Anticipated Backward Doubly Stochastic Differential Equations with Non-Lipschitz Coefficients

Abstract

The work presented in this paper focuses on a type of differential equations called anticipated backward doubly stochastic differential equations (ABDSDEs) whose generators not only depend on the anticipated terms of the solution ($Y_{·},Z_{·}$) but also satisfy one kind of non-Lipschitz assumption. Firstly, we give the existence and uniqueness theorem. Further, two comparison theorems for the solutions of these equations are obtained after finding a new comparison theorem for backward doubly stochastic differential equations (BDSDEs) with non-Lipschitz coefficients.

1. Introduction

In 1990, the pioneer research of Pardoux and Peng [1] proposed the theory of nonlinear backward stochastic differential Equations (BSDEs). Over the past 3 decades, BSDEs have attracted much attention from academia due to its wide application in lots of different fields of research, for example, financial mathematics (see El Karoui et al. [2]), stochastic optimal control, differential games and the theory of partial differential equations. Among others, a lot of effort has been made to relax the Lipschitz assumptions (see, e.g., [3,4,5,6]).

In order to obtain a probabilistic representation for a class of quasilinear stochastic partial differential equations, Pardoux and Peng [7] first presented a class of backward doubly stochastic differential Equations (BDSDEs in short) in the following

$$Y_t=\xi +{\int }_t^{T}f(s,Y_s,Z_s)ds+{\int }_t^{T}g(s,Y_s,Z_s)d{\overleftarrow{B}}_s-{\int }_t^T{Z}_{s}d{W}_s,t\in [0,T],$$

(1)

where the equations include a standard (forward) It$\widehat{\mathrm{o}}$ integral $d{W}_t$, and a backward It$\widehat{\mathrm{o}}$ integral $d{\overleftarrow{B}}_t$. They investigated the existence and uniqueness of solutions for BDSDE (1) under uniform Lipschitz generators. Then Shi et al. [8] gived a comparison theorem for BDSDEs with uniform Lipschitz condition on the generators. Refs. [9,10,11] have attempted to weaken the uniform Lipschitz assumption on the coefficients.

In 2009, Peng and Yang [12] introduced a new class of BSDEs called anticipated BSDEs (ABSDEs), whose generator involves not only the present values of the solutions but also the future situation. The authors proved ABSDEs have a unique solutions under uniform Lipschitz assumptions, obtained a comparison theorem for their solutions under some specific condition, and investigated the duality between anticipated BSDEs and delayed stochastic differential equations. Following the research of Peng and Yang [12], Zhang [13] studied the comparison theorems for one dimensional anticipated BSDEs under one kind of non-Lipschitz assumption. Zhou et al. [14] investigated the existence and uniqueness of this type equations under another non-Lipschitz conditions. Recently, Xu [15] and Zhang [16] introduced the following type of so-called anticipated BDSDES (ABDSDEs):

$$\left\{\begin{array}{cc}\hfill Y_t=& {\xi }_T+{\int }_t^{T}f(s,Y_s,Z_s,Y_{s+\delta (s)},Z_{s+\zeta (s)})\mathrm{d}s\hfill \\ \hfill & +{\int }_t^{T}g(s,Y_s,Z_s,Y_{s+\overline{\delta }(s)},Z_{s+\overline{\zeta }(s)})\mathrm{d}{\overleftarrow{B}}_s-{\int }_t^T{Z}_s\mathrm{d}{W}_s,t\in [0,T],\hfill \\ \hfill Y_t=& {\xi }_t,Z_t={\eta }_t,t\in [T,T+K],\hfill \end{array}\right.$$

(2)

where ${\xi }_{·},{\eta }_{·}$ are given stochastic processes, and $\delta (·),\overline{\delta }(·),\zeta (·)$ and $\overline{\zeta }(·)$ are four given nonnegative deterministic continuous functions and for $\gamma (·)=\delta (·),\overline{\delta }(·),\zeta (·),\overline{\zeta }(·)$ satisfying that:

(A1) there has a constant $K\ge 0$ such that, for each $t\in [0,T]$, $t+\gamma (t)⩽T+K$;

(A2) there has a constant $M⩾0$ such that, for each $t\in [0,T]$ and for any nonnegative integrable $h(·)$,

$${\int }_t^{T}h(s+\gamma (s))\mathrm{d}s⩽M{\int }_t^{T+K}h(s)\mathrm{d}s.$$

Xu [15] and Zhang [16] explored the existence and uniqueness of the solution for above equation, gave some comparison theorems, and investigated the duality between them and stochastic doubly differential equations with delay. Aidara [17,18] studied anticipated BDSDEs with one kind of non-Lipschitz coefficients, in which generator g does not depend on the anticipated term of $y,z$. They obtained the existence and uniqueness result and a comparison theorem in the one dimensional case. Recently, Wang and Yu [19] dealt with anticipated generalized backward doubly stochastic differential Equations (AGBDSDEs). Based on [15,16], we are concerned with anticipated BDSDEs under non-Lipschitz assumption. We will prove that under proper assumptions, the solution of the above ABDSDE with non-Lipschitz coefficients exists uniquely, and two comparison theorems are given for the one dimensional ABDSDEs with non-Lipschitz coefficients. These results are the cornerstones of ABDSDEs witn non-Lipschitz coefficients applied to some stochastic optimal control problems with delay effect.

This paper is divided into five sections as follows. In Section 2, some notations, assumptions and definition are given. We focus on the existence and uniqueness of the solutions of anticipated BDSDEs with non-Lipschitz coefficients in Section 3. In Section 4, we give two comparison theorems. Finally in Section 5, the conclusion and future work are presented.

2. Preliminaries

Let $(\Omega ,\mathcal{F},P)$ be a complete probability space. $T>0,K\ge 0$ are two fixed constants. Let $\{W_t;0\le t\le T\}$ and $\{B_t;0\le t\le T\}$ be two mutually independent standard Brownian motions with values, respectively, in ${\mathbb{R}}^d$ and ${\mathbb{R}}^l$. For any $x,y\in R^n$ and $z\in R^{n\times m}$, we use $\langle x,y\rangle$ to denote the inner product of x and y, and $\mid y\mid$ to represent for the vector norm of y and $\mid z\mid :=\sqrt{Tr(z{z}^{*})}$ means the matrix norm of z, where $z^{*}$ is the transpose of z. Set $\mathcal{N}$ to denote the class of P-null sets of $\mathcal{F}$. For all $t\in [0,T+K]$, we define

$${\mathcal{F}}_t\triangleq {\mathcal{F}}_{0,t}^W\vee {\mathcal{F}}_{t,T+K}^B,$$

where for each process $\left\{{\varphi }_t\right\},{\mathcal{F}}_{s,t}^{\eta }=\sigma \{{\varphi }_r-{\varphi }_s;s\le r\le t\}\vee \mathcal{N}$. Notice that $\{{\mathcal{F}}_{0,t}^W,t\in [0,T+K]\}$ is increasing and $\{{\mathcal{F}}_{t,T}^B,t\in [0,T+K]\}$ is decreasing, therefor $\{{\mathcal{F}}_t,t\in [0,T+K]\}$ do not constitute a filtration. The following notations will be used throughout the paper: for each $t\ge 0$, $n\in N$ and $b>a\ge 0$,

(i)

$L^2({\mathcal{F}}_t;{\mathbb{R}}^n)\triangleq \{\eta :\eta \in {\mathbb{R}}^n\mid \eta$ is a ${\mathcal{F}}_t$-measurable random variable with ${\mathbb{E}|\eta |}^2<\infty \}$;

(ii)

${\mathcal{M}}^2(a,b;{\mathbb{R}}^n)\triangleq \{\varphi :\Omega \times [a,b]\to {\mathbb{R}}^n\mid \varphi$ is a ${\mathcal{F}}_t$-progressively measurable processes such that ${\Vert \varphi \Vert }_{{\mathcal{M}}^2}^2=\mathbb{E}({\int }_a^b|{\varphi }_t{|}^{2}dt)<\infty \}$;

(iii)

${\mathcal{S}}^2([a,b];{\mathbb{R}}^n)\triangleq \{\varphi :\Omega \times [a,b]\to {\mathbb{R}}^n\mid \varphi$ is a continuous and ${\mathcal{F}}_t$- progressively measurable processes such that ${\Vert \varphi \Vert }_{{\mathcal{S}}^2}^2=\mathbb{E}\left(\underset{a\le t\le b}{sup}{|{\varphi }_t|}^2\right)<\infty$.

Let $f(t,·,·,·,·,·):\Omega \times {\mathbb{R}}^k\times {\mathbb{R}}^{k\times d}\times {\mathcal{M}}^2(t,T+K;{\mathbb{R}}^k)\times {\mathcal{M}}^2(t,T+K;{\mathbb{R}}^{k\times d})\to L^2({\mathcal{F}}_t;$${\mathbb{R}}^k)$, $g(t,·,·,·,·,·):\Omega \times {\mathbb{R}}^k\times {\mathbb{R}}^{k\times d}\times {\mathcal{M}}^2(t,T+K;{\mathbb{R}}^k)\times {\mathcal{M}}^2(t,T+K;{\mathbb{R}}^{k\times d})\to L^2({\mathcal{F}}_t;{\mathbb{R}}^{k\times l})$. We make the following assumptions about $(\xi ,f,g)$:

(H1)

$f(.,0,0,0,0)\in {\mathcal{M}}^2(0,T;{\mathbb{R}}^k),g(.,0,0,0,0)\in {\mathcal{M}}^2(0,T;{\mathbb{R}}^{k\times l})$.

(H2)

For each $t\in [0,T]$, $y,\overline{y}\in {\mathbb{R}}^k$, $z,\overline{z}\in {\mathbb{R}}^{k\times d}$, $\theta (·),\overline{\theta }(·)\in {\mathcal{M}}^2(t,T+K;{\mathbb{R}}^k)$, $\vartheta (·),\overline{\vartheta }(·)\in {\mathcal{M}}^2(t,T+K;{\mathbb{R}}^{k\times d})$, $r,r^{\prime }\in [t,T+K]$, we let

$$\left\{\begin{array}{cc}\hfill & |f(t,y,z,\theta (r),\vartheta (r^{\prime }))-f(t,\overline{y},\overline{z},\overline{\theta }(r),\overline{\vartheta }(r^{\prime })){|}^2\hfill \\ \hfill & ⩽C(|y-\overline{y}{|}^2+\parallel z-\overline{z}{\parallel }^2+{\mathbb{E}}^{{\mathcal{F}}_t}[|\theta (r)-\overline{\theta }{(r)|}^2+\parallel \vartheta (r^{\prime })-\overline{\vartheta }(r^{\prime }){\parallel }^2]),\hfill \\ \hfill & \parallel g(t,y,z,\theta (r),\vartheta (r^{\prime }))-g(t,\overline{y},\overline{z},\overline{\theta }(r),\overline{\vartheta }(r^{\prime })){\parallel }^2\hfill \\ \hfill & ⩽C(|y-\overline{y}{|}^2+{\mathbb{E}}^{{\mathcal{F}}_t}[|\theta (r)-\overline{\theta }{(r)|}^2])+{\alpha }_1\parallel z-\overline{z}{\parallel }^2+{\alpha }_2{\mathbb{E}}^{{\mathcal{F}}_t}[\parallel \vartheta (r^{\prime })-\overline{\vartheta }(r^{\prime }){\parallel }^2],\hfill \end{array}\right.$$

where $C>0,0<{\alpha }_1<1,0<{\alpha }_1+{\alpha }_{2}M<1$ are three given constants.

