CC BY
7
MSC 47D03, 47D06, 47D07, 60K25, 60J22
DOI: 10.14529/mmp 180303
LORD KELVIN AND ANDREY ANDREYEVICH MARKOV IN A QUEUE WITH SINGLE SERVER
A. Bobrowski, Lublin University of Technology, Lublin, Poland, a.bobrowski@pollub.pl
We use Lord Kelvin's method of images to show that a certain infinite system of equations with interesting boundary conditions leads to a Markovian dynamics in an L1-type space. This system originates from the queuing theory.
Keywords: queue; method of images; generation theorem; boundary conditions; Markovian dynamics.
Dedicated, to Professor Jan Kisyrlski on the occasion
of his 85th birthday.
Introduction Background
As shown by an increasing body of examples, an intelligent enrichment of an underlying state space may change the nature of a problem under consideration and allow for its elegant solution. For instance, even the simplest scalar delay-differential equation
x'(t) = ax(t) + bx(t — t), t > 0, (1)
where a,b E R and t > 0 are given, does not lead to a well-posed problem if we think of x(t) E R as a state of a process at time t and desire to describe dynamies in R. One should
x
ut :[-t, 0] ^ R, t > 0,
defined by ut(s) = x(s + t),s E [-t, 0] so that ut contains the entire information on x in the time interval [t — T,t]. Then, (1) may be rewritten as a differential equation in the space C[—t, 0] of continuous functions on [—t, 0] and the related initial value (Cauchy) problem is well-posed. As it transpires, this rather simple idea leads to a quite satisfying theory and has far-reaching consequences (see [1, Section VI.6] or [2] and references given there).
For another instance, a time-inhomogeneous Markov process (X(t))t>0 in a state S becomes time-homogeneous if extended to the pair (X(t),t)t>0 and considered in the Cartesian product space S x R+. Similarly, embedding the state-space of a non-Markovian process in a more suitable space may lead to Markovian dynamics (see e.g. [3]). Hidden Markov models, abounding in machine learning and biological sequence analysis [4,5], are the other side of the same coin. Here, although the process is in fact Markovian from the very beginning, what we observe is only a part or shadow of its state-space.
A similar trick to that described for time-inhomogeneous Markov process allows reducing non-autonomous Cauchy problems to autonomous ones (see [1, Section VI.9, Evolution Semigroups]). However, interestingly, in contrast to the examples presented
above, this transformation, presumably meant to simplify analysis, rarely helps in practice; in fact, I know of no particular instance were it is really useful.
The next example is at least intriguing. As developed by Samuel Goldstein [6] and Mark Kac [7], the stochastic process
£(t) = J(-l)Na(s) ds, t > 0, o
where Na(t),t > 0 is a Poisson process with expected value ENa(t) = at (a is a positive constant), lies behind a probabilistic formula for telegraph equation. In [8] Jan Kisyriski proves that the process
m> (-lfa(t)) ,
which is £(t) "enriched" by the coordinate (—l)Na(t), has independent increments in the non-commutative group R x { — l, 1} with the following multiplication rule:
(T,k) O (q,l) = (rl + S,kl).
In other words, £(t),t > 0 is a Levy process whereas (£(t), (—1)Na(t^ , t > 0 is a Markov process. While the former information suffices for a successful treatment of the telegraph equation (see e.g. [9,10]), the latter gives an additional insight and allows for natural generalizations [8].
