Hindawi Publishing Corporation
Advances in Operations Research
Volume 2016, Article ID 1925827, 56 pages
http://dx.doi.org/10.1155/2016/1925827
Research Article
Asymptotic Analysis of a Storage Allocation
Model with Finite Capacity: Joint Distribution
Eunju Sohn1 and Charles Knessl2
1
Department of Science and Mathematics, Columbia College Chicago, 623 South Wabash Avenue, Chicago, IL 60605, USA
Department of Mathematics, Statistics and Computer Science, University of Illinois at Chicago, Chicago, IL 60607-7045, USA
2
Correspondence should be addressed to Eunju Sohn; [email protected]
Received 7 August 2015; Accepted 26 January 2016
Academic Editor: Hsien-Chung Wu
Copyright © 2016 E. Sohn and C. Knessl. This is an open access article distributed under the Creative Commons Attribution
License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly
cited.
We consider a storage allocation model with a finite number of storage spaces. There are 𝑚 primary spaces and 𝑅 secondary spaces.
All of them are numbered and ranked. Customers arrive according to a Poisson process and occupy a space for an exponentially
distributed time period, and a new arrival takes the lowest ranked available space. We let 𝑁1 and 𝑁2 denote the numbers of occupied
primary and secondary spaces and study the joint distribution Prob[𝑁1 = 𝑘, 𝑁2 = 𝑟] in the steady state. The joint process (𝑁1 , 𝑁2 )
behaves as a random walk in a lattice rectangle. We study the problem asymptotically as the Poisson arrival rate 𝜆 becomes large,
and the storage capacities 𝑚 and 𝑅 are scaled to be commensurably large. We use a singular perturbation analysis to approximate
the forward Kolmogorov equation(s) satisfied by the joint distribution.
1. Introduction
We consider the following storage allocation model. There
are 𝑚 primary and 𝑅 secondary storage spaces. The primary
spaces are numbered {1, 2, . . . , 𝑚} and the secondary ones
are numbered {𝑚 + 1, 𝑚 + 2, . . . , 𝑚 + 𝑅}. Customers arrive
according to a Poisson process of rate 𝜆, and each customer
occupies a storage space for an exponentially distributed
amount of time, with the mean occupation time 1/𝜇. A new
arrival takes the lowest ranked available space. If all 𝑚 + 𝑅
spaces are filled, then a new arrival is turned away and lost.
The policy of taking the lowest ranked space is called “first-fit
allocation.”
We can consider the storage spaces as parking spaces of
a restaurant. The primary spaces are in a lot right next to the
restaurant, and the secondary spaces are located somewhere
further away from the restaurant. Lower ranked spaces will
be closer to the restaurant so it is natural for a customer
to use the first-fit policy. Since spaces are occupied and
emptied at random times, this model is called a dynamic
storage allocation model. Design and analysis of algorithms
for dynamic storage allocation are a fundamental part of
computer science [1]. In such applications we can consider
the customers as records, files, or lists and the storage device
as a memory device. As time evolves, items are inserted and
deleted, and the storage device, which is a linear array of
“cells,” will have regions of occupied cells alternating with
interior holes. This is referred to as memory fragmentation in
computers, and collapsing the holes corresponds to running
a defragmentation program.
In the language of queueing theory, the model with
finite secondary storage spaces can be called the 𝑀/𝑀/(𝑚 +
𝑅)/(𝑚+𝑅) queue (the Erlang loss model) with ranked servers.
The main contribution here is to study the effects of the finite
storage capacity, for systems with a large number of both
primary and secondary storage spaces and a commensurably
large traffic intensity, which we denote by 𝜌 = 𝜆/𝜇. Thus we
study the model asymptotically for 𝜌 → ∞ with 𝑚, 𝑅 = Θ(𝜌).
We let 𝑁1 and 𝑁2 be the numbers of occupied primary
and secondary spaces, and we will focus on the joint distribution of 𝑁1 and 𝑁2 , in the steady state. The distributions
of both 𝑁1 and 𝑁1 + 𝑁2 are readily computed, as these
processes behave as Erlang loss models, with 𝑚 and 𝑚 + 𝑅
servers, respectively. Thus their steady state distributions are
2
truncated Poisson distributions. However, the distribution of
the number 𝑁2 of occupied secondary spaces is much more
complicated, as is the joint distribution Prob[𝑁1 = 𝑘, 𝑁2 = 𝑟].
We focus here on only the steady state distribution but
comment that the transient behavior of the standard Erlang
loss model can be analyzed by singular perturbation methods
of the type employed here (see [2]). Thus we believe that, with
significant additional effort, the transient behavior of the joint
process (𝑁1 , 𝑁2 ) could also be ultimately analyzed.
There has been much past work on the model with an
infinite (secondary) storage capacity (𝑅 = ∞) since Kosten
[3]. Various aspects of the solution were also studied in [4–7],
but the solutions are in a complicated form, which is difficult
to evaluate asymptotically for 𝜌 → ∞, due to the presence
of an alternating sum. We derived the joint steady state
distribution of the (𝑁1 , 𝑁2 ) process in [8] using a discrete
version of the classic method of separation of variables. We
obtained the solution as a contour integral that involves
certain polynomials related to hypergeometric functions.
Such representations enabled us to obtain a complete set of
asymptotic results including the joint distribution Prob[𝑁1 =
𝑘, 𝑁2 = 𝑟], for 𝑅 = ∞ [9–12].
The solution of the finite capacity model with 𝑅 < ∞
seems more complicated than the solution of the model with
𝑅 = ∞. But we will show here that a singular perturbation
analysis is again fruitful, and we will obtain a complete
set of asymptotic results for 𝜋(𝑘, 𝑟), which depends also
parametrically on 𝜌, and the numbers 𝑚 and 𝑅 of primary and
secondary storage spaces. Most of the time we will scale all of
𝑘, 𝑟, 𝑚, and 𝑅 to be of the same order as the traffic intensity
𝜌. We will focus on understanding the effects of the finiteness
of the secondary storage capacity 𝑅.
The remainder of the paper is organized as follows. In
Section 2 we state the basic equations and briefly describe
their forthcoming analysis. In Section 3 we summarize all
of the main results, and the joint distribution will have
different asymptotic expansions in three main regions of the
state space, which is the lattice rectangle {(𝑘, 𝑟): 0 ⩽ 𝑘 ⩽
𝑚, 0 ⩽ 𝑟 ⩽ 𝑅}. Moreover, there are also various boundary,
corner, and transition curves where different expansions will
be needed. In Section 4 we derive the asymptotics of the joint
distribution in the three main regions, while in Sections 5–
7 we treat the boundary, corner, and transition ranges. In
Section 8 we will do some numerical comparisons to test
the accuracy and robustness of our asymptotic results. Some
discussion of our results also appears in Section 8. Since the
analysis is quite technical, we have written this paper so that
the derivations in Sections 4–7 can be omitted upon a first
(and perhaps even later) reading(s).
2. Statement of the Problem
We consider a system with 𝑚 primary and 𝑅 secondary
storage spaces (or servers). The primary spaces are ranked
and numbered 1, 2, . . . , 𝑚 while the secondary spaces are
numbered 𝑚+1, 𝑚+2, . . . , 𝑚+𝑅. Customers arrive according
to a Poisson process with rate parameter 𝜆 and a new arrival
takes the lowest ranked available space, if possible a primary
one. If all 𝑚+𝑅 spaces are occupied further arrivals are turned
Advances in Operations Research
away and lost. All of the storage spaces are identical and a
customer occupies a space for an exponentially distributed
amount of time, with the mean occupation (or service)
time being 1/𝜇. We then let 𝑁1 and 𝑁2 be the numbers of
occupied primary and secondary spaces, respectively. We also
introduce a dimensionless parameter
𝜌=
𝜆
𝜇
(1)
to denote the traffic intensity.
The joint process (𝑁1 , 𝑁2 ) corresponds to a continuous
time random walk in a lattice rectangle. Figure 1 indicates
transition rates. The steady state distribution
𝜋 (𝑘, 𝑟) = 𝜋 (𝑘, 𝑟; 𝜌, 𝑚, 𝑅) = lim Prob [𝑁1 (𝑡)
𝑡→∞
(0)
(0)
(2)
= 𝑘, 𝑁2 (𝑡) = 𝑟 | 𝑁1 (0) = 𝑘 , 𝑁2 (0) = 𝑟 ]
is independent of the initial values 𝑁1 (0) and 𝑁2 (0) and
satisfies the following balance equations:
(𝜌 + 𝑘 + 𝑟) 𝜋 (𝑘, 𝑟) = (𝑟 + 1) 𝜋 (𝑘, 𝑟 + 1)
+ (𝑘 + 1) 𝜋 (𝑘 + 1, 𝑟)
+ 𝜌𝜋 (𝑘 − 1, 𝑟) ,
(3)
1 ⩽ 𝑘 ⩽ 𝑚 − 1, 0 ⩽ 𝑟 ⩽ 𝑅 − 1,
(𝜌 + 𝑚 + 𝑟) 𝜋 (𝑚, 𝑟) = (𝑟 + 1) 𝜋 (𝑚, 𝑟 + 1)
+ 𝜌𝜋 (𝑚 − 1, 𝑟)
(4)
+ 𝜌𝜋 (𝑚, 𝑟 − 1) ,
1 ⩽ 𝑟 ⩽ 𝑅 − 1,
(𝜌 + 𝑘 + 𝑅) 𝜋 (𝑘, 𝑅) = (𝑘 + 1) 𝜋 (𝑘 + 1, 𝑅)
+ 𝜌𝜋 (𝑘 − 1, 𝑅) ,
(5)
1 ⩽ 𝑘 ⩽ 𝑚 − 1,
(𝜌 + 𝑟) 𝜋 (0, 𝑟) = (𝑟 + 1) 𝜋 (0, 𝑟 + 1) + 𝜋 (1, 𝑟) ,
0 ⩽ 𝑟 ⩽ 𝑅 − 1,
(𝜌 + 𝑅) 𝜋 (0, 𝑅) = 𝜋 (1, 𝑅) ,
(6)
(7)
(𝜌 + 𝑚) 𝜋 (𝑚, 0) = 𝜌𝜋 (𝑚 − 1, 0) + 𝜋 (𝑚, 1) ,
(8)
(𝑚 + 𝑅) 𝜋 (𝑚, 𝑅) = 𝜌𝜋 (𝑚, 𝑅 − 1) + 𝜌𝜋 (𝑚 − 1, 𝑅) .
(9)
The main balance equation (3) applies in the interior of
the lattice rectangle and along the boundary 𝑟 = 0, (4)–
(6) correspond to boundary conditions along three of the
four boundaries of the rectangle, and (7)–(9) are corner
conditions. Also, (6) applies at 𝑟 = 0 so the corner condition
at (0, 0) is 𝜌𝜋(0, 0) = 𝜋(0, 1) + 𝜋(1, 0). We also have the
normalization condition
𝑚
𝑅
∑ ∑𝜋 (𝑘, 𝑟) = 1.
𝑘=0 𝑟=0
(10)
Advances in Operations Research
N2
R
and in the present limit we have 0 < 𝑋0 , 𝑌0 < ∞, where we
view 𝑋0 and 𝑌0 as fixed as 𝜌 → ∞. We may then view the
process (𝑁1 , 𝑁2 ) on a “coarse” spatial scale, with
𝜆
𝜇k
𝜆
𝜇R
3
𝜇m
𝜇R
𝜇R
𝑋=
𝜆
r
𝜇k
𝜇r
𝜇r
0
𝜆
𝜇k
𝜇m
𝜆
k
(14)
𝜆
𝑟
𝑌 = , 0 ⩽ 𝑌 ⩽ 𝑌0 .
𝜌
𝜇r
On the (𝑋, 𝑌) scale the random walk takes small steps (=
1/𝜌) and the state space may be approximately viewed as the
continuous rectangle
𝜆
{(𝑋, 𝑌) : 0 ⩽ 𝑋 ⩽ 𝑋0 , 0 ⩽ 𝑌 ⩽ 𝑌0 } .
𝜇m
𝜆
𝑘
, 0 ⩽ 𝑋 ⩽ 𝑋0 ,
𝜌
m
N1
(15)
Setting
Figure 1: Steady state transition rates.
𝜋 (𝑘, 𝑟) = 𝜋 (𝑘, 𝑟; 𝜌, 𝑚, 𝑅) = 𝑃 (𝑋, 𝑌)
The process 𝑁1 by itself behaves precisely as the Erlang
loss model (𝑀/𝑀/𝑚/𝑚 queue with 𝑚 servers). This is well
known to have, in the steady state, a truncated Poisson
distribution; hence
the main balance equation (3) becomes
(11)
+ (𝑋 + 𝜌−1 ) 𝑃 (𝑋 + 𝜌−1 , 𝑌)
The total number, 𝑁1 + 𝑁2 , of occupied servers also
follows a truncated Poisson distribution. Therefore,
min{𝑚,𝐿}
∑ 𝜋 (𝑘, 𝑟) =
𝑘+𝑟=𝐿
∑
𝜋 (𝑘, 𝐿 − 𝑘)
𝑘=max{0,𝐿−𝑅}
=
(12)
𝜌𝐿 𝑒−𝜌 /𝐿!
ℓ −𝜌
∑𝑚+𝑅
ℓ=0 𝜌 𝑒 /ℓ!
, 0 ⩽ 𝐿 ⩽ 𝑚 + 𝑅.
𝑌0 =
𝑚
,
𝜌
𝑅
𝜌
(17)
+ 𝑃 (𝑋 − 𝜌−1 , 𝑌)
which is a difference equation with small differences, of order
𝑂(𝜌−1 ). The boundary condition along 𝑘 = 𝑚 in (4) may be
replaced by the “artificial boundary condition”
(𝑚 + 1) 𝜋 (𝑚 + 1, 𝑟) = 𝜌𝜋 (𝑚, 𝑟 − 1) , 1 ⩽ 𝑟 ⩽ 𝑅 − 1. (18)
We recently obtained in [13] explicit expressions for the
joint distribution 𝜋(𝑘, 𝑟), but they are not very insightful due
to their complexity. Thus we study the problem asymptotically, for 𝜌 → ∞ with 𝑚, 𝑅 = 𝑂(𝜌). This means that there
are many arrivals but the numbers of storage spaces, both
primary and secondary ones, are commensurately large. Note
that if 𝜌 → ∞ with 𝑚, 𝑅 = 𝑂(1) then the probability
distribution 𝜋(𝑘, 𝑟) would concentrate on a single lattice
point, with 𝜋(𝑘, 𝑟) → 𝛿𝑘,𝑚 𝛿𝑟,𝑅 as 𝜌 → ∞. Here 𝛿𝑘,𝑚 = 0,
𝑘 ≠ 𝑚, and 𝛿𝑘,𝑚 = 1, 𝑘 = 𝑚. Thus this limit would not
be particularly interesting. There are, however, certain cases
where either 𝑚 or 𝑅 is large but 𝑜(𝜌), that should lead to
interesting results, but we do not analyze them here.
We next introduce the parameters
𝑋0 =
(16)
(1 + 𝑋 + 𝑌) 𝑃 (𝑋, 𝑌) = (𝑌 + 𝜌−1 ) 𝑃 (𝑋, 𝑌 + 𝜌−1 )
𝑅
𝜌𝑘 𝑒−𝜌 /𝑘!
∑𝜋 (𝑘, 𝑟) = 𝑚 𝑘 −𝜌 , 0 ⩽ 𝑘 ⩽ 𝑚.
∑𝑘=0 𝜌 𝑒 /𝑘!
𝑟=0
= 𝑃 (𝑋, 𝑌; 𝜌, 𝑋0 , 𝑌0 )
The above is obtained by requiring that (3) holds also at 𝑘 = 𝑚
and comparing this to (4). Introducing 𝜋(𝑚 + 1, 𝑟) simplifies
some of the calculations, but this quantity has no physical
meaning.
The asymptotic structure of the joint distribution will be
very different for four main regions in the (𝑋0 , 𝑌0 ) parameter
space. We call these regions R1 –R4 and they are sketched in
Figure 2. They are defined by the inequalities
R1 = {(𝑋0 , 𝑌0 ) : 𝑋0 > 1, 𝑌0 > 0} ,
(19)
R2 = {(𝑋0 , 𝑌0 ) : 0 < 𝑋0 < 1, 𝑌0 + 𝑋0 > 1} ,
(20)
R3 = {(𝑋0 , 𝑌0 ) : 0 < 𝑋0 < 1, √𝑋0 − 𝑋0 < 𝑌0 < 1
(21)
− 𝑋0 } ,
(13)
R4 = {(𝑋0 , 𝑌0 ) : 0 < 𝑋0 < 1, 0 < 𝑌0 < √𝑋0 − 𝑋0 } . (22)
4
Advances in Operations Research
3. Summary of Results
Y0
1
In the analysis it proves sometimes useful to use the variables
(𝑛, 𝑝) where
ℛ2
𝑛 = 𝑚 − 𝑘,
ℛ1
𝑝 = 𝑅 − 𝑟,
ℛ3
so that 𝑛 (resp., 𝑝) measures the number of unoccupied
primary (resp., secondary) spaces. Then we also let
ℛ4
0
(26)
1
𝜋 (𝑘, 𝑟) = 𝜋 (𝑚 − 𝑛, 𝑅 − 𝑝) = 𝑄 (𝑛, 𝑝)
X0
Figure 2: Four regions of the (𝑋0 , 𝑌0 ) parameter space.
It will also prove useful to define as follows the curves that
separate these four regions:
R1 ∩ R2 = {(𝑋0 , 𝑌0 ) : 𝑋0 = 1, 𝑌0 > 0} ,
(23)
R2 ∩ R3 = {(𝑋0 , 𝑌0 ) : 𝑋0 + 𝑌0 = 1, 0 < 𝑋0 < 1} ,
(24)
so that 𝑄(0, 0) corresponds to the probability that all of
the storage spaces are full. In (27) we did not indicate the
dependence of 𝜋 and 𝑄 on the parameters 𝜌, 𝑚, and 𝑅.
We begin by giving asymptotic results for 𝑄(0, 0).
Proposition 1. For 𝜌 → ∞ and fixed 𝑋0 , 𝑌0 one has
𝜋 (𝑚, 𝑅) = 𝑄 (0, 0) ∼ 1 − 𝑋0 − 𝑌0 ,
(𝑋0 , 𝑌0 ) ∈ R3 ∪ R4 ,
(28)
2
R3 ∩ R4
= {(𝑋0 , 𝑌0 ) : 𝑋0 + 𝑌0 = √𝑋0 , 0 < 𝑋0 < 1} .
(27)
(25)
Note that the union of all the sets in (19)–(25) is the entire
open quarter plane in parameter space. We purposefully
exclude the coordinate axes 𝑋0 = 0 and 𝑌0 = 0, as they would
require entirely different asymptotic analyses. The separating
curves in (23)–(25) will also require separate analyses, and we
will obtain results that apply not only along the curves but also
in small neighborhoods of these curves, which will be defined
precisely later. This will produce results that asymptotically
match to those in the main regions.
The presence of the different regions can be explained
intuitively. If 𝑋0 > 1 (𝑚 > 𝜌) there are enough primary
spaces to service all storage requests and the secondary spaces
will generally not be needed. If 𝑋0 < 1 but 𝑋0 + 𝑌0 >
1 (𝑚 + 𝑅 > 𝜌) the primary spaces are insufficient but the
total number of spaces is adequate. Then we might expect
that typically all 𝑚 primary spaces and about 𝜌 − 𝑚 (< 𝑅)
secondary spaces will be occupied. If 𝑋0 + 𝑌0 < 1 then
typically all primary and secondary spaces will be occupied.
Then we might expect 𝜋(𝑘, 𝑟) to be concentrated near 𝑘 = 𝑚,
𝑟 = 𝑅. The further splitting of 𝑋0 +𝑌0 < 1 into the regions R3
and R4 is difficult to explain intuitively in terms of the basic
model, but we will explain this dichotomy via our asymptotic
analysis. We also note that the asymptotic behavior of the
distribution in (11) undergoes a transition when 𝑚/𝜌 passes
through 1, while (12) undergoes an analogous transition when
(𝑚 + 𝑅)/𝜌 passes through 1. However, neither (11) nor (12)
undergoes a transition along R3 ∩ R4 . In the analysis that
follows we will also need to, for each region of parameter
space, separately analyze several different regions of the state
space, which corresponds to the rectangle 0 ⩽ 𝑋 ⩽ 𝑋0 ,
0 ⩽ 𝑌 ⩽ 𝑌0 on the coarse spatial scale. It will sometimes prove
necessary to analyze boundary and corner regions where the
discrete nature of the model must be considered.
𝑄 (0, 0) ∼
𝑒−𝛾 /2
1
,
𝛾
√𝜌 ∫ 𝑒−𝑢2 /2 𝑑𝑢
−∞
𝑋0 + 𝑌0 = 1 +
𝛾
,
√𝜌
(29)
𝛾 = 𝑂 (1) (R2 ∩ R3 ) ,
𝑄 (0, 0)
∼
1
1
𝑒𝜌[𝑋0 +𝑌0 −1−(𝑋0 +𝑌0 )log(𝑋0 +𝑌0 )] ,
√𝜌 √2𝜋 (𝑋 + 𝑌 )
0
0
(30)
(𝑋0 , 𝑌0 ) ∈ R1 ∪ R2 .
Here and throughout the paper, we use the convention
that R1 ∪ R2 corresponds to the union of the open sets R1
and R2 and also the separating curve R1 ∩ R2 (cf. (23)).
Similar comments apply for R3 ∪ R4 and R2 ∪ R3 . We refer
to the asymptotic limit in (29) as corresponding to R2 ∩ R3 ,
where we now give the precise scaling, 𝑋0 +𝑌0 −1 = 𝑂(𝜌−1/2 ),
that applies near the separating curve in (24). The results in
(28)–(30) will follow from our asymptotic analysis of the joint
distribution, but we note that these also follow easily from
(12), by setting 𝑁 = 𝑚 + 𝑅 and expanding the result for
𝜌 → ∞ and different ranges of 𝑍0 = 𝑋0 +𝑌0 (thus 𝑍0 = 𝑁/𝜌).
It will prove convenient to express some of our results in
terms of the three constants 𝐶, 𝐶0 , and 𝐶1 ; these depend only
on the parameters 𝜌, 𝑋0 , and 𝑌0 . We summarize below the
leading order asymptotics of these constants.
Proposition 2. Define the constants 𝐶, 𝐶1 , and 𝐶0 by the
relations
𝐶 ∼ 𝜌−1/2 ,
𝐶∼√
(𝑋0 , 𝑌0 ) ∈ R1 ∪ R2 ,
−1
2𝜋 𝛾 −𝑢2 /2
𝑑𝑢] ,
[∫ 𝑒
𝜌
−∞
(31)
(𝑋0 , 𝑌0 ) ∈ R2 ∩ R3 , (32)
Advances in Operations Research
5
𝐶 ∼ √2𝜋 (𝑋0 + 𝑌0 ) (1 − 𝑋0 − 𝑌0 )
⋅ 𝑒𝜌[1−𝑋0 −𝑌0 +(𝑋0 +𝑌0 ) log(𝑋0 +𝑌0 )] ,
𝐶1 ∼ 𝜌−2/3 ,
𝐶1 ∼ 𝜌−2/3
(33)
(𝑋0 , 𝑌0 ) ∈ R3 ,
(𝑋0 , 𝑌0 ) ∈ R2 ,
𝛾
√2𝜋
(𝑋0 , 𝑌0 ) ∈ R2 ∩ R3 ,
,
2
∫−∞ 𝑒−𝑢 /2 𝑑𝑢
(34)
(35)
𝐶1 ∼ 𝜌1/6 √2𝜋 (𝑋0 + 𝑌0 ) (1 − 𝑋0 − 𝑌0 )
⋅ 𝑒𝜌[1−𝑋0 −𝑌0 +(𝑋0 +𝑌0 )log(𝑋0 +𝑌0 )] ,
𝐶1 ∼ 𝜌
−1/6
(𝑋0 , 𝑌0 ) ∈ R3 ,
(1 − 𝑋0 − 𝑌0 )
(37)
∞
𝑟0
𝑋0 + 𝑌0 − √𝑋0 = 𝜌−1/3 𝛿∗ = 𝑂 (𝜌−1/3 )
(38)
(R3 ∩ R4 ) ,
𝛿∗
𝑋01/6
1/3
(1 − √𝑋0 )
,
Φ (𝑋0 , 𝑌0 ) = (𝑌0 + 𝑊0 ) log (𝑌0 + 𝑊0 ) − 𝑌0 log 𝑌0
−
(39)
𝑊0
𝑋
log 𝑊0 − 0 log 𝑋0 + √𝑋0 − 1,
2
2
2
𝑊0 = (1 − √𝑋0 ) ,
where 𝐴𝑖(⋅) is the Airy function and 𝑟0 its maximal root
𝑟0 = max {𝑧 : 𝐴𝑖 (𝑧) = 0} = −2.3381 ⋅ ⋅ ⋅ .
(40)
𝐶1 ∼ 𝜌−5/6 √2𝜋 (1 − 𝑋0 − 𝑌0 ) 𝐴𝑖󸀠 (𝑟0 )
1/3
⋅ 𝑒−𝜌Φ(𝑋0 ,𝑌0 ) 𝑒−𝜌
Φ∗ (𝑌0 )
(41)
2/3
√𝑌0 𝑋01/3 (1 − √𝑋0 )
𝑌 + 𝑊0
⋅
( 0
)
2
(√𝑋0 − 𝑋0 − 𝑌0 ) √𝑊0 + 𝑌0 1 − √𝑋0
Φ∗ (𝑌0 ) = −𝑟0 (1 − √𝑋0 )
2/3
𝑋0−1/6 log (
1/2√𝑋0
,
𝑌0 + 𝑊0
),
1 − √𝑋0
𝐶0 ∼ 𝜌
(1 − 𝑋0 − 𝑌0 ) ,
(𝑋0 , 𝑌0 ) ∈ R3 ∪ R4 ,
𝛽
= 1 + 𝑂 (𝜌−1/2 ) .
√𝜌
(46)
Note that (46) can be predicted from the marginal distribution in (11), as the sum in the denominator undergoes a
transition for 𝑚 = 𝜌 + 𝑂(√𝜌), which is the same scaling as in
(46).
3.1. Joint Distribution and Its Limits. Now we consider the
joint distribution; 𝜋(𝑘, 𝑟) = 𝜋(𝑘, 𝑟; 𝜌, 𝑚, 𝑅) = 𝑃(𝑋, 𝑌) =
𝑃(𝑋, 𝑌; 𝜌, 𝑋0 , 𝑌0 ) for 𝜌 → ∞. We recall that 𝑋0 = 𝑚/𝜌 and
𝑌0 = 𝑅/𝜌 are the scaled numbers of primary and secondary
spaces. The state space of the random walk is the lattice
rectangle in Figure 1, and on the coarse (𝑋, 𝑌) spatial scale this
can be viewed as the continuous rectangle {(𝑋, 𝑌): 0 ⩽ 𝑋 ⩽
𝑋0 , 0 ⩽ 𝑌 ⩽ 𝑌0 }. Our goal is to give a complete asymptotic
description of the joint distribution for 𝜌 → ∞, including
ranges of the state space where there is appreciable mass
and also ranges where 𝜋(𝑘, 𝑟) is asymptotically small. This
corresponds to the tails of the distribution and in such ranges
𝜋(𝑘, 𝑟) = 𝑃(𝑋, 𝑌) is typically exponentially small for large 𝜌.
We first discuss the ranges where there is significant mass,
and this will lead to certain limiting distributions, which will
be very different for regions R1 –R4 of parameter space in
(19)–(25).
Proposition 3. For 𝜌 → ∞ one has the following limiting
distributions:
(i) 𝑋0 = 𝑚/𝜌 > 1 (thus (𝑋0 , 𝑌0 ) ∈ R1 )
𝜋 (𝑘, 𝑟) ≈ 𝛿0𝑟
𝑒−𝜌 𝜌𝑘
,
𝑘!
𝑘
= 𝑋 < 𝑋0 ,
𝜌
(47)
which can be recast as the limit
(42)
(𝑋0 , 𝑌0 ) ∈ R4 ,
−1/2
𝑋0 = 1 +
(36)
⋅ 𝑒−𝜌Φ(𝑋0 ,𝑌0 ) √2𝜋𝑋01/4 ∫ 𝑒𝛿1 𝑢 𝐴𝑖 (𝑢) 𝑑𝑢,
𝛿1 =
We note that the relation 𝐶0 ∼ 𝜌−1/2 𝑄(0, 0) holds for
all cases of the parameters. In (38) we have thus defined the
precise scaling near the separating curve R3 ∩ R4 , as 𝑌0 =
√𝑋0 − 𝑋0 + 𝑂(𝜌−1/3 ). Note that 𝐶 is not defined for region
R4 while 𝐶1 is not defined for R1 , as then the corresponding
constant will play no role in the analysis. We conclude by
giving the precise scaling for R1 ∩ R2 (near 𝑋0 = 1), which
will be
󸀠
√𝜌𝜋 (𝜌 + √𝜌𝑋 , 𝑟) 󳨀→ 𝛿0𝑟
1 −(𝑋󸀠 )2 /2
,
𝑒
√2𝜋
(48)
𝜌 󳨀→ ∞.
(43)
2
𝐶0 ∼ 𝜌−1
𝛾
𝑒−𝛾 /2
2
∫−∞ 𝑒−𝑢 /2 𝑑𝑢
,
𝐶0 ∼ 𝜌−1 [2𝜋 (𝑋0 + 𝑌0 )]
(𝑋0 , 𝑌0 ) ∈ R2 ∩ R3 ,
(44)
(ii) 0 < 𝑋0 < 1, 𝑋0 + 𝑌0 > 1 (thus (𝑋0 , 𝑌0 ) ∈ R2 )
𝜋 (𝑘, 𝑟) ∼
−1/2
⋅ 𝑒𝜌[𝑋0 +𝑌0 −1−(𝑋0 +𝑌0 )log(𝑋0 +𝑌0 )] ,
𝜌
1 − 𝑋0 𝑛
2
𝑋 exp {− [𝑌 − (1 − 𝑋0 )] }
2
√2𝜋𝜌 0
(49)
(45)
(𝑋0 , 𝑌0 ) ∈ R1 ∪ R2 .
and this applies for 𝑛 = 𝑚−𝑘 = 𝑂(1) and 𝑌−(1−𝑋0 ) =
𝑂(𝜌−1/2 ) (i.e., 𝑟 = 𝜌 − 𝑚 + 𝑂(√𝜌)).
6
Advances in Operations Research
(iii) 0 < 𝑋0 + 𝑌0 < 1 (thus (𝑋0 , 𝑌0 ) ∈ R3 ∪ R4 )
1 − 𝑋0 − 𝑌0 1
1−𝑤
−𝑛−1
[𝑧 (𝑤)]
∮
𝑋0
2𝜋𝑖 𝑧− (𝑤) − 𝑤 +
(50)
−𝑝−1
⋅𝑤
𝑑𝑤,
𝜋 (𝑘, 𝑟) ∼
𝑧± (𝑤) =
1
[1 + 𝑋0 + 𝑌0 − 𝑌0 𝑤
2𝑋0
(51)
2
± √(1 + 𝑋0 + 𝑌0 − 𝑌0 𝑤) − 4𝑋0 ] ,
which holds for 𝑛 = 𝑚−𝑘 = 𝑂(1) and 𝑝 = 𝑅−𝑟 = 𝑂(1).
When 𝑋0 > 1 we have 𝑚 > 𝜌 so the secondary storage
spaces will be rarely needed, and then 𝜋(𝑘, 𝑟) approximately
follows the Poisson distribution in (47), which has also the
Gaussian limit in (48). The results in (47) and (48) provide
no information on 𝜋(𝑘, 𝑟) for 𝑟 ⩾ 1, but later we will estimate
precisely these probabilities. We also note that when 𝑟 = 0,
(47) ceases to be valid for 𝑘 = 𝜌𝑋0 − 𝑂(1) = 𝑚 = 𝑂(1), for
then if almost all primary spaces are full there may well be
some secondary spaces also occupied, and thus 𝜋(𝑘, 0) may
become comparable to 𝜋(𝑘, 𝑟) for 𝑟 ⩾ 1, for this range of
𝑘. If 𝑋0 < 1 and 𝑋0 + 𝑌0 > 1 the primary storage spaces
are insufficient to meet the demand, but the total number of
spaces does suffice. Then (49) shows that 𝑚 − 𝑂(1) primary
spaces and 𝜌(1 − 𝑋0 ) + 𝑂(√𝜌) = 𝜌 − 𝑚 + 𝑂(√𝜌) secondary
spaces will tend to be occupied, with the joint distribution
being a product of a geometric and a Gaussian. This also
shows that, to leading order for large 𝜌, the processes 𝑁1
and 𝑁2 decouple. When 𝑋0 + 𝑌0 < 1 we have 𝑚 + 𝑅 < 𝜌
and the totality of storage spaces is not enough to meet the
demand. Then typically all but a few spaces, both primary and
secondary, will tend to be occupied, with the numbers 𝑛 and
𝑝 of available spaces following the discrete joint distribution
in (50). From (50) we can easily show that
∑ 𝜋 (𝑘, 𝑟) ∼ (1 − 𝑋0 − 𝑌0 ) (𝑋0 + 𝑌0 )
𝑘+𝑟=𝐿
𝐿
(52)
so that the total number of empty spaces is geometrically
distributed; this result also follows easily from the exact
expression in (12). We will later see that the tail behavior of
(50), for 𝑛 and/or 𝑝 → ∞, is quite different according as
(𝑋0 , 𝑌0 ) ∈ R3 or (𝑋0 , 𝑌0 ) ∈ R4 , which again will indicate
that the triangle 0 < 𝑋0 + 𝑌0 < 1 in parameter space needs to
be split into the two regions R3 and R4 .
We next study the transitions between the three limiting
results in Proposition 4.
Proposition 4. For 𝜌 → ∞ one has the limiting distributions:
̃ (𝑋) = 𝑌
̃ (𝑋; 𝑋0 , 𝑌0 ) = 𝑌0 +
𝑌=𝑌
2
(i) 𝑋0 − 1 = 𝛽/√𝜌 = 𝑂(𝜌−1/2 ), 𝑋 − 1 = 𝛼/√𝜌, and
𝑌 = Ω/√𝜌 (thus 𝑘 = 𝑚 + 𝛼√𝜌 and 𝑟 = √𝜌Ω)
2
𝜋 (𝑘, 𝑟) ∼ 𝜌−1 𝑒−𝛼 /4
𝜋 (𝑘, 𝑟) ∼ 𝜌−1 √
2
2
(Ω − 𝛼) 𝑒−(Ω−𝛼) /2 , 𝛼 < 0, Ω > 0. (54)
𝜋
(ii) 𝑋0 + 𝑌0 − 1 = 𝛾/√𝜌 = 𝑂(𝜌−1/2 ), 𝑛 = 𝑚 − 𝑘 = 𝑂(1),
̃ = 𝑂(𝜌−1/2 )
and 𝑌0 − 𝑌 = 𝜌−1/2 𝑦
𝜋 (𝑘, 𝑟)
∼ 𝜌−1/2 (1 − 𝑋0 ) 𝑋0𝑛 (∫
𝛾
−∞
2
−1
2
𝑒−𝑢 /2 𝑑𝑢) 𝑒−(̃𝑦−𝛾) /2 ,
(55)
̃ > 0.
𝑦
As 𝛾 → +∞ the truncated Gaussian distribution in (55)
approaches the free space Gaussian in (49), which applies for
1 − 𝑌0 < 𝑋0 < 1. For 𝛾 → −∞, (55) asymptotically matches
to (50), when the latter is expanded for 𝑋0 + 𝑌0 ↑ 1 and
simultaneously 𝑝 → ∞, with the product (1 − 𝑋0 − 𝑌0 )𝑝 =
−𝛾̃
𝑦 held fixed.
The complicated distribution in (53) is a necessary intermediate result since (47) and (49) do not asymptotically
match. The right-hand side of (53) is of the form 𝜌−1 × (density
in (𝛼, Ω)), with the density having support in the quarter
plane 𝛼 < 𝛽, Ω > 0. Thus if 𝑚 = 𝜌 + 𝑂(√𝜌) there will tend
to be 𝑂(√𝜌) empty primary spaces and 𝑂(√𝜌) full secondary
spaces, with now an intricate coupling between the processes
𝑁1 and 𝑁2 . Finally, we note that the results in items (i) and
(ii) in Proposition 3 and in item (i) of Proposition 4 are
independent of the secondary storage capacity 𝑌0 = 𝑅/𝜌,
while item (iii) in Proposition 3 and item (ii) in Proposition 4
do depend upon 𝑌0 .
3.2. Joint Distribution: Main Regions of State Space. The
asymptotic expansion of 𝜋(𝑘, 𝑟) = 𝑃(𝑋, 𝑌) will be different
for the four parameter ranges indicated in Figure 2 and also
for three main regions of the state space, which we call D0 ,
D+ , and D− , and we define/discuss these below.
First consider region R1 of parameter space, so that 𝑋0 >
1, and define the curve
2 (𝑋0 + 𝑌0 ) (𝑋0 + 𝑌0 − 1)
2
(53)
−∞ < 𝛼 < 𝛽, Ω > 0,
where 𝐷𝑧 (−𝛽) is the parabolic cylinder function of
order 𝑍 and argument −𝛽. When 𝛽 = 0 (𝑋0 = 1) the
above simplifies to
𝑌0
2
Ω𝑧−1 𝐷𝑧 (−𝛼)
1 𝑖∞
𝑑𝑧,
∫
2𝜋𝑖 −𝑖∞ 𝐷𝑧 (−𝛽) 𝐷𝑧−1 (−𝛽)
2
{𝑋0 − (𝑋0 + 𝑌0 ) + (𝑋0 + 𝑌0 ) (𝑋 − 𝑋0 )
(56)
2
+ √ [(𝑋0 + 𝑌0 ) + (𝑋0 + 𝑌0 ) (𝑋0 − 𝑋) − 𝑋0 ] − 4 (𝑋0 + 𝑌0 ) (𝑋0 + 𝑌0 − 1) (𝑋0 − 𝑋)} .
Advances in Operations Research
7
0.95
̃
(i) (𝑋, 𝑌) ∈ D+ (0 < 𝑌 < 𝑌(𝑋))
X0 = 1.5, Y0 = 1
ℛ1
1.00
𝜋 (𝑘, 𝑟) ∼ 𝐶 (𝜌) 𝐾+ (𝑋, 𝑌) 𝑒𝜌Ψ+ (𝑋,𝑌) ,
𝒟0
𝑤ℎ𝑒𝑟𝑒 𝐶 (𝜌) ∼ 𝜌−1/2 ,
(60)
Ψ+ (𝑋, 𝑌) = 𝑋0 log 𝑋0 + 𝑋0 − 1 + 𝑠 − 2𝑋0
Y 0.90
̃
Y
⋅ log (𝑠 + 𝑋0 ) +
0.85
−
𝒟+
0.80
0
0.5
1
𝑠
(1 − 𝑒𝑡 )
𝑠 + 𝑋0
𝑠
[𝑠 + 𝑋0 − 1 + 𝑒−𝑡 ]
𝑠 + 𝑋0
(61)
𝑠
)
𝑠 + 𝑋0
⋅ log [1 + (𝑠 + 𝑋0 − 1) 𝑒𝑡 ] + (𝑒−𝑡 −
1.5
⋅ (1 − 𝑋0 − 𝑠 − 𝑒𝑡 ) log [1 −
𝑠
𝑒𝑡 ] ,
𝑠 + 𝑋0
X
Figure 3: Region R1 .
where (𝑠, 𝑡) are related to (𝑋, 𝑌) via the mapping, for
0 < 𝑠 < 𝑌0 ,
This curve depends on both 𝑋0 and 𝑌0 and thus on both of
the total numbers, 𝑚 and 𝑅, of primary and secondary storage
spaces. The curve is defined for 0 ⩽ 𝑋 ⩽ 𝑋0 and we have
̃ (0) = 𝑌0 [1 −
𝑌
𝑋0
2
(𝑋0 + 𝑌0 )
̃ (𝑋0 ) = 𝑌0 .
], 𝑌
(57)
For region R1 (and indeed also for R2 and R3 ) we have
̃
𝑌(0)
> 0 so that (56) connects the point (0, Y0 [1 − 𝑋0 /(𝑋0 +
𝑌0 )2 ]) to the corner point (𝑋0 , 𝑌0 ) in the scaled state space.
̃
The curve 𝑌(𝑋)
divides the state space into the two regions
D0 and D+ , with
̃ (𝑋)} ,
D+ = {(𝑋, 𝑌) : 0 < 𝑋 ⩽ 𝑋0 , 0 < 𝑌 < 𝑌
(58)
̃ (𝑋) < 𝑌 < 𝑌0 } .
D0 = {(𝑋, 𝑌) : 0 < 𝑋 < 𝑋0 , 𝑌
(59)
Here we defined D0 as an open set, while D+ is bounded by
̃
the four curves 𝑋 = 0, 𝑌 = 0, 𝑋 = 𝑋0 , and 𝑌 = 𝑌(𝑋)
and
we include only the third of these (𝑋 = 𝑋0 ) as a part of D+ .
This is because the asymptotic expansion that will apply in the
interior of D+ will remain valid near 𝑋 = 𝑋0 , but not near the
other three bounding curves. The expansion valid in D0 will
̃
break down if either 𝑋 ≈ 0, 𝑌 ≈ 𝑌0 , or 𝑌 ≈ 𝑌(𝑋).
We sketch
̃
in Figure 3 the curve 𝑌(𝑋) and we recall that if 𝑋0 > 1 most
of the mass in 𝜋(𝑘, 𝑟) is concentrated in the range 𝑟 = 0 and
𝑘 = 𝜌 + 𝑂(√𝜌) (𝑋 = 1 + 𝑂(𝜌−1/2 )) (see Proposition 3), and
this corresponds to the lower bounding curve for D+ .
Proposition 5. For (𝑋0 , 𝑌0 ) ∈ R1 (𝑋0 > 1) the asymptotic
expansions of 𝜋(𝑘, 𝑟) = 𝜋(𝜌𝑋, 𝜌𝑌) are as follows:
𝑋 = [1 −
𝑠
𝑒𝑡 ] [1 + (𝑠 + 𝑋0 − 1) 𝑒−𝑡 ] ,
𝑠 + 𝑋0
𝑠
[𝑒−𝑡 − 1 + 𝑋0 + 𝑠] ,
𝑌=
𝑠 + 𝑋0
𝐾+ (𝑋, 𝑌) = [𝑒−𝑡 + 𝑠 + 𝑋0 − 1]
⋅ [𝑒−𝑡 −
2
⋅
−1/2
−1/2
𝑠
󵄨−1/2 √ 𝑋0
󵄨󵄨
]
󵄨󵄨Δ + (𝑠, 𝑡)󵄨󵄨󵄨
𝑠 + 𝑋0
2𝜋
[(𝑠 + 𝑋0 ) − 𝑋0 ]
(62)
(63)
3/2
3
(𝑠 + 𝑋0 )
,
where Δ + is the Jacobian associated with (62); that is,
Δ + (𝑠, 𝑡) = 𝑋𝑡 𝑌𝑠 − 𝑋𝑠 𝑌𝑡 =
1
[𝐷 (𝐸 − 𝐷) 𝑒𝑡
1−𝐸
− 2𝐷 (1 − 𝐷) − (𝐷 − 𝐸)2 𝑒−𝑡
+ (𝐸 + 𝐷 − 2𝐸𝐷) 𝑒−2𝑡 ] ,
𝐷=
(64)
𝑋0
𝑠
=1−
, 𝐸 = 1 − 𝑋0 − 𝑠.
𝑠 + 𝑋0
𝑠 + 𝑋0
̃
< 𝑌 < 𝑌0 )
(ii) (𝑋, 𝑌) ∈ D0 (𝑌(𝑋)
𝜋 (𝑘, 𝑟) ∼ 𝐶0 (𝜌) 𝐾 (𝑋, 𝑌) 𝑒𝜌Ψ(𝑋,𝑌) ,
(65)
8
Advances in Operations Research
where 𝐶0 (𝜌) is given by (45) for region R1
Ψ (𝑋, 𝑌) = 𝐴 (1 − 𝑒𝜏 ) + 𝑋0 log (1 − 𝐴) + 𝑌0
⋅ log (1 − 𝐵) − 𝑋 log (1 − 𝐴𝑒𝜏 ) − 𝑌 log (1 − 𝐵𝑒𝜏 )
= −𝐴𝑒𝜏 + [𝐴𝑒𝜏 − 1 + (𝑒−𝜏 − 𝐴) (1 −
⋅ log (1 − 𝐴𝑒𝜏 ) +
𝑋0
)]
1−𝐴
(66)
𝑌0 (𝑒−𝜏 − 𝐵)
log (1 − 𝐵𝑒𝜏 ) + 𝐴
𝐵−1
We can view (68) as representing a family of curves in
the (𝑋, 𝑌) plane, with 𝐴 indexing the family and 𝜏 increasing
along a curve. When 𝜏 = 0 the curves in (68) meet at the
corner point (𝑋, 𝑌) = (𝑋0 , 𝑌0 ) and we also note that the
Jacobian in (71) vanishes when 𝜏 = 0, indicating a singularity
in the transformation in (68). When 𝐴 = 𝐴 max the curve
becomes the horizontal segment 𝑌 = 𝑌0 . But then (69) shows
that (1 − 𝐴 max )(𝐴 max + 𝑋0 + 𝑌0 ) = 𝑋0 so that 𝐵 = ∞ in
(67). Thus near 𝑌 = 𝑌0 the expansion in (68) becomes invalid.
When 𝐴 = 𝑌0 /(𝑋0 + 𝑌0 ) the curve in (68) becomes
+ 𝑋0 log (1 − 𝐴) + 𝑌0 log (1 − 𝐵) ,
𝐵=
𝐴 (1 − 𝐴 − 𝑋0 )
,
(1 − 𝐴) (𝐴 + 𝑋0 + 𝑌0 ) − 𝑋0
(67)
𝑋0
− 1 + 𝑒𝜏 ) ,
1−𝐴
𝑋0
𝑌 = 𝑌0 − (1 − 𝑒 ) (𝐴 + 𝑋0 + 𝑌0 −
)
1−𝐴
(68)
𝐵 − 𝑒−𝜏
),
𝐵−1
where 𝐴 ∈ (𝐴 min , 𝐴 max ) with
𝐴 min = 1 − √𝑋0 ,
𝐴 max =
(69)
1
2
[1 − 𝑋0 − 𝑌0 + √(𝑋0 + 𝑌0 + 1) − 4𝑋0 ] ,
2
with 𝐴 ∈ (𝑌0 /(𝑋0 + 𝑌0 ), 𝐴 max ) corresponding to D0 ,
and
𝐾 (𝑋, 𝑌) = [(𝑒−𝜏 − 𝐴) (𝑒−𝜏 − 𝐵)]
⋅
−1/2
|Δ|−1/2
𝐴 (1 − 𝐴) √𝑌0 [(1 − 𝐴)2 − 𝑋0 ]
√2𝜋 [𝑌0 − 𝐴 (𝑋0 + 𝑌0 )] [𝑌0 + (1 − 𝑋0 − 𝑌0 ) 𝐴 − 𝐴2 ]
(70)
,
where Δ is the Jacobian associated with (68), so that
Δ = 𝑋𝜏 𝑌𝐴 − 𝑋𝐴 𝑌𝜏 = (𝑒−𝜏 − 1)
⋅ {[1 −
+
𝑋0
𝑋0
] [−𝐴𝑒𝜏 + (1 −
) 𝑒−𝜏 ]
2
1
−𝐴
(1 − 𝐴)
𝑌0
𝑋0
𝑒−𝜏 ]} = (𝑒−𝜏 − 1) {𝐴 + 𝑋0
[1 +
(71)
1−𝐵
(1 − 𝐴)2
+ 𝑌0 −
𝑋0
𝑋0
− 1]
+ 𝐴𝑒𝜏 [
1−𝐴
(1 − 𝐴)2
+ 𝑒−𝜏 [1 −
𝑋 (𝑋 + 𝑌 )
2𝑋0
+ 0 0 2 0 ]} .
