A generalization of the Newton method

Ion Păvăloiu
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Abstract

Let \(X\) be a Banach space, \(Y\) a normed space, \(G:X\rightarrow Y\) a nonlinear operator, and \(G\left( x\right) =0\) a nonlinear equation. We denote by \(F:X^{2}\rightarrow Y\) a nonlinear operator for which the restriction to the diagonal of \(X^{2}\) coincide with \(G\). We first prove a Taylor type formula for operators with two variables. Next we consider the following two-step Newton type method: \[F\left( x_{n},x_{n-1}\right) +F_{x}^{\prime}\left( x_{n},x_{n-1}\right) \left( x_{n+1}-x_{n}\right) +F_{y}^{\prime}\left( x_{n},x_{n-1}\right) \left( x_{n}-x_{n-1}\right)=0.\] We study the convergence to the solution of the above sequence.

Authors

Ion Păvăloiu
(Tiberiu Popoviciu Institute of Numerical Analysis)

Title

Original title (in French)

Une généralisation de methode de Newton

English translation of the title

A generalization of the Newton method

Keywords

Taylor polynomial with two variables; two-step Newton type method

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Cite this paper as:

I. Păvăloiu, Une généralisation de methode de Newton, Mathematica, 20(43) (1978) no. 1, pp. 45-52 (in French).

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References

[1] Kantorovici, L.V., Functionalnîi analiz i prikladnaia matematika, UMN, 28, 89-185 (1948).

[2] Pavaloiu, I., Sur les procédés iteratif à un ordere élevé de convergence. Mathematica, 12,  (35), 2 309-324 (1970).

[3] Weinisckhe, J. H., Über eine Klasse von Iterationsverfahren. Numeriche Mathematik , 6, 395-404, (1964).

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A generalization of the Newton method