(H3)

For each $t\in [0,T]$, $y,\overline{y}\in {\mathbb{R}}^k$, $z,\overline{z}\in {\mathbb{R}}^{k\times d}$, $\theta (·),\overline{\theta }(·)\in {\mathcal{M}}^2(t,T+K;{\mathbb{R}}^k)$, $\vartheta (·),\overline{\vartheta }(·)\in {\mathcal{M}}^2(t,T+K;{\mathbb{R}}^{k\times d})$, $r,r^{\prime }\in [t,T+K]$, we let

$$\left\{\begin{array}{cc}\hfill & |f(t,y,z,\theta (r),\vartheta (r^{\prime }))-f(t,\overline{y},\overline{z},\overline{\theta }(r),\overline{\vartheta }(r^{\prime })){|}^2\hfill \\ \hfill & ⩽{\rho }_1(t,|y-\overline{y}{|}^2)+{\rho }_2(r,{\mathbb{E}}^{{\mathcal{F}}_t}[|\theta (r)-\overline{\theta }{(r)|}^2])+C(\parallel z-\overline{z}{\parallel }^2+{\mathbb{E}}^{{\mathcal{F}}_t}[\parallel \vartheta (r^{\prime })-\overline{\vartheta }(r^{\prime }){\parallel }^2]),\hfill \\ \hfill & \parallel g(t,y,z,\theta (r),\vartheta (r^{\prime }))-g(t,\overline{y},\overline{z},\overline{\theta }(r),\overline{\vartheta }(r^{\prime })){\parallel }^2\hfill \\ \hfill & ⩽{\rho }_1(t,|y-\overline{y}{|}^2)+{\rho }_2(r,{\mathbb{E}}^{{\mathcal{F}}_t}[|\theta (r)-\overline{\theta }{(r)|}^2])+{\alpha }_1\parallel z-\overline{z}{\parallel }^2+{\alpha }_2{\mathbb{E}}^{{\mathcal{F}}_t}[\parallel \vartheta (r^{\prime })-\overline{\vartheta }(r^{\prime }){\parallel }^2],\hfill \end{array}\right.$$

where $C>0,0<{\alpha }_1<1,0<{\alpha }_1+{\alpha }_{2}M<1$ are three given constants and for $i=1,2$, ${\rho }_i:[0,T+K]\times {\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ satisfies:

• For fixed $t\in [0,T+K],{\rho }_i(t,.)$ is a concave and non-decreasing function such that ${\rho }_i(t,0)=0$.
• For fixed u, ${\int }_0^{T+K}{\rho }_i(t,u)dt<+\infty$.
• For any $L>0,S_2>S_1\ge 0$, the following ODE

  $$\left\{\begin{array}{cc}\hfill & u^{\prime }=-L({\rho }_1(s,u)+M{\rho }_2(s,u)),\hfill \\ \hfill & u(S_2)=0\hfill \end{array}\right.$$

  has a unique solution $u(s)\equiv 0,s\in [S_1,S_2]$.

(H4)

$({\xi }_{·},{\eta }_{·})\in {\mathcal{S}}^2([T,T+K];{\mathbb{R}}^k)\times {\mathcal{M}}^2(T,T+K;{\mathbb{R}}^{k\times d})$.

Remark 1.

1. 

It’s easy to check that ${\rho }_1(t,u)={\rho }_2(t,u)=Cu$ for $C>0$ is an example of the function ${\rho }_1$ and ${\rho }_2$, and in this case the assumption (H3) degenerates to the assumption (H2).

2. 

If for $i=1,2$, ${\rho }_i(t,u)$ has a linear growth that is ${\rho }_i(t,u)\le a_i(t)+b_i(t)u$ where $a_i(t)\ge 0,b_i(t)\ge 0$, with ${\int }_0^{T+K}{a}_i(t)dt<\infty and{\int }_0^{T+K}{b}_i(t)dt<\infty$, it’s easy to check that ${\rho }_i(t,u)$ satisfies assumption (H3). Similar assumptions were used in [3,4,5,6,9,11,14].

3. 

Similar non-Lipschitz assumption was also used in [14] when $g\not\equiv 0$ in the following form:

$$\begin{array}{cc}\hfill & |f(t,y,z,\theta (r),\vartheta (r^{\prime }))-f(t,\overline{y},\overline{z},\overline{\theta }(r),\overline{\vartheta }(r^{\prime })){|}^2\hfill \\ \hfill & ⩽\rho (t,|y-\overline{y}{|}^2)+\rho (r,{\mathbb{E}}^{{\mathcal{F}}_t}[|\theta (r)-\overline{\theta }{(r)|}^2])+C(\parallel z-\overline{z}{\parallel }^2+{\mathbb{E}}^{{\mathcal{F}}_t}[\parallel \vartheta (r^{\prime })-\overline{\vartheta }(r^{\prime }){\parallel }^2]).\hfill \end{array}$$

Definition 1.

A pair of processes $(Y_{·},Z_{·}):\Omega \times [0,T+K]\to {\mathbb{R}}^k\times {\mathbb{R}}^{k\times d}$ is a solution of ABDSDE (2) with non-Lipschitz coefficients, if $(Y_{·},Z_{·})\in {\mathcal{S}}^2([0,T+K],{\mathbb{R}}^k)\times {\mathcal{M}}^2(0,T+K,{\mathbb{R}}^{k\times d})$ and satisfies (2) and assumptions (H1), (H3) and (H4).

3. Existence and Uniqueness Theorem

We can obtain directly the following existence and uniqueness result for ABDSDEs with uniform Lipschitz condition through combing the results given by Xu [15] and Zhang [16].

Lemma 1.

Let (H1), (H2) and (H4) hold. Then there exists a unique solution $(Y_{·},Z_{·})\in {\mathcal{S}}^2([0,T+K],{\mathbb{R}}^k)\times {\mathcal{M}}^2(0,T+K,{\mathbb{R}}^{k\times d})$ of ADDSDE (2).

According to Lemma 1, we can construct the Picard-type iteration sequence of Equation (2) as follows:

$$\left\{\begin{array}{cc}\hfill Y_t^0=& 0,\hfill \\ \hfill Y_t^n=& {\xi }_T+{\int }_t^{T}f(s,Y_s^{n-1},Z_s^n,Y_{s+\delta (s)}^{n-1},Z_{s+\zeta (s)}^n)ds\hfill \\ \hfill & +{\int }_t^{T}g(s,Y_s^{n-1},Z_s^n,Y_{s+\overline{\delta }(s)}^{n-1},Z_{s+\overline{\zeta }(s)}^n)d{\overleftarrow{B}}_s-{\int }_t^T{Z}_s^{n}d{W}_s,t\in [0,T],\hfill \\ \hfill Y_t^n=& {\xi }_t,Z_t^n={\eta }_t,t\in [T,T+K].\hfill \end{array}\right.$$

(3)

In fact, for any $Y_{·}^{n-1}\in {\mathcal{S}}^2([0,T+K],{\mathbb{R}}^k)$, according to Lemma 1, the ABDSDE (3) admits a unique solution $(Y_{·}^n,Z_{·}^n)\in {\mathcal{S}}^2([0,T+K],{\mathbb{R}}^k)\times {\mathcal{M}}^2(0,T+K,{\mathbb{R}}^{k\times d})$. We want to find the unique solution of ABDSDEs (2) through proving that the sequence ${(Y_{·}^n,Z_{·}^n)}_{n\ge 0}$ converges in ${\mathcal{S}}^2([0,T+K],{\mathbb{R}}^k)\times {\mathcal{M}}^2(0,T+K,{\mathbb{R}}^{k\times d})$. In order to achieve this goal, we need the following two lemmas.

Lemma 2.

Assume (A1), (A2), (H1), (H3) and (H4). Then, for any $0\le t\le T,n,m\ge 1$, we get

$$\begin{array}{cc}\hfill \mathbb{E}|Y_t^{n+m}-Y_t^n{|}^2\le & e^{\frac{C(M+1)T}{1-{\alpha }_1-{\alpha }_{2}M}}\left(\frac{1-{\alpha }_1-{\alpha }_{2}M}{C(M+1)}+1\right)({\int }_t^T{\rho }_1(s,\mathbb{E}[|Y_s^{n+m-1}-Y_s^{n-1}{|}^2])ds\hfill \\ \hfill & +M{\int }_t^T{\rho }_2(s,\mathbb{E}[|Y_s^{n+m-1}-Y_s^{n-1}{|}^2])ds).\hfill \end{array}$$

Proof. 

In view of It$\widehat{\mathrm{o}}$’s formula, we have

$$\begin{array}{cc}\hfill & \mathbb{E}|Y_t^{n+m}-Y_t^n{|}^2+\mathbb{E}{\int }_t^T\parallel Z_s^{n+m}-Z_s^n{\parallel }^{2}ds=2\mathbb{E}{\int }_t^T\langle Y_s^{n+m}-Y_s^n,\hfill \\ \hfill & f(s,Y_s^{n+m-1},Z_s^{n+m},Y_{s+\delta (s)}^{n+m-1},Z_{s+\zeta (s)}^{n+m})-f(s,Y_s^{n-1},Z_s^n,Y_{s+\delta (s)}^{n-1},Z_{s+\zeta (s)}^n)\rangle ds\hfill \\ \hfill & +\mathbb{E}{\int }_t^T{\parallel g(s,Y_s^{n+m-1},Z_s^{n+m},Y_{s+\overline{\delta }(s)}^{n+m-1},Z_{s+\overline{\zeta }(s)}^{n+m})-g(s,Y_s^{n-1},Z_s^n,Y_{s+\overline{\delta }(s)}^{n-1},Z_{s+\overline{\zeta }(s)}^n)\parallel }^{2}ds.\hfill \end{array}$$

By the assumptions (H3), (A1), (A2), Young’s inequality $2ab\le \frac{1}{\theta }{a}^2+\theta b^2$, and Jensen’s inequality, for all $\theta >0$, we get

$$\begin{array}{cc}\hfill \mathbb{E}& |Y_t^{n+m}-Y_t^n{|}^2+\mathbb{E}{\int }_t^T{\parallel Z_t^{n+m}-Z_t^n\parallel }^{2}ds\hfill \\ \hfill \le & \frac{1}{\theta }\mathbb{E}{\int }_t^T|Y_t^{n+m}-Y_s^n{|}^{2}ds+(\theta +1){\int }_t^T{\rho }_1(s,\mathbb{E}[|Y_s^{n+m-1}-Y_s^{n-1}{|}^2])ds\hfill \\ \hfill & +(\theta +1)M{\int }_t^T{\rho }_2(s,\mathbb{E}[|Y_s^{n+m-1}-Y_s^{n-1}{|}^2])ds\hfill \\ \hfill & +(\theta C(M+1)+{\alpha }_1+{\alpha }_{2}M)\mathbb{E}{\int }_t^T{\parallel Z_s^{n+m}-Z_s^n\parallel }^{2}ds.\hfill \end{array}$$

Choosing $\theta =\frac{1-{\alpha }_1-{\alpha }_{2}M}{C(M+1)}>0$, it follows from Gronwall’s inequality that

$$\begin{array}{cc}\hfill \mathbb{E}|Y_t^{n+m}-Y_t^n{|}^2\le & e^{\frac{C(M+1)T}{1-{\alpha }_1-{\alpha }_{2}M}}\left(\frac{1-{\alpha }_1-{\alpha }_{2}M}{C(M+1)}+1\right)({\int }_t^T{\rho }_1(s,\mathbb{E}[|Y_s^{n+m-1}-Y_s^{n-1}{|}^2])ds\hfill \\ \hfill & +M{\int }_t^T{\rho }_2(s,\mathbb{E}[|Y_s^{n+m-1}-Y_s^{n-1}{|}^2])ds).\hfill \end{array}$$

□

Lemma 3.

Assume that (A1), (A2), (H1), (H3) and (H4) hold. Then, there exists $T_1\in [0,T]$ and a constant $L_1\ge 0$ such that for any $t\in [T_1,T],n\ge 1,E{|Y_t^n|}^2\le L_1$.

Proof. 

