doi:10.1155/2011/867136
Research Article
Asymptotic Behavior of a Discrete Nonlinear Oscillator with Damping Dynamical System
Hadi Khatibzadeh
1, 21Department of Mathematics, Zanjan University, P.O. Box 45195-313, Zanjan, Iran
2School of Mathematics, Institute for Research in Fundamental Sciences (IPM), P.O. Box 19395-5746, Tehran, Iran
Correspondence should be addressed to Hadi Khatibzadeh,[email protected] Received 24 December 2010; Accepted 10 February 2011
Academic Editor: Istvan Gyori
Copyrightq2011 Hadi Khatibzadeh. 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 propose a new discrete version of nonlinear oscillator with damping dynamical system governed by a general maximal monotone operator. We show the weak convergence of solutions and their weighted averages to a zero of a maximal monotone operatorA. We also prove some strong convergence theorems with additional assumptions onA. This iterative scheme gives also an extension of the proximal point algorithm for the approximation of a zero of a maximal monotone operator. These results extend previous results by Br´ezis and Lions1978, Lions1978 as well as Djafari Rouhani and H. Khatibzadeh2008.
1. Introduction
Let H be a real Hilbert space with inner product ·,· and norm | · |. We denote weak convergence in H by and strong convergence by →. Let A be a nonempty subset of H×Hwhich we will refer to as anonlinearpossibly multivalued operator inH.Ais called monotoneresp. strongly monotoneify2−y1, x2−x1≥0resp.y2−y1, x2−x1≥α|x1−x2|2 for someα >0for allxi, yi∈A,i1,2.Ais maximal monotone ifAis monotone andIA is surjective, whereIis the identity operator onH.
Nonlinear oscillator with damping dynamical system, ut γut Aut0,
u0 u0, u0 u1, 1.1
where A is a maximal monotone operator and γ > 0, has been investigated by many authors specially for asymptotic behavior. We refer the reader to1–6and references in there.
Following discrete version of1.1,
un1 IλnA−1unαnun−un−1 1.2
is called inertial proximal method and has been studied in3. This iterative algorithm gives a method for approximation of a zero of a maximal monotone operator. In this paper, we propose another discrete version of1.1and study asymptotic behavior of its solutions. By using approximations
ut uth−ut−h
2h oh,
ut uth−2ut ut−h
h2 oh,
1.3
for1.1, we get
un1−2unun−1
h2n γun1−un−1
2hn Aun1 0. 1.4
By lettingβγ/2,λn1h2n/1βhnandαn βhn−1/βhn1, we get
un1Jλn11−αnunαnun−1, n≥0,
u−10, u0x∈H, 1.5
whereαnresp.λnis nonnegativeresp. positivesequence andJλ IλA−1. This discrete version gives also an algorithm for approximation of a zero of maximal monotone operator A. This algorithm extends proximal point algorithm which was introduced by Martinet in7 withλn λandαn 0 and then generalized by Rockafellar8. We investigate asymptotic behavior of solutions of1.5as discrete version of1.1which also extend previous results of9–11on proximal point algorithm.
Letwn : n
k1λk−1n
k1λkuk. Under suitable assumptions, we investigate weak and strong convergence ofwnandunto an element ofA−10if and only if{un}is bounded.
Therefore, A−10/φ if and only if {un} is bounded provided∞
n1λn ∞. Our results extend previous results in2,3,5.
Throughout the paper, we denoteAun1 1−αnunαnun−1−un1/λn1, and we assume the following assumptions on the sequence{αn}:
0≤αn ≤1, {αn}is nonincreasing andαn−→0 asn−→∞. 1.6
2. Main Results
In this section, we establish convergence of the sequence{un}or its weighted average to an element ofA−10. First we recall the following elementary lemma without proof.
Lemma 2.1. Suppose that{αn}is a nonnegative sequence and{λn}is a positive sequence such that ∞
n1λn ∞. Ifαn/λn → 0 asn → ∞, thenn
k1αk/n
k1λk → 0 asn → ∞.
We start with a weak ergodic theorem which extends a theorem of Lions11 see also 12page 139 Theorem 3.1 as well as10Theorem 2.1.
Theorem 2.2. Assume thatunis a solution to1.5and{αn}satisfies1.6. If∞
k1λk ∞and αn/λn → 0, thenwn p∈A−10asn → ∞if and only ifunis bounded.
