Stochastic Processes and their Applications118(2008)864–895
https://www.doczj.com/doc/dc15630107.html,/locate/spa Stability of in?nite dimensional stochastic evolution
equations with memory and Markovian jumps$
Jiaowan Luo a,Kai Liu b,?
a School of Mathematics and Information Sciences,Guangzhou University,Guangzhou,Guangdong510405,PR China
b Division of Statistics and Probability,Department of Mathematical Sciences,The University of Liverpool,
Peach Street,Liverpool,L697ZL,United Kingdom
Received1August2005;received in revised form1January2007;accepted21June2007
Available online30June2007
Abstract
A strong solutions approximation approach for mild solutions of stochastic functional differential equations with Markovian switching driven by L′e vy martingales in Hilbert spaces is considered.The Razumikhin–Lyapunov type function methods and comparison principles are studied in pursuit of suf?cient conditions for the moment exponential stability and almost sure exponential stability of equations in which we are interested.The results of[A.V.Svishchuk,Yu.I.Kazmerchuk,Stability of stochastic delay equations of It?o form with jumps and Markovian switchings,and their applications in?nance,Theor.Probab.Math. Statist.64(2002)167–178]are generalized and improved as a special case of our theory.
c 2007Elsevier B.V.All rights reserved.
MSC:primary93E03;secondary60H10
Keywords:In?nite dimensional stochastic evolution equations with memory;L′e vy processes;Markovian jumps;Moment exponential stability;Almost sure exponential stability
1.Introduction
There exists an extensive literature dealing with stochastic differential equations with discontinuous paths incurred by L′e vy processes(for instance,see monographs[2,6,19]and references therein).These equations are used as models in the study of queues,insurance risks,
$This work was partially supported by NNSF of China(Grant No.10301036)and EPSRC of the UK(Grant No. GR/R37227).
?Corresponding author.Tel.:+441517944759;fax:+441517944754.
E-mail addresses:jwluo@https://www.doczj.com/doc/dc15630107.html,(J.Luo),k.liu@https://www.doczj.com/doc/dc15630107.html,(K.Liu).
0304-4149/$-see front matter c 2007Elsevier B.V.All rights reserved.
doi:10.1016/j.spa.2007.06.009
J.Luo,K.Liu/Stochastic Processes and their Applications118(2008)864–895865 dams,and more recently in mathematical?nance.On the other hand,some recent research in automatic control such as[7,10,17]has been devoted to stochastic differential equations with Markovian jumps.As a popular and important topic,the stability property of stochastic differential equations has always lain at the center of our understanding concerning stochastic models described by these equations.In particular,stability of stochastic differential equations with Markovian switching has recently received signi?cant attention.For example,Ji and Chizeck[10]and Mariton[17]studied the stability of a linear equation with jump coef?cient
d x(t)=A(r(t))x(t)d t,
where r(t)is a Markov chain taking values in S={1,2,...,N}.Basak et al.[5]investigated the stability of a semilinear stochastic differential equation with Markovian switching of the form
d x(t)=A(r(t))x(t)d t+σ(x(t),r(t))d w(t)
where w(t)is the usual?nite dimensional Brownian motion.Mao[14]considered the exponential stability of general nonlinear stochastic differential equations with Markovian switching of the form
d x(t)=f(x(t),t,r(t))d t+g(x(t),t,r(t))d w(t),
which can be regarded as the result of the following N equations:
d x(t)=f(x(t),t,i)d t+g(x(t),t,i)d w(t),1≤i≤N,
switching from one state to the others according to the movement of the Markov chain. Mao et al.[15,16]considered asymptotic and exponential stability of the following nonlinear stochastic delay differential equations with Markovian switching:
d x(t)=f(x(t),x(t?τ),t,r(t))d t+g(x(t),x(t?τ),t,r(t))d w(t)
whereτ>0is a constant.
Quite recently,in their paper[22]Svishchuk and Kazmerchuk made a?rst attempt to study the p th-moment exponential stability of solutions of linear It?o stochastic delay differential equations associated with Poisson jumps and Markovian switching which is motivated by some practical applications in mathematical?nance.
In the present paper we will be interested in the moment and almost sure stability property but content ourselves with some more general in?nite dimensional models,or to be precise, nonlinear stochastic functional differential equations with Markovian switching driven by L′e vy martingales in Hilbert spaces.One of the most remarkable advantages of studying this model is that it enables one to deal with stochastic partial functional differential equations with discontinuous paths,a case which the existing works such as those mentioned above fail to cover. However,it is worth mentioning that in comparison with stochastic stability in?nite dimensions, the theory presented in this paper is much more complicated due to at least three factors.The?rst is that the standard solution(strong solution)concept turns out to be too strong to apply for most stochastic partial functional differential equations in which we are especially interested.Actually, a natural generalization of this aspect is the mild solution(cf.[8]for its de?nition and relevant properties)which is more useful and also easy to formulate from both practical and theoretical viewpoints.The next factor which is closely related to the?rst one is that for the treatment of mild solutions,a lot of standard tools in stochastic calculus like It?o’s formula or Doob’s theorem could not be used any longer or in a straightforward way.In order to possibly take advantage of these powerful results,in our analysis of stability we would always require patience
866J.Luo,K.Liu/Stochastic Processes and their Applications118(2008)864–895
to overcome various dif?culties from calculus and probability so as to make our scheme move forward.For instance,our theory relies heavily on an appropriate use of a version of a Burkholder type of inequality for stochastic convolution driven by a compensated Poisson random measure which will be formulated properly for our stability purpose.We will also be introducing suitable approximation systems of strong solutions and using a limiting procedure technique so as to make the arguments involved with the use of It?o’s formula justi?able.The?nal factor is that, instead of the traditional Lyapunov functions in?nite dimensions,the corresponding Lyapunov functionals in in?nite dimensions should be constructed properly,a case which makes the usual construction approaches dif?cult to get through.Bearing this in mind,we shall employ an idea due to Razumikhin[20,21]to construct the so-called Razumikhin–Lyapunov functions to get round this dif?culty.
In this work,we shall derive some suf?cient conditions to ensure stability of in?nite dimensional stochastic systems with memory in the sense of both moment exponential stability and almost sure exponential stability.To the best of our knowledge,this problem has not been investigated in the existing literature.In particular,the results in[22]will be generalized and improved as a special case of our theory.In Section2,we shall?rst state some properties of mild solutions of the equations concerned and some results as regards approximating strong solution systems.The material of this part lays a good foundation not only for the stability analysis in this paper,but also for future research in connection with these models.We shall then discuss the moment,especially the mean square,and exponential stability of the equations studied in Section3,and then investigate the almost sure exponential stability for the same equations in Section4.In Section5,we shall present some comparison results which will clarify the relationships of stability between deterministic and stochastic time delay systems.Finally,we will give two illustrative applications of our theory in Section6.
2.Stochastic functional differential equations with Markovian switching driven by L′e vy martingales
Let{?,F,P}be a complete probability space equipped with some?ltration{F t}t≥0 satisfying the usual conditions,i.e.,the?ltration is right continuous and F0contains all P-null sets.Let H,K be two real separable Hilbert spaces and denote by ·,· H, ·,· K their inner products and by · H, · K their vector norms,respectively.We denote by L(K,H)the set of all linear bounded operators from K into H,equipped with the usual operator norm · .In this paper,we always use the same symbol · to denote norms of operators regardless of the spaces potentially involved when no confusion may arise.Letτ>0and D:=D([?τ,0];H)denote the family of all right-continuous functions with left-hand limits?from[?τ,0]to H.The space D([?τ,0];H)is assumed to be equipped with the norm ? D=sup?τ≤θ≤0 ?(θ) H.We
also use D b F
0([?τ,0];H)to denote the family of all almost surely bounded,F0-measurable,
D([?τ,0];H)-valued random variables.
Let{r(t),t∈R+},R+=[0,∞),be a right-continuous Markov chain on the probability space{?,F,P}taking values in a?nite state space S={1,2,...,N}with generatorΓ=
(γi j)N×N given by
P{r(t+h)=j|r(t)=i}= γ
i j h+o(h),if i=j, 1+γii h+o(h),if i=j,
for any t≥0and small h>0.Hereγi j≥0is the rate of transition from i to j if i=j,while
J.Luo,K.Liu /Stochastic Processes and their Applications 118(2008)864–895867
γii =?
j =i γi j .It is well known that almost every sample path of r (t )is a right-continuous step function with a ?nite number of jumps in any ?nite sub-interval of R +.
Let {W Q (t ),t ≥0}denote a K -valued {F t }t ≥0-Wiener process de?ned on {?,F ,P }with covariance operator Q ,i.e.,
E W Q (t ),x K W Q (s ),y K =(t ∧s ) Qx ,y K
for all x ,y ∈K ,
where Q is a positive,self-adjoint,trace class operator on K .In particular,we shall call such W Q (t ),t ≥0,a K -valued Q -Wiener process with respect to {F t }t ≥0.It is always assumed in the paper that the Markov chain r (·)is independent of the Q -Wiener process W Q (t ).
In order to de?ne stochastic integrals with respect to the Q -Wiener process W Q (t ),we introduce the subspace K 0=Q 1/2(K )of K which,endowed with the inner product
u ,v K 0= Q ?1/2u ,Q ?1/2v K ,
is a Hilbert space.Let L 02=L 2(K 0,H )denote the space of all Hilbert–Schmidt operators from K 0into H .It turns out to be a separable Hilbert space,equipped with the norm
Ψ 2L 02
=tr (ΨQ 1/2)(ΨQ 1/2)?
for any Ψ∈L 02.
Clearly,for any bounded operators Ψ∈L (K ,H ),this norm reduces to Ψ 2L 02
=tr (ΨQ Ψ?).For arbitrarily given T ≥0,let J (t ,ω),t ∈[0,T ],be an F t -adapted,
L 02-valued
process,and
we de?ne the following norm for arbitrary t ∈[0,T ]:
|J |t = E t 0
tr (J (s ,ω)Q 1/2)(J (s ,ω)Q 1/2)?
d s 12
.
In particular,we denote all L 02-valued predictable processes J satisfying |J |T <∞by U 2([0,T ];L 02).The stochastic integral t
0J (s ,ω)d W Q (s )∈H ,t ≥0,may be de?ned for all
J (t ,ω)∈U 2([0,T ];L 02)by
t 0
J (s ,ω)d W Q (s )=L 2?lim n →∞
n
i =1
t 0
λi J (s ,ω)e i d B i
s ,t ∈[0,T ],
where W Q (t )= ∞i =1√λi B i t e i .Here (λi ≥0,i ∈N )are the eigenvalues of Q and (e i ,i ∈N )
are the corresponding eigenvectors,(B i t ,i ∈N )are independent standard real-valued Brownian motions.The reader is referred to [8]for a systematic theory about stochastic integrals of this kind.
