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Existence of solution and algorithms for a class of bilevel variational inequalities with hierarchical nesting structure
Fixed Point Theory and Applications volume 2016, Article number: 41 (2016)
Abstract
In this paper we consider a class of bilevel variational inequalities with hierarchical nesting structure. We first of all get the existence of a solution for this problem by using the Himmelberg fixed point theorem. Then the uniqueness of the solution for an upperlevel variational inequality is given under some mild conditions. By using gap functions of the upperlevel and lowerlevel variational inequalities, we transform bilevel variational inequalities into a onelevel variational inequality. Moreover, we propose two iterative algorithms to find the solutions of the bilevel variational inequalities. Finally, the convergence of the proposed algorithm is derived under some mild conditions.
Introduction
The idea of bilevel programming problem may be considered to date back to 1934 when it had been formulated by Stackelberg in a monograph on market economy [1, 2]. Since it was introduced to the optimization community in the 1970s, a rapid development and intensive investigation of these problems begun in theoretical analysis [3–9]. Now it has been applied to industry [10], decision science [11], transportation [12], network [13], electricity [14], support chain management [15, 16], cloud computing market [17], and so on. Some mathematical programming problems can be transformed into variational inequality problems. In order to study the bilevel models better, it is necessary to study bilevel variational inequalities.
Bilevel variational inequalities models have been investigated in recent decades (see e.g. [18–24] and references therein). Many algorithms had been constructed to get the approximate solution of these models. These models can well be applied. An imperfection in these models is that the upperlevel variational inequality and lowerlevel variational inequality of all these models are not better embedded in each other. Sometimes the two levels of a bilevel problem are interactional; therefore, the bilevel variational inequalities models, of which the upper level’s variable is embedded into the lower level, and also the lower level’s variable is embedded into the upper level, need to be studied.
In [25], Wan and Chen introduced bilevel variational inequalities (shortly BVIs) which has a hierarchical nesting structure. They gave the existence theorem of a solution and constructed an algorithm, in the case that the solution set of the lowerlevel variational inequality \(M(x)\) is a singleton (\(M(x)\) is the solution set of a lowerlevel variational inequality, in regard to each parameter x from the upperlevel variational inequality). But sometimes, \(M(x)\) is not a singleton. In this case, one wonders whether the solution of the BVI exists. How to construct an algorithm? In this paper, our work will consider the two problems.
Now we review the model in [25]. Let K be nonempty subset of the ndimensional Euclidean space \(R^{n}\), and \(H:R^{m}\times R^{n}\rightarrow R^{n}\) and \(P:R^{m}\rightarrow R^{m}\) be two mappings. Let \(T:K\rightarrow 2^{R^{m}}\) be a setvalued mapping, where \(2^{R^{m}}\) is the family of all nonempty subsets of \(R^{m}\).
Consider the following bilevel variational inequalities (shortly BVIs): find \((x,z)\in K\times R^{m}\), such that
where \(z\in M(x)\), \(M(x)\) is the solution set of the following parametric variational inequality: find \(z\in T(x)\) such that
Inequations (1) and (2) are called the upperlevel variational inequality (shortly (UVI)) and the lowerlevel variation inequality (shortly (LVI)), respectively. The decision variables of the problem BVI are divided into two classes, namely, the upperlevel decision variable x and the lowerlevel decision variable z. Denote the optimal solution set of the BVI by Θ.
Obviously, the BVI involves two variational inequalities. The constraint region of the lowerlevel variational inequality \(T(x)\) is implicitly determined by the parameter x from upperlevel variational inequality. The lowerlevel decision variable z is embedded into upperlevel inequality. It is extremely difficult to solve globally, because of its nested structure. In a broad sense, it is similar to a quasivariational inequality [26–28].
Next we review some definitions that are referred in [25].
The solution set of the lowerlevel variational inequality in regard to the parameter x from the upperlevel variational inequality is
The solution set of the upperlevel variation inequality for every parameter z from the lowerlevel variational inequality is
We call the BVI wellposed bilevel variational inequalities, if \(M(x)\) is a singleton set. We call it illposed bilevel variational inequalities, if \(M(x)\) has more than one element. Wan and Chen gave the existence of the solution and an algorithm when the set \(M(x)\) is a singleton in [25]. But sometimes, \(M(x)\) is not a singleton set; this can be seen from the next example.
Example 1.1
Let \(R=(\infty,+\infty)\), \(K=[2,1]\) and \(T(x)=[2,x]\) for all \(x\in K\), \(z,v,x,y\in R\). Let \(\langle P(z),zv\rangle=\langle (z+1)(2z+3)(3z+4)e^{z},zv\rangle=(z+1)(2z+3)(3z+4)e^{z}(zv)\), \(\langle H(z,x),xy\rangle=\langle(z+2)(z+1)xe^{z},xy\rangle =(z+2)(z+1)xe^{z}(xy)\). The BVI is defined as follows: find \(x\in K\) such that
where z is a solution of the following variational inequality: find \(z\in T(x)\) such that
A simple computation shows that the solution set of lowerlevel variational inequality is
We can see that if \(x\in[\frac{4}{3},1)\), then \(M(x)=\{\frac {3}{2},\frac{4}{3}\}\); there are two elements in \(M(x)\). So it is not a singleton set. It is easy to verify that \(\{(2,x):x\in[2,\frac {3}{2}) \mbox{ or } x=1\}\) is the solution set of the BVI (3)(4). So it is meaningful to research the existence theorem of the solution and algorithm when the lowerlevel solution set \(M(x)\) is not a singleton. This paper will discuss these questions.
The rest of this paper is organized as follows. In Section 2, we recall some important results of [25] and give some preliminary definitions which are needed for our main results. In Section 3, we demonstrate the existence theorem of a solution when the lowerlevel solution set \(M(x)\) is not a singleton, and we investigate the unique solution condition of the upperlevel variational inequality for every parameter z from the lowerlevel variational inequality. In Section 4, we transform the BVI into a onelevel variational inequality by gap functions of the lowerlevel variational inequality and the upperlevel variational inequality. Based on Section 4, we proposed two iterative algorithms to compute the approximation solution of the BVI, and analyzed the convergence of the presented algorithm in Section 5. Some numerical examples are given in Section 6.
Preliminaries
We first review some definitions and lemmas which are needed for our main results.
Definition 2.1
([25])
Let Ξ be a nonempty subset of \(R^{m}\). A mapping \(P:\Xi\rightarrow R^{m}\) is said to be

