L Sylla
email: sylla.lamine@ugb.edu.sn

REFLECTED BACKWARD STOCHASTIC DIFFERENTIAL EQUATION WITH JUMPS AND VISCOSITY SOLUTION OF SECOND ORDER INTEGRO-DIFFERENTIAL EQUATION WITHOUT MONOTONICITY CONDITION: CASE WITH THE MEASURE OF LÉVY INFINITE

Keywords: Integro-partial differential equation, Reflected stochastic differential equations with jumps, Viscosity solution, Non-local operator MSC 2010 subject classifications: 35D40, 35R09, 60H30

We consider the problem of viscosity solution of integro-partial differential equation(IPDE in short) with one obstacle via the solution of reflected backward stochastic differential equations(RBSDE in short) with jumps. We show existence and uniqueness of a continuous viscosity solution of equation with non local terms, in case the generator is not monotonous and Levy's measure is infinite.

Introduction

We consider the following system of integro-partial differential equation with one-obstacle ℓ, which is a function of (t, x): ∀i ∈ {1, . . . , m},          min u i (t, x)ℓ(t, x); -∂ t u i (t, x)b(t, x) ⊤ D x u i (t, x) - 1 2 Tr(σσ ⊤ (t, x)D 2 xx u i (t, x)) -K i u i (t, x)h (i) (t, x, u i (t, x), (σ ⊤ D x u i )(t, x), B i u i (t, x)) = 0, (t, x) ∈ [0, T ] × R k ; u i (T, x) = g i (x);

(1.1)

where the operators B i and K i are defined as follows:

B i u i (t, x) = E γ i (t,
x, e)(u i (t, x + β(t, x, e))u i (t, x))λ(de);

(1.2)

K i u i (t, x) = E (u i (t, x + β(t,
x, e))u i (t, x)β(t, x, e) ⊤ D x u i (t, x))λ(de).

The resolution of (1.1) is in connection with the following system of backward stochastic differential equations with jumps and one-obstacle ℓ: (ii) Y i;t,x s ≥ ℓ(s, X t,x s ) and T 0 (Y i;t,x s ℓ(s, X t,x s ))dK i;t,x s = 0;

             (i) dY i;t,x s = -f (i) (s, X t,
(1.3) and the following standard stochastic differential equation of diffusion-jump type:

X t,x s = x + s t b(r, X t,x r ) dr + s t σ(r, X t,x r ) dB r + s t E β(r, X t,x r-, e) µ(dr, de),

for s ∈ [t, T ] and X t,x s = x if s ≤ t.

It is recalled that pioneering work was done for the resolution of (1.1), among these works we can mention those of Barles and al. [START_REF] Barles | Backward stochastic differential equations and integral-partial differential equations[END_REF] in case without obstacle, Harraj and al. [START_REF] Harraj | Double barriers Reflected BSDEs with jumps and viscosity solutions of parabolic Integro-differential PDEs[END_REF] in the case with two obstacles; with as common point the hypothesis of monotony on the generator and γ ≥ 0. But recently Hamadène and Morlais relaxed these conditions with λ(.) finite [START_REF] Hamadène | Viscosity solutions for second order integro-differential equations without monotonicity condition: The probabilistic Approach[END_REF].

In this work we propose to solve (1.1) by relaxing the monotonicity of the generator and the positivity of γ and assuming that λ = ∞.

Our paper is organized as follows: in the next section we give the notations and the assumptions of our objects; in section 3 we recall a number of existing results; in section 4 we build estimates and properties for a good resolution of our problem; section 5 is reserved to give our main result and the section 6 for doing an extension of our result.

And in the end, classical definition of the concept of viscosity solution is put in appendix.

Notations and assumptions

Let (Ω, F, (F t ) t≤T , P) be a stochastic basis such that F 0 contains all P-null sets of F, and F t = F t+ := ǫ>0 F t+ǫ , t ≥ 0, and we suppose that the filtration is generated by the two mutually independents processes:

(i) B := (B t ) t≥0 a d-dimensional Brownian motion and, (ii) a Poisson random measure µ on R + ×E where E := R ℓ -{0} is equipped with its Borel field E (ℓ ≥ 1). The compensator ν(dt, de) = dtλ(de) is such that { µ([0, t] × A) = (µλ)([0, t] × A)} t≥0 is a martingale for all A ∈ E satisfying λ(A) < ∞. We also assume that λ is a σ-finite measure on (E, E), integrates the function (1∧ | e | 2 ) and λ(E) = ∞.

Let's now introduce the following spaces:

(iii) P (resp. P) the field on [0, T ] × Ω of F t≤T -progressively measurable (resp. predictable) sets.

(iv) For κ ≥ 1, L 2 κ (λ) the space of Borel measurable functions ϕ := (ϕ(e)) e∈E from E into R κ such that ϕ 2

L 2 κ (λ) = E |ϕ(e)| 2
κ λ(de) < ∞; L 2 1 (λ) will be simply denoted by L 2 (λ); (v) S 2 (R κ ) the space of RCLL (for right continuous with left limits) P-measurable and R κvalued processes such that E[sup s≤T |Y s | 2 ] < ∞; A 2 c is its subspace of continuous non-decreasing processes (K t ) t≤T such that K 0 = 0 and E (K T ) 2 < ∞;

(vi) H 2 (R κ×d ) the space of processes Z := (Z s ) s≤T which are P-measurable, R κ×d -valued and satisfying

E T 0 |Z s | 2 ds < ∞;
(vii) H 2 (L 2 κ (λ)) the space of processes U := (U s ) s≤T which are P-measurable, L 2 κ (λ)-valued and satisfying

E T 0 U s (ω) 2 L 2 κ (λ) ds < ∞;
(viii) Π g the set of deterministics functions ̟ : (t, x) ∈ [0, T ] × R κ → ̟(t, x) ∈ R of polynomial growth, i.e., for which there exists two non-negative constants C and p such that for any (t, x)

∈ [0, T ] × R κ , |̟(t, x)| ≤ C(1 + |x| p ).
The subspace of Π g of continuous functions will be denoted by Π c g ; (ix) U the subclass of Π c g which consists of functions Φ : (t, x) ∈ [0, T ] × R κ → R such that for some non-negative constants C and p we have Φ(t, x) -Φ(t, x ′ ) ≤ C(1 + |x| p + x ′ p ) xx ′ , for any t, x, x ′ .

(x) For any process θ := (θ s ) s≤T and t ∈ (0, T ], θ t-= lim sրt θ s and ∆ t θ = θ tθ t-. Now let b and σ be the following functions:

b : (t, x) ∈ [0, T ] × R k → b(t, x) ∈ R k ; σ : (t, x) ∈ [0, T ] × R k → σ(t, x) ∈ R k×d .
We assume that they are jointly continuous in (t, x) and Lipschitz continuous w.r.t. x uniformly in t, i.e., there exists a constant C such that,

∀(t, x, x ′ ) ∈ [0, T ] × R k+k , b(t, x) -b(t, x ′ ) + σ(t, x) -σ(t, x ′ ) ≤ C x -x ′ .
(2.5)

Let us notice that by (2.5) and continuity, the functions b and σ are of linear growth, i.e., there exists a constant C such that

∀(t, x, x ′ ) ∈ [0, T ] × R k+k , |b(t, x)| + |σ(t, x)| ≤ C |1 + x| . (2.6) Let β : (t, x, e) ∈ [0, T ] × R k × E → β(t,
x, e) ∈ R k be a measurable function such that for some real constant C, and for all e ∈ E,

(i) |β(t, x, e)| ≤ C(1 ∧ |e|); (2.7) (ii) β(t, x, e) -β(t, x ′ , e) ≤ C x -x ′ (1 ∧ |e|); (iii) the mapping (t, x) ∈ [0, T ] × R k → β(t, x, e) ∈ R k is continuous for any e ∈ E.
We are now going to introduce the objects which are specifically connected to the RBSDE with jumps we will deal with. Let ℓ the barrier of (1.3); (g i ) i=1,m and (h (i) ) i=1,m be two functions defined as follows: for i = 1, . . . , m,

g i : R k -→ R m x -→ g i (x)
and

h (i) : [0, T ] × R k+m+d+1 -→ R (t, x, y, z, q) -→ h (i) (t, x, y, z, q).
Moreover we assume they satisfy:

(H1): The reflecting barrier ℓ is real valued and P-measurable process satisfying, ℓ ∈ U i.e., it is continuous and there exists constants C and p such that,

|ℓ(t, x) -ℓ(t, x ′ )| ≤ C(1 + |x| p + |x ′ | p ) |x -x ′ |, for any t ≥ 0, x, x ′ .
(H2): For any i ∈ {1, . . . , m}, the function g i belongs to U .

