Malliavin Calculus for Parabolic S.P.D.E.s with Jumps.

Nicolas FOURNIER1

October, 5, 1999

Abstract

We study a parabolic S.P.D.E. driven by a white noise and a compensated Poisson measure. We firstdefine the solutions in a weak sense, and we prove the existence and the uniqueness of a weak solution.Then we use the Malliavin calculus in order to show that under some non-degeneracy assumptions,the law of the weak solution admits a density with respect to the Lebesgue measure. To this aim, weintroduce two derivative operators associated with the white noise and the Poisson measure. The oneassociated with the Poisson measure is studied in details.

0 Introduction.

Let T > 0 be a positive time, let (Ω,F , (Ft)t∈[0,T ], P ) be a probability space, and let L be a positive realnumber. We consider [0, T ]× [0, L] and [0, T ]× [0, L]× IR endowed with their Borel σ-fields. Let W (dx, dt)be a space-time white noise on [0, L] × [0, T ] based on dxdt (see e.g. J.B. Walsh [9], p 269), and let Nbe a Poisson measure on [0, T ] × [0, L] × IR, independent of W , with intensity measure ν(dt, dx, dz) =dtdxq(dz), where q is a positive σ-finite measure on IR. The compensated Poisson measure is denoted byN = N − ν. Our purpose is to study the following one-dimensional stochastic partial differential equationon [0, L] × [0, T ] :

[

∂V

∂t(x, t) − ∂2V

∂x2(x, t)

]

dxdt = g(V (x, t))dxdt + f(V (x, t))W (dx, dt)

+

∫

IRh(V (x, t), z)N (dt, dx, dz) (0.1)

with Neumann boundary conditions

∀ t > 0,∂V

∂x(0, t) =

∂V

∂x(L, t) = 0 (0.2)

and with deterministic initial condition V (x, 0) = V0(x).In this paper, we first prove the existence and uniqueness of a weak solution V (x, t) to (0.1). Thenwe show, in the case where q(dz) admits a sufficiently regular density, and under some non-degeneracyconditions, that the law of V (x, t) is absolutely continuous with respect to the Lebesgue measure as soon

1Laboratoire de Probabilites, UMR 7599, Universite Paris VI, 4, Place Jussieu, Tour 56, 3o etage, F-75252 Paris Cedex

05, e-mail : fournier@proba.jussieu.fr.

1

as t > 0.Parabolic S.P.D.E.s driven by a white noise, i.e. equation (0.1) with h ≡ 0, have been introduced byWalsh, [8] and [9]. In [9], he defines his weak solutions, then he proves a theorem of existence, uniquenessand regularity. Since, various properties of Walsh’s equation have been investigated. In particular theMalliavin calculus has been developped by Pardoux, Zhang, [6], and Bally, Pardoux, [1].But Walsh builds his equation in order to model a discontinuous neurophysiological phenomenous. In [8],he explains that the white noise W approximates a Poisson point process. This approximation is realisticbecause there are many jumps, and the jumps are very small, but in any case, the observed phenomenousis discontinuous. However, S.P.D.E.s with jumps are much less known. In the case where f ≡ 0, SaintLoubert Bie has studied in [7] the existence, uniqueness, regularity, and stochastic variational calculus.We prove here a result of existence and uniqueness, because we define in a slighlty different way the weaksolutions, and because Saint Loubert Bie does not study exactly the same equation. Furthermore, hisresult about the absolute continuity does not extend to the present case.The Malliavin calculus for jump processes we will build extends the work of Bichteler, Gravereaux, Jacod,[2], who study diffusion processes with jumps. We can not apply directly their methods, essentially becausethe weak solution of (0.1) is not a semi-martingale. Bichteler et al., [2], use a ”scalar product of derivation”,which does not allow to obtain satisfying results in the present case (see Saint Loubert Bie, [7]). Thus wehave to introduce a real ”derivative operator”, which gives more information.Our method is also inspired by the paper of Bally, Pardoux, [1] who prove the existence of a smoothdensity in the case where h ≡ 0.The present work is organized as follows. In Section 1, we define the solutions of (0.1) in a weak sense,which is easy in the continuous case but slightly more difficult here, because there are ”predictability”problems. Then we state our main results. An existence and uniqueness result is proved in Section 2. Westudy the existence of a density for the law of the weak solution in Section 3. Finally, an appendix lies atthe end of the paper.

1 Statement of the main results.

In the whole work, we assume that (Ω,F , Ftt∈[0,T ], P ) is the canonical product probability space asso-ciated with W and N . In particular,

Ft = σ W (A) ; A ∈ B([0, L] × [0, t]) ∨ σ N(B) ; B ∈ B([0, t] × [0, L] × IR) (1.1)

Definition 1.1 Consider a process Y = Y (y, s)[0,L]×[0,T ]. We will say that Y is

• predictable if it is Pred ⊗ B([0, L])-measurable, where Pred is the predictable σ-field on Ω × [0, T ].

• bounded in L2 if sup[0,L]×[0,T ] E(Y 2(y, s)) < ∞.

• a version of X = X(y, s)[0,L]×[0,T ] if for all y ∈ [0, L], all s ∈ [0, T ], a.s., Y (y, s) = X(y, s).

• a weak version of X = X(y, s)[0,L]×[0,T ] if dP (ω)dyds-a.e., Y (y, s)(ω) = X(y, s)(ω).

• of class PV if it is bounded in L2 and if it is a weak version of a predictable process.

We now define the stochastic integrals we will use.

Definition 1.2 Let Y be a process that admits a predictable weak version Y−. Let Φ be a measurablefunction such that

∫ T

0

∫ L

0

∫

IRE(

φ2(Y (y, s), s, y, z))

q(dz)dyds < ∞ (1.2)

2

Then we set

∫ T

0

∫ L

0

∫

IRφ(Y (y, s), s, y, z)N (ds, dy, dz) =

∫ T

0

∫ L

0

∫

IRφ(Y−(y, s), s, y, z)N (ds, dy, dz) (1.3)

The obtained random variable does not depend on the choice of the predictable version, up to a P (dω)-negligible set. We define in the same way the stochastic integral against the white noise.

Using the classical stochastic integration theory (see Jacod, Shiryaev, [4] p 71-74, and Walsh, [9] p 292-298),we deduce, since Y− = Y dPdyds-a.e., that :

E

(

∫ T

0

∫ L

0

∫

IRΦ(Y (y, s), s, y, z)N (ds, dy, dz)

)2

=

∫ T

0

∫ L

0

∫

IRE(

Φ2(Y (y, s), s, y, z))

q(dz)dyds (1.4)

E

(

∫ T

0

∫ L

0Φ(Y (y, s), s, y)W (dy, ds)

)2

=

∫ T

0

∫ L

0E(

Φ2(Y (y, s), s, y))

dyds (1.5)

E

(

∫ T

0

∫ L

0Φ(Y (y, s), s, y)dyds

)2

≤ TL

∫ T

0

∫ L

0E(

Φ2(Y (y, s), s, y))

dyds (1.6)

We now would like to define the weak solutions of (0.1). First, we suppose the following conditions, whichin particular allow all the integrals below to be well-defined.Assumption (H) : f and g satisfy some global lipschitz conditions on IR, h is measurable on IR × IR, and

there exists a positive function η ∈ L2(IR, q) such that for all x, y, z ∈ IR,

|h(0, z)| ≤ η(z) and |h(x, z) − h(y, z)| ≤ η(z)|x − y| (1.7)

Assumption (D) : V0 is deterministic, B([0, L])-measurable, and bounded.

Following Walsh, [8], [9], or Saint Loubert Bie, [7], we define the weak solutions of (0.1) by using anevolution equation. Let Gt(x, y) be the Green kernel of the deterministic system :

∂u

∂t=

∂2u

∂x2;

∂u

∂x(0, t) =

∂u

∂x(L, t) = 0 (1.8)

This means that Gt(x, y) is the solution of the system with initial condition a Dirac mass at y. It iswell-known that

Gt(x, y) =1√4πt

∑

n∈Z

[

exp

(

−(y − x − 2nL)2

4t

)

+ exp

(

−(y + x − 2nL)2

4t

)]

(1.9)

All the properties of G that we will use can be found in the Appendix. Now we can define the weaksolutions of equation (0.1).

Definition 1.3 Assume (H) and (D). A process V of class PV is said to be a weak solution of (0.1) iffor all x in [0, L], all t > 0, a.s.

V (x, t) =

∫ L

0Gt(x, y)V0(y)dy +

∫ t

0

∫ L

0Gt−s(x, y)f(V (y, s))W (dy, ds) (1.10)

+

∫ t

0

∫ L

0Gt−s(x, y)g(V (y, s))dyds +

∫ t

0

∫ L

0

∫

IRGt−s(x, y)h(V (y, s), z)N (ds, dy, dz)

where we have used Definition 1.2

3

Let us finally state our first result.

Theorem 1.4 Assume (H) and (D). Equation (0.1) admits a unique solution V ∈ PV in the sense ofDefinition 1.3. The uniqueness holds in the sense that if V ′ ∈ PV is another weak solution, then V andV ′ are two versions of the same process, i.e. for each x, t, a.s., V (x, t) = V ′(x, t).

It is not standard to work with predictable weak versions. In the continuous case, no such problem appear,and the classical diffusion processes with jumps are a.s. cadlag. But here, the paths of a weak solutioncannot be cadlag in time. Indeed, this is even impossible in the much simpler case where V0 = 1, f = g = 0,h(x, z) = 1, where q(IR) < ∞, and where the Poisson measure is not compensated. In such a case, thePoisson measure is finite, thus it can be written as N =

∑µi=1 δTi,Xi,Zi, and hence the weak solution of

(0.1) is given by

V (x, t) = 1 +µ∑

i=1

Gt−Ti(x,Xi)1t>Ti

In this case, we see that for each ω ∈ Ω satisfying µ(ω) ≥ 1, the map t 7→ V (X1(ω), t)(ω) explodes when tdecreases to T1(ω).

We are now interested in the Malliavin calculus. We thus suppose some more conditions. First, the inten-sity measure of N has to be sufficiently ”regular”.

Assumption (M) : N has the intensity measure ν(ds, dy, dz) = ϕ(z)1O(z)dsdydz, where O is an open

subset of IR, and ϕ is a strictly positive C1 function on O.

The functions f , g, h also have to be regular enough.

Assumption (H ′) : f and g are C1 functions on IR, and their derivatives are bounded. The function h(x, z)on IR ×O admits the continuous partial derivatives h′

z, h′x, and h′′

zx = h′′xz. There exist a constant K and

a function η ∈ L2(O,ϕ(z)dz) such that for all x ∈ IR, all z ∈ O,

|h′z(0, z)| + |h′′

xz(x, z)| ≤ K ; |h(0, z)| + |h′x(x, z)| ≤ η(z) (1.11)

Notice that (H ′) is stronger than (H).Let ρ be a strictly positive C1 function on O such that ρ and ρ′ are bounded, and such that

ρ ∈ L1(O,ϕ(z)dz) (1.12)

This ”weight function” can be chosen according to the parameters of (0.1). The next condition is technical.

