R E S E A R C H
Open Access
Some results concerning the solution
mappings of mixed variational inequality
problems
Qi-Qing Song
1,2and Hui Yang
1**Correspondence:
1College of Science, Guizhou
University, Guiyang, 550025, China Full list of author information is available at the end of the article
Abstract
This paper shows some continuities of mappings between the space of mixed variational inequality problems and the graph space of their solution mappings. The space of mixed variational inequality problems is homeomorphic to the graph of a continuous mapping. These generalize the results in the corresponding references.
MSC: 49J53; 58E35; 47H04
Keywords: mixed variational inequality; monotone; homeomorphism
1 Introduction and preliminaries
Variational inequalities have become important methods to analyze many linear and non-linear problems; see [] and [], such as non-linear complementarity problems in [], convex optimization in [], imaging problems in [],etc.
Classical variational inequalities were introduced by Hartman and Stampacchia in the s; see [] and []. They are of the form suitable to findu∗such that (u–u∗)Tf(u∗)≥, ∀u∈K. The spaces consisting of nonlinear problems may have some interesting topolog-ical properties: some nonlinear problem spaces are homeomorphic to the graph spaces of their solution mappings, such as bimatrix games in [], normal game problems in [, ], game trees in [], classical variational inequality problems [],etc.
Duvaut and Lions considered a kind of mixed variational inequality, see [], which added a function to a classical variational inequality. The stability, algorithm, and gen-eralization of this kind of variational inequality has been studied in many forms [–] and applied to many fields; see [].
LetKbe a compact convex subset ofRn. We consider the following mixed variational
inequality problem: to find a pointu∗∈Ksuch thatu∗satisfies
θ(u) –θu∗+u–u∗Tfu∗≥, ∀u∈K, () wheref :K→Rnis a mapping andθ:K→Ris a real function. This was first introduced in []. Whenf≡ onK, this leads us to find a maximizeru∗∈Kof the functionθonK. Definition . A mappingf :K→Rnis said to be monotone if (f(x) –f(y))T(x–y)≥,
∀x,y∈K;f is said to be strictly monotone if,f is monotone and if (f(x) –f(y))T(x–y) = ,
thenx=y;fis strongly monotone if there exists a constantcsuch that (f(x) –f(y))T(x–y)≥ cx–y,∀x,y∈K.
Iff is monotone andθ is convex, we call the above variational inequality () a mixed monotone variational inequality problem. Denote byMthe set as follows:
M=
(θ,f): θ:K→Ris continuous and convex;
f :K→Rnis continuous and monotone
.
For any two (θ,f), (θ,f)∈M, define the metricρbetween (θ,f) and (θ,f) as
ρ(θ,f), (θ,f)
=sup
x∈K
f(x) –f(x)+sup
x∈K
θ(x) –θ(x).
ThenMis a metric space. For convenience, we denoteρ((θn,fn), (θ,f))→ by (θn,fn)
ρ
→ (θ,f). A pointu∗ is a solution of the mixed monotone variational inequality problem
(θ,f)∈M; we write it asu∗∈V(θ,f). Then a set-valued mappingV fromMtoKis de-fined. It is well known that, for each (θ,f)∈M, we haveV(θ,f) is nonempty. Iff is strictly monotone, thenV(θ,f) is a singleton set; for example,f +Iis strictly monotone, whereI
is the identity mapping onK, and iff is strongly monotone, thenf is strictly monotone. Denote byNthe graph of the set-valued mappingV, that is,
N=(θ,f,u)∈M×K|u∈V(θ,f) .
For each (θ,f,u∗)∈N, let
φθ,f,u∗= (θ,f–Cu∗),
whereCxdenotes the constant mapping withCx(u) =x,∀u∈K. Clearly,Cu∗∈Mis
mono-tone, thenφis mapping fromNtoM. For each (θ,f)∈M, defineψ(θ,f) such that
ψ(θ,f) = (θ,f +Cu¯,u¯),
whereu¯is the unique solution of the problem (θ,f+I).
