Electronic Journal of Differential Equations, Vol. 2007(2007), No. 164, pp. 1–18.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
A NEUMANN PROBLEM WITH THE q-LAPLACIAN ON A
SOLID TORUS IN THE CRITICAL OF SUPERCRITICAL CASE
ATHANASE COTSIOLIS, NIKOS LABROPOULOS
Abstract. Following the work of Ding [21] we study the existence of a nontrivial positive solution to the nonlinear Neumann problem
∆q u + a(x)uq−1 = λf (x)up−1 ,
u>0
on T,
∂u
∇u|q−2
+ b(x)uq−1 = λg(x)up̃−1 on ∂T ,
∂ν
2q
q
3
p=
> 6, p̃ =
> 4,
< q < 2,
2−q
2−q
2
on a solid torus of R3 . When data are invariant under the group G = O(2)×I ⊂
O(3), we find solutions that exhibit no radial symmetries. First we find the
best constants in the Sobolev inequalities for the supercritical case (the critical
of supercritical).
1. Introduction
In this paper we study the existence of positive solutions of the Neumann boundary problem
∆q u + a(x)uq−1 = λf (x)up−1 ,
u>0
on T,
∂u
+ b(x)uq−1 = λg(x)up̃−1 on ∂T ,
|∇u|q−2
∂ν
2q
q
3
p=
> 6, p̃ =
> 4,
< q < 2,
2−q
2−q
2
(1.1)
∂
where ∂ν
is the outer unit normal derivative, ∆q u = −div(|∇u|q−2 ∇u) is the qLaplacian and for q = 2, ∆2 = ∆ is the Laplace-Beltrami operator.
Let Ω be a bounded domain in Rn , n ≥ 3 with smooth boundary ∂Ω. A host of
literature exists, concerning problems of the same type with (1.1), when q = 2; see
e.g. [8, 41, 2, 3, 42, 34, 15, 31, 35, 25, 44, 37, 26, 10, 12, 11, 20, 13, 16, 9, 33, 38,
39, 36, 43] and the references therein.
2000 Mathematics Subject Classification. 35J65, 46E35, 58D19.
Key words and phrases. Neumann problem; q-Laplacian; solid torus; no radial symmetry;
critical of supercritical exponent.
c
2007
Texas State University - San Marcos.
Submitted May 18, 2006. Published November 30, 2007.
1
2
A. COTSIOLIS, N. LABROPOULOS
EJDE-2007/164
In [17] the first author proved (under symmetry assumptions on Ω) the existence
and the multiplicity of positive solutions and of nodal solutions for the problem
∆u + a(x)u = f (x)|u|p−2 u,
on Ω,
∂u
+ b(x)u = h(x)|u|p̃−2 u on ∂Ω,
∂ν
2n
2(n − 1)
p≥
, p̃ ≥
.
n−2
n−2
(1.2)
where a(x), f (x) are functions in C ∞ (Ω) and b(x), h(x) are functions in C ∞ (∂Ω).
In contrast to the case q = 2, the Neumann problem, for q 6= 2, has not been
studied so extensively; see e.g. [5, 7, 19, 32, 22, 6, 24]. In all the above mentioned
cases the supercritical exponent under consideration is not the highest possible one.
In problem (1.1) the main difficulty comes firstly from the dimension 3 of the
q
2q
> 6 and p̃ = 2−q
> 4,
domain and secondly because the exponents p = 2−q
3
<
q
<
2
of
the
equation
and
the
boundary
condition,
respectively,
are
both
2
the highest possible supercritital exponents (critical of supercritical). Also, the
boundary condition is more complicated than the one in the above problems with
q 6= 2. Additionally, we have to find solutions that exhibit no radial symmetries.
However, since the solid torus T ⊂ R3 is invariant under the group G = O(2) × I ⊂
O(3), the solutions inherit T ’s symmetry property.
Best constants in Sobolev inequalities are fundamental in the study of non-linear
PDEs on manifolds [1, 27, 30, 23, 29, 4] and the references therein. It is also well
known that Sobolev embeddings can be improved in the presence of symmetries
[30, 21, 17, 28, 19] and the references therein.
q
In our case for any q ∈ [1, 2) real, the embedding H1,G
(T ) ,→ LpG (T ) is compact
2q/(2−q)
q
for 1 ≤ p < 2q/(2 − q), while H1,G
(T ) ,→ L1,G
(T ), is only continuous [18].
q
We will prove that for any q ∈ [1, 2) real, the embedding H1,G
(T ) ,→ LpG (∂T ) is
q/(2−q)
q
compact for 1 ≤ p < q/(2 − q), while H1,G
(T ) ,→ LG
(∂T ), is only continuous.
In the spirit of [1, 4] we determine the best constants of the Sobolev trace inequality
kukqLp̃ (∂T ) ≤ Ak∇ukqLq (T ) + BkukqLq (∂T ) ,
where p̃ = q/(2 − q), 1 ≤ q < 2, which concern the supercritical case (the critical
of supercritical) and we use the above to solve the problem (1.1).
2. Notation and statement of results
Let us define the solid torus
p
T = {(x, y, z) ∈ R3 : ( x2 + y 2 − l)2 + z 2 ≤ r2 , l > r > 0}
and A = {(Ωi , ξi ) : i = 1, 2} be an atlas on T defined by
+
Ω1 = {(x, y, z) ∈ T : (x, y, z) ∈
/ HXZ
},
−
Ω2 = {(x, y, z) ∈ T : (x, y, z) ∈
/ HXZ
}
where
+
HXZ
= {(x, y, z) ∈ R3 : x > 0 , y = 0}
−
HXZ
= {(x, y, z) ∈ R3 : x < 0 , y = 0}
EJDE-2007/164
A NEUMANN PROBLEM
3
and ξi : Ωi → Ii × D, i = 1, 2, with
I1 = (0, 2π),
D = {(t, s) ∈ R2 : t2 + s2 < 1}
I2 = (−π, π),
and ξi (x, y, z) = (ωi , t, s), i = 1, 2 with
x
y
cos ωi = p
, sin ωi = p
,
2
2
2
x +y
x + y2
i = 1, 2
where

y

arctan x , x 6= 0
ω1 = π/2,
x = 0, y > 0


3π/2,
x = 0, y < 0

y

arctan x , x 6= 0
ω2 = π/2,
x = 0, y > 0


−π/2,
x = 0, y < 0
and
p
x2 + y 2 − l
z
t=
, s= .
r
r
The Euclidean metric g on (Ω, ξ) ∈ A can be expressed as
√
( g ◦ ξ −1 )(ω, t, s) = r2 (l + rt).
Consider the spaces of all G-invariant functions under the action of the group
G = O(2) × I ⊂ O(3)
q
H1,G
= {u ∈ H1q (T ) : u ◦ τ = u , ∀τ ∈ G},
where H1q (T ) is the completion of C ∞ (T ) with respect the to norm
kukH1q = k∇ukq + kukq .