By applying It$\widehat{\mathrm{o}}$’s formula, we get

$$\begin{array}{cc}\hfill & \mathbb{E}|Y_t^n{|}^2+\mathbb{E}{\int }_t^T\parallel Z_s^n{\parallel }^{2}ds=\mathbb{E}{|\xi |}^2+2\mathbb{E}{\int }_t^T\langle Y_s^n,f(s,Y_s^{n-1},Z_s^n,Y_{s+\delta (s)}^{n-1},Z_{s+\zeta (s)}^n)\rangle ds\hfill \\ \hfill & +\mathbb{E}{\int }_t^T{\parallel g(s,Y_s^{n-1},Z_s^n,Y_{s+\overline{\delta }(s)}^{n-1},Z_{s+\overline{\zeta }(s)}^n)\parallel }^{2}ds.\hfill \end{array}$$

From (H3), (A1), (A2) and Young’s inequality $2ab\le \frac{1}{\theta }{a}^2+\theta b^2$, for each $\theta >0$, we have

$$\begin{array}{cc}\hfill & 2\langle Y_s^n,f(s,Y_s^{n-1},Z_s^n,Y_{s+\delta (s)}^{n-1},Z_{s+\zeta (s)}^n)\rangle \hfill \\ \hfill & \le \frac{1}{\theta }|Y_s^n{|}^2+\theta {|f(s,Y_s^{n-1},Z_s^n,Y_{s+\delta (s)}^{n-1},Z_{s+\zeta (s)}^n)|}^2\hfill \\ \hfill & \le \frac{1}{\theta }|Y_s^n{|}^2+2\theta ({\rho }_1(s,|Y_s^{n-1}{|}^2)+{\rho }_2(s+\delta (s),{\mathbb{E}}^{{\mathcal{F}}_s}[|Y_{s+\delta (s)}^{n-1}{|}^2]))\hfill \\ \hfill & +2\theta C(\parallel Z_s^n{\parallel }^2+{\mathbb{E}}^{{\mathcal{F}}_s}[\parallel Z_{s+\zeta (s)}^n\parallel ])+{2\theta |f(s,0,0,0,0)|}^2,\hfill \\ \hfill & \parallel g(s,Y_s^{n-1},Z_s^n,Y_{s+\overline{\delta }(s)}^{n-1},Z_{s+\overline{\zeta }(s)}^n){\parallel }^2\le (1+\theta )({\rho }_1(s,|Y_s^{n-1}{|}^2)+{\rho }_2(s+\overline{\delta }(s),{\mathbb{E}}^{{\mathcal{F}}_s}[|Y_{s+\overline{\delta }(s)}^{n-1}{|}^2]))\hfill \\ \hfill & +(1+\theta ){\alpha }_1\parallel Z_s^n{\parallel }^2+(1+\theta ){\alpha }_2{\mathbb{E}}^{{\mathcal{F}}_s}[\parallel Z_{s+\overline{\zeta }(s)}^n{\parallel }^2])+\left(1+\frac{1}{\theta }\right){\parallel g(s,0,0,0,0)\parallel }^2.\hfill \end{array}$$

Therefore,

$$\begin{array}{cc}\hfill \mathbb{E}& |Y_t^n{|}^2+(1-2\theta C(1+M)-(1+\theta )({\alpha }_1+{\alpha }_{2}M))\mathbb{E}{\int }_t^T{\parallel Z_s^n\parallel }^{2}ds\hfill \\ \hfill & \le {\mathbb{E}|\xi |}^2+\frac{1}{\theta }\mathbb{E}{\int }_t^T|Y_s^n{|}^{2}ds+(3\theta +1){\int }_t^T({\rho }_1(s,\mathbb{E}|Y_s^{n-1}{|}^2)+M{\rho }_2(s,\mathbb{E}|Y_s^{n-1}{|}^2))ds\hfill \\ \hfill & +(3\theta +1)M{\int }_T^{T+K}{\rho }_2(s,\mathbb{E}|{\xi }_s{|}^2)ds+(2\theta CM+(1+\theta ){\alpha }_2)M\mathbb{E}{\int }_T^{T+K}{\parallel {\eta }_s\parallel }^{2}ds\hfill \\ \hfill & +\mathbb{E}{\int }_t^T{(2\theta |f(s,0,0,0,0)|}^2+\left(1+\frac{1}{\theta }\right){\parallel g(s,0,0,0,0)\parallel }^2)ds.\hfill \end{array}$$

We choose $\theta =\frac{1-{\alpha }_1-{\alpha }_{2}M}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}>0$, then

$$\begin{array}{cc}\hfill & \mathbb{E}|Y_t^n{|}^2\le {\mathbb{E}|\xi |}^2+\frac{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{1-{\alpha }_1-{\alpha }_{2}M}\mathbb{E}{\int }_t^T{|Y_s^n|}^{2}ds\hfill \\ \hfill & +\frac{3+2C(1+M)-2({\alpha }_1+{\alpha }_{2}M)}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{\int }_t^T({\rho }_1(s,\mathbb{E}|Y_s^{n-1}{|}^2)+M{\rho }_2(s,\mathbb{E}|Y_s^{n-1}{|}^2))ds\hfill \\ \hfill & +\frac{3+2C(1+M)-2({\alpha }_1+{\alpha }_{2}M)}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}M{\int }_T^{T+K}{\rho }_2(s,\mathbb{E}|{\xi }_s{|}^2)ds\hfill \\ \hfill & +(2\frac{1-{\alpha }_1-{\alpha }_{2}M}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}CM+\frac{2C(1+M)+1}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{\alpha }_2)\mathbb{E}{\int }_T^{T+K}{\parallel {\eta }_s\parallel }^{2}ds\hfill \\ \hfill & +\mathbb{E}{\int }_t^T(2\frac{1-{\alpha }_1-{\alpha }_{2}M}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{|f(s,0,0,0,0)|}^2+\frac{2C(1+M)+1}{1-{\alpha }_1-{\alpha }_{2}M}{\parallel g(s,0,0,0,0)\parallel }^2)ds.\hfill \end{array}$$

Now, in view of Gronwall’s inequality, we derive

$$\mathbb{E}|Y_t^n{|}^2\le {\delta }_t+L{\int }_t^T({\rho }_1(s,\mathbb{E}|Y_s^{n-1}{|}^2)+M{\rho }_2(s,\mathbb{E}|Y_s^{n-1}{|}^2))ds,$$

(4)

where

$$\begin{array}{cc}\hfill {\delta }_t^1=& e^{\frac{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{1-{\alpha }_1-{\alpha }_{2}M}T}{(\mathbb{E}|\xi |}^2+1+\frac{3+2C(1+M)-2({\alpha }_1+{\alpha }_{2}M)}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}M{\int }_T^{T+K}{\rho }_2(s,\mathbb{E}|{\xi }_s{|}^2)ds\hfill \\ \hfill & +(2\frac{1-{\alpha }_1-{\alpha }_{2}M}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}CM+\frac{2C(1+M)+1}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{\alpha }_2)\mathbb{E}{\int }_T^{T+K}{\parallel {\eta }_s\parallel }^{2}ds\hfill \\ \hfill & +\mathbb{E}{\int }_t^T(2\frac{1-{\alpha }_1-{\alpha }_{2}M}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{|f(s,0,0,0,0)|}^2+\frac{2C(1+M)+1}{1-{\alpha }_1-{\alpha }_{2}M}{\parallel g(s,0,0,0,0)\parallel }^2)ds),\hfill \end{array}$$

and

$$L=\frac{3+2C(1+M)-2({\alpha }_1+{\alpha }_{2}M)}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{e}^{\frac{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{1-{\alpha }_1-{\alpha }_{2}M}T}>0.$$

Let

$$\begin{array}{cc}\hfill L_1=& 2{\delta }_0^1=e^{\frac{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{1-{\alpha }_1-{\alpha }_{2}M}T}{(\mathbb{E}|\xi |}^2+1+\frac{3+2C(1+M)-2({\alpha }_1+{\alpha }_{2}M)}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}M{\int }_T^{T+K}{\rho }_2(s,\mathbb{E}|{\xi }_s{|}^2)ds\hfill \\ \hfill & +(2\frac{1-{\alpha }_1-{\alpha }_{2}M}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}CM+\frac{2C(1+M)+1}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{\alpha }_2)\mathbb{E}{\int }_T^{T+K}{\parallel {\eta }_s\parallel }^{2}ds\hfill \\ \hfill & +\mathbb{E}{\int }_0^T(2\frac{1-{\alpha }_1-{\alpha }_{2}M}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{|f(s,0,0,0,0)|}^2+\frac{2C(1+M)+1}{1-{\alpha }_1-{\alpha }_{2}M}{\parallel g(s,0,0,0,0)\parallel }^2)ds),\hfill \end{array}$$

By virtue of (H3),

$${\int }_0^T({\rho }_1(s,L_1)+M{\rho }_2(s,L_1)ds<\infty ,$$

so we can find $T_1$ such that

$${\int }_{T_1}^T({\rho }_1(s,L_1)+M{\rho }_2(s,L_1)ds\le \frac{{\delta }_0^1}{L}.$$

Indeed, if ${\int }_0^T({\rho }_1(s,L_1)+M{\rho }_2(s,L_1)ds\le \frac{{\delta }_0^1}{M}$, then we choose $T_1=0$. If

$${\int }_0^T({\rho }_1(s,L_1)+M{\rho }_2(s,L_1)ds>\frac{{\delta }_0^1}{L},$$

since $t\mapsto {\int }_t^T({\rho }_1(s,L_1)+M{\rho }_2(s,L_1)ds$ is continuous, there exists $T_1\in [0,T]$ such that

$${\int }_{T_1}^T({\rho }_1(s,L_1)+M{\rho }_2(s,L_1)ds=\frac{{\delta }_0^1}{L}.$$

Consequently, for any $t\in [T_1,T]$, from Equation (4) and ${\rho }_i(t,.)$ is increasing, we get

$$\begin{array}{cc}\hfill \mathbb{E}|Y_t^1{|}^2& \le {\delta }_t^1\le 2{\delta }_0^1=L_1,\hfill \\ \hfill \mathbb{E}|Y_t^2{|}^2& \le {\delta }_t^1+L{\int }_t^T({\rho }_1(s,\mathbb{E}|Y_s^1{|}^2)+M{\rho }_2(s,\mathbb{E}|Y_s^1{|}^2))ds\hfill \\ \hfill & \le {\delta }_0^1+L{\int }_t^T({\rho }_1(s,L_1)+M{\rho }_2(s,L_1)ds\le 2{\mu }_0^1=L_1,\hfill \\ \hfill \mathbb{E}|Y_t^3{|}^2& \le {\delta }_t^1+L{\int }_t^T({\rho }_1(s,\mathbb{E}|Y_s^2{|}^2)+M{\rho }_2(s,\mathbb{E}|Y_s^2{|}^2))ds\hfill \\ \hfill & \le {\delta }_0^1+L{\int }_t^T({\rho }_1(s,L_1)+M{\rho }_2(s,L_1)ds\le 2{\mu }_0^1=L_1.\hfill \end{array}$$

By induction, for all $n\ge 1,t\in [T_1,T]$,

$$\mathbb{E}|Y_t^n{|}^2\le L_1.$$

□

With the help of Lemmas 1 and 2, we can establish the existence and uniqueness theorem in the following.

Theorem 1.

Under (A1), (A2), (H1), (H3) and (H4). Then, ADDSDE (2) has a unique solution $(Y_{·},Z_{·})\in {\mathcal{S}}^2([0,T+K],{\mathbb{R}}^k)\times {\mathcal{M}}^2(0,T+K,{\mathbb{R}}^{k\times d})$.

Proof. 

