# Semilocal Convergence Theorem for the Inverse-Free Jarratt Method under New Hölder Conditions

Authors: Yueqing Zhao; Rongfei Lin; Zdenek Šmarda; Yasir Khan; Jinbiao Chen; Qingbiao Wu
Journal: The Scientific World Journal
 (2015)

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## Abstract

Under the new Hölder conditions, we consider the convergence analysis of the inverse-free Jarratt method in Banach space which is used to solve the nonlinear operator equation. We establish a new semilocal convergence theorem for the inverse-free Jarratt method and present an error estimate. Finally, three examples are provided to show the application of the theorem.


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## Body

## 1. Introduction

We consider the following boundary value problem:(1)x′′=-λG(x),x(a)=xa,x(b)=xb.
Those are equivalent to the following nonlinear integral equation (see [1, 2]):
(2)x(s)=α(s)+λ∫ab‍ks,tGxtdt,
where α(s)=1/b-a(xa(b-s)+xb(s-a)) and G:Ω⊂C[a,b]→C[a,b] is a twice Fréchet-differentiable operator. C[a,b] is the set of all continuous functions in [a,b]; k(s,t) is the Green function:
(3)k(s,t)=(b-s)(t-a)b-a,t≤s,(s-a)(b-t)b-a,s≤t.Instead of2, we can try to solve a nonlinear operator equation F(s)=0, where
(4)F:Ω⊂Ca,b⟶Ca,b,Fxs=xs-αs-λ∫ab‍ks,tGxtdt.
Solving the nonlinear operator equation is an important issue in the engineering and technology field as these kinds of problems appear in many real-world applications. Economics [3], chemistry [4], and physics [5–8] are some of the examples of the scientific and engineering technology areas applied to solve these type of equations. In this study, we consider to establish a new semilocal convergence theorem of the Jarratt method in Banach space which is used to solve the nonlinear operator equation
(5)F(x)=0,
where F is defined on an open convex Ω of a Banach space X with values in a Banach space Y.There are a lot of methods of finding a solution of equationF(x)=0. Particularly iterative methods are often used to solve this problem (see [1, 2, 9, 10]). If we use the famous Newton method, we can proceed as
(6)xn+1=xn-F′xn-1F(xn),(n≥0)(x0∈Ω).
Under a reasonable hypothesis, Newton’s method is the second-order convergence.To improve the convergence order, many modified methods have been presented. The famous Halley’s method and the supper-Halley method are the third-order convergence. References [11–22] give the convergence analysis for these methods. Now, we consider the following Jarratt method (see [23–25]):
(7)yn=xn-F′xn-1F(xn),Hxn,yn=32F′xn-1F′xn+23yn-xn-F′xn,xn+1=yn-12Hxn,ynI-Hxn,ynyn-xn.In this paper, we discuss the convergence of7 for solving nonlinear operator equations in Banach spaces and establish a new semilocal convergence theorem under the following condition (see [20, 21]):
(8)F′′′(x)-F′′′(y)≤ωx-y,
where ω:[0,+∞)→R is a nondecreasing continuous function. Finally, the corresponding error estimate is also given.