(i)
monotone on Ξ if
$$\bigl\langle P(x)P(y),yx\bigr\rangle \leq0,\quad \forall(x,y)\in\Xi\times\Xi; $$ 
(ii)
strictly monotone on Ξ if
$$\bigl\langle P(x)P(y),yx\bigr\rangle < 0,\quad \forall(x,y)\in\Xi\times\Xi, x \neq y; $$ 
(iii)
pseudomonotone if for any \((x,y)\in\Xi\times\Xi\)
$$\bigl\langle P(x),yx\bigr\rangle \geq0 \quad\Rightarrow\quad\bigl\langle P(y),xy\bigr\rangle \leq0. $$
It is easy to see that
Definition 2.2
Let Λ be a nonempty subset of \(R^{n}\). Let Ξ be a nonempty subset of \(R^{m}\), for all \(z\in\Lambda\), a function \(H:\Lambda\times R^{n}\rightarrow R^{n}\) is said to be

(i)
monotone on Ξ if
$$\bigl\langle H(z,x)H(z,y),yx\bigr\rangle \leq0, \quad\forall(x,y)\in\Xi\times \Xi; $$ 
(ii)
strictly monotone on Ξ if
$$\bigl\langle H(z,x)H(z,y),yx\bigr\rangle < 0, \quad\forall(x,y)\in\Xi\times\Xi ,x\neq y. $$
Definition 2.3
([29])
Let \(X\subseteq R^{n} \), \(Y\subseteq R^{m}\). A setvalued mapping \(T:X\rightarrow2^{Y}\) is said to be

(i)
upper semicontinuous (shortly, usc) at \(x_{0}\in X\) if, for each open set V with \(T(x_{0})\subset V\), there exists \(\delta>0\) such that
$$T(x)\subset V,\quad \forall x\in B(x_{0},\delta); $$ 
(ii)
lower semicontinuous (shortly, lsc) at \(x_{0}\in X\) if, for each open set V with \(T(x_{0})\cap V \neq\emptyset\), there exists \(\delta >0\) such that
$$T(x)\cap V \neq\emptyset,\quad \forall x\in B(x_{0},\delta); $$ 
(iii)
closed if the graph of T is closed, i.e., the set \(Gr(T)=\{(\zeta,x)\in P\times E:\zeta\in T(x)\}\) is closed in \(X\times Y\).
We say T is lsc (resp. usc) on X if it is lsc (resp. usc) at each \(x \in X\). T is called continuous at X if it is both lsc and usc on X.
Definition 2.4
([29])
Let K be a nonempty convex subset of \(R^{m}\). A setvalued mapping \(T:K\rightarrow2^{R^{m}}\) is said to be convexvalued (compactvalued, closedvalued) if, the images \(T(x)\) of all points \(x\in K\) are convex (compact, closed).
Definition 2.5
([30])
Let \(\Omega\in R^{n}\), \(E\in R^{m}\). A setvalued mapping \(G:\Omega \rightarrow2^{E}\) is said to be local intersection if, for any \(x\in \Omega\) there exists an open neighborhood \(U_{x}\) of x, such that \(\bigcap_{x^{\prime}\in U_{x}}G(x^{\prime})\neq\emptyset\). We denote \(G(\Omega)=\bigcup_{x\in\Omega}G(x)\).
Lemma 2.1
(Continuous selection theorem [30, 31])
Let \(X\subseteq R^{n} \), Y be a nonempty paracompact subset of \(R^{m}\). Let \(G,M :X\rightarrow2^{Y}\) be two setvalued mappings. Assume that the following conditions hold:

(i)
For any \(x\in X\), \(G(x)\) is nonempty, and \(co(G(x))\subset M(x)\);

(ii)
G is local intersection.
Then M has a continuous selection, namely, there is a continuous mapping \(f:X\rightarrow Y\), such that \(f(x)\in M(x)\), \(\forall x\in X\).
Remark 2.1
([31])
If X is a compact set, then \(coX\) is a paracompact set.
Lemma 2.2
([29])
Let \(X\subseteq R^{n} \), \(Y\subseteq R^{m}\), \(F,G:X\rightarrow2^{Y}\) be two setvalued mappings such that, for all \(x\in X\), \(F(x)\cap G(x)\neq\emptyset\). We suppose that:

(i)
F is upper semicontinuous at \(x_{0}\);

(ii)
\(F(x_{0})\) is compact;