For all G-invariants u we define the functions φ(t, s) = (u ◦ ξ −1 )(ω, t, s). Then we
have
Z
kukpLp (T ) = 2πr2
|φ(t, s)|p (l + rt) dt ds,
(2.1)
D
Z
k∇ukqLq (T ) = 2πr2−q
|∇φ(t, s)|q (l + rt) dt ds,
(2.2)
D
Z
kukpLp (∂T ) = 2πr
|φ(t, 0)|p (l + rt) dt,
(2.3)
∂D
where by φ we denote the extension of φ on ∂D.
Let K(2, q) be the best constant [1] of the Sobolev inequality
kϕkLp (R2 ) ≤ K(2, q)k∇ϕkLq (R2 )
for the Euclidean space R2 , where 1 ≤ q < 2, p = 2q/(2 − q) and K̃(2, q) be the
best constant [30] in the Sobolev trace embedding
kϕkLp̃ (∂R2+ ) ≤ K̃(2, q)k∇ϕkLq (R2+ )
for the Euclidean half-space R2+ , where 1 ≤ q < 2, p̃ = q/(2 − q).
Consider a point Pj (xj , yj , zj ) ∈ T , and by OPj denote the orbit of Pj under the
q
action of the group G. Let lj = x2j + yj2 be the horizontal distance of the orbit
OPj from the axis z 0 z. For ε > 0 given and δj = lj ε, consider a finite covering
(Tj )j=1,...N with
p
Tj = {(x, y, z) ∈ T : ( x2 + y 2 − lj )2 + (z − zj )2 < δj2 }
an open small solid torus (a tubular neighborhood of the orbit OPj ).
4
A. COTSIOLIS, N. LABROPOULOS
EJDE-2007/164
2.1. Best constants on the solid Torus.
Theorem 2.1. Let T be the solid torus and p̃, q be two positive real numbers such
that p̃ = q/(2 − q) with 1 < q < 2. Then for all ε > 0 there exists a real number Bε
q
such that for all u ∈ H1,G
the following inequality holds:
i
h K̃ q (2, q)
q
q
(2.4)
kukqLp̃ (∂T ) ≤
iq−1 + ε k∇ukLq (T ) + Bε kukLq (∂T )
[2π(l − r)
In addition the constant K̃ q (2, q)/[2π(l − r)]q−1 is the best constant for the above
inequality.
2.2. Resolution of the problem. Consider the set
Λ = {c = (α, β) ∈ R2 : α − β ≥ δ, q ≤ α ≤ p, q ≤ β ≤ p̃},
with δ ∈ (0, p − p̃) = 0, q/(2 − q) , and define the functionals
Z
Z
q
q
I(u) =
(|∇u| + a|u| )dV +
b|u|q dS
T
∂T
Z
Z
α
g|u|β dS
Ic (u) =
f |u|α dV +
β ∂T
T
q
and for any c ∈ Λ. I(u) and Ic (u) are well defined because the
for all u ∈ H1,G
q
imbeddings of H1,G
onto Lp and Lpe are continue according to the Sobolev theorem.
Define, also
q
Σc = {u ∈ H1,G
: Ic (u) = 1},
µc = inf{I(u) : u ∈ Σc },
c0 = (p, p̃),
t+ = sup(t, 0),
for all t ∈ R. Consequently for the problem (1.1) we have the following theorem.
Theorem 2.2. Let a, f , b and g be four smooth functions, G-invariant and q,
p, p̃ be three real numbers defined as in (1.1). Suppose that the function f has
q
constant sign (e.g. f ≥ 0). The problem (1.1) has a positive solution u ∈ H1,G
if
the following holds:
h K q (2, q)µ+ ip/2 p
h K
ipe/2
e q (2, q)µ+
c0
c0
(sup f )
+ (sup g)+
<1
(2.5)
q−1
q/2
pe ∂T
[2π(l − r)]
[π(l − r)]
T
and if
(1) f > 0 everywhere and g is arbitrary, or
(2) f ≥ 0, g > 0 everywhere and (−infT a)+ κ < 1, where
κ = inf{A > 0 : ∃B > 0 s.t.kψkqLq (T ) ≤ Ak∇ψkqLq (T ) + BkψkqLq (∂T ) }.
(2.6)
In the rest of this paper we denote K = K(2, q), K̃ = K̃(2, q) and L = 2π(l − r).
3. Proofs
Proof of Theorem 2.1. The proof is carried out in two steps.
q
Step 1. Suppose that there exist two real numbers A, B such that for all u ∈ H1,G
the following inequality holds:
kukqLp̃ (∂T ) ≤ Ak∇ukqLq (T ) + BkukqLq (∂T ) .
EJDE-2007/164
A NEUMANN PROBLEM
5
Then
K̃ q (2, q)
|2π(l − r)|q−1
A≥
Consider a transformation F : D → R2+ . Such a transformation, for example, is
F (t, s) = (
t2
4t
2(1 − t2 − s2 )
, 2
),
2
+ (1 + s) t + (1 + s)2
see [23]. Denote by (g̃ij ) the Euclidian metric on D, dx dy the Euclidian metric on
R2+ and dσ the induced on ∂R2+ . Choose a finite covering of D̄ consisting of disks
Dk , centered on pk , such that: If pk ∈ D, then entire Dk lies in D and if pk ∈ ∂D,
Dk is a Fermi neighborhood. In these neighborhoods we have
q
1 − ε0 6 det(g̃ij ) 6 1 + ε0
(3.1)
Suppose by contradiction that, there exists
A<
K̃ q
Lq−1
and B ∈ R
such that the inequality
kukqLp̃ (∂T ) 6 Ak∇ukqLq (T ) + BkukqLq (∂T )
G
G
(3.2)
G
q
(T ). Fix a point P0 ∈ ∂T , that belongs to the orbit of
holds for all u ∈ H1,G
minimum range l − r. For any ε0 > 0, we can choose δ = ε0 (l − r) < 1 and
Tδ = {Q ∈ R3 : d(Q, OP0 ) < δ}
such that, if I × U ⊂ I × D is the image of a neighborhood of P0 ∈ ∂T through the
chart ξ of T and V ⊂ R2+ the image of U through F , (3.1) holds. It follows that,
for any u ∈ C0∞ (Tδ ), we have successively:
Z
Z
Z
q/p̃
p̃
q
q
|u| dS
6A
|∇u| dV + B
|u| dS,
∂Tδ
Tδ
∂Tδ
Z
q/p̃
p̃
2πδ
|φ| (l − r + δt)dt
∂D
Z
Z
q
q
6 2πδ 2−q A
|∇φ| (l − r + δt) dt ds + +2πδB
|φ| (l − r + δt)dt,
D
∂D
Z
q/p̃
p
p̃
−1
(1 − ε0 )δL
(|φ| g̃) ◦ F dσ
F (∂D)
Z
p
≤ (1 + ε0 )δ 2−q AL
(|∇φ|q g̃) ◦ F −1 dx dy
F (D)
Z
p
+ (1 + ε0 )δLB
(|φ|q g̃) ◦ F −1 dσ,
F (∂D)
(1 − ε0 )2 δL
Z
p̃
|Φ| dσ
q/p̃
∂R2+
Z
6 (1 + ε0 )2 δL δ 1−q A
R2+
q
Z
|∇Φ| dx dy + B
∂R2+
q
|Φ| dσ ,
6
A. COTSIOLIS, N. LABROPOULOS
Z
p̃
|Φ| dσ
q/p̃
∂R2+
6 f (ε0 )Lq−1 A
Z
EJDE-2007/164
Z
q
|∇Φ| dx dy + B̃
R2+
q
|Φ| dσ,
(3.3)
∂R2+
where f (ε0 ) = (1 + ε0 )2 /(1 − ε0 )2q/p̃ , B̃ = f (ε0 )(δL)q−1 B and p̃ = q/(2 − q).
