RETRACTED ARTICLE: Anticipated backward doubly stochastic differential equations with non-Liphschitz coefficients

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)

$\begin{matrix} \{\begin{cases} - d Y_{t} = f (t, Y_{t}, Z_{t}, Y_{t + δ (t)}, Z_{t + ζ (t)}) d t - Z_{t} d W_{t}, 0 \leq t \leq T, \\ Y_{t} = ξ_{t}, Z_{t} = η_{t}, T \leq t \leq T + K, \end{cases} \end{matrix}$ $$\begin{array}{} \displaystyle \begin{cases} -dY_t= f\left(t, Y_t, Z_t,Y_ {t+\delta(t)}, Z_ {t+\zeta(t)}\right)dt - Z_t dW_t,\quad \quad 0\le t \le T,\\ Y_{t} = \xi_{t}, \quad Z_{t} =\eta_{t},\quad\quad T\le t \le T+K, \end{cases} \end{array}$$

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

$\begin{matrix} \{\begin{cases} Y_{t} = ξ_{T} + \int_{t}^{T} f (r, Y_{r}, Z_{r}, Y_{r + δ (r)}, Z_{r + ζ (r)}) d r + \int_{t}^{T} g (r, Y_{r}, Z_{r}) d B_{r} - \int_{t}^{T} Z_{r} d W_{r}, 0 \leq t \leq T, \\ Y_{t} = ξ_{t}, Z_{t} = η_{t}, T \leq t \leq T + K \end{cases} \end{matrix}$ $$\begin{array}{} \displaystyle \begin{cases} Y_t= \xi_{T} + \displaystyle \int_t^T f(r,Y_r, Z_r, Y_{r+\delta(r)}, Z_{r+\zeta(r)}) dr + \int_t^T g(r,Y_r, Z_r)dB_r - \displaystyle \int_t^T Z_r dW_r, \quad 0\le t \le T,\\ Y_{t} = \xi_{t}, \quad Z_{t} =\eta_{t}, \quad\quad T\le t \le T+K \end{cases} \end{array}$$(1)

where K is a positive constant, ξ., 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,

$\begin{matrix} \int_{t}^{T} h (r + ϕ (r)) d r \leq M \int_{t}^{T + K} h (r) d r, ϕ \in {δ, ζ} . \end{matrix}$ $$\begin{array}{} \displaystyle \int_t^T h (r+ \phi(r)) dr \le M \int_t^{T+K} h (r) dr, \quad \phi \in \{\delta, \zeta\}. \end{array}$$

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, ℱ a σ-algebra of sets of Ω and P a probability measure defined on ℱ. 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 (B_t)_0≤t≤T,

a d—dimensional Brownian motion (W_t)_0≤t≤T.

We consider the family (ℱ_t)_0≤t≤T given by

$\begin{matrix} F_{t} = F_{t}^{W} \lor F_{t, T}^{B}, 0 \leq t \leq T, G_{s} = F_{0, s}^{W} \lor F_{s, T + K}^{B}, 0 \leq s \leq T + K, \end{matrix}$ $$\begin{array}{} \displaystyle {\mathscr F}_{t} = {\mathscr F}^W_{t} \vee {\mathscr F}^B_{t, T}, \quad \; \;0\le t\le T,\quad \; \; {\mathscr G}_{s} = {\mathscr F}^W_{0,s} \vee {\mathscr F}^B_{s,T+K}, \quad 0\le s\le T+K, \end{array}$$

where for any process $\begin{matrix} {φ_{t}}_{t \geq 0}, F_{s, t}^{φ} = σ {φ_{r} - φ_{s}, s \leq r \leq t} \lor N, F_{t}^{φ} = F_{0, t}^{φ} . \end{matrix}$ $\begin{array}{} \displaystyle \{\varphi_t\}_{t\ge0}, \; {\mathscr F}^\varphi_{s,t} = \sigma\{\varphi_r-\varphi_s, \; s\le r\le t\}\vee {\mathscr N}, \; \; {\mathscr F}^\varphi_{t} = {\mathscr F}^\varphi_{0,t}. \end{array}$ Here 𝒩 denotes the class of P—null sets of ℱ. Note that (ℱ_t)_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(𝒢_T, R^k) the space of 𝒢_T-measurable random variable such that E[|ξ_T|²] < +∞.

$\begin{matrix} S_{[0, T]}^{2} \end{matrix}$ $\begin{array}{} \displaystyle {\mathscr S}_{[0, T]}^2 \end{array}$(𝒢, R^k) the space of 𝒢_t—adapted càdlàg processes

$\begin{matrix} Ψ : [0, T] \times Ω ⟶ R^{k}, | | Ψ | |_{S^{2} (R^{k})}^{2} = E (sup_{0 \leq t \leq T} | Ψ_{t} |^{2}) < \infty . \end{matrix}$ $$\begin{array}{} \displaystyle \Psi: [0, T]\times \Omega \longrightarrow \mathbf{R}^k, \; || \Psi ||_{\mathscr{S}^2(\mathbf{R}^k)}^2 = {\bf E} \left(\sup_{0\leq t\leq T} |\Psi_t|^2 \right) \lt \infty. \end{array}$$

$\begin{matrix} M_{[0, T]}^{2} \end{matrix}$ $\begin{array}{} \displaystyle \mathscr M_{[0, T]}^2 \end{array}$ (𝒢, R^k×d) the space of 𝒢_t—progressively measurable processes

$\begin{matrix} Ψ : [0, T] \times Ω ⟶ R^{k \times d}, | | Ψ | |_{M^{2} (R^{k \times d})}^{2} = E \int_{0}^{T} | Ψ_{t} |^{2} d t < \infty . \end{matrix}$ $$\begin{array}{} \displaystyle \Psi: [0, T]\times \Omega \longrightarrow \mathbf{R}^{k\times d},\; || \Psi ||_{\mathscr{M}^2(\mathbf{R}^{k\times d})}^2 = {\bf E} \int_0^T |\Psi_t|^2 \, dt \lt \infty. \end{array}$$

$\begin{matrix} C_{G}^{2} (0, T) = M_{[0, T]}^{2} (G, R^{k}) \times M_{[0, T]}^{2} (G, R^{k \times d}) \end{matrix}$ $\begin{array}{} \displaystyle {\mathscr C}_\mathscr{G}^2 (0, T) = \mathscr M_{[0, T]}^2 (\mathscr{G}, \mathbf{R}^k) \times \mathscr{M}_{[0, T]}^2 (\mathscr{G}, \mathbf{R}^{k\times d}) \end{array}$ endowed with the norm

$\begin{matrix} | | (Y, Z) | |_{C_{G}^{2} (0, T)}^{2} = | | Y | |_{M^{2} (R^{k})}^{2} + | | Z | |_{M^{2} (R^{k \times d})}^{2} . \end{matrix}$ $$\begin{array}{} \displaystyle ||(Y,Z)||^2_ {{\mathscr C}_\mathscr{G}^2 (0, T)} = || Y||_{\mathscr{M}^2(\mathbf{R}^k)}^2 + ||Z||_{\mathscr{M}^2(\mathbf{R}^{k\times d})}^2 . \end{array}$$

$\begin{matrix} B_{G}^{2} (0, T) = S_{[0, T]}^{2} (G, R^{k}) \times M_{[0, T]}^{2} (G, R^{k \times d}) \end{matrix}$ $\begin{array}{} \displaystyle {\mathscr B}_\mathscr{G}^2 (0, T) = \mathscr S_{[0, T]}^2 (\mathscr{G}, \mathbf{R}^k) \times \mathscr{M}_{[0, T]}^2 (\mathscr{G}, \mathbf{R}^{k\times d}) \end{array}$ endowed with the norm

$\begin{matrix} | | (Y, Z) | |_{B_{G}^{2} (0, T)}^{2} = | | Y | |_{S^{2} (R^{k})}^{2} + | | Z | |_{M^{2} (R^{k \times d})}^{2} . \end{matrix}$ $$\begin{array}{} \displaystyle ||(Y,Z)||^2_ {{\mathscr B}_\mathscr{G}^2 (0, T)} = || Y||_{\mathscr{S}^2(\mathbf{R}^k)}^2 + ||Z||_{\mathscr{M}^2(\mathbf{R}^{k\times d})}^2 . \end{array}$$

S be the set of all nondecreasing, continuous and concave function ρ(⋅) : R₊ → R₊ satisfying ρ(0) = 0, ρ(s) > 0 for s > 0 and $\begin{matrix} \int_{0 +} \frac{d u}{ρ (u)} = + \infty . \end{matrix}$ $\begin{array}{} \int_{0+}\frac{du}{\rho(u)}= +\infty. \end{array}$

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] × R^k × R^k×d, f(r, 0) = h(r, 0, 0, 0, 0), for all x, y ∈ R^k |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), if (Y, Z) ∈ $\begin{matrix} B_{G}^{2} \end{matrix}$ $\begin{array}{} \displaystyle {\mathscr B}_\mathscr{G}^2 \end{array}$ (0, T + K) and it satisfies eq.(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) with Lipschitz coefficients. For this purpose, we first make the following assumptions.

3.0.1

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

$\begin{matrix} \begin{aligned} | f (t, y, z, θ (r), φ (r)) - f (t, y^{'}, z^{'}, θ^{'} (r), φ^{'} (r)) |^{2} & \leq c (| y - y^{'} |^{2} + | z - z^{'} |^{2}) \\ + E^{F_{t}} [| θ (r) - θ^{'} (r) |^{2} + | φ (r) - φ^{'} (r) |^{2}], \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle |f(t, y,z, \theta(r), \varphi(r))- f(t, y',z', \theta'(r), \varphi'(r))|^{2} &\leq c (|y-y^\prime|^{2} + |z-z^\prime|^{2}) \\&+ {\bf E}^{\mathscr{F}_{t}}[|\theta(r) - \theta'(r)|^{2} + |\varphi(r) - \varphi'(r)|^{2}], \end{split} \end{array}$$

for all (r, r′) ∈ [t, T + K], (t, y, z, θ(r), φ(r)), (t, y′, z′, θ′(r), φ′(r)) ∈ 𝒜 × $\begin{matrix} C_{G}^{2} \end{matrix}$ $\begin{array}{} \displaystyle {\mathscr C}_\mathscr{G}^2 \end{array}$ (t, T + K).

