Electronic Journal of Qualitative Theory of Differential Equations
2005, No. 18, 1-12,http://www.math.u-szeged.hu/ejqtde/
EXISTENCE OF SOLUTIONS TO NONLOCAL AND SINGULAR VARIATIONAL ELLIPTIC INEQUALITY VIA GALERKIN
METHOD
FRANCISCO JULIO S. A. CORR ˆEA & SILVANO B. DE MENEZES
Abstract. In this article, we study the existence of solutions for nonlocal variational elliptic inequality
−M(kuk2)∆u≥f(x, u)
Making use of the penalized method and Galerkin approximations, we estab-lish existence theorems for both cases whenM is continuous and whenM is discontinuous.
1. Introduction
Let Ω⊂RN be a bounded smooth domain with boundary∂Ω. Let us consider
the Sobolev spacesL2(Ω), H1
0(Ω) whose inner products and norms will be denoted
by (,),|,|,((,)),k,k, respectively. We haveH1
0(Ω)⊂L2(Ω) dense and continuous.
Then the duals (L2(Ω))′ ⊂ (H1
0(Ω))′ also dense and continuous. If we identify
L2(Ω) = (L2(Ω))′ we have
H1
0(Ω)⊂L
2(Ω)⊂H−1(Ω),
whereH−1(Ω) is the dual ofH1
0(Ω). Throughout this paper, let us represent byK
a closed convex set ofL2(Ω), with 0∈K, which has the following property:
(H1) There exists a contractionρ:R−→R(e.g., |ρ(λ1)−ρ(λ2)|≤ |λ1−λ2|)
withρ(0) = 0 such that (PKv)(x) =ρ(v(x)),∀v ∈L2(Ω), wherePK is the
projec-tion operator fromL2(Ω) intoK.
For example, let us considerK={v∈L2(Ω); a≤v(x)≤b a.e. inΩ}
where−∞ ≤a≤0≤b≤ ∞. We defineρ(λ) as follows
ρ(λ) =
λ f or a < λ < b
b f or λ≥b (if b is f inite) a f or λ≤a (if a is f inite)
In this paper we study some questions related to the existence of solutions for the nonlocal elliptic variational inequality:
u∈K: (−M(kuk2)∆u, v)≥(f, v), f or all v∈K∩(H1
0(Ω)∩H
2(Ω)) (1.1)
where f ∈ H1
0(Ω) and M : R → R is a function whose behavior will be stated
problem (1.1), and its nonlinear counterpart, possesses a solution. This inequality has called our attention because the operator
Lu:=M(kuk2)∆u
contains the nonlocal term M(kuk2) which poses some interesting mathematical
questions. Also the operatorL appears in the Kirchhoff equation, which arises in nonlinear vibrations, namely
utt−M
Z
Ω
|∇u|2dx∆u=f(x, u) in Ω×(0, T),
u= 0 on∂Ω×(0, T), u(x,0) =u0(x), ut(x,0) =u1(x).
The mathematical aspects of this model were largely investigated. See, for example, Arosio-Spagnolo [2], Hazoya-Yamada [11], Lions [16], Pohozhaev [18]. A survey of the results about the mathematical aspects of Kirchhoff model can be found in Medeiros-Limaco-Menezes [14]. Unilateral problems for nonlinear operators of the Kirchhoff type were initially studied by Kludnev [12], Larkin [13], Medeiros-Milla Miranda [17] among others.
Recently, Alves-Corrˆea [1] focused their attention on problem related with (1.1) in case M(t) ≥ m0 > 0, for all t ≥ 0, where m0 is a constant. Among other
things they studied the aboveM-linear problem (1.1) whereM, besides the strict positivity mentioned before, satisfies the following assumption:
The functionH :R→Rwith
H(t) =M(t2)t
is monotone andH(R) =R.
In a previous paper, Menezes-Corrˆea [8] proved a similar results by allowingMto attain negative values andM(t)≥m0>0 only fortlarge enough. This is possible
thanks to a device explored by Alves-de Figueiredo [3], who use Galerkin method to attack a non-variational elliptic system. The technique can be conveniently adapted to problems such as (1.1). In this way we improve substantially the existence result on the above problem mainly because our assumptions onM are weakened. Indeed, we may also consider the case in whichM possesses a singularity. The methodology used in our proof consists in transforming, by penalty, the inequality (1.1) into a family of equations depending of a parameterǫ >0 and apply Galerkin’s method. In the application of Galerkin’s methods we use the sharp angle lemma(see Lions [15, p.53]). This paper is organized as follows: Section 2 is devoted to the study of (1.1) in the continuous case. In Section 3 the inequality (1.1) is studied in case M possesses a discontinuity. In Section 4 we analyze another type of variational inequality.
