# Two-Step Newton-Tikhonov Method for Hammerstein-Type Equations: Finite-Dimensional Realization

Authors: Santhosh George; Monnanda Erappa Shobha
Journal: ISRN Applied Mathematics
 (2012)

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## Abstract

Finite-dimensional realization of a Two-Step Newton-Tikhonov method is considered for obtaining a stable approximate solution to nonlinear ill-posed Hammerstein-type operator equationsKF(x)=f. Here F:D(F)⊆X→X is nonlinear monotone operator, K:X→Y is a bounded linear
operator, X is a real Hilbert space, and Y is a Hilbert space. The error analysis for this method is done under two general source conditions, the first one involves the operator K and the second one involves the Fréchet derivative of F at an initial approximation x0 of the the solution x̂: balancing principle of Pereverzev and Schock (2005) is employed in choosing the regularization parameter and order optimal error bounds are established. Numerical illustration is given to confirm the reliability of our approach.


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## Body

## 1. Introduction

Tikhonov’s regularization (e.g., [1]) method has been used extensively to stabilize the approximate solution of nonlinear ill-posed problems. In recent years, increased emphasis has been placed on iterative regularization procedures [2, 3] for obtaining the approximate solution of such problems. In this paper, we examine the use of iterative regularization procedures for Hammerstein-type [4, 5] equations of the form(1.1)KF(x)=f,
where F:D(F)⊆X→X, is nonlinear monotone operator, F′(·)-1 does not exists, and K:X→Y is a bounded linear operator. Throughout this paper, D(F) is the domain of F,F′(·) is the Fréchet derivative of F,X is a real Hilbert space, and Y is a Hilbert space. The inner product and the corresponding norm are denoted by 〈·,·〉 and ∥·∥, respectively.Recall that [6] the operator F is said to be a monotone operator if 〈F(x)-F(y),x-y〉≥0, for all x,y∈D(F).It is assumed throughout thatfδ∈Y are the available noisy data with(1.2)‖f-fδ‖≤δ,
and F possesses a uniformly bounded Fréchet derivative for each x∈D(F) (cf. [7]), that is,(1.3)‖F′(x)‖≤M,x∈D(F)
for some M.Observe that the solutionx of (1.1) with fδ in place of f can be obtained by first solving(1.4)Kz=fδ
for z and then solving the nonlinear problem(1.5)F(x)=z.
In [4, 5, 8] this was exploited. In [4], z is approximated with zαδ where(1.6)zαδ=(K*K+αI)-1K*fδ,α>0,δ>0,
and then solve (1.5) iteratively using the following Newton-type procedure:(1.7)xn+1,αδ=xn,αδ-F′(x0)-1(F(xn,αδ)-zαδ)
with x0,αδ:=x0 and obtained local linear convergence. Here and below x0∈D(F) is a known initial approximation of the solution x̂ of (1.1) such that ∥x0-x̂∥≤ρ.In [8], to solve (1.5), George and Kunhanandan used the iteration(1.8)xn+1,αδ=xn,αδ-F′(xn,αδ)-1(F(xn,αδ)-zαδ),x0,αδ:=x0,
where(1.9)zαδ=(K*K+αI)-1K*(fδ-KF(x0))+F(x0),
and obtained local quadratic convergence.A sequence(xn) in X with limxn=x* is said to be convergent of order p>1, if there exist positive reals β and γ such that, for all n∈N,(1.10)‖xn-x*‖≤βe-γpn.
If a sequence (xn) satisfies ∥xn-x*∥≤βqn, 0<q<1, then (xn) is said to be linearly convergent.As in [8], it is assumed that the solution x̂ of (1.1) satisfies(1.11)‖F(x̂)-F(x0)‖=min⁡{‖F(x)-F(x0)‖:KF(x)=f,x∈D(F)}.
The regularization parameter α is chosen from a finite set(1.12)DN={αi:α0<α1<α2<⋯<αN}
using the adaptive method considered by Pereverzev and Schock in [9].In [10], Argyros and Hilout considered a method called Two-Step Directional Newton Method (TSDNM) for approximating a zero x* of a differentiable function F defined on a convex subset 𝒟 of a Hilbert space H with values in ℝ. Motivated by TSDNM, in [11], we propose a Two-Step Newton-Tikhonov Methods (TSNTM) for solving (1.1).In fact, in [11] we consider two cases of F, in the first case we assume that F′(x0)-1 exist and in the second case we assume F is monotone. In this paper we consider the finite-dimensional realization of the second case, that is, F is monotone. The finite-dimensional realization of the method and associated algorithm are proposed for which local-cubic convergence is established theoretically and validated numerically.The organization of this paper is as follows. Section2 deals with Discretized Tikhonov regularization and Section 3 investigates the convergence of the Discretized TSNTM. Section 4 discusses the algorithm and finally the paper ends with a numerical example in Section 5.


