Pathwise Stationary Solutions of Stochastic Partial Differential

Pathwise Stationary Solutions of Stochastic Partial Differential Equations and Backward Doubly Stochastic Differential Equations on Infinite Horizon Qi Zhang1,2 , Huaizhong Zhao1 1

2

Department of Mathematical Sciences, Loughborough University, Loughborough, LE11 3TU, UK. [email protected], [email protected] School of Mathematics and System Sciences, Shandong University, Jinan, 250100, China.

Summary. The main purpose of this paper is to study the existence of stationary solution for stochastic partial differential equations. We establish a new connection between backward doubly stochastic differential equations on infinite time horizon and the stationary solution of the SPDEs. For this we study the existence of the solution of the associated BDSDEs on infinite time horizon and prove it is a stationary viscosity solution of the corresponding SPDEs.

Keywords: backward doubly stochastic differential equations, stochastic partial differential equations, pathwise stationary solution, random dynamical systems.

1 Intrduction As for the deterministic dynamical systems, the pathwise stationary solution of a stochastic dynamical system is one of the fundamental questions of basic importance and a generic phenomenon ([1], [9], [12], [13], [16], [18]-[20], [28], [29]). However, in contrast to the deterministic dynamical systems, the existence of stationary solutions of stochastic dynamical systems generated e.g. by stochastic differential equations (SDEs) or stochastic partial differential equations (SPDEs), is a difficult problem and results are only known in some special cases. In recent years, substantial results on the existence and uniqueness of invariant measures for SPDEs and weak convergence of the law of the solutions as time tends to ∞ have been proved ([4], [5], [11], [14], [15] to name, but a few). The invariant measure describes the invariance of a certain solution in law when time changes. As for the pathwise stationary solutions, some significant progresses have been made for certain SPDEs ([9], [12], [13], [16], [18]-[20], [28], [29]), however their existence still remained unsolved for many SPDEs. Their existence and stabilities are of great interests both in mathematics and physics. Let u : [0, ∞) × U × Ω → U be a

2

Q. Zhang and H.Z. Zhao

measurable random dynamical system on a measurable space (U, B) over a metric dynamical system (Ω, F, P , (θt )t≥0 ), then a stationary solution is a F-measurable random variable Y : Ω → X such that (Arnold [1]) u(t, Y (ω), ω) = Y (θt ω) t ≥ 0, a.s..

(1.1)

This ”one-force, one solution” setting is a natural extension of equilibrium in deterministic systems to stochastic counterparts. Different from the invariant measure, it describes the pathwise invariance of the stationary solution over time along the measurable and P -preserving transformation θt : Ω −→ Ω and the pathwise limit of the solutions of SPDEs. The simplest nontrivial example is the Ornstein-Uhlenbeck process defined by the stochastic differential eqaution du(t) = −u(t)dt + dBt . It defines a random dynamical sysRt tem u(t, u0 ) = u0 e−t + 0 e−(t−s) dBs and its stationary point is given by R0 Y (ω) = −∞ es dBs . But unlike deterministic cases, in general, the stationary solutions of the stochastic systems can not be given explicitly. It was even unthinkable to represent them as solutions of certain differential or functional equations. The existence of the stationary solutions for SPDEs is a subtle problem. There is no method which can be applied to SPDEs with great generalities. In [28], [29], the stationary strong solution of the stochastic Burgers’ equations with periodic or random forcing (C 3 in the space variable) was established by Sinai using the Hopf-Cole transformation. In [18], the existence of the H 1 [0, 1]-valued solution of the stochastic Burgers’ integral equation, as the stationary solution of the stochastic Burgers’ equation on [0, 1] with L2 [0, 1]-valued noise, was obtained by Liu and Zhao. In [12], the existence of stationary solution for certain SPDEs with linear one-dimensional noise was obtained by Duan, Lu and Schmalfuss. In [20], the stationary solution of the stochastic evolution equations was identified as a solution of a corresponding integral equation up to time +∞ and the existence was established for certain SPDEs by Mohammed, Zhang and Zhao. But the existence of solution of such a stochastic integral equation in general is far from clear. The local existence of stable/unstable manifold near a stationary solution for a stochastic dynamical system generated by SPDEs was studied by Mohammed, Zhang and Zhao in [19], [20], and by Duan, Lu and Schmalfuss in [12], [13]. In this paper, we use a new approach to study the stationary solutions of SPDEs via the solutions of backward doubly stochastic differential equations (BDSDEs) on infinite horizon. Backward stochastic differential equations (BSDEs) have been studied extensively in the last 15 years since the pioneering work of Pardoux and Peng [23]. The connection of BSDEs and quasilinear parabolic partial differential equations (PDEs) was discovered by Pardoux and Peng in [24] and Peng in [26]. The BDSDEs and their connections with the strong solutions of the SPDEs were studied by Pardoux and Peng in [25], also recently by Backdahn and Ma for the stochastic viscosity solutions in [6], [7], and by Bally and Matoussi for the weak solutions in [3]. On the other hand, the infinite horizon BSDE was first studied by Peng in

Stationary Solution of SPDEs and BDSDEs on Infinite Horizon

3

[26] and it was shown that the corresponding PDE is a Poisson equation (elliptic equation). This was studied systematically by Pardoux in [22]. Coupled forward-backward stochastic differential equations (FBSDEs) on infinite horizon and their connections with PDEs were recently studied by Shi and Zhao in [27]. Notice that the solutions of the Poisson equations can be regarded as the stationary solutions of the parabolic PDEs. Deepening this idea, it would not be unreasonable to conjecture that the solutions of infinite time horizon BDSDEs (if exists) be the stationary solutions of the corresponding SPDEs. Of course, we cannot write it as Poisson equation or stochastic Poisson equation like in the deterministic cases. However, it is very natural to describe the stationary solutions of SPDEs by infinite horizon BDSDEs. In this sense, BDSDEs (or BSDEs) can be regarded as more general SPDEs (or PDEs). We first study the existence of the solution to the BDSDE on the infinite time horizon in the space S p,−K ([0, ∞); R1 ) ∩ M 2,−K ([0, ∞); R1 ) × M 2,−K ([0, ∞); Rd ) (see Section 2 for definitions). The equation can be written in the following integral form: for any s ≥ t, Z ∞ Z ∞ 0 0 0 K − K 0 r t,x − K2 r t,x t,x t,x − K2 s t,x Ys = e f (Xr , Yr , Zr )dr + e 2 Yr dr e 2 s s Z ∞ Z ∞ K0 K0 + e− 2 r hg(Xrt,x , Yrt,x ), d† Br i − e− 2 r hZrt,x , dWr i. (1.2) s

s

Here f , g and K 0 satisfy conditions (H.1)0 , (H.2)0 , (H.3)0 , (H.4)0 in Section 3, {Bt } is a Rl -Brownian motion on a probability space (Ω 1 , F 1 , P 1 ) and {Wt } is a Rd -Brownian motion on a probability space (Ω 2 , F 2 , P 2 ); the stochastic integral Z ∞ Z n 0 K0 − K2 r t,x t,x † e hg(Xr , Yr ), d Br i = lim e− 2 r hg(Xrt,x , Yrt,x ), d† Br i n→∞

s

Rn

s

0 − K2

r in L2 (dP 1 × dP 2 ), where s e g(Xrt,x , Yrt,x )d† Br is a backward stochastic integral; Z ∞ Z n K0 K0 e− 2 r hZrt,x , dWr i = lim e− 2 r hZrt,x , dWr i n→∞

s

Rn

s

0 − K2

r in L2 (dP 1 ×dP 2 ), where s e hZrt,x , dWr i is a forward stochastic integral; t,x Xs is generated by the following stochastic differential equation dXst,x = b(Xst,x )ds + σ(Xst,x )dWs , s > t (1.3) Xst,x = x, 0 ≤ s ≤ t.

Here b(·), σ(·) satisfy the condition (H.5) in Section 3. Consider the following SPDEs Z t v(t, x) = v(0, x) + [Lv(s, x) + f x, v(s, x), σ ∗ (x)Dv(s, x) ]ds 0 Z t ˆs i. − hg x, v(s, x) , dB (1.4) 0

4


R ˆs is a two-sided Brownian motion, t hg x, v(s, x) , dB ˆs i is a forward Here B 0 stochastic integral; L is infinitesimal generator of the process Xst,x given by L=

n n X 1 X ∂2 ∂ aij (x) + bi (x) 2 i,j=1 ∂xi ∂xj ∂xi i=1

ˆT −s − B ˆT for all 0 ≤ with aij (x) = σσ ∗ (x). Take Bs in Eq.(1.2) to be B s < ∞. We then prove that, for arbitrary T > 0 and 0 ≤ t ≤ T , v(t, x, ω 1 ) = −t,x YTT−t (ω 1 ) is a stationary viscosity solution of Eq.(1.4). For this, we show ˆ· through time in Section 4 that the shift operator θ·1 of B· , defined from B 1 reversal at T , satisfies that θT −t B· is independent of T . This remarkable −t,x fact is used to show YTT−t (ω 1 ) is independent of T . It is proved that the stationary solution is a stochastic viscosity solution of Eq.(1.4). Stochastic viscosity solutions of such SPDEs were introduced by Buckdahn and Ma in [6]-[8]. As far as we know, the connection of the pathwise stationary solutions of the SPDEs and infinite horizon BDSDEs we study in this paper is new. We believe this new method can be used to many SPDEs such as those with quadratic or polynomial growth nonlinear terms, more types of noises, weak solutions etc. We don’t intend to include all these results in the current paper, but only study Lipschitz continuous nonlinear term and finite dimensional noise for simplicity in order to initiate this intrinsic method to the study of this basic problem in dynamics of SPDEs. We would like to point out that our BDSDE method depend on neither the continuity of the random dynamical system (contunuity means Φ(t, ·, ω) : U → U is a.s. continuous) nor on the method of the random attractors. The continuity problem for the SPDE (1.4) with the nonlinear noise considered in this paper still remains open mainly due to the failure of Kolmogorov’s continuity theorem in infinite dimensional setting as pointed out by many researchers (e.g. [12], [19]).

