(1)JJ J I II Go back Full Screen Close Quit ITERATIVE SOLUTIONS OF NONLINEAR EQUATIONS WITH φ-STRONGLY ACCRETIVE OPERATORS SHIN MIN KANG, CHI FENG and ZEQING LIU Abstract

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ITERATIVE SOLUTIONS OF NONLINEAR EQUATIONS WITH

φ

SHIN MIN KANG, CHI FENG and ZEQING LIU

Abstract. Suppose thatX is an arbitrary real Banach space andT :X→X is a Lipschitz continuousφ-strongly accretive operator or uniformly continuousφ-strongly accretive operator.

We prove that under different conditions the three-step iteration methods with errors converge strongly to the solution of the equationT x=f for a givenf∈X.

1. Introduction

LetX be a real Banach space with norm k · k and dualX^∗, and J denote the normalized duality mapping from X into 2^X∗ given by

J(x) ={f ∈X^∗:kfk²=kxk²=hx, fi}, x∈X,

whereh·,·iis the generalized duality pairing. In this paper,I denotes the identity operator onX,R⁺ andδ(K) denote the set of nonnegative real numbers and the diameter ofKfor anyK⊆X, respectively. An operatorT with domainD(T) and rangeR(T) inX is called φ-strongly accretiveif there exists a strictly increasing functionφ:R⁺→R⁺withφ(0) = 0 such that for anyx, y∈D(T) there existsj(x−y)∈J(x−y) such that

hT x−T y, j(x−y)i ≥φ(kx−yk)kx−yk.

(1.1)

Received April 9, 2007.

2000Mathematics Subject Classification. Primary 47H05, 47H10, 47H15.

Key words and phrases. φ-strongly accretive operators; three-step iteration method with errors; Banach spaces.

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If there exists a positive constant k >0 such that (1.1) holds with φ(kx−yk) replaced by kkx−yk, then T is called strongly accretive. The accretive operators were introduced independently in 1967 by Browder [1] and Kato [8]. An early fundamental result in the theory of accretive operator, due to Browder, states the initial value problem

du

dt +T u= 0, u(0) =u₀ (1.2)

is solvable ifT is locally Lipschitz and accretive onX. Martin [11] proved that ifT :X →X is strongly accretive and continuous, thenT is subjective so that the equation

T x=f (1.3)

has a solution for any given f ∈ X. Using the Mann and Ishikawa iteration methods with errors, Chang [3], Chidume [4], [5], Ding [7], Liu and Kang [10] and Osilike [12], [13]

obtained a few convergence theorems for Lipschitz φ-strongly accretive operators. Chang [2] and Yin, Liu and Lee [16] also got some convergence theorems for uniformly continuous φ-strongly accretive operators.

The purpose of this paper is to study the three-step iterative approximation of solution to equation (1.3) in the case when T is a Lipschitzφ-strongly accretive operator andX is a real Banach space. We also show that ifT :X →X is a uniformly continuousφ-strongly accretive operator, then the three-step iteration method with errors converges strongly to the solution of equation (1.3). Our results generalize, improve the known results in [2]–[7], [10], [12], [13] and [15].

2. Preliminaries

The following Lemmas play a crucial role in the proofs of our main results.

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Lemma 2.1 ([7]). Suppose that φ : R⁺ → R⁺ is a strictly increasing function with φ(0) = 0. Assume that{rn}^∞_n=0, {sn}^∞_n=0, {kn}^∞_n=0 and{tn}^∞_n=0 are sequences of nonnega- tive numbers satisfying the following conditions:

∞

X

n=0

kn <∞,

∞

X

n=0

tn<∞,

∞

X

n=0

sn=∞ (2.1)

and

r_n+1≤(1 +k_n)r_n−s_nr_n φ(r_n+1)

1 +rn+1+φ(rn+1)+t_n forn≥0.

(2.2)

Thenlim_n→∞rn= 0.

Lemma 2.2 ([10]). Suppose that X is an arbitrary Banach space andT :X →X is a continuous φ-strongly accretive operator. Then the equation T x=f has a unique solution for any f ∈X.

Lemma 2.3([9]).Let{αn}^∞_n=0,{βn}^∞_n=0and{γn}^∞_n=0be three nonnegative real sequences satisfying the inequality

α_n+1≤(1−ω_n)α_n+ω_nβ_n+γ_n forn≥0, where {ωn}^∞_n=0 ⊂ [0,1], P∞

n=0ωn = ∞, lim_n→∞βn = 0 and P∞

n=0γn < ∞. Then lim_n→∞αn= 0.

