AIMS’ Journals
Volume X, Number 0X, XX 200X pp. X–XX
POSITIVE AND NODAL SOLUTIONS FOR PARAMETRIC NONLINEAR ROBIN PROBLEMS WITH INDEFINITE
POTENTIAL
Genni Fragnelli
∗Department of Mathematics University of Bari
Via E. Orabona 4, 70125 Bari, Italy
Dimitri Mugnai
∗Department of Mathematics and Computer Sciences University of Perugia
Via Vanvitelli 1, 06123 Perugia, Italy
Nikolaos S. Papageorgiou
Department of Mathematics National Technical University Zografou Campus, Athens 15780, Greece
(Communicated by the associate editor name)
Abstract. We consider a parametric nonlinear Robin problem driven by the p−Laplacian plus an indefinite potential and a Carath´eodory reaction which is (p − 1)− superlinear without satisfying the Ambrosetti - Rabinowitz condition.
We prove a bifurcation-type result describing the dependence of the set of positive solutions on the parameter. We also prove the existence of nodal solutions. Our proofs use tools from critical point theory, Morse theory and suitable truncation techniques.
1. Introduction. Let Ω ⊂ R
Nbe a bounded domain with a C
2−boundary ∂Ω. In this paper, we study the following parametric nonlinear Robin problem:
(P
λ)
−∆
pu(z) + (ξ(z) + λ)u(z)
p−1= f (z, u(z)) in Ω,
∂u
∂n
p+ β(z)u(z)
p−1= 0 on ∂Ω, λ > 0, u > 0.
Here ∆
pdenotes the p−Laplace differential operator defined by
∆
pu = div(|Du|
p−2Du) for all u ∈ W
1,p(Ω), 1 < p < ∞.
The potential function ξ ∈ L
∞(Ω) is indefinite (that is, sign changing) and λ > 0 is a parameter. The reaction term f (z, x) is a Carath´ eodory function (that is, for all x ∈ R, z 7→ f (z, x) is measurable and for a.a. z ∈ Ω, x 7→ f (z, x) is continuous) which is (p − 1)− superlinear in the x−variable, but without satisfying the usual (in
2010 Mathematics Subject Classification. Primary: 35J20, 35J60; Secondary: 58E05.
Key words and phrases. indefinite potential, superlinear reaction, nonlinear regularity and maximum principle, Picone identity, bifurcation-type theorem for positive solutions, nodal solu- tions, critical groups.
∗Corresponding author: Dimitri Mugnai.
1
such cases) Ambrosetti - Rabinowitz condition. Indeed, we replace such a condition with a weaker one which lets us consider superlinear nonlinearities with slower growth near ∞, and not satisfying the Ambrosetti - Rabinowitz condition. Finally, in the boundary condition, ∂u
∂n
pdenotes the generalized normal derivative defined by
∂u
∂n
p= |Du|
p−2(Du, n)
RNfor all u ∈ W
1,p(Ω), (1) with n(·) being the outward unit normal on ∂Ω. This kind of normal derivative is dictated by the nonlinear Green’s identity (see, for example, Gasinski - Papageorgiou [11, p. 211]) and in an even more general form can be found also in the work of Lieberman [15]. The boundary weight function β(·) belongs to C
0,α(∂Ω) with α ∈ (0, 1) and β(z) ≥ 0 for all z ∈ ∂Ω. When β ≡ 0, we have the Neumann problem and in that case from (1) we see that the boundary condition becomes ∂u
∂n = 0 on
∂Ω (the usual normal derivative).
First, we look for positive solutions and our aim is to establish the precise de- pendence of the set of positive solutions on the parameter λ > 0. So, we prove a bifurcation-type result and show that there exists a critical parameter value λ
∗> 0 such that
for all λ > λ
∗problem (P
λ) admits at least two positive solutions, for all λ = λ
∗problem (P
λ) admits at least one positive solution, for all λ < λ
∗problem (P
λ) has no positive solutions.
We also show that for every λ ∈ [λ
∗, +∞) problem (P
λ) has a smallest positive solution u
∗λand we investigate the monotonicity and continuity properties of the map λ 7→ u
∗λin the relevant function space. Finally, in Section 4, we impose bilateral asymptotic conditions on f (z, ·) and prove the existence of nodal (sign changing) solutions.
Our work here continues and complements the papers of Motreanu - Motreanu - Papageorgiou [17] and Mugnai - Papageorgiou [20]. In [17] the authors exam- ine problem (P
λ) when ξ ≡ 0, β ≡ 0 and prove the existence of constant sign and nodal solutions when λ > 0 is big and the reaction term f (z, ·) is (p − 1)−
superlinear. However, they do not establish the precise dependence of the set of positive solutions on the parameter λ > 0 (bifurcation-type result). On the other hand Mugnai - Papageorgiou [20], study nonparametric Neumann problems (that is, β ≡ 0) and prove existence and multiplicity theorems for resonant problems.
Their work was extended recently to problems with a (p − 1)−superlinear reaction term by Fragnelli - Mugnai - Papageorgiou [10] (see also Mugnai - Papageorgiou [21]). We mention also the relevant works of Brock - Iturriaga - Ubilla [5] (nonlinear parametric Dirichlet problems with ξ ≡ 0), Cardinali - Papageorgiou - Rubbioni [6]
(nonlinear parametric Neumann problems with ξ ≡ 0 and a superdiffusive reaction term), Gasinski - Papageorgiou [11] (nonlinear parametric Dirichlet problems with ξ ≡ 0 and a logistic reaction term), Papageorgiou - Radulescu [24] (nonlinear para- metric Robin problems with ξ ≡ 0, the parameter λ > 0 multiplying the reaction term and the latter satisfying certain monotonicity properties) and Takeuchi [28], [29] (semilinear superdiffusive logistic equations driven by the Dirichlet Laplacian with zero potential). Finally, we recall also the work of Mugnai - Papageorgiou [22]
on logistic equations on R
Ndriven by the Dirichlet p−Laplacian with zero potential.
Our approach is variational, based on the critical point theory. In Section 4, in order to generate a nodal solution, we also use tools from Morse theory (critical groups).
2. Mathematical Background. Let X be a Banach space and X
∗its topological dual. By h·, ·i we denote the duality brackets for the pair (X
∗, X). Given ϕ ∈ C
1(X, R), we say that ϕ satisfies the Cerami condition (the “C-condition” for short), if the following property holds:
Every sequence {u
n}
n≥1⊂ X s.t. {ϕ(u
n)}
n≥1⊂ R is bounded and (1 + ku
nk)ϕ
0(u
n) → 0 in X
∗as n → +∞,
admits a strongly convergent subsequence.
This is a compactness-type condition on functional ϕ which compensates for the fact that the ambient space X need not be locally compact (X is in general infinite dimensional). The C-condition is a basic tool in probing a deformation theorem for the sublevel sets of ϕ, from which one can deduce the minimax theory for the critical values of ϕ. Prominent in that theory is the well-known “mountain pass theorem” due to Ambrosetti - Rabinowitz [3]. Here we state it in a slightly more general form (see, for example, Gasinski - Papageorgiou [11, p. 648]).
Theorem 2.1. If X is a Banach space, ϕ ∈ C
1(X, R), ϕ satisfies the C-condition, u
0, u
1∈ X, ku
1− u
0k > ρ > 0,
max n
ϕ(u
0), ϕ(u
1) o
< inf n
ϕ(u) : ku − u
0k < ρ o
= m
ρand c = inf
γ∈Γ
max
0≤t≤1
ϕ(γ(t)), with Γ = {γ ∈ C([0, 1], W
1,p(Ω)) : γ(0) = u
0, γ(1) = u
1}, then c ≥ m
ρand c is a critical value of ϕ.
