FLUID LIMITS FOR SHORTEST REMAININGPROCESSING TIME QUEUES

Douglas G. Down∗

Department of Computing and Software

McMaster University

1280 Main Street West

Hamilton, ON Canada L8S 4K1

H. Christian Gromoll†

Department of Mathematics

University of Virginia

Charlottesville, VA 22903

Amber L. Puha‡

Department of Mathematics

California State University, San Marcos

San Marcos CA 92096

June 20, 2008

Abstract

We consider a single server queue with renewal arrivals and i.i.d.service times, in which the server employs the shortest remaining pro-cessing time policy. To describe the evolution of this queue, we use ameasure valued process that keeps track of the residual service timesof all buffered jobs. We propose a fluid model (or formal law of largenumbers approximation) for this system and, under mild assumptions,prove existence and uniqueness of fluid model solutions. Furthermore,we prove a scaling limit theorem that justifies the fluid model as afirst order approximation of the stochastic model. The state descrip-tor of the fluid model is a measure valued function whose dynamics

∗Research supported in part by the Natural Sciences and Engineering Research Council

of Canada†Research supported in part by an NSF Mathematical Sciences Postdoctoral Research

Fellowship and NSF grant DMS 0707111‡Research supported in part by NSF Grant 03-05272 and an NSA Young Investigators

Award

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are governed by certain inequalities in conjunction with the standardworkload equation. In particular, these dynamics determine the evolu-tion of the left edge (infimum) of the state descriptor’s support, whichyields conclusions about sojourn times. We characterize the evolutionof this left edge as an inverse functional of the initial condition, arrivalrate, and service time distribution. This characterization reveals themanner in which the growth rate of the left edge depends on the servicetime distribution. It is shown that the rate can vary from logarithmicto polynomial by considering various examples. In addition, it is shownthat this characterization implies that critical fluid models for whichthe service time distribution has unbounded support converge to theempty queue as time approaches infinity, a perhaps unexpected resultsince the workload remains constant.

Contents

1 Introduction 31.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Stochastic and Fluid Models for an SRPT Queue 62.1 Stochastic Model . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 Fluid Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3 Results 113.1 Characterization of the Left Edge . . . . . . . . . . . . . . . . 113.2 Left Edge Asymptotics for Unbounded Support . . . . . . . . 133.3 Explicit Characterization of Fluid Model Solutions . . . . . . 153.4 Fluid Limit Theorem . . . . . . . . . . . . . . . . . . . . . . . 16

4 Proofs of Properties of Fluid Model Solutions 174.1 Proof of Theorem 3.1 . . . . . . . . . . . . . . . . . . . . . . . 174.2 Proof of Theorem 3.6 . . . . . . . . . . . . . . . . . . . . . . . 234.3 Proof of Corollaries 3.7 and 3.8 . . . . . . . . . . . . . . . . . 26

5 Proof of Fluid Limit Theorem 295.1 Foundation for the Proof . . . . . . . . . . . . . . . . . . . . . 30

5.1.1 Performance Processes . . . . . . . . . . . . . . . . . . 305.1.2 Dynamic Inequalities and Equations . . . . . . . . . . 315.1.3 Functional Weak Law of Large Numbers . . . . . . . . 38

5.2 Tightness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395.3 Characterization of Fluid Limit Points . . . . . . . . . . . . . 42

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5.3.1 Verification of (C2) . . . . . . . . . . . . . . . . . . . . 435.3.2 Verification of (C3) . . . . . . . . . . . . . . . . . . . . 44

1 Introduction

Consider a single server queue operating under the Shortest Remaining Pro-cessing Time (SRPT) scheduling policy. The SRPT scheduling policy givespreemptive priority to the job in the system with the shortest remainingprocessing time. Note that to implement this policy, it is assumed that theservice times of jobs are known upon arrival.

Interest in the SRPT policy stretches back to the first optimality resultdue to Schrage [16], who showed that SRPT minimizes the number of jobs inthe system at any point in time (see also Smith [19]). This was done with nodistributional assumptions on the underlying arrival and service processes.Expressions for the mean response time for a single server M/G/1/SRPTqueue were earlier developed by Schrage and Miller [17], with extended re-sults available in Schassberger [15] and Perera [12] (a nice survey from thesame time period is Schreiber [18]). Within these references, one can findexpressions for various performance measures, all of which depend on theentire service time distribution through nested integrals and thus are some-what difficult to work with, particularly if one wishes to make comparisonswith other policies.

Recently, there has been renewed interest in the SRPT policy, mainly incomputer science. For example, Bansal and Harchol-Balter [1] are interestedin the issue of fairness for SRPT ([1] is also a good source for a more extendedlist of prior work on SRPT). More recent work has attempted to provide aframework for comparing policies in the M/G/1 setting, see for exampleWierman and Harchol-Balter [20].

There has also been a recent body of work on the tail behavior of singleserver queues under SRPT; see for example Nunez Queija [10] and Nuyensand Zwart [11]. They discuss the advisability of implementing SRPT usinglarge deviations techniques.

In [3], Down and Wu employ diffusion limits to show certain optimalityproperties of a multi-layered round robin routing policy for a system of par-allel servers each operating under SRPT. This is done under the assumptionof a finitely supported service time distribution, mainly due to the absenceof such limits for more general service time distributions.

In this paper, the goal is to take first steps toward developing a generaldiffusion limit by characterizing the fluid limits (functional law of large num-

3

bers approximations) for a single server SRPT queue. Since SRPT is not ahead-of-the-line policy, we use a state descriptor that tracks the remainingservice times of all jobs in the system. Under mild conditions, we developfluid limits for the measure valued state descriptor that puts a unit of massat each element in the set of remaining service times. Such an approachis in the spirit of Gromoll et al. [6], Puha et al. [14], Doytchinov et al. [4],and Kruk et al. [9], who consider single server queues operating under theProcessor Sharing (PS) and Earliest Deadline First (EDF) policies respec-tively. The analysis here is more akin to that in [4, 9] for EDF. In part, thisis due to the observation that for both SRPT and EDF there is only onejob in service at any point in time, which contrasts with PS where all jobsreceive simultaneous service. There is some additional similarity betweenSRPT and EDF priority schemes, since they give preemptive priority to thejob that has the smallest residual service time and the smallest current leadtime respectively. Indeed the analysis here also makes use of a frontier pro-cess similar to the one considered in [4, 9]. However, since the lead timesin [4, 9] are required to be independent of the service times and decreaseconstantly at rate one, there are significant differences between the analysisin this paper and the work in [4, 9].

Under mild conditions that include a finite limiting arrival rate and afinite first moment for the limiting service time distribution, we prove thatthere is a unique fluid limit. This fluid limit is a measure valued function thathas a nondecreasing left edge, the infimum of the measure’s support. Thisis a direct reflection of the SRPT scheduling policy which gives preemptivepriority to the job with the shortest remaining service time. In particular,work in the fluid limit does not accumulate below the left edge. We analyzethe behavior of the fluid limit by defining a fluid model (see Section 2.2),proving that under mild conditions the fluid limit is a fluid model solution(Theorem 3.9), and analyzing the behavior of fluid model solutions (seeTheorems 3.1 and 3.6 and Corollaries 3.3, 3.7, and 3.8).

Of particular interest is the behavior of the left edge of a fluid modelsolution as a function of time. The results presented here include an ex-plicit description of the unique fluid model solution (Theorems 3.1 and 3.6)that characterizes the left edge as the right continuous inverse of a simplefunctional (18) of the fluid model data. This characterization allows us toprove that under mild conditions, critical fluid model solutions correspond-ing to service time distributions with unbounded support converge to thezero measure as time tends to infinity (Corollary 3.3). This is somewhatsurprising since the fluid analog of the workload is constant for critical fluidmodel solutions, i.e., the workload does not decrease with time. The charac-

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terization of the left edge is applied in some specific examples to determinethe rate at which the left edge increases as time increases, and hence the rateat which the critical fluid model empties (see Section 3.2). Interestingly, therate depends on the fluid model data and, in particular, on the tail behav-ior of the limiting service time distribution. Corollary 3.7 characterizes thelimiting behavior of critical fluid model solutions corresponding to servicetime distributions with bounded support.

The paper is organized as follows. In Section 2, we define our stochasticand fluid models for an SPRT queue. Section 3 contains the statementsof our main results. Section 4 contains the proofs of the results concerningfluid model solutions, and the remainder of the paper is devoted to the proofof the fluid limit theorem, Theorem 3.9.

1.1 Notation

The following notation will be used throughout the paper. Let N denotethe set of positive integers and let R denote the set of real numbers. Fora, b ∈ R, we write a∨ b for the maximum of a and b, a∧ b for the minimumof a and b, a+ and a− for the positive and negative parts of a respectively,⌊a⌋ for the largest integer less than or equal to a, and ⌈a⌉ for the smallestinteger greater than or equal to a. The nonnegative real numbers [0,∞) willbe denoted by R+. By convention, a sum of the form

∑mi=n with n > m, or

a sum over an empty set of indices equals zero. The sets (a, b), [a, b), and(a, b] are empty for a, b ∈ [0,∞] with a ≥ b and, unless otherwise specified,the infimum of the empty set equals ∞. For a function g : R+ → R, let‖g‖∞ = supx∈R+

|g(x)| and ‖g‖K = supx∈[0,K] |g(x)| for each K ≥ 0. Wedefine the positive and negative parts of such a function g by g+(x) = g(x)∨0and g−(x) = (−g(x)) ∨ 0 for all x ∈ R+.

For a Borel set B ⊂ R+, we denote the indicator of the set B by 1B . Inaddition, for ǫ > 0,

Bǫ = x ∈ R+ : infy∈B

|x − y| < ǫ. (1)

We also define the real valued function χ(x) = x, for x ∈ R+. For a topo-logical space A, denote by C+(A) the set of nonnegative, continuous, realvalued functions defined on A, and denote by C+

b (A) the functions in C+(A)that are bounded.

Let M denote the set of finite, nonnegative Borel measures on R+ andlet Ma denote those elements of M that do not charge the origin. Considerξ ∈ M and a Borel measurable function g : R+ → R which is integrable

5

with respect to ξ. We define 〈g, ξ〉 =∫

R+g(x)ξ(dx). The set M is endowed

with the weak topology, that is, for ξn, ξ ∈ M, n ∈ N, we have ξnw→ ξ if and

only if 〈g, ξn〉 → 〈g, ξ〉 as n → ∞, for all g : R+ → R that are bounded andcontinuous. With this topology, M is a Polish space [13]. We denote thezero measure in M by 0 and the measure in M that puts one unit of massat the point x ∈ R+ by δx. For x ∈ R+, the measure δ+

x is δx if x > 0 and0 otherwise.

We say that a measure ξ ∈ M has a finite first moment if 〈χ, ξ〉 < ∞.Let Mχ denote the set of all such measures and let M0 = Mχ ∩Ma. It willbe convenient to extend the notion of uniform integrability for random vari-ables (and their associated distributions) to elements of M. Call a sequenceξn ⊂ M uniformly integrable if 〈χ, ξn〉 < ∞ for all n, and

limx→∞

supn〈χ1[x,∞), ξn〉 = 0.

It is easy to show that if ξn ⊂ M is uniformly integrable and ξnw→ ξ, then

〈χ, ξ〉 < ∞ and 〈χ, ξn〉 → 〈χ, ξ〉.We use “⇒” to denote convergence in distribution of random elements

of a metric space. Following Billingsley [2], we use P and E respectivelyto denote the probability measure and expectation operator associated withwhatever space the relevant random element is defined on. Unless otherwisespecified, all stochastic processes used in this paper are assumed to havepaths that are right continuous with finite left limits (r.c.l.l.). For a Polishspace S, we denote by D([0,∞),S) the space of r.c.l.l. functions from [0,∞)into S, endowed with the Skorohod J1-topology [5].

2 Stochastic and Fluid Models for an SRPT Queue

2.1 Stochastic Model

Our stochastic model of an SRPT queue consists of the following: a randominitial condition Z(0) ∈ M specifying the state of the system at time zero,stochastic primitives E(·) and vkk∈N describing the arrival of jobs andtheir service times to the queue, and a measure valued state descriptor Z(·)describing the time evolution of the system. These are defined below.

Initial condition. The initial condition specifies the number Z(0) of jobsin the queue at time zero, as well as the initial service time of each job.Assume that Z(0) is a nonnegative integer valued random variable that is

6

finite almost surely. The initial service times are the first Z(0) elements of asequence vjj∈N of strictly positive, finite random variables. We sometimesrefer to jobs in the system at time zero as initial jobs. The initial job withservice time vj, j ≤ Z(0), is called job j.

A convenient way to express the initial condition is to define an initialrandom measure Z(0) ∈ M by

Z(0) =

Z(0)∑

j=1

δvj,

which equals 0 if Z(0) = 0. Our assumptions imply that Z(0) satisfies

P(〈1,Z(0)〉 ∨ 〈χ,Z(0)〉 < ∞) = 1. (2)

In particular, the number of initial jobs and the initial workload are finitealmost surely, and so Z(0) ∈ M0 almost surely.

Stochastic primitives. The stochastic primitives consist of an exogenousarrival process E(·) and a sequence of initial service times vkk∈N. Thearrival process E(·) is a rate α ∈ (0,∞) delayed renewal process. For t ∈[0,∞), E(t) represents the number of jobs that arrive to the queue duringthe time interval (0, t]. Jobs arriving after time 0 are indexed by integersj > Z(0). For t ∈ [0,∞), let

A(t) = Z(0) + E(t). (3)

Then job j ∈ N arrives at time Tj = inft ∈ [0,∞) : A(t) ≥ j. Hence, fori < j, Ti ≤ Tj and we say that job i arrives before job j.

For each k ∈ N, the random variable vk represents the initial servicetime of the (Z(0) + k)th job. That is, job j > Z(0) has initial service timevj−Z(0). Assume that the random variables vkk∈N are strictly positiveand form an independent and identically distributed sequence with commonBorel distribution ν on R+. Assume that the mean 〈χ, ν〉 ∈ (0,∞) and letβ = 〈χ, ν〉−1. Define the traffic intensity ρ = α/β.

It will be convenient to combine the stochastic primitives into a single,measure valued load process.

Definition 2.1 The load process is given by

V(t) =

E(t)∑

k=1

δvk, for t ∈ [0,∞).

Then V(·) ∈ D([0,∞),M).

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Evolution of the residual service times. In an SRPT queue, the small-est nonzero residual service time decreases at rate one until either it becomeszero or a job arrives that has a smaller initial service time, at which timethe rate changes to zero and the new smallest nonzero residual service timebegins decreasing at rate one. We adopt the convention that in case of a tie,the residual service time of the job that arrived first (that is, the job withsmaller index) begins decreasing at rate one. These dynamics are capturedby the unique solution to the following set of equations.

For x, y ∈ R+, let ϕ(x, y) = 1 if x = 0 and y = 1, and zero otherwise.For j ∈ N, let

wj =

vj , 1 ≤ j ≤ Z(0),

vj−Z(0), j > Z(0).(4)

For t ∈ [0,∞), let Z0(t) = 0 and, for all t ∈ [0,∞) and j ∈ N, define

wj(t) = wj −

∫ Tj∨t

Tj

ϕ(〈1[0,wj(s)),ZA(s)(s)〉, 〈1[0,wj (s)],Zj(s)〉)ds, (5)

Zj(t) =

j∑

ℓ=1

δ+wℓ(t)

. (6)

Since Z(0) < ∞ and E(t) < ∞ for all t ∈ [0,∞) almost surely, equations(4)–(6) have a unique right continuous solution wj(·)j∈N almost surely.For j ∈ N and t ∈ [0,∞), wj(t) is the residual service time at time t of jobj.

The unique solution of (4)–(6) satisfies the following properties. Firstly,since ϕ(·, ·) ≥ 0, wj(·) is continuous and nonincreasing for each j ∈ N.Furthermore, for all j ∈ N, 0 ≤ wj(t) ≤ wj for all t ∈ [0,∞). Hence, foreach j ∈ N, Zj(·) ∈ D ([0,∞),M), and Zj(·) does not charge the origin atany time.

We adopt the following terminology and definitions. For j ∈ N, we saythat job j is in the system at time t ∈ [0,∞) if t ≥ Tj and wj(t) > 0. Hence,if there are no jobs in the system at time t, ZA(t)(t) = 0 and the system isempty at time t. For t ∈ [0,∞) and j ∈ N, let

φj(t) =

0, if t ∈ [0, Tj),

ϕ(〈1[0,wj(t)),ZA(t)(t)〉, 〈1[0,wj (t)],Zj(t)〉), if t ∈ [Tj ,∞).(7)

For t ∈ [0,∞) and j ∈ N, we refer to φj(t) as the instantaneous rate ofservice allocated to job j at time t. If the system is not empty at timet ∈ [0,∞), φj(t) = 1 for exactly one j such that 1 ≤ j ≤ A(t), and is zero

8

for all other indices. Given t ∈ [0,∞) such that the system is not empty attime t, we refer to 1 ≤ j ≤ A(t) such that φj(t) = 1 as the job in service attime t. Thus, whenever the system is not empty there is exactly one job inservice and the server is busy. Otherwise, the system is empty, no jobs are inservice, and the server is idle. Once a job is in service, it remains in serviceuntil either its residual service time reaches zero, or a job enters the systemwith initial service time strictly smaller than the residual service time of thejob in service. Specifically, if φj(s) = 1 for some s ∈ [0,∞), then by rightcontinuity, there exists a time t > s such that φj(u) = 1 for all u ∈ [s, t).We say that a nonempty time interval [s, t) ⊂ [0,∞) is a busy period if foreach u ∈ [s, t), there exists 1 ≤ j ≤ A(u) such that φj(u) = 1. Finally, if attime t ∈ [0,∞), 0 < wi(t) < wj(t) for some 1 ≤ i, j ≤ A(t), then, since wj(·)is continuous and wi(·) is nonincreasing, φj(s) = 0 for all s ∈ [t,Di), whereDi = infu ∈ [Ti,∞) : wi(u) = 0. That is, job j cannot enter or resumeservice until job i departs the system.

