Sobolev solution for semilinear PDE with obsta
le under
monotoni
ity 
ondition
Anis MATOUSSI
Equipe �Statistique et Pro
essus�
Université du Maine
Avenue Olivier Messiaen
72085 LE MANS Cedex 9
FRANCE
anis.matoussi�univ-lemans.fr
Mingyu XU
Institute of Applied Mathemati
s
A
ademy of Mathemati
s and Systems S
ien
e
CAS, Beijing, 100190 China
xumy�amss.a
.
n
�rst revision version Mar
h 20, 2006, �nal a

epted version May 26, 2008
Abstra
t
We prove the existen
e and uniqueness of Sobolev solution of a semilinear PDE's and PDE's
with obsta
le under monotoni
ity 
ondition. Moreover we give the probabilisti
 interpretation of
the solutions in term of Ba
kward SDE and re�e
ted Ba
kward SDE respe
tively.
Key words: Ba
kward sto
hasti
 di�erential equation, Re�e
ted ba
kward sto
hasti
 di�erential
equation, monotoni
ity 
ondition, Sto
hasti
 �ow, partial di�erential equation with obsta
le.
AMS Classi�
ation: 35D05, 60H10, 60H30B
1 Introdu
tion
Our approa
h is based on Ba
kward Sto
hasti
 Di�erential Equations (in short BSDE's) whi
h were
�rst introdu
ed by Bismut [5℄ in 1973 as equation for the adjoint pro
ess in the sto
hasti
 version
of Pontryagin maximum prin
iple. Pardoux and Peng [14℄ generalized the notion in 1990 and were
the �rst to 
onsider general BSDE's and to solve the question of existen
e and uniqueness in the
non-linear 
ase. Sin
e then BSDE's have been widely used in sto
hasti
 
ontrol and espe
ially
in mathemati
al �nan
e, as any pri
ing problem by repli
ation 
an be written in terms of linear
BSDEs, or non-linear BSDEs when portfolios 
onstraints are taken into a

ount as in El Karoui,
Peng and Quenez [6℄.
The main motivation to introdu
e the non-linear BSDE's was to give a probabilisti
 interpreta-
tion (Feynman-Ka
's formula) for the solutions of semilinear paraboli
 PDE's. This result was �rst
obtained by Peng in [16℄, see also Pardoux and Peng [15℄ by 
onsidering the vis
osity and 
lassi
al
solutions of su
h PDE's. Later, Barles and Lesigne [2℄ studied the relation between BSDE's and
solutions of semi-linear PDE's in Soblev spa
es. More re
ently Bally and Matoussi [3℄ studied semi-
linear sto
hasti
 PDEs and ba
kward doubly SDE in Sobolev spa
e and their probabilisti
 method
is based on sto
hasti
 �ow.
The re�e
ted BSDE's was introdu
ed by the �ve authors El Karoui, Kapoudjian, Pardoux,
Peng and Quenez in [7℄, the setting of those equations is the following: let us 
onsider moreover an
adapted sto
hasti
 pro
ess L := (Lt)t 6 T whi
h stands for a barrier. A solution for the re�e
ted
http://arxiv.org/abs/0704.1414v2
BSDE asso
iated with (ξ, g, L) is a triple of adapted sto
hasti
 pro
esses (Yt, Zt,Kt)t 6 T su
h that
Yt = ξ +
g(s, ω, Ys, Zs)ds+KT −Kt −
ZsdBs, ∀ t ∈ [0, T ],
Yt > Lt and
(Yt − Lt)dKt = 0.
The pro
ess K is 
ontinuous, in
reasing and its role is to push upward Y in order to keep it above
the barrier L. The requirement
(Yt − Lt)dKt = 0 means that the a
tion of K is made with a
minimal energy.
The development of re�e
ted BSDE's (see for example [7℄, [10℄, [9℄) has been espe
ially motivated
by pri
ing Ameri
an 
ontingent 
laim by repli
ation, espe
ially in 
onstrained markets. A
tually
it has been shown by El Karoui, Pardoux and Quenez [8℄ that the pri
e of an Ameri
an 
ontingent

laim (St)t 6 T whose strike is γ in a standard 
omplete �nan
ial market is Y0 where (Yt, πt,Kt)t 6 T
is the solution of the following re�e
ted BSDE
−dYt = b(t, Yt, πt)dt+ dKt − πtdWt, YT = (ST − γ)+,
Yt > (St − γ)+ and
(Yt − (St − γ)+)dKt = 0
for an appropriate 
hoi
e of the fun
tion b. The pro
ess π allows to 
onstru
t a repli
ation strategy
and K is a 
onsumption pro
ess that 
ould have the buyer of the option. In a standard �nan
ial
market the fun
tion b(t, ω, y, z) = rty+zθt where θt is the risk premium and rt the spot rate to invest
or borrow. Now when the market is 
onstrained i.e. the interest rates are not the same whether
we borrow or invest money then the fun
tion b(t, ω, y, z) = rty + zθt − (Rt − rt)(y − (z.σ−1t .1))−
where Rt (resp. rt) is the spot rate to borrow (resp. invest) and σ the volatility.
Partial Di�erential Equations with obsta
les and their 
onne
tions with optimal 
ontrol prob-
lems have been studied by Bensoussan and Lions [4℄. They study su
h equations in the point of
view of variational inequalities. In a re
ent paper, Bally, Caballero, El Karoui and Fernandez [1℄
studied the the following semilinear PDE with obsta
le
(∂t + L)u + f(t, x, u, σ∗∇u) + ν = 0, u > h, uT = g,
where h is the obsta
le. The solution of su
h equation is a pair (u, ν) where u is a fun
tion in
2([0, T ],H) and ν is a positive measure 
on
entrated on the set {u = h}. The authors proved
the uniqueness and existen
e for the solution to this PDE when the 
oe�
ient f is Lips
hitz and
linear in
reasing on (y, z), and gave the probabilisti
 interpretation (Feynman-Ka
 formula) for
u and ∇u by the solution (Y, Z) of the re�e
ted BSDE (in short RBSDE). They prove also the
natural relation between Re�e
ted BSDE's and variational inequalities and prove uniqueness of the
solution for su
h variational problem by using the relation between the in
reasing pro
ess K and
the measure ν. This is also a point of view in this paper.
On the other hand, Pardoux [13℄ studied the solution of a BSDE with a 
oe�
ient f(t, ω, y, z),
whi
h satis�es only monotoni
ity, 
ontinuous and general in
reasing 
onditions on y, and a Lips
hitz

ondition on z, i.e. for some 
ontinuous, in
reasing fun
tion ϕ : R+ → R+, and real numbers µ ∈ R,
k > 0, ∀t ∈ [0, T ], ∀y, y′ ∈ Rn, ∀z, z′ ∈ Rn×d,
|f(t, y, 0)| 6 |f(t, 0, 0)|+ ϕ(|y|), a.s.; (1)
〈y − y′, f(t, y, z)− f(t, y′, z)〉 6 µ |y − y′|2 , a.s.;
|f(t, y, z)− f(t, y, z′)| 6 k |z − z′| , a.s..
In the same paper, he also 
onsidered the PDE whose 
oe�
ient f satis�es the monotoni
ity

ondition (1), proved the existen
e of a vis
osity solution u to this PDE and gave its probabilisti
interpretation via the solution of the 
orresponding BSDE. More re
ently, Lepeltier, Matoussi and
Xu [12℄ proved the existen
e and uniqueness of the solution for the re�e
ted BSDE under the
monotoni
ity 
ondition.
In our paper, we study the Sobolev solutions of the PDE and also the PDE with 
ontinuous
obsta
le under the monotoni
ity 
ondition (1). Using penalization method, we prove the existen
e
of the solution and give the probabilisti
 interpretation of the solution u and ∇u (resp.(u,∇u, ν))
by the solution (Y, Z) of ba
kward SDE (resp. the solution (Y, Z,K) of re�e
ted ba
kward SDE).
Furthermore we use equivalen
e norm results and a sto
hasti
 test fun
tion to pass from the solution
of PDE's to the one of BSDE's in order to get the uniqueness of the solution.
Our paper is organized as following: in se
tion 2, we present the basi
 assumptions and the
de�nitions of the solutions for PDE and PDE with obsta
le, then in se
tion 3, we re
all some useful
results from [3℄. We will prove the main results for PDE and PDE with 
ontinuous barrier under
monotoni
ity 
ondition in se
tion 4 and 5 respe
tively. Finally, we prove an analogue result to
Proposition 2.3 in [3℄ under the monotoni
ity 
ondition, and we also give a priori estimates for the
solution of the re�e
ted BSDE's.
2 Notations and preliminaries
Let (Ω,F , P ) be a 
omplete probability spa
e, and B = (B1, B2, · · · , Bd)∗ be a d-dimensional
Brownian motion de�ned on a �nite interval [0, T ], 0 < T < +∞. Denote by {F ts; t 6 s 6 T } the
natural �ltration generated by the Brownian motion B :
F ts = σ{Bs −Bt; t 6 r 6 s} ∪ F0,
where F0 
ontains all P−null sets of F .
We will need the following spa
es for studying BSDE or re�e
ted BSDE. For any given n ∈ N:
• L2n(F ts) : the set of n-dimensional F ts-measurable random variable ξ, su
h that E(|ξ|2) < +∞.
• H2n×m(t, T ) : the set of Rm×n-valued F ts-predi
table pro
ess ψ on the interval [t, T ], su
h that
‖ψ(s)‖2 ds < +∞.
• S2n(t, T ) : the set of n-dimensional F ts-progressively measurable pro
ess ψ on the interval [t, T ],
su
h that E(supt 6 s 6 T ‖ψ(s)‖
) < +∞.
• A2(t, T ) :={K : Ω× [t, T ] → R, F ts�progressively measurable in
reasing RCLL pro
esses
with Kt = 0, E[(KT )
2] <∞ }.
Finally, we shall denote by P the σ-algebra of predi
table sets on [0, T ]×Ω. In the real�valued

