# Semigroup Method on aMX/G/1 Queueing Model

Authors: Alim Mijit
Journal: Advances in Mathematical Physics
 (2013)

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## Abstract

By using the Hille-Yosida theorem, Phillips theorem, and Fattorini theorem in functional analysis we prove that theMX/G/1 queueing model with vacation times has a unique nonnegative time-dependent solution.


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## Body

## 1. Introduction

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 1.
The model has a nonnegative time-dependent solution.Hypothesis 2.
The time-dependent solution of the model converges to a nonzero steady-state solution.In this paper we investigate Hypothesis1. By using the Hille-Yosida theorem, Phillips theorem, and Fattorini theorem we prove that the model has a unique nonnegative time-dependent 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:
(1)dQ(t)dt=-λQ(t)+∫0∞‍υ(x)P0,0(x,t)dxdQ(t)dt=+∫0∞‍b(x)P1,1(x,t)dx,∂P0,0(x,t)∂t+∂P0,0(x,t)∂x=-[λ+υ(x)]P0,0(x,t),∂P0,n(x,t)∂t+∂P0,n(x,t)∂x=-[λ+υ(x)]P0,n(x,t)+λ∑k=1n‍ckP0,n-k(x,t),n≥1,∂P1,n(x,t)∂t+∂P1,n(x,t)∂x=-[λ+b(x)]P1,n(x,t)+λ∑k=1n‍ckP1,n-k+1(x,t),n≥1,P0,0(t)=λQ(t),P0,n(0,t)=0,n≥1,P1,n(0,t)=∫0∞‍υ(x)P0,n(x,t)dxP1,n(0,t)=+∫0∞‍b(x)P1,n+1(x,t)dx,n≥1,Q(0)=1,P0,n(x,0)=0,n≥0,P1,n(x,0)=0,n≥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
(2)υ(x)≥0,∫0∞‍υ(x)dx=∞.b(x) is the service rate of the server satisfying
(3)b(x)≥0,∫0∞‍b(x)dx=∞.


## 2. Problem Formulation

We first formulate the system (1) as an abstract Cauchy problem on a suitable state space. For convenience we take some notations as follows:
(4)Γ1=(e-x00⋯λe-x00⋯000⋯⋮⋮⋮⋱),Γ2=(00υ(x)0⋯000υ(x)⋯0000⋯⋮⋮⋮⋮⋱),Γ3=(0b(x)0⋯00b(x)⋯000⋯⋮⋮⋮⋱).
If we take state space
(5)X={(P0,P1)|P0=(Q,P0,0,P0,1,…)P0∈ℝ×L1[0,∞)×L1[0,∞)×⋯P1=(P1,1,P1,2,P1,3,…)P1∈L1[0,∞)×L1[0,∞)P1=×L1[0,∞)×⋯∥(P0,P1)∥=|Q|+∑i=0∞‍∥P0,i∥L1[0,∞)∥(P0,P1)∥=+∑i=1∞‍∥P1,i∥L1[0,∞)<∞},
then it is obvious that X is a Banach space. In the following we define operators and their domains;
(6)A(P0,P1)=((-ddx00⋯0-ddx0⋯00-ddx⋯⋮⋮⋮⋱)(-λ00⋯0-ddx0⋯00-ddx⋯⋮⋮⋮⋱)(QP0,0(x)P0,1(x)⋮),(-ddx00⋯0-ddx0⋯00-ddx⋯⋮⋮⋮⋱)(P1,1(x)P1,2(x)P1,3(x)⋮)),D(A)={(P0,P1)∈X|dP0,n(x)dx,dP1,k(x)dx∈L1[0,∞),P0,n(x),P1,k(x)(n≥0,k≥1)areabsolutelycontinuousfunctionsandsatisfyP0(0)=∫0∞‍Γ1P0(x)dx,P1(0)=∫0∞‍Γ2P0(x)dxP1(0)=+∫0∞‍Γ3P1(x)dx},U(P0,P1)=((0000⋯0υ~00⋯0λc1υ~0⋯0λc2λc1υ~⋯⋮⋮⋮⋮⋱)(QP0,0(x)P0,1(x)P0,2(x)⋮),(b~000⋯λc2b~00⋯λc3λc2b~0⋯λc4λc3λc2b~⋯⋮⋮⋮⋮⋱)(P1,1(x)P1,2(x)P1,3(x)P1,4(x)⋮)),
where υ~=-[λ+υ(x)], b~=-[λ+b(x)]+λc1,
(7)E(P0,P1)=((∫0∞‍υ(x)P0,0(x)dx+∫0∞‍b(x)P1,1(x)dx0⋮),(00⋮)),D(U)=D(E)=X.
Then the previous system of equations (1) can be rewritten as an abstract Cauchy problem in the Banach space X:
(8)d(P0,P1)(t)dt=(A+U+E)(P0,P1)(t),∀t∈[0,∞),(P0,P1)(0)=((10⋮),(00⋮)).