(H3): For any i ∈ {1, . . . , m}, (i) the function h (i) is Lipschitz in (y, z, q) uniformly in (t, x), i.e., there exists a real constant C such that for any

(t, x) ∈ [0, T ] × R k , (y, z, q) and (y ′ , z ′ , q ′ ) elements of R m+d+1 , h (i) (t, x, y, z, q) -h (i) (t, x, y ′ , z ′ , q ′ ) ≤ C( y -y ′ + z -z ′ + q -q ′ ); (2.8) (ii) the (t, x) → h (i) (t,
x, y, z, q), for fixed (y, z, q) ∈ R m+d+1 , belongs uniformly to U , i.e., it is continuous and there exists constants C and p (which do not depend on (y, z, q)) such that,

h (i) (t, x, y, z, q) -h (i) (t, x ′ , y, z, q) ≤ C(1 + |x| p + x ′ p ) x -x ′ , for any t ≥ 0, x, x ′ . (2.9) Next let γ i , i = 1, . . . , m be Borel measurable functions defined from [0, T ] × R k × E into R and satisfying: (i) γ i (t, x, e) ≤ C(1 ∧ |e|); (ii) γ i (t, x, e) -γ i (t, x ′ , e) ≤ C(1 ∧ |e|) x -x ′ (1 + |x| p + x ′ p ); (2.10) (iii) the mapping t ∈ [0, T ] → γ i (t, x, e) ∈ R is continuous for any (x, e) ∈ R k × E.
Finally we introduce the following functions (f (i) ) i=1,m defined by:

∀(t, x, y, z, ζ) ∈ [0, T ]×R k+m+d ×L 2 (λ), f (i) (t, x, y, z, ζ) := h (i) t, x, y, z, E γ i (t,
x, e)ζ(e)λ(de) .

(2.11)

The functions (f (i) ) i=1,m , enjoy the two following properties:

(a) The function f (i) is Lipschitz in (y, z, ζ) uniformly in (t, x), i.e.
, there exists a real constant C such that

f (i) (t, x, y, z, ζ) -f (i) (t, x, y ′ , z ′ , ζ ′ ) ≤ C( y -y ′ + z -z ′ + ζ -ζ ′ L 2 (λ) );
(2.12) since h (i) is uniformly Lipschitz in (y, z, q) and γ i verifies (2.8)-(i);

(b) The function (t, x) ∈ [0, T ] × R k → f (i) (t, x, 0, 0, 0) belongs to Π c g ;
and then

E T 0 f (i) (r, X t,x r , 0, 0, 0) 2 dr < ∞.
Having defined our data and put our assumptions, we can look at the state of the art.

Preliminaires

A class of diffusion processes with jumps

Let (t, x) ∈ [0, T ] × R d and (X t,x s ) s≤T be the stochastic process solution of (1.4). Under assumptions (2.5)-(2.7) the solution of Equation (1.4) exists and is unique (see [START_REF] Fujiwara | Stochastic differential equations of jump type and Lévy processes in differomorphism group[END_REF] for more details). We state some properties of the process {(X t,x s ), s ∈ [0, T ]} which can found in [START_REF] Fujiwara | Stochastic differential equations of jump type and Lévy processes in differomorphism group[END_REF].

Proposition 3.1 For each t ≥ 0, there exists a version of {(X t,x s ), s ∈ [t, T ]} such that s → X t s is a C 2 (R d )-valued rcll process. Moreover it satisfies the following estimates: ∀p ≥ 2, x, x ′ ∈ R d and s ≥ t, E[ sup t≤r≤s X t,x r -x p ] ≤ M p (s -t)(1 + |x| p ); E[ sup t≤r≤s X t,x r -X t,x ′ r -(x -x ′ ) p p ] ≤ M p (s -t)( x -x ′ p ); (3.13)
for some constant M p .

Existence and uniqueness for a RBSDE with jumps

Let (t, x) ∈ [0, T ] × R d and we consider the following m-dimensional RBSDE with jumps: Its proof is given in [START_REF] Hamadène | Reflected backward stochastic differential equation with jumps and random obstacle[END_REF] by using the penalization method (see p .5-12) and the Snell envelope method (see p. 14-16). Proposition 3.2 Assume that assumptions (H1), (H2) and (H3) hold. Then for any (t, x) ∈ [0, T ] × R d , the RBSDE (3.14) has an unique solution ( Y t,x , Z t,x , U t,x , K t,x ).

                       (i) Y t,x := (Y i,t,x ) i=1,m ∈ S 2 (R m ), Z t,x := (Z i,t,x ) i=1,m ∈ H 2 (R m×d ), K t,x := (K i,t,x ) i=1,m ∈ A 2 c , U t,x := (U i,t,x ) i=1,m ∈ H 2 (L 2 m (λ)); (ii) dY i;t,x s = -f (i) (s, X

Remark 1

The solution of this RBSDE with jumps exist and is unique since: (ii) for any fixed (t, x, y, z) ∈ [0, T ] × R k+m+d , the mapping

(i) E g(X t,x T ) 2 < ∞,
(q ∈ R) -→ h (i) (t, x, y, z, q) ∈ R is non-decreasing.
The function (u i ) i=1,m is a continuous viscosity solution (in Barles and al. 's sense, see Definition 6.1 in the Appendix) of (1.1).

For the proof see [START_REF] Harraj | Double barriers Reflected BSDEs with jumps and viscosity solutions of parabolic Integro-differential PDEs[END_REF] for the same way.

Finally, the solution (u i ) i=1,m of (1.1) is unique in the class Π c g .

Remark 2 (see [START_REF] Harraj | Double barriers Reflected BSDEs with jumps and viscosity solutions of parabolic Integro-differential PDEs[END_REF]) Under the assumptions (H1), (H2), (H3), there exists a unique viscosity solution of (1.1) in the class of functions satisfying

lim |x|→+∞ |u(t, x)| e -A[log(|x|)] 2 = 0 (3.16)
uniformly for t ∈ [0, T ], for some A > 0.

Estimates and properties

In this section we provide estimates for the functions (u i ) i=1,m defined in (3.15). Recall that, ( Y t,x , Z t,x , U t,x , K t,x ) := ((Y i;t,x ) i=1,m , (Z i;t,x ) i=1,m , (U i;t,x ) i=1,m , (K i;t,x ) i=1,m ) is the unique so- Next for any i = 1, . . . , m, the functions u i , g i and (t, x) → f (i) (t, x, 0, 0, 0) are of polynomial growth and finally y → f (i) (t, x, y, 0, 0) is Lipschitz uniformly w.r.t. (t, x). Then for some C and ρ ≥ 0,

Y i;t,x s = g i (X t,x T ) + T s f (i) (r, X t,x r , (u j (X t,x r )) j=1,m , Z i;t,x r , U i;t,x s )dr + K i;t,x T -K i;t,x s - T s Z i;t,x r dB r - T s E U i;t,
E g i (X t,x T ) p + T s f (i) (r, X t,x r , (u j (X t,x r )) j=1,m , 0, 0) 2 dr p 2 ≤ C (1 + |x| ρ ) . (4.20) Let us now fix i 0 ∈ {1, . . . , m}, ∀s ∈ [t, T ],                      (i) Y i 0 ,t,x s ∈ S 2 (R), Z i 0 ,t,x s ∈ H 2 (R d ), K i 0 ,t,x s ∈ A 2 c , U i 0 ,t,x s ∈ H 2 (L 2 (λ)); (ii) Y i 0 ,t,x s = g i 0 (X t,x T ) + T s f (i 0 ) (r, X t,x r , (u j (X t,x r )) j=1,m , Z i 0 ,t,x r , U i 0 ,t,x r )dr + K i 0 ,t,x T -K i 0 ,t,x s - T s Z i 0 ,t,x r dB r - T s E U i 0 ,t,x r (e) µ(dr, de). (iii) Y i 0 ,t,x s ≥ ℓ(s, X t,x s ) and T 0 (Y i 0 ,t,x s -ℓ(s, X t,x s ))dK i 0 ,t,x s = 0.
Applying Itô formula to Y i 0 ,t,x s 2 between s and T , we have

Y i 0 ,t,x s 2 + T s Z i 0 ,t,x r 2 dr + s≤r≤T (∆Y i 0 ,t,x r ) 2 (4.21) = g i 0 (X t,x T ) 2 + 2 T s Y i 0 ,t,x r f (i 0 ) (r, X t,x r , (u j (X t,x r )) j=1,m , Z i 0 ,t,x r , U i 0 ,t,x r ) dr + 2 T s Y i 0 ,t,x r dK i 0 ,t,x r -2 T s E Y i 0 ,t,x r U i 0 ,t,x r µ(dr, de) -2 T s Y i 0 ,t,x r Z i 0 ,t,x r dB r .
Notice that Y i 0 ,t,x r = u i 0 (r, X t,x r ) and we have that

|u i 0 (r, X t,x r )| ≤ C(1 + |X t,x r | q ). Next let us set Σ = 1 + sup s≤T |X t,x s |. Therefore |Y i 0 ,t,x r
| ≤ C q Σ q and g i 0 (X t,x T ) ≤ C q Σ q . By raising to the power p 2 and then taking expectation, it follows from (4.21) and the fact of, there exists C > 0 such that,

E T s U i 0 ,t,x r 2 L 2 (λ) dr p 2 ≤ CE s≤r≤T (∆Y i 0 ,t,x r ) 2 p 2
, (see [START_REF] Lenglart | Présentation unifiée de certaines inégalités de la théorie des martingales[END_REF] p. 28-45),

E Y i 0 ,t,x s p + E T s Z i 0 ,t,x r 2 dr p 2 + E T s U i 0 ,t,x r 2 L 2 (λ) dr p 2 ≤ 5 p 2 -1 E g i 0 (X t,x T ) p + 5 p 2 -1 E T s 2 Y i 0 ,t,x r f (i 0 ) (r, X t,x r , (u j (X t,x r )) j=1,m , Z i 0 ,t,x r , U i 0 ,t,x r ) dr p 2 +5 p 2 -1 E T s Y i 0 ,t,x r dK i 0 ,t,x r p 2 +5 p 2 -1 E T s E 2Y i 0 ,t,x r U i 0 ,t,x r µ(dr, de) p 2 + 5 p 2 -1 E T s 2Y i 0 ,t,x r Z i 0 ,t,x r dB r p 2 . ( 4.22)
For more comprehension, we adopt the following scripture for inequality (4.22);

E Y i 0 ,t,x s p + E T s Z i 0 ,t,x r 2 dr p 2 + E T s U i 0 ,t,x r 2 L 2 (λ) dr p 2 ≤ 5 p 2 -1 E g i 0 (X t,x T ) p +5 p 2 -1 T 1 (s) + 5 p 2 -1 T 2 (s) + 5 p 2 -1 T 3 (s) + 5 p 2 -1 T 4 (s) (4.23)
where,

T 1 (s) = E T s 2 Y i 0 ,t,x r f (i 0 ) (r, X t,x r , (u j (X t,x r )) j=1,m , Z i 0 ,t,x r , U i 0 ,t,x r ) dr p 2 ; T 2 (s) = E T s Y i 0 ,t,x r dK i 0 ,t,x r p 2 ; T 3 (s) = E T s E 2Y i 0 ,t,x r U i 0 ,t,x r µ(dr, de) p 2
;

T 4 (s) = E T s 2Y i 0 ,t,x r Z i 0 ,t,x r dB r p 2
.