1−𝐴
(1 − 𝐴)
−𝜏
𝑋 = (𝑋0 + 𝑌0 − 1 + 𝑒 ) (𝑒
−𝜏
= 𝑌0 (
1 − 𝑒−𝜏
),
𝑋0 + 𝑌0
𝜏
where (𝐴, 𝜏) are related to (𝑋, 𝑌) by
𝑋 = (𝑒−𝜏 − 𝐴) (
𝑌 = 𝑌0 (1 −
𝑌0
−
)
𝑋0 + 𝑌0
(72)
and eliminating 𝜏 we see that (72) is precisely the curve
̃
𝑌 = 𝑌(𝑋)
in (56) that separates D0 from D+ . For 𝐴 ∈
(𝐴 min , 𝑌0 /(𝑋0 + 𝑌0 )) the curves in (68) fill a portion of D+ ,
but then the leading term for 𝜋(𝑘, 𝑟) is given by (60), and (65)
corresponds to only an exponentially small correction to (60).
When 𝐴 = 𝐴 min = 1 − √𝑋0 the curve in (68) is tangent to the
line 𝑋 = 𝑋0 at the point 𝑌 = 𝑌0 , which will have significance
for the parameter region R4 . We also note that for 𝐴 ∈
(𝑌0 /(𝑋0 + 𝑌0 ), 𝐴 max ) when 𝜏 = − log 𝐴 the curves in (68)
hit the 𝑦-axis (then 𝑋 = 0 by (68)) and then the first factor in
(70) becomes singular, which indicates that the asymptotics
become invalid. Along 𝐴 = 𝑌0 /(𝑋0 + 𝑌0 ) corresponding to
̃
𝑌 = 𝑌(𝑋),
𝐾(𝑋, 𝑌) in (70) is again singular. Thus (70) is
singular when 𝐴 = 𝑌0 /(𝑋0 + 𝑌0 ), 𝐴 = 𝐴 max (𝐵 = ∞), and
̃
𝑒−𝜏 = 𝐴, corresponding to the three curves (𝑌 = 𝑌(𝑋),
𝑌 = 𝑌0 , and 𝑋 = 0) that bound the region D0 . We will
give the appropriate expansions near these bounding curves
in Section 3.3.
For 0 < 𝑠 < 𝑌0 the curves in (62) fill the entire region D+ ,
with 𝑠 = 0 corresponding to the line segment 𝑌 = 0, 0 < 𝑋 <
̃
𝑋0 , and 𝑠 = 𝑌0 corresponding to the curve 𝑌 = 𝑌(𝑋)
(then
(62) coincides with (72)). When 𝑋 = 𝑋0 we have 𝑡 = 0 and
the curves in (62) hit the line 𝑋 = 𝑋0 at finite and nonzero
slopes, for all 0 ⩽ 𝑠 ⩽ 𝑌0 . As 𝑡 increases each curve will hit
first either the 𝑥-axis or the 𝑦-axis. When 𝑠 = 0 the 𝑥-axis is
hit first. For 1 < 𝑋 < 𝑋0 this occurs at a finite value of 𝑡, when
𝑡 = log[(𝑋0 −1)/(𝑋−1)], but if 0 < 𝑋 < 1 in order to approach
the 𝑥-axis we must let 𝑠 → 0 and 𝑡 → ∞ in such a way that
𝑠𝑒𝑡 is held fixed. In this limit (62) may be approximated by
𝑋 ∼ 1 − 𝑠𝑒𝑡 /𝑋0 and 𝑌 ∼ (𝑋0 − 1)𝑠/𝑋0 . We discuss in more
detail the behavior of (60) as 𝑌 → 0 later, when we give the
asymptotic expansion(s) for 𝜋(𝑘, 𝑟) that apply for 𝑋0 > 1 and
𝑟 = 𝑂(1). When 𝑠 > 0 the curves in (62) hit the 𝑦-axis when
𝑡 = log(1 + 𝑋0 /𝑠), for then 𝑋 = 0. In particular if 𝑠 = 𝑌0
̃
the corresponding curve hits the 𝑦-axis at 𝑌(0)
in (57). Near
both the 𝑋- and 𝑌-axes, (60) will have singular behaviors and
other expansions must be constructed. Note, however, that
̃
(60) is not singular along the curve 𝑌 = 𝑌(𝑋),
whereas (65)
Advances in Operations Research
ℛ2
0.7
9
X0 = 0.7, Y0 = 0.7
ℛ3
0.3
X0 = 0.4, Y0 = 0.3
𝒟0
0.6
̃
Y
0.5
𝒟0
0.2
Y 0.4
Y
̃
Y
0.3
0.1
Y1
0.2
𝒟+
𝒟+
0.1
0
0.1
Y∗
Y∗
0.2
0.3
0.4
0.5
𝒟−
0.6
0.7
0
X
0
Figure 4: Region R2 .
𝜋 (𝑘, 𝑟)
∼ 𝜌−1/2 (
𝑋0
)
𝑋0 + 𝑌
(𝑋0 + 𝑌) − 𝑋0
5/2
√2𝜋
𝑒𝜌Ψ+ (𝑋0 ,𝑌) ,
Ψ+ (𝑋0 , 𝑌) = 𝑌 + 𝑋0 − 1 − (𝑋0 + 𝑌) log (𝑋0 + 𝑌) ,
0.3
0.4
We thus define D− as
1[
2√𝑋0 − 𝑋0 − 𝑋
2
[
and, for regions R2 and R3 , D+ now becomes
(74)
̃ (𝑋)} ,
D+ = {(𝑋, 𝑌) : 0 < 𝑋 < 𝑋0 , 𝑌∗ (𝑋) < 𝑌 < 𝑌
(75)
2
with D0 still defined by (59).
Proposition 6. For (𝑋0 , 𝑌0 ) ∈ R2 ∪ R3 the asymptotic
expansions of 𝜋(𝑘, 𝑟) are as follows:
(i) (𝑋, 𝑌) ∈ D− (0 < 𝑌 < 𝑌∗ (𝑋))
1/3
𝜋 (𝑘, 𝑟) ∼ 𝐶1 (𝜌) 𝐿 (𝑋, 𝑌) 𝑒𝜌Φ(𝑋,𝑌) 𝑒𝜌
Φ1 (𝑋,𝑌)
,
(79)
Φ (𝑋, 𝑌) = 𝑊0 log (𝑊0 + 𝑠1 ) + 𝑌 log (𝑌 + 𝑊0 ) − 𝑌
⋅ log 𝑌 − 𝑋 log [1 − √𝑊0
and thus, since now 𝑋0 < 1,
−
𝑌∗ (0) = 0,
(76)
(∈ (0, 𝑌0 )) .
(78)
where
− √𝑋0 − 𝑋√ (2 − √𝑋0 ) − 𝑋]
]
𝑌∗ (𝑋0 ) = √𝑋0 − 𝑋0
(77)
(73)
which holds for 𝑚 − 𝑘 = 𝜌(𝑋0 − 𝑋) = 𝑂(1) and 0 < 𝑌 < 𝑌0 .
However, we note that since 𝑋0 > 1 we have Ψ+ (𝑋0 , 𝑌) <
0 and thus 𝜋(𝑘, 𝑟) in (73) is exponentially small in 𝜌. This is
true for the entire domains D0 and D+ , as there is very little
probability mass in these ranges if 𝑋0 > 1.
We next consider regions R2 and R3 in parameter space,
where it will become necessary to break up the state space into
the three regions D+ , D− , and D0 . These regions are sketched
in Figures 4 and 5. The curve that separates D+ from D0 is
again given by (56), while the curve separating D+ from D−
will be
𝑌 = 𝑌∗ (𝑋) = 𝑌∗ (𝑋; 𝑋0 ) =
0.2
X
D− = {(𝑋, 𝑌) : 0 < 𝑋 < 𝑋0 , 0 < 𝑌 < 𝑌∗ (𝑋)}
2
(𝑋0 + 𝑌)
0.1
Figure 5: Region R3 .
is singular. We can simplify (60) near 𝑋 = 𝑋0 , and then we
obtain the more explicit form
𝑚−𝑘
𝒟−
𝑊0 + 𝑠1
]
𝑊0 + 𝑌
(80)
√𝑊0 (𝑊0 + 𝑠1 ) 1
− 𝑊0 log 𝑊0 ,
𝑊0 + 𝑌
2
𝑠1 = 𝑠1 (𝑋, 𝑌) = 𝑊0 +
𝑌 + 𝑊0
[𝑊0 + 1 − 𝑋
2√𝑊0
2
+ √(𝑊0 + 1 − 𝑋) − 4𝑊0 ] ,
(81)
10
Advances in Operations Research
Φ1 (𝑋, 𝑌) = Φ∗ (𝑠1 ) = −𝑟0
(1 − √𝑋0 )
2/3
𝑋01/6
0.20
X0 = 0.4, Y0 = 0.2
ℛ4
(82)
𝑠 + 𝑊0
⋅ log ( 1
),
1 − √𝑋0
0.15
3/2
𝐿 (𝑋, 𝑌) =
⋅
(𝑊0 + 𝑠1 )
1
√𝑌√𝑊0 + 𝑌 √𝑠1 − 𝑌
(𝑠1 + 𝑊0 ) √𝑋0 − 𝑠1
√2𝑊0 + 𝑠1 + 𝑌
⋅ 𝑋0−1/6 (1 − √𝑋0 )
⋅ exp [−
1 + √𝑋0
2√𝑋0
[1 − √𝑊0
−5/6
𝑊0 + 𝑠1 −1/2 1
]
𝑊0 + 𝑌
2𝜋
[𝐴𝑖󸀠 (𝑟0 )]
log (
𝒟0
Y 0.10
Yc
(83)
−2
𝑊0 + 𝑠1
)] ,
1 − √𝑋0
with 𝑊0 = (1−√𝑋0 )2 and 𝑟0 is the maximal root of the
Airy function 𝐴𝑖(⋅). In (79), 𝐶1 ∼ 𝜌−2/3 for (𝑋0 , 𝑌0 ) ∈
R2 , 𝐶1 is given by (35) for (𝑋0 , 𝑌0 ) ∈ R2 ∩ R3 (then
𝑋0 + 𝑌0 = 1 + 𝛾/√𝜌), and 𝐶1 is given by (36) for
(𝑋0 , 𝑌0 ) ∈ R3 .
̃
(ii) (𝑋, 𝑌) ∈ D+ (𝑌∗ (𝑋) < Y < 𝑌(𝑋)).
The expression in (60) applies with 𝐶 ∼ 𝜌−1/2 for
(𝑋0 , 𝑌0 ) ∈ R2 , 𝐶 is given by (32) for (𝑋0 , 𝑌0 ) ∈
R2 ∩ R3 , and 𝐶 is given by (33) for (𝑋0 , 𝑌0 ) ∈ R3 .
̃
< 𝑌 < 𝑌0 ).
(iii) (𝑋, 𝑌) ∈ D0 (𝑌(𝑋)
The expression in (65) applies with 𝐶0 given by (45) for
(𝑋0 , 𝑌0 ) ∈ R2 , 𝐶0 is given by (44) for (𝑋0 , 𝑌0 ) ∈ R2 ∩
R3 , and 𝐶0 ∼ 𝜌−1/2 (1 − 𝑋0 − 𝑌0 ) for (𝑋0 , 𝑌0 ) ∈ R3 .
In contrast to regions D+ and D0 , the expansion (79) in
D− is a completely explicit function of 𝑋, 𝑌, and 𝑋0 . We also
note that 𝑠1 (𝑋, 𝑌) has a simple linear dependence upon 𝑌,
and 𝑠1 (𝑋0 , 𝑌) = 𝑌. In Section 4 we give a more geometric
interpretation of this expansion, and we also observe that
the form of (79) is slightly different from the expansions
in D+ and D0 , as the former contains an additional factor
that is of order exp[𝑂(𝜌1/3 )], and thus gives an additional
subexponential dependence on 𝜌. While the forms of 𝐶,
𝐶0 , and 𝐶1 change according to whether (𝑋0 , 𝑌0 ) lies in the
regions R2 , R3 or R2 ∩ R3 of parameter space, the ratios
𝐶1 : 𝐶 : 𝐶0 remain the same for these three cases.
The expansion in (79) is valid only in the interior of D− .
As 𝑌 → 0 there is a singularity due to the factor 1/√𝑌 in
𝐿(𝑋, 𝑌) in (83). For 𝑋 → 𝑋0 there is also a singularity due
to the factor 1/√𝑠1 − 𝑌 in (83), and as 𝑋 → 𝑋0 , we find that
(𝑠1 − 𝑌)−1/2 = 𝑂[(𝑋0 − 𝑋)−1/4 ]. The curve 𝑌 = 𝑌∗ (𝑋) that
separates D+ from D− corresponds to 𝑠1 (𝑋, 𝑌) = √𝑋0 − 𝑋0 ,
and along 𝑌 = 𝑌∗ (𝑋) the factor (𝑠1 + 𝑊0 )√𝑋0 − 𝑠1 =
−(1 − √𝑋0 )[𝑠1 + 𝑋0 − √𝑋0 ] vanishes. Thus (83) shows that
𝐿(𝑋, 𝑌∗ (𝑋)) = 0 which also indicates a nonuniformity in the
asymptotics. Later we will give appropriate expansions near
the three bounding curves (𝑌 = 0, 𝑋 = 𝑋0 , and 𝑌 = 𝑌∗ (𝑋))
of region D− .
0.05
0
𝒟−
0.1
0.2
0.3
0.4
X
Figure 6: Region R4 .
In Figure 4 we also indicate the curve
2
𝑌 = 𝑌1 (𝑋) = 𝑌1 (𝑋; 𝑋0 ) =
(1 − 𝑋0 )
,
1−𝑋
(84)
which lies entirely within D+ (when (𝑋0 , 𝑌0 ) ∈ R2 ) and
corresponds to 𝜕Ψ+ /𝜕𝑌 = 0, where Ψ+ is in (61). We can also
show that Ψ+,𝑌𝑌(𝑋, 𝑌1 (𝑋)) < 0 so that, for a fixed 𝑋 ∈ (0, 𝑋0 ),
Ψ+ achieves a local maximum along 𝑌 = 𝑌1 (𝑋). Note that
𝑌 = 𝑌1 (𝑋) corresponds to 𝑠(𝑋, 𝑌) = 1 − 𝑋0 (> √𝑋0 − 𝑋0 ) in
(62), for then 𝑋 = 1 + (𝑋0 − 1)𝑒𝑡 and 𝑌 = (1 − 𝑋0 )𝑒−𝑡 .
For (𝑋0 , 𝑌0 ) ∈ R3 the curve in (84) plays no role, as it lies
outside of D+ , but now the curve
𝑌 = 𝑌2 (𝑋) = 𝑌2 (𝑋; 𝑋0 , 𝑌0 ) = 𝑌0
1 − 𝑋0
1−𝑋
(85)
lies within D0 , connecting the points (0, 𝑌0 (1 − 𝑋0 )) and
(𝑋0 , 𝑌0 ). Along 𝑌 = 𝑌2 (𝑋) we have Ψ𝑌 = 0 and then 𝐴 =
1 − 𝑋0 and 𝐵 = 0, so that (68) becomes 𝑋 = 1 − (1 − 𝑋0 )𝑒𝜏 and
𝑌 = 𝑌0 𝑒−𝜏 . Then for region R3 , Ψ will have a local maximum
along 𝑌 = 𝑌2 in D0 .
For (𝑋0 , 𝑌0 ) ∈ R3 most of the mass will lie near the
corner point (𝑋, 𝑌) = (𝑋0 , 𝑌0 ), where D0 and D+ meet,
but neither (60) nor (65) (with the appropriate 𝐶0 and 𝐶)
are valid there. For (𝑋0 , 𝑌0 ) ∈ R2 , 𝜋(𝑘, 𝑟) will be maximal
near the point (𝑋, 𝑌) = (𝑋0 , 1 − 𝑋0 ), and (73) applies for
𝑘 = 𝑚 − 𝑂(1) (or 𝑋 = 𝑋0 − 𝑂(𝜌−1 )) for both regions R1
and R2 . By expanding (73) about 𝑌 = 1 − 𝑋0 , which is where
Ψ+ (𝑋0 , 𝑌) is maximal, we obtain precisely the expression in
(49).
Next we consider region R4 of parameter space. Now
the state space will be split into D0 and D− , and D+ will be
absent. This is sketched in Figure 6, and we also observe that
̃
as 𝑋0 + 𝑌0 ↓ √𝑋0 in region R3 , the curves 𝑌∗ (𝑋) and 𝑌(𝑋)
in (75) and (56) become identical, and thus D+ shrinks to
Advances in Operations Research
11
this curve. For R4 there is a new curve that comes into play;
namely,
𝑌 = 𝑌𝑐 (𝑋) = 𝑌𝑐 (𝑋; 𝑋0 , 𝑌0 ) = −𝑊0 +
𝑌0 + 𝑊0
[𝑊0
2√𝑊0
(86)
2
+ 1 − 𝑋 − √(𝑊0 + 1 − 𝑋) − 4𝑊0 ] .
We have 𝑌𝑐 (𝑋0 ) = 𝑌0 and the curve hits the 𝑥-axis (𝑌 = 0)
when
𝑌𝑐 (𝑋) = 0 ⇐⇒
𝑋 = 𝑋0 −
𝑌02
1
.
𝑌0 + 𝑊0 √𝑊0
(87)
The curve 𝑌 = 𝑌𝑐 (𝑋) now separates D0 from D− and
corresponds to 𝐴 = 𝐴 min = 1 − √𝑋0 in (68), and also
𝑠1 (𝑋, 𝑌) = 𝑌0 in (81).
Proposition 7. For (𝑋0 , 𝑌0 ) ∈ R4 (thus 0 < 𝑌0 < √𝑋0 − 𝑋0 )
the asymptotic expansions of 𝜋(𝑘, 𝑟) are as follows:
(i) (𝑋, 𝑌) ∈ D− , 0 < 𝑌 < 𝑌𝑐 (𝑋), 𝑋0 − 𝑌02 /[√𝑊0 (𝑌0 +
𝑊0 )] < 𝑋 < 𝑋0 .
The expansion in (79) applies with now 𝐶1 (𝜌) given by
(41).
(ii) (𝑋, 𝑌) ∈ D0 (max{0, 𝑌𝑐 (𝑋)} < 𝑌 < 𝑌0 ).
The expansion in (65) applies for 𝐴 ∈ (𝐴 min , 𝐴 max )
with 𝐶0 (𝜌) ∼ 𝜌−1/2 (1 − 𝑋0 − 𝑌0 ).
Again, different expansions must be given near 𝑌 = 𝑌𝑐 (𝑋)
and near all boundaries of the state space. The curve 𝑌 =
𝑌2 (𝑋) in (85) still lies within D0 and is sketched in Figure 6,
with again Ψ𝑌 = 0 along this curve.
We conclude by noting that, for all regions of parameter
space R𝑗 , the expansion in D0 depends upon the secondary
storage capacity 𝑌0 (or 𝑅). For regions R1 and R2 the
expansions in D+ and D− (R2 only) are independent of
̃
𝑌0 , except through the curve 𝑌(𝑋)
that bounds D+ . Thus if
𝑋0 + 𝑌0 > 1 the effects of the finite storage capacity appear in
D0 only, and then letting 𝑌0 → ∞ will recover the results for
the storage model in [10, 11], which assumes that 𝑅 = ∞. For
(𝑋0 , 𝑌0 ) ∈ R3 , hence √𝑋0 < 𝑋0 + 𝑌0 < 1, the expansions in
D− and D+ depend upon 𝑌0 only through the multiplicative
constants 𝐶1 and 𝐶, which now depend on 𝑌0 in view of (33)
and (36). For (𝑋0 , 𝑌0 ) ∈ R4 , hence 0 < 𝑌0 < √𝑋0 − 𝑋0 , again
the D− result depends on 𝑌0 only through the constant 𝐶1 (cf.
(41)). For (𝑋0 , 𝑌0 ) ∈ R3 ∩R4 , that is, along and near the curve
̃
𝑋0 + 𝑌0 = √𝑋0 , we have 𝑌3 (𝑋) = 𝑌𝑐 (𝑋) = 𝑌(𝑋)
= 𝑌∗ (𝑋) so
all four of these curves coalesce. A special analysis is required
for 𝑋0 + 𝑌0 − √𝑋0 = 𝑂(𝜌−1/3 ) for values of (𝑋, 𝑌) near this
curve(s).
This completes our summary of the asymptotic expansions in the three main regions, D0 , D+ , and D− , of the state
space.
3.3. Joint Distribution: Boundary, Corner, and Transition
Regions. We next analyze the four boundary segments of the
state space rectangle, namely, the line segments {𝑌 = 0, 0 <
𝑋 < 𝑋0 }, {𝑋 = 𝑋0 , 0 < 𝑌 < 𝑌0 }, {𝑋 = 0, 0 < 𝑌 < 𝑌0 },
and {𝑌 = 𝑌0 , 0 < 𝑋 < 𝑋0 }. As we previously discussed,
the expansions in the three main regions become invalid near
the boundary of the state space, with the exception of the
D+ expansion in (60), which remains valid near 𝑋 = 𝑋0 ,
reducing to (73) when 𝑘 = 𝑚 − 𝑂(1) (𝑋 = 𝑋0 − 𝑂(𝜌−1 )), at
least for (𝑋0 , 𝑌0 ) ∈ R1 ∪R2 . For the four boundary segments
we will typically consider the respective scales 𝑟 = 𝑂(1),
𝑘 = 𝑚 − 𝑂(1), 𝑘 = 𝑂(1), and 𝑟 = 𝑅 − 𝑂(1), though at
times a different scaling must be considered in addition to
these discrete scales. Note also that a particular point on a
boundary segment may also require a separate expansion,
and this occurs in R2 ∪ R3 , when 𝑌∗ (𝑋) hits 𝑋 = 𝑋0
at the point (𝑋0 , √𝑋0 − 𝑋0 ). Also, in R1 ∪ R2 ∪ R3 , the
̃
curve 𝑌(𝑋)
that separates D0 from D+ hits the 𝑦-axis when
𝑌 = 𝑌0 [1 − 𝑋0 /(𝑋0 + 𝑌0 )2 ] (> 0) and this point requires a
separate analysis. Finally, in R4 the curve 𝑌𝑐 (𝑋) hits the 𝑥axis when 𝑋 = 𝑋0 − 𝑌02 /[(𝑌0 + 𝑊0 )√𝑊0 ].
After treating the four boundary segments we will give
asymptotic results that are valid near the four corner points,
(0, 0), (𝑋0 , 0), (𝑋0 , 𝑌0 ), and (0, 𝑌0 ), of the state space. Finally
we give results for points (𝑋, 𝑌) that lie on or near the
̃
transition curves 𝑌(𝑋)
(for R1 ∪ R2 ∪ R3 ), 𝑌∗ (𝑋) (for R2 ∪
R3 ), and 𝑌𝑐 (𝑋) (for R4 ). Unlike the previous subsections
where we listed the results by region R𝑗 of parameter space,
here we go by region of state space, and each proposition will
correspond to one such region and the different results for the
different R𝑗 will be collected in that proposition.
Proposition 8. For 𝑟 = 𝑂(1) and 𝑘 = 𝜌𝑋, 0 < 𝑋 < 𝑋0 , one
has the following expansions:
(i) R1 : 𝑋0 > 1, 0 < 𝑋 < 1
𝜋 (𝑘, 𝑟) ∼ 𝜌−𝑟
𝑋0 − 1
−2𝑟
(1 − 𝑋)−𝑟 (𝑋0 − 1)
2𝜋√𝑋𝑋0
𝜌(−2+𝑋+𝑋0 −𝑋log𝑋−𝑋0 log𝑋0 )
⋅ (𝑟 − 1)!𝑒
𝜋 (𝑘, 0) − 𝜌𝑘
,
(88)
𝑟 ⩾ 1,
𝑒−𝜌
−𝑋𝑒𝜌(−2+𝑋+𝑋0 −𝑋log𝑋−𝑋0 log𝑋0 )
.
∼ 𝜌−1
𝑘!
2𝜋√𝑋𝑋0 (𝑋0 − 1) (1 − 𝑋)
(89)
(ii) R1 : 𝑋0 > 1, 𝑋 = 1 + 𝛼/√𝜌, 𝛼 = 𝑂(1)
𝜋 (𝑘, 𝑟) ∼ 𝜌−𝑟/2
∞
𝑋0 − 1
−2𝑟
(𝑋0 − 1)
2𝜋√𝑋0
(90)
2
⋅ [∫ (𝛼 + 𝑢)𝑟−1 𝑒−𝑢 /2 𝑑𝑢] 𝑒𝜌(−1+𝑋0 −𝑋0 log𝑋0 ) ,
−𝛼
𝑟 ⩾ 1,
𝜋 (𝑘, 0) − 𝜌𝑘
𝑒−𝜌
∼ 𝜌−1/2
𝑘!
∞
2
−𝑒𝜌(−1+𝑋0 −𝑋0 log𝑋0 )
⋅
[∫ 𝑒−𝑢 /2 𝑑𝑢] .
2𝜋√𝑋0 (𝑋0 − 1) −𝛼
(91)
12
Advances in Operations Research
(92)
so that 𝑠10 corresponds to 𝑠1 (𝑋, 0) in (81), Φ∗ (⋅) is
defined in (82), and Φ(X, 0) can be computed from
(80) by setting 𝑌 = 0 and replacing 𝑠1 by 𝑠10 . The
value of 𝐶1 in (96) is given by (34) for R2 , by (35) for
R2 ∩ R3 (𝑋0 + 𝑌0 = 1 + 𝑂(𝜌−1/2 )), and by (36) for R3 .
(93)
(vi) R4 : 0 < 𝑌0 < √𝑋0 −𝑋0 , 𝑋0 −(𝑌02 /[√𝑊0 (𝑌0 +𝑊0 )]) <
𝑋 < 𝑋0 .
The expression in (96) applies with 𝐶1 now given by
(41). For region R3 ∩ R4 (hence 𝑋0 + 𝑌0 = √𝑋0 +
𝑂(𝜌−1/3 )), (96) holds for 0 < 𝑋 < 𝑋0 , where 𝐶1 can be
computed from (37).
(iii) R1 : 𝑋0 > 1, 1 < 𝑋 < 𝑋0
𝜋 (𝑘, 𝑟) ∼ 𝜌−1/2
3
(𝑋0 − 1)
√2𝜋𝑋0
(𝑋 − 1)𝑟−1
2 −𝑟−1 𝜌(−1+𝑋0 −𝑋0 log𝑋0 )
⋅ [𝑋 − 1 + (𝑋0 − 1) ]
𝑒
,
𝑟 ⩾ 1,
𝜋 (𝑘, 0) − 𝜌𝑘
⋅
𝑒−𝜌
∼ 𝜌−1/2
𝑘!
− (𝑋0 − 1) 𝑒𝜌(−1+𝑋0 −𝑋0 log𝑋0 )
2
√2𝜋𝑋0 [𝑋 − 1 + (𝑋0 − 1) ]
.
(vii) R4 : 0 < 𝑌0 < √𝑋0 −𝑋0 , 0 < 𝑋 < 𝑋0 −𝑌02 /[√𝑊0 (𝑌0 +
𝑊0 )] ≡ 𝑋𝐿
(iv) R1 ∩ R2 : 𝑋0 = 1 + 𝛽/√𝜌, 𝛽 = 𝑂(1), 0 < 𝑋 < 1
𝜋 (𝑘, 𝑟) ∼ 𝜌
−1/2
1 𝜌(−1+𝑋−𝑋log𝑋) 1
𝑒
√𝑋
2𝜋𝑖
𝜋 (𝑘, 𝑟) ∼ (1 − 𝑋0 − 𝑌0 )
(94)
Γ (𝑧 + 𝑟) (1 − 𝑋)𝑧
𝑑𝑧,
𝑟!𝐷𝑧 (−𝛽) 𝐷𝑧−1 (−𝛽)
⋅∫
𝐵𝑟+
Ψ (𝑋, 0)
= 𝐴 0 (1 −
where 𝐵𝑟+ is the imaginary axis in the 𝑧-plane if 𝑟 ⩾ 1,
and if 𝑟 = 0 the contour is to be indented to the right
of the pole at 𝑧 = 0. Here Γ(⋅) is the gamma function.
When 𝛽 = 0 the expression in (94) simplifies to
𝜋 (𝑘, 𝑟) ∼ √𝜌𝐶1 (𝜌) 𝑒𝜌Φ(𝑋,0) 𝑒𝜌
⋅
(1 − √𝑋0 )
1/6
Φ∗ (𝑠10 )
𝜌
𝑟
2
⋅ [𝑌0 (1 − 𝐴 0 ) + 𝑋0 𝐴 0 (1 − 𝑋0 − 𝐴 0 )] = (−𝑋)
𝑊0𝑟
(𝑊0 + 𝑠10 ) √√𝑋0 − 𝑋0 − 𝑠10
(96)
1/2√𝑋0
(
√𝑊0
)
𝑠10 + 𝑊0
,
that satisfies 𝐴 0 (0) = 1 − 𝑋0 − 𝑌0 and 𝐴 0 (𝑋𝐿 ) = 1 −
√𝑋0 = √𝑊0 ,
where 𝑊0 = (1 − √𝑋0 )2 ,
𝐵0 =
𝑠10
= −𝑊0
+
(100)
⋅ (1 − 𝑋0 − 𝐴 0 ) [(1 − 𝐴 0 ) (𝐴 0 + 𝑋0 + 𝑌0 ) − 𝑋0 ]
𝑟!
√2𝜋 [Ai󸀠 (𝑟0 )]2 𝑋01/6 √𝑠10 (𝑠10 + 2𝑊0 )
𝐴0
),
𝐵0
(𝐴 0 − 1 + 𝑋0 + 𝑌0 )
(v) R2 ∪ R3 : 0 < 𝑋0 < 1, 𝑋0 + 𝑌0 > √𝑋0 , 0 < 𝑋 < 𝑋0
1/3
(99)
where 𝐴 0 = 𝐴 0 (𝑋) = 𝐴 0 (𝑋; 𝑋0 , 𝑌0 ) is the solution of
the algebraic equation
(95)
2 (1 − 𝑋)
(2 − 𝑋)−𝑟−1 𝑒𝜌(−1+𝑋−𝑋log𝑋) .
𝜋 √𝑋
1
) + 𝑋0 log (1 − 𝐴 0 )
𝐵0
+ 𝑌0 log (1 − 𝐵0 ) − 𝑋 log (1 −
𝜋 (𝑘, 𝑟)
∼ 𝜌−1/2 √
𝜌𝑟 𝑌0 𝐵0 𝑟
) 𝑔 (𝑋) 𝑒𝜌Ψ(𝑋,0) , (98)
(
𝑟! 1 − 𝐵0
𝐴 0 (1 − 𝑋0 − 𝐴 0 )
,
(1 − 𝐴 0 ) (𝐴 0 + 𝑋0 + 𝑌0 ) − 𝑋0
(101)
(97)
√𝑊0
2
[𝑊0 + 1 − 𝑋 + √(𝑊0 + 1 − 𝑋) − 4𝑊0 ] ,
2
so that 𝐵0 (0) = 1−𝑋0 −𝑌0 and 𝐵0 (𝑋𝐿 ) = 𝑊0 /(𝑊0 +𝑌0 ),
and
2
𝑔 (𝑋) =
𝐴 0 (1 − 𝐴 0 ) [(1 − 𝐴 0 ) − 𝑋0 ] 𝑌0
(1 − 𝐵0 ) √𝐵0 − 𝐴 0 [𝑌0 − 𝐴 0 (𝑋0 + 𝑌0 )] [𝑌0 − 𝐴20 + (1 − 𝑋0 − 𝑌0 ) 𝐴 0 ]
−1/2
+
𝑋 (𝑋 + 𝑌 )
𝐴0
𝑋0
2𝑋0
[
− 1] + 𝐵0 [1 −
+ 0 0 20 ]}
𝐵0 (1 − 𝐴 0 )2
1 − 𝐴0
(1 − 𝐴 0 )
.
{𝐴 0 + 𝑋0 + 𝑌0 −
𝑋0
1 − 𝐴0
(102)
Advances in Operations Research
13
For 𝑋 → 0 and 𝑋 → 𝑋𝐿 one can obtain 𝑔(𝑋) more
explicitly, with
𝑔 (𝑋) ∼ √
𝑋0 + 𝑌0
, 𝑋 󳨀→ 0,
𝑋
(103)
5/2
3/2
𝑔 (𝑋) ∼
2 (𝑊0 + 𝑌0 ) (1 − √𝑋0 ) (𝑋𝐿 − 𝑋)
,
𝑌0 (2𝑊 + 𝑌 )3/2 (√𝑋 − 𝑋 − 𝑌 )3/2
0
0
0
0
0
(104)
𝑋 󳨀→ 𝑋𝐿 .
(viii) R4 : 0 < 𝑌0 < √𝑋0 − 𝑋0 , 𝑋 − 𝑋𝐿 = 𝜌−1/3 Λ = 𝑂(𝜌−1/3 )
𝜋 (𝑘, 𝑟) ∼ (1 − 𝑋0 − 𝑌0 )
𝜌𝑟 𝑊0𝑟 −1/3
𝜌
𝑟!
5/6
⋅
√𝑊0 + 𝑌0 𝑋01/6 (1 − √𝑋0 )
3/2
√2𝑊0 + 𝑌0 (√𝑋0 − 𝑋0 − 𝑌0 )
+𝜌
2/3
exp [𝜌Ψ (𝑋𝐿 , 0)
Λ3
1
Ψ𝑋 (𝑋𝐿 , 0) Λ + 𝜌1/3 Ψ𝑋𝑋 (𝑋𝐿 , 0) Λ2 ] exp [
2
6
[
(105)
(41) or (37). However, for 0 < 𝑋 < 𝑋𝐿 and 𝑋 ≈ 𝑋𝐿 , the
expressions in (98) and (105) show that now 𝜋(𝑘, 𝑟) depends
in an intricate way on 𝑌0 . Thus the finiteness of the secondary
storage capacity affects the probabilities that there are even a
few secondary spaces occupied.
For region R1 , 𝜋(𝑘, 𝑟) is exponentially small for 𝑟 ⩾ 1
and 𝜋(𝑘, 0) approximately follows a Poisson distribution, as
indicated previously in Proposition 3. The results in (88)–
(93) better quantify (47) and estimate the exponentially small
error from the Poisson approximation. For regions R2 , R3 ,
and R4 , 𝜋(𝑘, 𝑟) is always exponentially small for 𝑟 = 𝑂(1),
and the approximations do not distinguish 𝑟 = 0 from 𝑟 ⩾ 1.
By evaluating the contour integral in (105) for Λ → +∞ we
can verify that (105) asymptotically matches to (96) (with 𝐶1
given by (41)), as 𝑋 ↓ 𝑋𝐿 . Similarly, for Λ → −∞ (105) will
match to (98) as 𝑋 ↑ 𝑋𝐿 .
Next we consider points (𝑘, 𝑟) near 𝑋 = 𝑋0 , with 𝑋0 −𝑋 =
𝑜(1). Note that only regions D+ and D− can be bounded by
𝑋 = 𝑋0 , for 0 < 𝑌 < 𝑌0 .
Proposition 9. For 𝑋 ∼ 𝑋0 , one uses the scales 𝑘 = 𝑚 − 𝑛 and
𝑋 = 𝑋0 − 𝜌−2/3 ]1 , so that 𝑛 = 𝜌1/3 ]1 = 𝜌(𝑋0 − 𝑋) denotes the
number of empty primary storage spaces. The expansions are
now the following:
3
⋅
𝑊0 (𝑌0 + 𝑊0 ) (𝑌03 + 3𝑊0 𝑌02 + 4𝑊02 𝑌0 − 2√𝑋0 𝑊05/2 )
2
3
𝑌03 (𝑌0 + 2𝑊0 ) (√𝑋0 − 𝑋0 − 𝑌0 )
(i) R2 ∪ R3 : 0 < 𝑋0 < 1, 𝑋0 + 𝑌0 > √𝑋0 ; 𝑛 = 𝑂(1),
0 < 𝑌 < √𝑋0 − 𝑋0
]
5/3
⋅
]
(𝑌 + 𝑊0 ) (1 − √𝑋0 )
1 𝑖∞
1
exp [ 0
∫
2𝜋𝑖 −𝑖∞ 𝐴𝑖 (𝜃)
𝑌0 (𝑌0 + 2𝑊0 ) 𝑋01/6
[
Λ𝜃] 𝑑𝜃,
]
𝜌1/3
𝜋 (𝑘, 𝑟) ∼ 𝜌−1/6 𝐶1 (𝜌) 𝑒𝜌Φ(𝑋0 ,𝑌)𝑒
where
Ψ (𝑋𝐿 , 0) = 𝑌0 log (
⋅
𝑌0
𝑌
1
) − 0 + 𝑋0
𝑊0 + 𝑌0
√𝑊0 2
⋅ log (
√𝑋0 − 𝑋0 − 𝑌0
√𝑊0
Ψ𝑋 (𝑋𝐿 , 0) = − log (
(106)
⋅
),
√𝑋0 − 𝑋0 − 𝑌0
1 − √𝑋0
(107)
2
Ψ𝑋𝑋 (𝑋𝐿 , 0) = −
(𝑌0 + 𝑊0 ) √𝑊0
𝑌0 (𝑌0 + 2𝑊0 ) (√𝑋0 − 𝑋0 − 𝑌0 )
.
(108)
We note that the regions of validity of the expansions in
Proposition 8 are such that the four corner points of the state
space are excluded. For regions R1 and R2 the expansions for
𝑟 = 𝑂(1) are independent of the secondary storage capacity
𝑌0 . For region R3 the result in (96) does depend upon 𝑌0 ,
but only through the multiplicative constant 𝐶1 , which now
depends on 𝑌0 in view of (36). This is also true for region
R4 (and R3 ∩ R4 ) when 𝑋 > 𝑋𝐿 , as then 𝐶1 is given by
√𝑋0 (𝑌 + 𝑊0 )
(𝑌 + 𝑊0 ) √𝑋0 − 𝑌
]
(109)
(𝑌 + 𝑊0 ) √𝑋0 − 𝑌
√𝑌
⋅ exp [−
),
𝑛
(√𝑋0 )
−1/6
1
1
√
(1
−
𝑋
)
0
√2𝜋 𝐴𝑖󸀠 (𝑟0 )
⋅ 𝑋0−5/6 [𝑛 +
𝑌02
1
⋅ log 𝑋0 + (
− 𝑋0 )
𝑌0 + 𝑊0 √𝑊0
Φ∗ (𝑌)
1 + √𝑋0
2√𝑋0
log (
𝑊0 + 𝑌
)] ,
1 − √𝑋0
where 𝐶1 is given by (34), (35), and (36) for parameter
regions R2 , R2 ∩ R3 , and R3 , respectively, Φ∗ (𝑌) is
obtained by replacing 𝑠1 by 𝑠1 (𝑋0 , 𝑌) = 𝑌 in (82), and
Φ (𝑋0 , 𝑌) = (𝑌 + 𝑊0 ) log (𝑌 + 𝑊0 ) − 𝑌 log 𝑌
1
+ √𝑋0 − 1 − 𝑋0 log 𝑋0
2
1
− 𝑊0 log 𝑊0 .
2
(110)
14
Advances in Operations Research
(ii) R2 ∪ R3 : ]1 = 𝑂(1) with ]1 > 0, 0 < 𝑌 < √𝑋0 − 𝑋0
1/3
𝜋 (𝑘, 𝑟) ∼ 𝜌−1/6 𝐶1 (𝜌) 𝑒𝜌Φ(𝑋0 ,𝑌) 𝑒𝜌
⋅
[]1 log(√𝑋0 )+Φ∗ (𝑌)]
(𝑌 + 𝑊0 ) √𝑋0 − 𝑌
1
1
2
󸀠
√2𝜋 [𝐴𝑖 (𝑟0 )] [𝑌𝑋 (1 − √𝑋 )]1/2
0
0
⋅ exp [−
1 + √𝑋0
2√𝑋0
Proposition 10. For 𝑘 = 𝑂(1) and 𝑟 = 𝜌𝑌, 0 < 𝑌 < 𝑌0 , one
has the following expansions:
(i) R1 ∪ R2 ∪ R3 : 0 < 𝑌 < 𝑌0 [1 − 𝑋0 /(𝑋0 + 𝑌0 )2 ] ≡ 𝑌𝑈
𝜋 (𝑘, 𝑟)
(111)
𝜌Ψ+ (0,𝑌) 𝜌
∼ √𝜌𝐶 (𝜌) 𝑒
𝑊 +𝑌
log ( 0
)]
1 − √𝑋0
⋅ 𝐴𝑖 (] + 𝑟0 ) ,
Now the expressions in (109) and (111) apply over all
𝑌 (0 < 𝑌 < 𝑌0 ), with 𝐶1 given by (37) for R3 ∩ R4
and (41) for R4 .
𝜋 (𝑘, 𝑟) ∼ 𝜌−1/6 𝐶1 (𝜌) 𝑒𝜌[−1+√𝑋0 −√𝑋0 log(√𝑋0 )]
Ψ+ (0, 𝑌) = 𝐷0 +
The expansion of 𝐶 is given by (31)–(33) according to
subregions of R1 ∪ R2 ∪ R3 .
(112)
𝑘
𝐴 (𝐴 + 𝑋0 − 2) + 1
]
[ 1 1
𝑘!
1 − 𝐴1
𝑘
(117)
2
⋅
√𝑌0 𝐴 1 √𝐴 1 (𝐴 1 + 𝑋0 − 2) + 1 [(1 − 𝐴 1 ) − 𝑋0 ]
√𝑌 (𝑌0 − 𝑌)√󵄨󵄨󵄨󵄨Δ 𝐿 (𝑌)󵄨󵄨󵄨󵄨 [𝑌0 − 𝐴 1 (𝑋0 + 𝑌0 )]
,
where
𝐴𝑖 (] + 𝜃) −𝜔𝜃
𝑒 𝑑𝜃,
𝐴𝑖2 (𝜃)
𝐴 1 = 𝐴 1 (𝑌) = 𝐴 1 (𝑌; 𝑋0 , 𝑌0 ) =
where 𝜔 = (1 − √𝑋0 )−1/3 𝑋0−1/6 𝜔1 , so that ]1 𝜔1 =
√𝑋0 ]𝜔, and 𝐶1 is given again in Proposition 2 for the
three cases R2 , R2 ∩ R3 , and R3 .
(v) R1 ∪ R2 ∪ R3 : 0 < 𝑌 < 𝑌0 (R1 ) or √𝑋0 − 𝑋0 < 𝑌 <
𝑌0 (R2 ∪ R3 ), 𝑛 = 𝑂(1)
1
[1 − 𝑋0 − 𝑌0
2
(118)
2
+ √(𝑋0 + 𝑌0 − 1) + 4𝑌] ,
Ψ (0, 𝑌) = 𝐴 1 − 1 + 𝑋0 log (1 − 𝐴 1 ) + (𝑌0 − 𝑌)
⋅ log (1 − 𝐴 1 ) + 𝑌 log 𝐴 1 − 𝑌 log 𝑌 − (𝑌0 − 𝑌)
(119)
⋅ log (𝑌0 − 𝑌) + 𝑌0 log 𝑌0 ,
𝜋 (𝑘, 𝑟)
2
∼ 𝐶 (𝜌)
(116)
+ 2𝑋0 log (1 − 𝐷0 ) + 2𝑌 log 𝐷0 .
𝜋 (𝑘, 𝑟) ∼ 𝑄 (0, 0) 𝑒
1/3
(1 − √𝑋0 )
[]1 log (√𝑋0 ) −
]}
√2𝜋𝑋07/12
2√𝑋0
1 + √𝑋0
1
⋅ exp [
𝜔13 ]
2𝜋𝑖
6𝑋0 (1 − √𝑋0 )
−𝑖∞
𝑋0
− 2 − 𝑋0 log 𝑋0 − 𝑌 log 𝑌
1 − 𝐷0
𝜌Ψ(0,𝑌) 𝜌
𝜔12
(115)
(ii) R1 ∪ R2 ∪ R3 ∪ R4 : 𝑌𝑈 < 𝑌 < 𝑌0 (R1 ∪ R2 ∪ R3 )
or 0 < 𝑌 < 𝑌0 (R4 )
⋅ exp {−𝜌2/3 𝜔1 log (√𝑋0 )
𝑖∞
(114)
that satisfies 𝐷0 (0) = 1 − √𝑋0 = √𝑊0 and
(iv) R2 ∪ R3 : ]1 = 𝑂(1), 𝑌 − (√𝑋0 − 𝑋0 ) = 𝜌−1/3 𝜔1 =
𝑂(𝜌−1/3 )
⋅∫
𝑘! 𝐷 √𝑋 [𝑌 + 2𝐷2 (1 − 𝐷 )]
0
0
0
0
,
𝐷03 − 2𝐷02 + (1 − 𝑋0 − 𝑌) 𝐷0 + 𝑌 = 0
(iii) R3 ∩ R4 or R4 : 𝑛 = 𝑂(1) or ]1 = 𝑂(1).
+𝜌
𝑘
(1 + 𝑌 − 𝐷02 ) (1 − 𝐷0 ) 𝑌
where 𝐷0 = 𝐷0 (𝑌) is the root of the cubic equation
where ] = (1 − √𝑋0 )1/3 𝑋0−1/3 ]1 and 𝐶1 is again given
in Proposition 2 for the different ranges R𝑗 .
1/3
𝑘
(𝑋0 + 𝑌) − 𝑋0
5/2
√2𝜋 (𝑋0 + 𝑌)
(
𝑛
𝑋0
) 𝑒𝜌Ψ+ (𝑋0 ,𝑌) ,
𝑋0 + 𝑌
(113)
where Ψ+ (𝑋0 , 𝑌) is given in (74) and 𝐶 has the expansions in (31)–(33).
Note that the asymptotics are most complicated in (112),
which occurs when (𝑋, 𝑌) ≈ (𝑋0 , √𝑋0 − 𝑋0 ) and this is
where the curve 𝑌∗ (𝑋) hits the line 𝑋 = 𝑋0 . This intersection
occurs only for regions R2 and R3 . The integrand in (112) is
a meromorphic function of 𝜃, having double poles at all roots
of the Airy function Ai(⋅).
− Δ 𝐿 (𝑌) = (1 − 𝐴 1 ) {𝐴 1 + 𝑋0 + 𝑌0 +
−
𝑋0
𝑋0
−1
1 − 𝐴1
+ 𝐴 1 [1 −
2
(1 − 𝐴 1 )
(120)
𝑋 (𝑋 + 𝑌 )
2𝑋0
+ 0 0 20 ]}
1 − 𝐴1
(1 − 𝐴 1 )
and the expansion of 𝑄(0, 0) is given in Proposition 1
for the three cases R1 ∪ R2 , R2 ∩ R3 , and R3 ∪ R4
(when 𝑄(0, 0) ∼ 1 − 𝑋0 − 𝑌0 ).
Advances in Operations Research
15
(iii) R1 ∪ R2 ∪ R3 : 𝑌 − 𝑌𝑈 = 𝑦∗ /√𝜌 = 𝑂(𝜌−1/2 )
̃
̃
𝜋 (𝑘, 𝑟) ∼ 𝑄 (0, 0) 𝑒𝜌Ψ(0,𝑌(0)) 𝑒√𝜌𝑦∗ Ψ𝑌 (0,𝑌(0)) √𝜌
⋅
𝑘
𝜌𝑘
𝑋0
1
̃ (0))]
] exp [ 𝑦∗2 Ψ+,𝑌𝑌 (0, 𝑌
[𝑌0 +
𝑘!
