# A New Result Concerning the Solvability of a Class of General Systems of Variational Equations with Nonmonotone Operators: Applications to Dirichlet and Neumann Nonlinear Problems

Authors: Luisa Toscano; Speranza Toscano
Journal: International Journal of Differential Equations
 (2016)

---

## Abstract

A new result of solvability for a wide class of systems of variational equations depending on parameters and governed by nonmonotone operators is found in a Banach real and reflexive space with applications to Dirichlet and Neumann problems related to nonlinear elliptic systems.


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## Body

## 1. Introduction

Many researchers (e.g., [1–9]) have devoted (and are still devoting) their studies to the solvability and the investigation of multiple and positive solutions of nonlinear elliptic problems.A class of general systems of variational equations has been studied in [10]. The Dirichlet and Neumann problems investigated in [11] belong to this class.In this paper, we prove a further existence theorem related to the problem of [10] in the homogeneous case, by using the Lagrange multipliers and the “algebraic” approach which is based on the fibering method [12]. This theorem and the ones of [10] include the results of [8] and some results of [13–15]. Now, let us recall the problem studied in [10].LetW
1
,
·
1
,
…
,
W
n
,
·
n
n
≥
1 be Banach reflexive and real spaces. Let W
=
⨉
l
=
1
n
W
l with v
=
∑
l
=
1
n
v
l
l
∀
v
=
v
1
,
…
,
v
n
∈
W. Let ·
,
·
l [resp.,
·
,
·] be the duality between W
l
∗ dual space of W
l [resp.,
W
∗
dual  space  of
W] and W
l [resp.,
W]. Let us denote by “∂” Fréchet differential operator and by “∂
u
l” Fréchet differential operator with respect to u
l. Let A
≢
0 and D
j
≢
0 (j
=
1
,
…
,
m
;
m
≥
1) be real functionals defined in W and let B
l and B
^
l (l
=
1
,
…
,
n) be real functionals defined in W
l satisfying the following conditions:i
11
A is lower weakly semicontinuous in W and C
1
W
∖
0, B
l and B
^
l are weakly continuous in W
l and C
1
W
l, ∃
p
>
1
:
A
t
v
=
t
p
A
v  ∀
t
≥
0 and ∀
v
∈
W, B
l
t
v
l
=
t
p
B
l
v
l and B
^
l
t
v
l
=
t
p
B
^
l
v
l  ∀
t
≥
0 and ∀
v
l
∈
W
l.
i
12
D
j is weakly continuous in W and C
1
W
∖
0, ∃
q
j
>
1
:
D
j
t
v
=
t
q
j
D
j
v  ∀
t
≥
0, and ∀
v
∈
W, 1
<
q
1
<
⋯
<
q
m if m
>
1.Let us set the following:(1)Hλμv=Av-∑l=1nλlBlvl+μlB^lvl,Ev=Hλμv-∑j=1mDjv,∀v=v1,…,vn∈W,whereλ=λ1,…,λn,μ=μ1,…,μn∈Rn;Sλμ=v∈W:Hλμv=1,Vλμ-=v∈W:Hλμv<0,SD1=v∈W:D1v=-1,V+D1,…,Dm1=v∈W:∑j=1m1Djv>0asm1=1,…,m.Let us consider the following problem.Problem (P). Find u
=
u
1
,
…
,
u
n
∈
W
∖
0 such that(2)∂uiAu,vii=λi∂Biui,vii+μi∂B^iui,vii+∑j=1m∂uiDju,vii∀i∈1,…,n,∀vi∈Wi.In Section2, we present new cases c
1
–
c
4 in which Problem (P) is solvable. In these cases, we introduce one of the following hypotheses:i
13
∃
c
λ
,
μ
>
0
:
v
p
≤
c
λ
,
μ
H
λ
μ
v  ∀
v
∈
W.
i
14
∃
c
λ
,
μ
>
0
:
v
p
≤
c
λ
,
μ
H
λ
μ
v  ∀
v
∈
V
+
D
1 (if
V
+
D
1
≠
⌀).Theorem 1 assures the existence of at least one solution. It is possible to get the existence of a second solution from a result of [10, Theorem 2.2]. This result is based in particular on the following assumption:i
15
V
λ
μ
-
∩
S
D
1 is not empty and bounded in W.The applications to Dirichlet problems in Section3 [resp., Neumann problems in Section 4] (whose variational form is included in Problem (P)), for the sake of brevity, deal with the first case of solvability, since in this case we can use at the same time Theorems 1 and 4. When n
>
1, thanks to Propositions 2 and 5, we have got sufficient conditions so that the components of the found solutions are not identically equal to zero.


## 2. Solvability of Problem (P)

Let us consider the following cases:c
1
m
>
1, q
1
<
p, V
+
D
1
≠
⌀, D
j
≤
0  ∀
j
∈
2
,
…
,
m, and i
14 holds.
c
2
m
>
1, ∃
m
1
∈
2
,
…
,
m
:
q
m
1
<
p, D
j
≥
0  ∀
j
∈
1
,
…
,
m
1, D
j
≤
0  ∀
j
∈
m
1
+
1
,
…
,
m if m
1
<
m, and i
13 holds.
c
3
m
>
1, q
m
<
p, D
j
≥
0  ∀
j
∈
1
,
…
,
m
-
1, ∑
j
=
1
m
-
1
D
j
v
>
0  ∀
v
∈
W
∖
0, D
m changes sign, and i
13 holds.
c
4
m
>
1, p
<
q
1, D
1 changes sign, D
j
≥
0  ∀
j
∈
2
,
…
,
m, ∑
j
=
2
m
D
j
v
>
0  ∀
v
∈
W
∖
0, and i
13 holds.Let us introduce the open set A of the space W:(3)A=V+D1inc1,A=V+D1,…,Dm1inc2,A=W∖0inc3,c4,and let us set(4)E~t,v=Etv=tpHλμv-∑j=1mtqjDjv,(5)ψt,v=∂E~∂tt,v,∀t≥0,∀v∈A.It is easy to control that, for each v
∈
A, the equation(6)ψt,v=0has only one positive root t
v and it results in(7)∂ψ∂ttv,v>0,E~tv,v=mint≥0⁡E~t,v<0inc1–c3,(8)∂ψ∂ttv,v<0,E~tv,v=maxt≥0⁡E~t,v>0inc4.The implicit function theory assures the functional t
v belongs to C
1
A. Then, the functional E
~
~
v
=
E
~
t
v
,
v belongs to C
1
A and it results in(9)∂E~~u,v=ptup-1Hλμu-∑j=1mqjtuqj-1Dju·∂tu,v+tup∂Hλμu,v-∑j=1mtuqj∂Dju,v=tup·∂Hλμu,v-∑j=1mtuqj∂Dju,v∀u∈A,∀v∈W.It is important to remark that (9) can also be written as follows:(10)∂E~~u,v=tu·∂Hλμtuu,v-∑j=1m∂Djtuu,v∀u∈A,∀v∈W.Theorem 1.
Under assumptionsi
11, i
12, in cases c
1
–
c
4, one has(11)∃v0=v10,…,vn0∈Sλμ∩A:E~~v0=inf⁡E~~v:v∈Sλμ∩A,(12)u0=tv0v0issolutionofProblemP.Proof.
Let us set(13)e_=inf⁡E~~v:v∈Sλμ∩A,E~1t,v=tp-∑j=1mtqjDjv,ψ1t,v=∂E~1∂tt,v∀t≥0,∀v∈A,E~~1v=E~1t1v,v∀v∈A,where t
1
v is the positive root of equation ψ
1
t
,
v
=
0. Pointing out that E
~
~
1
∈
C
1
A and(14)E~~v=E~~1v∀v∈Sλμ∩A,let us prove that(15)∃v0∈A:E~~1v0=e_.Let v
k
⊆
S
λ
μ
∩
A such that E
~
~
1
v
k
→
e
_; that is,(16)t1vkp-∑j=1mt1vkqjDjvk⟶e_.We note that(17)pt1vkp-1-∑j=1mqjt1vkqj-1Djvk=0;moreover,(18)vkp≤cλ,μsince i
14 holds in c
1 and i
13 holds in c
2
–
c
4.
Inequality (18) implies that v
0
∈
W exists such that (within a subsequence) v
k
→
v
0 weakly in W. Then,(19)0≤Hλμv0≤1,(20)Djvk⟶Djv0asj=1,…,m.Let us verify that(21)sup⁡t1vk<+∞.In c
1
–
c
3, let δ
>
0 such that D
j
v
≤
δ
∀
v
∈
S
λ
μ
∩
A and ∀
j
∈
1
,
…
,
m. We set ∀
t
≥
0, Φ
t
=
p
t
p
-
1
-
δ
q
1
t
q
1
-
1, in c
1, and Φ
t
=
p
t
p
-
1
-
δ
∑
j
=
1
m
1
q
j
t
q
j
-
1 in c
2, Φ
t
=
p
t
p
-
1
-
δ
∑
j
=
1
m
q
j
t
q
j
-
1 in c
3.
Let us denote byt
0 the positive root of equation Φ
t
=
0. Since ψ
1
t
,
v
≥
Φ
t  ∀
t
≥
0 and ∀
v
∈
S
λ
μ
∩
A, we have t
1
v
k
≤
t
0  ∀
k
∈
N.
Inc
4, we note that from (17) t
1
v
k
q
1
D
1
v
k
=
p
/
q
1
t
1
v
k
p
-
∑
j
=
2
m
q
j
/
q
1
t
1
v
k
q
j
D
j
v
k; then,(22)t1vkp-∑j=1mt1vkqjDjvk=1-pq1t1vkp+∑j=2mqjq1-1t1vkqjDjvk≥1-pq1t1vkpfrom which (21) follows taking into account (16).
Relation (21) assures that ω
∈
0
,
+
∞ exists such that (within a subsequence) t
1
v
k
→
ω.