by
ION PAVALOIU
(Cluj-Napoca)
In this work we will study a Newton-type method for solving operational equations. For the construction of this method it is necessary to consider the generalized Taylor formula for applications of two variables.
Let X X XXXAnd Y Y YYYtwo normed linear spaces. We denote by X 2 == X × X X 2 == X × X X^(2)==X xx XX^{2}= =X \times XX2==X×Xthe Cartesian product of X X XXXby itself and consider an open set U ⊂ X 2 U ⊂ X 2 U subX^(2)U \subset X^{2}U⊂X2. Suppose that the elements ( x 0 , y 0 x 0 , y 0 x_(0),y_(0)x_{0}, y_{0}x0,y0) And ( x 0 + h , y 0 + k x 0 + h , y 0 + k x_(0)+h,y_(0)+kx_{0}+h, y_{0}+kx0+h,y0+k) belong to the set U U UUU, Or ( h , k ) ∈ X 2 ( h , k ) ∈ X 2 (h,k)inX^(2)(h, k) \in X^{2}(h,k)∈X2. Either f : U → Y f : U → Y f:U rarr Yf: U \rightarrow Yf:U→Yan application defined on U U UUUand values ​​in Y Y YYY.
THEOREM 1. If the application f : U → Y f : U → Y f:U rarr Yf: U \rightarrow Yf:U→Yadmits continuous partial derivatives, in the Fréchet sense, up to order 2 inclusive, for each element ( x , y ) ∈ U ( x , y ) ∈ U (x,y)in U(x, y) \in U(x,y)∈U, then we have the following inequality:
(1)
‖ f ( x 0 + h , y 0 + k ) − f ( x 0 , y 0 ) − f x ′ ( x 0 , y 0 ) h − f y ′ ′ ( x 0 , y 0 ) k ‖ ≦ ≤ 1 2 sup 0 ≤ θ ≤ 1 0 ≤ μ ≤ 1 ‖ f x 2 ′ ′ ( x 0 + θ h , y 0 + μ k ) ‖ ⋅ ‖ h ‖ 2 + + 1 2 sup 0 ≤ μ ≤ 1 0 ≤ μ ≤ 1 ‖ f y 2 ′ ′ ( x 0 + θ h , y 0 + μ k ) ‖ ⋅ ‖ k ‖ 2 + + sup 0 ≤ 0 ≤ 1 0 ≤ μ ≤ 1 ‖ f x y ′ ′ ( x 0 + θ h , y 0 + μ k ) ‖ ⋅ ‖ h ‖ ⋅ ‖ k ‖ f x 0 + h , y 0 + k − f x 0 , y 0 − f x ′ x 0 , y 0 h − f y ′ ′ x 0 , y 0 k ≦ ≤ 1 2 sup 0 ≤ θ ≤ 1 0 ≤ μ ≤ 1   f x 2 ′ ′ x 0 + θ h , y 0 + μ k ⋅ ‖ h ‖ 2 + + 1 2 sup 0 ≤ μ ≤ 1 0 ≤ μ ≤ 1   f y 2 ′ ′ x 0 + θ h , y 0 + μ k ⋅ ‖ k ‖ 2 + + sup 0 ≤ 0 ≤ 1 0 ≤ μ ≤ 1   f x y ′ ′ x 0 + θ h , y 0 + μ k ⋅ ‖ h ‖ ⋅ ‖ k ‖ {:[||f(x_(0)+h,y_(0)+k)-f(x_(0),y_(0))-f_(x)^(')(x_(0),y_(0))h-f_(y^('))^(')(x_(0),y_(0))k|| <= {: <= (1)/(2)su p_({:[0 <= theta <= 1],[0 <= mu <= 1]:})||f_(x^(2))^('')(x_(0)+theta h,y_(0)+mu k)||*||h||^(2)+:}+(1)/(2)su p_({:[0 <= mu <= 1],[0 <= mu <= 1]:})||f_(y^(2))^('')(x_(0)+theta h,y_(0)+mu k)||*||k||^(2)+],[+su p_({:[0 <= 0 <= 1],[0 <= mu <= 1]:})||f_(xy)^('')(x_(0)+theta h,y_(0)+mu k)||*||h||*||k||]:}\begin{aligned} \left\|f\left(x_{0}+h, y_{0}+k\right)-f\left(x_{0}, y_{0}\right)-f_{x}^{\prime}\left(x_{0}, y_{0}\right) h-f_{y^{\prime}}^{\prime}\left(x_{0}, y_{0}\right) k\right\| \leqq & \begin{array}{l} \leq \frac{1}{2} \sup _{\substack{0 \leq \theta \leq 1 \\ 0 \leq \mu \leq 1}}\left\|f_{x^{2}}^{\prime \prime}\left(x_{0}+\theta h, y_{0}+\mu k\right)\right\| \cdot\|h\|^{2}+ \\ \end{array}+\frac{1}{2} \sup _{\substack{0 \leq \mu \leq 1 \\ 0 \leq \mu \leq 1}}\left\|f_{y^{2}}^{\prime \prime}\left(x_{0}+\theta h, y_{0}+\mu k\right)\right\| \cdot\|k\|^{2}+ \\ & +\sup _{\substack{0 \leq 0 \leq 1 \\ 0 \leq \mu \leq 1}}\left\|f_{x y}^{\prime \prime}\left(x_{0}+\theta h, y_{0}+\mu k\right)\right\| \cdot\|h\| \cdot\|k\| \end{aligned}‖f(x0+h,y0+k)−f(x0,y0)−fx′(x0,y0)h−fy′′(x0,y0)k‖≦≤12sup0≤θ≤10≤μ≤1‖fx2′′(x0+θh,y0+μk)‖⋅‖h‖2++12sup0≤μ≤10≤μ≤1‖fy2′′(x0+θh,y0+μk)‖⋅‖k‖2++sup0≤0≤10≤μ≤1‖fxy′′(x0+θh,y0+μk)‖⋅‖h‖⋅‖k‖
or by f x ′ , f y ′ , f x ′ ′ , f x y ′ ′ , f y ′ ′ f x ′ , f y ′ , f x ′ ′ , f x y ′ ′ , f y ′ ′ f_(x)^('),f_(y)^('),f_(x)^(''),f_(xy)^(''),f_(y)^('')f_{x}^{\prime}, f_{y}^{\prime}, f_{x}^{\prime \prime}, f_{x y}^{\prime \prime}, f_{y}^{\prime \prime}fx′,fy′,fx′′,fxy′′,fy′′we have designated the partial derivatives of order 1 and 2 of the application f f fffwith respect to the specified variables.
Demonstration. We will give a demonstration analogous to that given in work [1] for Taylor's formula relating to applications of a single variable.
We designate by y y yyythe expression
(2) y = f ( x 0 + h , y 0 + k ) − f ( x 0 , y 0 ) − f x ′ ( x 0 , y 0 ) h − f y ′ ( x 0 , y 0 ) k (2) y = f x 0 + h , y 0 + k − f x 0 , y 0 − f x ′ x 0 , y 0 h − f y ′ x 0 , y 0 k {:(2)y=f(x_(0)+h,y_(0)+k)-f(x_(0),y_(0))-f_(x)^(')(x_(0),y_(0))h-f_(y)^(')(x_(0),y_(0))k:}\begin{equation*} y=f\left(x_{0}+h, y_{0}+k\right)-f\left(x_{0}, y_{0}\right)-f_{x}^{\prime}\left(x_{0}, y_{0}\right) h-f_{y}^{\prime}\left(x_{0}, y_{0}\right) k \tag{2} \end{equation*}(2)y=f(x0+h,y0+k)−f(x0,y0)−fx′(x0,y0)h−fy′(x0,y0)k
Either T : Y → R T : Y → R T:Y rarrRT: Y \rightarrow \mathbf{R}T:Y→Ra linear and continuous functional defined on Y Y YYYand values ​​in R R R\mathbf{R}R, which has the following properties:
(3) ‖ T ‖ = 1 T y = ‖ y ‖ (3) ‖ T ‖ = 1 T y = ‖ y ‖ {:[(3)||T||=1],[Ty=||y||]:}\begin{align*} & \|T\|=1 \tag{3}\\ & T y=\|y\| \end{align*}(3)‖T‖=1Ty=‖y‖
(The existence of such a functional is ensured by the HahnBanach theorem).
Now we consider the function φ : R × R → R φ : R × R → R varphi:RxxRrarrR\varphi: \mathbf{R} \times \mathbf{R} \rightarrow \mathbf{R}φ:R×R→Rgiven by the following equality:
(4) φ ( α , β ) = T ( f ( x 0 + α h , y 0 + β k ) ) . (4) φ ( α , β ) = T f x 0 + α h , y 0 + β k . {:(4)varphi(alpha","beta)=T(f(x_(0)+alpha h,y_(0)+beta k)).:}\begin{equation*} \varphi(\alpha, \beta)=T\left(f\left(x_{0}+\alpha h, y_{0}+\beta k\right)\right) . \tag{4} \end{equation*}(4)φ(α,β)=T(f(x0+αh,y0+βk)).
For the partial derivatives of the function φ φ varphi\varphiφwe have the following formulas:
φ α ′ ( α , β ) = T [ f x ′ ( x 0 + α h , y 0 + β k ) h ] ; φ β ′ ( α , β ) = T [ f y ′ ( x 0 + α h , y 0 + β k ) k ] ; φ α ′ ( α , β ) = T f x ′ x 0 + α h , y 0 + β k h ; φ β ′ ( α , β ) = T f y ′ x 0 + α h , y 0 + β k k ; {:[varphi_(alpha)^(')(alpha","beta)=T[f_(x)^(')(x_(0)+alpha h,y_(0)+beta k)h];],[varphi_(beta)^(')(alpha","beta)=T[f_(y)^(')(x_(0)+alpha h,y_(0)+beta k)k];]:}\begin{aligned} & \varphi_{\alpha}^{\prime}(\alpha, \beta)=T\left[f_{x}^{\prime}\left(x_{0}+\alpha h, y_{0}+\beta k\right) h\right] ; \\ & \varphi_{\beta}^{\prime}(\alpha, \beta)=T\left[f_{y}^{\prime}\left(x_{0}+\alpha h, y_{0}+\beta k\right) k\right] ; \end{aligned}φα′(α,β)=T[fx′(x0+αh,y0+βk)h];φβ′(α,β)=T[fy′(x0+αh,y0+βk)k];
(5)
φ α 2 ′ ′ ( α , β ) = T [ f x 2 ′ ′ ( x 0 + α h , y 0 + β k ) h 2 ] ; φ α β ′ ′ ( α , β ) = T [ f x y ′ ′ ( x 0 + α h , y 0 + β k ) h k ] ; φ β ′ ′ ′ ( α , β ) = T [ f y 0 ′ ′ ′ ( x 0 + α h , y 0 + β k ) k 2 ] . φ α 2 ′ ′ ( α , β ) = T f x 2 ′ ′ x 0 + α h , y 0 + β k h 2 ; φ α β ′ ′ ( α , β ) = T f x y ′ ′ x 0 + α h , y 0 + β k h k ; φ β ′ ′ ′ ( α , β ) = T f y 0 ′ ′ ′ x 0 + α h , y 0 + β k k 2 . {:[varphi_(alpha^(2))^('')(alpha","beta)=T[f_(x^(2))^('')(x_(0)+alpha h,y_(0)+beta k)h^(2)];],[varphi_(alpha beta)^('')(alpha","beta)=T[f_(xy)^('')(x_(0)+alpha h,y_(0)+beta k)hk];],[varphi_(beta^('))^('')(alpha","beta)=T[f_(y_(0)^('))^('')(x_(0)+alpha h,y_(0)+beta k)k^(2)].]:}\begin{aligned} & \varphi_{\alpha^{2}}^{\prime \prime}(\alpha, \beta)=T\left[f_{x^{2}}^{\prime \prime}\left(x_{0}+\alpha h, y_{0}+\beta k\right) h^{2}\right] ; \\ & \varphi_{\alpha \beta}^{\prime \prime}(\alpha, \beta)=T\left[f_{x y}^{\prime \prime}\left(x_{0}+\alpha h, y_{0}+\beta k\right) h k\right] ; \\ & \varphi_{\beta^{\prime}}^{\prime \prime}(\alpha, \beta)=T\left[f_{y_{0}^{\prime}}^{\prime \prime}\left(x_{0}+\alpha h, y_{0}+\beta k\right) k^{2}\right] . \end{aligned}φα2′′(α,β)=T[fx2′′(x0+αh,y0+βk)h2];φαβ′′(α,β)=T[fxy′′(x0+αh,y0+βk)hk];φβ′′′(α,β)=T[fy0′′′(x0+αh,y0+βk)k2].
The values ​​of the function φ φ varphi\varphiφand of these first-order partial derivatives at the point (0,0) are given by the following formulas:
φ ( 0 , 0 ) = T [ f ( x 0 , y 0 ) ] (6) φ α ′ ( 0 , 0 ) = T [ f x ′ ( x 0 , y 0 ) h ] φ β ′ ( 0 , 0 ) = T [ f y ′ , ( x 0 , y 0 ) k ] φ ( 0 , 0 ) = T f x 0 , y 0 (6) φ α ′ ( 0 , 0 ) = T f x ′ x 0 , y 0 h φ β ′ ( 0 , 0 ) = T f y ′ , x 0 , y 0 k {:[varphi(0","0)=T[f(x_(0),y_(0))]],[(6)varphi_(alpha)^(')(0","0)=T[f_(x)^(')(x_(0),y_(0))h]],[varphi_(beta)^(')(0","0)=T[f_(y)^('),(x_(0),y_(0))k]]:}\begin{gather*} \varphi(0,0)=T\left[f\left(x_{0}, y_{0}\right)\right] \\ \varphi_{\alpha}^{\prime}(0,0)=T\left[f_{x}^{\prime}\left(x_{0}, y_{0}\right) h\right] \tag{6}\\ \varphi_{\beta}^{\prime}(0,0)=T\left[f_{y}^{\prime},\left(x_{0}, y_{0}\right) k\right] \end{gather*}φ(0,0)=T[f(x0,y0)](6)φα′(0,0)=T[fx′(x0,y0)h]φβ′(0,0)=T[fy′,(x0,y0)k]
For the function φ φ varphi\varphiφwe have:
(7) φ ( 1 , 1 ) − φ ( 0 , 0 ) − φ α ′ ( 0 , 0 ) − φ β ′ ( 0 , 0 ) = = 1 2 [ φ α 2 ′ ′ ( θ , μ ) + 2 φ α β ′ ′ ( θ , μ ) + φ β 2 ′ ′ ( θ , μ ) ] (7) φ ( 1 , 1 ) − φ ( 0 , 0 ) − φ α ′ ( 0 , 0 ) − φ β ′ ( 0 , 0 ) = = 1 2 φ α 2 ′ ′ ( θ , μ ) + 2 φ α β ′ ′ ( θ , μ ) + φ β 2 ′ ′ ( θ , μ ) {:[(7)varphi(1","1)-varphi(0","0)-varphi_(alpha)^(')(0","0)-varphi_(beta)^(')(0","0)=],[=(1)/(2)[varphi_(alpha^(2))^('')(theta,mu)+2varphi_(alpha beta)^('')(theta,mu)+varphi_(beta^(2))^('')(theta,mu)]]:}\begin{align*} & \varphi(1,1)-\varphi(0,0)-\varphi_{\alpha}^{\prime}(0,0)-\varphi_{\beta}^{\prime}(0,0)= \tag{7}\\ & =\frac{1}{2}\left[\varphi_{\alpha^{2}}^{\prime \prime}(\theta, \mu)+2 \varphi_{\alpha \beta}^{\prime \prime}(\theta, \mu)+\varphi_{\beta^{2}}^{\prime \prime}(\theta, \mu)\right] \end{align*}(7)φ(1,1)−φ(0,0)−φα′(0,0)−φβ′(0,0)==12[φα2′′(θ,μ)+2φαβ′′(θ,μ)+φβ2′′(θ,μ)]
Or 0 ≦ θ ≦ 1 , 0 ≦ μ ≦ 1 0 ≦ θ ≦ 1 , 0 ≦ μ ≦ 1 0 <= theta <= 1,0 <= mu <= 10 \leqq \theta \leqq 1,0 \leqq \mu \leqq 10≦θ≦1,0≦μ≦1.
Taking into account formulas (2) - (7) we have:
(8)