(H1.2)

There exists a constant 0 < α₁ < 1 such that for any (t, y, z), (t, y′, z′) ∈ [0, T] × R^k × R^k×d

$\begin{matrix} | g (t, y, z) - g (t, y^{'}, z^{'}) |^{2} \leq c | y - y^{'} |^{2} + α_{1} | z - z^{'} |^{2} . \end{matrix}$ $$\begin{array}{} \displaystyle |g(t, y,z) - g(t, y',z')|^2 \leq c|y-y^\prime|^{2} + \alpha_{1}|z-z^\prime|^2. \end{array}$$

(H1.3)

For any (t, y, z) ∈ [0, T] × R^k × R^k×d,

$\begin{matrix} E [\int_{0}^{T} | f (s, 0) |^{2} d s] + E [\int_{0}^{T} | g (s, y, z) |^{2} d s] < \infty . \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E}\left[\int_0^T |f(s, 0)|^{2}ds\right] + {\bf E}\left[\int_0^T |g(s, y,z)|^{2}ds\right] \lt \infty. \end{array}$$

3.1

Existence and uniqueness of solution

Lemma 3.1

Suppose that (Y_t, Z_t)_0≤t≤T ∈ $\begin{matrix} C_{G}^{2} \end{matrix}$ $\begin{array}{} \displaystyle {\mathscr C}_\mathscr{G}^2 \end{array}$ (0, T + K) is the unique solution to the ABDSDE (1). Then Y ∈ $\begin{matrix} S_{[0, T]}^{2} \end{matrix}$ $\begin{array}{} \displaystyle \mathscr S_{[0, T]}^2 \end{array}$ (𝒢, R^k).

Proof

Itô’s formula applied to eq.(1) yields, for 0 ≤ t ≤ T

$\begin{matrix} \begin{aligned} | Y_{t} |^{2} + \int_{t}^{T} | Z_{r} |^{2} d r & = | ξ_{T} |^{2} + 2 \int_{t}^{T} 〈 Y_{r}, f (r, Y_{r}, Z_{r}, Y_{r + δ (r)}, Z_{r + ζ (r)}) 〉 d r + 2 \int_{t}^{T} 〈 Y_{r}, g (r, Y_{r}, Z_{r}) d B_{r} 〉 \\ - 2 \int_{t}^{T} 〈 Y_{r}, Z_{r} d W_{r} 〉 + \int_{t}^{T} | g (r, Y_{r}, Z_{r}) |^{2} d r . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle |Y_t|^2 + \int_t^T|Z_r|^2dr &= |\xi_{T}|^2+ 2\int_t^T\langle Y_r,f(r,Y_r, Z_r, Y_{r+\delta(r)}, Z_{r+\zeta(r)})\rangle dr + 2 \int_t^T\langle Y_r,g(r,Y_r, Z_r)dB_r\rangle \\ &-2\int_t^T\langle Y_r, Z_rdW_r\rangle + \int_t^T|g(r,Y_r, Z_r)|^2dr. \end{split} \end{array}$$(2)

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

$\begin{matrix} \begin{aligned} 2 E \int_{t}^{T} & 〈 Y_{r}, f (r, Y_{r}, Z_{r}, Y_{r + δ (r)}, Z_{r + ζ (r)}) 〉 d r \leq \frac{1}{ε} E \int_{t}^{T} | Y_{r} |^{2} d r + ε c E \int_{t}^{T} (| Y_{r} |^{2} + | Z_{r} |^{2}) d r \\ + ε E \int_{t}^{T} E^{F_{t}} [| Y_{r + δ (r)} |^{2} + | Z_{r + ζ (r)} |^{2}] d r + 2 E \int_{t}^{T} | Y_{r} | | f (r, 0) | d r . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle 2{\bf E}\int_t^T&\langle Y_r,f(r,Y_r, Z_r, Y_{r+\delta(r)}, Z_{r+\zeta(r)})\rangle dr \le \frac{1}{\varepsilon}{\bf E}\int_t^T|Y_r|^2dr + \varepsilon c{\bf E}\int_t^T(|Y_r|^2 +|Z_r|^2)dr \\ &+ \varepsilon{\bf E}\int_t^T{\bf E}^{\mathscr{F}_{t}}[|Y_{r+\delta(r)}|^{2} + |Z_{r+\zeta(r)}|^{2}]dr + 2{\bf E}\int_t^T|Y_r||f(r,0)|dr. \end{split} \end{array}$$

Applying (A2), we obtain finally

$\begin{matrix} \begin{aligned} 2 E \int_{t}^{T} & 〈 Y_{r}, f (r, Y_{r}, Z_{r}, Y_{r + δ (r)}, Z_{r + ζ (r)}) 〉 d r \leq (\frac{1}{ε} + ε (c + M)) E \int_{t}^{T} | Y_{r} |^{2} d r \\ + ε (c + M) E \int_{t}^{T} | Z_{r} |^{2} d r + 2 E \int_{t}^{T} | Y_{r} | | f (r, 0) | d r + ε M E \int_{T}^{T + K} (| ξ_{r} |^{2} + | η_{r} |^{2}) d r . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle 2{\bf E}\int_t^T&\langle Y_r,f(r,Y_r, Z_r, Y_{r+\delta(r)}, Z_{r+\zeta(r)})\rangle dr \le \left(\frac{1}{\varepsilon}+\varepsilon(c+M)\right){\bf E}\int_t^T|Y_r|^2dr\\ &+\varepsilon(c\!+\!M){\bf E}\int_t^T\!\!|Z_r|^2dr + 2{\bf E}\int_t^T|Y_r||f(r,0)|dr +\varepsilon M{\bf E}\int_T^{T\!+\!K}\!\!(|\xi_r|^2 \!+\!|\eta_r|^2)dr. \end{split} \end{array}$$

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

$\begin{matrix} \begin{aligned} E \int_{t}^{T} | g (r, Y_{r}, Z_{r}) |^{2} d r & \leq E \int_{t}^{T} | g (r, Y_{r}, Z_{r}) - g (r, 0, 0) |^{2} d r + E \int_{t}^{T} | g (r, 0, 0) |^{2} d r \\ \leq c E \int_{t}^{T} | Y_{r} |^{2} d r + α_{1} E \int_{t}^{T} | Z_{r} |^{2} d r + E \int_{t}^{T} g (r, 0, 0) |^{2} d r . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle {\bf E}\int_t^T|g(r,Y_r, Z_r)|^2dr &\le {\bf E}\int_t^T|g(r,Y_r, Z_r)-g(r,0,0)|^2dr +{\bf E}\int_t^T|g(r,0, 0)|^2dr\\ &\le c{\bf E}\int_t^T\!|Y_r|^2dr + \alpha_{1}{\bf E}\int_t^T\!\!|Z_r|^2dr +{\bf E}\int_t^Tg(r,0, 0)|^2dr. \end{split} \end{array}$$

Putting pieces together, we deduce from (2) that

$\begin{matrix} \begin{aligned} E \int_{t}^{T} | Z_{r} |^{2} d r & \leq E [| ξ_{T} |^{2}] + M ε E \int_{T}^{T + K} (| ξ_{r} |^{2} + | η_{r} |^{2}) d r + (\frac{1}{ε} + ε (c + M) + c) E \int_{t}^{T} | Y_{r} |^{2} d r \\ + 2 E \int_{t}^{T} | Y_{r} | | f (r, 0) | d r + E \int_{t}^{T} | g (r, 0, 0) |^{2} d r + (α_{1} + ε (c + M)) E \int_{t}^{T} | Z_{r} |^{2} d r . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle {\bf E}\int_t^T|Z_r|^2dr &\le {\bf E}[|\xi_{T}|^2] + M\varepsilon{\bf E}\int_T^{T\!+\!K}\!\!\!(|\xi_r|^2 \!+\!|\eta_r|^2)dr + \left(\frac{1}{\varepsilon}\!+\!\varepsilon(c\!+\!M)\!+\!c\right){\bf E}\int_t^T\!\!|Y_r|^2dr\\ &+ 2{\bf E}\int_t^T|Y_r||f(r,0)|dr + {\bf E}\int_t^T|g(r, 0,0)|^2dr +\left(\alpha_{1}\!+\!\varepsilon(c\!+\!M)\right){\bf E}\int_t^T\!\!|Z_r|^2\!dr. \end{split} \end{array}$$

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

$\begin{matrix} E \int_{t}^{T} | Z_{r} |^{2} d r \leq \frac{1}{C_{0}} E [X_{t} + 2 \int_{t}^{T} | Y_{r} | | f (r, 0) | d r + \int_{t}^{T} | g (r, 0, 0) |^{2} d r] \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E}\int_t^T|Z_r|^2dr \le \frac{1}{C_0}{\bf E} \left[ X_t + 2 \int_t^T|Y_r||f(r,0)|dr + \int_t^T|g(r, 0,0)|^2dr\right] \end{array}$$(3)

where putting $\begin{matrix} C_{1} = (\frac{1}{ε_{0}} + ε_{0} (c + M) + c), \end{matrix}$ $\begin{array}{} \displaystyle C_1=\left(\frac{1}{\varepsilon_0}\!+\!\varepsilon_0(c\!+\!M)\!+\!c\right), \end{array}$

$\begin{matrix} X_{t} = [| ξ_{T} |^{2} + M ε_{0} \int_{T}^{T + K} (| ξ_{r} |^{2} + | η_{r} |^{2}) d r + C_{1} \int_{t}^{T} | Y_{r} |^{2} d r] . \end{matrix}$ $$\begin{array}{} \displaystyle X_t= \left[|\xi_{T}|^2 + M\varepsilon_0\int_T^{T\!+\!K}\!\!\!(|\xi_r|^2 \!+\!|\eta_r|^2)dr + C_1\int_t^T\!\!|Y_r|^2dr\right]. \end{array}$$

By the same computations as before, we have

$\begin{matrix} \begin{aligned} E [| ξ_{T} |^{2}] & + 2 E \int_{t}^{T} 〈 Y_{r}, f (r, Y_{r}, Z_{r}, Y_{r + δ (r)}, Z_{r + ζ (r)}) 〉 d r + E \int_{t}^{T} | g (r, Y_{r}, Z_{r}) |^{2} d r \\ \leq \frac{1}{C_{0}} E [X_{t} + 2 \int_{t}^{T} | Y_{r} | | f (r, 0) | d r + \int_{t}^{T} | g (r, 0, 0) |^{2} d r] . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle {\bf E}[|\xi_{T}|^2] &+2{\bf E}\int_t^T\langle Y_r,f(r,Y_r, Z_r, Y_{r+\delta(r)}, Z_{r+\zeta(r)})\rangle dr +{\bf E}\int_t^T|g(r,Y_r, Z_r)|^2dr \\ &\le \frac{1}{C_0}{\bf E} \left[ X_t + 2 \int_t^T|Y_r||f(r,0)|dr + \int_t^T|g(r,0, 0 )|^2dr\right]. \end{split} \end{array}$$(4)

