on the eigenvalues of the m/m/1 queueing model with second optional service

J. Pseudo-Differ. Oper. Appl.DOI 10.1007/s11868-013-0087-8

On the eigenvalues of the M/M/1 queueing modelwith second optional service

Geni Gupur

Received: 18 October 2013 / Accepted: 18 November 2013© Springer Basel 2014

Abstract We study the eigenvalues of the underlying operator, which corresponds tothe M/M/1 queueing model with second optional service, on the left real line. We provethat if λ(μ1 +μ2) < μ1μ2 and λ[μ1(1 − r)−λ]2 > μ1μ2(μ1 −λ)+λμ2(2μ1 −λ),then all −λε are its eigenvalues with geometric multiplicity 1 when

0 < ε < min

{1 − r, 1 −

{2λμ1μ2 − λ2μ2

+√

(λ2μ2 − 2λμ1μ2)2 + 4λμ1μ2[μ1(1 − r) − λ]2(μ1 − λ)}/{

2λ[μ1(1 − r) − λ]2}}

.

Our results imply: (i) the underlying operator has uncountable eigenvalues in the leftreal line and therefore the C0-semigroup generated by the underlying operator is notcompact, even not eventually compact. (ii) The essential growth bound of the C0-semigroup is 0 and therefore it is not quasi-compact. (iii) It is impossible that thetime-dependent solution of the M/M/1 queueing model with second optional serviceexponentially converges to its steady-state solution. (iv) The essential spectral radiusof the C0-semigroup is equal to 1.

Keywords M/M/1 queueing model with second optional service · Eigenvalue ·Geometric multiplicity · Essential growth bound

This work was supported by the National Natural Science Foundation of China (No: 11371303).

G. Gupur (B)College of Mathematics and Systems Science, Xinjiang University,Urumqi 830046, People’s Republic of Chinae-mail: [email protected]

G. Gupur

Mathematics Subject Classification (2010) Primary 47A10 · 47D03;Secondary 60K25

1 Introduction

In real life, we always encounter such examples of queueing situations that all arriv-ing customers require the first essential service and only part of them may requirethe second optional service provided by the server. For example, at a barber’s shopeveryone may need a hair-cut but only a part of the customers may need a shaveafter their hair-cut. In 2000, Madan [10] considered these phenomena and describedsome of them by the M/G/1 queueing model with second optional service by usingthe supplementary variable technique. Then he studied the time-dependent solution ofthe model by using the probability generating functions and obtained the expressionof the Laplace–Stieltjes transform of the probability generating functions. Roughlyspeaking, he obtained the existence of the time-dependent solution of the model. In2009, by using C0-semigroup theory Zhao et al. [17] have proved the existence anduniqueness of the positive time-dependent solution of the M/G/1 queueing modelwith second optional service. When the first essential service rate and second optionalservice rate are constants, the M/G/1 queueing model with second optional serviceis called M/M/1 queueing model with second optional service. By studying spectraof the underlying operator which corresponds to the M/M/1 queueing model withsecond optional service and using Theorem 14 in Gupur et al. [9] or Theorem 1.96in Gupur [8], Xing [16] and Fang et al. [7] have obtained the asymptotic behaviorof the time-dependent solution of the model. In 2012, Wayit and Gupur [14] stud-ied eigenvalues of the underlying operator on the left half plane and found that −λ

is an eigenvalue of the underlying operator with geometric multiplicity 1. So far,any other results about dynamic analysis for the model have not been found in theliterature.

In Zhao et al. [17], Xing [16], Fang et al. [7] and Wayit et al. [14] the authorsobtained that all points on the right half plane belong to the resolvent set of the under-lying operator which corresponds to the M/M/1 queueing model with second optionalservice, all points on the imaginary axis except 0 belong to its resolvent set, 0 and −λ

are its eigenvalues. In this paper, by studying eigenvalues of the underlying operator,which is an infinite tridiagonal matrix that can be considered as a pseudo-differentialoperator, on the left half plane we investigate the structure of the corresponding C0-semigroup. We prove that all points in (−λε, 0) are its eigenvalues with geometricmultiplicity one when

0 < ε < min

{1 − r, 1 −

{2λμ1μ2 − λ2μ2

+√

(λ2μ2 − 2λμ1μ2)2 + 4λμ1μ2[μ1(1 − r) − λ]2(μ1 − λ)}/{

2λ[μ1(1 − r) − λ]2}}

.

Our results together with the spectral mapping theorem for the point spectrum (Engeland Nagel [5], p. 277) give that the C0-semigroup generated by the underlying operator

On the eigenvalues of the M/M/1 queueing model with second optional service

has uncountable eigenvalues and therefore it is not compact, even not eventually com-pact. The results in Zhao et al. [17], Xing [16] and Fang et al. [7] yield that the growthbound of the C0-semigroup is 0 and the spectral bound of the underlying operator is 0.

These together with our results in this paper and Corollary in [5], p. 258 imply that theessential growth bound of the C0-semigroup is 0 and therefore it is not quasi-compact.Moreover, by Nagel [11] and Arendt et al. [1,2] we show that the essential spectralradius of the C0-semigroup is 1 and it is impossible that the time-dependent solutionof the M/M/1 queueing model with second optional service exponentially convergesto its steady-state solution. These results indicate that queueing models are essentiallydifferent from the population equations (see Song and Yu [12], Webb [15]) and thereliability models that are described by a finite number of partial differential equationswith integral boundary equations (see Gupur [8]).

According to Madan [10] the M/M/1 queueing model with second optional servicecan be described by the following system of partial differential equations :

d Q(t)

dt= −λQ(t) + μ2 p(2)

0 (t) + (1 − r)μ1

∞∫0

p(1)0 (x, t)dx, (1.1)

∂p(1)0 (x, t)

∂x+ ∂p(1)

0 (x, t)

∂t= −(λ + μ1)p(1)

0 (x, t), (1.2)

∂p(1)n (x, t)

∂x+ ∂p(1)

n (x, t)

∂t= −(λ + μ1)p(1)

n (x, t) + λp(1)n−1(x, t), n ≥ 1, (1.3)

dp(2)0 (t)

dt= −(λ + μ2)p(2)

0 (t) + rμ1

∞∫0

p(1)0 (x, t)dx, (1.4)

dp(2)n (t)

dt= −(λ + μ2)p(2)

n (t) + λp(2)n−1(t) + rμ1

∞∫0

p(1)n (x, t)dx, n ≥ 1, (1.5)

p(1)0 (0, t) = (1 − r)μ1

∞∫0

p(1)1 (x, t)dx + μ2 p(2)

1 (t) + λQ(t), (1.6)

p(1)n (0, t) = (1 − r)μ1

∞∫0

p(1)n+1(x, t)dx + μ2 p(2)

n+1(t), n ≥ 1, (1.7)

Q(0) = 1, p(2)n (0) = 0, p(2)

n (x, 0) = 0, n ≥ 1. (1.8)

Where (x, t) ∈ [0,∞) × [0,∞).Q(t) represents the probability that at time t, thereis no customer in the system and the server is idle. p(1)

n (x, t) (n ≥ 0) represents

G. Gupur

the probability that at time t, there are n customers in the queue [not system (queue+ service)] excluding the one being provided the first essential service and elapsedservice time of this customer is x . p(2)

n (t) represents the probability that at time t,there are n customers in the queue excluding one customer being provided the secondoptional service. λ is the arrival rate of customers. μ1 is the service rate of comple-tion of the first essential service. μ2 is the service rate of completion of the secondoptional service. r is the probability that customers may opt to accept the secondservice.

In this paper, we use the notations in [7,14,16,17]. Choose a state space as follows:

X ={(

p(1), p(2)) ∣∣∣ ∥∥∥(p(1), p(2)

)∥∥∥ =∥∥∥p(1)

∥∥∥+∥∥∥p(2)

∥∥∥ < ∞}

,

here

p(1) = (Q, p(1)0 , p(1)

1 , p(1)2 , . . .) ∈ IR × L1[0,∞) × L1[0,∞) × · · · ,

p(2) = (p(2)0 , p(2)

1 , p(2)2 , . . .) ∈ l1,

∥∥∥p(1)∥∥∥ = |Q| +

∞∑n=0

∥∥∥p(1)n

∥∥∥L1[0,∞)

,

∥∥∥p(1)∥∥∥ =

∞∑n=0

|p(2)n |.

For simplicity, we introduce

�1 =

⎛⎜⎜⎜⎜⎜⎜⎝

e−x 0 0 0 · · ·λe−x 0 (1 − r)μ1 0 · · ·

0 0 0 (1 − r)μ1 · · ·...

......

.... . .

⎞⎟⎟⎟⎟⎟⎟⎠

,

�2 =

⎛⎜⎜⎜⎜⎜⎜⎝

0 0 0 · · ·0 μ2e−x 0 · · ·0 0 μ2e−x · · ·...

......

. . .

⎞⎟⎟⎟⎟⎟⎟⎠

.

In the following we define operators and their domains.


A(

p(1), p(2))

(x)

=

⎛⎜⎜⎜⎜⎜⎜⎝

⎛⎜⎜⎜⎜⎜⎝

−λ 0 0 · · ·0 − d

dx − (λ + μ1) 0 · · ·0 0 − d

dx − (λ + μ1) · · ·...

......

. . .

⎞⎟⎟⎟⎟⎟⎠

⎛⎜⎜⎜⎜⎜⎜⎝

Q

p(1)0 (x)

p(1)1 (x)

...

⎞⎟⎟⎟⎟⎟⎟⎠

,

⎛⎜⎜⎜⎜⎜⎝

−(λ + μ2) 0 0 · · ·0 −(λ + μ2) 0 · · ·0 0 −(λ + μ2) · · ·...

......

. . .

⎞⎟⎟⎟⎟⎟⎠

⎛⎜⎜⎜⎜⎜⎜⎝

p(2)0

p(2)1

p(2)2...

⎞⎟⎟⎟⎟⎟⎟⎠

⎞⎟⎟⎟⎟⎟⎟⎠

,

D(A) =

⎧⎪⎪⎨⎪⎪⎩(

p(1), p(2))

∈ X

∣∣∣∣∣∣∣∣

dp(1)n (x)dx ∈ L1[0,∞), p(1)

n (x) (n ≥ 0)

are absolutely continuous and satisfy

p(1)(0) = ∫∞0 �1 p(1)(x)dx + ∫∞

0 �2 p(2)dx

⎫⎪⎪⎬⎪⎪⎭

,

U(

p(1), p(2))

(x)=

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎛⎜⎜⎜⎜⎜⎝

0 0 0 · · ·0 0 0 · · ·0 λ 0 · · ·0 0 λ · · ·...

......

. . .

⎞⎟⎟⎟⎟⎟⎠

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

Q

p(1)0 (x)

p(1)1 (x)

p(1)2 (x)

...

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

,

⎛⎜⎜⎜⎜⎜⎝

0 0 0 · · ·λ 0 0 · · ·0 λ 0 · · ·0 0 λ · · ·...

......

. . .

⎞⎟⎟⎟⎟⎟⎠

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

p(2)0

p(2)1

p(2)2

p(2)3

...

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

,

E(

p(1), p(2))

(x)

=

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎛⎜⎜⎜⎜⎜⎝

μ2 p(2)0 + (1 − r)μ1

∫∞0 p(1)

0 (x)dx

0

0

...

⎞⎟⎟⎟⎟⎟⎠

,

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

rμ1∫∞

0 p(1)0 (x)dx

rμ1∫∞

0 p(1)1 (x)dx

rμ1∫∞

0 p(1)2 (x)dx

...

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

,

D(U ) = X, D(E) = X.

G. Gupur

Then the Eqs. (1.1)–(1.8) can be rewritten as an abstract Cauchy problem in the Banachspace X :

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

d(

p(1),p(2))(t)

dt = (A + U + E)(

p(1), p(2))(t), t ∈ (0,∞)

(p(1), p(2)

)(0) = (p(1)(0), p(2)(0)

)p(1)(0) = (1, 0, 0, 0, . . .) , p(2)(0) = (0, 0, 0, 0, · · · )

(1.9)

In [7,16,17] the authors obtained the following results:

Theorem 1.1 A+U + E generates a positive contraction C0-semigroup T (t). T (t) isisometric for the initial value. So, the system (1.9) has a unique positive time-dependentsolution

(p(1), p(2)

)(x, t) = T (t)

(p(1), p(2)

)(0) satisfying

∥∥∥(p(1), p(2))

(·, t)∥∥∥ = 1, ∀t ∈ [0,∞).

The set⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

γ ∈ C

∣∣∣∣∣∣∣∣∣∣

sup{

λ|γ+λ| ,

λReγ+λ+μ1

|γ+λ+μ1||γ+λ+μ2|+rμ1(Reγ+λ+μ1)|γ+λ+μ1||γ+λ+μ2|−[(1−r)μ1|γ+λ+μ2|+rμ1μ2] ,

λ|γ+λ+μ1| + μ2|γ+λ+μ2|(Reγ+λ+μ1)

+ rμ1μ2|γ+λ+μ1||γ+λ+μ2|2

× λ|γ+λ+μ1||γ+λ+μ2||γ+λ+μ1||γ+λ+μ2|−[(1−r)μ1|γ+λ+μ2|+rμ1μ2]

}< 1

⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭

belongs to the resolvent set of (A + U + E)∗, the adjoint operator of A + U + E .

In particular, all points on the imaginary axis except 0 belong to the resolvent set ofA+U +E . When λ

μ1+ rλ

μ2< 1, 0 is an eigenvalue of A+U +E and (A+U +E)∗ with

geometric multiplicity one. Therefore, the time-dependent solution(

p(1), p(2))(x, t)

of the system (1.9) strongly converges to its steady-state solution(

p(1), p(2))(x), i.e.,

limt→∞

∥∥∥(p(1), p(2))

(·, t) − α(

p(1), p(2))

(·)∥∥∥ = 0,

here α is decided by the eigenvector satisfying (A + U + E)∗(q(1), q(2)

) = 0 and theinitial value.

In 2012, Wayit and Gupur [14] have proved the following results:

Theorem 1.2 If λ(μ1 + μ2) < μ1μ2, then −λ is an eigenvalue of A + U + E withgeometric multiplicity one.

In this paper we will use the following results about cubic equation:

Theorem 1.3 [3,13] Let γ 3+aγ 2+bγ +c = 0 (a, b, c ∈ R). Define the discriminant

� = 18abc − 4a3c + a2b2 − 4b3 − 27c2.

We have the following results:


1. If � > 0, then the above equation has three distinct real roots.2. If � < 0, then the above equation has one real root and two complex conjugate

roots.3. If � = 0, then the above equation has a multiple root and its roots are real. This

case includes two special cases:

(1) If � = 0, a2 − 3b = 0, then the three roots are equal: γ1 = γ2 = γ3 = − a3 .

(2) If � = 0, a2 − 3b �= 0, then

γ1 = 4ab − 9c − a3

a2 − 3b, γ2 = γ3 = 9c − ab

2(a2 − 3b).

