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CONTROLLABILITY AND QUALITATIVE PROPERTIES OF THE

SOLUTIONS TO SPDES DRIVEN BY BOUNDARY L ´

EVY NOISE

ERIKA HAUSENBLAS, PAUL ANDR´

E RAZAFIMANDIMBY

Abstract. In the present paper we are interested in the qualitative properties of the Markov-

ian semigroups P= (Pt)t≥0associated to the solutions of certain stochastic partial diﬀerential

equations (SPDEs) with boundary noise. We assume that these problems can be written as

an abstract stochastic PDE on a Hilbert space Htaking the following form:

du(t, x) = Au(t, x)dt +B σ(u(t, x) ) dL(t), t > 0;

u(0, x) = x∈H.

(1)

Here Lis a real-valued L´evy process, A:D(H)⊂H→His an inﬁnitesimal generator of

a strongly continuous semigroup, σ:H→Ris a Lipschitz continuous map bounded from

below and above, and B:R→Ha possibly unbounded operator. As typical examples of

such stochastic evolution equation we mainly consider and treat the damped wave equation

and heat equation both driven by boundary L´evy noise.

In this article, we ﬁrst show that, if the system

du(t, x) = Au(t, x)dt +B v(t)dt, t > 0;

u(0, x) = x∈H

(2)

is approximate controllable at time T > 0 with control v, then, under some additional condi-

tions on B,Aand L, the probability measure on Hinduced by u(t, x) at a given time t > 0,

x∈H, is positive on open subsets of H. Secondly, we investigate under which conditions on

the L´evy process Land on the operators Aand Bthe solution to equation (1) is asymptotically

strong Feller. It follows from our results that the wave equation with boundary L´evy noise

has at most one invariant measure which is non–degenerate.

1. Introduction

To present the aim of this paper, let Hbe a real separable Hilbert space and ube the unique

H–valued solution of the inﬁnite dimensional system with L´evy noise, formally written as

du(t, x) = Au(t, x)dt +RRB σ(u(t, x)) z˜η(dz, dt), t > 0,

u(0, x) = x∈H.

(3)

In this equation, A:A:D(H)⊂H→His a linear operator generating a C0–semigroup on

H,B:R→Hand σ:H→Rare mappings satisfying some conditions which we will specify

later. The process η:B(R)× B(R+)→N0∪ {∞} is a compensated Poisson random measure

over a probability space A= (Ω,F,(Ft)t≥0,P) with intensity measure ν. Let P= (Pt)t≥0be

the Markovian semigroup induced on Hby the stochastic process u, i.e.

Ptφ(x) := Eφ(u(t, x)), x ∈H, t > 0, φ ∈Cb(H).(4)

Date: April 3, 2015.

1991 Mathematics Subject Classiﬁcation. 60H07, 60H10, 60H15, 60J75.

Key words and phrases. SPDEs, Poisson Random measures, support theorem, invariant measure, Asymptot-

ically Strong Feller Property.

1

2 E. HAUSENBLAS AND P. RAZAFIMANDIMBY APRIL 3, 2015

A typical example of such an equation is a stochastic partial diﬀerential equation (SPDE) with

boundary noise. The aim of this paper is to verify under which conditions on A,B,σand η

the Markovian semigroup deﬁned in (4) is irreducible. We also discuss the asymptotic strong

Feller property of this Markovian semigroup and its implications.

Regularity properties of the Markovian semigroup associated to a stochastic process play an

important role in studying the long time behavior of the process, especially for the investigation

of the uniqueness of invariant measure. In fact, if the Markovian semigroup is strong Feller

and satisﬁes an irreducible property, then it admits a unique invariant measure. To relax

these conditions Hairer and Mattingly introduced in [17] the so called asymptotic strong Feller

property. In particular, they proved that for the uniqueness of the invariant measure it is

suﬃcient to establish the existence of the invariant measure, some nondengeneracy property

and that the Markovian semigroup is asymptotically strong Feller.

Unlike the case of SPDEs driven by Wiener processes, there are very few works studying

support theorems and the uniqueness of the invariant measure for SPDEs driven by a L´evy

process. One of the ﬁrst results about the uniqueness of an invariant measure for L´evy driven

SPDEs were established in the articles of Chojnowska-Michalik [8, 9]. Fournier [16] investigated

SPDEs driven by space time L´evy noise. Applebaum analyzed in [3] the analytic property of

the generalized Mehler semigroup induced by L´evy noise and in [2] the self-decomposability

of a L´evy noise in Hilbert space. Further works are the two articles of Priola and Zabczyk

[29, 30]. We also refer to [20], [31], [32] for some recent results and review of progress for the

study of the ergodicity of the Markovian semigroup associated to the solution of a L´evy driven

SPDEs. The proofs of the results in [30, 31, 32] rely on the cylindrical and α-stability of the

noise, hence their approach does not cover the case we are treating in this paper.

In the present work we mainly prove that if a certain notion of null controllability is veriﬁed

for system (2), then the Markovian semigroup corresponding to equation (1) is irreducible

and enjoys the asymptotic strong Feller property. We apply our result to stochastic evolution

equations with L´evy boundary noise. For results related to SPDEs with white-noise boundary

condition we refer to [12],[22] and [7]. For stochastic evolution equations driven by Wiener

noise a similar result to ours was established long ago. Indeed, the Markovian semigroup of an

Ornstein-Uhlenbeck is irreducible and strong Feller if (2) is null controllable. For this result

we refer to the books of Da Prato and Zabczyk [12] and [14] and references therein. But their

result tells us nothing about the property of the Markovian semigroup when we consider an

Ornstein-Uhlenbeck driven by pure jump noise. Hence our work is an extension of some results

in [12] and [14] in the sense that we consider multiplicative and pure jump noise.

The structure of the paper is as follows. In Section 2 we give most of the hypotheses used

throughout the paper and prove an important relation between the irreducibility property and

(approximate) controllability. In particular, we prove in Section 2 that any ball centered at

x∈Hhas positive measure with respect to the law of the solution to (1) if (2) is approximate

controllable. Section 3 is devoted to the proof of the uniqueness of the invariant measure

of the Markovian semigroup associated to the solution of (3). In fact, we establish that the

Markovian semigroup satisﬁes the asymptotic strong Feller property if Aand Bgenerate an null

controllable system. The asymptotic strong Feller and the irreducibility of the aforementioned

semigroup implies the uniqueness of the invariant measure. To illustrate our results, we apply

them in Section 4 and Section 5 to a damped wave equation and a heat equation driven by

boundary L´evy noise, respectively. The last part of the paper consists of some appendices

collecting technical results about the change of measure. The proofs of our results are a

April 3, 2015 3

combination of the change of measure formula given by Bismuth, Graveraux and Jacod [5] and

Sato [33] (see also [18]) and the method used by Maslowski and Seidler [23].

Notation 1. Let R+:= (0,∞),R+

0:= [0,∞),R−:= (−∞,0),N0:= N∪ {0}and ¯

N:= N0∪

{∞}. Let (Z, Z)be a measurable space. By M+(Z)we denote the family of all positive measures

on Z, by M+(Z)we denote the σ-ﬁeld on M+(Z)generated by functions iB:M+(Z)∋µ7→

µ(B)∈R+

0,B∈ Z. By MI(Z)we denote the family of all σ–ﬁnite integer valued measures

on Z, by MI(Z)we denote the σ-ﬁeld on MI(Z)generated by functions iB:MI(Z)∋µ7→

µ(B)∈¯

N,B∈ Z. By M+

σ(Z)we denote the set of all σ–ﬁnite and positive measures on Z, by

M+

σ(Z)we denote the σ-ﬁeld on M+

σ(Z)generated by functions iB:M+

σ(Z)∋µ7→ µ(B)∈R,

B∈ Z. We denote by B(Z)the set of all Borel measurable, real-valued bounded functions.

For a Hilbert space Hwe denote by Cb(H)the space of all uniformly continuous and bounded

mappings φ:H→Rendowed with the norm |φ|∞:= supx∈H|φ(x)|.

The space of bounded linear maps from a Banach space Xinto another Banach space Yis

denoted by L(X, Y ).

Throughout the paper λwill denote the Lebesgue measure.

2. Irreducibility of the Markovian semigroup associated to equation (1)

In this section we will deﬁne the setting in which the results can be formulated. We begin

with the deﬁnition of a time homogenous (compensated) Poisson random measure.

Deﬁnition 2.1. Let (Z, Z, ν)be a measurable space with a σ-ﬁnite measure ν, and (Ω,F,F,P)

be a ﬁltered probability space with right continuous ﬁltration F= (Ft)t≥0. A time homogeneous

Poisson random measure ηon (Z, Z)over (Ω,F,F,P), is a measurable function η: (Ω,F)→

(MI(Z×[0,∞)),MI(Z×[0,∞))), such that

(i) η(∅ × I) = 0 a.s. for I∈ B([0,∞)) and η(A× ∅) = 0 a.s. for A∈ Z;

(ii) for each B×I∈ Z × B([0,∞)),η(B×I) := iB×I◦η: Ω →¯

Nis a Poisson random

variable with parameter1ν(B)λ(I).

(iii) ηis independently scattered, i.e. if the sets Bj×Ij∈ Z × B([0,∞)),j= 1,···, n,

are pairwise disjoint, then the random variables η(Bj×Ij),j= 1,···, n are mutually

independent.

(iv) for each U∈ Z, the ¯

N-valued process (N(t, U))t>0deﬁned by

N(t, U ) := η(U×(0, t]), t > 0

is (Ft)t≥0-adapted and its increments are independent of the past, i.e. if t > s ≥0, then

N(t, U )−N(s, U) = η((s, t]×U)is independent of Fs.

The measure νdeﬁned by

ν:Z ∋ A7→ Eη(A×(0,1]) ∈¯

N

is called the intensity of η.

We denote by ˜ηthe compensated Poisson random measure deﬁned by

˜η(U×I) = η(U×I)−ν(U)λ(I), U ×I∈ Z × B([0,∞)).

