# On Self-Adaptive Method for General Mixed Variational Inequalities

Authors: Abdellah Bnouhachem; Muhammad Aslam Noor; Eman H. Al-Shemas
Journal: Mathematical Problems in Engineering
 (2008)

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## Abstract

We suggest and analyze a new self-adaptive method for solving general mixed variational inequalities, which can be viewed as an improvement of the method of (Noor 2003). Global convergence of the new method is proved under the same assumptions as Noor's method. Some preliminary computational results are given to illustrate the efficiency of the proposed method. Since the general mixed variational inequalities include general variational inequalities, quasivariational inequalities, and nonlinear (implicit) complementarity problems as special cases, results proved in this paper continue to hold for these problems.


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## Body

## 1. Introduction

Variational inequalities introduced in the early
sixties have played a critical and significant part in the study of several
unrelated problems arising in finance, economics, network analysis,
transportation, elasticity, and optimization. Variational inequalities theory
has witnessed an explosive growth in theoretical advances, algorithmic
development, and applications across all disciplines of pure and applied
sciences, see [1–16]. A useful and important generalization of variational inequalities
is the mixed variational inequality containing a nonlinear term φ.
But the applicability of the projection method is limited due to the fact that
it is not easy to find the projection except in very special cases. Secondly,
the projection method cannot be applied to suggest iterative algorithms for
solving general mixed variational inequalities involving the nonlinear term φ. This fact has motivated many authors to
develop the auxiliary principle technique for solving the mixed variational
inequalities. In recent years, several techniques have been
developed to suggest and analyze various iterative methods for solving different
types of variational inequalities. It is worth mentioning that if the nonlinear
term in the variational inequalities is a proper, convex, and semilower
continuous function, then it is well known that the variational inequalities
involving the nonlinear term φ are equivalent to the fixed point problems and
the resolvent equations. In [11], Noor solved the general mixed
variational inequality problem by using the resolvent equations technique.
Inspired and motivated by the results of Noor [11], we propose a new method for
solving general mixed variational inequalities by using a new direction with a
new step size αk. We prove the global convergence of the
proposed method under the same assumptions as in [11]. An example is given to
illustrate the efficiency and its comparison with the results of Noor [11, 14].
This shows that the method is robust and efficient. This new method can be
viewed as an important and significant improvement of Noor and other methods.