Moreover using again eq.(2), we have

$\begin{matrix} \begin{aligned} E (sup_{t \leq r \leq T} | Y_{r} |^{2}) & \leq E [| ξ_{T} |^{2}] + 2 E sup_{t \leq s \leq T} (\int_{s}^{T} 〈 Y_{r}, f (r, Y_{r}, Z_{r}, Y_{r + δ (r)}, Z_{r + ζ (r)}) 〉 d r) \\ + 2 E sup_{t \leq s \leq T} |\int_{s}^{T} 〈 Y_{r}, g (r, Y_{r}, Z_{r}) d B_{r} 〉| + 2 E sup_{t \leq s \leq T} |\int_{s}^{T} 〈 Y_{r}, Z_{r} d W_{r} 〉| \\ + \int_{t}^{T} | g (r, Y_{r}, Z_{r}) |^{2} d r . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle {\bf E}\left(\sup_{t\le r\le T}|Y_r|^2\right) &\le {\bf E}[|\xi_{T}|^2] + 2{\bf E}\sup_{t\le s\le T}\left(\int_s^T\langle Y_r,f(r,Y_r, Z_r, Y_{r+\delta(r)}, Z_{r+\zeta(r)})\rangle dr\right)\\ &+ 2 {\bf E}\sup_{t\le s\le T}\left|\int_s^T\langle Y_r,g(r,Y_r, Z_r)dB_r\rangle\right| + 2{\bf E}\sup_{t\le s\le T}\left|\int_s^T\langle Y_r, Z_rdW_r\rangle\right|\\ &+ \int_t^T|g(r,Y_r, Z_r)|^2dr. \end{split} \end{array}$$(5)

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

$\begin{matrix} \begin{aligned} E sup_{t \leq s \leq T} |\int_{s}^{T} 〈 Y_{r}, g (r, Y_{r}, Z_{r}) d B_{r} 〉| & \leq \frac{1}{8} E (sup_{t \leq r \leq T} | Y_{r} |^{2}) + C \int_{t}^{T} | g (r, Y_{r}, Z_{r}) |^{2} d r \\ 2 E sup_{t \leq s \leq T} |\int_{s}^{T} 〈 Y_{r}, Z_{r} d W_{r} 〉| & \leq \frac{1}{8} E (sup_{t \leq r \leq T} | Y_{r} |^{2}) + C \int_{t}^{T} | Z_{r} |^{2} d r . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle {\bf E}\sup_{t\le s\le T}\left|\int_s^T\langle Y_r,g(r,Y_r, Z_r)dB_r\rangle\right| &\le \frac{1}{8}{\bf E}\left(\sup_{t\le r\le T}|Y_r|^2\right) + C\int_t^T|g(r,Y_r, Z_r)|^2dr \\ 2{\bf E}\sup_{t\le s\le T}\left|\int_s^T\langle Y_r, Z_rdW_r\rangle\right| &\le \frac{1}{8}{\bf E}\left(\sup_{t\le r\le T}|Y_r|^2\right) + C\int_t^T|Z_r|^2dr. \end{split} \end{array}$$

Using the above inequalities, we deduce from (5) that

$\begin{matrix} \begin{aligned} \frac{3}{4} E (sup_{t \leq r \leq T} | Y_{r} |^{2}) & \leq E [| ξ_{T} |^{2}] + 2 E sup_{t \leq s \leq T} (\int_{s}^{T} 〈 Y_{r}, f (r, Y_{r}, Z_{r}, Y_{r + δ (r)}, Z_{r + ζ (r)}) 〉 d r) \\ + C \int_{t}^{T} | g (r, Y_{r}, Z_{r}) |^{2} d r + C E \int_{t}^{T} | Z_{r} |^{2} d r \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle \frac{3}{4}{\bf E}\left(\sup_{t\le r\le T}|Y_r|^2\right) \!&\le {\bf E}[|\xi_{T}|^2] + 2{\bf E}\sup_{t\le s\le T}\left(\int_s^T\langle Y_r,f(r,Y_r, Z_r, Y_{r+\delta(r)}, Z_{r+\zeta(r)})\rangle dr\right) \\ &+ C\int_t^T|g(r,Y_r, Z_r)|^2dr + C{\bf E}\int_t^T\!|Z_r|^2dr \end{split} \end{array}$$

Applying (3) and (4), we deduce that

$\begin{matrix} \frac{3}{4} E (sup_{t \leq r \leq T} | Y_{r} |^{2}) \leq \frac{2 C}{C_{0}} E [X_{t} + 2 \int_{t}^{T} | Y_{r} | | f (r, 0) | d r + \int_{t}^{T} | g (r, 0, 0) |^{2} d r] . \end{matrix}$ $$\begin{array}{} \displaystyle \frac{3}{4}{\bf E}\left(\sup_{t\le r\le T}|Y_r|^2\right) \le \frac{2C}{C_0}{\bf E} \left[ X_t + 2 \int_t^T|Y_r||f(r,0)|dr + \int_t^T|g(r,0, 0 )|^2dr\right]. \end{array}$$(6)

Moreover, we have

$\begin{matrix} \frac{4 C}{C_{0}} E \int_{t}^{T} | Y_{r} | | f (r, 0) | d r \leq \frac{1}{4} E (sup_{t \leq r \leq T} | Y_{r} |^{2}) + 4 {(\frac{2 C}{C_{0}})}^{2} E {(\int_{t}^{T} | f (r, 0) | d r)}^{2} . \end{matrix}$ $$\begin{array}{} \displaystyle \frac{4C}{C_0}{\bf E}\int_t^T|Y_r||f(r,0)|dr \le \frac{1}{4}{\bf E}\left(\sup_{t\le r\le T}|Y_r|^2\right) + 4\left(\frac{2C}{C_0}\right)^2{\bf E}\left(\int_t^T|f(r,0)|dr\right)^2. \end{array}$$

Hence gathering (3) and (6) we obtain

$\begin{matrix} \frac{1}{2} E (sup_{t \leq r \leq T} | Y_{r} |^{2}) + E \int_{t}^{T} | Z_{r} |^{2} d r \leq C_{2} E [X_{t} + {(\int_{t}^{T} | f (r, 0) | d r)}^{2} + \int_{t}^{T} | g (r, 0, 0) |^{2} d r], \end{matrix}$ $$\begin{array}{} \displaystyle \frac{1}{2}{\bf E}\left(\sup_{t\le r\le T}|Y_r|^2\right) + {\bf E}\int_t^T|Z_r|^2dr \le C_2{\bf E} \left[X_t + \left(\int_t^T|f(r,0)|dr\right)^2 + \int_t^T|g(r, 0,0)|^2dr\right], \end{array}$$(7)

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

Then, applying the fubini’s theorem to (7), this leads to

$\begin{matrix} \begin{aligned} E (sup_{t \leq r \leq T} | Y_{r} |^{2}) & \leq C_{2} E [| ξ_{T} |^{2} + \int_{T}^{T + K} (| ξ_{r} |^{2} + | η_{r} |^{2}) d r + {(\int_{t}^{T} | f (r, 0) | d r)}^{2} + \int_{t}^{T} | g (r, 0, 0) |^{2} d r] \\ + C_{3} \int_{t}^{T} E [sup_{r \leq s \leq T} | Y_{s} |^{2}] d r, \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle {\bf E}\left(\sup_{t\le r\le T}|Y_r|^2\right) &\le C_2{\bf E}\bigg[|\xi_{T}|^2 + \int_T^{T\!+\!K}\!\!\!(|\xi_r|^2 \!+\!|\eta_r|^2)dr + \left(\int_t^T\!\!|f(r,0)|dr\right)^2 + \int_t^T\!\!|g (r, 0,0)|^2dr \bigg]\\ &+ C_3\int_t^T\!\!{\bf E}[\sup_{r\le s\le T}|Y_s|^2]dr, \end{split} \end{array}$$

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

$\begin{matrix} E (sup_{t \leq r \leq T} | Y_{r} |^{2}) \leq + \infty . \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E}\left(\sup_{t\le r\le T}|Y_r|^2\right) \le +\infty. \end{array}$$

This implies that Y ∈ $\begin{matrix} S_{[0, T]}^{2} \end{matrix}$ $\begin{array}{} \displaystyle \mathscr S_{[0, T]}^2 \end{array}$ (𝒢, R^k). 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

$\begin{matrix} Y_{t} = ξ_{T} + \int_{t}^{T} f (r) d r + \int_{t}^{T} g (r) d B_{r} - \int_{t}^{T} Z_{r} d W_{r}, 0 \leq t \leq T . \end{matrix}$ $$\begin{array}{} \displaystyle Y_t = \xi_{T} + \int_t^T f(r) dr + \int_t^T g(r)dB_r - \int_t^TZ_rdW_r, \quad 0\le t\le T. \end{array}$$(8)

where $\begin{matrix} f \in M_{[0, T]}^{2} (G, R^{k}), g \in M_{[0, T]}^{2} (G, R^{k \times ℓ}) and ξ_{T} \in L^{2} (G_{T}, R^{k}) . \end{matrix}$ $\begin{array}{} \displaystyle f\in\mathscr{M}_{[0, T]}^2 (\mathscr{G}, \mathbf{R}^k), \quad g\in\mathscr{M}_{[0, T]}^2 (\mathscr{G}, \mathbf{R}^{k\times \ell})~ \text{and}~ \xi_{T} \in L^{2}({\mathscr G}_{T}, \mathbf{R}^{k}). \end{array}$

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 ξ_T ∈ L²(𝒢_T, R^k), eq.(8) has a unique solution (Y_t, Z_t)_0≤t≤T ∈ $\begin{matrix} C_{G}^{2} \end{matrix}$ $\begin{array}{} \displaystyle {\mathscr C}_\mathscr{G}^2 \end{array}$ (0, T).