Assumption (S) : there exists a family of C1 positive functions Kǫ on O, with compact support (in O),bounded by 1, and such that

∀ z ∈ O,Kǫ(z) −→ǫ→0 1 ;

∫

O

(

K ′ǫ(z)

)2η2(z)ρ(z)ϕ(z)dz −→ǫ→0 0 (1.13)

We finally suppose one of the following non-degeneracy conditions :

Assumption (EW ) : for all x in IR, f(x) 6= 0orAssumption (EP1) : f = 0, and there exists η ∈ L1(O,ϕ(z)dz) such that 0 ≤ h′

x(x, z) ≤ η(z). For all xin IR,

∫

O1h′

z(x,z)6=0ϕ(z)dz = ∞ (1.14)

4

orAssumption (EP2) : we set H = z ∈ O /∀x ∈ IR, h′

z(x, z) 6= 0. There exist some constants C0 > 0,

r0 ∈]34 , 1[, and γ0 ≥ 0 such that for all γ ≥ γ0,

∫

H

(

1 − e−γρ(z))

ϕ(z)dz ≥ C0 × γr0 (1.15)

Our second main result is the next theorem.

Theorem 1.5 Assume (M), (D), (H ′), and (S). Let V be the unique weak solution of (0.1) in the senseof Definition 1.3, and let (x, t) ∈ [0, L]×]0, T ]. Then under one of the assumptions (EW ), (EP1) or(EP2), the law of V (x, t) admits a density with respect to the Lebesgue measure on IR.

We will use two derivative operators. The first one, associated with the white noise, is classical (see Nualart[5]). The second operator, associated with the Poisson measure, is inspired from Bichteler, Gravereaux,and Jacod, Chapter IV in [2]. They study the Malliavin calculus for diffusion processes with jumps, in thecase where the intensity measure of the Poisson measure is 1O(z)dsdz. Furthermore, they do not use anyderivative operator : they work with a ”scalar product of derivation”, which gives less information. Usingthis method, we could probably prove Theorem 1.5 only under (EP1).Our theorem gives in fact two results : the law of V (x, t) admits a density either thanks to W or thanksto N . It seems to be very difficult to state a “joint” non-degeneracy condition (see Subsection 3.5).Assumption (EW ) looks reasonnable : although Pardoux and Zhang prove this Theorem under a reallyless stringent assumption when h = 0 in [6] (it suffices that ∃y ∈ [0, L] such that f(V0(y)) 6= 0), they usethe continuity of their solution. The first condition in (EP1) (f = 0, h′

x ≥ 0, h′x ≤ η) is very stringent,

but the second one might be optimal : Bichteler et al. also have to assume this kind of condition. Finally,(EP2) is much more general, but it is an uniform non degeneracy assumption.

St Loubert Bie proves in [7] the existence of a density under the assumption f = 0, an hypothesis lessstringent than (M), an assumption quite similar to (H ′), and under (h1) or (h2) below (the notations areadapted to our context) :(h1) : h′

x = 0 and∫

O 1h′

z(z)=0ϕ(z)dz = ∞or(h2) : η ∈ L1(O,ϕ(z)dz), h′

x ≥ 0, and something like (EP1), but depending on the solution process V .

Condition (h1) is very restrictive, and (h2) is not very tractable : one has to know the behaviour of theweak solution. Saint Loubert Bie uses in both case the positivity of N (as in the proof of Theorem 1.5under (EP1)). But since the white noise is signed, this method can not be extended to the case wheref 6≡ 0. That is why in this work, the most interesting assumption is probably (EP2).

Let us finally give examples about assumptions (S) and (EP2).

Remark 1.6 Assume that O = IR. Then (S) is satisfied for any ϕ, η, and any choice of ρ.

Proof : it suffices to choose a family of C1 positive functions of the form

Kǫ(z) =

1 if |z| < 1/ǫ0 if |z| > 1/ǫ + 2

such that |K ′ǫ(z)| ≤ 1|z|∈[1/ǫ,1/ǫ+2. Using the Lebesgue Theorem and the fact that ρη2 ∈ L1(IR,ϕ(z)dz),

(1.13) is immediate.

Example 1 : assume that O = IR, and that ϕ is a C1 function on IR satisfying, for some K > a > 0,K > ϕ > a. We consider a function h(x, z) = c(x)η(z), where c is a strictly positive C1 function on IR

5

of which the derivative is bounded. η has to be C1 on IR, to belong to L2(IR,ϕ(z)dz), and η′ must bebounded. If for some b ∈ IR, [b,∞[⊂ η′ 6= 0, then (M), (H ′), (S), and (EP2) are satisfied :

thanks to Remark 1.6, it suffices to check (EP2). Choosing ρ(z) ≥ z−7

6 1z≥b∨1, we see that (1.15) issatisfied, since [b,∞[⊂ H, and using

∀ x ∈ [0, 1], 1 − e−x ≥ x/2 (1.16)

Example 2 : assume that O =]0, 1[, and that ϕ(z) = 1/zr, for some r > 7/4. We consider a functionh(x, z) = c(x)η(z), where c is a strictly positive C1 function on IR of which the derivative is bounded, andwhere η(z) = zα, for some α > 1 ∨ r−1

2 ∨ 7−r6 . Then (M), (H ′), (S), and (EP2) are satisfied :

(M) is met, and (H ′) holds, since α > 1 ∨ r−12 . It is clearly possible to choose ρ(z) of the form

ρ(z) =

zβ if z ≤ 1/4(1 − z)δ if z ≥ 3/4

(1.17)

with β > 1 ∨ (r − 1) and δ ≥ 1. Using (1.16), the facts that H =]0, 1[ and that ρ(z) ≥ zβ1]0,1/4[(z), we see

that (EP2) is satisfied if β < 43 (r − 1).

We now choose a family Kǫ of C1 positive functions on ]0, 1[, bounded by 1, and satisfying

Kǫ(z) =

0 if z < ǫ/21 if ǫ < z < 1 − ǫ0 if 1 − ǫ/2 < z < 1

; |K ′ǫ(z)| ≤ 4

ǫ1]ǫ/2,ǫ[∪]1−ǫ,1−ǫ/2[(z)

An explicit computation shows that (S) is satisfied if β > r + 1 − 2α and if δ > 1.Since α > 7−r

6 and r > 7/4, it is possible to choose β in ](r − 1), 43 (r − 1)[∩]1,∞[∩]r + 1− 2α,∞[, and the

conclusion follows.

Let us finally remark that (S) is satisfied for any O, η, φ, if ρ is of class C2b , and if ρ(z)+ |ρ′(z)| −→z→∂O 0.

Bichteler, Gravereaux and Jacod assume this kind of condition about ρ in [2].

2 Existence and uniqueness.

In this short section, we sketch the proof of Theorem 1.4. We begin with a fundamental lemma.

Lemma 2.1 Assume (H). Let Y be a process of class PV. Then the processes

U(x, t) =

∫ t

0

∫ L

0

∫

IRGt−s(x, y)h(Y (y, s), z)N (ds, dy, dz) (2.1)

X(x, t) =

∫ t

0

∫ L

0Gt−s(x, y)f(Y (y, s))W (dy, ds) (2.2)

Z(x, t) =

∫ t

0

∫ L

0Gt−s(x, y)g(Y (y, s))dyds (2.3)

belong to PV.

Proof : let us for example prove the lemma for U . First notice that U is bounded in L2 thanks to (1.4),(H), the fact that Y is bounded in L2, and the Appendix (4.3). We still have to prove that U admits apredictable weak version. We know from Walsh, [9] p 323, that

supx∈[0,L]

∫ T

0

∫ L

0

(

Gt(x, y) −N∑

k=0

φk(x)φk(y)e−λkt

)2

dydt −→N→∞ 0 (2.4)

6

where φ0 = 1√L, φk(x) =

√

2L cos

(

kπxL

)

, and λk =( π

L

)2k2. We thus can approximate U(x, t) by

UN (x, t) =N∑

k=0

φk(x)e−λkt∫ t

0

∫ L

0

∫

IRφk(y)eλksh(V (y, s), z)N (ds, dy, dz)

which clearly admits a predictable version since for each k, φk(x)e−λkt is deterministic and the process

t 7→∫ t

0

∫ L

0

∫

IRφk(y)eλksh(V (y, s), z)N (ds, dy, dz)

is a cadlag martingale. Using (1.4), (H), and (2.4), one easily checks that, when N goes to infinity,

supx,t

E

(

(

U(x, t) − UN (x, t))2)

−→ 0 (2.5)

Since for each N , UN admits a predictable version, and since there exists a subsequence of UN goingdPdxdt-a.e. to U , we deduce that U admits a predictable weak version.

Let us remark that even if Y is a predictable process, the process U defined by (2.1) is not a priori pre-dictable, but only admits a predictable weak version.

Proof of Theorem 1.4 : the uniqueness easily follows from Gronwall’s Lemma applied to the functionφ(t) = supx E((V (x, t) − V ′(x, t))2). Let us prove the existence. To this aim, we first build the followingPicard approximations.

V 0(x, t) =

∫ L

0Gt(x, y)V0(y)dy

V n+1(x, t) = V 0(x, t) +

∫ t

0

∫ L

0Gt−s(x, y)f(V n(y, s))W (dy, ds) (2.6)

+

∫ t

0

∫ L

0Gt−s(x, y)g(V n(y, s))dyds

+

∫ t

0

∫ L

0

∫

IRGt−s(x, y)h(V n(y, s), z)N (ds, dy, dz)

Due to (D) and the appendix (4.2), V 0 is deterministic and bounded. We deduce from Lemma 2.1 that foreach n, V n is well-defined and is of class PV . A simple computation using (1.4), (1.5), (1.6), assumption(H), and the appendix, (4.2), shows that for each n ≥ 1,

E(

(V n+1(x, t) − V n(x, t))2)

≤ K

∫ t

0

ds√t − s

supx

E(

(V n(x, s) − V n−1(x, s))2)

(2.7)

We now set φn(t) = supx E(

(V n+1(x, t) − V n(x, t))2)

. We obtain, iterating once (2.7), and using the factthat

∫ t0

ds√t−s

∫ s0

du√s−u

≤ 4,

φn(t) ≤ K

∫ t

0φn−1(s)

ds√t − s

≤ K

∫ t

0φn−2(s)ds (2.8)

Since φ0 is bounded (because of (D)), we deduce from the first inequality in (2.7) that φ1 is also bounded.Thus Picard’s Lemma allows to conclude that

∑

n

sup[0,T ]

(φ2n(t))1/2 < ∞ ;∑

n

sup[0,T ]

(φ2n+1(t))1/2 < ∞ (2.9)

7

and hence,∑

n

supx,t

E(

(V n+1(x, t) − V n(x, t))2)

< ∞ (2.10)

This clearly implies the existence of a process V bounded in L2 such that, when n tends to infinity,

sup[0,L]×[0,T ]

E(

(V n(x, t) − V (x, t))2)

−→ 0 (2.11)

This process belongs to PV : for each n, there exists a predictable process V n− which is a weak version of

V n, and it is clear that V n− goes to V dPdxdt-a.e.

Finally, making n go to infinity in (2.6), (by using (2.11)), we see that V satisfies (1.10). The proof isfinished.

3 The Malliavin calculus.

The aim of this main section is to prove Theorem 1.5. In Subsection 3.1, we will define some derivativeoperators. In Subsection 3.2, we will state the main properties of these operators, and derive a criterionof absolute continuity. We will say how to “differentiate” stochastic integrals in Subsection 3.3, and then“differentiate” the weak solution of (0.1) in Subsection 3.4. We will conclude in Subsection 3.5.

In the whole section, we will assume at least (M), (D), (H ′), and (S).

3.1 The derivative operators.

We denote by Cpc (IRd) (resp. Cp

b (IRd)) the set of Cp functions on IRd with compact support (resp. of whichthe derivatives of order 1 to p are bounded). As said previously, we will define two derivative operators.

