Hindawi Publishing Corporation
Advances in Mathematical Physics
Volume 2013, Article ID 893254, 10 pages
http://dx.doi.org/10.1155/2013/893254
Research Article
Semigroup Method on a MX/G/1 Queueing Model
Alim Mijit
Xinjiang Radio & TV University, Urumqi 830049, China
Correspondence should be addressed to Alim Mijit; alimjanmijit@yahoo.com.cn
Received 24 December 2012; Revised 20 March 2013; Accepted 21 March 2013
Academic Editor: B. G. Konopelchenko
Copyright © 2013 Alim Mijit. This is an open access article distributed under the Creative Commons Attribution License, which
permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
By using the Hille-Yosida theorem, Phillips theorem, and Fattorini theorem in functional analysis we prove that the MX /G/1
queueing model with vacation times has a unique nonnegative time-dependent solution.
1. Introduction
Hypothesis 1. The model has a nonnegative time-dependent
solution.
The queueing system when the server become idle is not
new. Miller [1] was the first to study such a model, where
the server is unavailable during some random length of
time for the M/G/1 queueing system. The M/G/1 queueing
models of similar nature have also been reported by a number
of authors, since Levy and Yechiali [2] included several
types of generalizations of the classical M/G/1 queueing
system. These generalizations are useful in model building
in many real life situations such as digital communication, computer network, and production/inventory system
[3–5].
At present, however, most studies are devoted to batch
arrival queues with vacation because of its interdisciplinary
character. Considerable efforts have been devoted to study
these models by Baba [6], Lee and Srinivasan [7], Lee et
al. [8, 9], Borthakur and Choudhury [10], and Choudhury
[11, 12] among others. However, the recent progress of MX /G/1
type queueing models of this nature has been served by Chae
and Lee [13] and Medhi [14].
In 2002, Choudhury [15] studied the MX /G/1 queueing
model with vacation times. By using the supplementary
variable technique [16] he established the corresponding
queueing model and obtained the queue size distribution at a
stationary (random) as well as a departure point of time under
multiple vacation policy based on the following hypothesis.
“The time-dependent solution of the model converges to a
nonzero steady-state solution.” By reading the paper we find
that the previous hypothesis, in fact, implies the following two
hypothesis.
Hypothesis 2. The time-dependent solution of the model
converges to a nonzero steady-state solution.
In this paper we investigate Hypothesis 1. By using the
Hille-Yosida theorem, Phillips theorem, and Fattorini theorem we prove that the model has a unique nonnegative timedependent solution, and therefore we obtain Hypothesis 1.
According to Choudhury [15], the MX /G/1 queueing
system with vacation times can be described by the following
system of equations:
∞
𝑑𝑄 (𝑡)
= −𝜆𝑄 (𝑡) + ∫ 𝜐 (𝑥) 𝑃0,0 (𝑥, 𝑡) 𝑑𝑥
𝑑𝑡
0
∞
+ ∫ 𝑏 (𝑥) 𝑃1,1 (𝑥, 𝑡) 𝑑𝑥,
0
𝜕𝑃0,0 (𝑥, 𝑡) 𝜕𝑃0,0 (𝑥, 𝑡)
+
= − [𝜆 + 𝜐 (𝑥)] 𝑃0,0 (𝑥, 𝑡) ,
𝜕𝑡
𝜕𝑥
𝜕𝑃0,𝑛 (𝑥, 𝑡) 𝜕𝑃0,𝑛 (𝑥, 𝑡)
+
𝜕𝑡
𝜕𝑥
= − [𝜆 + 𝜐 (𝑥)] 𝑃0,𝑛 (𝑥, 𝑡)
𝑛
+ 𝜆 ∑ 𝑐𝑘 𝑃0,𝑛−𝑘 (𝑥, 𝑡) ,
𝑘=1
𝜕𝑃1,𝑛 (𝑥, 𝑡) 𝜕𝑃1,𝑛 (𝑥, 𝑡)
+
𝜕𝑡
𝜕𝑥
𝑛 ≥ 1,
2
Advances in Mathematical Physics
= − [𝜆 + 𝑏 (𝑥)] 𝑃1,𝑛 (𝑥, 𝑡)
𝑛
+ 𝜆 ∑ 𝑐𝑘 𝑃1,𝑛−𝑘+1 (𝑥, 𝑡) ,
If we take state space
󵄨󵄨 𝑃 = (𝑄, 𝑃 , 𝑃 , . . .)
󵄨󵄨 0
0,0 0,1
{
{
󵄨󵄨
{
{
󵄨󵄨 ∈ R × 𝐿1 [0, ∞) × 𝐿1 [0, ∞) × ⋅ ⋅ ⋅
{
{
󵄨󵄨
{
{
󵄨󵄨 𝑃 = (𝑃 , 𝑃 , 𝑃 , . . .)
{
󵄨󵄨 1
{
1,1 1,2 1,3
{
󵄨󵄨
{
{
󵄨󵄨 ∈ 𝐿1 [0, ∞) × 𝐿1 [0, ∞)
{
{
󵄨󵄨
{
𝑋 = {(𝑃0 , 𝑃1 ) 󵄨󵄨󵄨󵄨
× 𝐿1 [0, ∞) × ⋅ ⋅ ⋅
{
󵄨
∞
{
{
󵄨󵄨󵄨 󵄩󵄩
{
{
󵄨󵄨 󵄩󵄩(𝑃0 , 𝑃1 )󵄩󵄩󵄩󵄩 = |𝑄| + ∑󵄩󵄩󵄩󵄩𝑃0,𝑖 󵄩󵄩󵄩󵄩 1
{
𝐿 [0,∞)
{
󵄨󵄨
{
󵄨󵄨
{
𝑖=0
{
󵄨
{
∞
{
󵄨󵄨󵄨
{
󵄩 󵄩
{
󵄨󵄨
{
+ ∑󵄩󵄩󵄩𝑃1,𝑖 󵄩󵄩󵄩𝐿1 [0,∞) < ∞
󵄨󵄨
󵄨
𝑖=1
{
𝑛 ≥ 1,
𝑘=1
𝑃0,0 (𝑡) = 𝜆𝑄 (𝑡) ,
𝑃0,𝑛 (0, 𝑡) = 0,
𝑛 ≥ 1,
∞
𝑃1,𝑛 (0, 𝑡) = ∫ 𝜐 (𝑥) 𝑃0,𝑛 (𝑥, 𝑡) 𝑑𝑥
0
∞
+ ∫ 𝑏 (𝑥) 𝑃1,𝑛+1 (𝑥, 𝑡) 𝑑𝑥,
0
𝑃0,𝑛 (𝑥, 0) = 0,
𝑄 (0) = 1,
𝑃1,𝑛 (𝑥, 0) = 0,
𝑛 ≥ 1,
𝑛 ≥ 0,
𝑛 ≥ 1,
(1)
where (𝑥, 𝑡) ∈ [0, ∞) × [0, ∞); 𝑄(𝑡) represents the probability
that there is no customer in the system and the server is idle
at time 𝑡; 𝑃0,𝑛 (𝑥, 𝑡)𝑑𝑥 (𝑛 ≥ 0) represents the probability that
at time 𝑡 there are 𝑛 customers in the system and the server is
on a vacation with elapsed vacation time of the server lying
in [𝑥, 𝑥 + 𝑑𝑥). 𝑃1,𝑛 (𝑥, 𝑡)𝑑𝑥 (𝑛 ≥ 1) represents the probability
that at time 𝑡 there are 𝑛 customers in the system with elapsed
service time of the customer undergoing service lying in
[𝑥, 𝑥 + 𝑑𝑥). 𝜆 is batch arrival rate of customers. 𝑐𝑘 (𝑘 ≥ 1)
represents the probability that at every arrival epoch a batch
of 𝑘 external customers arrives and satisfies ∑∞
𝑘=1 𝑐𝑘 = 1. 𝜐(𝑥)
is the vacation rate of the server, which satisfies
𝜐 (𝑥) ≥ 0,
then it is obvious that 𝑋 is a Banach space. In the following
we define operators and their domains;
𝐴 (𝑃0 , 𝑃1 )
−𝜆
0
0
𝑑
0 −
0
((
𝑑𝑥
(
= ((
𝑑
0
0 −
𝑑𝑥
..
..
..
.
.
.
((
−
(2)
0
..
( .
𝑏(𝑥) is the service rate of the server satisfying
𝑏 (𝑥) ≥ 0,
0
0
0
..
.
0 𝑏 (𝑥) 0
0 0 𝑏 (𝑥)
Γ3 = (0 0
0
..
..
..
.
.
.
0
..
.
⋅⋅⋅
𝑃1,1 (𝑥)
)
)
𝑃
0 ⋅ ⋅ ⋅)
1,2 (𝑥)
)
) (𝑃1,3 (𝑥))) ,
)
)
𝑑
−
⋅⋅⋅
..
𝑑𝑥
.
..
. d)
)
⋅⋅⋅
⋅⋅⋅
⋅ ⋅ ⋅) ,
d
𝑈 (𝑃0 , 𝑃1 )
(3)
0
⋅⋅⋅
0
0
⋅⋅⋅
⋅ ⋅ ⋅) , Γ2 = (0
..
.
d
𝑑
−
𝑑𝑥
0
𝑄
𝑃0,0 (𝑥)
)
) (𝑃0,1 (𝑥)) ,
⋅⋅⋅
..
