Steady-state analysis of the Join the Shortest Queue model in the Halfin-Whitt regime
SSteady-state analysis of the Join the Shortest Queue model inthe Halfin-Whitt regime
Anton Braverman Kellogg School of Management at Northwestern UniversityJune 12, 2019
Abstract
This paper studies the steady-state properties of the Join the Shortest Queue modelin the Halfin-Whitt regime. We focus on the process tracking the number of idleservers, and the number of servers with non-empty buffers. Recently, [10] proved thata scaled version of this process converges, over finite time intervals, to a two-dimensionaldiffusion limit as the number of servers goes to infinity. In this paper we prove that thediffusion limit is exponentially ergodic, and that the diffusion scaled sequence of thesteady-state number of idle servers and non-empty buffers is tight. Combined with theprocess-level convergence proved in [10], our results imply convergence of steady-statedistributions. The methodology used is the generator expansion framework basedon Stein’s method, also referred to as the drift-based fluid limit Lyapunov functionapproach in [36]. One technical contribution to the framework is to show how it canbe used as a general tool to establish exponential ergodicity.
We consider a system with n identical servers, where customers arrive according toa Poisson process with rate nλ , and service times are i.i.d. exponentially distributedwith rate 1. Each server maintains an individual buffer of infinite length. When acustomer arrives, he will either enter service immediately if an idle server is available,or be routed to the server with the smallest number of customers in its buffer; tiesare broken arbitrarily. Once a customer is routed to a server, he cannot switch to adifferent server. This model is known as the Join the Shortest Queue (JSQ) model.To describe the system, let Q i ( t ) be the number of servers with i or more customersat time t ≥
0, and let Q ( t ) = ( Q i ( t )) ∞ i =1 . Then { Q ( t ) } t ≥ is a continuous time Markovchain (CTMC), and it is positive recurrent provided nλ < Q i be the randomvariables having the stationary distributions of { Q i ( t ) } t .In this paper we work in the Halfin-Whitt regime [24], which assumes that λ = 1 − β/ √ n, (1.1) a r X i v : . [ m a t h . P R ] J un or some fixed β >
0. The first paper to study the JSQ model in this regime is [10],which shows that the scaled process (cid:110)(cid:16) Q ( t ) − n √ n , Q ( t ) √ n , Q ( t ) √ n , . . . (cid:17)(cid:111) t ≥ (1.2)converges to a diffusion limit as n → ∞ . The diffusion limit of (1.2) is essentially twodimensional, because Q i ( t ) / √ n becomes negligible for i ≥
3. The results of [10] arerestricted to the transient behavior of the JSQ model and steady-state convergenceis not considered, i.e. convergence to the diffusion limit is proved only for finite timeintervals.In the present paper, we study the steady-state properties of the JSQ system.Specifically, we prove the existence of an explicitly known constant C ( β ) > β such that n − E Q = n (1 − λ ) , E Q ≤ C ( β ) √ n, E Q i ≤ C ( β ) , i ≥ , n ≥ . (1.3)In other words, the expected number of idle servers is known, the expected number ofnon-empty buffers is at most of order √ n , and the expected number of buffers with twoor more waiting customers is bounded by a constant independent of n . A consequenceof (1.3) is tightness of the sequence of diffusion-scaled stationary distributions.In addition to (1.3), we also prove that the two-dimensional diffusion limit of theJSQ model is exponentially ergodic. Stability of this diffusion limit remained an openquestion until the present paper. Combining the process-level convergence of [10],tightness of the prelimit stationary distributions in (1.3), and stability of the diffusionlimit, we are able to justify convergence of the stationary distributions via a standardlimit-interchange argument.To prove our results, we use the generator expansion framework, which is a man-ifestation of Stein’s method [34] in queueing theory and was recently introduced tothe stochastic systems literature in [4, 20]; see [5] for an accessible introduction. Theidea is to perform Taylor expansion on the generator of a CTMC, and by looking atthe second-order terms, to identify a diffusion model approximating the CTMC. Onethen proves bounds on the steady-state approximation error of the diffusion, whichcommonly results in convergence rates to the diffusion approximation [4, 5, 11, 20, 28].In this paper, we use only the first-order terms of the generator expansion, which cor-respond to the generator of a related fluid model. We then carry out the machinery ofStein’s method to prove convergence rates to the fluid model equilibrium. The boundsin (1.3) are then simply an alternative interpretation of these convergence rates. Forother examples of Stein’s method for fluid, or mean-field models, see [16, 17, 40, 41].Specifically, [40] was the first to make the connection between Stein’s method andconvergence rates to the mean-field equilibrium.Our approach can also be tied to the drift-based fluid limit (DFL) Lyapunov func-tions used in [36], which appeared a few years before [40]. As we will explain in moredetail in Section 4, the DFL approach and Stein’s method for mean-field approxima-tions are essentially one and the same. his paper contributes another example of the successful application of the gen-erator expansion method to the queuing literature. Although the general frameworkhas already been laid out in previous work, examples of applying the framework tonon-trivial systems are the only way to display the power of the framework and pro-mote its adoption in the research community. Furthermore, tractable examples helpshowcase and expand the versatility of the framework and the type of results it canprove. The present paper contributes from this angle in two ways. First, the JSQmodel is an example where the dimension of the CTMC is greater than that of thediffusion approximation. To justify the approximation, one needs a way to show thatthe additional dimensions of the CTMC are asymptotically negligible; this is known asstate space collapse (SSC). Our way of dealing with SSC in Section 3 differs from thetypical solution of bounding the magnitude of the SSC terms [4, 9, 31, 32] (only [4] ofthe aforementioned papers uses the generator expansion framework, but the rest stilldeal with steady-state SSC in a conceptually similar way). Second, this paper presentsthe first working example of the generator expansion framework being used to proveexponential ergodicity of the diffusion approximation. The insight used is simple, butcan be easily generalized to prove exponential ergodicity for other models. Early work on the JSQ model appeared in the late 50’s and early 60’s [22, 30], followedby a number of papers in the 70’s–90’s [12, 13, 23, 25, 42]. This body of literature firststudied the JSQ model with two servers, and later considered heavy-traffic asymptoticsin the setting where the number of servers n is fixed, and λ →
1; see [10] for an itemizeddescription of the aforementioned works. A more recent paper [9] considers the steady-state behavior of the JSQ model, but again in the setting where n is fixed, and λ → n → ∞ has been untouched until very recently. In[35], the author studies a variant of the JSQ model where the routing policy is to joinan idle server if one is available, and otherwise join any buffer uniformly; this is knownas the Join the Idle Queue (JIQ) policy. In that paper, the arrival rate is nλ where λ < n → ∞ . The author shows that in this underloaded asymptoticregime, JIQ is asymptotically optimal on the fluid scale, and therefore asymptoticallyequivalent to the JSQ policy. We have already described [10], which is the first paper tostudy a non-underloaded regime. In [33], the authors work in the Halfin-Whitt regimeand show that JIQ is asymptotically optimal, and therefore asymptotically equivalentto JSQ, on the diffusion scale. Most recently, [19] studies the JSQ model in the non-degenerate slowdown (NDS) regime introduced in [1]. In this regime, λ = 1 − β/n forsome fixed β >
0, i.e. NDS is even more heavily loaded than the Halfin-Whitt regime.The authors of [19] establish a diffusion limit for the total customer count process. Fora recent overview of load balancing algorithms see [38]; that paper includes the JSQalgorithm and the closely related power-of- d class of policies.In the asymptotic regime where n → ∞ , all previous considerations of the diffusion-scaled model [10, 19, 33] have been in the transient setting. In particular, convergenceto the diffusion limit is only proved over finite time intervals. In contrast, the presentpaper deals with steady-state distributions. Since the seminal work of [15], justifyingconvergence of steady-state distributions has become the standard in heavy-traffic ap- roximations, and is recognized as being a non-trivial step beyond convergence overfinite-time intervals [4, 5, 6, 14, 20, 21, 28, 29, 36, 37, 39, 43].The methodology used in this paper can be discussed in terms of [16, 36, 40, 41].The main technical driver of our results are bounds on the derivatives of the solutionto a certain first order partial differential equation (PDE) related to the fluid modelof the JSQ system. In the language of [36], we need to bound the derivatives of theDFL Lyapunov function. These derivative bounds are a standard requirement to applyStein’s method, and [16, 40, 41] provide sufficient conditions to bound these derivativesfor a large class of PDEs. The bounds in [16, 40, 41] require continuity of the vector fielddefining the fluid model, but the JSQ fluid model does not satisfy this continuity dueto a reflecting condition at the boundary. To circumvent this, we leverage knowledgeof how the fluid model behaves to give us an explicit expression for the PDE solution,and we bound its derivatives directly using this expression. Using the behavior of thefluid model is similar to what was done in [36]. However, bounding the derivatives inthis way requires detailed understanding of the fluid model, and as such this is a case-specific approach that varies significantly from one model to another. Furthermore,unlike [36] where the dimension of the CTMC equals the dimension of the diffusionapproximation, our CTMC is infinite-dimensional whereas the diffusion process is two-dimensional. These additional dimensions in the CTMC create additional technicaldifficulties which we handle in Section 3.Regarding our proof of exponential ergodicity. The idea of using a fluid model Lya-punov function to establish exponential ergodicity of the diffusion model was initiallysuggested in Lemma 3.1 of [20]. However, the discussion in [20] is at a conceptual level,and it is only after the working example of the present paper that we have a simpleand general implementation of the idea. Indeed, our Lyapunov function in Section 5.1violates the condition in Lemma 3.1 of [20]. We use ⇒ to denote weak convergence, or convergence in distribution. We use 1( A ) todenote the indicator of a set A . We use D = D ([0 , ∞ ) , R ) to denote the space of rightcontinuous functions with left limits mapping [0 , ∞ ) to R . For any integer k ≥
2, welet D k = D ([0 , ∞ ) , R k ) be the product space D × . . . × D . LetΩ = ( −∞ , × [0 , ∞ ) . (1.4)Going forward, we adopt the convention that for any function f : Ω → R , partialderivatives are understood to be one-sided derivatives for those values x ∈ ∂ Ω wherethe derivative is not defined. For example, the partial derivative with respect to x isnot defined on the set { x = 0 , x ≥ } . In particular, for any integer k > C k (Ω) be the set of k -times continuously differentiable functions f : Ω → R obeyingthe notion of one-sided differentiability just described. We use f i ( x ) to denote df ( x ) dx i .The rest of the paper is structured as follows. We state our main results in Sec-tion 2, and provide a roadmap to prove them in Section 3. Section 4 is devoted tounderstanding the JSQ fluid model, and using this to prove the derivative bounds thatdrive the proof of our main results. Model and main results.
Consider the CTMC { Q ( t ) } t ≥ introduced in Section 1. The state space of the CTMCis S = (cid:8) q ∈ { , , , . . . , n } ∞ | q i ≥ q i +1 for i ≥ ∞ (cid:88) i =0 q i < ∞ (cid:9) . The requirement that (cid:80) ∞ i =0 q i < ∞ if q ∈ S means that we only consider states with afinite number of customers. Recall that Q i are random variables having the stationarydistributions of { Q i ( t ) } t , and let Q = ( Q i ) be the corresponding vector. Let us alsodefine the fluid-scaled CTMC { X ( t ) } t ≥ by X ( t ) = Q ( t ) − nn , X i ( t ) = Q i ( t ) n , i ≥ . Also, let X i be the random variables having the stationary distributions of { X i ( t ) } t ≥ ,and set X = ( X i ) ∞ i =1 . In addition to the fluid scaling, we refer to {√ nX ( t ) } t ≥ and {√ nX } ∞ n =1 as the diffusion-scaled CTMC and sequence of stationary distributions,respectively.As mentioned, convergence of the diffusion-scaled process was already proved. Thefollowing result is copied from [33] (but it was first proved in [10]). Theorem 1 (Theorem 1 of [33]) . Suppose Y (0) = ( Y (0) , Y (0)) ∈ R is a randomvector such that √ nX i (0) ⇒ Y i (0) for i = 1 , as n → ∞ and √ nX i (0) ⇒ for i ≥ as n → ∞ . Then the process {√ n ( X ( t ) , X ( t )) } t ≥ converges uniformly over boundedintervals to { ( Y ( t ) , Y ( t )) } t ≥ ∈ D , which is the unique solution of the stochasticintegral equation Y ( t ) = Y (0) + √ W ( t ) − βt + (cid:90) t ( − Y ( s ) + Y ( s )) ds − U ( t ) ,Y ( t ) = Y (0) + U ( t ) − (cid:90) t Y ( s ) ds, (2.1) where { W ( t ) } t ≥ is standard Brownian motion and { U ( t ) } t ≥ is the unique non-decreasing, non-negative process in D satisfying (cid:82) ∞ Y ( t ) < dU ( t ) = 0 . Although process-level convergence to the diffusion limit was proved, the question ofconvergence of stationary distributions remained open until the present paper. Estab-lishing steady-state convergence requires two ingredients: 1) proving tightness of thediffusion-scaled sequence {√ nX } ∞ n =1 and 2) showing that the diffusion model in (2.1)is positive recurrent. These ingredients are established via the following two theorems. Theorem 2.
For each β > , there exists a constant C ( β ) such that for all n ≥ , (cid:12)(cid:12) √ nX i (cid:12)(cid:12) ≤ C ( β ) , i = 1 , , (2.2) | nX i | ≤ C ( β ) , i ≥ . (2.3) Theorem 3.
The diffusion process { ( Y ( t ) , Y ( t )) } t ≥ defined in (2.1) is positive re-current. n addition to positive recurrence of the diffusion, we will actually prove that it isexponentially ergodic. Combining Theorems 1–3, we arrive at the following proposition.The proof is simple and relegated to Section A.5. Proposition 1.
Let Y = ( Y , Y ) have the stationary distribution of the diffusionprocess defined in (2.1). Then √ n ( X , X ) ⇒ Y as n → ∞ . (2.4)In the remainder of this paper, we prove Theorems 2 and 3. The former is provedin Section 3 while the latter is proved in Section 5. As we will see, both theorems willproved using very similar methodology. Introduction of this methodology is the topicof the next section. Let G Q be the generator of the CTMC { Q ( t ) } t ≥ , which acts on function f : S → R in the following way: G Q f ( q ) = nλ q < n ) (cid:0) f ( q + e (1) ) − f ( q ) (cid:1) + ∞ (cid:88) i =2 nλ q = . . . = q i − = n, q i < n ) (cid:0) f ( q + e ( i ) ) − f ( q ) (cid:1) + ∞ (cid:88) i =1 ( q i − q i +1 ) (cid:0) f ( q − e ( i ) ) − f ( q ) (cid:1) , where e ( i ) is the infinite dimensional vector where the i th element equals one, and therest equal zero. The generator of the CTMC encodes the stationary behavior of thechain. The relationship between the generator and the stationary distribution can beexploited via the following lemma, which is proved in Section A.1. Lemma 1.
For any function f : S → R such that E | f ( Q ) | < ∞ , E G Q f ( Q ) = 0 . (3.1)By choosing different test functions f ( q ), we can use (3.1) to obtain stationaryperformance measures of our CTMC. As an example of the idea, we are able to provethe following using simple test functions. Lemma 2.
For any n ≥ and λ ∈ (0 , , E Q = nλ, (3.2) E Q i = nλ P ( Q = . . . = Q i − = n ) , i > . (3.3)The lemma is proved in Section A.2. The main idea is to apply G Q to f ( q ) = (cid:80) ∞ j = i q j to get the equation involving E Q i . This type of analysis is also commonlycalled Lyapunov drift analysis, and so in this paper we will use the terms ‘test function’and ‘Lyapunov function’ interchangeably. Observe that (3.2) implies that n − E Q = (1 − λ ), which is what was claimed in (1.3) of Section 1. Furthermore, tightness of {√ nX } follows because √ n E | X | = n − E Q √ n = n (1 − λ ) √ n = β. Despite having (3.3) at our disposal, we do not prove the rest of Theorem 2 by analyzing P ( Q = . . . = Q i − = n ) directly. This is because we do not have a good handleto control these probabilities. One may continue to experiment by applying G Q tovarious test functions in the hope of getting more insightful results from (3.1), e.g. anexpression for E Q i that does not involve the complicated term P ( Q = . . . = Q i − = n ). In general, the more complicated the Markov chain, the less likely that ad-hocexperimentation with test functions will be a productive strategy.Let us begin to derive a more systematic approach to picking the test function. Fora function f : R → R , define the lifted version Af : S → R by( Af )( q ) = f ( x , x ) = f ( x ) , q ∈ S, where x = ( q − n ) /n , and x i = q i /n for i ≥
2. Keeping in mind the relationshipbetween x and q , we will abuse notation and sometimes write ( Af )( x ). The generator G X of the CTMC { X ( t ) } t ≥ acts on Af as follows: G X Af ( x ) = nλ q < n ) (cid:0) f ( x + e (1) /n ) − f ( x )+ nλ q = n, q < n ) (cid:0) f ( x + e (2) /n ) − f ( x ) (cid:1) + ( q − q ) (cid:0) f ( x − e (1) /n ) − f ( x ) (cid:1) + ( q − q ) (cid:0) f ( x − e (2) /n ) − f ( x ) (cid:1) . The essence of the generator comparison framework is that we can perform Taylorexpansion on G X above to extract a ‘generator’ for the associated fluid model. To beprecise, for any differentiable function f : R → R , let us define Lf ( x ) = ( − x + x − β/ √ n ) f ( x ) − x f ( x ) , x ∈ R , (3.4)where f i ( x ) = df ( x ) dx i . The following lemma expands G X into L plus additional terms,and is proved in Section A.3. Recall the set Ω introduced in (1.4). Lemma 3.