A Markovian Approach to a Single Server Queue
Our paper is devoted to an example of similar type originating from the queuing theory. It is well-known that unless quite restrictive conditions are imposed, a process of the form
N(t) = # of customers in a queue at time t
is non-Markovian [11]. However, as developed by e.g. D.E. Cox [12], in the case of M/G/l queue (with Markov-type arrivals, general distribution of service time and one server), the two-dimensional process
(N(t),x(t)),t > 0, (2)
where x(t) is the time the customer being served has spent at a service point up to time t, is Markov. Recently, the latter idea has been reinvestigated by P. Gwizdz in [13], where it was noted that the resulting process may be viewed as a piece-wise deterministic process of M.H.A. Davies [14-16] (see also [17]). Gwizdz's paper contains (see Theorem 2.1 there) a proof of the fact that the system
dptt) = -O0(t) + [ K*)Mt.x)dx (3)
8M'-X) 8pi(t-x) — (a + Kx))Mt,x), (4)
dt dx
dpn(t,x) dpn(t,x)
dt dx
— (a + /i(x))'pn(t,x) + apn-i(t,x), n > 2,x> 0, (5)
t
(where a > 0 is intensity of customers' arrivals, and fi(^), a non-negative, bounded function is the hazard rate function for service time) supplemented with the following boundary conditions:
has, for any initial data, a unique mild solution in a certain space of type L1 (see further on for details). Here, pn(t, x) is the probability density that at time t there are n > 1 customers and the one being served has spent time x at the service point; p0(t) is the probability that at that time there are no customers in the queue. From the perspective of the theory of semigroups of operators, this theorem says that the related operator is the generator of a semigroup in this space. The latter semigroup governs the evolution of probability distributions of the process (2), and the fact that the semigroup is composed of Markov operators is expressed in the relation
which holds for all i > 0 provided it hold s for t = 0. Our Goal
Theorem 2.1 in [13] is obtained by a Greiner-like [18] domain-perturbation technique, extended to include unbounded domain-perturbations in L1-type spaces, and designed in such a way that the perturbed semigroup remains positive if the original semigroup is positive (original Greiner's perturbation does not posses this feature). In this paper, we will show the same result more directly, using Lord Kelvin's method of images [19-24]. Here is the main idea of the proof.
If all the terms involving a or ^ are removed from system (3)-(5), and boundary conditions (6) are disregarded, the resulting equations may be solved explicitly on the entire real line by the very simple formula:
provided p0(0) and pn(0, x) are known for all x E R. Since restoring all removed terms is a matter of a bounded perturbation, the question of solving (3)-(6) reduces to that of existence of a procedure which, given a and ц, assigns to pn(0, •),n > 1 defined on R+ their unique extensions to the entire R in such a way that functions (7), as restricted to x> 0
conditions (6) are satisfied. Note that although in the latter equations a and ^ may be thought to be set to zero, they remain non-zero in the boundary conditions (6). To prove that this idea works well is the aim of this paper.
1. The Main Theorem
Throughout the paper we assume that ^ is a bounded, non-negative function on R+ and a > 0 is a positive constant.
(6)
оо
p0(t)= p0(0), pn(t,x)= pn(0, x - t), t > 0, x G R, n = 1, 2,.
(7)
Let L1(R+) be the space of absolutely continuous functions on R+, and let
L := R x l1(L1(R+))
be the space of sequences (fn)n>0 where /0 is a real number and fn,n > 1 are members of L1(R+), such that
II (fn)
n>0 11L := \fo\ + ^ 1 fn 1 L1(r+) < n=1
L
L
00
<X r.
£I (fn)n>0 = f0 + Y^ / fn(x) dx-
n=10
Let D be the set of (fn)n>0 G L such that
• each fn,n > 1 is absolutely continuous with fn G L1(R+),
• E(=1 IIfn11 L1(r+) < ro.
Also, let Fn : D ^ R be the linear functional given by
F1 (fm)m>0 = f1(0) - f ^(x)f2(x) dx - af0, (8)
0
X
Fn (fm)m>0 = fn(0) - j H(x)fn+1(x) dx, n >
0
Here is the main theorem of the paper. Theorem 1. The operator defined on the domain
D(A^a) = D n P| ker Fn
n> 1
by the formula where
,a ( fn)n>0 (gn)
n>n> 0
go = -afo + y(x)fi(x) dx,
91 = -f1 - (a + V)f1,
9n = -fn - (a + V)fn + afn-1, n >
L
To recall, a bounded linear operator P : L ^ L is said to be a Markov operator, if P (fn) n>0 is non-negative when (fn)n>0 is, and for such (fn)n>0 we have £ip (fn)n>0 = £I (fn)n>0- If the latter equality is replaced by the inequality £iP (fn)n>0 < £i (fn)n>0 the operator is said to be sub-Markov.
2. Reduction to a Simpler Case
We will show now that Theorem 1 may be deduced from the following result. Theorem 2. The operator B, a a defined on the do main D(B, a a) = D(A, a a) by the formula
B,,a (fn),n>0 = (gn)n>0 ,
where
go = gn = -fn, n >
generates a strongly continuous semigroup of positive operators in L.
