1 Introduction

In this paper, we shall consider the existence of mild solutions for impulsive neutral stochastic functional integro-differential inclusions with infinite delay of the following form:

d [ x ( t ) g ( t , x t , 0 t a ( t , s , x s ) d s ) ] d t [ A x ( t ) + f ( t , x t ) ] d t + F ( t , x t ) d w ( t ) , t J = [ 0 , b ] , t t k ,
(1.1)
Δx( t k )=x ( t k + ) x ( t k ) = I k ( x ( t k ) ) ,k=1,2,,m,
(1.2)
x(t)=ϕ(t) L 2 (Ω, B h )for a.e. t J 0 =(,0],
(1.3)

where the state x() takes values in a separable real Hilbert space H with inner product (,) and norm ||, A is the infinitesimal generator of a compact analytic resolvent operator S(t), t0, in the Hilbert space H. Suppose that {w(t):t0} is a given K-valued Brownian motion or Wiener process with a finite trace nuclear covariance operator Q0 and L(K,H) denotes the space of all bounded linear operators from K into H. Further a:D× B h H, g:J× B h ×HH, f:J× B h H and F:J× B h P( L Q (K,H)) are given functions, where D={(t,s)J×J:st}, P( L Q (K,H)) is the family of all nonempty subsets of L Q (K,H) and L Q (K,H) denotes the space of all Q-Hilbert-Schmidt operators from K into H, which will be defined in Section 2. Here, I k C(H,H) (k=1,2,,m) are bounded functions. Furthermore, the fixed times t k satisfies 0= t 0 < t 1 < t 2 << t m <b, x( t k + ) and x( t k ) denote the right and left limits of x(t) at t= t k . Δx( t k )=x( t k + )x( t k )= I k (x( t k )) represents the jump in the state x at time t k , where I k determines the size of jump. The histories x t :Ω B h , t0, which are defined by setting x t ={x(t+s):s(,0]}, belong to the abstract phase space B h , which will be defined in Section 2. The initial data ϕ={ϕ(t):<t0} is an F 0 -measurable, B h -valued random variables independent of {w(t):t0} with finite second moment.

The theory of impulsive integro-differential inclusions has become an active area of investigation due to their applications in the fields such as mechanics, electrical engineering, medicine biology, ecology, and so on (see [1, 2] and references therein).

The existence of impulsive neutral stochastic functional integro-differential equations or inclusions with infinite delays have attracted great interest of researchers. For example, Lin and Hu [3] consider the existence results for impulsive neutral stochastic functional integro-differential inclusions with nonlocal initial conditions. Hu and Ren [4] studied the existence results for impulsive neutral stochastic functional integro-differential equations with infinite delays.

Motivated by the previous mentioned papers, we prove the existence of solutions for impulsive neutral stochastic functional integro-differential inclusions with infinite delays.

2 Preliminaries

Throughout this paper, (H,||) and (K, | | K ) denote two real separable Hilbert spaces. Let (Ω,F,P;F) (F= { F t } t 0 ) be a complete filtered probability space satisfying the requirement that F 0 contains all P-null sets of ℱ. An H-valued random variable is an ℱ-measurable function x(t):ΩH and the collection of random variables S={x(t,w):ΩH|tJ} is called a stochastic process. Suppose that {w(t):t0} is a cylindrical K-valued Wiener process with a finite trace nuclear covariance operator Q0, denote T r Q= i = 1 λ i =λ<, which satisfies Q e i = λ i e i . So, actually, w(t)= i = 1 λ i w i (t) e i , where { w i ( t ) } i = 1 are mutually independent one-dimensional standard Wiener process. We assume that F t =σ{w(s):0st} is the σ-algebra generated by w and F T =F. Let ψL(K,H) and define

| ψ | Q 2 = T r ( ψ Q ψ ) = n = 1 | λ n ψ e n | 2 .

If | ψ | Q <, then ψ is called a Q-Hilbert-Schmidt operator. Let L Q (K,H) denote the space of all Q-Hilbert-Schmidt operator ψ:KH. The completion L Q (K,H) of L(K,H) with respect to the topology induced by the norm | | Q , where | ψ | Q 2 =ψ,ψ is a Hilbert space with the above norm topology.

Let A:D(A)H be the infinitesimal generator of a compact, analytic resolvent operator S(t), t0. Let 0ρ(A). Then it is possible to define the fractional power ( A ) α for 0<α1 as a closed linear operator with its domain D( ( A ) α ) being dense in H. We denote by H α the Banach space D( A α ) endowed with the norm x α = ( A ) α x, which is equivalent to the graph norm of ( A ) α .

Lemma 2.1 ([5])

The following properties hold:

  1. (i)

    If 0<β<α1, the H α H β and the embedding is continuous and compact whenever the resolvent operator of A is compact.

  2. (ii)

    For every 0<α<1, there exists a positive constant c α such that

    ( A ) α S ( t ) C α t α ,t>0.

Now, we define the abstract phase space B h . Assume that h:(,0](0,) is a continuous function with l= 0 h(t)dt<. For any a>0 we define

B h = { ψ : ( , 0 ] H : ( E | ψ ( θ ) | 2 ) 1 2 is a bounded and measurable function on [ a , 0 ] and 0 h ( s ) sup s θ 0 ( E | ψ ( θ ) | 2 ) 1 2 d s < } .

