On the stability of the linear Skorohod problem in an orthant

On the stability of the linear Skorohod problem in an orthant

Ahmed El Kharroubi, Abdelghani Ben Tahar and Abdelhak Yaacoubi

Universite´ Hassan II, Faculte´ des Sciences Ain Chock, B.P. 5366, Maarif, Casablanca, Maroc (e-mail: elkharroubi@facsc-achok.ac.ma)

Manuscript received: June 1999/Final version received: January 2002

Abstract. Dupuis and Williams proved that a su‰cient condition for the pos- itive recurrence for a semimartingale reflecting Brownian motion in an orthant (SRBM ) with data ðy; R; S; DÞ, is that the corresponding Linear Skorohod Problem LSPðyÞ is stable. In this paper we use the linear complementary prob- lem to give necessary conditions, on y A R

and matrix R, under which the linear Skorohod problem LSPðyÞ is stable. In the three dimensional case we characterize the vectors y A R

such that the LSPðyÞ is stable.

Key words: Skorohod problem, Stability, Linear complementary problem

1 Introduction

The solution of the Skorohod problem define a useful deterministic mapping of paths. One of the areas of application is the positive recurrence of semi- martingale reflecting Brownian motions in an orthant (SRBMs). Under suit- able conditions, the latter have been shown to arise as heavy tra‰c approxi- mations for multiclass queueing networks. Hence their recurrence behavior is of interest for applications. Let n be a positive integer, R be an n n matrix with column vectors r

; . . . ; r

, such that r

¼ 1 for all i ¼ 1; . . . ; n. S denotes the non negative orthant of R

: S ¼ fu A R

: u

b 0; i ¼ 1; . . . ; ng. Let x ¼ ðx

; t b 0Þ be a continuous function from R

into R

with x

A S (called tra- jectory with initial state in S). Then the pair ð y; zÞ of trajectories is said to solve the Skorohod problem for x, or simply, said to solve SPðxÞ, if they jointly satisfy

1) z

¼ x

þ Ry

A S for all t b 0.

2) for all i ¼ 1; . . . ; n, the i

component y

of y is nondecreasing with y

¼ 0,

and increases only on ft b 0 : z

¼ 0g.

As in [3], the Skorohod problem SPðxÞ is said to be a linear Skorohod problem with rate y if x ¼ ðx

þ yt; t b 0Þ; with x

A S and y A R

. This prob- lem is denoted by LSPðyÞ. In [2] it was shown that a necessary and su‰- cient condition for the existence of solution to SPðxÞ is that the matrix R be completely-S or S-matrix (see definition 2). Let D be a non degenerate co- variance matrix, according to the terminology of M. I. Rieman and R. J.

Williams [9], a semimartingale reflected Brownian Motion SRBM with data ðy; R; D; SÞ is a continuous n-dimensional process with state space S. On the interior of S this process behaves like a Brownian Motion with drift y and covariance matrix D, and reflected instantaneously at the boundary of S, where the direction of reflection on the i

face F

¼ fx A S : x

¼ 0g is given by r

, i ¼ 1; . . . ; n. ( for a precise mathematical definition see [9]). In [11] it was shown that a necessary and su‰cient condition for the existence and unique- ness of this process is that R be completely-S or S-matrix.

The results of this paper are motivated by the works of Dupuis and Wil- liams [8] and Chen [3]. The first authors proved that an SRBM with data ðy; R; D; SÞ is positive recurrent and has a unique stationary distribution if the corresponding linear Skorohod problem LSPðyÞ is stable, in other hand H.

Chen proved a su‰cient condition for the stability of LSPðyÞ, and hence posi- tive recurrence and the existence of a unique stationary distribution for the SRBM.