If the intensity of a Poisson random measure is a L´evy measure2, then one can construct from

the Poisson random measure a L´evy process. Conversely, tracing the jumps, one can ﬁnd a

1If ν(B)λ(I) = ∞, then obviously η(B×I) = ∞a.s..

2A L´evy measure on Ris a σ–ﬁnite measure such that ν({0}) = 0 and RR(|z|2∧1)ν(dz)<∞.

4 E. HAUSENBLAS AND P. RAZAFIMANDIMBY APRIL 3, 2015

Poisson random measure associated to each L´evy process. For more details on this relationship

we refer to [1, 6].

Let A= (Ω,F,F,P) be a complete probability space with right continuous ﬁltration F=

(Ft)t≥0,ηbe a time homogeneous Poisson random measure on Rover Awith intensity νbeing

a L´evy measure and compensator γνdeﬁned by

γν:B(R)× B([0,∞)) ∋(A×I)7→ γν(A×I) := ν(A)λ(I)∈R+

0.

Our assumptions are given below.

Hypothesis 1. We assume that the L´evy measure νis symmetric and has a density kwith

respect to the Lebesgue measure λ. Furthermore, we assume that for any ρ > 0the density kis

bounded on the interval (ρ, ∞), and that there exist an index α∈(1,2), constants K1, K2>0

and ρ0>0such that

K1|ρ|−α−1≤k(ρ)≤K2|ρ|−α−1,for all |ρ| ≤ ρ0.

Hypothesis 2. We assume that

Z|ρ|>ρ0|z|2ν(dz)<∞.

Remark 2.2. Note that Hypothesis 1 and Hypothesis 2 imply

(5) ZR|z|2ν(dz)<∞.

Hypothesis 3. Let H1be a separable Hilbert space such that the embedding H ֒→H1is

continuous. We assume that B:R→H1is a linear operator. We suppose that Ais the

inﬁnitesimal generator of a C0-semigroup {etA, t ≥0}on Hand for any T > 0there exists a

constant KTsuch that

(6) ZT

0ke(T−s)ABk2

L(R,H)ds < KT.

Hypothesis 4. Let σ:H→Rbe a Lipschitz continuous mapping which is bounded from below

and above, i.e., there exist positive constants Cσ,ℓsuch that

Cσ<|σ(u)| ≤ ℓ,

for any u∈H.

By u(t, x) we denote the solution of the following stochastic evolution equation

du(t, x) = Au(t, x) + RRBσ(u(t, x)) z˜η(dz, dt), t > 0,

u(0, x) = x∈H.

(7)

Typical examples of such system are SPDEs with boundary noise and are presented in the

following examples (for more details we refer to Section 4 and Section 5).

Example 2.3. We consider the vibration of a string of length 2πwhere one end is ﬁxed and

the other end is perturbed by a L´evy noise. To be more precise, let T > 0,˜ρ∈Rand consider

April 3, 2015 5

the system

(8)

utt(t, ξ)−uξξ (t, ξ ) + ˜ρut(t, ξ) = 0, t ∈(0, T ), ξ ∈(0,2π),

u(t, 0) = 0, t ∈(0, T ),

uξ(t, 2π) = 2 + sin(|u(t)|L2(O))˙

Lt, t ∈(0, T ),

u(0, ξ) = x0(ξ), ut(0, ξ) = x1(ξ), ξ ∈(0,2π),

where ˙

Lis the Radon Nikodym derivative of a real valued L´evy process with intensity measure

ν,x0∈H1

0(0,2π)and x1∈L2(0,2π).

Example 2.4. We consider the temperature of a one–dimensional rod of length 2πexposed

to a heat source with random ﬂuctuations on one side. To model this random ﬂuctuations a

L´evy noise is added at the boundary ξ= 2π, while the boundary ξ= 0 is ﬁxed to have zero

temperature. To be more precise, let T > 0and consider the system

(9)

ut(t, ξ)−uξξ (t, ξ) = 0, t ∈(0, T ), ξ ∈(0,2π),

u(t, 0) = 0, t ∈(0, T ),

uξ(t, 1) = ˙

Lt, t ∈(0, T ),

u(0, ξ) = x0(ξ), ξ ∈(0,2π).

Here x0∈L2(0,2π)and ˙

Lis the Radon–Nikodym derivative of a real valued L´evy process with

intensity measure ν.

The Example 2.3 and Example 2.4 are of diﬀerent type and we show their well-posedness,

i.e., the existence and uniqueness of their solutions in Section 4 and Section 5. In these sections

we also show how these examples can be written as an abstract evolution equation of type (7).

For the time being, we assume that they can be written as an abstract evolution equation of

the form (7) and satisfy the hypothesis listed before. Moreover, we assume that each of them

has a unique mild solution that is deﬁned in the next deﬁnition.

Deﬁnition 2.5. By a solution of (7) we mean a H-valued3and progressively measurable process

such that

PZt

0ZZe(t−s)ABσ(u(s, x))z

pν(dz)ds < ∞= 1,∀t≥0,

and satisfying the following equation P-a.s.

(10) u(t, x) = etAx+Zt

0ZZ

e(t−s)ABσ(u(s, x))z˜η(dz, ds),∀t≥0.

Before continuing we introduce some deﬁnitions from control theory.

Deﬁnition 2.6.

(i) We say that the system

˙uc(t, x, v) = Auc(t, x, v) + Bv(t), t ≥0,

uc(0, x, v) = x,

(11)

is null controllable at time Tiﬀ for any x∈Hthere exists a control v∈L2(0, T ;R)

such that uc(T , x, v) = 0.

3A process uis H–valued, iﬀ for all t≥0, u(t) is an H–valued random variable.

6 E. HAUSENBLAS AND P. RAZAFIMANDIMBY APRIL 3, 2015

(ii) We say that the system (11) is exactly controllable at time Tin x∈Hiﬀ for each

y∈Hthere exists a control v∈L2(0, T ;R)such that uc(T , x, v) = y.

(iii) We say that the system (11) is approximate controllable at time Tin x∈Hiﬀ for each

y∈Hand ǫ > 0there exists a control v∈L2(0, T ;R)such that |uc(T , x, v)−y| ≤ ǫ.

Remark 2.7.

•The system (11) associated to the wave equation with boundary control described in

Example 2.3 is exactly controllable at time T≥2π(Zuazua [35], or [34, Proposition

10.2.3, p. 325]);

•The system (11) associated to the heat equation with Neumann boundary control de-

scribed in Example 2.4 is approximate controllable at time T > 0(Laroche, Martin and

Rouchon [21], see also [4, Theorem 5.3, p. 337]). Moreover, it is null-controllable for

any T > 0(see [34, Example 11.2.5, p. 360]).

For C > 0 and y∈H, we set

DH(y, C ) := {z∈H:|z−y| ≤ C},

and, for simplicity, we just write DH(C) for DH(0, C ). Let ube the solution of the stochastic

evolution equation

du(t, x) = Au(t, x)dt +RRB z˜η(dz, dt)

u(0, x) = x∈H.

(12)

Then, we have the following theorems.

Theorem 2.8. Let us assume that Hypothesis 1 to Hypothesis 3 are satisﬁed. In addition,

we assume that the system (11) is approximate controllable at time T > 0. Let ube a unique

solution of Eq. (12). Then for any x, y ∈Hand δ > 0there exists a number κ=κ(x, y, δ)>0

such that

P(u(T, x)∈ DH(y, δ)) ≥κ.(13)

In case the system is exactly controllable the result of the above theorem can be strengthen

as follows.

Theorem 2.9. Let us assume that Hypothesis 1 to Hypothesis 3 are satisﬁed. In addition, we

assume that the system (11) is exactly controllable at time T > 0. Let ube a unique solution of

Eq. (12). Then for all C > 0, for all x0, y ∈H, and all δ > 0there exists κ=κ(x0, C, y, δ)>0

such that we have for all x∈ DH(x0, C)

P(u(T, x)∈ DH(y, δ)) ≥κ.

One gets a similar result, if the system (11) is only null controllable.

Theorem 2.10. Let us assume that Hypothesis 1 to Hypothesis 3 are satisﬁed. In addition,

we assume that the system (11) is null controllable. Let ube a unique solution of Eq. (12).

Then for all C > 0, all δ > 0there exists a number κ=κ(δ, C)>0such that all x∈ DH(C)

we have

P(u(T, x)∈ DH(δ)) ≥κ.

April 3, 2015 7

Let ube the solution of the stochastic evolution equation

du(t, x) = Au(t, x)dt +RZB σ(u(t, x)) z˜η(dz, dt)

u(0, x) = x.

(14)

We will prove the following result later.

Theorem 2.11. Let us assume that Hypothesis 1 to Hypothesis 4 are satisﬁed. In addition,

we assume that the system (11) is approximate controllable at time T > 0. Let ube a unique

solution of Eq. (14). If uis c`adl`ag in H, then for any δ > 0and for all x, y ∈H, there exists

a number κ=κ(x, y, δ)>0such that

P(u(T, x)∈ DH(y, δ)) ≥κ.(15)

Again, in the case when the system is exactly controllable the result of the above theorem

can be strengthen as follows.

Theorem 2.12. Let us assume that Hypothesis 1 to Hypothesis 4 are satisﬁed. In addition,

we assume that the system (11) is exactly controllable at time T > 0. Let ube a solution of

Eq. (14). If uis c`adl`ag in H, then for all C > 0, for all x0, y ∈H, and all δ > 0there exists

κ=κ(x0, C, y, δ)>0such that for all x∈ DH(x0, C )we have

P(u(T, x)∈ DH(y, δ)) ≥κ.

If the system is only null controllable at time T, then again the disk DH(y, C ) centered at

any point y∈Hhas to be replaced by a disc DH(C) centered at the point 0. Since this claim

is important for our analysis we state and prove it in the following theorem.

