Volume 2010, Article ID 975740,8pages doi:10.1155/2010/975740
Research Article
On a Max-Type Difference Inequality and Its Applications
Stevo Stevi ´c
1and Bratislav D. Iri ˇcanin
21Mathematical Institute of the Serbian Academy of Sciences, Knez Mihailova 36/III, 11000 Beograd, Serbia
2Faculty of Electrical Engineering, University of Belgrade, Bulevar Kralja Aleksandra 73, Belgrade 11120, Serbia
Correspondence should be addressed to Stevo Stevi´c,[email protected] Received 1 November 2009; Accepted 31 March 2010
Academic Editor: Leonid Berezansky
Copyrightq2010 S. Stevi´c and B. D. Iriˇcanin. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We prove a useful max-type difference inequality which can be applied in studying of some max- type difference equations and give an application of it in a recent problem from the research area.
We also give a representation of solutions of the difference equationxnmax{xan−11 , . . . , xan−kk }.
1. Introduction
The investigation of max-type difference equations attracted some attention recently; see, for example,1–20and the references therein. In the beginning of the study of these equations the following difference equation was investigated:
xnmax A1n
xn−1,A2n
xn−2, . . . ,Akn
xn−k
, n∈N0, 1.1
where k ∈ N,Ain , i 1, . . . , k, are real sequences mostly constant or periodic, and the initial valuesx−1, . . . , x−kare different from zerosee, e.g., monograph6or paper19and the references therein.
The study of the next difference equation
xnmax
Bn0, Bn1
xn−pr1 1 xsn−q1 1, Bn2
xrn−p2 2
xsn−q2 2, . . . , Bkn
xrn−pk k xsn−qk k
, n∈N0, 1.2
wherepi, qiare natural numbers such thatp1 < p2 <· · ·< pk,q1 < q2 <· · ·< qk,ri, si ∈R 0,∞, i 1, . . . , k, andk ∈ N, was proposed by the first author in numerous talks; see, for example,11,13. For some results in this direction see1,4,5,7,8,12,14–18,20.
A particular case of the difference equation
ynmax
A
yn−1· · ·yn−m1, 1
yn−m−1· · ·yn−2m1
, n∈N0, 1.3
arises naturally in certain models in automatic control see 9. By the change xn ynyn−1· · ·yn−m1the equation is transformed into the equation
xnmax
A, xn−1
xn−m
, n∈N0, 1.4
which is a special case of1.2and which is a natural prototype for the equation.
The following result, which extends the main result from the study in18was proved by the first author in17 see also16.
Theorem A. Every positive solution to the difference equation
xnmax A1
xαn−11 , A2
xαn−22 , . . . , Ak
xαn−kk
, n∈N0, 1.5
where−1< αi<1, Ai≥0, i1, . . . , k, converges to max1≤i≤k{A1/αi i1}.
Here we continue to study1.5by considering the cases when some ofαi’s are equal to one. We also give a representation of well-defined solutions of the difference equation xnmax{xan−11 , . . . , xan−kk },whereai∈R,i1, . . . , k.
2. Main Results
In this section we prove the main results of this note. Before this we formulate the following very useful auxiliary result which can be found in10and give a definition.
Lemma A. Letann∈Nbe a sequence of positive numbers which satisfy the inequality
ank≤qmax{ank−1, ank−2, . . . , an}, forn∈N, 2.1
whereq >0 andk∈Nare fixed. Then there exists anM >0 such that
an≤M k
qn
, n∈N. 2.2
Definition 2.1. For a sequencexn∞n−s,s∈N0, we say that it converges to zero geometrically if there is aq∈0,1andM >0 such that
|xn| ≤Mqn, 2.3
forn−s, . . . ,−1,0,1, . . . .
Now we are in a position to formulate and prove the main results of this note.
Proposition 2.2. Assume thatan∞n−k is a sequence of nonnegative numbers satisfying the difference inequality
an≤max{α1an−1−d1, . . . , αkan−k−dk}, n∈N0, 2.4
wherek ∈ N,αi ∈ 0,1,di ∈ R,i∈ {1, . . . , k}, and if, for somei,αi 1, thendi >0. Then the sequenceanconverges geometrically to zero asn → ∞.
