Stationary analysis of the shortest queue first service policy

Abstract

We analyze the so-called shortest queue first (SQF) queueing discipline whereby a unique server addresses queues in parallel by serving at any time, the queue with the smallest workload. Considering a stationary system composed of two parallel queues and assuming Poisson arrivals and general service time distributions, we first establish the functional equations satisfied by the Laplace transforms of the workloads in each queue. We further specialize these equations to the so-called “symmetric case,” with same arrival rates and identical exponential service time distributions at each queue; we then obtain a functional equation

$$\begin{aligned} M(z) = q(z) \cdot M \circ h(z) + L(z) \end{aligned}$$

for unknown function M, where given functions $q, L$, and $h$ are related to one branch of a cubic polynomial equation. We study the analyticity domain of function M and express it by a series expansion involving all iterates of function $h$. This allows us to determine empty queue probabilities along with the tail of the workload distribution in each queue. This tail appears to be identical to that of the head-of-line preemptive priority system, which is the key feature desired for the SQF discipline.

Appendices

Appendix 1: Proof for Assertion (B) of Proposition 3.1

Before proving Eqs. (3.7), we state preliminary expressions of $G_1$ and $G_2$.

Lemma 7.1

Given

$$\begin{aligned} E_{21}(s_1) = \int \limits _0^{+\infty } \mathrm{e}^{-s_1u_1} \varphi _2(u_1,0) \mathrm{d}u_1, \quad E_{12}(s_2) = \int \limits _0^{+\infty } \mathrm{e}^{-s_2u_2} \varphi _1(0,u_2) \mathrm{d}u_2, \end{aligned}$$

(7.1)

univariate transforms $G_1$ and $G_2$ satisfy

$$\begin{aligned} \left\{ \begin{array}{ll} G_1(s_1) = \displaystyle \frac{(1-\varrho )(\lambda - \lambda _1b_1(s_1)) - E_{21}(s_1) - \psi _2(0)}{s_1 - \lambda + \lambda _1b_1(s_1)}, \quad \mathfrak {R}(s_1) > 0,\\ G_2(s_2) = \displaystyle \frac{(1-\varrho )(\lambda - \lambda _2b_2(s_2)) - E_{12}(s_2) - \psi _1(0)}{s_2 - \lambda + \lambda _2b_2(s_2)}, \quad \mathfrak {R}(s_2) > 0. \end{array} \right. \end{aligned}$$

(7.2)

Proof

As $b_2$ and $F_1$ are transforms of distributions with no mass at the origin, we have $b_2(s_2) \rightarrow 0, F_1(s_1,s_2) \rightarrow 0$ when $s_2 \uparrow +\infty$ for fixed $s_1$ with $\mathfrak {R}(s_1) > 0$. Besides, we have $s_2F_2(s_1,s_2) \rightarrow E_{21}(s_1), s_2G_2(s_2) \rightarrow \psi _2(0)$ when $s_2 \uparrow +\infty$ with fixed $s_1, \; \mathfrak {R}(s_1) > 0$, where $E_{21}$ is the Laplace transform of the restriction of distribution $\varphi _2$ on the boundary $\delta _1$; as a consequence,

$$\begin{aligned} \lim _{s_2 \uparrow +\infty } s_2H_2(s_1,s_2) = \lim _{s_2 \uparrow +\infty } s_2(F_2(s_1,s_2) + G_2(s_2)) = E_{21}(s_1) + \psi _2(0) \end{aligned}$$

for fixed $s_1, \; \mathfrak {R}(s_1) > 0$. Now, letting $s_2 \uparrow +\infty$ in each side of (3.6), the above limit results entail $(s_1 - K(s_1,\infty ))G_1(s_1) + E_{21}(s_1) + \psi _2(0) = (1-\varrho )K(s_1,\infty )$ with $K(s_1,\infty ) = \lambda - \lambda _1b_1(s_1)$, which provides identity (7.2) for $G_1(s_1)$. Identity (7.2) for $G_2(s_2)$ is symmetrically deduced by letting $s_1 \uparrow +\infty$ in (3.7) with fixed $s_2, \; \mathfrak {R}(s_2) > 0$.$\square$