We will estimate T 1 (s), T 2 (s), T 3 (s) and T 4 (s), ∀s ∈ [t, T ].

(a) Before starting our estimations, let's linearize f with respect to (u j (X t,x r )) j=1,m and Z i 0 ,t,x r i.e.

f (i 0 ) (r, X t,x r , (u j (X t,x r )) j=1,m , Z i 0 ,t,x r , U i 0 ,t,x r ) = a t,x r Z i 0 ,t,x r +b t,x r (u j (X t,x r )) j=1,m +f (i 0 ) (r, X t,x r , 0, 0, U i 0 ,t,x r );
where a t,x r and b t,x r are progressively measurable processes respectively bounded by the Lipschitz constants of f in Z t,x r and (u j (X t,x r )) j=1,m i.e |a t,x r | ≤ C Z and |b t,x r | ≤ λ 1 . (b) We also take the fact that f is Lipschitz in U i 0 ,t,x r i.e there exists a constant Lipschitz λ 2 such that |f

(i 0 ) (r, X t,x r , 0, 0, U i 0 ,t,x r )| ≤ |f (i 0 ) (r, X t,x r , 0, 0, 0)| + λ 2 U i 0 ,t,x r L 2 m (λ)
. By combining (a) and (b) we have

|f (i 0 ) (r, X t,x r , (u j (X t,x r )) j=1,m , Z i 0 ,t,x r , U i 0 ,t,x r )| ≤ |a t,x r Z i 0 ,t,x r | + |b t,x r (u j (X t,x r )) j=1,m | + |f (i 0 ) (r, X t,x r , 0, 0, 0)| +λ 2 U i 0 ,t,x r L 2 m (λ) . (4.24)
Let's start our estimates.

For T 1 (s)

By using (4.24), it follows that;

T s 2 Y i 0 ,t,x r f (i 0 ) (r, X t,x r , (u j (X t,x r )) j=1,m , Z i 0 ,t,x r , U i 0 ,t,x r ) dr ≤ T s 2 Y i 0 ,t,x r a t,x r Z i 0 ,t,x r dr + T s 2 Y i 0 ,t,x r b t,x r (u j (X t,x r )) j=1,m dr + λ 2 T s 2 Y i 0 ,t,x r U i 0 ,t,x r dr + T s 2 Y i 0 ,t,x r f (i 0 ) (r, X t,x r , 0, 0, 0) dr ≤ C 2 q C Z T ǫ -1 1 Σ 2q + ǫ 1 C Z T s Z i 0 ,t,x r 2 dr + C 2 q λ 1 T ǫ -1 2 Σ 2q + ǫ 2 λ 1 C 2 q T Σ 2q + C 2 q T λ 2 ǫ -1 3 Σ 2q +λ 2 ǫ 3 T s U i 0 ,t,x r 2 L 2 m (λ) dr + C 2 q T ǫ -1 4 Σ 2q + C 2 q T ǫ 4 Σ 2q .
By raising to the power p 2 and then taking expectation it follows that,

T 1 (s) ≤ 8 p 2 -1 CC p q (C Z T ǫ -1 1 ) p 2 + (λ 1 T ǫ -1 2 ) p 2 + C p q (ǫ 2 λ 1 T ) p 2 + (T λ 2 ǫ -1 3 ) p 2 + (T ǫ -1 4 ) p 2 +(T ǫ 4 ) p 2 |1 + |x| pq | + 8 p 2 -1 (ǫ 1 C Z ) p 2 E T s Z i 0 ,t,x r 2 dr p 2 +8 p 2 -1 (λ 2 ǫ 3 ) p 2 E T s U i 0 ,t,x r 2 L 2 m (λ) dr p 2 . (4.25)
Before estimating T 2 (s), let us first give an estimate of

E |K i 0 ,t,x T -K i 0 ,t,x s
| p which will serve us in that of T 2 (s).

K i 0 ,t,x T -K i 0 ,t,x s = Y i 0 ,t,x s -g i 0 (X t,x T ) - T s f (i 0 ) (r, X t,x r , (u j (X t,x r )) j=1,m , Z i 0 ,t,x r , U i 0 ,t,x r )dr + T s Z i 0 ,t,x r dB r + T s E U i 0 ,t,x
r (e) µ(dr, de).

By (4.24) and Cauchy-Schwartz inequality; it follows that,

|K i 0 ,t,x T -K i 0 ,t,x s | ≤ 2C q Σ q + C q λ 1 T Σ q + C q T Σ q + C Z T s |Z i 0 ,t,x r | 2 dr 1 2 +λ 2 T s U i 0 ,t,x r 2 dr 1 2 + T s Z i 0 ,t,x r dB r + T s E U i 0 ,t,x
r (e) µ(dr, de).

By raising to the power p, taking expectation and BDG inequality we have,

E |K i 0 ,t,x T -K i 0 ,t,x s | p ≤ 5 p-1 CC p q {2 p + (T λ 1 ) p + T p } |1 + |x| pq | + 5 p-1 (C p Z + C p )E T 0 |Z i 0 ,t,x s | 2 ds p 2 +5 p-1 (λ p 2 + C p )E T 0 ds U i 0 ,t,x s 2 L 2 m (λ) p 2 . (4.26) For T 2 (s) T s |Y i 0 ,t,x s |dK i 0 ,t,x s ≤ T s |(Y i 0 ,t,x s -ℓ(s, X t,x s ))|dK i 0 ,t,x s + T s |ℓ(s, X t,x s )|dK i 0 ,t,x s ≤ sup s≤T |ℓ(s, X t,x s )|K i 0 ,t,x T ≤ ǫ -1 5 sup s≤T |ℓ(s, X t,x s )| 2 + ǫ 5 (K i 0 ,t,x T ) 2 ≤ ǫ -1 5 C 2 q Σ 2q + ǫ 5 (K i 0 ,t,x T ) 2 . ( 4
T 2 (s) ≤ 2 p 2 -1 CC p q (ǫ -1 5 ) p 2 |1 + |x| pq | + 2 p 2 -1 (ǫ 5 ) p 2 E (K i 0 ,t,x T ) p ≤ 2 p 2 -1 CC p q (ǫ -1 5 ) p 2 |1 + |x| pq | + (ǫ 5 ) p 2 2 p 2 -1 7 p-1 CC p q {2 p + (T λ 1 ) p + T p } |1 + |x| pq | + (ǫ 5 ) p 2 2 p 2 -1 7 p-1 (C p Z + C p )E T 0 |Z i 0 ,t,x s | 2 ds p 2 +(ǫ 5 ) p 2 2 p 2 -1 7 p-1 (λ p 2 + C p )E T 0 ds U i 0 ,t,x s 2 L 2 m (λ) p 2 T 2 (s) ≤ 2 p 2 -1 CC p q (ǫ -1 5 ) p 2 + (ǫ 5 ) p 2 2 p 2 -1 7 p-1 CC p q (2 p + (T λ 1 ) p + T p ) (1 + |x| pq ) +(ǫ 5 ) p 2 2 p 2 -1 7 p-1 (C p Z + C p )E T 0 |Z i 0 ,t,x s | 2 ds p 2 +(ǫ 5 ) p 2 2 p 2 -1 7 p-1 (λ p 2 + C p )E T 0 ds U i 0 ,t,x s 2 L 2 m (λ) p 2
.

For T 3 (s)

By BDG inequality,

T 3 (s) ≤ C p E T 0 |Y i 0 ,t,x s | 2 |Z i 0 ,t,x s | 2 ds p 4 ≤ C p E sup s≤T |Y i 0 ,t,x s | 2 T 0 |Z i 0 ,t,x s | 2 ds p 4 ≤ C p C p q ǫ -1 6 (1 + |x| pq ) + C p ǫ 6 E T 0 |Z i 0 ,t,x s | 2 ds p 2 . ( 4.28)
For T 4 (s)

By BDG inequality,

T 4 (s) ≤ C p E T 0 |Y i 0 ,t,x s | 2 U i 0 ,t,x s 2 L 2 m (λ) ds p 4 ≤ C p E sup s≤T |Y i 0 ,t,x s | 2 T 0 U i 0 ,t,x s 2 L 2 m (λ) ds p 4 ≤ C p C p q ǫ -1 7 (1 + |x| pq ) + C p ǫ 7 E T 0 U i 0 ,t,x s 2 L 2 m (λ) ds p 2 . (4.29)
Finally by taking estimation of T 1 (s), T 2 (s), T 3 (s), T 4 (s) and choosing ǫ 1 , ǫ 2 , ǫ 3 , ǫ 4 , ǫ 5 , ǫ 6 , ǫ 7 such that;

{(ǫ 5 ) p 2 2 p 2 -1 7 p-1 (C p Z + C p ) + C p ǫ 6 } < 1, {(ǫ 5 ) p 2 2 p 2 -1 7 p-1 (λ p 2 + C p ) + C p ǫ 7 } < 1
, and the sum of all coefficients of (1 + |x| pq ) was small than 1.

