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Multilinear estimates for periodic KdV equations and applications

a r X i v :m a t h /0110049v 3 [m a t h .A P ] 14 J u l 2006

MULTILINEAR ESTIMATES FOR PERIODIC KDV EQUATIONS,

AND APPLICATIONS

J.COLLIANDER,M.KEEL,G.STAFFILANI,H.TAKAOKA,AND T.TAO

Abstract.We prove an endpoint multilinear estimate for the X s,b spaces associated to the periodic Airy equation.As a consequence we obtain sharp local well-posedness results for periodic generalized KdV equations,as well as some global well-posedness results below the energy norm.

Contents

1.Introduction 21.1.Main Results

31.2.Remarks and possible extensions 41.3.Outline

52.A counter-example 63.Linear Estimates

74.Reduction to a multiplier bound 105.The non-endpoint case

126.Some elementary number theory 137.The proof of (4.4)in the endpoint case.

2J.COLLIANDER,M.KEEL,G.STAFFILANI,H.TAKAOKA,AND T.TAO

8.The proof of(4.5)in the endpoint case.17

9.Proof of Proposition121

10.Local well-posedness for small data24

https://www.doczj.com/doc/f39581528.html,rge periods25

12.An interpolation lemma29

13.Global well-posedness31

14.Proof of Lemma13.133

References38

1.Introduction

This paper studies the Cauchy problem for periodic generalized KdV equations of the form

(1.1) ?t u+1

4π2is convenient in order to make the

dispersion relationτ=ξ3,but it is inessential and we recommend that the reader ignore all powers of2πwhich appear in the sequel.We can assume that F has no constant or linear term since these can be removed by a Gallilean transformation.

The main result established in this paper is a sharp multilinear estimate which allows us to show the initial value problem(1.1)is locally well-posed in H s(T)for s≥1

MULTILINEAR ESTIMATES AND APPLICATIONS 3

The low-regularity study of the equation (1.1)on T has been based around

iteration in the spaces X s,

+ ξ s ?u L 2ξL 1τ

.We shall also need the companion spaces Z s de?ned by

(1.3) u Z s := u s,?1 τ?ξ3 L 2ξL 1

1.1.Main Results.The main technical result of this paper is the following mul-tilinear estimate.

Theorem 1.For any s ≥

i =1

u i Y s .

This improves on the results in [29],where this estimate was proven for s ≥1.By combining this estimate with an estimate of Kenig,Ponce,and Vega [24]we

shall obtain

Proposition 1.For any s ≥

In [24]the restriction s ≥

for k ≥4,and also for the k =3case if one allows the mean of the

u i to be non-zero.In the k =3case with mean zero the counter-example is a little trickier,and is discussed in Section 2.

The sharp estimate (1.5)of Proposition 1will be useful in studying the dynami-cal behavior of solutions of polynomial generalizations of the KdV equation.In this paper,we use Proposition 1to obtain some local and global well-posedness results for the Cauchy problem (1.1).In [29](see also [5],[22]),such equations were shown to be globally well-posed for H 1

data.

4J.COLLIANDER,M.KEEL,G.STAFFILANI,H.TAKAOKA,AND T.TAO Theorem2.If F(u)is a polynomial,then the Cauchy problem for the periodic

generalized KdV equation(1.1)is locally well-posed in H s(T)for all s≥1

2 [24](see also[5],[7]).When F is cubic we have the modi?ed KdV equation,for

which local well-posedness was already obtained for s≥1

2;this example can

be extended to general polynomials F of degree at least3.

On the real line with F(u)=u k+1,the equation(1.1)is known to be locally

well-posed down to the scaling exponent s≥1

k for k≥4[22](see also[4];

earlier results are in[18]).This was recently extended to k=3(except at the endpoint s=1

)in[19].Thus there is a loss of2

4loss one has in the k=1,2

cases.These observations resolve a problem2posed by Carlos Kenig.

There are some other consequences of Theorem1.It allows us to complete the proof(in[11])of global well-posedness of the KdV and mKdV equations down to s≥?1

.In particular in the symplectic space˙H?1

4π2u xxx+u3u x=0

u(x,0)=u0(x),x∈T is locally well-posed for large H s(T)data for s≥1

6.Moreover,the initial value problem(1.6)is analytically

ill-posed in H s(T)for s<1

6for global well-posedness

falls quite short of the local condition s≥1

2See the problem posed after Theorem 5.3in Kenig’s lecture notes at https://www.doczj.com/doc/f39581528.html,/publications/ln/msri/1997/ha/kenig/5/banner/03.html.

MULTILINEAR ESTIMATES AND APPLICATIONS5 on the?uctuation of our modi?ed Hamiltonian.From our experience with the KdV and mKdV equations in[11]however we believe it is reasonable to hope that global well-posedness should hold for all s≥1

8π2u2x?

and s≥1

2.Section3records linear esti-

mates between spaces associated to the Airy equation.Section4reduces Theorem 1to a multiplier bound which we establish in the non-endpoint case,s>1

2.Section9

establishes Proposition1.Theorem2is proven in Section10.Section11rescales various estimates to the setting of large periods4.Section12contains a general interpolation result revealing a certain?exibility in proofs of local well-posedness. The global result of Theorem3is proven in Sections13and14.

The authors thank Jorge Silva for a correction.

6J.COLLIANDER,M.KEEL,G.STAFFILANI,H.TAKAOKA,AND T.TAO

2.A counter-example

In this section we give an example which shows why the condition s≥1

2,and let N?1be a large integer.Let N0,N1,N2,N3,N4be integers

with distinct magnitudes such that

|N0|～1;|N1|,|N2|,|N3|,|N4|～N

and

(2.1)N0+N1+N2+N3+N4=0;N30+N31+N32+N33+N34=O(1).

For instance,we could choose

N0:=6;N1:=N?4;N2:=2N+1;N3:=?N?4;N4:=?2N+1.

Let u0be an H s function.We de?ne the iterates u(0),u(1)by

u(0)t+

4π2

u(1)xxx+(u(0))3u(0)x=0;u(1)(x,0)=u0(x).

In order for the Cauchy problem(1.6)to be locally analytically well-posed in H s,it is necessary that the non-linear map u0→u(1)maps H s to L∞t H s x,at least for short times t and small H s norm.This is because u(1)is the Taylor expansion of u to fourth order in terms of u0(see e.g.[7]for further discussion).

We now choose a speci?c choice of initial data u0,namely

u0(x):=ε

j=0N?s j cos(2πN j x)

for some0<ε?1.Clearly u0has H s norm O(ε)and mean zero.The zeroth

iterate u(0)is

u(0)(t,x)=ε

j=0N?s j cos(2π(N j x?N3j t)).

By(2.1),one can then see that the non-linear expression(u(0))3u(0)x contains a term of the form

Cε4N?s

0N?s

N?s

N4cos(2π(N4x?(N34+O(1))t))

for some absolute non-zero constant C.From this and a little Fourier analysis one can see that for all non-zero times|t|?1,the N4Fourier coe?cient norm of u(1)(t)is～c(t)ε4N1?3s for some non-zero quantity c(t)depending only on t.This implies that the H s norm of u(1)(t)is at least c(t)N1?2s.Since s<1

MULTILINEAR ESTIMATES AND APPLICATIONS7 more carefully we can show that the solution map is not even C4in the H s topology in the k=3case.For more general k,one can show C k+1ill-posedness.).

Note that the above example can be easily modi?ed to also show that Proposi-tion1fails for s<1

2and2≤p≤2

and2≤p≤∞.

