R E S E A R C H
Open Access
A fixed point operator for systems of
vector
p-Laplacian with singular weights
Xianghui Xu
1, Inbo Sim
2and Yong-Hoon Lee
1**Correspondence:
1Department of Mathematics,
Pusan National University, Busan, 609-735, Republic of Korea Full list of author information is available at the end of the article
Abstract
In this paper, after establishing a fixed point operator for a strongly coupled vector
p-Laplacian with a singular and sign-changing weight function, which may not be integrable, we investigate the existence for the Dirichlet boundary value problems of strongly coupled vectorp-Laplacian systems with a nonlinear term consisting of Hadamard product. The proofs are mainly based on topological degree arguments and the global continuation theorem.
MSC: 34B16; 34B18
Keywords: p-Laplacian system; sign-changing weight; existence; nontrivial solution
1 Introduction
We are concerned with the existence of nontrivial solutions for strongly coupled nonlinear differential systems of the form
(Pλ)
⎧ ⎨ ⎩
–p(u)=λh(t)·f(u), t∈(, ),
u() = =u(),
wherep> , p :RN →RN is defined by p(x) =|x|p–x,λ> is a parameter,h(t) =
(h(t), . . . ,hN(t)) withhi: (, )→R, andf(u) = (f(u), . . . ,fN(u)) with continuousfi:RN→
R. Here we denotex·y= (xy,xy, . . . ,xNyN) the Hadamard product ofxandyinRN.
Thus, problem (Pλ) can be rewritten as
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
–(|u(t)|p–u
(t))=λh(t)f(u), ..
.
–(|u(t)|p–u
N(t))=λhN(t)fN(u), t∈(, ),
ui() = =ui(), i= , . . . ,N.
Throughout the paper, we denote by| · |the absolute value onRor the Euclidean norm onRN and by·,·the inner product onRNand defineϕ
p:R→Rbyϕp(s) =|s|p–s. For a
weight functionh, we assume thathi∈H, where
H=
g∈Lloc(, ),R
ϕp–
s
g(τ) dτ
ds+
ϕp– s
g(τ) dτ
ds<∞
.
It is well known thatL(, )H. Thus, a function inHmay have stronger singularity at the boundary than a function inL(, ) (see examples in Section ). Ifh
i∈Hfor all
i= , , . . . ,N, then|h| ∈H. In this sense, we shall denoteh∈Hwheneverhi∈Hfor all
i= , , . . . ,N.
Scalar equations or systems ofp-Laplacian-like problem (Pλ) appear in various appli-cations, which describe reaction-diffusion systems, nonlinear elasticity, glaciology, pop-ulation biology, combustion theory, and non-Newtonian fluids (see [–]). The study on the existence of solutions forp-Laplacian scalar equations or systems or more generalized Laplacian systems has attracted much attention recently (see [–] and the references therein).
Among their general setup, a solution operator for nonlinearp-Laplacian systems was introduced in the pioneering works of Manásevich and Mawhin [, ]. They applied the solution operator to study the existence of solutions for systems of strongly coupled vector p-Laplacian-like operators withL-Carathéodory nonlinear perturbations.
We see that theL-Carathéodory condition in problem (Pλ) corresponds to the condi-tionh∈L((, ),RN). As a generalization of theL-Carathéodory condition, it is interest-ing to consider the caseh∈H. Since our problem involves systems of strongly coupled differential operators and the weight functionhmay change sign, related studies are not known yet, as far as the authors know. Recently, for a scalar equation of (Pλ), Sim and Lee [] established a new solution operator and proved an existence result by the global continuation theorem.
Thus, the goal of this paper is to get an existence result for (Pλ) where the differential operator is related to strongly coupled vector p-Laplacian and the weight function has stronger singularity at the boundary thanLand sign-changing. The novelty of the paper is providing a new solution operator, which is the most generalized so far.
This paper is organized as follows. In Section , we derive a solution operator for prob-lem (W)+(D) withg∈H. In Section , we prove the compactness of the solution operator for (Pλ) withλ= . In Section , we show the existence of solutions and give some illus-trative examples, which satisfy all assumptions in the paper and are not given in other studies.