In the analysis of problem (P
λ), in addition to the Sobolev space W
1,p(Ω), we will also use the Banach space C
1(Ω) and the boundary space L
τ(∂Ω) with τ ∈ [1, +∞].
In what follows by | · | we denote the norm on R
N, by (·, ·)
RNthe inner product of R
Nand by k · k the norm of the Sobolev space W
1,p(Ω) defined by
kuk = kuk
pp+ kDuk
ppp1for all u ∈ W
1,p(Ω).
The space C
1(Ω) is an ordered Banach space with positive cone C
+= n
u ∈ C
1(Ω) : u(z) ≥ 0 for all z ∈ Ω o . This cone has a nonempty interior given by
int C
+= n
u ∈ C
+: u(z) > 0 for all z ∈ Ω o .
On ∂Ω we consider the (N − 1)− dimensional Hausdorff (surface) measure σ(·).
With this measure on ∂Ω, we can define the Lebesgue spaces L
τ(∂Ω), 1 ≤ τ ≤
∞. From the theory of Sobolev spaces, we know that there exists a unique linear
continuous map γ
0: W
1,p(Ω) → L
p(∂Ω), known as the “trace map”, s.t. γ
0(u) =
u|
∂Ωfor all u ∈ W
1,p(Ω) ∩ C(Ω). So, we understand the trace map as representing
the “boundary values” of a Sobolev function u ∈ W
1,p(Ω). The trace map is
compact into L
r(∂Ω) for all r ∈
1, N p − p N − p
when p < N and into L
r(∂Ω) for all r ∈ [1, +∞) when p ≥ N . We know that
imγ
0= W
p01,p(∂Ω) 1 p + 1
p
0= 1
and kerγ
0= W
01,p(Ω).
In the rest of the work, for the sake of notational simplicity, we drop the use of the trace map γ
0. It is understood that all restrictions of Sobolev functions u ∈ W
1,p(Ω) on ∂Ω are defined in the sense of traces.
Let A : W
1,p(Ω) → W
1,p(Ω)
∗be the nonlinear map defined by hA(u), hi =
Z
Ω
|Du|
p−2(Du, Dh)
RNdz for all u, h ∈ W
1,p(Ω). (2) From Motreanu - Motreanu - Papageorgiou [18] (p. 40), we have:
Proposition 1. The nonlinear map A : W
1,p(Ω) → W
1,p(Ω)
∗defined by (2), is continuous, monotone, hence maximal monotone and of type (S)
+, that is, if u
n* u in W
1,p(Ω) and lim sup
n→∞hA(u
n), u
n− ui ≤ 0, then u
n→ u in W
1,p(Ω).
Let f
0: Ω × R → R be a Carath´eodory function s.t.
|f
0(z, x)| ≤ a
0(z)(1 + |x|
r−1) for a.a. z ∈ Ω and all x ∈ R, with a
0∈ L
∞(Ω)
+:= n
a ∈ L
∞(Ω) : a ≥ 0 in Ω o
and r ∈ (1, p
∗), where
p
∗=
N p
N − p , if p < N +∞, if N ≤ p.
Moreover, let g
0∈ C
0,η(∂Ω × R), η ∈ (0, 1), be such that
|g
0(z, x)| ≤ c
0|x|
τfor all (z, x) ∈ ∂Ω × R, with c
0> 0 and τ ∈ (1, p]. Finally, set F
0(z, x) = R
x0
f
0(z, s)ds, G
0(z, x) = R
x0
g
0(z, s)ds and consider the C
1− functional defined by ϕ
0(u) = 1
p kDuk
pp+ Z
∂Ω
G
0(z, u)dσ − Z
Ω
F
0(z, u)dz for all u ∈ W
1,p(Ω).
From Papageorgiou - Radulescu [23], we have
Proposition 2. If u
0∈ W
1,p(Ω) is a local C
1(Ω)− minimizer of ϕ
0, that is there exists ρ
0> 0 s.t.
ϕ
0(u
0) ≤ ϕ
0(u
0+ h) for all h ∈ C
1(Ω) with khk
C1(Ω)≤ ρ
0,
then u
0∈ C
1,α(Ω) for some α ∈ (0, 1) and u
0is also a local W
1,p(Ω)−minimizer of ϕ
0, that is there exists ρ
1> 0 s.t.
ϕ
0(u
0) ≤ ϕ
0(u
0+ h) for all h ∈ W
1,p(Ω) with khk ≤ ρ
1.
Remark 1. In fact the result is still true even when f
0(z, ·) has critical growth, namely r = p
∗, see Papageorgiou - Radulescu [25].
We will also need a strong comparison result which is of independent interest and for this reason is formulated in a more general setting.
So, let Θ ∈ C
1(0, ∞) be a function such that 0 < ˆ c ≤ Θ
0(t)t
Θ(t) ≤ ˆ c
0and ˆ c
1t
p−1≤ Θ(t) ≤ ˆ c
2(1 + t
p−1)
for all t > 0, with ˆ c
1, ˆ c
2> 0.
We consider a map a : R
N→ R
Nsatisfying the following hypotheses:
H(a) : a(y) = a
0(|y|)y for all y ∈ R
Nwith a
0(t) > 0 for all t > 0 and (i) a
0∈ C
1(0, ∞), t 7→ a
0(t)t is strictly increasing, lim
t→0+
a
0(t)t = 0 and lim
t→0+
a
00(t)t a
0(t) > −1;
(ii) |Da(y)| ≤ ˆ c
3Θ(|y|)
|y| for some ˆ c
3> 0, all y ∈ R
N\ {0};
(iii) (Da(y)Ξ, Ξ)
RN≥ Θ(|y|)
|y| |ξ|
2for all y ∈ R
N\ {0}, all Ξ ∈ R
N.
Remark 2. These hypotheses are motivated by the nonlinear regularity of Lieber- man [15] and the nonlinear maximum principle of Pucci-Serrin [27]. If a(y) =
|y|
p−2y with 1 < p < ∞, then the hypotheses above are satisfied with Θ(t) = t
p−1, and in this case, for every u ∈ W
1,p(Ω),
div a(Du) = ∆
pu.
Given two functions h
1, h
2∈ L
∞(Ω), we write h
1≺ h
2, if for every compact K ⊆ Ω, there exists = (K) > 0 such that
h
1(z) + ≤ h
2(z) for a.a. z ∈ K.
Evidently if h
1, h
2∈ C(Ω) and h
1(z) < h
2(z) for all z ∈ Ω, then h
1≺ h
2.
The next strong comparison principle extends analogous results for the Dirichlet p−Laplacian, by Guedda-Veron [13] and Arcoya-Ruiz [4]. In what follows by ∂u
∂n
awe denote the generalized directional derivative ∂u
∂n
a= (a(Du), n)
RNfor all u ∈ W
1,p(Ω). If a(y) = |y|
p−2y (1 < p < ∞), then we recover the generalized directional derivative from (1).