We begin with the derivative operator associated with the Poisson measure. We first denote by CL the setof measurable functions l(s, y, z) on [0, T ] × [0, L] × O, with compact support, C2 on O (in z), such thatl, l′z, and l′′zz are bounded on [0, T ] × [0, L] × O. Then we define the domain

SN =

X = F (N(g1), ..., N(gd)) + a/

d ≥ 1, F ∈ C2c (IRd), gi ∈ CL, a ∈ IR

(3.1)

where N(gi) stands for∫ T0

∫ LO

∫

O gi(s, x, z)N(ds, dx, dz). If X ∈ SN , if α ∈ [0, L], τ ∈ [0, T ], and ζ ∈ O, weset

DNα,τ,ζX =

d∑

i=1

∂iF (N(g1), ..., N(gd))(gi)′z(α, τ, ζ) (3.2)

In order to “close” (SN ,DN ), we introduce another operator. For X ∈ SN ,

LNX =1

2

d∑

i=1

∂iF (N(g1), ..., N(gd))N

(

ρ.(gi)′′zz +

(

ρϕ′

ϕ+ ρ′

)

.(gi)′z

)

+1

2

d∑

i,j=1

∂i∂jF (N(g1), ..., N(gd))N(

ρ.(gi)′z.(gj)

′z

)

(3.3)

We finally define a scalar product. If Sα,τ,ζ(ω) and Tα,τ,ζ(ω) are in L2(P (dω)ρ(ζ)N(ω, dτ, dα, dζ)), we set

〈S, T 〉ρN =

∫ T

0

∫ L

0

∫

OSα,τ,ζTα,τ,ζ ρ(ζ)N(dτ, dα, dζ) and 〈S〉ρN = 〈S, S〉ρN

8

Then one gets easily, for all X and Y in SN ,

⟨

DNX,DNY⟩

ρN= LNXY − XLNY − Y LNX (3.4)

By adapting Bichteler et al., [2] Proposition 9-3, p 113., we check in the lemma below that LN is well-defined : if X = F (N(f1), ..., N(fd)) = G(N(g1), ..., N(gq)), then using one expression or the other willgive the same LNX.

Lemma 3.1 If X = F (N(f1), ..., N(fk)) ∈ SN , and if X ≡ 0, then LNX ≡ 0, and thus⟨

DNX⟩

ρN≡ 0.

Proof : we assume that Ω is the set of the integer valued measures on [0, T ] × [0, L] × O. Let ω ∈ Ω and(t, x, z) ∈ supp ω be fixed. We set ω′ = ω − δ(t,x,z), and for λ ∈ Λ, ωλ = ω′ + δ(t,x,z+λ), where Λ is a

neighbourhood of 0 in IR such that z + Λ ⊂ O. Then ω′ and ωλ are in Ω. We set

Xt,x,z(λ) = X(ωλ) = F (ω′(f1) + f1(t, x, z + λ), ..., ω′(fk) + fk(t, x, z + λ))

Then Xt,x,z vanishes identically, and is C2 in λ. We deduce that

1

ϕ(z)

∂

∂λ

(

ρ(z + λ)ϕ(z + λ) × ∂

∂λXt,x,z(λ)

)∣

∣

∣

∣

λ=0= 0

Writing this explicitely, and summing the obtained expression on all the points (t, x, z) ∈ supp ω, we getLNX(ω) = 0.

We thus see that for each ω ∈ Ω, DNX(ω) is well-defined up to a N(ω)-negligible set : we could replaceDN

α,τ,ζX(ω) with anything when N(ω, α, τ, ζ) = 0. In order to understand this notion of derivative,assume that Ω is the canonical space associated with N . Then every ω ∈ Ω is a counting measure on[0, T ] × [0, L] × O, and

1ω(α,τ,ζ=1DNα,τ,ζX(ω) = 1ω(α,τ,ζ=1

∂

∂λX(ω − δ(α,τ,ζ) + δ(α,τ,ζ+λ))

∣

∣

∣

∣

λ=0(3.5)

The following lemma can be proved as Proposition 9-3, d), p 113 in [2].

Lemma 3.2 If X and Y are in SN , then

E(XLNY ) = E(Y LNX) = −1

2E

(

⟨

DNX,DNY⟩

ρN

)

(3.6)

We deduce from (3.6) the following lemma, which shows that DN is closable.

Lemma 3.3 Let Zn be a sequence of SN which goes to 0 in L2. Assume that there exists Sα,τ,ζ(ω) ∈L2(P (dω)ρ(ζ)N(ω, dτ, dα, dζ)) such that E

(

⟨

DNZn − S⟩

ρN

)

goes to 0. Then E(〈S〉ρN ) = 0.

Proof : Let X be in SN . The Cauchy-Schwarz inequality yields

E

(

⟨

S,DNX⟩

ρN

)

= limn

E

(

⟨

DNZn,DNX⟩

ρN

)

But, thanks to Lemma 3.2, E

(

⟨

DNZn,DNX⟩

ρN

)

= −2E(ZnLNX). Since Zn goes to 0 in L2, we deduce

that E

(

⟨

S,DNX⟩

ρN

)

= 0. Then we apply this with X = Zk, and we let k go to infinity.

9

We now define the derivative operator associated with the white noise. First we define the domain of the“smooth variables” :

SW =

X = F (W (f1), ...,W (fk)) + a/

k ≥ 1, F ∈ C2c (IRk), fi ∈ L2([0, L] × [0, T ]), a ∈ IR

(3.7)

Here, W (fi) =∫ T0

∫ L0 fi(s, x)W (dx, ds). If X is in SN , if α ∈ [0, L] and τ ∈ [0, T ], we set :

DWα,τX =

k∑

i=1

∂iF (W (f1), ...,W (fk))fi(α, τ) (3.8)

If Sα,τ (ω) and Tα,τ (ω) are in L2(P (dω)dαdτ), we set

〈S, T 〉leb =

∫ T

0

∫ L

0Sα,τTα,τdαdτ and 〈S〉leb = 〈S, S〉leb (3.9)

The following lemma can be found in Nualart, [5], p 26.

Lemma 3.4 Let Zn be a sequence of SW that goes to 0 in L2. Assume that there exists Sα,τ (ω) ∈L2(P (dω)dαdτ) such that E

(⟨

DW Zn − S⟩

leb

)

goes to 0. Then E(〈S〉leb) = 0.

Now we can build the operators on the product space. The smooth variables domain is

S =

Y = H(W (f1), , ...,W (fd), N(g1), ..., N(gk)) + a/

H ∈ C2c (IRd+k), k + d ≥ 1,

fi ∈ L2([0, L × [0, T ]), gj ∈ CL, a ∈ IR (3.10)

If Y belongs to S, we define DNY and LNY (resp. DW Y ) as previously, considering the variablesW (f1), ...,W (fd) (resp. N(g1), ..., N(gk)) as constants :

DWα,τY =

d∑

i=1

∂iH(W (f1), , ...,W (fd), N(g1), ..., N(gk))fi(α, τ) (3.11)

DNα,τ,ζY =

d+k∑

i=d+1

∂iH(W (f1), , ...,W (fd), N(g1), ..., N(gk))(gi)′z(α, τ, ζ) (3.12)

The scalar products are denoted as previously, and we see that if X and Z are in S, then X, Z, and⟨

DW X,DW Z⟩

lebare bounded ; and

⟨

DNX,DNZ⟩

ρNbelongs to ∩p<∞Lp.

If Z belongs to S, we set

|||Z|||2 =

[

E(

Z2)

+ E(⟨

DW Z⟩

leb

)

+ E

(

⟨

DNZ⟩

ρN

)] 1

2

(3.13)

We denote by D2 the closure of S for this norm. Because of Lemmas 3.3 and 3.4, the operators DWα,τ and

DNα,τ,ζ can be extended to the space D2.

Remark 3.5 We have extended DW and DN to D2, and the weak solution of (0.1) will belong to thisspace. But no integration by parts formula (like (3.6)) hold on D2, because LN can not be extended to thisspace. Nevertheless, the ”differentiability” of our solution will allow us to prove Theorem 1.5.

Notation 3.6 We will denote by Tnn≥0 a sequence of stopping times, by (Xn, Zn)n≥0 a sequence of[0, L] × O-valued random variables, such that for each n, (Xn, Tn) is FTn

-measurable, and such that

N(dt, dx, dz) =∞∑

n=0

δ(Tn,Xn,Zn)(dt, dx, dz) (3.14)

10

Remark 3.7 1. The way we have closed (SN ,DN ) shows that if X ∈ D2, and if Y = X a.s., thenY ∈ D2, and a.s.,

⟨

DW X − DW Y⟩

leb= 0 ;

⟨

DNX − DNY⟩

ρN= 0 (3.15)

2. Let Sα,τ,ζ(ω) and S′α,τ,ζ(ω) belong to L2(P (dω)ρ(ζ)N(ω, dτ, dα, dζ)). In the whole sequel, the notation

”Sα,τ,ζ = S′α,τ,ζ” or ”S = S′” will mean

a.s.,⟨

S − S′⟩ρN = 0 (3.16)

which is the same asa.s., ∀ n ∈ IN, SXn,Tn,Zn

= S′Xn,Tn,Zn

(3.17)

3. Let X be a random variable, eventually defined a.s. (X may be a stochastic integral,...). Inorder to prove that X ∈ D2 and that, for some Sα,τ (ω) ∈ L2(P (dω)dαdτ), some Tα,τ,ζ(ω) ∈L2(P (dω)ρ(ζ)N(ω, dτ, dα, dζ)),

DWα,τX = Sα,τ and DN

α,τ,ζX = Tα,τ,ζ (3.18)

it suffices to find a sequence Xn in S (or in D2) such that, when n goes to infinity,

E(

(X − Xn)2)

+ E(⟨

S − DW Xn

⟩

leb

)

+ E

(

⟨

T − DNXn

⟩

ρN

)

−→ 0 (3.19)

3.2 Properties of the derivative operators.

We now give the usual properties of our derivative operators. We omit the proofs of the two first ones,because the results are well known in the Gaussian context (we refer to Nualart [5]), and the adaptationsare easy.

Proposition 3.8 D2, endowed with the following scalar product, is Hilbert :

〈Y,Z〉D2= E(Y Z) + E

(⟨

DW Y,DW Z⟩

leb

)

+ E

(

⟨

DNY,DNZ⟩

ρN

)

(3.20)

Proposition 3.9 1. Let Y be in D2 and let F be in C1b (IR). Then Z = F (Y ) belongs to D2, DW Z =

F ′(Y )DW Y , and DNZ = F ′(Y )DNY .

2. If f0 is in L2([0, L] × [0, T ]), then W (f0) belongs to D2, DW W (f0) = f0, and DNW (f0) = 0.

3. If g0 is a measurable function on [0, T ] × [0, L] × O, of class C1 on O, with compact support, suchthat g0 and (g0)

′z are bounded, then N(g0) and N(g0) are in D2, DW N(g0) = DW N(g0) = 0, and

DNN(g0) = DN N(g0) = (g0)′z.

We carry on with a proposition which deals with the conditional expectations.

Proposition 3.10 1. Let Z be in S. Consider the cadlag martingale Zs = E(Z|Fs). Then, for eachs ∈ [0, T ], Zs belongs to D2, and for all α, τ , a.s.,

DWα,τZs = E

(

DWα,τZ

∣

∣

∣Fs

)

1τ≤s (3.21)

and for all n, a.s.,

DNXn,Tn,Zn

Zs = E(

DNXn,Tn,Zn

Z∣

∣

∣Fs

)

1Tn≤s (3.22)

Furthermore, |||Zs|||2 ≤ |||Z|||2.

