2. The Laplace–Stieltjes transforms

Note that by Definition 1, for $x\geq 0$,

(10) \begin{eqnarray}\mu(dy)^{(x)}_{ij}&\,=\,&\lim_{t\to\infty}\mathbb{P}(X(t)\in dy,\varphi(t)=j\ |\ \theta(0)>t,X(0)=x,\varphi(0)=i)\nonumber\\[4pt] &\,=\,&\lim_{t\to\infty}\frac{\mathbb{P}(X(t)\in dy,\varphi(t)=j,\theta(0)>t\ |\ X(0)=x,\varphi(0)=i)}{\mathbb{P}(\theta(0)>t\ |\ X(0)=x,\varphi(0)=i)}.\end{eqnarray}

For all $x\geq 0$ and $y>0$, define the matrix ${\textbf{E}}(dy)^{(x)}(s)=[E(dy)^{(x)}_{ij}(s)]_{i,j\in\mathcal{S}}$ and the vector ${\textbf{E}}^{(x)}(s)=[E^{(x)}_i(s)]_{i\in\mathcal{S}}$, which record the corresponding Laplace–Stieltjes transforms (LSTs),

(11) \begin{eqnarray}E(dy)^{(x)}_{ij}(s)&\,=\,&\int_0^\infty \mathbb{E}(e^{-s t} {\boldsymbol{1}}\{X(t)\in dy,\varphi(t)=j,\theta(0)>t\}\ |\ X(0)=x,\varphi(0)=i) dt,\nonumber\\[4pt]E^{(x)}_i(s)&\,=\,&\int_0^\infty \mathbb{E}(e^{-s t} {\boldsymbol{1}}\{\theta(0)>t\}\ |\ X(0)=x,\varphi(0)=i)dt,\end{eqnarray}

where ${\boldsymbol{1}}\{\cdot\}$ denotes an indicator function. We have

(12) \begin{eqnarray}{\textbf{E}}^{(x)}(s)&\,=\,&\int_{y=0}^{\infty}{\textbf{E}}(dy)^{(x)}(s){\boldsymbol{1}}.\end{eqnarray}

We partition ${\textbf{E}}(dy)^{(x)}(s)$, $x>0$, according to $\mathcal{S}\times \mathcal{S}$ for $\mathcal{S}=\mathcal{S}_1\cup\mathcal{S}_2\cup\mathcal{S}_0$ as

(13) \begin{equation}{\textbf{E}}(dy)^{(x)}(s)=\left[\begin{array}{c@{\quad}c@{\quad}c}{\textbf{E}}(dy)^{(x)}(s)_{11}&{\textbf{E}}(dy)^{(x)}(s)_{12}&{\textbf{E}}(dy)^{(x)}(s)_{10}\\[4pt]{\textbf{E}}(dy)^{(x)}(s)_{21}&{\textbf{E}}(dy)^{(x)}(s)_{22}&{\textbf{E}}(dy)^{(x)}(s)_{20}\\[4pt]{\textbf{E}}(dy)^{(x)}(s)_{01}&{\textbf{E}}(dy)^{(x)}(s)_{02}&{\textbf{E}}(dy)^{(x)}(s)_{00}\\[4pt]\end{array}\right],\end{equation}

and ${\textbf{E}}^{(x)}(s)$, $x>0$, as

(14) \begin{equation}{\textbf{E}}^{(x)}(s)=\left[\begin{array}{c}{\textbf{E}}^{(x)}(s)_1\\[4pt]{\textbf{E}}^{(x)}(s)_2\\[4pt]{\textbf{E}}^{(x)}(s)_0\end{array}\right].\end{equation}

We partition ${\textbf{E}}(dy)^{(0)}(s)$ according to $\mathcal{S}_1\times\mathcal{S}_1\cup\mathcal{S}_2\cup\mathcal{S}_0$ as

(15) \begin{equation}{\textbf{E}}(dy)^{(0)}(s)=\big[\begin{array}{c@{\quad}c@{\quad}c}{\textbf{E}}(dy)^{(0)}(s)_{11}&{\textbf{E}}(dy)^{(0)}(s)_{12}&{\textbf{E}}(dy)^{(0)}(s)_{10}\\[4pt]\end{array}\big]\end{equation}

and let

(16) \begin{equation}{\textbf{E}}(dy)^{(0)}(s)_{1}={\textbf{E}}(dy)^{(0)}(s){\boldsymbol{1}}.\end{equation}

Define ${\textbf{C}}_1=diag(c_i)_{i\in\mathcal{S}_1}$, ${\textbf{C}}_2=diag(|c_i|)_{i\in\mathcal{S}_2}$, and let ${\textbf{Q}}(s)$ be the key fluid generator matrix ${\textbf{Q}}(s)$ introduced in [Reference Bean, O’Reilly and Taylor16],

(17) \begin{equation}{\textbf{Q}}(s)=\left[\begin{array}{c@{\quad}c}{\textbf{Q}}_{11}(s) & {\textbf{Q}}_{12}(s) \\[4pt]{\textbf{Q}}_{21}(s) & {\textbf{Q}}_{22}(s)\end{array}\right],\end{equation}

where the block matrices are given by

(18) \begin{eqnarray}{\textbf{Q}}_{22}(s)&\,=\,&{\textbf{C}}^{-1}_{2}\left({\textbf{T}}_{22}-s{\textbf{I}}-{\textbf{T}}_{20}({\textbf{T}}_{00}-s{\textbf{I}})^{-1}{\textbf{T}}_{02}\right),\nonumber\\[4pt]{\textbf{Q}}_{11}(s)&\,=\,&{\textbf{C}}^{-1}_{1}\left({\textbf{T}}_{11}-s{\textbf{I}}-{\textbf{T}}_{10}({\textbf{T}}_{00}-s{\textbf{I}})^{-1}{\textbf{T}}_{01}\right),\nonumber\\[4pt]{\textbf{Q}}_{12}(s)&\,=\,&{\textbf{C}}^{-1}_{1}\left({\textbf{T}}_{12}-{\textbf{T}}_{10}({\textbf{T}}_{00}-s{\textbf{I}})^{-1}{\textbf{T}}_{02}\right),\nonumber\\[4pt]{\textbf{Q}}_{21}(s)&\,=\,&{\textbf{C}}^{-1}_{2}\left({\textbf{T}}_{21}-{\textbf{T}}_{20}({\textbf{T}}_{00}-s{\textbf{I}})^{-1}{\textbf{T}}_{01}\right).\end{eqnarray}

Here, ${\textbf{Q}}(s)$ exists for all real s such that $({\textbf{T}}_{00}-s{\textbf{I}})^{-1}=\int_{t=0}^{\infty}e^{-st}e^{{\textbf{T}}_{00}t}dt<\infty$, or for all real s when $S_0=\emptyset$.

Also, let ${\boldsymbol{\Psi}}(s)$ be the key matrix for SFMs [Reference Bean, O’Reilly and Taylor16] such that, for all $i\in\mathcal{S}_1, j\in\mathcal{S}_2$,

(19) \begin{equation}[{\boldsymbol{\Psi}}(s)]_{ij}=\mathbb{E}(e^{-s\theta( 0)} {\boldsymbol{1}}\{\theta(0)<\infty,\varphi(\theta(0))=j\}\ |\ \varphi(0)=i,X(0)=0)\end{equation}

is the LST of the first return time to the original level zero, with this occurring in phase j, given a start at level zero in phase i. Let $\boldsymbol{\psi}(t)$ be the corresponding density, so that ${\boldsymbol{\Psi}}(s)=\int_{t=0}^{\infty}e^{-st}\boldsymbol{\psi}(t)dt$. Clearly, ${\boldsymbol{\Psi}}(s)>0$ for real s such that ${\boldsymbol{\Psi}}(s)<\infty$ exists.

Define matrices

(20) \begin{eqnarray} {\textbf{K}}(s)&\,=\,&{\textbf{Q}}_{11}(s)+{\boldsymbol{\Psi}}(s){\textbf{Q}}_{21}(s),\nonumber\\[4pt]{\textbf{D}}(s)&\,=\,&{\textbf{Q}}_{22}(s)+{\textbf{Q}}_{21}(s){\boldsymbol{\Psi}}(s),\end{eqnarray}

and note that by the assumed stability of the process, the spectra of ${\textbf{K}}(s)$ and $({-}{\textbf{D}}(s))$ are separate for $s\geq 0$ by [Reference Bean, O’Reilly and Taylor15, Reference Bean, O’Reilly and Taylor16, Reference Bean, O’Reilly and Taylor17]; that is, $sp({\textbf{K}}(s))\cap sp({-}{\textbf{D}}(s))=\varnothing$.

We extend the result in [Reference Bean, O’Reilly and Taylor16, Equation (23)] from $Re(s)\geq 0$ to all real s such that ${\boldsymbol{\Psi}}(s)<\infty$ exists.

Lemma 1. For all real s such that ${\boldsymbol{\Psi}}(s)<\infty$ exists, the matrix ${\boldsymbol{\Psi}}(s)$ is a solution of the Riccati equation,

(21) \begin{equation}{\textbf{Q}}_{12}(s)+{\textbf{Q}}_{11}(s){\textbf{X}}+{\textbf{X}}{\textbf{Q}}_{22}(s)+{\textbf{X}}{\textbf{Q}}_{21}(s){\textbf{X}}={\boldsymbol{0}}.\end{equation}

Proof. Suppose s is real and ${\boldsymbol{\Psi}}(s)<\infty$. Then, by [Reference Bean, O’Reilly and Taylor16, Theorem 1] and [Reference Bean, O’Reilly and Taylor17, Algorithm 1 of Section 3.1],

(22) \begin{equation}\infty> {\boldsymbol{\Psi}}(s)=\int_{y=0}^{\infty}e^{{\textbf{Q}}_{11}(s)y}({\textbf{Q}}_{12}(s)+{\boldsymbol{\Psi}}(s){\textbf{Q}}_{21}(s){\boldsymbol{\Psi}}(s))e^{{\textbf{Q}}_{22}(s)y}dy.\end{equation}

Then, letting

(23) \begin{eqnarray}{\boldsymbol{\Psi}}(s,y)&\,=\,& e^{{\textbf{Q}}_{11}(s)y}({\textbf{Q}}_{12}(s)+{\boldsymbol{\Psi}}(s){\textbf{Q}}_{21}(s){\boldsymbol{\Psi}}(s)) e^{{\textbf{Q}}_{22}(s)y},\end{eqnarray}

we have

(24) \begin{eqnarray}\frac{\partial}{\partial y} {\boldsymbol{\Psi}}(s,y) &\,=\,&{\textbf{Q}}_{11}(s){\boldsymbol{\Psi}}(s,y)+{\boldsymbol{\Psi}}(s,y){\textbf{Q}}_{22}(s),\end{eqnarray}

(25) \begin{eqnarray}{\boldsymbol{\Psi}}(s,0)&\,=\,&{\textbf{Q}}_{12}(s)+{\boldsymbol{\Psi}}(s){\textbf{Q}}_{21}(s){\boldsymbol{\Psi}}(s),\,\,\end{eqnarray}

(26) \begin{eqnarray}\lim_{y\to\infty}{\boldsymbol{\Psi}}(s,y)&\,=\,&{\boldsymbol{0}},\qquad\qquad\qquad\qquad\qquad\qquad\end{eqnarray}

and so

(27) \begin{eqnarray}{\textbf{Q}}_{11}(s){\boldsymbol{\Psi}}(s)+{\boldsymbol{\Psi}}(s){\textbf{Q}}_{22}(s)&\,=\,&\int_{y=0}^{\infty}\frac{\partial}{\partial y}\left( {\boldsymbol{\Psi}}(s,y) \right)dy\nonumber\\[8pt] &\,=\,&\lim_{y\to\infty}{\boldsymbol{\Psi}}(s,y)-{\boldsymbol{\Psi}}(s,0)\end{eqnarray}

and

(28) \begin{equation}{\boldsymbol{0}}={\textbf{Q}}_{12}(s)+{\textbf{Q}}_{11}(s){\boldsymbol{\Psi}}(s)+{\boldsymbol{\Psi}}(s){\textbf{Q}}_{22}(s)+{\boldsymbol{\Psi}}(s){\textbf{Q}}_{21}(s){\boldsymbol{\Psi}}(s),\end{equation}

which implies that ${\boldsymbol{\Psi}}(s)$ is a solution of (21).

Below, we state expressions for ${\textbf{E}}(dy)^{(0)}(s)$ derived in [Reference Ahn and Ramaswami3, Theorem 3.1.1] and [Reference Bean and O’Reilly14, Theorem 2].

Lemma 2. We have

(29) \begin{eqnarray}{\textbf{E}}(dy)^{(0)}(s)_{11}&\,=\,&e^{{\textbf{K}}(s)y}{\textbf{C}}_1^{-1}dy,\end{eqnarray}

(30) \begin{eqnarray}{\textbf{E}}(dy)^{(0)}(s)_{12}&\,=\,&e^{{\textbf{K}}(s)y}{\boldsymbol{\Psi}}(s){\textbf{C}}_2^{-1}dy,\end{eqnarray}

(31) \begin{eqnarray}{\textbf{E}}(dy)^{(0)}(s)_{10}&\,=\,&\left[\begin{array}{c@{\quad}c}{\textbf{E}}(dy)^{(0)}(s)_{11}&{\textbf{E}}(dy)^{(0)}(s)_{12}\end{array}\right]\left[\begin{array}{c}{\textbf{T}}_{10}\\[4pt]{\textbf{T}}_{20}\end{array}\right] \times({-}({{\textbf{T}}}_{00}-s{\textbf{I}})^{-1})dy.\end{eqnarray}

In the next lemma, we derive expressions for ${\textbf{E}}(dy)^{(x)}(s)$, $x>0$. The formal proof of this lemma was already given by Ahn and Ramaswami [Reference Ahn and Ramaswami5]. However, since the lemma is crucial to all of the remaining analysis, we include its proof for completeness.

Lemma 3. For $x>0$ we have

(32) \begin{eqnarray}{\textbf{E}}(dy)^{(x)}(s)_{21}&\,=\,&\int_{z=0}^{\min\{x,y\}}e^{{\textbf{D}}(s)(x-z)}{\textbf{Q}}_{21}(s)e^{{\textbf{K}}(s)(y-z)}{\textbf{C}}_1^{-1}dzdy,\nonumber\\[7pt] {\textbf{E}}(dy)^{(x)}(s)_{22}&\,=\,&{\textbf{E}}(dy)^{(x)}(s)_{21}{\textbf{C}}_1{\boldsymbol{\Psi}}(s){\textbf{C}}_2^{-1}+e^{{\textbf{D}}(s)(x-y)}{\textbf{C}}_2^{-1}{\boldsymbol{1}}\{y<x\}dy,\nonumber\\[7pt] {\textbf{E}}(dy)^{(x)}(s)_{20}&\,=\,&\left[\begin{array}{c@{\quad}c}{\textbf{E}}(dy)^{(x)}(s)_{21}&{\textbf{E}}(dy)^{(x)}(s)_{22}\end{array}\right]\left[\begin{array}{c}{\textbf{T}}_{10}\\[7pt] {\textbf{T}}_{20}\end{array}\right] \times({-}({{\textbf{T}}}_{00}-s{\textbf{I}})^{-1})dy,\nonumber\\[7pt] {\textbf{E}}(dy)^{(x)}(s)_{11}&\,=\,&{\boldsymbol{\Psi}}(s){\textbf{E}}(dy)^{(x)}(s)_{21}+e^{{\textbf{K}}(s)(y-x)}{\textbf{C}}_1^{-1}{\boldsymbol{1}}\{y>x\}dy,\nonumber\\[7pt]{\textbf{E}}(dy)^{(x)}(s)_{12}&\,=\,&{\boldsymbol{\Psi}}(s){\textbf{E}}(dy)^{(x)}(s)_{22}+e^{{\textbf{K}}(s)(y-x)}{\boldsymbol{\Psi}}(s){\textbf{C}}_2^{-1}{\boldsymbol{1}}\{y> x\}dy,\nonumber\\[7pt]{\textbf{E}}(dy)^{(x)}(s)_{10}&\,=\,&\left[\begin{array}{c@{\quad}c}{\textbf{E}}(dy)^{(x)}(s)_{11}&{\textbf{E}}(dy)^{(x)}(s)_{12}\end{array}\right]\left[\begin{array}{c}{\textbf{T}}_{10}\\[7pt]{\textbf{T}}_{20}\end{array}\right] \times({-}({{\textbf{T}}}_{00}-s{\textbf{I}})^{-1})dy.\end{eqnarray}

Proof. The expressions for ${\textbf{E}}(dy)^{(x)}(s)_{10}$ and ${\textbf{E}}(dy)^{(x)}(s)_{20}$ follow from the argument in the proof of Lemma 2. Furthermore, by partitioning the sample paths, since the process may visit level y after returning to level x first, or without hitting level x at all, we have

(33) \begin{eqnarray}{\textbf{E}}(dy)^{(x)}(s)_{11}&\,=\,&{\boldsymbol{\Psi}}(s){\textbf{E}}(dy)^{(x)}(s)_{21}+{\textbf{E}}(d(y-x))^{(0)}(s)_{11}{\boldsymbol{1}}\{y>x\}dy,\nonumber\\[4pt]{\textbf{E}}(dy)^{(x)}(s)_{12}&\,=\,&{\boldsymbol{\Psi}}(s){\textbf{E}}(dy)^{(x)}(s)_{22}+{\textbf{E}}(d(y-x))^{(0)}(s)_{12}{\boldsymbol{1}}\{y>x\}dy,\end{eqnarray}

and so the expressions for ${\textbf{E}}(dy)^{(x)}(s)_{11}$ and ${\textbf{E}}(dy)^{(x)}(s)_{12}$ follow from Lemma 2.

Next, we consider ${\textbf{E}}(dy)^{(x)}(s)_{22}$. For $x>0$, define a matrix ${\textbf{G}}(x,t)=[G(x,t)_{kj}]$ such that for $k,j\in\mathcal{S}$,

(34) \begin{equation}G(x,t)_{kj}=\mathbb{P}(\theta(0)\leq t,\varphi(\theta(0))=j\ |\ X(0)=x,\varphi(0)=k)\end{equation}

is the probability that, given that the process starts from level x in phase k, it first hits level zero by time t, and does so in phase j. We partition ${\textbf{G}}(x,t)$ according to $\mathcal{S}_1\cup\mathcal{S}_2$ as

(35) \begin{equation}{\textbf{G}}(x,t)=\left[\begin{array}{c@{\quad}c}{\boldsymbol{0}}&{\textbf{G}}(x,t)_{12}\\[4pt]{\boldsymbol{0}}&{\textbf{G}}(x,t)_{22}\end{array}\right].\end{equation}

Also, define $\widetilde{\textbf{G}}(x,s)=\int_{t=0}^{\infty}e^{-s t}d{\textbf{G}}(x,t)$, which we partition in an analogous manner.

