Asymptotic Expansion of Markov Random Evolution
It is studied asymptotic expansion for solution of singularly perturbed equation for Markov random evolution in Rd. The views of regular and singular parts of solution are found.
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Інститут прикладної математики і механіки НАН України
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| Cite this: | Asymptotic Expansion of Markov Random Evolution / I.V. Samoilenko // Український математичний вісник. — 2006. — Т. 3, № 3. — С. 394-407. — Бібліогр.: 12 назв. — англ. |
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nasplib_isofts_kiev_ua-123456789-1245612025-02-09T18:01:56Z Asymptotic Expansion of Markov Random Evolution Samoilenko, I.V. It is studied asymptotic expansion for solution of singularly perturbed equation for Markov random evolution in Rd. The views of regular and singular parts of solution are found. The author thanks Acad. V. S. Korolyuk for the formulation of the problem studied. Acknowledgements also to the Institute of Applied Mathematics, University of Bonn for the hospitality and financial support by DFG project 436 UKR 113/70/0-1. 2006 Article Asymptotic Expansion of Markov Random Evolution / I.V. Samoilenko // Український математичний вісник. — 2006. — Т. 3, № 3. — С. 394-407. — Бібліогр.: 12 назв. — англ. 1810-3200 2000 MSC. 60J25, 35C20. https://nasplib.isofts.kiev.ua/handle/123456789/124561 en Український математичний вісник application/pdf Інститут прикладної математики і механіки НАН України |
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It is studied asymptotic expansion for solution of singularly perturbed equation for Markov random evolution in Rd. The views of regular and singular parts of solution are found. |
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Samoilenko, I.V. Asymptotic Expansion of Markov Random Evolution Український математичний вісник |
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Asymptotic Expansion of Markov Random Evolution |
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Asymptotic Expansion of Markov Random Evolution |
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Asymptotic Expansion of Markov Random Evolution |
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Asymptotic Expansion of Markov Random Evolution |
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Asymptotic Expansion of Markov Random Evolution |
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asymptotic expansion of markov random evolution |
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Інститут прикладної математики і механіки НАН України |
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Asymptotic Expansion of Markov Random Evolution / I.V. Samoilenko // Український математичний вісник. — 2006. — Т. 3, № 3. — С. 394-407. — Бібліогр.: 12 назв. — англ. |
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Український математичний вісник |
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AT samoilenkoiv asymptoticexpansionofmarkovrandomevolution |
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Український математичний вiсник
Том 3 (2006), № 3, 394 – 407
Asymptotic Expansion of Markov
Random Evolution
Igor V. Samoilenko
(Presented by V. S. Korolyuk)
Abstract. It is studied asymptotic expansion for solution of singularly
perturbed equation for Markov random evolution in R
d. The views of
regular and singular parts of solution are found.
2000 MSC. 60J25, 35C20.
Key words and phrases. Random evolution, singularly perturbed
equation, asymptotic expansion, diffusion approximation, estimate of
the remainder.
1. Introduction
A Markov random evolution (MRE) is created by a solution of the
evolutionary equation in Euclidean space R
d, d ≥ 1
duε(t)/dt = v(uε(t); æ(t/ε))
with the ergodic Markov switching process æ(t), t ≥ 0 on the standard
(Polish) phase-space (E, E) by the operator Q(x,B), x ∈ E, B ∈ E that
defines transition probabilities of a Markov chain æn, n ≥ 0
Q(x,B) = P{æn+1 ∈ B|æn = x}.
The operator of transition probabilities Q is defined by
Qf(x) =
∫
E
Q(x, dy)f(y), x ∈ E, (1.1)
for any bounded measurable real valued f defined on E.
Received 4.07.2006
ISSN 1810 – 3200. c© Iнститут математики НАН України
I. V. Samoilenko 395
We will see later that the equation for the regular and the singular
parts of a random evolution are defined by the generator (1.1) of a uni-
formly argodic Markov switching process. The Banach space B(E) is
splitted onto the two subspaces [7]:
B(E) = NQ
⊕
RQ,
where NQ := {ϕ : Qϕ = 0} is the null-space of Q, and RQ := {ψ : Qϕ =
ψ} is the range of Q.
We define the projector Π : NQ := ΠB(E), RQ := (I − Π)B(E);
Πϕ(x) := ϕ̂1, ϕ̂ :=
∫
E
ϕ(x)π(dx), where the stationary distribution
π(B), B ∈ E of the Markov process æ(t), t ≥ 0 satisfies the relations [4]
π(dx) = ρ(dx)m1(x)/m̂,
m̂ =
∫
E
m1(x)ρ(dx).