## 2. Discretized Tikhonov Regularization

This section deals with discretized Tikhonov regularized solutionzαh,δ of (1.4) and (an a priori and an a posteriori) error estimate for ∥F(x̂)-zαh,δ∥ using an error estimate for ∥F(x̂)-zα∥ from [8].The following assumption is used in [8] to obtain the error estimate.Assumption 2.1.
There exists a continuous, strictly monotonically increasing functionφ:(0,a]→(0,∞) with a≥∥K∥2 satisfying(i)
limλ→0φ(λ)=0,(ii)
(2.1)sup⁡λ>0αφ(λ)λ+α≤φ(α)∀λ∈(0,a],(iii)
(2.2)F(x̂)-F(x0)=φ(K*K)w
for somew∈X such that ∥w∥≤1.Remark 2.2.
The functions(2.3)φ(λ):=λν,λ>0,
for 0<ν≤1 and
(2.4)φ(λ)={[ln⁡1λ]-p,0<λ≤e-(p+1),0otherwise
for p≥0 satisfy the above assumption (see [12]).Theorem 2.3 (cf. [8], Theorem 4.3).
Letzα:=zα0 be as in (1.9) and Assumption 2.1 holds. Then
(2.5)‖F(x̂)-zα‖≤φ(α).
Let{Ph}h>0 be a family of orthogonal projections on X. Let
(2.6)ɛh:=‖K(I-Ph)‖,τh:=‖F′(x)(I-Ph)‖,∀x∈D(F)
and {bh:h>0} is such that limh→0(∥(I-Ph)x0∥/bh)=0,  limh→0(∥(I-Ph)F(x0)∥/bh)=0 and limh→0bh=0. We assume that ɛh→0 and τh→0 as h→0. The above assumption is satisfied if, Ph→I pointwise and if K and F′(x) are compact operators. Further we assume that  ɛh<ɛ0,τh≤τ0,bh≤b0 and δ∈(0,δ0] where δ0+ɛ0<(2/(2M+3))α0. The discretized Tikhonov regularization method for the regularized equation (1.4) consists of solving the equation
(2.7)(PhK*KPh+αPh)(zαh,δ-PhF(x0))=PhK*[fδ-KF(x0)].Theorem 2.4.
Suppose assumptions in Theorem2.3 hold. Let zαh,δ be as in (2.7) and bh≤(δ+ɛh)/α. Then
(2.8)‖F(x̂)-zαh,δ‖≤C(φ(α)+(δ+ɛhα)),
where C=(1/2)max{Mρ,1}+1.Proof .
Letzα=(K*K+αI)-1K*(f-KF(x0))+F(x0). Then
(2.9)‖zα-zαh‖=‖(K*K+αI)-1K*(f-KF(x0))-(PhK*KPh+αI)-1PhK*(f-KF(x0))+F(x0)-PhF(x0)‖≤‖(PhK*KPh+αPh)-1PhK*(KPh-K)(K*K+αI)-1K*K[F(x̂)-F(x0)]‖+‖(I-Ph)F(x0)‖≤‖F(x̂)-F(x0)‖ɛh2α+bh≤‖∫01F′(x0+t(x̂-x0))(x̂-x0)dt‖ɛh2α+bh≤Mρɛh2α+bh,‖zαh-zαh,δ‖=‖(PhK*KPh+αI)-1PhK*(f-fδ)‖≤δ2α.
Now the result follows from (2.9), Theorem 2.3 and the following triangle inequality:
(2.10)‖F(x̂)-zα,hδ‖≤‖F(x̂)-zα‖+‖zα-zαh‖+‖zαh-zαh,δ‖.


### 2.1. A Priori Choice of the Parameter

Note that the estimateφ(α)+(δ+ɛh)/√α in (2.8) is of optimal order for the choice α:=α(δ,h) which satisfies φ(α(δ,h))=(δ+ɛh)/α(δ,h). Let ψ(λ):=λφ-1(λ),0<λ≤a. Then we have δ+ɛh=α(δ,h)φ(α(δ,h))=ψ(φ(α(δ,h))) and(2.11)α(δ,h)=φ-1(ψ-1(δ+ɛh)).
So the relation (2.8) leads to ∥F(x̂)-zαh,δ∥≤2Cψ-1(δ+ɛh).


### 2.2. An Adaptive Choice of the Parameter

In this subsection, we consider the balancing principle established by Pereverzev and Shock [9] for choosing the parameter α. Let(2.12)DN={αi:0<α0<α1<α2<⋯<αN}
be the set of possible values of the parameter α.Let(2.13)l:=max⁡{i:φ(αi)≤δ+ɛhαi}<N,(2.14)k=max⁡{i:αi∈DN+},
where DN+={αi∈DN:∥zαiδ-zαjδ∥≤4C(δ+ɛh)/αj,j=0,1,2,…,i-1}.We use the following theorem, proof of which is analogous to the proof of Theorem  4.3 in [8], for our error analysis.Theorem 2.5 (cf. [8], Theorem 4.3).
Letl be as in (2.13), let k be as in (2.14), and let zαkh,δ be as in (2.7) with α=αk:=μkα0,μ>1. Then l≤k and
(2.15)‖F(x̂)-zαkh,δ‖≤C(2+4μμ-1)μψ-1(δ+ɛh).


## 2.1. A Priori Choice of the Parameter

Note that the estimateφ(α)+(δ+ɛh)/√α in (2.8) is of optimal order for the choice α:=α(δ,h) which satisfies φ(α(δ,h))=(δ+ɛh)/α(δ,h). Let ψ(λ):=λφ-1(λ),0<λ≤a. Then we have δ+ɛh=α(δ,h)φ(α(δ,h))=ψ(φ(α(δ,h))) and(2.11)α(δ,h)=φ-1(ψ-1(δ+ɛh)).
So the relation (2.8) leads to ∥F(x̂)-zαh,δ∥≤2Cψ-1(δ+ɛh).