We now address the derivation of Eqs. (3.7). Recall that subsets $\gamma _1, \delta _1$, etc. of state space $\mathcal {U}$ are defined in (2.6)–(2.7). Given $\varepsilon > 0$, define the function $Y_\varepsilon$ by $Y_\varepsilon (v) = \exp (-\varepsilon /v){\small 1}\!\!1_{\{v > 0\}}$; $Y_\varepsilon$ is twice continuously differentiable over $\mathbb {R}$, $\lim _{\varepsilon \downarrow 0} Y_\varepsilon (v) = {\small 1}\!\!1_{\{v > 0\}}$ for each $v \in \mathbb {R}$ and $\lim _{\varepsilon \downarrow 0} Y_\varepsilon ' = \delta _0$ (the Dirac mass at $v = 0$) for the weak convergence of distributions. For given $\mathfrak {R}(s_1) > 0, \mathfrak {R}(s_2) > 0$, let then be the test function: $\theta _\varepsilon (\mathbf{u}) = \mathrm{e}^{-\mathbf{s} \cdot \mathbf{u}}\chi _\varepsilon (\mathbf{u}), \mathbf{u} \in \mathcal {U}$, with

$$\begin{aligned} \chi _\varepsilon (\mathbf{u}) = Y_\varepsilon (u_1)Y_\varepsilon (u_2-u_1). \end{aligned}$$

(7.3)

Function $\theta _\varepsilon$ belongs to $\mathcal {C}_b^2(\mathcal {U})$ and is zero on the outside of $\gamma _1$; moreover, we have $\lim _{\varepsilon \downarrow 0} \chi _\varepsilon (\mathbf{u}) = {\small 1}\!\!1_{\{\mathbf{u} \in \gamma _1\}}$ so that $\lim _{\varepsilon \downarrow 0} \theta _\varepsilon = \theta$ pointwise in $\mathcal {U}$, with limit function $\theta$ defined by $\theta (\mathbf{u}) = \mathrm{e}^{-\mathbf{s} \cdot \mathbf{u}}{\small 1}\!\!1_{\{\mathbf{u} \in \gamma _1\}}, \mathbf{u} \in \mathcal {U}$.

By direct differentiation, we further calculate

$$\begin{aligned} \frac{\partial \theta _\varepsilon }{\partial u_1}(\mathbf{u}) = -s_1\theta _\varepsilon (\mathbf{u}) + \mathrm{e}^{-\mathbf{s}\cdot \mathbf{u}} \frac{\partial \chi _\varepsilon }{\partial u_1}(\mathbf{u}), \quad \; \displaystyle \frac{\partial \theta _\varepsilon }{\partial u_2}(\mathbf{u}) = -s_2\theta _\varepsilon (\mathbf{u}) + \mathrm{e}^{-\mathbf{s}\cdot \mathbf{u}} \frac{\partial \chi _\varepsilon }{\partial u_2}(\mathbf{u}) \end{aligned}$$

for $\mathbf{u} \in \mathcal {U}$, with

$$\begin{aligned} \frac{\partial \chi _\varepsilon }{\partial u_1}(\mathbf{u}) = Y_\varepsilon '(u_1)Y_\varepsilon (u_2-u_1) - Y_\varepsilon (u_1)Y_\varepsilon '(u_2-u_1) \end{aligned}$$

after (7.3); note that derivative $\partial \chi _\varepsilon / \partial u_1$ tends to $\delta _0(u_1) - \delta _0(u_2 - u_1)$ for the weak convergence of distributions as $\varepsilon \downarrow 0$.