Similarly,we have the energy estimate5

(3.3)X s,1

+,1

?δ,1

and2≤q,r<1

3?L4x,t

and

X0+,1

2?L q x,t

for all0<δ<1

1?2δ.In particular we may take q=6,6?,or6+.

If we interpolate with(3.6)instead we obtain

(3.8)X0+,1

5We use a+and a?to denote quantities a+ε,a?ε,whereε>0is arbitrarily small,and implicit constants are allowed to depend onε.

8J.COLLIANDER,M.KEEL,G.STAFFILANI,H.TAKAOKA,AND T.TAO

Now we give some embeddings for the Y s and Z s spaces.Since the Fourier transform of an L 1function is continuous and bounded,we have from (1.2)that (3.9)

Y s ?C t H s x ?L ∞t H s

x .

Let η(t )denote a bump function adapted to [?2,2]which equals one on [?1,1].It

is easy to see that multiplication by η(t )is a bounded operation on the spaces Y s ,Z s ,X s,b .

Let S (t )denote the evolution operator for the Airy equation:

S (t ):=exp(

(a (t ′)?a (t ?t ′))

for all t ∈[?2,2]and t ′∈[?3,3],we see that we may write η(t ) t

S (t ?t ′)F (t ′)dt ′

as a linear combination of

(3.11)η(t )S (t )

a (t ′)S (?t ′)F (t ′)dt ′

and (3.12)

η(t )

a (t ?t ′)S (t ?t ′)F (t ′)dt ′.

Consider the contribution of (3.11).By (3.10)it su?ces to show that

a (t ′)S (?t ′)F (t ′)dt ′

F Z s .Observe that the Fourier transform of

a (t ′)S (?t ′)F (t ′)dt ′at ξis given by

?a (τ?ξ3)?F

(ξ,τ)dτ.

MULTILINEAR ESTIMATES AND APPLICATIONS 9

Since one has the easily veri?ed bound (3.13)

?a (λ)=O ( λ ?1),

the claim then follows from (1.3).

Now consider the contribution of (3.12).

We may discard the η(t )cuto?.The spacetime Fourier transform of R a (t ?t ′)S (t ?t ′)F (t ′)dt ′at (τ,ξ)is equal to

?a (τ?ξ3)?F (τ,ξ).The claim then follows from (3.13),(1.2),(1.3). Finally,we shall need the following duality relationship between Y s and Z ?s .Lemma 3.2.We have

χ[0,1](t )u (x,t )v (x,t )dxdt u Y s v Z ?s for any s and any u ,v on T ×R .

Proof.Without the cuto?χ[0,1]this would be an immediate consequence of the duality of X s,

2.However,these spaces are not preserved by rough cuto?s,and one requires a little more care.

By writing χ[0,1]as the di?erence of two signum functions,it su?ces to show that |

sgn(t )uv dxdt | u Y s v Z ?s .By Plancherel we can write the left-hand side as

C |p.v. ?u (ξ,τ)?v (ξ,τ′)

dξdτdτ′

τ?τ′

First consider the contribution of the case |j ?k | 1.In this case we use the L 2

boundedness of the Hilbert transform and Cauchy-Schwarz to estimate the above by C

j,k ≥0:|j ?k | 1

ξ s ?u j 2 ξ ?s ?v k 2.

Since |j ?k | 1,we may estimate this by C j,k ≥0:|j ?k | 1

ξ s ξ?τ3

?v k 2,

which by another Cauchy-Schwarz is bounded by

C u s,1

u Y s v Z s

as desired.

10J.COLLIANDER,M.KEEL,G.STAFFILANI,H.TAKAOKA,AND T.TAO Now consider the contribution when|j?k|?1.In this case we observe that |τ?τ′| τ′?ξ3 ,so we may estimate(3.14)by

Cp.v. ξ s|?u(ξ,τ)| ξ ?s|?v(ξ,τ′)|

τ′?ξ3 L2

τ′

u Y s v Z s

as desired.

4.Reduction to a multiplier bound

In this section we make some preliminary reductions for Theorem1,exploit-ing the“denominator games”of Bourgain and reducing matters to a multilinear multiplier estimate.

Fix s,u i.Since the spaces Y s and X s?1,1

MULTILINEAR ESTIMATES AND APPLICATIONS 11

On the other hand,from Sobolev and H¨o lder,then another Sobolev and (3.9)we have

i =1

u i s ?1,0 k i =1

u i L 2t

L 1+x

u 1 L 2t

L k +

i =2 u i L ∞t

L k +

u 1 L 2t H s

x k

i =2 u i L ∞

t H s x

u 1 s,0

i =2

u i Y s

as desired.

From the above and symmetry,we may thus assume that (4.2)

τ?ξ3 ? τi ?ξ3

for all 1≤i ≤k .In particular we have

1? τ?ξ3

～|ξ3

k i =1

ξ3

i |.

It thus su?ces to show that

(4.3) ?

ξ s ?1|ξ3

i =1

ξ3i |12

We now rewrite (4.3)using the notation of [30].For all n ≥2and all symbols

m (ξ1,τ1,...,ξn ,τn )de?ned on the region

Γn :={(ξ1,...,τn )∈(Z ×R )n :ξ1+...+ξn =τ1+...+τn =0},

de?ne the norm

m [n ;Z ×R ]

to be the best constant such that one has the bound

Γn

m (ξ1,...,τn )n i =1

f i (ξi ,τi ) ≤ m [n ;Z ×R ]n i =1

f i L 2ξi L 2τi .By duality (4.3)can now be written as

| k +1i =1ξ3i |

1 ξk +1 1?s k i =1 ξi s τi ?ξ3i

12J.COLLIANDER,M.KEEL,G.STAFFILANI,H.TAKAOKA,AND T.TAO

Proof.From the estimate

ξ31+ξ32=(ξ1+ξ2)(ξ21?ξ1ξ2+ξ22)=O(|ξ1+ξ2||ξ1|2)

we see that

k+1

i=1ξ3i=O(|ξ1+ξ2||ξ1|2)+O(|ξ3|3).

Since|ξ1+ξ2|=O(|ξ3|)and|ξ3|≤|ξ2|～|ξ1|,the claim follows.

From Lemma4.1and symmetry of theξ1,...,ξk variables,it thus su?ces to show the estimates

(4.4)

|ξ1|12|ξk+1|1

ξk+1 1?s k i=1 ξi sλ1

2|ξ2|12

[k+1;Z×R] 1

where we adopt the notation

λi:= τi?ξ3i .

In the next section we prove these estimates in the non-endpoint case s>1

5.The non-endpoint case

We now prove(4.4),(4.5)in the non-endpoint case s>1

2|ξ2|12

ξ2 s?1

ξ2 s?12i

[k+1;Z×R] 1

which by duality becomes

u1...u k L2

x,t u1 0,1

However,by H¨o lder we may take u1in L4x,t,u2in L6x,t,and the other u i in L12(k?2)

x,t .

The claim then follows from(3.6),(3.7),and(3.5).

MULTILINEAR ESTIMATES AND APPLICATIONS13 Now consider(4.5).By duality as before it thus su?ces to show that

u1...u k

L2t H s?1

x u1 s?1

u2 s?1

u3 s?1

i=4 u i s,1

u2 0+,1

By standard Cauchy-Schwarz arguments(see[5],or apply[30]Proposition5.1and Lemma3.9)this result would obtain if we knew that the number of integer solutions of size O(N)to the equations

ξ=ξ1+ξ2+ξ3;τ=ξ31+ξ32+ξ33

was bounded by O(N0+)uniformly for non-zero(ξ,τ)and N≥1.But this follows from the identity(cf.[5])

τ?ξ3=3(ξ1+ξ2)(ξ2+ξ3)(ξ3+ξ1),

and the well-known observation(see e.g.[1])

(6.2)Every non-zero integerλhas at most O(|λ|0+)factors

from elementary number theory.(The caseτ?ξ3=0needs to be dealt with separately).