If B h is endowed with the norm

ψ B h = 0 h(s) sup s θ 0 ( E | ψ ( θ ) | 2 ) 1 2 dsfor allψ B h ,

then ( B h , B h ) is a Banach space [6]. Now, we consider the space

B b = { x : ( , b ] H such that x k C ( J k , H ) and there exist x ( t k + ) and x ( t k ) with x ( t k ) = x ( t k ) , x 0 = ϕ L 2 ( Ω , B h ) on ( , 0 ] , k = 1 , 2 , , m } ,

where x k is the restriction of x to J k =( t k , t k + 1 ], k=0,1,,m. Let b be a seminorm in B b defined by

x b = x 0 B h + sup 0 s b ( E | x ( s ) | 2 ) 1 2 ,x B b .

Lemma 2.2 ([7])

Assume that x B b , then for tJ, x t B h . Moreover

l ( E | x ( t ) | 2 ) 1 2 x t B h x 0 B h +l sup 0 s t ( E | x ( t ) | 2 ) 1 2 ,

where l= 0 h(s)ds<.

We use the notation P(H) for the family of all subsets H and denote

P c l ( H ) = { Y P ( H ) : Y  is closed } , P b d ( H ) = { Y P ( H ) : Y  is bounded } , P c v ( H ) = { Y P ( H ) : Y  is convex } , P c p ( H ) = { Y P ( H ) : Y  is compact } .

A multi-valued mapping Γ:HP(H) is called upper semicontinuous (u.s.c) if for any xH, the set Γ(x) is a nonempty closed subset of H and if for each open set G of H containing Γ(x), there exists an open neighborhood N of x such that Γ(N)G. Γ is said to be completely continuous if Γ(B) is relatively compact for every bounded subset of BH. If the multi-valued mapping Γ is completely continuous with nonempty compact values, then Γ is u.s.c. if and only if Γ has a closed graph, i.e., x n x, y n y, y n Γ( x n ) imply yΓ(x).

Definition 2.1 The multi-valued mapping F:J× B h P(H) is said to be L 2 -Carathéodory if

  1. (i)

    tF(t,v) is measurable for each v B h ,

  2. (ii)

    vF(t,v) is u.s.c. for almost all tJ and v B h ,

  3. (iii)

    for each q>0, there exists h q L 1 (J, R + ) such that

    F ( t , v ) 2 = sup f F ( t , v ) E ( | f | 2 ) h q (t),

for all v B h 2 q and for a.e. tJ.

The following lemma is crucial in the proof of our main result.

Lemma 2.3 ([8])

Let I be a compact interval and H be a Hilbert space. Let F be an L 2 -Carathéodory multi-valued mapping with N F , x ϕ and let Γ be a linear continuous mapping from L 2 (I,H) to C(I,H). Then the operator

Γ N F :C(I,H) P c p , c v (H),x(Γ N F )(x)=Γ( N F , x )

is a closed graph operator in C(I,H)×C(I,H), where N F , x is known as the selectors set from F; it is given by

σ N F , x = { σ L 2 ( L ( K , H ) ) : σ ( t ) F ( t , x ) for a.e. t J } .

Theorem 2.1 ([9])

Let X be a Banach space, Φ 1 :X P c l , c v , b d (X) and Φ 2 :X P c p , c v (X) be two multi-valued operators satisfying:

  1. (a)

    Φ 1 is a contraction,

  2. (b)

    Φ 2 is u.s.c. and completely continuous.

Then either

  1. (i)

    the operator inclusion λx Φ 1 x+ Φ 2 x has a solution for λ=1, or

  2. (ii)

    the set G={xX:λx Φ 1 x+ Φ 2 x,λ>1} is unbounded.

Lemma 2.4 ([10])

Let v,w:[0,b][0,) be continuous functions. If w is nondecreasing and there are constants θ>0, 0<α<1 such that

v(t)w(t)+θ 0 t v ( s ) ( t s ) 1 α ds,tJ,

then

v(t) e θ n Γ ( α ) n t n α Γ ( n α ) j = 0 n 1 ( θ b α α ) j w(t)

for every tJ and every nN such that nα>1 and Γ() is the Gamma function.

3 Main result

Let J 1 =(,b]. First, we present the definition of the mild solution of problem (1.1)-(1.3).

Definition 3.1 A stochastic process x: J 1 ×ΩH is called a mild solution of problem (1.1)-(1.3) if

  1. (i)

    x(t) is measurable and F t -adapted for each t0,

  2. (ii)

    Δx( t k )=x( t k + )x( t k ), k=1,2,,m,

  3. (iii)

    x(t)H has càdlàg paths on tJ a.e. and there exists a function σ N F , x such that

    x ( t ) = S ( t ) [ ϕ ( 0 ) g ( 0 , ϕ , 0 ) ] + g ( t , x t , 0 t a ( t , s , x s ) d s ) + 0 t A S ( t s ) g ( s , x s , 0 s a ( s , τ , x τ ) d τ ) d s + 0 t A S ( t s ) f ( s , x s ) d s + 0 t S ( t s ) σ ( s ) d w ( s ) + 0 < t k < t S ( t t k ) I k ( x ( t k ) ) , t J ,
  4. (iv)

    x 0 ()=ϕ L 2 (Ω, B h ) on J 0 =(,0] satisfies ϕ B h 2 <.