Existence. For any $n\ge 1$, and $t\in [0,T]$, we set

$${\phi }_0(t)=L{\int }_t^T({\rho }_1(s,L_1)+M{\rho }_2(s,L_1))ds,{\phi }_{n+1}(t)=L{\int }_t^T({\rho }_1(s,{\phi }_n(s))+M{\rho }_2(s,{\phi }_n(s)))ds.$$

Obviously, for all $t\in [T_1,T]$, we have

$$\begin{array}{cc}\hfill {\phi }_0(t)& =L{\int }_t^T({\rho }_1(s,L_1)+M{\rho }_2(s,L_1))ds\le L_1,\hfill \\ \hfill {\phi }_1(t)& =L{\int }_t^T({\rho }_1(s,{\phi }_0(s))+M{\rho }_2(s,{\phi }_0(s)))ds\hfill \\ \hfill & \le L{\int }_t^T({\rho }_1(s,L_1)+M{\rho }_2(s,L_1))ds={\phi }_0(t)\le L_1,\hfill \\ \hfill {\phi }_2(t)& =L{\int }_t^T({\rho }_1(s,{\phi }_1(s))+M{\rho }_2(s,{\phi }_1(s)))ds\hfill \\ \hfill & \le L{\int }_t^T({\rho }_1(s,{\phi }_0(s))+M{\rho }_2(s,{\phi }_0(s)))ds={\phi }_1(t)\le L_1.\hfill \end{array}$$

Through induction, we get ${\phi }_n(t)$ satisfies

$$0\le {\phi }_{n+1}(t)\le {\phi }_n(t)\le \cdots \le {\phi }_1(t)\le {\phi }_0(t)\le L_1$$

for any $n\ge 1,t\in [T_1,T]$. Further, for any $n\ge 1,t,t^{\prime }\in [T_1,T]$, we have

$$\begin{array}{cc}\hfill |{\phi }_n(t)-{\phi }_n(t^{\prime })|& =L{\int }_{t\wedge t^{\prime }}^{t\vee t^{\prime }}({\rho }_1(s,{\phi }_{n-1}(s))+M{\rho }_2(s,{\phi }_{n-1}(s)))ds\hfill \\ \hfill & \le L{\int }_{t\wedge t^{\prime }}^{t\vee t^{\prime }}({\rho }_1(s,L_1)+M{\rho }_2(s,L_1))ds.\hfill \end{array}$$

Since $t\mapsto {\int }_t^T({\rho }_1(s,L_1)+M{\rho }_2(s,L_1))ds$ is a continuous mapping, we can obtain

$$\underset{n}{sup}|{\phi }_n(t)-{\phi }_n(t^{\prime })|\to 0as|t-t^{\prime }|\to 0.$$

Consequently, ${\left\{{\phi }_n(t)\right\}}_{n\ge 1}$ is an equicontinuous family of function on $[T_1,T]$ and decreasing. Thus, by the Ascoli–Arzela theorem, ${\left\{{\phi }_n(t)\right\}}_{n\ge 1}$ converges to a limit $\phi (t)$, as $n\to 0$, satisfying

$$\phi (t)=L{\int }_t^T({\rho }_1(s,\phi (s))+M{\rho }_2(s,\phi (s)))ds.$$

From (H3), it follows that $\phi (t)=0,t\in [T_1,T]$.

According to Lemmas 2 and 3, the definition of ${\left\{{\phi }_n(t)\right\}}_{n\ge 1}$, and the fact that

$$e^{\frac{C(M+1)T}{1-{\alpha }_1-{\alpha }_{2}M}}\left(\frac{1-{\alpha }_1-{\alpha }_{2}M}{C(M+1)}+1\right)\le L,$$

for any $t\in [T_1,T],n,m\ge 1$, we get

$$\begin{array}{cc}\hfill \mathbb{E}|Y_t^{1,1+m}-Y_t^{1,1}{|}^2& \le e^{\frac{C(M+1)T}{1-{\alpha }_1-{\alpha }_{2}M}}\left(\frac{1-{\alpha }_1-{\alpha }_{2}M}{C(M+1)}+1\right){\int }_t^T({\rho }_1(s,\mathbb{E}[|Y_s^{1,m}{|}^2])+M{\rho }_2(s,\mathbb{E}[|Y_s^{1,m}{|}^2]))ds\hfill \\ \hfill & \le L{\int }_t^T({\rho }_1(s,L_1)+M{\rho }_2(s,L_1))ds={\phi }_0(t)\le L_1,\hfill \\ \hfill \mathbb{E}|Y_t^{1,2+m}-Y_t^{1,2}{|}^2& \le e^{\frac{C(M+1)T}{1-{\alpha }_1-{\alpha }_{2}M}}\left(\frac{1-{\alpha }_1-{\alpha }_{2}M}{C(M+1)}+1\right){\int }_t^T({\rho }_1(s,\mathbb{E}[|Y_s^{1,1+m}-Y_s^{1,1}{|}^2])\hfill \\ \hfill & +M{\rho }_2(s,\mathbb{E}[|Y_s^{1,1+m}-Y_s^{1,1}{|}^2]))ds\hfill \\ \hfill & \le L{\int }_t^T({\rho }_1(s,{\phi }_0(s))+M{\rho }_2(s,{\phi }_0(s)))ds={\phi }_1(t)\le L_1.\hfill \end{array}$$

Through induction, we can give that

$$\mathbb{E}|Y_t^{1,n+m}-Y_t^{1,n}{|}^2\le {\phi }_{n-1}(t)\le L_1.$$

(5)

Notice that ${\phi }_n(t)\to 0$ for any $t\in [T_1,T]$, as $n\to \infty$, thus we can know that $\left\{Y_{·}^{1,n}\right\}$ is a Cauchy sequence in ${\mathcal{M}}^2(T_1,T+K;{\mathbb{R}}^k)$. Consequently, it is simple to check that $\left\{Z_{·}^{1,n}\right\}$ is also a Cauchy sequence in ${\mathcal{M}}^2(T_1,T+K;{\mathbb{R}}^{k\times d})$. Let $Y_{·}^1=\underset{n\to +\infty }{lim}{Y}_{·}^{1,n},Z_{·}^1=\underset{n\to +\infty }{lim}{Z}_{·}^{1,n}$, we obtain

$$\left\{\begin{array}{cc}\hfill & Y_t^1=\xi +{\int }_t^{T}f(s,Y_s^1,Z_s,Y_{s+\delta (s)}^1,Z_{s+\zeta (s)}^1)\mathrm{d}s\hfill \\ \hfill & +{\int }_t^{T}g(s,Y_s^1,Z_s^1,Y_{s+\overline{\delta }(s)}^1,Z_{s+\overline{\zeta }(s)}^1)\mathrm{d}{\overleftarrow{B}}_s\hfill \\ \hfill & -{\int }_t^T{Z}_s^1\mathrm{d}{W}_s,t\in [T_1,T],\hfill \\ \hfill & Y_t^1={\xi }_t,Z_t^1={\eta }_t,t\in [T,T+K].\hfill \end{array}\right.$$

(6)

Applying It$\widehat{\mathrm{o}}$’s formula, and using the assumptions (A1), (A2), (H3) and (H4), Young’s inequality, Lemma 3 and Burkholder–Davis–Gundy’s inequality, we can derive the limit $(Y_{·}^1,Z_{·}^1)\in {\mathcal{S}}^2([T_1,T+K];{\mathbb{R}}^k)\times {\mathcal{M}}^2(T_1,T+K;{\mathbb{R}}^{k\times d})$.

Therefore, we have proved the existence of the solution to ABDSDE (2) on $[T_1,T]$. If $T_1=0$, then the existence is obtained.

If $T_1\ne 0$, we will consider the following equation

$$\left\{\begin{array}{cc}\hfill & Y_t^2=Y_{T_1}^1+{\int }_t^{T}f(s,Y_s^2,Z_s^2,Y_{s+\delta (s)}^2,Z_{s+\zeta (s)}^2)\mathrm{d}s\hfill \\ \hfill & +{\int }_t^{T}g(s,Y_s^2,Z_s^2,Y_{s+\overline{\delta }(s)}^2,Z_{s+\overline{\zeta }(s)}^2)\mathrm{d}{\overleftarrow{B}}_s\hfill \\ \hfill & -{\int }_t^T{Z}_s^2\mathrm{d}{W}_s,t\in [0,T_1],\hfill \\ \hfill & Y_t^2=Y_t^1,Z_t^2=Z_t^1,t\in [T_1,T+K].\hfill \end{array}\right.$$

(7)

Let us introduce the approximate sequence which is similar to (3) for Equation (7). Through the similar process as in Lemmas 2 and 3, for each $t\in [0,T_1],n,m\ge 1$, we can prove that

$$\begin{array}{cc}\hfill \mathbb{E}|Y_t^{2,n+m}-Y_t^{2,n}{|}^2\le & e^{\frac{C(M+1)T}{1-{\alpha }_1-{\alpha }_{2}M}}\left(\frac{1-{\alpha }_1-{\alpha }_{2}M}{C(M+1)}+1\right){\int }_t^{T_1}({\rho }_1(s,\mathbb{E}[|Y_s^{2,n+m-1}-Y_s^{2,n-1}{|}^2])\hfill \\ \hfill & +M{\rho }_2(s,\mathbb{E}[|Y_s^{2,n+m-1}-Y_s^{2,n-1}{|}^2]))ds.\hfill \end{array}$$

$$\mathbb{E}|Y_t^{2,n}{|}^2\le {\delta }_t^2+L{\int }_t^T({\rho }_1(s,\mathbb{E}|Y_s^{2,n-1}{|}^2)+M{\rho }_2(s,\mathbb{E}|Y_s^{2,n-1}{|}^2))ds,$$

where

$$\begin{array}{cc}\hfill {\delta }_t^2=& e^{\frac{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{1-{\alpha }_1-{\alpha }_{2}M}T}(\mathbb{E}|Y_{T_1}^1{|}^2+1+\frac{3+2C(1+M)-2({\alpha }_1+{\alpha }_{2}M)}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}M{\int }_{T_1}^{T+K}{\rho }_2(s,\mathbb{E}|Y_s^1{|}^2)ds\hfill \\ \hfill & +(2\frac{1-{\alpha }_1-{\alpha }_{2}M}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}CM+\frac{2C(1+M)+1}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{\alpha }_2)\mathbb{E}{\int }_{T_1}^{T+K}{|Z_s^1|}^{2}ds\hfill \\ \hfill & +\mathbb{E}{\int }_t^T(2\frac{1-{\alpha }_1-{\alpha }_{2}M}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{|f(s,0,0,0,0)|}^2+\frac{2C(1+M)+1}{1-{\alpha }_1-{\alpha }_{2}M}{\parallel g(s,0,0,0,0)\parallel }^2)ds),\hfill \end{array}$$

and

$$L=\frac{3+2C(1+M)-2({\alpha }_1+{\alpha }_{2}M)}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{e}^{\frac{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{1-{\alpha }_1-{\alpha }_{2}M}T}>0.$$

Let

$$\begin{array}{cc}\hfill L_2=& 2{\delta }_0^2=2e^{\frac{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{1-{\alpha }_1-{\alpha }_{2}M}T}(\mathbb{E}|Y_{T_1}^1{|}^2+1+\frac{3+2C(1+M)-2({\alpha }_1+{\alpha }_{2}M)}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}M{\int }_{T_1}^{T+K}{\rho }_2(s,\mathbb{E}|Y_s^1{|}^2)ds\hfill \\ \hfill & +(2\frac{1-{\alpha }_1-{\alpha }_{2}M}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}CM+\frac{2C(1+M)+1}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{\alpha }_2)\mathbb{E}{\int }_{T_1}^{T+K}{|Z_s^1|}^{2}ds\hfill \\ \hfill & +\mathbb{E}{\int }_0^T(2\frac{1-{\alpha }_1-{\alpha }_{2}M}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{|f(s,0,0,0,0)|}^2+\frac{2C(1+M)+1}{1-{\alpha }_1-{\alpha }_{2}M}{\parallel g(s,0,0,0,0)\parallel }^2)ds).\hfill \end{array}$$

