Absolute Continuity of the Laws of the Solutions to Parabolic SPDEs with Two Reflecting Walls

0.17 MB·

by  Wen Yue

Absolute Continuity of the Laws of the Solutions to Parabolic
SPDEs with Two Reflecting Walls
Wen Yue
6 InstituteofAnalysisandScientificComputing,TechnologyUniversityofVienna,WiednerHaupt. 8-10,Vienna,Austria
1
0 Abstract
2
In this paper, we focus on the existence of the density for the law of the solutions to
b
parabolic stochastic partial differential equations with two reflecting walls. The main tool
e
F is Malliavin Calculus.

8
Keywords: parabolic stochastic partial differential equations, two reflecting walls, absolute
1
continuity, Malliavin calculus.

]
R
P 1 Introduction
.
h
t Parabolic SPDEs with reflection are natural extension of the widely studied deterministic
a
m parabolic obstacle problems. It was proved by Funaki and Olla in [7] that the fluctuations
[ of a ∇φ interface model near a hard wall converge in law to the stationary solution of an

SPDE with reflection. In recent years, there is a growing interest on the study of SPDEs with
2
v reflection. Several works are devoted to the existence and uniqueness of the solutions, such as
1 [9]byNaulart andPardoux, [11]byXu andZhangand[14]by Otobe. Especially, theexistence
8
and uniqueness of the solution to a fully non-linear SPDE with two reflecting walls was proved
2
1 by Yang and Zhang [12].
0 We focus here on the existence of the density of the law of the solution, using Malliavin
.
1 calculus. Malliavin calculus associated with white noise was also used by Pardoux and Zhang
0
[10], Bally and Pardoux [3] to establish the existence of the density of the law of the solution
6
1 to parabolic SPDE. The case of parabolic stochastic partial differential equation with one
: reflecting wall was studied by Donati-martin and Pardoux [6]. For parabolic SPDEs with two
v
i reflecting walls, we construct a convergent sequence uǫ,δ with two indices, based on the case of
X
onereflecting wall. Itis moredemandingto provetheconvergence of uǫ,δ andidentify thelimit
r
a as the solution of the original equation. To prove the positivity of the Malliavin derivative of
the solution, we need more delicate partition of sample spaces.
This paper is organized as follows: Section 2 is devoted to fundamental knowledge of
parabolic stochastic partial differential equations with two reflecting walls and Malliavin cal-
culus associated with white noise. In Section 3, we recall some results obtained by Yang and
Zhang [12] about the existence and uniqueness of the solution to parabolic SPDEs with two
reflecting walls and we prove the Malliavin differentiability of the solution. Finally, we give
the existence of the density of the law of the solution.
1
2 Preliminaries
Notation: Let Q = [0,1] × R , Q = [0,1] ×[0,T], V = {u ∈ H1([0,1]),u(0) = u(1) = 0}
+ T
where H1([0,1]) denotes the usual Sobolev space of absolutely continuous funcitons defined on
[0,1] whose derivative belongs to L2([0,1]), and A= − ∂2 .
∂x2
Consider the following stochastic partial differential equation with two reflecting walls:
∂u ∂2u
= +f(x,t,u)+σ(x,t,u)W˙ (x,t)+η−ξ,
 ∂t ∂x2

 u(0,t) = 0,u(1,t) = 0, for t ≥ 0, (2.1)


u(x,0) = u (x) ∈ C([0,1]),
0
 h1(x,t) ≤ u(x,t) ≤ h2(x,t),for(x,t) ∈ Q,a.s.