.
d)
⋅⋅⋅
󵄨󵄨 𝑑𝑃 (𝑥) 𝑑𝑃 (𝑥)
1,𝑘
󵄨󵄨 0,𝑛
{
,
∈ 𝐿1 [0, ∞) ,
󵄨󵄨
{
{
󵄨
{
𝑑𝑥
𝑑𝑥
󵄨󵄨
{
{
󵄨󵄨 𝑃0,𝑛 (𝑥) , 𝑃1,𝑘 (𝑥) (𝑛 ≥ 0, 𝑘 ≥ 1)
{
{
󵄨󵄨
{
{
󵄨󵄨 are absolutely continuous
{
{
󵄨󵄨
{
{
󵄨󵄨
{
{
󵄨󵄨 functions and satisfy
{
{
󵄨󵄨
∞
= {(𝑃0 , 𝑃1 ) ∈ 𝑋 󵄨󵄨󵄨 𝑃 (0) = ∫ Γ 𝑃 (𝑥) 𝑑𝑥,
1 0
󵄨󵄨 0
{
{
󵄨󵄨
0
{
{
󵄨󵄨
{
∞
{
󵄨󵄨
{
{
󵄨󵄨 𝑃1 (0) = ∫ Γ2 𝑃0 (𝑥) 𝑑𝑥
{
{
󵄨󵄨
{
{
0
󵄨󵄨
{
{
󵄨󵄨
∞
{
{
󵄨󵄨
{
󵄨󵄨
+ ∫ Γ3 𝑃1 (𝑥) 𝑑𝑥
󵄨
󵄨
{
0
∫ 𝑏 (𝑥) 𝑑𝑥 = ∞.
We first formulate the system (1) as an abstract Cauchy
problem on a suitable state space. For convenience we take
some notations as follows:
0
0
0
..
.
0
⋅⋅⋅
𝐷 (𝐴)
∞
2. Problem Formulation
𝑒−𝑥
𝜆𝑒−𝑥
Γ1 = ( 0
..
.
𝑑
𝑑𝑥
( 0
(
(
(
0
∞
∫ 𝜐 (𝑥) 𝑑𝑥 = ∞.
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
,
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
(5)
0 𝜐 (𝑥) 0
0 0 𝜐 (𝑥)
0 0
0
..
..
..
.
.
.
⋅⋅⋅
⋅⋅⋅
⋅ ⋅ ⋅) .
d
(4)
0
0
= ((0
0
..
(( .
0 0
𝜐̃ 0
𝜆𝑐1 𝜐̃
𝜆𝑐2 𝜆𝑐1
..
..
.
.
0
0
0
𝜐̃
..
.
⋅⋅⋅
𝑄
⋅⋅⋅
𝑃0,0 (𝑥)
⋅ ⋅ ⋅) (𝑃0,1 (𝑥)) ,
⋅⋅⋅
𝑃0,2 (𝑥)
..
d) ( . )
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
,
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
Advances in Mathematical Physics
̃𝑏
0
0 0 ⋅⋅⋅
𝜆𝑐2
̃𝑏
0 0 ⋅⋅⋅
3
𝑑𝑃0,𝑛 (𝑥)
= −𝛾𝑃0,𝑛 (𝑥) + 𝑦0,𝑛 (𝑥) ,
𝑑𝑥
𝑑𝑃1,𝑛 (𝑥)
= −𝛾𝑃1,𝑛 (𝑥) + 𝑦1,𝑛 (𝑥) ,
𝑑𝑥
𝑃0,0 (0) = 𝜆𝑄,
𝑃1,1 (𝑥)
𝑃1,2 (𝑥)
(
)
)
(𝜆𝑐 𝜆𝑐 ̃𝑏 0 ⋅ ⋅ ⋅) (𝑃 (𝑥)))
2
( 3
) ( 1,3 )) ,
(
)
)
𝑃1,4 (𝑥)
𝜆𝑐4 𝜆𝑐3 𝜆𝑐2 ̃𝑏 ⋅ ⋅ ⋅
..
..
..
.. ..
. )
.
. . d) (
( .
)
𝑃0,𝑛 (0) = 0,
(10)
𝑛 ≥ 1,
(11)
(12)
𝑛 ≥ 1,
(13)
∞
(6)
𝑃1,𝑛 (0) = ∫ 𝜐 (𝑥) 𝑃0,𝑛 (𝑥) 𝑑𝑥
0
where 𝜐̃ = −[𝜆 + 𝜐(𝑥)], ̃𝑏 = −[𝜆 + 𝑏(𝑥)] + 𝜆𝑐1 ,
(14)
∞
+ ∫ 𝑏 (𝑥) 𝑃1,𝑛+1 (𝑥) 𝑑𝑥,
𝐸 (𝑃0 , 𝑃1 )
𝑛 ≥ 1.
0
∞
∞
0
0
Through solving (9)–(11), we have
∫ 𝜐 (𝑥) 𝑃0,0 (𝑥) 𝑑𝑥 + ∫ 𝑏 (𝑥) 𝑃1,1 (𝑥) 𝑑𝑥
= ((
𝑛 ≥ 0,
𝑄=
),
0
..
.
1
𝑦 ,
𝛾+𝜆 𝑄
(15)
𝑥
𝑃0,𝑛 (𝑥) = 𝑎0,𝑛 𝑒−𝛾𝑥 + 𝑒−𝛾𝑥 ∫ 𝑦0,𝑛 (𝜏) 𝑒𝛾𝜏 𝑑𝜏,
𝑛 ≥ 0,
(16)
𝑃1,𝑛 (𝑥) = 𝑎1,𝑛 𝑒−𝛾𝑥 + 𝑒−𝛾𝑥 ∫ 𝑦1,𝑛 (𝜏) 𝑒𝛾𝜏 𝑑𝜏,
𝑛 ≥ 1.
(17)
0
𝑥
0
(0)) ,
..
.
0
Combining (16) with (12) and (13), we obtain
𝐷 (𝑈) = 𝐷 (𝐸) = 𝑋.
𝑎0,0 = 𝑃0,0 (0) = 𝜆𝑄 =
(7)
Then the previous system of equations (1) can be rewritten as
an abstract Cauchy problem in the Banach space 𝑋:
𝑑 (𝑃0 , 𝑃1 ) (𝑡)
= (𝐴 + 𝑈 + 𝐸) (𝑃0 , 𝑃1 ) (𝑡) ,
𝑑𝑡
𝑎0,𝑛 = 𝑃0,𝑛 (0) = 0,
(18)
𝑛 ≥ 1.
(19)
Substituting (19) into (16), it follows that
𝑥
∀𝑡 ∈ [0, ∞) ,
𝑃0,𝑛 (𝑥) = 𝑒−𝛾𝑥 ∫ 𝑦0,𝑛 (𝜏) 𝑒𝛾𝜏 𝑑𝜏,
0
1
0
(𝑃0 , 𝑃1 ) (0) = ((0) , (0)) .
..
..
.
.
𝑛 ≥ 1.
(20)
By combining (15), (16), (17), and (20) with (14), we deduce
𝑎1,𝑛 = 𝑃1,𝑛 (0)
(8)
∞
𝑥
= ∫ 𝜐 (𝑥) 𝑒−𝛾𝑥 ∫ 𝑦0,𝑛 (𝜏) 𝑒𝛾𝜏 𝑑𝜏 𝑑𝑥
0
3. Well-Posedness of The System (8)
0
∞
+ 𝑎1,𝑛+1 ∫ 𝑏 (𝑥) 𝑒−𝛾𝑥 𝑑𝑥
0
Theorem 1. If 𝜐 = sup𝑥∈[0,∞) 𝜐(𝑥) < ∞, 𝑏 = sup𝑥∈[0,∞) 𝑏(𝑥)<
∞, then 𝐴 + 𝑈 + 𝐸 generates a positive contraction 𝐶0 semigroup 𝑇(𝑡).
Proof. We split the proof of the theorem into four steps.
Firstly, we prove that (𝛾𝐼 − 𝐴)−1 exists and is bounded for
some 𝛾. Secondly, we show that 𝐷(𝐴) is dense in 𝑋. Thirdly,
we verify that 𝑈 and 𝐸 are bounded linear operators. Thus by
using the Hille-Yosida theorem and the perturbation theorem
of 𝐶0 -semigroup we deduce that 𝐴 + 𝑈 + 𝐸 generates a 𝐶0 semigroup 𝑇(𝑡). Finally, we check that 𝐴 + 𝑈 + 𝐸 is dispersive,
and therefore we obtain the desired result.
For any given (𝑦0 , 𝑦1 ) ∈ 𝑋, we consider the equation (𝛾𝐼−
𝐴)(𝑃0 , 𝑃1 ) = (𝑦0 , 𝑦1 ); that is,
(𝛾 + 𝜆) 𝑄 = 𝑦𝑄,
𝜆
𝑦 ,
𝛾+𝜆 𝑄
(9)
∞
𝑥
0
0
+ ∫ 𝑏 (𝑥) 𝑒−𝛾𝑥 ∫ 𝑦1,𝑛+1 (𝜏) 𝑒𝛾𝜏 𝑑𝜏 𝑑𝑥,
𝑛 ≥ 1,
(21)
󳨐⇒
∞
𝑎1,𝑛 − 𝑎1,𝑛+1 ∫ 𝑏 (𝑥) 𝑒−𝛾𝑥 𝑑𝑥
0
∞
𝑥
= ∫ 𝜐 (𝑥) 𝑒−𝛾𝑥 ∫ 𝑦0,𝑛 (𝜏) 𝑒𝛾𝜏 𝑑𝜏 𝑑𝑥
0
0
∞
𝑥
0
0
+ ∫ 𝑏 (𝑥) 𝑒−𝛾𝑥 ∫ 𝑦1,𝑛+1 (𝜏) 𝑒𝛾𝜏 𝑑𝜏 𝑑𝑥.
