Semigroup Method on a M^X/G/1 Queueing Model

Abstract

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

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 M^X/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 M^X/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 Hypothesis 1. 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 M^X/G/1 queueing system with vacation times can be described by the following system of equations: 𝑑𝑄(𝑡)(𝑡)+𝑑𝑡=−𝜆𝑄^∞₀𝜐(𝑥)𝑃_0,0()+𝑥,𝑡𝑑𝑥_∞0𝑏(𝑥)𝑃_1,1(𝑥,𝑡)𝑑𝑥,𝜕𝑃_0,0(𝑥,𝑡)+𝜕𝑡𝜕𝑃_0,0(𝑥,𝑡)[(𝑥𝑃𝜕𝑥=−𝜆+𝜐)]_0,0(𝑥,𝑡),𝜕𝑃_0,𝑛(𝑥,𝑡)+𝜕𝑡𝜕𝑃_0,𝑛(𝑥,𝑡)[(𝑥𝑃𝜕𝑥=−𝜆+𝜐)]_0,𝑛(𝑥,𝑡)+𝜆^𝑛_𝑘=1𝑐_𝑘𝑃_{0,𝑛−𝑘}(𝑥,𝑡),𝑛≥1,𝜕𝑃_1,𝑛(𝑥,𝑡)+𝜕𝑡𝜕𝑃_1,𝑛(𝑥,𝑡)[(𝑥𝑃𝜕𝑥=−𝜆+𝑏)]_1,𝑛(𝑥,𝑡)+𝜆^𝑛_𝑘=1𝑐_𝑘𝑃_{1,𝑛−𝑘+1}()𝑃𝑥,𝑡,𝑛≥1,_0,0(𝑡)=𝜆𝑄(𝑡),𝑃_0,𝑛()𝑃0,𝑡=0,𝑛≥1,_1,𝑛(0,𝑡)=^∞₀𝜐(𝑥)𝑃_0,𝑛()+𝑥,𝑡𝑑𝑥_∞0𝑏(𝑥)𝑃_1,𝑛+1()𝑄(0)𝑥,𝑡𝑑𝑥,𝑛≥1,=1,𝑃_0,𝑛()𝑃𝑥,0=0,𝑛≥0,_1,𝑛(𝑥,0)=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; 𝑃_0,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). 𝑃_1,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. 𝑐_𝑘(𝑘≥1) represents the probability that at every arrival epoch a batch of 𝑘 external customers arrives and satisfies ∑^∞_𝑘=1𝑐_𝑘=1. 𝜐(𝑥) is the vacation rate of the server, which satisfies 𝜐(𝑥)≥0,^∞₀𝜐(𝑥)𝑑𝑥=∞. (2)𝑏(𝑥) is the service rate of the server satisfying 𝑏(𝑥)≥0,^∞₀𝑏(𝑥)𝑑𝑥=∞. (3)

