A note on the existence and uniqueness of the solutions to SFDES

  • Yeol JE Cho1,

    Affiliated with

    • Sever S Dragomir2 and

      Affiliated with

      • Young-Ho Kim3Email author

        Affiliated with

        Journal of Inequalities and Applications20122012:126

        DOI: 10.1186/1029-242X-2012-126

        Received: 26 March 2012

        Accepted: 7 June 2012

        Published: 7 June 2012

        Abstract

        In this paper, we establish a new proof of the existence and uniqueness of solution to stochastic functional differential equations with infinite delay at phase space BC((-∞, 0]: R d ) which denotes the family of bounded continuous R d -valued functions φ defined on (-∞, 0] with norm ||φ|| = sup-∞ < θ≤0|φ(θ)|.

        Mathematics Subject Classification (2000): 60H05, 60H10.

        Keywords

        existence uniqueness stochastic functional differential equations infinite delay.

        1. Introduction

        Stochastic differential equations (SDEs) are well known to model problems from many areas of science and engineering. For instance, in 2006, Henderson and Plaschko [1] published the SDEs in Science and Engineering, in 2007, Mao [2] published the SDEs, in 2010, Li and Fu [3] considered the stability analysis of stochastic functional differential equations with infinite delay and its application to recurrent neural networks.

        In recent years, there is an increasing interest in stochastic functional differential equations(SFDEs) with finite and infinite delay under less restrictive conditions than Lipschitz condition. For instance, in 2007, Wei and Wang [4] discussed the existence and uniqueness of the solution for stochastic functional differential equations with infinite delay, in 2008, Mao et al. [5] discussed almost surely asymptotic stability of neutral stochastic differential delay equations with Markovian switching, in 2008, Ren et al. [6] considered the existence and uniqueness of the solutions to SFDEs with infinite delay, and in 2009, Ren and Xia [7], discussed the existence and uniqueness of the solution to neutral SFDEs with infinite delay. Furthermore, on this topic, one can see Halidias [8], Henderson and Plaschko [1], Kim [9, 10], Ren [11], Ren and Xia [7], Taniguchi [12] and references therein for details.

        On the other hand, Mao [2] discussed d-dimensional stochastic functional differential equations with finite delay
        d x ( t ) = f ( x t , t ) d t + g ( x t , t ) d B ( t ) , t 0 t T , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equ1_HTML.gif
        (1.1)
        where x t = {x(t + θ): -τθ ≤ 0} could be considered as a C([-τ, 0]; R d )-value stochastic process. The initial value of (1.1) was proposed as follows:
        x t 0 = ξ = { ξ ( θ ) : - τ θ 0 } is an F t 0   - measurable C ( [ - τ , 0 ] ; R d ) - value random variable such that  E ξ 2 < . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equa_HTML.gif
        Furthermore, Ren et al. [6] also considered the stochastic functional differential equations with infinite delay at phase space BC((-∞, 0]; R d ) to be described below.
        d x ( t ) = f ( x t , t ) d t + g ( x t , t ) d B ( t ) , t 0 t T , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equ2_HTML.gif
        (1.2)
        where x t = {X(t + θ): -∞ ≤ θ ≤ 0} could be considered as a BC((-∞, 0]; R d )-value stochastic process. The initial value of (1.2) was proposed as follows:
        X t 0 = ξ = { ξ ( θ ) : - θ 0 } is an F t 0   - measurable BC ( ( - , 0 ] ; R d ) - value random variable such that ξ M 2 ( ( - , 0 ] ; R d ) . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equ3_HTML.gif
        (1.3)

        Following this way, now we recall the existence and uniqueness of the solutions to the Equation 1.2 with initial data (1.3) under the non-Lipschitz condition and the weakened linear growth condition. In this paper, we will give some new proof of the existence and uniqueness of the solutions to ISDEs under an alternative way.