## 2. Preliminaries

LetH be a real Hilbert space, whose inner product
and norm are denoted by 〈⋅,⋅〉 and ∥⋅∥, let I be the identity mapping on H, and T,g:H→H be two operators. Let ∂φ denotes the subdifferential of function φ, where φ:H→R∪{+∞} is a proper convex lower semicontinuous
function on H. It is well known that the subdifferential ∂φ is a maximal monotone operator. We consider
the problem of finding u*∈H such that
(2.1)〈T(u*),g(v)−g(u*)〉+φ(g(v))−φ(g(u*))≥0,∀g(v)∈H,which is known as the mixed
general variational inequality, see Noor [11]. We also note that the general
variational inequality can be written in the equivalent form as find u∗∈H such that(2.2)0∈Tu+∂φ(g(u)),which is known as the problem of
finding a zero of sum of two(more) monotone operators. It is well known that a
wide class of linear and nonlinear problems arising in pure and applied
sciences can be studied via the general mixed variational inequalities, see
[1–16] and the references therein.IfK is a closed convex set in H and φ(v)≡IK(v),∀v∈H, where(2.3)IK(v)={0ifv∈K,+∞,otherwiseis the indicator function of K, then the problem (2.1) is equivalent to
finding u*∈H such that g(u*)∈K and (2.4)〈T(u*),g(v)−g(u*)〉≥0,∀g(v)∈K.Problem (2.4) is called the
general variational inequality, which was first introduced and studied by
Noor [9] in 1988. For the applications, formulation, and numerical methods of
general variational inequalities (2.4), we refer the reader to the survey, see
[1–3, 7, 12, 13, 16].Ifg≡I, then the problem (2.4) is equivalent to
finding u*∈K such that
(2.5)〈T(u*),v−u*〉≥0,∀v∈K, which is known as the
classical variational inequality introduced and studied by Stampacchia [17].Lemma 2.1 (see [4]).
For a givenw∈H,z∈H,
and ρ>0, the inequality(2.6)〈w−z,z−v〉+ρφ(v)−ρφ(z)≥0,∀v∈Hholds if and only if z=Jφ(w), where Jφ=(I+ρ∂φ)−1 is the resolvent operator.It follows from
Lemma2.1 that (2.7)〈w−Jφ(w),Jφ(w)−v〉+ρφ(v)−ρφ(Jφ(w))≥0,∀v,w∈H.Ifφ is the indicator function of a closed convex
set K in H, then the resolvent operator Jφ(⋅)=PK, where PK is the projection of H onto the closed convex set K. It is well known that Jφ is nonexpansive, that is,(2.8)∥Jφ(u)−Jφ(v)∥≤∥u−v∥,∀u,v∈H.Lemma 2.2 (see [10]).
u* is solution of problem (2.1) if and only if u*∈H satisfies the relation:(2.9)g(u*)=Jφ[g(u*)−ρT(u*)], where Jφ=(I+ρ∂φ)−1 is the resolvent operator .From Lemma2.2, it is clear that u* is solution of (2.1) if and only if u* is a zero point of the function(2.10)r(u*,ρ):=g(u*)−Jφ[g(u*)−ρT(u*)].In [11], Noor used the
fixed-point formulation (2.9) and the resolvent equations to suggest and
analyze the following algorithm for solving problem (2.1).Algorithm 2.3.
For a givenu0∈H, compute the approximate solution uk+1 by the iterative schemes.
Predictor Step
(2.11)g(wk)=Jφ[g(uk)−ρkT(uk)],where ρk satisfies(2.12)ρk〈T(uk)−T(g−1(Jφ[g(uk)−ρkT(uk)])),r(uk,ρk)〉≤δ∥r(uk,ρk)∥2,δ∈(0,1).
Corrector Step
(2.13)g(uk+1)=Jφ[g(uk)−αkd(uk,ρk)],where(2.14)d(uk,ρk)=r(uk,ρk)−ρkT(uk)+ρkT(g−1(Jφ[g(uk)−ρkT(uk)])),αk=(1−δ)∥r(uk,ρk)∥2∥d(uk,ρk)∥2is the corrector step size.
Ifφ is an indicator function of a closed convex
set K in H, then Jφ≡PK [10], the projection of H onto K and consequently Algorithm 2.3 collapses.Algorithm 2.4 (see [11]).
For a givenu0∈H, compute the approximate solution uk+1 by the iterative schemes.
Predictor Step
(2.15)g(wk)=PK[g(uk)−ρkT(uk)],where ρk satisfies(2.16)ρk〈T(uk)−T(g−1(PK[g(uk)−ρkT(uk)])),r(uk,ρk)〉≤δ∥r(uk,ρk)∥2,δ∈(0,1).
Corrector Step
(2.17)g(uk+1)=PK[g(uk)−αkd(uk,ρk)],where(2.18)d(uk,ρk)=r(uk,ρk)−ρkT(uk)+ρkT(g−1(PK[g(uk)−ρkT(uk)])),αk=(1−δ)∥r(uk,ρk)∥2∥d(uk,ρk)∥2is the corrector step size.
Throughout this paper, we make following assumptions.Assumptions
(i)
H is finite dimension space.(ii)
g is homeomorphism on H,
that is, g is bijective, continuous and g−1 is continuous.(iii)
T is continuous and g-pseudomonotone operator on H,
that is,(2.19)〈T(u),g(u′)−g(u)〉≥0⟹〈T(u′),g(u′)−g(u)〉≥0∀u′,u∈H.(iv)
The solution set of problem (2.1) denoted by S* is nonempty.