ρ(B), B ∈ E is the stationary distribution of the Markov chain æn, n ≥ 0,
given by the equation
ρ(B) =
∫
E
Q(x,B)ρ(dx), ρ(E) = 1.
Let us consider the Banach space B(Rd) of real-valued test-functions
ϕ(u), u ∈ R
d which are bounded with all their derivatives equipped with
sup-norm
‖ϕ‖ := sup
u∈Rd
|ϕ(u)| < Cϕ.
The random evolution in B(Rd) is given by the relation
Φε
t (u, x) := E[ϕ(uε(t))|uε(0) = u,æε(0) = x]. (1.2)
The asymptotic behavior of MRE (1.2) as ε→ 0 is investigated under
the assumption of uniformly ergodicity of the Markov switching process
æ(t) described above and under the assumption of the existence of a
global solution of the deterministic equations
dux(t)/dt = v(ux(t);x), x ∈ E.
Let us consider the deterministic evolution
Φx(t, u) = ϕ(ux(t)), ux(0) = u.
396 Asymptotic Expansion...
It generates a corresponding semigroup
Vt(x)ϕ(u) := ϕ(ux(t)), ux(0) = u,
and its generator has the form:
V(x)ϕ(u) = v(u;x)ϕ′(u) :=
d∑
k=1
vk(u;x)ϕ
′
k(u),
ϕ′
k(u) := ∂ϕ(u)/∂uk, ϕ(u) ∈ C∞(Rd).
By the average principle [10] the weak convergence
uε(t) ⇒ û(t), ε→ 0 (1.3)
takes place. The average limit evolution û(t), t ≥ 0 is defined by a
solution of the average equation
dû(t)/dt = v̂(û(t)).
The average velocity v̂(u), u ∈ R
d is defined by
v̂(u) =
∫
E
v(u;x)π(dx)
(i.e. by the average of the initial velocity v(u;x) over the stationary
distribution π(B), B ∈ E).
The rate of convergence in (1.3) can be investigated in two directions:
i) asymptotic analysis of the fluctuations
ζε(t) = uε(t) − û(t); (1.4)
ii) asymptotic analysis of the average deterministic evolution (1.2).
The asymptotic analysis of fluctuations (1.4) leads to the diffusion
approximation of the random evolution [5, 10].
The asymptotic analysis of evolution (1.2) is realized in what follows
by constructing the asymptotic expansion in power of the small parameter
series ε→ 0(ε > 0) in the following form (τ = t/ε):
Φε
t (u, x) = u(0)(t) +
∞∑
k=1
εk[u(k)(t) + w(k)(τ)]. (1.5)
The asymptotic expansion (1.5) contains two parts:
i) the regular term uε(t) := u(0)(t) +
∑∞
k≥1 ε
ku(k)(t),
ii) the singular term (boundary layer) wε(τ) :=
∑∞
k≥1 ε
kw(k)(τ), τ = t/ε.
I. V. Samoilenko 397
In addition the initial condition:
u(0)(0) = ϕ(u)1
has to be valid for any x ∈ E, u ∈ R
d.
It’s well-known (see, e.g. [8]), that the evolution, determined by a
test-function ϕ(u) ∈ C∞(Rd) (here ϕ(u) is integrable on R
d): satisfy the
system of Kolmogorov backward differential equations:
∂
∂t
Φε
t (u, x) = [ε−1Q+ V]Φε
t (u, x),
Φε
0(u, x) = ϕ(u).
(1.6)
Asymptotic expansions with “boundary layers” were studied by many
authors (see [2, 3, 12]). In particular, functionals of Markov and semi-
Markov processes are investigated from this point of view in [6, 9, 11].
In this work we study system (1.6) with the first order singularity.
To find asymptotic expansion of the solution of (1.2) we use the method
proposed in [3,12]. The solution consists of two parts, regular terms and
singular terms, which are determined by different equations. Asymptotic
expansion lets not only determine the terms of asymptotic, but to see the
velocity of convergence in hydrodynamic limit.
Besides, when studying this problem, we improved the algorithm of
asymptotic expansion. Partially, the initial conditions for the regular
terms of asymptotic are determined without the use of singular terms,
i.e. the regular part of the solution may be found by a separate recursive
algorithm; scalar part of the regular term is found and without the use of
singular terms. These and other improves of the algorithm are pointed
later.