If we set
(22)
4
Advances in Mathematical Physics
∞
1 − ∫ 𝑏 (𝑥) 𝑒−𝛾𝑥 𝑑𝑥
0
0
0
∞
− ∫ 𝑏 (𝑥) 𝑒−𝛾𝑥 𝑑𝑥
(0
C=(
(
0
1
0
1
..
(.
..
.
..
.
0
⋅⋅⋅
0
⋅ ⋅ ⋅)
),
)
−𝛾𝑥
− ∫ 𝑏 (𝑥) 𝑒 𝑑𝑥 ⋅ ⋅ ⋅
∞
0
(23)
..
.
d)
𝑎1,1
𝑎1,2
𝑎1⃗ = (𝑎1,3 ) ,
..
.
then (21) can be rewritten as follows:
∞
𝑥
∞
𝑥
0
∞
0
𝑥
0
∞
0
𝑥
0
0
0
0
∫ 𝜐 (𝑥) 𝑒−𝛾𝑥 ∫ 𝑦0,1 (𝜏) 𝑒𝛾𝜏 𝑑𝜏 𝑑𝑥 + ∫ 𝑏 (𝑥) 𝑒−𝛾𝑥 ∫ 𝑦1,2 (𝜏) 𝑒𝛾𝜏 𝑑𝜏 𝑑𝑥
(24)
C𝑎1⃗ = (∫ 𝜐 (𝑥) 𝑒−𝛾𝑥 ∫ 𝑦0,2 (𝜏) 𝑒𝛾𝜏 𝑑𝜏 𝑑𝑥 + ∫ 𝑏 (𝑥) 𝑒−𝛾𝑥 ∫ 𝑦1,3 (𝜏) 𝑒𝛾𝜏 𝑑𝜏 𝑑𝑥) .
..
.
(
)
It is easy to calculate
∞
∞
0
0
∞
2
3
∞
1 ∫ 𝑏 (𝑥) 𝑒−𝛾𝑥 𝑑𝑥 (∫ 𝑏(𝑥)𝑒−𝛾𝑥 𝑑𝑥) (∫ 𝑏(𝑥)𝑒−𝛾𝑥 𝑑𝑥) ⋅ ⋅ ⋅
C−1
(
(0
(
=(
(
(
0
..
(.
1
∫ 𝑏 (𝑥) 𝑒−𝛾𝑥 𝑑𝑥
0
1
..
.
..
.
0
0
∞
2
)
(∫ 𝑏(𝑥)𝑒−𝛾𝑥 𝑑𝑥) ⋅ ⋅ ⋅)
)
0
).
)
∞
)
∫ 𝑏 (𝑥) 𝑒−𝛾𝑥 𝑑𝑥 ⋅ ⋅ ⋅
(25)
0
..
.
d)
From which together with (24), we derive
∞
∞
𝑘=0
0
∞
∞
1 󵄨󵄨 󵄨󵄨
󵄨 𝛾𝜏
󵄨
−𝛾𝑥
󵄨󵄨𝑎0,𝑛 󵄨󵄨 + ∫ 󵄨󵄨󵄨𝑦0,𝑛 (𝜏)󵄨󵄨󵄨 𝑒 ∫ 𝑒 𝑑𝑥 𝑑𝜏
𝛾
0
𝜏
1 󵄨 󵄨 1󵄩 󵄩
= 󵄨󵄨󵄨𝑎0,𝑛 󵄨󵄨󵄨 + 󵄩󵄩󵄩𝑦0,𝑛 󵄩󵄩󵄩𝐿1 [0,∞) , 𝑛 ≥ 0,
𝛾
𝛾
1 󵄨 󵄨 1󵄩 󵄩
󵄩󵄩 󵄩󵄩
󵄩󵄩𝑃1,𝑛 󵄩󵄩𝐿1 [0,∞) ≤ 󵄨󵄨󵄨𝑎1,𝑛 󵄨󵄨󵄨 + 󵄩󵄩󵄩𝑦1,𝑛 󵄩󵄩󵄩𝐿1 [0,∞) , 𝑛 ≥ 1.
𝛾
𝛾
=
𝑘
𝑎1,𝑛 = ∑ (∫ 𝑏 (𝑥) 𝑒−𝛾𝑥 𝑑𝑥)
∞
𝑥
0
0
× {∫ 𝜐 (𝑥) 𝑒−𝛾𝑥 ∫ 𝑦0,𝑘+𝑛 (𝜏) 𝑒𝛾𝜏 𝑑𝜏 𝑑𝑥
∞
𝑥
0
0
+ ∫ 𝑏 (𝑥) 𝑒−𝛾𝑥 ∫ 𝑦1,𝑘+𝑛+1 (𝜏) 𝑒𝛾𝜏 𝑑𝜏 𝑑𝑥} ,
𝑛 ≥ 1.
(26)
(27)
From (26) and Fubini theorem, we deduce
∞
By using Fubini theorem we estimate (16) and (17) as follows
(assume that 𝛾 > 𝜐 + 𝑏):
𝑘
∞
∞
󵄨
󵄨
× {𝜐 ∫ 󵄨󵄨󵄨𝑦0,𝑘+𝑛 (𝜏)󵄨󵄨󵄨 𝑒𝛾𝜏 ∫ 𝑒−𝛾𝑥 𝑑𝑥 𝑑𝜏
0
𝜏
󵄩󵄩 󵄩󵄩
󵄩󵄩𝑃0,𝑛 󵄩󵄩𝐿1 [0,∞)
∞
∞ ∞
𝑏
󵄨 󵄨
∑ 󵄨󵄨󵄨𝑎1,𝑛 󵄨󵄨󵄨 ≤ ∑ ∑ ( )
𝑛=1
𝑛=1𝑘=0 𝛾
∞
𝑥
󵄨
󵄨 󵄨
󵄨
≤ ∫ 󵄨󵄨󵄨𝑎0,𝑛 󵄨󵄨󵄨 𝑒−𝛾𝑥 𝑑𝑥 + ∫ 𝑒−𝛾𝑥 ∫ 󵄨󵄨󵄨𝑦0,𝑛 (𝜏)󵄨󵄨󵄨 𝑒𝛾𝜏 𝑑𝜏 𝑑𝑥
0
0
0
∞
∞
󵄨
󵄨
+𝑏 ∫ 󵄨󵄨󵄨𝑦1,𝑘+𝑛+1 (𝜏)󵄨󵄨󵄨 𝑒𝛾𝜏 ∫ 𝑒−𝛾𝑥 𝑑𝑥 𝑑𝜏}
0
𝜏
Advances in Mathematical Physics
=
5
𝜐 ∞ 𝑏 𝑘 ∞ 󵄩󵄩
󵄩󵄩
∑ ( ) ∑ 󵄩𝑦
󵄩1
𝛾 𝑘=0 𝛾 𝑛=1󵄩 0,𝑘+𝑛 󵄩𝐿 [0,∞)
it follows that, for any 𝜖 > 0, there exists a positive integer 𝑁
∞
such that ∑∞
𝑛=𝑁 ‖𝑃0,𝑛 ‖𝐿1 [0,∞) < 𝜖, ∑𝑛=𝑁 ‖𝑃1,𝑛 ‖𝐿1 [0,∞) < ∞. Let
󵄨󵄨 𝑃 (𝑥) = (𝑄, 𝑃 (𝑥) , 𝑃 (𝑥) , . . . ,
󵄨󵄨 0
0,0
0,1
{
{
󵄨󵄨
{
{
󵄨
𝑃
,
0,
. . .) ,
(𝑥)
{
󵄨󵄨
0,𝑁
{
󵄨󵄨
{
{
󵄨
{
󵄨󵄨 𝑃1 (𝑥) = (𝑃1,1 (𝑥) , 𝑃1,2 (𝑥) , . . . ,
{
{
󵄨󵄨
{
{
󵄨󵄨
{
𝑃1,𝑁 (𝑥) , 0, . . .) ,
L = {(𝑃0 , 𝑃1 ) 󵄨󵄨󵄨󵄨
{
󵄨󵄨 𝑄 ∈ R, 𝑃0,𝑖 (𝑥) ,
{
{
󵄨󵄨
{
{
󵄨󵄨 𝑃1,𝑗 (𝑥) ∈ 𝐿1 [0, ∞) , 𝑖 = 0, 1, . . . , 𝑁;
{
{
󵄨󵄨
{
󵄨󵄨
{
{
{
󵄨󵄨󵄨 𝑗 = 1, 2, . . . , 𝑁;
{
{
󵄨󵄨
󵄨󵄨 𝑁 is a finite positive integer
{
𝑏 𝑘+1 ∞ 󵄩
󵄩
+ ∑ ( ) ∑ 󵄩󵄩󵄩𝑦1,𝑘+𝑛+1 󵄩󵄩󵄩𝐿1 [0,∞)
𝛾
𝑛=1
𝑘=0
∞
≤
𝜐 ∞ 𝑏 𝑘 ∞ 󵄩󵄩 󵄩󵄩
∑ ( ) ∑ 󵄩𝑦 󵄩 1
𝛾 𝑘=0 𝛾 𝑛=0󵄩 0,𝑛 󵄩𝐿 [0,∞)
∞
𝑏 𝑘+1 ∞ 󵄩 󵄩
+ ∑ ( ) ∑ 󵄩󵄩󵄩𝑦1,𝑛 󵄩󵄩󵄩𝐿1 [0,∞)
𝛾
𝑛=1
𝑘=0
=
𝜐 ∞ 󵄩󵄩 󵄩󵄩
𝑏 ∞ 󵄩󵄩 󵄩󵄩
.