ase, i.e., n = 1, these spa
es will be simply denoted by L2(F ts), H2(t, T ) and S2(t, T ), respe
tively.
For the sake of the Sobolev solution of the PDE, the following notations are needed:
• Cmb (Rd,Rn) : the set of Cm-fun
tions f : Rd → Rn, whose partial derivatives of order less
that or equal to m, are bounded. (The fun
tions themselves need not to be bounded)
• C1,mc ([0, T ] × Rd,Rn) : the set of 
ontinuous fun
tions f : [0, T ] × Rd → Rn with 
ompa
t
support, whose �rst partial derivative with respe
t to t and partial derivatives of order less
or equal to m with respe
t to x exist.
• ρ : Rd → R, the weight, is a 
ontinuous positive fun
tion whi
h satis�es
ρ(x)dx <∞.
• L2(Rd, ρ(x)dx) : the weighted L2-spa
e with weight fun
tion ρ(x), endowed with the norm
L2(Rd,ρ) =
|u(x)|2 ρ(x)dx
We assume:
Assumption 2.1. g(·) ∈ L2(Rd, ρ(x)dx).
Assumption 2.2. f : [0, T ]× Rd × Rn×Rn×d → Rn is measurable in (t, x, y, z) and
|f(t, x, 0, 0)|2 ρ(x)dxdt <∞.
Assumption 2.3. f satis�es in
reasing and monotoni
ity 
ondition on y, for some 
ontinuous
in
reasing fun
tion ϕ : R+ → R+, real numbers k > 0, µ ∈ R su
h that ∀(t, x, y, y′, z, z′) ∈
[0, T ]× Rd × Rn × Rn × Rn×d × Rn×d
(i) |f(t, x, y, z)| 6 |f(t, x, 0, z)|+ ϕ(|y|),
(ii) |f(t, x, y, z)− f(t, x, y, z′)| 6 k |z − z′|,
(iii) 〈y − y′, f(t, x, y, z)− f(t, x, y′, z)〉 6 µ |y − y′|2,
(iv) y → f(t, x, y, z) is 
ontinuous.
For the PDE with obsta
le, we 
onsider that f satis�es assumptions 2.2 and 2.3, for n = 1.
Assumption 2.4. The obsta
le fun
tion h ∈ C([0, T ] × Rd;R) satis�es the following 
onditions:
there exists κ ∈ R, β > 0, su
h that ∀(t, x) ∈ [0, T ]× Rd
(i) ϕ(eµth+(t, x)) ∈ L2(Rd; ρ(x)dx),
(ii) |h(t, x)| 6 κ(1 + |x|β),
here h+ is the positive part of h.
Assumption 2.5. b : [0, T ]× Rd → Rd and σ : [0, T ]× Rd → Rd×d satisfy
b ∈ C2b (Rd;Rd) and σ ∈ C3b (Rd;Rd×d).
We �rst study the following PDE
(∂t + L)u + F (t, x, u,∇u) = 0, ∀ (t, x) ∈ [0, T ]× Rd
u(x, T ) = g(x), ∀x ∈ Rd
where F : [0, T ]× Rd × Rn × Rn×d → R, su
h that
F (t, x, u, p) = f(t, x, u, σ∗p)
i,j=1
∂xi∂xj
a := σσ∗. Here σ∗ is the transposed matrix of σ.
In order to study the weak solution of the PDE, we introdu
e the following spa
e
H := {u ∈ L2([0, T ]× Rd, ds⊗ ρ(x)dx)
∣∣ σ∗∇u ∈ L2(([0, T ]× Rd, ds⊗ ρ(x)dx)}
endowed with the norm
‖u‖2 :=
[|u(s, x)|2 + |(σ∗∇u)(s, x)|2]ρ(x)dsdx.
De�nition 2.1. We say that u ∈ H is the weak solution of the PDE asso
iated to (g, f), if
(i) ‖u‖2 <∞,
(ii) for every φ ∈ C1,∞c ([0, T ]× Rd)
(us, ∂tφ)ds + (u(t, ·), φ(t, ·)) − (g(·), φ(·, T )) +
E(us, φs)ds =
(f(s, ·, us, σ∗∇us), φs)ds.
where (φ, ψ) =
φ(x)ψ(x)dx denotes the s
alar produ
t in L2(Rd, dx) and
E(ψ, φ) =
((σ∗∇ψ)(σ∗∇φ) + φ∇((1
σ∗∇σ + b)ψ))dx
is the energy of the system of our PDE whi
h 
orresponds to the Diri
hlet form asso
iated to the
operator L when it is symmetri
. Indeed E(ψ, φ) = −(φ,Lψ).
The probabilisti
 interpretation of the solution of PDE asso
iated with g, f , whi
h satisfy As-
sumption 2.1-2.3 was �rstly studied by (Pardoux [13℄), where the author proved the existen
e of a
vis
osity solution to this PDE, and gave its probabilisti
 interpretation. In se
tion 4, we 
onsider
the weak solution to PDE (2) in Sobolev spa
e, and give the proof of the existen
e and uniqueness
of the solution as well as the probabilisti
 interpretation.
In the se
ond part of this arti
le, we will 
onsider the obsta
le problem asso
iated to the PDE
(2) with obsta
le fun
tion h, where we restri
t our study in the one dimensional 
ase (n = 1).
Formulaly, The solution u is dominated by h, and veri�es the equation in the following sense :
∀(t, x) ∈ [0, T ]× Rd
(i) (∂t + L)u+ F (t, x, u,∇u) 6 0, on u(t, x) > h(t, x),
(ii) (∂t + L)u+ F (t, x, u,∇u) = 0, on u(t, x) > h(t, x),
(iii) u(x, T ) = g(x) .
where L =
i=1 bi
i,j=1 ai,j
∂xi∂xj
, a = σσ∗. In fa
t, we give the following formulation of
the PDE with obsta
le.
De�nition 2.2. We say that (u, ν) is the weak solution of the PDE with obsta
le asso
iated to
(g, f, h), if
(i) ‖u‖2 <∞, u > h, and u(T, x) = g(x).
(ii) ν is a positive Radon measure su
h that
ρ(x)dν(t, x) <∞,
(iii) for every φ ∈ C1,∞c ([0, T ]× Rd)
(us, ∂sφ)ds+ (u(t, ·), φ(t, ·)) − (g(·), φ(·, T )) +
E(us, φs)ds (3)
(f(s, ·, us, σ∗∇us), φs)ds+
φ(s, x)1{u=h}dν(x, s).
3 Sto
hasti
 �ow and random test fun
tions
Let (Xt,xs )t 6 s 6 T be the solution of
dXt,xs = b(s,X
s )ds+ σ(s,X
s )dBs,
t = x,
where b : [0, T ]× Rd → Rd and σ : [0, T ]× Rd → Rd×d satisfy Assumption 2.5.
So {Xt,xs , x ∈ Rd, t 6 s 6 T } is the sto
hasti
 �ow asso
iated to the di�use {Xt,xs } and denote
by {X̂t,xs , t 6 s 6 T } the inverse �ow. It is known that x → X̂t,xs is di�erentiable (Ikeda and
Watanabe [?℄). We denote by J(Xt,xs ) the determinant of the Ja
obian matrix of X̂
s , whi
h is
positive, and J(X
t ) = 1.
For φ ∈ C∞c (Rd) we de�ne a pro
ess φt : Ω× [0, T ]× Rd → R by
φt(s, x) := φ(X̂
s )J(X̂
Following Kunita (See [11℄), we know that for v ∈ L2(Rd), the 
omposition of v with the sto
hasti
�ow is
(v ◦Xt,·s , φ) := (v, φt(s, ·)).
Indeed, by a 
hange of variable, we have
(v ◦Xt,·s , φ) =
v(y)φ(X̂t,ys )J(X̂
s )dy =
v(Xt,xs )φ(x)dx.
The main idea in Bally and Matoussi [3℄ and Bally et al. [1℄, is to use φt as a test fun
tion in (2)
and (3). The problem is that s→ φt(s, x) is not di�erentiable so that
(us, ∂sφ)ds has no sense.
However φt(s, x) is a semimartingale and they proved the following semimartingale de
omposition
of φt(s, x):
Lemma 3.1. For every fun
tion φ ∈ C2c(Rd),
φt(s, x) = φ(x)−
(σij(x)φt(r, x))
L∗φt(r, x)dr, (4)
where L∗ is the adjoint operator of L. So
dφt(r, x) = −
(σij(x)φt(r, x))
dBjr + L∗φt(r, x)dr, (5)
Then in (2) we may repla
e ∂sφds by the It� sto
hasti
 integral with respe
t to dφt(s, x), and
have the following proposition whi
h allows us to use φt as a test fun
tion. The proof will be given
in the appendix.
Proposition 3.1. Assume that assumptions 2.1, 2.2 and 2.3 hold. Let u ∈ H be a weak solution
of PDE (2), then for s ∈ [t, T ] and φ ∈ C2c (Rd),
u(r, x)dφt(r, x)dx − (g(·), φt(T, ·)) + (u(s, ·), φt(s, ·))−
E(u(r, ·), φt(r, ·))dr
f(r, x, u(r, x), σ∗∇u(r, x))φt(r, x)drdx. a.s. (6)
Remark 3.1. Here φt(r, x) is R-valued. We 
onsider that in (6), the equality holds for ea
h

omponent of u.
We need the result of equivalen
e of norms, whi
h play important roles in existen
e proof for
PDE under monotoni
 
onditions. The equivalen
e of fun
tional norm and sto
hasti
 norm is �rst
proved by Barles and Lesigne [2℄ for ρ = 1. In Bally and Matoussi [3℄ proved the same result
for weighted integrable fun
tion by using probabilisti
 method. Let ρ be a weighted fun
tion, we
take ρ(x) := exp(F (x)), where F : Rd → R is a 
ontinuous fun
tion. Moreover, we assume that
there exists a 
onstant R > 0, su
h that for |x| > R, F ∈ C2b (Rd,R). For instant, we 
an take
ρ(x) = (1 + |x|)−q or ρ(x) = expα |x|, with q > d+ 1, α ∈ R.
Proposition 3.2. Suppose that assumption 2.5 hold, then there exists two 
onstants k1, k2 > 0,
su
h that for every t 6 s 6 T and φ ∈ L1(Rd, ρ(x)dx), we have
|φ(x)| ρ(x)dx 6
∣∣φ(Xt,xs )
∣∣)ρ(x)dx 6 k1
|φ(x)| ρ(x)dx, (7)
Moreover, for every ψ ∈ L1([0, T ]× Rd, dt⊗ ρ(x)dx)
|ψ(s, x)| ρ(x)dsdx 6
∣∣ψ(s,Xt,xs )
∣∣)ρ(x)dsdx (8)
|ψ(s, x)| ρ(x)dsdx,
where the 
onstants k1, k2 depend only on T , ρ and the bounds of the �rst (resp. �rst and se
ond)
derivatives of b (resp. σ).
This proposition is easy to get from the follwing Lemma, see Lemma 5.1 in Bally and Matoussi
Lemma 3.2. There exist two 
onstants c1 > 0 and c2 > 0 su
h that ∀x ∈ Rd, 0 6 t 6 T
c1 6 E
ρ(t, X̂
t )J(X̂
6 c2.
4 Sobolev's Solutions for PDE's under monotoni
ity 
ondi-
In this se
tion we shall study the solution of the PDE whose 
oe�
ient f satis�es the monotoni
ity

ondition. For this sake, we introdu
e the BSDE asso
iated with (g, f): for t 6 s 6 T ,
Y t,xs = g(X
f(r,Xt,xr , Y
r , Z
r )dr −
Zt,xs dBs. (9)
Thanks to the equivalen
e of the norms result (3.2), we know that g(X
T ) and f(s,X
s , 0, 0) make
sense in the BSDE (9). Moreover we have
) ∈ L2n(FT ) and f(., Xt,x. , 0, 0) ∈ H2n(0, T ).
It follows from the results from Pardoux [13℄ that for ea
h (t, x), there exists a unique pair
(Y t,x, Zt,x) ∈ S2(t, T ) ×H2n×d(t, T ) of {F ts} progressively measurable pro
esses, whi
h solves this
BSDE(g, f).
The main result of this se
tion is
Theorem 4.1. Suppose that assumptions 2.1-2.3 and 2.4 hold. Then there exists a unique weak
solution u ∈ H of the PDE (2). Moreover we have the probabilisti
 interpretation of the solution:
u(t, x) = Y
t , (σ
∗∇u)(t, x) = Zt,xt , dt⊗ dx− a.e. (10)
and moreover Y t,xs = u(s,X
s ), Z
s = (σ
∗∇u)(s,Xt,xs ), dt⊗ dP ⊗ dx-a.e. ∀s ∈ [t, T ].
Proof : We start to prove the existen
e result.
a) Existen
e : We prove the existen
e in three steps. By integration by parts formula, we know
that u solves (2) if and only if
û(t, x) = eµtu(t, x)
is a solution of the PDE(ĝ, f̂), where
ĝ(x) = eµT g(x) and f̂(t, x, y, z) = eµtf(t, x, e−µty, e−µtz)− µy. (11)
Then the 
oe�
ient f̂ satis�es the assumption 2.3 as f , ex
ept that 2.3-(iii) is repla
ed by
(y − y′)(f(t, x, y, z)− f(t, x, y′, z)) 6 0. (12)
In the �rst two steps, we 
onsider the 
ase where f does not depend on ∇u, and write f(t, x, y)
for f(t, x, y, v(t, x)), where v is in L2([0, T ]× Rd, dt⊗ ρ(x)dx).
We assume �rst that f(t, x, y) satis�es the following assumption 2.3': ∀(t, x, y, y′) ∈ [0, T ] ×
d × Rn × Rn,
(i) |f(t, x, y)| 6 |f(t, x, 0)|+ ϕ(|y|),
(ii) 〈y − y′, f(t, x, y)− f(t, x, y′)〉 6 0,
(iii) y → f(t, x, y) is 
ontinuous, ∀(t, x) ∈ [0, T ]× Rd.
Step 1 : Suppose that g(x), f(t, x, 0) are uniformly bounded, i.e. there exists a 
onstant C,
su
h that
|g(x)| + sup
0 6 t 6 T
|f(t, x, 0)| 6 C (13)
where C as a 
onstant whi
h 
an be 
hanged line by line.
De�ne fn(t, y) := (θn ∗ f(t, ·))(y) where θn : Rn → R+ is a sequen
e of smooth fun
tions with