2 A fixed point operator
In this section, we construct a solution operator for a strongly coupled vectorp-Laplacian. Let us consider a problem of the form
(W) –p
w=g(t), t∈(, ), (D) w() = =w(),
where g∈H. Sinceg may not be inL((, ),RN), the solution of (W)+(D) may not be
inC([, ],RN). For an example of a simple scalar case, takeg(t) = (p– )t–| +lnt|p–, p> ; theng∈/L(, ), butg∈H, and the solutionuis given byu(t) = –tlnt, which is not inC[, ].
So by a solution to this problem we mean a functionw∈C([, ],RN)∩C((, ),RN)
withp(w) absolutely continuous that satisfies equations (W)+(D).
Remark . From the definition ofpandϕpwe get, for anyx,y∈RN,
p–(x+y) ≤ϕp–|x|+|y|≤Cp
ϕp–|x|+ϕp–|y|,
where
Cp= ⎧ ⎨ ⎩
, p> ,
–p–p, <p≤.
Remark . By the homogeneity ofϕ–
p we can deduce that ifh∈H, thenα·h∈Hfor all α∈C([, ],RN).
Letwbe a solution of (W)+(D). Then integrating both sides of (W) on the intervals [s,] and [,s] fors∈(,] ands∈[, ), respectively, we find that (W)+(D) is equivalent to
⎧ ⎨ ⎩
w(s) =–
p (a+
s g(τ)dτ), w() = , s∈(,
], w(s) =p–(a–s
g(τ)dτ), w() = , s∈[
, ),
(.)
wherea=p(w()). Applying Remark . withx=aandy=
s g(τ)dτ, we get
p–
a+
s
g(τ)dτ ≤ϕp–
|a|+
s
g(τ) dτ
≤Cpϕp–
|a|+Cpϕp–
s
g(τ) dτ
.
Sinceg∈H, we know that
p–
a+
s
g(τ)dτ
∈L
,
, p–
a–
s
g(τ)dτ
∈L
,
.
Thus, we may integrate both sides of (.) on the interval [,t] fort∈[,] and on the interval [t, ] fort∈[, ], and we get
w(t) =
⎧ ⎨ ⎩ t
p–(a+
s g(τ)dτ)ds, t∈[,],
t p–(–a+ s
g(τ)dτ)ds, t∈[
, ].
We need to check thatw( –
) =w( +
). Fora∈RN, define
Gg(a) =
p–
a+
s
g(τ)dτ
ds–
p–
–a+
s
g(τ)dτ
ds. (.)
Then the functionGg :RN →RN is well defined. IfGg has a unique zero, thenw(
– ) = w(
+
). For this, we give the following lemma.
Lemma . For given g∈H,the function Gg defined in(.)has a unique zero a=a(g)
Proof I. Existence. We claim that there exists r> such that Gg(a),a> for alla∈ ∂Br()⊂RN. If the claim is valid, then we consider the homotopy
h(λ,a) =λa+ ( –λ)Gg(a) forλ∈[, ].
By the claim,
h(λ,a),a=λa,a+ ( –λ)Gg(a),a
>
for anya∈∂Br(),λ∈[, ]. Taking=Br(), we see that the Brouwer degreedB(h(λ,a), , ) is well defined, and by the homotopy invariance property we get
dB
Gg(·),,
=dB
h(,a),, =dB
h(,a),, =dB(id,, ) =
since ∈. This completes the proof of the existence of a zero ofGg. We now prove the
claim. For convenience, we denote
Hg(a) p– a+ s
g(τ)dτ
ds, Wg(a) p– –a+ s
g(τ)dτ
ds.
Then it suffices to show that there existsr> such thatHg(a),a> andWg(a),a<
for alla∈∂Br()⊂RN. Indeed, we have
Hg(a),a = p– a+ s
g(τ)dτ ,a ds = δ p– a+ s
g(τ)dτ ,a ds+ δ p– a+ s
g(τ)dτ
,a
ds,
whereδ∈(,
) will be determined later. Sinceg∈H, both integrations are well defined, and we denote
H,δ δ p– a+ s
g(τ)dτ ,a ds, H,δ δ p– a+ s
g(τ)dτ
,a
ds.
