Stochastic PDEs and Infinite Horizon Backward Doubly Stochastic Differential Equations

Notation The Euclidean norm of a vector x ∈ R^k will be denoted by |x|, and, for a d × k matrix A, we define $$, where A^* is the transpose of A.

Let (Ω, ℱ, P) be a completed probability space and let {W_t} _t≥0 and {B_t} _t≥0 be two mutually independent standard Brownian motions, with values, respectively, in R^d and R^l, defined on (Ω, ℱ, P). Let 𝒩 denote the class of P-null sets of ℱ. For each t ∈ [0, ∞), we define

Note that $$ is an increasing filtration and $$ is a decreasing filtration, and the collection {ℱ_t, t ∈ [0, ∞)} is neither increasing nor decreasing.

Suppose

For any n ∈ N, let S²(R⁺; Rⁿ) denote the space of all {ℱ_t}-measurable n-dimensional processes v with norm of | | v | |_S≐[E(sup _s≥0 |v(s) | ) ²] ^1/2 < ∞.

We denote similarly by M²(R⁺; Rⁿ) the space of all (classes of dP ⊗ dt a.e. equal) {ℱ_t}-measurable n-dimensional processes v with norm of $$.

For any t ≥ 0, let L²(Ω, ℱ_t, P; Rⁿ) denote the space of all {ℱ_t}-measurable n-valued random variables ξ satisfying E | ξ|² < ∞.

We also denote

For each (X, Y) ∈ B², we define the norm of (X, Y) by Obviously B² is a Banach space.

Consider the following infinite horizon backward doubly stochastic differential equation:

where ξ ∈ L²(Ω, ℱ, P; R^k) is given. We note that the integral with respect to {B_t} is a backward Itô integral and the integral with respect to {W_t} is a standard forward Itô integral. These two types of integrals are particular cases of the Itô-Skorohod integral; see Boufoussi et al. [12]. Our aim is to find some conditions under which BDSDE (2.5) has a unique solution. Now we give the definition of solution of BDSDE (2.5).

Definition 2.1. A pair of processes (y, z) : Ω × R⁺ → R^k × R^k×d is called a solution of BDSDE (2.5), if (y, z) ∈ B² and satisfies BDSDE (2.5).

Let

satisfy the following assumptions:

• (H1)

  for any (y, z) ∈ R^k × R^k×d, f(·, y, z) and g(·, y, z) are {ℱ_t}-progressively measurable processes, such that

  $$ ()

• (H2)

  f and g satisfy Lipschitz condition with Lipschitz coefficient v : = {v(t)}; that is, there exists a positive nonrandom function {v(t)} such that

  $$ ()
  for all (t, y_i, z_i) ∈ R⁺ × R^k × R^k×d,  i = 1,2,

• (H3)

  $$.

Theorem 3.1. Under the above conditions, in particular (H1), (H2), and (H3), (2.5) has unique solution (y, z) ∈ B².

Lemma 3.2. Suppose (H1), (H2), and (H3) hold for f and g. For any T ∈ [0, ∞], let $$,  (Yⁱ, Zⁱ) and (yⁱ, zⁱ) ∈ B²  (i = 1,2) satisfy the following equation:

Then there is a constant C > 0, such that, for any τ ∈ [0, T], where $$ and I_[τ,T](·) is an indicator function.

Proof. Firstly, we assume that τ = 0,  T = ∞.

Set

Then We define the filtration {𝒢_t} _0≤t≤T by Obviously 𝒢_t is an increasing filtration. Since $$, $$ is a 𝒢_t-martingale, thus from (3.4) it follows that Note that Applying Doob inequality and B-D-G inequality, we can deduce where c₀ > 0 is a constant.

On the other hand, from (3.4) it follows that

where 〈M〉 is the variation process generated by the martingale M.

Consequently, (3.8) and (3.9) imply that

where C = (57 + 6c₀) is a constant, and $$.

For any τ, T ∈ [0, ∞], we set f₁(t, y_t, z_t) = f(t, y_t, z_t)I_[τ,T], and g₁(t, y_t, z_t) = g(t, y_t, z_t)I_[τ,T]. Then f₁, and g₁ satisfy the assumptions (H1), (H2), and (H3), and their Lipschitz constants are vI_[τ,T].

Obviously,

Since $$, $$ is a 𝒢_t-martingale, thus from (3.11) it follows that Note that Applying Doob inequality and B-D-G inequality, we can deduce where c₀ > 0 is a constant.

On the other hand, from (3.11) it follows that

Consequently, (3.14) and (3.15) imply that where C = (57 + 6c₀) is a constant, and $$.

Martingale Representation Theorem [4] Suppose Y is a random variable, such that E | Y² | < ∞. Note that M_t = E[Y∣𝒢_t] is a square integrable martingale with respect to 𝒢_t and can be represented using martingale representation theorem as $$, where $$.

Now we give the proof of the Theorem 3.1.

Proof. The proof of Theorem 3.1 is divided into two steps.

Step 1. We assume $$. For any (y, z) ∈ B², let

We will prove {M_t} is a square integrable 𝒢_t-martingale. From (H1)–(H3), it follows that which means {M_t} is a square integrable 𝒢_t-martingale. According to the martingale representation theorem, there exists a unique 𝒢_t-progressively measurable process Z_t with value in R^k×d such that Let So Then We show that {Y_t} and {Z_t} are in fact ℱ_t-measurable. For Y_t, this is obvious since, for each t, where $$ is indeed $$-measurable. Hence $$ is independent of ℱ_t⋁ σ(Θ), and Now and the right side is $$-measurable. Hence, from Itô’s martingale representation theorem, {Z_s, s > t} is $$-adapted. Consequently Z_s is $$-measurable, for any t < s, and, thus, Z_t is ℱ_t-measurable. So (Y, Z) ∈ B². Therefore (3.22) has constructed a mapping from B² to B², and we denote it by ϕ, that is, If ϕ is a contractive mapping with respect to the norm ∥·∥_B, by the fixed point theorem, there exists a unique (y, z) ∈ B², satisfying (3.22), which is just the unique solution to BDSDE (2.5).