where W˙ denotes the space-time white noise defined on a complete probability space
(Ω,F,{F } ,P), where F = σ(W(x,s) : x ∈ [0,1],0 ≤ s ≤ t), u is a continuous function
t t≥0 t 0
on [0,1], which vanishes at 0 and 1.
We assume that the reflecting walls h1(x,t),h2(x,t) are continuous functions satisfying
h1(0,t),h1(1,t) ≤ 0, h2(0,t),h2(1,t) ≥ 0, and
(H1)h1(x,t) < h2(x,t) for x ∈(0,1) and t ∈ R ;
+
(H2)∂hi + ∂2hi ∈ L2([0,1]×[0,T]), where ∂ and ∂2 are interpreted in a distributional sense;
∂t ∂x2 ∂t ∂x2
(H3)∂ hi(0,t) = ∂ hi(1,t) = 0 for t ≥ 0;
∂t ∂t
(H4)∂ (h2−h1) ≥ 0.
∂t
We also assume that the coefficients: f,σ(x,t,u(x,t)) : [0,1]×R ×R → R are measurable
+
and satisfy:
(F) : f,σ are of class of C1 with bounded derivatives with respect to the third element and σ
is bounded.
Thefollowing is thedefinition of the solution to a parabolic SPDE with two reflecting walls
h1, h2.
Definition 2.1 A triplet (u,η,ξ) defined on a filtered probability space
(Ω,P,F;{F }) is a solution to the SPDE(2.1), denoted by (u ;0,0;f,σ;h1,h2), if
t 0
(i) u ={u(x,t);(x,t) ∈ Q} is a continuous, adapted random field (i.e. u(x,t) is F -measuralbe
t
∀t ≥ 0, x ∈[0,1]) satisfying h1(x,t) ≤ u(x,t) ≤ h2(x,t), u(0,t) = 0 and u(1,t) = 0, a.s.;
(ii) η(dx,dt) and ξ(dx,dt) are positive and adapted (i.e. η(B) and ξ(B) are F -measurable if
t
B ⊂ (0,1)×[0,t]) random measures on (0,1)×R satisfying
+
η((θ,1−θ)×[0,T]) < ∞,ξ((θ,1−θ)×[0,T]) < ∞ a.s. (2.2)
for 0 < θ < 1 and T > 0;
2
(iii) for all t ≥ 0 and φ ∈ C∞((0,1) × (0,∞)) (the set of smooth functions with compact
k
supports) we have
t t t 1
′′
(u(t),φ)− (u(s),φ )ds− (f(y,s,u),φ)ds− φσ(y,s,u)W(dx,ds)
Z Z Z Z
0 0 0 0
2
t 1 t 1
= (u ,φ)+ φη(dxds)− φξ(dxds)a.s.;
0 Z Z Z Z
0 0 0 0
(2.3)
(iv) (u(x,t)−h1(x,t))η(dx,dt) = (h2(x,t)−u(x,t))ξ(dx,dt) = 0 a.s..
Q Q
R R
Remarks: We note that the stochastic integral in (2.3) is an Ito integral with respect to the
Brownian sheet {W(x,t);(x,t) ∈ [0,1]×R } defined on the canonical space Ω = C ([0,1]×
+ 0
R ) (the space of continuous functions on [0,1] ×R which are zero whenever one of their
+ +
arguments is zero). The Brownian sheet is equipped with its Borel σ-field F, the filtration
F = {σ(W(x,s)),x ∈ [0,1],s ≤ t} and the Wiener measure P.
t
Next, we recall Malliavin calculus associated with white noise:
Let S denote the set of ”simple random variables” of the form
F = f(W(h ),...,W(h )),n ∈ N,
1 n
where h ∈ H := L2([0,1]×R ) and W(h ) represent the Wiener integral of h , f ∈ C∞(Rn).
i + i i p
For such a variable F, we defineits derivative DF, a random variable with values in L2([0,1]×
R ) by
+
∂f
D F = Σn (W(h ),...,W(h ))·h (x,t).
x,t i=1∂x 1 n i
i
We denote by D1,2 the closure of S with respect to the norm:
||F||1,2 = (E(F2))12 +[E(||DF||2L2([0,1]×R+))]12.
D1,2 is a Hilbert space. It is the domain of the closure of derivation operator D.
We go back to consider the following parabolic SPDE:
∂u ∂2u
= +f(x,t,u)+σ(x,t,u)W˙ ,
 ∂t ∂x2
 u(0,t) = 0,u(1,t) = 0, for t ≥ 0, (2.4)