The expression for ${\textbf{E}}(dy)^{(x)}(s)_{22}$ then follows from partitioning the sample paths. The process can visit level y in some phase in $\mathcal{S}_2$ directly after a visit to level y in some phase in $\mathcal{S}_1$, or without visiting level y in some phase in $\mathcal{S}_1$ at all, and so we take the sum of expressions corresponding to these two possibilities, which gives

(36) \begin{eqnarray}{\textbf{E}}(dy)^{(x)}(s)_{22} = {\textbf{E}}(dy)^{(x)}(s)_{21}{\textbf{C}}_1{\boldsymbol{\Psi}}(s){\textbf{C}}_2^{-1}+\widetilde{\textbf{G}}(x-y,s){\textbf{C}}_2^{-1}{\boldsymbol{1}}\{y<x\}dy.\end{eqnarray}

The result follows since by [Reference Bean, O’Reilly and Taylor16], $\widetilde{\textbf{G}}(x-y,s)=e^{{\textbf{D}}(s)(x-y)}$.

Finally, we consider ${\textbf{E}}(dy)^{(x)}(s)_{21}$. Define

(37) \begin{equation}\underline{X}(t)=\inf_{u\in[0,t]}\{X(u)\}.\end{equation}

Note that, given that the process starts with $X(0)=x$, $\varphi(0)=i$, for the process to end with $X(t)\in dy$, $\varphi(t)=j$, with a taboo $\theta(0)>t$, one of the following two alternatives must hold.

The first alternative is that $y\geq x$. In this case, the following occurs:

• • First, given $X(0)=x$, $\varphi(0)=i$, the process must reach some infimum $\underline{X}(t)=z\in(0,x]$ at some time $u\in[0,t]$, in some phase in $\mathcal{S}_2$, with the corresponding density recorded by the matrix ${\textbf{G}}_{22}(x-z,u)$. This is followed by an instantaneous transition to some phase k in $\mathcal{S}_1$ according to the rate recorded by the block matrix ${\textbf{Q}}_{21}$ of the fluid generator Q, by the physical interpretation of Q in [Reference Bean, O’Reilly and Taylor16]. The corresponding density of this occurring is therefore $[{\textbf{G}}_{22}(x-z,u){\textbf{Q}}_{21}]_{ik}$.

• • Next, starting from level z in phase k at time u, the process must remain above level z during the time interval [u,t], ending in some level y in phase j at time t. The corresponding density of this occurring is $[\boldsymbol{\phi}(y-z,t-u)]_{kj}$.

Consequently, the LST of this alternative is

\begin{eqnarray*}&&{\int_{z=0}^x\int_{t=0}^{\infty}\int_{u=0}^te^{-s t}{\textbf{G}}_{22}(x-z,u){\textbf{Q}}_{21}(s)\boldsymbol{\phi}(y-z,t-u)_{11}dudtdz}\\[4pt] &&\,=\,\int_{z=0}^x\int_{u=0}^{\infty}\int_{t=u}^{\infty}e^{-s u}{\textbf{G}}_{22}(x-z,u){\textbf{Q}}_{21}(s)e^{-s (t-u)}\boldsymbol{\phi}(y-z,t-u)_{11}dudtdz\\[4pt] &&\,=\,\int_{z=0}^x\left(\int_{u=0}^{\infty}e^{-s u}{\textbf{G}}_{22}(x-z,u)du\right){\textbf{Q}}_{21}(s)\left(\int_{t=0}^{\infty}e^{-s t}\boldsymbol{\phi}(y-z,t)_{11}dt\right)dz\\[4pt] &&\,=\,\int_{z=0}^xe^{{\textbf{D}}(s)(x-z)}{\textbf{Q}}_{21}(s)e^{{\textbf{K}}(s)(y-z)}dz.\end{eqnarray*}

The second alternative is that $y<x$. The LST of this alternative, by an argument similar to the above, is

\begin{eqnarray*}\int_{z=0}^ye^{{\textbf{D}}(s)(x-z)}{\textbf{Q}}_{21}(s)e^{{\textbf{K}}(s)(y-z)}dz.\end{eqnarray*}

Taking the sum of the expressions corresponding to the two alternatives and right-multiplying by ${\textbf{C}}_1^{-1}$ results in the integral expression for ${\textbf{E}}(dy)^{(x)}(s)_{21}$.

Remark 2. Consider

(38) \begin{equation} {\textbf{E}}^{(x)}(s)_{21} =\int_{y=0}^{\infty}{\textbf{E}}(dy)^{(x)}(s)_{21}={\textbf{X}}{\textbf{C}}_1^{-1}, \end{equation}

where ${\textbf{X}}=\int_{y=0}^{\infty}{\textbf{X}}(y)dy$, and

(39) \begin{eqnarray} {\textbf{X}}(y)&\,=\,& \int_{z=0}^{\min\{x,y\}} e^{{\textbf{D}}(s)(x-z)}{\textbf{Q}}_{21}(s)e^{{\textbf{K}}(s)(y-z)}dz. \end{eqnarray}

Then, by integration by parts in (39), ${\textbf{X}}(y)$ is the solution of

(40) \begin{eqnarray} {\textbf{D}}(s){\textbf{X}}(y)+{\textbf{X}}(y){\textbf{K}}(s) &\,=\,&-\left[ e^{{\textbf{D}}(s)(x-z)}{\textbf{Q}}_{21}(s)e^{{\textbf{K}}(s)(y-z)} \right]_{z=0}^{\min\{x,y\}}, \end{eqnarray}

and by integrating (40), X is the solution of

(41) \begin{eqnarray} {\textbf{D}}(s){\textbf{X}}+{\textbf{X}}{\textbf{K}}(s)&\,=\,&e^{{\textbf{D}}(s)(x)}{\textbf{Q}}_{21}(s)({-}{\textbf{K}}(s)^{-1})-({-}{\textbf{D}}(s)^{-1}){\textbf{Q}}_{21}(s)\nonumber\\[4pt] &&+\,({-}{\textbf{D}}(s)^{-1})e^{{\textbf{D}}(s)(x)}{\textbf{Q}}_{21}(s)+{\textbf{Q}}_{21}(s){\textbf{K}}(s)^{-1}. \end{eqnarray}

3. Approach

The key idea is to write each of ${\textbf{E}}(dy)^{(x)}(s)$ and ${\textbf{E}}^{(x)}(s)$ in the form

(42) \begin{equation}\tilde f(s)=\tilde f(s^*) - C (s-s^*)^{1/2}+o((s-s^*)^{1/2}),\end{equation}

and then apply the Heaviside principle in order to evaluate (10). In this section, we summarise the relevant mathematical background required for this analysis.

Consider a function $f:\mathbb{R}\to\mathbb{R}$. Let $\tilde f(s):=\int_0^\infty e^{-s x}f(x)\ dx$ for $s\in \mathbb{R}$ be its Laplace transform. Consider the singularities of $\tilde{f}(s)$. We assume that one with the largest strictly negative real part is real, and we denote it by $s^*<0$. Notice that this yields the integrability of $\int_0^\infty|f(x)|\ dx$. The inversion formula reads

(43) \begin{equation}f(x)=\frac{1}{2\pi {\textrm{i}}}\int_{a-{\textrm{i}} \infty}^{a+{\textrm{i}}\infty}\tilde{f}(s) e^{sx}\ ds\end{equation}

for some (and then any) $a>s^*$.

We now focus on a class of theorems that infer the tail behaviour of a function from its Laplace transform, commonly referred to as Tauberian theorems. Importantly, the behaviour of the Laplace transform around the singularity $s^*$ plays a crucial role here. The following heuristic principle given in [Reference Abate and Whitt1] is often relied upon. Suppose that for $s^*$, some constants K and C, and a non-integer $q > 0$,

(44) \begin{equation}\tilde{f}(s)=K - C(s-s^*)^q+o((s-s^*)^q) \qquad\mbox{as $s\downarrow s^*$.}\end{equation}

Then

(45) \begin{equation}f(x)=\frac{C}{\Gamma({-}q)} x^{-q-1}e^{s^* x}(1+o(1)) \qquad \mbox{as $x\to\infty$,}\end{equation}

where $\Gamma(\cdot)$ is the gamma function. Below we specify conditions under which this relation can be rigorously proven. Later in our paper we apply it in the specific case that $q=1/2$; recall that $\Gamma({-}1/2)=-2\sqrt{\pi}$.

A formal justification of the above relation can be found in Doetsch [Reference Doetsch22, Theorem 37.1]. Following Miyazawa and Rolski [Reference Miyazawa and Rolski41], we consider the following specific form. For this we first recall the concept of the $\mathfrak{W}$-contour with a half-angle of opening $\pi/2<\psi\le \pi$, as depicted in [Reference Doetsch22, Figure 30, p. 240]; also, ${\mathscr G}_{\alpha}(\psi)$ is the region between the contour $\mathfrak W$ and the line $\Re(z)=0$. More precisely,

(46) \begin{equation}{\mathscr G}_{\alpha}(\psi) \equiv \{z \in \mathbb{C}; \Re(z) < 0, z \ne \alpha,\ |\ \arg (z - \alpha)| < \psi \},\end{equation}

where $\arg z$ is the principal part of the argument of the complex number z. In the following theorem, conditions are identified such that the above principle holds; we refer to this as Heaviside’s operational principle, or simply the Heaviside principle.

Theorem 1. (Heaviside principle) Suppose that for $\tilde f:\mathbb{C}\to \mathbb{C}$ and $s^* < 0$ the following three conditions hold:

1. (A1) $\tilde f(\cdot)$ is analytic in a region ${\mathscr G}_{s^*}(\psi)$ for some $\pi/2<\psi\le \pi$;

2. (A2) $\tilde f(s) \to 0$ as $|s| \to \infty$ with $s \in {\mathscr G}_{s^*}(\psi)$;

3. (A3) for some constants K and C, and a non-integer $q>0$,

   (47) \begin{eqnarray} \tilde f(s)=K - C (s-s^*)^{q}+o((s-s^*)^{q}), \end{eqnarray}
   where ${\mathscr G}_{s^*}(\psi)\ni s \to s^*$.

Then

\begin{eqnarray*} f(x)=\frac{C}{\Gamma({-}q)} x^{-q-1}e^{s^* x}(1+o(1)) \end{eqnarray*}

as $x\to\infty$.

We now discuss when the assumption (A1) is satisfied. To check that the Laplace transform $\tilde{f}(\cdot)$ is analytic in the region $\mathscr G_{s^*}(\psi)$, we can use the concept of semiexponentiality of f (see [30, p. 314]).

Definition 2. (Semiexponentiality) f is said to be semiexponential if for some $0< \phi\le\pi/2$ and all $-\phi\le {\vartheta}\le \phi$ there exists finite and strictly negative $\gamma({\vartheta})$, defined as the infimum of all a such that

\[\left|f(e^{{\textrm{i}}{\vartheta}}r)\right|<e^{ar}\]

for all sufficiently large r.

Relying on this concept, the following sufficient condition for (A1) applies.

Note that by Lemma 3, all assumptions of Proposition 1 are satisfied, and we can apply the Heaviside principle given in Theorem 1 for ${\textbf{E}}(dy)^{(x)}(s)$ and ${\textbf{E}}^{(x)}(s)$.

4. Application of the Heaviside principle

By Section 2, ${\textbf{E}}(dy)^{(x)}(s)$ and ${\textbf{E}}^{(x)}(s)$ are expressed in terms of ${\textbf{Q}}(s)$ and ${\boldsymbol{\Psi}}(s)$, and so we derive the expansion around $s^*$ for each of them first.

Consider ${\boldsymbol{\Psi}}(s)$ as defined in (19). We have ${\boldsymbol{\Psi}}(s)=\int_{t=0}^{\infty}e^{-st}\boldsymbol{\psi}(t)dt<\infty$ for all $s\geq 0$ by [Reference Bean, O’Reilly and Taylor16, Reference Bean, O’Reilly and Taylor17]. Define the singularity

(48) \begin{eqnarray}s^*&\,=\,&\max\{s\leq 0:{\boldsymbol{\Psi}}(s)<\infty, {\boldsymbol{\Psi}}(z)=\infty \mbox{ for all }z < s\},\end{eqnarray}

where the existence of $s^*$ follows from [Reference Doetsch22, Theorem 3.3, p. 15].

Consider matrices ${\textbf{K}}(s)$ and ${\textbf{D}}(s)$ as defined in (20), and recall that $sp({\textbf{K}}(s))\cap sp({-}{\textbf{D}}(s))=\varnothing$ for all $s\geq 0$. Define

(49) \begin{eqnarray}\delta^*&\,=\,&\max\{s\in[s^*,0):sp({\textbf{K}}(s))\cap sp({-}{\textbf{D}}(s))\not=\varnothing\},\end{eqnarray}

whenever the maximum exists. The definition implies that ${\textbf{K}}(\delta^*)$ and $({-}{\textbf{D}}(\delta^*))$ have a common eigenvalue.

Proof. Consider Equation (21), and for all s for which ${\textbf{Q}}(s)$ exists, define the following function of ${\textbf{X}}=[x_{ij}]_{i\in\mathcal{S}_1,j\in\mathcal{S}_2}$:

(50) \begin{equation} g_s({\textbf{X}})={\textbf{Q}}_{12}(s)+{\textbf{Q}}_{11}(s){\textbf{X}}+{\textbf{X}}{\textbf{Q}}_{22}(s)+{\textbf{X}}{\textbf{Q}}_{21}(s){\textbf{X}}. \end{equation}

For ${\textbf{X}}\geq {\boldsymbol{0}}$, ${\textbf{X}}\not= {\boldsymbol{0}}$, we have

(51) \begin{eqnarray} \frac{d}{ds}g_s({\textbf{X}})&\,=\,& -{\textbf{C}}^{-1}_1{\textbf{T}}_{10}({\textbf{T}}_{00}-s{\textbf{I}})^{-2}{\textbf{T}}_{02} {\textbf{X}} -{\textbf{C}}^{-1}_1\left({\textbf{I}}+{\textbf{T}}_{10}({\textbf{T}}_{00}-s{\textbf{I}})^{-2}{\textbf{T}}_{01}\right) {\textbf{X}} \nonumber\\[4pt] && -{\textbf{X}} {\textbf{C}}^{-1}_2\left({\textbf{I}}+{\textbf{T}}_{10}({\textbf{T}}_{00}-s{\textbf{I}})^{-2}{\textbf{T}}_{02}\right) - {\textbf{X}} {\textbf{C}}^{-1}_2{\textbf{T}}_{10}({\textbf{T}}_{00}-s{\textbf{I}})^{-2}{\textbf{T}}_{01} {\textbf{X}} \nonumber\\[4pt] &<&{\boldsymbol{0}}, \end{eqnarray}

since $({\textbf{T}}_{00}-s{\textbf{I}})^{-2}=\left(\int_{t=0}^{\infty}e^{-st}e^{{\textbf{T}}_{00}t}dy\right)^2>{\boldsymbol{0}}$, and so $g_s({\textbf{X}})$ is a decreasing function of s.

Also, define the functions $g_s^{(u,v)}({\textbf{X}})=[g_s({\textbf{X}})]_{uv}$, for $u\in\mathcal{S}_1$, $v\in\mathcal{S}_2$, by

(52) \begin{eqnarray} g_s^{(u,v)}({\textbf{X}})&\,=\,&[{\textbf{Q}}_{12}(s)]_{u,v} +\sum_{k\in\mathcal{S}_1} [{\textbf{Q}}_{11}(s)]_{uk}x_{kv} +\sum_{\ell\in\mathcal{S}_2} x_{u\ell}[{\textbf{Q}}_{22}(s)]_{\ell v} +\sum_{k\in\mathcal{S}_2,\ell\in\mathcal{S}_1} x_{uk}[{\textbf{Q}}_{21}(s)]_{k\ell}x_{\ell v};\nonumber\\ \end{eqnarray}

each of these corresponds to an $|\mathcal{S}_1|\times|\mathcal{S}_2|$-dimensional smooth quadratic surface. The matrix equation (21) is equivalent to the system of $|\mathcal{S}_1|\times|\mathcal{S}_2|$ quadratic polynomial equations given by

(53) \begin{eqnarray} g_s^{(u,v)}({\textbf{X}})&\,=\,&0 \quad \mbox{ for all }u\in\mathcal{S}_1,v\in\mathcal{S}_2, \end{eqnarray}

each corresponding to the (u,v)th level curve.

Now, by Lemma 1, for all $s\geq s^*$, ${\boldsymbol{\Psi}}(s)$ is a solution of $g_{s}({\textbf{X}})={\boldsymbol{0}}$ and so is an intersection point of all level curves (53).

Some other solutions to $g_{s}({\textbf{X}})={\boldsymbol{0}}$ may exist. For all real s, we denote by ${\textbf{X}}(s)$ the family of solutions that correspond to the intersection point ${\boldsymbol{\Psi}}(s)$. That is, when $s\geq s^*$, ${\textbf{X}}(s)={\boldsymbol{\Psi}}(s)$, and if ${\textbf{X}}(s)$ exists for $s<s^*$ in some neighbourhood of $s^*$, then ${\textbf{X}}(s)$ must be a continuous function of s in this neighbourhood, by the monotonicity and continuity of $g_s({\textbf{X}})$.

So suppose that there exist solutions ${\textbf{X}}(s)$ to $g_{s}({\textbf{X}})={\boldsymbol{0}}$ for $s<s^*$ in some neighbourhood of $s^*$, and that $\lim_{s\uparrow s^*}{\textbf{X}}(s)={\boldsymbol{\Psi}}(s^*)$. Then, since ${\boldsymbol{\Psi}}(s^*)>{\boldsymbol{0}}$, there exists ${\textbf{W}}>{\boldsymbol{0}}$ with $g_{s}({\textbf{W}})={\boldsymbol{0}}$ for some $s<s^*$ with $sp({\textbf{Q}}_{11}(s))\cap sp({-}{\textbf{Q}}_{22}(s))=\varnothing$ (since the spectra $sp({\textbf{Q}}_{11}(s))$ and $sp({-}{\textbf{Q}}_{22}(s))$ are discrete).

Therefore, by [Reference Guo27, Theorem 2.3] and [Reference Bean, O’Reilly and Taylor17, Algorithm 1], we have ${\boldsymbol{\Psi}}(s)<\infty$ for such $s<s^*$, and this contradicts the definition of $s^*$. Consequently, ${\textbf{X}}(s)$ does not exist for $s<s^*$, and so the level curves (53) must touch (have a common tangent line) at $s=s^*$, but not at $s>s^*$.