Suppose Y ={Y t },t ≥0,is a K -valued L′e vy process,so that Y has stationary and independent increments,is stochastically continuous and satis?es Y 0=0almost surely.Let p t be the law of Y t for each t ≥0;then (p t ,t ≥0)is a weakly continuous convolution semigroup of probability measures on K .We have the L′e vy–Khintchine formula (see e.g.[3])which yields for all t ≥0,u ∈K ,
E e i u ,Y t K
=e t η(u ),where
η(u )=exp
i b ,u K ?1
2
u ,Qu K
868J.Luo,K.Liu /Stochastic Processes and their Applications 118(2008)864–895
+
K ?{0}
e i u ,y K
?1?i u ,y K ·χ y K <1(y )
ν(d y )
,
where b ∈K ,Q is a positive,self-adjoint,trace class operator on K and νis a L′e vy measure
on K ?{0},i.e., K ?{0}( y 2K ∧1)ν(d y )<∞.Here we use χE to denote the characteristic
function on set E ?K .We can also de?ne the L′e vy measure on the whole of K via the assignment ν({0})=0.We call the triple (b ,Q ,ν)the characteristics of the process Y ,and the mapping ηthe characteristic exponent of Y .It can be shown that the L′e vy process has a c`
a dl`a g version which is always assumed to be this case in this paper.We will also strengthen the independent increments requirement on Y by assuming that Y t ?Y s is independent of F s for all 0≤s If Y is a L′ e vy process on K ,we write ?Y t =Y t ?Y t ?for all t ≥0where Y t ?:=lim s ↑t Y s .We obtain then a counting Poisson random measure N on (K ?{0})through N (t ,E )=#{0≤s ≤t ;?Y s ∈E }<∞, t ≥0, almost surely for any E ∈B (K ?{0})with 0∈ˉE ,the closure of E in K .Here B (K ?{0})denotes the Borel σ-?eld of K ?{0}.The associated compensated Poisson random measure ?N is de?ned by ?N (t ,d y )=N (t ,d y )?t ν(d y ).Let O ∈B (K ?{0})with 0∈ˉO and let νO denote the restriction of the measure νto O ,still denoted by ν,so that νis ?nite on O .Let P 2([0,T ]×O ;H )denote the space of all predictable mappings L :[0,T ]×O ×?→H for which T 0 O E L (t ,y ) 2H ν(d y )d t <∞. We may then de?ne T 0 O L (t ,y )N (d t ,d y )= 0≤t ≤T L (t ,?P t )χO (?P t ) where P t = O yN (t ,d y )= 0≤s ≤t ?Y s χO (?Y s ), as a random ?nite sum,which enables us to de?ne further T 0 O L (t ,y )?N (d t ,d y )= T 0 O L (t ,y )N (d t ,d y )? T 0 O L (t ,y )ν(d y )d t , and then by standard arguments (see e.g.[2]),we may see that t 0 O L (s ,y )?N (d s ,d y ),t ≥0,is actually an H -valued centered square integrable martingale with E T 0 O L (t ,y )?N (d t ,d y ) 2H = T 0 O E L (t ,y ) 2H ν(d y )d t ,for each t ≥0.We always assume in this paper that W ,r (·)and ?N are independent of F 0. J.Luo,K.Liu /Stochastic Processes and their Applications 118(2008)864–895869 Let T (t ),t ≥0,be some C 0-semigroup of bounded linear operators over H which has its in?nitesimal generator A with domain D (A )?H .Consider the following semilinear stochastic functional differential equation with Markovian switching driven by L′e vy processes:for any t ∈I =[0,T ],T ≥0, x (t )= t [Ax (s )+f (x s ,r (s ))]d s + t 0g (x s ,r (s ))d W Q (s )+ t 0 y K (d s ,d y )+ t 0 y K ≥c q (x s ,r (s ),y )N (d s ,d y ),x 0(·)=ξ∈D b F 0 ([?τ,0],H ),(2.1)for some c ∈(0,∞]where x t (θ):=x (t +θ),θ∈[?τ,0], t 0 y K h (x s ,r (s ),y )?N (d s ,d y ):=lim n →∞ t 0 1 n < y K h (x s ,r (s ),y )?N (d s ,d y ),and f :D ([?τ,0];H )×S →H , g :D ([?τ,0];H )×S →L (K ,H ), h :D ([?τ,0];H )×S ×K →H , q :D ([?τ,0];H )×S ×K →H are properly de?ned measurable functions such that the associated integrals make sense. The convenient parameter c ∈(0,∞]on the right-hand side of (2.1)allows us to specify what we mean by “small”and “large”jumps,respectively,in speci?c applications.If we want to put “small”and “large”jumps on the same footing we let c =∞or c →0so that the term involving q or h is absent in (2.1).In many situations,the term q in (2.1)involving “large jumps”may be handled by using an interlacing technique (see [2]).In the remainder of this paper,for the sake of simplicity,we proceed by omitting this term and concentrate on the study of the equation driven by continuous noise interspersed with “small jumps”.In other words,instead of (2.1)we wish to consider the stability of the following modi?ed equation: x (t )= t 0(Ax (s )+f (x s ,r (s )))d s + t 0g (x s ,r (s ))d W Q (s )+ t 0 y K h (x s ,r (s ),y )?N (d s ,d y )x 0(·)=ξ∈D b F 0 ([?τ,0],H ).(2.2)In particular,one can show by passing to the limit that if h ∈P 2([0,T ]×{ y K T 0 y K E L (t ,y ,ω) 2H ν(d y )d t <∞, the process t 0 y K (d s ,d y ),t ≥0,is an H -valued centered square integrable martingale with 870J.Luo,K.Liu/Stochastic Processes and their Applications118(2008)864–895 E t y K h(x s,r(s),y)?N(d s,d y) 2 H = t y K E h(x s,r(s),y) 2Hν(d y)d s, for each t≥0.Next,let us present the following de?nitions. De?nition2.1.A stochastic process x(t),t∈I,de?ned on(?,F,{F t}t≥0,P)is called a strong solution of(2.2)if (i)x(t)∈D(A),0≤t≤T,almost surely and is adapted to F t,t∈I; (ii)x(t)∈H has c`a dl`a g paths on t∈I almost surely,and for arbitrary0≤t≤T, x(t)=ξ(0)+ t 0[Ax(s)+f(x s,r(s))]d s+ t g(x s,r(s))d W Q(s) + t y K h(x s,r(s),y)?N(d s,d y), x0(·)=ξ(·)∈D b F ([?τ,0],H).(2.3) Generally speaking,this concept is quite strong and the much weaker one described below is more appropriate for practical purposes. De?nition2.2.A stochastic process x(t),t∈I,de?ned on(?,F,{F t}t≥0,P)is called a mild solution of(2.2)if (i)x(t)is adapted to F t,t≥0; (ii)x(t)∈H has c`a dl`a g paths on t∈I almost surely,and for arbitrary0≤t≤T, x(t)=T(t)ξ(0)+ t 0T(t?s)f(x s,r(s))d s+ t T(t?s)g(x s,x(s))d W Q(s) + t y K T(t?s)h(x s,r(s),y)?N(d s,d y), x0(·)=ξ(·)∈D b F ([?τ,0],H).(2.4) In order to establish the global existence and uniqueness of solutions of Eq.(2.2),for simplicity,let the coef?cients f(·,·),g(·,·)and h(·,·,·)of(2.2)be assumed to satisfy the following global Lipschitz continuous and linear growth conditions: (H)There exists a number L>0such that f(ξ,i)?f(η,i) 2 H + g(ξ,i)?g(η,i) 2≤L ξ?η 2 D ,(2.5) and y K H ν(d y)≤L ξ?η 2D,(2.6) for all i∈S andξ,η∈D,and there exists,moreover,an M>0such that f(ξ,i) 2 H + g(ξ,i) 2≤M(1+ ξ 2 D ),(2.7) and y K H ν(d y)≤M(1+ ξ 2D),(2.8) for allξ∈D and i∈S. J.Luo,K.Liu /Stochastic Processes and their Applications 118(2008)864–895871 As a direct application of the properties of semigroup theory,it may be easily proved that: Proposition 2.1.For arbitrary ξ(·)∈D b F 0 ([?τ,0],H )with ξ(θ)∈D (A ),θ∈[?τ,0],assume that x (t )∈D (A ),t ∈I ,is a strong solution of (2.2);then it is also a mild solution of (2.2). Note that the converse statement of Proposition 2.1is generally not true,i.e.,a mild solution of (2.2)is not necessarily a strong one.By a straightforward argument,it is possible to establish the following uniqueness result. Proposition 2.2.Assume the condition (H )holds;then there exists at most one mild solution of (2.2).In other words,under the condition (H )the mild solution of (2.2)is unique. Carrying out the usual Picard iteration or a probabilistic ?xed-point theorem type of procedure,we can follow the arguments in [9]to establish an existence theorem for mild solutions of (2.2)in the following form. Theorem 2.1.Assume the condition (H )holds and let ξ∈D b F 0 ([?