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Anticipated backward doubly stochastic differential equations with non-Liphschitz coefficients

   | Mar 04, 2019

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Introduction

Backward stochastic differential equations (BSDEs in short) were first introduced by Pardoux and Peng [4]. They proved an existence and uniqueness result under Lipschitz condition. Since then many efforts have been made in relaxing the Lipschitz assumption of the generator of the BSDEs (see among others Mao [3] and Wang and Huang [7]). Few years later, the same authors considered in [5] a new type of BSDEs, that is a class of backward doubly stochastic differential equations (BDSDEs in short) with two different directions of stochastic integrals. These equations are extensively used in the study of stochastic partial differential equations (SPDEs). Their link with SPDEs in the case of Lipchitzian drift was established in [5]. The key point of solvency of such equations is the martingale representation theorem. In this spirit, Bally and Matoussi [1] gave the probabilistic representation of the solution in Sobolev space of semilinear SPDEs in terms of BDSDEs.

On the other hand, Peng and Yang [6] introduced the following type of anticipated backward stochastic differential equations (ABSDEs in short)

{ dYt=f(t,Yt,Zt,Yt+δ(t),Zt+ζ(t))dtZtdWt,0tT,Yt=ξt,Zt=ηt,TtT+K, $$\left\{ \begin{array}{*{35}{l}}-d{{Y}_{t}}=f\left( t,{{Y}_{t}},{{Z}_{t}},{{Y}_{t+\delta \left( t \right)}},{{Z}_{t+\zeta \left( t \right)}} \right)dt-{{Z}_{t}}d{{W}_{t}},\ \ \ \ \ \ \ 0\le t\le T, \\{{Y}_{t}}={{\xi }_{t}},\ \ {{Z}_{t}}={{\eta }_{t}},\ \ \ \ \ \ \ \ \ \ T\le t\le T+K, \\\end{array} \right.$$

where δ and ζ are given nonnegative deterministic functions. In these equations, the generator includes not only the values of solutions of the present but also the future. In [6], the authors obtained the existence and uniqueness of the solution of ABSDE under Lipschitz assumption, gave the comparison theorem for one dimensional ABSDEs and finally they solved a stochastic control problem by showing the duality between linear stochastic differential delay equations and ABSDEs.

This paper is devoted to the following anticipated BDSDE

{ Yt=ξT+tTf(r,Yr,Zr,Yr+δ(r),Zr+ζ(r))dr+tTg(r,Yr,Zr)dBrtTZrdWr,0tT,Yt=ξt,Zt=ηt,TtT+K $$\left\{ \begin{array}{*{35}{l}}{{Y}_{t}}={{\xi }_{T}}+\int_{t}^{T}{f\left( r,{{Y}_{r}},{{Z}_{r}},{{Y}_{r+\delta \left( r \right)}},{{Z}_{r+\zeta \left( r \right)}} \right)dr+\int_{t}^{T}{g\left( r,{{Y}_{r}},{{Z}_{r}} \right)d{{B}_{r}}-\int_{t}^{T}{{{Z}_{r}}d{{W}_{r}},\ \ \ 0\le t\le T,}}} \\{{Y}_{t}}={{\xi }_{t}},\ \ \ {{Z}_{t}}={{\eta }_{t}},\ \ \ \ \ \ \ \ \ \ T\le t\le T+K \\\end{array} \right.$$

where K is a positive constant, x., η: are given stochastic processes and δ,ζ : [0,T]R+ are continuous functions satisfying:

(A1) : t +δ(t) ≤ T +K, t +ζ (t) ≤ T +K:

(A2) : There exists M ≥ 0 such that for 0 ≤ t ≤ T and non negative integrable function h,

tTh(r+ϕ(r))drMtT+Kh(r)dr,ϕ{ δ,ζ }. $$\int_{t}^{T}{h\left( r+\phi \left( r \right) \right)}dr\le M\int_{t}^{T+K}{h\left( r \right)dr,\ \ \ \phi \in \left\{ \delta ,\zeta \right\}}.$$

The paper is organized as follows. In section 2, we study first solvability of our equation in the case of Lipschtzian coefficients. Using this result, in section 3 we prove existence and uniqueness of solution with coefficients satisfying rather weaker conditions.

Preliminaries

Let Ω be a non-empty set, a σ‒algebra of sets of Ω and P a probability measure defined on . The triplet (Ω,ℱ, P) defines a probability space, which is assumed to be complete. For a fix real 0 < T ≤ ∞, we assume given two mutually independent processes:

an 𝓁‒dimensional Brownian motion (Bt)0≤t≤T,

a δ ‒dimensional Brownian motion (Wt)0≤t≤T.

We consider the family (t)0≤t≤T given by

Ft=FtWFt,TB,0tT,Gs=F0,sWFs,T+kB,0sT+K, $${{F}_{t}}=F_{t}^{W}\vee F_{t,T}^{B},\ \ \ \ \ \ 0\le t\le T,\ \ \ \ \ \ {{G}_{s}}=F_{0,s}^{W}\vee F_{s,T+k}^{B},\ \ \ \ \ \ 0\le s\le T+K,$$

where for any process { φt }t0,Fs,tφ=σ{ φrφs,srt }N,Ftφ=F0,tφ. ${{\left\{ {{\varphi }_{t}} \right\}}_{t\ge 0}},F_{s,t}^{\varphi }=\sigma \left\{ {{\varphi }_{r}}-{{\varphi }_{s}},s\le r\le t \right\}\vee N,F_{t}^{\varphi }=F_{0,t}^{\varphi }.$Here 𝒩 denotes the class of Pnull sets of F. Note that (Ft)0≤t≤T does not constitute a classical filtration.

For k ∈ N* we consider the following sets (where E denotes the mathematical expectation with respect to the probability measure P):

L2(T,Rk) the space of T -measurable random variable such that E[ | ξT |2 ]<+. $\mathbf{E}\left[ {{\left| {{\xi }_{T}} \right|}^{2}} \right]<+\infty .$

I[ 0,T ]2(G,Rk) $I_{\left[ 0,T \right]}^{2}\left( G,{{\text{R}}^{k}} \right)$the space of tadapted càdlàg processes

Ψ:[ 0,T ]×ΩRk, Ψ I2(Rk)2=E(sup0tT| Ψt |2)<. $$\Psi :\left[ 0,T \right]\times \Omega \to {{\text{R}}^{k}},\left\| \Psi \right\|_{{{I}^{2}}\left( {{\mathbf{R}}^{k}} \right)}^{2}=\mathbf{E}\left( \underset{0\le t\le T}{\mathop{\sup }}\,{{\left| {{\Psi }_{t}} \right|}^{2}} \right)<\infty .$$

M[ 0,T ]2(G,Rk×d) $M_{\left[ 0,T \right]}^{2}\left( G,{{\mathbf{R}}^{k\times d}} \right)$the space of tprogressively measurable processes

Ψ:[ 0,T ]×ΩRk×d, Ψ M2(Rk×d)2=EoT Ψt 2dt<. $$\Psi :\left[ 0,T \right]\times \Omega \to {{\mathbf{R}}^{k\times d}},\left\| \Psi \right\|_{{{M}^{2}}\left( {{\mathbf{R}}^{k\times d}} \right)}^{2}=\mathbf{E}\int_{o}^{T}{{{\left\| {{\Psi }_{t}} \right\|}^{2}}dt<\infty .}$$

CG2(0,T)=M[ 0,T ]2(G,Rk)×M[ 0.T ]2(G,Rk×d) $C_{G}^{2}\left( 0,T \right)=M_{\left[ 0,T \right]}^{2}\left( G,{{\mathbf{R}}^{k}} \right)\times M_{\left[ 0.T \right]}^{2}\left( G,{{\mathbf{R}}^{k\times d}} \right)$endowed with the norm

(Y,Z) CG2(0,T)2= Y M2(Rk)2+ Z M2(Rk×d)2. $$\left\| \left( Y,Z \right) \right\|_{C_{G}^{2}\left( 0,T \right)}^{2}=\left\| Y \right\|_{{{M}^{2}}\left( {{\mathbf{R}}^{k}} \right)}^{2}+\left\| Z \right\|_{{{M}^{2}}\left( {{\mathbf{R}}^{k\times d}} \right)}^{2}.$$

BG2(0,T)=I[ 0,T ]2(G,Rk)×M[ 0,T ]2(G,Rk×d) $B_{G}^{2}\left( 0,T \right)=I_{\left[ 0,T \right]}^{2}\left( G,{{\text{R}}^{k}} \right)\times M_{\left[ 0,T \right]}^{2}\left( G,{{\mathbf{R}}^{k\times d}} \right)$endowed with the norm

(Y,Z) BG2(0,T)2= Y S2(Rk)2+ Z 2M2(Rk×d). $$\left\| \left( Y,Z \right) \right\|_{B_{G}^{2}\left( 0,T \right)}^{2}=\left\| Y \right\|_{{{S}^{2}}\left( {{\mathbf{R}}^{k}} \right)}^{2}+{{\left\| Z \right\|}^{2}}_{{{M}^{2}}\left( {{\mathbf{R}}^{k\times d}} \right)}.$$

S be the set of all nondecreasing, continuous and concave function ρ(·) : R+ R+ satisfying ρ(0) = 0, ρ(s)>0fors>0and 0+duρ(u)=+. $\rho \left( s \right)>0\ \text{for}\ s\ >\ 0\ \text{and }\int_{0+}{\frac{du}{\rho \left( u \right)}=+\infty .}$

Remark 2.1. For any ρ ∈ S, we can find a pair of positive constants a and b such that ρ(ν) ≤ a +bν for all ν ≥ 0.

We denote A = Ω×[0,T]×Rk ×Rk×d, f(r,0) = h(r,0,0,0,0), for all x,y ∈ Rk |x| the Euclidean norm of x and denote by 〈x,y〉 the Euclidean inner product.

Definition 2.2. A pair of processes (Y,Z) is called a solution to ABDSDE (1.1), if(Y,Z)BG2(0,T+K) $if\left( Y,Z \right)\in \ B_{G}^{2}\left( 0,T+K \right)$and it satisfies eq (1.1).

First we investigate the case of lipschitz coefficients.

The case of Lipschitz coefficients

In this subsection, we will mainly study the existence and uniqueness of the solution to ABDSDE (1.1) with Lipschitz coefficients. For this purpose, we first make the following assumptions.

