Appendix A
Proof of Theorem 1
Let
$$\begin{aligned} Q(z,\tau )= & {} \sum \limits _{n=1}^{\infty }\pi _{n,1}(\tau )z^n. \end{aligned}$$
(7.1)
Using (7.1) in Eqs. (2.2), (2.3) and (2.4), we obtain
$$\begin{aligned} \frac{\partial Q (z,\tau )}{\partial \tau }= & {} \left[ -(\lambda _{1}+ \mu _{1})+\lambda _{1}z+\frac{ \mu _{1} }{z} \right] Q (z,\tau )+\eta \sum \limits _{n=1}^{k-1}\pi _{n,0}(\tau )z^n\nonumber \\&+\,\sigma \lambda _{0}\pi _{k-1,0}(\tau )z^{k}+\lambda _{1}z\pi _{0,1}(\tau )- \mu _{1} \pi _{1,1}(\tau ). \end{aligned}$$
Solving the above partial differential equation, we get
$$\begin{aligned} Q(z,\tau )= & {} \eta \int \limits _{0}^{\tau }\left[ \sum \limits _{n=1}^{k-1}\pi _{n,0}(\tau )z^n\right] \mathrm{e}^{-(\lambda _{1}+ \mu _{1})(\tau -u) } \mathrm{e}^{\left( \lambda _{1} z+ \frac{ \mu _{1}}{z}\right) (\tau -u) }\hbox {d}u \nonumber \\&+\,\sigma \lambda _{0}\int \limits _{0}^{\tau }\pi _{k-1,0}(u)z^{k}\mathrm{e}^{-(\lambda _{1}+ \mu _{1})(\tau -u) }\mathrm{e}^{(\lambda _{1} z+ \frac{ \mu _{1}}{z})(\tau -u) }\hbox {d}u \nonumber \\&+\,\lambda _{1} \int \limits _{0}^{\tau }\pi _{0,1}(u)z\mathrm{e}^{-(\lambda _{1}+ \mu _{1})(\tau -u) }\mathrm{e}^{(\lambda _{1} z+ \frac{ \mu _{1}}{z})(\tau -u) }\hbox {d}u\nonumber \\&-\,\mu _{1} \int \limits _{0}^{\tau }\pi _{1,1}(u)\mathrm{e}^{-(\lambda _{1}+ \mu _{1})(\tau -u) }\mathrm{e}^{(\lambda _{1} z+ \frac{ \mu _{1}}{z})(\tau -u) }\hbox {d}u. \end{aligned}$$
(7.2)
Let
$$\begin{aligned} \mathrm{e}^{(\lambda _{1} z+ \frac{ \mu _{1}}{z})\tau }= & {} \sum \limits _{-\infty \leqslant n \leqslant \infty }^{}(\beta z)^n I_{n}(\alpha \tau ). \end{aligned}$$
(7.3)
Using (7.3) in (7.2) and simplifying for \(n\geqslant 1\), we get
$$\begin{aligned} \pi _{n,1}(\tau )&= \eta \int \limits _{0}^{\tau }\left[ \sum \limits _{m=1}^{k-1}\pi _{m,0}(u)\right] \mathrm{e}^{-(\lambda _{1}+ \mu _{1})(\tau -u) }\beta ^{n-m} I_{n-m}(\alpha (\tau -u))\hbox {d}u \nonumber \\&\quad +\sigma \lambda _{0}\int \limits _{0}^{\tau }\pi _{k-1,0}(u)\mathrm{e}^{-(\lambda _{1}+ \mu _{1})(\tau -u) }\beta ^{n-k}I_{n-k}(\alpha (\tau -u))\hbox {d}u \nonumber \\&\quad +\lambda _{1}\int \limits _{0}^{\tau }\pi _{0,1}(u)\mathrm{e}^{-(\lambda _{1}+ \mu _{1})(\tau -u) }\beta ^{n-1}I_{n-1}(\alpha (\tau -u))\hbox {d}u \nonumber \\&\quad -\mu _{1}\int \limits _{0}^{\tau }\pi _{1,1}(u)\mathrm{e}^{-(\lambda _{1}+ \mu _{1})(\tau -u) }\beta ^{n}I_{n}(\alpha (\tau -u))\hbox {d}u. \end{aligned}$$
(7.4)
The above expression holds for \(n = -1, -2, -3, \cdots \). Using \( I_{-n}(y)= I_{n}(y)\), for \(n \geqslant 1,\) we have
