# On the Existence of Solutions of a Nonlocal Elliptic Equation with ap-Kirchhoff-Type Term

Authors: Francisco Julio S. A. Corrêa; Rúbia G. Nascimento
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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## Abstract

Questions on the existence of positive
solutions for the following class of elliptic problems are studied:−[M(∥u∥1,pp)]1,pΔpu=f(x,u), in Ω, u=0, on ∂Ω, where Ω⊂ℝN is a bounded smooth domain, f:Ω¯×ℝ+→ℝ and M:ℝ+→ℝ,ℝ+=[0,∞) are given functions.


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## Body

## 1. Introduction

In  this  paper,
we  are concerned with the  elliptic problem(1.1)−[M(∥u∥1,pp)]1,pΔpu=f(x,u)inΩ,u=0on∂Ω,where Ω⊂RN is a bounded smooth  domain, f:Ω¯×ℝ+→ℝ and M:ℝ+→ℝ,ℝ+=[0,∞) are given functions, Δp is  the p-Laplacian:
(1.2)Δpu=div(|∇u|p−2∇u),p>1,and ∥⋅∥1,p is the usual norm(1.3)∥u∥1,pp=∫Ω|∇u|pin the Sobolev space W01,p(Ω).Such a problem, which will be namedp-Kirchhoff
problem, is a generalization of the classical stationary Kirchhoff
equation(1.4)−[M(∥u∥2)]Δu=f(x,u)inΩ,u=0on∂Ω,where ∥u∥2=∫Ω|∇u|2 is the usual norm in H01(Ω).As it is well known, problem (1.4) is the stationary
counterpart of the hyperbolic Kirchhoff equation(1.5)ρ∂2u∂t2−(P0h+E2L∫0L|∂u∂x|2dx)∂2u∂x2=0that appeared at the first time
in the work of Kirchhoff [1],
in 1883. The equation in (1.5) is called Kirchhoff equation and it extends the
classical D'Alembert wave equation, by considering the effects of the changes
in the length of the strings during the vibrations.The interest of the mathematicians on the so-called
nonlocal problems like (1.1) (nonlocal because of the presence of the term M(∥u∥1,pp),
which implies that equations in (1.1) and (1.4) are no longer pointwise
equalities) has increased because they represent a variety of relevant physical
and engineering situations and requires a nontrivial apparatus to solve them.Particularly, problem (1.1) presents some combinations
that, at least to our knowledge, seem to be new. Indeed, in problem (1.1)
appears the nonlocal term M(∥u∥1,pp) motivated, among other things, by the above
physical situations. Furthermore, we have the presence of the p-Laplacian
operator that appears in several areas of the science such as astronomy,
glaciology, climatology, nonnewtonian fluids, petroleum extraction. Problems
that involve these two terms, M(∥u∥1,pp) and Δpu,
present several difficulties such as uniqueness, regularity, degeneracy, as we
will see throughout this paper.Beside these considerations, we also consider a case
with the presence of a singular term which poses an additional difficulty in
our study. Singular elliptic problems arise in chemical heterogeneous
catalysts, nonnewtonian fluids, nonlinear heat conduction, among other
phenomena.In casep=2,
problem (1.1) has been studied by several authors.
See [2–7], and the references therein. Particularly, this work
was motivated by [2–4, 6].We will establish existence results for problem (1.1)
by considering several classes of functions M and f.An outline of this work is as follows:In Section2, we recall some properties of the
p-Laplacian. In Section 3, we study the case in which f depends only on x∈Ω.
This is the Mp -Linear case. In Section 4, we attack problem
(1.1) when f is sublinear, that is, f(u)=uα,
for some 0<α<1.In both  Sections3 and 4, we
suitably adapt ideas developed in
[2, 3, 6].In Section5, we analyze the case in which f possesses a singular term. More precisely, f is of the following form:(1.6)f(x,u)=h(x)uγ−p+2+uα−p+2,for x∈Ω¯ and u>0,
with γ,α∈(p−2,p−1).
In this section, we have to use some arguments different from those in
[4].To end this introduction we recall thatu∈W01,p(Ω) is a weak  solution of problem (1.1) if				(1.7)[M(∥u∥1,pp)]p−1∫Ω|∇u|p−2∇u∇ϕ=∫Ωf(x,u)ϕ,∀ϕ∈W01,p(Ω).