Proposition 3. If Hypotheses H(a) hold, Υ, h
1, h
2∈ L
∞(Ω), Υ ≥ 0, h
1≺ h
2, u ∈ C
1(Ω) \ {0}, v ∈ intC
+, u ≤ v and they satisfy
− div a(Du(z)) + Υ(z)|u(z)|
p−2u(z) = h
1(z) in Ω,
− div a(Dv(z)) + Υ(z)v(z)
p−1= h
2(z) in Ω with ∂u
∂n
a∂Ω
< 0 or ∂v
∂n
a∂Ω
< 0, then (v−u)(z) > 0 for all z ∈ Ω and ∂(v − u)
∂n
G0<
0, where G
0= n
z ∈ ∂Ω : u(z) = v(z) o . Proof. To fix things we assume that ∂v
∂n
a∂Ω
< 0.
Let G := n
z ∈ Ω : u(z) = v(z) o
and E = n
z ∈ Ω : Du(z) = Dv(z) = 0 o . Claim: G ⊆ E .
Let z
0∈ G and set y = v − u. Then y attains its infimum at z
0and so Dv
0(z) = Du
0(z).
If Du(z
0) 6= 0, then we can find ρ > 0 small such that B
ρ(z
0) ⊆ Ω and
|Du(z)| > 0, |Dv(z)| > 0, (Du(z), Dv(z))
RN> 0
for all z ∈ B
ρ(z
0). By hypothesis, we have
− div a(Dv) − a(Du) = h
2(z) − h
1(z) − Υ(z)(v
p−1− u
p−1) in Ω. (3) If a = (a
k)
Nk=1, then using the mean value theorem, we have
a
k(Ξ) − a
k(Ξ
0) =
N
X
i=1
Z
1 0∂a
k∂y
i(Ξ
0+ t(Ξ − Ξ
0))(Ξ
i− Ξ
0i)dt
for all Ξ = (Ξ
i)
Ni=1, Ξ
0= (Ξ
0i)
Ni=1in R
N, k ∈ {1, ..., N }. On B
ρ(z
0) we define the continuous coefficients
θ
k,i(z) = Z
10
∂a
k∂y
i(Du(z) + t(Dv(z) − Du(z)))dt.
Then, we introduce the following linear differential operator:
L(w) = −div
N
X
i=1
θ
k,i(z) ∂w
∂z
i!
for all w ∈ W
1,p(B
ρ(z
0)).
From (3) we have
L(y) = h
2(z) − h
1(z) − Υ(z)(v
p−1− u
p−1) in B
ρ(z
0). (4) Recall that B
ρ(z
0) ⊆ Ω, h
1≺ h
2and ξ ∈ L
∞(Ω). So, from (4) and choosing ρ ∈ (0, 1) even smaller if necessary, we can have that L(·) is uniformly elliptic (see also [4] and [13]). Using the strong maximum principle (see, for example, Gasinski- Papageorgiou [11, p. 738]), we have
y(z) = (v − u)(z) > 0 for all z ∈ B
ρ(z
0).
But u(z
0) = v(z
0) since z
0∈ G , and thus a contradiction arises. This proves the Claim.
Since v ∈ intC
+satisfies ∂v
∂n
a∂Ω
< 0, we see that E ⊆ Ω is compact. So, we can find a smooth open set Ω
1⊆ Ω such that
G ⊆ Ω
1⊆ Ω
1⊆ Ω.
Then we can find > 0 such that
u(z) + ≤ v(z) for all z ∈ ∂Ω
1, (5) h
1(z) + ≤ h
2(z) for a.a. z ∈ Ω
1. (6) Let δ ∈ (0, min{, 1}) be such that
|Υ(z)|
|s|
p−2s − |t|
p−2t
≤ (7)
for a.a. z ∈ Ω, all s, t ∈ [−kuk
∞, kvk
∞] with |s − t| ≤ δ (recall that Υ ∈ L
∞(Ω)).
Then we have
− div a(D(u + δ)) + Υ(z)|u + δ|
p−2(u + δ)
= − div a(Du) + Υ(z)|u + δ|
p−2(u + δ)
≤ − div a(Du) + Υ(z)|u|
p−2u + (see(7))
=h
1(z) + ≤ h
2(z) (see(6))
= − div a(Dv) + Υ(z)v
p−1for a.a.z ∈ Ω
1.
(8)
From (8) and the weak comparison principle (see (5)) it follows that
u(z) + δ < v(z) for all z ∈ Ω
1.
But G ⊆ Ω
1. Hence G = ∅ and so
(v − u)(z) > 0 for all z ∈ Ω.
Moreover, from Hopf’s Lemma we have
∂(v − u)
∂n
G0
< 0.
Remark 3. Consider the following order cone in C
1(Ω):
C ˆ
+:=
y ∈ C
1(Ω) : y(z) ≥ 0 for all z ∈ Ω, ∂y
∂n
G0≤ 0
. This cone has a non empty interior given by
int ˆ C
+=
y ∈ C
1(Ω) : y(z) > 0 for all z ∈ Ω, ∂y
∂n
G0
< 0
.
According to Proposition 3, we have v − u ∈ int ˆ C
+. Of course if G
0= ∅ then C ˆ
+= C
+.
Let ξ ∈ L
∞(Ω) and β ∈ C
0,η(∂Ω) with η ∈ (0, 1), β(z) ≥ 0 for all z ∈ ∂Ω, and consider the following nonlinear eigenvalue problem:
−∆
pu(z) + ξ(z)|u(z)|
p−2u(z) = ˆ λ|u(z)|
p−2u(z), in Ω,
∂u
∂n
p+ β(z)|u|
p−2u = 0, on ∂Ω. (9)
As in Mugnai - Papageorgiou [20] (there β ≡ 0, Neumann problem), we can show that there exists a smallest eigenvalue ˆ λ
1(β) having the following properties:
ˆ λ
1(β) is isolated in the spectrum of (9), ˆ λ
1(β) is simple ,
ˆ λ
1(β) = inf ϑ(u) + R
∂Ω
β(z)|u|
pdσ kuk
pp: u ∈ W
1,p(Ω), u 6= 0
, (10) where
ϑ(u) = kDuk
pp+ Z
Ω
ξ(x)|u|
pdz for all u ∈ W
1,p(Ω).
The infimum in (10) is realized on the corresponding one dimensional eigenspace.
From (10) it is clear that the elements of this eigenspace do not change sign. In what follows, by ˆ u
1(β) we denote the positive L
p− normalized eigenfunction (that is kˆ u
1(β)k
p= 1) corresponding to ˆ λ
1(β). From the nonlinear regularity theory of Lieberman [15] and the nonlinear maximum principle (see Pucci - Serrin [27, p.
120]), we have that ˆ u
1(β) ∈ int C
+.
As we already mentioned in the Introduction, in Section 4 in order to produce a nodal solution, we will use tools from Morse theory (critical groups). So, let us recall the definition of critical groups at an isolated critical point.
Let X be a Banach space, ϕ ∈ C
1(X, R) and c ∈ R. We introduce the following sets:
ϕ
c:= {u ∈ X : ϕ(u) ≤ c}, K
ϕ= {u ∈ X : ϕ
0(u) = 0}
and
K
ϕc= {u ∈ K
ϕ: ϕ(u) = c}.
For every topological pair (Y
1, Y
2) with Y
1⊆ Y
2⊆ X and every k ∈ N
0, by H
k(Y
1, Y
2) we denote the k
threlative singular homology group with integer coeffi- cients. If u ∈ K
ϕcis isolated, then the critical groups of ϕ at u are defined by
C
k(ϕ, u) = H
k(ϕ
c∩ U, ϕ
c∩ U \ {u}) for all k ∈ N
0,
where U is a neighborhood of u s.t. K
ϕ∩ϕ
c∩U = {u}. The excision property of sin- gular homology theory implies that this definition of critical groups is independent of the particular choice of the neighborhood U as above.