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: Γ₁=𝑒^−𝑥00⋯𝜆𝑒^−𝑥00⋯000⋯⋮⋮⋮⋱,Γ₂=(𝑥)(𝑥)⋯,Γ00𝜐0⋯000𝜐0000⋯⋮⋮⋮⋮⋱₃=(𝑥)(𝑥)⋯.0𝑏0⋯00𝑏000⋯⋮⋮⋮⋱ (4) If we take state space 𝑋=𝑃₀,𝑃₁𝑃₀=𝑄,𝑃_0,0,𝑃_0,1,…∈ℝ×𝐿¹[0,∞)×𝐿¹[)𝑃0,∞×⋯₁=𝑃_1,1,𝑃_1,2,𝑃_1,3,…∈𝐿¹[0,∞)×𝐿¹[0,∞)×𝐿¹[)𝑃0,∞×⋯₀,𝑃₁=|𝑄|+∑^∞_𝑖=0𝑃_{0,𝑖𝐿1[0,∞)}+∑^∞_𝑖=1𝑃_{1,𝑖𝐿1[0,∞)},<∞ (5) then it is obvious that 𝑋 is a Banach space. In the following we define operators and their domains; 𝐴𝑃₀,𝑃₁=𝑑−𝜆00⋯0−𝑑𝑑𝑥0⋯00−⋯𝑄𝑃𝑑𝑥⋮⋮⋮⋱^0,0(𝑥)𝑃_0,1(𝑥)⋮,−𝑑𝑑𝑑𝑥00⋯0−𝑑𝑑𝑥0⋯00−⋯𝑃𝑑𝑥⋮⋮⋮⋱^1,1(𝑥)𝑃_1,2(𝑥)𝑃_1,3(𝑥)⋮,𝐷(𝐴)=𝑃₀,𝑃₁∈𝑋𝑑𝑃_0,𝑛(𝑥),𝑑𝑥𝑑𝑃_1,𝑘(𝑥)𝑑𝑥∈𝐿¹[),𝑃0,∞_0,𝑛(𝑥),𝑃_1,𝑘(𝑥)(𝑛≥0,𝑘≥1)areabsolutelycontinuousfunctionsandsatisfy𝑃₀(0)=∫^∞₀Γ₁𝑃₀(𝑥)𝑃𝑑𝑥,₁(0)=∫^∞₀Γ₂𝑃₀(𝑥)𝑑𝑥+∫^∞₀Γ₃𝑃₁(𝑥),𝑈𝑃𝑑𝑥₀,𝑃₁=00000⋯_∼𝜐00⋯0𝜆𝑐𝜐₁^∼0⋯0𝜆𝑐₂𝜆𝑐₁^∼𝑄𝑃𝜐⋯⋮⋮⋮⋮⋱^0,0(𝑥)𝑃_0,1(𝑥)𝑃_0,2(𝑥)⋮_∼,𝑏000⋯𝜆𝑐𝑏₂^∼00⋯𝜆𝑐₃𝜆𝑐𝑏₂^∼0⋯𝜆𝑐₄𝜆𝑐₃𝜆𝑐𝑏₂^∼⋯𝑃⋮⋮⋮⋮⋱^1,1(𝑥)𝑃_1,2(𝑥)𝑃_1,3(𝑥)𝑃_1,4(𝑥)⋮, (6) where 𝜐^∼=−[𝜆+𝜐(𝑥)], 𝑏^∼=−[𝜆+𝑏(𝑥)]+𝜆𝑐₁, 𝐸𝑃₀,𝑃₁=∫^∞₀𝜐(𝑥)𝑃_0,0(𝑥)𝑑𝑥+∫^∞₀𝑏(𝑥)𝑃_1,1(𝑥)0⋮,00⋮,𝐷(𝑈)(𝐸)𝑑𝑥=𝐷=𝑋. (7) Then the previous system of equations (1) can be rewritten as an abstract Cauchy problem in the Banach space 𝑋: 𝑑𝑃₀,𝑃₁(𝑡)=()𝑃𝑑𝑡𝐴+𝑈+𝐸₀,𝑃₁(𝑡)[),𝑃,∀𝑡∈0,∞₀,𝑃₁(0)=10⋮,00⋮. (8)

3. Well-Posedness of The System (8)

Theorem 1. If 𝜐=sup_{𝑥∈[0,∞)}𝜐(𝑥)<∞, 𝑏=sup_{𝑥∈[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+𝜆𝑄=𝑦_𝑄,𝑑𝑃_0,𝑛(𝑥)𝑑𝑥=−𝛾𝑃_0,𝑛(𝑥)+𝑦_0,𝑛(𝑥),𝑛≥0,𝑑𝑃_1,𝑛(𝑥)𝑑𝑥=−𝛾𝑃_1,𝑛(𝑥)+𝑦_1,𝑛(𝑥),𝑛≥1,𝑃_0,0(0)=𝜆𝑄,𝑃_0,𝑛(0)=0,𝑛≥1,𝑃_1,𝑛(0)=^∞₀𝜐(𝑥)𝑃_0,𝑛(𝑥)+𝑑𝑥_∞0𝑏(𝑥)𝑃_1,𝑛+1(𝑥)𝑑𝑥,𝑛≥1. (9)(10)(11)(12)(13)(14) Through solving (9)–(11), we have 1𝑄=𝑦𝛾+𝜆^𝑄,𝑃_0,𝑛(𝑥)=𝑎_0,𝑛𝑒^−𝛾𝑥+𝑒^{−𝛾𝑥𝑥}₀𝑦_0,𝑛(𝜏)𝑒^𝛾𝜏𝑑𝜏,𝑛≥0,𝑃_1,𝑛(𝑥)=𝑎_1,𝑛𝑒^−𝛾𝑥+𝑒^{−𝛾𝑥𝑥}₀𝑦_1,𝑛(𝜏)𝑒^𝛾𝜏𝑑𝜏,𝑛≥1. (15)(16)(17) Combining (16) with (12) and (13), we obtain 𝑎_0,0=𝑃_0,0(0)𝜆=𝜆𝑄=𝑦𝛾+𝜆^𝑄,𝑎_0,𝑛=𝑃_0,𝑛(0)=0,𝑛≥1. (18)(19) Substituting (19) into (16), it follows that 𝑃_0,𝑛(𝑥)=𝑒^{−𝛾𝑥𝑥}₀𝑦_0,𝑛(𝜏)𝑒^𝛾𝜏𝑑𝜏,𝑛≥1. (20) By combining (15), (16), (17), and (20) with (14), we deduce 𝑎_1,𝑛=𝑃_1,𝑛(0)=^∞₀𝜐(𝑥)𝑒^{−𝛾𝑥𝑥}₀𝑦_0,𝑛(𝜏)𝑒^𝛾𝜏𝑑𝜏𝑑𝑥+𝑎_1,𝑛+1^∞₀𝑏(𝑥)𝑒^−𝛾𝑥+𝑑𝑥_∞0𝑏(𝑥)𝑒^{−𝛾𝑥𝑥}₀𝑦_1,𝑛+1(𝜏)𝑒^𝛾𝜏𝑑𝜏𝑑𝑥,𝑛≥1,⟹𝑎_1,𝑛−𝑎_1,𝑛+1^∞₀𝑏(𝑥)𝑒^−𝛾𝑥=𝑑𝑥_∞0𝜐(𝑥)𝑒^{−𝛾𝑥𝑥}₀𝑦_0,𝑛(𝜏)𝑒^𝛾𝜏+𝑑𝜏𝑑𝑥_∞0𝑏(𝑥)𝑒^{−𝛾𝑥𝑥}₀𝑦_1,𝑛+1(𝜏)𝑒^𝛾𝜏𝑑𝜏𝑑𝑥. (21)(22) If we set 𝒞=1−∫^∞₀𝑏(𝑥)𝑒^−𝛾𝑥𝑑𝑥00⋯01−∫^∞₀𝑏(𝑥)𝑒^−𝛾𝑥𝑑𝑥0⋯001−∫^∞₀𝑏(𝑥)𝑒^−𝛾𝑥,𝑎𝑑𝑥⋯⋮⋮⋮⋮⋱₁=𝑎_1,1𝑎_1,2𝑎_1,3⋮, (23)then (21) can be rewritten as follows:𝒞𝑎₁=∫^∞₀𝜐(𝑥)𝑒^−𝛾𝑥∫^𝑥₀𝑦_0,1(𝜏)𝑒^𝛾𝜏𝑑𝜏𝑑𝑥+∫^∞₀𝑏(𝑥)𝑒^−𝛾𝑥∫^𝑥₀𝑦_1,2(𝜏)𝑒^𝛾𝜏∫𝑑𝜏𝑑𝑥_∞0𝜐(𝑥)𝑒^−𝛾𝑥∫^𝑥₀𝑦_0,2(𝜏)𝑒^𝛾𝜏𝑑𝜏𝑑𝑥+∫^∞₀𝑏(𝑥)𝑒^−𝛾𝑥∫^𝑥₀𝑦_1,3(𝜏)𝑒^𝛾𝜏⋮𝑑𝜏𝑑𝑥. (24)It is easy to calculate𝒞⁻¹=1∫^∞₀𝑏(𝑥)𝑒^−𝛾𝑥𝑑𝑥∫^∞₀𝑏(𝑥)𝑒^−𝛾𝑥𝑑𝑥²∫^∞₀𝑏(𝑥)𝑒^−𝛾𝑥𝑑𝑥³⋯01∫^∞₀𝑏(𝑥)𝑒^−𝛾𝑥𝑑𝑥∫^∞₀𝑏(𝑥)𝑒^−𝛾𝑥𝑑𝑥²⋯001∫^∞₀𝑏(𝑥)𝑒^−𝛾𝑥.𝑑𝑥⋯⋮⋮⋮⋮⋱ (25)From which together with (24), we derive 𝑎_1,𝑛=^∞_𝑘=0^∞₀𝑏(𝑥)𝑒^−𝛾𝑥𝑑𝑥^𝑘×^∞₀𝜐(𝑥)𝑒^{−𝛾𝑥𝑥}₀𝑦_0,𝑘+𝑛(𝜏)𝑒^𝛾𝜏+𝑑𝜏𝑑𝑥_∞0𝑏(𝑥)𝑒^{−𝛾𝑥𝑥}₀𝑦_{1,𝑘+𝑛+1}(𝜏)𝑒^𝛾𝜏𝑑𝜏𝑑𝑥,𝑛≥1. (26) By using Fubini theorem we estimate (16) and (17) as follows (assume that 𝛾>𝜐+𝑏): 𝑃_{0,𝑛𝐿1[0,∞)}≤^∞₀𝑎_0,𝑛𝑒^−𝛾𝑥𝑑𝑥+^∞₀𝑒^{−𝛾𝑥𝑥}₀𝑦_0,𝑛(𝜏)𝑒^𝛾𝜏=1𝑑𝜏𝑑𝑥𝛾𝑎_0,𝑛+^∞₀𝑦_0,𝑛(𝜏)𝑒^𝛾𝜏∞_𝜏𝑒^−𝛾𝑥=1𝑑𝑥𝑑𝜏𝛾𝑎_0,𝑛+1𝛾𝑦_{0,𝑛𝐿1[0,∞)}𝑃,𝑛≥0,_{1,𝑛𝐿1[0,∞)}≤1𝛾𝑎_1,𝑛+1𝛾𝑦_{1,𝑛𝐿1[0,∞)},𝑛≥1. (27) From (26) and Fubini theorem, we deduce ^∞_𝑛=1𝑎_1,𝑛≤^∞∞_𝑛=1𝑘=0𝑏𝛾^𝑘×𝜐^∞₀𝑦_0,𝑘+𝑛(𝜏)𝑒^𝛾𝜏∞_𝜏𝑒^−𝛾𝑥𝑑𝑥𝑑𝜏+𝑏^∞₀𝑦_{1,𝑘+𝑛+1}(𝜏)𝑒^𝛾𝜏∞_𝜏𝑒^−𝛾𝑥=𝜐𝑑𝑥𝑑𝜏𝛾^∞_𝑘=0𝑏𝛾^𝑘∞_𝑛=1𝑦_{0,𝑘+𝑛𝐿1[0,∞)}+^∞_𝑘=0𝑏𝛾^∞𝑘+1_𝑛=1𝑦_{1,𝑘+𝑛+1𝐿1[0,∞)}≤𝜐𝛾^∞_𝑘=0𝑏𝛾^𝑘∞_𝑛=0𝑦_{0,𝑛𝐿1[0,∞)}+^∞_𝑘=0𝑏𝛾^∞𝑘+1_𝑛=1𝑦_{1,𝑛𝐿1[0,∞)}=𝜐𝛾−𝑏^∞_𝑛=0𝑦_{0,𝑛𝐿1[0,∞)}+𝑏𝛾−𝑏^∞_𝑛=1𝑦_{1,𝑛𝐿1[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 𝑃₀,𝑃₁≤𝑦_𝑄+1𝛾+𝜆𝛾^∞_𝑛=0𝑎_0,𝑛+1𝛾^∞_𝑛=0𝑦_{0,𝑛𝐿1[0,∞)}+1𝛾^∞_𝑛=1𝑎_1,𝑛+1𝛾^∞_𝑛=1𝑦_{1,𝑛𝐿1[0,∞)}≤1𝛾𝑦_𝑄+𝛾+𝜐−𝑏𝛾𝛾−𝑏^∞_𝑛=0𝑦_{0,𝑛𝐿1[0,∞)}+1𝛾−𝑏^∞_𝑛=1𝑦_{1,𝑛𝐿1[0,∞)}≤1𝑦𝛾−𝜐−𝑏^𝑄+1𝛾−𝜐−𝑏^∞_𝑛=0𝑦_{0,𝑛𝐿1[0,∞)}+1𝛾−𝜐−𝑏^∞_𝑛=1𝑦_{1,𝑛𝐿1[0,∞)}=1𝑦𝛾−𝜐−𝑏⁰,𝑦₁. (29) Equation (29) shows that (𝛾𝐼−𝐴)⁻¹ exists for 𝛾>𝜐+𝑏 and 𝛾𝐼−𝐴⁻¹∶𝑋⟶𝐷(𝐴),𝛾𝐼−𝐴⁻¹≤1.𝛾−𝜐−𝑏 (30) As far as the second step is concerned, from |𝑄|+∑^∞_𝑛=0‖𝑃_0,𝑛‖_𝐿1[0,∞)+∑^∞_𝑛=1‖𝑃_1,𝑛‖_𝐿1[0,∞)<∞ for (𝑃₀,𝑃₁)∈𝑋 it follows that, for any 𝜖>0, there exists a positive integer 𝑁 such that ∑^∞_𝑛=𝑁‖𝑃_0,𝑛‖_𝐿1[0,∞)<𝜖, ∑^∞_𝑛=𝑁‖𝑃_1,𝑛‖_𝐿1[0,∞)<∞. Let 𝕃=𝑃₀,𝑃₁𝑃₀(𝑥)=𝑄,𝑃_0,0(𝑥),𝑃_0,1(𝑥)𝑃,…,_0,𝑁(𝑥),𝑃,0,…₁(𝑥)=𝑃_1,1(𝑥),𝑃_1,2(𝑥)𝑃,…,_1,𝑁(𝑥),0,…,𝑄∈ℝ,𝑃_0,𝑖(𝑥),𝑃_1,𝑗(𝑥)∈𝐿¹[)𝑁0,∞,𝑖=0,1,…,𝑁;𝑗=1,2,…,𝑁;isafinitepositiveinteger, (31) then 𝕃 is dense in 𝑋. If we set ℤ=𝑃₀,𝑃₁𝑃₀(𝑥)=𝑄,𝑃_0,0(𝑥),𝑃_0,1(𝑥)𝑃,…,_0,𝑙(𝑥),𝑃,0,…₁(𝑥)=𝑃_1,1(𝑥),𝑃_1,2(𝑥)𝑃,…,_1,𝑙(𝑥),𝑃,0,…_0,𝑖(𝑥),𝑃_1,𝑗(𝑥)∈𝐶^∞₀[0,∞),andthereexistspositivenumbers𝑐_0,𝑖>0,𝑐_1,𝑗>0,suchthat𝑃_0,𝑖(𝑥)=0,𝑥∈0,𝑐_0,𝑖,𝑃_1,𝑗(𝑥)=0,𝑥∈0,𝑐_1,𝑗;,𝑖=0,1,2,…,𝑙;𝑗=1,2,…,𝑙 (32) then by the relationship 𝐶^∞₀[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 𝑐_0,𝑖>0, 𝑐_1,𝑗>0(𝑖=0,1,…,𝑙; 𝑗=1,2,…,𝑙) such that 𝑃₀(𝑥)=𝑄,𝑃_0,0(𝑥),𝑃_0,1(𝑥),…,𝑃_0,𝑙(𝑥),𝑃,0,…₁(𝑥)=𝑃_1,1(𝑥),𝑃_1,2(𝑥),…,𝑃_1,𝑙(𝑥),𝑃,0,…_0,𝑖(𝑥)=0,for𝑥∈0,𝑐_0,𝑖𝑃,𝑖=0,1,2,…,𝑙,_1,𝑗(𝑥)=0,for𝑥∈0,𝑐_1,𝑗,𝑗=1,2,…,𝑙, (33) which