## 3. Well-Posedness of The System (8)

Theorem 1.
Ifυ=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,
(9)(γ+λ)Q=yQ,(10)dP0,n(x)dx=-γP0,n(x)+y0,n(x),n≥0,(11)dP1,n(x)dx=-γP1,n(x)+y1,n(x),n≥1,(12)P0,0(0)=λQ,(13)P0,n(0)=0,n≥1,(14)P1,n(0)=∫0∞‍υ(x)P0,n(x)dx+∫0∞‍b(x)P1,n+1(x)dx,n≥1.
Through solving (9)–(11), we have
(15)Q=1γ+λyQ,(16)P0,n(x)=a0,ne-γx+e-γx∫0x‍y0,n(τ)eγτdτ,n≥0,(17)P1,n(x)=a1,ne-γx+e-γx∫0x‍y1,n(τ)eγτdτ,n≥1.
Combining (16) with (12) and (13), we obtain
(18)a0,0=P0,0(0)=λQ=λγ+λyQ,(19)a0,n=P0,n(0)=0,n≥1.
Substituting (19) into (16), it follows that
(20)P0,n(x)=e-γx∫0x‍y0,n(τ)eγτdτ,n≥1.
By combining (15), (16), (17), and (20) with (14), we deduce
(21)a1,n=P1,n(0)a1,n=∫0∞‍υ(x)e-γx∫0x‍y0,n(τ)eγτdτdx+a1,n+1∫0∞‍b(x)e-γxdx+∫0∞‍b(x)e-γx∫0x‍y1,n+1(τ)eγτdτdx,n≥1,(22)⟹a1,n-a1,n+1∫0∞‍b(x)e-γxdx=∫0∞‍υ(x)e-γx∫0x‍y0,n(τ)eγτdτdx+∫0∞‍b(x)e-γx∫0x‍y1,n+1(τ)eγτdτdx.
If we set (23)𝒞=(1-∫0∞‍b(x)e-γxdx00⋯01-∫0∞‍b(x)e-γxdx0⋯001-∫0∞‍b(x)e-γxdx⋯⋮⋮⋮⋮⋱),a→1=(a1,1a1,2a1,3⋮),then (21) can be rewritten as follows:(24)𝒞a→1=(∫0∞‍υ(x)e-γx∫0x‍y0,1(τ)eγτdτdx+∫0∞‍b(x)e-γx∫0x‍y1,2(τ)eγτdτdx∫0∞‍υ(x)e-γx∫0x‍y0,2(τ)eγτdτdx+∫0∞‍b(x)e-γx∫0x‍y1,3(τ)eγτdτdx⋮).It is easy to calculate(25)𝒞-1=(1∫0∞‍b(x)e-γxdx(∫0∞‍b(x)e-γxdx)2(∫0∞‍b(x)e-γxdx)3⋯01∫0∞‍b(x)e-γxdx(∫0∞‍b(x)e-γxdx)2⋯001∫0∞‍b(x)e-γxdx⋯⋮⋮⋮⋮⋱).From which together with (24), we derive
(26)a1,n=∑k=0∞‍(∫0∞‍b(x)e-γxdx)k×{∫0∞‍υ(x)e-γx∫0x‍y0,k+n(τ)eγτdτdx+∫0∞‍b(x)e-γx∫0x‍y1,k+n+1(τ)eγτdτdx},n≥1.
By using Fubini theorem we estimate (16) and (17) as follows (assume that γ>υ+b):
(27)∥P0,n∥L1[0,∞)≤∫0∞‍|a0,n|e-γxdx+∫0∞‍e-γx∫0x‍|y0,n(τ)|eγτdτdx=1γ|a0,n|+∫0∞‍|y0,n(τ)|eγτ∫τ∞‍e-γxdxdτ=1γ|a0,n|+1γ∥y0,n∥L1[0,∞),n≥0,∥P1,n∥L1[0,∞)≤1γ|a1,n|+1γ∥y1,n∥L1[0,∞),n≥1.
From (26) and Fubini theorem, we deduce
(28)∑n=1∞‍|a1,n|≤∑n=1∞‍∑k=0∞‍(bγ)k×{υ∫0∞‍|y0,k+n(τ)|eγτ∫τ∞‍e-γxdxdτ+b∫0∞‍|y1,k+n+1(τ)|eγτ∫τ∞‍e-γxdxdτ}=υγ∑k=0∞‍(bγ)k∑n=1∞‍∥y0,k+n∥L1[0,∞)+∑k=0∞‍(bγ)k+1∑n=1∞‍∥y1,k+n+1∥L1[0,∞)≤υγ∑k=0∞‍(bγ)k∑n=0∞‍∥y0,n∥L1[0,∞)+∑k=0∞‍(bγ)k+1∑n=1∞‍∥y1,n∥L1[0,∞)=υγ-b∑n=0∞‍∥y0,n∥L1[0,∞)+bγ-b∑n=1∞‍∥y1,n∥L1[0,∞).
