Lemma 2.1
If
\(r(x)\)
and
\(s(x)\)
satisfy
$$ \lambda \int_{0}^{\infty } e^{- \lambda x - \int _{0}^{x} r ( \xi )\,d\xi }\,dx + \lambda \int_{0} ^{\infty } xs ( x ) e^{- \int_{0}^{x} s ( \xi )\,d\xi }\,dx < 1, $$
then 0 is an eigenvalue of
A
with geometric multiplicity one.
Proof
Consider the equation \(A ( p, Q )=0\), which is equivalent to
$$\begin{aligned}& \lambda p_{0} = \int_{0}^{\infty } Q_{0} ( x ) s ( x )\,dx, \end{aligned}$$
(2.1)
$$\begin{aligned}& \frac{d p_{n} ( x ) }{dx} =- \bigl[ \lambda + r ( x ) \bigr] p_{n} ( x ) , \quad \forall n \geq 1, \end{aligned}$$
(2.2)
$$\begin{aligned}& \frac{d Q_{0} ( x ) }{dx} =- \bigl[ \lambda + s ( x ) \bigr] Q_{0} ( x ) , \end{aligned}$$
(2.3)
$$\begin{aligned}& \frac{d Q_{n} ( x ) }{dx} =- \bigl[ \lambda + s ( x ) \bigr] Q_{n} ( x ) + \lambda Q_{n -1} ( x ) , \quad \forall n \geq 1, \end{aligned}$$
(2.4)
$$\begin{aligned}& p_{n} ( 0 ) = \int_{0}^{\infty } Q _{n} ( x ) s ( x )\,dx, \quad \forall n \geq 1, \end{aligned}$$
(2.5)
$$\begin{aligned}& Q_{0} ( 0 ) = \lambda p_{0} + \int_{0}^{\infty } p_{1} ( x ) r ( x )\,dx, \end{aligned}$$
(2.6)
$$\begin{aligned}& Q_{n} ( 0 ) = \lambda \int_{0}^{\infty } p_{n} ( x )\,dx + \int_{0}^{\infty } p_{n +1} ( x ) r ( x )\,dx, \quad \forall n \geq 1. \end{aligned}$$
(2.7)
Solving (2.2)∼(2.4), we have
$$\begin{aligned}& p_{n} ( x ) = a_{n} e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }, \quad \forall n \geq 1, \end{aligned}$$
(2.8)
$$\begin{aligned}& Q_{0} ( x ) = b_{0} e^{- \lambda x - \int_{0}^{x} s ( \xi )\,d\xi }, \end{aligned}$$
(2.9)
$$\begin{aligned}& Q_{n} ( x ) = b_{n} e^{- \lambda x - \int_{0}^{x} s ( \xi )\,d\xi } + \lambda e^{- \lambda x - \int_{0}^{x} s ( \xi )\,d\xi } \int_{0}^{x} e^{\lambda \tau + \int_{0} ^{\tau } s ( \xi )\,d\xi } Q_{n -1} ( \tau )\,d\tau , \quad \forall n \geq 1. \end{aligned}$$
(2.10)
Using (2.9) and (2.10) repeatedly, we deduce
$$ Q_{n} ( x ) = e^{- \lambda x - \int_{0}^{x} s ( \xi )\,d\xi } \sum_{k =0}^{n} b_{n - k} \frac{( \lambda x )^{k}}{k !}, \quad \forall n \geq 0. $$
(2.11)
It is hard to determine concrete expressions of all \(p_{n} ( x ) \ \mbox{and} \ Q_{n} ( x )\) and to prove that \(( p, Q )\in D ( A )\). We further use another method. We introduce the probability generating functions
$$ p ( x, z )= \sum_{n =1}^{\infty } p_{n} ( x ) z ^{n}, \qquad Q ( x, z )= \sum _{n =0}^{\infty } Q_{n} ( x ) z^{n} $$
for all complex variables \(\vert z \vert <1\). Theorem 1.1 ensures that \(p ( x, z )\) and \(Q ( x, z )\) are well-defined. Equation (2.2) gives
$$\begin{aligned} \frac{\partial \sum_{n =1}^{\infty } p_{n} ( x ) z^{n}}{\partial x} &=- \sum_{n =1}^{\infty } \bigl[ \lambda + r ( x ) \bigr] p _{n} ( x ) z^{n} \\ &\Rightarrow \quad \frac{\partial p ( x, z ) }{\partial x} =- \bigl[ \lambda + r ( x ) \bigr] p ( x, z ) \\ &\Rightarrow \quad p ( x, z ) = p ( 0, z ) e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }, \end{aligned}$$
(2.12)
where the summation and differential are interchangeable because of the convergence of \(\sum_{n =1}^{\infty } [ \lambda + r ( x )] p_{n} ( x ) z^{n}\) and the Lebesgue control theorem.
The convergence of \(\sum_{n =0}^{\infty } [\lambda + s ( x )] Q_{n} ( x ) z^{n}\) and \(\sum_{n =1}^{\infty } Q_{n -1} ( x ) z^{n}\) allows us to change the order of summation and differential, so (2.3) and (2.4) imply
$$\begin{aligned} &\frac{\partial \sum_{n =0}^{\infty } Q_{n} (x ) z^{n}}{\partial x} =- \sum_{n =0}^{\infty } \bigl[ \lambda + s ( x ) \bigr] Q _{n} ( x ) z^{n} + \lambda \sum_{n =1}^{\infty } Q_{n -1} ( x ) z^{n} \\ &\quad \Rightarrow \quad \frac{\partial Q ( x, z ) }{\partial x} =- \bigl[ \lambda + s ( x ) \bigr] Q ( x, z ) + \lambda zQ ( x, z ) = \bigl[ \lambda ( z -1 ) - s ( x ) \bigr] Q ( x, z ) \\ &\quad \Rightarrow \quad Q ( x, z ) = Q ( 0, z ) e^{\lambda ( z -1 ) x - \int_{0}^{x} s ( \xi )\,d\xi }. \end{aligned}$$
(2.13)
Applying (2.5), (2.13), and (2.1) and noting the convergence of \(\sum_{n =1}^{\infty } Q_{n} ( x ) z^{n}\) and \(\int_{0}^{\infty } \sum_{n =1}^{\infty } Q_{n} ( x ) z^{n}\,dx\), we have
$$\begin{aligned} p ( 0, z ) &= \sum_{n =1}^{\infty } p _{n} ( 0 ) z^{n} \\ &= \sum_{n =1}^{\infty } \biggl( \int_{0}^{\infty } Q _{n} ( x ) s ( x )\,dx \biggr) z^{n} \\ &= \int_{0}^{\infty } s ( x ) \sum _{n =1} ^{\infty } Q_{n} ( x ) z^{n} \,dx \\ &= \int_{0}^{\infty } s ( x ) \sum _{n=0} ^{\infty } Q_{n} ( x ) z^{n} \,dx - \int _{0}^{\infty } s ( x ) Q_{0} ( x )\,dx \\ &= \int_{0}^{\infty } s ( x ) Q ( x, z )\,dx - \lambda p_{0} \\ &= \int_{0}^{\infty } s ( x ) Q ( 0, z ) e^{\lambda ( z -1 ) x - \int_{0}^{x} s ( \xi )\,d\xi } \,dx - \lambda p_{0} \\ &= Q ( 0, z ) \int_{0}^{\infty } s ( x ) e^{\lambda ( z -1 ) x - \int_{0}^{x} s ( \xi )\,d\xi }\,dx - \lambda p_{0}. \end{aligned}$$
(2.14)
By combining (2.6) and (2.7) with (2.12) and (2.14) and noting the convergence of \(\int_{0}^{\infty } \sum_{n =1} ^{\infty } p_{n} ( x ) z^{n}\,dx\) and the Lebesgue control theorem we have
$$\begin{aligned} Q ( 0, z ) &= \sum_{n =0}^{\infty } Q _{n} ( 0 ) z^{n} \\ & = Q_{0} ( 0 ) + \sum_{n =1} ^{\infty } Q_{n} ( 0 ) z^{n} \\ &= Q_{0} ( 0 ) + \lambda \int_{0}^{\infty } \sum_{n =1}^{\infty } p_{n} ( x ) z^{n}\,dx + \int_{0}^{\infty } \frac{1}{z} r ( x ) \sum _{n =0}^{\infty } p_{n +1} ( x ) z^{n+1}\,dx \\ &\quad {}- \int_{0}^{\infty } r ( x ) p_{1} ( x )\,dx \\ &= \lambda p_{0} + \lambda \int_{0}^{\infty } p ( x, z )\,dx + \int_{0}^{\infty } \frac{1}{z} r ( x ) p ( x, z ) \,dx \\ &= \lambda p_{0} + \int_{0}^{\infty } \biggl[ \lambda + \frac{1}{z} r ( x ) \biggr] p ( x, z )\,dx \\ &= \lambda p_{0} + \int_{0}^{\infty } \biggl[ \lambda + \frac{1}{z} r ( x ) \biggr] p ( 0, z ) e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx \\ &= \lambda p_{0} + p ( 0, z ) \int_{0}^{\infty } \biggl[ \lambda + \frac{1}{z} r ( x ) \biggr] e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx \\ &= \lambda p_{0} + \biggl[ Q ( 0, z ) \int_{0}^{\infty } s ( x ) e^{\lambda ( z -1 ) x - \int_{0}^{x} s ( \xi )\,d\xi }\,dx - \lambda p_{0} \biggr] \\ &\quad {}\times \int_{0}^{\infty } \biggl[ \lambda + \frac{1}{z} r ( x ) \biggr] e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx \\ &= \lambda p_{0} - \lambda p_{0} \int_{0}^{\infty } \biggl[ \lambda + \frac{1}{z} r ( x ) \biggr] e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx \\ &\quad {}+ Q ( 0, z ) \int_{0}^{\infty } s ( x ) e^{\lambda ( z -1 ) x - \int_{0}^{x} s ( \xi )\,d\xi }\,dx \int_{0}^{\infty } \biggl[ \lambda + \frac{1}{z} r ( x ) \biggr] e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx \\ &\Rightarrow \quad \\ Q ( 0, z ) & = \frac{\lambda p_{0} - \lambda p_{0} \int_{0} ^{\infty } [ \lambda + \frac{1}{z} r ( x ) ] e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx}{1- \int_{0}^{\infty } s ( x ) e^{\lambda ( z -1 ) x - \int_{0}^{x} s ( \xi )\,d\xi }\,dx \int_{0}^{\infty } [ \lambda + \frac{1}{z} r ( x ) ] e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx} \\ &= \frac{z \lambda p_{0} - \lambda p_{0} \int_{0}^{\infty } [ z \lambda + r ( x ) ] e ^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx}{z - \int_{0}^{\infty } s ( x ) e^{\lambda ( z -1 ) x - \int_{0}^{x} s ( \xi )\,d\xi }\,dx \int_{0}^{\infty } [ z \lambda + r ( x ) ] e ^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx}. \end{aligned}$$
(2.15)
From (2.14), (2.15), \(\int_{0}^{\infty } s ( x ) e^{- \int_{0}^{x} s ( \xi )\,d\xi }\,dx =1\), \(\int _{0}^{\infty } [ \lambda + r ( x )] e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx =1\) (see (2.47), (2.48)), the Lebegue control theorem, and the l’Hospital rule it follows that
$$\begin{aligned}& p ( 0, z ) = \frac{ [ z \lambda p_{0} - \lambda p_{0} \int_{0}^{\infty } [ z \lambda + r ( x ) ] e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx ] \int_{0}^{\infty } s ( x ) e^{\lambda ( z -1 ) x - \int_{0}^{x} s ( \xi )\,d\xi }\,dx}{z - \int_{0}^{\infty } s ( x ) e^{ \lambda ( z -1 ) x - \int_{0}^{x} s ( \xi )\,d\xi }\,dx \int_{0}^{\infty } [ z \lambda + r ( x ) ] e^{- \lambda x - \int _{0}^{x} r ( \xi )\,d\xi }\,dx} \\& \hphantom{p ( 0, z ) =}{}- \lambda p_{0} \\& \hphantom{p ( 0, z ) }= \frac{- z \lambda p_{0} + z \lambda p_{0} \int_{0}^{\infty } s ( x ) e^{\lambda ( z -1 ) x - \int_{0}^{x} s ( \xi )\,d\xi }\,dx}{z - \int_{0}^{\infty } s ( x ) e^{\lambda ( z -1 ) x - \int_{0}^{x} s ( \xi )\,d\xi }\,dx \int_{0}^{\infty } [ z \lambda + r ( x ) ] e^{- \lambda x - \int _{0}^{x} r ( \xi )\,d\xi }\,dx} \\& \hphantom{p ( 0, z )}\Rightarrow \quad \\& \lim_{z \rightarrow 1} p ( 0, z ) = \frac{\lambda p_{0} \int_{0}^{\infty } \lambda xs ( x ) e ^{- \int_{0}^{x} s ( \xi )\,d\xi }\,dx}{1- \lambda \int_{0}^{\infty } e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx - \int_{0}^{\infty } \lambda xs ( x ) e ^{- \int_{0}^{x} s ( \xi )\,d\xi }\,dx}, \end{aligned}$$
(2.16)
$$\begin{aligned}& \lim_{z \rightarrow 1} Q ( 0, z ) = \frac{\lambda p_{0} - \lambda^{2} p_{0} \int_{0}^{\infty } e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx}{1- \lambda \int_{0}^{\infty } e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx - \int_{0}^{\infty } \lambda xs ( x ) e ^{- \int_{0}^{x} s ( \xi )\,d\xi }\,dx}. \end{aligned}$$
(2.17)
By using (2.12), (2.13), (2.16), (2.17), the condition of this lemma, Theorem 1.1, and the Lebesgue control theorem we derive
$$\begin{aligned}& \sum_{n =1}^{\infty } p_{n} ( x ) = \lim_{z \rightarrow 1} p ( x, z ) \\& \hphantom{\sum_{n =1}^{\infty } p_{n} ( x )}= \frac{\lambda p_{0} e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi } \int_{0}^{\infty } \lambda xs ( x ) e^{- \int_{0}^{x} s ( \xi )\,d\xi }\,dx}{1- \lambda \int _{0}^{\infty } e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx - \int_{0}^{ \infty } \lambda xs ( x ) e^{- \int _{0}^{x} s ( \xi )\,d\xi }\,dx} \\& \hphantom{\sum_{n =1}^{\infty } p_{n} ( x )}{}\Rightarrow \quad \\& \sum_{n =1}^{\infty } \int_{0}^{\infty } p_{n} ( x )\,dx = \frac{\lambda p_{0} \int_{0}^{\infty } e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx \int_{0}^{\infty } \lambda xs ( x ) e^{- \int_{0}^{x} s ( \xi )\,d\xi }\,dx}{1- \lambda \int _{0}^{\infty } e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx - \int_{0}^{ \infty } \lambda xs ( x ) e^{- \int_{0}^{x} s ( \xi )\,d\xi }\,dx} < \infty , \end{aligned}$$
(2.18)
$$\begin{aligned}& \sum_{n =0}^{\infty } Q_{n} ( x ) = \lim_{z \rightarrow 1} Q ( x, z ) = \frac{\lambda p_{0} e ^{- \int_{0}^{x} s ( \xi )\,d\xi } [ 1- \lambda \int_{0}^{\infty } e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx ] }{1- \lambda \int_{0}^{\infty } e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx - \int_{0}^{\infty } \lambda xs ( x ) e^{- \int_{0}^{x} s ( \xi )\,d\xi }\,dx} \\& \hphantom{\sum_{n =0}^{\infty } Q_{n} ( x ) }{}\Rightarrow \quad \\& \sum_{n =0}^{\infty } \int_{0}^{\infty } Q_{n} ( x )\,dx = \frac{\lambda p_{0} \int_{0} ^{\infty } e^{- \int_{0}^{x} s ( \xi )\,d\xi }\,dx [ 1- \lambda \int_{0}^{\infty } e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx ] }{1- \lambda \int_{0}^{\infty } e^{- \lambda x - \int _{0}^{x} r ( \xi )\,d\xi }\,dx - \int _{0}^{\infty } \lambda xs ( x ) e^{-\int_{0}^{x} s ( \xi )\,d\xi }\,dx} < \infty . \end{aligned}$$
(2.19)
Equations (2.18) and (2.19) show that 0 is an eigenvalue of A. Moreover, from (2.1), (2.5)–(2.8), and (2.11) we know that the eigenvectors corresponding to 0 span the following linear space:
$$ \varUpsilon := \left \{ \tilde{c} ( p, Q ) \left | \textstyle\begin{array}{l} p ( x )=( p_{0}, p_{1} ( x ), p_{2} ( x ),\ldots ), \qquad Q ( x )=( Q_{0} (x ), Q_{1} ( x ), Q_{2} ( x ),\ldots ), \\ p_{n} ( x )= a_{n} e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }, \quad \forall n \geq 1, \\ Q_{n} ( x )= e^{- \lambda x - \int_{0}^{x} s ( \xi )\,d\xi } \sum_{k =0}^{n} b_{n - k} \frac{( \lambda x )^{k}}{k !}, \quad \forall n \geq 0, \\ a_{n} = \sum_{k =0}^{n} b_{n - k} \frac{\lambda ^{k}}{k !} \int_{0}^{\infty } s ( x ) x^{k} e ^{- \lambda x - \int_{0}^{x} s ( \xi )\,d\xi }\,dx, \quad n \geq 1, \\ b_{n} = \lambda \sum_{k =0}^{n} b_{n - k} \frac{\lambda^{k}}{k !} \int_{0}^{\infty } s ( x ) x ^{k} e^{- \lambda x - \int_{0}^{x} s ( \xi )\,d\xi }\,dx \int_{0}^{\infty } e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx \\ \hphantom{b_{n} = }{}+ \sum_{k =0}^{n +1} b_{n +1- k} \frac{\lambda ^{k}}{k !} \int_{0}^{\infty } s ( x ) x^{k} e ^{- \lambda x - \int_{0}^{x} s ( \xi )\,d\xi }\,dx \int_{0}^{\infty } e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx, \\ n \geq 1, \quad b_{0} = \frac{\lambda p_{0}}{\int_{0}^{\infty } s ( x ) e^{- \lambda x - \int_{0}^{x} s ( \xi )\,d\xi }\,dx}, \quad \tilde{c}, p _{0}\in \mathbb{R}. \end{array}\displaystyle \right . \right \} $$
