1 Introduction
We consider here the problem given by the following system of equations and boundary conditions:
(1.1)
{
-
div
(
|
x
|
ζ
a
(
|
∇
u
|
)
∇
u
)
=
|
x
|
ν
f
11
(
u
)
f
12
(
v
)
in
Ω
,
-
div
(
|
x
|
ζ
~
b
(
|
∇
v
|
)
∇
v
)
=
|
x
|
ν
~
f
21
(
u
)
f
22
(
v
)
in
Ω
,
u
|
∂
Ω
=
0
,
v
|
∂
Ω
=
0
,
where Ω⊂ℝn, n>2, denotes a ball with center at the origin and radius R, and the functions a,b:(0,∞)↦[0,∞) are continuous. The exponents ζ,ν,ζ~,ν~ are suppose to be positive and such that ν≥ζ and ν~≥ζ~.
We will deal here with the existence of positive radially symmetric solutions of (1.1), and therefore we consider the problem
(1.2)
{
(
r
α
a
(
|
u
′
|
)
u
′
)
′
+
r
χ
f
11
(
u
(
r
)
)
f
12
(
v
(
r
)
)
=
0
,
r
∈
(
0
,
R
)
,
(
r
α
~
b
(
|
v
′
|
)
v
′
)
′
+
r
χ
~
f
21
(
u
(
r
)
)
f
22
(
v
(
r
)
)
=
0
,
r
∈
(
0
,
R
)
,
u
′
(
0
)
=
u
(
R
)
=
0
,
v
′
(
0
)
=
v
(
R
)
=
0
,
where r=|x|, R∈(0,∞), is fixed, and
(1.3)
α
=
n
-
1
+
ζ
,
χ
=
n
-
1
+
ν
,
α
~
=
n
-
1
+
ζ
~
,
χ
~
=
n
-
1
+
ν
~
.
Since (1.2) is a system of ordinary differential equations we can consider α, α~, ζ, and ζ~ as real numbers not necessarily tied up to the dimension of the space. Furthermore, we will replace the functions a(s)s and b(s)s by general homeomorphisms ϕ,ψ of ℝ, and thus we will consider henceforth the system
(1.4)
{
(
r
α
ϕ
(
u
′
)
)
′
+
r
χ
f
11
(
u
(
r
)
)
f
12
(
v
(
r
)
)
=
0
,
r
∈
(
0
,
R
)
,
(
r
α
~
ψ
(
v
′
)
)
′
+
r
χ
~
f
21
(
u
(
r
)
)
f
22
(
v
(
r
)
)
=
0
,
r
∈
(
0
,
R
)
,
u
′
(
0
)
=
u
(
R
)
=
0
,
v
′
(
0
)
=
v
(
R
)
=
0
,
where we assume χ≥α and χ~≥α~.
The proof of the existence of positive solutions to (1.4) that we give uses the blow-up method as a main ingredient for the search of a-priori bounds of solutions. This method originated in [12] was used later in [2] and [1] in connection with quasilinear operators. The blow-up argument is done by contradiction and uses a sort of scaling, reminiscent to the one used in the theory of minimal surfaces, see [12], and therefore the homogeneity of the operators, Laplacian or p-Laplacian, and second members powers or power like functions play a fundamental role.
Thus, when the differential operators are no longer homogeneous, and similarly for the second members, the application of the blow-up method to obtain a-priori bounds of solutions seems an almost impossible task. In spite of this fact, in [8], we were able to overcome this difficulty and obtain a-priori bounds for a certain type of problems. To do that we introduced a type of functions that we call asymptotically homogeneous functions (AH for short). They are extensively used in applied probability and statistics where they are known as regularly varying functions, see, for example, [19, 21]. They form an important class of non-homogeneous functions which without being necessarily asymptotic to any power have a suitable homogeneous asymptotic behavior. These asymptotically homogeneous functions provide, in some sense, a non-homogeneous rescaling, that allows us to apply the blow-up method to our situation.
In this paper we will generalize the results of [2, 8] and, in particular, to the more general asymptotically homogeneous functions framework [1]. Thus, most of the functions we will deal with in our hypotheses will belong to this class of functions.
A function h:[0,∞)↦[0,∞) is AH at +∞ of exponent δ>0 if for any σ>0,
(1.5)
lim
s
→
+
∞
h
(
σ
s
)
h
(
s
)
=
σ
δ
.
By replacing +∞ by 0 in (1.5), we obtain a similar equivalent definition for a function h to be AH of exponent δ at zero. From Karamata’s theorem (see [19, Theorem 0.6, p. 17]), it follows that for any σ>0,
(1.6)
lim
t
→
+
∞
h
(
σ
t
)
h
(
t
)
=
σ
ρ
⇔
lim
t
→
+
∞
H
(
t
)
t
h
(
t
)
=
1
ρ
+
1
,
where H(t)=∫0th(s)𝑑s. We will use primarily the right-hand side of (1.6) to define our functions.
Also for later use we recall the well-known fact that if h is AH at +∞ of exponent δ>0, then
lim
t
→
∞
t
-
ρ
-
ε
h
(
t
)
=
0
,
lim
t
→
∞
t
-
ρ
+
ε
h
(
t
)
=
∞
for each
ε
>
0
.
In this work we will assume that the functions ϕ,ψ:ℝ→ℝ are odd increasing homeomorphism on ℝ and the functions fi,j:ℝ→ℝ, for i,j∈{1,2} are odd continuous functions such that sfij(s)>0, for all s∈ℝ, with primitives
Φ
(
s
)
=
∫
0
s
ϕ
(
t
)
𝑑
t
,
Ψ
(
s
)
=
∫
0
s
ψ
(
t
)
𝑑
t
,
F
i
j
(
s
)
=
∫
0
s
f
i
j
(
t
)
𝑑
t
,
i
,
j
=
1
,
2
.
We will assume these functions satisfy the following hypotheses:
There exist p>1, q>1, δij>0 such that
lim
s
→
+
∞
s
ϕ
(
s
)
Φ
(
s
)
=
p
,
lim
s
→
+
∞
s
ψ
(
s
)
Ψ
(
s
)
=
q
,
lim
s
→
+
∞
s
f
i
j
(
s
)
F
i
j
(
s
)
=
δ
i
j
+
1
.
Furthermore, we assume that α>p-1 and α~>q-1. Notice that α and α~ are positive.
There exist p~>1,q~>1, μij>0 such that
lim
s
→
0
s
ϕ
(
s
)
Φ
(
s
)
=
p
~
,
lim
s
→
0
s
ψ
(
s
)
Ψ
(
s
)
=
q
~
,
lim
s
→
0
s
f
i
j
(
s
)
F
i
j
(
s
)
=
μ
i
j
+
1
,
and the following relationships among the exponents are determined by these limits:
δ
12
δ
21
>
(
p
-
δ
11
-
1
)
(
q
-
δ
22
-
1
)
,
p
-
δ
11
-
1
>
0
,
q
-
δ
22
-
1
>
0
,
and
μ
12
μ
21
>
(
p
~
-
μ
11
-
1
)
(
q
~
-
μ
22
-
1
)
,
p
~
-
μ
11
-
1
>
0
,
q
~
-
μ
22
-
1
>
0
.