Measure-valued state descriptor. For t ∈ [0,∞), define the state de-scriptor by

Z(t) =

A(t)∑

j=1

δ+wj(t)

. (8)

Note that Z(t) = ZA(t)(t) for all t ∈ [0,∞).

2.2 Fluid Model

The fluid model has two parameters, α ∈ (0,∞) and a Borel probabilitymeasure ν on R+ that does not charge the origin and satisfies 〈χ, ν〉 <∞. These parameters are limits of parameters in the stochastic model,where α corresponds to the rate at which jobs arrive to the system and theprobability measure ν corresponds to the distribution of the i.i.d. servicetimes for those jobs. The traffic intensity parameter ρ is given by ρ = α/β,where β = 1/〈χ, ν〉. The pair (α, ν) is referred to as the data for the fluidmodel. The adjectives strictly subcritical, subcritical, critical, supercritical,and strictly supercritical are used to refer to data that satisfy ρ < 1, ρ ≤ 1,ρ = 1, ρ ≥ 1, and ρ > 1, respectively.

Given a measure valued function ζ : [0,∞) → M, for each t ∈ [0,∞), let

l(ζ, t) = supx ∈ R+ : 〈1[0,x), ζ(t)〉 = 0, (9)

which is the infimum of the support of ζ(t). Note that for t ∈ [0,∞),l(ζ, t) equals infinity if ζ(t) = 0 and equals zero if 〈1[0,x), ζ(t)〉 > 0 for all

9

x ∈ (0,∞). When it is understood which measure valued function ζ(·)is under consideration, the dependence on ζ(·) is suppressed by using theabbreviated notation l(t) in place of l(ζ, t). We refer to l(·) as the left edgeof the measure valued function ζ(·).

Definition 2.2 Let (α, ν) be fluid model data and let ξ ∈ M0. A measurevalued function ζ : [0,∞) → M is a fluid model solution for the data (α, ν)and the initial measure ξ if each of the following hold:

(C1) ζ(·) is right continuous;

(C2) for all t ∈ [0,∞),

〈χ, ζ(t)〉 = [〈χ, ξ〉 + (ρ − 1)t]+ ; (10)

(C3) for all t ∈ [0,∞) and for all g ∈ C+b (R+),

〈g1(l(t),∞), ξ + αtν〉 ≤ 〈g, ζ(t)〉 ≤ 〈g1[l(t),∞), ξ + αtν〉, (11)

where [∞,∞) = (∞,∞) = ∅.

Note that the upper bound in (11) implies that ζ(t) can not have an atomat zero. That is,

supt∈[0,∞)

〈10, ζ(t)〉 = 0. (12)

This is immediate if l(t) > 0, and follows by bounded convergence if l(t) = 0,since neither ξ nor ν charges the origin. Together with (10) and (11), thisimplies that ζ(0) = ξ.

Condition (C1) is natural since a fluid model solution can be viewedas a formal functional law of large numbers approximation of the measurevalued state descriptor for the stochastic model, which is right continuous.Condition (C2) is the standard workload equation and is also natural sincethe SRPT policy is work conserving.

Condition (C3) is specific to SRPT. It implies that, for each t ∈ [0,∞),ζ(t) has no support below l(t) and agrees with the measure ξ + αtν abovel(t). If we intuitively regard the fluid model as a deterministic system thatreceives αt units of mass during each time interval (0, t] which, as it arrives,is instantaneously distributed over R+ according to the distribution ν, and isprocessed according to the SRPT discipline, then (C3) can be interpreted asfollows. Mass arriving below level l(t) at time t is instantaneously flushed outof the system, while mass arriving above l(t) by time t receives no processing

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by time t. Hence, the mass that is at l(t) at time t is being processed attime t. This reflects the fact that in an SRPT queue, jobs with the shortestremaining processing time are served first. The inequalities allow for thepossibility of atoms in ξ and ν. In particular, discrete distributions are alsoincluded in this analysis.

3 Results

3.1 Characterization of the Left Edge

Analysis of the left edge of fluid model solution depends on some details ofthe relationship between the data (α, ν) and the initial measure ξ. For data(α, ν), let

x1 = supx ∈ R+ : α〈χ1[0,x], ν〉 < 1, (13)

x2 = infx ∈ R+ : α〈χ1[0,x], ν〉 > 1. (14)

Since ν does not charge the origin, x1 > 0. In addition, x1 ≤ x2. If ρ < 1or ρ = 1 and ν has unbounded support, then x1 = x2 = ∞. If ρ > 1 orρ = 1 and ν has bounded support, x1 < ∞. In this case, it is possible thatα〈χ1[0,x1), ν〉 < 1 since ν may have an atom at x1. In fact, if ρ > 1, it maybe the case that α〈χ1[0,x1], ν〉 > 1. In such a case, x1 = x2. It is also possiblethat α〈χ1[0,x1], ν〉 = 1. Then, if 〈1(x1,x1+ǫ), ν〉 = 0 for some ǫ > 0, x1 < x2.In particular, if ρ = 1, x2 = ∞.

For ξ ∈ M, define the left edge of ξ by

lξ = supx ∈ R+ : 〈1[0,x), ξ〉 = 0. (15)

For data (α, ν), let

M1 = ξ ∈ M0 : 〈1[0,x1), ξ〉 > 0 = ξ ∈ M0 : lξ < x1. (16)

When x1 = ∞, M1 is simply M0 without 0. For data (α, ν), let

M2 = ξ ∈ M0 : ξ 6= 0 and 〈1[0,x1), ξ〉 = 0 = ξ ∈ M0 : x1 ≤ lξ < ∞.(17)

For data (α, ν) and ξ ∈ M0, let

s(x) =〈χ1[0,x], ξ〉

1 − α〈χ1[0,x], ν〉, for all x ∈ [0, x1), (18)

and definet1 = sups(x) : x ∈ [0, x1). (19)

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lξ x2x1

t

s−1

r(t)

t1

(t1 = ∞)

ρ > 1, or ρ = 1 and support of ν bounded

s(x)

lξ x1 = x2 = ∞

s

s−1

r(s)

t

s−1

r(t)

t1

(t1 = ∞)

ρ < 1, or ρ = 1 and support of ν unbounded

s(x)

Figure 1: Possible relationships among (α, ν), ξ, x1, x2, t1, and s(·) whenξ ∈ M1.

Note that s(·) is nondecreasing and right continuous, and may achieve thevalue t1 for some x ∈ [0, x1). For example, s(·) achieves the value t1 for somex ∈ [0, x1) if ρ < 1 and the union of the supports of ν and ξ is bounded.Also note that t1 < ∞ if and only if ρ < 1, or ρ ≥ 1 and either ξ 6∈ M1

or x1 < ∞ and α〈χ1[0,x1), ν〉 < 1, that is, ν has an atom at x1. Figure 1illustrates some of the possible relationships among (α, ν), ξ, x1, x2, t1, ands(·).

Let s−1r : [0, t1) → R+ be the right continuous inverse of s(·) on [0, t1),

which is given by

s−1r (t) = infx ∈ [0, x1) : s(x) > t, for all t ∈ [0, t1). (20)

If t1 < ∞, then for convenience we extend s−1r (·) to be defined on all of

[0,∞) by letting

s−1r (t) =

x1, if ξ ∈ M1 and t ∈ [t1,∞),

lξ, if ξ 6∈ M1 and t = 0,

x2 ∧ lξ, if ξ 6∈ M1 and t ∈ (0,∞).

(21)

Note that if ξ 6∈ M1, then t1 = 0. Also note that, when t1 < ∞, s−1r (t) is

not necessarily finite for t ∈ [t1,∞). For example, if either ρ < 1 or ρ = 1and ξ = 0, then t1 < ∞ and s−1

r (t) = ∞ for all t ∈ [t1,∞).The following result characterizes the left edge dynamics of fluid model

solutions. It is proved in Section 4.1.

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Theorem 3.1 Let (α, ν) be fluid model data and ξ ∈ M0. If ζ(·) is a fluidmodel solution for the data (α, ν) and initial measure ξ, then l(ζ, t) = s−1

r (t)for all t ∈ [0,∞).

Remark 3.2 In light of Theorem 3.1, for x ∈ [0, x1), s(x) can be viewed asthe fluid analog of the response time for a job of size x that is in the systemat time 0.

3.2 Left Edge Asymptotics for Unbounded Support

One corollary of Theorem 3.1 is that if (α, ν) are critical data and the supportof ν is unbounded, then all fluid model solutions for all initial measuresconverge to the zero measure as time tends to infinity (see Corollary 3.3).This happens despite the fact that the first moment, or workload, is constantfor the critical fluid model (cf. (C2)). In particular, for such data, the uniqueinvariant state is the zero measure and so the invariant manifold consists ofa single state. This presents challenges for developing a diffusion limit viastate space collapse arguments.

Corollary 3.3 Let (α, ν) be critical fluid model data such that x1 = ∞ andlet ξ ∈ M0. If ζ(·) is a fluid model solution for the data (α, ν) and initialmeasure ξ, then ζ(t)

w→ 0 as t → ∞.

Proof. If ξ = 0, then by (C2), 〈χ, ζ(t)〉 = 0 for all t ∈ [0,∞). Hence, by(12), ζ(·) ≡ 0 and the result follows. Otherwise, ξ 6= 0. Since x1 = ∞, (13)and Lemma 4.2 (iii) below imply that t1 = ∞ and limt→∞ s−1

r (t) = ∞. ByTheorem 3.1, for t ∈ [0,∞) such that s−1

r (t) > 0,

〈1, ζ(t)〉 = 〈1[s−1r (t),∞), ζ(t)〉 ≤

〈χ1[s−1r (t),∞), ζ(t)〉

s−1r (t)

=〈χ, ξ〉

s−1r (t)

,

where the final equality is by (C2). Letting t tend to infinity in the aboveinequality completes the proof.

The asymptotic behavior of critical fluid model solutions for data withx1 < ∞ is addressed in Corollary 3.7 below. At this point, it is instructiveto look at a few examples to see how the critical fluid model empties underdifferent distributional assumptions. In particular, we demonstrate that theasymptotic behavior of the left edge l(·) depends on the tail behavior of thedistribution ν.

13

Example 3.4 Let ξ ∈ M0 be such that ξ 6= 0. Suppose that ν is an expo-nential distribution with rate λ > 0 and let α = λ. Then ρ = 1, x1 = ∞,and for x ∈ R+,

1 − α〈χ1[0,x], ν〉 = λ

∫ ∞

x

λye−λydy = (λx + 1)e−λx.

Therefore,

s(x) = 〈χ1[0,x], ξ〉eλx

λx + 1, x ∈ R+.

Given ǫ ∈ (0, λ) and δ ∈ (0, 1), let y ∈ R+ be such that eǫx > λx + 1 and〈χ1[0,x], ξ〉 ≥ δ〈χ, ξ〉 for all x ≥ y. Then

δ〈χ, ξ〉e(λ−ǫ)x ≤ s(x) ≤ 〈χ, ξ〉eλx, for all x ≥ y.

Then the left edge l(ζ, ·) of any fluid model solution ζ(·) for data (α, ν) andinitial measure ξ satisfies

1

λln

(

t

〈χ, ξ〉

)

≤ l(ζ, t) ≤1

λ − ǫln

(

t

δ〈χ, ξ〉

)

, for all t ∈ [〈χ, ξ〉eλy ,∞).

Since ǫ ∈ (0, λ) is arbitrary,

l(ζ, t) ∼1

λln(t), as t → ∞.

Example 3.5 Let b, k > 0. Suppose that ν has Pareto density f(x) =(k + 1)bk+1/xk+2 for x ≥ b, which has mean (k + 1)b/k. Let α = k/(k + 1)b.Then ρ = 1, x1 = ∞, and for x ≥ b,

1 − α〈χ1[0,x], ν〉 = kbk

∫ ∞

x

y−k−1dy =

(

b

x

)k

.

Therefore,

s(x) = 〈χ1[0,x], ξ〉(x

b

)k

, for x ≥ b.

Let δ ∈ (0, 1) and let y > b be such that 〈χ1[0,y], ξ〉 ≥ δ〈χ, ξ〉. This impliesthat the left edge l(ζ, ·) of any fluid model solution ζ(·) for data (α, ν) andinitial measure ξ satisfies

b

(

t

〈χ, ξ〉

)1k

≤ l(ζ, t) ≤ b

(

t

δ〈χ, ξ〉

)1k

, t ∈ [〈χ, ξ〉(y/b)k ,∞).

Since δ ∈ (0, 1) is arbitrary,

l(ζ, t) ∼ b

(

t

〈χ, ξ〉

)1k

, as t → ∞.

14

3.3 Explicit Characterization of Fluid Model Solutions

The definition of fluid model solutions given in Section 2.2 arises naturallyby considering the limiting dynamics of the SRPT policy under fluid scalingof the stochastic model. However, this definition only determines fluid modelsolutions implicitly, as there is mutual dependence between ζ(t) and l(t) in(11). Although our main interest is in the left edge l(t), it is importantto have an explicit characterization of the entire fluid model solution ζ(t).This characterization is the key to proving existence and uniqueness of fluidmodel solutions, and also has some useful consequences described below.

Theorem 3.6 Let (α, ν) be fluid model data and ξ ∈ M0. A fluid modelsolution for the data (α, ν) and initial measure ξ exists and is unique. Theunique solution ζ(·) satisfies the following: for each t ∈ [0,∞) and g ∈C+

b (R+),

〈g, ζ(t)〉 =

g(s−1r (t))a(t) + 〈g1(s−1

r (t),∞), ξ + αtν〉, if s−1r (t) < ∞,

0, if s−1r (t) = ∞,

(22)

where for each t ∈ [0,∞) such that s−1r (t) < ∞,

a(t) =

0, if s−1r (t) = 0,

1s−1r (t)

[

〈χ1[0,s−1r (t)], ξ + αtν〉 − t

]

, if s−1r (t) > 0.

(23)

Theorem 3.6 is proved in Section 4.2.In the case of critical fluid model data (α, ν) such that ν has bounded

support, that is, x1 < ∞, M2 is nonempty. Furthermore, as a result ofTheorem 3.6, each measure in M2 is an invariant state. In particular, theset of invariant states is M2 ∪ 0. The following corollary of Theorem 3.6states that fluid model solutions with any initial measure ξ ∈ M0 convergeto the set of invariant states, and explicitly identifies the limiting invariantstate.

Corollary 3.7 Let (α, ν) be critical fluid model data such that x1 < ∞ andlet ξ ∈ M0. If ζ(·) is a fluid model solution for the data (α, ν) and initialmeasure ξ, then for all g ∈ C+

b (R+),

limt→∞

〈g, ζ(t)〉 =g(x1)〈χ1[0,x1), ξ〉

x1+ 〈g1[x1,∞), ξ〉,

and so the limiting invariant state is 〈χ1[0,x1), ξ〉x−11 δx1 + 1[x1,∞)ξ.

15

Here, 1[x1,∞)ξ is the measure in M such that 〈g, 1[x1,∞)ξ〉 = 〈1[x1,∞)g, ξ〉for all bounded continuous functions g. Corollary 3.7 is proved in Section4.3, as is the next corollary of Theorem 3.6.

Corollary 3.8 Let (α, ν) be fluid model data and ξ ∈ M0. The unique fluidmodel solution ζ(·) for data (α, ν) and initial measure ξ is continuous.

3.4 Fluid Limit Theorem

This section presents the limit theorem that rigorously justifies the fluidmodel discussed above as an approximation of the original stochastic model.We first define a sequence of systems over which the limit is taken. Let Rbe a sequence of positive real numbers increasing to infinity. Consider anR-indexed sequence of stochastic models, each defined as in Section 2.1. Foreach r ∈ R, there is an initial condition Zr(0); there are stochastic primi-tives Er(·) and vr

kk∈N with parameters αr, νr, βr, and ρr, and an arrivalprocess Ar(·) with arrival times T r

j j∈N; there is a corresponding measurevalued load process Vr(·); there is a state descriptor Zr(·). The stochasticelements of each model are defined on a probability space (Ωr,Fr,Pr) withexpectation operator Er. A fluid scaling (or law of large numbers scaling) isapplied to each model in the R-indexed sequence as follows. For each r ∈ Rand t ∈ [0,∞), let

Er(t) =1

rEr(rt), Vr(t) =

1

rVr(rt) and Zr(t) =

1

rZr(rt). (24)

Let α ∈ (0,∞) and define α(t) = αt for all t ∈ [0,∞). Let ν ∈ M0 be aprobability measure. Then

〈10, ν〉 = 0 and 0 < 〈χ, ν〉 < ∞. (25)

Let β = 〈χ, ν〉−1 and ρ = α/β. Define ρ(t) = ρt for all t ∈ [0,∞). For thesequence of exogenous arrival processes, assume that as r → ∞,

Er(·) ⇒ α(·). (26)

For the sequence of service time distributions, assume that

νr w→ ν and νr : r ∈ R is uniformly integrable. (27)

This implies that βr → β and thus ρr → ρ as r → ∞.For the sequence of fluid scaled initial conditions Zr(0) : r > 0, assume

that as r → ∞,(Zr(0), 〈χ, Zr(0)〉) ⇒ (Z0, 〈χ,Z0〉), (28)

16

where Z0 is a random measure satisfying

P(Z0 ∈ M0) = 1. (29)

In particular, Z0 has finite total mass, finite first moment, and does notcharge the origin almost surely. The following result establishes the fluidapproximation.

Theorem 3.9 Under the asymptotic assumptions (26)–(29), the sequenceZr(·) : r ∈ R converges in distribution on D([0,∞),M) to a measurevalued process Z∗(·) such that Z∗(0) is equal in distribution to Z0, andalmost surely, Z∗(·) is a fluid model solution for the data (α, ν) and initialcondition Z∗(0).

Proof. This follows from Theorems 5.15 and 5.16 in Section 5 below.

4 Proofs of Properties of Fluid Model Solutions

4.1 Proof of Theorem 3.1

This section begins with Lemma 4.1, which summarizes some basic prop-erties of the left edge of any fluid model solution. Then Lemma 4.2 andPropositions 4.3 and 4.4 summarize some basic relations satisfied by s−1

r (·).To conclude, these results are put together to prove Theorem 3.1.