3. Main Results

Theorem 3.1. Suppose that X is an arbitrary real Banach space and T : X → X is a Lipschitz φ-strongly accretive operator. Assume that {un}^∞_n=0, {vn}^∞_n=0, {wn}^∞_n=0 are se- quences inXand{an}^∞_n=0,{bn}^∞_n=0and{cn}^∞_n=0are sequences in[0,1]such that{kwnk}^∞_n=0

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is bounded and

∞

X

n=0

a²_n<∞,

∞

X

n=0

anbn <∞,

∞

X

n=0

kunk<∞,

∞

X

n=0

kvnk<∞, (3.1)

∞

X

n=0

an=∞.

(3.2)

For any given f ∈ X, define S : X → X by Sx= f +x−T x for all x∈ X. Then the three-step iteration sequence with errors{xn}^∞_n=0 defined for arbitraryx0∈X by

zn= (1−cn)xn+cnSxn+wn, y_n= (1−b_n)x_n+b_nSz_n+v_n,

xn+1= (1−an)xn+anSyn+un, n≥0 (3.3)

converges strongly to the unique solutionq of the equationT x=f. Moreover kxn+1−qk ≤[1 + (3 + 3L³+L⁴)a²_n+L(1 +L²)anbn]kxn−qk

−A(xn+1, q)ankxn−qk+anbnL²(3 +L)kwnk +a_nL(3 +L)kvnk+ (3 +L)kunk

(3.4)

forn≥0, where A(x, y) =1+kx−yk+φ(kx−yk)^φ(kx−yk) ∈[0,1) forx, y∈X.

Proof. It follows from Lemma2.2that the equationT x=f has a unique solutionq∈X. Let L⁰ denote the Lipschitz constant of T. From the definition of S we know that q is a fixed point of S and S is also Lipschitz with constantL= 1 +L⁰. Thus for any x, y∈X, there exists j(x−y)∈J(x−y) such that

h(I−S)x−(I−S)y, j(x−y)i ≥A(x, y)kx−yk².

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This implies that

h(I−S−A(x, y))x−(I−S−A(x, y))y, j(x−y)i ≥0 and it follows from Lemma 1.1 of Kato [8] that

kx−yk ≤ kx−y+r[(I−S−A(x, y))x−(I−S−A(x, y))y]k (3.5)

forx, y∈X andr >0. From (3.3) we conclude that for each n≥0 xn=xn+1+anxn−anSyn−un

= (1 +an)xn+1+an(I−S−A(xn+1, q))xn+1−(I−A(xn+1, q))anxn

+an(Sxn+1−Syn) + (2−A(xn+1, q))a²_n(xn−Syn)

−[1 + (2−A(x_n+1, q))a_n]u_n (3.6)

and

q= (1 +a_n)q+a_n(I−S−A(x_n+1, q))q−(I−A(x_n+1, q))a_nq.

(3.7)

It follows from (3.5)–(3.7) that

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kxn−qk

=k(1 +a_n)x_n+1+a_n(I−S−A(x_n+1, q))x_n+1−(I−A(x_n+1, q))a_nx_n +a_n(Sx_n+1−Sy_n) + (2−A(x_n+1, q))a²_n(x_n−Sy_n)

−[1 + (2−A(xn+1, q))an]un−(1 +an)q−an(I−S−A(xn+1, q))q + (I−A(xn+1, q))anqk

≥(1 +an)

xn+1−q+ an

1 +an

[(I−S−A(xn+1, q))xn+1

−(I−S−A(xn+1, q))q

−an(1−A(xn+1, q))kxn−qk

−(2−A(xn+1, q))a²_nkxn−Synk −ankSxn+1−Synk

−[1 + (2−A(xn+1, q))an]kunk

≥(1 +a_n)kxn+1−qk −a_n(1−A(x_n+1, q))kxn−qk

−(2−A(x_n+1, q))a²_nkxn−Sy_nk −a_nkSxn+1−Sy_nk

−[1 + (2−A(xn+1, q))an]kunk, which implies that

kxn+1−qk

≤1 + (1−A(xn+1, q))an

1 +a_n kx_n−qk+ (2−A(x_n+1, q))a²_nkx_n−Sy_nk +ankSxn+1−Synk+ [1 + (2−A(xn+1, q))an]kunk