2. Asymptotic Expansion of the Solution
Let P (t) = eQt = {pij(t); i, j ∈ E}. Put πj = limt→∞ pij(t) and
−R0 = {
∫ ∞
0 (pij(t) − πj) dt; i, j ∈ E} = {rij ; i, j ∈ E}.
Let Π be a projecting operator on the null-spaceNQ of the operatorQ.
For any vector g we have Πg = ĝ1, where ĝ = (g, π),1 = (1, . . . , 1). Then
for the operator Q the following correlations are true (see [7], chapter 3)
ΠQΠ = 0,
QR0 = R0Q = Π − I.
We put:
exp0(Qt) := eQt − Π.
398 Asymptotic Expansion...
Theorem 2.1. The solution of equation (1.6) with initial condition
Φε
0(u, x) = ϕ(u), where ϕ(u) ∈ C∞(Rd) and integrable on R
d has asymp-
totic expansion
Φε
t (u, x) = u(0)(t) +
∞∑
n=1
εn
(
u(n)(t) + w(n) (t/ε)
)
. (2.1)
Regular terms of the expansion are: u(0)(x, t), the solution of equation
∂
∂t
u(0)(t) − ΠVΠu(0)(t) = 0 (2.2)
with initial condition u(0)(0) = ϕ(u),
u(1)(t) = R0
[ d
dt
u(0)(t) − Vu(0)(t)
]
+ c(1)(t) := R0Lu
(0)(t) + c(1)(t),
for k ≥ 2 :
u(k)(t) = R0Lu
(k−1)(t) + c(k)(t)
where c(k)(t) ∈ NQ,
c(k)(u, t) = c(k)(V −1(t+ V (u)), 0) +
∫
Lk(V
−1(t+ V (u)), 0)
V −1(t+ V (u))
du
−
∫
Lk(u, t)
v(u)
du, k > 0,
here V (u) =
∫
du
v(u) , V
−1(w) is the backward function for V (u),
Lk(u, t) =
k−1∑
i=0
k−i∑
n=1
(−1)k(k − i− n+ 1)ΠVR0V
nΠ
dk−i−n
dtk−i−n
c(i)(t).
The singular terms of the expansion have the view:
w(1)(τ) = exp0(Qτ)Vϕ(u),
for k > 1 :
w(k)(τ) = exp0(Qτ)w
(k)(0) +
τ∫
0
exp0(Q(τ − s))Vw(k−1)(s) ds
− Π
∞∫
τ
Vw(k−1)(s) ds.
I. V. Samoilenko 399
Initial conditions:
c(0)(0) = ϕ(u),
w(1)(0) = −R0Lϕ(u),
c(1)(0) = 0,
for k > 1 :
w(k)(0) = −R0Lu
(k−1)(0),
c(k)(0) = Vw̃(k−1)(0),
where w̃(1)(0) = −R0Vϕ(u),
w̃(k)(0) = R0Lu
(k−1)(0) +R0Vw̃
(k−1)(0) + ΠV(w̃(k−1)(λ))′λ
∣∣
λ=0
,
(w̃(k)(λ))′λ
∣∣
λ=0
= R2
0Lu(k−1)(0)+R2
0Q1w̃
(k−1)(0)+R0V(w̃(k−1)(λ))′λ
∣∣
λ=0
.
Remark 2.1. The initial conditions for the regular terms of asymptotic
are determined without the use of singular terms, i.e. the regular part
of the solution may be found by a separate recursive algorithm (comp.
with [6]).
Proof of Theorem 2.1. Let us substitute the solution Φε
t (u, x) in the form
(2.1) to the equation (1.6) and equal the terms at ε degrees. We’ll have
the system for the regular terms of asymptotic:
{
Qu(0) = 0
Qu(k) = d
dt
u(k−1) − Vu(k−1) := Lu(k−1), k ≥ 1
(2.3)
and for the singular terms
dwε
dt
=
dwε
dτ
dτ
dt
= ε−1dw
ε
dτ
= (ε−1Q+ V)wε.
Thus, from dwε
dτ
= (Q+ εV)wε we obtain:
{
d
dτ
w(1) = Qw(1)
d
dτ
w(k) −Qw(k) = Vw(k−1), k > 1.
(2.4)
From (2.3) we have: u(0)(t) ∈ NQ.
The solvability condition for u(1)(t) has the view:
ΠQΠu(1)(t) = 0 =
∂
∂t
u(0)(t) − ΠVΠu(0)(t).