There exists $T_2\in [0,T_1[$ with

$$\mathbb{E}|Y_t^{2,n}{|}^2\le L_2,n\ge 1,t\in [T_2,T_1],$$

where $T_2=0$ or $T_2\in ]0,T_1[$ with ${\int }_{T_2}^{T_1}({\rho }_1(s,L_1)+M{\rho }_2(s,L_1)ds=\frac{{\delta }_0^2}{M}$. As proved above, ABDSDE (7) admits a solution on $[T_2,T+K]$. If $T_2=0$, the existence proof is complete. Otherwise, we can find a sequence $\{T_p,{\delta }_t^p,L_p,p\ge 1\}$ defined as

$$0\le T_p<T_{p-1}<\cdots <T_1<T_0=T,$$

$$\begin{array}{cc}\hfill {\delta }_t^p=& e^{\frac{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{1-{\alpha }_1-{\alpha }_{2}M}T}(\mathbb{E}|Y_{T_{p-1}}^{p-1}{|}^2+1+\frac{3+2C(1+M)-2({\alpha }_1+{\alpha }_{2}M)}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}M{\int }_{T_{p-1}}^{T+K}{\rho }_2(s,\mathbb{E}|Y_s^{p-1}{|}^2)ds\hfill \\ \hfill & +(2\frac{1-{\alpha }_1-{\alpha }_{2}M}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}CM+\frac{2C(1+M)+1}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{\alpha }_2)\mathbb{E}{\int }_{T_{p-1}}^{T+K}{|Z_s^{p-1}|}^{2}ds\hfill \\ \hfill & +\mathbb{E}{\int }_t^T(2\frac{1-{\alpha }_1-{\alpha }_{2}M}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{|f(s,0,0,0,0)|}^2+\frac{2C(1+M)+1}{1-{\alpha }_1-{\alpha }_{2}M}{\parallel g(s,0,0,0,0)\parallel }^2)ds),\hfill \end{array}$$

$$\begin{array}{cc}\hfill L_p=& 2{\delta }_0^p=2e^{\frac{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{1-{\alpha }_1-{\alpha }_{2}M}T}(\mathbb{E}|Y_{T_{p-1}}^{p-1}{|}^2+1+\frac{3+2C(1+M)-2({\alpha }_1+{\alpha }_{2}M)}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}M{\int }_{T_{p-1}}^{T+K}{\rho }_2(s,\mathbb{E}|Y_s^{p-1}{|}^2)ds\hfill \\ \hfill & +(2\frac{1-{\alpha }_1-{\alpha }_{2}M}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}CM+\frac{2C(1+M)+1}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{\alpha }_2)\mathbb{E}{\int }_{T_{p-1}}^{T+K}{|Z_s^{p-1}|}^{2}ds\hfill \\ \hfill & +\mathbb{E}{\int }_0^T(2\frac{1-{\alpha }_1-{\alpha }_{2}M}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{|f(s,0,0,0,0)|}^2+\frac{2C(1+M)+1}{1-{\alpha }_1-{\alpha }_{2}M}{\parallel g(s,0,0,0,0)\parallel }^2)ds),\hfill \end{array}$$

and

$${\int }_{T_p}^{T_{p-1}}({\rho }_1(s,L_p)+M{\rho }_2(s,L_p)ds\ge \frac{{\delta }_0^p}{L}$$

through repeating the above procedure. Therefore, we can derive a solution to ABDSDE (2) on $[T_p,T+K]$ through iteration method. In the following, we will check there exists a finite number $p\ge 1$ with $T_p=0$. Denote

$$\begin{array}{cc}\hfill A& =2e^{\frac{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{1-{\alpha }_1-{\alpha }_{2}M}T}(1+\mathbb{E}{\int }_0^T(2\frac{1-{\alpha }_1-{\alpha }_{2}M}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{|f(s,0,0,0,0)|}^2\hfill \\ \hfill & +\frac{2C(1+M)+1}{1-{\alpha }_1-{\alpha }_{2}M}{\parallel g(s,0,0,0,0)\parallel }^2)ds).\hfill \end{array}$$

Then, in view of (H1),

$$\begin{array}{cc}\hfill L_p& =2e^{\frac{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{1-{\alpha }_1-{\alpha }_{2}M}T}(\mathbb{E}|Y_{T_{p-1}}^{p-1}{|}^2+\frac{3+2C(1+M)-2({\alpha }_1+{\alpha }_{2}M)}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}M{\int }_{T_{p-1}}^{T+K}{\rho }_2(s,\mathbb{E}|Y_s^{p-1}{|}^2)ds\hfill \\ \hfill & +(2\frac{1-{\alpha }_1-{\alpha }_{2}M}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}CM+\frac{2C(1+M)+1}{2C(1+M)+{\alpha }_1+{\alpha }_{2}M}{\alpha }_2)\mathbb{E}{\int }_{T_{p-1}}^{T+K}|Z_s^{p-1}{|}^{2}ds)+A\hfill \\ \hfill & \ge A>0,\mathrm{for}\mathrm{all}p\ge 1.\hfill \end{array}$$

For any $p\ge 1$, since $\underset{T_p\le t\le T+K}{sup}\mathbb{E}{|Y_t^p|}^2<\infty$, $L_p$ is finite. So, we have

$$0<\frac{A}{L_p}\le 1,\mathrm{for}\mathrm{any}p\ge 1.$$

(8)

Since for $i=1,2$, ${\rho }_i(t,.)$ is a concave function, ${\rho }_i(t,0)=0$ while $t\in [0,T]$, we get

$${\rho }_i(t,\lambda x)\ge \lambda {\rho }_i(t,x),\mathrm{for}\mathrm{all}\mathrm{any}\lambda \in [0,1]\mathrm{and}\mathrm{any}x\in {\mathbb{R}}^{+}.$$

By the inequality (8), we have

$${\rho }_i(t,A)={\rho }_i\left(t,\frac{A}{L_p}{L}_p\right)\ge \frac{A}{L_p}{\rho }_i(t,L_p),\mathrm{for}\mathrm{each}p\ge 1.$$

Therefore, for each $p\ge 1$, we can derive that

$$\begin{array}{cc}\hfill {\int }_{T_p}^T({\rho }_1(s,A)+M{\rho }_2(s,A))ds& =\sum _{i=1}^p{\int }_{T_i}^{T_{i-1}}({\rho }_1(s,A)+M{\rho }_2(s,A))ds\hfill \\ \hfill & \ge \sum _{i=1}^p\frac{A}{L_i}{\int }_{T_i}^{T_{i-1}}({\rho }_1(s,L_i)+M{\rho }_2(s,L_i))ds\ge \sum _{i=1}^p\frac{A}{L_i}\frac{{\delta }_0^i}{L}.\hfill \end{array}$$

By definition, $\frac{{\delta }_0^i}{L_i}=\frac{1}{2}$, for any $i\ge 1$. So, for any $p\ge 1$,

$${\int }_{T_p}^T({\rho }_1(s,A)+M{\rho }_2(s,A))ds\ge \frac{pA}{2L}.$$

(9)

Notice that $A>0$, the right side of inequality (9) tends to $+\infty$ while $p\to +\infty$. Consequently, as ${\int }_0^T({\rho }_1(s,A)+M{\rho }_2(s,A))ds$ is finite, so we can obtain a sufficient large p such that

$${\int }_{T_p}^T({\rho }_1(s,A)+M{\rho }_2(s,A))ds\ge {\int }_0^T({\rho }_1(s,A)+M{\rho }_2(s,A))ds.$$

Therefore, there is a finite p with $T_p=0$, and we assert the existence of solution on $[0,T+K]$.

Uniqueness. Set $(Y_{·},Z_{·}),(Y_{·}^{\prime },Z_{·}^{\prime })\in {\mathbb{S}}^2([0,T+K];{\mathbb{R}}^k)\times {\mathbb{M}}^2(0,T+K;{\mathbb{R}}^{k\times d})$ be the two solutions of ABDSDE (2). Let $\beta >0$. Applying the It$\widehat{\mathrm{o}}$’s formula yields

$$\begin{array}{cc}\hfill & \mathbb{E}|Y_t-Y_t^{\prime }{|}^2{e}^{\beta t}+\beta \mathbb{E}{\int }_t^T|Y_s-Y_s^{\prime }{|}^2{e}^{\beta s}ds+\mathbb{E}{\int }_t^T\parallel Z_s-Z_s^{\prime }{\parallel }^2{e}^{\beta s}ds=2\mathbb{E}{\int }_t^T\langle Y_s-Y_s^{\prime },\hfill \\ \hfill & f(s,Y_s,Z_s,Y_{s+\delta (s)},Z_{s+\zeta (s)})-f(s,Y_s^{\prime },Z_s^{\prime },Y_{s+\delta (s)}^{\prime },Z_{s+\zeta (s)}^{\prime })\rangle e^{\beta s}ds\hfill \\ \hfill & +\mathbb{E}{\int }_t^T{\parallel g(s,Y_s,Z_s,Y_{s+\overline{\delta }(s)},Z_{s+\overline{\zeta }(s)})-g(s,Y_s^{\prime },Z_s^{\prime },Y_{s+\overline{\delta }(s)}^{\prime },Z_{s+\overline{\zeta }(s)}^{\prime })\parallel }^2{e}^{\beta s}ds.\hfill \end{array}$$

By the assumptions (A1), (A2), (H3), Young’s inequality $2ab\le \frac{1}{\beta }{a}^2+\beta b^2$, and Jensen’s inequality, we have

$$\begin{array}{cc}\hfill \mathbb{E}& |Y_t-Y_t^{\prime }{|}^2{e}^{\beta t}+(1-\frac{1}{\beta }C(M+1)-{\alpha }_1-{\alpha }_{2}M)\mathbb{E}{\int }_t^T{\parallel Z_s-Z_s^{\prime }\parallel }^2{e}^{\beta s}ds\hfill \\ \hfill \le & (\frac{1}{\beta }+1){\int }_t^T({\rho }_1(s,\mathbb{E}[|Y_s-Y_s^{\prime }{|}^2])+M{\rho }_2(s,\mathbb{E}[|Y_s-Y_s^{\prime }{|}^2]))e^{\beta s}ds.\hfill \end{array}$$

Choosing $\beta >\frac{C(M+1)}{1-{\alpha }_1-{\alpha }_{2}M}>0$, we get

$$\begin{array}{cc}\hfill \mathbb{E}& |Y_t-Y_t^{\prime }{|}^2+(1-\frac{1}{\beta }C(M+1)-{\alpha }_1-{\alpha }_{2}M)\mathbb{E}{\int }_t^T{\parallel Z_s-Z_s^{\prime }\parallel }^{2}ds\hfill \\ \hfill & \le (\frac{1}{\beta }+1)e^{\beta T}{\int }_t^T({\rho }_1(s,\mathbb{E}[|Y_s-Y_s^{\prime }{|}^2])+M{\rho }_2(s,\mathbb{E}[|Y_s-Y_s^{\prime }{|}^2]))ds.\hfill \end{array}$$

(10)

Therefore

$$\mathbb{E}|Y_t-Y_t^{\prime }{|}^2\le \left(\frac{1}{\beta }+1\right)e^{\beta T}{\int }_t^T({\rho }_1(s,\mathbb{E}[|Y_s-Y_s^{\prime }{|}^2])+M{\rho }_2(s,\mathbb{E}[|Y_s-Y_s^{\prime }{|}^2]))ds.$$

From the comparison Theorem for ODE, we can obtain

$$\mathbb{E}|Y_t-Y_t^{\prime }{|}^2\le \gamma (t),\mathrm{for}\mathrm{any}\mathrm{t}\in [0,\mathrm{T}],$$

where $\gamma (t)$ is the maximum left shift solution of the following equation:

$$\left\{\begin{array}{cc}\hfill u^{\prime }& =-(\frac{1}{\beta }+1)e^{\beta T}({\rho }_1(t,u)+M{\rho }_2(t,u));\hfill \\ \hfill u(T)& =0.\hfill \end{array}\right.$$

According to the assumption (H3), we have $\gamma (t)=0,t\in [0,T]$. So $\mathbb{E}|Y_t-Y_t^{\prime }{|}^2=0$, $t\in [0,T]$, this means $Y_t=Y_t^{\prime }$, a.s. for all $t\in [0,T]$. It immediately derives from (10) that $Z_t=Z_t^{\prime }$, a.s., for each $t\in [0,T]$. □

4. Comparison Theorems

In this part, we mainly focus on one dimensional ABDSDEs, that is, $k=1$. Firstly, we propose one comparison theorem for BDSDEs under non-Lipschitz coefficients, which acts as a starting point for the following investigation. For $i=1,2$, assume that ${\xi }^i\in L^2({\mathcal{F}}_T;\mathbb{R})$ and $f^i(t,y,z):[0,T]\times \Omega \times \mathbb{R}\times {\mathbb{R}}^d\to \mathbb{R}$ satisfies (H1) and (H3). Then, the following BDSDE:

$$Y_t^i={\xi }^i+{\int }_t^T{f}^i(s,Y_s^i,Z_s^i)\mathrm{d}s+{\int }_t^T{g}^i(s,Y_s^i,Z_s^i)\mathrm{d}{\overleftarrow{B}}_s-{\int }_t^T{Z}_s^i\mathrm{d}{W}_s,t\in [0,T],$$

(11)

has a unique solution $(Y_{·}^i,Z_{·}^i)\in {\mathcal{S}}^2([0,T];\mathbb{R})\times {\mathcal{M}}^2(0,T;{\mathbb{R}}^d)$ for $i=1,2$, according to Theorem 3.4 in [10]. We can assert the following comparison theorem.