For any q i ∈ Z + , let x = ( q − n ) /n and x i = q i /n for i ≥ . Suppose f ( y , y ) is defined on Ω , and f ( · , y ) , f ( y , · ) are absolutely continuous for all y ∈ Ω .Then for all q ∈ S , G X Af ( q ) = Lf ( x ) + ( f ( x ) − f ( x )) λ x = 0) + ε ( x ) , here ε ( x ) = − f ( x ) λ q = q = n ) + q (cid:90) x x − /n f ( x , u ) du + nλ q < n ) (cid:90) x +1 /nx ( x + 1 /n − u ) f ( u, x ) du + nλ q = n, q < n ) (cid:90) x +1 /nx ( x + 1 /n − u ) f ( x , u ) du + ( q − q ) (cid:90) x x − /n ( u − ( x − /n )) f ( u, x ) du + q (cid:90) x x − /n ( u − ( x − /n )) f ( x , u ) du. To make use of Lemma 3, observe that E | f ( X , X ) | < ∞ for any f : R → R because ( X , X ) can only take finitely many values. A variant of Lemma 1 then tellsus that E G X Af ( X ) = 0 . (3.5)Combining (3.5) with the Taylor expansion in Lemma 3 then yields E G X Af ( X ) = E Lf ( X ) + E (cid:0) ( f ( X ) − f ( X )) λ X = 0) (cid:1) + E ε ( X ) = 0 . (3.6)In other words, the expression above says that G X Af ( X ) can be decomposed intothree parts. The first term Lf ( X ) represents the first-order drift of the CTMC, whichis commonly referred to as the fluid model of the process. The higher order terms(corresponding to the diffusion approximation) are grouped into ε ( X ), which will actas an error term for our purposes. The final term, ( f ( X ) − f ( X )) λ X = 0), is areflection term that is present because X has to be non-positive. Unlike the terms in ε ( X ), this term cannot be treated as error.We will construct a Lyapunov function f ( x ) such that a) Lf ( x ) is well-understoodand the reflection term ( f ( x ) − f ( x )) λ x = 0) vanishes, and b) the derivatives of f ( x ) can be controlled to bound E ε ( X ). The following result tells us of the existenceof a Lyapunov function f ( x ) that satisfies both a) and b). Lemma 4.
Let Lf ( x ) be as in (3.4) and fix κ > β . The PDE Lf ( x ) = − (cid:0) ( x − κ/ √ n ) ∨ (cid:1) , x ∈ Ω , (3.7) f (0 , x ) = f (0 , x ) , x ≥ has a solution f ∗ ( x ) with f ∗ ( · , x ) , f ∗ ( x , · ) absolutely continuous for all x ∈ Ω , andthe second-order weak derivatives satisfy f ∗ ( x ) , f ∗ ( x ) , f ∗ ( x ) ≥ , x ∈ Ω , (3.9) f ∗ ( x ) = f ∗ ( x ) = 0 , x ∈ [0 , κ/ √ n ] , (3.10) f ∗ ( x ) ≤ √ nβ (cid:16) κκ − β + 1 (cid:17) , f ∗ ( x ) ≤ √ nβ (cid:16) κκ − β (cid:17) , x ≥ κ/ √ n. (3.11) he proof of Lemma 4 is postponed to Section 4, where we also comment on therelationship of f ∗ ( x ) to the DFL Lyapunov function of [36]. We are now in a positionto prove that {√ nX } is tight. Fix κ > β and let f ∗ ( x ) be as in Lemma 4. Going backto the Taylor expansion in Lemma 3, G X Af ∗ ( q ) = Lf ∗ ( x ) + (cid:0) ( f ∗ ( x ) − f ∗ ( x )) λ x = 0) (cid:1) + ε ( x )= − (cid:0) ( x − κ/ √ n ) ∨ (cid:1) + ε ( x ) , q ∈ S. Taking expected values on both sides and applying (3.5), we conclude that E (cid:0) ( X − κ/ √ n ) ∨ (cid:1) = E ε ( X ) . (3.12)We now prove that √ n | E ε ( X ) | is bounded by some constant C ( β ). Recall ε ( x ) = − f ∗ ( x ) λ q = q = n ) + q (cid:90) x x − /n f ∗ ( x , u ) du (3.13)+ nλ q < n ) (cid:90) x +1 /nx ( x + 1 /n − u ) f ∗ ( u, x ) du (3.14)+ nλ q = n, q < n ) (cid:90) x +1 /nx ( x + 1 /n − u ) f ∗ ( x , u ) du (3.15)+ ( q − q ) (cid:90) x x − /n ( u − ( x − /n )) f ∗ ( u, x ) du (3.16)+ q (cid:90) x x − /n ( u − ( x − /n )) f ∗ ( x , u ) du. (3.17)We first argue that lines (3.14)-(3.17) are all non-negative and provide an upper boundfor them. By (3.9) we know that (3.14)-(3.17) all equal zero when x < κ/ √ n − /n .Now suppose x ≥ κ/ √ n − /n . From (3.11) and the fact that q ≥ q if q ∈ S , wecan see that each of (3.14) and (3.16) is non-negative and bounded by β √ n (cid:16) κκ − β + 1 (cid:17) .Similarly, (3.11) tells us that each of (3.15) and (3.17) is non-negative and bounded by β √ n (cid:16) κκ − β (cid:17) . We conclude that (3.14)-(3.17) is bounded by1 β √ n (cid:16)
12 + 6 κκ − β (cid:17) x ≥ κ/ √ n − /n ) . Then (3.12) implies0 ≤ E (cid:0) ( X − κ/ √ n ) ∨ (cid:1) = E ε ( X ) ≤ β √ n (cid:16)
12 + 6 κκ − β (cid:17) P ( X ≥ κ/ √ n − /n ) − f ∗ (0 , λ P ( Q = Q = n ) + E (cid:20) Q (cid:90) X X − /n f ∗ ( X , u ) du (cid:21) . The term containing Q above is present because our CTMC is infinite dimensional,but the PDE (3.7)–(3.8) is two-dimensional. To deal with this error term, we invoke emma 2: − f ∗ (0 , λ P ( Q = Q = n ) + E (cid:20) Q (cid:90) X X − /n f ∗ ( X , u ) du (cid:21) = − f ∗ (0 ,
1) 1 n E Q + E (cid:20) Q (cid:90) X X − /n f ∗ ( X , u ) du (cid:21) = E (cid:20) Q (cid:90) X X − /n ( f ∗ ( X , u ) − f ∗ (0 , du (cid:21) ≤ , where in the last inequality we used f ( x ) , f ( x ) ≥ E (cid:0) ( X − κ/ √ n ) ∨ (cid:1) ≤ β √ n (cid:16)
12 + 6 κκ − β (cid:17) P ( X ≥ κ/ √ n − /n ) , (3.18)and hence E √ nX = κ + E ( √ nX − κ ) ≤ κ + 1 β (cid:16)
12 + 6 κκ − β (cid:17) P ( X ≥ κ/ √ n − /n ) , which establishes tightness of {√ nX } ∞ n =1 . The remainder of Theorem 2, namely (2.3),follows from a relatively simple bootstrapping argument involving (3.18). The proof ispresented in Section A.4. In the following section, we describe how to construct f ∗ ( x ). The aim of this section is to prove Lemma 4. The following informal discussion providesa roadmap of the procedure. Given x ∈ Ω, [10, Lemma 1] implies the existence anduniqueness of a solution v x ( t ) to the system of integral equations v ( t ) = x − β √ n t − (cid:90) t ( v ( s ) − v ( s )) ds − U ( t ) ,v ( t ) = x − (cid:90) t v ( s ) ds + U ( t ) , (cid:90) ∞ v ( s ) dU ( s ) = 0 , U ( t ) ≥ , t ≥ . (4.1)We refer to v x ( t ) as the fluid-model corresponding to the fluid-scaled CTMC { X ( t ) } t ≥ .The key idea is that f ∗ ( x ) = (cid:90) ∞ (cid:0) ( v ( x )2 ( s ) − κ/ √ n ) ∨ (cid:1) ds (4.2)will satisfy the PDE in Lemma 4. Our plan is to a) better understand the behavior ofthe fluid model and b) to use this knowledge to obtain a closed form representation of(4.2) and bound its derivatives. emark 1. Our choice of f ∗ ( x ) = (cid:82) ∞ (cid:0) ( v x ( s ) − κ/ √ n ) ∨ (cid:1) ds is a special case of aDFL Lyapunov function. More generally, a DFL Lyapunov function is any function ofthe form f ( h ) ( x ) = (cid:90) ∞ h ( v x ( s )) ds (4.3)and we expect it to satisfy Lf ( h ) ( x ) = − h ( x ) , (4.4)where L is the ‘generator’ of the fluid model. Section 2 of [36] proves rigorouslythat (4.4) is indeed true provided the fluid model satisfies ddt v ( t ) = F ( v ( t )) for somecontinuous vector field F ( · ). In our case the vector field is discontinuous because wedeal with a linear switching system (more on this in Section 4.1), and we would needto verify by hand that (4.4) is satisfied for any choice of h ( x ). The following is a heuristic description of the fluid model in (4.1). We refer to U ( t )as the regulator, because it prevents v x ( t ) from becoming positive. In the absence ofthis regulator, i.e. U ( t ) ≡
0, the system would have been a linear dynamical system˙ v = F ( v ) , where F ( v ) = ( − v + v − β/ √ n, − v ) . (4.5)However, due to the presence of the regulator, for values in the set { v = 0 , v ≥ β/ √ n } it is as if the vector field becomes F ( v ) = (0 , − β/ √ n ) . (4.6)Hence, we have a piece-wise linear system, whose dynamics are further illustrated inFigure 1.The fluid model can also be characterized analytically. Suppose the initial condition x <
0. Then the system behaves according to (4.5), meaning that until the verticalaxis is hit, i.e. for t ∈ [0 , inf s ≥ { v x ( s ) = 0 } ], its solution is (cid:18) v x ( t ) v x ( t ) (cid:19) = (cid:18) − β/ √ n + ( x + β/ √ n ) e − t + tx e − t x e − t (cid:19) . After the vertical axis is hit, the drift switches to (4.6) and v x ( t ) decreases linearly ata rate − β/ √ n until the point (0 , β/ √ n ) is reached, after which the system behavesaccording to (4.5) again.We require two elements to characterize v x ( t ). The first is the hitting timeinf { t ≥ v x ( t ) = 0 } (4.7)which is the first hitting time of the vertical axis given initial condition x . The secondis a curve Γ ( κ ) ⊂ Ω. The curve is defined such that for any point x ∈ Γ ( κ ) , thefluid path v x ( t ) first hits the vertical axis at the point (0 , κ/ √ n ). The following twolemmas present rigorous definitions of Γ ( κ ) and the hitting time. Lemma 5 is provedin Section B.1.1. β/ √ n β/ √ n − β/ √ n β/ √ n Figure 1: Dynamics of the fluid model. Any trajectory starting below the dahsed curvewill not hit the vertical axis, and anything starting above the curve will hit the axis andtravel down until reaching the point (0 , β/ √ n ). Lemma 5.
Fix κ ≥ β and x ≤ . The nonlinear system − β/ √ n + ( x + β/ √ n ) e − η + ηνe − η = 0 ,νe − η = κ/ √ n,ν ≥ κ/ √ n, η ≥ . (4.8) has exactly one solution ( ν ∗ ( x ) , η ∗ ( x )) . Furthermore, for every x ≤ , let us definethe curve Γ ( κ ) = { x ∈ Ω | x = ν ∗ ( x ) } and let γ ( κ ) ( x ) = (cid:110)(cid:0) − β/ √ n + ( x + β/ √ n ) e − t + tν ∗ ( x ) e − t , ν ∗ ( x ) e − t (cid:1) (cid:12)(cid:12)(cid:12) t ∈ [0 , η ∗ ( x )] (cid:111) . Then γ ( κ ) ( x ) ⊂ Γ ( κ ) for every x ≤ . Given κ ≥ β and x ∈ Ω, let Γ ( κ ) and ν ∗ ( x ) be as in Lemma 5. Let us adopt theconvention of writing x > Γ ( κ ) if x > ν ∗ ( x ) , (4.9)and define x ≥ Γ ( κ ) , x < Γ ( κ ) , and x ≤ Γ ( κ ) similarly. Observe that the sets { x ∈ Ω | x > Γ ( κ ) } , { x ∈ Ω | x < Γ ( κ ) } , and { x ∈ Ω | x ∈ Γ ( κ ) } are disjoint, and that their union equals Ω. Furthermore, { x ∈ Ω | x ≥ Γ ( κ ) } ∩ { x ∈ Ω | x ≤ Γ ( κ ) } = { x ∈ Ω | x ∈ Γ ( κ ) } . he next lemma characterizes the first hitting time of the vertical axis given initialcondition x , and shows that this hitting time is differentiable in x . It is proved inSection B.1.2. Lemma 6.