It is easy to see that \(a_{n}\) and \(b_{n}\) are decided by \(p_{0}\), and \(p_{n} ( x )\) and \(Q_{n} ( x )\) are decided by \(a_{n}\) and \(b_{n}\), respectively. Therefore, \(p_{n} ( x )\) and \(Q_{n} ( x )\) are determined by \(p_{0}\). Since \(p_{0} \in \mathbb{R} \) and \(\tilde{c}\in \mathbb{R}\), ϒ is a one-dimensional linear subspace of X, that is, the geometric multiplicity of 0 is one. □
According to Theorem 14 in Gupur, Li, and Zhu [11] or Theorem 1.96 in Gupur [12], we know that to obtain the asymptotic behavior of the time-dependent solution of system (1.9), we need to know the spectrum of A on the imaginary axis. By comparing to Zhang and Gupur [9] we find that the main difficult point is the boundary conditions and that there are infinitely many equations. In 1987, Greiner [16] put forward an idea to study the spectrum of A by perturbing boundary conditions when he studied a population equation which was described by a partial differential equation with an integral boundary condition. Using Greiner’s idea, Haji and Radl [18] obtained the resolvent set of the operator corresponding to the \(M / M^{B} /1\) queueing model where all parameters are constants and gave a result described by the Dirichlet operator. In the following, by applying the result, we deduce the resolvent set of A on the imaginary axis. To do this, define (\(A _{0}\), \(D ( A_{0} )\)) as
$$ A_{0} ( p, Q ) = A_{m} ( p, Q ) \quad \mbox{and} \quad D ( A_{0} )= \bigl\{ ( p, Q )\in D ( A_{m} ) \mid \varPsi (p, Q )=0 \bigr\} $$
and discuss the inverse of \(A_{0}\). For any given \(( y, z )\in X\), consider the equation \(( \gamma I - A_{0} )( p, Q )=( y, z )\), that is,
$$\begin{aligned}& ( \gamma + \lambda ) p_{0} = y_{0} + \int_{0}^{\infty } Q_{0} ( x ) s ( x )\,dx, \end{aligned}$$
(2.20)
$$\begin{aligned}& \frac{d p_{n} ( x ) }{dx} =- \bigl[ \gamma + \lambda + r ( x ) \bigr] p_{n} ( x ) + y_{n} ( x ) , \quad \forall n \geq 1, \end{aligned}$$
(2.21)
$$\begin{aligned}& \frac{d Q_{0} ( x ) }{dx} =- \bigl[ \gamma + \lambda + s ( x ) \bigr] Q_{0} ( x ) + z_{0} ( x ) , \end{aligned}$$
(2.22)
$$\begin{aligned}& \frac{d Q_{n} ( x ) }{dx} =- \bigl[ \gamma + \lambda + s ( x ) \bigr] Q_{n} ( x ) + \lambda Q_{n -1} ( x ) + z_{n} ( x ) , \quad \forall n \geq 1, \end{aligned}$$
(2.23)
$$\begin{aligned}& p_{n} ( 0 ) =0, \quad \forall n \geq 1, \qquad Q_{n} ( 0 ) =0, \quad \forall n \geq 0. \end{aligned}$$
(2.24)
By solving (2.20)–(2.24) we have
$$\begin{aligned}& p_{n} ( x ) = e^{- ( \gamma + \lambda ) x - \int _{0}^{x} r ( \xi )\,d\xi } \int_{0} ^{x} y_{n} ( \tau ) e^{ ( \gamma + \lambda ) \tau + \int_{0}^{\tau } r ( \xi )\,d\xi }\,d\tau , \quad \forall n \geq 1, \end{aligned}$$
(2.25)
$$\begin{aligned}& Q_{0} ( x ) = e^{- ( \gamma + \lambda ) x - \int _{0}^{x} s ( \xi )\,d\xi } \int_{0} ^{x} z_{0} ( \tau ) e^{ ( \gamma + \lambda ) \tau + \int_{0}^{\tau } s ( \xi )\,d\xi }\,d\tau , \end{aligned}$$
(2.26)
$$\begin{aligned}& Q_{n} ( x ) = e^{- ( \gamma + \lambda ) x - \int _{0}^{x} s ( \xi )\,d\xi } \int_{0} ^{x} z_{n} ( \tau ) e^{ ( \gamma + \lambda ) \tau + \int_{0}^{\tau } s ( \xi )\,d\xi }\,d\tau \\& \hphantom{Q_{n} ( x ) = }{}+ \lambda e^{- ( \gamma + \lambda ) x - \int_{0}^{x} s ( \xi )\,d\xi } \int_{0}^{x} Q_{n -1} ( \tau ) e^{ ( \gamma + \lambda ) \tau + \int_{0}^{\tau } s ( \xi )\,d\xi }\,d\tau , \quad \forall n \geq 1, \end{aligned}$$
(2.27)
$$\begin{aligned}& p_{0} = \frac{y_{0}}{\gamma + \lambda } + \frac{1}{\gamma + \lambda } \int_{0}^{\infty } Q_{0} ( x ) s ( x )\,dx \\& \hphantom{p_{0}}{}= \frac{y_{0}}{\gamma + \lambda } + \frac{1}{\gamma + \lambda } \int _{0}^{\infty } s ( x ) e^{- ( \gamma + \lambda ) x - \int_{0}^{x} s ( \xi )\,d\xi } \int_{0}^{x} z_{0} ( \tau ) e^{ ( \gamma + \lambda ) \tau + \int _{0}^{\tau } s ( \xi )\,d\xi }\,d\tau dx. \end{aligned}$$
(2.28)
If we set
$$\begin{aligned}& E_{r} f ( x ) = e^{- ( \gamma + \lambda ) x - \int_{0}^{x} r ( \xi )\,d\xi } \int _{0}^{x} f ( \tau ) e^{ ( \gamma + \lambda ) \tau + \int_{0}^{\tau } r ( \xi )\,d\xi }\,d\tau , \quad \forall f \in L^{1} [ 0, \infty ) , \\& E_{s} f ( x )= e^{-( \gamma + \lambda ) x - \int_{0}^{x} s ( \xi )\,d\xi } \int_{0}^{x} f ( \tau ) e^{( \gamma + \lambda ) \tau + \int _{0}^{\tau } s ( \xi )\,d\xi }\,d\tau , \quad \forall f \in L^{1} [0,\infty ), \end{aligned}$$
then Eqs. (2.25)–(2.28) give, provided that the resolvent of \(A_{0}\) exists,
$$\begin{aligned} ( \gamma I - A_{0} )^{-1} ( y, z )&= \left ( \left (\textstyle\begin{array}{c@{\quad }c@{\quad }c@{\quad }c} \frac{1}{\gamma + \lambda } & 0 & 0 & \cdots \\ 0 & E_{r} & 0 & \cdots \\ 0 & 0 & E_{r} & \cdots \\ \vdots & \vdots & \vdots & \ddots \end{array}\displaystyle \right ) \left (\textstyle\begin{array}{@{}c@{}} y_{0} \\ y_{1} ( x ) \\ y_{2} ( x ) \\ \vdots \end{array}\displaystyle \right ) \right . \\ &\quad {}+ \left (\textstyle\begin{array}{c@{\quad }c@{\quad }c} \frac{1}{\gamma + \lambda } \phi E_{s} & 0 & \cdots \\ 0 & 0 & \cdots \\ 0 & 0 & \cdots \\ \vdots & \vdots & \ddots \end{array}\displaystyle \right ) \left (\textstyle\begin{array}{@{}c@{}} z_{0} ( x ) \\ z_{1} ( x ) \\ z_{2} ( x ) \\ \vdots \end{array}\displaystyle \right ) , \\ &\quad {}\left . \left (\textstyle\begin{array}{c@{\quad }c@{\quad }c@{\quad }c} E_{s} & 0 & 0 & \cdots \\ \lambda E_{s}^{2} & E_{s} & 0 & \cdots \\ \lambda^{2} E_{s}^{3} & \lambda E_{s}^{2} & E_{s} & \cdots \\ \vdots & \vdots & \vdots & \ddots \end{array}\displaystyle \right ) \left (\textstyle\begin{array}{@{}c@{}} z_{0} ( x ) \\ z_{1} ( x ) \\ z_{2} ( x ) \\ \vdots \end{array}\displaystyle \right ) \right ) . \end{aligned}$$
From this, together with the definition of the resolvent set, we obtain the following result.