τ,0],H )be an arbitrarily given initial datum.Then there exists a unique global solution,x (t ),t ≥0,of (2.4)in the space L 2([0,∞)×?;H ). Remark 2.1.(1)For the purposes of the existence and uniqueness of a mild solution,it is possible to replace the global Lipschitz condition in (H)by a local one,that is with the condition holding with possibly different constants L k ,M k for ξ D , η D ≤k and each k ∈N . (2)The solution x (t )established in Theorem 2.1does not necessarily have almost sure c`a dl`a g paths at the point.However,by using the Lemma 2.2below we can ?nd under some circumstances a c`a dl`a g version for x (t )by a strong solution approximation procedure.The following stochastic Fubini theorem which was presented in [3]in a slightly different form is fundamental.Let P =P ([0,T ]×?)denote the predictable σ-algebra and (Z ,Z ,μ)be a ?nite measure space.Let O ∈B (K ?{0})and H 2(T ,O ,Z )be the real Hilbert space of all P ×B (O )×Z -measurable functions G from [0,T ]×?×O ×Z →H for which Z T 0 O E G (s ,y ,z ) 2H ν(d y )d s μ(d z )<∞. The space S (T ,O ,Z )is dense in H 2(T ,O ,Z ),where G ∈S (T ,O ,Z )if G = N 1 i =0N 2 j =0N 3 k =0 G i jk χA i χ(t j ,t j +1]χB k , where N 1,N 2,N 3∈N ,A 0,...,A N 1are disjoint sets in B (O ),0=t 0 B 0,...,B N 3is a partition of Z ,wherein each B k ∈Z and each G i jk is a bounded F t j -measurable random variable with values in H . Lemma 2.1.If G ∈H 2(T ,O ,Z ),then for each 0≤t ≤T , Z t 0 O G (s ,y ,z )?N (d s ,d y ) μ(d z )= t 0 O Z G (s ,y ,z )μ(d z ) ?N (d s ,d y )(2.9)almost surely. Proof.First note that both integrals in (2.9)are easily seen to exist in L 2(?,F ,P ).If G ∈ S (T ,O ,Z ),then the result holds with both sides of (2.9)equal to 872J.Luo,K.Liu /Stochastic Processes and their Applications 118(2008)864–895 N 1 i =0N 2 j =0N 3 k =0 G i jk ?N ((t j ,t j +1],A i )μ(B k )where ?N ((t j ,t j +1],A i ):=?N (t j +1,A i )??N (t j ,A i ).Now suppose that (G n ,n ∈N )is a sequence of mappings in S (T ,O ,Z )converging to G ∈H 2(T ,O ,Z );then E t 0 O Z [G (s ,y ,z )?G n (s ,y ,z )]μ(d z ) ?N (d s ,d y ) 2H = t 0 O E Z [G (s ,y ,z )?G n (s ,y ,z )]μ(d z ) 2H ν(d z )d s ≤μ(Z ) t 0 O Z E G (s ,y ,z )?G n (s ,y ,z ) 2 H μ(d z )ν(d y )d s →0, as n →∞. A similar argument shows that lim n →∞ E Z t 0 O [G n (s ,y ,z )?G (s ,y ,z )]?N (d s ,d y ) μ(d y ) 2H =0,and the result follows. The following result gives suf?cient conditions for a mild solution to be also a strong solution, which is quite useful in our stability analysis. Proposition 2.3.Suppose that the following conditions hold:for arbitrary η∈D,i ∈S,t ≥0, (1)ξ(·)∈D b F 0 ([?τ,0],H )with ξ(θ)∈D (A )for any θ∈[?τ,0];(2)T (t )f (η,i )∈D (A ),T (t )g (η,i )k ∈D (A ),T (t )h (η,i ,y )∈D (A )for any k ∈K ,and y ∈K ; (3) AT (t )f (η,i ) H ≤z 1(t ) η D ,z 1(·)∈L 1(0,T ;R +);(4) AT (t )g (η,i ) 2≤z 2(t ) η 2D ,z 2(·)∈L 1(0,T ;R +);(5) y K ([?τ,0],H )is also a strong solution such that x (t )∈D (A ),t ∈I ,almost surely. Proof.It suf?ces to prove that the mild solution x (t ),t ∈I ,takes values in D (A )and satis?es (2.3).By the above conditions,we have almost surely T 0 t AT (t ?s )f (x s ,r (s )) H d s d t <∞, T 0 t tr (AT (t ?s )g (x s ,r (s )))Q (AT (t ?s )g (x s ,r (s )))? d s d t <∞, and T 0 t 0 y K AT (t ?s )h (x s ,r (s ),y ) 2H ν(d y )d s d t <∞. J.Luo,K.Liu /Stochastic Processes and their Applications 118(2008)864–895873 Thus by the classic Fubini theorem,we have t 0 v 0AT (v ?s )f (x s ,r (s ))d s d v = t 0 t s AT (v ?s )f (x s ,r (s ))d v d s = t T (t ?s )f (x s ,r (s ))d s ? t 0 f (x s ,r (s ))d s .On the other hand,by the Fubini type of theorems for Q -Wiener processes in [8]and Lemma 2.1, we have t 0 v 0AT (v ?s )g (x s ,r (s ))d W Q (s )d v = t 0 t s AT (v ?s )g (x s ,r (s ))d v d W Q (s )= t 0 T (t ?s )g (x s ,r (s ))d W Q (s )? t g (x s ,r (s ))d W Q (s ),and t 0 v y K AT (v ?s )h (x s ,r (s ),y )?N (d s ,d y )d v = t 0 t s y K AT (v ?s )h (x s ,r (s ),y )d v ?N (d s ,d y )= t 0 y K T (t ?s )h (x s ,r (s ),y )?N (d s ,d y )? t 0 y K h (x s ,r (s ),y )?N (d s ,d y ).Hence,Ax (t )is integrable almost surely and t 0Ax (v)d v =T (t )ξ(0)?ξ(0)+ t 0T (t ?s )f (x s ,r (s ))d s ? t 0 f (x s ,r (s ))d s + t 0T (t ?s ) g (x s ,r (s ))d W Q (s )? t 0g (x s ,r (s ))d W Q (s ) + t 0 y K T (t ?s )h (x s ,r (s ),y )?N (d s ,d y )? t 0 y K h (x s ,r (s ),y )?N (d s ,d y )=x (t )?ξ(0)? t 0f (x s ,r (s ))d s ? t g (x s ,r (s ))d W Q (s ) ? t 0 y K h (x s ,r (s ),y )?N (d s ,d y ).In other words,X t ∈D (A ),t ∈I ,is a strong solution of (2.2). At the moment,we assume that A :D (A )?H →H is the in?nitesimal generator of a pseudo-contraction C 0-semigroup T (t ),t ≥0,of bounded linear operators in H ,i.e., T (t ) ≤e αt for some α≥0and any t ≥0.It is well known (see [12])that in this case we have Ax ,x H ≤α x 2H , ?x ∈D (A ). (2.10) Let O c ={y ∈K ?{0}: y K ν([0,T ]×O c ;H ),p ≥2,denote the space of H -valued mappings J (t ,y ),progressively measurable with respect to {F t }t ≥0such that 874J.Luo,K.Liu /Stochastic Processes and their Applications 118(2008)864–895 E T y K J (t ,y ) p H ν(d y )d t <∞.(2.11) We are interested in the stochastic convolution Z (t )= t 0 y K (d s ,d y ),de?ned for any ?xed t ∈[0,T ].In particular,we establish below a special case of Burkholder type of inequality for stochastic convolutions driven by the compensator ?N (·,·)of the Poisson random measure N (·,·): Lemma 2.2.Suppose J ∈M 2ν([0,T ]×O c ;H )∩M 4ν([0,T ]×O c ;H );then for any t ∈[0,T ],E sup 0≤s ≤t Z (s ) 2H ≤C E t y K J (s ,y ) 2H ν(d y )d s +E t 0 y K J (s ,y ) 4H ν(d y )d s 1/2 (2.12) for some number C =C (T )>0.In particular,if α=0,the number C (T )can be chosen independent of T. Proof.Step 1:First of all,let us suppose that J (·,·)∈M 2ν([0,T ]×O c ;D (A ))∩M 4 ν([0,T ]×O c ;H ) where D (A )is endowed with the usual graph norm.From Proposition 2.3,we know that the process Z (t )also satis?es that for any t ∈[0,T ], Z (t )= t AZ (s )d s + t 0 y K J (s ,y )?N (d s ,d y ).By applying It?o ’s formula to x 2H ,we get Z (t ) 2H = t 0 2 AZ (s ),Z (s ) H + y K J (s ,y ) 2H ν(d y ) d s + t 0 y K 2 Z (s ?),J (s ,y ) H + J (s , y ) 2H ?N (d s ,d y ).(2.13) Let us de?ne Z ?(t )=sup 0≤s ≤t Z (s ) H and J ?(t )= t 0 y K J (s ,y ) 2H ν(d y )d s . Taking into account (2.10),we obtain from (2.13)that Z ?(t )2≤2α t Z (s ) 2H d s +J ? (t ) +6sup 0≤s ≤t s 0 y K Z (v ?),J (v,y ) H ?N (d v,d y ) +3sup 0≤s ≤t s 0 y K J (v,y ) 2H ?N (d v,d y ) ,which,by the H¨o lder inequality and the usual Burkholder type of inequality,immediately yields J.Luo,K.Liu /Stochastic Processes and their Applications 118(2008)864–895875 E (Z ?(t )2 )≤2α t E (Z ?(s )2)d s +E (J ?