Moreover, the general formula for the roots is as follows (i2 = −1):

γ1 = −1

3

⎡⎣a + 3

√2a3 − 9ab + 27c + √−27�

2+ a2 − 3b

3√

2a3−9ab+27c+√−27�2

⎤⎦ ,

γ2 = −1

3

⎡⎣a+ −1+i

√3

23

√2a3−9ab+27c+√−27�

2+ (a2 − 3b)(1+i

√3)

3√

2a3−9ab+27c+√−27�2

⎤⎦ ,

γ3 = −1

3

⎡⎣a+ −1−i

√3

23

√2a3−9ab+27c+√−27�

2− (a2 − 3b)(−1+i

√3)

3√

2a3−9ab+27c+√−27�2

⎤⎦

Theorem 1.4 [6] Necessary and sufficient conditions that all roots of the cubic equa-tion γ 3 + aγ 2 + bγ + c = 0, (a, b, c ∈ R) have absolute value (modulus) less than1 are

1−a + b − c > 0, 1 + a + b + c > 0, 3−a−b + 3c > 0, 1−b + ac−c2 > 0.

2 Main Results

Theorem 2.1 If λ,μ1, μ2 and r satisfy λ(μ1 +μ2) < μ1μ2 and λ[μ1(1−r)−λ]2 >

μ1μ2(μ1 − λ) + λμ2(2μ1 − λ), then all −λε are eigenvalues of A + U + E withgeometric multiplicity 1 for

0 < ε < min

{1−r, 1−

{2λμ1μ2−λ2μ2

+√

(λ2μ2−2λμ1μ2)2+4λμ1μ2[μ1(1−r)−λ]2(μ1−λ)}/{

2λ[μ1(1−r)−λ]2}}

.

G. Gupur

Remark 2.2 1. By comparing with Theorems 1.1 and 1.2 it is easy to see that ourconditions are stronger than the conditions of that two theorems. The conditionλ(μ1 + μ2) < μ1μ2 implies that the service rates of the server are larger thanthe arrival rate of customers, which is necessary for the existence of the queue-ing system. If the server serves so fast that while a customer receives serviceand elapses, next customer does not arrive, i.e., the server becomes idle, theneconomically, it is not suitable. Our second condition λ[μ1(1 − r) − λ]2 >

μ1μ2(μ1 − λ) + λμ2(2μ1 − λ) restricts the service rates of the server by thearrival rate of customers. Hence, this condition is acceptable.

2. The conditions λ(μ1 + μ2) < μ1μ2 and λ[μ1(1 − r) − λ]2 > μ1μ2(μ1 − λ) +λμ2(2μ1 − λ) imply

λ(μ1 + μ2) < μ1μ2 ⇒ λ

μ1+ λ

μ2< 1 ⇒ λ < μ1, λ < μ2,

λ[μ1(1 − r) − λ]2 > μ1μ2(μ1 − λ) + λμ2(2μ1 − λ) > 0

⇒λ[μ1(1 − r) − λ]2 > μ2

1μ2 − λμ1μ2 − λ2μ2 + 2λμ1μ2

⇒λ[μ1(1 − r) − λ]2 + λ2μ2 − 2λμ1μ2 > μ2

1μ2 − λμ1μ2

⇒4λ[μ1(1 − r) − λ]2

{λ[μ1(1 − r) − λ]2 + λ2μ2 − 2λμ1μ2

}

> 4λ[μ1(1 − r) − λ]2(μ21μ2 − λμ1μ2)

⇒4λ2[μ1(1 − r) − λ]4 + 4λ[μ1(1 − r) − λ]2

(λ2μ2 − 2λμ1μ2

)

> 4λ[μ1(1 − r) − λ]2(μ21μ2 − λμ1μ2)

⇒4λ2[μ1(1 − r) − λ]4 + 4λ[μ1(1 − r) − λ]2

(λ2μ2 − 2λμ1μ2

)

+(λ2μ2 − 2λμ1μ2)2

> 4λ[μ1(1 − r) − λ]2(μ21μ2 − λμ1μ2) + (λ2μ2 − 2λμ1μ2)

2

⇒{2λ[μ1(1 − r) − λ]2 + λ2μ2 − 2λμ1μ2

}2

> (λ2μ2 − 2λμ1μ2)2 + 4λμ1μ2[μ1(1 − r) − λ]2(μ1 − λ)

⇒2λ[μ1(1 − r) − λ]2 + λ2μ2 − 2λμ1μ2

>√

(λ2μ2 − 2λμ1μ2)2 + 4λμ1μ2[μ1(1 − r) − λ]2(μ1 − λ)


⇒2λμ1μ2−λ2μ2+√(λ2μ2−2λμ1μ2)2+4λμ1μ2[μ1(1 − r)−λ]2(μ1−λ)

2λ[μ1(1 − r) − λ]2

< 1.

which means that (−λε, 0) is an interval.

Proof We consider (A+U + E)(

p(1), p(2)) = −λε

(p(1), p(2)

), which is equivalent

to

−λQ + μ2 p(2)0 + (1 − r)μ1

∞∫0

p(1)0 (x)dx = −λεQ, (2.1)

−dp(1)0 (x)

dx− (λ + μ1)p(1)

0 (x) = −λεp(1)0 (x), (2.2)

−dp(1)n (x)

dx− (λ + μ1)p(1)

n (x) + λp(1)n−1(x) = −λεp(1)

n (x), n ≥ 1, (2.3)

−(λ + μ2)p(2)0 + rμ1

∞∫0

p(1)0 (x)dx = −λεp(2)

0 , (2.4)

−(λ + μ2)p(2)n + λp(2)

n−1 + rμ1

∞∫0

p(1)n (x)dx = −λεp(2)

n (x), n ≥ 1, (2.5)

p(1)0 (0) = (1 − r)μ1

∞∫0

p(1)1 (x)dx + μ2 p(2)

1 + λQ, (2.6)

p(1)n (0) = (1 − r)μ1

∞∫0

p(1)n+1(x)dx + μ2 p(2)

n+1, n ≥ 1. (2.7)

By rearranging (2.1)–(2.5) we have

μ2 p(2)0 + (1 − r)μ1

∞∫0

p(1)0 (x)dx = (λ − λε)Q, (2.8)

dp(1)0 (x)

dx= −(λ − λε + μ1)p(1)

0 (x), (2.9)

dp(1)n (x)

dx= −(λ − λε + μ1)p(1)

n (x) + λp(1)n−1(x), n ≥ 1, (2.10)

G. Gupur

(λ − λε + μ2)p(2)0 = rμ1

∞∫0

p(1)0 (x)dx, (2.11)

(λ − λε + μ2)p(2)n = λp(2)

n−1 + rμ1

∞∫0

p(1)n (x)dx, n ≥ 1. (2.12)

By solving (2.9) and (2.10) we get

p(1)0 (x) = a0e−(λ−λε+μ1)x , (2.13)

p(1)n (x) = ane−(λ−λε+μ1)x + λe−(λ−λε+μ1)x

x∫0

p(1)n−1(τ )e(λ−λε+μ1)τ dτ, n ≥ 1.

(2.14)

By combining (2.8) with (2.11) we obtain

μ2 p(2)0 + (1 − r)

λ − λε + μ2

rp(2)

0 = (λ − λε)Q

⇒rμ2 + (1 − r)(λ − λε + μ2)

rp(2)

0 = (λ − λε)Q

⇒p(2)

0 = r(λ − λε)

rμ2 + (1 − r)(λ − λε + μ2)Q. (2.15)

By inserting (2.15) into (2.11) and using (2.13) and λ − λε + μ1 > 0 we deduce

(λ − λε + μ2)p(2)0 = rμ1

∞∫0

p(1)0 (x)dx = rμ1

λ − λε + μ1a0

⇒

a0 = (λ − λε + μ1)(λ − λε + μ2)

rμ1p(2)

0

= (λ − λε + μ1)(λ − λε + μ2)

rμ1

× r(λ − λε)

rμ2 + (1 − r)(λ − λε + μ2)Q

= (λ − λε)(λ − λε + μ1)(λ − λε + μ2)

μ1[rμ2 + (1 − r)(λ − λε + μ2)] Q. (2.16)


By using (2.13) and (2.14) repeatedly we derive

p(1)1 (x) = a1e−(λ−λε+μ1)x + λe−(λ−λε+μ1)x

x∫0

p(1)0 (τ )e(λ−λε+μ1)τ dτ

= a1e−(λ−λε+μ1)x + λe−(λ−λε+μ1)x

x∫0

a0dτ

v = [a1 + a0λx]e−(λ−λε+μ1)x . (2.17)


x∫0



x∫0

[a1 + a0λτ ]dτ

= a2e−(λ−λε+μ1)x + λe−(λ−λε+μ1)x[

a1x + a0λx2

2

]

=[

a2 + a1λx + a0(λx)2

2

]e−(λ−λε+μ1)x . (2.18)

. . . . . .


x∫0



x∫0

[a2 + a1λτ + a0

(λτ)2

2

]dτ

= a3e−(λ−λε+μ1)x + λe−(λ−λε+μ1)x[

a2x + a1λx2

2+ a0

λ2

2

x3

3

]

=[

a3 + a2λx + a1(λx)2

2+ a0

(λx)3

3!]

e−(λ−λε+μ1)x . (2.19)

. . . . . .

p(1)n (x) =

[an + an−1λx + an−2

(λx)2

2! + an−3(λx)3

3! + · · ·

+a1(λx)n−1

(n − 1)! + a0(λx)n

n!]e−(λ−λε+μ1)x

=n∑

k=0

ak(λx)n−k

(n − k)!e−(λ−λε+μ1)x , n ≥ 0. (2.20)

G. Gupur

(2.12) implies

(λ − λε + μ2)p(2)n = λp(2)

n−1 + rμ1

∞∫0

p(1)n (x)dx

⇒ p(2)n = λ

λ − λε + μ2p(2)

n−1 + rμ1

λ − λε + μ2

∞∫0

p(1)n (x)dx, n ≥ 1. (2.21)

(2.21) gives

p(1)1 = λ

λ − λε + μ2p(2)

0 + rμ1

λ − λε + μ2

∞∫0

p(1)1 (x)dx, (2.22)

p(2)2 = λ

λ − λε + μ2p(2)

1 + rμ1

λ − λε + μ2

∞∫0

p(1)2 (x)dx

= λ

λ − λε + μ2

⎡⎣ λ

λ − λε + μ2p(2)

0 + rμ1

λ − λε + μ2

∞∫0

p(1)1 (x)dx

⎤⎦

+ rμ1

λ − λε + μ2

∞∫0

p(1)2 (x)dx

=(

λ

λ − λε + μ2

)2

p(2)0 + λrμ1

(λ − λε + μ2)2

∞∫0

p(1)1 (x)dx

+ rμ1

λ − λε + μ2

∞∫0

p(1)2 (x)dx, (2.23)

p(2)3 = λ

λ − λε + μ2p(2)

2 + rμ1

λ − λε + μ2

∞∫0

p(1)3 (x)dx

= λ

λ − λε + μ2

⎡⎣( λ

λ − λε + μ2

)2

p(2)0 + λrμ1

(λ − λε + μ2)2

∞∫0

p(1)1 (x)dx

+ rμ1

λ − λε + μ2

∞∫0

p(1)2 (x)dx

⎤⎦+ rμ1

λ − λε + μ2

∞∫0

p(1)3 (x)dx

=(

λ

λ − λε + μ2

)3

p(2)0 + λ2rμ1

(λ − λε + μ2)3

∞∫0

p(1)1 (x)dx


+ λrμ1

(λ − λε + μ2)2

∞∫0

p(1)2 (x)dx

+ rμ1

λ − λε + μ2

∞∫0

p(1)3 (x)dx, (2.24)

p(2)4 = λ

λ − λε + μ2p(2)

3 + rμ1

λ − λε + μ2

∞∫0

p(1)4 (x)dx

= λ

λ − λε + μ2

⎡⎣(

λ

λ − λε + μ2

)3

p(2)0 + λ2rμ1

(λ − λε + μ2)3

∞∫0

p(1)1 (x)dx

+ λrμ1

(λ − λε + μ2)2

∞∫0

p(1)2 (x)dx + rμ1

λ − λε + μ2

∞∫0

p(1)3 (x)dx

⎤⎦

+ rμ1

λ − λε + μ2

∞∫0

p(1)4 (x)dx

=(

λ

λ − λε + μ2

)4

p(2)0 + λ3rμ1

(λ − λε + μ2)4

∞∫0

p(1)1 (x)dx

+ λ2rμ1

(λ − λε + μ2)3

∞∫0

p(1)2 (x)dx + λrμ1

(λ − λε + μ2)2

∞∫0

p(1)3 (x)dx

+ rμ1

λ − λε + μ2

∞∫0

p(1)4 (x)dx,

· · · · · · (2.25)

p(2)n =

(λ

λ − λε + μ2

)n

p(2)0 + λn−1rμ1

(λ − λε + μ2)n

∞∫0

p(1)1 (x)dx

+ λn−2rμ1

(λ − λε + μ2)n−1

∞∫0

p(1)2 (x)dx + λn−3rμ1

(λ − λε + μ2)n−2

∞∫0

p(1)3 (x)dx + · · ·

+ λrμ1

(λ − λε + μ2)2

∞∫0

p(1)n−1(x)dx + rμ1

λ − λε + μ2

∞∫0

p(1)n (x)dx

G. Gupur

=(

λ

λ − λε + μ2

)n

p(2)0 +

n−1∑k=0

λkrμ1

(λ − λε + μ2)k+1

∞∫0

p(1)n−k(x)dx

=(

λ

λ − λε + μ2

)n

p(2)0 +

n∑k=1

λn−krμ1

(λ − λε + μ2)n+1−k

∞∫0

p(1)k (x)dx, n ≥ 1.

(2.26)

By substituting (2.26) into (2.7) and using (2.14) we calculate

an = (1 − r)μ1

∞∫0

p(1)n+1(x)dx + μ2 p(2)

n+1

= (1 − r)μ1

∞∫0

p(1)n+1(x)dx + μ2

(λ

λ − λε + μ2

)n+1

p(2)0

+μ2

n+1∑k=1

λn+1−krμ1

(λ − λε + μ2)n+2−k

∞∫0

p(1)k (x)dx

= μ2

(λ

λ − λε + μ2

)n+1

p(2)0 + (1 − r)μ1

∞∫0

p(1)n+1(x)dx

+rμ1μ2

n+1∑k=1

λn+1−k

(λ − λε + μ2)n+2−k

∞∫0

p(1)k (x)dx, n ≥ 1. (2.27)

From (2.20), λ − λε + μ1 > 0 and∫∞

0 xke−(λ−λε+μ1)x dx = k!(λ−λε+μ1)k+1 for k =

0, 1, 2 · · · we determine

∞∫0

p(1)n (x)dx =

∞∫0

n∑k=0

ak(λx)n−k

(n − k)!e−(λ−λε+μ1)x dx

=n∑

k=0

akλn−k

(n − k)!∞∫

0

xn−ke−(λ−λε+μ1)x dx

=n∑

k=0

akλn−k

(n − k)!(n − k)!