u(x,0) = u (x) ∈ C([0,1]),
 0



where f,σ satisfy (F).
According to [11], we know u also satisfies the integral equation:
1 t 1
u(x,t) = G (x,y)u (y)dy+ G (x,y)f(u(y,s))dyds
Z t 0 Z Z t−s
0 0 0
t 1
+ G (x,y)σ(u(y,s))W(dyds)
Z Z t−s
0 0
And we have the following result from [10].
3
Proposition 2.1 [10]For all(x,t) ∈ (0,1)×R , u(x,t) isthe solutionto(2.4), Thenu(x,t) ∈
+
D1,2 and D u(x,t) is the solution of SPDE:
y,s
t 1
D u(x,t) = G (x,y)σ(u(y,s))+ G (x,z)f′(u(z,r))D u(z,r)dzdr
y,s t−s Z Z t−r y,s
s 0
t 1
+ G (x,z)σ′(u(z,r))D (u(z,r))W(dzdr).
Z Z t−r y,s
s 0
3 The Main Result and The Proof
We consider the penalized SPDE as follows:
∂uǫ,δ(x,t) ∂2uǫ,δ(x,t)
− +f(uǫ,δ(x,t)) = σ(uǫ,δ(x,t))W˙ (x,t)
 ∂t ∂x2

 1 1
 +δ(uǫ,δ(x,t)−h1(x,t))− − ǫ(uǫ,δ(x,t)−h2(x,t))+, (3.1)

uǫ,δ(0,t) = uǫ,δ(1,t) = 0,t ≥ 0,

 uǫ,δ(x,0) = u0(x),



and we can get the following proposition.
Proposition 3.1 If we have (H1),(H2), (H3), (H4) and (F). Then for any p ≥ 1,T > 0,
sup E(||uǫ,δ||T ) < ∞ and uǫ,δ converges uniformly on [0,1]×[0,T] to u as ǫ,δ → 0, where
ǫ,δ ∞
u,uǫ,δ are the solutions of SPDE (2.1) and the penalized SPDE (3.1).
Proof. Let uǫ,δ be the solution to the penalized SPDE (3.1).
Step 1: we prove that there exists u(x,t) such that
u:= limuǫ = limlimuǫ,δa.s. (3.2)
ǫ↓0 ǫ↓0 δ↓0
First fix ǫ,
let vǫ,δ be the solution of equation:
∂vǫ,δ(x,t) ∂2vǫ,δ(x,t)
− +f(vǫ,δ(x,t)) = σ(uǫ,δ(x,t))W˙ (x,t)
 ∂t ∂x2
 −1(uǫ,δ(x,t)−h2(x,t))+, (3.3)

ǫ
 vǫ,δ(x,0) = u0(x),vǫ,δ(0,t) = vǫ,δ(1,t) =0.



Then zǫ,δ = vǫ,δ −uǫ,δ is the unique solution in L2((0,T)×(0,1)) of
ǫ,δ
∂z 1
 ∂tt +Aztǫ,δ +f(vtǫ,δ)−f(uǫt,δ) = −δ(uǫ,δ(x,t)−h1(x,t))−, (3.4)


zǫ,δ(x,0) = 0,zǫ,δ(0,t) = zǫ,δ(1,t) = 0.


4
Multiplying Eq(3.4) by (zǫ,δ)+ and integrating it to obtain:
s
t ∂zǫ,δ(x,s) t ∂zǫ,δ(x,s) ∂(zǫ,δ(x,s))+
( ,(zǫ,δ(x,s))+)ds+ ( , )ds
Z ∂s Z ∂x ∂x
0 0
t
+ (f(vǫ,δ(x,s))−f(uǫ,δ(x,s)),(zǫ,δ(x,s))+)ds
Z
0
1 t
= − ((uǫ,δ(x,s)−h1(x,s))−,(zǫ,δ(x,s))+)ds. (3.5)
δ Z
0
According to Bensoussan and Lions [2] (Lemma 6.1, P132), (zǫ,δ)+ ∈ L2(0,T;V)∩C([0,T];H)
s
a.s.
t ∂ 1
( zǫ,δ,(zǫ,δ)+)ds = |(zǫ,δ)+|2
Z ∂s s s 2 t H
0
and similarly
t ∂ ∂ t ∂
( zǫ,δ, (zǫ,δ)+)ds = | (zǫ,δ)+|2ds ≥ 0,
Z ∂x s ∂x s Z ∂x s
0 0
and by Lipschitz continuity of f, we have
t t
(f(vǫ,δ(x,s))−f(uǫ,δ(x,s)),(zǫ,δ(x,s))+)ds ≥ −c |(zǫ,δ(x,s))+|2 ds,
Z Z H
0 0
and we deduce that
1 t ∂(zǫ,δ(x,s))+ t
0 ≥ |(zǫ,δ(x,t))+|2 + | |2 ds−c |(zǫ,δ(x,s))+|2 ds
2 H Z ∂x H Z H
0 0
1 t
≥ |(zǫ,δ(x,t))+|2 −c |zǫ,δ(x,s)+|2 ds.
2 H Z H
0
Hence,
t 1
c |(zǫ,δ(x,s))+|2 ds ≥ |zǫ,δ(x,t)+|2 (3.6)
Z H 2 H
0
From Gronwall’s Lemma: |(zǫ,δ(x,t))+|2 =0, ∀t≥ 0.a.s.
H
Then,
uǫ,δ(x,t) ≥ vǫ,δ(x,t),∀x ∈ [0,1],t ≥ 0.a.s. (3.7)
FromTheorem3.1in[5],wegetthatthefollowingequationhasauniquesolution{wǫ,δ(x,t);x ∈
[0,1],t ≥ 0}:
∂wǫ,δ(x,t) ∂2wǫ,δ(x,t)
− +f(wǫ,δ(x,t)+ sup (wǫ,δ(y,s)−h1(y,s))−)
 ∂t ∂x2
s≤t,y∈[0,1]