Fix κ ≥ β and x ∈ ( −∞ , × [ κ/ √ n, ∞ ) . Provided it exists, define τ ( x ) to be the smallest solution to β/ √ n − ( x + β/ √ n ) e − η − ηx e − η = 0 , η ≥ , and define τ ( x ) = ∞ if no solution exists. It follows immediately that τ (0 , x ) = 0 , x ≥ . (4.10) Furthermore, let Γ ( κ ) be as in Lemma 5.1. If x > Γ ( κ ) , then τ ( x ) < ∞ and x e − τ ( x ) > κ/ √ n, (4.11) and if x ∈ Γ ( κ ) , then τ ( x ) < ∞ and x e − τ ( x ) = κ/ √ n .2. If κ > β , then the function τ ( x ) is differentiable at all points x ≥ Γ ( κ ) with τ ( x ) = − e − τ ( x ) x e − τ ( x ) − β/ √ n ≤ , τ ( x ) = τ ( x ) τ ( x ) ≤ , x ≥ Γ ( κ ) , (4.12) where τ ( x ) is understood to be the left derivative when x = 0 .3. For any κ , κ with β < κ < κ , x ≥ Γ ( κ ) implies x > Γ ( κ ) , (4.13) i.e. the curve Γ ( κ ) lies strictly above Γ ( κ ) . Armed with Lemmas 5 and 6, we are now in a position to present the function f ∗ ( x ) that will satisfy the PDE in Lemma 4. f ∗ ( x ) Fix κ > β and partition the set Ω into three subdomains { x ∈ [0 , κ/ √ n ] } , { x ≤ Γ ( κ ) , x ≥ κ/ √ n } , and { x ≥ Γ ( κ ) } , where Γ ( κ ) is as in Lemma 5. From (4.11) we know that x ≥ Γ ( κ ) implies x ≥ κ/ √ n ,and therefore any point in Ω must indeed lie in one of the three subdomains. Thefollowing is an informal discussion of the intuition behind the form of f ∗ ( x ), which isgiven in (4.14) below.We already said that we will choose f ∗ ( x ) = (cid:90) ∞ (cid:0) ( v x ( s ) − κ/ √ n ) ∨ (cid:1) ds, nd so we aim to understand the integral on the right hand side. Recall that the fluidmodel is a piece-wise linear model satisfies (4.5) off the vertical boundary { x = 0 } ,and (4.6) on the vertical boundary. The simplest case to work with is if x ∈ [0 , κ/ √ n ].In both (4.5) and (4.6), the v component has a negative drift, meaning v x ( s ) ≤ v x (0)for all s ≥
0. Therefore, we let f ∗ ( x ) = (cid:90) ∞ (cid:0) ( v x ( s ) − κ/ √ n ) ∨ (cid:1) ds = 0 , if x ∈ [0 , κ/ √ n ] . Now suppose x ≥ κ/ √ n and x ≤ Γ ( κ ) . This means that the fluid model’s point ofcontact with the vertical axis is upper bounded by κ/ √ n . Therefore, the fluid modelsimply behaves like the linear system in (4.5) all the way until v x ( t ) = κ/ √ n , afterwhich time (cid:0) ( v x ( s ) − κ/ √ n ) ∨ (cid:1) becomes zero. This tells us that (cid:90) ∞ (cid:0) ( v x ( s ) − κ/ √ n ) ∨ (cid:1) ds = (cid:90) inf t ≥ { v x ( t )= κ/ √ n } (cid:0) ( v x ( s ) − κ/ √ n ) ∨ (cid:1) ds = (cid:90) inf t ≥ { v x ( t )= κ/ √ n } ( x e − s − κ/ √ n ) ds = (cid:90) log( x √ n/κ )0 ( x e − s − κ/ √ n ) ds = x (1 − κ/x √ n ) − κ/ √ n log( x √ n/κ ) , where in the second and third equalities we used the fact that ˙ v x ( t ) = − v x ( t ) or v x ( t ) = x e − t , and that the time until v x ( t ) = x e − t hits κ/ √ n is log( x √ n/κ ).Therefore, for x ≥ κ/ √ n and x ≤ Γ ( κ ) we set f ∗ ( x ) = x − κ/ √ n − κ/ √ n log( x √ n/κ ) . Lastly, if x ≥ Γ ( κ ) , then the fluid model behaves like (4.5) from time 0 until τ ( x ),at which point it hits the vertical axis (above (0 , κ/ √ n )). Once the fluid model hitsthe vertical axis, the v ( t ) component decreases linearly at a rate − β/ √ n until thepoint (0 , κ/ √ n ) is hit; afterwards, (cid:0) ( v x ( s ) − κ/ √ n ) ∨ (cid:1) = 0. Therefore, (cid:82) ∞ (cid:0) ( v x ( s ) − κ/ √ n ) ∨ (cid:1) ds is split into two parts: (cid:90) τ ( x )0 (cid:0) ( v x ( s ) − κ/ √ n ) ∨ (cid:1) ds + (cid:90) ∞ τ ( x ) (cid:0) ( v x ( s ) − κ/ √ n ) ∨ (cid:1) ds. The first term (before the vertical axis is hit) equals (cid:90) τ ( x )0 ( x e − s − κ/ √ n ) ds = x (1 − e − τ ( x ) ) − τ ( x ) κ/ √ n To evaluate the second term, we observe that once the vertical axis is hit at time τ ( x ),the v ∗ ( t ) component decreases linearly at a rate of β/ √ n until it hits the level κ/ √ n which occurs after ( v x ( τ ( x )) − κ/ √ n ) /β/ √ n time units). Therefore, (cid:90) ∞ τ ( x ) (cid:0) ( v x ( s ) − κ/ √ n ) ∨ (cid:1) ds = (cid:90) τ ( x )+( v x ( τ ( x )) − κ/ √ n ) /β/ √ nτ ( x ) ( v x ( s ) − κ/ √ n ) ds = (cid:90) ( v x ( τ ( x )) − κ/ √ n ) /β/ √ n ( v x ( τ ( x ) + s ) − κ/ √ n ) ds = (cid:90) ( v x ( τ ( x )) − κ/ √ n ) /β/ √ n ( x e − τ ( x ) − κ/ √ n − sβ/ √ n ) ds = 12 √ nβ (cid:0) x e − τ ( x ) − κ/ √ n (cid:1) . We conclude our above discussion by defining f ∗ ( x ) = , x ∈ [0 , κ/ √ n ] ,x − κ √ n − κ √ n log( √ nx /κ ) , x ≤ Γ ( κ ) and x ≥ κ/ √ n,x (1 − e − τ ( x ) ) − κ √ n τ ( x ) + √ nβ ( x e − τ ( x ) − κ/ √ n ) , x ≥ Γ ( κ ) . (4.14)Note that the above discussion is informal in the sense that we did not actually proveanything rigorous about the fluid model v x ( t ). Instead, we simply came up with acandidate PDE solution f ∗ ( x ) based on an intuitive grasp of the fluid model. Never-theless, the following Lemma confirms our intuition and shows that f ∗ ( x ) does indeedsolve the PDE; it is proved in Section B.2. Lemma 7.
The function f ∗ ( x ) in (4.14) is well-defined, has f ∗ ( · , x ) , f ∗ ( x , · ) abso-lutely continuous for all x ∈ Ω , satisfies the PDE (3.7) – (3.8) , and satisfies the deriva-tive bounds in (3.9) – (3.11) . Lemma 7 was the final piece in the proof of Theorem 2.
Theorem 1 proves that {√ n ( X ( t ) , X ( t )) } t ≥ converges to a diffusion limit. Conver-gence was established only over finite time intervals, but convergence of steady-statedistributions was not justified. In fact, it has not been shown that the process in (2.1)is even positive recurrent. We show that not only is this process positive recurrent(Theorem 3), but it is also exponentially ergodic. The proof involves a very similarapproach to that of Theorem 2. Namely, our proof will again revolve around comparingthe diffusion generator to its fluid model counterpart.Recall that the diffusion limit in Theorem 1 is Y ( t ) = Y (0) + √ W ( t ) − βt + (cid:90) t ( − Y ( s ) + Y ( s )) ds − U ( t ) ,Y ( t ) = Y (0) + U ( t ) − (cid:90) t Y ( s ) ds, here U ( t ) is the regulator and satisfies (cid:82) ∞ Y ( t ) < dU ( t ) = 0. To discuss geomet-ric ergodicity, we introduce the extended generator of this diffusion process. Denote by D ( G Y ) the set of all functions f : Ω → R for which there exists a measurable function g : Ω → R such that, for each x ∈ Ω, t ≥ E x f ( Y ( t ) , Y ( t )) − E x f ( Y (0) , Y (0)) = E x (cid:90) t g ( Y ( s ) , Y ( s )) ds, (5.1) (cid:90) t E x | g ( Y ( s ) , Y ( s )) | ds < ∞ . We write G Y f = g and call G Y the extended generator of { Y ( t ) } . An applicationof Ito’s lemma (see [26, Theorem 2] for an example of Ito’s lemma in the presence ofregulators) states that for any function f ( x ) ∈ C (Ω), E x f ( Y ( t ) , Y ( t )) − E x f ( Y (0) , Y (0))= E x (cid:90) t (cid:16) ( − Y ( s ) + Y ( s ) − β ) f ( Y ( s ) , Y ( s )) − Y ( s ) f ( Y ( s ) , Y ( s )) + f ( Y ( s ) , Y ( s )) (cid:17) ds + E x (cid:90) t (cid:0) − f (0 , Y ( s )) + f (0 , Y ( s )) (cid:1) dU ( s ) . Comparing the above expansion to (5.1), we see that for functions f ( x ) with f (0 , x ) = f (0 , x ), G Y f ( x ) = ( − x + x − β ) f ( x ) − x f ( x ) + f ( x ) , x ∈ Ω . The following theorem proves the existence of a function satsfying a Foster-Lyapunovcondition that is needed for exponential ergodicity.
Theorem 4.
Fix β > . There exist positive constants c and d , a compact set K , anda function V : Ω → [1 , ∞ ) with V ( x ) → ∞ as | x | → ∞ and V (0 , x ) = V (0 , x ) suchthat G Y V ( x ) ≤ − cV ( x ) + d x ∈ K ) , (5.2) The function V ( x ) , c , d , and K all depend on β . The proof of the theorem is given in Section 5.1. A consequence of Theorem 4 isexponential ergodicity of the diffusion in (2.1): the following corollary is an immediateconsequence of (5.2) and [8, Theorem 5.2].
Corollary 1.
The diffusion process { ( Y ( t ) , Y ( t )) } t ≥ defined in (2.1) is positive re-current. Furthermore, let Y = ( Y , Y ) be the vector having its stationary distribution,and let V ( x ) be the function from Theorem 4. There exist constants b < and B < ∞ such that sup | f |≤ V | E x f ( Y ( t )) − E f ( Y ) | ≤ BV ( x ) b t .1 Proving Theorem 4. The proof of Theorem 4 follows a similar line of reasoning as the proof of Theorem 2.Namely, we view the diffusion generator as G Y f ( x ) = Lf ( x ) + error , and we choose a function such that Lf ( x ) is well behaved. The error term will containderivatives of f ( x ), and so we want those to be controlled as well. Let us examine how G Y acts on Lyapunov functions of the form V ( x ) = e f ( x/ √ n ) (assuming for now that V (0 , x ) = V (0 , x )): G Y V ( x ) = ( − x + x − β ) V ( x ) − x V ( x ) + V ( x )= ( − x / √ n + x / √ n − β/ √ n )( f ( x/ √ n )) αV ( x ) − x √ n ( f ( x/ √ n )) αV ( x ) + V ( x )= ( Lf ( x/ √ n )) αV ( x ) + 1 n (cid:0) αf ( x/ √ n ) + α ( f ( x/ √ n )) (cid:1) V ( x ) . (5.3)Therefore, to satisfy a condition like (5.2), it suffices to choose a function f ( x ) suchthat Lf ( x/ √ n ) ≤ − c and both f ( x/ √ n ) and f ( x/ √ n ) are sufficiently under control.One candidate is to set V ( x ) equal toexp (cid:16) (cid:90) ∞ (cid:0) v x/ √ n ( t ) (cid:54)∈ [ − κ/ √ n, × [0 , κ/ √ n ] (cid:1)(cid:17) , (5.4)i.e. the exponential of the fluid hitting time to the set [ − κ/ √ n, × [0 , κ/ √ n ]. Theintegral in the exponent is a DFL Lyapunov function like in (4.3), and so we hope that L (cid:90) ∞ (cid:0) v x/ √ n ( t ) (cid:54)∈ [ − κ/ √ n, × [0 , κ/ √ n ] (cid:1) = − (cid:0) x/ √ n (cid:54)∈ [ − κ/ √ n, × [0 , κ/ √ n ] (cid:1) . However, we cannot use (5.4) directly because G Y acts on C (Ω) functions, and (5.4)does not have the required regularity; the indicator inside the integral is a discontinuousfunction. Instead, we will use a smoothed relative of (5.4). Let us define a smoothedindicator. For any (cid:96) < u , let φ ( (cid:96),u ) ( x ) = , x ≤ (cid:96), ( x − (cid:96) ) (cid:16) − ( x − (cid:96) )(( u + (cid:96) ) / − (cid:96) ) ( u − (cid:96) ) + u + (cid:96) ) / − (cid:96) )( u − (cid:96) ) (cid:17) , x ∈ [ (cid:96), ( u + (cid:96) ) / , − ( x − u ) (cid:16) ( x − u )(( u + (cid:96) ) / − u ) ( u − (cid:96) ) − u + (cid:96) ) / − u )( u − (cid:96) ) (cid:17) , x ∈ [( u + (cid:96) ) / , u ] , , x ≥ u. (5.5)It is straightforward to check that φ ( (cid:96),u ) ( x ) has an absolutely continuous first derivative,and that ( φ ( (cid:96),u ) ) (cid:48) ( (cid:96) ) = ( φ ( (cid:96),u ) ) (cid:48) ( u ) = 0 (5.6) (cid:12)(cid:12)(cid:12) ( φ ( (cid:96),u ) ) (cid:48) ( x ) (cid:12)(cid:12)(cid:12) ≤ u − (cid:96) , and (cid:12)(cid:12)(cid:12) ( φ ( (cid:96),u ) ) (cid:48)(cid:48) ( x ) (cid:12)(cid:12)(cid:12) ≤ u − (cid:96) ) . (5.7) ix κ > κ > β and α ∈ (0 , V ( κ ,κ ) ( x ) = exp (cid:0) α ( f (1) ( x/ √ n ) + f (2) ( x/ √ n )) (cid:1) (5.8)where f (1) ( x ) and f (2) ( x ) will be DFL Lyapunov functions constructed to satisfy Lf (1) ( x ) = − φ ( κ / √ n,κ / √ n ) ( − x ) , x ∈ Ω ,f (1)1 (0 , x ) = f (1)2 (0 , x ) , x ≥ , (5.9)and Lf (2) ( x ) = − φ ( κ / √ n,κ / √ n ) ( x ) , x ∈ Ω ,f (2)1 (0 , x ) = f (2)2 (0 , x ) , x ≥ . (5.10)Let us omit the superscript from V ( x ) for convenience and set f (Σ) ( x ) = f (1) ( x ) + f (2) ( x ) . Observe that V (0 , x ) = V (0 , x ) for all x ≥ Lf (Σ) ( x/ √ n ) = Lf (1) ( x/ √ n ) + Lf (2) ( x/ √ n ), G Y V ( x )= ( Lf (Σ) ( x/ √ n )) αV ( x ) + 1 n (cid:0) αf (Σ)11 ( x/ √ n ) + α ( f (Σ)1 ( x/ √ n )) (cid:1) V ( x )= ( − φ ( κ / √ n,κ / √ n ) ( − x / √ n ) − φ ( κ / √ n,κ / √ n ) ( x / √ n )) αV ( x )+ 1 n (cid:0) αf (Σ)11 ( x/ √ n ) + α ( f (Σ)1 ( x/ √ n )) (cid:1) V ( x ) ≤ − αV ( x )1( x (cid:54)∈ [ − κ , × [0 , κ ]) + 1 n (cid:0) αf (Σ)11 ( x/ √ n ) + α ( f (Σ)1 ( x/ √ n )) (cid:1) V ( x )= α (cid:16) − n (cid:0) f (Σ)11 ( x/ √ n ) + α ( f (Σ)1 ( x/ √ n )) (cid:1)(cid:17) V ( x ) (5.11)+ 1 n (cid:0) αf (Σ)11 ( x/ √ n ) + α ( f (Σ)1 ( x/ √ n )) (cid:1) V ( x )1( x ∈ [ − κ , × [0 , κ ]) , (5.12)where in the first inequality we used the fact that φ ( κ / √ n,κ / √ n ) ( x/ √ n ) = 1 for x ≥ κ .Let us compare (5.11)–(5.12) to our desired result in (5.2) to see that we need boundson f (Σ)1 ( x ), f (Σ)11 ( x ), and V ( x )1( x ∈ [ − κ , × [0 , κ ]). The following lemma presentsall the properties of f (1) ( x ) and f (2) ( x ) and their derivatives that we will need to proveTheorem 4. It is proved in Section C. Lemma 8.
Fix κ < κ such that κ > β . Then both PDE’s (5.9) and (5.10) havesolutions f (1) ( x ) and f (2) ( x ) , respectively. The solutions belong to C (Ω) , and forevery (cid:15) > , f (1) ( x ) ≤ log 2 , x ∈ [ − κ / √ n, × [0 , κ / √ n ] , (5.13) f (2) ( x ) ≤ log 2 + (cid:15)β , x ∈ [ − κ / √ n, × [0 , κ / √ n ] , (5.14) nd for all x ∈ Ω , (cid:12)(cid:12) f (1)1 ( x ) (cid:12)(cid:12) ≤ √ n(cid:15) log 2 , (cid:12)(cid:12) f (1)11 ( x ) (cid:12)(cid:12) ≤ n(cid:15) log 2 , (5.15) (cid:12)(cid:12) f (2)1 ( x ) (cid:12)(cid:12) ≤ √ nβ , (cid:12)(cid:12) f (2)11 ( x ) (cid:12)(cid:12) ≤ nβ(cid:15) (cid:16) β + (cid:15)(cid:15) (cid:17) . (5.16)With these derivative bounds, we are ready to prove Theorem 4. Proof of Theorem 4.