(iii)
G is closed.
Then the setvalued map \(F\cap G:x\rightarrow F(x)\cap G(x)\) is upper semicontinuous at \(x_{0}\).
Lemma 2.3
(Himmelberg fixed point theorem [32, 33])
Let X be a convex subset of \(R^{m}\). D is a nonempty compact subset of X, \(H:X\rightarrow2^{D}\) is an upper semicontinuous setvalued mapping. And for all \(x\in X\), \(H(x)\) is a nonempty closed convex subset of D. Then there is a point \(\bar{x}\in D\), such that \(\bar{x}\in H(\bar{x})\).
Lemma 2.4
([34])
Let K be a nonempty subset of \(R^{m}\). A setvalued mapping \(T:K\rightarrow2^{R^{m}}\) is compactvalued and continuous. \(f:K\times R^{n} \) is continuous. Then
defines a compactvalued, and \(n(x):=\max_{y\in T(x)}f(x,y)\) is continuous.
Existence of solution for the BVI
In this section, we investigate the existence of solution for the BVI when the lowerlevel solution set \(M(x)\) is not a singleton under some suitable conditions.
For simplicity, let \(\phi:R^{m}\times R^{n}\times R^{n}\rightarrow R^{n}\), \(\phi:R^{m}\times R^{n}\times R^{n}\rightarrow R^{n}\), we denote the following equations:
Remark 3.1
It is easy to verify that \(\phi(z,x,x)=\langle H(z,x),xx\rangle=0\), both \(v\mapsto\psi(\cdot,v)\) and \(y\mapsto\phi(\cdot,\cdot,y)\) are concave and continuous. For each \(w,v\in R^{m}\), \(z\mapsto\langle P(w),zv\rangle\) is continuous.
Lemma 3.1
([25])
Let K be nonempty convex subset of \(R^{n}\), \(T: K\rightarrow 2^{R^{m}}\) be closed convexvalued. Assume that the following conditions hold:

(i)
P is monotone;

(ii)
\(z\mapsto\psi(z,\cdot)\) is lowerhemicontinuous.
Then for each \(x\in K\), the solution set \(M(x)\) of (LVI) is closed and convex.
Since it is easy to verify that ψ and P in Example 3.2 satisfies all conditions of Lemma 3.1, and \(M(x)\) is a closed and convex set, from Example 3.2 we can see that Lemma 3.1 is applicable.
Lemma 3.2
([25])
Let K be nonempty convex subset of \(R^{n}\), \(T: K\rightarrow 2^{R^{m}}\) be convex and compactvalued. Assume that the following conditions hold:

(i)
P is pseudomonotone;

(ii)
\(z\mapsto\psi(z,\cdot)\) is lowerhemicontinuous.
Then for each \(x\in K\), the solution set \(M(x)\) of (LVI) is nonempty and compact.
Example 3.1 can be used to show that the result of Lemma 3.2 is also applicable.
Example 3.1
Let \(R=(\infty,+\infty)\), \(K=[1,\frac{1}{2}]\), \(T(x)=[x,1]\) for \(x\in K\), and let \(\langle P(z),zv\rangle=\langle e^{z}, zv\rangle\). It is obvious that all conditions of Lemma 3.2 are satisfied. By computation, for each \(x\in K\), the lowerlevel solution set \(M(x)=\{ x\}\). \(M(x)\) is a nonempty and compact set.
Remark 3.2
Since monotonicity ⇒ pseudomonotonicity, from the conditions of Lemma 3.2, we can further see that, for each \(x\in K\), the solution set \(M(x)\) of (LVI) is convex nonempty and compact.
Remark 3.3
Under the conditions of the Lemma 3.2, \(M(x)\) may be not a singleton set, this can be seen in Example 3.2. This example also shows that Lemma 3.2 is applicable.
Example 3.2
Let \(R=(\infty,+\infty)\), \(K=[0.5,6]\times[0.5,6]\), \(x=(x_{1},x_{2})^{\top}\), \(x=(y_{1},y_{2})^{\top}\), \(z=(z_{1},z_{2})^{\top}\), \(v=(v_{1},v_{2})^{\top}\), and \(T(x)=\{ y:0< y_{1}<1+x_{1},0<y_{2}<1+x_{2},y_{1}+y_{2}=1\}\) for \(x\in K\), \(z,v\in R^{2}\). Let \(P(z)^{\top}=(1,1)\).
Obviously, P satisfies the conditions of Lemma 3.2. By a simple computation one gets the solution set of (LVI),
It is obvious that \(M(x)\) is nonempty convex compact and not a singleton.
The next theorem shows the existence of solution when the solution set \(M(x)\) of (LVI) is not a singleton. That is, the BVI is an illposed bilevel variational inequality.
Theorem 3.1
Suppose that K is a nonempty compact convex subset of \(R^{m}\); \(T: K\rightarrow2^{R^{m}}\) is a convex and compactvalued setvalued mapping, and, for all \(x\in K\), \(K\in T(x)\). Assume that the following conditions hold:

(i)
P is pseudomonotone;

(ii)
\(z\mapsto\psi(z,\cdot)\) is lowerhemicontinuous;

(iii)
\(\inf_{z\in K} \sup_{v\in T(K)} \psi (z,v)\leq0\);