Lemma 2.2
Let
\(r ( x ) \), \(s(x):[0,\infty)\rightarrow [0,\infty )\)
be measurable, \(0\leq \inf_{x\in [ 0,\infty ) } r ( x ) \leq \sup_{x\in [ 0, \infty ) } r ( x ) <\infty \), and
\(0< \inf_{x\in [0,\infty )} s(x)\leq \sup_{x\in [0,\infty )} s(x)<\infty \). Then
$$ \left \{ \gamma \in \mathbb{C} \left | \textstyle\begin{array}{@{}l@{}} \mathfrak{R} \gamma + \lambda > 0, \\ \mathfrak{R} \gamma + \inf_{x \in [0,\infty )} s ( x ) >0 \end{array}\displaystyle \right .\right \} \subset \rho ( A_{0} ). $$
Proof. For any \(f \in L^{1} [0,\infty )\), using integration by parts, as in Gupur and Ehmet [17], we estimate
$$\begin{aligned}& \Vert E_{r} \Vert \leq \frac{1}{\mathfrak{R} \gamma + \lambda + \inf_{x \in [ 0,\infty ] } r ( x ) }, \end{aligned}$$
(2.29)
$$\begin{aligned}& \Vert E_{s} \Vert \leq \frac{1}{\mathfrak{R} \gamma + \lambda + \inf_{x \in [ 0,\infty ) } s ( x ) }. \end{aligned}$$
(2.30)
Since an absolutely convergent number series converges to the original limit if the original orders of its terms are changed, (2.29), (2.30), \(\Vert \phi \Vert \leq \sup_{x \in [0,\infty )} s ( x )\), and \(\Vert \psi \Vert \leq \sup_{x \in [0,\infty )} r ( x )\) imply, for \(( y, z )\in X\),
$$\begin{aligned}& \bigl\Vert ( \gamma I - A_{0} )^{-1} ( y, z ) \bigr\Vert \\& \quad = \biggl\vert \frac{y_{0}}{\gamma + \lambda } + \frac{1}{\gamma + \lambda } \phi E_{s} z_{0} \biggr\vert \\& \quad \quad {}+\Vert E_{r} y_{1} \Vert _{L^{1} [ 0,\infty ) } + \Vert E_{r} y_{2} \Vert _{L^{1} [ 0,\infty ) } + \Vert E_{r}y_{3} \Vert _{L^{1} [ 0,\infty ) } + \cdots \\& \quad \quad {}+\Vert E_{s} z_{0} \Vert _{L^{1} [ 0,\infty ) } + \bigl\Vert \lambda E_{s}^{2} z_{0} + E_{s} z_{1} \bigr\Vert _{L^{1} [ 0,\infty ) } + \bigl\Vert \lambda^{2} E_{s}^{3} z_{0} + \lambda E_{s}^{2} z_{1} + E_{s} z_{2} \bigr\Vert _{L^{1} [ 0, \infty ) } \\& \quad\quad {}+ \bigl\Vert \lambda^{3} E_{s}^{4} z_{0} + \lambda^{2} E_{s}^{3} z _{1} + \lambda E_{s}^{2} z_{2} + E_{s} z_{3} \bigr\Vert _{L^{1} [ 0,\infty ) } +\cdots \\& \quad \leq \frac{\vert y_{0} \vert }{\vert \gamma + \lambda \vert } + \frac{1}{\vert \gamma + \lambda \vert } \Vert \phi E_{s} z_{0} \Vert _{L^{1} [ 0,\infty ) } + \sum _{n =1}^{\infty } \Vert E_{r} y_{n} \Vert _{L^{1} [ 0,\infty ) } \\& \quad \quad {}+\Vert E_{s} z_{0} \Vert _{L^{1} [ 0,\infty ) } + \bigl\Vert \lambda E_{s}^{2} z_{0} \bigr\Vert _{L^{1} [ 0,\infty ) } +\Vert E_{s} z_{1} \Vert _{L^{1} [ 0,\infty ) } + \bigl\Vert \lambda^{2} E_{s}^{3} z_{0} \bigr\Vert _{L^{1} [ 0,\infty ) } \\& \quad \quad {}+ \bigl\Vert \lambda E_{s}^{2} z_{1} \bigr\Vert _{L^{1} [ 0,\infty ) } +\Vert E_{s} z_{2} \Vert _{L^{1} [ 0,\infty ) } + \bigl\Vert \lambda^{3} E_{s}^{4} z_{0} \bigr\Vert _{L^{1} [ 0,\infty ) } + \bigl\Vert \lambda^{2} E_{s}^{3} z_{1} \bigr\Vert _{L^{1} [ 0,\infty ) } \\& \quad \quad {}+ \bigl\Vert \lambda E_{s}^{2} z_{2} \bigr\Vert _{L^{1} [ 0,\infty ) } +\Vert E_{s} z_{3} \Vert _{L^{1} [ 0,\infty ) } +\cdots \\& \quad \leq \frac{\vert y_{0} \vert }{\vert \gamma + \lambda \vert } + \frac{1}{\vert \gamma + \lambda \vert } \Vert \phi \Vert \Vert E_{s} \Vert \Vert z_{0} \Vert _{L^{1} [ 0,\infty ) } + \sum_{n =1}^{\infty } \Vert E_{r} \Vert \Vert y_{n} \Vert _{L^{1} [ 0,\infty ) } \\& \quad \quad {}+\Vert E_{s} \Vert \Vert z_{0} \Vert _{L^{1} [ 0,\infty ) } + \lambda \Vert E_{s} \Vert ^{2} \Vert z_{0} \Vert _{L^{1} [ 0,\infty ) } +\Vert E_{s} \Vert \Vert z_{1} \Vert _{L^{1} [ 0,\infty ) } \\& \quad \quad {}+ \lambda^{2} \Vert E_{s} \Vert ^{3} \Vert z_{0} \Vert _{L^{1} [ 0,\infty ) } + \lambda \Vert E_{s} \Vert ^{2} \Vert z_{1} \Vert _{L^{1} [ 0,\infty ) } +\Vert E_{s} \Vert \Vert z_{2} \Vert _{L^{1} [ 0,\infty ) } \\& \quad \quad {}+ \lambda^{3} \Vert E_{s} \Vert ^{4} \Vert z_{0} \Vert _{L^{1} [ 0,\infty ) } + \lambda^{2} \Vert E_{s} \Vert ^{3} \bigl\Vert z_{1}\Vert _{L^{1} [ 0,\infty ) } +\lambda \Vert E_{s} \bigl\Vert ^{2} \bigr\Vert z_{2} \bigr\Vert _{L^{1} [ 0,\infty ) } \\& \quad \quad {}+\Vert E_{s} \Vert \Vert z_{3} \Vert _{L^{1} [ 0,\infty ) } +\cdots \\& \quad =\frac{\vert y_{0} \vert }{\vert \gamma + \lambda \vert } + \frac{1}{\vert \gamma + \lambda \vert } \Vert \phi \Vert \Vert E_{s} \Vert \Vert z_{0} \Vert _{L^{1} [ 0,\infty ) } + \sum_{n =1}^{\infty } \Vert E_{r} \Vert \Vert y_{n} \Vert _{L^{1} [ 0,\infty ) } \\& \quad \quad {}+ \sum_{n =0}^{\infty } \lambda^{n} \Vert E_{s}\Vert ^{n +1} \sum _{n =0}^{\infty } \Vert z_{n}\Vert _{L^{1} [ 0,\infty ) } \\& \quad = \frac{\vert y_{0} \vert }{\vert \gamma + \lambda \vert } + \frac{1}{\vert \gamma + \lambda \vert } \Vert \phi \Vert \Vert E_{s} \Vert \Vert z_{0} \Vert _{L^{1} [ 0,\infty ) } + \Vert E_{r} \Vert \sum_{n =1}^{\infty } \Vert y_{n} \Vert _{L^{1} [ 0,\infty ) } \\& \quad \quad {}+\Vert E_{s} \Vert \sum _{n =0}^{\infty } \lambda^{n} \Vert E_{s} \Vert ^{n} \sum_{n =0}^{\infty } \Vert z_{n} \Vert _{L^{1} [ 0,\infty ) } \\& \quad \leq \frac{\vert y_{0} \vert }{\vert \gamma + \lambda \vert } + \frac{1}{\vert \gamma + \lambda \vert } \frac{ \sup_{x \in [ 0,\infty ) } s ( x ) }{\mathfrak{R} \gamma + \lambda + \inf_{x \in [ 0,\infty ) } s ( x ) } \Vert z_{0} \Vert _{L^{1} [ 0,\infty ) } \\& \quad \quad {}+ \frac{1}{\mathfrak{R}\gamma + \lambda + \inf_{x \in [ 0,\infty ) } r ( x ) } \sum_{n =1} ^{\infty } \Vert y_{n} \Vert _{L^{1} [ 0,\infty ) } \\& \quad \quad {}+ \frac{1}{\mathfrak{R} \gamma + \lambda + \inf_{x \in [ 0,\infty ) } s ( x ) } \sum_{n =0} ^{\infty } \biggl( \frac{\lambda }{\mathfrak{R} \gamma + \lambda + \inf_{x \in [ 0,\infty ) } s ( x ) } \biggr) ^{n} \sum _{n =0}^{\infty } \Vert z_{n} \Vert _{L^{1} [ 0,\infty ) } \\& \quad \leq \frac{\vert y_{0} \vert }{\mathfrak{R} \gamma + \lambda } + \frac{1}{\mathfrak{R} \gamma + \lambda } \frac{ \sup_{x \in [ 0,\infty ) } s ( x ) }{\mathfrak{R} \gamma + \lambda + \inf_{x \in [ 0,\infty ) } s ( x ) } \Vert z_{0} \Vert _{L^{1} [ 0,\infty ) } \\& \quad \quad {} + \frac{1}{\mathfrak{R} \gamma + \lambda + \inf_{x \in [ 0,\infty ) } r ( x ) } \sum_{n =1} ^{\infty } \Vert y_{n} \Vert _{L^{1} [ 0,\infty ) } + \frac{1}{\mathfrak{R} \gamma + \inf_{x \in [ 0,\infty ) } s ( x ) } \sum_{n =0}^{\infty } \Vert z_{n} \Vert _{L^{1} [ 0,\infty ) } \\& \quad \leq \sup \biggl\{ \frac{1}{\mathfrak{R} \gamma + \lambda } + \frac{1}{ \mathfrak{R} \gamma + \lambda + \inf_{x \in [ 0,\infty ) } r ( x ) }, \\& \quad \quad {}\frac{1}{\mathfrak{R} \gamma + \lambda } \frac{ \sup_{x \in [0,\infty )} s ( x )}{\mathfrak{R} \gamma + \lambda + \inf_{x \in [0,\infty )} s ( x )} + \frac{1}{\mathfrak{R} \gamma + \inf_{x \in [0,\infty )} s ( x )} \biggr\} \bigl\Vert ( y, z ) \bigr\Vert , \end{aligned}$$
which means that the result of this lemma is right.