Remark 1. We stress again that, while a and are not featuring in the definition of the "action" of B,,a, both are involved in the definition of its domain. Hence, what we are facing here is a domain-changing perturbation; such perturbations were studied in detail by G. Greiner [18], and the approach presented in [13] follows Greiner's path. We are using a different method, i.e., Lord Kelvin's method of images that has been developed as a way to deal with boundary conditions in semigroup theory in [19-23] and [24]. For yet different ways of dealing with boundary conditions see [25] and [26].
Suppose thus that B,,a generates a semigroup of positive operators and then choose а к such that
к > sup ¡(x).
x>0
Then, for any (fn)n>0 E V(A„,a) = V(B^a),
,a (fn)n>0 = , a (fn)n>0 + , a, к (fn)n>0 — (a + K) (fn)n>0 , where C,,a,к is the bounded, non-negative, linear operator defined by
C,,а,к (fn)n>0 = (gn)n>0
with
<x
go = j p(x)fi(x) dx + Kfo, gi = (к - /i)fi,
o
gn = (к - ¡j)fn + afn-i, n > 2.
Thus, by the Phillips perturbation theorem, B, a + C, generates a semigroup of positive operators and this implies that so does A,,a, because the last two operators differ by a constant multiple of the identity operator.
Hence, we are left with showing that the semigroup generated by A,,a is composed of Markov operators. A well-known necessary and sufficient condition for that is for A (A — A,,a)-1 to be Markov operators for all sufficiently large A > 0. Sinee A,,a generates a semigroup, for sufficiently large A the resolvent equation
A (fn)n>0 — A,,a (fn)n>0 = (gn)n>0
Вестник ЮУрГУ. Серия «Математическое моделирование 22
и программирование» (Вестник ЮУрГУ ММП). 2018. Т. 11, № 3. С. 29-43
has a unique solution for all (gn)n> 0. Because the semigroup gene rated by is composed of positive operators, the map assigning solution (fn)n>0 to (gn)n>0 is positive also (since the resolvent is the Laplace transform of the semigroup). Applying the functional £/ to both sides of the resolvent equation we see thus that all we need to prove is that
SiA^a (fn)n>o = 0 (9)
for all non-negative (fn)n>0 G D(A^a).
To this end, we first use (8) to calculate (writing, for simplicity of notation, f ¡ifn
R+
oo
instead of f ¡i(x)fn(x) dx) 0
oo
E fn(0) = afo + Y, / f (fn)n>o G D(A^a).
n=l n=2R+
On the other hand, with similar notation,
SI,a (fn)n>0 = -af0 / Vnfn fn■
r^-O V - 1
n 2r+ n_1r+
Since J fn = -fn(0), these two relations combined imply (9), completing the proof.
r+
Remark 2. The last calculation shows that (9) holds in fact for all vectors (fn)n>0 in D(A^,a) whether positive or not. Hence, A (A — A^,a)-1 preserves the functional S;, i.e., SIA (A — A^,a)-1 (fn)n>0 = SI (fn)n>0 for all sufficiently large A. Therefore, the semigroup generated by A^a also preserves this functional, i.e., the value of S; does not change along this semigroup's trajectories (this is a stronger condition than that of being composed of Markov operators).
3. A Semigroup in a "Larger" Space
A first step in the procedure of the Lord Kelvin method of images is to construct a semigroup generated "by without boundary conditions" in a "larger" space. This is our goal in this section.
Let (fn)n>0 G L, and consider an n > 1 Any extension of fn to a function on R is determined by gn(x) = (—x), a function on R+. It would be nice to have gn G L1(R+) but this is rarely the case. Fortunately, for the particular extension we are looking for there is an w > 0 such that all functions x M e-uxgn(x) are in L1(R+). Hence, we introduce L^ (R+) as the space of functions g : R+ M R such that x M e-MXg(x) is a member of L1(R+). When equipped with the norm
oo
Wghi (r+) = J \g(x)\e-MX dx 0
L^ (R+) is a Banach space.