2. The M-linear Problem: Continuous Case
In this section we are concerned with the M-linear problem (1.1) where f ∈
H1
(M1) There exists a positive numberm0such thatM(t)≥m0, for allt≥0.
We have
Theorem 2.1. Assume that and (M1) and (H1) hold. Then for any choice of
06=f ∈H1
0(Ω) the problem (1.1) admits at least one solution.
The proof of Theorem 2.1 is given by the penalty method. In fact, let us represent byβ the operator from L2(Ω) intoL2(Ω) defined byβ =I−P
K, e.g., (βv)(x) =
v(x)−ρ(v(x)) for allv∈L2(Ω).The operatorβis monotone and Lipschitzian. The
next result can be found in Haraux [10] p. 58.
Lemma 2.2. Letg:R→Rbe a Lipschitzian and increasing function withg(0) =
0. Then(g(u),−∆u)≥0 for allu∈H1
0(Ω)∩H 2(Ω).
We have thatg(s) =s−ρ(s) is under the condition of the Lemma 2.2, then
(β(v),−∆v)≥0f or all v∈H01(Ω)∩H 2
(Ω). (2.1)
The penalized problem associated to the problem (1.1), consists in givenǫ >0, find uǫsolution in Ω of the problem
−M(kuǫk2)∆uǫ+
1
ǫβ(uǫ) =f in Ω, uǫ= 0 on∂Ω,
(2.2)
wheref is given in H1 0(Ω).
The existence of one solution of the penalized problem (2.2) is give by the
Theorem 2.3. Assume that(M1)hold. Then, for any06=f ∈H1
0(Ω)there exists at least oneuǫ∈H01(Ω)∩H2(Ω), solution of the problem (2.2).
Proof. We employ the Galerkin Method by using the sharp angle lemma. Let us consider the Hilbertian basis of spectral objects (ej)j∈Nand (λj)j∈Nfor the operator
−∆ inH1
0(Ω), cf. Brezis [5]. We know that the eigenvectors (ej)j∈Nare orthonormal
complete inL2(Ω) and complete inH1
0(Ω)∩H2(Ω). For eachm∈Nconsider Vm= span{e1, . . . , em},
the subspace ofH1
0(Ω)∩H2(Ω) generated bymeigenvectorse1, e2, ..., em. Ifuǫm ∈
Vm, then
uǫm = m
X
j=1
ξjej with real ξj, 1≤j ≤m.
We will consideruminstead ofuǫm . The approximate problem consists in finding
a solutionum∈Vmof the system of algebraic equations
M(kumk2)((um, ei)) +
1
ǫ(β(uǫ), ei) = (f, ei), i= 1, . . . , m. (2.3) We need to prove that (2.3) has a solutionum∈Vm. To this end, we will consider
the vectorη= (ηi)1≤i≤m ofRmdefined by
ηi =M(kumk2)((um, ei)) +
1
ǫ(β(um), ei)−(f, ei), i= 1, . . . , m.
Let ξ = (ξi)1≤i≤m be the components of the vector um of Vm. The mapping
P :Rm →Rm defined byP ξ =η is continuous. If we prove thathP ξ, ξi ≥0 for
kξkRn=r, with an appropriater, it will follow by the sharp angle lemma that there
exists aξ in the ballBr(0)⊂Rmsuch that P ξ= 0. This implies the existence of
a solution to (2.3). In fact, we have
hP ξ, ξi=
m
X
1
ξiηi=M(kumk)((um, um)) +
1
ǫ(β(um), um)−(f, um).
Using (M1), H¨older and Poincar´e inequalities and observing that (β(um), um)≥0,
we get
hP ξ, ξi ≥m0kumk2−Ckfkkumk
We can consider r large enough that (P ξ, ξ) ≥ 0 for kξkRm = r. Then P ξ = 0
for some ξ ∈ Br(0), which implies that system (2.3) has a solution um ∈ Vm
corresponding to thisξ. Thus, there isum∈Vm,
kumk ≤r, (2.4)
rdoes not depend onmandǫ, such that
M(kumk2)((um, ei)) +
1
ǫ(β(um), ei) = (f, ei), i= 1, . . . , m, which implies that
M(kumk2)((um, ω)) +
1
ǫ(β(um), ω) = (f, ω), for allω∈Vm. (2.5) Because (kumk2) is a bounded real sequence and M is continuous one has
kumk2→˜t0, (2.6)
for some ˜t0≥0, and
um⇀ uinH01(Ω), um→uin L2(Ω), M(kumk2)→M(˜t0), (2.7)
perhaps for a subsequence.
