ON THE NONLINEAR IMPLICIT COMPLEMENTARITY PROBLEM

In this paper, we consider a new class of implicit complementarity problem and study the existence of its solution. An iterative algorithm is also given to find the approximate solution of the new problem and prove that this approximate solution converges to the exact solution. Several special cases are also discussed.

problem which is called mildly nonlinear complementarity problem.
The recent research carried out in this field motivated us to introduce and study a new class of complementarity problem which includes implicit complementarity problem and mildly nonlinear cmplementarity problem as special cases.By using the technique of Isac [9] we prove an existence theorem.An algorithm is also given to find the approximate solution of the new complementarity problem and prove that this approximate solution converges to the exact solution.
2, PORELATION AND BASIC RESULTS.Let H be a Hilbert space with norm and inner product denoted by II" II and <.,.>, respectively and K be a closed convex Cone of H.If T and A are nonlinear operators from H into itself then the mildly nonlinear complementarity problem (M.N.C.P.) which is introduced and studied by Noor [12], is to find u E K such that T(X) + A(x) E K* and <x, T(x) + A() 0, (2.1) where we denote K* the polar cone of K, that is, K* x E K/<x,y> > 0 for all y E K}.We shall call it strongly nonlinear implicit complementarity problem (S.N.I.C.P.).
We recall that if P denotes the project,;on of H onto K that is for every is the unique element satisfying-.
then we have the following result.
DEFINITION 2.1 [9] Given a subset D H, we consider the mappings ,: R++ R+ and we say that.
(a) F is a -Lischitz mapping with respect to G if for all x,y e D, (b) F is a -strongly monotene mapping with respect to G if for all x,y e D.
is characterised by the x, for all x D then we say that F is a -Lipschitz mapping (respectively, F is a -strongly monotone mapping).
Obviously, if and are strictly positive constants, we obtain from Definition 2.1 (a) (respectively (b)) that F is a Lipschitz continuous (respectively, strongly monotone) mapping.

DEFINITION 2.2 [14]
A metric space (X,O) is said to be metrically convex, if for each x,y E X, (x # y) there is a z # x,y for which 0(x,y) 0(x,z) + O(z,y)- We denote, P {0(x,y)/x,y e X}.
TEOREM 2.1 [14] Let (X,O) be a complete metrically convex metric space.If for the mapping F:X X there is a mapping :p R satisfying, + 'i.0(F(x), F(y)) _<. (0(X,y)), for all x,y e. X, 2. (t) < t, for all t e Pk{o}, then F has a unique fixed point x and Tn(x)+ x for each x e X.
o o 3. EXISTENCE THEORY THEOREM 3.1.Assume that (i) T is -Lipschitz mapping with respect to G; (ii) T is -strongly monotone mapping with respect to G (iii) A is -Lipschitz mapping with respect to G; (iv) K C G(D); (v) there exists a real number > 0 such that, v2(t) for all t e R + Then the strongly nonlinear implicit complementarity problem has a solution.
Moreover, if G is one-to-one, then the problem (S.N.I.C.P.) has a unique solution.
PROOF.We consider the mappings p and q:K H (which are not unique) defined by p(u) T(x) and q(u) where x e G -l(u) {x e D/G(x) u} and u e K.
From this definition, we observe that p and q have the following properties: for all u,v e K v, < u-p., u-.> _> !lu-v l,llu-v I, for all u,v e K for all u,v e K, :R+/ R+.Now, we conclude that the S.N.I.C.P. is equivalent to the M.N.C.P.
Find u e K such that (M.N.C.P.) (3.1) p (u) +q (u) e/* and <u,p(u)+q(u)> 0 By Proposition 2.1, it is easy to prove that the S.N.C.P. has a solution if and only if the mapping F:K K defined by for all u e K, has a fixed point (where is the real number used in assumption We, now will show that T has a fixed point. < lu-(p(u) +q(u) )-v+ (p(v) +q(v)) II < lu-v-(p (u)-p (v))I + Since P is nonexpansive [15].K By (vi) (vii) and (viii) we obtain lu--cpcu)-pCv)) _< lu-".,I--:' lu-vl 1+'' cl lu-vl ! and lqCu-qCv I! _< lu-',,l I'cl lu-",,I !. Therefore, for all t e R+, we observe, by assumption (v) and the fact that a Hi lbert space is a complete metri- metric space, that all assumptions of Theorem 2.1 are satisfied.(=R+) cally convex Hence F has a unique fixed point u and Fn(u) u for every u K.

o o
Obviously, if G is one-to-one mapping then S.N.I.C.P. has a unique solution.
COROLLARY 3.2.Assume that (i) T is Y-strongly monotone mapping with respect to G; (ii) G is an expansive mapping, that is, there exists I > 1 such that for all x,y e D (iii) !-_< Ix-'s,l lcl IGCxCI! .
Then the S.N.I.C.P. has a unique solution.

ALGORITHM
In this section we give an iterative algorithm for finding the approximate solu- tion of the S.N.I.C.P. and prove that the approximate solution converges to the exact solution.For this, we need the following result.In view of this lemma, we suggest the following new unified algorithm for finding the approximate solution of the S.N.I.C.P.  where > 0 is a constant.
In order to discuss the convergence properties of the Algorithm 4.1, we need the following concepts.DEFINITION 4.1.An operator T:K H is called: (i) CoeP4ve if there exists a constant u > 0 such that < x> for all x e K; (ii) Cont4n2ou8 (boz,tded), if there is a constant 8 > 0 such that for all x,y e K.
Since @ < i, the fixed point problem (4.1) has a unique solution x and consequently the Picard iterates Xn+1 converges to x strongly in H.
REMARK 4.1. (i) If the nonlinear operator A is independent of x, that is, A(x) 0, then Algorithm 4.1 reduces to Algorithm 3.1 [16].
(ii) If G is the identity operator, that is, G(x) x, then Algorithm 4.1 reduces to Algorithm 2.1 [12].
If D C H is a subset of H and G:D H problem (I.C.P.) is to find x e D such that then the impl ic it complementarity G(x) e K, T(x) e K* and <G(x) T(x) > 0 (2.2) Now we consider a more genera] form of complementarity problem for which (M.N.C.P) and (I.C.P.) are special cases. (S .N .I .C .P .)Find x e D such that G(x) K, T(x)+A(x) e K* and (x) T(x)+A(x

LEMMA 4 .F
1.If K is a convex cone in H, x e H is a solut'ion of the S.N.I.C.P. if and only if it satisfies the relation x follows directly from Proposition 2.1.