Appendix
Proof of of Lemma 2.2
For any \(f\in L^{1}[0,\infty ),\) by using integration by parts, we have
$$\begin{aligned} \Vert E_0f\Vert _{L^1[0,\infty )}&=\int ^{\infty }_0|E_0f(x)|dx\nonumber \\&= \int ^{\infty }_0\left| e^{-\int ^x_0(\gamma +\lambda +\theta +\mu _0(\xi ))d\xi }\int ^x_0f(\tau )e^{\int ^{\tau }_0(\gamma +\lambda +\theta +\mu _0(\xi ))d\xi }d\tau \right| dx\nonumber \\&\le \int ^{\infty }_0e^{-\int ^x_0(\text {Re}\gamma +\lambda +\theta +\mu _0(\xi ))d\xi }\int ^x_0|f(\tau )|e^{\int ^{\tau }_0(\text {Re}\gamma +\lambda +\theta +\mu _0(\xi ))d\xi }d\tau dx\nonumber \\&= \int ^{\infty }_0\frac{-1}{\text {Re}\gamma +\lambda +\theta +\mu _0(x)}\int ^x_0|f(\tau )|e^{\int ^{\tau }_0(\text {Re}\gamma +\lambda +\theta +\mu _0(\xi ))d\xi }d\tau de^{-\int ^x_0(\text {Re}\gamma +\lambda +\theta +\mu _0(\xi ))d\xi }\nonumber \\&\le \int ^{\infty }_0\frac{-1}{\text {Re}\gamma +\lambda +\theta +\underline{\mu _0}}\int ^x_0|f(\tau )|e^{\int ^{\tau }_0(\text {Re}\gamma +\lambda +\theta +\mu _0(\xi ))d\xi }d\tau de^{-\int ^x_0(\text {Re}\gamma +\lambda +\theta +\mu _0(\xi ))d\xi }\nonumber \\&= -\frac{1}{\text {Re}\gamma +\lambda +\theta +\underline{\mu _0}}\Big \{e^{-\int ^x_0(\text {Re}\gamma +\lambda +\mu _0(\xi ))d\xi }\int ^x_0|f(\tau )|e^{\int ^{\tau }_0(\text {Re}\gamma +\lambda +\theta +\mu _0(\xi ))d\xi }d\tau \Big |^{x=\infty }_{x=0}\nonumber \\&\quad -\int ^{\infty }_0|f(x)|e^{\int ^x_0(\text {Re}\gamma +\lambda +\mu _0(\xi ))d\xi }e^{-\int ^x_0(\text {Re}\gamma +\lambda +\theta +\mu _0(\xi ))d\xi }dx\Big \}\nonumber \\&=\frac{1}{\text {Re}\gamma +\lambda +\theta +\underline{\mu _0}}\Vert f\Vert _{L^1[0,\infty )}\nonumber \\&\Longrightarrow \nonumber \\ \Vert E_0\Vert&\le \frac{1}{\text {Re}\gamma +\lambda +\theta +\underline{\mu _0}}. \end{aligned}$$
(A.1)
$$\begin{aligned} \Vert E_1f\Vert _{L^1[0,\infty )}&= \int ^{\infty }_0|E_1f(x)|dx\nonumber \\&=\int ^{\infty }_0\left| e^{-\int ^x_0(\gamma +\lambda +\mu _1(\xi ))d\xi }\int ^x_0f(\tau )e^{\int ^{\tau }_0(\gamma +\lambda +\mu _1(\xi ))d\xi }d\tau \right| dx\nonumber \\&\le \frac{1}{\text {Re}\gamma +\lambda +\underline{\mu _1}}\Vert f\Vert _{L^1[0,\infty )}\nonumber \\&\Longrightarrow \nonumber \\ \Vert E_1\Vert&\le \frac{1}{\text {Re}\gamma +\lambda +\underline{\mu _1}}. \end{aligned}$$
(A.2)
From (A.1) and (A.2) together with condition of this lemma and using \(\Vert \phi _0\Vert \le \overline{\mu _0},\;\Vert \phi _1\Vert \le \overline{\mu _1} \) we deduce, for any \((y_0,y_1)\in X\times Y\)