The following inequalities, that will be used later in this paper, are direct consequences of hypotheses (H1) and (H2): For all ε>0, there exist s0, c>0 and C>0 such that for all s>s0,
c
s
δ
i
j
+
1
-
ε
≤
F
i
j
(
s
)
≤
C
s
δ
i
j
+
1
+
ε
for all
i
,
j
=
1
,
2
,
c
s
p
-
ε
≤
Φ
(
s
)
≤
C
s
p
+
ε
and
c
s
q
-
ε
≤
Ψ
(
s
)
≤
C
s
q
+
ε
.
Similarly, for all ε>0, there exist 0<s1<1 such that for all 0<s<s1,
c
s
μ
i
j
+
1
+
ε
≤
F
i
j
(
s
)
≤
C
s
μ
i
j
+
1
-
ε
for all
i
,
j
=
1
,
2
,
c
s
p
~
+
ε
≤
Φ
(
s
)
≤
C
s
p
~
-
ε
and
c
s
q
~
+
ε
≤
Ψ
(
s
)
≤
C
s
q
~
-
ε
.
Furthermore, for all s>s0,
(1.7)
c
s
δ
i
j
-
ε
≤
f
i
j
(
s
)
≤
C
s
δ
i
j
+
ε
for all
i
,
j
=
1
,
2
,
(1.8)
c
s
p
-
1
-
ε
≤
ϕ
(
s
)
≤
C
s
p
-
1
+
ε
and
c
s
q
-
1
-
ε
≤
ψ
(
s
)
≤
C
s
q
-
1
+
ε
,
and for all 0<s<s1,
(1.9)
c
s
μ
i
j
+
ε
≤
f
i
j
(
s
)
≤
C
s
μ
i
j
-
ε
for all
i
,
j
=
1
,
2
,
(1.10)
c
s
p
~
-
1
+
ε
≤
ϕ
(
s
)
≤
C
s
p
~
-
1
-
ε
and
c
s
q
~
-
1
+
ε
≤
ψ
(
s
)
≤
C
s
q
~
-
1
-
ε
.
Some related works to this paper can be found in [9, 10, 4, 3, 5, 14, 6, 7, 11, 13, 17, 16, 20, 18, 22, 15].
This paper is organized as follows. In Section 2 we give some preliminary results. Here we first prove a lemma about positive supersolutions that is needed in a crucial form in Proposition 4.1. We continue by stating a lemma which says that a certain nonlinear algebraic system of equations has a solution of the form (α1(s),α2(s)) for s in an interval of the form (s*,∞) for a certain s*, and we give some additional properties of these solutions. These functions play the role of a nonlinear rescaling in the blow-up argument later. Since the proof of this lemma is very technical and long, due to space limitations, we do not provide it in this paper; it will appear somewhere else.
We end the section with a theorem for nonexistence of positive solutions for a quasilinear homogeneous system with weights. This result follows in the same way as a similar result in [1] for the non-weighted case. We will apply this theorem to a limiting system that comes from the blow-up method.
In Section 3 we consider a problem, see (3.1), depending on the product of two parameters λ and h. For λ=0, this problem gives problem (1.4), and we want to establish a homotopy between these two problems with a view of later Leray–Schauder degree calculations. We set an abstract formulation problem (3.1) and prove in this section that for λ=1 and all large h>0, problem (3.1) does not have solutions. This result will be combined with those of the next Section 4, where the a-priori bounds of solutions to system (3.1) is proved via the blow-up method. In this section we prove Proposition 4.1 which is the central result of the paper. Also in this section we introduce a new hypothesis (H5) that is needed to control the possible oscillations of the quotients introduced in hypotheses (H1) and (H2).
In Section 5 we establish and prove our main theorem about the existence of positive solutions to problem (1.4) via Leray–Schauder degree theory. In Section 6 we give some simple examples to which our results apply.
Throughout this paper C[0,R] will denote the usual Banach space of continuous functions defined on [0,R] with the sup norm, and ℬ(0,M) will denote a ball in the product space C[0,R]×C[0,R], with center at 0 and radius M>0.
2 Some Preliminary Results
We begin with a lemma concerning positive supersolutions.
Lemma 2.1.
Let ϕ satisfy assumptions (H1) and (H2) and let r0≥0 and R>r0. Let u∈C1([r0,R)), (rαϕ(u′))∈C1((r0,R)) be a nonnegative function satisfying
(2.1)
-
(
r
α
ϕ
(
u
′
)
)
′
≥
0
on
(
r
0
,
R
)
.
Assume u′(r0)≤0. Then for any r>r0 and any constant k>1 with kr<R, we have
(2.2)
u
(
r
)
≥
M
0
α
r
|
u
′
(
r
)
|
,
where M0 is a positive constant depending on k and α.
Proof.
Integrating (2.1) from r to t, with r0≤r≤t≤R, we find
-
t
α
ϕ
(
u
′
(
t
)
)
≥
-
r
α
ϕ
(
u
′
(
r
)
)
,
from which it follows immediately that u′(t)≤0 on (r0,R) (just let r→r0). Then we can write
-
t
α
ϕ
(
u
′
(
t
)
)
≥
r
α
ϕ
(
|
u
′
(
r
)
|
)
,
and hence
-
u
′
(
t
)
≥
ϕ
-
1
(
t
-
α
r
α
ϕ
(
|
u
′
(
r
)
|
)
)
.
Now we fix r>r0, with kr<R. We may assume that u′(r)<0, since if u′(r)=0, then (2.2) holds for every constant M0. Integrating the last inequality with respect to t, from r to kr<R, we find
u
(
r
)
≥
∫
r
k
r
ϕ
-
1
(
t
-
α
r
α
ϕ
(
|
u
′
(
r
)
|
)
)
𝑑
t
.
Making the change of variable τ=t-αrαϕ(|u′(r)|), we obtain
(2.3)
u
(
r
)
≥
r
ϕ
(
|
u
′
(
r
)
|
)
1
α
α
∫
ϕ
(
|
u
′
(
r
)
|
)
k
α
ϕ
(
|
u
′
(
r
)
|
)
τ
-
(
α
+
1
α
)
ϕ
-
1
(
τ
)
𝑑
τ
≥
r
α
ϕ
(
|
u
′
(
r
)
|
)
∫
ϕ
(
|
u
′
(
r
)
|
)
k
α
ϕ
(
|
u
′
(
r
)
|
)
ϕ
-
1
(
τ
)
𝑑
τ
.