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 ξ_T ∈ L²(𝒢_T, R^k). Then for any $\begin{matrix} (ξ, η) \in S_{[T, T + K]}^{2} (G, R^{k}) \times M_{[T, T + K]}^{2} (G, R^{k \times d}) \end{matrix}$ $\begin{array}{} \displaystyle (\xi,\eta)\in\mathscr{S}_{[T, T+K]}^2 (\mathscr{G}, \mathbf{R}^k)\times \mathscr{M}_{[T, T+K]}^2 (\mathscr{G}, \mathbf{R}^{k\times d}) \end{array}$ the ABDSDE (1) has a unique solution (Y_t, Z_t)_0≤t≤T ∈ $\begin{matrix} B_{G}^{2} \end{matrix}$ $\begin{array}{} \displaystyle {\mathscr B}_\mathscr{G}^2 \end{array}$ (0, T + K).

Proof

Existence. Let us consider the mapping

$\begin{matrix} \begin{aligned} Ψ : C_{G}^{2} (0, T + K) & \to C_{G}^{2} (0, T + K), \\ (y, z) & \to (Y, Z) \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle \Psi : {\mathscr C}_\mathscr{G}^2 (0, T+K) &\to {\mathscr C}_\mathscr{G}^2 (0, T+K), \\ (y, z) &\to (Y, Z) \end{split} \end{array}$$

where the pair (Y_t, Z_t)_0≤t≤T+K ∈ $\begin{matrix} C_{G}^{2} \end{matrix}$ $\begin{array}{} \displaystyle {\mathscr C}_\mathscr{G}^2 \end{array}$ (0, T + K) is s.t. (Y_t, Z_t)_T≤t≤T+K = (ξ_t, η_t) and it satisfies the equation

$\begin{matrix} \{\begin{cases} \forall 0 \leq t \leq T, \\ Y_{t} = ξ_{T} + \int_{t}^{T} f (r, y_{r}, z_{r}, y_{r + δ (r)}, z_{r + ζ (r)}) d r + \int_{t}^{T} g (r, y_{r}, z_{r}) d B_{r} - \int_{t}^{T} Z_{r} d W_{r}, \\ \forall t \in [T, T + K], Y_{t} = ξ_{t}, Z_{t} = η_{t} . \end{cases} \end{matrix}$ $$\begin{array}{} \displaystyle \begin{cases} \forall 0\le t \le T,\\ Y_t= \xi_{T} + \displaystyle \int_t^T f(r,y_r, z_r,y_{r+\delta(r)}, z_{r+\zeta(r)})dr + \int_t^T g(r,y_r, z_r)dB_r - \displaystyle \int_t^T Z_r dW_r, \\ \forall t\in [T, T+K], \quad Y_{t} = \xi_{t}, \quad Z_{t} = \eta_{t}. \end{cases} \end{array}$$(9)

Thanks to Proposition 3.2, the mapping Ψ is well defined. Let (Y, Z) and (Y͠, Z͠) be two solutions of eq.(9), i.e:

$\begin{matrix} (Y, Z) = Ψ (y, z) a n d (\tilde{Y}, \tilde{Z}) = Ψ (\tilde{y}, \tilde{z}) . \end{matrix}$ $$\begin{array}{} \displaystyle (Y, Z )=\Psi (y, z) \quad and \quad (\widetilde Y, \widetilde Z)= \Psi (\widetilde y, \widetilde z). \end{array}$$

Fix β ∈ R. The pair (Y, Z) solves the ABDSDE

$\begin{matrix} \{\begin{cases} {\bar{Y}}_{t} = \int_{t}^{T} Δ f (r) d r + \int_{t}^{T} Δ g (r) d B_{r} - \int_{t}^{T} {\bar{Z}}_{r} d W_{r}, \forall 0 \leq t \leq T, \\ \forall t \in [T, T + K], {\bar{Y}}_{t} = 0, {\bar{Z}}_{t} = 0, \end{cases} \end{matrix}$ $$\begin{array}{} \displaystyle \begin{cases} \overline Y_t= \displaystyle \int_t^T\!\! \Delta f(r) dr + \int_t^T \!\! \Delta g(r) dB_r - \displaystyle \int_t^T \overline Z_rdW_r,\quad \forall \, 0\le t \le T,\\ \forall t\in [T, T+K], \quad \overline Y_t = 0, \quad \overline Z_t = 0, \end{cases} \end{array}$$(10)

where for ρ ∈ {Y, Z}, ρ = ρ – ρ͠, Δ g(r) = g(r, y_r, z_r) – g(r, y͠_r, z͠_r) and

$\begin{matrix} Δ f (r) = f (r, y_{r}, z_{r}, y_{r + δ (r)}, z_{r + ζ (r)}) - f (r, {\tilde{y}}_{r}, {\tilde{z}}_{r}, {\tilde{y}}_{r + δ (r)}, {\tilde{z}}_{r + ζ (r)}) . \end{matrix}$ $$\begin{array}{} \displaystyle \Delta f(r) = f(r,y_r, z_r,y_{r+\delta(r)}, z_{r+\zeta(r)}) -f(r,\widetilde y_r, \widetilde z_r,\widetilde y_{r+\delta(r)}, \widetilde z_{r+\zeta(r)}). \end{array}$$

Applying Ito’s formula, we obtain

$\begin{matrix} E [e^{β t} | {\bar{Y}}_{t} |^{2} + β \int_{t}^{T} e^{β r} | {\bar{Y}}_{r} |^{2} d r + \int_{t}^{T} e^{β r} | {\bar{Z}}_{r} |^{2} d r] = 2 E \int_{t}^{T} e^{β r} 〈 {\bar{Y}}_{r}, Δ f (r) 〉 d r + E \int_{t}^{T} e^{β r} | Δ g (r) |^{2} d r . \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E}\bigg[e^{\beta t}|\overline Y_t|^2 + \beta\int_t^T\! e^{\beta r}|\overline Y_r|^2 dr + \int_t^T \!\!e^{\beta r} |\overline Z_r|^2 dr\bigg] = 2 {\bf E} \int_t^T \!\! e^{\beta r}\langle \overline Y_r, \Delta f(r)\rangle dr + {\bf E} \int_t^T\!\! e^{\beta r}|\Delta g(r)|^2 dr. \end{array}$$(11)

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

$\begin{matrix} E \int_{t}^{T} e^{β r} | Δ g (r) |^{2} d r \leq c E \int_{t}^{T + K} e^{β r} | {\bar{y}}_{r} |^{2} d r + α_{1} E \int_{t}^{T + K} e^{β r} | {\bar{z}}_{r} |^{2} d r . \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E} \int_t^{T} e^{\beta r}|\Delta g(r)|^2 dr \le c{\bf E} \int_t^{T+K} e^{\beta r}|\overline y_r|^2dr + \alpha_{1}{\bf E} \int_t^{T+K} e^{\beta r}|\overline z_r|^{2} dr. \end{array}$$

Similarly, we have

$\begin{matrix} 2 e^{β r} 〈 {\bar{Y}}_{r}, Δ f (r) 〉 \leq e^{β r} ε | {\bar{Y}}_{r} |^{2} + \frac{c}{ε} e^{β r} (| {\bar{y}}_{r} |^{2} + | {\bar{z}}_{r} |^{2}) + \frac{1}{ε} e^{β r} E^{F_{t}} [| {\bar{y}}_{r + δ (r)} |^{2} + | {\bar{z}}_{r + ζ (r)} |^{2}] . \end{matrix}$ $$\begin{array}{} \displaystyle 2e^{\beta r}\langle \overline Y_r,\Delta f(r)\rangle \le e^{\beta r}\varepsilon|\overline Y_r |^{2} + \frac{c}{\varepsilon}e^{\beta r}(|\overline y_r|^{2} +|\overline z_r|^{2}) + \frac{1}{\varepsilon}e^{\beta r}{\bf E}^{\mathscr{F}_{t}}\left[|\overline y_{r+\delta(r)}|^{2} + |\overline z_{r+\zeta(r)}|^{2}\right]. \end{array}$$

Which implies by virtue of condition (A2) that

$\begin{matrix} 2 E \int_{t}^{T} e^{β r} 〈 {\bar{Y}}_{r}, Δ f (r) 〉 d r \leq ε E \int_{t}^{T + K} e^{β r} | {\bar{Y}}_{r} |^{2} d r + \frac{1}{ε} (c + M) E \int_{t}^{T + K} e^{β r} (| {\bar{y}}_{r} |^{2} + | {\bar{z}}_{r} |^{2}) d r . \end{matrix}$ $$\begin{array}{} \displaystyle 2{\bf E}\int_t^T \!\! e^{\beta r}\langle \overline Y_r, \Delta f(r)\rangle dr \le \varepsilon {\bf E} \int_t^{T+K} \!\! e^{\beta r}|\overline Y_r |^{2} dr + \frac{1}{\varepsilon}(c+ M ){\bf E} \int_t^{T+K}\!e^{\beta r}(|\overline y_r|^{2} +|\overline z_r|^{2})dr. \end{array}$$

Therefore, we can write (where $\begin{matrix} γ = \frac{\frac{1}{ε} (c + M) + c}{\frac{1}{ε} (c + M) + α_{1}} \end{matrix}$ $\begin{array}{} \displaystyle \gamma = \frac{\frac{1}{\varepsilon}(c+ M )+c}{\frac{1}{\varepsilon}(c+ M )+\alpha_{1}} \end{array}$

$\begin{matrix} \begin{aligned} E (e^{β t} | {\bar{Y}}_{t} |^{2}) & + (β - ε) E \int_{t}^{T + K} e^{β r} | {\bar{Y}}_{r} |^{2} d r + E \int_{t}^{T + K} e^{β r} | {\bar{Z}}_{r} |^{2} d r \\ \leq (\frac{1}{ε} (c + M) + c) E \int_{t}^{T + K} e^{β r} | {\bar{y}}_{r} |^{2} d r + (\frac{1}{ε} (c + M) + α_{1}) E \int_{t}^{T + K} e^{β r} | {\bar{z}}_{r} |^{2} d r \\ = (\frac{1}{ε} (c + M) + α_{1}) E \int_{t}^{T + K} e^{β r} [γ | {\bar{y}}_{r} |^{2} + | {\bar{z}}_{r} |^{2}] d r . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle {\bf E}\left(e^{\beta t}|\overline Y_t|^2\right) &+ (\beta -\varepsilon){\bf E} \int_t^{T+K} \!\!\! e^{\beta r}|\overline Y_r|^2 dr + {\bf E}\int_t^{T+K} \!\!e^{\beta r} |\overline Z_r|^2 dr \\ &\le\left(\frac{1}{\varepsilon}(c+ M )+c\right){\bf E}\int_t^{T+K} \!\! e^{\beta r}|\overline y_r |^{2}dr + \left(\frac{1}{\varepsilon}(c+ M )+\alpha_{1}\right){\bf E}\int_t^{T+K} \!\! e^{\beta r}|\overline z_r|^{2}dr\\ &=\left(\frac{1}{\varepsilon}(c+ M )+\alpha_{1}\right){\bf E}\int_t^{T+K} \!\! e^{\beta r}\left[ \gamma |\overline y_r |^{2} + |\overline z_r|^{2}\right] dr. \end{split} \end{array}$$