Lemma 4.

Let $(Y_{·}^1,Z_{·}^1)$ and $(Y_{·}^2,Z_{·}^2)$ be solutions of BDSDEs (11), respectively. Assume that (1) ${\xi }_t^1⩾{\xi }_t^2,a.s.$; (2) $f^1(t,Y_t^1,Z_t^1)⩾f^2(t,Y_t^1,Y_t^1)or{f}^1(t,Y_t^2,Z_t^2)⩾f^2(t,Y_t^2,Y_t^2),a.s.,forallt\in [0,T]$; (3) $g^1(t,Y_t^1,Z_t^1)=g^2(t,Y_t^1,Y_t^1)or{g}^1(t,Y_t^2,Z_t^2)=g^2(t,Y_t^2,Y_t^2),a.s.,forallt\in [0,T]$. Then $Y_t^1⩾Y_t^2,a.s.,foranyt\in [0,T]$.

Proof. 

Let

$$\begin{array}{cc}\hfill & {\widehat{Y}}_t=Y_t^2-Y_t^1,{\widehat{Z}}_t=Z_t^2-Z_t^1,\widehat{\xi }={\xi }^2-{\xi }^1.\hfill \\ \hfill & {\widehat{f}}_t=f^2(t,Y_t^2,Z_t^2)-f^1(t,Y_t^1,Z_t^1),{\widehat{g}}_t=g^2(t,Y_t^2,Z_t^2)-g^1(t,Y_t^1,Z_t^1).\hfill \end{array}$$

In view of It$\widehat{\mathrm{o}}$-Meyer’s formula and ${\xi }^1⩾{\xi }^2$, a.s., we have

$$\mathbb{E}|{\widehat{Y}}_t^{+}{|}^2+\mathbb{E}{\int }_t^T{1}_{\{{\widehat{Y}}_s\ge 0\}}|{\widehat{Z}}_s{|}^{2}ds=2\mathbb{E}{\int }_t^T{\widehat{Y}}_s^{+}{\widehat{f}}_{s}ds+\mathbb{E}{\int }_t^T{1}_{\{{\widehat{Y}}_s\ge 0\}}{|{\widehat{g}}_s|}^{2}ds.$$

By the assumption (H3), Young’s inequality $2ab\le \frac{1}{\theta }{a}^2+\theta b^2$, and Jensen’s inequality, for each $\theta >0$, we get

$$\begin{array}{cc}\hfill 2\mathbb{E}{\int }_t^T{\widehat{Y}}_s^{+}{\widehat{f}}_{s}ds\le & 2\mathbb{E}{\int }_t^T{\widehat{Y}}_s^{+}(f^2(s,Y_s^2,Z_s^2)-f^2(t,Y_s^1,Z_s^1))ds\hfill \\ \hfill \le & \frac{2C}{1-{\alpha }_1}\mathbb{E}{\int }_t^T|{\widehat{Y}}_s^{+}{|}^{2}ds+\frac{1-{\alpha }_1}{2C}{\int }_t^T{1}_{\{{\widehat{Y}}_s\ge 0\}}{|f^2(s,Y_s^2,Z_s^2)-f^2(t,Y_s^1,Z_s^1)|}^{2}ds\hfill \\ \hfill \le & \frac{2C}{1-{\alpha }_1}\mathbb{E}{\int }_t^T|{\widehat{Y}}_s^{+}{|}^{2}ds+\frac{1-{\alpha }_1}{2C}{\int }_t^T{\rho }_1(s,\mathbb{E}[|{\widehat{Y}}_s^{+}{|}^2])ds+\frac{1-{\alpha }_1}{2}\mathbb{E}{\int }_t^T{1}_{\{{\widehat{Y}}_s\ge 0\}}{|{\widehat{Z}}_s|}^{2}ds,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \mathbb{E}{\int }_t^T{1}_{\{{\widehat{Y}}_s\ge 0\}}{|{\widehat{g}}_s|}^{2}ds=& \mathbb{E}{\int }_t^T{1}_{\{{\widehat{Y}}_s\ge 0\}}{|g^1(s,Y_s^2,Z_s^2)-g^1(t,Y_s^1,Z_s^1)|}^{2}ds\hfill \\ \hfill \le & {\int }_t^T{\rho }_1(s,\mathbb{E}[|{\widehat{Y}}_s^{+}{|}^2])ds+{\alpha }_1\mathbb{E}{\int }_t^T{1}_{\{{\widehat{Y}}_s\ge 0\}}{|{\widehat{Z}}_s|}^{2}ds.\hfill \end{array}$$

Then, thanks to the above inequalities, we obtain

$$\mathbb{E}|{\widehat{Y}}_t^{+}{|}^2\le \frac{2C}{1-{\alpha }_1}\mathbb{E}{\int }_t^T|{\widehat{Y}}_s^{+}{|}^{2}ds+(\frac{1-{\alpha }_1}{2C}+1){\int }_t^T{\rho }_1(s,\mathbb{E}[|{\widehat{Y}}_s^{+}{|}^2])ds.$$

From the Gronwall’s inequality, we have

$$\mathbb{E}|{\widehat{Y}}_t^{+}{|}^2⩽e^{\frac{2C}{1-{\alpha }_1}}(\frac{1-{\alpha }_1}{2C}+1){\int }_t^T{\rho }_1(s,\mathbb{E}[|{\widehat{Y}}_s^{+}{|}^2])ds.$$

Using the same proof method about the uniqueness in Theorem 1, we can obtain

$$\mathbb{E}|{\widehat{Y}}_t^{+}{|}^2=0,forallt\in [0,T].$$

Hence

$$Y_t^1⩾Y_t^2,a.s.,foranyt\in [0,T].$$

□

For $i=1,2$, we first study a comparison theorem of anticipated BDSDEs of the following generalized version:

$$\left\{\begin{array}{cc}\hfill & Y_t^i={\xi }_T^i+{\int }_t^T{f}^i(s,Y_s^i,Z_s^i,Y_{s+{\delta }^i(s)}^i,Z_{s+{\zeta }^i(s)}^i)\mathrm{d}s\hfill \\ \hfill & +{\int }_t^T{g}^i(s,Y_s^i,Z_s^i,Y_{s+{\overline{\delta }}^i(s)}^i,Z_{s+{\overline{\zeta }}^i(s)}^i)\mathrm{d}{\overleftarrow{B}}_s-{\int }_t^T{Z}_s^i\mathrm{d}{W}_s,t\in [0,T],\hfill \\ \hfill & Y_t^i={\xi }_t^i,Z_t^i={\eta }_t^i,t\in [T,T+K].\hfill \end{array}\right.$$

(12)

Let us assume that ${\delta }^i(·),{\overline{\delta }}^i(·),{\zeta }^i(·),{\overline{\zeta }}^i(·)$ satisfies (A1) and (A2), ${\xi }_{·}^i\in {\mathcal{S}}^2([T,T+K];\mathbb{R})$, ${\eta }_{·}^i\in {\mathcal{M}}^2(T,T+K;{\mathbb{R}}^d)$, and $(f^i,g^i)$ satisfies (H1) and (H3). Then, by Theorem 1, anticipated BDSDE (12) has a unique solution $(Y_{·}^i,Z_{·}^i)\in {\mathcal{S}}^2([0,T+K];\mathbb{R})\times {\mathcal{M}}^2(0,T+K;{\mathbb{R}}^d)$ for $i=1,2$.

Theorem 2.

Suppose $(Y_{·}^1,Z_{·}^1)$ and $(Y_{·}^2,Z_{·}^2)$ are solutions of ABDSDEs (12), respectively. Assume that (1) ${\xi }_t^1⩾{\xi }_t^2,a.s.,forallt\in [T,T+K]$; (2) $f^1(t,Y_t^1,Z_t^1,Y_{t+{\delta }^1(t)}^1,Z_{t+{\zeta }^1(t)}^1)⩾f^2(t,Y_t^1,Z_t^1,Y_{t+{\delta }^2(t)}^2,Z_{t+{\zeta }^2(t)}^2)or{f}^1(t,Y_t^2,Z_t^2,Y_{t+{\delta }^1(t)}^1,Z_{t+{\zeta }^1(t)}^1)⩾f^2(t,Y_t^2,Y_t^2,Y_{t+{\delta }^2(t)}^2,$$Z_{t+{\zeta }^2(t)}^2),a.s.,forallt\in [0,T]$; (3) $g^1(t,Y_t^1,Z_t^1,Y_{t+{\overline{\delta }}^1(t)}^1,Z_{t+{\overline{\zeta }}^1(t)}^1)=g^2(t,Y_t^1,Y_t^1,Y_{t+{\overline{\delta }}^2(t)}^2,$$Z_{t+{\overline{\zeta }}^2(t)}^2)or{g}^1(t,Y_t^2,Z_t^2,Y_{t+{\overline{\delta }}^1(t)}^1,Z_{t+{\overline{\zeta }}^1(t)}^1)=g^2(t,Y_t^2,Y_t^2,Y_{t+{\overline{\delta }}^2(t)}^2,Z_{t+{\overline{\zeta }}^2(t)}^2),a.s.,forallt\in [0,T]$. Then $Y_t^1⩾Y_t^2,a.s.,foranyt\in [0,T+K]$.

Proof. 

For $i=1,2$, set

$$F^i(t,y,z)=f^i(t,y,z,Y_{t+{\delta }^i(t)}^i,Z_{t+{\zeta }^i(t)}^i),G^i(t,y,z)=g^i(t,y,z,Y_{t+{\overline{\delta }}^i(t)}^i,Z_{t+{\overline{\zeta }}^i(t)}^i),$$

then $(Y_{·}^i,Z_{·}^i)$ is the unique solution of the following BDSDE,

$$Y_t^i={\xi }_T^i+{\int }_t^T{F}^i(s,Y_s^i,Z_s^i)\mathrm{d}s+{\int }_t^T{G}^i(s,Y_s^i,Z_s^i)\mathrm{d}{\overleftarrow{B}}_s-{\int }_t^T{Z}_s^i\mathrm{d}{W}_s,t\in [0,T].$$

According to Lemma 4, we can get

$$Y_t^1⩾Y_t^2,a.s.,forallt\in [0,T],$$

which implies

$$Y_t^1⩾Y_t^2,a.s.,forallt\in [0,T+K].$$

□

Example 1.

