The obstacle Problem for Quasilinear Stochastic PDEs - CiteSeerX

arXiv:1202.3296v1 [math.PR] 15 Feb 2012

The obstacle Problem for Quasilinear Stochastic PDEs: Analytical approach Laurent DENIS Université d’Evry-Val-d’Essonne-FRANCE e-mail: [email protected]

Anis MATOUSSI LUNAM Université, Université du Maine - FRANCE e-mail: [email protected]

Jing ZHANG Université d’Evry-Val-d’Essonne -FRANCE e-mail: [email protected] Abstract: We prove existence and uniqueness of the solution of quasilinear stochastic PDEs with obstacle. Our method is based on analytical technics coming from the parabolic potential theory. The solution is expressed as pair (u, ν) where u is a predictable continuous process which takes values in a proper Sobolev space and ν is a random regular measure satisfying minimal Skohorod condition. Keywords and phrases: parabolic potential, regular measure, stochastic partial differential equations, obstacle problem, penalization method, Ito’s formula, comparison theorem, space-time white noise. AMS 2000 subject classifications: Primary 60H15; 35R60; 31B150.

1. Introduction The starting point of this work is the following parabolic stochastic partial differential equation (in short SPDE) dut (x) = ∂i (ai,j (x)∂j ut (x) + gi (t, x, ut (x), ∇ut (x))) dt + f (t, x, ut (x), ∇ut (x))dt +∞ X hj (t, x, ut (x), ∇ut (x))dBtj , (1) + j=1

where a is a symmetric bounded measurable matrix which defines a second order operator on O ⊂ Rd , with null Dirichlet condition. The initial condition is given as u0 = ξ, a L2 (O)−valued random variable, and f , g = (g1 , ..., gd ) and h = (h1 , ...hi , ...) are non-linear random functions. Given an obstacle S : Ω×[0, T ]×O → R, we study the obstacle problem for the SPDE (1), i.e. we want to find a solution of (1) which satisfies "u ≥ S" where the ∗

The work of the first and third author is supported by the chair risque de crédit, Fédération bancaire Française † The research of the second author was partially supported by the Chair Financial Risks of the Risk Foundation sponsored by Société Générale, the Chair Derivatives of the Future sponsored by the Fédération Bancaire Française, and the Chair Finance and Sustainable Development sponsored by EDF and Calyon 1

2

L. Denis, A. Matoussi and J. Zhang

obstacle S is regular in some sense and controlled by the solution of an SPDE. Nualart and Pardoux [19] have studied the obstacle problem for a nonlinear heat equation on the spatial interval [0, 1] with Dirichlet boundary conditions, driven by an additive space-time white noise. They proved the existence and uniqueness of the solution and their method relied heavily on the results for a deterministic variational inequality. DonatiMartin and Pardoux [12] generalized the model of Nualart and Pardoux. They proved the existence of the solution by penalization method but they didn’t obtain the uniqueness result. Also, Xu and Zhang [25] have solved the problem of the uniqueness. However, in all their models, there isn’t the term of divergence and they do not consider the case where the coefficients depend on ∇u. The work of El Karoui et al [13] treats the obstacle problem for deterministic semi linear PDE’s within the framework of backward stochastic differential equations (BSDE in short). Namely the equation (1) is considered with f depending of u and ∇u, while the function g is null (as well h) and the obstacle v is continuous. They considered the viscosity solution of the obstacle problem for the equation (1) , they represented this solution stochastically as a process and the main new object of this BSDE framework is a continuous increasing process that controls the set {u = v}. Bally et al [3] (see also [16]) point out that the continuity of this process allows one to extend the classical notion of strong variational solution (see Theorem 2.2 of [5] p.238) and express the solution to the obstacle as a pair (u, ν) where ν is supported by the set {u = v}. Matoussi and Stoica [17] have proved an existence and uniqueness result for the obstacle problem of backward quasilinear stochastic PDE on the whole space Rd and driven by a finite dimentionnal Brownian motion. The method is based on the probabilistic interpretation of the solution by using the backward doubly stochastic differential equation (DBSDE). They have also proved that the solution is a pair (u, ν) where u is a predictable continuous process which takes values in a proper Sobolev space and ν is a random regular measure satisfying minimal Skohorod condition. In particular they gave for the regular measure ν a probabilistic interpretation in term of the continuous increasing process K where (Y, Z, K) is the solution of a reflected generalized BDSDE. Michel Pierre [20, 21] has studied the parabolic PDE with obstacle using the parabolic potential as a tool. He proved that the solution uniquely exists and is quasi-continuous. With the help of Pierre’s result, under suitable assumptions on f , g and h, our aim is to prove existence and uniqueness for the following SPDE with given obstacle S that we write formally as:  dut (x) = ∂i (ai,j (x)∂j ut (x) + gi (t, x, ut (x), ∇ut (x))) dt + f (t, x, ut (x), ∇ut (x))dt     +∞  X    hj (t, x, ut (x), ∇ut (x))dBtj , + j=1      ut ≥ St dt × dP − a.e. ,    u0 = ξ.

(2)

3

Theorem 1. Under the assumptions Lipschitz continuity and integrability of f , g and h, there exists a unique solution (u, ν) of the obstacle problem for the SPDE (2) associated to (ξ, f, g, h, S) where u is a predictable continuous process which takes values in a proper Sobolev space and ν is a random regular measure satisfying minimal Skohorod condition. In our paper, we will use the technics of parabolic potential theory developed by M. Pierre in the stochastic framework. We first prove a quasi-continuity result for the solution of the SPDE (1) with null Dirichlet condition on given domain O and driven by an infinite dimentionnal Brownian motion. This result is not obvious and its based on a mixing pathwise arguments and Mignot and Puel [18] existence result of the obstacle problem for some deterministic PDEs. Moreover, we prove in our context that the reflected measure ν is a regular random measure and we give the analytical representation of such measure in term of parabolic potential in the sense given by M. Pierre in [20]. This paper is divided as follows: in the second section, we set the assumptions then we introduce in the third section the notion of regular measure associated to parabolic potentials. The fourth section is devoted to prove the quasi-continuity of the solution of SPDE without obstacle. The fifth section is the main part of the paper in which we prove existence and uniqueness of the solution, to do that we begin with the linear case, and then by Picard iteration we get the result in the nonlinear case, we also establish the Ito’s formula. Finally, in the sixth section, we prove a comparison theorem for the solution of SPDE with obstacle. 2. Preliminaries We consider a sequence ((B i (t))t≥0 )i∈N of independent Brownian motions defined on a standard filtered probability space (Ω, F, (Ft )t≥0 , P ) satisfying the usual conditions. Let O ⊂ Rd be a bounded open domain and L2 (O) the set of square integrable functions with respect to the Lebesgue measure on O, it is an Hilbert space equipped with the usual scalar product and norm as follows Z Z u(x)v(x)dx, k u k= ( u2 (x)dx)1/2 . (u, v) = O

O

Let A be a symmetric second order differential operator, with domain D(A), given by A := −

d X

∂i (ai,j ∂j ).

i,j=1

We assume that a = (ai,j )i,j is a measurable symmetric matrix defined on O which satisfies the uniform ellipticity condition 2

λ|ξ| ≤

d X

ai,j (x)ξ i ξ j ≤ Λ|ξ|2 , ∀x ∈ O, ξ ∈ Rd ,

i,j=1

where λ and Λ are positive constants.


4

Let (F, E) be the associated Dirichlet form given by F := D(A1/2 ) = H01 (O) and E(u, v) := (A1/2 u, A1/2 v) and E(u) =k A1/2 u k2 , ∀u, v ∈ F, where H01 (O) is the first order Sobolev space of functions vanishing at the boundary. As usual we shall denote H −1 (O) its dual space. We consider the quasilinear stochastic partial differential equation (1) with initial condition u(0, ·) = ξ(·) and Dirichlet boundary condition u(t, x) = 0, ∀ (t, x) ∈ R+ × ∂O. We assume that we have predictable random functions f : R+ × Ω × O × R × Rd → R, g = (g1 , ..., gd ) : R+ × Ω × O × R × Rd → Rd , h = (h1 , ..., hi , ...) : R+ × Ω × O × R × Rd → RN , In the sequel, | · | will always denote the underlying Euclidean or l2 −norm. For example |h(t, ω, x, y, z)|2 =

+∞ X

|hi (t, ω, x, y, z)|2 .

i=1

Assumption (H): There exist non negative constants C, α, β such that for almost all ω, the following inequalities hold for all (x, y, z, t) ∈ O × R × R × R+ : 1. 2. 3. 4.

|f (t, ω, x, y, z) − f (t, ω, x, y ′ , z ′ )| ≤ C(|y − y ′ | + |z − z ′ |), P 1 ( di=1 |gi (t, ω, x, y, z) − gi (t, ω, x, y ′ , z ′ )|2 ) 2 ≤ C|y − y ′ | + α|z − z ′ |, 1 (|h(t, ω, x, y, z) − h(t, ω, x, y ′ , z ′ )|2 ) 2 ≤ C|y − y ′ | + β|z − z ′ |, the contraction property: 2α + β 2 < 2λ.

Remark 1. This last contraction property ensures existence and uniqueness for the solution of the SPDE without obstacle (see [8]). With the uniform ellipticity condition we have the following equivalent conditions: k f (u, ∇u) − f (v, ∇v) k≤ C k u − v k +Cλ−1/2 E 1/2 (u − v) k g(u, ∇u) − g(v, ∇v) kL2 (O;Rd ) ≤ C k u − v k +αλ−1/2 E 1/2 (u − v) k h(u, ∇u) − h(v, ∇v) kL2 (O;RN ) ≤ C k u − v k +βλ−1/2 E 1/2 (u − v) Assumption (I): Moreover we assume that for any T > 0, ξ ∈ L2 (Ω × O) is an F0 − measurable random variable f (·, ·, ·, 0, 0) := f 0 ∈ L2 ([0, T ] × Ω × O; R) g(·, ·, ·, 0, 0) := g 0 = (g10 , ..., gd0 ) ∈ L2 ([0, T ] × Ω × O; Rd ) h(·, ·, ·, 0, 0) := h0 = (h01 , ..., h0i , ...) ∈ L2 ([0, T ] × Ω × O; RN ). Now we introduce the notion of weak solution. For simplicity, we fix the terminal time T > 0. We denote by HT the space of H01 (O)-valued predictable continuous processes (ut )t≥0 which satisfy Z T 2 E(ut )dt < +∞. E sup k ut k +E t∈[0,T ]

0

5

It is the natural space for solutions. The space of test functions is denote by D = Cc∞ (R+ ) ⊗ Cc2 (O), where Cc∞ (R+ ) is the space of all real valued infinite differentiable functions with compact support in R+ and Cc2 (O) the set of C 2 -functions with compact support in O. Heuristquely, a pair (u, ν) is a solution of the obstacle problem for (1) if we have the followings: 1. u ∈ HT and u(t, x) ≥ S(t, x), dP ⊗ dt ⊗ dx − a.e. and u0 (x) = ξ, dP ⊗ dx − a.e.; 2. ν is a random measure defined on (0, T ) × O; 3. the following relation holds almost surely, for all t ∈ [0, T ] and ∀ϕ ∈ D, t

