# Characterisation of Exponential  Convergence to Nonequilibrium Limits for Stochastic Volterra Equations

Authors: John A. D. Appleby; Siobhán Devin; David W. Reynolds
Journal: Journal of Applied Mathematics and Stochastic Analysis
 (2008)

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## Abstract

This paper considers necessary and sufficient conditions for the solution
of a stochastically and deterministically perturbed Volterra equation to
converge exponentially to a nonequilibrium and nontrivial limit. Convergence
in an almost sure andpth mean sense is obtained.


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## Body

## 1. Introduction

In this paper, we study the exponential convergence of
the solution of(1.1a)dX(t)=(AX(t)+∫0tK(t−s)X(s)ds+f(t))dt+Σ(t)dB(t),t>0,(1.1b)X(0)=X0,
to a nontrivial random variable. Here the solution X is an n-dimensional vector-valued function on [0,∞), A is a real n×n-dimensional matrix, K is a continuous and integrable n×n-dimensional matrix-valued function on [0,∞), f is a continuous n-dimensional vector-valued function on [0,∞), Σ is a continuous n×d-dimensional matrix-valued function on [0,∞) and B(t)=(B1(t),B2(t),…,Bd(t)), where each component of the Brownian motion is
independent. The initial condition X0 is a deterministic constant vector.The solution of
(1.1a)-(1.1b) can be written in
terms of the solution of the resolvent equation
(1.2a)R′(t)=AR(t)+∫0tK(t−s)R(s)ds,t>0,(1.2b)R(0)=I,
where the matrix-valued function R is known as the resolvent or fundamental
solution. In [1], the
authors studied the asymptotic convergence of the solution R of (1.2a)-(1.2b) to a nontrivial limit R∞.
It was found that R−R∞ being integrable and the kernel being
exponentially integrable were necessary and sufficient for exponential
convergence. This built upon a result of Murakami [2] who considered the
exponential convergence of the solution to a trivial limit and a result of
Krisztin and Terjéki [3] who obtained necessary and sufficient conditions for
the integrability of R−R∞.
A deterministically perturbed version of (1.2a)-(1.2b),
(1.3a)x′(t)=Ax(t)+∫0tK(t−s)x(s)ds+f(t),t>0,(1.3b)x(0)=x0,was also studied in [1]. It was shown that the exponential decay of the tail
of the perturbation f combined with the integrability of R−R∞ and the exponential integrability of the
kernel were necessary and sufficient conditions for convergence to a nontrivial
limit.The case where (1.2a)-(1.2b) is stochastically
perturbed
(1.4a)dX(t)=(AX(t)+∫0tK(t−s)X(s)ds)dt+Σ(t)dB(t),t>0,(1.4b)X(0)=X0,
has been considered. Various authors including Appleby and Freeman
[4], Appleby and
Riedle [5], Mao
[6], and Mao and
Riedle [7] have
studied convergence to equilibrium. In particular the paper by Appleby and
Freeman [4] considered
the speed of convergence of solutions of (1.4a)-(1.4b) to equilibrium. It was shown
that under the condition that the kernel does not change sign on [0,∞) then (i) the almost sure exponential
convergence of the solution to zero, (ii) the pth mean exponential convergence of the solution
to zero, and (iii) the exponential integrability of the kernel and the
exponential square integrability of the noise are equivalent.Two papers by Appleby et al. [8, 9] considered the convergence
of solutions of (1.4a)-(1.4b) to a nonequilibrium limit in the mean square and almost
sure senses, respectively. Conditions on the resolvent, kernel, and noise for
the convergence of solutions to an explicit limiting random variable were
found. A natural progression from this work is the analysis of the speed of
convergence.This paper examines (1.1a)-(1.1b) and builds on the
results in [1, 8, 9]. The analysis of (1.1a)-(1.1b) is complicated, particularly in the almost sure
case, due to presence of both a deterministic and stochastic perturbation.
Nonetheless, the set of conditions which characterise the exponential
convergence of the solution of (1.1a)-(1.1b) to a nontrivial random variable is found. It can be
shown that the integrability of R−R∞,
the exponential integrability of the kernel, the exponential square
integrability of the noise combined with the exponential decay of the tail of
the deterministic perturbation, t↦∫t∞f(s)ds,
are necessary and sufficient conditions for exponential convergence of the
solution to a nontrivial random limit.


## 2. Mathematical Preliminaries

In this
section, we introduce some standard notation as well as giving a precise
definition of (1.1a)-(1.1b) and its solution.Letℝ denote the set of real numbers and let ℝn denote the set of n-dimensional vectors with entries in ℝ.
Denote by ei the ith   standard basis vector in ℝn.
Denote by ∥A∥ the standard Euclidean norm for a
vector A=(a1,…,an) given by(2.1)∥A∥2=∑i=1nai2=trAAT,where tr denotes the trace of a square matrix.Letℝn×n be the space of n×n matrices with real entries where I is the identity matrix. Let diag(a1,a2,…,an) denote the n×n matrix with the scalar entries a1,a2,…,an on the diagonal and 0 elsewhere. For A=(aij)∈ℝn×d the norm denoted by ∥⋅∥ is defined by(2.2)∥A∥2=∑i=1n∑j=1d|aij|2.The set of complex numbers is denoted byℂ;
the real part of z in ℂ being denoted by Rez.
The Laplace transform of the function A:[0,∞)→ℝn×d is defined as(2.3)A^(z)=∫0∞A(t)e−ztdt.If ϵ∈ℝ and ∫0∞∥A(t)∥e−ϵtdt<∞ then A^(z) exists for Rez≥ϵ and z↦A^(z) is analytic for Rez>ϵ.IfJ is an interval in ℝ and V a finite-dimensional normed space with norm ∥⋅∥ then C(J,V) denotes the family of continuous functions ϕ:J→V.
The space of Lebesgue integrable functions ϕ:[0,∞)→V will be denoted by L1([0,∞),V) where ∫0∞∥ϕ(t)∥dt<∞.
The space of Lebesgue square-integrable functions ϕ:[0,∞)→V will be denoted by L2([0,∞),V) where ∫0∞∥ϕ(t)∥2dt<∞.
When V is clear from the context, it is omitted it
from the notation.We now make our problem precise. We assume that the
functionK:[0,∞)→ℝn×n satisfies(2.4)K∈C([0,∞),ℝn×n)∩L1([0,∞),ℝn×n),the function f:[0,∞)→ℝn satisfies(2.5)f∈C([0,∞),ℝn)∩L1([0,∞),ℝn),and the function Σ:[0,∞)→ℝn×d satisfies(2.6)Σ∈C([0,∞),ℝn×d).Due to
(2.4) we may define K1 to be a function K1∈C([0,∞),ℝn×n) with(2.7)K1(t)=∫t∞K(s)ds,t≥0,where this function defines the
tail of the kernel K.
Similarly, due to (2.5), we may define f1 to be a function f1∈C([0,∞),ℝn) given by(2.8)f1(t)=∫t∞f(s)ds,t≥0.We let {B(t)}t≥0 denote d-dimensional Brownian motion on a complete
probability space (Ω,ℱ,{ℱB(t)}t≥0,ℙ) where the filtration is the natural one ℱB(t)=σ{B(s):0≤s≤t}.Under the hypothesis (2.4), it is well known that
(1.2a)-(1.2b) has a
unique continuous solution R,
which is continuously differentiable. We define the function t↦X(t;X0,f,Σ) to be the unique solution of the initial value
problem (1.1a)-(1.1b). If Σ and f are continuous then for any deterministic
initial condition X0 there exists an almost surely unique
continuous and ℱB-adapted solution to (1.1a)-(1.1b) given
by(2.9)X(t;X0,Σ,f)=R(t)X0+∫0tR(t−s)f(s)ds+∫0tR(t−s)Σ(s)dB(s),t≥0.When X0,f,
and Σ are clear from the context, we omit them from
the notation.The notion of convergence and integrability inpth mean and almost sure senses are now defined:
the ℝn-valued stochastic process {X(t)}t≥0converges inpth mean to X∞ if limt→∞𝔼∥X(t)−X∞∥p=0;
the process is pth mean exponentially convergent to X∞ if there exists a deterministic βp>0 such that(2.10)limsup⁡t→∞⁢1tlog⁡(E∥X(t)−X∞∥p)≤−βp;we say that the difference
between the stochastic process {X(t)}t≥0 and X∞ is integrable in thepth mean sense if(2.11)∫0∞E∥X(t)−X∞∥pdt<∞.If there exists a ℙ-null set Ω0 such that for every ω∉Ω0,
the following holds: limt→∞X(t,ω)=X∞(ω),
then Xconverges almost surely to X∞;
we say X is almost surely exponentially convergent to X∞ if there exists a deterministic β0>0 such that(2.12)limsup⁡t→∞⁢1tlog⁡∥X(t,ω)−X∞(ω)∥≤−β0,⁢a.s.Finally, the difference between
the stochastic process {X(t)}t≥0 and X∞ is square integrable in the almost sure
sense if(2.13)∫0∞∥X(t,ω)−X∞(ω)∥2dt<∞.Henceforth, 𝔼[Xp] will be denoted by 𝔼Xp except in cases where the meaning may be
ambiguous. A number of inequalities are used repeatedly in the sequel; they are
stated here for clarity. If, for p,q∈(0,∞),
the finite-dimensional random variables X and Y satisfy 𝔼∥X∥p<∞ and 𝔼∥Y∥q<∞,
respectively, then the Lyapunov inequality is useful when considering
the pth mean behaviour of random variables as any
exponent p>0 may be considered:(2.14)E[∥X∥p]1/p≤E[∥X∥q]1/q,0<p≤q.The following proves useful in
manipulating norms:(2.15)(∑i=1n|xi|)k≤nk−1∑i=1n|xi|k,n,k∈ℕ.