𝑋0 + 𝑌0
2
⋅√
𝑋0
𝑌0
(121)
2
⋅
∞
(𝑋0 + 𝑌0 ) − 𝑋0
√(𝑋0 + 𝑌0 )3 + 𝑋0 (𝑌0 − 𝑋0 )
2
(∫
𝑎𝑐 𝑦∗
𝑒−𝑢 /2 𝑑𝑢) ,
We see that for each of the three cases the dependence of
𝜋(𝑘, 𝑟) on 𝑘 is of the form 𝜌𝑘 [𝐴 ∗ (𝑌)]𝑘 /𝑘! (times a function of
𝑌 or 𝑦∗ ), where the geometric ratio 𝐴 ∗ depends upon 𝑌 = 𝑟/𝜌
̃
and undergoes a transition when 𝑌 increases past 𝑌𝑈 = 𝑌(0),
which can occur only for regions R1 , R2 , and R3 . We note
that 𝐴 1 (𝑌𝑈) = 𝑌0 /(𝑋0 + 𝑌0 ) and 𝐷0 (𝑌𝑈) = 𝑌0 /(𝑋0 + 𝑌0 ),
so that 𝐴 ∗ is continuous along 𝑌 = 𝑌𝑈, with 𝐴 ∗ (𝑌𝑈) = 1 +
𝑌0 − 𝑋0 /(𝑋0 + 𝑌0 ), as indicated in (121). The expansion in
(117) develops a singularity as 𝑌 ↓ 𝑌𝑈, in view of the factor
[𝑌0 − 𝐴 1 (𝑋0 + 𝑌0 )]−1 . The expansion is also singular as 𝑌 ↑ 𝑌0
(for all regions R𝑗 ) and as 𝑌 ↓ 0 (for region R4 only). Then
we are approaching the corner points (0, 𝑌0 ) or (0, 0) of the
state space. Using the expansions
where
̃ (0))
Ψ (0, 𝑌
= 𝑋0 log 𝑋0 + 𝑌0 log 𝑌0 − (𝑋0 + 𝑌0 ) log (𝑋0 + 𝑌0 ) −
+
𝑋0 𝑌0
2
(𝑋0 + 𝑌0 )
+ 𝑌0 [
𝑋0
𝑋0 + 𝑌0
∫
𝑎𝑐 𝑦∗
(122)
[2 log (𝑋0 + 𝑌0 ) − log 𝑌0 ]
𝑋0
(𝑋0 + 𝑌0 )
2
− 1] log [1 −
𝑋0
2
(𝑋0 + 𝑌0 )
],
̃ (0)) = Ψ+,𝑌 (0, 𝑌
̃ (0))
Ψ𝑌 (0, 𝑌
̃ (0))
Ψ+,𝑌𝑌 (0, 𝑌
2
2
(124)
(𝑋0 + 𝑌0 )
𝑋0
𝑌 (𝑋0 + 𝑌0 ) + 𝑋0
],
[1 − 0
3
2
𝑌0 (𝑋0 + 𝑌0 ) + 𝑋0 (𝑌0 − 𝑋0 )
𝑋0 (𝑋0 + 𝑌0 ) − 𝑋0
𝑎𝑐
5/2
=
(𝑋0 + 𝑌0 )
√(𝑋0 + 𝑌0 )2 − 𝑋0
3
2
√𝑋0 𝑌0 √(𝑋0 + 𝑌0 ) + 𝑋0 (𝑌0 − 𝑋0 )√(𝑋0 + 𝑌0 ) + 𝑌0 − 𝑋0
𝜋 (𝑘, 𝑟) ∼ 𝑄 (0, 0) 𝑒𝜌Ψ(𝑋,𝑌0 )
2
𝑒−𝑢 /2 𝑑𝑢
√2𝜋,
{
{
{
∼{
2 2
{
{(𝑎𝑐 𝑦∗ )−1 exp (− 𝑎𝑐 𝑦∗ ) ,
2
{
𝑦∗ 󳨀→ −∞
(126)
𝑦∗ 󳨀→ +∞
(123)
2
= log 𝑌0 − log [(𝑋0 + 𝑌0 ) − 𝑋0 ] ,
=
∞
.
(125)
we can easily show that (121) matches to (117) (for region R1 ∪
R2 ∪ R3 in the intermediate limit where 𝑦∗ → +∞ and 𝑌 ↓
𝑌𝑈) and to (114) (with now 𝑦∗ → −∞ and 𝑌 ↑ 𝑌𝑈). Note
also that the ratio 𝑄(0, 0)/𝐶(𝜌) is asymptotically the same for
each of the three regions R1 , R2 , and R3 , in view of (28)–
(33). For any region R𝑗 , for 𝑘 = 𝑂(1) and 𝑌 ∈ (0, 𝑌0 ), 𝜋(𝑘, 𝑟)
is exponentially small in 𝜌.
Proposition 11. For 𝑅 − 𝑟 = 𝑝 = 𝑂(1) and 0 < 𝑋 < 𝑋0 one
has (for all regions R𝑗 in parameter space)
2
(1 − 𝑋0 − 𝑌0 ) + 4𝑌0
𝜌𝑝
𝑝 √1 − 𝐴 max
]
[
[𝑌0 (𝑒𝜏𝑈 − 1)]
−𝜏
𝑝!
√𝑒 𝑈 − 𝐴 max (1 − 𝑌0 − 𝑋)2 − 4𝐴 max (1 − 𝑋0 / (1 − 𝐴 max ))
1/4
,
(127)
− 𝑌0 𝜏𝑈 − 𝑋 log (1 − 𝐴 max 𝑒𝜏𝑈 ) .
where 𝜏𝑈 = 𝜏𝑈(𝑋) = 𝜏𝑈(𝑋; 𝑋0 , 𝑌0 ) is given by
(129)
𝜏𝑈 = − log [1 − 𝑌0 − 𝑋
2
− √ (1 − 𝑌0 − 𝑋) − 4𝐴 max (1 −
+ log [2 (1 −
𝑋0
)]
1 − 𝐴 max
(128)
𝑋0
)] ,
1 − 𝐴 max
where 𝐴 max is given in (69), 𝑄(0, 0) has the respective expansions in (28)–(30), and
Ψ (𝑋, 𝑌0 ) = 𝐴 max (1 − 𝑒𝜏𝑈 ) + 𝑋0 log (1 − 𝐴 max )
Thus (127) gives the expansion when there are only a few
secondary spaces empty and a fraction 𝑋/𝑋0 = 𝑘/𝑚 ∈ (0, 1)
of primary spaces occupied. The expressions can be simplified
in the limits 𝑋 → 0 (then 𝜏𝑈(0) = −log(𝐴 max )) and 𝑋 → 𝑋0
(then 𝜏𝑈(𝑋0 ) = 0), but then other “corner” expansions will
apply. Note that (127) is a completely explicit expression in
terms of 𝑝 and 𝑋.
Next we examine the four corners of the state space, where
(𝑋, 𝑌) = (0, 𝑌0 ), (0, 0), (𝑋0 , 0), and (𝑋0 , 𝑌0 ). We recall that
these ranges are important in that for region R1 ∩ R2 (𝑋0 ≈
1) most of the mass concentrates near the corner (𝑋0 , 0),
while for region R3 ∪ R4 (𝑋0 + 𝑌0 < 1) most of the mass
16
Advances in Operations Research
occurs near (𝑋0 , 𝑌0 ). We will start with the corner (0, 𝑌0 )
and proceed about the perimeter of the state space in a
counterclockwise manner.
(iv) (𝑋0 , 𝑌0 ) ∈ R4 (0 < 𝑌0 < √𝑋0 − 𝑋0 )
𝜋 (𝑘, 𝑟) ∼ √2𝜋𝜌√𝑋0 + 𝑌0 (1 − 𝑋0 − 𝑌0 )
Proposition 12. For 𝑘 = 𝑂(1), 𝑝 = 𝑅 − 𝑟 = 𝑂(1), and any
parameter region R𝑗 one has
𝜋 (𝑘, 𝑟) ∼ √2𝜋𝜌𝑄 (0, 0)
⋅[
𝑌0 (1 − 𝐴 max )
]
𝐴 max
⋅ 𝑒𝜌Ψ(0,𝑌0 ) √
⋅ 𝑒𝜌[−𝑋0 −𝑌0 +(𝑋0 +𝑌0 )log(𝑋0 +𝑌0 )]
𝜌𝑘+𝑝
𝑘
(1 + 𝑌0 )
𝑘!𝑝!
⋅
𝑝
(130)
1/4
1 − 𝐴 max
2
[(1 − 𝑋0 − 𝑌0 ) + 4𝑌0 ] ,
𝐴 max
𝑘
𝑘+𝑟
𝜌
𝑋0
𝑌0
] [
− 𝑌0 ] .
[𝑌0 +
𝑘!𝑟!
𝑋0 + 𝑌0
𝑋0 + 𝑌0
(v) (𝑋0 , 𝑌0 ) ∈ R2 ∪ R3 ; 𝑘 = 𝜌2/3 𝑋󸀠 , 𝑟 = 𝜌2/3 𝑌󸀠 ; 𝑋 =
𝜌−1/3 𝑋󸀠 , 𝑌 = 𝜌−1/3 𝑌󸀠
𝜋 (𝑘, 𝑟) ∼ 𝜌2/3 𝐶1 (𝜌)
where 𝑄(0, 0) is given in Proposition 1, 𝐴 max is in (69), and
Ψ (0, 𝑌0 ) = 𝐴 max − 1 + 𝑋0 log (1 − 𝐴 max )
(131)
+ 𝑌0 log (𝐴 max ) .
This gives the expansion when there are but a few primary
spaces occupied and a few secondary spaces empty. Next
we consider the corner point (0, 0), and this will typically
correspond to 𝑘, 𝑟 = 𝑂(1). The results will be very different
for region R1 compared to those for the remaining regions.
Proposition 13. For 𝑘, 𝑟 = 𝑂(1) and 𝜌 → ∞ one has
𝑖∞
⋅∫
−𝑖∞
𝜌𝑘+𝑟 𝑟
𝑘 1
𝑊0 (1 − 𝑊0 )
𝑘!𝑟!
2𝜋𝑖
󸀠
𝑒−𝜉 𝑧
𝑑𝑧
[𝐴𝑖 (𝑧)]2
2
(𝑌󸀠 )
{1
2
𝑊0
⋅ exp { 𝜌1/3 [
−
(𝑋󸀠 ) ]
2
2
𝑊0
(1 − 𝑊0 )
[
]
{
(138)
3
󸀠
𝑊 (𝑊 + 2)
3 }
1 (𝑌 )
− [
+ 0 0 4 (𝑋󸀠 ) ]} ,
2
6
𝑊
(1 − 𝑊0 )
]}
[ 0
𝜉󸀠 =
(i) (𝑋0 , 𝑌0 ) ∈ R1 (𝑋0 > 1)
(137)
𝑟
2/3
(1 − √𝑋0 )
𝑋01/6
[
𝑌󸀠
𝑋󸀠
−
].
𝑊0 1 − 𝑊0
(139)
𝜋 (𝑘, 𝑟)
1−2𝑟
𝜌 (𝑟 − 1)! 𝑘−𝑟 (𝑋0 − 1)
∼√
𝜌
2𝜋 𝑘!
√𝑋0
𝜌(−2+𝑋0 −𝑋0 log𝑋0 )
𝑒
, (132)
𝑟 ⩾ 1,
𝜋 (𝑘, 0) −
𝜌𝑘 𝑒−𝜌
𝜌
𝜌𝑘 𝑒𝜌(−2+𝑋0 −𝑋0 log𝑋0 )
∼ −√
.
𝑘!
2𝜋 (𝑘 − 1)!
𝑋0 − 1
(133)
(ii) (𝑋0 , 𝑌0 ) ∈ R1 ∩ R2 (𝑋0 − 1 = 𝛽/√𝜌 = 𝑂(𝜌−1/2 )),
𝑟⩾0
𝜌𝑘 𝑒−𝜌 1
Γ (𝑧 + 𝑟)
𝜋 (𝑘, 𝑟) ∼
𝑑𝑧,
∫
𝑘!𝑟! √2𝜋𝑖 Br+ 𝐷𝑧 (−𝛽) 𝐷𝑧−1 (−𝛽)
(134)
We note that (136) and (137) are continuous along the
curve 𝑋0 + 𝑌0 = √𝑋0 , corresponding to R3 ∩ R4 , in view of
the expansion in (36) for 𝐶1 in R3 . Thus the leading term for
𝑘, 𝑟 = 𝑂(1) for (𝑋0 , 𝑌0 ) ∈ R3 ∩ 𝑅4 can be obtained by either
using (36) to compute 𝐶1 in (136) or replacing 𝑌0 by √𝑋0 −𝑋0
in (137). The scale 𝑘, 𝑟 = 𝑂(𝜌2/3 ) must be considered (cf.
(138)), as the approximation in (136) cannot directly match
to those for 𝑘 = 𝑂(1), 0 < 𝑌 < 𝑌𝑈 (cf. (114)) or 𝑟 = 𝑂(1),
0 < 𝑋 < 𝑋0 (cf. (96)). We can view (136) as a special case of
(138), letting 𝑋󸀠 , 𝑌󸀠 → 0 in the latter and noting that
(2𝜋𝑖)−1 ∫
𝑖∞
−𝑖∞
[Ai (𝑧)]−2 𝑑𝑧 = 1.
(140)
(136)
For region R4 we see from Figure 6 that only state space
region D0 meets the corner (𝑋, 𝑌) = (0, 0). Then (137)
matches directly to (98), in the limits 𝑘 → ∞, 𝑋 → 0, and to
(117), in the limits 𝑟 → ∞, 𝑌 → 0. Note that, for region R4 ,
𝐴 1 (0) = 1 − 𝑋0 − 𝑌0 , which follows from (118) (this is true
also for R3 , but then (117) applies only for 𝑌 > 𝑌𝑈 and thus
not near the corner (0, 0)). We can obtain a result analogous
to (138) for the transitional range R3 ∩R4 , but we do not give
that for the sake of brevity.
where 𝐶1 is given by (34), (35), or (36) for R2 , R2 ∩R3 ,
or R3 , respectively.
Proposition 14. For 𝑘 = 𝑚 − 𝑂(1) and 𝑟 = 𝑂(1), or for 𝑘 =
𝑚 − 𝑂(√𝜌) and 𝑟 = 𝑂(√𝜌), the expansions of 𝜋(𝑘, 𝑟) are as
follows:
where 𝐵𝑟+ is as in (94), and when 𝛽 = 0
𝜋 (𝑘, 𝑟) ∼
𝜌𝑘 𝑒−𝜌 −𝑟
2
𝑘!
(𝛽 = 0) .
(135)
(iii) (𝑋0 , 𝑌0 ) ∈ R2 ∪ R3 (𝑋0 < 1, 𝑌0 + 𝑋0 > √𝑋0 )
𝜋 (𝑘, 𝑟) ∼ 𝜌2/3 𝐶1 (𝜌)
𝜌𝑘+𝑟 𝑒−𝜌 𝑟
𝑘
𝑊0 (1 − 𝑊0 ) ,
𝑘!𝑟!
Advances in Operations Research
17
(i) (𝑋0 , 𝑌0 ) ∈ R1 , 𝑚 − 𝑘 = 𝑛 ⩾ 0
(vi) (𝑋0 , 𝑌0 ) ∈ R2 ∪ R3 ∪ R4 , 𝑚 − 𝑘 = 𝑛 ⩾ 0
1
𝑒𝜌(−1+𝑋0 −𝑋0 log𝑋0 ) (𝑋0 − 1) 𝑋0−𝑟−1 ,
𝜋 (𝑘, 𝑟) ∼
√2𝜋𝜌𝑋0
𝜋 (𝑘, 𝑟) ∼ 𝜌1/3 𝐶1 (𝜌)
𝑟 ⩾ 1 (141)
𝜋 (𝑘, 0) ∼
1
1
𝑒(−1+𝑋0 −𝑋0 log𝑋0 ) (𝑋0𝑛 −
).
𝑋0
√2𝜋𝜌𝑋0
1/3
⋅ 𝑒𝜌Φ(𝑋0 ,0) 𝑒𝜌
⋅ exp [−
(ii) (𝑋0 , 𝑌0 ) ∈ R1 ∩ R2 (𝑋0 − 1 = 𝛽/√𝜌), 𝑚 − 𝑘 = 𝑛 ⩾ 0
∼ 𝜌−1 𝜌−𝑧0 /2
2
(142)
𝑛𝑧0
𝑒−𝛽 /4
Γ (𝑧0 + 𝑟 + 1) [
+ 1] ,
𝑧0 + 𝑟
𝑟!Δ 0 (𝛽)
where 𝑧0 is the minimal root of the parabolic cylinder
function 𝐷𝑧 (⋅); that is, 𝑧0 = 𝑧0 (𝛽) = min{𝑧 :
𝐷𝑧 (−𝛽) = 0}, and Δ 0 (𝛽) = −(𝑑/𝑑𝑧)𝐷𝑧 (−𝛽)|𝑧=𝑧0 (𝛽) .
(iii) (𝑋0 , 𝑌0 ) ∈ R1 ∩ R2 (𝑋0 − 1 = 𝛽/√𝜌), 𝑘 = 𝜌 + √𝜌𝛼,
and 𝑟 = √𝜌Ω, with 𝛼 ⩽ 𝛽 and Ω > 0
2
𝜋 (𝑘, 𝑟) ∼ 𝜌−1
1 𝑖∞ 𝑒−𝛼 /4 𝐷𝑧 (−𝛼) Ω𝑧−1
𝑑𝑧
∫
2𝜋𝑖 −𝑖∞ 𝐷𝑧 (−𝛽) 𝐷𝑧−1 (−𝛽)
(143)
which when 𝛽 = 0 simplifies to
𝜋 (𝑘, 𝑟) ∼ 𝜌−1 √
2
2
(Ω − 𝛼) 𝑒−(Ω−𝛼) /2 ;
𝜋
(144)
𝛼 < 0, Ω > 0.
(iv) (𝑋0 , 𝑌0 ) ∈ R1 ∩ R2 (𝑋0 − 1 = 𝛽/√𝜌), 𝑘 = 𝜌 + √𝜌𝛼,
𝛼 < 𝛽, 𝑟 = 𝑂(1)
𝜋 (𝑘, 𝑟)
2
∼ 𝜌−1/2 𝜌−𝑧0 /2 𝑒−𝛼 /4
(145)
𝐷𝑧0 (−𝛼)
Γ (𝑟 + 𝑧0 )
.
𝑟!
Δ 0 (𝛽) 𝐷𝑧0 −1 (−𝛽)
−1/4
=
(v) (𝑋0 , 𝑌0 ) ∈ R1 ∩ R2 with now 𝑋0 − 1 = 𝛽∗ 𝜌
−1/4
𝑂(𝜌 ) and 𝛽∗ > 0, 𝑟 = 𝑂(1), 𝑘 = 𝜌+ √𝜌𝛼, 𝛼 = 𝑂(1)
𝜋 (𝑘, 𝑟) ∼
∞
2
𝛽∗3
(𝑢 + 𝛼)𝑟−1
⊗
𝑒−𝑢 /2 𝑑𝑢,
∫
∗
𝑟+1
2
2𝜋𝜌1/4
−𝛼 (𝑢 + 𝛼 + 𝛽 )
∗
(146)
𝑟 ⩾ 1,
2
𝜋 (𝑘, 0) −
∞
𝜌𝑘 𝑒−𝜌
𝛽
𝑒−𝑢 /2
𝑑𝑢,
∼ − ∗1/4 ⊗∗ ∫
2
𝑘!
2𝜋𝜌
−𝛼 𝑢 + 𝛼 + 𝛽∗
⊗∗ = exp [−
1
√𝜌 2 𝜌1/4 3
𝛽∗ +
𝛽∗ − 𝛽∗4 ] .
2
6
12
(147)
(148)
4/3
− √𝑋0 )
Φ∗ (0) (1
𝐴𝑖󸀠 (𝑟0 ) 𝑋01/3
(149)
1
log (1 − √𝑋0 )] ,
2√𝑋0
2/3
Φ∗ (0) = −𝑟0
𝜋 (𝑘, 𝑟)
𝑛
𝜌𝑟 𝑊0𝑟
(√𝑋0 ) (𝑛 + 1)
𝑟!
(1 − √𝑋0 )
𝑋01/6
log (1 − √𝑋0 ) ,
1
1
Φ (𝑋0 , 0) = 𝑊0 log (𝑋0 ) − 𝑋0 log (𝑋0 ) + √𝑋0
2
2
(150)
(151)
− 1,
and 𝐶1 is given in Proposition 2 for the different regions
R𝑗 .
For region R1 , (141) show that the dependence of 𝜋(𝑘, 𝑟)
on 𝑚 − 𝑘 and 𝑟 is quite simple, but we have to distinguish the
cases 𝑟 = 0 and 𝑟 ⩾ 1. The same is true for regions R2 , R3 ,
and R4 , where (149) applies for all 𝑟 ⩾ 0, except now the form
of 𝐶1 in (149) is different for the five cases in Proposition 2.
However, when 𝑋0 ≈ 1, the asymptotic structure of 𝜋(𝑘, 𝑟)
is quite complicated, and we must consider separately the
scales 𝑋0 − 1 = 𝑂(𝜌−1/2 ) (𝑚 − 𝜌 = 𝑂(√𝜌)) and 𝑋0 − 1 =
𝑂(𝜌−1/4 ) (𝑚 − 𝜌 = 𝑂(𝜌3/4 )). For X0 − 1 = 𝑂(𝜌−1/2 ) we
obtain the limiting density in (53) or (143), as the limit of
𝜌𝜋(𝜌 + √𝜌𝛼, √𝜌Ω) = 𝜌𝜋(𝑚 + √𝜌(𝛼 − 𝛽), √𝜌Ω) for 𝜌 → ∞.
This expansion applies for Ω > 0, for any 𝛼 ⩽ 𝛽, but becomes
invalid as Ω → 0. For Ω → 0 the asymptotic behavior of the
contour integral in (143) is determined by the singularity at
𝑧 = 𝑧0 (𝛽), and the density behaves as 𝑂(Ω𝑧0 −1 ) for Ω → 0 and
𝛼 < 𝛽. This corresponds to either an integrable singularity or
a zero of the density (unless 𝛽 = 0 then 𝑧0 = 1) and in either
case indicates a nonuniformity in the asymptotics. Thus we
need the expansion in (145) for 𝑟 = 𝑂(1). For 𝑟 → ∞ we
have, by Stirling’s formula,
Γ (𝑟 + 𝑧0 ) Γ (𝑟 + 𝑧0 )
𝑧 −1
=
∼ 𝑟𝑧0 −1 = (√𝜌Ω) 0
𝑟!
Γ (𝑟 + 1)
(152)
and then (145) matches to (143) in the intermediate limit
where 𝑟 → ∞ but Ω = 𝑟/√𝜌 → 0. The expansion in (145)
itself breaks down when 𝛼 → 𝛽, since by the definition of
𝑧0 (𝛽) we have 𝐷𝑧0 (𝛽) (−𝛼) → 0 as 𝛼 ↑ 𝛽. Then for 𝛼 ∼ 𝛽 we
have the expansion in (142), which holds for 𝑛 = 𝑂(1) and we
note that 𝑛 = 𝑚 − 𝑘 = (𝛽 − 𝛼)/√𝜌 so that 𝛼 = 𝛽 − 𝑛𝜌−1/2 .
We can show also that the expansions for 𝑛, 𝑟 = 𝑂(1)
(cf. (141), (142), and (149)) match in appropriate intermediate
limits. Consider (142) for 𝛽 → −∞ (then we are moving into
the range 𝑋0 < 1). We now have 𝑧0 → ∞ and more precisely
𝑧0 (𝛽) =
𝛽2
𝛽 2/3 1
− 𝑟0 (− ) − + 𝑜 (1) , 𝛽 󳨀→ −∞ (153)
4
2
2
18
Advances in Operations Research
so the approximation to the minimal root of the parabolic
cylinder function 𝐷𝑧 (−𝛽) involves the maximal root of the
Airy function Ai(⋅). Thus, for 𝛽 → −∞ we have
𝑧 /2
Δ 0 (𝛽) ∼ 𝑧00 𝑒−𝑧0 /2 (
1/3
2
)
−𝛽
√2𝜋Ai󸀠 (𝑟0 )
𝛽 −𝛽2 /2 −1 {1,
𝜌 {
𝑒
√2𝜋
𝑛 + 1,
{
(154)
which can be obtained by approximated 𝐷𝑧 (−𝛽) by Airy
functions in the double limit where 𝛽 → −∞ and 𝑧 →
∞, with 𝑧 − 𝛽2 /4 = 𝑂[(−𝛽)2/3 ]. Using (154) in (142) and
expanding Γ(𝑧0 (𝛽) + 𝑟 + 1) by Stirling’s formula (since now
𝑧0 → ∞) for a fixed 𝑟, (142) becomes
2
𝛽 4/3 𝑒−𝛽 /4 𝑧0 /2 −𝑧0 /2 𝑧0𝑟
𝑧 𝑒
𝜌−1 𝜌−𝑧0 /2 (𝑛 + 1) (− )
.
2
𝑟!
Ai󸀠 (𝑟0 ) 0
approximated by 1 if 𝑟 ⩾ 1 and is equal to 𝑛 + 1 if 𝑟 = 0. Now
also Γ(𝑧0 + 𝑟 + 1)/𝑟! → 1 and (142) becomes, for 𝛽 → +∞,
(155)
Then (155) must agree with the expansion of (149) as 𝑋0 ↑ 1.
Then we use the fact that 𝜌1/3 𝐶1 ∼ 𝜌−1/3 in region R2 . Also,
as 𝑋0 → 1, 𝑊0 = (1 − √𝑋0 )2 ∼ 𝛽2 /(4𝜌),
𝛽2
−𝛽
3
log (
) − 𝛽2 + 𝑜 (1) ,
4
2√𝜌
8
− 𝜌1/3 𝑟0 𝑋0−1/6 (1 − √𝑋0 )
2/3
𝜋 (𝑘, 𝑟) ∼
(𝑟 − 1)!𝛽∗1−2𝑟 𝜌−𝑟/2 ⊗∗ 𝜌(−1+𝑋−𝑋log𝑋)
𝑒
,
2𝜋√𝑋𝜌1/4 (1 − 𝑋)𝑟
for 𝑋 − 1 = 𝛼∗ 𝜌−1/4 = 𝑂(𝜌−1/4 ) with 0 < 𝛼∗ < 𝛽∗ we have
𝜋 (𝑘, 𝑟) ∼
(156)
log (1 − √𝑋0 )
∫
⋅ exp {[
2/3
𝛽
𝛽
− 𝑟0 (− )
4
2
𝛽
1
− ] log (−
)
2
2√𝜌
(𝑢 + 𝛼 +
∼
(157)
(161)
∞
−𝛼
But by using (153) in (155) we obtain again the expression in
(157), which verifies the matching.
Now consider (141) for 𝑋0 ↓ 1 and (142) for 𝛽 → +∞. We
now have
𝛽 −𝛽2 /2
,
𝑒
√2𝜋
𝛽∗−2−2𝑟
2
𝑟+1
𝛽∗2 )
𝑒−𝑢 /2 𝑑𝑢
(162)
−𝛼2 /2
(𝑟 − 1)!𝑒
−𝑟
(−𝛼) .
For 𝛼 → +∞ we have
∫
3
− 𝛽2 } .
8
𝑧0 (𝛽) ∼
(𝑢 + 𝛼)𝑟−1
∞
−∞
4/3
𝛽
1 𝛽2
(𝑛 + 1)
(−
( )
)
󸀠
𝑟! 4
2√𝜌
Ai (𝑟0 )
2
⊗∗ 𝛽∗3
, 𝑟 ⩾ 1;
√2𝜋𝜌3/4 𝛼∗2
and for the scale 𝑚−𝑘 = 𝑛 = 𝑂(1) we have 𝛼∗ ∼ 𝛽∗ so the last
factor in (161) may be approximated by 𝛽∗3 𝛼∗−2 ∼ 𝛽∗ . We can
easily verify that as 𝑋 ↑ 1, (159) matches to (146) as 𝛼 → −∞.
Note that in this limit we have
Using (156) we see that as 𝑋0 ↑ 1, (149) becomes
𝑟
(160)
𝑟 ⩾ 1;
𝛽 2/3
𝛽
∼ −𝑟0 (− ) log (−
).
2
2√𝜌
𝜌−1/3
(159)
𝑟 = 0,
and the above clearly agrees with (141), when we expand these
for 𝑋0 → 1.
The results in (146)–(148) assume the scaling 𝑋0 − 1 =
𝛽∗ 𝜌−1/4 , and these are needed to asymptotically connect the
parameter ranges 𝑋0 −1 = 𝑂(𝜌−1/2 ) and 𝑋0 > 1. It is only near
the corner (𝑋, 𝑌) = (𝑋0 , 0) that we must consider this scaling
(and indeed also the 𝛽-scale). For other ranges of 𝑋 and 𝑟 =
𝑂(1), we can get the expansions of 𝜋(𝑘, 𝑟) as limiting cases
of other expansions, as they lie in the asymptotic matching
range where 𝑋0 → 1 but (𝑋0 − 1)√𝜌(= 𝛽) → ∞. For a fixed
𝛽∗ > 0 and 0 < 𝑋 < 1 we have
1
𝜌 [−1 + √𝑋0 − 𝑋0 log 𝑋0 + 𝑊0 log (1 − √𝑋0 )]
2
=
𝑟⩾1
(𝑢 + 𝛼)𝑟−1
(𝑢 + 𝛼 +
2
𝑟+1
𝛽∗2 )
𝑒−𝑢 /2 𝑑𝑢 ∼
√2𝜋
√2𝜋
=
2
𝛼
√𝜌𝛼∗2
(163)
and then (146) clearly matches to (160) and in fact contains
the latter as a limiting case. When 𝑟 = 0 we have, for 0 < 𝑋 <
1,
𝜋 (𝑘, 0) −
𝜌𝑘 𝑒−𝜌
⊗∗
−√𝑋
𝑒𝜌(−1+𝑋−𝑋log𝑋) ;
∼
𝑘!
1 − 𝑋 2𝜋𝜌3/4 𝛽∗
(164)
for 𝑋 − 1 = 𝛼∗ 𝜌−1/4 , 𝛼∗ > 0, we have
√2𝜋 𝛽2 /4
Δ 0 (𝛽) ∼
𝑒 ,
𝛽
(158)
𝛽 󳨀→ +∞
so that 𝑧0 is exponentially small and Δ 0 is exponentially large.
The last factor in (142), namely, 𝑛𝑧0 /(𝑧0 + 𝑟) + 1, can be
𝜋 (𝑘, 0) −
𝜌𝑘 𝑒−𝜌
⊗∗ 𝛽∗
;
∼−
√2𝜋𝜌1/2 𝛼∗
𝑘!
(165)
and for 𝑛 = 𝑚 − 𝑘 = √𝜌(𝛽 − 𝛼) = 𝜌3/4 (𝛽∗ − 𝛼∗ ) = 𝑂(1)
𝜋 (𝑘, 0) ∼
⊗∗
𝛽 (𝑛 + 1) .
√2𝜋𝜌3/4 ∗
(166)
Advances in Operations Research
19
We can again easily verify that (147) matches to (164) for 𝑋 →
1 and 𝛼 → −∞ and to (165) for 𝛼 → +∞ and 𝛼∗ = 𝛼𝜌−1/4 →
0. Recalling that 𝑘 = 𝜌𝑋 = 𝜌 + 𝜌3/4 𝛼∗ = 𝜌 + 𝜌3/4 𝛽∗ − 𝑛, we
have
where
F (], 𝑇) =
𝑛𝛽
⊗∗
[exp ( 1/4∗ ) − 1 + 𝑂 (𝜌−3/4 )]
𝜌
√2𝜋𝜌
∼
⊗∗
𝛽 𝑛
√2𝜋𝜌3/4 ∗
𝐴𝑖 (] + 𝑟𝑗 )
𝑗=0
𝐴𝑖󸀠 (𝑟𝑗 )
(167)
𝑇1 =
Proposition 15. For 𝑘 = 𝑚 − 𝑜(𝜌) and 𝑟 = 𝑅 − 𝑜(𝜌) one uses
the variables 𝑚 = 𝑚 − 𝑘 = 𝜌1/3 ]1 and 𝑝 = 𝑅 − 𝑟 = 𝜌2/3 𝑇1 , and
the expansions are as follows:
(i) (𝑋0 , 𝑌0 ) ∈ R1 ∪ R2 ∪ R3 ∪ R4 , 𝑛 = 𝑂(1), 𝑝 = 𝑂(1)
𝑄 (0, 0) 1
1 − 𝑤 −𝑛−1 −𝑝−1
𝑤
𝑑𝑤,
𝑧
∮
𝑋0 2𝜋𝑖 𝑧− − 𝑤 +
(170)
𝑟𝑗 𝑇
𝑒
(168)
𝑋0
)
1 − √𝑋0
where 𝑧± (𝑤) are defined in (50) and the expansions
for 𝑄(0, 0) are in Proposition 1; in particular for region
R3 ∪ R4 one has 𝑄(0, 0) ∼ 1 − 𝑋0 − 𝑌0 > 0 and then
(168) is a proper discrete distribution over the range(s)
𝑛 ⩾ 0, 𝑝 ⩾ 0.
(ii) (𝑋0 , 𝑌0 ) ∈ R4 , 𝑚 − 𝑘 = 𝜌1/3 ]1 = 𝑂(𝜌1/3 ), ]1 > 0,
𝑅 − 𝑟 = 𝜌2/3 𝑇1 = 𝑂(𝜌2/3 )
𝜋 (𝑘, 𝑟) ∼ 𝜌
2/3
(1 − √𝑋0 )
5/3
⋅
(1 − √𝑋0 )
𝑋0−1/6
(√𝑋0 − 𝑋0 − 𝑌0 ) (𝑊0 + 𝑌0 )
⋅(
F (], 𝑇) (√𝑋0 )
𝑝
𝑌0
)
𝑌0 + 𝑊0
⋅
𝑊0 𝑋0−1/2
(√𝑋0 − 𝑋0 − 𝑌0 ) (𝑊0 + 𝑌0 )
⋅(
⋅ exp [−
𝑊 (𝑊 + 2𝑌0 ) 3
𝜌1/3 𝑊0
𝑇 ],
𝑇2 − 0 0
2𝑌0 (𝑊0 + 𝑌0 ) 1 6𝑌02 (W0 + 𝑌0 )2 1
(√𝑋0 )
𝑛
𝑝
√𝑋0 (𝑌0 + 𝑊0 )
𝑌0
) [𝑛 +
]
𝑌0 + 𝑊0
(𝑌0 + 𝑊0 ) √𝑋0 − 𝑌0
(172)
∞
𝑊0
⋅ [ ∑𝑒𝑟𝑗 𝑇 ] exp [−𝜌1/3
𝑇12
2𝑌
(𝑊
+
𝑌
)
0
0
0
[𝑗=0 ]
−
𝑊0 (𝑊0 + 2𝑌0 )
2
6𝑌02 (𝑊0 + 𝑌0 )
𝑇13 ] ,
𝑇 > 0.
(iv) (𝑋0 , 𝑌0 ) ∈ R3 ∩ R4 , 𝑚 − 𝑘 = 𝑛 = 𝜌1/3 ]1 = 𝑂(𝜌1/3 ),
]1 ⩾ 0, 𝑅 − 𝑟 = 𝑝 = 𝜌2/3 𝑇1 = 𝑂(𝜌2/3 )
⋅(
−
4/3
𝑋0−1/3 (√𝑋0 )
𝑛
𝑝
𝑊0
𝑌0
) exp [−𝜌1/3
𝑇2
𝑌0 + 𝑊0
2𝑌0 (𝑊0 + 𝑌0 ) 1
𝑊0 (𝑊0 + 2𝑌0 )
2
6𝑌02 (𝑊0 + 𝑌0 )
(173)
𝑇13 ] F (], 𝑇; 𝛿1 ) ,
where
𝛿1 𝑢
⋅ (∫ 𝑒
𝜃
(169)
𝑇,
𝜋 (𝑘, 𝑟) ∼ 𝜌−1 (1 − 𝑋0 − 𝑌0 )
∞
𝑛
(171)
and 𝑟𝑗 are the roots of 𝐴𝑖(⋅), ordered as 0 > 𝑟0 > 𝑟1 >
⋅⋅⋅.
(iii) (𝑋0 , 𝑌0 ) ∈ R4 , 𝑚−𝑘 = 𝑛 = 𝑂(1), 𝑅−𝑟 = 𝑝 = 𝜌2/3 𝑇1 =
𝑂(𝜌2/3 )
F (], 𝑇; 𝛿1 ) =
(1 − 𝑋0 − 𝑌0 )
,
(𝑊0 + 𝑌0 ) 𝑋01/6
𝜋 (𝑘, 𝑟) ∼ 𝜌−1/3 (1 − √𝑋0 )
−2/3
, 𝑇 > 0,
1/3
]1 = ] (
and this agrees with (166) for 𝑛 → ∞, which verifies the
matching between (165) and (166). Thus we have given 𝜋(𝑘, 𝑟)
for all ranges of 𝑘 for 𝑟 = 𝑂(1) and the scaling 𝑚 = 𝜌 + 𝜌3/4 𝛽∗ ,
𝛽∗ > 0. However, only for the range 𝑘 = 𝜌 + 𝑂(√𝜌) do we get
the new results in (146) and (147).
Next we examine the corner (𝑋, 𝑌) = (𝑋0 , 𝑌0 ), so both
primary and secondary spaces will be nearly full. For regions
R3 ∪ R4 most of the mass is in the range. We will need to
consider the scales 𝑚 − 𝑘, 𝑅 − 𝑟 = 𝑂(1) and also 𝑚 − 𝑘 =
𝑂(𝜌1/3 ). Consider 𝑅 − 𝑟 = 𝑂(𝜌2/3 ).
𝜋 (𝑘, 𝑟) ∼
∞
=∑
𝜌𝑘 𝑒−𝜌
𝛽∗
⊗∗ 𝛽∗ 𝜌𝑘 𝑒−𝜌
⊗∗
−
=
−
1/2
√2𝜋𝜌 𝑘!
𝑘!
𝑘!
√2𝜋𝜌 𝛽∗ − 𝜌−3/4 𝑛
=
𝑖∞
𝐴𝑖 (] + 𝜃) 𝜃𝑇
1 𝑑
𝑒 𝑑𝜃}
{∫
2𝜋𝑖 𝑑𝑇 −𝑖∞ 𝜃𝐴𝑖 (𝜃)
1 𝑖∞ 𝐴𝑖 (] + 𝜃) 𝜃(𝑇−𝛿1 )
𝑒
∫
2𝜋𝑖 −𝑖∞ [𝐴𝑖 (𝜃)]2
(174)
𝐴𝑖 (𝑢) 𝑑𝑢) 𝑑𝜃,
𝑋0 + 𝑌0 = √𝑋0 + 𝜌−1/3 𝛿∗ ,
𝛿∗ = 𝑋01/6 (1 − √𝑋0 )
1/3
(175)
𝛿1
and (], 𝑇) is given by (171) in terms of (]1 , 𝑇1 ).
20
Advances in Operations Research
When (𝑋0 , 𝑌0 ) ∈ R3 ∪ R4 most mass concentrates
on the (𝑛, 𝑝) scale so there tend to be but a few available
primary and secondary spaces. For (𝑋0 , 𝑌0 ) ∈ R1 ∪R2 (𝑋0 +
𝑌0 > 1), the result in (168) still applies but now 𝑄(0, 0) is
exponentially small (cf. (30)). For 𝑋0 + 𝑌0 − 1 = 𝑂(𝜌−1/2 )
we have 𝑄(0, 0) = 𝑂(𝜌−1/2 ). Later we will study the behavior
of the contour integral in (168) as 𝑛 and/or 𝑝 → ∞, and
we will see that for R1 ∪ R2 we can get exponential growth
in certain sectors, such as if 𝑝 → ∞ with 𝑛 = 𝑂(1). From
Proposition 15 we also see that the probabilities of finding
𝑂(𝜌1/3 ) empty primary spaces and 𝑂(𝜌2/3 ) secondary spaces
are quite complicated, and their estimation involves contour
integrals of Airy functions. These probabilities are however
1/3
quite small, in view of the factors (√𝑋0 )𝑛 = (√𝑋0 )𝜌 ]1 and
region(s) R1 ∪ R2 ∪ R3 , D+ meets D− for region R2 ∪ R3 ,
and D0 meets D− for region R4 (see also Figures 3–6).
̃
≡
Proposition 16. For (𝑋0 , 𝑌0 ) ∈ R1 ∪ R2 ∪ R3 , 𝑌 − 𝑌(𝑋)
𝜂∗ /√𝜌 = 𝑂(𝜌−1/2 ), and 0 < 𝑋 < 𝑋0 one has
̃ (𝑋))
𝜋 (𝑘, 𝑟) ∼ 𝐶 (𝜌) 𝐾+ (𝑋, 𝑌
⋅
̃ (𝑋)) 𝜂∗
+ √𝜌Ψ+,𝑌 (𝑋, 𝑌
(176)
1
̃ (𝑋)) 𝜂2 ] ,
+ Ψ+,𝑌𝑌 (𝑋, 𝑌
∗
2
2/3
[𝑌0 /(𝑌0 + 𝑊0 )]𝑝 = [𝑌0 /(𝑌0 + 𝑊0 )]𝜌 𝑇1 in (169) and (173). In
(173) we can use (175) and rewrite the result in terms of 𝑋0
and 𝛿∗ , thus eliminating 𝑌0 .
Finally we give results that apply near the transition
curves that separate D0 , D− , and D+ . Note D0 meets D+ for
∞
2
1
̃ (𝑋))
𝑒−𝑢 /2 𝑑𝑢) exp [𝜌Ψ+ (𝑋, 𝑌
(∫
√2𝜋 𝜂∗ /√𝑏(𝑡∗ )
where
̃ (𝑋)) = 𝑋0 log 𝑋0 + 𝑋0 + 𝑌0 − 1 − 2𝑋0 log (𝑋0 + 𝑌0 ) +
Ψ+ (𝑋, 𝑌
𝑌0
𝑌 𝑒𝑡∗
̃ (𝑋)
(1 − 𝑒𝑡∗ ) − 𝑋 log (1 − 0
)−𝑌
𝑋0 + 𝑌0
𝑋0 + 𝑌0
(177)
𝑡∗
⋅ log [1 − (1 − 𝑋0 − 𝑌0 ) 𝑒 ] ,
̃ (𝑋))] ,
𝑡∗ = log 𝑌0 − log [𝑌0 − (𝑋0 + 𝑌0 ) (𝑌0 − 𝑌
(178)
̃ (𝑋)) = log [
Ψ+,𝑌 (𝑋, 𝑌
̃ (𝑋))
𝑌0 − (𝑋0 + 𝑌0 ) (𝑌0 − 𝑌
],
̃ (𝑋)
(𝑋0 + 𝑌0 ) 𝑌
(179)
̃ (𝑋)) = −𝑒𝑡∗
Ψ+,𝑌𝑌 (𝑋, 𝑌
𝑠𝑌0 − (1 − 𝑋0 − 𝑌0 ) 𝑡𝑌0
,
1 − (1 − 𝑋0 − 𝑌0 ) 𝑒𝑡∗
(180)
𝑠𝑌0
=
(𝑋0 + 𝑌0 ) (𝑋0 + 𝑌0 + 𝑌0 𝑒−𝑡∗ 𝑡𝑌0 )
2
(𝑋0 + 𝑌0 ) + (𝑒−𝑡∗ − 1) 𝑋0
,
(181)
2
𝑡𝑌0 = 𝑒𝑡∗ (1 − 𝑒𝑡∗ ) (𝑋0 + 𝑌0 ) [𝑋0 𝑒𝑡∗ + (𝑋0 + 𝑌0 ) ]
2 2
2
−1
⋅ {(𝑋0 + 𝑌0 ) (𝑋02 − 𝑌02 − 𝑋0 ) + 𝑒𝑡∗ [𝑋0 − (𝑋0 + 𝑌0 ) ] + 𝑒2𝑡∗ (2 − 𝑒𝑡∗ ) 𝑋0 𝑌0 + 𝑒3𝑡∗ 𝑌0 (𝑋 + 𝑌0 ) } ,
2
𝑌0 √𝑋0 [(𝑋0 + 𝑌0 ) − 𝑋0 ]
̃ (𝑋)) =
𝐾+ (𝑋, 𝑌
3/2
̃ (𝑋)√𝑋0 𝑌0 − (𝑋0 + 𝑌0 )2 (𝑌0 − 𝑌
̃ (𝑋))
√2𝜋 󵄨󵄨󵄨󵄨Δ0+ 󵄨󵄨󵄨󵄨1/2 (𝑋0 + 𝑌0 )3 √𝑌
󵄨 󵄨󵄨 󵄨
−2
𝑋0 𝑌0 󵄨󵄨󵄨󵄨Δ0 󵄨󵄨󵄨󵄨 󵄨󵄨󵄨󵄨Δ0+ 󵄨󵄨󵄨󵄨
𝑌0 𝑒𝑡∗
−𝑡∗
𝑏 (𝑡∗ ) =
−
𝑌
)
𝑒
−
]
,
[(1
−
𝑋
0
0
2
𝑋0 + 𝑌0
(𝑋0 + 𝑌0 ) [(𝑋0 + 𝑌0 ) − 𝑋0 ]
Δ0+ = Δ0+ (𝑋) =
𝐷∗ =
(182)
,
1
2
[𝐷∗ (𝐸∗ − 𝐷∗ ) 𝑒𝑡∗ − 2𝐷∗ (1 − 𝐷∗ ) − (𝐷∗ − 𝐸∗ ) 𝑒−𝑡∗ + (𝐷∗ + 𝐸∗ − 2𝐷∗ 𝐸∗ ) 𝑒−2𝑡∗ ] ,
1 − 𝐸∗
𝑌0
,
𝑋0 + 𝑌0
(183)
(184)
(185)
(186)
𝐸∗ = 1 − 𝑋0 − 𝑌0 ,
2
0
0
−𝑡∗
Δ = Δ (𝑋) = (𝑒
𝑌0 ((𝑋0 + 𝑌0 ) − 𝑋0 ) 𝑡
𝑌0
3
+
𝑒 ∗ + 𝑒−𝑡∗ (1 − 2 (𝑋0 + 𝑌0 ) + (𝑋0 + 𝑌0 ) 𝑋0−1 )] .
𝑋0 + 𝑌0
𝑋0 (𝑋0 + 𝑌0 )
[
]
− 1) [
(187)
Advances in Operations Research
21
̃
𝑌(𝑋)
is given by (56), and 𝐶(𝜌) has the expansions in (31)–(33)
for the different regions R𝑗 .
For 𝜂∗ → −∞ we are moving into region D+ and the
integral in (176) approaches √2𝜋, and then (176) becomes
simply the expansion of the D+ result in (60), about Y =
̃
𝑌(𝑋).
Thus the matching of (176) to D+ is immediate, and
we can also verify the matching to (65), by expanding the
̃
latter as 𝑌 ↓ 𝑌(𝑋),
and (176) for 𝜂∗ → +∞. We recall that
Proposition 2 shows that the expansion of 𝐶0 (𝜌)/𝐶(𝜌) is the
same for regions R1 , R2 , and R3 . The function 𝑡∗ = 𝑡∗ (𝑋)
̃
in (178) is obtained by setting 𝑠 = 𝑌0 and then 𝑌 = 𝑌(𝑋)
in (62). The function Δ0+ (𝑋) is obtained by setting 𝑠 = 𝑌0
and 𝑡 = 𝑡∗ (𝑋) in (64) and corresponds to the values of this
̃
Jacobian along the curve 𝑌 = 𝑌(𝑋).