Z

(ut , ϕt ) − (ξ, ϕ0 ) −

(us , ∂s ϕs )ds +

0

=

Z

t

(fs (us , ∇us ), ϕs )ds + 0

+∞ Z t X j=1

4. Z

T 0

Z

0

Z

t

E(us , ϕs )ds + 0

d Z X i=1

(hjs (us , ∇us ), ϕs )dBsj +

0

t

(gsi (us , ∇us ), ∂i ϕs )ds

Z tZ 0

ϕs (x)ν(dx, ds); O

(u(s, x) − S(s, x))ν(dx, ds) = 0, a.s.. O

But, the random measure which in some sense obliges the solution to stay above the barrier is a local time so, in general, it is not absolutely continuous w.r.t Lebesgue measure. As a consequence, for example, the condition Z

T 0

Z

(u(s, x) − S(s, x))ν(dxds) = 0 O

makes no sense. Hence we need to consider precise version of u and S defined ν−almost surely. In order to tackle this difficulty, we introduce in the next section the notions of parabolic capacity on [0, T ]×O and quasi-continuous version of functions introduced by Michel Pierre in several works (see for example [20, 21]). Let us remark that these tools were also used by Klimsiak ([14]) to get a probabilistic interpretation to semilinear PDE’s with obstacle. Finally and to end this section, we give an important example of stochastic noise which is cover by our framework: Example 1. Let W be a noise white in time and colored in space, defined on a standard filtered probability space Ω, F, (Ft )t≥0 , P whose covariance function is given by: ˙ (x, s)W ˙ (y, t)] = δ(t − s)k(x, y), ∀s, t ∈ R+ , ∀x, y ∈ O, E[W

where k : O × O 7→ R+ is a symmetric and measurable function. Consider the following SPDE driven by W : dut (x) =

d X

i,j=1

∂i ai,j (x)∂j ut (x) + f (t, x, ut (x), ∇ut (x)) +

d X i=1

∂i gi (t, x, ut (x), ∇ut (x)) dt

˜ x, ut (x), ∇ut (x)) W (dt, x), + h(t,

(3)


6

˜ is a random real valued function. where f and g are as above and h We assume that the covariance function k defines a trace class operator denoted by K in L2 (O). It is well known (see [22]) that there exists an orthogonal basis (ei )i∈N∗ of L2 (O) consisting of eigenfunctions of K with corresponding eigenvalues (λi )i∈N∗ such that +∞ X

λi < +∞,

i=1

and k(x, y) =

+∞ X

λi ei (x)ei (y).

i=1

It is also well known that there exists a sequence ((B i (t))t≥0 )i∈N∗ of independent standard Brownian motions such that W (dt, ·) =

+∞ X

1/2

λi ei B i (dt).

i=1

So that equation (3) is equivalent to (1) with h = (hi )i∈N∗ where p ˜ x, y, z)ei (x). ∀i ∈ N∗ , hi (s, x, y, z) = λi h(s,

Assume as in [23] that for all i ∈ N∗ , kei k∞ < +∞ and +∞ X

λi kei k2∞ < +∞.

i=1

Since 1 2 = |h(t, ω, x, y, z) − h(t, ω, x, y ′ , z ′ )|2

+∞ X

λi kei k2∞

i=1

!

2 ˜ ˜ x, y ′ , z ′ ) , h(t, x, y, z) − h(t,

˜ satisfies a similar Lipschitz h satisfies the Lipschitz hypothesis (H)-(ii) if and only if h hypothesis. 3. Parabolic potential analysis 3.1. Parabolic capacity and potentials In this section we will recall some important definitions and results concerning the obstacle problem for parabolic PDE in [20] and [21]. K denotes L∞ ([0, T ]; L2 (O)) ∩ L2 ([0, T ]; H01 (O)) equipped with the norm: k v k2K = k v k2L∞ (0,T ;L2 (O)) + k v k2L2 (0,T ;H 1 (O)) 0 Z T = sup k vt k2 + k vt k2 +E(vt ) dt. t∈[0,T [

0

7

C denotes the space of continuous functions on compact support in [0, T [×O and finally: ∂ϕ ∈ L2 ([0, T ]; H −1 (O))}, ∂t

W = {ϕ ∈ L2 ([0, T ]; H01 (O));

∂ϕ 2 . k 2 −1 0 ∂t L ([0,T ];H (O)) It is known (see [15]) that W is continuously embedded in C([0, T ]; L2 (O)), the set of L2 (O)-valued continuous functions on [0, T ]. So without ambiguity, we will also consider WT = {ϕ ∈ W; ϕ(T ) = 0}, W + = {ϕ ∈ W; ϕ ≥ 0}, WT+ = WT ∩ W + . We now introduce the notion of parabolic potentials and regular measures which permit to define the parabolic capacity. endowed with the normk ϕ k2W =k ϕ k2L2 ([0,T ];H 1 (O)) + k

Definition 1. An element v ∈ K is said to be a parabolic potential if it satisfies: Z T Z T ∂ϕt + E(ϕt , vt )dt ≥ 0. , vt )dt + −( ∀ϕ ∈ WT , ∂t 0 0 We denote by P the set of all parabolic potentials. The next representation property is crucial: Proposition 1. (Proposition 1.1 in [21]) Let v ∈ P, then there exists a unique positive Radon measure on [0, T [×O, denoted by ν v , such that: Z TZ Z T Z T ∂ϕt ϕ(t, x)dν v . E(ϕt , vt )dt = , vt )dt + (− ∀ϕ ∈ WT ∩ C, ∂t 0 O 0 0 Moreover, v admits a right-continuous (resp. left-continuous) version vˆ (resp. v¯) : [0, T ] 7→ L2 (O) . Such a Radon measure, ν v is called a regular measure and we write: νv =

∂v + Av. ∂t

Remark 2. As a consequence, we can also define for all v ∈ P: vT = lim v¯t ∈ L2 (O). t↑T

Definition 2. Let K ⊂ [0, T [×O be compact, v ∈ P is said to be ν−superior than 1 on K, if there exists a sequence vn ∈ P with vn ≥ 1 a.e. on a neighborhood of K converging to v in L2 ([0, T ]; H01 (O)). We denote: SK = {v ∈ P; v is ν − superior to 1 on K}. Proposition 2. (Proposition 2.1 in [21]) Let K ⊂ [0, T [×O compact, then SK admits a v whose support is in K satisfies smallest vK ∈ P and the measure νK Z

T 0

Z

O

v dνK

Definition 3. (Parabolic Capacity)

Z = inf { v∈P

T 0

Z

O

dν v ; v ∈ SK }.


8

RT R v ; • Let K ⊂ [0, T [×O be compact, we define cap(K) = 0 O dνK • let O ⊂ [0, T [×O be open, we define cap(O) = sup{cap(K); K ⊂ O compact}; • for any borelian E ⊂ [0, T [×O, we define cap(E) = inf{cap(O); O ⊃ E open}. Definition 4. A property is said to hold quasi-everywhere (in short q.e.) if it holds outside a set of null capacity. Definition 5. (Quasi-continuous) A function u : [0, T [×O → R is called quasi-continuous, if there exists a decreasing sequence of open subsets On of [0, T [×O with: 1. for all n, the restriction of un to the complement of On is continuous; 2. limn→+∞ cap (On ) = 0. We say that u admits a quasi-continuous version, if there exists u ˜ quasi-continuous such that u ˜ = u a.e.. The next proposition, whose proof may be found in [20] or [21] shall play an important role in the sequel: Proposition 3. Let K ⊂ O a compact set, then ∀t ∈ [0, T [ cap({t} × K) = λd (K), where λd is the Lebesgue measure on O. As a consequence, if u : [0, T [×O → R is a map defined quasi-everywhere then it defines uniquely a map from [0, T [ into L2 (O). In other words, for any t ∈ [0, T [, ut is defined without any ambiguity as an element in L2 (O). Moreover, if u ∈ P, it admits version u ¯ 2 which is left continuous on [0, T ] with values in L (O) so that uT = u ¯T − is also defined without ambiguity. Remark 3. The previous proposition applies if for example u is quasi-continuous. Proposition 4. (Theorem III.1 in [21]) If ϕ ∈ W, then it admits a unique quasi-continuous version that we denote by ϕ. ˜ Moreover, for all v ∈ P, the following relation holds: Z

v

ϕdν ˜ = [0,T [×O

Z

T

(−∂t ϕ, v) + E(ϕ, v) dt + (ϕT , vT ) . 0

3.2. Applications to PDE’s with obstacle For any function ψ : [0, T [×O → R and u0 ∈ L2 (O), following M. Pierre [20, 21], F. Mignot and J.P. Puel [18], we define κ(ψ, u0 ) = ess inf{u ∈ P; u ≥ ψ a.e., u(0) ≥ u0 }.

(4)

This lower bound exists and is an element in P. Moreover, when ψ is quasi-continuous, this potential is the solution of the following reflected problem: κ ∈ P, κ ≥ ψ,

∂κ + Aκ = 0 on {u > ψ}, κ(0) = u0 . ∂t

9

Mignot and Puel have proved in [18] that κ(ψ, u0 ) is the limit (increasingly and weakly in L2 ([0, T ]; H01 (O))) when ǫ tends to 0 of the solution of the following penalized equation uǫ ∈ W, uǫ (0) = u0 ,

∂uǫ (uǫ − ψ)− + Auǫ − = 0. ∂t ǫ

Let us point out that they obtain this result in the more general case where ψ is only measurable from [0, T [ into L2 (O). For given f ∈ L2 ([0, T ]; H −1 (O)), we denote by κfu0 the solution of the following problem: κ ∈ W, κ(0) = u0 ,

∂κ + Aκ = f. ∂t

The next theorem ensures existence and uniqueness of the solution of parabolic PDE with obstacle, it is proved in [20], Theorem 1.1. The proof is based on a regularization argument of the obstacle, using the results of [6]. Theorem 2. Let ψ : [0, T [×O → R be quasi-continuous, suppose that there exists ζ ∈ P with |ψ| ≤ ζ a.e., f ∈ L2 ([0, T ]; H −1 (O)), and the initial value u0 ∈ L2 (O) with u0 ≥ ψ(0), then there exists a unique u ∈ κfu0 + P quasi-continuous such that: Z TZ f ˜ u−κu0 = 0 u(0) = u0 , u ˜ ≥ ψ, q.e.; (˜ u − ψ)dν 0

O

We end this section by a convergence lemma which plays an important role in our approach (Lemma 3.8 in [21]): Lemma 1. If v n ∈ P is a bounded sequence in K and converges weakly to v in L2 ([0, T ]; H01 (O)); if u is a quasi-continuous function and |u| is bounded by a element in P. Then Z TZ Z TZ n udν v . udν v = lim n→+∞ 0

O

0

O

Remark 4. For the more general case one can see [21] Lemma 3.8. 4. Quasi–continuity of the solution of SPDE without obstacle As a consequence of well-known results (see for example [8], Theorem 8), we know that under assumptions (H) and (I), SPDE (1) with zero Dirichlet boundary condition, admits a unique solution in HT , we denote it by U (ξ, f, g, h). The main theorem of this section is the following: Theorem 3. Under assumptions (H) and (I), u = U (ξ, f, g, h) the solution of SPDE (1) admits a quasi-continuous version denoted by u ˜ i.e. u = u ˜ dP × dt × dx a.e. and for almost all w ∈ Ω, (t, x) → u ˜t (w, x) is quasi-continuous. Before giving the proof of this theorem, we need the following lemmas. The first one is proved in [21], Lemma 3.3: Lemma 2. There exists C > 0 such that, for all open set ϑ ⊂ [0, T [×O and v ∈ P with v ≥ 1 a.e. on ϑ: capϑ ≤ C k v k2K .