Because of (3.2) and since the above function f : (0, 1) → (1, +∞) with
f (t) =
(1 + t)2
(1 − t)2q/p̃
is monotonically increasing, we can choose ε0 small enough, such that the following
inequality holds
K̃ q
A < f (ε0 )A < q−1
L
0
q
0
q−1
hence A < K̃ where A = f (ε0 )L A.
So for ε0 small enough and for all Φ ∈ C0∞ (D) we have
Z
Z
q/p̃
Z
q
q
p̃
6 A0
|∇Φ| dx dy + B̃
|Φ| dσ
(3.4)
|Φ| dσ
R2+
∂R2+
∂R2+
On the other hand by Hölder’s inequality, for all Φ ∈ C0∞ (Dδ ), where Dδ ⊂ D, we
have
Z
Z
q/p̃
q
p̃/q
1−(q/p̃)
|Φ| dσ0 6 [V ol(∂Dδ )]
(|Φ|q ) dσ0
∂Dδ
∂Dδ
and since p̃ = q/(2 − q), that is 1 − (q/p̃) = 1 − (2 − q) = q − 1, we have
Z
Z
q/p̃
q
p̃
q−1
|Φ| dσ0 6 V ol(∂Dδ )
|Φ| dσ0
∂Dδ
(3.5)
∂Dδ
Hence, choosing ε0 small enough, by (3.4) and (3.5), we get that there exists A00 <
K̃ q , such that for all Φ ∈ C0∞ (Dδ ),
Z
q/p̃
Z
p̃
q
00
|Φ| dσ
6A
|∇Φ| dx dy .
(3.6)
∂R2+
R2+
Let Ψ ∈ C0∞ (R2+ ) and set Ψλ (x) = λ1/p̃ Ψ(λx), λ > 0. For λ > 0, sufficiently
large, Ψλ ∈ C0∞ (D) and since kΨλ kLp̃ (∂R2+ ) = kΨkLp̃ (∂R2+ ) and k∇Ψλ kLq (R2+ ) =
k∇ΨkLq (R2+ ) , by (3.6), the following inequality
Z
Z
q/p̃
p̃
q
|Ψ| dσ
6 A00
|∇Ψ| dx dy
∂R2+
R2+
holds for all Ψ ∈ C0∞ (R2+ ). This is a contradiction since K̃ is the best constant for
the Sobolev inequality in R2+ .
q
Step 2. For all ε > 0 there exists a real number Bε such that for all u ∈ H1,G
the
following inequality holds:
kukqLp̃ (∂T ) ≤
K̃ q (2, q)
+ ε k∇ukqLq (T ) + Bε kukqLq (∂T )
q−1
[2π(l − r)]
Assume by contradiction that there exists ε0 > 0 such that for all α > 0 we can
q
find u ∈ H1,G
(T ) with
kukqLp̃ (∂T ) > (
G
K̃ q
+ ε0 )k∇ukqLq (T ) + αkukqLq (∂T )
G
G
Lq−1
EJDE-2007/164
or
A NEUMANN PROBLEM
k∇ukqLq (T ) + αkukqLq (∂T )
G
G
<
kukLp̃ (∂T )
7
−1
K̃ q
+ ε0
q−1
L
G
It follows that, the above inequality remains true for all ε ∈ (0, ε0 ) and setting
Iα =
k∇ukqLq (T ) + αkukqLq (∂T )
G
inf
1,q
G
kukqLp̃ (∂T )
u∈HG (T )\{0}
G
we conclude that for all α > 1, there exists θ0 > 0 independent of α such that
Iα <
As the quotient
−1
Lq−1
K̃ q
+
ε
=
− θ0
0
Lq−1
K̃ q
(3.7)
k∇ukqLq (T ) + αkukqLq (∂T )
G
G
kukqLp̃ (∂T )
G
q
is homogeneous, for any fixed α we can take a minimizing sequence (uk ) ⊂ H1,G
(T )
q
for it satisfying kuk kLp̃ (∂T ) = 1. As
G
k∇uk kqLq (T ) + αkuk kqLq (∂T ) → Iα ,
G
(3.8)
G
LqG (∂T )
LqG (∂T ).
we conclude that (uk ) is bounded in
and (∇uk ) is bounded in
By
the standard Sobolev trace inequality, we easily take, by contradiction, that there
exists a constant C such that
kukqLq (T ) ≤ C(k∇ukqLq (T ) + kukqLq (∂T ) ).