Now we are in the position to prove that ϕ is a contractive mapping. Supposing that (yⁱ, zⁱ) ∈ B², let (Yⁱ, Zⁱ) be the map ϕ of (yⁱ, zⁱ),   (i = 1,2), that is

We denote By Lemma 3.2, we have

Due to l_[0,∞] ≤ 1/2C, it follows that ϕ is a contractive mapping from B² to B².

Step 2. Since $$, then there exists a sufficiently large constant T such that

Let then (H1)–(H3) hold for f₁ and g₁, whose Lipschitz coefficients are $$. Obviously, By Step 1, there exists a unique $$ satisfying For $$ given as above, let us consider the following infinite BDSDE: According to the results of Pardoux and Peng [1], the above BDSDE has a unique solution $$ in [0, T], thus the above BDSDE has a unique solution such that $$ for every t > T. Let It is easy to check that (y_t, z_t) is the unique solution of (2.5).

Remark 3.3. Suppose v is a constant, if we choose v(t) = vI_[0,T](t), then Theorem 3.1 is the main theorem in the paper by Pardoux and Peng [1].

Remark 3.4. The condition (H3) is usually necessary. That is, if for any ξ ∈ L²(Ω, ℱ, P; R^k) and f, g hold in (H1) and (H2), BDSDE (2.5) has a unique solution in B², then the (H3) is necessary.

In fact, let us choose f(s, y_s, z_s) = (1/(1 + s))z_s,   g(s, y_s, z_s) = 0 and any ξ ∈ L²(Ω, ℱ, P; R^k), then the solution of BDSDE

should be where 〈y_t, W_t〉 is the variation process generated by the semimartingale y_t and Brownian motion W_t.

Thus the assumption (H3) is necessary.

Remark 3.5. The following example shows that if the coefficients f and g of BDSDE (2.5) satisfy the uniformly Lipschitz, the BDSDE (2.5) has no solution.

For all T > 0, let $$, then the BDSDE $$ has a unique solution pair $$,

When T → ∞, $$ and $$ in L²(Ω, ℱ, P), but z_t = 1/(1 + t) is not the solution of the following infinite horizon BDSDE:

because $$.

Theorem 4.1. Suppose ξ_i ∈ L²(Ω, ℱ, P; R^k),   (i = 1,2), and consider (H1)–(H3). Let (yⁱ, zⁱ) be the solutions of BDSDE (2.5) corresponding to the terminal data ξ = ξ₁,   ξ = ξ₂, respectively. Then there exists a constant $$ such that

Proof. Set $$,  $$. Since $$, we can choose a strictly increasing sequence 0 = t₀ < t₁ < ⋯<t_n < t_n+1 = ∞ such that

Applying Lemma 3.2, we have Thus In particular, we have From (4.4) and (4.5), it follows that Thus the desired result is obtained.

Theorem 4.2. Suppose ξ, ξ_i ∈ L²(Ω, ℱ, P; R^k),   (i = 1,2, …), (H1)–(H3) hold for f and g. Let (yⁱ, zⁱ) be the solutions of the following BDSDE:

If E | ξ_i − ξ|² → 0 as i → ∞, then there exists a pair (y, z) ∈ B² such that ||(yⁱ − y, zⁱ − z) | |_B → 0 as i → ∞. Furthermore, (y, z) is the solution of the following BDSDE:

Proof. For any n, m ≥ 1, let (yⁿ, zⁿ) and (y^m, z^m) be the solutions of (4.7) corresponding to ξ_n and ξ_m, respectively. Due to Theorem 4.1, there exists a constant $$ such that

which means that {(yⁱ, zⁱ),   i = 1,2, …} is a Cauchy sequence in B². Thus there exists a pair (y, z) ∈ B² such that ||(yⁱ − y, zⁱ − z) | |_B → 0 as i → ∞. Since Thus for any $$ and $$ in L²(Ω, ℱ, P). Taking the limit on both sides of (4.7), we deduce that (y, z) is the solution to BDSDE (4.8). The desired result is obtained.

Corollary 4.3. Assume ξ ∈ L²(Ω, ℱ, P; R^k), (H1)–(H3) hold for f and g. Let (y, z) be the solution of BDSDE (2.5). For any T > 0, let (y^T, z^T) be the solutions of the finite time interval BDSDE (4.11), then(y^T, z^T)→(y, z) in B² as T → ∞.

Proof. Note that $$ in L²(Ω, ℱ, P; R^k) as T → ∞. The proof is straightforward from Theorem 4.2.

Theorem 5.1. Let κ(t, x);   t ≥ 0,   x ∈ R^d be a random field such that κ(t, x) is $$-measurable for each (t, x),   κ ∈ C^0,2([0, ∞) × R^d; R^k) a.s., and κ satisfies (5.4). Then $$, where $$ solves the BDSDE (5.3).

Proof. We can apply the extension of the Itô formula [5] to the solution κ of (5.4):

We can see that $$ coincides with the unique solution of (5.3). It follows that $$.

Theorem 5.2. Let f and g satisfy (H1), (H2), and (H3). Then $$ is the unique classical solution of the system of backward SPDEs (5.3).

We can finish the proof exactly as in Theorem 3.2 of Hu and Ren [13].