Theorem 2.13. Let us assume that Hypothesis 1 to Hypothesis 4 are satisﬁed. In addition, we

assume that the system (11) is null controllable at time T > 0. Let ube a H-c`adl`ag solution

of Eq. (14). Then for all C > 0, and all δ > 0there exists a number κ=κ(C, δ)>0such that

for all x∈ DH(C)we have

P(u(T, x)∈ DH(δ)) ≥κ.(16)

As mentioned in the introduction, we illustrate the applicability of these theorems to the

Wave and Heat equations with L´evy boundary noises. In particular, we will give the detailed

mathematical treatment of Example 2.3 and Example 2.4 in Section 4 and Section 5, respec-

tively.

Proof of Theorem 2.8. For technical reasons we will switch to another representation of the

Poisson random measure. Let A= ( ¯

Ω,¯

F,¯

F,¯

P) be a ﬁltered probability space with ﬁltration

¯

F= ( ¯

Ft)t≥0and let µbe a Poisson random measure on R×[0, T ] over Ahaving the Lebesgue

measure λas its intensity measure. The compensator of µis denoted by γand given by

B(R)× B([0,∞)) ∋A×I7→ γ(A×I) := λ(A)λ(I).

Let U: (0,∞)→R+denote the tail integral U(r) = R∞

rk(s)ds. Observe that Uis a strictly

decreasing and continuous function. Let c:R+→R+be its inverse given by

c:R+∋r7→ c(r) := sup

κ>0{U(κ)≥r}if r > 0.(17)

8 E. HAUSENBLAS AND P. RAZAFIMANDIMBY APRIL 3, 2015

Since kis symmetric, ccan be extended to the negative numbers by putting c(−z) = −c(z),

z∈R+and c(0) = 0. Note, that hypothesis 1 implies, that there exists an index α∈(1,2),

constants K1, K2>0 and r0>0 such that

K1|z|−1

α≤c(z)≤K2|z|−1

α,for all |z| ≥ r0.

In what follows we assume that cis a function on R.

Let

L(t) := Zt

0ZR

z˜η(dz, ds), t ≥0,

and

Lc(t) := Zt

0ZR

c(z) ˜µ(dz, ds), t ≥0.

By [1, Theorem 2.3.7] we have

¯

Eexp hiy Zt

0ZR

c(z)˜µ(dz, ds)i= exptZR

(eiyz −1)λc(dz),

where λc=λ(c−1(B)) for each B∈ B(R). Thanks to the deﬁnition of c(·) we can deduce that

¯

Eexp hiy Zt

0ZR

c(z)˜µ(dz, ds)i= exptZR

(eiyz −1)k(z)dz

=Eexp hiy Zt

0ZR

z˜µ(dz, ds)i.

Thus, the processes L={L(t) : 0 ≤t < ∞} and Lc={Lc(t) : 0 ≤t < ∞} have the same law.

Now, the stochastic evolution equation given in (12) reads as follows on A

du(t, x) = Au(t, x)dt +RRB c(z)˜µ(dz, dt),

u(0, x) = x∈H.

(18)

Fix y∈H,δ > 0, T > 0 and x∈H. In order to prove Theorem 2.8 we need a result from

control theory. Given v∈L2([0,∞); R), let ucbe the solution to

duc(t, x, v) = Auc(t, x, v)dt +Bv(t)dt, t ≥0,

uc(0, x, v) = x.

(19)

Since the system (19) is supposed to be approximate controllable, there exists v∈L2(0, T ;R)

such that

|uc(T, x, v)−y| ≤ δ

3.(20)

Now, let ˜

R > 0 be as in Corollary A.4-(2) and choose R > ˜

R∨r0such that

3

δ2

C T R1−2

α≤1

2.(21)

Here Cis a generic constant which does not depend on δ,Tand R(see (30)). Next we set

gR=ZDR(R)

c(z)λ(dz).(22)

April 3, 2015 9

Let θ(R): [0, T ]×R→Rbe a transformation such that

−(v(s) + gR) = ZRhc(z)−c(θ(R)(s, z))iλ(dz), s ∈[0, T ].(23)

Observe, that for z∈ DR(R) we have c(θ(R)(s, z)) = c(z). The existence of such a transforma-

tion is given by Corollary A.3 and Corollary A.4. Now, Corollary A.3 implies that for any

R≥r0there exists ϑR:R×R\{0} → Rsuch that

−(v(s) + gR) = ZRc(z)−c(ϑR(−(v(s) + gR), z))λ(dz), s ∈[0, T ].

Setting

θ(R)(s, z) := ϑR(v(s) + gR, z),

it is easy to see that for any s∈[0, T ] and z∈R\{0}the transformation θ(R)satisﬁes (23).

In addition, due to Corollary A.4, we can assume that ρ(R)

z(x, z)| ≤ 1

2, for all z∈R\ {0}and

x∈R. For simplicity, we will write θinstead of θ(R)throughout this proof.

Let µθbe a random measure deﬁned by

µθ:¯

Ω× B(R)⊗ B([0, T ]) ∋(ω, A ×I)7→ ZRZI

1A(θ(s, z))µ(ω, dz , ds).

Remark 2.14. Note, that due to the fact that µis a (Poisson) random measure depending on

ω∈Ω, also µθdepends on ω. Moreover, µθis not necessarily time homogenous on A, i.e. it is

a random measure but not necessarily a Poisson random measure on A.

Let Qθbe the probability measure on Asuch that µθhas compensator γ. The existence of

such probability measure Qθis ensured by Lemma B.1. Under Qθ, the process uθ

µdeﬁned by

duθ

µ(t, x) = Auθ

µ(t, x)dt +RRBc(z) (µθ−γ)(dz, dt)

uθ

µ(0, x) = x,

(24)

has the same law as u. In particular, we have

E¯

P1[0,δ](|u(T, x)−y|)=EQθh1[0,δ](|uθ

µ(T, x)−y|)i.

On the other hand we know that under ¯

Pthe process uθ

µsolves the following stochastic

evolution equation

(25)

duθ

µ(t, x) = Auθ

µ(t, x)dt +RRB[c(θ(t, z)) −c(z)] (µ−γ)(dz, dt)

+RRB[c(θ(t, z)) −c(z)] γ(dz, dt) + RRBc(z)(µ−γ)(dz, dt),

uθ

µ(0, x) = x.

Hence we can write

P(|u(T, x)−y| ≤ δ) = Qθ(|uθ

µ(T, x)−y| ≤ δ) = EQθh1|uθ

µ(T,x)−y|≤δi

=E¯

PhGθ(T)1|uθ

µ(T,x)−y|≤δi,

where G(T) is such that dQθ=G(T)d¯

P. The existence of G(T) is again ensured by Lemma

B.1. Now, let X= 1|uθ

µ(T,x)−y|≤δand Y=Gθ(T). By H¨older’s inequality we have

E¯

Ph(X·Y)1

2Y−1

2i2≤E¯

P[X·Y]E¯

P[Y−1].

10 E. HAUSENBLAS AND P. RAZAFIMANDIMBY APRIL 3, 2015

Hence,

E¯

P[X·Y]≥E¯

P[X1

2]2

E¯

P[Y−1],

from which we infer that

P(|u(T, x)−y| ≤ δ)≥E¯

Ph1|uθ

µ(T,x)−y|≤δi2

E¯

P|G−1

θ(T)|.(26)

In what follows we will prove that the right hand side of the above inequality is bounded

from below by a positive constant. For this aim, we will ﬁrst give an upper estimate of

E¯

P|H(t)|=E¯

P|G−1

θ(T)|, and then give a lower estimate of E¯

Ph1|uθ

µ(T,x)−y|≤δi.

By Remark B.3 the process Hdeﬁned by H(t) := G−1(t), t≥0, solves the stochastic

diﬀerential equation

dH(t) = −H(t−)RR

θz(t,z)−1

θz(t,z)(µ−γ)(dz, dt) + H(t−)RR

[θz(t,z)−1]2

θz(t,z)γ(dz, dt),

H(0) = 1.

Let θzbe the partial derivative of θwith respect to to z. By Corollary A.4-(2) we have

1

2≤θz≤3

2for the choice of Ras above. Since His of ﬁnite variation we see that

E¯

P|H(t)| ≤ 1 + 2 Zt

0ZR

E¯

P|H(s)||θz(s, z)−1|dz ds + 2 Zt

0ZR

E¯

PH(s)|θz(s, z)−1|2dz ds

≤1 + 2 Zt

0ZR

E¯

P|H(s)||θz(s, z)−1|dz ds + 3 Zt

0ZR

E¯

PH(s)|θz(s, z)−1|dz ds.

Since, by Corollary A.4-(1), there exists C(R)>0 such that

Z∞

−∞ |θz(s, z)−1|dz ≤C(R)|v(s)|2,

we have

E¯

P|H(t)| ≤ 1 + 2C(R)Zt

0

E¯

P|H(s)||v(s)|2ds + 2C(R)Zt

0

E¯

P|H(s)||v(s)|2ds.(27)

Note that the function vis given by the control problem (19) and is not a random function.

Now, Gronwall’s inequality implies

(28) E¯

P|G−1

θ(T)|≤C′·exp C(R)ZT

0|v(s)|2ds.

Next, by inequality (20) we deduce that

E¯

Ph1|uθ

µ(T,x)−y|≤δi

≥E¯

Ph1|uc(T,x,v )−y|≤ δ

31|uc(T,x,v )−uθ

µ(T,x)|≤ δ

3i=E¯

Ph1|uc(T,x,v )−uθ

µ(T,x)|≤ δ

3i.(29)

By setting ∆(T) = uθ

µ(T, x)−uc(T , x, v) we have

E¯

Ph1|uθ

µ(T,x)−y|≤δi≥E¯

Ph1|∆(T)|≤ δ

3i.