Note also that 𝑡∗ (𝑋) can
̃
be obtained by setting 𝐴 = 𝑌0 /(𝑋0 + 𝑌0 ) and 𝑌 = 𝑌(𝑋)
in
0
(68), and Δ (𝑋) corresponds to the Jacobian in (71) along the
̃
curve 𝑌 = 𝑌(𝑋),
with 𝜏 = 𝑡∗ (𝑋).
Proposition 17. For (𝑋0 , 𝑌0 ) ∈ R2 ∪ R3 , 𝑌 − 𝑌∗ (𝑋) =
𝜌−1/3 𝜉1 = 𝑂(𝜌−1/3 ), and 0 < 𝑋 < 𝑋0 one has
𝜋 (𝑘, 𝑟) ∼ 𝜌−1/2 𝐶 (𝜌) exp [𝜌Φ (𝑋, 𝑌∗ (𝑋) + 𝜌−1/3 𝜉1 )]
⋅
1 𝑊0
2𝜋 𝑌∗ (𝑋) √
⋅∫
𝑖∞
−𝑖∞
1
2
𝑊0 − (𝑊0 + 𝑌∗ (𝑋))
1
2𝜋𝑖
2/3
(1 − √𝑋0 )
1
exp [− 1/6
2
[𝐴𝑖 (𝜃)]
𝑋 (𝑊0 + 𝑌∗ (𝑋))
[ 0
⋅ 𝜉1 𝜃] 𝑑𝜃,
]
(188)
Φ𝑌𝑌 (𝑋, 𝑌∗ ) = −
Φ𝑌𝑌𝑌 (𝑋, 𝑌∗ ) =
𝑊0 (𝑊0 + 2𝑌∗ )
2
𝑌∗2 (𝑊0 + 𝑌∗ )
,
(189)
and 𝐶(𝜌) has the expansions in (31)–(33) (one can replace
𝜌−1/2 𝐶(𝜌) by 𝜌−1/3 𝐶1 (𝜌) in (188)), and 𝑌∗ = 𝑌∗ (𝑋) =
𝑌∗ (𝑋; 𝑋0 ) is given in (75).
For 𝜉1 → +∞ we can expand the integral in (188) by the
saddle point method, after shifting the integration contour far
to the right, with Re(𝜃) ≫ 1. Then the Airy function in the
integrand may be approximated using
Ai (𝑧) ∼
1 −1/4 −(2/3)𝑧3/2
𝑧 𝑒
;
2√𝜋
󵄨
󵄨
𝑧 󳨀→ ∞, 󵄨󵄨󵄨arg 𝑧󵄨󵄨󵄨 < 𝜋
(190)
and we can verify that (188) for 𝜉1 → +∞ matches to the D+
result in (60), as 𝑌 ↓ 𝑌∗ (𝑋). For 𝜉1 → −∞ the behavior of
the integral in (188) is determined by the singularity with the
largest real part, which is the double pole at 𝜃 = 𝑟0 (< 0).
Then standard singularity analysis can be used to show that
(188) for 𝜉1 → −∞ agrees with the expansion of (79) as 𝑌 ↑
𝑌∗ (𝑋). Note that in this limit 𝜌1/3 Φ∗ (𝑠1 ) becomes 𝑂(1) and
proportional to 𝜉1 .
Proposition 18. For (𝑋0 , 𝑌0 ) ∈ R4 , 𝑌 − 𝑌𝑐 (𝑋) = 𝜌−1/3 𝜂∗∗ =
𝑂(𝜌−1/3 ), and 𝑋𝐿 < 𝑋 < 𝑋0 (with 𝑋𝐿 defined in item (vii) of
Proposition 8) one has
4/3
1/6
1 − 𝑋0 − 𝑌0 (1 − √𝑋0 ) 𝑋0
𝜋 (𝑘, 𝑟) ∼
𝜌5/6 √2𝜋 √𝑋0 − 𝑋0 − 𝑌0
where
⋅√
𝜌Φ (𝑋, 𝑌∗ + 𝜌−1/3 𝜉1 )
= 𝜌Φ (𝑋, 𝑌∗ ) + 𝜌2/3 Φ𝑌 (𝑋, 𝑌∗ ) 𝜉1
1
1
+ 𝜌1/3 Φ𝑌𝑌 (𝑋, 𝑌∗ ) 𝜉12 + Φ𝑌𝑌𝑌 (𝑋, 𝑌∗ ) 𝜉13
2
6
+ 𝑜 (1) ,
Φ (𝑋, 𝑌∗ )
= 𝑌∗ log (1 +
− 𝑋 log (
𝑊0
,
𝑌∗ (𝑊0 + 𝑌∗ )
𝑊0
𝑊0
)−
𝑌∗
𝑊0 + 𝑌∗
𝑌∗
),
𝑊0 + 𝑌∗
𝑊
Φ𝑌 (𝑋, 𝑌∗ ) = log (1 + 0 ) ,
𝑌∗
𝑌0
𝑊0 + 𝑌0
√
{[2𝑊0 + 𝑌0 + 𝑌𝑐 (𝑋)]
𝑌𝑐 (𝑋) 𝑌0 − 𝑌𝑐 (𝑋)
−1/2
⋅ [𝑊0 + 𝑌𝑐 (𝑋) − √𝑊0 (𝑊0 + 𝑌0 )]}
⋅ exp {𝜌 [Φ (𝑋, 𝑌𝑐 (𝑋) + 𝜌−1/3 𝜂∗∗ ) − Φ (𝑋0 , 𝑌0 )]}
(191)
2/3
⋅
(1 − √𝑋0 )
1
1 𝑖∞
exp [− 1/6
∫
2𝜋𝑖 −𝑖∞ 𝐴𝑖 (𝜃)
𝑋 (𝑊0 + 𝑌0 (𝑋))
[ 0
⋅ 𝜂∗∗ 𝜃] 𝑑𝜃,
]
where 𝑌𝑐 (𝑋) = 𝑌𝑐 (𝑋; 𝑋0 , 𝑌0 ) is defined in (86).
Thus the transition from D0 to D− involves a slightly
different integral (cf. (191)) compared to the transition from
D+ to D− (cf. (188)), as the former has simple poles at the
22
Advances in Operations Research
Airy roots. Using standard asymptotic analysis we can show
that
3
1 𝑖∞ 𝑒−⋆𝜃
𝑑𝜃 ∼ 2 ⋆ 𝑒−⋆ /3 , ⋆ 󳨀→ +∞,
∫
2𝜋𝑖 −𝑖∞ Ai (𝜃)
(192)
𝑒−⋆𝑟0
1 𝑖∞ 𝑒−⋆𝜃
𝑑𝜃 ∼ 󸀠
, ⋆ 󳨀→ −∞
∫
2𝜋𝑖 −𝑖∞ Ai (𝜃)
Ai (𝑟0 )
(193)
and (192) can be used to verify the matching between (191)
and the D0 result in (65), where 𝐶0 ∼ 𝜌−1/2 (1 − 𝑋0 − 𝑌0 ) in
R4 . Similarly (193) can be used to verify matching to the D−
result in (79), with 𝐶1 now given by (41). The factor involving
𝜌Φ(𝑋, 𝑌𝑐 +𝜌−1/3 𝜂∗∗ ) in (191) can be expanded in Taylor series,
similarly to (189), by replacing 𝑌∗ (𝑋) by 𝑌𝑐 (𝑋) and 𝜉1 by 𝜂∗∗ .
As we approach R3 ∩ R4 (where 𝑋0 + 𝑌0 = √𝑋0 ) from
within R4 we see that (191) develops a singularity, in view of
the factor (√𝑋0 − 𝑋0 − 𝑌0 )−1 . As we approach R3 ∩ R4 from
̃
in (183) vanishes
within R3 , the expression for 𝐾+ (𝑋, 𝑌(𝑋))
and that for 𝑏(𝑡∗ ) in (184) develops a singularity. Thus the
expansions in both Propositions 16 and 18 become invalid,
and below we give a new result that applies for R3 ∩R4 , where
D0 meets D− .
Proposition 19. For (𝑋0 , 𝑌0 ) ∈ R3 ∩ R4 , with 𝑋0 + 𝑌0 =
√𝑋0 + 𝛿∗ 𝜌−1/3 and 𝛿∗ = 𝑂(1),
3
𝜋 (𝑘, 𝑟) ∼
(1 − √𝑋0 ) 𝑋01/4 exp {𝜌 [Φ (𝑋, 𝑌𝑐 (𝑋) + 𝜌−1/3 𝜂∗∗ ) − Φ (𝑋0 , 𝑌0 )]}
1
2𝜋𝑖
√2𝜋𝜌𝑌∗ (𝑋) √√𝑋0 − 𝑋0 − 𝑌∗ (𝑋)√𝑌∗ (𝑋) + (1 − √𝑋0 ) (2 − √𝑋0 )
⋅∫
𝑖∞
−𝑖∞
−𝜃𝛿1
𝑒
∞
∫𝜃 𝑒𝛿1 𝑢 Ai (𝑢) 𝑑𝑢
exp [−
2
[Ai (𝜃)]
where 𝛿1 = 𝑋0−1/6 (1 − √𝑋0 )−1/3 𝛿∗ and 𝑌𝑐 (𝑋) is in (86) and
𝑌∗ (𝑋) in (75).
With the scaling 𝑋0 + 𝑌0 − √𝑋0 = 𝑂(𝜌−1/3 ), the curves
̃
nearly coincide, with the differences
𝑌𝑐 (𝑋), 𝑌∗ (𝑋), and 𝑌(𝑋)
−1/3
being 𝑂(𝜌 ). In (194) we can again replace Φ(𝑋, 𝑌𝑐 +
𝜌−1/3 𝜂∗∗ ) by its Taylor expansion and replace 𝑌0 everywhere
by √𝑋0 − 𝑋0 + 𝛿∗ 𝜌−1/3 , thus writing the result in terms of
𝑋0 and 𝛿∗ (or 𝛿1 ), along with 𝑋 and 𝜂∗∗ . Note that 𝜂∗∗ is
still defined by 𝜂∗∗ = 𝜌1/3 [𝑌 − 𝑌𝑐 (𝑋)], which differs from
𝜌1/3 [𝑌 − 𝑌∗ (𝑋)] by an amount that is 𝑂(1). Thus, for region
R3 ∩ R4 the transition range in state space, between D−
and D0 , involves a somewhat more complicated integral than
those in Propositions 18 and 19.
This completes our summary and discussion of the
various regions of state space, which carry zero volume in
(𝑋, 𝑌) space but are necessary since the results in D0 , D+ ,
and D− do not always apply.
4. Asymptotic Expansion in Region D0
In this section we will construct the expansion for (𝑋, 𝑌) ∈
D0 , that is, (65). The analysis for the complementary regions
D+ and D− is virtually identical to that for the model with
an infinite secondary capacity (𝑅 = ∞), and the detailed
calculations can be found in [10]. Here we will only discuss,
for regions D+ and D− , those aspects that change when 𝑅 <
∞. In order to uniquely determine the expansion in D0 , we
will need to use asymptotic matching to the corner expansion
that applies on the scale 𝑘 = 𝑚−𝑂(1), 𝑟 = 𝑅−𝑂(1), and this is
discussed in Section 4.2. For parameter region R4 (and also
for R3 ∩ R4 ) we will also need to carefully analyze the scale
𝑘 = 𝑚 − 𝑂(𝜌1/3 ) and 𝑟 = 𝑅 − 𝑂(𝜌2/3 ), which will be necessary
[
2/3
(1 − √𝑋0 )
𝑋01/6
(𝑊0 + 𝑌∗ (𝑋))
(194)
𝜂∗∗ 𝜃] 𝑑𝜃,
]
to determine the multiplicative constant 𝐶1 = 𝐶1 (𝜌; 𝑋0 , 𝑌0 )
that arises in the asymptotic expansion in D− ; this analysis is
done in Section 4.3.
4.1. Ray Expansion in the Interior of D0 . We analyze the
scaled equation in (17) using the ray method of geometrical
optics, where we assume an expansion of the form
𝑃 (𝑋, 𝑌) = 𝐶0 (𝜌) 𝑒𝜌Ψ(𝑋,𝑌) [𝐾 (𝑋, 𝑌) + 𝜌−1 𝐾(1) (𝑋, 𝑌)
+ 𝜌−2 𝐾(2) (𝑋, 𝑌) + ⋅ ⋅ ⋅] .
(195)
Then we have
𝑃 (𝑋 ± 𝜌−1 , 𝑌)
𝑃 (𝑋, 𝑌)
= 𝑒±Ψ𝑋 (𝑋,𝑌) [1
±𝐾
1
+ 𝜌 ( 𝑋 + Ψ𝑋𝑋 (𝑋, 𝑌)) + 𝑂 (𝜌−2 )]
𝐾
2
(196)
−1
and using (195) in (17) we obtain in the limit as 𝜌 → ∞ the
“eikonal” equation
1 + 𝑋 + 𝑌 = 𝑌𝑒Ψ𝑌 + 𝑋𝑒Ψ𝑋 + 𝑒−Ψ𝑋
(197)
and at the next order in 𝜌−1 the “transport” equation
1
𝑒Ψ (𝑌𝐾𝑌 + 𝑌Ψ𝑌𝑌𝐾 + 𝐾)
2
1
+ 𝑒Ψ𝑋 (𝑋𝐾𝑋 + 𝑋Ψ𝑋𝑋 𝐾 + 𝐾)
2
1
+ 𝑒−Ψ𝑋 (−𝐾𝑋 + Ψ𝑋𝑋 𝐾) = 0.
2
(198)
Advances in Operations Research
23
The first-order PDE in (197) can be solved by the method of
characteristics (see [14]), where one must solve the five ODEs
𝑑𝑋
= 𝑒−Ψ𝑋 − 𝑋𝑒Ψ𝑋 ,
𝑑𝜏
2
𝑑Ψ
𝑑𝑋
𝑑𝑌
= Ψ𝑋
+ Ψ𝑌
𝑑𝑡
𝑑𝜏
𝑑𝜏
= Ψ𝑋 (𝑒
(200)
Ψ𝑋
Ψ𝑌
− 𝑋𝑒 ) − 𝑌Ψ𝑌𝑒 ,
𝑑Ψ𝑋
= 𝑒Ψ𝑋 − 1,
𝑑𝜏
(201)
𝑑Ψ𝑌
= 𝑒Ψ𝑌 − 1.
𝑑𝜏
Here 𝜏 is a parameter along a given characteristic curve,
which is also called a “ray,” due to applications in optics.
To uniquely specify the solution Ψ(𝑋, 𝑌) to (197), we must
either specify Ψ along some curve, called the “initial manifold,” in the (𝑋, 𝑌) plane, or use a singular solution, where
all the rays emanate from a single point. The appropriate
solution to the eikonal equation must be determined for each
individual problem, and we will see that for the present model
we will need to use three different solutions Ψ to (197), with
two having the boundary 𝑋 = 𝑋0 as the initial manifold and
the third corresponding to all the rays emanating from the
corner point (𝑋0 , 𝑌0 ). The first two solutions will correspond
to regions D+ and D− and the third to D0 . Since D+ and D−
arose also in the infinite capacity model, where 𝑅 = ∞, we
discuss these only briefly, and the details of the corresponding
solutions to (197)–(201) can be found in [10].
Let us denote the solution in region D+ as
𝐶(𝜌)𝐾+ (𝑋, 𝑌)𝑒𝜌Ψ+ (𝑋,𝑌) to distinguish it from (195), which
will apply in D0 . If this expansion will satisfy the boundary
equation along 𝑋 = 𝑋0 (or 𝑘 = 𝑚) in (4), or, equivalently,
(18), we must have
𝑋0 𝑒Ψ+,𝑋 (𝑋0 ,𝑌) = 𝑒−Ψ+,𝑌 (𝑋0 ,𝑌) .
(202)
Requiring Ψ+ (𝑋, 𝑌) to satisfy (202) is equivalent, up to an
additive constant in Ψ+ which can be incorporated into 𝐶(𝜌),
to specifying Ψ+ along the initial manifold 𝑋 = 𝑋0 . Solving
(199)–(201) subject to (202) leads to
Ψ+,𝑋 = − log (1 −
(𝑌 + 𝑋0 ) − 𝑋0
𝑡 + 𝑂 (𝑡2 ) .
𝑋0 − 𝑋 =
𝑌 + 𝑋0
(199)
𝑑𝑌
= −𝑌𝑒Ψ𝑌 ,
𝑑𝜏
−Ψ𝑋
which is an explicit function of 𝑌. From the first expression
in (62) we find that for 𝑡 → 0
𝑠
𝑒𝑡 ) ,
𝑠 + 𝑋0
(203)
𝑡
Ψ+,𝑌 = − log [1 + (𝑠 + 𝑋0 − 1) 𝑒 ]
and then the rays are given in parametric form by (62), which
relates (𝑠, 𝑡) to (𝑋, 𝑌). When 𝑡 = 0 we have (𝑋, 𝑌) = (𝑋0 , 𝑠)
so that 𝑠 is the point where a ray hits the line 𝑋 = 𝑋0 . We use
now 𝑡 instead of 𝜏 as the parameter along a given ray, and 𝑠
is used to index the family. Finally, solving (200) with (Ψ, 𝜏)
replaced by (Ψ+ , 𝑡) leads to the expression in (61). We also note
that when 𝑡 = 0 (𝑋 = 𝑋0 ) we have
Ψ+ (𝑋0 , 𝑌) = 𝑋0 − 1 + 𝑌 − (𝑋0 + 𝑌) log (𝑋0 + 𝑌)
(204)
(205)
It follows that for 𝑡 > 0 the rays that start from 𝑋 = 𝑋0 enter
the state space, where 𝑋 ⩽ 𝑋0 , for 𝑡 > 0 only if (𝑌+𝑋0 )2 −𝑋0 >
0, or 𝑌+𝑋0 > √𝑋0 . If 𝑋0 > 1 this condition holds for all 𝑌 ⩾ 0
and hence the rays fill the region D+ indicated in Figure 3. If
𝑋0 < 1 and 𝑌0 + 𝑋0 > √𝑋0 then the condition holds only
for 𝑌 in the interval √𝑋0 − 𝑋0 < 𝑌 < 𝑌0 . Then these rays fill
the domain D+ indicated in Figures 4 and 5. But if 𝑋0 < 1
and 𝑋0 + 𝑌0 < √𝑋0 , which is true for parameter region R4 ,
the condition never holds and then this ray expansion plays
no role in the analysis (see also Figure 6). Once we compute
Ψ+ in D+ , we can integrate (198) to obtain 𝐾+ (𝑋, 𝑌) in (63).
Thus we have shown that (63), up to the constant 𝐶(𝜌) which
we have yet to determine, holds in the portion D+ of the state
space, for parameter regions R1 ∪ R2 ∪ R3 .
Now we observe that if 𝑋0 > 1, Ψ+ (𝑋0 , 𝑌) is maximal at
𝑌 = 0 and by evaluating also 𝐾+ at 𝑋 = 𝑋0 and 𝑌 = 0 we find
that near the corner (𝑋, 𝑌) = (𝑋0 , 0) with 𝑋 = 𝑋0 (𝑘 = 𝑚)
and 𝑌 = 𝑟/𝜌, and we have
𝐶𝐾+ 𝑒𝜌Ψ+ ∼ 𝐶
1 𝑋0 − 1 −𝑟 𝜌(𝑋0 −1−𝑋0 log𝑋0 )
𝑋 𝑒
.
√2𝜋 𝑋03/2 0
(206)
The scale 𝑟 = 𝑂(1) for 𝑋0 > 1 must be analyzed separately and
the details are carried out in [10]. For 𝑟 = 𝑂(1) and 𝑚 − 𝑘 =
𝑛 = 𝑂(1) the result in (141) applies, and by asymptotically
matching this to (206) (for 𝑟 ⩾ 1) we conclude that
𝐶 (𝜌) ∼ 𝜌−1/2 ,
(𝑋0 , 𝑌0 ) ∈ R1 .
(207)
If 𝑋0 < 1 and 𝑋0 + 𝑌0 > 1 then Ψ+ (𝑋0 , 𝑌) in (204) is maximal
at 𝑌 = 1 − 𝑋0 , which lies in the range (0, 𝑌0 ) precisely for
parameter region R2 . Expanding Ψ+ (𝑋, 𝑌) in (61) about 𝑋 =
𝑋0 and 𝑌 = 1 − 𝑋0 leads to
𝐶𝐾+ 𝑒𝜌Ψ+
∼ 𝐶 (1 − 𝑋0 ) 𝑋0𝑛
(208)
1
1
2
exp {− [𝑟 − (𝜌 − 𝑚)] } .
√2𝜋
2𝜌
For parameter region R2 the expansions in D− and D0 will
be uniformly exponentially small in 𝜌, as there is little mass
in these state space ranges. Thus the main contribution to the
double sum in the normalization condition (10) will come
from D+ and in particular from the scale 𝑚 − 𝑘 = 𝑛 =
𝑂(1) and 𝑟 = 𝜌 − 𝑚 + 𝑂(√𝜌) (corresponding to 𝑌 = 1 −
𝑋0 +𝑂(𝜌−1/2 )). Then normalizing the approximation in (208),
after approximating the sum over 𝑟 by an integral over 𝑌, we
conclude that 𝐶(𝜌) ∼ 𝜌−1/2 , and hence (207) applies also for
parameter region R2 (and, by continuity, the relation holds
in R1 ∩ R2 (𝑋0 − 1 = 𝑂(𝜌−1/2 )) also).
We have thus shown that the ray expansion in D+ has the
state space regions D− and D0 as “shadows,” and thus the
24
Advances in Operations Research
other solutions to (197) must apply. To construct the solution
in D− we must slightly modify the ansatz in (195) and now
expand the joint distribution as
1/3
𝑃 (𝑋, 𝑌) = 𝐶1 (𝜌) 𝑒𝜌Φ(𝑋,𝑌) 𝑒𝜌
Φ1 (𝑋,𝑌)
[𝐿 (𝑋, 𝑌)
+ 𝜌−1/3 𝐿(1) (𝑋, 𝑌) + 𝜌−2/3 𝐿(2) (𝑋, 𝑌) + ⋅ ⋅ ⋅] .
(209)
Now Φ will satisfy (197) and 𝐿 will satisfy (198), and the
subexponential term Φ1 will satisfy the PDE
𝑌𝑒Φ𝑌 Φ1,𝑌 + (𝑋𝑒Φ𝑋 − 𝑒−Φ𝑋 ) Φ1,𝑋 = 0.
(210)
Again the detailed analysis can be found in [10]. We now find
that
Φ𝑋 = − log [1 − (1 − √𝑋0 ) 𝑒𝑡1 ] ,
Φ𝑌 = − log [1 −
𝑊0 𝑡1
𝑒 ]
𝑊0 + 𝑠1
(211)
and the rays are given in parametric form by
𝑋 = [1 − (1 − √𝑋0 ) 𝑒𝑡1 ] [1 − (1 − √𝑋0 ) 𝑒−𝑡1 ] ,
−𝑡1
𝑌 = −𝑊0 + (𝑊0 + 𝑠1 ) 𝑒
(212)
,
where we recall that 𝑊0 = (1 − √𝑋0 )2 . When 𝑡1 = 0 we have
𝑋 = 𝑋0 and Y = 𝑠1 , but unlike the rays in D+ , those in D−
are all tangent to the boundary 𝑋 = 𝑋0 . Thus the boundary
is a “caustic boundary,” for 0 < 𝑌 < √𝑋0 − 𝑋0 for region
R2 ∪ R3 and for all 𝑌 with 0 < 𝑌 < 𝑌0 for region R4 . The
solution Φ to (197) is now given by
−𝑡1
Φ (𝑋, 𝑌) = [𝑊0 − (𝑊0 + 𝑠1 ) 𝑒
𝐶1 (𝜌)
∼ 𝜌−1/6 ,
𝐶 (𝜌)
(𝑋0 , 𝑌0 ) ∈ R2 ∪ R3 .
Ψ𝑌 = − log (1 − 𝐵𝑒𝜏 ) ,
(213)
⋅ log (1 − √𝑊0 𝑒𝑡1 ) − √𝑊0 𝑒𝑡1 + 𝑊0
1
⋅ log (𝑊0 + 𝑠1 ) − 𝑊0 log 𝑊0 .
2
For region D− we can invert transformation (212) and write
𝑠1 and 𝑡1 explicitly in terms of 𝑋 and 𝑌. This leads to the
expression in (81) for 𝑠1 = 𝑠1 (𝑋, 𝑌), and then (80) gives Φ
in terms of 𝑋 and 𝑌.
Equation (210) implies that Φ1 (𝑋, 𝑌) is constant along a
caustic ray, so we write Φ1 (𝑋, 𝑌) = Φ∗ (𝑠1 ). To determine
Φ∗ (⋅) and also completely determine 𝐿(𝑋, 𝑌) in (209), we
need to construct two “nested” boundary layer corrections
to the expansion in (209), near 𝑋 = 𝑋0 , corresponding
to the scales 𝑋0 − 𝑋 = 𝑂(𝜌−2/3 ) (𝑚 − 𝑘 = 𝑂(𝜌1/3 )) and
𝑋0 − 𝑋 = 𝑂(𝜌−1 ) (𝑚 − 𝑘 = 𝑂(1)). The boundary condition
in (18) can only be imposed on the expansion that applies for
𝑛 = 𝑚 − 𝑘 = 𝑂(1). Using asymptotic matching between the
expansions for 𝑋0 − 𝑋 = 𝑂(𝜌−1 ), 𝑋0 − 𝑋 = 𝑂(𝜌−2/3 ), and the
(214)
In R2 we have 𝐶(𝜌) ∼ 𝜌−1/2 so that 𝐶1 (𝜌) ∼ 𝜌−2/3 , but
we have yet to determine either 𝐶 or 𝐶1 for region R3 . The
conclusion in (214) can also be reached by constructing an
expansion near the point (𝑋, 𝑌) = (𝑋0 , √𝑋0 − 𝑋0 ), with the
scaling 𝑋0 − 𝑋 = 𝑂(𝜌−2/3 ) and 𝑌 − (√𝑋0 − 𝑋0 ) = 𝑂(𝜌−1/3 )
(see subsection 5.3 in [10] for that analysis). Note also that the
curve 𝑌 = 𝑌∗ (𝑋) corresponds to the ray with 𝑠1 = √𝑋0 − 𝑋0
in (212) and also the ray with 𝑠 = √𝑋0 − 𝑋0 in (62). Then
we have Ψ+ (𝑋, 𝑌∗ (𝑋)) = Φ(𝑋, 𝑌∗ (𝑋)) so that the exponential
factors in the expansions in (60) and (209) are continuous
along 𝑌 = 𝑌∗ (𝑋), which separates D+ from D− .
The region D0 is a shadow of both the D+ and D− rays
for parameter regions R2 ∪ R3 and a shadow of the D− rays
for parameter region R4 . For R4 , the caustic rays that fill D−
correspond to 0 < 𝑠1 < 𝑌0 in (212), and 𝑠1 = 𝑌0 corresponds
to the curve 𝑌 = 𝑌𝑐 (𝑋) in (86), which separates D− from D0 .
To fill the shadow D0 and thus obtain an approximation to
𝑃(𝑋, 𝑌) for (𝑋, 𝑌) ∈ D0 , we must use a singular solution to
(197), one that has all rays start from the corner point (𝑋, 𝑌) =
(𝑋0 , 𝑌0 ). To construct this solution we first integrate the two
ODEs in (201), which yields
Ψ𝑋 = − log (1 − 𝐴𝑒𝜏 ) ,
𝑊0 𝑒𝑡1
] log [1 −
]
𝑊0 + 𝑠1
− (1 − √𝑋0 − 𝑒−𝑡1 ) (1 − √𝑋0 − 𝑒𝑡1 )
ray expansion in (209) allows us to determine (209) up to the
multiplicative constant 𝐶1 .
To determine 𝐶1 we need to, for regions R2 and R3 , first
relate 𝐶1 (𝜌) to the constant 𝐶(𝜌) in the D+ expansion. This
can be done by analyzing the transition curve 𝑌 = 𝑌∗ (𝑋),
with the scaling 𝑌 − 𝑌∗ (𝑋) = 𝑂(𝜌−1/3 ). The details are again
presented in [10], and this leads to the conclusion that
(215)
where 𝐴 and 𝐵 are constant along a given ray. Evaluating the
eikonal equation in (197) along the corner 𝑋 = 𝑋0 , 𝑌 = 𝑌0
and using (215) with 𝜏 = 0 gives
1 + 𝑋0 + 𝑌0 =
𝑌0
𝑋0
+
+1−𝐴
1−𝐵 1−𝐴
(216)
and this leads to the relation between 𝐴 and 𝐵 in (67). Here
we take 𝜏 = 0 to correspond to when a ray starts from the
corner point 𝑋 = 𝑋0 , 𝑌 = 𝑌0 . Using (215) we solve the two
ODEs in (199), subject to 𝑋|𝜏=0 = 𝑋0 and 𝑌|𝜏=0 = 𝑌0 , and we
thus obtain the expressions in (68), which give the corner rays
in parametric form. Using (215) and (68) we then integrate
(200), and choosing Ψ(𝑋0 , 𝑌0 ) = 0 for convenience, we obtain
(66). Note that Ψ can only be determined from (197) up to
an additive constant, but such a constant can be incorporated
into 𝐶0 = 𝐶0 (𝜌; 𝑋0 , 𝑌0 ) in (195).
We can view 𝐴 as indexing this family of rays. If 𝐴 is such
that (𝐴 + 𝑋0 + 𝑌0 )(1 − 𝐴) − 𝑋0 = 0 then 𝐵 in (216) and
(67) becomes infinite, and this corresponds to 𝐴 = 𝐴 max in
(69). This critical ray corresponds to 𝑌 = 𝑌0 , and for 𝜏 ∈
(0, −log(𝐴 max )) we have 𝑋 ∈ (0, 𝑋0 ), so the ray is the upper
Advances in Operations Research
25
boundary of the state space rectangle, where all secondary
storage spaces are full. The ray 𝐴 = 𝐴 min = 1−√𝑋0 (for 𝑋0 <
1) corresponds to the curve 𝑌 = 𝑌𝑐 (𝑋) = 𝑌𝑐 (𝑋; 𝑋0 , 𝑌0 ) in
(86). For regions R1 ∪R2 ∪R3 we have 𝑌0 /(𝑋0 +𝑌0 ) > 1−√𝑋0
̃
and then the ray with 𝐴 = 𝑌0 /(𝑋0 + 𝑌0 ) is the curve 𝑌(𝑋)
in
(56). Thus for regions R1 ∪R2 ∪R3 it suffices to consider 𝐴 in
the range (𝑌0 /(𝑋0 + 𝑌0 ), 𝐴 max ) to fill region D0 in state space.
We also note that in order for the rays, which all start from
(𝑋, 𝑌) = (𝑋0 , 𝑌0 ), to enter the state space we need 𝑋𝜏 |𝜏=0 < 0
and 𝑌𝜏 |𝜏=0 < 0, which implies that 1 − 𝐴 − 𝑋0 /(1 − 𝐴) < 0
and 𝑋0 /(1 − 𝐴) − (𝐴 + 𝑋0 + 𝑌0 ) < 0. Adding the last two
inequalities implies that 𝐴 > (1/2)(1 − 𝑋0 − 𝑌0 ) and this is
certainly true if 𝐴 > 𝐴 min .
We next solve the transport equation (198) for 𝐾(𝑋, 𝑌).
This equation can be written as an ODE along a ray, with
𝐾𝜏
1
1
= ( 𝑌Ψ𝑌𝑌 + 1) 𝑒Ψ𝑌 + ( 𝑋Ψ𝑋𝑋 + 1) 𝑒Ψ𝑋
𝐾
2
2
1
+ Ψ𝑋𝑋 𝑒−Ψ𝑋
2
= 𝑒Ψ𝑋 + 𝑒Ψ𝑌 −
−
(217)
1
(𝑋 + 𝑋𝜏 𝑒Ψ𝑋 )
2 𝜏𝜏
1
(𝑌 + 𝑌𝜏 𝑒Ψ𝑌 ) ,
2 𝜏𝜏
𝐾𝜏 1
1
1 1
1 Δ𝜏
+
−
=
𝜏
𝜏
𝐾
2 1 − 𝐴𝑒
2 1 − 𝐵𝑒
2 Δ
= [(𝑒
− 𝐴) (𝑒
− 𝐵) |Δ (𝜏, 𝐴)|]
(221)
so that
𝑝 (𝐴 + 𝑋0 + 𝑌0 ) (1 − 𝐴) − 𝑋0
𝑌0 − 𝑌
,
= ∼
𝑋0 − 𝑋 𝑛
𝑋0 − (1 − 𝐴)2
(222)
𝜏 󳨀→ 0.
Here we recall that 𝑌 = 𝑌0 − 𝑝/𝜌 and 𝑋 = 𝑋0 − 𝑛/𝜌, and (222)
gives the slope at which the ray indexed by 𝐴 hits the corner
point (𝑋0 , 𝑌0 ).
We can thus invert the transformation in (68) locally, with
(222) corresponding to a quadratic equation for 𝐴 = 𝐴(𝑛/𝑝),
and hence
𝐴 ∼ 𝐴𝑠 ≡ 1 +
1 1
[(1 + 𝑋0 + 𝑌0 ) 𝑛
2𝑝−𝑛
(223)
𝐾0 (𝐴) .
Then we define 𝐵𝑠 by replacing 𝐴 by 𝐴 𝑠 in (67). Note that
the right-hand side of (223) approaches 𝐴 max if 𝑛 → ∞,
approaches 𝐴 min = 1 − √𝑋0 as 𝑝 → ∞, and is equal to
𝑌0 /(𝑋0 + 𝑌0 ) if 𝑛/𝑝 = [(𝑋0 + 𝑌0 )2 − 𝑋0 ]/𝑌0 . For 𝜏 → 0 we
also have Ψ(𝑋, 𝑌) → 0 with
Ψ (𝑋, 𝑌) = [(𝐴 +
+
𝑋0
− 1) log (1 − 𝐴)
1−𝐴
𝑌0
log (1 − 𝐵)] 𝜏 + 𝑂 (𝜏2 ) = (𝑋0 − 𝑋)
1−𝐵
(224)
⋅ log (1 − 𝐴 𝑠 ) + (𝑌0 − 𝑌) log (1 − 𝐵𝑠 ) + 𝑂 (𝜏2 )
𝐾 (𝑋, 𝑌)
−1/2
𝑋0
+ 𝐴 − 1) 𝜏,
1−𝐴
𝑌0
𝑌0 − 𝑌 ∼
𝜏
1−𝐵
(218)
and the most general solution to (218) is given by
−𝜏
𝑋0 − 𝑋 ∼ (
2
− √(1 + 𝑋0 + 𝑌0 ) 𝑛2 + 4 (𝑝2 − 𝑛2 ) 𝑋0 ] .
where we used (199), and the last equality in (217) follows
by differentiating (199) with respect to 𝜏. Introducing the
Jacobian Δ ≡ 𝑋𝜏 𝑌𝐴 −𝑋𝐴𝑌𝜏 associated with the mapping from
(𝜏, 𝐴) to (𝑋, 𝑌) variables, after some calculation we find that
(217) becomes
−𝜏
For 𝜏 → 0 we also have
(219)
Here 𝐾0 (𝐴) is an arbitrary function of the parameter that
indexes the rays and is thus constant along any particular
ray. We have thus determined the expansion in (195) up to
the constant 𝐶0 and the function 𝐾0 (⋅). To complete the D0
ray expansion we will need to use asymptotic matching to
a local expansion valid near the corner (𝑋, 𝑌) = (𝑋0 , 𝑌0 ).
This is constructed in Section 4.2. In order to accomplish
the matching we will need the behavior of (195) as 𝑋 ↑ 𝑋0 ,
𝑌 ↑ 𝑌0 , and this is examined next.
We note that (𝑋, 𝑌) → (𝑋0 , 𝑌0 ) corresponds to 𝜏 → 0
and in this limit the Jacobian in (71) vanishes, and
so that 𝑒𝜌Ψ(𝑋,𝑌) ∼ (1 − 𝐴 𝑠 )𝑛 (1 − 𝐵𝑠 )𝑝 . We have thus shown that
as 𝜏 → 0 we have
𝐶0 𝐾 (X, 𝑌) 𝑒𝜌Ψ(𝑋,𝑌) ∼ 𝐶0
𝑛
𝐾0 (𝐴 𝑠 )
√1 − 𝐴 𝑠 √1 − 𝐵𝑠
(1
𝑝
− 𝐴 𝑠 ) (1 − 𝐵𝑠 )
⋅
𝑋0
1
)
[(1 + 𝑋0 + 𝑌0 ) (1 +
2
√𝜏
(1 − 𝐴 𝑠 )
(225)
−1/2
4𝑋0
−
]
1 − 𝐴𝑠
,
where 𝐴 ∼ 𝐴 𝑠 was approximated by (223), and by (221) we
have
Δ
∼ (−𝜏) [(1 + 𝑋0 + 𝑌0 ) (1 +
𝑋0
4𝑋0
)−
] , (220)
2
1−𝐴
(1 − 𝐴)
𝜏 󳨀→ 0+ .
𝜏∼
(1 − 𝐵𝑠 ) (𝑌0 − 𝑌) 1 − 𝐵𝑠 𝑝
∼
,
𝑌0
𝑌0 𝜌
so that (225) becomes an explicit function of 𝑛 and 𝑝.
(226)
26
Advances in Operations Research
4.2. Analysis of the Scale 𝑘 = 𝑚 − 𝑂(1), 𝑟 = 𝑅 − 𝑂(1). We
use the variables 𝑛 = 𝑚 − 𝑘 and 𝑝 = 𝑅 − 𝑟 and obtain an
approximation to 𝜋(𝑘, 𝑟) that is valid when all but a few of
the primary and secondary spaces are occupied. This is likely
for parameter regions R3 and R4 but unlikely for R1 and
R2 . We define 𝑄 by
𝜋 (𝑘, 𝑟) = 𝜋 (𝑘, 𝑟; 𝜌, 𝑚, 𝑅) = 𝑄 (𝑛, 𝑝; 𝜌, 𝑋0 , 𝑌0 ) .
∞ ∞
̃ (𝑧, 𝑤) = ∑ ∑ 𝑄𝐿 (𝑛, 𝑝) 𝑧𝑛 𝑤𝑝
𝑄
(227)
We thus denote by 𝑄(𝑛, 𝑝) the exact probability that there are
𝑛 (resp., 𝑝) empty primary (resp., secondary) spaces, and we
will later denote by 𝑄𝐿 (𝑛, 𝑝) the leading term in an asymptotic
expansion of this probability, which is valid for the scale
𝑛, 𝑝 = 𝑂(1). Then clearly 𝑄(0, 0) ∼ 𝑄𝐿 (0, 0) so in asymptotic
relations involving 𝑄(0, 0) we can drop the subscript 𝐿.
Writing the balance equations in (3), (4), and (5) in terms
of (𝑛, 𝑝) leads to
[1 + 𝑋0 + 𝑌0 − 𝜌−1 (𝑛 + 𝑝)] 𝑄 (𝑛, 𝑝)
= [𝑌0 + 𝜌−1 (1 − 𝑝)] 𝑄 (𝑛, 𝑝 − 1)
Here 𝑄𝐿 (𝑛, 𝑝) in (232)–(234) is understood to be the leading
term in an asymptotic expansion of 𝑄(𝑛, 𝑝; 𝜌), for 𝜌 → ∞,
and the corner equation in (231) must also be satisfied by this
leading term.
We introduce the double generating function
and from (231)–(234) we obtain
1 ̃
[1 + 𝑋0 (1 − 𝑧) + 𝑌0 (1 − 𝑤) − ] 𝑄
(𝑧, 𝑤)
𝑧
1
1 1 ̃
= ( − )𝑄
(0, 𝑤) + (1 − ) 𝑄𝐿 (0, 0) .
𝑤 𝑧
𝑤
̃
̃ (𝑧, 1) = 𝑄 (0, 1)
𝑄
1 − 𝑧𝑋0
ℓ
∑ 𝑄𝐿 (𝑛, 𝑝) = 𝑄𝐿 (0, 0) (𝑋0 + 𝑌0 ) ,
[1 + 𝑋0 + 𝑌0 − 𝜌−1 𝑝] 𝑄 (0, 𝑝)
(229)
𝑝 ⩾ 1,
[1 + 𝑋0 + 𝑌0 − 𝜌−1 𝑛] 𝑄 (𝑛, 0)
= [𝑋0 + 𝜌−1 (1 − 𝑛)] 𝑄 (𝑛 − 1, 0) + 𝑄 (𝑛 + 1, 0) ,
(230)
𝑛 ⩾ 1,
and the corner condition in (9) becomes
(𝑋0 + 𝑌0 ) 𝑄 (0, 0) = 𝑄 (1, 0) + 𝑄 (0, 1) .
(238)
The expression in (238) shows that the total number of empty
spaces (= 𝑚 + 𝑅 − 𝑘 − 𝑟 = 𝑛 + 𝑝) follows asymptotically
a geometric distribution if 𝑋0 + 𝑌0 < 1, and this could be
also deduced from (12), by expanding the exact (truncated
Poisson) distribution for 𝜌 → ∞. Note, however, that (238)
holds also for 𝑋0 + 𝑌0 ⩾ 1. The factor in brackets in the lefthand side of (236) vanishes when 𝑧 = 𝑧± (𝑤), where 𝑧± are
̃ 𝑤) to be analytic at the smaller
given in (51). Requiring 𝑄(𝑧,
̃
root 𝑧− (𝑤) determines 𝑄(0, 𝑤) as
̃ (0, 𝑤) = 𝑧+ (𝑤) (1 − 𝑤) 𝑄𝐿 (0, 0) .
𝑄
(𝑧− (𝑤) − 𝑤) 𝑋0
(231)
Note that near this corner only the boundaries 𝑋 = 𝑋0 and
𝑌 = 𝑌0 of the state space rectangle are relevant, and the
problem in (228)–(231) corresponds to a random walk in a
quarter plane, in (𝑛, 𝑝) space. For 𝜌 → ∞ the problem in
(228)–(231) may be further approximated by
(239)
Then using (239) in (236) and partially inverting the transform in (236) lead to
∞
∑ 𝑤𝑝 𝑄𝐿 (𝑛, 𝑝)
𝑝=0
(1 + 𝑋0 + 𝑌0 ) 𝑄𝐿 (𝑛, 𝑝)
= 𝑌0 𝑄𝐿 (𝑛, 𝑝 − 1) + 𝑋0 𝑄𝐿 (𝑛 − 1, 𝑝)
(232)
=
+ 𝑄𝐿 (𝑛 + 1, 𝑝) ; 𝑛 ⩾ 1, 𝑝 ⩾ 1,
1−𝑤
−𝑛−1
[𝑧 (𝑤)]
𝑄𝐿 (0, 0) ,
(𝑧− (𝑤) − 𝑤) 𝑋0 +
(240)
𝑛 ⩾ 0,
(1 + 𝑋0 + 𝑌0 ) 𝑄𝐿 (0, 𝑝)
and hence by the Cauchy integral formula
= 𝑌0 𝑄𝐿 (0, 𝑝 − 1) + 𝑄𝐿 (1, 𝑝) + 𝑄𝐿 (0, 𝑝 + 1) ,
(233)
𝑝 ⩾ 1,
= 𝑋0 𝑄𝐿 (𝑛 − 1, 0) + 𝑄𝐿 (𝑛 + 1, 0) , 𝑛 ⩾ 1.
ℓ ⩾ 0.
𝑛+𝑝=ℓ
= [𝑌0 + 𝜌−1 (1 − 𝑝)] 𝑄 (0, 𝑝 − 1) + 𝑄 (1, 𝑝)
(1 + 𝑋0 + 𝑌0 ) 𝑄𝐿 (𝑛, 0)
(237)
̃ 𝑧) = 𝑄𝐿 (0, 0)/[1 − (𝑋0 + 𝑌0 )𝑧].
and if 𝑧 = 𝑤 we find that 𝑄(𝑧,
Setting 𝑧 = 𝑤 then in (235) leads to
+ 𝑄 (𝑛 + 1, 𝑝) ; 𝑛 ⩾ 1, 𝑝 ⩾ 1,
+ 𝑄 (0, 𝑝 + 1) ,
(236)
From (236), by setting 𝑤 = 1 we obtain
(228)
+ [𝑋0 + 𝜌−1 (1 − 𝑛)] 𝑄 (𝑛 − 1, 𝑝)
(235)
𝑛=0 𝑝=0
𝑄𝐿 (𝑛, 𝑝)
=
(234)
(241)
𝑄𝐿 (0, 0)
1−𝑤
−𝑛−1 −𝑝−1
𝑤
𝑑𝑤,
[𝑧+ (𝑤)]
∮
𝑋0 (2𝜋𝑖) 𝑧− (𝑤) − 𝑤
where the contour is over a small loop about 𝑤 = 0.
Advances in Operations Research
27
When 𝑋0 + 𝑌0 < 1 (𝑚 + 𝑅 < 𝜌, region R3 ∪ R4 ), most
primary and secondary spaces will be full, and then (241)
becomes a normalized discrete distribution with
𝑄 (0, 0) ∼ 𝑄𝐿 (0, 0) = 1 − 𝑋0 − 𝑌0 , R3 ∪ R4 ,
(242)
leading to the limit law in (50).
For regions R1 ∪ R2 , (241) still represents a local
approximation to 𝜋(𝑘, 𝑟), but now different arguments must
be used to determine 𝑄𝐿 (0, 0), and indeed now 𝑄𝐿 (0, 0) will
turn out to be exponentially small for large 𝜌. We proceed to
relate (241) to the ray expansions in D0 , D+ , and D− , noting
that, for parameter region R1 ∪ R2 ∪ R3 , D0 and D+ both
border the corner point, while, for R4 , D0 and D− border
this point. We will need to expand (241) asymptotically, for 𝑛
and/or 𝑝 → ∞. The integrand in (241) has branch points at
𝑤 = (1+𝑋0 +𝑌0 ±2√𝑋0 )/𝑌0 and possible poles at solutions of
𝑧− (𝑤) = 𝑤. Now, 𝑧− (1) = 1 for 𝑋0 > 1 but 𝑤 = 1 is not a pole
in view of the factor 1 − 𝑤 in the numerator. The only other
possible solution to 𝑧− (𝑤) = 𝑤 occurs at 𝑤 = 1/(𝑋0 + 𝑌0 ) and
this is a pole if
𝑌0
2𝑋0
= 1 + 𝑋0 + 𝑌0 −
𝑋0 + 𝑌0
𝑋0 + 𝑌0
− √ (1 + 𝑋0 + 𝑌0 −
(i) 𝑝 = 𝑂(1), 𝑛 → ∞
−𝑛
𝑄𝐿 (𝑛, 𝑝) ∼ 𝑄 (0, 0)
𝑝
⋅ 𝑛𝑝 (
2
𝑝−𝑛−2
.
(248)
(iii) 𝑝, 𝑛 → ∞ with 𝜃1 < 𝑛/𝑝 < ∞, for R1 ∪ R2 ∪ R3
𝑄𝐿 (𝑛, 𝑝)
∼
2
𝑌0
) − 4𝑋0
𝑋0 + 𝑌0
1
1
𝑤=1+
(1 + 𝑋0 − 𝑋0 𝑢 − )
𝑌0
𝑢
(243)
⋅
−1/4
𝑄 (0, 0) 2
2
[𝑛 (1 + 𝑋0 + 𝑌0 ) + 4 (𝑝2 − 𝑛2 ) 𝑋0 ]
√2𝜋
(249)
(1 − 𝑢𝑠 ) (𝑋0 𝑢𝑠2 − 1)
𝑢𝑠3/2 (𝑋0 + 𝑌0 − 𝑋0 𝑢𝑠 ) √(1 + 𝑋0 + 𝑌0 ) 𝑢𝑠 − 1 − 𝑋0 𝑢𝑠2
𝑝
⋅ 𝑢𝑠𝑝−𝑛 𝑌0 [(1 + 𝑋0 + 𝑌0 ) 𝑢𝑠 − 1 − 𝑋0 𝑢𝑠2 ]
−𝑝
,
where
𝑛
1
𝑢𝑠 = 𝑢𝑠 ( ) =
[(1 + 𝑋0 + 𝑌0 ) 𝑛
𝑝
2𝑋0 (𝑛 + 𝑝)
(250)
2
+ √(1 + 𝑋0 + 𝑌0 ) 𝑛2 + 4𝑋0 (𝑝2 − 𝑛2 )] .