Lemma 2.3
For
\(\gamma \in \{ \gamma \in \mathbb{C} \mid \mathfrak{R} \gamma + \lambda >0, \mathfrak{R} \gamma + \inf_{x\in [ 0,\infty ) } s ( x ) >0 \} \subset \rho ( A_{0} ) \), we have
$$\begin{aligned}& ( p, Q ) \in \ker ( \gamma I - A_{m} ) \quad \Leftrightarrow \quad p_{0} = \frac{b_{0}}{\gamma + \lambda } \int_{0}^{\infty } s ( x ) e^{- ( \gamma + \lambda ) x - \int_{0}^{x} s ( \xi )\,d\xi \,dx}, \end{aligned}$$
(2.31)
$$\begin{aligned}& p_{n} ( x ) = a_{n} e^{- ( \gamma + \lambda ) x - \int_{0}^{x} r ( \xi )\,d\xi }, \quad \forall n \geq 1, \end{aligned}$$
(2.32)
$$\begin{aligned}& Q_{n} ( x ) = e^{- ( \gamma + \lambda ) x - \int _{0}^{\infty } s ( \xi )\,d\xi } \sum_{k =0}^{n} \frac{( \lambda x )^{k}}{k !} b _{n - k}, \quad \forall n \geq 0, \end{aligned}$$
(2.33)
$$\begin{aligned}& \overrightarrow{a} = ( a_{1}, a_{2}, a_{3},\ldots ) \in l ^{1}, \qquad \overrightarrow{b} = ( b_{0}, b_{1}, b_{2},\ldots ) \in l^{1}. \end{aligned}$$
(2.34)
Proof. If \(( p, Q )\in \ker ( \gamma I - A_{m} )\), then \(( \gamma I - A_{m} )( p, Q )=0\), which is equivalent to
$$\begin{aligned}& ( \gamma + \lambda ) p_{0} = \int_{0}^{\infty } Q_{0} ( x ) s ( x )\,dx, \end{aligned}$$
(2.35)
$$\begin{aligned}& \frac{d p_{n} ( x ) }{dx} =- \bigl[ \gamma + \lambda + r ( x ) \bigr] p_{n} ( x ) , \quad \forall n \geq 1, \end{aligned}$$
(2.36)
$$\begin{aligned}& \frac{d Q_{0} ( x ) }{dx} =- \bigl[ \gamma + \lambda + s ( x ) \bigr] Q_{0} ( x ) , \end{aligned}$$
(2.37)
$$\begin{aligned}& \frac{d Q_{n} ( x )}{dx} =- \bigl[ \gamma + \lambda + s ( x ) \bigr] Q_{n} ( x ) + \lambda Q_{n -1} ( x ) , \quad \forall n \geq 1. \end{aligned}$$
(2.38)
By solving (2.36)–(2.38) we have
$$\begin{aligned}& p_{n} ( x ) = a_{n} e^{- ( \gamma + \lambda ) x - \int_{0}^{x} r ( \xi )\,d\xi }, \quad \forall n \geq 1, \end{aligned}$$
(2.39)
$$\begin{aligned}& Q_{0} ( x ) = b_{0} e^{- ( \gamma + \lambda ) x - \int_{0}^{x} s ( \xi )\,d\xi }, \end{aligned}$$
(2.40)
$$\begin{aligned}& Q_{n} ( x ) = b_{n} e^{- ( \gamma + \lambda ) x - \int_{0}^{x} s ( \xi )\,d\xi } \\& \hphantom{Q_{n} ( x ) = }{}+ \lambda e^{- ( \gamma + \lambda ) x - \int_{0}^{x} s ( \xi )\,d\xi } \int_{0}^{x} Q_{n -1} ( \tau ) e^{ ( \gamma + \lambda ) \tau + \int_{0}^{\tau } s ( \xi )\,d\xi }\,d\tau , \quad \forall n \geq 1. \end{aligned}$$
(2.41)
By inserting (2.40) into (2.35) and using (2.40) and (2.41) repeatedly we deduce
$$\begin{aligned}& p_{0} = \frac{b_{0}}{\gamma + \lambda } \int_{0}^{\infty } s ( x ) e^{- ( \gamma + \lambda ) x - \int_{0}^{x} s ( \xi )\,d\xi }\,dx, \end{aligned}$$
(2.42)
$$\begin{aligned}& Q_{n} ( x ) = e^{- ( \gamma + \lambda ) x - \int _{0}^{x} s ( \xi )\,d\xi } \sum_{k =0} ^{n} \frac{( \lambda x )^{k}}{k !} b_{n - k}, \quad \forall n \geq 0. \end{aligned}$$
(2.43)
Since \(( p, Q )\in \ker ( \gamma I - A_{m} )\), by Theorem 4.12 in Adams [19], which implies that \(W^{1,1} [0, \infty )\hookrightarrow \ L^{\infty } [0, \infty )\), we have
$$\begin{aligned}& \sum_{n =1}^{\infty } \vert a_{n} \vert = \sum_{n =1}^{\infty } \bigl\vert p _{n} ( 0 ) \bigr\vert \leq \sum_{n =1}^{\infty } \Vert p_{n} \Vert _{L^{\infty } [ 0,\infty ) } \\& \hphantom{\sum_{n =1}^{\infty } \vert a_{n} \vert }\leq \sum_{n =1}^{\infty } \biggl\{ \Vert p_{n} \Vert _{L^{1} [ 0,\infty ) } + \biggl\Vert \frac{d p _{n}}{dx} \biggr\Vert _{L^{1} [ 0,\infty ) } \biggr\} < \infty , \end{aligned}$$
(2.44)
$$\begin{aligned}& \sum_{n =0}^{\infty } \vert b_{n} \vert = \sum_{n =0}^{\infty } \bigl\vert Q _{n} ( 0 ) \bigr\vert \leq \sum_{n =0}^{\infty } \Vert Q_{n} \Vert _{L^{\infty } [ 0,\infty ) } \\& \hphantom{\sum_{n =0}^{\infty } \vert b_{n}\vert }\leq \sum_{n =0}^{\infty } \biggl\{ \Vert Q_{n} \Vert _{L^{1} [ 0,\infty ) } + \biggl\Vert \frac{d Q _{n}}{dx} \biggr\Vert _{L^{1} [ 0,\infty ) } \biggr\} < \infty . \end{aligned}$$
(2.45)
Relations (2.39) and (2.42)–(2.45) show that (2.31)–(2.34) are true.
Conversely, if (2.31)–(2.34) hold, then by using the formulas
$$\begin{aligned}& \int_{0}^{\infty } e^{- cx} x^{k}\,dx = \frac{k !}{c ^{k +1}},\quad c > 0, k \in \mathbb{N}, \end{aligned}$$
(2.46)
$$\begin{aligned}& \int_{0}^{\infty } r ( x ) e^{- \int _{0}^{x} r ( \xi )\,d\xi }\,dx =- e ^{- \int_{0}^{x} r ( \xi )\,d\xi } |_{0}^{\infty } =1, \end{aligned}$$
(2.47)
$$\begin{aligned}& \int_{0}^{\infty } s ( x ) e^{- \int _{0}^{x} s ( \xi )\,d\xi }\,dx =- e ^{- \int_{0}^{x} s ( \xi )\,d\xi } |_{0}^{\infty } =1, \end{aligned}$$
(2.48)
integration by parts, and the Cauchy product, we estimate
$$\begin{aligned}& \Vert p_{n} \Vert _{L^{1} [ 0,\infty ) } = \int_{0} ^{\infty } \bigl\vert a_{n} e^{- ( \gamma + \lambda ) x - \int_{0}^{x} r (\xi )\,d\xi } \bigr\vert \,dx \leq \vert a_{n} \vert \int_{0}^{\infty } e^{- ( \mathfrak{R} \gamma + \lambda ) x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx \\& \hphantom{\Vert p_{n} \Vert _{L^{1} [ 0,\infty ) }}\leq \vert a_{n} \vert \int_{0}^{\infty } e^{- [ \mathfrak{R} \gamma + \lambda + \inf_{x \in [ 0,\infty ) } r ( x ) ] x}\,dx = \frac{1}{\mathfrak{R} \gamma + \lambda + \inf_{x \in [ 0,\infty ) } r ( x ) } \vert a _{n} \vert , \quad \forall n \geq 1 \\& \hphantom{\Vert p_{n} \Vert _{L^{1} [ 0,\infty ) }}\Rightarrow \\& \vert p_{0} \vert + \sum_{n =1}^{\infty } \Vert p_{n} \Vert _{L^{1} [ 0,\infty ) } \\& \quad \leq \frac{\vert b_{0} \vert }{\vert \gamma + \lambda \vert } \int_{0}^{\infty } \bigl\vert s ( x ) e^{- ( \gamma + \lambda ) x - \int_{0} ^{x} s ( \xi )\,d\xi } \bigr\vert \,dx + \frac{1}{\mathfrak{R} \gamma + \lambda + \inf_{x \in [ 0,\infty ) } r ( x ) } \sum_{n =1} ^{\infty } \vert a_{n} \vert \\& \quad \leq \frac{\vert b_{0} \vert }{\vert \gamma + \lambda \vert } \int_{0}^{\infty } s ( x ) e^{- \int_{0}^{x} s ( \xi )\,d\xi }\,dx + \frac{1}{\mathfrak{R} \gamma + \lambda + \inf_{x \in [ 0,\infty ) } r ( x ) } \sum_{n =1} ^{\infty } \vert a_{n} \vert \\& \quad = \frac{\vert b_{0} \vert }{\vert \gamma + \lambda \vert } + \frac{1}{\mathfrak{R} \gamma + \lambda + \inf_{x \in [ 0,\infty ) } r ( x ) } \sum _{n =1} ^{\infty } \vert a_{n} \vert < \infty, \end{aligned}$$
(2.49)