Let us now calculate the limit $\mathcal {M} = \lim _{\varepsilon \downarrow 0} \mathcal {M}(\varepsilon )$ with $\mathcal {M}(\varepsilon )$ introduced in (3.9); to this end, we address successive terms of $\smallint _{\mathcal {U}}\mathcal {A}\theta _\varepsilon \mathrm{d}\Phi$ according to Definition (2.9). Integrating first $\partial \theta _\varepsilon /\partial u_1$ over $\gamma _1 \cup \delta _1$ against $\mathrm{d}\Phi$ reduces to

$$\begin{aligned} - \int \limits _{\gamma _1 \cup \delta _1} \frac{\partial \theta _\varepsilon }{\partial u_1} \mathrm{d}\Phi = - \int \limits _{\gamma _1} \frac{\partial \theta _\varepsilon }{\partial u_1}(\mathbf{u})\varphi _1(\mathbf{u})\mathrm{d}\mathbf{u} = \int \limits _{\gamma _1} \left[ s_1 \theta _\varepsilon (\mathbf{u}) - \mathrm{e}^{-\mathbf{s} \cdot \mathbf{u}} \frac{\partial \chi _\varepsilon }{\partial u_1}(\mathbf{u})\right] \varphi _1(\mathbf{u})\mathrm{d}\mathbf{u} \end{aligned}$$

since $\partial \theta _\varepsilon /\partial u_1$ vanishes on the outside of $\gamma _1$; on account of the above mentioned weak convergence properties, we then obtain

$$\begin{aligned} \lim _{\varepsilon \downarrow 0} - \int \limits _{\gamma _1 \cup \delta _1} \frac{\partial \theta _\varepsilon }{\partial u_1}(\mathbf{u})\mathrm{d}\Phi (\mathbf{u}) = s_1F_1(s_1,s_2) - E_{12}(s_2) + Z_1(s_1+s_2) \end{aligned}$$

(7.4)

with $E_{12}(s_2)$ defined as in Lemma 7.1 and where $Z_1(s) = \smallint _{u > 0} \; \mathrm{e}^{-su} \varphi _1(u,u) \mathrm{d}u$ defines the Laplace transform of distribution $\varphi _1$ restricted to the positive diagonal $\delta$ (function $Z_1$ is determined below). Besides, the integral of $\partial \theta _\varepsilon /\partial u_2$ over $\gamma _2 \cup \delta _2$ equals zero as this function vanishes on the outside of $\gamma _1$.

Further, we have $\lim _{\varepsilon \downarrow 0} \theta _\varepsilon (\mathbf{u} + \mathcal {T}_1\mathbf{e}_1) = \mathrm{e}^{-\mathbf{s} \cdot \mathbf{u}} \mathrm{e}^{-s_1\mathcal {T}_1} {\small 1}\!\!1_{\{\mathcal {T}_1 < u_2 - u_1\}}$ for given $\mathbf{u} \in \mathcal {U}$ and $\mathcal {T}_1 > 0$, therefore $\lim _{\varepsilon \downarrow 0} \mathbb {E}\theta _\varepsilon (\mathbf{u} + \mathcal {T}_1\mathbf{e}_1) = \mathrm{e}^{-\mathbf{s} \cdot \mathbf{u}} \mathbb {E}(\mathrm{e}^{-s_1\mathcal {T}_1} {\small 1}\!\!1_{\{\mathcal {T}_1 < u_2 - u_1\}})$ by the Dominated Convergence Theorem; hence

$$\begin{aligned}&\lim _{\varepsilon \downarrow 0} \; \mathbb {E}\theta _\varepsilon (\mathbf{U} + \mathcal {T}_1\mathbf{e}_1) = \int \limits _{\mathcal {U}} \mathrm{e}^{-\mathbf{s} \cdot \mathbf{u}} \mathbb {E} \left( \mathrm{e}^{-s_1\mathcal {T}_1} {\small 1}\!\!1_{\{\mathcal {T}_1 < u_2 - u_1\}} \right) \mathrm{d}\Phi (\mathbf{u}) \nonumber \\&\quad = \int \limits _{\gamma _1} \mathrm{e}^{-\mathbf{s} \cdot \mathbf{u}} \mathbb {E} \left( \mathrm{e}^{-s_1\mathcal {T}_1} {\small 1}\!\!1_{\{\mathcal {T}_1 < u_2 - u_1\}}\right) \varphi _1(\mathbf{u})\mathrm{d}\mathbf{u} \nonumber \\&\quad \quad + \int \limits _0^{+\infty } \mathrm{e}^{-s_2u_2}\mathbb {E} \left( \mathrm{e}^{-s_1\mathcal {T}_1} {\small 1}\!\!1_{\{\mathcal {T}_1 < u_2\}}\right) \psi _2(u_2)\mathrm{d}u_2 \end{aligned}$$

