**Lemma 2.1** *If*${\int}_{0}^{\mathrm{\infty}}\lambda xb(x){e}^{-{\int}_{0}^{x}b(\xi )\phantom{\rule{0.2em}{0ex}}d\xi}\phantom{\rule{0.2em}{0ex}}dx<1$, *then* 0 *is an eigenvalue of* *A* *with geometric multiplicity one*.

*Proof* We consider the equation

$A(p,Q)=0$,

*i.e.*,

By solving (2.2)-(2.5), we have

Through using (2.8)-(2.11) repeatedly, we deduce

By combining (2.10) and (2.11) with (2.7) and using (2.13), we deduce

It is difficult to determine directly all

${a}_{k}$ and to verify

${\sum}_{k=1}^{\mathrm{\infty}}{\parallel {p}_{k}\parallel}_{{L}^{1}[0,\mathrm{\infty})}<\mathrm{\infty}$. In the following, we use another method. We introduce the probability generating function

$P(x,z)={\sum}_{n=1}^{\mathrm{\infty}}{p}_{n}(x){z}^{n}$ for all complex variables

$|z|<1$. Theorem 1.1 ensures that

$P(x,z)$ is well defined. (2.2) and (2.3) give

By applying (2.6), (2.16), (2.14), (2.1),

${\int}_{0}^{\mathrm{\infty}}b(x){e}^{-{\int}_{0}^{x}b(\xi )\phantom{\rule{0.2em}{0ex}}d\xi}\phantom{\rule{0.2em}{0ex}}dx=1$,

${\int}_{0}^{\mathrm{\infty}}{b}_{0}(x){e}^{-{\int}_{0}^{x}{b}_{0}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi}\phantom{\rule{0.2em}{0ex}}dx=1$ and the L’Hospital rule it follows that

(2.16) and (2.17) give

$\begin{array}{rcl}\sum _{n=1}^{\mathrm{\infty}}{p}_{n}(x)& =& \underset{z\to 1}{lim}P(x,z)\\ =& \frac{{\int}_{0}^{\mathrm{\infty}}\lambda x{b}_{0}(x){e}^{-{\int}_{0}^{x}{b}_{0}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi}\phantom{\rule{0.2em}{0ex}}dx+{\int}_{0}^{\mathrm{\infty}}{b}_{0}(x){e}^{-\lambda x-{\int}_{0}^{x}{b}_{0}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi}\phantom{\rule{0.2em}{0ex}}dx}{1-{\int}_{0}^{\mathrm{\infty}}\lambda xb(x){e}^{-{\int}_{0}^{x}b(\xi )\phantom{\rule{0.2em}{0ex}}d\xi}\phantom{\rule{0.2em}{0ex}}dx}\lambda |{p}_{0}|\\ \times {\int}_{0}^{\mathrm{\infty}}{e}^{-{\int}_{0}^{x}b(\xi )\phantom{\rule{0.2em}{0ex}}d\xi}\phantom{\rule{0.2em}{0ex}}dx\\ <& \mathrm{\infty}.\end{array}$

(2.18)

(2.18) and (2.15) show that 0 is an eigenvalue of *A*. Moreover, from (2.12), (2.14), (2.1) and (2.6), it is easy to see that the eigenvectors corresponding to zero span one dimensional linear space, that is, the geometric multiplicity of 0 is one. □

According to Theorem 14 in Gupur, Li and Zhu [

7], we know that in order to obtain the asymptotic behavior of the time-dependent solution of the system (1.9), we need the spectrum of

*A* on the imaginary axis. Through investigating the system (1.9), we find that the infinite number of equations and the boundary conditions are the difficult points. Greiner [

11] put forward an idea to study the spectrum of

*A* by perturbing boundary conditions. And by using the Greiner idea, Haji and Radl [

12] gave a result which was 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

${A}_{0}p={A}_{m}p,\phantom{\rule{2em}{0ex}}D({A}_{0})=\{p\in D({A}_{m})\mid Lp=0\}$