Proof. Letβ∈0,1be chosen such that
max{a−k, . . . , a−1} ≤ cm
1−β, 2.5
where
cmmin
1,min
i:αi1{di}
. 2.6
Then from2.4and using the fact thatanare nonnegative numbers, we have that
an≤max
A max
i:αi∈0,1{an−i},max
j:αj1 an−j−cm
, 2.7
whereAmaxi:αi∈0,1{αi}.
From2.7,2.5and since 0<max{A, β}<1, we have that
a0≤max
A max
i:αi∈0,1{a−i},max
j:αj1 a−j−cm
≤max Acm
1−β, cm
1−β−cm,
max Acm
1−β, βcm
1−β
< cm
1−β.
2.8
Now assume thatan≤cm/1−β, for 0≤n≤n0−1.Then from2.4we get
an0≤max
A max
i:αi∈0,1{an0−i},max
j:αj1 an0−j−cm
≤max Acm
1−β, cm
1−β−cm,
max Acm
1−β, βcm
1−β
< cm
1−β.
2.9
Inequalities2.8and2.9along with the method of induction show that
0≤an≤ cm
1−β, forn∈ {−s, . . . ,−1} ∪N0. 2.10
Now note that from2.10we have that
an−cm≤βan, forn∈ {−s, . . . ,−1} ∪N0. 2.11
From2.7,2.11and the choice ofcm, it follows that forn∈N0
an≤max
A max
i:αi∈0,1{an−i}, βmax
j:αj1 an−j
≤max A, β
max1≤i≤k{an−i}.
2.12
Applying Lemma A in inequality2.12withqmax{A, β}, the result follows.
Remark 2.3. Note that the constant β in the proof of Proposition 2.2 depends on initial conditions of solutions to difference equation2.4, so that this is not a uniform constant.
Lemma 2.4. Consider the difference equation
znmin{C1−α1zn−1, C2−α2zn−2, . . . , Ck−αkzn−k}, n∈N0, 2.13
wherek∈N,Ci∈R, αj∈R,i1, . . . , k, and there isi0∈ {1, . . . , k}such thatCi0 0.Then
|zn| ≤max{|α1||zn−1| −C1,|α2||zn−2| −C2, . . . ,|αk||zn−k| −Ck}, n∈N0. 2.14
Proof. If all terms in the right-hand side of 2.13 are nonnegative then clearly 0 ≤ zn ≤
−αi0zn−i0, so that
|zn| ≤ |αi0||zn−i0||αi0||zn−i0| −Ci0. 2.15
Otherwise, the setS ⊆ {1, . . . , k} of all indices for which the terms in2.13are negative is nonempty, so thatznmini∈S{Ci−αizn−i}<0.
From this and since for suchi∈S,αizn−imust be positive, it follows that
|zn|max
i∈S {αizn−i−Ci}max
i∈S {|αi||zn−i| −Ci}. 2.16
From2.15and2.16inequality2.14easily follows.
ByProposition 2.2andLemma 2.4we obtain the following theorem.
Theorem 2.5. Consider the difference equation xnmax
A1
xαn−11 , A2
xαn−22 , . . . , Ak
xαn−kk
, n∈N0, 2.17
wherek ∈N, 0≤Ai≤1,−1≤αi ≤1,−1< αiAi<1 for eachi∈ {1, . . . , k}, andAi 1 for at least onei∈ {1, . . . , k}.Then every positive solution of 2.17converges to one.
Proof. Taking the logarithm of2.17and using the changeyn−lnxn, we obtain that
yn min
i:Ai/0
ln 1
Ai −αiyn−i
, n∈N0. 2.18
Now note that ln1/Ai ≥ 0 for thoseisuch that Ai/0, since Ai ∈ 0,1, and there is an S1 ⊂ {1, . . . , k}such that ln1/Ai 0 wheni ∈ S1.ByLemma 2.4we have that for every n∈N0
yn≤ max
i:Ai/0
|αi|yn−i−ln 1 Ai
. 2.19
From 2.19, noticing that if |αi| 1 and Ai/0, thenAi ∈ 0,1so that ln1/Ai > 0 and by applyingProposition 2.2we obtain that|yn| → 0 asn → ∞, from which it follows that xne−yn → 1 asn → ∞, as desired.