‖ y ‖ = T y = φ ( 1 , 1 ) − φ ( 0 , 0 ) − φ α ′ ( 0 , 0 ) − φ β ′ ( 0 , 0 ) = = 1 2 T [ f x 2 ′ ′ ( x 0 + θ h , y 0 + μ k ) h 2 + 2 f x y ′ ′ ( x 0 + θ h , y 0 + μ k ) h k + ‖ y ‖ = T y = φ ( 1 , 1 ) − φ ( 0 , 0 ) − φ α ′ ( 0 , 0 ) − φ β ′ ( 0 , 0 ) = = 1 2 T f x 2 ′ ′ x 0 + θ h , y 0 + μ k h 2 + 2 f x y ′ ′ x 0 + θ h , y 0 + μ k h k + {:[||y||=Ty=varphi(1","1)-varphi(0","0)-varphi_(alpha)^(')(0","0)-varphi_(beta)^(')(0","0)=],[=(1)/(2)T[f_(x^(2))^('')(x_(0)+theta h,y_(0)+mu k)h^(2)+2f_(xy)^('')(x_(0)+theta h,y_(0)+mu k)hk+:}]:}\begin{gathered} \|y\|=T y=\varphi(1,1)-\varphi(0,0)-\varphi_{\alpha}^{\prime}(0,0)-\varphi_{\beta}^{\prime}(0,0)= \\ =\frac{1}{2} T\left[f_{x^{2}}^{\prime \prime}\left(x_{0}+\theta h, y_{0}+\mu k\right) h^{2}+2 f_{x y}^{\prime \prime}\left(x_{0}+\theta h, y_{0}+\mu k\right) h k+\right. \end{gathered}‖y‖=Ty=φ(1,1)−φ(0,0)−φα′(0,0)−φβ′(0,0)==12T[fx2′′(x0+θh,y0+μk)h2+2fxy′′(x0+θh,y0+μk)hk+
≦ 1 2 ‖ T ‖ [ sup ‖ f x 2 ′ ′ ( x 0 + θ h , y 0 + μ k ) ‖ ⋅ ‖ h ‖ 2 + + 2 sup ‖ f x y ′ ′ ( x 0 + θ h , y 0 + μ k ) ‖ ⋅ ‖ h ‖ ⋅ ‖ k ‖ + + sup ‖ f y 2 ′ ′ ( x 0 + θ h , y 0 + μ k ) ) ‖ ⋅ ‖ k ‖ 2 ] , ≦ 1 2 ‖ T ‖ sup f x 2 ′ ′ x 0 + θ h , y 0 + μ k ⋅ ‖ h ‖ 2 + + 2 sup f x y ′ ′ x 0 + θ h , y 0 + μ k ⋅ ‖ h ‖ ⋅ ‖ k ‖ + + sup ‖ f y 2 ′ ′ x 0 + θ h , y 0 + μ k ‖ ⋅ ‖ k ‖ 2 , {:[ <= (1)/(2)||T||[s u p||f_(x^(2))^('')(x_(0)+theta h,y_(0)+mu k)||*||h||^(2)+:}],[+2s u p||f_(xy)^('')(x_(0)+theta h,y_(0)+mu k)||*||h||*||k||+],[{:+s u p||f_(y^(2))^('')(x_(0)+theta h,y_(0)+mu k))||*||k||^(2)]","]:}\begin{gathered} \leqq \frac{1}{2}\|T\|\left[\sup \left\|f_{x^{2}}^{\prime \prime}\left(x_{0}+\theta h, y_{0}+\mu k\right)\right\| \cdot\|h\|^{2}+\right. \\ +2 \sup \left\|f_{x y}^{\prime \prime}\left(x_{0}+\theta h, y_{0}+\mu k\right)\right\| \cdot\|h\| \cdot\|k\|+ \\ \left.\left.+\sup \| f_{y^{2}}^{\prime \prime}\left(x_{0}+\theta h, y_{0}+\mu k\right)\right)\|\cdot\| k \|^{2}\right], \end{gathered}≦12‖T‖[sup‖fx2′′(x0+θh,y0+μk)‖⋅‖h‖2++2sup‖fxy′′(x0+θh,y0+μk)‖⋅‖h‖⋅‖k‖++sup‖fy2′′(x0+θh,y0+μk))‖⋅‖k‖2],
Or 0 ≦ θ ≦ 1 , 0 ≦ μ ≦ 1 0 ≦ θ ≦ 1 , 0 ≦ μ ≦ 1 0 <= theta <= 1,0 <= mu <= 10 \leqq \theta \leqq 1,0 \leqq \mu \leqq 10≦θ≦1,0≦μ≦1.
From (2), (3) and (8) the inequality of the statement of the theorem results.
In what follows we consider the following equation
(9)
G ( x ) = 0 G ( x ) = 0 G(x)=0G(x)=0G(x)=0
Or G : X → Y G : X → Y G:X rarr YG: X \rightarrow YG:X→YAnd θ θ theta\thetaθis the neutral element of space Y Y YYY
We assume that X X XXXis a Banach space. We denote by F : X 2 → Y F : X 2 → Y F:X^(2)rarr YF: X^{2} \rightarrow YF:X2→Yan application and we assume that the application restriction F F FFFon the whole D D DDDcoincides with G G GGG, Or D = { ( x , y ) ∈ X 2 : x = y } D = ( x , y ) ∈ X 2 : x = y D={(x,y)inX^(2):x=y}D=\left\{(x, y) \in X^{2}: x=y\right\}D={(x,y)∈X2:x=y}.
Either ( x n , x n − 1 ) ∈ X 2 x n , x n − 1 ∈ X 2 (x_(n),x_(n-1))inX^(2)\left(x_{n}, x_{n-1}\right) \in X^{2}(xn,xn−1)∈X2any element of space X 2 X 2 X^(2)X^{2}X2. We assume that the application F F FFFadmits partial derivatives in the Fréchet sense at the point ( x n , x n − 1 ) ∈ X 2 x n , x n − 1 ∈ X 2 (x_(n),x_(n-1))inX^(2)\left(x_{n}, x_{n-1}\right) \in X^{2}(xn,xn−1)∈X2and we consider the linear equation
(10) F ( x n , x n − 1 ) + F x ′ ( x n , x n − 1 ) ( x n + 1 − x n ) + F y ′ ( x n , x n − 1 ) ( x n − x n − 1 ) = θ F x n , x n − 1 + F x ′ x n , x n − 1 x n + 1 − x n + F y ′ x n , x n − 1 x n − x n − 1 = θ F(x_(n),x_(n-1))+F_(x)^(')(x_(n),x_(n-1))(x_(n+1)-x_(n))+F_(y)^(')(x_(n),x_(n-1))(x_(n)-x_(n-1))=thetaF\left(x_{n}, x_{n-1}\right)+F_{x}^{\prime}\left(x_{n}, x_{n-1}\right)\left(x_{n+1}-x_{n}\right)+F_{y}^{\prime}\left(x_{n}, x_{n-1}\right)\left(x_{n}-x_{n-1}\right)=\thetaF(xn,xn−1)+Fx′(xn,xn−1)(xn+1−xn)+Fy′(xn,xn−1)(xn−xn−1)=θ,
where the unknown element is x n + 1 x n + 1 x_(n+1)x_{n+1}xn+1.
If we assume that the application F x ′ ( x n , x n − 1 ) ∈ L ( X , Y ) F x ′ x n , x n − 1 ∈ L ( X , Y ) F_(x)^(')(x_(n),x_(n-1))in L(X,Y)F_{x}^{\prime}\left(x_{n}, x_{n-1}\right) \in L(X, Y)Fx′(xn,xn−1)∈L(X,Y)is reversible (by L ( X , Y ) L ( X , Y ) L(X,Y)L(X, Y)L(X,Y)we denote the set of linear and continuous applications defined on X X XXXand values ​​in Y Y YYY), then the solution x n + 1 x n + 1 x_(n+1)x_{n+1}xn+1of equation (10) has the following representation:
(11)
x n + 1 = x n − Γ n [ F ( x n , x n − 1 ) + F y ′ ( x n , x n − 1 ) ( x n − x n − 1 ) ] x n + 1 = x n − Γ n F x n , x n − 1 + F y ′ x n , x n − 1 x n − x n − 1 x_(n+1)=x_(n)-Gamma_(n)[F(x_(n),x_(n-1))+F_(y)^(')(x_(n),x_(n-1))(x_(n)-x_(n-1))]x_{n+1}=x_{n}-\Gamma_{n}\left[F\left(x_{n}, x_{n-1}\right)+F_{y}^{\prime}\left(x_{n}, x_{n-1}\right)\left(x_{n}-x_{n-1}\right)\right]xn+1=xn−Γn[F(xn,xn−1)+Fy′(xn,xn−1)(xn−xn−1)]
Or
Γ n = [ F x ′ ( x n , x n − 1 ) ] − 1 Γ n = F x ′ x n , x n − 1 − 1 Gamma_(n)=[F_(x)^(')(x_(n),x_(n-1))]^(-1)\Gamma_{n}=\left[F_{x}^{\prime}\left(x_{n}, x_{n-1}\right)\right]^{-1}Γn=[Fx′(xn,xn−1)]−1
Using method (11) for each n = 0 , 1 , … n = 0 , 1 , … n=0,1,dotsn=0,1, \ldotsn=0,1,…, we get a sequence ( x n ) n = 0 ∞ x n n = 0 ∞ (x_(n))_(n=0)^(oo)\left(x_{n}\right)_{n=0}^{\infty}(xn)n=0∞of approximations for the solution of equation (9).