(t )) +6E t 0 y K Z (s ) 2H J (s , y ) 2H ν(d y )d s 1/2 +3E t 0 y K J (s ,y ) 4H ν(d y )d s 1/2 ≤2α t E (Z ?(s )2 )d s +E (J ? (t ))+6E Z ?(t )(J ?(t )) 1/2 +3E t 0 y K J (s , y ) 4H ν(d y )d s 1/2 . Let H ?(t ):=3 t 0 y K J (s ,y ) 4H ν(d y )d s 1/2 . We then have E (Z ? (t )2 )≤2α t E (Z ? (s )2 )d s +E (H ? (t )+J ? (t ))+6 E (Z ? (t )2 ) 1/2 E (J ?(t )) 1/2 which,by using the well-known Gronwall inequality,immediately implies E (Z ?(t )2) 1/2≤e 2αt E (Z ?(t )2) ?1 2 E (H ?(t )+J ?(t ))+6[E (J ?(t ))]1/2 .This is a second-order inequality in [E (Z ?(t ))2]1/2from which we obtain E (Z ?(t )2 ) ≤Const .·e 4αt E (J ?(t ))+E (H ?(t )+J ?(t )) ≤C 0 E (H ?(t ))+E (J ?(t )) , for some real number C 0=C 0(T )>0,dependent on T ≥0,i.e.,for any t ∈[0,T ], E sup 0≤s ≤t Z (s ) 2H ≤3C 0 E t y K J (s ,y ) 2H ν(d y )d s +E t 0 y K J (s ,y ) 4H ν(d y )d s 1/2 as desired. Step 2:In the general case,let J n (t ,y )=n R (n ,A )J (t ,y )where R (n ,A ),n ∈N ,is the resolvent of A ;then we have J n →J in M 4ν([0,T ]×O c ;H ),as n →∞,by a straightforward application of the dominated convergence theorem.Moreover,it is easy to see by Proposition 2.3 that J n ∈M 2ν([0,T ]×O c ;D (A ))∩M 4ν([0,T ]×O c ;H ).De?ning Z n (t )= t 0 y K T (t ?s )J n (s ,y )?N (d s ,d y ),we have that Z n (t )→Z (t )in L 2(?,F ,P ;H )for any t ∈[0,T ]as n →∞.On the other hand, we can apply the inequality (2.12)to the difference Z n (t )?Z m (t )with J (·,·)replaced by the 876J.Luo,K.Liu /Stochastic Processes and their Applications 118(2008)864–895 difference J n ?J m from which we deduce that E sup 0≤s ≤t Z (s )?Z n (s ) 2H →0,as n →∞, and hence (2.12)is true for any J ∈M 2ν([0,T ]×O c ;H )∩M 4ν ([0,T ]×O c ;H ). Since we mainly treat stability of mild solutions in this paper,a dif?culty encountered here is that we need strong solutions so as to use powerful tools like It?o ’s formula from stochastic calculus.We can deal with this problem,however,by introducing approximation systems of strong solutions and using a limiting type of argument.To this end,we introduce an approximation system of (2.2)as follows:for any t ≥0, x (t )= t 0[Ax (s )d s +R (l )f (x s ,r (s ))]d s + t 0R (l )g (x s ,r (s ))d W Q (s )+ t 0 y K R (l )h (x s ,r (s ),y )?N (d s ,d y ),x 0(θ)=R (l )ξ(θ),θ∈[?τ,0], (2.14)where l ∈ρ(A ),the resolvent set of A ,and R (l ):=l R (l ,A ),R (l ,A )is the resolvent of A . Proposition 2.4.Let ξ∈D b F ([?τ,0];H )be an arbitrarily given initial datum and assume that T (t )is a pseudo-contraction C 0-semigroup.Suppose that the terms f (·,·),g (·,·)and h (·,·,·)in (2.2)satisfy the conditions (2.5)–(2.8).Furthermore,we suppose that there exist numbers L 0>0and M 0>0such that y K h (ξ,i ,y )?h (η,i ,y ) 4H ν(d y )≤L 0 ξ?η 4 D , (2.15)for all i ∈S and ξ,η∈D,and y K h (ξ,i ,y ) 4H ν(d y )≤M 0(1+ ξ 4 D ), (2.16) for all ξ∈D and i ∈S.Then,for each l ∈ρ(A ),the stochastic differential equation (2.14)has a unique strong solution x l (t )∈D (A ),which lies in L 2(?,F ,P ;D (0,T ;H ))for all T >0.Moreover,there exists a subsequence,denote it by x n (t ),such that for arbitrary T >0,x n (t )→x (t )of (2.4)almost surely as n →∞,uniformly with respect to [0,T ]. Proof.The existence of a unique strong solution x l (t )of the kind we desire is an immediate consequence of Proposition 2.3and Theorem 2.1on noting the fact that AR (l )=Al R (l ,A )=l ?l 2R (l ,A )are bounded operators.To prove the remainder of the proposition,let us consider x (t )?x l (t )=T (t )(ξ(0)?R (l )ξ(0)) + t 0T (t ?s ) f (x s ,r (s ))?R (l )f (x l s ,r (s )) d s + t 0 T (t ?s ) g (x s ,r (s ))?R (l )g (x l s ,r (s )) d W Q (s ) J.Luo,K.Liu /Stochastic Processes and their Applications 118(2008)864–895877 + t 0 y K T (t ?s ) h (x s ,r (s ),y )? R (l )h (x l s ,r (s ), y ) ?N (d s ,d y )(2.17) for any t ≥0where x l t (θ)=x l (t +θ)for any θ∈[?τ,0].We thus have that for any T ≥0, E sup 0≤t ≤T x (t )?x l (t ) 2 H ≤42E sup 0≤t ≤T t 0T (t ?s )R (l ) f (x s ,r (s ))?f (x l s ,r (s )) d s 2 H +42E sup 0≤t ≤T t 0 T (t ?s )R (l ) g (x s ,r (s ))?g (x l s ,r (s )) d W Q (s ) 2H +42E sup 0≤t ≤T t 0 y K T (t ?s )R (l ) × h (x s ,r (s ),y )?h (x l s ,r (s ),y ) ?N (d s ,d y ) 2H +42 E sup 0≤t ≤T T (t )(ξ(0)?R (l )ξ(0))+ t T (t ?s )[I ?R (l )]f (x s ,r (s ))d s + t 0T (t ?s )[I ?R (l )]g (x s ,r (s ))d W Q (s ) + t 0 y K T (t ?s )[I ?R (l )]h (x s ,r (s ),y )?N (d s ,d y ) 2H :=16[I 1+I 2+I 3+I 4]. (2.18) Note that R (l ) ≤2for l >0large enough.The Lipschitz continuous conditions in (H)and H¨o lder’s inequality imply that I 1=E sup 0≤t ≤T t 0 T (t ?s )R (l ) f (x s ,r (s ))?f (x l s ,r (s )) H d s 2≤4T e 2αT E sup 0≤t ≤T t 0 f (x s ,r (s ))?f (x l s ,r (s )) 2H d s ≤4T e 2αT L E T sup 0≤r ≤s x r ?x l r 2 D d s +4T 2e 2αT L E sup ?r ≤θ≤0 (I ?R (l ))ξ(θ) 2H , (2.19) where L >0is the Lipschitz constant in (H).On the other hand,by virtue of the Burkholder–Davis–Gundy type of inequality for stochastic convolutions in [23],we have for l >0large enough that there exists a number C 1(T )>0such that I 2=E sup 0≤t ≤T t 0 T (t ?s )R (l ) g (x s ,r (s ))?g (x l s ,r (s )) d W Q (s ) 2H 878J.Luo,K.Liu /Stochastic Processes and their Applications 118(2008)864–895 ≤C 1(T )L T E sup 0≤r ≤s x r ?x l r 2 D d s +T C 1(T )L E sup ?r ≤θ≤0 (I ?R (l ))ξ(θ) 2H ,(2.20) and by Lemma 2.2and (H),it follows that there exists a number C 2(T )>0such that I 3=E sup 0≤t ≤T t 0 y K y K h (x s ,r (s ),y )?h (x l s ,r (s ),y ) 2H ν(d y )d s +E T y K h (x s ,r (s ),y )?h (x l s ,r (s ),y ) 4H ν(d y )d s 1/2 ≤C 2(T ) L E T 0 x s ?x l s 2 D d s + L 0E T 0 sup 0≤t ≤T x t ?x l t 2D · x s ?x l s 2D d s 1/2 ≤ LC 2(T )+ L 0C 2(T ) 2 T E x s ?x l s 2 D d s +12 E sup 0≤t ≤T x t ?x l t 2D .Also,it is easy to see that I 4≤16 E sup 0≤t ≤T T (t )(ξ(0)?R (l )ξ(0)) 2H +E sup 0≤t ≤T t 0T (t ?s )[I ?R (l )]f (x s ,r (s ))d s 2 H +E sup 0≤t ≤T t T (t ?s )[I ?R (l )]g (x s ,r (s ))d W Q (s ) 2H +E sup 0≤t ≤T t 0 y K T (t ?s )[I ?R (l )]h (x s ,r (s ),y )?N (d s ,d y ) 2H .(2.21) By using the dominated convergence theorem,we can obtain E sup 0≤t ≤T T (t )(ξ(0)?R (l )ξ(0)) 2H ≤e 2αT E (I ?R (l ))ξ(0) 2H →0,as l →∞, and E sup 0≤t ≤T t 0 T (t ?s )[I ?R (l )]f (x s ,r (s ))d s 2 H ≤e 2αT T E (I ?R (l ))f (x s ,r (s )) 2H d s →0, as l →∞.(2.22) In a similar manner,by using the Burkholder–Davis–Gundy type of inequality for stochastic J.Luo,K.Liu /Stochastic Processes and their Applications 118(2008)864–895879 convolutions in [23],it is easy to deduce that there exists a number C 3(T )>0such that E sup 0≤t ≤T t 0 T (t ?s )[I ?R (l )]g (x s ,r (s ))d W Q (s ) 2H ≤C 3(T ) T E (I ?R (l ))g (x s ,r (s )) 2H d s →0, as l →∞,(2.23) and by Lemma 2.2and the dominated convergence theorem,we deduce that there exists a number C 4(T )>0such that E sup 0≤t ≤T t 0 y K T (t ?s )[I ?R (l )]h (x s ,r (s ),y )?N (d s ,d y ) 2H ≤C 4(T ) E T 0 y K (I ?R (l ))h (x s ,r (s ),y ) 2H ν(d y )d s +E T y K (I ?R (l ))h (x s ,r (s ),y ) 4H ν(d y )d s 1/2 →0, as l →∞. (2.24) Hence,combining with (2.18)–(2.24),we can get that there exist numbers C (T )>0and ε(l )>0such that E sup 0≤t ≤T x t ?x l t 2D ≤E sup 0≤t ≤T x (t )?x l (t ) 2H +E sup θ∈[?