        2. Preliminary

        Let | · | denote Euclidean norm in R n . If A is a vector or a matrix, its transpose is denoted by A T ; if A is a matrix, its trace norm is represented by A = trace ( A T A ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq1_HTML.gif. Let t0 be a positive constant and ( Ω , F , P ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq2_HTML.gif, throughout this paper unless otherwise specified, be a complete probability space with a filtration { F t } t t 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq3_HTML.gif satisfying the usual conditions (i.e. it is right continuous and F t 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq4_HTML.gif contains all P-null sets). Assume that B(t) is a m-dimensional Brownian motion defined on complete probability space, that is B(t) = (B1(t), B2(t), ..., B m (t)) T . Let BC((-∞, 0]; R d ) denote the family of bounded continuous R d -value functions φ defined on (-∞, 0] with norm ||φ|| = sup-∞ < θ≤0(θ)|. We denote by M 2 ( ( - , 0 ] ; R d ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq5_HTML.gif the family of all F t 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq4_HTML.gif-measurable, ℝ d -valued process ψ(t) = ψ(t, w), t ∈ (-∞, 0], such that E - 0 ψ ( t ) 2 d t < http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq6_HTML.gif. And let L p ( [ a , b ] ; R d ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq7_HTML.gif is the family of R d -valued F t http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq8_HTML.gif-adapted processes {f(t)}atbsuch that a b f ( t ) p d t < http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq9_HTML.gif.

        With all the above preparation, consider a d-dimensional stochastic functional differential equations:
        d x ( t ) = f ( x t , t ) d t + g ( x t , t ) d B ( t ) , t 0 t T , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equ4_HTML.gif
        (2.1)
        where x t = {x(t + θ): -∞ < θ ≤ 0} can be considered as a BC((-∞, 0]; R d )-value stochastic process, where
        f : BC ( ( - , 0 ] ; R d ) × [ t 0 , T ] R d , g : BC ( ( - , 0 ] ; R d ) × [ t 0 , T ] R d × m http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equb_HTML.gif
        be Borel measurable. Next, we give the initial value of (2.1) as follows:
        x t 0 = ξ = { ξ ( θ ) : - < θ 0 } is an F t 0   - measurable BC ( ( - , 0 ] ; R d )  - value random variable such that  ξ M 2 ( ( - , 0 ] ; R d ) . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equ5_HTML.gif
        (2.2)

        To find out the solution, we give the definition of the solution of the Equation 2.1 with initial data (2.2).

        Definition 2.1. [6] R d -value stochastic process x(t) defined on -∞ < tT is called the solution of (2.1) with initial data (2.2), if x(t) has the following properties:
        1. (i)

          x(t) is continuous and { x ( t ) } t 0 t T http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq10_HTML.gif is F t http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq8_HTML.gif-adapted;

           
        2. (ii)

          { f ( x t , t ) } L 1 ( [ t 0 , T ] ; R d ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq11_HTML.gif and { g ( x t , t ) } L 2 ( [ t 0 , T ] ; R d × m ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq12_HTML.gif;

           
        3. (iii)
          x t 0 = ξ http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq13_HTML.gif, for each t0tT,
          x ( t ) = ξ ( 0 ) + t 0 t f ( x s , s ) d s + t 0 t g ( x s , s ) d B ( s ) a . s . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equ6_HTML.gif
          (2.3)
           
        A solution x(t) is called as a unique if any other solution x ̄ ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq14_HTML.gif is indistinguishable with x(t), that is
        P { x ( t ) = x ̄ ( t ) , for any - < t T } = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equc_HTML.gif

        The integral inequalities of Gronwall type have been applied in the theory of SDEs to prove the results on existence, uniqueness, and stability etc. [10, 1316]. Naturally, Gronwall's inequality will play an important role in next section.

        Lemma 2.1. (Gronwall's inequality). Let u(t) and b(t) be non-negative continuous functions for tα, and let
        u ( t ) a + α t b ( s ) u ( s ) d s , t α , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equd_HTML.gif
        where a ≥ 0 is a constant. Then
        u ( t ) a exp α t b ( s ) d s , t α . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Eque_HTML.gif
        Lemma 2.2. (Bihari's inequality). Let u and b be non-negative continuous functions defined on R+. Let g(u) be a non-decreasing continuous function on R+ and g(u) > 0 on (0, ∞). If
        u ( t ) k + 0 t b ( s ) g ( u ( s ) ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equf_HTML.gif
        for tR+, where k ≥ 0 is a constant. Then for 0 ≤ tt1,
        u ( t ) G - 1 G ( k ) + 0 t b ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equg_HTML.gif
        where
        G ( r ) = r 0 r d s g ( s ) , r > 0 , r 0 > 0 , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equh_HTML.gif
        and G-1 is the inverse function of G and t1R+ is chosen so that
        G ( k ) + 0 t b ( s ) d s Dom ( G - 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equi_HTML.gif

        for all tR+ lying in the interval 0 ≤ tt1.

        The following two lemmas are known as the moment inequality for stochastic integrals which will play an important role in next section.

