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SMOOTH DEPENDENCE ON PARAMETERS OF SOLUTION OF COHOMOLOGY EQUATIONS OVER ANOSOV SYSTEMS AND APPLICATIONS TO COHOMOLOGY EQUATIONS ON DIFFEOMORPHISM GROUPS R.DE LA LLAVE AND A.WINDSOR Abstract.We consider the dependence on parameters of the solutions of cohomology equations over Anosov di?eomorphisms.We show that the solutions depend on param-eters as smoothly as the data.As a consequence we prove optimal regularity results for the solutions of equations taking value in di?eomorphism groups.These results are motivated by applications to rigidity theory,dynamical systems,and geometry.In particular,in the context of di?eomorphism groups we show:Let f be a transitive Anosov di?eomorphism of a compact manifold M .Suppose that η∈C k +α(M,Di?r (N ))for a compact manifold N ,k,r ∈N ,r ≥1,and 0<α≤Lip.We show that if there exists a ?∈C k +α(M,Di?1(N ))solving ?f (x )=ηx ??x then in fact ?∈C k +α(M,Di?r (N )).1.Introduction Let f be a di?eomorphism of a compact manifold M .A cohomology equation over f is an equation of the form ?f (x )=ηx ·?x (1.1)where ηis given and we are to determine ?.The interpretation of ·varies depending on the context but is typically either a group operation or the composition of linear maps.Equations of the form (1.1)have been studied for many di?erent maps (including rotations,and horocycle ?ows).In this paper we will always consider f to be an Anosov
di?eomorphism,usually a transitive Anosov di?eomorphism.
It is relevant to point out that if f n (p )=p then,by applying (1.1)repeatedly,we obtain
?f n (p )=ηf n ?1(p )···ηp ·?p .
2R.DE LA LLAVE AND A.WINDSOR
Hence,if there is a solution?then,since?f n(p)=?p,we must have
ηf n?1p···ηp=Id.(1.2)
When f is an Anososv di?eomorphism a great deal of e?ort,starting with the pioneer-ing work[Liv71],has been devoted to showing that(1.2),supplemented with regularity and“localization”assumptions,is su?cient for the existence and regularity of?.
The main goal of this paper is to study the parameter dependence of these solutions given that the data depends smoothly on parameters.
In this paper,we will not be concerned with the existence of solutions of(1.1),rather, we will assume that a solution exists and establish smoothness with respect to parameters. Of course,there is an extensive literature establishing the existence of solution,see for example the references in[dW07].
There are several contexts in which to study cohomological equations(1.1).The most classical one is when?,ηtake values in a Lie group G.In dynamical systems and geometry,(1.1)also appears in some di?erent contexts.Given a bundle E over M,we take?to be an object de?ned on the?ber E x andηx is an action transporting objects in E x to objects in E f(x).For example,in[dW07],one can?nd an application where?x are conformal structures on E x(a subspace of T x M)andηx is the natural transportation of the conformal structure by the di?erential.We note that it is not necessary to assume that the bundle E is?nite dimensional.It su?ces to assume that the linear operators lie on a Banach algebra[BN98].Of course,when the bundle E is trivial,the linear operators on E x can be identi?ed with a matrix group,so that this geometric framework reduces to the Lie group framework with G a matrix group(or,more generally,a Banach algebra of operators).
One very interesting example,which indeed serves as the main motivation for this paper is when?x andηx are supposed to be C r di?eomorphisms of a compact manifold N and the operation in the right hand side of(1.1)is just composition.Though Di?r(N) is certainly a group under composition,it is not a Lie group since the group operation is not di?erentiable as a map from Di?r(N)to Di?r(N).The problem was considered in [NT96]where it was shown that results on(1.1)have implications for rigidity.In[NT96] one can?nd results on existence of solutions when N=T d and,in[dW07]for a general N.
Both in[NT96],[dW07],from the fact thatηx∈Di?r(N)one can only reach the conclusion that?x∈Di?r?R(N)(in[dW07]the regularity loss R=3,but in[NT96]R depends on the dimension of N).
SMOOTH DEPENDENCE3
The main goal of this paper is to overcome this loss of regularity and show that if ηx∈Di?r(N)for r≥1,and?∈Di?1(N),then?∈Di?r(N).See Theorem6for a more precise formulation.
As it turns out,the main technical tool in the proof of Theorem6is to establish results on the smooth dependence on parameters for the solutions of(1.1)which may be of independent interest.We formulate them as Theorem3,Theorem5.
When G is a commutative group smooth dependence on parameters is rather elemen-tary,see Proposition2,and our main technique is to reduce to the commutative case.
1.1.Sketch of the proofs.In this section we informally present the main ideas behind the proof omitting several of the details which we will treat later.
1.1.1.Some remarks on the relation between bootstrap of regularity and dependence on parameters.Going over the proofs in[NT96]and,more explicitly,in[dW07]it becomes apparent that,the loss of regularity of the solutions is closely related to the fact that the group operation is not di?erentiable.
Hence,we?nd it convenient to turn the tables.Rather than considering?:M×N→N as a function from M to a space of mappings on N,we consider?as a mapping from M to N with parameters from N.In this application,N plays two roles:as the ambient space for the di?eomorphisms and as the parameter.For the sake of clarity,we will develop our main technical results denoting the target space as a N and the set of parameters as U.This is,of course,more general,and in many cases,the parameters appear independently of the target space.
More precisely,if we?x u∈U and suppose that?is di?erentiable with respect to u then the chain rule tells us that
D u?f(x)(u)=D uηx??x(u)·D u?x(u)(1.3)
where D u denotes the derivative with respect to the parameter u∈U.Note that for typographical convenience we will often useηu x in place ofη(x,u),and?u x in place of ?(x,u).
We see that for each u∈U(1.3)is an equation of the form(1.1)for the cocycle D u?x(u)with generator
?ηu x=D uηx·?u x(1.4)
N.
In this case E is the bundle of linear maps from T u N to T?
x(u)
If?u x depends C1on u andηx depends C2on u,then?ηx is C1in https://www.doczj.com/doc/6018516165.html,ing the regularity theory for commutative cohomology equations we obtain that D u?u x is C1with respect to
4R.DE LA LLAVE AND A.WINDSOR
u.This means that?u x is C2with respect to u.This argument to improve the regularity can be repeated so long as we can di?erentiateηx.
1.1.
2.The dependence on parameters.The idea of the proof of the smooth dependence on parameters for(1.1)is more involved that the bootstrap of regularity since we need to justify the existence of the?rst derivative.
For simplicity of notation,we interpret(1.1)in the linear bundle maps framework. This will be the crucial technical result for the bootstrap of regularity in di?eomorphism groups.The case of Lie group valued cocycles will be dealt with in Section2.3.
We want to show that if?u,ηu solve(1.1)andηu depends smoothly on parameters, then?u depends smoothly on parameters.
However solutions of(1.1)are not unique.Taking advantage of this it is very easy to construct solutions which depend badly on parameters.It is therefore necessary to impose some additional condition such as that?u p is smooth in u∈U for some?xed p∈M.For convenience we will take p∈M to be a periodic point.
In order to prove di?erentiability we?rst?nd a candidate for the derivative using Livˇs ic methods and then argue that this candidate is the true derivative.The key observation is that if we formally di?erentiate(1.1)with respect to the parameter u we obtain
D u?u f(x)=D uηu x·?u x+ηu x·D u?u x
Using(1.1)we get
D u?u f(x)=D uηu x·?u x+?u f(x)·(?u x)?1·D u?u x(1.5) Multiplying on the right both sides of(1.5)by(?u
)?1,we obtain
f(x)
(?u f(x))?1·D u?u f(x)=(?u f(x))?1·D uηu x·?u x+(?u x)?1·D u?u x(1.6) where(?u x)?1refers to the inverse of the linear map?u x.This equation(1.6)is a coho-mology equation over a commutative group for the new unknown
ξx=(?u x)?1D u?u x.