It would be very convenient if one could replace0+by0in the above arguments; indeed,the endpoint estimate would then follow by a variant of the preceding arguments.Even with the epsilon loss in exponents,(3.7)is still strong enough to treat a large portion of(4.4)and(4.5)in the endpoint case.

We do not know how to prove the endpoint of(3.7)directly(see[5]for some further discussion of this issue and of a related L8x,t conjecture).However,we will be able to remove the epsilon in(6.1)when u3(for instance)has much smaller frequency than the other two functions,and this is enough to treat the remaining cases for(4.4)and(4.5).

To achieve this we shall rely on the following variant of(6.2).

Letξ,λ,N,L be integers such that0

(6.3)#{(l,n)∈Z2:|l?λ| L;|n?ξ| N;n|l},

14J.COLLIANDER,M.KEEL,G.STAFFILANI,H.TAKAOKA,AND T.TAO

where we use the notation a|b to denote that a divides b.In other words,for all l=λ+O(L),we count the divisors of l which lie in the intervalξ+O(N).From (6.2)we may clearly bound(6.3)by|λ|0+L.The purpose of the following lemma is to remove the|λ|0+under some additional assumptions.

Lemma6.1(Few divisors in small intervals).Letξ,λ,N,L be as above.Then

(6.3)≤N.

If we make the further assumptions

|λ| |ξ|3;0

6.Since l=O(|ξ|3),we obtain the desired contradiction.

From this lemma and the previous Cauchy-Schwarz argument we can obtain various partial endpoint versions of(6.1).For instance,we can prove

u1u2u3 L2

x,t u1 0,1

u3 0,1 3

1.We sketch the argument as follows.We repeat

the proof of(6.1)but observe that we may assumeξ=O(N1),ξ1+ξ2=?ξ?ξ3=?ξ+O(N3),andτ?ξ33=O(N31).If we discard the special caseτ?ξ33=0,we may invoke the second part of Lemma6.1(with L=1)and conclude that for?xed

MULTILINEAR ESTIMATES AND APPLICATIONS15τ,ξ,there are at most3values ofξ1+ξ2.Fixingξ1+ξ2determinesξ3,and hence ξ31+ξ32,which determinesξ1andξ2up to permutations.So we have at most6 integer solutions(ξ1,ξ2,ξ3)(rather than O(N0+

)),and the claim follows.

We will not use this partial endpoint result directly in the sequel,but arguments with the above?avor will be used to compensate for the0+loss in the L6Strichartz estimate at various junctures.

7.The proof of(4.4)in the endpoint case.

We now prove(4.4)in the endpoint case s=1

2 u2 0,1

2?δ

u2 0,1

?δ,1

2?1

+,1

100has no special signi?cance,and could be replaced by any

other small constant.

By(3.4),we need only show the trilinear estimate

u1u2u3 L2

x,t u1 0,1

100

u2 0,1

100

u3 1

100

By dyadic decomposition,it su?ces to show that

u1u2u3 L2

x,t N1100?

u1 0,1

100

u2 0,1

100

u3 0,1

100

for all N3≥1,where u3is supported on ξ3 ～N3.

100

in other indices,and so in the?nal estimate we will have some sort of gainδ>0throughout.

16J.COLLIANDER,M.KEEL,G.STAFFILANI,H.TAKAOKA,AND T.TAO Fix N3.By duality we reduce to

χ|ξ

3|～N3

v 2)we may assume thatξ1,ξ2have the same sign.By

symmetry we may thus takeξ1,ξ2non-negative.

Let us?rst deal with the case8whenξ2 (N3L1L2L3)10.In this case we may borrow some regularity from u3to place on u2,and it su?ces by duality to show that

u1u2u3 2 u1 0,1

100 u2 0+,1

100

u3 1

100

But this follows by(for instance)taking u1in L4x,t,u2in L5x,t,and u3in L20x,t,and

using(3.6),(3.8),(3.5).From this and symmetry we may restrict ourselves to the

caseξ1,ξ2?(N3L1L2L3)10.

By Cauchy-Schwarz([30],Lemma3.9)it su?ces to show that

χξ1,ξ2?(N3L1L2L3)10χ|ξ3|～N3χλ1～L1χλ2～L2χλ3～L3dξ1dξ2dτ1dτ2 N1?2100?for allξ4,τ4,whereξ3,τ3is given byξ1+ξ2+ξ3+ξ4=τ1+τ2+τ3+τ4=0.

Fixξ4,τ4.Performing theτintegrals9,we reduce to

L min L med#{(ξ1,ξ2,ξ3)∈Z3:ξ1+ξ2+ξ3+ξ4=0;ξ1,ξ2?(N3L1L2L3)10;|ξ3|～N3;

|ξ31+ξ32+ξ33+τ4| L max} N1?2100?

where L min≤L med≤L max are the minimum,median,and maximum of L1,L2,L3 respectively.It will su?ce to show that

#{(ξ1,ξ2,ξ3)∈Z3:ξ1+ξ2+ξ3+ξ4=0;ξ1,ξ2?(N3L max)10;|ξ3|～N3;

|ξ31+ξ32+ξ33+τ4| L max} N1?2100?

MULTILINEAR ESTIMATES AND APPLICATIONS17

Sinceξ1+ξ2+ξ3+ξ4=0,we have the identity

(7.3)ξ31+ξ32+ξ33+ξ34=3(ξ1+ξ2)(ξ2+ξ3)(ξ3+ξ1)

and so we reduce to showing

#{(ξ1,ξ2,ξ3)∈Z3:ξ1+ξ2+ξ3+ξ4=0;ξ1,ξ2?(N3L max)10;|ξ3|～N3; (7.4)

(ξ1+ξ2)(ξ2+ξ3)(ξ3+ξ1)∈I} N1?2100?

max

where I is the set

I:={l∈Z:|l+(τ4?ξ34)/3| L max}.

Sinceξ1,ξ2are positive and much larger thanξ3,we see that|ξ4| |ξ1|,|ξ2|,and that

|(ξ1+ξ2)(ξ2+ξ3)(ξ3+ξ1)| |ξ1+ξ2|3 |ξ4|3.

We may therefore assume that|τ4?ξ34| |ξ4|3,since(7.4)vanishes otherwise.

From Lemma6.1thus we see that

#{(n,l):|n+ξ4| N3;l∈I;n|l} min(N3,L max).

From this and the previous discussion we see that there are at most O(min(N3,L max)) possible values ofξ1+ξ2which can contribute to(7.4).But from elementary alge-bra we see that each value ofξ1+ξ2contributes at most O(1)elements to(7.4). The claim then follows.

8.The proof of(4.5)in the endpoint case.

It remains to prove(4.5),which we rewrite as

| u1...u k+1dxdt| u1 0,12 u3 0,12,0k i=4 u i 12.

We shall actually prove the stronger estimate

(8.1)

| u1...u k+1dxdt| u1 0,12?δ u3 0,12?δ,0k i=4 u i 12?δ.

for some0<δ?1.Again,this improved estimate shall be useful for large data applications.

We shall prove(8.1)in the case k≥4.The k=3case is slightly simpler,and can be obtained by a routine modi?cation of the following argument.

We?rst observe that

| u1...u k+1dxdt| u1 0,16 u2 0,16 u3 0,12+,0k i=4 u i 12+.