implies 𝑃_0,𝑖(𝑥)[]𝑃=0,𝑥∈0,2𝑠,𝑖=0,1,…,𝑙;_1,𝑗(𝑥)=0,𝑥∈[0,2𝑠],𝑗=1,…,𝑙, (34) where 0<2𝑠<min{𝑐_0,0,𝑐_0,1,…,𝑐_0,𝑙,𝑐_1,1,…,𝑐_1,𝑙}. Define 𝑓^𝑠₀(0)=𝑄,𝑓^𝑠_0,0(0),𝑓^𝑠_0,1(0),…,𝑓^𝑠_0,𝑙(0)=(),𝑓,0,…𝑄,𝜆𝑄,0,…,0,0,…_𝑠1(0)=𝑓^𝑠_1,1(0),𝑓^𝑠_1,2(0),…,𝑓^𝑠_1,𝑙(0),0,…, (35) where 𝑓^𝑠_1,𝑗(0)=^∞_2𝑠𝜐(𝑥)𝑃_0,𝑗(𝑥)𝑑𝑥+^∞_2𝑠𝑏(𝑥)𝑃_1,𝑗+1(𝑥)𝑓𝑑𝑥,𝑗=1,2,…,𝑙−1,_𝑠1,𝑙(0)=^∞_2𝑠𝜐(𝑥)𝑃_0,𝑙(𝑥)𝑓𝑑𝑥,_𝑠0(𝑥)=𝑄,𝑓^𝑠_0,0(𝑥),𝑓^𝑠_0,1(𝑥),…,𝑓^𝑠_0,𝑙(𝑥),𝑓,0,…_𝑠1(𝑥)=𝑓^𝑠_1,1(𝑥),𝑓^𝑠_1,2(𝑥),…,𝑓^𝑠_1,𝑙(𝑥),0,…, (36) where 𝑓^𝑠_0,𝑖(𝑥)=𝑓^𝑠_0,𝑖(0)𝑥1−𝑠²,𝑥∈[0,𝑠),−𝜇_0,𝑖(𝑥−𝑠)²(𝑥−2𝑠)²[),𝑃,𝑥∈𝑠,2𝑠_0,𝑖(𝑥)[),𝑓,𝑥∈2𝑠,∞_𝑠1,𝑗(𝑥)=𝑓^𝑠_1,𝑗(0)𝑥1−𝑠²,𝑥∈[0,𝑠),−𝜇_1,𝑗(𝑥−𝑠)²(𝑥−2𝑠)²[),𝑃,𝑥∈𝑠,2𝑠_1,𝑗(𝑥)[),𝜇,𝑥∈2𝑠,∞_0,𝑖=∫^𝑠₀𝜐(𝑥)𝑓^𝑠_0,𝑖(0)(1−(𝑥/𝑠))²𝑑𝑥∫_𝑠^2𝑠𝜐(𝑥)(𝑥−𝑠)²(𝑥−2𝑠)²,𝜇𝑑𝑥_1,𝑗=∫^𝑠₀𝑏(𝑥)𝑓^𝑠_1,𝑗(0)(1−(𝑥/𝑠))²𝑑𝑥∫_𝑠^2𝑠𝑏(𝑥)(𝑥−𝑠)²(𝑥−2𝑠)²,𝑑𝑥𝑖=0,1,…,𝑙;𝑗=1,…,𝑙. (37) Then it is not difficult to verify that (𝑓^𝑠₀,𝑓^𝑠₁)∈𝐷(𝐴). Moreover, 𝑃₀,𝑃₁−𝑓^𝑠₀,𝑓^𝑠₁=^𝑙_𝑖=0^∞₀𝑃_0,𝑖(𝑥)−𝑓^𝑠_0,𝑖(𝑥+)𝑑𝑥_{𝑙𝑗=1∞0}𝑃_1,𝑗(𝑥)−𝑓^𝑠_1,𝑗(𝑥=)𝑑𝑥_{𝑙𝑖=0𝑠0}𝑓^𝑠_0,𝑖(0𝑥)1−𝑠²+𝑑𝑥_𝑙𝑖=0𝜇_0,𝑖𝑠^2𝑠(𝑥−𝑠)²(𝑥−2𝑠)²+𝑑𝑥_{𝑙𝑗=1𝑠0}𝑓^𝑠_1,𝑗(0𝑥)1−𝑠²+𝑑𝑥_𝑙𝑗=1𝜇_1,𝑗𝑠^2𝑠(𝑥−𝑠)²(𝑥−2𝑠)²=𝑑𝑥_𝑙𝑖=0𝑓^𝑠_0,𝑖(0𝑠)3+^𝑙_𝑖=0𝜇_0,𝑖𝑠⁵+30^𝑙_𝑗=1𝑓^𝑠_1,𝑗(0𝑠)3+^𝑙_𝑗=1𝜇_1,𝑗𝑠⁵30⟶0as𝑠⟶0, (38) 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=1𝑐_𝑘=1 we have 𝑈𝑃₀,𝑃₁≤^∞_𝑛=0^∞₀[𝜆+𝜐(𝑥)]𝑃_0,𝑛(𝑥)𝑑𝑥+𝜆^∞_𝑘=1𝑐_𝑘^∞_𝑛=0^∞₀𝑃_0,𝑛(𝑥+)𝑑𝑥_{∞𝑛=1∞0}[𝜆+𝑏(𝑥)]𝑃_1,𝑛(𝑥)𝑑𝑥+𝜆^∞_𝑘=1𝑐_𝑘^∞_𝑛=1^∞₀𝑃_1,𝑛(𝑥≤())𝑑𝑥2𝜆+𝜐^∞_𝑛=0^∞₀𝑃_0,𝑛(𝑥+())𝑑𝑥2𝜆+𝑏^∞_𝑛=1^∞₀𝑃_1,𝑛(𝑥),(𝑃)𝑑𝑥≤max{(2𝜆+𝜐2𝜆+𝑏)}₀,𝑃₁,𝐸𝑃₀,𝑃₁≤^∞₀𝜐(𝑥)𝑃_0,0(𝑥+)𝑑𝑥_∞0𝑏(𝑥)𝑃_1,1(𝑥𝑃)𝑑𝑥≤𝜐_{0,0𝐿1[0,∞)}+𝑏𝑃_{1,1𝐿1[0,∞)}≤max{𝜐,𝑏}𝑃₀,𝑃₁. (39) 