        Lemma 2.3. [2]. If p ≥ 2, g M 2 ( [ 0 , T ] ; R d × m ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq15_HTML.gif such that
        E 0 T g ( s ) p d s < , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equj_HTML.gif
        then
        E 0 T g ( s ) d B ( s ) p p ( p - 1 ) 2 p 2 T p - 2 2 E 0 T g ( s ) p d s . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equk_HTML.gif

        In particular, for p = 2, there is equality.

        Lemma 2.4. [2]. If p ≥ 2, g M 2 ( [ 0 , T ] ; R d × m ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq15_HTML.gif such that
        E 0 T g ( s ) p d s < , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equl_HTML.gif
        then
        E sup 0 t T 0 t g ( s ) d B ( s ) p p 3 2 ( p - 1 ) p 2 T p - 2 2 E 0 T g ( s ) p d s . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equm_HTML.gif

        3. Existence and Uniqueness of the Solutions

        In order to obtain the existence and uniqueness of the solutions to (2.1) with initial data (2.2), we define x t 0 0 = ξ http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq16_HTML.gif and x0(t) = ξ(0), for t0tT. Let x t 0 n = ξ http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq17_HTML.gif, n = 1, 2, ... and define the Picard sequence
        x n ( t ) = ξ ( 0 ) + t 0 t f ( x s n - 1 , s ) d s + t 0 t g ( x s n - 1 , s ) d B ( s ) , t 0 t T . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equn_HTML.gif

        Now we begin to establish the theory of the existence and uniqueness of the solution. We first show that the non-Lipschitz condition and the weakened linea growth condition guarantee the existence and uniqueness.

        Theorem 3.1. Assume that there exist a positive number K such that
        1. (i)
          For any φ, ψ ∈ BC((-∞, 0]; R d ) and t ∈ [t0, T], it follows that
          f ( φ , t ) - f ( ψ , t ) 2 g ( φ , t ) - g ( ψ , t ) 2 κ ( φ - ψ 2 ) , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equ7_HTML.gif
          (3.1)
           
        where κ(·) is a concave non-decreasing function from ℝ+ to ℝ+ such that κ(0) = 0, κ(u) > 0 for u > 0 and 0 + d u / κ ( u ) = http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq18_HTML.gif.
        1. (ii)
          For any t ∈ [t0, T], it follows that f(0, t), g(0, t) ∈ L2 such that
          f ( 0 , t ) 2 g ( 0 , t ) 2 K . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equ8_HTML.gif
          (3.2)
           

        Then the initial value problem (2.1) has a solution x(t). Moreover, x t M 2 ( ( - , T ] ; R d ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq19_HTML.gif. We prepare a lemma to prove this theorem.

        Lemma 3.2. Let the assumption (3.1) and (3.2) of Theorem 3.1 hold. If x(t) is a solution of equation (2.1) with initial data (2.2), then
        E sup - < t T x ( t ) 2 E ξ 2 + c 2 e 6 b ( T - t 0 + 1 ) ( T - t 0 ) , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equo_HTML.gif

        where c2 = c1 + E ǁ ξ ǁ2, c1 = 3E ǁξ ǁ2 + 6(T - t0 + 1)(T - t0)[K + a]. In particular, x(t) belong to M 2 ( ( - , T ] ; R d ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq20_HTML.gif.