Define

(54) \begin{equation} \nabla g_{s^*}^{(u,v)}(x_{ij},[x_{ij}^*]) = \frac{\partial}{\partial x_{ij}}g^{(u,v)}([x_{ij}])\Big|_{[x_{ij}^*]}, \end{equation}

and note that

(55) \begin{eqnarray} && { \nabla g_{s^*}^{(u,v)}(x_{ij},[x_{ij}^*]) = \frac{\partial}{\partial x_{ij}}g_{s^*}^{(u,v)}([x_{ij}])\Big|_{[x_{ij}^*]} } \nonumber \\[2pt] &&\,=\, [{\textbf{Q}}_{11}(s^*)]_{ui}{\boldsymbol{1}}\{j=v\} +[{\textbf{Q}}_{22}(s^*)]_{jv}{\boldsymbol{1}}\{i=u\} +\sum_{k\in\mathcal{S}_2} [{\textbf{Q}}_{21}(s^*)]_{ki}\times x_{uk}^* {\boldsymbol{1}}\{j=v\} \nonumber\\ &&\quad \,+ \sum_{\ell\in\mathcal{S}_1}[{\textbf{Q}}_{21}(s^*)]_{j\ell}\times x_{\ell v}^*{\boldsymbol{1}}\{i=u\} \nonumber\\[2pt] &&\,=\, [{\textbf{K}}(s^*)]_{ui}{\boldsymbol{1}}\{j=v\} +[{\textbf{D}}(s^*)]_{jv}{\boldsymbol{1}}\{i=u\} . \end{eqnarray}

The tangent plane to the (u,v)th level curve (53) at ${\boldsymbol{\Psi}}(s^*)=[x_{ij}^*]$ is the solution to the equation

(56) \begin{eqnarray} 0&\,=\,& \sum_{i,j} \nabla g^{(u,v)}_{s^*}(x_{ij},[x_{ij}^*]) (x_{ij}-x_{ij}^*) \nonumber\\ &\,=\,&\sum_{i,j} \Big( [{\textbf{K}}(s^*)]_{ui}{\boldsymbol{1}}\{j=v\} +[{\textbf{D}}(s^*)]_{jv}{\boldsymbol{1}}\{i=u\} \Big) (x_{ij}-x_{ij}^*) \nonumber\\ &\,=\,& \sum_i[{\textbf{K}}(s^*)]_{ui}[{\textbf{X}}-{\boldsymbol{\Psi}}(s^*)]_{iv} +\sum_j [{\textbf{X}}-{\boldsymbol{\Psi}}(s^*)]_{uj}[{\textbf{D}}(s^*)]_{jv} \nonumber\\ &\,=\,& \left[ {\textbf{K}}(s^*)({\textbf{X}}-{\boldsymbol{\Psi}}(s^*)) +({\textbf{X}}-{\boldsymbol{\Psi}}(s^*)){\textbf{D}}(s^*)\right]_{uv}. \end{eqnarray}

From linear algebra, a matrix equation of the form ${\boldsymbol{0}}={\textbf{A}}{\textbf{X}}+{\textbf{X}}{\textbf{B}}$ has a nonzero solution if and only if A and $({-}{\textbf{B}})$ have a common eigenvalue (e.g. see [Reference Bhatia and Rosenthal18]). Therefore, the equation

(57) \begin{eqnarray} {\boldsymbol{0}}&\,=\,& {\textbf{K}}(s^*)({\textbf{X}}-{\boldsymbol{\Psi}}(s^*)) +({\textbf{X}}-{\boldsymbol{\Psi}}(s^*)){\textbf{D}}(s^*) \end{eqnarray}

has a solution ${\textbf{Z}}=[z_{ij}]\not= {\boldsymbol{\Psi}}(s^*)$ if and only if ${\textbf{K}}(s^*)$ and $({-}{\textbf{D}}(s^*))$ have a common eigenvalue, in which case the tangent planes (56) to all level curves (53) at ${\boldsymbol{\Psi}}(s^*)$ intersect one another at a tangent line that goes through Z and ${\boldsymbol{\Psi}}(s^*)$.

That is, the level curves (53) touch if and only if $sp({\textbf{K}}(s^*))\cap sp({-}{\textbf{D}}(s^*))\not=\varnothing$. Hence, $s^*=\delta^*$.

We now extend the result for $s>0$ in [Reference Bean, O’Reilly and Taylor16, Theorem 1] to all $s\geq s^*$.

Proof. Suppose $s\geq s^*$. Then ${\textbf{Q}}_{11}(s)\leq {\textbf{K}}(s)={\textbf{Q}}_{11}(s)+{\boldsymbol{\Psi}}(s){\textbf{Q}}_{21}(s)$ and ${\textbf{Q}}_{22}(s)\leq {\textbf{D}}(s)={\textbf{Q}}_{22}(s)+{\textbf{Q}}_{21}(s){\boldsymbol{\Psi}}(s)$, and so $sp({\textbf{Q}}_{11}(s))\cap sp({-}{\textbf{Q}}_{22}(s))=\varnothing$.

Therefore, by [Reference Guo27, Theorem 2.3] and [Reference Bean, O’Reilly and Taylor17, Algorithm 1], ${\boldsymbol{\Psi}}(s)$ is the minimum nonnegative solution of (21).

In order to illustrate the theory, we consider the following simple example, which we will analyse as we develop the results throughout the paper.

Example 1. Let $\mathcal{S}=\{1,2\}$, $\mathcal{S}_1=\{1\}$, $\mathcal{S}_2=\{2\}$, $c_1=1$, $c_2=-1$, and

(58) \begin{eqnarray} {\textbf{T}} &\,=\,&\left[ \begin{array}{c@{\quad}c} {\textbf{T}}_{11}&{\textbf{T}}_{12}\\[4pt] {\textbf{T}}_{21}&{\textbf{T}}_{22} \end{array} \right] =\left[ \begin{array}{c@{\quad}c} -a&a\\[4pt] b&-b \end{array} \right], \end{eqnarray}

(59) \begin{eqnarray} {\textbf{Q}}(s) &\,=\,&\left[ \begin{array}{c@{\quad}c} {\textbf{Q}}_{11}(s)&{\textbf{Q}}_{12}(s)\\[4pt] {\textbf{Q}}_{21}(s)&{\textbf{Q}}_{22}(s) \end{array} \right] =\left[ \begin{array}{c@{\quad}c} -a-s&a\\[4pt] b&-b-s \end{array} \right], \end{eqnarray}

with $a>b>0$ so that the process is stable.

Then ${\boldsymbol{\Psi}}(s)$ is the minimum nonnegative solution of (21), here equivalent to

(60) \begin{equation} b x^2 - (a+b+2s)x+a=0, \end{equation}

which has solutions provided $\Delta(s)=(a+b+2s)^2-4ab \geq 0$, that is, for all

(61) \begin{eqnarray} s\in \left({-}\infty, \frac{-(a+b)-2\sqrt{ab}}{2} \right] \cup\left[ \frac{-(a+b)+2\sqrt{ab}}{2}, +\infty\right) . \end{eqnarray}

Since

(62) \begin{eqnarray} (a+b+2s)-\sqrt{\Delta(s)}\geq 0 \iff s\leq \frac{2ab}{a+b}, \end{eqnarray}

it follows that ${\boldsymbol{\Psi}}(s)$ exists for all $s\geq \frac{-(a+b)+2\sqrt{ab}}{2}$, and

(63) \begin{eqnarray} {\boldsymbol{\Psi}}(s)&\,=\,&\frac{(a+b+2s)-\sqrt{\Delta(s)}}{2b},\qquad\quad\ \ \ \ \end{eqnarray}

(64) \begin{eqnarray} {\textbf{K}}(s)&\,=\,&-a-s+\frac{(a+b+2s)-\sqrt{\Delta(s)}}{2},\end{eqnarray}

(65) \begin{eqnarray} {\textbf{D}}(s)&\,=\,&-b-s+\frac{(a+b+2s)-\sqrt{\Delta(s)}}{2}. \end{eqnarray}

Therefore,

(66) \begin{equation} s^*=\frac{-(a+b)+2\sqrt{ab}}{2}<0 \end{equation}

and

(67) \begin{eqnarray} {\boldsymbol{\Psi}}(s^*)&\,=\,&\sqrt{\frac{a}{b}},\qquad\qquad\qquad\qquad\qquad \ \ \ \end{eqnarray}

(68) \begin{eqnarray} {\textbf{K}}(s^*)&\,=\,&-a-s^*+\sqrt{\frac{a}{b}}\ b = \frac{b-a}{2}<0 , \end{eqnarray}

(69) \begin{eqnarray} {\textbf{D}}(s^*)&\,=\,&-b-s^*+b\ \sqrt{\frac{a}{b}} =\frac{a-b}{2}>0. \end{eqnarray}

Note that $s^*=\delta^*$.

Lemma 5. For all $s>s^*$,

(70) \begin{eqnarray}{\textbf{Q}}_{22}(s)&\,=\,&{\textbf{Q}}_{22}(s^*)-{\textbf{A}}_{22}(s^*)(s-s^*)+o(s-s^*),\end{eqnarray}

(71) \begin{eqnarray}{\textbf{Q}}_{11}(s)&\,=\,&{\textbf{Q}}_{11}(s^*)-{\textbf{A}}_{11}(s^*)(s-s^*)+o(s-s^*),\end{eqnarray}

(72) \begin{eqnarray}{\textbf{Q}}_{12}(s)&\,=\,&{\textbf{Q}}_{12}(s^*)-{\textbf{A}}_{12}(s^*)(s-s^*)+o(s-s^*),\end{eqnarray}

(73) \begin{eqnarray}{\textbf{Q}}_{21}(s)&\,=\,&{\textbf{Q}}_{21}(s^*)-{\textbf{A}}_{21}(s^*)(s-s^*)+o(s-s^*),\end{eqnarray}

where, for all $s> s^*$,

(74) \begin{eqnarray}{\textbf{A}}_{22}(s)&\,=\,&-\frac{d}{ds}{\textbf{Q}}_{22}(s)={\textbf{C}}^{-1}_2\left({\textbf{I}}+{\textbf{T}}_{20}({\textbf{T}}_{00}-s{\textbf{I}})^{-2}{\textbf{T}}_{02} \right),\end{eqnarray}

(75) \begin{eqnarray}{\textbf{A}}_{11}(s)&\,=\,&-\frac{d}{ds}{\textbf{Q}}_{11}(s)={\textbf{C}}^{-1}_1\left({\textbf{I}}+{\textbf{T}}_{10}({\textbf{T}}_{00}-s{\textbf{I}})^{-2}{\textbf{T}}_{01} \right),\end{eqnarray}

(76) \begin{eqnarray}\ \ {\textbf{A}}_{12}(s)&\,=\,&-\frac{d}{ds}{\textbf{Q}}_{12}(s)={\textbf{C}}^{-1}_1{\textbf{T}}_{10}({\textbf{T}}_{00}-s{\textbf{I}})^{-2}{\textbf{T}}_{02},\qquad\qquad\!\!\!\end{eqnarray}

(77) \begin{eqnarray}\ \ {\textbf{A}}_{21}(s)&\,=\,&-\frac{d}{ds}{\textbf{Q}}_{21}(s)={\textbf{C}}^{-1}_2{\textbf{T}}_{20}({\textbf{T}}_{00}-s{\textbf{I}})^{-2}{\textbf{T}}_{01},\qquad\qquad\!\!\!\end{eqnarray}

and ${\textbf{A}}_{22}(s^*)=\lim_{s\downarrow s^*}{\textbf{A}}_{22}(s)<\infty$, ${\textbf{A}}_{11}(s^*)=\lim_{s\downarrow s^*}{\textbf{A}}_{11}(s)<\infty$, ${\textbf{A}}_{12}(s^*)=\lim_{s\downarrow s^*}$ ${\textbf{A}}_{12}(s)<\infty$, and ${\textbf{A}}_{21}(s^*)=\lim_{s\downarrow s^*}{\textbf{A}}_{21}(s)<\infty$.

Proof. For all $s>s^*$,

(78) \begin{eqnarray} -\frac{d}{ds}\left({-}({\textbf{T}}_{00}-s{\textbf{I}})^{-1} \right) &\,=\,&({\textbf{T}}_{00}-s{\textbf{I}})^{-2}, \end{eqnarray}

and so by (18),

(79) \begin{eqnarray}-\frac{d}{ds}{\textbf{Q}}_{22}(s)&\,=\,&{\textbf{C}}^{-1}_2\left({\textbf{I}}+{\textbf{T}}_{20}({\textbf{T}}_{00}-s{\textbf{I}})^{-2}{\textbf{T}}_{02} \right),\end{eqnarray}

which, with the notation ${\textbf{A}}_{22}(s)=-\frac{d}{ds}{\textbf{Q}}_{22}(s)$, implies

(80) \begin{eqnarray}{\textbf{Q}}_{22}(s+h)={\textbf{Q}}_{22}(s)-{\textbf{A}}_{22}(s)(h)+o(h).\end{eqnarray}

Next, since ${\boldsymbol{\Psi}}(s^*)<\infty$ by Lemma 4, we have $\left({-}({\textbf{T}}_{00}-s^*{\textbf{I}})^{-1} \right)<\infty$ and $({\textbf{T}}_{00}-s^*{\textbf{I}})^{-2}<\infty$, which implies that ${\textbf{A}}_{22}(s^*)<\infty$. Taking the limits as $s\downarrow s^*$ in (80) and substituting $h=(s-s^*)$ gives

(81) \begin{eqnarray}{\textbf{Q}}_{22}(s)&\,=\,&{\textbf{Q}}_{22}(s^*)-{\textbf{A}}_{22}(s^*)(s-s^*)+o(s-s^*).\end{eqnarray}

The proofs for the remaining expressions are analogous.

For $s>s^*$, let

(82) \begin{equation}{\boldsymbol{\Phi}} (s)=\frac{d}{ds}{\boldsymbol{\Psi}}(s)=\lim_{h\to 0}\frac{{\boldsymbol{\Psi}}(s+h)-{\boldsymbol{\Psi}}(s)}{h},\end{equation}

and, for $s\geq s^*$, let

(83) \begin{eqnarray}{\textbf{U}}(s)&\,=\,&{\textbf{A}}_{12}(s)+{\textbf{A}}_{11}(s){\boldsymbol{\Psi}}(s)+{\boldsymbol{\Psi}}(s){\textbf{A}}_{22}(s)+{\boldsymbol{\Psi}}(s){\textbf{A}}_{21}(s){\boldsymbol{\Psi}}(s),\end{eqnarray}

noting that ${\textbf{U}}(s^*)$ exists by Lemma 5.

Lemma 6. For $s>s^*$, ${\boldsymbol{\Phi}}(s)$ is the unique solution of the equation

(84) \begin{eqnarray} {\textbf{K}}(s){\textbf{X}}+{\textbf{X}}{\textbf{D}}(s) &\,=\,& {\textbf{U}}(s). \end{eqnarray}

Furthermore, ${\boldsymbol{\Phi}}(s^*)=\lim_{s\downarrow s^*}{\boldsymbol{\Phi}}(s)=-\infty$.

Proof. By Lemma 4, for all $s>s^*$, ${\textbf{K}}(s)$ and $({-}{\textbf{D}}(s))$ have no common eigenvalues, and so by [Reference Laub38, Theorem 13.18], Equation (84) has a unique solution. We now show that ${\boldsymbol{\Phi}} (s)$ is the solution of (84). Also see [Reference Bean, O’Reilly and Taylor16, Corollary 3]. Indeed, by taking derivatives with respect to s in Equation (21) for ${\boldsymbol{\Psi}}(s)$, we have

(85) \begin{eqnarray}{\boldsymbol{0}}&\,=\,&\frac{d}{ds}\Big({\textbf{Q}}_{12}(s)+{\textbf{Q}}_{11}(s){\boldsymbol{\Psi}}(s)+{\boldsymbol{\Psi}}(s){\textbf{Q}}_{22}(s)+{\boldsymbol{\Psi}}(s){\textbf{Q}}_{21}(s){\boldsymbol{\Psi}}(s)\Big)\nonumber\\[4pt]&\,=\,&-{\textbf{A}}_{12}(s)-{\textbf{A}}_{11}(s){\boldsymbol{\Psi}}(s)+{\textbf{Q}}_{11}(s){\boldsymbol{\Phi}}(s)-{\boldsymbol{\Psi}}(s){\textbf{A}}_{22}(s)+{\boldsymbol{\Phi}}(s){\textbf{Q}}_{22}(s)\nonumber\\[4pt]&&+{\boldsymbol{\Phi}}(s){\textbf{Q}}_{21}(s){\boldsymbol{\Psi}}(s)-{\boldsymbol{\Psi}}(s){\textbf{A}}_{21}(s){\boldsymbol{\Psi}}(s)+{\boldsymbol{\Psi}}(s){\textbf{Q}}_{21}(s){\boldsymbol{\Phi}}(s)\nonumber\\[4pt]&\,=\,&-{\textbf{U}}(s)+{\textbf{K}}(s){\boldsymbol{\Phi}} (s)+{\boldsymbol{\Phi}} (s){\textbf{D}}(s).\end{eqnarray}

Also, ${\boldsymbol{\Phi}}(s)<{\boldsymbol{0}}$, since

(86) \begin{equation}{\boldsymbol{\Phi}}(s)=\frac{d}{ds}{\boldsymbol{\Psi}}(s)=\frac{d}{ds}\int_0^{\infty}e^{-st}\boldsymbol{\psi}(t)dt=-\int_0^{\infty}te^{-st}\boldsymbol{\psi}(t)dt<{\boldsymbol{0}}.\end{equation}

When $s=s^*$, however, by Lemma 4, ${\textbf{K}}(s)$ and $({-}{\textbf{D}}(s))$ have a common eigenvalue, and so by [Reference Laub38, Theorem 13.18], Equation (84) does not have a unique solution.