(λ − λε + μ1)n−k+1

=n∑

k=0

akλn−k

(λ − λε + μ1)n−k+1 , n ≥ 0. (2.28)


By inserting (2.28) into (2.27) and using the Fubini theorem and (2.16) it follows

an = μ2

(λ

λ − λε + μ2

)n+1

p(2)0 + (1 − r)μ1

n+1∑k=0

akλn+1−k

(λ − λε + μ1)n+2−k

+ rμ1μ2

n+1∑k=1

λn+1−k

(λ − λε + μ2)n+2−k

k∑j=0

a jλk− j

(λ − λε + μ1)k+1− j

= μ2

(λ

λ − λε + μ2

)n+1

p(2)0 + (1 − r)μ1

n+1∑k=0

akλn+1−k

(λ − λε + μ1)n+2−k

+ rμ1μ2

n+1∑k=0

λn+1−k

(λ − λε + μ2)n+2−k

k∑j=0

a jλk− j

(λ − λε + μ1)k+1− j

− rμ1μ2λn+1

(λ − λε + μ2)n+2

1

λ − λε + μ1a0

= μ2

(λ

λ − λε + μ2

)n+1

p(2)0 + (1 − r)μ1

n+1∑k=0

akλn+1−k

(λ − λε + μ1)n+2−k

+ rμ1μ2

n+1∑j=0

n+1∑k= j

a jλn+1−k

(λ − λε + μ2)n+2−k

λk− j

(λ − λε + μ1)k+1− j

− rμ1μ2λn+1

(λ − λε + μ2)n+2

1

λ − λε + μ1a0

= μ2

(λ

λ − λε + μ2

)n+1

p(2)0 + (1 − r)μ1

n+1∑k=0

akλn+1−k

(λ − λε + μ1)n+2−k

+ rμ1μ2

n+1∑j=0

a jλn+1− j

(λ − λε + μ2)n+2

1

(λ − λε + μ1)1− j

×n+1∑k= j

(λ − λε + μ2

λ − λε + μ1

)k

− rμ1μ2λn+1

(λ − λε + μ2)n+2

1

λ − λε + μ1a0

= μ2

(λ

λ − λε + μ2

)n+1

p(2)0 + (1 − r)μ1

n+1∑k=0

akλn+1−k

(λ − λε + μ1)n+2−k

G. Gupur

+ rμ1μ2

n+1∑j=0

a jλn+1− j

(λ − λε + μ2)n+2

1

(λ − λε + μ1)1− j

× 1

1 − λ−λε+μ2λ−λε+μ1

[(λ − λε + μ2

λ − λε + μ1

) j

−(

λ − λε + μ2

λ − λε + μ1

)n+2]

− rμ1μ2λn+1

(λ − λε + μ2)n+2

1

λ − λε + μ1a0

= μ2

(λ

λ − λε + μ2

)n+1

p(2)0 + (1 − r)μ1

n+1∑k=0

akλn+1−k

(λ − λε + μ1)n+2−k

+ rμ1μ2

n+1∑j=0

a jλn+1− j (λ − λε + μ1)

j

(μ1 − μ2)(λ − λε + μ2)n+2

×[(

λ − λε + μ2

λ − λε + μ1

) j

−(

λ − λε + μ2

λ − λε + μ1

)n+2]

− rμ1μ2λn+1

(λ − λε + μ2)n+2

1

λ − λε + μ1a0

= μ2

(λ

λ − λε + μ2

)n+1

p(2)0 + (1 − r)μ1

n+1∑k=0

akλn+1−k

(λ − λε + μ1)n+2−k

+ rμ1μ2

μ1−μ2

n+1∑j=0

a jλn+1− j

(λ − λε + μ2)n+2− j− rμ1μ2

μ1 − μ2

n+1∑j=0

a jλn+1− j

(λ − λε + μ1)n+2− j

− rμ1μ2λn+1

(λ − λε + μ2)n+2

1

λ − λε + μ1a0

= μ2

(λ

λ − λε + μ2

)n+1

p(2)0

+[(1 − r)μ1 − rμ1μ2

μ1 − μ2

] n+1∑k=0

akλn+1−k

(λ − λε + μ1)n+2−k

+ rμ1μ2

μ1 − μ2

n+1∑k=0

akλn+1−k

(λ − λε + μ2)n+2−k


− rμ1μ2λn+1

(λ − λε + μ2)n+2

1

λ − λε + μ1a0

= μ2

(λ

λ − λε + μ2

)n+1

p(2)0

+ μ1[(1 − r)μ1 − μ2]μ1 − μ2

n+1∑k=0

akλn+1−k

(λ − λε + μ1)n+2−k

+ rμ1μ2

μ1 − μ2

n+1∑k=0

akλn+1−k

(λ − λε + μ2)n+2−k

− rμ1μ2λn+1

(λ − λε + μ2)n+2

1

λ − λε + μ1

(λ − λε + μ1)(λ − λε + μ2)

rμ1p(2)

0

= μ2

(λ

λ − λε + μ2

)n+1

p(2)0

+ μ1[(1 − r)μ1 − μ2]μ1 − μ2

n+1∑k=0

akλn+1−k

(λ − λε + μ1)n+2−k

+ rμ1μ2

μ1 − μ2

n+1∑k=0

akλn+1−k

(λ − λε + μ2)n+2−k− μ2

(λ

λ − λε + μ2

)n+1

p(2)0

= μ1[(1 − r)μ1 − μ2]μ1 − μ2

n+1∑k=0

akλn+1−k

(λ − λε + μ1)n+2−k

+ rμ1μ2

μ1 − μ2

n+1∑k=0

akλn+1−k

(λ − λε + μ2)n+2−k

={

μ1[(1 − r)μ1 − μ2](μ1 − μ2)(λ − λε + μ1)

+ rμ1μ2

(μ1 − μ2)(λ − λε + μ2)

}an+1

+{

λμ1[(1 − r)μ1 − μ2](μ1 − μ2)(λ − λε + μ1)2 + rμ1μ2λ

(μ1 − μ2)(λ − λε + μ2)2

}an

+ μ1[(1 − r)μ1 − μ2](μ1 − μ2)(λ − λε + μ1)

n−1∑k=0

ak

(λ

λ − λε + μ1

)n+1−k

+ rμ1μ2

(μ1 − μ2)(λ − λε + μ2)

n−1∑k=0

ak

(λ

λ − λε + μ2

)n+1−k

, n ≥ 1. (2.29)

G. Gupur

If we define

ξ = μ1[(1 − r)μ1 − μ2](μ1 − μ2)(λ − λε + μ1)

,

η = rμ1μ2

(μ1 − μ2)(λ − λε + μ2),

l1 = λ

λ − λε + μ1, l2 = λ

λ − λε + μ2,

then (2.29) becomes

an = (ξ + η)an+1+(ξ l1 + ηl2)an +ξ

n−1∑k=0

akln+1−k1 + η

n−1∑k=0

akln+1−k2

⇒

(ξ + η)an+1 = (1 − ξ l1 − ηl2)an − ξ

n−1∑k=0

akln+1−k1 − η

n−1∑k=0

akln+1−k2

⇒

an+1 = 1 − ξ l1 − ηl2ξ + η

an − ξ

ξ + η

n−1∑k=0

akln+1−k1

− η

ξ + η

n−1∑k=0

akln+1−k2 , n ≥ 1. (2.30)

⇒

an = 1 − ξ l1 − ηl2ξ + η

an−1 − ξ

ξ + η

n−2∑k=0

akln−k1

− η

ξ + η

n−2∑k=0

akln−k2 , n ≥ 2. (2.31)

an+2 = 1 − ξ l1 − ηl2ξ + η

an+1 − ξ

ξ + η

n∑k=0

akln+2−k1

− η

ξ + η

n∑k=0

akln+2−k2 , n ≥ 0. (2.32)

(2.30)–(2.31) ×l1 gives


an+1 − l1an = 1 − ξ l1 − ηl2ξ + η

an − ξ

ξ + η

n−1∑k=0

akln+1−k1

− η

ξ + η

n−1∑k=0

akln+1−k2 − l1(1 − ξ l1 − ηl2)

ξ + ηan−1

+ ξ

ξ + η

n−2∑k=0

akln+1−k1 + ηl1

ξ + η

n−2∑k=0

akln−k2

= 1 − ξ l1 − ηl2ξ + η

an − ξ l21

ξ + ηan−1 − ηl2

2

ξ + ηan−1

− l1 − ξ l21 − ηl1l2

ξ + ηan−1 + ηl1 − ηl2

ξ + η

n−2∑k=0

akln−k2

= 1 − ξ l1 − ηl2ξ + η

an − ξ l21 + ηl2

2 + l1 − ξ l21 − ηl1l2

ξ + ηan−1

+ηl1 − ηl2ξ + η

n−2∑k=0

akln−k2

= 1 − ξ l1 − ηl2ξ + η

an − ηl22 + l1 − ηl1l2

ξ + ηan−1

+η(l1 − l2)

ξ + η

n−2∑k=0

akln−k2

⇒η(l1−l2)

ξ + η

n−2∑k=0

akln−k2 = an+1 −

[l1+ 1 − ξ l1+ηl2

ξ + η

]an + l1−ηl1l2+ηl2

2

ξ + ηan−1

= an+1 − l1(ξ + η) + 1 − ξ l1 + ηl2ξ + η

an

+ l1 − ηl1l2 + ηl22

ξ + ηan−1

= an+1 − ηl1 + 1 − ηl2ξ + η

an + l1 − ηl1l2 + ηl22

ξ + ηan−1

⇒

G. Gupur

η

ξ + η

n−2∑k=0

akln−k2 = 1

l1 − l2an+1 − 1 + ηl1 − ηl2

(ξ + η)(l1 − l2)an

+ l1 − ηl1l2 + ηl22

(ξ + η)(l1 − l2)an−1. (2.33)

(2.30)–(2.31) ×l2 implies

an+1 − l2an = 1 − ξ l1 − ηl2ξ + η

an − ξ

ξ + η

n−1∑k=0

akln+1−k1

− η

ξ + η

n−1∑k=0

akln+1−k2 − l2(1 − ξ l1 − ηl2)

ξ + ηan−1

+ ξ l2ξ + η

n−2∑k=0

akln−k1 + η

ξ + η

n−2∑k=0

akln+1−k2

= 1 − ξ l1 − ηl2ξ + η

an − ξ l21

ξ + ηan−1

+ξ(l2 − l1)

ξ + η

n−2∑k=0

akln−k1 − ηl2

2

ξ + ηan−1

− l2(1 − ξ l1 − ηl2)

ξ + ηan−1

= 1 − ξ l1 − ηl2ξ + η

an − ξ l21 + ηl2

2 + l2 − ξ l1l2 − ηl22

ξ + ηan−1

+ξ(l2 − l1)

ξ + η

n−2∑k=0

akln−k1

= 1 − ξ l1 − ηl2ξ + η

an − ξ l21 + l2 − ξ l1l2

ξ + ηan−1

+ξ(l2 − l1)

ξ + η

n−2∑k=0

akln−k1

⇒ξ(l2 − l1)

ξ + η

n−2∑k=0

akln−k1 = an+1 −

[l2 + 1 − ξ l1 − ηl2

ξ + η

]an


+ξ l21 + l2 − ξ l1l2

ξ + ηan−1

= an+1 − ξ l2 + 1 − ξ l1ξ + η

an + l2 + ξ l21 − ξ l1l2

ξ + ηan−1

⇒ξ

ξ + η

n−2∑k=0

akln−k1 = 1

l2 − l1an+1 − ξ l2 + 1 − ξ l1

(ξ + η)(l2 − l1)an

+ l2 + ξ l21 − ξ l1l2

(ξ + η)(l2 − l1)an−1. (2.34)

By inserting (2.33) and (2.34) into (2.32) we have

an+2 = 1 − ξ l1 − ηl2ξ + η

an+1 − ξ

ξ + η

n∑k=0

akln+2−k1 − η

ξ + η

n∑k=0

akln+2−k2

= 1 − ξ l1 − ηl2ξ + η

an+1 − ξ l21

ξ + ηan − ξ l3

1

ξ + ηan−1 − ξ l2

1

ξ + η

n−2∑k=0

akln−k1

− ηl22

ξ + ηan − ηl3

2

ξ + ηan−1 − ηl2

2

ξ + η

n−2∑k=0

akln−k2

= 1 − ξ l1 − ηl2ξ + η

an+1 − ξ l21 + ηl2

2

ξ + ηan − ξ l3

1 + ηl32

ξ + ηan−1

−l21

[1

l2 − l1an+1 − ξ l2 + 1 − ξ l1

(ξ + η)(l2 − l1)an + l2 + ξ l2

1 − ξ l1l2(ξ + η)(l2 − l1)

an−1

]

−l22

[1

l1 − l2an+1 − 1 + ηl1 − ηl2

(ξ + η)(l1 − l2)an + l1 − ηl1l2 + ηl2

2

(ξ + η)(l1 − l2)an−1

]

=[

1 − ξ l1 − ηl2ξ + η

− l21

l2 − l1− l2

2

l1 − l2

]an+1

−[ξ l2

1 + ηl22

ξ + η− l2

1(1 − ξ l1 + ξ l2)

(ξ + η)(l2 − l1)− l2

2(1 + ηl1 − ηl2)

(ξ + η)(l1 − l2)

]an

−[ξ l3

1 + ηl32

ξ + η+ l2

1(l2 + ξ l21 − ξ l1l2)

(ξ + η)(l2 − l1)+ l2

2(l1 − ηl1l2 + ηl22)

(ξ + η)(l1 − l2)

]an−1

= (1 − ξ l1 − ηl2)(l2 − l1) + (l22 − l2

1)(ξ + η)

(ξ + η)(l2 − l1)an+1

G. Gupur

−[

ξ l21 + ηl2

2

ξ + η+ l2

2(1 + ηl1 − ηl2) − l21(1 − ξ l1 + ξ l2)

(ξ + η)(l2 − l1)

]an

−[

ξ l31 + ηl3

2

ξ + η+ l2

1(l2 + ξ l21 − ξ l1l2) − l2

2(l1 − ηl1l2 + ηl22)

(ξ + η)(l2 − l1)

]an−1

= 1 − ξ l1 − ηl2 + (l2 + l1)(ξ + η)

ξ + ηan+1

−[

ξ l21 + ηl2

2

ξ + η+ (l2

2 − l21) + ηl2

2(l1 − l2) + ξ l21(l1 − l2)

(ξ + η)(l2 − l1)

]an

−[

ξ l31 + ηl3

2

ξ + η+ l1l2(l1 − l2) + ξ l3

1(l1 − l2) + ηl32(l1 − l2)

(ξ + η)(l2 − l1)

]an−1

= 1 + ξ l2 + ηl1ξ + η

an+1 −[

ξ l21 + ηl2

2

ξ + η+ l2 + l1 − ηl2

2 − ξ l21

ξ + η

]an

−[

ξ l31 + ηl3

2

ξ + η+ −l1l2 − ξ l3

1 − ηl32

ξ + η

]an−1

= 1 + ξ l2 + ηl1ξ + η

an+1 − l1 + l2ξ + η

an + l1l2ξ + η

an−1, n ≥ 1. (2.35)

In the following we determine the coefficients in (2.35).

