Transport Equation on Semidiscrete Domains and Poisson-Bernoulli Processes

1 MB·English

by  Petr Stehlík

1
TRANSPORT EQUATION ON SEMIDISCRETE DOMAINS AND
POISSON-BERNOULLI PROCESSES
PETRSTEHL´IK,JONA´Sˇ VOLEK
Abstract. In this paper we consider a scalar transport equation with constant
coefficientsondomainswithdiscretespaceandcontinuous,discreteorgeneraltime.
2 We show that on all these underlying domains solutions of the transport equation
1 can conserve sign and integrals both in time and space. Detailed analysis reveals
0 that, under some initial conditions, the solutions correspond to counting stochastic
2 processesandrelatedprobabilitydistributions. Consequently,thetransportequation
couldgeneratevariousmodificationsoftheseprocessesanddistributionsandprovide
n
some insights into corresponding convergence questions. Possible applications are
a
suggestedanddiscussed.
J

4 Keywords. Transport Equation, Advection Equation, Conservation Law, Time
Scales,SemidiscreteMethod,Poissonprocess,Bernoulliprocess.
]
P
AMS Subject classification. 34N05,35F10,39A14,65M06.
A
. 1. Introduction
h
t Scalar transport equation with constant coefficients u +ku = 0 belongs among the simplest
a t x
m partial differential equations. Its importance is based on the following facts. Firstly, it describes
advectivetransportoffluids,aswellasone-waywavepropagation. Secondly,itservesasabasefor
[
astudyofhyperbolicpartialdifferentialequations(andisconsequentlyanalysedalsoinnumerical

1
analysis). Thirdly, its nonlinear modifications model complex transport of fluids, heat or mass.
v
Finally, its study is closely connected to conservation laws (see [9] or [13]).
8
5 Propertiesandsolutionsofpartialdifferenceequationshavebeenstudiedmainlyfromnumerical
8 (e.g. [13]) but also from analytical point of view (e.g. [5]). Meanwhile, in one dimension, there
0 has been a wide interest in the problems with mixed timing, which has been recently clustered
.
1 around the time scales calculus and the so-called dynamic equations (see [4], [11]). Nevertheless,
0 there is only limited literature on partial equations on time scales (see [1], [3], [18]). These papers
2 indicatethecomplexityofsuchsettingsandthenecessitytoanalyzebasicproblemsliketransport
1
equation. Our analysis is also closely related to numerical semidiscrete methods (e.g. [13, Section
:
v 10.4]) or analytical Rothe method (e.g. [17]).
Xi In this paper we consider a transport equation on domains with discrete space and general
(continuous, discreteand time scale) time (see Figure 1). We show that the solutions of transport
r
a equationdoesnotpropagatealongcharacteristicslinesasintheclassicalcaseandfeaturebehavior
close to the classical diffusion equation. Our analysis of sign and integral conservation discloses
interestingrelationshipbetweenthesolutionsonsuchdomainsandprobabilitydistributionsrelated
toPoissonandBernoullistochasticprocesses. Thesecountingprocessesareusedtomodelwaiting
timesforoccurenceofcertainevents(defects,phonecalls,customers’arrivals,etc.),see[2],[10]or
[15] for more details. Consequently, considering domains with general time, we are able not only
togeneralizethesestandardprocessesbutalsogeneratetransitionalprocessesofPoisson-Bernoulli
type and corresponding distributions. Moreover, our analysis provides a different perspective on
some numerical questions (numerical diffusion) and relate it to analytical problems (relationship
between the CFL condition and regressivity). Finally, it also establishes relationship between
the time scales calculus and heterogeneous and mixed probability distributions in the probability
theory.
In Section 3 we summarize well-known features of the classical transport equation. In Section
4weconsideratransportequationwithdiscretespaceandcontinuoustime. InSection5wesolve
1Thisisapreprint. ThefinalversionofthispaperwillappearinJournalofDifferenceEquationsandApplications
1
2 PETRSTEHL´IK,JONA´SˇVOLEK
Μt
Μx Μx Μx
Figure 1. Examples of various domains considered in this paper. We study
domains with discrete space and continuous (Section 4), discrete (Section 5) and
general time (Section 6).
the problem on domains with discrete time. In Section 6 we generalize those results to domains
with a general time and prove the necessary and sufficient conditions which ensure that the sign
and both time and space integrals are conserved (Theorem 18). Finally, in Section 7 we discuss
convergence issues, applications to probability distributions and stochastic processes and provide
two examples.