H1q (T ),
(3.9)
q
These two facts together imply that uk * u in
uk → u in L (T ) and uk → u
in Lq (∂T ). Since convergence in Lp spaces implies a.e. convergence, the function u
will be G-invariant. By theorem 4 of [4] we have uk → u in Lp̃G (∂T ), kukqLp̃ (∂T ) = 1
G
and
k∇ukqLq (T ) + αkukqLq (∂T ) = Iα ,
G
G
q
that is u is minimizing of Iα . Now, for each α > 0, let uα ∈ H1,G
(T ) satisfy
kuα kLp̃ (∂T ) = 1 and
G
Lq−1
− θ0
(3.10)
G
G
K̃ q
Following arguments similar to the ones that proved u is G-invariant minimizing of
q
Iα , we conclude that (uα ) is bounded in H1,G
(T ), thus we can take a subsequence
q
of (uα ), denoted (uα ) too, such that uα * u in H1,G
(T ), uα → u in LqG (T ) and
q
uα → u in LG (∂T ). Moreover, by (3.10) we obtain
k∇uα kqLq (T ) + αkuα kqLq (∂T ) = Iα 6
1 Lq−1
− θ0 ,
G
α K̃ q
and sending α to +∞ we have u = 0 on ∂T . Finally following the proof of theorem
4 of [4] we obtain that ∇uα → ∇u a.e. and so (∇uα ) is bounded in LqG (T ). Because
of (2.1), (2.2) and (2.3) and since 1 < q < 2, p̃ = q/ (2 − q) we have
kuα kqLq (∂T ) <
k∇uα kqLq (T ) + αkuα kqLq (∂T )
kuα kqLp̃ (∂T )
8
A. COTSIOLIS, N. LABROPOULOS
EJDE-2007/164
R
|∇uα |q dV + α ∂T |uα |q dS
R
( ∂T |uα |p̃ dS)q/p̃
2−q R
R
q
q
2π λδ
|∇φα | l − r + δ λt dtds + 2π λδ α ∂D |φα | l − r + δ λt dt
D
=
q/p̃
R
p̃
2π λδ ∂D |φα | l − r + δ λt dt
2−q R
R
q
q
2π λδ
|∇φα | l − r + δ λt dtds
2π λδ α ∂D |φα | l − r + δ λt dt
D
= +
q/p̃
q/p̃
R
R
p̃
p̃
2π λδ ∂D |φα | l − r + δ λt dt
2π λδ ∂D |φα | l − r + δ λt dt
R
2−q− p̃q
q
2πδ 2−q D |∇φα | l − r + δ λt dtds
1
=
q/p̃
R
λ
p̃
2πδ ∂D |φα | l − r + δ λt dt
R
1− p̃q
q
2πδα ∂D |φα | l − r + δ λt dt
1
+
q/p̃
R
λ
p̃
2πδ ∂D |φα | l − r + δ λt dt
R
R
2πδ 2−q D |∇φα |q (l − r + δ λt ) dt ds + 2πδα ∂D |φα |q (l − r + δ λt )dt
R
≤
(2πδ ∂D |φα |p̃ (l − r + δ λt )dt)q/p̃
R
=
T
and as λ → +∞ the above inequality yields
k∇uα kqLq (T ) + αkuα kqLq (∂T )
kuα kqLp̃ (∂T )
R
R
Lδ 2−q D |∇φα |q dt ds + Lδα ∂D |φα |q dt
R
≤
(Lδ ∂D |φα |p̃ dt)q/p̃
R
R
q
1−(q/p̃)
α ∂D |φα |q dt
q−1 D |∇φα | dtRds + δ
=L
( ∂D |φα |p̃ dt)q/p̃
R
R
|∇φα |q dt ds + α ∂D |φα |q dt
R
< Lq−1 D
( ∂D |φα |p̃ dt)q/p̃
From the above inequality and (3.10) we obtain
R
R
q
q
Lq−1
q−1 D |∇φα |R dt ds + α ∂D |φα | dt
− θ0
L
<
p̃
q/
p̃
( ∂D |φα | dt)
K̃
or
R
R
|∇φα |q dt ds + α ∂D |φα |q dt
1
D
R
<
−θ
(3.11)
( ∂D |φα |p̃ dt)q/p̃
K̃ q
According to [4, Theorem 4] such a function satisfying inequality (3.11) does not
exist, and the theorem is proved.
3.1. Proof of the main theorem.
Proof of Theorem 2.2. The proof is based on ideas from [14]. We recall, in this
point, some notation:
Λ = {c = (α, β) ∈ R2 : α − β ≥ δ, q ≤ α ≤ p, q ≤ β ≤ p̃},
where δ ∈ (0, p − p̃) = 0, q/(2 − q) ,
Z
Z
q
q
I(u) =
(|∇u| + a|u| )dV +
b|u|q dS,
T
∂T
EJDE-2007/164
A NEUMANN PROBLEM
Z
f |u|α dV +
Ic (u) =
T
α
β
Z
9
g|u|β dS
∂T
q
where u ∈ H1,G
and c ∈ Λ,
q
Σc = {u ∈ H1,G
: Ic (u) = 1},
µc = inf{I(u) : u ∈ Σc },
t+ = sup(t, 0), t ∈ R.
c0 = (p, p̃),
q
Because the imbeddings of H1,G
(T ) in LpG (T ) and Lp̃G (∂T ) are continuous but not
compact, we adopt the procedure of solving an approximating equation and then
we pass to the limit as in [1].
1. The proof of this part is carried out in six steps.
Step 1. A real tc ∈ Σc . Define on [0, +∞) the continuous function hc with
Z
Z
α β
α
gdS
hc (t) = t
f dV + t
β
∂T
T
Since α > β we have hc (0) = 0, limt→∞ hc (t) = +∞ and there exists tc > 0 such
that h(tc ) = 1. Hence the constant function, which in every point is equal to tc ,
belongs to Σc , and then Σc 6= ∅.
Step 2. sup{µc : c ∈ Λ} < +∞. We will prove that there exists t̃ ∈ R such that
tc 6 t̃ for all c ∈ Λ and the following holds
Z
Z
Z
Z
µc 6 I(tc ) =
adV +
bdS tqc 6
|a|dV +
|b|dS t̃q
T
• If
R
∂T
∂T
T
∂T
gdS > 0, since f > 0 and α > β, by equality
Z
Z
α β
1 = Ic (tc ) = tα
f
dV
+
t
gdS
c
β c ∂T
T
arises
tc =
Z
f dV +
T
However, if
R
T
α β−α
t
β c
Z
gdS
−1/α
∂T
while, if
T
f dV
−1/α
: q 6 α 6 p}
T
f dV < 1, since q 6 α 6 p, the following holds
Z
Z
−1/q
−1/α
f dV
6
f dV
<1
T
R
Z
> sup{
T
f dV > 1, we have the inequality
Z
Z
−1/q
−1/α
f dV
>
f dV
>1
T
T
Therefore, in this case, we set
Z
Z
−1/q
−1/α
t̃ = max{1,
f dV
} > sup{
f dV
:q6α6
T
• If
R
∂T
T
gdS < 0, let t̃0 ∈ R such that
R
| ∂T gdS| 1/δ t̃0 > max 1, p R
f dV
T
2q
}
2−q
10
A. COTSIOLIS, N. LABROPOULOS
EJDE-2007/164
When t > t̃0 , because of q 6 α 6 p, q 6 β 6 p̃ and β 6 α − δ, we get (α/β) 6 (p/q),
and then
Z
Z
α β
α
hc (t) = t
f dV + t
gdS
β
∂T
ZT
Z
p
> tα
f dV + tα−δ
gdS
q
T
∂T
R
Z
gdS| −δ p|
t
f dV
= tα 1 + R∂T
q T f dV
T
R
Z
p | ∂T gdS| −δ q
>t 1+ R
t̃0
f dV
q T f dV
T
Z
tq
>
f dV
q T
Hence, for
Z
−1/q
t̃ = max{q 1/q
f dV
, t̃0 }
T
we have hc (t̃) > 1 and then tc 6 t̃.
Step 3. inf{µc : c ∈ Λ} > −∞. Since f > 0 everywhere, m = inf T f > 0 and
q
because of H1,G
(T ) ,→ Lp̃G (∂T ) is continuous (see [18, lemma 2.1]) there exists
q
C > 1 such that for all ψ ∈ H1,G
(T ),
kψkp̃,∂T 6 C(k∇ψkq,T + kψkq,T )
We set
−1/q
m
C1 = sup
q6α6p
Z
f dV
(1/q)−(1/α) ,
T
C2 = sup 2β−1 q −1 p C β kgk∞ [V ol(∂T )]1−(β/p̃) ,
q6β6p̃
1/α
2 C1 (C2 + 1)1/α .
C3 = sup
q6α6p
We recall, for reals x, y ≥ 0, the elementary inequalities
(x + y)β 6 2β−1 (xβ + y β ),
1/α
(x + y)
62
1/α
1/α
(x
+y
1/α
(3.12)
).