2 BDSDE on infinite time interval Let (Ω 1 × Ω 2 , F 1 × F 2 , P 1 × P 2 ) be a product probability space, (Bt )t≥0 and (Wt )t≥0 be two mutually independent standard Brownian motion processes with values on Rl and Rd , defined on (Ω 1 , F 1 , P 1 ) and (Ω 2 , F 2 , P 2 ) respectively. Let N 1 × N 2 denote the class of P 1 × P 2 -null sets of F 1 × F 2 . For each t ≥ 0, we define _ B Ft , Ft,∞ × FtW N 1 × N 2. η η , Here forW any process {ηt }, σ{ηr − ηs ; 0 ≤ s ≤ r ≤ t}, Ftη = F0,t W Fs,t = η η η η F∞ = t≥0 Ft , Ft,∞ = T ≥0 Ft,T . Take K > 0, q ≥ 2, then we denote M q,−K ([0, ∞); Rd ) the set of jointly measurable random processes {ϕt , t ≥ 0} with values on Rd satisfying


5

(i) ϕt is Ft -measurable for t ≥ 0; R∞ (ii) E[ 0 e−Kt |ϕt |q dt] < ∞. Also we denote S q,−K ([0, ∞); R1 ) the set of jointly measurable random processes {ψt , t ≥ 0} with values on R1 satisfying (i) ψt is Ft -measurable for t ≥ 0 and continuous for P-a.s. ω; (ii) E[supt≥0 e−Kt |ψt |q ] < ∞. Similarly, M 2,0 ([0, T ]; Rd ) is the set of jointly measurable random processes {ϕ˜t , 0 ≤ t ≤ T } with values on Rd satisfying (i) ϕ˜t is Ft -measurable for 0 ≤ t ≤ T ; RT (ii) E[ 0 |ϕ˜t |q dt] < ∞. And S q,0 ([0, T ]; R1 ) is the set of jointly measurable random processes {ψ˜t , 0 ≤ t ≤ T } with values on R1 satisfying (i) ψ˜t is Ft -measurable for 0 ≤ t ≤ T and continuous for P-a.s. ω; (ii) E[sup0≤t≤T |ψ˜t |q ] < ∞. We assume that (H.1). f : Ω 1 × Ω 2 × [0, ∞) × R1 × Rd −→ R1 , g : Ω 1 × Ω 2 × [0, ∞) × R1 × Rd −→ Rl are jointly measurable and f· (0, 0) ∈ M p,−K ([0, ∞); R1 ), g· (0, 0) ∈ M p,−K ([0, ∞); Rl ), where p > d + 2. (H.2). There exist three constants C0 , C > 0, 0 < α < 13 , s.t. for any (ω 1 , ω 2 , t) ∈ Ω 1 × Ω 2 × [0, ∞), (y1 , z1 ), (y2 , z2 ) ∈ R1 × Rd , |ft (y1 , z1 ) − ft (y2 , z2 )| ≤ C0 |y1 − y2 | + C|z1 − z2 |, |gt (y1 , z1 ) − gt (y2 , z2 )| ≤ C|y1 − y2 | + α|z1 − z2 |. (H.3). There exists a constant µ > 0, s.t. for K < K 0 < 2K, 2µ − K 0 − p(2p − 1)C 2 > 0 and for any (ω 1 , ω 2 , t) ∈ Ω 1 × Ω 2 × [0, ∞), y1 , y2 ∈ R1 , z ∈ Rd , (y1 − y2 )(ft (y1 , z) − ft (y2 , z)) ≤ −µ|y1 − y2 |2 . The main purpose of this section is to prove the existence and uniqueness of the solutions of BDSDEs on infinite horizon. First, we briefly include the results obtained by Pardoux and Peng ([25]) and Buckdahn-Ma ([6]-[8]) on finite time BDSDE. The following BDSDE was first introduced by Pardoux and Peng in [25]: for 0 ≤ t ≤ T ,

6


Z

T

T

Z

Yt = ξ +

hg(s, Ys , Zs ), d† Bs i −

f (s, Ys , Zs )ds + t

t

Z

T

hZs , dWs i.(2.1) t

Here the integral w.r.t. {Bt } is a ”backward Itô’s integral” and the integral w.r.t. {Wt } is a standard ”forward Itô’s integral”. The backward integral is particular case of Itˆ o-Skorohod integral ([21]). An important fact is that, for 0 ≤ T ≤ T 0 and arbitrary a, b satisfying 0 ≤ a ≤ b ≤ T 0 , if a process h· with Rb values on Rl satisfies a |hs |ds < ∞ a.s., then T

Z

hhs , d† Bs i = −

Z

T 0 −t

ˆs i hhT 0 −s , dB

a.s..

(2.2)

T 0 −T

t

ˆs = BT 0 −s − BT 0 . Here B The following existence and uniqueness theorem for Ft -measurable solution (Yt , Zt )0≤t≤T of Eq.(2.1) was proved in [25]. It will be used in the proof of Theorem 2.3. Theorem 2.1 ([25]) Under conditions (H.1), (H.2), for any given Ft -measurable ξ ∈ L2 (dP 1 × dP 2 ), Eq.(2.1) has a unique Ft -measurable solution (Y· , Z· ) ∈ S 2,0 ([0, T ]; R1 ) × M 2,0 ([0, T ]; Rd ). Let (Xst,x )t≤s≤T be defined by Eq.(1.3), and consider Eq.(2.1) in the following form for t ≤ s ≤ T Z T Yst,x = h(XTt,x ) + f (Xrt,x , Yrt,x , Zrt,x )dr s

Z +

T

hg(Xrt,x , Yrt,x , Zrt,x ), d† Br i −

s

Z

T

hZrt,x , dWr i.

(2.3)

s

Pardoux and Peng also proved that under some strong smoothness conditions of h, b, σ, f and g (for details see [25]), u(t, x) = Ytt,x , (t, x) ∈ [0, T ] × Rd , is independent of ω 2 ∈ Ω 2 and is the unique classical solution of the following backward SPDE Z T u(t, x) = h(x) + [Lu(s, x) + f x, u(s, x), σ ∗ (x)Du(s, x) ]ds t

Z +

T

hg x, u(s, x), σ ∗ (x)Du(s, x) , d† Bs i,

0 ≤ t ≤ T. (2.4)

t

It is easy to see from (2.2) this equation is the time reversal version of a forward SPDE. We are particularly interested in the case that g(x, y, z) = g(x, y) where the stochastic viscosity solution can be defined. In such a case, v(t, x) = ˆs = BT −s − BT . The stochastic u(T − t, x) is a solution of Eq.(1.4) with B viscosity solution of SPDE (1.4) was studied by Buckdahn and Ma in [6]-[8] and it was proved that the solution Ytt,x of Eq.(2.3), (t, x) ∈ [0, T ] × Rd , is


7

a viscosity solution of Eq.(2.4) under conditions (H.1)0 , (H.2)0 , (H.4)0 , (H.5) stated in Sections 3 and 4. Therefore it gives the stochastic viscosity solution of Eq.(1.4) through the time reversal argument. To benefit readers, we include briefly Backdahn-Ma’s main idea of the stochastic viscosity solutions for SPDEs in [6]-[8]. Define λ(t, x, y) as the solution of the following SDE Z λ(t, x, y) = y −

t

ˆs i. hg(x, λ(s, x, y)), dB

0

Under the condition (H.4)0 , the equation can define a stochastic flow λ(t, x, y), i.e., for fixed x, the random field λ(·, x, ·) is continuously differentiable in the variable y, and the mapping y −→ λ(t, x, y) defines a diffeomorphism for all (t, x), a.s. ω 1 . Denote its inverse by ζ(t, x, y) = (λ(t, x, ·))−1 (y). Let v˜(t, x) = ζ(t, x, v(t, x)), the so-called Doss-Sussmann transformation, then v˜(t, x) satisfies the following random PDE Z v˜(t, x) = v˜(0, x) +

t

[L˜ v (s, x) + f˜ s, x, v˜(s, x), σ ∗ (x)D˜ v (s, x) ]ds. (2.5)

0

Here f˜(t, x, y, z) =

1 f (x, λ(t, x, y), σ ∗ (x)Dx λ(t, x, y) + Dy λ(t, x, y)z Dy λ(t, x, y) 1 +Lx λ(t, x, y) + hσ ∗ (x)Dxy λ(t, x, y), zi + Dyy λ(t, x, y)|z|2 . 2

Then the stochastic viscosity solution v(t, x) of Eq.(1.4) was defined in [6]-[8] through the ”deterministic” viscosity solution v˜(t, x) of Eq.(2.5) for a.s. ω 1 ∈ Ω 1 via the relation v(t, x) = λ(t, v˜(t, x)). The notion of viscosity solutions of partial differential equations was first introduced by Crandall and Lions in [10]. Buckdahn and Ma established the connection of the solution of BDSDE (2.3) and the stochastic viscosity solution of the SPDE (1.4). In particular, the following result of Buckdahn and Ma will be used in Section 4. Theorem 2.2 ([6]) Assume Conditions (H.1), (H.2) and that the function h : Rd → R is continuous and for some constants C, δ > 0, |h(x)| ≤ C(1 + −t,x |x|). Then v(t, x) = u(T − t, x) = YTT−t is a stochastic viscosity solution of Eq. (1.4). Now we consider the following BDSDE with infinite time horizon, for t ≥ 0, Z ∞ Z ∞ 0 0 0 K − K0 s − K2 t − K2 s Yt = e fs (Ys , Zs )ds + e 2 Ys ds e 2 t t Z ∞ Z ∞ K0 K0 + e− 2 s hgs (Ys , Zs ), d† Bs i − e− 2 s hZs , dWs i. (2.6) t

t

8


Note that the integral w.r.t. {Bt } and the integral w.r.t. {Wt } are the same as in Eq.(2.1). The main objective of this Section is to prove Theorem 2.3 Under the above conditions (H.1), (H.2), (H.3), Eq.(2.6) has a unique solution (Y· , Z· ) ∈ S p,−K ([0, ∞); R1 ) ∩ M 2,−K ([0, ∞); R1 ) × M 2,−K ([0, ∞); Rd ). Proof. Rewrite Eq.(2.6) in differential form ( dYt = −ft (Yt , Zt )dt − hgt (Yt , Zt ), d† Bt i + hZt , dWt i, K0 limT →∞ e− 2 T YT = 0 a.s.. So for arbitrary T > 0 and 0 ≤ t ≤ T , Eq.(2.6) is equivalent to ( RT RT RT Yt = YT + t fs (Ys , Zs )ds + t hgs (Ys , Zs ), d† Bs i − t hZs , dWs i, 0 (2.7) K limT →∞ e− 2 T YT = 0 a.s.. We are able to investigate Eq.(2.7) instead of Eq.(2.6). Uniqueness. Let (Yt1 , Zt1 ) and (Yt2 , Zt2 ) be two solutions of Eq.(2.7). Define Y¯t = Yt1 − Yt2 , Z¯t = Zt1 − Zt2 ,

t ≥ 0.