≤(1−A(xn+1, q)an+a²_n)kxn−qk+ 2a²_nkxn−Synk +a_nkSx_n+1−Sy_nk+ (1 + 2a_n)ku_nk

(3.8)

forn≥0. By (3.3) we get that

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kzn−qk ≤(1−cn)kxn−qk+cnkSxn−qk+kwnk

≤(1−c_n)kxn−qk+Lc_nkxn−qk+kwnk

≤Lkxn−qk+kwnk, (3.9)

kyn−qk ≤(1−bn)kxn−qk+bnkSzn−qk+kvnk

≤(1−b_n)kx_n−qk+Lb_nkz_n−qk+kv_nk, (3.10)

kxn−Sz_nk ≤ kxn−qk+kSzn−qk ≤ kxn−qk+Lkzn−qk, (3.11)

kxn−ynk ≤bnkxn−Sznk+kvnk (3.12)

and

kSyn−ynk ≤ kSyn−qk+kyn−qk ≤(1 +L)kyn−qk (3.13)

forn≥0. From (3.9)–(3.13) we obtain that

kxn−Sy_nk ≤(1 +L³)kxn−qk+L²b_nkwnk+Lkvnk (3.14)

and

kSxn+1−Synk ≤(Lbn+L³bn−Lanbn−L³anbn+L³an+L⁴an)kxn−qk + (L²bn+L³anbn)kwnk+ (L+L²an)kvnk+Lkunk (3.15)

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forn≥0. It follows from (3.8), (3.14) and (3.15) that

kx_n+1−qk ≤[1 + (3 + 3L³+L⁴)a²_n+L(1 +L²)a_nb_n]kx_n−qk

−A(x_n+1, q)a_nkxn−qk+a_nb_nL²(3 +L)kwnk + (3 +L)ankvnk+ (3 +L)kunk

(3.16)

forn≥0. Set

rn=kxn−qk, kn = (3 + 3L³+L⁴)a²_n+L(1 +L²)anbn, sn=an, tn =anbnL²(3 +L)kwnk+anL(3 +L)kvnk+ (3 +L)kunk forn≥0.

Then (3.16) yields that

r_n+1≤(1 +k_n)r_n−s_nr_n φ(rn+1)

1 +r_n+1+φ(r_n+1)+t_n forn≥0.

(3.17)

It follows from (3.1), (3.2), (3.17) and Lemma2.1that rn →0 asn→ ∞. That isxn →q

as n→ ∞. This completes the proof.

Remark 3.2. Theorem 3.1 extends Theorem 5.2 of [3], Theorem 1 of [4], Theorem 2 of [5], Theorem 1 of [6], Theorem 3.1 of [10], Theorem 1 of [12], Theorem 1 of [13] and Theorem 4.1 of [15].

Theorem 3.3. Let X, {un}^∞_n=0, {vn}^∞_n=0, {wn}^∞_n=0, {an}^∞_n=0, {bn}^∞_n=0 and {cn}^∞_n=0 be as in Theorem 3.1 and T : D(T) ⊂X → X be a Lipschitz φ-strongly accretive operator.

Suppose that the equation T x=f has a solution q∈D(T) for some f ∈X. Assume that the sequences {xn}^∞_n=0, {yn}^∞_n=0 and {zn}^∞_n=0 generated from an arbitrary x0 ∈ D(T) by (3.3) are contained in D(T). Then {xn}^∞_n=0, {yn}^∞_n=0 and {zn}^∞_n=0 converge strongly to q and satisfied (3.4).

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The proof of Theorem3.3uses the same idea as that of Theorem3.1. So we omit it.

Remark 3.4. Theorem 3.1 in [7] and Theorem 3.2 in [10] are special cases of our Theorem3.3.