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]⁺𝑄,𝑃_0,0(𝑥)⁺𝑃_0,0(𝑥),𝑃_0,1(𝑥)⁺𝑃_0,1(𝑥),𝜙,…₁(𝑥)=𝑃_1,1(𝑥)⁺𝑃_1,1(𝑥),𝑃_1,2(𝑥)⁺𝑃_1,2(𝑥),…, (40) where [𝑄]⁺=𝑄,if𝑄>0,0,if𝑃𝑄≤0,_0,𝑛(𝑥)⁺=𝑃_0,𝑛(𝑥),if𝑃_0,𝑛(𝑥)>0,0,if𝑃_0,𝑛(𝑥)𝑃≤0,𝑛≥0,_1,𝑛(𝑥)⁺=𝑃_1,𝑛(𝑥),if𝑃_1,𝑛(𝑥)>0,0,if𝑃_1,𝑛(𝑥)≤0,𝑛≥1. (41) If we define 𝑉_0,𝑖={𝑥∈[0,∞)∣𝑃_0,𝑖(𝑥)>0} and 𝑊_0,𝑖={𝑥∈[0,∞)∣𝑃_0,𝑖(𝑥)≤0} for 𝑖=0,1,2,…, then by a short argument we calculate ^∞₀𝑑𝑃_0,𝑖(𝑥)𝑃𝑑𝑥^0,𝑖(𝑥)⁺𝑃_0,𝑖(𝑥)=𝑑𝑥_𝑉0,𝑖𝑑𝑃_0,𝑖(𝑥)𝑃𝑑𝑥^0,𝑖(𝑥)⁺𝑃_0,𝑖(𝑥)+𝑑𝑥_𝑊0,𝑖𝑑𝑃_0,𝑖(𝑥)𝑃𝑑𝑥^0,𝑖(𝑥)⁺𝑃_0,𝑖(𝑥)𝑑𝑥=_𝑉0,𝑖𝑑𝑃_0,𝑖(𝑥)𝑃𝑑𝑥^0,𝑖(𝑥)⁺𝑃_0,𝑖(𝑥)𝑑𝑥=_𝑉0,𝑖𝑑𝑃_0,𝑖(𝑥)𝑑𝑥𝑑𝑥=^∞₀𝑑𝑃_0,𝑖(𝑥)⁺𝑃𝑑𝑥𝑑𝑥=−_0,𝑖(0)⁺,𝑖≥0. (42) Similar to (42), we get ^∞₀𝑑𝑃_1,𝑗(𝑥)𝑃𝑑𝑥^1,𝑗(𝑥)⁺𝑃_1,𝑗(𝑥)𝑑𝑥=−𝑃_1,𝑗(0)⁺,𝑗≥1. (43) By using boundary conditions on (𝑃₀,𝑃₁)∈𝐷(𝐴), (42), (43), and ∑^∞_𝑘=1𝑐_𝑘=1 for such (𝜙₀,𝜙₁), we derive (𝐴+𝑈+𝐸)𝑃₀,𝑃₁,𝜙₀,𝜙₁=−𝜆𝑄+^∞₀𝜐(𝑥)𝑃_0,0(𝑥)+𝑑𝑥_∞0𝑏(𝑥)𝑃_1,1(𝑥)[𝑄]𝑑𝑥⁺𝑄+^∞₀−𝑑𝑃_0,0(𝑥)−[(𝑥𝑃𝑑𝑥𝜆+𝜐)]_0,0(𝑥𝑃)^0,0(𝑥)⁺𝑃_0,0(𝑥)+𝑑𝑥_{∞𝑛=1∞0}−𝑑𝑃_0,𝑛(𝑥)−[(𝑥𝑃𝑑𝑥𝜆+𝜐)]_0,𝑛(𝑥)+𝜆^𝑛_𝑘=1𝑐_𝑘𝑃_{0,𝑛−𝑘}(𝑥𝑃)^0,𝑛(𝑥)⁺𝑃_0,𝑛(𝑥)+𝑑𝑥_{∞𝑛=1∞0}−𝑑𝑃_1,𝑛(𝑥)−[(𝑥𝑃𝑑𝑥𝜆+𝑏)]_1,𝑛(𝑥)+𝜆^𝑛_𝑘=1𝑐_𝑘𝑃_{1,𝑛−𝑘+1}(𝑥𝑃)^1,𝑛(𝑥)⁺𝑃_1,𝑛(𝑥)[𝑄]𝑑𝑥=−𝜆⁺+[𝑄]⁺𝑄^∞₀𝜐(𝑥)𝑃_0,0(𝑥)+[𝑄]𝑑𝑥₊𝑄^∞₀𝑏(𝑥)𝑃_1,1(𝑥)+𝑑𝑥_∞𝑛=0𝑃_0,𝑛(0)⁺−^∞_𝑛=0^∞₀[𝜆+𝜐(𝑥)]𝑃_0,𝑛(𝑥)⁺𝑑𝑥+𝜆^∞_𝑘=1𝑐_𝑘^∞_𝑛=𝑘^∞₀𝑃_{0,𝑛−𝑘}(𝑥)𝑃_0,𝑛(𝑥)⁺𝑃_0,𝑛(𝑥)+𝑑𝑥_∞𝑛=1𝑃_1,𝑛(0)⁺−^∞_𝑛=1^∞₀[𝜆+𝑏(𝑥)]𝑃_1,𝑛(𝑥)⁺𝑑𝑥+𝜆^∞_𝑘=1𝑐_𝑘^∞_𝑛=𝑘^∞₀𝑃_{1,𝑛−𝑘+1}(𝑥)𝑃_1,𝑛(𝑥)⁺𝑃_1,𝑛(𝑥)[𝑄]𝑑𝑥≤−𝜆⁺+[𝑄]⁺𝑄^∞₀𝜐(𝑥)𝑃_0,0(𝑥)⁺+[𝑄]𝑑𝑥₊𝑄^