Finally, we fix our notation. By | · |
Nwe denote the Lebesgue measure on R
N. If x ∈ R, then we set x
±= max{±x, 0}. For u ∈ W
1,p(Ω) we define u
±(·) = u(·)
±. We know that u
±∈ W
1,p(Ω), u = u
+− u
−, |u| = u
++ u
−. Also, if g : Ω × R → R is measurable, then we define the Nemitzskii operator N
g(u)(·) = g(·, u(·)) for all u ∈ W
1,p(Ω).
3. Positive solutions. In order to look for positive solutions, our hypotheses on the data of problem (P
λ) are the following:
H(ξ) : ξ ∈ L
∞(Ω),
H(β) : (a) β ∈ C
0,α(∂Ω) with α ∈ (0, 1) and β(z) > 0 for all z ∈ ∂Ω, or (b) β ≡ 0 (Neumann problem);
H(f ) : f : Ω × R → R is a Carath´eodory function such that f (z, 0) = 0 for a.a.
z ∈ Ω and
(i) |f (z, x)| ≤ a(z)(1 + x
r−1) for a.a. z ∈ Ω and all x ≥ 0, where a ∈ L
∞(Ω)
+and
p < r < p
∗=
N p
N − p , if p < N, +∞, if N ≤ p;
(ii) if F (z, x) = Z
x0
f (z, s)ds, then lim
x→+∞
F (z, x)
x
p−1= +∞ uniformly for a.a. z ∈ Ω;
(iii) there exist γ
0> 0 and µ ∈
max
(r − p) N p , 1
, p
∗such that 0 < γ
0≤ lim inf
x→+∞
f (z, x)x − pF (z, x)
x
µuniformly for a.a.z ∈ Ω;
(iv) there exists δ > 0 and q ∈ (1, p) such that c
0x
q−1≤ f (z, x) for a.a. z ∈ Ω and all x ∈ [0, δ].
Remark 4. The alternative in H(β) means that we exclude mixed problems.
Remark 5. Since we are looking for positive solutions and the above hypotheses concern the positive semiaxis R
+= [0, +∞), without any loss of generality, we may assume that f (z, x) = 0 for a.a. z ∈ Ω and all x ≤ 0.
Hypotheses H(f )(ii), (iii) imply that
x→+∞
lim f (z, x)
x
p−1= +∞ uniformly for a.a. z ∈ Ω,
that is f (z, ·) is (p − 1)− superlinear. We point out that we do not employ the Ambrosetti - Rabinowitz condition, which says that there exist τ > p and M > 0 such that
0 < τ F (z, x) ≤ f (z, x)x for a.a. z ∈ Ω, all x ≥ M,
essinf
ΩF (·, M ) > 0 (see Ambrosetti - Rabinowitz [3] and Mugnai [19]).
By direct integration of the Ambrosetti - Rabinowitz condition, we have ˆ
c
0x
τ≤ F (z, x) for a.a. z ∈ Ω, all x ≥ M and some ˆ c
0> 0.
Hence, F (z, ·) has at least τ − polynomial growth near +∞.
Example. The following functions satisfy hypotheses H(f ) (for the sake of sim- plicity, we drop the z− dependence):
f
1(x) = x
r−1+ x
q−1for all x ≥ 0 with 1 < q < p < r < p
∗, f
2(x) = x
p−1log x + 1 p
for all x ≥ 0.
Note that f
2does not satisfy the Ambrosetti - Rabinowitz condition.
Let L = {λ > 0 : problem (P
λ) has a positive solution} (this is the set of admis- sible parameters) and S(λ) the set of positive solutions of problem (P
λ) (for λ 6∈ L, we have S(λ) = ∅).
Proposition 4. If Hypotheses H(ξ), H(β), H(f ) hold, then S(λ) ⊆ int C
+for every λ > 0.
Proof. We assume that λ ∈ L (otherwise S(λ) = ∅). Let u ∈ S(λ). Then hA(u), hi +
Z
Ω
(ξ(z) + λ)u
p−1hdz + Z
∂Ω
β(z)u
p−1hdσ = Z
Ω
f (z, u)hdz for all h ∈ W
1,p(Ω), that is
−∆
pu(z) + (ξ(z) + λ)u(z)
p−1= f (z, u(z)) for a.a. z ∈ Ω,
∂u
∂n
p+ β(z)u
p−1= 0 on ∂Ω, (11)
see Papageorgiou - Radulescu [23].
From Winkert [30] and Papageorgiou - Radulescu [25], we have u ∈ L
∞(Ω). So, we can apply Theorem 2 of Lieberman [15] and have that u ∈ C
+\ {0}.
Let ρ = kuk
∞. Hypotheses H(f )(i), (iv) imply that we can find ˆ ξ
ρ> 0 such that f (z, x) + ˆ ξ
ρx
p−1≥ 0 for a.a. z ∈ Ω and all x ∈ [0, ρ]. (12) From (11) and (12), we have
∆
pu(z) ≤ (kξk
∞+ λ + ˆ ξ
ρ)u(z)
p−1for a.a. z ∈ Ω (see Hypothesis H(ξ)), which implies that u ∈ int C
+by the nonlinear maximum principle (see Pucci - Serrin [27, p. 120]).
Therefore S(λ) ⊆ int C
+for all λ > 0.
Proposition 5. If Hypotheses H(ξ), H(β), H(f ) hold, then L 6= ∅ and λ ∈ L implies that [λ, +∞) ⊆ L.
Proof. Let µ > kξk
∞(see Hypothesis H(ξ)), and consider the following nonlinear Robin problem:
−∆
pu(z) + (ξ(z) + µ)u(z)
p−1= 1, in Ω,
∂u
∂n
p+ β(z)u
p−1= 0, on ∂Ω, u > 0. (13) Let V : L
p(Ω) → L
p0(Ω)
1
p
+
p10= 1
be the nonlinear map defined by
V
p(u)(z) = (ξ(z) + µ)|u(z)|
p−2u(z).
Evidently V
p(·) is continuous, strictly monotone (recall µ > kξk
∞), hence maximal monotone, too. Moreover, let B
p: W
1,p(Ω) → L
p0(∂Ω) be the boundary map defined by
B
p(u)(z) = β(z)|u(z)|
p−2u(z) for all z ∈ ∂Ω.
Also this map is continuous, monotone (see Hypothesis H(β)), hence maximal monotone. If γ
0∈ L(W
1,p(Ω), L
p(∂Ω)) is the trace map, then γ
∗0∈ L(L
p0(∂Ω), W
1,p(Ω)
∗).
Finally, we introduce the map K : W
1,p(Ω) → W
1,p(Ω)
∗defined by K(u) = A(u) + V
p(u) + γ
0∗◦ B
p(u).
Then K is continuous, maximal monotone (see Gasinski - Papageorgiou [11, p.
326]), and we have
hK(u), ui = kDuk
pp+ Z
Ω
(ξ(z) + µ)|u|
pdz + Z
∂Ω
β(z)|u|
pdσ
≥ kDuk
pp+ (µ − kξk
∞)kuk
pp(see Hypothesis H(β)),
⇒ K(·) is coercive (recall that µ > kξk
∞).