Defining
h
(
r
)
=
1
|
u
′
(
r
)
|
ϕ
(
|
u
′
(
r
)
|
)
∫
ϕ
(
|
u
′
(
r
)
|
)
k
α
ϕ
(
|
u
′
(
r
)
|
)
ϕ
-
1
(
τ
)
𝑑
τ
,
it can be easily verified that h(r) is bounded below by a positive constant that we call M0. From this fact and (2.3), we obtain that (2.2) follows immediately. This proves the lemma. ∎
Lemma 2.2.
Assume that (H1) and (H3) hold and for s∈(0,∞), let us consider the system
(2.4)
F
11
(
x
1
)
F
12
(
x
2
)
-
τ
s
χ
-
α
x
2
Φ
(
s
x
1
)
=
0
,
(2.5)
F
21
(
x
1
)
F
22
(
x
2
)
-
σ
s
χ
~
-
α
~
x
1
Ψ
(
s
x
2
)
=
0
,
where τ=p(δ11+1)(δ12+1) and σ=q(δ21+1)(δ22+1). Then there exists s1>0 such that, for each s>s1, system (2.4)–(2.5) has a unique solution (x1,x2)=(α1(s),α2(s)) satisfying αi∈C1(s1,∞), αi′(s)>0 on (s1,∞) and αi(s)→∞ as s→∞ for i=1, 2. Furthermore, αi(s) is AH at ∞ of exponent Ei for i=1, 2, where
(2.6)
E
1
=
(
χ
+
p
-
α
)
(
q
-
δ
22
-
1
)
+
(
χ
~
+
q
-
α
~
)
δ
12
δ
12
δ
21
-
(
p
-
δ
11
-
1
)
(
q
-
δ
22
-
1
)
and
(2.7)
E
2
=
(
χ
~
+
q
-
α
~
)
(
p
-
δ
11
-
1
)
+
(
χ
+
p
-
α
)
δ
21
δ
12
δ
21
-
(
p
-
δ
11
-
1
)
(
q
-
δ
22
-
1
)
.
Finally, defining the functions R1(s) and R2(s) by
ℛ
1
(
s
)
=
f
11
(
α
1
(
s
)
)
f
12
(
α
2
(
s
)
)
s
χ
-
α
+
1
ϕ
(
s
α
1
(
s
)
)
,
ℛ
2
(
s
)
=
f
21
(
α
1
(
s
)
)
f
22
(
α
2
(
s
)
)
s
χ
~
-
α
~
+
1
ψ
(
s
α
2
(
s
)
)
,
then
lim
s
→
∞
ℛ
1
(
s
)
=
1
,
lim
s
→
∞
ℛ
2
(
s
)
=
1
.
The functions (α1(s),α2(s)), will play the role of a nonlinear rescaling (as mentioned in the introduction) which will permit to extend the blow-up method to our situation. As we pointed out in the introduction, the proof of the existence and asymptotic properties of the functions (α1(s),α2(s)) is lengthy and very technical, hence it will not be given here. The rest of the proof is straightforward.
In our final result in this section we consider the system
(2.8)
{
(
r
α
ϕ
p
(
u
′
)
)
′
+
r
χ
u
(
r
)
δ
11
v
(
r
)
δ
12
=
0
,
r
∈
(
0
,
∞
)
,
(
r
α
~
ϕ
q
(
v
′
)
)
′
+
r
χ
~
u
(
r
)
δ
21
v
(
r
)
δ
22
=
0
,
r
∈
(
0
,
∞
)
,
u
′
(
0
)
=
0
=
v
′
(
0
)
.
We have the following nonexistence of positive solutions result.
Theorem 2.3.
If 1<p,1<q, α>p-1, α~>q-1, condition (H3) holds, δij>0,i,j=1,2, and the condition
max
{
E
1
-
α
-
(
p
-
1
)
p
-
1
,
E
2
-
α
~
-
(
q
-
1
)
q
-
1
}
>
0
is satisfied, then problem (2.8) has no positive solution. Here E1 and E2 are given by (2.6) and (2.7), respectively.
The proof of this theorem can be derived as in [1] for a similar result in the non-weighted case and hence it will not be given here. We just notice at this point the interesting role played by the exponents E1 and E2 in the sense that they link asymptotic properties of the functions (α1(s),α2(s)) with the nonexistence result.
3 Homotopying with System (1.4)
We consider the system
(3.1)
{
(
r
α
ϕ
(
u
′
)
)
′
+
r
χ
(
f
11
(
|
u
(
r
)
|
)
f
12
(
|
v
(
r
)
|
)
+
λ
h
)
=
0
,
r
∈
(
0
,
R
)
,
(
r
α
~
ψ
(
v
′
)
)
′
+
r
χ
~
(
f
21
(
|
u
(
r
)
|
)
f
22
(
|
v
(
r
)
|
)
+
λ
h
)
=
0
,
r
∈
(
0
,
R
)
,
u
′
(
0
)
=
0
=
v
′
(
0
)
,
u
(
R
)
=
0
=
v
(
R
)
.
where λ∈[0,1], and h is a positive parameter that will be fixed later in this section. For the moment we assume it is fixed. We point out, as it is easy to check, that solutions (u,v) to this system satisfy u(r)>0, v(r)>0 for r∈[0,R).
This problem can be written in an equivalent operator form as
(3.2)
(
u
,
v
)
=
𝒯
(
(
u
,
v
)
,
λ
)
,
where 𝒯:C[0,R]×C[0,R]×[0,1]↦C[0,R]×C[0,R] is a completely continuous operator given by
𝒯
(
(
u
,
v
)
,
λ
)
=
(
∫
r
R
ϕ
-
1
(
1
s
α
∫
0
s
ξ
χ
(
f
11
(
|
u
(
ξ
)
|
)
f
12
(
|
v
(
ξ
)
|
)
+
λ
h
)
d
ξ
)
d
s
,
∫
r
R
ψ
-
1
(
1
s
α
~
∫
0
s
ξ
χ
~
(
f
21
(
|
u
(
ξ
)
|
)
f
22
(
|
v
(
ξ
)
|
)
+
λ
h
)
d
ξ
)
d
s
)
.
Solutions to (3.2), equivalently to (3.1), for λ=0, give solutions to the original problem (1.4), and hence solve our main problem.
We show next that for λ=1, there exists h0 such that for any h>h0, (3.2) does not have solutions. Indeed, for λ=1, equation (3.2) is equivalent to the problem
(3.3)
{
(
r
α
ϕ
(
u
′
)
)
′
+
r
χ
(
f
11
(
|
u
(
r
)
|
)
f
12
(
|
v
(
r
)
|
)
+
h
)
=
0
,
r
∈
(
0
,
R
)
,
(
r
α
~
ψ
(
v
′
)
)
′
+
r
χ
~
(
f
21
(
|
u
(
r
)
|
)
f
22
(
|
v
(
r
)
|
)
+
h
)
=
0
,
r
∈
(
0
,
R
)
,
u
′
(
0
)
=
u
(
R
)
=
0
,
v
′
(
0
)
=
v
(
R
)
=
0
.