τ,0] (I ?R (l ))ξ(θ) 2 H ≤C (T ) T 0 E sup 0≤r ≤s x r ?x l r 2 D d s +12 E sup 0≤t ≤T x t ?x l t 2D +ε(l ),where lim l →∞ε(l )=0.By the well-known Gronwall inequality,it is deduced that E sup 0≤t ≤T x (t )?x l (t ) 2H ≤E sup 0≤t ≤T x t ?x l t 2D ≤2ε(l )e 2C (T )T →0,as l →∞. (2.25) Now we are in a position to construct the desired sequence by using a standard diagonal sequence trick.Indeed,for the positive integer n =1,by virtue of (2.25),there exists a positive integer sequence {m 1(i )}∞i =1 in ρ(A )such that x l m 1(i )(t )→x (t )almost surely as i →∞,uniformly with respect to t ∈[0,1].Now for the positive integer n =2,consider the sequence x l m 1(i )(t );we can ?nd a subsequence x l m 2(i )(t ),{l m 2(i )}?{l m 1(i )},such that x l m 2(i )(t )→x (t )almost surely as i →∞,uniformly with respect to t ∈[0,2].Proceeding inductively,we ?nd successive subsequences,x l m n (i )(t ),so that (a)x l m n +1(i )(t )is a subsequence of x l m n (i )(t ),{l m n +1(i )}?{l m n (i )},and (b)x l m n (i )(t )→x (t )almost surely as i →∞,uniformly with respect to t ∈[0,n ].To get a subsequence converging for each n ,one may take the diagonal sequence l (n )=l m n (n ).Then we can obtain the sequence {x l (n )(t )}∞n =1,more simply denoted by {x n (t )}∞n =1,which has the desired properties. If we consider in (2.2)a special term h (ξ,i ,y )which is linear in the variable y ∈K ,then Proposition 2.4turns out to be a simpler form.To see this,assume that for any ξ∈D ,i ∈S and y ∈K , 880J.Luo,K.Liu/Stochastic Processes and their Applications118(2008)864–895 h(ξ,i,y)=P(ξ,i)y,(2.26) where P:D×S→L(K,H)satis?es that for some constant C>0, P(ξ,i)?P(η,i) 2≤C ξ?η 2 D , P(ξ,i) 2≤C(1+ ξ 2 D ), (2.27) for anyξ,η∈D and i∈S,and the L′e vy measureνsatis?es the relation y K K ν(d y)<∞.(2.28) Corollary2.1.Letξ∈C b F0([?τ,0];H)be an arbitrarily given initial datum and assume that T(t)is a pseudo-contraction C0-semigroup.Suppose that the terms f(·,·),g(·,·)and h(·,·,·) in(2.4)satisfy the conditions(2.5),(2.7)and(2.26),(2.27),(2.28),respectively.Then,for each l∈ρ(A),the stochastic differential equation(2.14)has a unique strong solution x l(t)∈D(A), which lies in L2(?,F,P;D(0,T;H))for any T≥0.Moreover,there exists a subsequence, denote it by x n(t),such that for arbitrary T≥0,x n(t)→x(t)of(2.4)almost surely as n→∞,uniformly with respect to[0,T]. 3.Moment exponential stability In this section,we will consider the moment exponential stability of mild solutions of(2.2) by means of the so-called Razumikhin–Lyapunov function techniques.The fundamental idea of this method was?rstly explored by Razumikhin in[20,21]to deal with stability of deterministic systems.For the purposes of stability,we shall assume that f(0,i)≡0,g(0,i)≡0and h(0,i,y)≡0for any i∈S,y∈K.(3.1) Then Eq.(2.2)obviously has a trivial solution whenξ≡0.The presentation in this section is closely related to the work[13]although some signi?cant dif?culties from calculus must be overcome to make our scheme proceed. We denote by C2,0(H×S;R+)the family of all non-negative function V(x,i)on H×S which are continuously twice differentiable with respect to x.For any(?,i)∈D([?τ,0];H)×S with ?(0)∈D(A),we introduce the following: (L V)(?,i)= V x(?(0),i),A?(0)+f(?,i) H +1 2 tr V xx(?(0),i)g(?,i)Qg(?,i)T + N j=1 γi j V(?(0),j) + y K V(?(0)+h(?,i,y),i)?V(?(0),i) ? V x (?(0),i),h(?,i,y) H ν(d y).(3.2) Theorem3.1.Suppose that T(t)is a pseudo-contraction C0-semigroup and the condi-tions(H)and(3.1)hold.Assume that there exist functions V(x,i)∈C2,0(H×S;R+)and w1(z),w2(z)∈C([0,∞);R+)such that for every t≥0,i∈S,the following two conditions hold: J.Luo,K.Liu /Stochastic Processes and their Applications 118(2008)864–895 881 w 1( x 2H )≤V (x ,i )≤w 2( x 2 H ),(3.3) E max 1≤i ≤N (L V )(φ,i ) ≤?λE max 1≤i ≤N V (φ(0),i ) for some λ>0, (3.4) for any random process φ(·,ω)∈D ([?τ,0];H )satisfying φ(0)∈D (A )and E min 1≤i ≤N V (φ(θ),i ) V (φ(0),i ) ,?θ∈[?τ,0], (3.5) for some q >1.Then,for arbitrarily given ξ∈D b F 0 ([?τ,0];H ),we have E w 1( x (t ,ξ) 2H )≤E w 2( ξ 2D )e ?γt ,t ≥0,(3.6) where γ=min {λ,ln q τ}provided E w 2( x (t ) 2H )<∞,t ≥0. Clearly,if w 1(z )=w 2(z )=z ,the trivial solution of (2.4)is mean square exponentially stable,and the corresponding Lyapunov exponent is not bigger than ?γ. Proof.For arbitrarily given initial datum ξ∈D b F 0 ([?τ,0];H ),let us denote the solution x (t ;ξ)of (2.4)by x (t ).If t ∈[?τ,0]we let r (t )=r (0).Since x (t )and r (t )are right continuous and E sup ?τ≤s ≤t x (s ) 2H <∞for t ≥0,then E V (x (t ),r (t ))is right continuous for t ≥τ.For suf?ciently small ε∈(0,γ),let ˉγ=γ?ε.De?ne U (t )=sup ?τ≤θ≤0 e ˉγ(t +θ) E V (x (t +θ),r (t +θ)) ,t ≥0.(3.7) We claim that D +U (t ) =lim sup h →0+U (t +h )?U (t ) h ≤0, t ≥0.(3.8) To show this,note that for each ?xed t 0≥0,if for all θ∈[?τ,0], U (t 0)>e ˉτ(t 0+θ) E V (x (t 0+θ),r (t 0+θ)), then,as E V (x (·),·,r (·))is right continuous,there exists δ>0suf?ciently small such that U (t 0)>e ˉγ(t 0+δ)E V (x (t 0+δ),t 0+δ,r (t 0+δ)). Hence U (t 0+δ)≤U (t 0), i.e., D +U (t 0)≤0. If there exists θ∈[?τ,0]such that U (t 0)=e ˉτ(t 0+θ)E V (x (t 0+θ),r (t 0 +θ)),then de?ne ˉθ=max θ∈[?τ,0]:U (t 0)=e ˉτ(t 0+θ)E V (x (t 0+θ),r (t 0+θ)) ,and we obviously have U (t 0)=e ˉγ(t 0+ˉθ)E V (x (t 0+ˉθ), r (t 0+ˉθ)).If ˉθ <0,one has e ˉγ(t 0+θ)E V (x (t 0+θ),r (t 0+θ)) r (t 0+ˉθ))for all ˉθ <θ≤0.It is therefore easy to observe that by the right continuity of E V (x (t ),r (t )), 882J.Luo,K.Liu /Stochastic Processes and their Applications 118(2008)864–895 for any h >0small enough, e ˉγ(t 0+h )E V (x (t 0+h ),r (t 0+h ))≤e ˉγ(t 0+ˉθ)E V (x (t 0+ˉθ), r (t 0+ˉθ)).Hence, U (t 0+h )≤U (t 0)and D +U (t 0)≤0. If ˉθ =0,then e ˉγ(t 0+θ)E V (x (t 0+θ),r (t 0+θ))≤e ˉγt 0 E V (x (t 0),r (t 0)) for any θ∈[?τ,0].Therefore, E V (x (t 0+θ),r (t 0+θ))≤e ?ˉγθE V (x (t 0),r (t 0))≤e ˉγτE V (x (t 0),r (t 0)) (3.9) for any θ∈[?τ,0].In the case of E V (x (t 0),r (t 0))=0,the relations (3.9)and (3.3)imply that x (t 0+θ)=0for all θ∈[?τ,0]almost surely.Recall that f (0,·)=0,g (0,·)=0and h (0,·,·)=0;it then follows that x (t 0+h )=0almost surely for all h >0,and hence U (t 0+h )=0and D +U (t 0)=0.On the other hand,in the case of E V (x (t 0),r (t 0))>0,the relation (3.9)implies E V (x (t 0+θ),r (t 0+θ))≤e ˉγτ E V (x (t 0),r (t 0)) (3.10) for any θ∈[?τ,0]owing to e γτ τ>0;it then follows from (3.11)and the right continuity of E V (x (t ),r (t ))that for some h >0small enough, E V (x (t 0+θ),r (t 0+θ))≤ e ˉγτ +ν2 E V (x (t 0),r (t 0)) for any θ∈[0,h ].Now we need to introduce the strong solution sequence {x n (t )}of (2.4)so that by Proposition 2.4,x n (t )→x (t )uniformly with respect to t ∈[0,T ]in the almost sure sense for any T ≥0as n →∞.Consequently,for some constant δ∈ 0, ν 4+2ν E V (x (t 0),r (t 0)) ,(3.11)there is a suf?ciently small constant h >0such that for any s ∈(t 0,t 0+h ], E V (x (s ),r (s ))>E V (x (t 0),r (t 0))?