 1
 = σ(uǫ,δ(x,t))W˙ (x,t)− (uǫ,δ(x,t)−h2(x,t))+,
ǫ
 wǫ,δ(·,0) = u0,wǫ,δ(0,t) = wǫ,δ(1,t) = 0.



5
We set
wǫ,δ(x,t) = wǫ,δ(x,t)+ sup (wǫ,δ(y,s)−h1(y,s))− = wǫ,δ(x,t)+Φǫ,δ (3.8)
t
s≥t,y∈[0,1]
wǫ,δ(x,t)−h1(x,t) ≥ 0 and Φǫ,δ is an increasing process.
t
For any T > 0, zǫ,δ = uǫ,δ −wǫ,δ is the unique solution in L2((0,T);H1(0,1)) of
∂zǫ,δ(x,t) dΦǫ,δ
+Azǫ,δ(x,t)+f(uǫ,δ(x,t))−f(wǫ,δ(x,t))+ t
 ∂t dt

 = 1(uǫ,δ(x,t)−h1(x,t))−,
 δ

zǫ,δ(·,0) = 0,

 zǫ,δ(0,t) = zǫ,δ(1,t) = −Φǫ,δ.
 t


Multiplying this equation by (zǫ,δ(x,s))+, we obtain by the same arguments as above:
t ∂zǫ,δ(x,s) t ∂zǫ,δ(x,s) ∂(zǫ,δ(x,s))+
( ,(zǫ,δ(x,s))+)ds+ ( , )ds
Z ∂s Z ∂x ∂x
0 0
t
+ (f(uǫ,δ(x,s))−f(wǫ,δ(x,s)),(zǫ,δ(x,s))+)ds
Z
0
t 1
+ (zǫ,δ(x,s))+dxdΦǫ,δ
Z Z s
0 0
1 t
= ((uǫ,δ(x,s)−h1(x,s))−,(zǫ,δ(x,s))+)ds (3.9)
δ Z
0
The right-hand side of the above equality is zero because (zǫ,δ(x,s))+ > 0 implies uǫ,δ(x,s)−
h1(x,s) > wǫ,δ(x,s)−h1(x,s) ≥ 0.
Hence we again deduce from Gronwall’s Lemma:
uǫ,δ(x,t) ≤ wǫ,δ(x,t) (3.10)
By (3.7),(3.10),
|uǫ,δ(x,t)| ≤ |vǫ,δ(x,t)|+|wǫ,δ(x,t)|+ sup (wǫ,δ(y,s)−h1(y,s))−
s≤t,y∈[0,1]
≤ |vǫ,δ(x,t)|+2 sup [|wǫ,δ(y,s)|+|h1(y,s)|]. (3.11)
s≤t,y∈[0,t]
From Lemma 6.1 in [5], for arbitrarily large p and any T > 0, consider that f′(vǫ,δ(x,t)) =
f(vǫ,δ(x,t))+1(uǫ,δ(x,t)−h2(x,t))+ isLipschitzcontinuouswithrespecttovǫ,δ andf′(wǫ,δ(x,t)+
ǫ
sup (wǫ,δ(y,s)−h1(y,s))−)=
s≥t,y∈[0,1]
f(wǫ,δ(x,t)+sup (wǫ,δ(y,s)−h1(y,s))−)+ 1(uǫ,δ(x,t)−h2(x,t))+ is Lipschitz contin-
s≥t,y∈[0,1] ǫ
uous with respect to wǫ,δ(x,t)+sup (wǫ,δ(y,s)−h1(y,s))−, we have that
s≥t,y∈[0,1]
6
sup E[sup |vǫ,δ(x,t)|p]< ∞ and sup E[sup |wǫ,δ(x,t)|p]< ∞,
δ (x,t)∈Q δ (x,t)∈Q
T T
which imply
supE[ sup |uǫ,δ(x,t)|p] < ∞. (3.12)
δ (x,t)∈Q
T
So uǫ = sup uǫ,δ is a.s. bounded on Q .
δ T
Let
(uǫ,δ(x,t)−h1(x,t))−
ηǫ = lim (3.13)
δ→0 δ
Similar as the proof of Th4.1 in [5], uǫ is continuous and uǫ is the solution to:
∂uǫ 1
+Auǫ+f(uǫ) = σ(uǫ)W˙ (x,t)+ηǫ(x,t)− (uǫ(x,t)−h2(x,t))+ (3.14)
∂t ǫ
Inaddition,bythedefinitionofuǫ,uǫ ≥ h1andusingTheorem1.2.6(ComparisonTheorem),uǫ