It follows then

E T 0 ds U i 0 ,t,x s 2 L 2 m (λ) p 2 ≤ C (1 + |x| ρ ) . (4.30)
Where ρ = pq.

Finally since i 0 ∈ {1, . . . , m} is arbitrary we then obtain the estimate (4.17).

Proposition 4.2 For any i = 1, . . . , m, u i belongs to U .

Proof. Let x and x ′ be elements of

R k . Let ( Y t,x , Z t,x , U t,x , K t,x ) (resp. ( Y t,x ′ , Z t,x ′ , U t,x ′ , K t,x ′ ))
be the solution of the RBSDE with jumps (3.14) associated with f (s,

X t,x s , y, η, ζ, g(X t,x T )) (resp. f (s, X t,x ′ s , y, η, ζ, g(X t,x ′ T ))). Applying Itô formula to Y t,x -Y t,x ′ 2
between s and T , we have

Y t,x s -Y t,x ′ s 2 + T s |∆Z r | 2 dr + s≤r≤T (∆ r Y t,x r ) 2 (4.31) = g(X t,x T ) -g(X t,x ′ T ) 2 + 2 T s < Y t,x s -Y t,x ′ s , ∆f (r) > dr + 2 T s Y t,x r -Y t,x ′ r d (∆K r ) -2 T s E Y t,x r -Y t,x ′ r (∆U r (e)) µ(dr, de) -2 T s Y t,x r -Y t,x ′ r (∆Z r ) dB r ;
and taking expectation we obtain: ∀s ∈ [t, T ],

E Y t,x s -Y t,x ′ s 2 + T s |∆Z r | 2 dr + T s ∆U r 2 L 2 (λ) dr (4.32) ≤ E g(X t,x T ) -g(X t,x ′ T ) 2 + 2 T s < Y t,x s -Y t,x ′ s , ∆f (r) > dr +E 2 T s Y t,x r -Y t,x ′ r d (∆K r ) ,
where the processes ∆X r , ∆Y r , ∆f (r), ∆K r , ∆Z r , ∆U r and ∆ℓ r are defined as follows: ∀r ∈

[t, T ], ∆f (r) := ((∆f (i) (r)) i=1,m = (f (i) (r, X i;t,x r , Y t,x r , Z i;t,x r , U i;t,x r )-f (i) (r, X i;t,x ′ r , Y i;t,x ′ r , Z i;t,x ′ r , U i;t,x ′ r )) i=1,m , ∆X r = X t,x r -X t,x ′ r , ∆Y (r) = Y t,x r -Y t,x ′ r = (Y j;t,x r -Y j;t,x ′ r ) j=1,m , ∆K r = K t,x r -K t,x ′ r , ∆Z r = Z t,x r -Z t,x ′ r , ∆U r = U t,x r -U t,x ′ r and ∆ℓ r = ℓ(r, X t,x r ) -ℓ(r, X t,x ′ r ) (< •, • > is the usual scalar product on R m
). Now we will give an estimation of each three terms of the second member of inequality (4.32).

• As for any i ∈ {1, . . . , m} g i belongs to U ; therefore

E g(X t,x T ) -g(X t,x ′ T ) 2 ≤ C X t,x T -X t,x ′ T 2 (1 + X t,x T 2p + X t,x ′ T 2p ) ≤ E x -x ′ 2 (1 + (X t,x T -x) + x 2p + (X t,x ′ T -x ′ ) + x ′ 2p ,
and by subsequently using the triangle inequality, the relation of proposition 3.1 and the fact that

(a + b) p ≤ 2 p-1 (a p + b p ). E g(X t,x T ) -g(X t,x ′ T ) 2 ≤ C x -x ′ 2 (1 + |x| 2p + x ′ 2p ), (4.33)
• using (iii) of (3.14

): E 2 T s Y t,x r -Y t,x ′ r d (∆K r ) can be replaced by E 2 T s ℓ(r, X t,x r ) -ℓ(r, X t,x ′ r ) d (∆K r ) .
Now by (H1) and Cauchy-Schwartz inequality we obtain:

E sup 0≤t≤T (∆ℓ t ) 2 × E (∆K T ) 2 ≤ 2CC ′ x -x ′ 2 (1 + |x| 2p + x ′ 2p ); (4.34)
where

C ′ = E (∆K T ) 2 .
• To complete our estimation of (4.32) we need to deal with E 2

T s < Y t,x s -Y t,x ′ s , ∆f (r) > dr .
Taking into account the expression of f (i) given by (2.11) we then split ∆f (r) in the follows way:

for r ≤ T ,

∆f (r) = (∆f (r)) i=1,m = ∆ 1 (r)+∆ 2 (r)+∆ 3 (r)+∆ 4 (r) = (∆ i 1 (r)+∆ i 2 (r)+∆ i 3 (r)+∆ i 4 (r)) i=1,m ,
where for any i = 1, . . . , m,

∆ i 1 (r) = h (i) r, X t,x r , Y t,x r , Z i;t,x r , E γ i (r, X t,x r , e)U i;t,x r (e)λ(de) -h (i) r, X t,x ′ r , Y t,x r , Z i;t,x r , E γ i (r, X t,x r , e)U i;t,x r (e)λ(de) ; ∆ i 2 (r) = h (i) r, X t,x ′ r , Y t,x r , Z i;t,x r , E γ i (r, X t,x r , e)U i;t,x r (e)λ(de) -h (i) r, X t,x ′ r , Y t,x ′ r , Z i;t,x r , E γ i (r, X t,x r , e)U i;t,x r (e)λ(de) ; ∆ i 3 (r) = h (i) r, X t,x ′ r , Y t,x ′ r , Z i;t,x r , E γ i (r, X t,x r , e)U i;t,x r (e)λ(de) -h (i) r, X t,x ′ r , Y t,x ′ r , Z i;t,x ′ r , E γ i (r, X t,x r , e)U i;t,x r (e)λ(de) ; ∆ i 4 (r) = h (i) r, X t,x ′ r , Y t,x ′ r , Z i;t,x ′ r , E γ i (r, X t,x r , e)U i;t,x r (e)λ(de) -h (i) r, X t,x ′ r , Y t,x ′ r , Z i;t,x ′ r , E γ i (r, X t,x ′ r , e)U i;t,x ′
r (e)λ(de) .

By Cauchy-Schwartz inequality, the inequality 2ab ≤ ǫa 2 + 1 ǫ b 2 , the relation (2.11) and the estimate (3.13) we have:

E 2 T s <∆Y (r),∆ 1 (r)> dr ≤ E 1 ǫ T s |∆Y (r)| 2 dr+C 2 ǫ T s |X t,x r -X t,x ′ r | 2 (1+|X t,x r | p +|X t,x ′ r | p ) 2 dr ≤ E 1 ǫ T s |∆Y (r)| 2 dr + C 2 ǫ|x -x ′ | 2 (1 + |x| p + |x ′ | p ) 2 . (4.35) Besides since h (i) is Lipschitz w.r.t. (y, z, q) then, E 2 T s < ∆Y (r), ∆ 2 (r) > dr ≤ 2CE T s |∆Y (r)| 2 dr , (4.36)
and

E 2 T s < ∆Y (r), ∆ 3 (r) > dr ≤ E 1 ǫ T s |∆Y (r)| 2 dr + C 2 ǫ T s |∆Z(r)| 2 dr . (4.37)
It remains to obtain a control of the last term. But for any s ∈ [t, T ] we have,

E 2 T s < ∆Y (r), ∆ 4 (r) > dr (4.38) ≤ 2CE T s |∆Y (r)| dr × E γ(r, X t,x r , e)U t,x r (e) -γ(r, X t,x ′ r , e)U t,x ′ r (e) λ(de) .
Next by splitting the crossing terms as follows γ(r, X t,x r , e)U t,x r (e)γ(r, X t,x ′ r , e)U t,x ′ r (e) = ∆U s (e)γ(s, X t,x s , e) + U t,x ′ s γ(s, X t,x s , e)γ(s, X t,x ′ s , e) and setting ∆γ s (e) := γ(s, X t,x s , e)γ(s, X t,x ′ s , e) , we obtain, Choosing now ǫ small enough we deduce the existence of a constant C ≥ 0 such that for any

E 2 T s < ∆Y (r), ∆ 4 (r) > dr ≤ 2CE T s |∆Y (r)|× E (|U t,
≤ CE {sup r∈[t,T ] |X t,x r -X t,x ′ r | 2 (1+sup r∈[t,T ] |X t,x r | p +|X t,x ′ r | p ) 2 dr} ×E T s dr E |U t,x ′ r (e)| 2 λ(de) ≤ C E {sup r∈[t,T ] |X t,x r -X t,x ′ r | 4 (1+sup r∈[t,T ] |X t,x r | p +|X t,x ′ r | p ) 4 } × E T s dr E |U t,x ′ r (e)| 2 λ(de) 2 ≤ C x -x ′ 2 (1 + |x| 2p + x ′ 2p ). ( 4
E Y t,x s -Y t,x ′ s 2 + T s |∆Z r | 2 dr + T s ∆U r 2 L 2 (λ) dr ≤ E g(X t,x T ) -g(X t,x ′ T ) 2 + 2 T s < Y t,x s -Y t,x ′ s , ∆f (r) > dr +E 2 T s Y t,x r -Y t,x ′ r d (∆K r ) ≤ x -x ′ 2 (1 + |x| 2p + x ′ 2p )(C + 2CC ′ + C 2 ǫ + C 3 ǫ) + 3 ǫ + 2C E T s |∆Y (
s ∈ [t, T ], E |∆Y (s)| 2 ≤ C |x -x ′ | 2 (1 + |x| 2p + |x ′ | 2p ) + E T s |∆Y (r)| 2 dr
and by Gronwall lemma this implies that for any s ∈ [t, T ],

E |∆Y (s)| 2 ≤ C x -x ′ 2 (1 + |x| 2p + x ′ 2p ).
Finally in taking s = t and considering (3.15) we obtain the desired result.