## 2. Main Results

In the section, we establish a new semilocal convergence theorem and present the error estimate. DenoteB(x,r)={y∈X∣y-x<r} and B(x,r)¯={y∈X∣y-x≤r}. Suppose that X and Y are the Banach spaces, Ω is an open convex of the Banach space X, and F:Ω⊂X→Y has continuous Fréchet derivative of the third-order. F′(x0)-1 exists, for some x0∈Ω, and F satisfies
(9)A1y0-x0=F′x0-1F(x0)≤η;(A2)F′x0-1F′′(x)≤M,x∈Ω,M≥0;(A3)F′x0-1F′′′(x)≤N,x∈Ω,N≥0;A4F′x0-1F′′′x-F′′′y≤ωx-y,x,y∈Ω.(A5)
ω(z) is a nondecreasing continuous real function for z>0 such that ω(0)≥0, and there exists a positive real number p∈(0,1] such that ω(tz)≤tpω(z) for t∈[0,1] and z∈[0,+∞).(A6)
DenoteA=∫01‍∫01‍t(1-t)(st)pdsdt=1/(p+1)(p+2)(p+3), B=1/3∫01‍ ∫01‍(2st/3)ptdsdt=2p/3p+1(p+1)(p+2). Let a0=Mη, b0=Nη2, c0=η2ω(η), an+1=anf2(an)g(an,bn,cn), bn+1=bnf3(an) g2(an,bn,cn), cn+1=f3+p(an)g2+p(an,bn,cn), where
(10)fx=22-2x-x2-x3,g(x,y,z)=5x3+2x4+x58+xy12+(A+B)z.First, we get some lemmas.Lemma 1.
Suppose thatf(x), g(x,y,z) are given by 10. Then∀x∈(0,1/2), f(x) is increasing and f(x)>1;∀x∈(0,1/2), y>0, g(x,y,z) is increasing;∀γ∈(0,1), x∈(0,1/2), p>0, f(γx)<f(x) and g(γx,γ2y,γ2+pz)< γ2+pg(x,y,z).Lemma 2.
Suppose thatf(x), g(x,y,z) are given by 10. If
(11)a0∈0,12,f2(a0)g(a0,b0,c0)<1,
then(i)
the sequences{an}, {bn}, {cn} are nonnegative and decreasing;(ii)
(1+an/2(1+an))an<1, ∀n≥0.Proof.
(i) Whenn=1,
(12)0≤a1=a0f2(a0)g(a0,b0,c0)≤a0,0≤b1=b0f3a0g2a0,b0,c0≤b0,0≤c1=c0f3+p(a0)g2+p(a0,b0,c0)≤c0.
Suppose aj≤aj-1,bj≤bj-1 for j=1,2,…,n. By Lemma 1, f and g are increasing; then
(13)an+1=anf2(an)g(an,bn,cn)≤anf2(a0)g(a0,b0,c0)≤an,bn+1=bnf3ang2an,bn,cn≤bnf3a0g2a0,b0,c0≤bn,cn+1=cnf3+p(an)g2+p(an,bn,cn)≤cnf2a0ga0,b0,c02+p≤cn.
(ii) By (i),{an} is decreasing and a0∈(0,1/2). So, for all n≥0,
(14)1+an21+anan≤1+a021+a0a0<1.
This completes the proof of Lemma 2.Lemma 3.
Suppose that the conditions of Lemma2 hold. Denote γ=a1/a0=f2(a0)g(a0,b0,c0)<1. Then (i)