A pair (f, g) E L1(R+) x LI (R+) may be identified with a function fr on R: it suffices to agree that h(x) = f (x),x > 0 and h(x) = g(-x) for x < 0 The space L^(R) of functions h : R ->■ R such that
LL(R) := \h(x)\ dx + e шx\h(-x)\dx
is finite, is a Banach space with this norm, and the spaces L1(R+) x Li(R+^d Li are isometrically isomorphic. All our extensions will be members of Li (R) and sequences of extensions will be members of
Ьш := R x l1(L1u
the space of sequences (hn)n>0 where h0 is a real number and hn,n > 1, are members of
<x
\\ (hn)n>0 \\lu := lho| + Whn\\Ll(r) < ro;
Li (R), such that
n=1
when equipped with this norm, LM is a Banach space. The formula
T(t)h(x) = h(x - t), t > 0,x E R, (10)
defines a strongly continuous semigroup in L^ (R), and since
t <x <x
\\T(t)h\\Ll(r) < i\h(x - t)\dx + i \h(x - t)\dx + i\h(-x - t)\e-ux dx
< eMt J e-MX\h(-x)\ dx + J \h(x)\ dx + eut J e-MX\h(-x)\ dx
0 o t
< ^WhL(r),
the operator norm of T(t) does not exceed eMt. A standard argument shows that the domain D(G) of the generator G of {T(t),t > 0} is composed of absolutely continuous members h of L^(R) such that h' E L^(R^^d Gh = -h' on this domain. It follows that
T(t)(K)n>o = (ho,T (t)hi,T (t)h2,...) (11)
defines a strongly continuous semigroup of operators in Lw. The operator norm of T(t) does not exceed eMt. The domain D(G) of the infinitesimal generator G of {T(t),t > 0} is composed of (hn )n> 0 such th at hn,n > 1 are absolutely continuous with h'n E L^ (R) and (0, h'1,h'2,...) E Lu. For such (hn)n>0,
G (hn)n> о = -{0,h'l,h'2,...).
t
4. Constructing the Images
As explained in the introduction to [20], given an (fm)m>0 G V(B^,a) it is a good idea to look for the sequence (fn) m>0 G LM of extensions of its terms satisfying
FnT(t) f) „>0 = 0, for all n > l,t > 0, (12)
where Fn given % (8) is now treated as a functional defined on D(G). This condition means that o
fn(-t) = J ti(x)f?(x — t)dx + af0 0
and o
fn—) = j n(x)fn+1(x — t) dx, for all n > 2, t > 0, 0
and thus may be rewritten as a requirement for
gn(x) := fn(—x), x > 0
as follows:
x o
g1(x) = J g2(x — y)v(v) dy ^y Mx + y)f2 (y) dy + af0,
00
x o
gn(x) = J gn+1(x — y)^(y)dy + J Mx + y)fn+1(y)dy, n > 2,x > 000
This in turn simply means that we need to have
g1 = n * g2 + Uf2 + af0, gn = n * gn+1 + Ufn+1, n > 2, (13)
o
where Uf (x) = f /i(x + y)f (y) dy, f G L1(R+). 0
We note that while condition (12) may be formulated only for (fn)n>0 in V(B^,a), (13) makes sense for all (fn)n>0 in L.
Lemma 1. Fix w > ||^||o := supx>0¡i(x). For each (fn)n>0 G L there is precisely one (gn)n>1 such that (13) holds and the sequence (fn)n>0 defined by
fo = f0, fn(x) = fn(x), x > 0,
fn(x) = gn(—x), x< °
is a member of Lu. Proof.
1. Let K := ll(L1^(R+)) be the space of sequences (kn)n> 1 such that kn G L^(R+)
and
W (kn)n>1 K :=Y Wknhl (r+) < n=1
Our task is thus to show that for any (fn)n>0 E L, there is precisely one (gn)n> 1 E K^ satisfying (13).