(iv)
the function ϕ is a continuous function.
Then there exist \(\bar{x}\in K\), \(\bar{z}\in M(\bar{x})\), such that \(\langle H(\bar{z},\bar{x}),\bar{x}y\rangle\leq0\), \(\forall y\in K\). That is, the BVI (1)(2) has at least one solution \((\bar{z},\bar{x})\).
Proof
By conditions (i), (ii), and Lemma 3.2, it is easy to show that the solution set \(M(x)\) of (LVI) is nonempty, convex, and compact. Since K is a nonempty convex compact set, from condition (ii) and (iii) we see that there exists \(z_{0}\in K\) satisfying
First, we will show that \(z_{0}\in\bigcap_{x\in K}M(x)\). Suppose by contradiction that \(z_{0}\notin\bigcap_{x\in K}M(x)\), then there exists \(x^{\prime}\in K\) such that \(z_{0}\notin M(x^{\prime})\). That is,
so there exists at least one \(v^{\prime}\in T(x^{\prime})\), such that \(\psi(z_{0},v^{\prime})>0\). In view of condition (iii), we get
which is a contradiction.
Second, we show that setvalued mapping M satisfies the conditions of Lemma 2.1 (continuous selection theorem). In fact, since, for any \(x\in K\), \(M(x)\) is nonempty convex compact set, it is obvious that there exists a nonempty convex compact subset \(N_{x}\) for x, satisfying \(z_{0}\in N_{x}\subset M(x)\), then we can find a setvalued mapping \(G:K\mapsto2^{R^{m}}\), for every \(x\in K\), such that \(G(x)=N_{x}\). Obviously, G satisfies Definition 2.5. It easy to see that the following two conditions are satisfied:

(i)
For any \(x\in K\), \(G(x)\) is nonempty, and \(co(G(x))=co(N_{x})\subset M(x)\);