But a maximal monotone, coercive map is surjective (see Gasinski - Papageorgiou [11, p. 319]). So, we can find ¯ u ∈ W
1,p(Ω), ¯ u 6= 0 such that
A(¯ u) + V
p(¯ u) + (γ
0∗◦ B
p)(¯ u) = 1. (14) On (14) we act with −¯ u
−∈ W
1,p(Ω). Then
kD¯ u
−k
pp+ (µ − kξk
∞)k¯ u
−k
pp≤ 0 (see Hypothesis H(β)),
⇒ ¯ u ≥ 0, ¯ u 6= 0.
From (14) we have Z
Ω
|D¯ u|
p−2(D ¯ u, Dh)
Rndz + Z
Ω
(ξ(z) + µ)¯ u
p−1hdz + Z
∂Ω
β(z)¯ u
p−1hdσ = Z
Ω
hdz
for all h ∈ W
1,p(Ω), that is
−∆
pu(z) + (ξ(z) + µ)¯ ¯ u(z)
p−1= 1 for a.a. z ∈ Ω,
∂ ¯ u
∂n
p+ β(z)¯ u
p−1= 0 on ∂Ω (15)
(see Papageorgiou - Radulescu [23]). As before (see the proof of Proposition 4) using the nonlinear regularity theory of Lieberman [15, Theorem 2], we have ¯ u ∈ C
+\{0}.
Then from (15), we have
∆
pu(z) ≤ (kξk ¯
∞+ µ)¯ u(z)
p−1for a.a. z ∈ Ω,
⇒ ¯ u ∈ int C
+(see Pucci - Serrin [27, p.120]). (16)
Let ¯ m = min
Ωu > 0 (see (16)) and let λ ¯
0= µ + kN
f(¯ u)k
∞¯
m
p−1(see Hypothesis H(f )(i)). For all h ∈ W
1,p(Ω) with h ≥ 0, we have
Z
Ω
|D¯ u|
p−2(D ¯ u, Dh)
RNdz + Z
Ω
(ξ(z) + λ
0)¯ u
p−1hdz + Z
∂Ω
β(z)¯ u
p−1hdσ
= Z
Ω
|D¯ u|
p−2(D ¯ u, Dh)
RNdz + Z
Ω
(ξ(z) + µ + kN
f(¯ u)k
∞¯
m
p−1)¯ u
p−1hdz +
Z
∂Ω
β(z)¯ u
p−1hdσ
≥ Z
Ω
|D¯ u|
p−2(D ¯ u, Dh)
RNdz + Z
Ω
(ξ(z) + µ)¯ u
p−1hdz + Z
Ω
f (z, ¯ u)hdz +
Z
∂Ω
β(z)¯ u
p−1hdσ (recall that h ≥ 0)
≥ Z
Ω
(1 + f (z, ¯ u))hdz (see Hypothesis H(β)and recall h ≥ 0)
≥ Z
Ω
f (z, ¯ u)hdz.
(17)
We consider the following truncations of the reaction term f (z, ·) and of the bound- ary term x 7→ β(z)|x|
p−2x:
f (z, x) = ˆ
( f (z, x) if x ≤ ¯ u(z)
f (z, ¯ u(z)) if ¯ u(z) < x (18) for all (z, x) ∈ Ω × R,
β(z, x) = ˆ
0 if x < 0
β(z)x
p−1if 0 ≤ x ≤ ¯ u(z) β(z)¯ u(z)
p−1if ¯ u(z) < x
(19)
for all (z, x) ∈ ∂Ω × R, and both are Carath´eodory functions.
We set ˆ F (z, x) = R
x0
f (z, s)ds, ˆ ˆ B(z, s) = R
x0
β(z, s)ds and consider the C ˆ
1− functional ˆ ϕ : W
1,p(Ω) → R defined by
ˆ ϕ(u) = 1
p kDuk
pp+ 1 p
Z
Ω
(ξ(z) + λ
0)|u|
pdz + Z
∂Ω
B(z, u)dσ − ˆ Z
Ω
F (z, u)dz ˆ
for all u ∈ W
1,p(Ω).
Since λ
0> µ > kξk
∞, from (18) and (19) it is clear that ˆ ϕ is coercive. Also, using the Sobolev embedding theorem and the trace theorem, we see that ˆ ϕ is sequentially weakly lower semicontinuous. So, by the Weierstrass theorem, we can find u
0∈ W
1,p(Ω) such that
ˆ
ϕ(u
0) = inf n ˆ
ϕ(u) : u ∈ W
1,p(Ω) o
. (20)
Let t ∈ (0, 1) be small such that
tˆ u
1(β) ≤ ¯ u and tˆ u
1(z) ∈ [0, δ] for all z ∈ Ω
(recall that ˆ u
1(β), ¯ u ∈ int C
+). We have ˆ
ϕ(tˆ u
1) = t
pp kDˆ u
1(β)k
pp+ t
pp
Z
Ω
(ξ(z) + λ
0)ˆ u
1(β)
pdz + t
pp
Z
∂Ω
β(z)ˆ u
1(β)
pdσ
− Z
Ω
F (z, ˆ u
1(β))dz (see (18) and (19))
≤ t
pp [ˆ λ
1(β) + λ
0] − t
qq c
0kˆ u
1(β)k
qq(see HypothesisH(f )(iv) and recall that kˆ u
1(β)k
p= 1).
(21)
Since q < p, see Hypothesis H(f )(iv), by choosing t ∈ (0, 1) even smaller if neces- sary, from (21) wee see that
ˆ
ϕ(tˆ u
1) < 0 = ˆ ϕ(0),
⇒ ˆ ϕ(u
0) < 0 = ˆ ϕ(0) (see (20)), hence u
06= 0.
From (20) we have ˆ
ϕ
0(u
0) = 0
⇒ hA(u
0), hi + Z
Ω
(ξ(z) + λ
0)|u
0|
p−2u
0hdz + Z
∂Ω
β(z, u ˆ
0)hdσ = Z
Ω
f (z, u ˆ
0)hdz (22) for all h ∈ W
1,p(Ω). In (22) we choose h = −u
−0∈ W
1,p(Ω). Using (18) and (19), we obtain
kDu
−0k
pp+ (λ
0− kξk
∞)ku
−0k
pp≤ 0,
⇒ u
0≥ 0, u
06= 0.
Next in (22) we choose h = (u
0− ¯ u)
+∈ W
1,p(Ω). Then hA(u
0), (u
0− ¯ u)
+i+
Z
Ω
(ξ(z) + λ
0)u
p−10(u
0− ¯ u)
+dz + Z
∂Ω
β(z)¯ u
p−1(u
0− ¯ u)
+dσ
= Z
Ω
f (z, ¯ u)(u
0− ¯ u)
+dz (see (18) and (19))
≤ hA(¯ u), (u
0− ¯ u)
+i + Z
Ω
(ξ(z) + λ
0)¯ u
p−1(u
0− ¯ u)
+dz +
Z
∂Ω
β(z)¯ u
p−1(u
0− ¯ u)
+dσ (see (17)), so that
hA(u
0) − A(¯ u), (u
0− ¯ u)
+i + Z
Ω
(λ
0− kξk
∞)(u
p−10− ¯ u
p−1)(u
0− ¯ u)
+dz ≤ 0, which implies that |{u
0> ¯ u}|
N= 0, and hence u
0≤ ¯ u (recall that λ
0> kξk
∞). So, we have proved that
u
0∈ [0, ¯ u] = n
u ∈ W
1,p(Ω) : 0 ≤ u(z) ≤ ¯ u(z) for a.a. z ∈ Ω o
, u
06= 0. (23) Then from (18), (19) and (23), equation (22) becomes
hA(u
0), hi + Z
Ω
(ξ(z) + λ
0)u
p−10hdz + Z
∂Ω
β(z)u
p−10hdσ = Z
Ω
f (z, u
0)hdz
for all h ∈ W
1,p(Ω). Hence
−∆
pu
0(z) + (ξ(z) + λ
0)u
0(z)
p−1= f (z, u
0(z)) for a.a. z ∈ Ω,
∂u
∂n
p+ β(z)u
p−10= 0 on ∂Ω
and u
0∈ S(λ
0) ⊆ int C
+(see Proposition 4). Hence λ
0∈ L 6= ∅.