$$\begin{aligned}& \Vert Q_{n} \Vert _{L^{1} [ 0,\infty ) } \\& \quad = \int_{0} ^{\infty } \Biggl\vert e^{- ( \gamma + \lambda ) x - \int_{0}^{x} s ( \xi )\,d\xi } \sum _{k =0}^{n} \frac{( \lambda x )^{k}}{k !} b_{n - k} \Biggr\vert \,dx \\& \quad \leq \sum_{k =0}^{n} \frac{\lambda^{k}}{k !} \vert b_{n - k} \vert \int_{0}^{\infty } x^{k} e^{- [ \mathfrak{R} \gamma + \lambda + \inf_{x \in [ 0,\infty ) } s ( x ) ] x}\,dx \\& \quad = \sum_{k =0}^{n} \frac{\lambda^{k}}{k !} \vert b _{n - k} \vert \frac{k !}{[\mathfrak{R} \gamma + \lambda + \inf_{x \in [ 0,\infty ) } s ( x ) ]^{k +1}} \\& \quad = \frac{1}{\mathfrak{R} \gamma + \lambda + \inf_{x \in [ 0,\infty ) } s ( x ) } \sum_{k =0} ^{n} \biggl( \frac{\lambda }{\mathfrak{R} \gamma + \lambda + \inf_{x \in [ 0,\infty ) } s ( x ) } \biggr) ^{k} \vert b_{n - k} \vert \\& \quad \Rightarrow \\& \sum_{n =0}^{\infty } \Vert Q_{n} \Vert _{L^{1} [ 0,\infty ) } \\& \quad \leq \sum_{n =0}^{\infty } \frac{1}{\mathfrak{R} \gamma + \lambda + \inf_{x \in [ 0,\infty ) } s ( x ) } \sum_{k =0} ^{n} \biggl( \frac{\lambda }{\mathfrak{R} \gamma + \lambda + \inf_{x \in [ 0,\infty ) } s ( x ) } \biggr) ^{k} \vert b_{n - k} \vert \\& \quad = \frac{1}{\mathfrak{R} \gamma + \lambda + \inf_{x \in [ 0,\infty ) } s ( x ) } \sum_{k =0} ^{\infty } \biggl( \frac{\lambda }{\mathfrak{R} \gamma + \lambda + \inf_{x \in [ 0,\infty ) } s ( x ) } \biggr) ^{k} \sum _{n =0}^{\infty } \vert b _{n} \vert \\& \quad = \frac{1}{\mathfrak{R} \gamma + \inf_{x \in [0,\infty )} s ( x )} \sum_{n =0}^{\infty } \vert b_{n} \vert < \infty . \end{aligned}$$
(2.50)
By direct calculation it is not difficult to verify (2.35)–(2.38). In addition, (2.36)–(2.38) imply
$$\begin{aligned}& \sum_{n =1}^{\infty } \biggl\Vert \frac{d p_{n}}{dx} \biggr\Vert _{L^{1} [ 0,\infty ) } \leq \Bigl[ \vert \gamma \vert + \lambda + \sup_{x \in [ 0,\infty ) } r ( x ) \Bigr] \sum _{n =1}^{\infty } \Vert p_{n} \Vert _{L^{1} [ 0,\infty ) } < \infty , \end{aligned}$$
(2.51)
$$\begin{aligned}& \sum_{n =0}^{\infty } \biggl\Vert \frac{d Q_{n}}{dx} \biggr\Vert _{L^{1} [0,\infty )} \leq \Bigl[ \vert \gamma \vert +2 \lambda + \sup_{x \in [ 0,\infty ) } s ( x ) \Bigr] \sum _{n =0}^{\infty } \Vert Q_{n} \Vert _{L^{1} [ 0,\infty ) } < \infty . \end{aligned}$$
(2.52)
Relations (2.49)–(2.52) mean that \(( p, Q )\in D ( A_{m} )\) and \(( \gamma I - A_{m} )( p, Q )=0\).
It is not difficult to see that Ψ is surjective. Moreover,
$$ \varPsi|_{\ker ( \gamma I - A_{m} )}:\ker ( \gamma I - A_{m} )\rightarrow \partial X $$
is invertible for \(\gamma \in \rho ( A_{0} )\). For \(\gamma \in \rho ( A_{0} )\) we define the Dirichlet operator as
$$ D_{\gamma }:=( \varPsi|_{\ker ( \gamma I - A_{m} )} )^{-1}:\partial X \rightarrow \ker ( \gamma I - A_{m} ). $$
Lemma 2.3 gives an explicit form of \(D_{\gamma }\) for \(\gamma \in \rho ( A_{0} ) \):
$$\begin{aligned} D_{\gamma } ( \overrightarrow{a}, \overrightarrow{b} ) &= \left ( \left (\textstyle\begin{array}{c@{\quad }c@{\quad }c@{\quad }c} 0 & 0 & 0 & \cdots \\ \varepsilon & 0 & 0 & \cdots \\ 0 & \varepsilon & 0 & \cdots \\ 0 & 0 & \varepsilon & \cdots \\ \vdots & \vdots & \vdots & \ddots \end{array}\displaystyle \right ) \left (\textstyle\begin{array}{@{}c@{}} a_{1} \\ a_{2} \\ a_{3} \\ a_{4} \\ \vdots \end{array}\displaystyle \right ) + \left (\textstyle\begin{array}{c@{\quad }c@{\quad }c} \frac{1}{\gamma + \lambda } \phi \delta_{0} & 0 & \cdots \\ 0 & 0 & \cdots \\ 0 & 0 & \cdots \\ 0 & 0 & \cdots \\ \vdots & \vdots & \ddots \end{array}\displaystyle \right ) \left (\textstyle\begin{array}{@{}c@{}} b_{0} \\ b_{1} \\ b_{2} \\ b_{3} \\ \vdots \end{array}\displaystyle \right ) \right . , \\ &\quad {}\left . \left (\textstyle\begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} \delta_{0} & 0 & 0 & 0 & \cdots \\ \delta_{1} & \delta_{0} & 0 & 0 & \cdots \\ \delta_{2} & \delta_{1} & \delta_{0} & 0 & \cdots \\ \delta_{3} & \delta_{2} & \delta_{1} & \delta_{0} & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array}\displaystyle \right ) \left (\textstyle\begin{array}{@{}c@{}} b_{0} \\ b_{1} \\ b_{2} \\ b_{3} \\ \vdots \end{array}\displaystyle \right ) \right ) , \end{aligned}$$
(2.53)
where
$$ \varepsilon = e^{-( \gamma + \lambda ) x - \int_{0}^{x} r ( \xi )\,d\xi }, \qquad \delta_{k} = \frac{( \lambda x )^{k}}{k !} e^{-( \gamma + \lambda ) x - \int_{0}^{x} s ( \xi )\,d\xi }, \quad k \geq 0. $$
From (2.53) and the definition of Φ it is easy to determine the expression of \(\varPhi D_{\gamma }\) by
$$\begin{aligned} \varPhi D_{\gamma } ( a, b ) &= \left ( \left (\textstyle\begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} \phi \delta_{1} & \phi \delta_{0} & 0 & 0 & \cdots \\ \phi \delta_{2} & \phi \delta_{1} & \phi \delta_{0} & 0 & \cdots \\ \phi \delta_{3} & \phi \delta_{2} & \phi \delta_{1} & \phi \delta_{0} & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array}\displaystyle \right ) \left (\textstyle\begin{array}{@{}c@{}} b_{0} \\ b_{1} \\ b_{2} \\ \vdots \end{array}\displaystyle \right ) \right . , \\ &\quad {}\left . \left (\textstyle\begin{array}{c@{\quad }c@{\quad }c@{\quad }c} \psi \varepsilon & 0 & 0 & \cdots \\ \lambda H \varepsilon & \psi \varepsilon & 0 & \cdots \\ 0 & \lambda H \varepsilon & \psi \varepsilon & \cdots \\ \vdots & \vdots & \vdots & \ddots \end{array}\displaystyle \right ) \left (\textstyle\begin{array}{@{}c@{}} a_{1} \\ a_{2} \\ a_{3} \\ \vdots \end{array}\displaystyle \right ) + \left (\textstyle\begin{array}{c@{\quad }c@{\quad }c} \frac{\lambda }{\gamma + \lambda } \phi \delta_{0} & 0 & \cdots \\ 0 & 0 & \cdots \\ 0 & 0 & \cdots \\ \vdots & \vdots & \ddots \end{array}\displaystyle \right ) \left (\textstyle\begin{array}{@{}c@{}} b_{0} \\ b_{1} \\ b_{2} \\ \vdots \end{array}\displaystyle \right ) \right ) . \end{aligned}$$
Haji and Radl [18] gave the following result, through which we deduce the resolvent set of A on the imaginary axis.
Lemma 2.4
If
\(\gamma \in \rho ( A_{0} )\)
and
\(1\notin \sigma ( \varPhi D_{\gamma } )\), then
$$ \gamma \in \sigma ( A )\quad \Leftrightarrow \quad 1\in \sigma ( \varPhi D_{\gamma } ). $$
By using Lemma 2.4 and p. 297 of Nagel [20] we derive the following result.
Lemma 2.5
Let
\(r(x)\), \(s(x):[0,\infty )\rightarrow [0,\infty )\)
be measurable,
$$ 0 < \inf_{x \in [ 0,\infty ) } r ( x ) \leq \sup_{x \in [ 0,\infty ) } r ( x ) < \infty ,\quad \textit{and}\quad 0< \inf_{x \in [0,\infty )} s ( x )\leq \sup_{x \in [0,\infty )} s ( x ) < \infty . $$
Then all points on the imaginary axis except zero belong to the resolvent set of A.
