥q)1/q∥Lpf∈S′(ℝn):Xp,qs(ℝn):551=∥f∥Xp,qs=∥(∑k∈ℤn‍2|k|sq|□kf|q)1/q∥Lp<∞}.
Moreover, we denote
(15)Lr(ℝ,Xp,qs):={f∈𝒮′:(∫ℝ‍∥f∥Xp,qsrdt)1/r<∞}.Main ResultsTheorem 1.
Assumef(u)=u|u|k, k=2m, m∈ℤ+, and r⩾k+2; then for any initial data u0∈X1,1s, 0⩽s<∞, there exists T*:=T*(∥u0∥X1,1s)>0 such that the initial value problem
(16)u(t)=S(t)u0-i∫0t‍S(t-τ)f(u(τ))dτ,(NLSequation)
has a unique solution
(17)u∈Llocr(0,T*;X1,1s).
Moreover, if T*<∞, then
(18)∥u∥Lr(0,T*;X1,1s)=∞,
where S(t)=F-1e-it|ξ|2F.Theorem 2.
Assumef(u)=(b+iβ)u|u|k+μu, k=2m, m∈ℤ+, and r⩾k+2. Assume also a,b>0, α,β,μ∈ℝ. Then, for any initial data u0∈Xp,1s, 1⩽p<∞, 0⩽s<∞, there exists T*:=T*(∥u0∥Xp,1s)>0 such that the initial value problem (complex Ginzburg-Landau equation)
(19)u(t)=U(t)u0-i∫0t‍U(t-τ)f(u(τ))dτ
has a unique solution
(20)u∈Llocr(0,T*;Xp,1s).
Moreover, if T*<∞, then
(21)∥u∥Lr(0,T*;Xp,1s)=∞,
where U(t)=F-1e-tP(ξ)F, P(ξ)=1+(a+iα)|ξ|2.This paper is organized as follows. In Section2, we will state some properties on Xp,qs, which is useful to establish the following inclusions. Section 3 is devoted to considering the multiplication algebra of Xp,qs. Some dispersive smooth effects for the Schrödinger and Ginzburg-Landau semigroups will be given in Section 4. Theorems 1 and 2 will be proved in Section 5.


## 2. The Properties onXp,qs

In order to study the Cauchy problem inXp,qs, we first give some properties on Xp,qs as follows.Proposition 3.
Letting0⩽s<∞, 0<p<∞, 0<q⩽∞, one has(1)
ifq1≤q2, then(22)Xp,q1s⊂Xp,q2s;(2)
ifε>0, then(23)Xp,q1s+ε⊂Xp,q2s;(3)
ifp<∞, then(24)Ep,p∧qs⊂Xp,qs⊂Ep,p∨qs,(p∧q=min(p,q),Ep,p∧qs⊂Xp,qs⊂Ep,p∨qs,p∨q=max(p,q)).Proof.
Becauseℓp⊂ℓp+a, a≥0, we can get the results of (1) directly. For (2), we know that
(25)(∑k∈ℤn‍2|k|sq2|ak|q2)1/q2=(∑k∈ℤn‍2-|k|εq22|k|(s+ε)q2|ak|q2)1/q2≲supk∈ℤn2|k|(s+ε)|ak|,(ε>0).
Then, lettingak=□kf, we get (2).
Finally, we prove (3). Letbk=2|k|s□kf. The proof can be classified into the following two kinds of cases.
Case  1
(
q
≤
p
)
. In this case, ℓq⊂ℓp. We have
(26)∥bk∥ℓp(Lp)≤∥bk∥Lp(ℓq)≤∥bk∥ℓq(Lp).
Actually, noticing that ℓq⊂ℓp, we have ∥bk∥ℓp(Lp)=∥bk∥Lp(ℓp)≤∥bk∥Lp(ℓq). So, the first part of the above inequality holds. On the other hand, by Minkowski’s inequality, we have
(27)∥bk∥Lp(ℓq)=∥∑k=0∞‍|bk|q∥Lp/q1/q⩽(∑k=0∞‍∥|bk|q∥Lp/q)1/q=∥bk∥ℓq(Lp).
This proves the second part.
Case  2
(
p
≤
q
)
. By Minkowski’s inequality and ℓp⊂ℓq, we have
(28)∥bk∥ℓq(Lp)=(∥∫ℝn‍|bk|pdx∥ℓq/p)1/p⩽(∫ℝn‍∥|bk|p∥ℓq/pdx)1/p=∥bk∥Lp(ℓq)⩽∥bk∥ℓp(Lp).
This finishes the proof of this proposition.Proposition 4 (completeness).