an≤γ3+pn-1an-1≤γ3+pn-1/2+pa0, bn≤γ3+pn-12bn-1 ≤ γ3+pn-1/2+p2b0, cn≤γ3+pn-12+pcn-1 ≤ γ3+pn-1c0∀n≥1;(ii)
f(an)g(an,bn,cn)≤γ3+pn-1f(a0)g(a0,b0,c0)=γ3+pn/f(a0), ∀n≥1.Proof.
First, by induction, we prove that (i) holds. Becausea1=γa0 and f(a0)>1, we have
(15)b1=b0f3(a0)g2(a0,b0,c0)≤γ2b0,c1=c0f3+p(a0)g2+p(a0,b0,c0)≤γ2+pc0.
Suppose that (i) holds for n≥1. Then we get
(16)an+1=anf2angan,bn,cnan+1≤γ3+pn-1an-1f2γ3+pn-1an-1an+1<×gγ3+pn-1an-1,γ3+pn-12bn-1,γ3+pn-12+pcn-1an+1≤γ3+pn-1an-1f2an-1γ3+pn-12+pgan-1,bn-1,cn-1an+1=γ3+pnan-1f2an-1gan-1,bn-1,cn-1=γ3+pnan,an+1≤γ3+pnan≤γ3+pnγ3+pn-1an-1an+1≤⋯≤γ3+pnγ3+pn-1⋯γ3+p0a0an+1=γ3+pn+1-1/2+pa0,bn+1=bnf3(an)g2(an,bn,cn)≤bnan+1an2≤γ3+pn2bnbn+1≤⋯≤γ3+pn2γ3+pn-12⋯γ3+p02b0bn+1=γ3+pn+1-1/2+p2b0,cn+1=cnf3+p(an)g2+p(an,bn,cn)cn+1≤cnf2angan,bn,cn2+p=cnan+1an2+pcn+1≤γ3+pn2+pcn≤⋯≤γ3+pn+1-1c0
and from (ii) we get
(17)fangan,bn,cn≤fγ3+pn-1/2+pa0×gγ3+pn-1/2+pa0,γ3+pn-1/2+p2b0,γ3n-1c0≤γ3+pn-1fa0ga0,b0,c0=γ3+pnfa0,n≥1.
This completes the proof of Lemma 3.Lemma 4.
Suppose thatX and Y are Banach spaces, Ω is an open convex of the Banach space X, F:Ω⊂X→Y has continuous Fréchet derivative of the second-order, and the sequences {xn}, {yn} are generated by 7. Then, for all natural numbers n≥0, the following formula holds:
(18)F(xn+1)=∫01‍F′′yn+txn+1-yn1-tdtxn+1-yn2+∫01‍F′′xn+tyn-xn1-tdt-12∫01‍F′′xn+23tyn-xndtyn-xn2-12∫01‍xn+23tyn-xnF′′xn+tyn-xn-F′′xn+23tyn-xndt×yn-xnHxn,ynyn-xn+12∫01‍F′′xn+tyn-xndt×yn-xnHxn,ynHxn,ynyn-xn.Proof.
Consider(19)Fyn=F(yn)-F(xn)-F′(xn)(yn-xn)=∫01‍F′′xn+tyn-xn1-tdtyn-xn2,F′(yn)(xn+1-yn)=-12[F′(yn)-F′(xn)]H(xn,yn)×[I-H(xn,yn)](yn-xn)-12F′xnHxn,ynI-Hxn,ynyn-xn=-12∫01‍F′′(xn+t(yn-xn))dt×(yn-xn)H(xn,yn)(yn-xn)+12∫01‍F′′(xn+t(yn-xn))dt×(yn-xn)H(xn,yn)H(xn,yn)(yn-xn)-12∫01‍F′′xn+23tyn-xndtyn-xn2+12∫01‍F′′xn+23tyn-xndt×yn-xnHxn,ynyn-xn,F(xn+1)=F(xn+1)-F(yn)-F′(yn)(xn+1-yn)+F(yn)+F′(yn)(xn+1-yn)=∫01‍F′′(yn+t(xn+1-yn))(1-t)dtxn+1-yn2+∫01‍F′′(xn+t(yn-xn))(1-t)dt-12∫01‍F′′xn+23tyn-xndtyn-xn2-12∫01‍xn+23tyn-xnF′′xn+tyn-xn-F′′xn+23tyn-xndt×(yn-xn)H(xn,yn)(yn-xn)+12∫01‍F′′xn+tyn-xndt×(yn-xn)H(xn,yn)H(xn,yn)(yn-xn).