$$\begin{aligned} 0&=\eta \int \limits _{0}^{\tau }\sum \limits _{m=1}^{k-1}\pi _{m,0}(u) \mathrm{e}^{-(\lambda _{1}+ \mu )(\tau -u) }\beta ^{-n-m} I_{n+m}(\alpha (\tau -u))\hbox {d}u \nonumber \\&\quad +\sigma \lambda _{0}\int \limits _{0}^{\tau }\pi _{k-1,0}(u)\mathrm{e}^{-(\lambda _{1}+ \mu )(\tau -u) }\beta ^{-n-k}I_{n+k}(\alpha (\tau -u))\hbox {d}u \nonumber \\&\quad +\lambda _{1}\int \limits _{0}^{\tau }\pi _{0,1}(u)\mathrm{e}^{-(\lambda _{1}+ \mu )(\tau -u) }\beta ^{-n-1}I_{n+1}(\alpha (\tau -u))\hbox {d}u\nonumber \\&\quad -\mu \int \limits _{0}^{\tau }\pi _{1,1}(u)\mathrm{e}^{-(\lambda _{1}+ \mu )(\tau -u) }\beta ^{-n}I_{n}(\alpha (\tau -u))\hbox {d}u. \end{aligned}$$
(7.5)
From (7.4) and (7.5), for \(n \geqslant 1\), we obtain (2.8). Taking Laplace transform of (2.1), we get
$$\begin{aligned} \hat{\pi }_{0,1}(s)=&\left( \frac{\eta }{s+ \lambda _{1}}\right) \hat{\pi }_{0,0}(s). \end{aligned}$$
(7.6)
On Laplace inversion of (7.6), we obtain (2.9).
Appendix B
Proof of Theorem 2
Taking Laplace transform of (2.7), we get
$$\begin{aligned} \hat{\pi }_{k-1,0}(s)=&\left( \frac{\sigma \lambda _{0}}{s+\sigma \lambda _{0}+\mu _{0}+\eta } \right) \hat{\pi }_{k-2,0}(s). \end{aligned}$$
(7.7)
On Laplace inversion of (7.7), we obtain (2.10).
Taking Laplace transform of (2.6), for \( 1 \leqslant n \leqslant k-2\), we get
$$\begin{aligned} \frac{\hat{\pi }_{n,0}(s)}{\hat{\pi }_{n-1,0}(s)}=&\frac{\sigma \lambda _{0}}{(s+\sigma \lambda _{0}+\mu _{0}+\eta )-\mu _{0}\frac{\hat{\pi }_{n+1,0}(s)}{\hat{\pi }_{n,0}(s)}}. \end{aligned}$$
Iterating the above equation, for \( 1 \leqslant n \leqslant k-2\), we get
$$\begin{aligned} \hat{\pi }_{n,0}(s)=&\beta _{0}^{n} \left( \frac{w_{0}-\sqrt{w_{0}^{2}-\alpha _{0}^{2}}}{\alpha _{0}} \right) ^{n}\hat{\pi }_{0,0}(s), \end{aligned}$$
(7.8)
where \(w_{0}=s+\sigma \lambda _{0}+\mu _{0}+\eta \). Laplace inversion of (7.8) leads to (2.11).
Taking Laplace transform of (2.8), for \(n=1\), and using (7.6), (7.7) and (7.8), we get
$$\begin{aligned} \hat{\pi }_{1,1} (s)=&\hat{A}(s)\hat{\pi }_{0,0}(s), \end{aligned}$$
(7.9)
where
$$\begin{aligned} \hat{A}(s)&=\frac{\eta }{\lambda _{1}} \sum \limits _{m=1}^{k-2}\beta _{0}^{m}\beta _{1}^{2-m}\left( \frac{w_{0}-\sqrt{w_{0}^{2}-\alpha _{0}^{2}}}{\alpha _{0}} \right) ^{m}\left( \frac{w_{1}-\sqrt{w_{1}^{2}-\alpha _{1}^{2}}}{\alpha _{1}} \right) ^{m}\\&\quad +\frac{\eta }{\lambda _{1}}\beta _{0}^{k-2}\beta _{1}^{3-k} \left( \frac{ \sigma \lambda _{0}}{s+\sigma \lambda _{0}+\mu _{0}+\eta }\right) \\&\quad \left( \frac{w_{0}-\sqrt{w_{0}^{2}-\alpha _{0}^{2}}}{\sigma \alpha _{0}} \right) ^{k-2}\left( \frac{w_{1}-\sqrt{w_{1}^{2}-\alpha _{1}^{2}}}{\alpha _{1}} \right) ^{k-1}\\&\quad +\beta _{1}\left( \frac{ \eta }{s+\lambda _{1}}\right) \left( \frac{w_{1}-\sqrt{w_{1}^{2}-\alpha _{1}^{2}}}{\alpha _{1}} \right) \\&\quad +\frac{(\sigma \lambda _{0})^{2}}{\lambda _{1}}\beta _{0}^{k-2}\beta _{1}^{2-k} \left( \frac{ 1}{s+\sigma \lambda _{0}+\mu _{0}+\eta }\right) \\&\quad \left( \frac{w_{0}-\sqrt{w_{0}^{2}-\alpha _{0}^{2}}}{\alpha _{0}} \right) ^{k-2}\left( \frac{w_{1}-\sqrt{w_{1}^{2}-\alpha _{1}^{2}}}{\alpha _{1}} \right) ^{k}. \end{aligned}$$