Assumptions

In the following, we assume that there exists ρ ∈ S such that f and g satisfy assumptions (H1).

(H1.1): There exists a constant c > 0 such that

f t , y , z , θ r , φ r f t , y , z , θ r , φ r 2 c y y 2 + z z 2 + E F t θ r θ r 2 + φ r φ r 2 , $$\begin{align} & {{\left| f\left( t,y,z,\theta \left( r \right),\varphi \left( r \right) \right)-f\left( t,{y}',{z}',{\theta }'\left( r \right),{\varphi }'\left( r \right) \right) \right|}^{2}}\le c\left( {{\left| y-{y}' \right|}^{2}}+{{\left| z-{z}' \right|}^{2}} \right) \\ & +{{\mathbf{E}}^{{{F}_{t}}}}\left[ {{\left| \theta \left( r \right)-{\theta }'\left( r \right) \right|}^{2}}+{{\left| \varphi \left( r \right){{-}\varphi'}\left( r \right) \right|}^{2}} \right], \\ \end{align}$$

for all r , r t , T + K , t , y , z , θ r , φ r , t , y , z , θ r , φ r A × C G 2 t , T + K . $\left( r,r' \right)\in \left[ t,T+K \right],\left( t,y,z,\theta \left( r \right),\varphi \left( r \right) \right),\left( t,{y}',{z}',{\theta }'\left( r \right),{\varphi }'\left( r \right) \right)\in A\times C_{G}^{2}\left( t,T+K \right).$

(H1.2): There exists a constant 0 < α1 < 1 such that for any (t,y, z), (t,y′, z′) [0,T]×Rk ×Rk×d

| g(t,y,z)g(t,y,z) |2c| yy |2+α1| zz |2. $${{\left| g\left( t,y,z \right)-g\left( t,{y}',{z}' \right) \right|}^{2}}\le c{{\left| y-{y}' \right|}^{2}}+{{\alpha }_{1}}{{\left| z-{z}' \right|}^{2}}.$$

(H1.3): For any (t,y,z) [0,T]×Rk ×Rk×d,

E[ oT| f(s,0) |2ds ]+E[ oT| g(s,y,z) |2ds ]<. $$\mathbf{E}\left[ \int_{o}^{T}{{{\left| f\left( s,0 \right) \right|}^{2}}ds} \right]+\mathbf{E}\left[ \int_{o}^{T}{{{\left| g\left( s,y,z \right) \right|}^{2}}ds} \right]<\infty .$$

Existence and uniqueness of solution
Lemma 3.1

Suppose that (Yt,Zt)0tTCG2(0,T+K) ${{\left( {{Y}_{t}},{{Z}_{t}} \right)}_{0\le t\le T}}\in C_{G}^{2}\left( 0,T+K \right)$is the unique solution to the ABDSDE (1.1). Then YS[ 0,T ]2(G,Rk). $Y\in S_{\left[ 0,T \right]}^{2}\left( G,{{\mathbf{R}}^{k}} \right).$

Proof. Itô’s formula applied to eq (1.1) yields, for 0 ≤ t ≤ T

| Yt |2+tT| Zr |2dr=| ξT |2+2tT Yr,f(r,Yr,Zr,Yr+δ(r),Zr+ζ(r)) dr+2tT Yr,g(r,Yr,Zr)dBr 2tT Yr,ZrdWr +tT| g(r,Yr,Zr) |2dr. $$\begin{align}& {{\left| {{Y}_{t}} \right|}^{2}}+\int_{t}^{T}{{{\left| {{Z}_{r}} \right|}^{2}}dr={{\left| {{\xi }_{T}} \right|}^{2}}+2\int_{t}^{T}{\left\langle {{Y}_{r}},f\left( r,{{Y}_{r}},{{Z}_{r}},{{Y}_{r+\delta \left( r \right)}},{{Z}_{r+\zeta \left( r \right)}} \right) \right\rangle dr+2\int_{t}^{T}{\left\langle {{Y}_{r}},g\left( r,{{Y}_{r}},{{Z}_{r}} \right)d{{B}_{r}} \right\rangle }}} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -2\int_{t}^{T}{\left\langle {{Y}_{r}},{{Z}_{r}}d{{W}_{r}} \right\rangle +\int_{t}^{T}{{{\left| g\left( r,{{Y}_{r}},{{Z}_{r}} \right) \right|}^{2}}dr.}} \\ \end{align}$$

Using the fact that 2ab ≤ εa2+b2 for ε > 0 and assumptionn (H1.1), we deduce that

2 E t T Y r , f r , Y r , Z r , Y r + δ r , Z r + ζ r d r 1 ε E t T Y r 2 d r + ε c E t T Y r 2 + Z r 2 d r + ε E t T E F t Y r + δ r 2 + Z r + ζ r 2 d r + 2 E t T Y r f r , 0 d r . $$\begin{align}& 2\mathbf{E}\int_{t}^{T}{\left\langle {{Y}_{r}},f\left( r,{{Y}_{r}},{{Z}_{r}},{{Y}_{r+\delta \left( r \right)}},{{Z}_{r+\zeta \left( r \right)}} \right) \right\rangle dr\le \frac{1}{\varepsilon }\mathbf{E}\int_{t}^{T}{{{\left| {{Y}_{r}} \right|}^{2}}dr+\varepsilon c\mathbf{E}\int_{t}^{T}{\left( {{\left| {{Y}_{r}} \right|}^{2}}+{{\left| {{Z}_{r}} \right|}^{2}} \right)}}dr} \\ & +\varepsilon \mathbf{E}\int_{t}^{T}{{{\mathbf{E}}^{{{F}_{t}}}}\left[ {{\left| {{Y}_{r+\delta \left( r \right)}} \right|}^{2}}+{{\left| {{Z}_{r+\zeta \left( r \right)}} \right|}^{2}} \right]}dr+2\mathbf{E}\int_{t}^{T}{\left| {{Y}_{r}} \right|\left| f\left( r,0 \right) \right|dr.} \\ \end{align}$$

Applying (A2), we obtain finally

2EtT Yr,f(r,Yr,Zr,Yr+δ(r),Zr+ζ(r)) dr(1ε+ε(c+M))EtT| Yr |2dr+ε(c+M)EtT| Zr |2dr+2EtT| Yr || f(r,0) |dr+εMETT+K(| ξr |2+| ηr |2)dr. $$\begin{align}& 2\mathbf{E}\int_{t}^{T}{\left\langle {{Y}_{r}},f\left( r,{{Y}_{r}},{{Z}_{r}},{{Y}_{r+\delta \left( r \right)}},{{Z}_{r+\zeta \left( r \right)}} \right) \right\rangle dr\le \left( \frac{1}{\varepsilon }+\varepsilon \left( c+M \right) \right)\mathbf{E}\int_{t}^{T}{{{\left| {{Y}_{r}} \right|}^{2}}dr}} \\ & \ \ \ \ \ \ \ \ \ +\varepsilon \left( c+M \right)\mathbf{E}\int_{t}^{T}{{{\left| {{Z}_{r}} \right|}^{2}}dr}+2\mathbf{E}\int_{t}^{T}{\left| {{Y}_{r}} \right|\left| f\left( r,0 \right) \right|dr+\varepsilon M\mathbf{E}\int_{T}^{T+K}{\left( {{\left| {{\xi }_{r}} \right|}^{2}}+{{\left| {{\eta }_{r}} \right|}^{2}} \right)dr.}} \\ \end{align}$$

In addition, for any 0 ≤ t ≤ T, we have

EtT| g(r,Yr,Zr) |2drEtT| g(r,Yr,Zr)g(r,0,0) |2dr+EtT| g(r,0,0) |2drcEtT| Yr |2dr+α1EtT| Zr |2dr+EtTg(r,0,0)|2dr. $$\begin{align}& \mathbf{E}\int_{t}^{T}{{{\left| g\left( r,{{Y}_{r}},{{Z}_{r}} \right) \right|}^{2}}dr\le \mathbf{E}\int_{t}^{T}{{{\left| g\left( r,{{Y}_{r}},{{Z}_{r}} \right)-g\left( r,0,0 \right) \right|}^{2}}dr+\mathbf{E}\int_{t}^{T}{{{\left| g\left( r,0,0 \right) \right|}^{2}}dr}}} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,\le c\mathbf{E}\int_{t}^{T}{{{\left| {{Y}_{r}} \right|}^{2}}dr+{{\alpha }_{1}}}\mathbf{E}\int_{t}^{T}{{{\left| {{Z}_{r}} \right|}^{2}}dr+\mathbf{E}\int_{t}^{T}{{{\left. g\left( r,0,0 \right) \right|}^{2}}}dr.} \\ \end{align}$$

Putting pieces together, we deduce from (3.1) that

E t T Z r 2 d r E ξ T 2 + M ε E T T + K ξ r 2 + η r 2 d r + 1 ε + ε c + M + c E t T Y r 2 d r + 2 E t T Y r f r , 0 d r + E t T g r , 0 , 0 2 d r + α 1 + ε c + M E t T Z r 2 d r . $$\begin{align}& \mathbf{E}\int_{t}^{T}{{{\left| {{Z}_{r}} \right|}^{2}}dr\le \mathbf{E}\left[ {{\left| {{\xi }_{T}} \right|}^{2}} \right]+M\varepsilon \mathbf{E}\int_{T}^{T+K}{\left( {{\left| {{\xi }_{r}} \right|}^{2}}+{{\left| {{\eta }_{r}} \right|}^{2}} \right)}dr+\left( \frac{1}{\varepsilon }+\varepsilon \left( c+M \right)+c \right)\mathbf{E}\int_{t}^{T}{{{\left| {{Y}_{r}} \right|}^{2}}dr}} \\ & +2\mathbf{E}\int_{t}^{T}{\left| {{Y}_{r}} \right|\left| f\left( r,0 \right) \right|dr+\mathbf{E}\int_{t}^{T}{{{\left| g\left( r,0,0 \right) \right|}^{2}}dr+\left( {{\alpha }_{1}}+\varepsilon \left( c+M \right) \right)\mathbf{E}\int_{t}^{T}{{{\left| {{Z}_{r}} \right|}^{2}}dr.}}} \\ \end{align}$$