Takek≤m,Vk ⊂Vm. Fixk and letm→ ∞in equation (2.5) to obtain
M(˜t0)((u, ω)) +
1
ǫ(β(u), ω) = (f, ω), for allω∈(H
1 0 ∩H
2
)(Ω). (2.8)
Since−∆ej=λjej we takeω=−∆um in (2.8). We obtain
M(te0)(∆um,∆um) +
1
ǫ(β(um),−∆um) = (∇f,∇um) (2.9) By (2.1), we obtain
|∆um|2≤C(|∇f|2+|∇um|2). (2.10)
Sincef ∈H1
0(Ω) and by (2.6), we have that
|∆um|2≤C (2.11)
whereCdoes not depend onmandǫ. By (2.7) and (2.11) and by compact imbed-ding of (H1
0(Ω)∩H2(Ω))⊂H01(Ω) we deduce that
The continuity ofM and convergence (2.7) and (2.12) permit to pass to the limit in (2.5). We obtain
M(kuk2)((u, v)) +1
ǫ(β(u), v) = (f, v), for allv∈H
1
0(Ω)∩H2(Ω). (2.13)
Thusuǫ is a weak solution of problem (2.2) and the proof of Theorem 2.3 is
com-plete.
Proof. of Theorem 2.1
Let (ǫn)n∈Nbe a sequence of real numbers such that
0< ǫn<1f or all n∈N and lim
n→∞ǫn= 0.
For eachn∈N, we get functionuǫnwhich satisfies Theorem 2.3. Since the estimates
were uniform onǫ and n, we can see that there exists a subsequence ofuǫn, again
calleduǫn, and a functionu∈H01(Ω)∩H2(Ω) such that
uǫm →uin H01(Ω). (2.14)
Considerv in H1
0(Ω)∩H
2(Ω) with vbelongs toK. By (2.5), we have
(−M(kuǫnk2)∆uǫn−f, v) = (−
1 ǫn
β(uǫn), v) (2.15)
and
(−M(kuǫnk2)∆uǫn−f,−uǫn) = (−
1 ǫn
β(uǫn),−uǫn). (2.16)
Follows of (2.15) and (2.16) that
(−M(kuǫnk2)∆uǫn−f, v−uǫn) =
1
ǫ(−β(uǫn), v−uǫn)
= 1
ǫ(β(v)−β(uǫn), v−uǫn)≥0,
(2.17)
becausev∈K and monotonicity ofβ. Hence
(−M(kuǫnk2)∆uǫn−f, v) + (f, uǫn)≥(−M(kuǫnk2)∆uǫn, uǫn), (2.18)
But
(−M(kuǫnk2)∆uǫn, uǫn) =M(kuǫnk2)(∇uǫn,∇uǫn) =
M(kuǫnk2)(∇(uǫn−u),∇(uǫn−u)) +M(kuǫnk2)(∇uǫn,∇u)+
M(kuǫnk2)(∇u,∇(uǫn−u))≥M(kuǫnk2)(∇uǫn,∇u)+
M(kuǫnk2)(∇u,∇(uǫn−u)),
(2.19)
becauseM(kuǫnk2)(∇(uǫn−u),∇(uǫn−u))≥0.Then,
lim inf[−M(kuǫnk2)(∆uǫn, uǫn)]≥ −M(kuk2)(∇u, u). (2.20)
By (2.18) and (2.20), we obtain (1.1).
In order to proveu∈K we observe that from (2.5), withω =uǫm, that
0≤(β(uǫn), uǫn)≤ǫC (2.21)
therefore
(β(uǫn), uǫn) converges to zero. (2.22)
Letv∈H1
0(Ω) be arbitrarily fixed; then the inequality
(β(uǫn)−β(v), uǫn−v)≥0
yields (β(v), u−v)≤0, hence (β(u−λω)), ω)≤0 with the choice of v=u−λω withλ >0 andω∈H1
0(Ω). By hemicontinuity ofβ we can letλ→0+ and obtain
(β(u), ω)≥0, hence β(u) = 0 by the arbitrariness ofω. Thereforeu∈K.