Fix κ < κ with κ > β and α >
0, let f (1) ( x ) and f (2) ( x ) be asin Lemma 8, and let V ( κ ,κ ) ( x ) = e α ( f (1) ( x/ √ n )+ f (2) ( x/ √ n )) . Our goal is to find positiveconstants c, d such that (5.2) is satisfied. It follows from (5.11)–(5.12) that G Y V ( x ) ≤ α (cid:16) − n (cid:0) f (Σ)11 ( x/ √ n ) + α ( f (Σ)1 ( x/ √ n )) (cid:1)(cid:17) V ( x )+ 1 n (cid:0) αf (Σ)11 ( x/ √ n ) + α ( f (Σ)1 ( x/ √ n )) (cid:1) V ( x )1( x ∈ [ − κ , × [0 , κ ])By (5.15)–(5.16),1 n (cid:12)(cid:12) f (Σ)11 ( x/ √ n ) + α ( f (Σ)1 ( x/ √ n )) (cid:12)(cid:12) = 1 n (cid:12)(cid:12) f (1)11 ( x/ √ n ) + f (2)11 ( x/ √ n ) + α ( f (1)1 ( x/ √ n ) + f (2)1 ( x/ √ n )) (cid:12)(cid:12) ≤ n (cid:16) n(cid:15) log 2 + nβ(cid:15) (cid:16) β + (cid:15)(cid:15) (cid:17) + α (cid:0) √ n(cid:15) log 2 + √ nβ (cid:1) (cid:17) . Note that the right hand side above is independent of n . Furthermore, by choosing (cid:15) large enough and α small enough, the term on the right hand side can be made to beless than one, meaning there exists a c > G Y V ( x ) ≤ − cV ( x ) + 1 n (cid:0) αf (Σ)11 ( x/ √ n ) + α ( f (Σ)1 ( x/ √ n )) (cid:1) V ( x )1( x ∈ [ − κ , × [0 , κ ])To bound the second term on the right hand side we use (5.13)–(5.14) and the factthat V ( x ) = e α ( f (1) ( x/ √ n )+ f (2) ( x/ √ n )) tell us that V ( x )1( x ∈ [ − κ , × [0 , κ ]) ≤ exp (cid:16) log 2 + log 2 + (cid:15)β (cid:17) x ∈ [ − κ , × [0 , κ ])Note that the upper bound does not depend on n . Therefore, G Y V ( x ) ≤ − cV ( x ) + d x ∈ K ) , where K = [ − κ , × [0 , κ ], and c, d are positive constants that depend only on β andno other system parameters such as λ or n . Remark 2.
In the proof of Theorem 3 we compare the generator of the diffusionprocess G Y to L , which can be thought of as the generator of the associated fluidmodel. One may wonder why we do not use a similar argument to compare L to G X ,and prove that the CTMC is also exponentially ergodic. The answer is that the CTMCis infinite dimensional, while the operator L acts on functions of only two variables. s a result, comparing G X to L leads to excess error terms that L does not accountfor, e.g. q in (3.13). Although we were able to get around this issue in the proof ofTheorem 2 by taking expected values, the same trick will not work now because (5.2)has to hold for every state. To prove exponential ergodicity, one needs to replace theoperator L and the PDE (3.7) by infinite-dimensional counterparts corresponding tothe infinite-dimensional fluid model of { ( X ( t ) , X ( t ) , X ( t ) , . . . ) } t ≥ . This is left as anopen problem to the interested reader, as Theorem 3 is sufficient for the purposes ofillustrating the proof technique. This paper contains a steady-state analysis of the JSQ model in the Halfin-Whittregime, using the generator expansion/DFL Lyapunov function methodology to provetightness of the customer count process. The proof procedure is to 1) write down theCTMC generator 2) perform Taylor expansion on it to extract a fluid model generatorand 3) set up a PDE related to the fluid model and bound the derivatives of thesolution to said PDE. The bottleneck of this methodology are the derivative bounds ofthe DFL Lyapunov function; this can only be done if the fluid model is relatively wellunderstood. In addition to proving tightness we saw in Section 5.1, that exponentialsof DFL Lyapunov functions can be used to prove exponential ergodicity of a process.One important open problem that this paper did not address is the following. WhenDFL Lyapunov functions were discussed in [36], the author considered fluid modelswith continuous vector fields (i.e. ddt v ( t ) = F ( v ( t )) where F ( · ) is continuous). In thatsetting, [36] showed by a simple argument that L (cid:90) ∞ h ( v x ( s )) ds = − h ( x ) , (6.1)where L is the ‘generator’ of the fluid model. Our JSQ model does not satisfy thecontinuity condition in [36]. Furthermore, our PDE has a reflecting boundary condition f (0 , x ) = f (0 , x ) , (6.2)which appears due to the presence of the regulator in the fluid model. In this paperwe must verify in a brute force manner that our DFL Lyapunov function satisfiesboth (6.1) and (6.2). It would be very useful to prove that (cid:82) ∞ h ( v x ( s )) automaticallysatisfies the aforementioned properties even in the presence of a discontinuous vectorfield and regulators in the fluid model. Acknowledgments.
This work was inspired by a talk given by David Gamarnik at Northwestern University’sKellogg School of Business in October 2017.
A Miscellaneous proofs.
This appendix contains proofs to a few miscellaneous lemmas used in the paper. .1 Lemma 1. Proof of Lemma 1.
A sufficient condition to ensure that E (cid:2) G Q f ( Q ) (cid:3) = 0is given by [27, Proposition 1.1] (alternatively, see [18, Proposition 3]). Namely, werequire that E (cid:104)(cid:12)(cid:12) G Q ( Q, Q ) f ( Q ) (cid:12)(cid:12)(cid:105) < ∞ , (A.1)where G Q ( q, q ) is the diagonal entry of the generator matrix G Q corresponding to state q ∈ S . It is not hard to check that in the JSQ system, | G Q ( q, q ) | < nλ + n for all states q ∈ S . Our assumption that E | f ( Q ) | < ∞ , is enough to ensure (A.1) is satisfied. A.2 Lemma 2
Proof of Lemma 2.
Fix
M > f ( q ) = min (cid:0) M, (cid:80) ∞ i =1 q i (cid:1) . Then G Q f ( q ) = nλ (cid:0) ∞ (cid:88) i =1 q i < M (cid:1) − q (cid:0) ∞ (cid:88) i =1 q i ≤ M (cid:1) . Using (3.1), nλ P (cid:0) T < M (cid:1) = E (cid:16) Q (cid:0) T ≤ M (cid:1)(cid:17) , where T = (cid:80) ∞ i =1 Q i is the total customer count. Although the infinite series in thedefinition of T may seem worrying at first, stability of the JSQ model in fact impliesthat T < ∞ almost surely. To see why this is true, observe that an alternative way todescribe the JSQ model is via the CTMC { ( S ( t ) , . . . , S n ( t )) } t ≥ , where S i ( t ) be thenumber of customers assigned to server i at time t ; we can view Q ( t ) as a deterministicfunction of ( S ( t ) , . . . , S n ( t )). This new CTMC is also positive recurrent, but nowthe total number of customers in the system at time t is the finite sum (cid:80) ni =1 S i ( t ).Therefore, T < ∞ almost surely, and we can take M → ∞ and apply the monotoneconvergence theorem to conclude that E Q = nλ. Repeating the argument above with f ( q ) = min (cid:0) M, (cid:80) ∞ j = i q j (cid:1) gives us nλ P ( Q = . . . = Q i − = n ) = E Q i . .3 Lemma 3. Proof of Lemma 3.
The CTMC generator satisfies G X Af ( q ) = nλ q < n ) (cid:0) f ( x + 1 /n, x ) − f ( x , x ) (cid:1) + nλ q = n, q < n ) (cid:0) f ( x , x + 1 /n ) − f ( x , x ) (cid:1) + ( q − q ) (cid:0) f ( x − /n, x ) − f ( x , x ) (cid:1) + ( q − q ) (cid:0) f ( x , x − /n ) − f ( x , x ) (cid:1) . (A.2)It is straightforward to verify that f ( x + e (1) /n ) − f ( x ) = 1 n f ( x ) + (cid:90) x +1 /nx ( x + 1 /n − u ) f ( u, x ) du,f ( x − e (1) /n ) − f ( x ) = − n f ( x ) + (cid:90) x x − /n ( u − ( x − /n )) f ( u ) du, (A.3)and that a similar expansion holds for f ( x + e (2) /n ) ± f ( x ). Applying (A.3) to (A.2)(but leaving the q term untouched), we see that G X Af ( q ) = f ( x ) 1 n (cid:0) nλ q < n ) − ( q − q ) (cid:1) + f ( x ) 1 n (cid:0) nλ q = n, q < n ) − q (cid:1) + nλ q < n ) (cid:90) x +1 /nx ( x + 1 /n − u ) f ( u ) du + nλ q = n, q < n ) (cid:90) x +1 /nx ( x + 1 /n − u ) f ( u ) du + ( q − q ) (cid:90) x x − /n ( u − ( x − /n )) f ( u ) du + q (cid:90) x x − /n ( u − ( x − /n )) f ( u ) du − q (cid:0) f ( x , x − /n ) − f ( x , x ) (cid:1) . (A.4)To conclude, we rewrite the first line of (A.4) as f ( x ) 1 n (cid:0) nλ − ( q − q ) (cid:1) − f ( x ) 1 n q + ( f ( x ) − f ( x )) λ q = n ) − f ( x ) λ q = q = n )= f ( x ) (cid:0) − β/ √ n − x + x (cid:1) − x f ( x )+ ( f ( x ) − f ( x )) λ q = n ) − f ( x ) λ q = q = n )= Lf ( x ) + ( f ( x ) − f ( x )) λ q = n ) − f ( x ) λ q = q = n ) . A.4 Proving (2.3)
Our goal is to prove (2.3), or that E nX i = E Q i ≤ C ( β ) for all i ≥
3. Since E Q i ≤ E Q for i ≥
3, it suffices to consider i = 3. Our starting point is (3.18), which we recall elow: E (cid:0) ( X − κ/ √ n ) ∨ (cid:1) ≤ β √ n (cid:16)
12 + 6 κκ − β (cid:17) P ( X ≥ κ/ √ n − /n ) . (A.5)Consider n such that max( β/ √ n, /n ) <
1, and fix ˜ κ ∈ (max( β/ √ n, /n ) , √ n ˜ κ in place of κ there to see that E (cid:16) ( X − ˜ κ )1( X ≥ ˜ κ ) (cid:17) ≤ β √ n (cid:16)
12 + 6˜ κ ˜ κ − β/ √ n (cid:17) P ( X ≥ ˜ κ − /n )= 1 βn (cid:16)
12 + 6˜ κ ˜ κ − β/ √ n (cid:17) √ n E (cid:16) X X X ≥ ˜ κ − /n ) (cid:17) ≤ βn (cid:16)
12 + 6˜ κ ˜ κ − β/ √ n (cid:17) κ − /n E √ nX ≤ βn (cid:16)
12 + 6˜ κ ˜ κ − β/ √ n (cid:17) κ − /n C ( β ) , (A.6)where in the last inequality we used (2.2). Therefore,1 βn (cid:16)
12 + 6˜ κ ˜ κ − β/ √ n (cid:17) κ − /n C ( β ) ≥ E (cid:16) ( X − ˜ κ )1( X ≥ ˜ κ ) (cid:17) ≥ (1 − ˜ κ ) P ( X = 1)= (1 − ˜ κ ) P ( Q = n ) ≥ (1 − ˜ κ ) 1 n E Q , where in the second inequality we used the fact that ˜ κ <
1, and in the last inequalitywe used Lemma 2.
Remark 3.
The bound in (2.3) will be sufficient for our purposes, but it is unlikelyto be tight. The argument in (A.6) can be modified by observing that for any integer m > P ( X ≥ ˜ κ − /n ) = n m n m E (cid:16) X m X m X ≥ ˜ κ − /n ) (cid:17) ≤ n m (˜ κ − /n ) m E ( √ nX ) m . Provided we have a bound on E ( √ nX ) m that is independent of n , it follows that E Q ≤ C ( β ) /n m − / . Although we have not done so, we believe the arguments usedin Theorem 2 can be extended to provide the necessary bounds on E ( √ nX ) m . A.5 Proposition 1
Proof of Proposition 1.
Lemma 2 and Theorem 2 imply that the sequence {√ n ( X , X ) } n is tight. It follows by Prohorov’s Theorem [2] that the sequence is also relativelycompact. We will now show that any subsequence of {√ n ( X , X ) } n has a furthersubsequence that converges weakly to Y .Fix n > { X ( t ) } t ≥ by letting √ nX (0) have the samedistribution as √ nX . Prohorov’s Theorem implies that for any subsequence {√ n (cid:48) X (0) } n (cid:48) ⊂ {√ nX (0) } n , here exists a further subsequence {√ n (cid:48)(cid:48) X (0) } n (cid:48)(cid:48) ⊂ {√ n (cid:48) X (0) } n (cid:48) that converges weakly to some random vector Y (0) = ( Y (0)1 , Y (0)2 , . . . ). Theorem 2implies that Y (0) i = 0 for i ≥
3. Now for any t ≥
0, let ( Y ( t ) , Y ( t )) solve the integralequation in (2.1) with intial condition ( Y (0) , Y (0)) = ( Y (0)1 , Y (0)2 ). Theorem 1 saysthat for any T > t ∈ [0 , T ], {√ n (cid:48)(cid:48) ( X ( t ) , X ( t )) , t ∈ [0 , T ] } ⇒ { ( Y ( t ) , Y ( t )) , t ∈ [0 , T ] } (A.7)as n → ∞ , where the convergence is uniform over bounded intervals. Furthermore,since { ( X ( t ) , X ( t )) } was initialized according to the stationary distribution,lim n →∞ √ n (cid:48)(cid:48) ( X ( t ) , X ( t )) d = lim n →∞ √ n (cid:48)(cid:48) ( X (0) , X (0)) = ( Y (0) , Y (0)) , t ∈ [0 , T ] . (A.8)It follows from (A.7) and (A.8) that( Y ( t ) , Y ( t )) d = ( Y (0) , Y (0)) = ( Y (0)1 , Y (0)2 ) , t ∈ [0 , T ] , meaning { ( Y ( t ) , Y ( t )) } is a stationary process, and must therefore be distributedaccording to its stationary distribution ( Y , Y ). To conclude, we have shown that √ n (cid:48)(cid:48) ( X , X ) converges in distribution to ( Y , Y ), which implies convergence of theoriginal sequence √ n ( X , X ). B Technical lemmas: Section 4.
In this appendix we prove the key technical lemmas from Section 4. Section B.1 hasthe proofs for Lemmas 5 and 6 and Section B.2 has the proof for Lemma 7.
B.1 Lemmas in Section 4.1.
A function known as the Lambert W function will play a central role here; the followingdiscussion is based on [7]. Define W ( x ) as the solution to x = W ( x ) e W ( x ) , x ∈ [ − e − , ∞ ) . (B.1)The function W ( x ) exists and is known as the Lambert W function. Taking logarithmson both sides of (B.1), W ( x ) = log x − log W ( x ) . (B.2)As is depicted in the plot of W ( x ) in Figure 2, W ( − e − ) = − W (0) = 0, and W ( x ) → ∞ as x → ∞ . Furthermore, W ( x ) is multi-valued for x ∈ ( − e − , W ( x ) and W − ( x ); the former is commonly calledthe principal branch. We will also need to use the fact that W ( x ) and W ( x ) are W ( x ) taken from [7]. For x ≤
0, the dashed line represents W − ( x )and the solid line represents W ( x ). differentiable for x > x ∈ ( − e − , W (cid:48) ( x ) = W ( x ) x (1 + W ( x )) > , x ∈ ( − e − , ∪ (0 , ∞ ); (B.3)c.f. section 3 of [7]. Going forward, we adopt the convention of using W ( x ) to mean W ( x ) for negative values of x . A useful property of W ( x ) is that x = W ( xe x ) , x ≥ − . (B.4)This can be seen by applying W ( x ) to both sides of (B.1) and using the fact thatthe range of W ( x ) is [ − , ∞ ] (again, we are using the convention W ( x ) = W ( x ) for x ∈ ( − e − , W ( x ) is invertible; indeed, W − ( x ) = xe x due to (B.4). B.1.1 Proving Lemma 5.
We first prove a technical result about W ( x ), and then prove Lemma 5. Lemma 9.
Fix κ ≥ β and x ≤ . The equation W (cid:16) − β/ √ nν e − ( x β/ √ n ) ν (cid:17) = − βκ (B.5) as a unique solution ν ∗ ≥ κ/ √ n . Furthermore, ddν (cid:18) − β/ √ nν e − ( x β/ √ n ) ν (cid:19) > , ν ≥ ν ∗ . (B.6) Proof of Lemma 9 .