Now, we assume the following hypotheses:

(H1) A is the infinitesimal generator of a compact analytic resolvent operator S(t), t0, in the Hilbert space H and there exist positive constants M and M 1 such that

S ( t ) 2 M, A β M 1 ,tJ.

(H2) a:D× B h H, D={(t,s)J×J:ts} is a continuous function and there exists a constant M a such that

E | 0 t [ a ( t , s , x ) a ( t , s , y ) ] d s | 2 M a x y B h 2 for all tJ,x,y B h .

(H3) There exist constants 0<β<1 and M g such that g is H β -valued, ( A ) β g is continuous and

E | ( A ) β g ( t , x 1 , y 1 ) ( A ) β g ( t , x 2 , y 2 ) | 2 M g [ x 1 x 2 B h 2 + E | y 1 y 2 | 2 ] .

(H4) The function f:J× B h H satisfies the following conditions:

  1. (i)

    tf(t,s) is measurable for each x B h ;

  2. (ii)

    xf(t,x) is continuous for almost all tJ;

  3. (iii)

    There exists a constant M f such that

    E | ( A ) β f ( t , x ) ( A ) β f ( t , y ) | 2 M f x y B h 2

    for all x,y B h , tJ and

    E | f ( t , x ) | 2 p(t)ψ ( x B h 2 )

    for almost all tJ, where p L 1 (J,R), ψ: R + (0,) is continuous and increasing with

    0 b μ ( s ) ¯ d s B 0 k 1 1 ψ ( s ) d s , μ ¯ ( t ) = B 0 k 3 p ( t ) , k 1 = 4 ϕ B h 2 + l 2 F 1 96 l 2 ( A ) β 2 M g ( 1 + 2 M a ) , k 2 = 96 b l 2 M g ( 1 + 2 M a ) c 1 β 2 1 96 l 2 ( A ) β 2 M g ( 1 + 2 M a ) , k 3 = 48 M b l 2 1 96 l 2 ( A ) β 2 M g ( 1 + 2 M a ) , L 0 = 3 l 2 [ M g ( 1 + M a ) ( ( A ) β 2 + ( C 1 β b β ) 2 2 β 1 ) + M f ( C 1 β b β ) 2 2 β 1 ] < 1 , B 0 = e k 2 n Γ ( β ) n b n β / Γ ( n β ) j = 1 n 1 ( k 2 b β β ) j , c 1 = b 2 sup ( t , s ) D a 2 ( t , s , 0 ) , c 2 = ( A ) β 2 sup t J g ( t , 0 , 0 ) 2

    and

    F = 4 M | ϕ ( 0 ) | 2 + 96 ( M + ( A ) β 2 ) c 2 + 192 ( A ) β 2 M g c 1 + 192 b 2 β C 1 β 2 2 β 1 ( c 2 + 2 M g c 1 ) + 48 M μ L loc 1 ( J , R + ) b 2 Tr ( Q ) + 48 M m 2 k = 1 m d k + 96 M ( A ) β 2 M g ϕ B h 2 .

    (H5) The multi-valued mapping F:J× B h P b d , c l , c v (L(K,H)) is an L 2 -Carathéodory function that satisfies the following conditions:

    1. (i)

      For each tJ, the function F(t,): B h P b d , c l , c v (L(K,H)) is u.s.c. and for each fixed x B h , the function F(,x) is measurable. For each x B h , the set

      N F , x = { σ L 2 ( K , H ) : σ ( t ) F ( t , x )  for a.e.  t J }

    is nonempty.

  4. (ii)

    There exists a positive function μ L loc 1 (J, R + ) such that

    F ( t , x ) 2 = sup σ F ( t , x ) E | σ | 2 μ(t).

(H6) I k C( H α , H α ) and there exist positive constants d k such that for each x H α ,

| I k ( x ) | 2 d k ,k=1,2,,m.

We consider the mapping Φ: B h P( B h ) defined by

Φx(t)={ ϕ ( t ) , t ( , 0 ] , S ( t ) [ ϕ ( 0 ) g ( 0 , ϕ , 0 ) ] + g ( t , x t , 0 t a ( t , s , x s ) d s ) + 0 t A S ( t s ) g ( s , x s , 0 s a ( s , τ , x τ ) d τ ) d s + 0 t A S ( t s ) f ( s , x s ) d s + 0 t S ( t s ) σ ( s ) d w ( s ) + 0 < t k < t S ( t t k ) I k ( x ( t k ) ) , t J ,

where σ N F , x . For each ϕ B h , we define

ϕ ˜ (t)={ ϕ ( t ) , t ( , 0 ] , S ( t ) ϕ ( 0 ) , t J ,

and then ϕ ˜ B h . Let x(t)=y(t)+ ϕ ˜ (t), t(,b]. Then it is easy to see that x satisfies (1.1)-(1.3) if and only if y satisfies y 0 =0 and