(244)
and (241) becomes
(iv) 𝑝, 𝑛 → ∞, 𝑛/𝑝 ≈ 𝜃1 (with 𝑛 − 𝑝𝜃1 = 𝑂(√𝑝)), for
R1 ∪ R2 ∪ R3
𝑄𝐿 (0, 0)
2𝜋𝑖
(245)
𝑝
(1 − 𝑢) (𝑋0 𝑢2 − 1) 𝑌0 𝑢𝑝−𝑛−2
(𝑋0 + 𝑌0 − 𝑋0 𝑢) [(1 + 𝑋0 + 𝑌0 ) 𝑢 − 𝑋0 𝑢2 − 1]
𝑑𝑢,
𝑝+1
𝜂
−∞
𝑧+ (0)
1
2
[1 + X0 + 𝑌0 + √(1 + 𝑋0 + 𝑌0 ) − 4𝑋0 ] ,
2𝑋0
𝑝−𝑛−2
𝑄𝐿 (𝑛, 𝑝) ∼ 𝑄 (0, 0) 𝑋0𝑛 (𝑋0 + 𝑌0 )
⋅∫
where Γ is a small loop about 𝑢 = 𝑧+ (0), with
=
) .
∼ 𝑄 (0, 0) [(𝑋0 + 𝑌0 ) − 𝑋0 ] 𝑋0𝑛 (𝑋0 + 𝑌0 )
But (243) holds precisely when (𝑋0 + 𝑌0 ) > 𝑋0 , which is
true for regions R1 ∪ R2 ∪ R3 . Thus for R4 the pole is
absent and along the transition curve R3 ∩ R4 (cf. (25))
the pole coalesces with the lower branch point, both being
at 𝑤 = 1/√𝑋0 (> 1). We can recast the integral in (241) by
using the conformal map 𝑢 = 𝑧+ (𝑤), so that the inverse is
Γ
√(1 + 𝑋0 + 𝑌0 )2 − 4𝑋0
(247)
𝑄𝐿 (𝑛, 𝑝)
2
⋅∫
𝑌0
(ii) 𝑝, 𝑛 → ∞ with 0 ⩽ 𝑛/𝑝 < 𝜃1 , 𝜃1 = [(𝑋0 + 𝑌0 )2 −
𝑋0 ]/𝑌0 , for (𝑋0 , 𝑌0 ) ∈ R1 ∪ R2 ∪ R3
󵄨󵄨
󵄨
󵄨󵄨(𝑋0 + 𝑌0 )2 − 𝑋0 󵄨󵄨󵄨
𝑋0
󵄨
󵄨.
= 𝑋0 + 𝑌0 +
−
𝑋0 + 𝑌0
𝑋0 + 𝑌0
𝑄𝐿 (𝑛, 𝑝) =
[𝑧+ (0)]
𝑝!
𝜂=
(246)
and the integrand in (245) has a pole of order 𝑝 + 1 at 𝑢 =
𝑧+ (0). Below we collect some asymptotic results for 𝑄𝐿 (𝑛, 𝑝)
that will be used in the matching calculations.
Proposition 20. For 𝑛 and/or 𝑝 → ∞ the function 𝑄𝐿 (𝑛, 𝑝)
has the following asymptotic expansions:
2
[(𝑋0 + 𝑌0 ) − 𝑋0 ]
1
√2𝜋
2
(251)
𝑒−V /2 𝑑V,
𝑌0 √𝑝
3
√𝑋0 + 𝑌0 √(𝑋0 + 𝑌0 ) + 𝑋0 + (𝑋0 + 𝑌0 ) (𝑌0 − 2𝑋0 )
𝑛
− ) = 𝑂 (1) .
𝑝
(𝜃1
(252)
(v) 𝑝, 𝑛 → ∞, 0 < 𝑛/𝑝 ⩽ ∞, for region R4 .
The expression in (249) holds for all 0 < 𝑛/𝑝 < ∞ and
(247) holds for 𝑛 → ∞ with 𝑝 = 𝑂(1).
28
Advances in Operations Research
(vi) 𝑛 = 𝑂(1), 𝑝 → ∞, for region R4
𝑄𝐿 (𝑛, 𝑝) ∼
⋅
and (257) is equivalent to the quadratic equation
𝑝
𝑌0
𝑄 (0, 0) 𝑛/2−1/4
𝑋0
(
)
𝑌0 + 𝑊0
2√𝜋
(
√𝑊0 + 𝑌0 (1 − √𝑋0 )
(253)
√𝑋0 − 𝑋0 − 𝑌0
⋅ 𝑝−3/2 [𝑛 +
√𝑋0
√𝑋0 − 1
+
√𝑋0
√𝑋0 − 𝑋0 − 𝑌0
].
(vii) 𝑛, 𝑝 → ∞ with 𝑛 = 𝑂(√𝑝), for region R3 ∩ R4 , with
𝑌 = √𝑋0 − 𝑋0 + 𝜌−1/2 𝛿, 𝛿 = 𝑂(1)
𝑝{
{ √1 − √𝑋0
𝑌0
𝑄 (0, 0) 𝑛/2
𝑋0 (
) { 1/4
√𝜋
𝑌0 + 𝑊0 { 𝑋0 √𝑝
{
𝑄𝐿 (𝑛, 𝑝) ∼
⋅ exp [
2
𝑛2
𝛿
1
√
)] +
(1 −
4𝑝
𝜌
𝑋
√𝑋0
√
0
}
∞
}
2
⋅ (∫ 𝑒−V /2 𝑑V)} ,
}
𝜂0
}
𝜂0 =
√1 − √𝑋0 𝑛
√2
√𝑝𝛿
−
.
1/4
√2𝑋0
√𝑝 √1 − √𝑋 𝑋1/4 √𝜌
0 0
−𝑛−1
[𝑧+ (𝑤)]
∼ [𝑧+ (0)]
−𝑛−1
−𝑛−1
[1 +
𝑤𝑧+󸀠 (0)
+ 𝑂 (𝑤2 )]
𝑧+ (0)
(256)
Note that (247) is not only asymptotically true but also an
exact expression when 𝑝 = 0, since 𝑄(𝑛, 0) satisfies the
boundary equation in (234). Turning to (245) we see that the
integrand has a simple pole at 𝑢 = 𝑢∗ ≡ (𝑋0 + 𝑌0 )/𝑋0 and
saddle point(s) where
(257)
2
1
< 𝑢∗ < 𝑢𝑠 < 𝑧+ (0)
√𝑋0
− 𝑝 log [(1 + 𝑋0 + 𝑌0 ) 𝑢 − 𝑋0 𝑢 − 1]} = 0,
(260)
and then the pole does not contribute as we dilate Γ to the
saddle point contour. Setting
𝐺 (𝑢) =
⋅
(1 − 𝑢) (𝑋0 𝑢2 − 1)
(𝑋0 + 𝑌0 ) 𝑢 − 1
1
,
𝑢2 [(1 + 𝑋0 + 𝑌0 ) 𝑢 − 𝑋0 𝑢2 − 1]
(261)
𝑛
𝑛
𝐹 (𝑢) = 𝐹 (𝑢; ) = log 𝑌0 + (1 − ) log 𝑢
𝑝
𝑝
− log [(1 + 𝑋0 + 𝑌0 ) 𝑢 − 𝑋0 𝑢2 − 1]
we use for (245) the standard saddle point estimate
𝑧󸀠 (0)
−𝑛−1
𝑤 ).
∼ [𝑧+ (0)]
exp (− +
𝑧+ (0) 1
𝑑
{(𝑝 − 𝑛) log 𝑢
𝑑𝑢
(259)
and we note that 𝑢∗ < 𝑧+ (0) is always true. Then we dilate
the small loop Γ about 𝑢 = 𝑧+ (0) to the saddle point contour
|𝑢 − 𝑧+ (0)| = |𝑢𝑠 − 𝑧+ (0)|, which is a circular contour that
traverses the saddle 𝑢𝑠 in the steepest descent directions,
which are arg(𝑢 − 𝑢𝑠 ) = ±𝜋/2. But in doing the dilation
we must take into account the contribution from the residue
at the pole 𝑢 = 𝑢∗ , in view of (259). It turns out that the
residue dominates the saddle point contribution, and we thus
obtain the expression in (248). For 𝑛/𝑝 ∈ (𝜃1 , ∞) we have the
ordering
(255)
The results in (247)–(255) may be obtained by expanding
the integrals in (241) or (245) by a combination of singularity
analysis and the saddle point method. Good references on
asymptotic expansion of integrals are the books [15–19], but
since these methods are now well established, we merely
sketch the proof of Proposition 20.
To obtain (247) we approximate the integrand in (241) for
𝑤 = 𝑂(𝑛−1 ), scaling 𝑤 = 𝑤1 /𝑛 and using
(258)
One root of (258) is given by 𝑢𝑠 = 𝑢𝑠 (𝑛/𝑝) in (250), and the
complementary root, with a minus sign in front of the square
root, will correspond to a second saddle which will not play
any role in the analysis. The pole 𝑢∗ and saddle 𝑢𝑠 , whose
location depends on the ratio 𝑛/𝑝, coalesce when 𝑛/𝑝 = 𝜃1 ,
for regions R1 , R2 , and R3 . Note also that as 𝑛/𝑝 → ∞ we
have 𝑢𝑠 → 𝑧+ (0) and that the integrand in (245) has a zero at
𝑢 = 1/√𝑋0 . For region R1 ∪ R2 ∪ R3 and 𝑛/𝑝 ∈ (0, 𝜃1 ) we
have the ordering
1
< 𝑢𝑠 < 𝑢∗ < 𝑧+ (0)
√𝑋0
(254)
𝑝𝛿2
𝑛𝛿
⋅ exp [
]
−
𝜌√𝑋0 (1 − √𝑋0 ) √𝜌√𝑋0
𝑛
𝑛
𝑛
+ 1) 𝑋0 𝑢2 − (1 + 𝑋0 + 𝑌0 ) 𝑢 + − 1 = 0.
𝑝
𝑝
𝑝
1
1
∫ 𝐺 (𝑢) 𝑒𝑝𝐹(𝑢) 𝑑𝑢 =
∫ 𝐺 (𝑢) 𝑒𝑝𝐹(𝑢) 𝑑𝑢
2𝜋𝑖 Γ
2𝜋𝑖 Γ𝑠
𝐺 (𝑢𝑠 ) 𝑝𝐹(𝑢𝑠 )
1
,
𝑒
∼
√2𝜋𝑝 √𝐹󸀠󸀠 (𝑢 )
(262)
𝑠
and this leads to the expression in (249). The transitional
result in (251) corresponds to 𝑛/𝑝 ≈ 𝜃1 and then we have 𝑢∗ ≈
𝑢𝑠 , so the saddle is close to a simple pole. Such situations are
discussed in detail in [15], and by expanding the integrand in
(245) about 𝑢∗ we ultimately obtain a simpler integrand that is
Advances in Operations Research
29
related to 𝐷−1 (⋅), the parabolic cylinder function of order −1,
which can in turn be expressed in terms of the standard error
function, leading to (251). Note that as 𝜂 → +∞ we approach
the region 𝑛/𝑝 < 𝜃1 and then (251) reduces to (248).
The pole at 𝑢∗ and zero at 1/√𝑋0 coalesce when 𝑋0 +𝑌0 =
√𝑋0 , which is precisely the curve R3 ∩ R4 which separates
R3 from R4 in parameter space. For (𝑋0 , 𝑌0 ) ∈ R4 we have
the ordering
𝑢∗ <
1
< 𝑢𝑠 < 𝑧+ (0) .
√𝑋0
(263)
Then for any 𝑛/𝑝 ∈ (0, ∞) we can deform Γ into the saddle
point contour and obtain (249) as the approximation to
𝑄𝐿 (𝑛, 𝑝). The limits 𝑛 = 𝑂(1), 𝑝 → ∞, and 𝑛 → ∞,
𝑝 = 𝑂(1) require a separate analysis. For the latter 𝑢𝑠 becomes
close to 𝑧+ (0) and we again obtain (247), while for the former
the saddle 𝑢𝑠 gets close to zero at 1/√𝑋0 . By expanding the
integrand near 𝑢 = 1/√𝑋0 , setting 𝑢 = 1/√𝑋0 + V/√𝑝, we
obtain
where Br+ is a vertical contour in the complex V-plane, which
is to the right of the pole at V = 𝛿/𝑋0 . The integral in (266)
may be evaluated as a combination of a Gaussian and an error
function, and this ultimately leads to (254) with (255). Since
the large parameter 𝜌 appears explicitly in (254) and (266)
we can define the asymptotic limit more precisely, as 𝜌 → ∞
with 𝑌0 + 𝑋0 − √𝑋0 = 𝑂(𝜌−1/2 ), 𝑝 = 𝑂(𝜌), and 𝑛 = 𝑂(√𝜌).
With Proposition 20 we are now ready to relate the corner
approximation 𝑄𝐿 (𝑛, 𝑝) to the various ray expansions, via
asymptotic matching. First consider parameter region R1 ∪
R2 ∪ R3 , so that D0 and D+ come together at the corner
point (𝑋0 , 𝑌0 ). In D+ the function Ψ+ (𝑋, 𝑌) can be expanded
in Taylor series about (𝑋, 𝑌) = (𝑋0 , 𝑌0 ) and as (𝑋, 𝑌) →
(𝑋0 , 𝑌0 ) we obtain
𝐶 (𝜌) 𝐾+ (𝑋, 𝑌) 𝑒𝜌Ψ+ (𝑋,𝑌) ∼ 𝐶 (𝜌) 𝐾+ (𝑋0 , 𝑌0 )
⋅ 𝑒𝜌Ψ+ (𝑋0 ,𝑌0 ) exp [𝜌 (𝑋 − 𝑋0 ) Ψ+,𝑋 (𝑋0 , 𝑌0 )
+ 𝜌 (𝑌 − 𝑌0 ) Ψ+,𝑌 (𝑋0 , 𝑌0 )] = 𝐶 (𝜌)
𝑝
𝑋0
𝑌0
𝑄𝐿 (𝑛, 𝑝) = 𝑄𝐿 (0, 0) (√𝑋0 ) (
)
𝑌0 + 𝑊0
𝑌0 + 𝑊0
𝑛
⋅
2
⋅
1 𝑖∞ V
1
[2√𝑋0 (√𝑋0 − 1)
∫
𝑋0 + 𝑌0 − √𝑋0 2𝜋𝑖 −𝑖∞ 𝑝
−
𝑋0 + 𝑌0 − √𝑋0
.
(265)
Evaluating the integral(s) in (264) we obtain (253). Note that
the coefficient of the 𝑂(𝑝−1 ) term in the integral is zero, so
the result is 𝑂(𝑝−3/2 ).
When 𝑌0 + 𝑋0 ≈ √𝑋0 the pole at 𝑢∗ and zero at 1/√𝑋0
are close together. If 𝑛, 𝑝 → ∞ so that 𝑛/𝑝 remains fixed and
positive, the saddle lies well to the right of these, and then
(249) holds. But if 𝑛, 𝑝 → ∞ with 𝑛 = 𝑂(√𝑝) then the
saddle, pole, and zero are all close. Then we must reexamine
the integrand in (245) and expand it about 𝑢 = 1/√𝑋0 .
Again setting V = √𝜌(𝑢 − 1/√𝑋0 ) = 𝑂(1) and replacing
𝑌0 by √𝑋0 − 𝑋0 + 𝛿/√𝜌 (with 𝛿 = 𝑂(1)), (245) becomes
asymptotically
𝑄𝐿 (𝑛, 𝑝) ∼ 𝑄 (0, 0) 𝑋0𝑛/2 (
⋅∫
Br+
+
𝑝
2√𝑋0 1
𝑌0
)
𝑌0 + 𝑊0
√𝜌 2𝜋𝑖
𝑋0 V
𝑛
√𝑋0 V
exp [−
𝑋0 V − 𝛿
√𝜌
3/2
𝑝 𝑋0
V2 ] 𝑑V,
𝜌 𝑌0 + 𝑊0
𝑛
𝑋0
) (𝑋0
𝑋0 + 𝑌0
𝑝
where
2𝑋03/2 (√𝑋0 − 1)
(𝑋0 + 𝑌0 )
(267)
√2𝜋
+ 𝑌0 ) ,
𝑋3/2 V2
𝑋02
V3
) 𝑑V,
] exp ( 0
𝑌0 + 𝑊0 √𝑝
𝑌0 + 𝑊0
B = −5𝑋03/2 + 3𝑋0 +
5/2
⋅ 𝑒𝜌[𝑋0 +𝑌0 −1−(𝑋0 +𝑌0 )log(𝑋0 +𝑌0 )] (
(264)
V
V
+B
+ 𝑂 (𝑝−1 )] [1 − √𝑋0 𝑛
] [1
√𝑝
√𝑝
(𝑋0 + 𝑌0 ) − 𝑋0
(266)
where we used 𝜌(𝑋 − 𝑋0 ) = −𝑛 and 𝜌(𝑌 − 𝑌0 ) = −𝑝. Near
̃
𝑋 = 𝑋0 , the curve 𝑌 = 𝑌(𝑋)
that separates D0 from D+ has
󸀠
̃
̃
the slope 𝑌 (𝑋0 ) = 𝑌0 /[(𝑋0 + 𝑌0 )2 − 𝑋0 ] so that 𝑌(𝑋)
can be
󸀠
̃ (𝑋0 )(𝑋 − 𝑋0 ),
approximated by the straight line 𝑌 − 𝑌0 = 𝑌
which is the same as 𝑝/𝑛 = 1/𝜃1 . Then D+ meets the corner
in the sector 𝑛/𝑝 ∈ [0, 𝜃1 ), and this corresponds precisely to
where the asymptotic result (248) applies. Comparing (267)
to (248) we see that the matching is possible, if 𝑄(0, 0) and
𝐶(𝜌) are related by
𝐶 (𝜌)
∼ √2𝜋√𝑋0 + 𝑌0 𝑒𝜌[1−𝑋0 −𝑌0 +(𝑋0 +𝑌0 )log(𝑋0 +𝑌0 )] 𝑄 (0, 0)
(268)
and this holds throughout R1 ∪ R2 ∪ R3 . For R1 ∪ R2
we have previously determined that 𝐶(𝜌) ∼ 𝜌−1/2 and thus
(268) determines 𝑄(0, 0), as in (30). For region R3 we have
𝑄(0, 0) ∼ 1 − 𝑋0 − 𝑌0 and then (268) leads to (33). For
R1 ∩R2 , corresponding to 𝑋0 +𝑌0 −1 = 𝛾/√𝜌 = 𝑂(𝜌−1/2 ), the
approximation in (208) holds for 𝑛 = 𝑂(1) and 𝑌 − (1 − 𝑋0 ) =
𝑂(𝜌−1/2 ), but now the upper boundary 𝑌 = 𝑌0 lies within this
range. Then applying the normalization condition in (10) to
(208) leads to
1
∞ ∞
𝜌
𝐶
2
(1 − 𝑋0 ) 𝑋0𝑛 exp [− (𝑌 + 𝑋0 − 1) ]
√
2
2𝜋
𝑝=0 𝑛=0
∼∑∑
30
Advances in Operations Research
2
𝜌 𝛾
𝑝
𝐶 ∞
=
− )]
∑ exp [− (
√2𝜋 𝑝=0
2 √𝜌 𝜌
∼
if we approach the corner along any straight line (excluding
slopes 0 and ∞) we are approaching from within D0 . By the
asymptotic matching principle, the expansion of 𝑄𝐿 (𝑛, 𝑝) for
𝑛, 𝑝 → ∞ should agree with the expansion of 𝐶0 𝐾𝑒𝜌Ψ as
𝑋 → 𝑋0 , 𝑌 → 𝑌0 . Thus in D0 we must compare (249) to
(225). But from (250) and (223) we see that (1 − 𝐴 𝑠 )𝑢𝑠 = 1
and from (67) we obtain
∞
𝛾
2
2
√𝜌
√𝜌
𝐶 ∫ 𝑒−(𝛾−V) /2 𝑑V =
𝐶 ∫ 𝑒−V /2 𝑑V
√2𝜋 0
√2𝜋 −∞
(269)
and then 𝐶 is given by (33). Then 𝑄𝐿 (0, 0) ∼ 𝑄(0, 0) can
be computed from (268), and since 𝜌[1 − 𝑋0 − 𝑌0 + (𝑋0 +
𝑌0 )log(𝑋0 + 𝑌0 )] ∼ 𝛾2 /2, we obtain the expression in (29). We
have thus determined both 𝑄(0, 0) and 𝐶(𝜌) for all cases of
the parameters 𝑋0 , 𝑌0 .
To match 𝑄(𝑛, 𝑝) to the ray expansion in D0 we note that,
for regions R1 ∪ R2 ∪ R3 , D0 meets the corner over the
sector 𝑛/𝑝 ∈ (𝜃1 , ∞), while for region R4 this holds for all
𝑛/𝑝 ∈ (0, ∞). We recall that for R4 the curve 𝑌 = 𝑌𝑐 (𝑋)
separates D0 from D− , but this curve has infinite slope at
𝑋 = 𝑋0 and is thus tangent to the line 𝑋 = 𝑋0 at 𝑌 = 𝑌0 . So
1 − 𝐵𝑠 =
𝑌0 𝑢𝑠
.
(𝑋0 + 𝑌0 − 1) 𝑢𝑠 − 1 − 𝑋0 𝑢𝑠2
It follows that
𝑛
(1 − 𝐴 𝑠 ) (1 − 𝐵𝑠 )
= 𝑢𝑠−𝑛 {
𝑝
𝑝
𝑢𝑠 𝑌0
}
[(1 + 𝑋0 + 𝑌0 ) 𝑢𝑠 − 1 − 𝑋0 𝑢𝑠2 ]
𝜌
𝑋0
4𝑋0
)−
]
√ [(1 + 𝑋0 + 𝑌0 ) (1 +
2
1 − 𝐴𝑠
√1 − 𝐴 𝑠 (1 − 𝐵𝑠 ) 𝑝
(1 − 𝐴 𝑠 )
−1/2
2
2
2
2
2
𝑄 (0, 0) (1 − 𝑢𝑠 ) (𝑋0 𝑢𝑠 − 1) [𝑛 (1 + 𝑋0 + 𝑌0 ) + 4 (𝑝 − 𝑛 ) 𝑋0 ]
∼
√2𝜋
𝑢3/2 (𝑋 + 𝑌 − 𝑋 𝑢 ) √(1 + 𝑋 + 𝑌 ) 𝑢 − 1 − 𝑋 𝑢2
0
0
Using (258) to express 𝑛/𝑝 in terms of 𝑢𝑠 , after some
calculation we find that
(1 + 𝑋0 + 𝑌0 )
=[
2
𝑛2
𝑛2
+
4𝑋
(1
−
)
0
𝑝2
𝑝2
(𝑋0 𝑢𝑠2 + 1) (1 + 𝑋0 + 𝑌0 ) − 4𝑋0 𝑢𝑠
𝑋0 𝑢𝑠2 + 1 − (1 + 𝑋0 + 𝑌0 ) 𝑢𝑠
2
(273)
] .
Then choosing 𝐶0 (𝜌) = 𝑄𝐿 (0, 0)/√𝜌 we have
𝐶0 (𝜌) ∼
1
𝑄 (0, 0)
√𝜌
(R1 ∪ R2 ∪ R3 ∪ R4 ) .
(274)
Then we use (273) to simplify (272) to
𝐾0 (𝐴 𝑠 ) √𝑌0
√1 − 𝐴 𝑠 (1 − 𝐵𝑠 )
=
𝑋0 𝑢𝑠2 − 1
1 1 − 𝑢𝑠
√2𝜋 𝑢𝑠3/2 𝑋0 + 𝑌0 − 𝑋0 𝑢𝑠
2
𝐴 𝑠 [(1 − 𝐴 𝑠 ) − 𝑋0 ]
1
=
√2𝜋 √1 − 𝐴 𝑠 [𝑌0 − (𝑋0 + 𝑌0 ) 𝐴 𝑠 ]
(275)
so we have determined the functional form of 𝐾0 (⋅), and
(275) along with (219) leads to (70). We have thus completely
determined the ray expansion in D0 , as 𝐶0 is known for all
parameter regions R𝑗 via (274).
(271)
and the matching is possible if
𝐶0 𝐾0 (𝐴 𝑠 ) √𝑌0
𝑠
(270)
0 𝑠
0
0
𝑠
(272)
−1/4
.
0 𝑠
It remains only to determine the constant 𝐶1 in the caustic
ray expansion, for regions R4 and R3 ∩ R4 (𝑋0 + 𝑌0 ≈
√𝑋0 ) (for R3 , 𝐶1 can be inferred from (214)). For this we
must relate 𝑄(0, 0) to 𝐶1 (𝜌) by asymptotic matching, but this
matching will require the analysis of another scale, which is
intermediate to the (𝑋, 𝑌) and (𝑛, 𝑝) scales. This analysis is
carried out in the next subsection.
4.3. Analysis of the Scale 𝑘 = 𝑚−𝑂(𝜌1/3 ), 𝑟 = 𝑅−𝑂(𝜌2/3 ). We
will consider 𝑚 − 𝑘 = ]1 𝜌1/3 = 𝑂(𝜌1/3 ) and 𝑅 − 𝑟 = 𝑇1 𝜌2/3 =
𝑂(𝜌2/3 ). Then 𝑋0 − 𝑋 = 𝑂(𝜌−2/3 ) and 𝑌0 − 𝑌 = 𝑂(𝜌−1/3 )
so we are examining the vicinity of the corner point (𝑋0 , 𝑌0 )
along parabolas, where (𝑌0 − 𝑌)2 /(𝑋0 − 𝑋) is constant. We
note that the scale 𝑘 = 𝑚 − 𝑂(𝜌1/3 ) (with 𝑟 = 𝜌𝑌 and
𝑌 > 0) was also important in the analysis of the model with
𝑅 = ∞ (see [10]). There, a more geometric interpretation
is given, in terms of caustic rays and caustic boundaries and
also in terms of sample paths of large deviations. Since the
caustic rays in region D− cannot fill the entire domain (for
parameter regions R4 and R3 ∩ R4 ) we would expect a
boundary effect near 𝑌 = 𝑌0 or 𝑟 = 𝑅. The corner scale
𝑘 = 𝑚 − 𝑂(1), 𝑟 = 𝑅 − 𝑂(1) that we analyzed in Section 4.2
is insufficient for fully understanding this boundary effect,
and hence we analyze the scaling 𝑟 = 𝑅 − 𝑂(𝜌2/3 ), which
will connect the cases 𝑟 = 𝑅 − 𝑂(1) and 𝑟 = 𝜌𝑌, 𝑌 < 𝑌0 .
We cannot give an a priori probabilistic argument of why this
scale is needed but mention that it leads to an interesting PDE,
namely, (285), and such problems arise in many other areas,
Advances in Operations Research
31
including queues with time-dependent arrival rates (see [20])
and steady two-dimensional convection-diffusion problems
past curved obstacles (see [21]).
We first set
𝑛
𝑄 (𝑛, 𝑝) = (√𝑋0 ) (
𝑝
𝑌0
) Q (𝑛, 𝑝)
𝑌0 + 𝑊0
[1 + 𝑋0 + 𝑌0 − 𝜌 (𝑛 + 𝑝)] Q (𝑛, 𝑝)
𝑊0 + 𝑌0
] Q (𝑛, 𝑝 − 1)
𝑌0
1
] Q (𝑛 − 1, 𝑝)
√𝑋0
(277)
𝑇1 =
⋅ Q (𝑛, 𝑝) − [𝑊0 + 𝑌0 + 𝜌
+ 𝑌0
(1 − 𝑝)]
𝑌0
5/3
𝑝2
𝑊0
] Q (𝑛, 𝑝)
2𝑌0 (𝑊0 + 𝑌0 ) 𝜌
𝑊0
𝜌1/3 𝑇12 ] Q (𝑛, 𝑝)
2𝑌0 (𝑊0 + 𝑌0 )
Q (]1 , 𝑇1 ) =
(278)
(279)
(281)
2
𝑌02 (𝑊0 + 𝑌0 )
𝜌
𝑎𝜌
Q𝑒
𝑄 (0, 0)
(𝑊0 + 𝑌0 )
(284)
,
where we isolated dominant term in (281) on the scale 𝑝 =
𝑂(𝜌2/3 ). Setting 𝑝 = 𝜌2/3 𝑇1 and 𝑛 = 𝜌1/3 ]1 , multiplying (278)
(285)
Including the exponential factor in (284) allows us to eliminate the last term in the left-hand side of (282), while the other
factors (such as 𝜌−2/3 and 𝑄(0, 0)) are purely for convenience.
We will show, by asymptotic matching between the (𝑛, 𝑝)
and (], 𝑇) scales, that Q must be proportional to 𝑄(0, 0),
and including the factor 𝜌−2/3 in (284) will lead to F being
asymptotically 𝑂(1).
Next we examine the boundary condition in (18), which
can be also written as (𝑋0 + 𝜌−1 )𝑃(𝑋0 + 𝜌−1 , 𝑌) = 𝑃(𝑋0 , 𝑌 −
𝜌−1 ), or, using the (𝑛, 𝑝) variables, as
(286)
In view of (276) and (279) we can also write (286) as
(√𝑋0 +
2
4𝑎2 𝑝2
4𝑎𝑝
2𝑎
=( 2 Q+
Q𝑝 + Q + Q𝑝𝑝 ) 𝑒𝑎𝑝 /𝜌
𝜌
𝜌
𝜌
𝑇12
(√𝑋0 − 𝑋0 −
𝑌0 ) 𝑋01/6
(𝑋0 + 𝜌−1 ) 𝑄 (−1, 𝑝) = 𝑄 (0, 𝑝 + 1) .
2
2𝑎𝑝
𝑊0
Q + Q𝑝 ) 𝑒𝑎𝑝 /𝜌 , 𝑎 = −
, (280)
𝜌
2𝑌0 (𝑊0 + 𝑌0 )
1/3
𝜌−2/3 (1 − √𝑋0 )
𝑇3 𝑊 (𝑊 + 2𝑌0 )
] F (], 𝑇)
⋅ exp [− 1 0 0
6 𝑌2 (𝑊0 + 𝑌0 )2
1 −1
𝜌 ) Q (−1, 𝑝)
√𝑋0
𝑌0
Q (0, 𝑝 + 1)
=
𝑌0 + 𝑊0
so that
∼
𝑇
F]] − ]F = F𝑇 ; ] > 0, 𝑇 > 0.
where the error term(s) involve derivatives of Q of order ⩾ 3
and terms of order 𝑂(𝜌−1 𝑝Q𝑝𝑝 ). We furthermore set
−2/3
2/3
(1 − √𝑋0 )
(283)
we obtain from (282) the separable, parabolic PDE
1
(𝑛 − 1) Q𝑛 (𝑛, 𝑝) + √𝑋0 Q𝑛𝑛 (𝑛, 𝑝) + ⋅ ⋅ ⋅ ,
√𝑋0
𝑊02 𝑇12
(𝑌0 + 𝑊0 ) 𝑋01/6
],
0
−1 𝑊0
1
⋅ Q𝑝 (𝑛, 𝑝) + (𝑊0 + 𝑌0 ) Q𝑝𝑝 (𝑛, 𝑝) + 𝜌−1
2
Q𝑝𝑝
𝑋0
)
1 − √X0
and letting
𝑊 + 𝑌0
1
(1 − 𝑝) +
0 = 𝜌 [𝑛 + 𝑝 + 0
(1 − 𝑛)]
𝑌0
√𝑋0
Q𝑝 = (
(282)
1/3
−1
= exp [−
𝜕2 Q
1
𝜕Q
+ ]1 (1 −
) Q − (𝑊0 + 𝑌0 )
𝜕𝑇1
𝜕]21
√𝑋0
𝑊 (𝑊 + 2𝑌0 ) 2
− 02 0
𝑇 Q = 0,
2𝑌0 (𝑊0 + 𝑌0 ) 1
To obtain a limiting PDE on the (]1 , 𝑇1 ) scale, we first
formally expand (277) using Q(𝑛±1, 𝑝) = Q(𝑛, 𝑝)±Q𝑛 (𝑛, 𝑝)+
(1/2)Q𝑛𝑛 (𝑛+𝑝)+⋅ ⋅ ⋅ , where we anticipate that 𝑛 will be scaled
to be large, with 𝑛 = 𝑂(𝜌1/3 ) and also 𝑝 = 𝑂(𝜌2/3 ). We thus
rewrite (277) as
Q (𝑛, 𝑝) = exp [−
, and using (279)–(281), we obtain the limiting
]1 = (
+ √𝑋0 Q (𝑛 + 1, 𝑝) .
⋅
𝑇12
which applies over the quarter plane ]1 > 0, 𝑇1 > 0, and
we now view Q as a function of (]1 , 𝑇1 ) instead of (𝑛, 𝑝). By
rescaling ]1 and 𝑇1 using
−1
+ [√𝑋0 + 𝜌−1 (1 − 𝑛)
√𝑋0
(276)
with which the main balance equation (228) becomes
= [𝑊0 + 𝑌0 + 𝜌−1 (1 − 𝑝)
1/3
by 𝜌2/3 𝑒−𝑎𝜌
PDE
(287)
or
(√𝑋0 +
𝑌0
1 −1
𝜌 ) Q (−1, 𝑝) =
𝑌0 + 𝑊0
√𝑋0
𝑌0
𝑎
⋅ exp [ (2𝑝 + 1)] Q (0, 𝑝 + 1) =
[1
𝜌
𝑌0 + 𝑊0
−
𝑊0
𝜌−1/3 𝑇1 + 𝑂 (𝜌−2/3 )] Q (0, 𝑝 + 1) ,
𝑌0 (𝑊0 + 𝑌0 )
where we used also (280) and (284).
(288)
32
Advances in Operations Research
]1 /√𝑇1 fixed. But 𝑛 = 𝜌1/3 ]1 , 𝑝 = 𝜌2/3 𝑇1 , and (283) shows
that
On the (]1 , 𝑇1 ) scale we have
Q (−1, 𝑝) = Q (−𝜌−1/3 , 𝑇1 )
= Q (0, 𝑇1 ) + 𝑂 (𝜌−1/3 ) ,
Q (0, 𝑝 + 1) = Q (0, 𝑇1 + 𝜌−2/3 )
2
√𝑋0 ]2
𝑛2 ] 1
=
=
,
𝑝
𝑇1 𝑊0 + 𝑌0 𝑇
(289)
2/3
(1 − √𝑋0 ) 𝑋01/12 ]
𝑛
=
.
3/2 𝑇3/2
𝑝3/2
𝜌2/3 (𝑊0 + 𝑌0 )
= Q (0, 𝑇1 ) + 𝑂 (𝜌−2/3 )
so as long as √𝑋0 ≠ 𝑌0 /(𝑌0 + 𝑊0 ) we conclude from (288)
that Q must vanish along ]1 = 0 for all 𝑇1 > 0, and thus, in
view of (284) so must F; hence
F (0, 𝑇) = 0, 𝑇 > 0.
(290)
The case where √𝑋0 = 𝑌0 /(𝑌0 + 𝑊0 ) corresponds to R3 ∩
R4 , and then the boundary condition will become more
complicated. First we analyze the interior of R4 where (290)
holds.
To analyze (285) with (290), we first derive the behavior of
F as ], 𝑇 → 0, by matching the expansions on the (𝑛, 𝑝) and
(], 𝑇) scales. To this end we need, for region R4 , the behavior
of 𝑄𝐿 (𝑛, 𝑝) for 𝑛, 𝑝 → ∞ with 𝑛 = 𝑂(√𝑝). This asymptotic
limit lies in the asymptotic matching region between (249)
(which applies for 0 < 𝑛/𝑝 < ∞ for region R4 ) and (253),
which applies for 𝑝 → ∞ but with 𝑛 = 𝑂(1). Setting 𝑢 =
1/√𝑋0 +V/√𝑝 in the integral in (245) and shifting the contour
Γ toward Re(𝑢) = 1/√𝑋0 we obtain
𝑛
𝑄𝐿 (𝑛, 𝑝) ∼ 𝑄 (0, 0) (√𝑋0 ) (
2𝑋03/2
(291)
𝑢
∼
𝑛
√𝑋0 V) (292)
√𝑝
𝑛
(√𝑋0 − 𝑋0 − 𝑌0 ) 𝑝3/2
⋅ exp (−
𝑇 > 0.
(296)
(297)
(298)
Introducing the Laplace transform
(299)
(296) becomes
̂ = −𝛿 (] − ]0 ) , ] > 0
̂ ]] − (] + 𝜃) F
F
(293)
2
𝑊0 + 𝑌0 𝑛
);
4√𝑋0 𝑝
𝑛, 𝑝 󳨀→ ∞ with 𝑛 = 𝑂 (√𝑝) .
By asymptotic matching the expansion on the (]1 , 𝑇1 ) scale
must behave as the right side of (293), when ]1 , 𝑇1 → 0 with
(300)
̂ 𝜃) = 0. The problem in (300) is a
and (298) implies that F(0,
standard Green’s function problem with solution
̂ = −𝜋 Ai (]0 + 𝜃) [Ai (] + 𝜃) Bi (𝜃)
F
Ai (𝜃)
− Bi (] + 𝜃) Ai (𝜃)] ,
(1 − √𝑋0 ) √𝑊0 + 𝑌0
𝑋01/4
]0 > 0
0
𝑝
𝑛
𝑌0
𝑄 (0, 0)
(√𝑋0 ) (
)
𝑌0 + 𝑊0
2√𝜋
⋅
F∗]] − ]F∗ = F∗𝑇 ; ] > 0, 𝑇 > 0
∞
and approximated Γ by a vertical contour. Evaluating the
integral in (291) explicitly we conclude that
𝑄𝐿 (𝑛, 𝑝) ∼
Note that (295) is consistent with the boundary condition in
(290) along ] = 0. The exponential factors in (279) and (284)
do not enter the matching condition, since 𝑇1 = 𝑂(1) and
𝜌1/3 𝑇12 = 𝜌−1 𝑝2 , and we can choose 𝑝 so that 𝑝 = 𝑜(𝜌1/2 ) in
the matching region.
To solve (285) subject to (295) it is useful to view the
function (𝜋𝑇)−1/2 exp[−]2 /(4𝑇)] as being an approximation
to the delta function 𝛿(]) for 𝑇 → 0, with the mass
concentrated in the range ] > 0. Then the right side of (295)
corresponds to the dipole −𝛿󸀠 (]). Consider the problem
̂=F
̂ (], 𝜃; ]0 ) = ∫ 𝑒−𝜃𝑇 F∗ (], 𝑇; ]0 ) 𝑑𝑇,
F
Here we used
𝑋0𝑛/2 exp (−
(295)
], 𝑇 󳨀→ 0 with ] = 𝑂 (√𝑇) .
F∗ (0, 𝑇) = 0,
𝑋03/2 2
𝑛
√𝑋0 V +
⋅ ∫ V exp [−
V ] 𝑑V.
𝑊0 + 𝑌0
√𝑝
−𝑖∞
−𝑛
]2
]
exp
(−
);
4𝑇
2√𝜋𝑇3/2
F∗ (], 0) = 𝛿 (] − ]0 ) ,
𝑖∞
1
V
=(
+
)
√𝑋0 √𝑝
F (], 𝑇) ∼
𝑌0
)
𝑊0 + 𝑌0
1
⋅
𝑝 (𝑊0 + 𝑌0 ) (𝑋0 + 𝑌0 − √𝑋0 ) 2𝜋𝑖
−𝑛
Then comparing (293) with (276) (using (279) and (284)) we
conclude that
𝑝
(√𝑋0 − 1)
(294)
(301)
0 < ] < ]0 ,
̂ = −𝜋 Ai (] + 𝜃) [Ai (]0 + 𝜃) Bi (𝜃)
F
Ai (𝜃)
(302)
− Bi (]0 + 𝜃) Ai (𝜃)] , ] > ]0 ,
where Ai(⋅) and Bi(⋅) are the Airy functions. Then the solution
F to (285) will be
󵄨󵄨
𝜕
󵄨
F (], 𝑇; ]0 )󵄨󵄨󵄨
.
F (], 𝑇) =
(303)
󵄨󵄨] =0
𝜕]0
0
Advances in Operations Research
33
By differentiating (302) with respect to ]0 , setting ]0 = 0, and
noting the Wronskian identity Ai(𝜃)Bi󸀠 (𝜃)−Ai󸀠 (𝜃)Bi(𝜃) = 𝜋−1
we find that
̂ 󵄨󵄨󵄨
𝜕F
Ai (] + 𝜃)
󵄨󵄨
, ] > 0.
=
(304)
󵄨
𝜕]0 󵄨󵄨󵄨]0 =0
Ai (𝜃)
Inverting the Laplace transform in (299) leads to F(], 𝑇) in
(170), so we have established the expansion on the (]1 , 𝑇1 )
scale. We can also easily verify that (295) is indeed satisfied,
since
1 𝑑 𝑖∞ Ai (] + 𝜃) 𝜃𝑇
𝑒 𝑑𝜃
F (], 𝑇) =
∫
2𝜋𝑖 𝑑𝑇 −𝑖∞ Ai (𝜃)
1 𝑑 𝑖∞ 𝑒𝜃𝑇−]
∼
∫
2𝜋𝑖 𝑑𝑇 −𝑖∞
𝜃
√𝜃
]2
)] 𝑑𝜃
[1 + 𝑂 (
√𝜃
1 𝑖∞ 𝜔2 𝑇−𝜔]
∼
𝜔 𝑑𝜔
∫ 𝑒
𝜋𝑖 −𝑖∞
=
(305)
𝑗=0
1 − √𝑋0
)
𝑋0
1/3
𝑛 ∞ −|𝑟𝑗 |𝑇
,
∑𝑒
𝜌1/3 𝑗=0
(306)
] 󳨀→ 0,
𝜋 (𝑘, 𝑟) ∼ 𝑄 (0, 0) 𝜌−1
0 = (𝑊0 − 1 − 𝑋0 ) Q (𝑛, 𝑇1 )
+ √𝑋0 [Q (𝑛 − 1, 𝑇1 ) + Q (𝑛 + 1, 𝑇1 )] .
(1 − √𝑋0 )
(𝑊0 + 𝑌0 ) √𝑋0 (√𝑋0 − 𝑋0 − 𝑌0 )
⋅(
𝑛
𝑝
√𝑋0 (𝑌0 + 𝑊0 )
𝑌0
) [𝑛 +
] 𝑓∗ (𝑇)
𝑌0 + 𝑊0
(𝑌0 + 𝑊0 ) √𝑋0 − 𝑌0
(309)
⋅ exp [−
−
(√𝑋0 )
𝑊0
𝜌1/3 𝑇12
2𝑌0 (𝑊0 + 𝑌0 )
𝑊0 (𝑊0 + 2𝑌0 )
2
𝑌0 (𝑊0 + 𝑌0 )
𝑇13 ] .
Here we made 𝑎(⋅) proportional to 𝑓∗ (⋅), which we will
determine shortly, and computed 𝑏(⋅) from (308). Now we
require (309), for 𝑛 → ∞, to match to the (]1 , 𝑇1 ) scale result,
for ]1 → 0. But in view of (306) we see that the matching is
possible provided that
∞
𝑓∗ (𝑇) = ∑ 𝑒−|𝑟𝑗 |𝑇 .
(310)
𝑗=0
Thus we now have three expansions valid for 𝑛 = 𝑂(1), on
the scales 𝑝 = 𝑂(1), 𝑝 = 𝑂(𝜌2/3 ), and 𝑝 = 𝑂(𝜌) with 𝑌 < 𝑌0 .
The matching between the 𝑝 = 𝑂(1) result and (309) follows
by letting 𝑝 → ∞ in (241) (corresponding to (253) for region
R4 ) and letting 𝑇 → 0. The matching condition is satisfied if
1 −3/2 √𝑊0 + 𝑌0 (1 − √𝑋0 )
𝑝
2√𝜋
𝑋01/4 (√𝑋0 − 𝑋0 − 𝑌0 )
(311)
2
and this holds for any fixed 𝑇 > 0. Consider the scale 𝑛 = 𝑂(1)
and 𝑇1 > 0. Viewing Q in (277) now as a function of 𝑛 and 𝑇1
and noting that 𝑝 → 𝑝 ± 1 corresponds to 𝑇1 → 𝑇1 ∓ 𝜌−2/3
the limiting form of (277) becomes
(307)
(308)
We have thus shown that the expansion on the (𝑛, 𝑇1 ) scale is
given by
2
Here we approximated the first integrand for 𝜃 → ∞ using
asymptotic properties of Ai(⋅).
We next relate the constant 𝐶1 in D− to 𝑄(0, 0) (∼ 1 −
𝑋0 − 𝑌0 in R4 ), having expressed the expansions near the
corner point (𝑋0 , 𝑌0 ) in terms of 𝑄(0, 0) on both the (𝑛, 𝑝)
and (]1 , 𝑇1 ) scales. As we discussed previously, the caustic
ray expansion in D− does not satisfy the boundary condition
along 𝑋 = 𝑋0 , and different expansions must be constructed
on the ]1 and 𝑛 scales, corresponding to 𝑋0 − 𝑋 = 𝑂(𝜌−2/3 )
and 𝑋0 − 𝑋 = 𝑂(𝜌−1 ). For region R4 , these expansions are
given, respectively, in (111) and (109) (with (110)), up to the
constant 𝐶1 . The details of their construction are given in [10].
We will need to carefully estimate 𝜋(𝑘, 𝑟) for 𝑛 = 𝑚−𝑘 = 𝑂(1).
For region R4 , (109) applies for 𝑛 = 𝑂(1) and all 𝑌 ∈ (0, 𝑌0 )
(but breaks down as either 𝑌 ↓ 0 or 𝑌 ↑ 𝑌0 ). For 𝑛 = 𝑂(1) and
𝑝 = 𝑂(1), (241) holds as then 𝜋(𝑘, 𝑟) ∼ 𝑄𝐿 (𝑛, 𝑝), and in (253)
we gave the asymptotics for 𝑄(𝑛, 𝑝) for 𝑝 → ∞ and 𝑛 = 𝑂(1).
However, we must derive yet another expansion on the scale
𝑇1 = 𝑂(1) (𝑌0 − 𝑌 = 𝑂(𝜌−1/3 )) with 𝑇1 > 0 and 𝑛 = 𝑂(1). The
expansion in (169) develops a nonuniformity as ] → 0, since
F(0, 𝑇) = 0. Since by definition of the Airy roots 𝑟𝑗 we have
Ai(𝑟𝑗 ) = 0, from the infinite series form in (170) we conclude
that
∞
𝑌0 𝑎 (𝑇1 ) = (𝑊0 + 𝑌0 ) √𝑋0 [𝑎 (𝑇1 ) − 𝑏 (𝑇1 )] .
⋅
]
]2
exp
(−
).