By inserting (18), (19), and (28) into (27) and using inequality (γ+υ-b)/γ(γ-b)<1/(γ-υ-b), we estimate
(29)∥(P0,P1)∥≤|yQ|γ+λ+1γ∑n=0∞‍|a0,n|+1γ∑n=0∞‍∥y0,n∥L1[0,∞)+1γ∑n=1∞‍|a1,n|+1γ∑n=1∞‍∥y1,n∥L1[0,∞)≤1γ|yQ|+γ+υ-bγ(γ-b)∑n=0∞‍∥y0,n∥L1[0,∞)+1γ-b∑n=1∞‍∥y1,n∥L1[0,∞)≤1γ-υ-b|yQ|+1γ-υ-b∑n=0∞‍∥y0,n∥L1[0,∞)+1γ-υ-b∑n=1∞‍∥y1,n∥L1[0,∞)=1γ-υ-b∥(y0,y1)∥.
Equation (29) shows that (γI-A)-1 exists for γ>υ+b and
(30)(γI-A)-1:X⟶D(A),∥(γI-A)-1∥≤1γ-υ-b.
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 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
(31)𝕃={(P0,P1)|P0(x)=(Q,P0,0(x),P0,1(x),…,P0(x)=P0,N(x),0,…),P1(x)=(P1,1(x),P1,2(x),…,P1(x)=P1,N(x),0,…),Q∈ℝ,P0,i(x),P1,j(x)∈L1[0,∞),i=0,1,…,N;j=1,2,…,N;Nisafinitepositiveinteger},
then 𝕃 is dense in X. If we set
(32)ℤ={(P0,P1)|P0(x)=(Q,P0,0(x),P0,1(x),…,P0(x)=P0,l(x),0,…),P1(x)=(P1,1(x),P1,2(x),…,P1(x)=P1,l(x),0,…),P0,i(x),P1,j(x)∈C0∞[0,∞),andthereexistspositivenumbersc0,i>0,c1,j>0,suchthatP0,i(x)=0,x∈[0,c0,i],P1,j(x)=0,x∈[0,c1,j];i=0,1,2,…,l;j=1,2,…,l},
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
(33)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,forx∈[0,c0,i],i=0,1,2,…,l,P1,j(x)=0,forx∈[0,c1,j],j=1,2,…,l,
which implies
(34)P0,i(x)=0,x∈[0,2s],i=0,1,…,l;P1,j(x)=0,x∈[0,2s],j=1,…,l,
where 0<2s<min{c0,0,c0,1,…,c0,l,c1,1,…,c1,l}. Define
(35)f0s(0)=(Q,f0,0s(0),f0,1s(0),…,f0,ls(0),0,…)=(Q,λQ,0,…,0,0,…),f1s(0)=(f1,1s(0),f1,2s(0),…,f1,ls(0),0,…),
where
(36)f1,js(0)=∫2s∞‍υ(x)P0,j(x)dx+∫2s∞‍b(x)P1,j+1(x)dx,P0,j(x)dx5+∫2s∞‍b(x)P1,j+1(x)dxj=1,2,…,l-1,f1,ls(0)=∫2s∞‍υ(x)P0,l(x)dx,f0s(x)=(Q,f0,0s(x),f0,1s(x),…,f0,ls(x),0,…),f1s(x)=(f1,1s(x),f1,2s(x),…,f1,ls(x),0,…),
where
(37)f0,is(x)={f0,is(0)(1-xs)2,x∈[0,s),-μ0,i(x-s)2(x-2s)2,x∈[s,2s),P0,i(x),x∈[2s,∞),f1,js(x)={f1,js(0)(1-xs)2,x∈[0,s),-μ1,j(x-s)2(x-2s)2,x∈[s,2s),P1,j(x),x∈[2s,∞),μ0,i=∫0s‍υ(x)f0,is(0)(1-(x/s))2dx∫s2s‍υ(x)(x-s)2(x-2s)2dx,μ1,j=∫0s‍b(x)f1,js(0)(1-(x/s))2dx∫s2s‍b(x)(x-s)2(x-2s)2dx,i=0,1,…,l;j=1,…,l.