## 2.2. An Adaptive Choice of the Parameter

In this subsection, we consider the balancing principle established by Pereverzev and Shock [9] for choosing the parameter α. Let(2.12)DN={αi:0<α0<α1<α2<⋯<αN}
be the set of possible values of the parameter α.Let(2.13)l:=max⁡{i:φ(αi)≤δ+ɛhαi}<N,(2.14)k=max⁡{i:αi∈DN+},
where DN+={αi∈DN:∥zαiδ-zαjδ∥≤4C(δ+ɛh)/αj,j=0,1,2,…,i-1}.We use the following theorem, proof of which is analogous to the proof of Theorem  4.3 in [8], for our error analysis.Theorem 2.5 (cf. [8], Theorem 4.3).
Letl be as in (2.13), let k be as in (2.14), and let zαkh,δ be as in (2.7) with α=αk:=μkα0,μ>1. Then l≤k and
(2.15)‖F(x̂)-zαkh,δ‖≤C(2+4μμ-1)μψ-1(δ+ɛh).


## 3. Discretized Two-Step Newton Method (DTSNM)

We need the following assumptions for the convergence of DTSNM and to obtain the error estimate.Assumption 3.1 (cf. [7], Assumption 3 (A3)).
There exists a constantk0≥0 such that for every x,u∈D(F) and v∈X there exists an element Φ(x,u,v)∈X such that [F′(x)-F′(u)]v=F′(u)Φ(x,u,v), ∥Φ(x,u,v)∥≤k0∥v∥∥x-u∥.Assumption 3.2.
There exists a continuous, strictly monotonically increasing functionφ1:(0,b]→(0,∞) with b≥∥F′(x0)∥ satisfying(i)
limλ→0φ1(λ)=0,(ii)
(3.1)sup⁡λ>0αφ1(λ)λ+α≤φ1(α)∀λ∈(0,b],(iii)
there existsv∈X with ∥v∥≤1 (cf. [6]) such that(3.2)x0-x̂=φ1(F′(x0))v,(iv)
for eachx∈Br(x0):={x:∥x-x0∥<r} there exists a bounded linear operator G(x,x0) (cf. [13]) such that
(3.3)F′(x)=F′(x0)G(x,x0)
with ∥G(x,x0)∥≤K1.First we consider a DTSNM for approximating the zero xc,αkh,δ of
(3.4)Ph(F(x)+αkc(x-x0))=Phzαkh,δ
and then we show that xc,αkh,δ is an approximation to the solution x̂ of (1.1) where c≤αk. For an initial guess x0∈X and for R(x):=PhF′(x)Ph+(αk/c)Ph, the DTSNM is defined as
(3.5)yn,αkh,δ=xn,αkh,δ-R(xn,αkh,δ)-1Ph[F(xn,αkh,δ)-zαkh,δ+αkc(xn,αkh,δ-x0,αkh,δ)],(3.6)xn+1,αkh,δ=yn,αkh,δ-R(xn,αkh,δ)-1Ph[F(yn,αkh,δ)-zαkh,δ+αkc(yn,αkh,δ-x0,αkh,δ)],
where x0,αkh,δ:=Phx0. Note that with the above notation
(3.7)‖R(xn,αkh,δ)-1PhF′(xn,αkh,δ)‖=‖(PhF′(xn,αkh,δ)Ph+αkcPh)-1PhF′(xn,αkh,δ)‖=‖(PhF′(xn,αkh,δ)Ph+αkcPh)-1PhF′(xn,αkh,δ)[Ph+I-Ph]‖≤‖(PhF′(xn,αkh,δ)Ph+αkcPh)-1PhF′(xn,αkh,δ)Ph‖+‖(PhF′(xn,αkh,δ)Ph+αkcPh)-1PhF′(xn,αkh,δ)(I-Ph)‖≤1+‖PhF′(xn,αkh,δ)(I-Ph)‖αk/c≤1+τh≤1+τ0.
Let
(3.8)en,αkh,δ:=‖yn,αkh,δ-xn,αkh,δ‖,∀n≥0.
and let k0 be such that
(3.9)k028(4+3k0(1+τ0))(1+τ0)2<1.Remark 3.3.
Note that the above assumption is satisfied if we choosek0<min{1,(1/(1+τ0))8/4+3(1+τ0)}.
Letg:(0,1)→(0,1) be the function defined by
(3.10)g(t)=k028(4+3k0(1+τ0)t)(1+τ0)2t2∀t∈(0,1).
Let ∥x̂-x0∥≤ρ, with
(3.11)ρ<1M(1-(32+M)δ0+ɛ0α0),γρ≔Mρ+(32+M)(ɛ0+δ0α0).Theorem 3.4.
Leten,αkh,δ and g be as in (3.8) and (3.10), respectively, and let xn,αkh,δ and yn,αkh,δ be as in (3.6) and (3.5), respectively, with δ∈(0,δ0],α=αk and ɛh∈(0,ɛ0]. Then the following holds:(a)