We first considerH,δ. Since
s
g(τ)dτ ≤
s
g(τ) dτ,
applying Remark ., we obtain
|H,δ| ≤
δ p– a+ s
g(τ)dτ
,a ds≤
δ p– a+ s
g(τ)dτ |a|ds
≤ δ
ϕp–
|a|+
s
g(τ)dτ
|a|ds≤
δ
ϕp–
|a|+
s
g(τ) dτ
≤ δ
Cp
ϕp–|a|+ϕ–p
s
g(τ) dτ
|a|ds
=Cpδ|a|p
∗ +Cp
δ
ϕ–p
s
g(τ) dτ
ds
|a|,
wherep∗=pp–. Thus, we get
H,δ≥–Cpδ|a|p
∗ –Cp
δ
ϕ–p
s
g(τ) dτ
ds
|a|
=|a|p∗
–Cpδ–Cp δ
ϕp–
s
g(τ) dτ
ds
|a|p∗–
. (.)
Now we considerH,δ. Sincep(x),x=|x|p,x∈RN, we see that
p–(x),x= p–(x) p=|x|(p∗–)p=|x|p∗.
Moreover, for s∈[δ,], |
s g(τ)dτ| ≤
δ |g(τ)|dτ <∞; thus, denoting
δ |g(τ)|dτ Mδ, we obtain
H,δ=
δ p– a+ s
g(τ)dτ
,a+
s
g(τ)dτ ds – δ p– a+ s
g(τ)dτ
,
s
g(τ)dτ ds ≥ δ a+ s
g(τ)dτ p∗
ds–Mδ
δ p– a+ s
g(τ)dτ ds.
Sincep∗> and
a+
s
g(τ)dτ ≥ |a|–
s
g(τ)dτ ≥ |a|–Mδ
fors∈[δ,], taking|a|large enough to satisfy|a|–Mδ> , we get
H,δ≥
δ
|a|–Mδ
p∗ ds–Mδ
δ
|a|+Mδ
p∗– ds
=
–δ
|a|–Mδ
p∗ –Mδ
|a|+Mδ
p∗–
=|a|p∗
–δ
–Mδ
|a|
p∗ –Mδ
+Mδ
|a|
p∗–
|a|
. (.)
Combining (.) and (.), we get that
Hg(a),a
≥ |a|p∗
–δ
–Mδ
|a|
p∗ –Mδ
·
+Mδ
|a|
p∗–
·|
a|
–Cpδ–Cp δ
ϕp–
s
g(τ) dτ
ds·
|a|p∗–
Sinceg∈H, we have thatϕ–
p (
s |g(τ)|dτ)∈L(,δ]. Choosingδ> sufficiently small and |a|=rsufficiently large, we can make the right-hand side of (.) strictly greater than . This implies that there existsr> such thatHg(a),a> for alla∈∂Br(). Applying
a similar argument, we can show thatWg(a), –a> for alla∈∂Br(). Therefore, we
conclude that there existsr> such thatGg(a),a> for alla∈∂Br(), and the claim is
proved.
II. Uniqueness. Suppose thataandaare two distinct zeros ofGg. Then
Gg(a) –Gg(a),a–a
= .
On the contrary,
Gg(a) –Gg(a),a–a
=Hg(a) –Hg(a),a–a
+W(a) –W(a),a–a
= p– a+ s
g(τ)dτ
–p–
a+
s
g(τ)dτ
,a–a
ds + p– –a+ s
g(τ)dτ
–p–
–a+
s
g(τ)dτ
,a–a
ds.
Therefore, we get
Gg(a) –Gg(a),a–a
= p– a+ s
g(τ)dτ
–p–
a+
s
g(τ)dτ , a+ s
g(τ)dτ – a+ s
g(τ)dτ ds + p– –a+ s
g(τ)dτ
–p–
–a+
s
g(τ)dτ , –a+ s
g(τ)dτ – –a+ s
g(τ)dτ
ds>
sincep–(x) –p–(y),x–y> for allx,y∈RN,x=y. This contradiction completes the
proof of uniqueness.