ξ + η = μ1[(1 − r)μ1 − μ2](μ1 − μ2)(λ − λε + μ1)

+ rμ1μ2

(μ1 − μ2)(λ − λε + μ2)

= μ1[(1 − r)μ1 − μ2](λ − λε + μ2) + rμ1μ2(λ − λε + μ1)

(μ1 − μ2)(λ − λε + μ1)(λ − λε + μ2)

= μ1(μ1 − μ2 − rμ1)(λ − λε + μ2) + rμ1μ2(λ − λε + μ1)

(μ1 − μ2)(λ − λε + μ1)(λ − λε + μ2)

= 1

(μ1 − μ2)(λ − λε + μ1)(λ − λε + μ2)

{μ1(μ1−μ2)(λ−λε+μ2)

−rμ21(λ − λε + μ2) + rμ1μ2(λ − λε + μ1)

}

= 1

(μ1 − μ2)(λ − λε + μ1)(λ − λε + μ2)

{μ1(μ1−μ2)(λ−λε+μ2)

+rμ1[μ2(λ − λε + μ1) − μ1(λ − λε + μ2)]}

= μ1(μ1 − μ2)(λ − λε + μ2) + rμ1(λ − λε)(μ2 − μ1)

(μ1 − μ2)(λ − λε + μ1)(λ − λε + μ2)


= μ1(λ − λε + μ2) − rμ1(λ − λε)

(λ − λε + μ1)(λ − λε + μ2)

= (μ1 − rμ1)(λ − λε) + μ1μ2

(λ − λε + μ1)(λ − λε + μ2)

= λμ1(1 − r)(1 − ε) + μ1μ2

(λ − λε + μ1)(λ − λε + μ2). (2.36)

1 + ξ l2 + ηl1 = 1 + λμ1[(1 − r)μ1 − μ2](μ1 − μ2)(λ − λε + μ1)(λ − λε + μ2)

+ λrμ1μ2

(μ1 − μ2)(λ − λε + μ1)(λ − λε + μ2)

= 1 + λμ1[(1 − r)μ1 − μ2 + rμ2](μ1 − μ2)(λ − λε + μ1)(λ − λε + μ2)

= 1 + λμ1[(1 − r)μ1 − (1 − r)μ2](μ1 − μ2)(λ − λε + μ1)(λ − λε + μ2)

= 1 + λμ1(1 − r)(μ1 − μ2)

(μ1 − μ2)(λ − λε + μ1)(λ − λε + μ2)

= 1 + λμ1(1 − r)

(λ − λε + μ1)(λ − λε + μ2)

= (λ − λε + μ1)(λ − λε + μ2) + λμ1(1 − r)

(λ − λε + μ1)(λ − λε + μ2). (2.37)

l1 + l2 = λ

λ − λε + μ1+ λ

λ − λε + μ2

= λ(2λ − 2λε + μ1 + μ2)

(λ − λε + μ1)(λ − λε + μ2)

= λ[2λ(1 − ε) + μ1 + μ2](λ − λε + μ1)(λ − λε + μ2)

. (2.38)

l1l2 = λ2

(λ − λε + μ1)(λ − λε + μ2). (2.39)

By combining (2.36) with (2.37), (2.38) and (2.39) we obtain

1 + ξ l2 + ηl1ξ + η

= (λ − λε + μ1)(λ − λε + μ2) + λμ1(1 − r)

λμ1(1 − r)(1 − ε) + μ1μ2, (2.40)

l1 + l2ξ + η

= λ[2λ(1 − ε) + μ1 + μ2]λμ1(1 − r)(1 − ε) + μ1μ2

, (2.41)

l1l2ξ + η

= λ2

λμ1(1 − r)(1 − ε) + μ1μ2. (2.42)

G. Gupur

By inserting (2.40)∼ (2.42) into (2.35) we deduce

an+2 = (λ − λε + μ1)(λ − λε + μ2) + λμ1(1 − r)

λμ1(1 − r)(1 − ε) + μ1μ2an+1

− λ[2λ(1 − ε) + μ1 + μ2]λμ1(1 − r)(1 − ε) + μ1μ2

an

+ λ2

λμ1(1 − r)(1 − ε) + μ1μ2an−1, n ≥ 1. (2.43)

(2.6), (2.13), (2.22), (2.28), (2.15) and (2.16) give

a0 = (1 − r)μ1

∞∫0

p(1)1 (x)dx + μ2 p(2)

1 + λQ

= (1 − r)μ1

∞∫0

p(1)1 (x)dx + λμ2

λ − λε + μ2p(2)

0

+ rμ1μ2

λ − λε + μ2

∞∫0

p(1)1 (x)dx + λQ

=[(1 − r)μ1 + rμ1μ2

λ − λε + μ2

] ∞∫0

p(1)1 (x)dx + λμ2

λ − λε + μ2p(2)

0 + λQ

= (1 − r)μ1(λ − λε + μ2) + rμ1μ2

λ − λε + μ2

∞∫0

p(1)1 (x)dx

+ λμ2

λ − λε + μ2p(2)

0 + λQ

= μ1(λ − λε + μ2) − rμ1[λ − λε + μ2 − μ2]λ − λε + μ2

∞∫0

p(1)1 (x)dx

+ λμ2

λ − λε + μ2p(2)

0 + λQ

= μ1(λ − λε + μ2) − λrμ1(1 − ε)

λ − λε + μ2

∞∫0

p(1)1 (x)dx

+ λμ2

λ − λε + μ2p(2)

0 + λQ


= λμ1(1 − ε) + μ1μ2 − λrμ1(1 − ε)

λ − λε + μ2

∞∫0

p(1)1 (x)dx

+ λμ2

λ − λε + μ2p(2)

0 + λQ

= λμ1(1 − r)(1 − ε) + μ1μ2

λ − λε + μ2

∞∫0

p(1)1 (x)dx + λμ2

λ − λε + μ2p(2)

0 + λQ

= λμ1(1 − r)(1 − ε) + μ1μ2

λ − λε + μ2

[λ

(λ − λε + μ1)2 a0 + 1

λ − λε + μ1a1

]

+ λμ2

λ − λε + μ2p(2)

0 + λQ

= λ[λμ1(1 − r)(1 − ε) + μ1μ2](λ − λε + μ2)(λ − λε + μ1)2 a0

+ λμ1(1 − r)(1 − ε) + μ1μ2

(λ − λε + μ1)(λ − λε + μ2)a1 + λμ2

λ − λε + μ2p(2)

0 + λQ

⇒λμ1(1 − r)(1 − ε) + μ1μ2

(λ − λε + μ1)(λ − λε + μ2)a1

={

1 − λ[λμ1(1 − r)(1 − ε) + μ1μ2](λ − λε + μ2)(λ − λε + μ1)2

}a0

− λμ2

λ − λε + μ2p(2)

0 − λQ

= (λ − λε + μ2)(λ − λε + μ1)2 − λ[λμ1(1 − r)(1 − ε) + μ1μ2]

(λ − λε + μ2)(λ − λε + μ1)2 a0

− λμ2

λ − λε + μ2p(2)

0 − λQ

⇒

a1 = (λ − λε + μ2)(λ − λε + μ1)2 − λ[λμ1(1 − r)(1 − ε) + μ1μ2]

(λ − λε + μ1)[λμ1(1 − r)(1 − ε) + μ1μ2] a0

− λμ2(λ − λε + μ1)

λμ1(1 − r)(1 − ε) + μ1μ2p(2)

0

−λ(λ − λε + μ1)(λ − λε + μ2)

λμ1(1 − r)(1 − ε) + μ1μ2Q

= (λ − λε + μ2)(λ − λε + μ1)2 − λ[λμ1(1 − r)(1 − ε) + μ1μ2]

λμ1(1 − r)(1 − ε) + μ1μ2

G. Gupur

× (λ − λε)(λ − λε + μ2)

μ1[rμ2 + (1 − r)(λ − λε + μ2)] Q

− λμ2r(λ − λε + μ1)(λ − λε)

[λμ1(1 − r)(1 − ε) + μ1μ2][rμ2 + (1 − r)(λ − λε + μ2)] Q

−λ(λ − λε + μ1)(λ − λε + μ2)

λμ1(1 − r)(1 − ε) + μ1μ2Q

={

(λ − λε + μ2)(λ − λε + μ1)2 − λ[λμ1(1 − r)(1 − ε) + μ1μ2]

λμ1(1 − r)(1 − ε) + μ1μ2

× (λ − λε)(λ − λε + μ2)

μ1[rμ2 + (1 − r)(λ − λε + μ2)]

− λμ2r(λ − λε + μ1)(λ − λε)

[λμ1(1 − r)(1 − ε) + μ1μ2][rμ2 + (1 − r)(λ − λε + μ2)]

−λ(λ − λε + μ1)(λ − λε + μ2)

λμ1(1 − r)(1 − ε) + μ1μ2

}Q. (2.44)

By combining (2.29) with (2.44) and (2.16) we derive

a1 = μ1[(1 − r)μ1 − μ2]μ1 − μ2

[λ2

(λ − λε + μ1)3 a0 + λ

(λ − λε + μ1)2 a1

+ 1

λ − λε + μ1a2

]+ rμ1μ2

μ1 − μ2

[λ2

(λ − λε + μ2)3 a0

+ λ

(λ − λε + μ2)2 a1 + 1

λ − λε + μ2a2

]

=[

λ2μ1[(1 − r)μ1 − μ2](μ1 − μ2)(λ − λε + μ1)3 + λ2rμ1μ2

(μ1 − μ2)(λ − λε + μ2)3

]a0

+[

λμ1[(1 − r)μ1 − μ2](μ1 − μ2)(λ − λε + μ1)2 + λrμ1μ2

(μ1 − μ2)(λ − λε + μ2)2

]a1

+[

μ1[(1 − r)μ1 − μ2](μ1 − μ2)(λ − λε + μ1)

+ rμ1μ2

(μ1 − μ2)(λ − λε + μ2)

]a2

= λ2μ1[(1 − r)μ1 − μ2](λ − λε + μ2)3 + λ2rμ1μ2(λ − λε + μ1)

3

(μ1 − μ2)(λ − λε + μ1)3(λ − λε + μ2)3 a0

+λμ1[(1 − r)μ1 − μ2](λ − λε + μ2)2 + λrμ1μ2(λ − λε + μ1)

2

(μ1 − μ2)(λ − λε + μ1)2(λ − λε + μ2)2 a1

+μ1[(1 − r)μ1 − μ2](λ − λε + μ2) + rμ1μ2(λ − λε + μ1)

(μ1 − μ2)(λ − λε + μ1)(λ − λε + μ2)a2



3

(μ1 − μ2)(λ − λε + μ1)3(λ − λε + μ2)3 a0


2

(μ1 − μ2)(λ − λε + μ1)2(λ − λε + μ2)2 a1

+ a2

(μ1 − μ2)(λ − λε + μ1)(λ − λε + μ2)

{μ1(μ1 − μ2)(λ − λε + μ2)

−rμ1[μ1(λ − λε + μ2) − μ2(λ − λε + μ1)]}


3

(μ1 − μ2)(λ − λε + μ1)3(λ − λε + μ2)3 a0


2

(μ1 − μ2)(λ − λε + μ1)2(λ − λε + μ2)2 a1

+μ1(μ1 − μ2)(λ − λε + μ2) − rμ1(λ − λε)(μ1 − μ2)

(μ1 − μ2)(λ − λε + μ1)(λ − λε + μ2)a2


3

(μ1 − μ2)(λ − λε + μ1)3(λ − λε + μ2)3 a0


2

(μ1 − μ2)(λ − λε + μ1)2(λ − λε + μ2)2 a1

+μ1(λ − λε + μ2) − rμ1(λ − λε)

(λ − λε + μ1)(λ − λε + μ2)a2

⇒λμ1(1 − r)(1 − ε) + μ1μ2

(λ − λε + μ1)(λ − λε + μ2)a2

=[

1− λμ1[(1 − r)μ1−μ2](λ − λε+μ2)2+λrμ1μ2(λ − λε + μ1)

2

(μ1 − μ2)(λ − λε + μ1)2(λ − λε + μ2)2

]a1

−λ2μ1[(1 − r)μ1 − μ2](λ − λε + μ2)3 + λ2rμ1μ2(λ − λε + μ1)

3

(μ1 − μ2)(λ − λε + μ1)3(λ − λε + μ2)3 a0

= 1

(μ1 − μ2)(λ − λε + μ1)2(λ − λε + μ2)2

×{(μ1 − μ2)(λ − λε + μ1)

2(λ − λε + μ2)2

−λμ1[(1 − r)μ1 − μ2](λ − λε + μ2)2 − λrμ1μ2(λ − λε + μ1)

2}

a1

−λ2μ1[(1 − r)μ1 − μ2](λ − λε + μ2)3 + λ2rμ1μ2(λ − λε + μ1)

3

(μ1 − μ2)(λ − λε + μ1)3(λ − λε + μ2)3 a0

G. Gupur

⇒

a2 = 1

(μ1 − μ2)(λ − λε + μ1)(λ − λε + μ2)[λμ1(1 − r)(1 − ε) + μ1μ2]

×{(μ1 − μ2)(λ − λε + μ1)

2(λ − λε + μ2)2


2}

a1

− λ2μ1[(1−r)μ1−μ2](λ − λε + μ2)3+λ2rμ1μ2(λ−λε+μ1)

3

(μ1−μ2)(λ−λε+μ1)2(λ−λε+μ2)2[λμ1(1 − r)(1 − ε) + μ1μ2]a0

= 1


×{(μ1 − μ2)(λ − λε + μ1)

2(λ − λε + μ2)2


2}

×{

(λ − λε + μ2)(λ − λε + μ1)2 − λ[λμ1(1 − r)(1 − ε) + μ1μ2]

λμ1(1 − r)(1 − ε) + μ1μ2

× (λ − λε)(λ − λε + μ2)

μ1[rμ2 + (1 − r)(λ − λε + μ2)]

− λμ2r(λ − λε + μ1)(λ − λε)

[λμ1(1 − r)(1 − ε) + μ1μ2][rμ2 + (1 − r)(λ − λε + μ2)]

−λ(λ − λε + μ1)(λ − λε + μ2)

λμ1(1 − r)(1 − ε) + μ1μ2

}Q

− λ2μ1[(1 − r)μ1 − μ2](λ − λε + μ2)3 + λ2rμ1μ2(λ − λε + μ1)

3


× λ − λε

μ1[rμ2 + (1 − r)(λ − λε + μ2)] Q. (2.45)

If we introduce

H =

⎛⎜⎜⎝

(λ−λε+μ1)(λ−λε+μ2)+λμ1(1−r)λμ1(1−r)(1−ε)+μ1μ2

− λ[2λ(1−ε)+μ1+μ2]λμ1(1−r)(1−ε)+μ1μ2

λ2

λμ1(1−r)(1−ε)+μ1μ2

1 0 0

0 1 0

⎞⎟⎟⎠,


then (2.43) is equivalent to

⎛⎝an+2

an+1an

⎞⎠ = H

⎛⎝an+1

an

an−1

⎞⎠= H2

⎛⎝ an

an−1an−2

⎞⎠= H3

⎛⎝an−1

an−2an−3

⎞⎠ = · · ·= Hn

⎛⎝a2

a1a0

⎞⎠ , n ≥ 1.