Let u ∈ Σc with kukq,T > 1. Since u ∈ Σc we have
Z
Z
α
α
g|u|β dS = 1
f |u| dV +
β ∂T
T
and then
Z
Z
α
α
f |u| dV =1 −
g|u|β dS
β ∂T
T
By (3.12), (3.13) and since q 6 α 6 p, q 6 β 6 p̃ we obtain
Z
1/q
q
kukq,T 6 m−1/q
f |u| dV
T
Z
(1/q)−(1/α) Z
1/α
α
≤ m−1/q
f dV
f |u| dV
T
T
Z
α
6 C1 (1 −
g|u|β dS)1/α
β ∂T
(3.13)
EJDE-2007/164
A NEUMANN PROBLEM
11
p
6 C1 (1 + kgk∞ [V ol(∂T )]1−(β/p̃) kukβp̃,∂T )1/α
q
p
6 C1 [1 + kgk∞ [V ol(∂T )]1−(β/p̃) C β (k∇ukq,T + kukq,∂T )β ]1/α
q
6 C1 [1 + 2β−1 q −1 pkgk∞ [V ol(∂T )]1−(β/p̃) C β (k∇ukβq,T + kukβq,T )]1/α
6 C1 [C2 k∇ukβq,T + (C2 + 1)kukβq,T ]1/α
1/α
6 21/α C1 [C2
β/α
β/α
β/α
k∇ukq,T + (C2 + 1)1/α kukq,T ]
β/α
6 C3 (k∇ukq,T + kukq,T ).
Since
δ
δ
β
61− 61− ,
α
α
p
if ε ∈ (0, 1) and C4 = C4 (ε, p, δ, C3 ) is a constant such that
ε
t1−(δ/p) 6
t + C4 ,
C3
for any t > 0, the latter inequality implies
1−(δ/p)
1−(δ/p)
kukq,T 6 C3 (1 + k∇ukq,T
+ kukq,T
)
6 ε(k∇ukq,T + kukq,T ) + C3 (1 + 2C4 )
Hence if ε1 = ε/(1 − ε) and C = C3 (1 + 2C4 )/(1 − ε) is a constant depending on
ε1 , but not on c ∈ Λ, by the last inequality we obtain
kukq,T 6 ε1 k∇ukq,T + C
(3.14)
Since we can take C > 1, (3.14) holds for c ∈ Λ and for all u ∈ Σc .
q
q
If b 6≡ 0, since H1,G
(T ) ,→ LqG (∂T ) and H1,G
(T ) ,→ LqG (T ) are compact, for any
ε0 ∈ (0, 1) we can find a constant C 0 = C 0 (ε0 , b) such that
q
q
0
0
kψkqq,∂T 6 kbk−1
∞ (ε k∇ψkq,T + C kψkq,T )
q
for all ψ ∈ H1,G
(T ), and then we obtain
Z
Z
q
q
I(ψ) =
(|∇ψ| + a|ψ| )dV +
T
b|ψ|q dS
∂T
≥ k∇ψkqq,T − kak∞,T kψkqq,T − kbk∞,T kψkqq,∂T
(3.15)
≥ (1 − ε0 )k∇ψkqq,T − kak∞,T kψkqq,T − C 0 kψkqq,T
= (1 − ε0 )k∇ψkqq,T − A(ε0 )kψkqq,T
where A(ε0 ) = kak∞,T + C 0 . An inequality of the type (3.15) is always true if b ≡ 0,
because of inequality
I(ψ) > k∇ψkqq,T − kak∞,T kψkqq,T .
By (3.14) because of (3.12) we obtain
kukqq,T ≤ 2q−1 εq1 k∇ukqq,T + 2q−1 C q
2q−1 A(ε0 )εq1
Thus if we choose ε1 , small enough, such that 1 − ε0 −
u ∈ Σc , by (3.15) and (3.16) we obtain
I(u) > 1 − ε0 − 2q−1 A(ε0 )εq1 k∇ukqq,T − 2q−1 A(ε0 )C q
> −2q−1 A(ε0 )C q
(3.16)
> 0, for all
12
A. COTSIOLIS, N. LABROPOULOS
EJDE-2007/164
Hence for all c ∈ Λ, µ ≥ −2q−1 A(ε0 )C q .
We observe that if Γ is a subset of ∪c∈Λ Σc , such that sup{I(u) : u ∈ Γ} = L <
q
+∞, Γ is bounded in H1,G
(T ). Indeed, according to the former, for any u ∈ Γ we
have
I(u) + 2q−1 A(ε0 )C q
L + 2q−1 A(ε0 )C q
k∇ukqq,T 6
=c
q 6
0
q−1
0
1 − ε − 2 A(ε )ε1
1 − ε0 − 2q−1 A(ε0 )εq1
and
kukq,T 6 ε1 k∇ukq,T + C 6 c0 ,
thus
sup(k∇ukq,T + kukq,T ) < +∞ .
u∈Γ
Step 4. µc = inf{ I(u) : u ∈ Σc } is attained. Suppose that c = (α, β) ∈ Λ, α <
p , β < p̃. Let a sequence (uj ) ∈ Σc such that limj→∞ I(uj ) = µc . Since
|γuj | = γ(|uj |), (γuj is the trace of uj on ∂T ), a.e. on ∂T and I(|uj |) = I(uj ),
Ic (|uj |) = Ic (uj ) a.e. in T , we conclude that |uj | ∈ Σc ; in a way similar to the
one that employs (|uj |) in place of (uj ), (γ(|uj |) in place of γuj respectively) or we
can consider uj ’s nonnegative a.e. (the same for γuj ’s). The sequence (I(uj )) is
q
bounded in H1,G
(T ) implies that supj∈N (kuj kH1q (T ) ) < +∞. Since the imbeddings
β
q
q
of H1,G (T ) in LG (T ), LqG (∂T ), Lα
G (T ) and LG (∂T ) are compacts, there is a funcq
q
tion uc ∈ H1,G
(T ) and a subsequence (uj ) of (uj ) such that (uj ) * uc in H1,G
(T ),
r
(uj ) → uc in everyone of the previous L spaces, (uj ) → uc a.e. in T . (The same
holds and for traces on ∂T ). Hence uc and γuc are nonnegative. We also have
Ic (uc ) = limj→∞ Ic (uj ) = 1 and then uc ∈ Σc and Ic (uc ) > µc . Moreover since
k∇uc kq 6 lim k∇uj kq
j→+∞
and
Z
auqc dV
Z
+
T
∂T
buqc dS
= lim
j→∞
Z
T
auqj dV
Z
+
buqj dS
∂T
holds, we conclude that I(uc ) 6 limj→∞ I(uj ) = µc and I(uc ) = µc .