∑ 󵄩󵄩𝑦0,𝑛 󵄩󵄩𝐿1 [0,∞) +
∑ 󵄩𝑦 󵄩 1
𝛾 − 𝑏 𝑛=0
𝛾 − 𝑏 𝑛=1󵄩 1,𝑛 󵄩𝐿 [0,∞)
(28)
By inserting (18), (19), and (28) into (27) and using inequality
(𝛾 + 𝜐 − 𝑏)/𝛾(𝛾 − 𝑏) < 1/(𝛾 − 𝜐 − 𝑏), we estimate
󵄨󵄨 󵄨󵄨
󵄨󵄨𝑦𝑄󵄨󵄨
1 ∞ 󵄨 󵄨 1 ∞󵄩 󵄩
󵄩
󵄩󵄩
+ ∑ 󵄨󵄨󵄨𝑎0,𝑛 󵄨󵄨󵄨 + ∑ 󵄩󵄩󵄩𝑦0,𝑛 󵄩󵄩󵄩𝐿1 [0,∞)
󵄩󵄩(𝑃0 , 𝑃1 )󵄩󵄩󵄩 ≤
𝛾 + 𝜆 𝛾 𝑛=0
𝛾 𝑛=0
+
1 ∞ 󵄨󵄨 󵄨󵄨 1 ∞ 󵄩󵄩 󵄩󵄩
∑ 󵄨𝑎 󵄨 + ∑ 󵄩𝑦 󵄩 1
𝛾 𝑛=1 󵄨 1,𝑛 󵄨 𝛾 𝑛=1󵄩 1,𝑛 󵄩𝐿 [0,∞)
1 󵄨 󵄨 𝛾 + 𝜐 − 𝑏 ∞ 󵄩󵄩 󵄩󵄩
≤ 󵄨󵄨󵄨𝑦𝑄󵄨󵄨󵄨 +
∑ 󵄩𝑦 󵄩 1
𝛾
𝛾 (𝛾 − 𝑏) 𝑛=0󵄩 0,𝑛 󵄩𝐿 [0,∞)
+
≤
1 ∞ 󵄩󵄩 󵄩󵄩
∑ 󵄩𝑦 󵄩 1
𝛾 − 𝑏 𝑛=1󵄩 1,𝑛 󵄩𝐿 [0,∞)
(29)
∞
1
1
󵄩 󵄩
󵄨󵄨 󵄨󵄨
∑ 󵄩󵄩󵄩𝑦0,𝑛 󵄩󵄩󵄩𝐿1 [0,∞)
󵄨󵄨𝑦𝑄󵄨󵄨 +
𝛾−𝜐−𝑏
𝛾 − 𝜐 − 𝑏 𝑛=0
∞
1
󵄩 󵄩
+
∑ 󵄩󵄩󵄩𝑦1,𝑛 󵄩󵄩󵄩𝐿1 [0,∞)
𝛾 − 𝜐 − 𝑏 𝑛=1
1
󵄩󵄩
󵄩
=
󵄩(𝑦 , 𝑦 )󵄩󵄩 .
𝛾−𝜐−𝑏󵄩 0 1 󵄩
Equation (29) shows that (𝛾𝐼 − 𝐴)−1 exists for 𝛾 > 𝜐 + 𝑏 and
(𝛾𝐼 − 𝐴)
−1
: 𝑋 󳨀→ 𝐷 (𝐴) ,
1
󵄩
󵄩󵄩
󵄩󵄩(𝛾𝐼 − 𝐴)−1 󵄩󵄩󵄩 ≤
󵄩 𝛾 − 𝜐 − 𝑏.
󵄩
(30)
then L is dense in 𝑋. If we set
󵄨󵄨 𝑃 (𝑥) = (𝑄, 𝑃 (𝑥) , 𝑃 (𝑥) , . . . ,
󵄨󵄨 0
0,0
0,1
{
{
󵄨󵄨
{
{
󵄨
𝑃
,
0,
.
. .) ,
(𝑥)
{
󵄨󵄨
0,𝑙
{
{
󵄨󵄨
{
{
󵄨󵄨 𝑃1 (𝑥) = (𝑃1,1 (𝑥) , 𝑃1,2 (𝑥) , . . . ,
{
{
󵄨󵄨
{
󵄨󵄨
{
𝑃1,𝑙 (𝑥) , 0, . . .) ,
{
󵄨󵄨
{
{
󵄨󵄨 𝑃 (𝑥) , 𝑃 (𝑥) ∈ 𝐶∞ [0, ∞) ,
{
{
󵄨
{
1,𝑗
0
󵄨 0,𝑖
Z = {(𝑃0 , 𝑃1 ) 󵄨󵄨󵄨󵄨 and there exists positive
{
󵄨󵄨
{
{
󵄨󵄨
{
{
󵄨󵄨 numbers 𝑐0,𝑖 > 0, 𝑐1,𝑗 > 0,
{
{
󵄨󵄨󵄨
{
{
󵄨󵄨 such that 𝑃0,𝑖 (𝑥) = 0, 𝑥 ∈ [0, 𝑐0,𝑖 ] ,
{
{
󵄨󵄨
{
{
󵄨󵄨
{
{
󵄨󵄨 𝑃1,𝑗 (𝑥) = 0, 𝑥 ∈ [0, 𝑐1,𝑗 ] ;
{
{
󵄨󵄨
{
󵄨󵄨 𝑖 = 0, 1, 2, . . . , 𝑙; 𝑗 = 1, 2, . . . , 𝑙
󵄨
{
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
,
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
(32)
then by the relationship 𝐶0∞ [0, ∞) and 𝐿1 [0, ∞) in Adams
[17], we know that Z is dense in L. Hence in order to prove
denseness of 𝐷(𝐴), it suffices to prove that 𝐷(𝐴) is dense in Z.
Take any (𝑃0 , 𝑃1 ) ∈ Z, then there are a finite positive integer
𝑙 and positive numbers 𝑐0,𝑖 > 0, 𝑐1,𝑗 > 0 (𝑖 = 0, 1, . . . , 𝑙; 𝑗 =
1, 2, . . . , 𝑙) such that
𝑃0 (𝑥) = (𝑄, 𝑃0,0 (𝑥) , 𝑃0,1 (𝑥) , . . . , 𝑃0,𝑙 (𝑥) , 0, . . .) ,
𝑃1 (𝑥) = (𝑃1,1 (𝑥) , 𝑃1,2 (𝑥) , . . . , 𝑃1,𝑙 (𝑥) , 0, . . .) ,
𝑃0,𝑖 (𝑥) = 0,
for 𝑥 ∈ [0, 𝑐0,𝑖 ] , 𝑖 = 0, 1, 2, . . . , 𝑙,
𝑃1,𝑗 (𝑥) = 0,
for 𝑥 ∈ [0, 𝑐1,𝑗 ] , 𝑗 = 1, 2, . . . , 𝑙,
𝑃0,𝑖 (𝑥) = 0,
𝑃1,𝑗 (𝑥) = 0,
(33)
𝑥 ∈ [0, 2𝑠] , 𝑖 = 0, 1, . . . , 𝑙;
𝑥 ∈ [0, 2𝑠] , 𝑗 = 1, . . . , 𝑙,
(34)
where 0 < 2𝑠 < min{𝑐0,0 , 𝑐0,1 , . . . , 𝑐0,𝑙 , 𝑐1,1 , . . . , 𝑐1,𝑙 }. Define
𝑠
𝑠
𝑠
𝑓0𝑠 (0) = (𝑄, 𝑓0,0
(0) , 𝑓0,1
(0) , . . . , 𝑓0,𝑙
(0) , 0, . . .)
= (𝑄, 𝜆𝑄, 0, . . . , 0, 0, . . .) ,
As far as the second step is concerned, from |𝑄| +
∞
∑∞
𝑛=0 ‖𝑃0,𝑛 ‖𝐿1 [0,∞) + ∑𝑛=1 ‖𝑃1,𝑛 ‖𝐿1 [0,∞) < ∞ for (𝑃0 , 𝑃1 ) ∈ 𝑋
}
}
}
}
}
}
}
}
}
}
}
}
}
}
,
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
(31)
𝑠
𝑠
𝑠
𝑓1𝑠 (0) = (𝑓1,1
(0) , 𝑓1,2
(0) , . . . , 𝑓1,𝑙
(0) , 0, . . .) ,
(35)
6
Advances in Mathematical Physics
where
∞
∞
2𝑠
2𝑠
𝑠
𝑓1,𝑗
(0) = ∫ 𝜐 (𝑥) 𝑃0,𝑗 (𝑥) 𝑑𝑥 + ∫ 𝑏 (𝑥) 𝑃1,𝑗+1 (𝑥) 𝑑𝑥,
𝑗 = 1, 2, . . . , 𝑙 − 1,
∞
𝑙
𝑙
𝑙
5
󵄨𝑠
󵄨𝑠
󵄨 𝑠
󵄨 𝑠
󵄨 󵄨𝑠
+ ∑ 󵄨󵄨󵄨󵄨𝑓1,𝑗
= ∑ 󵄨󵄨󵄨󵄨𝑓0,𝑖
(0)󵄨󵄨󵄨󵄨 + ∑ 󵄨󵄨󵄨𝜇0,𝑖 󵄨󵄨󵄨
(0)󵄨󵄨󵄨󵄨
3
30
3
𝑖=0
𝑖=0
𝑗=1
𝑙
󵄨 󵄨 𝑠5
+ ∑ 󵄨󵄨󵄨󵄨𝜇1,𝑗 󵄨󵄨󵄨󵄨
󳨀→ 0
30
𝑗=1
as 𝑠 󳨀→ 0,
𝑠
𝑓1,𝑙
(0) = ∫ 𝜐 (𝑥) 𝑃0,𝑙 (𝑥) 𝑑𝑥,
(38)
2𝑠
𝑠
𝑠
𝑠
𝑓0𝑠 (𝑥) = (𝑄, 𝑓0,0
(𝑥) , 𝑓0,1
(𝑥) , . . . , 𝑓0,𝑙
(𝑥) , 0, . . .) ,
𝑠
𝑠
𝑠
𝑓1𝑠 (𝑥) = (𝑓1,1
(𝑥) , 𝑓1,2
(𝑥) , . . . , 𝑓1,𝑙
(𝑥) , 0, . . .) ,
(36)
where
∞
𝑛=0
𝑠
∫0 𝜐 (𝑥) 𝑓0,𝑖
2𝑠
𝜇0,𝑖 =
∫𝑠 𝜐 (𝑥) (𝑥 − 𝑠)2 (𝑥 − 2𝑠)2 𝑑𝑥
𝑠
𝑠
∫0 𝑏 (𝑥) 𝑓1,𝑗
2𝑠
𝜇1,𝑗 =
2
2
∫𝑠 𝑏 (𝑥) (𝑥 − 𝑠) (𝑥 − 2𝑠) 𝑑𝑥
𝑖 = 0, 1, . . . , 𝑙;
𝑙
(37)
,
𝑛=1
(𝑓0𝑠 , 𝑓1𝑠 )
∞
∞
∞
󵄨
󵄨
+ (2𝜆 + 𝑏) ∑ ∫ 󵄨󵄨󵄨𝑃1,𝑛 (𝑥)󵄨󵄨󵄨 𝑑𝑥
0
𝑛=1
󵄩
󵄩
≤ max {(2𝜆 + 𝜐) , (2𝜆 + 𝑏)} 󵄩󵄩󵄩(𝑃0 , 𝑃1 )󵄩󵄩󵄩 ,
∈ 𝐷(𝐴).