In the proof of the next proposition, we will use the following lemma.
Lemma 3.1.
Assume that (H1) and (H2) hold. Then there exist a t0>0 and a constant K>0 such that if 0≤τ≤t and t≥t0, then
f
i
j
(
τ
)
≤
K
f
i
j
(
t
)
.
Proof.
Since
(
s
-
1
F
i
j
(
s
)
)
′
=
F
i
j
(
s
)
s
2
(
s
f
i
j
(
s
)
F
i
j
(
s
)
-
1
)
,
there exists t0>0 such that
(
s
-
1
F
i
j
(
s
)
)
′
>
0
,
s
≥
t
0
,
that is, s-1Fij(s) is increasing on [t0,∞). Moreover, from (H1), (H2), there exist d1>0 and d2>0 such that
d
1
≤
s
f
i
j
(
s
)
F
i
j
(
s
)
≤
d
2
,
s
≥
0
,
i
,
j
=
1
,
2
.
If t≥τ≥t0, then
f
i
j
(
τ
)
f
i
j
(
t
)
≤
d
2
τ
-
1
F
i
j
(
τ
)
d
1
t
-
1
F
i
j
(
t
)
≤
d
2
d
1
.
If t≥t0≥τ≥0, then
f
i
j
(
τ
)
f
i
j
(
t
)
≤
d
2
τ
-
1
F
i
j
(
τ
)
d
1
t
-
1
F
i
j
(
t
)
≤
d
2
sup
s
∈
[
0
,
t
0
]
|
s
-
1
F
i
j
(
s
)
|
d
1
t
0
-
1
F
i
j
(
t
0
)
.
Since lims→0s-1Fij(s)=0, we note that sups∈[0,s0]|s-1Fij(s)|<∞. The proof is complete. ∎
Proposition 3.2.
Assume that (H1) and (H2) hold. Then there exists h0 such that for every h>h0, system (3.3) (and hence (3.2) with λ=1) does not have solutions.
Proof.
By integration and taking into account that a solution (u,v) to (3.3) satisfies u(r)>0, v(r)>0, r∈[0,R), we have
u
(
r
)
=
∫
r
R
ϕ
-
1
(
1
s
α
∫
0
s
ξ
χ
(
f
11
(
u
(
ξ
)
)
f
12
(
v
(
ξ
)
)
+
h
)
𝑑
ξ
)
𝑑
s
and
v
(
r
)
=
∫
r
R
ψ
-
1
(
1
s
α
~
∫
0
s
ξ
χ
~
(
f
21
(
u
(
ξ
)
)
f
22
(
v
(
ξ
)
)
+
h
)
𝑑
ξ
)
𝑑
s
.
We continue by contradiction assuming there exists {hk} such that hk→∞ as k→∞, for which (3.3) has a positive solution (uk,vk) for each k. Then, from the last two equations, with (u,v,h) replaced by (uk,vk,hk), we obtain that
u
k
(
r
)
≥
∫
r
R
ϕ
-
1
(
1
s
α
∫
0
s
ξ
χ
h
k
𝑑
ξ
)
𝑑
s
≥
ϕ
-
1
(
1
χ
+
1
r
χ
-
α
+
1
h
k
)
(
R
-
r
)
,
v
k
(
r
)
≥
∫
r
R
ψ
-
1
(
1
s
α
~
∫
0
s
ξ
χ
~
h
k
𝑑
ξ
)
𝑑
s
≥
ψ
-
1
(
1
χ
~
+
1
r
χ
~
-
α
~
+
1
h
k
)
(
R
-
r
)
,
for all r∈[0,R]. Then for r∈[0,34R],
u
k
(
r
)
≥
u
k
(
3
4
R
)
≥
R
4
ϕ
-
1
(
1
χ
+
1
(
3
4
R
)
χ
-
α
+
1
h
k
)
,
v
k
(
r
)
≥
v
k
(
3
4
R
)
≥
R
4
ψ
-
1
(
1
χ
~
+
1
(
3
4
R
)
χ
~
-
α
~
+
1
h
k
)
.
Therefore, for every A>0, there exists k0, such that for all k>k0,
u
k
(
r
)
≥
A
,
v
k
(
r
)
≥
A
,
for all r∈[0,34R]. We also have
u
k
(
r
)
≥
∫
r
R
ϕ
-
1
(
1
R
α
∫
0
r
ξ
χ
f
11
(
u
k
(
ξ
)
)
f
12
(
v
k
(
ξ
)
)
𝑑
ξ
)
𝑑
s
=
(
R
-
r
)
ϕ
-
1
(
1
R
α
∫
0
r
ξ
χ
f
11
(
u
k
(
ξ
)
)
f
12
(
v
k
(
ξ
)
)
𝑑
ξ
)
≥
R
4
ϕ
-
1
(
1
R
α
∫
0
r
ξ
χ
f
11
(
u
k
(
ξ
)
)
f
12
(
v
k
(
ξ
)
)
𝑑
ξ
)
for all r∈[0,34R]. Similarly,
v
k
(
r
)
≥
R
4
ψ
-
1
(
1
R
α
~
∫
0
r
ξ
χ
~
f
21
(
u
k
(
ξ
)
)
f
22
(
v
k
(
ξ
)
)
𝑑
ξ
)
for all r∈[0,34R].