Indeed,this follows by taking u1,u2in L4x,t,u3in L∞t L2x,u k+1in L2t L∞x,and all other u i in L∞x,t,and then using(3.6),(3.3),(3.2),and(3.4).

18J.COLLIANDER,M.KEEL,G.STAFFILANI,H.TAKAOKA,AND T.TAO

By symmetry and interpolation10it thus su?ces to show that

| u1...u k+1dxdt| u1 0,12+ u3 0,12?12?12?12+,1

2?1

and so by(3.3)we have

(

i=5u i)u k+1 1100,0 u k+1 1100,0k i=5 u i 12+.

We thus reduce to the quintilinear estimate

| u1u2u3u4u5dxdt| u1 0,12+ u3 0,12?12?12?1

u2 0,1

u4 0,0 u5 0,0

for all N4,L4,N5≥1,where u4and u5have Fourier support in the regions ξ4 ～N4,λ4～L4and ξ5 ～N5respectively.

Let us?rst consider the contribution where u3is supported in the region ξ3 (N4N5L4)10.In this case it su?ces to show that

| u1u2u3u4u5dxdt| u1 0,12+ u3 0+,12?22?22?2

10Speci?cally,we permute{u

1,u2,u3}and{u4,...,u k+1}independently in this estimate and the previous one,and then use multilinear complex interpolation(see e.g.[3])to obtain the centroid of all the permuted estimates.The small losses of0+in some indices will be more than compensated for by the gains of1

in other indices,so in the interpolated estimate one will have some non-zero gainδ>0throughout.

MULTILINEAR ESTIMATES AND APPLICATIONS19 This quintilinear estimate is too complex to handle directly.The strategy will

be to reduce this estimate to a quartilinear estimate,and(in some cases)further

to a trilinear estimate.

By symmetry we may assume that|ξ1|≤|ξ2|≤|ξ3|;since|ξ4|+|ξ5|

(N4N5L4)10?|ξ1|,|ξ2|,|ξ3|,we thus see that|ξ2|～|ξ1+ξ2|.Thus we may freely

insert a factor of

χΣ(ξ1+ξ2,τ1+τ2)

in the previous expression,where

Σ:={(ξ,τ):(ξ,τ)=(ξ1+ξ2,τ1+τ2)for some(ξ1,τ1),(ξ2,τ2)∈?such that|ξ2|～|ξ1+ξ2|}.

We expand this out as

?χ?(ξ1,τ1)χ?(ξ2,τ2)χΣ(ξ1+ξ2,τ1+τ2)χ?(ξ3,τ3)χ|ξ

4| N4

χ|ξ

5| N5

χλ

4 L4 5

j=1?u j(ξj,τj) (N4N5L4)11005 j=1 u j L2x,t

where ?denotes integration over the regionξ1+...+ξ5=τ1+...+τ5=0.

We can write the left-hand side as11

(8.2) ??χΣ(ξ12,τ12)χ?(ξ3,τ3)χ|ξ4| N4χ|ξ5| N5χλ4 L4?F(ξ12,τ12) j=3,4,5?u j(ξj,τj) where ??integrates over the variablesξ12,ξ3,ξ4,ξ5,τ12,τ3,τ4,τ5such that

ξ12+ξ3+ξ4+ξ5=τ12+τ3+τ4+τ5=0,

and

?F(ξ

,τ12):= ξ1+ξ2=ξ12 τ1+τ2=τ12χ?(ξ1,τ1)χ?(ξ2,τ2)?u1(ξ1,τ1)?u2(ξ2,τ2). Observe that

F=(P?u1)(P?u2)

where P?is the space Fourier projection corresponding to the set?.From H¨o lder and(3.6)we thus have

F L2

x,t ≤ P?u1 L4

x,t

P?u2 L4

x,t

P?u1 0,1

u1 L2

x,t

u2 L2

x,t

It thus su?ces to show that the quantity(8.2)is bounded by

(N4N5L4)1100 F L2

x,t j=3,4,5 u j L2x,t.

If we relabelξ1,ξ23,ξ4,ξ5asξ′1,ξ′2,ξ′3,ξ′4and similarly for theτ,we thus see that it su?ces to show that

χΣ(ξ′1,τ′1)χ?(ξ′2,τ′2)χ|ξ′

3| N′

χ|ξ′

| N′

χλ′

L′3

[4;Z×R] (N′3N′4L′3)1100

20J.COLLIANDER,M.KEEL,G.STAFFILANI,H.TAKAOKA,AND T.TAO

where we have renamed N4,N5,L4as N′3,N′4,L′3to reduce confusion,andλ′3is short-hand for ξ′3?(τ′3)3 .(Of course we will de?ne the[4;Z×R]norm here to use the primed variablesξ′j,τ′j instead of the unprimed variables.) Since|ξ′3+ξ′4| N′3+N′4,we see that the variablesξ′1,ξ′2are constrained by the relationshipξ′1=?ξ′2+O(N′3+N′4).By Schur’s test([30],Lemma3.11)it thus su?ces to show that

χΣ(ξ′1,τ′1)χ?(ξ′2,τ′2)χξ′

1=?A+O(N′

+N′

)

χξ

2=A+O(N′3+N′4)

χ|ξ′

3| N′

χ|ξ′

| N′

χλ′

L′3

[4;Z×R] (N′3N′4L′3)1100

(8.3)

for all A∈Z.

Fix A.We may assume that|A|?(N′3N′4L′3)10since the above expression vanishes otherwise.

We split into two cases:L′3≤(N′3+N′4)3and L′3≥(N′3+N′4)3.

Case1:L′3≤(N′3+N′4)3(L′3not dominant).

We shall drop the primes from the variablesξ′j,τ′j.By Cauchy-Schwarz([30], Lemma3.9or Lemma3.14)we have

χ|ξ

2| N′3χ|ξ

3| N′4

χλ

2 L′3

[3;Z×R] min(N′3,N′4)12

so by the Composition Lemma([30],Lemma3.7;alternatively one can introduce an“F”as in the previous arguments)it su?ces to show that

χΣ(ξ1,τ1)χ?(ξ2,τ2)χξ

1=?A+O(N′3+N′4)χξ

2=A+O(N′3+N′4)

[3;Z×R] 1

since by hypothesis we have

min(N′3,N′4)12 (N′3N′4L′3)1100.

By Cauchy-Schwarz([30],Lemma3.9)it su?ces to show that

χΣ(ξ1,τ1)χ?(ξ2,τ2)χξ1=?A+O(N′3+N′4)χξ2=A+O(N′3+N′4)dξ1dτ1 1

for all(ξ3,τ3)∈Z×R,whereξ2,τ2are given by the formulaeξ1+ξ2+ξ3=τ1+τ2+τ3=0.

Fixξ3,τ3.We may assume that|ξ3| N′3+N′4since the integral vanishes otherwise.Performing theτintegral and expanding outΣ,we reduce to showing that

#{(ξ′1,ξ′′1,ξ2)∈Z3:ξ′1+ξ′′1+ξ2+ξ3=0;|ξ′1|～A;|ξ′′1|?(N′3+N′4)10;

ξ′1+ξ′′1=?A+O(N′3+N′4);ξ2:=A+O(N′3+N′4);

ξ′13+ξ′′13+ξ32+τ3=O(1)} 1.