∞₀𝑏(𝑥)𝑃_1,1(𝑥)⁺[𝑄]𝑑𝑥+𝜆⁺−^∞_𝑛=0^∞₀[𝜆+𝜐(𝑥)]𝑃_0,𝑛(𝑥)⁺𝑑𝑥+𝜆^∞_𝑘=1𝑐_𝑘^∞_𝑛=𝑘^∞₀𝑃_{0,𝑛−𝑘}(𝑥)⁺+𝑑𝑥_{∞𝑛=1∞0}𝜐(𝑥)𝑃_0,𝑛(𝑥)⁺𝑑𝑥+^∞_𝑛=2^∞₀𝑏(𝑥)𝑃_1,𝑛(𝑥)⁺−𝑑𝑥_{∞𝑛=1∞0}[𝜆+𝑏(𝑥)]𝑃_1,𝑛(𝑥)⁺𝑑𝑥+𝜆^∞_𝑘=1𝑐_𝑘^∞_𝑛=𝑘^∞₀𝑃_{1,𝑛−𝑘+1}(𝑥)⁺≤[𝑄]𝑑𝑥₊𝑄×−1_∞0𝜐(𝑥)𝑃_0,0(𝑥)⁺𝑑𝑥+^∞₀𝑏(𝑥)𝑃_1,1(𝑥)⁺−𝑑𝑥_{∞𝑛=0∞0}𝜆𝑃_0,𝑛(𝑥)⁺𝑑𝑥+𝜆^∞_𝑘=1𝑐_𝑘^∞_𝑛=0^∞₀𝑃_0,𝑛(𝑥)⁺−𝑑𝑥_{∞𝑛=1∞0}𝜆𝑃_1,𝑛(𝑥)⁺𝑑𝑥+𝜆^∞_𝑘=1𝑐_𝑘^∞_𝑛=1^∞₀𝑃_1,𝑛(𝑥)⁺𝑑𝑥≤0. (44) 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 that 𝑋^∗, the dual space of 𝑋, is 𝑋^∗=𝑞^∗₀,𝑞^∗₁𝑞^∗₀=𝑞^∗,𝑞^∗_0,0,𝑞^∗_0,1,…∈ℝ×𝐿^∞[0,∞)×𝐿^∞[)𝑞0,∞×⋯,_∗1=𝑞^∗_1,1,𝑞^∗_1,2,…∈𝐿^∞[0,∞)×𝐿^∞[)𝑞0,∞×⋯,_∗0,𝑞^∗₁=sup𝑞^∗,sup_𝑛≥0𝑞^∗_{0,𝑛𝐿∞[0,∞)},sup_𝑛≥1𝑞^∗_{1,𝑛𝐿∞[0,∞)}. (45)

It is easy to check that 𝑋^∗ is a Banach space. If we take a set 𝑆 in 𝑋 as 𝑆=𝑃₀,𝑃₁∈𝑋∣𝑄≥0,𝑃_0,𝑛(𝑥)𝑃≥0,𝑛≥0;_1,𝑛(𝑥)≥0,𝑛≥1,𝑥∈[0,∞), (46) then 𝑆 is a cone in 𝑋. For (𝑃₀,𝑃₁)∈𝐷(𝐴)∩𝑆, we take 𝑞^∗₀,𝑞^∗₁=𝑃₀,𝑃₁11⋮,11⋮∈𝑋^∗, (47) then we have 𝑃₀,𝑃₁,𝑞^∗₀,𝑞^∗₁=𝑃₀,𝑃₁|𝑄|+^∞_𝑛=0𝑃_0,𝑛(𝑥)𝑑𝑥+^∞_𝑛=1𝑃_1,𝑛(𝑥)=𝑑𝑥(𝑃₀,𝑃₁)²=(𝑞^∗₀,𝑞^∗₁)², (48) that is 𝑞^∗₀,𝑞^∗₁∈𝜃𝑃₀,𝑃₁=𝑞^∗₀,𝑞^∗₁∈𝑋^∗∣𝑃₀,𝑃₁,𝑞^∗₀,𝑞^∗₁=(𝑃₀,𝑃₁)²=(𝑞^∗₀,𝑞^∗₁)². (49) For such (𝑞^∗₀,𝑞^∗₁), by using boundary conditions on (𝑃₀,𝑃₁)∈𝐷(𝐴)∩𝑆 and ∑^∞_𝑘=1𝑐_𝑘=1, we have (𝐴+𝑈+𝐸)𝑃₀,𝑃₁,𝑞^∗₀,𝑞^∗₁=𝑃₀,𝑃₁×−𝜆𝑄+^∞₀𝜐(𝑥)𝑃_0,0(𝑥)𝑑𝑥+^∞₀𝑏(𝑥)𝑃_1,1(𝑥)+𝑑𝑥_∞0−𝑑𝑃_0,0(𝑥)−[(𝑥𝑃𝑑𝑥𝜆+𝜐)]_0,0(𝑥)𝑃₀,𝑃₁+𝑑𝑥_{∞𝑛=1∞0}−𝑑𝑃_0,𝑛(𝑥)−[(𝑥𝑃𝑑𝑥𝜆+𝜐)]_0,𝑛(𝑥)+𝜆^𝑛_𝑘=1𝑐_𝑘𝑃_{0,𝑛−𝑘}(𝑥)𝑃₀,𝑃₁+𝑑𝑥_{∞𝑛=1∞0}−𝑑𝑃_1,𝑛(𝑥)−[(𝑥𝑃𝑑𝑥𝜆+𝑏)]_1,𝑛(𝑥)+𝜆^𝑛_𝑘=1𝑐_𝑘𝑃_{1,𝑛−𝑘+1}(𝑥)𝑃₀,𝑃₁=𝑃𝑑𝑥₀,𝑃₁×−𝜆𝑄+^∞₀𝜐(𝑥)𝑃_0,0(𝑥)𝑑𝑥+^∞₀𝑏(𝑥)𝑃_1,1(𝑥)−𝑃𝑑𝑥₀,𝑃₁×^∞_𝑛=0^∞₀−𝑑𝑃_0,𝑛(𝑥)𝑑