Lemma 4.1 Suppose that ζ(·) is a fluid model solution for data (α, ν) andinitial measure ξ ∈ M0.

(i) If ξ ∈ M1, then l(t) < x1 for all t ∈ [0, t1) and l(t) = x1 for allt ∈ [t1,∞).

(ii) If ξ 6∈ M1, then t1 = 0, l(0) = lξ, and l(t) = x2 ∧ lξ for all t ∈ (0,∞).

(iii) l(·) is nondecreasing and right continuous for t ∈ [0, t1).

Proof. First, we derive some basic relationships satisfied by l(·). In (11),take a sequence gn

∞n=1 ⊂ C+

b (R+) such that gn ր χ as n → ∞. Thenan application of the monotone convergence theorem yields that for all t ∈[0,∞),

〈χ1(l(t),∞), ξ + αtν〉 ≤ 〈χ, ζ(t)〉 ≤ 〈χ1[l(t),∞), ξ + αtν〉. (30)

17

For t ∈ [0,∞) such that 〈χ, ζ(t)〉 > 0, (30) together with (10) implies that

〈χ1[0,l(t)), ξ + αtν〉 ≤ t ≤ 〈χ1[0,l(t)], ξ + αtν〉. (31)

Inequalities (31) have an intuitive interpretation as follows. In the fluidlimit, t units of work can be completed by time t. So (31) reflects the factthat the jobs with the shortest remaining processing time are served first.However, (31) by itself does not necessarily uniquely determine l(t) for eacht, since there may be intervals in R+ that do not intersect the union of thesupports of ν and ξ, e.g., when ν and ξ are discrete distributions. In anycase, from (31) we obtain that for t ∈ [0,∞) such that 〈χ, ζ(t)〉 > 0,

〈χ1[0,l(t)), ξ〉 ≤ t(

1 − α〈χ1[0,l(t)), ν〉)

(32)

t(

1 − α〈χ1[0,l(t)], ν〉)

≤ 〈χ1[0,l(t)], ξ〉. (33)

Parts (i) and (ii); ρ < 1. Suppose that ρ < 1. Note that in thiscase, x1 = x2 = ∞ and t1 = 〈χ, ξ〉/(1 − ρ) < ∞. Hence, for t ∈ [0, t1),〈χ, ξ〉 + (ρ − 1)t > 0 and for t ∈ [t1,∞), 〈χ, ξ〉 + (ρ − 1)t ≤ 0. This togetherwith (C2) implies that 〈χ, ζ(t)〉 > 0 for all t ∈ [0, t1) and 〈χ, ζ(t)〉 = 0 fort ∈ [t1,∞). Therefore, ζ(t) 6= 0 for all t ∈ [0, t1) and, by (12), ζ(t) = 0 forall t ∈ [t1,∞). Thus, if ρ < 1, then l(t) < ∞ for t ∈ [0, t1) and l(t) = ∞ fort ∈ [t1,∞). Hence, (i) and (ii) hold for ρ < 1.

Part (i); ρ ≥ 1. Suppose that ρ ≥ 1 and fix ξ ∈ M1. Then, by (C2),〈χ, ζ(t)〉 > 0 for all t ∈ [0,∞). In particular, l(t) < ∞ for all t ∈ [0,∞)and (32) and (33) hold for all t ∈ [0,∞). Fix t ∈ [0,∞). If x1 = ∞, thenρ = 1 and t1 = ∞, and since l(t) < ∞, l(t) < x1. So (i) holds if x1 = ∞.Otherwise, x1 < ∞. If l(t) > x1, then by (16), (32), and (13),

0 < 〈χ1[0,x1), ξ〉 ≤ 〈χ1[0,l(t)), ξ〉 ≤ t(

1 − α〈χ1[0,l(t)), ν〉)

≤ 0,

a contradiction. So l(t) ≤ x1. Hence, it suffices to show that l(t) = x1 ifand only if t1 < ∞ and t ∈ [t1,∞). First suppose that l(t) = x1. By (16)and (32),

0 < 〈χ1[0,x1), ξ〉 ≤ t(

1 − α〈χ1[0,x1), ν〉)

,

that is, 1 − α〈χ1[0,x1), ν〉 > 0. This together with (19) implies that t1 < ∞and so using (32),

t1 =〈χ1[0,x1), ξ〉

1 − α〈χ1[0,x1), ν〉=

〈χ1[0,l(t)), ξ〉

1 − α〈χ1[0,l(t)), ν〉≤ t.

18

This completes the proof of the ‘only if’ direction. For the ‘if’ direction,suppose now that t1 < ∞ and t ∈ [t1,∞). Then, by (19), 1−α〈χ1[0,x1), ν〉 >0 and

t1 =〈χ1[0,x1), ξ〉

1 − α〈χ1[0,x1), ν〉. (34)

Furthermore, since ρ ≥ 1 and 1 − α〈χ1[0,x1), ν〉 > 0, x1 < ∞. To obtaina contradiction, suppose that l(t) < x1. Then, s(l(t)) ≤ t1 by (34), andt ≤ s(l(t)) by (33). Hence t ≤ t1. But t ∈ [t1,∞), so t = t1 and we havel(t1) < x1. But then (33) and (34) imply that s(l(t1)) = t1. Hence the unionof the supports of ν and ξ does not intersect (l(t1), x1). So, by (C3), forsome a(t1) > 0,

〈g, ζ(t1)〉 = 〈g1[x1,∞), ξ + αt1ν〉 + g(l(t1))a(t1), for all g ∈ C+b (R+).

Then by (C2) and (30),

〈χ, ξ〉 + (ρ − 1)t1 = 〈χ1[x1,∞), ξ + αt1ν〉 + l(t1)a(t1),

so that〈χ1[0,x1), ξ + αt1ν〉 − t1 = l(t1)a(t1).

Then by (34), l(t1)a(t1) = 0 so that l(t1) = 0. Hence, by (C3) and the factthat neither ν nor ξ charges the origin, a(t1) = 0. This implies that theunion of the supports of ν and ξ do not intersect [l(t1), x1), a contradiction.Hence, l(t1) = x1 and (i) holds for ρ ≥ 1.

Part (ii); ρ ≥ 1. Suppose that ρ ≥ 1 and fix ξ 6∈ M1. Then by (18) and(19), t1 = 0. If ρ = 1 and ξ = 0, then by (C2) and (12), ζ(t) = 0 for allt ∈ [0,∞). Thus l(t) = ∞ for all t ∈ [0,∞), as desired. Otherwise, eitherρ > 1 or ξ 6= 0. Then by (C2), 〈χ, ζ(t)〉 > 0 for all t ∈ (0,∞). In particular,l(t) < ∞ and (32) and (33) hold for all t ∈ (0,∞). Since ζ(0) = ξ, it isimmediate that l(0) = lξ. Fix t ∈ (0,∞). If l(t) > x2 ∧ lξ, then the leftside of (32) is positive and the right side of (32) is negative, a contradiction.So l(t) ≤ x2 ∧ lξ. If l(t) < x1, then the left side of (33) is positive. Butξ 6∈ M1, so x1 ≤ lξ. Hence, if l(t) < x1, the right side of (33) is zero, acontradiction. So x1 ≤ l(t). If x1 = x2 ∧ lξ, it follows that l(t) = x2 ∧ lξ.Otherwise, x1 < x2 ∧ lξ. Then, since x2 ∧ lξ ≤ x2, α〈χ1[0,x2∧lξ), ν〉 = 1 sothat α〈χ1[x2∧lξ ,∞), ν〉 = ρ− 1. Suppose that x1 ≤ l(t) < x2 ∧ lξ. Then since

19

〈χ, ζ(t)〉 > 0, by (C2), (9), (C3), and (15),

〈χ, ξ〉 + αt〈χ1[x2∧lξ,∞), ν〉 = 〈χ, ξ〉 + (ρ − 1)t = 〈χ, ζ(t)〉

= 〈χ1[l(t),∞), ζ(t)〉 = 〈χ1[l(t),x2∧lξ), ζ(t)〉 + 〈χ1[x2∧lξ ,∞), ζ(t)〉

= 〈χ1[l(t),x2∧lξ), ζ(t)〉 + 〈χ1[x2∧lξ,∞), ξ〉 + αt〈χ1[x2∧lξ,∞), ν〉

= 〈χ1[l(t),x2∧lξ), ζ(t)〉 + 〈χ, ξ〉 + αt〈χ1[x2∧lξ,∞), ν〉.

But then 〈χ1[l(t),x2∧lξ), ζ(t)〉 = 0, which contradicts (9). Thus, l(t) ≥ x2 ∧ lξand so (ii) holds for ρ ≥ 1.

Part (iii). Note that [0, t1) is empty when t1 = 0. Henceforth we assumet1 > 0. So, by parts (i) and (ii), we may assume that ξ ∈ M1 and thusl(t) < ∞ for all t ∈ [0, t1). Hence, by (9), it follows that 〈χ, ζ(t)〉 > 0 for allt ∈ [0, t1). Suppose that there exists 0 ≤ s < t < t1 such that l(t) < l(s).Then by (31),

t ≤ 〈χ1[0,l(t)], ξ + αtν〉 ≤ 〈χ1[0,l(s)), ξ + αtν〉 ≤ s + (t − s)α〈χ1[0,l(s)), ν〉.

By part (i), l(s) < x1 and so by (13), α〈χ1[0,l(s)), ν〉 < 1, a contradiction.Thus l(·) is nondecreasing on [0, t1). For the verification that l(·) is rightcontinuous on [0, t1), let

z(t, x) = 〈1[0,x], ζ(t)〉, for all t ∈ [0,∞) and x ∈ R+.

For all t ∈ (0,∞), z(t, x) = 0 for x < l(t) and z(t, x) > 0 for x > l(t). Sinceζ(·) is right continuous, it follows that for each t ∈ [0, t1), limsցt z(s, x) =z(t, x) for all x ∈ R+ that are continuity points for z(t, ·). Since l(·) is nonde-creasing on [0, t1), we have l(t) ≤ l(t+) for all t ∈ [0, t1). Suppose that thereexists a t ∈ [0, t1) such that l(t) < l(t+). Then, since l(·) is nondecreasingon [0, t1), l(t+) ≤ l(s) for all s ∈ (t, t1). In particular, z(s, x) = 0 for alls ∈ (t, t1) and x ∈ (l(t), l(t+)). But, z(t, x) > 0 for all x ∈ (l(t), l(t+)).Hence given x ∈ (l(t), l(t+)) such that x is a continuity point for z(t, ·), itfollows that

0 < z(t, x) = limsցt

z(s, x) = 0,

a contradiction (since there are at most countably many points of disconti-nuity for z(t, ·)). This completes the proof of part (iii).

Inequalities (32) and (33) and right continuity of l(·) suggest that l(·)is related to the right continuous inverse s−1

r (·) of s(·). The next lemmastates some of the basic properties of s−1

r (·) on [0, t1). For this we need thefollowing definition:

x0 = infx ∈ [0, x1) : s(x) ≥ t1 ∧ x1. (35)

20

Lemma 4.2 Let (α, ν) be fluid model data and ξ ∈ M0. Then

(i) s−1r (t) < x1 for all t ∈ [0, t1);

(ii) s−1r (·) is nondecreasing and right continuous on [0, t1);

(iii) limtրt1 s−1r (t) = x0;

(iv) s(s−1r (t)) ≥ t for all t ∈ [0, t1);

(v) s(s−1r (t)−) ≤ t for all t ∈ [0, t1);

(vi) 〈1[s−1r (t),x), ξ + αtν〉 > 0 for all x ∈ (s−1

r (t), x1) and all t ∈ [0, t1).

Proof. Property (i) is an immediate consequence of (19) and (20). Property(ii) holds due to (20) and the fact that s(·) is nondecreasing and right con-tinuous on [0, x1). Property (iii) follows from (i), (ii), and (20). Properties(iv) and (v) follow from (20) and the fact that s(·) is nondecreasing and rightcontinuous on [0, x1). Property (vi) follows from (20) and right continuityof s(·). To see this, fix t ∈ [0, t1). If s(s−1

r (t)) = t, then, by (20), s(x) > tfor all x ∈ (s−1

r (t), x1), which implies (vi). Otherwise, s(s−1r (t)) > t and by

(v), (20), and the fact that s(·) is nondecreasing, s(·) has a discontinuity ats−1r (t), which corresponds to 〈1s−1

r (t), ξ + αtν〉 > 0.

Proposition 4.3 Let (α, ν) be fluid model data and ξ ∈ M0 \ 0.

(i) If either t ∈ [0, t1), or t ∈ [t1,∞) and ρ ≥ 1, then s−1r (t) < ∞.

(ii) If t ∈ [t1,∞) and ρ < 1, then s−1r (t) = ∞.

Proof. If t1 > 0 and t ∈ [0, t1), then ξ ∈ M1 and the result follows fromLemma 4.2 (i). Next, suppose that t1 < ∞, t ∈ [t1,∞), and ρ ≥ 1. Then by(19) and since ρ ≥ 1, x1 < ∞. If ξ ∈ M1, then by (21), s−1

r (t) = x1 < ∞.If ξ ∈ M2, then t1 = 0 and by (21), s−1

r (t) ≤ lξ < ∞. Hence (i) holds. Ift1 < ∞, t ∈ [t1,∞) and ρ < 1, then since ξ 6= 0, ξ ∈ M1 and by (13) and(21), s−1

r (t) = x1 = ∞. Hence (ii) holds.

Proposition 4.4 Let (α, ν) be fluid model data and ξ ∈ M0. Then, for allt ∈ [0,∞) such that s−1

r (t) < ∞,

〈χ1[0,s−1r (t)), ξ + αtν〉 ≤ t ≤ 〈χ1[0,s−1

r (t)], ξ + αtν〉. (36)

21

Proof. To verify this, fix t ∈ [0,∞) such that s−1r (t) < ∞. If ξ = 0, then

by (19), t1 = 0 and by (21), t ∈ (0,∞) and x2 = s−1r (t) < ∞. In particular,

ρ > 1. By (14), α〈χ1[0,x2), ν〉 ≤ 1 ≤ α〈χ1[0,x2], ν〉. In these inequalities,replace x2 with s−1

r (t) and multiply by t to obtain (36) for ξ = 0.Otherwise, ξ 6= 0 and by Proposition 4.3, either t ∈ [0, t1), or t ∈ [t1,∞)

and ρ ≥ 1. If t1 > 0 and t ∈ [0, t1), then by Lemma 4.2 (i) and (13),the denominator of s(s−1

r (t)) is positive. After rearranging terms in theinequalities in Lemma 4.2 (iv) and (v), we obtain (36).

If t1 < ∞, t ∈ [t1,∞), and ρ ≥ 1, then there are two subcases to consider:ξ ∈ M1 and ξ ∈ M2. If ξ ∈ M1, then t1 > 0 by (19) and s−1

r (t) = x1 by(21), i.e., x1 < ∞. Since (36) holds on [0, t1), Lemma 4.2 (iii) implies that

〈χ1[0,x0), ξ + αt1ν〉 ≤ t1 ≤ 〈χ1[0,x1], ξ + αt1ν〉.

If x0 = x1, then we can replace x0 with x1 in the above inequalities. Ifx0 < x1, then by (35), s(x) = t1 for all x ∈ (x0, x1). By right continuity ofs(·), s(x0) = t1 so that 〈χ1[0,x0], ξ + αt1ν〉 = t1 and 〈χ1(x0,x1), ξ + αt1ν〉 = 0.Hence, in both cases (x0 = x1 or x0 < x1),

〈χ1[0,x1), ξ + αt1ν〉 ≤ t1 ≤ 〈χ1[0,x1], ξ + αt1ν〉.

By (13),

〈χ1[0,x1), α(t − t1)ν〉 ≤ t − t1 ≤ 〈χ1[0,x1], α(t − t1)ν〉.

Adding the two previous displays and replacing x1 with s−1r (t) implies (36)

for the first subcase, ξ ∈ M1.For the second subcase, ξ ∈ M2, note that t1 = 0 by (19). Then (36) is

immediate for t = 0 since (21) implies that s−1r (0) = lξ. For t ∈ (0,∞), (21)

implies that s−1r (t) = x2 ∧ lξ. Then 〈χ1[0,s−1

r (t)), ξ〉 = 0 and s−1r (t) ∈ [x1, x2].

Hence by (13) and (14),

α〈χ1[0,s−1r (t)), ν〉 ≤ α〈χ1[0,x2), ν〉 ≤ 1 ≤ α〈χ1[0,x1], ν〉 ≤ α〈χ1[0,s−1

r (t)], ν〉.

In these inequalities, multiply by t, and add 〈χ1[0,s−1r (t)], ξ〉 or 〈χ1[0,s−1

r (t)), ξ〉

as appropriate to obtain (36).

Proof of Theorem 3.1. Suppose that ζ(·) is a fluid model solution forthe data (α, ν) and initial measure ξ ∈ M0. By Lemma 4.1 and (21),l(t) = s−1

r (t) for t ∈ [t1,∞). If t1 = 0, then the proof is complete. Otherwise,t1 > 0, and so ξ ∈ M1. Then it suffices to show that l(t) = s−1

r (t) for allt ∈ [0, t1). By Lemma 4.1 (i), l(t) < ∞ for all t ∈ [0, t1) so that, by (12),

22

〈χ, ζ(t)〉 > 0 for all t ∈ [0, t1). Furthermore, by Lemma 4.1 (i) and (13),1 − α〈χ1[0,l(t)], ν〉 > 0 for all t ∈ [0, t1). This together with (32) and (33)implies that

〈χ1[0,l(t)), ξ〉

1 − α〈χ1[0,l(t)), ν〉≤ t ≤

〈χ1[0,l(t)], ξ〉

1 − α〈χ1[0,l(t)], ν〉, for all t ∈ [0, t1). (37)

If l(t) = 0, then (37) implies that t = 0. Since s−1r (0) = 0, l(0) = s−1

r (0).Otherwise, l(t) > 0. By (18), (37) is equivalent to

s(l(t)−) ≤ t ≤ s(l(t)), for all t ∈ [0, t1). (38)

Fix t ∈ [0, t1). From (38) and Lemma 4.1 (i), for all ǫ ∈ (0, l(t) ∧ (t1 − t)),t+ǫ ≤ s(l(t+ǫ)) and s(l(t)−ǫ) ≤ t. Hence, by (20), l(t)−ǫ ≤ s−1

r (t) ≤ l(t+ǫ)for all ǫ ∈ (0, l(t) ∧ (t1 − t)). Then by Lemma 4.1 (iii), it follows thatl(t) ≤ s−1

r (t) ≤ l(t+) = l(t). Since t ∈ [0, t1) was arbitrary, the proof iscomplete.