Proof. Let \(\gamma = im\), \(m \in \mathbb{R}\backslash \{ 0 \}\), \(\overrightarrow{a} = ( a_{1}, a_{2},\ldots ) \in l^{1}\), and \(\overrightarrow{b} =( b_{0}, b_{1}, b_{2},\ldots )\in l^{1}\). The Riemann–Lebesgue lemma
$$ \lim_{m \rightarrow \infty } \int_{0}^{\infty } f ( x ) e^{imx}\,dx =0, \quad f \in L^{1} [ 0,\infty ) , f ( x ) >0, $$
implies that there exist \(\mathcal{M} >0\) and \(\theta_{1} \in (0,1)\) such that
$$ \biggl\vert \int_{0}^{\infty } f ( x ) e^{- imx}\,dx \biggr\vert < \theta_{1} \int_{0}^{\infty } f ( x )\,dx, \quad \vert m \vert \mathcal{>M}. $$
(2.54)
Replacing \(f ( x )\) in (2.54) with \(f ( x ) = e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi } \ \mbox{and} \ f ( x )= e^{- \lambda x - \int_{0}^{x} s ( \xi )\,d\xi }\) and using (2.47)–(2.48), we have
$$\begin{aligned}& \biggl\vert \int_{0}^{\infty } e^{- ( im + \lambda ) x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx \biggr\vert < \theta_{2} \int_{0}^{\infty } e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx, \quad \theta_{2} \in ( 0,1 ) , \\& \biggl\vert \int_{0}^{\infty } \frac{( \lambda x )^{n}}{n !} e^{-( im + \lambda ) x - \int_{0}^{x} s ( \xi )\,d\xi } \,dx \biggr\vert < \theta_{2} \int_{0}^{\infty } \frac{( \lambda x )^{n}}{n !} e^{- \lambda x - \int_{0}^{x} s ( \xi )\,d\xi } \,dx,\quad \forall n \geq 0, \theta_{2} \in (0,1), \end{aligned}$$
recalling that a convergent positive number series still converges to the original limit if the orders of its terms are changed we derive, for \(\vert m \vert > \mathcal{M}\) and \(( \overrightarrow{a},\overrightarrow{b} )\neq ( \overrightarrow{0}, \overrightarrow{0} )\),
$$\begin{aligned}& \bigl\Vert \varPhi D_{\gamma } ( \overrightarrow{a},\overrightarrow{b} ) \bigr\Vert \\& \quad = \vert \phi \delta_{1} b _{0} + \phi \delta_{0} b_{1} \vert + \vert \phi \delta_{2} b_{0} + \phi \delta_{1} b_{1} + \phi \delta_{0} b_{2} \vert \\& \quad \quad {}+ \vert \phi \delta_{3} b_{0} + \phi \delta_{2} b_{1} + \phi \delta_{1} b_{2} + \phi \delta_{0} b_{3} \vert \\& \quad \quad {}+ \vert \phi \delta_{4} b_{0} + \phi \delta_{3} b_{1} + \phi \delta_{2} b_{2} + \phi \delta_{1} b_{3} + \phi \delta_{0} b_{4} \vert + \cdots \\& \quad \quad {}+ \biggl\vert \psi \varepsilon a_{1} + \frac{\lambda }{\gamma + \lambda } \phi \delta_{0} b_{0} \biggr\vert + \vert \lambda H \varepsilon a_{1} + \psi \varepsilon a_{2} \vert \\& \quad \quad {}+ \vert \lambda H \varepsilon a_{2} + \psi \varepsilon a_{3} \vert + \vert \lambda H \varepsilon a_{3} + \psi \varepsilon a_{4} \vert +\cdots \\& \quad \leq \sum_{n =1}^{\infty } \vert \phi \delta _{n} \vert \vert b_{0} \vert + \sum _{k =1}^{\infty } \vert b_{k} \vert \sum _{n =0}^{\infty } \vert \phi \delta_{n} \vert \\& \quad \quad {}+ \biggl\vert \frac{\lambda }{\gamma + \lambda } \biggr\vert \vert \phi \delta_{0} \vert \vert b_{0} \vert + \bigl( \vert \psi \varepsilon \vert + \lambda \vert H \varepsilon \vert \bigr) \sum _{n =1}^{\infty } \vert a_{n} \vert \\& \quad < \vert b_{0} \vert \sum_{n =1}^{\infty } \vert \phi \delta_{n} \vert + \sum_{k =1}^{\infty } \vert b_{k} \vert \sum_{n =0}^{\infty } \vert \phi \delta_{n} \vert \\& \quad \quad {}+ \vert b_{0} \vert \vert \phi \delta_{0} \vert + \bigl( \vert \psi \varepsilon \vert + \lambda \vert H \varepsilon \vert \bigr) \sum_{n =1}^{\infty } \vert a_{n} \vert \\& \quad = \sum_{k =0}^{\infty } \vert b_{k} \vert \sum_{n =0}^{\infty } \vert \phi \delta_{n} \vert + \bigl( \vert \psi \varepsilon \vert + \lambda \vert H \varepsilon \vert \bigr) \sum _{n =1}^{\infty } \vert a_{n} \vert \\& \quad = \sum_{n =0}^{\infty } \biggl\vert \int_{0}^{ \infty } s ( x ) \frac{( \lambda x )^{n}}{n !} e^{- ( im + \lambda ) x - \int_{0}^{x} s ( \xi )\,d\xi }\,dx \biggr\vert \sum_{k =0}^{\infty } \vert b_{k} \vert \\& \quad \quad {}+ \biggl( \biggl\vert \int_{0}^{\infty } r ( x ) e^{- ( im + \lambda ) x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx \biggr\vert \\& \quad \quad {}+ \lambda \biggl\vert \int_{0}^{\infty } e ^{- ( im + \lambda ) x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx \biggr\vert \biggr) \sum_{n =1}^{ \infty } \vert a_{n} \vert \\& \quad < \sum_{n =0}^{\infty } \biggl( \theta_{2} \int _{0}^{\infty } s ( x ) \frac{( \lambda x )^{n}}{n !} e^{- \lambda x - \int_{0}^{x} s ( \xi )\,d\xi }\,dx \biggr) \sum_{k =0}^{\infty } \vert b_{k} \vert \\& \quad \quad {}+ \biggl[ \theta_{2} \int_{0}^{\infty } r ( x ) e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx + \theta_{2} \lambda \int_{0}^{\infty } e^{- \lambda x - \int_{0}^{x} r ( \xi )\,d\xi }\,dx \biggr] \sum _{n =1} ^{\infty } \vert a_{n} \vert \\& \quad = \theta_{2} \sum_{k =0}^{\infty } \vert b _{k} \vert \int_{0}^{\infty } s ( x ) \sum _{n =0}^{\infty } \frac{( \lambda x )^{n}}{n !} e^{- \lambda x - \int_{0}^{x} s ( \xi )\,d\xi }\,dx \\& \quad \quad {}+ \theta_{2} \sum_{n =1}^{\infty } \vert a _{n} \vert \int_{0}^{\infty } \bigl[ \lambda + r ( x ) \bigr] e^{- \int_{0}^{x} [\lambda + r ( \xi ) ]\,d\xi }\,dx \\& \quad = \theta_{2} \Biggl( \sum_{n =1}^{\infty } \vert a _{n} \vert + \sum_{k =0}^{\infty } \vert b _{k} \vert \int_{0}^{\infty } s ( x ) e^{\lambda x} e^{- \lambda x - \int_{0}^{x} s ( \xi )\,d\xi }\,dx \Biggr) \\& \quad = \theta_{2} \Biggl( \sum_{n =1}^{\infty } \vert a _{n} \vert + \sum_{k =0}^{\infty } \vert b _{k} \vert \int_{0}^{\infty } s ( x ) e^{- \int_{0}^{x} s ( \xi )\,d\xi }\,dx \Biggr) \\& \quad = \theta_{2} \Biggl( \sum_{n =1}^{\infty } \vert a _{n} \vert + \sum_{k =0}^{\infty } \vert b _{k} \vert \Biggr) \\& \quad = \theta_{2} \bigl\Vert ( \overrightarrow{a}, \overrightarrow{b} ) \bigr\Vert \\& \quad \Rightarrow \quad \\& \Vert \varPhi D_{\gamma } \Vert = \sup_{\Vert (\overrightarrow{a}, \overrightarrow{b} ) \Vert \neq 0} \frac{\Vert \varPhi D_{\gamma } (\overrightarrow{a},\overrightarrow{b} ) \Vert }{ \Vert ( \overrightarrow{a},\overrightarrow{b} ) \Vert } \leq \theta_{2} < 1. \end{aligned}$$
(2.55)
This means that when \(\vert m \vert > \mathcal{M}\), the spectral radius \(r ( \varPhi D_{\gamma } )\leq \Vert \varPhi D_{\gamma } \Vert \leq \theta_{2} < 1\), which implies \(1\notin \sigma ( \varPhi D_{ \gamma } )\) for \(\vert m \vert > \mathcal{M}\), and therefore by Lemma 2.4 we know that \(\gamma = im \notin \sigma ( A )\) for \(\vert m \vert > \mathcal{M}\), that is,
$$ \bigl\{ im \mid \vert m \vert >\mathcal{M} \bigr\} \subset\rho ( A ) , \qquad \bigl\{ im \mid \vert m \vert \leq\mathcal{ M} \bigr\} \subset \sigma ( A ) \cap i \mathbb{R}. $$
(2.56)
On the other hand, since \(T ( t )\) is positive uniformly bounded by Theorem 1.1, by Corollary 2.3 in Nagel [20], p. 297, we know that \(\sigma ( A )\cap i \mathbb{R}\) is imaginary additively cyclic, which states that \(im \in \sigma ( A )\cap i \mathbb{R}\Rightarrow imk \in \sigma ( A )\cap i \mathbb{R}\) for all integers k, from which, together with (2.56) and Lemma 2.1, we conclude that \(\sigma ( A )\cap i \mathbb{R} =\{0\}\), that is, all points on the imaginary axis except zero belong to the resolvent set of A.