2 Main results
Theorem . ψis a continuous mapping from M to N.φis continuous on N.
Proof For each (θ,f)∈M, sinceψ(θ,f) = (θ,f+Cu¯,u¯), whereu¯∈V(θ,f+I), we haveθ(u) –
θ(u¯) + (u–u¯)T(f(u¯) +u¯)≥,∀u∈K, that is,u¯∈V(θ,f+C
¯
u).ψmapsMontoN.
Next, we prove thatψ is continuous onM. Let{(θn,fn)}∞n=⊂Mwith (θn,fn)
ρ
→(θ,f).
We need to show that (θn,fn+Cu¯n)
ρ
→(θ,f+Cu¯) and¯un–u¯ →, whereu¯n∈V(θn,fn+ I) andu¯ ∈V(θ,f+I). Since
θn(u) –θn(u¯n) + (u–u¯n)T
fn(¯un) +u¯n
≥, ∀u∈K
and
θ(u) –θ(u¯ ) + (u–u¯ )T
we have
θn(¯u) –θn(¯un) + (¯u–u¯n)T
fn(¯un) +u¯n
≥ ()
and
θ(u¯n) –θ(¯u) + (u¯n–u¯ )T
f(¯u) +u¯ ≥. ()
Adding equation () and equation (), it follows that
θn(u¯) –θ(u¯) +θ(u¯n) –θn(u¯n) + (u¯–u¯n)T
fn(u¯n) –f(u¯) +u¯n–u¯
≥.
Then
θn(¯u) –θ(¯u) +θ(¯un) –θn(u¯n) + (¯u–u¯n)T
fn(¯un) –f(u¯ )
≥ ¯u–u¯n.
This is equivalent to the following inequality:
θn(¯u) –θ(¯u) +θ(¯un) –θn(u¯n)
– (u¯–u¯n)T
f(u¯) –f(u¯n) +f(u¯n) –fn(u¯n)
≥ ¯u–u¯n.
By transposition, we have
θn(¯u) –θ(¯u) +θ(¯un) –θn(u¯n) – (¯u–u¯n)T
f(¯un) –fn(¯un)
≥ ¯u–u¯n+ (u¯ –u¯n)T
f(u¯ ) –f(u¯n)
.
Sincefis monotone, we have (¯u–u¯n)T(f(u¯ ) –f(u¯n))≥. Then
θn(¯u) –θ(¯u) +θ(¯un) –θn(u¯n) – (¯u–u¯n)T
f(¯un) –fn(¯un)
≥ ¯u–u¯n.
Therefore, we have
¯u–u¯n ≤θn(u¯ ) –θ(¯u)+θ(¯un) –θn(u¯n)+(u¯ –u¯n)T
f(¯un) –fn(¯un)
≤θn(u¯ ) –θ(¯u)+θ(¯un) –θn(u¯n)+¯u–u¯nf(¯un) –fn(¯un).
Then
–f(¯un) –fn(¯un)¯u–u¯n ≤θn(¯u) –θ(u¯ )+θ(u¯n) –θn(¯un).
Sinceθn→θandmaxu∈Kf(u) –fn(u) →, we have¯u–u¯n →. Hence, it can be
checked that
fn(u) +Cu¯n(u) –f(u) –Cu¯(u)
≤fn(u) –f(u)+Cu¯n(u) –Cu¯(u)
=fn(u) –f(u)+¯un–u¯ →.
Then (θn,fn+Cu¯n)
ρ
For the part withφis continuous onN, let{(θn,fn,un)}∞n=⊂Nwith (θn,fn)
ρ
→(θ,f) and
un–u →; we show thatφ(θn,fn,un)
ρ
→φ(θ,f,u). In fact, for eachu∈K, we have θn(u) –θ(u)+fn(u) –f(u) –un+u
≤θn(u) –θ(u)+fn(u) –f(u)+un–u →,
thenφ(θn,fn,un)
ρ
→φ(θ,f,u).