This completes the proof of Lemma4.
By (A1)–(A6),10, and 11, if a0<1/2, then
(20)H(x0,y0)≤My0-x0=MF′x0-1F′(x0)y0-x0≤a0,x1-y0≤12H(x0,y0)I-H(x0,y0)y0-x0≤a02(1+a0)y0-x0,x1-x0≤x1-y0+y0-x0≤1+a021+a0y0-x0<Rη,
where R=[1+a0/2(1+a0)]1/1-f(a0)g(a0,b0,c0); hence, x1,y0∈S(x0,Rη). Consider
(21)F′x0-1F′x1-I≤Mx1-x0≤1+a021+a0a0<1.
By Banach lemma, F′x1-1 exists, and
(22)F′x1-1F′(x0)≤f(a0)=f(a0)F′x0-1F′(x0).By Lemma4, we have
(23)∫01‍F′′x0+23ty0-x0dtF′x0-1×∫01‍F′′x0+ty0-x01-tdt-12F′x0-1×∫01‍F′′x0+23ty0-x0dt=∫01‍F′′x0+23ty0-x0-F′′x0dtF′x0-1∫01‍F′′-F′′x0x0+ty0-x0-F′′x01-tdt-12F′x0-1×∫01‍F′′x0+23ty0-x0-F′′x0dt=F′x0-1∬01F′′′-F′′′(x0)(x0+st(y0-x0))-F′′′(x0)ds×t(1-t)dt(y0-x0)-13F′x0-1×∬01F′′′x0+23sty0-x0-F′′′(x0)F′′′x0+23sty0-x0dstdt(y0-x0)F′x0-1∫01‍∫01‍≤(A+B)ω(η)y0-x0,F′x0-1F(x1)≤M2x1-y02+N12a0y0-x03+M2a02y0-x02+(A+B)ω(η)y0-x03,y1-x1≤F′x1-1F′(x0)F′x0-1F(x1)≤f(a0)g(a0,b0,c0)y0-x0.
Hence,
(24)Hx1,y1≤MF′x1-1F′(x0)y1-x1≤Mf2(a0)g(a0,b0,c0)y0-x0=a1,NF′x1-1F′(x0)y1-x12≤Nf3a0g2a0,b0,c0y0-x02=b1,F′x1-1F′(x0)ωy1-x1y1-x12≤f3+pa0g2+pa0,b0,c0ωηy0-x02=c1.
Hence,
(25)x2-y1≤12a1(1+a1)y1-x1,x2-x1≤x2-y1+y1-x1≤1+12a11+a1y1-x1,x2-x0≤x2-x1+x1-x0≤1+a021+a0fa0ga0,b0,c0+1×y0-x0<Rη.
By
(26)F′x1-1F′(x2)-I≤MF′x1-1F′(x0)x2-x1≤a11+a121+a1<1,
hence F′x2-1F′(x0) exists, and F′x2-1F′(x0)≤f(a1)F′x1-1F′(x0). By induction, we can prove that the following Lemma 5 holds.Lemma 5.
Under the hypotheses of Lemma2, the following items are true for all n≥1:(I)
F′xn-1F′x0 exists and F′xn-1F′x0  ≤fan-1F′xn-1-1F′x0;(II)
yn-xn≤fan-1gan-1,bn-1,cn-1yn-1-xn-1;(III)
Hxn,yn≤MF′(xn)-1F′(x0)yn-xn≤an;(IV)
NF′xn-1F′(x0)yn-xn2≤bn;(V)
F′xn-1F′x0ωyn-xnyn-xn2≤cn;(VI)
xn+1-yn≤an/21+anyn-xn;(VII)
xn+1-xn≤1+an/21+anyn-xn;(VIII)
xn+1-x0≤Rη, where R=[1+a0/2(1+a0)] 1/1-f(a0)g(a0,b0,c0).Theorem 6.