Next, by Lemma 3.1, there exist t0>0 and a constant K>0 such that if 0≤τ≤t and t≥t0, then
f
i
j
(
τ
)
≤
K
f
i
j
(
t
)
for all i,j∈{1,2}. Then, by increasing A, if necessary, we first obtain
∫
0
r
ξ
χ
f
11
(
u
k
(
ξ
)
)
f
12
(
v
k
(
ξ
)
)
𝑑
ξ
≥
1
K
2
∫
0
r
ξ
χ
f
11
(
u
k
(
r
)
)
f
12
(
v
k
(
r
)
)
𝑑
ξ
=
1
K
2
(
χ
+
1
)
r
χ
+
1
f
11
(
u
k
(
r
)
)
f
12
(
v
k
(
r
)
)
≥
1
K
2
(
χ
+
1
)
(
R
4
)
χ
+
1
f
11
(
u
k
(
r
)
)
f
12
(
v
k
(
r
)
)
for all r∈[R4,34R], and hence
u
k
(
r
)
≥
R
4
ϕ
-
1
(
c
1
f
11
(
u
k
(
r
)
)
f
12
(
v
k
(
r
)
)
)
for all r∈[R4,34R], where c1 is a positive constant. Similarly, we have
v
k
(
r
)
≥
R
4
ψ
-
1
(
c
2
f
21
(
u
k
(
r
)
)
f
22
(
v
k
(
r
)
)
)
for all r∈[R4,3R4], where c2 is a positive constant. Then
ϕ
(
4
R
u
k
(
r
)
)
≥
c
1
f
11
(
u
k
(
r
)
)
f
12
(
v
k
(
r
)
)
and
ψ
(
4
R
v
k
(
r
)
)
≥
c
2
f
21
(
u
k
(
r
)
)
f
22
(
v
k
(
r
)
)
for all r∈[R4,34R]. Now, since uk(r)→∞ and vk(r)→∞ as k→∞ uniformly on r∈[R4,3R4], from (1.7) and (1.8), we find that
(
u
k
(
r
)
)
p
-
1
-
δ
11
+
2
ε
≥
c
1
′
(
v
k
(
r
)
)
δ
12
-
ε
and
(
v
k
(
r
)
)
q
-
1
-
δ
22
+
2
ε
≥
c
2
′
(
u
k
(
r
)
)
δ
21
-
ε
for all r∈[R4,34R]. From these two inequalities, it follows that
(3.4)
(
u
k
(
r
)
)
Γ
(
ε
)
≥
C
,
r
∈
[
R
4
,
3
4
R
]
,
where C is a positive constant and
Γ
(
ε
)
=
(
p
-
1
-
δ
11
+
2
ε
)
(
q
-
1
-
δ
22
+
2
ε
)
-
(
δ
12
-
ε
)
(
δ
21
-
ε
)
q
-
1
-
δ
22
+
2
ε
.
Clearly Γ(ε) is continuous, and we are interested in ε>0 small. Now
Γ
(
0
)
=
(
p
-
1
-
δ
11
)
(
q
-
1
-
δ
22
)
-
δ
12
δ
21
q
-
1
-
δ
22
<
0
,
and hence Γ(ε)<0 for ε>0 small, which gives a contradiction to (3.4), for k large. This ends the proof of the proposition. ∎
4 A-Priori Bounds via Blow-Up Method
4.1 Some Preliminaries
Before we state and prove our main result, we notice that from (H1) and (H2), there exists 1<p-≤p+, 0<q-≤q+, and -1<δij-≤δij+, for j=1, 2, such that
(4.1)
inf
s
∈
[
0
,
∞
)
s
ϕ
(
s
)
Φ
(
s
)
=
p
-
,
sup
s
∈
[
0
,
∞
)
s
ϕ
(
s
)
Φ
(
s
)
=
p
+
,
(4.2)
inf
s
∈
[
0
,
∞
)
s
ψ
(
s
)
Ψ
(
s
)
=
q
-
,
sup
s
∈
[
0
,
∞
)
s
ψ
(
s
)
Ψ
(
s
)
=
q
+
,
(4.3)
inf
s
∈
[
0
,
∞
)
s
f
i
j
(
s
)
F
i
j
(
s
)
=
δ
i
j
-
+
1
,
sup
s
∈
[
0
,
∞
)
s
f
i
j
(
s
)
F
i
j
(
s
)
=
δ
i
j
+
+
1
,
i
,
j
=
1
,
2
.
Notice that d1,d2 of Lemma 3.1 are redefined here as d1=δij-+1 and d2=δij++1 for notational convenience. To eliminate some unwanted solutions to the limiting system that comes out from the blow-up method we add here a new hypothesis that will be used in the second part of the proof of the proposition of this section. Thus, we assume
p
-
>
δ
11
+
+
1
and q->δ22++1.
From (H3) and since δij+≥δij for i, j=1, 2, p≥p- and q≥q-, we have that
δ
12
+
δ
21
+
>
(
p
-
-
δ
11
+
-
1
)
(
q
-
-
δ
22
+
-
1
)
.
Also, from (4.1)–(4.3), it is not difficult to show that the following inequalities hold true:
(4.4)
p
-
p
+
σ
p
+
-
1
≤
ϕ
(
σ
s
)
ϕ
(
s
)
≤
p
+
p
-
σ
p
-
-
1
,
s
∈
(
0
,
∞
)
,
0
<
σ
≤
1
,
q
-
q
+
σ
q
+
-
1
≤
ψ
(
σ
s
)
ψ
(
s
)
≤
q
+
q
-
σ
q
-
-
1
,
s
∈
(
0
,
∞
)
,
0
<
σ
≤
1
,
(4.5)
δ
i
j
-
+
1
δ
i
j
+
+
1
σ
δ
i
j
+
≤
f
i
j
(
σ
s
)
f
i
j
(
s
)
≤
δ
i
j
+
+
1
δ
i
j
-
+
1
σ
δ
i
j
-
,
s
∈
(
0
,
∞
)
,
0
<
σ
≤
1
.
For example, integrating (logΦ(s))′=ϕ(s)Φ(s)≤p+s on [σs,s], we have logΦ(s)Φ(σs)≤logσ-p+, that is,
Φ
(
σ
s
)
Φ
(
s
)
≥
σ
p
+
,
and hence
ϕ
(
σ
s
)
ϕ
(
s
)
=
σ
s
ϕ
(
σ
s
)
Φ
(
σ
s
)
Φ
(
s
)
s
ϕ
(
s
)
1
σ
Φ
(
σ
s
)
Φ
(
s
)
≥
p
-
p
+
σ
p
+
-
1
,
s
∈
(
0
,
∞
)
,
0
<
σ
≤
1
.
Now we state and prove the central result of this paper:
Proposition 4.1.
Assume that (H1), (H2), (H3) and (H5) hold. Then there exists M~>0 such that problem (3.2) does not have solutions ((u,v),λ) in the set ∂B(0,M)×[0,1], for any M≥M~.
The proof of this proposition in lengthy and technical and therefore it will be done in two steps. The first step, given in Section 4.2, explores the blow-up method to obtain a nontrivial solution to a system of limiting equations. The second step consists in proving that, thanks to (H5), such system cannot have a solution and is done in Section 4.3.
4.2 Proof of Step 1
The proof is based in the blow-up method and uses a contradiction argument. Thus, let us assume there are sequences {Mk}, Mk→∞, {(uk,vk)} and {λk∈[0,1]} such that ∥uk∥+∥vk∥=Mk, that satisfy
(
u
k
,
v
k
)
=
𝒯
(
(
u
k
,
v
k
)
,
λ
k
)
,
equivalently,
(4.6)
{
(
r
α
ϕ
(
u
k
′
)
)
′
+
r
χ
(
f
11
(
|
u
k
(
r
)
|
)
f
12
(
|
v
k
(
r
)
|
)
+
λ
k
h
)
=
0
,
r
∈
(
0
,
R
)
,
(
r
α
~
ψ
(
v
k
′
)
)
′
+
r
χ
~
(
f
21
(
|
u
k
(
r
)
|
)
f
22
(
|
v
k
(
r
)
|
)
+
λ
k
h
)
=
0
,
r
∈
(
0
,
R
)
,
u
k
′
(
0
)
=
0
=
v
k
′
(
0
)
,
u
k
(
R
)
=
0
=
v
k
(
R
)
.