Indeed, if the following ( x n ) n = 0 ∞ x n n = 0 ∞ (x_(n))_(n=0)^(oo)\left(x_{n}\right)_{n=0}^{\infty}(xn)n=0∞is convergent, and if x ¯ = lim x n x ¯ = lim x n bar(x)=limx_(n)\bar{x}=\lim x_{n}x¯=limxn, then it is obvious that F ( x ¯ , x ¯ ) = G ( x ¯ ) = θ F ( x ¯ , x ¯ ) = G ( x ¯ ) = θ F( bar(x), bar(x))=G( bar(x))=thetaF(\bar{x}, \bar{x})=G(\bar{x})=\thetaF(x¯,x¯)=G(x¯)=θ, that is to say that x ¯ x ¯ bar(x)\bar{x}x¯verifies equation (9).
To study the convergence of the proposed method, we consider the following system of inequalities:
(12) δ n + 1 ≦ a δ n 2 + b ⋅ δ n ⋅ δ n − 1 + c ⋅ δ n − 1 2 + d ⋅ δ n , n = 1 , 2 , … (12) δ n + 1 ≦ a δ n 2 + b ⋅ δ n ⋅ δ n − 1 + c ⋅ δ n − 1 2 + d ⋅ δ n , n = 1 , 2 , … {:(12)delta_(n+1) <= adelta_(n)^(2)+b*delta_(n)*delta_(n-1)+c*delta_(n-1)^(2)+d*delta_(n)","n=1","2","dots:}\begin{equation*} \delta_{n+1} \leqq a \delta_{n}^{2}+b \cdot \delta_{n} \cdot \delta_{n-1}+c \cdot \delta_{n-1}^{2}+d \cdot \delta_{n}, n=1,2, \ldots \tag{12} \end{equation*}(12)δn+1≦hasδn2+b⋅δn⋅δn−1+c⋅δn−12+d⋅δn,n=1,2,…
(oû̀) a ≧ 0 , b ≧ 0 , c ≧ 0 , c ≧ 0 , d ≧ 0 , δ 0 ≧ 0 et δ 1 ≧ 0 . (oû̀) a ≧ 0 , b ≧ 0 , c ≧ 0 , c ≧ 0 , d ≧ 0 , δ 0 ≧ 0  et  δ 1 ≧ 0 . {:(oû̀)a >= 0","b >= 0","c >= 0","quad c >= 0","quad d >= 0","quaddelta_(0) >= 0" et "delta_(1) >= 0.:}\begin{equation*} a \geqq 0, b \geqq 0, c \geqq 0, \quad c \geqq 0, \quad d \geqq 0, \quad \delta_{0} \geqq 0 \text { et } \delta_{1} \geqq 0 . \tag{oû̀} \end{equation*}(Or)has≧0,b≧0,c≧0,c≧0,d≧0,δ0≧0 And δ1≧0.
Relative to the system (12) we will demonstrate the following theorem: theorem, 2. If the elements of the sequence ( δ n ) n = 0 ∞ δ n n = 0 ∞ (delta_(n))_(n=0)^(oo)\left(\delta_{n}\right)_{n=0}^{\infty}(δn)n=0∞and real numbers u , b , c u , b , c u,b,cu, b, cu,b,cAnd d d dddmeet the following conditions:
(i) δ n ≧ 0 δ n ≧ 0 delta_(n) >= 0\delta_{n} \geqq 0δn≧0for each n = 0 , 1 n = 0 , 1 n=0,1n=0,1n=0,1,
(ii) the elements of the sequence ( δ n ) n aco ∞ δ n n  aco  ∞ (delta_(n))_(n" aco ")^(oo)\left(\delta_{n}\right)_{n \text { aco }}^{\infty}(δn)n aco ∞verify the inequalities (12);
(iii) δ 0 = k , d 1 = k t 1 δ 0 = k , d 1 = k t 1 delta_(0)=k,d_(1)=kt_(1)\delta_{0}=k, d_{1}=k t_{1}δ0=k,d1=kt1Or k > 0 k > 0 k > 0k>0k>0is a constant independent of n n nnn, And t 1 t 1 t_(1)t_{1}t1is the positive solution of the equation
(13) ( a k − 1 ) t 2 + ( d + b k ) t + c k = 0 (13) ( a k − 1 ) t 2 + ( d + b k ) t + c k = 0 {:(13)(ak-1)t^(2)+(d+bk)t+ck=0:}\begin{equation*} (a k-1) t^{2}+(d+b k) t+c k=0 \tag{13} \end{equation*}(13)(hask−1)t2+(d+bk)t+ck=0
(iv) the constants a , b , c , d a , b , c , d a,b,c,da, b, c, dhas,b,c,dAnd k k kkksatisfy the following inequality:
k ( a + b + c ) + d < 1 k ( a + b + c ) + d < 1 k(a+b+c)+d < 1k(a+b+c)+d<1k(has+b+c)+d<1
then:
(j) the elements of the sequence ( δ n ) n = 0 ∞ δ n n = 0 ∞ (delta_(n))_(n=0)^(oo)\left(\delta_{n}\right)_{n=0}^{\infty}(δn)n=0∞satisfy the inequalities δ n ≦ k l 1 n δ n ≦ k l 1 n delta_(n) <= kl_(1)^(n)\delta_{n} \leqq k l_{1}^{n}δn≦kL1n, n = 0 , 1 , … n = 0 , 1 , … n=0,1,dotsn=0,1, \ldotsn=0,1,…;
(dd) lim n → ∞ δ n = 0 lim n → ∞   δ n = 0 lim_(n rarr oo)delta_(n)=0\lim _{n \rightarrow \infty} \delta_{n}=0limn→∞δn=0;
(jjj) the series ∑ i = 0 ∞ δ i ∑ i = 0 ∞   δ i sum_(i=0)^(oo)delta_(i)\sum_{i=0}^{\infty} \delta_{i}∑i=0∞δiis convergent.
Proof. From (iv) we deduce that equation (13) has a root t 1 t 1 t_(1)t_{1}t1, 0 ≦ t 1 < 1 0 ≦ t 1 < 1 0 <= t_(1) < 10 \leqq t_{1}<10≦t1<1. Either φ ( t ) = ( a k − 1 ) t 2 + ( d + b k ) t + c k φ ( t ) = ( a k − 1 ) t 2 + ( d + b k ) t + c k varphi(t)=(ak-1)t^(2)+(d+bk)t+ck\varphi(t)=(a k-1) t^{2}+(d+b k) t+c kφ(t)=(hask−1)t2+(d+bk)t+ck, so we have φ ( 0 ) ≧ 0 φ ( 0 ) ≧ 0 varphi(0) >= 0\varphi(0) \geqq 0φ(0)≧0And φ ( 1 ) < 0 φ ( 1 ) < 0 varphi(1) < 0\varphi(1)<0φ(1)<0that is to say the equation considered admits a root t 1 t 1 t_(1)t_{1}t1which verifies the inequality 0 ≦ t 1 < 1 0 ≦ t 1 < 1 0 <= t_(1) < 10 \leqq t_{1}<10≦t1<1.
To demonstrate property (j) we will proceed by induction.
For n = 0 n = 0 n=0n=0n=0And n = 1 n = 1 n=1n=1n=1the inequalities of property (j) are verified by hypothesis. Suppose that property (j) holds for n = 2 , 3 , … , s n = 2 , 3 , … , s n=2,3,dots,sn=2,3, \ldots, sn=2,3,…,sand show that it takes place for n = s + 1 n = s + 1 n=s+1n=s+1n=s+1. Indeed, of 0 ≦ t 1 < 1 0 ≦ t 1 < 1 0 <= t_(1) < 10 \leqq t_{1}<10≦t1<1the following inequalities result:
(14) δ i ≦ k t 1 ≦ k , i = 2 , 3 , … , s . (14) δ i ≦ k t 1 ≦ k , i = 2 , 3 , … , s . {:(14)delta_(i) <= kt_(1) <= k","i=2","3","dots","s.:}\begin{equation*} \delta_{i} \leqq k t_{1} \leqq k, i=2,3, \ldots, s . \tag{14} \end{equation*}(14)δi≦kt1≦k,i=2,3,…,s.
We will now write inequalities (12) in the following form:
(15) δ n + 1 ≦ ( a δ n + d ) δ n + ( b δ n + c δ n − 1 ) δ n − 1 , n = 1 , 2 , … , (15) δ n + 1 ≦ a δ n + d δ n + b δ n + c δ n − 1 δ n − 1 , n = 1 , 2 , … , {:(15)delta_(n+1) <= (adelta_(n)+d)delta_(n)+(bdelta_(n)+cdelta_(n-1))delta_(n-1)","quad n=1","2","dots",":}\begin{equation*} \delta_{n+1} \leqq\left(a \delta_{n}+d\right) \delta_{n}+\left(b \delta_{n}+c \delta_{n-1}\right) \delta_{n-1}, \quad n=1,2, \ldots, \tag{15} \end{equation*}(15)δn+1≦(hasδn+d)δn+(bδn+cδn−1)δn−1,n=1,2,…,
(ou) δ n − 1 ≦ u n δ n + v n δ n − 1 , n = 1 , 2 , … (16) u n = a δ n + d et v n = b δ n + c δ n − 1 , n = 1 , 2 , … (ou) δ n − 1 ≦ u n δ n + v n δ n − 1 , n = 1 , 2 , … (16) u n = a δ n + d  et  v n = b δ n + c δ n − 1 , n = 1 , 2 , … {:[(ou)delta_(n-1) <= u_(n)delta_(n)+v_(n)delta_(n-1)","n=1","2","dots],[(16)u_(n)=adelta_(n)+d" et "v_(n)=bdelta_(n)+cdelta_(n-1)","n=1","2","dots]:}\begin{gather*} \delta_{n-1} \leqq u_{n} \delta_{n}+v_{n} \delta_{n-1}, n=1,2, \ldots \tag{ou}\\ u_{n}=a \delta_{n}+d \text { et } v_{n}=b \delta_{n}+c \delta_{n-1}, n=1,2, \ldots \tag{16} \end{gather*}(Or)δn−1≦unδn+vnδn−1,n=1,2,…(16)un=hasδn+d And vn=bδn+cδn−1,n=1,2,…
From (14), taking into account u n u n u_(n)u_{n}unAnd v n v n v_(n)v_{n}vnwe deduce
u n ≦ a k t 1 + d et v n ≦ b k t 1 + c k , n = 2 , 3 , … , s . u n ≦ a k t 1 + d  et  v n ≦ b k t 1 + c k , n = 2 , 3 , … , s . u_(n) <= akt_(1)+d" et "v_(n) <= bkt_(1)+ck,n=2,3,dots,s.u_{n} \leqq a k t_{1}+d \text { et } v_{n} \leqq b k t_{1}+c k, n=2,3, \ldots, s .un≦haskt1+d And vn≦bkt1+ck,n=2,3,…,s.
From (16), taking into account the above inequalities, we deduce
δ s + 1 ≦ ( a k t 1 + d ) k t 1 s + ( b k t 1 + c k ) k t 1 s − 1 = = k t 1 s − 1 ( a k t 1 2 + ( b k + d ) t 1 + c k ) = k t 1 s + 1 δ s + 1 ≦ a k t 1 + d k t 1 s + b k t 1 + c k k t 1 s − 1 = = k t 1 s − 1 a k t 1 2 + ( b k + d ) t 1 + c k = k t 1 s + 1 {:[delta_(s+1) <= (akt_(1)+d)kt_(1)^(s)+(bkt_(1)+ck)kt_(1)^(s-1)=],[=kt_(1)^(s-1)(akt_(1)^(2)+(bk+d)t_(1)+ck)=kt_(1)^(s+1)]:}\begin{aligned} & \delta_{s+1} \leqq\left(a k t_{1}+d\right) k t_{1}^{s}+\left(b k t_{1}+c k\right) k t_{1}^{s-1}= \\ & =k t_{1}^{s-1}\left(a k t_{1}^{2}+(b k+d) t_{1}+c k\right)=k t_{1}^{s+1} \end{aligned}δs+1≦(haskt1+d)kt1s+(bkt1+ck)kt1s−1==kt1s−1(haskt12+(bk+d)t1+ck)=kt1s+1
that's to say δ s + 1 ≦ k t 1 s + 1 δ s + 1 ≦ k t 1 s + 1 delta_(s+1) <= kt_(1)^(s+1)\delta_{s+1} \leqq k t_{1}^{s+1}δs+1≦kt1s+1, from which the properties (jj) and (jjj) result.
Now we can prove the following theorem:
THEOREM 3. If the application F F FFF, the initial elements x 0 , x 1 ∈ X x 0 , x 1 ∈ X x_(0),x_(1)in Xx_{0}, x_{1} \in Xx0,x1∈Xand the real number r > 0 r > 0 r > 0r>0r>0enjoy the following properties:
(i) the application F is derivable in the Fréchet sense, up to order 2 inclusive, with respect to x x xxxAnd y y yyyin each point of the set S = S = S=S=S=Int (S) where
S ¯ = { x ∈ X : ‖ x − x 0 ‖ ≦ r } ; S ¯ = x ∈ X : x − x 0 ≦ r ; bar(S)={x in X:||x-x_(0)|| <= r};\bar{S}=\left\{x \in X:\left\|x-x_{0}\right\| \leqq r\right\} ;S¯={x∈X:‖x−x0‖≦r};
(ii) for each point ( x , y ) ∈ S × S ( x , y ) ∈ S × S (x,y)in S xx S(x, y) \in S \times S(x,y)∈S×S, the application F x ′ ( x , y ) F x ′ ( x , y ) F_(x)^(')(x,y)F_{x}^{\prime}(x, y)Fx′(x,y)is invertible and the operator [ F x ′ ( x , y ) ] − 1 F x ′ ( x , y ) − 1 [F_(x)^(')(x,y)]^(-1)\left[F_{x}^{\prime}(x, y)\right]^{-1}[Fx′(x,y)]−1is bounded on S , c ′ S , c ′ S,c^(')S, c^{\prime}S,c′that is to say ‖ [ F x ′ ( x , y ) ] − 1 ‖ ≦≦ α F x ′ ( x , y ) − 1 ≦≦ α ||[F_(x)^(')(x,y)]^(-1)||≦≦alpha\left\|\left[F_{x}^{\prime}(x, y)\right]^{-1}\right\| \leqq \leqq \alpha‖[Fx′(x,y)]−1‖≦≦α, where a is a real and positive constant;
(iii) the application [ F x ′ ( x , y ) ] − 1 F y ′ ( x , y ) F x ′ ( x , y ) − 1 F y ′ ( x , y ) [F_(x)^(')(x,y)]^(-1)F_(y)^(')(x,y)\left[F_{x}^{\prime}(x, y)\right]^{-1} F_{y}^{\prime}(x, y)[Fx′(x,y)]−1Fy′(x,y)is uniformly bounded on S × S S × S S xx SS \times SS×S, that's to say ‖ [ F x ′ ( x , y ) ] − 1 F y ′ ( x , y ) ‖ ≦ β F x ′ ( x , y ) − 1 F y ′ ( x , y ) ≦ β ||[F_(x)^(')(x,y)]^(-1)F_(y)^(')(x,y)|| <= beta\left\|\left[F_{x}^{\prime}(x, y)\right]^{-1} F_{y}^{\prime}(x, y)\right\| \leqq \beta‖[Fx′(x,y)]−1Fy′(x,y)‖≦β, Or β β beta\betaβis a real and positive constant;
(iv) the following inequalities are verified
sup ( x , y ) ∈ S ‖ F x 2 ′ ′ ( x , y ) ‖ ≦ 2 p ; sup ( x , y ) ∈ S ‖ F x y ′ ′ ( x , y ) ‖ ≦ q et sup ( x , y ) ∈ S ‖ F y 2 ′ ′ ( x , y ) ‖ ≦ 2 r , sup ( x , y ) ∈ S   F x 2 ′ ′ ( x , y ) ≦ 2 p ; sup ( x , y ) ∈ S   F x y ′ ′ ( x , y ) ≦ q  et  sup ( x , y ) ∈ S   F y 2 ′ ′ ( x , y ) ≦ 2 r , {:[s u p_((x,y)in S)||F_(x^(2))^('')(x,y)|| <= 2p;s u p_((x,y)in S)||F_(xy)^('')(x,y)|| <= q" et "],[s u p_((x,y)in S)||F_(y^(2))^('')(x,y)|| <= 2r","]:}\begin{gathered} \sup _{(x, y) \in S}\left\|F_{x^{2}}^{\prime \prime}(x, y)\right\| \leqq 2 p ; \sup _{(x, y) \in S}\left\|F_{x y}^{\prime \prime}(x, y)\right\| \leqq q \text { et } \\ \sup _{(x, y) \in S}\left\|F_{y^{2}}^{\prime \prime}(x, y)\right\| \leqq 2 r, \end{gathered}sup(x,y)∈S‖Fx2′′(x,y)‖≦2p;sup(x,y)∈S‖Fxy′′(x,y)‖≦q And sup(x,y)∈S‖Fy2′′(x,y)‖≦2r,