Let f ( ν ) = − β/ √ nν e − ( x β/ √ n ) ν . Since κ ≥ β and W ( x ) is one-to-one (recall our convention that W ( x ) = W ( x ) for x ≤ f ( ν ) = − βκ e − β/κ . (B.7)If x = 0, then ν = κ/ √ n is the unique solution, and so we assume that x < W ( x ) is [ − e − , ∞ ), wecan only consider ν large enough such that f ( ν ) ≥ −
1. Observe that f ( κ/ √ n ) = − βκ e − β/κ e − x √ n/κ ≤ − βκ e − β/κ . Differentiating, f (cid:48) ( ν ) = β/ √ nν e − ( x β/ √ n ) ν − β/ √ nν e − ( x β/ √ n ) ν (cid:16) x + β/ √ nν (cid:17) = β/ √ nν e − ( x β/ √ n ) ν (cid:16) − x + β/ √ nν (cid:17) . We know that f ( ν ) → ν → ∞ . Case 1: x + β/ √ n ≤
0. In this case f (cid:48) ( ν ) > ν >
0, which implies thatthere exists a unique ν ∗ such that (B.7) is satisfied. Case 2: x + β/ √ n ≥
0. The form of f (cid:48) ( ν ) tells us that f ( ν ) is decreasing on(0 , x + β/ √ n ), but starts increasing after that. This again implies that a unique ν ∗ exists, and that ν ∗ ≥ x + β/ √ n , implying f (cid:48) ( ν ) > ν ≥ ν ∗ . Proof of Lemma 5.
Fix κ ≥ β and x ≤
0. We begin by showing that the system (4.8)has a unique solution. The first step is to write η in terms of ν . We rearrange − β/ √ n + ( x + β/ √ n ) e − η + ηνe − η = 0 (B.8)into − β/ √ nν = − ( x + β/ √ n ) ν e − η − ηe − η = (cid:16) − ( x + β/ √ n ) ν − η (cid:17) e − η , or (cid:16) − ( x + β/ √ n ) ν − η (cid:17) e − ( x β/ √ n ) ν − η = − β/ √ nν e − ( x β/ √ n ) ν . (B.9) bserve that the left hand side of (B.9) is in the form − xe − x , and so must lie in[ − e − , ∞ ). Therefore, existence of a solution to (4.8) imposes a natural constraint on ν that the right hand side above must lie in [ − e − , ∞ ). Assuming this is the case, weapply W ( x ) to both sides of (B.9) and using (B.4), we arrive at η = − ( x + β/ √ n ) ν − W (cid:18) − β/ √ nν e − ( x β/ √ n ) ν (cid:19) . (B.10)Since the Lambert W function is multivalued for x ≤
0, the above equation tells usthat given ν , there can be two potential choices for η . Plugging the above form of η back into (B.8), we see that − β/ √ n − W (cid:18) − β/ √ nν e − ( x β/ √ n ) ν (cid:19) νe − η = 0 , which, after using the fact that νe − η = κ/ √ n , becomes W (cid:18) − β/ √ nν e − ( x β/ √ n ) ν (cid:19) = − βκ . (B.11)Lemma 9 tells us that (B.11) does indeed have a unique solution ν ∗ ( x ). From (B.10)we know η can have up to two values, but we narrow this number down to one usingthe fact that νe − η = κ/ √ n . As an aside, it can be verified that ν ∗ (0) = κ/ √ n , and η ∗ (0) = 0.We now prove the second claim in the lemma that γ ( κ ) ( x ) ⊂ Γ ( κ ) for every x ≤ γ ( κ ) ( x ) = (cid:110)(cid:0) − β/ √ n + ( x + β/ √ n ) e − t + tν ∗ ( x ) e − t , ν ∗ ( x ) e − t (cid:1) (cid:12)(cid:12)(cid:12) t ∈ [0 , η ∗ ( x )] (cid:111) . Given x ≤ t ∈ [0 , η ∗ ( x )], define¯ x = − β/ √ n + ( x + β/ √ n ) e − t + tν ∗ ( x ) e − t . By uniqueness of ν ∗ (¯ x ) and η ∗ (¯ x ), it suffices to show that the pair ν = ν ∗ ( x ) e − t , η = η ∗ ( x ) − t solves (4.8) with ¯ x replacing x there. Indeed, ν ∗ ( x ) e − t e − ( η ∗ ( x ) − t ) = ν ∗ ( x ) e − η ∗ ( x ) = κ/ √ n, and − β/ √ n + (¯ x + β/ √ n ) e − ( η ∗ ( x ) − t ) + ( η ∗ ( x ) − t ) ν ∗ ( x ) e − t e − ( η ∗ ( x ) − t ) = − β/ √ n + (¯ x + β/ √ n ) e − ( η ∗ ( x ) − t ) + ( η ∗ ( x ) − t ) ν ∗ ( x ) e − η ∗ ( x ) = − β/ √ n + (( x + β/ √ n ) e − t + tν ∗ ( x ) e − t ) e − ( η ∗ ( x ) − t ) + ( η ∗ ( x ) − t ) ν ∗ ( x ) e − η ∗ ( x ) = − β/ √ n + ( x + β/ √ n ) e − η ∗ ( x ) + η ∗ ( x ) ν ∗ ( x ) e − η ∗ ( x ) = 0 . .1.2 Proving Lemma 6. Proof of Lemma 6 .
Fix κ ≥ β and x ∈ Ω. Assume that x ≥ Γ ( κ ) , which by definitionin (4.9) implies that there exists some δ ≥ x , x − δ ) ∈ Γ ( κ ) . Note that if x > Γ ( κ ) , then δ > β/ √ n − ( x + β/ √ n ) e − η − ηx e − η = 0 . (B.12)We first argue that η = − ( x + β/ √ n ) x − W (cid:18) − β/ √ nx e − ( x β/ √ n ) x (cid:19) . (B.13)Starting with (B.12), we can replicate the steps used to get (B.9) to see that (B.12) isequivalent to (cid:16) − ( x + β/ √ n ) x − η (cid:17) e − ( x β/ √ n ) x − η = − β/ √ nx e − ( x β/ √ n ) x . (B.14)Let us assume that x ≥ Γ ( κ ) implies that the right hand side of (B.14) is in the interval[ − e − , W ( · ) toboth sides of (B.14) and use (B.4) to conclude (B.13). Plugging (B.13) back into(B.12), − β/ √ n − W (cid:18) − β/ √ nx e − ( x β/ √ n ) x (cid:19) x e − η = 0 , (B.15)or x e − η = − β/ √ nW (cid:18) − β/ √ nx e − ( x β/ √ n ) x (cid:19) ≥ − β/ √ nW (cid:18) − β/ √ nx − δ e − ( x β/ √ n ) x − δ (cid:19) = κ/ √ n. Observe that the inequality above is strict if x > Γ ( κ ) , and that it becomes an equalityif δ = 0 (which means that x ∈ Γ ( κ ) ).To conclude the proof of (4.11), it remains verify our assumption that x ≥ Γ ( κ ) implies that the right hand side of (B.14) is in the interval [ − e − , x , x − δ ) ∈ Γ ( κ ) implies W (cid:18) − β/ √ nx − δ e − ( x β/ √ n ) x − δ (cid:19) = − βκ , or that − β/ √ nx − δ e − ( x β/ √ n ) x − δ = W − ( − β/κ ) ≥ W − ( −
1) = − e − , (B.16) here in the inequality above we used the fact that κ ≥ β and that W − ( · ) is anincreasing function. Now from (B.6) we know that ddν (cid:18) − β/ √ nν e − ( x β/ √ n ) ν (cid:19) > , ν ≥ x − δ, which implies − β/ √ nx e − ( x β/ √ n ) x ≥ − β/ √ nx − δ e − ( x β/ √ n ) x − δ = W − ( − β/κ ) ≥ − e − . This concludes the proof of (4.11).We now address the differentiability of τ ( x ) to prove (4.12). Fixing κ > β and x ≥ Γ ( κ ) , we see from (B.13) that τ ( x ) = − ( x + β/ √ n ) x − W (cid:18) − β/ √ nx e − ( x β/ √ n ) x (cid:19) . We know that W (cid:48) ( u ) exists for u ∈ ( − e − , − β/ √ nx e − ( x β/ √ n ) x > − e − , x ≥ Γ ( κ ) , which can be derived from (B.16). Therefore, τ ( x ) is differentiable at all points x ≥ Γ ( κ ) with x <
0. Only the one-sided derivative exists for x ∈ { x = 0 , x ≥ Γ ( κ ) } , i.e. those x that are on the vertical axis. To characterize the derivatives of τ ( x ), let us use theform τ ( x ) = − ( x + β/ √ n ) x + β/ √ nx e − τ ( x ) , which is implied by (B.13) and (B.15). Differentiating gives us τ ( x ) = − x (cid:16) − β/ √ nx e − τ ( x ) (cid:17) − = − x x e − τ ( x ) x e − τ ( x ) − β/ √ n = − e − τ ( x ) x e − τ ( x ) − β/ √ n , (B.17)where τ ( x ) is understood to be the left derivative when x = 0. Note that x ≥ Γ ( κ ) means the denominator in τ ( x ) is strictly positive due to our recently proved (4.11).Furthermore, τ ( x ) = x + β/ √ nx − β/ √ nx e − τ ( x ) + τ ( x ) β/ √ nx e − τ ( x ) = − x τ ( x ) + τ ( x ) β/ √ nx e − τ ( x ) , and so τ ( x ) = − x τ ( x ) (cid:16) − β/ √ nx e − τ ( x ) (cid:17) − = τ ( x ) τ ( x ) . (B.18) his proves (4.12), and we now prove the last claim of the lemma. Fix x = ( x , x )and assume that x ≥ Γ ( κ ) . By (4.11), we know that x e − τ ( x ) ≥ κ / √ n > κ / √ n .Now ddx x e − τ ( x ) = e − τ ( x ) − τ ( x ) x e − τ ( x ) > , x ≥ Γ ( κ ) , where the inequality follows from the form of τ ( x ) in (4.12). Therefore,( x + ε ) e − τ ( x ,x + ε ) ≥ κ / √ n > κ / √ n, ε ≥ . In other words, ( x , x + ε ) (cid:54)∈ Γ ( κ ) for all ε ≥ ( κ ) in Lemma 5.However, also by Lemma 5, there must exist some ¯ x ≥ x , ¯ x ) ∈ Γ ( κ ) ,which means that ¯ x = x − ¯ ε for some ¯ ε >
0, or that x > Γ ( κ ) . B.2 Lemmas in Section 4.2.
Proof of Lemma 7.
The proof proceeds as follows. We first show that f ( · , x ) , f ( x , · )are absolutely continuous for all x ∈ Ω. We then verify that f ∗ ( x ) satisfies the PDE(3.7) with the boundary condition (3.8). Lastly, we verify the bounds on the secondderivatives of f ∗ ( x ). B.2.1 First Derivatives
Recall the definition of f ∗ ( x ): f ∗ ( x ) = , x ∈ [0 , κ/ √ n ] ,x − κ √ n − κ √ n log( √ nx /κ ) , x ≤ Γ ( κ ) and x ≥ κ/ √ n,x (1 − e − τ ( x ) ) − κ √ n τ ( x ) + √ nβ ( x e − τ ( x ) − κ/ √ n ) , x ≥ Γ ( κ ) . Let us differentiate f ∗ ( x ) in the region x ≤ Γ ( κ ) and x ≥ κ/ √ n : f ∗ ( x ) = 0 , f ∗ ( x ) = 1 − κx √ n , f ∗ ( x ) = 1 x κ √ n , x ≤ Γ ( κ ) and x ≥ κ/ √ n. (B.19)Next, we differentiate f ∗ ( x ) when x ≥ Γ ( κ ) : f ∗ ( x ) = τ ( x ) x e − τ ( x ) − κ √ n τ ( x ) + √ nβ ( x e − τ ( x ) − κ/ √ n )( − x τ ( x ) e − τ ( x ) )= − τ ( x ) (cid:16) − x e − τ ( x ) + κ √ n + x e − τ ( x ) √ nβ − x e − τ ( x ) κβ (cid:17) = − τ ( x ) √ nβ (cid:16) x e − τ ( x ) − β √ n x e − τ ( x ) − x e − τ ( x ) κ √ n + κβn (cid:17) = − τ ( x ) √ nβ (cid:0) x e − τ ( x ) − β/ √ n (cid:1)(cid:0) x e − τ ( x ) − κ/ √ n (cid:1) = √ nβ e − τ ( x ) (cid:0) x e − τ ( x ) − κ/ √ n (cid:1) , (B.20) here in the last equality we used (4.12). Now we will prove that f ∗ ( x ) = 1 − x κ √ n + √ nβ (cid:0) x e − τ ( x ) − κ/ √ n (cid:1)(cid:16) x e − τ ( x ) − β/ √ nx + τ e − τ ( x ) (cid:17) , x ≥ Γ ( κ ) . (B.21)We begin by differentiating f ∗ ( x ): f ∗ ( x ) = (1 − e − τ ( x ) ) + τ ( x ) x e − τ ( x ) − κ √ n τ ( x )+ √ nβ ( x e − τ ( x ) − κ/ √ n )( e − τ ( x ) − τ ( x ) x e − τ ( x ) )= (1 − e − τ ( x ) ) + τ ( x ) x e − τ ( x ) − κ √ n τ ( x )+ √ nβ (cid:0) x e − τ ( x ) − τ ( x ) x e − τ ( x ) − e − τ ( x ) κ/ √ n + τ ( x ) x e − τ ( x ) κ/ √ n (cid:1) , which equals(1 − e − τ ( x ) ) + √ nβ x e − τ ( x ) − e − τ ( x ) κβ (B.22) − τ ( x ) √ nβ (cid:0) x e − τ ( x ) − β √ n x e − τ ( x ) − κ √ n x e − τ ( x ) + κβn (cid:1) . (B.23)We focus on (B.22), which equals1 + √ nβ x (cid:16) x e − τ ( x ) − x e − τ ( x ) κ √ n − x e − τ ( x ) β √ n (cid:17) = 1 − x κ √ n + √ nβ x (cid:16) x e − τ ( x ) − x e − τ ( x ) κ √ n − x e − τ ( x ) β √ n + κβn (cid:17) = 1 − x κ √ n + √ nβ x (cid:0) x e − τ ( x ) − β/ √ n (cid:1)(cid:0) x e − τ ( x ) − κ/ √ n (cid:1) . With the help of (4.12), we see that (B.23) equals − τ ( x ) √ nβ (cid:0) x e − τ ( x ) − β/ √ n (cid:1)(cid:0) x e − τ ( x ) − κ/ √ n (cid:1) = √ nβ τ e − τ ( x ) (cid:0) x e − τ ( x ) − κ/ √ n (cid:1) . Therefore, for all x ≥ Γ ( κ ) , f ∗ ( x ) = 1 − x κ √ n + √ nβ (cid:0) x e − τ ( x ) − κ/ √ n (cid:1)(cid:16) x e − τ ( x ) − β/ √ nx + τ e − τ ( x ) (cid:17) . We now verify continuity of the partial derivatives of f ∗ ( x ). Recall that the supportof f ∗ ( x ) is naturally partitioned into three subdomains: { x ∈ [0 , κ/ √ n ] } , { x ≤ Γ ( κ ) , x ≥ κ/ √ n } , and { x ≥ Γ ( κ ) } . (B.24) ontinuity of the partial derivatives on the interiors of these subdomains follows fromthe continuity of τ ( x ), and it remains to verify continuity on the intersections, whichare { x ∈ [0 , κ/ √ n ] } ∩ { x ≤ Γ ( κ ) , x ≥ κ/ √ n } = { x = κ/ √ n } , { x ≤ Γ ( κ ) , x ≥ κ/ √ n } ∩ { x ≥ Γ ( κ ) } = { x ∈ Γ ( κ ) } , { x ∈ [0 , κ/ √ n ] } ∩ { x ≥ Γ ( κ ) } = { (0 , κ/ √ n ) } ⊂ Γ ( κ ) . The fact that { (0 , κ/ √ n ) } ⊂ Γ ( κ ) follows from the definition of Γ ( κ ) in Lemma 5. When x = κ/ √ n , we see from (B.19) that f ∗ ( x ) = f ∗ ( x ) = 0, which confirms continuity on { x = κ/ √ n } . Now by definition, x ∈ Γ ( κ ) implies that x e − τ ( x ) = κ/ √ n , from whichwe see that (B.19) coincides with (B.20)-(B.21).Thus we have proved continuity of the derivatives of f ∗ ( x ) on Ω. It remains toprove that f ∗ ( · , x ) , f ∗ ( x , · ) are absolutely continuous for all x ∈ Ω.Fix x ≥
0. We will show that f ∗ ( · , x ) is differentiable almost everywhere. Fromthe form of f ∗ ( x ) and (B.19), we see that ddx f ∗ ( x ) = 0 on the set { x ≤ x < Γ ( κ ) } .Furthermore, from (B.20) we know that ddx f ∗ ( x ) exists on the interior of { x ≤ x ≥ Γ ( κ ) } (because τ ( x ) is differentiable). It may be that ddx f ∗ ( x ) does not exist on theset { x : x ∈ Γ ( κ ) } . However, Lemma 5 (and in particular the form of γ ( κ ) ( x )) tellsus that the curve Γ ( κ ) contains no horizontal segments because the second coordinateof γ ( κ ) ( x ) is always decreasing with t . Therefore, the set { x : x ∈ Γ ( κ ) } contains atmost one point, and for each fixed x ≥
0, the function f ∗ ( · , x ) is differentiable almosteverywhere, and therefore absolutely continuous.A similar argument holds for showing f ∗ ( x , · ) is absolutely continuous. For fixed x ≤