We will show that the periodic obstruction(1.2)corresponding to(1.6)is the derivative with respect to parameters of the periodic obstruction corresponding to(1.1). Hence,applying the results on Livˇs ic equations,we obtain a solution of(1.6).It remains to show that this candidate is the true derivative.
As mentioned before,the solutions of(1.6)are not unique,but,under the hypothesis that?p(u)=:γ(u)is smooth in u for some p,which we made at the outset,it is natural
SMOOTH DEPENDENCE5 to impose that the solution to(1.6)satisfy the normalization
D u?p=D uγ.
Once we have a candidate for a derivative,we will use a comparison argument,Propo-sition4to show that indeed it is a derivative.
Of course,once we have established the existence of the?rst derivative,we have also shown that it satis?es(1.6).Hence,to establish the existence of higher derivatives it su?ces to use the(much simpler)theory of dependence of parameters in commutative cohomology equations.
The above argument uses very heavily that we are dealing with the framework of linear operators on bundles.The adaptation of the above argument to general Lie groups requires some adaptations of the geometry,see Section2.3.
2.Smooth dependence on parameters of cohomology equation
In this section,we make precise the previous arguments and establish smooth depen-dence on parameters for the solutions of(1.1).In Section2.1we make precise what we mean by smooth dependence on parameters.In Section2.2we present the results when (1.1)is interpreted as an equation between linear bundle maps.This section will be crucial for the case of di?eomorphism groups.In Section2.3we present the results for cocycles taking values in a Lie group.
2.1.Notation on regularity.We will understand that a function is C r when it admits r continuous derivatives.When the spaces we consider are not compact,we will assume that all the derivatives are bounded.
For0<α≤Lip we will de?ne real valued Cαfunctions on a metric space N in the usual way
|f(x)?f(y)
|
Hα(N)=sup
x=y
6R.DE LA LLAVE AND A.WINDSOR
We will write h∈C k+α,r if
(1)for all0≤i≤r and all0≤j≤k the derivative D j x D i u?exists and is continuous
on M×U.
(2)for all0≤i≤r and all u∈U the derivative D i u?(·,u)∈Cα(M)and the H¨o lder
constant does not depend on u.
We also de?ne?C k+α,r(U,M)=C r U,C k+α(M) .That,is the set of functions h: M×U→N such that u→h(·,u)is C r when we give?(·,u)the C k+αtopology.
The reason to introduce these spaces is that C k+α,r is natural when performing geo-metric constructions involving derivatives.The space?C k+α,r is natural when we consider dependence on parameters of commutative Livˇs ic equations.Of course,the two spaces are closely related.The relation is formulated in the following proposition. Proposition1.For any0<α′<α≤Lip,k,r∈N,we have:
?C k+α,r?C k+α,r??C k+α′,r(2.1) The simple example?(u,x)=(x?u)k+αsatis?es?∈C k+α,0,?/∈?C k+α,0.Integrating with respect to u one can obtain similar examples for all r∈N.
Proof.The?rst inclusion of(2.1)is obvious.
Let?∈C k+α,r.Letα′=α·θfor0<θ< 1.We wish to show that?∈C r(U,C k+α′(M)).This is equivalent to?i u?∈C0(U,C k+α′(M)for all multi-indices i with0≤|i|≤r.Fix u∈U and consider a compact K with u∈K?U.Restricted to M×K the function?and all its partial derivatives are uniformly continuous.So we have
?j x?i u?(·,u)??j x?i u?(·,u′) 0
is controlled by d(u,u′),for all multi-indices j with0≤|j|≤k.It remains to show that for j=k we have
?j x?i u?(·,u)??j x?i u?(·,u′) α′
is controlled.For any multi-index j with|j|=k we have from the Kolmogorov-Hadamard inequalities[dlLO99]
?j x?i u?(·,u)??j x?i u?(·,u′) α′
≤C ?j x?i u?(·,u)??j x?i u?(·,u′) θα ?j x?i u?(·,u)??j x?i u?(·,u′) 1?θ
SMOOTH DEPENDENCE7
By de?nition of?∈C k+α,r we have that ?j x?i u?(·,u) αis bounded independent of u∈U and hence
?j x?i u?(·,u)??j x?i u?(·,u′) 0
is controlled since?j x?i u?is uniformly continuous.Consequently?∈C r(U,C k+α′(M)). Note that we do not claim uniform continuity in u.
Next,we discuss the regularity properties of the commutative cohomology equation. The results for k=0appear in[Liv71,Liv72]and the results for k>0appear in [dlLMM86].The following result will be one of our basic tools later.
Proposition2.Let M be a compact manifold and f a transitive Anosov di?eomorphism on M.Let p∈M be a periodic point for f.
Consider the commutative cohomology equation
?u?f(x)??u(x)=ηu(x)
(2.2)
?(p,·)∈C r(U)
Assume that for all values of u,the periodic orbit obstruction withηu vanishes.Then,
(1)ifη∈?C k+α,r,then?∈?C k+α,r,and
(2)ifη∈C k+α,r,then?∈C k+α,r.
Proof.The?rst statement is obvious since we are considering the solution of a linear equation and[dlLMM86]showed that the solution operator is continuous from C k+αto C k+α/R.
The second statement follows because,by the previous argument,we have that?∈?C k+α′,r.In particular for|i|≤r,?i
?(·,u)∈C k+α′but this function satis?es
u
?i u?u?f??i u?u=?i uηu
Since we assumed thatη∈C k+α,r we have that?i uη(·,u)∈C k+α(M).and,by the results of[dlLMM86],?i u?(·,u)∈C k+α(M).
2.2.Cohomology Equations for Bundle Maps.As we have mentioned before,coho-mology equations for bundle maps appear naturally when we consider the linearization of a dynamical system or cohomology equations for cocycles taking values in di?eomorphism groups.