## 3. Discussion of Results

We begin by stating the main result of this paper.
That is, we state the necessary and sufficient conditions required on the
resolvent, kernel, deterministic perturbation, and noise terms for the solution
of (1.1a)-(1.1b) to
converge exponentially to a limiting random variable. In this paper, we are
particularly interested in the case when the limiting random variable is nontrivial,
although the result is still true for the case when the limiting value is zero.Theorem 3.1.
LetK satisfy (2.4) and(3.1)∫0∞t2∥K(t)∥dt<∞,let Σ satisfy (2.6), and let f satisfy (2.5). If K satisfies(3.2)eachentryofKdoesnotchangesignon[0,∞),then the following are
equivalent.
(i)
There exists a constant matrixR∞ such that the solution R of (1.2a)-(1.2b) satisfies(3.3)R−R∞∈L2([0,∞),ℝn×n),and there exist constants α>0,γ>0,ρ>0,
and c1>0 such that K satisfies(3.4)∫0∞∥K(s)∥eαs⁢ds<∞,Σ satisfies(3.5)∫0∞∥Σ(s)∥2e2γs⁢ds<∞,and f1,
the tail of f,
defined by (2.8) satisfies(3.6)∥f1(t)∥≤c1e−γt,t≥0.(ii)
For all initial conditionsX0 and constants p>0 there exists an  a.s. finite ℱB(∞)-measurable random variable X∞(X0,Σ,f) with 𝔼∥X∞∥p<∞ such that the unique continuous adapted
process X(⋅;X0,Σ,f) which obeys (1.1a)-(1.1b)
satisfies(3.7)E[∥X(t)−X∞∥p]≤mp*e−βp*t,t≥0,where βp* and mp*=mp*(X0) are positive constants.(iii)
For all initial conditionsX0 there exists an  a.s. finite ℱB(∞)-measurable random variable X∞(X0,Σ,f) such that the unique continuous adapted
process X(⋅;X0,Σ,f) which obeys (1.1a)-(1.1b)
satisfies(3.8)limsup⁡t→∞⁢1tlog⁡∥X(t)−X∞∥≤−β0*⁢a.s.,where β0* is a positive constant.The proof of Theorem3.1 is complicated by the
presence of two perturbations so as an initial step the case when f:=0 is considered. That is we consider the
conditions required for exponential convergence of (1.4a)-(1.4b) to a limiting random
variable.Theorem 3.2.
LetK satisfy (2.4) and (3.1) and let Σ satisfy (2.6). If K satisfies (3.2) then the following are
equivalent.
(i)
There exists a constant matrixR∞ such that the solution R of (1.2a)-(1.2b) satisfies (3.3) and there exist constants α>0 and γ>0 such that K and Σ satisfy (3.4) and (3.5), respectively.(ii)
For all initial conditionsX0 and constants p>0 there exists an  a.s. finite ℱB(∞)-measurable random variable X∞(X0,Σ) with 𝔼∥X∞∥p<∞ such that the unique continuous adapted
process X(⋅;X0,Σ) which obeys (1.4a)-(1.4b) satisfies(3.9)E[∥X(t)−X∞∥p]≤mpe−βpt,t≥0,where βp and mp=mp(X0) are positive constants.(iii)
For all initial conditionsX0 there exists an  a.s. finite ℱB(∞)-measurable random variable X∞(X0,Σ) such that the unique continuous adapted
process X(⋅;X0,Σ) which obeys (1.4a)-(1.4b) satisfies(3.10)limsup⁡t→∞⁢1tlog⁡∥X(t)−X∞∥≤−β0⁢a.s.,where β0 is a positive constant.This result is interesting in its own right as it
generalises a result in [4] where necessary and sufficient conditions for
exponential convergence to zero are found. Theorem 3.2 collapses to this case
if R∞=0.It is interesting to note the relationship between the
behaviour of the solutions of (1.1a)-(1.1b), (1.2a)-(1.2b),
(1.3a)-(1.3b),
and (1.4a)-(1.4b) and the behaviour of the inputs K,f,
and Σ.
It is seen in [1] that K being exponentially integrable is the crucial
condition for exponential convergence when we consider the resolvent equation.
Each perturbed equation then builds on this resolvent case: for the
deterministically perturbed equation we require the exponential integrability
of K and the exponential decay of the tail of the
perturbation f (see [1]); for the stochastically perturbed case we require the
exponential integrability of K and the exponential square integrability of Σ.
In the stochastically and deterministically perturbed case it is seen that the
perturbations do not interact in a way that exacerbates or diminishes the
influence of the perturbations on the system: we can isolate the behaviours of
the perturbations and show that the same conditions on the perturbations are
still necessary and sufficient.Theorem3.1 has application in the analysis of initial
history problems. In particular this theoretical result could be used to
interpret the equation as an epidemiological model. Conditions under which a
disease becomes endemic (which is the interpretation that is given when
solutions settle down to a nontrivial limit) were studied in [9]. The theoretical results
obtained in this paper could be exploited to highlight the speed at which this
can occur within a population.The remainder of this paper deals with the proofs of
Theorems3.1 and
3.2. In Section  4 we
prove the sufficiency of conditions on R,K,
and Σ for the exponential convergence of the
solution of (1.4a)-(1.4b) while in
Section 5 we prove the necessity of these conditions. In
Section 6 we prove the sufficiency of conditions on R,K,Σ,
and f for the exponential convergence of the
solution of (1.1a)-(1.1b), while Section 7 deals with the necessity of the
condition on Σ.
In Section 8 we combine our results to prove the main theorems, namely,
Theorems 3.1 and
3.2.