Theorem 3.5. Suppose that X is an arbitrary real Banach space and T :X→X is a uniformly continuous φ-strongly accretive operator, and the range of either(I−T)orT is bounded. For any f ∈X, define S : X → X by Sx= f+x−T x for all x∈X and the three-step iteration sequence with errors{xn}^∞_n=0 by

x0, u0, v0, w0∈X,

z_n=a⁰⁰_nx_n+b⁰⁰_nSx_n+c⁰⁰_nw_n, y_n =a⁰_nx_n+b⁰_nSz_n+c⁰_nv_n,

xn+1=anxn+bnSyn+cnun, n≥0, (3.18)

where{un}^∞_n=0, {vn}^∞_n=0 and{wn}^∞_n=0 are arbitrary bounded sequences in X and{an}^∞_n=0, {bn}^∞_n=0,{cn}^∞_n=0, {a⁰_n}^∞_n=0,{b⁰_n}^∞_n=0, {c⁰_n}^∞_n=0,{a⁰⁰_n}^∞_n=0, {b⁰⁰_n}^∞_n=0 and{c⁰⁰_n}^∞_n=0 are real se- quences in[0,1]satisfying the following conditions

a_n+b_n+c_n= 1, a⁰_n+b⁰_n+c⁰_n= 1, a⁰⁰_n+b⁰⁰_n+c⁰⁰_n= 1, b_n+c_n∈(0,1), n≥0, (3.19)

∞

X

n=0

bn= +∞, lim

n→∞bn= lim

n→∞b⁰_n= lim

n→∞c⁰_n = lim

n→∞

cn

b_n+c_n = 0.

(3.20)

Then the sequence{xn}^∞_n=0converges strongly to the unique solution of the equationT x=f.

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Proof. It follows from Lemma2.2that the equationT x=f has a unique solutionq∈X. By (1.2) we have

hT x−T y, j(x−y)i=h(I−S)x−(I−S)y, j(x−y)i ≥A(x, y)kx−yk², whereA(x, y) = 1+kx−yk+φ(kx−yk)^φ(kx−yk) ∈[0,1) forx, y∈X. This implies that

h(I−S−A(x, y))x−(I−S−A(x, y))y, j(x−y)i ≥0 forx, y∈X. It follows from Lemma 1.1 of Kato [8] that

kx−yk ≤ kx−y+r[(I−S−A(x, y))x−(I−S−A(x, y))y]k (3.21)

forx, y∈X andr >0. Now we show thatR(S) is bounded. IfR(I−T) is bounded, then kSx−Syk=k(I−T)x−(I−T)yk ≤δ(R(I−T))

forx, y∈X.IfR(T) is bounded, we get that

kSx−Syk=k(x−y)−(T x−T y)k

≤φ⁻¹(kT x−T yk) +kT x−T yk

≤φ⁻¹(δ(R(T))) +δ(R(T)) forx, y∈X. HenceR(S) is bounded. Put

d_n=b_n+c_n, d⁰_n=b⁰_n+c⁰_n, d⁰⁰_n=b⁰⁰_n+c⁰⁰_n forn≥0 and

D= max{kx₀−qk,

sup{kx−qk:x∈ {un, vn, wn, Sxn, Syn, Szn:n≥0}}}.

(3.22)

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By (3.18) and (3.22) we conclude that

max{kxn−qk,kyn−qk,kzn−qk} ≤D forn≥0.

(3.23)

Using (3.18) we obtain that

(1−d_n)x_n =x_n+1−d_nSy_n−c_n(u_n−Sy_n)

= [1−(1−A(xn+1, q))dn]xn+1+dn(I−S−A(xn+1, q))xn+1

+dn(Sxn+1−Syn)−cn(un−Syn).

(3.24) Note that

(1−dn)q= [1−(1−A(xn+1, q))dn]q+dn(I−S−A(xn+1, q))q.

(3.25)

It follows from (3.21) and (3.23)–(3.25) that

(1−dn)kxn−qk ≥[1−(1−A(xn+1, q))dn]kxn+1−q

+ d_n

1−(1−A(xn+1, q))dn

[(I−S−A(xn+1, q))xn+1

−(I−S−A(xn+1, q))q]k −dnkSxn+1−Synk −cnkun−Synk

≥[1−(1−A(xn+1, q))dn]kxn+1−qk −dnkSxn+1−Synk −2Dcn. That is

kxn+1−qk ≤ 1−dn

1−(1−A(x_n+1, q))d_nkxn−qk

+ d_n

1−(1−A(xn+1, q))dn

kSxn+1−Sy_nk+ 2Dc_n

1−(1−A(xn+1, q))dn

≤[1−(1−A(xn+1, q))dn]kxn−qk+M dnkSxn+1−Synk+M cn

(3.26)