2. Preliminaries and Notation
The sets R, Z, N denote real, integer and natural numbers. Furthermore, let us introduce
N =N∪{0} and R+ =[0,∞). Finally, we use multiples of discrete number sets, e.g. a-multiple
0 0
of integers is denoted by aZ and defined by aZ={...,−2a,−a,0,a,2a,...}.
Partial derivatives are denoted by u (x,t) and u (x,t) and partial differences by
t x
u(x,t+µ )−u(x,t) u(x,t)−u(x−µ ,t)
(1) ∆ u(x,t)= t and ∇ u(x,t)= x ,
t µ x µ
t x
where µ and µ denote step sizes in time and space.
t x
In Section 6, we consider time to be a general time scale T, i.e. an arbitrary closed subset
of R. Time step could be variable, described by a graininess function µ : T → R+. We use
t 0
the partial delta derivative u∆t which reduces to ut in points in which µt(t) = 0 or to ∆tu in
those t in which µ(t)>0. Similarly, we work with the so-called delta-integral which corresponds
to standard integration if T = R or to summation if T = Z. Finally, the dynamic exponential
function e (x,x ) is defined as a solution of the initial value problem (under the regressivity
p 0
condition 1+p(t)µ(t)(cid:54)=0)
(cid:40)
x∆(t)=p(t)x(t),
x(0)=1.
For more details concerning the time scale calculus we refer the interested reader to the survey
monograph [4].
Given function u(x,t), by u(x,·) we mean functions of one variable having the form u(0,t),
u(1,t), etc. Similarly, by u(·,t) we understand one-dimensional sections of u(x,t) having the form
u(x,0), u(x,1), etc.
TRANSPORT EQUATION ON SEMIDISCRETE DOMAINS AND POISSON-BERNOULLI PROCESSES 3
3. Classical Transport Equation
Let us briefly summarize essential properties of the classical transport equation
(cid:40)
u (x,t)+ku (x,t)=0, t∈R+,x∈R,
(2) t x 0
u(x,0)=φ(x), x∈R.
with φ∈C1. Typical features whose counterparts are studied in this paper include:
• theuniquesolutionu(x,t)=φ(x−kt)couldbeobtainedviathemethodofcharacteristics,
the solution is constant on the characteristic lines where x−kt=C,
• consequently, the solution conserves sign, i.e. if φ(x)≥0 then u(x,t)≥0,
∞
(cid:82)
• moreover, the solution conserves integral in space sections, i.e. if φ(x)x. =K, then
−∞
∞
(cid:90)
u(x,t)x. =K, for all t≥0,
−∞
• finally, the solution conserves integral in time sections in the following sense. For k > 0
we have that
∞ x
(cid:90) 1 (cid:90)
u(x,t)t. = φ(s)s..
k
0 −∞
Consequently, if φ(x)=0 for x≥x , then the integral along time sections is constant for
0
all x≥x .
0
4. Discrete Space and Continuous Time
In contrast to the classical problem (2) we consider a domain with discrete space and the
problem
u (x,t)+k∇ u(x,t)=0, t∈R+,x∈Z,
 t (cid:40) x 0
(3) A, x=0,
u(x,0)=
 0, x(cid:54)=0,
where A>0, k >0 and ∇ u reduces to1
x
∇ u(x,t)=u(x,t)−u(x−1,t).
x
One could rewrite the equation in (3) into
u (x,t)=−ku(x,t)+ku(x−1,t),
t
which implies that the problem (3) could be viewed as an infinite system of differential equations.
lem 1. The unique solution of the problem (3) has the form:
 kx
 Ax!txe−kt, t∈R+0,x∈N0,
(4) u(x,t)=
 0, t∈R+,x∈Z,x<0.