## 4. Sufficient Conditions for Exponential Convergence of Solutions of (1.4a)-(1.4b)

In this section, sufficient conditions for exponential
convergence of solutions of (1.4a)-(1.4b) to a nontrivial limit are obtained.
Proposition 4.1 concerns convergence in the pth mean sense while Proposition  4.2 deals with the almost sure
case.Proposition 4.1.
LetK satisfy (2.4) and (3.1), let Σ satisfy (2.6) and R∞ be a constant matrix such that the solution R of (1.2a)-(1.2b) satisfies (3.3). If there exist constants α>0 and γ>0 such that (3.4) and (3.5) hold, then there
exist constants βp>0,
independent of X0,
and mp=mp(X0)>0,
such that statement (ii) of Theorem 3.2 holds.Proposition 4.2.
LetK satisfy (2.4) and (3.1), let Σ satisfy (2.6) and R∞ be a constant matrix such that the solution R of (1.2a)-(1.2b) satisfies (3.3). If there exist constants α>0 and γ>0 such that (3.4) and (3.5) hold, then there
exists a constant β0>0,
independent of X0 such that statement (iii) of Theorem 3.2 holds.In [8], the conditions which give mean square convergence to
a nontrivial limit were considered. So a natural progression in this paper is
the examination of the speed of convergence in the mean square case. Lemma 4.3
examines the case when p=2 in order to highlight this important case.
This lemma may be then used when generalising the result to all p>0.Lemma 4.3.
LetK satisfy (2.4) and (3.1), let Σ satisfy (2.6), and let R∞ be a constant matrix such that the solution R of (1.2a)-(1.2b) satisfies (3.3). If there exist constants α>0 and γ>0 such that (3.4) and (3.5) hold, then there
exist constants λ>0,
independent of X0,
and m=m(X0)>0,
such that(4.1)E∥X(t)−X∞∥2≤m(X0)e−2λt,t≥0.From [8, 9] it is evident that R−R∞∈L2([0,∞),ℝn×n) is a more natural condition on the resolvent
than R−R∞∈L1([0,∞),ℝn×n) when studying convergence of solutions of
(1.4a)-(1.4b). However, the deterministic results obtained in [1] are based on the assumption
that R−R∞∈L1([0,∞),ℝn×n).
Lemma 4.4 is required in order to make use of these results in this paper; this
result isolates conditions that ensure the integrability of R−R∞ once R−R∞ is square integrable.Lemma 4.4.
LetK satisfy (2.4) and (3.1) and let R∞ be a constant matrix such that the solution R of (1.2a)-(1.2b) satisfies (3.3). Then the solution R of (1.2a)-(1.2b) satisfies(4.2)R−R∞∈L1([0,∞),ℝn×n).We now state some supporting results. It is well known
that the behaviour of the resolvent Volterra equation influences the behaviour
of the perturbed equation. It is unsurprising therefore that an earlier result
found in [1]
concerning exponential convergence of the resolvent R to a limit R∞ in needed in the proof of Theorems 3.1 and
3.2.Theorem 4.5.
LetK satisfy (2.4) and (3.1). Suppose there exists
a constant matrix R∞ such that the solution R of (1.2a)-(1.2b) satisfies (4.2). If there exists a constant α>0 such that K satisfies (3.4) then there exist constants β>0 and c>0 such that(4.3)∥R(t)−R∞∥≤ce−βt,t≥0.In the proof of Propositions4.1 and
4.2, an explicit
representation of X∞ is required. In [8, 9] the asymptotic convergence of the solution of (1.4a)-(1.4b)
was considered. Sufficient conditions for convergence were obtained and an
explicit representation of X∞ was found:Theorem 4.6.
LetK satisfy (2.4) and(4.4)∫0∞t∥K(t)∥dt<∞,and let Σ satisfy (2.6) and(4.5)∫0∞∥Σ(t)∥2dt<∞.Suppose that the resolvent R of (1.2a)-(1.2b) satisfies (3.3). Then the solution X of (1.4a)-(1.4b) satisfies limt→∞X(t)=X∞ almost surely, where X∞ is an almost surely finite and ℱB(∞)-measurable random variable given
by(4.6)X∞=R∞(X0+∫0∞Σ(t)dB(t))⁢a.s.Lemma4.7 concerns the structure of X∞ in the almost sure case. It was proved in
[9].Lemma 4.7.
LetK satisfy (2.4) and (4.4). Suppose that for all
initial conditions X0 there is an almost surely finite random
variable X∞(X0,Σ) such that the solution t↦X(t;X0,Σ) of (1.4a)-(1.4b) satisfies(4.7)lim⁡t→∞⁢X(t;X0,Σ)=X∞(X0,Σ)a.s.,(4.8)X(⋅;X0,Σ)−X∞(X0,Σ)∈L2([0,∞),ℝn)a.s.Then(4.9)(A+∫0∞K(s)ds)X∞=0⁢a.s.It is possible to apply this lemma using oura
priori assumptions due to Theorem 4.8, which was proved in [9].Theorem 4.8.
LetK satisfy (2.4) and (4.4) and let Σ satisfy (2.6). If Σ satisfies (4.5) and there exists a constant matrix R∞ such that the solution R of (1.2a)-(1.2b) satisfies (3.3), then for all initial conditions X0 there is an almost surely finite ℱB(∞)-measurable random variable X∞(X0,Σ),
such that the unique continuous adapted process X(⋅;X0,Σ) which obeys (1.4a)-(1.4b) satisfies (4.7).
Moreover, if the functionΣ also satisfies(4.10)∫0∞t∥R∞Σ(t)∥2dt<∞,then (4.8) holds.Lemma4.9 below is required in the proof of Lemma
4.4.
It is proved in [8].
Before citing this result some notation is introduced. Let M=A+∫0∞K(t)dt and T be an invertible matrix such that J=T−1MT has Jordan canonical form. Let ei=1 if all the elements of the ith row of J are zero, and ei=0 otherwise. Let Dp=diag(e1,e2,…,en) and put P=TDpT−1 and Q=I−P.Lemma 4.9.
LetK satisfy (2.4) and (4.4). If there exists a
constant matrix R∞ such that the resolvent R of (1.2a)-(1.2b) satisfies (3.3), then(4.11)det⁡[I+F^(z)]≠0,Rez≥0,where F is defined by(4.12)F(t)=−e−t(Q+QA)−(e⁢*QK)(t)+P∫t∞K(u)du,t≥0.Lemma4.10 concerns the moments of a normally
distributed random variable. It can be extracted from [4, Theorem 3.3] and it is used
in Proposition 4.1.Lemma 4.10.
Suppose the functionσ∈C([0,∞)×[0,∞),ℝp×r) then(4.13)E‖∫abσ(s,t)dB(s)‖2m≤dm(p,r)(∫ab∥σ(s,t)∥2ds)m,where dm(p,r)=pm+1r2m+1(2m)!(m!2m)−1c2(p,r)m.The following lemma is used in Proposition4.2. A
similar result is proved in [4].Lemma 4.11.
Suppose thatK∈C([0,∞),ℝn×n)∩L1([0,∞),ℝn×n) and(4.14)∫0∞∥K(s)∥eαsds<∞.If λ>0 and η=2λ⋀α then(4.15)∫0te−2λ(t−s)e−αs∥K(s)∥ds≤ce−ηt,where c is a positive constant.The proofs of Propositions4.1 and
4.2 and Lemmas 4.3
and 4.4 are now given.Proof of Lemma4.3.
From Theorem4.6 we see that X(t)→X∞ almost surely where X∞ is given by (4.6), so we see
that(4.16)E∥X∞∥2=E[tr(X∞X∞T)]=∥R∞X0∥2+∫0∞∥R∞Σ(s)∥2ds<∞.Since(4.17)E[∥X(t)−X∞∥2]=E[tr(X(t)−X∞)(X(t)−X∞)T],we use (2.9) and (4.6) to expand
the right hand side of (4.17) to obtain(4.18)E[∥X(t)−X∞∥2]=∥(R(t)−R∞)X0∥2+∫0t∥(R(t−s)−R∞)Σ(s)∥2ds+∫t∞∥R∞Σ(s)∥2ds.In order to obtain an
exponential upper bound on (4.18) each term is considered individually. We
begin by considering the first term on the right-hand side of (4.18). Using
(3.1) and (3.3) we can apply Lemma 4.4 to obtain (4.2). Then using (3.1),
(4.2), and (3.4) we see from Theorem 4.5 that(4.19)∥(R(t)−R∞)X0∥2≤c1∥X0∥2e−2βt.We provide an argument to show