4𝑇
2√𝜋𝑇3/2
F (], 𝑇) ∼ ] ∑𝑒𝑟𝑗 𝑇 = (
But 𝑊0 = (1 − √𝑋0 )2 = 1 + 𝑋0 − 2√𝑋0 so that the general
solution of (307) is the linear function 𝑎(𝑇1 ) + 𝑛𝑏(𝑇1 ), where
𝑎(⋅) and 𝑏(⋅) are arbitrary functions of 𝑇1 . Note also that
(307) applies asymptotically whether the exponential factors
in (279) and (284) are included or excluded. The boundary
condition in (287) can be applied on the 𝑛 = 𝑂(1) scale and
leads asymptotically to (𝑊0 + 𝑌0 )√𝑋0 Q(−1, 𝑇1 ) = 𝑌0 Q(0, 𝑇1 ).
This restricts the functions 𝑎(⋅) and 𝑏(⋅) in the linear solution
to (307), by
∼
(1 − √𝑋0 ) 𝜌−1
(𝑊0 + 𝑌0 ) √𝑋0 (√𝑋0 − 𝑋0 − 𝑌0 )
𝑓∗ (𝑇) ,
in some intermediate limit where 𝑝 → ∞ and 𝑇1 = 𝜌−2/3 𝑝 →
0. Simplifying (311) by using (283) we must show that
𝑓∗ (𝑇) ∼
1 −3/2
𝑇 , 𝑇 󳨀→ 0.
2√𝜋
(312)
34
Advances in Operations Research
The Airy roots 𝑟𝑗 are well known to satisfy −𝑟𝑗 ∼ (3𝑗𝜋/2)2/3 ,
𝑗 → ∞, so for 𝑇 → 0 we approximate the sum in (310) by an
integral to get
∞
2/3
∞
3
∑𝑒−|𝑟𝑗 |𝑇 ∼ ∫ exp [− ( 𝜉𝜋) 𝑇] 𝑑𝜉
2
0
𝑗=0
=
∞
2 −3/2
3
(∫ 𝑒−𝑢 √𝑢 𝑑𝑢)
𝑇
3𝜋
2
0
=
1 −3/2 3
𝑇 Γ( )
𝜋
2
(313)
which verifies (312).
Now we match (309) for 𝑇 → ∞ with (109) as 𝑌 → 𝑌0 .
For 𝑇 → ∞ we have 𝑓∗ (𝑇) ∼ 𝑒−|𝑟0 |𝑇 as the zeroth term in the
sum in (310) dominates. By expanding (109) for 𝑌 → 𝑌0 , after
canceling some common factors the matching implies that
𝑝
𝜌−1 𝑄 (0, 0) 𝑊0
𝑌0
) 𝑒𝑟𝑇
(
(𝑊0 + 𝑌0 ) √𝑋0 (√𝑋0 − 𝑋0 − 𝑌0 ) 𝑌0 + 𝑊0
⋅ exp [−
−
𝑇3 ]
2 1
𝑌02 (𝑊0 + 𝑌0 )
∼𝜌
−1/6
1/3
⋅ 𝑒𝜌Φ(𝑋0 ,𝑌0 ) 𝑒𝜌Φ𝑌 (𝑋0 ,𝑌0 )(𝑌−𝑌0 ) 𝑒𝜌
−
(314)
+ 𝜌1/3 Φ󸀠∗ (𝑌0 ) (𝑌 − 𝑌0 )]
⋅
2√𝑋0
log (
(𝑌0 + 𝑊0 ) √𝑋0 − 𝑌0
𝑌0 + 𝑊0
1
1
)]
󸀠
√2𝜋 Ai (𝑟0 )
1 − √𝑋0
1/6
√𝑌0 𝑋05/6 (1 − √𝑋0 )
.
𝑒
=𝑒
(315)
2/3
2
3
(318)
𝑌0
[1
𝑌0 + 𝑊0
𝑊0
𝜌−1/3 𝑇1 + 𝑂 (𝜌−2/3 )] F (0, 𝑇1
𝑌0 (𝑊0 + 𝑌0 )
(319)
(𝑌0 + 𝑊0 ) √𝑋0 − 𝑌0
= (1 − √𝑋0 ) (√𝑋0 − 𝑋0 − 𝑌0 )
By using (110) to compute Φ𝑌𝑌 and Φ𝑌𝑌𝑌 and noting that
𝜌 (𝑌 − 𝑌0 ) = −𝑇13
𝑇13 ] F (], 𝑇)
But now
(1 − √𝑋0 ) ]
= 𝑒𝑟0 𝑇 .
= exp [𝑟0 𝑇1
1/6
(𝑊0 + 𝑌0 ) 𝑋0
[
]
𝜌 (𝑌 − 𝑌0 ) = 𝜌1/3 𝑇12 ,
2
6𝑌02 (𝑊0 + 𝑌0 )
+ 𝜌−2/3 ) .
𝑝
𝑌0
=(
) ,
𝑌0 + 𝑊0
exp [𝜌1/3 Φ󸀠∗ (𝑌0 ) (𝑌 − 𝑌0 )] = exp [−𝑇1 Φ󸀠∗ (𝑌0 )]
𝑊0 (2𝑌0 + 𝑊0 )
(317)
Note that in (317) we included the factor 𝜌−1/3 whereas in
(284) we have the two factors 𝜌−2/3 and [√𝑋0 − 𝑋0 − 𝑌0 ]−1 , as
the product of these becomes, in region R3 ∩ R4 , 𝑂(𝜌−2/3 ) ×
𝑂(𝜌1/3 ) = 𝑂(𝜌−1/3 ). Including the factor 𝜌−1/3 will insure that
F is 𝑂(1) for 𝜌 → ∞, and the factor (1 − √𝑋0 )1/3 𝑋0−1/3
is purely for convenience. But for region R3 ∩ R4 we must
reexamine the boundary condition in (288), which in terms
of F becomes
−
−𝑝Φ𝑌 (𝑋0 ,𝑌0 )
𝑛
𝑝
𝑌0
𝑊0
) exp [−
𝜌1/3 𝑇12
𝑌0 + 𝑊0
2𝑌0 (𝑊0 + 𝑌0 )
[√𝑋0 + 𝑂 (𝜌−1 )] F (−𝜌1/3 , 𝑇1 ) =
Now,
𝜌Φ𝑌 (𝑋0 ,𝑌0 )(𝑌−𝑌0 )
(√𝑋0 )
F]] − 𝜌F = F𝑇 ; ] > 0, 𝑇 > 0.
Φ∗ (𝑌0 )
1
3
+ 𝜌Φ𝑌𝑌𝑌 (𝑋0 , 𝑌0 ) (𝑌 − 𝑌0 )
6
1 + √𝑋0
1/3
1 − √𝑋0
)
𝑋0
since the factors in (317) correspond to the same sequence of
transformations we made for the analysis of the interior of
R4 , and F will again satisfy the PDE
𝐶1 (𝜌)
1
2
⋅ exp [ 𝜌Φ𝑌𝑌 (𝑋0 , 𝑌0 ) (𝑌 − 𝑌0 )
2
⋅ exp [−
𝜋 (𝑘, 𝑟) ∼ 𝑄 (0, 0) 𝜌−1/3 (
⋅(
𝑊0
𝜌1/3 𝑇12
2𝑌0 (𝑊0 + 𝑌0 )
𝑊0 (𝑊0 + 2𝑌0 )
we see that the dependence on 𝑇1 is exactly the same in the
left- and right-hand sides of (314). Thus we have a relation
between the constants 𝑄(0, 0) and 𝐶1 for region R4 , and since
𝑄(0, 0) ∼ 1 − 𝑋0 − 𝑌0 we have derived the expression for 𝐶1
in (41).
The analysis for region R4 breaks down as 𝑌0 → √𝑋0 −
𝑋0 , as then the expressions in (284), (293), (309), and (314)
all become singular. We thus examine the transitional case
R3 ∩ R4 , where 𝑋0 + 𝑌0 = √𝑋0 + 𝑂(𝜌−1/3 ). Then we certainly
have 𝑄(0, 0) ∼ 1 − 𝑋0 − 𝑌0 and (241) applies for any R𝑗 . But
now on parabolic scales, where 𝑛, 𝑝 → ∞ with 𝑛2 /𝑝 fixed,
𝑄(𝑛, 𝑝) has the expansion in (254). We now define F(], 𝑇)
by setting
(316)
(320)
= − (1 − √𝑋0 ) 𝛿∗ 𝜌−1/3
so that the limiting boundary condition from (319) becomes
√𝑋0
𝜕F
+ (𝛿∗ − 𝑇1 ) F = 0, ]1 = 0.
𝜕]1
(321)
Advances in Operations Research
35
Then scaling 𝛿∗ = 𝑋01/6 (1 − √𝑋0 )1/3 𝛿1 , in terms of (], 𝑇) (321)
becomes
F] (0, 𝑇) + (𝛿1 − 𝑇) F (0, 𝑇) = 0, 𝑇 > 0.
(322)
A matching condition for F, as ], 𝑇 → 0 with ] = 𝑂(√𝑇), is
obtained by comparing (317) to (254). From the definitions of
𝛿 and 𝛿∗ we have 𝛿𝜌−1/2 = 𝛿∗ 𝜌−1/3 so that 𝛿 = 𝛿∗ 𝜌1/6 . Then
also
𝑝𝛿2
= 𝑇1 𝛿∗2 ,
𝜌
𝑛𝛿
= ]1 𝛿∗ ,
√𝜌
(323)
1/3
F (], 𝑇)
1/2
1 (1 − √𝑋0 )
∼
√𝜋 𝑋01/4 √𝑇1
1 − √𝑋0 ]21
exp (−
)
4√𝑋0 𝑇1
(324)
]2
1
exp (− ) ;
√𝜋𝑇
4𝑇
(325)
], 𝑇 󳨀→ 0 with ] = 𝑂 (√𝑇) .
𝑖∞
⋅∫
−𝑖∞
1
2𝜋𝑖
∞
Ai (] + 𝜃) 𝜃(𝑇−𝛿1 )
𝑒
(∫
𝑒𝛿1 𝑢 Ai (𝑢) 𝑑𝑢) 𝑑𝜃.
Ai2 (𝜃)
𝜃
(326)
The right-hand side of (326) clearly satisfies (318), since this
PDE is separable and any function of the form 𝑔(𝜃) Ai(] +
𝜃)𝑒𝜃𝑇 is a solution, for any function 𝑔(⋅) and any complex 𝜃.
We can also verify that (322) holds along ] = 0, by using
󸀠
Ai (𝜃) 𝜃(𝑇−𝛿1 ) 𝛿1 − 𝑇 𝜃(𝑇−𝛿1 )
𝑒
+
𝑒
Ai (𝜃)
[Ai (𝜃)]2
=−
𝜃(𝑇−𝛿1 )
𝑑 𝑒
[
]
𝑑𝜃 Ai (𝜃)
0
3/2
⋅ 𝜃−1/4 [1 −
∞
1
𝜉2
)]
∫ exp [−√𝜃𝜉 + 𝑂 (
√4𝜋 0
√𝜃
3/2
𝜉
𝜉2
+ 𝑂 ( 2 )] 𝑑𝜉 = 𝑒−(2/3)𝜃
4𝜃
𝜃
(327)
(328)
𝛿
1 −1/4 1
1
+ 1 + 𝑜 (𝜃−1 )] = Ai (𝜃) [
[
𝜃
√4𝜋
√𝜃 𝜃
√𝜃
+
𝛿1
+ 𝑜 (𝜃−1 )] .
𝜃
Here we also expanded the integrand for small 𝜉, since by
Watson’s lemma the main contribution comes from the range
𝜉 = 𝑂(𝜃−1/2 ). Using (328) in (326) and noting that
1 𝑖∞ 𝜃𝑇−√𝜃] 𝑑𝜃
1
]2
=
exp (− ) ,
∫ 𝑒
√𝜃 √𝜋𝑇
2𝜋𝑖 −𝑖∞
4𝑇
(329)
we obtain, for ], 𝑇 → 0 with ] = 𝑂(√𝑇),
F (], 𝑇) =
Here we also used 𝑌0 + 𝑊0 ∼ 1 − √𝑋0 in region R3 ∩ R4 , so
that 𝑇1 ∼ (1 − √𝑋0 )1/3 𝑋01/6 𝑇.
The matching condition in (325) may be replaced by the
condition F|𝑇=0 = 𝛿(]) and we analyzed such problems in
[20] where it was found that the solution to (318) with (322)
is given by
F (], 𝑇) =
𝜃
∞
2
1 𝑖∞ 𝜃𝑇−√𝜃] 𝑑𝜃
1
𝑒−𝑢 /2 𝑑𝑢
=
∫
∫ 𝑒
√2𝜋 ]/√2𝑇
2𝜋𝑖 −𝑖∞
𝜃
and in terms of (], 𝑇) this simplifies to
F (], 𝑇) ∼
∞
∼ 𝑒−(2/3)𝜃
⋅
are all 𝑂(1), while 𝑝−1/2 = 𝜌−1/3 𝑇1−1/2 and 𝛿𝜌−1/2 = 𝛿∗ 𝜌−1/3 ,
so in the matching region, where 𝑇1 is small but 𝛿∗ = 𝑂(1),
the Gaussian first term in the braces in (254) dominates the
error function second term, and the matching will imply that
1 − √𝑋0
)
𝑋0
∞
𝑒−𝜃𝛿1 ∫ 𝑒𝛿1 𝑢 Ai (𝑢) 𝑑𝑢 = ∫ 𝑒𝛿1 𝜉 Ai (𝜃 + 𝜉) 𝑑𝜉
⋅ [1 + 𝛿1 𝜉 + 𝑂 (𝜉2 )]
2
𝑛2 ] 1
=
𝑝
𝑇1
(
and some integration by parts in (326). To verify the matching
condition in (325) we expand the integrand in (326) for ] →
0, 𝑇 → 0, and 𝜃 → ∞, using (190) to approximate the Airy
√
function. We have Ai(] + 𝜃)/Ai(𝜃) ∼ 𝑒−] 𝜃 and
]2
1
exp (− )
√𝜋𝑇
4𝑇
∞
2
2
𝑒−𝑢 /2 𝑑𝑢 + 𝑜 (1) .
+ √ 𝛿1 ∫
𝜋
]/√2𝑇
(330)
The expression in (330) is a two-term asymptotic approximation in this limit, and the leading term verifies the matching
condition in (325). The correction term in (330), which is
proportional to 𝛿1 , will asymptotically match to the term
proportional to the error function in (254), as in the matching
region 𝑝𝛿2 /𝜌 = 𝑇1 𝛿∗2 and 𝑛𝛿/√𝜌 = ]1 𝛿∗ are 𝑜(1), and
𝜂0 ∼
(1 − √𝑋0 )
√2𝑋01/4
1/2
]1
]
,
∼
√𝑇1 √2𝑇
1/3
1 − √𝑋0
2 𝛿
√
)
= 𝜌−1/3 (
𝑋0 √𝜌
𝑋0
(331)
√2𝛿1 .
We proceed to determine 𝐶1 . Now the approximation in
(209) (with (326)) remains valid on the scale 𝑛 = 𝑂(1), since
F(0, 𝑇) ≠ 0. Thus for 𝑛 = 𝑂(1) 𝜋(𝑘, 𝑟) can be approximated
by (317) with F(], 𝑇) replaced by F(0, 𝑇) and for 𝑇 → ∞
36
Advances in Operations Research
the expansion of (326) is determined by the pole at 𝜃 = 𝑟0 < 0,
which is a simple pole if ] = 0. Hence,
F (0, 𝑇) ∼
∞
𝑒𝑟0 (𝑇−𝛿1 )
(∫
𝑒𝛿1 𝑢 Ai (𝑢) 𝑑𝑢) ,
Ai󸀠 (𝑟0 ) 𝑟0
(332)
𝑇 󳨀→ ∞.
For 𝑛 = 𝑂(1) and 0 < 𝑌 < 𝑌0 , the result in (109) holds,
and this remains finite as 𝑌 ↑ 𝑌0 even if 𝑋0 + 𝑌0 = √𝑋0 . In
parameter region R3 ∩ R4 we have 𝑌0 + 𝑊0 ∼ 1 − √𝑋0 and
then as 𝑌 ↑ 𝑌0 , (109) becomes
𝜌−1/6 𝐶1 (𝜌) (√𝑋0 )
𝑛
1/3
⋅ 𝑒𝜌Φ(𝑋0 ,𝑌0 ) 𝑒−𝑝Φ𝑌 (𝑋0 ,𝑌0 ) 𝑒𝜌
Φ∗ (𝑌0 ) 𝑟0 𝑇
𝑒
1
⋅ exp [ 𝜌1/3 Φ𝑌𝑌 (𝑋0 , 𝑌0 ) 𝑇12
2
F (], 𝑇) ∼
− √𝑋0 )
𝑋0−7/12 .
𝜌1/3 Φ∗ (𝑌0 )
= −𝜌
𝑟0
= −𝜌1/3 𝑟0
(1 − √𝑋0 )
2/3
𝑋01/6
(1 − √𝑋0 )
𝑋01/6
∼ −𝑟0 (1 − √𝑋0 )
−1/3
log (1 +
−1/3
𝜌 𝛿∗
)
1 − √𝑋0
2
4𝑇2
=
1
1
1
F∼
−
]
[
√2𝜋 2√𝜃𝑠 2√𝜃𝑠 + ]
∞
𝑄 (0, 0) 𝜌−1/3 𝑋0−1/3 (∫ 𝑒𝛿1 𝑢 Ai (𝑢) 𝑑𝑢)
𝐶1 (𝜌) −1/6 −7/12 𝜌Φ(𝑋0 ,𝑌0 )
.
𝜌 𝑋0 𝑒
√2𝜋
(] − 𝑇2 )
]2
] 𝑇2
−
+ .
4𝑇2 2
4
(338)
Taking ]/𝑇2 > 1 then a standard saddle point estimate leads
to
Thus (333) agrees with (317), with ] = 0 and 𝑇 → ∞, if
∼
𝜃 = 𝜃𝑠 ≡
(334)
𝑋0−1/6 𝛿∗ = −𝑟0 𝛿1 .
𝑟0
𝑑
2
2
[𝜃𝑇 + 𝜃3/2 − (𝜃 + ])3/2 ] = 𝑇 + √𝜃 − √𝜃 + ]
𝑑𝜃
3
3
(337)
=0
and this occurs when
𝑌 + 𝑊0
log ( 0
)
1 − √𝑋0
2/3
(336)
Here we can view 𝜃 as being scaled to be 𝑂(]) and 𝑇 to be
𝑂(√]). The integrand in (336) has a saddle point where
Comparing (333) to (317) with (332), the exponential parts
agree as for region R4 , and we now have
1/3
𝜃 1/4 𝜃𝑇
1
) 𝑒
∫ (
2𝜋𝑖 Br ] + 𝜃
2
⋅ exp { [𝜃3/2 − (𝜃 + ])3/2 ]} 𝑑𝜃.
3
(333)
1
1
1
(1
− Φ𝑌𝑌𝑌 (𝑋0 , 𝑌0 ) 𝑇13 ]
󸀠
√2𝜋 Ai (𝑟0 )
6
1/3
(𝑋, 𝑌) → (𝑋0 , 𝑌0 ) along parabolic scales with (𝑌0 −𝑌)2 /(𝑋0 −
𝑋) (= 𝑇12 /]1 ) held fixed and letting ]1 , 𝑇1 → ∞ in (169). We
have already verified that along linear scales, where (𝑋, 𝑌) →
(𝑋0 , 𝑌0 ) with (𝑌0 − 𝑌)/(𝑋0 − 𝑋) (= 𝑝/𝑛) fixed, the D0 ray
expansion matches directly to (168). The matching of the D0
and (], 𝑇) scale expansions will show that there are no “gaps”
in the asymptotics, which would require the analysis of yet
other scales.
We first expand F(], 𝑇) in (170) for ], 𝑇 → ∞, and we
note that, along parabolic scales with ] = 𝑂(𝑇2 ), ]/𝑇2 > 1
corresponds to moving from the corner (𝑋0 , 𝑌0 ) in region D0 ,
while ]/𝑇2 < 1 corresponds to moving into D− . By using
(190) to approximate the integrand in (170) and shifting the
contour to the right, to the range Re(𝜃) ≫ 1, we are led to
(335)
But now 𝑄(0, 0) ∼ 1 − 𝑋0 − 𝑌0 ∼ 1 − √𝑋0 and thus (335) leads
to the result in (37).
We have now determined the values of 𝑄(0, 0), 𝐶0 (𝜌),
𝐶(𝜌), and 𝐶1 (𝜌) for all possible regions R𝑗 of parameter
space, thus establishing Propositions 1 and 2. We also have
analyzed all of the relevant scales near the corner point
(𝑋0 , 𝑌0 ) of the state space and thus established all parts
of Proposition 15. As a final step we briefly discuss the
asymptotic matching between the ray expansion in D0 and
the expansion on the (], 𝑇) scale in (169). We only consider
region R4 and note that this matching will involve letting
⋅ exp {𝜃𝑠 𝑇 +
⋅ exp (
−1/2
(
1/4
𝜃𝑠
)
𝜃𝑠 + ]
2 3/2
] − 𝑇2
3/2
[𝜃𝑠 − (𝜃𝑠 + ]) ]} =
3
2√𝜋𝑇3/2
(339)
1 3 1
1 ]2
𝑇 − ]𝑇 −
) ; ], 𝑇 󳨀→ ∞.
12
2
4𝑇
By using (339) in (169) we match the result to the D0 ray
expansion, which is 𝜌−1/2 𝑄(0, 0)𝐾(𝑋, 𝑌)𝑒𝜌Ψ(𝑋,𝑌) , as (𝑋, 𝑌) →
(𝑋0 , 𝑌0 ). By separating the exponential factors from the
algebraic ones, the matching will hold if both
𝜌−1/2 𝐾 (𝑋, 𝑌) ∼ 𝜌−2/3
5/3
⋅
𝑋0−1/6
] − 𝑇2
,
2√𝜋 (𝑊0 + 𝑌0 ) (√𝑋0 − 𝑋0 − 𝑌0 ) 𝑇3/2
(1 − √X0 )
(340)
Advances in Operations Research
𝑛
𝑒𝜌Ψ(𝑋,𝑌) ∼ (√𝑋0 ) (
⋅ exp [−𝜌1/3
⋅ exp [−
−
37
𝑝
𝑌0
)
𝑌0 + 𝑊0
Now
𝑊0
𝑇2 ]
2𝑌0 (𝑊0 + 𝑌0 ) 1
𝑊0 (𝑊0 + 2𝑌0 )
𝑇3
2 1
6𝑌02 (𝑊0 + 𝑌0 )
𝐾 (𝑋, 𝑌)
(341)
1
1
+ 𝑇3 − ]𝑇
12
2
1/2
We thus proceed to evaluate 𝐾 and Ψ as (X, 𝑌) → (𝑋0 , 𝑌0 )
along parabolic scales, where (𝑌0 − 𝑌)2 /(𝑋0 − 𝑋) is fixed. In
this limit we have 𝐴 → 𝐴 min = 1 − √𝑋0 and 𝐵 → 𝐵min =
𝑊0 /(𝑌0 + 𝑊0 ), and it is useful to rewrite (68) as
𝑋0
+ 𝐴𝑒𝜏 − 1) ,
1−𝐴
exp [𝜌 (𝑋 − 𝑋0 ) Ψ𝑋 (𝑋0 , 𝑌0 )]
𝜏 󳨀→ 0, 𝐴 󳨀→ 1 − √𝑋0 .
(344)
1/3
1 (1 − √𝑋0 )
,
√2𝑇 𝑋01/12 √𝑌0 + 𝑊0
(345)
where we wrote the result in terms of 𝑇 rather than 𝑇1 . From
the first expression in (342) we have
𝑋0 − 𝑋 ∼ 𝜏 [2 (𝐴 − 𝐴 min ) + 𝐴 min 𝜏]
(346)
Then peeling off the linear part of Ψ near 𝑋 = 𝑋0 , 𝑌 = 𝑌0 ,
(341) is equivalent to showing that
+ (𝑌0 − 𝑌) Ψ𝑌 (𝑋0 , 𝑌0 )] ∼ −𝜌1/3
⋅ 𝑇12 −
𝑊0 (𝑊0 + 2𝑌0 )
6𝑌02 (𝑊0 + 𝑌0 )
1
[𝑋0 − 𝑋 − (1 − √𝑋0 ) 𝜏2 ]
2𝜏
1/3 ] − 𝑇2
1
= 𝜌−1/3 X01/6 (1 − √𝑋0 )
.
2
𝑇
1 3 1
1]
𝑇 − ]𝑇 −
.
12
2
4𝑇
+ (𝑌0 − 𝑌) Ψ𝑌 (𝑋0 , 𝑌0 )
2
− 𝑋0 log (
(347)
𝑇13 +
(350)
2
Ψ (𝑋, 𝑌) + (𝑋0 − 𝑋) Ψ𝑋 (𝑋0 , 𝑌0 )
1 − 𝐴𝑒𝜏
)
1 − 𝐴 min
+ (𝑌0 − 𝑌) log (
(𝑌 − 𝑌)
1 𝑊0 + 𝑌0
]
[𝑋0 − 𝑋 − (1 − √𝑋0 ) 0
2
2 𝑌0 − 𝑌
(𝑊0 + 𝑌)
1 − √𝑋0 2
𝑊 + 𝑌0
1
= 𝜌−1/3 0
[]1 −
𝑇]
2 1
2
𝑇1
(𝑊0 + 𝑌0 )
2
𝑊0
2𝑌0 (𝑊0 + 𝑌0 )
We can rewrite Ψ in (66) as
= (𝑋0 − 𝑋) log (
so that
∼
𝑌0
] .
(𝑌0 + 𝑊0 )
𝜌 [Ψ (𝑋, 𝑌) + (𝑋0 − 𝑋) Ψ𝑋 (𝑋0 , 𝑌0 )
But from (342) we have 𝑌0 − 𝑌 = 𝜌−1/3 𝑇1 ∼ (𝑊0 + 𝑌0 )𝜏 so that
𝐴 − 𝐴 min ∼
(349)
𝑝
(343)
Expanding the Jacobian Δ in (71) for 𝜏 → 0 and 𝐴 → 𝐴 min
leads to
∼𝜌
so using (347) and (345) we can easily verify that (340) is
satisfied.
To verify (341) we first note that Ψ(𝑋0 , 𝑌0 ) = 0,
exp [𝜌 (𝑌 − 𝑌0 ) Ψ𝑌 (𝑋0 , 𝑌0 )] = [
𝐴 󳨀→ 1 − √𝑋0 .
|Δ|
(348)
𝑌0−1/2 |Δ|−1/2 𝐾0 (𝐴) ,
(342)
√2𝑌0 (1 − √𝑋0 ) (𝐴 − 𝐴 min )
𝐾0 (𝐴) ∼
,
√𝜋 (√𝑋0 − 𝑌0 − 𝑋0 ) (𝑊0 + 𝑌0 )
1/6
|Δ|−1/2 𝐾0 (𝐴)
𝑛
The function 𝐾0 (𝐴) in (275) vanishes in this limit, in view of
the factor (1 − 𝐴)2 − 𝑋0 , and we have
−1/2
−1/2
= exp [𝜌 (𝑋0 − 𝑋) log (1 − 𝐴 min )] = (√𝑋0 ) ,
𝑌0
𝑌0 − 𝑌 =
(1 − 𝑒−𝜏 ) .
1−𝐵
Δ ∼ −2𝜏 (𝑊0 + 𝑌0 ) ;
(𝑒−𝜏 − 𝐵)
∼ 𝑋0−1/4 (𝑌0 + 𝑊0 )
1 ]2
].
4𝑇
𝑋0 − 𝑋 = (1 − 𝑒−𝜏 ) (
−1/2
= (𝑒−𝜏 − 𝐴)
1 − 𝐵𝑒𝜏
)
1 − 𝐵min
(351)
1 − 𝐴𝑒𝜏
1 − 𝐵𝑒𝜏
) − 𝑌0 log (
)
1−𝐴
1−𝐵
+ 𝐴 (1 − 𝑒𝜏 ) ,
where we have partially expressed the right-hand side of (351)
in terms of 𝑋0 − 𝑋 = 𝜌−2/3 ]1 and 𝑌0 − 𝑌 = 𝜌−1/3 𝑇1 . Next we
38
Advances in Operations Research
expand (351) as a triple Taylor series about 𝜏 = 0, 𝐴 = 𝐴 min ,
and 𝐵 = 𝐵min . We have
𝐴 (1 − 𝑒𝜏 ) − 𝑋0 log [1 −
− 𝑌0 log [1 −
+ 𝐴 min 𝜏] ∼ −
𝐴 (𝑒𝜏 − 1)
]
1−𝐴
− √𝑋0 )
𝐵 (𝑒𝜏 − 1)
]
1−𝐵
⋅
𝑌𝐵
𝑋𝐴
= [ 0 + 0 − 𝐴] (𝑒𝜏 − 1)
1−𝐴 1−𝐵
+[
1 − 𝐵𝑒𝜏
) = (𝑌0 − 𝑌) log [1
1 − 𝐵min
= 𝜌−1/3 𝑇1 [−
(355)
From (342) we have
𝑌0
1
2
∼−
(𝐴 − 𝐴 min ) .
2
√𝑋0 (𝑌0 + 𝑊0 )
(353)
By (347), on the (], 𝑇) scale, 𝐴 − 𝐴 min is 𝑂(𝜌 ) so that 𝐵 −
𝐵min = 𝑂(𝜌−2/3 ), and thus (352) may be further approximated
by (using (𝑒𝜏 − 1)2 = 𝜏2 + 𝜏3 + 𝑂(𝜏4 ))
𝑋0 𝐴2min
+
2
2 (1 − 𝐴 min )
+
+
+
2
𝑌0 𝐵min
2
2 (1 − 𝐵min )
+
𝑋0 𝐴3min
(𝑌0 − 𝑌) (1 − 𝐵) = 𝜌−1/3 𝑇1 [1 − 𝐵min + 𝑂 (𝜌−2/3 )]
= 𝑌0 (1 − 𝑒−𝜏 )
3
3 (1 − 𝐴 min )
𝑋0 𝐴2min
3
]
𝜏
+
[
3
2
3 (1 − 𝐵min )
2 (1 − 𝐴 min )
2
2 (1 − 𝐵min )
𝑋0 𝐴2min
3
(1 − 𝐴 min )
𝑋0 𝐴 min
2
]𝜏 + [
(354)
(356)
1
= 𝑌0 [𝜏 − 𝜏2 + 𝑂 (𝜏3 )]
2
so we can refine the estimate above (345) to
𝜏 = 𝜌−1/3
3
𝑌0 𝐵min
2
𝑌0 𝐵min
+ 𝑂 (𝜏3 )] .
2
2 (1 − 𝐵min )
−1/3
[
𝐵 − 𝐵min
𝐵 𝜏
− min
1 − 𝐵min 1 − 𝐵min
𝐵min 𝜏2
−
But the relation in (216) implies that the coefficient of (𝑒𝜏 − 1)
in (352) vanishes and also that
𝐵 − 𝐵min
(1 − √𝑋0 )
1
[ 𝑋01/6 (1
2
𝐵 − 𝐵min 𝐵min (𝑒𝜏 − 1)
2
−
+ 𝑂 ((𝐵 − 𝐵min ) )]
1 − 𝐵min
1 − 𝐵min
−
+ 𝑂 (𝜏4 ) .
1/3
] + 𝑇2
,
𝑇
(352)
𝑋0 𝐴3
𝑌0 𝐵3
3
+
] (𝑒𝜏 − 1)
3
3
3 (1 − 𝐴)
3 (1 − 𝐵)
𝑋01/6
] − 𝑇2 (1 − √𝑋0 )
]
𝑇1 ] = −
+
𝑇
𝑊0 + 𝑌0
2𝜌
1/3
(𝑌0 − 𝑌) log (
𝑋0 𝐴2
𝑌0 𝐵2
2
+[
+
] (𝑒𝜏 − 1)
2 (1 − 𝐴)2 2 (1 − 𝐵)2
𝜌−1 ]
𝑇12
𝑇1
1
+ 𝜌−2/3
+ 𝑂 (𝜌−1 ) . (357)
2
𝑊0 + 𝑌0 2
(𝑊0 + 𝑌0 )
Adding the expressions in (354) and (355) and using the
estimates in (347), (353), and (357), (351) becomes, when
multiplied by 𝜌,
2
2
(1 − 𝐴 min )
] (𝐴 − 𝐴 min ) 𝜏2
[
⋅
with an 𝑂(𝜏4 ) error, which is not needed for the matching
verification since exp[𝑂(𝜌𝜏4 )] = exp[𝑂(𝜌−1/3 )] ∼ 1. We also
have
3
𝑊3
𝑊2 (1 − √𝑋0 )
(1 − √𝑋0 )
+ 02 ]
+ 0 +
2
2𝑌0
3𝑌0
3√𝑋0
𝑇13
2
3
(𝑊0 + 𝑌0 )
+ [1 − √𝑋0 +
1/3
⋅
𝑋01/6 (1 − √𝑋0 )
(1 − √𝑋0 )
√𝑋0
]
𝑇12 (] − 𝑇2 )
2
𝑇
2 (𝑊0 + 𝑌0 )
2
(𝑋0 − 𝑋) log (
−
𝜏
1 − 𝐴𝑒
) = 𝜌−2/3 ]1 log [1
1 − 𝐴 min
𝐴 min 𝜏
𝐴 − 𝐴 min
−
1 − 𝐴 min 1 − 𝐴 min
+ 𝑂 (𝜏2 , 𝜏 (𝐴 − 𝐴 min ))] ∼ −
(1 − √𝑋0 )
𝑊2
+[
+ 0]
2
2𝑌0
⋅ [𝜌1/3
𝜌−2/3 ]1
[𝐴 − 𝐴 min
√𝑋0
+
𝑇12
(𝑊0 + 𝑌0 )
2
+
𝑇13
]2 ]𝑇
]
−
−
3
2𝑇
2
(𝑊0 + 𝑌0 )
𝑌0
𝑊0 + 𝑌0
1
𝑇1
2
𝑌0
(𝑊0 + 𝑌0 ) √𝑋0
Advances in Operations Research
2/3
𝑋1/3 (1 − √𝑋0 )
⋅ 0
4
⋅ 𝑇1 (
−
(
39
2
𝑊 (𝑊 + 𝑌 )
]
− 𝑇) − 0 0 2 0
𝑇
𝑤𝑌0
2
𝑇1
)
𝑊0 + 𝑌0
𝑇13
𝑊0 1/3 𝑇12
[𝜌
+
].
𝑌0
𝑊0 + 𝑌0 2 (𝑊0 + 𝑌0 )2
(358)
Comparing (358) to (350) we find, after some simplification,
that they are identical. The comparison is facilitated by
separately comparing terms proportional to 𝜌1/3 𝑇12 , ]2 /𝑇,
]𝑇, and 𝑇3 . This verifies the matching between the D0 ray
expansion and that on the (], 𝑇) scale. The Airy functions
that arose on the (], 𝑇) scale disappear in the limit where
], 𝑇 → ∞. If we let ], 𝑇 → ∞ with ]/𝑇2 < 1, then the
expansion of F(], 𝑇) is much different, and now we have
F (], 𝑇)
∼
(359)
1
2
1
exp [− ]3/2 + 𝑟0 (𝑇 − √])] ,
󸀠
1/4
3
2√𝜋Ai (𝑟0 ) ]
as the pole at 𝜃 = 𝑟0 determines the asymptotic behavior.
We can show that then (169) with (359) agrees with the ray
1/3
expansion in D− , that is, 𝐶1 𝐿(𝑋, 𝑌)𝑒𝜌Φ(𝑋,𝑌) 𝑒𝜌 Φ∗ (𝑠1 ) , as the
latter is expanded for (𝑋, 𝑌) → (𝑋0 , 𝑌0 ), for parameter
region R4 . The case where ]/𝑇2 ≈ 1 will be important in
determining the transition layer expansion that applies where
D0 meets D− , which is along the curve 𝑌 = 𝑌𝑐 (𝑋). Then
neither (339) nor (359) apply.
5. Asymptotic Expansions
near State Space Boundaries
We discuss the four boundary segments, 𝑋 = 0, 𝑋 = 𝑋0 ,
𝑌 = 0, and 𝑌 = 𝑌0 , of the state space rectangle, avoiding for
now the corner points. From Figures 3–6 we see that D+ will
border 𝑋 = 0 and 𝑋 = 𝑋0 for parameter regions R2 and
R3 and also border 𝑌 = 0 for region R1 . As we previously
discussed the ray expansion in D+ remains valid near 𝑋 = 𝑋0
but breaks down near 𝑌 = 0 (R1 ) and 𝑋 = 0 (R2 ∪ R3 ). The
analysis of 𝑌 = 0 (corresponding to the scales 𝑟 = 𝑂(1) and
𝑟 = 𝑂(√𝜌)) is identical to that of the infinite capacity model,
and we show in [10] how to construct appropriate boundary
layer corrections; this leads to (88)–(95) in Proposition 8. The
analysis near 𝑋 = 0 (with 0 < 𝑌 < [(𝑋0 + 𝑌0 )2 − 𝑋0 ]/(𝑋0 +
𝑌0 )2 ) is also very similar to that in [10], except that if 𝑋0 +
𝑌0 < 1 (R2 ∪ R3 ) we must use the values of 𝐶(𝜌) in (32)
and (33). Thus both the D+ ray expansion and the boundary
layer expansion, which applies for 𝑘 = 𝑂(1), are rescaled by a
constant. We thus obtain (114).
For parameter region(s) R2 ∪ R3 ∪ R4 , D− meets the
line 𝑋 = 𝑋0 , and there the expansions in (109)–(112) apply.
Their derivation for R2 is identical to that in [10], while
for the other subcases we must simply multiply both the
D− ray expansion and the boundary layer corrections by
the appropriate 𝐶1 (𝜌), from among (35), (36), (37), and (41).
Similarly, the boundary layer correction along 𝑌 = 0 is given
by (96), with the appropriate value of 𝐶1 (𝜌). For R2 ∪ R3 this
covers the entire range 0 < 𝑋 < 𝑋0 , but for R4 only the range
𝑋𝐿 < 𝑋 < 𝑋0 (with 𝑋𝐿 defined in item (vii) in Proposition 8),
as in the range 0 < 𝑋 < 𝑋𝐿 , D0 meets 𝑌 = 0.
Here we will only examine where D0 meets the state space
boundaries, and this occurs along 𝑌 = 𝑌0 for all regions R𝑗 ,
along 𝑋 = 0 with [(𝑋0 + 𝑌0 )2 − 𝑋0 ]/(𝑋0 + 𝑌0 )2 < 𝑌 < 𝑌0 for
R1 ∪ R2 ∪ R3 , and all 0 < 𝑌 < 𝑌0 for R4 , and along 𝑌 = 0
with 0 < 𝑋 < 𝑋𝐿 for R4 only. We thus proceed to construct
boundary layer corrections to the D0 ray expansion for these
three boundary segments.
5.1. The Boundary Segment 𝑋 = 0. We consider the scale 𝑘 =
𝑂(1) and let
𝜋 (𝑘, 𝑟) = 𝐶2 (𝜌) 𝑒𝜌Ψ(0,𝑌) 𝜌𝑘 Π𝑘 (𝑌; 𝜌) .
(360)
Then the main balance equation (3), after dividing by
exp[𝜌Ψ(0, 𝑌)], becomes
[𝜌 (1 + 𝑌) + 𝑘] Π𝑘 (𝑌; 𝜌) = (𝜌𝑌 + 1) Π𝑘 (𝑌 + 𝜌−1 ; 𝜌)
⋅ exp {𝜌 [Ψ (0, 𝑌 + 𝜌−1 ) − Ψ (0, 𝑌)]} + 𝜌 (𝑘 + 1)
(361)
⋅ Π𝑘+1 (𝑌; 𝜌) + Π𝑘−1 (𝑌; 𝜌) .
Taking Π𝑘 (𝑌) to be the leading order approximation to
Π𝑘 (𝑌; 𝜌), dividing (361) by 𝜌, and letting 𝜌 → ∞ lead to the
limiting equation
[1 + 𝑌 − 𝑌𝑒Ψ𝑌 (0,𝑌) ] Π𝑘 (𝑌) = (𝑘 + 1) Π𝑘+1 (𝑌) ,
𝑘 ⩾ 1.
(362)
By examining the boundary condition in (6) we conclude that
(362) holds also if 𝑘 = 0, and thus
Π𝑘 (𝑌) =
𝑘
Π0 (𝑌)
[1 + 𝑌 − 𝑌𝑒Ψ𝑌 (0,𝑌) ] .
𝑘!
(363)
To determine 𝐶2 (𝜌) and Π0 (𝑌) we asymptotically
match (360) to the D0 ray expansion. We thus expand
𝜌−1/2 𝑄(0, 0)𝐾(𝑋, 𝑌)𝑒𝜌Ψ(𝑋,𝑌) as 𝑋 → 0 and compare the result
to the large 𝑘 expansion of (360). By using Stirling’s formula
we see that the matching is possible if, as 𝑋 → 0,
Ψ (𝑋, 𝑌)
= Ψ (0, 𝑌) − 𝑋 log 𝑋
(364)
+ 𝑋 [1 + log (1 + 𝑌 − 𝑌𝑒Ψ𝑌 (0,𝑌) )] + 𝑜 (𝑋) ,
𝐶 (𝜌)
𝑄 (0, 0)
Π0 (𝑌) ,
𝐾 (𝑋, 𝑌) ∼ 2
√2𝜋𝜌𝑋
√𝜌
𝑋 󳨀→ 0.
(365)
Note that the exponential factor in (360) must be included
for the matching to be possible. Thus we could set 𝐶2 (𝜌) =
𝑄(0, 0) and then Π0 (𝑌) will be the limit of √2𝜋𝑋𝐾(𝑋, 𝑌) as
𝑋 → 0, which we show below to be finite and nonzero.
40
Advances in Operations Research
From (68) we see that 𝑋 = 0 corresponds to 𝐴 = 𝑒−𝜏
and we set 𝐴 1 (𝑌) = 𝐴(0, 𝑌), where we view 𝐴 as a function
of (𝑋, 𝑌) via the mapping in (68). Then we also set 𝐵1 (𝑌) =
𝐵(0, 𝑌) and from (68) find that
𝐴 1 − 𝐵1 =
𝑌 (1 − 𝐴 1 )
.
𝑌0 − 𝑌
(366)
Using (366) in (67) leads to a quadratic equation for 𝐴 1 (𝑌),
whose solution is given by (118). Also, replacing (𝑒𝜏 , 𝐴, 𝐵)
by (1/𝐴 1 , 𝐴 1 , 𝐵1 ) in (66) leads to the expression in (119) for
Ψ(0, 𝑌), which is an explicit function of 𝑌. As 𝑋 → 0,
𝑋/(𝑒−𝜏 − 𝐴) is finite and from (68) we obtain
𝑒−𝜏 − 𝐴 ∼
𝐴 1 (1 − 𝐴 1 )
𝑋, 𝑋 󳨀→ 0.
𝐴 1 (𝐴 1 + 𝑋0 − 2) + 1
(367)
Evaluating the Jacobian Δ in (71) along 𝑋 = 0 leads to Δ 𝐿 (𝑌)
in (120). Thus as 𝑋 → 0 we have
(𝑒−𝜏 − 𝐴)
−1/2
(𝑒−𝜏 − 𝐵)
−1/2
∼ (𝐴 1 − 𝐵1 )
⋅√
−1/2
|Δ|−1/2 𝐾0 (𝐴)
󵄨󵄨 󵄨󵄨−1/2
𝐾0 (𝐴 1 )
󵄨󵄨Δ 𝐿 󵄨󵄨
⋅√
⋅
(369)
𝐴 1 (𝐴 1 + 𝑋0 − 2) + 1
1 − 𝐴1
=1+𝑌−
𝑌𝐴 1
𝐴 1 − 𝐵1
(𝑌0 − 𝑌)
𝐴 .
1 − 𝐴1 1
=
(370)
∼ log [
𝑋0
,
𝑋0 + 𝑌0
(375)
Ψ+,𝑌 (0, 𝑌𝑈) = Ψ𝑌 (0, 𝑌𝑈)
(𝑋 + 𝑌0 ) − 𝑋0
= − log [ 0
].
𝑌0
1 − 𝐷0 (𝑌𝑈)
𝐷0 (𝑌𝑈)
⋅
𝑋0
− 1 + 𝑒𝜏 )]
1−𝐴
𝑋0 𝐴 1
+ 1 − 𝐴 1] .
1 − 𝐴1
𝐴 1 (𝑌𝑈) (𝐴 1 (𝑌𝑈) + 𝑋0 − 2) + 1
1 − 𝐴 1 (𝑌𝑈)
After some calculation we find that
Ψ𝑋 + log 𝑋 = − log (1 − 𝐴𝑒𝜏 )
𝑋0
− 1) 𝑒−𝜏 + 1]
1−𝐴
(374)
2
But from (118) we see that
+ (𝑋0 + 𝑌0 − 1)𝐴 1 − 𝑌 = 0 so
that the right-hand sides of (370) and (371) agree.
Using (68) we have, as 𝑋 → 0,
∼ log [(
𝑘
𝜌𝑘
𝑌0
) Π∗ (𝑦∗ ) ,
(1 + 𝑌0 −
𝑘!
𝑋0 + 𝑌0
= 𝑌0 +
(371)
𝐴21
+ log [(𝑒−𝜏 − 𝐴) (
𝑦∗ )
1 + 𝑌𝑈 − 𝐷02 (𝑌𝑈)
and Ψ has the small 𝑋 behavior indicated in (364). Now, Ψ𝑌 =
−log(1 − 𝐵𝑒𝜏 ) so that Ψ𝑌(0, 𝑌) = −log(1 − 𝐵1 /𝐴 1 ) and using
(366) leads to
1 + 𝑌 − 𝑌𝑒Ψ𝑌 (0,𝑌) = 1 + 𝑌 −
(373)
where we note that 1+𝑌−𝑌𝑒Ψ𝑌 (0,𝑌) = 1+𝑌0 −𝑌0 /(𝑋0 +𝑌0 ) when
𝑌 = 𝑌𝑈, since 𝐴 1 (𝑌𝑈) = 𝑌0 /(𝑋0 + 𝑌0 ). To determine Π∗ (⋅)
we first infer its behaviors as 𝑦∗ → ±∞ by asymptotically
matching (374) to the expansions that apply for 𝑌 < 𝑌𝑈 and
𝑌 > 𝑌𝑈.
We expand the result in (114) as 𝑌 ↑ 𝑌𝑈. We have
̃
= 𝑌0 /(𝑋0 + 𝑌0 ) so that
𝐷0 (𝑌𝑈) = 𝐷0 (𝑌(0))
Using (366) and 𝐾0 (⋅) in (275) we see that (360) agrees with
(117) as 𝑋 → 0, if
1 + 𝑌 − 𝑌𝑒Ψ𝑌 (0,𝑌) =
𝐴21 + (𝑋0 − 2) 𝐴 1 + 1
) 𝑂 (𝑋2 )
1 − 𝐴1
and this verifies (364). We have thus shown that the
asymptotic matching holds and also determined Π0 (𝑌). This
completes the derivation of (117)–(120).
For region R4 , (117) holds for all 0 < 𝑌 < 𝑌0 , and then
the expansion matches to the corner expansions that apply for
𝑌 = 𝑂(𝜌−1 ) (𝑟 = 𝑂(1)) and 𝑌 = 𝑌0 − 𝑂(𝜌−1 ) (𝑝 = 𝑅 − 𝑟 =
𝑂(1)). But for region R1 ∪ R2 ∪ R3 and 𝑘 = 𝑂(1), (117) holds
only for 𝑌𝑈 < 𝑌 < 𝑌0 , while (114) applies for 0 < 𝑌 < 𝑌𝑈. To
connect these it is necessary to construct another expansion
for 𝑌 − 𝑌𝑈 = 𝑂(𝜌−1/2 ). Using the variables 𝑘 and 𝑦∗ ≡ √𝜌(𝑌 −
𝑌𝑈) = 𝑂(1) we conclude from (3) and (6) that
(368)
󵄨󵄨 󵄨󵄨−1/2 √
2𝜋𝐾0 (𝐴 1 )
󵄨󵄨Δ 𝐿 󵄨󵄨
𝐴 1 (𝐴 1 + 𝑋0 − 2) + 1
.