Lemma . implies that ifg∈H, then the solutionwof (W)+(D) can be represented by
w(t) =
⎧ ⎨ ⎩ t
p–(a(g) +
s g(τ)dτ)ds, t∈[,],
t p–(–a(g) + s
g(τ)dτ)ds, t∈[
, ],
(.)
wherea(g)∈RN satisfies
p– a(g) + s
g(τ)dτ ds= p– –a(g) + s
g(τ)dτ
We note thata(g) is determined uniquely up tog, and from this uniqueness property the following corollary is obvious.
Corollary . Let g∈H,Then,as a function of g,a is homogeneous,that is,
a(λg) =λa(g) for allλ∈R.
On the other hand, it is not hard to see that the functionwdefined in (.) satisfies w∈C([, ],RN)∩C((, ),RN),
p(w) is absolutely continuous on (, ), andwsatisfies
(W)+(D). Therefore, we conclude that ifg∈H, thenwis a solution of (W)+(D) if and only ifwsatisfies (.).
3 Compactness of the fixed point operator
Consider a nonlinear problem of the form
(P)
⎧ ⎨ ⎩
–p(u)=h(t)·f(u), t∈(, ),
u() = =u(),
whereh∈Handf ∈C(RN,RN). We note that, by Remark .,h·f(u)∈H. Let us apply the solution representation for (W)+(D) given in (.) replacinggwithh·f(u). Then we may rewrite problem (P) equivalently as
u=T(u),
whereT:C([, ],RN)→C([, ],RN) is defined by
T(u)(t) =
⎧ ⎨ ⎩ t
–
p (a(h·f(u)) +
s h(τ)·f(u(τ))dτ)ds, t∈[,
],
t p–(–a(h·f(u)) + s
h(τ)·f(u(τ))dτ)ds, t∈[
, ].
In this section, we prove that the solution operatorTis completely continuous. For this, we need two lemmas about the properties ofa(h·f(u)). Sincehandf are fixed, we regard a(h·f(u)) as a function ofu∈C([, ],RN).
Lemma . The function a sends bounded sets in C([, ],RN)into bounded sets inRN.
Proof Assume that a sequence{un}is bounded inC([, ],RN). Let us denoteana(h·
f(un)) andGnGh·f(un). Suppose that{an}is unbounded inRN. Then there exists a
sub-sequence{ank}such that|ank| → ∞ask→ ∞. Since eachankis a zero ofGnk, we see that
Gnk(ank),ank= for allk. On the other hand, by the same calculation as in the proof of Lemma . we obtain
Hnk(ank),ank
≥ |ank|
p∗
–δ
–MHδ
|ank|
p∗ –MHδ
·
+MHδ
|ank|
p∗–
·
|ank|
–Cpδ–Cpϕp–(M) δ
ϕp–
s
h(τ) dτ
ds·
|ank|p ∗–
,
whereM=supk∈Nf(unk)∞andHδ=
Apply-ing a similar argument forWnk, we conclude thatGnk(ank),ank> for sufficiently largek,
and this contradiction completes the proof.
Remark . IfBis a bounded set inC([, ],RN), then{a(h·v)|v∈B}is also bounded inRN. The proof is similar to that of Lemma . by replacingMwithsupv∈Bv∞.
Lemma . The function a:C([, ],RN)→RN is continuous.
Proof Assume thatun→uinC([, ],RN). Then for the continuity ofa, we need to show
thata(h·f(un))→a(h·f(u)) inRN asn→ ∞. Denote againana(h·f(un)). We know
that{an}is bounded inRN by Lemma .; thus, it has a convergent subsequence{ank}, which converges to, say,aˆ∈RN. We first claim that
p–
ˆ
a+
s
h(τ)·fu(τ)dτ
ds
=
p–
–aˆ+
s
h(τ)·fu(τ)dτ
ds. (.)