10

Let κ = κ(u, u+ (0)) be defined by relation (4). One has to note that κ is a random function. From now on, we always take for κ the following measurable version κ = sup v n , n

where (v n ) is the non-decreasing sequence of random functions given by  n  ∂vt = Lv n + n(v n − u )− t t t ∂t  n v0 = u+ (0).

(5)

Using the results recalled in Subsection 3, we know that for almost all w ∈ Ω, v n (w) converges weakly to v(w) = κ(u(w), u+ (0)(w)) in L2 (0, T ; H01 (O)) and that v ≥ u. Lemma 3. We have the following estimate: Ekκ

k2K

≤C Ek

u+ 0

2

2

k +E k u0 k +E

Z

T

k

0

ft0

2

k +k

|gt0 |

2

k +k

|h0t |

2

k dt ,

where C is a constant depending only on the structure constants of the equation. Proof. All along this proof, we shall denote by C or Cǫ some constant which may change from line to line. The following estimate for the solution of the SPDE we consider is well-known: E sup k ut k2 +E t∈[0,T ]

Z

T

E(ut )dt ≤ CE(k u0 k2 +

T

Z

0

0

(k ft0 k2 + k |gt0 | k2 + k |h0t | k2 )dt) (6)

where C is a constant depending only on the structure constants of the equation. Consider the approximation (v n )n defined by (5), P -almost surely, it converges weakly to v = κ(u, u+ (0)) in L2 (0, T ; H01 (O)). We remark that v n − u satisfies the following equation: d(vtn −ut )+A(vtn −ut )dt = −ft (ut , ∇ut )dt−

d X

∂i gti (ut , ∇ut )dt−

+∞ X

hjt (ut , ∇ut )dBtj +n(vtn −ut )− dt,

j=1

i=1

applying the Itô’s formula to (v n − u)2 , see Lemma 7 in [9], we have k vtn − ut k2 +2 +2

d Z X

0

−2

t

i=1 +∞ XZ t j=1

0

Z

t 0

2 E(vsn − us )ds =k u+ 0 − u0 k −2

(∂i (vsn − us ), gsi (us , ∇us ))ds +

Z

(vsn − us , hjs (us , ∇us ))dBsj + 2

Z

t

Z

t 0

(vsn − us , fs (us , ∇us ))ds

k |hs (us , ∇us )| k2 ds

0 t

0

(n(vsn − us )− , vsn − us )ds.

(7)

11

The last term in the right member of (7) is obviously non-positive so Z t Z t + n 2 n 2 E(vs − us )ds ≤k u0 − u0 k −2 (vsn − us , fs (us , ∇us ))ds k vt − ut k +2 0

0

+

Z

t

k |hs (us , ∇us )| k2 ds + 2

0

−2

i=1

+∞ Z t X j=1

d Z t X

0

0

(∂i (vsn − us ), gsi (us , ∇us ))ds

(vsn − us , hjs (us , ∇us ))dBsj .

(8)

Then taking expectation and using Cauchy-Schwarz’s inequality, we get Z t Z t ǫ + n 2 n 2 k vsn − us k2 ds E(vs − us )ds ≤ E k u0 − u0 k +E E k vt − ut k +(2 − )E λ 0 0 Z t Z t Z t k |hs (us , ∇us )| k2 ds. k |gs (us , ∇us )| k2 ds + E k fs (us , ∇us ) k2 ds + Cǫ E +E 0

0

0

Therefore, by using the Lipschitz conditions on the coefficients we have: Z t Z t ǫ n 2 2 n + E k vt − ut k +(2 − )E k vsn − us k2 ds E(vs − us )ds ≤ E k u0 − u0 k +E λ 0 0 Z t Z t Z t α β2 C E(us )ds. k us k2 ds + ( + + )E (k fs0 k2 + k |gs0 | k2 + k |h0s | k2 )ds + CE +CE λ λ λ 0 0 0 Combining with (6), this yields Z t Z t ǫ n + 2 Ek − ut k +(2 − )E k vsn − us k2 ds E(vs − us )ds ≤ E k u0 − u0 k +E λ 0 0 Z T 2 (k ft0 k2 + k |gt0 | k2 + k |h0t | k2 )dt). + CE(k u0 k + vtn

2

0

We take now ǫ small enough such that (2− λǫ ) > 0, then, with Gronwall’s lemma, we obtain for each t ∈ [0, T ] Z T n 2 c′ T + 2 2 k ft0 k2 + k |gt0 | k2 + k |h0t | k2 dt). E k vt − ut k ≤ Ce (E k u0 − u0 k +E k u0 k +E 0

As we a priori know that P -almost surely, (vn )n tends to κ strongly in L2 ([0, T ] × O), the previous estimate yields, thanks to the dominated convergence theorem, that (vn )n converges to κ strongly in L2 (Ω × [0, T ] × O) and Z T 2 c′ T + 2 2 sup E k κt − ut k ≤ Ce (E k u0 − u0 k +E k u0 k +E k ft0 k2 + k |gt0 | k2 + k |h0t | k2 dt). 0

t∈[0,T ]

Moreover, as (v n )n tends to κ weakly in L2 ([0, T ]; H01 (O)) P -almost-surely, we have for all t ∈ [0, T ]: Z T Z T E(κs − us )ds ≤ lim inf E E(vsn − us )ds E n 0 0 Z T ′ + cT 2 2 k ft0 k2 + k |gt0 | k2 + k |h0t | k2 dt). ≤ T Ce (E k u0 − u0 k +E k u0 k +E 0


12

Let us now study the stochastic term in (8). Let define the martingales +∞ Z t +∞ Z t X X n j j n (κs − us , hjs )dBsj . (vs − us , hs )dBs and Mt = Mt = j=1

0

0

j=1

Then E[|MTn

2

− MT | ] = E

Z

0

+∞ T X

(κs −

vsn , hs )2 ds

≤E

Z

T

k κs − vsn k2 k |hs | k2 ds.

0

j=1

Using the strong convergence of (v n ) to κ we conclude that M n tends to M in L2 sense. Passing to the limit in (8), we get: 2

k κt − ut k +2

Z

t

E(κs − us )ds ≤ k 0

+ 2

u+ 0

2

− u0 k −2

d Z X

0

i=1

− 2

t

+∞ Z X

t

(κs − us , fs (us , ∇us ))ds 0

(∂i (κs − us ), gsi (us , ∇us ))ds

t

(κs −

0

j=1

Z

us , hjs (us , ∇us ))dBsj

+

Z

t

k |hs (us , ∇us )| k2 ds.

0

As a consequence of the Burkholder-Davies-Gundy’s inequalities, we get Z TX +∞ +∞ Z t X j j (κs − us , hs (us , ∇us ))dBs | ≤ CE[ (κs − us , hjs (us , ∇us ))2 ds]1/2 E sup | t∈[0,T ] j=1

0

0

Z ≤ CE[

0

j=1

+∞ T X

sup k κt − ut k2 k hjs (us , ∇us ) k2 ds]1/2

j=1 t∈[0,T ]

Z ≤ CE[ sup k κt − ut k ( t∈[0,T ]

T

k |ht (ut , ∇ut )| k2 dt)1/2 ]

0

≤ ǫE sup k κt − ut k2 +Cǫ E t∈[0,T ]

Z

T

k |ht (ut , ∇ut )| k2 dt.

0

By Lipschitz conditions on h and (6) this yields +∞ Z t X (κs − us , hs (us , ∇us ))dBs | ≤ ǫE sup k κt − ut k2 +C(E k u0 k2 E sup | t∈[0,T ] j=1

0

t∈[0,T ]

+ E

Z

0

T

(k ft0 k2 + k |gt0 | k2 + k |h0t | k2 )dt)

Hence, (1 − ǫ)E sup k κt − ut k2 + (2 − t∈[0,T ]

+ E

Z

ǫ )E λ T

0

Z

0

T

2 2 E(κt − ut )dt ≤ C(E k u+ 0 − u0 k +E k u0 k

k ft0 k2 + k |gt0 | k2 + k |h0t | k2 dt).

13

We can take ǫ small enough such that 1 − ǫ > 0 and 2 − 2

E sup k κt − ut k

+ E

Z

+ E

Z

T

0

t∈[0,T ]

T

0

ǫ λ

> 0, hence,

2 2 E(κt − ut )dt ≤ C(E k u+ 0 − u0 k +E k u0 k

k ft0 k2 + k |gt0 | k2 + k |h0t | k2 dt).

Then, combining with (6), we get the desired estimate: Z Z T + 2 2 2 E(κt )dt ≤ C(E k u0 k +E k u0 k +E E sup k κt k +E

0

0

t∈[0,T ]

T

k ft0 k2 + k |gt0 | k2 + k |h0t | k2 dt).

Proof of Theorem 3: For simplicity, we put ft (x) = f (t, x, ut (x), ∇ut (x)), gt (x) = g(t, x, ut (x), ∇ut (x)) and ht (x) = h(t, x, ut (x), ∇ut (x)). We introduce (Pt ) the semi-group associated to operator A and put for each n ∈ N∗ , i ∈ {1, · · · , d} and each j ∈ N∗ : f n = P 1 f,

un0 = P 1 u0 , n

gin = P 1 gi ,

n

and hnj = P 1 hj .

n

n

Then (un0 )n converges to u0 in L2 (Ω; L2 (O)), (f n )n , (gn )n and (hn )n are sequences of elements in L2 (Ω × [0, T ]; D(A)) which converge respectively to f , g and h in L2 (Ω × [0, T ]; L2 (O)). For all n ∈ N we define unt

=

Pt un0

+

Z

t 0

= Pt+ 1 u0 + n

Pt−s fsn ds

+

d Z X i=1

Z

0

t

Pt+ 1 −s fs ds + n

t 0

n Pt−s ∂i gi,s ds

j=1

d Z t X i=1

+

+∞ Z X

0

t 0

Pt−s hnj,s dBsj

Pt+ 1 −s ∂i gi,s ds + n

+∞ Z X j=1

t 0

Pt+ 1 −s hj,s dBsj . n

We denote by G(t, x, s, y) the kernel associated to Pt , then Z tZ Z 1 1 n G(t + , x, s, y)f (s, y)dyds G(t + , x, 0, y)u0 (y)dy + u (t, x) = n n 0 O O Z Z +∞ d XZ tZ X t 1 1 G(t + , x, s, y)hjs (y)dydBsi . G(t + , x, s, y)∂i gsi (y)dyds + + n n 0 O 0 O j=1

i=1

But, as A is strictly elliptic, G is uniformly continuous in space-time variables on any compact away from the diagonal in time ( see Theorem 6 in [1]) and satisfies Gaussian estimates (see Aronson [2]), this ensures that for all n, un is P -almost surely continuous in (t, x). We consider a sequence of random open sets ϑn = {|un+1 − un | > ǫ},

Θp =

+∞ [

n=p

ϑn .