𝐴 1 (1 − 𝐴 1 )
= 𝑋 log (
−1/2
We have thus identified Π0 (𝑌) as
−1/2
Ψ (𝑋, 𝑌) + 𝑋 log 𝑋 − 𝑋
𝜋 (𝑘, 𝑟) ∼ 𝐶3 (𝜌) 𝑒𝜌Ψ(0,𝑌𝑈 +𝜌
𝐴 1 (𝐴 1 + 𝑋0 − 2) + 1 1
.
√𝑋
𝐴 1 (1 − 𝐴 1 )
Π0 (𝑌) = (𝐴 1 − 𝐵1 )
Integrating the above, noting that Ψ𝑋 +log 𝑋 is an analytic
function of 𝑋, we conclude that
(372)
−1/2
𝑌𝑈
[𝑌𝑈 + 2𝐷02 (𝑌𝑈) (1 − 𝐷0 (𝑌𝑈))]
√𝑋0
=√
(376)
𝑋0
𝑋0
[1 −
]
2
𝑌0
(𝑋0 + 𝑌0 )
3/2
⋅
(𝑋0 + 𝑌0 )
√(𝑋0 + 𝑌0 )3 + 𝑋0 (𝑌0 − 𝑋0 )
.
Advances in Operations Research
41
1
Π∗ (𝑦∗ ) exp [ Ψ𝑌𝑌 (0, 𝑌𝑈) 𝑦∗2 ]
2
Furthermore,
Ψ (0, 𝑌𝑈) = Ψ+ (0, 𝑌𝑈) + 1 − 𝑋0 − 𝑋0
+ (𝑋0 + 𝑌0 ) log (𝑋0 + 𝑌0 )
2
∼
(377)
√(𝑋0 + 𝑌0 )3 + 𝑋0 (𝑌0 − 𝑋0 )
√
𝑋0
𝑌0
(379)
1
⋅ exp [ Ψ+,𝑌𝑌 (0, 𝑌𝑈) 𝑦∗2 ] .
2
and we note that, in view of Propositions 1 and 2,
𝐶 (𝜌) ∼ √2𝜋√𝑋0 + 𝑌0 𝑄 (0, 0) 𝑒𝜌[Ψ(0,𝑌𝑈 )−Ψ+ (0,𝑌𝑈 )]
(𝑋0 + 𝑌0 ) − 𝑋0
Next we expand (117) as 𝑌 ↓ 𝑌𝑈. Since (𝑋0 + 𝑌0 )𝐴 1 (𝑌𝑈) =
𝑌0 , (117) is singular in this limit, and we use
(378)
which holds in regions R1 ∪ R2 ∪ R3 . Using (375)–(378)
to infer the behavior of (114) as 𝑌 ↑ 𝑌𝑈 and comparing the
result to (374) we can take 𝐶3 (𝜌) = √2𝜋𝜌𝑄(0, 0) and then, as
𝑦∗ → −∞,
𝑌0 − 𝐴 1 (𝑌) (𝑋0 + 𝑌0 )
∼ −𝐴󸀠1 (𝑌𝑈) (𝑋0 + 𝑌0 ) (𝑌 − 𝑌𝑈)
(380)
= −𝐴󸀠1 (𝑌𝑈) (𝑋0 + 𝑌0 ) 𝑦∗ 𝜌−1/2
to obtain
2
√𝐴 1 (𝐴 1 + 𝑋0 − 2) + 1 [(1 − 𝐴 1 ) − 𝑋0 ] √𝐴 1
𝑌0
√
󵄨1/2
󵄨󵄨
𝑌 (𝑌0 − 𝑌)
󵄨󵄨Δ 𝐿 (𝑌)󵄨󵄨󵄨 [𝑌0 − 𝐴 1 (𝑋0 + 𝑌0 )]
2
∼
√𝜌 √𝐴 1 (𝑌𝑈) (𝐴 1 (𝑌𝑈) + 𝑋0 − 2) + 1 [(1 − 𝐴 1 (𝑌𝑈)) − 𝑋0 ] √𝐴 1 (𝑌𝑈)
󵄨1/2
󵄨󵄨
𝑦∗
󵄨󵄨Δ 𝐿 (𝑌𝑈)󵄨󵄨󵄨 [− (𝑋0 + 𝑌0 )] 𝐴󸀠1 (𝑌𝑈)
(381)
𝑋0
√𝜌
√(𝑋0 + 𝑌0 )2 − 𝑋0 √(𝑋0 + 𝑌0 )2 + 𝑌0 − 𝑋0
=
𝑦∗ (𝑋 + 𝑌 )5/2
0
0
=
−1/2 1
√𝜌 𝑋0
2
3
√
[(𝑋0 + 𝑌0 ) − 𝑋0 ] [(𝑋0 + 𝑌0 ) + 𝑋0 (𝑌0 − 𝑋0 )]
,
𝑦∗ 𝑌0
𝑎𝑐
where 𝑎𝑐 was defined in (125). To obtain (381) we also used
𝐴󸀠1 (𝑌𝑈) =
𝑋0 + 𝑌0
2
(𝑋0 + 𝑌0 ) + 𝑌0 − 𝑋0
Ψ+,𝑌𝑌 (0, 𝑌𝑈) − 𝑎𝑐2 = Ψ𝑌𝑌 (0, 𝑌𝑈)
,
(382)
2
=
(𝑋0 + 𝑌0 )
2
[(𝑋0 + 𝑌0 ) 𝑌0 + 𝑋0 ] ,
which follow from (118) and (120). Denoting the ratio of the
left- and right-hand sides of (379) by 𝑓(𝑦∗ ), the matching
conditions as 𝑦∗ → ±∞ may be written as
𝑓 (𝑦∗ ) 󳨀→ 1,
(384)
so that the error function
󵄨󵄨
󵄨
󵄨󵄨Δ 𝐿 (𝑌𝑈)󵄨󵄨󵄨
(𝑋0 + 𝑌0 ) + 𝑌0 − 𝑋0
After some calculation we can show that
𝑓 (𝑦∗ ) =
∞
2
1
∫ 𝑒−𝑢 /2 𝑑𝑢
√2𝜋 𝑎𝑐 𝑦∗
satisfies the conditions in (383). To determine 𝑓(𝑦∗ ) completely we need to match (374), with 𝐶3 = √2𝜋𝜌𝑄(0, 0), to
the expansion in the transition layer where D+ and D0 meet,
which is given by (176). This matching has 𝑋 → 0 in (176)
and 𝑘 → ∞ in (374), with 𝜂∗ fixed. Noting that as 𝑋 → 0
̃ (𝑋)] = √𝜌 [𝑌 − 𝑌𝑈 + 𝑂 (𝑋)]
𝜂∗ = √𝜌 [𝑌 − 𝑌
𝑦∗ 󳨀→ −∞
1
𝑓 (𝑦∗ ) ∼
√2𝜋𝑎𝑐 𝑦∗
= 𝑦∗ + 𝑂 (𝜌−1/2 )
𝑦2
⋅ exp { ∗ [Ψ+,𝑌𝑌 (0, 𝑌𝑈) − Ψ𝑌𝑌 (0, 𝑌𝑈)]} ,
2
𝑦∗ 󳨀→ ∞.
(383)
(385)
(386)
we can show, using (178) and (184)–(187), that 𝑏(𝑡∗ ) → 𝑎𝑐−2 as
𝑋 → 0 and thus (385) determines 𝑓(𝑦∗ ) for all 𝑦∗ = 𝑂(1).
5.2. The Boundary Segment 𝑌 = 0. We consider 𝑟 = 𝑂(1), in
ranges where D0 meets the 𝑥-axis, which can occur only for
42
Advances in Operations Research
R4 . The analysis for where D+ (region R1 ) or D− (regions
R2 and R3 ) meet the 𝑥-axis is very similar to that in [10] so
we omit it. By examining (3), noting that this equation holds
even if 𝑟 = 0, on the scale 𝑟 = 𝑂(1) with 𝑘 = 𝜌𝑋, 𝑋 > 0, we
find that
̃ 𝑟 (𝑋) 𝑒𝜌Ψ(𝑋,0) ,
𝜋 (𝑘, 𝑟) ∼ 𝐶4 (𝜌) 𝜌𝑟 Π
(387)
̃ 𝑟 (𝑋) will satisfy
where Π
̃ 𝑟 (𝑋)
[1 + 𝑋 − 𝑒−Ψ𝑋 (𝑋,0) − 𝑋𝑒Ψ𝑋 (𝑋,0) ] Π
̃ 𝑟+1 (𝑋) .
= (𝑟 + 1) Π
(388)
Noting that Ψ𝑋 (𝑋, 0) = −log(1 − 𝐴𝑒𝜏 ), we define 𝐴 0 (⋅) by
𝐴 0 (𝑋) = 𝐴(𝑋, 0). A ray in D0 reaches the 𝑥-axis when 𝑒−𝜏 =
𝐵 so we also let 𝐵0 (𝑋) = 𝐵(𝑋, 0). Then evaluating (197) along
𝑌 = 0 leads to
1 + 𝑋 − 𝑋𝑒Ψ𝑋 (𝑋,0) − 𝑒−Ψ𝑋 (𝑋,0) = lim [𝑌𝑒Ψ𝑌 ]
𝑌→0
𝐵 − 𝑒−𝜏
= −𝜏
lim {𝑌0
exp [− log (1 − 𝐵𝑒𝜏 )]}
𝑒 →𝐵
𝐵−1
=
we simply replace 𝐴 by 𝐴 0 (𝑋) in the factors that are not
singular and use
−1/2
(𝑒−𝜏 − 𝐵)
(395)
𝑔 (𝑋)
𝑊 + 𝑌0 √1 − √𝑋0 𝑋0 − (1 − 𝐴 0 )
∼√ 0
,
2𝑊0 + 𝑌0 √𝑋0 (√𝑋 − 𝑋 − 𝑌 )3/2
2
0
0
(396)
0
𝑋 󳨀→ 𝑋𝐿 .
(389)
𝑌0 𝐵0
1 − 𝐵0
By implicit differentiation of (100) we find that
𝐴󸀠0 (𝑋𝐿 ) = −
𝑟
̃ 𝑟 (𝑋) = Π
̃ 0 (𝑋) 1 [ 𝑌0 𝐵0 (𝑌) ] .
Π
𝑟! 1 − 𝐵0 (𝑌)
(390)
We proceed to determine the constant 𝐶4 (𝜌) and function
̃ 0 (⋅) by asymptotically matching (387) to the D0 ray expanΠ
sion.
First we note that setting 𝑒−𝜏 = 𝐵0 in (66) and replacing
(𝐴, 𝐵) by (𝐴 0 , 𝐵0 ) lead to the expression in (99) for Ψ(𝑋, 0).
As 𝑌 → 0 we also have, in view of (389),
Ψ𝑌 (𝑋, 𝑌) = − log 𝑌 + log (
𝑌0 𝐵0
) + 𝑜 (1) .
1 − 𝐵0
(391)
By integrating (391) we see that the D0 ray expansion behaves
for 𝑌 → 0 as
𝑌𝐵
𝑄 (0, 0)
𝐾 (𝑋, 𝑌) 𝑒−𝜌𝑌log𝑌 𝑒𝜌𝑌 exp [𝜌𝑌 log ( 0 0 )]
𝜌
1
− 𝐵0
√
𝑟
𝑟
(1 − 𝑋0 − 𝑌0 )
𝜌 𝑟 𝑌0 𝐵0
𝑒 (
)
𝐾 (𝑋, 𝑌) .
𝑟
𝑟
1 − 𝐵0
√𝜌
(392)
Here we used 𝑄(0, 0) ∼ 1 − 𝑋0 − 𝑌0 (in R4 ) and 𝜌𝑌 =
𝑟. Expanding 𝑟! in (390) by Stirling’s formula we see the
matching is possible if 𝐶4 = 1 − 𝑋0 − 𝑌0 and
𝐾 (𝑋, 𝑌) ∼
1 ̃
Π (𝑋) , 𝑌 󳨀→ 0.
√2𝜋𝑌 0
(393)
We denote by 𝑔(𝑋) the limit of [√2𝜋𝑌𝐾(𝑋, 𝑌)] as 𝑌 → 0.
Writing
𝐾 (𝑋, 𝑌)
= (𝑒−𝜏 − 𝐵)
−1/2
𝑌
𝑌
1
√ 0 .
∼
(1 − 𝐵)]
√𝑌 1 − 𝐵0
𝑌0
Then 𝑔(𝑋) becomes precisely the expression in (102). We have
̃ 0 (𝑋) as 𝑔(𝑋) and thus derived (98)–(102).
thus determined Π
The approximation in (387) applies only for 0 < 𝑋 < 𝑋𝐿 ,
and (96) applies for 𝑋𝐿 < 𝑋 < 𝑋0 . But for 𝑋 ≈ 𝑋𝐿 a new
approximation is needed, as 𝑔(𝑋) vanishes as 𝑋 ↑ 𝑋𝐿 , due
to the factor (1 − 𝐴 0 )2 − 𝑋0 , since 1 − 𝐴 0 (𝑋𝐿 ) = √𝑋0 . Since
𝐵0 (𝑋𝐿 ) = 𝑊0 /(𝑊0 + 𝑌0 ) we find from (102) that
so that the solution to (388) is
=
=[
−1/2
(𝑒−𝜏 − 𝐴)
−1/2
|Δ|−1/2 𝐾0 (𝐴)
(394)
𝑊0 (𝑊0 + 𝑌0 )
.
𝑌0 (2𝑊0 + 𝑌0 )
(397)
Using
2
𝑋0 − (1 − 𝐴 0 ) ∼ 2 (1 − 𝐴 0 ) 𝐴󸀠0 (𝑋𝐿 ) (𝑋 − 𝑋𝐿 )
= 2√𝑋0 𝐴󸀠0 (𝑋𝐿 ) (𝑋 − 𝑋𝐿 )
(398)
and (397) we obtain from (396) the expression in (104), which
will be used for asymptotic matching verifications.
We consider the scale 𝑋 − 𝑋𝐿 = 𝜌−1/3 Λ = 𝑂(𝜌−1/3 ),
retaining 𝑟 = 𝑂(1). From the balance equations we can
conclude that the expansion has the form
𝐶5 (𝜌) 𝑒𝜌Ψ(𝑋,0) [
⋅ 𝑒𝜌Ψ(𝑋𝐿 ,0)
𝑟
𝑌0 𝐵0 (𝑋𝐿 )
𝜌𝑟
]
𝑔 (𝑋𝐿 ) ∼ 𝐶5 (𝜌)
1 − 𝐵0 (𝑋𝐿 ) 𝑟!
𝜌𝑟 𝑊0𝑟
𝑟!
(399)
1
⋅ exp [𝜌2/3 Ψ𝑋 (𝑋𝐿 , 0) Λ + 𝜌1/3 Ψ𝑋𝑋 (𝑋𝐿 , 0) Λ2 ]
2
̂ (Λ) ,
⋅Π
where the last exponential factor follows from expanding
Ψ(𝑋, 0) about 𝑋 = 𝑋𝐿 , and these are necessary to have
a chance of matching to the expansion for 𝑋 < 𝑋𝐿 . By
comparing (399) to the behavior of (98) as 𝑋 ↑ 𝑋𝐿 we
immediately conclude that 𝐶5 (𝜌) = 𝑂(𝜌−1/3 ) so we set
𝐶5 (𝜌) = 𝜌−1/3 (1 − 𝑋0 − 𝑌0 ), and then
̂ (Λ) ∼ exp [ 1 Ψ𝑋𝑋𝑋 (𝑋𝐿 , 0) Λ3 ]
Π
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Advances in Operations Research
3/2
⋅
2 (𝑊0 + 𝑌0 )
3/2
𝑌0 (2𝑊0 + 𝑌0 )
43
𝑊05/4
̂
We next derive a matching condition for Π(Λ)
as Λ → ∞.
0
We first note from (97) that if we write 𝑠1 + 𝑊0 = √𝑊0 𝑓(𝑋)
then
(−Λ) ,
3/2
(√𝑋0 − 𝑋0 − 𝑌0 )
Λ 󳨀→ −∞.
(400)
𝑓 (𝑋)
=
From (101) we find that
𝑌0
𝐵󸀠
2 0
(1 − 𝐵0 )
(𝑋) = [1 −
𝑓 (𝑋𝐿 ) =
𝑋0
] 𝐴󸀠0
2
(𝑋)
(1 − 𝐴 0 )
(401)
𝑊0 + 𝑌0
𝑊 + 𝑌0
= 0
󳨐⇒ 𝑠10 (𝑋𝐿 ) = 𝑌0 .
√𝑊0
1 − √𝑋0
𝐴󸀠0
1
1
−
) 𝐵0󸀠 +
𝐵0 𝐵0 − 𝐴 0
𝐵0 − 𝐴 0
𝑓󸀠 (𝑋𝐿 ) = −
(402)
and thus Ψ𝑋𝑋 (𝑋𝐿 , 0) = 𝐴󸀠0 (𝑋𝐿 )/[𝐵0 (𝑋𝐿 ) − 1 + √𝑋0 ] is as in
(107). From (402) we also conclude that
(408)
3
󸀠󸀠
𝑓 (𝑋𝐿 ) = −
2𝑊05/2 (𝑌0 + 𝑊0 )
3
𝑌03 (𝑌0 + 2𝑊0 )
.
From (80) with 𝑌 = 0 we have Φ𝑋 (𝑋, 0) = −log[1 − 𝑓(𝑋)]
from which we can show that
𝐴󸀠󸀠0 (𝑋𝐿 )
𝐵0 (𝑋𝐿 ) − 𝐴 0 (𝑋𝐿 )
Φ𝑋𝑋𝑋 (𝑋𝐿 , 0)
(403)
2
𝐴󸀠0 (𝑋𝐿 )
+[
] .
𝐵0 (𝑋𝐿 ) − 𝐴 0 (𝑋𝐿 )
𝐴󸀠󸀠0 (𝑋𝐿 ) = −
=
3
𝑊0 (𝑊0 + 𝑌0 ) [𝑌03 + 3𝑊0 𝑌02 + 4𝑊02 𝑌0 − 2√𝑋0 𝑊05/2 ] (410)
.
2
3
𝑌03 (2𝑊0 + 𝑌0 ) (√𝑋0 − 𝑋0 − 𝑌0 )
By using the expression for 𝐶1 in (41) we have
Φ (𝑋𝐿 , 0) − Φ (𝑋0 , 𝑌0 ) = Ψ (𝑋𝐿 , 0) ,
By further differentiation of (100) and (401) we find that
𝐵0󸀠󸀠 (𝑋𝐿 ) = −
𝜌1/3 [Φ∗ (𝑠10 ) − Φ∗ (𝑌0 )] = −𝜌1/3 𝑟0
𝑊02
2
,
𝑌0 √𝑋0 (2𝑊0 + 𝑌0 )2
2𝑊03/2 (𝑌0 + 𝑊0 )
𝑌0 √𝑋0 (2𝑊0 + 𝑌0 )
2𝑊03 (𝑊0 + 𝑌0 )
(1 − √𝑋0 )
𝑠10 + 𝑊0
) = −𝜌1/3 𝑟0
𝑌0 + 𝑊0
𝑋01/6
⋅ log (
(1 − √𝑋0 )
𝑓 (𝑋)
) ∼ −𝑟0 Λ
𝑓 (𝑋𝐿 )
𝑋01/6
(404)
3
𝑋01/6
2/3
2
√𝑋0 𝑌02 (2𝑊0 + 𝑌0 )
2/3
(1 − √𝑋0 )
⋅ log (
2/3
2
−
(409)
Φ𝑋𝑋 (𝑋𝐿 , 0) = Ψ𝑋𝑋 (𝑋𝐿 , 0) ,
1
1
−
] 𝐵󸀠󸀠 (𝑋𝐿 )
𝐵0 (𝑋𝐿 ) 𝐵0 (𝑋𝐿 ) − 𝐴 0 (𝑋𝐿 ) 0
+
2
(𝑌0 + 𝑊0 )
,
𝑌0 (𝑌0 + 2𝑊0 )
Φ𝑋 (𝑋𝐿 , 0) = Ψ𝑋 (𝑋𝐿 , 0) ,
Ψ𝑋𝑋𝑋 (𝑋𝐿 , 0)
=[
,
𝑌0 + 𝑊0 (1 − √𝑋0 )
= 𝑟0 Λ
𝑌0 (𝑌0 + 2𝑊0 )
𝑋01/6
and thus
𝑓󸀠 (𝑋𝐿 )
𝑓 (𝑋𝐿 )
5/3
,
1/6
√𝜌𝐶1
Ψ𝑋𝑋𝑋 (𝑋𝐿 , 0)
3
=−
5/2
2 𝑊0 (𝑊0 + 𝑌0 )
1
3
2
√𝑋0 𝑌0 (2𝑊0 + 𝑌0 ) √𝑋0 − 𝑋0 − 𝑌0
4
+
𝑊0 (𝑊0 + 𝑌0 )
𝑌02
2
(407)
From (406) we then obtain
so that 𝐵0󸀠 (𝑋𝐿 ) = 0. Then from (99) we find that
Ψ𝑋𝑋 (𝑋, 0) = (
(406)
1
2
[𝑊0 + 1 − 𝑋 + √(𝑊0 + 1 − 𝑋) − 4𝑊0 ] ,
2
1
2
(2𝑊0 + 𝑌0 ) (√𝑋0 − 𝑋0 − 𝑌0 )
.
(1 − √𝑋0 )
𝑋01/6 [Ai󸀠 (𝑟0 )]
2
1/2√𝑋0
𝑊0 + 𝑠10
(405)
√𝑊
( 0 0 )
⋅
𝑠
0
0
√𝑠 (𝑠 + 2𝑊 ) 1 + 𝑊0
1
1
0
⋅ √ √𝑋0 − 𝑋0 − 𝑠10 ∼ 𝜌−1/3
(1 − 𝑋0 − 𝑌0 )
Ai󸀠 (𝑟0 )
44
Advances in Operations Research
⋅ √2𝜋√
5/6
1/6
𝑊0 + 𝑌0 (1 − √𝑋0 ) 𝑋0
𝑌0 + 2𝑊0 (√𝑋 − 𝑋 − 𝑌 )3/2
0
0
⋅√
0
⋅ exp [−𝜌Φ (𝑋0 , 𝑌0 ) − 𝜌1/3 Φ∗ (𝑌0 )] .
1
1 𝑖∞ 𝑒⋆𝜃
𝑑𝜃,
⋅ exp [ Λ3 Φ𝑋𝑋𝑋 (𝑋𝐿 , 0)]
∫
6
2𝜋𝑖 −𝑖∞ Ai (𝜃)
(411)
With (409)–(411) we have obtained the behavior of (96) as
𝑋 ↓ 𝑋𝐿 and thus derived the matching condition
5/3
𝑌0 + 𝑊0 (1 − √𝑋0 )
𝑌0 (𝑌0 + 2𝑊0 )
𝑋01/6
]
]
1
Ai󸀠 (𝑟0 )
(412)
5/6
0
𝜂∗∗ = 𝜌1/3 [𝑌 − 𝑌𝑐 (𝑋)]
𝑟
− 𝑌𝑐󸀠 (𝑋𝐿 ) (𝑋 − 𝑋𝐿 )] ∼ −𝑌𝑐󸀠 (𝑋𝐿 ) Λ
𝜌
(413)
𝑊3/2 (𝑌0 + 𝑊0 )
𝑌𝑐󸀠 (𝑋𝐿 ) = 0
.
𝑌0 (𝑌0 + 2𝑊0 )
(414)
We can expand 𝜌[Φ(𝑋, 𝑌𝑐 + 𝜌−1/3 𝜂∗∗ ) − Φ(𝑋0 , 𝑌0 )] in (191)
as 𝑋 → 𝑋𝐿 which is equivalent to expanding 𝜌[Φ(𝑋, 𝑌) −
Φ(𝑋0 , 𝑌0 )] for (𝑋, 𝑌) → (𝑋𝐿 , 0) and show that the exponential factors in (191) agree with those in (399), after 𝑟! is
approximated by Stirling’s formula, so that
𝜌𝑟 𝑊0𝑟
∼ (2𝜋)−1/2 𝑟−1/2
𝑟!
𝑌0 + 𝑊0 (1 − √𝑋0 )
𝑌0 (𝑌0 + 2𝑊0 )
𝑋01/6
.
(415)
⋅ exp {−𝜌𝑌 log 𝑌 + 𝜌𝑌 [1 + log (𝑊0 )]} .
Then using 𝐶5 = 𝜌−1/3 (1 − 𝑋0 − 𝑌0 ) we compare algebraic
factors in (399) and (191), which yields
In this limit 𝑌 ∼ 𝑌𝑐󸀠 (𝑋𝐿 )(𝑋 − 𝑋𝐿 ) and Λ is fixed, so that (416)
̂ as
determines Π(⋅)
5/6
(1 − √𝑋0 )
𝑋01/6
3/2
(√𝑋0 − 𝑋0 − 𝑌0 )
𝑊0 + 𝑌0
1
1
exp [ Λ3 Φ𝑋𝑋𝑋 (𝑋𝐿 , 0)]
2𝑊0 + 𝑌0
6
2𝜋𝑖
𝑖∞
⋅∫
−𝑖∞
(418)
𝑒⋆𝜃
𝑑𝜃.
Ai (𝜃)
We have thus derived the result in (105). By using the
asymptotic results in (192) and (193) (with ⋆ replaced by −⋆)
we see immediately that (412) holds. The matching condition
in (400) will also hold since
1 ⋆ 3 1
1
Φ𝑋𝑋𝑋 (𝑋𝐿 , 0) + ( ) = Ψ𝑋𝑋𝑋 (𝑋𝐿 , 0) ,
6
3 Λ
6
(419)
5.3. The Boundary Segment 𝑌 = 𝑌0 . We consider the scale
𝑝 = 𝜌(𝑌0 − 𝑌) = 𝑂(1) and 0 < 𝑋 < 𝑋0 . Now the
analysis will be the same for any region R𝑗 , and the expansion
we construct will hold everywhere except near the corner
points (𝑋, 𝑌) = (0, 𝑌0 ) and (𝑋0 , 𝑌0 ). By expanding (3) and
the boundary condition in (5) along 𝑟 = 𝑅, we find that an
asymptotic solution in this range is given by
𝜋 (𝑘, 𝑟) ∼ 𝐶6 (𝜌)
𝜌𝑝
𝑝
[𝑌 (𝑒𝜏𝑈 − 1)] 𝑒𝜌Ψ(𝑋,𝑌0 ) 𝑔 (𝑋) , (420)
𝑝! 0
where 𝜏𝑈 = 𝜏𝑈(𝑋) is given in (128). Again, (420) must
contain the factor exp[𝜌Ψ(𝑋, 𝑌0 )] in order to have a chance of
matching to the D0 ray expansion, and this factor determines
the geometric factor in 𝑝. We must only determine 𝐶6 (𝜌) and
𝑔(𝑋) by asymptotic matching.
As 𝑌 → 𝑌0 we have 𝐴 → 𝐴 max and 𝐵 → −∞ in (68).
Also,
Ψ𝑌 = − log (−𝐵) − 𝜏 + 𝑜 (1) , 𝑌 󳨀→ 𝑌0
1
−1/2 ̂
Π (Λ)
(1 − 𝑋0 − 𝑌0 ) 𝜌−1/3 (𝜌𝑌)
√2𝜋
(417)
by (417), (405), and (410). This completes our analysis of the
scale 𝑟 = 𝑂(1).
remains also 𝑂(1). From (86) we have
∼ (1 − 𝑋0 − 𝑌0 ) 𝜌−5/6
5/3
⋅√
0
̂
Thus (400) and (412) yield the behavior of Π(Λ)
as
Λ → ±∞, but to determine this function for all Λ we must
use a third matching condition, to the transition layer in
Proposition 18, which applies for 𝑌 − 𝑌𝑐 (𝑋) = 𝜌−1/3 𝜂∗∗ =
𝑂(𝜌−1/3 ). Thus we let 𝑋 → 𝑋𝐿 in (191) and let 𝑟 → ∞ in
(399), in such a way that Λ is fixed. Since Λ = 𝜌1/3 (𝑋 − 𝑋𝐿 )
this means that
∼ 𝜌1/3 [
where
̂ (Λ) =
Π
1/6
𝑊 + 𝑌0 (1 − √𝑋0 ) 𝑋0
, Λ 󳨀→ +∞.
⋅√ 0
𝑌0 + 2𝑊0 (√𝑋 − 𝑋 − 𝑌 )3/2
0
(416)
⋆=Λ
̂ (Λ) ∼ exp [ 1 Φ𝑋𝑋𝑋 (X𝐿 , 0) Λ3
Π
6
[
+ 𝑟0 Λ
𝑊0 + 𝑌0
1
2𝑊0 + 𝑌0 √𝑌󸀠 (𝑋 ) (𝑋 − 𝑋 )
𝐿
𝐿
𝐶
(421)
and from (68) we obtain
5/6
1 (1 − √𝑋0 )
√2𝜋 (√𝑋 − 𝑋 −
0
0
𝑋01/6
3/2
𝑌0 )
−𝐵 ∼
𝑌0 (1 − 𝑒−𝜏𝑈 )
,
𝑌0 − 𝑌
(422)
Advances in Operations Research
45
where 𝜏𝑈 = 𝜏𝑈(𝑋) is obtained by solving the first equation
in (68) with 𝐴 replaced by 𝐴 max . Using (421) and (422) we
conclude that
Ψ (𝑋, 𝑌) = Ψ (𝑋, 𝑌0 ) + (𝑌 − 𝑌0 ) log (𝑌0 − 𝑌) + 𝑌0
− 𝑌 + (𝑌0 − 𝑌) log [𝑌0 (𝑒𝜏𝑈 − 1)]
(423)
+ 𝑜 (𝑌0 − 𝑌) , 𝑌 󳨀→ 𝑌0 .
𝑄 (0, 0) 𝜌−1/2 𝐾 (𝑋, 𝑌) 𝑒𝜌Ψ(𝑋,𝑌)
(424)
as 𝑌 → 𝑌𝑈 we see that we can take 𝐶6 = 𝑄(0, 0) ∼ 1 − 𝑋0 − 𝑌0
and the matching will hold if
1
1
𝑔 (𝑋) = 𝜌−1/2
𝑔 (𝑋)
√2𝜋𝑝
√2𝜋 (𝑌0 − 𝑌)
∼𝜌
(425)
𝐾 (𝑋, 𝑌) , 𝑌 󳨀→ 𝑌0 .
To expand (70) as 𝑌 → 𝑌0 we note that in this limit (𝑒−𝜏 −
𝐴)−1/2 and |Δ|−1/2 are finite, while (𝑒−𝜏 − 𝐵)−1/2 ∼ (−𝐵)−1/2
vanishes and 𝐾0 (𝐴) is singular, due to the factor 𝑌0 + (1 −
𝑋0 − 𝑌0 )𝐴 − 𝐴2 in the denominator. We thus have
1
−1/2 󵄨
󵄨󵄨Δ 𝑈 (𝑋)󵄨󵄨󵄨−1/2
(𝑒−𝜏𝑈 − 𝐴 max )
𝐾 (𝑋, 𝑌) ∼
󵄨
󵄨
√−𝐵
⋅√
2
𝑌0 (1 − 𝐴 max ) − 𝑋0
−1
,
2𝜋 1 − 𝑋0 − 𝑌0 − 2𝐴 max 𝐴 − 𝐴 max
(426)
𝑌 󳨀→ 𝑌0 .
Here Δ 𝑈(𝑋) is the Jacobian in (71) evaluated along 𝑌 = 𝑌0 ,
and we also used the identity 𝐴 max (𝐴 max −1) = 𝑌0 −𝐴 max (𝑋0 +
𝑌0 ). From (67) we see that
𝐵∼
𝐴 max (1 − 𝐴 max − 𝑋0 )
1
1 − 𝑋0 − 𝑌0 − 2𝐴 max 𝐴 − 𝐴 max
1
1
1 − 𝑒−𝜏𝑈
√ −𝜏
√2𝜋 √𝑌0 − 𝑌 𝑒 𝑈 − 𝐴 max
2
𝑛
= − 𝑌0 𝜏𝑈󸀠 (𝑋0 )
𝜌
𝑛
𝜌√
(429)
𝑌0
2
(1 + 𝑋0 + 𝑌0 ) − 4𝑋0
and thus, for 𝑋 → 𝑋0 , (127) reduces to (247). This completes
the analysis of the scale 𝑝 = 𝑅 − 𝑟 = 𝑂(1).
6. Approximations near State Space Corners
We consider the four state space corners. The upper right
corner (𝑋, 𝑌) = (𝑋0 , 𝑌0 ) was analyzed already in Section 4,
since this was necessary to completely determine the ray
expansion in D0 . The analysis of the lower right corner
(𝑋, 𝑌) = (𝑋0 , 0) is essentially the same as that for the infinite
capacity model in [10]. For regions R1 and R2 the leading
terms for 𝜋(𝑘, 𝑟) near this corner are the same as for the
infinite capacity model, while for regions R3 and R4 the
analysis differs only through the multiplicative constant 𝐶1 .
Thus we omit the analysis of these two corners, focusing on
the lower and upper left corners, (0, 0) and (0, 𝑌0 ).
6.1. The Corner (𝑋, 𝑌) = (0, 0). For regions R1 ∪R2 ∪R3 the
analysis is very similar to that in [10], leading to (132)–(136)
in Proposition 13. For region R4 we set
𝜋 (𝑘, 𝑟) = 𝐶7 (𝜌) 𝑒𝜌Ψ(0,0) 𝜌𝑘+𝑟 𝑄∗ (𝑘, 𝑟; 𝜌) ,
(427)
𝑄∗ (𝑘, 𝑟) = (𝑟 + 1) 𝑄∗ (𝑘, 𝑟 + 1)
𝑋 − (1 − 𝐴 max ) 󵄨󵄨 󵄨󵄨−1/2
⋅ 0
.
󵄨󵄨Δ 𝑈󵄨󵄨
1 − 𝐴 max
Using (428) in (425) we can identify 𝑔(𝑋) and then (420)
becomes the same as (127).
Finally we note that as 𝑋 ↑ 𝑋0 , (127) will asymptotically
match to (168), the approximation valid for 𝑛, 𝑝 = 𝑂(1). For
𝑛 → ∞ with 𝑝 = 𝑂(1), (168) can be approximated by (247).
(431)
Note that (6) implies that (431) holds along 𝑘 = 0, 𝑟 ⩾ 0.
Equation (431) has many different solutions, and anything of
the form 𝛼0𝑘 𝛽0𝑟 /(𝑘!𝑟!) will be a solution provided that 𝛼0 +𝛽0 =
1. Let us write
𝑄∗ (𝑘, 𝑟) =
(428)
(430)
recalling that Ψ(0, 0) = −𝑋0 − 𝑌0 + (𝑋0 + 𝑌0 )log(𝑋0 + 𝑌0 ).
Dividing the main balance equation in (3) by 𝜌, using (430),
and letting 𝑄∗ (𝑘, 𝑟) denote the leading term approximation
for 𝑄∗ (𝑘, 𝑟; 𝜌), we obtain in the limit 𝜌 → ∞
+ (𝑘 + 1) 𝑄∗ (𝑘 + 1, 𝑟) ; 𝑘 ⩾ 1, 𝑟 ⩾ 0.
so that with (422) we have, as 𝑌 ↑ 𝑌0 ,
𝐾 (𝑋, 𝑌) ∼
𝑌0 (𝑒𝜏𝑈 (𝑋) − 1) ∼ 𝑌0 𝜏𝑈󸀠 (𝑋0 ) (𝑋 − 𝑋0 )
=
Expanding 𝑝! by Stirling’s formula and comparing the result
to
−1/2
As 𝑋 → 𝑋0 we have 𝜏𝑈(𝑋) → 0, Ψ(𝑋0 , 𝑌0 ) = 0, Ψ𝑋 (𝑋0 , 𝑌0 ) =
−log(1 − 𝐴 max ), and 𝑧+ (0)(1 − 𝐴 max ) = 1. Then
1 𝑘
𝑟
𝛼 (1 − 𝛼0 )
𝑘!𝑟! 0
(432)
and we will show by asymptotic matching that only one value
of 𝛼0 is needed, and the matching will also determine the
appropriate value. For regions R1 ∪ R2 ∪ R3 the same
argument was used in [10] to determine 𝛼0 as 1−𝑊0 = 2√𝑋0 −
𝑋0 , leading to (136). For region R4 we can asymptotically
match (430) to (117) (then 𝑟 → ∞ with 𝑘 = 𝑂(1)), to (98)
(then 𝑘 → ∞ with 𝑟 = 𝑂(1)), or to the D0 ray expansion
(then 𝑘, 𝑟 → ∞ with 𝑘/𝑟 fixed). We discuss only the first two
matchings, as the third will lead to the same conclusion.
46
Advances in Operations Research
For fixed 𝑘 and 𝑟 → ∞, (430) and (432) become
𝜋 (𝑘, 𝑟) ∼ 𝐶7 (𝜌)
⋅
that 𝐴 1 (0) = 1 − 𝑋0 − 𝑌0 . Then comparing the geometric
factors in 𝑘, in (117) and (433), we must have
𝐶 (𝜌)
𝜌𝑘+𝑟 𝜌Ψ(0,0) 𝑘
𝑟
𝛼0 (1 − 𝛼0 ) ∼ 7
𝑒
𝑘!𝑟!
√2𝜋𝜌𝑌
𝜌𝑘 𝑘
𝛼 exp {𝜌 [Ψ (0, 0) − 𝑌 log 𝑌 + 𝑌
𝑘! 0
𝛼0 =
(433)
𝐴 1 (0) (𝐴 1 (0) + 𝑋0 − 2) + 1
𝑋0
= 𝑌0 +
. (434)
1 − 𝐴 1 (0)
𝑋0 + 𝑌0
From (119) we find that as 𝑌 → 0
Ψ (0, 𝑌) = Ψ (0, 0) + 𝑌 − 𝑌 log 𝑌
+ 𝑌 log (1 − 𝛼0 )]} ,
+ 𝑌 log [
where the last formula holds in the matching region where
𝑟 → ∞ but 𝑌 = 𝑟/𝜌 → 0, and we wrote the expression in
terms of 𝑌. To expand (117) as 𝑌 → 0, we observe from (118)
𝐶7 (𝜌)
√2𝜋𝜌𝑌
𝑌0 (1 − 𝑋0 − 𝑌0 )
] + 𝑜 (𝑌)
𝑋0 + 𝑌0
which implies that 1−𝛼0 = 𝑌0 (1−𝑋0 −𝑌0 )/(𝑋0 +𝑌0 ), consistent
with (434). The matching also implies that
2
∼ 𝑄 (0, 0)
√𝐴 1 (0)√𝐴 1 (0) (𝐴 1 (0) + 𝑋0 − 2) + 1 {[1 − 𝐴 1 (0)] − 𝑋0 }
√𝑌√󵄨󵄨󵄨󵄨Δ 𝐿 (0)󵄨󵄨󵄨󵄨 [𝑌0 − 𝐴 1 (0) (𝑋0 + 𝑌0 )]
After some calculation we find from (120) that
𝑋 + 𝑌0 (𝑋0 + 𝑌0 )
−Δ 𝐿 (0) = (1 − 𝑋0 − 𝑌0 ) 0
𝑋0 + 𝑌0
(437)
𝐶7 (𝜌) ∼ √𝜌√2𝜋 (𝑋0 + 𝑌0 )𝑄 (0, 0) .
(438)
We have thus derived (137), since 𝑄(0, 0) ∼ 1 − 𝑋0 − 𝑌0 for
region R4 .
The same conclusions follow by matching to (98), as then,
for fixed 𝑟 and 𝑘 → ∞, (430) with (432) becomes
𝐶7 (𝜌) 𝜌𝑟
𝑟
(1 − 𝛼0 )
√2𝜋𝜌𝑋 𝑟!
(439)
⋅ exp {𝜌 [Ψ (0, 0) − 𝑋 log 𝑋 + 𝑋 + 𝑋 log (𝛼0 )]} .
We expand (98) as 𝑋 → 0. We have 𝐴 0 (0) = 1 − 𝑋0 − 𝑌0 and
from (100) find that
𝑌0
.
𝐴󸀠0 (0) = −
(440)
𝑋0 + 𝑌0 (𝑋0 + 𝑌0 )
(441)
We have, as 𝑋 → 0,
∼𝑋
−1/2
[𝐵0󸀠
−1/2
(0) −
𝐴󸀠0
(0)]
−1/2
(443)
.
Then after some calculation, using (440) and (441) we find
that 𝑔(𝑋) has the expansion in (103) as 𝑋 → 0. Expanding
(98) as 𝑋 → 0 and comparing the result to (439), noting that
𝑌0 𝐵0 (0)/[1 − 𝐵0 (0)] = 1 − 𝛼0 , regain the expression in (438)
for 𝐶7 (𝜌). We can also easily verify that (137) matches to the
D0 ray expansion, by expanding the latter for 𝑋, 𝑌 → 0 along
lines of constant slope 𝑌/𝑋.
6.2. The Corner (𝑋, 𝑌) = (0, 𝑌0 ). Now we use the variables 𝑘
and 𝑝 = 𝑅 − 𝑟, with
(444)
𝐵0 (𝑋) − 𝐴 0 (𝑋)
] 󳨀→
𝑋
log [𝐵0 (0)] − log [𝐵0󸀠 (0) − 𝐴󸀠0 (0)]
𝑋0
].
𝑋0 + 𝑌0
Near this corner only the balance equations in (3), (5), (6),
and (7) apply. Using (444) in (3) we obtain
̂ (𝑘, 𝑝; 𝜌)
(𝜌 + 𝜌𝑌0 + 𝑘 − 𝑝) 𝑄
𝐴 (𝑋)
Ψ𝑋 (𝑋, 0) + log 𝑋 = log 𝑋 − log [1 − 0
]
𝐵0 (𝑋)
= log [𝑌0 +
[𝐵0 (𝑋) − 𝐴 0 (𝑋)]
2
(𝑋0 + 𝑌0 ) − 𝑋0
.
𝑋0 + 𝑌0 (𝑋0 + 𝑌0 )
= log [𝐵0 (𝑋)] − log [
(436)
̂ (𝑘, 𝑝; 𝜌) .
𝜋 (𝑘, 𝑟) = 𝐶8 (𝜌) 𝑒𝜌Ψ(0,𝑌0 ) 𝜌𝑘+𝑝 𝑄
Then also 𝐵0 (0) = 1 − 𝑋0 − 𝑌0 and, using (101),
𝐵0󸀠 (0) = −
.
By using (442) to infer the small 𝑋 behavior of Ψ(𝑋, 0) and
comparing the result to the exponential part of (439), we
again conclude that 𝛼0 is as in (434). To expand 𝑔(𝑋) in (102)
as 𝑋 → 0 we note that it is singular due to the factor
and thus
𝜋 (𝑘, 𝑟) ∼
(435)
̂ (𝑘, 𝑝 − 1; 𝜌)
= [𝑌0 + (1 − 𝑝) 𝜌−1 ] 𝑄
(442)
(445)
̂ (𝑘 + 1, 𝑝; 𝜌) + 𝑄
̂ (𝑘 − 1, 𝑝; 𝜌) .
+ 𝜌 (𝑘 + 1) 𝑄
̂ 𝑝) be the limiting form of 𝑄(𝑘,
̂ 𝑝; 𝜌) as 𝜌 → ∞,
Letting 𝑄(𝑘,
(445) leads to
̂ (𝑘 + 1, 𝑝)
̂ (𝑘, 𝑝) = (𝑘 + 1) 𝑄
(1 + 𝑌0 ) 𝑄
(446)
Advances in Operations Research
47
whose most general solution is
̂ (𝑘, 𝑝) =
𝑄
Thus,
𝑘
(1 + 𝑌0 )
𝑄 (𝑝) .
𝑘!
(447)
Equations (5)–(7) provide, in the limit 𝜌 → ∞, no additional
information. To determine 𝑄(𝑝) in (447) we use asymptotic
matching to the boundary layer expansion in (127), which
applies for 𝑝 = 𝑂(1) and 0 < 𝑋 < 𝑋0 .
As 𝑋 → 0 we have 𝜏𝑈(𝑋) → 𝜏𝑈(0) = −log(𝐴 max ),
and then Ψ(0, 𝑌0 ) is as in (131). Also, from (68) with (𝐴, 𝜏)
replaced by (𝐴 max , 𝜏𝑈) we find that
𝜏𝑈 (𝑋)
1 − 𝐴 max 𝑒
𝑋
∼
, 𝑋 󳨀→ 0
1 + 𝑌0
𝑒𝜌Ψ(0,𝑌) ∼ 𝑒𝜌Ψ(0,𝑌0 ) [
After some calculation we find from (120) that
−Δ 𝐿 (𝑌0 ) = (1 + 𝑌0 ) [
⋅√
(449)
2
√𝐴 max (𝐴 max + 𝑋0 − 2) + 1
√𝐴 max (1 − 𝐴 max )
2
[(1 − 𝐴 max ) − 𝑋0 ]
(456)
1 − 𝐴 max √
2
(1 − 𝑋0 − 𝑌0 ) + 4𝑌0 .
𝐴 max
With (454)–(456) the expansion of (117) agrees precisely with
the large 𝑝 behavior of (444), with (447), (450), and (451).
This completes the matching verification.
(450)
7. Approximations near Transition Layers
1 − 𝐴 max
1
)
(𝑌
𝑝! 0 𝐴 max
(451)
We have thus established the result in (130).
We conclude by showing that (130), for 𝑝 → ∞ and fixed
𝑘, matches asymptotically to (117), for 𝑌 ↑ 𝑌0 and fixed 𝑘.
In this limit, (117) is singular due to the factor 1/√𝑌0 − 𝑌 =
√𝜌/𝑝. First we note from (118) that 𝐴 1 (𝑌0 ) = 𝐴 max and from
(119) we get
(𝑌0 − 𝑌) 𝐴 1 (𝑌)
]
𝑌 (1 − 𝐴 1 (𝑌))
1/2
√𝐴 max √𝐴 max (𝐴 max + 𝑋0 − 2) + 1 [(1 − 𝐴 max ) − 𝑋0 ]
= √1 + 𝑌0 √
𝑝
Ψ𝑌 (0, 𝑌) = log [
(455)
2
=
and then 𝑄(⋅) is determined as
1/4
1 − 𝐴 max
2
⋅√
[(1 − 𝑋0 − 𝑌0 ) + 4𝑌0 ] .
𝐴 max
− 1] (1 − 𝐴 max )
𝑌0 − (𝑋0 + 𝑌0 ) 𝐴 max
By comparing (449) to the large 𝑘 expansion of (444), with
(447), we can take
𝑄 (𝑝) =
2
(1 − 𝐴 max )
and also
1/4
1 − 𝐴 max
2
[(1 − 𝑋0 − 𝑌0 ) + 4𝑌0 ] .
𝐴 max
𝐶8 (𝜌) = √2𝜋𝜌𝑄 (0, 0)
𝑋0
= (1 + 𝑌0 ) [(1 − 𝑋0 − 𝑌0 ) + 4𝑌0 ]
𝜋 (𝑘, 𝑟) ∼ 𝑄 (0, 0) 𝑒𝜌Ψ(0,𝑌0 ) 𝑒𝜌[𝑋log𝑋−𝑋] 𝑒𝜌𝑋log(1+𝑌0 )
𝑝
1
1
− 1)]
[𝑌0 (
𝑝!
𝐴 max
(454)
𝑌 󳨀→ 𝑌0 .