For any0⩽s<∞, 0<p<∞, 0<q⩽∞, one has the following:(1)
Xp,qsis a quasi-Banach space. Moreover, if1⩽p<∞,1⩽q⩽∞, thenXp,qsis a Banach space;(2)
S(ℝn)⊂Xp,q0⊂S′(ℝn);(3)
if0<p,q<∞, then𝒮(ℝn)is dense inXp,q0.Proof
Step  1. In this step, we prove that S(ℝn)⊂Xp,q0⊂S′(ℝn). Actually, by S(ℝn)⊂Ep,q0⊂S′(ℝn) [12] and Ep,p∧qs⊂Xp,qs⊂Ep,p∨qs (Proposition 3), we immediately obtain the result.
Step 2.
X
p
,
q
s is a quasinormed space. We prove the completeness. Let {fℓ}ℓ=1∞ be a Cauchy sequence in Xp,qs (with respect to a fixed quasinorm in Xp,qs). Proposition 3 shows that {fℓ}ℓ=1∞ is also a Cauchy sequence in Ep,p∨qs. Because Ep,p∨qs is a complete topological linear space, we can find a limit element f∈Ep,p∨qs. Then, F-1σkFfℓ converges to F-1σkFf in Ep,p∨qs if ℓ→∞. On the other hand, {F-1σkFfℓ}ℓ=1∞ is a Cauchy sequence in Lp(ℝn) (Xp,qs⊂Ep,p∨qs, s⩾0.). By Appendix Theorem B, it is also a Cauchy sequence in L∞(ℝn). This shows that the limiting element of {F-1σkFfℓ}ℓ=1∞ in Lp(ℝn) (which is the same as in L∞(ℝn)) coincides with {F-1σkFf}. Now it follows by standard arguments that f belongs to Xp,qs and that fℓ converges in Xp,qs to f. Hence, Xp,qs is complete.
Step 3. We prove that 𝒮(ℝn) is dense in Xp,q0(ℝn) if 0<p<∞ and 0<q<∞. Let f∈Xp,q0; then we put
(29)fN(x)=∑k=0NF-1σkFf.
Of course, fN∈Xp,q0(ℝn). Consequently (by Appendix Theorem F),
(30)∥f-fN∥Xp,q0(ℝn)⩽c∥(∑k=N∞∑r=-11|F-1σkσk+rFf|q)1/q∥Lp(ℝn)⩽c∥(∑k=N∞|F-1σkFf|q)1/q∥Lp(ℝn).
Lebesgue’s bounded convergence theorem proves that the right-hand side of the above inequality tends to zero if N→∞. Hence, fN approximates f in Xp,q0(ℝn). Next, we let φ∈𝒮 with φ(0)=1 and suppFφ⊂{y:|y|⩽1}. Let fδ(x)=φ(δx)f(x) with 0<δ<1. Then, (fN)δ∈𝒮(ℝn) approximates fN in LpΩ:={f∈Lp:suppf^⊂Ω} with Ω={y:|y|⩽2N+2} if δ→0. However this is also an approximation of fN in Xp,q0(ℝn). This proves that 𝒮(ℝn) is dense in Xp,q0(ℝn).Proposition 5 (equivalent norm).
Assume{σk}k∈ℤn, {ϕk}k∈ℤn∈Υn, 0<p<∞, 0<q⩽∞. Then, {σk}k∈ℤn and {ϕk}k∈ℤn generate equivalent norms on Xp,qs.Proof.
Firstly, we have the following translation equality:(31)(F-1mFf)(x)=eixk[F-1m(·+k)F(e-ikyf(y))](x).
For the sake of convenience, we denote
(32)□kσ=F-1σkF,□kϕ=F-1ϕkF.
Then, we have
(33)□kσ=∑|ℓ|∞≤1‍□kσ□k+ℓϕ,∥□kσf∥Lp(ℓq)≤∥∑|ℓ|∞≤1‍□kσ(□k+ℓϕf)∥Lp(ℓq).
By Appendix Theorem F, we have
(34)∥□kσ(□k+ℓϕf)∥Lp(ℓq)≲supk∥σk∥Hs∥□k+ℓϕf∥Lp(ℓq),s>n(11∧p-12).
Then, by ∥σk∥Hs<C, we have
(35)∥□kσf∥Lp(ℓq)≲∑|ℓ|∞≤1‍∥□k+ℓϕf∥Lp(ℓq).
By the above discussion, we get ∥f∥Xp,q0{σk}≲∥f∥Xp,q0{ϕk}.The inverse inequality can be proved similarly. Based on the above observations, one can also obtain that∥f∥Xp,qs{σk}~∥f∥Xp,qs{ϕk}.Theorem 6.
Assume1⩽p2⩽p1<∞, 1⩽q⩽∞, 0⩽s<∞; then one has
(36)∥u∥Xp1,qs≲∥u∥Xp2,qs.Proof.