(3.12)‖xn,αkh,δ-yn-1,αkh,δ‖≤(1+τ0)k0en-1,αkh,δ2‖yn-1,αkh,δ-xn-1,αkh,δ‖;(b)
(3.13)‖xn,αkh,δ-xn-1,αkh,δ‖≤(1+(1+τ0)k0en-1,αkh,δ2)‖yn-1,αkh,δ-xn-1,αkh,δ‖;(c)
(3.14)‖yn,αkh,δ-xn,αkh,δ‖≤g(en-1,αkh,δ)‖yn-1,αkh,δ-xn-1,αkh,δ‖;(d)
(3.15)g(en,αkh,δ)≤g(γρ)3n,∀n≥0;(e)
(3.16)en,αkh,δ≤g(γρ)(3n-1)/2γρ,∀n≥0.Proof.
Observe that(3.17)xn,αkh,δ-yn-1,αkh,δ=yn-1,αkh,δ-xn-1,αkh,δ-R(xn-1,αkh,δ)-1×Ph(F(yn-1,αkh,δ)-F(xn-1,αkh,δ)+αkc(yn-1,αkh,δ-xn-1,αkh,δ))=R(xn-1,αkh,δ)-1[R(xn-1,αkh,δ)(yn-1,αkh,δ-xn-1,αkh,δ)-Ph(F(yn-1,αkh,δ)-F(xn-1,αkh,δ))-αkc(yn-1,αkh,δ-xn-1,αkh,δ)]=R(xn-1,αkh,δ)-1[Ph(F(yn-1,αkh,δ)-F(xn-1,αkh,δ))-αkc(yn-1,αkh,δ-xn-1,αkh,δ)(PhF′(xn-1,αkh,δ)Ph+αkcPh)(yn-1,αkh,δ-xn-1,αkh,δ)-Ph(F(yn-1,αkh,δ)-F(xn-1,αkh,δ))-αkc(yn-1,αkh,δ-xn-1,αkh,δ)]=R(xn-1,αkh,δ)-1Ph×∫01[F′(xn-1,αkh,δ)-F′(xn-1,αkh,δ+t(yn-1,αkh,δ-xn-1,αkh,δ))]Ph(yn-1,αkh,δ-xn-1,αkh,δ)dt.
Now by Assumption 3.1 and (3.7) we have
(3.18)‖xn,αkh,δ-yn-1,αkh,δ‖≤(1+τ0)‖∫01Φ(xn-1,αkh,δ,xn-1,αkh,δ+t(yn-1,αkh,δ-xn-1,αkh,δ),yn-1,αkh,δ-xn-1,αkh,δ)dt‖≤(1+τ0)k02‖yn-1,αkh,δ-xn-1,αkh,δ‖2.
This proves (a). Now (b) follows from (a) and the triangle inequality
(3.19)‖xn,αkh,δ-xn-1,αkh,δ‖≤‖xn,αkh,δ-yn-1,αkh,δ‖+‖yn-1,αkh,δ-xn-1,αkh,δ‖.
To prove (c) we observe that
(3.20)en,αkh,δ=‖xn,αkh,δ-yn-1,αkh,δ-R(xn,αkh,δ)-1Ph(F(xn,αkh,δ)-zαkh,δ+αkc(xn,αkh,δ-x0,αkh,δ))+R(xn-1,αkh,δ)-1Ph(F(yn-1,αkh,δ)-zαkh,δ+αkc(yn-1,αkh,δ-x0,αkh,δ))‖=‖xn,αkh,δ-yn-1,αkh,δ-R(xn,αkh,δ)-1Ph(F(xn,αkh,δ)-F(yn-1,αkh,δ)+αkc(xn,αh,δ-yn-1,αkh,δ))+[R(xn-1,αkh,δ)-1-R(xn,αkh,δ)-1]×Ph(F(yn-1,αkh,δ)-zαkh,δ+αkc(yn-1,αkh,δ-x0,αkh,δ))‖:=Γ1+Γ2,
where
(3.21)Γ1:=‖xn,αkh,δ-yn-1,αkh,δ-R(xn,αkh,δ)-1Ph(F(xn,αkh,δ)-F(yn-1,αkh,δ)+αkc(xn,αh,δ-yn-1,αkh,δ))‖,Γ2:=‖[R(xn-1,αkh,δ)-1-R(xn,αkh,δ)-1]Ph(F(yn-1,αkh,δ)-zαkh,δ+αkc(yn-1,αkh,δ-x0,αkh,δ))‖.
Note that
(3.22)Γ1=‖R(xn,αkh,δ)-1[PhF′(xn,αkh,δ)Ph(xn,αkh,δ-yn-1,αkh,δ)-Ph(F(xn,αkh,δ)-F(yn-1,αkh,δ))]‖=‖R(xn,αkh,δ)-1Ph∫01[F′(xn,αkh,δ)-F′(yn-1,αkh,δ+t(xn,αkh,δ-yn-1,αkh,δ))]×Ph(xn,αkh,δ-yn-1,αkh,δ)dt‖≤(1+τ0)‖∫01Φ(xn,αkh,δ,yn-1,αkh,δ+t(xn,αkh,δ-yn-1,αkh,δ),xn,αkh,δ-yn-1,αkh,δ)dt‖≤(1+τ0)k02‖xn,αkh,δ-yn-1,αkh,δ‖2.
The last but one step follows from Assumption 3.1 and (3.7). Similarly one can prove that
(3.23)Γ2≤(1+τ0)k0‖xn,αkh,δ-xn-1,αkh,δ‖‖xn,αkh,δ-yn-1,αkh,δ‖.
Thus from (3.20), (3.22), (3.23), (a) and (b) we have
(3.24)en,αkh,δ≤(1+τ0)2(k022+3k03(1+τ0)8‖yn-1,αkh,δ-xn-1,αkh,δ‖)×‖yn-1,αkh,δ-xn-1,αkh,δ‖3≤g(en-1,αkh,δ)en-1,αkh,δ.
Again since for μ∈(0,1),g(μt)≤μ2g(t), for all t∈(0,1), by (3.24) we have
(3.25)g(en,αkh,δ)≤g(e0,αkh,δ)3n,(3.26)en,αkh,δ≤g3(en-2,αkh,δ)en-1,αkh,δ≤g3(en-2,αkh,δ)g3(en-3,αkh,δ)en-2,αkh,δ⋯g(e0,αkh,δ)e0,αkh,δ≤g(e0,αkh,δ)3n-1+3n-2+⋯+1e0,αkh,δ≤g(e0,αkh,δ)(3n-1)/2e0,αkh,δ
provided en,αkh,δ<1, for alln≥0. From (3.26) it is clear that, en,αkh,δ≤1 if e0,αkh,δ≤1. This can be seen as follows:
(3.27)e0,αkh,δ=‖y0,αkh,δ-x0,αkh,δ‖=‖y0,αkh,δ-Phx0‖=‖(PhF′(Phx0)Ph+αkc)-1Ph(F(Phx0)-zαkh,δ)‖≤‖F(Phx0)-zαkh‖+‖zαkh-zαkh,δ‖,‖F(Phx0)-zαkh‖≤‖F(Phx0)-F(x0)‖+‖F(x0)-zαk‖+‖zαk-zαkh‖≤‖∫01F′(x0+t(Phx0-x0))(Phx0-x0)dt‖+‖(K*K+αkI)-1K*K(F(x̂)-F(x0))‖+‖zαk-zαkh‖≤Mbh+‖F(x̂)-F(x0)‖+‖zαk-zαkh‖≤Mbh+Mρ+‖zαk-zαkh‖.
Therefore by (3.27) and (2.9) we have
(3.28)e0,αkh,δ≤(M+1)bh+(1+ɛh2αk)Mρ+δ2αk≤(M+1)ɛh+δαk+Mρ+12max⁡{Mρ,1}ɛ0+δ0α0≤(M+1)ɛ0+δ0α0+Mρ+ɛ0+δ02α0≤γρ<1.
Now since g is monotonic increasing and e0,αkh,δ≤γρ, we have g(e0,αkh,δ)≤g(γρ). This completes the proof of the theorem.Theorem 3.5.
Letr=(1/(1-g(γρ))+(1+τ0)(k0/2)(γρ/(1-g(γρ)2)))γρ and the assumptions of Theorem 3.4 hold. Then xn,αkh,δ,yn,αkh,δ∈Br(Phx0), for all n≥0.Proof.
Note that by (b) of Theorem3.4 we have
(3.29)‖x1,αkh,δ-Phx0‖≤[1+(1+τ0)k02e0,αkh,δ]e0,αkh,δ≤[1+(1+τ0)k02γρ]γρ≤r,
that is, x1∈Br(Phx0). Again note that by (3.29) and (c) of Theorem 3.4 we have
(3.30)‖y1,αkh,δ-Phx0‖≤‖y1,αkh,δ-x1,αkh,δ‖+‖x1,αkh,δ-Phx0‖≤(1+g(e0,αkh,δ)+(1+τ0)k02e0,αkh,δ)e0,αkh,δ≤(1+g(γρ)+(1+τ0)k02γρ)γρ≤r,
that is, y1,αkh,δ∈Br(Phx0). Further by (3.29) and (b) of Theorem 3.4 we have
(3.31)‖x2,αkh,δ-Phx0‖≤‖x2,αkh,δ-x1,αkh,δ‖+‖x1,αkh,δ-Phx0‖≤(1+(1+τ0)k02e1,αkh,δ)e1,αkh,δ+(1+(1+τ0)k02e0,αkh,δ)e0,αkh,δ≤(1+g(e0,αkh,δ)+(1+τ0)k02e0,αkh,δ(1+g(e0,αkh,δ)2))e0,αkh,δ≤r.
The last but one step follows from the monotonicity of g and (3.28).
And by (3.31) and (c) of Theorem 3.4 we have
(3.32)‖y2,αkh,δ-Phx0‖≤‖y2,αkh,δ-x2,αkh,δ‖+‖x2,αkh,δ-Phx0‖≤g(e1,αkh,δ)e1,αkh,δ+(1+g(e0,αkh,δ)+(1+τ0)k02e0,αkh,δ(1+g(e0,αkh,δ)2))e0,αkh,δ≤g(e0,αkh,δ)4e0,αkh,δ+(1+g(e0,αkh,δ)+(1+τ0)k02e0,αkh,δ(1+g(e0,αkh,δ)2))e0,αkh,δ≤(1+g(γρ)+g(γρ)2+(1+τ0)k02γρ(1+g(γρ)2))γρ≤r,
that is, x2,αkh,δ,y2,αkh,δ∈Br(Phx0). Continuing this way one can prove that xn,αkh,δ,yn,αkh,δ∈Br(Phx0), for all n≥0. This completes the proof.The main result of this section is the following theorem.Theorem 3.6.
Letyn,αkh,δ and xn,αkh,δ be as in (3.5) and (3.6), respectively, and let assumptions of Theorem 3.5 hold. Then (xn,αkh,δ) is a Cauchy sequence in Br(Phx0) and converges to xc,αkh,δ∈Br(Phx0)¯. Further Ph[F(xc,αkh,δ)+(αk/c)(xc,αkh,δ-x0)]=Phzαkh,δ and
(3.33)‖xn,αkh,δ-xc,αkh,δ‖≤C0¯e-γ13n,
where C0¯=(1/(1-g(γρ)3)+(1+τ0)(k0γρ/2)(1/(1-(g(γρ)2)3))g(γρ)3n)γρ and γ1=-logg(γρ).Proof.
Using the relation (b) and (e) of Theorem3.4 and (3.28), we obtain
(3.34)‖xn+m,αkh,δ-xn,αkh,δ‖≤∑i=0i=m-1‖xn+i+1,αkh,δ-xn+i,αkh,δ‖≤∑i=0i=m-1(1+(1+τ0)k0en+i,αkh,δ2)en+i,αkh,δ≤∑i=0i=m-1(1+(1+τ0)k0e0,αkh,δ2g(e0,αkh,δ)3n+i)g(e0,αkh,δ)3n+ie0,αkh,δ≤[(1+g(γρ)3+g(γρ)32+⋯+g(γρ)3m)+(1+τ0)k0γρ2×(1+(g(γρ)2)3+(g(γρ)2)32+⋯+(g(γρ)2)3m)×g(γρ)3nk0γρ2]g(γρ)3nγρ≤C0¯g(γρ)3n≤C0¯e-γ13n.
Thus (xn,αkh,δ) is a Cauchy sequence in Br(Phx0) and hence it converges, say, to xc,αkh,δ∈Br(Phx0)¯. Observe that from (3.5)