Then for 0 ≤ t ≤ T ,  RT 2 2 1 1   Y¯t = Y¯TR+ t fs (Ys , Zs ) − fs (Ys , Zs ) ds R T T + t hgs (Ys1 , Zs1 ) − gs (Ys2 , Zs2 ), d† Bs i − t hZ¯s , dWs i,  0 K  limT →∞ e− 2 T Y¯T = 0 a.s.. 0

Rewrite it by differential form and note that K2 ≤ K,   dY¯s = − fs (Ys1 , Zs1 ) − fs (Ys2 , Zs2 ) ds −hgs (Ys1 , Zs1 ) − gs (Ys2 , Zs2 ), d† Bs i + hZ¯s , dWs i,  limT →∞ e−KT Y¯T = 0 a.s.. 2 Apply Itˆ o’s formula to e−Ks |Y¯s | , then 2 d(e−Ks |Y¯s | ) 2 = −Ke−Ks |Y¯s | ds + 2e−Ks Y¯s , dY¯s

−e−Ks |gs (Ys1 , Zs1 ) − gs (Ys2 , Zs2 )|2 ds + e−Ks |Z¯s |2 ds.


9

Therefore, Z 2 E[e−Kt |Y¯t | ] + E[

T

Z e−Ks |Z¯s |2 ds] − E[

t

T

2 Ke−Ks |Y¯s | ds]

t

Z 2 = E[e−KT |Y¯T | ] + E[

T

2e−Ks Y¯s fs (Ys1 , Zs1 ) − fs (Ys2 , Zs1 ) ds]

t

Z +E[

T

2e−Ks Y¯s

fs (Ys2 , Zs1 ) − fs (Ys2 , Zs2 ) ds]

t

Z +E[

T

e−Ks |gs (Ys1 , Zs1 ) − gs (Ys2 , Zs2 )|2 ds]

t

Z 2 ≤ E[e−KT |Y¯T | ] − E[

T

Z 2 2µe−Ks |Y¯s | ds] + E[

2e−Ks C|Y¯s ||Z¯s |ds]

t

t

Z +E[

T

T

e−Ks (C|Y¯s | + α|Z¯s |)2 ds].

t

That is, T

Z 2 E[e−Kt |Y¯t | ] + E[

(1 −

t

Z +E[

T

1 − 2α2 )e−Ks |Z¯s |2 ds] 2

2 (2µ − K − 4C 2 )e−Ks |Y¯s | ds]

t 2

≤ E[e−KT |Y¯T | ].

(2.8)

So 2

2

E[e−Kt |Y¯t | ] ≤ E[e−KT |Y¯T | ]. With K replaced by K 0 it is also valid, so 0

2

E[e−K t |Y¯t | ] ≤ e(K−K

0

)T

2

E[e−KT |Y¯T | ].

Since Yt1 , Yt2 ∈ S p,−K ([0, ∞); R1 ) and S p,−K ([0, ∞); R1 ) ⊂ S 2,−K ([0, ∞); R1 ), 2 E[e−KT |Y¯T | ] ≤ Cp for all T ≥ 0. Here and in the following, Cp is a generic constant. Taking the limit of T, we have 0

2

E[e−K t |Y¯t | ] = 0. That is Y¯t = 0 From (2.8), for a.e. t,

a.s..

10


Z¯t = 0

a.s..

Existence. For each n ∈ N, we define a sequence of BDSDEs as follows, Z n Z n Z n hgs (Ysn , Zsn ), d† Bs i − hZsn , dWs i. (2.9) fs (Ysn , Zsn )ds + Ytn = t

t

t

Also let (Ytn , Ztn )t≥n = (0, 0), and according to [25] (or Theorem 2.1), Eq.(2.9) has a unique solution (Y·n , Z·n ) ∈ S 2,−K ([0, ∞); Rn )∩M 2,−K ([0, ∞); Rn )× M 2,−K ([0, ∞); Rn×d ). Here we want to prove under conditions of Theorem 2.3 that Y·n ∈ S p,−K ([0, ∞); Rn ). We first prove Lemma 2.4 below. Lemma 2.4 Let (Ytn )t≥0 be the solution of Eq.(2.9), then under the conditions of Theorem 2.3, Y·n ∈ S p,−K ([0, ∞); R1 ). Proof. Apply Itˆ o’s formula to e−Kr |Yrn |p , we have d e−Kr |Yrn |p p = −Ke−Kr |Yrn |p + e−Kr |Yrn |p−2 − 2Yrn fr (Yrn , Zrn )dr 2 −|gr (Yrn , Zrn )|2 dr + |Zrn |2 dr − 2Yrn hgr (Yrn , Zrn ), d† Br i p(p − 2) −Kr n p−4 +2Yrn hZrn , dWr i + e |Yr | − 4|Yrn |2 |gr (Yrn , Zrn )|2 dr 8 +4|Yrn |2 |Zrn |2 dr . Then e

−Ks

|Ysn |p

p(p − 1) + 2

Z

n

e

−Kr

|Yrn |p−2 |Zrn |2 dr

Z −K

s

n

Z

e−Kr |Yrn |p−2 Yrn fr (Yrn , Zrn )dr

=p s

Z

n

e−Kr |Yrn |p−2 Yrn hgr (Yrn , Zrn ), d† Br i s Z p(p − 1) n −Kr n p−2 e |Yr | |gr (Yrn , Zrn )|2 dr + 2 s Z n −p e−Kr |Yrn |p−2 Yrn hZrn , dWr i s Z n ≤p e−Kr |Yrn |p−2 Yrn fr (Yrn , Zrn ) − fr (0, Zrn ) dr s Z n +p e−Kr |Yrn |p−2 Yrn fr (0, Zrn ) − fr (0, 0) dr Zs n +p e−Kr |Yrn |p−2 Yrn fr (0, 0)dr +p

s

s

n

e−Kr |Yrn |p dr


Z

11

n

e−Kr |Yrn |p−2 Yrn hgr (Yrn , Zrn ), d† Br i Z n e−Kr |Yrn |p−2 |gr (Yrn , Zrn ) − gr (0, 0)|2 dr +p(p − 1) s Z n Z n −Kr n p−2 2 +p(p − 1) e |Yr | |gr (0, 0)| dr − p e−Kr |Yrn |p−2 Yrn hZrn , dWr i s s Z n Z n ≤ −pµ e−Kr |Yrn |p dr + p e−Kr |Yrn |p−2 C|Yrn ||Zrn |dr s Z ns −Kr n p−2 e |Yr | |fr (0, 0)||Yrn |dr +p s Z n +p e−Kr |Yrn |p−2 Yrn hgr (Yrn , Zrn ), d† Br i s Z n +p(p − 1) e−Kr |Yrn |p−2 (2C 2 |Yrn |2 + 2α2 |Zrn |2 )dr s Z n Z n −Kr n p−2 2 +p(p − 1) e |Yr | |gr (0, 0)| dr − p e−Kr |Yrn |p−2 Yrn hZrn , dWr i s s Z Z n 1 p n −Kr n p−2 e |Yr | (2C 2 |Yrn |2 + |Zrn |2 )dr ≤ −pµ e−Kr |Yrn |p dr + 2 2 s Zs p n −Kr n p−2 3p 2ε + e |Yr | ( |fr (0, 0)|2 + |Yrn |2 )dr 2 s 2ε 3p Z n +p e−Kr |Yrn |p−2 Yrn hgr (Yrn , Zrn ), d† Br i s Z n +2p(p − 1) e−Kr |Yrn |p−2 (C 2 |Yrn |2 + α2 |Zrn |2 )dr s Z n Z n +p(p − 1) e−Kr |Yrn |p−2 |gr (0, 0)|2 dr − p e−Kr |Yrn |p−2 Yrn hZrn , dWr i s Z Z n s Z n p n −Kr n p−2 n 2 −Kr n p 2 −Kr n p e |Yr | |Zr | dr ≤ −pµ e |Yr | dr + pC e |Yr | dr + 4 s s s Z Z 3p2 n −Kr n p−2 ε n −Kr n p + e |Yr | |fr (0, 0)|2 dr + e |Yr | dr 4ε s 3 s Z n +p e−Kr |Yrn |p−2 Yrn hgr (Yrn , Zrn ), d† Br i s Z n Z n 2 −Kr n p 2 +2p(p − 1)C e |Yr | dr + 2p(p − 1)α e−Kr |Yrn |p−2 |Zrn |2 dr s s Z n Z n −Kr n p−2 2 +p(p − 1) e |Yr | |gr (0, 0)| dr − p e−Kr |Yrn |p−2 Yrn hZrn , dWr i. +p

s

s

s

Here ε is an arbitrary positive real number. Just as above, using the Hölder inequality and Young inequality, we have

12


p 10 19 e−Ks |Ysn |p + ( p − ) 4 9 9

Z

n

e−Kr |Yrn |p−2 |Zrn |2 dr Z n 2 e−Kr |Yrn |p dr +(pµ − K − p(2p − 1)C − ε) s Z n Z n −Kr p ≤ Cp e |fr (0, 0)| dr + Cp e−Kr |gr (0, 0)|p dr s s Z n +p e−Kr |Yrn |p−2 Yrn hgr (Yrn , Zrn ), d† Br i Zs n −p e−Kr |Yrn |p−2 Yrn hZrn , dWr i. s

(2.10)

s

Due to the arbitrariness of ε and (H.3), we can fix ε > 0 s.t. pµ − K − p(2p − 1)C 2 − ε > 0. Note here Cp depends on ε, but for fixed ε, Cp is a fixed number. Taking expectation on (2.10), we have Z ∞ Z ∞ E[ e−Kr |Yrn |p−2 |Zrn |2 dr] + E[ e−Kr |Yrn |p dr] (2.11) 0 0 Z ∞ Z ∞ ≤ Cp E[ e−Kr |fr (0, 0)|p dr] + Cp E[ e−Kr |gr (0, 0)|p dr] < ∞. 0

0

By B-D-G inequality, we get from (2.10) that E[sup e−Kt |Ytn |p ] t≥0

Z ∞ ≤ Cp E[ e−Kr (|fr (0, 0)|p + |gr (0, 0)|p )dr] 0 s Z ∞

e−2Kr |Yrn |2p−2 |gr (Yrn , Zrn )|2 dr]

+pE[ 0

sZ +pE[ Z ≤ Cp E[

∞

e−2Kr |Yrn |2p−2 |Zrn |2 dr]

0 ∞

e−Kr (|fr (0, 0)|p + |gr (0, 0)|p )dr] 0 Z ∞ 1 + E[sup e−Kt |Ytn |p ] + Cp E[ e−Kr |Yrn |p−2 |gr (Yrn , Zrn )|2 dr] 3 t≥0 0 Z ∞ 1 e−Kr |Yrn |p−2 |Zrn |2 dr]. (2.12) + E[sup e−Kt |Ytn |p ] + Cp E[ 3 t≥0 0

That is 1 E[sup e−Kt |Ytn |p ] 3 t≥0


Z ≤ Cp E[

13

∞

e−Kr (|fr (0, 0)|p + |gr (0, 0)|p )dr] Z Z ∞ −Kr n p−2 n 2 e |Yr | |Zr | dr] + Cp E[ +Cp E[ 0

∞

e−Kr |Yrn |p dr]. (2.13)

0

0

By (H.1) and (2.11), Y·n ∈ S p,−K ([0, ∞); R1 ). Lemma 2.4 is proved.