Proof. Suppose thatwn p∈A−10by1.5; we get
un1−p≤Jλn11−αnunαnun−1−p≤1−αnun−pαnun−1−p. 2.1
This implies that
un1−p≤maxu1−p,u0−p. 2.2
Then{un}is bounded and this proves necessity. Now, we prove sufficiency. By monotonicity ofA, we have
Aun1, um1 Aum1, un1≤Aum1, um1 Aun1, un1 2.3
for allm, n ≥0. Multiplying both sides of the above inequality byλm1λn1and using1.5, we deduce
1−αnun−un1, λm1um1 αnun−1−un1, λm1um1
1−αmum−um1, λn1un1 αmum−1−um1, λn1un1
≤λm11−αnun−un1, un1 λm1αnun−1−un1, un1 λn11−αmum−um1, um1 λn1αmum−1−um1, um1.
2.4
Summing both sides of this inequality fromm0 tomk−1, we get
1−αn
un−un1,k−1
m0
λm1um1
αn
un−1−un1,k−1
m0
λm1um1
≤λn1|un1|k−1
m0
αm|um−1−um|k−1
m0
um1−um, λn1un1
k−1
m0
λm1
1−αnun−un1, un1
k−1
m0
λm1
αnun−1−un1, un1
λn1
k−1 m0
1−αm
2 |um|2− 1−αm 2 |um1|2
λn1
k−1 m0
αm
2 |um−1|2−αm
2 |um1|2 λn1|un1|k−1
m0
αm|um−1−um| uk−u0, λn1un1
k−1
m0
λm1
1−αnun−un1, un1
k−1
m0
λm1
αnun−1−un1, un1
λn1k−1
m0
1
2|um|2− 1 2|um1|2
λn1k−1
m0
αm
2 |um−1|2− αm 2 |um|2
.
2.5
Divide both sides of the above inequality byk−1
m0λm1and suppose thatknjandwnj p asj → ∞. By assumptions on{αn},{λn}andLemma 2.1, we have
1−αn
un−un1, p αn
un−1−un1, p
≤1−αnun−un1, un1 αnun−1−un1, un1. 2.6
This implies that
1−αnunαnun−1−un1, un1−p
≥0. 2.7
From1.6, we get
un1−pαnun−p≤un−pαn−1un−1−p. 2.8 By 1.6 and boundedness of{un}, we get limn→∞|un −p| exists. If wnk q, we obtain again limn→∞|un − q| exists. Therefore, limn→∞1/2|un − p|2 − |un − q|2, and hence limn→∞un, p − q exists. This follows that limn→∞wn, p − q exists. It implies that
q, p−q p, p−q and hence p q and wn p ∈ H asn → ∞. Now we prove p∈A−10. Suppose thatx, y∈A. By monotonicity ofAand Assumption1.6, we get
⎛
⎝x− n−1
i0
λi1 −1n−1
i0
λi1ui1, y
⎞
⎠
n−1
i0
λi1 −1n−1
i0
λi1
x−ui1, y
≥ n−1
i0
λi1 −1n−1
i0
λi1x−ui1, Aui1
n−1
i0
λi1
−1n−1
i0
x−ui1,1−αiuiαiui−1−ui1
n−1
i0
λi1
−1n−1
i0
−1−αiui1−x, ui−x−αiui1−x, ui−1−x |ui1−x|2
≥ n−1
i0
λi1 −1n−1
i0
1 2
|ui1−x|2− |ui−x|2 1
2
αi|ui−x|2−αi−1|ui−1−x|2 .
2.9
Lettingn → ∞, we get:x−p, y≥0. By maximality ofA, we getp∈A−10.
Remark 2.3. Since range of Jλn is DA the domain of A, as a trivial consequence of Theorem 2.2, we have that IfDAis bounded thenA−10/φ.
In the following, we prove a weak convergence theorem. Since the necessity is obvious, we omit the proof of necessity in the next theorems.
Theorem 2.4. Letunbe a solution to1.5andλn ≥λ0 >0. If{αn}satisfies1.6, thenun p∈ A−10asn → ∞if and only if{un}is bounded.
Proof. Since assumption on{λn}implies that∞
n1λn ∞, from1.5and2.7, we get λ2n1|Aun1|2un1−pλn1Aun12−un1−p2−2λn1
Aun1, un1−p
≤1−αn un−p
αn
un−1−p2−un1−p2
≤1−αnun−p2αnun−1−p2−un1−p2
≤αn−1un−1−p2−αnun−p2un−p2−un1−p2.
2.10
The last inequality follows from Assumption1.6. Summing both sides of this inequality fromn1 tomand lettingm → ∞, since{αn}satisfies1.6, we have
∞ n1
λ2n1|Aun1|2<∞. 2.11
By assumption on {λn}, we have |Aun| → 0 asn → ∞. Assumeunj qasj → ∞, by the monotonicity of A, we haveAum−Aunj, um −unj ≥ 0. Lettingj → ∞, we get Aum, um−q≥0. Similar to the proof ofTheorem 2.2, limm→∞|um−q|exists. This implies thatun qp∈A−10asn → ∞.