δ>0, (3.12) E V (x (s +θ),r (s +θ)) ?θ∈[?τ,0], (3.13)e ˉγτE V (x (t 0),r (t 0)) (3.14) and by the strong solution approximation,there is an integer N 0>0large enough such that for any n ≥N 0and s ∈(t 0,t 0+h ], E V (x n (s ),r (s ))>E V (x (s ),r (s ))?δ>0, (3.15)e ˉγτE V (x (s ),r (s )) (3.16) E V (x n (s +θ),r (s +θ)) (3.17) Thus,we have by using (3.17),(3.13)and (3.10)that for any s ∈(t 0,t 0+h ],n ≥N 0, E V (x n (s +θ),r (s +θ)) ≤E V (x (t 0+θ),r (t 0+θ))+2δ ≤e ˉγτE V (x (t 0),r (t 0))+2δ. (3.18) J.Luo,K.Liu /Stochastic Processes and their Applications 118(2008)864–895883 Using (3.14)and (3.15),this yields that E V (x n (s +θ),r (s +θ)) E V (x (s ),r (s ))+3δ ≤e ˉγτ E V (x n (s ),r (s ))+4δ, (3.19) which,together with (3.11)and (3.15),immediately implies that E V (x n (s +θ),r (s +θ)) E V (x n (s ),r (s ))+ν(E V (x (s ),r (s ))?δ) E V (x n (s ),r (s ))+νE V (x n (s ),r (s )) =q E V (x n (s ),r (s )), ?θ∈[?τ,0].(3.20) Hence, E min 1≤i ≤N V (x n s (θ),i ) ≤q E max 1≤i ≤N V (x n (s ),i ) , ?τ≤θ≤0. Hence,by the conditions of the theorem,(3.20)implies that for some λ>0, E max 1≤i ≤N (L V )(x n s ,i ) V (x n (s ),i ) , which immediately yields that E (L V )(x n s ,r (s ))≤?λE V (x n (s ),r (s )), ?s ∈[t 0,t 0+h ]. (3.21) Applying It?o ’s formula to the function e ˉγ t V (·,i )along the strong solutions x n (t )of (2.2),we can derive,by using (3.21),that for any ˉh ∈(0,h ],e ˉγ(t 0+ˉh )E V (x n (t 0+ˉh ),r (t 0+ˉh ))≤e ˉγt 0 E V (x n (t 0),r (t 0))+(ˉγ?λ) t 0+ˉh t 0 e ˉγs E V (x n (s ),r (s ))d s + t 0+ˉh t 0e ˉγs E V x (x n (s ),r (s )),(R (n )?I )f (x n s ,r (s )) H d s ? t 0+ˉh t 0 y K E V x (x n (s ),r (s )),(R (n )?I )h (x n s ,r (s ),y ) H ν(d y )d s + t 0+ˉh t 0 y K E V (x n (s )+R (n )h (x n s ,r (s ),y ),r (s ))ν(d y )d s ? t 0+ˉh t 0 y K +1 2 t 0+ˉh t 0E e ˉγs tr V xx (x n (s ),r (s ))R (n )g (x n s ,r (s ))Q R (n )g (x n s ,r (s ))? d s ?12 t 0+ˉh t 0 E e ˉγs tr V xx (x n (s ),r (s ))g (x n s ,r (s ))Qg (x n s ,r (s ))? d s , (3.22) which,letting n →∞,immediately yields e ˉγ(t 0+ˉh )E V (x (t 0+ˉh ),r (t 0+ˉh )) 分 类变化方法 举例 规则变化单数名词词尾直接加-s boy — boys pen — pens 以s，x ，ch，sh结尾 的单词一般加-es glass—glasses box—boxes watch—watches brush—brushes class— classes 特例：stomach — stomachs 以“辅音字母+y”结尾 的变“y”为“i”再加 “-es” baby—babies lady —ladies family—families 注意：penny的两种复数形式含义有所不同： pence（便士的钱数）pennies(便士的枚数) 以“o”结尾的多数加 -s radios zoos photos pianos kilos tobaccos bamboos studios 而下列名词的复数却要加-es： tomato —tomatoes potato —potatoes hero — heroes echo—echoes 以“f”或“fe”结尾 的名词复数形式变 “f”或“fe”为 “v”，之后再加-es wife—wives self — selves 特例：handkerchief—handkerchiefs gulf—gulfs belief—beliefs chief—chiefs roof—roofs scarf—scarfs 不规则变化改变名词中的元音字母 或其他形式 man-men woman-women foot-feet goose-geese mouse-mice tooth—teeth 特例：child-children 单复数相同 s heep deer means（方法）works（作品、工厂、 著作）crossroads species Chinese Japanese 合成名词变成复数的情 况： ●将主体名词变为复数 ●无主体名词时将最后 一部分变为复数 ●将两部分都变为复数 ●sons-in-law lookers-on passers-by story-tellers boy friends ●grown-ups housewives stopwatches ●women singers men servants 儿童歌谣大全 各位读友大家好，此文档由网络收集而来，欢迎您下载，谢谢 篇一：儿童歌谣歌词大全 [小螺号]儿歌歌词 小螺号，嘀嘀嘀吹海鸥听了展翅飞小螺号，嘀嘀嘀吹浪花听了笑微微小螺号，嘀嘀嘀吹 声声唤船归啰 小螺号，嘀嘀嘀吹 阿爸听了快快回啰。 茫茫的海洋，蓝蓝的海水 吹起了小螺号，心里美吔 [小鸟小鸟] 蓝天里有阳光，树林里有花香 小鸟小鸟，你自由地飞翔 在田野，在草地，在湖边，在山冈 小鸟小鸟迎着春天歌唱 啦啦啦啦啦。爱春天，爱阳光，爱湖水，爱花香小鸟小鸟，我的好朋友让我们一起飞翔歌唱 一起飞翔歌唱 啦啦啦啦啦 [让我们荡起双桨] 让我们荡起双桨，小船儿推开波浪海面倒映着白塔，四周环绕着绿树红墙小船儿轻轻飘荡在水中 迎面吹来了凉爽的风 红领巾迎着太阳，阳光洒在海面上 水中鱼儿望着我们 悄悄听我们愉快歌唱 小船儿轻轻飘荡在水中 迎面吹来了凉爽的风 做完了一天的功课，我们来尽情欢乐 我问你亲爱的伙伴 谁给我们安排下幸福的生活？ 小船儿轻轻飘荡在谁中 迎面吹来了凉爽的风 [小红花]儿歌歌词 金波词尚疾曲 花园里，篱笆下，我种下一朵小红花 春天的太阳当头照，春天的小雨沙沙下 啦啦啦啦啦，啦啦啦啦啦 小红花张嘴笑哈哈 花园里，篱笆下，我种下一朵小红花 春天的太阳当头照，春天的小雨沙沙下 啦啦啦啦啦，啦啦啦啦啦 小红花张嘴笑哈哈[我家几口] 金苗苓词曲我家有几口？ 让我扳指头 爸爸，妈妈，还有我 再加一个布娃娃 哟！有四口我家有几口？ 让我扳指头 爸爸，妈妈，还有我 再加一个布娃娃 哟！有四口 [谁会这样] 少数民族儿歌潘振声曲 谁会飞呀，鸟会飞 鸟儿鸟儿怎样飞？ 拍拍翅膀飞呀飞 谁会游呀，鱼会游 鱼儿鱼儿怎样游？ 摇摇尾巴点点头 谁会跑呀，马会跑马儿马儿怎样跑？四脚离地身不摇。[我叫轻轻] 张友珊词汪玲曲 走路轻轻轻轻 上夜班的阿姨还没醒呀 敲门轻轻轻轻 给邻居叔叔送呀送封信 说话轻轻轻轻 姐姐灯下看书多用心呀 大家夸我是好孩子 给我取个名字叫呀叫轻轻 走路轻轻轻轻 上夜班的阿姨还没醒呀 敲门轻轻轻轻 给邻居叔叔送呀送封信 说话轻轻轻轻 姐姐灯下看书多用心呀 关于国内工程保理业务介绍 根据我行最新国内保理业务管理办法，现将本业务作如下介绍： 一、本业务涉及到的几个定义 （一）国内保理业务是指境内卖方将其向境内买方销售货物或提供服务所产生的应收账款转让给我行，由我行为卖方提供应收账款管理、保理融资及为买方提供信用风险担保的综合性金融业务。其中，我行为卖方提供的应收账款管理和保理融资服务为卖方保理，为买方提供的信用风险担保服务为买方保理。 工程保理是指工程承包商或施工方将从事工程建设服务形成 的应收账款转让给卖方保理行，由保理行提供综合性保理服务。（二）卖方（供应商、债权人）：对所提供的货物或服务出具发票 的当事方，即保理业务的申请人。 （三）买方（采购商、债务人）：对由提供货物或服务所产生的应 收账款负有付款责任的当事方。 （四）应收账款：本办法所称应收账款指卖方或权利人因提供货物、服务或设施而获得的要求买方或义务人付款的权利，包括现有的金钱债权及其产生的收益，包括但不限于应收账款本金、利息、违约金、损害赔偿金，以及担保权利、保险权益等所有主债权、从债权以及与主债权相关的其他权益，但不包括因票据或其他有价证券而产生的付款请求权。 （四）应收账款发票实有金额：是指发票金额扣除销售方已回笼货款后的余额。 （五）保理监管专户：指卖方保理行为卖方开立的用于办理保理项下应收账款收支业务的专用账户。该账户用途包括：收取保理项下应收账款；收取其他用以归还我行保理融资的款项；支付超过保理融资余额以外的应收账款等。卖方不能通过应收账款专户以网上银行或自助设备等形式开展任何业务。 二、保理业务分类 （一）按是否对保理融资具有追索权，我行国内保理分为有追索权保理和无追索权保理。 有追索权保理是指卖方保理行向卖方提供保理项下融资后，若买方在约定期限内不能足额偿付应收账款，卖方保理行有权按照合同约定向卖方追索未偿还融资款的保理业务；无追索权保理是指卖方保理行向卖方提供保理项下融资后，若买方因财务或资信原因在约定期限内不能足额偿付应收账款，卖方保理行无权向卖方追索未偿还融资款。 三、应收账款的范围、条件及转让方式 (一）可办理国内保理业务的应收账款范围： 1、因向企事业法人销售商品、提供服务、设施而形成的应收 账款； 2、由地市级（含）以上政府的采购部门统一组织的政府采购 行为而形成的应收账款； 3、由军队军级（含）以上单位的采购部门统一组织的军队采 购行为而形成的应收账款； 《樱桃小丸子》 小小年纪谈起理想一串串 想当专家、想做博士、想出唱片老爸老妈老师老友都夸赞 想来容易，说来简单，做做就难要数一百，先数一二三 要过明天先过好今天 瞄准目标看齐 噼里啪啦，噼里啪啦 做事不偷懒 噼里啪啦，噼里啪啦 学习不怕难 我们脚踏实地地干 瞄准目标看齐 噼里啪啦，噼里啪啦 读书真勤快 噼里啪啦，噼里啪啦 今天学得好 噼里啪啦，噼里啪啦 明天理想能实现 《多啦A梦》 如果我有仙女棒变大变小变漂亮 还要变个都是漫画巧克力和玩具的家 如果我有机器猫我要叫他小叮当 竹蜻蜓和时光隧道能去任何的地方 让小孩大人坏人都变成好人 (hi 大家好,我是小叮当) ang ang ang小叮当帮我实现所有的愿望 躺在草地上幻想想动想西想玩耍 想到老师还有考试一个头就变成两个大好在我有小叮当困难时候求求他 万能笔和时间机器能做任何的事情 让我的好朋友一齐分享他 (啊!救命啊!有老鼠!) ang ang ang 小叮当帮我实现所有的愿望 躺在草地上幻想想动想西想玩耍 想到老师还有考试一个头就变成两个大好在我有小叮当困难时候求求他 万能笔和时间机器能做任何的事情 让我的好朋友一齐分享他 (小叮当永远是你们的好朋友喔!) ang ang ang 小叮当帮我实现所有的愿望 ang ang ang 小叮当帮我实现所有的愿望 《铁臂阿童木》 越过辽阔天空，啦啦啦飞向遥远群星，来吧！阿童木，爱科学的好少年。善良勇敢的啦啦啦铁臂阿童木，十万马力七大神力，无私无畏的阿童木。穿过广阔大地，啦啦啦潜入深深海洋，来吧！阿童木，爱和平的好少年。善良勇敢的啦啦啦铁臂阿童木，我们的好朋友啊， 无私无畏的阿童木。 《小龙人之歌》 天上有，无数颗星星，那颗最小的就是我，我不知道我从哪里来，也不知道我在哪里生。地上有，无数个龙人，那个最小的就是我，我不知道我从哪里来，也不知道我在哪里生。啊----这是我将在妈妈怀 抱里，啊寻遍天涯，去找他 《我是一条小青龙》我头上有只角，我身后有尾巴，谁也不知道，我有多少秘密？我头上有只角，我身后有尾巴，谁也不知道，我有多少秘密。我是一条小青龙（小青龙，小青龙）我有许多小秘密（小秘 名词的所有格 名词所有格在句中主要用来表示所有关系，如：Tom’s book, China’s capital。也可以用来表示类别：children’s books儿童读物，women’s clothes。它还可以表示动作的执行者或承受者，前者如the teacher’s praise;后者如children’s education。 名词所有格构成法分为三种： 1.-’s所有格 (1)表示有生命东西的名词所有格，一般在名词后加-’s 单数名词在词尾加’s eg. Betty’s phone number,cow’s milk 以s结尾的复数名词在词尾加’eg: a teachers’ college,a workers’ rest home 不以s结尾的复数名词在词尾加-’s eg. The men’s room,children’s books 以s结尾的专有名词可以在词尾加-’s也可以只加’ eg. Charles’s/Charles’ letter,Yeats’s/Yeats’ poems 短语，在短语最后加-’s eg.the woman next door’s husband An hour and a half’s talk 如果一个事物为两人所共有，只在后一个名词的词尾加-’s;如分别为各人所有，则两个名词词尾都要加-’s Eg. Tom and Bill’s desk, Jane’s and Helen’s bikes (2) 表示时间，价值，度量，国家，城市，天体等无生命名词的所有格，也可以在词尾加-’s Eg. A day’s work, a moment’s rest,today’s newspaper,a pound’s weight,Shanghai’s industries (3)在一些固定习语中，用’s所有格 For God’s sake, be at death’s door,be at one’s wits’ end (4)-’s所有格后面名词的省略 A.表示店铺机构或某家名词所有格后面的shop, house等名词常省略。