∞
󵄨
󵄩󵄩
󵄨
󵄩
󵄩󵄩𝐸 (𝑃0 , 𝑃1 )󵄩󵄩󵄩 ≤ ∫ 𝜐 (𝑥) 󵄨󵄨󵄨𝑃0,0 (𝑥)󵄨󵄨󵄨 𝑑𝑥
0
∞
󵄨
󵄨
+ ∫ 𝑏 (𝑥) 󵄨󵄨󵄨𝑃1,1 (𝑥)󵄨󵄨󵄨 𝑑𝑥
0
󵄩 󵄩
󵄩 󵄩
≤ 𝜐󵄩󵄩󵄩𝑃0,0 󵄩󵄩󵄩𝐿1 [0,∞) + 𝑏󵄩󵄩󵄩𝑃1,1 󵄩󵄩󵄩𝐿1 [0,∞)
󵄩
󵄩
≤ max {𝜐, 𝑏} 󵄩󵄩󵄩(𝑃0 , 𝑃1 )󵄩󵄩󵄩 .
𝑗=1
(39)
𝑠
2
𝑥
󵄨
󵄨 𝑠
= ∑ ∫ 󵄨󵄨󵄨󵄨𝑓0,𝑖
(0)󵄨󵄨󵄨󵄨 (1 − ) 𝑑𝑥
𝑠
𝑖=0 0
2𝑠
󵄨 󵄨
+ ∑ 󵄨󵄨󵄨𝜇0,𝑖 󵄨󵄨󵄨 ∫ (𝑥 − 𝑠)2 (𝑥 − 2𝑠)2 𝑑𝑥
𝑠
𝑙
𝑠
𝑥 2
󵄨
󵄨 𝑠
+ ∑ ∫ 󵄨󵄨󵄨󵄨𝑓1,𝑗
(0)󵄨󵄨󵄨󵄨 (1 − ) 𝑑𝑥
𝑠
𝑗=1 0
󵄨 󵄨 2𝑠
+ ∑ 󵄨󵄨󵄨󵄨𝜇1,𝑗 󵄨󵄨󵄨󵄨 ∫ (𝑥 − 𝑠)2 (𝑥 − 2𝑠)2 𝑑𝑥
𝑠
𝑗=1
𝑘=1
∞
𝑛=0
,
𝑙
∞
󵄨
󵄨
𝑠
+ ∑ ∫ 󵄨󵄨󵄨󵄨𝑃1,𝑗 (𝑥) − 𝑓1,𝑗
(𝑥)󵄨󵄨󵄨󵄨 𝑑𝑥
0
𝑙
∞
∞
𝑖=0
𝑖=0
∞
󵄨
󵄨
≤ (2𝜆 + 𝜐) ∑ ∫ 󵄨󵄨󵄨𝑃0,𝑛 (𝑥)󵄨󵄨󵄨 𝑑𝑥
0
󵄨
󵄨
𝑠
= ∑ ∫ 󵄨󵄨󵄨󵄨𝑃0,𝑖 (𝑥) − 𝑓0,𝑖
(𝑥)󵄨󵄨󵄨󵄨 𝑑𝑥
0
𝑙
∞
󵄨
󵄨
+ 𝜆 ∑ 𝑐𝑘 ∑ ∫ 󵄨󵄨󵄨𝑃1,𝑛 (𝑥)󵄨󵄨󵄨 𝑑𝑥
0
∞
𝑙
𝑛=0
∞
𝑛=1 0
𝑗 = 1, . . . , 𝑙.
Then it is not difficult to verify that
Moreover,
󵄩󵄩
𝑠
𝑠 󵄩
󵄩󵄩(𝑃0 , 𝑃1 ) − (𝑓0 , 𝑓1 )󵄩󵄩󵄩
𝑘=1
∞
2
(0) (1 − (𝑥/𝑠)) 𝑑𝑥
∞
󵄨
󵄨
+ ∑ ∫ [𝜆 + 𝑏 (𝑥)] 󵄨󵄨󵄨𝑃1,𝑛 (𝑥)󵄨󵄨󵄨 𝑑𝑥
2
(0) (1 − (𝑥/𝑠)) 𝑑𝑥
∞
󵄨
󵄨
+ 𝜆 ∑ 𝑐𝑘 ∑ ∫ 󵄨󵄨󵄨𝑃0,𝑛 (𝑥)󵄨󵄨󵄨 𝑑𝑥
0
𝑠
{
{
{𝑓1,𝑗
𝑠
∞
󵄨
󵄨
󵄩
󵄩󵄩
󵄩󵄩𝑈 (𝑃0 , 𝑃1 )󵄩󵄩󵄩 ≤ ∑ ∫ [𝜆 + 𝜐 (𝑥)] 󵄨󵄨󵄨𝑃0,𝑛 (𝑥)󵄨󵄨󵄨 𝑑𝑥
0
𝑥 2
𝑠
{
𝑥 ∈ [0, 𝑠) ,
𝑓0,𝑖
(0) (1 − ) ,
{
{
𝑠
𝑠
2
2
𝑓0,𝑖 (𝑥) = {−𝜇 (𝑥 − 𝑠) (𝑥 − 2𝑠) , 𝑥 ∈ [𝑠, 2𝑠) ,
{
{ 0,𝑖
𝑥 ∈ [2𝑠, ∞) ,
{𝑃0,𝑖 (𝑥) ,
𝑥 2
𝑥 ∈ [0, 𝑠) ,
(0) (1 − ) ,
𝑠
𝑠
𝑓1,𝑗
(𝑥) = {−𝜇 (𝑥 − 𝑠)2 (𝑥 − 2𝑠)2 , 𝑥 ∈ [𝑠, 2𝑠) ,
{
{ 1,𝑗
𝑥 ∈ [2𝑠, ∞) ,
{𝑃1,𝑗 (𝑥) ,
which shows that 𝐷(𝐴) is dense in Z. Hence, 𝐷(𝐴) is dense in
𝑋. From the first step, the second step, and the Hille-Yosida
theorem [18] we know that 𝐴 generates a 𝐶0 -semigroup.
Next we will verify that 𝑈 and 𝐸 are bounded linear
operators. From the definition of 𝑈 and 𝐸 and ∑∞
𝑘=1 𝑐𝑘 = 1
we have
The previous two formulas show that 𝑈 and 𝐸 are bounded
operators. It is easy to check that 𝑈 and 𝐸 are linear operators.
Hence from the perturbation theorem of 𝐶0 -semigroup [18],
we obtain that 𝐴 + 𝑈 + 𝐸 generates a 𝐶0 -semigroup 𝑇(𝑡).
Lastly, we will prove that 𝐴+𝑈+𝐸 is a dispersive operator.
For (𝑃0 , 𝑃1 ) ∈ 𝐷(𝐴) we take (𝜙0 , 𝜙1 ) as
+
𝜙0 (𝑥) = (
+
[𝑄]+ [𝑃0,0 (𝑥)] [𝑃0,1 (𝑥)]
,
, . . .) ,
,
𝑄
𝑃0,0 (𝑥)
𝑃0,1 (𝑥)
+
+
[𝑃1,1 (𝑥)] [𝑃1,2 (𝑥)]
𝜙1 (𝑥) = (
,
, . . .) ,
𝑃1,1 (𝑥)
𝑃1,2 (𝑥)
(40)
Advances in Mathematical Physics
7
𝑄,
[𝑄]+ = {
0,
+𝜆 ∑ 𝑐𝑘 𝑃1,𝑛−𝑘+1 (𝑥)}
if 𝑄 > 0,
if 𝑄 ≤ 0,
𝑘=1
= −𝜆[𝑄]+ +
𝑃 (𝑥) ,
+
[𝑃0,𝑛 (𝑥)] = { 0,𝑛
0,
if 𝑃0,𝑛 (𝑥) > 0,
if 𝑃0,𝑛 (𝑥) ≤ 0,
𝑛 ≥ 0,
𝑃 (𝑥) ,
+
[𝑃1,𝑛 (𝑥)] = { 1,𝑛
0,
if 𝑃1,𝑛 (𝑥) > 0,
if 𝑃1,𝑛 (𝑥) ≤ 0,
𝑛 ≥ 1.