## 3. Iterative Method and Basic Results

In this section, we suggest and analyze a new method
for solving mixed general variational inequality (2.1) by using a new direction
with a new step size αk, and this is the main motivation of this paper.Algorithm 3.1
Step 1.
Givenγ∈[1,2),ρ0>0,δ∈(0,1),u0∈H,
and k=0.Step 2.
Set(3.1)wk=g−1(Jφ[g(uk)−ρkT(uk)]).Step 3.
If(3.2)rk:=∥ρk(T(uk)−T(wk))∥∥g(uk)−g(wk)∥≤δ, then set (3.3)εk=ρk(T(wk)−T(uk)),(3.4)d1(uk,ρk):=g(uk)−g(wk)+εk,(3.5)d(uk,ρk):=g(uk)−g(wk)+ρkT(wk),(3.6)ϕ(uk,ρk):=〈g(uk)−g(wk),d1(uk,ρk)〉
the stepsize
(3.7)αk:=ϕ(uk,ρk)∥d(uk,ρk)∥2and the next iterate(3.8)g(uk+1)=Jφ[g(uk)−γαkd(uk,ρk)],ρk:={1.5⁢*ρkifrk≤0.5,ρkotherwise,ρk+1=ρk and k:=k+1; go to Step 2.Step 4.
Reduce the value ofρk by ρk:=2/3*ρk*min{1,1/rk};
setwk=g−1(PK[g(uk)−ρkT(uk)]) and go to Step 3.
Ifφ is an indicator function of a closed convex
set K in H, then Jφ≡PK [10], the projection of H onto K. Consequently, Algorithm 3.1 reduces to
Algorithm 3.2 for solving the general variational inequalities (2.4).Algorithm 3.2
Step 1.
Givenγ∈[1,2),ρ0>0,δ∈(0,1),u0∈H,
and k=0.Step 2.
Set(3.9)wk=g−1(PK[g(uk)−ρkT(uk)]).Step 3.
If(3.10)rk:=∥ρk(T(uk)−T(wk))∥∥g(uk)−g(wk)∥≤δ,
then set(3.11)εk:=ρk(T(wk)−T(uk)),d1(uk,ρk):=g(uk)−g(wk)+εk,d(uk,ρk):=g(uk)−g(wk)+ρkT(wk),ϕ(uk,ρk):=〈g(uk)−g(wk),d1(uk,ρk)〉
the stepsize(3.12)αk:=ϕ(uk,ρk)∥d(uk,ρk)∥2
and the next iterate(3.13)g(uk+1)=PK[g(uk)−γαkd(uk,ρk)],ρk:={1.5⁢*ρkifrk≤0.5,ρkotherwise,
ρk+1=ρk and k:=k+1; go to Step 2.Step 4.
Reduce the value ofρk by ρk:=2/3*ρk*min{1,1/rk};
setwk=g−1(Jφ[g(uk)−ρkT(uk)]) and go to Step 3.Remark 3.3.
Equation (3.2) implies that
(3.14)|〈g(uk)−g(wk),εk〉|≤δ∥g(uk)−g(wk)∥2,0<δ<1.
The next lemma shows thatαk and ϕ(uk,ρk) are lower bounded away from zero, whenever uk≠wk.Lemma 3.4.
For givenuk∈H and ρk>0,
let wk and εk satisfy to (3.1) and (3.3), then (3.15)ϕ(uk,ρk)≥(1−δ)∥g(uk)−g(wk)∥2,αk≥(1−δ)∥g(uk)−g(wk)∥2∥d(uk,ρk)∥2.Proof.
It follows from (3.4) and (3.14)
that(3.16)ϕ(uk,ρk):=〈g(uk)−g(wk),d1(uk,ρk)〉=∥g(uk)−g(wk)∥2+〈g(uk)−g(wk),εk〉≥(1−δ)∥g(uk)−g(wk)∥2.
Otherwise, we have(3.17)αk:=〈g(uk)−g(wk),d1(uk,ρk)〉∥d(uk,ρk)∥2≥(1−δ)∥g(uk)−g(wk)∥2∥d(uk,ρk)∥2,we can get the assertion of this
lemma.The next lemma shows that∥r(u,ρ)∥ is a nondecreasing function with respect to ρ, which can be proved using the techniques as in
[1].Lemma 3.5.
For allu∈H and ρ′>ρ>0, it holds that(3.18)∥r(u,ρ′)∥≥∥r(u,ρ)∥.Next lemma has already been studied in [11].Lemma 3.6.
For allu∈H,u*∈S*,
and ρ>0, we have
(3.19)〈g(uk)−g(u*),d(uk,ρk)〉≥ϕ(uk,ρk), where d(uk,ρk) and ϕ(uk,ρk) are defined in (3.5) and (3.6), respectively.