Theorem3.Let M and N be compact smooth manifolds,U?R n open,and E a bundle over N.Let0 8R.DE LA LLAVE AND A.WINDSOR Let f:M→M be a transitive C r Anosov di?eomorphism,σ:U→N withσ∈C r(N) andτ:M×U→N withτ∈C k+α,r(resp.τ∈?C k+α,r). Suppose that we have linear maps ?u x:Eσ(u)→Eτ(x,u), ηu x:Eτ(x,u)→Eτ(f(x),u). such that for all u∈U we have ?u f(x)=ηu x·?u x.(2.3) Ifη∈C k+α,r(resp.η∈?C k+α,r)and there exists a periodic point p∈M such that ?(p,·)∈C r(U)then?∈C k+α,r(resp.?∈?C k+α,r). The most di?cult case of Theorem3is the case when r=1.The higher di?erentia-bility cases follow from this one rather straightforwardly.The case r=1of Theorem3 will be the inductive step for the bootstrap of regularity for cohomology equations on di?eomorphism groups. The proof we present of the r=1case uses that f is transitive but the subsequent bootstrap argument does not require that f is transitive. As indicated before,we will obtain a candidate for a derivative and,then,we argue it is indeed a true derivative. Smooth dependence is a local question.For that reason we consider a local trivial-ization of the bundle E about the pointsσ(u)andτ(x,u).The base space portion of the map?u x is as smooth as the minimum smoothness ofσandτ.We will concentrate therefore on the smoothness of the linear operator in the?bers. Proof.We note that if we could take derivatives in(2.3)then D u?u x,the derivative of?u x with respect to u,would satisfy D u?u f(x)=D uηu x·?u x+ηu x·D u?u x(2.4) Observing thatηu x=?u ·(?u x)?1we get f(x) (?u f(x))?1·D u?u f(x)=(?u f(x))?1·D uηu x·?u x+(?u x)?1·D u?u x.(2.5) Writing ξu x:=(?u x)?1·D u?u x(2.6) we see thatξu x would satisfy ξu f(x)?ξu x=(?u f(x))?1·D uηu x·?u x(2.7) SMOOTH DEPENDENCE9 Note that(2.7)is a commutative cohomology equation with the group operation being addition on a vector space. In this case,as we will show,the methods of[Liv72,dlLMM86]to obtain existence and regularity forξ.Using(2.6)we will obtain a candidate for D u?u x. Since we are assuming that there are solutions of(2.3)for all u in an open set U,we have that for any p,f n(p)=p ηu f n?1(p)·····ηu p=Id.(2.8) Di?erentiating(2.8)with respect to u we obtain 0=D uηu f n?1(p)·ηu f n?2(p)···ηu p+ηu f n?1(p)·D uηu f n?2(p)·ηu f n?3(p)···ηu p +···+ηu f n?1(p)·ηu f n?2(p)···ηu f(p)·D uηu p. Using the(2.3)to express the products ofη,we obtain 0=D uηu f n?1(p)·?u f n?1(p)·(?u p)?1 +?u p·(?u f n?1(p))?1·D uηu f n?2(p)·?u f n?2(p)·(?u p)?1 (2.9) +···+?u p·(?u f(p))?1·D uηu p. Multiplying(2.9)on the left by(?u p)?1and multiplying by?u p on the right we obtain 0=(?u p)?1·D uηu f n?1(p)·?u f n?1(p)+(?u f n?1(p))?1·D uηu f n?2(p)·?u f n?2(p) (2.10) +···+(?u f(p))?1·D uηu p·?u p which shows that the periodic orbit obstruction for the existence of a solution to the cohomology equation(2.7)vanishes.Hence we have a solutionξu x to(2.7). Using(2.6)we see that we have a candidate for the derivative, D u?u x=?u x·ξu x. It remains to show that this candidate,which we obtained by a formal argument is actually the derivative. We start by proving that all the solutions of(2.3)are di?erentiable in the stable manifold of a periodic point.For the sake of notation,and without any loss of generality, we will consider a?xed point p. Applying(2.3)repeatedly we have ?u f n+1(x)=ηu f n(x)···ηu x·?u x.(2.11) For x∈W s p we have d(f n(x),p)≤C xλn for some0<λ<1,n≥0.Sinceηu x is H¨o lder in x,andηu p=Id is continuous,in any smooth local trivialization around p we obtain ηu f n(x)?Id ≤Cλαn. 10R.DE LA LLAVE AND A.WINDSOR Therefore we can pass to the limit in(2.11)and obtain for x∈W s p ηu f n(x)···ηu x·?u x(2.12) ?u p=lim n→∞ where the convergence in(2.12)is uniform in bounded sets of W s p(in the topology of W s p). We will show that(2.12)can be di?erentiated with respect to u.By the product rule D u ηu f n(x)···ηu x =D uηu f n(x)·ηu f n?1(x)···ηu x+ηu f n(x)·D uηu f n?1(x)·ηu f n?2(x)···ηu x +···+ηu f n(x)···ηu f(x)D uηu x =?u f n+1(x)·(?u f n+1(x))?1·D uηu f n(x)·?u f n(x)·(?u x)?1+?u f n+1(x)·(?u f n(x))?1· )?1·D uηu x·?u x·(?u x)?1 D uηu f n?1(x)·?u f n?1(x)·(?u x)?1+···+?u f n+1(x)·(?u f(x) =?u f n+1(x)· n?1 i=0(?u f i+1(x))?1·D uηu f i(x)·?u f i(x) ·(?u x)?1 (2.13) where we have used again(2.3).Since p is a?xed point using(2.9)we obtain that (?u p)?1·D uηu p·?u p=0. Because of the assumed regularity on?x,D uηu x and the exponentially fast convergence of f n(x)to p,we obtain that the general term in(2.13)converges to zero exponentially. By the Weierstrass M-test we obtain that D u ηu f n(x)···ηu x converges uniformly on compact subsets of W s p.Therefore,the limit is the derivative of lim n→∞ηf n(x)···ηx and from that,it follows immediately that?u x is di?erentiable with respect to u for x∈W s p. Of course,the argument so far allows only to conclude that D u?u x is bounded on bounded sets of W s p.This does not show that it is bounded on M since W s p is unbounded. Nevertheless,we realize that the derivative solves(2.4)on W s p.Of course,so does the candidate for the derivative that we produced earlier.From the fact that these two functions satisfy the same functional equation in W s p,and that they agree at zero,we will show that they agree.This will allow us to conclude that the candidate is indeed the derivative in W s https://www.doczj.com/doc/6018516165.html,ing that it is a continuous function on the whole manifold,there is a standard argument that shows it is the derivative everywhere. Proposition4.Let p∈M be a?xed point of f.Letξ1,ξ2be continuous functions on W s p.Assume thatξ1(p)=ξ2(p)and that both of them satisfy(2.7).Thenξ1(x)=ξ2(x) for all x∈W s p. SMOOTH DEPENDENCE11 Proof.Note that A(x)≡ξ1(x)?