that the second term decays exponentially. Using (3.5) and the fact that R decays exponentially quickly to R∞ we can choose 0<λ<min(β,γ) such that eλΣ and eλ(R−R∞)∈L2[0,∞) where the function eλ is defined by eλ(t)=eλt.
Since the convolution of an L2[0,∞) function with an L2[0,∞) function is itself an L2[0,∞) function we get(4.20)e2λt∫0t∥(R(t−s)−R∞)Σ(s)∥2ds≤∫0te2λ(t−s)∥R(t−s)−R∞∥2e2λs∥Σ(s)∥2ds≤c2,and so the second term of (4.18)
decays exponentially quickly.
We can show that the third term on the right hand side
of (4.18) decays exponentially using (3.5) and the following
argument:(4.21)Σ¯:=∫0∞∥Σ(s)∥2e2γsds≥∫t∞∥Σ(s)∥2e2γsds≥e2γt∫t∞∥Σ(s)∥2ds.
Combining these facts we see that(4.22)E[∥X(t)−X∞∥2]≤m(X0)e−2λt,where m(X0)=c1∥X0∥2+c2+Σ¯∥R∞∥2 and λ<min(β,γ).Proof of Proposition4.1.
Consider
the case where0<p≤2 and p>2 separately. We begin with the case where 0<p≤2.
The argument given by (4.16) shows that 𝔼[∥X∞∥2]<∞.
Now applying Lyapunov's inequality we see that(4.23)E∥X∞∥p≤E[∥X∞∥2]p/2<∞.We now show that (3.9) holds for 0≤p≤2.
Lyapunov's inequality and Lemma 4.3 can be applied as follows:(4.24)E[∥X(t)−X∞∥p]≤E[∥X(t)−X∞∥2]p/2≤mp(X0)e−βpt,t≥0,where mp(X0)=m(X0)p/2 and βp=λp.
Now consider the case wherep>2.
In this case, there exists a constant m∈ℕ such that 2(m−1)<p≤2m.
We now seek an upper bound on 𝔼∥X∞∥2m and 𝔼[∥X(t)−X∞∥2m],
which will in turn give an upper bound on 𝔼∥X∞∥p and 𝔼[∥X(t)−X∞∥p] by using Lyapunov's inequality. By applying
Lemma 4.10 we see that(4.25)E∥X∞∥2m≤c∥R∞X0∥2m+c(∫0∞∥R∞Σ(s)∥2ds)m<∞,where c is a positive constant, so 𝔼∥X∞∥p≤𝔼[∥X∞∥2m]p/2m<∞.
Now consider𝔼[∥X(t)−X∞∥2m].
Using the variation of parameters representation of the solution and the
expression obtained for X∞,
taking norms, raising both sides of the equation to the 2mth power, then taking expectations across the
inequality, we arrive at(4.26)E[∥X(t)−X∞∥2m]≤32m−1(∥(R(t)−R∞)X0∥2m+E[‖∫0t(R(t−s)−R∞)Σ(s)dB(s)‖2m]+E[‖∫t∞R∞Σ(s)dB(s)‖2m]).We consider each term on the
right hand side of (4.26). By Theorem 4.5 we have(4.27)∥(R(t)−R∞)X0∥2m≤c1∥X0∥2me−2mβt.Now, consider the second term on
the right-hand side of (4.26). By (4.20) we see that ∫0t∥(R(t−s)−R∞)Σ(s)∥2ds≤c2e−2λt where λ<min(β,γ).
Using this and Lemma 4.10 we see that(4.28)E[∥∫0t(R(t−s)−R∞)Σ(s)dB(s)∥2m]≤dm(n,d)(∫0t∥R(t−s)−R∞∥2∥Σ(s)∥2ds)m≤dm(n,d)c2me−2mλt.
Using (4.21) combined with Lemma 4.10 and Fatou's
lemma, we show that the third term decays exponentially quickly:(4.29)E[‖∫t∞Σ(s)dB(s)‖2m]≤dm(n,d)(∫t∞∥Σ(s)∥2ds)m≤dm(n,d)Σ¯me−2mγt.Combining (4.27), (4.28), and
(4.29) the inequality (4.26) becomes(4.30)E[∥X(t)−X∞∥2m]≤32m−1(c1∥X0∥2me−2mβt+dm(n,d)c2me−2mλt+dm(n,d)∥R∞∥2mΣ¯me−2mγt).Using Lyapunov's inequality, the
inequality (4.30) implies(4.31)E[∥X(t)−X∞∥p]≤mp(X0)e−βpt,where mp(X0)=3p((2m−1)/2m)(c1∥X0∥2m+dm(n,d)c2m+dm(n,d)∥R∞∥2mΣ¯m)p/2m and βp=λp.Proof of Proposition4.2.
In order
to prove this proposition we show that(4.32)E[sup⁡n−1≤t≤n∥X(t)−X∞∥2]≤m(X0)e−2η(n−1),η>0.For each t>0 there exists n∈ℕ such that n−1≤t<n.
Define Δ(t)=X(t)−X∞.
Integrating (1.4a)-(1.4b) over [n−1,t],
then adding and subtracting X∞ on both sides we get(4.33)X(t)−X∞=(X(n−1)−X∞)+∫n−1tA(X(s)−X∞)ds+∫n−1t∫0sK(s−u)(X(u)−X∞)du⁢ds+∫n−1tΣ(s)dB(s)+∫n−1t(A+∫0∞K(u)du)X∞⁢ds−∫n−1t∫s∞K(v)X∞⁢dv⁢ds.By applying Theorem 4.8, we see
that (4.7) and (4.8) hold so Lemma 4.7 may be applied to obtain(4.34)Δ(t)=Δ(n−1)+∫n−1t(AΔ(s)+(K*Δ)(s))ds+∫n−1tΣ(s)dB(s)−∫n−1tK1(s)dsX∞.Taking norms on both sides of
(4.34), squaring both sides, taking suprema, before finally taking expectations
yields:(4.35)E[sup⁡n−1≤t≤n∥Δ(t)∥2]≤4{E∥Δ(n−1)∥2+E[(∫n−1n∥A∥∥Δ(s)∥+(∥K∥*∥Δ∥)(s)ds)2]+E[sup⁡n−1≤t≤n‖∫n−1tΣ(s)dB(s)‖2]+(∫n−1n∥K1(s)∥ds)2E∥X∞∥2}.We now consider each term on the
right hand side of (4.35). From Lemma 4.3 we see that the first term
satisfies(4.36)E[∥Δ(n−1)∥2]≤m(X0)e−2λ(n−1).In order to obtain an
exponential bound on the second term on the right hand side of (4.26) we make
use of the Cauchy-Schwarz inequality as follows:(4.37)(∫n−1n∥A∥∥Δ(s)∥+(∥K∥*∥Δ∥)(s)ds)2≤2∫n−1n(∥A∥2∥Δ(s)∥2+(∥K∥*∥Δ∥)2(s))ds≤2∫n−1n[∥A∥2∥Δ(s)∥2+(∫0seα(s−u)/2∥K(s−u)∥1/2e−α(s−u)/2∥K(s−u)∥1/2∥Δ(s)∥du)2]ds≤2∫n−1n[∥A∥2∥Δ(s)∥2+K¯α∫0se−α(s−u)∥K(s−u)∥∥Δ(s)∥2du]ds,where K¯α=∫0∞eαt∥K(t)∥dt.
Take expectations and examine the two terms within the integral. Using Lemma
4.3 we obtain(4.38)E[∫n−1n∥A∥2∥Δ(s)∥2ds]≤∥A∥2m(X0)∫n−1ne−2λsds≤c1(X0)e−2λ(n−1).In order to obtain an
exponential upper bound for the second term within the integral we apply Lemma
4.11 with K=K, α=α,λ=λ and η=η:(4.39)E[∫n−1nK¯α∫0se−α(s−u)∥K(s−u)∥∥Δ(s)∥2du⁢ds]≤m(X0)K¯α∫n−1n∫0se−αu∥K(u)∥e−2λ(s−u)⁢du⁢ds≤c2(X0)e−η(n−1).Next, we obtain an exponential
upper bound on the third term. Using (4.21) and the Burkholder-Davis-Gundy
inequality, there exists a constant c3>0 such that(4.40)E[sup⁡n−1≤t≤n‖∫n−1tΣ(s)dB(s)‖2]≤c3Σ¯e−2γ(n−1).Now consider the last term on
the right hand side of (4.35). Using (3.4) we see that(4.41)K¯α:=∫0∞∥K(s)∥eαsds≥eαt∫t∞∥K(s)∥ds≥eαt∥K1(t)∥.Using this and the fact that 𝔼∥X∞∥2<∞ (see (4.16)) we obtain(4.42)(∫n−1n∥K1(s)∥ds)2E∥X∞∥2≤E∥X∞∥2(∫n−1nK¯αe−αsds)2≤c4e−2α(n−1).Combining (4.36), (4.38),
(4.39), (4.40), and (4.42) we obtain(4.43)E[sup⁡n−1≤t≤n∥X(t)−X∞∥2]≤m(X0)e−2η(n−1),where m(X0)=4(m(X0)+c1(X0)+c2(X0)+c3Σ¯+c4) and η<min(2λ,α).
We can now apply the line of reasoning used in
[10, Theorem 4.4.2] to
obtain (3.10).Proof of Lemma4.4.
We use a reformulation of (1.2a)-(1.2b) in the proof of
this result. It is obtained as follows: multiply both sides of R′(s)=AR(s)+(K*R)(s) across by the function Φ(t−s),
where Φ(t)=P+e−tQ,
integrate over [0,t],
use integration by parts, add and subtract R∞ from both sides to obtain(4.44)Y(t)+(F*Y)(t)=G(t),t≥0,where Y=R−R∞, F is defined by (4.12) and G is defined by(4.45)G(t)=e−tQ−e−t(QR∞+QAR∞)+∫t∞∫s∞PK(u)R∞⁢du⁢ds−∫t∞QK(u)R∞⁢du−(e⁢*QKR∞)(t),t≥0.
Consider the reformulation of (1.2a)-(1.2b) given by (4.44).
It is well known that Y can be expressed as(4.46)Y(t)=G(t)−∫0tr(t−s)G(s)ds,where the function r satisfies r+F*r=F and r+r*F=F.
We refer the reader to [11] for details. Consider the first term on the right hand
side of (4.46). As (3.1) holds it is clear that the function G is integrable. Now consider the second term.
Since (3.3) and (4.4) hold we may apply Lemma 4.9 to obtain (4.11). Now we may
apply a result of Paley and Wiener (see [11]) to see that r is integrable. The convolution of an
integrable function with an integrable function is itself integrable. Now
combining the arguments for the first and second terms we see that (4.2) must
hold.