decreases when ǫ → 0.
Hence, there exists u(x,t) such that
u:= limuǫ = limlimuǫ,δa.s. (3.15)
ǫ↓0 ǫ↓0 δ↓0
Step 2: Next we prove u(x,t) is continuous.
Let v˜ǫ,δ be the solution of
∂v˜ǫ,δ
+Av˜ǫ,δ = σ(uǫ,δ)W˙ , (3.16)
∂t
and let vˆ be the solution of
∂vˆ
+Avˆ= σ(u)W˙ . (3.17)
∂t
Remember
∂uǫ,δ(x,t) ∂2uǫ,δ(x,t)
− +f(uǫ,δ(x,t)) = σ(uǫ,δ(x,t))W˙ (x,t)
∂t ∂x2
1 1
+ (uǫ,δ(x,t)−h1(x,t))− − (uǫ,δ(x,t)−h2(x,t))+,
δ ǫ
Let z˜ǫ,δ = uǫ,δ −v˜ǫ,δ, then z˜ǫ,δ is the solution of
∂z˜ǫ,δ
+Az˜ǫ,δ +f(z˜ǫ,δ +v˜ǫ,δ)
∂t
1 1
= (z˜ǫ,δ +v˜ǫ,δ −h1)−− (z˜ǫ,δ +v˜ǫ,δ −h2)+. (3.18)
δ ǫ
Let zˆǫ,δ be the solution of
∂zˆǫ,δ 1 1
+Azˆǫ,δ +f(zˆǫ,δ +vˆ) = (zˆǫ,δ +vˆ−h1)− − (zˆǫ,δ +vˆ−h2)+. (3.19)
∂t δ ǫ
7
We have
||z˜ǫ,δ −zˆǫ,δ|| ≤ ||v˜ǫ,δ −vˆ|| . (3.20)
T,∞ T,∞
zˆǫ,δ is continuous. According to proof of Theorem 2.1 in [14] , zˆǫ,δ → zˆ (continuous).
It means
zˆ= limzˆǫ = lim limzˆǫ,δ.
ǫ→0 ǫ→0δ→0
Fix ǫ, zˆǫ,δ ↑ zˆǫ(continuous), and from Dini theorem, zˆǫ,δ uniformly converges to zˆǫ. i.e.||zˆǫ,δ−
zˆǫ|| → 0, δ → 0.
T,∞
Since zˆǫ ↓zˆ, and from Dini theorem, zˆǫ uniformly converges to zˆ. i.e.||zˆǫ −zˆ|| → 0.
T,∞
Then we get
||zˆǫ,δ −zˆ|| = ||zˆǫ,δ −zˆǫ+zˆǫ−zˆ|| ≤ ||zˆǫ,δ −zˆǫ|| +||zˆǫ −zˆ|| → 0
T,∞ T,∞ T,∞ T,∞
(δ → 0,ǫ → 0). (3.21)
i.e. zˆǫ,δ → zˆ uniformly.
Next we prove v˜ǫ,δ → vˆ uniformly with respect to s,t as ǫ → 0,δ → 0:
Let I(x,t) = v˜ǫ,δ(x,t)−vˆ(x,t) = t 1G (x,y)(σ(uǫ,δ)−σ(u))W(dyds), from the proof of
0 0 t−s
Corollary 3.4 in [?], R R
tWs 1
E|I(x,t)−I(y,s)|)p ≤ CTE (|σ(uǫ,δ)−σ(u)|)pdzdr|(x,t)−(y,s)|p4−3,
Z Z
0 0
and following the same calculation as in the proof of Theorem 2.1 in Xu and Zhang [11], we
deduce
T 1
E( sup |I(x,t)|)p ≤ C E (|σ(uǫ,δ)−σ(u)|)pdxdt.
T
Z Z
x∈[0,1],t∈[0,T] 0 0
Again according to u := lim lim uǫ,δ and σ(x,t,u(x,t)) is Lipschitz continuous and
ǫ→0 δ→0
bounded, we can have
T 1
E( sup |I(x,t)|)p ≤ C E (|σ(uǫ,δ)−σ(u)|)pdtdx
T
Z Z
x∈[0,1],t∈[0,T] 0 0
→ 0
Then we have that v˜ǫ,δ → vˆ uniformly a.s. and again from (3.20) and (3.21) we deduce that
z˜ǫ,δ → zˆ uniformly a.s..