## 4. Convergence Analysis

In this
section, we prove the global convergence of the proposed method. The following
result plays a crucial role in the convergence analysis of the proposed method.Theorem 4.1.
Letu*∈H be a solution of problem (2.1) and let uk+1 be the sequence obtained from Algorithm 3.1.
Then uk is bounded and
(4.1)∥g(uk+1)−g(u*)∥2≤∥g(uk)−g(u*)∥2−γ(2−γ)(1−δ)2∥g(uk)−g(wk)∥4∥d(uk,ρk)∥2.Proof.
Letu*∈H be a solution of problem (2.1).
Then(4.2)∥g(uk+1)−g(u*)∥2≤∥g(uk)−g(u*)−γαkd(uk,ρk)∥2=∥g(uk)−g(u*)∥2−2γαk〈g(uk)−g(u*),d(uk,ρk)〉+γ2αk2∥d(uk,ρk)∥2≤∥g(uk)−g(u*)∥2−2γαkϕ(uk,ρk)+γ2αkϕ(uk,ρk)≤∥g(uk)−g(u*)∥2−γ(2−γ)(1−δ)2∥g(uk)−g(wk)∥4∥d(uk,ρk)∥2,where the first inequality
follows from the nonexpansiveness of the resolvent operator, the second
inequality follows from (3.7) and (3.19), and the third inequality follows from
(3.15). Since γ∈[1,2) and δ∈(0,1),
we have(4.3)∥g(uk+1)−g(u*)∥≤∥g(uk)−g(u*)∥≤⋯≤∥g(u0)−g(u*)∥.
Sinceg is homeomorphism, it is easy to verify that
the sequence uk is bounded.We now prove the convergence of Algorithm3.1.Theorem 4.2.
The sequence{uk} generated by the Algorithm 3.1 converges to a
solution of problem (2.1).Proof.
It follows from (4.1) that(4.4)∑k=0∞∥g(uk)−g(wk)∥4∥d(uk,ρk)∥2<∞,which means that (4.5)lim⁡k→∞∥g(uk)−g(wk)∥=0. Since g is homeomorphism, we have(4.6)lim⁡k→∞∥uk−wk∥=0.This implies that {wk} is bounded. Since ∥r(uk,ρ)∥ is a nondecreasing function of ρ, it follows from ρk≥ρmin that(4.7)∥r(wk,ρmin)∥≤∥r(wk,ρk)∥=∥g(wk)−Jφ[g(wk)−ρkT(wk)]∥(using(3.1)and(3.3))=∥Jφ[g(uk)−ρkT(wk)+εk]−Jφ[g(wk)−ρkT(wk)]∥≤∥g(uk)−g(wk)+εk∥(using(3.2))≤(1+δ)∥g(uk)−g(wk)∥and from (4.5), we get(4.8)lim⁡k→∞r(wk,ρmin)=0. Let u¯ be a cluster point of {wk} and the subsequence {wkj} converges to u¯. Since r(u,ρ) is a continuous function of u,
it follows from (4.8) that(4.9)r(u¯,ρmin)=lim⁡j→∞r(wkj,ρmin)=0.From Lemma 2.2, it follows that u¯ is a solution point of problem (2.1). Note
that inequality (4.1) is true for all solution point of problem (2.1), hence we
have (4.10)∥g(uk+1)−g(u¯)∥≤∥g(uk)−g(u¯)∥,∀k≥0.Since {g(wkj)}→g(u¯) and g(uk)−g(wk)→0,
for any given ε>0,
there is an l>0,
such that
(4.11)∥g(wkl)−g(u¯)∥<ε2,∥g(ukl)−g(wkl)∥<ε2.Therefore, for any k≥kl,
it follows from (4.10) and (4.11) that(4.12)∥g(uk)−g(u¯)∥≤∥g(ukl)−g(u¯)∥≤∥g(ukl)−g(wkl)∥+∥g(wkl)−g(u¯)∥<εand thus the sequence {g(uk)} converges to g(u¯).
Using g is homeomorphism, we see that the sequence {uk} converges to u¯.
We now prove that the sequence{uk} has exactly one cluster point. Assume that u is another cluster point and satisfies(4.13)δ:=∥g(u)−g(u¯)∥>0.Since u¯ is a cluster point of the sequence {uk} and g is homeomorphism, there is a k0>0 such that(4.14)∥g(uk0)−g(u¯)∥≤δ2.On the other hand, since u¯∈S* and from (4.1), we have(4.15)∥g(uk)−g(u¯)∥≤∥g(uk0)−g(u¯)∥∀k≥k0,it follows that(4.16)∥g(uk)−g(u)∥≥∥g(u)−g(u¯)∥−∥g(uk)−g(u¯)∥≥δ2∀k≥k0.This contradicts the assumption
that u is cluster point of {uk}. Thus the sequence {uk} converges to u¯∈S*.