By the definition ofXp,qs, we have
(37)∥u∥Xp1,qs=∥(∑i∈ℤn‍2|i|qs|□iu|q)1/q∥Lp1=∥(∑i∈ℤn‍2|i|qs|□i∑|ℓ|∞⩽1‍□i+ℓu|q)1/q∥Lp1=∥(∑i∈ℤn‍2|i|qs|σi∨*(∑|ℓ|∞⩽1‍□i+ℓu)|q)1/q∥Lp1⩽∥(∑i∈ℤn‍2|i|qs∥σi∨|*|(∑|ℓ|∞⩽1‍□i+ℓu)∥q)1/q∥Lp1=∥(∑i∈ℤn‍2|i|qs∥σ0∨|*|(∑|ℓ|∞⩽1‍□i+ℓu)∥q)1/q∥Lp1≲∥(∑i∈ℤn‍2|i|qs|(∑|ℓ|∞⩽1‍□i+ℓu)|q)1/q∥Lp2≲∥(∑i∈ℤn‍2|i|qs|□iu|q)1/q∥Lp2.
In this proof, we used∥|σ0∨|*f∥Lp1⩽∥f∥Lp2 (p1⩾p2, Young’s inequality) and Appendix Theorem G.Theorem 7.
Assumef∈L2; then one has
(38)∥f∥L2~∥f∥X2,20.Proof.
We first prove∥f∥L2≲∥f∥X2,20. Consider the following:
(39)∥f∥L2=(∫ℝn‍|∑i∈ℤn□if|2dx)1/2=(∫ℝn‍|∑i∈ℤnσiFf|2dx)1/2≲(∫ℝn‍∑i∈ℤn|σiFf|2dx)1/2=(∫ℝn‍∑i∈ℤn|F-1σiFf|2dx)1/2=(∫ℝn‍((∑i∈ℤn|F-1σiFf|2)1/2)2dx)1/2=∥f∥X2,20.
The inverse inequality can be proved similarly (∑i∈ℤn‍|σiFf|2≲|∑i∈ℤn‍σiFf|2). Moreover, by the similar discussion as mentioned previously, we can also obtain that ∥(∑k∈ℤn‍|F-1(1+|ξ|2)s/2σkFf|2)1/2∥L2~∥f∥H2s.Theorem 8.
Let0<s<∞, 0<p<∞, 0<q⩽∞. Then, one has
(40)Xp,qs(ℝn)⊂C∞(ℝn).Proof.
One can obtain the result directly by Appendix Theorem D and Proposition3.Theorem 9.
Let0<p<∞, 0<q⩽1. One has
(41)Xp,q0(ℝn)⊂Fp,q0(ℝn).Proof.
Letak=max(0,2k-1-n), bk=2k+1+n. We can easily find that if |i|∈-[ak,bk], then Δk□if=0. For p>0, q⩽1, by Appendix Theorem F, we have
(42)∥Δkf∥Lp(ℓq)=∥(∑k|Δkf|q)1/q∥Lp⩽∥(∑k|∑i∈ℤn,|i|∈[ak,bk]Δk□i□if|q)1/q∥Lp≲∥(∑k∑i∈ℤn,|i|∈[ak,bk]|F-1(φkσi)F□if|q)1/q∥Lp≲∥(∑i∈ℤn∑k,|i|∈[ak,bk]|F-1(φkσi)F□if|q)1/q∥Lp≲∥(∑i∈ℤn|□if|q)1/q∥Lp.
This proves the theorem.Theorem 10.
LetFp,qs(ℝn) be Triebel space; then one has
(43)F2,qs(ℝn)⊂X2,q0(ℝn),s>n(1q-12),F2,qs(ℝn)5.5⊂X2,q0(ℝn),0<q<2,X2,q0(ℝn)⊂F2,qs(ℝn),s<n(1q-12),X2,q0(ℝn)5.5⊂F2,qs(ℝn),2<q⩽∞.Proof.
By Appendix Theorem A, we know that(44)Hs⊂E2,q0;s>n(1q-12),0<q<2.
Then, by the embedding between Ep,qs and Xp,qs (Proposition 3), we have (for any given ε>0)
(45)F2,qn(1/q-1/2)+ε⊂F2,2n(1/q-1/2)+ε=Hn(1/q-1/2)+ε⊂E2,q0⊂X2,q0.
This proves the first result (F2,2s~Hs, [11]).
On the other hand, by Appendix Theorem A, we have(46)E2,q0⊂Hs;s<n(1q-12),2<q⩽∞.