Eg. Uncle Fred’s（house）the station’ s文具店the barber’s 理发店the baker’s面包店the butcher’s 肉店 B.名词所有格修饰的词，如前面已提到，可省略。 My book is here. Where is Mike’s ? 2.Of所有格 (1)表示无生命东西的名词所有格一般用of短语 The side of the road,the top of the hill, the leg of chair. (2)表示有生命东西的名词所有格，也可以用of短语，特别是当该名词有较长的后置定语时 The plays of Shakespeare,the name of the man over there,the mother of the girl playing by the lake. **the wife of John强调John,John’s wife强调wife 3.双重所有格 用‘of短语+’s’表示所属关系，称为双重所有格。只用于指人。(1)表示部分观念：名词前有不定冠词（a/an）不定代词 56首经典儿歌歌词大全 1、做早操 早上空气真叫好，我们都来做早操。 伸伸臂，弯弯腰，踢踢腿，蹦蹦跳，天天锻炼身体好。 2、饭前要洗手 小脸盆，水清请，小朋友们笑盈盈，小手儿，伸出来， 洗一洗，白又净，吃饭前，先洗手，讲卫生，不得病。 3、小手绢 小手绢，四方方，天天带在我身上。 又擦鼻涕又擦汗，干干净净真好看。 4、搬鸡蛋 小老鼠，搬鸡蛋，鸡蛋太大怎么办?一只老鼠地上躺， 紧紧抱住大鸡蛋。一只老鼠拉尾巴，拉呀拉呀拉回家。 5、大骆驼 骆驼骆驼志气大，风吹日晒都不怕。 走沙漠，运盐巴，再苦再累不讲话。 6、螳螂 螳螂哥，螳螂哥，肚儿大，吃得多。飞飞能把粉蝶捕， 跳跳能把蝗虫捉。两把大刀舞起来，一只害虫不放过 7、大蜻蜓 大蜻蜓，绿眼睛，一对眼睛亮晶晶， 飞一飞，停一停，飞来飞去捉蚊蝇。 8、小鸭子 小鸭子，一身黄，扁扁嘴巴红脚掌。 嘎嘎嘎嘎高声唱，一摇一摆下池塘。 9、拍手歌 你拍一，我拍一，天天早起练身体。 你拍二，我拍二，天天都要带手绢。 你拍三，我拍三，洗澡以后换衬衫。 你拍四，我拍四，消灭苍蝇和蚊子。 你拍五，我拍五，有痰不要随地吐。 你拍六，我拍六，瓜皮果核不乱丢。 你拍七，我拍七，吃饭细嚼别着急。 你拍八，我拍八，勤剪指甲常刷牙。 你拍九，我拍九，吃饭以前要洗手。 你拍十，我拍十，脏的东西不要吃。 10 、小螃蟹 小螃蟹，真骄傲，横着身子到处跑， 吓跑鱼，撞倒虾，一点也不懂礼貌 11 、庆六一 儿童节，是六一，小朋友们真欢喜。 又唱歌来又跳舞，高高兴兴庆六一。 12、花猫照镜子 小花猫，喵喵叫，不洗脸，把镜照， 左边照，右边照，埋怨镜子脏，气得胡子翘。 13、蚂蚁搬虫虫 小蚂蚁，搬虫虫，一个搬，搬不动，两个搬，掀条缝， 三个搬，动一动，四个五个六七个，大家一起搬进洞。 14、小青蛙 小青蛙，呱呱呱，水里游，岸上爬， 吃害虫，保庄稼，人人都要保护它。 15、花儿好看我不摘 公园里，花儿开，红的红，白的白， 花儿好看我不摘，人人都说我真乖。 16 、红绿灯 大马路，宽又宽，警察叔叔站中间， 红灯亮，停一停，绿灯亮，往前行。 17 、七个果果 一二三四五六七，七六五四三二一。 七个阿姨来摘果，七个篮子手中提。七个果子摆七样。 苹果、桃儿、石榴、柿子、李子、栗子、梨。 18、睡午觉 枕头放放平，花被盖盖好。 小枕头，小花被，跟我一起睡午觉，看谁先睡着。 19 、吃荸荠 荸荠有皮，皮上有泥。洗掉荸荠皮上的泥，削去荸荠外面的皮，荸荠没了皮和泥，干干净净吃荸荠。 20 、小云骑牛去打油 小云骑牛去打油，遇着小友踢皮球，皮球飞来吓了牛，摔下小云撒了油。 21 、盆和瓶 车上有个盆，盆里有个瓶，乒乒乒，乓乓乓，不知是瓶碰盆，还是盆碰瓶。 名词所有格的形式和用法。 （1）名词所有格一般是词尾加′s构成，如：the boy’s bag；our teacher’s room 等。如果原词已经 有复数词尾-s，则仅仅加一个（′）即可，如boys′ school等。词尾无s的复数名词 则仍要加′s，如： men’s clothes等。 （2）表示无生命东西的名词的所有格不可用词尾加（′s）或（′），而是用of 属格， 如：the window of the room等。但在表示时间、距离以及其他习惯用语中，则需用（′s）或（′）表 示所有格，如： ten minutes′ walk等。 （3）如果一样东西为两人共有，则只在后一个名词后加“'s”。 如：We visited Xiao Li and Xiao Zhang's room. 我们参观了小和小的房间。 （4）名词的双重所有格。（本部分只出现在教师版中） 物主代词不可与a, an, this, that, these, those, some, any, several, no, each, every, such, another, which等词一起修饰一个名词，而必须用双重所有格。 公式为：a, an, this, that +名词+of +名词性物主代词。 如：a friend of mine 我朋友中的一个 each brother of his 他的每个哥哥 名词 名词是人类认识事物所使用的基本词汇，它主要用来指人或各种事物具体的名称，也可 以指抽象的概念。 名词在句中可以作主语、宾语、表语、定语、状语、称呼语等。 名词可以分为专有名词和普通名词。专有名词是某个（些）人，地方，机构等专有的名 称，如Beijing， 名词所有格's和of的用法和区别一、's所有格 有生命的人或物的所有格用-'s表示，有时也可用of表示。如a man's voice=the voice of a man。此外，使用时还需注意以下几点： 1. 单数名词词尾加“-'s”，复数名词词尾没有“s”，也要加“-'s”。例： the boy's bag男孩的书包 men's room男厕所 2. 以“-s”结尾的单数普通名词后加“-'s”。例： The boss's son was arrogant to all the employees. 老板的儿子对所有雇佣人员都很傲慢无礼。 3. 以“-s”结尾的复数名词，其后只加“'”。例： the workers' struggle工人的斗争 4. 表示时间、距离、金额、天体、国家或城市等的名词也用“-'s”表示。例：two hours' drive 两个小时的车程 the city's scenic spots 这座城市的一些风景区 5. 如果两个名词并列，并且分别有“-'s”，则表示“分别有”；只有一个“-'s”，则表示“共有”。例： John's and Mary's room（两间） John and Mary's room（一间） 6. 作为一个整体的词组，一般在最后一个词的词尾加“-'s”。例： an hour and a half's walk（步行一个半小时的路程） 7. 不定代词后接else，所有格放在else 上。例： somebody else's bag 另外某人的书包 8. 下列情况可以将“-'s”所有格中的名词省略： 1)名词所有格所修饰的词，如果前面已经提到，往往可以省略，以免重复。例：This notebook is not mine, nor John's, nor Peter's. 这个笔记本不是我的，也不是约翰和彼得的。 2) -'s所有格后的名词若是不言而喻时，或者是某人的住所、店铺、诊所等时，通常省略。例： the doctor's (office) 医生的诊所 We had a great evening at Paul's. 我们在保罗家度过了一个愉快的夜晚。 9. 若是以“-s”结尾的专有名词，则既可只加撇号，也可加“-'s”。例：Dickens' “A Tale of Two Cities”is a literary classic. 狄更斯的《双城记》是一部古典文学作品。 二、of所有格 “名词+of+名词”构成of所有格。主要用法如下： 1. 表示无生命东西的所有关系。例： the window of the room 房间的窗户 2. 表示名词化的词的所有关系。例： the problem of the poor 穷人的问题 三、双重所有格 名词的所有格”构成双重所有格。主要用法如下：of+“ 幼儿儿歌歌词大全 1、小白兔 小白兔乖乖，把门开开，快点开开，我要进来，不开不开就不开，妈妈没回来，谁叫也不开。小白兔乖乖，把门开开，妈妈回来，我要进来，快开快开快快开，妈妈回来了，我来把门开。 2、小花狗 小花狗，真叫脏，不洗脚丫就上床。问它为什么？它说忘，忘，忘！ 3、吃饭 小宝宝，坐坐好，妈妈盛饭喂宝宝。细细嚼，慢慢咽，宝宝吃得直叫好。 4、干净 鼻涕擦干净，指甲剪干净，脸儿洗干净，妈妈和和亲一亲。 5、乒乓球 乒乓球,两人打,他给我,我给他,谁都不愿落在自己家. 6、踢毽子 小毽子,小毽子,飞上天, 落下地，我们都来踢踢它,踢不好儿没关系. 7、走夜路 夜色黑,星星闪,小朋友,把家还,突然窜出一只兔,吓得它,满处窜. 8、小盒子 小盒子,作用大,铅笔橡皮装的下, 还有一支小钢笔,装在里面心欢喜, 9、大马路 马路长又宽,我站在中间, 老师过来拜拜手,让我赶快走,说是有危险. 10、木头人 三三三，我们都是木头人， 不许哭来不许笑，还有一个不许动。 11、小皮球,小小篮,落地开花二十一,二五六,二五七, 二八二九三十一;三五六,三五七,三八三九四十一…… 12、小蜜蜂 小蜜蜂，嗡嗡嗡，嗡嗡嗡，大家一起勤劳动， 来匆匆，去匆匆，走得兴味浓，春暖花开不做工， 将来哪里好过冬？嗡嗡嗡，嗡嗡嗡，不学懒惰虫。 13、请你唱个歌吧 小杜鹃,小杜鹃，我们请你唱个歌，快来呀,大家来呀， 我们静听你的歌，咕咕!咕咕!歌声使我们快乐 14、谁会飞 谁会飞呀，鸟会飞，鸟儿鸟儿怎样飞？拍拍翅膀飞呀飞 谁会游呀，鱼会游，鱼儿鱼儿怎样游？摇摇尾巴点点头 谁会跑呀，马会跑，马儿马儿怎样跑？四脚离地身不摇。 15、好朋友 你帮我来梳梳头，我帮你来扣纽扣 团结友爱手拉手，我们都是好朋友 嘿嘿！ 16、数鸭子 门前大桥下，游过一群鸭， 快来快来数一数，二四六七八。 17、小雪花 小雪花，小雪花，飘在空中像朵花，小雪花，小雪花 飘在窗上变窗花，小雪花，小雪花，飘在手上不见了 名词所有格's和of的用法和区别 名词所有格是名词的语法范畴之一。它是名词和代词的一种变化形式，在句中表示与其它词的关系。名词有三个格：主格、宾格和所有格。在英语中有些名词可以加“'s”来表示所有关系，带这种词尾的名词形式称为该名词的所有格，如：a teacher's book。它有三种不同的形式。 一、用's表示 主要表示有生命的事物或自然界独一无二的事物以及时间距离等所有格，如the world's，the sun's，the earth's，today's，yesterday's 等。有生命的人或物的所有格用's表示，有时也可用of表示。如a man's voice=the voice of a man。 1. 单数名词词尾加“'s”，复数名词词尾没有s，也要加“'s”。 例the boy's bag 男孩的书包 men's room 男厕所 2. 若名词已有复数词尾又是s ，只加“'”。 例the workers' struggle工人的斗争 3. 凡不能加“'s”的名词，都可以用“名词+of +名词”的结构来表示所有关系。 例the title of the song 歌的名字 4. 在表示店铺或教堂的名字或某人的家时，名词所有格的后面常常不出现它所修饰的名词。 例the barber's 理发店 5. 如果两个名词并列，并且分别有's，则表示“分别有”;只有一个's，则表示“共有”。 例John's and Mary's room(两间) John and Mary's room(一间) 6. 在复合名词或短语中，'s 加在最后一个词的词尾。 例 a month or two's absence 7. 作为一个整体的词组，一般在最后一个词的词尾加's。 例an hour and a half's walk (步行一个半小时的路程) Carol and Charles' boat (卡咯和查尔斯两人共有的船) “ s ”构成名词所有格的用法 在英语中，有些名词可以加 's 来表示所有关系，带这种词尾的名词形式称为该名词的所有格。 一、 's 构成名词所有格的方法： 1. 单数名词直接在词尾加 's 。例如： This is my mother's bag. （这是我妈妈的包。） Where are Tom's books? （汤姆的书在哪儿？） 2. 在不规则复数名词的词尾加 's 。例如： Mrs Li is good at writing children's books. （李女士擅长写儿童书籍。） 3. 若名词已有复数形式的词尾 -s ，则仅在词尾加 ' 。例如： She is in the teachers' reading-room. （她在教师阅览室里。） Can you tell me how to get to the Workers' Stadium? （你能告诉我怎样才能到达工人体育馆去吗？） 4. 以 -s 结尾的专有名词（尤其是人名），后面可加 's ，也可仅加 ' ，但均读作 [iz] 。例如： On the shelf over there are Engels's （或 Engels' ） works. （那边书架上都是恩格斯的著作。） Have you ever read Burns's （或 Burns' ） poems? （你读过彭斯的诗吗？） 5. 复合名词的所用格和某些短语的所有格是在最后的那个词的词尾上加 's 。例如： This is her brother-in-law's bike. （这是她姐夫的自行车。） I don't know the editor-in-chief's telephone number. （我不知道总编辑的电话号码。） Here comes the Premier of France's car. （法国总理的汽车来了。） This work took us almost half a year's time. （这项工作花了我们几乎半年的时间。） 6. 在并列名词表示共同所有时，在后一个名词词尾上加 's 。例如： Mr Smith is Mary and Tom's father. （史密斯先生是玛丽和汤姆的爸爸。） 《一闪一闪亮晶晶》 一闪一闪亮晶晶。漫天都是小星星。挂在天空放光明。好像千万小眼睛。太阳慢慢向西沉。乌鸦回家一群群。星星眨着小眼睛。光辉照耀到天明。一闪一闪亮晶晶。满天都是小星星。 《春天在哪里》 春天在哪里呀。春天在哪里。春天在那青翠的山林里。这里有红花呀这里有绿草。还有那会唱歌的小黄鹂。嘀哩哩嘀哩嘀哩哩。嘀哩哩嘀哩嘀哩哩。春天在青翠的山林里。还有那会唱歌的小黄鹂。春天在哪里呀。春天在哪里。春天在那湖水的倒影里。映出红的花呀映出绿的草。还有那会唱歌的小黄鹂。 《卖报歌》 卖报歌。啦啦啦，啦啦啦，我是卖报的小行家，不等天明去等派报，一面走一面叫，今天的新闻真正好，七个铜板就买两份报。啦啦啦，啦啦啦，我是卖报的小行家，大风大雨里满街跑，走不好滑一跤，满身的泥水惹人笑，饥饿寒冷只有我知道。 《两只老虎》 两只老虎。两只老虎。跑的块跑的快。一只没有耳朵。一只没有尾巴。真奇怪。真奇怪。两只老虎。两只老虎。跑的块跑的快。一只没有鼻子。一只没有头发。真奇怪。真奇怪。 《葫芦娃》 葫芦娃葫芦娃。一根藤上七朵花。风吹雨打都不怕。啦啦啦啦。叮当咚咚当当。葫芦娃。叮当咚咚当当。本领大啦啦啦啦。葫芦娃葫芦娃。本领大。葫芦娃葫芦娃。一根藤上七朵花。风吹雨打都不怕。啦啦啦啦。叮当咚咚当当。葫芦娃。叮当咚咚当当。本领大啦啦啦啦。 《歌声与微笑》 请把我的歌带回你的家。请把你的微笑留下。请把我的歌带回你的家。请把你的微笑留下。明天明天这歌声飞遍海角天涯。飞遍海角天涯。明天明天这微笑将是遍野春花。将是遍野春花。 《小螺号》 小螺号嘀嘀嘀吹。海鸥听了展翅飞。小螺号嘀嘀嘀吹。