ompa
t support, whi
h approximate the Dira
 distribution at 0, and satisfy
θn(z)dz = 1. Let
{(Y n,t,xs , Zn,t,xs ), t 6 s 6 T } be the solution of BSDE asso
iated to (g(X
T ), fn), namely,
n,t,x
s = g(X
T ) +
f(r,Xt,xr , Y
n,t,x
r )dr −
n,t,x
r dBr, P-a.s.. (14)
Then for ea
h n ∈ N, we have ∣∣Y n,t,xs
∣∣ 6 eTC,
and ∣∣fn(s,Xt,xs , Y n,t,xs )
∣∣2 6 2
∣∣fn(s,Xt,xs , 0)
∣∣2 + 2ψ2(e
where ψ(r) := supn sup|y| 6 r
ϕ(|y|)θn(y − z)dz. So there exists a 
onstant C > 0, s.t.
∣∣Y n,t,xs
∣∣2 +
∣∣fn(s,Xt,xs , Y n,t,xs )
∣∣2 +
∣∣Zn,t,xs
∣∣2)ρ(x)dsdx 6 C. (15)
Then let n → ∞ on the both sides of (14), we get that the limit (Y t,xs , Zt,xs ) of (Y n,t,xs , Zn,t,xs ),
satis�es
Y t,xs = g(X
T ) +
f(r,Xt,xr , Y
r )dr −
Zt,xr dBr, P-a.s.. (16)
Moreover we obtain from the estimate (15) that
∣∣Y t,xs
∣∣2 +
∣∣Zt,xs
∣∣2)ρ(x)dsdx <∞. (17)
Noti
e that (Y
t , Z
t ) are F tt measurable, whi
h implies they are deterministi
. De�ne u(t, x) :=
t , and v(t, x) := Z
t . By the �ow property of X
r and by the uniqueness of the solution of the
BSDE (16), we have that Y t,xs = u(s,X
s ) and Z
s = v(s,X
The terminal 
ondition g and f(., ., 0, 0) are not 
ontinuous in t and x, and assumed to belong in
a suitable weighted L2 spa
e, so the solution u and for instan
e v are not in general 
ontinuous,
and are only de�ned a.e. in [0, T ]× Rd. So in order to give meaning to the expression u(s,Xt,xs )
(resp. v(s,Xt,xs )), and following Bally and Matoussi [3℄, we apply a regularization pro
edure on the
�nal 
ondition g and the 
oe�
ient f . A
tually, a

ording to Pardoux and Peng ([15℄, Theorem
3.2), if the 
oe�
ient (g, f) are smooth, then the PDE (2) admits a unique 
lassi
al solution
u ∈ C1,2([0, T ] × Rd). Therefore the approximated expression u(s,Xt,xs ) (resp. v(s,Xt,xs )) has a
meaning and then pass to the limit in L2 spa
es like us in Bally and Matoussi [3℄.
Now, the equivalen
e of norm result (8) and estimate (17) follow that u, v ∈ L2([0, T ] × Rd, dt ⊗
ρ(x)dx). Finally, let F (r, x) = f(r,Xt,xr , Y
r ), we know that F (s, x) ∈ L2([0, T ]×Rd, dt⊗ ρ(x)dx),
in view of
|F (s, x)|2 ρ(x)dsdx 6 1
∣∣F (s,Xt,xs )
∣∣2 ρ(x)dsdx
∣∣f(s,Xt,xs , Y t,xs )
∣∣2 ρ(x)dsdx <∞.
So that from theorem 2.1 in [3℄, we get that v = σ∗∇u and that u ∈ H solves the PDE asso
iated
to (g, f) under the bounded assumption.
Step 2 : We assume g ∈ L2(Rd, ρ(x)dx), f satis�es the assumption 2.3' and f(t, x, 0) ∈ L2([0, T ]×
d, dt⊗ ρ(x)dx). We approximate g and f by bounded fun
tions as follows :
gn(x) = Πn(g(x)), (18)
fn(t, x, y) = f(t, x, y)− f(t, x, 0) + Πn(f(t, x, 0)),
where
Πn(y) :=
min(n, |y|)
|y| y.
Clearly, the pair (gn, fn) satis�es the assumption (13) of step 1, and
gn → g in L2(Rd, ρ(x)dx), (19)
fn(t, x, 0) → f(t, x, 0) in L2([0, T ]× Rd, dt⊗ ρ(x)dx).
Denote (Y n,t,xs , Z
n,t,x
s ) ∈ S2n(t, T ) × H2n×d(t, T ) the solution of the BSDE(ξn, fn), where ξn =
T ), i.e.
n,t,x
s = gn(X
T ) +
fn(r,X
r , Y
n,t,x
r )dr −
n,t,x
r dBr.
Then from the results in step 1, un(t, x) = Y
n,t,x
t and un(t, x) ∈ H, is the weak solution of the
PDE(gn, fn), with
n,t,x
s = un(s,X
s ), Z
n,t,x
s = (σ
∗∇un)(s,Xt,xs ), a.s. (20)
For m,n ∈ N, applying It�'s formula to |Y m,t,xs − Y n,t,xs |
, we get
∣∣Y m,t,xs − Y n,t,xs
∣∣2 + E
∣∣Zm,t,xr − Zn,t,xr
∣∣2 dr 6 E
∣∣gm(Xt,xT )− gn(X
∣∣Y m,t,xr − Y n,t,xr
∣∣2 dr + E
∣∣fm(r,Xt,xr , 0)− fn(r,Xt,xr , 0)
∣∣2 dr.
From the equivalen
e of the norms (7) and (8), it follows
∣∣Y m,t,xs − Y n,t,xs
∣∣2 ρ(x)dx 6
∣∣gm(Xt,xT )− gn(X
∣∣2 ρ(x)dx
∣∣Y m,t,xr − Y n,t,xr
∣∣2 drρ(x)dx +
∣∣fm(r,Xt,xr , 0)− fn(r,Xt,xr , 0)
∣∣2 drρ(x)dx
∣∣Y m,t,xr − Y n,t,xr
∣∣2 drρ(x)dx + k1
E |gm(x)− gn(x)|2 ρ(x)dx
|fm(r, x, 0)− fn(r, x, 0)|2 ρ(x)drdx,
and by Gronwall's inequality and (19), we get as m,n→ ∞
t 6 s 6 T
∣∣Y m,t,xs − Y n,t,xs
∣∣2 ρ(x)dx → 0.
It follows immediately as m,n→ ∞
∣∣Y m,t,xr − Y n,t,xr
∣∣2 ρ(x)drdx +
∣∣Zm,t,xr − Zn,t,xr
∣∣2 ρ(x)drdx → 0.
Using again the equivalen
e of the norms (8), we get:
|um(s, x)− un(s, x)|2 + |σ∗∇um(s, x)− σ∗∇un(s, x)|2 ρ(x)dxds
∣∣um(s,Xt,xs )− un(s,Xt,xs )
∣∣2 +
∣∣σ∗∇um(s,Xt,xs )− σ∗∇un(s,Xt,xs )
∣∣2)ρ(x)dsdx
∣∣Y m,t,xs − Y n,t,xs
∣∣2 +
∣∣Zm,t,xs − Zn,t,xs
∣∣2)ρ(x)dsdx → 0.
as m,n→ ∞, i.e. {un} is Cau
hy sequen
e in H. Denote its limit as u, so u ∈ H, and satis�es for
every φ ∈ C1,∞c ([0, T ]× Rd),
(us, ∂tφ)ds + (u(t, ·), φ(t, ·)) − (g(·), φ(·, T )) +
E(us, φs)ds =
(f(s, ·, us), φs)ds. (22)
On the other hand, (Y
n,t,x
· , Z
n,t,x
· ) 
onverges to (Y
· , Z
· ) in S
n(0, T )×H2n×d(0, T ), whi
h is the
solution of the BSDE with parameters (g(X
), f); by the equivalen
e of the norms, we dedu
e
s = u(s,X
s ), Z
s = σ
∗∇u(s,Xt,xs ), a.s. ∀s ∈ [t, T ],
spe
ially Y
t = u(t, x), Z
t = σ
∗∇u(t, x).
Now, it's easy to the generalize the result to the 
ase when f satis�es assumption 2.2 .
Step 3: In this step, we 
onsider the 
ase where f depends on ∇u. Assume that g, f satisfy
the assumptions 2.1 - 2.3, with assumption 2.3-(iii) repla
ed by (12). From the result in step 2, for
any given n×d-matrix-valued fun
tion v ∈ L2([0, T ]×Rd, dt⊗ρ(x)dx), f(t, x, u, v(t, x)) satis�es the
assumptions in step 2. So the PDE(g, f(t, x, u, v(t, x))) admits a unique solution u ∈ H satisfying
(i) and (ii) in the de�nition 2.1.
Set V t,xs = v(s,X
s ), then V
s ∈ H2n×d(0, T ) in view of the equivalen
e of the norms. We