Consequently, from (16), (17), and (20), we obtain(23)ωp-∑j=1mωqjDjv0=e_,(24)pωp-1-∑j=1mqjωqj-1Djv0=0.Let us add that(25)ω>0,v0∈A.In c
1
–
c
3, since from (7) e
_
<
0, (23) implies that ω
>
0 and v
0
≠
0; then, v
0
∈
A in c
3, while (24) ⇒ v
0
∈
A in c
1 and c
2. In case c
4 (where from (8) e
_
≥
0), since from (17) ∑
j
=
1
m
q
j
t
1
v
k
q
j
-
p
D
j
v
k
=
p, we have ∑
j
=
1
m
q
j
ω
q
j
-
p
D
j
v
0
=
p from which we obtain (25).
Evidently, ((24), (25)) ⇒ ω
=
t
1
v
0; then, (23) can be written in the form t
1
v
0
p
-
∑
j
=
1
m
t
1
v
0
q
j
D
j
v
0
=
e
_; that is, E
~
~
1
v
0
=
e
_. Then, (15) holds.
Let us prove that(26)Hλμv0=1in19.In fact, taking into account that i
14 holds in c
1 and i
13 holds in c
2
–
c
4, we obtain(27)v0∈A⟹v0>0⟹Hλμv0>0.On the other hand, since v
0
∈
A
⇒
θ
v
0
∈
A  ∀
θ
>
0, we have(28)
d
d
θ
E
~
~
1
θ
v
0
=
p
t
1
θ
v
0
p
-
1
-
∑
j
=
1
m
q
j
t
1
θ
v
0
q
j
-
1
D
j
θ
v
0
d
d
θ
t
1
θ
v
0
-
∑
j
=
1
m
q
j
t
1
θ
v
0
q
j
θ
q
j
-
1
D
j
v
0
=
-
θ
-
1
p
t
1
θ
v
0
p
∀
θ
>
0
.Then, if H
λ
μ
v
0
<
1, we get the contradiction(29)Hλμv0-1/pv0∈Sλμ∩A,E~~1Hλμv0-1v0<E~~1v0=e_.Relations (14), (15), and (26) allow (11).
From (11), a Lagrange multiplier τ exists such that(30)∂E~~v0,v=τ∂Hλμv0,v∀v∈W.Setting in (30) v
=
v
0, we get τ
=
0, since ∂
H
λ
μ
v
0
,
v
0
=
p
H
λ
μ
v
0
=
p, and from (9) ∂
E
~
~
v
0
,
v
0
=
t
v
0
p
∂
H
λ
μ
v
0
,
v
0 − ∑
j
=
1
m
t
v
0
q
j
∂
D
j
v
0
,
v
0
=
p
t
v
0
p − ∑
j
=
1
m
q
j
t
v
0
q
j
D
j
v
0
=
0. Then, (30), taking into account (10), implies ∂
H
λ
μ
t
v
0
v
0
,
v
=
∑
j
=
1
m
∂
D
j
t
v
0
v
0
,
v  ∀
v
∈
W; then, (12) holds.Proposition 2.
Letn
>
1. Let v
0 as in Theorem 1 and F
⊆
S
λ
μ
∩
A. Let us suppose the following:i
0
h
∀
v
∈
F
v
~
h
∈
W
h
∖
0 and a function Φ
·
,
v
,
v
~
h
→
W belonging to C
o
s
0
,
1
∩
C
1
s
0
,
1 (0
≤
s
0
<
1) [resp., C
o
1
,
s
0
∩
C
1
1
,
s
0 (1
<
s
0
≤
+
∞)]  exist  such  that Φ
1
,
v
,
v
~
h
=
v, v
~
s
=
Φ
s
,
v
,
v
~
h
∈
S
λ
μ
∀
s
∈
s
0
,
1 [resp., ∀
s
∈
1
,
s
0] and for some s
1
∈
s
0
,
1 [r
e
s
p
.
,
s
1
∈
1
,
s
0] ∑
j
=
1
m
t
v
~
s
q
j
d
/
d
s
D
j
v
~
s
<
0 [resp., >
0] ∀
s
∈
s
1
,
1 [resp., ∀
s
∈
1
,
s
1].Then, v
0
∉
F.Proof.
In fact, withv and v
~
s as in i
0
h
s
2
∈
s
1
,
1 [resp.,
s
2
∈
1
,
s
1] exists such that v
~
s
∈
S
λ
μ
∩
A and d
/
d
s
E
~
~
v
~
s
=
-
∑
j
=
1
m
t
v
~
s
q
j
d
/
d
s
D
j
v
~
s
>
0 [resp., <
0] ∀
s
∈
s
2
,
1 [resp., ∀
s
∈
1
,
s
2]; then, E
~
~
v
~
s
<
E
~
~
v  ∀
s
∈
s
2
,
1 [resp., ∀
s
∈
1
,
s
2].Remark 3.
Conditioni
0
h with F
⊆
S
λ
μ includes condition i
16
h introduced in [10] and it implies the conclusion of Theorem 2.2 of [10].In the casec
5
m
>
1, i
15 holds, q
m
<
p, and D
j
≤
0  ∀
j
∈
2
,
…
,
m,for each v
∈
V
λ
μ
-
∩
S
D
1, (6) has only one positive root t
v and we have ∂
ψ
/
∂
t
t
v
,
v
≠
0.SetE
~
~
v
=
E
~
t
v
,
v
∀
v
∈
V
λ
μ
-
∩
S
D
1, in [10, Theorem 4.2], the following result has been proved.Theorem 4.
Under conditionsi
11, i
12, in case c
5, one has the following:(31)∃v-=v-1,…,v-n∈Vλμ-∩SD1:E~~v-=inf⁡E~~v:v∈Vλμ-∩SD1,u-=tv-v-issolutionofProblemP.We add the following proposition.Proposition 5.
Letn
>
1. Let v
- as in Theorem 4 and F
⊆
V
λ
μ
-
∩
S
D
1. Let one suppose the following:i
_
h
∀
v
∈
F
v
~
h
∈
W
h
∖
0 and a function Φ
·
,
v
,
v
~
h
→
W belonging to C
o
s
0
,
1
∩
C
1
s
0
,
1  (0
≤
s
0
<
1)  [resp., C
o
1
,
s
0
∩
C
1
1
,
s
0 (1
<
s
0
≤
+
∞)] exist such that Φ
1
,
v
,
v
~
h
=
v, v
~
s
=
Φ
s
,
v
,
v
~
h
∈
S
D
1
∀
s
∈
s
0
,
1  [resp., ∀
s
∈
1
,
s
0] and for some s
1
∈
s
0
,
1  [resp., s
1
∈
1
,
s
0] t
v
~
s
p
d
/
d
s
H
λ
μ
v
~
s
-
∑
j
=
1
m
t
v
~
s
q
j
d
/
d
s
D
j
v
~
s
>
0  [resp., <
0] ∀
s
∈
s
1
,
1 [resp., ∀
s
∈
1
,
s
1].Then, v
-
∉
F.Proof.
Letv and v
~
s be as in i
_
h. Since t
v
>
0, ψ
t
v
,
v
=
0 and ∂
ψ
/
∂
t
t
v
,
v
≠
0; then, an open ball B
~ exists with center v included in V
λ
μ
- and a unique functional t
∗
w belonging to C
1
B
~ such that t
∗
w
>
0 and ψ
t
∗
w
,
w
=
0
∀
w
∈
B
~. Evidently, t
∗
w
=
t
w
∀
w
∈
B
~
∩
S
D
1. Set s
2
∈
s
1
,
1 [resp., s
2
∈
1
,
s
1] such that v
~
s
∈
B
~
∀
s
∈
s
2
,
1 [resp., ∀
s
∈
1
,
s
2]; we have(32)
d
d
s
t
∗
v
~
s
p
H
λ
μ
v
~
s
-
∑
j
=
1
m
t
∗
v
~
s
q
j
D
j
v
~
s
=
t
v
~
s
p
d
d
s
H
λ
μ
v
~
s
-
∑
j
=
1
m
t
v
~
s
q
j
d
d
s
D
j
v
~
s
>
0
resp.,
<
0
∀
s
∈
s
2
,
1
resp.,
∀
s
∈
1
,
s
2
.Consequently, E
~
~
v
~
s
<
E
~
~
v
∀
s
∈
s
2
,
1 [resp.,
∀
s
∈
1
,
s
2].Remark 6.
LetW
l
l
=
1
,
…
,
n be a vector lattice. Let v
0[resp., v
-] be as in Theorem 1 [resp., Theorem 4]. If H
λ
μ
v
1
,
…
,
v
n
=
H
λ
μ
v
1
,
…
,
v
n and D
j
v
1
,
…
,
v
n
=
D
j
v
1
,
…
,
v
n
∀
v
=
v
1
,
…
,
v
n
∈
W and ∀
j
∈
1
,
…
,
m, then v
1
0
,
…
,
v
n
0
∈
S
λ
μ
∩
A, t
v
1
0
,
…
,
v
n
0
=
t
v
0, E
~
~
v
1
0
,
…
,
v
n
0
=
E
~
~
v
0 [resp., v
-
1
,
…
,
v
-
n
∈
V
λ
μ
-
∩
S
D
1
,
t
v
-
1
,
…
,
v
-
n
=
t
v
-
,
E
~
~
v
-
1
,
…
,
v
-
n
=
E
~
~
v
-].Then, reasoning as in Theorem1 proof’s final step [resp., Theorem 4.2 of [10]], we see that ±
t
v
0
v
1
0
,
…
,
v
n
0 [resp.,
±
t
v
-
v
-
1
,
…
,
v
-
n] are solutions of Problem (P). Consequently, we can assume that v
l
0
≥
0; that is, u
l
0
≥
0 [resp., v
-
l
≥
0
i.e.,
u
-
l
≥
0]as
l
=
1
,
…
,
n.