Indeed, let us takeK=supn∈N|an|,M=supn∈Nf(un)∞and fixs∈(,]. Then we get h(τ)·funk(τ) ≤M h(τ)
for allτ ∈[s,]. Moreover,hi∈Lloc(, ) implies|h| ∈L[s,]. Thus, by the continuity of
p–and applying the Lebesgue dominated convergence theorem componentwise, we get
lim
k→∞
–
p
ank+
s
h(τ)·funk(τ)
dτ
=p–
ˆ
a+
s
h(τ)·fu(τ)dτ
.
Similarly, fork∈N,
p–
ank+
s
h(τ)·funk(τ)
dτ ≤A+Bϕp–
s
h(τ) dτ
,
whereA=Cpϕ–p (K) andB=Cpϕp–(M). Sinceh∈H, the right-hand side of the last
in-equality is inL(,]. Thus, applying the Lebesgue dominated convergence theorem com-ponentwise again, we have
lim
k→∞
p–
ank+
s
h(τ)·funk(τ)
dτ
ds
=
p–
ˆ
a+
s
h(τ)·fu(τ)dτ
ds. (.)
By the same argument, for fixeds∈[, ), we also get
lim
k→∞
p–
–ank+
s
h(τ)·funk(τ)
dτ
ds
=
p–
–aˆ+
s
h(τ)·fu(τ)dτ
Moreover, by the definition ofank given in (.), we know that
p–
ank+
s
h(τ)·funk(τ)
dτ
ds
=
p–
–ank+
s
h(τ)·funk(τ)
dτ
ds.
This implies that both limits in (.) and (.) are the same, and thus (.) is valid. Equation (.) implies thataˆ=a(h·f(u)) by the uniqueness ofa. So we conclude thatˆ limk→∞ank(= a(h·f(unk))) =a(h·f(u)) inR
N. It is not hard to see by the standard subsequence argument
thatlimn→∞an(=a(h·f(un))) =a(h·f(u)), and the proof is done. Remark . Ifvn∈C([, ],RN) withvn→vasn→ ∞, thena(h·vn)→a(h·v) asn→ ∞. In particular, if v= , thena(h·vn)→ asn→ ∞. The proof is similar to that of
Lemma . by replacingMwithsupv∈Bv∞.
Lemma . The operator T:C([, ],RN)→C([, ],RN)is completely continuous.
Proof The continuity ofT is easily verified mainly by Lemma . and the Lebesgue dom-inated convergence theorem. LetB be a bounded subset ofC([, ],RN). Then by the
Arzelà-Ascoli theorem, it suffices to show thatT(B) is uniformly bounded and equicon-tinuous. TakeMB=supu∈Bf(u)∞,KB=supu∈B|a(h·f(u))|, and denoteaua(h·f(u)).
Then, fort∈(,],
T(u)(t) ≤
t
p–
au+
s
h(τ)·fu(τ)dτ ds
≤ t
ϕp–
KB+MB
s
h(τ) dτ
ds
≤
Cpϕ –
p (KB) +Cpϕp–(MB)
ϕp–
s
h(τ) dτ
ds.
Sinceh∈H, we see that the last bound is independent ofu∈Bandt∈(,
]. The bound on the interval [, ) can be obtained similarly, and thusT(B) is uniformly bounded.
To show the equicontinuity ofT(B), lett,t∈[, ] witht<t. Case .t,t∈[,] ort,t∈[, ]. We have
T(u)(t) –T(u)(t)
≤ t
t
p–
au+
s
h(τ)·fu(τ)dτ ds
≤Cpϕp–(KB)(t–t) +Cpϕ–p (MB) t
t
ϕp–
s
h(τ) dτ
ds.
The bound is independent ofu∈Bandϕ–p (
s |h(τ)|dτ)∈L
(,
Case . <t≤ <t< . Sincetandtcan be considered sufficiently close, without loss of generality, we assume that ≤t≤<t≤. Then, by the definition ofT,
T(u)(t) =
t
p–
au+
s
h(τ)·fu(τ)dτ ds = p–
au+
s
h(τ)·fu(τ)dτ ds – t p–
au+
s
h(τ)·fu(τ)dτ
ds
and
T(u)(t) =
t
p–
–au+ s
h(τ)·fu(τ)dτ ds = p–
–au+ s
h(τ)·fu(τ)dτ ds – t p–
–au+ s
h(τ)·fu(τ)dτ
ds.