Next, let λ ∈ L and pick θ > λ. Let u
λ∈ S(λ) ⊆ int C
+(see Proposition 4). We have
− ∆
pu
λ(z) + (ξ(z) + θ)u
λ(z)
p−1≥ −∆
pu
λ(z) + (ξ(z) + λ)u
λ(z)
p−1= f (z, u
λ(z)) for a.a. z ∈ Ω. (24) In this case, we truncate the reaction term and the boundary term at u
λ(z) (instead of truncating at ¯ u(z), as in (18), (19)). Repeating the argument above with λ
0replaced by θ using this time (24) instead of (17), via the direct method of the calculus of variations, we obtain
u
θ∈ S(θ) ∩ [0, u
λ] ⊆ [0, u
λ] ∩ int C
+(see Proposition 4),
⇒ θ ∈ L and so [λ, +∞) ⊆ L.
Remark 6. In fact a careful reading of the above proof, reveals that the solutions of (P
λ) exhibit the following “monotonicity” property. Given λ ∈ L and θ > λ, then θ ∈ L and there exists u
θ∈ S(θ) ⊆ int C
+such that u
θ≤ u
λ. This property can be improved provided we strengthen the Hypotheses on f (z, ·).
H(f )
1: f : Ω × R → R is a Carath´eodory function such that f (z, 0) = 0 for a.a. z ∈ Ω, Hypotheses H(f )
0(i), (ii), (iii), (iv) are the same as the corresponding Hypotheses H(f )(i), (ii), (iii), (iv) and
(v) for every ρ > 0, there exists ˆ ξ
ρ> 0 such that for a.a. z ∈ Ω, x 7→ f (z, x) + ˆ ξ
ρx
p−1is nondecreasing on [0, ρ].
Remark 7. Clearly this is true if for a.a. z ∈ Ω, f (z, ·) is nondecreasing. More generally, suppose that f (z, x) = c
0x
q−1+ f
0(z, x) with f
0(z, ·) differentiable and (f
0)
0x(z, x) ≥ −ˆ cx
d−2for a.a. z ∈ Ω, all x ≥ 0, with d > q. For this setting, Hypothesis H(f )
1(v) is satisfied.
With this stronger condition on f (z, ·), we can prove the following “strong mono- tonicity” property.
Proposition 6. If Hypotheses H(ξ), H(β), H(f )
1hold, λ ∈ L, u
λ∈ S(λ) ⊆ int C
+and θ > λ, then we can find u
θ∈ S(θ) ⊆ int C
+such that u
λ− u
θ∈ int ˆ C
+with G
0:= {z ∈ ∂Ω : u
λ(z) = u
θ(z)}, if H(β)(a) holds, and u
λ− u
θ∈ int C
+, if H(β)(b) is in force.
Proof. As we already mentioned in the previous Remark, we have:
θ ∈ L and we can find u
θ∈ S(θ) ⊆ int C
+s.t.u
θ≤ u
λ.
For δ > 0, we set u
δθ= u
θ+ δ ∈ int C
+. Let ρ = ku
λk
∞and let ˆ ξ
ρ> 0 be
as postulated by Hypothesis H(f )
1(v). We can always take ˆ ξ
ρ> 0 such that
ξ ˆ
ρ+ λ > kξk
∞. Being u
θbounded, we have
− ∆
pu
δθ+ (ξ(z) + λ + ˆ ξ
p)(u
δθ)
p−1≤ −∆
pu
θ+ (ξ(z) + θ)u
p−1θ− (θ − λ)u
p−1θ+ ˆ ξ
ρu
p−1θ+ χ(δ), with χ(δ) → 0
+as δ → 0
+,
≤ −∆
pu
θ+ (ξ(z) + θ)u
p−1θ− (θ − λ)m
p−1θ+ ˆ ξ
ρu
p−1θ+ χ(δ), with m
θ= min
Ω
u
θ> 0 (recall u
θ∈ int C
+)
= f (z, u
θ) + ˆ ξ
ρu
p−1θ+ (χ(θ) − (θ − λ)m
p−1θ)
≤ f (z, u
λ) + ˆ ξ
ρu
p−1λ+ (χ(δ) − (θ − λ)m
p−1θ) (see Hypothesis H(f )
1(v) and recall u
θ≤ u
λ)
= −∆
pu
λ+ (ξ(z) + λ + ˆ ξ
ρ)u
p−1λ+ (χ(δ) − (θ − λ)m
p−1θ)
< −∆
pu
λ+ (ξ(z) + λ + ˆ ξ
ρ)u
p−1λfor δ > 0 small.
(25)
If β(z) > 0 for all z ∈ ∂Ω, then from (25) and Proposition 3, we infer that u
λ− u
θ∈ int ˆ C
+with G
0:= {z ∈ ∂Ω : u
λ(z) = u
θ(z)}.
If β ≡ 0 (Neumann problem), multiplying (25) by (u
δθ− u
λ)
+∈ W
1,p(Ω), inte- grating over Ω and using the nonlinear Green’s identity, we obtain for δ > 0 small enough,
hA(u
δθ) − A(u
λ), (u
δθ− u
λ)
+i + ( ˆ ξ
ρ+ λ − kξk
∞)
Z
Ω
((u
δθ)
p−1− u
p−1λ)(u
δθ− u
λ)
+dz ≤ 0, and so u
δθ≤ u
λfor δ > 0 small, which implies that
u
λ− u
θ∈ intC
+.
Now, let λ
∗= inf L.
In what follows, for every λ > 0, let ϕ
λ: W
1,p(Ω) → R be the energy functional for problem (P
λ) defined by
ϕ
λ(u) = 1
p kDuk
pp+ 1 p
Z
Ω
(ξ(z) + λ)|u|
pdz + 1 p
Z
∂Ω
β(z)|u|
pdσ − Z
Ω
F (z, u)dz for all u ∈ W
1,p(Ω). Here F (z, x) = R
x0
f (z, s)ds. Evidently ϕ
λ∈ C
1(W
1,p(Ω)).
To be able to fix λ
∗more precisely, we need to strengthen the condition on f (z, ·) near zero.
H(f )
2: f : Ω × R → R is a Carath´eodory function such that f (z, 0) = 0 for a.a. z ∈ Ω, Hypotheses H(f )
2(i), (ii), (iii), (iv) are the same as Hypotheses H(f )(i), (ii), (iii), (iv) and
(v) there exists D ⊆ Ω with |D|
N> 0 such that
λ ˆ
1(β)x
p−1≤ f (z, x) for a.a. z ∈ Ω and all x ≥ 0,
λ ˆ
1(β)x
p−1< f (z, x) for a.a. z ∈ D and all x > 0
(recall ˆ λ
1(β) is the first eigenvalue of (9)).