Then it is not difficult to verify that (f0s,f1s)∈D(A). Moreover,
(38)∥(P0,P1)-(f0s,f1s)∥=∑i=0l‍∫0∞‍|P0,i(x)-f0,is(x)|dx+∑j=1l‍∫0∞‍|P1,j(x)-f1,js(x)|dx=∑i=0l‍∫0s‍|f0,is(0)|(1-xs)2dx+∑i=0l‍|μ0,i|∫s2s‍(x-s)2(x-2s)2dx+∑j=1l‍∫0s‍|f1,js(0)|(1-xs)2dx+∑j=1l‍|μ1,j|∫s2s‍(x-s)2(x-2s)2dx=∑i=0l‍|f0,is(0)|s3+∑i=0l‍|μ0,i|s530+∑j=1l‍|f1,js(0)|s3+∑j=1l‍|μ1,j|s530⟶0ass⟶0,
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 thatU and E are bounded linear operators. From the definition of U and E and ∑k=1∞‍ck=1 we have
(39)∥U(P0,P1)∥≤∑n=0∞‍∫0∞‍[λ+υ(x)]|P0,n(x)|dx+λ∑k=1∞‍ck∑n=0∞‍∫0∞‍|P0,n(x)|dx+∑n=1∞‍∫0∞‍[λ+b(x)]|P1,n(x)|dx+λ∑k=1∞‍ck∑n=1∞‍∫0∞‍|P1,n(x)|dx≤(2λ+υ)∑n=0∞‍∫0∞‍|P0,n(x)|dx+(2λ+b)∑n=1∞‍∫0∞‍|P1,n(x)|dx≤max{(2λ+υ),(2λ+b)}∥(P0,P1)∥,∥E(P0,P1)∥≤∫0∞‍υ(x)|P0,0(x)|dx+∫0∞‍b(x)|P1,1(x)|dx≤υ∥P0,0∥L1[0,∞)+b∥P1,1∥L1[0,∞)≤max{υ,b}∥(P0,P1)∥.
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 thatA+U+E is a dispersive operator. For (P0,P1)∈D(A) we take (ϕ0,ϕ1) as
(40)ϕ0(x)=([Q]+Q,[P0,0(x)]+P0,0(x),[P0,1(x)]+P0,1(x),…),ϕ1(x)=([P1,1(x)]+P1,1(x),[P1,2(x)]+P1,2(x),…),
where
(41)[Q]+={Q,ifQ>0,0,ifQ≤0,[P0,n(x)]+={P0,n(x),ifP0,n(x)>0,0,ifP0,n(x)≤0,n≥0,[P1,n(x)]+={P1,n(x),ifP1,n(x)>0,0,ifP1,n(x)≤0,n≥1.
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
(42)∫0∞‍dP0,i(x)dx[P0,i(x)]+P0,i(x)dx=∫V0,i‍dP0,i(x)dx[P0,i(x)]+P0,i(x)dx+∫W0,i‍dP0,i(x)dx[P0,i(x)]+P0,i(x)dx=∫V0,i‍dP0,i(x)dx[P0,i(x)]+P0,i(x)dx=∫V0,i‍dP0,i(x)dxdx=∫0∞‍d[P0,i(x)]+dxdx=-[P0,i(0)]+,i≥0.
Similar to (42), we get
(43)∫0∞‍dP1,j(x)dx[P1,j(x)]+P1,j(x)dx=-[P1,j(0)]+,j≥1.
By using boundary conditions on (P0,P1)∈D(A), (42), (43), and ∑k=1∞‍ck=1 for such (ϕ0,ϕ1), we derive