If we choose ε = ε0 satisfying 1/C0 = (1[α1+ε0(c+M)])1 > 0, we deduce that

EtT| Zr |2dr1C0E[ Xt+2tT| Yr || f(r,0) |dr+tT| g(r,0,0) |2dr ] $$\mathbf{E}\int_{t}^{T}{{{\left| {{Z}_{r}} \right|}^{2}}dr}\le \frac{1}{{{C}_{0}}}\mathbf{E}\left[ {{X}_{t}}+2\int_{t}^{T}{\left| {{Y}_{r}} \right|\left| f\left( r,0 \right) \right|dr+\int_{t}^{T}{{{\left| g\left( r,0,0 \right) \right|}^{2}}dr}} \right]$$

where putting C1=(1ε0+ε0(c+M+c)), ${{C}_{1}}=\left( \frac{1}{{{\varepsilon }_{0}}}+{{\varepsilon }_{0}}\left( c+M+c \right) \right),$

Xt=[ | ξT |2+Mε0TT+k(| ξr |2+| ηr |2)dr+C1tT| Yr |2dr ]. $${{X}_{t}}=\left[ {{\left| {{\xi }_{T}} \right|}^{2}}+M{{\varepsilon }_{0}}\int_{T}^{T+k}{\left( {{\left| {{\xi }_{r}} \right|}^{2}}+{{\left| {{\eta }_{r}} \right|}^{2}} \right)dr+{{C}_{1}}\int_{t}^{T}{{{\left| {{Y}_{r}} \right|}^{2}}dr}} \right].$$

By the same computations as before, we have

E[ | ξT |2 ]+2EtT Yr,f(r,Yr,Zr,Yr+δ(r),Zr+ζ(r)) dr+EtT| g(r,Yr,Zr) |2dr1C0E[ Xt+2tT| Yr || f(r,0) |dr+tT| g(r,0,0) |2dr ]. $$\begin{align}& \mathbf{E}\left[ {{\left| {{\xi }_{T}} \right|}^{2}} \right]+2\mathbf{E}\int_{t}^{T}{\left\langle {{Y}_{r}},f\left( r,{{Y}_{r}},{{Z}_{r}},{{Y}_{r+\delta \left( r \right)}},{{Z}_{r+\zeta \left( r \right)}} \right) \right\rangle }dr+E\int_{t}^{T}{{{\left| g\left( r,{{Y}_{r}},{{Z}_{r}} \right) \right|}^{2}}dr} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \le \frac{1}{{{C}_{0}}}\mathbf{E}\left[ {{X}_{t}}+2\int_{t}^{T}{\left| {{Y}_{r}} \right|\left| f\left( r,0 \right) \right|dr+{{\int_{t}^{T}{\left| g\left( r,0,0 \right) \right|}}^{2}}dr} \right]. \\ \end{align}$$

Moreover using again eq (3.1), we have

E sup t r T Y r 2 E ξ T 2 + 2 E sup t r T s T Y r , f r , Y r , Z r , Y r + δ r , Z r + ζ r d r + 2 E sup t s T s T Y r , g r , Y r , Z r d B r + 2 E sup t s T s T Y r , Z r d W r + t T g r , Y r , Z r 2 d r . $$\begin{align} & \mathbf{E}\left( \underset{t\le r\le T}{\mathop{\sup }}\,{{\left| {{Y}_{r}} \right|}^{2}} \right)\le \mathbf{E}\left[ {{\left| {{\xi }_{T}} \right|}^{2}} \right]+2\text{E}\underset{t\le r\le T}{\mathop{\sup }}\,\left( \int_{s}^{T}{\left\langle {{Y}_{r}},f\left( r,{{Y}_{r}},{{Z}_{r}},{{Y}_{r+\delta \left( r \right),}}{{Z}_{r+\zeta \left( r \right)}} \right) \right\rangle dr} \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+2\mathbf{E}\underset{t\le s\le T}{\mathop{\sup }}\,\left| \int_{s}^{T}{\left\langle {{Y}_{r}},g\left( r,{{Y}_{r}},{{Z}_{r}} \right)d{{B}_{r}} \right\rangle } \right|+2\mathbf{E}\underset{t\le s\le T}{\mathop{\sup }}\,\left| \int_{s}^{T}{\left\langle {{Y}_{r}},{{Z}_{r}}d{{W}_{r}} \right\rangle } \right| \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\int_{t}^{T}{{{\left| g\left( r,{{Y}_{r}},{{Z}_{r}} \right) \right|}^{2}}}dr. \\ \end{align}$$

By Burkhölder-Davis-Gundy inequality, there exists a constant C > 0 which may vary from line to line such that

E sup t s T s T Y r , g r , Y r , Z r d B r 1 8 E sup t r T Y r 2 + C t T g r , Y r , Z r 2 d r 2 E sup t s T s T Y r , Z r d W r f 1 8 E sup t r T Y 1 2 + C t T Z r 2 d r $$\begin{align}& \mathbf{E}\underset{t\le s\le T}{\mathop{\sup }}\,\left| \int_{s}^{T}{\left\langle {{Y}_{r}},g\left( r,{{Y}_{r}},{{Z}_{r}} \right)d{{B}_{r}} \right\rangle } \right|\le \frac{1}{8}\mathbf{E}\left( \underset{t\le r\le T}{\mathop{\sup }}\,{{\left| {{Y}_{r}} \right|}^{2}} \right)+C\int_{t}^{T}{{{\left| g\left( r,{{Y}_{r}},{{Z}_{r}} \right) \right|}^{2}}}dr \\ & 2\mathbf{E}\underset{t\le s\le T}{\mathop{\sup }}\,\left| \int_{s}^{T}{\left\langle {{Y}_{r}},{{Z}_{r}}d{{W}_{r}} \right\rangle } \right|\le f\frac{1}{8}\mathbf{E}\left( \underset{t\le r\le T}{\mathop{\sup }}\,{{\left| {{Y}_{1}} \right|}^{2}} \right)+C\int_{t}^{T}{{{\left| {{Z}_{r}} \right|}^{2}}dr} \\ \end{align}$$

Using the above inequalities, we deduce from (3.4) that

34E(suptrT| Yr |2)E[ | ξT |2 ]+2EsuptsT(sT Yr,f(r,Yr,Zr,Yr+δ(r),Zr+ζ(r)) dr)+CtT| g(r,Yr,Zr) |2dr+CEtT| Zr |2dr $$\begin{align}& \frac{3}{4}\mathbf{E}\left( \underset{t\le r\le T}{\mathop{\sup }}\,{{\left| {{Y}_{r}} \right|}^{2}} \right)\le \mathbf{E}\left[ {{\left| {{\xi }_{T}} \right|}^{2}} \right]+2\mathbf{E}\underset{t\le s\le T}{\mathop{\sup }}\,\left( \int_{s}^{T}{\left\langle {{Y}_{r}},f\left( r,{{Y}_{r}},{{Z}_{r}},{{Y}_{r+\delta \left( r \right)}},{{Z}_{r+\zeta \left( r \right)}} \right) \right\rangle dr} \right) \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,+C\int_{t}^{T}{{{\left| g\left( r,{{Y}_{r}},{{Z}_{r}} \right) \right|}^{2}}dr+C\mathbf{E}\int_{t}^{T}{{{\left| {{Z}_{r}} \right|}^{2}}dr}} \\ \end{align}$$

Applying (3.2) and (3.3), we deduce that

34E(suptrT| Yr |2)2CC0E[ Xt+2tT| Yr || f(r,0) |dr+tT| g(r,0,0) |2dr ]. $$\frac{3}{4}\mathbf{E}\left( \underset{t\le r\le T}{\mathop{\sup }}\,{{\left| {{Y}_{r}} \right|}^{2}} \right)\le \frac{2C}{{{C}_{0}}}\mathbf{E}\left[ {{X}_{t}}+2\int_{t}^{T}{\left| {{Y}_{r}} \right|\left| f\left( r,0 \right) \right|dr+\int_{t}^{T}{{{\left| g\left( r,0,0 \right) \right|}^{2}}dr}} \right].$$

Moreover, we have

4CC0EtT| Yr || f(r,0) |dr14E(suptrT| Yr |2)+4(2CC0)2E(tT| f(r,0) |dr)2. $$\frac{4C}{{{C}_{0}}}\mathbf{E}\int_{t}^{T}{\left| {{Y}_{r}} \right|\left| f\left( r,0 \right) \right|dr\le \frac{1}{4}\mathbf{E}\left( \underset{t\le r\le T}{\mathop{\sup }}\,{{\left| {{Y}_{r}} \right|}^{2}} \right)+4{{\left( \frac{2C}{{{C}_{0}}} \right)}^{2}}\mathbf{E}{{\left( \int_{t}^{T}{\left| f\left( r,0 \right) \right|dr} \right)}^{2}}.}$$

Hence gathering (3.2) and (3.5) we obtain

12E(suptrT| Yr |2)+EtT| Zr |2drC2E[ Xt+(tT| f(r,0) |dr)2+tT| g(r,0,0) |2dr ], $$\frac{1}{2}\mathbf{E}\left( \underset{t\le r\le T}{\mathop{\sup }}\,{{\left| {{Y}_{r}} \right|}^{2}} \right)+\mathbf{E}\int_{t}^{T}{{{\left| {{Z}_{r}} \right|}^{2}}dr\le {{C}_{2}}\mathbf{E}\left[ {{X}_{t}}+{{\left( \int_{t}^{T}{\left| f\left( r,0 \right) \right|}dr \right)}^{2}}+\int_{t}^{T}{{{\left| g\left( r,0,0 \right) \right|}^{2}}dr} \right]},$$

where C2 is a positive constant (which may change from line to line).