Laplace transform of (2.5) yields
$$\begin{aligned} \hat{\pi }_{0,0}(s)=&\frac{1}{(s+\sigma \lambda _{0}+\eta )-\mu _{0}\frac{\hat{\pi }_{1,0}(s)}{\hat{\pi }_{0,0}(s)}-\mu _{1}\frac{\hat{\pi }_{1,1}(s)}{\hat{\pi }_{0,0}(s)}}. \end{aligned}$$
(7.10)
Using (7.8), for \(n=1\), and (7.9) in (7.10), we get
$$\begin{aligned} \hat{\pi }_{0,0}(s)=&\sum \limits _{k=0}^{\infty }\sum \limits _{j=0}^{k}\left( {\begin{array}{c}{k} \\ j \\ \end{array}} \right) \frac{(\mu _{0}\beta _{0})^{j}\mu _{1}^{k-j}{\hat{A}(s)}^{k-j}}{(s+\sigma \lambda _{0}+\eta )^{k+1}}\left( \frac{w_{0}-\sqrt{w_{0}^{2}-\alpha _{0}^{2}}}{\alpha _{0}} \right) ^{j}. \end{aligned}$$
(7.11)
On Laplace inversion of (7.11), we obtain (2.12). It is difficult to derive the steady-state distributions from the transient counterpart as the results in time-dependent state are obtained as finite and infinite summations.
Appendix C
Proof of Theorem 3
Applying (7.1) in Eqs. (3.1) to (3.4), we obtain
$$\begin{aligned} \frac{\partial Q (z,\tau )}{\partial \tau }&=\left[ -(\lambda _{1}+ \mu _{1})+\lambda _{1}z+\frac{ \mu _{1} }{z} \right] Q (z,\tau )\\&\quad +\eta \sum \limits _{n=1}^{k-1}\pi _{n,0}(\tau )z^n+\sigma \lambda _{0}\pi _{k-1,0}(\tau )z^{k-1}- \mu _{1} \pi _{1,1}(\tau ). \end{aligned}$$
Solving the above partial differential equation, we obtain
$$\begin{aligned} Q (s,\tau )&=\eta \int \limits _{0}^{\tau }\left[ \sum \limits _{n=1}^{k-1}\pi _{n,0}(\tau )z^n\right] \mathrm{e}^{-(\lambda _{1}+ \mu _{1})(\tau -u) } \mathrm{e}^{\left( \lambda _{1} z+ \frac{ \mu _{1}}{z}\right) (\tau -u) }\hbox {d}u \nonumber \\&\quad +\sigma \lambda _{0}\int \limits _{0}^{\tau }\pi _{k-1,0}(u)z^{k-1}\mathrm{e}^{-(\lambda _{1}+ \mu _{1})(\tau -u) }\mathrm{e}^{(\lambda _{1} z+ \frac{ \mu _{1}}{z})(\tau -u) }\hbox {d}u\nonumber \\&\quad -\mu _{1} \int \limits _{0}^{\tau }\pi _{1,1}(u)\mathrm{e}^{-(\lambda _{1}+ \mu _{1})(\tau -u) }\mathrm{e}^{(\lambda _{1} z+ \frac{ \mu _{1}}{z})(\tau -u)}\hbox {d}u. \end{aligned}$$
(7.12)
Using (7.3) in (7.12) and simplifying for \(n\geqslant 1\), we get
$$\begin{aligned} \pi _{n,1}(\tau )&= \eta \int \limits _{0}^{\tau }\left[ \sum \limits _{m=1}^{k-1}\pi _{m,0}(u)\right] \mathrm{e}^{-(\lambda _{1}+ \mu )(\tau -u) }\beta ^{n-m} I_{n-m}(\alpha (\tau -u))\hbox {d}u \nonumber \\&\quad +\sigma \lambda _{0}\int \limits _{0}^{\tau }\pi _{k-1,0}(u)\mathrm{e}^{-(\lambda _{1}+ \mu )(\tau -u) }\beta ^{n-k+1}I_{n-k+1}(\alpha (\tau -u))\hbox {d}u \nonumber \\&\quad -\mu \int \limits _{0}^{\tau }\pi _{1,1}(u)\mathrm{e}^{-(\lambda _{1}+ \mu )(\tau -u) }\beta ^{n}I_{n}(\alpha (\tau -u))\hbox {d}u. \end{aligned}$$