$$\begin{aligned}&\left\| (\gamma I-A_0)^{-1}(y_0,y_1)\right\| \\&\quad =\bigg |\frac{1}{\gamma +\lambda }y_{0,0}+\frac{1}{\gamma +\lambda }\phi _0E_0 y_{1,0}+\frac{1}{\gamma +\lambda }\phi _1E_1 y_{1,1}\bigg |\\&\qquad +\Vert E_0 y_{1,0}\Vert _{L^1[0,\infty )}+\left\| \lambda E_0^2 y_{1,0}+E_0 y_{2,0}\right\| _{L^1[0,\infty )}\\&\qquad +\left\| \lambda ^2 E_0^3 y_{1,0}+\lambda E_0^2 y_{2,0}+E_0 y_{3,0}\right\| _{L^1[0,\infty )}+\left\| \sum ^3_{k=0}\lambda ^k E_0^{k+1} y_{4-k,0}\right\| _{L^1[0,\infty )}\\&\qquad +\cdots +\left\| \sum ^{n-1}_{k=0}\lambda ^k E_0^{k+1} y_{n-k,0}\right\| _{L^1[0,\infty )}+\cdots \\&\qquad +\Vert E_1 y_{1,1}\Vert _{L^1[0,\infty )}+\left\| \lambda E_1^2 y_{1,1}+E_1 y_{2,1}\right\| _{L^1[0,\infty )}\\&\qquad +\left\| \lambda ^2 E_1^3 y_{1,1}+\lambda E_1^2 y_{2,1}+E_1 y_{3,1}\right\| _{L^1[0,\infty )}+\left\| \sum ^3_{k=0}\lambda ^k E_1^{k+1} y_{4-k,1}\right\| _{L^1[0,\infty )}\\&\qquad +\cdots +\left\| \sum ^{n-1}_{k=0}\lambda ^k E_1^{k+1} y_{n-k,1}\right\| _{L^1[0,\infty )}+\cdots \\&\quad \le \frac{1}{|\gamma +\lambda |}|y_{0,0}|+\frac{1}{|\gamma +\lambda |}\Vert \phi _0\Vert \Vert E_0\Vert \Vert y_{1,0}\Vert _{L^1[0,\infty )}\\&\qquad +\frac{1}{|\gamma +\lambda |}\Vert \phi _1\Vert \Vert E_1\Vert \Vert y_{1,1}\Vert _{L^1[0,\infty )}+\sum ^{\infty }_{n=1}\sum ^{n-1}_{k=0}\lambda ^k \Vert E_0\Vert ^{k+1} \Vert y_{n-k,0}\Vert _{L^1[0,\infty )}\\&\qquad +\sum ^{\infty }_{n=1}\sum ^{n-1}_{k=0}\lambda ^k \Vert E_1\Vert ^{k+1} \Vert y_{n-k,1}\Vert _{L^1[0,\infty )}\\&\quad \le \frac{1}{|\gamma +\lambda |}|y_{0,0}|+\frac{\overline{\mu _0}}{|\gamma +\lambda |(\text{ Re }\gamma +\lambda +\theta +\underline{\mu _0})}\Vert y_{1,0}\Vert _{L^1[0,\infty )}\\&\qquad +\frac{\overline{\mu _1}}{|\gamma +\lambda |(\text{ Re }\gamma +\lambda +\underline{\mu _1})} \Vert y_{1,1}\Vert _{L^1[0,\infty )}\\&\qquad +\frac{1}{\text{ Re }\gamma +\lambda +\theta +\underline{\mu _0}}\sum ^{\infty }_{k=0} \left( \frac{\lambda }{\text{ Re }\gamma +\lambda +\theta +\underline{\mu _0}}\right) ^k \sum ^{\infty }_{n=1} \Vert y_{n,0}\Vert _{L^1[0,\infty )}\\&\qquad +\frac{1}{\text{ Re }\gamma +\lambda +\underline{\mu _1}}\sum ^{\infty }_{k=0} \left( \frac{\lambda }{\text{ Re }\gamma +\lambda +\underline{\mu _1}}\right) ^k \sum ^{\infty }_{n=1} \Vert y_{n,1}\Vert _{L^1[0,\infty )}\\&\quad \le \sup \bigg \{\frac{1}{|\gamma +\lambda |},\frac{\overline{\mu _0}}{|\gamma +\lambda |(\text{ Re }\gamma +\lambda +\theta +\underline{\mu _0})}\\&\qquad + \frac{1}{\text{ Re }\gamma +\theta +\underline{\mu _0}} \bigg \} \left( |y_0|+\sum ^{\infty }_{n=1} \Vert y_{n,0}\Vert _{L^1[0,\infty )} \right) \\&\qquad +\sup \bigg \{\frac{1}{|\gamma +\lambda |},\frac{\overline{\mu _1}}{|\gamma +\lambda |(\text{ Re }\gamma +\lambda +\underline{\mu _1})}\\&\qquad + \frac{1}{\text{ Re }\gamma +\underline{\mu _1}} \bigg \} \sum ^{\infty }_{n=1} \Vert y_{n,1}\Vert _{L^1[0,\infty )} \\&\quad \le \sup \bigg \{\frac{1}{|\gamma +\lambda |},\frac{\overline{\mu _0}}{|\gamma +\lambda |(\text{ Re }\gamma +\lambda +\theta +\underline{\mu _0})}+ \frac{1}{\text{ Re }\gamma +\theta +\underline{\mu _0}},\\&\qquad +\frac{\overline{\mu _1}}{|\gamma +\lambda |(\text{ Re }\gamma +\lambda +\underline{\mu _1})}+\frac{1}{\text{ Re }\gamma +\underline{\mu _1}} \bigg \} \Vert (y_0,y_1)\Vert <\infty . \end{aligned}$$
This shows that the result of this lemma is right. \(\square \)
Proof of Lemma 2.3
If \((p_{0},p_{1})\in \ker (\gamma I-A_m),\) then \((\gamma I-A_m)(p_{0},p_{1})=0,\) which is equivalent to
$$\begin{aligned} (\gamma +\lambda ) p_{0,0}&=\int ^{\infty }_0 \mu _0(x) p_{1,0}(x)dx+\int ^{\infty }_0\mu _0(x)p_{1,1}(x)dx, \end{aligned}$$
(A.3)
$$\begin{aligned} \frac{dp_{1,0}(x)}{dx}&=-(\gamma +\lambda +\theta +\mu _0(x))p_{1,0}(x), \end{aligned}$$
(A.4)
$$\begin{aligned} \frac{dp_{n,0}(x)}{dx}&=-(\gamma +\lambda +\theta +\mu _0(x))p_{n,0}(x)+\lambda p_{n-1,0}(x),\quad n\ge 2, \end{aligned}$$
(A.5)
$$\begin{aligned} \frac{dp_{1,1}(x)}{dx}&=-(\gamma +\lambda +\mu _0(x))p_{1,1}(x), \end{aligned}$$
(A.6)
$$\begin{aligned} \frac{dp_{n,1}(x)}{dx}&=-(\gamma +\lambda +\mu _0(x))p_{n,1}(x)+\lambda p_{n-1,1}(x),\quad n\ge 2. \end{aligned}$$
(A.7)
By solving (A.4), (A.5) and (A.6), (A.7), we have
$$\begin{aligned} p_{1,0}(x)&=a_{1,0}e^{-\int ^x_0(\gamma +\lambda +\theta +\mu _0(\xi ))d\xi }, \end{aligned}$$
(A.8)
$$\begin{aligned} p_{n,0}(x)&=a_{n,0}e^{-\int ^x_0(\gamma +\lambda +\theta +\mu _0(\xi ))d\xi }\nonumber \\&\quad +\lambda e^{-\int ^x_0(\gamma +\lambda +\theta +\mu _0(\xi ))d\xi }\int ^x_0 p_{n-1,0}(\tau )e^{\int ^{\tau }_0(\gamma +\lambda +\theta +\mu _0(\tau ))d\xi }d\tau ,\; n\ge 2, \end{aligned}$$
(A.9)
$$\begin{aligned} p_{1,1}(x)&=a_{1,1}e^{-\int ^x_0(\gamma +\lambda +\mu _1(\xi ))d\xi }, \end{aligned}$$
(A.10)
$$\begin{aligned} p_{n,1}(x)&=a_{n,1}e^{-\int ^x_0(\gamma +\lambda +\mu _1(\xi ))d\xi }\nonumber \\&\quad +\lambda e^{-\int ^x_0(\gamma +\lambda +\mu _1(\xi ))d\xi }\int ^x_0p_{n-1,1}(\tau ) e^{\int ^{\tau }_0(\gamma +\lambda +\mu _1(\xi ))d\xi }d\tau ,\; n\ge 2. \end{aligned}$$