Since, by the definition ofau,
p–
au+
s
h(τ)·fu(τ)dτ ds = p–
–au+ s
h(τ)·fu(τ)dτ
ds,
we get
T(u)(t) –T(u)(t)
= t p–
–au+ s
h(τ)·fu(τ)dτ ds – t p–
au+
s
h(τ)·fu(τ)dτ ds ≤ t
ϕp–
KB+MB s
h(τ) dτ
ds+
t
ϕp–
KB+MB
s
h(τ) dτ
ds
≤ t
ϕp–
KB+MB
h(τ) dτ
ds+
t
ϕ–p
KB+MB
h(τ) dτ
ds.
Thus, using Remark ., we obtain
T(u)(t) –T(u)(t)
≤Cp t
ϕp–(KB)ds+Cp t
ϕp–(MB)ϕp–
h(τ) dτ
+Cp
t
ϕp–(KB)ds+Cp
t
ϕp–(MB)ϕp–
h(τ) dτ
ds
≤
Cpϕp–(KB) +Cpϕp–(MB)ϕp–
h(τ) dτ
(t–t).
Since the coefficient att–tis a constant independent onu∈B, the proof of the
equicon-tinuity ofT(B) is complete.
4 Applications
In this section, we apply the solution operator obtained in Section and use the com-pactness of the operator in Section to show the existence of nontrivial solutions for the problem
(Pλ)
⎧ ⎨ ⎩
–p(u)=λh(t)·f(u), t∈(, ),
u() = =u().
For this, we first give one assumption onf.
(F) fi(, . . . , ) > andlim|s|→∞fi(s)/|s|p–= fors∈RN,i= , . . . ,N.
LetXbe a Banach space, andG:R×X→Xbe completely continuous withG(,u) = . Consider
u=G(λ,u). (.)
Denote bySthe set of solutions of (.),R+= [,∞), andR–= (–∞, ]. As the basic tool for the proof of our main theorem, we introduce the following theorem known as the global continuation theorem.
Theorem .([]) Let X be a Banach space,and G:R×X→X be continuous and
compact with G(,u) = .ThenScontains a pair of unbounded componentsC+andC–in R+×X andR–×X,respectively,andC+∩C–={(, )}.
For our fitting, let us takeX=C([, ],RN). Then the usual norm forXto be a Banach
space is defined byu∞=Ni=ui∞. In this paper, for the convenience of computation,
we establish an equivalent norm, which is defined by
uX=max
≤t≤
u(t), . . . ,uN(t) =max
≤t≤
u(t) +· · ·+uN(t)/.
Indeed, it is easy to see that
uX≤ u∞≤NuX.
We are ready to state our main existence theorem.
We know that to solve (Pλ) is equivalent to solve
u=G(λ,u),
whereG: (,∞)×X→Xis defined by
G(λ,u)(t) =
⎧ ⎨ ⎩ t
p–(a(λh·f(u)) +
s λh(τ)·f(u(τ))dτ)ds, t∈[,
],
t p–(–a(λh·f(u)) + s
λh(τ)·f(u(τ))dτ)ds, t∈[
, ].
By Remark . and Lemma . we can easily show thatGis continuous and compact with G(,u) = . Since Theorem . guarantees an unbounded continuumC+, if we provide the a priori boundedness of solutions for (Pλ), then the unbounded continuum allows the existence of solutions for allλ> .
Lemma . Assume that h∈Hand that f satisfies(F).Let any> be given,and let(λ,u) be a solution for(Pλ)withλ∈(,].Then there exists a constant C() > ,depending only on,such thatuX≤C().
Proof Assume that there exists a sequence (λn,un)∈(,]×Xsuch that, for anyn∈N,
un=G(λn,un)
withunX→ ∞asn→ ∞.