It is then not difficult to show that ∥uk∥→∞ and ∥vk∥→∞ as k→∞. Let α1,α2:[t0,∞)→ℝ be the functions determined in Section 2 and for k>k0, k0 large, set
χ
k
=
α
1
-
1
(
∥
u
k
∥
)
+
α
2
-
1
(
∥
v
k
∥
)
and
t
1
,
k
=
α
1
(
χ
k
)
,
t
2
,
k
=
α
2
(
χ
k
)
,
and introduce the change of variables
(4.7)
y
=
χ
k
r
,
w
k
(
y
)
=
u
k
(
r
)
t
1
,
k
,
z
k
(
y
)
=
v
k
(
r
)
t
2
,
k
in system (4.6), which becomes
(4.8)
{
-
(
y
α
ϕ
(
t
1
,
k
χ
k
w
k
′
)
)
′
=
χ
k
-
(
χ
-
α
+
1
)
y
χ
(
f
11
(
t
1
,
k
|
w
k
(
y
)
|
)
f
12
(
t
2
,
k
|
z
k
(
y
)
|
)
+
λ
k
h
)
,
-
(
y
α
~
ψ
(
t
2
,
k
χ
k
z
k
′
)
)
′
=
χ
k
-
(
χ
~
-
α
~
+
1
)
y
χ
~
(
f
21
(
t
1
,
k
|
w
k
(
y
)
|
)
f
22
(
t
2
,
k
|
z
k
(
y
)
|
)
+
λ
k
h
)
,
w
k
′
(
0
)
=
0
=
z
k
′
(
0
)
,
w
k
(
χ
k
R
)
=
0
=
z
k
(
χ
k
R
)
=
0
.
where y∈(0,χkR). Let R1>0 be arbitrary. We take k so large that R1<χkR and we suppose, in the following argument, that y∈[0,R1]. By integrating system (4.8) and using the boundary conditions at zero, we get
(4.9)
-
w
k
′
(
y
)
=
ϕ
-
1
(
ρ
1
,
k
I
1
,
k
(
y
)
ℛ
1
,
k
)
ϕ
-
1
(
ρ
1
,
k
)
,
-
z
k
′
(
y
)
=
ψ
-
1
(
ρ
2
,
k
I
2
,
k
(
y
)
ℛ
2
,
k
)
ψ
-
1
(
ρ
2
,
k
)
,
where ℛ1,k=ℛ1(χk), ℛ2,k=ℛ2(χk), ρ1,k=ϕ(t1,kχk),ρ2,k=ψ(t2,kχk),
(4.10)
I
1
,
k
(
y
)
=
y
-
α
∫
0
y
s
χ
(
f
11
(
t
1
,
k
|
w
k
(
s
)
|
)
f
12
(
t
2
,
k
|
z
k
(
s
)
|
)
f
11
(
t
1
,
k
)
f
12
(
t
2
,
k
)
+
λ
k
h
f
11
(
t
1
,
k
)
f
12
(
t
2
,
k
)
)
𝑑
s
,
and
I
2
,
k
(
y
)
=
y
-
α
∫
0
y
s
χ
(
f
21
(
t
1
,
k
|
w
k
(
s
)
|
)
f
22
(
t
2
,
k
|
z
k
(
s
)
|
)
f
21
(
t
1
,
k
)
f
22
(
t
2
,
k
)
+
λ
k
h
f
21
(
t
1
,
k
)
f
22
(
t
2
,
k
)
)
𝑑
s
for y∈(0,R1]. Clearly, ρ1,k, ρ2,k→∞ as k→∞.
Now, by Lemma 3.1, there exists a positive constant K>0 such that
(4.11)
f
11
(
t
1
,
k
|
w
k
(
s
)
|
)
f
12
(
t
2
,
k
|
z
k
(
s
)
|
)
f
11
(
t
1
,
k
)
f
12
(
t
2
,
k
)
≤
K
2
,
f
21
(
t
1
,
k
|
w
k
(
s
)
|
)
f
22
(
t
2
,
k
|
z
k
(
s
)
|
)
f
21
(
t
1
,
k
)
f
22
(
t
2
,
k
)
≤
K
2
,
for all s∈[0,R1], and since χ-α+1>0 and χ~-α~+1>0, we have
I
1
,
k
(
0
)
=
lim
y
→
0
I
1
,
k
(
y
)
=
0
,
I
2
,
k
(
0
)
=
lim
y
→
0
I
2
,
k
(
y
)
=
0
.
The bounds (4.11) together with the fact that {ℛ1,k} and {ℛ2,k} are bounded sequences imply that there exists a positive constant C such that
|
w
k
′
(
y
)
|
≤
ϕ
-
1
(
ρ
1
,
k
C
)
ϕ
-
1
(
ρ
1
,
k
)
,
|
z
k
′
(
y
)
|
≤
ψ
-
1
(
ρ
2
,
k
C
)
ψ
-
1
(
ρ
2
,
k
)
.
Finally, since ϕ-1 and ψ-1 are AH functions at infinite (see [21, Lemma 1.8]), it follows that {wk′} and {zk′} are bounded sequences, implying that the sequences {wk} and {zk} are equicontinuous and uniformly bounded. Then, by the Arzela-Ascoli theorem, they possess subsequences (renamed the same) converging uniformly in [0,R1], say
w
k
→
w
^
,
z
k
→
z
^
uniformly in
[
0
,
R
1
]
,
and notice that w^≥0, z^≥0 on [0,R1]. Then, an application of the Lebesgue Dominated Convergence Theorem gives that for each y∈[0,R1],
(4.12)
I
1
,
k
(
y
)
→
I
1
,
∞
(
y
)
,
I
2
,
k
(
y
)
→
I
2
,
∞
(
y
)
as
k
→
∞
,
where
I
1
,
∞
(
y
)
=
y
-
α
∫
0
y
s
χ
w
^
(
s
)
δ
11
z
^
(
s
)
δ
12
𝑑
s
,
I
2
,
∞
(
y
)
=
y
-
α
∫
0
y
s
χ
w
^
(
s
)
δ
21
z
^
(
s
)
δ
22
𝑑
s
,
with
I
1
,
∞
(
0
)
=
0
,
I
2
,
∞
(
0
)
=
0
.