Then, applying the fubini’s theorem to (3.6) , this leads to

E(suptrT| Yr |2)C2E[ | ξT |2+TT+K(| ξr |2+| ηr |2)dr+(tT| f(r,0) |dr)2+tT| g(r,0,0) |2dr ]+C3tTE[suprsT| Ys |2]dr, $$\begin{align}& \mathbf{E}\left( \underset{t\le r\le T}{\mathop{\sup }}\,{{\left| {{Y}_{r}} \right|}^{2}} \right)\le {{C}_{2}}\mathbf{E}\left[ {{\left| {{\xi }_{T}} \right|}^{2}}+\int_{T}^{T+K}{\left( {{\left| {{\xi }_{r}} \right|}^{2}}+{{\left| {{\eta }_{r}} \right|}^{2}} \right)dr+{{\left( \int_{t}^{T}{\left| f\left( r,0 \right) \right|dr} \right)}^{2}}+\int_{t}^{T}{{{\left| g\left( r,0,0 \right) \right|}^{2}}dr}} \right] \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,+{{C}_{3}}\int_{t}^{T}{\mathbf{E}[\underset{r\le s\le T}{\mathop{\sup }}\,{{\left| {{Y}_{s}} \right|}^{2}}]dr,} \\ \end{align}$$

where C3 = 2C1C2. Hence Gronwall’s inequality yields

E(suptrT| Yr |2)+. $$E\left( \underset{t\le r\le T}{\mathop{\sup }}\,{{\left| {{Y}_{r}} \right|}^{2}} \right)\le +\infty .$$

This implies that YS[ 0,T ]2(G,Rk). $Y\in S_{\left[ 0,T \right]}^{2}\left( G,{{\mathbf{R}}^{k}} \right).$This completes the proof.

To solve our equations, we examine first the cases where the coefficients do not depend on the variables. Namely, we consider the stochastic equation

Yt=ξT+tTf(r)dr+tTg(r)dBrtTZrdWr,0tT. $${{Y}_{t}}={{\xi }_{T}}+\int_{t}^{T}{f\left( r \right)dr+\int_{t}^{T}{g\left( r \right)d{{B}_{r}}-\int_{t}^{T}{{{Z}_{r}}d{{W}_{r}},\ \ \ \ 0\le t\le T.}}}$$

where fM[ 0,T ]2(G,Rk),gM[ 0,T ]2(G,Rk×l)and ξTL2(GT,Rk). $f\in M_{\left[ 0,T \right]}^{2}\left( G,{{\mathbf{R}}^{k}} \right),\ \ \ \ g\in M_{\left[ 0,T \right]}^{2}\left( G,{{\mathbf{R}}^{k\times l}} \right)\ \text{and }{{\xi }_{T}}{{L}^{2}}\left( {{G}_{T}},{{\mathbf{R}}^{k}} \right).$

Let us recall the following result which will be useful in the sequel (the proof is omitted since it is an adaptation of Theorem 3.1 in Xu [9]).

Proposition 3.2. Given ξT ∈ L2(GT,Rk), eq (3.7) has a unique solution (Yt,Zt)0tTCG2(0,T). ${{\left( {{Y}_{t}},{{Z}_{t}} \right)}_{0\le t\le T}}\in C_{G}^{2}\left( 0,T \right).$

We are now in position to give our main results of this section.

Theorem 3.3

Assume that the assumptions (A1), (A2) and (H1) are true and let ξT ∈ L2(T,Rk). Then for any (ξ,η)S[ T,T+K ]2(G,Rk)×M[ T,T+K ]2(G,Rk×d) $\left( \xi ,\eta \right)\in S_{\left[ T,T+K \right]}^{2}\left( G,{{\mathbf{R}}^{k}} \right)\times M_{\left[ T,T+K \right]}^{2}\left( G,{{\mathbf{R}}^{k\times d}} \right)$the ABDSDE (1.1) has a unique solution (Yt ,Zt)0≤t≤T BG2(0,T+K). $B_{G}^{2}\left( 0,T+K \right).$

Proof. (i) Existence. Let us consider the mapping

Ψ:CG2(0,T+K)CG2(0,T+K),(y,z)(Y,Z) $$\begin{align}& \Psi :C_{G}^{2}\left( 0,T+K \right)\to C_{G}^{2}\left( 0,T+K \right), \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \,\,\,\,\left( y,z \right)\to \left( Y,Z \right) \\ \end{align}$$

where the pair (Yt,Zt)0tT+KCG2(0,T+K)is s.t(Yt,Zt)TtT+K=(ξn,ηt) ${{\left( {{Y}_{t}},{{Z}_{t}} \right)}_{0\le t\le T+K}}\in C_{G}^{2}\left( 0,T+K \right)\ \text{is s}\text{.t}\ \ {{\left( {{Y}_{t}},{{Z}_{t}} \right)}_{T\le t\le T+K}}=\left( {{\xi }_{n}},{{\eta }_{t}} \right)$and it satisfies the equation

0 t T , Y t = ξ T + t T f r , y r , z r , y r + δ r , z r + ζ r d r + t T g r , y r , z r d B r t T Z r d W r , t T , T + K , Y t = ξ t , Z t = n t $$\left\{ \begin{array}{*{35}{l}} \forall 0\le t\le T, \\ {{Y}_{t}}={{\xi }_{T}}+\int_{t}^{T}{f\left( r,{{y}_{r}},{{z}_{r}},{{y}_{r+\delta \left( r \right)}},{{z}_{r+\zeta \left( r \right)}} \right)dr+\int_{t}^{T}{g\left( r,{{y}_{r}},{{z}_{r}} \right)d{{B}_{r}}-\int_{t}^{T}{{{Z}_{r}}d{{W}_{r}},}}} \\ \forall t\in \left[ T,T+K \right], {{Y}_{t}}={{\xi }_{t}}, {{Z}_{t}}={{n}_{t}} \\ \end{array} \right.$$ .

Thanks to Proposition 3.2, the mapping Ψ is well defined. Let (Y,Z) and (Y˜,Z˜) $\left( \tilde{Y},\tilde{Z} \right)$be two solutions of eq.(3.8), i.e :

(Y,Z)=Ψ(y,z)and(Y˜,Z˜)=Ψ(y˜,z˜). $\left( Y,Z \right)=\Psi \left( y,z \right)\ \ \ \ \text{and}\ \ \ \ \left( \tilde{Y},\tilde{Z} \right)=\Psi \left( \tilde{y},\tilde{z} \right).$

Fix β ∈ R. The pair (Y¯,Z¯) $\left( \overline{Y},\overline{Z} \right)$solves the ABDSDE

{ Y¯t=tTΔf(r)dr+tTΔg(r)dBrtTZ¯rdWr,0tT,t[ T,T+K ],Y¯t=0,Z¯t=0, $$\left\{ \begin{array}& {{{\bar{Y}}}_{t}}=\int_{t}^{T}{\Delta f\left( r \right)dr+}\int_{t}^{T}{\Delta g\left( r \right)d{{B}_{r}}-\int_{t}^{T}{{{{\bar{Z}}}_{r}}d{{W}_{r}},\,\,\,\,\,}}\forall 0\le t\le T, \\ \forall t\,\in \left[ T,T+K \right],\,\,\,\,{{{\bar{Y}}}_{t}}=0,\,\,\,{{{\bar{Z}}}_{t}}=0,\\ \end{array} \right.$$

where for ρ{ Y,Z },ρ¯=ρρ˜,Δg(r)=g(r,yr,zr)g(r,y˜r,z˜r) $\rho \in \left\{ Y,Z \right\},\bar{\rho }=\rho -\tilde{\rho },\Delta g\left( r \right)=g\left( r,{{y}_{r}},{{z}_{r}} \right)-g\left( r,{{{\tilde{y}}}_{r}},{{{\tilde{z}}}_{r}} \right)$and

Δf(r)=f(r,yr,zr,yr+δ(r),zr+ζ(r))f(r,y˜r,z˜r,y˜r+δ(r)). $$\Delta f\left( r \right)=f\left( r,{{y}_{r}},{{z}_{r}},{{y}_{r+\delta \left( r \right)}},{{z}_{r+\zeta \left( r \right)}} \right)-f\left( r,{{{\tilde{y}}}_{r}},{{{\tilde{z}}}_{r}},{{{\tilde{y}}}_{r+\delta \left( r \right)}} \right).$$

Applying Ito’s formula, we obtain

E e β t Y ¯ t 2 + β t T e β r Y ¯ r 2 d r + t T e β r Z ¯ r 2 d r = 2 E t T e β r Y ¯ r , Δ f r d r + E t T e β r Δ g r 2 d r . $$\text{E}\left[ {{e}^{\beta t}}{{\left| {{{\bar{Y}}}_{t}} \right|}^{2}}+\beta \int_{t}^{T}{{{e}^{\beta r}}}{{\left| {{{\bar{Y}}}_{r}} \right|}^{2}}dr+\int_{t}^{T}{{{e}^{\beta r}}}{{\left| {{{\bar{Z}}}_{r}} \right|}^{2}}dr \right]=2\text{E}\int_{t}^{T}{{{e}^{\beta r}}}\left\langle {{{\bar{Y}}}_{r}},\Delta f\left( r \right) \right\rangle dr+\text{E}\int_{t}^{T}{{{e}^{\beta r}}}{{\left| \Delta g\left( r \right) \right|}^{2}}dr.$$

Using the inequality 2ab ≤ εa2+ b2 (where ε > 0 will be chosen later) and assumption (H1.1), we obtain

EtTeβr| Δg(r) |2drcEtT+Keβr| y¯r |2dr+α1EtT+Keβr| z¯r |2dr. $$\text{E}\int_{t}^{T}{{{e}^{\beta r}}{{\left| \Delta g\left( r \right) \right|}^{2}}dr\le c\text{E}}\int_{t}^{T+K}{{{e}^{\beta r}}{{\left| {{{\bar{y}}}_{r}} \right|}^{2}}dr+}{{\alpha }_{1}}\text{E}\int_{t}^{T+K}{{{e}^{\beta r}}{{\left| {{{\bar{z}}}_{r}} \right|}^{2}}dr}.$$

Similarly, we have

2eβr Y¯r,Δf(r) eβrε| Y¯r |2+cεeβr(| y¯r |2+| z¯r |2)+1εeβrEFt[ | y¯r+δ(r) |2+| z¯r+ζ(r) |2 ]. $$2{{e}^{\beta r}}\left\langle {{{\bar{Y}}}_{r}},\Delta f\left( r \right) \right\rangle \le {{e}^{\beta r}}\varepsilon {{\left| {{{\bar{Y}}}_{r}} \right|}^{2}}+\frac{c}{\varepsilon }{{e}^{\beta r}}\left( {{\left| {{{\bar{y}}}_{r}} \right|}^{2}}+{{\left| {{{\bar{z}}}_{r}} \right|}^{2}} \right)+\frac{1}{\varepsilon }{{e}^{\beta r}}{{\mathbf{E}}^{{{F}_{t}}}}\left[ {{\left| {{{\bar{y}}}_{r+\delta \left( r \right)}} \right|}^{2}}+{{\left| {{{\bar{z}}}_{r+\zeta \left( r \right)}} \right|}^{2}} \right].$$