In this article we are also interested in the study of the LSPðyÞ. We seek to establish necessary and su‰cient conditions for its stability. Our approach is based on the existing link between the linear Skorohod problem LSPðyÞ and the linear complementary problem LCP associated with the matrix R and the vector y. This last problem is stated as follows:

Find v and w in R

such that:

w ¼ y þ Rv w b 0; v b 0

v

w

¼ 0 for all i A f1; . . . ; ng : 8 <

:

In the literature (see for example [4]) the matrix R is said to be a Q-matrix if the LCP has a solution for every y A R

. In this case the complementary cones fG

; J H Ig is a collection whose union contains R

, where G

is the closed convex cone generated by the vectors fðr

Þ

; ðe

Þ

g. (ðe

Þ

is the canonical euclidean basis). In section 3 we show that if the matrix R is completely-S and if the LSPðyÞ is stable then we have

y A G

6

JHI;J0q

G

R is invertible:

8 >

<

> : ð1Þ

The first condition means that y A G

and y B G

for all nonempty subset J J I.

(G

is the interior of the cone G generated by r

; . . . ; r

). In section 4 we are concerned with the stability of LSPðyÞ in the three dimensional case. First, we show by an example that the condition (1) is not su‰cient for the stability of LSPðyÞ (see example 4.1). Next we characterize y A G

6

JHI;J0q

G

for which

the LSPðyÞ is stable (Theorem 1). The general case remains an open problem.

2 Definitions and preliminaries

Let n be a positive integer, let I ¼ f1; . . . ; ng, let K J I, jKj denotes the car- dinality (size) of K. For an n n matrix R and K; J J I, R

is an jJ j jKj matrix whose elements are from R with row indices in J and column indices in K. R

is short for the principal submatrix R

. A vector u A R

will be treated as a column vector with components u

for i ¼ 1; . . . ; n. We write u > 0 (resp u b 0) if and only if each component u

is positive (resp non- negative), and u

denotes the vector whose components are those of u with indices in K. If K ¼ fig we write u

instead of u

. Let r

; . . . ; r

be the column vectors of the matrix R, and ðe

; i A IÞ be the canonical euclidean basis. For K J J J I, G

is the closed convex cone generated by the whole of the vectors fðe

Þ

; ðr

Þ

g.

G

¼ X

kAK

l

e

X

kAJnK

l

r

nl

b 0 for all j A J 8 <

:

9 =

; :

If J ¼ I, then G

is a complementary cone which is simply denoted by G

, and G is short for G

. CðR

; R

Þ denotes the set of continuous mapping from R

into R

.

Definition 1. Let x A CðR

; R

Þ (called trajectory) with x

A S then the pair ðy; zÞ A CðR

; R

Þ CðR

; R

Þ is said to solve the Skorohod problem for x or simply said to solve SPðxÞ if they jointly satisfy

i) z

¼ x

þ Ry

A S for all t b 0.

ii) For all i A I , the i

component y

of y is nondecreasing with y

¼ 0, and increases only when z

¼ 0.

Definition 2. Let R be an n n matrix. R is said to be completely-S (or S- matrix) if and only if for each principal submatrix R R of R there exists a non- ~ negative vector u u such that ~ R R~ ~ u u > 0.

Throughout this paper the pair ðy; zÞ solving SPðxÞ is referred to as a R- regulation of x, and ðz

; t b 0Þ as a R-regulated trajectory of x. It was shown in Bernard and El Kharroubi [2] that R is completely-S is a neces- sary and su‰cient condition for the existence of a solution to SP, but the uniqueness does not always hold. If the trajectory x ¼ ðx

; t b 0Þ has the form ðx

¼ x

þ yt; t b 0Þ with x

A S and y A R

, the SPðxÞ will be called as in [3], linear Skorohod problem, and is denoted by LSPðyÞ. We recall the following properties of a solution to the linear Skorohod problem. Let ðy; zÞ be a solution to SPðxÞ with x

A S. Then we have the well known properties:

Shift. For every t > 0 the pair of trajectories ð y y; ~ zzÞ ~ defined by y ~

y

¼ y

y

~ zz

¼ z

is a solution to SPðxÞ with initial state z

.

Scaling. For every e > 0, the pair of trajectories ðy; zÞ defined by y

¼

z

¼

(

is a solution to SPðxÞ with initial state

. Definition 3 [8].

i) A function z A CðR

; R

Þ is said to be attracted to the origin if for every e > 0 there exists a T < þ y such that kz

k < e for all t b T .

ii) A linear Skorohod problem LSPðyÞ with initial state x

A S is said to be stable if the z component of all of its solutions is attracted to the origin. If LSPðyÞ is stable for every initial state x

A S, then we simply say that LSPðyÞ is stable.