11

2. Let Y be a Fs-measurable element of D2. Then for each α, τ , each n,

DWα,τY = DW

α,τY 1τ≤s and DNXn,Tn,Zn

Y = DNXn,Tn,Zn

Y 1Tn≤s (3.23)

and these random variables are Fs-measurable.

3. Let X and Z be in D2. Assume that X and Z are independent. Then XZ belongs to D2, and

DW XZ = XDW Z + ZDW X ; DNXZ = XDNZ + ZDNX. (3.24)

Proof : 1. Let s ∈ [0, T ] and Z = F (W (f1), ...,W (fm), N(g1), ..., N(gd)) ∈ S. We set

fi(α, τ) = fi(α, τ)1τ≤s ; fi(α, τ) = fi(α, τ)1τ>s

gi(α, τ, ζ) = gi(α, τ, ζ)1τ≤s ; gi(α, τ, ζ) = gi(α, τ, ζ)1τ>s

Then for each i, W (fi) and N(gi) are Fs-measurable, and W (fi), N(gi) are independent of Fs. Thus, if µ

denotes the law of (W (f1), ..., N(gd)) and if H(X1, ..., Yd) =

∫

IRm+d

F (X1 + x1, ..., Yd + yd))µ(dx1, ..., dyd),

thenZs = E (Z|Fs) = H(W (f1), ..., N(gd)) (3.25)

Since F is of class C1c , it is easy to check that H is in C1

b . It is not difficult to deduce that Zs belongs toD2, and that for all α, τ , all n,

DWα,τZs =

m∑

i=1

∂iH(W (f1), ..., N(gd))fi(α, τ)

DNXn,Tn,Zn

Zs =m+d∑

i=m+1

∂iH(W (f1), ..., N(gd))(gi)′z(Xn, Tn, Zn)

But ∂iH(W (f1), ..., N(gd)) = E (∂iF (W (f1), ..., N(gd)) | Fs). We easily deduce (3.21), and on the otherhand,

DNXn,Tn,Zn

Zs =m+d∑

i=m+1

E [∂iF (W (f1), ..., N(gd)) | Fs] (gi)′z(Xn, Tn, Zn)1Tn≤s

= E

m+d∑

i=m+1

∂iF (W (f1), ..., N(gd))(gi)′z(Xn, Tn, Zn)1Tn≤s

∣

∣

∣

∣

∣

∣

Fs

= E(

DNXn,Tn,Zn

Z∣

∣

∣Fs

)

1Tn≤s (3.26)

and (3.22) follows. The inequality |||Zs|||2 ≤ |||Z|||2 is a consequence of (3.21), (3.22) and of Jensen’sinequality : for example,

E

(

⟨

DNZs

⟩

ρN

)

= E

(

∑

n

E(

DNXn,Tn,Zn

Z∣

∣

∣Fs

)21Tn≤sρ(Zn)

)

≤ E

(

∑

n

E

(

(

DNXn,Tn,Zn

Z)2∣

∣

∣

∣

Fs

)

1Tn≤sρ(Zn)

)

= E

(

∑

n

(

DNXn,Tn,Zn

Z)2

1Tn≤sρ(Zn)

∣

∣

∣

∣

∣

Fs

)

≤ E

(

⟨

DNZ⟩

ρN

)

(3.27)

12

2. is a straightforward consequence of 1. : let Zk be a sequence of S going to Y in D2. Then it is clearthat Zk

s = E(Zk|Fs) goes to Y in L2. Furthermore, we know from 1. that for each k, Zks belongs to D2,

and|||Zk+1

s − Zks |||2 ≤ |||Zk+1 − Zk|||2

Thus Zks is Cauchy in D2, and thus tends to Y in D2. One concludes easily by using 1.

3. Let G (resp. A) be the σ-field generated by X (resp. Z). Let X ′k (resp. Z ′k) be a sequence of Sgoing to X (resp. Z) in D2. Using the same arguments as in 1., one can check that Xk = E(X ′k|G)(resp. Xk = E(X ′k|G), still belongs to D2 and converges to X (resp. Z) in D2. Furthermore, the variables

X, Xk,⟨

DNX⟩

leb,⟨

DNXk⟩

leb,⟨

DN (X − Xk)⟩

leb, and

⟨

DNX⟩

ρN,⟨

DNXk⟩

ρN,⟨

DN (X − Xk)⟩

ρNare

G-measurable. The same list of random variables, replacing X and Xk with Z and Zk are A-measurable.On the other hand, it is easy to check that for each k, XkZk ∈ D2, and that DW XkZk = XkDW Zk +ZkDW Xk and DNXkZk = XkDNZk +ZkDNXk. The convergence of XkZk to XZ in D2 is easily proved,using repeatedly the same independence argument.

We finally state the absolute continuity criterion that we will use. This is adapted from Nualart [5], p 87.

Theorem 3.11 Assume that Z belongs to D2, and set σ =⟨

DW Z⟩

leb+⟨

DNZ⟩

ρN. Then, if σ > 0 a.s.,

the law of Z admits a density with respect to the Lebesgue measure on IR.

Proof : Suppose first that |Z| < 1. Let φ be a Lebesgue-negligible function from ] − 1, 1[ to [0, 1]. We

have to show that φ(Z) = 0 a.s. Let Ψ(y) =

∫ y

−1φ(x)dx. Following Nualart, [5] p 87, one can show that

Ψ(Z) ∈ D2, that DW Ψ(Z) = φ(Z)DW Z, and that DNΨ(Z) = φ(Z)DNZ.On the other hand, Ψ(Z) = 0. So its derivatives vanish, and the uniqueness of the derivatives yields thata.s.,

⟨

φ(Z)DW Z⟩

leb+⟨

φ(Z)DNZ⟩

ρN= 0

It follows that φ2(Z)σ = 0 a.s., and thus that φ(Z) vanishes almost surely. The first step is finished.If Z is not bounded any more, it suffices to apply what precedes with Φ(Z), where Φ is a bijective C1

b

function from IR to ] − 1, 1[, with a strictly positive derivative.

3.3 Derivation and stochastic integrals.

In the evolution equation (1.10), one can see three random integrals. In order to apply Theorem 3.11, wehave to compute their derivatives. We begin with a remark which might avoid confusions.

Remark 3.12 1. Let Y and Y ′ be two weak versions of the same process. Assume that for each y,each s, Y (y, s) ∈ D2. Then for almost all y, s,

Y ′(y, s) ∈ D2 ;⟨

DW Y (y, s) − DW Y ′(y, s)⟩

leb+⟨

DNY (y, s) − DNY ′(y, s)⟩

ρN= 0 a.s.

2. Let Y be a process such that for each y, s, Y (y, s) ∈ D2. Assume that for each α, τ , DWα,τY (y, s) =

Sα,τ (y, s) dPdyds-a.e. (i.e. Sα,τ is a weak version of DWα,τY ), and that for each n, DN

Xn,Tn,ZnY (y, s) =

S′Xn,Tn,Zn

(y, s) dPdyds-a.e. (i.e. S′Xn,Tn,Zn

is a weak version of DNXn,Tn,Zn

Y ). Then for almost ally, s, a.s.,

⟨

DW Y (y, s) − S(y, s)⟩

leb+⟨

DNY (y, s) − S′(y, s)⟩

ρN= 0

13

Proof : 1. This is immediate, since for almost all y, s, a.s., Y (y, s) = Y ′(y, s).2. For example, thanks to Fubini’s Theorem,

∫ T

0

∫ L

0E(⟨

DW Y (y, s) − S(y, s)⟩

leb

)

dyds

=

∫ T

0

∫ L

0dαdτE

(

∫ T

0

∫ L

0

(

DWα,τY (y, s) − Sα,τ (y, s)

)2dyds

)

= 0

Let us now define a class of processes of which the integrals will belong to D2.

Definition 3.13 Let Y be a process of class PV. We will say that Y is D2-PV if the following conditionshold :

• For every y, s, Y (y, s) belongs to D2, and supy,s

|||Y (y, s)|||2 < ∞.

• For each α, τ fixed, the process DWα,τY (y, s) admits a predictable weak version (and vanishes when

τ > s). The map ω,α, τ, y, s −→ DWα,τY (y, s)(ω) is globally measurable.

• For each n ≥ 0, the process DNXn,Tn,Zn

Y (y, s) admits a predictable weak version (and vanishes whenTn > s).

Remark 3.14 Let Z ∈ S. Consider the cadlag martingale Zs = E(Z|Fs). This process is in D2-PV, andfor all s, |||Zs|||2 ≤ |||Z|||2.

This is a straightforward consequence of Proposition 3.10-1. The following Remark is a well-known factabout Hilbert spaces, and will allow to ”separate” the variables y, s and ω in the stochastic integrals.

Remark 3.15 Let Zkk≥0 be an orthonormal (for 〈 〉D2) basis of S. Then every element Y of D2 can be

written asY =

∑

k≥0

λkZk with

∑

k≥0

λ2k < ∞ where λk =

⟨

Y,Zk⟩

D2

(3.28)

Let us apply this to a D2-PV process.

Lemma 3.16 Let Y belong to D2-PV.

1. Thanks to Remark 3.15, we can write Y as

Y (y, s) =∑

k≥0

φk(y, s)Zk with supy,s

∑

k≥0

φ2k(y, s) < ∞ (3.29)

where φk(y, s) =⟨

Y (y, s), Zk⟩

D2

is B([0, L] × [0, T ])-measurable.

2. If Zks = E(Zk|Fs) (see Remark 3.14), then for every N ,

|||Y (y, s) −N∑

k=0

φk(y, s)Zks |||2 ≤ |||Y (y, s) −

N∑

k=0

φk(y, s)Zk|||2 (3.30)

Proof : 1. The only problem is to prove that φk is measurable, which follows from the properties of aD2-PV process, since

φk(y, s) = E(

Y (y, s)Zk)

+ E

(

∫ T

0

∫ L

0DW

α,τY (y, s)DWα,τZkdαdτ

)

+∑

n

E(

DNXn,Tn,Zn

Y (y, s)DNXn,Tn,Zn

Zk)

14

2. can be proved in the same way as Proposition 3.10.

We will also need the following definition.

Definition 3.17 Let Sα,τ,ζ(ω, y, s) be function on Ω×([0, L]×[0, T ]×O)×([0, L]×[0, T ]). We will say thatS belongs to the class DN if supy,s E(〈S(y, s)〉ρN ) < ∞ and if for each n ≥ 0, the process SXn,Tn,Zn

(y, s)admits a predictable weak version, and vanishes when Tn > s.

These conditions, which are satisfied by the derivative related to N of a D2-PV process, will allow to provethe technical (but fundamental) lemma 3.18. Let us notice that if S is such a function, then 〈S(y, s)〉ρN

admits a predictable weak version (this is immediate by using the Notation 3.6). The following lemmatakes the place of the L2-isometry which is constantly used in the Gaussian case. The functions f, g, h arethe parameters of (0.1), and satisfy (H ′).