ξ2(x)satis?es A(x)=A(f(x)).For every x∈W s p we have lim n→∞f n(x)=p.Since A is continuous,and A(p)=0we see that A(x)=A f n(x) =lim n→∞A f n(x) =A lim n→∞f n(x) =A(p) =0. Thusξ1(x)=ξ2(x)for all x∈W s x. We have now shown that the function is di?erentiable in a leaf of the foliation which is dense.Furthermore,D u?u x extends continuously–indeed H¨o lder continuously–to the whole manifold.In this case,an elementary real analysis argument,which we will present now,shows that?u x is everywhere di?erentiable. Since we just want to show that the continuous function is indeed a derivative in the stable leaf,we can just take a trivialization of the bundles in a small neighborhood.We will use the same notation for the objects in the trivialization and the corresponding one in the bundles.This allows us to subtract objects in di?erent?bers. Ifψ(s)is any smooth path in U contained in the trivializing neighborhood withψ(0)= u0,ψ1=u1we have,for x∈W s p ?u1x??u2x= 10D u?ψ(s)xψ′(s)ds(2.14) Both sides of the equation are H¨o lder in x.Therefore,we can pass to the limit in(2.14) and conclude that the set of points x for which(2.14)holds is closed.But we had already shown that(2.14)held for x∈W s p,which is a dense set since f is transitive.Thus(2.14) holds for every x∈M.This shows immediately that our candidate is indeed the true derivative and concludes the proof of the case r=1of Theorem3. It is now relatively simple to obtain higher derivatives.The auxiliary functionξsatis?es the commutative cohomology equation(2.7).By assumption D uη∈C k+α,r?1(resp. D uη∈?C k+α,r?1)hence,if?∈C k+α,m(resp.?∈?C k+α,m)for m≤r?1then the right hand side of(2.7)is in C k+α,m(resp.?C k+α,m).Applying Proposition2we then obtain D u?∈C k+α,m(resp.D u?∈?C k+α,m)and thus?∈C k+α,m+1(resp.?∈C k+α,m+1).The induction stops at m=r?1when we have exhausted the regularity of D uη.At this point?∈C k+α,r(resp.?∈?C k+α,r)as required. 12R.DE LA LLAVE AND A.WINDSOR 2.3.Cohomology Equations for Lie Group Valued Cocycles.In this section we consider the dependence on parameters of the solution to(1.1)whenη,and?are func-tions taking values in a Lie group G.We denote the Lie algebra of the Lie group G by g and we denote the identity in G by e. The proof follows along the same lines as the proof of Theorem3.We derive a func-tional equation for a candidate for the?rst derivative,show that there is a solution for this equation,and that the candidate is a true derivative.Finally we use a bootstrapping argument to get full regularity.The main di?erence with the previous section is that we have to deal with the fact that the group operation is not just the product of linear operations,so that the derivatives of the functional equation involve the derivatives of the group operation. We introduce the notation L g h=g·h R g h=h·g (2.15) where·denotes the group operation. It follows directly from the de?nitions of L,R that L g?L h=L g·h R g?R h=R h·g (2.16) It is immediate to show that if g u,h u are smooth families D u(g u·h u)=DL g u(h u)D u h u+DR h u(g u)D u g u(2.17) Theorem 5.Let M be a compact manifold,U?R d open,G be a Lie group,f a transitive Anosov di?eomorphism of M,and p∈M a periodic point for f.Suppose that η:M×U→G withη∈C k+α,r(resp.η∈?C k+α,r)for0<α≤Lip and k,r∈N with r≥1. If?:M×U→G solves ?u f(x)=ηu x·?u x and?p∈C r(U,G)then?∈C k+α,r(resp.?∈?C k+α,r). The de?nition of the spaces C k+α,r and?C k+α,r appears in Section2.1. Proof.Taking derivatives of(1.1)and applying the product rule we obtain that if there is a derivative of D u?u x,it should satisfy: D u?u f(x)=DLηu x (?u x)D u?u x+DR?u x (ηu x)D uηu x(2.18) SMOOTH DEPENDENCE13 We introduce a functionξ:M×U→g by D u?u x=DL?u x (e)ξu x(2.19) Introducing the notation(2.19)is geometrically natural because we want to transport all the in?nitesimal derivatives to the identity,so thatξu x takes values in the Lie algebra g. In terms ofξ,the equation(2.18)becomes DL?u f(x)(e)ξu f(x)=DLηu x (?u x)DL?u x (e)ξu x+DR?u x (ηu x)D uηu x.(2.20) The?rst factor of the?rst term in(2.20)can be simpli?ed DLηu x (?u x)·DL?u x (e)=D u Lηu x?L?u x (e)=D u Lηu x·?u x (e)=DL?u f(x)(e).(2.21) Substituting(2.21)into(2.20)we obtain DL?u f(x)(e)ξu f(x)=DL?u f(x) (e)ξu x+DR?u x (ηu x)D uηu x.(2.22) Multiplying(2.22)in the left by DL?u f(x)(e) ?1we obtain: ξu f(x)=ξu x+ DL?u f(x)(e) ?1DR?u x(ηu x)D uηu x(2.23) Equation(2.23)can be simpli?ed further;from(2.16),we have Id=L(?u f(x) )?1?L?u f(x) where Id is the identity map on G.Thus,by the chain rule,we have Id=DL(?u f(x) )?1(?u f(x))DL?u f(x)(e) where Id is the identity map on g.Hence DL?u f(x)(e) ?1=DL(?u f(x))?1(?u f(x)).(2.24) Therefore,(2.23)can be rewritten as ξu f(x)=ξu x+DL(?u f(x) )?1(?u f(x))DR?u x(ηu x)D uηu x.(2.25) We are assuming?u p is a C r function of u.Thus,taking derivatives,and using(2.19) we obtain D u?u p=DL?u p (e)ξu p(2.26) so we see thatξp:U→g is C r. In summary,if the function?u x is di?erentiable then the g valued functionξintroduced in(2.19)would satisfy(2.25).Equation(2.25)is a cohomology equation for functions taking values in g.It is therefore a commutative cohomology equation.Thus a necessary and su?cient condition for the existence of a smooth solution is the vanishing of the periodic orbit obstruction. 