onsider the following BSDE with solution (Y
· , Z
Y t,xs = g(X
T ) +
f(s,Xt,xs , Y
s , V
s )ds−
Zt,xs dBs,
then Y t,xs = u(s,X
s ), Z
s = σ
∗∇u(s,Xt,xs ), a.s. ∀s ∈ [t, T ].
Now we 
an 
onstru
t a mapping Ψ from H into itself. For any u ∈ H, u = Ψ(u) is the weak
solution of the PDE with parameters g(x) and f(t, x, u, σ∗∇u).
Symmetri
ally we introdu
e a mapping Φ from H2n(t, T ) × H2n×d(t, T ) into itself. For any
(U t,x, V t,x) ∈ H2n(t, T )×H2n×d(t, T ), (Y t,x, Zt,x) = Φ(U t,x, V t,x) is the solution of the BSDE with
parameters g(X
T ) and f(s,X
s , Y
s , V
s ). Set V
s = σ
∗∇u(s,Xt,xs ), then Y t,xs = u(s,Xt,xs ),
Zt,xs = σ
∗∇u(s,Xt,xs ), a.s.a.e..
Let u1, u2 ∈ H, and u1 = Ψ(u1), u2 = Ψ(u2), we 
onsider the di�eren
e △u := u1 − u2, △u :=
u1−u2. Set V t,x,1s := σ∗∇u1(s,Xt,xs ), V t,x,2s := σ∗∇u2(s,Xt,xs ). We denote by (Y t,x,1, Zt,x,1)(resp.
(Y t,x,2, Zt,x,2)) the solution of the BSDE with parameters g(X
) and f(s,Xt,xs , Y
s , V
t,x,1
s ) (resp.
f(s,Xt,xs , Y
s , V
t,x,2
s )); then for a.e. ∀s ∈ [t, T ],
Y t,x,1s = u1(s,X
s ), Z
t,x,1
s = σ
∗∇u1(s,Xt,xs ),
t,x,2
s = u2(s,X
s ), Z
t,x,2
s = σ
∗∇u2(s,Xt,xs ),
Denote △Y t,xs := Y t,x,1s −Y t,x,2s , △Zt,xs := Zt,x,1s −Zt,x,2s , △V t,xs := V t,x,1s −V t,x,2s . By It�'s formula
applied to eγtE |△Y t,xs |
, for some α and γ ∈ R, we have
∣∣△Y t,xs
∣∣2 + E
eγs(γ
∣∣△Y t,xr
∣∣2 +
∣∣△Zt,xr
∣∣2)dr 6 E
∣∣△Y t,xr
∣∣2 + α
∣∣△V t,xr
∣∣2)dr,
Using the equivalen
e of the norms, we dedu
e that
eγs(γ |△u(s, x)|2 + |σ∗∇(△u)(s, x)|2)ρ(x)dsdx
eγsE(γ
∣∣△Y t,xr
∣∣2 +
∣∣△Zt,xr
∣∣2)ρ(x)drdx
∣∣△Y t,xr
∣∣2 + α
∣∣△V t,xr
∣∣2)ρ(x)drdx
|△u(s, x)|2 + α |σ∗∇(△u)(s, x)|2)ρ(x)dsdx.
Set α = k2
, γ = 1 +
k2, then we get
γs(|△u(s, x)|2 + |σ∗∇(△u)(s, x)|2)ρ(x)dsdx
γs |σ∗∇(△u)(s, x)|2 ρ(x)dsdx,
γs(|△u(s, x)|2 + |σ∗∇(△u)(s, x)|2)ρ(x)dsdx.
Consequently, Ψ is a stri
t 
ontra
tion on H equipped with the norm
eγs(|u(s, x)|2 + |σ∗∇u(s, x)|2)ρ(x)dsdx.
So Ψ has �xed point u ∈ H whi
h is the solution of the PDE (2) asso
iated to (g, f). Moreover,
for t 6 s 6 T ,
Y t,xs = u(s,X
s ), Z
s = σ
∗∇u(s,Xt,xs ), .a.e.
and spe
ially Y
t = u(t, x), Z
t = σ
∗∇u(t, x), a.e.
b) Uniqueness : Let u1 and u2 ∈ H be two solutions of the PDE(g, f). From Proposition 3.1,
for φ ∈ C2c (Rd) and i = 1, 2
ui(r, x)dφt(r, x)dx + (u
i(s, ·), φt(s, ·))− (g(·), φt(·, T ))−
E(ui(r, ·), φt(r, ·))dr
φt(r, x)f(r, x, u
i(r, x), σ∗∇ui(r, x))drdx. (23)
By (4), we get
dφt(r, x)dx =
(σ∗∇ui)(r, x)φt(r, x)dx)dBr
(σ∗∇ui)(σ∗∇φr) + φ∇((
σ∗∇σ + b)uir)
dxdr.
We substitute this in (23), and get
ui(s, x)φt(s, x)dx = (g(·), φt(·, T ))−
(σ∗∇ui)(r, x)φt(r, x)dxdBr
φt(r, x)f(r, x, u
i(r, x), σ∗∇ui(r, x))drdx.
Then by the 
hange of variable y = X̂t,xr , we obtain
ui(s,Xt,ys )φ(y)dy =
T )φ(y)dy +
φ(y)f(s,Xt,ys , u
i(s,Xt,ys ), σ
∗∇ui(s,Xt,ys ))dyds
(σ∗∇ui)(r,Xt,yr )φ(y)dydBr .
Sin
e φ is arbitrary, we 
an prove this result for ρ(y)dy almost every y. So (ui(s,Xt,ys ), (σ
∗∇ui)(s,Xt,ys ))
solves the BSDE(g(X
T ), f), i.e. ρ(y)dy a.s., we have
ui(s,Xt,ys ) = g(X
T ) +
f(s,Xt,ys , u
i(s,Xt,ys ), σ
∗∇ui(s,Xt,ys ))ds−
(σ∗∇ui)(r,Xt,yr )dBr .
Then by the uniqueness of the BSDE, we know u1(s,Xt,ys ) = u
2(s,Xt,xs ) and (σ
∗∇u1)(s,Xt,ys ) =
(σ∗∇u2)(s,Xt,ys ). Taking s = t we dedu
e that u1(t, y) = u2(t, y), dt⊗ dy-a.s. �
5 Sobolev's solution for PDE with obsta
le under monotoni
-
ity 
ondition
In this se
tion we study the PDE with obsta
le asso
iated with (g, f, h), whi
h satisfy the assump-
tions 2.1-2.4 for n = 1. We will prove the existen
e and uniqueness of a weak solution to the
obsta
le problem. We will restri
t our study to the 
ase when ϕ is polynomial in
reasing in y, i.e.
Assumption 5.1. We assume that for some κ1 ∈ R, β1 > 0, ∀y ∈ R,
|ϕ(y)| 6 κ1(1 + |y|β1).
For the sake of PDE with obsta
le, we introdu
e the re�e
ted BSDE asso
iated with (g, f, h),
like in El Karoui et al. [7℄:


Y t,xs = g(X
f(r,Xt,xr , Y
r , Z
r )dr +K
−Kt,xt −
Zt,xs dBs, P -a.s ∀ s ∈ [t, T ]
Y t,xs > L
s , P -a.s
(Y t,xs − Lt,xs )dKt,xs = 0, P -a.s.
where Lt,xs = h(s,X
s ) is a 
ontinuous pro
ess. Moreover following Lepeltier et al [12℄, we shall
need to estimate
E[ sup
t 6 s 6 T
ϕ2(eµt(Lt,xs )
+)] = E[ sup
t 6 s 6 T
ϕ2(eµth(s,Xt,xs )
6 Ce2β1µTE[ sup
t 6 s 6 T
∣∣Xt,xs
∣∣2β1β)]
6 C(1 + |x|2β1β),
where C is a 
onstant whi
h 
an be 
hanged line by line. By assumption 2.4-(ii), with same
te
hniques we get for x ∈ R, E[supt 6 s 6 T ϕ2((Lt,xs )+)] < +∞. Thanks to the assumption 2.1 and
2.2, by the equivalen
e of norms 7 and 8, we have
) ∈ L2(FT ) and f(s,Xt,xs , 0, 0) ∈ H2(0, T ).
By the existen
e and uniqueness theorem for the RBSDE in [12℄, for ea
h (t, x), there exists a
unique triple (Y t,x, Zt,x,Kt,x) ∈ S2(t, T ) ×H2d(t, T ) ×A2(t, T ) of {F ts} progressively measurable
pro
esses, whi
h is the solution of the re�e
ted BSDE with parameters (g(X
), f(s,Xt,xs , y, z),
h(s,Xt,xs ))We shall give the probabilisti
 interpretation for the solution of PDE with obsta
le (3).
The main result of this se
tion is
Theorem 5.1. Assume that assumptions 2.1-2.5 hold and ρ(x) = (1 + |x|)−p with p > γ where
γ = β1β + β + d+ 1. There exists a pair (u, ν), whi
h is the solution of the PDE with obsta
le (3)
asso
iated to (g, f, h) i.e. (u, ν) satis�es De�nition 2.2-(i) -(iii). Moreover the solution is given by:
u(t, x) = Y
t , a.e. where (Y
s , Z
s )t 6 s 6 T is the solution of RBSDE (24), and
Y t,xs = u(s,X
s ), Z
s = (σ
∗∇u)(s,Xt,xs ). (25)
Moreover, we have for every measurable bounded and positive fun
tions φ and ψ,
φ(s, X̂t,xs )J(X̂
s )ψ(s, x)1{u=h}(s, x)dν(s, x) =
φ(s, x)ψ(s,Xt,xs )dK
s , a.s.. (26)
If (u, ν) is another solution of the PDE (3) su
h that ν satis�es (26) with some K instead of K,
where K is a 
ontinuous pro
ess in A2F(t, T ), then u = u and ν = ν.
Remark 5.1. The expression (26) gives us the probabilisti
 interpretation (Feymamn-Ka
's for-
mula) for the measure ν via the in
reasing pro
ess Kt,x of the RBSDE. This formula was �rst
introdu
ed in Bally et al. [1℄, where the authors prove (26) when f is Lips
hitz on y and z uni-
formly in (t, ω). Here we generalize their result to the 
ase when f is monotoni
 in y and Lips
hitz
in z.
Proof. As in the proof of theorem 4.1 in se
tion 4, we �rst noti
e that (u, ν) solves (3) if and
only if
(û(t, x), dν̂(t, x)) = (eµtu(t, x), eµtdν(t, x))
is the solution of the PDE with obsta
le (ĝ, f̂ , ĥ), where ĝ, f̂ are de�ned as in (12) with
ĥ(t, x) = eµth(t, x).
Then the 
oe�
ient f̂ satis�es the same assumptions in assumption 2.3 with (iii) repla
ed by (12),
whi
h means that f is de
reasing on y in the 1-dimensional 
ase. The obsta
le ĥ still satis�es
assumption 2.4, for µ = 0. In the following we will use (g, f, h) instead of (ĝ, f̂ , ĥ), and suppose
that (g, f, h) satis�es assumption 2.1, 2.2, 2.4, 2.5 and 2.3 with (iii) repla
ed by (12).
a) Existen
e : The existen
e of a solution will be proved in 4 steps. From step 1 to step 3,
we suppose that f does not depend on ∇u, satis�es assumption 2.3' for n = 1, and f(t, x, 0) ∈
2([0, T ]× Rd, dt⊗ ρ(x)dx). In the step 4, we study the 
ase when f depend on ∇u.
Step 1 : Suppose g(x), f(t, x, 0), h+(t, x) uniformly bounded i.e. that there exists a 
onstant C
su
h that
|g(x)|+ sup
0 6 t 6 T
|f(t, x, 0)|+ sup
0 6 t 6 T
h+(t, x) 6 C.
We will use the penalization method. For n ∈ N, we 
onsider for all s ∈ [t, T ],
n,t,x
s = g(X
T ) +
f(r,Xt,xr , Y
n,t,x
r )dr + n
(Y n,t,xr − h(r,Xt,xr ))−dr −
n,t,x
r dBr.
From Theorem 4.1 in se
tion 3, we know that un(t, x) := Y
n,t,x
t , is solution of the PDE(g, fn),
where fn(t, x, y, x) = f(t, x, y, z) + n(y − h(t, x))−, i.e. for every φ ∈ C1,∞c ([0, T ]× Rd)
(uns , ∂sφ)ds+ (u
n(t, ·), φ(t, ·)) − (g(·), φ(·, T )) +
E(uns , φs)ds
(f(s, ·, uns ), φs)ds+ n
((un − h)−(s, ·), φs)ds.
Moreover
Y n,t,xs = un(s,X
s ), Z
n,t,x
s = σ
∗∇un(s,Xt,xs ), (27)
Set Kn,t,xs = n
(Y n,t,xr − h(r,Xt,xr ))−dr. Then by (27), we have that Kn,t,xs = n
(un −
h)−(r,Xt,xr )dr.
Following the estimates and 
onvergen
e results for (Y n,t,x, Zn,t,x) in the step 1 of the proof of
Theorem 2.2 in [12℄, for m, n ∈ N, we have,as m,n→ ∞
∣∣Y n,t,xs − Y m,t,xs
∣∣2 ds+ E
∣∣Zn,t,xs − Zm,t,xs
∣∣2 ds+ E sup
t 6 s 6 T
∣∣Kn,t,xs −Km,t,xs
∣∣2 → 0,
∣∣Y n,t,xs
∣∣2 +
∣∣Zn,t,xs
∣∣2 + (Kn,t,xT )
2) 6 C.
By the equivalen
e of the norms (8), we get
ρ(x)(|un(s, x)− um(s, x)|2 + |σ∗∇un(s, x)− σ∗∇um(s, x)|2)dsdx
ρ(x)E
∣∣Y n,t,xs − Y m,t,xs
∣∣2 +
∣∣Zn,t,xs − Zm,t,xs
∣∣2)dsdx→ 0.
Thus (un) is a Cau
hy sequen
e in H, and the limit u = limn→∞ un belongs to H.
Denote νn(dt, dx) = n(un − h)−(t, x)dtdx and πn(dt, dx) = ρ(x)νn(dt, dx), then by (7)
πn([0, T ]× Rd) =
ρ(x)νn(dt, dx) =
ρ(x)n(un − h)−(t, x)dtdx
ρ(x)E
∣∣∣Kn,0,xT
∣∣∣ dx 6 C
ρ(x)dx <∞.
It follows that
πn([0, T ]× Rd) <∞. (28)
In the same way like in the existen
e proof step 2 of theorem 14 in [1℄, we 
an prove that πn([0, T ]×
d) is bounded and then πn is tight. So we may pass to a subsequen
e and get πn → π where π is
a positive measure. De�ne ν = ρ−1π; ν is a positive measure su
h that
ρ(x)dν(t, x) < ∞,
and so we have for φ ∈ C1,∞c ([0, T ]× Rd) with 
ompa
t support in x,
φdνn =
dπn →
Now passing to the limit in the PDE(g, fn), we 
he
k that (u, ν) satis�es the PDE with obsta
le
(g, f, h), i.e. for every φ ∈ C1,∞c ([0, T ]× Rd), we have
(us, ∂sφ)ds+ (u(t, ·), φ(t, ·)) − (g(·), φ(·, T )) +
E(us, φs)ds
(f(s, ·, us), φs)ds+
φ(s, x)1{u=h}(s, x)dν(x, s). (29)
The last is to prove that ν satis�es the probabilisti
 interpretation (26). Sin
e Kn,t,x 
onverges
to Kt,x uniformly in t, the measure dKn,t,x → dKt,x weakly in probability.
Fix two 
ontinuous fun
tions φ, ψ : [0, T ]× Rd → R+ whi
h have 
ompa
t support in x and a