4.2 Proof of Theorem 3.6

In this section, Theorem 3.1 is used to prove Theorem 3.6.Proof of Theorem 3.6. Given fluid model data (α, ν) and ξ ∈ M0, for eacht ∈ [0,∞), define ζ(t) to be the unique finite Borel measure that satisfies(22). Note that by (23) and Proposition 4.4, a(t) ≥ 0 for all t ∈ [0,∞) suchthat s−1

r (t) < ∞. Thus, ζ(t) ∈ M for each t ∈ [0,∞). Furthermore, ζ(0) = ξsince s−1

r (0) = lξ. We begin by proving that ζ(·) is a fluid model solution,which implies the existence of a fluid model solution that also satisfies (22).For this, we need to verify that ζ(·) satisfies (C1)–(C3).

We begin by verifying that ζ(·) satisfies (C1). First, suppose that ξ 6∈M1. Then it suffices to show that ζ(t)

w→ ξ as t → 0. This is immediate by

(21) if lξ ≤ x2. Otherwise, x2 < lξ. Then the support of ξ does not intersect[0, x2] or (x2, lξ) and the result follows from (22) and (23). For ξ ∈ M1, rightcontinuity of ζ(·) on [t1,∞) follows from (21), (22), and (23). For ξ ∈ M1,right continuity of ζ(·) on [0, t1) follows from Lemma 4.2 (ii), (22), and (23),once we show that a(·) is right continuous on [0, t1). Since ξ ∈ M1, t1 > 0.Due to Lemma 4.2 (ii) and (23), the only issue is to verify right continuity ofa(t) at t ∈ [0, t1) such that s−1

r (t) = 0. Since s(0) = 0, t ∈ [0, t1) and Lemma4.2 (iv) imply that s−1

r (t) = 0 only if t = 0. Therefore, in order to completethe verification of right continuity of ζ(·), it suffices to show that if t1 > 0and s−1

r (0) = 0, then limtց0 a(t) = 0. Note that if t1 > 0 and s−1r (0) = 0,

23

then by right continuity of s−1r (·), limtց0 s−1

r (t) = 0. Then since neither ξnor ν charges the origin, it follows that

0 ≤ lim inftց0

a(t) ≤ lim suptց0

a(t) ≤ lim suptց0

s−1r (t)〈1[0,s−1

r (t)], ξ + αtν〉

s−1r (t)

= 0.

Therefore ζ(·) is right continuous.Next, we verify that ζ(·) satisfies (C2). For this, take a sequence,

gn∞n=1 ⊂ C+

b (R+) in (22) such that gn ր χ as n → ∞. Then an applica-tion of the monotone convergence theorem yields that for each t ∈ [0,∞),

〈χ, ζ(t)〉 =

〈χ1[0,s−1r (t)], ξ + αtν〉 − t + 〈χ1(s−1

r (t),∞), ξ + αtν〉, if s−1r (t) < ∞,

0, otherwise.

(39)For t ∈ [0,∞) such that s−1

r (t) < ∞, (39) together with Proposition 4.4implies that (10) holds at time t. For t ∈ [0,∞) such that s−1

r (t) = ∞,(22) implies that ζ(t) = 0. Hence, in order to verify that (10) holds fort ∈ [0,∞) such that s−1

r (t) = ∞, we must show that 〈χ, ξ〉 + (ρ − 1)t ≤ 0.By Proposition 4.3, there are two cases to consider: t ∈ [t1,∞) and ρ < 1,and ξ = 0. If t ∈ [t1,∞) and ρ < 1, then x1 = ∞ and t1 < ∞. By (18)and (19), 〈χ, ξ〉 + (ρ − 1)t1 = 0. So for t ∈ [t1,∞), 〈χ, ξ〉 + (ρ − 1)t ≤ 0and (10) holds at time t. If ξ = 0, then t1 = 0 by (19). For t = 0, (10) isimmediate since ξ = 0. For t ∈ (0,∞), x2 ∧ lξ = s−1

r (t) = ∞ and so ρ ≤ 1.This together with ξ = 0 implies that (10) holds at time t. Therefore (C2)holds.

Next, we verify (C3). For t ∈ [0,∞) such that s−1r (t) = ∞, (C3) is

immediate. Fix t ∈ [0,∞) such that s−1r (t) < ∞. We first show that

l(t) = s−1r (t). For this by (22), it suffices to show that either a(t) > 0 or

〈1(s−1r (t),x], ξ + αtν〉 > 0 for all x > s−1

r (t). Since s−1r (t) < ∞, Proposition

4.3 implies that either t ∈ [0, t1), or t ∈ [t1,∞) and ρ ≥ 1. If t ∈ [0, t1), thenby Lemma 4.2 (i) s−1

r (t) < x1. Hence, by (20) and right continuity of s(·),either s(s−1

r (t)) > t or s(s−1r (t)) = t. If s(s−1

r (t)) > t, then by (23) a(t) > 0.Otherwise, s(s−1

r (t)) = t (and so a(t) = 0). Then, by (20) s(x) > t forall x ∈ (s−1

r (t), x1]. In particular, s(x) > s(s−1r (t)) for all x ∈ (s−1

r (t), x1].Hence, 〈1(s−1

r (t),x], ξ+αtν〉 > 0 for all x ∈ (s−1r (t), x1]. Therefore, if t ∈ [0, t1),

l(t) = s−1r (t). Otherwise t ∈ [t1,∞) and ρ ≥ 1. Then, by (21), s−1

r (t) ≥ x1.Hence α〈χ1[0,s−1

r (t)], ν〉 ≥ 1. If 〈χ1[0,s−1r (t)], ξ〉 > 0 or α〈χ1[0,s−1

r (t)], ν〉 > 1,

then by (23) a(t) > 0. Otherwise, 〈χ1[0,s−1r (t)], ξ〉 = 0 and α〈χ1[0,s−1

r (t)], ν〉 =

1 (and so a(t) = 0). Since s−1r (t) ≥ x1 and 〈χ1[0,s−1

r (t)], ξ〉 = 0, ξ 6∈ M1.

24

Hence, by (21), (14), and (15), 〈1(s−1r (t),x], ξ + αtν〉 > 0 for all x > s−1

r (t).

Thus, if t ∈ [t1,∞) and ρ ≥ 1, l(t) = s−1r (t), which completes the proof that

l(t) = s−1r (t).

Having established that l(t) = s−1r (t) for t ∈ [0,∞) such that s−1

r (t) < ∞,we return to the proof of (C3) for t ∈ [0,∞) such that s−1

r (t) < ∞. Again,fix t ∈ [0,∞) such that s−1

r (t) < ∞. Then, the first inequality in (11) followsfrom l(t) = s−1

r (t), (22), and the fact that a(t) ≥ 0 (cf. Proposition 4.4). Toverify that the second inequality in (11) holds, it suffices to show that

a(t) ≤ 〈1s−1r (t), ξ + αtν〉. (40)

By (23), (40) holds if s−1r (t) = 0. So assume that s−1

r (t) > 0. The firstinequality in (36) together with (23) implies that

a(t) =1

s−1r (t)

[

〈χ1[0,s−1r (t)], ξ + αtν〉 − t

]

≤1

s−1r (t)

[

〈χ1[0,s−1r (t)], ξ + αtν〉 − 〈χ1[0,s−1

r (t)), ξ + αtν〉]

= 〈1s−1r (t), ξ + αtν〉.

This completes the proof that ζ(·) is a fluid model solution for (α, ν) andξ ∈ M0.

It remains to prove uniqueness of fluid model solutions. For this, let ζ1(·)be a fluid model solution for the data (α, ν) and initial measure ξ ∈ M0.Then by Theorem 3.1, l1(·) = l(ζ1, ·) = s−1

r (·). We show that ζ1(·) = ζ(·),where ζ(·) is given by (22). For this, fix t ∈ [0,∞). Since l1(t) = s−1

r (t), ifs−1r (t) = ∞, ζ1(t) = 0 = ζ(t). Otherwise, s−1

r (t) < ∞. Then (C3) impliesthat ζ1(t) has no support below s−1

r (t) and agrees with ξ +αtν above s−1r (t).

Thus, for some a1(t) ∈ R+,

〈g, ζ1(t)〉 = g(s−1r (t))a1(t) + 〈g1(s−1

r (t),∞), ξ + αtν〉, for all g ∈ C+b (R+).

(41)In order to completely characterize ζ1(t), one simply needs to determinea1(t). To verify uniqueness, we must show that a1(t) = a(t). If l1(t) =s−1r (t) = 0, then by (C3), a1(t) = 0 since neither ν nor ξ charges the origin.

Otherwise, 0 < l1(t) = s−1r (t) < ∞. Then take a sequence gn

∞n=1 ⊂

C+b (R+) in (41) such that gn ր χ as n → ∞, to obtain

〈χ, ζ1(t)〉 = s−1r (t)a1(t) + 〈χ1(s−1

r (t),∞), ξ + αtν〉. (42)

Since 0 < l1(t) = s−1r (t) < ∞, 〈χ, ζ1(t)〉 > 0. Thus, by combining (42) and

(C2), one can solve for a1(t) to verify that a1(t) = a(t).

25

4.3 Proof of Corollaries 3.7 and 3.8

Proof of Corollary 3.7. Fix g ∈ C+b (R+). If ξ 6∈ M1, then by (16),

lξ ≥ x1 and so by (19), t1 = 0. Since ρ = 1, x2 = ∞. Hence, by (21),s−1r (t) = lξ for all t ∈ [0,∞). If lξ = ∞, then ξ = 0 and by Theorem 3.6,

ζ(t) = 0 for all t ∈ [0,∞), and the result follows. Otherwise, lξ < ∞. Then,by Theorem 3.6 and the equalities 〈1(x1,∞), ν〉 = 0 and α〈χ1[0,lξ ], ν〉 = 1, forall t ∈ [0,∞),

〈g, ζ(t)〉 =g(lξ)〈χ1[0,lξ ], ξ〉

lξ+ 〈g1(lξ ,∞), ξ〉 = 〈g1[lξ ,∞), ξ〉 = 〈g1[x1,∞), ξ〉,

and the result follows.Otherwise, ξ ∈ M1. Since x1 < ∞, Lemma 4.2 (i) and (21) imply that

s−1r (t) < ∞ for all t ∈ [0,∞). Consider the case t1 < ∞. By Theorem 3.6,

(21), and the equalities 〈1(x1,∞), ν〉 = 0 and α〈χ1[0,x1], ν〉 = 1, we have forall t ≥ t1,

〈g, ζ(t)〉 =g(x1)〈χ1[0,x1], ξ〉

x1+〈g1(x1,∞), ξ〉 =

g(x1)〈χ1[0,x1), ξ〉

x1+〈g1[x1,∞), ξ〉.

Hence the result holds if t1 < ∞. Next consider the case t1 = ∞. By Lemma4.2 (i) and (iii), we have the two equalities

limt→∞

〈g1(s−1r (t),∞), ξ〉 = 〈g1[x1,∞), ξ〉,

limt→∞

g(s−1r (t))〈χ1[0,s−1

r (t)], ξ〉

s−1r (t)

=g(x1)〈χ1[0,x1), ξ〉

x1.

Hence, by Theorem 3.6 and the fact 〈1[x1,∞), ν〉 = 0, it suffices to show that

limt→∞

[

α〈g1(s−1r (t),x1), ν〉 −

g(s−1r (t))

s−1r (t)

(

1 − α〈χ1[0,s−1r (t)], ν〉

)

]

t = 0.

Firstly, note that for all t ∈ [0,∞), we have the two inequalities

〈g1(s−1r (t),x1), ν〉 ≥

infg(x) : x ∈ (s−1r (t), x1)

x1〈χ1(s−1

r (t),x1), ν〉,

〈g1(s−1r (t),x1), ν〉 ≤

supg(x) : x ∈ (s−1r (t), x1)

s−1r (t)

〈χ1(s−1r (t),x1), ν〉.

In addition, x1 is not an atom of ν since t1 = ∞. So for all t ∈ [0,∞),

α〈χ1(s−1r (t),x1), ν〉 = 1 − α〈χ1[0,s−1

r (t)], ν〉.

26

Thus, for all t ∈ [0,∞),

[

infg(x) : x ∈ (s−1r (t), x1)

x1−

g(s−1r (t))

s−1r (t)

]

[

1 − α〈χ1[0,s−1r (t)], ν〉

]

t

≤

[

α〈g1(s−1r (t),x1), ν〉 −

g(s−1r (t))

s−1r (t)

(

1 − α〈χ1[0,s−1r (t)], ν〉

)

]

t

≤

[

supg(x) : x ∈ (s−1r (t), x1)

s−1r (t)

−g(s−1

r (t))

s−1r (t)

]

[

1 − α〈χ1[0,s−1r (t)], ν〉

]

t.

Since

limt→∞

[

infg(x) : x ∈ (s−1r (t), x1)

x1−

g(s−1r (t))

s−1r (t)

]

= 0, and

limt→∞

[

supg(x) : x ∈ (s−1r (t), x1)

s−1r (t)

−g(s−1

r (t))

s−1r (t)

]

= 0,

it suffices to show that 0 ≤[

1 − α〈χ1[0,s−1r (t)], ν〉

]

t ≤ 〈χ, ξ〉 for all t ∈ [0,∞).

For this, note that by Lemma 4.2 (iv), for all t ∈ [0,∞),

0 ≤[

1 − α〈χ1[0,s−1r (t)], ν〉

]

t =〈χ1[0,s−1

r (t)], ξ〉t

s(s−1r (t))

≤ 〈χ, ξ〉.

In order to prove Corollary 3.8, it suffices by (C1) to verify left continuity.For this it will be necessary to consider the left continuous inverse s−1

ℓ (·) ofs(·) defined as follows. Given fluid model data (α, ν) and ξ ∈ M0,

s−1ℓ (t) = infx ∈ [0, x1) : s(x) ≥ t, for all t ∈ [0, t1). (43)

For ease of notation, we extend the definition of s−1ℓ (·) to all of [0,∞). For

this, recall the definition of x0 = infx ∈ [0, x1) : s(x) ≥ t1 ∧ x1 given in(35). If t1 < ∞, then let

s−1ℓ (t) =

x0, if t = t1,

x0, if ρ < 1, ξ 6= 0, and t ∈ (t1,∞),

x1, if ρ ≥ 1 or ξ = 0, and t ∈ (t1,∞).

(44)

Note that s−1ℓ (·) is left continuous on R+. Also note that if ξ 6∈ M1, then

t1 = 0 and x0 = 0.

Lemma 4.5 Let (α, ν) be fluid model data and ξ ∈ M0. Then

27

(i) s−1ℓ (t) ≤ s−1

r (t) for all t ∈ [0,∞);

(ii) s−1r (s) ≤ s−1

ℓ (t) for s ∈ [0, t), t ∈ [0, t1), and, if t1 < ∞, t = t1;

(iii) s−1r (t−) = s−1

ℓ (t) for all t ∈ (0, t1) and, if t1 < ∞, for t = t1;

(iv) 〈1(s−1ℓ

(t),s−1r (t)), ξ + αtν〉 = 0 for all t ∈ [0,∞);

(v) s(s−1ℓ (t)) ≥ t for all t ∈ [0, t1);

(vi) 〈χ1[0,s−1ℓ

(t)], ξ +αtν〉− t = 0 for all t ∈ [0, t1) such that s−1ℓ (t) < s−1

r (t)

and, if t1 < ∞ and s−1ℓ (t1) < s−1

r (t1), for t = t1 as well.

Proof. Properties (i)-(iii) and (v) follow directly from the definitions, usingLemma 4.2 (iii) for (iii). We now verify (iv). First consider t ∈ [0, t1). Sinces(x) = t for all x ∈ [s−1

ℓ (t), s−1r (t)), it is immediate that 〈1(s−1

ℓ(t),s−1

r (t)), ξ +

αtν〉 = 0. The proof is complete if t1 = ∞. Otherwise, t1 < ∞ and itsuffices to consider t ∈ [t1,∞). If t1 > 0, then ξ ∈ M1 and s(x) = t1 forall x ∈ [x0, x1) and so 〈1(x0,x1), ξ + αtν〉 = 0, which implies (iv). If t1 = 0,

then ξ 6∈ M1. Then we have s−1ℓ (0) = 0 and s−1

r (0) = lξ, which implies(iv) for t = 0. For t ∈ (0,∞), s−1

ℓ (t) = x1 and s−1r (t) = x2 ∧ lξ, which also

implies (iv) for t ∈ (0,∞). Hence (iv) holds. To verify (vi), note that fort ∈ [0, t1) such that s−1

ℓ (t) < s−1r (t), (v) and Lemma 4.2 (v) combine to yield

t ≤ s(s−1ℓ (t)) ≤ s(s−1

r (t)−) ≤ t. Hence, s(s−1ℓ (t)) = t, which in turn implies

the desired result. If t1 < ∞ and s−1ℓ (t1) < s−1

r (t1), we have x0 < x1 ands(x) = t1 for all x ∈ [x0, x1) and in particular for x = x0.