Step 5. There exists a week solution uc0 ≥ 0. We observe that the deferential
DIc (u)
is 6= 0 for all u ∈ Σc . (Because if DIc (u) = 0 for all ψ ∈ C0∞ (T )
R of Ic α−2
then T f u|u|
ψdV = 0. This implies that f u|u|α−2 ψ = 0 in (C0∞ (T ))0 and since
q
f > 0, u = 0. Then u = 0 in H1,G
(T ), which is impossible since Ic (u) = 1). After
q
this a Lagrange multiplicand λc exists such that, for all ψ ∈ H1,G
(T ), it satisfies
the next Euler equation
Z
Z
q−2
q−1
(|∇uc | ∇uc ∇ψ + auc ψ)dV +
buq−1
ψdS
c
T
∂T
Z
Z
(3.17)
β−1
= λc
f uα−1
ψdV
+
gu
ψdS
c
c
T
∂T
In the following we suppose that c = (α, β) → c0 = (p, p̃) and we will prove that
there are a real λc0 and a function uc0 such that
Z
Z
(|∇uc0 |q−2 ∇uc0 ∇ψ + auq−1
ψ)dV
+
buq−1
c0
c0 ψdS
T
∂T
Z
Z
= λc0 ( f up−1
gup̃−1
c0 ψdV +
c0 ψdS)
T
∂T
EJDE-2007/164
A NEUMANN PROBLEM
13
that is uc0 is a week solution of (1.1). Substituting ψ = uc in (3.17) we obtain
Z
Z
Z
Z
q
q
q
α
(|∇uc | + auc )dV +
buc dS = λc
f uc dV +
guβc dS
T
∂T
T
∂T
or
λc
Z
f uα
c dV +
Z
T
guβc dS = I(uc ) = µc
∂T
Moreover, we have
Z
α
guβc dS
f uα
dV
+
c
β
T
Z
Z
Z∂T
α
α
α
= ( f uc dV +
f uα dV
guβc dS) + (1 − )
β T
β T c
∂T
Z
1 = Ic (uc ) =
and since
α
(1 − )
β
Z
f uα
c dV < 0,
T
we obtain
Z
Z
β
q
> >0
α
p
T
∂T
Hence λc and µc have the same sign and since the set {µc }c ∈Λ is bounded a constant
C exists, such that
p
|λc | 6 |µc | 6 C
q
Since sup{I(uc ) : c ∈ Λ} < ∞, we have (step 2)
f uα
c dV +
guβc dS >
sup{kuc kH1q : c ∈ Λ} < ∞.
q
Moreover, because the embeddings of H1,G
(T ) in LpG (T ) and Lp̃G (∂T ) are continuous, we have
sup{kuc kp,T : c ∈ Λ} < ∞,
sup{kuc kp̃,∂T : c ∈ Λ} < ∞
We observe that
β−1
p̃−1
< 1 and then
β−1
kucβ−1 kp̃/(p̃−1),∂T 6 [V ol(∂T )]1−(β/p̃) kuc kp̃,∂T
6 [max(1, [V ol(∂T )]1−(β/p̃) )][max(1, kuc kp̃−1
p̃,∂T )] 6 ct
that is, the sets {uα−1
}, (respectively {uβ−1
}) are bounded in the Banach reflexive
c
c
spaces Lp/(p−1) (T ), (respectively Lp̃/(p̃−1) (∂T )), of which the dual Lp (T ) (respecq
tively Lp̃ (∂T )) contain H1,G
(T ).
The above implies the existence of a sequence cj = (αj , βj ) ∈ Λ, which converges
to c0 , of a real λc0 which is the limit of (λcj ) and of a function uc0 with the following
properties:
(a) ucj * uc0 on H1q (T ), (by Banach’s theorem),
(b) ucj → uc0 on Lq (T ) (resp. Lq (∂T ), (by Kondrakov’s theorem),
(c) ucj → uc0 a.e. in T , (resp. ∂T ) (by proposition 3.43 of [1]) and
α −1
β −1
(d) ucjj
* up−1
in Lp/(p−1) (T ) (resp. ucjj
* up̃−1
in Lp̃/(p̃−1) (∂T )) (by
c0
c0
Banach theorem).
14
A. COTSIOLIS, N. LABROPOULOS
EJDE-2007/164
From (a) arises that
Z
Z
q−2
|∇ucj | ∇ucj ∇ψdV →
|∇uc0 |q−2 ∇uc0 ∇ψdV
T
T
for all ψ ∈ H1q (T ).
From (c) arises that uc0 is G-invariant and uc0 > 0 on T (resp. γuc0 > 0 in ∂T ).
We may also assume that the sequence (µcj ) converges, with limit µ0 6 µc0 .
Indeed, let u ∈ Σc0 be a non-negative function such that µc0 6 I(u) 6 µc0 + ε, with
ε > 0. Then for all j, there exists a unique real number tj > 0 such that Icj (tj u) = 1
and limj→∞ tj = 1. If this is not the case, there will exist a subsequence tjk with
limj→∞ tjk = l 6= 1. But in T the following holds,
0 6 f uαjk 6 f (1 + up ) ∈ L1 (T )
and on ∂T the following also holds,
0 6 |g|uαjk 6 |g|(1 + up̃ ) ∈ L1 (∂T ).
According to the dominated convergence theorem we have
Z
Z
p
Icjk (tjk u) → lp
f up dV + lp̃
gup̃ dS
p̃
∂T
T
namely, Ic0 (lu) = 1. This is contradiction since l 6= 1 and Ic0 (u) = 1, whereas
we know that there exists unique real number r > 0 such that Ic0 (ru) = 1. Since
µcj 6 I(tj u) = tqj I(u) we conclude that lim supj→+∞ µcj 6 µc0 .
Now we can write equation (3.14) for ucj , and as j → +∞ by Lebesgue’s theorem,
q
we find that for all ψ ∈ H1,G
(T ) the following holds
Z
Z
(|∇uc0 |q−2 ∇uc0 ∇ψ + aucq−1
ψ)dV
+
bucq−1
ψdS
0
0
T
∂T
Z
Z
p̃−1
= λc0
f ucp−1
ψdV
+
gu
ψdS
c
0
0
T
∂T
which implies that (λc0 , uc0 ) is a week solution of the problem (1.1).
Step 6. uc0 > 0 everywhere. We proved in step 5 that uc0 ≥ 0. By the maximum
principle [40], the function uc0 is identically equal to 0 or uc0 > 0 everywhere in
T , and finally in T̄ : since every point P , where uc0 attains it’s minimum in T̄ ,
belongs to ∂T , assume that uc0 is regular and that there exists P0 ∈ ∂T such that
uc0 (P0 ) = 0. By
Hopf’s lemma [40], we have that the normal derivative has strict
sign, ∂uc0 /∂ν (P0 ) < 0, but the boundary condition imposes
∂uc0
p̃−1
(P0 ) = (−buq−1
c0 + λc0 guc0 )(P0 ) = 0,
∂ν
a contradiction which proves that uc0 (P ) > 0 in T̄ .