Then, by the embedding betweenEp,qs and Xp,qs (Proposition 3), we have
(47)X2,q0⊂E2,q0⊂Hn(1/q-1/2)-ε⊂F2,qn(1/q-1/2)-ε.
This implies the theorem.Theorem 11.
Lets>n(1/q-1/2), 0<q<2. Then, one has
(48)∥f∥X2,q0⩽Cs(1+[ln(1+∥f∥Hs)](1/q-1/2)∥f∥Hn(1/q-1/2)).Proof.
By the definition ofXp,qs, we have
(49)∥f∥X2,q0=∥(∑k∈ℤn‍|□kf|q)1/q∥L2≲∥(∑|k|⩽T‍|□kf|q)1/q∥L2+∥(∑|k|>T‍|□kf|q)1q∥L2=A+B.
Then, by Appendix Theorem F, Theorem 7 and Hölder’s inequality, the term A can be dominated by
(50)A=∥×(1+|ξ|2(1+|k|)2)n(1/q-1/2)/2σkFf|q)1/q∥L2(∑|k|⩽T‍|F-1((1+|k|)21+|ξ|2)n(1/q-1/2)/2=(∑|k|⩽T‍|F-151×(1+|ξ|2(1+|k|)2)n(1/q-1/2)/2σkFf|q)1/q∥L2≲∥(∑|k|⩽T‍|F-1(1+|ξ|2(1+|k|)2)n(1/q-1/2)/2σkFf|q)1/q∥L2≲(∑|k|⩽T‍(1+|k|)-n)(1/q-1/2)×∥(∑|k|⩽T|F-1(1+|ξ|2)n(1/q-1/2)/2σkFf|2‍)1/2∥L2≲(∑0⩽i⩽T‍(1+i)-1)(1/q-1/2)∥f∥Hn(1/q-1/2)≲[ln(1+T)](1/q-1/2)∥f∥H(n(1/q-1/2)).
On the other hand, by the similar discussion as the above, we have
(51)B≲(∑|k|>T‍(1+|k|)-2sq/(2-q))(1/q-1/2)×∥(∑|k|>T‍|F-1(1+|ξ|2)s/2σkFf|2)1/2∥L2≲(∑i>T‍∑|k|=i‍(1+i)-2sq/(2-q))(1/q-1/2)∥f∥Hs≲(∑i>T‍(1+i)-2sq/(2-q)+n-1)(1/q-1/2)∥f∥Hs⩽CsT-s+n(1/q-1/2)∥f∥Hs.
Let
(52)T=max(1,∥f∥Hs1/(s-n(1/q-1/2))).
From the above discussion, we get the result.Remark 12.
By Theorem9, we have
(53)Xp,10(ℝn)⊂Fp,10(ℝn).
Then, by Theorem 11, we have
(54)Hs(ℝn)⊂X2,10(ℝn)⊂F2,10(ℝn),s>n2.
So, X2,10 is an intermediate space between Hs(s>n/2) and F2,10.


## 3. Multilinear Estimate inXp,qs

It is well known thatBp,qs is a multiplication algebra if s>n/p, see [2]. The regularity indices, for which Xp,qs constitutes a multiplication algebra, are quite different from those of Besov space.Theorem 13.
Let0⩽s<∞, 0<p⩽p1, p2<∞, 0<q⩽∞. If 1/p=1/p1+1/p2, then one has
(55)∥fg∥Xp,qs≲C2Cs∥f∥Xp1,q∧1s∥g∥Xp2,q∧1s,
where C is independent of s,q.Proof.
Recall that we denotek∈ℤn and |k|=|k1|+⋯+|kn|. We have
(56)∥fg∥Xp,qs=∥(∑k∈ℤn‍2|k|sq|□k(fg)|q)1/q∥Lp.
By ∑k∈ℤn‍σk=1, we can obtain the decomposition of fg as follows:
(57)fg=∑i,j∈ℤn‍(□if)(□jg).
Then, we have
(58)□k(fg)=∑i,j∈ℤn‍□k(□if)(□jg).
It is easy to see that F(□if□jg)=(σif^)*(σjg^) and
(59)supp(σif^)*(σjg^)⊂{ξ:|ξ-i-j|⩽22n}.
Then, we have
(60)□k(□if□jg)=□k(∑i∈ℤn‍□if□j(i,k)g),|k-i-j(i,k)|⩽C(n).