𝑥𝑑𝑥+^∞_𝑛=0^∞₀[𝜆+𝜐(𝑥)]𝑃_0,𝑛(𝑥)𝑑𝑥−𝜆^∞_𝑘=1𝑐_𝑘^∞_𝑛=0^∞₀𝑃_0,𝑛(𝑥)𝑑𝑥+^∞_𝑛=1^∞₀𝑑𝑃_1,𝑛(𝑥)+𝑑𝑥𝑑𝑥_{∞𝑛=1∞0}[𝜆+𝑏(𝑥)]𝑃_1,𝑛(𝑥)𝑑𝑥−𝜆^∞_𝑘=1𝑐_𝑘^∞_𝑛=1^∞₀𝑃_1,𝑛(𝑥)=𝑃𝑑𝑥₀,𝑃₁×−𝜆𝑄+^∞₀𝜐(𝑥)𝑃_0,0(𝑥)𝑑𝑥+^∞₀𝑏(𝑥)𝑃_1,1(𝑥)−𝑃𝑑𝑥₀,𝑃₁×−𝜆𝑄+^∞_𝑛=0^∞₀𝜐(𝑥)𝑃_0,𝑛(𝑥)−𝑑𝑥_{∞𝑛=1∞0}𝜐(𝑥)𝑃_0,𝑛(𝑥)−𝑑𝑥_{∞𝑛=2∞0}𝑏(𝑥)𝑃_1,𝑛(𝑥)𝑑𝑥+^∞_𝑛=1^∞₀𝑏(𝑥)𝑃_1,𝑛(𝑥)𝑑𝑥=0. (50) 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)

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,𝑃₁0)+_𝑡𝑡+ℎ((𝑡+ℎ−𝑠)𝑇(𝑠)(𝐴+𝑈+𝐸)²𝑃₀,𝑃₁(0)(𝑡𝑃𝑑𝑠=𝑇)₀,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()𝑃)𝑡+ℎ₀,𝑃₁(0),𝑇(𝑡)𝑃₀,𝑃₁(0)^∗≥Re𝑇(𝑡+ℎ)𝑃₀,𝑃₁(0),𝑇(𝑡)𝑃₀,𝑃₁(0)^∗=𝑇(𝑡)𝑃₀,𝑃₁(0)²+ℎRe𝜌(𝑡,ℎ),𝑇(𝑡)𝑃₀,𝑃₁(0)^∗. (53) 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≥𝑇(𝑡)𝑃₀),𝑃₁(0+ℎ)𝑇(𝑡)𝑃₀,𝑃₁(0𝜌(),𝑇(𝑡)𝑃)×Re𝑡,ℎ₀,𝑃₁(0)^∗≥𝑇(𝑡)𝑃₀,𝑃₁(0)−ℎ𝛽(𝑡,ℎ), (55) 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<⋯<𝑡_𝑚=𝑡₀+𝜂 be partition of the interval 𝑡₀≤𝑡≤𝑡₀+𝜂 such that 𝑡_𝑗−𝑡_𝑗−1≤𝛿(1≤𝑗≤𝑚). Then by (55), one has0≤𝑇𝑡₀𝑃₀,𝑃₁(0)−𝑇𝑡₀+𝜂𝑃₀,𝑃₁(0≤_𝑚𝑗=1)𝑇𝑡_𝑗−1𝑃₀,𝑃₁(0)−𝑇𝑡_𝑗𝑃₀,𝑃₁(0≤_𝑚𝑗=1)𝑡_𝑗−𝑡_𝑗−1𝛽𝑡_𝑗,𝑡_𝑗−𝑡_𝑗−1≤𝜂𝜖. (56) 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 Theorems 1 and 2 we obtain the main result in this paper.

Theorem 3. If 𝜐=sup_{𝑥∈[0,∞)}𝜐(𝑥)<∞, 𝑏=sup_{𝑥∈[0,∞)}𝑏(𝑥)<∞, then the system (8) has a unique nonnegative time-dependent 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) From which together with Theorem 2 (i.e., (51)) we have 𝑃₀,𝑃₁(⋅,𝑡)=𝑇(𝑡)𝑃₀,𝑃₁(0)=𝑃₀,𝑃₁(0[),)=1,∀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).