14

Let κn = κ( 1ǫ (un+1 − un ), 1ǫ (un+1 − un )+ (0)) + κ(− 1ǫ (un+1 − un ), 1ǫ (un+1 − un )− (0)), from the definition of κ and the relation (see [21]) κ(|v|) ≤ κ(v, v + (0)) + κ(−v, v − (0)) we know that κn satisfy the conditions of Lemma 2, i.e. κn ∈ P et κn ≥ 1 a.e. on ϑn , thus we get the following relation cap (Θp ) ≤

+∞ X

cap (ϑn ) ≤

+∞ X

k κn k2K .

n=p

n=p

Thus, remarking that un+1 − un = U (un+1 − un0 , f n+1 − f n , gn+1 − g n , hn+1 − hn ), we apply 0 1 n+1 1 n+1 n n Lemma 3 to κ( ǫ (u − u ), ǫ (u − u )+ (0)) and κ(− 1ǫ (un+1 − un ), 1ǫ (un+1 − un )− (0)) and obtain: E[cap (Θp )] ≤

+∞ X

Ek

κn k2K

n=p

Z T +∞ X 1 n+1 n 2 k ftn+1 − ftn k2 ≤ 2C (E k u0 − u0 k +E 2 ǫ 0 n=p − hnt | k2 dt) + k |gtn+1 − gtn | k2 + k |hn+1 t

Then, by extracting a subsequence, we can consider that Ek

un+1 0

−

un0

Z

2

k +E

Then we take ǫ =

1 n2

T 0

k ftn+1 − ftn k2 + k |gtn+1 − gtn | k2 + k |hn+1 − hnt | k2 dt ≤ t

1 2n

to get E[cap (Θp )] ≤

+∞ X 2Cn4

n=p

2n

Therefore lim E[cap (Θp )] = 0.

p→+∞

For almost all ω ∈ Ω, un (ω) is continuous in (t, x) on (Θp (w))c and (un (ω))n converges uniformly to u on (Θp (w))c for all p, hence, u(ω) is continuous in (t, x) on (Θp (w))c , then from the definition of quasi-continuous, we know that u(ω) admits a quasi-continuous version since cap (Θp ) tends to 0 almost surely as p tends to +∞. 5. Existence and uniqueness of the solution of the obstacle problem 5.1. Weak solution Assumption (O): The obstacle S is assumed to be an adapted process, quasi-continuous, such that S0 ≤ ξ P -almost surely and controlled by the solution of an SPDE, i.e. ∀t ∈ [0, T ], St ≤ St′

(9)

15

where S ′ is the solution of the linear SPDE P P+∞ ′ j ′ dt + dSt′ = LSt′ dt + ft′ dt + di=1 ∂i gi,t j=1 hj,t dBt S ′ (0) = S0′ ,

(10)

where S0′ ∈ L2 (Ω × O) is F0 -measurable, f ′ , g′ and h′ are adapted processes respectively in L2 ([0, T ] × Ω × O; R), L2 ([0, T ] × Ω × O; Rd ) and L2 ([0, T ] × Ω × O; RN ). Remark 5. Here again, we know that S ′ uniquely exists and satisfies the following estimate: Z T Z T 2 2 ′ ′ ′ 2 ′ 2 ′ ′ 2 (k ft k + k |gt | k + k |ht | k )dt E(St )dt ≤ CE k S0 k + E sup k St k +E 0

0

t∈[0,T ]

(11)

Moreover, from Theorem 3,

S′

admits a quasi-continuous version.

We now are able to define rigorously the notion of solution to the problem with obstacle we consider. Definition 6. A pair (u, ν) is said to be a solution of the obstacle problem for (1) if 1. u ∈ HT and u(t, x) ≥ S(t, x), dP ⊗ dt ⊗ dx − a.e. and u0 (x) = ξ, dP ⊗ dx − a.e.; 2. ν is a random regular measure defined on [0, T ) × O; 3. the following relation holds almost surely, for all t ∈ [0, T ] and ∀ϕ ∈ D, (ut , ϕt ) − (ξ, ϕ0 ) −

Z

t

(us , ∂s ϕs )ds +

0

=

Z

t

(fs (us , ∇us ), ϕs )ds + 0

+∞ Z t X j=1

0

Z

t

E(us , ϕs )ds + 0

d Z X i=1


0

t

(gsi (us , ∇us ), ∂i ϕs )ds

Z tZ 0

ϕs (x)ν(dx, ds)(12) O

4. u admits a quasi-continuous version, u ˜, and we have Z TZ ˜ x))ν(dx, ds) = 0, a.s.. (˜ u(s, x) − S(s, 0

O

The main result of this paper is the following: Theorem 4. Under assumptions (H), (I) and (O), there exists a unique weak solution of the obstacle problem for the SPDE (1) associated to (ξ, f, g, h, S). We denote by R(ξ, f, g, h, S) the solution of SPDE (1) with obstacle when it exists and is unique. As the proof of this theorem is quite long, we split it in several steps: first we prove existence and uniqueness in the linear case then establish an Ito’s formula and finally prove the Theorem thanks to a fixed point argument. 5.2. Proof of Theorem 4 in the linear case All along this subsection, we assume that f , g and h do not depend on u and ∇u, so we consider that f , g and h are adapted processes respectively in L2 ([0, T ] × Ω × O; R),


16

L2 ([0, T ] × Ω × O; Rd ) and L2 ([0, T ] × Ω × O; RN ). For n ∈ N, let un be the solution of the following SPDE dunt

=

Lunt dt

+ ft dt +

d X

∂i gi,t dt +

+∞ X

hj,t dBtj + n(unt − St )− dt

(13)

j=1

i=1

with initial condition un0 = ξ and null Dirichlet boundary condition. We know from Theorem 8 in [DS04] that this equation admits a unique solution in HT and that the solution admits L2 (O)−continuous trajectories. Lemma 4. un satisfies the following estimate: Z Z T n n 2 E(ut )dt + E E sup k ut k +E

0

0

t∈[0,T ]

T

n k (unt − St )− k2 dt ≤ C,

where C is a constant depending only on the structure constants of the SPDE. Proof. From (13) and (10), we know that un − S ′ satisfies the following equation: d(unt − St′ ) = L(unt − St′ )dt + f˜t dt +

d X

∂i g˜ti dt +

+∞ X

˜hj dB j + n(un − St )− dt t t t

j=1

i=1

˜ = h − h′ . Applying Itô’s formula to (un − S ′ )2 , we have: where f˜ = f − f ′ , g˜ = g − g ′ and h k

unt

−2

−

St′

2

k +2

d Z t X 0

i=1

Z

(∂i (uns

t 0

E(uns

−

Ss′ )ds

− S˜s ), g˜si )ds + 2

=2

Z

Z tZ 0

t 0

O

((uns

−

Ss′ ), f˜s )ds

+2

+∞ Z X j=1

(uns

−

Ss′ )n(uns

−

− Ss ) ds +

t 0

˜ j )dB j ((uns − Ss′ ), h s s t

Z

˜ s | k2 ds. k |h

0

We remark first: Z tZ Z tZ n ′ n − (us − Ss )n(us − Ss ) ds = (uns − Ss + Ss − Ss′ )n(uns − Ss )− ds 0 O 0 O Z tZ Z tZ n − 2 (Ss − Ss′ )n(uns − Ss )− dxds n((us − Ss ) ) ds + = − 0

0

O

O

the last term in the right member is non-positive because St ≤ St′ , thus, Z t Z t Z t n − 2 n ′ n ′ 2 (uns − Ss′ , f˜s )ds n k (us − S) k ds ≤ 2 E(us − Ss )ds + 2 k ut − St k +2 0

−2

d Z t X i=1

0

(∂i (uns − Ss′ ), g˜si )ds + 2

0 Z +∞ X t j=1

0

0

˜ j )dB j + (uns − Ss′ , h s s

Z

t

˜ s | k2 ds. k |h

0

Then using Cauchy-Schwarz’s inequality, we have ∀t ∈ [0, T ], 2|

Z

t 0

(uns

−

Ss′ , f˜s )ds|

≤ǫ

Z

0

T

k

uns

−

Ss′

1 k ds + ǫ 2

Z

T 0

k f˜s k2 ds

17

and 2|

d Z X i=1

t 0

(∂i (uns

−

Ss′ ), g˜si )ds|

≤ǫ

Z

T

k 0

∇(uns

−

Ss′ )

1 k ds + ǫ 2

Z

T

k |˜ g | k2 ds.