April 3, 2015 11

It is clear that

∆(T) = ZT

0ZR

e(T−s)AB[c(θ(s, z)) −c(z)] (µ−γ)(dz, ds)

+ZT

0ZR

e(T−s)ABc(z) (µ−γ)(dz, ds)

+ZT

0ZR

e(T−s)AB[c(θ(s, z)) −c(z)] γ(dz, dt)−ZT

0

e(T−s)Av(s)ds

=ZT

0ZR

e(T−s)AB[c(θ(s, z)) −c(z)] γ(dz, dt)−ZT

0

e(T−s)Av(s)ds

+ZT

0ZR

e(T−s)ABc(θ(s, z))(µ−γ)(dz, ds).

From the last line and (23) we deduce that

∆(T) = ZT

0ZR

e(T−s)ABc(θ(s, z))(µ−γ)(dz, ds) + ZT

0

e(T−s)ABgRds.

Now we proceed by giving a lower estimate of ¯

P|∆(T)| ≤ δ

3. For this aim, we apply the Bayes

Theorem and get

¯

P|∆(T)| ≤ δ

3=¯

P(µ(DR(R)×[0, T ]) = 0)

×¯

P ZT

0ZR\DR(R)

e(T−s)ABc(θ(s, z))(µ−γ)(dz, ds)

+ZT

0ZDR(R)

e(T−s)ABc(θ(s, z))(µ−γ)(dz, ds)

+ZT

0

e(T−s)ABgRds≤δ

3µ(DR(R)×[0, T ]) = 0!

+¯

P(µ(DR(R)×[0, T ]) >0)

×¯

PZT

0ZR

e(T−s)ABc(θ(s, z))(µ−γ)(dz, ds)

+ZT

0

e(T−s)ABgRds≤δ

3µ(DR(R)×[0, T ]) >0

Since c(θ(s, z)) = c(z) on DR(R)×[0, T ], we have

ZT

0ZDR(R)

e(T−s)ABc(θ(s, z))(µ−γ)(dz, ds) = −ZT

0ZDR(R)

e(T−s)ABc(θ(s, z))λ(dz)ds

=−ZT

0

e(T−s)ABgRds,

12 E. HAUSENBLAS AND P. RAZAFIMANDIMBY APRIL 3, 2015

on the event µ(DR(R)×[0, T ]) = 0. Thus,

¯

P|∆(T)| ≤ δ

3

≥¯

P ZT

0ZR\DR(R)

e(T−s)ABc(θ(s, z))(µ−γ)(dz, ds)≤δ

3µ(DR(R)×[0, T ]) = 0!

×¯

P(µ(DR(R)×[0, T ]) = 0) .

Since the random variables µ(DR(R)×[0, T ]) and µ(A×[0, T ]) for all A∈ B(R\ DR(R)) are

independent, it follows that

¯

P|∆(T)| ≤ δ

3≥¯

P ZT

0ZR\DR(R)

e(T−s)ABc(θ(s, z))(µ−γ)(dz, ds)≤δ

3!

×¯

P(µ(DR(R)×[0, T ]) = 0) .

By the Chebyscheﬀ inequality we know that

¯

PRT

0RR\DR(R)e(T−s)ABc(θ(s, z))(µ−γ)(dz, ds)≥δ

3

≤3

δ2×E¯

PRT

0RR\DR(R)e(T−s)ABc(θ(s, z))(µ−γ)(dz, ds)

2.

It follows from the Burkholder inequality and the inequality |c(θ(t, z))|≤ |c(z)|(see the con-

struction of θ) that

E¯

P

ZT

0ZR\DR(R)

e(T−s)ABc(θ(s, z))(µ−γ)(dz, ds)

2

≤"ZT

0ZR\DR(R)ke(T−s)ABk2

L(R,H)|c(θ(s, z))|λ(dz)ds#

≤"ZT

0ZR\DR(R)ke(T−s)ABk2

L(R,H)|c(z)|λ(dz)ds#≤C T R1−2

α.

Therefore, collecting all together we obtain that

E¯

Ph1|uc(T,x,v )−uθ

µ(T,x)|≤ δ

3i≥1

C(R) 1−3

δ2

C T R1−2

α!.

Since Ris chosen in such a way that

3

δ2

C T R1−2

α≤1

2

(30)

and using the fact that

¯

P(µ(DR(R)×[0, T ]) = 0) = e−λ(DR(R))T

April 3, 2015 13

we get

E¯

Ph1|uθ

µ(T,x)−y|≤δi≥E¯

Ph1|∆(T)|≤ δ

3i≥¯

P(µ(DR(R)×[0, T ]) = 0)

× 1−¯

P ZT

0ZR\DR(R)

e(T−s)ABc(θ(s, z))(µ−γ)(dz, ds)≥δ

3µ(DR(R)×[0, T ]) = 0!!

≥e−λ(DR(R))TC(R) 1−3

δ2

C T R1−2

α!≥1

2e−λ(DR(R))TC(R).

Hence, we have shown that

¯

P(|u(T, x)−y| ≤ δ)≥1

4e−2λ(DR(R))TC(R) exp(−|v|2

L2([0,T ];R)),(31)

which completes the proof of Theorem 2.8.

Proof of Theorem 2.9. Let x∈ DH(x0, C ) and ucthe solution to

˙uc(t, x, v) = Auc(t, x, v) + Bv(t), t ≥0,

uc(0, x, v) = x,

(32)

In order to prove Theorem 2.9, we have to show that the RHS of inequality (31) can be

estimated from below for all x∈ DH(x0, C). That is, we have to check that for any y∈H,

there exists a constant K > 0 such that for any x∈ DH(x0, C) and control vx,y ∈L2(0, T ;R)

with uc(T, x, vx,y ) = y, we have

|vx,y|L2(0,T ;R)≤K, x ∈ DH(x0, C ).

To be more precise, let us deﬁne the linear mapping ΦT:L2(0, T ;R)→Hby

ΦT(v) = ZT

0

e(T−s)ABv(s)ds.

Now, the exact controllability of system (32) is equivalent to the fact that Ran(ΦT) = H

(see [34, Deﬁnition 11.1.1]). By [34, Proposition 12.1.3 ] it follows that ΦTis invertible. In

particular, there exists a constant c > 0 such that

|ΦT(v)|H≥c|v|L2(0,T ;R), v ∈L2(0, T ;R).

Since u(t, x1, vx1,y) = u(t, x2, vx2,y ) and u(t, xi, vxi,y ) = eT A xi+RT

0e(T−s)ABvxi,y (s)ds, we have

c|vx1,y −vx2,y|2

L2([0,T ];R)≤ZT

0

e(T−s)AB(vx1,y(s)−vx2,y (s)) ds

2

H

≤ZT

0

e(T−s)ABvx1,y (s)ds −ZT

0

e(T−s)ABvx2,y (s)ds

2

H

=eT Ax1−eT A x2

2

H≤˜

C|x1−x2|2

H.

Hence, for all x∈ DH(x0, C) there exists a vx,y ∈ DL2([0,T ];R)(x0,˜

CC

c) such that uc(T, x, vx,y ) =

y. Since, by the choice of R(see (21)), Rdepends on δ, we see by estimate (31) that κcan be

chosen such that it depends only on δand C. The constant ˜

Cand care given by the system.

This concludes the proof of Theorem 2.9

14 E. HAUSENBLAS AND P. RAZAFIMANDIMBY APRIL 3, 2015

Proof of Theorem 2.10. Let x∈Hand ucthe solution to

˙uc(t, x, v) = Auc(t, x, v) + Bv(t), t ≥0,

uc(0, x, v) = x,

(33)

Again, in order to prove Theorem 2.10, we have to show that the RHS of inequality (31) can be

estimated from below for all x∈ DH(C). That is, we have to check that there exists a constant

K > 0 such that for any x∈ DH(C) and control vx∈L2(0, T ;R) with uc(T , x, vx) = 0 we have

|vx|L2(0,T ;R)≤K, x ∈ DH(x0, C).

However, due to Deﬁnition 11.1.1 [34, p. 356] the pair (A, B) is null controllable at time Tiﬀ

Ran ΦT⊃Ran eT A, where

ΦT:L2(0, T ;R)∋v7→ ZT

0

e(T−s)ABv(s)ds ∈H.

In addition, by the closed–graph Theorem (see e.g. [34, p.385, Proposition 12.1.2]), it follows

that there exists an operator L∈ L(H, L2(0, T ;R)) such that eT A = ΦTL. In particular,

vx=Lx, where vxis the control which steers xto 0 in system (33). Since L∈ L(H, L2(0, T ;R)),

there exists a constant K > 0 such that kvxkL2(0,T;R)≤K|x|H. Since for all x∈ DH(C) we

have |x| ≤ 2C, we infer that

|vx|L2(0,T ;R)≤2C K,

from which and inequality (31) we derive that

¯

P(|u(T, x)−y| ≤ δ)≥1

4e−2λ(DR(R))TC(R) exp(−2CK).

This completes the proof of Theorem 2.10.

Proof of Theorem 2.11. The beginning of the proof uses similar argument as used in the

proof of Theorem 2.8. Here, we only list the points where the proofs diﬀer. Let us consider the

following SPDE

du(t, x) = Au(t, x) + RRBσ(u(s, x)) c(z)˜µ(dz, ds),

u(0, x) = x∈H.

(34)

Let ˜

R > 0 be as in Corollary A.4-(2). We start by choosing a number R > ˜

R∨r0such that

3

δ2

C T R1−2

αCσ(1 + K)≤1

2,(35)

where

K:= ZR\DR(R)|c(z)|2λ(dz)!ZT

0

e(T−s)AB

2

L(R,H)ds.

By the assumptions on B,νand σ,Kis ﬁnite. The next diﬀerence is that one has to ﬁnd a

predictable transformation θ(R): Ω ×[0, T ]→Rsuch that

−v(s)

σ(u(s−, x)) +gR=ZR\BR(ρ)hc(z)−c(θ(R)(s, z))iλ(dz), s ∈[0, T ].