For the solution uc0 to be strictly positive it suffices uc0 6≡ 0, which implies that
k uc0 kq,T = limj→∞ k ucj kq,T > 0. By [18, theorem 3.1] and theorem 2.1 of this
p
paper, we conclude that for any K > K/ L/2 and for any K̃ > K̃/L(q−1)/q there
q
exists a constant C(K, K̃) such that for all ψ ∈ H1,G
(T ) the following inequalities
hold
|∇uc0 |q−2
kψkqp,T 6 Kq k∇ψkqq,T + Ckψkqq,T ,
(3.18)
kψkqp̃,∂T
(3.19)
6 K̃
q
k∇ψkqq,T
+
Ckψkqq,T
EJDE-2007/164
A NEUMANN PROBLEM
15
By (3.15) we have
k∇uc kqq,T 6 (1 + ε)I(uc ) + A(ε)kuc kqq,T
0
0
(3.20)
0
where 1 + ε = 1/(1 − ε ), A(ε) = A(ε )/(1 − ε ).
Let ε > 0 and D(ε, p) be a constant such that for any x, y > 0 and ρ ∈ [0, p] the
following holds
(x + y)ρ/q 6 (1 + ε)xρ/q + Dy ρ/q
(3.21)
By (3.18) because of (3.20), (3.21) and since I(uc ) = µc for α < p, β < p̃ we can
write
q
q
q
α/q
kuc kα
p,T 6 (K k∇uc kq,T + Ckuc kq,T )
q
q
α/q
6 [(1 + ε)q/p Kq µ+
c + (AK + C)kuc kq,T ]
q
α
α/q
(µ+
c )
q
α/q
6 (1 + ε) K
+ D(AK + C)
Similarly by (3.19) because of (3.20), (3.21) we can write,
(3.22)
kuc kα
q,T
β/q
kuc kβp̃,∂T 6 (1 + ε)q K̃β (µ+
+ D(AK̃q + C)β/q kuc kβq,T
c )
(3.23)
So by (3.22), (3.23) and Hölder inequality we have
Z
Z
α
1 = Ic (uc ) =
f uα
dV
+
guβc dS
c
β
T
∂T
6 (sup f )[V ol(T )]1−(a/p) kuc kα
p,T
T
α
+ (sup g )+ [V ol(∂T )]1−(β/p̃) kuc kβp̃,∂T
β ∂T
q
1−(a/p)
6 (1 + ε) [(sup f )[V ol(T )]
K
T
α
(3.24)
α/q
(µ+
]
c )
α
β/q
+ (1 + ε)q [ (sup g )+ [V ol(∂T )]1−(β/p̃) K̃β (µ+
]
c )
β ∂T
β
+ C1 kuc kα
q,T + C2 kuc kq,T
where the constants
C1 = D(AKq + C)α/q (sup f )[V ol(T )]1−(a/p) ,
T
C2 = (α/β)D(AK̃q + C)β/q (sup g )+ [V ol(∂T )]1−(β/p̃)
∂T
are bounded by a constant C̃(ε, K, K̃) > 0 independent of α and β. By (3.24) for
c = cj , since limj→∞ µcj 6 µc0 for j → +∞, we obtain
p
p̃/q
p/q
1 6 (1 + ε)q (sup f )Kp (µ+
+ (sup g )+ K̃p̃ (µ+
c0 )
c0 )
p̃ ∂T
T
+ C̃(kuc0 kpq,T + kuc0 kp̃q,T )
because the sequence (ucj ) → uc0 strongly into Lq (T ). Thus if the condition (2.5)
of the theorem
pis satisfied and if we choose ε > 0 small enough and K, K̃ close
enough to K/ L/2, K̃/L(q−1)/q , respectively, we obtain kuc0 kq,T > 0 and hence
we proved the first part of the theorem.
2. If f becomes 0, f > 0, we impose the condition g > 0, namely ν = inf ∂T g > 0,
the proof follows along similar lines as in case 1. Thus we have to find two positive
constants A1 , A2 > 0 such that if u ∈ ∪c ∈Λ Σc the following will hold,
I(u) > A1 k∇ukqq − A2
16
A. COTSIOLIS, N. LABROPOULOS
EJDE-2007/164
Since g > 0 we have supu∈∪c∈Λ Σc kukq,∂T = D < +∞, because for such a u the
following holds
Z
1 = Ic (u) >
guβ dS > νkukββ,∂T > ν[V ol(∂T )]1−(β/q) kukβq,∂T
∂T
Let κ be the best constant of the inequality (2.6), namely of
q
kψkqq,T 6 Ak∇ψkqq,T + Bkψkqq,∂T , ψ ∈ H1,G
If ăκ < 1 where ă = (− inf a)+ we choose A close to κ such that A1 = 1 − Aă > 0
and then we have
Z
Z
I(u) ≥
(|∇u|q − ăuq )dV +
buq dS
T
∂T
≥ k∇ukqq − ăkukqq,T − kbk∞ kukqq,∂T
≥ k∇ukqq − ăAk∇ukqq − ăBkukqq,∂T − kbk∞ kukqq,∂T
= (1 − ăA)k∇ukqq − (ăB + kbk∞ )kukqq,∂T
≥ A1 k∇ukqq − A2
where A1 = 1 − ăA, A2 = (ăB + kbk∞ )Dq and the theorem is proved.
Acknowledgments. The authors would like to thank the anonymous referee for
the valuable suggestions that improved this article.
References
[1] Th. Aubin, Some non linear problems in Riemannian Geometry, Springer, Berlin, 1998.
[2] Adimurthi and S.L. Yadava, Existence of a nonradial positive solution for the critical exponent
problem with Neumann boundary condition. J. Differential Equations, 104, no. 2, (1993),
298-306.
[3] Adimurthi, F. Pacella and S. L. Yadava, Interaction between the geometry of the boundary
and positive solutions of a semilinear Neumann problem with critical nonlinearity, J. Func.
Anal., 113, (1993), 318-350.
[4] R. J. Biezuner, Best constants in Sobolev trace inequalities, Nonlinear Analysis., 54 (2003),
575-589.
[5] P. Binding, P.Drabek Y. X. Huang, On Neumann boundary value problems for some quasilinear elliptic equations, Electronic Journal of Differential Equations, Vol 1997, No 05, (1997),
1-11.
[6] Gabriella Bognar, Pavel Drabek, The p-Laplacian equation with superlinear and supercritical
growth, multiplicity of radial solutions Nonlinear Analysis.60 (2005), 719-728.
[7] J. Fernández Bonder, J. Rossi, Existence results for the p−Laplacian with nonlinear boundary
conditions, Journal of Mathematical Analysis and Applications 263 (2001), 195-223.
[8] C. Budd, M. C. Knaap and L. A. Peletier, Asymptotic behavior of solutions of elliptic equations with critical exponent and Neumann boundary conditions, Proc. Roy. Soc. Edinburgh,
117A (1991), 225-250.
[9] Daomin Cao and Pigong Han, A note on the Positive Energy Solutions for Elliptic Equations
Involving Critical Sobolev Exponents, Applied Mathematics Letters, 16 (2003), 1105-1113.
[10] D. Cao, E.S. Noussair, S. Yan, Existence and nonexistence of interior-peaked solution for a
nonlinear Neumann problem, Pacific J. Math. 200 (1) (2001), 19-41.