(44)〈(A+U+E)(P0,P1),(ϕ0,ϕ1)〉={-λQ+∫0∞‍υ(x)P0,0(x)dx+∫0∞‍b(x)P1,1(x)dx}[Q]+Q+∫0∞‍{-dP0,0(x)dx-[λ+υ(x)]P0,0(x)}[P0,0(x)]+P0,0(x)dx+∑n=1∞‍∫0∞‍{∑k=1n‍-dP0,n(x)dx-[λ+υ(x)]P0,n(x)+λ∑k=1n‍ckP0,n-k(x)}[P0,n(x)]+P0,n(x)dx+∑n=1∞‍∫0∞‍{∑k=1n‍-dP1,n(x)dx-[λ+b(x)]P1,n(x)+λ∑k=1n‍ckP1,n-k+1(x)}[P1,n(x)]+P1,n(x)dx=-λ[Q]++[Q]+Q∫0∞‍υ(x)P0,0(x)dx+[Q]+Q∫0∞‍b(x)P1,1(x)dx+∑n=0∞‍[P0,n(0)]+-∑n=0∞‍∫0∞‍[λ+υ(x)][P0,n(x)]+dx+λ∑k=1∞‍ck∑n=k∞‍∫0∞‍P0,n-k(x)[P0,n(x)]+P0,n(x)dx+∑n=1∞‍[P1,n(0)]+-∑n=1∞‍∫0∞‍[λ+b(x)][P1,n(x)]+dx+λ∑k=1∞‍ck∑n=k∞‍∫0∞‍P1,n-k+1(x)[P1,n(x)]+P1,n(x)dx≤-λ[Q]++[Q]+Q∫0∞‍υ(x)[P0,0(x)]+dx+[Q]+Q∫0∞‍b(x)[P1,1(x)]+dx+λ[Q]+-∑n=0∞‍∫0∞‍[λ+υ(x)][P0,n(x)]+dx+λ∑k=1∞‍ck∑n=k∞‍∫0∞‍[P0,n-k(x)]+dx+∑n=1∞‍∫0∞‍υ(x)[P0,n(x)]+dx+∑n=2∞‍∫0∞‍b(x)[P1,n(x)]+dx-∑n=1∞‍∫0∞‍[λ+b(x)][P1,n(x)]+dx+λ∑k=1∞‍ck∑n=k∞‍∫0∞‍[P1,n-k+1(x)]+dx≤([Q]+Q-1)×(∫0∞‍υ(x)[P0,0(x)]+dx+∫0∞‍b(x)[P1,1(x)]+dx)-∑n=0∞‍∫0∞‍λ[P0,n(x)]+dx+λ∑k=1∞‍ck∑n=0∞‍∫0∞‍[P0,n(x)]+dx-∑n=1∞‍∫0∞‍λ[P1,n(x)]+dx+λ∑k=1∞‍ck∑n=1∞‍∫0∞‍[P1,n(x)]+dx≤0.
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).It is not difficult to see thatX*, the dual space of X, is (45)X*={(q0*,q1*)|q0*=(q*,q0,0*,q0,1*,…)q0*∈ℝ×L∞[0,∞)×L∞[0,∞)×⋯,q1*=(q1,1*,q1,2*,…)q0*∈L∞[0,∞)×L∞[0,∞)×⋯,∥|(q0*,q1*)|∥=sup{|q*|,supn≥0∥q0,n*∥L∞[0,∞),∥(q0*,q1*)∥=supn≥1∥q1,n*∥L∞[0,∞)}}.It is easy to check thatX* is a Banach space. If we take a set S in X as
(46)S={(P0,P1)∈X∣Q≥0,P0,n(x)≥0,n≥0;P1,n(x)≥0,n≥1,x∈[0,∞)},
then S is a cone in X. For (P0,P1)∈D(A)∩S, we take
(47)(q0*,q1*)=∥(P0,P1)∥((11⋮),(11⋮))∈X*,
then we have
(48)〈(P0,P1),(q0*,q1*)〉=∥(P0,P1)∥(|Q|+∑n=0∞‍P0,n(x)dx+∑n=1∞‍P1,n(x)dx)=∥(P0,P1)∥2=∥|(q0*,q1*)|∥2,
that is
(49)(q0*,q1*)∈θ((P0,P1))={∥(P0,P1)∥2(q0*,q1*)∈X*∣〈(P0,P1),(q0*,q1*)〉=∥(P0,P1)∥2=∥|(q0*,q1*)|∥2}.
For such (q0*,q1*), by using boundary conditions on (P0,P1)∈D(A)∩S and ∑k=1∞‍ck=1, we have
(50)〈(A+U+E)(P0,P1),(q0*,q1*)〉=∥(P0,P1)∥×{-λQ+∫0∞‍υ(x)P0,0(x)dx+∫0∞‍b(x)P1,1(x)dx}+∫0∞‍{-dP0,0(x)dx-[λ+υ(x)]P0,0(x)}∥(P0,P1)∥dx+∑n=1∞‍∫0∞‍{∑n=1∞‍-dP0,n(x)dx-[λ+υ(x)]P0,n(x)+λ∑k=1n‍ckP0,n-k(x)}∥(P0,P1)∥dx+∑n=1∞‍∫0∞‍{∑n=1∞‍-dP1,n(x)dx-[λ+b(x)]P1,n(x)+λ∑k=1n‍ckP1,n-k+1(x)}∥(P0,P1)∥dx=∥(P0,P1)∥×{-λQ+∫0∞‍υ(x)P0,0(x)dx+∫0∞‍b(x)P1,1(x)dx}-∥(P0,P1)∥×{∑n=0∞‍∫0∞‍-dP0,n(x)dxdx+∑n=0∞‍∫0∞‍[λ+υ(x)]P0,n(x)dx-λ∑k=1∞‍ck∑n=0∞‍∫0∞‍P0,n(x)dx+∑n=1∞‍∫0∞‍dP1,n(x)dxdx+∑n=1∞‍∫0∞‍[λ+b(x)]P1,n(x)dx-λ∑k=1∞‍ck∑n=1∞‍∫0∞‍P1,n(x)dx}=∥(P0,P1)∥×{-λQ+∫0∞‍υ(x)P0,0(x)dx+∫0∞‍b(x)P1,1(x)dx}-∥(P0,P1)∥×{-λQ+∑n=0∞‍∫0∞‍υ(x)P0,n(x)dx-∑n=1∞‍∫0∞‍υ(x)P0,n(x)dx-∑n=2∞‍∫0∞‍b(x)P1,n(x)dx+∑n=1∞‍∫0∞‍b(x)P1,n(x)dx}=0.