Remark 2.5 The proof of Lemma 2.4 also works with p replaced by 2. Note that if f· (0, 0) ∈ M p,−K ([0, ∞); R1 ), then by H¨ older inequality Z ∞ E[ e−Ks |fs (0, 0)|2 ds] 0 Z ∞ Z ∞ p−2 2 2 e− 2 Ks ds) p ( ≤ E[( e−Ks |fs (0, 0)|p ds) p ] < ∞, 0

0

therefore f· (0, 0) ∈ M 2,−K ([0, ∞); R1 ). By the same method we have g· (0, 0) ∈ M 2,−K ([0, ∞); R1 ) as well. So it can be easily seen in (2.11) with p replaced by 2 or follows from [25] that (Y·n , Z·n ) ∈ M 2,−K ([0, ∞); R1 ) × M 2,−K ([0, ∞); Rd ). Then back to the proof of Theorem 2.3. We want to show that (Y·n , Z·n ) is a Cauchy sequence in the space of S p,−K ([0, ∞); R1 ) ∩ M 2,−K ([0, ∞); R1 ) × M 2,−K ([0, ∞); Rd ) with the norm Z ∞ Z ∞ 1 2 (E[sup e−Kt | · |p ]) p + E[ e−Kr | · |2 dr] + E[ e−Kr | · |2 dr] 2 , t≥0

0

0

as in Pardoux [22]. Firstly we show that, for m, n ∈ N and m ≥ n, lim E[sup e−Kt |Ytm − Ytn |p ] = 0.

n,m→∞

t≥0

Define Y¯tm,n = Ytm − Ytn , Z¯tm,n = Ztm − Ztn . (i) When n ≤ t ≤ m, Z m Z m,n m m m ¯ Yt = Yt = fs (Ys , Zs )ds + t

t

m

hgs (Ysm , Zsm ), d† Bs i

Z

m

−

By (2.10), we have Z m Z m −Kr m p−2 m 2 E[ e |Yr | |Zr | dr] + E[ e−Kr |Yrm |p dr] n n Z m ≤ Cp E[ e−Kr (|fr (0, 0)|p + |gr (0, 0)|p )dr]. n

hZsm , dWs i.

t

(2.14)

14


As n, m → ∞, the right hand side converges to 0. Similar method as in (2.12) and (2.13), by (2.10) and (2.14), we have E[ sup e−Kt |Ytm |p ] n≤t≤m Z m

≤ Cp E[

(2.15)

e−Kr (|fr (0, 0)|p + |gr (0, 0)|p )dr] −→ 0, as n, m −→ ∞.

n

(ii) When 0 ≤ t ≤ n, Z Y¯tm,n = Ynm + t

n

f¯r dr +

Z t

n

h¯ gr , d† Br i −

Z

n

hZ¯rm,n , dWr i.

t

Here f¯r = fr (Yrm , Zrm ) − fr (Yrn , Zrn ), g¯r = gr (Yrm , Zrm ) − gr (Yrn , Zrn ). Applying Itˆ o’s formula to e−Kr |Y¯rm,n |p , Hölder inequality and Young inequality, we have for s ≤ n, Z p(p − 1) n −Kr ¯ m,n p−2 ¯ m,n 2 e |Yr | |Zr | dr e−Ks |Y¯sm,n |p + 2 s Z n −K e−Kr |Y¯rm,n |p dr s Z n −Kn m p =e |Yn | + p e−Kr |Y¯rm,n |p−2 Y¯rm,n f¯r dr s Z n +p e−Kr |Y¯rm,n |p−2 Y¯rm,n h¯ gr , d† Br i s Z p(p − 1) n −Kr ¯ m,n p−2 e |Yr | |¯ gr |2 dr + 2 s Z n −Kr ¯ m,n p−2 ¯ m,n ¯ m,n −p e |Yr | Yr hZr , dWr i s Z n ≤ e−Kn |Ynm |p + p e−Kr |Y¯rm,n |p−2 Y¯rm,n fr (Yrm , Zrm ) − fr (Yrn , Zrm ) dr s Z n +p e−Kr |Y¯rm,n |p−2 Y¯rm,n fr (Yrn , Zrm ) − fr (Yrn , Zrn ) dr Zs n +p e−Kr |Y¯rm,n |p−2 Y¯rm,n h¯ gr , d† Br i s Z p(p − 1) n −Kr ¯ m,n p−2 e |Yr | (C|Y¯rm,n | + α|Z¯rm,n |)2 dr + 2 s


Z

15

n

e−Kr |Y¯rm,n |p−2 Y¯rm,n hZ¯rm,n , dWr i Z n −Kn m p e−Kr |Y¯rm,n |p dr ≤e |Yn | − pµ s Z n +p e−Kr |Y¯rm,n |p−2 C|Y¯rm,n ||Z¯rm,n |dr s Z n +p e−Kr |Y¯rm,n |p−2 Y¯rm,n h¯ gr , d† Br i s Z p(p − 1) n −Kr ¯ m,n p−2 e |Yr | (2C 2 |Y¯rm,n |2 + 2α2 |Z¯rm,n |2 )dr + 2 s Z n −p e−Kr |Y¯rm,n |p−2 Y¯rm,n hZ¯rm,n , dWr i s Z n −Kn m p e−Kr |Y¯rm,n |p dr ≤e |Yn | − pµ s Z p n −Kr ¯ m,n p−2 1 + e |Yr | (2C 2 |Y¯rm,n |2 + |Z¯rm,n |2 )dr 2 s 2 Z n +p e−Kr |Y¯rm,n |p−2 Y¯rm,n h¯ gr , d† Br i s Z p(p − 1) n −Kr ¯ m,n p−2 + e |Yr | (2C 2 |Y¯rm,n |2 + 2α2 |Z¯rm,n |2 )dr 2 s Z n −p e−Kr |Y¯rm,n |p−2 Y¯rm,n hZ¯rm,n , dWr i. −p

s

s

That is Z n p (14p − 23) e−Kr |Y¯rm,n |p−2 |Z¯rm,n |2 dr 36 s Z n 2 2 −Kr ¯ m,n p +(pµ − K − p C ) e |Yr | dr s Z n ≤ e−Kn |Ynm |p + p e−Kr |Y¯rm,n |p−2 Y¯rm,n h¯ gr , d† Br i s Z n −p e−Kr |Y¯rm,n |p−2 Y¯rm,n hZ¯rm,n , dWr i. (2.16) e−Kr |Y¯sm,n |p +

s

So Z E[

n

e

−Kr

Z m,n p−2 ¯ m,n 2 ¯ |Yr | |Zr | dr] + E[

0

≤ Cp E[e−Kn |Ynm |p ].

n

e−Kr |Y¯rm,n |p dr]

0

(2.17)

From (i), the right hand side converges to 0, as n, m −→ ∞. By (2.16) and (2.17), similar method as in (2.12) and (2.13), we have

16


1 E[ sup e−Kt |Y¯tm,n |p ] 3 0≤t≤n Z n ≤ E[e−Kn |Ynm |p ] + Cp E[ e−Kr |Y¯rm,n |p−2 |Z¯rm,n |2 dr] 0 Z n +Cp E[ e−Kr |Y¯rm,n |p dr] 0

≤ Cp E[e−Kn |Ynm |p ] −→ 0,

as n, m −→ ∞.

From (i) (ii), we have for m, n ∈ N, lim E[sup e−Kt |Ytm − Ytn |p ] = 0.

n,m→∞

t≥0

Furthermore the above arguments also hold for p = 2 in (2.15) and (2.17), by Remark 2.5, as n, m −→ ∞ we have Z ∞ Z ∞ E[ e−Kr |Y¯rm,n |2 dr] + E[ e−Kr |Z¯rm,n |2 dr] −→ 0. 0

0

Therefore, (Y·n , Z·n ) is a Cauchy sequence in the Banach space S p,−K ([0, ∞); Rn ) ∩ M 2,−K ([0, ∞); Rn ) × M 2,−K ([0, ∞); Rn×d ). We take (Yt , Zt )t≥0 as the limit of (Ytn , Ztn )t≥0 in the space mentioned above. We will show that (Yt , Zt )t≥0 is the solution of Eq.(2.6). Since for t ≤ n, Z n Z n Z n Ytn = fs (Ysn , Zsn )ds + hgs (Ysn , Zsn ), d† Bs i − hZsn , dWs i, t

t

t

it turns out that for t ≤ n, Z n Z n 0 K0 K0 K − K0 s n e− 2 s fs (Ysn , Zsn )ds + e 2 Ys ds e− 2 t Ytn = 2 t t Z n K0 + e− 2 s hgs (Ysn , Zsn ), d† Bs i Zt n K0 − e− 2 s hZsn , dWs i.