        Proof. For each number n ≥ 1, define the stopping time
        τ n = T inf { t [ t 0 , T ] : | | x ( t ) | | n } . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equp_HTML.gif
        Obviously, as n → ∞, τ n T a.s. Let x n (t) = x(tτ n ), t ∈ (-∞, T]. Then, for t0tT, x n (t) satisfy the following equation
        x n ( t ) = ξ ( 0 ) + t 0 t f x s n , s I [ t 0 , τ n ] ( s ) d s + t 0 t g ( x s n , s ) I [ t 0 , τ n ] ( s ) d B ( s ) . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equq_HTML.gif
        Using the elementary inequality x i p n p - 1 x i p http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq21_HTML.gif when p ≥ 1, we have
        x n ( t ) 2 3 ξ ( 0 ) 2 + 3 t 0 t f ( x s n , s ) I [ t 0 , τ n ] ( s ) d s 2 + 3 t 0 t g ( x s n , s ) I [ t 0 , τ n ] ( s ) d B ( s ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equr_HTML.gif
        By the Hölder's inequality and the moment inequality, we have
        E x n ( t ) 2 3 E ξ 2 + ( t - t 0 ) E t 0 t f ( x s n , s ) 2 d s + E t 0 t g ( x s n , s ) 2 d s . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equs_HTML.gif
        Hence, by the condition (3.1) and (3.2) one can further show that
        E x n ( t ) 2 3 E ξ 2 + 6 ( t - t 0 + 1 ) E t 0 t κ ( x s n 2 ) d s + E t 0 t K d s . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equt_HTML.gif
        Given that κ(·) is concave and κ(0) = 0, we can find a positive constants a and b such that κ(u) ≤ a + bu for u ≥ 0. So, we obtains that
        E sup t 0 s t x n ( s ) 2 c 1 + 6 b ( t - t 0 + 1 ) t 0 t E x s n 2 d s , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equu_HTML.gif
        where c1 = 3E||ξ||2 + 6(T - t0 + 1)(T - t0)[K + a]. Noting the fact that sup - < s t x n ( s ) 2 ξ 2 + sup t 0 s t x n ( s ) 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq22_HTML.gif, we obtain
        E sup - < s t x n ( s ) 2 c 2 + 6 b ( t - t 0 + 1 ) t 0 t E sup - < r s x n ( r ) 2 d s , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equv_HTML.gif
        where c2 = c1 + E||ξ||2. So, by the Gronwall inequality yields that
        E sup - < s t x n ( s ) 2 c 2 exp ( 6 b ( T - t 0 + 1 ) ( T - t 0 ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equw_HTML.gif
        Letting t = T, it then follows that
        E sup - < s T x ( s τ n ) 2 E ξ 2 + c 2 e 6 b ( T - t 0 + 1 ) ( T - t 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equx_HTML.gif
        Thus
        E sup - < s τ n x ( s ) 2 E ξ 2 + c 2 e 6 b ( T - t 0 + 1 ) ( T - t 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equy_HTML.gif

        Consequently the required result follows by letting n → ∞. □

        Proof of Theorem 3.1. Let x(t) and x ̄ ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq14_HTML.gif be any tow solutions of (2.1). By Lemma 3.2, x(t), x ̄ ( t ) M 2 ( ( - , T ] ; R d ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq23_HTML.gif. Note that
        x ( t ) - x ̄ ( t ) = t 0 t [ f ( x s , s ) - f ( x ̄ s , s ) ] d s + t 0 t [ g ( x s , s ) - g ( x ̄ s , s ) ] d B ( s ) . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equz_HTML.gif
        By the elementary inequality (u + v)2 ≤ 2(u2 + v2), one then gets
        x ( t ) - x ̄ ( t ) 2 = 2 t 0 t [ f ( x s , s ) - f ( x ̄ s , s ) ] d s 2 + 2 t 0 t [ g ( x s , s ) - g ( x ̄ s , s ) ] d B ( s ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equaa_HTML.gif
        By the Hölder's inequality, the moment inequality, and (3.1) we have
        E sup t 0 s t x ( s ) - x ̄ ( s ) 2 2 ( T - t 0 + 1 ) E t 0 t κ ( x s - x ̄ s 2 ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equab_HTML.gif
        Since κ(·) is concave, by the Jensen inequality, we have
        E κ ( x s - x ̄ s 2 ) κ ( E x s - x ̄ s 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equac_HTML.gif
        Consequently, for any ϵ > 0,
        E sup t 0 s t x ( s ) - x ̄ ( s ) 2 ϵ + 2 ( T - t 0 + 1 ) t 0 t κ E sup t 0 r s x ( r ) - x ̄ ( r ) 2 d s . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equad_HTML.gif
        By the Bihari inequality, one deduces that, for all sufficiently small ϵ > 0,
        E sup t 0 s t x ( s ) - x ̄ ( s ) 2 G - 1 [ G ( ϵ ) + 2 ( T - t 0 + 1 ) ( T - t 0 ) ] , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equ9_HTML.gif
        (3.3)
        where
        G ( r ) = 1 r 1 κ ( u ) d u http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equae_HTML.gif
        on r > 0, and G-1(·) be the inverse function of G(·). By assumption 0 + d u κ ( u ) = http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq24_HTML.gif and the definition of κ(·), one sees that limϵ↓0G(ϵ) = -∞ and then
        lim ϵ 0 G - 1 [ G ( ϵ ) + 2 ( T - t 0 + 1 ) ( T - t 0 ) ] = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equaf_HTML.gif
        Therefore, by letting ϵ → 0 in (3.3), gives
        E sup t 0 t T x ( t ) - x ̄ ( t ) 2 = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equag_HTML.gif

        This implies that x ( t ) = x ̄ ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq25_HTML.gif for t0tT, Therefore, for all -∞ < tT, x ( t ) = x ̄ ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq25_HTML.gif a.s. The uniqueness has been proved.