Lemma 3.18 Let Y be a process of class PV, and let S be in DN . We set, for τ ≤ t :

T aα,τ,ζ(x, t) =

∫ t

0

∫ L

0Gt−s(x, y)f ′(Y (y, s))Sα,τ,ζ(y, s)W (dy, ds)

T bα,τ,ζ(x, t) =

∫ t

0

∫ L

0Gt−s(x, y)g′(Y (y, s))Sα,τ,ζ(y, s)dyds (3.31)

T cα,τ,ζ(x, t) =

∫ t

0

∫ L

0

∫

OGt−s(x, y)h′

x(Y (y, s), z)Sα,τ,ζ(y, s)N (ds, dy, dz)

and T aα,τ,ζ(x, t) = T b

α,τ,ζ(x, t) = T cα,τ,ζ(x, t) = 0 if τ > t. These functions belong to DN , and

E[

〈T a(x, t)〉ρN

]

=

∫ t

0

∫ L

0G2

t−s(x, y)E[

f ′(Y (y, s))2 〈S(y, s)〉ρN

]

dyds (3.32)

E

[

⟨

T b(x, t)⟩

ρN

]

≤ TL

∫ t

0

∫ L

0G2

t−s(x, y)E[

g′(Y (y, s))2 〈S(y, s)〉ρN

]

dyds (3.33)

E[

〈T c(x, t)〉ρN

]

=

∫ t

0

∫ L

0

∫

OG2

t−s(x, y)E[

h′x(Y (y, s), z)

2 〈S(y, s)〉ρN

]

ϕ(z)dzdyds (3.34)

Notice here again that the integrals in (3.31) are not well-defined for each α, τ, ζ. One more time, we meanhere that α, τ, ζ have to be replaced by Xn, Tn, Zn.

Proof : we first notice that if (3.32), (3.33), and (3.34) hold, the lemma will follow easily, by using an easyadaptation of Lemma 2.1. Let us for example check (3.34). Using Notation 3.6, and applying Fubini’sTheorem (everything is positive), we obtain

E(

〈T c(x, t)〉ρN

)

= E

∑

n≥0

ρ(Zn)[

T cXn,Tn,Zn

(x, t)]2

=∑

n≥0

E

(

ρ(Zn)[

T cXn,Tn,Zn

(x, t)]2)

=∑

n≥0

E

(

∫ t

0

∫ L

0

∫

OGt−s(x, y)h′

x(Y (y, s), z)√

ρ(Zn)SXn,Tn,Zn(y, s)N(ds, dy, dz)

)2

But we know that√

ρ(Zn)SXn,Tn,Zn(y, s) =

√

ρ(Zn)SXn,Tn,Zn(y, s)1Tn≤s belongs to PV . We thus can

apply (1.4) :

E(

〈T c(x, t)〉ρN

)

=∑

n≥0

∫ t

0

∫ L

0

∫

OG2

t−s(x, y)E(

h′x(Y (y, s), z)

2ρ(Zn)S2

Xn,Tn,Zn(y, s)

)

ϕ(z)dzdyds

We conclude by using again Fubini’s Theorem.

Now we can state the main proposition of this section :

15

Proposition 3.19 Let Y belong to D2-PV. Let us consider the following processes

U1(x, t) =

∫ t

0

∫ L

0Gt−s(x, y)f(Y (y, s))W (dy, ds)

U2(x, t) =

∫ t

0

∫ L

0Gt−s(x, y)g(Y (y, s))dyds

U3(x, t) =

∫ t

0

∫ L

0

∫

IRGt−s(x, y)h(Y (y, s), z)N (ds, dy, dz)

Then U1, U2, U3 are in D2-PV. If Y− is a predictable weak version of Y , then :

DWα,τU1(x, t) = Gt−τ (x, α)f(Y−(α, τ))1τ≤t +

∫ t

0

∫ L

0Gt−s(x, y)f ′(Y (y, s))DW

α,τ Y (y, s)W (dy, ds)

DNα,τ,ζU1(x, t) =

∫ t

0

∫ L

0Gt−s(x, y)f ′(Y (y, s))DN

α,τ,ζY (y, s)W (dy, ds)

DWα,τU2(x, t) =

∫ t

0

∫ L

0Gt−s(x, y)g′(Y (y, s))DW

α,τY (y, s)dyds

DNα,τ,ζU2(x, t) =

∫ t

0

∫ L

0Gt−s(x, y)g′(Y (y, s))DN

α,τ,ζY (y, s)dyds

DWα,τU3(x, t) =

∫ t

0

∫ L

0

∫

IRGt−s(x, y)h′

x(Y (y, s), z)DWα,τ Y (y, s)N (ds, dy, dz)

DNα,τ,ζU3(x, t) = Gt−τ (x, α)h′

z(Y−(α, τ), ζ)1τ≤t

+

∫ t

0

∫ L

0

∫

IRGt−s(x, y)h′

x(Y (y, s), z)DNα,τ,ζY (y, s)N (ds, dy, dz)

Notice that the obtained derivatives do not depend on the choice for the predictable weak version Y− ofY . Indeed, if Y − is another predictable weak version of Y , then it is clear that for each x, t, a.s.,

∫ t

0

∫ L

0

Gt−τ (x, α)f(Y−(α, τ)) − Gt−τ (x, α)f(Y −(α, τ))2

dαdτ

+

∫ t

0

∫ L

0

∫

IR

Gt−τ (x, α)h′z(Y−(α, τ), ζ) − Gt−τ (x, α)h′

z(Y−(α, τ), ζ)

2ρ(ζ)N(dτ, dα, dζ) = 0

We will only prove the proposition for U3(x, t), because the other cases are simpler and proved similarly.We begin with a lemma :

Lemma 3.20 Let φ be a measurable function on [0, T ]×[0, L]×O, of class C1 on O (in z), with φ′z bounded,

and such that |φ(s, y, z)| ≤ Kη(z) (where η ∈ L2(O,ϕ(z)dz) is defined in (H ′)). Let (t, x) ∈]0, T ] × [0, L]be fixed, and

G(s, y, z) = Gt−s(x, y)φ(s, y, z)1s≤t (3.35)

Then N(G) =∫ T0

∫ L0

∫

O G(s, y, z)N (ds, dy, dz) belongs to D2 , and its derivatives are given by :

DWα,τ N(G) = 0 and DN

α,τ,ζN(G) = Gt−τ (x, α)φ′z(τ, α, ζ)1τ≤t (3.36)

Proof : this lemma is an easy extension of Proposition 3.9-3. Let Tǫ be a family of C∞ functions on IR,such that |T ′

ǫ | ≤ 1 such that

Tǫ(u) =

u if |u| ≤ 1/ǫ1 + 1/ǫ if u ≥ 2 + 1/ǫ−1 − 1/ǫ if u ≤ −2 − 1/ǫ

and |Tǫ(u)| ≤ |u|

16

On the other hand, using assumption (S), we consider a family Kǫ of C1 positive functions on O,bounded by 1, with compact support (in O), and satisfying

∀ z ∈ O,Kǫ(z) −→ǫ→0 1 ;

∫

O

(

K ′ǫ(z)

)2η2(z)ρ(z)ϕ(z)dz −→ǫ→0 0 (3.37)

We setGǫ(s, y, z) = Tǫ(Gt−s(x, y))Tǫ(φ(s, y, z))Kǫ(z)1s≤t

Then Gǫ satisfies the conditions of Proposition 3.9-3 :

(Gǫ)′z(s, y, z) = Tǫ(Gt−s(x, y))1s≤t

[T ′ǫ (φ(s, y, z))φ′

z(s, y, z)Kǫ(z) + Tǫ(φ(s, y, z))K ′ǫ(z)

]

Thus, N(Gǫ) ∈ D2, andDW N(Gǫ) = 0 ; DN N(Gǫ) = (Gǫ)

′z

One easily checks, by using the Lebesgue Theorem and (3.37), that N(Gǫ) goes to N(G) in L2, and that

E(

⟨

G′z − (Gǫ)

′z

⟩

ρN

)

tends to 0. This yields the result.

Proof of Proposition 3.19 for U = U3

Step 1 : if z is fixed, h( . , z) is C1b on IR. Hence, using Proposition 3.9-1., for every (x, t, z) in [0, L] ×

[0, T ] × IR, h(Y (x, t), z) ∈ D2, and we have

DW h(Y (x, t), z) = h′x(Y (x, t), z)DW Y (x, t) ; DNh(Y (x, t), z) = h′

x(Y (x, t), z)DNY (x, t)

Furthermore, due to (H ′), we see that supx,t |||h(Y (x, t), z)|||2 ≤ Kη(z). Let us write h(Y, z) in D2 (thanksto Lemma 3.16) as :

h(Y (x, t), z) =∑

k≥0

φk(x, t, z)Zk

We know that for all k, φk is measurable. Furthermore,

supx,t

∑

k

φ2k(x, t, z) = sup

x,t|||h(Y (x, t), z)|||22 ≤ Kη2(z)

Using Lebesgue’s Theorem and (H ′), we see that

φk(y, s, z) =⟨

h(Y (y, s), z), Zk⟩

D2

= E[

h(Y (y, s), z)Zk]

+ E[

h′x(Y (y, s), z)

⟨

DW Y (y, s),DW Zk⟩

leb

]

+E

[

h′x(Y (y, s), z)

⟨

DNY (y, s),DNZk⟩

ρN

]

(3.38)

is of class C1 in z, and that its derivative is bounded. On the other hand, since h′z( . , z) is of class C1

b ,Proposition 3.9-1. shows that h′

z(Y (y, s), z) belongs to D2, allows to compute its derivatives, and to seethat (due to (H ′)) :

supx,t,z

∑

k

(

(φk)′z(x, t, z)

)2= sup

x,t,z|||h′

z(Y (x, t), z)|||22 < ∞

We see also that h′z(Y (y, s), z) =

∑

k≥0(φk)′z(y, s, z)Zk in D2. Because of Remark 3.14, setting Zk

s =

E(Zk|Fs) and

ΨN (y, s, z) =N∑

k=0

φk(y, s, z)Zks (3.39)

17

we know that :∀y, s, z |||ΨN (y, s, z) − h(Y (y, s), z)|||2 −→ 0 (3.40)

supN

supy,s

|||ΨN (y, s, z) − h(Y (y, s), z)|||22 ≤ Kη2(z) (3.41)

∀y, s, z |||(ΨN )′z(y, s, z) − h′z(Y (y, s), z)|||2 −→ 0 (3.42)

supN

supy,s,z

|||(ΨN )′z(y, s, z) − h′z(Y (y, s), z)|||22 ≤ sup

y,s,z|||h′

z(Y (y, s), z)|||22 < ∞ (3.43)

Step 2 : let us first show that for each k,

Uk(x, t) =

∫ t

0

∫ L

0

∫

OGt−s(x, y)φk(y, s, z)Zk

s N(ds, dy, dz)

belongs to D2, and let us compute its derivatives. It is really useful to use the sequence Zks , because the

processes φk(y, s, z)Zk do not a priori admit predictable weak versions. We use a Peano approximation for

Zks : if 0 ≤ s ≤ T , we set sn = sup

inT

/

inT < s

∨ 0. Then we consider

Ukn(x, t) =

∫ t

0

∫ L

0

∫

OGt−s(x, y)φk(y, s, z)Zk

snN(ds, dy, dz)

=n−1∑

i=0

Zki

nT×∫

[0,t]∩] i

nT, i+1

nT ]

∫ L

0

∫

OGt−s(x, y)φk(y, s, z)N (ds, dy, dz) (3.44)

Since Zki

nT

belongs to D2 and is F i

nT -measurable, since φk satisfies the assumptions of Lemma 3.20, this

Lemma and Proposition 3.10-3 allows us to say that Ukn(x, t) ∈ D2, and

DNα,τ,ζU

kn(x, t) =

n−1∑

i=0

∫

[0,t]∩] i

nT, i+1

nT ]

