Convergence of solutions for the stochastic porous media

J. Evol. Equ. 11 (2011), 339–370© 2010 Springer Basel AG1424-3199/11/020339-32, published online January 13, 2011DOI 10.1007/s00028-010-0094-7

Journal of EvolutionEquations

Convergence of solutions for the stochastic porous media equationsand homogenization

Ioana Ciotir

Abstract. This work addresses the stochastic porous media equation with multiplicative noise and diffusiv-ity function depending on the space variable. The first part of the paper proves an existence and uniquenessresult for this type of equation, the second part proves the convergence of the solutions in the case of graphconvergence of the porous media operator, and this result is used in the third part for an homogenizationtheorem.

1. Introduction

Let O be an open bounded domain of Rd (1 ≤ d ≤ 3) with smooth boundary ∂O.

Recall the distribution space H−1(O) (the dual of H10 (O)) which is endowed with

the scalar product given by 〈x, z〉H−1(O)not= 〈x, z〉−1 = ∫

O (−�)−1 xzdξ, for allx, z ∈ H1

0 (O) and the norm

|x |−1 =∣∣∣(−�)−1 x

∣∣∣

H10 (O)

, (1)

where (−�)−1 x = y is the solution to Dirichlet problem −�y = x in O, y ∈H1

0 (O). For the linear operator −� in L2(O) defined on H2(O) ∩ H10 (O), there

exists a complete orthogonal system of eigenfunctions {ek} in L2(O) and the corre-

sponding sequence of eigenvalues denoted by {λk}, i.e.,

{−�ek = λkek, in Oek = 0, on ∂O ,

with {ek} ∈ H10 (O) ∩ C∞(O).

Consider a filtered probability space{�,F , {Ft }t≥0 , P

}and a cylindrical Wiener

process in L2(O) of the form W (t) =∞∑

k=1γk(t)ek, t ≥ 0, where {γk} is a sequence

of mutually independent standard Brownian motions.

Mathematics Subject Classification (2000): 60H15, 47H05, 35B27Keywords: Stochastic porous media equation, Diffusivity function depending on the space variable,

Yosida approximation, Trotter-type result, Graph convergence, Homogenization.

340 I. Ciotir J. Evol. Equ.

We also consider σ (x) h = ∑∞k=1 μk (h, ek)L2(O) xek, for all x ∈ H−1(O) and

h ∈ L2(O), and we assume that

∞∑

k=1

μ2kλ

2k = C < ∞. (2)

Since d ≤ 3, we have by the Sobolev embedding theorem that

∞∑

k=1

μ2k |xek |2−1 ≤ c1

∞∑

k=1

μ2kλ

2k |x |2−1 ≤ C1 |x |2−1 , for all x ∈ H−1(O),

for some constants c1, C1 > 0. We obtain that σ (x) is a Hilbert Schmidt operatorfrom L2(O) to H−1(O). Note that, since σ is linear, it follows that x → σ (x) isLipschitz from H−1(O) to L2

(L2 (O) ; H−1(O)

).

The stochastic porous media equation with the diffusivity function independent ofthe space variable was studied in [2,8–11,15,16,22,24].

In this work, we shall consider the following stochastic partial differential equation

⎧⎨

⎩

dX (t, ξ) − � (ξ, X (t, ξ)) dt = σ (X (t, ξ)) dW (t), in (0, T ) × O,

(ξ, X (t, ξ)) = 0, on (0, T ) × ∂O,

X (0) = x, in O.

(3)

Some results concerning the variational convergence of the nonlinear partial differ-ential operators were given in [27]. In [14], it was proved that we have convergenceof the solutions in the situation of graph convergence of the operator in R.

In this work, we prove convergence of the solution for a more general equation(with the diffusivity function depending on the space variable), for a different typeof graph convergence (the porous media operator −� in H−1(O)) and using a dif-ferent main argument in the proof. This new result, unlike the one in [14], is suitableto prove homogenization which is carried out in the last part of the paper, for the casewhen has separable variables, i.e. (ξ, x) = a (ξ) (x). This particular form ofthe operator describes the isotropic materials (in our case the soil), where the thefunction a represents the hydraulic conductivity coefficient. From the physical pointof view, the limiting case corresponds to perfectly isolating inclusions if a = 0 andto highly conducting inclusions in the case a = ∞. In our case, the function a takesonly finite and strictly positive values. We assume that we have also an incompressiblefluid with no chemical reaction with the soil. In this paper, we are interested only inthe hydraulic process so we disregard all possible interactions that water may havewith chemical substances from the soil particles. Moreover, we shall assume that tem-perature variations are small enough not to influence the process such that we shallnot associate thermic laws.

However, most natural phenomena exhibit variability which cannot be modeled byusing deterministic approaches. More precisely, in natural systems, flows in porousmedia can be represented as stochastic models and the deterministic description can

Vol. 11 (2011) Convergence of solutions for the stochastic porous media equations 341

be considered as a subset of the pertinent stochastic models. This was thoroughlyexplained in [21]. We should however mention that the purpose of making a fluidelement follows a random path through the porous medium is well achieved by addinga Wiener process that varies stochastically with the time and space.

The homogenization for the case of a porous media equation describing soil is inter-esting form the point of view of the damage produced in the porous structure by a fluidflow. The flowing fluid transports eroded particles by diffusion and deposits them inother areas of the soil. Finally, this induces the change of permeability to a constantwhich is the mean value of a, and that in our case describes the homogenization ofthe soil. Homogenization theory exploits the fact that transport phenomena in a peri-odic porous media with period α transform the equation on the microscopic scale toeffective equation as α → 0. The limit corresponds to the size of the cells becomingprogressively smaller and greater in number until the structure becomes homogeneousand conductivity becomes constant in all points.

The problem of stochastic homogenization is discussed in the case of periodic lineardegenerate PDE’s in [20], but our case is not covered by this work.

The paper is organized in five sections as follows: the second section proves anexistence and uniqueness result for equations of the type (3), the third section proves aTrotter-type theorem for this equation,and the fourth section shows the homogeniza-tion in the particular case when the diffusivity function has separable variables usingthe Trotter-type theorem. The fifth section is an appendix with some useful lemmas.

NOTATION 1. We shall denote by β : O×R −→ R a function such that, for all ξ

fixed, z → β (ξ, z) is the inverse of x → (ξ, x).

• For β (ξ, z), we denote by βξ the gradient of β with respect to the first variableand by βz the derivative of β with respect to the second variable.

• For (ξ, x), we denote by ξ the gradient of with respect to the first variableand by x the derivative of with respect to the second variable;

• We also denote ∇ (ξ, x (ξ)) = ξ (ξ, x (ξ)) + x (ξ, x (ξ)) ∇x (ξ) , for allx ∈ L2(O).

The following spaces will be used throughout this paper.Lq

W (0, T ; L p (�; H)) (H a Hilbert space and p, q ∈ [1,+∞]) denotes the spaceof all q-integrable processes u : [0, T ] → L p (�; H) which are adapted to the filtra-tion {Ft }t≥0 . By CW

([0, T ] ; L2 (�; H)

)we shall denote the space of all H -valued

adapted processes which are mean square continuous. This space is endowed with thenorm

‖X‖2CW ([0,T ];L2(�,F ,P;H))

= supt∈[0,T ]

E |X (t)|2H ,

(see [5,17,18,23]).We shall use the symbol C to denote several positive constants independent of t, x ,

and ξ .


2. Existence and uniqueness of the solution

The following hypotheses will be assumed everywhere in this paper.

H1 For ξ ∈ O fixed, the function (ξ, .) : R → R is a continuous and monotoni-cally increasing function and for x ∈ R fixed the function (., x) : O −→ R

is measurable. Assume that (ξ, 0) = 0 and β (ξ, 0) = 0.

H2 There exist Ci > 0, i = 1, 3 independent of ξ and m ≥ 1 such that

g (ξ, x) =∫ x

0 (ξ, y) dy ≥ C1 |x |m+1 + C2x2, for all (ξ, x) ∈ O×R, (4)

and

| (ξ, x)| ≤ C3(|x |m + 1

), for all (ξ, x) ∈ O×R.

H3 The function β (ξ, z) is continuously differentiable in z, piecewise continuouslydifferentiable in ξ , and there exists C4 > 0, independent of ξ, such that

∣∣βξ (ξ, z)

∣∣ ≤ C4 (|z| + 1) , for a.e. (ξ, z) ∈ O×R.

There exist x0, y0 ∈ H−1(O) ∩ L1(O) such that g (., x0 (.)) ∈ L1(O) andg∗ (., y0 (.)) ∈ L1(O), where g∗ is the conjugate of g with respect to the secondvariable.

REMARK 1. The function g defined above is a normal convex integrand and,by [[25] Lemma 5], g∗ (ξ, z) is also a normal convex integrand. Then β (ξ, z) =∂zg∗ (ξ, z) , for a.e. (ξ, z) ∈ O×R since ∂g∗ = (∂g)−1 .

REMARK 2. Note that by assumption H2, using (ξ, x) = ∂x g (ξ, x) and thedefinition of the subdifferential, we get

g (ξ, x) ≤ C3

(|x |m+1 + |x |

), for all (ξ, x) ∈ O × R, (5)

and, since (ξ, x) x ≥ g (ξ, x) − g (ξ, 0), we have

(ξ, x) x ≥ C1 |x |m+1 + C2x2, for all (ξ, x) ∈ O×R.