(7.5)

where random variable $\mathbf{U} = (U_1,U_2)$ has distribution $\Phi$. For given $\mathbf{u} \in \mathcal {U}$, we similarly have $\lim _{\varepsilon \downarrow 0} \mathbb {E}\theta _\varepsilon (\mathbf{u} + \mathcal {T}_2\mathbf{e}_2) = \mathrm{e}^{-\mathbf{s} \cdot \mathbf{u}} \mathbb {E}(\mathrm{e}^{-s_2\mathcal {T}_2} {\small 1}\!\!1_{\{\mathcal {T}_2 > u_1 - u_2, u_1 > 0\}})$ and therefore

$$\begin{aligned}&\lim _{\varepsilon \downarrow 0} \; \mathbb {E}\theta _\varepsilon (\mathbf{U} + \mathcal {T}_2\mathbf{e}_2) = \int \limits _{\mathcal {U}} \mathrm{e}^{-\mathbf{s} \cdot \mathbf{u}} \mathbb {E} \left( \mathrm{e}^{-s_2\mathcal {T}_2} {\small 1}\!\!1_{\{\mathcal {T}_2 > u_1 - u_2, u_1 > 0\}}\right) \mathrm{d}\Phi (\mathbf{u}) \nonumber \\&\quad = b_2(s_2)F_1(s_1,s_2) + \int \limits _{\gamma _2} \mathrm{e}^{-\mathbf{s} \cdot \mathbf{u}} \mathbb {E} \left( \mathrm{e}^{-s_2\mathcal {T}_2} {\small 1}\!\!1_{\{\mathcal {T}_2 > u_1 - u_2\}}\right) \varphi _2(\mathbf{u}) \mathrm{d}\mathbf{u} \nonumber \\&\qquad + \int \limits _0^{+\infty } \mathrm{e}^{-s_1u_1}\mathbb {E} \left( \mathrm{e}^{-s_2\mathcal {T}_2} {\small 1}\!\!1_{\{\mathcal {T}_2 > u_1\}}\right) \psi _1(u_1)\mathrm{d}u_1. \end{aligned}$$

(7.6)

Finally, noting that $\lim _{\varepsilon \downarrow 0} \smallint _{\mathcal {U}} \theta _\varepsilon (\mathbf{u}) \mathrm{d}\Phi (\mathbf{u}) = F_1(s_1,s_2)$ and adding limit terms (7.4), (7.5), (7.6) according to (2.9) gives limit $\mathcal {M} = \lim _{\varepsilon \downarrow 0} \mathcal {M}(\varepsilon ) = 0$ the final expression

$$\begin{aligned}&s_1 F_1(s_1,s_2) - E_{12}(s_2) + Z_1(s_1 + s_2) \nonumber \\&\quad +\,\, \lambda _1 \left[ b_1(s_1)F_1(s_1,s_2) - \int \limits _{\gamma _1} \mathrm{e}^{-\mathbf{s} \cdot \mathbf{u}} \mathbb {E} \left( \mathrm{e}^{-s_1\mathcal {T}_1} {\small 1}\!\!1_{\{\mathcal {T}_1 > u_2 - u_1\}}\right) \varphi _1(\mathbf{u})\mathrm{d}\mathbf{u}\right] \\&\quad + \,\,\lambda _1 \left[ b_1(s_1)G_2(s_2) - \int \limits _0^{+\infty } \mathrm{e}^{-s_2u_2}\mathbb {E} \left( \mathrm{e}^{-s_1\mathcal {T}_1} {\small 1}\!\!1_{\{\mathcal {T}_1 > u_2\}}\right) \psi _2(u_2)\mathrm{d}u_2\right] \\&\quad + \,\,\lambda _2 b_2(s_2) F_1(s_1,s_2) + \lambda _2 \; \bigg [ \int \limits _{\gamma _2} \mathrm{e}^{-\mathbf{s} \cdot \mathbf{u}} \mathbb {E} \left( \mathrm{e}^{-s_2\mathcal {T}_2} {\small 1}\!\!1_{\{\mathcal {T}_2 > u_1 - u_2\}}\right) \varphi _2(\mathbf{u}) \mathrm{d}\mathbf{u}\\&\quad + \,\,\int \limits _0^{+\infty } \mathrm{e}^{-s_1u_1}\mathbb {E} \left( \mathrm{e}^{-s_2\mathcal {T}_2} {\small 1}\!\!1_{\{\mathcal {T}_2 > u_1\}} \right) \psi _1(u_1)\mathrm{d}u_1 \bigg ] - \lambda F_1(s_1,s_2) = 0. \end{aligned}$$