Remark 3 This result give also estimate of U where we use the function h (i) ∀ i = 1, . . . , m contrary in estimate (4.17). Proof. The main point to notice is that λ integrates (1 ∧ |e| p ) ∀p ≥ 2.

We have that

|B i u i (t, x)| ≤ E |γ i (t, x, e)| × |(u i (t, x + β(t, x, e)) -u i (t, x))| λ(de) ≤ E C(1 ∧ |e|)|β(t, x, e)|(1 + |x + β(t, x, e)| p + |x| p ) λ(de) ≤ C 2 (1 + |x| p (1 + 2 p-1 )) E C(1 ∧ |e| 2 ) λ(de) + (2 p-1 p C p+2 p ) p E C(1 ∧ |e| p ) λ(de).
Which finish the proof.

Now by remark 3, the last estimate of U confirm the following result;

Proposition 4.4 For any i = 1, . . . , m, (t, x) ∈ [0, T ] × R k , U i;t,x s (e) = u i (s, X t,x s-+ β(s, X t,x s-, e)) -u i (s, X t,x s-), dP ⊗ ds ⊗ dλ -a.e. on Ω × [t, T ] × E. (4.42)
Proof. First note that since the measure λ is note finite, then we cannot use the same technique as in [START_REF] Hamadène | Viscosity solutions for second order integro-differential equations without monotonicity condition: The probabilistic Approach[END_REF] where the authors use the jumps of processes and (3.15).

In our case U i;t,x is only square integrable and not necessarily integrable w.r.t. dP ⊗ ds ⊗ dλ.

Therefore we first begin by truncating the Lévy measure as the same way in [START_REF] Hamadène | Viscosity solutions of second order integral-partial differential equations without monotonicity condition: A new result[END_REF].

Step 1: Truncation of the Lévy measure

For any k ≥ 1, let us first introduce a new Poisson random measure µ k (obtained from the truncation of µ) and its associated compensator ν k as follows:

µ k (ds, de) = 1 {|e|≥ 1 k } µ(ds, de) and ν k (ds, de) = 1 {|e|≥ 1 k } ν(ds, de).

Which means that, as usual, µ k (ds, de) := (µ kν k )(ds, de), is the associated random martingale measure.

The main point to notice is that

λ k (E) = E λ k (de) = E 1 {|e|≥ 1 k } λ(de) = {|e|≥ 1 k } λ(de) = λ({|e| ≥ 1 k }) < ∞. (4.43)
As in [START_REF] Hamadène | Viscosity solutions of second order integral-partial differential equations without monotonicity condition: A new result[END_REF], let us introduce the process k X t,x solving the following standard SDE of jump-diffusion type:

k X t,x s = x + s t b(r, k X t,x r ) dr + s t σ(r, k X t,x r ) dB r + s t E β(r, k X t,x r-, e) µ k (dr, de), t ≤ s ≤ T ; k X t,x r = x if s ≤ t.
(4.44)

Note that thanks to the assumptions on b, σ, β the process k X t,x exists and is unique. Moreover it satisfies the same estimates as in (3.13) since λ k is just a truncation at the origin of λ which integrates (1 ∧ |e| 2 ) e∈E .

On the other hand let us consider the following Markovian RBSDE with jumps

                                   (i) E sup s≤T k Y t,x s 2 + T s k Z t,x r 2 dr + T s k U t,x r 2 L 2 (λ k ) dr < ∞ (ii) k Y t,x := ( k Y i,t,x ) i=1,m ∈ S 2 (R m ), k Z t,x := ( k Z i,t,x ) i=1,m ∈ H 2 (R m×d ), k K t,x := ( k K i,t,x ) i=1,m ∈ A 2 c , k U t,x := ( k U i,t,x ) i=1,m ∈ H 2 (L 2 m (λ k )); (iii) k Y t,x s = g( k X t,x T ) + T s f µ k (r, k X t,x r , k Y t,x r , k Z t,x r , k U t,x r ) dr + k K t,x T -k K t,x s - T s k Z t,x r dB r + k E U t,x r (e) µ k (dr, de) , s ≤ T ; (iv) k Y i;t,x s ≥ ℓ(s, k X t,x s ) and T 0 ( k Y i;t,x s -ℓ(s, k X t,x s )) d( k K i;t,x s ) = 0. (4.45)
Finally let us introduce the following functions (f (i) ) i=1,m defined by: ∀

(t, x, y, z, ζ) ∈ [0, T ] × R k × R m × R m×d × L 2 m (λ k ), f µ k (t, x, y, z, ζ) = (f (i) µ k (t, x, y, z i , ζ i )) i=1,m := h (i) t, x, y, z, E γ i (t, x, e)ζ i (e)λ k (de) i=1,m .
First let us emphasize that this latter RBSDE is related to the filtration (F k s ) s≤T generated by the Brownian motion and the independent random measure µ k . However this point does not raise major issues since for any s ≤ T , F k s ⊂ F s and thanks to the relationship between µ and µ k .

Next by the properties of the functions b, σ, β and by the same opinions of proposition 3.2 and proposition 3.3, there exists an unique quadriple ( k Y t,x , k K t,x , k Z t,x , k U t,x ) solving (4.45) and there also exists a function

u k from [0, T ] × R k into R m of Π c g such that ∀s ∈ [t, T ], k Y t,x := u k (s, k X t,x ), P -a.s. (4.46)
Moreover as in proposition 4.2, there exists positive constants C and p wich do not depend on k such that:

∀t, x, x ′ , |u k (t, x) -u k (t, x ′ )| ≤ C x -x ′ (1 + |x| p + x ′ p ). (4.47)
Finally as λ k is finite then we have the following relationship between the process k U t,x := ( k U i;t,x ) i=1,m and the deterministics functions u k := (u k i ) i=1,m (see [START_REF] Hamadène | Viscosity solutions for second order integro-differential equations without monotonicity condition: The probabilistic Approach[END_REF]): ∀i = 1, . . . , m;

k U i;t,x s (e) = u k i (s, k X t,x s-+ β(s, k X t,x s-, e)) -u k i (s, k X t,x s-), dP ⊗ ds ⊗ dλ k -a.e. on Ω × [t, T ] × E.
This is mainly due to the fact that k U t,x belongs to

L 1 ∩ L 2 (ds ⊗ dP ⊗ dλ k ) since λ k (E) < ∞
and then we can split the stochastic integral w.r.t. µ k in (4.45). Therefore for all i = 1, . . . , m,

k U i;t,x s (e)1 {|e|≥ 1 k } = (u k i (s, k X t,x s-+β(s, k X t,x s-, e))-u k i (s, k X t,x s-))1 {|e|≥ 1 k } , dP⊗ds⊗dλ k -a.e. on Ω×[t, T ]×E. (4.48)
Step 2: Convergence of the auxiliary processes Let's now prove the following convergence result;

E sup s≤T Y t,x s -k Y t,x s 2 + (K t,x T -k K t,x T ) 2 + T 0 Z t,x s -k Z t,x s 2 ds + T 0 ds E λ(de) U t,x s (e) -k U t,x s (e)1 {|e|≥ 1 k } 2 -→ k-→+∞ 0; (4.49)
where (Y t,x , K t,x , Z t,x , U t,x ) is solution of the RBSDE with jumps (3.14).