Hence if we choose ε = ε₀ satisfying $\begin{matrix} \bar{c} = (\frac{1}{ε_{0}} (c + M) + α_{1}) < 1, \end{matrix}$ $\begin{array}{} \displaystyle \overline c=\left(\frac{1}{\varepsilon_0}(c+ M )+\alpha_{1}\right) \lt 1, \end{array}$ choose β = ε₀ + γ, then we deduce

$\begin{matrix} E \int_{t}^{T + K} e^{β r} [γ | {\bar{Y}}_{r} |^{2} + | {\bar{Z}}_{r} |^{2}] d r \leq \bar{c} E \int_{t}^{T + K} e^{β r} [γ | {\bar{y}}_{r} |^{2} + | {\bar{z}}_{r} |^{2}] d r . \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E}\int_t^{T+K}\!\!\! e^{\beta r}\left[\gamma|\overline Y_r|^2 + |\overline Z_r|^2\right] dr\le\overline c{\bf E}\int_t^{T+K}\!\!\! e^{\beta r}\left[\gamma|\overline y_r|^2 + |\overline z_r|^2\right]dr. \end{array}$$

Thus, the mapping Ψ is a strict contraction on $\begin{matrix} C_{G}^{2} \end{matrix}$ $\begin{array}{} \displaystyle {\mathscr C}_\mathscr{G}^2 \end{array}$ (0, T + K) and it has a unique fixed point

$\begin{matrix} (Y, Z) \in C_{G}^{2} (0, T + K) . \end{matrix}$ $$\begin{array}{} \displaystyle (Y, Z)\in {\mathscr C}_\mathscr{G}^2 (0, T+K). \end{array}$$

It remains to prove that the above solution is in $\begin{matrix} B_{G}^{2} \end{matrix}$ $\begin{array}{} \displaystyle {\mathscr B}_\mathscr{G}^2 \end{array}$ (0, T + K). Indeed, by Lemma 3.1, we have Y ∈ $\begin{matrix} S_{[0, T]}^{2} \end{matrix}$ $\begin{array}{} \displaystyle \mathscr S_{[0, T]}^2 \end{array}$ (𝒢, R^k). Thus, we obtain (Y_t, Z_t)_0≤t≤T ∈ $\begin{matrix} B_{G}^{2} \end{matrix}$ $\begin{array}{} \displaystyle {\mathscr B}_\mathscr{G}^2 \end{array}$ (0, T + K).

Uniqueness. Let (Y, Z) and (Y͠, Z͠) two solutions of eq.(1). Itô’s formula applied to eq.(10) yields, for 0 ≤ t ≤ T

$\begin{matrix} E [| {\bar{Y}}_{t} |^{2}] + E \int_{t}^{T} | {\bar{Z}}_{r} |^{2} d r \leq 2 E \int_{t}^{T} 〈 {\bar{Y}}_{r}, Δ f (r) 〉 d r + E \int_{t}^{T} | Δ g (r) |^{2} d r . \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E}[|\overline Y_t|^2] + {\bf E}\int_t^T|\overline Z_r|^2dr\le 2{\bf E} \int_t^T \!\!\langle \overline Y_r ,\Delta f(r)\rangle dr + {\bf E}\int_t^T\!\!|\Delta g(r)|^2 dr. \end{array}$$(12)

Using assumption (H1), we have:

$\begin{matrix} \begin{aligned} 2 E \int_{t}^{T} 〈 {\bar{Y}}_{r}, Δ f (r) 〉 d r & \leq (\frac{1}{ε} (c + M) + ε) E \int_{t}^{T} | {\bar{Y}}_{r} |^{2} d r + \frac{1}{ε} (c + M) E \int_{t}^{T} | {\bar{Z}}_{r} |^{2} d r, \\ E \int_{t}^{T} | Δ g (r) |^{2} d r & \leq c E \int_{t}^{T} | {\bar{Y}}_{r} |^{2} d r + α_{1} E \int_{t}^{T} | {\bar{Z}}_{r} |^{2} d r . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle 2{\bf E}\int_t^T\langle \overline Y_r,\Delta f(r)\rangle dr &\le(\frac{1}{\varepsilon}(c+M)+\varepsilon){\bf E}\int_t^T|\overline Y_r |^{2}dr + \frac{1}{\varepsilon}(c+M){\bf E}\int_t^T|\overline Z_r|^{2}dr, \\ {\bf E}\int_t^T |\Delta g(r)|^2dr &\le c{\bf E}\int_t^T|\overline Y_r|^2dr + \alpha_{1}{\bf E}\int_t^T|\overline Z_r|^{2}dr. \end{split} \end{array}$$

Hence if we choose ξ = ξ₀ satisfying $\begin{matrix} \bar{α} = \frac{1}{ε_{0}} (c + M) + α_{1} < 1 \end{matrix}$ $\begin{array}{} \displaystyle \overline\alpha=\frac{1}{\varepsilon_0}(c+M)+\alpha_{1} \lt 1 \end{array}$ and denote $\begin{matrix} \bar{c} = \frac{c (ε_{0} + 1) + M}{ε_{0}} + ε_{0}, \end{matrix}$ $\begin{array}{} \displaystyle \overline c=\frac{c(\varepsilon_0+1)+M}{\varepsilon_0}+\varepsilon_0, \end{array}$ then using the above inequalities, from (12), we obtain

$\begin{matrix} E (| {\bar{Y}}_{t} |^{2}) + (1 - \bar{α}) E \int_{t}^{T} | {\bar{Z}}_{r} |^{2} d r \leq \bar{c} E \int_{t}^{T} | {\bar{Y}}_{r} |^{2} d r . \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E}\left(|\overline Y_t|^2\right) + (1-\overline\alpha){\bf E}\int_t^T \!\!|\overline Z_r|^2 dr \le \overline c{\bf E}\int_t^T|\overline Y_r|^2dr. \end{array}$$

Then we can use Gronwall’s inequality to deduce Y = 0 and Z = 0. 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) with non-Lipschitz coefficients. For this purpose, we first make the following assumptions.

4.0.1

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

$\begin{matrix} \begin{aligned} | f (t, y, z, θ (r), φ (r)) - f (t, y^{'}, z^{'}, θ^{'} (r), φ^{'} (r)) |^{2} & \leq c (ρ (| y - y^{'} |^{2}) + | z - z^{'} |^{2}) \\ + E^{F_{t}} [ρ (| θ (r) - θ^{'} (r) |^{2}) + | φ (r) - φ^{'} (r) |^{2}], \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle |f(t, y,z, \theta(r), \varphi(r))- f(t, y',z', \theta'(r), \varphi'(r))|^{2} \!&\leq c (\rho(|y-y^\prime|^{2}) + |z-z^\prime|^{2}) \\&+ {\bf E}^{\mathscr{F}_{t}}[\rho(|\theta(r) - \theta'(r)|^{2}) + |\varphi(r) - \varphi'(r)|^{2}], \end{split} \end{array}$$

(H2.2)

There exists a constant 0 < α₁ < 1 such that for any (t, y, z), (t, y′, z′) ∈ [0, T] × R^k × R^k×d

$\begin{matrix} | g (t, y, z) - g (t, y^{'}, z^{'}) |^{2} \leq ρ (| y - y^{'} |^{2}) + α_{1} | z - z^{'} |^{2} . \end{matrix}$ $$\begin{array}{} \displaystyle |g(t, y,z) - g(t, y',z')|^2 \leq \rho(|y-y^\prime|^{2}) + \alpha_{1}|z-z^\prime|^2. \end{array}$$

(H2.3)

(H1.3) holds.

4.1

Existence and uniqueness of solution

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

$\begin{matrix} \{\begin{cases} Y_{t}^{0} = 0, Z_{t}^{0} = 0, 0 \leq t \leq T + K, \\ Y_{t}^{n} = ξ_{T} + \int_{t}^{T} f (r, Y_{r}^{n - 1}, Z_{r}^{n}, Y_{r + δ (r)}^{n - 1}, Z_{r + ζ (r)}^{n}) d r + \int_{t}^{T} g (r, Y_{r}^{n - 1}, Z_{r}^{n}) d B_{r} - \int_{t}^{T} Z_{r}^{n} d W_{r}, 0 \leq t \leq T, \\ Y_{t}^{n} = ξ_{t}, Z_{t}^{n} = η_{t}, T \leq t \leq T + K . \end{cases} \end{matrix}$ $$\begin{array}{} \displaystyle \begin{cases} Y_t^0=0, \; Z_t^0=0, \quad 0\le t \le T+K,\\ Y_t^{n} = \xi_{T} + \displaystyle \int_t^T\!\! f(r,Y_r^{n-1}, Z_r^{n},Y_{r+\delta(r)}^{n-1}, Z_{r+\zeta(r)}^{n}) dr + \int_t^T \!\! g(r, Y_r^{n-1}, Z_r^{n})dB_r - \displaystyle \int_t^T Z_r^{n} dW_r, \quad\quad 0\le t \le T,\\ Y_{t}^{n} = \xi_{t}, \quad Z_{t}^{n} = \eta_{t}, \quad\quad T\le t \le T+K. \end{cases} \end{array}$$(13)

Thanks to Theorem 3.3, this sequence is well defined since the generators $\begin{matrix} f (r, Y_{r}^{n - 1}, \cdot, Y_{r + δ (r)}^{n - 1}, \cdot) \end{matrix}$ $\begin{array}{} \displaystyle f(r, Y_{r}^{n-1},\cdot, Y_{r+\delta(r)}^{n-1}, \cdot) \end{array}$ and $\begin{matrix} g (r, Y_{r}^{n - 1}, \cdot) \end{matrix}$ $\begin{array}{} \displaystyle g(r, Y_{r}^{n-1},\cdot) \end{array}$ are Γ-Lipschitz. Let us state the following previous result