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,

By (2.19)-(2.24) it is easy to calculate

then the above equations (

2.25)-(2.29) give, if the resolvent of

${A}_{0}$ exists,

$\begin{array}{rcl}{(\gamma I-{A}_{0})}^{-1}(y,z)& =& \left(\right(\begin{array}{ccccc}\frac{1}{\gamma +\lambda}& \frac{1}{\gamma +\lambda}\psi E& 0& 0& \cdots \\ 0& E& 0& 0& \cdots \\ 0& \lambda {E}^{2}& E& 0& \cdots \\ 0& {\lambda}^{2}{E}^{3}& \lambda {E}^{2}& E& \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array}\left)\right(\begin{array}{c}{y}_{0}\\ {y}_{1}(x)\\ {y}_{2}(x)\\ {y}_{3}(x)\\ \vdots \end{array})\\ +\left(\begin{array}{cccc}\frac{1}{\gamma +\lambda}\varphi {E}_{0}& 0& 0& \cdots \\ 0& 0& 0& \cdots \\ 0& 0& 0& \cdots \\ \vdots & \vdots & \vdots & \ddots \end{array}\right)\left(\begin{array}{c}{z}_{1}(x)\\ {z}_{2}(x)\\ {z}_{3}(x)\\ \vdots \end{array}\right),\\ \left(\begin{array}{cccc}{E}_{0}& 0& 0& \cdots \\ \lambda {E}_{0}^{2}& {E}_{0}& 0& \cdots \\ {\lambda}^{2}{E}_{0}^{3}& \lambda {E}_{0}^{2}& {E}_{0}& \cdots \\ \vdots & \vdots & \vdots & \ddots \end{array}\right)\left(\begin{array}{c}{z}_{1}(x)\\ {z}_{2}(x)\\ {z}_{3}(x)\\ \vdots \end{array}\right)).\end{array}$

From which together with the definition of the resolvent set we have the following result.

**Lemma 2.2** *Let*$b(x),{b}_{0}(x):[0,\mathrm{\infty})\to [0,\mathrm{\infty})$*be measurable*,

$0<{inf}_{x\in [0,\mathrm{\infty})}b(x)\le {sup}_{x\in [0,\mathrm{\infty})}b(x)<\mathrm{\infty}$*and*$0<{inf}_{x\in [0,\mathrm{\infty})}{b}_{0}(x)\le {sup}_{x\in [0,\mathrm{\infty})}{b}_{0}(x)<\mathrm{\infty}$.

*Then*$\{\gamma \in \mathbb{C}\phantom{\rule{0.25em}{0ex}}|\phantom{\rule{0.25em}{0ex}}\begin{array}{l}Re\gamma +\lambda >0,\\ Re\gamma +{inf}_{x\in [0,\mathrm{\infty})}b(x)>0,\\ Re\gamma +{inf}_{x\in [0,\mathrm{\infty})}{b}_{0}(x)>0\end{array}\}\subset \rho ({A}_{0}).$

*Proof* For any

$f\in {L}^{1}[0,\mathrm{\infty})$, by using integration by parts, we estimate

Similarly,

$\parallel {E}_{0}\parallel \le \frac{1}{Re\gamma +\lambda +{inf}_{x\in [0,\mathrm{\infty})}{b}_{0}(x)}.$

(2.31)

From (2.30), (2.31),

$\parallel \varphi \parallel \le {sup}_{x\in [0,\mathrm{\infty})}{b}_{0}(x)$ and

$\parallel \psi \parallel \le {sup}_{x\in [0,\mathrm{\infty})}b(x)$ we deduce, for

$(y,z)\in X$,

which means that the result of this lemma is right. □

**Lemma 2.3**
*For*
$\gamma \in \rho ({A}_{0})$
*we have*
*Proof* If

$(p,Q)\in ker(\gamma I-{A}_{m})$, then

$(\gamma I-{A}_{m})(p,Q)=0$, which is equivalent to

By solving (2.37)-(2.40) we have

Through inserting (2.41) and (2.43) into (2.36), it follows that

$\begin{array}{rcl}{p}_{0}& =& \frac{{a}_{1}}{\gamma +\lambda}{\int}_{0}^{\mathrm{\infty}}b(x){e}^{-(\gamma +\lambda )x-{\int}_{0}^{x}b(\xi )\phantom{\rule{0.2em}{0ex}}\mathrm{d}\xi}\phantom{\rule{0.2em}{0ex}}\mathrm{d}x\\ +\frac{{b}_{1}}{\gamma +\lambda}{\int}_{0}^{\mathrm{\infty}}{b}_{0}(x){e}^{-(\gamma +\lambda )x-{\int}_{0}^{x}{b}_{0}(\xi )\phantom{\rule{0.2em}{0ex}}\mathrm{d}\xi}\phantom{\rule{0.2em}{0ex}}\mathrm{d}x.\end{array}$

(2.45)

By using (2.41), (2.42), (2.43) and (2.44) repeatedly, we deduce

Since

$(p,Q)\in ker(\gamma I-{A}_{m})$, by the imbedding theorem in Adams [

15],

(2.45)-(2.54) show that (2.32)-(2.35) are true.