(2.46)

In the following we first determine Hn by discussing the characteristic equation of H,

then we calculate an+1. The characteristic equation of H is

|γ I − H | = 0

⇒γ 2[γ − (λ − λε + μ1)(λ − λε + μ2) + λμ1(1 − r)

λμ1(1 − r)(1 − ε) + μ1μ2

]

− λ2

λμ1(1 − r)(1 − ε) + μ1μ2+ λ[2λ(1 − ε) + μ1 + μ2]

λμ1(1 − r)(1 − ε) + μ1μ2γ = 0

⇒γ 3 − (λ − λε + μ1)(λ − λε + μ2) + λμ1(1 − r)

λμ1(1 − r)(1 − ε) + μ1μ2γ 2

+ λ[2λ(1 − ε) + μ1 + μ2]λμ1(1 − r)(1 − ε) + μ1μ2

γ − λ2

λμ1(1 − r)(1 − ε) + μ1μ2= 0. (2.47)

By comparing with Theorem 1.3 we find

a = − (λ − λε + μ1)(λ − λε + μ2) + λμ1(1 − r)

λμ1(1 − r)(1 − ε) + μ1μ2,

b = λ[2λ(1 − ε) + μ1 + μ2]λμ1(1 − r)(1 − ε) + μ1μ2

,

c = − λ2

λμ1(1 − r)(1 − ε) + μ1μ2,

� = 18abc − 4a3c + a2b2 − 4b3 − 27c2

= 18λ3[2λ(1 − ε) + μ1 + μ2][λμ1(1 − r)(1 − ε) + μ1μ2]3

×[(λ − λε + μ1)(λ − λε + μ2) + λμ1(1 − r)]

−4λ2[(λ − λε + μ1)(λ − λε + μ2) + λμ1(1 − r)]2

[λμ1(1 − r)(1 − ε) + μ1μ2]4

+ λ2[2λ(1 − ε) + μ1 + μ2]2

[λμ1(1 − r)(1 − ε) + μ1μ2]4

×[(λ − λε + μ1)(λ − λε + μ2) + λμ1(1 − r)]2

G. Gupur

− 4λ3[2λ(1 − ε) + μ1 + μ2]3

[λμ1(1 − r)(1 − ε) + μ1μ2]3 − 27λ4

[λμ1(1 − r)(1 − ε) + μ1μ2]2

= λ2

[λμ1(1 − r)(1 − ε) + μ1μ2]4

{18λ[2λ(1 − ε) + μ1 + μ2][(λ − λε + μ1)(λ − λε + μ2) + λμ1(1 − r)]

×[λμ1(1 − r)(1−ε)+μ1μ2]−4[(λ − λε+μ1)(λ − λε+μ2) + λμ1(1 − r)]3

+[2λ(1 − ε) + μ1 + μ2]2[(λ − λε + μ1)(λ − λε + μ1) + λμ1(1 − r)]2

−4λ[2λ(1 − ε) + μ1 + μ2]3[λμ1(1 − r)(1 − ε) + μ1μ2]

−27λ2[λμ1(1 − r)(1 − ε) + μ1μ2]2}

= λ2

[λμ1(1 − r)(1 − ε) + μ1μ2]4

×{

2λ[2λ(1 − ε) + μ1 + μ2][λμ1(1 − r)(1 − ε) + μ1μ2]

×{

9(λ − λε + μ1)(λ − λε + μ2) + 9λμ1(1 − r)−2[2λ(1 − ε) + μ1 + μ2]2}

+[(λ − λε + μ1)(λ − λε + μ1) + λμ1(1 − r)]2

×{[2λ(1 − ε) + μ1 + μ2]2 − 4[(λ − λε + μ1)(λ + λε + μ2) + λμ1(1 − r)]

}

−27λ2[λμ1(1 − r)(1 − ε) + μ1μ2]2}

= λ2

[λμ1(1 − r)(1 − ε) + μ1μ2]4

×{

2λ[2λ(1 − ε) + μ1 + μ2][λμ1(1 − r)(1 − ε) + μ1μ2]

×{

9[λ(1 − ε) + μ1][λ(1 − ε) + μ2] + 9λμ1(1 − r)

−2[4λ2(1 − ε)2 + 4λμ1(1 − ε) + 4λμ2(1 − ε) + 2μ1μ2 + μ2

1 + μ22

]}

+[(λ − λε + μ1)(λ − λε + μ1) + λμ1(1 − r)]2

×{

4λ2(1 − ε)2 + 4λμ1(1 − ε) + 4λμ2(1 − ε)

+2μ1μ2 + μ21 + μ2

2 − 4[λ(1 − ε) + μ1][λ(1 − ε) + μ2] − 4λμ1(1 − r)}

−27λ2[λμ1(1 − r)(1 − ε) + μ1μ2]2}


= λ2

[λμ1(1 − r)(1 − ε) + μ1μ2]4

×{

2λ[2λ(1 − ε) + μ1 + μ2][λμ1(1 − r)(1 − ε) + μ1μ2]

×{

9λ2(1 − ε)2 + 9λμ1(1 − ε) + 9λμ2(1 − ε) + 9μ1μ2 + 9λμ1(1 − r)

−8λ2(1 − ε)2 − 8λμ1(1 − ε) − 8λμ2(1 − ε) − 4μ1μ2 − 2μ21 − 2μ2

2

}

+[(λ − λε + μ1)(λ − λε + μ1) + λμ1(1 − r)]2

×{

4λ2(1 − ε)2 + 4λμ1(1 − ε) + 4λμ2(1 − ε) + 2μ1μ2 + μ21 + μ2

2

−4λ2(1 − ε)2 − 4λμ1(1 − ε) − 4λμ2(1 − ε) − 4μ1μ2 − 4λμ1(1 − r)}

−27λ2[λμ1(1 − r)(1 − ε) + μ1μ2]2}

= λ2

[λμ1(1 − r)(1 − ε) + μ1μ2]4

×{

2λ[2λ(1 − ε)+μ1+μ2][λμ1(1−r)(1 − ε)+μ1μ2]

×{λ2(1 − ε)2 + λμ1(1−ε)+λμ2(1 − ε)+5μ1μ2+9λμ1(1−r)−2μ2

1−2μ22

}

+[(λ − λε + μ1)(λ − λε + μ1) + λμ1(1 − r)]2

×[μ2

1 − 2μ1μ2 + μ22 − 4λμ1(1 − r)

]− 27λ2[λμ1(1 − r)(1 − ε) + μ1μ2]2

}

= λ2

[λμ1(1 − r)(1 − ε) + μ1μ2]4

×{

2λ[2λ(1 − ε) + μ1 + μ2][λμ1(1 − r)(1 − ε) + μ1μ2]

×{λ2(1 − ε)2 + λμ1(1 − ε) + λμ2(1 − ε)

+5μ1μ2 + 9λμ1(1 − r) − 2μ21 − 2μ2

2

}

+(μ1 − μ2)2[(λ − λε + μ1)(λ − λε + μ1) + λμ1(1 − r)]2

−4λμ1(1 − r)[(λ − λε + μ1)(λ − λε + μ1) + λμ1(1 − r)]2

−λ[λμ1(1 − r)(1 − ε) + μ1μ2][27λ2μ1(1 − r)(1 − ε) + 27λμ1μ2]}

G. Gupur

= λ2

[λμ1(1 − r)(1 − ε) + μ1μ2]4

×{λ[λμ1(1 − r)(1 − ε) + μ1μ2]

×[4λ3(1 − ε)3 + 4λ2μ1(1 − ε)2 + 4λ2μ2(1 − ε)2 + 20λμ1μ2(1 − ε)

+36λ2μ1(1 − r)(1 − ε) − 8λμ21(1 − ε) − 8λμ2

2(1 − ε) + 2λ2μ1(1 − ε)2

+2λμ21(1 − ε) + 2λμ1μ2(1 − ε) + 10μ2

1μ2 + 18λμ21(1 − r) − 4μ3

1 − 4μ1μ22

+2λ2μ2(1 − ε)2 + 2λμ1μ2(1 − ε) + 2λμ22(1 − ε) + 10μ1μ

22

+18λμ1μ2(1 − r) − 4μ21μ2 − 4μ3

2 − 27λ2μ1(1 − r)(1 − ε) − 27λμ1μ2

]

+(μ1 − μ2)2[(λ − λε + μ1)(λ − λε + μ1) + λμ1(1 − r)]2

−4λμ1(1 − r)[(λ − λε + μ1)(λ − λε + μ1) + λμ1(1 − r)]2}

= λ2

[λμ1(1 − r)(1 − ε) + μ1μ2]4

×{λ[λμ1(1 − r)(1 − ε) + μ1μ2]

×[4λ3(1 − ε)3 + 6λ2μ1(1 − ε)2 + 6λ2μ2(1 − ε)2 + 24λμ1μ2(1 − ε)

+9λ2μ1(1 − r)(1 − ε) − 6λμ21(1 − ε) − 6λμ2

2(1 − ε) + 6μ21μ2

+18λμ21(1 − r) − 4μ3

1 + 6μ1μ22 + 18λμ1μ2(1 − r) − 4μ3

2 − 27λμ1μ2

]

+(μ1 − μ2)2[(λ − λε + μ1)(λ − λε + μ1) + λμ1(1 − r)]2

−4λμ1(1 − r){[λ(1 − ε) + μ1][λ(1 − ε) + μ1] + λμ1(1 − r)

}2}

= λ2

[λμ1(1 − r)(1 − ε) + μ1μ2]4

×{λ[4λ4μ1(1−r)(1 − ε)4+6λ3μ2

1(1−r)(1−ε)3+6λ3μ1μ2(1 − r)(1 − ε)3

+24λ2μ21μ2(1 − r)(1 − ε)2+9λ3μ2

1(1 − r)2(1−ε)2−6λ2μ31(1 − r)(1 − ε)2

−6λ2μ1μ22(1 − r)(1 − ε)2 + 6λμ3

1μ2(1−r)(1 − ε)+18λ2μ31(1 − r)2(1 − ε)

−4λμ41(1 − r)(1 − ε) + 6λμ2

1μ22(1 − r)(1 − ε) + 18λ2μ2

1μ2(1 − r)2(1 − ε)

−4λμ1μ32(1 − r)(1 − ε) − 27λ2μ2

1μ2(1 − r)(1 − ε) + 4λ3μ1μ2(1 − ε)3

+6λ2μ21μ2(1 − ε)2 + 6λ2μ1μ

22(1 − ε)2 + 24λμ2

1μ22(1 − ε)


+9λ2μ21μ2(1 − r)(1 − ε) − 6λμ3

1μ2(1 − ε) − 6λμ1μ32(1 − ε) + 6μ3

1μ22

+18λμ31μ2(1−r)−4μ4

1μ2+6μ21μ

32+18λμ2

1μ22(1−r)−4μ1μ

42−27λμ2

1μ22

]

+(μ1 − μ2)2[(λ − λε + μ1)(λ − λε + μ1) + λμ1(1 − r)]2 − 4λμ1(1 − r)

×[λ2(1 − ε)2 + λμ1(1 − ε) + λμ2(1 − ε) + μ1μ2 + λμ1(1 − r)

]2}

= λ2

[λμ1(1 − r)(1 − ε) + μ1μ2]4

×{λ[4λ4μ1(1−r)(1−ε)4+6λ3μ2

1(1−r)(1−ε)3+6λ3μ1μ2(1−r)(1 − ε)3

+24λ2μ21μ2(1−r)(1 − ε)2+9λ3μ2

1(1−r)2(1 − ε)2−6λ2μ31(1 − r)(1 − ε)2

−6λ2μ1μ22(1 − r)(1 − ε)2+6λμ3

1μ2(1−r)(1 − ε)+18λ2μ31(1 − r)2(1 − ε)

−4λμ41(1 − r)(1 − ε) + 6λμ2

1μ22(1 − r)(1 − ε) + 18λ2μ2

1μ2(1 − r)2(1 − ε)

−4λμ1μ32(1 − r)(1 − ε) − 18λ2μ2

1μ2(1 − r)(1 − ε) + 4λ3μ1μ2(1 − ε)3

+6λ2μ21μ2(1 − ε)2 + 6λ2μ1μ

22(1 − ε)2 + 24λμ2

1μ22(1 − ε)

−6λμ31μ2(1 − ε) − 6λμ1μ

32(1 − ε) + 6μ3

1μ22 + 18λμ3

1μ2(1 − r)

−4μ41μ2 + 6μ2

1μ32 + 18λμ2

1μ22(1 − r) − 4μ1μ

42 − 27λμ2

1μ22

]

+(μ1 − μ2)2[(λ − λε + μ1)(λ − λε + μ1) + λμ1(1 − r)]2

−4λμ1(1 − r)[λ4(1 − ε)4 + λ2μ2

1(1 − ε)2 + λ2μ22(1 − ε)2 + μ2

1μ22

+λ2μ21(1 − r)2 + 2λ3μ1(1 − ε)3 + 2λ3μ2(1 − ε)3 + 2λ2μ1μ2(1 − ε)2

+2λ3μ1(1 − r)(1 − ε)2 + 2λ2μ1μ2(1 − ε)2 + 2λμ21μ2(1 − ε)

+2λ2μ21(1 − r)(1 − ε) + 2λμ1μ

22(1 − ε) + 2λ2μ1μ2(1 − r)(1 − ε)

+2λμ21μ2(1 − r)

]}

= λ2

[λμ1(1 − r)(1 − ε) + μ1μ2]4

×{λ[4λ4μ1(1 − r)(1 − ε)4 + 6λ3μ2

1(1 − r)(1 − ε)3

+6λ3μ1μ2(1−r)(1−ε)3+24λ2μ21μ2(1−r)(1−ε)2+9λ3μ2

1(1 − r)2(1−ε)2

−6λ2μ31(1 − r)(1 − ε)2 − 6λ2μ1μ

22(1 − r)(1 − ε)2 + 6λμ3

1μ2(1 − r)(1 − ε)

+18λ2μ31(1 − r)2(1 − ε) − 4λμ4

1(1 − r)(1 − ε) + 6λμ21μ

22(1 − r)(1 − ε)

+18λ2μ21μ2(1 − r)2(1 − ε)−4λμ1μ

32(1−r)(1−ε)−18λ2μ2

1μ2(1−r)(1−ε)

G. Gupur

+4λ3μ1μ2(1 − ε)3 + 6λ2μ21μ2(1 − ε)2 + 6λ2μ1μ

22(1 − ε)2

+24λμ21μ

22(1 − ε) − 6λμ3

1μ2(1 − ε) − 6λμ1μ32(1 − ε) + 6μ3

1μ22

+18λμ31μ2(1−r)−4μ4

1μ2+6μ21μ

32+18λμ2

1μ22(1−r)−4μ1μ

42−27λμ2

1μ22

]

+(μ1 − μ2)2[(λ − λε + μ1)(λ − λε + μ1) + λμ1(1 − r)]2

−λ[4λ4μ1(1−r)(1−ε)4+4λ2μ3

1(1−r)(1 − ε)2 + 4λ2μ1μ22(1 − r)(1 − ε)2

+4μ31μ

22(1 − r) + 4λ2μ3

1(1 − r)3 + 8λ3μ21(1 − r)(1 − ε)3

+8λ3μ1μ2(1 − r)(1 − ε)3+8λ2μ21μ2(1−r)(1−ε)2+8λ3μ2

1(1 − r)2(1−ε)2

+8λ2μ21μ2(1 − r)(1 − ε)2 + 8λμ3

1μ2(1 − r)(1 − ε) + 8λ2μ31(1 − r)2(1 − ε)