So, we have equation (2.2) for u(0)(t).
400 Asymptotic Expansion...
For u(1)(t) we have:
u(1)(t) = R0Lu
(0)(t) + c(1)(t).
Using the second equation from (2.3) we obtain:
u(k)(t) = R0Lu
(k−1)(t) + c(k)(t),
where c(k)(t) ∈ NQ.
To find c(k)(t) we’ll use the fact that u(0)(t) ∈ NQ. Let us put c(0)(t) =
u(0)(t).
For the equation
Qu(2)(t) =
∂
∂t
u(1)(t) − Vu(1)(t) =
d
dt
R0
[ d
dt
c(0)(t) − Vc(0)(t)
]
+
d
dt
c(1)(t) − VR0
[ d
dt
c(0)(t) − Vc(0)(t)
]
− Vc(1)(t)
we use the solvability condition
ΠQΠu(2)(t) = 0 =
d
dt
c(1)(t) − Vc(1)(t) + ΠR0Π
d2
dt2
c(0)(t)
− ΠR0VΠ
d
dt
c(0)(t) − ΠVR0Π
d
dt
c(0)(t) + ΠVR0VΠc(0)(t).
We find:
d
dt
c(1)(t) − Vc(1)(t) = −ΠVR0VΠc(0)(t).
By induction:
d
dt
c(k)(t) − Vc(k)(t)
=
k−1∑
i=0
k−i∑
n=1
(−1)k(k − i− n+ 1)ΠVR0V
nΠ
dk−i−n
dtk−i−n
c(i)(t), k > 0.
So, we have the following equation for c(k)(u, t):
d
dt
c(k)(u, t) − v(u)
d
du
c(k)(t) = Lk(u, t),
here
Lk(u, t) =
k−1∑
i=0
k−i∑
n=1
(−1)k(k − i− n+ 1)ΠVR0V
nΠ
dk−i−n
dtk−i−n
c(i)(t).
I. V. Samoilenko 401
To find a solution we should write down a system
dt
1
= −
du
v(u)
=
dc(k)
Lk(u, t)
.
The independent integrals of this system are:
t+
∫
du
v(u)
= C1,
c(k)(u, t) +
∫
Lk(u, t)
v(u)
du = C2.
As soon as c(k)(u, t) is only in one of the first integrals, we may present
the solution in the form:
c(k)(u, t) = fk
(
t+
∫
du
v(u)
)
−
∫
Lk(u, t)
v(u)
du, k > 0,
where fk is any differentiable function. Using initial condition for c(k)(u, t)
we find a condition for fk:
fk
(∫
du
v(u)
)
= c(k)(u, 0) +
∫
Lk(u, 0)
v(u)
du.
We may put now V (u) =
∫
du
v(u) and make a change of variables
w = V (u). So, u = V −1(w) and we have:
fk(w) = c(k)(V −1(w), 0) +
∫
Lk(V
−1(w), 0)
v(V −1(w))
dw
w
.
Thus, we obtain
c(k)(u, t) = c(k)(V −1(t+ V (u)), 0)
+
∫
Lk(V
−1(t+ V (u)), 0)
V −1(t+ V (u))
du−
∫
Lk(u, t)
v(u)
du, k > 0.
Initial conditions for c(k)(u, 0) are found later through Laplace trans-
form for the singular terms of asymptotic.
For the singular terms we have from (10):
w(1)(τ) = exp0(Qτ)w
(1)(0).
Here we should note that the ordinary solution w(1)(τ) = exp(Qτ)×
w(1)(0) is corrected by the term −Πw(1)(0) in order to receive the fol-
lowing limτ→∞w(1)(τ) = 0. We choose this limit to be equal 0 for all
402 Asymptotic Expansion...
singular terms, that may done due to uniform ergodicity of switching
Markovian process.
The following statements are made using a method proposed in [2].
For the second equation of the system the corresponding solution should
be
w(k)(τ) = exp0(Qτ)w
(k)(0) +
τ∫
0
exp0(Q(τ − s))Vw(k−1)(s) ds,
where the homogenous part has the following solution
w(k)(τ) = exp0(Qτ)w
(k)(0).
But here we should again correct the solution, in order to receive the
limit lim
τ→∞
w(k)(τ) = 0, by the term −Π
∫ ∞
τ
Vw(k−1)(s) ds.