∫ L

0

∫

OGt−s(x, y)φk(y, s, z)N (ds, dy, dz) × DN

α,τ,ζZki

nT

+n−1∑

i=0

Zki

nT× Gt−τ (x, α)(φk)′z(α, τ, ζ)1τ∈] i

nT, i+1

nT ]∩[0,t]

=

∫ t

0

∫ L

0

∫

OGt−s(x, y)φk(y, s, z)DN

α,τ,ζZksn

N(ds, dy, dz)

+Gt−τ (x, α)(φk)′z(α, τ, ζ)Zkτn

1τ≤t (3.45)

and, by the same way,

DWα,τ U

kn(x, t) =

∫ t

0

∫ L

0

∫

OGt−s(x, y)φk(y, s, z)DW

α,τ Zksn

N(ds, dy, dz)

Thus we set

DNα,τ,ζU

k(x, t) =

∫ t

0

∫ L

0

∫

OGt−s(x, y)φk(y, s, z)DN

α,τ,ζZks N(ds, dy, dz)

+Gt−τ (x, α)(φk)′z(α, τ, ζ)Zkτ−1τ≤t (3.46)

and

DWα,τ U

k(x, t) =

∫ t

0

∫ L

0

∫

OGt−s(x, y)φk(y, s, z)DW

α,τ Zks N(ds, dy, dz)

We still have to check the convergence in D2. First, because of (1.4), and since |φk| ≤ Kη ∈ L2(O,ϕ(z)dz),

E[

(Uk(x, t) − Ukn(x, t))2

]

=

∫ t

0

∫ L

0

∫

OG2

t−s(x, y)φ2k(y, s, z)E

(

(Zks − Zk

sn)2)

ϕ(z)dzdyds

≤ K

∫ t

0

∫ L

0G2

t−s(x, y)E(

(Zks − Zk

sn)2)

dyds (3.47)

18

This goes to 0 by using (twice) the Lebesgue Theorem : for each s, |Zks −Zk

sn| goes to 0 a.s., and is smaller

than 2 supω |Zk(ω)| < ∞. Thus for each s, E(

(Zks − Zk

sn)2)

goes to 0. This expectation is also bounded,

and G2t−s(x, y) belongs to L1(dyds) : Lebesgue’s Theorem (for dyds) yields the convergence.

On the other hand,

E

[

⟨

DN Uk(x, t) − DN Ukn(x, t)

⟩

ρN

]

≤ C

∫ t

0

∫ L

0

∫

OG2

t−τ (x, α)(

(φk)′z(α, τ, ζ)

)2E(

(Zkτn

− Zkτ−)2

)

ρ(ζ)ϕ(ζ)dζdαdτ

+CE[

∫ t

0

∫ L

0

∫

O

(

∫ t

0

∫ L

0

∫

OGt−s(x, y)φk(y, s, z)

(

DNα,τ,ζZ

ksn

− DNα,τ,ζZ

ks

)

N(ds, dy, dz)

)2

ρ(ζ)N(dτ, dα, dζ)]

The first part in this expression tends to 0 as above. Lemma 3.18 allows us to upperbound the second onewith

∫ t

0

∫ L

0

∫

OG2

t−s(x, y)φ2k(y, s, z)E

(

⟨

DNZksn

− DNZks

⟩

ρN

)

ϕ(z)dzdyds

which goes to 0 by the same way (here⟨

DNZksn

− DNZks

⟩

ρNis not upperbounded by a constant, but by

the random variable

Xk = 4 sup[0,T ]

E

[

⟨

DNZk⟩

ρN

∣

∣

∣

∣

Fs

]

which belongs to L1(Ω) due to Doob’s inequality, since⟨

DNZk⟩

ρN∈ L2(Ω), because Zk ∈ S).

Finally, one can prove as well that E[⟨

DW Uk(x, t) − DW Ukn(x, t)

⟩

leb

]

tends to 0.

Step 3 : we now approximate U(x, t) with

UN (x, t) =N∑

k=0

Uk(x, t)

Using the first step, we know that UN (x, t) belongs to D2, and that (ΨN (y, s, z) is defined by equation(3.39)) :

DWα,τU

N (x, t) =

∫ t

0

∫ L

0

∫

OGt−s(x, y)DW

α,τΨN (y, s, z)N (ds, dy, dz)

DNα,τ,ζU

N (x, t) = Gt−τ (x, α)(ΨN )′z(α, τ, ζ)1τ≤t

+

∫ t

0

∫ L

0

∫

OGt−s(x, y)DN

α,τ,ζΨN (y, s, z)N (ds, dy, dz)

We now denote by DWα,τU(x, t) and DN

α,τ,ζU(x, t) the expressions of the statement, even if we still do notknow if these are really the derivatives of U(x, t). First, using (1.4),

E[

(U(x, t) − UN (x, t))2]

≤∫ t

0

∫ L

0

∫

OG2

t−s(x, y)E(

(h(Y (y, s), z) − ΨN (y, s))2)

ϕ(z)dydzds

This goes to 0 by the Lebesgue Theorem, and thanks to (3.40) and (3.41). Furthermore,

E

[

⟨

DNU(x, t) − DNUN (x, t)⟩

ρN

]

≤ K

∫ t

0

∫ L

0

∫

OG2

t−τ (x, α)E(

(h′z(Y (α, τ), ζ) − (ΨN )′z(α, τ, ζ))2

)

ρ(ζ)ϕ(ζ)dζdαdτ

+KE[

∫ T

0

∫ L

0

∫

O

(

∫ t

0

∫ L

0

∫

OGt−s(x, y)(DN

α,τ,ζh(Y (y, s), z) − DWα,τΨN (y, s, z))

N(ds, dy, dz))2

ρ(ζ)N(dτ, dα, dζ)]

19

The first term tends to 0 as above (because of (3.42) and (3.43)). We upperbound the second one, byusing Lemma 3.18, with

∫ t

0

∫ L

0

∫

OG2

t−s(x, y)E

(

⟨

DNΨN (y, s, z) − DNh(Y (y, s), z)⟩

ρN

)

ϕ(z)dzdyds

which goes also to 0 thanks to the Lebesgue Theorem. The third convergence can be checked by the sameway.

Step 4 : we still have to prove that U(x, t) belongs to D2-PV . First, U is predictable, as DWα,τU(x, t) if α, τ

are fixed. The global measurability of DWα,τU(x, t)(ω) is obvious, as the predictability of DW

α,τU(x, t) (for α, τ

fixed) and of DNXn,Tn,Zn

U(x, t) (for n ≥ 0 fixed). U is classically bounded in L2, and E

(

⟨

DNU(x, t)⟩

ρN

)

is bounded by Lemma 3.18, (H ′), and the Appendix (4.3). Furthermore, using Fubini’s Theorem and (H ′),

E(⟨

DW U(x, t)⟩

leb) =

∫ T

0

∫ L

0E

(

∫ t

0

∫ L

0

∫

OGt−s(x, y)h′

x(Y (y, s), z)DWα,τ Y (y, s)N(ds, dy, dz)

)2

dαdτ

≤∫ T

0

∫ L

0

∫ t

0

∫ L

0

∫

OG2

t−s(x, y)η2(z)E

(

DWα,τY (y, s)

2)

ϕ(z)dzdydsdαdτ

=

∫ t

0

∫ L

0

∫

OG2

t−s(x, y)η2(z)E(⟨

DW Y (y, s)⟩

leb

)

ϕ(z)dzdyds

This is clearly bounded, because Y is D2-PV , since η ∈ L2(O,ϕ(z)dz), and due to the Appendix (4.3).

3.4 Derivation of the solution.

In order to apply Theorem 3.11, we have to prove that V is in D2, and to compute its derivatives.

Theorem 3.21 Assume (H ′), (M) and (D). Let V be the unique weak solution of (0.1). Then V belongsto D2-PV, and if V− is a predictable weak version of V ,

DWα,τV (x, t) = Gt−τ (x, α)f(V−(α, τ))1t≥τ

+

∫ t

0

∫ L

0Gt−s(x, y)f ′(V (y, s))DW

α,τ V (y, s)W (dy, ds)

+

∫ t

0

∫ L

0Gt−s(x, y)g′(V (y, s))DW

α,τV (y, s)dyds

+

∫ t

0

∫ L

0

∫

OGt−s(x, y)h′

x(V (y, s), z)DWα,τ V (y, s)N (ds, dy, dz) (3.48)

DNα,τ,ζV (x, t) = Gt−τ (x, α)h′

z(V−(α, τ), ζ)1t≥τ

+

∫ t

0

∫ L

0Gt−s(x, y)f ′(V (y, s))DN

α,τ,ζV (y, s)W (dy, ds)

+

∫ t

0

∫ L

0Gt−s(x, y)g′(V (y, s))DN

α,τ,ζV (y, s)dyds

+

∫ t

0

∫ L

0

∫

OGt−s(x, y)h′

x(V (y, s), z)DNα,τ,ζV (y, s)N (ds, dy, dz) (3.49)

In order to prove this Theorem, we will denote by Aα,τ (x, t) and Bα,τ,ζ(x, t) the solutions of (3.48) and(3.49), then we will check that they are really the derivatives of V .Equations (3.48) and (3.49) are in fact “systems”. In particular, in the case of equation (3.49), we do notwant to solve the equation for each α, τ, ζ fixed, but rather for each n, replacing α, τ, ζ with Xn, Tn, Zn.The solution B(x, t) will be considered as taking its values in L2(P (dω)ρ(ζ)N(ω, dτ, dα, dζ)), for each x, t.

20

Lemma 3.22 1. Equation (3.48) admits a unique solution

x, t 7→ A(x, t)

[0, L] × [0, T ] 7→ L2(P (dω)dαdτ) (3.50)

such that for each fixed α, τ , the process Aα,τ (x, t) admits a predictable weak version and such that

supx,t

E (〈A(x, t)〉leb) < ∞

The uniqueness holds in the sense that, if A′ is another solution, then

supx,t

E[⟨

A(x, t) − A′(x, t)⟩

leb

]

= 0

2. Equation (3.49) admits a unique solution

x, t 7→ B(x, t)

[0, L] × [0, T ] 7→ L2(P (dω)ρ(ζ)N(ω, dτ, dα, dζ)) (3.51)

belonging to DN . The solution is unique in the sense that if B′ is another solution, then

supx,t

E[

⟨

B(x, t) − B′(x, t)⟩

ρN

]

= 0

Proof : Let us for example prove 2. The uniqueness follows easily from Lemma 3.18 and assumption (H ′).We prove the existence by using a Picard iteration : we set

B0α,τ,ζ(x, t) = Gt−τ (x, α)h′

z(V−(α, τ), ζ)1t≥τ

Bn+1α,τ,ζ(x, t) = B0

α,τ,ζ(x, t) +

∫ t

0

∫ L

0Gt−s(x, y)f ′(V (y, s))Bn

α,τ,ζ(y, s)W (dy, ds)

+

∫ t

0

∫ L

0Gt−s(x, y)g′(V (y, s))Bn

α,τ,ζ(y, s)dyds

+

∫ t

0

∫ L

0

∫

OGt−s(x, y)h′

x(V (y, s), z)Bnα,τ,ζ(y, s)N(ds, dy, dz)

One can check recursively, by using Lemma 3.18, that for every n, Bn belongs to DN . Then Lemma 3.18,assumption (H ′), the Appendix (4.2), and Picard’s Lemma allows us to say that the series with generalterm

[

supx,t

E

(

⟨

Bn+1(x, t) − Bn(x, t)⟩

ρN

)

]1

2

does converge. We conclude easily.