Recall the definition of a solution to Eq. 3 (see [9,11,15,16]).

DEFINITION 1. Let x ∈ H−1(O). An H−1 (O) valued continuous Ft -adaptedprocess X = X (t, x) is called a solution to (3) on [0, T ] if

X ∈ Lm+1 (� × (0, T ) × O) ∩ L2(

0, T ; L2(�; H−1(O)

))

and there exists η ∈ Lm+1

m (� × (0, T ) × O) such that P−a.s.⟨X (t), e j

⟩2 = ⟨

x, e j⟩2 + ∫ t

0

∫O η (s, ξ)�e j (ξ) dξds

+∞∑

k=1μk∫ t

0

⟨X (s) ek, e j

⟩2 dβk(s),


for all j ∈ N, and all t ∈ [0, T ] , and

η (ω, t, ξ) = (ξ, X (ω, t, ξ)), a.e. in � × (0, T ) × O.

Here, m is the exponent arising in assumption H2, and {ek} is the above orthonormalbasis in L2 (O). Taking into account that −�ek = λkek in O, we may equivalentlywrite the above equation as follows

⟨X (t), e j

⟩−1 = ⟨

x, e j⟩2 −

∫ t

0

∫

Oη (s, ξ) e j (ξ) dξds

+∞∑

k=1

μk

∫ t

0

⟨X (s) ek, e j

⟩−1 dβk(s),

for all j ∈ N, and all t ∈ [0, T ].Now we can formulate the existence result of this paper. We have to say that the

proof follows the strategy first presented in [9], the only difference being the depen-dence of on the space variable ξ . However, we shall include this theorem to makethe paper self-contained.

THEOREM 1. Assume that Hi , i = 1, 3 and (2) hold. Then for each x ∈ H−1 (O),there is a unique solution

X ∈ Lm+1 (� × (0, T ) × O) ∩ CW

([0, T ] ; L2

(�; H−1(O)

))

for the equation (3).

3. A Trotter convergence result for the stochastic porous media equation

For x ∈ L p(O), where p ≥ max{2m, 4}, we consider the following equations⎧⎨

⎩

dX (t, ξ) − � (ξ, X (t, ξ)) dt = σ (X (t, ξ)) dW (t), in (0, T ) × O, (ξ, X (t, ξ)) = 0, on (0, T ) × ∂O,X (0) = x, in O,

(6)

and⎧⎨

⎩

dXα (t, ξ) − �α (ξ, Xα (t, ξ)) dt = σ (Xα (t, ξ)) dW (t), in (0, T ) × O,α (ξ, Xα (t, ξ)) = 0, on (0, T ) × ∂O,Xα (0) = x, in O,

(7)

for α ∈ (0,∞). Throughout all of this section,we shall assume that and α satisfyhypotheses Hi , i = 1, 3 and (2), such that Ci , i = 1, 3 are independent of α. Notethat C4 from H3 may depend on α since this assumption is used only in the existenceresult and in Lemma 1, and in both cases, we do not need it to be independent of α.In addition to the hypotheses used for the existence result, we have to assume that


H4 For all ξ ∈ O fixed, functions x → g (ξ, x) and x → gα (ξ, x) definedby (4) belong to C2 (R). Moreover, for all x ∈ R fixed, the functions ξ → (ξ, x) and ξ → α (ξ, x) are piecewise continuously differentiable, withx (ξ, x) ,α

x (ξ, x) bounded in ξ on O. We also suppose that

∫

O

(ξ (ξ, x(ξ)) ,∇x (ξ)

)dξ ≥ 0, for all x ∈ H1

0 (O)

∫

O

(α

ξ (ξ, x(ξ)) ,∇x (ξ))

dξ ≥ 0, for all x ∈ H10 (O)

and, for all α > 0, there exists a constant C5 > 0 and a constant m ∈ (1, 5),independent of ξ and α, such that

|x (ξ, x)| + ∣∣α

x (ξ, x)∣∣ ≤ C5

(|x |m−1 + 1

), for all x ∈ R and all α > 0.

Consider the approximating equations

{dXλ (t, ξ) + Aλ Xλ (t, ξ) dt = σ (Xλ (t, ξ)) dW (t), t ≥ 0,

Xλ (0) = x,(8)

and{

dXαλ (t, ξ) + Aα

λ Xαλ (t, ξ) dt = σ

(Xα

λ (t, ξ))

dW (t), t ≥ 0,

Xαλ (0) = x,

(9)

where the operators A and Aα are defined as A : D (A) ⊂ H−1(O) → H−1 (O) ofthe form

⎧⎨

⎩

Au = {−�v; v ∈ H10 (O), v (ξ) ∈ (ξ, u (ξ)) a.e.ξ ∈ O},

D (A) = {u ∈ L1(O) ∩ H−1(O); ∃v ∈ H1

0 (O) ,

v (ξ) ∈ (ξ, u (ξ)) a.e.ξ ∈ O}.(10)

and Aλ, Aαλ are the Yosida approximations of A and respectively Aα.

THEOREM 2. Let x ∈ L p(O) where p ≥ max {2m, 4}. Assume that Hi , i = 1, 4and (2) hold. If Aα and A satisfy

(1 + λAα

)−1w → (1 + λA)−1 w, strongly in H−1(O) (11)

for all λ > 0 and all w ∈ H−1 (O) , as α → 0, then the solution Xα to equa-tion (7) is convergent in L2

W

(�; C

([0, T ] ; H−1 (O)

))to the solution X to (6), i.e.,

E |Xα (t) − X (t)|2−1 → 0 uniformly on [0, T ]

In order to prove this result, we need a more precise preparation, i.e. some estimatesfor Xα

λ , Xλ, Aα (Xα), and A (X). In the following two subsections, we shall presentthe necessary evaluations.


3.1. A priori estimates

In this subsection, we assume that α is fixed. We also assume here that α and

are strongly monotone in the second variable, i.e. there exists M > 0, independent ofξ and α such that

(α (ξ, z) − α (ξ, z)

)(z − z) ≥ M (z − z)2 , for all z, z ∈ R, (12)

( (ξ, z) − (ξ, z)) (z − z) ≥ M (z − z)2 , for all z, z ∈ R. (13)

We mention however that for the main result, we shall dispense with these hypotheses.The following lemmas can be used for the two equations above, but we shall presentthe proof only in the case of Eq. 7.

LEMMA 1. We denote by Jαλ (x) the resolvent of Aα . Then

∥∥Jα

λ (x)∥∥

L2(O)≤ ‖x‖L2(O) (14)

and for p ≥ max {4, m + 1}∥∥Jα

λ (x)∥∥

L p(O)≤ ‖x‖L p(O) . (15)

Proof. Fix λ > 0. For x ∈ L p(O) fixed, we set yα = (1 + λAα)−1 x, i.e.

yα (ξ) − λ�α(ξ, yα (ξ)

) = x (ξ) , a.e. ξ ∈ O. (16)

Then yα ∈ D (A) and α (., yα (.)) ∈ H10 (O).

Step IWe firstly prove that

∫O (∇α (ξ, yα (ξ)) ,∇ yα (ξ)) dξ ≥ 0. To this purpose,

we have to show that yα ∈ H10 (O). Denoting α (., yα (.)) = zα , we have

yα = βα (., zα (.)) and using (12), we get that z → βα (ξ, z) is a Lipschitz con-tinuous function. Since for zα ∈ H1

0 (O), we have

∇βα(ξ, zα (ξ)

) = βαξ

(ξ, zα (ξ)

)+ βαz

(ξ, zα (ξ)

)∇zα (ξ), for a.e. ξ ∈ O.

then |βα (., zα (.))|2H1

0 (O)≤ C

(|zα|2L2(O)

+ |zα|2H1

0 (O)+ 1

)and this leads to

∣∣βα

(., zα (.)

)∣∣2H1

0 (O)≤ C

(∣∣zα∣∣2

H10 (O)

+ 1).

We obtain that∣∣yα

∣∣

H10 (O)

≤ C(∣∣α

(., yα (.)

)∣∣H1

0 (O)+ 1

)(17)

with C a constant that, for each α, is independent of yα, and consequently, we havethat yα ∈ H1

0 (O). Now we have that∫

O∇α

(ξ, yα (ξ)

)∇ yα (ξ) dξ

=∫

Oα

ξ

(ξ, yα (ξ)

)∇ yα (ξ) dξ +∫

Oα

y

(ξ, yα (ξ)

) ∣∣∇ yα∣∣2 dξ


and, using the monotonicity of α and assumption H4, we obtain that

∫

O

(∇α(ξ, yα (ξ)

),∇ yα (ξ)

)dξ ≥ 0. (18)

Step IIAs in the proof of Lemma 7 in the Appendix, it follows by (16) and ( 18) that

∥∥yα

∥∥

L2(O)≤ ‖x‖L2(O) .

To prove (15), we firstly need that

∫

O

(∇α(ξ, yα (ξ)

),∇h

(yα (ξ)

))dξ ≥ 0

for h : R → R an increasing Lipschitz continuous function. Considering the mollifi-ers sequence hk ∈ C1 (R) , h′

k ≥ 0, k ∈ N, we get that ∇h (x) = limk→∞ h′k (x) ∇x

in L2(O) and so, it suffices to have (∇α (., yα) ,∇ yα) ≥ 0, a.e. on O. On the other

hand, by multiplying both side of (16) by h (y) = (yα)p−1

1+τ |yα |p−2 and integrating over O,

we obtain

∫

O(yα)p

1 + τ |yα|p−2 dξ ≤∫

O(yα)p−1 x

1 + τ |yα|p−2 dξ.

and consequently we have that ‖yα‖L p(O) ≤ ‖x‖L p(O) .