3 Stability of the linear Skorohod problem:necessary conditions

In this section we give necessary conditions for the stability of linear Skorohod problem LSPðyÞ.

Proposition 1. Assume that R is completely-S. Let y A R

. Suppose that the linear Skorohod problem LSPðyÞ is stable. Then the following tow conditions hold:

i) R is invertible.

ii) y A L ¼ G

6

JHI;J0q

G

.

Proof. i) Since R is completely-S, it is also Q-matrix [9]. By Theorem 1 in cottle [4], R is a strictly semi-monotone matrix and thus by Theorem 7 in Cottle and Dantzig [5] the LCP

w ¼ y þ Rv w b 0; v b 0

v

w

¼ 0 for all i A f1; . . . ; ng;

8 <

:

has a basic complementary feasible solution ðv; wÞ for every vector y. Then the linear function ðyt; t b 0Þ has the following R-regulation:

y

¼ vt z

¼ wt:

From the stability of LSPðyÞ we have necessary w ¼ 0 and then y ¼ Rv.

Thus

y A G : ð2Þ

We have to show that

v > 0: ð3Þ

Denote J ¼ fi A I : v

¼ 0g and suppose that J 0 q . For x

A S such that x

> 0 and x

¼ 0 ðJ ¼ InJÞ, the function ðx

þ yt; t b 0Þ with initial state x

has a R-regulation ðy; zÞ given by:

y

¼ vt z

¼ x

:

But this contradicts the stability of LSPðyÞ. Thus ðv; 0Þ is a basic comple- mentary feasible solution to LCP, hence R is invertible.

ii) We conclude from (2) and (3) that y A G

. It remains to prove that y B G

for all J J I with J 0 q . Suppose that there is a nonempty subset J

J I such that y A G

, it follows from the definition of G

that there are l

; . . . ; l

b 0 such that

y ¼ X

jAJ0

l

e

X

jAJ0

l

r

:

For x

A S such that x

> 0 and x

¼ 0, the z component of the following solution to LSPðyÞ defined by

y

¼ 0 for j A J

l

t for j A J

and z

¼ x

þ l

t for j A J

0 for j A J

is not attracted to the origin. r

Remark 1. The proof of part i) of this proposition can be proved by quoting existing known results. First if the linear Skorohod problem LSPðyÞ is sta- ble then by Theorem 2.6 of Dupuis and Williams [8] the SRBM with data ðy; R; D; SÞ is positive recurrent and has a unique stationary distribution. Next, by Proposition 2 in Dai and Harrison [6] the matrix R is invertible.

The following corollary is an immediate consequence of the Proposition 1 and the Corollary 2.8 in Chen [3]. Recall that an n n matrix R is Schur-S if all its principal submatrices are non singular and there exists a positive vector v such that

ðv

Þ

ðR

R

ðR

Þ

R

Þ > 0;

for any J J I (J ¼ InJ,

is transpose).

Corollary 1. Suppose that R is both completely-S and Schur-S, then the LSPðyÞ is stable if and only if

y A G

:

The next proposition establish the same result for the class of admissible ma- trices. This class of matrices is used in the study of some queueing networks (see for example the article of Dai and Harrison [7]). A matrix R is said to be admissible if there is a positive diagonal matrix D such that DR þ R

D is positive definite. It is well known that admissible matrices is a subclass of P- matrices (see Theorem 2.3, p. 134, in Berman and Plemmons [1]).

Proposition 2. Assume that R is an n n admissible matrix. Then the LSPðyÞ is stable if and only if

y A G

:

Proof. The if is obtained from Proposition 1. It remains the only if. Since R is admissible, there exists a positive diagonal matrix D such that DR is positive definite. By Theorem 2.5 in Chen [3] it su‰ces to find a vector h such that, given a subset J of I

ðh

Þ

ðy

R

ðR

Þ

y

Þ < 0 whenever ðR

Þ

y

a 0:

Let R R ~ ¼ DR and y y ~ ¼ Dy. Let J J I such that ðR

Þ

y

a 0. From Schur’s formula:

ð R R ~

y yÞ ~

¼ ð R R ~

R R ~

J

ð R R ~

Þ

R R ~

Þ

ð y y ~

R R ~

J

ð R R ~

Þ

y y ~

Þ : ð4Þ One can check that if an n n matrix R is positive definite then so is R

and all its principal submatrices. Therefore the matrix ð R R ~

R R ~

J

ð R R ~

Þ

R R ~

Þ is posi- tive definite. Then

ðð R R ~

y yÞ ~

Þ

ð R R ~

R R ~

J

ð R R ~

Þ

R R ~

Þð R R ~

y yÞ ~

¼ ðð R R ~

y yÞ ~

Þ

ð y y ~

R R ~

J

ð R R ~

Þ

y y ~

Þ

¼ ððR

yÞ

Þ

D

ðy

R

ðR

Þ

y

Þ

> 0:

Thus with h ¼ DR

y we have ðh

Þ

ðy

R

J

ðR

Þ

y

Þ < 0 which proves the stability. r

Remark 2. It was claimed in Chen [3] that in two dimensional case all P- matrices are Schur-S. We have the converse in the following sense, if R is both completely-S and Schur-S, then it is a P-matrix. Thus in two dimensional case the LSPðyÞ is stable if and only if y A G

and R is a P-matrix.

4 Stability of the linear Skorohod problem in three dimensional case

The purpose of this section is to investigate the following problem: For which vectors y A G

is the LSPðyÞ stable?

We begin by studying the following example.

4.1 Example

Let a be a real number, let RðaÞ be the matrix:

0 B @

1 a 0

0 1 a

a 0 1 1 C A

One verifies easily that the matrix RðaÞ is completely-S if and only if a > 1.

In fact it is a P-matrix in this case. We verify also that RðaÞ is a Minkowski matrix if and only if 1 < a a 0, Schur-S if and only if 1 < a < 1 and ad- missible if and only if 1 < a < 2. Let y ¼ ð1; 1; 1Þ

and x

> 0. Con- sider the linear Skorohod problem LSPðyÞ with initial state x

¼ ð0; 0; x

Þ

.

. _{For 1} < a a 1 this problem has the unique solution ðy; zÞ defined by:

for 0 a t a t

¼

y

¼ 0 B @

ð1 aÞt t 0

1 C A ; z

¼

0 B @

0 0

x

þ ðað1 aÞ 1Þt 1 C A

and for t b t

y

¼ t t

1 þ a y þ y

; z

¼ 0:

. For a > 1, the problem has the unique solution ðy; zÞ defined by:

On the interval ½t

; t

with t

t

¼ ða 1Þ

x

and p b 0

y

¼ 0 B @

0 t t

0 1

C A þ y

; z

¼ 0 B @

ða 1Þðt t

Þ 0

ða 1Þ

x

ðt t

Þ 1 C A ;

on the interval ½t

; t

with t

t

¼ ða 1Þ

x

y

¼ 0 B @

0 0 t t

1

C A þ y

; z

¼ 0 B @

ða 1Þ

x

ðt t

Þ ða 1Þðt t

Þ

0

1 C A

and on the interval ½t

; t

with t

t

¼ ða 1Þ

x

y

¼

t t

0 0 0

@

1

A þ y

; z

¼ 0 B @

0

ða 1Þ

x

ðt t

Þ ða 1Þðt t

Þ

1

C A :

So we have z

¼ ð0; 0; z

Þ

where z

¼ ða 1Þ

x

. Then the R- regulated trajectory ðz

; t b 0Þ behaves like:

i) either a spiral going to the origin, if 1 < a < 2:

ii) or a spiral going to infinite, if a > 2:

iii) or a loop if a ¼ 2:

Fig. 1

Fig. 2

Fig. 3

Then the linear Skorohod problem LSPðyÞ for y ¼ ð1; 1; 1Þ

is stable if and only if 1 < a < 2.

Remark 3. Taking account of the result of the Corollary 1 above, this example shows that in d-dimensional case ðd b 3Þ, any P-matrix is not necessarily Schur-S. Indeed one has for a > 2, RðaÞ is a P-matrix, y ¼ ð1; 1; 1Þ

belongs to G

and yet LSPðyÞ is not stable. This answers to the question in Chen ([3] Remark 3, pp 762). However we have the following proposition in the three dimensional case.