## 2. Preliminaries on Thep-Laplacian

We will briefly expose some properties of thep-Laplacian operator defined by(2.1)Δpu≡div(∇u|p−2∇u),1<p<∞.First, we consider the problem(2.2)−Δpu=f(x),inΩ,u=0,on∂Ω,where f∈W−1,p′(Ω),p′=p/(p−1),p>1,
and the boundary condition will be understood as u∈W01,p(Ω).The following results holds as a simple consequence of
a minimization of a suitable functional.Theorem 2.1.
Iff∈W−1,p′(Ω),
then problem (2.2) has only a solution u∈W01,p(Ω) in the weak sense, namely,(2.3)[M(∥u∥1,pp)]p−1∫Ω|∇u|p−2∇u∇ϕ=∫Ωf(x)ϕ,∀ϕ∈W01,p(Ω).So, we have defined an operator(−Δp)−1:W−1,p′(Ω)→W01,p(Ω),
the inverse of −Δp,
which satisfies the following.(a)−Δp:W01,p(Ω)→W−1,p′(Ω) is uniformly continuous on bounded sets, where
such operator is defined as(2.4)−Δp:W01,p(Ω)→W−1,p′(Ω),u↦−Δpu,where(2.5)〈−Δpu,ϕ〉=∫Ω|∇u|p−2∇u∇ϕ,∀ϕ∈W01,p(Ω).Here, we denote by 〈,〉 the duality pairing between W−1,p′(Ω) and W01,p(Ω).(b) Iff∈C0(Ω¯),
then the weak solution of (2.2) belongs to C1,α(Ω¯),
for some 0<α<1,
and the mapping (−Δp)−1:C0(Ω¯)→C1(Ω¯) is compact.Theorem 2.2 (weak comparison principle).
Letu1,u2∈W01,p(Ω) satisfy(2.6)−Δpu1≤−Δpu2,inΩ,(intheweaksense)u1≤u2,on∂Ω.Then, u1≤u2 a.e. in Ω.Theorem 2.3 (a Hopf-type maximum principle).
Ifu∈C1(Ω¯)∩W01,p(Ω) and verifies(2.7)−Δpu≥0inΩ,u>0inΩ,u=0on∂Ω,then ∂u/∂η<0 on ∂Ω,
where η is the outward normal to ∂Ω.Theorem 2.4 (a strong maximum principle).
Assume thatk∈ℝ is a nonnegative number 1<p≤2 and Ω is a bounded domain ℝN.
Suppose that u∈C1(Ω¯) satisfies(2.8)−Δpu+ku≥0inΩ(intheweaksense),u≥0 and u≢0 in Ω.
Then, u>0 in Ω.
The conclusion is still true for all p>1 when k=0.We now consider the eigenvalue problem(2.9)−Δpu=λ|u|p−2uinΩ,u=0on∂Ω,where λ∈ℝ is a parameter. We say that λ∈ℝ is an eigenvalue of (2.9) if there exists a
function u∈W01,p(Ω),u≢0,
satisfying (2.9) in the weak sense. Such a function is called an eigenfunction
of (2.9) associated to the eigenvalue λ.There exists the first positive eigenvalueλ1 of problem (2.9) which is characterized as the
minimum of the Rayleigh quotient(2.10)λ1=min0≠u∈W01,p(Ω)∫Ω|∇u|p∫Ω|u|p>0.Moreover, λ1 is simple (i.e., all the associated first
eigenfunctions u are merely constant multiples of each other)
and isolated (i.e., there are no eigenvalues less
than λ1 and no eigenvalues in some right reduced
neighborhood of λ1). There is a positive (in Ω) eigenfunction φ1 corresponding to λ1.For more informations on thep-Laplacian the reader
may consult [8–11].


## 3. TheMp-Linear Case

This section is devoted to the study of the so-calledMp-Linear problem(3.1)−[M(∥u∥1,pp)]p−1Δpu=finΩ,u=0on∂Ω,where p>1,f∈W−1,p′(Ω),p′ is the conjugate exponent of p,
that is, p′=p/(p−1) and W−1,p′(Ω) is the topological dual of W01,p(Ω).The next result is an adaptation of some ideas
contained in [12, 13] (which were used
for another nonlocal problem) for problem (3.1).Theorem 3.1.
For each0≠f∈W−1,p′(Ω),
problem (3.1) possesses as many solutions as the folloing
equation:(3.2)M(t)t1/p=∥ω∥1,p,t>0,
where ω∈W01,p(Ω) is the only solution of problem (2.2).Proof.
First, let us suppose thatu∈W01,p(Ω) is a solution of (3.1), where 0≠f∈W−1,p′(Ω) is fixed. Hence(3.3)div(|∇(M(∥u∥1,pp)u)|p−2∇(M(∥u∥1,pp)u))=fand so ω=M(∥u∥1,pp)u is the solution of(3.4)−Δpω=finΩ,ω=0on∂Ω,and observe that ∥ω∥1,p=M(∥u∥1,pp)∥u∥1,p from which we conclude that t=∥u∥1,pp is a solution of (3.2).
Conversely, letω be the solution of (3.4) and suppose that t>0 is a solution of (3.2).
Define(3.5)u=t1/pω∥ω∥1,p,and so ∥u∥1,p=t1/p.
A straightforward calculation shows that(3.6)−[M(∥u∥1,pp)]p−1Δpu=−[M(t)t1/p∥ω∥1,p]p−1ΔpωinΩ,and, because t>0 is a solution of M(t)t1/p=∥ω∥1,p,
one has (M(t)t1/p/∥ω∥1,p)=1.
Hence,(3.7)−[M(∥u∥1,pp)]p−1Δpu=finΩ,u=0on∂Ω,which concludes the proof of the
theorem.Remark 3.2.
Let us suppose thatM:ℝ+→ℝ+ is a continuous function. So,(3.8)limt→0+M(t)t1/p=0.Assume, in addition, that there aret0,m0>0 such that M(t)≥m0>0 for t≥t0.
In this case,(3.9)limt→+∞M(t)t1/p=+∞.Because of the intermediate
value theorem, for ∥ω∥1,p,
there is t>0 such that(3.10)M(t)t1/p=∥ω∥1,p.In particular, if M(t)t1/p is an increasing function for t>0,
problem (3.1) possesses only a solution for each 0≠f∈W−1,p′(Ω).
As an example, we take M(t)=e−t for t≥0.
Thus, problem (3.1) possesses as many solutions as the following equation:(3.11)e−tt1/p=∥ω∥1,p.A simple exercise shows that the functiong(t)=e−tt1/p attains its maximum in 1/p and g(1/p)=1/e1/ppp.
Consequently, if ∥ω∥1,p>1/e1/ppp,
problem (3.1) does not possess any solution; if ∥ω∥1,p=1/e1/ppp,
problem (3.1) possesses exactly one solution; and if 0<∥ω∥1,p<1/e1/ppp,
problem (3.1) possesses exactly two solutions.As we see, the presence of the term[M(∥u∥1,pp)]p−1 produces great difference between problems (2.2)
and (3.1).Remark 3.3.
Let us consider the casef=0,
that is,(3.12)−[M(∥u∥1,pp)]p−1Δpu=0inΩ,u=0on∂Ω.IfM(t)>0 for all t≥0,
problem (3.12) possesses only the null solution. If M(t0)=0,
for some t0>0,
problem (3.12) possesses infinitely many solutions. Indeed, if u∈W01,p(Ω),u≠0,
we have that(3.13)v=t01/pu∥u∥1,pis a solution of (3.12).