(ii)
\(\bigcap_{x\in K}G(x)=\bigcap_{x\in K}N_{x}\neq\emptyset\),
that is, M satisfies Lemma 2.1. Then, from Lemma 2.1 and Remark 2.1 we know that there is a continuous mapping \(f:K\mapsto R^{m}\) such that \(f(x)\in M(x)\), \(\forall x\in K\).
For every \(x\in K\), suppose
Clearly, \(F:K\rightarrow2^{K}\) is a setvalued mapping. According to the continuous property of f and ϕ, it follows that \(y\rightarrow\phi(f(x),x,y)\) is continuous. By the compactness property of K, we know that there exists at least one \(\hat{y}\in K\) such that
so, \(F(x)\) is a nonempty set.
Next, we will show that the setvalued mapping \(F:K\rightarrow2^{K}\) is convexvalued, compactvalued, and uppersemicontinuous.
First of all, we show that F is convexvalued. For every \(x\in K\), suppose \(y_{1} , y_{2}\in F(x)\), that is,
Let \(y_{0}=\lambda y_{1}+(1\lambda)y_{2}\), \(\forall\lambda\in(0,1) \), Since \(y\mapsto\phi(\cdot,\cdot,y)\) is concave, we have
so
that is, \(F(x)\) is a convex set.
Second, we prove that F is compactvalued. For every \((x,y)\in K\times K\), let \(\eta(x,y)=\phi(f(x),x,y)\). It is easy to see that \(y\rightarrow\eta(x,y)\) is continuous, and
For every sequence \(\{y_{k}\}\subset F(x)\), \(y_{k}\longrightarrow y_{0}\), we can get the following equation:
therefore, \(y_{0}\in F(x)\), F is compactvalued.
Finally, we show that F is upper semicontinuous. Let \(Q:K\rightarrow 2^{K}\) be setvalued mapping satisfying \(Q(x)=K\), \(\forall x\in K\). It is obvious that Q is a continuous and compactvalued mapping.
We denote
where \(J(x)=\{y: \eta(x,\hat{y})=\eta(x,y)\}\). Due to \(J(x)\) is closed. According to Lemma 2.2 we know that H is upper semicontinuous.
From the above discussion, it is obvious to know that F satisfies the conditions of Lemma 2.3 (Himmelberg fixed point theorem), so there exists a point \(\bar{x}\in K\), such that \(\bar{x}\in F (\bar {x})\). Then the point \(\bar{x}\in K\) satisfies
Let \(\bar{z}=f(\bar{x})\), then \(\bar{z}\in M(\bar{x})\), and \(\phi (\bar {z},\bar{x},\bar{x})=\max_{y\in K} \phi(\bar{z},\bar{x},y)\). So, for every \(y\in K\)
that is, \((\bar{z},\bar{x})\in\Theta\), here Θ is the solution set of the BVI (1)(2). □
Example 3.3
Let \(R=(\infty,+\infty)\), \(K=[0.5,6]\times[0.5,6]\), \(x=(x_{1},x_{2})^{\top}\), \(z=(z_{1},z_{2})^{\top}\), \(v=(v_{1},v_{2})^{\top}\), \(y=(y_{1},y_{2})^{\top}\), \(u=(u_{1},u_{2})^{\top}\), and \(T(x)=\{ u:0< u_{1}<1+x_{1},0<u_{2}<1+x_{2},u_{1}+u_{2}=1\}\) for \(x\in K\), \(z,v,y\in R^{2}\). Let \(H(z,x)^{\top}=(1,1)z(\frac{1}{2},\frac{1}{2})\), \(P(z)^{\top}=(1,1)\).
It is not hard to see that all conditions of Theorem 3.1 are satisfied.
Simple computation we get the solution set of the lowerlevel problem
It is obvious that \(M(x)\) is not a singleton set. Further, we can see that \(M(x)\) is a nonempty convex and compact set, and \(\{(z,x)\in R_{+}^{2}\times R_{+}^{2}: z_{1}+z_{2}=1, x_{1}+x_{2}=1\}\) is the solution set of the BVI.
Remark 3.4
From Example 3.3 we can verify that M satisfies Lemma 2.1. We can choose the continuous function \(f(x)=z\), \(z\in M(x)\). It is obvious that \(f(x)\) is a constant function.
We can find the following phenomenon from Example 3.3.
If the upperlevel decision maker feedback parameter \(x=(0.5,0.5)\) to the lowerlevel decision maker, then we can get the lowerlevel solution set \(M((0.5,0.5))=\{z: 0\leq z_{1}, 0\leq z_{2}, z_{1}+z_{2}=1\}\). Here, if the lowerlevel decision maker feedback parameter \(z=(0.5,0.5)\) to the upperlevel decision maker, then it is well known that \(U((0.5,0.5))=\{(z,x)\in R_{+}^{2}\times R_{+}^{2}: z_{1}+z_{2}=1, x_{1}+x_{2}=1\}\) is the solution set of the upperlevel variational inequality, and \(U((0.5,0.5))\) is not a singleton set. If the conditions of Theorem 3.1 are strengthened, we will get the unique solution of the upperlevel variational inequality, in regard to every parameter z from the lowerlevel variational inequality. That is, \(U(z)\) is a singleton set.
Theorem 3.2
(unique solution of (UVI))
Under the hypotheses of Theorem 3.1, if, moreover, \(H(z,x)\) is strictly monotone, for every \(z\in R^{m}\), then, for every \(z\in M(x_{0})\), the solution set of the upperlevel variation inequality
is a singleton.
Proof
For each \(x_{0}\in K\) from Theorem 3.1, we know that \(M(x_{0})\neq\emptyset\). Using a proof by contradiction, suppose that there exists a \(z\in M(x_{0})\) such that \(U(z)\) is not a singleton. That is, there exist \(x_{1}, x_{2}\in K\), \(x_{1}\neq x_{2}\) such that
and
moreover, we have
and
Since for every \(z\in R^{m}\), \(H(z,x)\) is strictly monotone, from (ii) of Definition 2.2, we have
that is,
which contradicts (6). Therefore, \(U(z)\) is a singleton. □
Equivalence transformation of the BVI
In this section, we shall transform the bilevel variational inequalities (1) and (2) into a onelevel variational inequality by the gap functions of (LVI) and (UVI).
We now define the functions \(\alpha: R^{m}\times R^{n}\rightarrow R\), \(\beta:R^{m}\times R^{n}\rightarrow R\), \(g: R^{m}\times R^{n}\rightarrow R\), by
and
Remark 4.1
For each \(x\in K\), \(z\in M(x)\) if \(\alpha(z,x)=0\). Moreover, we obtain that \(\alpha(z,x)=0\) from \(\alpha(z,x)\geq0\) for all \(z\in T(x)\). The same holds for \(\beta(z,x)\).
We consider the parametric variational inequality defined by the bifunction PVI: find \(x\in K\), \(z\in T(x)\) such that