3. The M-linear Problem: A Discontinuous Case
In this section we concentrate our atention on problem (1.1) whenM possesses a discontinuity. More precisely, we study problem (1.1) with M : R/{θ} → R
continuous such that
(M2) limt→θ+M(t) = limt→θ−M(t) = +∞
(M3) lim supt→+∞M(t2)t= +∞.
Theorem 3.1. Assume that (M1)–(M3) and (H1) hold. Then for any choice of
f ∈H1
0(Ω)the problem (1.1)admits at least one solution solution.
The proof of Theorem 3.1 is given as in the proof of Theorem 2.1. We will formulate the penalized problem, associated with the variational inequality (1.1), as follows. Givenǫ >0, find a functionuǫ∈H01(Ω) solution of the problem
−M(kuǫk2)∆uǫ+
1
ǫβ(uǫ) =f in Ω, uǫ= 0 on∂Ω,
(3.1)
wheref is given in H1 0(Ω).
The existence of one solution of the penalized problem (3.1) is given by the
Theorem 3.2. Assume that (M1)−(M3) hold. Then, for any 0 6=f ∈ H1 0(Ω) there exists at least oneuǫ∈L2(Ω), solution of the problem (3.1).
Proof. We first consider the sequence of functionsMn :R→Rgiven by
Mn(t) =
(
n, θ−δ′
n≤t≤θ+δ′′n,
M(t), t≤θ−δ′
n or t≥θ+δn′′,
forn > m0, whereθ−δn′ andθ+δ′′n,δn′,δ′′n>0, are, respectively, the points closest
toθ, at left and at right, so that
M(θ−δ′n) =M(θ+δ ′′ n) =n.
We point out that, in this case,δ′
n, δn′′→0 asn→ ∞.
Taken > m0 and observe that the horizontal linesy=ncross the graph ofM.
Hence Mn is continuous and satisfies (M1), for each n > m0. In view of this, for
eachnlike above, there isun ∈H01(Ω)
T
H2(Ω) satisfying
Mn(kunk2)((un, ω)) +
1
ǫ(β(un), ω) = (f, ω), for allω∈H01(Ω)∩H
2
(Ω).
Takingω=un in the above equation one has
Mn(kunk2)kunk2+
1
ǫ(β(un), un) = (f, un), and so
Mn(kunk2)kunk ≤ kfk,
since 0 ≤ (β(un), un). Because of (M3) the sequence (kunk) must be bounded.
Hence
un⇀ u inH01(Ω),
un→u inL2(Ω),
kunk2→θ0, for someθ0,
(3.3)
perhaps for subsequences. We note that if (Mn(kunk2) converges its limit is different
of zero. Suppose thatkunk2→θ. Ifkunk2> θ+δn′′orkunk2< θ−δ′n, for infinitely
manyn, we would getMn(kunk2) =M(kunk2), for suchn, and so
M(kunk2)kunk2≤(f, un) ⇒ +∞ ≤(f, u)
which is a contradiction. On the other hand, if there are infinitely manynso that θ−δ′
n≤ kunk2 ≤θ+δ′′n ⇒ Mn(kunk2) =n and sonkunk2 ≤(f, un) ⇒ +∞ ≤
(f, u) and we arrive again in a contradiction.
Consequentlykunk2→θ06=θwhich implies that fornlarge enough
kunk2< θ−δn′ or kunk2> θ+δn′′
and soMn(kunk2) =M(kunk2) which yields
M(kunk2)((un, ω)) +
1
ǫ(β(un), ω) = (f, ω), ∀ω∈H
1
0(Ω)∩H
2(Ω). (3.4)
Taking ω = −∆un in (3.4), using(2.1) and arguments of compactness as in the
proof of theorem 2.1, we obtain that
un→u ∈H01(Ω) (3.5)
The estimates (3.3) and (3.5) are sufficient to pass to the limit in the approximate equation and to obtain
M(kuk2)((u, ω)) +1
ǫ(β(u), ω) = (f, ω), ∀ω∈H
1
0(Ω)∩H
2(Ω). (3.6)
Proof. of Theorem 3.1
As in the proof of Theorem 2.1, let (ǫn)n∈N be a sequence of real numbers such,
that
0< ǫn<1f or all n∈N and lim
n→∞ǫn= 0.