Lemma 4.1

Assume that the assumptions (A1), (A2) and (H2) are true and let ξ_T ∈ L²(𝒢_T, R^k). Then for any $\begin{matrix} (ξ, η) \in S_{[T, T + K]}^{2} (G, R^{k}) \times M_{[T, T + K]}^{2} (G, R^{k \times d}) \end{matrix}$ $\begin{array}{} \displaystyle (\xi, \eta)\in \mathscr S_{[T, T+K]}^2 (\mathscr{G}, \mathbf{R}^k) \times \mathscr{M}_{[T, T+K]}^2 (\mathscr{G}, \mathbf{R}^{k\times d}) \end{array}$ there exists a positive constant C^′ such that

$\begin{matrix} sup_{n \geq 0} E | Y_{t}^{n} |^{2} \leq C^{'} (1 + E [X]), 0 \leq t \leq T + K \end{matrix}$ $$\begin{array}{} \displaystyle \sup_{{n\ge 0}} {\bf E}|Y_{t}^{n}|^{2} \le C'(1 + {\bf E} [X]), \quad \quad 0\le t\le T+K \end{array}$$(14)

where

$\begin{matrix} X = | ξ_{T} |^{2} + \int_{T}^{T + K} (| ξ_{r} |^{2} + | η_{r} |^{2}) d r + \int_{0}^{T} | f (r, 0) |^{2} d r + \int_{0}^{T} | g (r, 0, 0) |^{2} d r . \end{matrix}$ $$\begin{array}{} \displaystyle X = |\xi_{T}|^{2} + \int_{T}^{T+K}(|\xi_r|^{2} +|\eta_{r}|^{2})dr + \int_{0}^{T} |f(r, 0)|^{2}dr + \int_{0}^{T} |g(r, 0,0)|^{2}dr. \end{array}$$

Proof

For β > 0, apply Itô’s formula to $\begin{matrix} e^{β t} | Y_{t}^{n} |^{2}, \end{matrix}$ $\begin{array}{} \displaystyle e^{\beta t}|Y_t^n|^2, \end{array}$

$\begin{matrix} \begin{aligned} E [e^{β t} | Y_{t}^{n} |^{2}] & + β E \int_{t}^{T} e^{β r} | Y_{r}^{n} |^{2} d r + E \int_{t}^{T} e^{β r} | Z_{r}^{n} |^{2} d r = E [e^{β T} | ξ_{T} |^{2}] \\ + 2 E \int_{t}^{T} e^{β r} 〈 Y_{r}^{n}, f (r, Y_{r}^{n - 1}, Z_{r}^{n}, Y_{r + δ (r)}^{n - 1}, Z_{r + ζ (r)}^{n}) 〉 d r + E \int_{t}^{T} e^{β r} | g (r, Y_{r}^{n - 1}, Z_{r}^{n}) |^{2} d r . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle {\bf E}[ e^{\beta t}|Y_t^n|^2] &+ \beta {\bf E} \int_t^T\!\!e^{\beta r} |Y_r^n|^2 dr + {\bf E}\int_t^T\!\!e^{\beta r} |Z_r^n|^2 dr = {\bf E}[e^{\beta T}|\xi_T|^2] \\ &+ 2{\bf E}\int_t^T\!\!e^{\beta r}\langle Y_{r}^n, f(r,Y_r^{n-1}, Z_r^{n},Y_{r+\delta(r)}^{n-1}, Z_{r+\zeta(r)}^{n})\rangle dr + {\bf E}\int_t^T\!\!e^{\beta r} |g(r,Y_r^{n-1}, Z_r^{n})|^2dr. \end{split} \end{array}$$

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

$\begin{matrix} E \int_{t}^{T} e^{β r} | g (r, Y_{r}^{n - 1}, Z_{r}^{n}) |^{2} d r \leq E \int_{t}^{T} e^{β r} | g (r, 0, 0) |^{2} d r + E \int_{t}^{T} e^{β r} [ρ (| Y_{r}^{n - 1} |^{2}) + α_{1} | Z_{r}^{n} |^{2}] d r \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E}\int_t^T\!\!e^{\beta r} |g(r,Y_r^{n-1}, Z_r^{n})|^2dr \le {\bf E}\int_t^T\!\!e^{\beta r}|g(r,0,0)|^2dr + {\bf E}\int_t^T\!\!e^{\beta r}\left[\rho(|Y_r^{n-1}|^{2}) + \alpha_{1}|Z_r^n|^2 \right]dr \end{array}$$

and

$\begin{matrix} \begin{aligned} 2 E \int_{t}^{T} e^{β r} & 〈 Y_{r}^{n}, f (r, Y_{r}^{n - 1}, Z_{r}^{n}, Y_{r + δ (r)}^{n - 1}, Z_{r + ζ (r)}^{n}) 〉 d r \leq \frac{ε}{2} E \int_{t}^{T} e^{β r} | Y_{r}^{n} |^{2} d r + \frac{2}{ε} E \int_{t}^{T} e^{β r} | f (r, 0) |^{2} d r \\ + \frac{2}{ε} c E \int_{t}^{T} e^{β r} (ρ (| Y_{r}^{n - 1} |^{2}) + | Z_{r}^{n} |^{2}) d r + \frac{2}{ε} E \int_{t}^{T} e^{β r} E^{F_{t}} [ρ (| Y_{r + δ (r)}^{n - 1} |^{2}) + | Z_{r + ζ (r)}^{n} |^{2}] d r . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle 2{\bf E}\int_t^T\!\!e^{\beta r}&\langle Y_{r}^n, f(r,Y_r^{n-1}, Z_r^{n},Y_{r+\delta(r)}^{n-1}, Z_{r+\zeta(r)}^{n})\rangle dr\le \frac{\varepsilon}{2}{\bf E}\int_t^T\!\!e^{\beta r}| Y_{r}^n |^2 dr +\frac{2}{\varepsilon}{\bf E}\int_t^T\!\!e^{\beta r} |f(r,0)|^2 dr \\ &+ \frac{2}{\varepsilon}c{\bf E}\int_t^T\!\!e^{\beta r}\left( \rho(|Y_r^{n-1}|^{2}) + |Z_r^n|^2\right)dr+ \frac{2}{\varepsilon}{\bf E}\int_t^T\!\!e^{\beta r} {\bf E}^{\mathscr{F}_{t}}\left[\rho(|Y_{r+\delta(r)}^{n-1}|^{2}) + |Z_{r+\zeta(r)}^{n}|^2 \right]dr. \end{split} \end{array}$$(15)

Applying condition (A2), the last term on the right-hand side of (15) is less than

$\begin{matrix} \frac{2}{ε} M E \int_{t}^{T} e^{β r} [ρ (| Y_{r}^{n - 1} |^{2}) + | Z_{r}^{n} |^{2}] d r + \frac{2}{ε} M E \int_{T}^{T + K} e^{β r} [ρ (| ξ_{r} |^{2}) + | η_{r} |^{2}] d r . \end{matrix}$ $$\begin{array}{} \displaystyle \frac{2}{\varepsilon}M{\bf E}\int_t^T\!\!e^{\beta r}\left[\rho(|Y_r^{n-1}|^{2}) + |Z_r^{n}|^2 \right]dr + \frac{2}{\varepsilon}M{\bf E}\int_T^{T+K}\!\!e^{\beta r}\left[\rho(|\xi_r|^{2}) + |\eta_r|^2 \right]dr. \end{array}$$

Putting pieces together, we obtain finally

$\begin{matrix} \begin{aligned} E [e^{β t} & | Y_{t}^{n} |^{2}] + β E \int_{t}^{T} e^{β r} | Y_{r}^{n} |^{2} d r + E \int_{t}^{T} e^{β r} | Z_{r}^{n} |^{2} d r \leq E [e^{β T} | ξ_{T} |^{2}] + \frac{ε}{2} E \int_{t}^{T} e^{β r} | Y_{r}^{n} |^{2} d r \\ + (\frac{2}{ε} (c + M) + 1) E \int_{t}^{T} e^{β r} ρ (| Y_{r}^{n - 1} |^{2}) d r + (\frac{2}{ε} (c + M) + α_{1}) E \int_{t}^{T} e^{β r} | Z_{r}^{n} |^{2} d r \\ + \frac{2}{ε} M E \int_{T}^{T + K} e^{β r} [ρ (| ξ_{r} |^{2}) + | η_{r} |^{2}] d r + \frac{2}{ε} E \int_{t}^{T} e^{β r} | f (r, 0) |^{2} d r + E \int_{t}^{T} e^{β r} | g (r, 0, 0) |^{2} d r . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle {\bf E}[ e^{\beta t}&|Y_t^n|^2] + \beta {\bf E} \int_t^T\!\!e^{\beta r} |Y_r^n|^2 dr + {\bf E}\int_t^T\!\!e^{\beta r} |Z_r^n|^2 dr \le {\bf E}[e^{\beta T}|\xi_T|^2] + \frac{\varepsilon}{2}{\bf E} \int_t^T\!\!e^{\beta r}|Y_{r}^n|^2dr\\ &+ \left(\frac{2}{\varepsilon}(c+M)+1\right){\bf E}\int_t^T\!\!e^{\beta r}\rho(|Y_{r}^{n-1}|^2)dr + \left(\frac{2}{\varepsilon}(c+M)+\alpha_{1}\right){\bf E}\int_t^T\!\!e^{\beta r}|Z_r^n|^2 dr \\ &+\frac{2}{\varepsilon}M{\bf E}\int_T^{T+K}\!\!e^{\beta r}\left[\rho(|\xi_r|^{2})+ |\eta_r|^2 \right]dr + \frac{2}{\varepsilon} {\bf E}\int_t^T\!\!e^{\beta r}|f(r,0)|^2dr + {\bf E}\int_t^T\!\!e^{\beta r}|g(r,0,0)|^2dr. \end{split} \end{array}$$