Case  1
(
q
⩽
1
)
. Based on the previous observations and Appendix Theorem F, we have
(61)∥fg∥Xp,qs=∥(∑k∈ℤn‍2|k|sq|□k(fg)|q)1/q∥Lp⩽∥(∑k∈ℤn‍2|k|sq|□k(∑i∈ℤn‍□if□j(i,k)g)|q)1/q∥Lp⩽∥(∑k∈ℤn‍∑i∈ℤn‍2|k|sq|□if□j(i,k)g|q)1/q∥Lp⩽∥(∑k∈ℤn‍∑i∈ℤn‍2|k-i-j(i,k)|sq|(2|i|s□if)(2|j(i,k)|s□j(i,k)g)|q)1/q∥Lp≲∑t=0C(n)∥(∑k∈ℤn‍∑i∈ℤn,j=k-i±t‍2|k-i-j|sq‍≲∑t=0C(n)∥(∑k∈ℤn‍∑i∈ℤn,j=k-i±t‍2‍×|(2|i|s□if)(2|j|s□jg)|q∑k∈ℤn‍∑i∈ℤn,j=k-i±t‍2|k-i-j|sq)1/q∥Lp⩽C2Cs∥(∑i∈ℤn‍2|i|sq|□if|q)1/q(∑j∈ℤn‍2|j|sq|□jg|q)1/q∥Lp⩽C2Cs∥f∥Xp1,qs∥g∥Xp2,qs,
where we used |k-i-j(i,k)|<C(n) and the following Young’s inequality:
(62)∥ai*bj∥ℓp⩽∥ai∥ℓp1∥bj∥ℓp2,1+1p=1p1+1p2.
Case  2
(
q
>
1
)
. By the definition of Xp,qs and Appendix Theorem F, we have
(63)∥fg∥Xp,qs=∥(∑k∈ℤn‍2|k|sq|□k(fg)|q)1/q∥Lp⩽∥∑k∈ℤn‍2|k|s|□k(fg)|∥Lp,(ℓ1⊂ℓq,ifq⩾1)⩽∥∑k∈ℤn‍2|k|s|□k(∑i∈ℤn‍□if□j(i,k)g)|∥Lp⩽∥∑k∈ℤn‍2|k|s|∑i∈ℤn‍□if□j(i,k)g|∥Lp≲∑t=0C(n)‍∥∑k∈ℤn‍2|k|s|∑i∈ℤn,j=k-i±t‍(□if□jg)|∥Lp⩽∑t=0C(n)‍∥∑k∈ℤn‍2|k-i-j|s|∑i∈ℤn,j=k-i±t‍(2|i|s□if)(2|j|s□jg)|∥Lp⩽C2Cs∥∑i∈ℤn‍|(2is□if)|∑j∈ℤn‍|(2js□jg)|∥Lp⩽C2Cs∥f∥Xp1,1s∥g∥Xp2,1s.
By the above discussion, we can obtain the claimed results.Corollary 14.
Let0⩽s<∞, 0<p⩽pi<∞, i=1,…,N, 0<q⩽∞. If 1/p=1/p1+⋯+1/pN, then one has
(64)∥∏i=1N‍ui∥Xp,qs≲CN2CNs∏i=1N‍∥ui∥Xpi,q∧1s,
where C is independent of s,q.Proof.
Let1/p=1/p1+1/p*, where 1/p*=1/p2+⋯+1/pN. Then, we have
(65)∥∏i=1N‍ui∥Xp,qs⩽C2Cs∥u1∥Xp1,q∧1s∥∏i=2N‍ui∥Xp*,q∧1s.
By induction, we can easily see that the claimed result holds.


## 4. Smooth Effects of the Schrödinger and Ginzburg-Landau Semigroups

In this section, we will discuss a kind of Strichartz’s estimates. This kind of estimates is first introduced by Strichartz in [20], then developed by [18, 21, 22], and so forth.Put(66)S(t)=F-1e-it|ξ|2F,U(t)=F-1e-tP(ξ)F,P(ξ)=1+(a+iα)|ξ|2,U(t)=F-1e-tP(ξ)F,5551P(ξ)a>0,α∈ℝ.
Our aim is to derive the estimates of S(t), U(t) in the spaces Xp,qs.Theorem 15.
Assume2⩽p<∞, p′⩽q⩽p, 1/p+1/p′=1, 0⩽s<∞; then one has
(67)∥S(t)f∥Xp,qs≲(1+|t|)-n(1/2-1/p)∥f∥Xp′,qs.Proof
Step  1. By Appendix Theorem C, for 2⩽p<∞, we have
(68)∥□kS(t)f∥Lp≲(1+|t|)-n(1/2-1/p)∥□kf∥Lp′.
Thus, for0⩽s<∞, we have
(69)∥S(t)f∥Ep,qs≲(1+|t|)-n(1/2-1/p)∥f∥Ep′,qs.
Step  2. By Proposition 3, for p′⩽q⩽p, we have
(70)∥S(t)f∥Xp,qs⩽∥S(t)f∥Ep,qs⩽(1+t)-n(1/2-1/p)∥f∥Ep′,qs⩽(1+t)-n(1/2-1/p)∥f∥Xp′,qs.