Defining $H(s_1,s_2)$ as in (3.8) to gather all remaining integrals, the latter identity reads

$$\begin{aligned} K_1(s_1,s_2)F_1(s_1,s_2) + \lambda _1b_1(s_1) G_2(s_2) = E_{12}(s_2) - Z_1(s_1 + s_2) + H(s_1,s_2)\qquad \quad \end{aligned}$$

(7.7)

with $K_1(s_1,s_2) = s_1-K(s_1,s_2), E_{12}(s_2)$ being defined by (7.1) and where $Z_1$ defines the Laplace transform of distribution $\varphi _1$ restricted to the diagonal. Changing index 1 into 2, and noting that $H(s_1,s_2)$ changes into $-H(s_1,s_2)$, symmetrically yields second equation:

$$\begin{aligned} K_2(s_1,s_2)F_2(s_1,s_2) + \lambda _2b_2(s_2) G_1(s_1) = E_{21}(s_1) - Z_2(s_1 + s_2) - H(s_1,s_2)\qquad \quad \end{aligned}$$

(7.8)

with $K_2(s_1,s_2) = s_2 - K(s_1,s_2), E_{21}(s_1)$ being defined by (7.1) and where $Z_2$ defines the Laplace transform of distribution $\varphi _2$ restricted to the diagonal. To conclude the proof, we prove the following technical lemma:

Lemma 7.2

Functions $Z_1$ and $Z_2$ are identically zero.

Proof

Adding Eqs. (7.7) and (7.8) (and omitting arguments for notation simplicity) yields $K_1F_1 + K_2F_2 = -\lambda _1b_1G_2 - \lambda _2b_2G_1 + E_{12} - Z_1 + E_{21} - Z_2$. On the other hand, Eq. (3.6) gives $K_1F_1 + K_2F_2 = (1-\varrho )K - K_1G_1 - K_2G_2$; equating right-hand sides of the latter equations then provides the identity

$$\begin{aligned}&(1-\varrho )K(s_1,s_2) - (s_1-\lambda +\lambda _1b_1(s_1))G_1(s_1) - (s_2-\lambda +\lambda _2b_2(s_2))G_2(s_2) \\&\quad = E_{12}(s_2) + E_{21}(s_1) - Z_1(s_1+s_2) - Z_2(s_1+s_2). \end{aligned}$$

Using expressions (7.2) for $G_1(s_1)$ and $G_2(s_2)$, the latter identity simply reduces to $Z_1(s_1+s_2) + Z_2(s_1 + s_2) = \lambda (1-\varrho ) - \psi _1(0) - \psi _2(0)$, showing that function $Z_1 + Z_2$ is constant. As both $Z_1$ and $Z_2$ vanish at $+\infty$, this constant is 0 and since these functions are non-negative by definition, this entails that $Z_1 = Z_2 = 0$.$\square$

After using Eqs. (7.2) to express $E_{12}(s_2)$ and $E_{21}(s_1)$ in terms of $G_2(s_2)$ and $G_1(s_1)$, respectively, Lemma 7.2 finally enables us to reduce (7.7) and (7.8) to Eqs. (3.7). This concludes the proof of Proposition 3.1.