浪花听了笑微微。小螺号嘀嘀嘀吹。声声唤船归啰。小螺号嘀嘀嘀吹。阿爸听了快快回啰。茫茫的海滩。蓝蓝的海水。吹起了螺号。心里美吔。 《小毛驴》 我有一只小毛驴我从来也不骑。有一天我心血来潮骑着去赶集。我手里拿着小皮鞭我心里正得意。不知怎么哗拉拉拉拉我摔了一身泥。我有一只小毛驴我从来也不骑。有一天我心血来潮骑着去赶集。我手里拿着小皮鞭我心里正得意。不知怎么哗拉拉拉拉我摔了一身泥。 《数鸭子》 门前大桥下游过一群鸭。快来快来数一数。二四六七八。嘎嘎嘎嘎。真呀真多呀。数不清到底多少鸭。数不清到底多少鸭。赶鸭老爷爷胡子白花花。唱呀唱着家乡戏。还会说笑话。小孩小孩。快快上学校。别考个鸭蛋抱回家。别考个鸭蛋抱回家。 《一只哈巴狗》 一只哈巴狗。站在大门口。眼睛黑油油。想吃肉骨头。一只哈巴狗。吃完肉骨头。尾巴摇一摇。向我点点头。 ’s所有格用法 1. 以s结尾的复数名词直接加“ ' ” 其余加“ 's ” 2 . 以s结尾的人名加“ ' ”或加“ 's ” 3.人，国家，动物，表示船只，飞机，火车，汽车，以及其它车辆或飞行器及其部件的所属关系，时间等 例如： Have you read Robert Browning’s poems？你读过罗伯特-勃郎宁的诗吗？ It’s made from mare’s，cow’s or ewe’s milk．它是用马奶、牛奶或者羊奶制成的。 但也可用于表示时间、城市、地域、团体、机构等非生命的事物。例如： We accepted the invitation without a moment’s hesitation． 我们一点也没有犹豫就接受了邀请。 New York’s population is much larger than Washington’s，though it is not the capital city． 纽约的人口比华盛顿多得多，虽然它并不是首都城市。 They are holding conferences to discuss the Europe’s future． 他们正召开各种会议来讨论欧洲的前景。 We heartily applauded the delegation’s successful visit． 我们热烈欢呼代表团访问成功。 Professor Smith is teaching at Yale’s Department of Literature． 史密斯教授在耶鲁大学文学系任教。 在某些习惯用语中，尽管是表示无生命的名词，也需要’s的所有格。例如： The driver escaped the death by a hair’s breadth．那个司机这回真是九死一生。 Now you may sing to your heart’s content．你现在可以尽情地唱了。 另外，for friendship’s sake（为了友情），at a stone’s throw（一箭之远），at one’s finger’s tip（手头上有），at arm’s length（保持距离），at one’s wits’end（黔驴技穷）等都属此类。 也可用于无生命的东西的名词之后：表示时间的名词，today's paper.今天的报纸。表示国家的名词，England's shore.英国的海岸。一些表示车，船，用具的名词，I like the car's design.我喜欢这辆车的设计 （一）表示诊所、店铺或某人的家等地点名词，其名词所有格后的被修饰语常常省略。如： 幼儿园儿歌歌词 1《花仙子之歌》 lu lu lu lu lu lu lu lu lu lu lu lu lu lu lu lu lu 能给人们带来幸福的花儿啊。你在哪里悄悄地开放。我到处把你找。脚下的路伸向远方。大波斯菊是我的帽子。蒲公英在我在我枕边飘荡。穿过那阴森的针槐林。奋勇向前向前。幸福的花仙子就是我。名字叫Lulu不寻常。说不定说不定有那么一天。就来到来到你身旁。lu lu lu lu lu。lu lu lu lu lu。lu lu lu lu lu lu lu。大波斯菊是我的帽子。蒲公英在我在我枕边飘荡。穿过那阴森的针槐林。奋勇向前向前。幸福的花仙子就是我。名字叫Lulu不寻常。说不定说不定有那么一天。就来到来到你身旁。lu lu lu lu lu。lu lu lu lu lu。lu lu lu lu lu lu lu。 2《哪咤传奇》 是他。是他。就是他。我们的朋友。小哪咤。就是他。是他。就是他。少年英雄。小哪咤上天他比天要高。下海他比海更大。智斗妖魔勇降鬼怪。少年英雄就是小哪咤。有时他也聪明有时他也犯傻。他的个头和我一般高。有时他很努力有时他也贪玩。他的年纪和我一般大。上天他比天要高。下海他比海更大。智斗妖魔勇降鬼怪。少年英雄就是小哪咤。 3《两个小娃娃打电话》 两个小娃娃呀正在打电话呀。喂喂喂你在哪里呀。诶诶诶我在幼儿园。两个小娃娃呀正在打电话呀。喂喂喂你在做什么。诶诶诶我在学唱歌。两个小娃娃呀正在打电话呀。喂喂喂你在哪里呀。诶诶诶我在幼儿园。两个小娃娃呀正在打电话呀。喂喂喂你在做什么。诶诶诶我在学唱歌。两个小娃娃呀正在打电话呀。喂喂喂你在哪里呀。诶诶诶我在幼儿园。两个小娃娃呀正在打电话呀。喂喂喂你在做什么。诶诶诶我在学唱歌。 4《山猫和吉米》 蓝蓝的天空，青青的草地 快乐的山猫，快乐的吉咪 来到了大自然，呼吸新空气 快乐成长，健康成长， 这里是快乐的，快乐的天地 来吧。。来吧来吧。。。来吧。。来吧来吧。。。来吧。。来到我的身边 来吧。。来吧来吧。。。来吧。。来吧来吧。。。和我一起唱歌跳舞 蓝蓝的天空，青青的草地 帅气的山猫，漂亮的吉咪 来到了我身边，快乐无极限快乐成长，健康成长 这里是快乐的，快乐的天地 5《大头儿子小头爸爸》 大头儿子,小头爸爸.。一对好朋友,快乐父子俩.。儿子的头大手儿小,。爸爸的头小手儿大.。大手牵小手,走路不怕滑.。走着走着走走走走,转眼儿子就长大.。 啦啦啦啦啦啦拉啦,转眼儿子就长大! 6《小羊儿》 小羊咩咩叫妈妈。母羊咩咩也叫他。跟着妈妈一道去。吃饱早回家。小羊咩咩叫妈妈。 母羊咩咩也叫他。跟着妈妈一道去。吃饱早回家。小羊咩咩叫妈妈。母羊咩咩也叫他。 跟着妈妈一道去。吃饱早回家。小羊咩咩叫妈妈。母羊咩咩也叫他。跟着妈妈一道去。 吃饱早回家。跟着妈妈一道去。吃饱早回家。啦......啦...... 幼儿儿歌歌词20首 1、小白兔 小白兔乖乖，把门开开，快点开开，我要进来，不开不开就不开，妈妈没回来，谁叫也不开。 小白兔乖乖，把门开开，妈妈回来，我要进来，快开快开快快开，妈妈回来了，我来把门开。 2、小花狗 小花狗，真叫脏，不洗脚丫就上床。问它为什么？它说忘，忘，忘！ 3、吃饭 小宝宝，坐坐好，妈妈盛饭喂宝宝。细细嚼，慢慢咽，宝宝吃得直叫好。 4、干净 鼻涕擦干净，指甲剪干净，脸儿洗干净，妈妈和和亲一亲。 5、乒乓球 乒乓球,两人打,他给我,我给他, 谁都不愿落在自己家. 6、踢毽子 小毽子,小毽子,飞上天, 落下地，我们都来踢踢它,踢不好儿没关系. 7、走夜路 夜色黑,星星闪,小朋友,把家还, 突然窜出一只兔,吓得它,满处窜. 8、小盒子 小盒子,作用大,铅笔橡皮装的下, 还有一支小钢笔,装在里面心欢喜, 9、大马路 马路长又宽,我站在中间, 老师过来拜拜手, 让我赶快走,说是有危险. 10、木头人 三三三，我们都是木头人， 不许哭来不许笑，还有一个不许动。 11、小皮球,小小篮,落地开花二十一,二五六,二五七, 二八二九三十一;三五六,三五七,三八三九四十一…… 12、小蜜蜂 小蜜蜂，嗡嗡嗡，嗡嗡嗡，大家一起勤劳动， 来匆匆，去匆匆，走得兴味浓，春暖花开不做工， 将来哪里好过冬？嗡嗡嗡，嗡嗡嗡，不学懒惰虫。 13、请你唱个歌吧 小杜鹃,小杜鹃，我们请你唱个歌，快来呀,大家来呀， 我们静听你的歌，咕咕!咕咕!歌声使我们快乐 14、谁会飞 谁会飞呀，鸟会飞，鸟儿鸟儿怎样飞？拍拍翅膀飞呀飞 谁会游呀，鱼会游，鱼儿鱼儿怎样游？摇摇尾巴点点头 谁会跑呀，马会跑，马儿马儿怎样跑？四脚离地身不摇。 15、好朋友 你帮我来梳梳头，我帮你来扣纽扣 团结友爱手拉手，我们都是好朋友 嘿嘿！ 16、数鸭子 门前大桥下，游过一群鸭， 名词所有格用法口诀英语名词所有格，表示物品所有权。 名词后加’s，这种情况最常见。 两者共有添最后，各有各添记心间。 复数名词有s, 后面只把’来添。 名词若为无生命，我们常把of用。 说明： 在英语中，表示名词所有关系的形式叫名词所有格。其构成方法如下： 1. 单数名词词尾加’s。如： Maria’s hair 玛丽亚的头发 My father’s pen 我爸爸的钢笔 2. 表示两者或两者以上共同的所有关系，仅在最后一个名词词尾加’s。如：Lily and Lucy’s mother 莉莉和露西的妈妈 3. 表示两者或两者以上各自的所有关系，每个名词词尾均需 加’s。如： Lily’s and Lucy’s bag 莉莉和露西的书包 4. 规则复数名词后只加’。如： teachers’office 老师们的办公室 students’books 学生们的书 5. 名词若是无生命的东西，还可以用of 构成名词所有格。翻译时需注意英汉语序的不同。如： a map of China 一幅中国地图 名词所有格用法 【速记口诀】 名词所有格，表物是“谁的”， 若为生命词，加“’s”即可行， 词尾有s，仅把逗号择； 并列名词后，各自和共有， 前者分别加，后者最后加； 若为无生命词，of所有格， 前后须倒置，此是硬规则。 【妙语诠释】①有生命的名词所有格一般加s，但如果名词以s结尾，则只加“’”；②并列名词所有格表示各自所有时，分别加“’s”，如果是共有，则只在最后名词加“’s”；③如果是无生命的名词则用of表示所有格，这里需要注意它们的顺序与汉语不同，A of B要翻译为B的A。 英语名词所有格 百首儿歌童谣歌词大全 1、做早操 ?早上空气真叫好，我们都来做早操。 伸伸臂，弯弯腰，踢踢腿，蹦蹦跳，天天锻炼身体好。 2、饭前要洗手 小脸盆，水清请，小朋友们笑盈盈，小手儿，伸出来，洗一洗，白又净，吃饭前，先洗手，讲卫生，不得病。 3、小手绢 小手绢，四方方，天天带在我身上。 又擦鼻涕又擦汗，干干净净真好看。 4、搬鸡蛋 小老鼠，搬鸡蛋，鸡蛋太大怎么办？一只老鼠地上躺，紧紧抱住大鸡蛋。一只老鼠拉尾巴，拉呀拉呀拉回家。 5、大骆驼 骆驼骆驼志气大，风吹日晒都不怕。 走沙漠，运盐巴，再苦再累不讲话。 6、螳螂 螳螂哥，螳螂哥，肚儿大，吃得多。飞飞能把粉蝶捕，跳跳能把蝗虫捉。两把大刀舞起来，一只害虫不放过7、大蜻蜓 大蜻蜓，绿眼睛，一对眼睛亮晶晶， 飞一飞，停一停，飞来飞去捉蚊蝇。 8、小鸭子 小鸭子，一身黄，扁扁嘴巴红脚掌。 嘎嘎嘎嘎高声唱，一摇一摆下池塘。 9、拍手歌 你拍一，我拍一，天天早起练身体。 你拍二，我拍二，天天都要带手绢。 你拍三，我拍三，洗澡以后换衬衫。 你拍四，我拍四，消灭苍蝇和蚊子。 你拍五，我拍五，有痰不要随地吐。 你拍六，我拍六，瓜皮果核不乱丢。 你拍七，我拍七，吃饭细嚼别着急。 你拍八，我拍八，勤剪指甲常刷牙。 你拍九，我拍九，吃饭以前要洗手。 你拍十，我拍十，脏的东西不要吃。 10 、小螃蟹 小螃蟹，真骄傲，横着身子到处跑， 吓跑鱼，撞倒虾，一点也不懂礼貌 11 、庆六一 儿童节，是六一，小朋友们真欢喜。 又唱歌来又跳舞，高高兴兴庆六一。 12、花猫照镜子 小花猫，喵喵叫，不洗脸，把镜照， 左边照，右边照，埋怨镜子脏，气得胡子翘。 13、蚂蚁搬虫虫 小蚂蚁，搬虫虫，一个搬，搬不动，两个搬，掀条缝，三个搬，动一动，四个五个六七个，大家一起搬进洞。 14、小青蛙 小青蛙，呱呱呱，水里游，岸上爬， 吃害虫，保庄稼，人人都要保护它。 15、花儿好看我不摘 公园里，花儿开，红的红，白的白， 花儿好看我不摘，人人都说我真乖。 16 、红绿灯 大马路，宽又宽，警察叔叔站中间， 红灯亮，停一停，绿灯亮，往前行。 17 、七个果果 一二三四五六七，七六五四三二一。 七个阿姨来摘果，七个篮子手中提。七个果子摆七样。苹果、桃儿、石榴、柿子、李子、栗子、梨。 18、睡午觉 枕头放放平，花被盖盖好。 小枕头，小花被，跟我一起睡午觉，看谁先睡着。 19 、吃荸荠 荸荠有皮，皮上有泥。 洗掉荸荠皮上的泥，削去荸荠外面的皮， 荸荠没了皮和泥，干干净净吃荸荠。 20 、小云骑牛去打油 小云骑牛去打油，遇着小友踢皮球， 皮球飞来吓了牛，摔下小云撒了油。 21 、盆和瓶 车上有个盆，盆里有个瓶，乒乒乒，乓乓乓， 不知是瓶碰盆，还是盆碰瓶。 22、小脏手，长指甲 保理业务 保理是指卖方、供应商或出口商与保理商之间存在的一种契约关系。根据该契约，卖方、供应商或出口商将其现在或将来的基于其与买方（债务人）订立的货物销售或服务合同所产生的应收账款转让给保理商，由保理商为其提供贸易融资、销售分户账管理、应收账款的催收、信用风险控制与坏账担保等服务中的至少两项。 目录 1发展历程 2详细情况 3业务分类 ?有追索权的保理 ?无追索权的保理 ?明保理 ?暗保理 ?折扣保理 ?到期保理 4业务费用 5保理业务流程 1发展历程编辑 近年来，无论国内贸易还是国际贸易，赊销结算方式日渐盛行，以国际贸易为例，信用证的使用率已经降至16%，在发达国家已降至10%以下，赊销基本上取代了信用证成为主流结算方式。在赊销贸易下，企业对应收账款的管理和融资需求正是保理业务发展的基础。 近年来随着国际贸易竞争的日益激烈，国际贸易买方市场逐渐形成。对进口商不利的信用证结算的比例逐年下降，赊销日益盛行。由于保理业务能够很好地解决赊销中出口商面临的资金占压和进口商信用风险的问题，因而在欧美、东南亚等地日渐流行。在世界各地发展迅速。据统计，98年全球保理业务量已达5000亿美元。 2详细情况编辑 国际统一私法协会《国际保理公约》对保理的定义 保理是指卖方/供应商/出口商与保理商间存在一种契约关系。根据该契约，卖方/供应商/出口商将其现在或将来的基于其与买方(债务人)订立的货物销售/服务合同所产生的应收帐款转让给保理商，由保理商为其提供下列服务中的至少两项： 贸易融资 销售分户账管理 销售分户账是出口商与债务人（进口商）交易的记录。在保理业务中，出口商可将其管理权授予保理公司，从而可集中力量进行生产、经营管理和销售，并减少了相应的财务管理人员和办公设备，从而缩小了办公占用面积。在卖方叙做保理业务后，保理商会根据卖方的要求，定期/不定期向其提供关于应收帐款的回收情况、逾期帐款情况、信用额度变化情况、对帐单等各种财务和统计报表，协助卖方进行销售管理。 应收帐款的催收 保理商一般有专业人员和专职律师进行帐款追收。保理商会根据应收帐款逾期的时间采取信函通知、打电话、上门催款直至采取法律手段。 信用风险控制与坏账担保 卖方与保理商签订保理协议后，保理商会为债务人核定一个信用额度，并且在协议执行过程中，根据债务人资信情况的变化对信用额度进行调整。对于卖方在核准信用额度内的发货所产生的应收帐款，保理商提供100%的坏帐担保。 3业务分类 在实际的运用中，保理业务有多种不同的操作方式。一般可以分为：有追索权和无追索权的保理；明保理和暗保理；折扣保理和到期保理 有追索权的保理 有追索权的保理是指供应商将应收账款的债权转让银行（即保理商），供应商在得到款项之后，如果购货商拒绝付款或无力付款，保理商有权向供应商进行追索，要求偿还预付的货币资金。当前银行出于谨慎性原则考虑，为了减少日后可能发生的损失，通常情况下会为客户提供有追索权的保理。 无追索权的保理 无追索权的保理则相反，是由保理商独自承担购货商拒绝付款或无力付款的风险。供应商在与保理商开展了保理业务之后就等于将全部的风险转嫁给了银行。因为风险过大，银行名词单复数名词所有格
儿童歌谣大全
关于国内工程保理业务介绍
儿歌歌词大全
名词的所有格
56首经典儿歌歌词大全
名词所有格地形式和用法
名词所有格's和of的用法和区别
幼儿儿歌歌词大全
名词所有格's和of的用法和区别
“ s ”构成名词所有格的用法 和upto的用法
幼儿园儿歌歌词大全
名词所有格
幼儿园儿歌歌词大全
幼儿儿歌歌词20首
名词所有格口诀
百首儿歌童谣歌词大全(可自编手指操)
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