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,
(51)∥T(t)(P0,P1)(0)∥=∥(P0,P1)(0)∥,∀t∈[0,∞).Proof.
SinceA+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
(52)T(t+h)(P0,P1)(0)=T(t)(P0,P1)(0)+h(A+U+E)T(t)(P0,P1)(0)+∫tt+h‍(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),
where ∥ρ(t,h)∥→0 as h→0, uniformly in t≥0. Then
(53)∥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))*〉=∥T(t)(P0,P1)(0)∥2+hRe〈ρ(t,h),(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
(54)Ω={t∈[0,∞)∣∥T(t)(P0,P1)(0)∥=∥(P0,P1)(0)∥}.
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
(55)∥T(t+h)(P0,P1)(0)∥≥∥T(t)(P0,P1)(0)∥+h∥T(t)(P0,P1)(0)∥×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 nowϵ 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(56)0≤∥T(t0)(P0,P1)(0)∥-∥T(t0+η)(P0,P1)(0)∥≤∑j=1m(∥T(tj-1)(P0,P1)(0)∥-∥T(tj)(P0,P1)(0)∥)≤∑j=1m‍(tj-tj-1)β(tj,tj-tj-1)≤ηϵ.
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.From Theorems1 and 2 we obtain the main result in this paper.Theorem 3.
Ifυ=supx∈[0,∞)υ(x)<∞, b=supx∈[0,∞)b(x)<∞, then the system (8) has a unique nonnegative time-dependent solution (P0,P1)(x,t), which satisfies
(57)∥(P0,P1)(·,t)∥=1,∀t∈[0,∞).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
(58)(P0,P1)(x,t)=T(t)(P0,P1)(0),∀t∈[0,∞).
From which together with Theorem 2 (i.e., (51)) we have
(59)∥(P0,P1)(·,t)∥=∥T(t)(P0,P1)(0)∥=∥(P0,P1)(0)∥=1,=∥T(t)(P0,P1)(0)∥=∥(P0,P1)(0)∥=1,5∀t∈[0,∞),
this just reflects the physical background of the problem.


## 4. Concluding Remarks

If we know the spectrum ofA+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.



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*Source: 893254-2013-05-13.xml*