From (4.9), by integrating, we first get
(4.13)
w
k
(
y
)
=
w
k
(
0
)
-
∫
0
y
ϕ
-
1
(
ρ
1
,
k
I
1
,
k
(
s
)
ℛ
1
,
k
)
ϕ
-
1
(
ρ
1
,
k
)
𝑑
s
,
(4.14)
z
k
(
y
)
=
z
k
(
0
)
-
∫
0
y
ψ
-
1
(
ρ
2
,
k
I
2
,
k
(
s
)
ℛ
2
,
k
)
ψ
-
1
(
ρ
2
,
k
)
𝑑
s
.
Since ϕ-1 is AH of exponent 1p-1 (see [21, Lemma 1.8]), by using (4.12), the fact that
lim
k
→
∞
ℛ
1
,
k
=
1
,
lim
k
→
∞
ℛ
2
,
k
=
1
,
and by taking the limit as k→∞ in (4.13) and (4.14), a new application of Lebesgue’s dominated theorem, yields
w
^
(
y
)
=
w
^
(
0
)
-
∫
0
y
(
I
1
,
∞
(
s
)
)
1
p
-
1
𝑑
s
and
z
^
(
y
)
=
z
^
(
0
)
-
∫
0
y
(
I
2
,
∞
(
s
)
)
1
q
-
1
𝑑
s
,
hence
-
w
^
′
(
y
)
=
(
I
1
,
∞
(
y
)
)
1
p
-
1
=
(
y
-
α
∫
0
y
s
χ
w
^
(
s
)
δ
11
z
^
(
s
)
δ
12
𝑑
s
)
1
p
-
1
,
-
z
^
′
(
y
)
=
(
I
2
,
∞
(
y
)
)
1
q
-
1
=
(
y
-
α
∫
0
y
s
χ
w
^
(
s
)
δ
21
z
^
(
s
)
δ
22
𝑑
s
)
1
q
-
1
.
Differentiating both sides of the two equations, we get that for all y∈(0,R1], w^(y) and z^(y) satisfy
(4.15)
{
(
y
α
ϕ
p
(
w
^
′
)
)
′
+
y
χ
w
^
(
y
)
δ
11
z
^
(
y
)
δ
12
=
0
,
y
∈
[
0
,
R
1
]
,
(
r
α
~
ϕ
q
(
z
^
′
)
)
′
+
r
χ
~
w
^
(
y
)
δ
21
z
^
(
y
)
δ
22
=
0
,
y
∈
[
0
,
R
1
]
,
w
^
′
(
0
)
=
0
=
z
^
′
(
0
)
.
Arguing like in [2] or [8], one obtains that w^ and z^ can be extended to (0,∞) as solutions of system (2.8) that satisfy w^(y)≥0 and z^(y)≥0 for all y in [0,∞), and arguing like in [1], one can show that either w^(0)>0 or z^(0)>0, and hence (w^,z^) is not the trivial solution.
An important observation at this point is that system (4.15) admits a solution of the form (w^,z^)=(C,0) or (w^,z^)=(0,C), where C is a positive constant. We will show next that these solutions are not possible under (H5).
4.3 Proof of Step 2
We come back to system (4.8), more specifically to equation (4.9), that we rewrite as
(4.16)
-
ϕ
(
t
1
,
k
χ
k
w
k
′
(
y
)
)
ϕ
(
t
1
,
k
χ
k
)
=
I
1
,
k
(
y
)
ℛ
1
,
k
,
-
ψ
(
t
2
,
k
χ
k
z
k
′
(
y
)
)
ψ
(
t
2
,
k
χ
k
)
=
I
2
,
k
(
y
)
ℛ
2
,
k
.
Applying the change of variables (4.7) to (2.2) in Lemma 2.1, we obtain that
(4.17)
w
k
(
y
)
≥
y
|
w
k
′
(
y
)
|
M
1
α
for y≤Rk2, where we have taken k=2 in Lemma 2.1 and Rk=χkR.
Next, let T a fixed but otherwise arbitrary positive number. Let also ε>0 be a small positive number such that T>ε. Let k0∈ℕ be large enough so that T≤Rk2 for all k≥k0.
Then, for all y∈[ε,T], from the first of (4.16) and (4.17), we find
(4.18)
ϕ
(
t
1
,
k
χ
k
α
M
1
y
w
k
(
y
)
)
ϕ
(
t
1
,
k
χ
k
)
≥
I
1
,
k
(
y
)
ℛ
1
,
k
≥
1
2
I
1
,
k
(
y
)
.
We study now I1,k(y) given by (4.10). Recalling that 0≤wk(s)≤1 and 0≤zk(s)≤1, from (4.5), it follows that
f
11
(
t
1
,
k
w
k
(
s
)
)
f
11
(
t
1
,
k
)
≥
δ
11
-
+
1
δ
11
+
+
1
(
w
k
(
s
)
)
δ
11
+
and
f
12
(
t
2
,
k
z
k
(
s
)
)
f
12
(
t
2
,
k
)
≥
δ
12
-
+
1
δ
12
+
+
1
(
z
k
(
s
)
)
δ
12
+
.
Since wk(s) and zk(s) are nonnegative and nonincreasing on [0,y], we have
f
11
(
t
1
,
k
w
k
(
s
)
)
f
11
(
t
1
,
k
)
≥
δ
11
-
+
1
δ
11
+
+
1
(
w
k
(
y
)
)
δ
11
+
and
f
12
(
t
2
,
k
z
k
(
s
)
)
f
12
(
t
2
,
k
)
≥
δ
12
-
+
1
δ
12
+
+
1
(
z
k
(
y
)
)
δ
12
+
for s∈[0,y]. Then
I
1
,
k
(
y
)
≥
(
c
1
(
w
k
(
y
)
)
δ
11
+
(
z
k
(
y
)
)
δ
12
+
+
ε
1
,
k
)
y
χ
-
α
+
1
χ
+
1
,
where
c
1
=
(
δ
11
-
+
1
)
(
δ
12
-
+
1
)
(
δ
11
+
+
1
)
(
δ
12
+
+
1
)
,
ε
1
,
k
=
λ
k
h
f
11
(
t
1
,
k
)
f
12
(
t
2
,
k
)
.
Hence,
I
1
,
k
(
y
)
≥
c
1
χ
+
1
y
χ
-
α
+
1
(
w
k
(
y
)
)
δ
11
+
(
z
k
(
y
)
)
δ
12
+
.
Therefore, from (4.18), it follows that
(4.19)
ϕ
(
t
1
,
k
χ
k
α
M
1
y
w
k
(
y
)
)
ϕ
(
t
1
,
k
χ
k
)
≥
c
1
2
(
χ
+
1
)
y
χ
-
α
+
1
(
w
k
(
y
)
)
δ
11
+
(
z
k
(
y
)
)
δ
12
+
.
Without loss of generality, we may assume that T>α/M1, and then for α/M1≤y≤T, we have that
0
<
α
M
1
y
≤
1
.