0
Proof. First, let us observe that u(x,t) = 0 for all t ∈ R+, x < 0. The uniqueness of the trivial
0
solutionforx<0followse.g. from[16,Corollary1]ormoregenerallyfrom[6,Theorem3.1.3]. Let
us prove the rest (i.e. x≥0 by mathematical induction. Obviously, we have that u(0,t)=Ae−kt,
since u (0,t)=−ku(0,t)+ku(−1,t)=−ku(0,t) and u(0,0)=A.
t
1We assume that k > 0 so that the solution is bounded and does not vanish. Moreover, we use the nabla
differenceinsteadofdeltadifference. Thesinglereasonisthesimplerformofthesolution(4). Ifweusedthedelta
difference,wewouldconsiderk<0andthesolutionwouldpropagatetothequadrantwitht>0andx<0. This
appliesalsototheproblemswhichwestudyinthefollowingsections.
4 PETRSTEHL´IK,JONA´SˇVOLEK
ux,(cid:42) u(cid:42),t
1.0
(cid:72) (cid:76) (cid:72) (cid:76)
0.8
ux,1
ux,4
ux,8
ux,t 0.6 u(cid:72)x,12(cid:76)
(cid:72) (cid:76)
(cid:72) (cid:76)
(cid:72) (cid:76)
0.4
(cid:72) (cid:76)
0.2
x
t
1 2 3 4 5 t x
Figure 2. Solutionofthetransportequationwithdiscretespaceandcontinuous
time (3) with A=1 and k =1.
Moreover, if we assume that x∈N and u(x,t)=Akxtxe−kt, then u(x+1,t) satisfies
0 x!
(cid:40)
u (x+1,t)=−ku(x+1,t)+Akxtxe−kt, t∈R+,
t x! 0
u(x,0)=0.
One could use the variation of parameters to show that the unique solution is u(x + 1,t) =
A kx+1 tx+1e−kt, which proves the inductive step and consequently finishes the proof. (cid:3)
(x+1)!
Let us analyze the sign and integral preservation of (3).
lem 2. The solution of the problem (3) conserves the sign, the integral in time and the sum in
the space variable.
Proof. The sign preservation follows from the positivity of all terms in (4). Next, we could use
integration by parts to obtain (we skip the details since we prove this result in more general
settings in Theorem 15)
(cid:90) ∞ A
(5) u(x,t)t. = .
k
0
Similarly, summing over x we get
(cid:88)∞ kx (cid:88)∞ (kt)x
(6) A txe−kt =Ae−kt =Ae−ktekt =A.
x! x!
x=0 x=0
(cid:3)
If we go deeper and analyze values obtained in (5) and (6) we get the first indication of the
relationship of the semidiscrete transport equation with stochastic processes.
Remark3. IfA=kthentimesectionsofthesolution (4)generatetheprobabilitydensityfunction
of Erlang distributions (note that for x = 0 we get the exponential distribution and that Erlang
distributions are special cases of Gamma distributions).
Similarly, if A = 1 the space sections of (4) form the probability mass functions of Poisson
distributions.
Consequently, if A = k = 1 the solution u(x,t) describes Poisson process. All these facts are
further discussed in Section 7.
We conclude this section with two natural extensions. Firstly, we mention possible generaliza-
tions to other discrete space structures.
TRANSPORT EQUATION ON SEMIDISCRETE DOMAINS AND POISSON-BERNOULLI PROCESSES 5
Remark 4. If we consider the problem (3) on a domain with a discrete space having the constant
step µ >0, not necessarily µ =1, we obtain qualitatively equivalent problem, since
x x
u(x,t)−u(x−µ ,t) k
u (x,t)+k x =u (x,t)+ (u(x,t)−u(x−µ ,t))
t µ t µ x
x x
=u (x,t)+kˆ∇ u(x,t).
t x
In contrast to the rest of this paper, the value of µ does not play essential role here. Therefore,
x
for presentation purposes, we restricted our attention to µ =1.
x
Finally,wediscussmoregeneralinitialconditionandshowthatthesolutionisthesumofpoint
initial conditions which justifies their use not only in this section but also in the remainder of this
paper.