## 4. on A Sublinear Problem

In this
section, we are going to study the sublinear problem(4.1)−[M(∥u∥1,pp)]p−1Δpu=uαinΩ,u>0inΩ,u=0on∂Ω,where α satisfies 0<α<p−1 with p>1.Such a kind of problem belongs to a class of problems
known as sublinear whose prototype is(4.2)Δu=uαinΩ,u>0inΩ,u=0on∂Ω,with 0<α<1 (note that, in this case, p=2), which has been vastly studied. See
[14–17]. For the nonlocal problem, with p=2,
we cite [2, 3, 6, 7], and the references therein.In particular, Díaz and Saá [16] study the
problem(4.3)Δpu=uαinΩ,u>0inΩ,u=0on∂Ω,1<p<∞,0<α<p−1,
and show that it possesses only a positive solution u∈W01,p(Ω)∩L∞(Ω).
Indeed, such a solution also belongs to Wloc2,2(Ω).
Our next result describes what happens with the nonlocal sublinear problem
(4.1).Theorem 4.1.
Suppose thatM:ℝ+→ℝ is a continuous function satisfying M(t)>0 for all t≥0.
Then problem (4.1) has at least as many solutions as the
equation(4.4)[M(t)]p−1⋅t(p−1−α)/p=∥v∥1,pp−1−α,t>0,where v is the solution of (4.1).Proof.
We adapt for problem (4.1) the ideas developed in
[3, 6]. Let us suppose that t>0 is a solution of the (4.4) and set γ=t1/p/∥v∥1,p which implies ∥γv∥1,p=t1/p.
So, a simple calculation shows that(4.5)[M(∥γv∥1,pp)]p−1=γα−(p−1).
Defineu=γv and let us show that such a function u is a solution of problem (4.1). Indeed, this
follows from the calculation below:(4.6)−[M(∥u∥1,pp)]p−1Δpu=−[M(∥γv∥1,pp)]p−1Δp(γv)=−[M(∥γv∥1,pp)]p−1div(|∇(γv)|p−2∇(γv))=−[M(∥γv∥1,pp)]p−1div(γp−2γ|∇v|p−2∇v)=[M(∥γv∥1,pp)]p−1γp−1(−Δpv)=γα−(p−1)γp−1(−Δpv)=γαvα=(γv)α=uαinΩ.Hence, u is a solution of problem (4.1).Remark 4.2.
Remark3.2, mutatis mutandis, remains valid for problem (4.1).