2. We start the proof by noting that Uf defined below (13) belongs to Ll(R+) and
\\Uf\\Ll(R+) ||bi(R+). (14)
Ш
This is showed by the following calculation:
oo oo
Li(r+) < e шх ¡(x + y)\f (y)\ dy dx <
oo
oo
<j е~шхЫooj \f(y)\dy dx = ^\\f \\li(r+). oo
Inequality (14) implies that
U (fn)n>0 := (U f2 + af0,Uf3,UU,...)
is a member of K^ with
\U (fn)n>0 < jt \\Ufn\\Ll(r+) + аШ< m^(\\n\Ua) \\ (fn)n>0 \l. (15)
Ш Ш
n=2
3. Finally, for a fixed (fn)n>0 E L, let M : Kw — Kw be the map given by
M (kn)n>1 = (V * kn+1)n>1 + U (fn)n>0 • (16)
For any k E Ll (R+),
\\v * k\\Lh (r+) < ML (r+) \\k\\Lh (r+) < - \\k\\Lh (r+) • (17)
u
It follows that
\\ (^ * kn+1)n>1 \\ku < — \\ (kn)n>1 K• (18)
_ u ~
Since < u, M is a contraction mapping and Banach's fixed point theorem implies
that there is precisely one (gn)n> 1 E Ku such that M (gn)n> 1 = (gn)n> 1, i-e. precisely one (gn) n> 1 E Ki satisfying (13). □
The sequence (gn)n> 15 whose existence has just been established, will be called the (^,a)-image of (fn)n>0, and f^) n> 0 will be called the a)-extension of (fn)n>0■ We note that for non-negative (fn)n>0, the vector U (fn)n>0 is also non-negative, and thus the map M defined in (16) leaves the non-negative cone in invariant. Since the fix-point of M may be obtained as lim Ml (kn)n> 1 for any (kn)n> 1 E Kl and in particular we may
i^m > >
take (kn)n>1 = U (fn)n>0 it follows that (gn)n>1 is non-negative provided (fn)n>0 is. It is worth noting that (18) forces
\\M (kn)n>l \\Ki < —\\ (kn)n>1 К + \U (fn)n>0 K . Ш
Thus, by induction
)
\\M1U (fn)n>0 \\Ku < ¿ (MkY \\U (fn)n>0 \\Ku, i > 1,
i=0 ^ U '
and so, using (15) and (gn)n>1 = lim^^ MlU (fn)
n)n>0''
\\ (gn)n>1 K < -^V\\U (fn)n>0 < \\ (fn)n>0 \\l. (19)
- u — U ™ - u —
We also need information on regularity of (y, a)-images, contained in Lemma 3 (see later on) which in turn is based on the following Sobolev type inequality.
Lemma 2. Let u > 0. Suppo se f E L^ (R+) is absolutely continuo us with f' E L^ (R+). Then
\f (0)\< u\\f \\li (r+ ) + \\f '\\li (r+).
Proof. The function x M e-uxf (x) is also absolutely continuous with derivative equal to e-<¿xf' (x) — ue-ux f (x) .For x > 0 we have thus
x
e-MXf (x) — f (0) = j [e-^ f (y)]'dy. o
Since the limit as x M to of the right-hand side exists, so must the limit limx^^ e-ux f (x). But the latter must be zero, x M e-ux f (x) being integrable. Therefore,
\f (0)\
< u\\f hi(r+) + \\f'hi(r+),
[we-ujy f (y) — e-uy f' (y)]dy
0
as desired. n
Lemma 3. Fix (fn)n> 1 G D (see Section 1). Then gn,n > 1 are absolutely continuous with g'n G L^ (R+) and
\\ (g'n)n>i \\ki = e \\gn\L¿(r+)
n=1
Proof.
1. Suppose f G L1(R+) is absolutely continuous with f' G L1(R+^. Then Uf (defined right after (13)) is absolutely continuous also, and
—(Uf )' = Uf' + f (0)».
2. Let (kn)n> 1 := U (fn)n>0 G Kw. By point 1, each kn is absolutely continuous with — k'n = Ufn+1 + fn+1(0)». Therefore, by Lemma 2 with w = 0 and (14),
o o o
e Wk'nWLl(r+) < e WUfnULl(r+) + ||»lk(r+) e \fn(0)\ <
n=1 n=2 n=2
OO II II oo
\\ fn \\ L1(r+) + \fn\L!(r+) < TO.
uu
n=2 n=2
Combining this with (15) we see that
\\ (kn)n>1 \\ku + \\ (k'n)n>1 \\ku < *>• (20)
i
3. An induction argument shows that the n-th coordinate of Ml (kn)n> 1 is ji* *
> i=0
ki+n,l > 1 Since (gn)n> 1 = lim Ml (kn)n> 1 we have
gn
* kt+n, n > 1. (21)
i=0
4. If a k E Ll(R+) is absolutely continuous with k' E Ll(R+), then so is j * k and (j * k)' = j * k' + k(0)j. Since (j1* * k)(0) = 0 for i > 1, an induction argument shows that
(ji* * k)' = ji* * k' + k(0)ji*, i > 1.