Again, the existence of such a transformation is given by Corollary A.3. In fact, letting

G(s) := v(s)

σ(u(s−, x)) +gR,

April 3, 2015 15

we see from Corollary A.3 that there exists ϑR:R×R\{0} → Rsuch that

−G(s) = ZR\BR(ρ)c(z)−c(ϑR(G(s), z))λ(dz), s ∈[0, T ],

so we take θ(R): [0, T ]×R\{0} → Rsuch that for any s∈[0, T ] and z∈R\{0}

θ(R)(s, z) := ϑR−v(s)

σ(u(s−, x)) +gR, z.

Again, we will write θinstead of θ(R)for the rest of the proof.

Let µθa random measure deﬁned by

µθ: Ω × B(R)⊗ B([0, T ]) ∋(A×I)7→ ZR2ZI

1A(θ(s, z))µ(ω, dz , ds),

and Qbe a probability measure on Asuch that µθhas compensator γ. The process uθ

µdeﬁned

by

duθ

µ(t, x) = Auθ

µ(t, x)dt +RRBc(z)σ(u(s−, x)) (µθ−γ)(dz, dt)

uθ

µ(0, x) = x.

(36)

has under Qthe same law as u. In particular, we have

EP1[0,δ](|u(T , x0)−y|)=EQh1[0,δ](|uθ

µ(T, x0)−y|)i.

On the other hand we know that under ¯

Pthe process uθ

µsolves

(37)

duθ

µ(t, x) = Auθ

µ(t, x)dt +RRBσ(uθ

µ(t−, x)) [c(θ(t, z)) −c(z)] (µ−γ)(dz, dt)

+RRBσ(uθ

µ(t−, x)) [c(θ(t, z)) −c(z)] γ(dz, dt) + RRBσ(uθ

µ(t−, x))c(z)(µ−γ)(dz, dt),

uθ

µ(0, x) = x.

Hence as before we can again write

P(|u(T, x)−y| ≤ δ) = Q(|uθ

µ(T, x)−y| ≤ δ) = EQh1|uθ

µ(T,x)−y|≤δi=E¯

PhGθ(T)1|uθ

µ(T,x)−y|≤δi.

Similarly to (26) we obtain

¯

P(|u(T, x)−y| ≤ δ)≥

E¯

Ph1|uθ

µ(T,x)−y|≤δi2

E¯

P|G−1

θ(T)|.

As in proof of Theorem 2.8 we want to ﬁnd a positive lower bound for the right hand side of

the above inequality. First note that thanks to Hypothesis 4 we can show by arguing as in

the proof of the inequality (28) that the denominator E¯

P|G−1

θ(T)|satisﬁes an estimate which

is very similar to (28). In fact, to estimate (27), we used that vis deterministic. But using

the fact that σis bounded from below and above, estimate (28) can be shown similarly. To

estimate the numerator we closely follow the lines of the proof of Theorem 2.9. To be more

precise, we obtain ﬁrst

E¯

Ph1|uθ

µ(T,x)−y|≤δi

≥E¯

Ph1|uc(T,x,v )−y|≤ δ

31|uc(T,x,v )−uθ

µ(T,x)|≤ δ

3i=E¯

Ph1|uc(T,x,v )−uθ

µ(T,x)|≤ δ

3i.(38)

16 E. HAUSENBLAS AND P. RAZAFIMANDIMBY APRIL 3, 2015

By rewriting the diﬀerence ∆(T) = uθ

µ(T, x)−uc(T , x, v) as follows

∆(T) = ZT

0ZR

e(T−s)ABσ(u(s−, x))c(θ(t, z))(µ−γ)(dz, ds)

+ZT

0

e(T−s)ABσ(u(s−, x))gRds,

we infer that

E¯

Ph1|uθ

µ(T,x)−y|≤δi≥E¯

Ph1|∆(T)|≤ δ

3i.

Let us deﬁne

∆(T) := ZT

0ZR\DR(R)

e(T−s)ABσ(u(s−, x))c(θ(s, z))(µ−γ)(dz, ds)

,

and

I(T) := ¯

P ∆(T)>δ

3µ(DR(R)×[0, T ]) = 0!

Arguing as in proof of Theorem 2.8 we also obtain

¯

P|∆(T)| ≤ δ

3≥¯

Pµ(DR(R)×[0, T ]) = 0×¯

P ∆(T)≤δ

3µ(DR(R)×[0, T ]) = 0!

≥¯

Pµ(DR(R)×[0, T ]) = 0×[1 −I(T)].

Now we need to prove that due to the choice of R, we have I(T)≤1

2. To do so we apply the

Chebyscheﬀ inequality to I(T) and use the fact that the random variables µ(DR(R)×[0, T ])

and µ(A×[0, T ]) for all A∈ B(R\ DR(R)) are independent. Thus we obtain

I(T)≤3

δ2

×E¯

PZT

0ZR\DR(R)

e(T−s)ABσ(u(s−, x))c(θ(s, z))(µ−γ)(dz, ds)

2

.

The Burkholder inequality gives

I(T)≤3

δ2

×E¯

P"ZT

0ZR\DR(R)e(T−s)ABσ(u(s−, x))c(z)

2λ(dz)ds#

Since c(θ(t, z)) ≤c(z) and σis bounded (see Hypothesis 4), we get

... ≤C3

δ2

×T R1−2

αCσ 1 + E¯

P"ZT

0ZR\DR(R)ke(T−s)ABk2

L(R,H)|c(z)|2λ(dz)ds#!

≤C3

δ2

×T R1−2

αCσ(1 + ZR\DR(R)|c(z)|2λ(dz)!ZT

0ke(T−s)ABk2

L(R,H)ds).

Therefore, collecting all together

E¯

Ph1|uc(T,x,v )−uθ

µ(T,x)|≤ δ

3i≥1

C(R) 1−3

δ2

C T R1−2

αCσ(1 + K)!.

April 3, 2015 17

Since Rsatisﬁes inequality (35) we get

E¯

Ph1|uθ

µ(T,x)−y|≤δi≥1

2e−λ(DR(R))TC(R).

Hence, we have shown that

P(|u(T, x)−y| ≤ δ)≥1

2e−λ(DR(R))TC(R) exp(−|v|2

L2(0,T ;R)),(39)

which concludes the proof of Theorem 2.11.

Proof of Theorem 2.13. Let x∈ DH(C) and ucthe solution to

˙uc(t, x, v) = Auc(t, x, v) + Bv(t), t ≥0,

uc(0, x, vx) = x,

(40)

In order to show Theorem 2.13 we use the same arguments as we have used in the proof of

Theorem 2.9. That is, we ﬁrstly choose Rin such a way that inequality (35) holds for all

initial conditions x∈ DH(C). Secondly, we have to show that the RHS of inequality (39) can

be estimated from below for all x∈ DH(C). That is, we have to show that there exists a

constant C′>0 such that for any x∈ DH(C) there exists a control vx∈L2(0, T ;R) such that

uc(T, x, vx) = 0 and

kvxkL2(0,T ;R)≤C′.

However, again, this constant is given by continuity properties of system (40). Now, one can

proceed as in the proof of Theorem 2.8.

3. Uniqueness of the invariant measure and the asymptotic strong Feller

property

Let ube the solution of the following stochastic evolution equation

du(t, x) = Au(t, x) + RRBσ(u(t, x)) z˜η(dz, ds),

u(0, x) = x∈H.

(41)

If (41) admits a unique H–valued c`adl`ag solution, then the Markovian semigroup (Pt)t≥0

deﬁned by

Ptφ(x) := Eφ(u(t, x)), φ ∈Cb(H), t ≥0,

is a stochastically continuous Feller semigroup on Cb(H). That is, (Pt)t≥0satisﬁes (see [11])

(1) Pt◦Ps=Pt+s;

(2) for all φ∈Cb(H) and for all x∈Hwe have limt→0Ptφ(x) = φ(x).

Since Item (1) is clear we only check Item (2). For this purpose let ε > 0 and φ∈Cb(H) with

|φ|∞= 1. Item (2) follows from the fact that limt→0Eφ(u(t, x)) = φ(x). Equivalently, for all

ǫ > 0 there exists δ > 0 such that |Eφ(u(t, x)) −φ(x)| ≤ ǫfor all 0 ≤t < δ. Since φis uniformly

continuous on all compact sets Kof Hand the law of uis tight in D([0, T ]; H), there exists a

compact set Kǫsuch that P(u(t)6∈ Kǫfor all 0 ≤t≤T)≤ǫ

3(compare [15, Remark 7.3]) and

there exists a δ1>0 such that |φ(x)−φ(y)| ≤ ǫ

2for all x, y ∈Kǫsuch that |x−y|H≤δ1.

18 E. HAUSENBLAS AND P. RAZAFIMANDIMBY APRIL 3, 2015

Then for t≤δ:= ǫ

6δ2

1we know by the Chebyscheﬀ inequality that P(|u(t, x)−x|H≥δ1)≤ǫ

2.

Hence,

E|φ(u(t, x)) −φ(x)| ≤ E|φ(u(t, x)) −φ(x)|1u(t,x)6∈Kǫ

+E|φ(u(t, x)) −φ(x)|1u(t,x)∈Kǫ1{|u(t,x)−x|≥δ1}P(|u(t, x)−x|H≥δ1)

+E|φ(u(t, x)) −φ(x)|1u(t,x)∈Kǫ1{|u(t,x)−x|<δ1}≤ǫ

3+ǫ

3+ǫ

3,

from which it follows that the Markovian semigroup (Pt)t≥0on Cb(H) is stochastically contin-

uous.

Since in this section we are interested in the existence of an invariant measure, we assume, in

addition to the assumption Hypothesis 1-Hypothesis 4, that the semigroup generated by Ais of

contractive type. To be more precise, we suppose that we have the following set of conditions.

Hypothesis 5. We assume that Ais the inﬁnitesimal generator of a contraction type C0-

semigroup {etA, t ≥0}on H, i.e., there exists ω > 0and a constant M > 0such that for any

t≥0

ketAkL(H,H )≤M e−ωt.

In addition, we suppose that there exists a Hilbert space H1such that the embedding H1֒→H

is compact and Agenerates a contraction type C0-semigroup {etA, t ≥0}on H1. Moreover,

assume that there exists a constant KBsuch that

(42) Z∞

0kesABk2

L(R,H1)ds < KB.