## 3. Dirichlet Problems

We assumeW
=
W
0
1
,
p
Ω
n (n
≥
1) with v
=
∑
l
=
1
n
∫
Ω
∇
v
l
p
d
x
1
/
p  ∀
v
=
v
1
,
…
,
v
n
∈
W, B
l
v
l
=
p
-
1
∫
Ω
b
l
v
l
p
d
x
∀
v
l
∈
W
0
1
,
p
Ω, and B
^
l
≡
0, where 1
<
p
<
∞, Ω
⊆
R
N is open, bounded, connected and C
2
,
β set 0
<
β
≤
1, b
l
∈
L
∞
Ω
∖
0 with b
l
≥
0.Let us consider the functionalA (as in i
11) such that A
v
≥
p
-
1
c
~
v
p
∀
v
∈
W  (
c
~
=
const.
>
0
).Let us use the following notations:p
~
=
N
p
/
N
-
p if N
>
p, p
~
=
∞ else; ·
N is the Lebesgue measure in R
N; H
λ, S
λ, V
λ
- are instead of H
λ
μ, S
λ
μ, V
λ
μ
-; λ
l
∗ is the first eigenvalue and u
l
∗ is the first eigenfunction of the problem u
l
∈
W
0
1
,
p
Ω
:
-
c
~
div
⁡
∇
u
l
p
-
2
∇
u
l
=
θ
b
l
u
l
p
-
2
u
l in Ω [16].We present some results about the validity of assumptionsi
13
–
i
15. To this aim, let us set I
=
1
,
…
,
n and, for each I
∗
≠
⌀
⊆
I, V
∗
=
v
∈
W
:
v
l
≡
0
if
l
∈
I
∖
I
∗
,
v
l
=
c
l
u
l
∗
if
l
∈
I
∗
with
c
l
∈
R
and
c
l
≠
0
as  some
l
∈
I
∗, let us introduce the following hypothesis:i
3
I
∗
⊆
I exists: D
1
v
<
0 for every v
∈
V
∗.
i
_
3
I
∗
⊆
I exists such that D
1
v
<
0 and A
v
=
c
~
p
-
1
∑
l
∈
I
∗
∫
Ω
∇
v
l
p
d
x for every v
∈
V
∗
.Proposition 7.
Ifλ
l
<
λ
l
∗ for each l
∈
1
,
…
,
n, then i
13 holds. When V
+
D
1
≠
⌀ and i
3 holds with I
∗
≠
I, then, with λ
l
<
λ
l
∗ as l
∈
I
∖
I
∗, ∃
δ
∗
>
0
:
i
14 holds if λ
l
<
λ
l
∗
+
δ
∗ as l
∈
I
∗. When V
+
D
1
≠
⌀ and i
3 holds with I
∗
=
I, then ∃
δ
∗
>
0
:
i
14 holds if λ
l
<
λ
l
∗
+
δ
∗ as l
∈
I.Proof.
The first statement is obvious. We can prove the second and the third ones as in [11, Propositions 2.3 and 2.4].Proposition 8.
Wheni
_
3 holds with I
∗
≠
I, then with λ
l
<
λ
l
∗ as l
∈
I
∖
I
∗
∃
δ
∗
>
0
:
i
15 holds if λ
l
∈
λ
l
∗
,
λ
l
∗
+
δ
∗  ∀
l
∈
I
∗ and λ
l
>
λ
l
∗ as some l. When i
_
3 holds with I
∗
=
I, then ∃
δ
∗
>
0
:
i
15 holds if λ
l
∈
λ
l
∗
,
λ
l
∗
+
δ
∗  ∀
l
∈
I and λ
l
>
λ
l
∗ as some l.Proof.
See [11, Propositions 2.5 and 2.6].Let us now investigate Problem (P) in two concrete cases wherec
~
=
1.Application 1. Let n
>
1 and, for each v
=
v
1
,
…
,
v
n
∈
W,(33)Av=p-1∫Ω∑l=1n∇vlγp/γdx,D1v=∫Ωρ1v1q11dx∫Ω∑l=1nvlγ1q12/γ1dx,Djv=qj-1∫Ωρjdj1v1γj+djjvjγjqj/γjdxasj=2,…,n,where(34)1<γ<p,q11>1,1<γ1<q12,1<γj<qjasj=2,…,n,q1=q11+q12<q2<⋯<qn<p~,q1<p;ρ1∈L∞Ω;ρj∈L∞Ω,ρj<0,dj1,djj∈L∞Ω∖0asj=2,…,n.Let us consider the following system:(35)-div⁡∑l=1n∇ulγp/γ-1∇u1γ-2∇u1=λ1b1u1p-2u1+q11∫Ω∑l=1nulγ1q12/γ1dxρ1u1q11-2u1+q12∫Ωρ1u1q11dx∑l=1nulγ1q12/γ1-1·u1γ1-2u1+∑j=2nρjdj1u1γj+djjujγjqj/γj-2·dj1u1γj+djjujγjdj1u1γj-2u1inΩ,-div⁡∑l=1n∇ulγp/γ-1∇uiγ-2∇ui=λibiuip-2ui+q12∫Ωρ1u1q11dx∑l=1nulγ1q12/γ1-1·uiγ1-2ui+ρidi1u1γi+diiuiγiqi/γi-2·di1u1γi+diiuiγidiiuiγi-2uiinΩasi=2,…,n,ui=0on∂Ωasi=1,…,n.Let us introduce the following conditions:(36)ρ1+≢0⟹V+D1≠⌀,(37)∫Ωρ1u1∗q11dx<0⟹D1c1u1∗,…,cnun∗<0iffc1≠0.Then (Propositions 7 and 8),(38)
36
⟹
with
λ
l
<
λ
l
∗
as
l
=
1
,
…
,
n
i
14
holds
,
(39)
36
,
37
⟹
with
λ
l
<
λ
l
∗
as
l
=
2
,
…
,
n
∃
δ
1
∗
>
0
:
i
14
holds  if
λ
1
<
λ
1
∗
+
δ
1
∗
,
(40)
37
⟹
with
λ
l
<
λ
l
∗
as
l
=
2
,
…
,
n
∃
δ
2
∗
>
0
:
i
15
holds  if
λ
1
∈
λ
1
∗
,
λ
1
∗
+
δ
2
∗
.Proposition 9 (Theorems1 and 4, Remark 6,).
Under conditions (34), one has the following:When (36) holds [resp., (36) and (37) hold], with λ
1
,
…
,
λ
n as in (38) [resp., (39)], system (35) has at least two weak solutions u
0 and -
u
0
u
0
=
τ
0
v
0
,
τ
0
=
c
o
n
s
t
.
>
0
,
v
0
∈
S
λ
∩
V
+
D
1, and one has u
i
0
≥
0 as i
=
1
,
…
,
n, u
1
0
≢
0.
Whenq
n
<
p and (37) holds, with λ
1
,
…
,
λ
n as in (40), system (35) has at least two weak solutions u
- and -
u
-
u
-
=
τ
-
v
-
,
τ
-
=
c
o
n
s
t
.
>
0
,
v
-
∈
V
λ
-
∩
S
D
1, and one has u
-
i
≥
0 as i
=
1
,
…
,
n, u
-
1
≢
0.Consequently, when q
n
<
p and (36), (37) hold, with λ
l
<
λ
l
∗ as l
≥
2 and λ
1
∈
λ
1
∗
,
λ
1
∗
+
min
⁡
δ
1
∗
,
δ
2
∗, system (35) has at least four different weak solutions.Proposition 10.
If(41)eitherγ1<γ≤γjasj=2,…,n(42)ordj1djj<0,γj<γasj=2,…,n,then u
i
0
≢
0 as i
=
2
,
…
,
n.Proof.
It is sufficient (Proposition2) to verify the following.
Ash
=
2
,
…
,
n
i
0
h holds with F
=
v
∈
S
λ
∩
V
+
D
1
:
v
l
≥
0
as
l
=
1
,
…
,
n
,
v
1
≢
0
,
v
h
≡
0
≠
⌀.
Let, for example,h
=
2. Let v
∈
F. First of all,(43)∫Ωρ1v1q11dx>0,λ2<λ2∗⟹δ=∫Ω∇v1pdx-λ2∫Ωb2v1pdx>0.Let us introduce the function g
s
,
τ
=
H
λ
s
v
1
,
τ
v
1
,
…
,
s
v
n = p
-
1
[
∫
Ω
τ
γ
∇
v
1
γ
+
s
γ
∑
l
≠
2
∇
v
l
γ
p
/
γ
d
x − τ
p
λ
2
∫
Ω
b
2
v
1
p
d
x
-
s
p
∑
l
≠
2
λ
l
∫
Ω
b
l
v
l
p
d
x
]  ∀
s
,
τ
≥
0. Let us note that(44)g1,0=1,∂g∂τs,τ≥δτp-1>0∀s≥0,∀τ>0,gs,0=sp<1∀s∈0,1,limτ→+∞⁡gs,τ=+∞∀s≥0;then,(45)only oneτs>0τs>0ifs<1,τ1=0exists such thatgs,τs=1,∀s∈0,1.About the function τ
s, we have(46)limτ→1-⁡τs=0.In fact, with s
n
⊆
0,1 and lim
⁡
s
n
=
1, since g
s
n
,
τ
s
n
=
1, τ
s
n is necessarily bounded. Then (within a subsequence), lim
⁡
τ
s
n
=
ω with g
1
,
ω
=
1, from which ω
=
0. Then, τ
s belongs to C
0
0,1.
Let us add thatτ
s belongs to C
1
0,1 and it results in(47)τ′s=-1τsγ-1g~s,τs∀s∈0,1,lims→1-⁡g~s,τs∈0,+∞.Then, with v
~
s
=
s
v
1
,
τ
s
v
1
,
…
,
s
v
n, we have(48)v~1=v,v~s∈Sλ∀s∈0,1,and, moreover, taking into account the first one of (43),(49)lims→1-⁡ddsD1v~s=-∞,lims→1-⁡ddsDjv~s∈Rasj≥2when  (41)  holds,lims→1-⁡ddsD1v~s<+∞,lims→1-⁡ddsD2v~s=-∞,lims→1-⁡ddsDjv~s∈Rasj>2ifn>2when  (42)  holds,from which lim
s
→
1
-
⁡
∑
j
=
1
n
t
v
~
s
q
j
d
/
d
s
D
j
v
~
s
=
-
∞.Proposition 11.