0, the function f ∗ ( x , · ) is differentiable everywhere except the point x = κ/ √ n and the set { x : x ∈ Γ ( κ ) } . However, the latter contains only a single point becausegiven x , Lemma 5 (namely, uniqueness of ν ∗ ( x )) tells us the set { x : x ∈ Γ ( κ ) } contains only a single point (i.e. Γ ( κ ) contains no vertical lines). Therefore, f ∗ ( x , · )is differentiable almost everywhere and is therefore absolutely continuous. B.2.2 Satisfying the PDE
We now verify that f ∗ ( x ) satisfies the PDE (3.7) and the boundary condition (3.8).For x ∈ [0 , κ/ √ n ], Lf ∗ ( x ) = 0 = − (cid:0) ( x − κ/ √ n ) ∨ (cid:1) , and f ∗ (0 , x ) = f ∗ (0 , x ) = 0 , and so both (3.7) and (3.8) are trivially satisfied. When x ≤ Γ ( κ ) and x ≥ κ/ √ n , Lf ∗ ( x ) = ( − x + x − β/ √ n ) f ∗ ( x ) − x f ∗ ( x ) = − x (cid:0) − κx √ n (cid:1) = − (cid:0) ( x − κ/ √ n ) ∨ (cid:1) , nd the only intersection of x ≤ Γ ( κ ) and x ≥ κ/ √ n with the vertical axis is the point(0 , κ/ √ n ), meaning f ∗ (0 , x ) = 0 f ∗ (0 , x ) = 1 − κx √ n = 0 . The last case is x ≥ Γ ( κ ) . Using (B.20) and (B.21): Lf ∗ ( x ) = ( − x + x − β/ √ n ) f ∗ ( x ) − x f ∗ ( x )= ( − x + x − β/ √ n ) √ nβ e − τ ( x ) (cid:0) x e − τ ( x ) − κ/ √ n (cid:1) − x (cid:16) − x κ √ n + √ nβ (cid:0) x e − τ ( x ) − κ/ √ n (cid:1)(cid:16) x e − τ ( x ) − β/ √ nx + τ e − τ ( x ) (cid:17)(cid:17) = − (cid:0) x − κ √ n (cid:1) + √ nβ (cid:0) x e − τ ( x ) − κ/ √ n (cid:1)(cid:16) β/ √ n + x τ ( x ) e − τ ( x ) + ( − x − β/ √ n ) e − τ ( x ) (cid:17) = − (cid:0) ( x − κ/ √ n ) ∨ (cid:1) , where the last equality follows from the definition of τ ( x ) in Lemma 6. Verifying theboundary condition: f ∗ (0 , x ) = √ nβ e − τ (0 ,x ) (cid:0) x e − τ (0 ,x ) − κ/ √ n (cid:1) = √ nβ (cid:0) x − κ/ √ n (cid:1) , and f ∗ (0 , x ) = 1 − x κ √ n + √ nβ (cid:0) x e − τ (0 ,x ) − κ/ √ n (cid:1)(cid:16) x e − τ (0 ,x ) − β/ √ nx + τ (0 , x ) e − τ (0 ,x ) (cid:17) = 1 − x κ √ n + √ nβ x (cid:0) x − κ/ √ n (cid:1)(cid:16) x − β/ √ n (cid:17) = 1 − x κ √ n + √ nβ x (cid:0) x − x κ/ √ n − x β/ √ n + κβn (cid:1) = √ nβ (cid:16) β √ n − x κβn (cid:17) + √ nβ (cid:0) x − κ/ √ n − β/ √ n + κβn x (cid:1) = √ nβ ( x − κ/ √ n ) = f ∗ (0 , x ) . Therefore, our f ∗ ( x ) satisfies (3.7)–(3.8). B.2.3 Second Derivatives
It remains to prove the bounds on the second derivatives (3.9)–(3.11). When x ∈{ x ∈ [0 , κ/ √ n ] } , f ∗ ( x ) = f ∗ ( x ) = f ∗ ( x ) = 0 and when x ∈ { x ≤ Γ ( κ ) , x ≥ κ/ √ n } , f ∗ ( x ) = f ∗ ( x ) = 0 and f ∗ ( x ) = 1 x κ √ n ≤ √ nκ ≤ √ nβ , hich satisfies (3.9)–(3.11).Therefore, we are left to deal with the case when x ≥ Γ ( κ ) . First we take on f ∗ ( x ).Differentiating (B.20) and using (4.12) one arrives at f ∗ ( x ) = √ nβ e − τ ( x ) x e − τ ( x ) + (cid:0) x e − τ ( x ) − κ/ √ n (cid:1)(cid:0) x e − τ ( x ) − β/ √ n (cid:1) , from which we conclude that0 ≤ f ∗ ( x ) ≤ √ nβ (cid:16) x e − τ ( x ) x e − τ ( x ) − β/ √ n + 1 (cid:17) = √ nβ (cid:16) − β/ √ nx e − τ ( x ) + 1 (cid:17) ≤ √ nβ (cid:16) − βκ + 1 (cid:17) , where all three inequalities above follow from the fact that x e − τ ( x ) ≥ κ/ √ n > β/ √ n ;c.f. (4.11) in Lemma 6. Taking the derivative in (B.20) with respect to x , we see that f ∗ ( x ) = √ nβ (cid:0) − τ ( x ) e − τ ( x ) (cid:1)(cid:0) x e − τ ( x ) − κ/ √ n (cid:1) + √ nβ e − τ ( x ) (cid:0) e − τ ( x ) − τ ( x ) x e − τ ( x ) (cid:1) . The quantity above is non-negative because x e − τ ( x ) ≥ κ/ √ n and − τ ( x ) ≥
0; thelatter follows from (4.12). Lastly, we can differentiate (B.21) and use τ ( x ) = τ ( x ) τ ( x )from (4.12) to see that f ∗ ( x ) = 1 x κ √ n + √ nβ (cid:0) e − τ ( x ) − τ ( x ) τ ( x ) x e − τ ( x ) (cid:1)(cid:16) x e − τ ( x ) − β/ √ nx + τ ( x ) e − τ ( x ) (cid:17) + √ nβ (cid:0) x e − τ ( x ) − κ/ √ n (cid:1)(cid:16) β/ √ nx − τ ( x ) τ ( x ) e − τ ( x ) (cid:17) . Again, f ∗ ( x ) ≥ x e − τ ( x ) ≥ κ/ √ n and − τ ( x ) ≥
0. Let us now bound f ∗ ( x ). The first term on the right hand side above is bounded by √ n/κ , because x ≥ Γ ( κ ) implies x ≥ κ/ √ n . For the second term, note that x e − τ ( x ) − β/ √ nx + τ ( x ) e − τ ( x ) ≤ e − ≤ , and using the form of τ ( x ) from (4.12), e − τ ( x ) − τ ( x ) τ ( x ) x e − τ ( x ) = e − τ ( x ) + e − τ ( x ) x e − τ ( x ) − β/ √ n τ x e − τ ( x ) = e − τ ( x ) + τ e − τ ( x ) − β/ √ nx e − τ ( x ) ≤ e − − βκ ≤ κκ − β . or the third term, observe that (cid:0) x e − τ ( x ) − κ/ √ n (cid:1)(cid:16) β/ √ nx − τ ( x ) τ ( x ) e − τ ( x ) (cid:17) ≤ (cid:0) x e − τ ( x ) − κ/ √ n (cid:1)(cid:16) x βκ − τ ( x ) τ ( x ) e − τ ( x ) (cid:17) = x e − τ ( x ) − κ/ √ nx βκ + x e − τ ( x ) − κ/ √ nx e − τ ( x ) − β/ √ n τ ( x ) e − τ ( x ) ≤ βκ + 1 ≤ , where we use x ≥ κ/ √ n in the first inequality, the form of τ ( x ) in (4.12) in the firstequation, and the fact that β < κ in the last two inequalities. Combining the boundson all three terms, we conclude that f ∗ ( x ) ≤ √ nκ + √ nβ (cid:16) κκ − β (cid:17) + 2 √ nβ ≤ √ nβ (cid:16) κκ − β (cid:17) , where in the last inequality we used the fact that √ n/κ < √ n/β . C Proving Lemma 8.
In this section we prove Lemma 8 by constructing solutions to (5.9) and (5.10). Theintuition behind the forms of these solutions is the same as in Section 4. Namely, that f (1) ( x ) = (cid:90) ∞ φ ( κ / √ n,κ / √ n ) ( − v x ( t )) dt, and f (2) ( x ) = (cid:90) ∞ φ ( κ / √ n,κ / √ n ) ( v x ( t )) dt solves (5.9) and (5.10), respectively. For the remainder of this section, we fix κ < κ with κ > β , and let us write φ ( x ) instead of φ ( κ / √ n,κ / √ n ) ( x ) to simplify notation. C.1 Solving the first PDE.
We begin by constructing a candidate solution to (5.9), and proving the associatedproperties in Lemma 8. Recall the definition of W ( x ) from Section B.1, and for any κ > β , define˜ τ ( κ ) ( x ) = − ( x + β/ √ n ) x − W (cid:16) ( κ − β ) / √ nx e − ( x β/ √ n ) x (cid:17) , x ≤ − κ/ √ n, x > . (C.1)The quantity in (C.1) is well defined because the argument of W ( · ) is positive. Fur-thermore, differentiability of W ( · ) implies differentiability of ˜ τ ( κ ) ( x ). One can checkthat − β/ √ n + ( x + β/ √ n ) e − ˜ τ ( κ ) ( x ) + x ˜ τ ( κ ) ( x ) e − ˜ τ ( κ ) ( x ) = − κ/ √ n (C.2) or those x where ˜ τ ( κ ) ( x ) is defined by repeating the arguments used to show theequivalence of (B.8) and (B.10) in Section B.1.1. Intuitively, ˜ τ ( κ ) ( x ) is the time thefluid model hits the set { x = − κ/ √ n } . The following lemma tells us that we canextend the definition of ˜ τ ( κ ) ( x ) to x = 0; it is proved in Section C.1.1. Lemma 10.
For any κ > β and x ≤ − κ/ √ n , lim x ↓ ˜ τ ( κ ) ( x ) = log (cid:16) −√ nx − βκ − β (cid:17) , (C.3) meaning that the function in (C.1) can be extended to x ≥ . Furthermore, ˜ τ ( κ ) ( − κ/ √ n, x ) = 0 , x ≥ and − ˜ τ ( κ )1 ( x ) = e − ˜ τ ( κ ) ( x ) x e − ˜ τ ( κ ) ( x ) + ( κ − β ) / √ n , ˜ τ ( κ )2 ( x ) = − ˜ τ ( κ )1 ( x )˜ τ ( κ ) ( x ) (C.5) for x ≤ − κ/ √ n and x ≥ , where the derivatives at { x = − κ/ √ n } and { x = 0 } are interpreted as the one-sided derivatives. The following lemma presents the candidate solution to (5.9); it is proved in Sec-tion C.1.2.
Lemma 11.
The function f (1) : Ω → R + defined as f (1) ( x ) = ˜ τ ( κ ) ( x ) + (cid:82) ˜ τ ( κ ( x )˜ τ ( κ ( x ) φ (cid:0) β/ √ n − ( x + β/ √ n ) e − t − x te − t (cid:1) dt, x ≤ − κ / √ n, (cid:82) ˜ τ ( κ ( x )0 φ (cid:0) β/ √ n − ( x + β/ √ n ) e − t − x te − t (cid:1) dt, x ∈ (cid:104) − κ √ n , − κ √ n (cid:105) , , x ∈ [ − κ / √ n, . belongs to C (Ω) . Furthermore, f (1)1 ( x ) = − (cid:82) ˜ τ ( κ ( x )˜ τ ( κ ( x ) e − t φ (cid:48) (cid:0) β/ √ n − ( x + β/ √ n ) e − t − x te − t (cid:1) dt, x ≤ − κ / √ n, − (cid:82) ˜ τ ( κ ( x )0 e − t φ (cid:48) (cid:0) β/ √ n − ( x + β/ √ n ) e − t − x te − t (cid:1) dt, x ∈ (cid:104) − κ √ n , − κ √ n (cid:105) , , x ∈ [ − κ / √ n, . and f (1)2 ( x ) = − (cid:82) ˜ τ ( κ ( x )˜ τ ( κ ( x ) te − t φ (cid:48) (cid:0) β/ √ n − ( x + β/ √ n ) e − t − x te − t (cid:1) dt, x ≤ − κ / √ n, − (cid:82) ˜ τ ( κ ( x )0 te − t φ (cid:48) (cid:0) β/ √ n − ( x + β/ √ n ) e − t − x te − t (cid:1) dt, x ∈ (cid:104) − κ √ n , − κ √ n (cid:105) , , x ∈ [ − κ / √ n, . Let us now verify that f (1) ( x ) from Lemma 11 satisfies (5.9). The boundary con- ition f (1)1 (0 , x ) = f (1)2 (0 , x ) is trivially satisfied. For x ≤ − κ / √ n ,( − x + x − β/ √ n ) f (1)1 ( x ) − x f (1)2 ( x )= (cid:90) ˜ τ ( κ ( x )˜ τ ( κ ( x ) (cid:0) ( x − x + β/ √ n ) e − t + x te − t (cid:1) φ (cid:48) (cid:0) β/ √ n − ( x + β/ √ n ) e − t − x te − t (cid:1) dt = (cid:90) κ / √ nκ / √ n φ (cid:48) ( u ) du = φ ( κ / √ n ) − φ ( κ / √ n ) = − − φ ( − x ) , and a similar argument works when x ∈ (cid:104) − κ √ n , − κ √ n (cid:105) . Therefore, f (1) ( x ) solves (5.9).We now bound f (1)1 ( x ) and f (1)11 ( x ) to prove (5.15), and then bound f (1) ( x ) to prove(5.13). Since φ (cid:48) ( κ / √ n ) = φ (cid:48) ( κ / √ n ) = 0 (see (5.6)), differentiating f (1)1 ( x ) gives us f (1)11 ( x ) = − (cid:82) ˜ τ ( κ ( x )˜ τ ( κ ( x ) e − t φ (cid:48)(cid:48) (cid:0) β/ √ n − ( x + β/ √ n ) e − t − x te − t (cid:1) dt, x ≤ − κ / √ n, − (cid:82) ˜ τ ( κ ( x )0 e − t φ (cid:48)(cid:48) (cid:0) β/ √ n − ( x + β/ √ n ) e − t − x te − t (cid:1) dt, x ∈ (cid:104) − κ √ n , − κ √ n (cid:105) , , x ∈ [ − κ / √ n, . Using (5.7), (cid:12)(cid:12)(cid:12) f (1)1 ( x ) (cid:12)(cid:12)(cid:12) ≤ √ n (cid:12)(cid:12) ˜ τ ( κ ) ( x ) − ˜ τ ( κ ) ( x ) (cid:12)(cid:12) κ − κ , x ≤ − κ / √ n, (cid:12)(cid:12)(cid:12) f (1)11 ( x ) (cid:12)(cid:12)(cid:12) ≤ n (cid:12)(cid:12) ˜ τ ( κ ) ( x ) − ˜ τ ( κ ) ( x ) (cid:12)(cid:12) ( κ − κ ) , x ≤ − κ / √ n, (cid:12)(cid:12)(cid:12) f (1)1 ( x ) (cid:12)(cid:12)(cid:12) ≤ √ n (cid:12)(cid:12) ˜ τ ( κ ) ( x ) (cid:12)(cid:12) κ − κ , (cid:12)(cid:12)(cid:12) f (1)11 ( x ) (cid:12)(cid:12)(cid:12) ≤ n (cid:12)(cid:12) ˜ τ ( κ ) ( x ) (cid:12)(cid:12) ( κ − κ ) , x ∈ (cid:104) − κ √ n , − κ √ n (cid:105) . When x ≤ − κ / √ n ,˜ τ ( κ ) ( x ) − ˜ τ ( κ ) ( x ) = W (cid:16) ( κ − β ) / √ nx e − ( x β/ √ n ) x (cid:17) − W (cid:16) ( κ − β ) / √ nx e − ( x β/ √ n ) x (cid:17) = log (cid:16) κ − βκ − β (cid:17) − log (cid:32) W (cid:16) ( κ − β ) / √ nx e − ( x β/ √ n ) x (cid:17) W (cid:16) ( κ − β ) / √ nx e − ( x β/ √ n ) x (cid:17) (cid:33) ≤ log (cid:16) κ − βκ − β (cid:17) , x ≤ − κ / √ n, (C.6)where the second equation follows from (B.2) and the inequality follows from the factthat W ( · ) is an increasing function and κ > κ . The first equation in (C.6) andmonotonicity of W ( · ) means that ˜ τ ( κ ) ( x ) − ˜ τ ( κ ) ( x ) >
0, and therefore (cid:12)(cid:12)(cid:12) ˜ τ ( κ ) ( x ) − ˜ τ ( κ ) ( x ) (cid:12)(cid:12)(cid:12) ≤ log (cid:16) κ − βκ − β (cid:17) , x ≤ − κ / √ n. hen x ∈ (cid:104) − κ √ n , − κ √ n (cid:105) ,0 = ˜ τ ( κ ) ( − κ / √ n, x ) ≤ ˜ τ ( κ ) ( x ) ≤ ˜ τ ( κ ) ( − κ / √ n, x )= ˜ τ ( κ ) ( − κ / √ n, x ) − ˜ τ ( κ ) ( − κ / √ n, x ) ≤ log (cid:16) κ − βκ − β (cid:17) . (C.7)The two equalities above are due to (C.4), the first two inequalities follow from thefact that ˜ τ ( κ )1 ( x ) ≤ (cid:12)(cid:12)(cid:12) f (1)1 ( x ) (cid:12)(cid:12)(cid:12) ≤ √ nκ − κ log (cid:16) κ − βκ − β (cid:17) , (cid:12)(cid:12)(cid:12) f (1)11 ( x ) (cid:12)(cid:12)(cid:12) ≤ n ( κ − κ ) log (cid:16) κ − βκ − β (cid:17) . Choosing κ = β + (cid:15) and κ = β + 2 (cid:15) proves (5.15). To prove (5.13), note that (cid:12)(cid:12)(cid:12) f (1) ( x ) (cid:12)(cid:12)(cid:12) ≤ ˜ τ ( κ ) ( x ) ≤ log (cid:16) κ − βκ − β (cid:17) , x ∈ (cid:104) − κ √ n , − κ √ n (cid:105) , where the first inequality follows from the form of f (1) ( x ) and the second inequalityfollows from (C.7). For x ≤ − κ / √ n , (cid:12)(cid:12)(cid:12) f (1) ( x ) (cid:12)(cid:12)(cid:12) ≤ ˜ τ ( κ ) ( x ) + (cid:12)(cid:12)(cid:12) ˜ τ ( κ ) ( x ) − ˜ τ ( κ ) ( x ) (cid:12)(cid:12)(cid:12) = ˜ τ ( κ ) ( x ) + ˜ τ ( κ ) ( x ) − ˜ τ ( κ ) ( x )= ˜ τ ( κ ) ( x ) ≤ log (cid:16) κ − βκ − β (cid:17) , where the last inequality follows from (C.6) and the fact that ˜ τ ( κ ) ( x ) ≥