(A.11)
By using (A.8) and (A.9) repeatedly, we obtain
$$\begin{aligned} p_{2,0}(x)&=a_{2,0}e^{-\int ^x_0(\gamma +\lambda +\theta +\mu _0(\xi ))d\xi }+\lambda e^{-\int ^x_0(\gamma +\lambda +\theta +\mu _0(\xi ))d\xi }\int ^x_0a_{1,0}d\tau \nonumber \\&=e^{-\int ^x_0(\gamma +\lambda +\theta +\mu _0(\xi ))d\xi }[a_{2,0}+\lambda xa_{1,0}],\nonumber \\ p_{3,0}(x)&=a_{3,0}e^{-\int ^x_0(\gamma +\lambda +\theta +\mu _0(\xi ))d\xi }+\lambda e^{-\int ^x_0(\gamma +\lambda +\theta +\mu _0(\xi ))d\xi }\int ^x_0[a_{2,0}+\lambda \tau a_{1,0}]d\tau \nonumber \\&=e^{-\int ^x_0(\gamma +\lambda +\theta +\mu _0(\xi ))d\xi }\left[ a_{3,0}+\lambda xa_{2,0}+\frac{(\lambda x)^2}{2}a_{1,0}\right] ,\nonumber \\&\cdots \cdots \nonumber \\ p_{n,0}(x)&=e^{-\int ^x_0(\gamma +\lambda +\theta +\mu _0(\xi ))d\xi }\left[ a_{n,0}+\lambda xa_{n-1,0}+\frac{(\lambda x)^2}{2}a_{n-2,0}+\cdots +\frac{(\lambda x)^n}{n!}a_{1,0}\right] \nonumber \\&=e^{-\int ^x_0(\gamma +\lambda +\theta +\mu _0(\xi ))d\xi }\sum ^{n-1}_{k=0}\frac{(\lambda x)^k}{k!}a_{n-k,0},\quad n\ge 1. \end{aligned}$$
(A.12)
Similarly, by applying (A.10) and (A.11) repeatedly, we deduce
$$\begin{aligned} p_{n,1}(x)=e^{-\int ^x_0(\gamma +\lambda +\mu _1(\xi ))d\xi }\sum ^{n-1}_{k=0}\frac{(\lambda x)^k}{k!}a_{n-k,1},\quad n\ge 1. \end{aligned}$$
(A.13)
Through inserting (A.8) and (A.10) into (A.3), we derive
$$\begin{aligned} p_{0,0}&=\frac{1}{\gamma +\lambda }a_{1,0}\int ^{\infty }_0\mu _0(x)e^{-\int ^x_0(\gamma +\lambda +\theta +\mu _0(\xi ))d\xi }dx\nonumber \\&\quad +\frac{1}{\gamma +\lambda }a_{1,1}\int ^{\infty }_0\mu _1(x)e^{-\int ^x_0(\gamma +\lambda +\mu _1(\xi ))d\xi }dx. \end{aligned}$$
(A.14)
Since \((p_{0},p_{1})\in \ker (\gamma I-D(A_m)),\)
\((p_{0},p_{1})\in D(A_m)\) implies by the imbedding theorem in Adams [1],
$$\begin{aligned} \sum ^{\infty }_{n=1}|a_{n,0}|&=\sum ^{\infty }_{n=1}|p_{n,0}(0)|\le \sum ^{\infty }_{n=1}\Vert p_{n,0}\Vert _{L^{\infty }[0,\infty )}\\&\le \sum ^{\infty }_{n=1}\Vert p_{n,0}\Vert _{L^1[0,\infty )}+\sum ^{\infty }_{n=1}\left\| \frac{dp_{n,0}}{dx}\right\| _{L^1[0,\infty )}<\infty ,\\ \sum ^{\infty }_{n=1}|a_{n,1}|&=\sum ^{\infty }_{n=1}|p_{n,1}(0)|\le \sum ^{\infty }_{n=1}\Vert p_{n,1}\Vert _{L^{\infty }[0,\infty )}\\&\le \sum ^{\infty }_{n=1}\Vert p_{n,1}\Vert _{L^1[0,\infty )}+\sum ^{\infty }_{n=1}\left\| \frac{dp_{n,1}}{dx}\right\| _{L^1[0,\infty )}<\infty . \end{aligned}$$
From which together with (A.12)–(A.14) we know that (2.33) hold.