First note that the following convergence result was established in [START_REF] Hamadène | Viscosity solutions of second order integral-partial differential equations without monotonicity condition: A new result[END_REF] E sup s≤T

X t,x s -k X t,x s 2 -→ k-→+∞ 0. (4.50)
We now focus on (4.49). Note that we can apply Ito's formula, even if the RBSDEs are related to filtrations and Poisson random measures which are not the same, since:

(i) F k s ⊂ F s , ∀s ≤ T ; (ii) for any s ≤ T , s 0 k E U i;t,x (e) µ k (dr, de) = s 0 k E U i;t,x (e)1 {|e|≥ 1
k } µ (dr, de) and then the first (F k s ) s≤T -martingale is also an (F s ) s≤T -martingale. ∀s ∈ [0, T ],

Y t,x s -k Y t,x s 2 + T 0 Z t,x s -k Z t,x s 2 , ds + s≤r≤T ( k ∆ r Y t,x r ) 2 = g(X t,x T ) -g( k X t,x T ) 2 + 2 T s Y t,x r -k Y t,x r × k ∆f (r) dr + 2 T s Y t,x r -k Y t,x r d k ∆K r -2 T s E Y t,x r -k Y t,x r k ∆U r (e) µ(dr, de) -2 T s Y t,x r -k Y t,x r k ∆Z r dB r ;
and taking expectation we obtain: ∀s ∈ [t, T ],

E Y t,x s -k Y t,x s 2 + k ∆K T 2 + T 0 Z t,x s -k Z t,x s 2 + E U t,x s -k U t,x s 1 {|e|≥ 1 k } 2 λ(de) ds ≤ E g(X t,x T ) -g( k X t,x T ) 2 + 2 T s Y t,x r -k Y t,x r × k ∆f (r) dr + E sup s≤T k ∆ℓ s 2 ; (4.51)
where the processes k ∆X r , k ∆Y r , k ∆f (r), k ∆K r , k ∆Z r , k ∆U r and k ∆ℓ r are defined as follows:

∀r ∈ [0, T ], k ∆f (r) := (( k ∆f (i) (r)) i=1,m = (f (i) (r, X t,x r , Y t,x r , Z i;t,x r , U i;t,x r )-f (i) k (r, k X t,x r , k Y t,x r , k Z t,x r , k U t,x r )) i=1,m , k ∆X r = X t,x r -k X t,x r , k ∆Y (r) = Y t,x r -k Y t,x r = (Y j;t,x r -k Y j;t,x r ) j=1,m , k ∆K r = K t,x r -k K t,x r , k ∆Z r = Z t,x r -k Z t,x r , k ∆U r = U t,x r -k U t,x s 1 {|e|≥ 1 k } and k ∆ℓ r = ℓ(r, X t,x r ) -ℓ(r, k X t,x r ) .
Next let us set for r ≤ T ,

k ∆f (r) = (f (r, X t,x r , Y t,x r , Z t,x r , U t,x r )-f k (r, k X t,x r , k Y t,x r , k Z t,x r , k U t,x r )) = A(r)+B(r)+C(r)+D(r);
where for any i = 1, . . . , m,

A(r) = h (i) r, X t,x r , Y t,x r , Z i;t,x r , E γ i (r, X t,x r , e)U i;t,x r (e)λ(de) -h (i) r, k X t,x r , Y t,x r , Z i;t,x r , E γ i (r, X t,x r , e)U i;t,x r (e)λ(de) i=1,m ; B(r) = h (i) r, k X t,x r , Y t,x r , Z i;t,x r , E γ i (r, X t,x r , e)U i;t,x r (e)λ(de) -h (i) r, k X t,x r , k Y t,x r , Z i;t,x r , E γ i (r, X t,x r , e)U i;t,x r (e)λ(de) i=1,m ; C(r) = h (i) r, k X t,x r , k Y t,x r , Z i;t,x r , E γ i (r, X t,x r , e)U i;t,x r (e)λ(de) -h (i) r, k X t,x r , k Y t,x r , k Z i;t,x r , E γ i (r, X t,x r , e)U i;t,x r (e)λ(de) i=1,m
;

D(r) = h (i) r, k X t,x r , k Y t,x r , k Z i;t,x r , E γ i (r, X t,x r , e)U i;t,x r (e)λ(de) -h (i) r, k X t,x r , k Y t,x r , k Z i;t,x r , E γ i (r, k X t,x r , e) k U i;t,x r (e)λ k (de) i=1,m
.

By (4.50) and the of g ∈ U and ℓ ∈ U we have,

E g(X t,x T ) -g( k X t,x T ) 2 -→ 0 k→+∞ (4.52) and E sup s≤T ℓ(X t,x s ) -ℓ( k X t,x s ) 2 -→ 0 k→+∞ . (4.53) Now we will interest to E T s Y t,x r -k Y t,x r
× k ∆f (r) dr for found (4.49).

By (2) and (2.8), we have:

∀r ∈ [0, T ] |A(r)| ≤ C X t,x r -k X t,x r (1 + X t,x r p + k X t,x r p ); (4.54) |B(r)| ≤ C Y t,x r -k Y t,x r and |C(r)| ≤ Z t,x r -k Z t,x r ;
where C is a constant. Finally let us deal with D(r) which is more involved. First note that

D(r) = (D i (r)) i=1,m
where

D i (r) = h (i) r, k X t,x r , k Y t,x r , k Z i;t,x r , E γ i (r, X t,x r , e)U i;t,x r (e)λ(de) -h (i) r, k X t,x r , k Y t,x r , k Z i;t,x r , E γ i (r, k X t,x r , e) k U i;t,x r (e)λ k (de) .
But as h (i) is Lipschitz w.r.t to the last component q then,

|D(r)| 2 ≤ C E γ i (r, X t,x r , e)U i;t,x r (e) -γ i (r, k X t,x r , e) k U i;t,x r (e)1 {|e|≥ 1 k } 2 λ(de) ≤ C E γ i (r, X t,x r , e) -γ i (r, k X t,x r , e) U i;t,x r (e) λ(de) 2 + E γ i (r, X t,x r , e) U i;t,x r (e) -k U i;t,x r (e)1 {|e|≥ 1 k } λ(de) 2 ≤ C X t,x r -k X t,x r (1 + X t,x r p + k X t,x r p ) E U i;t,x r (e) λ(de) 2 +C E (1 ∧ |e|) U i;t,x r (e) -k U i;t,x r (e)1 {|e|≥ 1 k } 2 λ(de), (4.55)
and (4.51) become by using the majorations obtain in (4.54) and in (4.55);

E Y t,x s -k Y t,x s 2 + k ∆K T 2 + T 0 Z t,x s -k Z t,x s 2 + E U t,x s -k U t,x s 1 {|e|≥ 1 k } 2 λ(de) ds ≤ E g(X t,x T ) -g( k X t,x T ) 2 + E sup s≤T ℓ(X t,x s ) -ℓ( k X t,x s ) 2 + CE T s Y t,x s -k Y t,x s 2 +CE T 0 X t,x r -k X t,x r 2 (1 + X t,x r p + k X t,x r p ) 2 dr +CE T 0 dr X t,x r -k X t,x r (1 + X t,x r p + k X t,x r p ) E U i;t,x r (e)λ(de) 2 .
(4.56)

The two first terms converge to 0 by (4.52) and (4.53).

For the fourth term we have:

E T 0 X t,x r -k X t,x r 2 (1 + X t,x r p + k X t,x r p ) 2 dr ≤ E sup r≤T X t,x r -k X t,x r 2 T 0 (1 + X t,x r p + k X t,x r p ) 2 dr ≤ E sup r≤T X t,x r -k X t,x r 2 1 2 E T 0 (1 + X t,x r p + k X t,x r p ) 2 X t,x r -k X t,x r 2 dr 1 2
.

The first factor in the right-hand side of this inequality goes to 0 when k → ∞ due to (4.50) and the second factor is uniformly bounded by the uniform estimates (3.13) of X t,x and k X t,x .

Note also the last term converge to 0 when k → ∞, it is a consequence of (4.50), the fact that k X t,x verifies estimates (3.13) uniformly, the Cauchy-Schwartz inequality (used twice) and finally (4.17) of lemma 4.1. Then by Gronwall's lemma we deduce first that for any s ≤ T ,

E Y t,x s -k Y t,x s 2 -→ 0 k→+∞ (4.57) and in taking s = t we obtain u k (t, x) -→ u(t, x) k→+∞ . As (t, x) ∈ [0, T ]×R k is arbitrary then u k -→ u k→+∞ pointwisely.
Next going back to (4.56) take the limit w.r.t k and using the uniform polynomial growth of u k and the Lebesgue dominated convergence theorem as well, to obtain:

E T t E U t,x s -k U t,x s 1 {|e|≥ 1 k } 2 λ(de) ds -→ 0 k→+∞ . (4.58)
Step 3: Conclusion

First note that by (4.47) and the pointwise convergence of

(u k ) k to u, if (x k ) k is a sequence of R k which converge to x then ((u k (t, x k )) k ) converge to u(t, x).
Now let us consider a subsequence which we still denote by {k} such that sup s≤T X t,x s -k X t,x s 2 -→ 0 k→+∞ , P-a.s. (and then

X t,x s--k X t,x s- -→ 0 k→+∞ since X t,x s--k X t,x s-≤ sup s≤T X t,x s -k X t,x s 2
). By (4.50), this subsequence exists. As the mapping x → β(t, x, e) is Lipschitz then the sequence

k U t,x s (e)1 {|e|≥ 1 k } k = (u k i (s, k X t,x s-+ β(s, k X t,x s-, e)) -u k i (s, k X t,x s-))1 {|e|≥ 1 k } k≥1 -→ k→+∞ (u i (s, X t,x s-+ β(s, X t,x s-, e)) -u i (s, X t,x s-)), dP ⊗ ds ⊗ dλ -a.e. on Ω × [t, T ] × E (4.59)
for any i = 1, . . . , m. Finally from (4.58) we deduce that

U t,x s (e) = (u i (s, X t,x s-+ β(s, X t,x s-, e)) -u i (s, X t,x s-)), on Ω × [t, T ] × E (4.60)
which is the desired result.

The main result

First we give the definition of viscosity solution of IPDEs as given in [START_REF] Hamadène | Viscosity solutions of second order integral-partial differential equations without monotonicity condition: A new result[END_REF] and [START_REF] Hamadène | Viscosity solutions for second order integro-differential equations without monotonicity condition: The probabilistic Approach[END_REF]. Our main result deal with this definition.