## 5. A Singular Problem Via The Galerkin Method

In this section, we will study problem(5.1)−[M(∥u∥1,pp)]p−1Δpu=h(x)uγ−p+2+uα−p+2inΩ,u>0inΩ,u=0on∂Ω,which is a singular perturbation
of problem (4.1), where Ω⊂ℝN is a bounded smooth domain, 2≤p<N and h is a suitable function defined in Ω.We will attack problem (5.1) by using the Galerkin
method which rests heavily on the following result, which is a variant of the
well-known Brouwer fixed point theorem, whose proof may be found in Lions
[18].Proposition 5.1.
SupposeF:ℝm→ℝm is a continuous function such that 〈F(ξ),ξ〉≥0 with |ξ|=r,
for some r>0,
where 〈⋅,⋅〉 denotes the usual norm in ℝm.
Then, there is ξ0∈B¯r(0) such that F(ξ0)=0.We will suppose thatM:ℝ+→ℝ+ satisfies(M1) there are m0>0 and θ1>0 such thatM(t)≥m0 if    t≥θ1;(M2)θ2=sup{t>0;M(t)=0}>0.IfM(t)≥m0>0 for all t≥0,
condition (M2) is vacuous.Theorem 5.2.
Leth:Ω¯→ℝ be a continuous function with h>0 on Ω¯,γ,α∈(p−2,p−1),2≤p<N and M:ℝ+→ℝ+ is a continuous function satisfying (M1)-(M2).
Then problem (5.1) possesses a positive solution.We will split the proof of this theorem in several
lemmas. First, for each fixedϵ>0,
we will consider the following problem:(5.2)−[M(∥u∥1,pp)]p−1Δpu=h(x)(ϵ+u)γ−p+2+uα−p+2,inΩ,u>0,inΩ,u=0,on∂Ω.In what follows, throughout Section5, we are always
supposing that M,h,α, and γ enjoy assumptions of Theorem 5.2Lemma 5.3.
For each fixedϵ>0,
problem (5.2) possesses a solution uϵ.Proof.
Let us  consider the problem(5.3)−[M(∥u∥1,pp)]p−1Δpu=h(x)(ϵ+|u|)γ−p+2+|u|α−p+2,inΩ,u=0,on∂Ω,where M+:ℝ+→ℝ is given by(5.4)M+(t)={0,if0≤t≤θ2,M(t),ift>θ2.
We now consider𝔹={e1,e2,e3,…} a Schauder's basis for W01,p(Ω) (we recall that a Schauder's basis for a
Banach space X is a sequence (en) such that to each x∈X there exists a unique sequence of scalars (αn) for which the partial sums of ∑αnen converge to x in the norm of X). For more informations on Schauder's basis
see [19] and the
references therein.
For eachm∈ℕ,
let 𝔹m=span{e1,e2,…} be the finite dimensional vector space spanned
by the functions e1,…,em.
So, each u∈𝔹m is written as u=∑j=1mξjej.
We will use on 𝔹m the norm(5.5)∥u∥m=∑j=1m|ξj|.
We note that(𝔹m,∥⋅∥m) and (ℝm,|⋅|s) are isomorphic through the following map:(5.6)T:(𝔹,∥⋅∥m)→(ℝ,|⋅|s),u=∑j=1mξjej↦T(u)=ξ=(ξ1,ξ2,…,ξm),with ∥u∥m=|ξ|s=|T(u)|s where |ξ|s=∑j=1m|ξj|.
Since, for eachm∈ℕ,𝔹m  is a finite dimensional vector space the norms ∥⋅∥m and ∥⋅∥1,p,
induced from W01,p(Ω) on 𝔹m,
are equivalent to each other and so, there exist positive constants c(m) and k(m) such that(5.7)c(m)∥u∥m≤∥u∥1,p≤k(m)∥u∥m.
We now consider, for eachm∈ℕ,
the following application(5.8)F:ℝm→ℝm,ξ↦F(ξ)=(F1(ξ),F2(ξ),…,Fm(ξ)),where(5.9)Fj(ξ)=[M+(∥u∥1,pp)]p−1∫Ω|∇u|p−2∇u∇ej−∫Ωh(x)ej(ϵ+|u|)γ−p+2−∫Ω|u|α−p+2ej,j=1,…,m.
A simple calculation leads us to(5.10)〈F(ξ),ξ〉=[M+(∥u∥1,pp)]p−1∫Ω|∇u|p−2∇u∇u−∫Ωh(x)u(ϵ+|u|)γ−p+2−∫Ω|u|α−p+2u,=[M+(∥u∥1,pp)]p−1∥u∥1,pp−∫Ωh(x)u(ϵ+|u|)γ−p+2−∫Ω|u|α−p+2u.We now note that(5.11)∫Ωh(x)u(ϵ+|u|)γ−p+2≤∥h∥∞∫Ω|u|ϵγ−p+2≤Cϵ∥u∥1,p,and because p−2<α<p−1,
we have 1<α−p+3<2,
and so(5.12)∫Ω|u|α−p+2u≤∫Ω|u|α−p+3≤C∥u∥1,pα−p+3.Hence,(5.13)〈F(ξ),ξ〉≥[M(∥u∥1,pp)]p−1∥u∥1,pp−Cϵ∥u∥1,p−C∥u∥1,pα−p+3,and from (M1),