This proves the theorem.Theorem 16.
Assumer⩾1, 0⩽s<∞, p′⩽q⩽p, 2⩽p<∞, Af=∫0t‍S(t-τ)f(τ)dτ; then one has
(71)∥Af∥Lr(-T,T;Xp,qs)≲T2/r∥f∥Lr′(-T,T;Xp′,qs).Proof.
By Theorem15, we have
(72)∥Af∥Lr(-T,T;Xp,qs)⩽∥∫0t‍∥S(t-τ)f(τ)∥Xp,qsdτ∥Lr(-T,T)⩽∥∫0t‍(1+|t-τ|)-n(1/2-1/p)∥f(τ)∥Xp′,qsdτ∥Lr(-T,T)≲∥∫0∞‍χτ∈[0,t]∥f(τ)∥Xp′,qsdτ∥Lr(-T,T)≲T1/r(∫-TT‍∥f(τ)∥Lr′(-T,T;Xp′,qs)rdt)1/r≲T2/r∥f∥Lr′(-T,T;Xp′,qs).
This proves the theorem.Theorem 17.
Letting0⩽s<∞, one has
(73)∥S(t)f∥X1,1s⩽(1+|t|)n/2∥f∥X1,1s,
which holds for all f∈X1,1s.Proof.
By [18] (Appendix Theorem H), we have
(74)∥□kS(t)f∥Lp≲(1+|t|)n|1/2-1/p|∥□kf∥Lp,1⩽p⩽∞.
Then, we have
(75)∥S(t)f∥X1,1s=∥∑k∈ℤn‍2|k|s|□kS(t)f|∥L1=∑k∈ℤn‍2|k|s∥□kS(t)f∥L1≲(1+|t|)n/2∑k∈ℤn‍2|k|s∥□kf∥L1≲(1+|t|)n/2∥f∥X1,1s.
This proves the theorem.Theorem 18.
Assumer⩾1,0⩽s<∞, Af=∫0t‍S(t-τ)f(τ)dτ; then one has
(76)∥Af∥Lr(-T,T;X1,1s)≲(1+T)n/2T2/r∥f∥Lr′(-T,T;X1,1s).Proof.
By Theorem17, we have
(77)∥Af∥Lr(-T,T;X1,1s)⩽∥∫0t∥S(t-τ)f(τ)∥X1,1sdτ‍∥Lr(-T,T)⩽∥∫0t‍(1+|t-τ|)n/2∥f(τ)∥X1,1sdτ∥Lr(-T,T)≲(1+T)n/2∥∫0∞‍χτ∈[0,t]∥f(τ)∥X1,1sdτ∥Lr(-T,T)≲(1+T)n/2T1/r(∫-TT‍∥f(τ)∥Lr′(-T,T;X1,1s)rdt)1/r≲(1+T)n/2T2/r∥f∥Lr′(-T,T;X1,1s).
This proves the theorem.Theorem 19.
Assumer⩾k+2, 0⩽s<∞. Assume also f(u)=u|u|k, k=2m, m∈ℤ+, A(f)=∫0t‍S(t-τ)f(u(τ))dτ. Then, one has
(78)∥A(f)∥Lr(-T,T;X1,1s)≲(1+T)n/2T1-k/r∥u∥Lr(-T,T;X1,1s)k+1.Proof.
By the definition, Theorems18 and 13, we have
(79)∥A(f)∥Lr(-T,T;X1,1s)=∥∫0t‍S(t-τ)f(u(τ))dτ∥Lr(-T,T;X1,1s)≲(1+T)n/2T2/r∥f∥Lr′(-T,T;X1,1s)≲(1+T)n/2T2/r(∫-TT‍∥f(u(τ))∥X1,1sr′dτ)1/r′≲(1+T)n/2T2/r(∫-TT‍∥u(τ)∥Xk+1,1s(k+1)r′dτ)1/r′≲(1+T)n/2T2/r(∫-TT‍∥u(τ)∥X1,1srdτ)(k+1)/rT(r-k-2)/r≲(1+T)n/2T1-k/r∥u∥Lr(-T,T;X1,1s)k+1.
This proves the theorem.Theorem 20.
Let0<p<∞, 0<q⩽∞, 0⩽s<∞; then
(80)∥U(t)f∥Xp,qs⩽C∥f∥Xp,qs,
which holds for all f∈Xp,qs.Proof
Step  1 (see [12]). Let L>n(n/min(1,p,q)-1/2) and L∈ℤ+; then we have ∥σke-tP(ξ)∥HL≲1. Actually, in view of the Leibnitz rule, we have
(81)|DL(σke-tP(ξ))|⩽C∑L1+L2=L|DL1σkDL2e-tP(ξ)|⩽C∑0⩽s⩽L|Dse-tP(ξ+k)|χ|ξ|∞⩽1‍.