LetXbe the set of (θ,f) satisfying: (i)θ:K→Ris proper convex lower semi-continuous; (ii)f :K→Rnis continuous and monotone. Then each variational inequality (θ,f)∈Xhas
a solution (see []) and further iff is strictly monotone, then the corresponding solution is unique. Then (X,ρ) is a metric space. From Theorem ., there can be obtained some stability results for solution sets of mixed monotone variational inequalities.
Corollary . The set-valued mapping V is upper semi-continuous from M toK.
Proof From the proof of Theorem .,ψ :X→N is continuous, noting thatM⊂Xis closed, thenN, the graph ofV, is closed. Therefore,Vis upper semi-continuous.
Remark . By Corollary .,Vis upper semi-continuous onM. Then, whenf+εnI→f
as positiveεn→ andθn→θ, there is a convergent subsequence{xnk}of{xn}with{xn}=
V(θn,f+εnI) such thatxnk→x∈V(θ,f). Alternatively, from Theorem ., letεn> , if we defineψεn(θ,f) = (θ,f +εnCu¯n,u¯n) for each (θ,f)∈M, where{¯un}=V(θ,f +εnI), then
ψεnis also a continuous mapping fromMtoN. Note thatu¯n∈V(θ,f +εnCu¯n), therefore, ifεn→, we assert that there exists a subsequence{¯unk}of{¯un}withu¯nk→u∈V(θ,f).
Theorem . The mappingsφ◦ψ andψ◦φare identity mappings on M and N, respec-tively.
Proof For each (θ,f)∈M, from the definition ofψandφ, we have
φ◦ψ(θ,f) =φ(θ,f+Cu¯,u¯)
= (θ,f +Cu¯–Cu¯,u¯)
= (θ,f),
whereu¯∈V(θ,f +I).
For each (θ,f,u∗)∈N, we have
ψ◦φθ,f,u∗=ψ(θ,f–Cu∗)
= (θ,f –Cu∗+Cu¯,u¯),
whereu¯∈V(θ,f –Cu∗+I). Then
Particularly, we have
θu∗–θ(u¯) +u∗–u¯Tf(u¯) –u∗+u¯≥. () Note that (θ,f,u∗)∈N, we have
θ(u¯) –θu∗+u¯–u∗Tfu∗≥. () Then, by adding equation () and equation (), we obtain
u∗–u¯Tf(¯u) –fu∗+u¯–u∗≥, hence,
u∗–u¯Tfu∗–f(u¯)+u∗–u¯Tu∗–u¯≤.
From the monotonicity off, we getu∗–u¯ ≤. Therefore,u∗=u¯, hence,ψ◦φ(θ,f,u∗) =
(θ,f,u∗).
Theorem . The spaces N and M are homeomorphic.
Proof It follows immediately from Theorems . and ..
Remark . The homeomorphism results were shown between game spaces and the graphs of solutions mappings in games [–]. For any population gameF:X→Rn, each
Nash equilibrium pointx∗of the gameFis equivalent tox∗being a solution of variational inequality (y–x∗)TF(x∗)≤; see []. A population gameFbelongs to the class of stable
population games whenF is monotone, then, from Theorem ., we can assert that the space of stable population games is homeomorphic to the graph space of their solution mappings. Theorem . generalizes the homeomorphism result for classical variational inequalities in [].
From Theorem .,Nis homeomorphic to the spaceM. Furthermore, we see thatNis homeomorphic to the graph of a continuous mapping fromMtoK.
Letπ be the projection fromN to K andK˜ be a compact convex subset ofRn with K⊂int(f¯). Then there exists a retractionrfromK˜ toK, that is,r(x) =x,∀x∈K. From Urysohn lemma, there is a continuous mappingsfromK˜ to the closed interval [, ] such thats(x) = ,∀x∈K, ands(x) = ,∀x∈Bd(Kε), whereBd(Kε) denotes the boundary ofK˜.