This implies thanks to Remark 2.1 that

$\begin{matrix} \begin{aligned} E [e^{β t} | Y_{t}^{n} |^{2}] & + β E \int_{t}^{T} e^{β r} | Y_{r}^{n} |^{2} d r + E \int_{t}^{T} e^{β r} | Z_{r}^{n} |^{2} d r \leq E [e^{β T} | ξ_{T} |^{2}] + \frac{ε}{2} E \int_{t}^{T} e^{β r} | Y_{r}^{n} |^{2} d r \\ + (\frac{2}{ε} (c + M) + 1) b E \int_{t}^{T} e^{β r} | Y_{r}^{n - 1} |^{2} d r + (\frac{2}{ε} (c + M) + α_{1}) E \int_{t}^{T} e^{β r} | Z_{r}^{n} |^{2} d r \\ + \frac{2}{ε} M E \int_{T}^{T + K} e^{β r} [b | ξ_{r} |^{2} + | η_{r} |^{2}] d r + \frac{2}{ε} E \int_{t}^{T} e^{β r} | f (r, 0) |^{2} d r + E \int_{t}^{T} e^{β r} | g (r, 0, 0) |^{2} d r + C_{ε} \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle {\bf E}[ e^{\beta t}|Y_t^n|^2] &+ \beta {\bf E} \int_t^T\!\!e^{\beta r} |Y_r^n|^2 dr + {\bf E}\int_t^T\!\!e^{\beta r} |Z_r^n|^2 dr \le {\bf E}[e^{\beta T}|\xi_T|^2] + \frac{\varepsilon}{2}{\bf E} \int_t^T\!\!e^{\beta r}|Y_{r}^n|^2dr\\ &+ \left(\frac{2}{\varepsilon}(c+M)+1\right)b{\bf E}\int_t^T\!\!e^{\beta r}|Y_{r}^{n-1}|^2dr + \left(\frac{2}{\varepsilon}(c+M)+\alpha_{1}\right){\bf E}\int_t^T\!\!e^{\beta r}|Z_r^n|^2 dr \\ &+\frac{2}{\varepsilon}M{\bf E}\int_T^{T+K}\!\!e^{\beta r}\left[b|\xi_r|^{2} + |\eta_r|^2 \right]dr + \frac{2}{\varepsilon} {\bf E}\!\!\int_t^T\!\!e^{\beta r}|f(r,0)|^2dr + {\bf E}\!\!\int_t^T\!\!e^{\beta r}|g(r,0,0)|^2dr + C_\varepsilon \end{split} \end{array}$$

where $\begin{matrix} \begin{aligned} C_{ε} = \frac{a}{β} [\frac{2}{ε} (c + 2 M) + 1] e^{β (T + K)} . \end{aligned} \end{matrix}$ $\begin{array}{} \begin{split}{} \displaystyle C_\varepsilon= \frac{a}{\beta}\left[\frac{2}{\varepsilon}(c+2M)+1\right]e^{\beta(T+K)}. \end{split} \end{array}$ Choose ε = ε₀ such that $\begin{matrix} β = \frac{1}{2} (ε_{0} + 1), C_{0} = C_{ε_{0}} a n d \frac{2}{ε_{0}} (M + c) + α_{1} \end{matrix}$ $\begin{array}{} \displaystyle \beta=\frac{1}{2}(\varepsilon_0 + 1), C_0= C_{\varepsilon_0}\;\; \rm{and}\; \frac{2}{\varepsilon_0}(M+c)+ \alpha_{1} \end{array}$ = 1/2. Therefore, we obtain

$\begin{matrix} \begin{aligned} E [e^{β t} | Y_{t}^{n} |^{2}] & + \frac{1}{2} E \int_{t}^{T} e^{β r} | Y_{r}^{n} |^{2} d r + \frac{1}{2} E \int_{t}^{T} e^{β r} | Z_{r}^{n} |^{2} d r \leq (\frac{2}{ε_{0}} (c + M) + 1) b E \int_{t}^{T} e^{β r} | Y_{r}^{n - 1} |^{2} d r \\ + E [e^{β T} | ξ_{T} |^{2}] + \frac{2}{ε_{0}} M E \int_{T}^{T + K} e^{β r} [b | ξ_{r} |^{2} + | η_{r} |^{2}] d r + \frac{2}{ε_{0}} E \int_{0}^{T} e^{β r} | f (r, 0) |^{2} d r \\ + E \int_{0}^{T} e^{β r} | g (r, 0, 0) |^{2} d r + C_{0}, 0 \leq t \leq T . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split}{} \displaystyle {\bf E}[ e^{\beta t}|Y_t^n|^2] &+ \frac{1}{2} {\bf E} \int_t^T\!\!e^{\beta r} |Y_r^n|^2 dr + \frac{1}{2} {\bf E}\int_t^T\!\!e^{\beta r} |Z_r^n|^2 dr \le \left(\frac{2}{\varepsilon_0}(c+M)+1\right)b{\bf E}\int_t^T\!\!e^{\beta r}|Y_{r}^{n-1}|^2dr\\ &+ {\bf E}[e^{\beta T}|\xi_T|^2] + \frac{2}{\varepsilon_0}M{\bf E}\int_T^{T+K}\!\!e^{\beta r}\left[b|\xi_r|^{2} + |\eta_r|^2 \right]dr + \frac{2}{\varepsilon_0} {\bf E}\int_0^T\!\!e^{\beta r}|f(r,0)|^2dr\\ &+ {\bf E}\int_0^T\!\!e^{\beta r}|g(r,0,0)|^2dr + C_0, \quad 0\leq t\leq T. \end{split} \end{array}$$

This leads to

$\begin{matrix} E [| Y_{t}^{n} |^{2}] + \frac{1}{2} E \int_{t}^{T} | Y_{r}^{n} |^{2} d r + \frac{1}{2} E \int_{t}^{T} | Z_{r}^{n} |^{2} d r \leq C^{'} (E [X] + 1) + C^{'} E \int_{t}^{T} | Y_{r}^{n - 1} |^{2} d r, 0 \leq t \leq T \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E}[|Y_t^n|^2]+ \frac{1}{2}{\bf E} \int_t^T\!\!|Y_r^n|^2 dr + \frac{1}{2}{\bf E}\int_t^T\!\!|Z_r^n|^2 dr \le C'\left( {\bf E}[X]+ 1\right) + C'{\bf E}\int_t^T\!\! |Y_r^{n-1}|^{2} dr,\quad 0\leq t\leq T \end{array}$$

where C′ is a positive constant(which may vary from line to line).

In particular, putting $\begin{matrix} q_{n} (t) = sup_{n \in N} E [| Y_{t}^{n} |^{2}], \end{matrix}$ $\begin{array}{} q_n(t)= \sup_{n\in\mathbb{N}}{\bf E}[|Y_t^n|^2] , \end{array}$ we have

$\begin{matrix} q_{n} (t) \leq C^{'} (E [X] + 1) + C^{'} \int_{t}^{T} q_{n} (r) d r, 0 \leq t \leq T . \end{matrix}$ $$\begin{array}{} \displaystyle q_n(t) \le C'\left( {\bf E}[X]+ 1\right) + C'\int_t^T\!\!q_n(r)dr,\quad 0\leq t\leq T. \end{array}$$

Gronwall’s inequality yields

$\begin{matrix} sup_{n \in N} E [| Y_{t}^{n} |^{2}] \leq C^{'} (1 + E [X]), 0 \leq t \leq T . \end{matrix}$ $$\begin{array}{} \displaystyle \sup_{n\in\mathbb{N}}{\bf E}[|Y_t^n|^2] \le C'(1 + {\bf E}[X]), \quad 0\leq t\leq T. \end{array}$$

This immediately gives (14).□

Now we establish the main result of this section.

Theorem 4.2

Proof

Existence. We consider the sequence defined in eq.(13). For a process ρ ∈ {Y, Z}, and n ∈ N, m ∈ N, $\begin{matrix} {\bar{ρ}}_{t}^{n, m} = ρ_{t}^{n} - ρ_{t}^{m}, Δ g^{(n, m)} (r) = f (r, Y_{r}^{n - 1}, Z_{r}^{n}) - g (r, Y_{r}^{m - 1}, Z_{r}^{m}) \end{matrix}$ $\begin{array}{} \displaystyle \overline \rho_t^{n, m}= \rho_t^{n}-\rho_t^{m}, \Delta g^{(n, m)}(r)=f(r,Y_r^{n-1}, Z_r^{n})- g(r,Y_r^{m-1}, Z_r^{m}) \end{array}$ and $\begin{matrix} Δ f^{(n, m)} (r) = f (r, Y_{r}^{n - 1}, Z_{r}^{n}, Y_{r + δ (r)}^{n - 1}, Z_{r + ζ (r)}^{n}) - f (r, Y_{r}^{m - 1}, Z_{r}^{m}, Y_{r + δ (r)}^{m - 1}, Z_{r + ζ (r)}^{m}) . \end{matrix}$ $\begin{array}{} \displaystyle \Delta f^{(n, m)}(r)=f(r,Y_r^{n-1}, Z_r^{n},Y_{r+\delta(r)}^{n-1}, Z_{r+\zeta(r)}^{n})- f(r,Y_r^{m-1}, Z_r^{m},Y_{r+\delta(r)}^{m-1}, Z_{r+\zeta(r)}^{m}). \end{array}$

Note that the pair (Y^n,m, Z^n,m) solves the following equation

$\begin{matrix} \{\begin{cases} {\bar{Y}}_{t}^{n, m} = \int_{t}^{T} Δ f^{(n, m)} (r) d r + \int_{t}^{T} Δ g^{(n, m)} (r) d B_{r} - \int_{t}^{T} {\bar{Z}}_{r}^{n, m} d W_{r}, 0 \leq t \leq T, \\ {\bar{Y}}_{t}^{n, m} = 0, {\bar{Z}}_{t}^{n, m} = 0, T \leq t \leq T + K . \end{cases} \end{matrix}$ $$\begin{array}{} \displaystyle \begin{cases} \overline Y_t^{n, m} = \displaystyle \int_t^T\!\! \Delta f^{(n, m)}(r) dr + \int_t^T \!\! \Delta g^{(n, m)}(r) dB_r - \displaystyle \int_t^T \overline Z_r^{n, m} dW_r, \quad \quad 0\le t \le T,\\ \overline Y_t^{n, m} = 0, \quad \overline Z_t^{n, m} = 0, \quad\quad T\le t \le T +K. \end{cases} \end{array}$$