0

Moreover, thanks to the Burkholder-Davies-Gundy inequality, we get E sup |

+∞ Z X

t∈[0,T ] j=1

t 0

(uns

−

˜ j )dB j | Ss′ , h s s

Z ≤ c1 E[ Z ≤ c1 E[

0

0

+∞ T X

˜ j )2 ds]1/2 (uns − Ss′ , h s

j=1

+∞ T X

˜ j k2 ds]1/2 sup k uns − Ss′ k2 k h s

j=1 s∈[0,T ]

≤ c1 E[ sup k s∈[0,T ]

≤ ǫE sup k s∈[0,T ]

uns

uns

−

−

Ss′

Ss′

Z k(

T

˜ s | k2 ds)1/2 ] k |h

0

c1 k + E 4ǫ 2

Z

T

˜ s | k2 ds. k |h

0

Then using the strict ellipticity assumption and the inequalities above, we get Z T Z T n k (uns − Ss )− k2 ds E(uns − Ss′ )ds + 2E (1 − 2ǫ(T + 1))E sup k unt − St′ k2 +(2λ − ǫ)E 2 ≤ C(E k ξ k2 + E ǫ

Z

T

0

0

0

t∈[0,T ]

2 c1 ˜ s | k2 ds). k f˜s k2 + k |˜ gs | k2 +( + 1) k |h ǫ 2ǫ

We take ǫ small enough such that (1 − 2ǫ(T + 1)) > 0, this yields (2λ − ǫ) > 0 Z T Z T n ′ n ′ 2 n k (unt − St )− k2 dt ≤ C. E(ut − St )dt + E E sup k ut − St k +E t∈[0,T ]

0

0

Then with (11), we obtain the desired estimate. End of the proof of Theorem 4. We now introduce z, the solution of the corresponding SPDE without obstacle: dzt + Azt dt = ft dt +

d X i=1

∂i gi,t dt +

+∞ X

hj,t dBtj ,

j=1

starting from z0 = ξ, with null Dirichlet condition on the boundary. As a consequence of Theorem 3, we can take for z a quasi-continuous version. For each n ∈ N, we put v n = un − z. Clearly, v n satisfies dvtn + Avtn dt = n(vtn − (St − zt ))− dt = n(unt − St )− dt. Since S −z is quasi-continuous almost-surely, by the results established by Mignot and Puel in [18], we know that P -almost surely, the sequence (vn )n is increasing and converges in n L2 ([0, T ]×O) P -almost surely to v and that the sequence of random measures ν v = n(unt − St )− dtdx converges vaguely to a measure associated to v: ν = ν v . As a consequence of the


18

previous lemma, (un ) and (v n ) are bounded sequences in L2 (Ω × [0, T ]; H01 (O)) which is an RT Hilbert space ( equipped the norm E 0 k ut kH01 (O) dt ), by a double extraction argument, we can construct subsequences (unk )k and (v nk )k such that the first one converges weakly in L2 (Ω × [0, T ]; H01 (O)) to an element that we denote u and the second one to an element which necessary is equal to v since (v n )n is increasing. Moreover, we can construct sequences (ˆ un ) and (ˆ v n ) of convex combinations of elements of the form u ˆn =

Nn X

αnk unk and vˆn =

k=1

Nn X

αnk v nk

k=1

converging strongly to u an v respectively in L2 (Ω × [0, T ]; H01 (O)). From the fact that un is the weak solution of (13), we get (unt , ϕt ) − (ξ, ϕ0 ) − =

Z

t

(fs , ϕs )ds + 0

t

Z

0

+∞ Z t X 0

j=1

t

Z

(uns , ∂s ϕs )ds +

E(uns , ϕs )ds +

0

0

O

t

0

i=1

Z tZ

(hjs , ϕs )dBsj +

d Z X

(gsi , ∂i ϕs )ds

ϕs (x)n(uns − Ss )− dxds a.s.

(14)

Hence (ˆ unt , ϕt ) − (ξ, ϕ0 ) − =

Z

t

(fs , ϕs )ds + 0

j=1

We have Z tZ 0

+∞ Z X

ϕs (x)

O

Nn X

0

t

t

Z

0

(hjs , ϕs )dBsj +

nk (uns k

−

− Ss )

k=1

!

0

Z tZ 0

t

Z

(ˆ uns , ∂s ϕs )ds +

E(ˆ uns , ϕs )ds +

d Z X

0

i=1

Nn X

ϕs (x) O

t

(gsi , ∂i ϕs )ds

nk (uns k − Ss )−

k=1

T

Z

dxds =

−(

0

∂ϕt n , vˆ )dt + ∂t t

Z

T

0

!

dxds

(15)

E(ϕt , vˆtn )dt

so that we have almost-surely, at least for a subsequence: ! Z T Z tZ Z T Nn X ∂ϕt nk − E(ϕt , vt )dt , vt )dt + ϕs (x) −( lim dxds = nk (us − Ss ) n→+∞ 0 ∂t 0 O 0 k=1 Z TZ ϕt (x)ν(dx, dt). = 0

(ˆ un )n

As obtain:

converges to u in

L2 (Ω

(ut , ϕt ) − (ξ, ϕ0 ) − =

Z

t

(fs , ϕs )ds + 0

Z

×

[0, T ]; H01 (O)),

t

(us , ∂s ϕs )ds + 0

+∞ Z t X j=1

0

Z

O

by making n tend to +∞ in (15), we

t

E(us , ϕs )ds + 0

(hjs , ϕs )dBsj +

d Z X i=1

Z tZ 0

0

t

(gsi , ∂i ϕs )ds

ϕs (x)ν(dx, ds), a.s.. O

19

In the next subsection, we’ll show that u satisfies an Itô’s formula, as a consequence by applying it to u2t , using standard arguments we get that u ∈ HT so for almost all ω ∈ Ω, u(ω) ∈ K. And from Theorem 9 in [8], we know that for almost all ω ∈ Ω, z(ω) ∈ K. Therefore, for almost all ω ∈ Ω, v(ω) = u(w) − z(w) ∈ K. Hence, ν = ∂t v + Av is a regular measure by definition. Moreover, by [20, 21] we know that v admits a quasi continuous version v˜ which satisfies the minimality condition Z Z (˜ v − S + z˜)ν(dxdt) = 0. (16) z is quasi-continuous version hence u ˜ = z + v˜ is a quasi-continuous version of u and we can write (16) as Z Z (˜ u − S)ν(dxdt) = 0.

The fact that u ≥ S comes from the fact that v ≥ u − z, so at this stage we have proved that (u, ν) is a solution to the obstacle problem we consider. Uniqueness comes from the fact that both z and v are unique, which ends the proof of Theorem 4.

5.3. Itô’s formula The following Itô’s formula for the solution of the obstacle problem is fundamental to get all the results in the non linear case. Let us also remark, that any solution of the non-linear equation (1) may be viewed as the solution of a linear one so that it satisfies also the Itô’s formula. Theorem 5. Under assumptions of the previous subsection 5.2, let u be the solution of SPDE(1) with obstacle and Φ : R+ ×R → R be a function of class C 1,2 . We denote by Φ′ and Φ” the derivatives of Φ with respect to the space variables and by ∂Φ ∂t the partial derivative with respect to time. We assume that these derivatives are bounded and Φ′ (t, 0) = 0 for all t ≥ 0. Then P − a.s. for all t ∈ [0, T ], t

Z tZ

∂Φ (s, us (x))dxds 0 O ∂s O 0 O Z t +∞ Z t d Z tZ X X (Φ′ (s, us ), hj )dBsj Φ′′ (s, us (x))∂i us (x)gi (x)dxds + + (Φ′ (s, us ), fs )ds − Z

Φ(t, ut (x))dx +

0

+

1 2

′

E(Φ (s, us ), us )ds =

i=1

+∞ Z t Z X j=1

Z

0

O

0

Z

Φ(0, ξ(x))dx +

O

Φ′′ (s, us (x))(hj,s (x))2 dxds +

j=1

Z tZ 0

0

Φ′ (s, u ˜s (x))ν(dxds).

O

Proof. We keep the same notations as in the previous subsection and so consider the n ) approximating u and also (ˆ sequence un ) the sequence of convex combinations u ˆn = n PNn n (u nk converging strongly to u in L2 (Ω × [0, T ]; H 1 (O)). 0 k=1 αk u Moreover, by standard arguments such as the Banach-Saks theorem, since (un )n is nondecreasing, we can choose the convex combinations such that (ˆ un )n is also a non-decreasing sequence. We start by a key lemma:


20

Lemma 5. Let t ∈ [0, T ], then lim E

n→+∞

Z tZ 0

O

(ˆ uns − Ss )−

Nn X

αnk nk (usnk − Ss )− dxds = 0.

k=1

P n n − nk Proof. We write as above un = v n + z and we denote νˆn = N k=1 αk nk (us − Ss ) so that Z tZ Z tZ Z tZ (ˆ uns − Ss )ˆ (zs − Ss )ˆ ν n (dxds) vˆsn νˆn (dxds) + ν n (dxds) = 0

O

0

0

O

O

From Lemma 1, we know that Z tZ Z tZ (zs − Ss )ν(dxds). (zs − Ss )ˆ ν n (dxds) → 0

0

O

O

Moreover, by Lemma II.6 in [20] we have for all n: Z TZ Z T 1 n n 2 vˆsn νˆn (dxds), E(ˆ vs )ds = k vˆT k + 2 0 O 0 and

1 k vT k2 + 2

Z

T

E(vs )ds =

T

Z

0

0

As (ˆ vn )n tends to v in L2 ([0, T ], H01 (O)), Z Z T n E(ˆ vs )ds = lim n→+∞ 0

Z

v˜s ν(dxds).

O

T

E(vs )ds. 0

Let us prove that k vˆTn k tends to k vT k. Since, (ˆ vTn ) is non-decreasing and bounded in L2 (O) it converges in L2 (O) to m = supn vˆTn . Let ρ ∈ H01 (O) then the map defined by ϕ(t, x) = ρ(x) belongs to W hence as a consequence of Proposition 4 Z T Z E(ρ, vˆs n ) ds + (ρ, vˆTn ), ρdˆ νn = 0

[0,T [×O

and Z

ρ˜dν = [0,T [×O

Z

T

E(ρ, vs ) ds + (ρ, vT ), 0

making n tend to +∞ and using one more time Lemma 1, we get lim (ρ, vˆTn ) = (ρ, m) = (ρ, vT ).

n→+∞

Since ρ is arbitrary, we have vT = m and so limn→+∞ k vˆTn k=k vT k and this yields Z TZ Z TZ Z TZ n n (Ss − zs )ν(dxds). v˜s ν(dxds) = vˆs νˆ (dxds) = lim n→+∞ 0

0

O

This proves that lim

Z tZ

n→+∞ 0

O

O

0

O

(ˆ uns − Ss )ˆ ν n (dxds) = 0,

21

we conclude by remarking that Z tZ Z tZ Z tZ (˜ us −Ss )ν(dxds) = 0. (us −Ss )ˆ ν n (dxds) = (ˆ uns −Ss )+ νˆn (dxds) ≤ lim lim n→+∞ 0

n→+∞ 0

O

0

O

O

Proof of Theorem 5: We consider the penalized solution (un ), we know that its convex ˆn satisfies the combination u ˆn converges strongly to u in L2 (Ω × [0, T ]; H01 (O)). And u following SPDE dˆ unt + Aˆ unt dt = ft dt +

d X

∂i gti dt +

+∞ X

hjt dBtj +

j=1

i=1

Nn X

αnk nk (uns k − Ss )− dt

k=1

From the Itô’s formula for the solution of SPDE without obstacle (see Lemma 7 in [9]), we have, almost surely, for all t ∈ [0, T ], t

Z tZ

∂Φ (s, uˆns )dxds ∂s 0 O O 0 O Z t +∞ Z t d Z tZ X X (Φ′ (s, u ˆns ), hj )dBsj Φ′′ (s, u ˆns (x))∂i u ˆns (x)gi (x)dxds + ˆns ), fs )ds − + (Φ′ (s, u Z

Φ(t, u ˆnt (x))dx +

Z

0

E(Φ′ (s, u ˆns ), uˆns )ds =

i=1

+∞ Z t Z

1X + 2 j=1

0

Φ

′′

O

0

Z

Φ(0, ξ(x))dx +

O

0

j=1

(s, u ˆns (x))(hj (x))2 dxds

+

Z tZ 0

O

Φ′ (s, u ˆns )

Nn X

αnk nk (uns k − Ss )− dxds.