        Next to check the existence. Define x t 0 0 = ξ http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq16_HTML.gif and x0(t) = ξ(0) for t0tT. For each n = 1, 2, ..., set x t 0 n = ξ http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq17_HTML.gif and define, by the Picard iterations,
        x n ( t ) = ξ ( 0 ) + t 0 t f ( x s n - 1 , s ) d s + t 0 t g ( x s n - 1 , s ) d B ( s ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equ10_HTML.gif
        (3.4)
        for t0tT. Obviously, x 0 ( t ) M 2 ( [ t 0 , T ] : R d ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq26_HTML.gif. Moreover, it is easy to see that x n ( t ) M 2 ( ( - , T ] : R d ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq27_HTML.gif, in fact
        x n ( t ) 2 3 ξ ( 0 ) 2 + 3 t 0 t f ( x s n - 1 , s ) d s 2 + 3 t 0 t g ( x s n - 1 , s ) d B ( s ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equah_HTML.gif
        Taking the expectation on both sides and using the Hölder inequality and moment inequality, we have
        E x n ( t ) 2 3 E ξ 2 + 3 ( t - t 0 ) E t 0 t f ( x s n - 1 , s ) 2 d s + 3 E t 0 t g ( x s n - 1 , s ) d B ( s ) 2 3 E ξ 2 + 3 ( t - t 0 ) E t 0 t f ( x s n - 1 , s ) - f ( 0 , s ) + f ( 0 , s ) 2 d s + 3 E t 0 t g ( x s n - 1 , s ) - g ( 0 , s ) + g ( 0 , s ) 2 d s . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equai_HTML.gif
        Using the elementary inequality (u + v)2 ≤ 2u2 + 2v2, (3.1), and (3.2), we have
        E x n ( t ) 2 3 E ξ 2 + 3 ( t - t 0 + 1 ) E t 0 t 2 κ ( x s n - 1 2 )  + 2 K d s 3 E ξ 2 + 6 K ( T - t 0 ) ( T - t 0 + 1 ) + 6 ( T - t 0 + 1 ) E t 0 t κ ( x s n - 1 2 ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equaj_HTML.gif
        Given that κ(·) is concave and κ(0) = 0, we can find a positive constants a and b such that κ(u) ≤ a + bu for u ≥ 0. So, we have
        E x n ( t ) 2 c 1 + 6 b ( T - t 0 + 1 ) t 0 t E x s n - 1 2 d s , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equak_HTML.gif
        where c1 = 3E||ξ||2 + 6(T - t0)(T - t0+ 1)[K + a]. It also follows from the inequality that for any k ≥ 1,
        max 1 n k E x n ( t ) 2 c 1 + 6 b ( T - t 0 + 1 ) t 0 t max 1 n k E x n - 1 ( s ) 2 d s c 1 + 6 b ( T - t 0 + 1 ) t 0 t ( E ξ 2 + max 1 n k E x n ( s ) 2 ) d s c 2 + 6 b ( T - t 0 + 1 ) t 0 t max 1 n k E x n ( s ) 2 d s , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equal_HTML.gif
        where c2 = c1 = 6b(T - t0)(T - t0+ 1) E||ξ||2. The Gronwall inequality implies
        max 1 n k E x n ( t ) 2 c 2 exp ( 6 b ( T - t 0 ) ( T - t 0 + 1 ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equam_HTML.gif
        Since k is arbitrary, we must have
        E x n ( t ) 2 c 2 exp ( 6 b ( T - t 0 + 1 ) ( T - t 0 ) ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equ11_HTML.gif
        (3.5)

        for all t0tT, n ≥ 1.