Letβ=(β1,...,βq), and let Λqs={β:β1+⋯+βq=s,β1,...,βq⩾1}. It is easy to see that
(82)Dsef(η)=ef(η){∑q=1s‍∑Λqs‍Cβ∏i=1qDβif(ξ)}.
Letf(ξ)=-tP(ξ); one easily sees that
(83)|Dβif(ξ)|⩽Ct(1+|ξ+k|2),k∈ℤ.
Hence, we have(84)|Dsef(η)|⩽e-ct(1+|ξ+k|2)∑q=1s‍(t(1+|ξ+k|2))q⩽e-ct(1+|ξ+k|2).
Combining the previous discussion, we have(85)∥σke-tP(ξ)∥HL≲Ce-ct(1+|k|2).
Step  2. By Step 1, we have supk∥σke-tP(ξ)∥HL≲1. Then by Appendix Theorem F, we can obtain that
(86)∥U(t)f∥Xp,qs⩽Csupk∥σke-tP(ξ)∥Hs∥f∥Xp,qs≲∥f∥Xp,qs.
This proves the theorem.Theorem 21.
Assume1⩽p<∞, 1⩽q⩽∞, r⩾1,0⩽s<∞, Af=∫0t‍U(t-τ)f(τ)dτ; then one has
(87)∥Af∥Lr(-T,T;Xp,qs)≲T2/r∥f∥Lr′(-T,T;Xp,qs).Proof.
By Theorem20, we have
(88)∥Af∥Lr(-T,T;Xp,qs)⩽∥∫0t‍∥U(t-τ)f(τ)∥Xp,qsdτ∥Lr(-T,T)⩽∥∫0t‍∥f(τ)∥Xp,qsdτ∥Lr(-T,T)≲∥∫0∞‍χτ∈[0,t]∥f(τ)∥Xp,qs∥Lr(-T,T)≲T2/r∥f∥Lr′(-T,T;Xp,qs).
This proves the theorem.Theorem 22.
Assumeb>0, β,μ∈ℝ, f(u)=(b+iβ)u|u|k+μu,A(f)=∫0t‍U(t-τ)f(u(τ))dτ. Assume also k=2m, r⩾k+2, m∈ℤ+, 1⩽p<∞, 1⩽q⩽∞. Then, one has
(89)∥A(f)∥Lr(0,T;Xp,qs)≲T1-k/r∥u∥Lr(0,T;Xp,1s)k+1+T∥u∥Lr(0,T;Xp,qs).Proof.
By Theorems20 and 21 and the multilinear estimate, we have
(90)∥Af∥Lr(-T,T;Xp,qs)≲T2/r∥f∥Lr′(-T,T;Xp,qs)=T2/r(∫-TT∥f(u(τ))∥Xp,qsr′‍dτ)1/r′≲T2/r(∫-TT‍∥u|u|k∥Xp,qsr′dτ)1/r′+T2/r(∫-TT∥u(τ)∥Xp,qsr′‍dτ)1/r′≲T2/r(∫-TT‍∥u(τ)∥X(k+1)p,q∧1s(k+1)r′dτ)1/r′+T2/r(∫-TT‍∥u(τ)∥Xp,qsr′dτ)1/r′≲T2/r(∫-TT‍∥u(τ)∥Xp,1srdτ)(k+1)/rT(r-k-2)/r+T∥u∥Lr(0,T;Xp,qs)≲T1-k/r∥u∥Lr(0,T;Xp,1s)k+1+T∥u∥Lr(0,T;Xp,qs).
This proves the theorem.


## 5. Well-Posedness of Nonlinear Schrödinger Equation and Ginzburg-Landau Equation

In this section, we first study the well-posedness of the following Schrödinger equation:(91)iut+Δu=f(u);u(0,x)=u0(x).
The solution, u(x,t), of the above Cauchy problem is given by:
(92)u(t)=S(t)u0-i∫0t‍S(t-τ)f(u(τ))dτ,
where S(t)=F-1eit|ξ|2F. Then, we study the well-posedness of the following complex Ginzburg-Landau equation:
(93)ut-(a+iα)Δu+(b+iβ)|u|ku+(μ+1)u=0,u(0,x)=u0(x).
The solution, u(x,t), of the above Cauchy problem is given by
(94)u(t)=U(t)u0-i∫0t‍U(t-τ)f(u(τ))dτ,
where a,b>0, α,β,μ∈ℝ, U(t)=F-1e-tP(ξ)F, and P(ξ)=1+(a+iα)|ξ|2.