## 5. On the Necessity of (3.5) for Exponential Convergence of Solutions of (1.4a)-(1.4b)

In this section, the necessity of condition (3.5) for
exponential convergence in the almost sure and pth mean senses is shown. Proposition 5.1 concerns
the necessity of the condition in the almost sure case while Proposition 5.2
deals with the pth mean case.Proposition 5.1.
LetK satisfy (2.4) and (4.4) and Σ satisfy (2.6). If there exists a constant α>0 such that (3.4) holds, and if for all X0 there is a constant vector X∞(X0,Σ) such that the solution t↦X(t;X0,Σ) of (1.4a)-(1.4b) satisfies statement (iii) of Theorem
3.2, then there exists a constant γ>0,
independent of X0,
such that (3.5) holds.Proposition 5.2.
LetK satisfy (2.4) and (4.4) and Σ satisfy (2.6). If there exists a constant α>0 such that (3.4) holds, and if for all X0 there is a constant vector X∞(X0,Σ) such that the solution t↦X(t;X0,Σ) of (1.4a)-(1.4b) satisfies statement (ii) of Theorem
3.2, then there exists a constant γ>0,
independent of X0,
such that (3.5) holds.In order to prove these propositions the integral
version of (1.4a)-(1.4b) is considered. By reformulating this version of the equation
an expression for a term related to the exponential integrability of the
perturbation is found. Using various arguments, including the Martingale
Convergence Theorem in the almost sure case, this term is used to show that
(3.5) holds.Some supporting results are now stated. Lemma5.3 is the analogue of
Lemma 4.7 in the mean
square case. It was proved in [8].Lemma 5.3.
LetK satisfy (2.4) and (4.4). Suppose that for all
initial conditions X0 there is a ℱB(∞)-measurable and almost surely finite random
variable X∞(X0,Σ) with 𝔼∥X∞∥2<∞ such that the solution t↦X(t;X0,Σ) of (1.4a)-(1.4b) satisfies(5.1)lim⁡t→∞⁢E∥X(t;X0,Σ)−X∞(X0,Σ)∥2=0,E∥X(⋅;X0,Σ)−X∞(X0,Σ)∥2∈L1([0,∞),ℝ).Then X∞ obeys(5.2)(A+∫0∞K(s)ds)X∞=0⁢a.s.Lemma5.4 may be extracted from [4]; it is required in the proof
of Proposition 5.2.Lemma 5.4.
LetN=(N1,…,Nn) where Ni∼𝒩(0,vi2) for i=1,…,n.
Then there exists a {vi}i=1n-independent constant d1>0 such that(5.3)E[∥N∥2]≤d1E[∥N∥]2.Proof of Proposition5.1.
In order to prove this result we follow the argument
used in [4, Theorem 4.1]. Let 0<γ<α⋀β0.
By defining the process Z(t)=eγtX(t) and the matrix κ(t)=eγtK(t) we can rewrite (1.4a)-(1.4b) as(5.4)dZ(t)=((γI+A)Z(t)+∫0tκ(t−s)Z(s)ds)dt+eγtΣ(t)dB(t),the integral form of which
is(5.5)Z(t)−Z(0)=(γI+A)∫0tZ(s)ds+∫0t∫0sκ(s−u)Z(u)du⁢ds+∫0teγsΣ(s)dB(s).Using Z(t)=eγtX(t) and rearranging this becomes(5.6)∫0teγsΣ(s)dB(s)=eγtX(t)−X0−(γI+A)∫0teγsX(s)ds−∫0teγs∫0sK(s−u)X(u)du⁢ds.Adding and subtracting X∞ from the right hand side and applying Lemma
4.7 we obtain:(5.7)∫0teγsΣ(s)dB(s)=eγt(X(t)−X∞)−(X0−X∞)−(γI+A)∫0teγs(X(s)−X∞)ds−∫0teγs∫0sK(s−u)(X(u)−X∞)du⁢ds+∫0teγsK1(s)dsX∞.Consider each term on the right
hand side of (5.7). We see that the first term tends to zero as
(3.10) holds
and γ<β0.
The second term is finite by hypothesis. Again, using the fact that γ<β0 and that assumption (3.10)
holds we see that eγ(X−X∞)∈L1[0,∞),
so the third term tends to a limit as t→∞.
Now consider the fourth term. Since 0<γ<α⋀β0,
we can choose γ1>0 such that γ<γ1<α⋀β0.
So the functions t↦eγ1tK(t) and t↦eγ1t(X(t)−X∞) are both integrable. The convolution of these
two integrable functions is itself an integrable function, so(5.8)‖∫0sK(s−u)(X(u)−X∞)du‖≤ce−γ1s.Thus, it is clear that the
fourth term has a finite limit as t→∞.
Finally, the fifth term on the right hand side of (5.7) has a finite limit at
infinity, using (4.41).
Each term on the right hand side of the inequality has
a finite limit ast→∞,
so therefore(5.9)lim⁡t→∞⁢∫0teγsΣ(s)dB(s)existsandisalmostsurelyfinite.The Martingale Convergence
Theorem [12,
Proposition 5.1.8] may now be applied component by component to obtain (3.5).Proof of Proposition5.2.
By
Lemma5.3, (5.7) still holds. Define γ<α⋀β1,
take norms and expectations across (5.7) to obtain(5.10)E[‖∫0teγsΣ(s)dB(s)‖]≤E[eγt∥X(t)−X∞∥]+E[∥X0−X∞∥]+∥γI+A∥∫0tE[eγs∥X(s)−X∞∥]ds+∫0teγs∫0s∥K(u)∥E[∥X(s−u)−X∞∥]du⁢ds+∫0teγs∥K1(s)∥dsE∥X∞∥.There exists m1 such that(5.11)E[eγt∥X(t)−X∞∥]≤m1e−(β1−γ)t,thus the first, second and third
terms on the right hand side of (5.10) are uniformly bounded on [0,∞).
Now consider the fourth term. Since 0<γ<α⋀β1,
we can choose γ1>0 such that γ<γ1<α⋀β1 so that the functions t↦eγ1tK(t) and t↦eγ1t𝔼∥X(t)−X∞∥ are both integrable. The convolution of two
integrable functions is itself an integrable function, so(5.12)∫0s∥K(s−u)∥E∥X(u)−X∞∥du≤ce−γ1s,so it is clear that the fourth
term is uniformly bounded on [0,∞).
Finally, we consider the final term on the right hand side of (5.10). Using
(4.41) we obtain(5.13)∫0teγs∥K1(s)∥dsE∥X∞∥≤K¯αE∥X∞∥∫0te−(α−γ)s⁢ds<∞,since γ<α.
Thus there is a constant c>0 such that(5.14)E[‖∫0teγsΣ(s)dB(s)‖]≤c.The proof now follows the line
of reasoning found in [4, Theorem 4.3]: observe that(5.15)‖∫0teγsΣ(s)dB(s)‖2=∑i=1nNi(t)2,where(5.16)Ni(t)=∑j=1d∫0teγsΣij(s)dBj(s).It is clear that Ni(t) is normally distributed with zero mean and
variance given by(5.17)vi(t)2=∑j=1d∫0te2γsΣij(s)2ds.Lemma 5.4 and (5.14) may now be
applied to obtain:(5.18)∫0te2γs∥Σ(s)∥2ds=∑i=1n∑j=1d∫0te2γs|Σij(s)|2⁢ds=∑i=1nvi(t)2=E[‖∫0teγsΣ(s)dB(s)‖2]≤d1E[‖∫0teγsΣ(s)dB(s)‖]2≤d1c2.Allowing t→∞ on both sides of this inequality yields the
desired result.