Example. The following function satisfies Hypotheses H(f )
2above (as before, for the sake of simplicity we drop the z− dependence):
f (x) = ηx
p−1log x + 1 p
+ ξx
q−1for all x ≥ 0 with 1 < q < p, η > pˆ λ
1(β) and ξ > ˆ λ
1(β).
Proposition 7. If Hypotheses H(ξ), H(β), H(f )
2hold, then λ
∗> 0.
Proof. We argue indirectly. So, suppose that λ
∗= 0 and let λ
n↓ 0 as n → ∞.
Then {λ
n}
n≥1⊆ L (see Proposition 5). Moreover, from the last part of the proof of Proposition 5, we know that we can find {u
n}
n≥1⊆ W
1,p(Ω) such that u
n∈ S(λ
n) ⊆ int C
+(see Proposition 4) and
ϕ
λn(u
n) < 0 for all n ∈ N. (26) From (26) we have
− Z
Ω
pF (z, u
n)dz < −kDu
nk
pp− Z
Ω
(ξ(z) + λ
n)u
pndz − Z
∂Ω
β(z)u
pndσ for all n ∈ N.
(27) In addition, since u
n∈ S(λ
n) for all n ∈ N, we have
hA(u
n), hi + Z
Ω
(ξ(z) + λ
n)u
p−1nhdz + Z
∂Ω
β(z)u
p−1nhdσ = Z
Ω
f (s, u
n)hdz (28) for all h ∈ W
1,p(Ω) and all n ∈ N.
In (28) we choose h = u
n∈ W
1,p(Ω). Then Z
Ω
f (z, u
n)u
ndz = kDu
nk
pp+ Z
Ω
(ξ(z) + λ
n)u
pndz + Z
∂Ω
β(z)u
pndσ for all n ∈ N.
(29) We add (27) and (29). Then
Z
Ω
[f (z, u
n)u
n− pF (z, u
n)]dz < 0 for all n ∈ N. (30) Hypotheses H(f )
2(i), (iii) imply that we can find γ ∈ (0, γ
0) and c
1> 0 such that
γx
µ− c
1≤ f (z, x)x − pF (z, x) for a.a. z ∈ Ω, all x ≥ 0.
Using this estimate in (30) we obtain tat
{u
n}
n≥1⊆ L
µ(Ω) is bounded. (31) It is clear from Hypothesis H(f )
2(iii), that we may assume, without loss of generality, that µ < r.
First assume N 6= p and let t ∈ (0, 1) be such that 1
r = 1 − t µ + t
p
∗. (32)
From the interpolation inequality (see, for example, Gasinski - Papageorgiou [11, p.
905]), we have
ku
nk
r≤ ku
nk
1−tµku
nk
tp∗, so that
ku
nk
rr≤ c
2ku
nk
trfor some c
2> 0, all n ∈ N, see ( 31). (33) From Hypothesis H(f )
1(i) we have
f (z, u
n)u
n≤ c
3(1 + u
rn) for a.a. z ∈ Ω, all n ∈ N, some c
3> 0.
So, if in (28), we choose h = u
n∈ W
1,p(Ω), then
kDu
nk
pp≤ c
4(1 + ku
nk
rr) for some c
4> 0, all n ∈ N (see Hypotheses H(ξ), H(β))
≤ c
5(1 + ku
nk
tr) for some c
5> 0, all n ∈ N, see ( 33).
(34)
Recall that u 7→ kDuk
p+ kuk
µis an equivalent norm on W
1,p(Ω) (see, for example, Gasinski - Papageorgiou [11, p. 227]). So, from (31) and (34) we have
ku
nk
p≤ c
6(1 + ku
nk
tr) for some c
6> 0, all n ∈ N. (35) From (32) we get
tr = (r − µ)p
∗p
∗− µ < p by Hypothesis H(f )
2(iii), so from (35) we infer that
{u
n}
n≥1⊆ W
1,p(Ω) is bounded.
If N = p, then W
1,p(Ω) ,→ L
θ(Ω) for all θ ∈ [1, ∞). So, for the argument above to work, it is enough to replace p
∗by s > r large enough. More precisely, from (32) we have
tr = (r − µ)s s − µ and
(r − µ)s
s − µ → r − µ < p as s → +∞ (see Hypothesis H(f )
2(iii))
and so for s > r big, we have tr < p. Then the previous argument works and again we have that
{u
n}
n≥1⊆ W
1,p(Ω) is bounded.
So, we may assume that
u
n* u
∗in W
1,p(Ω) and u
n→ u
∗in L
p(Ω) and in L
p(∂Ω) as n → ∞. (36) In (28) we choose h = u
n− u
∗∈ W
1,p(Ω), pass to the limit as n → ∞ and use (36).
Then
n→∞
lim hA(u
n), u
n− u
∗i = 0.
Hence
u
n→ u
∗in W
1,p(Ω) (see Proposition 1). (37) So, if in (28) we pass to the limit as n → ∞ and use (37), then
hA(u
∗), hi + Z
Ω
ξ(z)u
p−1∗hdz + Z
∂Ω
β(z)u
p−1∗hdσ = Z
Ω
f (z, u
∗)hdz (38) for all h ∈ W
1,p(Ω) (recall that λ
n↓ 0). Hence
−∆u
∗(z) + ξ(z)u
∗(z)
p−1= f (z, u
∗(z)) for a.a. z ∈ Ω,
∂u
∗∂n
p+ β(z)u
p−1∗= 0 on ∂Ω,
see Papageorgiou - Radulescu [23]. It follows that u
∗∈ C
+by Lieberman [15, Theorem 2].
We will show that u
∗6= 0. To this end, note that given r ∈ (p, p
∗) and using Hypotheses H(f )
2(i), (ii), (iv), we can find c
7> 0 such that
f (z, x) ≥ c
0x
q−1− c
7x
r−1for a.a. z ∈ Ω, all x ≥ 0. (39)
For every λ > λ
∗= 0, we consider the following auxiliary nonlinear Robin problem:
(P)
λ
−∆
pu(z) + (ξ(z) + λ)u(z)
p−1= c
0u(z)
q−1− c
7u(z)
r−1in Ω,
∂u
∂n
p+ β(z)u(z)
p−1= 0 on ∂Ω,
u > 0.
As before, let µ > kξk
∞(see Hypothesis H(ξ)) and consider the C
1− functional ψ
λ: W
1,p(Ω) → R defined by
ψ
λ(u) = 1
p kDuk
pp+ 1 p
Z
Ω
(ξ(z) + λ)|u|
pdz + 1 p
Z
∂Ω
β(z)(u
+)
pdσ + µ
p ku
−k
pp+ c
7r ku
+k
rr− c
0q ku
+k
qq≥ 1
p kDu
+k
pp+ c
8ku
+k
rp− kξk
∞ku
+k
pp− c
9ku
+k
qq+ 1
p kDu
−k
pp+ 1
p [λ + µ − kξk
∞]ku
−k
ppform some c
8, c
9> 0 (see Hypotheses H(ξ), H(β) and recall that p < r)
= 1
p kDu
+k
pp+ (c
8ku
+k
r−qp− kξk
∞ku
+k
p−qp− c
9)ku
+k
qp+ 1
p kDu
−k
pp+ 1
p [λ + µ − kξk
∞]ku
−k
pp.
(40)
Since q < p < r and µ > kξk
∞, from (40) we infer that ψ
λis coercive. Also, it is sequentially weakly lower semicontinuous. So, by the Weierstrass Theorem, we can find ˜ u
λ∈ W
1,p(Ω) such that
ψ
λ(˜ u
λ) = inf n
ψ
λ(u) : u ∈ W
1,p(Ω) o
. (41)
As before (see the proof of Proposition 5), since q < p < r, we see that ψ
λ(˜ u
λ) < 0 = ψ
λ(0),
and hence ˜ u
λ6= 0.