Let $f^1(t,\tilde{y},\tilde{z},\tilde{\theta }(r),\tilde{\vartheta }(\overline{r}))=|\tilde{y}|+|\tilde{z}|+{\mathbb{E}}^{{\mathcal{F}}_t}[|\tilde{\theta }(r)|]+{\mathbb{E}}^{{\mathcal{F}}_t}[cos\tilde{\vartheta }(\overline{r})]$, $f^2(t,\tilde{y},\tilde{z},\tilde{\theta }(r),$$\tilde{\vartheta }(\overline{r}))=\tilde{y}+|\tilde{z}|+{\mathbb{E}}^{{\mathcal{F}}_t}[sin\tilde{\theta }(r)]-1$, $g^1(t,\tilde{y},\tilde{z},\tilde{\theta }(r),\tilde{\vartheta }(\overline{r}))=g^2(t,\tilde{y},\tilde{z},\tilde{\theta }(r),\tilde{\vartheta }(\overline{r}))=\tilde{y}+\frac{|\tilde{z}|}{2}+{\mathbb{E}}^{{\mathcal{F}}_t}[\tilde{\theta }(r)\tilde{\vartheta }(\overline{r})]$. Then by Theorem 3, we can derive $Y_t^1⩾Y_t^2$, a.s., $\mathrm{for}\mathrm{any}t\in [0,T+K]$ when the assumption (1) is satisfied.

Secondly, for $i=1,2$, we will study a comparison theorem of anticipated BDSDEs of the following type:

$$\left\{\begin{array}{cc}\hfill & Y_t^i={\xi }_T^i+{\int }_t^T{f}^i(s,Y_s^i,Z_s^i,Y_{s+\delta (s)}^i,Z_{s+{\zeta }^i(s)}^i)\mathrm{d}s\hfill \\ \hfill & +{\int }_t^T{g}^i(s,Y_s^i,Z_s^i,Y_{s+{\overline{\delta }}^i(s)}^i,Z_{s+{\overline{\zeta }}^i(s)}^i)\mathrm{d}{\overleftarrow{B}}_s-{\int }_t^T{Z}_s^i\mathrm{d}{W}_s,t\in [0,T],\hfill \\ \hfill & Y_t^i={\xi }_t^i,Z_t^i={\eta }_t^i,t\in [T,T+K].\hfill \end{array}\right.$$

(13)

Assume that $(\delta (·),{\zeta }^i(·),{\overline{\delta }}^i(·),{\overline{\zeta }}^i(·))$ satisfies (A1) and (A2), ${\xi }_{·}^i\in {\mathcal{S}}^2([T,T+K];\mathbb{R})$, ${\eta }_{·}^i\in {\mathcal{M}}^2(T,T+K;{\mathbb{R}}^d)$ and $(f^i,g^i)$ satisfies (H1) and (H3). Then, by Theorem 1, anticipated BDSDE (13) has a unique solution $(Y_{·}^i,Z_{·}^i)\in {\mathcal{S}}^2([0,T+K];\mathbb{R})\times {\mathcal{M}}^2(0,T+K;{\mathbb{R}}^d)$ for $i=1,2$. In order to obtain the following comparison theorem, we further assume ${\rho }_i(t,u)$ with linear growth, that is, for $i=1,2$, ${\rho }_i(t,u)\le a_i(t)+b_i(t)u$ where $a_i(t)\ge 0,b_i(t)\ge 0$, with

$${\int }_0^{T+K}{a}_i(t)dt<\infty and{\int }_0^{T+K}{b}_i(t)dt<\infty .$$

Theorem 3.

Suppose $(Y_{·}^1,Z_{·}^1)$ and $(Y_{·}^2,Z_{·}^2)$ are solutions of ABDSDEs (13), respectively. Assume that (1) ${\xi }_t^1⩾{\xi }_t^2,a.s.,forallt\in [T,T+K]$; (2) $f^1(t,Y_t^1,Z_t^1,Y_{t+\delta (t)}^1,Z_{t+{\zeta }^1(t)}^1)⩾f^2(t,Y_t^1,Y_t^1,Y_{t+\delta (t)}^1,Z_{t+{\zeta }^2(t)}^2),a.s.,forallt\in [0,T]$; (3) for any $(t,y,z)\in [0,T]\times R\times R^d,\gamma \in L^2({\mathcal{F}}_s;{\mathbb{R}}^d)$ and $s\in [t,T+K],f^2(t,y,z,·,\gamma )$ is increasing, i.e., $f^2(t,y,z,y_2,\gamma )\ge f^2(t,y,z,y_1,\gamma )$, if $y_2\ge y_1$ with $y_1,y_2\in R$; (4) $g^1(t,Y_t^1,Z_t^1,Y_{t+{\overline{\delta }}^1(t)}^1,Z_{t+{\overline{\zeta }}^1(t)}^1)=g^2(t,Y_t^1,Z_t^1,Y_{t+{\overline{\delta }}^2(t)}^2,Z_{t+{\overline{\zeta }}^2(t)}^2),a.s.,forallt\in [0,T]$. Then $Y_t^1⩾Y_t^2,a.s.,foranyt\in [0,T+K]$.

Proof. 

For $i=1,2$, set

$$F^i(t,y,z)=f^i(t,y,z,Y_{t+\delta (t)}^i,Z_{t+{\zeta }^i(t)}^i),G^i(t,y,z)=g^i(t,y,z,Y_{t+{\overline{\delta }}^i(t)}^i,Z_{t+{\overline{\zeta }}^i(t)}^i),$$

then $(Y_{·}^i,Z_{·}^i)$ be the unique solution of the following BDSDE,

$$\left\{\begin{array}{cc}\hfill & Y_t^i={\xi }_T^i+{\int }_t^T{F}^i(s,Y_s^i,Z_s^i)\mathrm{d}s+{\int }_t^T{G}^i(s,Y_s^i,Z_s^i)\mathrm{d}{\overleftarrow{B}}_s-{\int }_t^T{Z}_s^i\mathrm{d}{W}_s,t\in [0,T],\hfill \\ \hfill & Y_t^i={\xi }_t^i,Z_t^i={\eta }_t^i,t\in [T,T+K].\hfill \end{array}\right.$$

Let ${\overline{f}}^2(t,y,z)=f^2(t,y,z,Y_{t+\delta (t)}^1,Z_{t+{\zeta }^i(t)}^2)$, then the following BDSDE admits a unique solution $(Y^3,Z^3)$,

$$\left\{\begin{array}{cc}\hfill & Y_t^3={\xi }_T^2+{\int }_t^T{\overline{f}}^2(s,Y_s^3,Z_s^3)\mathrm{d}s+{\int }_t^T{G}^2(s,Y_s^3,Z_s^3)\mathrm{d}{\overleftarrow{B}}_s-{\int }_t^T{Z}_s^3\mathrm{d}{W}_s,t\in [0,T],\hfill \\ \hfill & Y_t^3={\xi }_t^2,Z_t^3={\eta }_t^2,t\in [T,T+K].\hfill \end{array}\right.$$

According to the assumptions (1), (2), (4) and Lemma 4, we can get $Y_t^1⩾Y_t^3,a.s.,\mathrm{for}\mathrm{each}\mathrm{t}\in [0,\mathrm{T}],$ which implies $Y_t^1⩾Y_t^3,a.s.,\mathrm{for}\mathrm{any}\mathrm{t}\in [0,\mathrm{T}+\mathrm{K}].$ The following BDSDE,

$$\left\{\begin{array}{cc}\hfill & Y_t^4={\xi }_T^2+{\int }_t^T{f}^2(s,Y_s^4,Z_s^4,Y_{t+\delta (t)}^3,Z_{t+{\zeta }^2(t)}^2)\mathrm{d}s+{\int }_t^T{G}^2(s,Y_s^4,Z_s^4)\mathrm{d}{\overleftarrow{B}}_s-{\int }_t^T{Z}_s^4\mathrm{d}{W}_s,t\in [0,T],\hfill \\ \hfill & Y_t^4={\xi }_t^2,Z_t^4={\eta }_t^2,t\in [T,T+K]\hfill \end{array}\right.$$

admits a unique solution $(Y_{·}^4,Z_{·}^4)$. From Lemma 4 and assumptions (3), we can get $Y_t^3⩾Y_t^4,a.s.,\mathrm{for}\mathrm{all}\mathrm{t}\in [0,\mathrm{T}],$ which implies $Y_t^3⩾Y_t^4,a.s.,\mathrm{for}\mathrm{any}\mathrm{t}\in [0,\mathrm{T}+\mathrm{K}].$ For $j\ge 5$, let

$$\left\{\begin{array}{cc}\hfill & Y_t^j={\xi }_T^2+{\int }_t^T{f}^2(s,Y_s^j,Z_s^j,Y_{t+\delta (t)}^{j-1},Z_{t+{\zeta }^2(t)}^2)\mathrm{d}s+{\int }_t^T{G}^2(s,Y_s^j,Z_s^j)\mathrm{d}{\overleftarrow{B}}_s-{\int }_t^T{Z}_s^j\mathrm{d}{W}_s,t\in [0,T],\hfill \\ \hfill & Y_t^j={\xi }_t^2,Z_t^j={\eta }_t^2,t\in [T,T+K].\hfill \end{array}\right.$$

According to Lemma 4 and by induction, we can get $Y_t^4⩾Y_t^5\ge \cdots \ge Y_t^{j-1}\ge Y_t^j,a.s.,\mathrm{for}\mathrm{all}\mathrm{t}\in [0,\mathrm{T}+\mathrm{K}],$ hence, for all $j\ge 3$

$$Y_t^1⩾Y_t^j,a.s.,\mathrm{for}\mathrm{all}\mathrm{t}\in [0,\mathrm{T}+\mathrm{K}].$$

(14)

For any $j\ge 4,\beta >0$, apply It$\widehat{\mathrm{o}}$’s formula to $e^{\beta t}{|Y_t^j|}^2$, in view of (H3), Young’s inequality, Jensen’s inequality and for $i=1,2$, ${\rho }_i(t,u)$ has a linear growth, we have

$$\begin{array}{cc}\hfill \mathbb{E}|Y_t^j{|}^2{e}^{\beta t}+(1-{\alpha }_1-\frac{{\alpha }_1+2C}{\beta })\mathbb{E}{\int }_t^T{|Z_s^j|}^2{e}^{\beta s}ds\le & C_1+(1+\frac{3}{\beta })\mathbb{E}{\int }_t^T{b}_1(s)e^{\beta s}\mathbb{E}[|Y_s^j{|}^2]ds\hfill \\ \hfill & +\frac{2M}{\beta }\mathbb{E}{\int }_t^T{b}_2(s)e^{\beta s}\mathbb{E}[|Y_s^{j-1}{|}^2]ds,\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill C_1=& \mathbb{E}[e^{\beta T}|{\xi }_T^2{|}^2]+(1+\frac{3}{\beta }){\int }_0^T{a}_1(s)e^{\beta s}\mathrm{d}s+({\alpha }_2+\frac{{\alpha }_2+2C}{\beta })M\mathbb{E}{\int }_0^{T+K}{|Z_s^2|}^2{e}^{\beta s}ds\hfill \\ \hfill & +\frac{2M}{\beta }\mathbb{E}{\int }_T^{T+K}{b}_2(s)e^{\beta s}\mathbb{E}[|{\xi }_s^2{|}^2]ds+(1+\frac{1}{\beta })M{\int }_0^{T+K}{\rho }_2(s,\mathbb{E}[|Y_s^2{|}^2])e^{\beta s}ds\hfill \\ \hfill & +\frac{2M}{\beta }{\int }_0^{T+K}{a}_2(s)e^{\beta s}ds+\frac{2}{\beta }\mathbb{E}{\int }_0^T|f^2{(s,0,0,0,0)|}^2{e}^{\beta s}\mathrm{d}s+(1+\beta ){\int }_0^T{|g^2(s,0,0,0,0)|}^2{e}^{\beta s}ds.\hfill \end{array}$$