for ∥u∥1,pp≥θ1,
we have M+(∥u∥1,pp)=M(∥u∥1,pp)≥m0>0.
Thus,(5.14)〈F(ξ),ξ〉≥m0p−1∥u∥1,pp−Cϵ∥u∥1,p−C∥u∥1,pα−p+3≥m0p−1[c(m)]p∥u∥mp−Cϵk(m)∥u∥m−[k(m)]α−p+3C∥u∥mα−p+3≥m0p−1C(m)|ξ|sp−Nϵ(m)|ξ|s−Q(m)|ξ|sα−p+3.
We now fixρm>0 and for ξ∈ℝm,|ξ|s=ρm and noting that 1<α−p+3<2≤p,
we obtain(5.15)〈F(ξ),ξ〉≥m0p−1C(m)ρmp−Nϵ(m)ρm−Q(m)ρmα−p+3.So, if ρm is large enough, we have(5.16)〈F(ξ),ξ〉>0if|ξ|s=ρm,which implies(5.17)〈F(ξ),ξ〉>0if|ξ|=ρm∗,where ρm∗>0,
because |⋅| and |⋅|s are equivalent in ℝm.
From Proposition5.1 there is ξm∈ℝm,
with |ξ|≤ρm∗,
such that F(ξm)=0 and note that |ξm|s≤ρm for some ρm is large enough.
Through the isometric identification of(ℝm,|⋅|s) with (𝔹m,∥⋅∥m),
we find (um)⊂𝔹m,∥um∥m≤ρm,
that is, ∥um∥1,p≤ρ¯m,
such that(5.18)[M+(∥um∥1,pp)]p−1∫Ω|∇um|p−2∇um∇ej−∫Ωh(x)ej(ϵ+|um|)γ−p+2−∫Ω|um|α−p+2ej=0,for each j=1,…,m,
which gives us(5.19)[M+(∥um∥1,pp)]p−1∫Ω|∇um|p−2∇um∇ω−∫Ωh(x)ω(ϵ+|um|)γ−p+2−∫Ω|um|α−p+2ω=0,for all ω∈𝔹m.
Since um∈𝔹m,(5.20)[M+(∥um∥1,pp)]p−1∥um∥1,pp−∫Ωh(x)um(ϵ+|um|)γ−p+2−∫Ω|um|α−p+2um=0.
Because(5.21)∫Ωh(x)um(ϵ+|um|)γ−p+2≤∥u∥∞ϵγ−p+2∫Ω|um|≤Cϵ∥um∥1,p,∫Ω|um|α−p+2um≤C∥um∥1,pα−p+3,
we obtain(5.22)[M(∥um∥1,pp)]p−1∥um∥1,pp≤Cϵ∥um∥1,p+C∥um∥1,pα−p+3.
Let us show that the sequence(∥um∥1,p) is bounded. Indeed, suppose, on the contrary,
that (∥um∥1,p) is not bounded. So, up to a subsequence, we
may suppose that ∥um∥1,p→+∞.
From condition (M1) and inequality (5.22) we obtain(5.23)0<m0p−1∥um∥1,pp≤Cϵ∥um∥1,p+C∥um∥1,pα−p+3,which give us(5.24)0<m0p−1≤Cϵ∥um∥1,pp−1+C∥um∥1,p2p−α−3.Since ∥um∥1,p→+∞ we arrive in(5.25)0<m0p−1≤0,which is impossible. Thus, (∥um∥1,p) is bounded. Consequently, up to a subsequence,(5.26)∥um∥1,pp→t0,um⇀uinW01,p(Ω),um→uinLq(Ω),1≤q<p∗,um(x)→u(x)a.e.Ω.And in view of continuity of M(5.27)[M+(∥um∥1,pp)]p−1→[M+(t0)]p−1.
We now fixl≤m and so 𝔹l⊂𝔹m.
For φ∈𝔹l we have(5.28)[M+(∥um∥1,pp)]p−1∫Ω|∇um|p−2∇um∇φ=∫Ωh(x)φ(ϵ+|um|)γ−p+2+∫Ω|um|α−p+2φ,for all φ∈𝔹l.
We now remark that(5.29)|h(x)φ(ϵ+|um|)γ−p+2|≤Cϵγ−p+2|φ|∈L1(Ω),h(x)φ(ϵ+|um(x)|)γ−p+2→h(x)φ(ϵ+|u(x)|)γ−p+2a.e.Ω.By the Lebesgue dominated
convergence theorem(5.30)∫Ωh(x)φ(ϵ+|u|)γ−p+2→∫Ωh(x)φ(ϵ+|u|)γ−p+2.Furthermore, because um→u∈Lp(Ω),
we have(5.31)|um|α−p+2→|u|α−p+2inLp/(α−p+2)(Ω),and because Ω is a bounded domain and p<p/(α−p+2) we have Lp/(α−p+2)(Ω)⊂Lp(Ω) and so(5.32)|um|α−p+2→|u|α−p+2inLp(Ω).Consequently,(5.33)|um(x)|α−p+2→|u(x)|α−p+2a.e.inΩ,and there is g∈Lp(Ω) such that(5.34)|um(x)|α−p+2≤g(x)a.e.inΩ∀m∈ℕ.Thus,(5.35)||um|α−p+2φ|≤g|φ|∈L1(Ω),and using again the Lebesgue
dominated convergence theorem(5.36)∫Ω|um|α−p+2φ→∫Ω|u|α−p+2φ∀φ∈𝔹l.
We will now consider the following claim, whose proof
will be postponed to Section6.
Claim 1.
The following  convergence holds(5.37)[M+(∥um∥1,pp)]p−1∫Ω|∇um|p−2∇um∇φ→[M+(t0)]p−1∫Ω|∇u|p−2∇u∇φ,for all φ∈W01,p(Ω).
In the inequality (5.28) we make m→∞ and we use (5.27),
(5.30), and (5.36) to obtain(5.38)[M+(t0)]p−1∫Ω|∇u|p−2∇u∇φ=∫Ωh(x)φ(ϵ+|u|)γ−p+2+∫Ω|u|α−p+2,for all φ∈𝔹l.
Sincel∈ℕ is arbitrary, the above inequality holds true
for all φ∈W01,p(Ω).
Because of this M+(t0)>0 and so M+(t0)=M(t0),
which implies(5.39)[M(t0)]p−1∫Ω|∇u|p−2∇u∇φ=∫Ωh(x)φ(ϵ+|u|)γ−p+2+∫Ω|u|α−p+2,for all φ∈W01,p(Ω).