## 5. Numerical Results

In this section,
we present some numerical results for the proposed method. In order to verify
the theoretical assertions, we consider the following problems:(5.1)minh(u)=∑j=1nujlog(ujpj),s.t.Au∈Π,u≥0,where A is an n×n matrix, Π is a simple closed convex set in Rn,0<p∈Rn is a parameter vector. Here, the statement
that the set Π is simple means that the projection onto Π is simple to carry out. For the same reason
given in Fletcher (see  [5, page 222]), each element of the optimal solution of problem
(5.1) is positive. Thus the bounds u≥0 are inactive and can be ignored, therefore
problem (5.1) can be written as (5.2)minh(u)=∑j=1nujlog⁡(ujpj),s.t.Au−v=0,v∈Π.By attaching the Lagrange multiplier y∈Rn to the equality constraints Au−v=0,
the Lagrange function of problem (5.2) is(5.3)L(u,v,y)=h(u)−yT(Au−v),which is defined on Rn×Π×Rn.
If (u*,v*,y*)∈Rn×Π×Rn is a KKT point of problem (5.2), then we
have (5.4)∇h(u*)−ATy*=0,(v−v*)Ty*≥0,∀v∈Π,Au*−v*=0. Note that problem (5.1) is invariant under
multiplication h by some positive scalar ρ. Denoting f(u)=∇h(u) and eliminating v and v* in (5.4), we see that problem (5.1) is
equivalent to a general variational inequality problem. Find w* such that(5.5)Q(w*)∈Ω,[Q(v1)−Q(w*)]TF(w*)≥0,∀Q(v1)∈Ω,where (5.6)w=(uy),Q(w)=(uAu),F(w)=(ρf(u)−ATyy),Ω=Rn×Π.It is
well known (see [7, Theorem 1]) that solving (5.5)-(5.6) is
equivalent to finding a zero point of the function(5.7)e(w):=Q(w)−PΩ[Q(w)−F(w)]=(ρf(u)−ATyAu−PΠ[Au−y]).Then, solving (5.5)-(5.6) is
equivalent to find a pair (u*,y*),
such that(5.8)ρf(u*)=ATy*,(5.9)g(u*)∈Π,(g(v)−g(u*))Ty*≥0,∀g(v)∈Π, where(5.10)g(u)=Au.In this case Algorithms 2.3 and 3.1 collapse to Algorithms 2.4 and 3.2, respectively.In the test, we letv′∈Rn be a randomly generated vector, vj′∈(−0.5,0.5), and A=I−2(v′v′T)/(v′Tv′) be an n×n Householder matrix. Let(5.11)uj*∈(0.1,1.1),yi*∈(−0.5,0.5).Note that(5.12)fj(u*)=(∇h(u*))j=log⁡(uj*)−log⁡(pj)+1.Since(5.13)f(u*)=ATy*,we set(5.14)pj=uj*exp⁡(1−ejTATy*),Π={z∣lB≤z≤uB},where(5.15)(lB)i={(Au*)iifyi*≥0,(Au*)i+yi*otherwise,(uB)i={(Au*)iifyi*<0,(Au*)i+yi*otherwise. In this way, we have(5.16)Au*∈Π,Au*=PΠ[Au*−y*].In all the tests, we takeδ=0.95 and γ=1.95. The calculations are started with a vector u0, whose elements are randomly chosen in (0,1)