where p , q p , q p,qp, qp,qAnd r r rrrare real and non-negative constants;
(v)
‖ x 1 − x 0 ‖ ′ ≦ k , ‖ x 2 − x 1 ‖ ≦ k t 1 x 1 − x 0 ′ ≦ k , x 2 − x 1 ≦ k t 1 ||x_(1)-x_(0)||^(') <= k,||x_(2)-x_(1)|| <= kt_(1)\left\|x_{1}-x_{0}\right\|^{\prime} \leqq k,\left\|x_{2}-x_{1}\right\| \leqq k t_{1}‖x1−x0‖′≦k,‖x2−x1‖≦kt1
Or k k kkkis a real and positive constant, which does not depend on n , x 2 n , x 2 n,x_(2)n, x_{2}n,x2is given by (11) for n = 1 n = 1 n=1n=1n=1And t 1 t 1 t_(1)t_{1}t1is the positive root of the equation
( α p k − 1 ) t 2 + ( β + α q k ) t + α r k = 0 ( α p k − 1 ) t 2 + ( β + α q k ) t + α r k = 0 (alpha pk-1)t^(2)+(beta+alpha qk)t+alpha rk=0(\alpha p k-1) t^{2}+(\beta+\alpha q k) t+\alpha r k=0(αpk−1)t2+(β+αqk)t+αrk=0
(vi) the following inequalities hold:
(17) k ( p + q + r ) + β < 1 (17) k ( p + q + r ) + β < 1 {:(17)k(p+q+r)+beta < 1:}\begin{equation*} k(p+q+r)+\beta<1 \tag{17} \end{equation*}(17)k(p+q+r)+β<1
4 - Mathematica - Volume 20 (43) No. 1/1978
and
(18)
γ ≧ k / ( 1 − t 1 ) , γ ≧ k / 1 − t 1 , gamma >= k//(1-t_(1)),\gamma \geqq k /\left(1-t_{1}\right),γ≧k/(1−t1),
then equation (9) and procedure (11) enjoy the following properties:
(j) the sequence ( x n ) n = 0 ∞ x n n = 0 ∞ (x_(n))_(n=0)^(oo)\left(x_{n}\right)_{n=0}^{\infty}(xn)n=0∞is convergent and if x ¯ = lim n → ∞ x n x ¯ = lim n → ∞   x n bar(x)=lim_(n rarr oo)x_(n)\bar{x}=\lim _{n \rightarrow \infty} x_{n}x¯=limn→∞xn, SO x ¯ n x ¯ n bar(x)_(n)\bar{x}_{n}x¯nis a solution to equation (9);
(jj) the following inequalities hold:
‖ x n − x ¯ ‖ ≦ k t 1 n / ( 1 − t 1 ) , x n − x ¯ ≦ k t 1 n / 1 − t 1 , ||x_(n)-( bar(x))|| <= kt_(1)^(n)//(1-t_(1)),\left\|x_{n}-\bar{x}\right\| \leqq k t_{1}^{n} /\left(1-t_{1}\right),‖xn−x¯‖≦kt1n/(1−t1),
for each n = 0 , 1 , … n = 0 , 1 , … n=0,1,dotsn=0,1, \ldotsn=0,1,….
Demonstration. By hypothesis we have x 1 ∈ S x 1 ∈ S x_(1)in Sx_{1} \in Sx1∈SAnd x 2 ∈ S x 2 ∈ S x_(2)in Sx_{2} \in Sx2∈S. Suppose the following properties hold:
a) δ k = ‖ x k + 1 − x k ‖ ≦ k t 1 k δ k = x k + 1 − x k ≦ k t 1 k delta_(k)=||x_(k+1)-x_(k)|| <= kt_(1)^(k)\delta_{k}=\left\|x_{k+1}-x_{k}\right\| \leqq k t_{1}^{k}δk=‖xk+1−xk‖≦kt1k, Or k = 0 , 1 , … , n − 1 k = 0 , 1 , … , n − 1 k=0,1,dots,n-1k=0,1, \ldots, n-1k=0,1,…,n−1
b) x k ∈ S x k ∈ S x_(k)in Sx_{k} \in Sxk∈S, Or k = 0 , 1 … , n k = 0 , 1 … , n k=0,1dots,nk=0,1 \ldots, nk=0,1…,n.
From the hypotheses of the theorem it follows that the properties a a aahas) And b b bbb) take place for k = 0 k = 0 k=0k=0k=0And k = 1 k = 1 k=1k=1k=1. Let us now show that these properties hold for k = n k = n k=nk=nk=nrespectively k = n + 1 k = n + 1 k=n+1k=n+1k=n+1
Indeed, we deduce from Theorem 1
‖ F ( x s − 1 , x s ) ‖ ≦ ‖ F ( x s + 1 , x s ) − F ( x s , x s − 1 ) − F x ′ ( x s , x s − 1 ) ( x s − 1 − x s ) − − F y ′ ( x s , x s − 1 ) ( x s − 1 − x s − 1 ) ‖ ≦ ≦ p ‖ x s + 1 − x s ‖ 2 + q ‖ x s + 1 − x s ‖ ⋅ ‖ x s − x s − 1 ‖ + + r ‖ x s − x s − 1 ‖ 2 . F x s − 1 , x s ≦ ‖ F x s + 1 , x s − F x s , x s − 1 − F x ′ x s , x s − 1 x s − 1 − x s − − F y ′ x s , x s − 1 x s − 1 − x s − 1 ‖ ≦ ≦ p x s + 1 − x s 2 + q x s + 1 − x s ⋅ x s − x s − 1 + + r x s − x s − 1 2 . {:[||F(x_(s-1),quadx_(s))|| <= ||F(x_(s+1),x_(s))-F(x_(s),x_(s-1))-F_(x)^(')(x_(s),x_(s-1))(x_(s-1)-x_(s))-],[-F_(y)^(')(x_(s),x_(s-1))(x_(s-1)-x_(s-1))|| <= ],[ <= p||x_(s+1)-x_(s)||^(2)+q||x_(s+1)-x_(s)||*||x_(s)-x_(s-1)||+],[+r||x_(s)-x_(s-1)||^(2).]:}\begin{aligned} \left\|F\left(x_{s-1}, \quad x_{s}\right)\right\| \leqq & \| F\left(x_{s+1}, x_{s}\right)-F\left(x_{s}, x_{s-1}\right)-F_{x}^{\prime}\left(x_{s}, x_{s-1}\right)\left(x_{s-1}-x_{s}\right)- \\ & -F_{y}^{\prime}\left(x_{s}, x_{s-1}\right)\left(x_{s-1}-x_{s-1}\right) \| \leqq \\ & \leqq p\left\|x_{s+1}-x_{s}\right\|^{2}+q\left\|x_{s+1}-x_{s}\right\| \cdot\left\|x_{s}-x_{s-1}\right\|+ \\ & +r\left\|x_{s}-x_{s-1}\right\|^{2} . \end{aligned}‖F(xs−1,xs)‖≦‖F(xs+1,xs)−F(xs,xs−1)−Fx′(xs,xs−1)(xs−1−xs)−−Fy′(xs,xs−1)(xs−1−xs−1)‖≦≦p‖xs+1−xs‖2+q‖xs+1−xs‖⋅‖xs−xs−1‖++r‖xs−xs−1‖2.
Now taking into account (ii) and (iii) we have:
(20) ‖ x s + 1 − x s ‖ ≦ α ‖ F ( x s , x s − 1 ) ‖ + β ‖ x s − x s − 1 ‖ ; s = 1 , 2 , … , n x s + 1 − x s ≦ α F x s , x s − 1 + β x s − x s − 1 ; s = 1 , 2 , … , n quad||x_(s+1)-x_(s)|| <= alpha||F(x_(s),x_(s-1))||+beta||x_(s)-x_(s-1)||;s=1,2,dots,n\quad\left\|x_{s+1}-x_{s}\right\| \leqq \alpha\left\|F\left(x_{s}, x_{s-1}\right)\right\|+\beta\left\|x_{s}-x_{s-1}\right\| ; s=1,2, \ldots, n‖xs+1−xs‖≦α‖F(xs,xs−1)‖+β‖xs−xs−1‖;s=1,2,…,n. Either ρ s = ‖ F ( x s − 1 , x s ) ‖ ρ s = F x s − 1 , x s rho_(s)=||F(x_(s-1),x_(s))||\rho_{s}=\left\|F\left(x_{s-1}, x_{s}\right)\right\|ρs=‖F(xs−1,xs)‖And δ s = ‖ x s + 1 − x s ‖ δ s = x s + 1 − x s delta_(s)=||x_(s+1)-x_(s)||\delta_{s}=\left\|x_{s+1}-x_{s}\right\|δs=‖xs+1−xs‖For s = 0 , 1 , … s = 0 , 1 , … s=0,1,dotss=0,1, \ldotss=0,1,…