Applying Theorem 3.1, for each n∈ N, we get a function uǫn ∈ H01(Ω)∩H2(Ω)
which satisfies
M(kuǫnk2)((uǫn, ω)) +
1 ǫn
(β(uǫn), ω) = (f, ω), ∀ω∈H01(Ω)∩H
2(Ω). (3.7)
Since the estimates were uniform onǫ and n, we can see that there exists a sub-sequence of (uǫn), again called (uǫn), and a function u ∈ H01(Ω)∩H2(Ω) such
that
uǫn→uinH01(Ω). (3.8)
We note that kuǫnk does not converges to θ. In fact, we assume by contradiction
thatkuǫnk →θ. By (3.7) and (3.8), withω=uǫn, we get
M(kuǫnk2)kuǫnk ≤C
By hyphotese (M2), lettingn→ ∞, we have a contradiction. Thus M(kuǫnk2)→M(kuk2),
sincekuǫnk → kuk andM is continuous in an neighboard ofkuk 6=θ.
Considerv in H1
0(Ω) withv belonging toK. By (3.7), we have
(−M(kuǫnk2)∆uǫn−f, v) = (−
1
ǫβ(uǫn), v) (3.9) and
(−M(kuǫnk2)∆uǫn−f,−uǫn) = (−
1
ǫβ(uǫn),−uǫn). (3.10) Follows of (3.9) and (3.10) that
(−M(kuǫnk2)∆uǫn−f, v−uǫn) =
1
ǫ(−β(uǫn), v−uǫn)
= 1
ǫ(β(v)−β(uǫn), v−uǫn)≥0,
(3.11)
becausev∈K and monotonicity ofβ. Hence
(−M(kuǫnk2)∆uǫn−f, v) + (f, uǫn)≥(−M(kuǫnk2)∆uǫn, uǫn). (3.12)
As in the proof of Theorem 2.1, letting n → ∞, we obtain that u satisfies the conditions of Theorem 3.2.
4. Another Nonlocal Problem
Next, we make some remarks on a nonlocal problem which is a slight generaliza-tion of one studied by Chipot-Lovat [6] and Chipot-Rodrigues [7]. More precisely, the above authors studied the problem
−a Z
Ω
u∆u=f in Ω,
u= 0 on∂Ω,
(4.1)
where Ω ⊂ RN is a bounded domain, N ≥ 1, and a : R → (0,+∞) is a given
function. Equation (4.1) is the stationary version of the parabolic problem
ut−a
Z
Ω
u(x, t)dx∆u=f in Ω×(0, T),
Here T is some arbitrary time and u represents, for instance, the density of a population subject to spreading. See [6, 7] for more details. In particular, [6] studies problem (4.1), withf ∈H−1(Ω), and proves the following result.
Proposition 4.1. Let a:R→(0,+∞)be a positive function,f ∈H−1(Ω). Then problem (4.1) has as many solutionsµ as the equation
a(µ)µ=hhf, ϕii,
whereϕis the function(unique) satisfying
−∆ϕ= 1 inΩ, ϕ= 0 on∂Ω.
In the present work, we consider the variational inequality
u∈K: (−a Z
Ω
|u|q∆u, ω)≥(f, ω) f or all v∈K∩(H1
0(Ω)∩H
2(Ω)) (4.2)
where K and f are as before and 1 < q <2N/(N−2), N ≥3. When q = 2 we have the well known Carrier model.
Theorem 4.2. If t 7→ a(t) is a decreasing and continuous function, for t ≥ 0,
limt→+∞a(tq)t= +∞andt7→a(tq)t is injective, fort≥0, then, for each06=f ∈
H−1(Ω), problem (4.2) possesses a weak solution.
The proof of Theorem 4.2 is given as in the proof of Theorem 2.1. We will formulate the penalized problem, associated with the variational inequality (4.2), as follows: Givenǫ >0, find a functionuǫ∈H01(Ω) solution of the problem
−a Z
Ω
|uǫ|q∆uǫ+
1
ǫβ(uǫ) =f in Ω, uǫ= 0 on∂Ω,
(4.3)
wheref is given in H1 0(Ω).