The proof of this lemma is now complete. �

REMARK 3. We have ‖Aλ (x)‖L p(O) = 1λ

‖x − yα‖L p(O) ≤ 2λ

‖x‖L p(O) .

LEMMA 2. The solutions to (8) and (9) resp. belongs to CW([

0, T]; L2

(�; L2

(O))).

Proof. We have by [[9], Lemma 3.2], that Jαλ = (1 + λAα)−1 is Lipschitz continuous

in L2(O) and consequently Aαλ = 1

λ

(I − Jα

λ

)is Lipschitz continuous in L2(O). Since

‖σ (x)‖L2(L2(O),L2(O)) ≤ C∑∞

k=1 λ2kμ

2k |x |2

L2(O), it follows that for all x ∈ L2(O),

we have Xλ, Xαλ ∈ CW

([0, T ] ; L2

(�; L2(O)

)). �

LEMMA 3. The solutions Xλ, Xαλ to (8) and resp. (9) are bounded in

L∞ (0, T ; L p (� × O)

).

Proof. The proof is similar to Lemma 3.3 from [9]. �

LEMMA 4. For each α, we have

Xαλ → Xα, strongly in L∞ (

0, T ; L2(�; H−1(O)

))as λ → 0.


Proof. We shall only sketch the proof because it is similar to Lemma 3.4 from [9].Using Itô’s formula and Gronwall’s lemma, we obtain that

E∣∣Xα

λ (t) − Xα(t)∣∣2−1

≤∫ t

0

∫

O

∣∣∣α

(ξ,(1 + λAα

)−1Xα (s)

)− α

(ξ, Xα (s)

)∣∣∣∣∣Xα

λ (s) − Xα (s)∣∣ dξds.

Since by Proposition 2 the operator Aα is m-accretive in L1(O), we have that

(1 + λAα

)−1Xα → Xα, strongly in L1(O) for a.e. (ω, t) ∈ � × [0, T ] .

On the other hand, by Lemma 1, it follows that

∣∣∣(1 + λAα

)−1Xα∣∣∣L p

≤ ∣∣Xα

∣∣L p .

Arguing as in [9], we get

(1 + λAα

)−1Xα → Xα, strongly in L p (O × � × [0, T ]) , as λ → 0. (19)

We also have∣∣∣α

(ξ,(1 + λAα

)−1Xα)

− α(ξ, Xα

)∣∣∣

≤∫ 1

0α

x

(ξ, γ

(1 + λAα

)−1Xα + (1 − γ ) Xα

) ∣∣∣(1 + λAα

)−1Xα − Xα

∣∣∣ dγ

≤ C

(∣∣∣(1 + λAα

)−1Xα∣∣∣m−1 + ∣

∣Xα∣∣m−1 + 1

) ∣∣∣(1 + λAα

)−1Xα − Xα

∣∣∣ ,

a.e. on O × � × [0, T ]. By (19 ) and Lemma 3, we obtain that

Xαλ → Xα, strongly in L∞ (

0, T ; L2(�; H−1(O)

)), as λ → 0.

�

3.2. A priori estimates independent of α

As in the previous subsection, we assume in addition to all the previous hypothesesthat α and satisfy (12) and resp. ( 13). The following lemmas holds for the twoequations above, but we shall present the proof only in the case of Eq. 7 in order toassure the independence of α.

LEMMA 5. The solutions{

Xαλ

}λ

and Xα to (9) and resp. (7) are bounded inL∞

W ([0, T ] ; L p (�; L p(O))) uniformly in α.

Proof. The proof is similar to Lemma 3.1 from [11] but we shall sketch it to assurethat the property is uniform in α. We shall apply Itô’s formula to (9) for the function


ϕρ (x) = 1

p

∣∣(1 + ρ A0)

−1 x∣∣p

p, ρ > 0, with A0 = −� and D (A0) = H2(O) ∩H1

0 (O). After letting ρ → 0, we get

E∣∣Xα

λ (t)∣∣p

p + E

∫ t

0

⟨Aα

λ

(Xα

λ (s)),∣∣Xα

λ (s)∣∣p−2

Xαλ (s)

⟩

2ds

= |x |pp + p − 1

2E

∞∑

k=1

μk

∫ t

0

∫

O

∣∣Xα

λ (s)∣∣p−2 ∣∣Xα

λ (s) ek∣∣2 dξds

≤ |x |pp + p − 1

2CE

∫ t

0

∫

O

∣∣Xα

λ (s)∣∣p dξds.

By Lemma 1, we have that∣∣Y α

λ

∣∣

p ≤ ∣∣Xα

λ

∣∣

p and then⟨Aα

λ

(Xα

λ

),∣∣Xα

λ

∣∣p−2

Xαλ

⟩

2≥ 0.

Finally, via Gronwall’s lemma, we obtain that

E∣∣Xα

λ (t)∣∣p

p ≤ |x |pp exp

(

Cp − 1

2

)

, ∀ t ∈ [0, T ],

where C is independent of x, t , and α.Then,

{Xα

λ

}λ

is bounded in L∞W (0, T ; L p (�; L p(O))) , uniformly in α, and con-

sequently ,{Xα}α is bounded in L∞W (0, T ; L p (�; L p(O))). �

LEMMA 6. We have E∫ t

0

∣∣Aα

λ

(Xα

λ

)∣∣2−1 ds ≤ C, for all λ > 0 and also that

E

∫ t

0

∣∣Aα

(Xα)∣∣2−1 ds ≤ C, (20)

with C independent of λ and α.

Proof. Let gα be defined as in H2. We consider the Moreau-Yosida approximation ofgα, i.e.,

gαε (ξ, x) = min

{

gα (ξ, y) + |x − y|22ε

; y ∈ R

}

,

for all ξ ∈ O and x ∈ R. By classical theory, we have that

gαε (ξ, x) =

∫ x

0α

ε (ξ, r) dr,

for all ξ ∈ O and x ∈ R, where αε is the Yosida approximation of α, with respect

to the second variable. �We define

gαε (ξ, x) = gα

ε (ξ, x) + ε

2x2 and α

ε (ξ, x) = αε (ξ, x) + εx,

for all ξ ∈ O and x ∈ R. Note that αε is Lipschitz and strictly monotone in x . For

gαε defined above, we consider

ϕαε : H−1(O) → R

ϕαε (u) =

{∫O gα

ε (ξ, u (ξ)) dξ, if u ∈ L1(O) and gαε (ξ, u (ξ)) ∈ L1(O),

+∞, otherwise,


and the corresponding Moreau-Yosida approximation

ϕαε,λ (u) = inf

v∈H−1(O)

{

ϕαε (v) +

|u − v|2H−1(O)

2λ

}

, for u ∈ H−1(O).

We know that ϕαε,λ ∈ C1

(H−1(O)

)and ∇ϕα

ε,λ = Aαε,λ, where Aα

ε,λ is the Yosidaapproximation of Aα

ε : D(

Aαε

) ⊂ H−1(O) → H−1(O), Aαε = −�α

ε . (see e.g. [[4]Theorem 2.2]).

Consider the approximating equation{

dXαε,λ(t) + Aα

ε,λ

(Xα

ε,λ(t))

dt = σ(

Xαε,λ(t)

)dW (t),

Xαε,λ (0, x) = x .

We intend to apply Itô’s formula for this equation, using ϕαε,λ, but, in order to do

that, both, the first order and the second order derivatives of ϕαε,λ have to exist and to

be continuous. Since ϕαε,λ ∈ C1

(H−1 (O)

), we only need to prove that the second

derivative exists and that it is continuous for every ε, λ > 0.Fix ε, λ > 0.

By the definition of the Fréchet derivative, we have for every u ∈ H−1(O) that

(A′

ε,λ (u) h, w) =

∫

O

(ε

)′(ξ, Jλ (u)) [D Jλ (u) h] wdξ, for all h, w ∈ H−1(O),

where(ε

)′is the derivative of ε with respect to the second variable, Jλ (u) =

(1 + λAα

ε

)−1(u) and D Jλ (u) h is the Fréchet derivative of Jλ (u).

We shall firstly prove that the Fréchet derivative of Jλ (u) exists as the solutionz = D Jλ (u) h of the following equation

z − λ�[(

ε

)′(ξ, Jλ (u)) z

]= h, (21)

which is equivalent to

(−�)−1 z + λ[(

ε

)′(ξ, Jλ (u)) z

]= (−�)−1 h. (22)

Since (−�)−1 is maximal monotone operator in L2(O)×L2(O) and � : L2(O) →L2(O), � (x) = (

ε

)′(ξ, Jλ (u)) x is linear, continuous and bounded we have by

[4, Corollary 1.1, p. 44] that B = � + (−�)−1 is a maximal monotone operator inthe same space.

We can easily verify that B satisfies the hypotheses of Corollary 1.2 from Chap-ter 2 of [4], and then, we get that it is also surjective. Consequently, for all h ∈H−1(O) ((−�)−1 h ∈ L2(O)), Eq. 21 has an unique solution z ∈ L2(O), such that(ε

)′(ξ, Jλ (u)) z ∈ H1

0 (O).