Definition 5.1 We say that a family of deterministics functions u = (u i ) i=1,m which belongs to U ∀i ∈ {1, . . . , m} is a viscosity sub-solution (resp. super-solution) of the IPDE (1.1) if:

(i) ∀x ∈ R k , u i (x, T ) ≤ g i (x) (resp. u i (x, T ) ≥ g i (x)); (ii) For any (t, x) ∈ [0, T ] × R k and any function φ of class C 1,2 ([0, T ] × R k ) such that (t, x) is
a global maximum point of u iφ (resp. global minimum point of u iφ) and (u iφ)(t, x) = 0 one has min u i (t, x)ℓ(t, x); -∂ t φ(t, x) -L X φ(t, x)h i (t, x, (u j (t, x)) j=1,m , σ ⊤ (t, x))D x φ(t, x), B i u i (t, x)) ≤ 0

(5.61)

(resp. min u i (t, x) -ℓ(t, x); -∂ t φ(t, x) -L X φ(t, x) -h i (t, x, (u j (t, x)) j=1,m , σ ⊤ (t, x))D x φ(t, x), B i u i (t, x)) ≥ 0 .
(5.62)

The family u = (u i ) i=1,m is a viscosity solution of (1.1) if it is both a viscosity sub-solution and viscosity super-solution.

Note that L X φ(t, x) = b(t, x) ⊤ D x φ(t, x) + 1 2 Tr(σσ ⊤ (t, x)D 2 xx φ(t, x)) + Kφ(t, x); where Kφ(t, x) = E (φ(t, x + β(t, x, e))φ(t, x)β(t, x, e) ⊤ D x φ(t, x))λ(de).

Proof.

Step 1: Existence Assume that assumptions (H1), (H2) and (H3) are fulfilled, then the following multi-dimensional

RBSDEs with jumps

                           (i) Y t,x := (Y i;t,x ) i=1,m ∈ S 2 (R m ), Z t,x := (Z i;t,x ) i=1,m ∈ H 2 (R m×d ), K t,x := (K i;t,x ) i=1,m ∈ A 2 c , U t,x := (U i;t,x ) i=1,m ∈ H 2 (L 2 m (λ)); (ii) Y i;t,x s = g i (X t,x T ) + K i;t,x T -K i;t,x s - T s Z i;t,x dB r - T s E U i;t,x
r (e) µ(dr, de).

+ T s h (i) (r, X t,x r , Y i;t,x r , Z i;t,x r , E γ i (t, X t,x r , e){(u i (t, X t,x r-+ β(t, X t,x r-, e)) -u i (t, X t,x r-))} λ(de))dr (iii) Y i;t,x s ≥ ℓ(s, X t,x s ) and T 0 (Y i;t,x s -ℓ(s, X t,x s ))dK i;t,x s = 0;
(5.63) has unique solution (Y , Z, K, U ). Next as for any i = 1, . . . , m, u i belongs to U , then by proposition 3.3 the (3.15), there exists a family of deterministics continuous functions of polynomial growth (u i ) i=1,m that fact for any

(t, x) ∈ [0, T ] × R k , ∀s ∈ [t, T ], Y i;t,x s = u i (s, X t,x s ).
Such that by the same proposition, the family (u i ) i=1,m is a viscosity solution of the following system:

         min u i (t, x) -ℓ(t, x); -∂ t u i (t, x) -b(t, x) ⊤ D x u i (t, x) -1 2 Tr(σσ ⊤ (t, x)D 2 xx u i (t, x)) -K i u i (t, x) -h (i) (t, x, (u j (t, x)) j=1,m , (σ ⊤ D x u i )(t, x), B i u i (t, x)) = 0, (t, x) ∈ [0, T ] × R k ; u i (T, x) = g i (x).
(5.64)

Now we have the family (u i ) i=1,m is a viscosity solution, our main objective is to found relation between (u i ) i=1,m and (u i ) i=1,m which is defined in (3.15).

For this, let us consider the system of RBSDE with jumps

                           (i) Y t,x := (Y i;t,x ) i=1,m ∈ S 2 (R m ), Z t,x := (Z i;t,x ) i=1,m ∈ H 2 (R m×d ), K t,x := (K i;t,x ) i=1,m ∈ A 2 c , U t,x := (U i;t,x ) i=1,m ∈ H 2 (L 2 m (λ)); (ii) Y i;t,x s = g i (X t,x T ) + K i;t,x T -K i;t,x s - T s Z i;t,x dB r - T s E U i;t,x
r (e) µ(dr, de). Therefore u i = u i , such that by (4.58) we obtain U t,x s (e) = (u i (s, X t,x s-+β(s, X t,x s-, e))-u i (s, X t,x s-)), on Ω× [t, T ]×E, which give the viscosity solution in the sense of definition 5.1 (see [START_REF] Hamadène | Viscosity solutions of second order integral-partial differential equations without monotonicity condition: A new result[END_REF]) by pluging (4.59) in h (i) of (5.64).

+ T s h (i) (r, X t,x r , Y i;t,x r , Z i;t,x r , E γ i (t, X t,
Step 2: Uniqueness For uniqueness, let (u i ) i=1,m be another family of U which is solution viscosity of the system (1.1) in the sense of definition 5.1 and we consider RBSDE with jumps defined with

u i .                            (i) Y t,x := (Y i;t,x ) i=1,m ∈ S 2 (R m ), Z t,x := (Z i;t,x ) i=1,m ∈ H 2 (R m×d ), K t,x := (K i;t,x ) i=1,m ∈ A 2 c , U t,x := (U i;t,x ) i=1,m ∈ H 2 (L 2 m (λ)); (ii) Y i;t,x s = g i (X t,x T ) + K i;t,x T -K i;t,x s - T s Z i;t,x dB r - T s E U i;t,x r
(e) µ(dr, de).

+ T s h (i) (r, X t,x r , Y i;t,x r , Z i;t,x r , E γ i (t, X t,x r , e)(u i (s, X t,x s-+ β(s, X t,x s-, e)) -u i (s, X t,x s-)) λ(de))dr; (iii) Y i;t,x s ≥ ℓ(s, X
) ∈ [0, T ] × R k , ∀s ∈ [t, T ], Y i;t,x s = v i (s, X t,x s ).
Such that by the same proposition, the family (v i ) i=1,m is a viscosity solution of the following system:

         min v i (t, x) -ℓ(t, x); -∂ t v i (t, x) -b(t, x) ⊤ D x v i (t, x) -1 2 Tr(σσ ⊤ (t, x)D 2 xx v i (t, x)) -K i v i (t, x) -h (i) (t, x, (v j (t, x)) j=1,m , (σ ⊤ D x v i )(t, x), B i u i (t, x)) = 0, (t, x) ∈ [0, T ] × R k ; u i (T, x) = g i (x).
(5.67)

By uniqueness of solution of (5.65) u i is viscosity solution of (5.66); and by proposition 3.3

v i = u i ∀i ∈ {1, . . . , m}.
Now for completing our proof we show that on Ω×[t, T ]×E, ds⊗dP⊗dλ-a.e. ∀i ∈ {1, . . . , m};

U i;t,x s (e) = (v i (s, X t,x s-+ β(s, X t,x s-, e)) -v i (s, X t,x s-))
= (u i (s, X t,x s-+ β(s, X t,x s-, e))u i (s, X t,x s-)).

(5.68) By Remark 3.4 in [START_REF] Hamadène | Viscosity solutions of second order integral-partial differential equations without monotonicity condition: A new result[END_REF]; let us considere (x k ) k≥1 a sequence of R k which converges to x ∈ R k and the two following RBSDE with jumps (adaptation is w.r.t. F k ):

                                   (i) Y k,t,x := (Y i;k,t,x ) i=1,m ∈ S 2 (R m ), Z k,t,x := (Z i;k,t,x ) i=1,m ∈ H 2 (R m×d ), K k,t,x := (K i;k,t,x ) i=1,m ∈ A 2 c , U k,t,x := (U i;k,t,x ) i=1,m ∈ H 2 (L 2 m (λ)); (ii) Y i;k,t,x s = g i (X k,t,x T ) + K i;k,t,x T -K i;k,t,x s - T s Z i;k,t,x dB r - T s E U i;k,t,x r (e) µ(dr, de) + T s h (i) r, X k,t,x r , Y i;k,t,x r , Z i;k,t,x r , E γ i (t, X k,t,x k r , e)(u i (s, X k,t,x k s- + β(s, X k,t,x k s- , e)) -u i (s, X k,t,x k s- )) λ(de) dr; (iii) Y i;k,t,x k s ≥ ℓ(s, X k,t,x k s ) and T 0 (Y i;k,t,x k s -ℓ(s, X k,t,x k s ))dK i;k,t,x k s = 0; (5.69) and                                    (i) Y k,t,x k := (Y i;k,t,x k ) i=1,m ∈ S 2 (R m ), Z k,t,x k := (Z i;k,t,x k ) i=1,m ∈ H 2 (R m×d ), K k,t,x k := (K i;k,t,x k ) i=1,m ∈ A 2 c , U k,t,x k := (U i;k,t,x k ) i=1,m ∈ H 2 (L 2 m (λ)); (ii) Y i;k,t,x k s = g i (X k,t,x k T ) + K i;k,t,x k T -K i;k,t,x k s - T s Z i;k,t,x k dB r - T s E U i;k,t,x k r (e) µ(dr, de) + T s h (i) r, X k,t,x k r , Y i;k,t,x k r , Z i;k,t,x k r , E γ i (t, X k,t,x k r , e)(u i (s, X k,t,x k s- + β(s, X k,t,x k s- , e)) -u i (s, X k,t,x k s-
)) λ(de) dr;