14R.DE LA LLAVE AND A.WINDSOR In the next paragraphs we will show that indeed this periodic orbit obstruction is met, so that indeed one can?nd aξsolving(2.25).Since(1.1)holds for all u,if f n(p)=p, we have for all u e=ηu f n?1(p)···ηu p =(?u p)?1·ηu f n?1(p)···ηu p·?u p Di?erentiating with respect to u,we obtain: 0=DRηu f n?1(p) ···ηu p·?u p (?u p)?1 D u(?u p)?1 +DL(?u p )?1(ηu f n?1(p)···ηu p·?u p)DRηu f n?2(p)···ηu p·?u p(ηu f n?1(p))D uηu f n?1(p) +···+DL(?u p )?1·ηu f n?1(p) ···ηu f(p) (ηu p?u p)DR?u p (ηu p)D uηu p +DL(?u p )?1·ηu f n?1(p) ···ηu p (?u p)D u?u p. (2.27) Using(1.1)to reduce the products,we transform(2.27)into 0=DR?u p (?u p)?1 D u(?u p)?1 +DL(?u p )?1(?u p)DR?u f n?1(p)(ηu f n?1(p))D uηu f n?1(p) +DL(?u f n?1(p) )?1(?u f n?1(p))DR?u f n?2(p)(ηu f n?2(p))D uηu f n?2(p) +···+DL(?u f(p) )?1(?u f(p))DR?u p(ηu p)D uηu p +DL(?u p )?1(?u p)D u?u p. (2.28) Finally we observe that DR?u p (?u p)?1 D u(?u p)?1+DL(?u p)?1(?u p)D u?u p=D u (?u p)?1·?u p =0 and hence(2.28)becomes 0=DL(?u p )?1(?u p)DR?u f n?1(p)(ηu f n?1(p))D uηu f n?1(p) +DL(?u f n?1(p) )?1(?u f n?1(p))DR?u f n?2(p)(ηu f n?2(p))D uηu f n?2(p) +···+DL(?u f2(p) )?1(?u f2(p))DR?u f(p)(ηu f(p))D uηu f(p) +DL(?u f(p) )?1(?u f(p))DR?u p(ηu p)D uηu p which shows the vanishing of the periodic orbit obstruction for(2.25). The rest of the argument does not need any modi?cation from the argument in the previous case and we just refer to the previous section.We argue that?is di?erentiable in W s p for some p∈M and that the candidate for the drivative is indeed the true derivative. SMOOTH DEPENDENCE15 3.Cohomology equations on diffeomorphism groups In this section,we consider(1.1)whenηand?take values in the di?eomorphism group of a compact manifold.As before f is a transitive Anosov di?eomorphism.In order to emphasize that the parameter space and target space are the same smooth manifold N and to match the notation in[dW07]we use y in place of u. Theorem6.Letη∈C k+α M,Di?r(N) and p is a periodic point of f. If?∈C k+α M,Di?1(N) solves ?f(x)=ηx??x(3.1) and?p∈Di?r(N)then?∈C k+α M,Di?r(N) . Notice that questions of di?erentiability is entirely local and hence the global structure of the di?emorphism group,which is quite complicated,does not enter into the argument. This should be compared with the existence argument in[dW07]for k=0which devotes considerable e?ort to de?ning the metric on Di?r(N).The local di?erential structure on Di?r(N)can be found in[Ban97].We remark that to show that?:M→Di?r(N)is C k+αit su?ces to show that all partial derivatives in N are C k+αuniformly. Proof.Taking derivatives of(1.1)with respect to the variable y in the manifold N we obtain D y?f(x)(y)=D yηx??x(y)·D y?x(y). This is possible since for all x we have?x∈Di?1(N).Now we writeψy x=D y?x(y)and ?ηy x=D yηx??x(y).The equation is then ψy f(x) =?ηy x·ψy x We have ψy x:T y N→T? x(y) N ?ηy x:T? x(y)N→T? f(x) (y) N. Thus we have the set up of Theorem3withσ(y)=y andτ(x,y)=?y x.Clearlyσ∈C∞(N). If we assume that?∈C k+α(M,Di?m(N))thenτ∈C k+α,m.By hypothesisη∈ C k+α(M,Di?r(N))thus for m≤r?1?η∈C k+α,m.Applying Theorem3we obtain that D y?∈C k+α,m.Thus?∈C k+α,m+1.We proceed by induction until m=r?1at which point we have?∈C k+α,r. 16R.DE LA LLAVE AND A.WINDSOR We have addressed on?x and not(?x)?1however since we know that?x∈Di?1(N) the inverse function theorem shows that(?x)?1is as smooth as?x and depends on parameters with the same smoothness. Remark.Note that the main result of[dW07]is that if the periodic orbit obstruction is met,f is hyperbolic enough,andη∈Cα M,Di?r(N) ,r≥4is close enough to the identity,then the cohomology equation(3.1)has a solution?∈Cα M,Di?1(N) .Hence we can apply Theorem6to show that in fact?∈Cα M,Di?r(N) . Remark.Though the existence result requires both a localization assumption and a rela-tion between the H¨o lder exponent and the hyperbolicity of f the bootstrap result requires none of these assumptions. Acknowledgements Alistair Windsor gratefully acknowledges the hospitality of the University of Texas at Austin.R.L.was partially supported by NSF grants. References [Ban97]Augustin Banyaga.The structure of classical di?eomorphism groups.Kluwer Academic Pub-lishers Group,Dordrecht,1997. [BN98]Hari Bercovici and Viorel Nit?ic?a.A Banach algebra version of the Livsic theorem.Discrete Contin.Dynam.Systems,4(3):523–534,1998. [CFdlL03]Xavier Cabr′e,Ernest Fontich,and Rafael de la Llave.The parameterization method for invariant manifolds.II.Regularity with respect to parameters.Indiana Univ.Math.J., 52(2):329–360,2003. [dlLMM86]R.de la Llave,J.M.Marco,and R.Moriy′o n.Canonical perturbation theory of Anosov systems and regularity results for the Livˇs ic cohomology equation.Ann.of Math.(2), 123(3):537–611,1986. [dlLO99]R.de la Llave and R.Obaya.Regularity of the composition operator in spaces of H¨o lder functions.Discrete Contin.Dynam.Systems,5(1):157–184,1999. [dW07]Rafael de la Llave and Alistair Windsor.Livˇs ic theorems for non-commutative groups includ-ing di?eomorphism groups and results on the existence of conformal structures for Anosov systems,2007. [HP69]Morris W.Hirsch and Charles C.Pugh.Stable manifolds for hyperbolic sets.Bull.Amer. Math.Soc.,75:149–152,1969. [Liv71] A.N.Livˇs ic.Certain properties of the homology of Y-systems.Mat.Zametki,10:555–564, 1971. [Liv72] A.N.Livˇs ic.Cohomology of dynamical systems.Izv.Akad.Nauk SSSR Ser.Mat.,36:1296–1320,1972. [NT96]Viorel Nit?ic?a and Andrei T¨o r¨o k.Regularity results for the solutions of the Livsic cohomology equation with values in di?eomorphism groups.Ergodic Theory Dynam.Systems,16(2):325– 333,1996. 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My father was in the Xinjiang half a year ago. 10.My grandfather was a famer six years ago. 11.He burned in 1991 有限元分析中的网格划分好坏直接关系到模型计算的准确性。