By using Remark . withx=a(λnh·f(un)),y=
s λnh(τ)·f(un(τ))dτ and the
homo-geneity ofϕ–
p andawe can estimate the solutionunas follows:
un(t) = t
p–
aλnh·f(un)
+
s
λnh(τ)·f
un(τ)
dτ
ds
≤ t
p–
aλnh·f(un)
+
s
λnh(τ)·f
un(τ)
dτ ds
≤ t
ϕ–p aλnh·f(un) +
s
λnh(τ)·f
un(τ)
dτ
ds
≤ϕ–p (λn)
ϕp– ah·f(un) +
s
h(τ)·fun(τ)
dτ
ds
≤ϕ–p ()
ϕp– |
a(h·f(un))| unpX–
+|
s h(τ)·f(un(τ))dτ| unpX–
dsunX
for allt∈[,]. By the homogeneity ofaagain, we get
un(t) ≤ϕ–p ()
ϕp– a
h· f(un)
unpX–
+
s
h(τ) |f(un(τ))|
unpX–
dτ
dsunX.
By (F), for anyε> , there existsl> such that for alls∈RN with|s| ≥l,
Sincefiis continuous on{s∈RN| |s| ≤l}, there exists a constantM> such that
fi(s) ≤M
on{s∈RN | |s| ≤l
}fori= , . . . ,N. Thus, we have
fi(s) ≤ε|s|p–+M for alls∈RN,i= , . . . ,N. (.)
SinceunX→ ∞asn→ ∞, there existsn∈Nsuch that for anyn≥n, we have
unX≥
M
p–
,
that is,
unpX–
≤
M .
Using (.), we get that, for anyn≥n andt∈[, /],
|fi(un(t))| unpX–
≤·|un(t)| p–
unpX–
+ M
unpX–
≤+M·
M =
and
f(un)X unpX–
≤f(un)∞ unpX–
=
N
i=fi(un)∞ unpX–
≤N·= N. (.)
Take
B=
f(un) unpX–
n≥n
.
ThenBis a bounded subset inX. Thus, by Remark . we see that the set{a(h·v)|v∈B} is bounded inRN. Moreover, by (.) and Remark . we may choose a constantC
= C(N) > satisfyingC→ as→ such that
a
h· f(un)
unpX–
≤C for anyn≥n.
Therefore, fort∈[,], we obtain
un(t) ≤
ϕ–p ()
ϕp–
C+
s
h(τ) dτ
ds
unX
≤
ϕ
–
p ()Cpϕp–(C)
+ϕp–()Cpϕp–()
ϕp–
s
h(τ) dτ
ds
By similar arguments, fort∈[, ], we obtain
un(t) ≤
ϕ–p ()
ϕ–p
C+
s
h(τ) dτ
ds
unX
≤
ϕ
–
p ()Cpϕp–(C)
+ϕp–()Cpϕp–()
ϕp– s
h(τ) dτ
ds
unX. (.)
DenotingChmax{
ϕ–p (
s |h(τ)|dτ)ds,
ϕ
–
p ( s
|h(τ)|dτ)ds}, we can choose>
small enough such that
ϕ
–
p ()Cpϕp–(C) +ϕp–()Cpϕp–()Ch≤
.
Consequently, combining (.) and (.), we obtain, fort∈[, ],
un(t) ≤
unX.
This implies that
unX≤ forn≥n,
which contradicts
unX≥
M
p–
> forn≥n
and this completes the proof.
Example Consider the followingp-Laplacian system:
(E)
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
–(|u|p–u)=λh(t)[(u+v) p–
+ ],
–(|u|p–v)=λh(t)e–v[ + (u)p– ], t∈(, ),
u() =v() = =u() =v(),
where u = (u,v),λ> is a parameter, andh(t) = (h(t),h(t)) is given by
h(t) =
⎧ ⎨ ⎩
t–α, t∈(, ], –, t∈(
, ), <α<p,
h(t) = –, t∈(, ).
We note thath∈Llocbuth∈/L. We now show thath∈H. Indeed,
s
τ–αdτ = –
α– τ –(α–)
s
= –
α–
–(α–) –s–(α–)
=
α–
s–(α–)– α–≤
Since <α<p, we have α–s–(α–)> fors∈(, ) and
ϕp–
s
τ–αdτ
ds≤
ϕp–
α– s –(α–)
ds=
s–(α–)
α–
p–
ds
= p– (α– )p–(p–α)
spp––α
<∞.