## 6. Sufficient Conditions for Exponential Convergence of Solutions of (1.1a)-(1.1b)

In this section, sufficient conditions for exponential
convergence of solutions of (1.1a)-(1.1b) to a nontrivial limit are found. Proposition
6.1
concerns the pth mean sense while Proposition 6.2 deals with
the almost sure case.Proposition 6.1.
LetK satisfy (2.4) and (3.1), let Σ satisfy (2.6), let f satisfy (2.5), and let R∞ be a constant matrix such that the solution R of (1.2a)-(1.2b) satisfies (3.3). If there exist constants α>0, ρ>0 and γ>0 such that  (3.4), (3.6) and (3.5)
hold, then there exist constants βp*>0,
independent of X0,
and mp*=mp*(X0)>0,
such that statement (ii) of Theorem 3.1 holds.Proposition 6.2.
LetK satisfy (2.4) and (3.1), let Σ satisfy (2.6), let f satisfy (2.5), and let R∞ be a constant matrix such that the solution R of (1.2a)-(1.2b) satisfies (3.3). If there exist constants α>0,ρ>0  and γ>0  such that (3.4), (3.6) and (3.5)
hold, then there exists constant β0*>0,
independent of X0 such that statement (iii) of Theorem 3.1
holds.As in the case wheref:=0 we require an explicit formulation for X∞.
The proof of this result follows the line of reasoning used in the proof of
Theorem 4.6 and is therefore omitted.Theorem 6.3.
LetK satisfy (2.4) and (4.4), let Σ satisfy (2.6) and (4.5), and let f satisfy
(2.5). Suppose that the resolvent R of (1.2a)-(1.2b) satisfies (3.3), then the solution X of (1.1a)-(1.1b) satisfies X→X∞(X0,Σ,f) almost surely, where(6.1)X∞(X0,Σ,f)=X∞(X0,Σ)+R∞∫0∞f(t)dt,⁢a.s.and X∞(X0,Σ,f) is almost surely finite.Proof of Proposition6.1.
We
begin by showing that𝔼∥X∞(X0,Σ,f)∥p is finite. Clearly, we see
that(6.2)E∥X∞(X0,Σ,f)∥p≤2p−1(E∥X∞(X0,Σ)∥p+‖∫0∞R∞f(s)ds‖p)<∞.Now, consider the difference
between the solution X(⋅;X0,Σ,f) of (1.1a)-(1.1b) and its limit X∞(X0,Σ,f) given by (6.1):(6.3)X(t;X0,Σ,f)−X∞(X0,Σ,f)=(X(t;X0,Σ)−X∞(X0,Σ))+∫0t(R(t−s)−R∞)f(s)ds−∫t∞R∞f(s)ds.Using integration by parts this
expression becomes(6.4)X(t;X0,Σ,f)−X∞(X0,Σ,f)=(X(t;X0,Σ)−X∞(X0,Σ))−f1(t)+(R(t)−R∞)f1(0)−∫0tR′(t−s)f1(s)ds.Taking norms on both sides of
equation (6.4), raising the power to p on both sides, and taking expectations across
we obtain(6.5)E∥X(t;X0,Σ,f)−X∞(X0,Σ,f)∥p≤4p−1(E∥X(t;X0,Σ)−X∞(X0,Σ)∥p+∥f1(t)∥p+∥R(t)−R∞∥p∥f1(0)∥p+(∫0t∥R′(t−s)∥∥f1(s)∥ds)p).Now consider the right hand side
of (6.5). The first term decays exponentially quickly due to Theorem 3.2. The
second term decays exponentially quickly due to assumption (3.6). By applying
Lemma 4.4 we see that (4.2) holds so we can apply Theorem 4.5 to show that the
third term must decay exponentially. In the sequel, an argument is provided to
show that R′ decays exponentially; thus the final term must
decay exponentially. Combining these arguments we see that (3.7) holds, where βp*<min(βp,β,ρ).
It is now shown thatR′ decays exponentially. It is clear from the
resolvent equation (1.2a)-(1.2b) that(6.6)R′(t)=A(R(t)−R∞)+∫0tK(t−s)(R(s)−R∞)ds−K1(t)R∞+(A+∫0∞K(s)ds)R∞.Consider each term on the right
hand side of (6.6). We can apply Theorem 4.5 to obtain that R decays exponentially quickly to R∞.
In order to show that the second term decays exponentially we proceed as
follows: since R−R∞ decays exponentially and (3.4) holds it is
possible to choose μ such that the functions t↦eμtK(t) and t↦eμt(R(t)−R∞) are both in L1([0,∞),ℝn×n).
The convolution of two integrable functions is itself an integrable function,
so(6.7)eμt‖∫0tK(t−s)(R(s)−R∞)ds‖=‖∫0teμ(t−s)K(t−s)eμs(R(s)−R∞)ds‖≤c.To see that the third term
decays exponentially we use (4.41). Finally, we consider the fourth term. By
Lemma 4.4 and (3.3) we have that (4.2) holds. In [1, Theorem 6.1] it was shown that (A+∫0∞K(s)ds)R∞=0 under this hypothesis and (3.1). Combining the
above we see that R′ decays exponentially quickly to 0.Proof of Proposition6.2.
Take norms across (6.4) to obtain(6.8)∥X(t;X0,Σ,f)−X∞(X0,Σ,f)∥≤∥X(t;X0,Σ)−X∞(X0,Σ)∥+∥f1(t)∥+∥R(t)−R∞∥∥f1(0)∥+‖∫0tR′(t−s)f1(s)ds‖.Using Theorem 3.2, we see that
the first term on the right hand side of (6.8) decays exponentially. The second
term on the right hand side decays exponentially as (3.6) holds. We can apply
Theorem 4.5 to show that the third term must decay exponentially. An argument
was provided in Proposition 6.1 to show that R′ decays exponentially. Combining this with
(3.6) enables us to show that the fourth term decays exponentially. Using the
above arguments we obtain (3.8), where β*≤min(β0,β,ρ).


