Hongmei XU Yan LIANG
In this paper,we are interested in the global existence and time-asymptotic behavior of solutions to generalized Benjamin-Bona-Mahony(GBBM)equations in all space dimensions.The GBBM equation is defined as
whereu∈R1,ηis a positive constant,andβis a real constant vector.f(u)=(f1(u),···,fn(u))T,andfi(u)=u2,wherenis the space dimension.In this paper,n≥1.The initial data is given by
The well-known Benjamin-Bona-Mahony(BBM)equation is of the form
It was proposed and studied in[1]by Benjamin,Bona and Mahony for the special physical situations in the long wave limit for nonlinear dispersive media.Since then,the existence and uniqueness of solutions to various generalized BBM equaitons have been proved by many authors(see[1–4]).The decays of solutions were also studied in[5–9].However,most of these studies are in low space dimensions and the decay estimates are inLpnorm.The aim of this paper is to give the global existence and pointwise decay rates of solutions to Cauchy problems of the GBBM equation in all space dimensions.
First we introduce some notations.As usual,Fourier transformation to the variablex∈Rnis
and the inverse Fourier transformation to the variableξis
We also useF−1()to denote the inverse Fourier transformation of functionDαf=for multi-indexα=s∈Z+,p∈[1,∞],denotes the usual Sobelev space with the norm
In particular,We denote the generic constant byC.All the convolutions are about the spatial variablexin this paper.
We arrange this paper as follows.In Section 2,we derive the solution formula of the Cauchy problem.We need many inequalities in our analysis.We list the inequalities and their proofs in Section 3.We construct a solution sequence due to the solution formula of(1.1)–(1.2),and then prove that the sequence is a Cauchy sequence in a Banach space.Thus it converges to the solution of our problem.We leave these treatment processes to Section 4 and Section 5.Finally,we give our conclusion in Section 6.
The aim of this section is to derive the solution formula from the problem(1.1)–(1.2).The linearized form of(1.1)is
Taking Fourier transform to variablexof(2.1),we have
The corresponding initial data is given by
The solution to the problem(2.2)–(2.3)is given by
Set(ξ,t)=By the Duhamel principle,we get the solution formula for(1.1)–(1.2):
Set(ξ,t)=Then
Due to(2.5),we define a solution sequencesatisfying
wherem≥1.Next we will prove thatis a Cauchy sequence in a Banach space,and then it converges to the solution to(1.1)–(1.2).To do so,we shall need many inequalities.We collect them in the following section.
In order to estimatewe must analyse the decay property forG,Hfirst.Set
whereχ1,χ2are smooth cut-off functions andχ1(ξ)+χ2(ξ)=1.
Setfori=1,2.
ForG1,H1,we have the decay property as follows.
Lemma 3.1There exists a constantdepending on N,αsuch that
where N is a positive integer.Throughout this paper N>
Conveniently,we next denoteBN(x,t)=
ProofWhen|ξ|is bounded,using the Taylor expansion,we have
Then
We know thatis a parallel operator.It can not contribute to the decay factor,but it has a physical meaning.So next we will not neglect the effect of the operator.From(3.3)–(3.4),we get
where(|α|−|β|)+:=
Then
Similarly we get(3.2).
G2,H2have the construction as follows.
Lemma 3.2There exists a positive constant b and distributionsfor i=1,2such that
where δ(x)is Dirac function and
withbeing sufficiently small.
ProofWhen|ξ|is large enough,using Taylor expansions,we have
Thus we have
Then there exists a positive constantbsuch that
with|γ|≥1.From[10,Lemma 3.2],we get our results.
When dealing with the convolution with the nonlinearized part,we need the following four lemmas.
Lemma 3.3For positive constants b,N,when t is large enough,we have
ProofNoticing thatifN,tare large enough,we have
Thus we get our lemma.
Lemma 3.4Whenand n3=min{n1,n2},we have
ProofWe just prove(3.8).The proof of(3.9)is similar.
When|x−y|(3.8)is easily got.
When|x−y|we have|y|Thus
If|x|≤we haveBn3(x,t)≥C.(3.8)is easily got.
If|x|>we have
Thus
Then(3.8)is proved.
Set
Lemma 3.5
ProofWe divide the proof into two different cases.
Case 1|x−
In this case,(x−βt,t)≥C.From[11,Lemma 5.2],we have
Case 2|x−βt|≥
We also divide the proof into two different cases.
Case 2.1s
If|y−βs|then
From[11,Lemma 5.2],we have
If|y−βs|≤then|x−y−β(t−s)|≥We have
Case 2.2s≥.
The proof of this part is similar to that of Case 2.1,so we omit it here.
Lemma 3.6There exists a constant depending only on n such that
ProofSince|x−βt|2≤2(|x−βs|2+|β|2(t−s)2),we have
Thus we get our lemma.
We can now enter into the estimate of the sequence
We first give an estimate for(x,t),and then use mathematical induction to get the estimate for
Lemma 4.1If u0∈Hl,l>1+with E small enough,then we have
ProofFrom(2.6),we have
From(3.1)and Lemma 3.4,we have
From(3.7),when|γ|>,we have
If|α|<l−1−,we have
From Lemma 3.2 and(4.4),we have
Taking|γ|=2Nin(4.3),from(4.3)and(4.5),we have
From Lemma 3.3,we get
From(4.2)and(4.6),we get our result.
Lemma 4.2For m>1,if
then
ProofBecause(u)=,we have
We still denote
From Lemmas 3.1,3.5 and(4.7),we have
From Lemma 3.2,we have
From Lemma 3.6 and(4.7),we have
From(3.5),(4.7),Lemmas 3.4 and 3.6,we have
From(3.6),(4.7)and Lemma 3.6,we have
Together with(4.9)–(4.12),we get
From(2.6),we know that
From(4.8),(4.13)–(4.14)and Lemma 4.1,we get our result.
From Lemmas 4.1 and 4.2,using mathematical induction,we know that for allm≥1 and|α|<l−1−
In this section,we will prove thatis a Cauchy sequence in a Banach space,and thus it converges to the solution of(1.1)–(1.2).
From[11,Lemma 5.2],it follows that
From(4.15)and(5.1),we know(x,t)∈L∞(0,∞;Thusis in a Banach space.We next prove that it is a Cauchy sequence.
Lemma 5.1is a Cauchy sequence in L∞(0,∞;
ProofFrom(2.6),satisfies the following equation
Thus
Set=−(x,t).Then
Multiplyingin the two sides of(5.3)and integrating with respect toxin Rn,we get
Whenm≥2,from(2.6)we have(x,0)=0.Thus form≥3,we have
From(4.15),we know
IfCE<1,from(5.4)we know thatis a Cauchy sequence in Banach spaceL∞(0,∞;and thus it converges to the solution to(1.1)–(1.2).
Theorem 6.1If=E,l>1+≤CE(1+with E small enough,then(1.1)–(1.2)have a global solution in time u(x,t),for|α|≤l−−2,satisfying
Remark 6.1The solution has the same decay rate as the heat kernel,so our estimation must be optimal.
Remark 6.2The solution decays much faster away along the characteristic linex=βt,so we can say that the solution propagates along the characteristic line.It coincides with the physical phenomenon.
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Chinese Annals of Mathematics,Series B2014年4期