+8λμ21μ

22(1 − r)(1 − ε) + 8λ2μ2

1μ2(1 − r)2(1 − ε) + 8λμ31μ2(1 − r)2

]}

= λ2

[λμ1(1 − r)(1 − ε) + μ1μ2]4

×{λ[4λ4μ1(1−r)(1−ε)4+6λ3μ2

1(1−r)(1−ε)3+6λ3μ1μ2(1 − r)(1 − ε)3

+24λ2μ21μ2(1 − r)(1 − ε)2+9λ3μ2

1(1 − r)2(1 − ε)2 − 6λ2μ31(1 − r)(1 − ε)2

−6λ2μ1μ22(1 − r)(1 − ε)2 + 6λμ3

1μ2(1 − r)(1 − ε) + 18λ2μ31(1 − r)2(1 − ε)

−4λμ41(1 − r)(1 − ε) + 6λμ2

1μ22(1 − r)(1 − ε) + 18λ2μ2

1μ2(1 − r)2(1 − ε)

−4λμ1μ32(1 − r)(1 − ε) − 18λ2μ2

1μ2(1 − r)(1 − ε) + 4λ3μ1μ2(1 − ε)3

+6λ2μ21μ2(1 − ε)2 + 6λ2μ1μ

22(1 − ε)2 + 24λμ2

1μ22(1 − ε) − 6λμ3

1μ2(1 − ε)

−6λμ1μ32(1 − ε) + 6μ3

1μ22 + 18λμ3

1μ2(1 − r) − 4μ41μ2 + 6μ2

1μ32

+18λμ21μ

22(1 − r) − 4μ1μ

42 − 27λμ2

1μ22 − 4λ4μ1(1 − r)(1 − ε)4

−4λ2μ31(1 − r)(1 − ε)2 − 4λ2μ1μ

22(1 − r)(1 − ε)2 − 4μ3

1μ22(1 − r)

−4λ2μ31(1 − r)3 − 8λ3μ2

1(1 − r)(1 − ε)3 − 8λ3μ1μ2(1 − r)(1 − ε)3

−8λ2μ21μ2(1 − r)(1 − ε)2−8λ3μ2

1(1 − r)2(1−ε)2−8λ2μ21μ2(1 − r)(1 − ε)2

−8λμ31μ2(1 − r)(1 − ε) − 8λ2μ3

1(1 − r)2(1 − ε) − 8λμ21μ

22(1 − r)(1 − ε)

−8λ2μ21μ2(1 − r)2(1 − ε) − 8λμ3

1μ2(1 − r)2]

+(μ1 − μ2)2[(λ − λε + μ1)(λ − λε + μ1) + λμ1(1 − r)]2

}

= λ2

[λμ1(1 − r)(1 − ε) + μ1μ2]4


×{λ[

− 2λ3μ21(1 − r)(1 − ε)3 − 2λ3μ1μ2(1 − r)(1 − ε)3

+8λ2μ21μ2(1 − r)(1 − ε)2 + λ3μ2

1(1 − r)2(1 − ε)2 − 10λ2μ31(1 − r)(1 − ε)2

−10λ2μ1μ22(1 − r)(1 − ε)2−2λμ3

1μ2(1−r)(1−ε) + 10λ2μ31(1 − r)2(1 − ε)

−4λμ41(1 − r)(1 − ε) − 2λμ2

1μ22(1 − r)(1 − ε) + 10λ2μ2

1μ2(1 − r)2(1 − ε)

−4λμ1μ32(1 − r)(1 − ε) − 18λ2μ2

1μ2(1 − r)(1 − ε) + 4λ3μ1μ2(1 − ε)3

+6λ2μ21μ2(1 − ε)2 + 6λ2μ1μ

22(1 − ε)2 + 24λμ2

1μ22(1 − ε) − 6λμ3

1μ2(1 − ε)

−6λμ1μ32(1 − ε) + 6μ3

1μ22 + 18λμ3

1μ2(1 − r) − 4μ41μ2 + 6μ2

1μ32

+18λμ21μ

22(1 − r) − 4μ1μ

42 − 27λμ2

1μ22 − 4μ3

1μ22(1 − r) − 4λ2μ3

1(1 − r)3

−8λμ31μ2(1−r)2

]+(μ1−μ2)

2[(λ−λε+μ1)(λ−λε+μ1)+λμ1(1−r)]2}

= λ2

[λμ1(1 − r)(1 − ε) + μ1μ2]4

×{λ{ [

4λ3μ1μ2 − 2λ3μ21(1 − r) − 2λ3μ1μ2(1 − r)

](1 − ε)3

+[8λ2μ21μ2(1 − r) + λ3μ2

1(1 − r)2 − 10λ2μ31(1 − r) − 10λ2μ1μ

22(1 − r)

+6λ2μ21μ2 + 6λ2μ1μ

22

](1 − ε)2 + [10λ2μ3

1(1 − r)2 − 2λμ31μ2(1 − r)

−4λμ41(1 − r) − 2λμ2

1μ22(1 − r) + 10λ2μ2

1μ2(1 − r)2 − 4λμ1μ32(1 − r)

−18λ2μ21μ2(1 − r) − 24λμ2

1μ22 − 6λμ3

1μ2 − 6λμ1μ32

](1 − ε) + 6μ3

1μ32

+18λμ31μ2(1 − r) − 4μ4

1μ2 + 6μ21μ

32 + 18λμ2

1μ22(1 − r) − 4μ1μ

42

−27λμ21μ

22 − 4μ3

1μ22(1 − r) − 4λ2μ3

1(1 − r)3 − 8λμ31μ2(1 − r)2

}

+(μ1 − μ2)2[(λ − λε + μ1)(λ − λε + μ1) + λμ1(1 − r)]2

}

= λ2

[λμ1(1 − r)(1 − ε) + μ1μ2]4

×{λ{ [

2λ3μ1μ2(1 + r) − 2λ3μ21(1 − r)

](1 − ε)3

+[2λ2μ21μ2(7 − 4r) + λ3μ2

1(1 − r)2 − 10λ2μ31(1 − r)

+2λ2μ1μ22(5r − 2)

](1 − ε)2 + [10λ2μ3

1(1 − r)2 − 2λμ31μ2(4 − r)

−4λμ41(1 − r) − 2λμ2

1μ22(13 − r) − 2λ2μ2

1μ2(1 − r)(4 + 5r)

−2λμ1μ32(5 − 2r)

](1 − ε) + 6μ3

1μ32 + 2λμ3

1μ2(1 − r)(5 + 4r) − 4μ41μ2

G. Gupur

+6μ21μ

32 − 9λμ2

1μ22(1 + 2r) − 4μ1μ

42 − 4μ3

1μ22(1 − r) − 4λ2μ3

1(1 − r)3}

+(μ1 − μ2)2[(λ − λε + μ1)(λ − λε + μ1) + λμ1(1 − r)]2

}

= λ2

[λμ1(1 − r)(1 − ε) + μ1μ2]4

×{λ{

2λ3μ1[μ2(1 + r) − μ1(1 − r)](1 − ε)3

+λ2μ1[2μ1μ2(7−4r) + λμ1(1 − r)2 − 10μ2

1(1 − r)+2μ22(5r − 2)

](1−ε)2

+λμ1[10λμ2

1(1 − r)2 − 2μ21μ2(4 − r) − 4μ3

1(1 − r) − 2μ1μ22(13 − r)

−2λμ1μ2(1 − r)(4 + 5r) − 2μ32(5 − 2r)

](1 − ε)

+μ1[6μ2

1μ32 + 2λμ2

1μ2(1 − r)(5 + 4r) − 4μ31μ2 + 6μ1μ

32

−9λμ1μ22(1 + 2r) − 4μ4

2 − 4μ21μ

22(1 − r) − 4λ2μ2

1(1 − r)3]}

+(μ1 − μ2)2[(λ − λε + μ1)(λ − λε + μ1) + λμ1(1 − r)]2

}

= λ2

[λμ1(1 − r)(1 − ε) + μ1μ2]4

×{λμ1

{2λ3[μ2(1 + r) − μ1(1 − r)](1 − ε)3

+λ2[2μ1μ2(7 − 4r) + λμ1(1 − r)2 − 10μ21(1 − r) + 2μ2

2(5r − 2)](1 − ε)2

+λ[10λμ2

1(1 − r)2 − 2μ21μ2(4 − r) − 4μ3

1(1 − r) − 2μ1μ22(13 − r)

−2λμ1μ2(1 − r)(4 + 5r) − 2μ32(5 − 2r)

](1 − ε) + 6μ2

1μ32

+2λμ21μ2(1 − r)(5 + 4r) − 4μ3

1μ2 + 6μ1μ32

−9λμ1μ22(1 + 2r) − 4μ4

2 − 4μ21μ

22(1 − r) − 4λ2μ2

1(1 − r)3}

+(μ1 − μ2)2[(λ − λε + μ1)(λ − λε + μ1) + λμ1(1 − r)]2

}. (2.48)

In the following, by Theorem 1.3 and (2.48) we discuss the roots of the equation (2.47)and therefore an+1. 1. If � > 0, then the equation (2.47) has three distinct real roots,say γ1, γ2, γ2. From the related knowledge of higher algebra we know that there existsan invertible matrix


L =⎛⎝γ 2

1 γ 22 γ 2

3γ1 γ2 γ31 1 1

⎞⎠ ,

L−1 = 1

(γ1 − γ2)(γ2 − γ3)(γ1 − γ3)

⎛⎜⎜⎝

γ2 − γ3 γ 23 − γ 2

2 γ2γ3(γ2 − γ3)

γ3 − γ1 γ 21 − γ 2

3 γ1γ3(γ3 − γ1)

γ1 − γ2 γ 22 − γ 2

1 γ1γ2(γ1 − γ2)

⎞⎟⎟⎠

such that

L−1 H L =⎛⎝γ1 0 0

0 γ2 00 0 γ3

⎞⎠ .

Which implies

H = L

⎛⎝γ1 0 0

0 γ2 00 0 γ3

⎞⎠ L−1

⇒

Hn = L

⎛⎝γ1 0 0

0 γ2 00 0 γ3

⎞⎠

n

L−1 = L

⎛⎝γ n

1 0 00 γ n

2 00 0 γ n

3

⎞⎠ L−1. (2.49)

This together with (2.46) give

an+2 = 1

(γ1 − γ2)(γ2 − γ3)(γ1 − γ3)

×{[

γ n+21 (γ2 − γ3) + γ n+2

2 (γ3 − γ1) + γ n+23 (γ1 − γ2)

]a2

+[γ n+2

1 (γ 23 − γ 2

2 ) + γ n+22 (γ 2

1 − γ 23 ) + γ n+2

3 (γ 22 − γ 2

1 )]a1

+[γ n+2

1 γ2γ3(γ2 − γ3) + γ n+22 γ1γ3(γ3 − γ1)

+γ n+23 γ1γ2(γ1 − γ2)

]a0

}

= 1

(γ1 − γ2)(γ2 − γ3)(γ1 − γ3)

×{γ n+2

1

[(γ2 − γ3)a2 + (γ 2

3 − γ 22 )a1 + γ2γ3(γ2 − γ3)a0

]

+γ n+22

[(γ3 − γ1)a2 + (γ 2

1 − γ 23 )a1 + γ1γ3(γ3 − γ1)a0

]

+γ n+23

[(γ1 − γ2)a2 + (γ 2

2 − γ 21 )a1 + γ1γ2(γ1 − γ2)a0

]}, n ≥ 1

⇒

G. Gupur

∞∑n=0

|an| = |a0| + |a1| + |a2| +∞∑

n=1

|an+2|

≤ |a0| + |a1| + |a2| + 1

|(γ1 − γ2)(γ2 − γ3)(γ1 − γ3)|

×{[

|γ2−γ3||a2|+∣∣∣γ 2

3 − γ 22

∣∣∣ |a1|+|γ2γ3(γ2−γ3)| |a0|] ∞∑

n=1

|γ1|n+2

+[|γ3 − γ1||a2| +

∣∣∣γ 21 − γ 2

3

∣∣∣ |a1| + |γ1γ3(γ3 − γ1)||a0|] ∞∑

n=1

|γ2|n+2

+[|γ1−γ2||a2|+

∣∣∣γ 22 − γ 2

1

∣∣∣ |a1|+|γ1γ2(γ1−γ2)||a0|] ∞∑

n=1

|γ3|n+2}.

(2.50)

2. If � < 0, then the Eq. (2.47) has one real root and two complex conjugate roots,say γ1, γ2, γ3 (γ 2 = γ3), that is to say, these three roots are different from each other.So, it is the same as the former step, we construct L and obtain (2.50) again.3. If � = 0, then we discuss the roots of the equation (2.47) by a2 − 3b.

a2 − 3b = [(λ − λε + μ1)(λ − λε + μ2) + λμ1(1 − r)]2

[λμ1(1 − r)(1 − ε) + μ1μ2]2

− 3λ[2λ(1 − ε) + μ1 + μ2]λμ1(1 − r)(1 − ε) + μ1μ2

= 1

[λμ1(1 − r)(1 − ε) + μ1μ2]2

×{[(λ − λε + μ1)(λ − λε + μ2) + λμ1(1 − r)]2

−3λ[2λ(1 − ε) + μ1 + μ2][λμ1(1 − r)(1 − ε) + μ1μ2]}

= 1

[λμ1(1 − r)(1 − ε) + μ1μ2]2

×{{

[λ(1 − ε) + μ1][λ(1 − ε) + μ2] + λμ1(1 − r)}2

−3λ[2λ2μ1(1 − r)(1 − ε)2 + 2λμ1μ2(1 − ε)

+λμ21(1 − r)(1 − ε) + μ2

1μ2

+λμ1μ2(1 − r)(1 − ε) + μ1μ22

]}

= 1

[λμ1(1 − r)(1 − ε) + μ1μ2]2

×{[

λ2(1 − ε)2 + λμ1(1 − ε) + λμ2(1 − ε)


−λ2μ21(1 − r)(1 − ε) + 2λμ1μ

22(1 − ε)

−λ2μ1μ2(1 − r)(1 − ε) + 2λμ21μ2(1 − r)

+μ1μ2 + λμ1(1 − r)]2 − 6λ3μ1(1 − r)(1 − ε)2

−6λ2μ1μ2(1 − ε) − 3λ2μ21(1 − r)(1 − ε)

−3λμ21μ2 − 3λ2μ1μ2(1 − r)(1 − ε) − 3λμ1μ

22

}

= 1

[λμ1(1 − r)(1 − ε) + μ1μ2]2

×{λ4(1 − ε)4 + λ2μ2

1(1 − ε)2 + λ2μ22(1 − ε)2

+μ21μ

22 + λ2μ2

1(1 − r)2 + 2λ3μ1(1 − ε)3

+2λ3μ2(1 − ε)3 + 2λ2μ1μ2(1 − ε)2

+2λ3μ1(1 − r)(1 − ε)2 + 2λ2μ1μ2(1 − ε)2

+2λμ21μ2(1 − ε) + 2λ2μ2

1(1 − r)(1 − ε)

+2λμ1μ22(1 − ε) + 2λ2μ1μ2(1 − r)(1 − ε)

+2λμ21μ2(1 − r) − 6λ3μ1(1 − r)(1 − ε)2

−6λ2μ1μ2(1 − ε) − 3λ2μ21(1 − r)(1 − ε)

−3λμ21μ2 − 3λ2μ1μ2(1 − r)(1 − ε) − 3λμ1μ

22

}

= 1

[λμ1(1 − r)(1 − ε) + μ1μ2]2

×{λ4(1 − ε)4 + λ2μ2

1(1 − ε)2 + λ2μ22(1 − ε)2

+μ21μ

22 + λ2μ2

1(1 − r)2 + 2λ3μ1(1 − ε)3

+2λ3μ2(1 − ε)3 + 4λ2μ1μ2(1 − ε)2

−4λ3μ1(1 − r)(1 − ε)2 + 2λμ21μ2(1 − ε)

−6λ2μ1μ2(1 − ε) − 3λμ21μ2 − 3λμ1μ

22

}. (2.51)

(1) If a2 − 3b = 0, then the Eq. (2.47) has a multiple real root

γ1 = γ2 = γ3 = (λ − λε + μ1)(λ − λε + μ1) + λμ1(1 − r)

3[λμ1(1 − r)(1 − ε) + μ1μ2] .