Which implies by virtue of condition (A2) that

2EtTeβr Y¯r,Δf(r) drεEtT+Keβr| Y¯r |2dr+1ε(c+M)EtT+Keβr(| y¯r |2+| z¯r |2)dr. $$2\mathbf{E}\int_{t}^{T}{{{e}^{\beta r}}\left\langle {{{\bar{Y}}}_{r}},\Delta f\left( r \right) \right\rangle }\,dr\le \varepsilon \mathbf{E}\int_{t}^{T+K}{{{e}^{\beta r}}{{\left| {{{\bar{Y}}}_{r}} \right|}^{2}}dr}+\frac{1}{\varepsilon }\left( c+M \right)\mathbf{E}\int_{t}^{T+K}{{{e}^{\beta r}}\left( {{\left| {{{\bar{y}}}_{r}} \right|}^{2}}+{{\left| {{{\bar{z}}}_{r}} \right|}^{2}} \right)dr}.$$

Therefore, we can write (Whereγ=1ε(c+M)+c1ε(c+M)+α1) $\left( \text{Where}\,\gamma =\frac{\frac{1}{\varepsilon }\left( c+M \right)+c}{\frac{1}{\varepsilon }\left( c+M \right)+{{\alpha }_{1}}} \right)$

E(eβt| Y¯t |2)+(βε)EtT+Keβr| Y¯r |2dr+ErT+Keβr| Z¯r |2dr(1ε(c+M)+c)EtT+Keβr| y¯r |2dr+(1ε(c+M)+α1)EtT+Keβr| z¯r |2dr=(1ε(c+M)+α1)EtT+Keβr[ γ| y¯r |2+| z¯r |2 ]dr. $$\begin{align}& \mathbf{E}\left( {{e}^{\beta t}}{{\left| {{{\bar{Y}}}_{t}} \right|}^{2}} \right)+\left( \beta -\varepsilon \right)\mathbf{E}\int_{t}^{T+K}{{{e}^{\beta r}}}{{\left| {{{\bar{Y}}}_{r}} \right|}^{2}}dr+\mathbf{E}\int_{r}^{T+K}{{{e}^{\beta r}}{{\left| {{{\bar{Z}}}_{r}} \right|}^{2}}}dr \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\le \left( \frac{1}{\varepsilon }\left( c+M \right)+c \right)\mathbf{E}\int_{t}^{T+K}{{{e}^{\beta r}}{{\left| {{{\bar{y}}}_{r}} \right|}^{2}}dr+\left( \frac{1}{\varepsilon }\left( c+M \right)+{{\alpha }_{1}} \right)\mathbf{E}\int_{t}^{T+K}{{{e}^{\beta r}}{{\left| {{{\bar{z}}}_{r}} \right|}^{2}}dr}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( \frac{1}{\varepsilon }\left( c+M \right)+{{\alpha }_{1}} \right)\mathbf{E}\int_{t}^{T+K}{{{e}^{\beta r}}\left[ \gamma {{\left| {{{\bar{y}}}_{r}} \right|}^{2}}+{{\left| {{{\bar{z}}}_{r}} \right|}^{2}} \right]}\,dr. \\ \end{align}$$

Hence if we choose ε = ε0 satisfying c¯=(1ε0(c+M)+α1)<1, $\bar{c}=\left( \frac{1}{{{\varepsilon }_{0}}}\left( c+M \right)+{{\alpha }_{1}} \right)<1,$choose β = ε0+ γ, then we deduce

EtT+Keβr[ γ| Y¯r |2+| Z¯r |2 ]drc¯EtT+Keβr[ γ| y¯|2+| z¯r |2 ]dr. $$\mathbf{E}\int_{t}^{T+K}{{{e}^{\beta r}}\left[ \gamma {{\left| {{{\bar{Y}}}_{r}} \right|}^{2}}+{{\left| {{{\bar{Z}}}_{r}} \right|}^{2}} \right]}\,dr\le \bar{c}\mathbf{E}\int_{t}^{T+K}{{{e}^{\beta r}}}\left[ \gamma {{\left| {\bar{y}} \right|}^{2}}+{{\left| {{{\bar{z}}}_{r}} \right|}^{2}} \right]\,dr.$$

Thus, the mapping Ψ is a strict contraction on CG2(0,T+K) $C_{G}^{2}\left( 0,T+K \right)$and it has a unique fixed point

(Y,Z)CG2(0,T+K). $\left( Y,Z \right)\in C_{G}^{2}\left( 0,T+K \right).$

It remains to prove that the above solution is in BG2(0,T+K). $B_{G}^{2}\left( 0,T+K \right).$Indeed, by Lemma 3.1, we have YS[ 0,T ]2(G,Rk). $Y\in S_{\left[ 0,T \right]}^{2}\left( G\text{,}{{\mathbf{R}}^{k}} \right).$Thus, we obtain (Yt,Zt)0tTBG2(0,T+K). ${{\left( {{Y}_{t}},{{Z}_{t}} \right)}_{0\le t\le T}}B_{G}^{2}\left( 0,T+K \right).$

(ii) Uniqueness. Let (Y,Z) and (Y˜,Z˜) $\left( \tilde{Y},\tilde{Z} \right)$two solutions of eq.(1.1). Itô’s formula applied to eq.(3.9) yields, for 0 ≤ t ≤ T

E[ | Y¯t |2 ]+EtT| Z¯r |2dr2EtT Y¯r,Δ(r) dr+EtT| Δg(r) |2dr. $$\mathbf{E}\left[ {{\left| {{{\bar{Y}}}_{t}} \right|}^{2}} \right]+\mathbf{E}\int_{t}^{T}{{{\left| {{{\bar{Z}}}_{r}} \right|}^{2}}dr\le 2\mathbf{E}}\int_{t}^{T}{\left\langle {{{\bar{Y}}}_{r}},\Delta \left( r \right) \right\rangle dr+\mathbf{E}\int_{t}^{T}{{{\left| \Delta g\left( r \right) \right|}^{2}}}}dr.$$

Using assumption (H1), we have :

2EtT Y¯r,Δf(r) dr(1ε(c+M)+ε)EtT| Y¯r |2dr+1ε(c+M)EtT| Z¯r |2dr,EtT| Δg(r) |2drcEtT| Y¯r |2dr+α1EtT| Z¯r |2dr. $$\begin{align}& 2\mathbf{E}\int_{t}^{T}{\left\langle {{{\bar{Y}}}_{r}},\Delta f\left( r \right) \right\rangle dr\le \left( \frac{1}{\varepsilon }\left( c+M \right)+\varepsilon \right)\mathbf{E}}\int_{t}^{T}{{{\left| {{{\bar{Y}}}_{r}} \right|}^{2}}}dr+\frac{1}{\varepsilon }\left( c+M \right)\mathbf{E}\int_{t}^{T}{{{\left| {{{\bar{Z}}}_{r}} \right|}^{2}}dr}, \\ & \,\,\mathbf{E}\int_{t}^{T}{{{\left| \Delta g\left( r \right) \right|}^{2}}dr}\le c\mathbf{E}\int_{t}^{T}{{{\left| {{{\bar{Y}}}_{r}} \right|}^{2}}}dr+{{\alpha }_{1}}\mathbf{E}\int_{t}^{T}{{{\left| {{{\bar{Z}}}_{r}} \right|}^{2}}dr}. \\ \end{align}$$

Hence if we choose ξ = ξ0 satisfying α¯=1ε0(c+M)+α1<1 $\bar{\alpha }=\frac{1}{{{\varepsilon }_{0}}}\left( c+M \right)+{{\alpha }_{1}}<1$and denote c¯=c(ε0+1)+Mε0+ε0, $\bar{c}=\frac{c\left( {{\varepsilon }_{0}}+1 \right)+M}{{{\varepsilon }_{0}}}+{{\varepsilon }_{0}},$then using the above inequalities, from (3.11), we obtain

E(| Y¯t |2)+(1α¯)EtT| Z¯r |2drc¯EtT| Y¯r |2dr. $$\mathbf{E}\left( {{\left| {{{\bar{Y}}}_{t}} \right|}^{2}} \right)+\left( 1-\bar{\alpha } \right)\mathbf{E}\int_{t}^{T}{{{\left| {{{\bar{Z}}}_{r}} \right|}^{2}}dr\le \bar{c}\mathbf{E}}\int_{t}^{T}{{{\left| {{{\bar{Y}}}_{r}} \right|}^{2}}dr}.$$

Then we can use Gronwall’s inequality to deduce Y ¯ = 0 a n d Z ¯ = 0 . $\bar{Y}=0\,\text and \,\bar{Z}=\text{0}\text{.}$This completes the proof.

The case of non-Lipschitz coefficients

In this subsection, we will mainly study the existence and uniqueness of the solution to ABDSDE (1.1) with non-Lipschitz coefficients. For this purpose, we first make the following assumptions.

Assumptions

In the following, we assume that there exists ρ ∈ S such that f and g satisfy assumptions (H2).

(H2.1): There exists a constant c > 0 such that

| f(t,y,z,θ(r),φ(r))f(t,y,z,θ(r),φ(r)) |2c(ρ(| yy |2)+| zz |2)+EFt[ ρ(| θ(r)θ(r) |2)+| φ(r)φ(r) |2 ], $$\begin{align}& {{\left| f\left( t,y,z,\theta \left( r \right),\varphi \left( r \right) \right)-f\left( t,{y}',{z}',{\theta }'\left( r \right),{\varphi }'\left( r \right) \right) \right|}^{2}}\le c\left( \rho \left( {{\left| y-{y}' \right|}^{2}} \right)+{{\left| z-{z}' \right|}^{2}} \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+{{E}^{{{F}_{t}}}}\left[ \rho \left( {{\left| \theta \left( r \right)-{\theta }'\left( r \right) \right|}^{2}} \right)+{{\left| \varphi \left( r \right)-{\varphi }'\left( r \right) \right|}^{2}} \right], \\ \end{align}$$

forall(r,r)[ t,T+K ],(t,y,z,θ(r),φ(r)),(t,y,z,θ(r),φ(r))A×CG2(t,T+K). $$\text{for}\,\text{all}\left( r,{r}' \right)\in \left[ t,T+K \right],\left( t,y,z,\theta \left( r \right),\varphi \left( r \right) \right),\,\left( t,{y}',{z}',{\theta }'\left( r \right),{\varphi }'\left( r \right) \right)\in A\times C_{G}^{2}\left( t,T+K \right).$$

(H2.2): There exists a constant 0 < α1 < 1 such that for any (t,y,z),(t,y,z)[ 0,T ]×Rk×Rk×d $\left( t,y,z \right),\,\left( t,{y}',{z}' \right)\in \left[ 0,T \right]\times {{\mathbf{R}}^{k}}\times {{\mathbf{R}}^{k\times d}}$

| g(t,y,z)g(t,y,z) |2ρ(| yy |2)+α1| zz |2. $${{\left| g\left( t,y,z \right)-g\left( t,{y}',{z}' \right) \right|}^{2}}\le \rho \left( {{\left| y-{y}' \right|}^{2}} \right)+{{\alpha }_{1}}{{\left| z-{z}' \right|}^{2}}.$$

(H2.3): (H1.3) holds.