Denote the solution set of the PVI by S.
Theorem 4.1
\((x,z)\in\Theta\) if and only if \((x,z)\in S\).
Proof
First, we verify that \((x,z)\in\Theta\Rightarrow(x,z)\in S\). Assume that \((x,z)\in\Theta\), then \(x\in K\), \(z\in T(x)\),
and
further, we have
therefore,
this implies that \((z,x)\) is a solution of the PVI, so \((z,x)\in S\).
The next thing to do in the proof is to show that \((x,z)\in S \Rightarrow(x,z)\in\Theta\). Let \((z,x)\in S\), then
from Remark (4.1) we have
from (9), (10), and (11), we have \(\alpha (z,x)=0 \), \(\beta(z,x)=0\), so \((z,x) \) is the solution of the BVI, that is, \((z,x)\in\Theta\). □
Corollary 4.1
If all conditions of Theorem 3.1 are satisfied, then S is nonempty and
Algorithm and convergence analysis
In this section, we first present Algorithm 1 for the BVI (1) and (2) from the theoretical point of view, and then consider the limiting behavior of the sequence generated by the algorithm. Algorithm 1 can be used to solve the BVI which has only one solution. In order to solve the BVI which has more than one solution we modify Algorithm 1 to obtain Algorithm 2.
We have transformed the BVI (1) and (2) into a singlelevel variational inequality (PVI) (8), and discussed the relationship between the two problems in the previous section. From Theorem 4.1, we know that the two problems are equivalent. So we can find the solution of the BVI (1) and (2) by solving the PVI problem (8). In Algorithm 1, we will solve the BVI (1)(2) by solving a sequence of the approximation problem PVI (8).
Algorithm 1
Step 1. Take \(\{\mu_{k}\}_{k\in Z^{+}}\subset(1,+\infty )\), where \(Z^{+}\) is the set of all nonnegative integers. Choose \(x_{0}\in K\) arbitrarily, let \(m=x_{0}\), compute \(z_{0}=\arg\min_{z\in T(x_{0})}g(x_{0},z)\). Let the tolerance \(\varepsilon\geq 0\), \(k=0\). If \(g(z_{0},x_{0})\leq\varepsilon\), stop, put out \((z_{0},x_{0})\). Otherwise, go to Step 2.
Step 2. Let \(K=K\backslash m\), take \(m \in K\). Compute \(n=\arg\min_{z\in T(m)}g(z,m)\), go to Step 3.
Step 3. If
let \(z_{k+1}=n\), \(x_{k+1}=m\) go to Step 4. Otherwise, go to Step 2.
Step 4. If
stop, put out \((z_{k+1},x_{k+1})\). Otherwise, let \(k=k+1\) go to Step 2.
Remark 5.1
In Step 2, we discretize K within acceptable error range of the solution. We take different discrete points m of K in every iterative process. For example, if the error band of solution is 2δ, \(K\in R\), then in every iterative process, \(m=m+\delta\) (or \(m=m\delta\)). If with the algorithm we have not found a solution until m has equality to one of the boundary points of K, put the other boundary point in m. Algorithm 1 is effective for the bilevel variational inequalities which have only one solution.
Remark 5.2
In order to guarantee the existence of a solution for \(\min_{z\in T(m)}g(z,m)\) in Step 2, we need to suppose that ψ and ϕ are continuous and T is compactvalued and continuous. Then according to Lemma 2.4, it follows that \(z\mapsto\alpha(z,x)\) and \(z\mapsto\beta(z,x)\) are continuous; further, \(z\mapsto g(z,x)\) is continuous. Thus \(\min_{z\in T(m)}g(z,m)\) has a solution.
To explain the Algorithm 1, we give the next example.
Example 5.1
Let \(R=(\infty,+\infty)\), \(K=[2,1]\), and \(T(x)=[2,x]\) for \(x\in K\), \(z,v,y\in R\). \(\langle H(z,x),xy\rangle=\langle zxe^{z},xy\rangle =zxe^{z}(xy)\), \(\langle P(z),zv\rangle= \langle (2z+3)e^{z},zv\rangle=(2z+3)e^{z}(zv)\). The BVI is defined as follows: find \(x\in K\) such that
where z is the solution of the following variational inequality: find \(z\in T(x)=[2,x]\) such that
By a simple computation we get the lowerlevel solution set,
Observe that M(x) is not a singleton, and \((2,2)\) is the unique solution of the BVI (13)(14). By the proposed Algorithm 1, first we should transform the BVI into a onelevel variational inequality.
The gap function of the lowerlevel variational inequality is
Moreover,
The gap function of upperlevel variational inequality is
By equality (9), we have
Next, we illustrate the process of solving the BVI (13)(14) by the proposed Algorithm 1.
Step 1. Choose \(x_{0}=\frac{3}{2}\) and the tolerance \(\varepsilon =10^{5}\), \(\mu_{k}=\mu=2\).
Step 2. Verify whether there exists \(z_{0}\in T(x_{0})=[2,\frac {3}{2}]\) such that inequality (12) holds.
From equality (15) we have
In order to find \(z_{0}\in T(x_{0})\), we need to solve the following optimization problem:
Applying the descent direction method to solve (16), we conclude that \(z_{0}=2\), which is an optimal solution of (16), but
Go to Step 2, set \(K=K\backslash\{\frac{3}{2}\}\). For simplicity, take \(m=2\in K\). In numerical experiments, we take m by the method of Remark 5.1. Solve the following optimization problem:
Applying the descent direction method to solve (17), we get \(n=2\), and go to Step 3.
Step 3. We have
Let \(z_{1}=n\), \(x_{1}=m\), go to Step 4.
Step 4. As \(g(z_{1},x_{1})=0\leq\varepsilon\), stop, put out \((z_{1},x_{1})\). So \((2,2)\) is the solution of the BVI (13)(14).
Convergence analysis
Let us consider the behavior of a sequence \(\{(x_{k},z_{k})\}\) generated by the proposed Algorithm 1.
Theorem 5.1
Suppose K is a nonempty compact convex subset of \(R^{n}\); \(T:K\rightarrow2^{R^{m}}\) is a continuous, convex, and compactvalued setvalued mapping. Assume that the following conditions hold:

(i)
\(z\mapsto\psi(z,\cdot)\) is lowerhemicontinuous;

(ii)
ϕ is a continuous function.
If the solution of the BVI (1)(2) exists, then the sequence \(\{(x_{k},x_{k})\}\) generated by the proposed Algorithm 1 converges to a solution of the BVI (1)(2).
Proof
According to Algorithm 1 we have
This shows that
Observe that \(\mu_{k}>1\), then \(k\nearrow+\infty\), \(\prod^{k}_{i=0}\frac {1}{\mu_{j}}\searrow0\). Since \(T: K\rightarrow2^{R^{m}}\) is convex compactvalued, from condition (i), we know that \(\alpha(z_{0},x_{0})\) is finite. Similarly, since K is nonempty compact convex subset of \(R^{n}\), from condition (ii), we see that \(\beta(z_{0},x_{0})\) is finite. Further,
is finite. This combined with \(k\nearrow+\infty\) shows that we have \(\prod^{k}_{j=1}\frac{1}{\mu_{j}} g(z_{0},x_{0})\searrow0\).
Observe that \(\{x_{k}\}_{k\in Z^{+}}\subseteq K\) is compact. Without loss of generality, let \(x_{k}\rightarrow x^{*}\in K\). Owing to \(T: K\rightarrow2^{ R^{m}}\) being convex compactvalued and continuous, there exists \(z^{*}\in T(x^{*})\) such that \(z_{k}\rightarrow z^{*}\) and \(\lim_{k\rightarrow\infty}T(x_{k})=T(x^{*})\). By condition (i), we have
Similarly, by condition (ii), we have
By equality (18), we have
and combining inequality (21) and (22) it is easy to see that
that is, \((z^{*},x^{*})\in S\). According to Theorem 4.1, it follows that \((z^{*},x^{*})\in\Theta\). So, the sequence \(\{(z_{k},x_{k})\}\) generated by Algorithm 1 converges to a solution of the BVI. □
In order to solve the bilevel variational inequalities which has more than one solution, we modify Algorithm 1 to obtain Algorithm 2.
Algorithm 2
Step 1. Take \(\{\mu_{k}^{j}\}_{k\in Z^{+}}\subset (1,+\infty)\), \(j\in Z^{+}\), where \(Z^{+}\) is the set of all nonnegative integers. Let the tolerance \(\varepsilon\geq0\), \(k=0\), \(j=1\). Choose \(x_{0}^{1}\in K\) arbitrarily, compute \(z_{0}^{1}=\arg\min_{z\in T(x_{0}^{1})}g(x_{0}^{1},z)\), such that \(g(z_{0}^{1},x_{0}^{1})>\varepsilon\), let \(K=K\backslash x_{0}^{1}\). Go to Step 2.
Step 2. Take \(m \in K\), let \(K=K\backslash m\). Compute \(n=\arg\min_{z\in T(m)}g(z,m)\), go to Step 3.
Step 3. If
let \(z_{k+1}^{j}=n\), \(x_{k+1}^{j}=m\), go to Step 4. Otherwise, go to Step 2.
Step 4. If
put out \((z_{k+1}^{j},x_{k+1}^{j})\), \(j=j+1\), choose \(x_{0}^{j}\in K\), compute \(z_{0}^{j}=\arg\min_{z\in T(x_{0}^{j})}g(x_{0}^{j},z)\) such that \(g(z_{0}^{j},x_{0}^{j})>\varepsilon\) (automatically initialize \(x_{0}^{j}\) and \(z_{0}^{j}\) again), take \(\{ \mu_{k}^{j}\}_{k\in Z^{+}}\subset(1,+\infty)\), let \(K=K\backslash x_{0}^{j}\), go to Step 5. Otherwise, let \(k=k+1\), go to Step 5.
Step 5. If \(K=\emptyset\), stop. Otherwise, go to Step 2.
Remark 5.3
In Step 2 we discretize K within an acceptable error range of solution. m takes different discrete points m of K in every iterative process. It is not hard to see that the major part of Algorithm 2 is similar to Algorithm 1. If we use Algorithm 1 to solve the BVI which has more than one solution, Algorithm 1 will stop when we get one solution. In order to get all the solutions, we should choose another initial point again, and reiterate the process of Algorithm 1. So, Step 4 in Algorithm 2 is important. m needs to take all the discrete points of K. Since we discretize K within an acceptable error range of solution, the stopping criterion \(K=\emptyset\) is reasonable.
Theorem 5.2
Suppose all the conditions of Theorem 5.1 are satisfied. If the BVI (1)(2) has n solutions, then, for all \(j\in[1,n]\) the sequence \(\{(x_{k}^{j},x_{k}^{j})\}\) generated by the proposed Algorithm 2 converges to one solution of the BVI (1)(2). Further, Algorithm 2 can be used to find all the solution of the BVI (1)(2).
Proof
The proof of Theorem 5.1 is similar to the proof of Theorem 5.2, so it is omitted here. □
Numerical illustrative examples
Next, we will given two numerical examples of the proposed Algorithm 1.
Example 6.1
Let \(R=(\infty,+\infty)\), \(K=[1,2]\) and \(T(x)=[1,x]\) for \(x\in K\), \(z,v,y\in R\), \(\langle H(z,x),xy\rangle=\langle z x,xy\rangle=z x(xy)\), \(\langle P (z),zv\rangle=\langle z,zv\rangle=z(zv)\). The BVI is defined as follows: find \(x\in K\) such that
where z is the solution of the following variational inequality: find \(z\in T(x)=[1,x]\) such that
It is easy to see that the lowerlevel solution set is \(M(x)=\{1\}\), \(\forall x\in K\), and \((1,1)\) is the unique solution of the BVI (23)(24).
The gap function of the lower level variational inequality is
The gap function of the upperlevel variational inequality is
Further,
Example 6.2
Let \(R=(\infty,+\infty)\), \(K=[0.5,5]\), and \(T(x)=[0.5,2x]\) for \(x\in K\), \(z,v,y\in R\), \(\langle H(z,x),xy\rangle=\langle 3zx^{2},xy\rangle = 3zx^{2}(xy)\), \(\langle P(z),zv\rangle=\langle3z,zv\rangle =3z(zv)\). The BVI is defined as follows: find \(x\in K\) such that
where z is the solution of the following variational inequality: find \(z\in T(x)=[0.5,2x]\) such that
It is easy to see that the rational reaction set is \(M(x)=\{0.5\}\), \(\forall x\in K\), and \((0.5,5)\) is the unique solution of the BVI (25)(26).
The gap function of the lower level variational inequality is
The gap function of the upperlevel variational inequality is
Further,
Using the proposed Algorithm 1, we obtain the results for solving Example 6.1 and Example 6.2. In Table 1, the fifth column denotes the solutions of the BVI (23)(24) and the BVI (25)(26); the sixth column denotes the value of \(g(z,x)\), which corresponds to equality to the BVI; the third column denotes the value of \(\mu_{k}\), here we let \(\mu_{k}\) be a fixed value μ; the last column denotes the operation time. From the numerical results, we observe that the proposed algorithm can solve the BVI well.
Next, we will given two numerical examples for the proposed Algorithm 2.
Example 6.3
Let \(R=(\infty,+\infty)\), \(K=[1,2]\), and \(T(x)=[1,2+x]\) for \(x\in K\), \(z,v,y\in R\), \(\langle H(z,x),xy\rangle=\langle z(z1.2)x,xy\rangle =z(z1.2)x(xy)\), \(\langle P (z),zv\rangle=\langle z(z1.5)^{2},zv\rangle=z(z1.5)^{2}(zv)\). The BVI is defined as follows: find \(x\in K\) such that