Proof of Corollary 3.8. By (C1), it suffices to show that ζ(·) is leftcontinuous, i.e., that for all t ∈ (0,∞) and g ∈ C+

b (R+),

limuրt

〈g, ζ(u)〉 = 〈g, ζ(t)〉. (45)

By (22), (23), and (21), (45) holds for t ∈ (t1,∞). Fix t ∈ (0, t1) andg ∈ C+

b (R+). By Lemma 4.5 (ii), there are two cases to consider: s−1r (u) =

s−1ℓ (t) for some u < t and s−1

r (u) < s−1ℓ (t) for all u < t. We first show that,

in both cases,

limuրt

〈g, ζ(u)〉 = 〈g1(s−1ℓ

(t),∞), ξ + αtν〉

+ g(s−1ℓ (t))

1

s−1ℓ (t)

[

〈χ1[0,s−1ℓ

(t)], ξ + αtν〉 − t]

. (46)

28

If s−1r (u) = s−1

ℓ (t) for some u < t, then since it is nondecreasing, s−1r (·) is

constant on [u, t) by Lemma 4.5 (ii). In this case, (46) follows immediatelyfrom (22) and (23). Otherwise, (22), (23), and Lemma 4.5 (iii) imply that

limuրt

〈g, ζ(u)〉 = 〈g1[s−1ℓ

(t),∞), ξ + αtν〉

+ g(s−1ℓ (t))

1

s−1ℓ (t)

[

〈χ1[0,s−1ℓ

(t)), ξ + αtν〉 − t]

. (47)

Then (46) follows by subtracting g(s−1ℓ (t))〈1s−1

ℓ(t), ξ + αtν〉 from the first

term on the right side of (47) and adding the equivalent expression

g(s−1ℓ (t))

1

s−1ℓ (t)

〈χ1s−1ℓ

(t), ξ + αtν〉

to the second term on the right side of (47).With (46) established, there are two cases to consider: s−1

ℓ (t) = s−1r (t)

and s−1ℓ (t) < s−1

r (t). If s−1ℓ (t) = s−1

r (t), then (46) implies (45) immediatelyby (22) and (23). Otherwise, (46), Lemma 4.5 (iv), and Lemma 4.5 (vi)imply that

limuրt

〈g, ζ(u)〉 = 〈g1[s−1r (t),∞), ξ + αtν〉

= 〈g1(s−1r (t),∞), ξ + αtν〉 + g(s−1

r (t))1

s−1r (t)

〈χ1s−1r (t), ξ + αtν〉.

Since s−1ℓ (t) < s−1

r (t) and s(·) is nondecreasing, Lemma 4.2 (v) and Lemma4.5 (v) imply that s(s−1

r (t)−) = t. It follows that 〈χ1[0,s−1r (t)), ξ+αtν〉−t = 0,

and so (45) holds by (22) and (23).If t1 = 0 or t1 = ∞, the proof is complete. Otherwise, 0 < t1 < ∞,

and we must show that limtրt1〈g, ζ(t)〉 = 〈g, ζ(t1)〉. If s−1ℓ (t1) = ∞, then

by Lemma 4.5 (i), s−1r (t1) = ∞, and so ζ(t1) = 0 by (22). By Lemma 4.5

(iii), limtրt1 s−1r (t) = ∞. This together with (22) and (23) implies (45) for

t = t1. Otherwise, s−1ℓ (t1) < ∞ and (46) also holds for t = t1 by the same

argument given for t ∈ (0, t1). Then equation (45) follows by an argumentsimilar to that used for t ∈ (0, t1).

5 Proof of Fluid Limit Theorem

Theorem 3.9 is proved in this section. We lay the foundation for the proofin Section 5.1. In Section 5.2 we verify tightness. Fluid limit points arecharacterized in Section 5.3.

29

5.1 Foundation for the Proof

In this section, we set up the framework necessary for proving tightness andcharacterizing fluid limit points. We begin by defining some performanceprocesses for the stochastic model. We then establish some relationshipsamong these processes, arising from the dynamics of SRPT. We also statea weak law of large numbers for the measure valued load process.

5.1.1 Performance Processes

Here we introduce the performance processes that play a key role in theanalysis of the stochastic model of an SRPT queue. Denote by L(·) the leftedge process of the measure valued state descriptor Z(·), which is given by

L(t) = supx ∈ R+ : 〈1[0,x),Z(t)〉 = 0, for all t ∈ [0,∞). (48)

Denote by C(·) the current residual service time process, which is given by

C(t) =

L(t), if Z(t) 6= 0,

0, otherwise,for all t ∈ [0,∞). (49)

Define the frontier process F (·) by

F (t) = sup0≤s≤t

C(s), for all t ∈ [0,∞). (50)

The queue length process is given by

Z(t) = 〈1,Z(t)〉, for all t ∈ [0,∞).

For each x ∈ R+, the truncated arrival process E(·, x) and truncated queuelength process Z(·, x) are respectively given by

E(t, x) = 〈1[0,x],V(t)〉 and Z(t, x) = 〈1[0,x],Z(t)〉, for all t ∈ [0,∞).

The immediate workload process W (·) is given by

W (t) = 〈χ,Z(t)〉, for all t ∈ [0,∞).

Three related processes are defined as follows: for t ∈ [0,∞), let

V (t) = 〈χ,V(t)〉, X(t) = W (0)+V (t)−t, and Y (t) = inf0≤s≤t

X(s).

30

Our analysis of the SRPT queue also exploits properties of the truncatedimmediate workload processes W (·, x), x ∈ R+, where for each x ∈ R+,

W (t, x) = 〈χ1[0,x],Z(t)〉, for all t ∈ [0,∞).

For each x ∈ R+, three more truncated processes are defined as follows: forall t ∈ [0,∞),

V (t, x) = 〈χ1[0,x],V(t)〉, X(t, x) = W (0, x) + V (t, x) − t,

Y (t, x) = inf0≤s≤t

X(s, x).(51)

5.1.2 Dynamic Inequalities and Equations

Next, the SRPT dynamics are used to obtain equations and bounds thatthe performance processes satisfy. For future reference, these equations andbounds are collected and written out at the end for the fluid scaled, R-indexed sequence of systems.

Queue Length

Lemma 5.1 Almost surely, for all s, t ∈ [0,∞) such that s ≤ t, all x, y ∈R+ such that x ≤ y, all a > 0, and all closed B ⊂ R+,

Z(t) ≤ Z(0) + E(t), (52)

〈1[x,∞),Z(t)〉 ≤ 〈1[x,∞),Z(0) + V(t)〉, (53)

〈1[x,y],Z(t)〉 ≤ 〈1[x,y],Z(0) + V(t)〉 + 1, (54)

〈1B ,Z(t)〉 ≤ 〈1B ,Z(s)〉 + E(t) − E(s) + 1, (55)

〈1B ,Z(s)〉 ≤ 〈1B ,Z(t)〉 + 〈1[0,a],Z(s)〉 +t − s

a+ 1. (56)

Proof. Inequalities (52) and (53) follow from Definition 2.1 and (8) becausethe residual service times are nonincreasing.

Inequality (54) holds because the residual service times are continuousand nonincreasing, and at most one job with initial service time not in[x, y] has positive residual service in [x, y] at time t. To prove this, fixt ∈ [0,∞) and x, y ∈ R+. Let It

x,y = 1 ≤ j ≤ A(t) : wj 6∈ [x, y] and wj(t) ∈[x, y] ∩ (0,∞), where A(·), w·, and w·(·) are defined by (3), (4), and (5).Then

〈1[x,y],Z(t)〉 ≤ 〈1[x,y],Z(0) + V(t)〉 +∣

∣Itx,y

∣

∣ ,

where | · | denotes cardinality of the set. Suppose that∣

∣Itx,y

∣

∣ ≥ 2, and leti, j ∈ It

x,y with i < j. If wi(Tj) ≤ wj, then φj(u) = 0 for all u ∈ [Tj ,Di]

31

(see (7) and the discussion at the end of Section 2.1). Since Di > t, thisimplies that wj(t) = wj 6∈ [x, y], a contradiction. Alternatively, if wi(Tj) >wj , then φi(u) = 0 for all u ∈ [Tj ,Dj ]. Since Dj > t, this implies thatwi(t) = wi(Tj) > wj . Since the residual service times are continuous andnonincreasing, wj > y, which yields another contradiction. Thus,

∣

∣Itx,y

∣

∣ ≤ 1.Inequality (55) is proved similarly, but with a slight modification. It

holds because the residual service times are continuous and nonincreasing,and at most one job with residual service time not in B at time s has positiveresidual service in B at time t. To prove this, fix s, t ∈ [0,∞) with s ≤ t andB ⊂ R+. Let Is,t

B = 1 ≤ j ≤ A(s) : wj(s) 6∈ B and wj(t) ∈ B ∩ (0,∞).Then

〈1B ,Z(t)〉 ≤ 〈1B ,Z(s)〉 + E(t) − E(s) +∣

∣

∣Is,tB

∣

∣

∣.

Suppose that∣

∣

∣Is,tB

∣

∣

∣≥ 2, and let i, j ∈ Is,t

B with i < j. If wi(s) ≤ wj(s),

then φj(u) = 0 for all u ∈ [s,Di]. Since Di > t, this implies that wj(t) =wj(s) 6∈ B, a contradiction. One arrives at the analogous contradiction if

wi(s) > wj(s). Thus,∣

∣

∣Is,tB

∣

∣

∣≤ 1.

For (56), fix s, t ∈ [0,∞) with s ≤ t, a > 0, B ⊂ R+, and let

Js,tB = 1 ≤ j ≤ A(s) : wj(s) ∈ B ∩ (a,∞) and wj(t) 6∈ B ∩ (0,∞).

Then〈1B ,Z(s)〉 ≤ 〈1B ,Z(t)〉 + 〈1[0,a],Z(s)〉 +

∣

∣

∣Js,t

B

∣

∣

∣.

Assume Js,tB has at least two elements, as the statement is trivial otherwise.

Let wJ = minwj(s) : j ∈ Js,tB and let i = minj ∈ Js,t

B : wj(s) = wJ.

Then for all j ∈ Js,tB \ i, φj(u) = 0 for all u ∈ [s, s + a], since Di > s + a.

That is, job j ∈ Js,tB \ i cannot enter or resume service until job i has exited

the system, which requires at least a units of time. So job j ∈ Js,tB \ i enters

service in (s + a, t). Iterating this argument and using the fact that all jobsin Js,t

B \ i either enter or resume service in (s, t) implies that the job in Js,tB

to enter or resume service last does so during the nonempty time interval(s + (|Js,t

B | − 1)a, t), which implies that |Js,tB | ≤ (t − s)a−1 + 1.

Immediate Workload

Lemma 5.2 Almost surely, for all t ∈ [0,∞),

W (t) = X(t) + [Y (t)]−. (57)

32

Almost surely, for all t ∈ [0,∞) and x ∈ R+,

〈χ1(x,∞),Z(t)〉 ≤ 〈χ1(x,∞),Z(0) + V(t)〉. (58)

In addition, almost surely, for all busy periods [s, t) ⊂ [0,∞),

W (t) = W (s) + V (t) − V (s) − (t − s). (59)

Finally, almost surely for each x ∈ R+ and t ∈ [0,∞),

W (t, x) ≤ W (0, x) + V (t, x) − V (τ(t, x)−, x) + x − (t − τ(t, x)), (60)

where τ(t, x) = sup0 ≤ s ≤ t : W (s, x) = 0 with the supremum of theempty set taken to be zero.

Proof. Since the server idles if and only if the system is empty, (57) is merelythe classical characterization of the workload process for a work conservingservice discipline. Also, (58) follows from the fact that χ(·) is nondecreasingand each residual service time is nonincreasing. In addition, (59) follows bysubtracting the sum of (5) at time s over 1 ≤ k ≤ A(s) from the sum of (5)

at time t over 1 ≤ k ≤ A(t), and using the fact that∑A(u)

k=1 φk(u) = 1 for allu ∈ [s, t).

To verify (60), we consider three cases. Case 1 is when W (t, x) = 0.Then τ(t, x) = t and (60) holds. Case 2 is when W (t, x) > 0 and τ(t, x) < t.Then [τ(t, x), t) is a busy period. For all s ∈ [τ(t, x), t), the job in service attime s has residual service time less than or equal to x. Hence, by the samereasoning used to verify (59),

W (t, x) = W (τ(t, x), x) + V (t, x) − V (τ(t, x), x) − (t − τ(t, x)).

At this point Case 2 splits into two subcases. Case 2a is when τ(t, x) = 0.Then W (τ(t, x), x) = W (0, x), and so, since V (τ(t, x)−, x) ≤ V (τ(t, x), x),(60) holds. Case 2b is when τ(t, x) > 0. Then either there exists ǫ > 0such that the residual service time of the job in service on the time interval(τ(t, x) − ǫ, τ(t, x)] decreases to x as time approaches τ(t, x), or jobs arriveat time τ(t, x) with initial service time in (0, x]. Thus, W (τ(t, x), x) ≤V (τ(t, x), x)−V (τ(t, x)−, x)+x, and (60) holds. Case 3 is when W (t, x) > 0and τ(t, x) = t. Then by the last argument, W (t, x) ≤ V (t, x)−V (t−, x)+xand (60) holds.

33

Behavior Above the Frontier

Lemma 5.3 Almost surely, for all t ∈ [0,∞),

(i) 〈1[0,C(t)),Z(t)〉 = 0;

(ii) for all j ∈ N such that F (t) < wj , wj(s) = wj for all s ∈ [0, t].

Proof. Property (i) follows from (48) and (49). For a proof of property(ii), suppose that there exist t ∈ [0,∞) and j ∈ N such that F (t) < wj andwj(t) < wj (recall that wj(·) is nonincreasing). Then by (5), t > 0 and thereexist 0 ≤ a < b ≤ t such that φj(s) = 1 for all s ∈ [a, b) and wj(a) = wj . Sofor all s ∈ [a, b),

〈1[0,wj(s)),Z(s)〉 = 0.

Hence, L(s) = wj(s) for s ∈ [a, b), and so C(s) = wj(s) for s ∈ [a, b). Butthen F (t) ≥ C(a) = wj, a contradiction. Hence (ii) holds.

The following result is an immediate consequence of Lemma 5.3 and(4)–(8).

Corollary 5.4 Almost surely, for all measurable functions g : R+ → R+

and t ∈ [0,∞),

〈g1(F (t),∞),Z(t)〉 = 〈g1(F (t),∞),Z(0) + V(t)〉, (61)

〈g1[F (t),∞),Z(t)〉 ≤ 〈g1[F (t),∞),Z(0) + V(t)〉. (62)

Behavior Below the Frontier

Lemma 5.5 Almost surely, for each t ∈ [0,∞),

〈χ1[0,F (t)),Z(t)〉 ≤ V (t, F (t)) − V (τ(t)−, F (t)) + F (t) − (t − τ(t)), (63)

where τ(t) = sup0 ≤ s ≤ t : C(s) = F (s) ∨ sup0 ≤ s ≤ t : W (s) = 0,with the supremum of the empty set taken to be zero.

Proof. Let t ∈ [0,∞). If τ(t) = t, (63) is trivial since 〈χ1[0,F (t)),Z(t)〉 = 0almost surely in that case. Henceforth, we assume that τ(t) < t. ThenC(s) < F (s) for all s ∈ (τ(t), t] and W (s) > 0 for all s ∈ [τ(t), t]. Hence,F (t) = F (τ(t)) and [τ(t), t) is a busy period. Therefore, by (59), (61), andcombining like terms,

W (t) = W (τ(t)) + V (t) − V (τ(t)) − (t − τ(t))

= W (τ(t), F (t)) + 〈χ1(F (t),∞),Z(0)〉 + 〈χ1(F (t),∞),V(τ(t))〉

+ V (t) − V (τ(t)) − (t − τ(t))

= W (τ(t), F (t)) + 〈χ1(F (t),∞),Z(0)〉 − V (τ(t), F (t))

+ V (t, F (t)) + 〈χ1(F (t),∞),V(t)〉 − (t − τ(t)).

34

Hence, by subtracting 〈χ1(F (t),∞),Z(0)〉+ 〈χ1(F (t),∞),V(t)〉 from both sidesand then using (61), we obtain

W (t, F (t)) = W (τ(t), F (t)) + V (t, F (t)) − V (τ(t), F (t)) − (t − τ(t)).

We have W (τ(t), F (t)) = 〈χ1[0,F (t)),Z(τ(t))〉+F (t)〈1F (t) ,Z(τ(t))〉. Clearly,〈χ1[0,F (t)),Z(τ(t)−)〉 = 0. Hence, 〈χ1[0,F (t)),Z(τ(t))〉 ≤ V (τ(t), F (t)) −V (τ(t)−, F (t)). Then,

W (t, F (t)) ≤ V (t, F (t)) − V (τ(t)−, F (t))

+F (t)〈1F (t),Z(τ(t))〉 − (t − τ(t)).

If we subtract F (t)〈1F (t),Z(t)〉 from both sides of this inequality, theresult follows provided that 〈1F (t),Z(τ(t))〉 − 〈1F (t),Z(t)〉 ≤ 1. SinceC(s) < F (s) and W (s) > 0 for all s ∈ (τ(t), t], 〈1[0,F (t)),Z(s)〉 > 0 forall s ∈ (τ(t), t]. Hence, any job with residual service time equal to F (t)at time s ∈ (τ(t), t] is not in service at time s. At time τ(t), at mostone job with residual service time equal to F (t) is in service. Therefore,〈1F (t),Z(τ(t))〉 ≤ 〈1F (t),Z(t)〉 + 1, and the desired bound follows.

Bounds for the Frontier Process

Lemma 5.6 Almost surely, for all x ∈ R+,

W (t, x) = X(t, x), for all t ∈ [0, τx),

where τx = supt ∈ [0,∞) : Y (t, x) ≥ 0.

Proof. For x ∈ R+, let τx = infs ∈ [0,∞) : W (s, x) = 0. Note that ifτx = 0, then by (51), τx = 0 and there is nothing to prove. Suppose τx > 0.For all x ∈ R+ and t ∈ [0, τx), W (t, x) > 0, and so 0 < L(t) ≤ x. Thus, forall x ∈ R+ and t ∈ [0, τx), C(t) ≤ x, and consequently F (t) ≤ x. Therefore,by (61), almost surely for all x ∈ R+ and t ∈ [0, τx),

W (t) = W (t, x) + 〈χ1(x,∞),Z(0)〉 + 〈χ1(x,∞),V(t)〉.