本文简述了网格划分应用的基本理论，并以ANSYS限元分析中的网格划分为实例对象，详细讲述了网格划分基本理论及其在工程中的实际应用，具有一定的指导意义。 1 引言 ANSYS有限元网格划分是进行数值模拟分析至关重要的一步，它直接影响着后续数值计算分析结果的精确性。网格划分涉及单元的形状及其拓扑类型、单元类型、网格生成器的选择、网格的密度、单元的编号以及几何体素。从几何表达上讲，梁和杆是相同的，从物理和数值求解上讲则是有区别的。同理，平面应力和平面应变情况设计的单元求解方程也不相同。在有限元数值求解中，单元的等效节点力、刚度矩阵、质量矩阵等均用数值积分生成，连续体单元以及壳、板、梁单元的面内均采用高斯（Gauss）积分，而壳、板、梁单元的厚度方向采用辛普生（Simpson）积分。辛普生积分点的间隔是一定的，沿厚度分成奇数积分点。由于不同单元的刚度矩阵不同，采用数值积分的求解方式不同，因此实际应用中，一定要采用合理的单元来模拟求解。 2 ANSYS网格划分的指导思想 ANSYS网格划分的指导思想是首先进行总体模型规划，包括物理模型的构造、单元类型的选择、网格密度的确定等多方面的内容。在网格划分和初步求解时，做到先简单后复杂，先粗后精，2D单元和3D单元合理搭配使用。为提高求解的效率要充分利用重复与对称等特征，由于工程结构一般具有重复对称或轴对称、镜象对称等特点，采用子结构或对称模型可以提高求解的效率和精度。利用轴对称或子结构时要注意场合，如在进行模态分析、屈曲分析整体求解时，则应采用整体模型，同时选择合理的起点并设置合理的坐标系，可以提高求解的精度和效率，例如，轴对称场合多采用柱坐标系。有限元分析的精度和效率与单元的密度和几何形状有着密切的关系，按照相应的误差准则和网格疏密程度，避免网格的畸形。在网格重划分过程中常采用曲率控制、单元尺寸与数量控制、穿透控制等控制准则。在选用单元时要注意剪力自锁、沙漏和网格扭曲、不可压缩材 一个网格划分实例的详解 该题目条件如下图所示： Part 1：本部分将平台考虑成蓝色的虚线 1. 画左边的第一部分，有多种方案。 方法一：最简单的一种就是不用布置任何初始的2dmesh直接用one volume 画，画出来的质量相当不错。 One volume是非常简单而且强大的画法，只要是一个有一个方向可以 mapped的实体都可以用这个方法来画网格，而事实上，很多不能map的单元也都可以用这个命令来画，所以在对三维实体进行网格划分的时候，收件推荐用one volume来试下效果，如果效果不错的话，就没有必要先做二维单元后再来画。 方法二：先在其一个面上生成2D的mesh，在来利用general选项，这样的优点是可以做出很漂亮的网格。 相比之下：方法二所做出来的网格质量要比一要高。 2. 画第二段的网格，同样演示两种方法： 方法一：直接用3D>solid map>one volume 方法二：从该段图形来看，左端面实际上由3个面组成，右端面由一个部分组成，故可以先将左端面的另两个部分的面网格补齐，再用general选项来拉伸，但是，问题是左面砖红色的部分仅为3D单元，而没有可供拉伸的源面网格，故，应该先用face命令生成二维网格后，再来拉伸，其每一步的结果分见下： 在用general选项时，有个问题需要注意：在前面我们说过，source geom和elemes to drag二选一都可以，但是这里就不一样了，因为source geom选面的话，只能选择一个面，而此处是3个面，所以这里只能选elemes to drag而不能选择source geom. ●I wonder if it’s because I have been at school for so long that I’ve grown so crazy about going home. ●It is because she wasn’t well that she fell far behind her classmates this semester. ●I can well remember that there was a time when I took it for granted that friends should do everything for me. ●In order to make a difference to society, they spent almost all of their spare time in raising money for the charity. ●It’s no pleasure eating at school any longer because the food is not so tasty as that at home. ●He happened to be hit by a new idea when he was walking along the riverbank. ●I wonder if I can cope with stressful situations in life independently. ●It is because I take things for granted that I make so many mistakes. ●The treasure is so rare that a growing number of people are looking for it. ●He picks on the weak mn in order that we may pay attention to him. ●It’s no pleasure being disturbed whena I settle down to my work. ●I can well remember that when I was a child, I always made mistakes on purpose for fun. ●It’s no pleasure accompany her hanging out on the street on such a rainy day. ●I can well remember that there was a time when I threw my whole self into study in order to live up to my parents’ expectation and enter my dream university. ●I can well remember that she stuck with me all the time and helped me regain my confidence during my tough time five years ago. ●It is because he makes it a priority to study that he always gets good grades. ●I wonder if we should abandon this idea because there is no point in doing so. ●I wonder if it was because I ate ice-cream that I had an upset student this morning. ●It is because she refused to die that she became incredibly successful. ●She is so considerate that many of us turn to her for comfort. ●I can well remember that once I underestimated the power of words and hurt my friend. ●He works extremely hard in order to live up to his expectations. ●I happened to see a butterfly settle on the beautiful flower. ●It’s no pleasure making fun of others. ●It was the first time in the new semester that I had burned the midnight oil to study. ●It’s no pleasure taking everything into account when you long to have the relaxing life. ●I wonder if it was because he abandoned himself to despair that he was killed in a car accident when he was driving. ●Jack is always picking on younger children in order to show off his power. ●It is because he always burns the midnight oil that he oversleeps sometimes. ●I happened to find some pictures to do with my grandfather when I was going through the drawer. ●It was because I didn’t dare look at the failure face to face that I failed again. ●I tell my friend that failure is not scary in order that she can rebound from failure. ●I throw my whole self to study in order to pass the final exam. ●It was the first time that I had made a speech in public and enjoyed the thunder of applause. ●Alice happened to be on the street when a UFO landed right in front of her. ●It was the first time that I had kept myself open and talked sincerely with my parents. ●It was a beautiful sunny day. The weather was so comfortable that I settled myself into the 高中英语~词性~句子成分~语法构成 第一章节：英语句子中的词性 1.名词：n. 名词是指事物的名称，在句子中主要作主语.宾语.表语.同位语。 2.形容词;adj. 形容词是指对名词进行修饰~限定~描述~的成份，主要作定语.表语.。形容词在汉语中是（的）.其标志是: ous. Al .ful .ive。. 3.动词：vt. 动词是指主语发出的一个动作，一般用来作谓语。 4.副词：adv. 副词是指表示动作发生的地点. 时间. 条件. 方式. 原因. 目的. 结果.伴随让步. 一般用来修饰动词. 形容词。副词在汉语中是（地）.其标志是：ly。 5.代词：pron. 代词是指用来代替名词的词，名词所能担任的作用，代词也同样.代词主要用来作主语. 宾语. 表语. 同位语。 6.介词：prep.介词是指表示动词和名次关系的词，例如：in on at of about with for to。其特征： 介词后的动词要用—ing形式。介词加代词时，代词要用宾格。例如：give up her（him）这种形式是正确的，而give up she（he）这种形式是错误的。 7.冠词：冠词是指修饰名词，表名词泛指或特指。冠词有a an the 。 8.叹词：叹词表示一种语气。例如：OH. Ya 等 9.连词：连词是指连接两个并列的成分，这两个并列的成分可以是两个词也可以是两个句子。例如：and but or so 。 10.数词：数词是指表示数量关系词，一般分为基数词和序数词 第二章节：英语句子成分 主语：动作的发出者，一般放在动词前或句首。由名词. 代词. 数词. 不定时. 动名词. 或从句充当。 谓语：指主语发出来的动作，只能由动词充当，一般紧跟在主语后面。 宾语：指动作的承受着，一般由代词. 名词. 数词. 不定时. 动名词. 或从句充当. 介词后面的成分也叫介词宾语。 定语：只对名词起限定修饰的成分，一般由形容 虽然做出来了.但是我还是有一个问题想请教大家: vsweep和mapp分网后形成网格各有什么规律?如何结合两种方法划分出整齐规则的网格呢. 