This in turn yields
(i \ ' i i
e Ji* * ki+n = e Ji* * k[+n + e ki+n(0)ji*. i=0 i=0 i=1
5. We claim that
<x / <x <x \
e e i * k'+n\\ Li(r+) + e \\ki+n(0)i%*\\Ll(r+) < (22)
n=1 \i=0 i=1 /
Since (use (17))
oo ^ ^ ^ у \ г
ee \\il* * ki+n \ Li (r+)
n=1 i=0 n=1 i=0 ^ '
< e г e и*ik(r+)
г=0 V / n=1
¿(r+)
i=0 4 ' n=1
\\ (kn)n>1 \\Ki
id —
is finite by (20), to prove (22) we need to estimate
m m
Y,Y,\ki+n(0)ji* wn (r+).
n=1 i=1
By Lemma 2, however, this quantity does not exceed
m /11.11 \i m m
n
e (^v iHknwLi (r+) + \\k'n \\li (r+)] i=1 v ' n=1
- П (kn)n>1 \\Ki + \\ (kn)n>1 \\k
n
Ш — \\i\\^
and this is finite by (20). This completes the proof of (22).
6. Let D(D) c Ll(R+) be the set of absolutely continuous f E Ll(R+) such that f' E Ll (R+), and let Df = f. A straightforward argument shows that D is closed. Inequality (22) implies that for n > 1, the series
<x
kn + ^^ \_y% * ki+n + ki+n(0)^*} (23)
i=1
converges absolutely in Ll(R+). ft follows that D(^l=0 y%* * ki+n) (calculated in point 4.) converges to the sum of this series. Since D is closed, gn (given by (21)) belongs to V(D), i.e., it is absolutely continuous with gn equal to the sum of the series (23). Combining
\\g'n\\Ll (R+) < E * ki+nhl (R+) + E Wki+n^^W^ (R+)
i=0 i=1
and (22),
we complete the proof. ^
5. Abstract Kelvin Formula for {etB^a ,t > 0}
The stage is now ready for the proof of Theorem 2. In fact, we will show the following result, giving a somewhat deeper insight into the nature of the semigroup generated by
B^,a ■
Theorem 3. Let R : Ll(R) ^ L1(R+) map an f E Ll(R) to its restriction f\[0,ro), and let R : Ll ^ L be defined by
R (fn)n>0 =(fo,Rfl,Rf2,Rf3,... ).
Also, let E : L ^ Ll map elements of L to their (y, a)-extensions. The abstract Kelvin formula
5(t) = RT(t)E, t > 0, (24)
where {T(t),t > 0} is the translation semigroup of (11), defines a strongly continuous semigroup of operators in L, and the infinitesimal generator of {S(t),t > 0^5
To see that Theorem 2 is a direct consequence of Theorem 3 it suffices to recall from the previous section that (y, a)-extensions of non-negative (fn)n>0 are non-negative, and that {T(t),t > 0} is a semigroup of non-negative operators.
The proof of Theorem 3 will become more clear if we extract from it the following lemma.
Lemma 4. Let (fn)n>0 E V(B^a) be fixed.
(a) fn) n> 0, the (y, a)-extension of (fn)n>0, belongs to the domain D(G) of the infinitesimal generator of the translation semigroup (11),
(b) For all t > 0, T(t)(fn)n>0 is the (y, a)-extension of its own restriction RT(t)(fn)n>i
Proof, (a) By Lemma 3, our assumption implies that each gn,n > 1 is absolutely continuous with g'n E L^(R+). Moreover, by definition of gn,
9i(0) = Uf2(0) + afo = J ß(y)f2(y) dy + afo
o
<x
gn(0) = Ufn+1(0) = i ß(y)fn+i(y) dy, n > 2. (25)
o
Since (fm)m>о E Пn>i ker Fn, fn(0) = gn(0) and so f? is absolutely continuous for all n > 1. Moreover, by
>)fWli (r) = llfniLi(r+) + Wg'nWbi (r+) ,
Lemma 3 implies
completing the proof of (a).