y ( t ) = S ( t ) g ( 0 , ϕ , 0 ) + g ( t , y t + ϕ ˜ t , 0 t a ( t , s , y s + ϕ ˜ s ) d s ) + 0 t A S ( t s ) g ( s , y s + ϕ ˜ s , 0 s a ( s , τ , y τ + ϕ ˜ τ ) d τ ) d s + 0 t A S ( t s ) f ( s , y s + ϕ s ) d s + 0 t S ( t s ) σ ( s ) d w ( s ) + 0 < t k < t S ( t t k ) I k ( y ( t k ) + ϕ ˜ ( t k ) ) , t J ,

where σ N F , y . Let B h ={y B h : y 0 =0 B h }. For any y B h ,

y b = y 0 B h + sup 0 s b ( E | y ( s ) | 2 ) 1 2 = sup 0 s b ( E | y ( s ) | 2 ) 1 2

and thus ( B h , b ) is a Banach space. Set B q ={y B h : y b 2 q} for some q0. Then B q B h is uniformly bounded and for any y B q , from Lemma 2.2, we see that

y t + ϕ ˜ t B h 2 2 y t B h 2 + 2 ϕ ˜ t B h 2 4 l 2 sup 0 s t E | y ( s ) | 2 + 4 y 0 B h 2 + 4 l 2 sup 0 s t ϕ ˜ ( s ) 2 + 4 ϕ ˜ 0 B h 2 4 l 2 ( q + M | ϕ ( 0 ) | 2 ) + 4 ϕ ˜ B h 2 : = q .

Define the operator Φ ˜ : B h P( B h ) by

Φ ˜ y(t)={ 0 , t ( , 0 ] , S ( t ) g ( 0 , ϕ , 0 ) + g ( t , y t + ϕ ˜ t , 0 t a ( t , s , y s + ϕ ˜ s ) d s ) + 0 t A S ( t s ) g ( s , y s + ϕ ˜ , 0 s a ( s , τ , y τ + ϕ ˜ τ ) d τ ) d s + 0 t A S ( t s ) f ( s , y s + ϕ ˜ s ) d s + 0 t S ( t s ) σ ( s ) d w ( s ) + 0 < t k < t S ( t t k ) I k ( y ( t k ) + ϕ ˜ ( t k ) ) , t J ,

where σ N F , y . Obviously, the operator Φ has a fixed point is equivalent to proving that Φ ˜ has a fixed point. Now, we decompose Φ ˜ as Φ ˜ 1 + Φ ˜ 2 , where

Φ ˜ 1 y ( t ) = S ( t ) g ( 0 , ϕ , 0 ) + g ( t , y t + ϕ ˜ t , 0 t a ( t , s , y s + ϕ ˜ s ) d s ) + 0 t A S ( t s ) g ( s , y s + ϕ ˜ s , 0 s a ( s , τ , y τ + ϕ ˜ τ ) d τ ) d s + 0 t A S ( t s ) f ( s , y s + ϕ ˜ s ) d s

and

Φ ˜ 2 y(t)= 0 s S(ts)σ(s)dw(s)+ 0 < t k < t S(t t k ) I k ( y ( t k ) + ϕ ˜ ( t k ) ) ,tJ,

where σ N F , y . In what follows, we show that the operators Φ ˜ 1 and Φ ˜ 2 satisfy all the conditions of Theorem 2.1.

Lemma 3.1 Assume that the assumptions (H1)-(H6) hold. Then Φ ˜ 1 is a contraction and Φ ˜ 2 is u.s.c. and completely continuous.

Proof We give the proof in several steps:

Step 1. Φ ˜ 1 is a contraction.

Let u,v B h . Then we have

E | ϕ ˜ 1 u ( t ) ϕ ˜ 1 v ( t ) | 2 3 E | g ( t , u t + ϕ ˜ t , 0 t a ( t , s , u s + ϕ ˜ s ) d s ) g ( t , v t + ϕ ˜ t , 0 t a ( t , s , v s + ϕ ˜ s ) d s ) | 2 + 3 b E ( 0 t | A S ( t s ) [ g ( s , u s + ϕ ˜ s , 0 s a ( s , τ , u τ + ϕ ˜ τ ) d τ ) g ( s , v s + ϕ ˜ s , 0 s a ( s , τ , u τ + ϕ ˜ τ ) d τ ) ] | 2 d s ) + 3 b E ( 0 t | A S ( t s ) [ f ( s , u s + ϕ ˜ s ) f ( s , v s + ϕ ˜ s ) ] | 2 d s ) 3 ( A ) β 2 M g ( u t v t B h 2 + M a u t v t B h 2 ) + 3 b 0 t C 1 β 2 ( t s ) 2 ( 1 β ) M g ( u s v s B h 2 + M a u s v s B h 2 ) d s + 3 b 0 t C 1 β 2 ( t s ) 2 ( 1 β ) M f u s v s B h 2 d s 3 ( A ) β 2 M g ( 1 + M a ) u t v t B h 2 + 3 M g ( 1 + M a ) ( C 1 β b β ) 2 2 β 1 u t v t B h 2 + 3 M f ( C 1 β b β ) 2 2 β 1 u t v t B h 2 3 [ M g ( 1 + M a ) ( ( A ) β 2 + ( C 1 β b β ) 2 2 β 1 ) + M f ( C 1 β b β ) 2 2 β 1 ] × [ l 2 sup s [ 0 , b ] E | u ( s ) v ( s ) | 2 + u 0 B h 2 + v 0 B h 2 ] = 3 l 2 [ M g ( 1 + M a ) ( ( A ) β 2 + ( C 1 β b β ) 2 2 β 1 ) + M f ( C 1 β b β ) 2 2 β 1 ] sup s [ 0 , b ] E | u ( s ) v ( s ) | 2 = L 0 sup s [ 0 , b ] E | u ( s ) v ( s ) | 2 ,

where L 0 =3 l 2 [ M g (1+ M a )( ( A ) β 2 + ( C 1 β b β ) 2 2 β 1 )+ M f ( C 1 β b β ) 2 2 β 1 ]<1 and we have used the fact that u 0 B h 2 =0 and v 0 B h 2 =0. Taking the supremum over t, we obtain