比如:为什么图中的(1)部分用MAPP划分,(2)部分用SWEEP划分呢就可以出现上图中的那种整齐规则的网格?反过来(1)部分用SWEEP,(2)部分用MAPP划分就不会出现整齐规则的网格呢? 部分(1)和部分(2)不可看成一个整体划分吗? 我试了一个,如果把两个部分看成整体,可以分网但是不会出现那种整齐的网格. 只有掌握了生成网格规律才容易得到合理,整齐,规则的网格,总不能分网时把各种方法都试一遍吧. 恳请各位谈点自己的在分网方面的经验.谢谢 1的三个边如果都设了分段数则sweep和map是一样的 et,1,42 et,2,45 cyl4,,,20 lsel,all lesize,all,,,10 esize,,10 vext,1,,,,,20 aclear,all amesh,1 不过好象中间不大好的!!!还望高手指点! 命令流; et,1,42 et,2,45 blc4,,,10,5 lesize,1,,,5 lesize,2,,,10 mshape,0,2d mshkey,1 amesh,1 esize,,5 vrotat,1,,,,,,1,4 aclear,all 用map也可以,,取其四分之一,单元大小可控制!做了一个! /PREP7 CYL4, , ,5 RECTNG,-1,1,-1,1, FLST,2,2,5,ORDE,2 FITEM,2,1 FITEM,2,-2 AOVLAP,P51X wpro,,90.000000, wpro,,,45.000000 ASBW, 3 wpro,,,-45.000000 wpro,,,-45.000000 FLST,2,2,5,ORDE,2 FITEM,2,1 FITEM,2,4 ASBW,P51X wpro,,,45.000000 wpro,,90.000000, ET,1,PLANE42 ESIZE,1,0, AMAP,6,12,9,7,8 WPSTYLE,,,,,,,,0 AMAP,7,12,10,5,8 AMAP,3,10,11,6,5 AMAP,5,6,7,9,11 AMAP,2,8,5,6,7 TYPE, 1 EXTOPT,ESIZE,10,0, EXTOPT,ACLEAR,0 EXTOPT,ATTR,0,0,0 MAT,_Z2 REAL,_Z4 ESYS,0 ET,2,SOLID45 TYPE, 2 EXTOPT,ESIZE,10,0, EXTOPT,ACLEAR,0 EXTOPT,ATTR,0,0,0 MAT,_Z2 REAL,_Z4 ESYS,0 VOFFST,2,5, , VOFFST,6,5, , VOFFST,7,5, , M A: Has the case been closed yet? B: No, the magistrate still needs to decide the outcome. magistrate n.地方行政官，地方法官，治安官 A: I am unable to read the small print in the book. B: It seems you need to magnify it. magnify vt.１．放大，扩大；２．夸大，夸张 A: That was a terrible storm. B: Indeed, but it is too early to determine the magnitude of the damage. magnitude n.１．重要性，重大；２．巨大，广大 A: A young fair maiden like you shouldn’t be single. B: That is because I am a young fair independent maiden. maiden n.少女，年轻姑娘，未婚女子 a.首次的，初次的 A: You look majestic sitting on that high chair. B: Yes, I am pretending to be the king! majestic a.雄伟的，壮丽的，庄严的，高贵的 A: Please cook me dinner now. B: Yes, your majesty, I’m at your service. majesty n.１．［Ｍ－］陛下（对帝王，王后的尊称）；２．雄伟，壮丽，庄严 A: Doctor, I traveled to Africa and I think I caught malaria. B: Did you take any medicine as a precaution? malaria n.疟疾 A: I hate you! B: Why are you so full of malice? malice n.恶意，怨恨 A: I’m afraid that the test results have come back and your lump is malignant. B: That means it’s serious, doesn’t it, doctor? malignant a.１．恶性的，致命的；２．恶意的，恶毒的 A: I’m going shopping in the mall this afternoon, want to join me? B: No, thanks, I have plans already. mall n.（由许多商店组成的）购物中心 A: That child looks very unhealthy. B: Yes, he does not have enough to eat. He is suffering from malnutrition. 一、前言 有限元网格划分是进行有限元数值模拟分析至关重要的一步，它直接影响着后续数值计算分析结果的精确性。网格划分涉及单元的形状及其拓扑类型、单元类型、网格生成器的选择、网格的密度、单元的编号以及几何体素。从几何表达上讲，梁和杆是相同的，从物理和数值求解上讲则是有区别的。同理，平面应力和平面应变情况设计的单元求解方程也不相同。在有限元数值求解中，单元的等效节点力、刚度矩阵、质量矩阵等均用数值积分生成，连续体单元以及壳、板、梁单元的面内均采用高斯（Gauss）积分，而壳、板、梁单元的厚度方向采用辛普生（Simpson）积分。辛普生积分点的间隔是一定的，沿厚度分成奇数积分点。由于不同单元的刚度矩阵不同，采用数值积分的求解方式不同，因此实际应用中，一定要采用合理的单元来模拟求解。 CAD软件中流行的实体建模包括基于特征的参数化建模和空间自由曲面混合造型两种 方法。Pro/E和SoildWorks是特征参数化造型的代表，而CATIA与Unigraphics等则将特征参数化和空间自由曲面混合造型有机的结合起来。现有CAD软件对表面形态的表示法已经大大超过了CAE软件，因此，在将CAD实体模型导入CAE软件的过程中，必须将CAD 模型中其他表示法的表面形态转换到CAE软件的表示法上，转换精度的高低取决于接口程序的好坏。在转换过程中，程序需要解决好几何图形（曲线与曲面的空间位置）和拓扑关系（各图形数据的逻辑关系）两个关键问题。其中几何图形的传递相对容易实现，而图形间的拓扑关系容易出现传递失败的情况。数据传递面临的一个重大挑战是，将导入CAE程序的CAD模型改造成适合有限元分析的网格模型。在很多情况下，导入CAE程序的模型可能包含许多设计细节，如细小的孔、狭窄的槽，甚至是建模过程中形成的小曲面等。这些细节往往不是基于结构的考虑，保留这些细节，单元数量势必增加，甚至会掩盖问题的主要矛盾，对分析结果造成负面影响。 CAD模型的“完整性”问题是困扰网格剖分的障碍之一。对于同一接口程序，数据传递的品质取决于CAD模型的精度。部分CAD模型对制造检测来说具备足够的精度，但对有限元网格剖分来说却不能满足要求。值得庆幸的是，这种问题通常可通过CAD软件的“完整性检查”来修正。改造模型可取的办法是回到CAD系统中按照分析的要求修改模型。一方面检查模型的完整性，另一方面剔除对分析无用的细节特征。但在很多情况下，这种“回归”很难实现，模型的改造只有依靠CAE软件自身。CAE中最直接的办法是依靠软件具有的“重构”功能，即剔除细部特征、缝补面和将小面“融入”大曲面等。有些专用接口在模型传递过程中甚至允许自动完成这种工作，并且通过网格剖分器检验模型的“完整性”，如发现“完整性”不能满足要求，接口程序可自动进行“完整性”修复。当几何模型距CAE分析的要求相差太大时，还可利用CAE程序的造型功能修正几何模型。“布尔运算”是切除细节和修理非完整特征的有效工具之一。 目前数据传递一般可通过专用数据接口，CAE程序可与CAD程序“交流”后生成与CAE 程序兼容的数据格式。另一种方式是通过标准图形格式如IGES、SAT和ParaSolid传递。现有的CAD平台与通用有限元平台一般通过IGES、STL、Step、Parasolid等格式来数据 意见应以事实为根据. 3 来自辞典例句 192. The bombers swooped ( down ) onthe air base. 轰炸机 突袭 空军基地. 来自辞典例句 193. He mounted their engines on a rubber base. 他把他们的发动机装在一个橡胶垫座上. 14 来自辞典例句 194. The column stands on a narrow base. 柱子竖立在狭窄的地基上. 14 来自辞典例句 195. When one stretched it, it looked like grey flakes on the carvas base. 你要是把它摊直, 看上去就象好一些灰色的粉片落在帆布底子上. 18 来自辞典例句 196. Economic growth and human well - being depend on the natural resource base that supports all living systems. 经济增长和人类的福利依赖于支持所有生命系统的自然资源. 12 1 来自辞典例句 197. The base was just a smudge onthe untouched hundred - mile coast of Manila Bay. 那基地只是马尼拉湾一百英里长安然无恙的海岸线上一个硝烟滚滚的污点. 6 来自辞典例句 198. You can't base an operation on the presumption that miracles are going to happen. 你不能把行动计划建筑在可能出现奇迹的假想基础上. 各种网格划分方法 1．输入实体模型尝试用映射、自由网格划分，并综合利用多种网格划分控制方法 本题提供IGES 文件 1． 以轴承座为例,尝试对其进行映射，自由网格划分，并练习一般后处理的多种技术，包 括等值图、云图等图片的获取方法，动画等。 2． 一个瞬态分析的例子 练习目的：熟悉瞬态分析过程 瞬态(FULL)完全法分析板-梁结构实例 如图所示板-梁结构，板件上表面施加随时间变化的均布压力，计算在下列已知条件下结构的瞬态响应情况。 全部采用A3钢材料，特性： 杨氏模量=2e112/m N 泊松比=0.3 密度=7.8e33 /m Kg 板壳： 厚度=0.02m 四条腿（梁）的几何特性： 截面面积=2e-42m 惯性矩=2e-84m 宽度=0.01m 高度=0.02m 压力载荷与时间的 关系曲线见下图所示。 图 质量梁-板结构及载荷示意图 0 1 2 4 6 时间（s ） 图 板上压力-时间关系 分析过程 第1步：设置分析标题 1. 选取菜单途径Utility Menu>File>Change Title 。 2. 输入“ The Transient Analysis of the structure ”，然后单击OK 。 第2步：定义单元类型 单元类型1为SHELL63，单元类型2为BEAM4 第3步：定义单元实常数 实常数1为壳单元的实常数1，输入厚度为0.02（只需输入第一个值，即等厚度壳） 实常数2为梁单元的实常数，输入AREA 为2e-4惯性矩IZZ=2e-8，IYY ＝2e-8宽度TKZ=0.01，高度TKY=0.02。 第5步：杨氏模量EX=2e112/m N 泊松比NUXY=0.3 密度DENS=7.8e33 /m Kg 第6步：建立有限元分析模型 1． 创建矩形，x1=0,x2=2,y1=0,y2=1 2． 将所有关键点沿Z 方向拷贝，输入DZ ＝－1 3． 连线。将关键点1，5；2，6；3，7；4，8分别连成直线。 4． 设置线的分割尺寸为0.1，首先给面划分网格；然后设置单元类型为2，实常数为2， 对线5到8划分网格。 第7步：瞬态动力分析 1. 选取菜单途径Main Menu>Solution>-Analysis Type-New Analysis ，弹出New Analysis 对话框。 2. 选择Transient ，然后单击OK ，在接下来的界面仍然单击OK 。 3. 