Corollary 5. The unique solution of
(cid:40)
u (x,t)+k∇ u(x,t)=0, t∈R+,x∈Z,
(7) t x 0
u(x,0)=C .
x
is given by
(cid:88)x (kt)x−i
(8) u(x,t)= C e−kt.
i(x−i)!
i=−∞
Proof. One couldsplit (7)intoproblems withpoint initialconditions, use Lemma 1to solve them
and then employ linearity of the equation to get (8). (cid:3)
5. Discrete Space and Discrete Time
In this section, we assume that both time and space are homogenously discrete with steps
µ >0 and µ >0 respectively. In other words, we consider a discrete domain
t x
Ω={(x,t)=(mµ ,nµ ), with m∈Z,n∈N }.
x t 0
The transport equation and the corresponding problem then have the form

∆ u(x,t)+k∇ u(x,t)=0, (x,t)∈Ω,
 t (cid:40) x
(9) A, x=0,
u(x,0)=
 0, x(cid:54)=0,
where A > 0, k > 0. Using the definition of partial differences in (1), we can easily rewrite the
equation in (9) into
(cid:18) (cid:19)
kµ kµ
(10) u(x,t+µ )= 1− t u(x,t)+ tu(x−µ ,t)
t µ µ x
x x
and derive the unique solution.
lem 6. Let m∈Z and n∈N . The unique solution of (9) has the form:
0
 (cid:18)n(cid:19)(cid:18) kµ (cid:19)n−m(cid:18)kµ (cid:19)m
 A 1− t t , n≥m≥0,
m µ µ
(11) u(mµ ,nµ )= x x
x t
 0, 0≤n<m, or m<0.
Proof. First, let us show that the solution vanishes uniquely for u(−mµ ,nµ )=0, for all m,n∈
x x
N. Consulting(10),weobservethatthevalueofu(−mµ ,nµ )isobtainedasalinearcombination
x x
ofinitialconditionsu(−mµ ,0),u(−(m+1)µ ,0),...,u(−(m+n)µ ,0),i.e. alinearcombination
x x x
of n+1 zeros.
We prove the rest of the statement by induction. Apparently,
(cid:18) kµ (cid:19)n (cid:18) kµ (cid:19)n
u(0,nµ )= 1− t u(0,0)=A 1− t .
t µ µ
x x
6 PETRSTEHL´IK,JONA´SˇVOLEK
Next, let us assume that u(mµ ,nµ ) satisfies (11), then
x t
(cid:18) kµ (cid:19)n
u((m+1)µ ,nµ )= 1− t u((m+1)µ ,0)
x t µ x
x
+ n(cid:80)−1 (cid:16)1− kµt(cid:17)n−1−rm+1A(cid:16)1− kµt(cid:17)rm+1−m(cid:16)kµt(cid:17)m+1rm(cid:80)+1−1...r(cid:80)3−1r(cid:80)2−11
rm+1=0 µx µx µx rm=0 r2=0r1=0
=A(cid:16)1− kµt(cid:17)n−(m+1)(cid:16)kµt(cid:17)m+1 n(cid:80)−1 ...r(cid:80)3−1r(cid:80)2−11.
µx µx rm+1=0 r2=0r1=0
At this stage, let us observe the properties of the falling factorials (see e.g. [8, Section 2.1] or
[12, Section 2.1]) to get that
n(cid:88)−1 r(cid:88)3−1r(cid:88)2−1 nm+1 (cid:18) n (cid:19)
... 1= = ,
(m+1)! m+1
rm+1=0 r2=0r1=0
which finishes the proof. (cid:3)
The closed-form solution enables us to analyze sign and integral conservation.
lem 7. If the inequality
kµ
(D1) 1− t >0,
µ
x
holds then the solution of (9) satisfies
(i) u(x,t)≥0,
∞
(cid:80)
(ii) u(mµ ,t) is constant for all t={0,µ ,2µ ,...},
x t t
m=−∞
∞
(cid:80)
(iii) u(x,nµ ) is constant for all x={0,µ ,2µ ,...}.
t x x
n=0
Proof. (i) The inequality follows immediately from Lemma 6 .