0. Choosing κ = β + (cid:15) and κ = β + 2 (cid:15) proves (5.13). C.1.1 Proof of Lemma 10.
Proof of Lemma 10.
Let u = ( κ − β ) / √ nx e − ( x β/ √ n ) x . Using (B.2),˜ τ ( κ ) ( x ) = − ( x + β/ √ n ) x − W ( u ) = log (cid:18) √ nκ − β x W ( u ) (cid:19) , x ≤ − κ/ √ n, x > . Therefore, it remains to evaluatelim x ↓ W (cid:16) ( κ − β ) / √ nx e − ( x β/ √ n ) x (cid:17) /x , hich we do using L’Hopital’s rule. The derivative of the numerator with respect to x is W (cid:48) ( u ) u (cid:18) − x + x + β/ √ nx (cid:19) = W ( u )1 + W ( u ) 1 x ( − x + x + β/ √ n ) , where we used (B.3) to get the equality above. Thereforelim x ↓ W (cid:16) ( κ − β ) / √ nx e − ( x β/ √ n ) x (cid:17) /x = lim x ↓ ddx W (cid:16) ( κ − β ) / √ nx e − ( x β/ √ n ) x (cid:17) − /x = − x − β/ √ n, and (C.3) follows. We now prove (C.4), or that ˜ τ ( κ ) ( − κ/ √ n, x ) = 0 for x ≥
0. Theclaim is true when x = 0 by (C.3). For x > τ ( κ ) ( − κ/ √ n, x ) = − ( − κ/ √ n + β/ √ n ) x − W (cid:16) ( κ − β ) / √ nx e − ( − κ/ √ n + β/ √ n ) x (cid:17) = 0 , where the second equality comes from the fact that W ( xe x ) = x ; c.f. (B.4). ThatDifferentiability of ˜ τ ( κ ) ( x ) follows from the differentiability of W ( · ). To verify (C.5),plug in (C.1) into (C.2) to see that W (cid:16) ( κ − β ) / √ nx e − ( x β/ √ n ) x (cid:17) = ( κ − β ) / √ nx e − ˜ τ ( κ ) ( x ) , and therefore ˜ τ ( κ ) ( x ) = − ( x + β/ √ n ) x − ( κ − β ) / √ nx e − ˜ τ ( κ ) ( x ) . The proof of (C.5) is then identical to the arguments used to prove (B.17)-(B.18).
C.1.2 Proof of Lemma 11.
Proof of Lemma 11 .
For convenience, we recall that f (1) ( x )= ˜ τ ( κ ) ( x ) + (cid:82) ˜ τ ( κ ( x )˜ τ ( κ ( x ) φ (cid:0) β/ √ n − ( x + β/ √ n ) e − t − x te − t (cid:1) dt, x ≤ − κ / √ n, (cid:82) ˜ τ ( κ ( x )0 φ (cid:0) β/ √ n − ( x + β/ √ n ) e − t − x te − t (cid:1) dt, x ∈ (cid:104) − κ √ n , − κ √ n (cid:105) , , x ∈ [ − κ / √ n, . ontinuity of f (1) ( x ) on the sets { x = − κ / √ n } and { x = − κ / √ n } follows from(C.4). Let us differentiate f (1) ( x ) on the set x ≤ − κ / √ n . Using the Leibniz rule, f (1)1 ( x ) = ˜ τ ( κ )1 ( x ) + φ (cid:0) β/ √ n − ( x + β/ √ n ) e − ˜ τ ( κ ( x ) − x ˜ τ ( κ ) ( x ) e − ˜ τ ( κ ( x ) (cid:1) ˜ τ ( κ )1 ( x ) − φ (cid:0) β/ √ n − ( x + β/ √ n ) e − ˜ τ ( κ ( x ) − x ˜ τ ( κ ) ( x ) e − ˜ τ ( κ ( x ) (cid:1) ˜ τ ( κ )1 ( x ) − (cid:90) ˜ τ ( κ ( x )˜ τ ( κ ( x ) e − t φ (cid:48) (cid:0) β/ √ n − ( x + β/ √ n ) e − t − x te − t (cid:1) dt = ˜ τ ( κ )1 ( x ) + φ (cid:0) κ / √ n (cid:1) ˜ τ ( κ )1 ( x ) − φ (cid:0) κ / √ n (cid:1) ˜ τ ( κ )1 ( x ) − (cid:90) ˜ τ ( κ ( x )˜ τ ( κ ( x ) e − t φ (cid:48) (cid:0) β/ √ n − ( x + β/ √ n ) e − t − x te − t (cid:1) dt = − (cid:90) ˜ τ ( κ ( x )˜ τ ( κ ( x ) e − t φ (cid:48) (cid:0) β/ √ n − ( x + β/ √ n ) e − t − x te − t (cid:1) dt, where the second and third equalities follow from (C.2), or φ (cid:0) β/ √ n − ( x + β/ √ n ) e − ˜ τ ( κ ( x ) − x ˜ τ ( κ ) ( x ) e − ˜ τ ( κ ( x ) (cid:1) = φ ( κ / √ n ) = 1 ,φ (cid:0) β/ √ n − ( x + β/ √ n ) e − ˜ τ ( κ ( x ) − x ˜ τ ( κ ) ( x ) e − ˜ τ ( κ ( x ) (cid:1) = φ ( κ / √ n ) = 0 . Repeating the same argument on the set x ∈ (cid:2) − κ √ n , − κ √ n (cid:3) , we conclude that f (1)1 ( x )= − (cid:82) ˜ τ ( κ ( x )˜ τ ( κ ( x ) e − t φ (cid:48) (cid:0) β/ √ n − ( x + β/ √ n ) e − t − x te − t (cid:1) dt, x ≤ − κ / √ n, − (cid:82) ˜ τ ( κ ( x )0 e − t φ (cid:48) (cid:0) β/ √ n − ( x + β/ √ n ) e − t − x te − t (cid:1) dt, x ∈ (cid:104) − κ √ n , − κ √ n (cid:105) , , x ∈ [ − κ / √ n, . and f (1)2 ( x )= − (cid:82) ˜ τ ( κ ( x )˜ τ ( κ ( x ) te − t φ (cid:48) (cid:0) β/ √ n − ( x + β/ √ n ) e − t − x te − t (cid:1) dt, x ≤ − κ / √ n, − (cid:82) ˜ τ ( κ ( x )0 te − t φ (cid:48) (cid:0) β/ √ n − ( x + β/ √ n ) e − t − x te − t (cid:1) dt, x ∈ (cid:104) − κ √ n , − κ √ n (cid:105) , , x ∈ [ − κ / √ n, . Continuity of the first order derivatives follows from (C.4). Furthermore, existenceand continuity of f (1)11 ( x ) , f (1)12 ( x ), and f (1)22 ( x ) follows from existence and continuity of φ (cid:48)(cid:48) ( x ). .2 Solving the second PDE. Recall the definitions of Γ ( κ ) and τ ( x ) from Lemmas 5 and 6, respectively. PartitionΩ into four subdomains: S = { x ∈ Ω | x ≤ κ / √ n } ,S = { x ∈ Ω | x ≥ κ / √ n, x ≤ Γ ( κ ) } ,S = { x ∈ Ω | Γ ( κ ) ≤ x ≤ Γ ( κ ) } ,S = { x ∈ Ω | x ≥ Γ ( κ ) } . Figure 3 helps to visualize the four sets. Let us verify that S ∪ S ∪ S ∪ S does indeedequal Ω. Fix x ≤ x ≥
0. Then x must lie in one of { x ≤ κ / √ n } , { x ≥ κ / √ n, x ≤ Γ ( κ ) } , { x ≥ κ / √ n, Γ ( κ ) ≤ x ≤ Γ ( κ ) } , or { x ≥ κ / √ n, x ≥ Γ ( κ ) , x ≥ Γ ( κ ) } . From (4.11) we see that x ≥ Γ ( κ ) implies x ≥ κ / √ n , or { x ≥ κ / √ n, Γ ( κ ) ≤ x ≤ Γ ( κ ) } = S , and { x ≥ κ / √ n, x ≥ Γ ( κ ) , x ≥ Γ ( κ ) } = { x ≥ Γ ( κ ) , x ≥ Γ ( κ ) } . From (4.13) we know that x ≥ Γ ( κ ) implies x > Γ ( κ ) , or { x ≥ Γ ( κ ) , x ≥ Γ ( κ ) } = S . Therefore S ∪ S ∪ S ∪ S equals Ω. The following lemma is proved at the end of thissection. Lemma 12.
The function f (2) ( x ) = , x ∈ S , (cid:82) log( √ nx /κ )0 φ ( x e − t ) dt, x ≤ κ / √ n, x ∈ S , log( √ nx /κ ) + (cid:82) log( κ /κ )0 φ (cid:0) κ √ n e − t (cid:1) dt, x ≥ κ / √ n, x ∈ S , (cid:82) τ ( x )0 φ ( x e − t ) dt + √ nβ (cid:82) x e − τ ( x ) κ / √ n φ ( t ) dt, x ≤ κ / √ n, x ∈ S , log( √ nx /κ ) + (cid:82) τ ( x )log( √ nx /κ ) φ (cid:0) x e − t (cid:1) dt + √ nβ (cid:82) x e − τ ( x ) κ / √ n φ ( t ) dt, x ≥ κ / √ n, x ∈ S ,τ ( x ) + x e − τ ( x ) − κ / √ nβ/ √ n + √ nβ (cid:82) κ / √ nκ / √ n φ ( t ) dt, x ∈ S , is well-defined for x ∈ Ω and belongs to C (Ω) . Its derivatives are f (2)1 ( x ) = , x ∈ S , , x ∈ S , √ nβ φ ( x e − τ ( x ) ) e − τ ( x ) , x ∈ S , √ nβ e − τ ( x ) , x ∈ S , (C.8) / √ nκ / √ n Γ ( κ ) Γ ( κ ) x x Figure 3: An aide to visualize S , S , S , S . and f (2)2 ( x ) = , x ∈ S , x φ ( x ) , x ∈ S , x (cid:0) φ ( x ) − φ ( x e − τ ( x ) ) (cid:1) + φ ( x e − τ ( x ) ) √ nβ e − τ ( x ) ( τ ( x ) + 1) , x ∈ S , √ nβ e − τ ( x ) ( τ ( x ) + 1) , x ∈ S . (C.9)In the remainder of this section, we 1) verify Lf (2) ( x ) = − φ ( x ), 2) verify f (2)1 (0 , x ) = f (2)2 (0 , x ) 3) bound f (2) ( x ) , f (2)1 ( x ) , and f (2)11 ( x ), and 4) prove Lemma 12.We now verify that Lf (2) ( x ) = − φ ( x ). When x ∈ S and x ∈ S , this fact is rivial. For x ∈ S ,( − x + x − β/ √ n ) f (2)1 ( x ) − x f (2)2 ( x )= ( − x + x − β/ √ n ) √ nβ φ ( x e − τ ( x ) ) e − τ ( x ) − x (cid:16) x (cid:0) φ ( x ) − φ ( x e − τ ( x ) ) (cid:1) + φ ( x e − τ ( x ) ) √ nβ e − τ ( x ) ( τ ( x ) + 1) (cid:17) = − ( x + β/ √ n ) √ nβ φ ( x e − τ ( x ) ) e − τ ( x ) − (cid:0) φ ( x ) − φ ( x e − τ ( x ) ) (cid:1) − x φ ( x e − τ ( x ) ) √ nβ e − τ ( x ) τ ( x )= √ nβ φ ( x e − τ ( x ) ) e − τ ( x ) (cid:16) − ( x + β/ √ n ) − x τ ( x ) (cid:17) − (cid:0) φ ( x ) − φ ( x e − τ ( x ) ) (cid:1) = φ ( x e − τ ( x ) ) − (cid:0) φ ( x ) − φ ( x e − τ ( x ) ) (cid:1) = − φ ( x ) , x ∈ Ω , where in the second last inequality we used the fact that − ( x + β/ √ n ) e − τ ( x ) − x τ ( x ) e − τ ( x ) = − β/ √ n from Lemma 6. For x ∈ S ,( − x + x − β/ √ n ) f (2)1 ( x ) − x f (2)2 ( x )= ( − x + x − β/ √ n ) √ nβ e − τ ( x ) − x √ nβ e − τ ( x ) ( τ ( x ) + 1)= √ nβ e − τ ( x ) (cid:0) − x − β/ √ n − x τ ( x ) (cid:1) = − − φ ( x ) , where in the last equality we used the fact that x ∈ S implies x ≥ κ / √ n . Therefore, Lf (2) ( x ) = − φ ( x ) for all x ∈ Ω. Let us now verify that f (2)1 (0 , x ) = f (2)2 (0 , x ) . (C.10)From Lemma 6 we know that τ (0 , x ) = 0, which suggests that (C.10) holds for x ∈ S ∪ S . Furthermore, the only point in S with x = 0 is the point (0 , κ / √ n ), whichmeans that (C.10) holds for x ∈ S as well. Therefore, f (2) ( x ) solves (5.10).We now bound f (2)1 ( x ) and f (2)11 ( x ) to prove (5.16). The bound on f (2)1 ( x ) is straight-forward and the details are omitted. Differentiating f (2)1 ( x ), we see that f (2)11 ( x ) = , x ∈ S , , x ∈ S , − τ ( x ) √ nβ φ ( x e − τ ( x ) ) e − τ ( x ) − τ ( x ) x e − τ ( x ) √ nβ φ (cid:48) ( x e − τ ( x ) ) , x ∈ S , − τ ( x ) √ nβ e − τ ( x ) , x ∈ S . ecall from (4.12) that − τ ( x ) = e − τ ( x ) x e − τ ( x ) − β/ √ n . From (4.11), we know that x e − τ ( x ) ≥ κ / √ n for x ≥ Γ ( κ ) , which means that (cid:12)(cid:12)(cid:12) f (2)11 ( x ) (cid:12)(cid:12)(cid:12) ≤ √ nβ κ / √ n − β/ √ n = nβ ( κ − β ) , x ∈ S . Similarly, (cid:12)(cid:12)(cid:12) f (2)11 ( x ) (cid:12)(cid:12)(cid:12) ≤ nβ ( κ − β ) + e − τ ( x ) x e − τ ( x ) − β/ √ n x e − τ ( x ) √ nβ (cid:12)(cid:12)(cid:12) φ (cid:48) ( x e − τ ( x ) ) (cid:12)(cid:12)(cid:12) ≤ nβ ( κ − β ) + x e − τ ( x ) x e − τ ( x ) − β/ √ n √ nβ √ nκ − κ ≤ nβ ( κ − β ) + κ κ − β nβ ( κ − κ ) , x ∈ S , where in the second inequality we used (5.7) and in the last inequality we used the factthat x e − τ ( x ) ≥ κ / √ n for x ≥ Γ ( κ ) . This proves (5.16) and we now prove the boundon f (2) ( x ) in (5.14). From the form of f (2) ( x ), we see that f (2) ( x ) ≤ log( √ nx /κ ) ≤ log( κ /κ ) , x ≤ κ / √ n, x ∈ S ,f (2) ( x ) ≤ τ ( x ) + √ nβ ( x e − τ ( x ) − κ / √ n ) ≤ τ ( x ) + κ − κ β , x ≤ κ / √ n, x ∈ S . We do not need to bound f (2) ( x ) for x ∈ S , because from (4.11) we know that x ∈ S implies x ≥ κ / √ n . To bound τ ( x ) in the second line above, note from (4.11) that x e − τ ( x ) ≥ κ / √ n, x ≤ κ / √ n, x ∈ S , which means that τ ( x ) ≤ log( κ /κ ) , x ≤ κ / √ n, x ∈ S . This proves (5.14).