Finally, we show that $\lim_{s\downarrow s^*}{\boldsymbol{\Phi}}(s)=-\infty$. By standard methodology [Reference Laub38, Section 13.3], for $s>s^*$, the unique solution to Equation (84) can be written in the form

(87) \begin{eqnarray} vec({\boldsymbol{\Phi}}(s)) &\,=\,& ({\textbf{Z}}(s))^{-1}vec({\textbf{U}}(s)) = \frac{adj({\textbf{Z}}(s))}{det({\textbf{Z}}(s))}vec({\textbf{U}}(s)), \end{eqnarray}

where $vec({\boldsymbol{\Phi}}(s))$ and $vec({\textbf{U}}(s))$ are column vectors obtained by stacking the columns (from left to right) of the original matrices one under another,

(88) \begin{equation} {\textbf{Z}}(s)=({\textbf{I}}\otimes {\textbf{K}}(s))+({\textbf{D}}(s)^T \otimes {\textbf{I}}), \end{equation}

and the eigenvalues of ${\textbf{Z}}(s)$ are $(\lambda_i-\mu_j)$, where $\lambda_i$ are eigenvalues of ${\textbf{K}}(s)$ and $\mu_j$ are eigenvalues of $({-}{\textbf{D}}(s))$. Because $det({\textbf{Z}}(s))$ is the product of the eigenvalues of ${\textbf{Z}}(s)$, and because, as $s\downarrow s^*$, one of the eigenvalues will approach zero since $s^*=\delta^*$ by Lemma 4, we have $\lim_{s\downarrow s^*}det({\textbf{Z}}(s))=0$ and so ${\boldsymbol{\Phi}}(s^*)=\lim_{s\downarrow s^*}{\boldsymbol{\Phi}}(s)=-\infty$, where the negative sign is due to the fact that ${\boldsymbol{\Phi}}(s)<{\boldsymbol{0}}$ for all $s>s^*$.

We now state the key result of this paper.

Theorem 2. For all $s>s^*$,

(89) \begin{eqnarray} {\boldsymbol{\Psi}}(s)&\,=\,&{\boldsymbol{\Psi}}(s^*)-{\textbf{B}}(s^*)\sqrt{s-s^*}+o(\sqrt{s-s^*}), \end{eqnarray}

where ${\boldsymbol{0}}<{\textbf{B}}(s^*)<\infty$ solves

(90) \begin{eqnarray} {\textbf{B}}(s^*){\textbf{Q}}_{21}(s^*){\textbf{B}}(s^*) &\,=\,&{\textbf{U}}(s^*)-{\textbf{Y}}(s^*),\end{eqnarray}

(91) \begin{eqnarray} {\textbf{K}}(s^*){\textbf{B}}(s^*)+{\textbf{B}}(s^*){\textbf{D}}(s^*)&\,=\,&{\boldsymbol{0}}, \end{eqnarray}

and

(92) \begin{eqnarray} {\textbf{Y}}(s^*)&\,=\,&\lim_{s\downarrow s^*} \left( {\textbf{K}}(s^*){\boldsymbol{\Phi}}(s)+{\boldsymbol{\Phi}}(s){\textbf{D}}(s^*) \right) . \end{eqnarray}

Proof. Note that for any function $h(\cdot)$ with $h(s-s^*)=o(s-s^*)$ or $h(s-s^*)=c\cdot (s-s^*)$ for some constant c, we have

(93) \begin{equation} -\lim_{s\downarrow s^*} \left( \frac{{\boldsymbol{\Psi}}(s)-{\boldsymbol{\Psi}}(s^*)}{s-s^*}h(s-s^*) \right) ={\boldsymbol{0}}.\end{equation}

Consider $h(s-s^*)=(s-s^*)/|| {\boldsymbol{\Psi}}(s)-{\boldsymbol{\Psi}}(s^*) ||$. We have

(94) \begin{equation} \lim_{s\downarrow s^*}\frac{s-s^*}{h(s-s^*)} =\lim_{s\downarrow s^*}|| {\boldsymbol{\Psi}}(s)-{\boldsymbol{\Psi}}(s^*) || =0,\end{equation}

which implies $(s-s^*)=o(h(s-s^*))$, and

(95) \begin{equation} \lim_{s\downarrow s^*} \Bigg|\Bigg| \frac{{\boldsymbol{\Psi}}(s)-{\boldsymbol{\Psi}}(s^*)}{s-s^*}h(s-s^*) \Bigg|\Bigg| =1\not= 0.\end{equation}

Therefore, there exists a continuous, positive-valued function $h(\cdot)$ such that $(s-s^*)=$ $o(h(s-s^*))$ and

(96) \begin{equation} -\lim_{s\downarrow s^*} \left( \frac{{\boldsymbol{\Psi}}(s)-{\boldsymbol{\Psi}}(s^*)}{s-s^*}h(s-s^*) \right) ={\textbf{B}}(s^*)\end{equation}

for some constant matrix ${\boldsymbol{0}}<{\textbf{B}}(s^*)<\infty$. For such $h(\cdot)$, define the function $g(\cdot)$ by

(97) \begin{equation} g(s-s^*)=\frac{s-s^*}{h(s-s^*)};\end{equation}

clearly $\lim_{s\downarrow s^*}g(s-s^*)=0$ since $(s-s^*)=o(h(s-s^*))$.

Consequently, we have

(98) \begin{equation} -\lim_{s\downarrow s^*} \left( \frac{{\boldsymbol{\Psi}}(s)-{\boldsymbol{\Psi}}(s^*)}{g(s-s^*)} \right) ={\textbf{B}}(s^*),\end{equation}

which implies that

(99) \begin{eqnarray} {\boldsymbol{\Psi}}(s)&\,=\,&{\boldsymbol{\Psi}}(s^*)-{\textbf{B}}(s^*)g(s-s^*) +o(g(s-s^*)).\end{eqnarray}

We now solve for ${\textbf{B}}(s^*)$ and $g(s-s^*)$. By (21) and Lemma 5, since

(100) \begin{eqnarray} {\boldsymbol{0}}&\,=\,& {\textbf{Q}}_{12}(s^*) + {\textbf{Q}}_{11}(s^*){\boldsymbol{\Psi}}(s^*) +{\boldsymbol{\Psi}}(s^*){\textbf{Q}}_{22}(s^*) +{\boldsymbol{\Psi}}(s^*){\textbf{Q}}_{21}(s^*){\boldsymbol{\Psi}}(s^*),\end{eqnarray}

we have

(101) \begin{eqnarray}{\boldsymbol{0}}&\,=\,& {\textbf{Q}}_{12}(s) +{\textbf{Q}}_{11}(s){\boldsymbol{\Psi}}(s)+{\boldsymbol{\Psi}}(s){\textbf{Q}}_{22}(s) +{\boldsymbol{\Psi}}(s){\textbf{Q}}_{21}(s){\boldsymbol{\Psi}}(s)\nonumber \\[4pt] &\,=\,&\left({\textbf{Q}}_{12}(s^*)-{\textbf{A}}_{12}(s^*)(s-s^*)\right) \nonumber\\[4pt] && + \left( {\textbf{Q}}_{11}(s^*)-{\textbf{A}}_{11}(s^*)(s-s^*) \right) \left( {\boldsymbol{\Psi}}(s^*)-{\textbf{B}}(s^*)g(s-s^*) \right) \nonumber\\[4pt] && + \left( {\boldsymbol{\Psi}}(s^*)-{\textbf{B}}(s^*)g(s-s^*) \right) \left( {\textbf{Q}}_{22}(s^*)-{\textbf{A}}_{22}(s^*)(s-s^*) \right) \nonumber\\[4pt] && + \left( {\boldsymbol{\Psi}}(s^*) - {\textbf{B}}(s^*)g(s-s^*) \right) \left( {\textbf{Q}}_{21}(s^*)-{\textbf{A}}_{21}(s^*)(s-s^*) \right) \left( {\boldsymbol{\Psi}}(s^*) - {\textbf{B}}(s^*)g(s-s^*) \right) \nonumber\\[4pt] && +o(s-s^*)+o(g(s-s^*)),\end{eqnarray}

and so

(102) \begin{eqnarray} {\boldsymbol{0}} &\,=\,& -(s-s^*){\textbf{U}}(s^*)+g(s-s^*){\textbf{W}}(s^*) +g^2(s-s^*){\textbf{V}}(s^*) +o(s-s^*) +o(g(s-s^*)), \nonumber\\[4pt]\end{eqnarray}

where ${\textbf{U}}(s^*)$ is as defined in (83), and

(103) \begin{eqnarray} {\textbf{W}}(s^*)&\,=\,& \left( {\textbf{Q}}_{11}(s^*)+{\boldsymbol{\Psi}}(s^*){\textbf{Q}}_{21}(s^*) \right) {\textbf{B}}(s^*) +{\textbf{B}}(s^*) \left( {\textbf{Q}}_{22}(s^*)+{\textbf{Q}}_{21}(s^*){\boldsymbol{\Psi}}(s^*) \right) \nonumber\\[4pt] &\,=\,& {\textbf{K}}(s^*){\textbf{B}}(s^*)+{\textbf{B}}(s^*){\textbf{D}}(s^*) , \nonumber\\[4pt] {\textbf{V}}(s^*) &\,=\,& {\textbf{B}}(s^*){\textbf{Q}}_{21}(s^*){\textbf{B}}(s^*) .\end{eqnarray}

We now use (102) to solve for ${\textbf{B}}(s^*)$ and $g(s-s^*)$. We note that ${\textbf{V}}(s^*)\not= {\boldsymbol{0}}$ and ${\textbf{U}}(s^*)\not= {\boldsymbol{0}}$. Indeed, ${\textbf{V}}(s^*)\not= {\boldsymbol{0}}$, since ${\textbf{V}}(s^*)> {\boldsymbol{0}}$ because ${\textbf{B}}(s^*)>{\boldsymbol{0}}$, ${\textbf{Q}}_{21}(s^*)\geq {\boldsymbol{0}}$, and ${\textbf{Q}}_{21}(s^*)\not= {\boldsymbol{0}}$. Furthermore, ${\textbf{U}}(s^*)\not= {\boldsymbol{0}}$ since ${\textbf{U}}(s^*)> {\boldsymbol{0}}$. Indeed, in the case $\mathcal{S}_0=\varnothing$, since ${\textbf{C}}_1,{\textbf{C}}_2>{\boldsymbol{0}}$ and ${\boldsymbol{\Psi}}(s^*)>{\boldsymbol{0}}$, we have ${\textbf{U}}(s^*)={\textbf{C}}_1^{-1}{\boldsymbol{\Psi}}(s^*)+{\boldsymbol{\Psi}}(s^*){\textbf{C}}_2^{-1}>{\boldsymbol{0}}$. In the case $\mathcal{S}_0\not=\varnothing$, we have $-({\textbf{T}}_{00}-s^*{\textbf{I}})^{-1}=\int_{t=0}^{\infty}e^{-s^*t}e^{{\textbf{T}}_{00}t}dt>{\boldsymbol{0}}$, and $({\textbf{T}}_{00}-s^*{\textbf{I}})^{-2}=({-}({\textbf{T}}_{00}-s^*{\textbf{I}})^{-1})^2>0$. Therefore ${\textbf{A}}_{11}(s^*),{\textbf{A}}_{22}(s^*)>{\boldsymbol{0}}$, ${\textbf{A}}_{12}(s^*),{\textbf{A}}_{21}(s^*)\geq 0$, and ${\boldsymbol{\Psi}}(s^*)>{\boldsymbol{0}}$, so that ${\textbf{U}}(s^*)={\textbf{A}}_{12}(s^*)+{\textbf{A}}_{11}(s^*){\boldsymbol{\Psi}}(s^*)+{\boldsymbol{\Psi}}(s^*){\textbf{A}}_{22}(s^*)+{\boldsymbol{\Psi}}(s^*){\textbf{A}}_{21}(s^*){\boldsymbol{\Psi}}(s^*)>{\boldsymbol{0}}$.

Consequently, below we consider the two cases ${\textbf{W}}(s^*)\not= {\boldsymbol{0}}$ and ${\textbf{W}}(s^*)= {\boldsymbol{0}}$, respectively labelled Case I and Case II.

Case I. Suppose ${\textbf{W}}(s^*)\not= {\boldsymbol{0}}$. Then

(104) \begin{eqnarray} {\boldsymbol{0}} &\,=\,& -(s-s^*){\textbf{U}}(s^*)+g(s-s^*){\textbf{W}}(s^*) +g^2(s-s^*){\textbf{V}}(s^*) +o(s-s^*) +o(g(s-s^*)). \nonumber\\[4pt]\end{eqnarray}

Consider $(s-s^*)$ and $g(s-s^*)$. Either one of them dominates the other, or one is a multiple of the other.

1. (i) If $g(s-s^*)=o(s-s^*)$, then dividing Equation (104) by $(s-s^*)$ and taking limits as $s\downarrow s^*$ gives ${\boldsymbol{0}}={\textbf{U}}(s^*)$, a contradiction.

2. (ii) If $(s-s^*)=o(g(s-s^*))$, then dividing Equation (104) by $g(s-s^*)$ and taking limits as $s\downarrow s^*$ gives ${\boldsymbol{0}}={\textbf{W}}(s^*)$, a contradiction.

3. (iii) If $g(s-s^*)=c\cdot (s-s^*)$ for some constant $c>0$, then without loss of generality we may assume $c=1$, since ${\textbf{B}}(s^*)g(s-s^*)=({\textbf{B}}(s^*)c )(s-s^*)$ suggests the substitution $\tilde {\textbf{B}}(s^*)\equiv {\textbf{B}}(s^*)c$. We then have

   (105) \begin{eqnarray} {\boldsymbol{\Psi}}(s)&\,=\,&{\boldsymbol{\Psi}}(s^*)-\tilde{\textbf{B}}(s^*)(s-s^*) +o(s-s^*),\end{eqnarray}
   with $\tilde {\textbf{B}}(s^*)<\infty$. However, dividing Equation (105) by $(s-s^*)$ and taking limits as $s\downarrow s^*$ gives, by Lemma 6,
   (106) \begin{eqnarray} \tilde{\textbf{B}}(s^*)=-\lim_{s\downarrow s^*}\frac{{\boldsymbol{\Psi}}(s)-{\boldsymbol{\Psi}}(s^*)}{s-s^*}=\infty,\end{eqnarray}
   a contradiction.

That is, the assumption ${\textbf{W}}(s^*)\not= {\boldsymbol{0}}$ leads to a contradiction.

Case II. By the above, we must have ${\textbf{W}}(s^*)= {\boldsymbol{0}}$, or equivalently,

(107) \begin{equation}{\textbf{K}}(s^*){\textbf{B}}(s^*)+{\textbf{B}}(s^*){\textbf{D}}(s^*)={\boldsymbol{0}},\end{equation}

and so,

(108) \begin{eqnarray} {\boldsymbol{0}}&\,=\,&-(s-s^*){\textbf{U}}(s^*)+g^2(s-s^*){\textbf{V}}(s^*)+o(s-s^*) +o(g(s-s^*)).\end{eqnarray}

We note that $g^2(s-s^*)=o(g(s-s^*))$, and consider the following.

(i) First, we show that $(s-s^*)=o(g(s-s^*))$. Indeed, if $g(s-s^*)=o(s-s^*)$ or $g(s-s^*)=c\cdot (s-s^*)$ for some $c\not= 0$, then dividing Equation (108) by $(s-s^*)$ and taking limits as $s\downarrow s^*$ gives ${\textbf{U}}(s^*)={\boldsymbol{0}}$, a contradiction. Therefore we must have $(s-s^*)=o(g(s-s^*))$. That is, $g(s-s^*)$ dominates both $g^2(s-s^*)$ and $(s-s^*)$.

Then,

(109) \begin{eqnarray} \lim_{s\to s^*}\frac{o(s-s^*)}{g(s-s^*)}=\lim_{s\to s^*}\frac{o(s-s^*)}{(s-s^*)}\frac{(s-s^*)}{g(s-s^*)} =0,\end{eqnarray}

which gives $o(s-s^*)=o(g(s-s^*))$, and so we write (108) in the form

(110) \begin{eqnarray} {\boldsymbol{0}}&\,=\,&-(s-s^*){\textbf{U}}(s^*)+g^2(s-s^*){\textbf{V}}(s^*) +o(g(s-s^*)).\end{eqnarray}

Since $(s-s^*)=o(g(s-s^*))$, we consider two cases, $o(g(s-s^*))={\boldsymbol{0}}$ and $o(g(s-s^*))\not ={\boldsymbol{0}}$, respectively labelled (A) and (B) below.

(A) Suppose $o(g(s-s^*))={\boldsymbol{0}}$. Then (110) reduces to

(111) \begin{eqnarray}{\boldsymbol{0}}&\,=\,&-(s-s^*){\textbf{U}}(s^*)+g^2(s-s^*){\textbf{V}}(s^*).\end{eqnarray}

If $(s-s^*)=o(g^2(s-s^*))$, we divide (111) by $g^2(s-s^*)$ and take limits as $s\downarrow s^*$ to get ${\textbf{V}}(s^*)={\boldsymbol{0}}$, a contradiction. If $g^2(s-s^*)=o(s-s^*)$, we divide (111) by $(s-s^*)$ and take limits as $s\downarrow s^*$ to get ${\textbf{U}}(s^*)={\boldsymbol{0}}$, a contradiction. So we must have $(s-s^*)=c\cdot g^2(s-s^*)$ for some constant $c>0$, and without loss of generality we may assume $c=1$. Then dividing (111) by $(s-s^*)$ and taking limits as $s\downarrow s^*$ gives ${\textbf{U}}(s^*)={\textbf{V}}(s^*)$, or equivalently,

(112) \begin{equation}{\textbf{B}}(s^*){\textbf{Q}}_{21}(s^*){\textbf{B}}(s^*)={\textbf{U}}(s^*).\end{equation}

That is, Case (A) gives $g(s-s^*)=\sqrt{s-s^*}$.

(B) Suppose $o(g(s-s^*))\not={\boldsymbol{0}}$. Then we write the term $o(g(s-s^*))$ in the form

(113) \begin{eqnarray} o(g(s-s^*))&\,=\,& L(s-s^*){\textbf{Y}}(s^*) +o(L(s-s^*))\end{eqnarray}

for some function $L(\cdot)\not= 0$ such that $L(s-s^*)=o(g(s-s^*))$ and some constant ${\textbf{Y}}(s^*)\not= {\boldsymbol{0}}$.

We then have

(114) \begin{eqnarray} {\boldsymbol{0}}&\,=\,&-(s-s^*){\textbf{U}}(s^*)+g^2(s-s^*){\textbf{V}}(s^*) +L(s-s^*){\textbf{Y}}(s^*) +o(L(s-s^*)). \end{eqnarray}

In relation to the terms $(s-s^*)$, $g^2(s-s^*)$, and $L(s-s^*)$, we consider three cases, labelled (B)(ii)–(B)(iv). We will show that Case (B)(ii) gives a contradiction and Cases (B)(iii) and (B)(iv) give $g(s-s^*)=\sqrt{s-s^*}$.

(B)(ii) Suppose one of $(s-s^*)$, $g^2(s-s^*)$, and $L(s-s^*)$ dominates the other two.

If $(s-s^*)$ dominates the other terms, that is, $g^2(s-s^*)=o(s-s^*)$ and $L(s-s^*)=$ $o(s-s^*)$, then dividing Equation (114) by $(s-s^*)$ and taking limits as $s\downarrow s^*$ gives ${\textbf{U}}(s^*)={\boldsymbol{0}}$, a contradiction.