Define two mappingsα,βfrom [, ]×M× ˜KtoM× ˜Ksuch that, for each (t,θ,f,x˜)∈ [, ]×M× ˜K,
αt=α(t,θ,f,x˜) = (θ,f¯,x˜) withf¯=f–s(˜x)tCx
and
βt=β(t,θ,f¯,x˜) = (θ,f,x˜) withf=f¯+s(˜x)tCx,
Theorem . Let h=π◦ψ :M→K.Then: (i)for each t∈[, ],βt◦αt andαt ◦βt are identity mappings on M× ˜K; (ii)for each t∈[, ],x˜∈Bd(K˜),αt(·,·,x˜)is a constant mapping; (iii)αis an identity on M× ˜K;αis a homeomorphism between N and the graph of h with its inverseβ.
Proof (i) For each (θ,f,x˜)∈M× ˜K, we haveβt◦αt(θ,f,x˜) =βt(θ,f –s(˜x)tCx,x˜) = (θ,f,x˜)
andαt◦βt(θ,f,x˜) =αt(θ,f+s(˜x)tCx,x˜) = (θ,f,x˜).
(ii) For eachx˜∈Bd(K˜) and (θ,f,x˜)∈M× ˜K,αt(θ,f,x˜) = (θ,f –s(x˜)tCx,x˜). Noting that
˜
x∈Bd(K˜), we haves(˜x) = , thenαt(θ,f,x˜) = (θ,f,x˜).
(iii) It is clear thatαis an identity onM× ˜K. Next, for each (θ,f,x˜)∈N, we havex˜∈ V(θ,f), thenx˜∈K. Hences(˜x) = andr(˜x) =x=x˜. Therefore,α(θ,f,x˜) = (θ,f –Cx,x˜) =
(φ(θ,f,x),x). One needs to show thatx=π◦ψ(φ(θ,f,x)). Sinceψ◦φis an identity onN, we haveπ◦ψ(φ(θ,f,x)) =π(θ,f,x) =x. Conversely, for each point (θ,f,x˜) on the graph of theh, we need to show (θ,f,x˜)∈α(N). Sincex˜=π◦ψ(θ,f), we haveψ(θ,f) = (θ,f+Cx˜,x˜),
then (θ,f+Cx˜,x˜)∈N. Hence,α(θ,f+Cx˜,x˜) = (θ,fˆ,x˜) withˆf=f+Cx˜–s(˜x)Cxandx=r(˜x).
Noting that (θ,f+Cx˜,x˜)∈N, we havex˜∈K, thenx=r(˜x) =x˜ands(˜x) = . Therefore,fˆ=f.
The proof is completed.
Remark . From Theorems . and ., the graph of set-valued mappingVis homeo-morphic toMand the graph of a continuous mapping. Easily, we see that the graph of a continuous mapping can be homeomorphic to the graph of a constant mapping. There-fore,NandMare all homeomorphic to the graph of a constant mapping. Like (un)knots and Nash dynamics [], this may contribute to variational dynamics like [].
In the following part, we generalize the results (Theorems .-.) to Hilbert spaces in relation to linear mappings.
LetKbe a compact convex subset in a Hilbert space (X,·,·), where·,·represents the inner product onX. Denote byX∗the dual space ofX(all continuous linear mappings onX). A mapping T fromK to X∗ is said to be monotone if (T(u) –T(v),u–v)≥, ∀u,v∈K, where (·,·) is the pairing ofX∗andX. A monotone mappingTis called strictly monotone if (T(u) –T(v),u–v) = impliesu=v.
LetT:K→X∗ andf ∈X∗, we consider the mixed variational inequality problem: to find au∈Ksuch that
T(u),v–u+θ(v) –θ(u)≥(f,v–u), ∀v∈K. ()
Denote byMthe set
M=
⎧ ⎪ ⎨ ⎪ ⎩(θ,T,f):
θ:K→Ris continuous and convex;
T:K→X∗is continuous onK;
f ∈X∗
⎫ ⎪ ⎬ ⎪ ⎭.
For each (θ,T,f), it is well known that this kind of mixed variational inequality problem with (θ,T,f) has a solution; ifTis strictly monotone, then the solution is unique; denote byV(θ,T,f) the set of all solutions of the problem (θ,T,f), then a set-valued mappingV
For twoα= (θ,T,f),α= (θ,T,f)∈M, measure the metric between them by
ρ(α,α) =sup x∈K
θ(x) –θ(x)+sup x∈K
T(x) –T(x)+f–f,
where, for eachg,g∈X∗, g–g= sup
y=,y∈X
(g,y) – (g,y).