(2.18)

t

We will show that Eq.(2.18) converges to Eq.(2.6) as m −→ ∞. We verify this term by term. For the first term, E[ |e−

K0 2

t

Ytn − e−

K0 2

t

Yt |2 ] ≤ E[sup e−Kt |Ytn − Yt |2 ] −→ 0. t≥0


For the second term, by H¨ older inequality, Z n Z ∞ 0 K0 − K2 s n n E[ | e fs (Ys , Zs )ds − e− 2 s fs (Ys , Zs )ds|2 ] t t Z n K0 ≤ 2E[ | e− 2 s fs (Ysn , Zsn ) − fs (Ys , Zs ) ds|2 ] t Z ∞ K0 e− 2 s fs (Ys , Zs )ds|2 ] +2E[ | Z n n Z n 0 ≤ 2E[ e−(K −K)s ds e−Ks |fs (Ysn , Zsn ) − fs (Ys , Zs )|2 ds] t t Z ∞ Z ∞ 0 +2E[ e−(K −K)s ds e−Ks |fs (Ys , Zs )|2 ds] n n Z ∞ Z ∞ −Ks n 2 e−Ks |Zsn − Zs |2 ds] e |Ys − Ys | ds] + Cp E[ ≤ Cp E[ 0 0 Z ∞ Z ∞ −Ks 2 +Cp E[ e |Ys | ds] + Cp E[ e−Ks |Zs |2 ds] n n Z ∞ +Cp E[ e−Ks |fs (0, 0)|2 ds] −→ 0. n

The third term can be seen similarly Z n 0 Z ∞ 0 K − K0 s n K − K0 s 2 e e 2 Ys ds|2 ] Ys ds − E[ | 2 2 t t Z ∞ 0 Z n 0 K − K0 s K − K0 s n e 2 (Ys − Ys )ds|2 ] + 2E[ | e 2 Ys ds|2 ] ≤ 2E[ | 2 2 n t Z n 02 Z n K −(K 0 −K)s ≤ 2E[ e ds e−Ks |Ysn − Ys |2 ds] 4 t t Z ∞ 02 Z ∞ K −(K 0 −K)s +2E[ e ds e−Ks |Ys |2 ds] 4 n n Z ∞ Z ∞ −Ks n 2 ≤ Cp E[ e |Ys − Ys | ds] + Cp E[ e−Ks |Ys |2 ds] −→ 0. 0

n

For the fourth term, noting K 0 > K and by Itô’s isometry, we have Z ∞ Z n K0 K0 e− 2 s hgs (Ys , Zs ), d† Bs i|2 ] e− 2 s hgs (Ysn , Zsn ), d† Bs i − E[ | t t Z n K0 ≤ 2E[ | e− 2 s hgs (Ysn , Zsn ) − gs (Ys , Zs ), d† Bs i|2 ] t Z ∞ K0 +2E[ | e− 2 s hgs (Ys , Zs ), d† Bs i|2 ] Z n n 0 = 2E[ e−K s |gs (Ysn , Zsn ) − gs (Ys , Zs )|2 ds] t

17

18


Z ∞ 0 +2E[ e−K s |gs (Ys , Zs )|2 ds] Z ∞ Z n∞ −Ks n 2 e−Ks |Zsn − Zs |2 ds] e |Ys − Ys | ds] + Cp E[ ≤ Cp E[ 0 0 Z ∞ Z ∞ −Ks 2 +Cp E[ e |Ys | ds] + Cp E[ e−Ks |Zs |2 ds] n n Z ∞ +Cp E[ e−Ks |gs (0, 0)|2 ds] −→ 0. n

The last term is Z n Z ∞ 0 K0 − K2 s n E[ | e hZs , dWs i − e− 2 s hZs , dWs i|2 ] t t Z Z ∞ n K0 K0 − 2 s n 2 e e− 2 s hZs , dWs i|2 ] ≤ 2E[ | hZs − Zs , dWs i| ] + 2E[ | n Z nt Z ∞ −K 0 s n 2 −K 0 s = 2E[ e |Zs − Zs | ds] + 2E[ e |Zs |2 ds] Zt n Z n∞ −Ks n 2 ≤ 2E[ e |Zs − Zs | ds] + 2E[ e−Ks |Zs |2 ds] −→ 0. t

n

That is to say (Yt , Zt )t≥0 is the solution of Eq.(2.6). The proof of Theorem 2.3 is completed.

3 The stationary solution of BDSDE In Section 2, we proved the existence and uniqueness of the solution of Eq.(1.1). In this section we will show that the solution is stationary under some conditions. Let θt = (θt1, θt2 ): Ω 1 × Ω 2 −→ Ω 1 × Ω 2 , t ≥ 0, be defined by θt (ω 1 , ω 2 ) = 1 B θt (ω 1 ), θt2 (ω 2 ) . Here θt1 and θt2 , t ≥ 0, are shifts on (Ω 1 , F∞ , P 1 ) and 2 2 W (Ω , F∞ , P ) respectively, s.t. for arbitrary s, t, u ≥ 0, (i) (P 1 × P 2 ) · (θt1 , θt2 )−1 = P 1 × P 2 ; (ii) θ01 = I 1 , θ02 = I 2 , where I 1 and I 2 are the identity transformation on Ω 1 and Ω 2 respectively; 1 2 (iii) θs1 ◦ θt1 = θs+t , θs2 ◦ θt2 = θs+t ;

(iv) θt1 ◦ Bs = Bs+t − Bt , θt2 ◦ Ws = Ws+t − Wt ;


19

B B W W (v) (θt1 )−1 (Fs,u ) = Fs+t,u+t , (θt2 )−1 (Fs,u ) = Fs+t,u+t .

Also for arbitrary measurable φ : (Ω 1 × Ω 2 , F 1 × F 2 , P 1 × P 2 ) −→ R, we define θ ◦ φ(ω 1 , ω 2 ) = φ θ(ω 1 , ω 2 ) . Then we give the added condition which makes f· (y, z) and g· (y, z) stationary w.r.t. θ· , (H.4). θr ◦fs (y, z) = fs+r (y, z), θr ◦gs (y, z) = gs+r (y, z) for arbitrary r, s ≥ 0 and fixed y, z. We will prove Proposition 3.1 Under conditions (H.1), (H.2), (H.3), (H.4), Eq.(2.6) has a stationary solution (Yt , Zt )t≥0 , s.t. for fixed r ≥ 0, θr ◦ Yt = Yt+r , θr ◦ Zt = Zt+r a.s.. ˆs = BT 0 −s − BT 0 for arbitrary T 0 > 0 and −∞ ≤ s ≤ T 0 . Then Proof. Let B ˆs , we ˆ ˆ0 = 0. For fixed r ≥ 0, acting θr1 on B Bs is a Brownian motion with B have ˆs = θr1 ◦ (BT 0 −s − BT0 ) = BT 0 −s+r − BT 0 +r θr1 ◦ B ˆs−r − B ˆ−r . = (BT 0 −s+r − BT0 ) − (BT 0 +r − BT0 ) = B So for 0 ≤ t ≤ T ≤ T 0 and a process h· which satisfies the condition in (2.2), Z T Z T 0 −t † ˆs i θr ◦ hhs , d Bs i = −θr ◦ hhT 0 −s , dB T 0 −T

t

Z

0

T −t

ˆs−r i hθr ◦ hT 0 −s , dB

=− T 0 −T

Z

T 0 −t−r

ˆs i hθr ◦ hT 0 −s−r , dB

=− Z

T 0 −T −r T +r

=

hθr hs−r , d† Bs i.

(3.1)

t+r

As T 0 can be chosen arbitrarily, so we can get for arbitrary T ≥ 0, 0 ≤ t ≤ T , r ≥ 0, Z T Z T +r † θr ◦ hhs , d Bs i = θr hhs−r , d† Bs i. t

t+r

By (H.1) and (H.2), it is easy to see that g· (Y· , Z· ) satisfies the condition in (2.2), hence by (H.4) and (3.1)

20


Z θr ◦

T

hgs (Ys , Zs ), d† Bs i =

Z

t

T +r

hgs (θr Ys−r , θr Zs−r ), d† Bs i a.s.. (3.2)

t+r

We consider Eq.(2.7) which is equivalent to Eq.(2.6). Pass θr on both sides of Eq.(2.7) and by (3.2), θr ◦ Yt satisfies the following equation R T +r  θr ◦ Yt = θr ◦ YT + t+r fs (θr ◦ Ys−r , θr ◦ Zs−r )ds   R T +r   + t+r hgs (θr ◦ Ys−r , θr ◦ Zs−r ), d† Bs i R T +r  − t+r hθr ◦ Zs−r , dWs i,    K0 limT →∞ e− 2 (T +r) (θr ◦ YT ) = 0 a.s.. On the other hand, from Eq.(2.7), it follows that  R T +r R T +r †   Yt+r = YTR+r + t+r fs (Ys , Zs )ds + t+r hgs (Ys , Zs ), d Bs i T +r − t+r hZs , dWs i,   K0 limT →∞ e− 2 (T +r) YT +r = 0 a.s..

(3.3)

(3.4)

Let Yˆ· = θr ◦Y·−r , Zˆ· = θr ◦Z·−r . Compare (3.3) with (3.4), by the uniqueness of solution of Eq.(2.7), Yˆ· = Y· , Zˆ· = Z·

a.s..

We proved the desired result.

W , P 2) Before we give a type of FBDSDE, we first consider the SDE on (Ω 2 , F∞ below. For 0 ≤ t ≤ s, Z s Z s Xst,x = x + b(Xut,x )du + σ(Xut,x )dWu . (3.5) t

t

The coefficients in Eq(3.5) satisfy (H.5) . b(·) : Rd −→ Rd , σ(·) : Rd −→ Rd×d are globally Lipschitz continuous with Lipschitz constant L which satisfies K − pL − p(p−1) L2 > 0 for p, 2 K in conditions of considered BDSDE. We regulate that Xst,x is defined on [0, ∞) by defining (Xst,x )s 0, 0 < α < (x1 , y1 , z1 ), (x2 , y2 , z2 ) ∈ Rd × R1 × Rd ,

1 3,

s.t. for any

|f (x1 , y1 , z1 ) − f (x2 , y2 , z2 )| ≤ C1 |x1 − x2 | + C0 |y1 − y2 | + C|z1 − z2 |, |g(x1 , y1 , z1 ) − g(x2 , y2 , z2 )| ≤ C1 |x1 − x2 | + C|y1 − y2 | + α|z1 − z2 |. (H.3)0 . There exists a constant µ > 0, s.t. for p > d + 2, K < K 0 < 2K, 2µ − p2 K 0 − p(2p − 1)C 2 > 0 and for any (ω 1 , ω 2 , t) ∈ Ω 1 × Ω 2 × [0, ∞), y 1 , y 2 ∈ R1 , z ∈ Rd , (y1 − y2 )(ft (y1 , z) − ft (y2 , z)) ≤ −µ|y1 − y2 |2 . Now we give FBDSDE on infinite time interval below. For s ≥ 0, Z ∞ Z ∞ 0 0 0 K − K 0 r t,x − K2 s t,x − K2 r t,x t,x t,x e Ys = e f (Xr , Yr , Zr )dr + e 2 Yr dr 2 s s Z ∞ Z ∞ K0 K0 e− 2 r hZrt,x , dWr i, + e− 2 r hg(Xrt,x , Yrt,x , Zrt,x ), d† Br i − s

s

which is equivalent to for 0 ≤ s ≤ T ,  RT t,x t,x t,x t,x t,x   Ys = YT R + s f (Xu , Yu , Zu )du R T T + s hg(Xut,x , Yut,x , Zut,x ), d† Bu i − s hZut,x , dWu i,  0 K  limT →∞ e− 2 T YTt,x = 0 a.s..