Proposition 3. Let R be an 3 3 matrix. If R is both completely S and Schur- S, then it is a P-matrix.

Proof. R is Schur-S, so there exists a positive vector u ¼ ðu

; u

; u

Þ

such that ðu

Þ

ðR

R

J

ðR

Þ

R

Þ > 0 ð5Þ

for any J J I. For J ¼ fig, we have 1 r

ðR

Þ

r

> 0:

Thus

ðR

Þ

¼ det R

det R ¼ ð1 r

ðR

Þ

r

Þ

> 0:

Which means that all minors of second order have the same mark as det R. If we assume that R is not P-matrix, then all minors of second order are negative (since all minors of first order are positive). Hence, from the characterization of S-matrices in tow dimensional case R is a nonnegative matrix. From (5) with fi; j; kg ¼ f1; 2; 3g we have

u

ð1 r

r

Þ þ u

ðr

r

r

Þ > 0

and then r

> r

r

. Multiplying this inequality by the positive number r

, we obtain

r

r

> r

r

r

> r

;

which is impossible. r

4.2 R-regulation of an a‰ne trajectory

Let R be an 3 3 completely-S matrix, let y be a vector in R

and let ðy; zÞ

be a solution to linear Skorohod problem LSPðyÞ. We purpose in this para-

graph to characterize vectors y in G

for which the R-regulated trajectories

behave like a spiral ( figure 1 or figure 2) or like a loop ( figure 3). We begin by

introducing the following tow cones:

C

¼ fu A R

: u < 0; R Ru ^ < 0; R Ru ~ > 0g : C

¼ fu A R

: u < 0; R Ru ^ > 0; R Ru ~ < 0g:

Where R R ^ and R R ~ are tow matrices deduced from R

R ~ R ¼

1 r

0

0 1 r

r

0 1 0

B @

1 C A ; R R ^ ¼

1 0 r

r

1 0 0 r

1 0

B @

1 C A

If R is a Minkowski matrix then C

and C

are empty. But when R is a P- matrix, one can check easily that C

(resp C

) is non empty if and only if the matrix ðI R RÞ ~ (resp ðI R RÞ) is nonnegative and det ^ R R ~ < 0 (resp det R R ^ < 0Þ. (I is the identity matrix). For the matrix RðaÞ in the previous example, the cone C

is non empty if and only if a > 1, ðy ¼ ð1; 1; 1Þ

A C

Þ, while C

is empty. Before stating the main result of this paragraph, remind that

L ¼ G

6

JHI;J0q

G

:

Proposition 4. Let y A L and let x

A qSnf0g. Assume that R is an 3 3 in- vertible S-matrix. Then the following tow assertions are equivalent:

i) y A C

W C

,

ii) Every R-regulated trajectory ðz

; t b 0Þ of ðx

þ yt; t b 0Þ behaves like a spiral (figure 1, and figure 2), or a loop (figure 3).

The proof of this proposition is based on the following pair of lemmas:

Lemma 1. The following conditions are equivalent:

i) y A C

W C

,

ii) y A L and y

B G

, for all subset J H I with jJ j ¼ 2.

Proof. We have

. for every subset J ¼ fi; jg H I ¼ f1; 2; 3g, with i 0 j

y

A G

) ðy

r

y

Þðy

r

y

Þ b 0 ð6Þ

. and

y A C

W C

,

ðy

r

y

Þðy

r

y

Þ < 0 for all i; j A I; i 0 j y < 0

y A L 8 <

: ð7Þ

The only if follows immediately from (6) and (7).

For the if, since y B 6

JHI;J0q

G

and y

B G

for every J H I and jJ j ¼ 2,

one deduces easily that y A C

W C

. r

Lemma 2. Assume that y A LnC

W C

. Then there exists a subset J

H I , jJ

j ¼ 2 such that the principal submatrix R

is a P-matrix and y

A G

. Proof. Let y A LnC

W C

. From Lemma 1, there exists J

H I with jJ

j ¼ 2 such that y

A G

. We have the result if R

is a P-matrix. Otherwise, denoting J

¼ fi

; i

g, J

¼ fi

; i

g and J

¼ fi

; i

g with fi

; i

; i

g ¼ f1; 2; 3g.