ϕ ˜ 1 u ϕ ˜ 1 v b 2 L 0 u v b 2

and so ϕ ˜ 1 is a contraction.

Now, we show that the operator Φ ˜ 2 is completely continuous.

Step 2. Φ ˜ 2 y is convex for each y B h .

In fact, if u 1 , u 2 Φ ˜ 2 (y), then there exist σ 1 , σ 2 N F , y such that

u i (t)= 0 t S(ts) σ i (s)dw(s)+ 0 < t k < t S(t t k ) I k ( y ( t k ) + ϕ ˜ ( t k ) )

for i=1,2 and tJ. Let λ[0,1]. Then for each tJ, we have

λ u 1 ( t ) + ( 1 λ ) u 2 ( t ) = 0 t S ( t s ) [ λ σ 1 ( s ) + ( 1 λ ) σ 2 ( s ) ] d w ( s ) + 0 < t k < t S ( t t k ) I k ( y ( t k ) + ϕ ˜ ( t k ) ) .

Since N F , y is convex (because F has convex values), we obtain

λ u 1 (t)+(1λ) u 2 (t) Φ ˜ 2 (y).

Step 3. Φ ˜ 2 maps bounded sets into bounded sets in B h .

It is enough to show that there exists a positive constant Λ such that for each u Φ ˜ 2 y, y B q ={y B h : y b q} one has u b Λ. If u Φ ˜ 2 (y), there exists σ N F , y such that for each tJ

u(t)= 0 t S(st)σ(s)dw(s)+ 0 < t k < t S(t t k ) I k ( y ( t k ) + ϕ ˜ ( t k ) )

and so

E | u ( t ) | 2 = E | 0 t S ( t s ) σ ( s ) d w ( s ) + 0 < t k < t S ( t t k ) I k ( y ( t k ) + ϕ ˜ ( t k ) ) | 2 2 E | 0 t S ( t s ) σ ( s ) d w ( s ) | 2 + 2 E | 0 < t k < t S ( t t k ) I k ( y ( t k ) + ϕ ˜ ( t k ) ) | 2 2 Tr ( Q ) M b 0 b μ ( s ) d s + 2 M m 2 k = 1 m d k 2 Tr ( Q ) M b 2 μ L loc 1 ( J , R + ) + 2 M m 2 k = 1 m d k : = Λ .

Thus, for each y B h , we get u b 2 Λ.

Step 4. Φ ˜ 2 maps bounded sets into equicontinuous sets of B h .

Let 0< τ 1 < τ 2 b. For each y B q ={y B h : y b q} and u Φ ˜ 2 (y). Let τ 1 , τ 2 J{ t 1 , t 2 ,, t m }. Then there exists σ N F , y such that for each tJ,

u(t)= 0 t S(ts)σ(s)dw(s)+ 0 < t k < t S(t t k ) I k ( y ( t k ) + ϕ ˜ ( t k ) ) .

Thus we have

E | u ( τ 2 ) u ( τ 1 ) | 2 = E | 0 τ 2 S ( τ 2 s ) σ ( s ) d w ( s ) + 0 < t k < τ 2 S ( τ 2 s ) I k ( y ( t k ) + ϕ ˜ ( t k ) ) 0 τ 1 S ( τ 1 s ) σ ( s ) d w ( s ) 0 < t k < τ 1 S ( τ 1 t k ) I k ( y ( t k ) + ϕ ˜ ( t k ) ) | 2 2 E | 0 τ 1 ε ( S ( τ 2 s ) σ ( s ) S ( τ 1 s ) σ ( s ) ) d w ( s ) + τ 1 ε τ 1 ( S ( τ 2 s ) σ ( s ) S ( τ 1 s ) σ ( s ) ) d w ( s ) + τ 1 τ 2 S ( τ 2 s ) σ ( s ) d w ( s ) | 2 + 2 E | 0 < t k < τ 1 [ S ( τ 2 t k ) S ( τ 1 t k ) ] I k ( y ( t k ) + ϕ ˜ ( t k ) ) + τ 1 < t k < τ 2 S ( τ 2 t k ) I k ( y ( t k ) + ϕ ˜ ( t k ) ) | 2 6 ε Tr ( Q ) 0 τ 1 ε μ ( s ) S ( τ 2 s ) S ( τ 1 s ) 2 d s + 6 ε Tr ( Q ) τ 1 ε τ 1 μ ( s ) S ( τ 2 s ) S ( τ 1 s ) 2 d s + 6 ( τ 2 τ 1 ) Tr ( Q ) τ 1 τ 2 μ ( s ) S ( τ 2 s ) 2 d s + 4 m 2 0 < t k < τ 1 S ( τ 2 s ) S ( τ 1 s ) 2 d k + 4 m 2 M τ 1 < t k < τ 2 d k .