Remark 2.6. Recently Gelis¸ken and C¸ inar in the paper: “On the global attractivity of a max- type difference equation,” Discrete Dynamics in Nature and Society, vol. 2009, Article ID 812674, 5 pages, 2009, have studied the asymptotic behavior to positive solutions of the difference equation
xnmax A
xn−1, 1 xn−3α
, n∈N0, 2.20
whereα ∈ 0,1 andA > 0. They claim that ifA ∈ 0,1, then every positive solution to 2.20converges to one. However the proof given there cannot be regarded as complete one.
Namely, they first formulated the following lemma.
Lemma 2.7. Letynbe a solution to the difference equation
ynmax 1−yn−1,−αyn−3
, n∈N0. 2.21
Then for alln∈N0, the following inequality holds:
yn≤max yn−1−1, αyn−3. 2.22 Then they tried to show thatyn → 0 asn → ∞. Note that2.21is obtained by the changexn Ayn from2.20, so that if it is proved thatyn → 0 asn → ∞then xn → 1 asn → ∞,from which the claim follows. In the beginning of the proof of the theorem they choose a numberβsuch that 0 < |yn−1| −1 ≤β|yn|,but do not say if these inequalities hold for allnor not, which is a bit confusing. Note that for differentnthe chosen numberβcan be different, which means that in this caseβmight be a function ofn. Hence it is important that these inequalities hold for everyn∈N0∪ {−2,−1},which was not proved. This motivated us to proveProposition 2.2which, among others, removes the gap.
Now we present a representation of solutions of a particular case of1.5. The first author would like to express his sincere thanks to Professor L. Berg for a nice communication regarding this2.
Theorem 2.8. Consider the equation
xnmax xan−11 , . . . , xan−kk
, n∈N0, 2.23
where k ∈ N,ai ∈ R, i 1, . . . , k.Then every well-defined solution of equation 2.23 has the following form:
xnd
k j1ai
jn j
n , 2.24
where
nk k
≤i1n · · ·ikn ≤n1, n∈N0, 2.25
ijn ≥0,j 1, . . . , k,and wherednis equal to one of the initial valuesx−k, . . . , x−1. Moreover, if−1< ai <1,i1, . . . , k,thenxn → 1 asn → ∞.
Proof. The casek1 is well known and simple. Just note thatxnxa−1n11 . Hence assume that k≥2.We prove the result by induction. Forn0 we have
x0max xa−11, . . . , xa−kk
. 2.26
Note thatx0can be equal to one of the numbersxa−11, . . . , xa−kk and that
xa−ii xai
i /ja0j
−i , i1, . . . , k, 2.27
which is nothing but formula2.24in this case. From this we also have that 0k
k
1i11 · · ·ik1 110, 2.28
which is2.25in this case.
Now assume that we have proved2.24and2.25forl≤n−1.Then
xnmax xan−11 , . . . , xan−kk
max
⎧⎨
⎩d
k j1ai
jn−1δj 1 j
n−1 , . . . , d
k j1ai
jn−kδj k j
n−k
⎫⎬
⎭, 2.29
whereδijis the Kronecker symbol andlk/k≤i1l · · ·ikl ≤l,forl≤n−1.Thus
i1n−s· · ·ikn−sδsj≤n−s11≤n1, 2.30
fors1, . . . , kand
i1n−s· · ·ikn−sδjs≥
n−sk k
1
nk−sk k
≥ nk
k
, 2.31
s1, . . . , k.Hence the first statement follows by induction.
Now assume that max1≤j≤k{|aj|}<1. From this and2.25we have
k
j1
aijjn ≤
max1≤j≤k ajnk/k
. 2.32
Inequality2.32, the assumption max1≤j≤k{|aj|} < 1, and2.24imply thatxn tends to 1 as n → ∞,finishing the proof of the theorem.
Remark 2.9. Note that formula2.24holds for each value of parametersaj, j 1, . . . , k, and for all solutions whose initial values are different from zero if one of these exponents is negative.
Remark 2.10. The second statement inTheorem 2.8follows easily also from Lemma A.
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