(41)
+
∞
0
∞
∞
𝑘=1
𝑛=𝑘 0
=∫
∞
0
∫
(42)
𝑑𝑃0,𝑖 (𝑥) [𝑃0,𝑖 (𝑥)]
𝑑𝑃0,𝑖 (𝑥)
𝑑𝑥 = ∫
𝑑𝑥
𝑑𝑥
𝑃0,𝑖 (𝑥)
𝑑𝑥
𝑉0,𝑖
𝑃1,𝑗 (𝑥)
+
∞
∞
𝑘=1
𝑛=𝑘 0
+
+
𝑗 ≥ 1.
(43)
∞
𝑘=1
𝑛=𝑘 0
∞
+
∞
∞
+
∞
+
𝑛=1 0
∞
𝑛=2 0
∞
+
𝑛=1 0
∞
∞
𝑘=1
𝑛=𝑘 0
∞
+
+ 𝜆 ∑ 𝑐𝑘 ∑ ∫ [𝑃1,𝑛−𝑘+1 (𝑥)] 𝑑𝑥
[𝑄]+
+ ∫ 𝑏 (𝑥) 𝑃1,1 (𝑥) 𝑑𝑥}
𝑄
0
≤(
+
[𝑃0,0 (𝑥)]
𝑑𝑃0,0 (𝑥)
𝑑𝑥
+ ∫ {−
− [𝜆 + 𝜐 (𝑥)] 𝑃0,0 (𝑥)}
𝑑𝑥
𝑃0,0 (𝑥)
0
∞
𝑑𝑃0,𝑛 (𝑥)
+∑∫ {−
− [𝜆 + 𝜐 (𝑥)] 𝑃0,𝑛 (𝑥)
𝑑𝑥
𝑛=1 0
∞
𝑑𝑃1,𝑛 (𝑥)
− [𝜆 + 𝑏 (𝑥)] 𝑃1,𝑛 (𝑥)
𝑑𝑥
∞
+
0
∞
+
0
∞
+
∞
∞
𝑘=1
𝑛=0 0
∞
∞
𝑘=1
𝑛=1 0
∞
+
− ∑ ∫ 𝜆[𝑃0,𝑛 (𝑥)] 𝑑𝑥 + 𝜆 ∑ 𝑐𝑘 ∑ ∫ [𝑃0,𝑛 (𝑥)] 𝑑𝑥
𝑛=0 0
+
[𝑃0,𝑛 (𝑥)]
𝑑𝑥
𝑃0,𝑛 (𝑥)
[𝑄]+
− 1)
𝑄
× (∫ 𝜐 (𝑥) [𝑃0,0 (𝑥)] 𝑑𝑥 + ∫ 𝑏 (𝑥) [𝑃1,1 (𝑥)] 𝑑𝑥)
∞
𝑛=1 0
+
+ ∑ ∫ 𝜐 (𝑥) [𝑃0,𝑛 (𝑥)] 𝑑𝑥 + ∑ ∫ 𝑏 (𝑥) [𝑃1,𝑛 (𝑥)] 𝑑𝑥
∞
+∑∫ {−
∞
− ∑ ∫ [𝜆 + 𝑏 (𝑥)] [𝑃1,𝑛 (𝑥)] 𝑑𝑥
0
∞
∞
∞
= {−𝜆𝑄 + ∫ 𝜐 (𝑥) 𝑃0,0 (𝑥) 𝑑𝑥
∞
[𝑄]+ ∞
+
∫ 𝑏 (𝑥) [𝑃1,1 (𝑥)] 𝑑𝑥
𝑄 0
+ 𝜆 ∑ 𝑐𝑘 ∑ ∫ [𝑃0,𝑛−𝑘 (𝑥)] 𝑑𝑥
∞
𝑘=1
[𝑄]+ ∞
+
∫ 𝜐 (𝑥) [𝑃0,0 (𝑥)] 𝑑𝑥
𝑄 0
𝑛=0 0
𝑑𝑥 = −[𝑃1,𝑗 (0)] ,
+𝜆 ∑ 𝑐𝑘 𝑃0,𝑛−𝑘 (𝑥)}
[𝑃1,𝑛 (𝑥)]
𝑑𝑥
𝑃1,𝑛 (𝑥)
+ 𝜆[𝑄]+ − ∑ ∫ [𝜆 + 𝜐 (𝑥)] [𝑃0,𝑛 (𝑥)] 𝑑𝑥
⟨(𝐴 + 𝑈 + 𝐸) (𝑃0 , 𝑃1 ) , (𝜙0 , 𝜙1 )⟩
𝑛
+
+
∞
∞
𝑖 ≥ 0.
By using boundary conditions on (𝑃0 , 𝑃1 ) ∈ 𝐷(𝐴), (42), (43),
and ∑∞
𝑘=1 𝑐𝑘 = 1 for such (𝜙0 , 𝜙1 ), we derive
∞
∞
𝑛=1 0
≤ −𝜆[𝑄]+ +
+
𝑑[𝑃0,𝑖 (𝑥)]
+
𝑑𝑥 = −[𝑃0,𝑖 (0)] ,
𝑑𝑥
𝑑𝑥
0
[𝑃0,𝑛 (𝑥)]
𝑑𝑥
𝑃0,𝑛 (𝑥)
+ 𝜆 ∑ 𝑐𝑘 ∑ ∫ 𝑃1,𝑛−𝑘+1 (𝑥)
𝑑𝑃0,𝑖 (𝑥) [𝑃0,𝑖 (𝑥)]
𝑑𝑥
𝑑𝑥
𝑃0,𝑖 (𝑥)
𝑑𝑃1,𝑗 (𝑥) [𝑃1,𝑗 (𝑥)]
∞
𝑛=1
Similar to (42), we get
∞
+
∞
+
+
𝑉0,𝑖
+
𝑛=0 0
∞
+
=∫
∞
+ 𝜆 ∑ 𝑐𝑘 ∑ ∫ 𝑃0,𝑛−𝑘 (𝑥)
+
𝑊0,𝑖
∞
+
+ ∑ [𝑃1,𝑛 (0)] − ∑ ∫ [𝜆 + 𝑏 (𝑥)] [𝑃1,𝑛 (𝑥)] 𝑑𝑥
𝑑𝑃0,𝑖 (𝑥) [𝑃0,𝑖 (𝑥)]
𝑑𝑥
𝑑𝑥
𝑃0,𝑖 (𝑥)
+∫
[𝑄]+ ∞
∫ 𝑏 (𝑥) 𝑃1,1 (𝑥) 𝑑𝑥
𝑄 0
𝑛=0
+
𝑉0,𝑖
[𝑄]+ ∞
∫ 𝜐 (𝑥) 𝑃0,0 (𝑥) 𝑑𝑥
𝑄 0
∞
𝑑𝑃0,𝑖 (𝑥) [𝑃0,𝑖 (𝑥)]
𝑑𝑥
𝑑𝑥
𝑃0,𝑖 (𝑥)
=∫
[𝑃1,𝑛 (𝑥)]
𝑑𝑥
𝑃1,𝑛 (𝑥)
+ ∑ [𝑃0,𝑛 (0)] − ∑ ∫ [𝜆 + 𝜐 (𝑥)] [𝑃0,𝑛 (𝑥)] 𝑑𝑥
If we define 𝑉0,𝑖 = {𝑥 ∈ [0, ∞) | 𝑃0,𝑖 (𝑥) > 0} and 𝑊0,𝑖 =
{𝑥 ∈ [0, ∞) | 𝑃0,𝑖 (𝑥) ≤ 0} for 𝑖 = 0, 1, 2, . . ., then by a short
argument we calculate
∫
+
𝑛
where
∞
∞
+
∞
+
− ∑ ∫ 𝜆[𝑃1,𝑛 (𝑥)] 𝑑𝑥 + 𝜆 ∑ 𝑐𝑘 ∑ ∫ [𝑃1,𝑛 (𝑥)] 𝑑𝑥
𝑛=1 0
≤ 0.
(44)
8
Advances in Mathematical Physics
Equation (44) shows that 𝐴 + 𝑈 + 𝐸 is a dispersive operator.
From which together with the first step, the second step, and
the Phillips theorem, we know that 𝐴 + 𝑈 + 𝐸 generates a
positive contraction 𝐶0 -semigroup [18]. By the uniqueness of
a 𝐶0 -semigroup we conclude that this semigroup is just 𝑇(𝑡).
∞
𝑛=1 0
It is not difficult to see that 𝑋 , the dual space of 𝑋, is
𝑋
𝑛
𝑘=1
∞
∞
+∑∫ {−
𝑛=1 0
󵄨󵄨
󵄨󵄨
{
󵄨󵄨
{
{
󵄨󵄨
{
{
󵄨󵄨
{
{
󵄨󵄨
{
{
󵄨󵄨
{
{
󵄨󵄨
{
{
󵄨
{
{ ∗ ∗ 󵄨󵄨󵄨
= {(𝑞0 , 𝑞1 ) 󵄨󵄨󵄨
󵄨󵄨
{
{
{
󵄨󵄨󵄨
{
{
󵄨󵄨
{
{
󵄨󵄨
{
{
󵄨󵄨
{
{
󵄨󵄨
{
{
{
󵄨󵄨󵄨
󵄨󵄨
{
󵄨
𝑞0∗
∗
= (𝑞
∗
∗
, 𝑞0,0
, 𝑞0,1
, . . .)