Now let us choose $\beta =\frac{{\alpha }_1+2C}{1-{\alpha }_1}$. Then, there exists $C_2>0$, which is independent of j, such that for $j⩾4$,

$$\mathbb{E}[|Y_t^j{|}^2]⩽C_2+C_2\mathbb{E}{\int }_t^T{b}_1(s)|Y_s^j{|}^2\mathrm{d}s+C_2\mathbb{E}{\int }_t^T{b}_2(s){|Y_s^{j-1}|}^2\mathrm{d}s,$$

which leads to

$$\underset{j⩾4}{sup}\mathbb{E}[|Y_t^j{|}^2]⩽C_2+C_2\mathbb{E}{\int }_0^T{b}_2(s)|Y_t^3{|}^2\mathrm{d}t+C_2{\int }_t^T(b_1(s)+b_2(s))\underset{j⩾4}{sup}\mathbb{E}[|Y_s^j{|}^2]\mathrm{d}s,$$

Then Gronwall’s inequality yields

$$\underset{j⩾4}{sup}\mathbb{E}[|Y_t^j{|}^2]⩽C_2{e}^{C_2{\int }_t^T(b_1(s)+b_2(s))\mathrm{d}s}[1+\mathbb{E}{\int }_0^T{b}_2(s)|Y_t^3{|}^2\mathrm{d}t],$$

which implies

$$\underset{0⩽t⩽T}{sup}\underset{j⩾4}{sup}\mathbb{E}[|Y_t^j{|}^2]⩽C_2{e}^{C_2{\int }_0^T(b_1(s)+b_2(s))\mathrm{d}s}[1+{\int }_0^T{b}_2(s)\mathbb{E}[\underset{0⩽t⩽T}{sup}|Y_t^3{|}^2]\mathrm{d}t]<\infty .$$

(15)

For $j⩾4,p⩾1$, Set

$$\begin{array}{cc}\hfill & {\widehat{Y}}_t^j=Y_t^{j+p}-Y_t^j,{\widehat{Z}}_t^j=Z_t^{j+p}-Z_t^j,\hfill \\ \hfill & {\widehat{F}}_t^j=f^2(t,Y_t^{j+p},Z_t^{j+p},Y_{t+\delta (t)}^{j+p-1},Z_{t+{\zeta }^2(t)}^2)-f^2(t,Y_t^j,Z_t^j,Y_{t+\delta (t)}^{j-1},Z_{t+{\zeta }^2(t)}^2),\hfill \\ \hfill & {\widehat{G}}_t^j=G^2(t,Y_t^{j+p},Z_t^{j+p})-G^2(t,Y_t^j,Z_t^j),\hfill \end{array}$$

Then for $j\ge 4$, $({\widehat{Y}}_{·}^j,{\widehat{Z}}_{·}^j)$ satisfies

$$\left\{\begin{array}{cc}\hfill & {\widehat{Y}}_t^j={\int }_t^T{\widehat{F}}_s^j\mathrm{d}s+{\int }_t^T{\widehat{G}}_s^j\mathrm{d}{\overleftarrow{B}}_s-{\int }_t^T{Z}_s^j\mathrm{d}{W}_s,t\in [0,T],\hfill \\ \hfill & {\widehat{Y}}_t^j=0,{\widehat{Z}}_t^j=0,t\in [T,T+K].\hfill \end{array}\right.$$

Applying It$\widehat{\mathrm{o}}$’s formula to $e^{\beta t}{|{\widehat{Y}}_t^j|}^2$ where $\beta >0$, and from the assumption (H3), Young’s inequality, Jensen’s inequality, we can obtain

$$\begin{array}{cc}\hfill \mathbb{E}|{\widehat{Y}}_t^j{|}^2{e}^{\beta t}+\mathbb{E}{\int }_t^T{|{\widehat{Z}}_s^j|}^2{e}^{\beta s}ds\le & \frac{1}{\beta }{\int }_t^T({\rho }_1(s,\mathbb{E}[|{\widehat{Y}}_s^j{|}^2])+M{\rho }_2(s,\mathbb{E}[|{\widehat{Y}}_s^{j-1}{|}^2]))e^{\beta s}ds\hfill \\ \hfill & +(\frac{C}{\beta }+{\alpha }_1)\mathbb{E}{\int }_t^T|{\widehat{Z}}_s^j{|}^2{e}^{\beta s}ds+{\int }_t^T{\rho }_1(s,\mathbb{E}[|{\widehat{Y}}_s^j{|}^2])e^{\beta s}ds.\hfill \end{array}$$

Now let us choose $\beta =\frac{2C}{1-{\alpha }_1}$. Then there exists $C_3>0$, which is independent of $j,p$, such that

$$\mathbb{E}[|{\widehat{Y}}_t^j{|}^2]+\mathbb{E}{\int }_t^T|{\widehat{Z}}_s^j{|}^2\mathrm{d}s⩽C_3{\int }_t^T({\rho }_1(s,\mathbb{E}[|{\widehat{Y}}_s^j{|}^2])+M{\rho }_2(s,\mathbb{E}[|{\widehat{Y}}_s^{j-1}{|}^2]))ds.$$

(16)

Denote $h(t)={lim\; sup}_{k\to \infty }\mathbb{E}[|{\widehat{Y}}_t^j{|}^2]$. Then in view of (15) we have $0⩽h(t)<\infty$. Then, apply Fatou’s Lemma to the right-hand side of (16), we obtain

$$h(t)⩽C_3{\int }_t^T({\rho }_1(s,h(s))+M{\rho }_2(s,h(s)))\mathrm{d}s.$$

Using the same proof method about the uniqueness in Theorem 1, we can obtain $h(t)=0$, that is ${lim\; sup}_{j\to \infty }\mathbb{E}[|{\widehat{Y}}_t^j{|}^2]=0$, for all $t\in [0,T]$. Moreover, from (16) it follows that

$$\underset{0⩽t⩽T}{sup}\mathbb{E}[|{\widehat{Y}}_t^j{|}^2]⩽C_3{\int }_0^T(\rho (\mathbb{E}[|{\widehat{Y}}_s^j{|}^2])+\rho (\mathbb{E}[|{\widehat{Y}}_s^{j-1}{|}^2]))\mathrm{d}s$$

Applying Fatou’s Lemma again leads to

$$\underset{0⩽t⩽T}{sup}\mathbb{E}[|{\widehat{Y}}_t^j{|}^2]\to 0\mathrm{as}j\to \infty .$$

By taking $j\to \infty$ in (16), we use Fatou’s Lemma again to obtain

$$\mathbb{E}{\int }_0^T{|{\widehat{Z}}_t^j|}^2\mathrm{d}t\to 0\mathrm{as}j\to \infty .$$

Hence, we have shown that $(Y_{·}^k,Z_{·}^k)$ is a Cauchy sequence in $S^2(0,T;\mathbb{R})\times M^2(0,T;{\mathbb{R}}^d)$, thus it is also a Cauchy sequence in $S^2(0,T+K;\mathbb{R})\times M^2(0,T+K;{\mathbb{R}}^d)$. As a consequence, we can find $(Y_{·},Z_{·})\in S^2(0,T+K;\mathbb{R})\times M^2(0,T+K;{\mathbb{R}}^d)$ such that $Y_t={\xi }_t^2$, $Z_t={\eta }_t^2$ for $T⩽t⩽T+K$ and

$$\underset{0⩽t⩽T}{sup}\mathbb{E}[|Y_t^j-Y_t{|}^2]+\mathbb{E}[{\int }_0^T|Z_t^j-Z_t{|}^2\mathrm{d}t]\to 0\mathrm{as}j\to \infty .$$

Therefore, it’s easy to verify that, as $j\to \infty$,

$$\mathbb{E}{\int }_t^T{|f^2(s,Y_s^j,Z_s^j,Y_{s+\delta (s)}^{j-1},Z_{s+{\zeta }^2(s)}^2)-f^2(s,Y_s,Z_s,Y_{s+\delta (s)},Z_{s+{\zeta }^2(s)}^2)|}^2\mathrm{d}s\to 0,$$

$$\mathbb{E}{\int }_t^T{|g^2(s,Y_s^j,Z_s^j,Y_{s+{\overline{\delta }}^2(s)}^2,Z_{s+{\overline{\zeta }}^2(s)}^2)-g^2(s,Y_s,Z_s,Y_{s+{\overline{\delta }}^2(s)}^2,Z_{s+{\overline{\xi }}^2(s)}^2)|}^2\mathrm{d}s\to 0,$$

$$\mathbb{E}[|{\int }_t^T{Z}_s^j·\mathrm{d}{W}_s-{\int }_t^T{Z}_s·\mathrm{d}{W}_s{|}^2]\to 0.$$

So, we conclude that $(Y_{·},Z_{·})$ solves the following ABDSDE:

$$\left\{\begin{array}{cc}\hfill & Y_t={\xi }_T^2+{\int }_t^T{f}^2(s,Y_s,Z_s,Y_{s+\delta (s)},Z_{s+{\zeta }^2(s)}^i)\mathrm{d}s\hfill \\ \hfill & +{\int }_t^T{g}^2(s,Y_s,Z_s,Y_{s+{\overline{\delta }}^2(s)}^2,Z_{s+{\overline{\zeta }}^2(s)}^2)\mathrm{d}{\overleftarrow{B}}_s-{\int }_t^T{Z}_s\mathrm{d}{W}_s,t\in [0,T],\hfill \\ \hfill & Y_t={\xi }_t^2,Z_t={\eta }_t^2,t\in [T,T+K].\hfill \end{array}\right.$$

Therefore, we can infer that $Y_t=Y_t^2,a.s.,\mathrm{for}\mathrm{all}t\in [0,T+K]$ from the uniqueness part of Theorem 1. Let $j\to \infty$ in (14) yields $Y_t^1\ge Y_t^2,a.s.,\mathrm{for}\mathrm{all}t\in [0,T+K]$. □

Example 2.

Let $f^1(t,\tilde{y},\tilde{z},\tilde{\theta }(r),\tilde{\vartheta }(\overline{r}))=|\tilde{y}|+|\tilde{z}|+{\mathbb{E}}^{{\mathcal{F}}_t}[|\tilde{\theta }(r)|]+{\mathbb{E}}^{{\mathcal{F}}_t}[\arctan \tilde{\vartheta }(\overline{r})]$, $f^2(t,\tilde{y},\tilde{z},$$\tilde{\theta }(r),\tilde{\vartheta }(\overline{r}))=\tilde{y}-|\tilde{z}|+{\mathbb{E}}^{{\mathcal{F}}_t}[\tilde{\theta }(r)]-\frac{\pi }{2}$, $g^1(t,\tilde{y},\tilde{z},\tilde{\theta }(r),\tilde{\vartheta }(\overline{r}))=g^2(t,\tilde{y},\tilde{z},\tilde{\theta }(r),\tilde{\vartheta }(\overline{r}))=\mathrm{arccot}\tilde{y}+|\tilde{z}|+{\mathbb{E}}^{{\mathcal{F}}_t}[\tilde{\theta }(r)+\tilde{\vartheta }(\overline{r})]$. Then by Theorem 3, we can derive $Y_t^1⩾Y_t^2$, a.s., $\mathrm{for}\mathrm{any}t\in [0,T+K]$ when the assumption (1) is satisfied.

5. Conclusions

The purpose of this paper is to introduce and study a type of anticipated BDSDEs with non-Lipschitze coefficients. We first show that the adapted solution of this kind of ABDSDEs is existent and unique. Furthermore, we give two comparison theorems one dimensional situation. In our future publications, we will concentrate on investigating this interesting problem and pay much attention to the application of this kind of equation, especially in control such as [20,21,22,23,24,25].