According to Zhang and Gupur [9] and Jiang and Gupur [10], \(X^{*}\), the dual space of X, is as follows:
$$ X^{*} = \left \{ ( p^{*}, Q^{*}) \left | \textstyle\begin{array}{@{}l@{}} p^{*} \in \mathbb{R}\times L^{\infty } [0,\infty )\times L^{\infty } [0,\infty )\times \cdots , \\ Q^{*} \in L^{\infty } [0,\infty )\times L^{\infty } [0,\infty )\times \cdots , \\ \Vert ( p^{*}, Q^{*} ) \Vert =\max \{ \vert p _{0}^{*} \vert , \\ \sup_{n \geq 1} \Vert p_{n}^{*} \Vert _{L^{\infty } [ 0,\infty ) }, \sup_{n \geq 0} \Vert Q_{n}^{*} \Vert _{L^{\infty } [0,\infty )} \} < \infty \end{array}\displaystyle \right .\right \} . $$
It is obvious that \(X^{*}\) is a Banach space. Zhang and Gupur [9] gave the following expression of \(A^{*}\), the adjoint operator of A:
$$ A^{*} \bigl( p^{*}, Q^{*} \bigr)=( L + N + R ) \bigl( p^{*}, Q^{*} \bigr), \quad \bigl( p^{*}, Q^{*} \bigr) \in D ( L ), $$
where
$$\begin{aligned}& L \bigl( p^{*}, Q^{*} \bigr) = \left ( \left (\textstyle\begin{array}{c@{\quad }c@{\quad }c@{\quad }c} - \lambda & 0 & 0 & \cdots \\ 0 & \frac{d}{dx} - [ \lambda + r ( x ) ] & 0 & \cdots \\ 0 & 0 & \frac{d}{dx} - [ \lambda + r ( x ) ] & \cdots \\ \vdots & \vdots & \vdots & \ddots \end{array}\displaystyle \right ) \left (\textstyle\begin{array}{@{}c@{}} p_{0}^{*} \\ p_{1}^{*} ( x ) \\ p_{2}^{*} ( x ) \\ \vdots \end{array}\displaystyle \right ) , \right . \\& \hphantom{L ( p^{*}, Q^{*} ) =}{}\left . \left (\textstyle\begin{array}{c@{\quad }c@{\quad }c@{\quad }c} \frac{d}{dx} -[ \lambda + s ( x )] & 0 & 0 & \cdots \\ 0 & \frac{d}{dx} -[ \lambda + s ( x )] & 0 & \cdots \\ 0 & 0 & \frac{d}{dx} -[ \lambda + s ( x )] & \cdots \\ \vdots & \vdots & \vdots & \ddots \end{array}\displaystyle \right ) \left (\textstyle\begin{array}{@{}c@{}} Q_{0}^{*} ( x ) \\ Q_{1}^{*} ( x ) \\ Q_{2}^{*} ( x ) \\ \vdots \end{array}\displaystyle \right ) \right ) , \\& D ( L )= \left \{ \bigl( p^{*}, Q^{*} \bigr) \left | \textstyle\begin{array}{@{}l@{}} \frac{d p_{i}^{*} ( x ) }{dx} \ ( i \geq 1 ) , \frac{dQ_{n}^{*} ( x ) }{dx} \ ( n \geq 0 )\mbox{ exist and } \\ p_{i}^{*} (\infty )= K_{0} \ ( i \geq 1), Q_{n}^{*} (\infty )= K_{0} \ ( n\geq 0) \end{array}\displaystyle \right .\right \} , \end{aligned}$$
where \(K_{0}\) is a constant which is irrelevant to i, n in \(D ( L )\).
$$\begin{aligned}& N \bigl( p^{*}, Q^{*} \bigr) = \left ( \left (\textstyle\begin{array}{c@{\quad }c@{\quad }c@{\quad }c} \lambda & 0 & 0 & \cdots \\ 0 & \lambda & 0 & \cdots \\ 0 & 0 & \lambda & \cdots \\ \vdots & \vdots & \vdots & \ddots \end{array}\displaystyle \right ) \left (\textstyle\begin{array}{@{}c@{}} Q_{0}^{*} ( 0 ) \\ Q_{1}^{*} ( 0 ) \\ Q_{2}^{*} ( 0 ) \\ \vdots \end{array}\displaystyle \right ) , \left (\textstyle\begin{array}{c@{\quad }c@{\quad }c@{\quad }c} 0 & \lambda & 0 & \cdots \\ 0 & 0 & \lambda & \cdots \\ 0 & 0 & 0 & \cdots \\ \vdots & \vdots & \vdots & \ddots \end{array}\displaystyle \right ) \left (\textstyle\begin{array}{@{}c@{}} Q_{0}^{*} ( x ) \\ Q_{1}^{*} ( x ) \\ Q_{2}^{*} ( x ) \\ \vdots \end{array}\displaystyle \right ) \right ) , \\& R \bigl( p^{*}, Q^{*} \bigr) = \left ( \left (\textstyle\begin{array}{c@{\quad }c@{\quad }c@{\quad }c} 0 & 0 & 0 & \cdots \\ r ( x ) & 0 & 0 & \cdots \\ 0 & r ( x ) & 0 & \cdots \\ 0 & 0 & r ( x ) & \cdots \\ \vdots & \vdots & \vdots & \ddots \end{array}\displaystyle \right ) \left (\textstyle\begin{array}{@{}c@{}} Q_{0}^{*} ( 0 ) \\ Q_{1}^{*} ( 0 ) \\ Q_{2}^{*} ( 0 ) \\ Q_{3}^{*} ( 0 ) \\ \vdots \end{array}\displaystyle \right ) , \right . \\& \hphantom{N ( p^{*}, Q^{*} ) =}{}\left . \left (\textstyle\begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} s ( x ) & 0 & 0 & 0 & \cdots \\ 0 & s ( x ) & 0 & 0 & \cdots \\ 0 & 0 & s ( x ) & 0 & \cdots \\ 0 & 0 & 0 & s ( x ) & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array}\displaystyle \right ) \left (\textstyle\begin{array}{@{}c@{}} p_{0}^{*} \\ p_{1}^{*} (0) \\ p_{2}^{*} (0) \\ p_{3}^{*} (0) \\ \vdots \end{array}\displaystyle \right ) \right ) . \end{aligned}$$
Since \(T ( t )\) is uniformly bounded, by Lemma 2.3 in Arendt and Batty [21] and Lemma 2.1 we know that 0 is an eigenvalue of \(A^{*}\). Furthermore, replacing α and β in [9] with \(r ( x )\) and \(s ( x )\), respectively, we deduce the following result.
Lemma 2.6
If
$$ \lambda \int_{0}^{\infty } e^{- \lambda x - \int _{0}^{x} r ( \xi )\,d\xi }\,dx + \int_{0}^{\infty } \lambda xs ( x ) e^{- \int_{0}^{x} s ( \xi )\,d\xi }\,dx < 1, $$
then 0 is an eigenvalue of
\(A^{*}\)
with geometric multiplicity one.
Since by combining Lemmas 2.1 and 2.6 and Lemma 5 in Gupur [22] we know that 0 is an eigenvalue of \(A^{*}\) with algebraic multiplicity one, Theorem 1.1, Lemma 2.1, Lemma 2.5, and Lemma 2.6 satisfy the conditions of Theorem 14 in Gupur, Li, and Zhu [11] (Theorem 1.96 in Gupur [12]). Thus, we obtain the desired result in this section, which is the direct result of Theorem 14 in Gupur, Li, and Zhu [11].
Theorem 2.1
Let
\(r(x),s(x):[0,\infty )\rightarrow [0,\infty )\)
be measurable,
$$ 0 < \inf_{x \in [0,\infty )} r ( x )\leq \sup_{x \in [0,\infty )} r ( x ) < \infty\quad \textit{and}\quad 0< \inf_{x \in [0,\infty )} s ( x )\leq \sup_{x \in [0,\infty )} s ( x ) < \infty . $$
If
$$ \lambda \int_{0}^{\infty } e^{- \lambda x - \int _{0}^{x} r ( \xi )\,d\xi }\,dx + \int_{0}^{\infty } \lambda xs ( x ) e^{- \int_{0}^{x} s ( \xi )\,d\xi }\,dx < 1, $$
then the time-dependent solution of system (1.9) converges strongly to its steady-state solution, that is,
$$ \lim_{t \rightarrow \infty } \bigl\Vert ( p, Q ) (\cdot , t )- \bigl\langle \bigl(p^{*},Q^{*} \bigr), \bigl( p (0), Q (0) \bigr) \bigr\rangle ( p, Q ) (\cdot ) \bigr\Vert =0, $$
where (\(p^{*}\), \(Q^{*}\)) and (p, Q) are the eigenvectors in Lemmas 2.6
and
2.1, respectively.
Remark 2.1
Gomez-Corral [4] obtained that the Markov process is ergodic if and only if \(\lambda \tilde{\beta }_{1} < \tilde{\alpha } ( \lambda ) \). This condition is quite different from the conditions in Theorem 2.1. Hence, it is worth studying the relation between the ergodicity of the Markov process and the conditions in Theorem 2.1. This is our future research topic.