By the related knowledge of higher algebra we know that there exists an invertiblematrix

L =⎛⎝1 2γ1 γ 2

10 1 γ10 0 1

⎞⎠

G. Gupur

such that

L−1 H L =⎛⎝γ1 0 0

1 γ1 00 1 γ1

⎞⎠ ⇒ H = L

⎛⎝γ1 0 0

1 γ1 00 1 γ1

⎞⎠ L−1

⇒ Hn = L

⎛⎝γ1 0 0

1 γ1 00 1 γ1

⎞⎠

n

L−1

=⎛⎝1 2γ1 γ 2

10 1 γ10 0 1

⎞⎠⎛⎜⎜⎝

γ n1 0 0

nγ n−11 γ n

1 0

n(n−1)2 γ n−2

1 nγ n−11 γ n

1

⎞⎟⎟⎠

×⎛⎝1 −2γ1 γ 2

10 1 −γ10 0 1

⎞⎠ . (2.52)

(2.52) and (2.46) imply

an+2 = γ n1

[1 + 2n + n(n − 1)

2

]a2 − γ n+1

1 n(n + 2)a1

+γ n+21

[n + n(n − 1)

2

]a0, n ≥ 1

⇒sum∞

n=0|an| = |a0| + |a1| + |a2| +∞∑

n=1

|an+2|

≤ |a0| + |a1| + |a2| + |a2|∞∑

n=1

|γ1|n[

1 + 2n + n(n − 1)

2

]

+|a1|∞∑

n=1

|γ1|n+1n(n + 2) + |a0|∞∑

n=1

|γ1|n+2[

n + n(n − 1)

2

].

(2.53)

(2) If a2 − 3b �= 0, then the Eq. (2.47) has three real roots as follows:

γ1 = 4ab − 9c − a3

a3 − 3b, γ2 = γ3 = 9c − ab

2(a3 − 3b).

By the related knowledge of higher algebra we know that there is an invertible matrix

L =⎛⎝γ 2

1 2γ2 γ 22

γ1 1 γ21 0 1

⎞⎠


L−1 = 1

(γ1 − γ2)2

⎛⎜⎜⎝

1 −2γ2 γ 22

γ2 − γ1 γ 21 − γ 2

2 γ1γ2(γ2 − γ1)

−1 2γ2 γ1(γ1 − 2γ2)

⎞⎟⎟⎠

such that

L−1 H L =⎛⎜⎝

γ1 0 0

0 γ2 0

0 1 γ2

⎞⎟⎠

⇒

H = L

⎛⎜⎝

γ1 0 0

0 γ2 0

0 1 γ2

⎞⎟⎠ L−1

⇒

Hn = L

⎛⎜⎝

γ n1 0 0

0 γ n2 0

0 nγ n−12 γ n

2

⎞⎟⎠

n

L−1

= 1

(γ1 − γ2)2

⎛⎜⎝

γ 21 2γ2 γ 2

2

γ1 1 γ2

1 0 1

⎞⎟⎠⎛⎜⎝

γ n1 0 0

0 γ n2 0

0 nγ n−12 γ n

2

⎞⎟⎠

×⎛⎜⎝

1 −2γ2 γ 22

γ2 − γ1 γ 21 − γ 2

2 γ1γ2(γ2 − γ1)

−1 2γ2 γ1(γ1 − 2γ2)

⎞⎟⎠ . (2.54)

From which together with (2.46) we derive

an+1 =[γ n+2

1 + (γ2 − γ1)(n + 2)γ n+12 − γ n+2

2

]a2

+[

− 2γ2γn+21 + (γ 2

1 − γ 22 )(n + 2)γ n+1

2 + 2γ n+32

]a1

+[γ 2

2 γ n+21 + γ1γ2(γ2 − γ1)(n + 2)γ n+1

2 + γ1(γ1 − 2γ2)γn+22

]a0

⇒∞∑

n=0

|an| = |a0| + |a1| + |a2| +∞∑

n=1

|an+2|

≤ |a0| + |a1| + |a2|

G. Gupur

+ 1

|γ1 − γ2|2[|a2| + 2|γ2||a1| + |γ2|2

] ∞∑n=1

|γ1|n+2

+ 1

|γ1 − γ2|2[|γ2 − γ1||a2| +

∣∣∣γ 21 − γ 2

2

∣∣∣ |a1|

+|γ1γ2(γ2 − γ1)||a0|] ∞∑

n=1

(n + 2)|γ2|n+1

+ 1

|γ1 − γ2|2[|a2| + 2|a1| + |γ1(γ1 − 2γ2)||a0|

] ∞∑n=1

|γ2|n+2.

(2.55)

From (2.55), (2.53) and (2.50) we know that convergence of∑∞

n=0 |an| is decidedby the convergence of

∑∞n=1 |γ j |n ( j = 1, 2, 3), which involve absolute values or

modulus of γ j ( j = 1, 2, 3). Since

f (γ ) := γ 3 − (λ − λε + μ1)(λ − λε + μ2) + λμ1(1 − r)

λμ1(1 − r)(1 − ε) + μ1μ2γ 2

+ λ[2λ(1 − ε) + μ1 + μ2]λμ1(1 − r)(1 − ε) + μ1μ2

γ − λ2

λμ1(1 − r)(1 − ε) + μ1μ2

= γ 3 + aγ 2 + bγ + c

satisfies

f (0) = c < 0, f (1) = 1 + a + b + c > 0 (see (2.57) below),

from the intermediate value theorem one may know that there exists a real root in(0, 1). But, from the general formula for the roots of the equation (2.47) (see Theorem1.3) we know that the estimation of absolute values or modulus of all roots is not easy.Hence, in the following by Theorem 1.4 we will show that absolute values or modulusof all the roots of the Eqs. (2.47) are less than 1. By applying the conditions

λ(μ1 + μ2) < μ1μ2 ⇒ λ

μ1+ λ

μ2< 1 ⇒ λ < μ1, λ < μ2,

r ∈ (0, 1), 0 < ε < 1 − r,

we obtain

1 − a + b − c = 1 + (λ − λε + μ1)(λ − λε + μ2) + λμ1(1 − r)

λμ1(1 − r)(1 − ε) + μ1μ2

+ λ[2λ(1 − ε) + μ1 + μ2]λμ1(1 − r)(1 − ε) + μ1μ2

+ λ2

λμ1(1 − r)(1 − ε) + μ1μ2

> 0. (2.56)


1 + a + b + c = 1 − (λ − λε + μ1)(λ − λε + μ2) + λμ1(1 − r)

λμ1(1 − r)(1 − ε) + μ1μ2

+ λ[2λ(1 − ε) + μ1 + μ2]λμ1(1 − r)(1 − ε) + μ1μ2

− λ2

λμ1(1 − r)(1 − ε) + μ1μ2

= 1

λμ1(1 − r)(1 − ε) + μ1μ2

{λμ1(1 − r)(1 − ε)

+μ1μ2 − [(λ − λε + μ1)(λ − λε + μ2)

+λμ1(1 − r)] + λ[2λ(1 − ε) + μ1 + μ2] − λ2}

= 1

λμ1(1 − r)(1 − ε) + μ1μ2

{λμ1(1 − r)(1 − ε)

+μ1μ2 − [λ(1 − ε) + μ1][λ(1 − ε) + μ2]−λμ1(1 − r) + 2λ2(1 − ε) + λμ1 + λμ2 − λ2

}

= 1

λμ1(1 − r)(1 − ε) + μ1μ2

{λμ1(1 − r)(1 − ε)

+μ1μ2 − λ2(1 − ε)2 − λμ1(1 − ε) − λμ2(1 − ε)

−μ1μ2 − λμ1(1 − r) + 2λ2(1 − ε) + λμ1 + λμ2 − λ2}

= 1

λμ1(1 − r)(1 − ε) + μ1μ2

{λμ1(1 − r)(1 − ε − 1)

−λ2(1 − ε)2 + λμ1ε + λμ2ε + 2λ2(1 − ε) − λ2}

= 1

λμ1(1 − r)(1 − ε) + μ1μ2

{− λμ1(1 − r)ε

−λ2[(1 − ε)2 − 2(1 − ε) + 1] + λμ1ε + λμ2ε}

= 1

λμ1(1 − r)(1 − ε) + μ1μ2

{− λμ1(1 − r)ε

−λ2(1 − ε − 1)2 + λμ1ε + λμ2ε}

= 1

λμ1(1 − r)(1 − ε) + μ1μ2

{− λμ1(1 − r)ε

λ2ε2 + λμ1ε + λμ2ε}

= 1

λμ1(1 − r)(1 − ε) + μ1μ2

{λμ1ε[−(1 − r) + 1]

+λε(μ2 − λε)}

G. Gupur

= 1

λμ1(1 − r)(1 − ε) + μ1μ2

{λμ1εr + λε(μ2 − λε)

}

> 0. (2.57)

3 − a − b + 3c = 3 + (λ − λε + μ1)(λ − λε + μ2) + λμ1(1 − r)

λμ1(1 − r)(1 − ε) + μ1μ2

− λ[2λ(1 − ε) + μ1 + μ2]λμ1(1 − r)(1 − ε) + μ1μ2

− 3λ2

λμ1(1 − r)(1 − ε) + μ1μ2

= 1

λμ1(1 − r)(1 − ε) + μ1μ2

×{

3λμ1(1 − r)(1 − ε) + 3μ1μ2 + (λ − λε + μ1)(λ − λε + μ2)

+λμ1(1 − r) − λ[2λ(1 − ε) + μ1 + μ2] − 3λ2}

= 1

λμ1(1 − r)(1 − ε) + μ1μ2

×{

3λμ1(1−r)(1 − ε)+3μ1μ2+[λ(1−ε)+μ1][λ(1−ε)+μ2]

+λμ1(1 − r) − 2λ2(1 − ε) − λμ1 − λμ2 − 3λ2}

= 1

λμ1(1 − r)(1 − ε) + μ1μ2

×{

3λμ1(1−r)(1−ε)+3μ1μ2+λ2(1−ε)2+λμ1(1−ε)

+λμ2(1−ε)+μ1μ2+λμ1(1−r)−2λ2(1−ε)−λμ1−λμ2−3λ2}

= 1

λμ1(1 − r)(1 − ε) + μ1μ2

×{λμ1(1 − r)(3 − 3ε + 1) + 4μ1μ2

+λ2(1 − ε)(1 − ε − 2) − λμ1ε − λμ2ε − 3λ2}

= 1

λμ1(1 − r)(1 − ε) + μ1μ2

×{λμ1(1 − r)(4 − 3ε) + 4μ1μ2

−λ2(1 − ε)(1 + ε) − λμ1ε − λμ2ε − 3λ2}

= 1

λμ1(1 − r)(1 − ε) + μ1μ2


×{λμ1(1 − r)(4 − 3ε) + 4μ1μ2 + λ2ε2 − λμ1ε − λμ2ε − 4λ2

}

>1

λμ1(1 − r)(1 − ε) + μ1μ2

×{λμ1(1 − r)(4 − 3ε) + 4λμ1 + 4λμ2

+λ2ε2 − λμ1ε − λμ2ε − 4λ2}

= 1

λμ1(1 − r)(1 − ε) + μ1μ2

×{λμ1(1 − r)(4 − 3ε) + λμ1(1 − ε) + λμ2(1 − ε)

+λ2ε2 + 3λμ1 − 3λ2 + 3λμ2 − λ2}

= 1

λμ1(1 − r)(1 − ε) + μ1μ2

×{λμ1(1 − r)(4 − 3ε) + λμ1(1 − ε) + λμ2(1 − ε)

+λ2ε2 + 3λ(μ1 − λ) + 2λμ2 + λ(μ2 − λ)}

> 0. (2.58)

Since the condition

0 < ε < min

{1 − r, 1 −

{2λμ1μ2 − λ2μ2

+√

(λ2μ2 − 2λμ1μ2)2 + 4λμ1μ2[μ1(1 − r) − λ]2(μ1 − λ)}/{

2λ[μ1(1 − r) − λ]2}}

implies

1 − r > ε ⇒ 1 − r − ε > 0, ε ∈ (0, 1),

1 − ε >{

2λμ1μ2 − λ2μ2

+√

(λ2μ2 − 2λμ1μ2)2 + 4λμ1μ2[μ1(1 − r) − λ]2(μ1 − λ)}/{

2λ[μ1(1 − r) − λ]2}

⇒λ[μ1(1 − r) − λ]2(1 − ε)2 + (λ2μ2 − 2λμ1μ2)(1 − ε)

−μ21μ2 + λμ1μ2 > 0,

which together with the conditions

λ(μ1 + μ2) < μ1μ2 ⇒ λ

μ1+ λ

μ2< 1 ⇒ μ1 > λ, μ2 > λ, r ∈ (0, 1)

G. Gupur

give

1 − b + ac − c2 = 1 − λ[2λ(1 − ε) + μ1 + μ2]λμ1(1 − r)(1 − ε) + μ1μ2

+λ2[(λ − λε + μ1)(λ − λε + μ2) + λμ1(1 − r)][λμ1(1 − r)(1 − ε) + μ1μ2]2

− λ4

[λμ1(1 − r)(1 − ε) + μ1μ2]2

= 1

[λμ1(1 − r)(1 − ε) + μ1μ2]2

{[λμ1(1 − r)(1 − ε) + μ1μ2]2

−λ[2λ(1 − ε) + μ1 + μ2][λμ1(1 − r)(1 − ε) + μ1μ2]+λ2[(λ − λε + μ1)(λ − λε + μ2) + λμ1(1 − r)] − λ4

}

= 1

[λμ1(1 − r)(1 − ε) + μ1μ2]2

{λ2μ2

1(1 − r)2(1 − ε)2 + 2λμ21μ2(1 − r)(1 − ε)

+μ21μ

22 − λ

[2λ2μ1(1 − r)(1 − ε)2 + 2λμ1μ2(1 − ε)

+λμ21(1 − r)(1 − ε) + μ2

1μ2 + λμ1μ2(1 − r)(1 − ε)

+μ1μ22

]+ λ2[λ(1 − ε) + μ1][λ(1 − ε) + μ2]

+λ3μ1(1 − r) − λ4}

= 1

[λμ1(1 − r)(1 − ε) + μ1μ2]2

{λ2μ2

1(1 − r)2(1 − ε)2 + 2λμ21μ2(1 − r)(1 − ε)

+μ21μ

22 − 2λ3μ1(1 − r)(1 − ε)2 − 2λ2μ1μ2(1 − ε)

−λ2μ21(1 − r)(1 − ε) − λμ2

1μ2 − λ2μ1μ2(1 − r)(1 − ε)