Then from (19) and (20) we deduce
ρ s ≦ p δ s 2 + q ⋅ δ s ⋅ δ s − 1 + r ⋅ δ s − 1 2 , s = 1 , 2 , … , n − 1 ρ s ≦ p δ s 2 + q ⋅ δ s ⋅ δ s − 1 + r ⋅ δ s − 1 2 , s = 1 , 2 , … , n − 1 rho_(s) <= pdelta_(s)^(2)+q*delta_(s)*delta_(s-1)+r*delta_(s-1)^(2),s=1,2,dots,n-1\rho_{s} \leqq p \delta_{s}^{2}+q \cdot \delta_{s} \cdot \delta_{s-1}+r \cdot \delta_{s-1}^{2}, s=1,2, \ldots, n-1ρs≦pδs2+q⋅δs⋅δs−1+r⋅δs−12,s=1,2,…,n−1
(21) δ s ≦ α ⋅ ρ s − 1 + β δ s − 1 , s = 1 , 2 , … , n (21) δ s ≦ α ⋅ ρ s − 1 + β δ s − 1 , s = 1 , 2 , … , n {:(21)delta_(s) <= alpha*rho_(s-1)+betadelta_(s-1)","s=1","2","dots","n:}\begin{equation*} \delta_{s} \leqq \alpha \cdot \rho_{s-1}+\beta \delta_{s-1}, s=1,2, \ldots, n \tag{21} \end{equation*}(21)δs≦α⋅ρs−1+βδs−1,s=1,2,…,n
from which it results
(22) δ s + 1 ≦ α p ⋅ δ s 2 + α ⋅ q ⋅ δ s ⋅ δ s − 1 + α r ⋅ δ s − 1 2 + β δ s ′ (22) δ s + 1 ≦ α p ⋅ δ s 2 + α ⋅ q ⋅ δ s ⋅ δ s − 1 + α r ⋅ δ s − 1 2 + β δ s ′ {:(22)delta_(s+1) <= alpha p*delta_(s)^(2)+alpha*q*delta_(s)*delta_(s-1)+alpha r*delta_(s-1)^(2)+betadelta_(s^(')):}\begin{equation*} \delta_{s+1} \leqq \alpha p \cdot \delta_{s}^{2}+\alpha \cdot q \cdot \delta_{s} \cdot \delta_{s-1}+\alpha r \cdot \delta_{s-1}^{2}+\beta \delta_{s^{\prime}} \tag{22} \end{equation*}(22)δs+1≦αp⋅δs2+α⋅q⋅δs⋅δs−1+αr⋅δs−12+βδs′
From (22), (17) and taking into account Theorem 2 we deduce
(23)
δ n = ‖ x n + 1 − x n ‖ ⩽ k t 1 n , δ n = x n + 1 − x n ⩽ k t 1 n , delta_(n)=||x_(n+1)-x_(n)|| <= kt_(1)^(n),\delta_{n}=\left\|x_{n+1}-x_{n}\right\| \leqslant k t_{1}^{n},δn=‖xn+1−xn‖⩽kt1n,
that is to say it follows that property a) takes place for k = n k = n k=nk=nk=n.
For b) we have:
‖ x n + 1 − x 0 ‖ ≦ ∑ i = 0 n ‖ x i + 1 − x i ‖ = ∑ i = 0 n δ i ≦ k ∑ i = 0 n t 1 i < k 1 − t 1 ≦ r , x n + 1 − x 0 ≦ ∑ i = 0 n   x i + 1 − x i = ∑ i = 0 n   δ i ≦ k ∑ i = 0 n   t 1 i < k 1 − t 1 ≦ r , ||x_(n+1)-x_(0)|| <= sum_(i=0)^(n)||x_(i+1)-x_(i)||=sum_(i=0)^(n)delta_(i) <= ksum_(i=0)^(n)t_(1)^(i) < (k)/(1-t_(1)) <= r,\left\|x_{n+1}-x_{0}\right\| \leqq \sum_{i=0}^{n}\left\|x_{i+1}-x_{i}\right\|=\sum_{i=0}^{n} \delta_{i} \leqq k \sum_{i=0}^{n} t_{1}^{i}<\frac{k}{1-t_{1}} \leqq r,‖xn+1−x0‖≦∑i=0n‖xi+1−xi‖=∑i=0nδi≦k∑i=0nt1i<k1−t1≦r,
SO x n + 1 ∈ S x n + 1 ∈ S x_(n+1)in Sx_{n+1} \in Sxn+1∈S.
Let us now show that the sequence ( x n ) n = 0 ∞ x n n = 0 ∞ (x_(n))_(n=0)^(oo)\left(x_{n}\right)_{n=0}^{\infty}(xn)n=0∞is convergent. Indeed, from (23) we deduce
‖ x n + p − x n ‖ ≦ ∑ k = n n + p − 1 ‖ x k + 1 − x k ‖ = ∑ k = n n + p − 1 δ k ≦ k t 1 n ( 1 + t 1 + … + t 1 n + p − 1 ) ≦ k t 1 n / ( 1 − t 1 ) x n + p − x n ≦ ∑ k = n n + p − 1   x k + 1 − x k = ∑ k = n n + p − 1   δ k ≦ k t 1 n 1 + t 1 + … + t 1 n + p − 1 ≦ k t 1 n / 1 − t 1 {:[||x_(n+p)-x_(n)|| <= sum_(k=n)^(n+p-1)||x_(k+1)-x_(k)||=sum_(k=n)^(n+p-1)delta_(k) <= kt_(1)^(n)(1+t_(1)+dots+t_(1)^(n+p-1))],[ <= kt_(1)^(n)//(1-t_(1))]:}\begin{gathered} \left\|x_{n+p}-x_{n}\right\| \leqq \sum_{k=n}^{n+p-1}\left\|x_{k+1}-x_{k}\right\|=\sum_{k=n}^{n+p-1} \delta_{k} \leqq k t_{1}^{n}\left(1+t_{1}+\ldots+t_{1}^{n+p-1}\right) \\ \leqq k t_{1}^{n} /\left(1-t_{1}\right) \end{gathered}‖xn+p−xn‖≦∑k=nn+p−1‖xk+1−xk‖=∑k=nn+p−1δk≦kt1n(1+t1+…+t1n+p−1)≦kt1n/(1−t1)
for each p = 1 , 2 , … ; n = 1 , 2 , … p = 1 , 2 , … ; n = 1 , 2 , … p=1,2,dots;n=1,2,dotsp=1,2, \ldots ; n=1,2, \ldotsp=1,2,…;n=1,2,….
From the above inequality, taking into account that t 1 < 1 t 1 < 1 t_(1) < 1t_{1}<1t1<1, it follows that the following ( x n ) n = 0 ∞ x n n = 0 ∞ (x_(n))_(n=0)^(oo)\left(x_{n}\right)_{n=0}^{\infty}(xn)n=0∞is convergent.
If we pass to limit for gans the inequality (24) and we write x ¯ = lim x n x ¯ = lim x n bar(x)=limx_(n)\bar{x}=\lim x_{n}x¯=limxnwe have
(25) ‖ x ¯ − x n ‖ ≦ k t 1 n / ( 1 − t 1 ) (25) x ¯ − x n ≦ k t 1 n / 1 − t 1 {:(25)||( bar(x))-x_(n)|| <= kt_(1)^(n)//(1-t_(1)):}\begin{equation*} \left\|\bar{x}-x_{n}\right\| \leqq k t_{1}^{n} /\left(1-t_{1}\right) \tag{25} \end{equation*}(25)‖x¯−xn‖≦kt1n/(1−t1)
Because F F FFFis a derivable application in the sense of Fréchet it follows that F F FFFis continuous; then passing to limit in (11) for n → ∞ n → ∞ n rarr oon \rightarrow \inftyn→∞, we have
θ = F ( x ¯ , x ¯ ) = G ( x ¯ ) θ = F ( x ¯ , x ¯ ) = G ( x ¯ ) theta=F( bar(x), bar(x))=G( bar(x))\theta=F(\bar{x}, \bar{x})=G(\bar{x})θ=F(x¯,x¯)=G(x¯)
that's to say x ¯ x ¯ bar(x)\bar{x}x¯is a solution to equation (9).

BIBLIOGRAPHY

[1] Kantorovici, IV, Funktionaljnîi analiz i prikladnaia matematika, UMN 28, 89185, (1948).
[2] Păvalo1u, I., On iterative processes at a high order of convergence, Mathematica, (1970).
[3] Weinischke, JH, Über eine Klasse von Iterationsverfahren, Numerische Mathematik, 6, 395-404, (1964).
1978

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