## 7. On the Necessity of (3.6) and (3.5) for Exponential Convergence of Solutions of (1.1a)-(1.1b)

In this section, the necessity of (3.6) and (3.5) for
exponential convergence of solutions of (1.1a)-(1.1b) in the almost
sure and pth mean senses is shown. Proposition 7.1 concerns
the necessity of the conditions in the pth mean case while Proposition 7.2 deals with the
almost sure case.Proposition 7.1.
LetK satisfy (2.4) and (4.4), let Σ satisfy (2.6), and let f satisfy (2.5). If there exists constant α>0 such that (3.4) holds, and if for all X0 there is constant vector X∞(X0,Σ,f) such that the solution t↦X(t;X0,Σ,f) of (1.1a)-(1.1b) satisfies statement (ii) of
Theorem 3.1, then there
exist constants ρ>0 and γ>0,
independent of X0,
such that (3.6) and (3.5) hold.Proposition 7.2.
LetK satisfy (2.4) and (4.4), let Σ satisfy (2.6), and let f satisfy (2.5). If there exists constant α>0 such that (3.4) holds, and if for all X0 there is a constant vector X∞(X0,Σ,f) such that the solution t↦X(t;X0,Σ,f) of (1.1a)-(1.1b) satisfies statement (iii) of
Theorem 3.1, then
there exist constants ρ>0 and γ>0,
independent of X0,
such that (3.6) and (3.5) hold.The following lemma is used in the proof of
Proposition7.2. This lemma allows us to separate the behavior of the
deterministic perturbation from the stochastic perturbation in the almost sure
case. It is interesting to note that we can prove this lemma without any
reference to the integro-differential equation.Lemma 7.3.
Supposec>0 is an almost surely finite random variable
and(7.1)∥f1(t)+μ1(t,ω)∥≤c(ω)e−λt,where λ>0, ω∈Ω*, ℙ[Ω*]=1 and the functions f1 and μ1 are defined by (2.8) and(7.2)μ1(t)=∫t∞Σ(s)dB(s),t≥0,respectively. Then (3.5) and
(3.6) hold.In order to prove Lemma7.3 we require Lemmas
7.4 and
7.5 below. Lemma 7.5 was proved in [13]. The proof of
Lemma 7.4 makes use of Kolmogorov's
Zero-One Law. It follows the proof of Theorem 2 in [14, Chapter IV, Section 1] and so is omitted.Lemma 7.4.
Let{ξi}i=1∞ be a sequence of independent Gaussian random
variables with 𝔼[ξi]=0 and 𝔼[ξi2]=vi2≥1.
Then(7.3)limsup⁡m→∞⁢∑i=1mξi=∞,liminf⁡m→∞⁢∑i=1mξi=−∞,⁢a.s.Lemma 7.5.
If there is aγ>0 such that σ∈C([0,∞),ℝ) and(7.4)∫0∞σ(s)2e2γs⁢ds<∞,then(7.5)limsup⁡t→∞⁢1tlog⁡|∫t∞σ(s)dB(s)|≤−γ,⁢a.s.where {B(t)}t≥0 is a one-dimensional standard Brownian
motion.Lemmas7.6 and 7.7 are used in the proofs of
Propositions 7.1 and
7.2, respectively and are the analogues of Lemmas 5.3 and
4.7. Their proofs are identical in all important aspects and so are
omitted.Lemma 7.6.
LetK satisfy (2.4) and (4.4). Suppose that for all
initial conditions X0 there is an ℱB(∞)-measurable and almost surely finite random
variable X∞(X0,Σ) with 𝔼∥X∞∥2<∞ such that the solution t↦X(t;X0,Σ) of (1.1a)-(1.1b) satisfies(7.6)lim⁡t→∞⁢E∥X(t;X0,Σ,f)−X∞(X0,Σ,f)∥2=0,E∥X(⋅;X0,Σ,f)−X∞(X0,Σ,f)∥2∈L1([0,∞),ℝ).Then X∞ obeys(7.7)(A+∫0∞K(s)ds)X∞=0⁢a.s.Lemma 7.7.
LetK satisfy (2.4) and (4.4). Suppose that for all
initial conditions X0 there is an ℱB(∞)-measurable and almost surely finite random
variable X∞(X0,Σ,f) such that the solution t↦X(t;X0,Σ,f) of (1.1a)-(1.1b) satisfies(7.8)lim⁡t→∞⁢X(t;X0,Σ,f)=X∞(X0,Σ,f)a.s.,X(⋅;X0,Σ,f)−X∞(X0,Σ,f)∈L2([0,∞),ℝn)a.s.Then X∞ obeys (7.7).Proof of Proposition7.1.
Since (3.7) holds for every initial condition we can choose X0=0:
this simplifies calculations. Moreover using (3.7) in Lemma 7.6 it is clear
that assumption (7.7) holds. Consider the integral form of (1.1a)-(1.1b). Adding and
subtracting X∞ from both sides and applying Lemma 7.6 we
obtain(7.9)Δ(t)=−X∞+∫0tδ(s)ds+∫0tf(s)ds+μ(t)−∫0tK1(s)dsX∞,where Δ(t)=X(t)−X∞,
the function δ is defined by(7.10)δ(t)=AΔ(t)+(K*Δ)(t),and μ(t)=∫0tΣ(s)dB(s).
Taking expectations across (7.9) and allowing t→∞ we obtain(7.11)−E[X∞]=−∫0∞E[δ(s)]ds−∫0∞f(s)ds+∫0∞K1(s)dsE[X∞],where 𝔼[δ(t)]=A𝔼[Δ(t)]+(K*𝔼[Δ])(t).
Using this expression for 𝔼[X∞] we obtain(7.12)f1(t)=−E[Δ(t)]−∫t∞E[δ(s)]ds+∫t∞K1(s)dsE[X∞],The first term on the right-hand
side of (7.12) decays exponentially due to (3.7). Assumptions (3.4) and (3.7)
imply that 𝔼[δ(⋅)] decays exponentially so the second term decays
exponentially. The third term on the right-hand side of (7.12) decays
exponentially due to the argument given by (4.41). Hence, f1 decays exponentially.
Proving that (3.5) holds breaks into two steps. We
begin by showing that(7.13)‖∫0∞eγtf(t)dt‖<∞,where γ>0.
By choosing γ<α⋀β1 we can obtain the following reformulation of
(1.1a)-(1.1b) using
methods applied in [15, Proposition 5.1]
(7.14)eγtΔ(t)=Δ(0)+(γI+A)∫0teγsΔ(s)ds+∫0teγs∫0sK(s−u)Δ(u)du⁢ds−∫0teγsK1(s)du⁢dsX∞+∫0teγsf(s)ds+∫0teγsΣ(s)dB(s).Rearranging (7.14), taking
expectations and then norms on both sides we can obtain(7.15)‖∫0teγsf(s)ds‖≤E∥eγtΔ(t)∥+E∥Δ(0)∥+∥γI+A∥∫0teγsE∥Δ(s)∥ds+∫0teγs∫0s∥K(s−u)∥E∥Δ(u)∥du⁢ds+∫0teγs∥K1(s)∥dsE∥X∞∥.Since (3.7) holds this implies
that both the first and third terms on the right-hand side of (7.15) are
bounded. The second term is bounded due to our assumptions. Since 0<γ<α⋀β1,
we can choose γ1>0 such that γ<γ1<α⋀β1.
It can easily be shown that(7.16)∫0s∥K(s−u)∥E[∥Δ(u)∥]du≤ce−γ1s.Finally, we see that the fifth
term is bounded using (4.41). So, (7.13) holds.
We now return to (7.14). Again rearranging the
equation and taking norms and then expectations across both sides, we
obtain(7.17)E‖∫0teγsΣ(s)dB(s)‖≤E∥eγtΔ(t)∥+E∥Δ(0)∥+∥γI+A∥∫0teγsE∥Δ(s)∥ds+∫0teγs∫0s∥K(s−u)∥E∥Δ(u)∥du⁢ds+∫0teγs∥K1(u)∥du⁢dsE∥X∞∥+‖∫0teγsf(s)ds‖.We already provided an argument
above to show that the first five terms on the right hand side of this
expression are bounded. Also, we know that (7.13) holds. Thus,(7.18)E‖∫0teγsΣ(s)dB(s)‖≤C.The proof is now identical to
Proposition 5.2.Proof of Proposition7.2.
Since Lemma7.7 holds we can obtain (7.9). Thus, as t→∞,
we obtain(7.19)−X∞=−∫0∞δ(s)ds−∫0∞f(s)ds−μ(∞)+∫0∞K1(s)dsX∞,where δ is defined by (7.10). Using this expression for X∞,
(7.9) becomes(7.20)Δ(t)=−∫t∞δ(s)ds−f1(t)−μ1(t)+∫t∞K1(s)dsX∞,where μ1(t)=∫t∞Σ(s)dB(s).
Rearranging the equation and taking norms yields(7.21)∥f1(t)+μ1(t)∥≤∥Δ(t)∥+‖∫t∞δ(s)ds‖+‖∫t∞K1(s)dsX∞‖.The first term on the right hand
side of (7.21) decays exponentially due to (3.8). Using the argument given in
(4.41) we see that the third term on the right hand side of (7.21) decays
exponentially. Finally, we consider the second term. Clearly ∫t∞AΔ(s)ds decays exponentially due to (3.8). In order to
show that ∫t∞(K*Δ)(s)ds decays exponentially we use an argument
similar to that applied in the proof of Proposition 7.1. So there is an almost
surely finite random variable c>0 such that(7.22)∥f1(t)+μ1(t)∥≤ce−λt∀t≥0,a.s.,where λ<min(β0*,α).
We can now apply Lemma 7.3 to obtain (3.6) and (3.5).Proof of Lemma7.3.
We
suppose that there exists a constantγ such that (3.5) holds. Using the equivalence
of norms we see that for all 1≤i≤n and 1≤j≤d assumption (3.5) implies that(7.23)∫0∞Σij(s)2e2γs⁢ds<∞.Applying Lemma 7.5 we
obtain(7.24)limsup⁡t→∞⁢1tlog⁡|∫t∞Σij(s)dBj(s)|≤−γ,ω∈Ωij,⁢ℙ[Ωij]=1.Choose any ϵ∈(0,γ).
For each ω∈Ωij we can choose a constant cij(ω,ϵ)≥1 such that(7.25)|∫t∞Σij(s)dBj(s)|≤cij(ω,ϵ)e−(γ−ϵ)t.Now, summing over j we see that(7.26)|μ1i(t)|≤ci(ω,ϵ)e−(γ−ϵ)t,where ω∈Ωi=∩j=1dΩij, ci=∑j=1dcij and μ1i(t)=∑j=1d∫t∞Σij(s)dBj(s).
Now, since(7.27)|f1i(t)+μ1i(t)|2≤∑i=1n|f1i(t)+μ1i(t)|2=∥f1(t)+μ1(t)∥2we see that(7.28)|f1i(t)+μ1i(t)|≤c(ω)e−λt,ω∈Ω*.So for ω∈Ωi∩Ω* we see that |f1i(t)|≤|f1i(t)+μ1i(t)|+|μ1i(t)|≤c(ω)e−λt+ci(ω,ϵ)e−(γ−ϵ)t.
This gives(7.29)|f1i(t)|≤c¯i(ω)e−ρt,ω∈Ω*∩Ωi,where c¯i>0 is finite and ρ≤max(λ,γ−ϵ).
Now summing over i we obtain (3.6), by picking out any ω∈∩i(Ω*∩Ωi).
This concludes the case when (3.5) holds.
Now, consider the case where assumption (3.5) fails to
hold. We choose a constant γ such that 0<γ<λ and define the function d as(7.30)d(t)=∫0t1γ(eγs−1)f(s)ds,and the vector martingale M as(7.31)M(t)=∫0t1γ(eγs−1)Σ(s)dB(s).We let Mi denote the ith component of M and 〈Mi〉 denote the quadratic variation of Mi given by(7.32)〈Mi〉(t)=∑j=1d∫0t1γ2(eγs−1)2Σij(s)2⁢ds.We show at the end of this proof
that(7.33)∥d(t)+M(t)∥<c*(ω),ω∈Ω*,ℙ[Ω*]=1,and therefore assume it for the
time being.
Since (3.5) fails to hold there exists an entry i,1≤i≤n,
of the martingale M such that(7.34)lim⁡t→∞〈Mi〉(t)=∞.It follows that liminft→∞Mi(t)=−∞ and limsupt→∞Mi(t)=∞ a.s. Consider the corresponding ith entry of d,
denoted di;
it is either bounded or unbounded. If di is bounded then Mi is bounded and so, by the Martingale
Convergence Theorem, 〈Mi〉(t) is bounded: this contradicts (7.34). So, we
suppose the latter, that di is unbounded, and proceed to show this is also
contradictory. Since |di(t)+Mi(t)|<c*(ω),
for ω∈Ω* it is clear that −c*−Mi(t)<di(t).
Taking the limit superior on both sides of the inequality
yields(7.35)∞=−c*−liminf⁡t→∞⁢Mi(t)≤limsup⁡t→∞⁢di(t).As d is deterministic, there exists an increasing
sequence of deterministic times {tm}m=0∞ with t0=0 such that di(tm)→∞ as m→∞.
Consequently, Mi(tm)→−∞ as m→∞.
We choose a subsequence of these times {τm}m=0∞ with τ0=t0 such that(7.36)vl2:=∑j=1d∫τl−1τl1γ2(eγs−1)2Σij2(s)ds≥1.Define Si(m)=Mi(τm).
Obviously(7.37)Si(m)=∑l=1mξl(i)where(7.38)ξl(i)=∑j=1d∫τl−1τl1γ(eγs−1)Σij(s)dBj(s).It is clear that {ξl(i)}l=1∞ is an indepenendent normally distributed
sequence with the variance of each ξl(i) given by vl2≥1 so we may apply Lemma 7.4.
We now show that assumption (7.33) holds. By changing
the order of integration we can show that(7.39)d(t)=∫0teγs(f1(s)−f1(t))ds,M(t)=∫0teγs(μ1(s)−μ1(t))ds.Thus, as 0<γ<λ,(7.40)∥d(t)+M(t)∥≤∫0teγs(∥f1(t)+μ1(t)∥+∥f1(s)+μ1(s)∥)ds≤c(ω)∫0teγs(e−λt+e−λs)ds<c1(ω),ω∈Ω*.