By(7.3)we may rewrite this as

#{(ξ′1,ξ′′1,ξ2)∈Z3:ξ′1+ξ′′1+ξ2+ξ3=0;|ξ′1|～A;|ξ′′1|?(N′3+N′4)10;

ξ′1+ξ′′1=?A+O(N′3+N′4);ξ2:=A+O(N′3+N′4);

(ξ′1+ξ′′1)(ξ′1+ξ2)(ξ′′1+ξ2)∈I} 1

(8.4)

to与for的用法和区别

to与for的用法和区别一般情况下, to后面常接对象; for后面表示原因与目的为多。 Thank you for helping me. Thanks to all of you. to sb.表示对某人有直接影响比如，食物对某人好或者不好就用to; for表示从意义、价值等间接角度来说，例如对某人而言是重要的，就用for. for和to这两个介词，意义丰富，用法复杂。这里仅就它们主要用法进行比较。 1. 表示各种“目的” 1. What do you study English for? 你为什么要学英语？ 2. She went to france for holiday. 她到法国度假去了。 3. These books are written for pupils. 这些书是为学生些的。 4. hope for the best, prepare for the worst. 作最好的打算，作最坏的准备。 2．对于 1．She has a liking for painting. 她爱好绘画。 2．She had a natural gift for teaching. 她对教学有天赋/ 3．表示赞成同情，用for不用to. 1. Are you for the idea or against it? 你是支持还是反对这个想法？ 2. He expresses sympathy for the common people.. 他表现了对普通老百姓的同情。 3. I felt deeply sorry for my friend who was very ill. 4 for表示因为，由于（常有较活译法） 1 Thank you for coming. 谢谢你来。 2. France is famous for its wines. 法国因酒而出名。 5．当事人对某事的主观看法，对于（某人），对…来说（多和形容词连用）用介词to，不用for.. He said that money was not important to him. 他说钱对他并不重要。 To her it was rather unusual. 对她来说这是相当不寻常的。 They are cruel to animals. 他们对动物很残忍。 6．for和fit, good, bad, useful, suitable 等形容词连用，表示适宜，适合。 Some training will make them fit for the job. 经过一段训练，他们会胜任这项工作的。 Exercises are good for health. 锻炼有益于健康。 Smoking and drinking are bad for health. 抽烟喝酒对健康有害。 You are not suited for the kind of work you are doing. 7. for表示不定式逻辑上的主语，可以用在主语、表语、状语、定语中。 1．It would be best for you to write to him. 2．The simple thing is for him to resign at once. 3．There was nowhere else for me to go. 4．He opened a door and stood aside for her to pass.

of与for的用法以及区别

of与for的用法以及区别 for 表原因、目的 of 表从属关系介词of的用法（1）所有关系 this is a picture of a classroom （2）部分关系 a piece of paper a cup of tea a glass of water a bottle of milk what kind of football，American of soccer？（3）描写关系 a man of thirty 三十岁的人 a man of shanghai 上海人（4）承受动作 the exploitation of man by man．人对人的剥削。（5）同位关系 It was a cold spring morning in the city of London in England．（6）关于，对于 What do you think of Chinese food？你觉得中国食品怎么样？介词 for 的用法小结 1. 表示“当作、作为”。如: I like some bread and milk for breakfast. 我喜欢把面包和牛奶作为早餐。What will we have for supper? 我们晚餐吃什么?

2. 表示理由或原因,意为“因为、由于”。如: Thank you for helping me with my English. 谢谢你帮我学习英语。 Thank you for your last letter. 谢谢你上次的来信。 Thank you for teaching us so well. 感谢你如此尽心地教我们。 3. 表示动作的对象或接受者,意为“给……”、“对…… (而言)”。如: Let me pick it up for you. 让我为你捡起来。 Watching TV too much is bad for your health. 看电视太多有害于你的健康。 4. 表示时间、距离,意为“计、达”。如: I usually do the running for an hour in the morning. 我早晨通常跑步一小时。We will stay there for two days. 我们将在那里逗留两天。 5. 表示去向、目的,意为“向、往、取、买”等。如: let’s go for a walk. 我们出去散步吧。 I came here for my schoolbag.我来这儿取书包。 I paid twenty yuan for the dictionary. 我花了20元买这本词典。 6. 表示所属关系或用途,意为“为、适于……的”。如: It’s time for school. 到上学的时间了。 Here is a letter for you. 这儿有你的一封信。 7. 表示“支持、赞成”。如: Are you for this plan or against it? 你是支持还是反对这个计划? 8. 用于一些固定搭配中。如: Who are you waiting for? 你在等谁? For example, Mr Green is a kind teacher. 比如,格林先生是一位心地善良的老师。

延时子程序计算方法

学习MCS-51单片机，如果用软件延时实现时钟，会接触到如下形式的延时子程序：delay:mov R5,#data1 d1:mov R6,#data2 d2:mov R7,#data3 d3:djnz R7,d3 djnz R6,d2 djnz R5,d1 Ret 其精确延时时间公式：t=（2*R5*R6*R7+3*R5*R6+3*R5+3）*T （“*”表示乘法，T表示一个机器周期的时间）近似延时时间公式：t=2*R5*R6*R7 *T 假如data1,data2,data3分别为50，40，248，并假定单片机晶振为12M，一个机器周期为10-6S，则10分钟后，时钟超前量超过1.11秒，24小时后时钟超前159.876秒（约2分40秒）。这都是data1,data2,data3三个数字造成的，精度比较差，建议C描述。

上表中e=-1的行（共11行）满足（2*R5*R6*R7+3*R5*R6+3*R5+3）=999，999 e=1的行（共2行）满足（2*R5*R6*R7+3*R5*R6+3*R5+3）=1，000，001 假如单片机晶振为12M，一个机器周期为10-6S，若要得到精确的延时一秒的子程序，则可以在之程序的Ret返回指令之前加一个机器周期为1的指令（比如nop指令）， data1,data2,data3选择e=-1的行。比如选择第一个e=-1行，则精确的延时一秒的子程序可以写成： delay:mov R5,#167 d1:mov R6,#171 d2:mov R7,#16 d3:djnz R7,d3 djnz R6,d2

djnz R5,d1 nop ；注意不要遗漏这一句 Ret 附： #include"iostReam.h" #include"math.h" int x=1,y=1,z=1,a,b,c,d,e(999989),f(0),g(0),i,j,k; void main() { foR(i=1;i<255;i++) { foR(j=1;j<255;j++) { foR(k=1;k<255;k++) { d=x*y*z*2+3*x*y+3*x+3-1000000; if(d==-1) { e=d;a=x;b=y;c=z; f++; cout<<"e="<

常用介词用法(for to with of)

For的用法 1. 表示“当作、作为”。如: I like some bread and milk for breakfast. 我喜欢把面包和牛奶作为早餐。 What will we have for supper? 我们晚餐吃什么? 2. 表示理由或原因,意为“因为、由于”。如: Thank you for helping me with my English. 谢谢你帮我学习英语。 3. 表示动作的对象或接受者,意为“给……”、“对…… (而言)”。如: Let me pick it up for you. 让我为你捡起来。 Watching TV too much is bad for your health. 看电视太多有害于你的健康。 4. 表示时间、距离,意为“计、达”。如: I usually do the running for an hour in the morning. 我早晨通常跑步一小时。 We will stay there for two days. 我们将在那里逗留两天。 5. 表示去向、目的,意为“向、往、取、买”等。如: Let’s go for a walk. 我们出去散步吧。 I came here for my schoolbag.我来这儿取书包。 I paid twenty yuan for the dictionary. 我花了20元买这本词典。 6. 表示所属关系或用途,意为“为、适于……的”。如: It’s time for school. 到上学的时间了。 Here is a letter for you. 这儿有你的一封信。 7. 表示“支持、赞成”。如: Are you for this plan or against it? 你是支持还是反对这个计划? 8. 用于一些固定搭配中。如: Who are you waiting for? 你在等谁? For example, Mr Green is a kind teacher. 比如,格林先生是一位心地善良的老师。尽管for 的用法较多，但记住常用的几个就可以了。 to的用法: 一：表示相对，针对 be strange (common, new, familiar, peculiar) to This injection will make you immune to infection. 二：表示对比，比较 1：以-ior结尾的形容词，后接介词to表示比较，如：superior ,inferior,prior,senior,junior 2: 一些本身就含有比较或比拟意思的形容词，如equal，similar，equivalent，analogous A is similar to B in many ways.