k=1

Because of the strong convergence of u ˆn , the convergence of all the terms except the last one are clear. To obtain the convergence of the last term, we do as follows: Z tZ 0

Φ

O

′

(s, uˆns )

Nn X

αnk nk (uns k

Z tZ

−

− Ss ) dxds =

0

k=1

O

Z tZ

+

0

(Φ

′

(s, uˆns ) −

′

Φ (s, Ss ))

Nn X

αnk nk (uns k − Ss )− dxds

k=1

Φ′ (s, Ss )

O

Nn X

αnk nk (uns k − Ss )− dxds.

k=1

For the first term in the right member, we have: |

Z tZ 0

(Φ O

≤C

Z tZ

=C

Z tZ

0

0

=C

Z tZ 0

′

(s, u ˆns )

′

− Φ (s, Ss ))

Nn X

αnk nk (uns k − Ss )− dxds|

k=1

O

O

O

|ˆ uns − Ss | ·

Nn X


k=1

((ˆ uns − Ss )+ + (ˆ uns − Ss )− )

Nn X


k=1

(ˆ uns

+

− Ss )

Nn X k=1

αnk nk (uns k

−

− Ss ) dxds + C

Z tZ 0

O

(ˆ uns

−

− Ss )

Nn X k=1



22

We have the following inequality because (ˆ un ) converges to u increasingly: Z tZ Z tZ Nn Nn X X (ˆ uns − Ss )− (us − Ss )+ αnk nk (uns k − Ss )− dxds ≤ αnk nk (uns k − Ss )− dxds 0

O

0

k=1

=

Z tZ 0

O

k=1

(us − Ss ) O

Nn X


k=1

With Lemma 1, we know that Z tZ Z tZ Nn X nk − n (˜ us − S˜s )ν(dxds) = 0. αk nk (us − Ss ) dxds → (us − Ss ) lim n→∞ 0

O

0

k=1

O

And from Lemma 5, we have Z tZ Nn X (ˆ uns − Ss )− αnk nk (uns k − Ss )− dxds → 0. 0

O

k=1

Therefore,

Z tZ 0

O

(Φ′ (s, u ˆns ) − Φ′ (s, Ss ))

Nn X

αnk nk (uns k − Ss )− dxds → 0.

k=1

Moreover, with Lemma 1, we have Z tZ Z tZ Nn X ′ n nk − Φ (s, Ss ) Φ′ (s, Ss )ν(dxds) αk nk (us − Ss ) dxds → 0

O

k=1

0

O

and Z tZ Z tZ Z tZ ′ ′ | Φ (s, us )ν(dxds) − Φ (s, Ss )ν(dxds)| ≤ C |˜ us − Ss |ν(dxds) 0 O 0 O 0 O Z tZ = C (˜ us − Ss )ν(dxds) = 0 0

O

Therefore, taking limit, we get the desired Itô’s formula.

5.4. Itô’s formula for the difference of the solutions of two RSPDEs We still consider (u, ν) solution of the linear equation as in Subsection 5.2 P P j j dut + Aut dt = ft dt + di=1 ∂i gti dt + +∞ j=1 ht dBt + ν(dt, x) u ≥ S

¯ respectively in and consider another linear equation with adapted coefficients f¯, g¯, h 2 2 d 2 L ([0, T ] × Ω × O; R), L ([0, T ] × Ω × O; R ) and L ([0, T ] × Ω × O; RN ) and obstacle S¯ which satisfies the same hypotheses (O) as S i.e; S¯0 ≤ ξ and S¯ is dominated by the solution of an SPDE (not necessarily the same as S). We denote by (y, ν¯) the unique solution to the associated SPDE with obstacle with initial condition y0 = u0 = ξ. P P ¯ j j ¯(dt, x) dyt + Ayt dt = f¯t dt + di=1 ∂i g¯ti dt + +∞ j=1 ht dBt + ν y ≥ S¯

23

Theorem 6. Let Φ as in Theorem 5, then the difference of the two solutions satisfy the following Itô’s formula for all t ∈ [0, T ]: Z t Z t Z ′ (Φ′ (s, us − ys ), fs − f¯s )ds E(Φ (s, us − ys ), us − ys )ds = Φ(t, ut (x) − yt (x))dx + −

d Z tZ X 0

i=1

+ +

0

0

O

Φ”(s, us − ys )∂i (us − ys )(gsi − g¯si )dxds +

O

j=1

+∞ Z t Z X

1 2 j=1 Z tZ 0

0

+∞ Z t X

O

¯ j )2 dxds + Φ”(s, us − ys )(hjs − h s

Φ′ (s, u˜s − y˜s )(ν − ν¯)(dx, ds)

Z tZ 0

O

0

¯ j )dB j (Φ′ (s, us − ys ), hjs − h s s

∂Φ (s, us − ys )dxds ∂s (17)

a.s.

O

Proof. We begin with the penalized solutions. The corresponding penalization equations are +∞ d X X i n n hjt dBtj + n(unt − St )− dt ∂i gt dt + dut + Aut dt = ft dt + j=1

i=1

and

dytm + Aytm dt = f¯t dt +

d X

∂i g¯ti dt +

+∞ X

¯ j dB j + m(y m − S¯t )− dt h t t t

j=1

i=1

from the proofs above, we know that the penalized solution converges weakly to the soluPNn′ n n′ P n n ni and yˆn = i=1 βi y i such tion and we can take convex combinations u ˆn = N i=1 αi u n n that (ˆ u )n and (ˆ y )n are non-decreasing and converge strongly to u and y respectively in L2 ([0, T ], H01 (O)). As in the proof of Theorem 5, we first establish a key lemma: Lemma 6. For all t ∈ [0, T ], lim E

n→+∞

Z tZ 0

′

O

u ˆns

lim E

n→+∞

0

Proof. We put for all n: ν n (ds, dx) =

Nn X

O

βkn n′k (ys k − S¯s )− dxds = E

Z tZ

αnk nk (uns k

Z tZ

n′

k=1

and Z tZ

Nn X

yˆsn

Nn X

−

− Ss ) dxds = E

0

0

k=1

u ˜ν¯(ds, dx),

O

y˜ν(ds, dx).

O

′

αnk nk (uns k − Ss )− dxds and ν¯n (ds, dx) =

k=1

Nn X k=1

As in the proof of Lemma 5, we write for all n: un = z + v n . In the same spirit, we introduce z¯ the solution of the linear spde: d¯ zt + A¯ zt = f¯t dt +

d X i=1

∂i g¯ti dt +

+∞ X j=1

′

n βkn n′k (ys k − S¯s )− dxds.

¯ j dB j , h t t


24

with initial condition z¯0 = ξ and put ∀n ∈ N, v¯n = y n − z¯, vˆ¯n = yˆn − z¯ and v¯ = y − z¯. As a consequence of Lemma II.6 in [21], we have for all n, P -almost surely: Z

t

1 k vt − v¯t k2 + 2

Z

1 k vˆtn − vˆ ¯tn k2 + 2

0

E(ˆ vsn − vˆ¯sn )ds =

and

Z tZ 0

t

E(vs − v¯s )ds =

O

Z tZ 0

0

(ˆ vsn − vˆ¯sn )(ν n − ν¯n )(dx, ds),

(˜ vs − v˜¯s )(ν − ν¯)(dx, ds).

O

But, as in the proof of Lemma 5, we get that vˆtn − vˆ¯tn tends to vt − v¯t in L2 (O) almost surely and Z tZ Z tZ n n v˜s ν(dx, ds), vˆs ν (dx, ds) = lim n

lim n

0

0

O

Z tZ 0

ˆsn ν¯n (dx, ds) = v¯

O

O

Z tZ 0

v˜¯s ν¯(dx, ds). O

This yields: Z tZ Z t Z Z tZ Z tZ n n n n ˆ v˜¯s ν(dx, ds). v˜s ν¯(dx, ds) + v¯s ν (dx, ds) = vˆs ν¯ (dx, ds) + lim n

0

0

O

0

O

0

O

O

But, we have lim sup n

Z tZ 0

O

vˆsn ν¯n (dx, ds) ≤ lim sup n

Z tZ 0

vs ν¯n (dx, ds) =

Z tZ 0

O

v˜s ν¯(dx, ds),

O

and in the same way: lim sup n

Z tZ 0

O

vˆ¯sn ν n (dx, ds) ≤

Z tZ 0

v˜¯s ν(dx, ds). O

Let us remark that these inequalities also hold for any subsequence. From this, it is easy to deduce that necessarily: lim n

and lim n

Z tZ

vˆsn ν¯n (dx, ds)

Z tZ

vˆ ¯sn ν n (dx, ds)

0

0

O

O

=

Z tZ 0

=

Z tZ 0

v˜s ν¯(dx, ds), O

v˜¯s ν(dx, ds). O

We end the proof of this lemma by using similar arguments as in the proof of Lemma 5. End of the proof of Theorem 6: We begin with the equation which u ˆn − yˆn satisfies: d(ˆ unt − yˆtn ) + A(ˆ unt − yˆtn )dt = (ft − f¯t )dt +

d X i=1

n

n

+ (ν − ν¯ )(x, dt)

∂i (gti − g¯ti )dt +

+∞ X j=1

¯ j )dB j (hjt − h t t

25

Applying Itô’s formula to Φ(ˆ un − yˆn ), we have Z

O

−

Φ(t, u ˆnt (x)

d Z tZ X 0

i=1

+ +

−

O

0

0

O

+

Z

t

E(Φ 0

′

(s, u ˆns

−

yˆsn ), uˆns

−

yˆsn )ds

=

+∞ Z X

t

Φ”(s, u ˆns − yˆsn )∂i (ˆ uns − yˆsn )(gsi − g¯si )dxds +

j=1

+∞ Z t Z X

1 2 j=1 Z tZ

yˆtn (x))dx

O

¯ j )2 dxds + Φ”(s, uˆns − yˆsn )(hjs − h s

Z tZ 0

O

Z

0

0

t

(Φ′ (s, uˆns − yˆsn ), fs − f¯s )ds

¯ j )dB j (Φ′ (s, uˆns − yˆsn ), hjs − h s s

∂Φ (s, u ˆns − yˆsn )dxds ∂s

Φ′ (s, uˆns − yˆsn )(ν n − ν¯n )(dx, dt)

Because that u ˆn and yˆn converge strongly to u and y respectively, the convergence of all the terms except the last term are clear. For the convergence of the last term, we do as follows: Z t Z Z tZ ′ n n ′ n n ′ n ′ n [Φ (s, u ˆs − yˆs ) − Φ (s, us − yˆs )]ν (dxds) + [Φ (s, us − yˆs ) − Φ (s, us − ys )]ν (dxds) 0 O 0 O Z tZ Z tZ |ˆ ysn − ys |ν n (dx, ds) |ˆ uns − us |ν n (dxds) + ≤C 0

0

O

O

As a consequence of Lemma 5 and using the fact that u ˆn ≤ u: Z tZ Z tZ n n (us − u ˆns )ν n (dxds) = 0. |ˆ us − us |ν (dxds) = lim lim n

0

n

O

0

O

By Lemma 6 and the fact that yˆn ≤ y: Z tZ Z tZ n n (ys − yˆsn )ν n (dx, ds) = 0, |ˆ ys − ys |ν (dx, ds) = lim lim n

0

n

O

this yields: lim n

Z tZ 0

O

0

O

(Φ′ (s, u ˆns − yˆsn ) − Φ′ (s, us − ys )ν n (dx, dt) = 0,

but by Lemma 1, we know that Z tZ Z tZ ′ n lim Φ (s, us − ys )ν (dx, dt) = Φ′ (s, u ˜s − y˜s )¯ ν (dx, dt), n

0

lim

Z tZ

so n

0

O

0

Φ O

′

(s, u ˆns

−

yˆsn )ν n (dx, dt)

=

O

Z tZ 0

Φ′ (s, u ˜s − y˜s )ν(dx, dt). O

In the same way, we prove: Z tZ Z tZ ′ n n n Φ′ (s, u ˜s − y˜s )¯ ν (dx, dt). Φ (s, u ˆs − yˆs )¯ ν (dx, dt) = lim n

0

O

The proof is now complete.