Ifd
j
1
d
j
j
<
0 and γ
j
<
γ
1
≤
γ as j
=
2
,
…
,
n, then u
-
i
≢
0 as i
=
2
,
…
,
n.Proof.
It is sufficient (Proposition5) to prove the following.
Ash
=
2
,
…
,
n
i
_
h holds with F
=
v
∈
V
λ
-
∩
S
D
1
:
v
l
≥
0
as
l
=
1
,
…
,
n
,
v
1
≢
0
,
v
h
≡
0
≠
⌀.
Let, for example,h
=
2. Let v
∈
F. Firstly, we have(50)∫Ωρ1v1q11dx<0.Let us consider the function g
1
s
,
τ
=
D
1
s
v
1
,
τ
v
1
,
…
,
s
v
n
=
s
q
11
∫
Ω
ρ
1
v
1
q
11
d
x
∫
Ω
τ
γ
1
v
1
γ
1
+
s
γ
1
∑
l
≠
2
v
l
γ
1
q
12
/
γ
1
d
x
∀
s
,
τ
≥
0. Since(51)g11,0=-1,∂g1∂τs,τ<0∀s>0,∀τ>0,g1s,0=-sq1>-1∀s∈0,1,limτ→+∞⁡g1s,τ=-∞∀s>0,we have the following: (52)only oneτs≥0τs>0ifs<1,τ1=0exists such thatg1s,τs=-1,∀s∈0,1,the function τ
s belongs to C
0
0,1
∩
C
1
0,1, τ
′
s
=
-
1
/
τ
s
γ
1
-
1
g
~
1
s
,
τ
s, and(53)lims→1-⁡g~1s,τs∈0,+∞,∀s∈0,1.Then, set v
~
s
=
s
v
1
,
τ
s
v
1
,
…
,
s
v
n; we get(54)v~1=v,D1v~s=-1∀s∈0,1,lims→1-⁡ddsHλv~s∈R,lims→1-⁡ddsD2v~s=-∞,lims→1-⁡ddsDjv~s∈Rasj>2ifn>2;in particular, lim
s
→
1
-
⁡
t
v
~
s
p
d
/
d
s
H
λ
v
~
s
-
∑
j
=
1
n
t
v
~
s
q
j
d
/
d
s
D
j
v
~
s
=
+
∞.Application 2. Let n
>
2 and, for each v
=
v
1
,
…
,
v
n
∈
W,(55)
A
v
=
p
-
1
∑
l
=
1
n
∫
Ω
∇
v
l
p
d
x
,
D
1
v
=
q
1
-
1
∫
Ω
d
1
v
1
γ
1
+
d
∑
l
=
2
n
v
l
γ
1
q
1
/
γ
1
d
x
-
∫
Ω
d
~
∑
l
=
2
n
v
l
γ
1
q
1
/
γ
1
d
x
,
D
j
v
=
q
j
-
1
∑
l
=
1
l
≠
j
n
∫
Ω
ρ
j
d
j
j
v
j
γ
j
+
d
j
l
v
l
γ
j
q
j
/
γ
j
d
x
as
j
=
2
,
…
,
n
,where(56)1<γj<qjasj=1,…,n,q1<⋯<qn,q1<p,qn<p~;d1,d,d~∈L∞Ω,withd1,d>0,dq1/γ1<d~;ρj∈L∞Ω,withρj<0,djl∈L∞Ω∖0asj=2,…,n,l=1,…,n.Let us consider the system:(57)
-
div
⁡
∇
u
1
p
-
2
∇
u
1
=
λ
1
b
1
u
1
p
-
2
u
1
+
d
1
d
1
u
1
γ
1
+
d
∑
l
=
2
n
u
l
γ
1
q
1
/
γ
1
-
1
u
1
γ
1
-
2
u
1
+
∑
j
=
2
n
ρ
j
d
j
j
u
j
γ
j
+
d
j
1
u
1
γ
j
q
j
/
γ
j
-
2
d
j
j
u
j
γ
j
+
d
j
1
u
1
γ
j
d
j
1
u
1
γ
j
-
2
u
1
in
Ω
,
-
div
⁡
∇
u
i
p
-
2
∇
u
i
=
λ
i
b
i
u
i
p
-
2
u
i
+
d
d
1
u
1
γ
1
+
d
∑
l
=
2
n
u
l
γ
1
q
1
/
γ
1
-
1
u
i
γ
1
-
2
u
i
-
d
~
∑
l
=
2
n
u
l
γ
1
q
1
/
γ
1
-
1
·
u
i
γ
1
-
2
u
i
+
∑
l
=
1
l
≠
i
n
ρ
i
d
i
i
u
i
γ
i
+
d
i
l
u
l
γ
i
q
i
/
γ
i
-
2
·
d
i
i
u
i
γ
i
+
d
i
l
u
l
γ
i
d
i
i
u
i
γ
i
-
2
u
i
+
∑
j
=
2
j
≠
i
n
ρ
j
d
j
j
u
j
γ
j
+
d
j
i
u
i
γ
j
q
j
/
γ
j
-
2
d
j
j
u
j
γ
j
+
d
j
i
u
i
γ
j
d
j
i
u
i
γ
j
-
2
u
i
in
Ω
as
i
=
2
,
…
,
n
,
u
i
=
0
on
∂
Ω
as
i
=
1
,
…
,
n
.Evidently, V
+
D
1
≠
⌀ and D
1
0
,
c
2
u
2
∗
,
…
,
c
n
u
n
∗
<
0
∀
c
2
,
…
,
c
n
∈
R
n
-
1
∖
0. Then, set I
∗
≠
⌀
⊆
2
,
…
,
n; we have (Propositions 7 and 8)(58)withλl<λl∗asl∈I∖I∗∃δ1∗>0:i14holds  ifλl<λl∗+δ1∗∀l∈I∗,(59)withλl<λl∗asl∈I∖I∗∃δ2∗>0:i15holds  ifλl∈λl∗,λl∗+δ2∗∀l∈I∗,λl>λl∗as  somel.Proposition 12 (see Theorems1 and 4, Remark 6).
Under conditions (56), one has the following:Withλ
1
,
…
,
λ
n as in (58), system (57) has at least two weak solutions u
0 and -
u
0
u
0
=
τ
0
v
0
,
τ
0
=
c
o
n
s
t
.
>
0
,
v
0
∈
S
λ
∩
V
+
D
1, and one has u
i
0
≥
0 as i
=
1
,
…
,
n
,
u
1
0
≢
0.
Whenq
n
<
p, with λ
1
,
…
,
λ
n as in (59), system (57) has at least two weak solutions u
- and -
u
-
u
-
=
τ
-
v
-
,
τ
-
=
c
o
n
s
t
.
>
0
,
v
-
∈
V
λ
-
∩
S
D
1, and one has u
-
i
≥
0 as i
=
1
,
…
,
n.Consequently, when q
n
<
p, with λ
l
<
λ
l
∗ as l
∈
I
∖
I
∗, λ
l
∈
λ
l
∗
,
λ
l
∗
+
min
⁡
δ
1
∗
,
δ
2
∗  ∀
l
∈
I
∗, and λ
l
>
λ
l
∗ as some l, system (57) has at least four different weak solutions.Proposition 13.
Letd
j
j
d
j
l
<
0 as j
=
2
,
…
,
n and l
=
1
,
…
,
n with l
≠
j. Let γ
j
<
γ
1 as j
=
2
,
…
,
n. Then,(60)ui0≢0asi=2,…,n;u-i≢0asi=1,…,n.Proof.
It is sufficient (Propositions2 and 5) to prove that(61)
As
h
=
2
,
…
,
n
i
0
h
holds  with
F
=
v
∈
S
λ
∩
V
+
D
1
:
v
l
≥
0
as
l
=
1
,
…
,
n
,
v
1
≢
0
,
v
h
≡
0
≠
⌀
;
(62)
i
_
1
holds  with
F
=
v
∈
V
λ
-
∩
S
D
1
:
v
l
≥
0
as
l
=
1
,
…
,
n
,
v
1
≡
0
≠
⌀
;
(63)
As
h
=
2
,
…
,
n
i
_
h
holds  with
F
=
v
∈
V
λ
-
∩
S
D
1
:
v
l
≥
0
as
l
=
1
,
…
,
n
,
v
1
≢
0
,
v
h
≡
0
,
if
F
≠
⌀
.
About (61), let v
∈
F. Let K
⊆
Ω be a compact set such that K
N
>
0 and v
1
>
0 in K. Let [11, Proposition A
.
1] φ
ε
0
<
ε
<
ε
0
⊆
C
0
∞
Ω with 0
≤
φ
ε
≤
1 such that(64)φε⟶χstrongly  inLsΩ,∫Ω∇φεsdx⟶+∞asε⟶0+∀s∈1,+∞,where χ is the characteristic function of K. We choose ε such that(65)δ=p-1∫Ω∇φεpdx-λh∫Ωbhφεpdx>0,∫Ωρhdh1qh/γh-2v1qh-γhdh1dhhφεγhdx>0and let us set v
~
h
=
δ
-
1
/
p
φ
ε. With v
~
s
=
s
1
/
p
v
1
,
…
,
1
-
s
1
/
p
v
~
h
,
…
,
s
1
/
p
v
n, we have(66)v~1=v,v~s∈Sλ∀s∈0,1.It is easy to verify that(67)ddsD1v~s=-1-sγ1/p-1f1s+f~1s∀s∈0,1withlims→1-⁡f1s∈R,lims→1-⁡f~1s∈R,ddsDhv~s~-ch1-sγh/p-1ass⟶1-ch=const.>0,ddsDjv~s=cj1-sqj∖p-1+f~jsasj∈2,…,n∖h∀s∈0,1withcj=const.>0,lims→1-⁡f~js∈Rifvjv~h≡0,lims→1-⁡ddsDjv~s=-∞ifvjv~h≢0.Then, lim
s
→
1
-
⁡
∑
j
=
1
n
t
v
~
s
q
j
d
/
d
s
D
j
v
~
s
=
-
∞.