Proof of Theorem 3.21 : we consider the Picard approximations of V defined in Section 2 by (2.6). It isimmediate, by using recursively Proposition 3.19, that for each n, V n belongs to D2-PV , and we also havean expression of its derivatives. For example, if V n

− is a predictable weak version of V n,

DNα,τ,ζV

n+1(x, t) = Gt−τ (x, α)h′z(V

n− (α, τ), ζ)1t≥τ

+

∫ t

0

∫ L

0Gt−s(x, y)f ′(V n(y, s))DN

α,τ,ζVn(y, s)W (dy, ds)

+

∫ t

0

∫ L

0Gt−s(x, y)g′(V n(y, s))DN

α,τ,ζVn(y, s)dyds

+

∫ t

0

∫ L

0

∫

OGt−s(x, y)h′

x(V n(y, s), z)DNα,τ,ζV

n(y, s)N(ds, dy, dz) (3.52)

21

We already know (see (2.11) in the proof of Theorem 1.4), that V n(x, t) goes to V (x, t) uniformly in L2.

Thus we just have to check that E(⟨

A(x, t) − DW V n(x, t)⟩

leb

)

and E

(

⟨

B(x, t) − DNV n(x, t)⟩

ρN

)

go to

0. We set

Gn(x, t) = E

[

⟨

B(x, t) − DNV n(x, t)⟩

ρN

]

and φn(t) = supx

Gn(x, t)

One can check that Gn+1(x, t) ≤ K(In1 (x, t) + ... + In

7 (x, t)), where :

In1 (x, t) =

∫ t

0

∫ L

0

∫

OG2

t−τ (x, α)E(

[

h′z(V

n(α, τ), ζ) − h′z(V (α, τ), ζ)

]2)

ρ(ζ)ϕ(ζ)dζdαdτ

In4 (x, t) = E

[

∫ T

0

∫ L

0

∫

O

(

∫ t

0

∫ L

0

∫

OGt−s(x, y)[h′

x(V (y, s), z) − h′x(V n(y, s), z)]

B(y, s)N(ds, dy, dz))2

ρ(ζ)N(dτ, dα, dζ)]

In7 (x, t) = E

[

∫ T

0

∫ L

0

∫

O

(

∫ t

0

∫ L

0

∫

OGt−s(x, y)h′

x(V n(y, s), z)

×[B(y, s) − DNα,τ,ζV

n(y, s)]N(ds, dy, dz))2

ρ(ζ)N(dτ, dα, dζ)]

and where In2 and In

3 (resp. In5 and In

6 ) correspond to the same term as In4 (resp. In

7 ) but with the whitenoise and the Lebesgue measure.First, h′′

zx is bounded, hence

[

h′z(V

n(α, τ), ζ) − h′z(V (α, τ), ζ)

]2 ≤ K(V n(α, τ) − V (α, τ))2

Since supα,τ E(

(V n(α, τ) − V (α, τ))2)

tends to 0 (see (2.11), and using the Appendix (4.3), we see thatIn1 (x, t) ≤ K1

n −→ 0.Lemma 3.18 shows that In

4 (x, t) equals :

∫ t

0

∫ L

0

∫

OG2

t−s(x, y)E[

(h′x(V (y, s), z) − h′

x(V n(y, s), z))2 × 〈B(y, s)〉ρN

]

ϕ(z)dzdyds

Applying Holder’s inequality (for the measure dyds, with p = 5/4 and q = 5), we upperbound In4 (x, t)

with :[

∫ t

0

∫ L

0(Gt−s(x, y))5/2dyds

]4/5

×[

∫ T

0

∫ L

0

[∫

OE(

(

h′x(V (y, s), z) − h′

x(V n(y, s), z))2 〈B(y, s)〉ρN

)

ϕ(z)dz

]5

dyds

]1/5

The first part in the product is bounded (see (4.3) in the Appendix), and the second one does not dependany more on x, t, so we denote it by K2

n. Then one can show by using three times the Lebesgue Theorem(for the measures P , ϕ(z)dz, then dyds), by using (H ′), that K2

n goes to 0.After a simple computation using Lemma 3.18 and (H ′), we see that

In7 (x, t) ≤ K

∫ t

0

∫ L

0G2

t−s(x, y)Gn(y, s)dyds

We finally obtain

Gn+1(x, t) ≤ Kn + K

∫ t

0

∫ L

0Gn(y, s)G2

t−s(x, y)dyds ≤ K ′n + K ′

∫ t

0φn(s)

ds√t − s

22

where K ′n −→ 0. Hence

φn+1(x, t) ≤ K ′n + K ′

∫ t

0φn(s)

ds√t − s

Since φ0 is easily bounded, it is standard to deduce that sup[0,T ] φn(t) goes to 0.

One can show in the same way that supx,t E[⟨

A(x, t) − DW V n(x, t)⟩

leb

]

tends to 0, and Theorem 3.21 is

proved.

3.5 Existence of the density.

We have now enough information to prove Theorem 1.5. We consider (x, t) ∈ [0, L]×]0, T ], and we assumethat (M), (D), and (H ′) hold. Using Theorems 3.11 and 3.21, we just have to show that a.s.,

σ(x, t) =⟨

DW V (x, t)⟩

leb+⟨

DNV (x, t)⟩

ρN= σW (x, t) + σN (x, t)

is strictly positive under one of the assumptions (EW ), (EP1), or (EP2).We did not manage to compute explicitely σ(x, t). That is why we have to write three proofs : we willshow that under (EW ), σW (x, t) > 0 a.s., and that under (EP1) or (EP2), σN (x, t) > 0 a.s.

3.5.1 Existence of the density under (EP1).

We begin with the standard remark :

Remark 3.23 It suffices to prove the result when g′ ≥ c, for an arbitrary c > 0.

Proof : let a ∈ IR be fixed. Notice that the Green kernel associated with the system

u′t = u′′

xx − au , u′x(0, t) = u′

x(L, t) = 0

is given by Ht(x, y) = e−atGt(x, y). Since V is a weak solution of equation (0.1), it also is weak solution of

[

∂V

∂t(x, t)dxdt − ∂2V

∂x2(x, t) + aV (x, t)

]

dxdt = (g(V (x, t)) + aV (x, t)) dxdt + f(V (x, t))W (dx, dt)

+

∫

IRh(V (x, t), z)N (dt, dx, dz) (3.53)

Hence,

V (x, t) =

∫ L

0V0(y)Ht(x, y) +

∫ t

0

∫ L

0Ht−s(x, y)f(V (y, s))W (dy, ds)

+

∫ t

0

∫ L

0Ht−s(x, y) [g(V (y, s)) + aV (y, s)] W (dy, ds)

+

∫ t

0

∫ L

0

∫

IRHt−s(x, y)h(V (y, s), z)N (ds, dy, dz)

Since g′ is bounded, since a is arbitrary, and since H behaves in the same way as G, in the sense that0 < Ht(x, y) ≤ Gt(x, y), the Remark is proved.

Then we see that since V is in D2-PV , DNα,τ,ζV (x, t) = 0 as soon as τ > t. Furthermore, we know (by

(EP1)) that f = 0, and that |h′x| ≤ η ∈ L1(O,ϕ(z)dz). Thus setting G′(x) = g′(x) − ∫O h′

x(x, z)ϕ(z)dz,

23

we obtain :

DNα,τ,ζV (x, t) = Gt−τ (x, α)h′

z(V−(α, τ), ζ)

+

∫ t

τ

∫ L

0Gt−s(x, y)G′(V (y, s))DN

α,τ,ζV (y, s)dyds

+

∫ t

τ

∫ L

0

∫

OGt−s(x, y)h′

x(V (y, s), z)DNα,τ,ζV (y, s)N(ds, dy, dz) if τ ≤ t

DNα,τ,ζV (x, t) = 0 if τ > t

Let Sα,τ (x, t) be the unique solution (in the sense of Lemma 3.22-2.) of the following equation :

Sα,τ (x, t) = Gt−τ (x, α) +

∫ t

τ

∫ L

0Gt−s(x, y)G′(V (y, s))Sα,τ (y, s)dyds

+

∫ t

τ

∫ L

0

∫

OGt−s(x, y)h′

x(V (y, s), z)Sα,τ (y, s)N(ds, dy, dz) if τ ≤ t

Sα,τ (x, t) = 0 if τ > t

(3.54)

A uniqueness argument yields that DNα,τ,ζV (x, t) = h′

z(V−(α, τ), ζ))Sα,τ (x, t) in the sense where

supx,t

E

(

⟨

DNV (x, t) − h′z(V−( . , . ), . )S(x, t)

⟩

ρN

)

= 0

which of course implies that for each x, t, a.s., σN (x, t) = 〈h′z(V−( . , . ), . )S(x, t)〉ρN . Using Remark 3.23,

we can assume that G′ ≥ 0. Since h′x ≥ 0, it is obvious that for each x, t, P (dω)N(ω, dτ, dα, dζ)-a.e.,

Sα,τ (x, t) ≥ Gt−τ (x, α)1τ≤t

and we just have to check that for any t > 0, any x ∈ [0, L], a.s.,

∫ t

0

∫ L

0

∫

OG2

t−τ (x, α)(h′z(V−(α, τ), ζ))2ρ(ζ)N(dτ, dα, dζ) > 0

Since ρ > 0, it suffices to show that for every t > 0, a.s.,

∫ t

0

∫ L

0

∫

O1h′

z(V−(α,τ),ζ)6=0N(dτ, dα, dζ) > 0

To this aim, we consider the stopping time

R = inf

s > 0

/

∫ s

0

∫ L

0

∫

O1h′

z(V−(α,τ),ζ)6=0N(dτ, dα, dζ) > 0

and we prove that R = 0 a.s. : since V− is predictable, and since N is a counting measure,

E

(

∫ R

0

∫ L

0

∫

O1h′

z(V−(α,τ),ζ)6=0ϕ(ζ)dζdαdτ

)

= E

(

∫ R

0

∫ L

0

∫

O1h′

z(V−(α,τ),ζ)6=0N(dτ, dα, dζ)

)

≤ 1

which implies that a.s.,∫ R

0

∫ L

0

∫

O1h′

z(V−(α,τ),ζ)6=0ϕ(ζ)dζdαdτ < ∞

This contradicts (EP1), except if R = 0 a.s., and Theorem 1.5 is proved under (EP1).

24

3.5.2 Existence of the density under (EP2).

As under (EP1), we write DNα,τ,ζV (x, t) = h′

z(V−(α, τ), ζ)Sα,τ (x, t), where Sα,τ (x, t) is the unique solution,in the sense of Lemma 3.22-2, of

Sα,τ (x, t) = Gt−τ (x, α) +

∫ t

τ

∫ L

0

∫

OGt−s(x, y)f ′(V (y, s))Sα,τ (y, s)W (dy, ds)

+

∫ t

τ

∫ L

0Gt−s(x, y)g′(V (y, s))Sα,τ (y, s)dyds

+

∫ t

τ

∫ L

0

∫

OGt−s(x, y)h′

x(V (y, s), z)Sα,τ (y, s)N(ds, dy, dz) if τ ≤ t

Sα,τ (x, t) = 0 if τ > t

(3.55)

A uniqueness argument shows that

σN (x, t) =

∫ T

0

∫ L

0

∫

OS2

α,τ (x, t)(

h′z(V−(α, τ), ζ)

)2ρ(ζ)N(dτ, dα, dζ) a.s.