Then det R

¼ 1 r

r

a 0. R

is completely-S so r

> 0 and r

> 0.

Then y

A G

leads to y

r

y

b 0 y

r

y

b 0 y

a 0 8 >

<

> : ð8Þ

join those relations (8) with the fact that y B ðG

W G

Þ we obtain:

y

r

y

< 0 y

r

y

< 0 (

ð9Þ

so we have the result if the principal submatrix R

is a P-matrix and y

A G

. Otherwise, thanks to relations (8), (9) and the fact that R

is completely-S, we have necessarily y

A G

. Which gives

y

a 0 y

r

y

b 0 (

ð10Þ

In the end, from the relations (8)–(10) one has necessarily y

B ðG

W G

W G

Þ ;

and since R

is completely-S, then y

A G

and det R

> 0. r Proof of Proposition 4. Fix y A C

W C

and x

A qSnf0g. Since C

is disjoint from C

, we suppose that y A C

. It follows from the definition of C

that

ð

Þ

y

r

y

> 0 y

r

y

> 0 y

r

y

> 0 8 >

<

> :

y

r

y

< 0 y

r

y

< 0 y

r

y

< 0 8 >

<

> :

y

< 0 y

< 0 y

< 0 8 <

:

let I ðx

Þ ¼ fi A I : x

¼ 0g, x

belongs to, either a face ðjIðx

Þj ¼ 1Þ or a line ðjIðx

Þj ¼ 2Þ. We distinguish two cases

Case 1. jI ðx

Þj ¼ 2. We suppose without loss of generality that Iðx

Þ ¼ f1; 2g.

Let ðy; zÞ be a solution to linear Skorohod problem LSPðyÞ. On an enough small period of time we have

z

¼ y

t þ R

y

z

¼ x

þ y

t þ r

y

þ r

y

(

Because y

B G

, y

and y

may not increase simultaneously. The fact that y

r

y

< 0 prevents y

to increase during this period of time, then only y

increases on ½0; t

with t

¼

y³r₂³y²

. Thus for 0 a t a t

y

¼ 0 B @

0 y

t

0 1 C A ; z

¼

ðy

r

y

Þt 0

x

þ ðy

r

y

Þt 0

B @

1 C A

and z

¼ ðz

; 0; 0Þ, where z

¼ ðy

r

y

Þ

ðy

r

y

Þ x

: ð11Þ

We have Iðz

Þ ¼ fi A I : z

¼ 0g ¼ f2; 3g. Since y

B G

1Þ;qÞ

and y

r

y

< 0, only y

increases on ½t

; t

with t

¼ t

1t1

y¹r¹₃y³

. Thus on ½t

; t

: y

¼

0 B @

0 y

t

ðt t

Þy

1 C A ; z

¼

0 B @

z

þ ðy

r

y

Þðt t

Þ ðy

r

y

Þðt t

Þ

0

1 C A

and z

¼ ð0; z

; 0Þ, where z

¼ ðy

r

y

Þ

ðy

r

y

Þ z

: ð12Þ Similarly since Iðz

Þ ¼ fi A I : z

¼ 0g ¼ f1; 3g, y

B G

2Þ;qÞ

and y

r

y

< 0, only y

increases on ½t

; t

where t

¼ t

2 t2

y²r₁²y¹

. Therefore

y

¼

ðt t

Þy

y

t

ðt

t

Þy

0

B @

1 C A ; z

¼

0

z

þ ðy

r

y

Þðt t

Þ ðy

r

y

Þðt t

Þ 0

B @

1 C A

and z

¼ ð0; 0; z

Þ with z

¼ ðy

r

y

Þ

ðy

r

y

Þ z

: ð13Þ One deduces from (11)–(13) that

z

¼ b

ðyÞ x

; where

b

ðyÞ ¼ ðy

r

y

Þðy

r

y

Þðy

r

y

Þ

ðy

r

y

Þðy

r

y

Þðy

r

y

Þ : ð14Þ

By iterating this procedure, we obtain at the p

iteration

z

¼ ð b

ðyÞÞ

x

: ð15Þ

Thus the R-regulated trajectory ðz

; t b 0Þ behaves like, either a spiral if b

ðyÞ 0 1 ( figure 1, figure 2), or a loop if b

ðyÞ ¼ 1 ( figure 3).