LetNbe the graph ofV, that is,
N=(θ,T,f,u)∈M×K|u∈V(θ,T,f) .
Define a mappingφ:N→Msuch that, for each (θ,T,f,u)∈N,
φ(θ,T,f,u) = (θ,T–lu,f),
wherelu:K→X∗ is a mapping withlu(x) =u,·,∀x∈K. For eachT:K→X∗, letRT: K→X∗such that, for eachx∈K,
RT(x),z
=T(x),z+x,z, ∀z∈X.
Then we can check thatRTis strictly monotone. For each (θ,T,f)∈M, define a mapping
ψonMsuch that
ψ(θ,T,f) = (θ,T+lu,f,u),
whereuis the unique solution of the problem (θ,RT,f), that is,V(θ,RT,f) ={u}.
Theorem . ψis a continuous mapping from Mto N.
Proof For each (θ,T,f)∈M, we haveψ(θ,T,f) = (θ,T +lu,u), where u∈V(θ,RT,f).
Then
θ(v) –θ(u) +T(u),v–u+u,v–u
=θ(v) –θ(u) +T(u) +lu,v–u
≥(f,v–u), ∀v∈K,
it follows thatu∈V(T+lu), that is,ψmapsMontoN.
For the part thatψ is continuous onM: let{(θn,Tn,fn)}∞n= ⊂M with (θn,Tn,fn)
ρ
→ (θ,T,f)∈M. It is sufficient to show thatψ(θn,Tn,fn)→ψ(θ,T,f), that is, (θn,Tn+ lun,fn)
ρ
→(θ,T+lu,f) andun–u →, whereun∈V(θn,RTn,fn) andu∈V(θ,RT,
f), this leads to the fact that, for eachx∈K,
θn(v) –θn(un) +
Tn(un),v–un
and
θ(v) –θ(u) +
T(u),v–u+u,v–u ≥(f,v–u), ∀v∈K.
Particularly, we have
θn(u) –θn(un) +
Tn(un),u–un
+un,u–un ≥(fn,u–un) ()
and
θ(un) –θ(u) +
T(u),un–u
+u,un–u ≥(f,un–u). ()
Adding equation () and equation (), we get
θn(u) –θ(u) +θ(un) –θn(un) +
Tn(un) –T(u),u–un
≥ un–u+ (fn–f,u–un).
This is equivalent to the following:
θn(u) –θ(u) +θ(un) –θn(un) +
(Tn–T)(un) +T(un–u),u–un
≥ un–u+ (fn–f,u–un).
SinceTis monotone, we have (T(un–u),u–un)≤. Then
θn(u) –θ(u) +θ(un) –θn(un)
+(Tn–T)(un),u–un
– (fn–f,u–un)
≥ un–u.
Hence,
un–u
≤θn(u) –θ(u) +θ(un) –θn(un)
+un–u
(Tn–T)(un),
u–un
u–un
–un–u
fn–f,
u–un
u–un
,
thus,
un–u ≤sup x∈K
θn(x) –θ(x)
+un–usup x∈K
(Tn–T)(x)+un–ufn–f,
consequently, we obtain
–sup
x∈K
(Tn–T)(x)–fn–f
un–u ≤sup x∈K
θn(x) –θ(x).
Since (θn,Tn,fn)
ρ
In addition, for eachx∈K,
θn(x) –θ(x)+(Tn+lun)(x) – (T+lu)(x)+fn–f
≤θn(x) –θ(x)+(Tn–T)(x)+(lun–lu)(x)+fn–f
≤θn(x) –θ(x)+(Tn–T)(x)+un–u+fn–f →.
Note thatun–u → and (θn,Tn,fn)
ρ
→(θ,T,f), we have (θn,Tn+lun,fn)
ρ
→(θ,T+
lu,f). The proof is completed.