(7.13)
The above expression holds for \(n = -1, -2, -3, \cdots \). Using \( I_{-n}(y)= I_{n}(y)\), for \(n \geqslant 1,\) we have
$$\begin{aligned} 0&=\eta \int \limits _{0}^{\tau }\sum \limits _{m=1}^{k-1}\pi _{m,0}(u) \mathrm{e}^{-(\lambda _{1}+ \mu )(\tau -u) }\beta ^{-n-m} I_{n+m}(\alpha (\tau -u))\hbox {d}u \nonumber \\&\quad +\sigma \lambda _{0}\int \limits _{0}^{\tau }\pi _{k-1,0}(u)\mathrm{e}^{-(\lambda _{1}+ \mu )(\tau -u) }\beta ^{-n-k+1}I_{n+k-1}(\alpha (\tau -u))\hbox {d}u \nonumber \\&\quad -\mu \int \limits _{0}^{\tau }\pi _{1,1}(u)\mathrm{e}^{-(\lambda _{1}+ \mu )(\tau -u) }\beta ^{-n}I_{n}(\alpha (\tau -u))\hbox {d}u. \end{aligned}$$
(7.14)
From (7.13) and (7.14), for \(n \geqslant 1\), we obtain (3.8).
Appendix D
Proof of Theorem 4
Taking Laplace transform of (3.8), for \(n=1\), and using (7.7) and (7.8), we get
$$\begin{aligned} \hat{\pi }_{1,1} (s)=&\hat{B}(s)\hat{\pi }_{0,0}(s), \end{aligned}$$
(7.15)
where
$$\begin{aligned} \hat{B}(s)&=\frac{\eta }{\lambda _{1}} \sum \limits _{m=1}^{k-2}\beta _{0}^{m}\beta _{1}^{2-m}\left( \frac{w_{0}-\sqrt{w_{0}^{2}-\alpha _{0}^{2}}}{\alpha _{0}} \right) ^{m}\left( \frac{w_{1}-\sqrt{w_{1}^{2}-\alpha _{1}^{2}}}{\alpha _{1}} \right) ^{m}\\&\quad +\frac{\eta }{\lambda _{1}}\beta _{0}^{k-2}\beta _{1}^{3-k} \left( \frac{ \sigma \lambda _{0}}{s+\sigma \lambda _{0}+\mu _{0}+\eta }\right) \\&\quad \left( \frac{w_{0}-\sqrt{w_{0}^{2}-\alpha _{0}^{2}}}{\alpha _{0}} \right) ^{k-2}\left( \frac{w_{1}-\sqrt{w_{1}^{2}-\alpha _{1}^{2}}}{\alpha _{1}} \right) ^{k-1}\\&\quad +\frac{(\sigma \lambda _{0})^{2}}{\lambda _{1}}\beta _{0}^{k-2}\beta _{1}^{2-k} \left( \frac{ 1}{s+\sigma \lambda _{0}+\mu _{0}+\eta }\right) \\&\quad \left( \frac{w_{0}-\sqrt{w_{0}^{2}-\alpha _{0}^{2}}}{\alpha _{0}} \right) ^{k-2}\left( \frac{w_{1}-\sqrt{w_{1}^{2}-\alpha _{1}^{2}}}{\alpha _{1}} \right) ^{k}. \end{aligned}$$
Laplace transform of (3.5) yields
$$\begin{aligned} \hat{\pi }_{0,0}(s)=&\frac{1}{(s+\sigma \lambda _{0})-\mu _{0}\frac{\hat{\pi }_{1,0}(s)}{\hat{\pi }_{0,0}(s)}-\mu _{1}\frac{\hat{\pi }_{1,1}(s)}{\hat{\pi }_{0,0}(s)}}. \end{aligned}$$
(7.16)
Using (7.8), for \(n=1\), and (7.15) in (7.16), we get
$$\begin{aligned} \hat{\pi }_{0,0}(s)=&\sum \limits _{k=0}^{\infty }\sum \limits _{j=0}^{k}\left( {\begin{array}{c}{k} \\ j \\ \end{array}} \right) \frac{(\mu _{0}\beta _{0})^{j}\mu _{1}^{k-j}{\hat{B}(s)}^{k-j}}{(s+\sigma \lambda _{0})^{k+1}}\left( \frac{w_{0}-\sqrt{w_{0}^{2}-\alpha _{0}^{2}}}{\alpha _{0}} \right) ^{j}. \end{aligned}$$
(7.17)
On Laplace inversion of (7.17), we obtain (3.9).