−λμ1μ22 + λ2[λ2(1 − ε)2 + λμ1(1 − ε)

+λμ2(1 − ε) + μ1μ2]+ λ3μ1(1 − r) − λ4

}

= 1

[λμ1(1 − r)(1 − ε) + μ1μ2]2


{λ2μ2

1(1 − r)2(1 − ε)2 + 2λμ21μ2(1 − r)(1 − ε)

+μ21μ

22 − 2λ3μ1(1 − r)(1 − ε)2 − 2λ2μ1μ2(1 − ε)

−λ2μ21(1 − r)(1 − ε) − λμ2

1μ2 − λ2μ1μ2(1 − r)(1 − ε)

−λμ1μ22 + λ4(1 − ε)2 + λ3μ1(1 − ε)

+λ3μ2(1 − ε) + λ2μ1μ2 + λ3μ1(1 − r) − λ4}

= 1

[λμ1(1 − r)(1 − ε) + μ1μ2]2

{λ2μ2

1(1 − r)2(1 − ε)2 − 2λ3μ1(1 − r)(1 − ε)2

+λ4(1 − ε)2 − 2λ2μ1μ2(1 − ε) + λ3μ2(1 − ε)

−λμ21μ2 + λ2μ1μ2 + 2λμ2

1μ2(1 − r)(1 − ε) + μ21μ

22

−λ2μ21(1 − r)(1 − ε) − λ2μ1μ2(1 − r)(1 − ε)

−λμ1μ22 + λ3μ1(1 − ε) + λ3μ1 − λ3μ1r − λ4

}

= 1

[λμ1(1 − r)(1 − ε) + μ1μ2]2

{λ2[μ2

1(1 − r)2 − 2λμ1(1 − r) + λ2](1 − ε)2

+(−2λ2μ1μ2 + λ3μ2)(1 − ε) − λμ21μ2 + λ2μ1μ2

+λμ21(1 − r)(1 − ε)(μ2 − λ)

+λμ1μ2(1 − r)(1 − ε)(μ1 − λ) + μ1μ22(μ1 − λ)

+λ3μ1(1 − r − ε) + λ3(μ1 − λ)}

= 1

[λμ1(1 − r)(1 − ε) + μ1μ2]2

{λ2[μ1(1 − r) − λ]2(1 − ε)2

+λ(−2λμ1μ2 + λ2μ2)(1 − ε) + λ(−μ21μ2 + λμ1μ2)

+λμ21(1 − r)(1 − ε)(μ2 − λ)

+λμ1μ2(1 − r)(1 − ε)(μ1 − λ) + μ1μ22(μ1 − λ)

+λ3μ1(1 − r − ε) + λ3(μ1 − λ)}

= 1

[λμ1(1 − r)(1 − ε) + μ1μ2]2

G. Gupur

{λ{λ[μ1(1 − r) − λ]2(1 − ε)2

+(−2λμ1μ2 + λ2μ2)(1 − ε) − μ21μ2 + λμ1μ2

}

+λμ21(1 − r)(1 − ε)(μ2 − λ)

+λμ1μ2(1 − r)(1 − ε)(μ1 − λ) + μ1μ22(μ1 − λ)

+λ3μ1(1 − r − ε) + λ3(μ1 − λ)

}

> 0. (2.59)

By combining (2.56)–(2.59) with Theorem 1.4 we know that all the roots of theequation (2.47), say γ1, γ2, γ3, satisfy

|γ1| < 1, |γ2| < 1, |γ3| < 1. (2.60)

This together with the basic knowledge of power series imply

∞∑n=1

|γ j |n < ∞,

∞∑n=1

n|γ j |n < ∞,

∞∑n=1

n2|γ j |n < ∞, j = 1, 2, 3. (2.61)

From which together with (2.16), (2.44), (2.45), (2.50), (2.53) and (2.55) we deduce

∞∑n=0

|an| < ∞. (2.62)

By using (2.20), (2.62), the Cauchy product of series and

λ − λε + μ1 > λ > 0,

∞∫0

xn−ke−(λ−λε+μ1)x dx = (n − k)!(λ − λε + μ1)n+1−k

we derive

∞∫0

∣∣∣p(1)n (x)

∣∣∣ dx =∞∫

0

∣∣∣∣∣n∑

k=0

ak(λx)n−k

(n − k)!e−(λ−λε+μ1)x

∣∣∣∣∣ dx

≤n∑

k=0

|ak | λn−k

(n − k)!∞∫

0

xn−ke−(λ−λε+μ1)x dx


=n∑

k=0

|ak | λn−k

(n − k)!(n − k)!

(λ − λε + μ1)n+1−k

=n∑

k=0

|ak | λn−k

(λ − λε + μ1)n+1−k, n ≥ 0

⇒∞∑

n=0

∥∥∥p(1)n

∥∥∥L1[0,∞)

=∞∑

n=0

∞∫0

∣∣∣p(1)n (x)

∣∣∣ dx

=∞∑

n=0

n∑k=0

|ak | λn−k

(λ − λε + μ1)n+1−k

=∞∑

k=0

∞∑n=k

|ak | λn−k

(λ − λε + μ1)n+1−k

=∞∑

k=0

|ak | 1

λ − λε + μ1

∞∑n=k

(λ

λ − λε + μ1

)n−k

=∞∑

k=0

|ak | 1

λ − λε + μ1

1

1 − λλ−λε+μ1

= 1

μ1 − λε

∞∑k=0

|ak | < ∞. (2.63)

By using (2.26), the Fubini theorem, λ−λε+μ2 > λ > 0, (2.15) and (2.63) it follows

∞∑n=0

∣∣∣p(2)n

∣∣∣ =∣∣∣p(2)

0

∣∣∣+∞∑

n=1

∣∣∣p(2)n

∣∣∣

≤∣∣∣p(2)

0

∣∣∣+∞∑

n=1

(λ

λ − λε + μ2

)n ∣∣∣p(2)0

∣∣∣

+∞∑

n=1

n∑k=1

λn−krμ1

(λ − λε + μ2)n+1−k

∞∫0

∣∣∣p(1)k (x)

∣∣∣ dx

=∣∣∣p(2)

0

∣∣∣+ ∣∣∣p(2)0

∣∣∣ 1


λ

λ − λε + μ2

+∞∑

k=1

∞∑n=k

λn−krμ1

(λ − λε + μ2)n+1−k

∥∥∥p(1)k

∥∥∥L1[0,∞)

G. Gupur

=∣∣∣p(2)

0

∣∣∣+ ∣∣∣p(2)0

∣∣∣ λ

μ2 − λε

+∞∑

k=1

∥∥∥p(1)k

∥∥∥L1[0,∞)

rμ1

λ − λε + μ2

1


= λ − λε + μ2

μ2 − λε

∣∣∣p(2)0

∣∣∣+ rμ1

μ2 − λε

∞∑k=1

∥∥∥p(1)k

∥∥∥L1[0,∞)

< ∞. (2.64)

(2.63) and (2.64) give our desired result:

∥∥∥(p(1), p(2))∥∥∥ =

∞∑n=0

∣∣∣p(2)n

∣∣∣+∞∑

n=0

∥∥∥p(1)n

∥∥∥L1[0,∞)

< ∞.

This means that −λε are eigenvalues of A + U + E . Moreover, (2.16), (2.20), (2.26),(2.28), (2.43), (2.44) and (2.45) imply that the eigenvectors’ spaces are one dimensionallinear spaces, that is to say, geometric multiplicity of −λε is 1. ��

Remark 2.3 It seems that (2.48) does not equal to 0. If so, we can shorten (2.51)∼(2.55). But we could not prove it. It is also seems that (2.51) does not equal to 0. Ifso, the equation (2.47) has no multiple roots. But we failed to prove this point.

Theorem 2.1 implies that the whole interval (−λε, 0) belongs to the point spectrumof A + U + E, i.e., A + U + E has uncountable eigenvalues in the left real line.This together with the spectral mapping theorem for the point spectrum (see Engeland Nagel [5], p. 277)

σp(T (t)) = etσp(A+U+E) ∪ {0},

here σp(T (t)) and σp(A + U + E) are the point spectra of T (t) and A + U +E respectively, imply that the C0-semigroup T (t) has an uncountable number ofeigenvalues. Hence, T (t) is not compact, even not eventually compact ([5], p. 330).

Corollary 2.11 in Engel and Nagel [5] (p. 258) states that let T (t) be a stronglycontinuous semigroup on a Banach space with generator A + U + E, then

I. ω0 = max{ωess, s(A + U + E)}.II. σ(A + U + E) ∩ {γ ∈ C | Reγ ≥ w} is finite for every w > ωess .

where ω0 is the growth bound of T (t), ωess is the essential growth bound of T (t), ands(A + U + E) is the spectral bound of A + U + E .

Theorem 1.1 implies that ω0 = 0 and s(A + U + E) = 0. These together with theabove I show that

ωess ≤ 0. (2.65)


Since Theorem 1.1 and Theorem 2.1 show that (−λε, 0] ⊂ σp(A + U + E). Whichtogether with the above II give

ωess ≥ 0. (2.66)

(2.65) and (2.66) indicate that ωess = 0, which together with Proposition 3.5 in Engeland Negel [5], p. 332 yield that T (t) is not quasi-compact. Hence, we know that thequeueing models are essentially different from the population equations (see Songand Yu [12], Webb [15]) and reliability models that are described by a finite numberof partial differential equations with integral boundary conditions (see Gupur [8], p.104). Since ω0 = ωess = 0, by Nagel [11], p. 74 we know

r(T (t)) = ress(T (t)) = etωess = e0 = 1,

here r(T (t)) is the spectral radius of T (t) and ress(T (t)) is the essential spectral radiusof T (t). Since Theorems 1.1 and 2.1 imply that all points in the interval (−λε, 0] belongto the point spectral set of A+U + E, i.e., 0 is an accumulation point, the convergenceof the time-dependent solution of the system (1.9) in Theorem 1.1 is optimal, i.e.,it is impossible that the time-dependent solution of the system (1.9) exponentiallyconverges to its steady-state solution. In fact, Theorem 1.1 together with Proposition4.3.14 in Arendt et al. [2], p. 268 give the decomposition

(p(1), p(2)

)(x) =

(p(1), p(2)

)0(x) +

(y(1), y(2)

)(x), (2.67)

here(

p(1), p(2)) ∈ X; (p(1), p(2)

)0 (x) is the eigenvector with respect to 0,

i.e., (A + U + E)(

p(1), p(2))

0 (x) = 0; (y(1), y(2)) ∈ Range(A + U + E) and

limt→∞∥∥T (t)

(y(1), y(2)

)∥∥ = 0 by ABLV theorem [1]. Let(

p(1), p(2))ε(x) be

eigenvectors with respect to −λε in Theorem 2.1, then by using (A + U +E)(

p(1), p(2))ε(x) = −λε

(p(1), p(2)

)ε(x) we derive

T (t)((

p(1), p(2))

0(x) + (A + U + E)

(p(1), p(2)

)ε(x))

=(

p(1), p(2))

0(x) + T (t)(A + U + E)

(p(1), p(2)

)ε(x)

=(

p(1), p(2))

0(x) − λεT (t)

(p(1), p(2)

)ε(x)

=(

p(1), p(2))

0(x) − λεe−λεt

(p(1), p(2)

)ε(x)

⇒∥∥∥T (t)

((p(1), p(2)

)0(·)+(A+U +E)

(p(1), p(2)

)ε(·))−(

p(1), p(2))

0(·)∥∥∥

= λεe−λεt∥∥∥(p(1), p(2)

)ε

∥∥∥ , ∀t ≥ 0. (2.68)

This shows that there are no positive constants ϒ > 0 and � > 0 such that

∥∥∥T (t)((

p(1), p(2))

0(·) + (A + U + E)

(p(1), p(2)

)(·))

−(

p(1), p(2))

0(·)∥∥∥

G. Gupur

≤ ϒe−�t∥∥∥(p(1), p(2)

)∥∥∥ , ∀t ≥ 0, ∀(

p(1), p(2))

∈ D(A),

that is, it is impossible that the time-dependent solution of the system (1.9) exponen-tially converges to its steady-state solution.

Theorems 1.1, 1.2 and 2.1 mean that {−λ}∪(−λε, 0] belongs to the point spectrumof A + U + E . According to the definition of the essential spectrum (see Davies [4],p. 113) we know that A + U + E has no essential spectrum in the right half plane,imaginary axis and in {−λ}�(−λε, 0]. Naturally, there are a lot of works to be studiedin the future: whether the interval (−λ,−λε] belongs to the point spectral set ofA + U + E ? Can we obtain the results in Theorem 1.2 and Theorem 2.1 under onecondition λ

μ1+ λr

μ2< 1 ? Where the essential spectrum of A + U + E locate?

Acknowledgments The author completed this work during his stay at the Department of Mathematics,Louisiana State University.

References

1. Arendt, W., Batty, C.J.: Tauberian theorems and stabaility of one-parameter semigroups. Trans. Am.Math. Soc. 306, 837–852 (1988)

2. Arendt, W., Batty, C.J.K., Hieber, M., Neubrander, F.: Vector-Valued Laplace Transforms and CauchyProblems. Springer, Basel (2011)

3. Cubic function.: http://en.wikipedia.org/wiki/Cubic_function4. Davies, E.B.: Linear Operators and Their Spectra. Cambridge University Press, London (2008)5. Engel, K.J., Nagel, R.: One-Parameter Semigroups for Linear Evolution Equations. Springer, New

York (2000)6. Faddeev, D.K., Simonskii, I.S.: Problems in Higher Algebra. W. H. Freeman and Company, San

Francisco (1965)7. Fang, G.S., Gupur, G.: An eigenvalue of the M/M/1 queueing model with second optional service and

its application. Acta Anal. Funct. Appl. 11, 334–345 (2009)8. Gupur, G.: Functional Analysis Methods for Reliability Models. Springer, Basel (2011)9. Gupur, G., Li, X.Z., Zhu, G.T.: Functional Analysis Method in Queueing Theory. Research Information

Ltd, Hertfordshire (2001)10. Madan, K.C.: An M/G/1 queue with second optional service. Queueing Syst. 34, 37–46 (2000)11. Nagel, R.: One-Parameter Semigroups of Positive Operators. Springer, Berlin (1986)12. Song, J., Yu, J.: Population System Control. Springer, Berlin (1988)13. Sheng Jin formula.: http://www.86wiki.com/view/606391.htm14. Wayit, Z., Gupur, G.: Another eigenvalue of the M/M/1 queueing model with second optional service.

Acta Anal. Funct. Appl. 14, 419–430 (2012)15. Webb, G.F.: Theory of Nonlinear Age-Dependent Population Dynamics. Marcel Dekker, New York

(1985)16. Xing, X.M.: The resolvent set of M/M/1 queueing model with second optional service. J. Xinjiang

Uni. Nat. Sci. Edn. 25, 403–415 (2008)17. Zhao, H.B., Gupur, G., Helil, M.: Well-posedness of M/G/1 queueing model with second optional

service. Int. J. Pure Appl. Math. 50, 465–482 (2009)

http://en.wikipedia.org/wiki/Cubic_function

http://www.86wiki.com/view/606391.htm

on the eigenvalues of the m/m/1 queueing model with second optional service

Documents