Now we have the existence of D Jλ (u) h and this assures the existence of A′ε,λ. We

still need to prove the continuity of A′ε,λ, i.e., for un → u, strongly in H−1(O), as

n → ∞, we have⟨A′

ε,λ (un) h, w⟩−1 → ⟨

A′ε,λ (u) h, w

⟩−1,


which is equivalent to

∫

O

(ε

)′(ξ, Jλ (un)) [D Jλ (un) h] wdξ →

∫

O

(ε

)′(ξ, Jλ (u)) [D Jλ (u) h] wdξ,

as n → ∞, for all h, w ∈ H−1(O).

We firstly prove that, we have

Jλ (un) → Jλ (u) , strongly in L2(O), as n → ∞. (23)

By taking the scalar product in H−1(O) of

Jλ (un) − Jλ (um) − λ�[ε (ξ, Jλ (un)) − ε (ξ, Jλ (um))

] = un − um,

with −�[ε (ξ, Jλ (un)) − ε (ξ, Jλ (um))

]and using in the first term the strong

monotonicity of ε, we get that

ε |Jλ (un) − Jλ (um)|22 + λ∣∣ε (ξ, Jλ (un)) − ε (ξ, Jλ (um))

∣∣2

H10 (O)

= 1

4λ|un − um |2−1 + λ

∣∣�(ε (ξ, Jλ (un)) − ε (ξ, Jλ (um))

)∣∣2−1 .

Since |�x |−1 = |x |H10 (O) for all x ∈ H1

0 (O), we can conclude that (23) holds.

Since using H4 we get that(ε

)′is continuous on R in the second variable, we

obtain now that

(ε

)′(ξ, Jλ (un)) → (

ε

)′(ξ, Jλ (u)) , a.e. in O, as n → ∞.

Then for every w ∈ L2(O), we have

∣∣∣(ε

)′(ξ, Jλ (un)) w − (

ε

)′(ξ, Jλ (u)) w

∣∣∣2 → 0, a.e. in O, as n → ∞.

Since ε is Lipschitz in the second variable with the constant1

ε+ ε we have that

∣∣∣(ε

)′(ξ, Jλ (un)) w − (

ε

)′(ξ, Jλ (u)) w

∣∣∣2 ≤ 4

(1

ε+ ε

)2

|w|2 , a.e. in O.

Using Lebesgue’s dominated convergence theorem, we get that

(ε

)′(ξ, Jλ (un)) w → (

ε

)′(ξ, Jλ (u)) w, strongly in L2(O) (24)

as n → ∞.On the other hand, we have from Eq. 21 that

zn − λ�[(

ε

)′(ξ, Jλ (un)) zn

]= h.


By taking the scalar product with(ε

)′(ξ, Jλ (un)) zn in L2(O) and using the strong

monotonicity of ε, we get that {zn} is bounded in L2(O) and that{(

ε

)′(ξ, Jλ (un)) zn

}

nis bounded in H1

0 (O). (25)

Then there exists η ∈ L2(O) such that

zn = D Jλ (un) h ⇀ η, weakly in L2(O), as n → ∞.

From the definition of the subdifferential, we have that

D Jλ (un) h → D Jλ (u) h, strongly in H−1(O), as n → ∞and then

D Jλ (un) h ⇀ D Jλ (u) h, weakly in L2(O), as n → ∞. (26)

By (24) and (26), we have that∫

O

(ε

)′(ξ, Jλ (un)) [D Jλ (un) h] wdξ →

∫

O

(ε

)′(ξ, Jλ (u)) [D Jλ (u) h] wdξ,

(27)

as n → ∞, for all h ∈ H−1(O) and all w ∈ L2(O).

Then{(

ε

)′(ξ, Jλ (un)) [D Jλ (un) h]

}

n

is weakly convergent in L2(O) to(ε

)′(ξ, Jλ (u)) [D Jλ (u) h] as n → ∞.

We know by (25) that {(ε

)′(ξ, Jλ (un)) zn}n is weakly convergent in H1

0 (O) andconsequently (27) holds also for all w ∈ H−1(O). We can conclude that {A′

ε,λ} iscontinuous on H−1(O). Arguing as in [[20], Lemma 4.3], we conclude that we canapply Itô’s formula.

We obtain that

E[ϕα

ε,λ

(Xα

ε,λ(t))]+ 2E

∫ t

0

∣∣Aα

ε,λ

(Xα

ε,λ

)∣∣2−1ds (28)

= ϕαε,λ (x) + E

∫ t

0

∞∑

k=1

∫

O

(α)′ (

ξ, Jε

(ξ, Jλ

(Xα

ε,λ

)))D Jε

(ξ, Jλ

(Xα

ε,λ

))

× [D Jλ

(Xα

ε.λ

)Xα

ε.λei]

Xαε.λei dξds,

where D Jε is the derivative of Jε (ξ, x) = (1 + εα (ξ, .)

)−1(x) with respect to the

second variable.From Eq. 21 in our case, we have that

z − λ�[(

α)′ (

ξ, Jε

(ξ, Jλ

(Xα

ε,λ

)))D Jε

(ξ, Jλ

(Xα

ε,λ

))z]

= h.

By taking the scalar product of the equation above with(α)′ (

ξ, Jε

(ξ, Jλ

(Xα

ε,λ

)))D Jε

(ξ, Jλ

(Xα

ε,λ

))z


in L2(O) and using the strict monotonicity of(α), the fact that D Jε is positive

(because Jε is a monotonically increasing function in the second variable) and |D Jε| ≤1 (by [4, Chapter 2, Proposition 1.4, i])), we get that

M

2

∣∣D Jε

(ξ, Jλ

(Xα

ε,λ

))z∣∣22

≤ 1

2M

∣∣∣(α)′ (

ξ, Jε

(ξ, Jλ

(Xα

ε,λ

)))h∣∣∣2

2,

where M is the constant arising in (12).Note that in our case z = D Jλ

(Xα

ε.λ

)Xα

ε.λei and h = Xαε.λei ∈ L2(O).

Replacing in the last term of (28), we get by Hölder’s inequality that

E[ϕα

ε,λ

(Xα

ε,λ(t))]+ 2E

∫ t

0

∣∣Aα

ε,λ

(Xα

ε,λ

)∣∣2−1ds

= ϕαε,λ (x) + CE

∫ t

0

∫

O

∣∣∣(α)′ (

ξ, Jε

(ξ, Jλ

(Xα

ε,λ

)))Xα

ε,λ

∣∣∣2

dξds,

with C independent of α, ε and λ.Since ϕα

ε,λ is positive, ϕαε,λ (x) is bounded uniformly in α, ε, λ and by (H4), we get

that

E

∫ t

0

∣∣Aα

ε,λ

(Xα

ε,λ

)∣∣2−1ds ≤ C, (29)

with C independent of α, ε, and λ.We firstly shall pass to the limit for λ > 0 fixed and for ε → 0, as follows.Recall that Aα

ε = −�αε and Aα = �α .

Since αε (ξ, x) → α (ξ, x) pointwise as ε → 0, we can easily see that

(1 + λAα

ε

)−1(x) → (

1 + λAα)−1

(x) , strongly in H−1(O) (30)

as ε → 0, for all x ∈ L2(O).

Indeed, if we denote zε = (1 + λAα

ε

)−1(x) and z = (1 + λAα)−1 (x) and taking

the scalar product of the difference

zε − z − λ�(α

ε (ξ, zε) − α (ξ, z)) = 0

with zε − z in H−1(O), we get that

|zε − z|2−1 + λ

∫

O

(α

ε (ξ, zε) − α (ξ, z))(zε − z) dξ + λε

∫

O(zε − z)2 dξ = 0.

Then, by the monotonicity of αε , we have that

|zε − z|2−1 ≤ λ

∫

O

(α

ε (ξ, z) − α (ξ, z))(z − zε) dξ (31)

Arguing as in Lemma 1, we have that {zε} is bounded in L2(O), and by using Lebes-gue’s dominated convergence theorem, we can easily see that α

ε (ξ, z) → α (ξ, z)strongly in L2(O). Passing to the limit in (31) for ε → 0, we get (30).


Then, arguing as in Step III from Theorem 2 below, we get that

Xαε,λ → Xα

λ , strongly in L2(�; C

([0, T ] ; H−1(O)

)),

as ε → 0, for every λ fixed. Note that Lemma 6 is not used in Step III of Theorem 2.Consequently,

Xαε,λ → Xα

λ, strongly in H−1(O)

a.e. in [0, T ] × �, as ε → 0.

We obtain now that

(1 + λAα

ε

)−1 (Xα

ε,λ

) → (1 + λAα

)−1 (Xα

λ

), strongly in H−1(O) (32)

a.e. in [0, T ] × �, as ε → 0, by using the same techniques as we did for (30).

Since Aαε,λ

(Xα

ε,λ

)= Aα

(Jε

(ξ, Jλ

(Xα

ε,λ

)))where

Jε (ξ, x) = (1 + εα (ξ, .)

)−1(x) and Jλ (x) = (

1 + λAαε

)−1(x),

we only need to prove that

Jε

(ξ, Jλ

(Xα

ε,λ

)) → Jλ

(Xα

λ

), strongly in H−1(O) (33)

a.e. in [0, T ] × �, as ε → 0.

We denote by

qε,λ = (1 + εα (ξ, .)

)−1 (1 + λAα

ε

)−1 (Xα

ε,λ

)

wε,λ = (1 + λAα

ε

)−1 (Xα

ε,λ

).