If $g^2(s-s^*)$ dominates the other terms, that is, $(s-s^*)=o(g^2(s-s^*))$ and $L(s-s^*)=o(g^2(s-s^*))$, then dividing Equation (114) by $g^2(s-s^*)$ and taking limits as $s\downarrow s^*$ gives ${\textbf{V}}(s^*)={\boldsymbol{0}}$, a contradiction.

If $L(s-s^*)$ dominates the other terms, that is, $(s-s^*)=o(L(s-s^*))$ and $g^2(s-s^*)=o(L(s-s^*))$, then dividing Equation (114) by $L(s-s^*)$ and taking limits as $s\downarrow s^*$ gives ${\textbf{Y}}(s^*)={\boldsymbol{0}}$, a contradiction.

That is, Case (B)(ii) gives a contradiction. Therefore at least two of $(s-s^*)$, $g^2(s-s^*)$, and $L(s-s^*)$ must be multiples of each other.

(B)(iii) Suppose each of $(s-s^*)$, $g^2(s-s^*)$ and $L(s-s^*)$ is a multiple of the others. Then $(s-s^*)=c\cdot g^2(s-s^*)=d\cdot L(s-s^*)$, and without loss of generality we may assume $c=1$, $d=1$, by an argument analogous to the one used before. Therefore, dividing Equation (114) by $(s-s^*)$ and taking limits as $s\downarrow s^*$ gives ${\boldsymbol{0}}=-{\textbf{U}}(s^*)+{\textbf{V}}(s^*)+{\textbf{Y}}(s^*)$, or equivalently,

(115) \begin{eqnarray}{\textbf{B}}(s^*){\textbf{Q}}_{21}(s^*){\textbf{B}}(s^*) &\,=\,&{\textbf{U}}(s^*)-{\textbf{Y}}(s^*).\end{eqnarray}

That is, Case (B)(iii) gives $g(s-s^*)=\sqrt{s-s^*}$.

(B)(iv) Suppose exactly two of $(s-s^*)$, $g^2(s-s^*)$, and $L(s-s^*)$ are multiples of one another. Then these two terms must dominate the third term, or we have a contradiction by Part (i) of Case II above.

If $(s-s^*)=c\cdot g^2(s-s^*)$ for some $c>0$, then without loss of generality we may assume $c=1$. Also, we must have $L(s-s^*)=o(s-s^*)$. Therefore, $g(s-s^*)=\sqrt{s-s^*}$, and dividing Equation (114) by $(s-s^*)$ and taking limits as $s\downarrow s^*$ gives ${\textbf{U}}(s^*)={\textbf{V}}(s^*)$, or equivalently,

(116) \begin{equation}{\textbf{B}}(s^*){\textbf{Q}}_{21}(s^*){\textbf{B}}(s^*)={\textbf{U}}(s^*).\end{equation}

If $L(s-s^*)=c\cdot (s-s^*)$ for some $c\not= 0$, then dividing Equation (114) by $(s-s^*)$ and taking limits as $s\downarrow s^*$ gives ${\textbf{V}}(s^*)={\boldsymbol{0}}$, a contradiction.

If $L(s-s^*)=c\cdot g^2(s-s^*)$ for some $c> 0$, then without loss of generality we may assume $c=1$. Also, we must have $L(s-s^*)=o(s-s^*)$. Therefore, Equation (114) becomes

(117) \begin{eqnarray} {\boldsymbol{0}}&\,=\,&-(s-s^*){\textbf{U}}(s^*)+g^2(s-s^*)({\textbf{V}}(s^*)+{\textbf{Y}}(s^*)) +o(g^2(s-s^*)).\end{eqnarray}

In this case, if $g^2(s-s^*)=o(s-s^*)$, then dividing Equation (117) by $(s-s^*)$ and taking limits as $s\downarrow s^*$ gives ${\textbf{U}}(s^*)={\boldsymbol{0}}$, a contradiction. If $(s-s^*)=o(g^2(s-s^*))$, then dividing Equation (117) by $g^2(s-s^*)$ and taking limits as $s\downarrow s^*$ gives ${\textbf{U}}(s^*)+{\textbf{Y}}(s^*)={\boldsymbol{0}}$, a contradiction. Therefore, we must have $g^2(s-s^*)=c\cdot (s-s^*)$ for some $c> 0$. Without loss of generality we may assume $c=1$. Therefore, $g(s-s^*)=\sqrt{s-s^*}$, and dividing Equation (117) by $g^2(s-s^*)$ and taking limits as $s\downarrow s^*$ gives

(118) \begin{eqnarray} {\textbf{B}}(s^*){\textbf{Q}}_{21}(s^*){\textbf{B}}(s^*)&\,=\,& {\textbf{U}}(s^*)- {\textbf{Y}}(s^*).\end{eqnarray}

That is, Case (B)(iv) gives $g(s-s^*)=\sqrt{s-s^*}$.

By the above arguments, we must have

(119) \begin{equation} g(s-s^*)=\sqrt{s-s^*},\end{equation}

and

(120) \begin{eqnarray} {\textbf{B}}(s^*){\textbf{Q}}_{21}(s^*){\textbf{B}}(s^*)&\,=\,& {\textbf{U}}(s^*)- {\textbf{Y}}(s^*), \end{eqnarray}

(121) \begin{eqnarray} {\textbf{K}}(s^*){\textbf{B}}(s^*)+{\textbf{B}}(s^*){\textbf{D}}(s^*)&\,=\,&{\boldsymbol{0}},\end{eqnarray}

and $-\infty<{\textbf{Y}}(s^*)\leq {\textbf{U}}(s^*)$. Here, ${\textbf{Y}}(s^*)= {\boldsymbol{0}}$ whenever the term $o(g(s-s^*))$ in (110) satisfies $o(g(s-s^*))=o(s-s^*)$, and ${\textbf{Y}}(s^*)\not= {\boldsymbol{0}}$ when $o(g(s-s^*))=(s-s^*){\textbf{Y}}(s^*)+o(s-s^*)$.

Finally, we show (92). By L’Hospital’s rule,

(122) \begin{eqnarray}{\textbf{B}}(s^*)&\,=\,& -\lim_{s\downarrow s^*}\frac{{\boldsymbol{\Psi}}(s)-{\boldsymbol{\Psi}}(s^*)}{\sqrt{s-s^*}}\ = \ \lim_{s\downarrow s^*}\left({\boldsymbol{\Phi}}(s)2\sqrt{s-s^*}\right),\end{eqnarray}

and so, by taking limits as $s\downarrow s^*$ in (85),

(123) \begin{eqnarray}&&{{\textbf{B}}(s^*){\textbf{Q}}_{21} (s^*){\textbf{B}}(s^*)+{\textbf{Y}}(s^*)= {\textbf{U}}(s^*)}\nonumber\\[4pt] &&\,=\, \lim_{s\downarrow s^*} \Big[ {\textbf{K}}(s){\boldsymbol{\Phi}}(s) +{\boldsymbol{\Phi}}(s){\textbf{D}}(s) \Big]\nonumber\\[4pt] &&\,=\,\lim_{s\downarrow s^*}\Big[({\textbf{Q}}_{11}(s)+{\boldsymbol{\Psi}}(s){\textbf{Q}}_{21}(s)){\boldsymbol{\Phi}}(s)+{\boldsymbol{\Phi}}(s)({\textbf{Q}}_{22}(s)+{\textbf{Q}}_{21}(s){\boldsymbol{\Psi}}(s))\Big]\nonumber &&\,=\,\lim_{s\downarrow s^*}\Bigg[\frac{1}{2}\left(\frac{{\boldsymbol{\Psi}}(s)-{\boldsymbol{\Psi}}(s^*)}{\sqrt{s-s^*}}\right){\textbf{Q}}_{21}(s)\left({\boldsymbol{\Phi}}(s)2\sqrt{s-s^*}\right)\nonumber\\[4pt] &&+\frac{1}{2}\left({\boldsymbol{\Phi}}(s)2\sqrt{s-s^*}\right){\textbf{Q}}_{21}(s)\left(\frac{{\boldsymbol{\Psi}}(s)-{\boldsymbol{\Psi}}(s^*)}{\sqrt{s-s^*}}\right)\nonumber\\[4pt]&&+{\textbf{Q}}_{11}(s){\boldsymbol{\Phi}}(s)+{\boldsymbol{\Phi}}(s){\textbf{Q}}_{22}(s)+{\boldsymbol{\Psi}}(s^*){\textbf{Q}}_{21}(s){\boldsymbol{\Phi}}(s)+{\boldsymbol{\Phi}}(s){\textbf{Q}}_{21}(s){\boldsymbol{\Psi}}(s^*)\Bigg]\nonumber\\[4pt]&\,=\,&\frac{1}{2}{\textbf{B}}(s^*){\textbf{Q}}_{21} (s^*){\textbf{B}}(s^*)+\frac{1}{2}{\textbf{B}}(s^*){\textbf{Q}}_{21} (s^*){\textbf{B}}(s^*)\nonumber\\[4pt]&&+\lim_{s\downarrow s^*}\Bigg[\left(\frac{{\textbf{Q}}_{11}(s)-{\textbf{Q}}_{11}(s^*)}{\sqrt{s-s^*}}\right)({\boldsymbol{\Phi}}(s)\sqrt{s-s^*})+(\sqrt{s-s^*}{\boldsymbol{\Phi}}(s))\left(\frac{{\textbf{Q}}_{22}(s)-{\textbf{Q}}_{22}(s^*)}{\sqrt{s-s^*}} \right)\nonumber\\[4pt]&&+{\textbf{Q}}_{11}(s^*){\boldsymbol{\Phi}}(s)+{\boldsymbol{\Phi}}(s){\textbf{Q}}_{22}(s^*)\nonumber\\[4pt]&&+{\boldsymbol{\Psi}}(s^*)\left(\frac{{\textbf{Q}}_{21}(s)-{\textbf{Q}}_{21}(s^*)}{\sqrt{s-s^*}}\right)({\boldsymbol{\Phi}}(s)\sqrt{s-s^*})+({\boldsymbol{\Phi}}(s)\sqrt{s-s^*})\left(\frac{{\textbf{Q}}_{21}(s)-{\textbf{Q}}_{21}(s^*)}{\sqrt{s-s^*}}\right){\boldsymbol{\Psi}}(s^*)\nonumber\\[4pt]&&+{\boldsymbol{\Psi}}(s^*){\textbf{Q}}_{21}(s^*){\boldsymbol{\Phi}}(s)+{\boldsymbol{\Phi}}(s){\textbf{Q}}_{21}(s^*){\boldsymbol{\Psi}}(s^*)\Bigg]\nonumber\\[4pt]&\,=\,&{\textbf{B}}(s^*){\textbf{Q}}_{21} (s^*){\textbf{B}}(s^*)+ {\boldsymbol{0}}+\lim_{s\downarrow s^*}\Big[{\textbf{K}}(s^*){\boldsymbol{\Phi}}(s)+{\boldsymbol{\Phi}}(s){\textbf{D}}(s^*)\Big],\end{eqnarray}

which completes the proof.

The next result follows immediately by Theorem 1.

Corollary 1. We have

(124) \begin{equation}\boldsymbol{\psi}(t)={\textbf{B}}(s^*)\frac{1}{2\sqrt{\pi}}t^{-3/2}e^{s^* t}(1+o(1)).\end{equation}

Example 1. Since $\lim_{s\downarrow s^*} \Delta(s)=\Delta(s^*)=0$, we have

(125) \begin{eqnarray}\lim_{s\downarrow s^*}\frac{d}{ds}{\boldsymbol{\Psi}}(s)&\,=\,& \lim_{s\downarrow s^*}\frac{d}{ds} \frac{(a+b+2s)-\sqrt{\Delta(s)}}{2b}\nonumber\\[4pt]&\,=\,& \lim_{s\downarrow s^*}\frac{d}{ds}\left(\frac{1}{b}-\frac{1}{4b \sqrt{\Delta(s)}}(8s+4(a+b))\right)\nonumber\\[4pt]&\,=\,&-\infty,\end{eqnarray}

as expected. Furthermore,

(126) \begin{eqnarray}\lim_{s\downarrow (s^*)}\frac{{\boldsymbol{\Psi}}(s)-{\boldsymbol{\Psi}}(s^*)}{\sqrt{s-s^*}}&\,=\,&\lim_{s\downarrow (s^*)}\left(\frac{(a+b+2s)-\sqrt{\Delta(s)}}{2b\sqrt{s-s^*}}-\frac{(a+b+2s^*)-\sqrt{\Delta(s^*)}}{2b\sqrt{s-s^*}}\right)\nonumber\\[4pt]&\,=\,& \frac{\sqrt{2\sqrt{ab}}}{-b},\end{eqnarray}

which implies

(127) \begin{equation}{\boldsymbol{\Psi}}(s)={\boldsymbol{\Psi}}(s^*)-{\textbf{B}}(s^*)\sqrt{s-s^*}+o(\sqrt{s-s^*}),\end{equation}

where

(128) \begin{equation}{\textbf{B}}(s^*)=\frac{\sqrt{2\sqrt{ab}}}{b}.\end{equation}

Therefore, by Theorem 1,

(129) \begin{equation}\boldsymbol{\psi}(t)=\frac{\sqrt{2\sqrt{ab}}}{2b\sqrt{\pi}}t^{-3/2}\exp\left(\left(\frac{-(a+b)+2\sqrt{ab}}{2}\right)t\right)(1+o(1)).\end{equation}

Also, ${\textbf{A}}_{12}(s^*)=0$, ${\textbf{A}}_{21}(s^*)=0$, ${\textbf{A}}_{11}(s^*)=1$, ${\textbf{A}}_{22}(s^*)=1$, so

(130) \begin{eqnarray}{\textbf{B}}(s^*){\textbf{Q}}_{21}(s^*){\textbf{B}}(s^*)&\,=\,&b\left(\frac{\sqrt{2\sqrt{ab}}}{b}\right)^2= \ 2\sqrt{\frac{a}{b}}\ = \ {\textbf{U}}(s^*),\end{eqnarray}

and

(131) \begin{eqnarray}{\textbf{K}}(s^*){\textbf{B}}(s^*)+{\textbf{B}}(s^*){\textbf{D}}(s^*)&\,=\,&\left( \frac{b-a}{2}+\frac{a-b}{2} \right)\left(\frac{\sqrt{2\sqrt{ab}}}{b}\right)={\boldsymbol{0}},\end{eqnarray}

(132) \begin{eqnarray} \lim_{s\downarrow s^*} \left( {\textbf{K}}(s^*){\boldsymbol{\Phi}}(s)+{\boldsymbol{\Phi}}(s){\textbf{D}}(s^*) \right)&\,=\,&\lim_{s\downarrow s^*}\left(\left( \frac{b-a}{2}+\frac{a-b}{2} \right)2 {\boldsymbol{\Phi}}(s)\right)={\boldsymbol{0}}. \end{eqnarray}

For $n\geq 1$, define the matrices

(133) \begin{eqnarray}{\textbf{H}}_{1,n}(s^*)&\,=\,&\sum_{i=0}^{n-1}\left({\textbf{K}}(s^*) \right)^i\times{\textbf{B}}(s^*){\textbf{Q}}_{21}(s^*)\times\left({\textbf{K}}(s^*) \right)^{n-1-i}\end{eqnarray}

and

(134) \begin{equation}{\textbf{H}}(s^*,y)=\sum_{n=1}^{\infty}\frac{y^n}{n!}{\textbf{H}}_{1,n}(s^*),\ \ {\textbf{H}}(s^*)=\int_{y=0}^{\infty}{\textbf{H}}(s^*,y)dy,\end{equation}

and define a column vector

(135) \begin{eqnarray}\widetilde{\textbf{H}}(s^*)&\,=\,&{\textbf{H}}(s^*){\textbf{C}}_1^{-1}{\boldsymbol{1}}+\left({-}({\textbf{K}}(s^*))^{-1}{\textbf{B}}(s^*)+{\textbf{H}}(s^*){\boldsymbol{\Psi}}(s^*)\right){\textbf{C}}_2^{-1}{\boldsymbol{1}}\nonumber\\[4pt]&&+\left[\begin{array}{c@{\quad}c}{\textbf{H}}(s^*)&-({\textbf{K}}(s^*))^{-1}{\textbf{B}}(s^*)+{\textbf{H}}(s^*){\boldsymbol{\Psi}}(s^*)\end{array}\right]\left[\begin{array}{c}{\textbf{C}}_1^{-1}{\textbf{T}}_{10}\\[4pt]{\textbf{C}}_2^{-1}{\textbf{T}}_{20}\end{array}\right]({-}({{\textbf{T}}}_{00}-s^*{\textbf{I}})^{-1}){\boldsymbol{1}}.\nonumber\\[4pt]\end{eqnarray}

Below, we derive the expressions for $\boldsymbol{\mu}(dy)^{(0)}$.