Under Hypothesis 1-Hypothesis 4, and the additional Hypothesis 5, the existence of the

invariant measure can be shown by an application of the Krylov-Bogoliubov Theorem (see [14,

Theorem 3.1.1] ). First, we will deﬁne for T > 0 and x∈Hthe following probability measure

on (H, B(H))

B(H)∋Γ7→ RT(x, Γ) := 1

TZT

0

Pt(x, Γ) dt.(43)

In addition, for any ρ∈M1(H), let R∗

Tρbe deﬁned by

B(H)∋Γ7→ ZH

RT(x, Γ) ρ(dx).

Corollary 3.1.2 in [14] says that, if for some probability measure ρon (H, B(H)) and some

sequence Tn↑ ∞, the sequence {R∗

Tnρ:n∈N}is tight, and, in addition the Markovian (Pt)t≥0

semigroup is also a Feller semigroup, then there exists an invariant measure for (Pt)t≥0. That

means, for the existence of an invariant measure it is suﬃcient to show that for all ǫ > 0 for

all x∈H, there exists a compactly embedded subspace E ֒→H, a R > 0 such that we have

for all T > 0

1

TZT

0

P(|u(t, x)|E≥R)dt < ǫ.(44)

Observe that, if Agenerates a contraction type C0-semigroup (etA)t≥0on E, then

Z∞

0

etA

2

L(E,E )dt < ∞.

April 3, 2015 19

By Hypothesis 5 we can take E=H1. Thanks to the boundedness of σ(see Hypothesis 4)

and Eq. (42) of Hypothesis 5 we can show by direct calculation that the following estimate

holds

sup

0≤t≤T

E|u(t, x)|2

H1≤C, T ≥0.(45)

Then, the Markovian semigroup (Pt)t≥0admits an invariant measure. In fact, the Chebyshev

inequality implies that for any R > 0

1

TZT

0

P(|u(t, x)|H1≥R)dt ≤1

TZT

0

E|u(t, x)|2

H1

R2ds

≤C

R2,

from which we easily deduce inequality (44) by taking R > (C

ǫ)1

2.

By [17, Corollary 3.17] we know that if the semigroup is asymptotically strong Feller (for

the deﬁnition of asymptotic strong Feller we refer to [17]) and there exists a point x∈Hsuch

that x∈supp(ϕ), whenever ϕis an invariant measure of the Markovian semigroup (Pt)t≥0,

then the semigroup (Pt)t≥0admits at most one invariant measure.

Assume for the time being that the semigroup (Pt)t≥0is asymptotically strong Feller. Hence,

it remains to prove that there exists a x∈Hsuch that x∈supp(ϕ). Now, the two following

properties imply that 0 ∈supp(ϕ)4whenever ϕis an invariant measure.

•There exists a constant C > 0 such that

inf

{ρis an invariant measure}ρ(DH(C)) >0.(46)

•For all δ > 0 and for all x∈ DH(C) there exists a time Tδ>0 and some κ=κ(C, δ)>0

depending only on Cand δ, such that

P(u(Tδ, x)∈ DH(δ)) ≥κ.(47)

Indeed since ϕis invariant we have

ϕ(DH(δ)) ≥κ·inf

{ρis invariant measure}ρ(¯

DH(C)).

Now, estimates (46) and (47) imply 0 ∈supp(ϕ).

Inequality (47) follows from Theorem 2.13. Estimate (46) follows from the fact that for any

invariant measure ρof the semigroup (Pt)t≥0, there exists a constant C > 0 such that

ZH|u|2

Hρ(du)≤C.(48)

Estimate (48) follows by the same lines as in [24, Lemma B.1] and the a priori estimate (45).

Now, an application of the Chebyscheﬀ Inequality leads to inequality (46).

It remains to investigate under which conditions the semigroup satisﬁes the asymptotic

strong Feller property. But, before continuing further we introduce an additional concept of

controllability.

4Note, that x∈supp(ϕ) iﬀ for all δ > 0, ϕ(DH(x, δ )) >0.

20 E. HAUSENBLAS AND P. RAZAFIMANDIMBY APRIL 3, 2015

Deﬁnition 3.1. We say that the system

˙uc(t, x, v) = Auc(t, x, v) + Bv(t), t ≥0,

uc(0, x, v) = x,

is null controllable with vanishing energy (see [26, 28]), if it is null controllable and for any

x∈Hthere exists a sequence of times {tn≥0 : n∈N}and a sequence of controls {vn:n∈

N} ⊂ L2([0, tn]; R)such that u(tn, x, vn) = 0 for any n∈Nand

lim

n→∞ Ztn

0|vn(s)|2

Rds = 0.

Now we are ready to present our result about the asymptotically strong Feller property.

Theorem 3.2. We assume that Hypothesis 1 to Hypothesis 5 are satisﬁed. If the system (11)

is null controllable with vanishing energy, and (41) admits an H–valued c`adl`ag solution, then

the Markovian semigroup of system (41) is asymptotically strong Feller.

Due to the null controllability, one can prove the following Corollary.

Corollary 3.3. Assume that all the assumptions of Theorem 3.2 are satisﬁed. Then, the Mar-

kovian semigroup (Pt)t≥0generated by the solution to system (41) admits a unique invariant

measure.

Some similar theorem holds also for the system (12), if the solution is not c`adl`ag. Albeit,

the Markovian semigroup is not Feller, and therefore, the Krylov-Bogoliubov Theorem is not

applicable. However, one can prove a similar result to the one stated in Theorem 3.2.

Theorem 3.4. We assume that Hypothesis 1 to Hypothesis 4 are satisﬁed. If the system (11)

is null controllable with vanishing energy, then the Markovian semigroup of system (12) is

asymptotically strong Feller.

Corollary 3.5. Assume that all assumptions of Theorem 3.4 are satisﬁed. Then, the Markov-

ian semigroup (Pt)t≥0generated by the solution to system (12), admits at most one invariant

measure.

Proof of Theorem 3.2:Again, we will switch for technical reasons to another representation

of the Poisson random measure. Let A= (¯

Ω,¯

F,¯

F,¯

P) be a ﬁltered probability space with

ﬁltration ¯

F= ( ¯

Ft)t≥0and let µbe a Poisson random measure on Rover Ahaving intensity

measure λ. The compensator of µis denoted by γand given by

B(R)× B([0,∞)) ∋A×I7→ γ(A×I) := λ(A)λ(I).

Let

c:R+∋r7→ sup

ρ>0Z∞

ρ

k(s)ds ≥rif r > 0.(49)

Now, the stochastic evolution equation given in (7) reads as follows

du(t, x) = Au(t, x) + RRBσ(u(s−, x)) c(z)˜µ(dz, ds),

u(0, x) = x∈H.

(50)

We split the proof into several steps.

April 3, 2015 21

Step I:. Fix x∈H. In order to show the asymptotic strong Feller property of (Pt)t≥0in x,

we have to show that there exist an increasing sequence {tn:n∈N}and a totally separating

sequence of pseudometrics {dn:n∈N}5such that6

lim

ǫ→0+lim sup

n→∞ sup

y∈D(x,ǫ)kPtn(x, ·)− Ptn(y, ·)kdn= 0.(51)

Let {an:n∈N}be a sequence of positive real numbers converging to zero. Let

dn(y, z) := 1 ∧(|z−y|H/an), z, y ∈H, n ∈N.

Fix h∈Hwith |h|= 1. Since system (11) is linear and null controllable with vanishing

energy, we can ﬁnd a sequence of times {tn:n∈N}and controls {vn:n∈N}such that

a2

n≥Rtn

0|vn(s)|2ds,n∈N, and the solution ucto

duc(t, x, vn) = Auc(t, x, vn)dt +Bvn(t)dt, 0≤t≤tn

uc(0, x, vn) = x,

satisfy uc(tn, x, vn) = uc(tn, x +h, 0). For simplicity, put y=x+ǫh and vn

ǫ:= ǫ·vn. Then,

it follows by the linearity that uc(tn, x, ǫvn) = uc(tn, x +ǫh, 0) = uc(tn, y, 0). In order to give

an estimate of kPtn(x, ·)− Ptn(y, ·)kdnin terms of ǫand n, we deﬁne the following sequence

of continuous functions. Let φ∈Cb(H), there exists a sequence {φn:n∈N},φn∈C∞

b(H),

such that φn→φpointwise, kφnk∞≤ kφk∞and kφnkdn≤1 for all n∈N. Now, the following

identity holds

kPtn(x, ·)− Ptn(y, ·)kdn=E¯

P[φn(u(tn, x +ǫh))] −E¯

P[φn(u(tn, x))].

Hence, we have to show that

lim

ǫ→0+lim sup

n→∞ sup

y∈D(x,ǫ)E¯

P[φn(u(tn, x +ǫh))] −E¯

P[φn(u(tn, x))]= 0.(52)

This will be the sub ject of the second step of the proof.

Step II:. Let us introduce a transformation ϑǫ

n: [0,∞)×R→R, such that we have

ZR

(c(z)−c(ϑn

ǫ(s, z)))λ(dz) = vn

ǫ(s)

σ(u(s−, x)) for all s∈[0, tn].

Note, that hypothesis 1 implies, that there exists an index α∈(1,2), constants K1, K2>0

and r0>0 such that

K1|z|−1

α≤c(z)≤K2|z|−1

α,for all |z| ≥ r0.

5An increasing sequence {dn:n∈N}of pseudo metrics is called a totally separating system of pseudo metrics

for X, if limn→∞ dn(z, y) = 1 for all z, y ∈ X ,z6=y.

6Let dbe a pseudo–metric on X, we denote by L(X, d) the space of d-Lipschitz functions from Xinto R.

That is, the function φ:X → Ris an element of L(X, d) if

kφkd:= sup

z,y∈X

z6=y

|φ(z)−φ(y)|

d(z, y )<∞.