In two following, theorems we show strong convergence of {un} under suitable assumptions on operatorAand the sequence{λn}.
Theorem 2.5. Assume that I A−1 is compact and ∞
n1λ2n ∞. Ifαn satisfies 1.6, then un → p∈A−10asn → ∞if and only if{un}is bounded.
Proof. By2.11and assumption on{λn}, we get lim infn→∞|Aun| 0 andun pasn →
∞. Therefore, there exists a subsequence{Aunj}of{Aun}such that|Aunj| → 0 asj → ∞ and{unjAunj} is bounded. The compacity ofIA−1 implies that{unj}has a strongly convergent subsequencewe denote again by{unj}top. By the monotonicity ofA, we have Aun−Aunj, un−unj≥ 0. Lettingj → ∞, we obtainAun, un−p ≥0. Now, the proof of Theorem 2.2shows that limn→∞|un−p|2exists. This implies thatun → pasn → ∞.
Theorem 2.6. Assume thatAis strongly monotone operator and∞
n1λn ∞. If{αn}satisfies 1.6, thenun → p∈A−10asn → ∞if and only if{un}is bounded.
Proof. By the proof ofTheorem 2.2,wn p∈A−10asn → ∞, and limn→∞|un−p|2exists.
SinceAis strongly monotone, we have
Aun1, un1−p
≥αun1−p2. 2.12 Multiplying both sides of2.12byλn1and summing fromn1 tom, we have
αm
n1
λn1un1−p2≤m
n1
1−αnunαnun−1−un1, un1−p
m
n1
1−αn
un−p, un1−p αn
un−1−p, un1−p
−un1−p2
≤ 1 2
m n1
1−αnun−p2αnun−1−p2−un1−p2
≤ 1 2
m n1
un−p2−un1−p2αn−1un−1−p2−αnun−p2 .
2.13
The last inequality follows from Assumption1.6. Lettingm → ∞, we get:
∞ n1
λn1un1−p2<∞. 2.14 So, lim infn→∞|un−p|20. This implies thatun → pasn → ∞.
In the following theorem, we assume that A ∂ϕ, where ϕ is a proper, lower semicontinuous and convex function and Argminϕ / φ.
Theorem 2.7. LetA∂ϕ, whereϕis a proper, lower semicontinuous, and convex function. Assume thatA−10is nonempty (i.e.,ϕhas at least one minimum point) and∞
n1λn ∞. If{αn}satisfies 1.6, thenun p∈A−10asn → ∞.
Proof. SinceAis subdifferential ofϕandp∈A−10, by Assumption1.6, we have ϕun1−ϕ
p
≤ 1 λn1
1−αnunαnun−1−un1, un1−p
≤ 1 λn1
1−αn 2
un−p2−un1−p2 αn
2
un−1−p2−un1−p2
≤ 1 λn1
1 2
un−p2−un1−p2 1
2
αn−1un−1−p2−αnun−p2 . 2.15
Multiplying both sides of the above inequality byλn1 and summing fromn 1 to mand lettingm → ∞, we get
∞ n1
λn1
ϕun1−ϕ p
<∞. 2.16
By assumption on{λn}, we deduce lim inf
n→∞ϕun ϕ p
. 2.17
By convexity ofϕ, we have
ϕun1−1−αnϕun−αnϕun−1
≤ϕun1−ϕ1−αnunαnun−1
≤ 1
λn11−αnunαnun−1−un1, un1−1−αnun−αnun−1
≤0.
2.18
Therefore,
ϕun1≤1−αnϕun αnϕun−1. 2.19
From2.19, by Assumption1.6, we get
ϕun1 αnϕun≤ϕun αn−1ϕun−1. 2.20
Again by2.19, we get
ϕun≤max
ϕu0, ϕu1
2.21
for alln >1. By2.20and2.21, we have that
nlim→∞
ϕun1 αnϕun
2.22
exists. From Assumptions1.6,2.17, and2.21, we get
nlim→∞ϕun ϕ p
. 2.23
Ifunj q, thenϕp lim infj→∞ϕunj≥ ϕq. This implies thatq∈A−10. On the other hand, for eachp ∈A−10by1.5, we get2.7. The proof ofTheorem 2.2implies that there exists limn→∞|un−p|. Then the theorem is concluded by Opial’s Lemmasee13.
Acknowledgment
This research was in part supported by a Grant from IPMno. 89470017.
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