ontinuous fun
tion with 
ompa
t support θ : Rd → R+, we have
φ(s, X̂t,xs )J(X̂
s )ψ(s, x)θ(x)dν(s, x)
= lim
φ(s, X̂t,xs )J(X̂
s )ψ(s, x)θ(x)n(un − h)−(t, x)dtdx
= lim
φ(s, x)ψ(s,Xt,xs )θ(X
s )n(un − h)−(t,Xt,xs )dtdx
= lim
φ(s, x)ψ(s,Xt,xs )θ(X
s )dK
n,t,x
φ(s, x)ψ(s,Xt,xs )θ(X
s )dK
s dx.
We take θ = θR to be the regularization of the indi
ator fun
tion of the ball of radius R and
pass to the limit with R→ ∞, it follows that
φ(s, X̂t,xs )J(X̂
s )ψ(s, x)dν(s, x) =
φ(s, x)ψ(s,Xt,xs )dK
s dx. (30)
Sin
e (Y n,t,xs , Z
n,t,x
n,t,x
s ) 
onverges to (Y
s , Z
s ) as n → ∞ in S2(t, T )×H
2(t, T ) ×
2(t, T ), and (Y t,xs , Z
s ) is the solution of RBSDE(g(X
), f, h), then we have
(Y t,xs − Lt,xs )dKt,xs =
(u− h)(t,Xt,xs )dKt,xs = 0, a.s.
it follows that dKt,xs = 1{u=h}(s,X
s )dK
s . In (30), setting ψ = 1{u=h} yields
φ(s, X̂t,xs )J(X̂
s )1{u=h}(s, x)dν(s, x) =
φ(s, X̂t,xs )J(X̂
s )dν(s, x), a.s.
Note that the family of fun
tions A(ω) = {(s, x) → φ(s, X̂t,xs ) : φ ∈ C∞c } is an algebra whi
h
separates the points (be
ause x → X̂t,xs is a bije
tion). Given a 
ompa
t set G, A(ω) is dense in
C([0, T ] × G). It follows that J(X̂t,xs )1{u=h}(s, x)dν(s, x) = J(X̂t,xs )dν(s, x) for almost every ω.
While J(X̂t,xs ) > 0 for almost every ω, we get dν(s, x) = 1{u=h}(s, x)dν(s, x), and (26) follows.
Then we get easily that Y t,xs = u(s,X
s ) and Z
s = σ
∗∇u(s,Xt,xs ), in view of the 
onvergen
e
results for (Y n,t,xs , Z
n,t,x
s ) and the equivalen
e of the norms. So u(s,X
s ) = Y
s > h(t, x). Spe-

ially for s = t, we have u(t, x) > h(t, x)
Step 2 : As in the proof of the RBSDE in Theorem 2.2 in [12℄, step 2, we relax the bounded