About (62), let v
∈
F. Set(68)
g
1
s
,
τ
=
D
1
τ
u
1
∗
,
s
v
2
,
…
,
s
v
n
=
q
1
-
1
∫
Ω
d
1
τ
γ
1
u
1
∗
γ
1
+
d
s
γ
1
∑
l
=
2
n
v
l
γ
1
q
1
∖
γ
1
d
x
-
∫
Ω
d
~
s
q
1
∑
l
=
2
n
v
l
γ
1
q
1
∖
γ
1
d
x
∀
s
,
τ
≥
0
;we have(69)g11,0=-1,∂g1∂τs,τ>0∀s≥0,∀τ>0,g1s,0=-sq1<-1∀s>1,limτ→+∞⁡g1s,τ=+∞∀s≥0.Relations (69) imply that(70)only oneτs>0τs>0ifs>1,τ1=0exists such thatg1s,τs=-1,∀s≥1and the function τ
s belongs to C
0
1
,
+
∞
∩
C
1
1
,
+
∞ and it results in(71)τ′s=-1τsγ1-1g~1s,τs∀s>1,lims→1+⁡g~1s,τs∈-∞,0.Then, with v
~
s
=
τ
s
u
1
∗
,
s
v
2
,
…
,
s
v
n, we have(72)v~1=v,v~s∈SD1∀s≥1;moreover,(73)lims→1+⁡ddsHλv~s∈-∞,0,lims→1+⁡ddsDjv~s=+∞ifvj≢0,lims→1+⁡ddsDjv~s∈Rifvj≡0,asj=2,…,nfrom which lim
s
→
1
+
⁡
t
v
~
s
p
d
/
d
s
H
λ
v
~
s
-
∑
j
=
1
n
t
v
~
s
q
j
d
/
d
s
D
j
v
~
s
=
-
∞ since v
j
≢
0 as some j
≥
2.
Finally, let us verify (63), for example, when h
=
2. Let v
∈
F. Since(74)v1≢0⟹vj≢0as  somej>2,we can suppose v
3
~
0. Set v
~
s
=
v
1
,
1
-
s
1
/
γ
1
v
3
,
s
1
/
γ
1
v
3
,
…
,
v
n; we have(75)v~1=v,v~s∈SD1∀s∈0,1;we add that(76)lims→1-⁡ddsHλv~s∈R,lims→1-⁡ddsD2v~s=-∞,lims→1-⁡ddsD3v~s=-∞,lims→1-⁡ddsDjv~s<+∞,asj>3ifn>3from which lim
s
→
1
-
⁡
t
v
~
s
p
d
/
d
s
H
λ
v
~
s
-
∑
j
=
1
n
t
v
~
s
q
j
d
/
d
s
D
j
v
~
s
=
+
∞.


## 4. Neumann Problems

We assumeW
=
W
1
,
p
Ω
n (n
≥
1
,
1
<
p
<
∞
,
Ω
⊆
R
N is an open, bounded, connected and C
0,1 set, v
=
∑
l
=
1
n
∫
Ω
∇
v
l
p
+
v
l
p
d
x
1
/
p
∀
v
=
v
1
,
…
,
v
n
∈
W), B
l
v
l
=
p
-
1
∫
Ω
b
l
v
l
p
d
x
∀
v
l
∈
W
1
,
p
Ω with b
l
∈
L
∞
Ω
∖
0, and B
^
l
v
l
=
p
-
1
∫
∂
Ω
b
^
l
v
l
p
d
σ
∀
v
l
∈
W
1
,
p
Ω with b
^
l
∈
L
∞
∂
Ω
∖
0 (σ is the measure on ∂
Ω, v
l
=
γ
0
v
l, where γ
0
:
W
1
,
p
Ω
→
W
p
-
1
/
p
,
p
∂
Ω is the trace operator).Let us consider the functionalA (as in i
11) such that A
v
≥
p
-
1
c
~
∑
l
=
1
n
∫
Ω
∇
v
l
p
d
x
∀
v
∈
W (c
~
=
const.
>
0). Moreover, ·
N and p
~ have been settled as in Section 3, ν is the outward orthogonal unitary vector to ∂
Ω, p
^
=
N
-
1
p
/
N
-
p if p
<
N, and p
^
=
∞ if p
≥
N.About the validity of assumptionsi
13
–
i
15, we set C
∗
=
{
c
=
c
1
,
…
,
c
n
∈
R
n
:
c
l
=
0
if
l
∈
I
∖
I
∗, c
l
≠
0 as  some l
∈
I
∗
}  I
=
1
,
…
,
n
,
I
∗
≠
⌀
⊆
I and we introduce the following assumptions:i
4
I
∗
⊆
I exists: D
1
c
<
0 for every c
∈
C
∗.
i
_
4
I
∗
⊆
I exists: D
1
c
<
0 and A
c
=
0 for every c
∈
C
∗.Proposition 14.
Whenb
l
≥
0 and b
^
l
≥
0 as l
∈
I, then i
13 holds if λ
l
≤
0, μ
l
≤
0, and λ
l
+
μ
l
<
0 as l
∈
I. When i
4 holds with I
∗
≠
I, b
l
≥
0 and b
^
l
≥
0 as l
∈
I
∖
I
∗, V
+
D
1
≠
⌀, then with λ
l
≤
0, μ
l
≤
0 and λ
l
+
μ
l
<
0 as l
∈
I
∖
I
∗
∃
δ
∗
>
0
:
i
14 holds if λ
l
,
μ
l
≤
δ
∗
∀
l
∈
I
∗. When i
4 holds with I
∗
=
I and V
+
D
1
≠
⌀, then ∃
δ
∗
>
0
:
i
14 holds if λ
l
,
μ
l
≤
δ
∗
∀
l
∈
I.Proof.
We reason by contradiction as in [11, Propositions 3.2 and 3.3].Proposition 15.
Wheni
_
4 holds with I
∗
≠
I, b
l
≥
0 and b
^
l
≥
0 as l
∈
I
∖
I
∗, ∫
Ω
b
l
d
x
>
0 and ∫
∂
Ω
b
^
l
d
σ
>
0 as l
∈
I
∗, then with λ
l
≤
0, μ
l
≤
0, and λ
l
+
μ
l
<
0 as l
∈
I
∖
I
∗
∃
δ
∗
>
0
:
i
15 holds if λ
l
,
μ
l
∈
0
,
δ
∗
∀
l
∈
I
∗ and λ
l
+
μ
l
>
0 for some l. When i
_
4 holds with I
∗
=
I, ∫
Ω
b
l
d
x
>
0 and ∫
∂
Ω
b
^
l
d
σ
>
0 as l
∈
I, then ∃
δ
∗
>
0
:
i
15 holds if λ
l
,
μ
l
∈
0
,
δ
∗
∀
l
∈
I and λ
l
+
μ
l
>
0 for some l.Proof.
See [11, Propositions 3.4 and 3.5].Passing to the applications, we state in advance that the results of [8] for problem(77)-div⁡∇up-2∇u=bxus-1inΩ,∇up-2∂u∂ν=axuq-1on∂Ωcan be obtained by using Theorem 1 and Proposition 14, Theorems 2.1 and 2.2 of [10], and Propositions 3.3 and A
.
4 of [11] (taking into account [17] too).Application 3. Let n
>
1 and, for each v
=
v
1
,
…
,
v
n
∈
W.We set(78)
A
v
=
p
-
1
∑
l
=
1
n
∫
Ω
∇
v
l
p
d
x
,
D
1
v
=
q
1
-
1
∫
∂
Ω
∑
l
=
1
n
d
l
v
l
γ
q
1
/
γ
d
σ
-
∑
l
=
1
n
∫
∂
Ω
d
^
l
v
l
q
1
d
σ
,
D
2
v
=
-
∏
l
=
1
n
∫
∂
Ω
ρ
^
l
v
l
γ
l
d
σ
γ
l
′
/
γ
l
,where(79)1<γ<q1<p,1<γl<γl′,γl<p^,q1<q2=∑l=1nγl′;dl,d^l∈L∞∂Ω,dl,d^l>0;ρ^l∈L∞∂Ω∖0.Let us consider the system:(80)-div⁡∇uip-2∇ui=λibiuip-2uiinΩ,∇uip-2∂ui∂ν=μib^iuip-2ui+∑l=1ndlulγq1/γ-1diuiγ-2ui-d^iuiq1-2ui-∏l≠i∫∂Ωρ^lulγldσγl′/γl·γi′∫∂Ωρ^iuiγidσγl′/γl-2∫∂Ωρ^iuiγidσ·ρ^iuiγi-2uion∂Ωasi=1,…,n.Proposition 16.
Letu
=
u
1
,
…
,
u
n be a weak solution of system (80). Then,(81)λi≠0,ui≢0forsomei∈1,…,n⟹uiisnotconstant;(82)ui≥0∀i∈1,…,n⟹ui∈L∞Ω∩Cloc1,αiΩ∀i∈1,…,n,ui>0ifui≢0.Proof.
Relation (81) is evident. About (82), it is easy to verify that(83)∑i=1n∫Ω∇uip-2∇ui·∇vidx≤c1∫Ω∑i=1nui+1p-1·∑i=1nvidx+∫∂Ωc2+c3∑l=1nul+1γ0-p·∑i=1nui+1p-1∑i=1nvidσ∀v=v1,…,vn∈W1,pΩ∩L∞Ωnwithvi≥0,where c
1
–
c
3 are positive constants and γ
0
=
max
⁡
γ
1
,
…
,
γ
n. Consequently [11, Proposition A
.