In addition, sincehandhare constants on (, ) and (, ), respectively, by Remark . we geth∈H.
Next, we need to check that bothf(u,v) = (u+v)p– + andf(u,v) =e–v[ + (u)
p– ]
satisfy assumption (F). In fact,f(, ) =f(, ) = > , and
lim
|(u,v)|→∞ f(u,v)
|(u,v)|p–=|(u,limv)|→∞
(u+v)p– +
(u+v)p–
= lim
|(u,v)|→∞
(u+v)p–
+
(u+v)p–
= ,
≤ lim
|(u,v)|→∞
f(u,v)
|(u,v)|p–=|(u,limv)|→∞
e–v[ + (u)p– ]
(u+v)p–
≤ lim
|(u,v)|→∞
ev
(u+v)p–
+
ev
(u+v)p–
= .
that is,lim|(u,v)|→∞ f(u,v)
|(u,v)|p– = . Consequently, by Theorem . we see that problem (E) has at least one nontrivial solution for allλ> .
Example Consider the followingp-Laplacian system withp= :
(E)
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
–(|u|u)=λh(t)[ – (u+v)],
–(|u|v)=λh(t)[ –e–(u+v)], t∈(, ), u() =v() = =u() =v(),
where u = (u,v),λ> is a parameter, andh(t) = (h(t),h(t)) is given by
h(t) =
⎧ ⎨ ⎩
t–, t∈(, ], –, t∈(, ),
and
h(t) =
⎧ ⎨ ⎩
t–, t∈(, ], , t∈(
, ).
Example Consider the followingp-Laplacian system:
(E)
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
–(|u|p–u
)=λh(t)ln((u+· · ·+uN)
+ ),
.. .
–(|u|p–u
N)=λhN(t)ln((u+· · ·+uN)
+N+ ), t∈(, ),
ui() = =ui(), i= , . . . ,N,
where u = (u, . . . ,uN),λ> is a parameter,h(t) = (h(t), . . . ,hN(t)) is defined by
hi(t) =
tα( –t)α –
p, t∈(, ), <α<p,i= , . . . ,N,
and
fi(u, . . . ,uN) =ln
u+· · ·+uN
+i+ , i= , . . . ,N.
We note that eachhiis not inL(, ),hi() = α– p< for <α<p, andh: (, )→RN
is locally integrable. By similar arguments as in Example , we can easily check thath∈H. Next, let us check (F) forfi(u, . . . ,uN) =ln((u+· · ·+uN)
+i+ ). In fact,fi(, . . . , ) =
ln(i+ ) > , and settingx:= (u
+· · ·+uN)
, we have
≤ lim
|(u,...,uN)|→∞
fi(u, . . . ,uN) |(u, . . . ,uN)|p–
= lim
|(u,...,uN)|→∞
ln((u+· · ·+uN) +i+ )
(u
+· · ·+uN)
p–
= lim
x→+∞
ln(x+i+ ) xp–
= lim
x→+∞
x+i+ ·
(p– )xp–
≤ lim
x→+∞
(p– )xp–= ,
that is,lim|(u,...,uN)|→∞
fi(u,...,uN)
|(u,...,uN)|p– = fori= , . . . ,N. Consequently, by Theorem . we see that problem (E) has at least one nontrivial solution for allλ> .
Competing interests
The authors declare that they have no competing interests for this paper.
Authors’ contributions
All authors have equally contributed in obtaining new results in this article and also read and approved the final manuscript.
Author details
1Department of Mathematics, Pusan National University, Busan, 609-735, Republic of Korea.2Department of
Mathematics, University of Ulsan, Ulsan, 680-749, Republic of Korea.
Acknowledgements
The second author was supported by the National Research Foundation of Korea, Grant funded by the Korea Government (MEST) (NRF2012R1A1A2000739). The third author was supported by the National Research Foundation of Korea, Grant funded by the Korea Government (MEST) (NRF2014R1A1A2056339).
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