LetX and Y be two Banach spaces and F:Ω⊂X→Y has continuous Fréchet derivative of the third-order on a nonempty open convex Ω. One supposes that Γ0=F′x0-1∈L(Y,X) exists for some x0∈Ω and conditions (A1)–(A6) and 11 hold. If S(x0,Rη)¯⊂Ω, then the sequence {xn} generated by 7 is well defined and converges to a unique solution x* of 2 in S(x0,2/M-Rη)∩Ω. Furthermore, the following error estimate is obtained:
(27)x*-xn≤1+γ3+pn-1/2+pa021+γ3+pn-1/2+pa0×11-γ3+pnΔγ3+pn-1/2+pΔnη,
where γ=f2(a0)g(a0,b0,c0)=a1/a0 and Δ=1/f(a0), R=(1+a0/2(1+a0)) 1/1-γΔ.Proof.
Firstly, we prove that the sequence{xn} is a Cauchy one. From (II) and by Lemma 3, we have
(28)yn-xn≤fan-1gan-1,bn-1,cn-1yn-1-xn-1≤⋯≤∏i=0n-1‍faigai,bi,ciη≤∏i=0n-1‍γ3+piΔη=γ3+pn-1/2+pΔnη.
For n≥0, m≥1,
(29)xn+m-xn≤xn+m-xn+m-1+xn+m-1-xn+m-2+⋯+xn+1-xn≤1+an21+an×yn+m-1-xn+m-1+⋯+yn+1-xn≤1+an21+an×γ3+pn+m-1-1/2+pΔn+m-1γ3+pn+m-1-1/2+pΔn+m-1+⋯+γ3+pn-1/2+pΔnη=1+an21+anγ3+pn-1/2+pΔnη×γ3+pn3+pm-1-1/2+pΔm-1+⋯+1.
By the Bernoulli inequality 1+xk-1>kx, so 3+pk-1>k(2+p). Hence, we have
(30)xn+m-xn<1+an21+an11-γ3+pnΔγ3+pn-1/2+pΔnη.
Hence, {xn} is a Cauchy sequence and x*=lim⁡n→∞xn. Obviously, xm∈B(x0,Rη), for all m≥1, as if n=0 in 30; we obtain
(31)xm-x0<1+a021+a011-γΔη=Rη.
Following a similar procedure, we have yn∈B(x0,Rη), for all n≥0.
Now, letn→∞ in 28. It follows that F′xn-1F(xn)→0. Besides F(xn)→0, since F(xn)≤F′xnF′xn-1F(xn) and {F′(xn)} is a bounded sequence, therefore F(x*)=0 by the continuity of F in S(x0,Rη)¯.
By lettingm→∞ in 30, we obtain error estimate 28.
To show uniqueness, let us assume that there exists a second solutiony* of 2 in S(x0,2/M-Rη)∩Ω. Then
(32)∫01‍F′x0-1F′x*+ty*-x*-F′x0dt≤M∫01‍x*+t(y*-x*)-x0dt≤M∫01‍1-tx*-x0+ty*-x0dt<M2Rη+2M-Rη=1.
By Banach lemma, we can obtain that the inverse of the linear operator ∫01‍F′(x*+t(y*-x*))dt exists and
(33)∫01‍F′x*+ty*-x*dty*-x*=Fy*-Fx*=0.
We get that x*=y*.
This completes the proof of Theorem6.