4], u
i
∈
L
∞
Ω from which, since -
div
⁡
∇
u
i
p
-
2
∇
u
i
=
λ
i
b
i
u
i
p
-
2
u
i in Ω, we deduce that u
i
∈
C
l
o
c
1
,
α
i
Ω [17] and u
i
>
0 if u
i
≢
0 [18].Let us introduce the conditions:(84)d1q1/γ>d^1⟹V+D1≠⌀,(85)∑l=2ndlq1/γ<min⁡d^2,…,d^n⟹D10,c2,…,cn<0∀c2,…,cn∈Rn-1∖0,(86)b1≥0,b^1≥0,(87)∫Ωbldx>0,∫∂Ωb^ldσ>0asl=2,…,n.Then (Propositions 14 and 15),(88)
84
–
86
⟹
with
λ
1
≤
0
,
μ
1
≤
0
,
λ
1
+
μ
1
<
0
∃
δ
1
∗
>
0
:
i
14
holds  if
λ
l
,
μ
l
≤
δ
1
∗
∀
l
∈
2
,
…
,
n
,
(89)
85
–
87
⟹
with
λ
1
≤
0
,
μ
1
≤
0
and
λ
1
+
μ
1
<
0
∃
δ
2
∗
>
0
:
i
15
holds  if
λ
l
,
μ
l
∈
0
,
δ
2
∗
∀
l
∈
2
,
…
,
n
and
λ
l
+
μ
l
>
0
as  some
l
.Proposition 17 (Theorems1 and 4, Remark 6, Proposition 16).
Under conditions (79), one has the following:When (84)–(86) hold, with λ
l
,
μ
l as in (88), system (80) has at least two weak solutions u
0 and -
u
0
u
0
=
τ
0
v
0
,
τ
0
=
c
o
n
s
t
.
>
0
,
v
0
∈
S
λ
μ
∩
V
+
D
1, and one has u
i
0
∈
L
∞
Ω
∩
C
l
o
c
1
,
α
i
0
Ω, u
i
0
≥
0 as i
=
1
,
…
,
n
,
u
i
0
>
0 if u
i
0
≢
0.
Whenq
2
<
p and (85)–(87) hold, with λ
l
,
μ
l as in (89), system (80) has at least two weak solutions u
- and -
u
-
u
-
=
τ
-
v
-
,
τ
-
=
c
o
n
s
t
.
>
0
,
v
-
∈
V
λ
μ
-
∩
S
D
1, and one has u
-
i
∈
L
∞
Ω
∩
C
l
o
c
1
,
α
-
i
Ω, u
-
i
≥
0 as i
=
1
,
…
,
n
,
u
-
i
>
0 if u
-
i
~
0.Consequently, when q
2
<
p and (84)–(87) hold, with λ
l
,
μ
l as in (89) and min
⁡
δ
1
∗
,
δ
2
∗ instead of δ
2
∗, system (80) has at least four different weak solutions.Proposition 18.
One gets the following:(90)u10≢0on∂Ω;(91)eitherγh′≥porγ≤γh′⟹uh0≢0assomeh∈2,…,n.Proof.
Relation (90) is evident. About (91), it is sufficient (Proposition 2) to verify thati
0
h
holds withF
=
v
∈
S
λ
μ
∩
V
+
D
1
:
v
l
≥
0
as
l
=
1
,
…
,
n
,
v
h
≡
0
≠
⌀
.Let us check, for example, the case h
=
2. Let v
∈
F; then, Γ
⊆
∂
Ω exists such that σ
Γ
>
0 and ∑
l
≠
2
d
l
v
l
γ
>
0 on Γ. Let k
⊆
Ω be a compact set with positive measure and Ω
′ an open set such that Ω
-
′
⊆
Ω, k
⊆
Ω
′. Thanks to [11, Propositions A
.
1, A
.
2], a compact set Γ
^
⊆
Γ with σ
Γ
^
>
0, φ
1
ε
0
<
ε
<
ε
0
⊆
C
0
∞
Ω, and φ
2
ε
0
<
ε
<
ε
0
⊆
C
0
∞
R
N exist such that(92)0≤φ1ε≤1,suppφ1ε⊆Ω′,φ1ε⟶χstrongly  inLsΩ,∫Ω∇φ1εsdx⟶+∞asε⟶0+∀s∈1,+∞,0≤φ2ε≤1,suppφ2ε⊆RN∖Ω-′,φ2ε⟶χ^strongly  inLs∂Ω,∫RNφ2εsdx⟶0asε⟶0+∀s∈1,+∞,where χ
resp.,
χ
^ is the characteristic function of K
resp.,
Γ
^. Set φ
ε
=
φ
1
ε
+
φ
2
ε; let us fix ε such that(93)
δ
=
p
-
1
∫
Ω
∇
φ
ε
p
d
x
-
λ
2
∫
Ω
b
2
φ
ε
p
d
x
-
μ
2
∫
∂
Ω
b
^
2
φ
ε
p
d
σ
>
0
,
∫
∂
Ω
∑
l
≠
2
d
l
v
l
γ
q
1
/
γ
-
1
d
2
φ
ε
γ
d
σ
>
0
.Then, with v
~
s
=
s
1
/
p
v
1
,
1
-
s
1
/
p
δ
-
1
/
p
φ
ε
,
…
,
s
1
/
p
v
n, we have(94)v~1=v,v~s∈Sλμ∀s∈0,1,ddsD1v~s~-c^11-sγ/p-1ass⟶1-c^1=const.>0,D2v~s=-c^2sq2-γ2′/p1-sγ2′/p∀s∈0,1c^2=const.≥0⟹lims→1-⁡∑j=12tv~sqjddsDjv~s=-∞.Proposition 19.
Let for someh
∈
2
,
…
,
n either γ
h
′
≥
p or γ
≤
γ
h
′. Let b
h
≥
0, λ
h
>
0, and μ
h
∫
∂
Ω
b
^
h
d
σ
≥
0. Then, u
h
0
≢
0 on ∂
Ω.Proof.
Since in Proposition18  u
h
0
>
0, it is sufficient (Proposition 2) to prove thati
0
h
holds withF
=
v
∈
S
λ
μ
∩
V
+
D
1
:
v
l
≥
0
as
l
=
1
,
…
,
n
,
v
h
>
0
,
v
h
≡
0
on
∂
Ω
≠
⌀.Let us suppose, for example, h
=
2. Let v
∈
F and c
2
=
const.
>
0. Set g
s
,
τ
=
H
λ
μ
(
s
v
1
,
s
v
2
+
τ
c
2
,
…
,
s
v
n
) = p
-
1
s
p
∫
Ω
∇
v
2
p
d
x
-
λ
2
∫
Ω
b
2
s
v
2
+
τ
c
2
p
d
x
-
μ
2
τ
p
c
2
p
∫
∂
Ω
b
^
2
d
σ
+
s
p
H
λ
μ
v
1
,
0
,
…
,
v
n  ∀
s
,
τ
≥
0; we have(95)g1,0=1,∂g∂τs,τ<0∀s≥0,∀τ>0,gs,0=sp>1∀s>1,limτ→+∞⁡gs,τ=-∞∀s≥0;then, τ
:
1
,
+
∞
→
0
,
+
∞ belonging to C
0
1
,
+
∞
∩
C
1
1
,
+
∞ exists such that g
s
,
τ
s
=
1
∀
s
≥
1, τ
s
>
0 if s
>
1, τ
1
=
0, and lim
s
→
1
+
⁡
τ
′
s
∈
0
,
+
∞. Consequently, set v
~
s
=
s
v
1
,
s
v
2
+
τ
s
c
2
,
…
,
s
v
n; it results in(96)v~1=v,v~s∈Sλμ∀s≥1,lims→1+⁡ddsD1v~s=q1D1v>0,lims→1+⁡ddsD2v~s=0⟹lims→1+⁡∑j=12tv~sqjddsDjv~s∈0,+∞.Proposition 20.
If eitherγ
1
′
>
γ or ∫
∂
Ω
ρ
^
1
d
σ
=
0, it results in(97)u-1≢0;(98)u-1≢0on∂Ωwhenλ1=0.Proof.
About (97) [resp., (98)], it is sufficient (Proposition 5) to prove thati
_
1
holds withF
=
v
∈
V
λ
μ
-
∩
S
D
1
:
v
l
≥
0
as
l
=
1
,
…
,
n
,
v
1
≡
0
resp.,
v
1
≡
0
on
∂
Ω
≠
⌀
.Let v
∈
F and c
1
=
const.
>
0. Let g
1
s
,
τ
=
D
1
τ
c
1
,
s
v
2
,
…
,
s
v
n = q
1
-
1
∫
∂
Ω
d
1
τ
γ
c
1
γ
+
s
γ
∑
l
≠
1
d
l
v
l
γ
q
1
/
γ
d
σ
-
τ
q
1
c
1
q
1
∫
∂
Ω
d
^
1
d
σ
-
s
q
1
∑
l
≠
1
∫
∂
Ω
d
^
l
v
l
q
1
d
σ  ∀
s
≥
0 and ∀
τ
≥
0. We note that(99)g11,0=-1,∂g1∂τs,τ>0∀s≥0,∀τ>0,g1s,0=-sq1<-1∀s>1,limτ→+∞⁡g1s,τ=+∞∀s≥0.Then, τ
:
1
,
+
∞
→
0
,
+
∞
τ
s
>
0
if
s
>
1
,
τ
1
=
0 belonging to C
0
1
,
+
∞
∩
C
1
1
,
+
∞ exists such that(100)g1s,τs=-1∀s≥1,τ′s=-1τsγ-1g~1s,τs∀s>1,lims→1+⁡g~1s,τs∈-∞,0,from which with v
~
s
=
τ
s
c
1
,
s
v
2
,
…
,
s
v
n [resp.,v
~
s
=
s
v
1
+
τ
s
c
1
,
s
v
2
,
…
,
s
v
n] we get(101)v~1=v,v~s∈SD1∀s≥1,lims→1+⁡ddsD2v~s=0,lims→1+⁡tv~spddsHλμv~s∈-∞,0.Evidently,u
-
2
≢
0 on ∂
Ω if n
=
2.Proposition 21.
Whenn
>
2 and for some h
∈
2
,
…
,
n
∫
∂
Ω
ρ
^
h
d
σ
=
0, b
h
≥
0, μ
h
>
0, then(102)u-h≢0on∂Ω.Proof.
Let us prove (102) as h
=
2. Reasoning by contradiction, let u
-
2
≡
0 on ∂
Ω; that is, v
-
2
≡
0 on ∂
Ω. Let c
2
=
const.