Conversely, if (2.32)-(2.35) hold, then by using the formulas

integration by parts and the Fubini theorem, we estimate

By combining (2.57), (2.58), (2.59) and (2.60) with (2.55) and (2.56), we derive

(2.55)-(2.62) mean that $(p,Q)\in D({A}_{m})$ and $(\gamma I-{A}_{m})(p,Q)=0$. □

It is not difficult to see that

*L* is surjective. Moreover,

$L{|}_{ker(\gamma I-{A}_{m})}:ker(\gamma I-{A}_{m})\to \partial X$

is invertible for

$\gamma \in \rho ({A}_{0})$. For

$\mathrm{\forall}\gamma \in \rho ({A}_{0})$ we define the Dirichlet operator as

${D}_{\gamma}:={(L{|}_{ker(\gamma I-{A}_{m})})}^{-1}:\partial X\to ker(\gamma I-{A}_{m}).$

Lemma 2.3 gives the explicit form of

${D}_{\gamma}$ for

$\gamma \in \rho ({A}_{0})$$\begin{array}{rcl}{D}_{\gamma}(\overrightarrow{a},\overrightarrow{b})& =& \left(\right(\begin{array}{cccc}\frac{1}{\gamma +\lambda}\psi {\u03f5}_{0}& 0& 0& \cdots \\ {\u03f5}_{0}& 0& 0& \cdots \\ {\u03f5}_{1}& {\u03f5}_{0}& 0& \cdots \\ {\u03f5}_{2}& {\u03f5}_{1}& {\u03f5}_{0}& \cdots \\ \vdots & \vdots & \vdots & \ddots \end{array}\left)\right(\begin{array}{c}{a}_{1}\\ {a}_{2}\\ {a}_{3}\\ {a}_{4}\\ \vdots \end{array})+(\begin{array}{cccc}\frac{1}{\gamma +\lambda}\varphi {\delta}_{0}& 0& 0& \cdots \\ 0& 0& 0& \cdots \\ 0& 0& 0& \cdots \\ 0& 0& 0& \cdots \\ \vdots & \vdots & \vdots & \ddots \end{array}\left)\right(\begin{array}{c}{b}_{1}\\ {b}_{2}\\ {b}_{3}\\ {b}_{4}\\ \vdots \end{array}),\\ \left(\begin{array}{ccccc}{\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}\right)\left(\begin{array}{c}{b}_{1}\\ {b}_{2}\\ {b}_{3}\\ {b}_{4}\\ \vdots \end{array}\right)),\end{array}$

(2.63)

From (2.63) and the definition of Φ, it is easy to determine the expression of

$\mathrm{\Phi}{D}_{\gamma}$ for

$\gamma \in \rho ({A}_{0})$.

$\begin{array}{rcl}\mathrm{\Phi}{D}_{\gamma}(\overrightarrow{a},\overrightarrow{b})& =& \left(\right(\begin{array}{ccccc}\psi {\u03f5}_{1}& \psi {\u03f5}_{0}& 0& 0& \cdots \\ \psi {\u03f5}_{2}& \psi {\u03f5}_{1}& \psi {\u03f5}_{0}& 0& \cdots \\ \psi {\u03f5}_{3}& \psi {\u03f5}_{2}& \psi {\u03f5}_{1}& \psi {\u03f5}_{0}& \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array}\left)\right(\begin{array}{c}{a}_{1}\\ {a}_{2}\\ {a}_{3}\\ \vdots \end{array})\\ +\left(\begin{array}{ccccc}\varphi {\delta}_{1}& \varphi {\delta}_{0}& 0& 0& \cdots \\ \varphi {\delta}_{2}& \varphi {\delta}_{1}& \varphi {\delta}_{0}& 0& \cdots \\ \varphi {\delta}_{3}& \varphi {\delta}_{2}& \varphi {\delta}_{1}& \varphi {\delta}_{0}& \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array}\right)\left(\begin{array}{c}{b}_{1}\\ {b}_{2}\\ {b}_{3}\\ \vdots \end{array}\right),\\ \left(\begin{array}{cccc}\frac{\lambda}{\gamma +\lambda}\psi {\u03f5}_{0}& 0& 0& \cdots \\ 0& 0& 0& \cdots \\ 0& 0& 0& \cdots \\ \vdots & \vdots & \vdots & \ddots \end{array}\right)\left(\begin{array}{c}{a}_{1}\\ {a}_{2}\\ {a}_{3}\\ \vdots \end{array}\right)+\left(\begin{array}{cccc}\frac{\lambda}{\gamma +\lambda}\varphi {\delta}_{0}& 0& 0& \cdots \\ 0& 0& 0& \cdots \\ 0& 0& 0& \cdots \\ \vdots & \vdots & \vdots & \ddots \end{array}\right)\left(\begin{array}{c}{b}_{1}\\ {b}_{2}\\ {b}_{3}\\ \vdots \end{array}\right)).\end{array}$