And so the solution is:
w(k)(τ) = exp0(Qτ)w
(k)(0) +
∫ τ
0
exp0(Q(τ − s))Vw(k−1)(s)ds
− Π
∞∫
τ
Vw(k−1)(s) ds.
We should finally find the initial conditions for the regular and sin-
gular terms.
We put c(0)(t) = u(0)(t), so c(0)(0) = u(0)(0) = ϕ(u).
From the initial condition for the solution uε(0) = u(0)(0) = ϕ(u),
we have to determine u(k)(0) + w(k)(0) = 0, k ≥ 1. Let us rewrite this
equation for the null-space NQ of matrix Q:
Πu(k)(0) + Πw(k)(0) = 0, k ≥ 1, (2.5)
and the space of values RQ:
(I − Π)u(k)(0) + (I − Π)w(k)(0) = 0, k ≥ 1. (2.6)
So, for k = 1 we obtain:
u(1)(0) = R0Lu
(0)(0) + c(1)(0) = (I − Π)R0Lϕ(u) + Πc(1)(0),
w(1)(0) = (I − Π)w(1)(0).
Thus, c(1)(0) = 0, w(1)(0) = −R0Lϕ(u).
I. V. Samoilenko 403
By analogy, for k > 1:
u(k)(0) = R0Lu
(k−1)(0) + c(k)(0) = (I − Π)R0Lu
(k−1)(0) + Πc(k)(0),
w(k)(0) = (I − Π)w(k)(0) − Π
∞∫
0
Vw(k−1)(s) ds.
Functions w(k−1)(s), u(k−1)(0) are known from the previous steps of
induction. So, we’ve found Πw(k)(0) in (2.5) and (I −Π)u(k)(0) in (2.6).
Now we may use the correlations (2.5), (2.6) to find the unknown
initial conditions:
c(k)(0) =
∞∫
0
Vw(k−1)(s) ds,
w(k)(0) = −R0Lu
(k−1)(0).
In [6] an analogical correlation was found for c(k)(0). To find c(k)(0)
explicitly and without the use of singular terms we’ll find Laplace trans-
form for the singular term. The following lemma is true.
Lemma 2.1. Laplace transform for the singular term of asymptotic ex-
pansion
w̃(k)(λ) =
∞∫
0
e−λsw(k)(s) ds
has the view:
w̃(1)(λ) = (λ− Π + (R0 + Π)−1)−1[−R0Vϕ(u)],
w̃(k)(λ) = (λ− Π + (R0 + Π)−1)−1
Lu(k−1)(0)
+ (λ− Π + (R0 + Π)−1)−1
Vw̃(k−1)(λ)
+
1
λ
ΠV[w̃(k−1)(λ) − w̃(k−1)(0)],
where
w̃(1)(0) = −R0Vϕ(u),
(w̃(1)(λ))′λ
∣∣
λ=0
= −R2
0VΠϕ(u),
w̃(k)(0) = R0Lu
(k−1)(0) +R0Vw̃
(k−1)(0) + ΠV(w̃(k−1)(λ))′λ
∣∣
λ=0
,
(w̃(k)(λ))′λ
∣∣
λ=0
= R2
0Lu(k−1)(0)+R2
0Q1w̃
(k−1)(0)+R0V(w̃(k−1)(λ))′λ
∣∣
λ=0
.
404 Asymptotic Expansion...
Proof.
w̃(1)(λ) =
∞∫
0
e−λsw(1)(s) ds =
∞∫
0
e−λs[eQs − Π] ds · w(1)(0)
= (λ− Π + (R0 + Π)−1)−1[−Vϕ(u)],
where the correlation for the resolvent was found in [7].
w̃(1)(0) = −R0Vϕ(u),
(w̃(1)(λ))′λ
∣∣
λ=0
= lim
λ→0
R(λ) −R0
λ
[−Vϕ(u)] = −R2
0Vϕ(u).
For the next terms we have:
w̃(k)(λ) = (λ− Π + (R0 + Π)−1)−1
Lu(k−1)(0)
+ (λ− Π + (R0 + Π)−1)−1
Vw̃(k−1)(λ)
+
1
λ
ΠV[w̃(k−1)(λ) − w̃(k−1)(0)],
here the last term was found using the following correlation:
∞∫
0
e−λs
∞∫
s
Vw(k−1)(θ) dθ ds =
∞∫
0
θ∫
0
e−λs
Vw(k−1)(θ) ds dθ
=
∞∫
0
(
−
1
λ
)
(e−λθ − 1)Vw(k−1)(θ) dθ =
1
λ
V[w̃(k−1)(λ) − w̃(k−1)(0)].