>
0. As g
1
s
,
τ
=
D
1
s
v
-
1
,
v
-
2
+
τ
c
2
,
…
,
s
v
-
n = q
1
-
1
∫
∂
Ω
d
2
τ
γ
c
2
γ
+
s
γ
∑
l
≠
2
d
l
v
-
l
γ
q
1
/
γ
d
σ
-
τ
q
1
c
2
q
1
∫
∂
Ω
d
^
2
d
σ
-
s
q
1
∑
l
≠
2
∫
∂
Ω
d
^
l
v
-
l
q
1
d
σ  ∀
s
,
τ
≥
0, we have(103)g1s,0=-sq1>-1∀s∈0,1,(104)limτ→+∞⁡g1s,τ=-∞∀s≥0.Thanks to (103), (104), for each s
∈
0,1, it is possible to choose τ
s
>
0 such that g
1
s
,
τ
s
=
-
1. Then, set v
~
s
=
s
v
-
1
,
v
-
2
+
τ
s
c
2
,
…
,
s
v
-
n; it results in(105)D1v~s=-1∀s∈0,1;moreover, since(106)
H
λ
μ
v
~
s
=
p
-
1
∫
Ω
∇
v
-
2
p
d
x
-
λ
2
∫
Ω
b
2
v
-
2
+
τ
s
c
2
p
d
x
-
μ
2
τ
s
p
c
2
p
∫
∂
Ω
b
^
2
d
σ
+
s
p
H
λ
μ
v
-
1
,
0
,
…
,
v
-
n
<
p
-
1
∫
Ω
∇
v
-
2
p
d
x
-
λ
2
∫
Ω
b
2
v
-
2
p
d
x
+
s
p
H
λ
μ
v
-
1
,
0
,
…
,
v
-
n
⟶
H
λ
μ
v
-
<
0
as
s
⟶
1
-
,
s
-
∈
0,1 exists such that(107)Hλμv~s<0∀s∈s-,1.Relations (105), (107) imply that (Theorem 4)(108)E~~v~s≥E~~v-∀s∈s-,1.Since D
2
v
-
=
D
2
v
~
s
=
0, we have(109)ptv-p-1Hλμv-+q1tv-q1-1=0,ptv~sp-1Hλμv~s+q1tv~sq1-1=0∀s∈0,1,from which(110)E~~v-=δHλμv--q1/p-q1,E~~v~s=δHλμv~s-q1/p-q1∀s∈0,1,where δ
=
q
1
/
p
q
1
/
p
-
q
1
-
q
1
/
p
p
/
p
-
q
1
>
0.
From (108), (110), we get(111)Hλμv-≤Hλμv~s∀s∈s-,1.We add that since(112)∂g1∂τs,τ=τγ-1g~s,τ∀s,τ≥0,lims→1-τ→0+⁡g~s,τ∈0,+∞,then s
0
∈
s
-
,
1 and τ
0
∈
0,1 exist such that(113)∂g1∂τs,τ>0∀s∈s0,1,∀τ∈0,τ0,from which(114)τs≥τ0∀s∈s0,1thanks to (103). Relations (111), (114) imply that(115)Hλμv-≤-p-1τspc2pμ2∫∂Ωb^2dσ+p-1∫Ω∇v-2pdx-λ2∫Ωb2v-2pdx+spHλμv-1,0,…,v-n≤-p-1τ0pc2pμ2∫∂Ωb^2dσ+p-1∫Ω∇v-2pdx-λ2∫Ωb2v-2pdx+spHλμv-1,0,…,v-n∀s∈s0,1,from which, as s
→
1
-, we get the contradiction(116)Hλμv-≤-p-1τ0pc2pμ2∫∂Ωb^2dσ+Hλμv-<Hλμv-.Application 4. Let n
>
1 and, for each v
=
v
1
,
…
,
v
n
∈
W, we set(117)
A
v
=
p
-
1
∫
Ω
∑
l
=
1
n
∇
v
l
γ
p
/
γ
d
x
,
D
1
v
=
q
1
-
1
∫
Ω
ρ
1
∑
l
=
1
n
d
l
v
l
γ
1
q
1
/
γ
1
d
x
,
D
j
v
=
q
j
-
1
∫
Ω
∑
l
=
1
n
d
j
l
v
l
γ
j
q
j
/
γ
j
d
x
-
∑
l
=
1
n
∫
Ω
d
~
j
l
v
l
q
j
d
x
as
j
=
2
,
…
,
m
,where(118)1<γ<p,1<γj<qjasj=1,…,m,q1<p,q1<q2<⋯<qm<p~;ρ1∈C0Ω∩L∞Ω,dl=const.>0;djl,d~jl∈L∞Ω,djl,d~jl>0,∑l=1ndjlqj/γj≤min⁡d~j1,…,d~jn.Let us consider the system:(119)
-
div
⁡
∑
l
=
1
n
∇
u
l
γ
p
/
γ
-
1
∇
u
i
γ
-
2
∇
u
i
=
λ
i
b
i
u
i
p
-
2
u
i
+
ρ
1
∑
l
=
1
n
d
l
u
l
γ
1
q
1
/
γ
1
-
1
d
i
u
i
γ
1
-
2
u
i
+
∑
j
=
2
m
∑
l
=
1
n
d
j
l
u
l
γ
j
q
j
/
γ
j
-
1
d
j
i
u
i
γ
j
-
2
u
i
-
d
~
j
i
u
i
q
j
-
2
u
i
in
Ω
,
∑
l
=
1
n
∇
u
l
γ
p
/
γ
-
1
∇
u
i
γ
-
2
∂
u
i
∂
ν
=
μ
i
b
^
i
u
i
p
-
2
u
i
on
∂
Ω
as
i
=
1
,
…
,
n
.Proposition 22.
Let eitherq
m
<
p
^ or μ
l
=
0
∀
l
∈
1
,
…
,
n. If u
=
u
1
,
…
,
u
n is a weak solution of system (119) with all nonnegative components, then u
i
∈
L
∞
Ω
∀
i
∈
1
,
…
,
n.Proof.
The statement is true in virtue of [11, Proposition A
.
4, Remark A
.
5], and of relation(120)∑i=1n∫Ω∑l=1n∇ulγp/γ-1∇uiγ-2∇ui·∇vidx≤∫Ωc1+c2∑l=1nul+1qm-p∑i=1nui+1p-1·∑i=1nvidx+c3maxl⁡μl∫∂Ω∑i=1nui+1p-1·∑i=1nvidσ∀v=v1,…,vn∈W1,pΩ∩L∞Ωnwithvi≥0,where c
1
–
c
3 are positive constants.Let us introduce the conditions:(121)ρ1+≢0⟹V+D1≠⌀,(122)∫Ωρ1dx<0⟹D1c<0∀c∈Rn∖0,(123)∫Ωbldx>0,∫∂Ωb^ldσ>0asl=1,…,n.We have (Propositions 14 and 15)(124)
121
,
122
⟹
∃
δ
1
∗
>
0
:
i
14
holds  if
λ
l
,
μ
l
≤
δ
1
∗
∀
l
∈
1
,
…
,
n
,
(125)
122
,
123
⟹
∃
δ
2
∗
>
0
:
i
15
holds  if
λ
l
,
μ
l
∈
0
,
δ
2
∗
∀
l
∈
1
,
…
,
n
,
λ
l
+
μ
l
>
0
as  some
l
.Proposition 23 (see Theorems1 and 4, Remark 6).
Under conditions (118), one has the following:When (121) and (122) hold, with λ
l, μ
l as in (124), system (119) has at least two weak solutions u
0 and -
u
0
u
0
=
τ
0
v
0
,
τ
0
=
c
o
n
s
t
.
>
0
,
v
0
∈
S
λ
μ
∩
V
+
D
1, and one has u
i
0
≥
0 as i
=
1
,
…
,
n.
Whenq
m
<
p and (122), (123) hold, with λ
l
,
μ
l as in (125), system (119) has at least two weak solutions u
- and -
u
- (u
-
=
τ
-
v
-
,
τ
-
=
c
o
n
s
t
.
>
0
,
v
-
∈
V
λ
μ
-
∩
S
D
1), and one has u
-
i
≥
0 as i
=
1
,
…
,
n.Consequently, when q
m
<
p and (121)–(123) hold, with λ
l
,
μ
l as in (125) and min
⁡
δ
1
∗
,
δ
2
∗ instead of δ
2
∗, system (119) has at least four different weak solutions.From Propositions22 and 23, we deduce the following.Proposition 24.
We haveu
-
i
∈
L
∞
Ω as i
=
1
,
…
,
n. If either q
m
<
p
^ or μ
l
=
0
∀
l
∈
1
,
…
,
n, then u
i
0
∈
L
∞
Ω as i
=
1
,
…
,
n.Proposition 25.
LetΩ
0
=
x
∈
Ω
:
ρ
1
x
=
0
. If γ
>
γ
1 and Ω
0
N
=
0, then u
i
0
≢
0 as i
=
1
,
…
,
n.Proof.
It is sufficient (Proposition2) to show thatash
=
1
,
…
,
n
i
0
h holds  with F
=
v
∈
S
λ
μ
∩
V
+
D
1
:
v
l
≥
0
as
l
=
1
,
…
,
n
,
v
h
≡
0
≠
⌀.Let us consider, for example, the case h
=
1. Let v
∈
F. Firstly, let us prove that v
~
1
∈
W
1
,
p
Ω exists with v
~
1
≥
0 such that(126)δ=∫Ω∇v~1pdx-λ1∫Ωb1v~1pdx-μ1∫∂Ωb^1v~1pdσ>0,(127)∫Ω∑l≠1∇vlγp/γ-1∇v~1γdx>0,(128)∫Ωρ1∑l≠1dlvlγ1q1/γ1-1v~1γ1dx>0.Let Ω
+
=
x
∈
Ω
:
ρ
1
x
>
0 and Ω
l
=
x
∈
Ω
:
∇
v
l
x
>
0  l
=
1
,
…
,
n
. Since v
∈
V
+
D
1, then l
0
∈
2
,
…
,
n exists such that Ω
l
0
N
>
0. In fact, Ω
l
N
=
0  ∀
l
∈
2
,
…
,
n
⇒
v
l
=
const.  ∀
l
∈
2
,
…
,
n
⇒
D
1
v
<
0.