(3.7)

We have the following proposition. Proposition 3.2 Under conditions (H.1)0 , (H.2)0 , (H.3)0 , (H.5), Eq.(3.7) has a unique solution (Y·t,x , Z·t,x ) ∈ S p,−K ([0, ∞); R1 ) ∩ M 2,−K ([0, ∞); R1 ) × M 2,−K ([0, ∞); Rd ). Proof. Let

22


fˆs (y, z) = f (Xst,x , y, z),

gˆs (y, z) = g(Xst,x , y, z).

We just need to verify that fˆ, gˆ satisfy conditions (H.1), (H.2), (H.3) in Theorem 1.1. It is obvious that fˆ, gˆ satisfy (H.2), (H.3), and we show that fˆ, gˆ satisfy (H.1) as well, i.e. Z ∞ Z ∞ −Ks ˆ p e−Ks |ˆ gs (0, 0)|p ds] < ∞. e |fs (0, 0)| ds] < ∞, E[ E[ 0

0

Notice Z ∞ e−Ks |fˆs (0, 0)|p ds] E[ Z0 ∞ = E[ e−Ks |f (Xst,x , 0, 0)|p ds] 0 Z ∞ Z ∞ ≤ Cp E[ e−Ks |f (Xst,x , 0, 0) − f (0, 0, 0)|p ds] + Cp E[ e−Ks |f (0, 0, 0)|p ds] 0 0 Z ∞ Z ∞ −Ks p t,x p ≤ Cp E[ e C1 |Xs | ds] + Cp E[ e−Ks |f (0, 0, 0)|p ds]. 0

0

Now applying Itˆ o’s formula to e−Kr |Xrt,x |p , we have d e−Kr |Xrt,x |p p = −Ke−Kr |Xrt,x |p dr + e−Kr |Xrt,x |p−2 2hXrt,x , b(Xst,x )idr 2 +kσ(Xrt,x )k2 dr + 2hXst,x , σ(Xrt,x )dWr i p(p − 2) −Kr t,x p−4 e |Xr | hσ(Xrt,x )σ ∗ (Xrt,x )Xrt,x , Xrt,x idr. + 2 Then by Lipschitz condition and Young inequality, for 0 ≤ t ≤ s, Z s e−Ks |Xst,x |p + Ke−Kr |Xrt,x |p dr t Z s ≤ e−Kt |x|p + pe−Kr |Xrt,x |p−2 hXrt,x , b(Xst,x )idr t Z s p(p − 1) −Kr t,x p−2 e |Xr | kσ(Xrt,x )k2 dr + 2 t Z s + pe−Kr |Xrt,x |p−2 hXrt,x , σ(Xrt,x )dWr i t Z s ≤ e−Kt |x|p + pe−Kr |Xrt,x |p−2 |Xrt,x ||b(Xst,x )|dr t

Stationary Solution of SPDEs and BDSDEs on Infinite Horizon s

Z +

Zt s

23

p(p − 1) −Kr t,x p−2 e |Xr | (L|Xrt,x | + kσ(0)k)2 dr 2

pe−Kr |Xrt,x |p−2 hXrt,x , σ(Xrt,x )dWr i Z s −Kt p pe−Kr |Xrt,x |p−1 (L|Xrt,x | + |b(0)|)dr ≤e |x| + t Z s p(p − 1) −Kr t,x p−2 + e |Xr | (L|Xrt,x | + kσ(0)k)2 dr 2 Zt s + pe−Kr |Xrt,x |p−2 hXrt,x , σ(Xrt,x )dWr i t Z s Z s ≤ e−Kt |x|p + pL e−Kr |Xrt,x |p dr + p e−Kr |Xrt,x |p−1 |b(0)|dr t t Z s 2ε p(p − 1) t,x p−2 |Xr | (1 + )L2 |Xrt,x |2 + Cp kσ(0)k2 dr + e−Kr 2 2 3p(p − 1)L Zt s + pe−Kr |Xrt,x |p−2 hXrt,x , σ(Xrt,x )dWr i t Z s Z ε s −Kr t,x p ≤ e−Kt |x|p + pL e−Kr |Xrt,x |p dr + e |Xr | dr 3 t t Z s Z Z p(p − 1) 2 s −Kr t,x p ε s −Kr t,x p +Cp e−Kr |b(0)|p dr + L e |Xr | dr + e |Xr | dr 2 3 t t t Z s Z s +Cp e−Kr |Xrt,x |p−2 kσ(0)k2 dr + p e−Kr |Xrt,x |p−2 hXrt,x , σ(Xrt,x )dWr i t t Z s p(p − 1) 2 L + ε) e−Kr |Xrt,x |p dr ≤ e−Kt |x|p + (pL + 2 t Z s Z s −Kr p p +Cp e (|b(0)| + kσ(0)k )dr + pe−Kr |Xrt,x |p−2 hXrt,x , σ(Xrt,x )dWr i. +

t

t

t

Therefore, Z s p(p − 1) 2 L − ε) e−Kr |Xrt,x |p dr e−Ks |Xst,x |p + (K − pL − 2 t Z s ≤ e−Kt |x|p + Cp e−Kr (|b(0)|p + kσ(0)kp )dr t Z s +p e−Kr |Xrt,x |p−2 hXrt,x , σ(Xrt,x )dWr i. t

Due to the arbitrariness of ε, we have Z s Z −Kr t,x p −Kt p E[ e |Xr | dr] ≤ e |x| + Cp E[ t

s

e−Kr (|b(0)|p + kσ(0)kp )dr] < ∞.

t

Taking the limit of s and noting that (Xst,x )s 0, we will show that under conditions (H.1)0 , (H.2)0 , (H.3)0 , (H.4)0 , (H.5) and for arbitrary t ∈ [0, T ], x belonging to an arbitrary −t,x bounded set in Rd , the solution YTT−t (ω 1 ) of Eq.(4.1) is the viscosity solution ([6]) of the following SPDE Z t v(t, x) = v(t, v0 )(x) = v0 (x) + [Lv(s, x) + f x, v(s, x), σ ∗ (x)Dv(s, x) ]ds 0 Z t ˆs i, − hg x, v(s, x) , dB t ≥ 0. (4.2) 0

ˆs } is a two-sided Brownian motion. Denote by U the Banach space Here {B d 1 of C 0;δ (0 < δ < 1 − d+2 p ) functions f : R −→ R for which the norm |f | = sup x∈Rd

|f (x)| |f (x) − f (y)| + sup 1 + |x| x6=y, x,y∈Rd |x − y|δ

is finite and consider its Borel σ-field B. If we assume that Eq. (4.2) has a unique stochastic viscosity solution, then it is not difficult to see that (4.2) defines a measurable random dynamical system v : [0, T ] × U × Ω 1 −→ U for any T > 0. This can be seen from (2.5), the Doss-Sussmann transformation, the cocycle property of the stochastic flow λ and some standard arguments. The uniqueness of the stochastic viscosity solution was studied in Buckdahn and Ma [7]. Note here we don’t need the continuity of random dynamical system. In fact the continuity of the map v(t, ·, ω 1 ) : U −→ U a.s. remains open for Eq. (4.2). ˆT −s − B ˆT For arbitrary T > 0, define u(t, x) = v(T − t, x) and Bs = B for 0 ≤ s < ∞. Then {Bs } is a standard Brownian motion and g x, u(·, x) satisfies condition in (2.2). Hence by (2.2) and integral transformation in Eq.(4.2), we can see that u satisfies the backward SPDE of the following form Z T u(t, x) = u(T, x) + [Lu(s, x) + f x, u(s, x), σ ∗ (x)Du(s, x) ]ds t

Z + t

T

hg x, u(s, x) , d† Bs i,

0 ≤ t ≤ T.

(4.3)

26


We first show that under conditions (H.1)0 , (H.2)0 , (H.3)0 , (H.4)0 , (H.5) and for t ∈ [0, T ], x belonging to an arbitrary bounded set in Rd , the solution Ytt,x (ω 1 ) of Eq.(4.1) is the viscosity solution of Eq.(4.3), then by the relation−t,x ship between u and v, we obtain that YTT−t (ω 1 ) is the viscosity solution of Eq.(4.2). We need Lemma 4.1 Under condition (H.5), let (Xst,x )s≥0 be the solution of Eq.(3.5), then for arbitrary T > 0, t, t0 ∈ [0, T ] and x, x0 belonging to an arbitrary bounded set in Rd , Z ∞ p 0 0 E[ e−Kr |Xrt ,x − Xrt,x |p dr] ≤ Cp (|x0 − x|p + |t0 − t| 2 ) a.s.. 0

Proof. Without loss of generality, we assume t0 ≥ t ≥ 0. For r ≥ 0, let ¯ r = Xrt0 ,x0 − Xrt,x , X ¯br = br (X t0 ,x0 ) − br (X t,x ), σ ¯r =

r 0 0 σr (Xrt ,x )

r

− σr (Xrt,x ).

Then for r ≥ t0 ,

¯ r = ¯br dr + σ dX ¯r dWr , t,x 0 ¯ 0 Xt = x − Xt0 .