Conversely, if (2.33) is right, then by using \(\int ^{\infty }_0x^ke^{-bx}dx=\frac{k!}{b^{k+1}},\quad k\ge 1,\; b>0,\) integration by parts and the Fubini theorem we estimate
$$\begin{aligned} \Vert p_{n,0}\Vert _{L^1[0,\infty )}=&\int ^{\infty }_0\left| e^{-\int ^x_0(\gamma +\lambda +\theta +\mu _0(\xi ))d\xi }\sum ^{n-1}_{k=0}\frac{(\lambda x)^k}{k!}a_{n-k,0}\right| dx\nonumber \\ \le&\sum ^{n-1}_{k=0}|a_{n-k,0}|\frac{\lambda ^k}{k!}\int ^{\infty }_0 x^k e^{-\left( \text{ Re }\gamma +\lambda +\theta +\underline{\mu _0} \right) x}dx\nonumber \\ =&\sum ^{n-1}_{k=0}\frac{\lambda ^k}{(\text {Re}\gamma +\lambda +\theta +\underline{\mu _0})^{k+1}}|a_{n-k,0}|\nonumber \\&\Longrightarrow \nonumber \\ |p_{0,0}|+\sum ^{\infty }_{n=1}\Vert p_{n,0}\Vert _{L^1[0,\infty )}&=\frac{\overline{\mu _0}}{|\gamma +\lambda +\theta |(\text{ Re }\gamma +\lambda +\theta +\underline{\mu _0})}|a_{1,0}|\nonumber \\&\quad +\frac{\overline{\mu _1}}{|\gamma +\lambda +\theta |(\text{ Re }\gamma +\lambda +\underline{\mu _1})}|a_{1,1}|\nonumber \\&\quad +\frac{1}{\text {Re}\gamma +\theta +\underline{\mu _0}}\sum ^{\infty }_{n=1}|a_{n,0}|<\infty . \end{aligned}$$
(A.15)
Similarly, we get
$$\begin{aligned} \sum ^{\infty }_{n=1}\Vert p_{n,1}\Vert _{L^1[0,\infty )}&=\frac{\overline{\mu _1}}{|\gamma +\theta |(\text{ Re }\gamma +\lambda +\theta +\underline{\mu _1})}|a_{1,0}|\nonumber \\&\quad +\frac{\overline{\mu _1}}{|\gamma +\lambda |(\text{ Re }\gamma +\lambda +\underline{\mu _1})}|a_{1,1}|\nonumber \\&\quad +\frac{1}{\text {Re}\gamma +\theta +\underline{\mu _1}}\sum ^{\infty }_{n=1}|a_{n,1}|<\infty . \end{aligned}$$
(A.16)
(A.16) and (A.17) give
$$\begin{aligned} |p_{0,0}|+\sum ^{\infty }_{n=1}\Vert p_{n,0}\Vert _{L^1[0,\infty )}+\sum ^{\infty }_{n=1}\Vert p_{n,1}\Vert _{L^1[0,\infty )}<\infty . \end{aligned}$$
Since
$$\begin{aligned} \frac{dp_{1,0}(x)}{dx}&=-(\gamma +\lambda +\theta +\mu _0(x))p_{1,0}\\ \frac{dp_{n,0}(x)}{dx}&=-(\gamma +\lambda +\theta +\mu _0(x))p_{n,0}(x)+\lambda p_{n-1,0}(x),\quad n\ge 2, \\ \frac{dp_{1,1}(x)}{dx}&=-(\gamma +\lambda +\mu _1(x))p_{1,1}(x),\\ \frac{dp_{n,1}(x)}{dx}&=-(\gamma +\lambda +\mu _1(x))p_{n,1}(x)+\lambda p_{n-1,1}(x),\quad n\ge 2. \end{aligned}$$
It is immediately obtained
$$\begin{aligned} \sum ^{\infty }_{n=1}\left\| \frac{dp_{n,0}}{dx}\right\| _{L^1[0,\infty )}&\le (\text {Re}\gamma +\lambda +\theta +\overline{\mu _0})\sum ^{\infty }_{n=0}\Vert p_{n,0}\Vert _{L^1[0,\infty )}\nonumber \\&\quad +\lambda \sum ^{\infty }_{n=2}\Vert p_{n-1,0}\Vert _{L^1[0,\infty )}\nonumber \\&\le \left( \frac{\text {Re}\gamma +\lambda +\theta +\overline{\mu _0}}{\text {Re}\gamma +\theta +\underline{\mu _0}}+\frac{\lambda }{\text {Re}\gamma +\theta +\underline{\mu _0}}\right) \sum ^{\infty }_{n=1}|a_{n,0}|<\infty , \end{aligned}$$
(A.17)
$$\begin{aligned} \sum ^{\infty }_{n=1}\left\| \frac{dp_{n,1}}{dx}\right\| _{L^1[0,\infty )}&\le (\text {Re}\gamma +\lambda +\overline{\mu _1})\sum ^{\infty }_{n=1}\Vert p_{n,1}\Vert _{L^1[0,\infty )}\nonumber \\&\quad +\lambda \sum ^{\infty }_{n=2}\Vert p_{n-1,1}\Vert _{L^1[0,\infty )}\nonumber \\&\le \left( \frac{\text {Re}\gamma +\lambda +\overline{\mu _1}}{\text {Re}\gamma +\underline{\mu _1}}+\frac{\lambda }{\text {Re}\gamma +\underline{\mu _1}}\right) \sum ^{\infty }_{n=0}|a_{n,1}| <\infty . \end{aligned}$$
(A.18)
(A.15)–(A.18) show that \((p_{0},p_{1})\in \ker (\gamma I-A_m).\)
\(\square \)