}
}
}
}
}
}
}
}
}
}
∗
∗
∗
}
𝑞1 = (𝑞1,1 , 𝑞1,2 , . . .)
}
}
}
} (45)
∞
∞
∈ 𝐿 [0, ∞) × 𝐿 [0, ∞) × ⋅ ⋅ ⋅ ,
.
}
}
}
󵄩󵄩 ∗ 󵄩󵄩
}
󵄨󵄨 ∗ 󵄨󵄨
󵄩󵄩󵄨󵄨 ∗ ∗ 󵄨󵄨󵄩󵄩
}
󵄩󵄩󵄨󵄨(𝑞0 , 𝑞1 )󵄨󵄨󵄩󵄩 = sup {󵄨󵄨𝑞 󵄨󵄨 , sup󵄩󵄩󵄩𝑞0,𝑛 󵄩󵄩󵄩𝐿∞ [0,∞) , }
}
}
}
𝑛≥0
}
}
}
}
}
󵄩󵄩 ∗ 󵄩󵄩
sup󵄩󵄩󵄩𝑞1,𝑛 󵄩󵄩󵄩𝐿∞ [0,∞) } }
}
𝑛≥1
𝑆 = {(𝑃0 , 𝑃1 ) ∈ 𝑋 | 𝑄 ≥ 0, 𝑃0,𝑛 (𝑥) ≥ 0, 𝑛 ≥ 0;
𝑃1,𝑛 (𝑥) ≥ 0, 𝑛 ≥ 1, 𝑥 ∈ [0, ∞)} ,
𝑛
𝑘=1
󵄩
󵄩
= 󵄩󵄩󵄩(𝑃0 , 𝑃1 )󵄩󵄩󵄩
0
0
∞
∞ ∞
𝑑𝑃0,𝑛 (𝑥)
𝑑𝑥 + ∑ ∫ [𝜆 + 𝜐 (𝑥)] 𝑃0,𝑛 (𝑥) 𝑑𝑥
𝑑𝑥
𝑛=0 0
∞
× {∑ ∫ −
𝑛=0 0
∞
∞
𝑘=1
𝑛=0 0
∞
∞
− 𝜆 ∑ 𝑐𝑘 ∑ ∫ 𝑃0,𝑛 (𝑥) 𝑑𝑥 + ∑ ∫
∞
∞
𝑛=1 0
𝑑𝑃1,𝑛 (𝑥)
𝑑𝑥
𝑑𝑥
∞
+ ∑ ∫ [𝜆 + 𝑏 (𝑥)] 𝑃1,𝑛 (𝑥) 𝑑𝑥
𝑛=1 0
(47)
∞
∞
𝑘=1
𝑛=1 0
∞
−𝜆 ∑ 𝑐𝑘 ∑ ∫ 𝑃1,𝑛 (𝑥) 𝑑𝑥}
󵄩
󵄩
= 󵄩󵄩󵄩(𝑃0 , 𝑃1 )󵄩󵄩󵄩
⟨(𝑃0 , 𝑃1 ) , (𝑞0∗ , 𝑞1∗ )⟩
∞
∞
𝑛=0
𝑛=1
󵄩
󵄩
= 󵄩󵄩󵄩(𝑃0 , 𝑃1 )󵄩󵄩󵄩 (|𝑄| + ∑ 𝑃0,𝑛 (𝑥) 𝑑𝑥 + ∑ 𝑃1,𝑛 (𝑥) 𝑑𝑥) (48)
󵄩
󵄩2 󵄩󵄨
󵄨󵄩2
= 󵄩󵄩󵄩(𝑃0 , 𝑃1 )󵄩󵄩󵄩 = 󵄩󵄩󵄩󵄨󵄨󵄨(𝑞0∗ , 𝑞1∗ )󵄨󵄨󵄨󵄩󵄩󵄩 ,
that is
∗
∈𝑋 |
∞
∞
0
0
× {−𝜆𝑄 + ∫ 𝜐 (𝑥) 𝑃0,0 (𝑥) 𝑑𝑥 + ∫ 𝑏 (𝑥) 𝑃1,1 (𝑥) 𝑑𝑥}
󵄩
󵄩
− 󵄩󵄩󵄩(𝑃0 , 𝑃1 )󵄩󵄩󵄩
∞
∞
× {−𝜆𝑄 + ∑ ∫ 𝜐 (𝑥) 𝑃0,𝑛 (𝑥) 𝑑𝑥
𝑛=0 0
(𝑞0∗ , 𝑞1∗ ) ∈ 𝜃 ((𝑃0 , 𝑃1 ))
=
∞
󵄩
󵄩
− 󵄩󵄩󵄩(𝑃0 , 𝑃1 )󵄩󵄩󵄩
(46)
then we have
{ (𝑞0∗ , 𝑞1∗ )
∞
× {−𝜆𝑄 + ∫ 𝜐 (𝑥) 𝑃0,0 (𝑥) 𝑑𝑥 + ∫ 𝑏 (𝑥) 𝑃1,1 (𝑥) 𝑑𝑥}
then 𝑆 is a cone in 𝑋. For (𝑃0 , 𝑃1 ) ∈ 𝐷(𝐴) ∩ 𝑆, we take
1
1
󵄩󵄩
󵄩󵄩
∗ ∗
1
(𝑞0 , 𝑞1 ) = 󵄩󵄩(𝑃0 , 𝑃1 )󵄩󵄩 (( ) , (1)) ∈ 𝑋∗ ,
..
..
.
.
𝑑𝑃1,𝑛 (𝑥)
− [𝜆 + 𝑏 (𝑥)] 𝑃1,𝑛 (𝑥)
𝑑𝑥
󵄩
󵄩
+𝜆 ∑ 𝑐𝑘 𝑃1,𝑛−𝑘+1 (𝑥)} 󵄩󵄩󵄩(𝑃0 , 𝑃1 )󵄩󵄩󵄩 𝑑𝑥
∈ R × 𝐿∞ [0, ∞) × 𝐿∞ [0, ∞) × ⋅ ⋅ ⋅ ,
It is easy to check that 𝑋∗ is a Banach space. If we take a
set 𝑆 in 𝑋 as
𝑑𝑃0,𝑛 (𝑥)
− [𝜆 + 𝜐 (𝑥)] 𝑃0,𝑛 (𝑥)
𝑑𝑥
󵄩
󵄩
+𝜆 ∑ 𝑐𝑘 𝑃0,𝑛−𝑘 (𝑥)} 󵄩󵄩󵄩(𝑃0 , 𝑃1 )󵄩󵄩󵄩 𝑑𝑥
∗
∗
∞
+∑∫ {−
∞
⟨(𝑃0 , 𝑃1 ) , (𝑞0∗ , 𝑞1∗ )⟩
(49)
𝑛=1 0
∞
󵄩
󵄩2 󵄩󵄨
󵄨󵄩2
= 󵄩󵄩󵄩(𝑃0 , 𝑃1 )󵄩󵄩󵄩 = 󵄩󵄩󵄩󵄨󵄨󵄨(𝑞0∗ , 𝑞1∗ )󵄨󵄨󵄨󵄩󵄩󵄩 } .
For such (𝑞0∗ , 𝑞1∗ ), by using boundary conditions on (𝑃0 , 𝑃1 ) ∈
𝐷(𝐴) ∩ 𝑆 and ∑∞
𝑘=1 𝑐𝑘 = 1, we have
∞
− ∑ ∫ 𝜐 (𝑥) 𝑃0,𝑛 (𝑥) 𝑑𝑥
∞
∞
∞
− ∑ ∫ 𝑏 (𝑥) 𝑃1,𝑛 (𝑥) 𝑑𝑥 + ∑ ∫ 𝑏 (𝑥) 𝑃1,𝑛 (𝑥) 𝑑𝑥}
𝑛=2 0
𝑛=1 0
= 0.
⟨(𝐴 + 𝑈 + 𝐸) (𝑃0 , 𝑃1 ) , (𝑞0∗ , 𝑞1∗ )⟩
(50)
󵄩
󵄩
= 󵄩󵄩󵄩(𝑃0 , 𝑃1 )󵄩󵄩󵄩
∞
∞
0
0
× {−𝜆𝑄 + ∫ 𝜐 (𝑥) 𝑃0,0 (𝑥) 𝑑𝑥 + ∫ 𝑏 (𝑥) 𝑃1,1 (𝑥) 𝑑𝑥}
∞
+∫ { −
0
𝑑𝑃0,0 (𝑥)
󵄩
󵄩
− [𝜆 + 𝜐 (𝑥)] 𝑃0,0 (𝑥)} 󵄩󵄩󵄩(𝑃0 , 𝑃1 )󵄩󵄩󵄩 𝑑𝑥
𝑑𝑥
Which shows that 𝐴 + U + 𝐸 is a conservative operator. So we
can use the Fattorini theorem [19] and state it as follows.
Theorem 2. 𝑇(𝑡) is isometric for the initial value of the system
(8); that is,
󵄩 󵄩
󵄩
󵄩󵄩
󵄩󵄩𝑇 (𝑡) (𝑃0 , 𝑃1 ) (0)󵄩󵄩󵄩 = 󵄩󵄩󵄩(𝑃0 , 𝑃1 ) (0)󵄩󵄩󵄩 ,
∀𝑡 ∈ [0, ∞) .