By taking the scalar product of the difference

qε,λ − qμ,λ + εα(ξ, qε,λ

)− μα(ξ, qμ,λ

) = wε,λ − wμ,λ

with qε,λ − qμ,λ in H−1 (O) and using Hölder’s inequality, we get that

3

4

∣∣qε,λ − qμ,λ

∣∣2−1 (34)

≤ ∣∣wε,λ − wμ,λ

∣∣2−1 + ⟨

εα(ξ, qε,λ


), qμ,λ − qε,λ

⟩−1 .

In the second term of the right side, we have

⟨εα

(ξ, qε,λ


), (−�)−1 (qμ,λ − qε,λ

)⟩

2

≤ 1

4

∣∣∣(−�)−1 (qμ,λ − qε,λ

)∣∣∣2

Lm+1(O)+ ∣∣εα

(ξ, qε,λ


)∣∣2L

m+1m (O)

.


From H4, we have that m < 5, and the by the Sobolev imbedding,we get H10 (O) ⊂

Lm+1 (O). Since∣∣(−�)−1 x

∣∣

H10 (O)

= |x |−1, then ( 34) becomes

1

2

∣∣qε,λ − qμ,λ

∣∣2−1 ≤ ∣

∣wε,λ − wμ,λ

∣∣2−1 + ∣

∣εα(ξ, qε,λ


)∣∣2L

m+1m (O)

.

Using Lemma 1 and assumption H4, we have that{α

(ξ, qμ,λ

)}μ

is bounded in

Lm+1

m (O). Then, considering (32), we can pass to the limit for ε, μ → 0 and weobtain (33).

We know that Aα is maximal monotone in H−1 (O)× H−1(O) and Aαε,λ

(Xα

ε,λ

)=

Aα(

Jε

(ξ, Jλ

(Xα

ε,λ

))). Then (33) and (29) lead to

Aαε,λ

(Xα

ε,λ

) → Aαλ

(Xα

λ

), weakly in H−1(O),

a.e. in [0, T ] × � as ε → 0.

On the other hand, by Fatou’s lemma, we have that

E

∫ t

0

∣∣Aα

λ

(Xα

λ

)∣∣2−1 ds ≤ E

∫ t

0lim inf

ε→0

∣∣Aα

ε,λ

(Xα

ε,λ

)∣∣2−1ds

≤ lim infε→0

E

∫ t

0

∣∣Aα

ε,λ

(Xα

ε,λ

)∣∣2−1ds ≤ C

and also that

E

∫ t

0

∣∣Aα

(Xα)∣∣2−1 ds ≤ C,

with C independent of α, ε, and λ. This concludes the proof of the lemma.Proof of Theorem2. We firstly assume that z −→ α (ξ, z) and z −→ (ξ, z)

are strongly monotone with the same constant M, independent of ξ and α. We applyTheorem 1 for Eqs. 6 and 7 and we get that each has a unique solution X, Xα ∈L2

W

(�; C

([0, T ] ; H−1(O)

)) ∩ Lm+1 (� × (0, T ) × O).Let Xλ and Xα

λ be the solutions to the approximating Eq. 8 and resp. (9). We have

E∣∣X − Xα

∣∣2−1 ≤ 3E

(|X − Xλ|2−1 + ∣

∣Xλ − Xαλ

∣∣2−1 + ∣

∣Xαλ − Xα

∣∣2−1

).

Step IWe firstly prove that, for λ → 0, we have

(Xα

λ − Xα) → 0 strongly in L2

W

(�; C

([0, T ] ; H−1(O)

))

uniformly in α. We formally apply Itô’s formula to the difference

d(Xα (t, ξ) − Xα

λ (t, ξ))+ (

Aα(Xα (t, ξ)

)− Aαλ

(Xα

λ (t, ξ)))

dt

= σ(Xα (t, ξ) − Xα

λ (t, ξ))

dW (t).


and we get for all t ∈ [0, T ] that

1

2

∣∣Xα(t) − Xα

λ (t)∣∣2−1 e−ηt

+∫ t

0

⟨Aα(Xα (s)

)− Aαλ

(Xα

λ (s)),(Xα (s) − Xα

λ (s))⟩

−1 e−ηsds

≤∫ t

0e−ηs ⟨Xα (s) − Xα

λ (s), σ(Xα (s) − Xα

λ (s))

dW (s)⟩−1 ds

+(

C∞∑

k=1

μ2kλ

2k − 1

2η

)∫ t

0

∣∣Xα (s) − Xα

λ (s)∣∣2−1 e−ηsds, P−a.s..

On the other hand by the strong monotonicity of Aα , we have via standard argumentsas e.g. in [14, p. 5609], that

t∫

0

⟨Aα(Xα (s)

)− Aαλ

(Xα

λ (s)),(Xα (s) − Xα

λ (s))⟩

−1 e−ηsds

≥ −λC∫ t

0

∣∣Aα Xα (s)

∣∣2−1 e−ηsds − λC

∫ t

0

∣∣Aα

λ Xαλ (s)

∣∣2−1 e−ηsds,

for all t ∈ [0, T ] , P−a.s., with C independent of λ and α.By the Burkholder-Davis-Gundy inequality, for r ∈ [0, T ] and t ≤ r we get that

E

∫ r

0e−ηs ⟨Xα (s) − Xα

λ (s) , σ(Xα (s) − Xα

λ (s))

dW (s)⟩−1 ds

≤ E sups∈[0,r ]

∣∣Xα (s) − Xα

λ (s)∣∣−1 e− η

2 s(

C∫ r

0

∣∣Xα (s) − Xα

λ (s)∣∣2−1 e−ηsds

)1/2

≤ 1

4E sup

s∈[0,r ]

∣∣Xα (s) − Xα

λ (s)∣∣2−1 e−ηs + CE

(∫ r

0

∣∣Xα (s) − Xα


)

.

Hence, for η > 0 large enough, we obtain that

1

4E sup

t∈[0,r ]


λ (t)∣∣2−1 e−ηt ≤ λCE

∫ r

0

∣∣Aα

(Xα (s)

)∣∣2−1 e−ηsds

+λCE

∫ r

0

∣∣Aα

λ

(Xα

λ (s))∣∣2−1 e−ηsds + CE

(∫ r

0

∣∣Xα (s) − Xα


)

(35)

for all λ ∈ (0, 1) and r ∈ [0, T ].Using Gronwall’s lemma for (35), we get that

1

4E sup

t∈[0,r ]


λ (t)∣∣2−1 e−ηt

≤ λCE

∫ r

0

∣∣Aα

(Xα (s)

)∣∣2−1 e−ηsds + λCE

∫ r

0

∣∣Aα

λ

(Xα

λ (s))∣∣2−1 e−ηsds

and by Lemma 6, letting λ → 0 uniformly in α, for large η, we conclude Step I.


Step IIIn order to obtain

(Xλ − X) → 0, strongly in L2(�; C

([0, T ] ; H−1 (O)

))

as λ → 0 we argue as in Step I.Step IIITo complete the proof, it suffices to show that

(Xα

λ − Xλ

) → 0 strongly in L2(�; C

([0, T ] ; H−1(O)

)),

for all fixed λ > 0, as α → 0. Applying Itô’s formula for equation

d(Xα

λ (t) − Xλ(t))− �

(α

λ

(Xα

λ (t))− λ (Xλ(t))

)dt

= σ(Xα

λ (t) − Xλ(t))

dW (t)

for η > 0, large enough, and after some calculation involving Burkholder-Davis-Gundy inequality as above, we get that

1

4E sup

t∈[0,r ]

∣∣Xα

λ (t) − Xλ(t)∣∣2−1 e−ηt

+E

∫ r

0

⟨Aα

λ

(Xα

λ (s))− Aλ (Xλ (s)) , Xα

λ (s) − Xλ (s)⟩−1 e−ηsds

≤ CE

∫ r

0

∣∣Xα

λ (s) − Xλ (s)∣∣2−1 e−ηsds. (36)

We have by the monotonicity of Aαλ that

E

∫ r

0

⟨Aα

λ

(Xα

λ (s))− Aλ (Xλ (s)), Xα

λ (s) − Xλ (s)⟩−1 e−ηsds

≥ E

∫ r

0

⟨Aα

λ (Xλ (s)) − Aλ (Xλ (s)) , Xαλ (s) − Xλ (s)

⟩−1 e−ηsds.

By Gronwall’s lemma, we get that

E supt∈[0,r ]

∣∣Xα

λ (t) − Xλ(t)∣∣2−1 e−ηt (37)

≤ CE

∫ r

0

∣∣∣⟨Aα

λ (Xλ (s)) − Aλ (Xλ (s)) , Xλ (s) − Xαλ (s)

⟩−1

∣∣∣ e−ηsds.


Moreover, from the last term of the inequality above, we have

E

∫ r

0

∣∣∣⟨Aα

λ (Xλ (s)) − Aλ (Xλ (s)), Xλ (s) − Xαλ (s)

⟩−1

∣∣∣ e−ηsds

≤ E

∫ r

0

∣∣Aα

λ (Xλ (s)) − Aλ (Xλ (s))∣∣−1

∣∣Xλ (s) − Xα

λ (s)∣∣−1 e−ηsds

=⟨∣∣Aα

λ (Xλ) − Aλ (Xλ)∣∣−1 ,

∣∣(Xλ − Xα

λ

)∣∣−1 e−η·⟩

L2(�×[0,r ])

≤(

E

∫ r

0

∣∣Aα

λ (Xλ (s)) − Aλ (Xλ (s))∣∣2−1 ds

)1/2

×(

E

∫ r

0

∣∣(Xλ (s) − Xα

λ (s))∣∣2−1 e−ηsds

)1/2

. (38)

Using Lemma 5, it follows that∣∣(Xλ − Xα

λ

)e−η·∣∣

L2(�×[0,r ];H−1(O))≤ ∣∣(Xλ − Xα

λ

)e−η·∣∣

L p(�×[0,r ]×O)≤ C |x |p

p ,

(39)

with C independent of x, t, λ and α. Now, replacing (38) and (39) in (37), we obtainthat

1

4E sup

t∈[0,r ]

∣∣Xα

λ (t)−Xλ(t)∣∣2−1 ≤C |x |p

p

(

E

∫ r

0

∣∣Aα

λ (Xλ (s))− Aλ (Xλ (s))∣∣2−1 ds

)1/2

.

(40)

We denote

Y αλ = (

1 + λAα)−1

Xλ = Jαλ (Xλ) and Yλ = (1 + λA)−1 Xλ = Jλ (Xλ) ,

i.e.,

Y αλ (ξ) + λAα

(Y α

λ (ξ)) = Xλ (ξ) and Yλ (ξ) + λA (Yλ (ξ)) = Xλ (ξ).

By taking the difference and the scalar product in H−1 (O) with Aα(Y α

λ

)− A (Yλ),we get that

⟨Y α

λ − Yλ, Aα(Y α

λ

)− A (Yλ)⟩−1 + λ

∣∣Aα

(Y α

λ

)− A (Yλ)∣∣2−1 = 0,

a.e. in � × [0, T ].This leads to

λ∣∣Aα

(Y α

λ

)− A (Yλ)∣∣2−1 ≤ λ

2

∣∣Aα

(Y α

λ

)− A (Yλ)∣∣2−1 + 1

2λ

∣∣Y α

λ − Yλ

∣∣2−1 ,

a.e. in � × [0, T ].Using Lemma 1, we have now that

∣∣Aα

λ (Xλ) − Aλ (Xλ)∣∣2−1 ≤ 1

λ2|Xλ|2L2(O)

, a.e. in � × [0, T ] . (41)


On the other hand, by (11) and from

(Aλ (Xλ) − Aα

λ (Xλ)) = 1

λ

((1 + λA)−1 Xλ − (

1 + λAα)−1

Xλ

)

it follows that∣∣Aα

λ (Xλ) − Aλ (Xλ)∣∣−1 → 0 as α → 0, a.e. on � × [0, T ] (42)

(see [1]). From (41) and (42) we obtain, by Lebesgue’s dominated convergence theo-rem, that

E

∫ r

0

∣∣Aα

λ (Xλ (s)) − Aλ (Xλ (s))∣∣−1 ds → 0,

as α → 0, for all λ > 0 fixed. Finally, from (40), we get that

E supt∈[0,r ]

∣∣Xα

λ (t) − Xλ(t)∣∣2−1 → 0, as α → 0, for all λ > 0

and this concludes the proof of Step III.We can now go back to

E∣∣Xα − X

∣∣2−1 ≤ 3E

(∣∣Xα − Xα

λ

∣∣2−1 + ∣

∣Xαλ − Xλ

∣∣2−1 + |Xλ − X |2−1

).

For all ε > 0, we first choose λ0, independent of α, such that the first and the third

term are less thenε

3. Having fixed λ0 this way, we can choose α such that the second

term is less thenε

3and finally we obtain

E∣∣Xα(t) − X (t)

∣∣2−1 ≤ ε, uniformly on [0, T ].

The proof is now complete in the case and α are strictly monotone in the secondvariable. We show below that this assumption can be dispensed with. Consider theapproximating equations corresponding to

Aαδ (x) = −�

(α (., x) + δx

), δ > 0,

D(

Aαδ

) ={

x ∈ H−1 (O) ∩ L1(O) : α (., x) + δx ∈ H10 (O)

}

and

Aδ (x) = −�( (., x) + δx) , δ > 0,

D (Aδ) ={

x ∈ H−1(O) ∩ L1(O) : (., x) + δx ∈ H10 (O)

}.

We have that∣∣Xα − X

∣∣2−1 ≤ 3

(∣∣Xα − Xα

δ

∣∣2−1 + ∣

∣Xαδ − Xδ

∣∣2−1 + |Xδ − X |2−1

),

where Xαδ and Xδ are solutions corresponding to Aα

δ and resp. Aδ.


The main part of the proof is the convergence of the first term, uniformly in α.Arguing as in Step I, we have

1

4E sup

t∈[0,r ]


δ (t)∣∣2−1 e−ηt

≤ δCE

∫ r

0

∫

O

∣∣Xα (s)

∣∣2 e−ηsdξds + δCE

∫ r

0

∫

O

∣∣Xα

δ (s)∣∣2 e−ηsdξds.

Since Aαδ satisfy the hypotheses required in Lemma 5, we get by using it that

E

∫ t

0

∫

O

∣∣Xα

δ (s)∣∣2 dξds ≤ C,

with C independent of α and δ. Passing to the limit for δ → 0, we get

E

∫ t

0

∫

O

∣∣Xα (s)

∣∣2 dξds ≤ C,

with C independent of α.The proof of the theorem is now complete. �

4. The homogenization of the stochastic porous media equation

Consider the following stochastic partial differential equation⎧⎨

⎩

dX (t, ξ) − �(a (ξ) (X (t, ξ))) dt = σ (X (t, ξ)) dW (t), in (0, T ) × O (ξ, X (t, ξ)) = 0, on (0, T ) × ∂OX (0) = x, in O

.

Throughout this section, we assume that the following hypotheses hold

A1 The function a : Rd → R is Y−periodic for Y = [0, l]d , for l ∈ (0,∞) such

that Y ⊂ O and satisfies(1) a ∈ C2 (intY ) ∩ C (Y );(2) �a ≤ 0 on intY and ∇a belongs to L∞ (Y );(3) a ≥ ρ > 0, in Y for a constant ρ.

A2 The function : R → R is continuous, differentiable, monotonically increasingand (0) = 0. There exists Ci > 0, i = 1, 3, and 1 ≤ m < 5 such that

γ (x) =∫ x

0(y) dy ≥ C1 |x |m+1 + C2x2, for all x ∈ R,

′ (x) ≤ C3

(|x |m−1 + 1

), for all x ∈ R. (43)

A3 The function −1 : R → R is continuous, differentiable and there exists C4 >

0, such that(−1

)′(z) ≤ C4

(|z| 1−m

m + 1), for all |z| < 1,

(−1

)′(z) ≤ C4, for all |z| ≥ 1.


There exists x0, y0 ∈ H−1(O) ∩ L1(O) for which a (.) γ (x0 (.)) ∈ L1(O) and

γ ∗(

y0(.)a(.)

)∈ L1(O).

In this section, we shall assume also that (2) holds.

REMARK 4. We note that, denoting by β : Rd × R → R the inverse of x →

a (ξ) (x), we have

β (ξ, z) = −1(

z

a (ξ)

)

= (∂γ )−1(

z

a (ξ)

)

= ∂γ ∗(

z

a (ξ)

)

.

By assumption A2, we have that

(x) x ≥ C1 |x |m+1 + C2x2, for all x ∈ R,

and

γ (x) ≤ C3

(|x |m+1 + |x |

), for all x ∈ R.

Define

α (ξ, x) = aα (ξ) (x) = a

(ξ

α

)

(x)

and Aα : H−1(O) → H−1(O), Aα (X) = −�α (., X (.)) with

D(

Aα) =

{x ∈ H−1(O) ∩ L1(O);α (., X (.)) ∈ H1

0 (O)}

.

We intend to use the Trotter-type theorem from the previous section, and to thispurpose, we have to verify that hypotheses Hi , i = 1, 4 are satisfied. One can easilysee that Hi , i = 1, 2, the second part of H3 and the second part of H4 hold. It remainsnow to prove that α satisfies the first parts of H3 and H4.

Regarding assumption H3, we have two situations. Firstly, for |z| < 1 we have that∣∣∣βα

ξ (ξ, z)∣∣∣ =

∣∣∣∣

(−1

)′ ( z

aα (ξ)

)(

− z

(aα (ξ))2

)

∇aα (ξ)

∣∣∣∣

≤ C4

∣∣∣∇a

(ξα

)∣∣∣

α |aα (ξ)|2 ·(∣∣∣∣

z

aα (ξ)

∣∣∣∣

1−mm + 1

)

|z|

≤ C(|z| 1

m + |z|)

≤ C (|z| + 1), a.e. on O.

Secondly, for |z| ≥ 1 we have that∣∣∣βα

ξ (ξ, z)∣∣∣ =

∣∣∣∣

(−1

)′ ( z

aα (ξ)

)(

− z

(aα (ξ))2

)

∇aα (ξ)

∣∣∣∣

≤ C4

∣∣∣∇a

(ξα

)∣∣∣

α |aα (ξ)|2 · |z| ≤ C (|z| + 1) , a.e. on O.

It is obvious now that there exists C4 > 0, depending on α and satisfying∣∣∣βα

ξ (ξ, z)∣∣∣ ≤ C4 (|z| + 1), a.e. on O.


REMARK 5. The fact that C4 depends on α is not a problem since the relationabove is used only in the existence result and in Lemma 1, where we do not needindependence of α.

Regarding assumption H4, we consider∫

O

(α

ξ (ξ, x (ξ)) ,∇x (ξ))

dξ = 1

α

∫

O

(

∇a

(ξ

α

)

,∇γ (x (ξ))

)

dξ

= − 1

α2

∫

O�a

(ξ

α

)

γ (x (ξ)) dξ ≥ 0.

Thus, all the hypotheses that we mentioned before are satisfied for α and conse-quently, the equation

⎧⎨

⎩

dXα (t, ξ) − �α (ξ, Xα (t, ξ)) dt = σ (Xα (t, ξ)) dW (t), in (0, T ) × Oα (ξ, Xα (t, ξ)) = 0, on (0, T ) × ∂OXα (0) = x, in O

.

(44)

with x ∈ L p(O), p ≥ max {2m, 4}has a unique solution and, moreover, the assump-tion of Theorem 2 is satisfied. Before that we need similar assumptions for the limitequations. To this aim we define

hom (x) = MY (a) (x), ∀x ∈ R.

where MY (a) = 1

|Y |∫

Y a (y) dy is the mean value of a over Y, |Y | is the Lebesgue

measure of Y and

Ahom : H−1(O) → H−1(O), Ahom (X) = −�hom (X)

D(

Ahom) = {

x ∈ H−1(O) ∩ L1(O);hom (X) ∈ H10 (O)

}.

We can easily see that hom satisfy Hi , i = 1, 4, and therefore, the homogenizedequation

⎧⎨

⎩

dXhom (ξ) − �hom(Xhom (ξ)

)dt = σ

(Xhom (ξ)

)dW (t), in (0, T ) × O

hom(Xhom (ξ)

) = 0, on (0, T ) × ∂OXhom (0) = x, in O

(45)

with x ∈ L p(O), where p ≥ max {2m, 4}has a unique solution. We can now formulatethe main result of this section.

THEOREM 3. Consider the functions a and satisfying assumptions Ai , i = 1, 3and α,hom defined above and x ∈ L p(O), p ≥ max {2m, 4}. Then the solutionXα to Eq. 44 is convergent for α → 0 in L2

W

(�; C

([0, T ] ; H−1(O)

))to the solution

Xhom to the homogenized Eq. 45.


Proof. We intend to prove this theorem by applying the Trotter-type theorem from theprevious section of this paper. In order to do that, since all the other assumptions hold,we just have to show, for all x ∈ H−1(O) and all λ > 0, arbitrary fixed, that we have

(1 + λAα

)−1x →

(1 + λAhom

)−1x, in H−1(O),

as α → 0.Fix λ > 0. We firstly assume that is strictly monotone, i.e., there exists a constant

M > 0 such that ( (x) − (y)) (x − y) ≥ M (x − y)2 , ∀x, y ∈ R.

Fix x ∈ L2(O) and denote (1 + λAα)−1 x = yα and(1 + λAhom

)−1x = yhom.

We can easily see that

1

2

∣∣yα

∣∣2−1 + λ

∫

Oaα (ξ)

(yα (ξ)

)yα (ξ) dξ ≤ 1

2|x |2−1 .

By assumption A2, we have

∣∣yα

∣∣2−1 + 2λρC1

∣∣yα

∣∣m+1Lm+1(O)

≤ |x |2−1

and we obtain that

yα → y, weakly in Lm+1 (O),

yα → y, weakly in H−1 (O).

On the other hand, we have∫

O

∣∣aα (ξ)

(yα (ξ)

)∣∣m+1

m dξ ≤ C(

1 + |x |2−1

)

and consequently

aα (.) (yα (.)

) → χ, weakly in Lm+1

m (O) .

Similarly for equation

{yhom − λ�hom

(yhom

) = x,

hom(yhom

) ∈ H10 (O),

we have that

2λρC1

∣∣∣yhom

∣∣∣m+1

Lm+1(O)≤ |x |2−1

and also that∫

O

∣∣∣hom

(yhom (ξ)

)∣∣∣

m+1m

dξ ≤ C(

1 + |x |2−1

).


Since is strictly monotone, we have that {yα} is bounded in H10 (O). Indeed, we

can define α

(ξ, z) = α (ξ, z) − Mρ2 z, for all (ξ, z) ∈ O×R, which is strictly

monotone with respect to z. Multiplying the equation

yα − λ�α (

ξ, yα)+ �

Mρ

2yα = x

by yα in L2(O), we get that {yα} is bounded in H10 (O). Hence by the Rellich-

Kondrachov theorem, {yα}α is relatively compact in Lq(O), with m + 1 < q < 6 andthus

yα → y strongly in Lq(O), as α → 0.

Now we have to prove that

lim supα→0

∫

Oα

(ξ, yα

)yαdξ ≤

∫

Oχydξ. (46)

We consider the equation

yα − x − λ�α(ξ, yα

) = 0. (47)

By taking the scalar product in H−1(O) with yα we get

⟨yα − x, yα

⟩−1 + λ

∫

Oα

(ξ, yα

)yαdξ = 0.

Passing to the limit as α → 0, we obtain

〈y − x, y〉−1 + λ lim supα→0

∫

Oα

(ξ, yα

)yαdξ ≤ 0.

On the other hand, if we take the scalar product of (47) with y and pass to the limit,we get

〈y − x, y〉−1 + λ

∫

Oχydξ = 0

and this leads to (46).Using [4, Chapter 2, Lemma 1.3], we obtain

limα→0

∫

Oα

(ξ, yα

)yαdξ =

∫

Oχydξ.

Recall from [3, Lemma 2.1] or [26, Chapter 5, Lemma 4.1] that

aα → MY (a) in L∞(O) weak *.

In order to conclude the proof, we still have to show that χ = hom (y). By thedefinition of the subdifferential, we have

∫

Oa

(ξ

α

)(γ(yα)− γ (z)

)dξ ≤

∫

Oα

(ξ, yα

) (yα − z

)dξ, (48)


for all z ∈ Lm+1(O), (see [4, Proposition 2.9]). We prove now that

∫

Oa

(ξ

α

)

γ(yα)

dξ →∫

OMY (a) γ (y) dξ, as α → 0. (49)

Because yα → y, strongly in Lq (O) , as α → 0, we get on a subsequence that

yα → y, a.e. in O, as α → 0

and using the continuity of γ , we obtain

γ(yα) → γ (y) , a.e. in O, as α → 0.

From the Egorov theorem, we have that for every δ > 0, there exists a set Eδ, withthe Lebesgue norm |Eδ| < δ, such that

γ(yα) → γ (y), uniformly on O \ Eδ, as α → 0.

On the other hand, since aα is bounded and since∫

O

∣∣γ(yα)∣∣

qm+1 dξ ≤ C3

(∫

O

∣∣yα

∣∣q dξ + 1

)

≤ C

we have that∫

Eδ

a

(ξ

α

)(γ(yα (ξ)

)− γ (y (ξ)))

dξ

≤ C |Eδ|q−m−1

q

(∫

O

∣∣γ(yα (ξ)

)− γ (y (ξ))∣∣

qm+1 dξ

)m+1q ≤ Cδ.

We consider now∫

Oa

(ξ

α

)

γ(yα (ξ)

)dξ

=∫

O\Eδ

a

(ξ

α

)(γ(yα)− γ (y)

)dξ

+∫

Eδ

a

(ξ

α

)(γ(yα)− γ (y)

)dξ +

∫

Oa

(ξ

α

)

γ (y) dξ.

We obtain that

limα→0

∫

Oa

(ξ

α

)

γ(yα (ξ)

)dξ =

∫

OMY (a) γ (y) dξ.

Passing to the limit for α → 0 in (48), we have∫

OMY (a) (γ (y) − γ (z)) dξ ≤

∫

Oχ (y − z) dξ, for all z ∈ Lm+1(O).


Let D be an arbitrary measurable subset of O and let z be defined by

z (ξ) ={

y0, for ξ ∈ Dy (ξ), for ξ ∈ O\D

.

We obtain∫

D

[MY (a) γ (y (ξ)) − MY (a) γ (y0) − χ (ξ) (y (ξ) − y0)

]dξ ≤ 0,

for all y0 ∈ R and since D is arbitrary, this implies that for a.e. ξ ∈ O, we haveMY (a) γ (y (ξ)) ≤ MY (a) γ (y0) + χ (ξ) (y (ξ) − y0) , for all y0 ∈ R and conse-quently that χ (ξ) = MY (a) ∂γ (y (ξ)) = hom (y (ξ)) for a.e. ξ ∈ O. Now it iseasily seen that the limit y of the sequence {yα} is the solution of the equation

{yhom − λ�hom

(yhom

) = x,

hom(yhom

) ∈ H10 (O),

i.e. y = yhom, and we finally obtain that

yα → y, strongly in H−1(O) as α → 0,

since Lq(O) ⊂ L2(O) ⊂ H−1(O) continuously.The proof is now complete in the case x ∈ L2(O) and strongly monotone.We shall prove now that those additional assumptions can be dispensed with. In

order to do that we first consider the following strictly monotone approximations ofα and hom

αδ (ξ, z) = α (ξ, z) + δz and hom

δ (z) = hom (z) + δz,

for (ξ, z) ∈ O × R and δ > 0. Denote by

Aαδ (z) = −�α

δ (ξ, z) = −�(α (ξ, z) + δz), for z ∈ D (Aα),

Ahomδ (z) = −�hom

δ (z) = −�(hom (z) + δz

), for z ∈ D

(Ahom

),

and

D(

Aαδ

) = {x ∈ H−1 (O) ∩ L1(O); α (., z) + δz ∈ H1

0 (O)},

D(

Ahomδ

)= {

x ∈ H−1 (O) ∩ L1(O); hom (z) + δz ∈ H10 (O)

}.