Firstly, let us supposeΩ
l
0
∩
Ω
+
N
>
0. Then, a compact set K
0
⊆
Ω
+ exists such that K
0
N
>
0 and ∇
v
l
0
>
0 in K
0. Let [11, Proposition A
.
1] φ
ε
0
<
ε
<
ε
0
⊆
C
0
∞
Ω
+ with 0
≤
φ
ε
≤
1 such that(129)φε⟶χstrongly  inLsΩ,∫Ω∇φεsdx⟶+∞asε⟶0+∀s∈1,+∞,where χ is the characteristic function of K
0. Set Ω
ε
=
x
∈
Ω
l
0
:
∇
φ
ε
x
>
0 and J
=
ε
∈
0
,
ε
0
:
Ω
ε
N
>
0; we can consider two cases:(130)inf⁡J=0,(131)eitherJ=⌀orinf⁡J>0.In case (130), thanks to (129), it is possible to choose ε such that (126)–(128) hold with v
~
1
=
φ
ε. In case (131), also from (129), we can find ε such that v
~
1
=
φ
v
l
0
+
φ
ε satisfying (126)–(128), where φ
∈
C
0
∞
Ω
+, 0
≤
φ
≤
1, and φ
=
1 in K
0.
Let us suppose nowΩ
l
0
∩
Ω
+
N
=
0. Since Ω
0
N
=
0, we have(132)∃a  compact  setK0⊆Ω∖Ω-+:K0N>0,∇vl0>0inK0.
Let us add that(133)v∈V+D1⟹∃l1∈2,…,n,a  compact  setK1⊆Ω+:K1N>0,vl1>0inK1.Let φ
ε
0
<
ε
<
ε
0
⊆
C
0
∞
Ω
+ such that 0
≤
φ
ε
≤
1, φ
ε
→
χ strongly in L
s
Ω and ∫
Ω
∇
φ
ε
s
d
x
→
+
∞ as ε
→
0
+
∀
s
∈
1
,
+
∞, where χ is the characteristic function of K
1. Evidently, v
~
1
=
η
φ
v
l
0
+
φ
ε satisfies (126)–(128) with suitable η
>
0 and ε
,
φ
∈
C
0
∞
Ω
∖
Ω
-
+, 0
≤
φ
≤
1, and φ
=
1 in K
0.
Let us introduce the functiong
s
,
τ
=
H
λ
μ
τ
v
~
1
,
s
v
2
,
…
,
s
v
n
=
p
-
1
[
∫
Ω
τ
γ
∇
v
~
1
γ
+
s
γ
∑
l
≠
1
∇
v
l
γ
p
/
γ
d
x
-
τ
p
λ
1
∫
Ω
b
1
v
~
1
p
d
x
+
μ
1
∫
∂
Ω
b
^
1
v
~
1
p
d
σ
-
s
p
∑
l
≠
1
λ
l
∫
Ω
v
l
p
d
x
+
μ
l
∫
∂
Ω
b
^
l
v
l
p
d
σ
]
∀
s
,
τ
≥
0. Equation (126) implies that(134)
lim
τ
→
+
∞
⁡
g
s
,
τ
=
+
∞
∀
s
≥
0
,
∂
g
∂
τ
s
,
τ
≥
δ
τ
p
-
1
>
0
∀
s
≥
0
,
∀
τ
>
0
.Consequently, since g
1,0
=
1 and g
s
,
0
=
s
p
<
1
∀
s
∈
0,1, ∃
τ
:
0,1
→
0
,
+
∞
τ
1
=
0 and τ
s
>
0
if
s
<
1 belonging to C
0
0,1
∩
C
1
0,1 such that g
s
,
τ
s
=
1
∀
s
∈
0,1, and we have(135)
τ
′
s
=
-
1
τ
s
γ
-
1
g
~
s
,
τ
s
∀
s
∈
0,1
;and then, lim
s
→
1
-
⁡
g
~
s
,
τ
s
∈
0
,
+
∞ by (127).
In conclusion, setv
~
s
=
τ
s
v
~
1
,
s
v
2
,
…
,
s
v
n and, taking into account (128), (135), it results in(136)
v
~
1
=
v
,
v
~
s
∈
S
λ
μ
∀
s
∈
0,1
,
lim
s
→
1
-
⁡
d
d
s
D
1
v
~
s
=
-
∞
,
lim
s
→
1
-
⁡
d
d
s
D
j
v
~
s
<
+
∞
as
j
≠
1
⟹
lim
s
→
1
-
⁡
∑
j
=
1
m
t
v
~
s
q
j
d
d
s
D
j
v
~
s
=
-
∞
.Proposition 26.
If for somej
0
∈
2
,
…
,
m
γ
j
0
<
γ
≤
γ
1, then u
i
0
≢
0 as i
=
1
,
…
,
n.Proof.
It is sufficient (Proposition2) to show thatash
=
1
,
…
,
n
i
0
h holds  with F
=
v
∈
S
λ
μ
∩
V
+
D
1
:
v
l
≥
0
as
l
=
1
,
…
,
n
,
v
h
≡
0
≠
⌀.Let us suppose h
=
1 and let v
∈
F. Let us verify that v
~
1
∈
W
1
,
p
Ω exists with v
~
1
≥
0 such that (126) and (127) hold. At this aim, we note that(137)
v
∈
V
+
D
1
⟹
∃
l
0
∈
2
,
…
,
n
:
Ω
l
0
N
>
0
,where Ω
l
0
=
x
∈
Ω
:
∇
v
l
0
x
>
0. Let K
⊆
Ω be a compact set with positive measure. Let φ
ε
0
<
ε
<
ε
0
⊆
C
0
∞
Ω such that 0
≤
φ
ε
≤
1, φ
ε
→
χ strongly in L
s
Ω and ∫
Ω
∇
φ
ε
s
d
x
→
+
∞ as ε
→
0
+
∀
s
∈
1
,
+
∞ (χ = characteristic function of K). With Ω
ε and J as in Proposition 25, it is possible to find ε such that v
~
1
=
φ
ε [resp., v
~
1
=
v
l
0
+
φ
ε] satisfying (126) and (127) in case (130) [resp., (131)]. Then, if τ
s and v
~
s are as in Proposition 25, we have(138)
v
~
1
=
v
,
v
~
s
∈
S
λ
μ
∀
s
∈
0,1
,
lim
s
→
1
-
⁡
d
d
s
D
j
0
v
~
s
=
-
∞
,
lim
s
→
1
-
⁡
d
d
s
D
j
v
~
s
<
+
∞
as
j
≠
j
0
⟹
lim
s
→
1
-
⁡
∑
j
=
1
m
t
v
~
s
q
j
d
d
s
D
j
v
~
s
=
-
∞
.Proposition 27.
If for somej
0
∈
2
,
…
,
m
γ
j
0
<
γ
1
≤
γ, then u
-
i
≢
0 as i
=
1
,
…
,
n.Proof.
It is sufficient (Proposition5) to show thatash
=
1
,
…
,
n
i
_
h holds with F
=
v
∈
V
λ
μ
-
∩
S
D
1
:
v
l
≥
0
as
l
=
1
,
…
,
n
,
v
h
≡
0, if F
≠
⌀.Let us suppose, for example, h
=
1 and let us set Ω
-
=
x
∈
Ω
:
ρ
1
x
<
0. Let K
⊆
Ω
- be a compact set such that K
N
>
0 and v
l
0
>
0 in K for some l
0
∈
2
,
…
,
n. Let φ
∈
C
0
∞
Ω
- with 0
≤
φ
≤
1 and φ
=
1 in K. Set g
1
s
,
τ
=
D
1
τ
φ
,
s
v
2
,
…
,
s
v
n
=
q
1
-
1
∫
Ω
ρ
1
d
1
τ
γ
1
φ
γ
1
+
s
γ
1
∑
l
≠
1
d
l
v
l
γ
1
q
1
/
γ
1
d
x
∀
s
,
τ
≥
0; it results in(139)
g
1
1,0
=
-
1
,
∂
g
1
∂
τ
s
,
τ
<
0
∀
s
≥
0
,
∀
τ
>
0
,
g
1
s
,
0
=
-
s
q
1
>
-
1
∀
s
∈
0,1
,
lim
τ
→
+
∞
⁡
g
1
s
,
τ
=
-
∞
∀
s
≥
0
.Then, ∃
τ
:
0,1
→
0
,
+
∞
(
τ
1
=
0
,
τ
s
>
0
if
s
<
1
) belonging to C
0
0,1
∩
C
1
0,1 such that g
1
s
,
τ
s
=
-
1
∀
s
∈
0,1, and we have(140)
τ
′
s
=
-
1
τ
s
γ
1
-
1
g
~
1
s
,
τ
s
∀
s
∈
0,1
,
lim
s
→
1
-
⁡
g
~
1
s
,
τ
s
∈
0
,
+
∞
.Consequently, with v
~
s
=
τ
s
φ
,
s
v
2
,
…
,
s
v
n, we get(141)
v
~
1
=
v
,
v
~
s
∈
S
D
1
∀
s
∈
0,1
,
lim
s
→
1
-
⁡
d
d
s
H
λ
μ
v
~
s
∈
-
∞
,
0
,
lim
s
→
1
-
⁡
d
d
s
D
j
0
v
~
s
=
-
∞
,
lim
s
→
1
-
⁡
d
d
s
D
j
v
~
s
<
+
∞
as
j
≠
j
0
⟹
l
i
m
s
→
1
-
t
v
~
s
p
d
d
s
H
λ
μ
v
~
s
-
∑
j
=
1
m
t
v
~
s
q
j
d
d
s
D
j
v
~
s
=
+
∞
.



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*Source: 1683759-2016-01-19.xml*