(r)
< OO,
n=1
fn ) n> 0
(b) Fix t > 0. By (a) fn?) n>0 is a member of D(G). ft follows that so is fn T(t) )fn)n>0 and th at FnT(s) ftm )m>0 = 0 for all s > 0 and n > 1 (on D(G), (12) and~(13) are equivalent). Therefore, for all s and n > 1, FnT(s) (f'*) m>0 = FnT(s+1) f) m>0 = 0. This means, by definition, that fm)m>0 is the (y, a)-extension (of to own restriction). □
Proof of Theorem 3
1. Fix ш > ||y||^, an(l ^ Еш С Ьш be the space of (y, a)-extensions of members of L. Inequality (19) shows that E mapping L onto Еш is bounded. Since E has a bounded inverse R Еш is closed in L^, and hence is a Banach space (with norm inherited from Lu). The spaces L and Еш are isomorphic with the isomorphism E : L — Еш and its inverse R : Еш — L.
2. Since V(B^a) is dense in L (as a straightforward argument shows), so is its image EV(B^a) in Еш. Lemma 4 now says that EV(B^a) is invariant for the translation semigroup {T(t),t > 0}. It follows that so is Еш. Hence, {T(t),t > 0} restricted to Еш is a strongly continuous semigroup. The semigroup defined by the abstract Kelvin formula (24) is thus the isomorphic image of {T(t),t > 0} restricted to Еш, and it is obviously strongly continuous.
3. We are left with showing that the generator of {S(t),t > 0} is B^,a. To this end, we recall that the generator of {T(t),t > 0} restricted to Еш is the part Gp of G in Еш (G was defined in Section 3). Thus (f?)n>0 E Еш is a member of V(GP) (= D(G) П Еш) iff
fn,n > 1 are absolutely continuous and (0, (f', (f',...) E Lu] then
GP )f?)n>0 = -(0, (fi)', (f?)',...);
the vector on the right-hand side here automatically belongs to Еш since Еш is invariant for the translation semigroup.
On the other hand, (fn)n>0 belongs to the domain of the generator, say Gi, of {S(t),t > 0} iff n>0 = E (fn)n>0 belongs to D(GP). Lemma 4 tells us that for (fn)n>0 E V(B^,a)
the latter condition holds. Conversely, if (fn) n> 0 is a member of D(G) n Ethen each fn (being the restrict ion of ff) must be absolutely continu ous with f G L1(R+), and we must have
J2fn Hli(r+) Wn, (r) <
n=1 n=1
Also, absolute continuity of ff implies fn(0) = gn(0) for all n, and then a look at (25) reveals that (fn)n>0 G f|n>1 ker Fn, thus showing that V(Gi) = V(B^a). For such (fn)n>o,
G1 (fn)n>0 = RGE (fn)n>o = RG (ff)n>0 =
= -R(0, (ff)', (ff)',... ) = -(0,f1,f2,... ) = B^a (fn)n>0 .
This completes the proof. □
Acknowledgment. I would like to thank Lukasz St§pien for careful reading of the
manuscript and for many remarks that helped to improve the presentation of the material
discussed here.
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Received June 21, 2018
УДК 517.9+519.21+517.958 Б01: 10.14529/ттр180303
ЛОРД КЕЛЬВИН И АНДРЕЙ АНДРЕЕВИЧ МАРКОВ К ОЧЕРЕДИ С ОДИНОЧНОГО СЕРВЕРА
А. Бобровский, Люблинский технологический университет, г. Люблин, Польша
Мы используем метод изображений лорда Кельвина, чтобы показать, что некоторая бесконечная система уравнений с интересными граничными условиями приводит к марковской динамике в пространстве Ь1-типа. Эта система берет свое начало в теории массового обслуживания.
Ключевые слова: очередь; метод изображений; теорема генерации; граничные условия; Марковская динамика.
Адам Бобровский, профессор, кафедра математики, Люблинский технологический университет (г. Люблин, Польша), a.bobrowski@pollub.pl.
Поступила в редакцию 21 июня 2018 г.