The right-hand side of the above inequality tends to zero as τ 1 τ 2 with ε sufficiently small, since S(t) is strongly continuous and the compactness of S(t) for t>0 implies the continuity in the uniform operator topology. Thus, the set { Φ ˜ 2 y:y B q } is equicontinuous. Here we consider the case 0< τ 1 < τ 2 b, since the case τ 1 < τ 2 0 or τ 1 0 τ 2 b is simple.

Step 5. Φ ˜ 2 maps B q into a precompact set in H.

Let 0<tb and 0<ε<t. For y B q and u Φ ˜ 2 (y), there exists σ N F , y such that

u ( t ) = 0 t ε S ( t s ) σ ( s ) d w ( s ) + t ε t S ( t s ) σ ( s ) d w ( s ) + 0 < t k < t S ( t t k ) I k ( y ( t k ) + ϕ ˜ ( t k ) ) .

Define

u ε (t)=S(ε) 0 t ε S(tεs)σ(s)dw(s)+ 0 < t k < t ε S(t t k ) I k ( y ( t k ) + ϕ ˜ ( t k ) ) .

Since S(t) is a compact operator, the set V ε (t)={ u ε (t): u ε Φ ˜ 2 ( B q )} is relatively compact in H for each ε, 0<ε<t. Moreover,

E | u ( t ) u ε ( t ) | 2 = E | 0 t ε S ( t s ) σ ( s ) d w ( s ) + t ε t S ( t s ) σ ( s ) d w ( s ) + 0 < t k < t S ( t t k ) I k ( y ( t k ) + ϕ ˜ ( t k ) ) S ( ε ) 0 t ε S ( t ε s ) σ ( s ) d w ( s ) 0 < t k < t ε S ( t t k ) I k ( y ( t k ) + ϕ ˜ ( t k ) ) | 2 4 M b Tr ( Q ) ε μ L loc 1 ( J , R + ) + 4 m 2 M t ε < t k < t d k .

Therefore letting ε0, we can see that there are relative compact sets arbitrarily close to the set {u(t):u Φ ˜ 2 ( B q )}. Thus, the set {u(t):u Φ ˜ 2 ( B q )} is relatively compact in H. Hence, the Arzelá-Ascoli theorem shows that Φ ˜ 2 is a compact multi-valued mapping.

Step 6. Φ ˜ 2 has a closed graph.

Let y n y , u n Φ ˜ 2 ( y n ) and u n u . We prove that u Φ ˜ 2 ( y ).

Indeed, u n Φ ˜ 2 ( y n ) means that there exists σ n N F , y n such that

u n (t)= 0 t S(ts) σ n (s)dw(s)+ 0 < t k < t S(t t k ) I k ( y n ( t k ) + ϕ ˜ ( t k ) ) ,tJ.

Thus we must prove that there exists σ N F , y such that

u (t)= 0 t S(ts) σ (s)dw(s)+ 0 < t k < t S(t t k ) I k ( y ( t k ) + ϕ ˜ ( t k ) ) ,tJ.

Since I k , k=1,2,,m, are continuous, we see that

0 < t k < t S ( t t k ) I k ( y n ( t k ) + ϕ ˜ ( t k ) ) 0 < t k < t S ( t t k ) I k ( y ( t k ) + ϕ ˜ ( t k ) ) b 2 0

as n. Consider the linear continuous operator Γ: L 2 (J,H)C(J,H) with Γ(σ)(t)= 0 t S(ts)σ(s)dw(s), where σ N F , y . From Lemma 2.3, it follows that Γ N F is a closed graph operator. Moreover, we have

u n (t) 0 < t k < t S(t t k ) I k ( y n ( t k ) + ϕ ˜ ( t k ) ) Γ( N F , y n ).

Since y n y , from Lemma 2.3, we obtain

u (t) 0 < t k < t S(t t k ) I k ( y ( t k ) + ϕ ˜ ( t k ) ) Γ( N F , y ).

That is, there exists a σ N F , y such that

u ( t ) 0 < t k < t S ( t t k ) I k ( y ( t k ) + ϕ ˜ ( t k ) ) = Γ ( σ ( t ) ) = 0 t S ( t s ) σ ( s ) d w ( s ) .

Therefore Φ ˜ 2 has a closed graph and Φ ˜ 2 is u.s.c. This completes the proof. □

Lemma 3.2 Assume that the assumptions (H1)-(H2) hold. Then there exists a constant K>0 such that y t + ϕ ˜ t B h 2 K for all tJ, where K is depends only on b and the functions ψ and μ ¯ .

Proof Let y be a possible solution of yλ Φ ˜ (y) for some 0<λ<1. Then there exists σ N F , y such that for tJ we have

y ( t ) = λ S ( t ) g ( 0 , ϕ , 0 ) + λ g ( t , y t + ϕ ˜ t , 0 t a ( t , s , y s + ϕ ˜ s ) d s ) + λ 0 t A S ( t s ) g ( s , y s + ϕ ˜ s , 0 s a ( s , τ , y τ + ϕ ˜ τ ) d τ ) d s + λ 0 t S ( t s ) f ( s , y s + ϕ ˜ s ) d s + 0 t S ( t s ) σ ( s ) d w ( s ) + λ 0 < t k < t S ( t t k ) I k ( y ( t k ) + ϕ ˜ ( t k ) ) .