of和for的用法

of 1....的,属于 One of the legs of the table is broken. 桌子的一条腿坏了。 Mr.Brown is a friend of mine. 布朗先生是我的朋友。 2.用...做成的;由...制成 The house is of stone. 这房子是石建的。 3.含有...的;装有...的 4....之中的;...的成员 Of all the students in this class,Tom is the best. 在这个班级中,汤姆是最优秀的。 5.(表示同位) He came to New York at the age of ten. 他在十岁时来到纽约。 6.(表示宾格关系) He gave a lecture on the use of solar energy. 他就太阳能的利用作了一场讲演。 7.(表示主格关系) We waited for the arrival of the next bus. 我们等待下一班汽车的到来。

I have the complete works of Shakespeare. 我有莎士比亚全集。 8.来自...的;出自 He was a graduate of the University of Hawaii. 他是夏威夷大学的毕业生。 9.因为 Her son died of hepatitis. 她儿子因患肝炎而死。 10.在...方面 My aunt is hard of hearing. 我姑妈耳朵有点聋。 11.【美】(时间)在...之前 12.(表示具有某种性质) It is a matter of importance. 这是一件重要的事。 For 1.为,为了 They fought for national independence. 他们为民族独立而战。 This letter is for you. 这是你的信。

单片机C延时时间怎样计算

C程序中可使用不同类型的变量来进行延时设计。经实验测试，使用unsigned char类型具有比unsigned int更优化的代码，在使用时应该使用unsigned char作为延时变量。以某晶振为12MHz的单片机为例，晶振为12M H z即一个机器周期为1u s。一. 500ms延时子程序程序: void delay500ms(void) { unsigned char i,j,k; for(i=15;i>0;i--) for(j=202;j>0;j--) for(k=81;k>0;k--); } 计算分析: 程序共有三层循环一层循环n:R5*2 = 81*2 = 162us DJNZ 2us 二层循环m:R6*(n+3) = 202*165 = 33330us DJNZ 2us + R5赋值 1us = 3us 三层循环: R7*(m+3) = 15*33333 = 499995us DJNZ 2us + R6赋值 1us = 3us

循环外: 5us 子程序调用 2us + 子程序返回2us + R7赋值 1us = 5us 延时总时间 = 三层循环 + 循环外 = 499995+5 = 500000us =500ms 计算公式:延时时间=[(2*R5+3)*R6+3]*R7+5 二. 200ms延时子程序程序: void delay200ms(void) { unsigned char i,j,k; for(i=5;i>0;i--) for(j=132;j>0;j--) for(k=150;k>0;k--); } 三. 10ms延时子程序程序: void delay10ms(void) { unsigned char i,j,k; for(i=5;i>0;i--) for(j=4;j>0;j--) for(k=248;k>0;k--);

for和to区别

1.表示各种“目的”，用for （1）What do you study English for 你为什么要学英语？（2）went to france for holiday. 她到法国度假去了。（3）These books are written for pupils. 这些书是为学生些的。（4）hope for the best, prepare for the worst. 作最好的打算，作最坏的准备。 2．“对于”用for （1）She has a liking for painting. 她爱好绘画。（2）She had a natural gift for teaching. 她对教学有天赋/ 3．表示“赞成、同情”，用for （1）Are you for the idea or against it 你是支持还是反对这个想法？（2）He expresses sympathy for the common people.. 他表现了对普通老百姓的同情。（3）I felt deeply sorry for my friend who was very ill. 4. 表示“因为，由于”（常有较活译法），用for （1）Thank you for coming. 谢谢你来。

（2）France is famous for its wines. 法国因酒而出名。 5．当事人对某事的主观看法，“对于（某人），对…来说”，（多和形容词连用），用介词to，不用for. （1）He said that money was not important to him. 他说钱对他并不重要。（2）To her it was rather unusual. 对她来说这是相当不寻常的。（3）They are cruel to animals. 他们对动物很残忍。 6．和fit, good, bad, useful, suitable 等形容词连用，表示“适宜，适合”，用for。（1）Some training will make them fit for the job. 经过一段训练，他们会胜任这项工作的。（2）Exercises are good for health. 锻炼有益于健康。（3）Smoking and drinking are bad for health. 抽烟喝酒对健康有害。（4）You are not suited for the kind of work you are doing. 7. 表示不定式逻辑上的主语，可以用在主语、表语、状语、定语中。（1）It would be best for you to write to him. （2） The simple thing is for him to resign at once.

51单片机延时时间计算和延时程序设计

一、关于单片机周期的几个概念 ●时钟周期时钟周期也称为振荡周期，定义为时钟脉冲的倒数（可以这样来理解，时钟周期就是单片机外接晶振的倒数，例如12MHz的晶振，它的时间周期就是1/12 us），是计算机中最基本的、最小的时间单位。在一个时钟周期内，CPU仅完成一个最基本的动作。 ●机器周期完成一个基本操作所需要的时间称为机器周期。以51为例,晶振12M，时钟周期(晶振周期)就是(1/12)μs，一个机器周期包执行一条指令所需要的时间，一般由若干个机器周期组成。指令不同，所需的机器周期也不同。对于一些简单的的单字节指令，在取指令周期中，指令取出到指令寄存器后，立即译码执行，不再需要其它的机器周期。对于一些比较复杂的指令，例如转移指令、乘法指令，则需要两个或者两个以上的机器周期。 1.指令含义 DJNZ：减1条件转移指令这是一组把减1与条件转移两种功能结合在一起的指令，共2条。 DJNZ Rn，rel ；Rn←（Rn）-1 ；若（Rn）=0，则PC←（PC）+2 ；顺序执行；若（Rn）≠0，则PC←（PC）+2+rel，转移到rel所在位置DJNZ direct，rel ；direct←（direct）-1 ；若（direct）= 0，则PC←（PC）+3；顺序执行；若（direct）≠0，则PC←（PC）+3+rel，转移到rel 所在位置 2.DJNZ Rn,rel指令详解例：

MOV R7，#5 DEL:DJNZ R7,DEL; rel在本例中指标号DEL 1.单层循环由上例可知，当Rn赋值为几，循环就执行几次，上例执行5次，因此本例执行的机器周期个数=1（MOV R7，#5）+2（DJNZ R7,DEL）×5=11，以12MHz的晶振为例，执行时间（延时时间）=机器周期个数×1μs=11μs，当设定立即数为0时，循环程序最多执行256次，即延时时间最多256μs。 2.双层循环 1）格式： DELL:MOV R7，#bb DELL1:MOV R6，#aa DELL2:DJNZ R6,DELL2; rel在本句中指标号DELL2 DJNZ R7,DELL1; rel在本句中指标号DELL1 注意：循环的格式，写错很容易变成死循环，格式中的Rn和标号可随意指定。 2）执行过程