Takingu as a test function in the inequality (5.39), we
get(5.40)[M(t0)]p−1∥u∥1,pp=∫Ωh(x)u(ϵ+|u|)γ−p+2+∫Ω|u|α−p+2u.
We now takeφ=um in (5.28) to obtain(5.41)[M(∥um∥1,pp)]p−1∥um∥1,pp=∫Ωh(x)um(ϵ+|um|)γ−p+2+∫Ω|um|α−p+2um.
Since(5.42)|h(x)um(ϵ+|um|)γ−p+2|≤Cϵγ−p+2|um|and because um→u∈Lq(Ω),1≤q<p∗,
it follows that |um(x)|→|u(x)| a.e. in Ω and there is g∈L1(Ω) such that |um|≤g a.e. in Ω for all m∈ℕ.
Consequently,(5.43)|h(x)um(ϵ+|um|)γ−p+2|≤Cϵγ−p+2|um|≤Cϵγ−p+2g∈L1(Ω).
Reasoning as before, we obtain(5.44)∫Ωh(x)um(ϵ+|um|)γ−p+2→∫Ωh(x)u(ϵ+|u|)γ−p+2,∫Ω|um|α−p+2um→∫Ω|u|α−p+2u.
Taking limits on both sides of the equality
(5.41) and
using ∥um∥1,pp→t0,
(5.27), (5.44), we obtain(5.45)[M+(t0)]p−1t0=∫Ωh(x)u(ϵ+|u|)γ−p+2+∫Ω|u|α−p+2u.
Comparing this last equality with the one in (5.40) we
obtain t0=∥u∥1,pp because M+(t0)>0.
Then, from (5.39)(5.46)[M(∥u∥1,pp)]p−1∫Ω|∇u|p−2∇u∇φ=∫Ωh(x)φ(ϵ+|u|)γ−p+2+∫Ω|u|α−p+2φfor all φ∈W01,p(Ω).
This shows thatu∈W01,p(Ω) is a weak solution of the auxiliary problem
(5.2) which is positive by virtue of the maximum principle. This proves the
lemma.In what follows, for eachn∈ℕ,
we set ϵ=1/n and u1/n=un where u1/n is the solution obtained in the last
lemma.Lemma 5.4.
There existsδ>0 such that [M(∥un∥1,pp)]p−1≥δ>0,
for all n∈ℕ.Proof.
We will reason by contradiction. Suppose
that(5.47)liminf[M(∥un∥1,pp)]p−1=0.If it is the case, the sequence (∥un∥1,pp) is bounded because, on the contrary, we would
have ∥un∥1,pp>θ1,
perhaps for a subsequence, and so(5.48)[M(∥un∥1,pp)]p−1≥m0p−1>0,and this would imply(5.49)liminf[M(∥un∥1,pp)]p−1≥m0p−1>0,which is impossible in view of
(5.47). Therefore, up to  subsequences,(5.50)∥un∥1,pp→θ0,un⇀inW01,p(Ω),un(x)→u(x)a.e.inΩ.Since M is continuous, we have(5.51)0=liminf[M(∥un∥1,pp)]p−1=lim[M(∥un∥1,pp)]p−1=[M(θ0)]p−1.
We now note that(5.52)h(x)(1/n+t)γ−p+2+tα−p+2≥h(x)(1+t)γ−p+2+tα−p+2≥C(1+t)γ−p+2+tα−p+2≥C˜[1(1+t)γ−p+2+tα−p+2]≥C0>0,for all x∈Ω¯ and t≥0,
because the function t↦1/(1+t)γ−p+2+tα−p+2,t≥0,
attains a positive minimum.
Since(5.53)−[M(∥un∥1,pp)]p−1=h(x)(1/n+un)γ−p+2+unα−p+2≥C0>0,inΩ,un=0,on∂Ω.Taking φ∈C01(Ω¯),φ>0 in Ω,
as a test function in the  last equality, we obtain(5.54)[M(∥un∥1,pp)]p−1∫Ω|∇un|p−2∇un∇φ≥C0∫Ωφ>0,and so M(∥un∥1,pp)]p−1>0.
Taking limit on the above expression we get(5.55)0=[M(θ0)]p−1∫Ω|∇u|p−2∇u∇φ≥C0∫Ωφ>0,which is impossible. This
completes the proof of the lemma.Lemma 5.5.
The sequence(∥un∥1,pp) is bounded, where un is as above.Proof.
First of all we note that(5.56)[M(∥un∥1,pp)]p−1∥un∥1,pp=∫Ωh(x)un(1/n+un)γ−p+2+∫Ωunα−p+3,∫Ωh(x)un(1/n+un)γ−p+2≤C1∥un∥1,pp−γ−1,∫Ωunα−p+3≤C2∥un∥1,pα−p+3,where C1 and C2 are positive constants independent of n.
Consequently,(5.57)δ∥un∥1,pp≤[M(∥un∥1,pp)]p−1∥un∥1,pp≤C1∥un∥1,pp−γ−1+C2∥un∥1,pα−p+3.Since p−γ−1<1,α−p+3<2≤p,
we conclude that (∥un∥1,pp) is bounded.As a consequence of the preceding lemma we
have(5.58)0<δ≤[M(∥un∥1,pp)]p−1≤M∞,∀n∈ℕ.Lemma 5.6.
The sequence(un),
obtained in the last lemma, converges to a solution of problem (5.1).Proof.
As(un) is a bounded sequence in W01,p(Ω) we have, up to subsequence, that(5.59)∥un∥1,pp→t0>0,un⇀uinW01,p(Ω),un→inLq(Ω),1≤q<p∗,un(x)→u(x)a.e.Ω.From the continuity of M,(5.60)M(∥un∥1,pp)→M(t0).