and stopped, whenever ∥r(u,ρ)∥∞≤10−7.Sinceu* is known, we also report the distance ∥uk−u*∥ after ∥r(u,ρ)∥∞≤10−7. All codes are written in Matlab and run on a
P4-2.00G note book computer. We test the problem with dimensions n=200 and n=300.
The iteration numbers and the computational time for Algorithms 2.4 and 3.2 with different dimensions and initial parameter ρ are given in the Tables 1-2, and for Algorithm 3.2 and the method
of Noor [14] in Tables 3-4Table 1
Numerical results for problem (5.9)
with n=200.
ρAlgorithm3.2Algorithm2.4No. It.CPU(Sec.)∥uk−u*∥No. It.CPU(Sec.)∥uk−u*∥105372.171.4*10−153074.311.12*10−7104321.801.4*10−153774.331.16*10−7102201.051.4*10−152932.811.22*10−7190.461.4*10−153212.661.22*10−710−130.751.4*10−152802.251.21*10−710−361.621.4*10−151148451.239.54*10−8Table 2
Numerical results for problem (5.9)
with n=300.
ρAlgorithm3.2Algorithm2.4No. It.CPU(Sec.)∥uk−u*∥No. It.CPU(Sec.)∥uk−u*∥105392.471.88*10−155398.399.36*10−8104341.951.88*10−156459.539.71*10−8102221.371.88*10−155276.819.5*10−81110.661.88*10−155726.439.45*10−810−150.841.88*10−155166.219.18*10−810−351.751.88*10−1511532120.019.56*10−8Table 3
Numerical results for problem (5.9)
with n=200.
ρAlgorithm3.2The method in [14]No. It.CPU(Sec.)∥uk−u*∥No. It.CPU(Sec.)∥uk−u*∥104321.801.4*10−15423.171.4*10−15102201.051.4*10−15322.561.4*10−1510151.511.4*10−15251.561.4*10−1510−130.751.4*10−15151.151.4*10−1510−5162.561.4*10−157157.789.49*10−810−6222.861.4*10−15665144.999.18*10−8Table 4
Numerical results for problem (5.9)
with n=300.
ρAlgorithm3.2The method in [14]No. It.CPU(Sec.)∥uk−u*∥No. It.CPU(Sec.)∥uk−u*∥104341.951.88*10−15463.451.88*10−15102221.371.88*10−15362.161.88*10−1510171.561.88*10−15291.751.88*10−1510−150.841.88*10−15161.791.88*10−1510−5152.781.88*10−1540810.511.88*10−1510−6212.311.88*10−15405782.821.88*10−15From Tables1–4, we could see that Algorithms 2.4
and the method in [14] work well, if ρ is sufficient large. If the parameter ρ is too small, then the iteration numbers and
the computational time can increase significantly. Also these tables show that
Algorithm 3.2 is very efficient for the problem tested. In addition, for our
method, it seems that the computational time and the iteration numbers are not
very sensitive to the problem size.



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*Source: 280956-2008-05-18.xml*