(448)
and then Ψ𝑋 (𝑋, 𝑌0 ) + log𝑋 → log(1 + 𝑌0 ) as 𝑋 → 0. Using
the above in (127) we find that for 𝑋 → 0 we have
⋅
𝑝
𝜌 𝑝
1 − 𝐴 max
𝑌0 ] ( ) 𝑒𝑝 ,
𝐴 max
𝑝
(452)
We analyze the vicinities of the curves where the regions D0 ,
D+ , and D− meet. From Figures 3–6 we see that, for R1 , D0
̃
meets D+ along the curve 𝑌 = 𝑌(𝑋).
For regions R2 ∪R3 , D0
also meets D+ , while D+ and D− meet along 𝑌 = 𝑌∗ (𝑋). For
region R4 , D0 and D− meet along 𝑌 = 𝑌𝑐 (𝑋). The analysis for
𝑌 = 𝑌∗ (𝑋), with the scaling 𝑌 − 𝑌∗ (𝑋) = 𝑂(𝜌−1/3 ), is carried
out in [10] and we omit it here. For R2 the analysis is exactly
as in [10], while for R2 ∩ R3 and R3 the D+ and D− ray
expansions must be multiplied by the appropriate constants
𝐶 and 𝐶1 , but the analysis is otherwise unchanged. We thus
obtain the result in (188).
7.1. Transition Layer near 𝑌 = 𝑌𝑐 (𝑋), Region R4 . This layer
arises only for parameter region R4 . We set 𝜌1/3 [𝑌−𝑌𝑐 (𝑋)] =
𝜂∗∗ = 𝑂(1) and use (𝑋, 𝜂∗∗ ) as the variables. Let us write
𝐿(𝑋, 𝑌) in the ray expansion in (79) as
2
and thus
𝐿 (𝑋, 𝑌) =
Ψ (0, 𝑌) − Ψ (0, 𝑌0 )
= (𝑌 − 𝑌0 ) log (𝑌0 − 𝑌) + 𝑌0 − 𝑌
+ (𝑌 − 𝑌0 ) log [
𝐴 max 1
] + 𝑜 (𝑌 − 𝑌0 ) .
1 − 𝐴 max 𝑌0
(453)
(𝑊0 + 𝑠1 )
1
√𝑌√𝑊0 + 𝑌 √𝑠1 − 𝑌
√𝑊0 (𝑊0 + 𝑠1 )
1
⋅
[1 −
]
𝑊0 + 𝑌
√2𝑊0 + 𝑠1 + 𝑌
⋅ 𝐿 0 (𝑠1 ) ≡ 𝐿 1 (𝑋, 𝑌) 𝐿 0 (𝑠1 ) ,
−1/2
(457)
48
Advances in Operations Research
where we used 𝐿(𝑋, 𝑌) ∼ 𝐿(𝑋, 𝑌𝑐 (𝑋)). If (465) is to
asymptotically match to the D− ray expansion we must have
where
−2
1
[Ai󸀠 (𝑟0 )] 𝑋0−1/6 (1 − √𝑋0 )
2𝜋
𝐿 0 (𝑠1 ) =
⋅
−5/6
(𝑠1 + 𝑊0 ) √𝑋0 − 𝑠1
L0 (−∞) = 1.
(458)
√𝑊0 + 𝑠1
2/3
= 𝜌1/3 𝑟0 (1 − √𝑋)
The scaling 𝑌 − 𝑌𝑐 (𝑋) = 𝑂(𝜌−1/3 ) corresponds to rays that
have 𝑠1 ≈ 𝑌0 and more precisely 𝑠1 − 𝑌0 = 𝑂(𝜌−1/3 ). We thus
set
(459)
In view of (75) and (86) we have
𝑌𝑐 (𝑋) + 𝑊0 𝑌∗ (𝑋) + 𝑊0
=
𝑌0 + 𝑊0
√𝑊0
(460)
which relates the two curves 𝑌𝑐 and 𝑌∗ . From the definition
of 𝑠1 in (81) we then have
𝑠1 + 𝑊0 = 𝑌0 + 𝑊0 + 𝜌−1/3̃𝑠1 =
√𝑊0 (𝑌 + 𝑊0 )
𝑌∗ (𝑋) + 𝑊0
(461)
= 𝜌−1/3
=
𝑌 + 𝑊0
𝑌 (𝑋) + 𝑊0
− 𝑐
]
𝑌∗ (𝑋) + 𝑊0 𝑌∗ (𝑋) + 𝑊0
√𝑊0
[𝑌 − 𝑌𝑐 (𝑋)]
𝑌∗ (𝑋) + 𝑊0
Φ1 (𝑋,𝑌)
𝐿 (𝑋, 𝑌) L (𝑋, 𝜂∗∗ ) ,
(463)
where we can replace 𝑌 by 𝑌𝑐 (𝑋) + 𝜌−1/3 𝜂∗∗ , so the expansion
is in terms of 𝑋 and 𝜂∗∗ . We can also view L as being a
function of the ray variables 𝑡 and ̃𝑠1 . But then the product
𝐿L will satisfy the transport equation in (198). Since 𝐿 is a
particular solution, L must be constant along a ray and thus
a function of ̃𝑠1 but not 𝑡, so in view of (462) we write
L (𝑋, 𝜂∗∗ ) = L0 (
𝑌0 + 𝑊0
𝜂 ).
𝑌𝑐 (𝑋) + 𝑊0 ∗∗
⋅ exp [𝜌Φ (𝑋, 𝑌𝑐 (𝑋) + 𝜌
+𝜌
1/3
Φ (𝑋, 𝑌𝑐 (𝑋) + 𝜌
−1/3
(464)
−1/3
𝜂∗∗ )] ,
𝑋01/6 [𝑊0 + 𝑌𝑐 (𝑋)]
.
Ψ (𝑋, 𝑌𝑐 (𝑋)) = Φ∗ (𝑋, 𝑌𝑐 (𝑋)) − Φ (𝑋0 , 𝑌0 ) ,
Ψ𝑌 (𝑋, 𝑌𝑐 (𝑋)) = Φ𝑌 (𝑋, 𝑌𝑐 (𝑋)) ,
(468)
Ψ𝑌𝑌 (𝑋, 𝑌𝑐 (𝑋)) = Φ𝑌𝑌 (𝑋, 𝑌𝑐 (𝑋)) ,
𝑌
]
𝑌0
𝑌
− log [(1 − 𝐵) ] ,
𝑌0
(469)
where we expressed 𝜏 in terms of 𝐵 and 𝑌 using (68). Along
𝑌 = 𝑌𝑐 we have 𝜕𝐵/𝜕𝑌 = 0 so that from (469) we obtain
Ψ𝑌𝑌 =
𝑌0
1
𝜕𝐵
1−𝐵
+
1 − 𝐵 𝑌0 𝐵 + (1 − 𝐵) 𝑌 𝜕𝑌 𝑌0 𝐵 + (1 − 𝐵) 𝑌
1
−
𝑌
(470)
and hence
Ψ𝑌𝑌 (𝑋, 𝑌𝑐 (𝑋)) = −
𝑊0
.
𝑌𝑐 (𝑌𝑐 + 𝑊0 )
(471)
From the relation between 𝐵 and 𝐴 in (67) we have
𝑋0
𝑌0 𝜕𝐵
𝜕𝐴
]
= [1 −
2 𝜕𝑌
2 𝜕𝑌
(1 − 𝐵)
(1 − 𝐴)
𝑌0 + 𝑊0
𝜂 )
𝑌𝑐 (𝑋) + 𝑊0 ∗∗
𝜂∗∗ )
𝑟0 𝜂∗∗ (1 − √𝑋0 )
Next we examine the ray expansion in D0 near the curve
𝑌 = 𝑌𝑐 (𝑋). This will yield a matching condition for L0 (⋅) as
𝜂∗∗ → +∞. We can easily establish the following continuity
conditions between Ψ(𝑋, 𝑌) and Φ(𝑋, 𝑌) across 𝑌 = 𝑌𝑐 (𝑋)
We thus write the expansion in the transition layer as
𝜋 (𝑘, 𝑟) ∼ 𝐶1 𝐿 (𝑋, 𝑌𝑐 (𝑋)) L0 (
(467)
2/3
∼−
(462)
Consider the balance equation (17) on the scale 𝜂∗∗ =
𝑂(1), with 0 < 𝑋 < 𝑋0 . An asymptotic solution is given by
𝐶1 𝑒𝜌Φ(𝑋,𝑌) 𝑒𝜌
𝑊0 + 𝑌0
)
𝑊0 + 𝑠1
Ψ𝑌 (𝑋, 𝑌) = log [𝐵 + (1 − 𝐵)
𝑌0 + 𝑊0
𝜂 .
𝑌𝑐 (𝑋) + 𝑊0 ∗∗
1/3
𝑋0−1/6 log (
where we recall that 𝑌 = 𝑌𝑐 (𝑋) corresponds to the D0 ray
with 𝐴 = 1 − √𝑋0 , 𝐵 = 𝑊0 /(𝑊0 + 𝑌0 ). To this end we note
that
so that
̃𝑠1 = 𝜌−1/3 √𝑊0 [
Using the fact that Φ1 (𝑋, 𝑌) = Φ∗ (𝑠1 ) we have
𝜌1/3 [Φ∗ (𝑠1 ) − Φ (𝑌0 )]
𝑊 +𝑠
1
1
⋅ exp [− ( +
) log ( 0 1 )] .
2 2√𝑋0
1 − √𝑋0
𝑠1 = 𝑌1 + 𝜌−1/3̃𝑠1 .
(466)
(472)
and thus
(465)
2𝑌0
𝜕2 𝐵
𝜕𝐴 2
=−
( ) ,
2
2
𝜕𝑌
√𝑋0 (𝑌0 + 𝑊0 ) 𝜕𝑌
(473)
at 𝑌 = 𝑌𝑐 (𝑋) .
Advances in Operations Research
49
The function 𝐾0 (𝐴) vanishes as 𝐴 → 1 − √𝑋0 , in view of the
factor [(1 − 𝐴)2 − 𝑋0 ], and we have
It also follows from (470) that
Ψ𝑌𝑌𝑌 (𝑋, 𝑌𝑐 (𝑋))
=
𝐾0 (𝐴)
1
1
−
𝑌𝑐2 (𝑊0 + 𝑌𝑐 )2
(474)
2
(𝑌0 + 𝑊0 ) 𝜕2 𝐵
+
(𝑋, 𝑌𝑐 (𝑋)) .
𝑌0 (𝑌𝑐 + 𝑊0 ) 𝜕𝑌2
∼
2√𝑌0 (1 − √𝑋0 )
√2𝜋 (𝑌0 + 𝑊0 ) (√𝑋0 − 𝑋0 − 𝑌0 ) (𝑊0 + 𝑌𝑐 )
Combining (479) and (480) leads to, for 𝑌 → 𝑌𝑐 (𝑠),
(𝑋 + 𝑌 − 1) [𝐵 + (1 − 𝐵)
𝐶0 𝐾 (𝑋, 𝑌) ∼ 2 (1 − 𝑋0 − 𝑌0 ) 𝜌−5/6 𝜂∗∗
𝑌
]+𝐴
𝑌0
= (𝐴 + 𝑋0 + 𝑌0 − 1) [𝐵 + (1 − 𝐵)
2
(475)
2
𝑌
] .
𝑌0
By implicit differentiation of (475) we obtain
√𝑊0
𝜕𝐴
(𝑋, 𝑌𝑐 (𝑋)) =
𝜕𝑌
𝑊0 + 𝑌𝑐 (𝑋)
(477)
= Φ𝑌𝑌𝑌 (𝑋, 𝑌𝑐 (𝑋))
2𝑊0
1
.
√𝑋0 (𝑊0 + 𝑌𝑐 )3
√𝑋0
(
1
3
(478)
3
𝜂∗∗
) .
𝑊0 + 𝑌𝑐 (𝑋)
2
(𝑌0 + 𝑊0 )
−Δ∼
[𝑌𝑐 + √𝑋0 𝑊0 − √𝑊0 𝑌0 ] ,
𝑌0 − 𝑌𝑐
(𝑌 + 𝑌𝑐 + 2𝑊0 ) .
𝑊0 + 𝑌0 0
⋅ exp [𝑟0
[
2/3
(1 − √𝑋0 )
𝑋01/6
2/3
𝑋0−1/6
.
⋆
𝑌0 + 𝑊0
⋆
𝑌0 + 𝑊0
(482)
Thus (466) and (482) give the behaviors of L0 (⋆) as ⋆ →
±∞. But to determine L0 (⋅) completely we must use asymptotic matching to the corner approximation in (169), which
applies on the (], 𝑇) scale. We thus expand (465) for 𝑋 → 𝑋0
and asymptotically match this to (169), expanding the latter
for ] → ∞, 𝑇 → ∞ but with 𝑇 − √] = 𝑂(1). In this limit we
have
3/2
1 𝑖∞ 𝑒(𝑇−√])𝜃
𝑒−(2/3)]
[
∼
𝑑𝜃] .
∫
2√𝜋]1/4 2𝜋𝑖 −𝑖∞ Ai (𝜃)
(𝑒−𝜏 − 𝐴) (𝑒−𝜏 − 𝐵)
𝑌𝑐
√𝑌𝑐 + 𝑊0 − √𝑊0 (𝑊0 + 𝑌0 )
1 𝑖∞ Ai (] + 𝜃) 𝜃𝑇
𝑒 𝑑𝜃
∫
2𝜋𝑖 −𝑖∞ Ai (𝜃)
As 𝑌 → 𝑌𝑐 (𝑋), 𝑒−𝜏 → (𝑊0 + 𝑌𝑐 )/(𝑊0 + 𝑌0 ),
∼
1
(481)
2
𝜌 {Ψ (𝑋, 𝑌𝑐 + 𝜌−1/3 𝜂∗∗ )
(1 − √𝑋)
1
1
𝑊0 + 𝑌𝑐 √𝑋0 − 𝑋0 − 𝑌0
3
1 (1 − √𝑋0 )
⋆
−
(
) ] , ⋆ 󳨀→ +∞.
3
𝑌0 + 𝑊0
√𝑋0
]
Combining (468) with (477) we have
⋅
⋅
L0 (⋆) ∼ 2Ai󸀠 (𝑟0 ) (1 − √𝑋0 )
2𝑊0
1
−
√𝑋0 (𝑊0 + 𝑌𝑐 )3
2
(1 − √𝑋0 ) √𝑌0 √𝑊0 + 𝑌0 1
√𝑌0 − 𝑌𝑐 √𝑌𝑐 + 𝑌0 + 2𝑊0 √2𝜋
Using the expression in (41) for 𝐶1 and the continuity
conditions in (468) and (477), we compare (465) to the D0
ray expansion as 𝑌 → 𝑌𝑐 (𝑋) to conclude that
1
1
−
𝑌𝑐2 (𝑊0 + 𝑌𝑐 )2
− [Φ (𝑋, 𝑌𝑐 + 𝜌−1/3 𝜂∗∗ ) − Φ (𝑋0 , 𝑌0 )]} ∼ −
⋅
⋅
(476)
and then using (473) in (474) leads to
−
(480)
⋅ 𝜌−1/3 𝜂∗∗ , 𝑌 󳨀→ 𝑌𝑐 .
Eliminating 𝜏 in (68) leads to
Ψ𝑌𝑌𝑌 (𝑋, 𝑌𝑐 (𝑋)) =
2
(479)
(483)
Recalling that 𝑋0 − 𝑋 = 𝜌−2/3 ]1 we have
𝑌0 − 𝑌𝑐 (𝑋) ∼
𝑌0 + 𝑊0
𝑊01/4
𝜌−1/3 √]1
(484)
50
Advances in Operations Research
U → +∞ a standard saddle point calculation shows that
(490) is asymptotic to
and then
𝜂∗∗ = 𝜌1/3 [𝑌 − 𝑌𝑐 (𝑋)] = −𝑇1 + 𝜌1/3 [𝑌0 − 𝑌𝑐 (𝑋)]
∼
(𝑌0 + 𝑊0 ) 𝑋01/6
2/3
(1 − √𝑋0 )
(√] − 𝑇) ,
3
Ai󸀠 (𝑟0 ) 𝑒𝑟0 U (2U) 𝑒−U /3 ,
(485)
(𝑋, 𝑌) 󳨀→ (𝑋0 , 𝑌0 ) .
Apart from the exponential factor
exp {𝜌 [Φ (𝑋, 𝑌𝑐 (𝑋) + 𝜌−1/3 𝜂∗∗ ) − Φ (𝑋0 , 𝑌0 )]} ,
(486)
as 𝑌 → 𝑌𝑐 the expression in (465) becomes
(1 − 𝑋0 − 𝑌0 ) 𝜌−5/6
4/3
⋅
𝑋0−1/12
√2 (𝑊0 + 𝑌0 ) (√𝑋0 − 𝑋0 − 𝑌0 )
[
𝑌0 + 𝑊0
⋅ L0 (
2/3
𝑊01/4
(487)
2/3
(1 − √𝑋0 )
⋅
(√] − 𝑇)) .
Here we note that 𝐿 1 (𝑋, 𝑌𝑐 (𝑋)) is singular as 𝑋 → 𝑋0 , in
view of the factor (𝑠1 − 𝑌𝑐 )−1/2 , and in this limit
𝑠1 − 𝑌𝑐 = 𝑠1 − 𝑌𝑐 + 𝑌0 − 𝑌𝑐 (𝑋)
∼
𝑌0 + 𝑊0
𝑊01/4
√𝑋0 − 𝑋.
(1 − √𝑋0 )
5/3
(488)
]−1/4 1 𝑖∞ 𝑒(𝑇−√])𝜃
𝑑𝜃] .
⋅
[
∫
2√𝜋 2𝜋𝑖 −𝑖∞ Ai (𝜃)
(489)
=
(𝑌0 + 𝑊0 ) 𝑋01/6
2/3
(1 − √𝑋0 )
(𝑋)
1
√𝑌𝑐0 (𝑋) + (1 − √𝑋0 ) (2 − √𝑋0 )
5/3
𝑋01/12
(1
−2
1
[Ai󸀠 (𝑟0 )]
√2𝜋
∞
𝑟0
The factors in (492) that precede L1 come from 𝑒𝜌Φ(𝑋,𝑌) , 𝐶1
1/3
(now given by (37)), 𝑒𝜌 Φ1 (𝑋,𝑌) , 𝐿 1 (𝑋, 𝑌), and 𝐿 0 (𝑠1 ), except
we exclude from 𝐿 0 (⋅) the factor (𝑠1 + 𝑊0 )√𝑋0 − 𝑠1 , which
vanishes if 𝑠1 → 𝑌0 and 𝑌0 → √𝑋0 − 𝑋0 . Instead we include
an extra factor of 𝜌−1/3 , and then 𝜌−1/2 = 𝜌−1/3 𝜌−1/6 , where
𝜌−1/6 comes from 𝐶1 in (37). Also, in (492)
󵄨
𝑌𝑐0 (𝑋) ≡ 𝑌𝑐 (𝑋)󵄨󵄨󵄨𝑌0 =√𝑋0 −𝑋0 .
(493)
In region R3 ∩ R4 we have 𝑊0 + 𝑌0 − √𝑊0 (𝑊0 + 𝑌0 ) ∼ 𝑌𝑐
and 𝑌0 + 2𝑊0 ∼ 2 + 𝑋0 − 3√𝑋0 , and also
Comparing (487) with (489) determines L0 (⋅) as
L0 (
√√𝑋0 − 𝑋0 −
(492)
𝑌𝑐0
⋅ (∫ 𝑒𝛿1 𝑢 Ai (𝑢) 𝑑𝑢) .
(1 − 𝑋0 − 𝑌0 )
(𝑌0 + 𝑊0 ) (√𝑋0 − 𝑋0 − 𝑌0 )
⋅
𝑌0 + 𝑊0
1
𝜂∗∗ ) 0
𝑌𝑐 + 𝑊0
𝑌𝑐 (𝑋)
1
− √𝑋0 )
The limit of (169) as 𝑇, ] → ∞, using (483), apart from some
exponential factors, which will automatically match those
from (486), is given by
𝜌−2/3
(1 − √𝑋0 )
𝜂∗∗ ]
⋅ exp [−𝑟0 1/6
𝑋0 (𝑊0 + 𝑌𝑐 )
[
]
⋅ 𝜌−1/2 L1 (
exp [−𝑟0 (√] − 𝑇)]
(𝑌0 + 𝑊0 ) 𝑋01/6
7.2. Transition Layer near 𝑌 = 𝑌𝑐 (𝑋), Region R3 ∩ R4 . We
̃
take 𝑋0 + 𝑌0 = √𝑋0 + 𝜌−1/3 𝛿∗ . Now the curves 𝑌(𝑋),
𝑌∗ (𝑋),
and 𝑌𝑐 (𝑋) are all close to each other, coinciding if 𝛿∗ = 0.
Most of the analysis closely parallels that in R4 , so we include
here fewer of the details. We again use the variables 𝑋 and 𝜂∗∗
to find that in the transition layer
∼ exp {𝜌 [Φ (𝑋, 𝑌𝑐 + 𝜌−1/3 𝜂∗∗ ) − Φ (𝑋0 , 𝑌0 )]}
−1/2
⋅ 𝜌−1/3 √]1 ]
which is consistent with (482).
𝜋 (𝑘, 𝑟)
−1
1
[Ai󸀠 (𝑟0 )]
√2𝜋
(1 − √𝑋0 )
(491)
U)
𝐿 0 (𝑠1 )
(490)
Ai󸀠 (𝑟0 ) 𝑒𝑟0 U 𝑖∞ 𝑒−𝜃U
𝑑𝜃.
∫
2𝜋𝑖
−𝑖∞ Ai (𝜃)
With (490), (465) becomes the same as (191), so we have
established Proposition 8. Note also that (490) is consistent
with (466), since for U → −∞ the asymptotics of the contour
integral are determined by the pole at 𝜃 = 𝑟0 < 0. For
(𝑠1 + 𝑊0 ) √𝑋0 − 𝑠1
󳨀→
−4/3
−2
1
,
[Ai󸀠 (𝑟0 )] 𝑋0−1/6 (1 − √𝑋0 )
2𝜋
(494)
𝑠1 󳨀→ 𝑌0 .
Thus apart from the factors 𝜌−1/3 L1 (⋅), (492) is just the D−
ray expansion expanded near 𝑌 → 𝑌𝑐 (𝑋) (or 𝑠1 → 𝑌0 ),
Advances in Operations Research
51
divided by (𝑠1 + 𝑊0 )√𝑋0 − 𝑠1 . Thus (492) matches to the D−
ray expansion if
󵄨󵄨
𝑌 + 𝑊0
󵄨
𝜂∗∗ )󵄨󵄨󵄨
𝜌−1/3 L1 ( 0
󵄨󵄨𝜂 →−∞
𝑌𝑐 + 𝑊0
∗∗
(495)
󵄨󵄨
󵄨
∼ [(𝑠1 + 𝑊0 ) √𝑋0 − 𝑠1 ]󵄨󵄨󵄨
.
󵄨𝑠1 →𝑌0
But the left side of (495), in view of (461) and (462), becomes
(𝑠1 + 𝑊0 ) √𝑋0 − 𝑠1
2
∼ (1 − √𝑋0 ) 𝜌−1/3 (
−𝜂∗∗
).
𝑌𝑐 + 𝑊0
(497)
Next we match (492) to the D0 ray expansion and thus
infer the behavior of L1 (𝑈) as 𝑈 → +∞. Now 𝐶0 ∼ (1 −
𝑋0 − 𝑌0 )𝜌−1/2 ∼ (1 − √𝑋0 )𝜌−1/2 . First evaluating 𝐾(𝑋, 𝑌) at
𝑌0 = √𝑋0 − 𝑋0 and then letting 𝑌 → 𝑌𝑐0 we get
2
󵄨
[ 𝐾 (𝑋, 𝑌)|𝑌0 =√𝑋0 −𝑋0 ]󵄨󵄨󵄨󵄨𝑌→𝑌0 ∼ √
𝑐
𝜋
⋅
2
(1 − √𝑋0 )
𝑌𝑐0 (𝑋)
1
1
√√𝑋0 − 𝑋0 − 𝑌𝑐0 (𝑋) √𝑊0 + √𝑊0 + 𝑌𝑐0 (𝑋)
(498)
.
−2
[Ai󸀠 (𝑟0 )] (∫ 𝑒𝛿1 𝑢 Ai (𝑢) 𝑑𝑢) (1 − √𝑋0 )
5/3
2/3
2
∞
𝛿1 𝑢
⋅ (∫ 𝑒
𝑟0
=
2
1[
𝑋 − 𝑋0 + √𝑋0 − 𝑋√ (2 − √𝑋0 ) − 𝑋]
2
]
[
∼ (1 − √𝑋0 )
1/2
(503)
𝜌−1/3 √]1 .
𝜂∗∗ ∼ 𝑋01/6 (1 − √𝑋0 )
1/3
(√] − 𝑇) .
(504)
With (503) and (504), the algebraic part of (492) becomes
−2
1 −1/6 −1/4 −1/4
𝜌 ] 𝑋0 [Ai󸀠 (𝑟0 )]
2√𝜋
(505)
1/3
⋅ L1 (𝑋01/6 (1 − √𝑋0 )
(√] − 𝑇)) .
Comparing (502) to (505) determines L1 (⋅) as
3
𝜂
1 (1 − √𝑋0 )
( ∗∗ ) ] .
3
𝑊0 + 𝑌𝑐
√𝑋0
]
We thus have the matching condition
4/3
√𝑋0 − 𝑋0 − 𝑌∗ (𝑋)
𝑟0
L1 (𝑈) = 𝑋01/6 (1 − √𝑋0 )
−
L1 (𝑈) ∼ 2𝑋1/6 (1 − √𝑋0 )
(502)
Now we evaluate the algebraic part of (492), for 𝑋 → 𝑋0 .
From (460), since 𝑌0 + 𝑊0 → 1 − √𝑋0 = √𝑊0 , we see that
𝑌𝑐0 (𝑋) is the same as 𝑌∗ (𝑋), which is independent of 𝑌0 . We
also have 𝑌∗ (𝑋0 ) = √𝑋0 − 𝑋0 so that (498) becomes singular
as 𝑋 → 𝑋0 . We have
⋅ (∫ 𝑒𝛿1 𝑢 Ai (𝑢) 𝑑𝑢)
(499)
𝜂∗∗
⋅ exp [𝑟0 1/6
𝑋0 (𝑊0 + 𝑌0 )
[
1 1
2√𝜋 2𝜋𝑖
∞
3
𝑌0 + 𝑊0
𝜂∗∗ ) ∼ 2𝑋01/4 (1 − √𝑋0 )
𝑌𝑐 + 𝑊0
(1 − √𝑋0 )
𝑋0−1/3 ]−1/4
∞
𝑒𝜃(𝑇−√]−𝛿1 )
(∫
𝑒𝛿1 𝑢 Ai (𝑢) 𝑑𝑢) 𝑑𝜃.
2
𝜃
[Ai (𝜃)]
𝜌−1/2 𝑒−𝑟0 (𝑇−√])
𝑟0
⋅ 𝑋01/12 L1 (
⋅∫
𝑖∞
4/3
The result in (485) still applies, now simplifying to
The estimate in (478) still holds so if 𝐶0 𝐾(𝑋, 𝑌)𝑒𝜌Ψ(𝑋,𝑌) is to
agree, for 𝑌 → 𝑌𝑐 , with the large 𝜂∗∗ asymptotics of (492), we
must have
∞
𝜌−1/3 (1 − √𝑋0 )
−𝑖∞
𝑈 󳨀→ −∞.
𝑋01/4
2
(501)
Ai (] + 𝜃) ∼ 2−1 𝜋−1/2 ]−1/4 exp (− ]3/2 − √]𝜃)
3
in the integral we expand (173) for ], 𝑇 → ∞ with 𝑇 − √] =
𝑂(1). The result contains some exponential factors and some
algebraic ones, with the latter being
(496)
Now 𝑌0 + 𝑊0 ∼ 1 − √𝑋0 so that (495) and (496) lead to
L1 (𝑈) ∼ (1 − √𝑋0 ) (−𝑈) ,
To determine L1 (⋅) completely we need to match (492) to the
corner layer expansion in (173). Using
∞
⋅ [∫ 𝑒𝛿1 𝑢 Ai (𝑢) 𝑑𝑢]
𝑟0
[Ai󸀠 (𝑟0 )]
2
⋅∫
𝑖∞
−𝑖∞
−1
Ai (𝑢) 𝑑𝑢)
(500)
𝑟0 𝑈
𝑈3
],
⋅ exp [
−
1/3
1/6
√𝑋
3√𝑋
(1
−
)
0
0
[ 𝑋0 (1 − √𝑋0 )
]
𝑈 󳨀→ +∞.
−1
4/3
[Ai󸀠 (𝑟0 )]
2
1
2𝜋𝑖
∞
𝑒−𝜃𝛿1
𝑒𝛿1 𝑢 Ai (𝑢) 𝑑𝑢)
(∫
Ai2 (𝜃) 𝜃
𝑈𝜃
] 𝑑𝜃
⋅ exp [−
1/3
1/6
√𝑋
𝑋
(1
−
)
0
[ 0
]
𝑈
].
⋅ exp [𝑟0
1/3
1/6
[ 𝑋0 (1 − √𝑋0 ) ]
(506)
52
Advances in Operations Research
We have thus established the result in (194). By asymptotically
expanding the contour integral in (506) in the limits 𝑈 →
±∞ we can easily verify that (497) and (500) are satisfied.
Changing variables from (𝑋, 𝜂∗ ) to (𝑡∗ , 𝜂∗ ) the PDE in (510)
becomes
̃
7.3. Transition Layer near 𝑌 = 𝑌(𝑋),
Region R1 ∪ R2 ∪ R3 .
We consider the curve that separates D0 from D+ . We use the
scaling
(515)
̃ (𝑋) + 𝜌−1/2 𝜂∗ ,
𝑌=𝑌
𝜂∗ = 𝑂 (1)
(507)
and expand the joint distribution as
̃ exp [𝜌Ψ+ (𝑋, 𝑌)
̃
𝜋 (𝑘, 𝑟) ∼ 𝐶 (𝜌) 𝐹 (𝑋, 𝜂∗ ) 𝐾+ (𝑋, 𝑌)
(508)
̃ 𝜂∗ + 1 Ψ+,𝑌𝑌 (𝑋, 𝑌)
̃ 𝜂2 ] .
+ √𝜌Ψ+,𝑌 (𝑋, 𝑌)
∗
2
The exponential factor in (508) corresponds to the expansions
̃
If (508) as 𝜂∗ → −∞ is to match
of Ψ+ (𝑋, 𝑌) about 𝑌 = 𝑌(𝑋).
to the D+ ray expansion we must have
𝐹 (𝑋, −∞) = 1,
∀0 < 𝑋 < 𝑋0 .
1 ̃󸀠 2
̃ Ψ+,𝑌 ] 𝐹𝜂 𝜂
[(𝑌 ) (𝑋𝑒Ψ+,𝑋 + 𝑒−Ψ+,𝑋 ) + 𝑌𝑒
∗ ∗
2
+ 𝜂∗ 𝑒
Ψ+,𝑋
𝐹𝜂∗ + (𝑋𝑒
−Ψ+,𝑋
−𝑒
for 𝑋 ∈ (0, 𝑋0 ) and 𝜂∗ ∈ (−∞, ∞). In (510), Ψ+ is understood
̃
to be evaluated at (𝑋, 𝑌) = (𝑋, 𝑌(𝑋)).
We can write the curve
̃
𝑌(𝑋)
in parametric form, as
𝑌0
𝑋 = (1 −
𝑒𝑡∗ ) [1 + (𝑋0 + 𝑌0 − 1) 𝑒−𝑡∗ ] ,
𝑋0 + 𝑌0
̃ (𝑋) = 𝑌0 [𝑋0 + 𝑌0 − 1 + 𝑒−𝑡∗ ] ,
𝑌
𝑋0 + 𝑌0
(511)
̃
𝑑𝑋
𝑑𝑡∗
(512)
and define 𝑀∗ from
󸀠
2
̃
̃
̃ (𝑋)] [𝑋𝑒Ψ+,𝑋 (𝑋,𝑌(𝑋)) + 𝑒−Ψ+,𝑋 (𝑋,𝑌(𝑋)) ]
𝑀∗ = [𝑌
̃
̃ (𝑋) 𝑒Ψ+,𝑌 (𝑋,𝑌(𝑋))
+𝑌
.
(513)
We can view 𝑀∗ as being a function of either 𝑋 or 𝑡∗ . In
terms of 𝑡∗ we have
𝑌 𝑒−𝑡∗
𝑀∗ (𝑡∗ ) = 0
𝑋0 + 𝑌0
+ [2 + (𝑋0 + 𝑌0 − 1) 𝑒−𝑡∗
𝑎󸀠 (𝑡∗ )
1
] ⋆ 𝐹0󸀠 (⋆)
𝑀∗ 𝑎2 (𝑡∗ ) 𝐹0󸀠󸀠 (⋆) + [𝐵 (𝑡∗ ) −
2
𝑎 (𝑡∗ )
(516)
= 0,
where
𝐹0 (𝜂∗ 𝑎 (𝑡∗ )) = 𝐹 (𝑡∗ , 𝜂∗ ) ,
−1
𝐵 (𝑡∗ ) = [1 + (𝑋0 + 𝑌0 − 1) 𝑒𝑡∗ ] .
(517)
Such a similarity solution is possible if 𝑎(𝑡∗ ) satisfies the
nonlinear ODE
𝐵 (𝑡∗ ) −
𝑎󸀠 (𝑡∗ ) 1
= 𝑀∗ (𝑡∗ ) 𝑎2 (𝑡∗ ) .
2
𝑎 (𝑡∗ )
(518)
Setting 𝑎(𝑡∗ ) = [𝑏(𝑡∗ )]−1/2 , the Bernoulli equation in (518)
becomes the linear equation
𝑏󸀠 (𝑡∗ ) + 2𝐵 (𝑡∗ ) 𝑏 (𝑡∗ ) = 𝑀∗ (𝑡∗ ) .
(519)
Solving (519) subject to 𝑏(0) = 0 leads to the expression in
(184). With (518), (516) becomes
𝐹0󸀠󸀠 (⋆) + ⋆𝐹0󸀠 (⋆) = 0
with the latter equation corresponding to (178). Since 𝑡∗ =
𝑡∗ (𝑋) we note that
̃
̃ 0 ) = 𝑌0 .
and note that 𝑡∗ = 0 corresponds to 𝑋 = 𝑋0 , as 𝑌(𝑋
Next we assume that 𝐹(𝑡∗ , 𝜂∗ ) will be a function of a single
“similarity” variable, which we call ⋆ = 𝜂∗ 𝑎(𝑡∗ ) and with
which (515) becomes the ordinary differential equation
(510)
) 𝐹𝑋 = 0,
𝑋𝑒Ψ+,𝑋 (𝑋,𝑌(𝑋)) − 𝑒−Ψ+,𝑋 (𝑋,𝑌(𝑋)) = −
𝑡∗ > 0
(509)
We use (508) in the main balance equation (3) and after
a lengthy calculation we find that 𝐹(𝑋, 𝜂∗ ) satisfies the
parabolic PDE
Ψ+,𝑌
𝜂∗
1
𝐹 = 𝐹𝑡∗ ,
𝑀∗ 𝐹𝜂∗ 𝜂∗ +
2
1 + (𝑋0 + 𝑌0 − 1) 𝑒𝑡∗ 𝜂∗
(520)
and (509) implies that 𝐹(−∞) = 1, and thus
𝐹0 (
𝜂∗
√𝑏 (𝑡∗ )
)=
∞
2
1
𝑒−𝑢 /2 𝑑𝑢.
∫
√2𝜋 𝜂∗ /√𝑏(𝑡∗ )
(521)
We have thus derived (176).
With (521) substituted for 𝐹(𝑋, 𝜂∗ ) in (508), we can show
that as 𝜂∗ → +∞, (508) asymptotically matches to the D0 ray
expansion. As 𝜂∗ → ∞ from (521) we have
𝐹0 (
𝜂∗
√𝑏 (𝑡∗ )
)∼
√𝑏 (𝑡∗ ) 1
𝜂2
exp [− ∗ ] ,
√2𝜋 𝜂∗
2𝑏 (𝑡∗ )
(522)
𝜂∗ 󳨀→ ∞.
𝑌0
−
𝑒𝑡∗ ]
𝑋0 + 𝑌0
(514)
2
𝑌0 𝑒−𝑡∗
⋅[ 𝑡
] .
𝑌0 𝑒 ∗ + (𝑋0 + 𝑌0 ) (𝑋0 + 𝑌0 − 1) 𝑒−𝑡∗
We can easily establish the continuity conditions
̃ (𝑋)) = Ψ+ (𝑋, 𝑌
̃ (𝑋)) − Ψ+ (𝑋0 , 𝑌0 ) ,
Ψ (𝑋, 𝑌
̃ (𝑋)) = Ψ+,𝑌 (𝑋, 𝑌
̃ (𝑋))
Ψ𝑌 (𝑋, 𝑌
(523)
Advances in Operations Research
53
and, after a lengthy calculation, show that
̃ (𝑋)) =
Ψ+,𝑌𝑌 (𝑋, 𝑌
1
̃ (𝑋)) .
+ Ψ𝑌𝑌 (𝑋, 𝑌
𝑏 (𝑡∗ )
But
(524)
Thus expanding the D0 ray expansion, 𝐶0 𝐾(𝑋, 𝑌)𝑒𝜌Ψ(𝑋,𝑌) , as
̃
𝑌 ↓ 𝑌(𝑋)
and comparing this to the large 𝜂∗ expansion of
(508), using (523) and (524), lead to
𝐾+ (𝑋0 , 𝑌0 )
2
=−
√𝑏 (𝑡∗ )
𝑒𝜌Ψ+ (𝑋0 ,𝑌0 )
,
̃ (𝑋)]
√2𝜋 √𝜌 [𝑌 − 𝑌
(525)
̃ (𝑋) .
𝑌 󳨀→ 𝑌
3
∼
(526)
Ψ+ (𝑋0 , 𝑌0 ) = −1 + 𝑋0 + 𝑌0 − (𝑋0 + 𝑌0 ) log (𝑋0
+ 𝑌0 ) .
̃
The curve 𝑌 = 𝑌(𝑋)
corresponds to the ray 𝐴 = 𝑌0 /(𝑋0 +𝑌0 ),
and then 𝐵 = 1 − 𝑋0 − 𝑌0 . From (70) we see that 𝐾 becomes
̃
singular as 𝑌 → 𝑌(𝑋),
due to the factor [𝑌0 − 𝐴(𝑋0 + 𝑌0 )]−1 .
This singularity precisely matches that in the right-hand side
of (525), and yet another lengthy calculation shows that
̃ (𝑋)] 𝐾 (𝑋, 𝑌)}
lim {[𝑌 − 𝑌
(527)
̃ (𝑋)) ,
= √𝑋0 + 𝑌0 √𝑏 (𝑡∗ )𝐾+ (𝑋, 𝑌
which establishes the matching.
We note that the PDE (515) contains many solutions other
than the similarity solution. An initial condition as 𝑡∗ →
0 can be obtained by asymptotically matching (508) to the
corner layer valid on the (𝑛, 𝑝) scale. This will determine the
solution to (515) uniquely, but the matching will again lead
to the conclusion that 𝐹 is given by (521). Below we only
briefly verify that the matching holds. Near the corner the
̃
curve 𝑌 = 𝑌(𝑋)
can be approximated by the straight line
󸀠
̃
𝑌 − 𝑌0 = 𝑌 (𝑋0 )(𝑋 − 𝑋0 ), which corresponds to
𝑝
𝑌0 − 𝑌
𝑌0
1
= ,
= =
𝑋0 − 𝑋 𝑛 (𝑋0 + 𝑌0 )2 − 𝑋0 𝜃1
(528)
where 𝜃1 was defined in Proposition 20. Thus we must show
that (508) for 𝑋 → 𝑋0 is the same as (251). The exponential
parts agree automatically and 𝐶(𝜌) ∼ 𝜌−1/2 𝑄(0, 0), so we must
only show that, for 𝑋 → 𝑋0 ,
∞
̃ √𝑋0 + 𝑌0 ∫
𝐾+ (𝑋, 𝑌)
𝜂∗ /√𝑏(𝑡∗ )
2
2
𝑒−𝑢 /2 𝑑𝑢
𝜂
2
(𝑋0 + 𝑌0 ) − 𝑋0 1
∼
∫ 𝑒−𝑢 /2 𝑑𝑢.
2
√
2𝜋 −∞
(𝑋0 + 𝑌0 )
𝑌0
1
𝑛] .
[−𝑝 +
2
√𝜌
(𝑋0 + 𝑌0 ) − 𝑋0
𝑏 (𝑡∗ )
𝐶
∼ √2𝜋𝜌√𝑋0 + 𝑌0 exp {𝜌 [1 − 𝑋0 − 𝑌0
𝐶0
̃
𝑌→𝑌(𝑋)
∼
(530)
𝑝
̃ (𝑋)]
+ √𝜌 [𝑌0 − 𝑌
√𝜌
From (184)–(187) we obtain, as 𝑋 → 𝑋0 or 𝑡∗ → 0,
But from Proposition 2 we see that, for R1 ∪ R2 ∪ R3 ,
+ (𝑋0 + 𝑌0 ) log (𝑋0 + 𝑌0 )]} ,
,
̃ (𝑋)] = √𝜌 [𝑌 − 𝑌0 + 𝑌0 − 𝑌
̃ (𝑋)]
𝜂∗ = √𝜌 [𝑌 − 𝑌
𝐶0 𝐾 (𝑋, 𝑌)
̃ (𝑋))
∼ 𝐶𝐾+ (𝑋, 𝑌
−5/2
= (2𝜋)−1/2 [(𝑋0 + 𝑌0 ) − 𝑋0 ] (𝑋0 + 𝑌0 )
(529)
𝑌0 (𝑋0 + 𝑌0 ) [(𝑋0 + 𝑌0 ) + (𝑋0 + 𝑌0 ) (𝑌0 − 2𝑋0 ) + 𝑋0 ] 𝑛 (531)
2
[(𝑋0 + 𝑌0 ) − 𝑋0 ]
3
𝜌
since 𝑛/𝑝 ∼ 𝜃1 , as 𝑡∗ → 0 𝜂∗ /√𝑏(𝑡∗ ) approaches −𝜂 in
(252), which verifies (529). This completes the analysis of the
transition from D+ to D0 .
8. Numerical Studies and Discussion
Next we show that our asymptotic results can also be used to
accurately estimate the probabilities 𝜋(𝑘, 𝑟) in ranges where
there is little mass. We will consider 𝜋(0, 𝑅) which is the (very
unlikely) situation where no primary spaces are occupied but
all of the secondary spaces are full. To estimate 𝜋(0, 𝑅) we
must use the expansion that is valid for 𝑘 = 𝑂(1) and 𝑝 = 𝑅 −
𝑟 = 𝑂(1), and this corresponds to (130). Thus we set 𝑘 = 𝑝 = 0
in (130) and note that 𝑄(0, 0) has the different expansions
in Proposition 1, according to regions in parameter space. In
Table 1 we take 𝑋0 = 2, 𝑌0 = 1 and use the expression for
𝑄(0, 0) that holds in parameter region R1 . We see that the
exact and asymptotic results agree to three decimal places
even if 𝜌 is as small as 2. In Table 2 we have 𝑋0 = 0.8 and
𝑌0 = 0.4, and 𝑄(0, 0) is again computed from (30). Now the
agreement is not quite as good as in Table 1, and we typically
have errors of about 10%. This is probably due to the fact that
numerically 𝑋0 + 𝑌0 = 1.2, which exceeds one only slightly,
so it may be preferable to use (29) to approximate 𝑄(0, 0). In
Tables 3 and 4 we take, respectively, 𝑋0 = 0.4, 𝑌0 = 0.4 and
𝑋0 = 0.4, 𝑌0 = 0.2. Table 3 corresponds to region R3 and
Table 4 to region R4 . The agreement is certainly better for
region R4 , as again the sum 𝑋0 + 𝑌0 is further away from the
critical value of one. Here we used 𝑄(0, 0) ∼ 1 − 𝑋0 − 𝑌0 for
region(s) R3 ∪ R4 .
Tables 1–4 show that the very small values of 𝜋(0, 𝑅) are
well predicted by the asymptotic formula(s).
To summarize, we have done a rather thorough asymptotic analysis for this storage allocation model. We have
shown that as long as 𝑋0 + 𝑌0 > 1 (𝑚 + 𝑅 > 𝜌) the effects
of the finiteness of the secondary storage capacity occur only
for (𝑋, 𝑌) ∈ D0 and for those state space regions that border
D0 . However, for 𝑋0 + 𝑌0 < 1 (𝑚 + 𝑅 < 𝜌) the entire state
54
Advances in Operations Research
Table 1: 𝜋(0, 𝑅) for region R1 , 𝑋0 = 2, 𝑌0 = 1.
𝜌
2
4
6
8
10
Exact
5.41 (10−4 )
2.54 (10−7 )
1.19 (10−10 )
5.58 (10−14 )
2.62 (10−17 )
Asymptotic
5.41 (10−4 )
2.54 (10−7 )
1.19 (10−10 )
5.58 (10−14 )
2.62 (10−17 )
Table 2: 𝜋(0, 𝑅) for region R2 , 𝑋0 = 0.8, 𝑌0 = 0.4.
𝜌
5
10
15
20
25
30
35
Exact
1.47 (10−3 )
1.70 (10−6 )
1.96 (10−9 )
2.27 (10−12 )
2.65 (10−15 )
3.09 (10−18 )
3.63 (10−21 )
Asymptotic
1.13 (10−3 )
1.35 (10−6 )
1.61 (10−9 )
1.92 (10−12 )
2.29 (10−15 )
2.73 (10−18 )
3.25 (10−21 )
Table 3: 𝜋(0, 𝑅) for region R3 , 𝑋0 = 0.4, 𝑌0 = 0.4.
𝜌
10
20
30
40
50
60
70
Exact
1.86 (10−4 )
2.29 (10−8 )
2.64 (10−12 )
2.96 (10−16 )
3.25 (10−20 )
3.53 (10−24 )
3.81 (10−28 )
Asymptotic
1.08 (10−4 )
1.56 (10−8 )
1.94 (10−12 )
2.28 (10−16 )
2.59 (10−20 )
2.89 (10−24 )
3.18 (10−28 )
Table 4: 𝜋(0, 𝑅) for region R4 , 𝑋0 = 0.4, 𝑌0 = 0.2.
𝜌
10
20
30
40
50
60
70
Exact
5.13 (10−4 )
1.33 (10−7 )
3.12 (10−11 )
6.99 (10−15 )
1.53 (10−18 )
3.29 (10−22 )
6.98 (10−26 )
Asymptotic
4.17 (10−4 )
1.17 (10−7 )
2.83 (10−11 )
6.48 (10−15 )
1.43 (10−18 )
3.11 (10−22 )
6.66 (10−26 )
space is affected by the finite capacity, as then there are not
enough storage spaces to satisfy the demand for them. For
𝑋0 + 𝑌0 < 1 it proves useful to consider the numbers, 𝑚 − 𝑁1
and 𝑅 − 𝑁2 , of empty primary and secondary spaces, and
then the steady state distribution 𝜋(𝑘, 𝑟) = 𝜋(𝑚 − 𝑛, 𝑅 − 𝑝)
has for 𝜌 → ∞ the limiting form in (50). Even though
the marginal distributions of 𝑁1 and the sum 𝑁1 + 𝑁2 are
particularly simple, the joint distribution of (𝑁1 , 𝑁2 ) is quite
complicated, as the analysis involves many different ranges
of the parameter space (cf. Figure 2) and the state space (cf.
Figures 3–6). In some ranges special functions such as Airy
and parabolic cylinder functions play a key role.
Competing Interests
The authors declare that they have no competing interests.
Acknowledgments
This work is partially supported by a Faculty Development
Grant from Columbia College Chicago.
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