Theorem 3. The matrix $\boldsymbol{\mu}(dy)^{(0)}$ is unique, and

(136) \begin{eqnarray}\boldsymbol{\mu}(dy)^{(0)}_{11}&\,=\,&diag({\widetilde{\textbf{H}}(s^*)})^{-1}{ {\textbf{H}}(s^*,y){\textbf{C}}_1^{-1}}dy,\nonumber\\[4pt]\boldsymbol{\mu}(dy)^{(0)}_{12}&\,=\,&diag({\widetilde{\textbf{H}}(s^*)})^{-1}\left(e^{{\textbf{K}}(s^*)y}{\textbf{B}}(s^*)+{\textbf{H}}(s^*,y){\boldsymbol{\Psi}}(s^*)\right){\textbf{C}}_2^{-1}dy,\nonumber\\[4pt]\boldsymbol{\mu}(dy)^{(0)}_{10}&\,=\,& \left[\begin{array}{c@{\quad}c}\boldsymbol{\mu}(dy)^{(0)}_{11} &\boldsymbol{\mu}(dy)^{(0)}_{12}\end{array}\right]\left[\begin{array}{c}{\textbf{T}}_{10}\\[4pt]{\textbf{T}}_{20}\end{array}\right]({-}({{\textbf{T}}}_{00}-s^*{\textbf{I}})^{-1}).\end{eqnarray}

Proof. By Lemma 2, Lemma 5, and Theorem 2, we have

(137) \begin{eqnarray}e^{{\textbf{K}}(s)y}&\,=\,&\lim_{K\to\infty}\sum_{n=0}^{K}\frac{y^n}{n!}\left({\textbf{Q}}_{11}(s)+{\boldsymbol{\Psi}}(s){\textbf{Q}}_{21}(s)\right)^n\nonumber\qquad\ \\[4pt] &\,=\,&\lim_{K\to\infty}\sum_{n=0}^{K}\frac{y^n}{n!}\left({\textbf{Q}}_{11}(s^*)+\left( {\boldsymbol{\Psi}}(s^*)-{\textbf{B}}(s^*)\sqrt{s-s^*} \right){\textbf{Q}}_{21}(s^*)+o(\sqrt{s-s^*})\right)^n\nonumber\\[4pt] &\,=\,&\lim_{K\to\infty}\sum_{n=0}^{K}\frac{y^n}{n!}\left({\textbf{Q}}_{11}(s^*)+{\boldsymbol{\Psi}}(s^*){\textbf{Q}}_{21}(s^*)\right)^n-\lim_{K\to\infty}\sqrt{s-s^*}\sum_{n=1}^{K }\frac{y^n}{n!}{\textbf{H}}_{1,n}+o(\sqrt{s-s^*})\nonumber\\[4pt] &\,=\,&e^{{\textbf{K}}(s^*)y}-\sqrt{s-s^*}{\textbf{H}}(s^*,y)+o(\sqrt{s-s^*}),\end{eqnarray}

which gives

(138) \begin{eqnarray}{\textbf{E}}(dy)^{(0)}(s)_{11}&\,=\,&e^{{\textbf{K}}(s)y}{\textbf{C}}_1^{-1}dy\nonumber\\[4pt]&\,=\,&{\textbf{E}}(dy)^{(0)}(s^*)_{11}-\sqrt{s-s^*}{\textbf{H}}(s^*,y){\textbf{C}}_1^{-1}dy+o(\sqrt{s-s^*})\end{eqnarray}

and

(139) \begin{eqnarray} {\textbf{E}}(dy)^{(0)}(s)_{12}& = & e^{{\textbf{K}}(s)y}{\boldsymbol{\Psi}}(s){\textbf{C}}_2^{-1}dy\nonumber\\[4pt] &=\,&\left(e^{{\textbf{K}}(s^*)y}-\sqrt{s-s^*}{\textbf{H}}(s^*,y)\right)({\boldsymbol{\Psi}}(s^*)-{\textbf{B}}(s^*)\sqrt{s-s^*}){\textbf{C}}_2^{-1}dy+o(\sqrt{s-s^*})\nonumber\\[4pt] &=\,&{\textbf{E}}(dy)^{(0)}(s^*)_{12}-\sqrt{s-s^*}\left(e^{{\textbf{K}}(s^*)y}{\textbf{B}}(s^*)+{\textbf{H}}(s^*,y){\boldsymbol{\Psi}}(s^*)\right){\textbf{C}}_2^{-1}dy+o(\sqrt{s-s^*}).\nonumber\\[5pt]\end{eqnarray}

Also, noting that $({\textbf{T}}_{00}-s{\textbf{I}})^{-1}-({\textbf{T}}_{00}-s^*{\textbf{I}})^{-1}=(s-s^*)({\textbf{T}}_{00}-s^*{\textbf{I}})^{-2}+o(s-s^*)$, which gives $({{\textbf{T}}}_{00}-s{\textbf{I}})^{-1}=({{\textbf{T}}}_{00}-s^*{\textbf{I}})^{-1}+o(\sqrt{s-s^*})$, we have

(140) \begin{eqnarray}&& {{\textbf{E}}(dy)^{(0)}(s)_{10}=\left[\begin{array}{c@{\quad}c} e^{{\textbf{K}}(s)y} {\textbf{C}}_1^{-1}&e^{{\textbf{K}}(s)y}{\boldsymbol{\Psi}}(s){\textbf{C}}_2^{-1}\end{array}\right]\left[\begin{array}{c}{\textbf{T}}_{10}\\[4pt] {\textbf{T}}_{20}\end{array}\right]({-}({{\textbf{T}}}_{00}-s{\textbf{I}})^{-1})}\nonumber\\[4pt] &&\,=\,{\textbf{E}}(dy)^{(0)}(s^*)_{10}-\sqrt{s-s^*}\left[\begin{array}{c@{\quad}c}{\textbf{H}}(s^*,y)&\left(e^{{\textbf{K}}(s^*)y}{\textbf{B}}(s^*)+{\textbf{H}}(s^*,y){\boldsymbol{\Psi}}(s^*)\right)\end{array}\right]\nonumber\\[4pt] &&\quad\times\left[\begin{array}{c}{\textbf{C}}_1^{-1}{\textbf{T}}_{10}\\[4pt]{\textbf{C}}_2^{-1}{\textbf{T}}_{20}\end{array}\right]({-}({{\textbf{T}}}_{00}-s^*{\textbf{I}})^{-1})dy+o(\sqrt{s-s^*}).\end{eqnarray}

Furthermore,

(141) \begin{eqnarray}{\textbf{E}}^{(0)}(s)_1&\,=\,&\int_{y=0}^{\infty}{\textbf{E}}(dy)^{(0)}(s)_{11}{\boldsymbol{1}}+\int_{y=0}^{\infty}{\textbf{E}}(dy)^{(0)}(s)_{12}{\boldsymbol{1}}+\int_{y=0}^{\infty}{\textbf{E}}(dy)^{(0)}(s)_{10}{\boldsymbol{1}}\nonumber\\[8pt] &\,=\,&{\textbf{E}}^{(0)}(s^*)_1-\sqrt{s-s^*}{\textbf{H}}(s^*){\textbf{C}}_1^{-1}{\boldsymbol{1}}-\sqrt{s-s^*}\left({-}({\textbf{K}}(s^*))^{-1}{\textbf{B}}(s^*)+{\textbf{H}}(s^*){\boldsymbol{\Psi}}(s^*)\right){\boldsymbol{1}}\nonumber\\[8pt] &&-\sqrt{s-s^*}\left[\begin{array}{c@{\quad}c}{\textbf{H}}(s^*)&\left({-}({\textbf{K}}(s^*))^{-1}{\textbf{B}}(s^*)+{\textbf{H}}(s^*){\boldsymbol{\Psi}}(s^*)\right)\end{array}\right]\nonumber\\[9pt] &&\times\left[\begin{array}{c}{\textbf{C}}_1^{-1}{\textbf{T}}_{10}\\[10pt] {\textbf{C}}_2^{-1}{\textbf{T}}_{20}\end{array}\right]({-}({{\textbf{T}}}_{00}-s^*{\textbf{I}})^{-1}){\boldsymbol{1}}+o(\sqrt{s-s^*})\nonumber\\[4pt]&\,=\,&{\textbf{E}}^{(0)}(s^*)_1-\sqrt{s-s^*}\widetilde{\textbf{H}}(s^*)+o(\sqrt{s-s^*}).\end{eqnarray}

The result follows from Theorem 1 and (10), since the relevant terms cancel out. Indeed, for $i,j\in\mathcal{S}_1$, by (138)–(141), Theorem 1, and (10),

(142) \begin{eqnarray} \mu(dy)^{(0)}_{ij} &\,=\,& \lim_{t\to\infty} \frac{ \mathbb{P}(X(t)\in dy,\varphi(t)=j,\theta(0)>t\, |\, X(0)=0,\varphi(0)=i) } { \mathbb{P}(\theta(0)>t\, |\, X(0)=0,\varphi(0)=i) } \nonumber\\[8pt] &\,=\,& \frac{ \lim_{t\to\infty} ([{\textbf{H}}(s^*,y) {\textbf{C}}_1^{-1}]_{ij}\Gamma(1/2)^{-1}t^{-1/2-1}e^{s^*t}(1+o(1)) } { \lim_{t\to\infty} ([\widetilde{\textbf{H}}(s^*)]_{i}\Gamma(1/2)^{-1}t^{-1/2-1}e^{s^*t}(1+o(1)) }dy \nonumber\\[8pt] &\,=\,& \frac{[{\textbf{H}}(s^*,y){\textbf{C}}_1^{-1}]_{ij} } { [\widetilde{\textbf{H}}(s^*)]_{i} }dy, \end{eqnarray}

which gives the result for $\boldsymbol{\mu}(dy)^{(0)}_{11}$. The expressions for $\boldsymbol{\mu}(dy)^{(0)}_{12}$ and $\boldsymbol{\mu}(dy)^{(0)}_{10}$ follow in a similar manner.

Example 1. Finally,

(143) \begin{eqnarray}{\textbf{H}}(s^*,y)&\,=\,&\sum_{n=1}^{\infty}\frac{y^n}{n!}{\textbf{H}}_{1,n}(s^*)\nonumber\\[4pt]&\,=\,&\sum_{n=1}^{\infty}\frac{y^n}{n!}\sum_{i=0}^{n-1}\left({\textbf{K}}(s^*) \right)^i\times{\textbf{B}}(s^*){\textbf{Q}}_{21}(s^*)\times\left({\textbf{K}}(s^*) \right)^{n-1-i}\nonumber\\[4pt]&\,=\,&\sum_{n=1}^{\infty}\frac{y^n}{n!}\sum_{i=0}^{n-1}\left( -(a-b)/2\right)^{n-1}\sqrt{2\sqrt{ab}}\nonumber\\[4pt]&\,=\,&ye^{\left( -(a-b)/2 \right)y}\sqrt{2\sqrt{ab}},\end{eqnarray}

and

(144) \begin{eqnarray}{\textbf{H}}(s^*)&\,=\,&\int_{y=0}^{\infty}ye^{\left( -(a-b)/2 \right)y}\sqrt{2\sqrt{ab}}\ dy\nonumber\\[4pt] &\,=\,&\frac{\sqrt{2\sqrt{ab}}}{(a-b)^2/4},\nonumber\\[4pt] \widetilde{\textbf{H}}(s^*)&\,=\,&{\textbf{H}}(s^*)+\left({-}({\textbf{K}}(s^*))^{-1}{\textbf{B}}(s^*)+{\textbf{H}}(s^*){\boldsymbol{\Psi}}(s^*)\right)\nonumber\\[4pt] &\,=\,&\frac{\sqrt{2\sqrt{ab}}}{(a-b)^2/4} \left(1+\sqrt{\frac{a}{b}} \right)+\frac{2}{a-b}\left(\frac{\sqrt{2\sqrt{ab}}}{b}\right),\end{eqnarray}

so

(145) \begin{eqnarray}\boldsymbol{\mu}(dy)^{(0)}_{11}&\,=\,& diag({\widetilde{\textbf{H}}(s^*)})^{-1} { {\textbf{H}}(s^*,y)} dy \nonumber\\[4pt] &\,=\,&\left(\frac{1}{(a-b)^2/4}\, \left(1+\sqrt{\frac{a}{b}} \right) +\frac{2}{a-b} \left( \frac{1}{b} \right)\right)^{-1}ye^{({-}(a-b)/2)y}dy,\nonumber\\[4pt] \boldsymbol{\mu}(dy)^{(0)}_{12}&\,=\,&diag({\widetilde{\textbf{H}}(s^*)})^{-1}\left(e^{{\textbf{K}}(s^*)y}{\textbf{B}}(s^*)+{\textbf{H}}(s^*,y){\boldsymbol{\Psi}}(s^*)\right)dy\nonumber\\[4pt] &\,=\,& \left(\frac{1}{(a-b)^2/4}\, \left(1+\sqrt{\frac{a}{b}} \right) +\frac{2}{a-b} \left( \frac{1}{b} \right) \right)^{-1} \left( \frac{1}{b}+y\sqrt{\frac{a}{b}} \right) e^{({-}(a-b)/2)y} dy.\nonumber\\[4pt] \end{eqnarray}

We plot the values of $\boldsymbol{\mu}(dy)^{(0)}_{11}$ and $\boldsymbol{\mu}(dy)^{(0)}_{12}$ in Figure 1.

Figure 1:

The values of $\boldsymbol{\mu}(dy)^{(0)}_{11}/dy$ and $\boldsymbol{\mu}(dy)^{(0)}_{12}/dy$ in Example 4 for $b=1$, $a=4,3,2$ (dotted, solid, and dashed lines, respectively).

We will now find the Yaglom limit for a strictly positive initial position of $X(0)=x>0$. For $n\geq 1$, define the matrices

(146) \begin{eqnarray} \mathbf{W}(s^*, x-z)&\,=\,&\sum_{n=1}^\infty \frac{(x-z)^n}{n!}\sum_{i=1}^{n-1}\mathbf{D}(s^*)^i \times\mathbf{Q}_{21}(s^*) \mathbf{B}(s^*) \times\mathbf{D}(s^*)^{n-1-i}, \nonumber\\[4pt] \mathbf{W}_x(s^*)&\,=\,&\int_{z=0}^x \mathbf{W}(s^*, x-z)dz, \nonumber\\[4pt] \mathbf{Z}_x(s^*,y)&\,=\,&\int_{z=0}^{\min\{x,y\}}\left(\mathbf{W}(s^*, x-z){\textbf{Q}}_{21}(s^*)e^{{\textbf{K}}(s^*)(y-z)}\right. \nonumber\\[4pt] &&\left.+e^{{\textbf{D}}(s^*)(x-z)}{\textbf{Q}}_{21}(s^*)\mathbf{H}(s^*, y-z) \right)dz, \nonumber\\[4pt] {\mathbf{Z}}_x(s^*)&\,=\,&\int_{y=0}^\infty \mathbf{Z}_x(s^*,y)\;dy, \end{eqnarray}

and define the column vectors

(147) \begin{eqnarray*} \widetilde{\mathbf{Z}}_x(s^*) &\,=\,&\widetilde{\mathbf{Z}}_x(s^*)_{11}{\boldsymbol{1}} + \widetilde{\mathbf{Z}}_x(s^*)_{12}{\boldsymbol{1}} \nonumber\\[4pt] && + \left[ \begin{array}{c@{\quad}c} \widetilde{\mathbf{Z}}_x(s^*)_{11} & \widetilde{\mathbf{Z}}_x(s^*)_{12} \end{array} \right] \left[ \begin{array}{c} {\textbf{T}}_{10}\\[4pt] {\textbf{T}}_{20} \end{array} \right] ({-}({{\textbf{T}}}_{00}-s^*{\textbf{I}})^{-1}) {\boldsymbol{1}}, \widetilde{\widetilde{\mathbf{Z}}}_x(s^*) &\,=\,&\widetilde{\widetilde{\mathbf{Z}}}_x(s^*)_{21}{\boldsymbol{1}} + \widetilde{\widetilde{\mathbf{Z}}}_x(s^*)_{22}{\boldsymbol{1}} \nonumber\\[4pt] && + \left[ \begin{array}{c@{\quad}c} \widetilde{\widetilde{\mathbf{Z}}}_x(s^*)_{21} & \widetilde{\widetilde{\mathbf{Z}}}_x(s^*)_x(s^*)_{22} \end{array} \right] \left[ \begin{array}{c} {\textbf{T}}_{10}\\[4pt] {\textbf{T}}_{20} \end{array} \right] ({-}({{\textbf{T}}}_{00}-s^*{\textbf{I}})^{-1}) {\boldsymbol{1}},\end{eqnarray*}

where

(148) \begin{eqnarray} \widetilde{\mathbf{Z}}_x(s^*)_{11}&\,=\,& {\mathbf{Z}}_x(s^*){\textbf{C}}_1^{-1}, \nonumber\\[4pt] \widetilde{\mathbf{Z}}_x(s^*)_{12}&\,=\,& \left( {\textbf{E}}^{(x)}(s^*)_{21}{\textbf{C}}_1 \mathbf{B}(s^*) +{\mathbf{Z}}_x(s^*){\boldsymbol{\Psi}}(s^*)+\mathbf{W}_x(s^*) \right) {\textbf{C}}_2^{-1}, \nonumber\\[4pt] \widetilde{\widetilde{\mathbf{Z}}}_x(s^*)_{21} &\,=\,& \mathbf{B}(s^*){\textbf{E}}^{(x)}(s^*)_{21} +{\boldsymbol{\Psi}}(s^*)\mathbf{Z}_x(s^*){\textbf{C}}_1^{-1} +\mathbf{H}(s^*){\textbf{C}}_1^{-1}, \nonumber\\[4pt] \widetilde{\widetilde{\mathbf{Z}}}_x(s^*)_{22}&\,=\,& \mathbf{B}(s^*){\textbf{E}}^{(x)}(s^*)_{22} +{\boldsymbol{\Psi}}(s^*) \Big( {\textbf{E}}^{(x)}(s^*)_{21}{\textbf{C}}_1\mathbf{B}(s^*){\textbf{C}}_2^{-1} +\mathbf{Z}_x(s^*){\boldsymbol{\Psi}}(s^*){\textbf{C}}_2^{-1} + \mathbf{W}(s^*){\textbf{C}}_2^{-1} \Big) \nonumber\\[4pt] && + \left( ({-}{{\textbf{K}}(s^*))^{-1}}\mathbf{B}(s^*) + {\textbf{H}}(s^*){\boldsymbol{\Psi}}(s^*) \right){\textbf{C}}_2^{-1}, \end{eqnarray}

with ${\textbf{E}}^{(x)}(s^*)_{21}=\int_{y=0}^{\infty}{\textbf{E}}(dy)^{(x)}(s^*)_{21}$ as considered in Remark 2, and

(149) \begin{eqnarray}{\textbf{E}}^{(x)}(s^*)_{22}&\,=\,&{\textbf{E}}^{(x)}(s^*)_{21}{\textbf{C}}_1{\boldsymbol{\Psi}}(s^*){\textbf{C}}_2^{-1}+\int_{y=0}^xe^{{\textbf{D}}(s^*)(x-y)}{\textbf{C}}_2^{-1}dy\nonumber\\[4pt]&\,=\,&{\textbf{E}}^{(x)}(s^*)_{21}{\textbf{C}}_1{\boldsymbol{\Psi}}(s^*){\textbf{C}}_2^{-1}-\int_{w=0}^xe^{{\textbf{D}}(s^*)w}{\textbf{C}}_2^{-1}dw\nonumber\\[4pt]&\,=\,&{\textbf{E}}^{(x)}(s^*)_{21}{\textbf{C}}_1{\boldsymbol{\Psi}}(s^*){\textbf{C}}_2^{-1}-({\textbf{D}}(s^*))^{-1}\left(e^{{\textbf{D}}(s^*)x}-{\textbf{I}}\right){\textbf{C}}_2^{-1}.\end{eqnarray}