(ii) If we fix t and sum up the equation (10) over x we get
∞ (cid:18) (cid:19) ∞ ∞
(cid:88) u(mµ ,t+µ )= 1− kµt (cid:88) u(mµ ,t)+ kµt (cid:88) u((m−1)µ ,t).
x t µ x µ x
x x
m=−∞ m=−∞ m=−∞
The assumption (D1) implies that the sum on the left hand side is a linear combination of
two sums on the right hand side. Since these sums are equal, we get that
∞ ∞
(cid:88) (cid:88)
u(mµ ,t+µ )= u(mµ ,t).
x t x
m=−∞ m=−∞
(iii) Similarly, one could sum up the equation (10) over t to get for a fixed x>0
∞ (cid:18) (cid:19) ∞ ∞
(cid:88)u(x,nµ )= 1− kµt (cid:88)u(x,nµ )+ kµt (cid:88)u(x−µ ,nµ ).
t µ t µ x t
x x
n=1 n=0 n=0
Since u(x,0)=0 for x>0 we have that
∞ ∞
(cid:88) (cid:88)
u(x,nµ )= u(x−µ ,nµ ).
t x t
n=0 n=0
(cid:3)
Once again, we could study the solutions’ relationship to probability distributions.
thm 8. Let u(x,t) be a solution of (9). Then the space and time sections µ u(x,·) and µ u(·,t)
x t
form probability mass functions if and only if the assumptions (D1),
Aµ
(D2) x =1,
k
TRANSPORT EQUATION ON SEMIDISCRETE DOMAINS AND POISSON-BERNOULLI PROCESSES 7
ux,(cid:42) ux,(cid:42)
(cid:72) (cid:76) (cid:72) (cid:76)
u1,t u1,t
u2,t u2,t
u3,t u3,t
u(cid:72)4,t(cid:76) u(cid:72)4,t(cid:76)
ux,t (cid:72) (cid:76) (cid:72) (cid:76)
(cid:72) (cid:76) (cid:72) (cid:76)
(cid:72) (cid:76) (cid:72) (cid:76)
(cid:72) (cid:76)
x
t
t t
Figure 3. Solution of the transport equation with discrete space and discrete
time (9) with A=1, k =1, µ =.25 and µ =1.
t x
and
(D3) Aµ =1,
x
hold.
Proof. Lemma7yieldsthatthesolutionsarenonnegativeandconservesums. Itsufficestoinclude
(cid:80) (cid:80)
step-lengths µ and µ and identify conditions under which µ u(x,0)=1 and µ u(0,t)=
x t x x t t
1. Giventheinitialcondition,theformersumisequaltoAµ . Hencetheassumption(D3). Finally,
x
(cid:16) (cid:17)n
since u(−µ ,t)=0, the equation (10) implies that u(0,nµ )=A 1− kµt . Consequently,
x t µx
1=Aµ (cid:88)∞ (cid:18)1− kµt(cid:19)n = Aµx.
t µ k
x
n=0
(cid:3)
Corollary 9. Let u(x,t) be a solution of (9). Then the space and time sections µ u(x,·) and
x
µ u(·,t) form probability mass functions if and only if k =1, µ <µ and A= 1 .
t t x µx
Proof. (D2) and (D3) hold if and only if k =1. Consequently, (D1) could be satisfied if and only
if µ <µ . (cid:3)
t x
Closer examination again reveals that the sections form probability mass functions of discrete
probability distributions related to Bernoulli counting processes.
Remark 10. Let us consider the solution (11). If we put A=k =µ =1 and µ =p we get
x t
(cid:18) (cid:19)
n
u(n,m·p)= (1−p)n−mpm, n≥m.
m
which forms, for each fixed n ∈ N , a probability mass function of the binomial distribution.
0
Similarly, for each fixed m∈N , p=µ -multiple forms a probability mass function of a version of
0 t
the negative binomial distribution (the value p·u(n,m·p) describes a probability that for m failures
we need n trials). Consequently, the solution of (9) describes a counting Bernoulli stochastic
process (see [2]).
8 PETRSTEHL´IK,JONA´SˇVOLEK
6. Discrete Space and General Time
Letusextendtheresultsfromthelasttwosectionsbyconsideringmoregeneraltimestructures.
