A NEWTON-TYPE METHOD AND ITS APPLICATION

We prove an existence and uniqueness theorem for solving the operator equation F(x) + G(x)= 0, where F is a continuous and Gâteaux differentiable operator and the operator G satisfies Lipschitz condition on an open convex subset of a Banach space. As corollaries, a recent theorem of Argyros (2003) and the classical convergence theorem for modified Newton iterates are deduced. We further obtain an existence theorem for a class of nonlinear functional integral equations involving the Urysohn operator.


Introduction
This paper considers the problem of approximating a locally unique solution x * of the equation F(x) + G(x) = 0, where F and G are continuous operators defined on an open convex subset D of a Banach space X with values in a Banach space Y .The solution is obtained as the limit of a sequence of Newton-type iterates given by It may be noted that for G ≡ 0, (1.1) reduces to Newton's method of iterates whose convergence is proved under the usual hypotheses that F is Fréchet differentiable and F x is invertible.Recently, Argyros [3] obtained an interesting generalization of Newton's method which is further extended in this paper along with an earlier result of Vijesh and Subrahmanyam [1].More specifically, we prove the convergence of a sequence of Newton-type iterates under mild conditions on F. In particular, the Gâteaux differentiable operator F is assumed to satisfy the inequality F x0 −1 F x − F x0 ≤ in a certain neighbourhood N x 0 of x 0 while G is required to be a contraction in N x 0 .Using the main theorem, the existence of a unique continuous solution to a nonlinear functional integral equation involving the Urysohn operator is deduced from a corollary to the main theorem.This seems to be the first application of Newton-type algorithm to such nonlinear equations.Illustrative examples are also discussed.

Convergence analysis
Theorem 2.1 is a general theorem on the convergence of Newton-type iterates, proved under mild assumptions.It generalizes the theorems in [1,3].
Then the sequence x n (n ≥ 0) generated recursively by (1.1) is well defined, remains in U(x 0 ,r) for all n ≥ 0, and converges to a unique solution x * ∈ U(x 0 ,r) of the equation F(x) + G(x) = 0.Moreover, the following error bounds hold for all n ≥ 2: Proof.Clearly by (ii) Thus x 2 ∈ U(x 0 ,r).Again by Banach's lemma (see [4]) In view of hypotheses (iii) and (iv), it follows as before from Banach's lemma that (F xn ) −1 exists and By hypotheses (iii) and (iv) and (2.5), we obtain (2.6) Thus by induction hypothesis (2.4), (2.7) Hence x n ∈ U(x 0 ,r) for all n ≥ 0. For k ≥ 2, m ≥ 0.
x n is a Cauchy sequence in the closed subset U(x 0 ,r) of the Banach space X; it hence converges to an element x * in U(x 0 ,r).From hypothesis (iii) using triangle inequality, it follows that (2.10) Proceeding to the limit in (2.10) as n tends to infinity and using the continuity of F and G it follows from the convergence of (x n ) to x * that F(x * ) + G(x * ) = 0.If x * and y * are two solutions of F(x) + G(x) = 0 in U(x 0 ,r), then (2.11) This contradiction implies that x * = y * .Hence the theorem holds.(iii) for some ∈ (0,1/3), there exists δ > 0 such that , where c 0 = /(1 − ) and c = 2c 0 .Then the Newton iterates (x n ) generated by (1.1) are well defined, remain in U(x 0 ,δ) for all n ≥ 0, and converge to a solution x * ∈ U(x 0 ,δ) of equation F(x) = 0.Moreover, for all n ≥ 2, the following error bounds hold: (2.12) Since F is continuously Fréchet differentiable, F x is hemicontinuous at each x ∈ U(x 0 ,δ) and thus all the conditions of Theorem 2.1 are satisfied.So F has a unique zero in U(x 0 ,δ).(iii) for some r > 0, (F x0 ) −1 [F x − F xo ] ≤ whenever x ∈ U(x 0 ,r).Suppose (1 + c 0 / (1 − c))η < r and 0 < 3 < 1, where c 0 = /(1 − ) and c = 2c 0 ; (iv) F x is piecewise hemi continuous at each x ∈ U(x 0 ,r).Then the modified Newton iterates (x n ), n ≥ 0, generated by are well defined, remain in U(x 0 ,r) for all n ≥ 0, and converge to a solution x * ∈ U(x 0 ,r) of F(x) = 0.
To prove Corollary 2.3, set G ≡ 0 in Theorem 2.1 and proceed as in Theorem 2.1.

Solutions of a class of nonlinear functional integral equations
In this section, some existence theorems for an operator equation involving the Urysohn operator are proved.More specifically, given a compact subset Ω of R, g ∈ C(Ω × R) and K ∈ C(Ω 2 × R), Theorem 3.2 gives sufficient conditions for the existence of a unique solution of the nonlinear functional integral equation of the form Hereafter we denote x(t) + Ω K(t,s,x(s))ds by Fx(t).
the Lipschitz condition with respect to the third variable.Then the Fréchet derivative of F exists for x ∈ U(x 0 ,r) and x 0 ∈ C(Ω) and the derivative at x is given by For the proof, see [2].
and let Ω be a compact subset of R whose Lebesgue measure is equal to d > 0. Suppose that (i) for some m ∈ (0,1), 2 and let r 1 and r 2 with r 1 ≤ r 2 be the positive roots of q Then the functional integral equation Fx(t) + g(t,x(t)) = 0 for all t ∈ Ω has a unique solution in U(x 0 ,r 0 ) for all r 0 ∈ r 1 ,min r 2 ,(1 − β − * )/3md and the sequence of iterates given in (1.1) converges to this solution.

Illustrative examples
The example below illustrates Theorem 3.2.
Example 4.2.The next example shows that Theorem 2.1 is more general than Corollary 2.2, the main theorem obtained by Argyros [3].

Corollary 2 . 2 (
see Argyros[3, Theorem 1]).Let F be a continuous operator defined on an open convex subset D of a Banach space X with values in a Banach space Y and continuously Fréchet differentiable at some x 0 ∈ D. Assume that(i) (F xo ) −1 ∈ L(Y ,X);(ii) there exists a parameter η such that 0 ≤ (F xo ) −1 F(x 0 ) ≤ η;