So
lim limuǫ,δ = u= zˆ+vˆ
ǫ→0δ→0
is continuous.
Step 3: Next we prove u(x,t) is the solution of
∂u
+Au+f(u)= σ(u)W˙ (x,t)+η(x,t)−ξ(x,t). (3.22)
∂t
8
For ψ ∈ C∞((0,1)×[0,∞)),
0
t t t
− (uǫ(x,s),ψ (s))ds− (uǫ(x,s),Aψ)ds+ (f(uǫ),ψ)ds
s
Z Z Z
0 0 0
t 1 t 1
= (σ(uǫ),ψ)W(dx,ds)+ ψ(x,t)(ηǫ(dx,dt)−ξǫ(dx,dt))
Z Z Z Z
0 0 0 0
(3.23)
(uǫ,δ −h1)− (uǫ−h2)+
ηǫ = lim ,ξǫ = .
δ→0 δ ǫ
Let ǫ → 0,
t t t
− (u(x,s),ψ )ds− (u(x,s),Aψ)ds+ (f(u),ψ)ds
s
Z Z Z
0 0 0
t 1 t 1
= (σ(u),ψ)W(dx,ds)+lim ψ(x,t)(ηǫ(dx,dt)−ξǫ(dx,dt)).
Z0 Z0 ǫ→0Z0 Z0
Then it is clear that, under the limit ǫ → 0, lim (ηǫ −ξǫ) exists in the sense of Schwartz
ǫ→0
distribution a.s..
Because uǫ uniformly converges to u, similarly as Theorem 3.1 in [12] we get ηǫ → η and
ξǫ → ξ. Let ǫ → 0 to see that (u,η,ξ) satisfies condition (iii) of Def 3.2.1.
Multiplying both sides of Eq(3.23) by ǫ and letting ǫ → 0,
t 1 (uǫ,δ −h1)−
lim ψ(x,t)(ǫlim −(uǫ−h2)+)(dx,dt) = 0 (3.24)
ǫ→0Z0 Z0 δ→0 δ
then t 1ψ(x,t)(u − h2)+(dx,dt) = 0, and we can get u ≤ h2. And since uǫ ≥ h1, then
0 0
u≥ hR1. RCombining these two inequalities, we have h1 ≤ u≤ h2.
Finally, we can show that (u−h1)dη = (h2−u)dξ = 0.
QT QT
For ǫ ≤ ǫ′, uǫ ≥ uǫ′, therRefore supp(ηǫ) ⊂Rsupp(ηǫ′), we get supp(η) ⊂ supp(ηǫ). we know
uǫ − h1 ≤ 0 on suppηǫ. So (uǫ − h1)dη ≤ 0. Then (u − h1)dη = 0. Because ξǫ =
QT QT
1(uǫ−h2)+, then 0 ≥ (uǫR−h2)dξǫ ≥ 0. And since ξǫ →R ξ, then (u−h2)dξ = 0.
ǫ QT QT
By taking ψ ∈ C∞((0,1R)×(0,∞)) such that ψ = 1 on (suppη)∩((δ,R1−δ)×[0,T]) and ψ = 0
0
on suppξ. Hence, in view of (2.3),
T 1 T
η([δ,1−δ]×[0,T]) = ψ(x,t)η(dx,dt)− φ(x,t)ξ(dx,dt) < ∞
Z Z Z
0 0 0
for all 0 < δ < 1 and T > 0. Similarly we can get ξ([δ,1−δ]×[0,T]) < ∞ for all 0 < δ < 1
2 2
and T > 0. ✷
9
Setk (uǫ,δ−h1(x,t)) = arctan[(uǫ,δ−h1(x,t))∧0]2 andk (uǫ,δ−h2(x,t)) = arctan[(h2(x,t)−
1 2
uǫ,δ)∧0]2. Consider the following penalized SPDE:
∂uǫ,δ(x,t) ∂2uǫ,δ(x,t)
− +f(uǫ,δ(x,t)) = σ(uǫ,δ(x,t))W˙ (x,t)
 ∂t ∂x2
 +1k (uǫ,δ −h1(x,t))− 1k (uǫ,δ −h2(x,t)), (3.25)
 1 2
δ ǫ
 uǫ,δ(x,0) = u0(x).