LetTbeatimescalesuchthatminT=0andsupT=+∞. Inthisparagraphweconsiderdomains
Ω={(x,t):x∈µ Z,t∈T},
x
and the problem:

u∆t(x,t)+(cid:40)k∇xu(x,t)=0, (x,t)∈Ω,
(12) A, x=0,
u(x,0)=
 0, x(cid:54)=0,
where A>0, k >0 and ∇xu(x,t) is the backward difference defined in (1) and u∆t is the delta-
derivative in time variable. Since the space is discrete, we could again rewrite the equation in 12
into
k
(13) u∆t(x,t)=− (u(x,t)−u(x−µ ,t)).
µ x
x
In order to conserve the sign of solutions we assume that
kµ (t)
(TS1) 1− t >0,
µ
x
i.e. the condition which is similar to the positive regressivity in the time scale theory (e.g. [4,
Section 2.2]) or the so-called CFL condition in the discretization of the transport equation (e.g.
[13, Section 4.4]).
Let u be a solution of (12). One could use [14, Proposition 5.2] to show that u(x,t)=0 for all
x<0 is the unique solution there. Since u(−µx,t)=0, we could see that u∆t(0,t)=−µkxu(0,t).
Given the initial condition and assumption (TS1), we get u(0,t)=Ae (t;0), where e (t;0)
− k − k
µx µx
is a time scale exponential function (see [4, Section 2]).
lem 11. The solution of (12) satisfies
(i) lim u(0,t)=0,
t→∞
∞
(ii) (cid:82) u(0,t)∆t=Aµx.
k
0
Proof. (i) Follows directly from the assumption (TS1) and the properties of the exponential
functions [4, Section 2.2],
(ii)
∞ ∞
(cid:90) (cid:90) µ (cid:16) (cid:17) µ
u(0,t)∆t=A e (t;0)∆t= lim −A x e (t;0)−1 =A x.
−µkx t→∞ k −µkx k
0 0
(cid:3)
Unique solutions of u(mµ ,t) could be found using the variation of constants (see e.g. [4,
x
Theorem 2.77]). However, these computations depend critically on a particular time scale and
cannot be performed in general. For example, one could compute that the second branch of the
solution has the form
k (cid:90) t ∆τ
u(µ ,t)=A e (t;0) .
x µx −µkx 0 1− kµt(τ)
µx
This implies that we can’t derive closed-form solutions as in previous sections. Formally, these
solutions can be expressed as Taylor-like series with generalized polynomials whose form depends
onparticulartimescale(see[14]and[4,Section1.6]). Wedeterminethesesolutionsinspecialcases
(see Lemmata 1, 6 and 19). Therefore, we are forced use another means to show the properties of
solutions we are interested in.
TRANSPORT EQUATION ON SEMIDISCRETE DOMAINS AND POISSON-BERNOULLI PROCESSES 9
lem12. Letx∈µ N. If (TS1)issatisfiedandu(x−µ ,t)≥0andforallt∈Tandu(x−µ ,t)>
x x x
0 at least for one t∈T then u(x,t)≥0 for all t∈T.
Proof. First, note that u(x,0)=0 for all x>0. Consequently, (13) implies that u∆t(x,t)>0 at
the beginning of the support of u(x−µ ,t) and u(x,t) is strictly increasing there.
x
• If t is right-scattered then we can rewrite the equation (13) into
(cid:18) (cid:19)
kµ kµ
u(x,t+µ )= 1− t u(x,t)+ tu(x−µ ,t).
t µ µ x
x x
If u(x,t) ≥ 0, then this is the weighted average of two nonnegative values and thus
nonnegative as well.
• If t is right-dense then the equation (13) has the form
k k
u (x,t)=− u(x,t)+ u(x−µ ,t).
t µ µ x
x x
Since both u(x−µ ,t) ≥ 0 and u(x,t) ≥ 0, we have that u (x,t) ≥ − k u(x,t) and thus
x t µx
u(x,t) cannot become negative.