Theorem 4. For $x>0$, the matrix $\boldsymbol{\mu}(dy)^{(x)}$ is unique, and

\begin{eqnarray*}\boldsymbol{\mu}(dy)^{(x)}_{21}&\,=\,&diag({\widetilde{\textbf{Z}}_x(s^*)})^{-1}{{\textbf{Z}}_x(s^*,y){\textbf{C}}_1^{-1}}dy,\nonumber\\[8pt] \boldsymbol{\mu}(dy)^{(x)}_{22}&\,=\,&diag({\widetilde{\textbf{Z}}_x(s^*)})^{-1}\Big({\textbf{E}}(dy)^{(x)}(s^*)_{21}{\textbf{C}}_1\mathbf{B}(s^*){\textbf{C}}_2^{-1}+\mathbf{Z}_x(s^*,y){\boldsymbol{\Psi}}(s^*){\textbf{C}}_2^{-1}dy\nonumber\\[8pt] &&+ \mathbf{W}(s^*, x-y){\boldsymbol{1}}\{y<x\}{\textbf{C}}_2^{-1}dy\Big),\\[8pt] \boldsymbol{\mu}(dy)^{(x)}_{20}&\,=\,&\left[\begin{array}{c@{\quad}c}\boldsymbol{\mu}(dy)^{(x)}_{21}&\boldsymbol{\mu}(dy)^{(x)}_{22}\end{array}\right]\left[\begin{array}{c}{\textbf{T}}_{10}\\[4pt] {\textbf{T}}_{20}\end{array}\right]({-}({{\textbf{T}}}_{00}-s^*{\textbf{I}})^{-1}),\nonumber\\[8pt] \boldsymbol{\mu}(dy)^{(x)}_{11}&\,=\,&diag({\widetilde{\widetilde{\textbf{Z}}}_x(s^*)})^{-1}\Big(\mathbf{B}(s^*){\textbf{E}}(dy)^{(x)}(s^*)_{21}+{\boldsymbol{\Psi}}(s^*)\mathbf{Z}_x(s^*,y){\textbf{C}}_1^{-1} dy\\[4pt] &&+\mathbf{H}(s^*, y-x){\textbf{C}}_1^{-1}{\boldsymbol{1}}\{y>x\}dy\Big),\boldsymbol{\mu}(dy)^{(x)}_{12}&\,=\,&diag({\widetilde{\widetilde{\textbf{Z}}}_x(s^*)})^{-1}\Big\{\Big(\mathbf{B}(s^*){\textbf{E}}(dy)^{(x)}(s^*)_{22}+{\boldsymbol{\Psi}}(s^*)\Big({\textbf{E}}(dy)^{(x)}(s^*)_{21}{\textbf{C}}_1\mathbf{B}(s^*){\textbf{C}}_2^{-1}\nonumber\\[4pt]&&+\mathbf{Z}_x(s^*,y){\boldsymbol{\Psi}}(s^*){\textbf{C}}_2^{-1}dy+ \mathbf{W}(s^*, x-y){\boldsymbol{1}}\{y<x\}{\textbf{C}}_2^{-1}dy\Big)\Big)\nonumber\\[4pt]&&+\left(e^{{\textbf{K}}(s^*)(y-x)}\mathbf{B}(s^*)+{\textbf{H}}(s^*,y-x){\boldsymbol{\Psi}}(s^*)\right){\textbf{C}}_2^{-1}{\boldsymbol{1}}\{y>x\}dy\Big\},\nonumber\\[4pt]\boldsymbol{\mu}(dy)^{(x)}_{10}&\,=\,&\left[\begin{array}{c@{\quad}c}\boldsymbol{\mu}(dy)^{(x)}_{11}&\boldsymbol{\mu}(dy)^{(x)}_{12}\end{array}\right]\left[\begin{array}{c}{\textbf{T}}_{10}\\[4pt]{\textbf{T}}_{20}\end{array}\right]({-}({{\textbf{T}}}_{00}-s^*{\textbf{I}})^{-1}). \end{eqnarray*}

The next corollary follows from Theorems 3 and 4.

Proof. Our proof is again based on Theorem 1 and (10). Note that

(150) \begin{align}e^{\mathbf{D}(s)(x-z)}& =\lim_{K\to+\infty}\sum_{n=0}^K\frac{(x-z)^n}{n!}\left(\mathbf{Q}_{22}(s)+\mathbf{Q}_{21}(s){\boldsymbol{\Psi}}(s)\right)^n\nonumber\\[4pt] & =\lim_{K\to+\infty}\sum_{n=0}^k\frac{(x-z)^n}{n!}\left(\mathbf{Q}_{22}(s^*)+\mathbf{Q}_{21}(s^*)({\boldsymbol{\Psi}}(s^*)-\mathbf{B}(s^*)\sqrt{s-s^*}+o(\sqrt{s-s^*})))\right)^n\nonumber\\[4pt] & = e^{\mathbf{D}(s^*)(x-z)}- \sqrt{s-s^*}\mathbf{W}(s^*, x-z)+o(\sqrt{s-s^*}).\end{align}

By (137), (150), Lemmas 5 and 3, and Theorem 2, we have

(151) \begin{eqnarray}{\textbf{E}}(dy)^{(x)}(s)_{21}&\,=\,&\int_{z=0}^{\min\{x,y\}}e^{{\textbf{D}}(s)(x-z)}{\textbf{Q}}_{21}(s)e^{{\textbf{K}}(s)(y-z)}{\textbf{C}}_1^{-1}dzdy\nonumber\\[4pt]&\,=\,&\int_{z=0}^{\min\{x,y\}}\left(e^{\mathbf{D}(s^*)(x-z)}- \sqrt{s-s^*}\mathbf{W}^*(s^*, x-z)\right)\mathbf{Q}_{21}(s^*)\nonumber\\[4pt]&&\times \left(e^{\mathbf{K}(s^*)(y-z)}- \sqrt{s-s^*}\mathbf{H}(s^*, y-z)\right){\textbf{C}}_1^{-1}dz dy+o(\sqrt{s-s^*})\nonumber\\[4pt]&\,=\,&{\textbf{E}}(dy)^{(x)}(s^*)_{21}- \sqrt{s-s^*}\mathbf{Z}_x(s^*,y){\textbf{C}}_1^{-1} dy +o(\sqrt{s-s^*}),\end{eqnarray}

(152) \begin{align}& {\textbf{E}}(dy)^{(x)}(s)_{22}={\textbf{E}}(dy)^{(x)}(s)_{21}{\textbf{C}}_1{\boldsymbol{\Psi}}(s){\textbf{C}}_2^{-1}+e^{{\textbf{D}}(s)(x-y)}{\textbf{C}}_2^{-1}{\boldsymbol{1}}\{y<x\}dy\nonumber\\[4pt] & = \left({\textbf{E}}(dy)^{(x)}(s^*)_{21}{\textbf{C}}_1- \sqrt{s-s^*}\mathbf{Z}_x(s^*,y) dy\right)\left({\boldsymbol{\Psi}}(s^*)-{\textbf{B}}(s^*)\sqrt{s-s^*}\right){\textbf{C}}_2^{-1}\nonumber\\[4pt] & \quad +\left(e^{\mathbf{D}(s^*)(x-y)}- \sqrt{s-s^*}\mathbf{W}(s^*, x-y)\right){\textbf{C}}_2^{-1}{\boldsymbol{1}}\{y<x\}dy+o(\sqrt{s-s^*})\nonumber\\[4pt] & ={\textbf{E}}(dy)^{(x)}(s^*)_{22}- \sqrt{s-s^*}\Big({\textbf{E}}(dy)^{(x)}(s^*)_{21}{\textbf{C}}_1({-}\mathbf{B}(s^*)){\textbf{C}}_2^{-1}+\mathbf{Z}_x(s^*,y){\boldsymbol{\Psi}}(s^*){\textbf{C}}_2^{-1}dy\nonumber\\[4pt] & \quad + \mathbf{W}(s^*, x-y){\boldsymbol{1}}\{y<x\}{\textbf{C}}_2^{-1}dy\Big)+o(\sqrt{s-s^*}),\end{align}

(153) \begin{align}& {{\textbf{E}}(dy)^{(x)}(s)_{20}=\left[\begin{array}{c@{\quad}c}{\textbf{E}}(dy)^{(x)}(s)_{21}&{\textbf{E}}(dy)^{(x)}(s)_{22}\end{array}\right]\left[\begin{array}{c}{\textbf{T}}_{10}\\[4pt] {\textbf{T}}_{20}\end{array}\right]({-}({{\textbf{T}}}_{00}-s{\textbf{I}})^{-1})}\nonumber\\[4pt] & = {\textbf{E}}(dy)^{(x)}(s^*)_{20}\nonumber\\[4pt] & \quad - \sqrt{s-s^*} \left[ \begin{array}{c@{\quad}c} {\mathbf{Z}}_x(s^*,y) & {\textbf{E}}(dy)^{(x)}(s^*)_{21}{\textbf{C}}_1 \mathbf{B}(s^*) +{\mathbf{Z}}_x(s^*){\boldsymbol{\Psi}}(s^*)dy+\mathbf{W}(s^*,x-y){\boldsymbol{1}}\{y<x\}dy \end{array} \right] \nonumber\\[4pt] & \quad\times \left[ \begin{array}{c} {\textbf{C}}_1^{-1}{\textbf{T}}_{10}\\[4pt] {\textbf{C}}_2^{-1}{\textbf{T}}_{20} \end{array} \right] ({-}({{\textbf{T}}}_{00}-s^*{\textbf{I}})^{-1})+o(\sqrt{s-s^*}),\end{align}

and

(154) \begin{eqnarray}{\textbf{E}}^{(x)}(s)_2&\,=\,&\int_{y=0}^{\infty}{\textbf{E}}(dy)^{(x)}(s)_{21}{\boldsymbol{1}}+\int_{y=0}^{\infty}{\textbf{E}}(dy)^{(x)}(s)_{22}{\boldsymbol{1}}+\int_{y=0}^{\infty}{\textbf{E}}(dy)^{(x)}(s)_{20}{\boldsymbol{1}}\nonumber\\[4pt]&\,=\,&{\textbf{E}}^{(x)}(s^*)_1- \sqrt{s-s^*}\widetilde{\textbf{Z}}_x(s^*)+o(\sqrt{s-s^*}).\end{eqnarray}

Thus, the expressions for $\boldsymbol{\mu}(dy)^{(x)}_{21}$, $\boldsymbol{\mu}(dy)^{(x)}_{22}$, and $\boldsymbol{\mu}(dy)^{(x)}_{20}$ follow from arguments similar to those in the proof of Theorem 3.

Furthermore, by (137), Lemmas 5 and 3, and Theorem 2, we have

(155) \begin{align}&{{\textbf{E}}(dy)^{(x)}(s)_{11}={\boldsymbol{\Psi}}(s){\textbf{E}}(dy)^{(x)}(s)_{21}+e^{{\textbf{K}}(s)(y-x)}{\textbf{C}}_1^{-1}{\boldsymbol{1}}\{y>x\}dy}\nonumber\\[4pt] & = ({\boldsymbol{\Psi}}(s^*)-\mathbf{B}(s^*)\sqrt{s-s^*})({\textbf{E}}(dy)^{(x)}(s^*)_{21}- \sqrt{s-s^*}\mathbf{Z}_x(s^*,y){\textbf{C}}_1^{-1} dy)\nonumber\\[4pt] & \quad +\left(e^{\mathbf{K}(s^*)(y-x)}- \sqrt{s-s^*}\mathbf{H}(s^*, y-x)\right){\textbf{C}}_1^{-1}{\boldsymbol{1}}\{y>x\}dy+o(\sqrt{s-s^*})\nonumber\\[4pt] & = {\textbf{E}}(dy)^{(x)}(s^*)_{11}- \sqrt{s-s^*}\Big(\mathbf{B}(s^*){\textbf{E}}(dy)^{(x)}(s^*)_{21}+{\boldsymbol{\Psi}}(s^*)\mathbf{Z}_x(s^*,y){\textbf{C}}_1^{-1} dy\nonumber\\[4pt] & \quad +\mathbf{H}(s^*, y-x){\textbf{C}}_1^{-1}{\boldsymbol{1}}\{y>x\}dy\Big)+o(\sqrt{s-s^*})\end{align}

and

(156) \begin{align}& {{\textbf{E}}(dy)^{(x)}(s)_{12}={\boldsymbol{\Psi}}(s){\textbf{E}}(dy)^{(x)}(s)_{22}+e^{{\textbf{K}}(s)(y-x)}{\boldsymbol{\Psi}}(s){\textbf{C}}_2^{-1}{\boldsymbol{1}}\{y> x\}dy}\nonumber\\[4pt] & = \left({\boldsymbol{\Psi}}(s^*)-\mathbf{B}(s^*)\sqrt{s-s^*}\right)\Big( {\textbf{E}}(dy)^{(x)}(s^*)_{22}- \sqrt{s-s^*}\Big({\textbf{E}}(dy)^{(x)}(s^*)_{21}{\textbf{C}}_1\mathbf{B}(s^*){\textbf{C}}_2^{-1}\nonumber\\[4pt] & \quad +\mathbf{Z}_x(s^*,y){\boldsymbol{\Psi}}(s^*){\textbf{C}}_2^{-1}dy+ \mathbf{W}(s^*, x-y){\boldsymbol{1}}\{y<x\}{\textbf{C}}_2^{-1}dy\Big)\nonumber\\[4pt] & \quad +\Big(e^{{\textbf{K}}(s^*)(y-x)}{\boldsymbol{\Psi}}(s^*){\textbf{C}}_2^{-1}{\boldsymbol{1}}\{y>x\}dy\nonumber\\[4pt] & \quad - \sqrt{s-s^*}\left(e^{{\textbf{K}}(s^*)(y-x)}\mathbf{B}(s^*)+{\textbf{H}}(s^*,y-x){\boldsymbol{\Psi}}(s^*)\right){\textbf{C}}_2^{-1}dy\Big)+o(\sqrt{s-s^*})\nonumber\\[4pt] & = {\textbf{E}}(dy)^{(x)}(s^*)_{12}- \sqrt{s-s^*}\Big(\mathbf{B}(s^*){\textbf{E}}(dy)^{(x)}(s^*)_{22}+{\boldsymbol{\Psi}}(s^*)\Big({\textbf{E}}(dy)^{(x)}(s^*)_{21}{\textbf{C}}_1\mathbf{B}(s^*){\textbf{C}}_2^{-1}\nonumber\\[4pt] & \quad +\mathbf{Z}_x(s^*,y){\boldsymbol{\Psi}}(s^*){\textbf{C}}_2^{-1}dy+ \mathbf{W}(s^*, x-y){\boldsymbol{1}}\{y<x\}{\textbf{C}}_2^{-1}dy\Big)\Big)\nonumber\\[4pt] & \quad - \sqrt{s-s^*}\left(e^{{\textbf{K}}(s^*)(y-x)}\mathbf{B}(s^*)+{\textbf{H}}(s^*,y-x){\boldsymbol{\Psi}}(s^*)\right){\textbf{C}}_2^{-1}{\boldsymbol{1}}\{y>x\}dy+o(\sqrt{s-s^*}).\nonumber\\[4pt] \end{align}

Thus, the expressions for $\boldsymbol{\mu}(dy)^{(x)}_{11}$, $\boldsymbol{\mu}(dy)^{(x)}_{12}$, and $\boldsymbol{\mu}(dy)^{(x)}_{10}$ follow from a similar argument, with

(157) \begin{eqnarray}{\textbf{E}}^{(x)}(s)_1&\,=\,&\int_{y=0}^{\infty}{\textbf{E}}(dy)^{(x)}(s)_{11}{\boldsymbol{1}}+\int_{y=0}^{\infty}{\textbf{E}}(dy)^{(x)}(s)_{12}{\boldsymbol{1}}+\int_{y=0}^{\infty}{\textbf{E}}(dy)^{(x)}(s)_{10}{\boldsymbol{1}}\nonumber\\[4pt]&\,=\,&{\textbf{E}}^{(x)}(s^*)_1- \sqrt{s-s^*}\widetilde{\widetilde{\textbf{Z}}}_x(s^*)+o(\sqrt{s-s^*}).\end{eqnarray}

5. Example with non-scalar ${\boldsymbol{\Psi}}(s)$

Below we construct an example where, unlike in Example 1, key quantities are matrices, rather than scalars. We derive expressions for this example analytically and illustrate these results with some numerical output as well.

Example 2. Consider a system with $N=2$ sources based on an example analysed in [Reference Anick, Mitra and Sondhi6]. Let $\mathcal{S}=\{1,2,3\}$, $\mathcal{S}_1=\{1\}$, $\mathcal{S}_2=\{2,3\}$, $c_1=1$, $c_2=c_3=-1$, and

\begin{eqnarray}{\textbf{T}}&\,=\,&\left[\begin{array}{c@{\quad}c}{\textbf{T}}_{11}&{\textbf{T}}_{12}\\[4pt] {\textbf{T}}_{21}&{\textbf{T}}_{22}\end{array}\right]=\left[\begin{array}{c|cc}-2\lambda&2\lambda&0\\[4pt] \hline 1&-(1+\lambda)&\lambda\\[4pt] 0&2&-2\end{array}\right],\nonumber\\[4pt]{\textbf{Q}}(s)&\,=\,&\left[\begin{array}{c@{\quad}c}{\textbf{Q}}_{11}(s)&{\textbf{Q}}_{12}(s)\\[4pt]{\textbf{Q}}_{21}(s)&{\textbf{Q}}_{22}(s)\end{array}\right]=\left[\begin{array}{c|cc}-(2\lambda +s)&2\lambda&0\\[4pt]\hline1&-(1+\lambda +s)&\lambda\\[4pt]0&2&-(2+s)\end{array}\right],\nonumber\end{eqnarray}

with some parameter $\lambda>\sqrt{2}-1$ such that the process is stable. In our plots of the output below, we will assume the value $\lambda=2.5$.

Denote by $[x\ z]={\boldsymbol{\Psi}}(s)=\int_{t=0}^{\infty}e^{-st}\boldsymbol{\psi}(t)dt$ the minimum nonnegative solution of (21), here equivalent to

(158) \begin{eqnarray}[0\ 0]&\,=\,&[2\lambda\ 0]-(2\lambda+s)[x\ z]+[x\ z]\left[\begin{array}{c@{\quad}c}-(1+\lambda+s)&\lambda\\[4pt]2&-(2+s)\end{array}\right]\nonumber\\[4pt]&&+[x\ z]\left[\begin{array}{c}1\\[4pt]0\end{array}\right][x\ z],\end{eqnarray}

which we write as a system of equations

(159) \begin{eqnarray} 0&\,=\,&x^2-(1+3\lambda+2s)x+2z+2\lambda,\end{eqnarray}

(160) \begin{eqnarray}0&\,=\,&-(2+2\lambda+2s-x)z+\lambda x.\quad\ \ \ \ \end{eqnarray}

The minimum nonnegative solution $[x\ z]$ of (159)–(160) must be strictly positive, satisfy $2+2\lambda+2s-x> 0$, and occur at the intersection of the two curves

(161) \begin{eqnarray}z=z_1(x,s)&\,=\,&-\frac{1}{2}x^2+\frac{1}{2}(1+3\lambda+2s)x-\lambda,\end{eqnarray}

(162) \begin{eqnarray}z=z_2(x,s)&\,=\,&\lambda x/(2+2\lambda+2s-x).\qquad\ \ \ \ \ \ \end{eqnarray}

To facilitate the analysis that follows, we consider the shape of the curves in (161)–(162); see Figure 2. It is a straightforward exercise to verify that, when $s=0$, we have $z_1(x,0)<-\lambda<z_2(x,0)$ for all $x< 0$, and so the two curves may only intersect at some point (x, z) with $x>0$.