## 6. Proof of Theorem1

Proof.
FixT>0, δ>0 to be chosen later. Let
(95)𝒟={u∈Lr(0,T;X1,1s):∥u∥Lr(0,T;X1,1s)<δ}
be equipped with the distance(96)d(u,υ)=∥u-υ∥Lr(0,T;X1,1s).
It is easy to know that(𝒟,d) is a complete metric space. Now we consider the following map:
(97)ℑ:u(t)⟶S(t)u0-i∫0t‍S(t-τ)f(u(τ))dτ.
We will prove that there existsT,δ>0, such that ℑ:(𝒟,d)→(𝒟,d) is a strict contraction map.
By the nonlinear term estimate and the semigroup estimate,(98)∥|u|ku∥X1,1s≲∥u∥X(k+1),1sk+1,∥S(t)u0∥X1,1s≲(1+|t|)n/2∥u0∥X1,1s,∥A(f)∥Lr(0,T;X1,1s)≲(1+T)n/2T1-k/r∥u∥Lr(0,T;X1,1s)k+1,
we have
(99)∥ℑu∥Lr(0,T;X1,1s)≲(1+T)n/2T1/r∥u0∥X1,1s+(1+T)n/2T1-k/r∥u∥Lr(0,T;X1,1s)k+1.
Then, we let
(100)δ=2C∥u0∥X1,1s,T<1.
Then, we choose T such that
(101)2CT(1-k/r)δk⩽12.
Then, we have
(102)∥ℑu∥Lr(0,T;X1,1s)<δ.
Moreover, we have that
(103)∥ℑu-ℑυ∥Lr(0,T;X1,1s)⩽12∥u-υ∥Lr(0,T;X1,1s).
In this way, we prove that ℑ:(𝒟,d)→(𝒟,d) is a strict contraction map. By Banach’s fixed-point theorem, there exists a unique solution u∈𝒟 that satisfies the conditions in the theorem. Using standard argument, we can extend the solution, considering the map
(104)ℑ:u(t)⟶S(t-T)uT-i∫Tt‍S(t-τ)f(u(τ))dτ,
where u(T) is the solution that we have chosen. We claim u(T)∈X1,1s; one can refer to [23] for the details. In this way, we can find a maximum T*>0 which satisfies the conditions in the theorem.


## 7. Proof of Theorem2

Proof.
FixT>0, δ>0 to be chosen later. Let
(105)𝒟={u∈Lr(0,T;Xp,1s):∥u∥Lr(0,T;Xp,1s)<δ}
be equipped with the distance(106)d(u,υ)=∥u-υ∥Lr(0,T;Xp,1s).
It is easy to know that(𝒟,d) is a complete metric space. Now, we consider the following map:
(107)ℑ:u(t)⟶U(t)u0-i∫0t‍U(t-τ)f(u(τ))dτ.
We shall prove that there existsT,δ>0, such that ℑ:(𝒟,d)→(𝒟,d) is a strict contraction map.
By the nonlinear term estimate and the semigroup estimate,(108)∥|u|ku∥Xp,1s≲∥u∥Xp(k+1),1sk+1,∥U(t)u0∥Xp,1s≲∥u0∥Xp,1s,∥A(f)∥Lr(0,T;Xp,1s)≲T1-k/r∥u∥Lr(0,T;Xp,1s)k+1+T∥u∥Lr(0,T;Xp,1s),
we have(109)∥ℑu∥Lr(0,T;Xp,1s)≲T1/r∥u0∥Xp,1s+T1-k/r∥u∥Lr(0,T;Xp,1s)k+1+T∥u∥Lr(0,T;Xp,1s).
Then, we let(110)δ=3C∥u0∥Xp,1s,T<13.
Then, we chooseT such that
(111)CT(1-k/r)δk⩽13.
Then, we have(112)∥ℑu∥Lr(0,T;Xp,1s)<δ.
Moreover, we have that(113)∥ℑu-ℑυ∥Lr(0,T;Xp,1s)⩽23∥u-υ∥Lr(0,T;Xp,1s).
In this way, we prove thatℑ:(𝒟,d)→(𝒟,d) is a strict contraction map. By Banach’s fixed-point theorem, there exists a unique solution u∈𝒟 that satisfies the conditions in the theorem. Using standard argument, we can extend the solution considering the map
(114)ℑ:u(t)⟶U(t-T)uT-i∫Tt‍U(t-τ)f(u(τ))dτ,
where u(T) is the solution that we have chosen. We claim u(T)∈Xp,1s; one can refer to [23] for the details. In this way, we can find a maximum T*>0 which satisfies the conditions in the theorem.



---

*Source: 176596-2013-07-10.xml*