双宾语 to for的用法

1.两者都可以引出间接宾语，但要根据不同的动词分别选用介词to 或for：(1) 在give, pass, hand, lend, send, tell, bring, show, pay, read, return, write, offer, teach, throw 等之后接介词to。如：请把那本字典递给我。正：Please hand me that dictionary. 正：Please hand that dictionary to me. 她去年教我们的音乐。正：She taught us music last year. 正：She taught music to us last year. (2) 在buy, make, get, order, cook, sing, fetch, play, find, paint, choose,prepare, spare 等之后用介词for 。如：他为我们唱了首英语歌。正：He sang us an English song. 正：He sang an English song for us. 请帮我把钥匙找到。正：Please find me the keys. 正：Please find the keys for me. 能耽搁你几分钟吗(即你能为我抽出几分钟吗)? 正：Can you spare me a few minutes? 正：Can you spare a few minutes for me? 注：有的动词由于搭配和含义的不同，用介词to 或for 都是可能的。如：do sb a favour＝do a favour for sb 帮某人的忙 do sb harm＝do harm to sb 对某人有害

双宾语tofor的用法

1. 两者都可以引出间接宾语，但要根据不同的动词分别选用介词to 或for： (1) 在give, pass, hand, lend, send, tell, bring, show, pay, read, return, write, offer, teach, throw 等之后接介词to。如：请把那本字典递给我。正：Please hand me that dictionary. 正：Please hand that dictionary to me. 她去年教我们的音乐。正：She taught us music last year. 正：She taught music to us last year. (2) 在buy, make, get, order, cook, sing, fetch, play, find, paint, choose,prepare, spare 等之后用介词for 。如：他为我们唱了首英语歌。正：He sang us an English song. 正：He sang an English song for us. 请帮我把钥匙找到。正：Please find me the keys. 正：Please find the keys for me. 能耽搁你几分钟吗(即你能为我抽出几分钟吗)? 正：Can you spare me a few minutes? 正：Can you spare a few minutes for me? 注：有的动词由于搭配和含义的不同，用介词to 或for 都是可能的。如： do sb a favou r do a favour for sb 帮某人的忙 do sb harnn= do harm to sb 对某人有害

for和of的用法

for的用法： 1. 表示“当作、作为”。如: I like some bread and milk for breakfast. 我喜欢把面包和牛奶作为早餐。 What will we have for supper? 我们晚餐吃什么? 2. 表示理由或原因,意为“因为、由于”。如: Thank you for helping me with my English. 谢谢你帮我学习英语。 Thank you for your last letter. 谢谢你上次的来信。 Thank you for teaching us so well. 感谢你如此尽心地教我们。 3. 表示动作的对象或接受者,意为“给……”、“对…… (而言)”。如: Let me pick it up for you. 让我为你捡起来。 Watching TV too much is bad for your health. 看电视太多有害于你的健康。 4. 表示时间、距离,意为“计、达”。如:

I usually do the running for an hour in the morning. 我早晨通常跑步一小时。 We will stay there for two days. 我们将在那里逗留两天。 5. 表示去向、目的,意为“向、往、取、买”等。如: Let’s go for a walk. 我们出去散步吧。 I came here for my schoolbag.我来这儿取书包。 I paid twenty yuan for the dictionary. 我花了20元买这本词典。 6. 表示所属关系或用途,意为“为、适于……的”。如: It’s time for school. 到上学的时间了。 Here is a letter for you. 这儿有你的一封信。 7. 表示“支持、赞成”。如: Are you for this plan or against it? 你是支持还是反对这个计划? 8. 用于一些固定搭配中。如:

英语形容词和of for 的用法

加入收藏夹主题：介词试题It’s + 形容词 + of sb. to do sth.和It’s + 形容词 + for sb. to do sth.的用法区别。内容： It's very nice___pictures for me. A.of you to draw B.for you to draw C.for you drawing C.of you drawing 提交人：杨天若时间：1/23/2008 20:5:54 主题：for 与of 的辨别内容：It's very nice___pictures for me. A.of you to draw B.for you to draw C.for you drawing C.of you drawing 答：选A 解析：该题考查的句型It’s + 形容词+ of sb. to do sth.和It’s +形容词+ for sb. to do sth.的用法区别。 “It’s + 形容词+ to do sth.”中常用of或for引出不定式的行为者，究竟用of sb.还是用for sb.，取决于前面的形容词。 1) 若形容词是描述不定式行为者的性格、品质的，如kind，good，nice，right，wrong，clever，careless，polite，foolish等，用of sb. 例： It’s very kind of you to help me. 你能帮我，真好。 It’s clever of you to work out the maths problem. 你真聪明，解出了这道数学题。 2) 若形容词仅仅是描述事物，不是对不定式行为者的品格进行评价，用for sb.，这类形容词有difficult，easy，hard，important，dangerous，（im）possible等。例： It’s very dangerous for children to cross the busy street. 对孩子们来说，穿过繁忙的街道很危险。 It’s difficult for u s to finish the work. 对我们来说，完成这项工作很困难。 for 与of 的辨别方法：用介词后面的代词作主语，用介词前边的形容词作表语，造个句子。如果道理上通顺用of，不通则用for. 如： You are nice.(通顺，所以应用of)。 He is hard.(人是困难的，不通，因此应用for.) 由此可知，该题的正确答案应该为A项。提交人：f7_liyf 时间：1/24/2008 11:18:42

to和for的用法有什么不同(一)

to和for的用法有什么不同（一）一、引出间接宾语时的区别两者都可以引出间接宾语，但要根据不同的动词分别选用介词to 或for，具体应注意以下三种情况： 1. 在give, pass, hand, lend, send, tell, bring, show, pay, read, return, write, offer, teach, throw 等之后接介词to。如：请把那本字典递给我。正：Please hand me that dictionary. 正：Please hand that dictionary to me. 她去年教我们的音乐。正：She taught us music last year. 正：She taught music to us last year. 2. 在buy, make, get, order, cook, sing, fetch, play, find, paint, choose, prepare, spare 等之后用介词for 。如：他为我们唱了首英语歌。正：He sang us an English song. 正：He sang an English song for us. 请帮我把钥匙找到。正：Please find me the keys. 正：Please find the keys for me. 能耽搁你几分钟吗(即你能为我抽出几分钟吗)? 正：Can you spare me a few minutes?

正：Can you spare a few minutes for me? 3. 有的动词由于用法和含义不同，用介词to 或for 都是可能的。如： do sb a favor＝do a favor for sb 帮某人的忙 do sb harm＝do harm to sb 对某人有害在有的情况下，可能既不用for 也不用to，而用其他的介词。如： play sb a trick＝play a trick on sb 作弄某人请比较： play sb some folk songs＝play some folk songs for sb 给某人演奏民歌有时同一个动词，由于用法不同，所搭配的介词也可能不同，如leave sbsth 这一结构，若表示一般意义的为某人留下某物，则用介词for 引出间接宾语，即说leave sth for sb；若表示某人死后遗留下某物，则用介词to 引出间接宾语，即说leave sth to sb。如： Would you like to leave him a message? / Would you like to leave a message for him? 你要不要给他留个话？ Her father left her a large fortune. / Her father left a large fortune to her. 她父亲死后给她留下了一大笔财产。二、表示目标或方向的区别两者均可表示目标、目的地、方向等，此时也要根据不同动词分别对待。如： 1. 在come, go, walk, move, fly, ride, drive, march, return 等动词之后通常用介词to 表示目标或目的地。如： He has gone to Shanghai. 他到上海去了。 They walked to a river. 他们走到一条河边。