Existence and uniqueness of solution

We consider now the sequence (Θn)n∈N = (Yn,Zn)n∈N given by

Y t 0 = 0 , Z t 0 = 0 , 0 t T + K , Y t n = ξ T + t T f r , Y r n 1 , Z r n , Y r + δ r n 1 , Z r + ζ r n d r + t T g r , Y r n 1 , Z r n d B r t T Z r n d W r , 0 t T , Y t n = ξ t , Z t n = η t , T t T + K . $$\left\{ \begin{array} \text Y_{t}^{0}=0,\,Z_{t}^{0}=0,\,\,\,\,\,\,\,\,0\le t\le T+K, \\ Y_{t}^{n}={{\xi }_{T}}+\int_{t}^{T}{f\left( r,Y_{r}^{n-1},Z_{r}^{n},Y_{r+\delta \left( r \right)}^{n-1},Z_{r+\zeta \left( r \right)}^{n} \right)dr+}\int_{t}^{T}{g\left( r,Y_{r}^{n-1},Z_{r}^{n} \right)d{{B}_{r}}-}\int_{t}^{T}{Z_{r}^{n}d{{W}_{r}},}\,\,\,\,\,\,\,0\le t\le T, \\ Y_{t}^{n}={{\xi }_{t}},\,\,\,Z_{t}^{n}={{\eta }_{t}},\,\,\,\,\,\,\,\,\,\,\,\,\,T\le t\le T+K. \\ \end{array} \right.$$

Thanks to Theorem 3.3, this sequence is well defined since the generators f(r,Yrn1,,Yr+δ(r)n1,)andg(r,Yrn1,) $f\left( r,Y_{r}^{n-1},\cdot ,Y_{r+\delta \left( r \right)}^{n-1},\cdot \right)\,\text{and}\,\text{g}\left( r,Y_{r}^{n-1},\cdot \right)$are Г-Lipschitz. Let us state the following previous result

Lemma 4.1

Assume that the assumptions (A1), (A2) and (H2) are true and let ξT ∈ L2(T,Rk). Then for any (ξ,η)S[ T,T+K ]2(G,Rk)×M[ T,T+K ]2(G,Rk×d) $\left( \xi ,\eta \right)\in S_{\left[ T,T+K \right]}^{2}\left( G,{{\mathbf{R}}^{k}} \right)\times M_{\left[ T,T+K \right]}^{2}\left( G,{{\mathbf{R}}^{k\times d}} \right)$there exists a positive constant C′ such that

supn0E| Ytn |2C(1+E[ X ]),0tT+K $$\underset{n\ge 0}{\mathop{\sup }}\,\mathbf{E}{{\left| Y_{t}^{n} \right|}^{2}}\le {C}'\left( 1+\mathbf{E}\left[ X \right] \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\le t\le T+K$$

where

X=| ξT |2+TT+K(| ξr |2+| ηr |2)dr+0T| f(r,0) |2dr+0T| g(r,0,0) |2dr. $$X={{\left| {{\xi }_{T}} \right|}^{2}}+\int_{T}^{T+K}{\left( {{\left| {{\xi }_{r}} \right|}^{2}}+{{\left| {{\eta }_{r}} \right|}^{2}} \right)dr+\int_{0}^{T}{{{\left| f\left( r,0 \right) \right|}^{2}}dr}}+\int_{0}^{T}{{{\left| g\left( r,0,0 \right) \right|}^{2}}dr}.$$

Proof. For β > 0, apply Itô’s formula to eβt| Ytn |2, ${{e}^{\beta t}}{{\left| Y_{t}^{n} \right|}^{2}},$

E[ eβt| Ytn |2 ]+βEtTeβr| Yrn |2dr+EtTeβr| Zrn |2dr=E[ eβT| ξT |2 ]+2EtTeβr Yrn,f(r,Yrn1,Zrn,Yr+δ(r)n1,Zr+ζ(r)n) dr+EtTeβr| g(r,Yrn1,Zrn) |2dr. $$\begin{align}& \mathbf{E}\left[ {{e}^{\beta t}}{{\left| Y_{t}^{n} \right|}^{2}} \right]+\beta \mathbf{E}\int_{t}^{T}{{{e}^{\beta r}}{{\left| Y_{r}^{n} \right|}^{2}}}dr+\mathbf{E}\int_{t}^{T}{{{e}^{\beta r}}{{\left| Z_{r}^{n} \right|}^{2}}}dr=\mathbf{E}\left[ {{e}^{\beta T}}{{\left| {{\xi }_{T}} \right|}^{2}} \right] \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+2\mathbf{E}\int_{t}^{T}{{{e}^{\beta r}}\left\langle Y_{r}^{n},f\left( r,Y_{r}^{n-1},Z_{r}^{n},Y_{r+\delta \left( r \right)}^{n-1},Z_{r+\zeta \left( r \right)}^{n} \right) \right\rangle }dr+\mathbf{E}\int_{t}^{T}{{{e}^{\beta r}}{{\left| g\left( r,Y_{r}^{n-1},Z_{r}^{n} \right) \right|}^{2}}}dr. \\ \end{align}$$

Using the inequality 2ab ≤ εa2+ b2 (where ε > 0 will be chosen later), we deduce from assumptions (H2.1) and (H2.2)

EtTeβr| g(r,Yrn1,Zrn) |2drEtTeβr| g(r,0,0) |2dr+EtTeβr[ ρ(| Yrn1 |2)+α1| zrn |2 ]dr $$\mathbf{E}\int_{t}^{T}{{{e}^{\beta r}}{{\left| g\left( r,Y_{r}^{n-1},Z_{r}^{n} \right) \right|}^{2}}dr}\le \mathbf{E}\int_{t}^{T}{{{e}^{\beta r}}{{\left| g\left( r,0,0 \right) \right|}^{2}}dr+\mathbf{E}\int_{t}^{T}{{{e}^{\beta r}}\left[ \rho \left( {{\left| Y_{r}^{n-1} \right|}^{2}} \right)+{{\alpha }_{1}}{{\left| z_{r}^{n} \right|}^{2}} \right]}}\,dr$$

and

2EtTeβr Yrn,f(r,Yrn1,Zrn,Yr+δ(r)n1,Zr+ζ(r)n) drε2EtTeβr| Yrn |2dr+2εEtTeβr| f(r,0) |2dr+2εcEtTeβr(ρ(| Yrn1 |2)+| Zrn |2)dr+2εEtTeβrEFt[ ρ(| Yr+δ(r)n1 |2)+| Zr+ζ(r)n |2 ]dr. $$\begin{align}& 2\mathbf{E}\int_{t}^{T}{{{e}^{\beta r}}\left\langle Y_{r}^{n},f\left( r,Y_{r}^{n-1},Z_{r}^{n},Y_{r+\delta \left( r \right)}^{n-1},Z_{r+\zeta \left( r \right)}^{n} \right) \right\rangle }dr\le \frac{\varepsilon }{2}\mathbf{E}\int_{t}^{T}{{{e}^{\beta r}}{{\left| Y_{r}^{n} \right|}^{2}}dr}+\frac{2}{\varepsilon }\mathbf{E}\int_{t}^{T}{{{e}^{\beta r}}{{\left| f\left( r,0 \right) \right|}^{2}}dr} \\ & \,\,\,\,\,\,\,\,+\frac{2}{\varepsilon }c\mathbf{E}\int_{t}^{T}{{{e}^{\beta r}}\left( \rho \left( {{\left| Y_{r}^{n-1} \right|}^{2}} \right)+{{\left| Z_{r}^{n} \right|}^{2}} \right)}dr+\frac{2}{\varepsilon }\mathbf{E}\int_{t}^{T}{{{e}^{\beta r}}{{\mathbf{E}}^{{{F}_{t}}}}\left[ \rho \left( {{\left| Y_{r+\delta \left( r \right)}^{n-1} \right|}^{2}} \right)+{{\left| Z_{r+\zeta \left( r \right)}^{n} \right|}^{2}} \right]}\,dr. \\ \end{align}$$

Applying condition (A2), the last term on the right-hand side of (4.3) is less than

2εMEtTeβr[ ρ(| Yrn1 |2)+| Zrn |2 ]dr+2εMETT+Keβr[ ρ(| ξr |2)+| ηr |2 ]dr. $$\frac{2}{\varepsilon }ME\int_{t}^{T}{{{e}^{\beta r}}\left[ \rho \left( {{\left| Y_{r}^{n-1} \right|}^{2}} \right)+{{\left| Z_{r}^{n} \right|}^{2}} \right]}\,dr+\frac{2}{\varepsilon }ME\int_{T}^{T+K}{{{e}^{\beta r}}}\left[ \rho \left( {{\left| {{\xi }_{r}} \right|}^{2}} \right)+{{\left| {{\eta }_{r}} \right|}^{2}} \right]\,dr.$$