The existence of one solution of the penalized problem (4.3) is given by the
Theorem 4.3. Under the hypotheses of Theorem 4.2, then for each 0 6= f ∈
H−1(Ω)there exists at least one u
ǫ∈H01(Ω) solution of the problem (4.3). Proof. As in the proof of Theorem 2.1, letP={e1, . . . , em, . . .}be an orthonormal
basis of the Hilbert space H1
0(Ω). For each m∈Nconsider the finite dimensional
Hilbert space
Vm= span{e1, . . . , em},
P :Rm→Rm be the functionP(ξ) = (P1(ξ), . . . , Pm(ξ)), where
Pi(ξ) =a(kukqq)((u, ei)) +
1
ǫ(β(u), ei)− hhf, eiii, i= 1, . . . , m withu=Pmj=1ξjej and the identifications ofRmandVmmentioned before. So
Pi(ξ) =a(kukqq)ξi+
1
ǫ(β(u), ei)− hhf, eiii, i= 1, . . . , m and then
< P(ξ), ξ >=a(kukqq)kuk 2
+1
ǫ((β(u), u))− hhf, uii
We have to show that there isr >0 so thathP(ξ), ξi ≥0, for allkξkRm =rinVm.
Suppose, on the contrary, that for eachr >0 there isur∈Vm such thatkurk=r
and
hP(ξr), ξri<0, ξr↔ur.
Takingr=n∈Nwe obtain a sequence (un),kunk=n,un∈Vmand
hP(un), uni=a(kunkqq)kunk2+
1
ǫ(β(un), un)− hhf, unii<0 and so
a(kunkqq)kunk< Ckfk, ∀n= 1,2, . . . ,
since(β(un), un) > 0. Because of the continuous immersion H01(Ω) ⊂ Lq(Ω) one
getskukq≤Ckukand the monotonicity ofayieldsa(kukqq)≥a(Ckukq) and so
a(Ckunkq)kunk< Ckfk.
In view of limt→+∞a(tq)t= +∞one has thatkunk ≤C,∀n∈N, which contradicts
kunk=n. So, there isrm>0 such thathF(ξ), ξi ≥0, for all|ξ|=rm. In view of
the sharp angle lemma there isum ∈Vm, kumk ≤ rm such that Pi(um) = 0, i=
1, . . . , m, that is,
a kumkqq
((um, ω)) +
1
ǫ(β(um), ω) =hhf, ωii, ∀ω∈Vm. (4.4) Reasoning as before, by using the facts that t → a(t) is decreasing for t ≥ 0 , limt→+∞a(tq)t= +∞and
R
Ωβ(u)u >0 we conclude thatkumk ≤C,∀m= 1,2. . .
for some constantCthat does not depend onm. Hence,um⇀ uinH01(Ω),um→u
in Lq(Ω), 1 < q < 2N
N−2, and so kumkq → kukq. Taking limits on both sides of
equation (4.4) we conclude that the function uis a solution of problem (4.3).
Remark 4.4. The function
a(t) = 1 t2β+ 1,
whereβ andq are related by 2βq <1, satisfies the assumptions of Theorem 4.2
Proof. of Theorem 4.2
Let (ǫn)n∈Nbe a sequence of real numbers such, that
0< ǫn<1f or all n∈N and lim
n→∞ǫn= 0.
For eachn∈N, we get functionuǫnwhich satisfies Theorem 4.3. Since the estimates
were uniform onǫ and n, we can see that there exists a subsequence ofuǫn, again
calleduǫn, and a functionu∈H01(Ω)∩L
q(Ω), 1≤q≤ 2N
N−2 such that
uǫn⇀ uin H01(Ω)∩H
2(Ω) (4.5)
uǫn→uinLq(Ω), 1≤q≤
2N
N−2 (4.6)
kuǫnkq → kukq (4.7)
Considerv in H1
(−a(kuǫnkpq)∆uǫn−f, v−uǫn) = (−
1
ǫβ(uǫn), v−uǫn) = 1
ǫ(β(v)−β(uǫn), v−uǫn)≥0
(4.8)
becausev∈K and monotonicity ofβ.
As in the proof of Theorem 2.1, whenǫn →0 we obtain thatusatisfies the
condi-tions of Theorem 4.2.
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(Received December 12, 2004)
Francisco Julio S. A. Corrˆea
Departamento de Matem´atica-CCEN, Universidade Federal do Par´a, 66.075-110 Bel´em Par´a Brazil
E-mail address: [email protected] Silvano D. B. Menezes
Departamento de Matem´atica-CCEN, Universidade Federal do Par´a, 66.075-110 Bel´em Par´a Brazil