0

O


26

5.5. Proof of Theorem 4 in the nonlinear case Let γ and δ 2 positive constants. On L2 (Ω × [0, T ]; H01 (O)), we introduce the norm Z k u kγ,δ = E(

T

e−γs (δ k us k2 + k ∇us k2 )ds),

0

which clearly defines an equivalent norm on L2 (Ω × [0, T ]; H01 (O)). Let us consider the Picard sequence (un ) defined by u0 = ξ and for all n ∈ N we denote by (un+1 , ν n+1 the solution of the linear SPDE with obstacle (un+1 , ν n+1 ) = R(ξ, f (un , ∇un ), g(un , ∇un ), h(un , ∇un ), S). Then, by the Itô’s formula (17), we have −γT

k

Z

T

e

+2 +

Z

un+1 T −γs

e 0

−

unT

2

k +2

Z

T

−γs

e 0

E(un+1 s

(fˆs , un+1 − uns )ds − 2 s

d Z X i=1

T

ˆ s | k2 ds + 2 e−γs k |h

0

Z

T 0

Z

uns )ds

−

T

−γs

e

0

= −γ

Z

T 0

(ˆ gsi , ∂i (un+1 s

e−γs k un+1 − uns k2 ds s

−

uns ))ds

+2

+∞ Z X j=1

0

T

ˆ j , un+1 − un )dB j e−γs (h s s s s

e−γs (un+1 − uns )(ν n+1 − ν n )(dxds) s

O

ˆ = where fˆ = f (un , ∇un ) − f (un−1 , ∇un−1 ), gˆ = g(un , ∇un ) − g(un−1 , ∇un−1 ) and h n n n−1 n−1 h(u , ∇u )−h(u , ∇u ).Clearly, the last term is non-positive so using Cauchy-Schwarz’s inequality and the Lipschitz conditions on f , g and h, we have 2

Z

T

−γs

e 0

(un+1 s

−

uns , fˆs )ds

Z T Z 1 T −γs n+1 n 2 ≤ k fˆs k2 ds e k us − us k ds + ǫ ǫ 0 0 Z T Z 1 T −γs n 2 e−γs k uns − usn−1 k2 ds e k un+1 − u k ds + Cǫ ≤ s s ǫ 0 0 Z T e−γs k ∇(uns − usn−1 ) k2 ds + Cǫ 0

and 2

d Z X i=1

T

−γs

e 0

(ˆ gsi , ∂i (un+1 s

−

uns ))ds

≤2

Z

T 0

e−γs k ∇(un+1 − uns )(C k uns − usn−1 k s

T

C e−γs k ∇(un+1 − uns ) k2 ds + +α k ∇(uns − usn−1 ) k)ds ≤ Cǫ s ǫ 0 Z T Z T e−γs k uns − usn−1 k2 ds e−γs k ∇(un+1 − uns ) k2 ds + α +α s Z

0

−γs

e

ˆ s | k2 ds ≤ C(1 + 1 ) k |h ǫ

T

0

e−γs k uns − usn−1 k2 ds

0

0

and Z T

Z

Z

T

−γs

e 0

k

uns

−

usn−1

2

2

k ds + β (1 + ǫ)

Z

0

T

e−γs k ∇(uns − usn−1 ) k2 ds

27

where C, α and β are the constants in the Lipschitz conditions. Using the elliptic condition and taking expectation, we get: Z T Z T 1 −γs n+1 n 2 e−γs k ∇(un+1 − uns ) k2 ds ≤ e k us − us k ds + (2λ − α)E (γ − )E s ǫ 0 0 Z T Z T 2 −γs n n−1 2 2 C(1 + ǫ + ) e−γs k ∇(uns − usn−1 ) k2 ds e k us − us k ds + (Cǫ + α + β (1 + ǫ))E ǫ 0 0

We choose ǫ small enough and then γ such that Cǫ + α + β 2 (1 + ǫ) < 2λ − α and If we set δ =

γ−1/ǫ 2λ−α ,

k un+1 −un kγ,δ ≤

C(1 + ǫ + 2/ǫ) γ − 1/ǫ = 2λ − α Cǫ + α + β 2 (1 + ǫ)

we have the following inequality:

Cǫ + α + β 2 (1 + ǫ) n Cǫ + α + β 2 (1 + ǫ) k un −un−1 kγ,δ ≤ ... ≤ ( ) k u1 kγ,δ 2λ − α 2λ − α 2

(1+ǫ) n when n → ∞, ( Cǫ+α+β ) → 0, we deduce that un converges strongly to u in L2 (Ω × 2λ−α [0, T ]; H01 (O)). Moreover, as (un+1 , ν n+1 ) = R(ξ, f (un , ∇un ), g(un , ∇un ), h(un , ∇un ), S), we have for any ϕ ∈ D: Z t Z t d Z t X n+1 n (gsi (uns , ∇uns ), ∂i ϕs )ds E(u , ϕ )ds + (u , ∂ ϕ )ds + (un+1 , ϕ ) − (ξ, ϕ ) − s s s t 0 s s t 0

0

=

Z

t

0

(fs (uns , ∇uns ), ϕs )ds +

+∞ Z t X j=1

0

0

i=1

(hjs (uns , ∇uns ), ϕs )dBsj +

Z tZ 0

ϕs (x)ν n+1 (dxds). a.s.

O

Let v n+1 the random parabolic potential associated to ν n+1 : ν n+1 = ∂t v n+1 + Av n+1 . We denote z n+1 = un+1 − v n+1 , so z n+1 = U (ξ, f (un , ∇un ), g(un , ∇un ), h(un , ∇un )) converges strongly to z in L2 (Ω×[0, T ]; H01 (O)). As a consequence of the strong convergence of un+1 , we deduce that v n+1 converges strongly to v in L2 (Ω × [0, T ]; H01 (O)). Therefore, for fixed ω, Z t Z t Z t Z t ∂s ϕs ∂s ϕs n+1 (− E(ϕs , vs )ds = lim (− E(ϕs , vsn+1 )ds ≥ 0 , vs )ds + , vs )ds + ∂s ∂s 0 0 0 0

i.e. v(ω) ∈ P. Then from Proposition 1, we obtain a regular measure associated with v, and ν n+1 converges vaguely to ν. Taking the limit, we obtain Z t Z t d Z t X (gsi (us , ∇us ), ∂i ϕs )ds E(us , ϕs )ds + (us , ∂s ϕs )ds + (ut , ϕt ) − (ξ, ϕ0 ) − 0

0

=

Z

0

t

(fs (us , ∇us ), ϕs )ds +

+∞ Z X j=1

t

0

i=1


0

Z tZ 0

O

ϕs (x)ν(dx, ds), a.s..


28

From the fact that u and z are in HT , we know that v is also in HT , by definition, ν is a random regular measure. 6. Comparison theorem 6.1. A comparison Theorem in the linear case We first establish a comparison theorem for the solutions of linear SPDE with obstacle in the case where the obstacles are the same, this gives a comparison between the regular measures. So, for this part only, we consider the same hypotheses as in the Subsection 5.2. So we consider adapted processes f , g, h respectively in L2 ([0, T ] × Ω × O; R), L2 ([0, T ] × Ω × O; Rd ) and L2 ([0, T ]×Ω×O; RN ), an obstacle S which satisfies assumption (O) and ξ ∈ L2 (Ω×O) is an F0 -measurable random variable such that ξ ≤ S0 . We denote by (u, ν) be the solution of R(ξ, f, g, h, S). We are given another ξ ′ ∈ L2 (Ω × O) is F0 -measurable and such that ξ ′ ≤ S0 and another adapted process f ′ in L2 ([0, T ] × Ω × O; R). We denote by (u′ , ν ′ ) the solution of R(ξ ′ , f ′ , g, h, S). We have the followingg comparison theorem: Theorem 7. Assume that the following conditions hold 1. ξ ≤ ξ ′ , dx ⊗ dP − a.e.; 2. f ≤ f ′ , dt ⊗ dx ⊗ dP − a.e.. Then for almost all ω ∈ Ω, u ≤ u′ , q.e. and ν ≥ ν ′ in the sense of distribution. Proof. We consider the following two penalized equations: dunt

=

Aunt dt

+ ft dt +

d X

∂i gti dt

+

′

′

d X i=1

hjt dBtj + n(unt − St )− dt

j=1

i=1

dutn = Autn dt + ft′ dt +

+∞ X

∂i gti dt +

+∞ X

′

hjt dBtj + n(utn − St )− dt

j=1

we denote Ft (x, unt ) = ft (x) + n(unt − St )− Ft′ (x, unt ) = ft′ (x) + n(unt − St )− with assumption 2 we have that Ft (x, unt ) ≤ Ft′ (x, unt ), dt ⊗ dx ⊗ dP − a.e., therefore, from the comparison theorem for SPDE (without obstacle, see [D]), we know that ∀t ∈ [0, T ], ′ ′ unt ≤ utn , dx ⊗ dP − a.e., thus n(unt − St )− ≥ n(utn − St )− . The results are immediate consequence of the construction of (u, ν) and (u′ , ν ′ ) given in Subsection 5.2.

29

6.2. A comparison theorem in the general case We now come back to the general setting and still consider (u, ν) = R(ξ, f, g, h, S) the solution of the SPDE with obstacle  d  X    ∂i gi (t, x, ut (x), ∇ut (x))dt du (x) = Lu (x)dt + f (t, x, u (x), ∇u (x))dt + t t t t     i=1  +∞ X  hj (t, x, ut (x), ∇ut (x))dBtj + ν(x, dt) +     j=1     u ≥ S , u0 = ξ,

where we assume hypotheses (H), (I) and (O). We consider another coefficients f ′ which satisfies the same assumptions as f , another obstacle S ′ which satisfies (O) and another initial condition ξ ′ belonging to L2 (Ω × O) and F0 adapted such that ξ ′ ≥ S0′ . We denote by (u′ , ν ′ ) = R(ξ ′ , f ′ , g, h, S ′ ). Theorem 8. Assume that the following conditions hold 1. ξ ≤ ξ ′ , dx ⊗ dP − a.e. 2. f (u, ∇u) ≤ f ′ (u, ∇u), dtdx ⊗ P − a.e. 3. S ≤ S ′ , dtdx ⊗ P − a.e. Then for almost all ω ∈ Ω, u(t, x) ≤ u′ (t, x), q.e..