For a pseudo-metric don Xwe deﬁne the distance between two probability measures P1and P2with respect

to dby

kP1− P2kd:= sup

φ∈L(X,d)

kφkd=1

ZX

φ(x) (P1− P2)(dx).

22 E. HAUSENBLAS AND P. RAZAFIMANDIMBY APRIL 3, 2015

Now, let ˜

R > 0 be as in Corollary A.4-(2) and put R=˜

R∧r0. In fact, by Corollary A.3 such

a transformation ϑn

ǫexists and is given by

ϑn

ǫ: [0,∞)×R∋(s, z)7→ ϑR(−vn

ǫ,σ(s)),(53)

where

vn

ǫ,σ(s) = vn

ǫ(s)

σ(u(s−, x)).

In addition, let µn

ǫbe a random measure deﬁned by

B(Z)× B([0, t]) ∋A×I7→ µn

ǫ(A×I) = ZIZA

1A(ϑn

ǫ(s, z))µ(dz, ds)(54)

Let Qǫ,n be the probability measure on Asuch that the intensity measure of µn

ǫis the Lebesgue

measure λ. Let un

ǫbe the solution to

dun

ǫ(t, x) = Aun

ǫ(t, x)dt +RRBσ(u(t−, x))[c(ϑn

ǫ(t, z)) −c(z)]µ(dz, dt)

+RRBσ(u(t−, x))c(z)(µ−γ)(dz, dt),

un

ǫ(0, x) = x,

and let uc

µ,n,ǫ be the solution to

duc

µ,n,ǫ(t, x, vn

ǫ,σ) = Auc

µ,n,ǫ(t, x, vn

ǫ)dt +Bvn

ǫ,σ(t)dt +RRBσ(u(t−, x))c(z)(µ−γ)(dz, dt),

uc

µ,n,ǫ(0, x, vn

ǫ) = x.

Observe, that, by the choice of the transformation ϑǫ

n, under Qǫ,n , the random variable un

ǫ(tn, x)

is identical in law to the process u(tn, x). In particular, we have

EQǫ,n

tn[φn(un

ǫ(tn, x))] = E¯

P[φn(u(tn, x))] .

From the choice of the control and the linearity of Awe also have uc

µ,n,ǫ(tn, x, vn

ǫ) = u(tn, x+ǫh).

For t≥0 let Qǫ,n

tbe the restriction of Qǫ,n onto ¯

Ftand ¯

Ptbe the restriction of ¯

Ponto ¯

Ft. Now

we are ready to give an estimate of kPtn(x, ·)− Ptn(y, ·)kdn.First, we write

kPtn(x, ·)− Ptn(y, ·)kdn=E¯

P[φn(u(tn, x +ǫh))] −E¯

P[φn(u(tn, x))],(55)

and

E¯

P[φn(u(tn, x +ǫh))] −E¯

P[φn(u(tn, x))]

=E¯

Pφnuc

µ,n,ǫ(tn, x, vn

ǫ)−E¯

P[φn(un

ǫ(tn, x))] + E¯

P[φn(un

ǫ(tn, x))] −E¯

P[φn(u(tn, x))]

=E¯

Pφnuc

µ,n,ǫ(tn, x, vn

ǫ)−φn(un

ǫ(tn, x))+E¯

P1−dQǫ,n

tn

d¯

Ptnφn(un

ǫ(tn, x))

+EQǫ,n [φn(un

ǫ(tn, x))] −E¯

P[φn(u(tn, x))]

=E¯

Pφnuc

µ,n,ǫ(tn, x, vn

ǫ)−φn(un

ǫ(tn, x))+E¯

P1−dQǫ,n

tn

d¯

Ptnφn(un

ǫ(tn, x)).

April 3, 2015 23

We have the following inequality

E¯

P[φn(u(tn, x +ǫh))] −E¯

P[φn(u(tn, x))]

≤1

an

E¯

Puc

µ,n,ǫ(tn, x, vn

ǫ)−un

ǫ(tn, x)+||φn||∞E¯

P

1−dQǫ,n

tn

d¯

Ptn

,

=: 1

an

I1+||φn||∞I2.

By the construction of un

ǫ(t, x) and uc

µ,n,ǫ(t, x, vn

ǫ) we see that

un

ǫ(tn, x)−uc

µ,n,ǫ(tn, x, vn

ǫ) = Ztn

0ZR

e(tn−s)ABσ(u(s−, x)) [c(z)−c(ϑǫ

n(s, z))] (γ−µ)(dz, ds).

Applying Hypothesis 4

E¯

Pun

ǫ(tn, x)−uc

µ,n,ǫ(tn, x, vn

ǫ)

2

≤E¯

PZtn

0ZRe(tn−s)ABσ(u(s−, x)) [c(z)−c(ϑǫ

n(s, z))]λ(dz)ds

≤1

Cσ

E¯

PZtn

0ke(tn−s)ABkL(R,H)|vn

ǫ(s)|ds.

By Applying Hypothesis 3 we derive that

|Iǫ

1| ≤ E¯

Pu(tn, x)−uc

µ,n,ǫ(tn, x, vn

ǫ)

2≤C

CσZtn

0ke(tn−s)ABk2

L(R,H)ds1

2

E¯

PZtn

0|vn

ǫ(s)|2ds1

2

≤CK2

B

CσZtn

0|vn

ǫ(s)|2ds1

2

≤C(γ, ρ, B )

CσZtn

0|vn

ǫ(s)|2

Rds1

2

.

To give an estimate of the second term Iǫ

2we apply [18, Theorem 1] to get an exact represen-

tation of the Radon Nikodym derivative. In particular, we obtain

Iǫ

2≤E¯

Ph(1 −dQǫ,n

tn

d¯

Ptn)φ(un

ǫ(tn, x))i

≤E¯

P[(1 − Gn

ǫ(tn)) φ(un

ǫ(tn, x))]

where Gn

ǫis deﬁned by (see Lemma B.1 and (A.3))

(56)

dGn

ǫ(t) = Gn

ǫ(t)RR+[−ςz(κ+(|vn

ǫ,σ(s)|), z) sgn(vn

ǫ,σ(s))] (µ−γ)(dz, ds)

Gn

ǫ(0) = 1.

The H¨older inequality gives

Iǫ

2≤E¯

P|1− Gn

ǫ(tn)| kφk∞.

24 E. HAUSENBLAS AND P. RAZAFIMANDIMBY APRIL 3, 2015

First we will give an estimate of Esup0≤s≤tn|Gn

ǫ(s)|. By the choice of ϑ, the process Gis of

ﬁnite variation, thus we easily derive that for 0 ≤t≤tn

E¯

Psup

0≤s≤t|Gn

ǫ(s)| ≤ 1 + Zt

0ZR|Gn

ǫ(s−)||1−ϑz(vn

ǫ,σ(s))|dz ds

≤1 + CZt

0|Gn

ǫ(s−)|ZR|1−ϑz(vn

ǫ,σ(s))|dzds.(57)

From Corollary A.4 it follows

ZR|1−ϑz(vn

ǫ,σ(s))|dz ≤C

Cσ|vn

ǫ(s)|2.

Substituting this last estimate in (57) we obtain

E¯

Psup

0≤s≤t|Gn

ǫ(s)| ≤ 1 + C

CσZt

0|Gn

ǫ(s−)||vn

ǫ(s)|2ds(58)

≤1 + C

Cσ

Esup

0≤s≤t|Gn

ǫ(s)|Zt

0|vn

ǫ(s)|2ds.(59)

Since Rtn

0|vn

ǫ(s)|2ds ≤a2

nand an→0, there exists a n0∈Nsuch that Ca2

n<1/2. Therefore,

for n≥n0we obtain

E¯

Psup

0≤s≤t|Gn

ǫ(s)| ≤ 2.

Arguing as above above we obtain that

E¯

P|Gn

ǫ(tn)−1| ≤ C(R)

Cσ

E¯

PZtn

0|Gn

ǫ(s−)||vn

ǫ(s)|2ds

≤C

Cσ

E¯

Psup

0≤s≤tn|Gn

ǫ(s)|Ztn

0|vn

ǫ(s)|2ds,

≤C

Cσ

2ǫa2

n.

Going back to equation (55) and taking the limit, it follows that there exists some constants

C1, C2>0 and n0∈N, such that for all n≥n0

E¯

P[φn(u(tn, x +ǫh))] −E¯

P[φn(u(tn, x))]

≤(C1

anZtn

0|vn

ǫ(s)|2ds1

2

+C2kφk∞2ǫ2a2

n).

Hence, we have

≤C1ǫan

an

+C2ǫan.

Taking the limit n→ ∞ we get

lim sup

n→∞ sup

y∈D(x,ǫ)kPtn(x, ·)− Ptn(y, ·)kdn≤Cǫ.

We ﬁnish the proof of Theorem 3.2 by passing to limit as ǫ→0 in the last estimate.

April 3, 2015 25

Proof of Theorem 3.4:The proof works along the lines of the proof of Theorem 3.2, only

one has to put σ= 1. Since one does not need that the process [0,∞)∋t7→ u(t) is c`adl`ag,

but only one needs that the process [0,∞)∋t7→ σ(u(t)) is c`adl`ag, the proof of Theorem 3.4

remains the same.

4. An Example - the damped wave equation with boundary noise

As mentioned in the introduction, we consider as example an elastic string, ﬁxed at one end

and perturbed by a L´evy noise at the other end. Mathematically, the system can be formulated

as a damped wave equation with boundary L´evy noise.