Haji and Radl [12] 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*$1\notin \sigma (\mathrm{\Phi}{D}_{\gamma})$,

*then*$\gamma \in \sigma (A)\phantom{\rule{1em}{0ex}}\u27fa\phantom{\rule{1em}{0ex}}1\in \sigma (\mathrm{\Phi}{D}_{\gamma}).$

By using Lemma 2.4 and Nagel [14], page 297, we derive the following result.

**Lemma 2.5** *Let*$b(x),{b}_{0}(x):[0,\mathrm{\infty})\to [0,\mathrm{\infty})$*be measurable*, $0<{inf}_{x\in [0,\mathrm{\infty})}b(x)\le {sup}_{x\in [0,\mathrm{\infty})}b(x)<\mathrm{\infty}$*and*$0<{inf}_{x\in [0,\mathrm{\infty})}{b}_{0}(x)\le {sup}_{x\in [0,\mathrm{\infty})}{b}_{0}(x)<\mathrm{\infty}$. *Then all points on the imaginary axis except zero belong to the resolvent set of* *A*.

*Proof* Take

$\gamma =im$,

$m\in \mathbb{R}\setminus \{0\}$,

$\overrightarrow{a}=({a}_{1},{a}_{2},\dots )\in {l}^{1}$ and

$\overrightarrow{b}=({b}_{1},{b}_{2},\dots )\in {l}^{1}$. Then by the Riemann-Lebesgue lemma,

we know there exists

$\mathcal{M}>0$ such that

$|m|>\mathcal{M}$By replacing

$f(x)$ in (2.64) with

$f(x)={e}^{-\lambda x-{\int}_{0}^{x}b(\xi )\phantom{\rule{0.2em}{0ex}}d\xi}$,

$f(x)={e}^{-\lambda x-{\int}_{0}^{x}{b}_{0}(\xi )\phantom{\rule{0.2em}{0ex}}d\xi}$ and using the fact

we derive, for

$|m|>\mathcal{M}$,

(2.65) means that when

$|m|>\mathcal{M}$, the spectral radius

$r(\mathrm{\Phi}{D}_{\gamma})\le \parallel \mathrm{\Phi}{D}_{\gamma}\parallel <1$, which implies

$1\notin \sigma (\mathrm{\Phi}{D}_{\gamma})$ for

$|m|>\mathcal{M}$, and therefore by Lemma 2.4, we know

$\gamma =im\notin \sigma (A)$ for

$|m|>\mathcal{M}$, that is,

$\{im\mid |m|>\mathcal{M}\}\subset \rho (A),\phantom{\rule{2em}{0ex}}\{im\mid |m|\le \mathcal{M}\}\subset \sigma (A)\cap i\mathbb{R}.$

(2.66)

On the other hand, since $T(t)$ is positive uniformly bounded by Theorem 1.1, by Corollary 2.3 in Nagel [14], page 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 integer *k*, from which together with (2.66) and Lemma 2.1 we conclude $\sigma (A)\cap i\mathbb{R}=\{0\}$. □

It is not difficult to prove

${X}^{\ast}$, dual space of

*X*, is as follows:

${X}^{\ast}=\{({p}^{\ast},{Q}^{\ast})\phantom{\rule{0.25em}{0ex}}|\phantom{\rule{0.25em}{0ex}}\begin{array}{l}{p}^{\ast}\in \mathbb{R}\times {L}^{\mathrm{\infty}}[0,\mathrm{\infty})\times {L}^{\mathrm{\infty}}[0,\mathrm{\infty})\times \cdots ,\\ {Q}^{\ast}\in {L}^{\mathrm{\infty}}[0,\mathrm{\infty})\times {L}^{\mathrm{\infty}}[0,\mathrm{\infty})\times {L}^{\mathrm{\infty}}[0,\mathrm{\infty})\times \cdots ,\\ \u2980({p}^{\ast},{Q}^{\ast})\u2980=max\left\{\begin{array}{l}sup\{|{p}_{0}^{\ast}|,{sup}_{n\ge 1}{\parallel {p}_{n}^{\ast}\parallel}_{{L}^{\mathrm{\infty}}[0,\mathrm{\infty})}\},\\ {sup}_{n\ge 1}{\parallel {Q}_{n}^{\ast}\parallel}_{{L}^{\mathrm{\infty}}[0,\mathrm{\infty})}\end{array}\right\}<\mathrm{\infty}\end{array}\}.$

It is obvious that

${X}^{\ast}$ is a Banach space. Gupur [

4] gave the expression of

${A}^{\ast}$, the adjoint operator of

*A* as follows:

${A}^{\ast}({p}^{\ast},{Q}^{\ast})=(G+F+\mathrm{\Re})({p}^{\ast},{Q}^{\ast}),\phantom{\rule{1em}{0ex}}({p}^{\ast},{Q}^{\ast})\in D(G),$

Since $T(t)$ is uniformly bounded, by Arendt and Batty [16] and Lemma 2.1, we know that 0 is an eigenvalue of ${A}^{\ast}$. Furthermore, by replacing *μ* and *η* in Lemma 3 in Gupur [4] with $b(x)$ and ${b}_{0}(x)$, respectively, we deduce the following result.

**Lemma 2.6** *If*${\int}_{0}^{\mathrm{\infty}}\lambda xb(x){e}^{-{\int}_{0}^{x}b(\xi )\phantom{\rule{0.2em}{0ex}}d\xi}\phantom{\rule{0.2em}{0ex}}dx<1$, *then* 0 *is an eigenvalue of*${A}^{\ast}$*with geometric multiplicity one*.

Since Theorem 1.1, Lemma 2.1, Lemma 2.5 and Lemma 2.6 satisfy the conditions of Theorem 14 in Gupur, Li and Zhu [7], the following conclusion is the direct result of Theorem 14 in Gupur, Li and Zhu [7].

**Theorem 2.7** *Let*$b(x),{b}_{0}(x):[0,\mathrm{\infty})\to [0,\mathrm{\infty})$*be measurable*,

$0<{inf}_{x\in [0,\mathrm{\infty})}b(x)\le {sup}_{x\in [0,\mathrm{\infty})}b(x)<\mathrm{\infty}$*and*$0<{inf}_{x\in [0,\mathrm{\infty})}{b}_{0}(x)\le {sup}_{x\in [0,\mathrm{\infty})}{b}_{0}(x)<\mathrm{\infty}$.

*If*${\int}_{0}^{\mathrm{\infty}}\lambda xb(x)\times {e}^{-{\int}_{0}^{x}b(\xi )\phantom{\rule{0.2em}{0ex}}d\xi}\phantom{\rule{0.2em}{0ex}}dx<1$,

*then the time*-

*dependent solution of the system* (1.9)

*converges strongly to its steady*-

*state solution*,

*that is*,

$\underset{t\to \mathrm{\infty}}{lim}\parallel (p,Q)(\cdot ,t)-\u3008({p}^{\ast},{Q}^{\ast}),(p(0),Q(0))\u3009(p,Q)(\cdot )\parallel =0,$

*where*$({p}^{\ast},{Q}^{\ast})$*and*$(p,Q)$*are the eigenvectors in Lemma* 2.6 *and Lemma* 2.1, *respectively*.

When $b(x)=\mu $ and ${b}_{0}(x)=\eta $, Lin and Gupur [9] proved that if
, then
are eigenvalues of *A* with geometric multiplicity one for all $\theta \in (0,1)$. Which means that the result in Theorem 2.7 is optimal, that is to say, it is impossible that the time-dependent solution of the system (1.9) exponentially converges to its steady-state solution.