Proof of Lemma 12 .
Recall for convenience that f (2) ( x ) = , x ∈ S , (cid:82) log( √ nx /κ )0 φ ( x e − t ) dt, x ≤ κ / √ n, x ∈ S , log( √ nx /κ ) + (cid:82) log( κ /κ )0 φ (cid:0) κ √ n e − t (cid:1) dt, x ≥ κ / √ n, x ∈ S , (cid:82) τ ( x )0 φ ( x e − t ) dt + √ nβ (cid:82) x e − τ ( x ) κ / √ n φ ( t ) dt, x ≤ κ / √ n, x ∈ S , log( √ nx /κ ) + (cid:82) τ ( x )log( √ nx /κ ) φ (cid:0) x e − t (cid:1) dt + √ nβ (cid:82) x e − τ ( x ) κ / √ n φ ( t ) dt, x ≥ κ / √ n, x ∈ S ,τ ( x ) + x e − τ ( x ) − κ / √ nβ/ √ n + √ nβ (cid:82) κ / √ nκ / √ n φ ( t ) dt, x ∈ S , nd S = { x ∈ Ω | x ≤ κ / √ n } ,S = { x ∈ Ω | x ≥ κ / √ n, x ≤ Γ ( κ ) } ,S = { x ∈ Ω | Γ ( κ ) ≤ x ≤ Γ ( κ ) } ,S = { x ∈ Ω | x ≥ Γ ( κ ) } . Let us first verify the continuity of f (2) ( x ) on each of S i ∩ S j for i, j ∈ { , , , } ; theonly non-empty intersections are S ∩ S , S ∩ S , and S ∩ S . Continuity on S ∩ S is straightforward because S ∩ S = { x = κ / √ n } . Now S ∩ S = { x ∈ Γ ( κ ) } , and recall from (4.11) that x e − τ ( x ) = κ/ √ n or τ ( x ) = log( √ nx /κ ) , x ∈ Γ ( κ ) . (C.11)Therefore, we can compare the definitions of f (2) ( x ) on x ≤ κ / √ n, x ∈ S vs. x ≤ κ / √ n, x ∈ S and x ≥ κ / √ n, x ∈ S vs. x ≥ κ / √ n, x ∈ S to see that they coincide. Lastly, S ∩ S = { x ∈ Γ ( κ ) } , and recall that x ∈ Γ ( κ ) implies x ≥ κ / √ n . Therefore, we need only to compare the definitions of f (2) ( x ) on x ≥ κ / √ n, x ∈ S vs. x ∈ S and use (C.11) to conclude that f (2) ( x ) is continuous.Differentiating f (2) ( x ) and using the Leibniz integration rule, we get f (2)1 ( x ) = , x ∈ S , , x ∈ S ,τ ( x ) φ ( x e − τ ( x ) ) (cid:0) − √ nβ x e − τ ( x ) (cid:1) , x ∈ S ,τ ( x ) (cid:0) − √ nβ x e − τ ( x ) (cid:1) , x ∈ S , and f (2)2 ( x ) = , x ∈ S , x φ ( x ) , x ∈ S , x (cid:0) φ ( x ) − φ ( x e − τ ( x ) ) (cid:1) + τ ( x ) φ ( x e − τ ( x ) )+ √ nβ φ ( x e − τ ( x ) ) (cid:0) e − τ ( x ) − τ ( x ) x e − τ ( x ) (cid:1) , x ∈ S ,τ ( x ) + √ nβ (cid:0) e − τ ( x ) − τ ( x ) x e − τ ( x ) (cid:1) , x ∈ S . rom (4.12), we know that − τ ( x ) = e − τ ( x ) x e − τ ( x ) − β/ √ n , and τ ( x ) = τ ( x ) τ ( x ) , which proves the forms of f (2)1 ( x ) and f (2)2 ( x ) as stated in the Lemma. Let us now provethat the first order derivatives are continuous. We begin with f (2)1 ( x ), for which we haveto verify continuity on S ∩ S = { x ∈ Γ ( κ ) } and S ∩ S = { x ∈ Γ ( κ ) } . On the formerset, φ ( x e − τ ( x ) ) = φ ( κ / √ n ) = 0, and on the latter set, φ ( x e − τ ( x ) ) = φ ( κ / √ n ) = 1,which proves continuity of f (2)1 ( x ). We now prove continuity of f (2)2 ( x ). On S ∩ S = { x = κ / √ n } , f (2)2 ( x ) = x φ ( κ / √ n ) = 0. On S ∩ S = { x ∈ Γ ( κ ) } , continuityfollows from the fact that φ ( x e − τ ( x ) ) = φ ( κ / √ n ) = 0 and on S ∩ S = { x ∈ Γ ( κ ) } continuity follows both from the fact that both φ ( x e − τ ( x ) ) = φ ( κ / √ n ) = 1 and1 ≥ φ ( x ) ≥ φ ( x e − τ ( x ) ) = 1. Continuity of the second order partial derivatives isstraightforward to check using similar arguments, and the proof is omitted. References [1]
Atar, R. (2012). A diffusion regime with nondegenerate slowdown.
OperationsResearch , http://dx.doi.org/10.1287/opre.1110.1030 .[2] Billingsley, P. (1999).
Convergence of probability measures . 2nd ed. Wiley,New York.[3]
Bramson, M. (2011). Stability of join the shortest queue networks.
Ann. Appl.Probab. , https://doi.org/10.1214/10-AAP726 .[4] Braverman, A. and
Dai, J. G. (2017). Stein’s method for steady-state diffusionapproximations of M/ Ph /n + M systems. Annals of Applied Probability , http://dx.doi.org/10.1214/16-AAP1211 .[5] Braverman, A. , Dai, J. G. and
Feng, J. (2016). Stein’s method for steady-state diffusion approximations: an introduction through the Erlang-A and Erlang-C models.
Stochastic Systems , .[6] Budhiraja, A. and
Lee, C. (2009). Stationary distribution convergence forgeneralized Jackson networks in heavy traffic.
Math. Oper. Res. , http://dx.doi.org/10.1287/moor.1080.0353 .[7] Corless, R. M. , Gonnet, G. H. , Hare, D. E. G. , Jeffrey, D. J. and
Knuth, D. E. (1996). On the Lambert W function.
Advances in ComputationalMathematics , https://doi.org/10.1007/BF02124750 .[8] Down, D. , Meyn, S. P. and
Tweedie, R. L. (1995). Exponential and uniformergodicity of markov processes.
Ann. Probab. , https://doi.org/10.1214/aop/1176987798 . Eryilmaz, A. and
Srikant, R. (2012). Asymptotically tight steady-state queuelength bounds implied by drift conditions.
Queueing Systems , http://dx.doi.org/10.1007/s11134-012-9305-y .[10] Eschenfeldt, P. and
Gamarnik, D. (2015). Join the shortest queue withmany servers. the heavy traffic asymptotics. URL https://arxiv.org/abs/1502.00999 .[11]
Feng, J. and
Shi, P. (2017). Steady-state diffusion approximations for discrete-time queue in hospital inpatient flow management. URL https://arxiv.org/abs/1612.00790 .[12]
Flatto, L. and
McKean, H. P. (1977). Two queues in parallel.
Communi-cations on Pure and Applied Mathematics , https://dx.doi.org/10.1002/cpa.3160300206 .[13] Foschini, G. J. and
Salz, J. (1978). A basic dynamic routing problem anddiffusion.
IEEE Transactions on Communications , https://dx.doi.org/10.1109/TCOM.1978.1094075 .[14] Gamarnik, D. and
Stolyar, A. L. (2012). Multiclass multiserver queueingsystem in the Halfin-Whitt heavy traffic regime: asymptotics of the stationarydistribution.
Queueing Systems , http://dl.acm.org/citation.cfm?id=2339029 .[15] Gamarnik, D. and
Zeevi, A. (2006). Validity of heavy traffic steady-stateapproximation in generalized Jackson networks.
Ann. Appl. Probab. , http://dx.doi.org/10.1214/105051605000000638 .[16] Gast, N. (2017). Expected values estimated via mean-field approximation are1/n-accurate.
Proc. ACM Meas. Anal. Comput. Syst. , http://doi.acm.org/10.1145/3084454 .[17] Gast, N. and
Van Houdt, B. (2017). A Refined Mean Field Approximation.
Proceedings of the ACM on Measurement and Analysis of Computing Systems , . URL https://hal.inria.fr/hal-01622054 .[18] Glynn, P. W. and
Zeevi, A. (2008). Bounding stationary expectations ofMarkov processes. In
Markov processes and related topics: a Festschrift forThomas G. Kurtz , vol. 4 of
Inst. Math. Stat. Collect.
Inst. Math. Statist., Beach-wood, OH, 195–214. URL http://dx.doi.org/10.1214/074921708000000381 .[19]
Gupta, V. and
Walton, N. (2017). Load balancing in the non-degenerateslowdown regime. URL https://arxiv.org/abs/1707.01969 .[20]
Gurvich, I. (2014). Diffusion models and steady-state approximations for ex-ponentially ergodic Markovian queues.
The Annals of Applied Probability , http://dx.doi.org/10.1214/13-AAP984 . Gurvich, I. (2014). Validity of heavy-traffic steady-state approximations in mul-ticlass queueing networks: the case of queue-ratio disciplines.
Mathematics of Op-erations Research , http://dx.doi.org/10.1287/moor.2013.0593 .[22] Haight, F. A. (1958). Two queues in parallel.
Biometrika , http://dx.doi.org/10.1093/biomet/45.3-4.401 .[23] Halfin, S. (1985). The shortest queue problem.
Journal of Applied Probability , .[24] Halfin, S. and
Whitt, W. (1981). Heavy-traffic limits for queues with manyexponential servers.
Oper. Res. , Hanqin, Z. and
Rongxin, W. (1989). Heavy traffic limit theorems for a queueingsystem in which customers join the shortest line.
Advances in Applied Probability , .[26] Harrison, J. M. and
Reiman, M. I. (1981). Reflected Brownian motion on anorthant.
Ann. Probab. , http://links.jstor.org/sici?sici=0091-1798(198104)9:2<302:RBMOAO>2.0.CO;2-P&origin=MSN .[27] Henderson, S. G. (1997).
Variance reduction via an approximating Markovprocess . Ph.D. thesis, Department of Operations Research, Stanford University. http://people.orie.cornell.edu/shane/pubs/thesis.pdf .[28]
Huang, J. and
Gurvich, I. (2016). Beyond heavy-traffic regimes: universalbounds and controls for the single-server queue. Submitted for publication, URL http://papers.ssrn.com/sol3/papers.cfm?abstract_id=2784752 .[29]
Katsuda, T. (2010). State-space collapse in stationarity and its application to amulticlass single-server queue in heavy traffic.
Queueing Syst. , http://dx.doi.org/10.1007/s11134-010-9178-x .[30] Kingman, J. F. C. (1961). Two similar queues in parallel.
The Annals ofMathematical Statistics , Maguluri, S. T. , Burle, S. K. and
Srikant, R. (2016). Optimal heavy-trafficqueue length scaling in an incompletely saturated switch.
SIGMETRICS Perform.Eval. Rev. , http://doi.acm.org/10.1145/2964791.2901466 .[32] Maguluri, S. T. and
Srikant, R. (2016). Heavy traffic queue length behaviorin a switch under the maxweight algorithm.
Stochastic Systems , http://dx.doi.org/10.1214/15-SSY193 .[33] Mukherjee, D. , Borst, S. C. , van Leeuwaarden, J. S. H. and Whiting,P. A. (2016). Universality of load balancing schemes on the diffusion scale.
J.Appl. Probab. , https://projecteuclid.org:443/euclid.jap/1481132840 . Stein, C. (1986). Approximate computation of expectations.
Lecture Notes-Monograph Series , . URL .[35] Stolyar, A. L. (2015). Pull-based load distribution in large-scale heterogeneousservice systems.
Queueing Systems , https://doi.org/10.1007/s11134-015-9448-8 .[36] Stolyar, A. L. (2015). Tightness of stationary distributions of a flexible-serversystem in the Halfin-Whitt asymptotic regime.
Stoch. Syst. , http://dx.doi.org/10.1214/14-SSY139 .[37] Tezcan, T. (2008). Optimal control of distributed parallel server systems underthe Halfin and Whitt regime.
Mathematics of Operations Research , http://search.proquest.com/docview/212618995?accountid=10267 .[38] van der Boor, M. , Borst, S. C. , van Leeuwaarden, J. S. H. and Mukher-jee, D. (2017). Scalable load balancing in networked systems: Universality prop-erties and stochastic coupling methods. URL https://arxiv.org/abs/1712.08555 .[39]
Ye, H.-Q. and
Yao, D. D. (2012). A stochastic network under proportional fairresource control—diffusion limit with multiple bottlenecks.
Operations Research , http://dx.doi.org/10.1287/opre.1120.1047 .[40] Ying, L. (2016). On the approximation error of mean-field models. In
Proceedingsof the 2016 ACM SIGMETRICS International Conference on Measurement andModeling of Computer Science . ACM, Antibes Juan-les-Pins, France, 285–297.URL http://dx.doi.org/10.1145/2964791.2901463 .[41]
Ying, L. (2017). Stein’s method for mean field approximations in light andheavy traffic regimes.
Proc. ACM Meas. Anal. Comput. Syst. , http://doi.acm.org/10.1145/3084449 .[42] Zhang, H. , Hsu, G. H. and
Wang, R. (1995). Heavy traffic limit theorems fora sequence of shortest queueing systems.
Queueing Systems , https://doi.org/10.1007/BF01158582 .[43] Zhang, J. and
Zwart, B. (2008). Steady state approximations of limited proces-sor sharing queues in heavy traffic.
Queueing Systems: Theory and Applications , http://dx.doi.org/10.1007/s11134-008-9095-4 ..