ondition on the barrier h in step 1, and prove the existen
e of the solution under assumption 2.4.
Similarly to step 2 in the proof of theorem 2.2 in [12℄, after some transformation, we know
that it is su�
ient to prove the existen
e of the solution for the PDE with obsta
le (g, f, h), where
(g, f, h) satis�es
g(x), f(t, x, 0) 6 0.
Let h(t, x) satisfy assumption 2.4 for µ = 0, i.e. ∀(t, x) ∈ [0, T ]×,Rd
ϕ(h(t, x)+) ∈ L2(Rd; ρ(x)dx),
|h(t, x)| 6 κ(1 + |x|β).
hn(t, x) = h(t, x) ∧ n,
then the fun
tion hn(t, x) are 
ontinuous, sup0 6 t 6 T h
n (t, x) 6 n, and hn(s,X
s ) → h(s,Xt,xs ) in
F (t, T ), in view of Dini's theorem and dominated 
onvergen
e theorem.
We 
onsider the PDE with obsta
le asso
iated with (g, f, hn). By the results of step 1, there
exists (un, νn), whi
h is the solution of the PDE with obsta
le asso
iated to (g, f, hn), where un ∈ H
and νn is a positive measure su
h that
ρ(x)dνn(t, x) <∞. Moreover
Y n,t,xs = un(s,X
s ), Z
n,t,x
s = σ
∗∇un(s,Xt,xs ),
φ(s, X̂t,xs )J(X̂
s )ψ(s, x)1{un=hn}(s, x)dνn(s, x) =
φ(s, x)ψ(s,Xt,xs )dK
n,t,x
s dx,
Here (Y n,t,x, Zn,t,x,Kn,t,x) is the solution of the RBSDE(g(X
T ), f, hn). Thanks to proposition
6.1 in Appendix, and the bounded assumption of g and f , we know that
[ ∫ T
( ∣∣Y n,t,xs
∣∣2 +
∣∣Zn,t,xs
∣∣2 )ds+ (Kn,0,x
1 + E
ϕ2( sup
0 6 t 6 T
h+(t,X
+ sup
0 6 t 6 T
(h+(t,X
6 C(1 + |x|2β1β + |x|2β).
By the Lemma 2.3 in [12℄, Y n,t,xs → Y t,xs in S2(0, T ), Zn,t,xs → Zt,xs in H2d(0, T ) and Kn,t,xs → Kt,xs
2(0, T ), as n→ ∞. Moreover (Y t,xs , Zt,xs ,Kt,xs ) is the solution of RBSDE(g(X
T ), f, h).
By the 
onvergen
e result of (Y n,t,xs , Z
n,t,x
s ) and the equivalen
e of the norms (8), we get
(|un(t, x)− um(t, x)|2 + |σ∗∇un(s, x)− σ∗∇um(s, x)|2)dsdx
ρ(x)E
∣∣Y n,t,xs − Y m,t,xs
∣∣2 +
∣∣Zn,t,xs − Zm,t,xs
∣∣2)dsdx → 0.
So {un} is a Cau
hy sequen
e in H, and admits a limit u ∈ H. Moreover Y t,xs = u(s,Xt,xs ), Zt,xs =
σ∗∇u(s,Xt,xs ). In parti
ular u(t, x) = Y
t > h(t, x).
Set πn = ρνn, like in step 1, we �rst need to prove that πn([0, T ]× Rd) is uniformly bounded.
In (31), let φ = ρ, ψ = 1, then we have
ρ(X̂0,xs )J(X̂
s )dνn(s, x) =
ρ(x)dKn,0,xs dx.
Re
all Lemma 3.2: there exist two 
onstants c1 > 0 and c2 > 0 su
h that ∀x ∈ Rd, 0 6 t 6 T
c1 6 E
ρ(t, X̂
t )J(X̂
6 c2.
Applying Hölder's inequality and S
hwartz's inequality, we have
πn([0, T ]× Rd)
ρ(x)νn(dt, dx)
2 (x)
2 (t, X̂
2 (X̂
2 (x)ρ
2 (t, X̂
2 (X̂
t )νn(dt, dx)
ρ(t, X̂
t )J(X̂
ρ(x)νn(dt, dx)
ρ(t, X̂
t )J(X̂
t )νn(dt, dx)
ρ(t, X̂
t )J(X̂
ρ(x)νn(dt, dx)
ρ(t, X̂
t )J(X̂
t )νn(dt, dx)
ρ(t, X̂
t )J(X̂
ρ(x)νn(dt, dx)
n,0,x
t ρ(x)dx
ρ(x)νn(dt, dx)
ρ(x)E[K
n,0,x
T ]dx
So by (32) and (7), we get
πn([0, T ]× Rd) 6 C
ρ(x)E[K
n,0,x
T ]dx (33)
ρ(x)(1 + |x|β1β + |x|β)dx <∞.
Using the same arguments as in step 1, we dedu
e that πn is tight. So we may pass to a subsequen
e
and get πn → π where π is a positive measure.
De�ne ν = ρ−1π, then ν is a positive measure su
h that
ρ(x)dν(t, x) < ∞. Then for
φ ∈ C([0, T ]× Rd) with 
ompa
t support in x, we have as n→ ∞,
φdνn =
dπn →
Now passing to the limit in the PDE(g, f, hn), we 
he
k that (u, ν) satis�es the PDE with
obsta
le asso
iated to (g, f, h), i.e. for every φ ∈ C1,∞c ([0, T ]× Rd)
(us, ∂sφ)ds+ (u(t, ·), φ(t, ·)) − (g(·), φ(·, T )) +
E(us, φs)ds
(f(s, ·, us), φs)ds+
φ(s, x)1{u=h}dν(x, s). (34)
Then we will 
he
k if the probabilisti
 interpretation (26) still holds. Fix two 
ontinuous fun
-
tions φ, ψ : [0, T ]× Rd → R+ whi
h have 
ompa
t support in x. With the 
onvergen
e result of
Kn,t,x, whi
h implies dKn,t,x → dKt,x weakly in probability, in the same way as step 1, passing to
the limit in (31) we have
φ(s, X̂t,xs )J(X̂
s )ψ(s, x)dν(s, x) =
φ(s, x)ψ(s,Xt,xs )dK
Sin
e (Y t,xs , Z
s ) is the solution of RBSDE(g(X
T ), f, h), then by the integral 
ondition, we
dedu
e the dKt,xs = 1{u=h}(s,X
s )dK
s . In (35), setting ψ = 1{u=h} yields
φ(s, X̂t,xs )J(X̂
s )1{u=h}(s, x)dν(s, x) =
φ(s, X̂t,xs )J(X̂
s )dν(s, x).
With the same arguments, we get that dν(s, x) = 1{u=h}(s, x)dν(s, x), and (26)holds for ν and K.
Step 3 : Now we will relax the bounded 
ondition on g(x) and f(t, x, 0). Then for m,n ∈ N, let
gm,n(x) = (g(x) ∧ n) ∨ (−m),
fm,n(t, x, y) = f(t, x, y)− f(t, x, 0) + (f(t, x, 0) ∧ n) ∨ (−m).
So gm,n(x) and fm,n(t, x, 0) are bounded and for �xed m ∈ N, as n→ ∞, we have
gm,n(x) → gm(x) in L2(Rd, ρ(x)dx),
fm,n(t, x, 0) → fm(t, x, 0) in L2([0, T ]×Rd, dt⊗ ρ(x)dx),
where
gm(x) = g(x) ∨ (−m),
fm(t, x, y) = f(t, x, y)− f(t, x, 0) + f(t, x, 0) ∨ (−m).
Then as m→ ∞, we have
gm(x) → g(x) in L2(Rd, ρ(x)dx),
fm(t, x, 0) → f(t, x, 0) in L2([0, T ]×Rd, dt⊗ ρ(x)dx),
in view of assumption 2.1 and f(t, x, 0) ∈ L2([0, T ]× Rd, dt⊗ ρ(x)dx).
Now we 
onsider the PDE with obsta
le asso
iated to (gm,n, fm,n, h). By step 2, there exists a
(um,n, νm,n) whi
h is the solution of the PDE with obsta
le asso
iated to (gm,n, fm,n, h). In parti
-
ular the representation formulas (25) and (26) are satis�ed. Denote by (Y m,n,t,x, Zm,n,t,x,Km,n,t,x)
the solution of the RBSDE (gm,n(X
T ), fm,n, h).
Re
all the 
onvergen
e results in step 3 of theorem 2.2 in [12℄, we know that for �xed m ∈ N, as
n→ ∞, (Y m,n,t,xs , Zm,n,t,xs ,Km,n,t,xs ) → (Y m,t,xs , Zm,t,xs ,Km,t,xs ) in S2(0, T )×H2d(0, T )×A2(0, T ),
and that (Y m,t,xs , Z
m,t,x
m,t,x
s ) is the solution of RBSDE(gm(X
T ), fm, h).
By It�'s formula, we have for n, p ∈ N,
∣∣Y m,n,t,xs − Y m,p,t,xs
∣∣2 +
∣∣Zm,n,t,xs − Zm,p,t,xs
∣∣2)ds
∣∣gm,n(Xt,xT )− gm,p(X
∣∣2 + CE
∣∣fm,n(s,Xt,xs , 0)− fm,p(s,Xt,xs , 0)
∣∣2 ds,
so by the equivalen
e of the norms (7) and (8), it follows that as n→ ∞,
ρ(x)(|um,n(t, x)− um,p(t, x)|2 + |σ∗∇um,n(s, x) − σ∗∇um,p(s, x)|2)dsdx
ρ(x) |gm,n(x)− gm,p(x)|2 dx+
ρ(x) |fm,n(s, x, 0)− fm,p(s, x, 0)|2 dsdx→ 0.
i.e. for ea
h �xed m ∈ N, {um,n} is a Cau
hy sequen
e in H, and admits a limit um ∈ H. Moreover
Y m,t,xs = um(s,X
s ), Z
m,t,x
s = σ
∗∇um(s,Xt,xs ), a.s., in parti
ular um(t, x) = Y
m,t,x
t > h(t, x).
Then we �nd the measure νm by the sequen
e {νm,n}. Set πm,n = ρνm,n, by proposition 6.1 in
Appendix, we have for ea
h m,n ∈ N, 0 6 t 6 T
∣∣Km,n,t,xT
∣∣2) 6 CE[g2m,n(X
T ) +
m,n(s,X
s , 0, 0)ds+ ϕ
2( sup
t 6 s 6 T
(h+(s,Xt,xs )))
+ sup
t 6 s 6 T
(h+(s,Xt,xs ))
2 + 1 + ϕ2(2T )]
6 CE[g(X
2(s,Xt,xs , 0, 0)ds+ ϕ
2( sup
0 6 s 6 T
(h+(s,Xt,xs )))
+ sup
0 6 s 6 T
(h+(s,Xt,xs ))
2 + 1 + ϕ2(2T )]
6 C(1 + |x|2β1β + |x|2β). (35)
By the same way as in step 2, we dedu
e that for ea
h �xed m ∈ N, πm,n is tight, we may pass to
a subsequen
e and get πm,n → πm where πm is a positive measure. If we de�ne νm = ρ−1πm, then
νm is a positive measure su
h that
ρ(x)dνm(t, x) <∞. So we have for all φ ∈ C([0, T ]×Rd)
with 
ompa
t support in x,
φdνm,n =
dπm,n →
dπm =
φdνm.
Now for ea
h �xed m ∈ N, let n → ∞, in the PDE(gm,n, fm,n, h), we 
he
k that (um, νm)
satis�es the PDE with obsta
le asso
iated to (gm, fm, h), and by the weak 
onvergen
e result of
dKm,n,t,x, we have easily that the probabilisti
 interpretation (26) holds for νm and K
Then let m → ∞, by the 
onvergen
e results in step 4 of theorem 2.2 in [12℄, we apply the
same method as before. We dedu
e that limm→∞ um = u is in H and Y t,xs = u(s,Xt,xs ), Zt,xs =
σ∗∇u(s,Xt,xs ), a.s., where (Y t,x, Zt,x,Kt,x) is the solution of the RBSDE(g, f, h), in parti
ular,
setting s = t, u(t, x) = Y
t > h(t, x).
From (35), it follows that
m,t,x
2] 6 C(1 + |x|2ββ1 + |x|2β).
By the same arguments, we 
an �nd the measure ν by the sequen
e {νm}, whi
h satis�es that for
all φ and ψ with 
ompa
t support,
φ(s, X̂t,xs )J(X̂
s )ψ(s, x)1{u=h}(s, x)dν(s, x) =
φ(s, x)ψ(s,Xt,xs )dK
s dx.
Finally we �nd a solution (u, ν) to the PDE with obsta
le (g, f, h), when f does not depend on
∇u. So for every φ ∈ C1,∞c ([0, T ]× Rd)
(us, ∂sφ)ds+ (u(t, ·), φ(t, ·)) − (g(·), φ(·, T )) +
E(us, φs)ds
(f(s, ·, us), φs)ds+
φ(s, x)1{u=h}dν(x, s). (36)
Step 4 : Finally we study the 
ase when f depends on ∇u, and satis�es a Lips
hitz 
ondition
on ∇u. We 
onstru
t a mapping Ψ from H into itself. For some u ∈ H, de�ne
u = Ψ(u),
where (u, ν) is a weak solution of the PDE with obsta
le (g, f(t, x, u, σ∇u), h). Then by this
mapping, we denote a sequen
e {un} in H, beginning with a fun
tion v0 ∈ L2([0, T ] × Rd, dt ⊗
ρ(x)dx). Sin
e f(t, x, u, v0(t, x)) satis�es the assumptions of step 3, the PDE(g, f(t, x, u, v0(t, x)), h)
admits a solution (u1, v1) ∈ H. For n ∈ N, set un(t, x) = Ψ(un−1(t, x)).
Symmetri
ally we introdu
e a mapping Φ from H2(t, T ) × H2d(t, T ) into itself. For V t,x,0 =
v0(s,Xt,xs )), then V
s ∈ H2d(t, T ) in view of the equivalen
e of the norms. Set
(Y t,x,n, Zt,x,n) = Φ(Y t,x,n−1, Zt,x,n−1),
where (Y t,x,n, Zt,x,n,Kt,x,n)is the solution of the RBSDE with parameters g(X
T ), f(s,X
s , Y
s , Z
t,x,n−1
and h(s,Xt,xs ).Then Y
t,x,n
s = un(s,X
s ), Z
t,x,n
s = σ
∗∇un(s,Xt,xs ) a.s. and
φ(s, X̂t,xs )J(X̂
s )ψ(s, x)1{u=h}(s, x)dνn(s, x) =
φ(s, x)ψ(s,Xt,xs )dK
t,x,n
s dx.
Set ũn(t, x) := un(t, x) − un−1(t, x). To deal with the di�eren
e ũn, we need the di�eren
e of
the 
orresponding BSDE, denote Ỹ t,x,ns := Y
t,x,n
s − Y t,x,n−1s , Z̃t,x,ns := Zt,x,ns − Zt,x,n−1s , K̃t,x,ns :=
Kt,x,ns −Kt,x,n−1s . It follows from It�'s formula, for some α, γ ∈ R,
∣∣∣Ỹ t,x,ns
eγr(γ
∣∣∣Ỹ t,x,nr
∣∣∣Z̃t,x,nr
∣∣∣Ỹ t,x,nr
∣∣∣Z̃t,x,n−1r
sin
e
eγrỸ t,x,nr dK̃
t,x,n
eγr(Y t,x,ns − h(r,Xt,xr ))dKt,x,n +
eγr(Y t,x,n−1s − h(r,Xt,xr ))dKt,x,n−1
eγr(Y t,x,ns − h(r,Xt,xr ))dKt,x,n−1 +
eγr(Y t,x,n−1s − h(r,Xt,xr ))dKt,x,n
then by the equivalen
e of the norms, for γ = 1 +
k2, we have
eγs(|ũn(s, x)|2 + |σ∗∇(ũn)(s, x)|2)ρ(x)dsdx
eγs(|ũ2(s, x)|2 + |σ∗∇(ũ2)(s, x)|2)ρ(x)dsdx
)n−1(‖u1(s, x)‖2γ + ‖u2(s, x)‖
where ‖u‖2
eγs(|u(s, x)|2 + |σ∗∇u(s, x)|2)ρ(x)dsdx, whi
h is equivalent to the norm ‖·‖
of H. So {un} is a Cau
hy sequen
e in H, it admits a limit u in H, whi
h is the solution to
the PDE with obsta
le (2). Then 
onsider σ∗∇u as a known fun
tion by the result of step 3,
we know that there exists a positive measure ν su
h that
ρ(x)dν(t, x) < ∞, and for every
φ ∈ C1,∞c ([0, T ]× Rd),
(us, ∂sφ)ds+ (u(t, ·), φ(t, ·)) − (g(·), φ(·, T )) +
E(us, φs)ds
(f(s, ·, us, σ∗∇us), φs)ds+
φ(s, x)1{u=h}dν(x, s). (37)
Moreover, for t 6 s 6 T ,
s = u(s,X
s ), Z
s = σ
∗∇u(s,Xt,xs ), a.s.a.e.,
φ(s, X̂t,xs )J(X̂
s )ψ(s, x)1{u=h}(s, x)dν(s, x)
φ(s, x)ψ(s,Xt,xs )dK
b) Uniqueness : Set (u, ν) to be another solution of the PDE with obsta
le (3) asso
iated
to (g, f, h); with ν veri�es (26) for an in
reasing pro
ess K. We �x φ : Rd → R, a smooth fun
tion
in C2c (R
d) with 
ompa
t support and denote φt(s, x) = φ(X̂
s )J(X̂
s ). From proposition 3.1, one
may use φt(s, x) as a test fun
tion in the PDE(g, f, h) with ∂sφ(s, x)ds repla
ed by a sto
hasti
integral with respe
t to the semimartingale φt(s, x). Then we get, for t 6 s 6 T
u(r, x)dφt(r, x)dx + (u(s, ·), φt(s, ·))− (g(·), φt(·, T )) +
E(ur, φr)dr (38)
f(r, x, u(r, x), σ∗∇u(r, x))φt(r, ·)dr +
φt(r, x)1{u=h}dν(x, r).
By (5) in Lemma 3.1, we have
udrφt(r, x)dx =
(σ∗∇u)(r, x)φt(r, x)dx)dBr
(σ∗∇u)(σ∗∇φr) + φt∇((
σ∗∇σ + b)u)
dxdr.
Substitute this equality in (38), we get
u(s, x)φt(s, x)dx = (g(·), φt(·, T ))−
(σ∗∇u)(r, x)φt(r, x)dx)dBr
f(r, x, u(r, x), σ∗∇u(r, x))φt(s, ·)dr +
φt(r, x)1{u=h}dν(x, r).
Then by 
hanging of variable y = X̂t,xr and applying (26) for ν, we obtain
u(s,Xt,ys )φ(y)dy
T )φ(y)dy +
φ(y)f(s,Xt,ys , u(s,X
s ), σ
∗∇u(s,Xt,ys )ds
φ(y)1{u=h}(r,X
s )dK
r dy −
(σ∗∇u)(r,Xt,yr )φ(y)dy)dBr .
Sin
e φ is arbitrary, we 
an prove that for ρ(y)dy almost every y, (u(s,Xt,ys ), (σ
∗∇u)(s,Xt,ys ), K̂t,xs )
solves the RBSDE(g(X
), f, h). Here K̂t,xs =
1{u=h}(r,X
r )dK
r . Then by the uniqueness
of the solution of the re�e
ted BSDE, we know u(s,Xt,ys ) = Y
s = u(s,X
s ), (σ
∗∇u)(s,Xt,ys ) =
Zt,ys = (σ
∗∇u)(s,Xt,ys ) and K̂t,ys = Kt,ys . Taking s = t we dedu
e that u(t, y) = u(t, y), ρ(y)dy-a.s.
and by the probabilisti
 interpretation (26), we obtain
φt(r, x)1{u=h}(r, x)dν(x, r) =
φt(r, x)1{u=h}(r, x)dν(x, r).
So 1{u=h}(r, x)dν(x, r) = 1{u=h}(r, x)dν(x, r). �
6 Appendix
6.1 Proof of proposition 3.1
First we 
onsider the 
ase when f does not depend on z and satis�es assumption 2.3'. As in step
2 of the proof of theorem 4.1, we approximate g and f as in (18), then gn → g in L2(Rd, ρ(x)dx)
and fn(t, x, 0) → f(t, x, 0) in L2([0, T ]× Rd, dt⊗ ρ(x)dx), as n→ ∞.
Sin
e for ea
h n ∈ N, |gn| 6 n and |fn(t, x, 0)| 6 n, by the result of the step 1 of theorem 4.1,
the PDE(gn, fn) admits the weak solution un ∈ H and sup0 6 t 6 T |un(t, x)| 6 Cn. So we know
|fn(t, x, un(t, x))|2 6 |fn(t, x, 0)|2 + ϕ( sup
0 6 t 6 T
|un(t, x)|) 6 Cn.
Set Fn(t, x) := fn(t, x, un(t, x)), then Fn(t, x) ∈ L2([0, T ]× Rn, dt⊗ ρ(x)dx).
From proposition 2.3 in Bally and Matoussi [3℄, for φ ∈ C2c (Rd), we get, for t 6 s 6 T
un(r, x)dφt(r, x)dx + (un(s, ·), φt(s, ·))− (gn(·), φt(·, T )) +
E(un(r, ·), φt(r, ·))dr
f(r, x, un(r, x))φt(r, x)drdx +
(fn(r, x, 0)− f(r, x, 0))φt(r, x)drdx.
By step 2, we know that as n→ ∞, un → u in H, where u is a weak solution of the PDE(g, f), i.e.
un → u in L2([0, T ]× Rd, dt⊗ ρ(x)dx),
∗∇un → ∇u in L2([0, T ]× Rd, dt⊗ ρ(x)dx).
Then there exists a fun
tion u∗ in L2([0, T ] × Rd, dt ⊗ ρ(x)dx), su
h that for a subsequen
e of
{un}, |unk | 6 |u∗| and unk → u, dt ⊗ dx-a.e. Thanks to assumption 2.3'-(iii), we have that
f(r, x, un(r, x)) → f(r, x, u(r, x)), dt ⊗ dx-a.e. Now, for all 
ompa
t support fun
tion φ ∈ C2c (Rd),
the se
ond term in the right hand side of (39) 
onverge to 0 as n→ ∞ and it is not hard to prove by
using the dominated 
onvergen
e theorem the term in the left hand side of (39) 
onverges. Thus,
we 
on
lude that limn→∞
f(r, x, un(r, x))φt(r, x)drdx exists. Moreover by the monotono
ity