Appendix 2: Proof of Proposition 3.3

For an exponentially distributed service time $\mathcal {T}_1$ with parameter $\mu _1$, the factor of $\lambda _1$ in Definition (3.8) of $H(s_1,s_2)$ reads

$$\begin{aligned}&{\mathbb {E}}\left[ \mathrm{e}^{-s_1U_1-s_2U_2}{\small 1}\!\!1_{\{0 \le U_1 < U_2\}}\mathrm{e}^{-s_1\mathcal {T}_1}{\small 1}\!\!1_{\{\mathcal {T}_1 > U_2-U_1\}}\right] \\&\quad = \int \limits _0^{+\infty } \int \limits _0^{+\infty } \left[ \int \limits _{u_2-u_1}^{+\infty } \mathrm{e}^{-s_1x_1} \mu _1 \mathrm{e}^{-\mu _1x_1} \mathrm{d}x_1\right] \mathrm{e}^{-s_1u_1-s_2u_2}{\small 1}\!\!1_{\{0 \le u_1 < u_2\}}\mathrm{d}\Phi (u_1,u_2), \end{aligned}$$

for $\mathfrak {R}(s_1)\ge$ and $\mathfrak {R}(s_2)\ge 0$. By Definition (2.8), the latter term is equal to

$$\begin{aligned}&\int \limits _0^{+\infty } \int \limits _0^{+\infty } \frac{\mu _1}{\mu _1+s_1}\mathrm{e}^{\mu _1u_1-(s_1+s_2+\mu _1)u_2}{\small 1}\!\!1_{\{0 \le u_1 < u_2\}} \\&\qquad \times \,\,\Big [ \psi _2(u_2){\small 1}\!\!1_{\{u_2 > 0\}}\mathrm{d}u_2 \otimes \delta _0(u_1) + \varphi _1(u_1,u_2) \mathrm{d}u_1\mathrm{d}u_2 \Big ] \\&\qquad = \frac{\mu _1}{\mu _1+s_1} \big [ G_2(s_1+s_2+\mu _1) + F_1(-\mu _1,s_1+s_2+\mu _1) \big ] \end{aligned}$$

where, by Corollary 3.2, each term inside brackets is analytically defined for $(s_1,s_2)$ such that $\mathfrak {R}(s_1+s_2 + \mu _1) > \widetilde{s}_2$ and $\mathfrak {R}(s_1+s_2+\mu _1) > \max (\widetilde{s}_2,\widetilde{s}_2+\mu _1) = \widetilde{s}_2 + \mu _1$, respectively, that is at least for $\mathfrak {R}(s_1 + s_2) > \widetilde{s}_2$. Similarly, for an exponentially distributed service time $\mathcal {T}_2$ with parameter $\mu _2$, the factor of $\lambda _2$ in Definition (3.8) of $H(s_1,s_2)$ reads

$$\begin{aligned}&- {\mathbb {E}}\left[ \mathrm{e}^{-s_1U_1-s_2U_2}{\small 1}\!\!1_{\{0 \le U_2 < U_1\}}\mathrm{e}^{-s_2\mathcal {T}_2}{\small 1}\!\!1_{\{\mathcal {T}_2 > U_1-U_2\}}\right] \\&\quad = - \frac{\mu _2}{\mu _2+s_2}\left[ G_1(s_1+s_2+\mu _2) + F_2(s_1+s_2+\mu _2,-\mu _2)\right] \end{aligned}$$

where, by Corollary 3.2, each term inside brackets is analytically defined for $(s_1,s_2)$ such that $\mathfrak {R}(s_1+s_2 + \mu _2) > \widetilde{s}_1$ and $\mathfrak {R}(s_1+s_2+\mu _2) > \max (\widetilde{s}_1,\widetilde{s}_1+\mu _2) = \widetilde{s}_1 + \mu _2$, respectively, hence for $\mathfrak {R}(s_1 + s_2) > \widetilde{s}_1$. Adding up the two above expressions, we obtain claimed expressions.