Throughout this section suppose that we are given a ﬁltered probability space (Ω,F,F,P)

such that the ﬁltration F= (Ft)t≥0satisﬁes the usual condition. On this probability space we

assume that we are given a real valued L´evy process L. Let T > 0 and ˜ρ∈R. We consider the

system

utt(t, ξ)−Λu(t, ξ) + ˜ρ uξ(t, ξ) = 0, t ∈(0, T ), ξ ∈(0,2π),

u(t, 0) = 0, t ∈(0, T ),

uξ(t, 2π) = σ(u(t)) ˙

Lt, t ∈(0, T ),

u(0, ξ) = x0(ξ), ut(0, ξ) = x1(ξ),

(60)

where Λ = ∆ is the Laplacian and ˙

Lis the Radon Nikodym derivative of a real valued L´evy

process with characteristic measure νsatisfying Hypothesis 1 and Hypothesis 2. Let x0∈

H1

0(0,2π) and x1∈L2(0,2π). Here we have set σ(u(t, ξ)) = (2 + 2 + sin(|u(t, ξ )|L2(O))for

any t∈(0, T ) and ξ∈(0,2π).

Equation (60) can be reformulated as an evolution equation of order one. In fact, let H0=

L2([0,2π]) and D(Λ) = {u∈H0,Λu∈H0and u(0) = 0, ux(2π) = 0}. Observe, that for θ≥0

we have

D(Λθ

2) =

Hθ,2(0,2π) for θ < 1

2

{u∈Hθ,2(0,2π) : u(0) = 0}for 1

2≤θ < 3

2

{u∈Hθ,2(0,2π) : u(0) = 0, u′(2π) = 0}for 3

2≤θ.

Finally, to reformulate Equation (60) as an evolution equation of order one let us deﬁne the

Hilbert space H=D(Λ 1

2)⊗H0equipped with the scalar product

hw, ziH=hΛ1

2w1,Λ1

2z1i+hw1, w2i, w =w1

w2∈ H and z=z1

z2∈ H,

where h·,·i denotes the scalar product on L2([0,2π]). Deﬁne an operator Awith domain

D(A) = D(Λ) ×D(Λ1

2)→ H by

(61) Az1

z2=0I

Λ 0z1

z2,

and B˜ρ:H → H by

B˜ρz1

z2=0

˜ρz1.

It is not diﬃcult to prove that for ˜ρ > 0A+B˜ρgenerates a C0semigroup (S(t))t≥0of

contractional type on H. First, Λ has eigenfunctions en(ξ) = sin((2n−1)ξ/4), n∈N, with

eigenvalues λn= (2n−1)2/16. Now, if {λn:n∈N}are the eigenvalues and {φn:n∈N}the

26 E. HAUSENBLAS AND P. RAZAFIMANDIMBY APRIL 3, 2015

eigenfunction of A, then {µn:n∈R}with µn=−˜ρ±p˜ρ2−4λn,µ−n= ¯µn,n∈N, are the

eigenvalues and

ψ=1

√21

iµnφn

φn:n∈Z,

are the eigenfunction of A(see [34, Proposition]). The semigroup S={S(t), t ≥0}can be

written as

S(t)f

g=1

√2X

n∈R

eµnµnhdf

dx,dφn

dx iL2([0,1]) +hg, φniL2([0,1])ψn,f

g∈R.

To rewrite (60) as a stochastic evolution equations on the Hilbert space Hwe need to ﬁnd

a way of transforming the nonhomogeneous boundary conditions in (60) to homogeneous one.

Therefore, we introduce an operator DB,γ as follows. For every a∈R,DB,γ a=vis a solution

to the problem

v′′(ξ) = γv(ξ), ξ ∈(0,2π),

vξ(2π) = a,

v(0) = 0.

By a short calculation it follows that given a∈R,

v=DB,0(ξ) = aξ/(2π), ξ ∈[0,2π].

Moreover, for all θ < 3

2, the mapping DB,0:R→D(Λθ

2) is bounded. Also, observe that for

any θ < 1

4the operator

A0

DB,0:R→D(Λ θ+1

2)×D(Λθ

2)

is bounded. Following the approach in [22] and [27] we see that (60) can be transformed into

the following

(62)

dX(t, X0) = (A+B˜ρ)X(t, X0)dt +A 0

DB,0σ◦Π1(X(t, X0))dL!,

X(0, X0) = X0,

where X= (u, ˙u)T,X0= (x0, x1)T. Here Π1X=udenotes the projection from Honto D(Λ1

2).

From now on we will work with (62). Since D(Λθ+1

2)×D(Λθ

2)֒→ H continuously, one can

show by mimicking the proof of [27, Theorem 15.7.2] that the problem (62) is well posed.

Moreover, the solution uis c`adl`ag in Hand is a Markov-Feller process. In particular, the

family of operators (Pt)t≥0deﬁned by

Ptφ(x) := Eφ(X), φ ∈Cb(H), t ≥0.(63)

is a semigroup on Cb(H). By means of Remark 2.7 and Theorem 2.9 the following results can

be obtained.

Theorem 4.1. Assume that the L´evy measure νof the L´evy noise Lsatisﬁes Hypothesis 1 and

Hypothesis 2. Then, there exists a time T > 0such that for any C > 0and δ > 0, there exists

aκ=κ(C, δ)>0with

(P∗

Tδx) (DH(δ)) = P(|X(T, x)−y|H≤δ)≥κ, x, y ∈ DH(C).

April 3, 2015 27

Proof of Theorem 4.1: System (60) is exactly controlable at any time T > 2π(see [10, Section

2.4], or [34, Example 11.2.6]). Hence, the system satisﬁes the assumptions of Theorem 2.9,

which implies Theorem 4.1.

The following Corollary is a consequence of Theorem 4.1 and Theorem 3.2.

Corollary 4.2. Assume that the L´evy measure νof the L´evy noise Lsatisﬁes Hypothesis 1

and Hypothesis 2. If ˜ρ > 0, then the Markovian semigroup (Pt)t≥0deﬁned by (63) has at most

one invariant measure µ. Moreover, the invariant measure is nondegenerate. In particular, for

any x∈ H and δ > 0

µ(DH(x, δ)) >0.

We need to show some facts which are essential for the results in the previous sections to be

applicable in for our example. First, in [28, Section 6.3] it is shown that the following system

utt(t, ξ)−Λu(t, ξ) + ˜ρ u(t, ξ) = 0, t ∈(0, T ), ξ ∈(0,2π),

u(t, 0) = 0, t ∈(0, T ),

uξ(t, 2π) = v(t), t ∈(0, T ),

u(0, ξ) = u0(ξ), ut(0, ξ) = u1(ξ),

(64)

with control v∈L2([0,∞); R) and ˜ρ≥0 is null controllable with vanishing energy. Moreover,

it is shown (see Remark 2.7), that (64) is exact controllable for T≥2π.

Now we are ready to prove the existence and uniqueness of the invariant measure.

Proof of Corollary 4.2: It was proved in [10, Section 2.4] (see also e.g. [34, Example 11.2.6])

that (60) is controllable at any time T > 2π, hence it is null controllable. Since approximate

controllability is weaker than exact contractibility, it is also null controllable. Now it remains

to prove that it is null controllable with vanishing energy. For this purpose we mainly follow

the idea in [28]. Let us write Has the direct sum H=Hs⊕Huwhere Hu={0}and Hs=H.

Therefore we see that [28, Hypothesis 1.1] are satisﬁed in our case. Moreover, since Ais the

inﬁnitesimal generator of a strongly continuous semigroup of contractions we can deduce from

[25, Chapter 1, Corollary 3.6] that the spectrum σ(A) is contained in {λ∈C: Re(λ)≤0}.

This fact implies that S(A) = sup{Re(λ) : λ∈σ(A)} ≤ 0. Therefore we can deduce from [28,

Theorem 1.1] that (64) is null controllable with vanishing energy.

In the next step we will proof the existence of an invariant measure by an application of the

Krylov-Bogoliubov Theorem. Let 0 < θ < 1

4and

H1=D(Λθ+1

2)×D(Λθ

2).

First, note that the embeddings D(Λθ+1

2)×D(Λθ

2)֒→ H are compact. Now, to establish the

existence of the invariant measure we have to show inequality (45) with H1deﬁned above. By

standard arguments (see [19]) we get

Zt

0ZRS(t)A0

DB,0(σ◦Π1(X(t))) z˜η(dz, ds)

2

H1

≤Zt

0ZRS(t)A0

DB,0(σ◦Π1(X(t))) z

2

H1

ν(dz)ds.

28 E. HAUSENBLAS AND P. RAZAFIMANDIMBY APRIL 3, 2015

Since

A0

DB,0:R→D(Λ θ+1

2)×D(Λθ

2),

and Sis a semigroup of contractive type, we get

... ≤CZt

0ZR

e−˜ρ(t−s)A0

DB,0(σ◦Π1(X(t))) z

2

H1

ν(dz)ds

≤CZt

0ZR

e−˜ρ(t−s)|z|2ν(dz)ds.

Due to the fact that σis bounded, estimate (45) is satisﬁed. Thus, the damped wave equation

with L´evy noise boundary condition (60) has an invariant measure (compare also with [14,

p. 181 ]).

It remains to show the uniqueness of the invariant measure. Owing to the Remark 2.7 and

Theorem 3.2 the semigroup Ptis asymptotically strong Feller. By [17, Corollary 3.17] we

know that if the semigroup is asymptotically strong Feller and there exists a point x∈ H

such that x∈supp(ρ), whenever ρis an invariant measure of (Pt)t≥0, then the Markovian

(Pt)t≥0semigroup admits at most one invariant measure. Therefore, we have to show that

there exists a point x∈Hsuch that for any invariant measure ρ,x∈supp(ρ), i.e. for all δ > 0,

ρ(DH(δ)) >0. Since null controllability implies approximate null controllability Theorem 2.8

can be applied to our example. In particular, there exists a time T > 0 such that for all C > 0

and δ > 0 there exists a κ > 0 with

P(u(T, x)∈ DH(δ)) ≥κ, x ∈¯

DH(C).(65)

It remains to show (46). In particular, we should check that there exists a constant C > 0 such

that

inf

{ϕis an invariant measure}ϕ(¯

DH(C)) >0.(66)

It follows that 0 ∈supp(ρ) by the following observations. First, since ρis an invariant measure

we have

ρ(DH(δ)) ≥ρ¯

DH(C)·inf

{ϕis invariant measure}ϕ(¯

DH(C)).