We put u ˆ = u − u′ , ξˆ = ξ − ξ ′ , fˆt = f (t, ut , ∇ut ) − f ′ (t, u′t , ∇u′t ), gˆt = g(t, ut , ∇ut ) − ′ 2 ˆ t = h(t, ut , ∇ut ) − h(t, u′ , ∇u′ ). The main idea is to evaluate E k u g(t, ut , ∇u′t ) and h ˆ+ t t t k , thanks to Itô’s formula and then apply Gronwall’s inequality. Therefore, we start by the following lemma Lemma 7. For all t ∈ [0, T ], we have Z t Z t + ˆ + 2 ˆ (∇ˆ u+ ˆs )ds (ˆ us , fs )ds − 2E = E k ξ k +2E k +2E Ek s ,g 0 0 0 Z t Z tZ + ′ ˆ s | k2 ds a.s.. k I{ûs >0} |h (18) u ˆs (x)(ν − ν )(dxds) + E +2E u ˆ+ t

Z

2

0

O

t

E(ˆ u+ s )ds

0

Proof. We approximate the function ψ : y ∈ R → (y + )2 by a sequence (ψn ) of regular functions: let ϕ be a C ∞ increasing function such that ∀y ∈] − ∞, 1], ϕ(y) = 0 and ∀y ∈ [2, +∞[, ϕ(y) = 1. We set for all n: ∀y ∈ R, ψn (y) = y 2 ϕ(ny). It is easy to verify that (ψn ) converges uniformly to the function ψ and that moreover we have the estimates: ∀y ∈ R+ , ∀n, 0 ≤ ψn (y) ≤ ψ(y), 0 ≤ ψn′ (y) ≤ Cy, |ψ”n (y)| ≤ C.


30

Thanks to Theorem 6, for all n and t ∈ [0, T ] we have dx ⊗ dP −almost everywhere Z t ˆ (ψn′ (ˆ us ), fˆs )ds ψn (ξ)dx + E =E ψn (ˆ us )dx + E E 0 O 0 O Z tZ Z t Z tZ 1 ˆ 2 dxds ψn′ (ˆ us )ˆ ν (dxds) + E (∇ψn′ (ˆ us ), gˆs )ds + E −E ψ”n (ˆ us )h s 2 0 O 0 0 O Z

Z

t

Z

E(ψn′ (ˆ us ), u ˆs )ds

we take the limit and get the desired result. 2 Proof of Theorem 8: Applying Itô’s formula (18) to (ˆ u+ t ) , we have

Ek

u ˆ+ t

+E

Z

t

Z

t

k +2E

0

0

t

Z

t

Z

ˆ (ˆ u+ s , fs )ds

(ˆ u+ ˆs )ds + 2E I{uˆs >0} E(ˆ us )ds = 2E s ,g 0 0 0 Z tZ 2 ˆ (us − u′s )+ (x)(ν − ν ′ )(dx, ds). k I{ûs >0} |hs | k ds + 2E 2

O

As we assume that f (u, ∇u) ≤ f ′ (u, ∇u), ′ ′ ′ ′ ′ ˆ u ˆ+ ˆ+ ˆ+ s fs = u s {f (s, us , ∇us ) − f (s, us , ∇us )} + u s {f (s, us , ∇us ) − f (s, us , ∇us )} ′ ′ ′ ′ ≤ u ˆ+ s {f (s, us , ∇us ) − f (s, us , ∇us )}.

then with the Lipschitz condition, using Cauchy-Schwartz’s inequality, we have the following relations: E

Z

0

E

t 0

ˆ (ˆ u+ s , fs )ds t

C ≤ (C + )E ǫ

t

ˆ s | k2 ds ≤ CE k I{ûs >0} |h

Z

0

(∇ˆ u+ ˆs ) s ,g

k

0

Z

Z

t

Z

ǫ+α E ≤ λ

E Z

t

0

0

t

u ˆ+ s

Cǫ k ds + E λ 2

E(ˆ u+ s )ds

C + E ǫ

2 ku ˆ+ s k ds +

t

Z

β2

0

Z

0

t

E(ˆ u+ s )ds.

2 ku ˆ+ s k ds

+ǫ E λ

Z

t 0

E(ˆ u+ s )ds.

RtR

The last term is equal to −2E 0 O (us − u′s )+ (x)ν ′ (dx, ds) ≤ 0, because that on {u ≤ u′ }, (u − u′ )+ = 0 and on {u > u′ }, ν(dx, ds) = 0. Thus we have the following inequality Ek

u ˆ+ t

2α + 2ǫ 2Cǫ β 2 + ǫ k +(2 − − − )E λ λ λ 2

we can take ǫ small enough such that 2 −

2α+2ǫ λ

2 Eku ˆ+ t k ≤ CE

Z

0

− t

Z

t 0

E(ˆ u+ s )ds

2Cǫ λ

−

β 2 +ǫ λ

≤ CE

Z

t 0

2 ku ˆ+ s k ds.

> 0, we have

2 ku ˆ+ s k ds,

then we deduce the result from the Gronwall’s lemma.

Remark 6. Applying the comparison theorem to the same obstacle gives another proof of the uniqueness of the solution.

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References [1] Aronson D.G.: On the Green’s function for second order parabolic differential equations with discontinuous coefficients. Bulletin of the American Mathematical Society, 69, 841-847 (1963). [2] Aronson, D.G. Non-negative solutions of linear parabolic equations, Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 3, tome 22 (4), pp. 607-694 (1968):. [3] Bally V., Caballero E., El-Karoui N. and Fernandez, B. : Reflected BSDE’s PDE’s and Variational Inequalities. preprint INRIA report (2004). [4] Bally V. and Matoussi, A. : Weak solutions for SPDE’s and Backward Doubly SDE’s. J. of Theoret.Probab. 14, 125-164 (2001). [5] Bensoussan A. and Lions J.-L. Applications des Inéquations variationnelles en contrôle stochastique. Dunod, Paris (1978). [6] Charrier P. and Troianiello G.M. Un résultat d’existence et de régularité pour les solutions fortes d’un problème unilatéral d’évolution avec obstacle dépendant du temps. C.R.Acad. Sc. Paris, 281, série A, p. 621 (1975). [7] Denis L.: Solutions of SPDE considered as Dirichlet Processes. Bernoulli Journal of Probability, 10(5), 783-827 (2004). [8] Denis L. and Stoïca L.: A general analytical result for non-linear s.p.d.e.’s and applications.Electronic Journal of Probability, 9, p. 674-709 (2004). [9] Denis L., Matoussi A. and Stoïca L.: Lp estimates for the uniform norm of solutions of quasilinear SPDE’s. Probability Theory Related Fields, 133, 437-463 (2005). [10] Denis L., Matoussi A. and Stoïca L.: Maximum principle for parabolic SPDE’s: first approach, Stohcastic Partial Differential Equations and Applications VIII, Levico, Jan. 6-12 (2008). [11] Denis L., Matoussi A. and Stoïca L.: Maximum Principle and Comparison Theorem for Quasi-linear Stochastic PDE’s. Electronic Journal of Probability, 14, p. 500-530 (2009). [12] Donati-Martin C. and Pardoux E.: White noise driven SPDEs with reflection. Probability Theory and Related Fields, 95, 1-24 (1993). [13] El Karoui N., Kapoudjian C., Pardoux E., Peng S., and Quenez M.C.:Reflected Solutions of Backward SDE and Related Obstacle Problems for PDEs. The Annals of Probability, 25 (2), 702-737 (1997). [14] Klimsiak T.: Reflected BSDEs and obstacle problem for semilinear PDEs in divergence form. Stochastic Processes and their Applications, 122 (1), 134-169 (2012). [15] Lions J.L. and Magenes E.: Problèmes aux limites non homogènes et applications. 1 , Dunod, Paris (1968). [16] Matoussi, A. Xu, M. : Sobolev solution for semilinear PDE with obstacle under monotonicity condition. Electronic Journal of Probability 13, 1035-1067 (2008). [17] Matoussi A. and Stoïca L.:The Obstacle Problem for Quasilinear Stochastic PDE’s. The Annals of Probability, 38, 3, 1143-1179 (2010). [18] Mignot F. and Puel J.P. : Inéquations d’évolution paraboliques avec convexes dépendant du temps. Applications aux inéquations quasi-variationnelles d’évolution. Arch. for Rat. Mech. and Ana., 64, No.1, 59-91 (1977). [19] Nualart D. and Pardoux E.: White noise driven quasilinear SPDEs with reflection. Probability Theory and Related Fields, 93, 77-89 (1992).

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[20] Pierre M.: Problèmes d’Evolution avec Contraintes Unilaterales et Potentiels Parabolique. Comm. in Partial Differential Equations, 4(10), 1149-1197 (1979). [21] Pierre M. : Représentant Précis d’Un Potentiel Parabolique. Séminaire de Théorie du Potentiel, Paris, No.5, Lecture Notes in Math. 814, 186-228 (1980). [22] Riesz, F. and Nagy, B. . Functional Analysis. Dover, New York, 1990. [23] Sanz M. , Vuillermot P. (2003) : Equivalence and Hölder Sobolev regularity of solutions for a class of non-autonomous stochastic partial differential equations. Ann. I. H. Poincaré , 39 (4) 703-742. [24] Walsh, J.B. : An introduction to stochastic partial differential equations. Ecole d’Eté de St-Flour XIV, 1984, Lect. Notes in Math, Springer Verlag , 1180 , 265-439 (1986). [25] Xu T.G. and Zhang T.S.: White noise driven SPDEs with reflection: Existence, uniqueness and large deviation principles Stochatic processes and their applications, 119, 3453-3470 (2009).

Laurent DENIS Laboratoire d’Analyse et Probabilités Université d’Evry Val d’Essonne Rue du Père Jarlan F-91025 Evry Cedex, FRANCE e-mail: [email protected]

Anis MATOUSSI LUNAM Université, Université du Maine Fédération de Recherche 2962 du CNRS Mathématiques des Pays de Loire Laboratoire Manceau de Mathématiques Avenue Olivier Messiaen F-72085 Le Mans Cedex 9, France email : [email protected] and CMAP, Ecole Polytechnique, Palaiseau

Jing ZHANG Laboratoire d’Analyse et Probabilités Université d’Evry Val d’Essonne Rue du Père Jarlan F-91025 Evry Cedex, FRANCE Email: [email protected]