Then, by the assumptions, we deduce that

E | y ( t ) | 2 E | S ( t ) g ( 0 , ϕ , 0 ) + g ( t , y t + ϕ ˜ t , 0 t a ( t , s , y s + ϕ ˜ s ) d s ) + 0 t A S ( t s ) g ( s , y s + ϕ ˜ s , 0 s a ( s , τ , y τ + ϕ ˜ τ ) d τ ) d s + 0 t S ( t s ) f ( s , y s + ϕ ˜ s ) d s + 0 t S ( t s ) σ ( s ) d w ( s ) + 0 < t k < t S ( t t k ) I k ( y ( t k ) + ϕ ˜ ( t k ) ) | 2 12 { 2 M ( ( A ) β 2 M g ϕ B h 2 + c 2 ) + 2 ( A ) β 2 [ M g ( y s + ϕ ˜ s B h 2 + 2 M a y s + ϕ ˜ s B h 2 + 2 c 1 ) + c 2 ] + 2 b 0 t c 1 β 2 ( t s ) 2 ( 1 β ) [ M g ( y s + ϕ ˜ s B h 2 + 2 M a y s + ϕ ˜ s B h 2 + 2 c 1 ) + c 2 ] d s + M b 0 t p ( s ) ψ ( y s + ϕ ˜ s B h 2 ) d s + M μ L loc 1 ( J , R + ) b 2 Tr ( Q ) + M m 2 k = 1 m d k } = 24 ( M + ( A ) β 2 ) c 2 + 48 ( A ) β 2 M g c 1 + 48 b 2 β c 1 β 2 2 β 1 ( c 2 + 2 M g c 1 ) + 12 M μ L loc 1 ( J , R + ) b 2 Tr ( Q ) + 12 M m 2 k = 1 m d k + 24 M ( A ) β 2 M g ϕ B h 2 + 24 ( A ) β 2 M g ( 1 + 2 M a ) y s + ϕ ˜ s B h 2 + 24 b M g ( 1 + 2 M a ) c 1 β 2 0 t y s + ϕ ˜ s B h 2 ( t s ) 2 ( 1 β ) d s + 12 M b 0 t p ( s ) ψ ( y s + ϕ ˜ s B h 2 ) d s .

From Lemma 2.2 we see that

y t + ϕ ˜ t B h 2 4 l 2 sup 0 s t E | y ( s ) | 2 +4 l 2 M | ϕ ˜ ( 0 ) | 2 +4 ϕ ˜ B h 2 .

Thus, for any tJ, we have

y t + ϕ ˜ t B h 2 4 l 2 M | ϕ ˜ ( 0 ) | 2 + 4 ϕ ˜ B h 2 + 96 l 2 ( M + ( A ) β 2 ) c 2 + 192 l 2 ( A ) β 2 M g c 1 + 192 l 2 b 2 β C 1 β 2 2 β 1 ( c 2 + 2 M g c 1 ) + 48 M μ L loc 1 ( J , R + ) b 2 l 2 Tr ( Q ) + 48 M l 2 m 2 k = 1 m d k + 96 M l 2 ( A ) β 2 M g ϕ B h 2 + 96 l 2 ( A ) β 2 M g ( 1 + 2 M a ) y s + ϕ ˜ s B h 2 + 96 b l 2 M g ( 1 + 2 M a ) C 1 β 2 0 t y s + ϕ ˜ s B h 2 ( t s ) 2 ( 1 β ) d s + 48 M b l 2 0 t p ( s ) ψ ( y s + ϕ ˜ s B h 2 ) d s = 4 ϕ B h 2 + l 2 F + 96 l 2 ( A ) β 2 M g ( 1 + 2 M a ) sup 0 s t y s + ϕ ˜ s B h 2 + 96 b l 2 M g ( 1 + 2 M a ) C 1 β 2 0 t y s + ϕ ˜ s B h 2 ( t s ) 2 ( 1 β ) d s + 48 M b l 2 0 t p ( s ) ψ ( y s + ϕ ˜ s B h 2 ) d s .

Let v(t)= sup 0 s t y s + ϕ ˜ s B h 2 . Then the function v(t) is nondecreasing in J. Thus, we obtain

v ( t ) 4 ϕ B h 2 + l 2 F + 96 l 2 ( A ) β 2 M g ( 1 + 2 M a ) v ( t ) + 96 b l 2 M g ( 1 + 2 M a ) C 1 β 2 0 t v ( s ) ( t s ) 2 ( 1 β ) d s + 48 M b l 2 0 t p ( s ) ψ ( v ( s ) ) d s .

