i The sets D
i
are closures of domains with non-zero volumes and boundaries at least C
2
in D : this
implies the existence of a unique unit normal vector n
i
x at each point x ∈ D ∩ ∂ D
i
. ii Each set
D
i
has the Uniform Exterior Sphere property restricted to D, i.e.
∃α
i
0, ∀x ∈ D ∩ ∂ D
i
B x
− α
i
n
i
x, α
i
∩ D
i
= ; iii Each set
D
i
has the Uniform Normal Cone property restricted to D, i.e.
for some β
i
∈ [0, 1[ and δ
i
0 and for each x ∈ D ∩ ∂ D
i
there is a unit vector l
i x
s.t. ∀y ∈
D ∩ ∂ D
i
∩ Bx, δ
i
n
i
y.l
i x
≥ p
1 − β
2 i
iv compatibility assumption There exists β
p 2 max
1 ≤i≤p
β
i
satisfying
∀x ∈ ∂ D, ∃l
x
∈ S
m
, ∀i s.t. x ∈ ∂ D
i
l
x
. n
i
x ≥ β
Under the above assumptions, D ∈ U ESα and D ∈ UN Cβ, δ hold with α = β
min
1 ≤i≤p
α
i
, δ = min
1 ≤i≤p
δ
i
2 and β = p
1 − β
− 2 max
1 ≤i≤p
β
i
β
2
. Moreover, the vectors normal to the boundary
∂ D are convex combinations of the vectors normal to the boundaries ∂ D
i
:
∀x ∈ ∂ D N
D
x
=
n ∈ S
m
, n =
X
∂ D
i
∋x
c
i
n
i
x with each c
i
≥ 0
Remark 3.5 : Thanks to the compatibility assumption i v, n =
P
∂ D
i
∋x
c
i
n
i
x with non-negatives
c
i
’s and
|n| = 1 implies that
X
∂ D
i
∋x
c
i
≤ X
∂ D
i
∋x
c
i
n
i
x.l
x
β =
n.l
x
β ≤
1 β
so that the last equality in proposition 3.4 can be rewritten as
N
D
x
=
n ∈ S
m
, n =
X
∂ D
i
∋x
c
i
n
i
x with ∀i c
i
≥ 0 and X
∂ D
i
∋x
c
i
≤ 1
β
Corollary 3.6. of theorem 3.3
If D =
p
\
i=1
D
i
satisfies assumptions i · · · iv and if σ and b are bounded Lipschitz continuous func-
tions, then
Xt = X0 +
Z
t
σ XsdWs +
Z
t
bXsds +
p
X
i=1
Z
t
n
i
Xsd L
i
s 2
has a unique strong solution with local times L
i
satisfying L
i
· = Z
·
1I
∂ D
i
Xs d L
i
s
150
Proof of corollary 3.6
Thanks to proposition 3.4, D satisfies the assumptions of theorem 3.3. Thus equation 2 has a
unique strong solution X with local time L and reflection direction n. Using proposition 3.4 again,
there are non-unique coefficients c
i
ω, s ∈ [0,
1 β
] for each normal vector in the reflection term to be written as a convex combination :
n ω, s =
X
∂ D
i
∋Xω,s
c
i
ω, sn
i
Xω, s 3
Let us prove that there exists a measurable choice of the c
i
’s : The map
n resp. n
i
X is only defined for ω, s such that Xω, s ∈ ∂ D resp. Xω, s ∈ ∂ D
i
. We extend these maps by zero to obtain measurable maps on Ω
× [0, T ] for an arbitary positive T . Note that equality 3 holds for the extended maps too.
For each ω, s, we define the map f
ω,s
on R
p
by f
ω,s
c = | P
p i=1
c
i
n
i
Xω, s − nω, s|. For
a positive integer k, let R
k
= {0,
1 k
,
2 k
, . . . ,
1 k
⌊
k β
⌋}
p
denote the
1 k
-lattice on [0,
1 β
]
p
endowed with lexicographic order. The smaller point in R
k
for which f
ω,s
reaches its minimum value is c
k
ω, s = X
c ∈R
k
c Y
c
′
∈R
k
,c
′
6=c
1I
f
ω,s
c
′
f
ω,s
c
+ 1I
f
ω,s
c
′
= f
ω,s
c1I
c
′
c
c
k
is a measurable map on Ω × [0, T ] and 3 implies that | f
ω,s
c
k
ω, s| ≤
p k
. Taking coordi- nate after coordinate the limsup of the sequence of c
k k
, we obtain a measurable process c
∞
satisfying 3. Finally, let L
i
= Z
·
1I
∂ D
i
Xsc
i
s dLs. L
i
has bounded variations on each [0, T ] since the c
i
’s are bounded, and it is a local time in the sense of remark 2.1.
Here, strong uniqueness holds for process X and for the reflection term
p
X
i=1
Z
t
n
i
Xsd L
i
s. The uniqueness of this term does not imply uniqueness of the L
i
’s, because the c
i
’s are not unique.
Proof of proposition 3.4
Let x
∈ ∂ D. Since D = T
p i=1
D
i
, the set {i s.t. x ∈ ∂ D
i
} is not empty. Since D
i
satisfies U ES α
i
restricted to D, for each y ∈ D
i
y − x.n
i
x +
1 2
α
i
|y − x|
2
≥ 0, and consequently : ∀i s.t. ∂ D
i
∋ x ∀y ∈ D y − x.n
i
x +
1 2 min
∂ D
j
∋x
α
j
|y − x|
2
≥ 0 Summing over i for non-negative c
i
’s such that P
∂ D
i
∋x
c
i
≤
1 β
we obtain :
∀y ∈ D
X
∂ D
j
∋x
c
i
n
i
x.y − x +