¯ r |p , we have Apply Itˆ o’s formula to e−Kr |X Z s ¯ s |p + ¯ r |p dr e−Ks |X Ke−Kr |X t0 Z s t,x p −Kt0 0 ¯ r |p−2 hX ¯ r , ¯br idr =e |x − Xt0 | + pe−Kr |X t0 Z s Z s p(p − 1) −Kr ¯ p−4 ∗ ¯ ¯ ¯ r |p−2 hX ¯r , σ e |Xr | h¯ σr σ ¯r Xr , Xr idr + pe−Kr |X ¯r dWr i + 2 t0 t0 Z s 0 p ¯ r |p−2 hX ¯ r , ¯br idr ≤ e−Kt |x0 − Xtt,x pe−Kr |X 0 | + 0 t Z s Z s p(p − 1) −Kr ¯ p−2 ¯ r |p−2 hX ¯r , σ e |Xr | k¯ σr k2 dr + pe−Kr |X ¯r dWr i + 2 t0 t0 Z s Z s 0 p(p − 1) 2 −Kr ¯ p p ¯ r |p dr + L e |Xr | dr ≤ e−Kt |x0 − Xtt,x pLe−Kr |X 0 | + 2 t0 t0 Z s ¯ r |p−2 hX ¯r , σ + pe−Kr |X ¯r dWr i t0 Z s p(p − 1) 2 t,x p 0 ¯ r |p dr ≤ |x − Xt0 | + (pL + L ) e−Kr |X 2 t0 Z s ¯ r |p−2 hX ¯r , σ +p e−Kr |X ¯r dWr i. t0


27

By (3.8), then E[e

−Ks

Z s p(p − 1) 2 p ¯ ¯ r |p dr] L )E[ e−Kr |X |Xs | ] + (K − pL − 2 0 t

p ≤ E[ |x0 − Xtt,x 0 | ].

Now for arbitrary T > 0, we consider t, t0 ∈ [0, T ] and x, x0 belonging to an arbitrary bounded set in Rd . Referring to P184 in [17], we have p

p 0 p 0 2 E[ |x0 − Xtt,x 0 | ] ≤ Cp (|x − x| + |t − t| ).

Noting that K − pL − p(p−1) L2 > 0, we have 2 Z ∞ ¯ r |p dr] ≤ Cp (|x0 − x|p + |t0 − t| p2 ). e−Kr |X E[

(4.4)

t0

Also referring to the same formula in [17], we have p

p

E[ |x0 − Xrt,x |p ] ≤ Cp (|x0 − x|p + |t0 − t| 2 + |t0 − r| 2 ). Therefore t0

Z E[

¯ r |p dr] e−Kr |X

0

Z t Z = E[ e−Kr |x0 − x|p dr] + E[ 0

t0

e−Kr |x0 − Xrt,x |p dr]

t

≤ Cp |x0 − x|p +

Z

t0

e−Kr E[ |x0 − Xrt,x |p ]dr

t 0

p

Z

≤ Cp |x − x| +

t0

p

p

e−Kr Cp (|x0 − x|p + |t0 − t| 2 + |t0 − r| 2 )dr

t p

p

≤ Cp |x − x| + Cp |x0 − x|p + Cp |t0 − t| 2 + Cp |t0 − t| 2 +1 . 0

p

(4.5)

By (4.4) and (4.5), together with bounded |t0 − t|, it follows that Z ∞ ¯ r |p dr] ≤ Cp (|x0 − x|p + |t0 − t| p2 ) a.s., E[ e−Kr |X 0 0

for t, t ∈ [0, T ] and x, x0 belonging to an arbitrary bounded set in Rd . Lemma 4.1 is proved. Proposition 4.2 Under conditions (H.1)0 , (H.2)0 , (H.3)0 , (H.5), let (Yst,x )s≥0 be the solution of Eq.(4.1), then for arbitrary T > 0, t ∈ [0, T ] and x belonging to an arbitrary bounded set in Rd , (t, x) −→ Ytt,x is a.s. continuous. In particular, Ytt,x is a.s. δ-H¨ older continuous w.r.t. x for any 0 < δ < 1 − d+2 p t,· and of linear growth, so Yt ∈ U .

28


0 0 0 0 Proof. For t, t0 r ≥ 0, let Y¯r = Yrt ,x − Yrt,x , Z¯r = Zrt ,x − Zrt,x , 0 0 0 0 0 0 f¯r = f (Xrt ,x , Yrt ,x , Zrt ,x ) − f (Xrt,x , Yrt,x , Zrt,x ), 0

0

0

0

g¯r = g(Xrt ,x , Yrt ,x ) − g(Xrt,x , Yrt,x ). Then (

dY¯r = −f¯r dr − h¯ gr , d† Br i + hZ¯r , dWr i, K0 limT →∞ e− 2 T Y¯T = 0 a.s..

Apply Itˆ o’s formula to e−

pK 0 2

r

|Y¯r |p , for 0 ≤ s ≤ T , we have

Z Z p(p − 1) T − pK 0 r ¯ p−2 ¯ 2 pK 0 T − pK 0 r ¯ p e 2 |Yr | |Zr | dr − e 2 |Yr | dr 2 2 s s Z T pK 0 pK 0 e− 2 r |Y¯r |p−2 Y¯r f¯r dr = e− 2 T |Y¯T |p + p e−

pK 0 2

s

Z

T

|Y¯s |p +

s

+p

e

0 − pK 2

r

s T

Z

e−

−p

pK 0 2

p(p − 1) |Y¯r |p−2 Y¯r h¯ gr , d† Br i + 2

Z

T

e−

pK 0 2

r

|Y¯r |p−2 |¯ gr |2 dr

s

r

|Y¯r |p−2 Y¯r hZ¯r , dWr i

pK 0 2

r

0 0 0 0 0 0 0 0 0 0 |Y¯r |p−2 Y¯r f (Xrt ,x , Yrt ,x , Zrt ,x ) − f (Xrt ,x , Yrt,x , Zrt ,x ) dr

pK 0 2

r

0 0 0 0 |Y¯r |p−2 Y¯r f (Xrt ,x , Yrt,x , Zrt ,x ) − f (Xrt,x , Yrt,x , Zrt,x ) dr

s

≤ e−

pK 0 2

T

Z

T

|Y¯T |p e−

+p s T

Z

e−

+p s

Z p(p − 1) T − pK 0 r ¯ p−2 ¯ r | + C|Y¯r |)2 dr + e 2 |Yr | (C1 |X 2 s Z T Z T pK 0 pK 0 +p e− 2 r |Y¯r |p−2 Y¯r h¯ gr , d† Br i − p e− 2 r |Y¯r |p−2 Y¯r hZ¯r , dWr i s

≤e

0 − pK 2

s T

|Y¯T |p − pµ

T

Z

e

0 − pK 2

r

|Y¯r |p dr

s

p + 2

T

Z

e−

pK 0 2

r

s

Z

C2 ¯ 2 ¯ 2 |Y¯r |p−2 (2C 2 |Y¯r |2 + 12 |X r | + |Zr | )dr C

T

e−

+p(p − 1)

pK 0 2

r

¯ r |2 + C 2 |Y¯r |2 )dr |Y¯r |p−2 (C12 |X

s

Z +p

T

e−

pK 0 2

r

|Y¯r |p−2 Y¯r h¯ gr , d† Br i − p

s

Therefore, for 0 ≤ s ≤ T ,

Z s

T

e−

pK 0 2

r

|Y¯r |p−2 Y¯r hZ¯r , dWr i.


29

Z p(p − 2) T − pK 0 r ¯ p−2 ¯ 2 e 2 |Yr | |Zr | dr 2 s Z T pK 0 pK 0 2 2 +(pµ − −p C ) e− 2 r |Y¯r |p dr 2 s Z T 2 pK 0 pK 0 pC ¯ r |2 dr ≤ e− 2 T |Y¯T |p + ( 12 + p(p − 1)C12 ) e− 2 r |Y¯r |p−2 |X 2C s Z T Z T pK 0 pK 0 e− 2 r |Y¯r |p−2 Y¯r h¯ pe− 2 r |Y¯r |p−2 Y¯r hZ¯r , dWr i. +p gr , d† Br i − p e−

pK 0 2

s

|Y¯s |p +

s

≤ e−

pK 0

s

2

T

Z

T

|Y¯T |p + ε

T

Z

e−

pK 0 2

r

|Y¯r |p dr + Cp

s

e

+p

0 − pK 2

r

T

Z

e−

pK 0 2

r

¯ r |p dr |X

s

|Y¯r |p−2 Y¯r h¯ gr , d† Br i − p

Z

s

T

e−

pK 0 2

r


s

Equivalently, for 0 ≤ s ≤ T , Z pK 0 p(p − 2) T − pK 0 r ¯ p−2 ¯ 2 e 2 |Yr | |Zr | dr e− 2 s |Y¯s |p + 2 s Z T pK 0 pK 0 +(pµ − − p2 C 2 − ε) e− 2 r |Y¯r |p dr 2 s Z T Z T 0 pK 0 pK 0 T p − pK − ¯ r |p dr + p gr , d† Br i ≤ e 2 |Y¯T | + Cp e 2 r |X e− 2 r |Y¯r |p−2 Y¯r h¯ s

Z −p

T

e

0 − pK 2

r

s


(4.6)

s

Then, for 0 ≤ s ≤ T , Z T pK 0 p(p − 2) p ¯ E[ e− 2 r |Y¯r |p−2 |Z¯r |2 dr] |Ys | ] + E[e 2 s Z T pK 0 pK 0 +(pµ − − p2 C 2 − ε)E[ e− 2 r |Y¯r |p dr] 2 s Z T pK 0 pK 0 ¯ r |p dr]. ≤ E[e− 2 T |Y¯T |p ] + Cp E[ e− 2 r |X 0

− pK 2 s

(4.7)

s

Since (Y·t,x ) ∈ S p,−K ([0, ∞); R1 ), for arbitrary T ≥ 0, E[e−

pK 0 2

T

|Y¯T |p ] ≤ E[sup e−Ks |Y¯s |p ] < ∞. s≥0

Therefore we can apply Lebesgue’s dominated convergence theorem by taking the limit of T to have lim E[e

T →∞

pK 0 2

T

|Y¯T |p ] = E[( lim e− T →∞

K0 2

T

|Y¯T |)p ] = 0.