From this we derive that

v ( t ) 4 ϕ B h 2 + l 2 F 1 96 l 2 ( A ) β 2 M g ( 1 + 2 M a ) + 96 b l 2 M g ( 1 + 2 M a ) C 1 β 2 1 96 l 2 ( A ) β 2 M g ( 1 + 2 M a ) 0 t v ( s ) ( t s ) 1 β d s + 48 M b l 2 1 96 l 2 ( A ) β 2 M g ( 1 + 2 M a ) 0 t p ( s ) ψ ( v ( s ) ) d s k 1 + k 2 0 t v ( s ) ( t s ) 1 β d s + k 3 0 t p ( s ) ψ ( v ( s ) ) d s .

By Lemma 2.4, we get

v(t) B 0 ( k 1 + k 3 0 t p ( s ) ψ ( v ( s ) ) d s ) ,

where

B 0 = e k 2 n Γ ( β ) n b n β / Γ ( n β ) j = 1 n 1 ( k 2 b β β ) j .

Let us take the right-hand side of the above inequality as μ(t). Then μ(0)= B 0 k 1 , v(t)μ(t), tJ and

μ (t) B 0 k 3 p(t)ψ ( v ( t ) ) .

Since ψ is nondecreasing, we have

μ ( t ) B 0 k 3 p ( t ) ψ ( μ ( t ) ) = μ ¯ ( t ) ψ ( μ ( t ) ) .

It follows that

μ ( 0 ) μ ( t ) 1 ψ ( s ) d s 0 b μ ( s ) ¯ d s B 0 K 1 1 ψ ( s ) d s ,

which indicates that μ(t)<. Thus, there exists a constant K such that μ(t)K, tJ. Furthermore, we see that y t + ϕ ˜ t B h 2 v(t)μ(t)K, tJ. □

Theorem 3.1 Assume that the assumptions (H1)-(H6) hold. The problem (1.1)-(1.3) has at least one mild solution on J.

Proof Let us take the set

G(Φ)= { x B h : x λ Φ ( x )  for some  λ ( 0 , 1 ) } .

Then for any xG(Φ), we have

x t B h 2 = y t + ϕ ˜ t B h 2 K,tJ,

where K>0 is a constant in Lemma 3.2. This show that G is bounded on J. Hence from Theorem 2.1 there exists a fixed point x(t) for Φ on B h , which is a mild solution of (1.1)-(1.3) on J. □

4 An example

As an application of Theorem 3.1, we consider the impulsive neutral stochastic functional integro-differential inclusion of the following form:

t ( z ( t , x ) + g ( t , z ( t h , x ) , 0 t a ( t , s , z ( s h , x ) ) d s ) ) 2 x 2 z ( t , x ) + ( f ( t , z ( t h , x ) ) + [ Q 1 ( t , z ( t h , x ) ) , Q 2 ( t , z ( t h , x ) ) ] ) d w ( t ) ,
(4.1)
0 x π , t J , t t k , Δ z ( t k , x ) = z ( t k + , x ) z ( t k , x ) = I k ( z ( t k , x ) ) , k = 1 , 2 , , m ,
(4.2)
z(t,0)=z(t,π)=0,tJ,
(4.3)
z(t,x)=ρ(t,x),<t0,0xπ,
(4.4)

where J=[0,b], k=1,2,,m, z( t k + ,x)= lim h 0 + z( t k +h,x), z( t k ,x)= lim h 0 z( t k +h,x), Q 1 , Q 2 :J×RR are two given functions and w(t) is a one-dimensional standard Wiener process. We assume that for each tJ, Q 1 (t,) is lower semicontinuous and Q 2 (t,) is upper semicontinuous. Let J 1 =(,b] and H= L 2 ([0,π]) with norm . Define A:HH by Av= v with domain D(A)={vH:v, v  are absolutely continuous, v H,v(0)=v(π)=0}. Then

Av= n = 1 n 2 (v, v n ),vD(A),

where v n = 2 π sin(ns), n=1,2, , is the orthogonal set of eigenvectors in A. It is well known that A is the infinitesimal generator of an analytic semigroup S(t), t0 in H given by

S(t)v= n = 1 e n 2 t (v, v n ) v n ,vH.

For every vH, ( A ) 1 2 v= n = 1 1 n (v, v n ) v n and ( A ) 1 2 =1. The operator ( A ) 1 2 is given by

( A ) 1 2 v= n = 1 n(v, v n ) v n

on the space D( ( A ) 1 2 )={vH: n = 1 n(v, v n ) v n H}. Since the analytic semigroup S(t) is compact [10], there exists a constant M>0 such that S(t)M and satisfies (H1). Now, we give a special B h -space. Let h(s)= e 2 s , s<0. Then l= 0 h(s)ds= 1 2 and let

φ B h = 0 h(s) sup s θ 0 ( E | φ ( θ ) | 2 ) 1 2 ds.

It follows from [5] that ( B h , B h ) is a Banach space. Hence for (t,ϕ)[0,b]× B h , let

ϕ ( θ ) x = ϕ ( θ , x ) , ( θ , x ) ( , 0 ] × [ 0 , π ] , z ( t ) ( x ) = z ( t , x )

and

F(t,ϕ)(x)= [ Q 1 ( t , ϕ ( θ , x ) ) , Q 2 ( t , ϕ ( θ , x ) ) ] ,<θ0,x[0,π].

Then (4.1)-(4.4) can be rewritten as the abstract form as the system (1.1)-(1.3). If we assume that (H2)-(H6) are satisfied, then the system (4.1)-(4.4) has a mild solution on [0,b].