Since, for each x ∈ R+, [0, τx) is a busy period, (59) implies that almostsurely for all x ∈ R+ and t ∈ [0, τx),

W (t) = W (0) + V (t) − t = 〈χ,Z(0)〉 + 〈χ,V(t)〉 − t.

Subtracting the first of the preceding two displays from the second yieldsthat almost surely for all x ∈ R+ and t ∈ [0, τx),

W (t, x) = X(t, x).

35

Thus, almost surely X(t, x) > 0 for all x ∈ R+ and t ∈ [0, τx). Hence,almost surely Y (t, x) ≥ 0 for all x ∈ R+ and t ∈ [0, τx). Therefore, almostsurely τx ≤ τx for all x ∈ R+. If τx = ∞, the proof is complete. If τx < ∞,then W (τx, x) = 0 by right continuity. Thus, no job arrives at time τx withservice time less than or equal to x and, consequently, X(τx, x) = 0. So by(51) and since V (·, x) is piecewise constant almost surely, Y (s, x) < 0 for alls > τx. So almost surely, τx ≥ τx.

Corollary 5.7 Almost surely, for all t ∈ [0,∞),

F (t) ≤ infx ∈ R+ : Y (t, x) > 0. (64)

Proof. Fix t ∈ [0,∞). If x ∈ R+ : Y (t, x) > 0 = ∅, (64) is trivial sinceinf ∅ = ∞. Otherwise, there exists x ∈ R+ such that Y (t, x) > 0. Then by(51), τx > t and so by Lemma 5.6, W (s, x) = X(s, x) ≥ Y (t, x) > 0 for alls ∈ [0, t]. Hence, for all s ∈ [0, t], L(s) ≤ x, which implies that C(s) ≤ x,and thus F (t) ≤ x.

Corollary 5.8 Let τ = sups ∈ [0,∞) : Y (s) > 0. Almost surely, for allt ∈ [0, τ),

F (t) ≥ supx ∈ R+ : Y (t, x) < 0. (65)

Proof. Note that if s ∈ [0,∞) : Y (s) > 0 = ∅, then τ = 0 and thereis nothing to prove. Suppose τ > 0 and fix t ∈ [0, τ). If x ∈ R+ :Y (t, x) < 0 = ∅, (65) is trivial. Otherwise, there exists an x ∈ R+ suchthat Y (t, x) < 0. Then τx < t < τ . Almost surely, W (τx, x) = 0 sinceτx = τx (see the proof of Lemma 5.6). Since τx < τ , W (τx) > 0. Therefore,x ≤ L(τx) < ∞, so that x ≤ C(τx) ≤ F (τx). Since τx < t, F (t) ≥ x.

Application to the Fluid Scaled Sequence of Systems Recall thatwe append a superscript r to each object associated with the rth modelin the R-indexed sequence, including the performance processes defined inSection 5.1.1 and the random times τ(·, x), τ(·), and τ defined in Lemmas5.2, 5.5, and Corollary 5.8 respectively. A fluid scaling is applied to eachmodel in the sequence. In addition to the scaling already defined in (24),

36

define for each r ∈ R, t ∈ [0,∞), and x ∈ R+,

Lr(t) = Lr(rt), Cr(t) = Cr(rt), F r(t) = F r(rt),

Er(t, x) =1

rEr(rt, x),

Zr(t) =1

rZr(rt), Zr(t, x) =

1

rZr(rt, x),

W r(t) =1

rW r(rt), W r(t, x) =

1

rW r(rt, x),

V r(t) =1

rV r(rt), V r(t, x) =

1

rV r(rt, x),

Xr(t) =1

rXr(rt), Xr(t, x) =

1

rXr(rt, x),

Y r(t) =1

rY r(rt), Y r(t, x) =

1

rY r(rt, x),

τ r =1

rτ r, τ r(t) =

1

rτ r(rt), τ r(t, x) =

1

rτ r(rt, x).

(66)

Then, (2), Lemmas 5.1 and 5.2, Corollary 5.4, Lemma 5.5, and Corollaries5.7 and 5.8, imply that, for each r ∈ R, almost surely, for all s, t ∈ [0,∞)such that s ≤ t, all x, y ∈ R+ such that x ≤ y, all a > 0, all closed B ⊂ R+,

37

and all measurable g : R+ → R+,

Zr(0) < ∞, (67)

W r(0) < ∞, (68)

Zr(t) ≤ Zr(0) + Er(t), (69)

〈1[x,∞), Zr(t)〉 ≤ 〈1[x,∞), Z

r(0) + Vr(t)〉, (70)

〈1[x,y], Zr(t)〉 ≤ 〈1[x,y], Z

r(0) + Vr(t)〉 +1

r, (71)

〈1B , Zr(t)〉 ≤ 〈1B , Zr(s)〉 + Er(t) − Er(s) +1

r, (72)

〈1B , Zr(s)〉 ≤ 〈1B , Zr(t)〉 + 〈1[0,a], Zr(s)〉 +

t − s

a+

1

r, (73)

W r(t) = Xr(t) +[

Y r(t)]−

, (74)

〈χ1(x,∞), Zr(t)〉 ≤ 〈χ1(x,∞), Z

r(0) + Vr(t)〉, (75)

W r(t, x) ≤ W r(0, x) + V r(t, x) − V r(τ r(t, x)−, x)

+x

r− (t − τ r(t, x)), (76)

〈g1(F r(t),∞), Zr(t)〉 = 〈g1(F r(t),∞), Z

r(0) + Vr(t)〉, (77)

〈g1[F r(t),∞), Zr(t)〉 ≤ 〈g1[F r(t),∞), Z

r(0) + Vr(t)〉, (78)

〈χ1[0,F r(t)), Zr(t)〉 ≤ V r(t, F r(t)) − V r(τ r(t)−, F r(t))

+F r(t)

r− (t − τ r(t)), (79)

F r(t) ≤ infx ∈ R+ : Y r(t, x) > 0, (80)

and, for all t ∈ [0, τ r),

F r(t) ≥ supx ∈ R+ : Y r(t, x) < 0. (81)

5.1.3 Functional Weak Law of Large Numbers

The following result gives the limiting behavior under fluid scaling of thestochastic primitives. It is a special case, for example, of Lemma 5.1 in [7]for the case of one route (I = 1).

Proposition 5.9 As r → ∞,

(Vr(·), V r(·)) ⇒ (V∗(·), V ∗(·)),

where V∗(t) = αtν and V ∗(t) = ρt for all t ∈ [0,∞).

38

5.2 Tightness

To prove the fluid limit result, we first need to establish that Zr(·) : r ∈ Ris tight. There are two main steps: verify a compact containment conditionand verify that sample path oscillations are uniformly small.

Compact Containment

Lemma 5.10 Let T > 0 and η > 0. There exists a compact K ⊂ M suchthat

lim infr→∞

Pr(

Zr(t) ∈ K for all t ∈ [0, T ])

≥ 1 − η.

Proof. By (29), there exists an M > 0 such that

P (〈1,Z0〉 ∨ 〈χ,Z0〉 ≥ M) ≤ η.

Fix such an M and let K = (α + ρ)T + 1. For each r ∈ R, let

Ωr1 = Zr(0) ∨ W r(0) < M,

Ωr2 = Er(T ) ∨ V r(T ) < K,

Ωr3 = Zr(t) ≤ Zr(0) + Er(T ) and W r(t) ≤ W r(0) + V r(T ) for all t ∈ [0, T ].

Since ξ 7→ 〈1, ξ〉 is continuous on M, (28) implies that

(

Zr(0), W r(0))

⇒ (〈1,Z0〉, 〈χ,Z0)〉) , as r → ∞.

The set (z,w) ∈ R+ × R+ : z ∨ w < M is open, so by the Portmanteautheorem,

lim infr→∞

Pr (Ωr1) ≥ P (〈1,Z0〉 ∨ 〈χ,Z0〉 < M) ≥ 1 − η.

Similarly, (Er(T ), V r(T )) ⇒ (αT, ρT ) as r → ∞, by Proposition 5.9. So bychoice of K, lim infr→∞ Pr(Ωr

2) = 1. By (69) and (75), Pr(Ωr3) = 1 for all

r ∈ R. Hence,lim infr→∞

Pr (Ωr1 ∩ Ωr

2 ∩ Ωr3) ≥ 1 − η.

Let K be the closure in M of the set ξ ∈ M : 〈1, ξ〉∨〈χ, ξ〉 ≤ M +K. Theset K is compact by Theorem 15.7.5 in [8]. Furthermore, on Ωr

1 ∩ Ωr2 ∩ Ωr

3,Zr(t) ≤ M + K and W r(t) ≤ M + K for all t ∈ [0, T ]. In particular,Ωr

1 ∩ Ωr2 ∩ Ωr

3 ⊂ Zr(t) ∈ K for all t ∈ [0, T ].

39

Asymptotic Regularity Near the Origin In order to control the os-cillations of the measure valued state descriptors, it is necessary to controlthe number of departures in a short period of time. A large number of de-partures can only occur if a large number of jobs build up arbitrarily closeto the origin. Lemma 5.11 below implies that, with high probability, thenumber of jobs in a sufficiently small region around the origin is uniformlysmall over a compact time interval.

Lemma 5.11 Let T > 0. For each ǫ, η ∈ (0, 1) there exists an a > 0 suchthat

lim infr→∞

Pr

(

supt∈[0,T ]

Zr(t, a) ≤ ǫ

)

≥ 1 − η.

Proof. Almost surely, Z0 ∈ M0 and so 〈10,Z0〉 = 0. Hence, there existsa1 > 0 such that

P(

〈1[0,a1],Z0〉 <ǫ

4

)

≥ 1 − η. (82)

Furthermore, since ν does not have an atom at the origin, there is an a2 > 0such that

αT 〈1[0,a2], ν〉 <ǫ

4. (83)

Let a = a1 ∧ a2 and define

Ωr1 =

Zr(0, a) <ǫ

4

,

Ωr2 =

Er(T, a) <ǫ

4

,

Ωr3 =

Zr(t, a) ≤ Zr(0, a) + Er(T, a) +1

rfor all t ∈ [0, T ]

.

The set ξ ∈ M : 〈1[0,a1], ξ〉 < ǫ/4 is open in the weak topology. So by(28), (82), the Portmanteau theorem, and since a ≤ a1,

lim infr→∞

Pr (Ωr1) ≥ 1 − η.

Similarly, by Proposition 5.9, (83), the Portmanteau theorem, and sincea ≤ a2,

lim infr→∞

Pr (Ωr2) = 1.

In addition, by (71) with x = 0 and y = a, and since 〈1[0,a], Vr(t)〉 ≤ Er(T, a)

for all t ∈ [0, T ], Pr (Ωr3) = 1 for all r ∈ R. Hence,

lim infr→∞

Pr (Ωr1 ∩ Ωr

2 ∩ Ωr3) ≥ 1 − η.

40

On Ωr1 ∩ Ωr

2 ∩ Ωr3,

supt∈[0,T ]

Zr(t, a) ≤ǫ

4+

ǫ

4+

1

r,

which is bounded above by ǫ for sufficiently large r.

Oscillations

Definition 5.12 Let d be the metric on M given by

d[ζ, ξ] = infǫ > 0 : 〈1B , ζ〉 ≤ 〈1Bǫ , ξ〉 + ǫ and

〈1B , ξ〉 ≤ 〈1Bǫ , ζ〉 + ǫ for all closed B ⊂ R+.

Definition 5.13 Let T > 0 and δ ∈ [0, T ]. For each ζ(·) ∈ D ([0,∞),M),define a modulus of continuity on [0, T ] by

wT (ζ(·), δ) = supd[ζ(t), ζ(s)] : 0 ≤ s, t ≤ T, |t − s| < δ.

Lemma 5.14 Let T > 0. For all ǫ, η ∈ (0, 1) there exists δ > 0 such that

lim infr→∞

Pr(

wT (Zr(·), δ) ≤ ǫ)

≥ 1 − η.

Proof. Fix ǫ, η ∈ (0, 1). Let a > 0 be as in Lemma 5.11 with ǫ replaced byǫ/3 and define

Ωr1 =

supt∈[0,T ]

∣

∣Er(t) − αt∣

∣ ≤ǫ

6

,

Ωr2 =

supt∈[0,T ]

Zr(t, a) ≤ǫ

3

,

Ωr3 =

〈1B , Zr(t)〉 ≤ 〈1B , Zr(s)〉 + Er(t) − Er(s) +1

r

and 〈1B , Zr(s)〉 ≤ 〈1B , Zr(t)〉 + 〈1[0,a], Zr(s)〉 +

t − s

a+

1

r

for all closed B ⊂ R+ and all 0 ≤ s ≤ t ≤ T

.

By (26), lim infr→∞ Pr(Ωr1) = 1. Together with Lemma 5.11, (72), and (73),

this implies thatlim infr→∞

Pr(Ωr1 ∩ Ωr

2 ∩ Ωr3) ≥ 1 − η.

41

Thus, in order to complete the proof, it suffices to specify δ > 0 such that

Ωr1 ∩ Ωr

2 ∩ Ωr3 ⊂ wT (Zr(·), δ) ≤ ǫ,

for all sufficiently large r.Set δ = ǫ

3α∧ ǫa

3 . Fix r > 3/ǫ, ω ∈ Ωr1 ∩ Ωr

2 ∩ Ωr3, and 0 ≤ s ≤ t ≤ T such

that t − s < δ. Once we verify that d[Zr(t)(ω), Zr(s)(ω)] ≤ ǫ the proof iscomplete. By the definitions of δ, Ωr

1, Ωr2, and Ωr

3, for each closed B ⊂ R+,

〈1B , Zr(t)〉 ≤ 〈1B , Zr(s)〉 + α(t − s) +ǫ

6+

ǫ

6+

1

r≤ 〈1Bǫ , Zr(s)〉 + ǫ.

Similarly, for all closed B ⊂ R+,

〈1B , Zr(s)〉 ≤ 〈1B , Zr(t)〉 +ǫ

3+

δ

a+

1

r≤ 〈1Bǫ , Zr(t)〉 + ǫ.

Proof of Tightness

Theorem 5.15 The sequence Zr(·) : r ∈ R is C-tight.

Proof. For each T > 0 and δ ∈ [0, T ], let w′T (·, δ) be the modulus of

continuity on D ([0,∞),M) used in Corollary 3.7.4 of [5]. By Definition 5.13,w′

T (ζ(·), δ) ≤ wT+δ(ζ(·), 2δ) for all ζ(·) ∈ D ([0,∞),M). So by Lemmas 5.10and 5.14, Zr(·) : r ∈ R satisfies the compact containment and oscillationconditions of Corollary 3.7.4 in [5]. Thus, Zr(·) : r ∈ R is tight. Moreover,Definition 5.13 and Lemma 5.14 imply that any weak limit point Z∗(·) iscontinuous almost surely.

5.3 Characterization of Fluid Limit Points

Let Z∗(·) be a weak limit of Zr(·) : r ∈ R and let Q ⊂ R be a subsequencesuch that

Zq(·) ⇒ Z∗(·), as q → ∞.

By Theorem 5.15, Z∗(·) is continuous almost surely. Let W ∗(0) = 〈χ,Z∗(0)〉.By (28) and Proposition 5.9, and since (Vr(·), V r(·)) : r ∈ R converges indistribution to a deterministic process, we have the joint convergence

(

Vq(·), V q(·), W q(0), Zq(·))

⇒ (V∗(·), V ∗(·),W ∗(0),Z∗(·)) , as q → ∞.

42

By the Skorohod representation theorem, we assume that all random ele-ments are defined on a common probability space (Ω∗,P∗,F∗) such that,almost surely as q → ∞,

(

Vq(·), V q(·), W q(0), Zq(·))

→ (V∗(·), V ∗(·),W ∗(0),Z∗(·)) , (84)

uniformly on compact time intervals. Note that since the Skorohod repre-sentations in (84) are equal in distribution to the original processes, theystill satisfy (67)–(81) almost surely, with the various functionals there de-fined analogously as for the original processes. In this section, we prove thefollowing theorem.

Theorem 5.16 Z∗(·) is almost surely a fluid model solution for the data(α, ν) and initial measure Z∗(0).

In order to prove Theorem 5.16, we will need to verify that Z∗(·) almostsurely satisfies (C1), (C2), and (C3). For this, let Ω∗ ∈ F∗ be such that

P∗(

Ω∗)

= 1 and, on Ω∗, (84) holds, Z∗(·) is continuous, Z∗(0) ∈ M0 (which

is possible due to (29)), and finally, that (67)–(81) hold for all q ∈ Q. Then(C1) holds on Ω∗. It remains to show that Z∗(·) satisfies (C2) and (C3) onΩ∗. This is accomplished in Sections 5.3.1 and 5.3.2 respectively. To simplifythe presentation, ω ∈ Ω∗ is fixed throughout Sections 5.3.1 and 5.3.2 andall random elements are evaluated at this fixed ω. Let ξ = Z∗(0)(ω). Thenξ ∈ M0. Using this ξ and the data (α, ν) given in (26) and (27), define x1,x2, lξ, M1, M2, s(·), t1, s−1

r (·), and s−1ℓ (·) respectively by (13), (14), (15),

(16), (17) (18), (19), (20), (21), (43), and (44).

5.3.1 Verification of (C2)

We begin with an observation. The convergence of the second and thirdcomponents in (84) implies uniform integrability (for finite measures) ofthe sequences of measures Zq(0)q∈Q and Vq(t)q∈Q for each t ∈ [0,∞).Together with (75), this implies that Zq(t)q∈Q is uniformly integrable foreach t ∈ [0,∞). Thus, for each t ∈ [0,∞),

〈χ,Z∗(t)〉 = limMր∞

〈χ ∧ M,Z∗(t)〉

= limMր∞

limq→∞

〈χ ∧ M, Zq(t)〉

≤ lim supq→∞

〈χ, Zq(t)〉 < ∞.