ondition of f and the same arguments as in step 2 of the proof of theorem 4.1, we get for all 
ompa
t
support fun
tion φ ∈ C2c (Rd)
u(r, x)dφt(r, x)dx + (u(s, ·), φt(s, ·)) − (g(·), φt(·, T )) +
E(u(r, ·), φt(r, ·))dr
f(r, x, u(r, x))φt(r, x)drdx .
Now we 
onsider the 
ase when f depends on ∇u and satis�es the assumption 2.3 with (iii)
repla
ed by (12). Like in the step 3 of the proof of theorem 4.1, we 
onstru
t a mapping Ψ from H
into itself. Then by this mapping, we de�ne a sequen
e {un} in H, beginning with a matrix-valued
fun
tion v0 ∈ L2([0, T ] × Rn×d, dt ⊗ ρ(x)dx). Sin
e f(t, x, u, v0(t, x)) satis�es the assumptions of
step 2, the PDE(g, f(t, x, u, v0(t, x))) admits a unique solution u1 ∈ H. For n ∈ N, denote
un(t, x) = Ψ(un−1(t, x)),
i.e. un is the weak solution of the PDE(g, f(t, x, u, σ
∗∇un−1(t, x))). Set ũn(t, x) := un(t, x) −
un−1(t, x). In order to estimate the di�eren
e, we introdu
e the 
orresponding BSDE(g, fn) for
n = 1, where fn(t, x, u) = f(t, x, u,∇un−1(t, x)). So we have Y n,t,xs = un(s,Xt,xs ), Zn,t,xs =
σ∇un(s,Xt,xs ). Then we apply the It�'s formula to |Ỹ n,t,x|2, where Ỹ n,t,xs := Y n,t,xs − Y n−1,t,xs .
With the equivalen
e of the norms, similarly as in step 3, for γ = 1 +
k2, we have
eγs(|ũn(s, x)|2 + |σ∗∇(ũn)(s, x)|2)ρ(x)dsdx
eγs(|ũ2(s, x)|2 + |σ∗∇(ũ2)(s, x)|2)ρ(x)dsdx
)n−1(‖u1(s, x)‖2γ + ‖u2(s, x)‖
where ‖u‖2
eγs(|u(s, x)|2 + |σ∗∇u(s, x)|2)ρ(x)dsdx, whi
h is equivalent to the norm ‖·‖
of H. So {un} is a Cau
hy sequen
e in H, it admits a limit u in H, and by the �xed point theorem,
u is a solution of the PDE(g, f).
Then for ea
h n ∈ N, we have for φ ∈ C2c (Rd)
un(r, x)dφt(r, x)dx + (un(s, ·), φt(s, ·))− (g(·), φt(·, T )) +
E(un(r, ·), φr(r, ·))dr
f(r, x, un(r, x), σ
∗∇un−1(r, x))φt(r, x)drdx
f(r, x, un(r, x), σ
∗∇u(r, x))φt(r, x)drdx
[f(r, x, un(r, x), σ
∗∇un−1(r, x)) − f(r, x, un(r, x), σ∗∇u(r, x))]φt(r, x)drdx.
Noti
ing that f is Lips
hitz in z, we get
|f(r, x, un(r, x), σ∗∇un−1(r, x)) − f(r, x, un(r, x), σ∗∇u(r, x))| 6 k |σ∗∇un−1(r, x) − σ∗∇u(r, x)| .
So the last term of the right side 
onverges to 0, sin
e {σ∗∇un} 
onverges to σ∗∇u in L2([0, T ]×
d, dt ⊗ ρ(x)dx). Now we are in the same situation as in the �rst part of proof, and in the same
way, we dedu
e that the following holds: for φ ∈ C2c (Rd)
u(r, x)dφt(r, x)dx + (u(s, ·), φt(s, ·)) − (g(·), φt(·, T )) +
E(u(r, ·), φt(r, ·))dr
f(r, x, u(r, x), σ∗∇u(r, x))φt(r, x)drdx, dt ⊗ dx, a.s..
Now if f satis�es assumption 2.3, we know that u is solution of the PDE(g, f) if and only if
û = eµtu is solution of the PDE(ĝ, f̂), where
ĝ(x) = eµT g(x), f̂(t, x, y, x) = eµtf(t, x, e−µty, e−µtz)− µy,
and f̂ satis�es assumption 2.3-(iii) repla
ed by (12). So we know now: for φ ∈ C2c (Rd),
û(r, x)dφt(r, x)dx + (û(s, ·), φt(s, ·)) − (ĝ(·), φt(·, T )) +
E(û(r, ·), φt(r, ·))dr
f̂(r, x, û(r, x),∇û(r, x))φt(r, x)drdx, dt ⊗ dx, a.s..
Noti
e that d(eµru(r, x)) = µeµru(r, x)dr + eµrd(u(r, x)), so by the integration by parts formula
(for sto
hasti
 pro
ess), we get
u(r, x)dφt(r, x)dx
e−µrû(r, x)dφt(r, x)dxdr
= e−µT (ĝ(·), φt(·, T ))− e−µs(û(s, ·), φt(s, ·)) + µ
û(r, x)φt(r, x)dxdr
e−µrφt(r, x)[Lû(r, x) + f̂(r, x, û(r, x),∇û(r, x))]drdx.
Using (11), we get that for φ ∈ C2c (Rd),
u(r, x)dφt(r, x)dx = (g(·), φt(·, T ))− (u(s, ·), φt(s, ·))−
φt(r, x)Lu(r, x)drdx
φt(r, x)f(r, x, u(r, x),∇u(r, x))drdx
= (g(·), φt(·, T ))− (u(s, ·), φt(s, ·))−
E(u(r, ·), φt(r, ·))dr
φt(r, x)f(r, x, u(r, x),∇u(r, x))drdx,
and �nally, the result follows. �
6.2 Some a priori estimates
In this subse
tion, we 
onsider the non-markovian Re�e
ted BSDE asso
iated to (ξ, f, L) :