(iii) Y i;k,t,x k s ≥ ℓ(s, X k,t,x k s ) and T 0 (Y i;k,t,x k s -ℓ(s, X k,t,x k s ))dK i;k,t,x k s = 0.
(5.70)

By proof of step 2 of proposition 4.4, (Y i;k,t,x , K i;k,t,x , Z i;k,t,x , U i;k,t,x 1 {|e|≥ 1 k } ) k converge to (Y i;t,x , K i;t,x , Z i;t,x , U i;t,x ) in S 2 (R) × A 2 c × H 2 (R κ×d ) × H 2 (L 2 (λ)). Let ((v k i=1,m
)) k≥1 be the sequence of continuous deterministics functions such that for any t ≤ T and s ∈ [t, T ],

Y i;k,t,x s = v k i (s, k X t,x s ) and Y i;k,t,x k s = v k i (s, k X t,x k s ) ∀i = 1, . . . , m.
Such that we have respectively by proof of proposition 4.4 in step 1 and step 2:

(i) U i;k,t,x s (e) = (v i (s, k X t,x s-+ β(s, k X t,x s-, e)) -v i (s, k X t,x s-)), ds ⊗ dP ⊗ dλ k -a.e on [t, T ] × Ω × E; (ii) the sequence ((v k i=1,m
)) k≥1 converge to v i (t, x) by using (4.57). So that x k -→ k x we take the following estimation which is obtaining by Ito's formula and by the properties of h (i) .

E Y k,t,x k s -Y k,t,x s 2 + K k,t,x k T -K k,t,x T 2 + T 0 Z k,t,x k s -Z k,t,x s 2 + E U k,t,x k s -U k,t,x s 2 λ k (de) ds ≤ E g( k X t,x k T ) -g( k X t,x T ) 2 + E sup s≤T ℓ( k X t,x k s ) -ℓ( k X t,x s ) 2 + CE T s Y k,t,x k r -Y k,t,x r 2 dr +CE T 0 k X t,x k r -k X t,x r 2 (1 + k X t,x k r p + k X t,x r p ) 2 dr +C i=1,m E T s B i u i (r, k X t,x k r ) -B i u i (r, k X t,x r ) 2 dr .
(5.71)

Next using (4.52) and (4.53), the continuty of the function (t, x) → B i u i (t, x) and the fact of it belong to Π g and in the other hand the majoration of the fourth term of (4.56); we can use Gronwall's lemma for s = t ∀i = 1, . . . , m,

v k i (t, x k ) -→ k v k i (t, x).
Therefore by (i)-(ii) we have, for any i = 1, . . . , m, U i;t,x s (e) = (v i (s, X t,x s-+β(s, X t,x s-, e))-v i (s, X t,x s-)) ds⊗dP⊗dλ-a.e. in [t, T ]×Ω×E, ∀i ∈ {1, . . . , m}.

(5.72)

By this result we can replace (u i (s, X t,x s-+ β(s, X t,x s-, e))u i (s, X t,x s-)) by U i;t,x s (e) in (5.68), we deduce that the quadriple (Y t,x , K t,x , Z t,x , U t,x ) verifies: ∀i ∈ {1, . . . , m}

                           (i) Y t,x
:= (Y i;t,x ) i=1,m ∈ S 2 (R m ), Z t,x := (Z i;t,x ) i=1,m ∈ H 2 (R m×d ), K t,x := (K i;t,x ) i=1,m ∈ A With the uniqueness of solution (5.66), we have u i = u i = v i which means that the solution of (1.1) in the sense of Definition 5.1 is unique inside the class U .

Extension

In this section, we will redefine the function h (i) as a function of U i,t,x L 2 (λ) ∀i ∈ {1, . . . , m}. And to show that the results of the previous section remain valid. Let us consider for any i ∈ {1, . . . , m} the functions f (i) , defined by

∀(t, x, y, z, ζ) ∈ [0, T ] × R k × R m+d × L 2 (λ); f (i) (t, x, y, z, ζ) = h (i) (t, x, y, z, ζ L 2 (λ) );
where the functions (h (i) ) i=1,m are the sames defined in section 2.

We recall that the result of Theorem 5.2 is obtained by having mainly U t,x s (e) = (u i (s, X t,x s-+ β(s, X t,x s-, e))u i (s, X t,x s-)); this makes it possible to have the definition 4.1 by passing through a modification of the expression of B i u i ∀i ∈ {1, . . . , m}.

We show that U i,t,x s (e) 2 L 2 (λ) = (u i (s, X t,x s-+ β(s, X t,x s-, e))u i (s, X t,x s-))| 2 L 2 (λ) and that in this case B i u i is well ∀i ∈ {1, . . . , m}.

Let now (t, x) ∈ [0, T ] × R d and let us consider the following m-dimensional RBSDE with jumps:

                            
(i) Y t,x := (Y i,t,x ) i=1,m ∈ S 2 (R m ), Z t,x := (Z i,t,x ) i=1,m ∈ H 2 (R m×d ), K t,x := (K i,t,x ) i=1,m ∈ A 2 c , U t,x := (U i,t,x ) i=1,m ∈ H 2 (L 2 m (λ)); ∀i ∈ {1, . . . , m} Y i;t,x T = g i (X t,x T ) and; (ii) dY i;t,x s = -f (i) (s, X = w i (s, X t,x s ).

Such that by the same proposition, the family (w i ) i=1,m is a viscosity solution of the following system:          min w i (t, x)ℓ(t, x); -∂ t w i (t, x)b(t, x) ⊤ D x w i (t, x) -1 2 Tr(σσ ⊤ (t, x)D 2 xx w i (t, x)) -K i w i (t, x)h (i) (t, x, (w j (t, x)) j=1,m , (σ ⊤ D x w i )(t, x), B i w i (t, x)) = 0, (t, x) ∈ [0, T ] × R k ; w i (T, x) = g i (x).

(6.75) Indeed, using Lemma 4.1 and the fact that U i;t,x k converges to U i;t,x ∀i ∈ {1, . . . , m} when

x k -→ k x, we deduce that U i,t,x k s (e) L 2 (λ) -→ k U i,t,x s (e) L 2 (λ) .

Moreover, from property of h (i) and the proof of theorem 5.2 step 2 (viscosity solution uniqueness), U i,t,x s (e) 2 L 2 (λ) = (w i (s, X t,x s-+ β(s, X t,x s-, e))w i (s, X t,x s-)) 2 L 2 (λ) ; from where

B i w i = E
|(w i (s, X t,x s-+ β(s, X t,x s-, e))w i (s, X t,x s-))| 2 λ(de) .

Thanks to corollary 4.3, we deduce that B i w i is well defined ∀i ∈ {1, . . . , m}.

(

  |∆U r (e)| 2 λ(de) .

Corollary 4 . 3

 43 For u i ∈ U ∀i = 1, . . . , m B i u i defined in (1.2) is well posed since the functions β and (γ i ) i=1,m verify (2.7) and (2.10) respectively.

Theorem 5 . 2

 52 Under assumptions (H1), (H2) and (H3), the IPDE (1.1) has unique solution which is the m-tuple of functions (u i ) i=1,m defined in proposition 3.3 (3.15).

  Since that the RBSDE with jumps (5.64) has solution and it is unique by assumed that (H1), (H2) and (H3) are verified. By proposition 3.3 the (3.15), there exists a family of deterministic continuous functions of polynomial growth (v i ) i=1,m that fact for any (t, x

	associated to IPDE (1.1).
	t,x s ) and	0	T	(Y	i;t,x s	-ℓ(s, X t,x s ))dK	i;t,x s	= 0.	(5.66)
	By Feynman Kac formula u i (s, X t,x s ) = Y i;t,x s	where Y i;t,x s	satisfies the RBSDE with jumps (1.3)

  By assumed that (H1), (H2) and (H3) are verified and by proposition 3.3 the (3.15), there exists a family of deterministics continuous functions of polynomial growth (w i ) i=1,m that fact for any (t, x) ∈ [0, T ] × R

		t,x s , (Y i;t,x s	) i=1,m , Z i;t,x s	, U t,x s (e) L 2 (λ) )ds -dK i;t,x s	(6.74)
	(iii) Y i;t,x s	+Z i;t,x s ≥ ℓ(s, X t,x s ) and dB s + T 0	U i;t,x s (Y i;t,x E s -ℓ(s, X t,x (e) µ(ds, de), s ≤ T ; s ))dK i;t,x s = 0.

k , ∀s ∈ [t, T ],

Y i;t,x s

Appendix. Barles et al.'s definition for viscosity solution of IPDE (1.1)

In the paper by Barles et al. [START_REF] Barles | Backward stochastic differential equations and integral-partial differential equations[END_REF], the definition of the viscosity solution of the system (1.1) is given as follows. Definition 6.1 We say that a family of deterministics functions u = (u i ) i=1,m which is continuous ∀i ∈ {1, . . . , m}, is a viscosity sub-solution (resp. super-solution) of the IPDE (1.1) if:

The family u = (u i ) i=1,m is a viscosity solution of (1.1) if it is both a viscosity sub-solution and viscosity super-solution.

Note that L X φ(t, x) = b(t, x) ⊤ D x φ(t, x) + 1 2 Tr(σσ ⊤ (t, x)D 2 xx φ(t, x)) + Kφ(t, x); where Kφ(t, x) = E (φ(t, x + β(t, x, e))φ(t, x)β(t, x, e) ⊤ D x φ(t, x))λ(de).