[11] J. Chabrowski, On the nonlinear Neumann problem involving the critical Sobolev exponent
on the boundary, J. Math. Anal. Appl., 290 (2004), 605-619.
[12] J. Chabrowski, P. Girao, Symmetric solutions of the Neumann problem involving a critical
Sobolev exponent, Topological Methods in Nonlinear Analysis, 19 (2002), 1-27.
[13] J. Chabrowski, E. Tonkes, On the nonlinear Neumann problem with critical and supercritical
nonlinearities, Dissertationes Math. Rozprawy Mat., 417 (2003), 59pp.
[14] P. Cherrier, Probèmes de Neumann non lièaires sur les variètes riemanniennes J Functional.,
57 2 (1984), 154-206.
EJDE-2007/164
A NEUMANN PROBLEM
17
[15] M. Chipot, I. Shafrir and Fila, On Solutions to some elliptic equations with nonlinear Neumann boundary conditions, Advances in Differential Equations, Vol1, No 1,(1996), 91-110.
[16] D. Costa, and P. Girao, Existence and nonexistence of least energy solutions of the Neumann
problem for a semilinear elliptic equation with critical Sobolev exponent and a critical lowerorder perturbation, J. Differential Equations 188, (2003), 164202.
[17] A. Cotsiolis, Problème de Neumann avec exposant sur-critique et une non-linéarité surcritique sur le bord, C. R. Acad. Sci. Paris, 322 (1996), 139-142.
[18] A. Cotsiolis and N. Labropoulos, Dirichlet problem on a solid torus in the critical of supercritical case, Bull. Greek Math. Soc., To appear.
[19] F. Demengel and B. Nazaret, On some nonlinear partial differential equations involving the
p−Laplacian and critical Sobolev traces maps, Asymptotic Analysis,23, (2000), 135-156.
[20] Yinbin Deng and Shuangjie Peng Existence of multiple positive solutions for inhomogeneous
Neumann problem J. Math. Anal. Appl., 271 (2002), 155-174.
[21] W. Ding, On a Conformally Invariant Elliptic Equation on Rn , Commun. Math. Phys., 107
(1986), 331-335.
[22] E. Lami Dozo, O. Torne, Symmetry and Symmetry breaking for minimizers in the trace
inequality, Communications in Contemporary Mathematics, Vol 7, No 6 (2005), 727-746.
[23] J. Escobar, Sharp constant in a Sobolev trace inequality. Indiana Univ. Math. J. 37 no. 3,
(1988), 687-698.
[24] Michael Filippakis, Leszek Gasinski and Nikolaos S. Papageorgiou, Multiplicity Results for
Nonlinear Neumann Problems Canad. J. Math. Vol. 58 1, (2006) 64-92.
[25] C. Gui and N. Ghoussoub, Multi-peak solutions for a semilinear Neumann problem involving
the critical exponent, Math. Z. 229, (1998), 443-474.
[26] Zheng-Chao Han and YanYan Li, The Yamabe problem on manifolds with boundary: Existence and compactness results, Duke Mathematical Journal, Vol 99, No 3, (1999) 489-542.
[27] E. Hebey, Nonlinear Analysis on Manifolds: Sobolev Spaces and Inequalities, Courant Institute of Mathematical Sciences, Lectures Notes in Mathematics, 5, 1999.
[28] E. Hebey, M. Vaugon, Sobolev spaces in the presence of symmetries, J.Math.Pures Appl., 76
(1997), 859-881.
[29] YY Li, M. Zhu., Sharp Sobolev inequalities on Riemannian manifolds with boundaries,
Comm. Pure Appl.Math. 50 no 5 (1997), 449-487.
[30] P. L. Lions, The concentration compactness principle in the calculus of variations. The limit
case, I, II. Rev. Mat. Iberoamericana 1 no 1,2 (1985), 145-201, 45-121.
[31] S. Maier-Paape, K. Schmitt, Zhi-Qiang Wang, On Neumann problems for semilinear elliptic equations with critical nonlinearity. Existence and symmetry of multi-peaked solutions,
Commun. in Partial Differential Equations, Vol. 22, (9 and 10), (1997), 1493-1527.
[32] M. Motron, Méthode des sur et sous solutions pour la résolution d’un problème de type
Neumann faisant intervenir le p−Laplacian, C.R.Acad. Sci. Paris, Ser. I 335, (2002), 341344.
[33] Shuangjie Peng, The uniqueness of symmetric single peak solutions for a Neumann problem
involving critical exponent, Nonlinear Analysis 56, (2004), 19-42.
[34] D. Pierotti, S. Terracini, On a Neumann problem with critical Sobolev exponent and critical
nonlinearity on the boundary, Commun. in Partial Differential Equations 20(7 and 8) (1995),
1155-1187.
[35] D. Pierotti, S. Terracini, On a Neumann problem involving two critical Sobolev exponents:
remarks on geometrical and topological aspects, Calc. Var. 5, (1997), 271-291.
[36] Manuel del Pino, Monica Musso, Angela Pistoia, Super-critical boundary bubbling in a semilinear Neumann problem Ann. I. H. Poincare AN 22 (2005), 45-82.
[37] O. Rey, An elliptic Neumann problem with critical nonlinearity in three dimensional domains,
Comm. Contemp. Math. 1 (1999), 405-449.
[38] Olivier Rey and Juncheng Wei, Blowing up solutions for an elliptic Neumann problem with
sub- or supercritical nonlinearity Part I:N = 3, Journal of Functional Analysis, 212 (2004),
472-499.
[39] Olivier Rey, Juncheng Wei, Blowing up solutions for an elliptic Neumann problem with subor supercritical nonlinearity. Part II: N ≥ 4, Ann. I. H. Poincarè, AN 22 (2005), 459-484.
[40] J. L. Vazquez. A strong maximum principle for some quasilinear elliptic equations. Appl.
Math. Optim., 12, (1984), 191-202.
18
A. COTSIOLIS, N. LABROPOULOS
EJDE-2007/164
[41] Xu-Jia Wang, Neumann problems os semilinear elliptic equations involving critical Sobolev
exponents, Journal of Differential equations, 93 No 4, (1991), 283-310.
[42] Zhi-Qiang Wang, High-energy and multi-peaked solutions for a nonlinear Neumann problem
with critical exponents, Proceedings of the Royal Society of Edindurgh, 125A, (1995), 10031029.
[43] Juncheng Wei , Shusen Yan, Solutions with interior bubble and boundary layer for an elliptic
Neumann problem with critical nonlinearity C. R. Acad. Sci. Paris, Ser. I 343, (2006), 311316.
[44] Meijun Zhu, Uniqueness Results through a Priori Estimates I. A Three Dimensional Neumann
Problem Journal of differential equations 154, (1999), 284-317.
Athanase Cotsiolis
Department of Mathematics, University of Patras, Patras 26110, Greece
E-mail address: cotsioli@math.upatras.gr
Nikos Labropoulos
Department of Mathematics, University of Patras, Patras 26110, Greece
E-mail address: nal@upatras.gr