(3.35)‖Ph(F(xn,αkh,δ)-zαkh,δ)+αkc(xn,αkh,δ-Phx0)‖=‖R(xn,αkh,δ)(xn,αkh,δ-yn,αkh,δ)‖≤‖R(xn,αkh,δ)‖‖yn,αkh,δ-xn,αkh,δ‖≤(‖PhF′(xn,αkh,δ)Ph‖+αkc)en,αkh,δ≤(‖PhF′(xn,αkh,δ)Ph‖+αkc)×g(e0,αkh,δ)3ne0,αkh,δ≤(M+αkc)g(γρ)3nγρ.
Now by letting n→∞ in (3.35) we obtain PhF(xc,αkh,δ)+(αk/c)(xc,αkh,δ-Phx0)=Phzαkh,δ. This completes the proof.Hereafter we assume thatr<1/k0 and K1<(1-k0r)/(1-c). The proof of the following theorem is analogous to the proof of Theorem  3.14 in [11] but for the sake of completeness we give the proof.Theorem 3.7 (cf. [11], Theorem 3.14).
Supposexc,αkδ is the solution of
(3.36)F(x)+αkc(x-x0)=zαkδ
and Assumptions 3.1 and 3.2 holds, then
(3.37)‖x̂-xc,αkδ‖≤φ1(αk)+(2+4μ/(μ-1))μψ-1(δ+ɛh)1-(1-c)K1-k0r.Proof.
Note thatc(F(xc,αkδ)-zαkδ)+αk(xc,αkδ-x0)=0, so
(3.38)(F′(x0)+αkI)(xc,αkδ-x̂)=(F′(x0)+αkI)(xc,αkδ-x̂)-c(F(xc,αkδ)-zαkδ)-αk(xc,αδ-x0)=αk(x0-x̂)-c(F(x̂)-zαkδ)+F′(x0)(xc,αkδ-x̂)-c[F(xc,αkδ)-F(x̂)].
Thus
(3.39)‖xc,αkδ-x̂‖≤‖αk(F′(x0)+αkI)-1(x0-x̂)‖+‖(F′(x0)+αkI)-1c(F(x̂)-zαkδ)‖+‖(F′(x0)+αkI)-1[F′(x0)(xc,αkδ-x̂)-c(F(xc,αkδ)-F(x̂))]‖≤‖αk(F′(x0)+αkI)-1(x0-x̂)‖+‖F(x̂)-zαkδ‖+Γ,
where Γ:=∥(F′(x0)+αkI)-1∫01[F′(x0)-cF′(x̂+t(xc,αkδ-x̂)](xc,αkδ-x̂)dt∥. So by Assumption 3.2, we obtain
(3.40)Γ≤‖(F′(x0)+αkI)-1∫01[F′(x0)-F′(x̂+t(xc,αkδ-x̂))]×(xc,αkδ-x̂)dt‖+(1-c)‖(F′(x0)+αkI)-1F′(x0)×∫01G(x̂+t(xc,αkδ-x̂),x0)(xc,αkδ-x̂)dt‖≤k0r‖xc,αkδ-x̂‖+(1-c)K1‖xc,αkδ-x̂‖
and hence by (3.39) and (3.40) we have
(3.41)‖xc,αkδ-x̂‖≤‖αk(F′(x0)+αkI)-1(x0-x̂)‖+‖F(x̂)-zαkδ‖1-(1-c)K1-k0r≤φ1(αk)+(2+4μ/(μ-1))μψ-1(δ+ɛh)1-(1-c)K1-k0r.
This completes the proof of the theorem.Theorem 3.8.
Supposexc,αkh,δ is the solution of (3.4) and Assumption 2.1 and Theorem 3.7 hold. In addition if τ0<1, then
(3.42)‖xc,αkh,δ-xc,αkδ‖≤21-τ0(δ+ɛhαk).Proof.
Supposexc,αkδ and xc,αkh,δ are the solutions of (3.36) and (3.4), respectively, then by (3.36) we have,
(3.43)PhF(xc,αkδ)+αkc(Phxc,αkh,δ-Phx0)=Phzαkδ.
So from (3.4) and (3.43),
(3.44)Ph[F(xc,αkh,δ)-F(xc,αkδ)]+αkcPh(xc,αkh,δ-xc,αkδ)=Ph(zαkh,δ-zαkδ).
Let Mf=∫01F′(xc,αkδ+t(xc,αkh,δ-xc,αkδ))dt. Then by (3.44) we have
(3.45)Ph[Mf(xc,αkh,δ-xc,αkδ)]+αkcPh(xc,αkh,δ-xc,αkδ)=Ph(zαkh,δ-zαkδ)
and hence
(3.46)‖xc,αkh,δ-xc,αkδ‖≤‖zαkh,δ-zαkδ‖+‖Mf(Ph-I)‖‖xc,αkh,δ-xc,αkδ‖≤‖zαkh,δ-zαkδ‖+τ0‖xc,αkh,δ-xc,αkδ‖.
Thus
(3.47)‖xc,αkh,δ-xc,αkδ‖≤11-τ0‖zαkh,δ-zαkδ‖≤11-τ0[‖zαkh,δ-zαkh‖+‖zαkh-zαk‖+‖zαk-zαkδ‖].
Now the result follows from (2.9), (3.47) and the relation
(3.48)‖zαk-zαkδ‖≤δ2αk.
The following theorem is a consequence of Theorems3.6, 3.7, and 3.8.Theorem 3.9.
Letxn,αkh,δ be as in (3.6) and let assumptions in Theorems 3.6, 3.7 and 3.8 hold. Then
(3.49)‖x̂-xnk,αkh,δ‖≤C0¯e-γ13n+φ1(αk)+(2+4μ/(μ-1))μψ-1(δ+ɛh)1-(1-c)K1-k0r+21-τ0(δ+ɛhαk),
where C0¯ and γ1 are as in Theorem 3.6.Theorem 3.10.
Letxn,αkh,δ be as in (3.6) and let assumptions in Theorem 3.9 hold. Further let φ1(αk)≤φ(αk) and
(3.50)nk:=min⁡{n:e-γ13n≤δ+ɛhαk}.
Then
(3.51)‖x̂-xn,αkh,δ‖=O(ψ-1(δ+ɛh)).