Thus, inequality (4.4) implies that
ϕ
(
t
1
,
k
χ
k
α
M
1
y
w
k
(
y
)
)
ϕ
(
t
1
,
k
χ
k
)
≤
p
+
p
-
(
α
M
1
y
w
k
(
y
)
)
p
-
-
1
=
p
+
p
-
(
α
M
1
)
p
-
-
1
y
-
(
p
-
-
1
)
(
w
k
(
y
)
)
p
-
-
1
for α/M1≤y≤T. Hence, from (4.19), it follows that
p
+
p
-
(
α
M
1
)
p
-
-
1
y
-
(
p
-
-
1
)
(
w
k
(
y
)
)
p
-
-
1
≥
c
1
2
(
χ
+
1
)
y
χ
-
α
+
1
(
w
k
(
y
)
)
δ
11
+
(
z
k
(
y
)
)
δ
12
+
,
that is,
(4.20)
(
w
k
(
y
)
)
p
-
-
1
-
δ
11
+
≥
C
1
y
χ
-
α
+
p
-
(
z
k
(
y
)
)
δ
12
+
,
α
/
M
1
≤
y
≤
T
,
where C1>0 is a constant.
Similarly, there exist constants M2>0 and C2>0 such that
(4.21)
(
z
k
(
y
)
)
q
-
-
1
-
δ
22
+
≥
C
2
y
χ
~
-
α
~
+
q
-
(
w
k
(
y
)
)
δ
21
+
,
α
~
/
M
2
≤
y
≤
T
.
Set T0=max{αM1,α~M2}. From (4.20) and (4.21), it follows that
(
w
k
(
y
)
)
p
-
-
1
-
δ
11
+
δ
12
+
≥
C
1
1
δ
12
+
y
χ
-
α
+
p
-
δ
12
+
z
k
(
y
)
and
z
k
(
y
)
≥
C
2
1
q
-
-
1
-
δ
22
+
y
χ
~
-
α
~
+
q
-
q
-
-
1
-
δ
22
+
(
w
k
(
y
)
)
δ
21
+
q
-
-
1
-
δ
22
+
for T0≤y≤T. Combining these inequalities, we find that
(
w
k
(
y
)
)
p
-
-
1
-
δ
11
+
δ
12
+
≥
C
1
1
δ
12
+
y
χ
-
α
+
p
-
δ
12
+
C
2
1
q
-
-
1
-
δ
22
+
y
χ
~
-
α
~
+
q
-
q
-
-
1
-
δ
22
+
(
w
k
(
y
)
)
δ
21
+
q
-
-
1
-
δ
22
+
,
that is,
C
3
y
-
χ
-
α
+
p
-
δ
12
+
-
χ
~
-
α
~
+
q
-
q
-
-
1
-
δ
22
+
≥
(
w
k
(
y
)
)
-
p
-
-
1
-
δ
11
+
δ
12
+
+
δ
21
+
q
-
-
1
-
δ
22
+
,
which implies
w
k
(
y
)
≤
C
~
1
y
-
Γ
1
for T0≤y≤T, where C~1 is a positive constant and
Γ
1
=
(
χ
-
α
+
p
-
)
(
q
-
-
1
-
δ
22
+
)
+
δ
12
+
(
χ
~
-
α
~
+
q
-
)
δ
12
+
δ
21
+
-
(
p
-
-
1
-
δ
11
+
)
(
q
-
-
1
-
δ
22
+
)
>
0
.
Similar to these calculations, from (4.20) and (4.21), we obtain that
w
k
(
y
)
≥
C
1
1
p
-
-
1
-
δ
11
+
y
χ
-
α
+
p
-
p
-
-
1
-
δ
11
+
(
z
k
(
y
)
)
δ
12
+
p
-
-
1
-
δ
11
+
and
(
z
k
(
y
)
)
q
-
-
1
-
δ
22
+
δ
21
+
≥
C
2
1
δ
21
+
y
χ
~
-
α
~
+
q
-
δ
21
+
w
k
(
y
)
for T0≤y≤T, which implies that
z
k
(
y
)
≤
C
~
2
y
-
Γ
2
for T0≤y≤T, where C~2 is a positive constant, and
Γ
2
=
(
χ
~
-
α
~
+
q
-
)
(
p
-
-
1
-
δ
11
+
)
+
δ
21
+
(
χ
-
α
+
p
-
)
δ
12
+
δ
21
+
-
(
p
-
-
1
-
δ
11
+
)
(
q
-
-
1
-
δ
22
+
)
>
0
.
Consequently, we have that
0
≤
w
k
(
T
)
≤
C
~
1
T
-
Γ
1
,
0
≤
z
k
(
T
)
≤
C
~
2
T
-
Γ
2
.
Letting k→∞, we obtain that
0
≤
w
^
(
T
)
≤
C
~
1
T
-
Γ
1
,
0
≤
z
^
(
T
)
≤
C
~
2
T
-
Γ
2
for all T>0 sufficiently large. Then we conclude that neither (w^(y),z^(y))≡(C,0) nor (w^(y),z^(y))≡(0,C) for some C>0.
Thus, any solution to system (4.15) satisfies w^(y)>0, z^(y)>0, for all y∈(0,∞). But by Theorem 2.3, these type of solutions cannot exist, ending the contradiction argument and thus finishing the proof of Proposition 4.1.
5 Main Theorem and Its Proof
We are now in a position to establish and prove our main result.
Theorem 5.1.
Under hypotheses (H1)–(H5), and
max
{
E
1
-
α
-
(
p
-
1
)
p
-
1
,
E
2
-
α
~
-
(
q
-
1
)
q
-
1
}
>
0
,
where E1 and E2 are defined in (2.6) and (2.7), respectively, problem (1.4) has a solution (u,v) such that u(r)>0, v(r)>0 for all r∈[0,R).
Proof.
Let us denote by B(0,ℛ) the ball with center 0 and radius ℛ in C[0,R], and by ℬ(0,ℛ) the ball in the product space, i.e.,
ℬ
(
0
,
ℛ
)
:=
B
(
0
,
ℛ
)
×
B
(
0
,
ℛ
)
⊂
C
[
0
,
R
]
×
C
[
0
,
R
]
.
In what follows we consider a fixed h>h0 that satisfies Proposition 3.2. Then, for any ℛ>0,
(5.1)
d
LS
(
I
-
𝒯
(
⋅
,
1
)
,
ℬ
(
0
,
ℛ
)
,
0
)
=
0
,
where dLS denotes the Leray–Schauder degree and I is the identity in the product space.
By Proposition 4.1, there exists M~>0 such that for any fixed M≥M~, the equation
(
u
,
v
)
=
𝒯
(
(
u
,
v
)
,
λ
)
,
does not have solutions for ((u,v),λ)∈∂ℬ(0,M)×[0,1]. