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Electronic Journal of Differential Equations, Vol. 2004(2004), No. 05, pp. 1–30.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu (login: ftp)

PERIODIC AND INVARIANT MEASURES FOR STOCHASTIC WAVE EQUATIONS

JONG UHN KIM

Abstract. We establish the existence of periodic and invariant measures for a semilinear wave equation with random noise. These are counterparts of time-periodic and stationary solutions of a deterministic equation. The key element in our analysis is to prove that the family of probability distributions of a solution is tight.

1. Introduction

We consider the semi-linear wave equation with random noise utt+ 2αut−∆u+βu=f(t, x, u) +

X

k=1

gk(t, x, u)dBk

dt (1.1)

where (t, x)∈(0,∞)×R3,α >0,β >0 are constants, andf, gk’s are given nonlin- ear functions. Bk’s are mutually independent standard Brownian motions. Later on, we will make precise assumptions onf and gk’s. When gk ≡0 for all k, and f(t, x, u) =f1(t, x)− |u|p−1uwith 1≤p≤3, (1.1) is a model equation in nonlinear meson theory. The Cauchy problem and the initial-boundary value problem associ- ated with this deterministic equation have been completely investigated. See Lions [12], Reed and Simon [16], Temam [17], and references therein. It has been also investigated as a model equation whose solutions converge to a global attractor.

For an extensive discussion of the dynamical system associated with this equation, see [17]. On the other hand, the Cauchy problem for the stochastic equation with random noise was discussed by Chow [2], Garrido-Atienza and Real [8] and Pardoux [14]. Crauel, Debussche and Flandoli [4] proved the existence of random attractors for an initial-boundary value problem associated with (1.1). Here our goal is to obtain a periodic measure for (1.1) when the given functions are time-periodic, and an invariant measure when the given functions are independent of time. A proba- bility measureµon the natural function class for (1.1) is called a periodic measure if the initial probability distribution equal toµgenerates time-periodic probability distributions of the solution, and is called an invariant measure if it results in time- invariant probability distributions of the solution. We can handle both the Cauchy

2000Mathematics Subject Classification. 35L65, 35R60, 60H15.

Key words and phrases. Wave equation, Brownian motion, periodic measure, invariant measure, probability distribution, tightness.

c

2004 Texas State University - San Marcos.

Submitted December 15, 2002. Published January 2, 2004.

1

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problem in the whole spaceR3and an initial-boundary value problem in a bounded domain. But we will present full details only for the case of the whole spaceR3, and give a sketch of the procedure for a bounded domain. The case of the whole space is more challenging for lack of compact imbedding of usual Sobolev spaces. For the Cauchy problem for (1.1), Chow [2] used a basic result of Da Prato and Zabczk [6]

for evolution equations. The main difficulty arises from polynomial nonlinearity in the equation. This type of nonlinearity can be handled by the truncation method.

For its use for parabolic equations, see Gy¨ongy and Rovira [9]. The result in [6] is based on the analysis of stochastic convolutions in the frame of semigroup theory.

Our goal is to obtain periodic and invariant measures. For this, we need some basic estimates to ensure tightness of the probability laws. Such estimates can be ob- tained most conveniently in the frequency domain via the Fourier transform. These new estimates can be derived through the representation formula for solutions in the frequency domain. Hence, it seems natural to obtain solutions in the same context, and we will present the proof of existence independently of the previous works. But we will borrow a truncation device from [2].

Da Prato and Zabczyk [6, 7] present some general results on the existence of invariant measures for stochastic semilinear evolution equations, which cover basi- cally two different cases. The first case is the equations with suitable dissipation.

By means of translation of the time variable and two-sided Brownian motions, dissi- pation of the energy results in invariant measures. The second case is the equations associated with compact semigroups. This compactness of semigroups can be used to prove tightness of the probability distributions of a solution, which in turn yields invariant measures. Parabolic equations fall in this category, and so far most of the works on invariant measures for nonlinear equations have been concerned with equations of parabolic type. However, there are other types of equations which are not covered by either of these cases. The equation (1.1) is one of them. The result of [4] combined with that of [5] yields invariant measures for (1.1) in a bounded space domain. According to their method, only additive noise with sufficiently regular coefficients can be handled. Our method can relax such restrictive assumptions; see remarks in Section 6 below. Our method is based upon the works of Khasminskii [11] and Parthasarathy [15]. The idea of [11] was used in [3] for quasilinear para- bolic equations. The main task is to prove tightness of the probability distributions of a solution. We borrow an essential idea from Parthasarathy [15, Theorem 2.2].

But substantial technical adaptation is necessary for our problem. When the space domain is unbounded, we cannot use compact imbedding of usual Sobolev spaces.

For reaction-diffusion equations, Wang [18] overcame this difficulty by approximat- ing the whole space through expanding balls. This idea was also used in Lu and Wang [13]. We will adopt this for our problem.

In section 2, we introduce notation and present some preliminaries for stochastic processes. In Section 3, we prove the existence of a solution and establish some estimates which will be used later. In Section 4, we prove that the family of probability distributions of a solution is tight, and establish the existence of a periodic measure and an invariant measure in Section 5. Finally, we explain how our method can be used for initial-boundary value problem in Section 6.

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2. Notation and Preliminaries

WhenG is a subset ofRn, C(G) is the space of continuous functions on G, and C0(G) is the space of continuous functions with compact support contained in G.

Hm(R3) stands for the usual Sobolev space of order m. For a function in R3, its Fourier transform is given by

f(ξ) =ˆ 1 (2π)3/2

Z

R3

f(x)e−iξ·xdx and the inversion formula is

f(x) = 1 (2π)3/2

Z

R3

fˆ(ξ)eiξ·xdξ, which will be also expressed byf =Fξ−1

. In this context, it is convenient to define the convolution by

f ∗g

(x) = 1 (2π)3/2

Z

R3

f(x−y)g(y)dy.

For function spaces with respect to the variableξin the frequency domain, we use the notationLp(ξ) andHm(ξ) to denoteLp(R3) andHm(R3), respectively.

We recall some properties of Sobolev spaces. For 2 ≤ q ≤ 6, there is some positive constantCq such that

kψkLq(R3)≤CqkψkH1(R3), for allψ∈H1(R3). (2.1) We also have

Lemma 2.1. Suppose0≤s <1 andψ∈L2(R3). If

Z

R3

ψ(x) I−∆

φ(x)dx

≤CkφkH1+s(R3) (2.2) holds for allφ∈C0(R3), for some positive constant C, thenψ∈H1−s(R3).

Proof. By the Parseval’s identity, Z

R3

ψ(x) I−∆

φ(x)dx= Z

R3

ψ(ξ) 1 +ˆ |ξ|2(1−s)/2φ(ξ) 1 +ˆ |ξ|2(1+s)/2 dξ which, with (2.2), yields ˆψ(ξ) 1 +|ξ|2(1−s)/2

∈L2(R3), andkψkH1−s(R3)≤C.

Lemma 2.2. Let 1≤p <3 andq= 3−p2 . Then,

kψ|ψ|p−1kHq(R3)≤CpkψkpH1(R3) (2.3) holds for allψ∈H1(R3), for some positive constant Cp.

Proof. For anyφ∈C0(R3), we see, by (2.1), ψ|ψ|p−1

L2(

R3)≤CkψkpH1(R3) (2.4) and

Z

R3

ψ|ψ|p−1∆φ dx =

p

Z

R3

|ψ|p−1∇ψ· ∇φ dx

(2.5)

≤Cpk∇ψkL2(R3)

|ψ|p−1

L6/(p−1)(R3)k∇φkL6/(4−p)(R3)

≤CpkψkpH1(R3)kφkH(1+p)/2(R3).

(2.3) follows from (2.4), (2.5) and Lemma 2.1.

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Throughout this paper, {Bk(t)}k=1 is a set of mutually independent standard Brownian motions over the stochastic basis{Ω,F,Ft, P} whereP is a probability measure over the σ-algebra F, {Ft} is a right-continuous filtration over F, and F0 contains allP-negligible sets. E(·) denotes the expectation with respect toP. When X is a Banach space, B(X) denotes the set of all Borel subsets of X. For 1 ≤p < ∞, Lp Ω;X

denotes the set of all X-valued F-measurable functionsh such that

Z

khkpXdP <∞.

L Ω;X

is the set of all X-valued F-measurable functions hsuch that khkX is essentially bounded with respect to the measure P. For general information on stochastic processes, see Karatzas and Shreve [10]. We need the following fact due to Berger and Mizel [1].

Lemma 2.3. Leth(t, s;ω)beB([0, T]×[0, T])⊗ F-measurable and adapted to{Fs} ins for eacht. Suppose that for almost allω ∈Ω,his absolutely continuous int, and

Z T 0

Z t 0

∂h

∂t(t, s)

2

dsdt <∞, for almost allω, Z t

0

|h(t, s)|2ds <∞, for almost allω, for eacht. Let

zk(t) = Z t

0

h(t, s)dBk(s), k= 1,2, . . . . Then, it holds that

dzk(t) =h(t, t)dBk(t) +Z t 0

∂h

∂t(t, s)dBk(s) dt.

3. The Cauchy problem We start from the linear problem.

utt+ 2αut−∆u+βu=f +

X

k=1

gk

dBk

dt , (3.1)

u(0) =u0, ut(0) =u1. (3.2) We suppose that (u0, u1) isH1(R3)×L2(R3)-valuedF0-measurable,

(u0, u1)∈L2 Ω;H1(R3)×L2(R3)

(3.3) and thatf, gk’s areL2(R3)-valued predictable processes such that

f, gk ∈L2 Ω;L2(0, T;L2(R3))

, (3.4)

EX

k=1

Z T 0

kgkk2L2(R3)dt

<∞ (3.5)

for each T > 0. By taking the Fourier transform, this problem is transformed in the frequency domain as follows.

ˆ

utt+ 2αˆut+|ξ|2uˆ+βuˆ= ˆf+

X

k=1

ˆ gk

dBk

dt , (3.6)

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ˆ

u(0) = ˆu0, uˆt(0) = ˆu1. (3.7) Let us first consider more restrictive data. So we assume that (ˆu0,uˆ1) is F0- measurable, and

ˆ

u0,uˆ1∈L2(Ω;C0(K)) (3.8) and that ˆf ,gˆk’s are predictable, and

f ,ˆgˆk∈L Ω;C0([0,∞)×K)

(3.9)

X

k=1

kˆgkk2L(Ω;C0([0,∞)×K))<∞ (3.10) where K is a compact subset of R3. Then, the solution ˆuof (3.6) - (3.7) is given by, for eachξ∈R3,

ˆ

u(t, ξ) =e−αtcos(p

|ξ|2+γt)ˆu0(ξ) +e−αtsin(p

|ξ|2+γt) p|ξ|2+γ uˆ1(ξ) +

Z t 0

e−α(t−s)sin p

|ξ|2+γ(t−s) p|ξ|2

fˆ(s, ξ)ds (3.11)

+

X

k=1

Z t 0

e−α(t−s)sin p

|ξ|2+γ(t−s)

p|ξ|2+γ ˆgk(s, ξ)dBk(s)

for allt≥0, for almost allω∈Ω, whereγ=β−α2. Since|ξ|2+γ≤0 is possible, we note that

cos p

|ξ|2+γt

=

X

k=0

(−1)k |ξ|2k t2k (2k)!

sin p

|ξ|2+γt p|ξ|2+γ =

X

k=0

(−1)k |ξ|2k t2k+1 (2k+ 1)! .

One can apply Ito’s formula to (3.6) for each fixedξ, but we first have to find the manner in which ˆudepends onξ. It is easy to see that

e−αtcos p

|ξ|2+γt

, e−αtsin p

|ξ|2+γt p|ξ|2

and their time derivatives are continuous and uniformly bounded for t ≥ 0 and ξ∈K. This fact is used to estimate the terms in the right-hand side of (3.11). But the last term needs some manipulation for a necessary estimate, because it is not a martingale. Thus, we use Lemma 2.3 to write

Jk:=

Z t 0

e−α(t−s)sin p

|ξ|2+γ(t−s)

p|ξ|2+γ gˆk(s, ξ)dBk(s)

= Z t

0

Z s 0

(−α)e−α(s−η)sin p

|ξ|2+γ(s−η)

p|ξ|2+γ ˆgk(η, ξ)dBk(η)ds +

Z t 0

Z s 0

e−α(s−η)cos p

|ξ|2+γ(s−η) ˆ

gk(η, ξ)dBk(η)ds, k= 1,2, . . . ,

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and

tJk= Z t

0

Z s 0

α2e−α(s−η)sin p

|ξ|2+γ(s−η)

p|ξ|2+γ gˆk(η, ξ)dBk(η)ds +

Z t 0

Z s 0

(−α)e−α(s−η)cos p

|ξ|2+γ(s−η) ˆ

gk(η, ξ)dBk(η)ds +

Z t 0

Z s 0

(−α)e−α(s−η)cos p

|ξ|2+γ(s−η) ˆ

gk(η, ξ)dBk(η)ds

− Z t

0

Z s 0

e−α(s−η)p

|ξ|2+γsin p

|ξ|2+γ(s−η) ˆ

gk(η, ξ)dBk(η)ds +

Z t 0

ˆ

gk(s, ξ)dBk(s), k= 1,2, . . . .

These integrals are easy to estimate, and we find that for eachT >0,

ξ1lim→ξ2

E

Jk1)−Jk2)

2 C1([0,T])

= 0. (3.12)

By (3.10) and (3.12), the last term of (3.11) belongs to C0 K;L2(Ω;C1([0, T])) for eachT >0. It is easy to see that other terms also belong to the same function class, and we conclude

ˆ

u∈C0 K;L2(Ω;C1([0, T]))

for all T > 0. Let ˜K be another compact subset whose interior contains K. By partition of unity, ˆucan be approximated inC0 K;˜ L2(Ω;C1([0, T]))

by a sequence of functions of the form

ˆ uN =

N

X

j=1

aN j(ξ)bN j(ω, t) where eachaN j∈C0( ˜K) andbN j∈L2 Ω;C1([0, T])

, andbN j(t) isFt-measurable for everyt. Since each ˆuN ∈L2 Ω;C1([0, T];L2( ˜K))

and kˆuNkL2(Ω;C1([0,T];L2( ˜K)))≤MK˜kˆuNkC

0( ˜K;L2(Ω;C1([0,T])))

where MK˜ is a positive constant depending only on ˜K. It follows that{uˆN} is a Cauchy sequence inL2 Ω;C1([0, T];L2( ˜K))

. Hence, ˆu∈L2 Ω;C1([0, T];L2( ˜K)) . Meanwhile, each ˆuN(t) isL2( ˜K)-valuedFt-measurable and so is ˆu(t). Also, for each t, ˆuN(t) and ˆu(t) areB( ˜K)⊗ Ft-measurable. By Ito’s formula, we find that for each ξ∈K,

|ˆut|2(t) + (|ξ|2+β+ 2α)|ˆu(t)|2+ 2Re uˆt(t)ˆu(t)

=|ˆu1|2+ (|ξ|2+β+ 2α)|ˆu0|2+ 2Re uˆ10

(3.13)

−(4α−2) Z t

0

|ˆut(s)|2ds−2 Z t

0

(|ξ|2+β)|ˆu(s)|2ds +

Z t 0

2Re

fˆ(s)ˆut(s) +fˆ(s)ˆu(s) ds+

X

k=1

Z t 0

|ˆgk(s)|2ds

+

X

k=1

Z t 0

2Re ˆ

gk(s)ˆut(s) +ˆgk(s)ˆu(s) dBk(s)

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holds fort≥0, for almost allω∈Ω. In fact, (3.13) holds inC0 K;L1(Ω;C([0, T]) , for allT >0, because each term belongs to C0 K;L1(Ω;C([0, T])

, for allT >0.

By (3.10), we can apply the stochastic Fubini theorem to the last term, and find that

kˆut(t)k2L2(ξ)+

p|ξ|2+β+ 2αˆu(t)

2

L2(ξ)+ 2 Z

R3

Re uˆt(t)ˆu(t) dξ

=kˆu1k2L2(ξ)+

p|ξ|2+β+ 2αˆu0

2

L2(ξ)+ 2 Z

R3

Re ˆu10

−(4α−2) Z t

0

Z

R3

|ˆut(s)|2dξ ds−2 Z t

0

Z

R3

(|ξ|2+β)|ˆu(s)|2dξds +

Z t 0

Z

R3

2Re

fˆ(s)ˆut(s) +fˆ(s)ˆu(s) dξds+

X

k=1

Z t 0

kˆgk(s)k2L2(ξ)ds

+

X

k=1

Z t 0

Z

R3

2Re ˆ

gk(s)ˆut(s) +ˆgk(s)ˆu(s)

dξ dBk(s) (3.14) holds for allt≥0, for almost all ω. We now choosesuch that

0< < α . (3.15)

By the Burkholder-Davis-Gundy inequality, we have E

sup

0≤s≤t

X

k=1

Z s 0

Z

R3

2Re ˆ

gk(η)ˆut(η) +ˆgk(η)ˆu(η)

dξ dBk(η)

≤M EX

k=1

Z t 0

Z

R3

2Re ˆ

gk(s)ˆut(s) +ˆgk(s)ˆu(s) dξ

2

ds1/2

(3.16)

≤ρE sup

0≤s≤t

ku(s)kˆ 2L2(ξ)+kˆut(s)k2L2(ξ)

+M ρ

X

k=1

EZ t 0

kˆgk(s)k2L2(ξ)ds

, for allρ >0.

Thus, by using (3.16) with suitably smallρ, we can derive from (3.14) E

sup

0≤s≤t

kˆut(s)k2L2(ξ)+kp

|ξ|2+βu(s)kˆ 2L2(ξ)

≤M E

kˆu1k2L2(ξ)+

p|ξ|2+βuˆ0

2 L2(ξ)

(3.17) +M tEZ t

0

kfˆ(s)k2L2(ξ)ds +M

X

k=1

EZ t 0

kˆgk(s)k2L2(ξ)ds

whereM denotes positive constants independent ofK andt≥0. We now consider the general data satisfying (3.3) - (3.5). Let us fix any T > 0. We can choose sequences{ˆu0,n},{ˆu1,n},{fˆn} and{ˆgk,n}such that

p|ξ|2+βˆu0,n→p

|ξ|2+βuˆ0 inL2 Ω;L2(ξ) , ˆ

u1,n→uˆ1 in L2 Ω;L2(ξ) , fˆn →fˆ inL2 Ω;L2(0, T;L2(ξ))

,

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X

k=1

EZ T 0

kˆgk,n(s)−gˆk(s)k2L2(ξ)ds

→0,

and each ˆu0,n, ˆu1,n satisfy (3.8), and each ˆfn,gˆk,n satisfy (3.9), (3.10) with some compact subset Kn. It follows from (3.17) that un, ∂tun

corresponding to ˆ

u0,n,uˆ1,n,fˆn,ˆgk,n forms a Cauchy sequence inL2 Ω;C([0, T];H1(R3)×L2(R3)) , and its limit u, ut

is the solution which also satisfies (3.17). In addition, if we

have

X

k=1

gk

2

L(Ω;L2(0,T;L2(R3)))<∞, for allT >0, (3.18) then (3.14) is also valid. We now suppose that there is another solution u, ut

∈ L2 Ω;C([0, T];H1(R3)×L2(R3))

. Then, u−u, ut−ut

is a solution of the de- terministic wave equation, and the pathwise uniqueness of solution follows directly from the uniqueness result for the deterministic wave equation.

We now summarize what has been established.

Lemma 3.1. Suppose that(u0, u1)satisfies (3.3)and thatf andgk’s satisfy (3.4) and (3.5). Then, there is a pathwise unique solution of (3.1) and (3.2)such that (u(t), ut(t))is aH1(R3)×L2(R3)-valued predictable process and

(u, ut)∈L2

Ω;C [0, T];H1(R3)×L2(R3)

for all T >0. Furthermore, it satisfies (3.17), and if (3.18) holds, (3.14) is also valid.

Next we will consider the semi-linear case.

utt+ 2αut−∆u+βu=f(t, x, u) +

X

k=1

gk(t, x, u)dBk

dt , (3.19)

u(0) =u0, ut(0) =u1. (3.20) Let us suppose that

f(t, x, u) =f1(t, x) +f2(u); (3.21) gk(t, x, u) =gk,1(t, x) +φ(x)gk,2(u); (3.22) φ∈H1(R3)∩L(R3); (3.23) f1∈C([0,∞);L2(R3)), f1(t) =f1(t+L), for allt≥0, (3.24) whereLis a fixed positive number;

f2(0) = 0, kf2(v)−f2(w)kL2(R3)≤Mkv−wkH1(R3) (3.25) for allv, w∈H1(R3), for some positive constantM;

gk,1∈C([0,∞);L2(R3)), gk,1(t) =gk,1(t+L), for allt≥0; (3.26) kgk,1(t)kL2(R3))≤Mk, for allt≥0; (3.27)

|gk,2(y)| ≤M˜k, |gk,2(y)−gk,2(z)| ≤αk|y−z|, for ally, z∈R (3.28)

with

X

k=1

(Mk2+ ˜Mk22k)<∞. (3.29)

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We employ the standard iteration scheme. Let us setu(0)=u0 and letu(n)be the solution of

utt+ 2αut−∆u+βu=f(t, x, u(n−1)) +

X

k=1

gk(t, x, u(n−1))dBk dt , u(0) =u0, ut(0) =u1.

Fix anyT >0. By subtraction, we can obtain an equation satisfied byu(n+1)−u(n). By treatingf(t, x, u(n))−f(t, x, u(n−1)) andgk(t, x, u(n))−gk(t, x, u(n−1)) as given functions, we interpret u(n+1)−u(n), u(n+1)t −u(n)t

as a solution of the linear problem. By the pathwise uniqueness of solution for the linear problem, we can apply the estimate (3.17) with help of (3.25), (3.28) and (3.29) to derive

E sup

0≤s≤t

ku(n+1)t (s)−u(n)t (s)k2L2(R3)+ku(n+1)(s)−u(n)(s)k2H1(R3)

≤M Z t

0

E

ku(n)(s)−u(n−1)(s)k2H1(R3)

ds (3.30)

+MX

k=1

α2kZ t 0

E

ku(n)(s)−u(n−1)(s)k2L2(R3)

ds, for all 0≤t≤T . By induction, we have, for all 0≤t≤T and alln≥1,

E sup

0≤s≤t

ku(n+1)t (s)−u(n)t (s)k2L2(R3)+ku(n+1)(s)−u(n)(s)k2H1(R3)

≤Kntn/n!

for some constantK independent of n andt. Thus, the sequence{(u(n), u(n)t )} is a Cauchy sequence inL2

Ω;C [0, T];H1(R3)×L2(R3)

. The limit u, ut is the solution, and u(t), ut(t)

isH1(R3)×L2(R3)-valuedFt-measurable. Suppose that u, ut

∈L2 Ω;C([0, T];H1(R3)×L2(R3))

is another solution. By subtraction, we can obtain an equation satisfied by v =u−u. By the same argument as for (3.30), we can derive

E sup

0≤s≤t

kvt(s)k2L2(R3)+kv(s)k2H1(R3)

≤M Z t

0

E kv(s)k2H1(R3)

ds+MX

k=1

α2kZ t 0

E kv(s)k2L2(R3)

ds,

for all 0≤t≤T. This yields the pathwise uniqueness. Since T can be arbitrarily large, it follows from the pathwise uniqueness that for almost allω∈Ω,

u, ut

∈C [0,∞);H1(R3)×L2(R3) . We now drop the assumption (3.25) and consider the case

f2(v) =−v|v|p−1, 1≤p <3. (3.31) Borrowing a truncation device from [2], we set, for a positive integerN,

f2,N(v) =−ηN kvkH1(R3)

v|v|p−1

whereηN(y) =η(y/N), η∈C0(R) such that 0≤η(y)≤1, for ally, η(y) = 1, for

|y| ≤2, andη(y) = 0, for |y| ≥3. Then, it follows from (2.1) that f2,N(v1)−f2,N(v2)

L2(

R3)≤CNkv1−v2kH1(R3). (3.32)

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Hence, there is a solution uN of (3.19) - (3.20) withf =f1+f2,N such that for eachT >0,

uN, ∂tuN

∈L2

Ω;C [0, T];H1(R3)×L2(R3) , and for almost allω∈Ω,

uN, ∂tuN

∈C [0,∞);H1(R3)×L2(R3) . Now define

τN = (inf

t:kuN(t)kH1(R3)> N ,

∞, if{t:kuN(t)kH1(R3)> N}=∅ (3.33) so that uN satisfies the original equation withf =f1+f2 for 0≤t ≤τN(ω),for almost allω. ForN1< N2, we set

v(t) =uN1(t∧τN1∧τN2)−uN2(t∧τN1∧τN2).

We then note thatv(t) is the solution of

vtt+ 2αvt−∆v+βv =F(t, x) +

X

k=1

Gk(t, x)dBk dt on the interval [0, τN1∧τN2) satisfyingv(0) = 0, vt(0) = 0 where

F(t, x) =f2,N1 uN1(t∧τN1∧τN2)

−f2,N2 uN2(t∧τN1∧τN2)

, (3.34) Gk(t, x) =φ(x)

gk,2 uN1(t∧τN1∧τN2)

−gk,2 uN2(t∧τN1∧τN2)

. (3.35) We also note that

f2,N1 uN1(t∧τN1∧τN2)

−f2,N2 uN2(t∧τN1∧τN2) L2(

R3)

=

f2 uN1(t∧τN1∧τN2)

−f2 uN2(t∧τN1∧τN2) L2(

R3)

≤CN2p−1kv(t)kH1(R3)

for all 1≤N1 < N2 and all t≥0, for almost all ω. We can treat v as a solution of the linear equation where F and Gk’s are given functions. By the pathwise uniqueness of solution of the linear problem,v must satisfy

E sup

0≤s≤t

kvt(s)k2L2(R3)+kv(s)k2H1(R3)

≤M Z t

0

E kv(s)k2H1(R3)

ds+MX

k=1

α2kZ t 0

E kv(s)k2L2(R3)

ds,

for allt≥0. It follows from the Gronwall inequality thatv(t)≡0, for allt≥0, for almost allω. Thus,τN1≤τN2, for almost allω. Let

τ= lim

N→∞τN and define

u(t) = lim

N→∞uN(t), for 0≤t < τ.

Apparently,u(t∧τN) =uN(t∧τN), for allt≥0 and allN ≥1, and consequently, u, ut

∈C [0, τ);H1(R3)×L2(R3)

, for almost allω.

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On account of (3.22) - (3.23) and (3.26) - (3.29), we can use (3.14) so that for each N ≥1,

kut(t∧τN)k2L2(R3)+k∇u(t∧τN)k2L2(R3)+ (β+ 2α)ku(t∧τN)k2L2(R3)

+ 2hut(t∧τN), u(t∧τN)i

=ku1k2L2(R3)+k∇u0k2L2(R3)+ (β+ 2α)ku0k2L2(R3)+ 2∠u1, u0i

−(4α−2) Z t∧τN

0

kut(s)k2L2(R3)ds−2 Z t∧τN

0

k∇u(s)k2L2(R3)+βku(s)k2L2(R3)

ds + 2

Z t∧τN

0

hf1(s) +f2,N(u(s)), ut(s) +u(s)ids +

X

k=1

Z t∧τN

0

gk,1(s) +φgk,2(u(s))

2 L2(R3)ds +

X

k=1

Z t∧τN

0

2hgk,1(s) +φgk,2(u(s)), ut(s) +u(s)idBk(s)

for allt≥0, for almost allω, whereh·,·iis the inner product inL2(R3). We write Q(t) =kut(t)k2L2(R3)+k∇u(t)k2L2(R3)+ (β+ 2α)ku(t)k2L2(R3)

+ 2hut(t), u(t)i+ 2 p+ 1

Z

R3

|u(t)|p+1dx.

We then have

Q(t∧τN) =Q(0)−(4α−2) Z t∧τN

0

kut(s)k2L2(R3)ds

−2 Z t∧τN

0

k∇u(s)k2L2(R3)+βku(s)k2L2(R3)

ds

−2 Z t∧τN

0

Z

R3

|u(s)|p+1dx ds+ 2 Z t∧τN

0

hf1(s), ut(s) +u(s)ids +

X

k=1

Z t∧τN

0

gk,1(s) +φgk,2(u(s))

2

L2(R3)ds (3.36) +

X

k=1

Z t∧τN

0

2hgk,1(s) +φgk,2(u(s)), ut(s) +u(s)idBk(s) for allt≥0, for almost allω∈Ω. In addition to (3.3), we assume

u0∈Lp+1 Ω;Lp+1(R3) so thatE Q(0)

<∞.

By the same argument as for (3.17), we can derive from (3.36) E

sup

0≤s≤t

Q(s∧τN)

≤E Q(0)

+M tEZ t∧τN

0

kf1(s)k2L2(R3)ds +M t

X

k=1

Mk2+ ˜Mk2kφk2L2(R3)

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for allt≥0 and allN ≥1, for some positive constantM. Thus, for each T >0, E

sup

0≤t≤T

Q(t∧τN)

≤MT, for allN ≥1,

whereMT is a positive constant independent ofN. By the same argument as in [2], this implies thatτN ↑ ∞asN → ∞, for almost allω. Using this fact and Fatou’s lemma, we passN → ∞to arrive at

E sup

0≤t≤T

Q(t)

≤MT. (3.37)

Hence, for eachT >0, we have obtained a solution u, ut

∈L2

Ω;C [0, T];H1(R3)×L2(R3) . Suppose that there is another solution

u, ut

∈L2

Ω;C [0, T];H1(R3)×L2(R3) . We defineτN by (3.33) in terms ofu. Then, by means of

f2 u(t∧τN ∧τN)

−f2 u(t∧τN ∧τN) L2(

R3)

≤CNp−1

u(t∧τN∧τN)−u(t∧τN ∧τN) H1(

R3), we can derive

E sup

0≤t≤T

u(t∧τN∧τN)−u(t∧τN ∧τN)

2 H1(R3)

+

ut(t∧τN∧τN)−ut(t∧τN ∧τN)

2 L2(R3)

= 0 for allN≥1. SinceτN ∧T ↑T, as N → ∞, for almost allω, we have

u, ut

= u, ut

in C [0, T];H1(R3)×L2(R3)

for almost allω. This proves the pathwise uniqueness. Next by passingN → ∞in (3.36), we find

Q(t) =Q(0)−(4α−2) Z t

0

kut(s)k2L2(R3)ds

−2 Z t

0

k∇u(s)k2L2(R3)+βku(s)k2L2(R3)

ds

−2 Z t

0

Z

R3

|u(s)|p+1dx ds

+ 2 Z t

0

hf1(s), ut(s) +u(s)ids+

X

k=1

Z t 0

kgk,1(s) +φgk,2(u(s))k2L2(R3)ds

+

X

k=1

Z t 0

2hgk,1(s) +φgk,2(u(s)), ut(s) +u(s)idBk(s) for allt≥0,for almost allω∈Ω. Thus,Q(t) is a solution of

dQ(t)

dt =−(4α−2)kut(t)k2L2(R3)−2 k∇u(t)k2L2(R3)+βku(t)k2L2(R3)

−2 Z

R3

|u(t)|p+1dx+ 2hf1(t), ut(t) +u(t)i

(13)

+

X

k=1

gk,1(t) +φgk,2(u(t))

2

L2(R3) (3.38)

+

X

k=1

2hgk,1(t) +φgk,2(u(t)), ut(t) +u(t)idBk(t) dt .

Recalling that was chosen by (3.15), we can choose δ = δ(, α, β) > 0, κ = κ(, α, β)>0 so that

−(4α−2)kut(t)k2L2(R3)−2 k∇u(t)k2L2(R3)+βku(t)k2L2(R3)

−2 Z

R3

|u(t)|p+1dx+ 2hf1(t), ut(t) +u(t)i (3.39)

≤ −δQ(t) +κkf1(t)k2L2(R3)

for allt≥0, for almost allω.

LetY(t) be the solution of the following initial value problem.

dY(t)

dt =−δY(t) +κkf1(t)k2L2(R3)+

X

k=1

gk,1(t) +φgk,2(u(t))

2 L2(R3)

+

X

k=1

2hgk,1(t) +φgk,2(u(t)), ut(t) +u(t)idBk(t)

dt , (3.40)

Y(0) =Q(0). (3.41)

It follows from (3.38) and (3.40) thatQ(t)−Y(t) is continuously differentiable in tfor almost allω, and, by (3.39) and (3.40),

d

dt Q(t)−Y(t)

≤ −δ Q(t)−Y(t)

for allt≥0, for almost allω. Hence, by (3.41),Q(t)≤Y(t) for allt, for almost all ω. SinceY(t) can be given by

Y(t) =Q(0)e−δt+κ Z t

0

e−δ(t−s)kf1(s)k2L2(R3)ds +

X

k=1

Z t 0

e−δ(t−s)

gk,1(s) +φgk,2(u(s))

2 L2(R3)ds +

X

k=1

Z t 0

e−δ(t−s)2< gk,1(s) +φgk,2(u(s)), ut(s) +u(s)> dBk(s), we have

Q(t)≤Q(0)e−δt+κ Z t

0

e−δ(t−s)kf1(s)k2L2(R3)ds +

X

k=1

Z t 0

e−δ(t−s)

gk,1(s) +φgk,2(u(s))

2

L2(R3)ds (3.42) +

X

k=1

Z t 0

e−δ(t−s)2hgk,1(s) +φgk,2(u(s)), ut(s) +u(s)idBk(s)

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for allt≥0, for almost allω. By means of (3.23), (3.24), (3.26) - (3.29) and EX

k=1

Z t 0

e−δ(t−s)2hgk,1(s) +gk,2(u(s)), ut(s) +u(s)idBk(s)

= 0, we can derive from (3.42) that

E Q(t)

≤M, for allt≥0. (3.43)

By virtue of (3.15), this implies that E

ut(t)

2

L2(R3)+ u(t)

2 H1(R3)

≤M, for allt≥0. (3.44) For later use, we can further derive

E |Q(t)|2

≤M, for allt≥0, (3.45)

provided (u0, u1)∈ L4 Ω;H1(R3)×L2(R3)

and u0 ∈L2p+2 Ω;Lp+1(R3) . This follows from (3.42) and

E

X

k=1

Z t 0

e−δ(t−s)2hgk,1(s) +gk,2(u(s)), ut(s) +u(s)idBk(s)

2

≤M EX

k=1

Z t 0

e−2δ(t−s) Mk2+ ˜Mk2kφk2L2(R3)

kutk2L2(R3)+2kuk2L2(R3)

ds

≤M, for allt≥0, where (3.44) has been used.

We now state the existence result we have obtained.

Lemma 3.2. Suppose that (u0, u1) is the same as in Lemma 3.1 with additional assumption u0 ∈Lp+1 Ω;Lp+1(R3)

and that f, gk satisfy the conditions (3.21) - (3.24), (3.26) - (3.29) and (3.31). Then, there is a pathwise unique solution of (3.19) and (3.20) such that u(t), ut(t)

is a H1(R3)×L2(R3)-valued predictable process and

(u, ut)∈L2

Ω;C [0, T];H1(R3)×L2(R3)

for allT >0.Furthermore,(3.43)is valid, and if(u0, u1)∈L4 Ω;H1(R3)×L2(R3) andu0∈L2p+2 Ω :Lp+1(R3)

, then (3.45) is also valid.

4. Tightness of probability laws Let u, ut

be the solution of (3.19) - (3.20) in Lemma 3.2. In this section, the goal is to establish tightness of the probability laws for u, ut

.

We will present basic estimates which are necessary for tightness of the prob- ability distributions of a solution. For this, we suppose that the initial value

u0, u1

belongs to L4 Ω;H1(R3)×L2(R3)

and u0 ∈ L2p+2 Ω;Lp+1(R3) . We also retain all other conditions in Lemma 3.2 so that a unique solution u, ut

∈ L4

Ω;C [0, T];H1(R3)×L2(R3)

may exist for all T >0.

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Estimate I. We choose a function χR such that χR(x) = χ(x/R), χ ∈C(R3), and

χ(x) =

(1, |x| ≤1

0, |x| ≥2, 0≤χ(x)≤1, for allx∈R3. We setv= 1−χR

u.Then,v is the solution of vtt+ 2αvt−∆v+βv=F+

X

k=1

Gk

dBk dt v(0) =v0= 1−χR

u0, vt(0) =v1= 1−χR

u1

where

F= 1−χR

f+ 2∇χR· ∇u+u∆χR

Gk= 1−χR gk

By the uniqueness of the solution, v can be obtained as a solution of the linear problem of the form (3.1)–(3.2). Hence, we can apply (3.14) tovso that

kvt(t)k2L2(R3)+k∇v(t)k2L2(R3)+ (β+ 2α)kv(t)k2L2(R3)+ 2hvt(t), v(t)i

=kv1k2L2(R3)+k∇v0k2L2(R3)+ (β+ 2α)kv0k2L2(R3)+ 2hv1, v0i

−(4α−2) Z t

0

kvt(s)k2L2(R3)ds−2 Z t

0

k∇v(s)k2L2(R3)+βkv(s)k2L2(R3)

ds

+ Z t

0

2hF(s), vt(s) +v(s)ids+

X

k=1

Z t 0

kGk(s)k2L2(R3)ds (4.1)

+

X

k=1

Z t 0

2< Gk, vt(s) +v(s)> dBk(s)

for allt≥0, for almost allω, whereh·,·iis the inner product inL2(R3). We write QR(t) =

1−χR ut(t)

2

L2(R3)+

∇ (1−χR)u(t)

2 L2(R3)

+ (β+ 2α)

1−χR

u(t)

2

L2(R3)+ 2h(1−χR)ut(t),(1−χR)u(t)i

+ 2

p+ 1 Z

R3

1−χR2

|u(t)|p+1dx

Then, by the same argument as for (3.42), we derive from (4.1) that QR(t)≤QR(0)e−δt+M

Z t 0

e−δ(t−s)

(1−χR)f1(s)

2 L2(R3)ds +M

R2 Z t

0

e−δ(t−s) k∇uk2L2(R3)+kuk2L2(R3)

ds

+

X

k=1

Z t 0

e−δ(t−s)

1−χR

gk,1(s) +φgk,2(u(s))

2

L2(R3)ds (4.2) +

X

k=1

Z t 0

e−δ(t−s)2

(1−χR) gk,1(s) +φgk,2(u(s)) , (1−χR) ut(s) +u(s)

dBk(s)

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