Letψ1>0 be an eigenfunction of (−Δp,W01,p(Ω)) associated to first eigenvalue λ1 and satisfying(5.61)C0>λ1M∞p−1ψ1p−1,∀x∈Ω¯,where C0 is the constant obtained in (5.54).
Consequently,(5.62)−[M(∥un∥1,pp)]p−1Δpun=h(x)(1/n+un)γ−p+2+unα−p+2inΩ,≥C0>λ1M∞p−1ψ1p−1inΩ,un=ψ1=0on∂Ω.Since(5.63)−[M(∥un∥1,pp)]p−1Δpun≥C0>λ1M∞p−1ψ1p−1inΩ,it follows that(5.64)−Δp([M(∥un∥1,pp)]un)>−Δp(M∞ψ1)inΩ,and as [M(∥un∥1,pp)]un=M∞ψ1=0 on ∂Ω.
By the comparison principle,(5.65)[M(∥un∥1,pp)]un>M∞ψ1inΩ,and from (5.58)(5.66)un(x)>M∞ψ1(x)M∞1/(p−1)inΩ.From which we conclude that un(x)↛0 for each x∈Ω.
Next, we will use the following.
Hardy-Sobolev Inequality. If u∈W01,p(Ω) and 1<p≤N,
then u/ψ1r∈Lr(Ω),
where 1/r=1/p−(1−τ)/N,0≤τ≤1,
and(5.67)∥uψ1r∥Lr≤∥∇u∥Lp,where C>0 is a constant and is an eigenfunction of (−Δp,W01,p(Ω)) associated to the first eigenvalue λ1.Whose proof may
be found in [20].
As(5.68)|h(x)φ(1/n+un)γ−p+2|≤h(x)|φ||un|γ−p+2,and using (5.66) it follows(5.69)h(x)|φ||un|γ−p+2<∥h∥∞C|φ|ψ1γ−p+2.From the Hardy-Sobolev
inequality(5.70)∥h∥∞C|φ|ψ1γ−p+2∈L1(Ω).Furthermore,(5.71)h(x)φ(1/n+un(x))γ−p+2→h(x)φu(x)γ−p+2inΩ,and so, by the Lebesgue
dominated convergence theorem(5.72)∫Ωh(x)φ(1/n+un(x))γ−p+2→∫Ωh(x)φu(x)γ−p+2,∀φ∈W01,p(Ω).
SinceΩ is bounded and p<p/(α−p+2),
we have Lp/(α−p+2)↪Lp(Ω) and so(5.73)unα−p+2→uα−p+2inLp(Ω).Since Ω is bounded and p<p/(α−p+2),
we have Lp/(α−p+2)↪Lp(Ω) and so(5.74)un(x)α−p+2→u(x)α−p+2a.e.inΩ,and there exists g∈Lp(Ω) such that(5.75)unα−p+2≤ga.e.inΩ,∀n∈ℕ.Hence,(5.76)|unα−p+2φ|≤g|φ|∈L1(Ω),and using again the Lebesgue
dominated convergence theorem, we obtain(5.77)∫Ωunα−p+2φ→∫Ωuα−p+2φ,for all φ∈W01,p(Ω).
As(un) is solution of the auxiliary
problem(5.78)[M(∥un∥1,pp)]p−1∫Ω|∇un|p−2∇un∇φ=∫Ωh(x)φ(1/n+un)γ−p+2+∫Ωunα−p+2φ,for all φ∈W01,p(Ω),
taking limits on both sides of the above expression, using the same reasoning
as in the proof of the claim and the convergences in (5.60),
(5.72), and
(5.77)
we obtain(5.79)[M(t0)]p−1∫Ω|∇u|p−2∇u∇φ=∫Ωh(x)φuγ−p+2+∫Ωuα−p+2,for all φ∈W01,p(Ω).
Using u as a test function in the last
expression(5.80)[M(t0)]p−1∥u∥1,pp=∫Ωh(x)up−γ−1+∫Ωuα−p+3.Taking φ=un in the expression (5.78), it follows
that(5.81)[M(∥un∥1,pp)]p−1∥un∥1,pp=∫Ωh(x)un(1/n+un)γ−p+2+∫Ωunα−p+3.Noticing that un→u in Lq(Ω),1≤q<p∗,
and 1<1/(p−γ−1) we have L1/(p−γ−1)(Ω)↪L1(Ω) and so(5.82)un→uinL1/(p−γ−1)(Ω),un(x)→u(x)a.e.inΩ,and there exists ω∈L1/(p−γ−1)(Ω) such that(5.83)0≤un(x)≤ω(x)a.e.inΩ,∀n∈ℕ.Thus,(5.84)|h(x)un(1/n+un)γ−p+2|≤∥h∥∞unp−γ−1∈L1(Ω),h(x)un(1/n+un)γ−p+2→h(x)u(x)p−γ−1a.e.inΩ.By the Lebesgue dominated
convergence theorem(5.85)∫Ωh(x)un(1/n+un)γ−p+2→∫Ωh(x)u(x)p−γ−1,and because 1<α−p+3<2,
it follows that(5.86)∫Ωunα−p+3→∫Ωuα−p+3.
We now take
limits on both sides of (5.81), by using
(5.60),
(5.85), and
(5.86), to obtain(5.87)[M(t0)]p−1t0=∫Ωh(x)uuγ−p+2+∫Ωuα−p+3,that is,(5.88)[M(t0)]p−1t0=∫Ωh(x)up−γ−1+∫Ωuα−p+3.
Comparing (5.80),
(5.88) and using [M(t0)]p−1>0,
we get ∥u∥1,pp=t0.
Therefore, from (5.79)(5.89)[M(∥u∥1,pp)]p−1∫Ω|∇u|p−2∇u∇φ=∫Ωh(x)φuγ−p+2+∫Ωuα−p+2φ,for all φ∈W01,p(Ω),
which shows that u is a weak solution of (5.1).