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

$\begin{matrix} E | {\bar{Y}}_{t}^{n, m} |^{2} + \frac{1}{2} E \int_{t}^{T} | {\bar{Y}}_{r}^{n, m} |^{2} d r + \frac{1}{2} E \int_{t}^{T} | {\bar{Z}}_{r}^{n, m} |^{2} d r \leq C^{'} E \int_{t}^{T} ρ (| {\bar{Y}}_{r}^{n - 1, m - 1} |^{2}) d r, 0 \leq t \leq T . \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E} |\overline Y_t^{n,m}|^2 + \frac{1}{2}{\bf E} \int_t^T |\overline Y_r^{n,m}|^2dr + \frac{1}{2}{\bf E} \int_t^T |\overline Z_r^{n,m}|^2dr \le C'{\bf E}\int_t^T \!\!\rho(|\overline Y_r^{n-1,m-1}|^2)dr,\quad 0\le t \le T. \end{array}$$(16)

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

$\begin{matrix} q (t) \leq C^{'} \int_{t}^{T + K} ρ (q (r)) d r, 0 \leq t \leq T + K \end{matrix}$ $$\begin{array}{} \displaystyle q(t)\le C'\int_t^{T+K} \!\!\rho(q(r))dr, \quad 0\le t \le T +K \end{array}$$

where $\begin{matrix} q (t) = lim_{n, m \to \infty} sup E | {\bar{Y}}_{t}^{n, m} |^{2}, 0 \leq t \leq T + K . \end{matrix}$ $\begin{array}{} q(t)=\lim_{n,m \to\infty}\sup{\bf E}|\overline Y_t^{n,m}|^2,\quad 0\le t \le T+K. \end{array}$ Therefore, we can use Bihari’s inequality to get q(t) = 0, i.e. $\begin{matrix} lim_{n, m \to \infty} sup E | {\bar{Y}}_{t}^{n, m} |^{2} = 0 \end{matrix}$ $\begin{array}{} \lim_{n,m \to\infty}\sup{\bf E}|\overline Y_t^{n,m}|^2=0 \end{array}$ for all 0 ≤ t ≤ T + K.

So, from inequality (16), we obtain

$\begin{matrix} lim_{n, m \to \infty} E (| Y_{t}^{n} - Y_{t}^{m} |^{2} + \int_{t}^{T + K} | Z_{r}^{n} - Z_{r}^{m} |^{2} d r) = 0, 0 \leq t \leq T + K . \end{matrix}$ $$\begin{array}{} \displaystyle \lim_{n,m\to\infty}{\bf E}\left(|Y_t^n - Y_t^m|^2 + \int_t^{T+K}\!\!\!|Z_r^n - Z_r^m|^2dr \right)=0,\quad 0\le t \le T +K. \end{array}$$

Then, there exists (Y, Z) ∈ $\begin{matrix} B_{G}^{2} \end{matrix}$ $\begin{array}{} \displaystyle {\mathscr B}_\mathscr{G}^2 \end{array}$ (0, T + K) such that

$\begin{matrix} lim_{n \to \infty} E (| Y_{t}^{n} - Y_{t} |^{2} + \int_{t}^{T + K} | Z_{r}^{n} - Z_{r} |^{2} d r) = 0, 0 \leq t \leq T + K . \end{matrix}$ $$\begin{array}{} \displaystyle \lim_{n\to\infty}{\bf E}\left(|Y_t^n - Y_t|^2 + \int_t^{T+K}\!\!\!|Z_r^n - Z_r|^2dr \right)=0, \quad 0\le t \le T +K. \end{array}$$

Finally, taking limit in eq.(13) as n → +∞, we conclude that (Y, Z) solves

$\begin{matrix} \{\begin{cases} 0 \leq t \leq T, \\ Y_{t} = ξ_{T} + \int_{t}^{T} f (r, Y_{r}, Z_{r}, Y_{r + δ (r)}, Z_{r + ζ (r)}) d r + \int_{t}^{T} g (r, Y_{r}, Z_{r}) d B_{r} - \int_{t}^{T} Z_{r} d W_{r}, \\ Y_{t} = ξ_{t}, Z_{t} = η_{t}, T \leq t \leq T + K . \end{cases} \end{matrix}$ $$\begin{array}{} \displaystyle \begin{cases} 0\le t \le T,\\ Y_t= \xi_{T} + \displaystyle \int_t^T f(r, Y_r, Z_r,Y_{r+\delta(r)}, Z_{r+\zeta(r)}) dr + \int_t^T g(r, Y_r, Z_r)dB_r- \displaystyle\int_t^T Z_r dW_r,\\ Y_{t} = \xi_{t}, \quad Z_{t} =\eta_{t}, \quad\quad T\le t \le T+K. \end{cases} \end{array}$$

This shows that (Y, Z) ∈ $\begin{matrix} B_{G}^{2} \end{matrix}$ $\begin{array}{} \displaystyle {\mathscr B}_\mathscr{G}^2 \end{array}$ (0, T + K) solves ABDSDE (1). The proof of existence is complete.

Uniqueness. Let (Yⁱ, Zⁱ) ∈ $\begin{matrix} B_{G}^{2} \end{matrix}$ $\begin{array}{} \displaystyle {\mathscr B}_\mathscr{G}^2 \end{array}$ (0, T + K), i = 1, 2 be two solutions of ABDSDE (1). Define $\begin{matrix} {\bar{Y}}_{t} = Y_{t}^{1} - Y_{t}^{2}, {\bar{Z}}_{t} = Z_{t}^{1} - Z_{t}^{2}, Δ g (r) = g (r, Y_{r}^{1}, Z_{r}^{1}) - g (r, Y_{r}^{2}, Z_{r}^{2}) \end{matrix}$ $\begin{array}{} \displaystyle \overline Y_t= Y_t^1 - Y_t^2,\quad \overline Z_t= Z_t^1 - Z_t^2, \Delta g(r)=g(r, Y_r^1, Z_r^{1})- g(r, Y_r^{2}, Z_r^{2}) \end{array}$ and $\begin{matrix} Δ f (r) = f (r, Y_{r}^{1}, Z_{r}^{1}, Y_{r + δ (r)}^{1}, Z_{r + ζ (r)}^{1}) - f (r, Y_{r}^{2}, Z_{r}^{2}, Y_{r + δ (r)}^{2}, Z_{r + ζ (r)}^{2}) . \end{matrix}$ $\begin{array}{} \displaystyle \Delta f(r) = f(r, Y_r^1, Z_r^{1},Y_{r+\delta(r)}^{1}, Z_{r+\zeta(r)}^{1}) -f(r, Y_r^{2}, Z_r^{2},Y_{r+\delta(r)}^{2}, Z_{r+\zeta(r)}^{2}). \end{array}$

We obtain the following equation

$\begin{matrix} \{\begin{cases} {\bar{Y}}_{t} = \int_{t}^{T} Δ f (r) d r + \int_{t}^{T} Δ g (r) d B_{r} - \int_{t}^{T} {\bar{Z}}_{r} d W_{r}, 0 \leq t \leq T, \\ {\bar{Y}}_{t} = 0, {\bar{Z}}_{t} = 0, T \leq t \leq T + K . \end{cases} \end{matrix}$ $$\begin{array}{} \displaystyle \begin{cases} \overline Y_t = \displaystyle \int_t^T\!\! \Delta f(r) dr + \int_t^T \!\! \Delta g(r)dB_r - \displaystyle\int_t^T \overline Z_r dW_r, \quad \quad 0\le t \le T,\\ \overline Y_t = 0, \quad \overline Z_t = 0, \quad\quad T\le t \le T +K. \end{cases} \end{array}$$(17)

By the same computations as in Lemma 4.1, we obtain

$\begin{matrix} E [| {\bar{Y}}_{t} |^{2}] + \frac{1}{2} E \int_{t}^{T} | {\bar{Y}}_{r} |^{2} d r + \frac{1}{2} E \int_{t}^{T} | {\bar{Z}}_{r} |^{2} d r \leq C^{'} E \int_{t}^{T} ρ (| {\bar{Y}}_{r} |^{2}) d r, 0 \leq t \leq T . \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E} [|\overline Y_t|^2] + \frac{1}{2}{\bf E} \int_t^T |\overline Y_r|^2 dr + \frac{1}{2}{\bf E} \int_t^T |\overline Z_r|^2 dr \le C'{\bf E}\int_t^T \!\!\rho(|\overline Y_r|^2)dr, \quad 0\le t \le T. \end{array}$$

This leads to

$\begin{matrix} E [| {\bar{Y}}_{t} |^{2}] \leq C^{'} E \int_{t}^{T + K} ρ (| {\bar{Y}}_{r} |^{2}) d r, 0 \leq t \leq T + K . \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E}[|\overline Y_t|^2]\le C'{\bf E}\int_t^{T+K} \!\!\rho(|\overline Y_r|^2)dr, \quad 0\le t \le T+K. \end{array}$$

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

$\begin{matrix} E | {\bar{Y}}_{t} |^{2} \leq C^{'} \int_{t}^{T + K} ρ (E | {\bar{Y}}_{r} |^{2}) d r, 0 \leq t \leq T + K . \end{matrix}$ $$\begin{array}{} \displaystyle {\bf E}|\overline Y_t|^2 \le C'\int_t^{T+K} \rho({\bf E}|\overline Y_r|^2)dr, \quad 0\le t \le T+K. \end{array}$$

Then we can use Bihari’s inequality to obtain $\begin{matrix} E | {\bar{Y}}_{t} |^{2} = 0, 0 \leq t \leq T + K . \end{matrix}$ $\begin{array}{} \displaystyle {\bf E}|\overline Y_t|^2=0, 0\le t\le T+K. \end{array}$ This implies Z_t = 0.□

eISSN:: 2444-8656
Language:: English

Publication timeframe:: Volume Open
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RETRACTED ARTICLE: Anticipated backward doubly stochastic differential equations with non-Liphschitz coefficients

Published Online: Jun 24, 2019

Page range: 101 - 112

Received: Jan 02, 2019

Accepted: Mar 19, 2019

DOI: https://doi.org/10.2478/AMNS.2019.1.00011

KeywordsAnticipated backward doubly stochastic differential equation, non-lipschitz coefficients, Itô’s representation formula and Gronwall lemma

© 2019 Sadibou Aidara, published by Sciendo

This work is licensed under the Creative Commons Attribution 4.0 International License.

Keywords
Anticipated backward doubly stochastic differential equation, non-lipschitz coefficients, Itô’s representation formula and Gronwall lemma