## 8. On the Necessary and Sufficient Conditions for Exponential Convergence of Solutions of (1.1a)-(1.1b) and (1.4a)-(1.4b)

We now combine
the results from Sections4 and 5 to prove Theorem 3.2 and combine the results from
Sections 6 and 7 to prove Theorem 3.1.We showed the necessity of (3.5) for the exponential
convergence of the solution of (1.4a)-(1.4b) in
Section 5. In order to prove the necessity of
the exponential integrability of the kernel we require the following result
which was extracted from [1].Theorem 8.1.
LetK satisfy (2.4) and (3.1). Suppose that there
exists a constant matrix R∞ and constants β>0 and c>0 such that the solution R of (1.2a)-(1.2b) satisfies (4.3). If the kernel K satisfies (3.2) then there exists a constant α>0 such that K satisfies (3.4).Proof of Theorem3.2.
We begin by proving the equivalence between (i) and (ii).
The implication (i) implies (ii) is the subject of Proposition4.1. We can
demonstrate that (ii) implies (i) as follows. We begin by proving that (3.9)
implies (3.4). We consider the following n solutions of (1.4a)-(1.4b); {Xj(t)}j=1,…,n,
where Xj(0)=ej.
Since (3.9) holds we obtain(8.1)m1(ej)e−β1t≥E∥Xj(t)−Xj(∞)∥≥∥E[Xj(t)−Xj(∞)]∥=∥(R(t)−R(∞))ej∥.for each j=1,…,n.
Thus, the resolvent R of (1.2a)-(1.2b) decays exponentially to R∞.
We can apply Theorem 8.1 to obtain (3.4) after which Proposition
5.2 can be
applied to obtain (3.5). As (8.1) holds it is clear that (3.3) holds.
We now prove the equivalence between (i) and (iii).
The implication (i) implies (iii) is the subject of Proposition4.2. We now
demonstrate that (iii) implies (i). We begin by proving that (3.10) implies
(3.4). As (3.10) holds for all X0 we can consider the following n+1 solutions of (1.4a)-(1.4b); Xj(t)j=1,…,n+1 where(8.2)Xj(0)=ejforj=1,…,n,Xn+1(0)=0.We know that Xj(t) approaches Xj(∞) exponentially quickly in the almost sure
sense. Introduce(8.3)Sj(t)=Xj(t)−Xn+1(t),and notice Sj(0)=ej.
Let S=[S1,…,Sn]∈Rn×n.
Then(8.4)S′(t)=AS(t)+(K*S)(t),t>0,S(0)=I.If we define Sj(∞)=Xj(∞)−Xn+1(∞) then S(t)→S∞ exponentially quickly so we can apply Theorem
8.1 to obtain (3.4). As (3.4) and (3.10)
hold we can apply Proposition
5.1 to
obtain (3.5). Also evident from this argument is that (3.3) holds. This proves
that (iii) implies (i).Proof of Theorem3.1.
We begin by proving the equivalence between (i) and (ii).
The implication that (i) implies (ii) is the subject of Proposition6.1. Now
consider the implication (ii) implies (i). Using (3.7) we see
that(8.5)∥E[X(t)−X∞]∥≤E∥X(t)−X∞∥≤m*e−β1*t.Consider the n+1 solutions Xj of (1.1a)-(1.1b) with initial conditions Xj(0)=ej for j=1,…,n and Xn+1(0)=0.
Since R(t)ej=Xj(t)−Xn+1(t) we see that(8.6)E[Xj(t)−Xj(∞)]+E[Xn+1(t)−Xn+1(∞)]=R(t)ej−E[cj],where cj=Xj(∞)−Xn+1(∞) is an almost surely finite constant. As both
terms on the left hand side of this expression are decaying exponentially to
zero, t↦R(t)ej must decay exponentially to 𝔼[cj] as t→∞.
Thus R must satisfy (4.3). Now, apply Theorem 8.1 to
obtain (3.4) and Proposition 7.1 to obtain (3.6) and (3.5).
We now prove the equivalence between (i) and (iii).
The implication (i) implies (iii) is the subject of Proposition6.2. Once
again, consider the n+1 solutions Xj(t) with initial conditions Xj(0)=ej for j=1,…,n and Xn+1(0)=0.
Since R(t)ej=Xj(t)−Xn+1(t) for j=1,…,n,
we can write(8.7)(Xj(t)−Xj(∞))−(Xn+1(t)−Xn+1(∞))=R(t)ej−cj,where cj=Xj(∞)−Xn+1(∞) is an almost surely finite random variable.
From (3.8) we know that Xj decays exponentially quickly to Xj(∞),
similarly Xn+1 decays exponentially quickly to Xn+1(∞).
Thus, R decays exponentially to a limit. As a
result (4.3) must hold. Now apply Theorem 8.1 to obtain (3.4) and Proposition
7.2 to obtain (3.6) and (3.5).



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