(4.8)

30


So we take the limit of T in (4.7) and by Lemma 4.1 and monotone convergence theorem to have Z ∞ pK 0 pK 0 p(p − 2) e− 2 r |Y¯r |p−2 |Z¯r |2 dr] E[e− 2 s |Y¯s |p ] + E[ 2 Z ∞0 pK 0 pK 0 − pC 2 − ε)E[ e− 2 r |Y¯r |p dr] +(pµ − 2 0 Z ∞ ¯ r |p dr] e−Kr |X ≤ Cp E[ 0 p

≤ Cp (|x0 − x|p + |t0 − t| 2 ). By the arbitrariness of ε, it follows that Z ∞ Z 0 − pK r ¯ p−2 ¯ 2 2 e E[ |Yr | |Zr | dr] + E[ 0

∞

e−

pK 0 2

r

|Y¯r |p dr]

0 p

≤ Cp (|x0 − x|p + |t0 − t| 2 ).

(4.9)

From (4.6), E[ sup e−

pK 0 2

s

|Y¯s |p ]

0≤s≤T

≤ E[e−

pK 0 2

T

Z |Y¯T |p ] + Cp E[

∞

e−

pK 0 2

r

¯ r |p dr] |X

0

sZ

∞

e−pK 0 r |Y¯r |2p−2 |¯ gr |2 dr]

+pE[ 0

sZ +pE[

∞

e−pK 0 r |Y¯r |2p−2 |Z¯r |2 dr].

(4.10)

0

Noting (4.8), Taking the limit of T on both sides of (4.10) and using monotone convergence theorem we have E[sup e−

pK 0 2

s

|Y¯s |p ]

s≥0

Z ≤ Cp E[

∞

e−

pK 0 2

0

Z +Cp E[

∞

e−

r

0 ¯ r |p dr] + 1 E[sup e− pK2 s |Y¯s |p ] |X 3 s≥0

pK 0 2

r

pK 0 2

r

0

Z +Cp E[

∞

e−

pK 0 1 |Y¯r |p dr] + E[sup e− 2 s |Y¯s |p ] 3 s≥0

|Y¯r |p−2 |Z¯r |2 dr].

0

From this inequality, by Lemma 4.1 and (4.9), for arbitrary T > 0, t, t0 ∈ [0, T ] and x, x0 belonging to an arbitrary bounded set in Rd , we have

Stationary Solution of SPDEs and BDSDEs on Infinite Horizon p E[sup e−pKs |Y¯s |p ] ≤ Cp (|x0 − x|p + |t0 − t| 2 ).

31

(4.11)

s≥0

(·,·)

Noting p > d+2 in (4.11), by Kolmogorov Lemma (see [17]), we have Ys has a continuous modification for t ∈ [0, T ] and x belonging to an arbitrary bounded set in Rd under the norm sup e−Ks |Ys(·,·) |. s≥0

In particular, 0

0

lim e−Kt |Ytt ,x − Ytt,x | = 0.

t→t0 x→x0

Then we have a.s. 0

0

0

lim |e−Kt Ytt0 ,x − e−Kt Ytt,x |

t→t0 x→x0

0

0

0

0

0

0

0

≤ lim0 (|e−Kt Ytt0 ,x − e−Kt Ytt ,x | + |e−Kt Ytt ,x − e−Kt Ytt,x |) = 0. t→t x→x0

0

0

The convergence of the first term follows from the continuity of Yst ,x in s. That is to say e−Kt Ytt,x is a.s. continuous, therefore Ytt,x is continuous in t ∈ [0, T ] and x belonging to an arbitrary bounded set in Rd . Furthermore, from (4.11) by Kolmogorov Lemma, Ytt,x is a.s. δ-Hölder continuous w.r.t. x for any 0 < δ < 1 − d+2 p and of linear growth for x belonging to an arbitrary d bounded set in R . Proposition 4.2 is proved. With the above results, we claim that Theorem 4.3 Assume conditions (H.1)0 , (H.2)0 , (H.3)0 , (H.4)0 , (H.5) and Eq. (4.3) has a unique stochastic viscosity solution. Define u(t, x, ω 1 ) = Ytt,x (ω 1 ) for t ∈ [0, T ] and x ∈ Rd , where Ytt,x (ω 1 ) is the solution of Eq.(4.1). Then u(t, x, ω 1 ) is a stationary stochastic viscosity solution of Eq. (4.3) w.r.t. θ·1 defined in Section 3, i.e. there exists Ω 1,∗ ⊂ Ω 1 , s.t. P 1 (Ω 1,∗ ) = 1 and for all ω 1 ∈ Ω 1,∗ , t, r ≥ 0, x ∈ Rd , u(t + r, x, ω 1 ) = u(t, x, θr1 ω 1 ). Proof. By Proposition 4.2, u(·, ·, ω 1 ) is a.s. continuous for t ∈ [0, T ] and x belonging to an arbitrary bounded set in Rd and of linear growth in x. In particular, u(T, x) is continuous in x and of linear growth in x. So, referring to [6] (or Theorem 2.2) for the time reveral version, we see that u(t, x, ω 1 ) is a stochastic viscosity solution of Eq.(4.3). To prove the pefection version of the ¯ N ) the closed ball in Rd of radius N centred stationary solution, define B(0, S∞ ¯ N ) = Rd . First consider when x is in a at 0. It is obvious that N =1 B(0, t,· ¯ N ). Since Yt ∈ C(B(0, ¯ N )) and C(B(0, ¯ N )) is a separable space, ball B(0,

32


by [2] and (3.9), we know there exist a modification of Ytt,· , still denoted 1 by Ytt,· , and a full measure set Ω 1,N ⊂ Ω 1 such that for all Ω 1,N , r, T∞ω ∈ 1,N t+r,x t,x 1 1 1 1,∗ ¯ t ≥ 0, x ∈ B(0, N ), Yt+r (ω ) = Yt (θr ω ). Take Ω = N =1 Ω , then ¯ N ), but P 1 (Ω 1,∗ ) = 1. Now for any x ∈ Rd , there exists an N s.t. x ∈ B(0, 1 1,∗ 1 1,N for all ω ∈ Ω , it is obvious that ω ∈ Ω , so for any t, r ≥ 0, x ∈ Rd , u(t + r, x, ω 1 ) = u(t, x, θr1 ω 1 ). −t,x Note that v(t, x) = u(T − t, x), hence v(t, x)(ω 1 ) = YTT−t (ω 1 ) is the T −t,x solution of Eq.(4.2) for arbitrary T > 0. In fact YT −t (ω 1 ) does not depend −t,x on T . For example, if we take T 0 ≥ T , then we can show that YTT−t (ω 1 ) = 0 −t,x ˆT −s − B ˆT , 0 ≤ s < ∞, and YTT0 −t (˜ ω 1 ) when 0 ≤ t ≤ T , where ω 1 (s) = B ˆT 0 −s − B ˆT 0 , 0 ≤ s < ∞. Let θ· and θ˜· are the shifts of ω 1 (·) and ω ˜ 1 (s) = B ω ˜ 1 (·) respectively. Since by (3.9) we have −t,x YTT−t (ω 1 ) = θT1 −t Y00,x (ω 1 ) = Y00,x (θT1 −t ω 1 ) 0

−t,x YTT0 −t (˜ ω 1 ) = θ˜T1 0 −t Y00,x (˜ ω 1 ) = Y00,x (θ˜T1 0 −t ω ˜ 1 ),

˜ 1 . In fact we have we just need to show that θT1 −t ω 1 = θ˜T1 0 −t ω (θT1 −t ω 1 )(s) = ω 1 (T − t + s) − ω 1 (T − t) ˆT −(T −t+s) − B ˆT − B ˆT −(T −t) − B ˆT = B ˆt−s − B ˆt . =B The right side of the above formula does not depend on T , therefore ˆt−s − B ˆt . That is to say Y T −t,x (ω 1 ) does not ˜ 1 (s) = B θT1 −t ω 1 (s) = θ˜T1 0 −t ω T −t depend on the choice of T . ˆ B , P 1 ), we define θˆt1 = (θt1 )−1 , t ≥ 0. Note On probability space (Ω 1 , F∞ that {Bs } is a two-sided Brownian motion due to the two-sidedness of Brow1 ˆs }, so (θt1 )−1 = θ−t is well defined (see [1]). It is easy to see nian motion {B 1 ˆ ˆ that θt is a shift w.r.t. {Bt } satisfying (i) P 1 · (θˆt1 )−1 = P 1 ; (ii) θˆ01 = I 1 ; 1 (iii) θˆs1 ◦ θˆt1 = θˆs+t ;

ˆs = B ˆs+t − B ˆt ; (iv) θˆt1 ◦ B ˆ ˆ B B (v) (θˆt1 )−1 (Fs,u ) = Fs+r,u+r . −t,x Since v(t, x)(ω 1 ) = u(T − t, x)(ω 1 ) = YTT−t (ω 1 ) a.s.,


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θˆr1 v(t, x)(ω 1 ) 1 = θ−r u(T − t, x)(ω 1 ) = u(T − t − r, x)(ω 1 ) = v(t + r, x)(ω 1 ), for r ≥ 0 and T > t + r a.s.. In particular, let Y (ω 1 ) = v0 (ω 1 ) = YTT,· (ω 1 ) which can be regarded as a point in U . Then above formula implies that θˆt1 Y (ω 1 ) = Y (θˆt1 ω 1 ) = v(t, ω 1 ) = v(t, v0 (ω 1 ), ω 1 ) = v(t, Y (ω 1 ), ω 1 ), t ≥ 0, a.s.. Here as we have seen before, v is a measurable random dynamical system. −t,x That is to say v(t, ω 1 )(x) = Y (θˆt1 ω 1 )(x) = YTT−t (ω 1 ) is a stationary solution of the SPDE (4.2). Therefore we have the following theorem Theorem 4.4 Assume conditions (H.1)0 , (H.2)0 , (H.3)0 , (H.4)0 , (H.5) and Eq. (4.2) has a unique stochastic viscosity solution. Define v(t, x, ω 1 ) = −t,x YTT−t (ω 1 ) for t ∈ [0, T ] a.s. and x belonging to an arbitrary bounded set in Rd , where Ytt,x (ω 1 ) is the solution of Eq.(4.1). Then v(t, x, ω 1 ) is a stationary stochastic viscosity solution of Eq.(4.2) w.r.t. θˆ·1 . Acknowledgements We have discussed this result with many researchers. In particular, it is our great pleasure to thank Z. Brzezniak, Z. Dong, J. Duan, K. D. Elworthy, F. Gong, R. Hudson, Y. Liu, S. Mohammed, S. Peng, A. Truman, T. Zhang and Z. Zheng for useful conversations and comments. We acknowledge financial support from the Loughborough Development Fund and the EPSRC Grant GR/R69518.

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