Figure 2:

Plots of (161)–(162) for $s=0$ (left) and $s=-2$ (right), when $\lambda=2.5$.

Furthermore, when $2+2\lambda+2s-x> 0$, we have

(163) \begin{eqnarray}\frac{\partial z_2(x,s)}{\partial x}=\frac{\lambda(2+2\lambda+2s-x)+\lambda x}{(2+2\lambda+2s-x)^2} > 0,\end{eqnarray}

and so, when $s=0$, the minimum nonnegative solution $[x\ z]$ of (159)–(160) is in fact the minimum real-valued solution of (159)–(160).

Also, when $x>0$ and $2+2\lambda+2s-x> 0$, we have

(164) \begin{eqnarray}\frac{\partial z_1(x,s)}{\partial s}=x&\ >\ &0,\nonumber\\[4pt] \frac{\partial z_2(x,s)}{\partial s}=\frac{-2\lambda x}{(2+2\lambda+2s-x)^2}&\ <\ &0,\end{eqnarray}

and so as $s\downarrow s^*$ we have $z_1(x,s)\downarrow$ while $z_2(x,s)\uparrow$, until the two curves touch when $s=s^*$, and then move apart when $s<s^*$. Therefore, by the argument about the continuity of ${\boldsymbol{\Psi}}(s)$ which was used in the proof of Lemma 4, for all $s\in [s^*,0]$, ${\boldsymbol{\Psi}}(s)=[x\ z]$ is the minimum real-valued solution of (159)–(160).

Instead of looking at the problem in terms of two intersecting curves $z_1(x,s)$ and $z_2(x,s)$, we now look at it in terms of one cubic curve $g_s(x)$. Substitute (162) into (159) and multiply by $(2+2\lambda+2s-x)$ to get

(165) \begin{eqnarray}0&\,=\,&-x^3+(3+5\lambda+4s)x^2-(2+2\lambda+2s)(1+3\lambda+2s)x+(2+2\lambda+2s)2\lambda\nonumber\\[4pt]&\,=\,&g_s(x),\end{eqnarray}

which is of the form

(166) \begin{equation}ax^3+bx^2+cx+d=0,\end{equation}

with $g_s(0)=d>0$ (we have $d>0$ because $0<x<2+2\lambda+2s$, since $z>0$ in (162)). (See the plots of $g_s(x)$ in Figure 3 for the case $\lambda=2.5$.) Noting that $a=-1<0$, we conclude that when $s=s^*$, the solution $[x\ z]$ corresponds to the local minimum,

(167) \begin{equation}x=\min \left\{ \frac{-b+ \sqrt{b^2-3ac}}{3a},\frac{-b - \sqrt{b^2-3ac}}{3a} \right\}=\frac{-b+ \sqrt{b^2-3ac}}{3a},\end{equation}

where

(168) \begin{equation}b^2-3ac>0.\end{equation}

Figure 3:

The plot of (165) for $s=0$ (top left), $s=-2$ (top right), and $s=-1.1178$ (bottom), when $\lambda=2.5$.

We transform the cubic equation (166) into the form

(169) \begin{equation}y^3+py+q=0\end{equation}

using the substitution

(170) \begin{eqnarray}x&\,=\,&y-\frac{b}{3a}\end{eqnarray}

with

(171) \begin{eqnarray}p&\,=\,&\frac{3ac-b^2}{3a^2}= s\times c_p^{(1)}+s^2\times c_p^{(2)}+c_p\end{eqnarray}

for suitable $c_p^{(1)}$, $c_p^{(2)}$, $c_p$, and

(172) \begin{eqnarray}q&\,=\,&\frac{2b^3+27a^2d-9abc}{27a^3}= s\times c_q^{(1)}+s^2\times c_q^{(2)}+s^3\times c_q^{(3)}+c_q\end{eqnarray}

for suitable $c_q^{(1)}$, $c_q^{(2)}$, $c_q^{(3)}$, $c_q$.

Below, we choose the convention of writing p(s) to demonstrate that p is a function of s, with similar notation applied for other quantities such as q, x, y, and so on. Observe that

\begin{eqnarray*}s^3-(s^*)^3&\,=\,&(s-s^*)(s^2+ss^*+(s^*)^2)=C_3\times (s-s^*)+o(s-s^*),\\[4pt]s^2-(s^*)^2&\,=\,&(s-s^*)(s+s^*)=C_2\times (s-s^*)+o(s-s^*),\end{eqnarray*}

where $C_3=3(s^*)^2$ and $C_2=2s^*$, and so by (171)–(172),

(173) \begin{eqnarray}p(s)-p(s^*)&\,=\,&C_p\times (s-s^*)+o(s-s^*),\end{eqnarray}

(174) \begin{eqnarray}q(s)-q(s^*)&\,=\,&C_q\times (s-s^*)+o(s-s^*),\end{eqnarray}

where the constants $C_p$ and $C_q$ are given by

(175) \begin{eqnarray}C_p&\,=\,& c_p^{(1)}+C_2\times c_p^{(2)},\nonumber\\[4pt]C_q&\,=\,&c_q^{(1)}+C_2\times c_q^{(2)}+C_3\times c_q^{(3)}.\end{eqnarray}

Consider (169) and apply Vieta’s substitution,

(176) \begin{equation}y=u-\frac{p}{3u},\end{equation}

where $u^3$ solves the quadratic equation

(177) \begin{equation}(u^3)^2+qu^3-\frac{p^3}{27}=0.\end{equation}

The two solutions are

(178) \begin{equation}u^3(s)=\frac{-q(s)\pm \sqrt{\Delta (s)}}{2}\end{equation}

with

(179) \begin{equation}\Delta (s)= q^2(s)+4\times \frac{p^3(s)}{27},\end{equation}

where $\Delta(s)<0$ for $s>s^*$, and the repeated root requires

(180) \begin{equation}\Delta(s^*)=q^2(s^*)+4\times \frac{p^3(s^*)}{27}=0.\end{equation}

When $s>s^*$, the three (real) solutions of (169) are the three cube roots

(181) \begin{eqnarray}y_0,y_1,y_2&\,=\,&\left(\frac{-q(s) + \sqrt{\Delta (s)}}{2}\right)^{1/3},\end{eqnarray}

and we choose the minimum,

(182) \begin{equation}y(s)=\min\{y_0(s),y_1(s),y_2(s)\},\end{equation}

which corresponds to the minimum $x(s)={\boldsymbol{\Psi}}(s)_1$, where ${\boldsymbol{\Psi}}(s)_i$ denotes the ith element of ${\boldsymbol{\Psi}}(s)$.

Therefore, by (174),

(183) \begin{eqnarray}u^3(s)-u^3(s^*)&\,=\,&\frac{-q(s) + \sqrt{\Delta (s)}}{2}+ \frac{q(s^*)}{2}\nonumber\\[4pt]&\,=\,&-\frac{1}{2}C_q\times (s-s^*) + \frac{1}{2}\sqrt{\Delta(s)}+o(s-s^*).\end{eqnarray}

Now,

(184) \begin{align}\Delta(s) & =\Delta(s)-\Delta(s^*)\nonumber\\[4pt] & = \frac{1}{2}(q(s)-q(s^*))(q(s)+q(s^*))+\frac{4}{27} (p(s)-p(s^*))(p(s)^2+p(s)p(s^*)+p(s^*)^2),\nonumber\\[4pt]\end{align}

and so by (173)–(174),

(185) \begin{equation}\Delta(s)= C_\Delta\times (s-s^*)+o(s-s^*),\end{equation}

where the constant $C_\Delta<0$ is given by

(186) \begin{eqnarray}C_\Delta &\,=\,&\frac{1}{2}C_q\times 2q(s^*)+\frac{4}{27} C_p\times 3p^2(s^*).\end{eqnarray}

Therefore,

\begin{eqnarray*}\lim_{s\downarrow s^*}\left(\frac{u^3(s)-u^3(s^*)}{\sqrt{s-s^*}}\right)&\,=\,&\lim_{s\downarrow s^*}\left({-}\frac{1}{2} C_q\sqrt{s-s^*}+\frac{1}{2}\sqrt{\frac{C_\Delta\times (s-s^*)+o(s-s^*)}{s-s^*}}\right)\\[4pt]&\,=\,&\frac{1}{2}\sqrt{C_\Delta},\end{eqnarray*}

and

\begin{eqnarray*}\lim_{s\downarrow s^*}\left(\frac{u(s)-u(s^*)}{\sqrt{s-s^*}}\right)&\,=\,&\lim_{s\downarrow s^*}\left(\frac{u(s)-u(s^*)}{\sqrt{s-s^*}}\times\frac{u^2(s)+u(s)u(s^*)+u^2(s^*)}{u^2(s)+u(s)u(s^*)+u^2(s^*)}\right)\nonumber\\[4pt]&\,=\,&\lim_{s\downarrow s^*}\left(\frac{u^3(s)-u^3(s^*)}{\sqrt{s-s^*}}\times\frac{1}{u^2(s)+u(s)u(s^*)+u^2(s^*)}\right)\nonumber\\[4pt]&\,=\,&\frac{1}{6u^2(s^*)}\sqrt{C_\Delta},\end{eqnarray*}

where $u(s^*)\neq 0$ by (177), since $p(s^*)\neq 0$ by (168) and (171).

From the above we conclude that by (176),

\begin{align*} & {\lim_{s\downarrow s^*}\left(\frac{y(s)-y(s^*)}{\sqrt{s-s^*}}\right)=\lim_{s\downarrow s^*}\left(\frac{u(s)-u(s^*)}{\sqrt{s-s^*}} -\frac{1}{3\sqrt{s-s^*}}\left(\frac{p(s)}{u(s)}-\frac{p(s^*)}{u(s^*)}\right)\right)}\\[4pt] & = \lim_{s\downarrow s^*}\left(\frac{u(s)-u(s^*)}{\sqrt{s-s^*}}-\frac{(p(s)-p(s^*))u(s^*)}{3\sqrt{s-s^*}u(s)u(s^*)}+\frac{p(s^*)(u(s)-u(s^*))}{3\sqrt{s-s^*}u(s)u(s^*)}\right)\\[4pt] & = \pm \frac{1}{6u^2(s^*)}\sqrt{C_\Delta}-0 \frac{1}{6u^2(s^*)}\sqrt{C_\Delta}\frac{p(s^*)}{3u^2(s^*)}\\[4pt] & = \frac{1}{6u^2(s^*)}\sqrt{C_\Delta}\left(1+\frac{p(s^*)}{3u^2(s^*)}\right). \end{align*}

Therefore, by (170), we have

(187) \begin{eqnarray}\lim_{s\downarrow s^*}\left(\frac{{\boldsymbol{\Psi}}(s)_1-{\boldsymbol{\Psi}}(s^*)_1}{\sqrt{s-s^*}}\right)&\,=\,&\lim_{s\downarrow s^*}\left(\frac{x(s)-x(s^*)}{\sqrt{s-s^*}}\right)\nonumber\\[4pt]&\,=\,&\lim_{s\downarrow s^*}\left(\frac{y(s)-y(s^*)}{\sqrt{s-s^*}} +o(1)\right)\nonumber\\[4pt]&\,=\,&\frac{1}{6u^2(s^*)}\sqrt{C_\Delta}\left(1+\frac{p(s^*)}{3u^2(s^*)}\right)\nonumber\\[4pt]&\,=\,&-\mathbf{B}(s^*)_1.\end{eqnarray}

Furthermore, by (162),

(188) \begin{align}& { \lim_{s\downarrow s^*} \left(\frac{{\boldsymbol{\Psi}}(s)_2-{\boldsymbol{\Psi}}(s^*)_2}{\sqrt{s-s^*}}\right)= \lim_{s\downarrow s^*} \left( \frac{z(s)-z(s^*)}{\sqrt{s-s^*}} \right)}\nonumber\\[4pt] & = \lim_{s\downarrow s^*} \left(\frac{1}{\sqrt{s-s^*}}\left(\frac{\lambda x(s)}{2+2\lambda+2s-x(s)}-\frac{\lambda x(s^*)}{2+2\lambda+2s^*-x(s^*)}\right)\right)\nonumber\\[4pt] & =\frac{2\lambda(1+\lambda+s^*)}{(2+2\lambda+2s^*-x(s^*))^2}\mathbf{B}(s^*)_1\nonumber\\[4pt] & = -\mathbf{B}(s^*)_2,\end{align}

which gives

(189) \begin{equation}\lim_{s\downarrow s^*}\left(\frac{{\boldsymbol{\Psi}}(s)-{\boldsymbol{\Psi}}(s^*)}{\sqrt{s-s^*}}\right)=-[\mathbf{B}(s^*)_1\ \ \ \mathbf{B}(s^*)_2]=-{\textbf{B}}(s^*),\end{equation}

as expected (cf. (89)).

Now, assuming $\lambda=2.5$, we solve (180) numerically, obtaining

(190) \begin{eqnarray}s^* \approx -1.1178.\end{eqnarray}

We evaluate $[x\ z]={\boldsymbol{\Psi}}(s^*)$, using (167) to get x and then (162) to get z,

(191) \begin{eqnarray}{\boldsymbol{\Psi}}(s^*) \approx [1.7878\quad 1.5016].\end{eqnarray}

We obtain ${\textbf{K}}(s^*)$ and $({-}{\textbf{D}}(s^*))$ using (20):

(192) \begin{eqnarray}{\textbf{K}}(s^*) \approx [-2.0944],\qquad\qquad\end{eqnarray}

(193) \begin{eqnarray}-{\textbf{D}}(s^*) \approx \left[\begin{array}{c@{\quad}c}-0.5944 & 4.0016\\[4pt] 2.0000 & -0.8822\end{array}\right],\end{eqnarray}

which have common eigenvalue $\gamma \approx -2.0944$. Also, we use (83) to obtain

(194) \begin{eqnarray}{\textbf{U}}(s^*) \approx [3.5756\quad 3.0031].\end{eqnarray}

Finally, we evaluate ${\textbf{B}}(s^*)$ using (187) and ${\textbf{Y}}(s^*)$ using (90):

(195) \begin{eqnarray}\qquad\qquad\quad\ \ {\textbf{B}}(s^*) \approx [1.6416 \quad\quad\quad 2.2069],\end{eqnarray}

(196) \begin{eqnarray}{\textbf{B}}(s^*){\textbf{Q}}_{21}(s^*){\textbf{B}}(s^*) \approx [2.6948\quad\quad\quad 3.6228],\end{eqnarray}

(197) \begin{eqnarray}\qquad\qquad\ \ \ \ \ {\textbf{Y}}(s^*) \approx [0.8808 \quad\quad -0.6197].\end{eqnarray}

This yields

(198) \begin{eqnarray}{\textbf{K}}(s^*){\textbf{B}}(s^*)+{\textbf{B}}(s^*){\textbf{D}}(s^*) \approx 10^{-14}\times [0.7550 \quad -0.0888],\end{eqnarray}

which is approximately zero, as expected (cf. (91)).

Finally,

(199) \begin{eqnarray}{\textbf{H}}(s^*,y)&\,=\,&\sum_{n=1}^{\infty}\frac{y^n}{n!}{\textbf{H}}_{1,n}(s^*)\nonumber\\[4pt]&\,=\,&\sum_{n=1}^{\infty}\frac{y^n}{n!}\sum_{i=0}^{n-1}\left({\textbf{K}}(s^*) \right)^i\times\mathbf{B}(s^*){\textbf{Q}}_{21}(s^*)\times\left({\textbf{K}}(s^*) \right)^{n-1-i}\nonumber\\[4pt]&\,=\,&\sum_{n=1}^{\infty}\frac{y^n}{n!}\sum_{i=0}^{n-1}\left( {\textbf{K}}(s^*)\right)^{n-1}\left(\mathbf{B}(s^*){\textbf{Q}}_{21}(s^*)\right)\nonumber\\[4pt]&\,=\,&ye^{{\textbf{K}}(s^*)y}\mathbf{B}(s^*){\textbf{Q}}_{21}(s^*)\nonumber\\[4pt]&\approx&1.6416 ye^{-2.0944\times y},\end{eqnarray}

and

(200) \begin{eqnarray}{\textbf{H}}(s^*)&\,=\,&\int_{y=0}^{\infty}ye^{{\textbf{K}}(s^*)y}{\textbf{B}}(s^*){\textbf{Q}}_{21}(s^*)dy\nonumber\\[4pt]&\,=\,&({\textbf{K}}(s^*))^{-2}{\textbf{B}}(s^*){\textbf{Q}}_{21}(s^*),\nonumber\\[4pt]\widetilde{\textbf{H}}(s^*)&\,=\,&{\textbf{H}}(s^*)+\left({-}({\textbf{K}}(s^*))^{-1}\mathbf{B}(s^*)+{\textbf{H}}(s^*){\boldsymbol{\Psi}}(s^*)\right){\boldsymbol{1}}\nonumber\\[4pt]&\approx& 3.4428,\end{eqnarray}

so

(201) \begin{eqnarray} \boldsymbol{\mu}(dy)^{(0)}_{11}&\,=\,& diag({\widetilde{\textbf{H}}(s^*)})^{-1} { {\textbf{H}}(s^*,y)} dy \nonumber\\[4pt] &\approx& 0.2905\times 1.6416 ye^{-2.0944\times y} dy ,\nonumber\\[4pt] \boldsymbol{\mu}(dy)^{(0)}_{12}&\,=\,& diag({\widetilde{\textbf{H}}(s^*)})^{-1} \left( e^{{\textbf{K}}(s^*)y}{\textbf{B}}(s^*) + {\textbf{H}}(s^*,y){\boldsymbol{\Psi}}(s^*) \right) dy \nonumber\\[4pt] &\approx& 0.2905\times \left( [1.6416\quad 2.2069] +1.6416 y [1.7878\quad 1.5016] \right) e^{-2.0944\times y} dy , \nonumber\\[4pt] \end{eqnarray}

where $\mu(dy)^{(0)}_{ij}$ is the limiting conditional distribution (Yaglom limit) of observing the process in level y and phase j, given that the process started from level zero in phase i at time zero, and has been evolving without hitting level zero.