Following the induction principle (e.g. [4, Theorem 1.7]), we could see that u(x,t) ≥ 0 for all
t∈T. (cid:3)
Lemma 12 serves as the inductive step in the proof of the sign-conservation.
thm 13. If (TS1) holds then u(x,t)≥0 for all (x,t)∈Ω.
Proof. We prove the statement by mathematical induction. Firstly, u(0,t) = Ae (t;0) > 0.
− k
Secondly, if u(x,t)≥0 then Lemma 12 implies that u(x+µ ,t)≥0 which finishes thµex proof. (cid:3)
x
The following auxiliary lemma shows that the variation of constant formula which generates
further branches of solutions conserve zero-limits at infinity.
lem 14. Let us consider a time scale T, a constant K such that 1−µK > 0 and a function
∞
f :T→[0,∞) such that the integral (cid:82) f(t)∆t is finite. If we define g :T→[0,∞) by
0
t
(cid:90)
g(t)= e (t,σ(τ))f(τ)∆τ,
−K
0
then lim g(t)=0.
t→∞
∞
(cid:82)
Proof. Since f(t)∆(t) is finite we know that for each (cid:15) > 0 there exists T > 0 such that for all
0
t∈T, t>T the inequality
∞
(cid:90) (cid:15)
(14) f(τ)∆τ < ,
2
t
holds. Similarly,propertiesoftimescaleexponentialfunctionimplythatforeach(cid:15)>0andT >0
there exists R>T such that for all t∈T, t>R the following inequality is satisfied
T
(cid:90) (cid:15)
(15) e (t;σ(τ))∆τ < ,
−K 2F
0
10 PETRSTEHL´IK,JONA´SˇVOLEK
with F = maxf(t). Consequently, inequalities (14) and (15) imply that for each for each (cid:15) > 0
t∈T
there exists T >0 and R>T such that for all t>R
t
(cid:90)
g(t)= e (t,σ(τ))f(τ)∆τ
−K
0
T t
(cid:90) (cid:90)
= e (t;σ(τ))f(τ)∆τ + e (t;σ(τ))f(τ)∆τ
−K −K
0 T
T t
(cid:90) (cid:90)
≤F e (t;σ(τ))∆τ + f(τ)∆τ
−K
0 T
(cid:15) (cid:15)
<F +
2F 2
=(cid:15),
which implies that lim g(t)=0. (cid:3)
t→∞
Consequently, we are able to show that the integrals are constant for each fixed x≥0.
thm 15. If (TS1) holds and u(x,t) is a solution of (12) then
∞ ∞
(cid:90) (cid:90) µ
u(x,t)∆t= u(0,t)∆t=A x,
k
0 0
for all x∈µ N .
x 0
Proof. We proceed by mathematical induction.
• For x=0 the convergence of the integral to Aµx follows from Lemma 11 (ii).
k
• Let us fix x∈µ N and assume that the statement holds for a function u(x−µ ,t). If we
x x
integrate (13) we get
(cid:90) ∞ k (cid:18)(cid:90) ∞ (cid:90) ∞ (cid:19)
(16) u∆t(x,τ)∆τ =− u(x,τ)∆τ − u(x−µ ,τ)∆τ .
µ x
0 x 0 0
Let us concentrate on the left-hand side term. The variation of constants formula ([4,
Theorem 2.77]) implies that
t
(cid:90)
u(x,t)= e (t,σ(τ))u(x−µ ,τ)∆τ.
− k x
µx
0
Consequently,Lemma14impliesthat lim u(x,t)=0. Usingtheinitialconditionu(x,0)=
t→∞
0, we could rewrite the left-hand side of (16) into
(cid:90) ∞
u∆t(x,τ)∆τ = lim u(x,t)−u(x,0)=0.
t→∞
0
This implies that (16) could be rewritten into
k (cid:18)(cid:90) ∞ (cid:90) ∞ (cid:19)
0=− u(x,τ)∆τ − u(x−µ ,τ)∆τ ,
µ x
x 0 0
or equivalently into
(cid:90) ∞ (cid:90) ∞
u(x,τ)∆τ = u(x−µ ,τ)∆τ,
x
0 0
which finishes the proof.
(cid:3)
Finally, we show that the integrals (sums in this case) remains constant in time as well.