## 4. Algorithm

Note that fori,j∈{0,1,2,…,N},(4.1)zαih,δ-zαjh,δ=(αj-αi)(PhK*KPh+αjI)-1(PhK*KPh+αiI)-1PhK*(fδ-KF(x0)).Therefore the balancing principle algorithm associated with the choice of the parameter specified in Section2 involves the following steps:Step 1.
Chooseα0 such that δ0+ɛ0<2α0/(2M+3) and μ>1.Step 2.
αi=μ2iα0.Step 3.
Solve forwi:
(4.2)(PhK*KPh+αiI)wi=PhK*(fδ-KF(x0)).Step 4.
Solve forj<i,   zijh,δ:(PhK*KPh+αjI)zijh,δ=(αj-αi)wi.Step 5.
If∥zijh,δ∥>4C(δ+ɛh)/αj, then take k=i-1.Step 6.
Otherwise, repeat withi+1 in place of i.Step 7.
Choosenk=min{n:e-γ13n≤(δ+ɛh)/αk}.Step 8.
Solvexnk,αkh,δ using the iteration (3.6).In the next section we consider an example to illustrate the above algorithm. The computational results provided endorse the reliability and effectiveness of our method.


## 5. Example

In this section we consider an example satisfying the assumptions made in this paper and give the numerical illustration. Consider the operatorKF:L2(0,1)→L2(0,1) where K:L2(0,1)→L2(0,1) is defined by(5.1)K(x)(t)=∫01k(t,s)x(s)ds
and F:D(F)⊆L2(0,1)→L2(0,1) defined by(5.2)F(u):=∫01k(t,s)u3(s)ds,
where(5.3)k(t,s)={(1-t)s,0≤s≤t≤1,(1-s)t,0≤t≤s≤1.
Then for all x(t),y(t):x(t)>y(t) (see [7], Section 4.3):(5.4)〈F(x)-F(y),x-y〉=∫01[∫01k(t,s)(x3-y3)(s)ds](x-y)(t)dt≥0.Thus the operatorF is monotone. The Fréchet derivative of F is given by(5.5)F′(u)w=3∫01k(t,s)(u(s))2w(s)ds.LetVn be a sequence of finite-dimensional subspaces of X and let Ph=P1/n denote the orthogonal projection on X with range R(Ph)=Vn. We assume that dimVn=n+1 and ∥Phx-x∥→0 as h→0 for all x∈X. We choose the linear splines {v1,v2,…,vn+1} in a uniform grid of n+1 points in [0,1] as a basis of Vn.Sincewi∈Vn,wi is of the form ∑i=1n+1λivi for some scalars λ1,λ2,…,λn+1. It can be seen that wi is a solution of (4.2) if and only if λ¯=(λ1,λ2,…,λn+1)T is the unique solution of(5.6)(Mn+αiBn)λ¯=a¯,
where(5.7)Mn=(〈Kvi,Kvj〉),i,j=1,2,…,n+1,Bn=(〈vi,vj〉),i,j=1,2,…,n+1,a¯=(〈PhK*(fδ-KF(x0)),vi〉)T,i=1,2,…,n+1.Observe thatzijh,δ in Step 4 of algorithm is again in Vn and hence zijh,δ=∑k=1n+1μkijvk for some μkij,k=1,2,…,n+1. One can see that for j<i,zijh,δ is a solution of(5.8)(PhK*KPh+αjI)zijh,δ=(αj-αi)wi
if and only if μij¯=(μ1ij,μ2ij,…,μn+1ij)T is the unique solution of(5.9)(Mn+αjBn)μij¯=b¯,
where(5.10)b¯=(〈(αj-αi)wi,vi〉)T.
Compute zijh,δ till ∥zijh,δ∥>4C(δ+ɛh)/αj and fix k=i-1. Now choose nk=min{n:e-γ13n≤(δ+ɛh)/(αk)}.Letξn=(ξ1n,ξ2n,…,ξn+1n),  ηn=(η1n,η2n,…,ηn+1n),  yn,αkh,δ=∑i=1n+1ξinvi and xn,αkh,δ=∑i=1n+1ηinvi. Then from (3.5) we have(5.11)(PhF′(xn,αkh,δ)+αkc)∑i=1n+1(ξin-ηin)vi=∑i=1n+1λivi-∑i=1n+1PhF(xn,αkh,δ)vi+αkc∑i=1n+1(x0(ti)-ηin)vi,
where t1,t2,…,tn+1 are the grid points.Observe that(yn,αkh,δ-xn,αkh,δ) is a solution of (3.5) if and only if (ξn-ηn¯)=(ξ1n-η1n,ξ2n-η2n,…,ξn+1n-ηn+1n)T is the unique solution of(5.12)(Qn+αkcBn)(ξn-ηn¯)=Bn[λ¯-Fh1+αkc(X0-ηn¯)],
where Qn=〈F′(xn,αkh,δ)vi,vj〉,i,j=1,2,…,n+1(5.13)Fh1=[F(xn,αkh,δ)(t1),F(xn,αkh,δ)(t2),…,F(xn,αkh,δ)(tn+1)]T
and X0=[x0(t1),x0(t2),…,x0(tn+1)]T.Further from (3.6) it follows that(5.14)(PhF′(xn,αkh,δ)+αkc)(xn+1,αkh,δ-yn,αkh,δ)=Ph[zαkh,δ-F(yn,αkh,δ)+αkc(x0,αkh,δ-yn,αkh,δ)].
Thus (xn+1,αkh,δ-yn,αkh,δ) is a solution of (5.14) if and only if (ηn+1-ξn¯)=(η1n+1-ξ1n,η2n+1-ξ2n,…,ηn+1n+1-ξn+1n)T is the unique solution of(5.15)(Qn+αkcBn)(ηn+1-ξn¯)=Bn[λ¯-Fh2+αkc(X0-ξn¯)],
where Fh2=[F(yn,αkh,δ)(t1),F(yn,αkh,δ)(t2),…,F(yn,αkh,δ)(tn+1)]T.