So,
w̃(k)(0) = R0Lu
(k−1)(0) +R0Vw̃
(k−1)(0) + ΠV(w̃(k−1)(λ))′λ
∣∣
λ=0
,
(w̃(k)(λ))′λ
∣∣
λ=0
=R2
0Lu(k−1)(0)+R2
0Q1w̃
(k−1)(0)+R0V(w̃(k−1)(λ))′λ
∣∣
λ=0
− lim
λ→0
{ 1
λ2
ΠV[w̃(k−1)(λ) − w̃(k−1)(0)] −
1
λ
ΠV(w̃(k−1)(λ))′λ
}
,
where the last limit tends to 0.
Lemma is proved.
So, the obvious view of the initial condition for the c(k)(t) is:
c(k)(0) = Vw̃(k−1)(0).
Theorem is proved.
I. V. Samoilenko 405
3. Estimate of the Remainder
Let function ϕ(u) in the definition of the functional Φε
t belongs to
Banach space of twice continuously differentiable by u functions C2(Rd).
Let us write (1.6) in the view
Φ̃ε(t) = Φε(t) − Φε
2(t) (3.1)
where Φε
2(t) = u(0)(t)+ε(u(1)(t)+w(1)(t))+ε2(u(2)(t)+w(2)(t)), and the
explicit view of the functions u(i)(t), w(j)(t), i = 0, 2, j = 1, 2 is given in
Theorem 2.1.
By Theorem 3.2.1 from [7] in Banach space C2(Rd × E) for the gen-
erator of Markovian evolution Lε = ε−1Q + V, exists bounded inverse
operator (Lε)−1.
Let us substitute the function (3.1) into equation (1.6):
d
dt
Φ̃ε − LεΦ̃ε =
d
dt
Φε
2 − LεΦε
2 := εθε. (3.2)
Here εθε = ε[ d
dt
u(1) − εV(u(2) + w(2))].
The initial condition has the order ε, so we may write it in the view:
Φ̃ε(0) = εΦ̃ε(0).
Let Lε
tϕ(u) = E[ϕ(uε(t))|uε(0) = u,æε(0) = x] be the semigroup
corresponding to the operator Lε.
Theorem 3.1. The following estimate is true for the remainder (3.1)
of the solution of equation (1.6):
‖Φ̃ε(t)‖ ≤ ε‖Φ̃ε(0)‖ exp
{
εL‖θε‖
}
,
where L ≥ 2||(Lε)−1||.
Proof. The solution of equation (3.2) is:
Φ̃ε(t) = ε
[
Lε
t Φ̃
ε(0) +
t∫
0
Lε
t−sθ
ε(s) ds
]
.
For the semigroup we have Lε
t = I+Lε
∫ t
0 L
ε
sds, so
∫ t
0 L
ε
sds = (Lε)−1×
(Lε
t − I).
Using Gronwell–Bellman inequality [1], we receive
‖Φ̃ε(t)‖ ≤ εLε
t‖Φ̃
ε(0)‖ exp
{
ε
t∫
0
Lε
sθ
ε(t− s) ds
}
≤ εLε
t‖Φ̃
ε(0)‖ exp{εL‖θε‖},
406 Asymptotic Expansion...
where L ≥ 2‖(Lε)−1‖.
Theorem is proved.
Remark 3.1. For the remainder of asymptotic expansion (1.5) of the
view
Φ̃ε
N+1(t) := Φε(t) − Φε
N+1(t),
where Φε
N+1(t) = u(0)(t)+
∑N+1
k=1 ε
k(u(k)(t)+w(k)(t)) we have analogical
estimate:
‖Φ̃ε
N+1(t)‖ ≤ εN‖Φ̃ε(0)‖ exp{εNL‖θε
N‖},
where d
dt
Φε
N+1 − LεΦε
N+1 := εNθε
N .
Acknowledgements. The author thanks Acad. V. S. Korolyuk for
the formulation of the problem studied. Acknowledgements also to the
Institute of Applied Mathematics, University of Bonn for the hospitality
and financial support by DFG project 436 UKR 113/70/0-1.
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I. V. Samoilenko 407
Contact information
Igor V. Samoilenko Institute of Mathematics,
Ukrainian National Academy of Sciences,
3 Tereshchenkivs’ka,
Kyiv, 01601,
Ukraine
E-Mail: isamoil@imath.kiev.ua
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