Notice that the corresponding penalized elements in Proposition 3.3.1 are (uǫ,δ − h1(x,t))−
and(uǫ,δ −h2(x,t))+. It was shown in [4](also in [6]) that the choice of k ,k does not change
1 2
the limit of uǫ,δ, but makes k ,k differentiable with respect to uǫ,δ.
1 2
Proposition 3.2 For all (x,t) ∈ [0,1]×R+, u(x,t) ∈ D and there exists a subsequence of
1,p
Duǫ,δ(x,t) that converges to Du(x,t) inthe weak topology of Lp(Ω;H) and H = L2([0,1]×R+).
Proof. Let uǫ,δ be the solution to the following SPDE:
∂uǫ,δ(x,t) ∂2uǫ,δ(x,t)
− +f(uǫ,δ(x,t)) = σ(uǫ,δ(x,t))W˙ (x,t)
 ∂t ∂x2
 +1k (uǫ,δ −h1(x,t))− 1k (uǫ,δ −h2(x,t)), (3.26)
 1 2
δ ǫ
 uǫ,δ(x,0) = u0(x).



Then it can be expressed as,
t t 1
uǫ,δ(x,t) = G (x,y)u (y)dy+ G (x,y)σ(uǫ,δ(x,t))W(dyds)
Z t 0 Z Z t−s
0 0 0
t 1 1 1
+ G (x,y)[−f(uǫ,δ(x,t))+ k − k ]dyds,
Z Z t−s δ 1 ǫ 2
0 0
where G (x,y) is the heat kernel.
t
And we also know from Section 3.2 that:
D uǫ,δ(x,t) = G (x,y)σ(uǫ,δ(y,s))
y,s t−s
t 1
+ G (x,z)σ′(uǫ,δ(z,r))D (uǫ,δ(z,r))W(dzdr)
Z Z t−r y,s
s 0
+ t 1G (x,z)[−f′ + 1k′ − 1k′]D (uǫ,δ(z,r))dzdr
Z Z t−r δ 1 ǫ 2 y,s
s 0
Let
D uǫ,δ(x,t) = σ(uǫ,δ(y,s))Sǫ,δ(x,t) (3.27)
y,s y,s
ǫ,δ
and then S (x,t) is the solution of
y,s
t 1
Sǫ,δ(x,t) = G (x,y)+ G (x,z)σ′(uǫ,δ(z,r))Sǫ,δ(z,r)W(dzdr)
y,s t−s Z Z t−r y,s
s 0
10