## 3. Application

In this section, we apply the convergence theorem and show three numerical examples.Example 1.
Consider the root of the equationF(x)=x10/3+x7/2-x-1=0 on x∈(0,+∞). Then, we easily get that
(34)F′′′(x)=28027x1/3+1058x1/2
does not satisfy (K,p) Hölder condition
(35)F′′′(x)-F′′′(y)≤Kx-yp
because, for all p∈(0,1],
(36)sup⁡x,y∈(0,+∞)280/27x-y1/3+105/8x-y1/2x-yp=+∞.
Let
(37)ωz=28027z1/3+1058z1/2,z>0;
then ω(tz)≤t1/3ω(z) for t∈[0,1] and z∈[0,+∞);
(38)F′′′(x)-F′′′(y)≤ωx-y.
Let us consider a particular case of2 from the operator given by the following nonlinear integral equation of mixed Hammerstein type (see [26]):
(39)x(s)=α(s)-∑i=1m‍∫ab‍ks,tφixtdt,
where -∞<a<b<+∞, u, φi, for i=1,2,…,m, are known functions and x is a solution to be determined. If φ′′′ is (Li,pi) Hölder continuous in Ω, for i=1,2,…,m, the corresponding operator F:Ω⊆C[a,b]→C[a,b],
(40)Fxs=xs+∑i=1m‍∫ab‍ks,tφixtdt-αs,s∈a,b,
does not satisfy (K,p) Hölder condition; for instance, the max-norm is considered. In this case,
(41)F′′′(x)-F′′′(y)≤∑i=1m‍Lix-ypi,Li>0,pi∈0,1,x,y∈Ω.
To solve this type of equations, we can consider
(42)F′′′(x)-F′′′(y)≤ω(x-y),x,y∈Ω,
where ω(z)=Σi=1mLizpi satisfy ω(tz)≤tqω(z), where q=min⁡{pi,p2,…,pm}.Remark 7.
Observe that ifF′′′ is Lipschitz continuous in Ω, we can choose ω(z)=Kz, K>0, so that Jarratt’s method is of R-order, at least four order. If F′′′ is (L,p) Hölder continuous in Ω, then we can choose ω(z)=Lzp, L<0, p∈(0,1], and Jarratt’s method is of R-order, at least 3+p.Example 2.
Consider the case as follows:(43)xs=1+132∫01‍k(s,t)xt16/5dt+130∫01‍k(s,t)xt10/3dt,
where the space is X=C[0,1] with the norm
(44)x=max⁡0≤s≤1x(s),k(s,t)=t(1-s),t≤s,s(1-t),s≤t.
This equation arises in the theory of the radiative transfer and neutron transport and in the kinetic theory of gasses. Let us define the operatorF on X by
(45)Fx=x(s)-132∫01‍k(s,t)xt16/5dt-130∫01‍k(s,t)xt10/3dt-1.
The first, the second, and the third derivatives of F are defined by
(46)F′xus=u(s)-110∫01‍k(s,t)xt11/5u(t)dt-19∫01‍k(s,t)xt7/3u(t)dt,u∈X,F′′xuvs=-1150∫01‍k(s,t)xt6/5u(t)v(t)dt-727∫01‍k(s,t)xt43u(t)v(t)dt,u∈X,F′′′xuvws=-66250∫01‍k(s,t)xt1/5u(t)v(t)w(t)dt-2881∫01‍k(s,t)xt1/3u(t)v(t)w(t)dt,
and we have
(47)[F′′′(x)-F′′′(y)]uvw≤66250max⁡s∈[0,1]×∫01‍k(s,t)xt1/5-yt1/5u(t)v(t)w(t)dt+2881max⁡s∈[0,1]×∫01‍k(s,t)xt1/3-yt1/3u(t)v(t)w(t)dt≤66250×18x-y1/5uvw+2881×18x-y1/3uvw.
To apply Theorem 6, we choose x0=x0(s)=1 and we look for a domain in the form
(48)Ω=B(1,2)⊆C([0,1]).
In this case, we have
(49)I-F′(x0)≤19720<1
and from the Banach lemma, we obtain
(50)F′x0-1≤720701,F′x0-1F(x0)≤720701×18132+130=η=9311216,M=0.148766⋯,N=0.0948511⋯,ω(z)=33100z1/5+7162z1/3,p=15.
Then a0=Mη=0.00123353<1/2, b0=6.52127×10-6, c0=1.47132×10-6, γ=f2(a0)g(a0,b0,c0)=3.48167×10-7<1, Δ=0.998766⋯, and R=1.00062⋯. This means that the hypothesis of Theorem 6 is satisfied. Then, the error bound becomes
(51)x*-xn≤1+γ3.2n-1/2.2a021+γ3.2n-1/2.2a0×11-γ3.2nΔγ(3.2n-1/2.2)Δnη.
For n=1,2,3,4, we get
(52)x1-x*≤4.28944×10-10,x2-x*≤5.76451×10-16,x3-x*≤6.63209×10-23,x4-x*≤2.86064×10-32.Example 3.
Let us consider the system of equationsF(u,v)=0, where
(53)F(u,v)=(u7/2-uv-v10/3+1,u7/2+uv-v10/3-1)T.
Then, we have
(54)F′(u,v)=72u5/2-v-103v7/3-u72u5/2+v-103v7/3+u,F′u,v-1=114/2u7/2+20/3v10/3×-103v73+u103v73+u-72u52-v72u52-v,,F′′(u,v)=354u3/2-1-1-709v4/3354u3/211-709v4/3.,F′′′(u,v)s,t3=1058u1/228027v1/31058u1/228027v1/3s3t3.
Now, we choose x0=(u0,v0)=(1.5,1.5) and Ω={x∣x-x0≤1.5}. We take the max-norm in R2 and the norm A=max{a11+a12,a21+a22} for A=a11a12a21a22. We define the norm of a bilinear operator B on R2 by
(55)B=sup⁡u=1max⁡i∑j=12‍∑k=12‍bijkuk,
where u=u1,u2T and B=b111b112b121b122b211b212b221b222.Then, we get the following results:η=F′x0-1F(x0)=0.09598⋯, M=9.20456⋯, N=10.7635⋯, and p=1/3.We get that the hypotheses of Theorem6 are satisfied.



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*Source: 346571-2015-03-25.xml*