### 5.1. Numerical Example

Example 5.1.
To illustrate the method discussed in the above section, we consider the spaceX=Y=L2[0,1] and the Fredholm integral operator K:L2[0,1]→L2[0,1]. The algorithm in Section 5 is applied by choosing Vn as the space of linear splines in a uniform grid of n+1 points in [0,1].
In our computation, we takef(t)=(1/36π2)(27sinπt-sin3πt)+(1/36π)(27t2cosπt-3t2cos3πt+6tcos3πt-3cos3πt-27tcosπt) and fδ=f+δ. Then the exact solution
(5.16)x̂(t)=sin⁡πt.
We use
(5.17)x0(t)=sin⁡πt+34π2(1+tπ2-t2π2-cos⁡2(πt))
as our initial guess, so that the function x0-x̂ satisfies the source condition
(5.18)x0-x̂=F′(x̂)1=φ1(F′(x0))G(x0,x̂),
where φ1(λ)=λ. Thus we expect to have an accuracy of order at least O((δ+εh)1/2).
We chooseα0=(1.5)δ2,μ=1.5,δ=0.0667=c,ɛh=1/10n2,ρ=0.19,γρ=0.8173, and g(γρ)=0.54 approximately. For all n the number of iteration nk=3 in this example. The results of the computation are presented in Table 1. The plots of the exact and the approximate solution obtained are given in Figures 1 and 2.Table 1nkαk∥xk-x̂∥∥xk-x̂∥/(δ+ɛh)1/2840.17900.03630.13881640.17290.04320.16693240.17140.04500.17426440.17100.04550.176112840.17090.04560.176525640.17090.04560.176751240.17090.04560.1767102440.17090.04560.1767Curve of the exact and approximate solutions.
(a)(b)(c)(d)Curve of exact and approximate solutions.
(a)(b)(c)(d)


## 5.1. Numerical Example

Example 5.1.
To illustrate the method discussed in the above section, we consider the spaceX=Y=L2[0,1] and the Fredholm integral operator K:L2[0,1]→L2[0,1]. The algorithm in Section 5 is applied by choosing Vn as the space of linear splines in a uniform grid of n+1 points in [0,1].
In our computation, we takef(t)=(1/36π2)(27sinπt-sin3πt)+(1/36π)(27t2cosπt-3t2cos3πt+6tcos3πt-3cos3πt-27tcosπt) and fδ=f+δ. Then the exact solution
(5.16)x̂(t)=sin⁡πt.
We use
(5.17)x0(t)=sin⁡πt+34π2(1+tπ2-t2π2-cos⁡2(πt))
as our initial guess, so that the function x0-x̂ satisfies the source condition
(5.18)x0-x̂=F′(x̂)1=φ1(F′(x0))G(x0,x̂),
where φ1(λ)=λ. Thus we expect to have an accuracy of order at least O((δ+εh)1/2).
We chooseα0=(1.5)δ2,μ=1.5,δ=0.0667=c,ɛh=1/10n2,ρ=0.19,γρ=0.8173, and g(γρ)=0.54 approximately. For all n the number of iteration nk=3 in this example. The results of the computation are presented in Table 1. The plots of the exact and the approximate solution obtained are given in Figures 1 and 2.Table 1nkαk∥xk-x̂∥∥xk-x̂∥/(δ+ɛh)1/2840.17900.03630.13881640.17290.04320.16693240.17140.04500.17426440.17100.04550.176112840.17090.04560.176525640.17090.04560.176751240.17090.04560.1767102440.17090.04560.1767Curve of the exact and approximate solutions.
(a)(b)(c)(d)Curve of exact and approximate solutions.
(a)(b)(c)(d)



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*Source: 783579-2012-05-08.xml*