Case 2. jIðx

Þj ¼ 1. We suppose that I ðx

Þ ¼ f1g. Since y

< 0, it follows from (*) that

y

¼ y

t

0 0 0 B @

1 C A ; z

¼

0

x

þ ðy

r

y

Þt x

þ ðy

r

y

Þt 0

B @

1 C A

for t a t

¼

y²r²₁y¹

. Because jIðz

Þj ¼ 2, then we end up in the first case with initial state z

¼ ð0; 0; z

Þ. By the same argument, if y A C

, the R-regulated trajectory ðz

; t b 0Þ is, either a spiral if b

ðyÞ 0 1, or a loop if b

ðyÞ ¼ 1, where

b

ðyÞ ¼ 1

b

ðyÞ : ð16Þ

Conversely, assume y B C

W C

. From lemma 2 there exists J

H I; jJ

j ¼ 2, such that y

A G

and R

is a P-matrix. Let x

A qSnf0g with x

> 0 and x

¼ 0; the pair ðy; zÞ defined by

on the interval ½0; t

, with t

¼

y^J0R_J^J0

0ðRJ0Þ¹y^J0

y

¼ ðR

Þ

y

t y

¼ 0

(

; and z

¼ 0

z

¼ x

þ ðy

R

ðR

Þ

y

Þt (

and on ½t

; þ y

y

¼ R

yt; and z

¼ 0

is a solution to LSPðyÞ, which is neither a spiral nor a loop. r

In the following proposition, we show that if y A LnðC

W C

Þ, all R-regulated trajectories reach the origin in finite time.

Proposition 5. Suppose y A LnðC

W C

Þ and let T ¼ inf ft : z

¼ 0g. Then for every x

A Snf0g and every R-regulated trajectory z ¼ ðz

; t b 0Þ of ðx

þ yt;

t b 0Þ there exists a positive constant C depending only on R and y such that

T a Ckx

k: ð17Þ

Proof. Recall that t

b 0 is a time of face change for the R-regulated trajectory

z, if t

¼ 0; or for all e > 0 , there exist s A t

e; t

½ and i A f1; 2; 3g such that

z

0 0 and z

¼ 0 (see [2]). Let x

A Snf0g, let ð y; zÞ be a solution to LSPðyÞ, and let t

and t

be tow consecutive times of face change for the R-regulated trajectory z such that z

0 0. First, we show that there exists a positive con- stant k

such that

t

t

a k

kz

k: ð18Þ

Denote Iðz

Þ ¼ fi A I n : z

¼ 0g. Let K J Iðz

Þ such that y

A G

, hence for t

a t a t

the R-regulated trajectory is written by

z

z

z

0 B B

@

1 C C A ¼

0 y

ðt t

Þ þ R

K

y

z

þ y

ðt t

Þ þ R

K

y

0

B B

@

1 C C

A ; ð19Þ

where K ¼ Iðz

ÞnK, Iðz

Þ ¼ InIðz

Þ and t

¼ infft > t

: z

¼ 0 for some i A Iðz

Þg. Since y B G

tpÞWK

there exists i A Iðz

Þ such that y

ðt t

Þ þ r

K

y

< 0 for all t > t

; consequently t

is finite. Define S

¼ fv A R

jKj þ

: y

þ R

K

v b 0; y

þ R

v ¼ 0g, if K 0 q ; and S

¼ f0g. Let e

be the maximum of ðy

þ r

K

vÞ over all v A S

and all indices i A Iðz

Þ such that y

þ r

K

v < 0. Because y A L and the matrix R

is completely-S, we have e

> 0. Now let i

A Iðz

Þ such that

z

þ y

ðt

t

Þ þ r

K

y

¼ 0;

it follows that

ðt

t

Þ ¼ 1 y

þ r

K y^K_tpþ1 ðtpþ1tpÞ

z

:

According to the first two equations in (19) one has

pþ1tpÞ

y

A S

. Then (18) holds with k

¼

K

. Since y A LnðC

W C

Þ the trajectory z is neither a spiral nor a loop (Proposition 4). Then there exist q A f1; . . . ; 4g and T > t

such that z is a‰ne on ½t

; þ y ½ and z

¼ 0. By Lemma 1 in Bernard and El Kharroubi [2], there exists a constant c such that

kz

z

k a cðt

t

Þ

for all p A f1; . . . ; 4g. From the last inequality and (18), one deduces the con- stant C in (17). r

4.3 Main result

Theorem 1. Let R be an 3 3 matrix invertible and completely-S. We have:

1) If y A C

(resp y A C

), then the LSPðyÞ is stable if and only if b

ðyÞ < 1 (resp b

ðyÞ < 1).

2) If y A LnðC

W C

Þ, then the LSPðyÞ is stable.

Proof. 1) Let y A C

such that b

ðyÞ < 1. We show that the condition of The- orem 2.5 in Chen [3] holds, i.e. there exists a strictly positive vector h such that, for all J J I, we have

ðh

Þ

ðy

þ R

J

uÞ < 0 ð20Þ

for all u A fv A R

jJj

þ

: y

þ R

v ¼ 0g. From Lemma 1, fv A R

jJj þ

: y

þ R

v ¼ 0g ¼ q for all J H I; jJ j ¼ 1 and y < 0. Then it su‰ces to find h > 0 such that the following inequalities hold

h

ðy

r

y

Þ þ h

ðy

r

y

Þ < 0 h

ðy

r

y

Þ þ h

ðy

r

y

Þ < 0 h

ðy

r

y

Þ þ h

ðy

r

y

Þ < 0:

8 >

<

> : ð21Þ

The matrix

M ¼

1

2r₃²y³

y¹r₃¹y³

0

0 1

3r₁³y¹

y²r₁²y¹ y¹r₂¹y²

y³r₂³y²

0 1 0

B B B B B B

@

1 C C C C C C A

is M-matrix (I-M is positive with spectral radius ðb

ðyÞÞ

< 1). Hence there exists h ¼ ðh

; h

; h

Þ

> 0 such that Mh > 0 and (21) holds. Conversely, let x

¼ ð0; 0; x

Þ

with x

> 0. From (15) it appears clear that the stability of LSPðyÞ implies b

ðyÞ < 1.

2) Let y A LnðC

W C

Þ, x

A Snf0g, then T ¼ infft : z

¼ 0g is finite by Proposition 5. We will show that

z

¼ 0

for all t b T. Using the shift and scaling properties, it su‰ces to consider x

¼ 0. Assume that there exist a R-regulated trajectory z of ðyt; t b 0Þ and h > 0 such that

kz

k ¼ d > 0:

Let e < d and let h

¼ maxfs < h : kz

k ¼ eg. Thanks to shift and scaling prop- erties the pair ð y y; ~ ~ zzÞ of trajectories defined by

~ y

y

¼

~ zz

¼

(

is a solution to LSPðyÞ with initial state

. Let t

¼ inf ft : ~ zz

¼ 0g, by the Proposition 5 there exists k > 0 such that

t

a k z

e

¼ k:

First, we have ~ zz

¼ z

¼ 0 and from lemma 1 in [2] there exists a constant c > 0 such that

kz

z

k a ckykðt h

Þ:

In particular, we have for all t A ½h

; h

þ et

kz

k a eð1 þ ckykkÞ :

Choose e such that eð1 þ c k kkÞ y < d. Then h

þ et

< h and therefore there exists s A h

þ et

; h½ such that kz

k ¼ e, which contradicts the definition of h

. r

The following corollary is an immediate consequence of the previous the- orem and the Theorem in Samelson et al. [10].

Corollary 2. Assume R is an 3 3 P-matrix, then we have:

i) If y A C

(resp C

), then the LSPðyÞ is stable if and only if b

ðyÞ < 1 (resp b

ðyÞ < 1).

ii) If y A G

nðC

W C

Þ, then the LSPðyÞ is stable.

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