Yt = ξ +
f(t, Ys, Zs)ds+KT −Kt −
ZsdBs,
Yt > Lt,
(Ys − Ls)dKs = 0
under the following assumptions :
(H1) a �nal 
ondition ξ ∈ L2(FT ),
(H2) a 
oe�
ient f : Ω× [0, T ]× R× Rd → R, whi
h is su
h that for some 
ontinuous in
reasing
fun
tion ϕ : R+ −→ R+, a real numbers µ and C > 0:
(i) f(·, y, z) is progressively measurable, ∀(y, z) ∈ R× Rd;
(ii) |f(t, y, 0)| 6 |f(t, 0, 0)|+ ϕ(|y|), ∀(t, y) ∈ [0, T ]× R, a.s.;
(iii) E
|f(t, 0, 0)|2 dt <∞;
(iv) |f(t, y, z)− f(t, y, z′)| 6 C |z − z′| , ∀(t, y) ∈ [0, T ]× R, z, z′ ∈ Rd, a.s.
(v) (y − y′)(f(t, y, z)− f(t, y′, z)) 6 µ(y − y′)2, ∀(t, z) ∈ [0, T ]× Rd, y, y′ ∈ R, a.s.
(vi) y → f(t, y, z) is 
ontinuous, ∀(t, z) ∈ [0, T ]× Rd, a.s.
(H3) a barrier (Lt)0 6 t 6 T , whi
h is a 
ontinuous progressively measurable real-valued pro
ess,
satisfying
E[ϕ2( sup
0 6 t 6 T
(eµtL+t ))] <∞,
and (L+t )0 6 t 6 T ∈ S2(0, T ), LT 6 ξ, a.s.
We shall give an a priori estimate of the solution (Y, Z,K) with respe
t to the terminal 
ondition
ξ, the 
oe�
ient f and the barrier L. Unlike the Lipshitz 
ase, we have in addition the term
Eϕ2(sup0 6 t 6 T (L
t )) and a 
onstant, whi
h only depends on ϕ, µ, k and T :
Proposition 6.1. There exists a 
onstant C, whi
h only depends on T , µ and k, su
h that
0 6 t 6 T
|Yt|2 +
|Zs|2 ds+ |KT |2
f2(t, 0, 0)dt+ ϕ2( sup
0 6 t 6 T
(L+t ))
+ CE[ sup
0 6 t 6 T
(L+t )
2 + 1 + ϕ2(2T )].
Proof. Applying It�'s formula to |Yt|2, and taking expe
tation, then
E[|Yt|2 +
|Zs|2 ds] = E[|ξ|2 + 2
Ysf(s, Ys, Zs)ds+ 2
LsdKs
6 E[|ξ|2 + 2
Ysf(s, 0, 0)ds+ 2
(µ |Ys|2 + k |Ys| |Zs|)ds+ 2
LsdKs].
It follows that
E[|Yt|2 +
|Zs|2 ds] 6 E[|ξ|2 + 2
f2(s, 0, 0)ds+ (2µ+ 1 + 2k2)
|Ys|2 ds+ 2
LsdKs].
Then by Gronwall's inequality, we have
E |Yt|2 6 CE[|ξ|2 +
f2(s, 0, 0)ds+
LsdKs], (40)
|Zs|2 ds 6 CE[|ξ|2 +
f2(s, 0, 0)ds+
LsdKs], (41)
where C is a 
onstant only depends on µ, k and T , in the following this 
onstant 
an be 
hanged
line by line.
Now we estimate K by approximation. By the existen
e of the solution, theorem 2.2 in Lepeltier
et al. [12℄, we take the pro
ess Z as a known pro
ess. Without losing generality we write f(t, y)
for f(t, y, Zt), here f(t, 0) = f(t, 0, Zt) is a pro
ess in H
2(0, T ). Set
m,n = (ξ ∨ (−n)) ∧m,
fm,n(t, y) = f(t, y)− f(t, 0) + (f(t, 0) ∨ (−n)) ∧m.
Form, n ∈ N, ξm,n and sup0 6 t 6 T fm,n(t, 0) are uniformly bounded. Consider the RBSDE(ξm,n, fm,n, L),
t = ξ
m,n +
fm,n(t, Y m,ns )ds+K
Zm,ns dBs,
t > Lt,
(Y m,ns − Ls)dKm,ns = 0.
if we re
all the transform in step 2 of the proof of theorem 2.2 in Lepeltier et al. [12℄, sin
e ξm,n,
fm,n(t, 0) 6 m, we know that (Y
t , Z
t ) is the solution of this RBSDE, if and only if
(Y m,n′, Zm,n′,Km,n′) is the solution of RBSDE(ξm,n′, fm,n′, L′), where
t , Z
t ) = (Y
t +m(t− 2(T ∨ 1)), Z
ξm,n′ = ξm,n + 2mT −m(T ∨ 1),
fm,n′(t, y) = fm,n(t, y −m(t− 2(T ∨ 1)))−m,
t = Lt +m(t− 2(T ∨ 1)).
Without losing generality we set T > 1. Then ξm,n′ 6 0 and fm,n′(t, 0) 6 0. Sin
e (Y m,n′, Zm,n′,Km,n′)
is the solution of RBSDE(ξm,n′, fm,n′, L′), then we have
T = Y
0 − ξ
m,n′ −
m,n′(s, Y m,n′s , Zs)ds+
s dBs,
whi
h follows
)2] 6 4E[
∣∣Y m,n′0
∣∣2 + |ξm,n′|2 + (
fm,n′(s, Y m,n′s )ds)
|Zm,n′s |
ds]. (42)
Applying It�'s formula to |Y m,n|2, like (40) and (41), we have
E |Y m,nt |
|Zm,ns |
ds 6 CE[|ξm,n|2 +
(fm,n(s, 0))2ds+
∣∣Y m,n′0
∣∣2 +
|Zm,n′s |
ds = 2 |Y m,n0 |
+ 8m2T 2 + E
|Zm,ns |
6 CE[|ξm,n|2 +
(fm,n(s, 0)ds)2 +
s ] + 8m
2T 2.
For the third term on the right side of (42), from Lemma 2.3 in Lepeltier et al. [12℄, we remember
fm,n′(s, Y m,n′s )ds)
6 max{(
fm,n′(s, Ỹ m,ns )ds)
fm,n′(s, Y
s )ds)
2}, (43)
where (Ỹ m,n, Z̃m,n) is the solution the following BSDE
t = ξ
m,n′ +
m,n′(s, Ỹ m,ns )ds−
s dBs, (44)
s = ess sup
τ∈Tt,T
+1{τ<T} + (ξ
m,n)+1{τ=T}|Ft] = ess sup
τ∈Tt,T
+|Ft].
From (44), and proposition 2.2 in Pardoux [13℄, we have
fm,n′(s, Ỹ m,ns )ds)
6 CE[|ξm,n′|2 + (
fm,n′(s, 0)ds)2]
6 CE[|ξm,n|2 +
(fm,n(s, 0))2ds] + Cϕ2(2mT ) + Cm2.
For the other term in (43), with sup0 6 t 6 T Y
s = sup0 6 t 6 T (L
, we get
m,n′(s, Y
s )ds)
2(fm,n′(s, 0))2ds+ 2Tϕ2( sup
0 6 t 6 T
(L′t)
6 E[4
fm,n(s, 0)2ds+ 2Tϕ2( sup
0 6 t 6 T
+)] + 2m2T + 4Tϕ2(2mT ).
Consequently, we dedu
e that
2] = E[(K
6 CE[|ξm,n|2 +
(fm,n(s, 0))2ds+
s + ϕ
2( sup
0 6 t 6 T
+) +m2 + ϕ2(2mT )]
6 CE[|ξ|2 +
(f(s, 0, Zs))
2ds+ ϕ2( sup
0 6 t 6 T
+) + sup
0 6 t 6 T
((Lt)
+)2] +
+C(m2 + ϕ2(2mT )).
Moreover using (41) and the fa
t that f is Lips
hitz on z, it follows that
)2] 6 CE[|ξ|2 +
(f(s, 0, 0))2ds+ ϕ2( sup
0 6 t 6 T
+) + sup
0 6 t 6 T
((Lt)
+)2 (45)
LsdKs] + C(m
2 + ϕ2(2mT )).
Let m→ ∞, then
E[|ξm,n − ξn|2] → 0, E
|fm,n(t, 0)− fn(t, 0)|2 → 0,
where ξn = ξ ∨ (−n) and fn(t, y) = f(t, y)− f(t, 0) + f(t, 0) ∨ (−n).
Thanks to the 
onvergen
e result of step 3 of the proof for theorem 2.2 in [12℄, we know that
(Y m,n, Zm,n,Km,n) → (Y n, Zn,Kn) in S2(0, T )×H2d(0, T )×A2(0, T ), where (Y n, Zn,Kn) is the
soultion of the RBSDE(ξn, fn, L). MoreoverK
T ց KnT in L2(FT ), so we haveKnT 6 K
T , whi
h
implies for ea
h n ∈ N,
E[(KnT )
2] 6 E[(K
2] (46)
Then, letting n→ ∞, by the 
onvergen
e result in step 4, sin
e
E[|ξn − ξ|2] → 0, E
|fn(t, 0)− f(t, 0)|2 → 0,
the sequen
e (Y n, Zn,Kn) → (Y, Z,K) in S2(0, T )× H2d(0, T ) ×A2(0, T ), where (Y, Z,K) is the
solution of the RBSDE(ξ, f, L). From (46), and (45) for m = 1, we get
E[(KT )
2] 6 CE[|ξ|2 +
(f(s, 0, 0))2ds+ ϕ2( sup
0 6 t 6 T
+) + sup
0 6 t 6 T
((Lt)
LsdKs] + C(1 + ϕ
2(2T ))
6 CE[|ξ|2 +
(f(s, 0, 0))2ds+ ϕ2( sup
0 6 t 6 T
+) + sup
0 6 t 6 T
((Lt)
E[(KT )
2] + C(1 + ϕ2(2T )).
Then it follows that for ea
h t ∈ [0, T ],
E[|Yt|2 +
|Zs|2 ds+ (KT )2] 6 CE[|ξ|2 +
(f(s, 0, 0))2ds+ ϕ2( sup
0 6 t 6 T
+ sup
0 6 t 6 T
((Lt)
+)2] + C(1 + ϕ2(2T )).
Finally we get the result, by applying BDG inequality. �
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Springer Verlag, New York (1999).
	Introduction
	Notations and preliminaries
	Stochastic flow and random test functions
	Sobolev's Solutions for PDE's under monotonicity condition
	Sobolev's solution for PDE with obstacle under monotonicity condition
	Appendix
	Proof of proposition 3.1
	Some a priori estimates

  We prove the existence and uniqueness of the solution of a semilinear PDE's
and also PDE's with obstacle under monotonicity condition. Moreover we give the
probabilistic interpretation of the Sobolev's solutions in term of Backward SDE
and reflected Backward SDE respectively.

Introduction
	Notations and preliminaries
	Stochastic flow and random test functions
	Sobolev's Solutions for PDE's under monotonicity condition
	Sobolev's solution for PDE with obstacle under monotonicity condition
	Appendix
	Proof of proposition 3.1
	Some a priori estimates