        Next, we that the sequence {x n (t)} is Cauchy sequence. For all n ≥ 0 and t0tT, we have
        x n ( t ) - x m ( t ) = t 0 t [ f ( x s n - 1 , s ) - f ( x s m - 1 , s ) ] d s + t 0 t [ g ( x s n - 1 , s ) - g ( x s m - 1 , s ) ] d B ( s ) . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equan_HTML.gif
        Next, using an elementary inequality ( u + v ) 2 1 α u 2 + 1 1 - α v 2 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq28_HTML.gif and the condition (H3), we derive that
        x n ( t ) - x m ( t ) 2 1 α t 0 t [ f ( x s n - 1 , s ) - f ( x s m - 1 , s ) ] d s 2 + 1 1 - α t 0 t [ g ( x s n - 1 , s ) - g ( x s m - 1 , s ) ] d B ( s ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equao_HTML.gif
        On the other hand, by Hölder's inequality, Lemma (2.4), and the condition, one can show that
        E sup t 0 < s t x n ( s ) - x m ( s ) 2 β t 0 t κ E sup t 0 u s x n - 1 ( u ) - x m - 1 ( u ) 2 d s , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equ12_HTML.gif
        (3.6)
        where β = (T - t0)/α + 4/(1 - α). Let
        Z ( t ) = lim n , m sup  E sup t 0 s t x n ( s ) - x m ( s ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equap_HTML.gif
        From (3.6), for any ϵ > 0, we get
        Z ( t ) ϵ + β t 0 t κ ( Z ( s ) ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equaq_HTML.gif
        By the Bihari inequality, one deduces that, for all sufficiently small ϵ > 0,
        Z ( t ) G - 1 [ G ( ϵ ) + β ( T - t 0 ) ] , http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equar_HTML.gif
        where
        G ( r ) = 1 r 1 κ ( u ) d u http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equas_HTML.gif
        on r > 0, and G-1(·) be the inverse function of G(·). By assumption, we get Z(t) = 0. This shows the sequence {x n (t), n ≥ 0} is a Cauchy sequence in L2. Hence, as n → ∞, x n (t) - x(t), that is E |x n (t) - x(t)|2 → 0. Letting n → ∞ in (3.5) then yields that
        E sup t 0 s t x ( s ) 2 c 2 exp ( 6 b ( T - t 0 + 1 ) ( T - t 0 ) ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equat_HTML.gif
        for all t0tT. Therefore, x t M 2 ( ( - , T ] ; R d ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_IEq19_HTML.gif. It remains to show that x(t) satisfies Equation 2.3. Note that
        E t 0 t ( f ( x s n , s ) - f ( x s , s ) ) d s 2 + E t 0 t ( g ( x s n , s ) - g ( x s , s ) ) d B ( s ) 2 ( t - t 0 ) E t 0 t ( f ( x s n , s ) - f ( x s , s ) ) 2 d s + E t 0 t ( g ( x s n , s ) - g ( x s , s ) ) 2 d s ( t - t 0 + 1 ) t 0 T κ ( E ( sup t 0 u s x n ( u ) - x ( u ) 2 ) ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equau_HTML.gif
        Noting that sequence x n (t) is uniformly converge on (-∞, T ], it means that
        E sup t 0 u s x n ( u ) - x ( u ) 2 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equav_HTML.gif
        as n → ∞, further
        κ E sup t 0 u s x n ( u ) - x ( u ) 2 0 http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equaw_HTML.gif
        as n → ∞. Hence, taking limits on both sides in the Picard sequence, we obtain that
        x ( t ) = x 0 + t 0 t f ( x s , s ) d s + t 0 t g ( x s , s ) d B ( s ) http://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-126/MediaObjects/13660_2012_Article_266_Equax_HTML.gif

        on t0tT. The above expression demonstrates that x(t) is the solution of (2.3). So, the existence of theorem is complete.

        Remark 3.1. In the proof of Theorem 3.1, the solution is constructed by the successive approximation. It shows that how to get the approximate solution of (2.1) and how to construct Picard sequence x n (t). For SFDEs, we know that a weakened linear growth condition imposed on Theorem 3.1 is rigorous for our discussion. In papers [6, 7], the proofs of the assertions are based on some function inequalities. Recall that the procedures has become more and more complicated in those proofs. For this reason, although analogous problem is studied here, the proof of the assertion in the Theorem is completely different with respect to the ones from [7]. Moreover, our new proof in this paper is completed by Bihari's inequality and more simple than that reported in [7].

        Declarations

        Acknowledgements

        The authors wish to thank the anonymous referees for their endeavors and valuable comments. Also, this research was supported by Basic Science Research Program through the National Research Foundation of Korea(NRF) funded by the Ministry of Education, Science and Technology(2011-0023547)

        Authors’ Affiliations

        (1)
        Department of Mathematics Education, Gyeongsang National University
        (2)
        School of Computer Science and Mathematics, Victoria University of Technology
        (3)
        Department of Mathematics, Changwon National University

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        © Cho et al; licensee Springer. 2012

        This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://​creativecommons.​org/​licenses/​by/​2.​0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.