选取菜单途径Main Menu>Solution>-Load Step Opts-Time/Frequenc> Damping ， 弹出Damping Specifications 窗口。 4. 在Mass matrix multiplier 处输入5。单击OK 。 5. 选取菜单途径Main Menu > Solution > -Loads-Apply > -Structural- Displacement>On Nodes 。弹出拾取（Pick ）窗口，在有限元模型上点取节点232、242、252和262，单击OK ，弹出Apply U,ROT on Nodes 对话框。 6. 在DOFS to be constrained 滚动框中，选种“All DOF ”(单击一次使其高亮度显示， 确保其它选项未被高亮度显示）。单击OK 。 7. 选取菜单途径Utility Menu>Select>Everything 。 8. 选取菜单途径Main Menu>Solution>-Load Step Opts-Output Ctrls>DB/Results File ，弹出Controls for Database and Results File Writing 窗口。 9. 在Item to be controlled 滚动窗中选择All items ，下面的File write frequency 中选择Every substep 。单击OK 。 10. 选取菜单途径Main Menu>Solution>-Load Step Opts-Time/Frequenc> Time – Time Step ，弹出Time – Time Step Options 窗口。 11. 在Time at end of load step 处输入1；在Time step size 处输入0.2；在Stepped or ramped b.c 处单击ramped ；单击Automatic time stepping 为on ；在Minimum time step size 处输入0.05；在Maximum time step size 处输入0.5。单击OK 。 12. 选取菜单途径Main Menu>Solution>-Loads-Apply>-Structure-Pressure>On Areas 。弹出Apply PRES on Areas 拾取窗口。 13. 单击Pick All ，弹出Apply PRES on Areas 对话框。 14. 在pressure value 处输入10000。单击OK 15. 选取菜单途径Main menu>Solution>Write LS File ，弹出Write Load Step File 对 话框。 16. 在Load step file number n 处输入1，单击OK 。 17. 选取菜单途径Main Menu>Solution>-Load Step Opts-Time/Frequenc> Time – Time Step ，弹出Time – Time Step Options 窗口。 英语造句大全English sentence 在句子中，更好的记忆单词！ 1、(1)、able adj. 能 句子：We are able to live under the sea in the future. (2)、ability n. 能力 句子：Most school care for children of different abilities. (3)、enable v. 使。。。能句子：This pass enables me to travel half-price on trains. 2、(1)、accurate adj. 精确的句子：We must have the accurate calculation. (2)、accurately adv. 精确地 句子：His calculation is accurately. 3、(1)、act v. 扮演 句子：He act the interesting character. （2）、actor n. 演员 句子：He was a famous actor. (3)、actress n. 女演员 句子：She was a famous actress. (4)、active adj. 积极的 句子：He is an active boy. 4、add v. 加 句子：He adds a little sugar in the milk. 5、advantage n. 优势 句子：His advantage is fight. 6、age 年龄n. 句子：His age is 15. 7、amusing 娱人的adj. 句子：This story is amusing. 8、angry 生气的adj. 句子：He is angry. 9、America 美国n. 主语+谓语 1. 理解主谓结构 1) The students arrived. The students arrived at the park. 2) They are listening. They are listening to the music. 3) The disaster happened. 2．体会状语的位置 1) Tom always works hard. 2) Sometimes I go to the park at weekends.. 3) The girl cries very often. 4) We seldom come here. The disaster happened to the poor family. 3. 多个状语的排列次序 1) He works. 2) He works hard. 3) He always works hard. 4) He always works hard in the company. 5) He always works hard in the company recently. 6) He always works hard in the company recently because he wants to get promoted. 4. 写作常用不及物动词 1. ache My head aches. I’m aching all over. 2. agree agree with sb. about sth. agree to do sth. 3. apologize to sb. for sth. 4. appear (at the meeting, on the screen) 5. arrive at / in 6. belong to 7. chat with sb. about sth. 8. come (to …) 9. cry 10. dance 11. depend on /upon 12. die 13. fall 14. go to … 15. graduate from 16. … happen 17. laugh 18. listen to... 19. live 20. rise 21. sit 22. smile 23. swim 24. stay (at home / in a hotel) 25. work 26. wait for 汉译英： 1.昨天我去了电影院。 2.我能用英语跟外国人自由交谈。 3.晚上7点我们到达了机场。 4.暑假就要到了。 5.现在很多老人独自居住。 6.老师同意了。 7.刚才发生了一场车祸。 8.课上我们应该认真听讲。9. 我们的态度很重要。 10. 能否成功取决于你的态度。 11. 能取得多大进步取决于你付出多少努力。 12. 这个木桶能盛多少水取决于最短的一块板子的长度。 【it's time to和it's time for】 ——————这其实是一个句型,只不过后面要跟不同的东西. ——————It's time to跟的是不定式（to do）.也就是说,要跟一个动词,意思是“到做某事的时候了”.如： It's time to go home. It's time to tell him the truth. ——————It's time for 跟的是名词.也就是说,不能跟动词.如： It's time for lunch.(没必要说It's time to have lunch) It's time for class.(没必要说It's time to begin the class.) They can't wait to see you Please ask liming to study tonight. Please ask liming not to play computer games tonight. Don’t make/let me to smoke I can hear/see you dance at the stage You had better go to bed early. You had better not watch tv It’s better to go to bed early It’s best to run in the morning I am enjoy running with music. With 表伴随听音乐 I already finish studying You should keep working. You should keep on studying English Keep calm and carry on 保持冷静继续前行二战开始前英国皇家政府制造的海报名字 I have to go on studying I feel like I am flying I have to stop playing computer games and stop to go home now I forget/remember to finish my homework. I forget/remember cleaning the classroom We keep/percent/stop him from eating more chips I prefer orange to apple I prefer to walk rather than run I used to sing when I was young What’s wrong with you There have nothing to do with you I am so busy studying You are too young to na?ve I am so tired that I have to go to bed early 1. 网格划分 1.1 Hypermesh 中六面体网格划分的功能介绍 ?六面体网格划分的工具主要有： ?Drag ?Spin ?Line drag ?element offset ?solid map ?其中solid map集成了部分其它功能； 1.1.1：drag 面板 此面板的功能是在二维网格接触上沿着一个线性路径挤压拉伸而形成三维实体单元。要求： 1）有初始的二维网格； 2）截面保持不变：相同尺寸，相同曲率和空间中的相同方向； 3）线性路径。 1.1.2：spin 面板 -1- 此面板的功能是在二维网格基础上沿着一个旋转轴旋转一定角度形成三维实体单元。要求： 1）有初始的二维网格； 2）界面保持不变； 3）圆形路径； 4）不能使用在没有中心孔的实体部件上。 1.1.3：line drag 面板 此面板的功能上在二维网格的基础上沿着一条线拉伸成三维实体单元。 要求： 1）初始的二维网格； 2）截面保持不变； 3）有一条定义的曲线或直线路径。 1.1.4：element offset 面板 此面板的功能是在二维网格的基础上沿着法线方向偏置挤压形成三维实体单元。 要求： 1）初始的二维网格； 2）截面可以是非平面的； -2- -3- 3） 常厚度或者近似常厚度。 1.1.5：soild map 面板 此面板的功能是在二维网格基础上，首先挤压网格，然后将挤压的网格映射到一个由几何要 素定义的实体中，从而形成三维实体单元。 1.2 drag 面板网格划分指导 导入几何，drag 实体之前必须先生成2D 网格，如下图 拉伸的距离 定义方向 需要拉伸的层数on the contrary的解析
英语造句
ANSYS网格划分总结大全
网格划分实例详细步骤
学生造句--Unit 1
英语句子结构和造句
ANSYS网格划分的一些例子
六级单词解析造句记忆MNO
ANSYS结构有限元分析中的网格划分技术及其应用实例
base on的例句
各种网格划分方法
英语造句大全
(完整版)主谓造句
初中英语造句
hypermesh六面体网格划分指导(含实例)
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