From (41) we have
ψ
λ0(˜ u
λ) = 0.
It implies
hA(˜ u
λ), hi + Z
Ω
(ξ(z) + λ)|˜ u
λ|
p−2u ˜
λhdz − µ Z
Ω
(˜ u
−λ)
p−1hdz +
Z
∂Ω
β(z)(˜ u
+λ)
p−1hdσ
= c
0Z
Ω
(˜ u
+λ)
p−1hdz − c
7Z
Ω
(˜ u
+λ)
r−1hdz, for all h ∈ W
1,p(Ω).
(42)
In (42) we choose h = −˜ u
−λ∈ W
1,p(Ω). Then
kD˜ u
−λk
pp+ λ + µ − kξk
∞k˜ u
−λk
pp≤ 0,
so that ˜ u
λ≥ 0, ˜ u
λ6= 0 (recall that µ > kξk
∞). Then (42) becomes hA(˜ u
λ), hi +
Z
Ω
(ξ(z) + λ)˜ u
p−1λhdz + Z
∂Ω
β(z)(˜ u
λ)
p−1hdσ
= c
0Z
Ω
˜
u
q−1λhdz − c
7Z
Ω
˜ u
p−1λhdz
for all h ∈ W
1,p(Ω). Thus ˜ u
λis a positive solution of (P)
λ(see Papageorgiou - Radulescu [23]).
The nonlinear regularity theory implies that ˜ u
λ∈ C
+\ {0}. In addition, we have
∆
pu ˜
λ(z) ≤ (kξk
∞+ λ + c
7k˜ u
λk
r−p∞)˜ u
λ(z)
p−1for a.a. z ∈ Ω, so that ˜ u
λ∈ int C
+(see Pucci - Serrin [27, p.120]).
We will show that ˜ u
λ∈ int C
+is the unique positive solution of (P)
λ. Indeed, suppose that ˜ v
λis another positive solution of (P)
λ. As above, we can show that
˜
v
λ∈ int C
+.
Let j : L
1(Ω) → R = R ∪ {+∞} be the integral functional defined by
j(u) =
Z
Ω
|Du
1p|
pdz + Z
∂Ω
β(z)udσ if u ≥ 0, u
1p∈ W
1,p(Ω),
+∞ otherwise.
From Diaz - Saa [8] we know that j(·) is convex. Set domj = u ∈ L
1(Ω) : j(u) <
+∞ (the effective domain of j). Since ˜ u
λ, ˜ v
λ∈ int C
+, for h ∈ C
1(Ω) and |t| < 1 small, we have
˜
u
pλ+ th, ˜ v
λp+ th ∈ dom j.
Then the Gateaux derivative of j(·) at ˜ u
pλand ˜ v
λpin the direction h ∈ C
1(Ω) exists and using the chain rule and the nonlinear Green’s identity, we obtain
j
0(˜ u
pλ)(h) = 1 p Z
Ω
−∆
pu ˜
λ˜
u
p−1λhdz j
0(˜ v
pλ)(h) = 1
p Z
Ω
−∆
pv ˜
λ˜
v
λp−1hdz.
The convexity of j(·) implies the monotonicity of j
0(·). Hence
0 ≤ Z
Ω
−∆
pu ˜
λ˜
u
p−1λ− −∆
pv ˜
λ˜ v
λp−1!
(˜ u
pλ− ˜ v
λp)dz
= Z
Ω
"
c
01
˜
u
p−qλ− 1
˜ v
λp−q!
− c
7(˜ u
r−pλ− ˜ v
r−pλ)
#
(˜ u
pλ− ˜ v
λp)dz ≤ 0.
Thus ˜ u
λ= ˜ v
λand ˜ u
λ∈ int C
+is the unique positive solution of (P)
λ.
Let u ∈ S(λ) and consider the Carath´ eodory functions g(z, x) for (z, x) ∈ Ω × R and β
0(z, x) for (z, x) ∈ ∂Ω × R defined by
g(z, x) =
0 if x < 0,
c
0x
q−1− c
7x
r−1+ µx
p−1if 0 ≤ x ≤ u(z), c
0u(z)
q−1− c
7u(z)
r−1+ µu(z)
p−1if u(z) < x,
(43)
β
0(z, x) =
0 if x < 0, β(z)x
p−1if 0 ≤ x ≤ u(z), β(z)u(z)
−1if u(z) < x
(44)
(as before µ > kξk
∞). We set G(z, x) = R
x0
g(z, s)ds, B
0(z, x) = R
x0
β
0(z, s)ds and consider the C
1− functional ˆ γ : W
1,p(Ω) → R defined by
ˆ γ(u) = 1
p kDuk
pp+ 1 p
Z
Ω
(ξ(z) + λ + µ)|u|
pdz + Z
∂Ω
B
0(z, u)dσ − Z
Ω
G(z, u)dz for all u ∈ W
1,p(Ω). Since µ > kξk
∞, from (43) and (44) it is clear that ˆ γ is coercive.
Also, it is sequentially weakly lower semicontinuous. So, by the Weierstrass theorem, we can find ˜ u
0λ∈ W
1,p(Ω) such that
ˆ
γ(˜ u
0λ) = inf n ˆ
γ(u) : u ∈ W
1,p(Ω) o
. (45)
Since q < p < r, we have
ˆ
γ(˜ u
0λ) < 0 = ˆ γ(0).
Hence ˜ u
0λ6= 0. From (45) we have ˆ
γ
0(˜ u
0λ) = 0, thus
hA(˜ u
0λ), hi+
Z
Ω
(ξ(z)+λ+µ)|˜ u
0λ|
p−2u ˜
0λhdz+
Z
∂Ω
β
0(z, ˜ u
0λ)hdσ = Z
Ω
g(z, ˜ u
0λ)hdz (46) for all h ∈ W
1,p(Ω).
In (46), we choose h = −(˜ u
0λ)
−∈ W
1,p(Ω) and using (43), (44) we obtain kD(˜ u
0p)
−k
pp+ λ + µ − kξk
∞k(˜ u
0λ)
−k
pp≤ 0,
hence ˜ u
0λ≥ 0, ˜ u
0λ6= 0.
Next in (46) we choose (˜ u
0λ− u)
+∈ W
1,p(Ω). Using (43) and (44) we have hA(˜ u
0λ), (˜ u
0λ− u)
+i +
Z
Ω
(ξ(z) + λ + µ)(˜ u
0λ)
p−1(˜ u
0λ− u)
+dz +
Z
∂Ω
β(z)u
p−1(˜ u
0λ− u)
+dσ
= Z
Ω
[c
0u
q−1− c
7u
r−1+ µu
p−1](˜ u
0λ− u)
+dz
≤ Z
Ω
[f (z, u) + µu
p−1](˜ u
0λ− u)
+dz (see (39))
= hA(u), (˜ u
0λ− u)
+i + Z
Ω
(ξ(z) + λ + µ)u
p−1(˜ u
0λ− u)
+dz +
Z
∂Ω
β(z)u
p−1(˜ u
0λ− u)
+dσ (since u ∈ S(λ)).
Hence
hA(˜ u
0λ) − A(u), (˜ u
0λ− u)
+i + (λ + µ − kξk
∞) Z
Ω