Considering the approximating equations

yαδ + λAα

δ

(yαδ

) = x

yδ + λAhomδ (yδ) = x

we have for y = yhom

∣∣yα − y

∣∣−1 ≤ ∣

∣yα − yαδ

∣∣−1 + ∣

∣yαδ − yδ

∣∣−1 + |yδ − y|−1, (50)

where yαδ and yδ are solutions corresponding to Aα

δ and resp. Aδ.


The main step of the proof is the convergence of the first term to 0, for δ → 0,

uniformly in α. We take the H−1(O)-scalar product of the difference(yα − yα

δ

)− λ�(α

(ξ, yα

)− α(ξ, yα

δ

)− δyαδ

) = 0

with yα − yαδ and we get that

∣∣yα − yα

δ

∣∣2−1 ≤ δλ

∫

Oyαδ

(yα − yα

δ

)dξ.

Since, by using assumption A2, we have that {yα} and{

yαδ

}are bounded in L2(O)

uniformly in α, then∣∣yα − yα

δ

∣∣−1 → 0, as δ → 0, uniformly in α.

The convergence of the third term of (50) is similar to the one we just showed, andthe second term is in the case with strict monotonicity. So, for all ε > 0, we can choose

δ0 such that the first and the third terms to be less thenε

3. For this δ0 fixed, we choose

α such that the second term is less thenε

3. We can conclude that |yα − y|2−1 < ε.

We consider now x ∈ H−1(O) and using the density of L2(O) in H−1 (O), weapproximate x by a sequence {xn} ⊂ L2(O) such that xn → x strongly in H−1(O).Consider the equations

yαn − λ�α

(ξ, yα

n

) = xn and yn − λ�hom (yn) = xn .

We have∣∣yα − y

∣∣−1 ≤ ∣

∣yα − yαn

∣∣−1 + ∣

∣yαn − yn

∣∣−1 + |yn − y|−1. (51)

By taking the H−1(O)-scalar product of

yαn − yα − λ�

(α

(ξ, yα

n

)− α(ξ, yα

)) = xn − x

with yαn − yα , we obtain that

yαn → yα, strongly in H−1(O)

as n → ∞ uniformly in α. Similarly, we get yn → y, strongly in H−1(O) as n → ∞.So, for all ε > 0, we can choose n0 such that the first and the third term of (51) to beless then ε

3 . For this n0 fixed, we choose α such that the second term is less than ε3 .

We finally obtain

yα → y, strongly in H−1(O) as α → 0.

By the Trotter-type result from the previous section, we get

Xα → Xhom

in L2W

(�; C

([0, T ] ; H−1 (O)

))and the proof is now complete. �


5. Appendix

These results are known, but since we could not find any bibliographic references,we shall briefly present them. In this section, hypotheses Hi , i = 1, 4 and (2) are stillin force.

LEMMA 7. For each f ∈ L2(O), equation

u (ξ) − � (ξ, u (ξ)) = f (ξ) (52)

has a unique solution u ∈ L2(O), such that (., u (.)) ∈ H10 (O) ∩ H2(O).

Proof. For g(ξ, x) = ∫ x0 (ξ, y) dy, for all (ξ, x) ∈ O×R we define Ig : L2(O) →

R by

Ig (u) ={∫

O g (ξ, u (ξ)) dξ, if g (., u (.)) ∈ L1(O),

+∞, otherwise.

From [4, Chapter 2, Proposition 2.9], we know that Ig is well defined, proper, l.s.c.,convex function, and the subdifferential ∂ Ig : L2(O) → L2(O) satisfies

∂ Ig(u) = {v ∈ L2(O); v (ξ) ∈ ∂g (ξ, u (ξ)) , a.e.inO}.

We note that ∂g (ξ, u (ξ)) = (ξ, u (ξ)) , for almost all (ξ, x) ∈ O×R. Using[4, Chapter 2, Theorem 2.1], we have also that ∂ Ig is a maximal monotone subsetof L2(O) × L2(O). Equation 52 can be equivalently written as

u (ξ) − �∂ Ig (u (ξ)) = f (ξ) in O, a.e. ξ ∈ O

and since(∂ Ig

)−1 = ∂ I ∗g where I ∗

g is the conjugate of Ig , we have

∂ I ∗g (v (ξ)) − �v (ξ) = f (ξ) in O, a.e. ξ ∈ O.

By [25, Theorem 2] we get ∂ I ∗g (v (ξ)) = ∂ Ig∗ (v (ξ)) = ∂g∗ (ξ, v (ξ)) = β (ξ, v (ξ))

and so Eq. 52 can be equivalently written as

β (ξ, v (ξ)) − �v (ξ) = f (ξ) in O, a.e. ξ ∈ O.

We consider the Yosida approximation βλ (ξ, v) = 1λ

(v − yλ) (see [12,13]) for v →β (ξ, v) where yλ = (1 + λβ (ξ, .))−1 v, i.e., the solution of

yλ + λβ (ξ, yλ) = v, for a fixed v ∈ L2(O).

According to Corollary 1.1 and Corollary 1.2 from Chapter 2 of [4] this equationhas a solution yλ for all v ∈ L2(O). It is easily seen that the approximating equation

βλ (ξ, vλ (ξ)) − �vλ (ξ) = f (ξ) (53)


has a unique solution vλ ∈ H10 (O) ∩ H2(O) (arguing as in [7] Chapter 3

Theorem 3.5.1).By multiplying (53) by vλ (ξ) and integrating over O, we get that

∫

O|∇vλ (ξ)|2 dξ ≤ 1

2C

∫

O|vλ (ξ)|2 dξ + C

2

∫

O| f (ξ)|2 dξ

where C depends only on O (from |.|L2(O) ≤ C |.|H10 (O)) and this leads to

|vλ|2H10 (O)

=∫

O|∇vλ (ξ)|2 dξ ≤ C (54)

where C is independent of λ. Assumption H3 assures that βλ (., vλ (.)) ∈ H10 (O). By

multiplying (53) by βλ (ξ, vλ (ξ)) and integrating over O we get that

1

2

∫

O|βλ (ξ, vλ)|2 dξ +

∫

O〈∇βλ (ξ, vλ) ,∇vλ〉 dξ = 1

2

∫

O| f |2 dξ.

For the second term of the left side we have∫

O〈∇βλ (ξ, vλ) ,∇vλ〉 dξ =

∫

O

⟨(βλ)ξ (ξ, vλ) ,∇vλ

⟩dξ

+∫

O(βλ)z (ξ, vλ) |∇vλ|2 dξ ≥

∫

O(βλ)ξ (ξ, vλ) ∇vλdξ

(see e.g. [7, (5.11) p. 154 and p. 155, l. 4]]).Using assumption H3 and (54) we get that

∫

O〈∇βλ (ξ, vλ (ξ)) ,∇vλ (ξ)〉 dξ ≥

∫

O

⟨(βλ)ξ (ξ, vλ (ξ)) ,∇vλ (ξ)

⟩dξ

≥ −∣∣∣∣

∫

O

⟨(βλ)ξ (ξ, vλ (ξ)) ,∇vλ (ξ)

⟩dξ

∣∣∣∣ ≥ −C

(|vλ|2H1

0 (O)+ 1

)≥ −C1.

and we obtain∫

O|βλ (ξ, vλ (ξ))|2 dξ ≤ C2

with C, C1 and C2 independent of λ.We have that βλ (., vλ (.)) → χ, weakly in L2(O). It is easily seen that {vλ}λ is

a Cauchy sequence in H10 (O). Then

vλ → v, strongly in H10 (O)

and also that χ = β (., v (.)). The proof of v ∈ H2(O) can be found in [4, Chapter 2,Remark 2.2], and the uniqueness of the solution is obtained by classical arguments(see [4,7,6]). We can now conclude that Eq. 52 has a unique solution u ∈ L2(O) suchthat (., u (.)) ∈ H1

0 (O) ∩ H2 (O). �


PROPOSITION 1. The operator A ⊂ H−1(O) × H−1(O) defined by Au =−� (., u (.)) where

D (A) = {u ∈ L1(O) ∩ H−1(O); ∃ v ∈ H1

0 (O),

v (ξ) ∈ (ξ, u (ξ)) a.e. ξ ∈ O}

is maximal monotone in H−1(O) × H−1(O).

This result can be proved arguing as in Proposition 2.9 and Proposition 2.12 fromChapter 2 of [4].

PROPOSITION 2. The operator A ⊂ L1(O) × L1(O) defined by

Au ={−�v; v ∈ W 1,1

0 (O), v (ξ) = (ξ, u (ξ)) a.e. ξ ∈ O}

where

D (A) ={

u ∈ L1(O); ∃v ∈ W 1,10 (O),

such that �v ∈ L1(O), v (ξ) = (ξ, u (ξ)) a.e. ξ ∈ O}

is m-accretive in L1(O) × L1(O).

This result can be proved arguing as in Proposition 3.10 from Chapter 2 of [4].

Acknowledgments

The author thanks the referee for the constructive comments and suggestions.

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I. CiotirDepartment of Mathematics,Faculty of Economics and BusinessAdministration,“Al.I. Cuza” UniversityBd. Carol no. 9-11, Iasi, România,E-mail: [email protected]

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Convergence of solutions for the stochastic porous media

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