## 6. Proof of The Claim

We recall that
we have to prove that(6.1)[M+(∥u∥1,pp)]p−1∫Ω|∇um|p−2∇um∇φ→[M+(t0)]p−1∫Ω|∇u|p−2∇u∇φfor all φ∈W01,p(Ω).First of all, we claim that(−Δpum)⊂(W01,p(Ω))′ is a bounded sequence. Indeed,(6.2)|〈−Δpum,v〉|=|∫Ω|∇um|p−2∇um∇v|≤∫Ω|∇um|p−1|∇v|,and because um∈𝔹m⊂W01,p(Ω),
we have |∇um|∈Lp(Ω) and |∇um|p−1∈Lp/(p−1)(Ω).
Applying the Hölder inequality, by using the exponents p and p/(p−1),
it follows that(6.3)|〈−Δpum,v〉|≤(∫Ω|∇um|p)(p−1)/p(∫Ω|∇v|p)1/p=∥um∥1,pp∥v∥1,p,which implies(6.4)∥−Δpum∥(W01,p(Ω))′≤∥u∥1,pp≤Cϵ,where Cϵ>0 is a constant, because the boundedness of (∥um∥1,p) was proved in Lemma 5.5.SinceW01,p(Ω) is a separable Banach space and (−Δum) is a bounded sequence in (W01,p(Ω))′ then, up to a subsequence,(6.5)−Δpum⇀∗χin(W01,p(Ω))′,that is,(6.6)〈−Δpum,ψ〉→〈χ,ψ〉∀ψ∈W01,p(Ω),which is equivalent
to(6.7)∫Ω|∇um|p−2∇um∇ψ→〈χ,ψ〉∀ψ∈W01,p(Ω).Since(6.8)[M+(∥um∥1,pp)]p−1→[M+(t0)]p−1,it follows from (6.7)
that(6.9)[M+(∥um∥1,pp)]p−1∫Ω|∇um|p−2∇um∇ψ→[M+(t0)]p−1〈χ,ψ〉,for all ψ∈W01,p(Ω).
Taking limits on both sides of (5.28) and
using the convergences (5.30) and
(5.36), we obtain(6.10)[M+(∥um∥1,pp)]p−1∫Ω|∇um|p−2∇um∇ψ→∫Ωh(x)ψ(ϵ+|u|)γ−p+2+∫Ω|u|α−p+2ψ,for all ψ∈W01,p(Ω).
So, from (6.9),
(6.10) and the uniqueness of the limit(6.11)[M+(t0)]p−1〈χ,ψ〉=∫Ωh(x)ψ(ϵ+|u|)γ−p+2+∫Ω|u|α−p+2ψ,for all ψ∈W01,p(Ω).Using the monotonicity of the operator(−Δp),
that is,(6.12)〈−Δpω−(−Δpv),ω−v〉≥0∀ω,v∈W01,p(Ω),we have(6.13)[M+(∥ω∥1,pp)]p−1∫Ω(|∇ω|p−2∇ω−|∇v|p−2∇v,∇ω−∇v)dx≥0.Taking ω=um and v=ψ in the last expression, we obtain(6.14)[M+(∥um∥1,pp)]p−1∫Ω(|∇um|p−2∇um−|∇ψ|p−2∇ψ,∇um−∇ψ)dx≥0,and so(6.15)[M+(∥um∥1,pp)]p−1∥um∥1,pp−[M+(∥um∥1,pp)]p−1∫Ω|∇um|p−2∇um∇ψ−[M+(∥um∥1,pp)]p−1∫Ω|∇ψ|p−2∇ψ∇um+[M+(∥um∥1,pp)]p−1∫Ω|∇ψ|p≥0.In (5.28) we take φ=um to obtain(6.16)[M+(∥um∥1,pp)]p−1∥um∥1,pp=∫Ωh(x)um(ϵ+|um|)γ−p+2+∫Ω|um|α−p+2um.Consequently,(6.17)∫Ωh(x)um(ϵ+|um|)γ−p+2+∫Ω|um|α−p+2um−[M+(∥um∥1,pp)]p−1∫Ω|∇um|p−2∇um∇ψ−[M+(∥um∥1,pp)]p−1∫Ω|∇ψ|p−2∇ψ∇um+[M+(∥um∥1,pp)]p−1∫Ω|∇ψ|p≥0,using the definition of −Δp(6.18)∫Ωh(x)um(ϵ+|um|)γ−p+2+∫Ω|um|α−p+2um−[M+(∥um∥1,pp)]p−1〈−Δpum,ψ〉−[M+(∥um∥1,pp)]p−1〈−Δpψ,um〉+[M+(∥um∥1,pp)]p−1〈−Δpψ,ψ〉≥0,taking limits as m→∞ and using (5.44)  we have(6.19)∫Ωh(x)u(ϵ+|u|)γ−p+2+∫Ω|u|α−p+2u−[M+(∥u∥1,pp)]p−1〈χ,ψ〉−[M+(t0)]p−1〈−Δpψ,u〉+[M+(t0)]p−1〈−Δpψ,ψ〉≥0.We note that(6.20)[M+(t0)]p−1〈χ,ψ〉=∫Ωh(x)ψ(ϵ+|u|)γ−p+2+∫Ω|u|α−p+2,for all ψ∈𝔹l⊂W01,p(Ω),
and so(6.21)[M+(t0)]p−1〈χ,u〉=∫Ωh(x)u(ϵ+|u|)γ−p+2+∫Ω|u|α−p+2u.Therefore,(6.22)[M+(t0)]p−1〈χ,u〉−[M+(t0)]p−1〈χ,ψ〉−[M+(t0)]p−1〈−Δpψ,u〉+[M+(t0)]p−1〈−Δpψ,ψ〉≥0,that is,(6.23)〈χ−(−Δpψ),u−ψ〉≥0.Setting(6.24)ψ=u−λφ,λ>0,φ∈𝔹l,we obtain(6.25)〈χ,−[−Δp(u−λφ)],φ〉≥0,∀λ>0,φ∈𝔹l,taking limits on both sides of
the above expression as λ→0 and using the continuity of the operator −Δp(6.26)〈χ,−(−Δpu),φ〉≥0,∀φ∈𝔹l.So,(6.27)〈χ−(−Δpu),−φ〉≥0,and we obtain(6.28)〈χ−(−Δpu),φ〉≤0,∀φ∈𝔹l.Hence,(6.29)〈χ−(−Δpu),φ〉=0∀φ∈𝔹l,which implies χ=−Δpu,
because l is arbitrary. This means that the above
equality is valid for all φ∈W01,p(Ω).
From (6.10) and
(6.11) we conclude that(6.30)[M+(∥um∥1,pp)]p−1∫Ω|∇um|p−2∇um∇ψ→[M+(t0)]p−1〈−Δu,ψ〉,for all ψ∈W01,p(Ω).
Hence,(6.31)[M+(∥um∥1,pp)]p−1∫Ω|∇um|p−2∇um∇ψ→[M+(t0)]p−1∫Ω|∇u|p−2∇u∇ψfor all ψ∈W01,p(Ω).
This concludes the proof of the claim.



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*Source: 364085-2009-01-05.xml*