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forn≥0, whereM is some constant. In view of (3.18)–(3.20) we infer that kxn+1−ynk ≤ kxn+1−xnk+kyn−xnk

≤bnkSyn−xnk+cnkun−xnk+b⁰_nkSzn−xnk+c⁰_nkvn−xnk

≤bnkSyn−xnk+cnkun−xnk+b⁰_nkSzn−znk+c⁰_nkvn−xnk +b⁰_n(b⁰⁰_nkSxn−xnk+c⁰⁰_nkwn−xnk)

≤2D(dn+d⁰_n+b⁰_nd⁰⁰_n)→0 as n→ ∞. SinceS is uniformly continuous, we have

kSx_n+1−Sy_nk →0 asn→ ∞.

(3.27)

Set inf{A(xn+1, q) :n≥0}=r. We claim thatr= 0. If not, thenr >0. It is easy to check that

kxn+1−qk ≤(1−rd_n)kxn−qk+M d_nkSxn+1−Sy_nk+M c_n forn≥0.

Put

cn =tndn, αn=kxn−qk, ωn=rdn,

βn =M r⁻¹(kSxn+1−Synk+tn), γn= 0 forn≥0.

(3.2) ensures that tn → 0 as n→ ∞. It follows from (3.20), (3.27) and Lemma 2.3 that ωn∈(0,1] withP∞

n=0ωn=∞,lim_n→∞βn = 0,P∞

n=0γn<∞. Sokxn−qk →0 asn→ ∞, which means thatr= 0. This is a contradiction. Thusr= 0 and there exists a subsequence {kxn_i+1−qk}^∞_i=0 of{kxn+1−qk}^∞_n=0 satisfying

kxn_i+1−qk →0 asi→ ∞.

(3.28)

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From (3.28) and (3.29) we conclude that for given ε >0 there exists a positive integerm such that forn≥m,

kxn_m+1−qk< ε (3.29)

and

MkSxn+1−Sy_nk+Mc_n dn

<min 1

2ε, φ(ε)ε 1 +φ(³₂ε) +³₂ε

. (3.30)

Now we claim that

kxn_m+j−qk< ε forj≥1.

(3.31)

In fact (3.29) means that (3.31) holds for j = 1. Assume that (3.31) holds for j =k. If kxnm+k+1−qk> ε, we get that

kxn_m+k+1−qk

≤ kxn_m+k−qk+M dn_m+kkSxn_m+k+1−Syn_m+kk+M cn_m+k

≤ε+ min 1

2ε, φ(ε)ε 1 +φ(³₂ε) +³₂ε

dn_m+k

≤ 3 2ε.

(3.32)

Note thatφ(kx_nm+k+1−qk)> φ(ε). From (3.32) we get that A(xnm+k+1, q)≥ φ(ε)

1 +φ(³₂ε) +³₂ε. (3.33)

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By virtue of (3.26) (3.30) and (3.33) we obtain that kx_nm+k+1−qk

≤

1− φ(ε)ε

1 +φ(³₂ε) +³₂εdn_m+k

kxn_m+k−qk +M d_nm+kkSx_nm+k+1−Sy_nm+kk+M c_nm+k

≤

1− φ(ε)ε

1 +φ(³₂ε) +³₂εdn_m+k

ε+ min 1

2ε, φ(ε)ε 1 +φ(³₂ε) +³₂ε

dn_m+k

≤ε.

That is

ε <kxnm+k+1−qk ≤ε,

which is a contradiction. Hence kxn_m+k+1−qk ≤ε. By induction (3.29) holds for j ≥1.

Thus (3.31) yields that xn→qasn→ ∞. This completes the proof.

Remark 3.6. Theorem 3.5extends and improves Theorem 3.4 in [2] and Theorem 3.1 in [16].

Acknowledgement. This work was supported by the Science Research Foundation of Educational Department of Liaoning Province (20060467).

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Shin Min Kang, Department of Mathematics and the Research Institute of Natural Science, Gyeongsang National University, Jinju 660-701, Korea,e-mail:[email protected]

Chi Feng, Department of Science, Dalian Fisheries College, Dalian, Liaoning, 116023, People’s Republic of China,e-mail:[email protected]

Zeqing Liu, Department of Mathematics, Liaoning Normal University, P.O. Box 200, Dalian, Liaoning, 116029, People’s Republic of China,e-mail:[email protected]