Putting pieces together, we obtain finally

E[ eβt| Ytn |2 ]+βEtTeβr| Yrn |2dr+EtTeβr| Zrn |2drE[ eβT| ξT |2 ]+ε2EtTeβr| Yrn |2dr+(2ε(c+M)+1)EtTeβrρ(| Yrn1 |2)dr+(2ε(c+M)+α1)EtTeβr| Zrn |2dr+2εMETT+Keβr[ ρ(| ξr |2)+| ηr |2 ]dr+2εEtTeβr| f(r,0) |2dr+EtTeβr| g(r,0,0) |2dr. $$\begin{align}& \mathbf{E}\left[ {{e}^{\beta t}}{{\left| Y_{t}^{n} \right|}^{2}} \right]+\beta \mathbf{E}\int_{t}^{T}{{{e}^{\beta r}}{{\left| Y_{r}^{n} \right|}^{2}}dr+\mathbf{E}}\int_{t}^{T}{{{e}^{\beta r}}{{\left| Z_{r}^{n} \right|}^{2}}dr}\le \mathbf{E}\left[ {{e}^{\beta T}}{{\left| {{\xi }_{T}} \right|}^{2}} \right]+\frac{\varepsilon }{2}\mathbf{E}\int_{t}^{T}{{{e}^{\beta r}}{{\left| Y_{r}^{n} \right|}^{2}}dr} \\ & \,\,\,\,\,\,\,\,+\left( \frac{2}{\varepsilon }\left( c+M \right)+1 \right)\mathbf{E}\int_{t}^{T}{{{e}^{\beta r}}\rho \left( {{\left| Y_{r}^{n-1} \right|}^{2}} \right)dr}+\left( \frac{2}{\varepsilon }\left( c+M \right)+{{\alpha }_{1}} \right)\mathbf{E}\int_{t}^{T}{{{e}^{\beta r}}{{\left| Z_{r}^{n} \right|}^{2}}dr} \\ & \,\,\,\,\,\,\,\,\,+\frac{2}{\varepsilon }ME\int_{T}^{T+K}{{{e}^{\beta r}}\left[ \rho \left( {{\left| {{\xi }_{r}} \right|}^{2}} \right)+{{\left| {{\eta }_{r}} \right|}^{2}} \right]}\,dr+\frac{2}{\varepsilon }\mathbf{E}\int_{t}^{T}{{{e}^{\beta r}}}{{\left| f\left( r,0 \right) \right|}^{2}}dr+\mathbf{E}\int_{t}^{T}{{{e}^{\beta r}}}{{\left| g\left( r,0,0 \right) \right|}^{2}}dr. \\ \end{align}$$

This implies thanks to Remark 2.1 that

E e β t Y t n 2 + β E t T e β r Y r n 2 d r + E t T e β r Z r n 2 d r E e β T ξ T 2 + ε 2 E t T e β r Y r n 2 d r + 2 ε c + M + 1 b E t T e β r ρ Y r n 1 2 d r + 2 ε c + M + α 1 E t T e β r Z r n 2 d r + 2 ε M E T T + K e β r b ξ r 2 + η r 2 d r + 2 ε E t T e β r f r , 0 2 d r + E t T e β r g r , 0 , 0 2 d r + C ε $$\begin{align} & \mathbf{E}\left[ {{e}^{\beta t}}{{\left| Y_{t}^{n} \right|}^{2}} \right]+\beta \mathbf{E}\int_{t}^{T}{{{e}^{\beta r}}{{\left| Y_{r}^{n} \right|}^{2}}dr+\mathbf{E}}\int_{t}^{T}{{{e}^{\beta r}}{{\left| Z_{r}^{n} \right|}^{2}}dr}\le \mathbf{E}\left[ {{e}^{\beta T}}{{\left| {{\xi }_{T}} \right|}^{2}} \right]+\frac{\varepsilon }{2}\mathbf{E}\int_{t}^{T}{{{e}^{\beta r}}{{\left| Y_{r}^{n} \right|}^{2}}dr} \\ & +\left( \frac{2}{\varepsilon }\left( c+M \right)+1 \right)\,b\mathbf{E}\int_{t}^{T}{{{e}^{\beta r}}\rho \left( {{\left| Y_{r}^{n-1} \right|}^{2}} \right)dr}+\left( \frac{2}{\varepsilon }\left( c+M \right)+{{\alpha }_{1}} \right)\mathbf{E}\int_{t}^{T}{{{e}^{\beta r}}{{\left| Z_{r}^{n} \right|}^{2}}dr} \\ & +\frac{2}{\varepsilon }ME\int_{T}^{T+K}{{{e}^{\beta r}}\left[ b\left( {{\left| {{\xi }_{r}} \right|}^{2}} \right)+{{\left| {{\eta }_{r}} \right|}^{2}} \right]}dr+\frac{2}{\varepsilon }\mathbf{E}\int_{t}^{T}{{{e}^{\beta r}}}{{\left| f\left( r,0 \right) \right|}^{2}}dr+\mathbf{E}\int_{t}^{T}{{{e}^{\beta r}}}{{\left| g\left( r,0,0 \right) \right|}^{2}}dr+{{C}_{\varepsilon }} \\ \end{align}$$

where Cε=aβ[ 2ε(c+2M)+1 ]eβ(T+K). ${{C}_{\varepsilon }}=\frac{a}{\beta }\left[ \frac{2}{\varepsilon }\left( c+2M \right)+1 \right]{{e}^{\beta \left( T+K \right)}}.$Choose ε = ε0 such that β=12(ε0+1),C0=Cε0and2ε0(M+c)+α1= $\beta = \frac{1}{2}\left( {{\varepsilon }_{0}}+1 \right),\,\,{{C}_{0}}={{C}_{{{\varepsilon }_{0}}}}and\,\frac{2}{{{\varepsilon }_{0}}}\left( M+c \right)+{{\alpha }_{1}}=$1/2. Therefore, we obtain

E e β t Y t n 2 + 1 2 E t T e β r Y r n 2 d r + 1 2 E t T e β r Z r n 2 d r 2 ε 0 c + M + 1 b E t T e β r Y r n 1 2 d r + E e β T ξ T 2 + 2 ε 0 M E T T + K e β r b ξ r 2 + η r 2 d r + 2 ε 0 E 0 T e β r f r , 0 2 d r + E 0 T e β r g r , 0 , 0 2 d r + C 0 , 0 t T . $$\begin{align} & \mathbf{E}\left[ {{e}^{\beta t}}{{\left| Y_{t}^{n} \right|}^{2}} \right]+\frac{1}{2}\mathbf{E}\int_{t}^{T}{{{e}^{\beta r}}{{\left| Y_{r}^{n} \right|}^{2}}dr+\frac{1}{2}\mathbf{E}}\int_{t}^{T}{{{e}^{\beta r}}{{\left| Z_{r}^{n} \right|}^{2}}dr}\le \left( \frac{2}{{{\varepsilon }_{0}}}\left( c+M \right)+1 \right)b\mathbf{E}\int_{t}^{T}{{{e}^{\beta r}}{{\left| Y_{r}^{n-1} \right|}^{2}}dr} \\ & +\mathbf{E}\left[ {{e}^{\beta T}}{{\left| {{\xi }_{T}} \right|}^{2}} \right]+\frac{2}{{{\varepsilon }_{0}}}M\mathbf{E}\int_{T}^{T+K}{{{e}^{\beta r}}\left[ b{{\left| {{\xi }_{r}} \right|}^{2}}+{{\left| {{\eta }_{r}} \right|}^{2}} \right]dr+\frac{2}{{{\varepsilon }_{0}}}\mathbf{E}\int_{0}^{T}{{{e}^{\beta r}}{{\left| f\left( r,0 \right) \right|}^{2}}dr}} \\ & \,\,\,\,+E\int_{0}^{T}{{{e}^{\beta r}}{{\left| g\left( r,0,0 \right) \right|}^{2}}dr+{{C}_{0}},\,\,\,\,0}\le t\le T. \\ \end{align}$$

This leads to

E[ | Ytn |2 ]+12EtT| Yrn |2dr+12EtT| Zrn |2drCEtT| Yrn1 |2dr,0tT $$\mathbf{E}\left[ {{\left| Y_{t}^{n} \right|}^{2}} \right]+\frac{1}{2}\mathbf{E}\int_{t}^{T}{{{\left| Y_{r}^{n} \right|}^{2}}}dr+\frac{1}{2}\mathbf{E}\int_{t}^{T}{{{\left| Z_{r}^{n} \right|}^{2}}dr}\le {C}'E{{\int_{t}^{T}{\left| Y_{r}^{n-1} \right|}}^{2}}dr,\,\,0\le t\le T$$

where C′ is a positive constant(which may vary from line to line). In particular, putting qn(t)=supnE[ | Ytn |2 ], ${{q}_{n}}\left( t \right)={{\sup }_{n\in \mathbb{N}}}\mathbf{E}\left[ {{\left| Y_{t}^{n} \right|}^{2}} \right],$we have

qn(t)C(E[ X ]+1)+CtTqn(r)dr,0tT. $${{q}_{n}}\left( t \right)\le {C}'\left( E\left[ X \right]+1 \right)+{C}'\int_{t}^{T}{{{q}_{n}}\left( r \right)dr},\,\,\,\,\,0\le t\le T.$$

Gronwall’s inequality yields

supnE[ | Ytn |2 ]C(1+E[ X ]),0tT. $$\underset{n\in \mathbb{N}}{\mathop{\sup }}\,\mathbf{E}\left[ {{\left| Y_{t}^{n} \right|}^{2}} \right]\le {C}'\left( 1+\mathbf{E}\left[ X \right] \right),\,\,\,\,\,\,\,\,\,\,0\le t\le T.$$

This immediately gives (4.2).

Now we establish the main result of this section.