Using (EP2), we see that σN (x, t) > 0 as soon as

Σ(x, t) =

∫ T

0

∫ L

0

∫

OS2

α,τ (x, t)1H(ζ)ρ(ζ)N(dτ, dα, dζ) > 0

We thus split Sα,τ (x, t) = Gt−τ (x, α)1τ≤t + Qα,τ (x, t), where

Qα,τ (x, t) =

∫ t

τ

∫ L

0Gt−s(x, y)f ′(V (y, s))Sα,τ (y, s)W (dy, ds)

+

∫ t

τ

∫ L

0Gt−s(x, y)g′(V (y, s))Sα,τ (y, s)dyds

+

∫ t

τ

∫ L

0

∫

OGt−s(x, y)h′

x(V (y, s), z)Sα,τ (y, s)N (ds, dy, dz) if τ ≤ t

Qα,τ (x, t) = 0 if τ > t

Hence, for every ǫ > 0 small enough,

Σ(x, t) ≥ 23

∫ t

t−ǫ

∫ L

0

∫

HG2

t−τ (x, α)ρ(ζ)N(dτ, dα, dζ) − 2

∫ t

t−ǫ

∫ L

0

∫

HQ2

α,τ (x, t)ρ(ζ)N(dτ, dα, dζ)

= 23Aǫ(x, t) − 2Bǫ(x, t)

The following lemma shows that Bǫ(x, t) is small.

Lemma 3.24 There exists C1 > 0 such that for any ǫ > 0,

E (Bǫ(x, t)) ≤ C1ǫ

Proof : Using Lemma 3.18 then (H ′), we easily obtain

E (Bǫ(x, t)) ≤ K

∫ t

t−ǫ

∫ L

0G2

t−s(x, y)E

(

⟨

Sα,τ (y, s)1H(ζ)1[t−ǫ,s](τ)⟩

ρN

)

dyds

But, for s ∈ [t − ǫ, t],

E

(

⟨

Sα,τ (y, s)1H(ζ)1[t−ǫ,s](τ)⟩

ρN

)

≤ E

(

⟨

Sα,τ (y, s)1H(ζ)1[s−ǫ,s](τ)⟩

ρN

)

≤ KE

(

∫ s

s−ǫ

∫ L

0

∫

HG2

s−τ (y, α)ρ(ζ)N(dτ, dα, dζ)

)

+KE

(

∫ s

s−ǫ

∫ L

0

∫

HQ2

α,τ (y, s)ρ(ζ)N(dτ, dα, dζ)

)

= K [Iǫ1(y, s) + Iǫ

2(y, s)]

25

Lemma 3.18 and (H ′) yield

Iǫ2(y, s) ≤ K

∫ s

s−ǫ

∫ L

0G2

s−s′(y, y′)E(

⟨

Sα,τ (y′, s′)1[s−ǫ,s′](τ)1H(ζ)

⟩

ρN

)

dy′ds′ ≤ K√

ǫ

by using the Appendix (4.4), since S is defined as satisfying supy,s E(

〈S(y, s)〉ρN

)

< ∞. Furthermore,

Iǫ1(y, s) =

∫ s

s−ǫ

∫ L

0

∫

HG2

s−τ (y, α)ρ(ζ)ϕ(ζ)dζdαdτ ≤ K√

ǫ

since ρ ∈ L1(O,ϕ(ζ)dζ), and thanks to the Appendix (4.4). We thus get

E (Bǫ(x, t)) ≤ K√

ǫ

∫ t

t−ǫ

∫ L

0G2

t−s(x, y)dyds

and it suffices to use one more time (4.4) to conclude.

The next lemma will allow to prove that E(

e−λAǫ(x,t))

is small (when λ is large).

Lemma 3.25 There exist λ0 ≥ 0, ǫ0 > 0, and K0 > 0, such that for all λ ≥ λ0, for all ǫ ≤ ǫ0,

∫ ǫ

0

∫ L

0

∫

H

(

1 − e−λG2s(x,y)ρ(z)

)

ϕ(z)dzdyds ≥ K0λr0ǫ

3−2r02 (3.56)

Proof : let us first notice, using the Appendix (4.1), that for every s ∈ [ǫ/2, ǫ], y ∈ [x − √ǫ, x +

√ǫ],

G2s(x, y) ≥ C/ǫ, where C > 0 is a constant. The left member of (3.56) is thus greater than

∫

H

∫ ǫ

ǫ/2

∫ x+√

ǫ

x−√ǫ

(

1 − e−λG2s(x,y)ρ(z)

)

dyds ϕ(z)dz ≥ Kǫ√

ǫ

∫

H

(

1 − e−C λ

ǫρ(z))

ϕ(z)dz ≥ K0ǫ3

2

(

λ

ǫ

)r0

where the last inequality, which holds as soon as Cλ/ǫ ≥ γ0, comes from assumption (EP2).

Now we can check that Σ(x, t) > 0 a.s. We notice that for all η > 0, ǫ > 0, and λ > 0,

P (Σ(x, t) > 0) ≥ P(

23Aǫ(x, t) > η

)

+ P (2Bǫ(x, t) < η) − 1

≥ 1 − eληE(

e−2

3λAǫ(x,t)

)

− 2ηE (Bǫ(x, t))

But E (Bǫ(x, t)) ≤ C1ǫ, and if ǫ < ǫ0, if λ ≥ 32λ0, Lemma 3.25 yields

E(

e−2

3λAǫ(x,t)

)

= exp

(

−∫ t

t−ǫ

∫ L

0

∫

H

(

1 − e−2

3λG2

t−τ(x,α)ρ(ζ)

)

ϕ(ζ)dζdαdτ

)

≤ exp(

−C2λr0ǫ

3

2−r0

)

Hence for all η > 0, ǫ < ǫ0, λ ≥ 32λ0,

P (Σ(x, t) > 0) ≥ 1 − exp(

−C2λr0ǫ

3

2−r0 + λη

)

− 2C1ǫ

η

We choose λ = η−1 = ǫ−α where α > 0. We obtain, for all ǫ > 0 small enough :

P (Σ(x, t) > 0) ≥ 1 − exp

(

1 − C2

(

1

ǫ

)αr0− 3

2+r0

)

− 2C1ǫ1−α

26

Since r0 > 34 , we can choose α > 0 such that αr0 − 3

2 + r0 > 0 and 1−α > 0. Letting ǫ go to 0, we deducethat Σ(x, t) > 0 a.s., and Theorem 1.5 is proved under (EP2).

Comparing the proofs of Theorem 1.5 under (EP1) and under (EP2), we see how useful are the local

derivatives. Under (EP1), we only need to consider⟨

DNV (x, t)⟩

ρN, and we do not really use the expres-

sion of DNα,τ,ζV (x, t) for each α, τ, ζ. Saint Loubert Bie works in a quite similar way. But under (EP2),

we need the local expressions of the derivatives, which allow us to take into account the ”explosion” of theGreen kernel.

In [2], Bichteler et al. do not define the local derivatives, they work directly with⟨

DNXt

⟩

ρN(where Xt

is a diffusion process) : since this scalar product satisfies a linear S.D.E., they can use the Doleans-Dadeformula in order to study its positivity. Here, we can not use such a method, because of the Green kernelGt(x, y).

3.5.3 Existence of the density under (EW ).

We show here that σW (x, t) > 0 a.s. The next proof is inspired by Bally and Pardoux in [1], althoughthey use the Holder regularity of their solution.The proof is quite similar (but easier) to that of Subsection 3.5.2. We first use a uniqueness argument, inorder to write

σW (x, t) =

∫ t

0

∫ L

0(Sα,τ (x, t))2 f2(V−(α, τ))dαdτ

where Sα,τ satisfies equation (3.54) in the sense of Lemma 3.22-1 (this in not the same object as in theprevious paragraphs). Using (EW ), we just have to prove that

Σ(x, t) =

∫ t

0

∫ L

0(Sα,τ (x, t))2 dαdτ > 0 a.s.

We split Sα,τ (x, t) = Gt−τ (x, α)1τ≤t + Qα,τ (x, t), and we obtain, for all ǫ > 0 :

Σ(x, t) ≥ 2

3

∫ t

t−ǫ

∫ L

0G2

t−τ (x, α)dαdτ − 2

∫ t

t−ǫ

∫ L

0(Qα,τ (x, t))2dαdτ ≥ 2

3Jǫ(x, t) − 2Iǫ(x, t)

In the Appendix (4.5), one can see that Jǫ(x, t) ≥ C√

ǫ. Furthermore, an easy computation (as in Lemma3.24) shows that E(Iǫ(x, t)) ≤ Kǫ. The conclusion follows.

4 Appendix.

The results below are elementary properties of the Green kernel Gt(x, y) defined in Section 1. In all theequations below, the inequalities remain true for (x, t) ∈ [0, L]×]0, T ], and CT is a constant dependingonly on T . The three first equations are proved in Walsh, [9], p 311-323.

1√4πt

exp

−(y − x)2

4t

≤ Gt(x, y) ≤ CT√t

exp

−(y − x)2

4t

(4.1)

If r > 0,

∫ L

0Gr

t (x, y)dy ≤ CT t1−r

2 (4.2)

If r ∈]0, 3[,

∫ t

0

∫ L

0Gr

s(x, y)dyds ≤ CT (4.3)

27

The two following inequalities are proved by Bally and Pardoux in the Appendix of [1].

If r ∈]0, 3[, then ∀ ǫ > 0,

∫ t

t−ǫ

∫ L

0Gr

t−s(x, y)dyds ≤ CT × ǫ3−r

2 (4.4)

For all ǫ > 0,

∫ t

t−ǫ

∫ x+√

ǫ

x−√ǫ

G2t−s(x, y)dyds ≥ K ×√

ǫ (4.5)

Acknowledgements : I wish to thank Sylvie Meleard for her help and support during the preparationof this paper. I am very grateful to the anonymous referee for his careful reading and fruitful remarks.

References

[1] V. Bally, E. Pardoux, Malliavin Calculus for white noise driven SPDEs, Potential Analysis, vol. 9,no 1, p 27-64, 1998.

[2] K. Bichteler, J.B. Gravereaux, J. Jacod, Malliavin calculus for processes with jumps, Number 2 inStochastic monographs, Gordon and Breach, 1987.

[3] K. Bichteler, J. Jacod, Calcul de Malliavin pour les diffusions avec sauts, existence d’une densite dansle cas unidimensionel, Seminaire de Probabilites XVII, L.N.M. 986, p 132-157, Springer, 1983.

[4] J. Jacod, A.N. Shiryaev, Limit Theorems for Stochastic Processes, Springer, 1987.

[5] D. Nualart, Malliavin Calculus and related topics, Springer, 1995.

[6] E. Pardoux, T. Zhang, Absolute continuity for the law of the solution of a parabolic S.P.D.E., J. ofFunct. Anal. 112, 447-458, 1993.

[7] E. Saint Loubert Bie, Etude d’une EDPS conduite par un bruit Poissonnien, Manuscrit de these,1998.

[8] J.B. Walsh, A stochastic model for neural response, Advances in applied probability, vol. 13, 231-281,1981.

[9] J.B. Walsh, An introduction to stochastic partial differential equations, Ecole d’ete de Probabilite deSaint Flour 14, L.N.M 1180 p 265-439, Springer, 1986.

28