43

Furthermore, for all q ∈ Q, t ∈ [0,∞) and M > 0,

∣

∣〈χ, Zq(t)〉 − 〈χ,Z∗(t)〉∣

∣ ≤∣

∣〈χ ∧ M, Zq(t)〉 − 〈χ ∧ M,Z∗(t)〉∣

∣

+ 〈χ1(M,∞), Zq(t)〉 + 〈χ1(M,∞),Z

∗(t)〉.

Hence, for each t ∈ [0,∞),

limq→∞

W q(t) = limq→∞

〈χ, Zq(t)〉 = 〈χ,Z∗(t)〉. (85)

We wish to show that 〈χ,Z∗(t)〉 = [W ∗(0) + (ρ − 1)t]+ for all t ∈ [0,∞).For this, it suffices to show that, for all t ∈ [0,∞),

limq→∞

W q(t) = [W ∗(0) + (ρ − 1)t]+ . (86)

For this, define X∗(·) and Y ∗(·) as follows: for t ∈ [0,∞),

X∗(t) = W ∗(0) + V ∗(t) − t = W ∗(0) + (ρ − 1)t, (87)

and

Y ∗(t) = inf0≤s≤t

X∗(s) =

W ∗(0), if ρ ≥ 1,

W ∗(0) + (ρ − 1)t, if ρ < 1.(88)

The following lemma is an immediate consequence of (84).

Lemma 5.17 For each t ∈ [0,∞),

limq→∞

Xq(t) = X∗(t) and limq→∞

Y q(t) = Y ∗(t).

Combining Lemma 5.17 with (74) proves (86). Hence (C2) holds.

5.3.2 Verification of (C3)

Verifying that Z∗(·) satisfies (C3) presents the greatest challenge. First, wedevelop some elementary results that will facilitate this. Then we deriveasymptotic bounds for the sequence of fluid scaled frontier processes. Thisleads to the proof that Z∗(·) satisfies (C3).

44

Fluid Limits for Truncated Processes For x ∈ R+, let ρx = α〈χ1[0,x], ν〉and W ∗(0, x) = 〈χ1[0,x],Z

∗(0)〉. For x ∈ R+ and t ∈ [0,∞), let

V ∗(t, x) = ρxt, (89)

X∗(t, x) = W ∗(0, x) + V ∗(t, x) − t = W ∗(0, x) + (ρx − 1)t, (90)

Y ∗(t, x) = inf0≤s≤t

X∗(s, x). (91)

Lemma 5.18

(i) For each x ∈ R+ such that ξ does not have an atom at x, W q(0, x)converges to W ∗(0, x) as q → ∞.

(ii) For each x ∈ R+ such that ν does not have an atom at x, V q(·, x) con-verges to V ∗(·, x) uniformly on compact time intervals as q → ∞.

Proof. Note that (i) follows from the fact that Zq(0) → ξ ∈ M0 as q → ∞.For a proof of (ii), fix x ∈ R+ such that ν does not have an atom at x.Since Vq(t)

w→ V∗(t), limq→∞ V q(t, x) = V ∗(t, x) for each t ∈ [0,∞). Since

V ∗(·, x) is continuous and nondecreasing, the convergence is uniform oncompact time intervals.

Corollary 5.19 For each t ∈ [0,∞) and x ∈ R+ such that neither ξ nor νhas an atom at x,

limq→∞

Xq(t, x) = X∗(t, x),

and

limq→∞

Y q(t, x) = Y ∗(t, x) =

W ∗(0, x) + (ρx − 1)t, x < x1,

W ∗(0, x), x ≥ x1.

Asymptotics for the Frontier We wish to analyze the behavior of thefrontier process F q(·) as q → ∞. This is slightly delicate since it may not bethe case that the frontier of Zq(·) converges to the frontier F ∗(·) of Z∗(·),where for all t ∈ [0,∞),

L∗(t) = supx ∈ R+ : 〈1[0,x),Z∗(t)〉 = 0,

C∗(t) =

L∗(t), if Z∗(t) 6= 0,

0, otherwise,,

F ∗(t) = sup0≤s≤t

C∗(s).

45

To see this, consider the following example. Suppose Q = N and, for eachq ∈ Q, suppose that [0, 4] does not intersect the support of νq, and

Zq(0) = qδ1 + δ2 + qδ3.

Then, Z∗(0) = δ1 + δ3,

F q(t) =

1, 0 ≤ t < 1,

2, 1 ≤ t < 1 + 2q,

3, 1 + 2q≤ t < 4 + 2

q,

and F ∗(t) =

1, 0 ≤ t < 1,

3, 1 ≤ t < 4.

Thus, F q(·) does not converge to F ∗(·) in the Skorohod J1-topology. Thefact that Z∗(0) does not have any support in the interval (1, 3) allows theprelimits to have a very small, but positive mass in (1, 3), which delaysthe time at which the frontier process achieves the value 3. In general,the prelimits for the frontier process can exhibit similar behavior wheneverthere are open intervals that do not intersect the union of the supports ofξ and ν. It is sufficient for our purposes to bound lim infq→∞ F q(·) andlim supq→∞ F q(·).

Theorem 5.20

(i) If ξ ∈ M1, then for all t ∈ [0,∞),

s−1ℓ (t) ≤ lim inf

q→∞F q(t) ≤ lim sup

q→∞F q(t) ≤ s−1

r (t). (92)

(ii) If ξ ∈ M2, then for all t ∈ [0,∞),

s−1ℓ (t) ≤ lim inf

q→∞F q(t) ≤ lim sup

q→∞F q(t) ≤ lξ. (93)

Remark 5.21 If ξ ∈ M2, then by (21), s−1r (t) = x2 ∧ lξ for all t ∈ (0,∞).

Hence, if lξ ≤ x2, then s−1r (t) = lξ for all t ∈ [0,∞). Therefore, the upper

bound in (93) is consistent with that in (92). However, if x2 < lξ, then, forall t ∈ (0,∞), s−1

r (t) < lξ and the upper bound in (93) is not equal to s−1r (t).

This results from the fact that, at time zero, the prelimit frontier processescan jump above x2, and thereby remain larger than x2 for all time. Hence,the frontier process fails to characterize the left edge of the fluid limit whenξ 6= 0 and x2 < lξ.

Proof. We begin with a proof of (i). If ξ ∈ M1, then t1 > 0. First, fixt ∈ [0, t1). By Lemma 4.2 (i), there is an ǫ > 0 such that s−1

r (t)+ ǫ < x1 and

46

neither ξ nor ν has an atom at s−1r (t) + ǫ. By (20), s(s−1

r (t) + ǫ) > t. Hence,by (18) and (90), X∗(t, s−1

r (t)+ǫ) > 0, which implies that Y ∗(t, s−1r (t)+ǫ) >

0. By Corollary 5.19, there exists Q such that q > Q implies Y q(t, s−1r (t) +

ǫ) > 0. By (80), q > Q implies F q(t) ≤ s−1r (t) + ǫ. Since ǫ may be chosen

arbitrarily small, the third inequality in (92) holds at time t. If s−1ℓ (t) = 0,

the proof that (92) holds at time t is complete. Otherwise, s−1ℓ (t) > 0. In

this case, choose ǫ > 0 such that 0 < s−1ℓ (t) − ǫ and neither ξ nor ν has an

atom at s−1ℓ (t)− ǫ. By (43), s(s−1

ℓ (t)− ǫ) < t. Hence, X∗(t, s−1ℓ (t)− ǫ) < 0,

which implies that Y ∗(t, s−1ℓ (t)−ǫ) < 0. Since ξ ∈ M1, W ∗(0) > 0. If ρ ≥ 1,

then Y ∗(t) = W ∗(0) > 0. Otherwise, ρ < 1 and t < t1 = W ∗(0)/(1 − ρ),which also implies that Y ∗(t) > 0. Since Y ∗(t, s−1

ℓ (t)−ǫ) < 0 and Y ∗(t) > 0,by Lemma 5.17 and Corollary 5.19, there exists Q such that q > Q impliesY q(t, s−1

ℓ (t)−ǫ) < 0 and Y q(t) > 0. Note that Y q(t) > 0 implies that t < τ q.So by (81), q > Q implies s−1

ℓ (t) − ǫ ≤ F q(t). Hence, (92) holds at time t,for t ∈ [0, t1). If t1 = ∞, the proof of (i) is complete.

Otherwise, t1 < ∞ and we must verify that (92) holds for t ∈ [t1,∞).Fix t ∈ [t1,∞). We begin by verifying that the first inequality in (92) holdsat time t. If t = t1, this follows because (92) holds for t ∈ [0, t1), F q(·) isnondecreasing, and s−1

ℓ (·) is left continuous. Otherwise, t ∈ (t1,∞). If ρ < 1or x0 = x1, then s−1

ℓ (t) = s−1ℓ (t1) by (44). Since F q(·) is nondecreasing, the

first inequality in (92) holds at time t. Otherwise, ρ ≥ 1 and x0 < x1. Thens(x) = t1 for all x ∈ [x0, x1). Hence, X∗(t1, x) = 0 for all x ∈ [x0, x1). Sincet ∈ (t1,∞) and ρx−1 < 0 for all x ∈ [x0, x1), X∗(t, x) < 0 for all x ∈ [x0, x1).Hence, Y ∗(t, x) < 0 for all x ∈ [x0, x1). Since ρ ≥ 1, Y ∗(t) = W ∗(0) > 0. Fixx ∈ [x0, x1). By Lemma 5.17 and Corollary 5.19, there exists a Q such thatq > Q implies Y q(t, x) < 0 and Y q(t) > 0. So then by (81), q > Q impliesF q(t) ≥ x. Since s−1

ℓ (t) = x1 by (44), this implies that the first inequalityin (92) holds at time t. Next, we verify that the third inequality in (92)holds at time t ∈ [t1,∞). Note that s−1

r (t) = x1. Hence, if x1 = ∞, thereis nothing to prove. Suppose that x1 < ∞. Since ξ ∈ M1, W ∗(0, x1) > 0.By definition of x1, ρx1 − 1 ≥ 0. Thus, Y ∗(t, x1) = W ∗(0, x1) > 0. Then byCorollary 5.19, there exists Q such that q > Q implies Y q(t, x1) > 0. Thistogether with (80) implies that F q(t) ≤ x1 for all q > Q. Thus, the thirdinequality in (92) holds at time t. This completes the proof of (i).

In order to prove (ii), suppose that ξ ∈ M2. Then t1 = 0, x1 < ∞, andρ ≥ 1. We begin by verifying the first inequality in (93). For t = 0, thefirst inequality in (93) is trivial since by (44), s−1

ℓ (0) = 0. Fix t ∈ (0,∞).Then by (44), s−1

ℓ (t) = x1. Note that X∗(t) = W ∗(0) + (ρ − 1)t and soY ∗(t) = W ∗(0) > 0, since ρ ≥ 1 and ξ ∈ M2. Given x ∈ [0, x1), sinceξ 6∈ M1, W ∗(0, x) = 0. Thus, X∗(t, x) = (ρx − 1)t for all x ∈ [0, x1). Since

47

(ρx−1) < 0 for all x ∈ [0, x1), Y ∗(t, x) < 0 for all x ∈ [0, x1). Fix x ∈ [0, x1).By Lemma 5.17 and Corollary 5.19, there exists Q such that q > Q impliesY q(t, x) < 0 and Y q(t) > 0. Hence, by (81), for all q > Q, F q(t) ≥ x. Sincex ∈ [0, x1) was arbitrary, the first inequality in (93) holds at time t. Toverify the third inequality in (93), note that since ξ ∈ M2, x1 ≤ lξ < ∞.Fix t ∈ [0,∞). Thus, Y ∗(t, x) = W ∗(0, x) > 0 for all x > lξ. Fix x > lξ.By Corollary 5.19, there exists Q such that q > Q implies that Y q(t, x) > 0.By (80), F q(t) ≤ x for all q > Q. Thus, (93) holds.

Analysis of Z∗(·) We are now prepared to prove that Z∗(·) satisfies (C3).The idea is that for q ∈ Q, g ∈ C+

b (R+), and t ∈ [0,∞),

〈g, Zq(t)〉 = 〈g1[0,F q(t)), Zq(t)〉 + 〈g1[F q(t),∞), Z

q(t)〉.

To obtain a lower bound, we can remove the first term on the right side,replace the closed interval in the second term with an open interval, and use(77). Then by combining this with Theorem 5.20 and (84), a preliminarylower bound is obtained for the fluid limit (see Lemma 5.22 below). Toobtain a suitable upper bound, we show that the first term on the rightside tends to zero (see Lemma 5.23 below). With that accomplished, (78),Theorem 5.20, and (84) are used to obtain a preliminary upper bound (seeLemma 5.24). The preliminary bounds stated in Lemmas 5.22 and 5.24 areused in the proof of Lemma 5.25, where it is verified that Z∗(·) satisfies(C3).

Lemma 5.22 Let g ∈ C+b (R+). Then

(i) if lξ < x2, then for all t ∈ [0,∞),

〈g,Z∗(t)〉 ≥ 〈g1(s−1r (t),∞), ξ + αtν〉, and (94)

(ii) if lξ ≥ x2, then for all t ∈ [0,∞),

〈g,Z∗(t)〉 ≥ 〈g1(lξ ,∞), ξ + αtν〉. (95)

Proof. If ξ = 0, then lξ = ∞, lξ ≥ x2, and (95) holds. Otherwise, ξ 6= 0and lξ < ∞. Fix g ∈ C+

b (R+) and t ∈ [0,∞). If t ∈ [t1,∞) and ρ < 1,then x2 = ∞ and so lξ < x2. Also, by Propositon 4.3, s−1

r (t) = ∞. Hence,(94) holds at time t. Therefore, it suffices to consider the case where eithert ∈ [0, t1), or t ∈ [t1,∞) and ρ ≥ 1. Then, by Proposition 4.3, s−1

r (t) < ∞.

48

Let ǫ > 0 be such that neither ν nor ξ has an atom at either s−1r (t) + ǫ or

lξ + ǫ, and define

b(t, ǫ) =

s−1r (t) + ǫ, if lξ < x2,

lξ + ǫ, if lξ ≥ x2.

Recall that if ξ ∈ M2 and lξ < x2, then s−1r (t) = lξ (see (21)). Hence, by

Theorem 5.20, there exists Q such that q > Q implies that F q(t) < b(t, ǫ).Since g is nonnegative, (77) implies that, for all q ∈ Q,

〈g, Zq(t)〉 ≥ 〈g1(F q(t),∞), Zq(t)〉 = 〈g1(F q(t),∞), Z

q(0)〉 + 〈g1(F q(t),∞), Vq(t)〉.

Hence, for all q > Q,

〈g, Zq(t)〉 ≥ 〈g1[b(t,ǫ),∞), Zq(0)〉 + 〈g1[b(t,ǫ),∞), V

q(t)〉.

Since neither ν nor ξ has an atom at b(t, ǫ), (84) implies that

limq→∞

(

〈g1[b(t,ǫ),∞), Zq(0)〉 + 〈g1[b(t,ǫ),∞), V

q(t)〉)

= 〈g1[b(t,ǫ),∞), ξ + αtν〉.

It follows that〈g,Z∗(t)〉 ≥ 〈g1[b(t,ǫ),∞), ξ + αtν〉.

Since ǫ > 0 can be made arbitrarily small, the result follows.

Lemma 5.23 Let g ∈ C+b (R+).

(i) If lξ < x2 and either t ∈ [0, t1), or t ∈ [t1,∞) and ρx1 ≤ 1, then

limq→∞

〈g1[0,F q(t)), Zq(t)〉 = 0.

(ii) If ξ 6∈ M1, then for all t ∈ [0,∞) and x ∈ [0, x2 ∧ lξ),

limq→∞

〈g1[0,x], Zq(t)〉 = 0.

Proof. We begin by proving (i). For this, fix t ∈ [0,∞). We first show that

limq→∞

〈χ1[0,F q(t)), Zq(t)〉 = 0. (96)

If t ∈ [t1,∞) and ρ < 1, then (96) is an immediate consequence of (85),(C2), (18), and (19). Henceforth, we assume that either t ∈ [0, t1), ort ∈ [t1,∞) and ρ ≥ 1. Then, since ξ 6= 0, Proposition 4.3 implies thats−1r (t) < ∞. Let ǫ > 0 be such that neither ν nor ξ has an atom at

49

s−1r (t) + ǫ; if t ∈ [0, t1), by Lemma 4.2 (i), we may further require that

s−1r (t)+ǫ < x1. Let x = s−1

r (t)+ǫ. By (79), Theorem 5.20, and monotonicityof V q(t, ·) − V q(s−, ·) for each fixed 0 ≤ s < t < ∞, there exists Q1 suchthat q > Q1 implies

〈χ1[0,F q(t)), Zq(t)〉 ≤ V q(t, x) − V q(τ q(t)−, x) − (t − τ q(t)) +

x

q.

By Lemma 5.18 (ii), there exists Q2 ≥ Q1 such that q > Q2 implies

〈χ1[0,F q(t)), Zq(t)〉 ≤ ρx(t − τ q(t)) + ǫ − (t − τ q(t)) +

x

q.

If t ∈ [0, t1), then x < x1 and so ρx − 1 < 0. Otherwise, t ∈ [t1,∞) andρ ≥ 1. Then since lξ < x2, (21) implies that x > lξ ∨ x1. Hence, ρx − 1 ≥ 0.Then, for q > Q2,

〈χ1[0,F q(t)), Zq(t)〉 ≤

ǫ + xq, t ∈ [0, t1),

(ρx − 1)t + ǫ + xq, t ∈ [t1,∞).

If t ∈ [t1,∞), then since lξ < x2 and ρx1 ≤ 1, it follows by (21) thatlimǫց0 ρx − 1 = ρlξ∨x1 − 1 = 0. Thus, letting q tend to infinity and thenletting ǫ decrease to zero completes the proof of (96).

To complete the proof of (i), fix g ∈ C+b (R+) and η > 0. Since neither ν

nor ξ has an atom at the origin, there exists 0 < δ < 1 such that

〈1[0,δ], ξ + αtν〉 <η

3‖g‖∞.