Theorem 4.2

Assume that the assumptions (A1), (A2) and (H2) are true and let ξT ∈ L2(GT,Rk). Then for any (ξ,η)S[ T,T+K ]2(G,Rk)×M[ T,T+K ]2(G,Rk×d) $\left( \xi ,\eta \right)\in S_{\left[ T,T+K \right]}^{2}\left( G,{{\mathbf{R}}^{k}} \right)\times M_{\left[ T,T+K \right]}^{2}\left( G,{{\mathbf{R}}^{k\times d}} \right)$the ABDSDE (1.1) has a unique solution (Y,Z)BG2(0,T+K). $\left( Y,Z \right)\in B_{G}^{2}\left( 0,T+K \right).$

Proof. (i) Existence. We consider the sequence defined in eq.(4.1). For a process ρ ∈ {Y,Z}, and n ∈ N,m ∈ N, ρtn,m=ρtnρtm,Δg(n,m)(r)=f(r,Yrn1,Zrn)g(r,Yrm1,Zrm)andΔf(n,m)(r)=f(r,Yrn1,Zrn,Yr+δ(r)n1,Zr+ζ(r)n)f(r,Yrm1,Zrm,Yr+δ(r)m1,Zr+ζ(r)m). $\begin{align}& \rho _{t}^{-n,m}=\rho _{t}^{n}-\rho _{t}^{m},\Delta {{g}^{\left( n,m \right)}}\left( r \right)=f\left( r,Y_{r}^{n-1},Z_{r}^{n} \right)-g\left( r,Y_{r}^{m-1},Z_{r}^{m} \right)\text{and} \\ & \Delta {{f}^{\left( n,m \right)}}\left( r \right)=f\left( r,Y_{r}^{n-1},Z_{r}^{n},Y_{r+\delta \left( r \right)}^{n-1},Z_{r+\zeta \left( r \right)}^{n} \right)-f\left( r,Y_{r}^{m-1},Z_{r}^{m},Y_{r+\delta \left( r \right)}^{m-1},Z_{r+\zeta \left( r \right)}^{m} \right). \\ \end{align}$

Note that the pair (Yn,m,Zn,m) solves the following equation

{ Y¯tn,m=tTΔf(n,m)(r)dr+tTΔg(n,m)(r)dBrtTZ¯rn,mdWr,0tT,Y¯tn,m=0,Z¯tn,m=0,TtT+K. $$\left\{ \begin{align}& \bar{Y}_{t}^{n,m}=\int_{t}^{T}{\Delta {{f}^{\left( n,m \right)}}\left( r \right)dr+\int_{t}^{T}{\Delta {{g}^{\left( n,m \right)}}\left( r \right)d{{B}_{r}}-\int_{t}^{T}{\bar{Z}_{r}^{n,m}d{{W}_{r}},}}}\,\,\,0\le t\le T, \\ & \bar{Y}_{t}^{n,m}=0,\,\,\,\,\bar{Z}_{t}^{n,m}=0,\,\,\,\,\,T\le t\le T+K. \\ \end{align} \right.\,$$

By the same computations as in the proof of Lemma 4.1, we have

E| Y¯tn,m |2+12EtT| Y¯rn,m |2dr+12EtT| Z¯rn,m |2drCEtTρ(| Y¯rn1,m1 |2)dr,0tT. $$\mathbf{E}{{\left| \bar{Y}_{t}^{n,m} \right|}^{2}}+\frac{1}{2}\mathbf{E}\int_{t}^{T}{{{\left| \bar{Y}_{r}^{n,m} \right|}^{2}}dr+\frac{1}{2}\mathbf{E}}\int_{t}^{T}{{{\left| \bar{Z}_{r}^{n,m} \right|}^{2}}dr\le {C}'E}\int_{t}^{T}{\rho \left( {{\left| \bar{Y}_{r}^{n-1,m-1} \right|}^{2}} \right)}dr,\,\,\,\,0\le t\le T.$$

Applying Fatou’s lemma and the fact that ρ ∈ S, we deduce that

q(t)CtT+Kρ(q(r))dr,0tT+K $$q\left( t \right)\le {C}'\int_{t}^{T+K}{\rho \left( q\left( r \right) \right)dr},\,\,\,0\le t\le T+K$$

where q(t)=limn,msupE| Y¯tn,m |2,0tT+K. $q\left( t \right)={{\lim }_{n,m\to \infty }}\sup \mathbf{E}{{\left| \bar{Y}_{t}^{n,m} \right|}^{2}},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\le t\le T+K.$Therefore, we can use Bihari’s inequality to get q(t) = 0, i. limn,msupE| Y¯tn,m |2=0 ${{\lim }_{n,m\to \infty }}\sup \mathbf{E}{{\left| \bar{Y}_{t}^{n,m} \right|}^{2}}=0$for all 0 ≤ t ≤ T + K.

So, from inequality (4.4), we obtain

limn,mE(| YtnYtm |2+tT+K| ZrnZrm |2dr)=0,0tT+K. $$\underset{n,m\to \infty }{\mathop{\lim }}\,\mathbf{E}\left( {{\left| Y_{t}^{n}-Y_{t}^{m} \right|}^{2}}+\int_{t}^{T+K}{{{\left| Z_{r}^{n}-Z_{r}^{m} \right|}^{2}}dr} \right)=0,\,\,\,\,\,\,\,\,0\le t\le T+K.$$

Then, there exists (Y,Z)BG2(0,T+K) $\left( Y,Z \right)\in B_{G}^{2}\left( 0,T+K \right)$such that

limnE(| YtnYt |2+tT+K| ZrnZr |2dr)=0,0tT+K. $$\underset{n\to \infty }{\mathop{\lim }}\,\mathbf{E}\left( {{\left| Y_{t}^{n}-{{Y}_{t}} \right|}^{2}}+{{\int_{t}^{T+K}{\left| Z_{r}^{n}-{{Z}_{r}} \right|}}^{2}}dr \right)=0,\,\,\,\,\,\,0\le t\le T+K.$$

Finally, taking limit in eq (1.1) as n ➝+∞, we conclude that (Y,Z) solves

{ 0tT,Yt=ξT+tTf(r,Yr,Zr,Yr+δ(r),Zr+ζ(r))dr+tTg(r,Yr,Zr)dBrtTZrdWr,Yt=ξt,Zt=ηt,TtT+K. $$\left\{ \begin{align}& 0\le t\le T, \\ & {{Y}_{t}}={{\xi }_{T}}+\int_{t}^{T}{f\left( r,{{Y}_{r}},{{Z}_{r}},{{Y}_{r+\delta \left( r \right)}},{{Z}_{r+\zeta \left( r \right)}} \right)dr+}\int_{t}^{T}{g\left( r,{{Y}_{r}},{{Z}_{r}} \right)d{{B}_{r}}-\int_{t}^{T}{{{Z}_{r}}d{{W}_{r}}}}, \\ & {{Y}_{t}}={{\xi }_{t}},\,\,\,{{Z}_{t}}={{\eta }_{t}},\,\,\,\,T\le t\le T+K. \\ \end{align} \right.$$

This shows that (Y,Z)BG2(0,T+K) $\left( Y,Z \right)\in B_{G}^{2}\left( 0,T+K \right)$solves ABDSDE (1.1). The proof of existence is complete.

(ii) Uniqueness. Let (Yi,Zi)BG2(0,T+K),i=1,2 $\left( {{Y}^{i}},{{Z}^{i}} \right)\in B_{G}^{2}\left( 0,T+K \right),\,\,\,i=1,2$be two solutions of ABDSDE (1.1).

Define Y¯t=Yt1Yt2,Z¯t=Zt1Zt2,Δg(r)=g(r,Yr1,Zr1)g(r,Yr2,Zr2) ${{\bar{Y}}_{t}}=Y_{t}^{1}-Y_{t}^{2},\,\,\,\,\,\,{{\bar{Z}}_{t}}=Z_{t}^{1}-Z_{t}^{2},\Delta g\left( r \right)=g\left( r,Y_{r}^{1},Z_{r}^{1} \right)-g\left( r,Y_{r}^{2},Z_{r}^{2} \right)$and Δf(r)=f(r,Yr1,Zr1,Yr+δ(r)1,Zr+ζ(r)1)f(r,Yr2,Zr2,Yr+δ(r)2,Zr+ζ(r)2). $\Delta f\left( r \right)=f\left( r,Y_{r}^{1},Z_{r}^{1},Y_{r+\delta \left( r \right)}^{1},Z_{r+\zeta \left( r \right)}^{1} \right)-f\left( r,Y_{r}^{2},Z_{r}^{2},Y_{r+\delta \left( r \right)}^{2},Z_{r+\zeta \left( r \right)}^{2} \right).$

We obtain the following equation

{ Y¯t=tTΔf(r)dr+tTΔg(r)dBrtTZ¯rdWr,0tT,Y¯t=0,Z¯t=0,TtT+K. $$\left\{ \begin{array}& {{{\bar{Y}}}_{t}}=\int_{t}^{T}{\Delta f\left( r \right)dr+\int_{t}^{T}{\Delta g\left( r \right)d{{B}_{r}}-\int_{t}^{T}{{{{\bar{Z}}}_{r}}d{{W}_{r}},\,\,\,\,\,\,\,0\le t\le T,}}} \\ {{{\bar{Y}}}_{t}}=0,\,\,\,\,{{{\bar{Z}}}_{t}}=0,\,\,\,\,\,\,\,\,\,\,T\le t\le T+K. \\ \end{array}\right.$$

By the same computations as in Lemma 4.1 , we obtain

E[ | Y¯t |2 ]+12EtT| Y¯r |2dr+12EtT| Z¯r |2drCEtTρ(| Y¯r |2)dr,0tT. $$E\left[ {{\left| {{{\bar{Y}}}_{t}} \right|}^{2}} \right]+\frac{1}{2}\mathbf{E}\int_{t}^{T}{{{\left| {{{\bar{Y}}}_{r}} \right|}^{2}}dr+\frac{1}{2}\mathbf{E}}\int_{t}^{T}{{{\left| {{{\bar{Z}}}_{r}} \right|}^{2}}}dr\le {C}'\mathbf{E}\int_{t}^{T}{\rho }\left( {{\left| {{{\bar{Y}}}_{r}} \right|}^{2}} \right)dr,\,\,\,\,0\le t\le T.$$

This leads to

E[ | Y¯t |2 ]CEtT+Kρ(| Y¯r |2)dr,0tT+K. $$\mathbf{E}\left[ {{\left| {{{\bar{Y}}}_{t}} \right|}^{2}} \right]\le {C}'\mathbf{E}\int_{t}^{T+K}{\rho \left( {{\left| {{{\bar{Y}}}_{r}} \right|}^{2}} \right)}dr,\,\,\,\,\,\,0\le t\le T+K.$$

Using Fubini’s theorem and Jensen’s inequality, we deduce that

E| Y¯t |2CtT+Kρ(E| Y¯r |2)dr,0tT+K. $$\mathbf{E}{{\left| {{{\bar{Y}}}_{t}} \right|}^{2}}\le {C}'\int_{t}^{T+K}{\rho \left( \mathbf{E}{{\left| {{{\bar{Y}}}_{r}} \right|}^{2}} \right)}\,dr,\,\,\,\,\,\,0\le t\le T+K.$$

Then we can use Bihari’s inequality to obtain E| Y¯t |2=0,0tT+K. $\mathbf{E}{{\left| {{{\bar{Y}}}_{t}} \right|}^{2}}=0,0\le t\le T+K.$This implies Z¯t=0. ${{\bar{Z}}_{t}}=0.$

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