where z is the solution of the following variational inequality: find \(z\in T(x)=[1,2+x]\) such that
It is easy to see that the lowerlevel solution set is \(M(x)=\{1,1.5\}\), \(\forall x\in K\). \(M(x)\) is not a singleton, and \(\{(1,2), (1.5,1)\}\) is the solution set of the BVI (27)(28).
The gap function of the lower level variational inequality is
The gap function of the upperlevel variational inequality is
Moreover,
Further,
Example 6.4
Let \(R=(\infty,+\infty)\), \(K=[1,4]\), and \(T(x)=[1,4+2x]\) for \(x\in K\), \(z,v,y\in R\), \(\langle H(z,x),xy\rangle=\langle (z1.5)x,xy\rangle =(z1.5)x(xy)\), \(\langle P(z),zv\rangle=\langle z^{2}(z2)^{2},zv\rangle= z^{2}(z2)^{2}(zv)\). The BVI is defined as follows: find \(x\in K\) such that
where z is the solution of the following variational inequality: find \(z\in T(x)=[1,4+2x]\) such that
It is easy to see that the lowerlevel solution set is \(M(x)=\{1,2\}\), \(\forall x\in K\). \(M(x)\) is not a singleton, and \(\{(1,4), (2,1)\}\) is the solution set of the BVI (29)(30).
The gap function of the lower level variational inequality is
The gap function of the upperlevel variational inequality is
Moreover,
Further,
Example 6.5
Let \(R=(\infty,+\infty)\), \(K=[1,2]\times[1,2]\), and \(T(x)=[1,2+x_{1}]\) for \(x\in K\), \(z,v,y\in R\), \(\langle H(z,x),xy\rangle= ((1.2z)x_{1},8)(x_{1}y_{2},x_{2}y_{2})^{\top}\), \(\langle P(z),zv\rangle= (z1.4)^{2}(zv)\). The BVI is defined as follows: find \(x\in K\) such that
where z is the solution of the following variational inequality: find \(z\in T(x)=[1,2+x_{1}]\) such that
It is easy to see that the lowerlevel solution set is \(M(x)=\{1,1.4\}\), \(\forall x\in K\). \(M(x)\) is not a singleton, and \(\{((2,1),1.4), ((1,1),1)\}\) is the solution set of the BVI (29)(30).
The gap function of the lower level variational inequality is
The gap function of the upperlevel variational inequality is
Moreover,
Let \(\varepsilon=0.01 \), use the proposed Algorithm 2, we obtain the results for solving Example 6.3, Example 6.4, and Example 6.5. In Table 2, the fourth column denotes the solutions of the BVI (27)(28), the BVI (29)(30), and the BVI (31)(32); the fifth column denotes the value of \(g(z,x)\) which is equivalent to the BVI; the second column denotes the initial value of \(x_{0}^{j}\). From this column we can see that, in order to get the approximate solution of Example 6.3, Algorithm 2 only needs to initialize \(x_{0}\) two times, that is, \(x_{0}^{1}=1.9\), \(x_{0}^{2}=1.91\); the third column denotes the value of \(\mu_{k}^{j}\), here we let \(\mu_{k}^{j}=\mu^{j}\) be a fixed value 1.1; the last column denotes the operation time. From the numerical results, we observe that the proposed Algorithm 2 can get all solutions of the BVI.
Next, we solve the BVI (23)(24) which has only one solution in Example 6.1 with the two proposed algorithms, respectively; see the results detailed in Table 3.
In Table 3, we can see that the two algorithms can solve the BVI (23)(24) well. The last column denotes the operation time. Obviously, Algorithm 1 is faster than Algorithm 2. Because the two algorithms have different termination conditions, if the BVI has only one solution, we should solve it by Algorithm 1.
Conclusions
In this paper, we have investigated a class of bilevel variational inequalities (BVIs) with hierarchical nesting structure, which was introduced in [25] first. In the case that the solution set \(M(x)\) of the lowerlevel variational inequality is not a singleton, we call it Illposed bilevel variational inequality. We first obtained the existence theorem of a solution for illposed bilevel variational inequalities by the Himmelberg fixed point theorem. Then the uniqueness of a solution for the upperlevel variational inequality in regard to every feedback parameter is given under some mild condition. We transform the BVI into a onelevel variational inequality by gap functions of the upperlevel and lowerlevel function, and we proved their equivalence. Two algorithms to find the solutions of the BVI were constructed. Finally, we have proved the convergence of the iterative sequence generated by the proposed algorithm under some mild conditions. We said that the two algorithms can also be used to solve wellposed bilevel variational inequalities. Furthermore, we can see that for the two algorithms \(M(x)\) does not need to be compact or convex. In the future, we will consider the following questions. We know that one of the conditions for a solution to exist is that \(M(x)\) is a convex set. In the future we will consider the solution existence conditions in the case that \(M(x)\) is not a convex set. Since the BVI is a very complicated model, it is quite difficulty to design an algorithm that is efficient for all examples. From the numerical examples, we can see that the two algorithms are efficient in R. In the future, we will design algorithms that are more efficient in \(R^{d}\) for \(d> 2\).
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Acknowledgements
This work of Z Wan was partially supported by the Natural Science Foundation of China, No. 71471140 and No. 71171150. The work of Jw Chen was partially supported by the Natural Science Foundation of China, No. 11401487. The work of G Li was partially supported by Graduate Students Independent Research Foundation of Wuhan University, No. 2015201020204.
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Li, G., Wan, Z., Chen, J. et al. Existence of solution and algorithms for a class of bilevel variational inequalities with hierarchical nesting structure. Fixed Point Theory Appl 2016, 41 (2016). https://doi.org/10.1186/s1366301605245
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MSC
 90C33
 49J40
 90C30
Keywords
 bilevel variational inequalities
 Himmelberg fixed point theorem
 existence
 gap function
 iterative algorithm
 convergence