(51)
Advances in Mathematical Physics
9
Proof. Since 𝐴 + 𝑈 + 𝐸 is conservative with respect to 𝜃 and
(𝑃0 , 𝑃1 )(0) ∈ 𝐷(𝐴2 ) ∩ 𝑆, from the Taylor expansion of 𝑇(𝑡 + ℎ)
for 𝑡, ℎ ≥ 0 we have
𝑇 (𝑡 + ℎ) (𝑃0 , 𝑃1 ) (0)
= 𝑇 (𝑡) (𝑃0 , 𝑃1 ) (0) + ℎ (𝐴 + 𝑈 + 𝐸) 𝑇 (𝑡) (𝑃0 , 𝑃1 ) (0)
+∫
𝑡+ℎ
𝑡
(𝑡 + ℎ − 𝑠) 𝑇 (𝑠) (𝐴 + 𝑈 + 𝐸)2 (𝑃0 , 𝑃1 ) (0) 𝑑𝑠
= 𝑇 (𝑡) (𝑃0 , 𝑃1 ) (0) + ℎ (𝐴 + 𝑈 + 𝐸) 𝑇 (𝑡) (𝑃0 , 𝑃1 ) (0)
+ ℎ𝜌 (𝑡, ℎ) ,
(52)
where ‖𝜌(𝑡, ℎ)‖ → 0 as ℎ → 0, uniformly in 𝑡 ≥ 0. Then
󵄩󵄩
󵄩󵄩 󵄩󵄩
󵄩󵄩
󵄩󵄩𝑇 (𝑡 + ℎ) (𝑃0 , 𝑃1 ) (0)󵄩󵄩 󵄩󵄩𝑇 (𝑡) (𝑃0 , 𝑃1 ) (0)󵄩󵄩
󵄨
∗ 󵄨
≥ 󵄨󵄨󵄨󵄨⟨𝑇 (𝑡 + ℎ) (𝑃0 , 𝑃1 ) (0) , (𝑇 (𝑡) (𝑃0 , 𝑃1 ) (0)) ⟩󵄨󵄨󵄨󵄨
∗
≥ Re ⟨𝑇 (𝑡 + ℎ) (𝑃0 , 𝑃1 ) (0) , (𝑇 (𝑡) (𝑃0 , 𝑃1 ) (0)) ⟩
(53)
󵄩2
󵄩
= 󵄩󵄩󵄩𝑇 (𝑡) (𝑃0 , 𝑃1 ) (0)󵄩󵄩󵄩
In view of (52) and the fact that 𝐴 + 𝑈 + 𝐸 is conservative with
respect to 𝜃, consider the set
󵄩 󵄩
󵄩
󵄩
Ω = {𝑡 ∈ [0, ∞) | 󵄩󵄩󵄩𝑇 (𝑡) (𝑃0 , 𝑃1 ) (0)󵄩󵄩󵄩 = 󵄩󵄩󵄩(𝑃0 , 𝑃1 ) (0)󵄩󵄩󵄩} .
(54)
Since 0 ∈ Ω, Ω is nonempty, moreover Ω is obviously a closed
interval because of continuity of 𝑇(𝑡). If Ω ≠ [0, ∞), then let
𝑡0 be the right end point of Ω and 𝜂 > 0 is so small that
‖𝑇(𝑡)(𝑃0 , 𝑃1 )(0)‖ is bounded away from zero in 𝑡0 ≤ 𝑡 ≤ 𝑡0 +𝜂.
For any such 𝑡 we divide (53) by ‖𝑇(𝑡)(𝑃0 , 𝑃1 )(0)‖ and get
󵄩󵄩
󵄩󵄩
󵄩󵄩𝑇 (𝑡 + ℎ) (𝑃0 , 𝑃1 ) (0)󵄩󵄩
ℎ
󵄩
󵄩
≥ 󵄩󵄩󵄩𝑇 (𝑡) (𝑃0 , 𝑃1 ) (0)󵄩󵄩󵄩 + 󵄩󵄩
󵄩
󵄩󵄩𝑇 (𝑡) (𝑃0 , 𝑃1 ) (0)󵄩󵄩󵄩
(55)
∗
× Re ⟨𝜌 (𝑡, ℎ) , (𝑇 (𝑡) (𝑃0 , 𝑃1 ) (0)) ⟩
󵄩
󵄩
≥ 󵄩󵄩󵄩𝑇 (𝑡) (𝑃0 , 𝑃1 ) (0)󵄩󵄩󵄩 − ℎ𝛽 (𝑡, ℎ) ,
where 𝛽(𝑡, ℎ) is positive and 𝛽(𝑡, ℎ) → 0 when ℎ → 0,
uniformly in 𝑡0 ≤ 𝑡 ≤ 𝑡0 + 𝜂.
Let now 𝜖 be are small positive number and 𝛿 > 0 such
that ‖𝛽(𝑡, ℎ)‖ ≤ 𝜖 for 0 ≤ ℎ ≤ 𝛿 and 𝑡0 ≤ 𝑡 ≤ 𝑡0 + 𝜂. Let
𝑡0 < 𝑡1 < 𝑡2 < ⋅ ⋅ ⋅ < 𝑡𝑚 = 𝑡0 + 𝜂 be partition of the interval
𝑡0 ≤ 𝑡 ≤ 𝑡0 + 𝜂 such that 𝑡𝑗 − 𝑡𝑗−1 ≤ 𝛿 (1 ≤ 𝑗 ≤ 𝑚). Then by
(55), one has
󵄩
󵄩 󵄩
󵄩
0 ≤ 󵄩󵄩󵄩𝑇 (𝑡0 ) (𝑃0 , 𝑃1 ) (0)󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑇 (𝑡0 + 𝜂) (𝑃0 , 𝑃1 ) (0)󵄩󵄩󵄩
𝑚
󵄩 󵄩
󵄩
󵄩
≤ ∑ (󵄩󵄩󵄩󵄩𝑇 (𝑡𝑗−1 ) (𝑃0 , 𝑃1 ) (0)󵄩󵄩󵄩󵄩 − 󵄩󵄩󵄩󵄩𝑇 (𝑡𝑗 ) (𝑃0 , 𝑃1 ) (0)󵄩󵄩󵄩󵄩)
𝑚
𝑗=1
Theorem 3. If 𝜐 = sup𝑥∈[0,∞) 𝜐(𝑥) < ∞, 𝑏 = sup𝑥∈[0,∞) 𝑏(𝑥) <
∞, then the system (8) has a unique nonnegative timedependent solution (𝑃0 , 𝑃1 )(𝑥, 𝑡), which satisfies
󵄩
󵄩󵄩
󵄩󵄩(𝑃0 , 𝑃1 ) (⋅, 𝑡)󵄩󵄩󵄩 = 1,
∀𝑡 ∈ [0, ∞) .
(57)
Proof. Since (𝑃0 , 𝑃1 )(0) ∈ 𝐷(𝐴2 ) ∩ 𝑆, by Theorem 1 and Theorem 11 in Gupur et al. [18], we know that the system (8) has
a unique nonnegative time-dependent solution (𝑃0 , 𝑃1 )(𝑥, 𝑡)
which can be expressed as
(𝑃0 , 𝑃1 ) (𝑥, 𝑡) = 𝑇 (𝑡) (𝑃0 , 𝑃1 ) (0) ,
∀𝑡 ∈ [0, ∞) .
(58)
󵄩 󵄩
󵄩 󵄩
󵄩
󵄩󵄩
󵄩󵄩(𝑃0 , 𝑃1 ) (⋅, 𝑡)󵄩󵄩󵄩 = 󵄩󵄩󵄩𝑇 (𝑡) (𝑃0 , 𝑃1 ) (0)󵄩󵄩󵄩 = 󵄩󵄩󵄩(𝑃0 , 𝑃1 ) (0)󵄩󵄩󵄩 = 1,
∗
≤ ∑ (𝑡𝑗 − 𝑡𝑗−1 ) 𝛽 (𝑡𝑗 , 𝑡𝑗 − 𝑡𝑗−1 ) ≤ 𝜂𝜖.
From Theorems 1 and 2 we obtain the main result in this
paper.
From which together with Theorem 2 (i.e., (51)) we have
+ ℎ Re ⟨𝜌 (𝑡, ℎ) , (𝑇 (𝑡) (𝑃0 , 𝑃1 ) (0)) ⟩ .
𝑗=1
Since 𝜖 is arbitrary, it follows that ‖𝑇(𝑡0 + 𝜂)(𝑃0 , 𝑃1 )(0)‖ =
‖𝑇(𝑡0 )(𝑃0 , 𝑃1 )(0)‖, which contradicts the fact that 𝑡0 is
the right endpoint of Ω. Hence Ω = [0, ∞). That is,
‖𝑇(𝑡)(𝑃0 , 𝑃1 )(0)‖ = ‖(𝑃0 , 𝑃1 )(0)‖ for 𝑡 ∈ [0, ∞). The proof of
the theorem is complete.
(56)
∀𝑡 ∈ [0, ∞) ,
(59)
this just reflects the physical background of the problem.
4. Concluding Remarks
If we know the spectrum of 𝐴 + 𝑈 + 𝐸 on the imaginary
axis, then by Theorem 1 and Theorem 14 in Gupur et al. [18],
we obtain the asymptotic behavior of the time-dependent
solution of the system (8), which describes Hypothesis 2. It
is our next research work.
Acknowledgments
The author would like to thank the reviewer for his/her
valuable suggestions and comments which helped to improve
the quality and clarity of the paper. This work was supported
by The Open University of China General Foundation (no:
Q4301E-Y) and Xinjiang Radio & TV University Foundation
(no: 2013xjddkt001).
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