Theorem . φis a continuous mapping from Nto M.
Proof Let{(θn,Tn,fn,un)}∞n=⊂Nwith (θn,Tn,fn,un)→(θ,T,f,u)∈N, that is, (θn,Tn, fn)
ρ
→(θ,T,f), andun–u →. We need to show that (θn,Tn–lun,fn)
ρ
→(θ,T– lu,f). For eachx∈K, we have
ρ(θn,Tn–lun,fn), (θ,T–lu,f)
=sup
x∈K
θn(x) –θ(x)+sup x∈K
(Tn–T)(x) – (lun–lu)(x)+fn–f
≤sup
x∈K
θn(x) –θ(x)+sup x∈K
(Tn–T)(x)+fn–f
+sup
x∈K
(lun–lu)(x)
=ρ(θn,Tn,fn), (θ,T,f)
+ sup
y=,y∈X
un–u,y
≤ρ(θn,Tn,fn), (θ,T,f)
+ sup
y=,y∈X
un–uy
=ρ(θn,Tn,fn), (θ,T,f)
+un–u →,
thenφ(θn,Tn–lun,fn)→φ(θ,T–lu,f).
Theorem . φ◦ψandψ◦φare identity mappings on Mand N,respectively.
Proof (a)φ◦ψis an identity mapping onM. For each (θ,T,f)∈M, by Theorem ., we have
φ◦ψ(θ,T,f) =φ(θ,T+lu,f,u) = (θ,T+lu–lu,f) = (θ,T,f).
(b)ψ◦φis an identity mapping onN. For each (θ,T,f,u∗)∈N, we have
ψ◦φθ,T,f,u∗=ψ(θ,T–lu∗,f) = (θ,T–lu∗+lu¯,f,u¯),
whereu¯∈V(θ,RT–lu∗,f), andRT–lu∗means that, for eachx∈K,
RT–lu∗(x),z
SinceψmapsMontoN, we haveu¯∈V(θ,T–lu∗+lu¯,f), then
θ(v) –θ(u¯) +T(u¯) –lu∗(u¯) +lu¯(u¯),v–u¯
≥(f,v–u¯), ∀v∈K.
Hence,
θ(v) –θ(¯u) +(T–lu∗+lu¯)(¯u),v–u¯
≥(f,v–u¯), ∀v∈K. ()
Letv=u∗in (), then
θu∗–θ(¯u) +(T–lu∗+lu¯)(¯u),u∗–u¯
≥f,u∗–u¯. ()
Sinceu∗∈V(θ,T,f), we have
θ(u¯) –θu∗+Tu∗,u¯–u∗≥f,u¯–u∗. () Add¯u–u∗,u¯–u∗to the left-hand side of equation (),
θ(u¯) –θu∗+Tu∗,u¯–u∗+u¯–u∗,u¯–u∗≥f,u¯–u∗, that is,
θ(u¯) –θu∗+(T–lu∗+lu¯)
u∗,u¯–u∗≥f,u¯–u∗. () Adding equation () and equation (), we get
(T–lu∗+lu¯)
¯
u–u∗,u∗–u¯≥. Note thatT–lu∗+lu¯ is monotone, we have
(T–lu∗+lu¯)
¯
u–u∗,u¯–u∗= .
Furthermore,Tis strictly monotone, it follows thatu¯=u∗. Theorem . The spaces Mand Nare homeomorphic.
Proof It follows from Theorems ., ., and ..
Remark . Theorem . generalizes the homeomorphism result for the variational in-equalities to find a pointu∈Ksuch that (T(u),v–u)≥,∀v∈K, in [].
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.
Author details
1College of Science, Guizhou University, Guiyang, 550025, China.2College of Science, Guilin University of Technology,
Acknowledgements
This project is funded by National Natural Science Fund of China (11271098, 11661030, 11561013), Guangxi Natural Science Fund (2016GXNSFAA380059), and China Postdoctoral Science Foundation (2016M590905).
Received: 1 June 2016 Accepted: 13 September 2016 References
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