Appendix 3: Proof of Proposition 5.2

With $s = z - \alpha (z)$ and $\xi ^-(s) = 2z - s = z + \alpha (z)$, second equation (5.3) reads

$$\begin{aligned} M(z)&= \frac{z + \alpha (z)+\mu }{\mu } \Big [ \frac{2(z-\alpha (z)+\mu )}{\lambda }G(z-\alpha (z)) \nonumber \\&+ \; (1-\varrho )\frac{(z+\alpha (z)+\mu )}{2\alpha (z)} \Big ]. \end{aligned}$$

(9.1)

We successively make the following points:

• By Lemma 4.2, function $z \mapsto \alpha (z)$ is analytic on the cut plane $\mathbb {C} {\setminus } [\eta _2,\eta _1]$, where ramification points $\eta _2, \eta _1$ are determined as the real negative roots of discriminant $\Delta (z)$. As $\eta _2 =-\mu < \eta _1 < 0$, function $z \mapsto \alpha (z)$ is, in particular, analytic in the half-plane $\{z \in \mathbb {C} \mid \; \mathfrak {R}(z) > \eta _1\}$.

• By Definition (4.12), we may have $\alpha (z) = 0$ only if $z(z+\mu )(z+\mu -\lambda )= 0$, that is, $z = 0$ or $z = -\mu$ or $z=\sigma _0=-\mu (1-\varrho )$; in the case $z=0$, we have

  $$\begin{aligned} \alpha (0) = \frac{\lambda -\sqrt{\lambda ^2+4\mu ^2}}{2}< \beta (0)=0 < \gamma (0) = \frac{\lambda +\sqrt{\lambda ^2+4\mu ^2}}{2}, \end{aligned}$$

  and in the case $z=\sigma _0$,

  $$\begin{aligned} \alpha (\sigma _0) = \frac{-\mu -\sqrt{\mu ^2+4\lambda ^2}}{2}< \beta (\sigma _0)=0 < \gamma (\sigma _0) = \frac{-\mu +\sqrt{\mu ^2+4\lambda ^2}}{2}; \end{aligned}$$

  hence, we conclude that we cannot have $\alpha (z)=0$ if $\mathfrak {R}(z)>\eta _1$.

• By Corollary 3.2, transform $G$ is analytic on $\widetilde{\omega } = \{s \in \mathbb {C} \mid \; \mathfrak {R}(s) > \widetilde{s} \}$ where $\widetilde{s} = \sigma _0 = -\mu (1-\varrho )$ if $\varrho > 1/2$ and $\widetilde{s} = \zeta ^+$ if $\varrho < 1/2$.

From expression (9.1) and the latter observations, we deduce that $M$ is analytic at any point $z$ with $\mathfrak {R}(z) > \eta _1$ and

$$\begin{aligned} \mathfrak {R}(A(z)) > \widetilde{s} \end{aligned}$$

(9.2)

where $A(z) = z - \alpha (z)$.

• (a) Assume first that $\varrho >1/2$. In the $(O,z,s)$ plane, the diagonal $z=s$ intersects the curve $z = X^+(s) = (s+\xi ^+(s))/2$ at $s=\sigma _0$ (see Fig. 5). Further, we easily verify that $A(z) = z-\alpha (z)>\sigma _0$ for

  $$\begin{aligned} z > \frac{\sigma _0+\xi ^-(\sigma _0)}{2} = \frac{1}{2} \left( \sigma _0-\frac{\mu }{2} \right) \end{aligned}$$

  and condition (9.2) is therefore fulfilled in this first case. We then conclude that function $M$ is analytic for $z > \frac{1}{2}(\sigma _0 - \frac{\mu }{2})$, and thus for $\mathfrak {R}(z) > \frac{1}{2}(\sigma _0 - \frac{\mu }{2})$ (recall by Definition (4.4) that $M$ is the sum of two non-negative Laplace transforms).

• (b) Assume now that $\varrho \le 1/2$ (see Fig. 6). We have shown above that we cannot have $\sigma _0=\sigma _0-\alpha (\sigma _0)$, which would otherwise imply $\alpha (\sigma _0)=0$. We thus necessarily have $\sigma _0 < s^*$, which entails that $A(z) = z-\alpha (z) > \sigma _0$ for $z>\eta _1$ and condition (9.2) is therefore fulfilled in this second case. We then conclude that function $M$ is analytic for $z>\eta _1$, hence for $\mathfrak {R}(z) > \eta _1$.

Fig. 5

Case $\varrho >1/2$ ($\lambda =1.8, \mu =2$; $\varrho =0.9$)

Fig. 6

Case $\varrho <1/2$ ($\lambda =0.6, \mu =2$; $\varrho =0.3$)