Fix such a δ that, in addition, has the property that neither ν nor ξ has anatom at δ. Then there exists Q such that q > Q implies that

〈1[0,δ], Zq(0)+Vq(t)〉 <

η

3‖g‖∞,

1

q<

η

3‖g‖∞, and 〈χ1[0,F q(t)), Z

q(t)〉 <ηδ

3‖g‖∞.

By (71), with x = 0 and y = δ, for all q > Q,

〈g1[0,F q(t)), Zq(t)〉 ≤ ‖g‖∞

(

〈1[0,δ], Zq(t)〉 +

1

δ〈χ1[δ,F q(t)), Z

q(t)〉

)

≤ ‖g‖∞

(

〈1[0,δ], Zq(0) + Vq(t)〉 +

1

q+

1

δ〈χ1[0,F q(t)), Z

q(t)〉

)

< η.

Hence (i) holds.

50

Next we prove (ii). For this, suppose that ξ 6∈ M1. Fix t ∈ [0,∞),x ∈ [0, x2 ∧ lξ), and ǫ > 0. Assume first that neither ν nor ξ has an atomat x. Then by (76), Lemma 5.18 (i), the inequality x < lξ, and Lemma 5.18(ii), there exists Q such q > Q implies

W q(t, x) ≤ ǫ +x

q+ ρx(t − τ q(t, x)) + ǫ − (t − τ q(t, x)).

Since x < x2 ∧ lξ, ρx ≤ 1. Hence, for all q > Q,

W q(t, x) ≤ ǫ +x

q+ ǫ.

Letting q → ∞ and then ǫ → 0 implies that limq→∞ W q(t, x) = 0. Sincethere are at most countably many atoms for ν and ξ in [0, x2 ∧ lξ), andW q(t, ·) is nonnegative and nondecreasing,

limq→∞

W q(t, x) = 0 for all x ∈ [0, x2 ∧ lξ).

The result in (ii) follows from this by an argument analogous to that usedin the previous paragraph to obtain (i) from (96).

Lemma 5.24 Let g ∈ C+b (R+).

(i) If lξ < x2 and either t ∈ [0, t1), or t ∈ [t1,∞) and ρx1 ≤ 1, then

〈g,Z∗(t)〉 ≤ 〈g1[s−1ℓ

(t),∞), ξ + αtν〉. (97)

(ii) If lξ ≥ x2, then for all t ∈ [0,∞),

〈g,Z∗(t)〉 ≤ 〈g1[x2,∞), ξ + αtν〉. (98)

Proof. We begin by proving (i). Fix t ∈ [0,∞). Since lξ < x2, it followsthat ξ 6= 0. If t ∈ [t1,∞) and ρ < 1, then (18), (19), (C2), and (12) implythat Z∗(t) = 0, so (97) holds. Otherwise, by Proposition 4.3 (i) and Lemma4.5 (i), s−1

ℓ (t) < ∞. For all q ∈ Q,

〈g, Zq(t)〉 = 〈g1[0,F q(t)), Zq(t)〉 + 〈g1[F q(t),∞), Z

q(t)〉.

This, together with (84) and Lemma 5.23 (i) implies that

〈g,Z∗(t)〉 ≤ lim supq→∞

〈g1[F q(t),∞), Zq(t)〉. (99)

51

By (78), for all q ∈ Q,

〈g1[F q(t),∞), Zq(t)〉 ≤ 〈g1[F q(t),∞), Z

q(0)〉 + 〈g1[F q(t),∞), Vq(t)〉. (100)

If t = 0, (97) follows since s−1ℓ (0) = 0. Otherwise, t ∈ (0,∞). Then

s−1ℓ (t) > 0. Let 0 < ǫ < s−1

ℓ (t) be such that neither ν nor ξ has an atom ats−1ℓ (t)− ǫ and define a(t, ǫ) = s−1

ℓ (t)− ǫ. By (100) and Theorem 5.20, thereexists Q such that q > Q implies

〈g1[F q(t),∞), Zq(t)〉 ≤ 〈g1(a(t,ǫ),∞), Z

q(0)〉 + 〈g1(a(t,ǫ),∞), Vq(t)〉. (101)

Hence, since neither ν nor ξ has an atom at a(t, ǫ), (84) and (101) implythat

lim supq→∞

〈g1[F q(t),∞), Zq(t)〉 ≤ 〈g1(a(t,ǫ),∞), ξ + αtν〉.

Letting ǫ ց 0, we obtain

lim supq→∞

〈g1[F q(t),∞), Zq(t)〉 ≤ 〈g1[s−1

ℓ(t),∞), ξ + αtν〉. (102)

Combining (99) and (102) yields (97).Next, we prove (ii). Fix t ∈ [0,∞). Since lξ ≥ x2, ξ 6∈ M1. Hence, by

(84) and Lemma 5.23 (ii), for each x < x2,

〈g,Z∗(t)〉 ≤ lim supq→∞

〈g1(x,∞), Zq(t)〉.

Fix x < x2 such that neither ξ nor ν has an atom at x. Given ǫ > 0, let M >x be such that neither ξ nor ν has an atom at M and ‖g‖∞〈1[M,∞), ξ+αtν〉 <

ǫ. Let y11 = x and y1

2 = M . For n = 2, 3, 4, . . . , let ynk

m(n)k=1 ⊂ [x,M ] be

such that yn−1k

m(n−1)k=1 ⊂ yn

km(n)k=1 , neither ξ nor ν has an atom at yn

k for allk = 1, . . . ,m(n), and 0 < yn

k+1−ynk < 2(M−x)/n for all k = 1, . . . ,m(n)−1.

Given n ∈ N and y ∈ [x,M), let k be such that y ∈ [ynk , yn

k+1) and set

gn(y) = sup

g(z) : z ∈[

ynk , yn

k+1

]

.

Then, for each q and n ∈ N,

〈g1(x,∞), Zq(t)〉 ≤

m(n)∑

k=1

gn(ynk )〈1[yn

k,yn

k+1), Zq(t)〉 + ‖g‖∞〈1[M,∞), Z

q(t)〉.

52

By (84) and (71), for each k = 1, . . . ,m(n) − 1,

lim supq→∞

〈1[ynk

,ynk+1)

, Zq(t)〉 ≤ lim supq→∞

〈1[ynk

,ynk+1]

, Zq(0) + Vq(t)〉 +1

q

= 〈1[ynk

,ynk+1]

, ξ + αtν〉

= 〈1[ynk

,ynk+1)

, ξ + αtν〉.

Also, by (70) and (84),

lim supq→∞

〈1[M,∞), Zq(t)〉 ≤ lim sup

q→∞〈1[M,∞), Z

q(0) + Vq(t)〉 = 〈1[M,∞), ξ + αtν〉.

It follows that for all n ∈ N,

〈g,Z∗(t)〉 ≤ 〈gn1[x,M), ξ + αtν〉 + ǫ.

Letting n tend to infinity and ǫ tend to zero,

〈g,Z∗(t)〉 ≤ 〈g1[x,∞), ξ + αtν〉.

Finally, letting x increase to x2 completes the proof.

The following definitions are needed for the proof of the final lemma,Lemma 5.25.

Time Shifted Fluid Limits For s ∈ [0,∞) and q ∈ Q, define the times shifted stochastic processes Eq

s (t) = Eq(s + t) − Eq(s), Vqs (t) = Vq(s +

t) − Vq(s), V qs (t) = V q(s + t) − V q(s), Zq

s (t) = Zq(s + t), and W qs (t) =

W q(s + t) for all t ∈ [0,∞). In addition, for s ∈ [0,∞) define the time sshifted limit process V∗

s (t) = V∗(s + t) − V∗(s), V ∗s (t) = V ∗(s + t) − V ∗(s),

Z∗s (t) = Z∗(s+ t), and W ∗

s (t) = W ∗(s+ t) for all t ∈ [0,∞). It is immediatethat for s ∈ [0,∞), the time s shifted stochastic processes satisfy (25)–(27).In addition, by (84) and (85) for each s ∈ [0,∞), as q → ∞,

(

Vqs (·), V q

s (·), W qs (0), Zq

s (·))

→ (V∗s (·), V ∗

s (·),W ∗s (0),Z∗

s (·)) . (103)

Then for each s ∈ [0,∞) such that Z∗s (0) does not charge the origin, it

follows that the time s shifted stochastic processes satisfy (25)–(29). Inparticular, if s ∈ [0,∞) is such that Z∗

s (0) does not charge the origin, thenany result proved for Z∗(·) also holds for Z∗

s (·).

Lemma 5.25 Z∗(·) satisfies (C3).

53

Proof. For t = 0, (C3) is immediate since Z∗(0) = ξ and L∗(0) = lξ.Therefore it suffices to consider t ∈ (0,∞). This is proved in three cases.Case One: Assume that lξ ≥ x2. By Lemmas 5.22 (ii) and 5.24 (ii), for allt ∈ (0,∞) and g ∈ C+

b (R+),

〈g1(lξ ,∞), ξ + αtν〉 ≤ 〈g,Z∗(t)〉 ≤ 〈g1[x2,∞), ξ + αtν〉. (104)

If x2 = ∞, then by (104), Z∗(t) = 0 and L∗(t) = ∞ for all t ∈ (0,∞) so that(C3) holds. Otherwise, x2 < ∞. Then ρ > 1 and by (104), x2 ≤ L∗(t) forall t ∈ (0,∞). Hence (104) implies that for all t ∈ (0,∞) and g ∈ C+

b (R+),

〈g,Z∗(t)〉 ≤ 〈g1[L∗(t),∞), ξ + αtν〉. (105)

Letting g ր χ in (105) and invoking the monotone convergence theoremyields, for all t ∈ (0,∞),

〈χ,Z∗(t)〉 ≤ 〈χ1[L∗(t),∞), ξ + αtν〉.

This, together with (C2) and the inequality ρ > 1 implies that α〈χ1[0,L∗(t)), ν〉 ≤1 for all t ∈ (0,∞). Hence, by (14), L∗(t) ≤ x2 for all t ∈ (0,∞). Thus, forall t ∈ (0,∞),

L∗(t) = x2. (106)

Case One (a): Assume that lξ = x2. Then by (104) and (106), (C3) holdsfor all t ∈ (0,∞).Case One (b): Assume that lξ > x2. For each ǫ > 0, consider the timeshifted fluid limit point Z∗

ǫ (·) = Z∗(· + ǫ). Then by (106), for each ǫ > 0,L∗

ǫ (t) = x2 for all t ∈ [0,∞); in particular, L∗ǫ(0) = x2. Since x2 > 0, Z∗

ǫ (0)does not charge the origin for each ǫ > 0. Hence by the commentary onTime Shifted Fluid Limit Points (see page 53) and Case One (a), (C3) holdsfor Z∗

ǫ (·) for each ǫ > 0. Then, for all t ∈ (0,∞), ǫ ∈ (0, t), and g ∈ C+b (R+),

we obtain

〈g1(x2 ,∞),Z∗(ǫ) + α(t− ǫ)ν〉 ≤ 〈g,Z∗

ǫ (t− ǫ)〉 ≤ 〈g1[x2,∞),Z∗(ǫ) + α(t− ǫ)ν〉.

Using the fact that for each 0 < ǫ < t < ∞ and g ∈ C+b (R+), 〈g,Z∗

ǫ (t−ǫ)〉 =〈g,Z∗(t)〉, letting ǫ decrease to zero, and using continuity of Z∗(·) completesthe proof of (C3) for t ∈ (0,∞) when lξ > x2.Case Two: Assume that lξ < x2 and t ∈ (0, t1), or t ∈ [t1,∞) and ρx1 ≤ 1.Fix t ∈ (0,∞). By Lemmas 5.22 (i) and 5.24 (i), for all g ∈ C+

b (R+),

〈g1(s−1r (t),∞), ξ + αtν〉 ≤ 〈g,Z∗(t)〉 ≤ 〈g1[s−1

ℓ(t),∞), ξ + αtν〉. (107)

54

In the upper bound, we wish to replace s−1ℓ (t) with s−1

r (t). Since s−1ℓ (t) ≤

s−1r (t), it suffices to show that

〈1[0,s−1r (t)),Z

∗(t)〉 = 0. (108)

Then, to complete the proof of (C3) in Case Two, we must show that

L∗(t) = s−1r (t). (109)

Indeed the combination of (107), (108), and (109) imply (C3) at time t. Wenow proceed to verify (108) and (109).Case Two (a): Assume that t ∈ (0, t1). Then by Lemma 4.5 (ii) s−1

r (t) ≤s−1ℓ (s) for t < s < t1. This together with (107) at time s, for t < s < t1,

implies that [0, s−1r (t)) does not intersect the support of Z∗(s) for all t <

s < t1. Hence, by continuity of Z∗(·), (108) holds. Then (107) and (108)imply that for all g ∈ C+

b (R+),

〈g1(s−1r (t),∞), ξ + αtν〉 ≤ 〈g,Z∗(t)〉 ≤ 〈g1[s−1

r (t),∞), ξ + αtν〉. (110)

Hence s−1r (t) ≤ L∗(t). Suppose that s−1

r (t) < L∗(t). Then by (110), for allg ∈ C+

b (R+),〈g,Z∗(t)〉 = 〈g1[L∗(t),∞), ξ + αtν〉. (111)

Letting g ր χ in (111) and using the monotone convergence theorem andthen (C2) together with the fact that 〈χ, ξ〉 + (ρ − 1)t > 0 since t ∈ (0, t1),it follows that

〈χ, ξ〉 + (ρ − 1)t = 〈χ1[L∗(t),∞), ξ + αtν〉. (112)

Hence,〈χ1[0,L∗(t)), ξ〉 = t(1 − 〈χ1[0,L∗(t)), αν〉). (113)

If L∗(t) > x1, then the left side of (113) is positive since t1 > 0 impliesξ ∈ M1. However, the right side is nonpositive, a contradiction. Hence,s−1r (t) < L∗(t) ≤ x1. Then both sides of (113) are necessarily positive since

the right side is positive if L∗(t) < x1 and the left side is positive if L∗(t) = x1

since ξ ∈ M1. Rearranging (113) gives s(L∗(t)−) = t. By (20), s(s−1r (t)) ≥

t. Hence by monotonicity of s(·), s(x) = t for all x ∈ [s−1r (t), L∗(t)), which

contradicts (20). So s−1r (t) = L∗(t) and (109) holds.

Case Two (b): Assume that t ∈ [t1,∞) and ρ < 1. Then t1 < ∞ and sinceρ < 1, (C2) implies that Z∗(t) = 0 and (108) holds. Moreover L∗(t) = ∞.Since ρ < 1 and lξ < x2 = x1, by (21), s−1

r (t) = x1 = ∞. Hence, (109)holds.

55

Case Two (c): Assume that t ∈ [t1,∞), ρx1 ≤ 1, and ρ ≥ 1. If t1 > 0, thenξ ∈ M1. If t ∈ (t1,∞), then by (21) and (44), s−1

ℓ (t) = s−1r (t) = x1 and

so by (107), (108) holds. Since s−1r (t1) = x1, right continuity of Z∗(·) and

(108) on (t1,∞) together imply (108) for t = t1. If t1 = 0, then ξ 6∈ M1 andt ∈ (0,∞). Hence, s−1

r (t) = x2 ∧ lξ = lξ and by Lemma 5.23 (ii) and (84),(108) holds. Thus, in Case Two (c), (108) holds.

Next we verify (109). By (108), L∗(t) ≥ s−1r (t). Suppose that L∗(t) >

s−1r (t) = x1 ∨ lξ ≥ lξ. Then we may replace s−1

ℓ (t) with L∗(t) in the upperbound in (107), let g ր χ, invoke the monotone convergence theorem, apply(C2) and the inequality ρ ≥ 1, and rearrange terms to obtain

〈χ1[0,L∗(t)), ξ + αtν〉 ≤ t.

But then since L∗(t) > lξ, 〈χ1[0,L∗(t)), ξ〉 > 0. Hence,

α〈χ1[0,L∗(t)), ν〉 < 1.

But then L∗(t) ≤ x1, a contradiction. Hence (109) holds.Case Three: Assume that lξ < x2, t ∈ [t1,∞), and ρx1 > 1. Then x1 = x2,ξ ∈ M1, ρ > 1, and t1 ∈ (0,∞).Case Three (a): Assume that t = t1. By Case Two (a), continuity of Z∗(·),and the fact that limtրt1 s−1

r (t) = x0 ≤ x1 = x2 (see Lemma 4.2 (iii)), itfollows that for all g ∈ C+

b (R+),

〈g1(x0,∞), ξ + αt1ν〉 ≤ 〈g,Z∗(t1)〉 ≤ 〈g1[x0,∞), ξ + αt1ν〉.

Hence L∗(t1) ≥ x0. In addition, by the same reasoning used in case one toargue from (105), L∗(t1) ≤ x2. Hence, if x0 = x2, (C3) follows at time t1.Otherwise, x0 < x2 = x1. Since s(x) = t1 for all x ∈ [x0, x1), it follows thatthe union of the supports of ν and ξ does not intersect (x0, x1). Hence, forsome a ≥ 0 and for all g ∈ C+

b (R+),

〈g,Z∗(t1)〉 = ag(x0) + 〈g1[x2,∞), ξ + αt1ν〉.

Since ξ ∈ M1, x0 6= 0. Letting g ր χ, applying the monotone convergencetheorem, using (C2) and the fact that ν and ξ do not charge (x0, x1) impliesthat a = (〈χ1[0,x0], ξ + αt1ν〉 − t1)/x0. Hence, by (21), (44) and Lemma 4.5(vi), a = 0. Then L∗(t1) = x2 and (C3) follows at time t1.Case Three (b): Assume that t ∈ (t1,∞). Then consider the time shiftedfluid limit point Z∗

t1(·) = Z∗(· + t1). By Case Three (a), L∗

t1(0) = L∗(t1) =

x2. Since x2 > 0, Z∗t1

(0) does not charge the origin and so by the commen-tary on Time Shifted Fluid Limit Points (see page 53) and Case One, (C3)follows.

56

Acknowledgment. The authors gratefully acknowledge the stimulatingenvironment at EURANDOM, where this work was begun.

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