延时计算

t=n*(分频/f) t：是你所需的延时时间 f：是你的系统时钟(SYSCLK) n：是你所求，用于设计延时函数的程序如下： void myDelay30s() reentrant { unsigned inti,k; for(i=0;i<4000;i++) /*系统时钟我用的是24.576MHZ，分频是12分频，达到大约10s延时*/ for(k=0;k<8000;k++); } //n=i*k |评论 2012-2-18 20:03 47okey|十四级 debu(g调试），左侧有运行时间。在你要测试的延时子函数外设一断点，全速运行到此断点。记下时间，再单步运行一步，跳到下一步。再看左侧的运行时间，将这时间减去上一个时间，就是延时子函数的延时时间了。不知能不能上图。追问在delayms处设置断点，那么对应的汇编语言LCALL是否被执行呢？还有，问问您，在C8051F020单片机中，MOV指令都是多少指令周期呢？我在KEIL下仿真得出的结果，与我通过相应的汇编语言分析的时间，总是差了很多。回答 C编译时，编译器都要先变成汇编。只想知道延时时间，汇编的你可以不去理会。只要看运行时间就好了。 at8051单片机12m晶振下，机器周期为1us，而c8051 2m晶振下为1us。keil 调试里频率默认为24m，你要设好晶振频率。

202X中考英语：to和for的区别与用法.doc

202X中考英语：to和for的区别与用法中考栏目我为考生们整理了“202X中考英语：to和for的区别与用法”，希望能帮到大家，想了解更多考试资讯，本网站的及时更新哦。 202X中考英语：to和for的区别与用法 to和for的区别与用法是什么一般情况下, to后面常接对象; for后面表示原因与目的为多。 Thank you for helping me. Thanks to all of you. to sb. 表示对某人有直接影响比如，食物对某人好或者不好就用to; for 表示从意义、价值等间接角度来说，例如对某人而言是重要的，就用for. for和to这两个介词，意义丰富，用法复杂。这里仅就它们主要用法进行比较。 1. 表示各种“目的” 1. What do you study English for? 你为什么要学英语? 2. She went to france for holiday. 她到法国度假去了。 3. These books are written for pupils. 这些书是为学生些的。 4. hope for the best, prepare for the worst. 作最好的打算，作最坏的准备。

2.对于 1.She has a liking for painting. 她爱好绘画。 2.She had a natural gift for teaching. 她对教学有天赋。 3.表示赞成同情，用for不用to. 1. Are you for the idea or against it? 你是支持还是反对这个想法? 2. He expresses sympathy for the common people.. 他表现了对普通老百姓的同情。 3. I felt deeply sorry for my friend who was very ill. 4 for表示因为，由于(常有较活译法) 1.Thank you for coming. 谢谢你来。 2. France is famous for its wines. 法国因酒而出名。 5.当事人对某事的主观看法，对于(某人)，对?来说(多和形容词连用)用介词to，不用for.. He said that money was not important to him. 他说钱对他并不重要。 To her it was rather unusual. 对她来说这是相当不寻常的。 They are cruel to animals. 他们对动物很残忍。

keep的用法及of 、for sb.句型区别

keep的用法 1. 用作及物动词 ①意为"保存；保留；保持；保守"。如： Could you keep these letters for me, please? 你能替我保存这些信吗？ ②意为"遵守；维护"。如： Everyone must keep the rules. 人人必须遵守规章制度。 The teacher is keeping order in class.老师正在课堂上维持秩序。 ③意为"使……保持某种(状态、位置或动作等)"。这时要在keep的宾语后接补足语，构成复合宾语。其中宾语补足语通常由形容词、副词、介词短语、现在分词和过去分词等充当。如：例：We should keep our classroom clean and tidy.(形容词) 我们应保持教室整洁干净。 You'd better keep the child away from the fire.(副词)你最好让孩子离火远一点。 The bad weather keeps us inside the house.(介词短语)坏天气使我们不能出门。 Don't keep me waiting for long.(现在分词)别让我等太久。 The other students in the class keep their eyes closed.(过去分词) 班上其他同学都闭着眼睛。 2. 用作连系动词构成系表结构：keep＋表语，意为"保持，继续(处于某种状态)"。其中表语可用形容词、副词、介词短语等充当。如：例：You must look after yourself and keep healthy.(形容词) 你必须照顾好自己，保持身体健康。 Keep off the grass.(副词)请勿践踏草地。 Traffic in Britain keeps to the left.(介词短语)英国的交通是靠左边行驶的。注意：一般情况下，keep后接形容词较为多见。再如： She knew she must keep calm.她知道她必须保持镇静。 Please keep silent in class.课堂上请保持安静。 3. ①keep doing sth. 意为"继续干某事"，表示不间断地持续干某事，keep后不能接不定式或表示静止状态的v-ing形式，而必须接延续性的动词。例：He kept working all day, because he wanted to finish the work on time. 他整天都在不停地工作，因为他想准时完成工作。 Keep passing the ball to each other, and you'll be OK.坚持互相传球，你们就

to of和for的区别

to , of 和for的区别 1.to有到的意思，常常和go，come，get连用引出地点。Go to school , go to the shop , go to the cinema. 常见的短语：the way to 去---的路 On one’s way to 在某人去---的路上以上的用法中，当地点是副词home，here，there等是to 要去掉。如：get home，the way here To后跟动词原形，是不定式的标志 It is +形容词+(for/of +人+)to do sth.(括号内部分可以省略) It is easy for me to learn English. It is very kind of you to lend me your money. 当形容词表示人的行为特征时用of表示to do的性质时用for Want, hope ,decide, plan , try , fail等词后跟to do I want to join the swimming club. Would like to do I’d like to play basketball with them. It is time to have a break. Next to , close to , from ---to--- 2.for 为，表示目的。 Thank you for Buy sth for sb =buy sb sth It is time for bed. Here is a letter for you.

I will study for our country. 3.of表示所属关系意思是：---的 a map of the world a friend of mine

for和of引导的不定式结构的区别

for和of引导的不定式结构的区别不定式是一种非谓语动词，不能单独作谓语，因此没有语法上的主语。但由于不定式表示的是动作，在意义上可以有它的主体。我们称之为逻辑主语。提起不定式逻辑主语，人们首先想到的会是“for+名词（宾格代词）+不定式”的复合结构。如：It is important for us to study English well．然而，有时不定式的逻辑主语须要用“of+名词（代词宾格）”才行。例如：It is kind of you to help me．而不能说：It is kind for you to help me．在选择介词“for”还是“of”时，人们往往总是凭感觉而定。有时受习惯影响，多选介词“for”。于是常出现这样的错误：It was careless for him to lose his way．It is cruel for you to do so．由于众多语法书对这种结构中使用“for”与“of”的区别介绍甚少，一些人对其概念认识尚不完全清楚，笔者认为有必要就这一问题作些探讨与介绍。一、在句中的语法作用不同 a．不定式for结构在句中可以作主、宾、表、定、状、同位语： 1．It is easy for Tom to do this work．（主语）汤姆做此工作是容易的。 2．I'd like for him to come here．（宾语）我喜欢他来这里。 3．His idea is for us to travel in two different groups．（表语）他的想法是：我们分成两组旅行。 4．Have you heard about the plan for you to go abroad．（定语）你听到让你出国的计划吗？ 5．The word is too difficult for him to pronounce well．（状语）这单词太难，他念不准。 6．In the most schools，it is the custom for the headmaster to declare the newterm start．在大部分学校，校长宣布新学期开始是一个习惯。 b．不定式of结构只能在句中作主语。 1．It was careless of him to leave his umbrella in the train．他把伞丢在火车上真是太粗心了。 2．It is awfully good of you to come to see me off at the station．谢谢你来车站送我。二、逻辑主语的名词有所不同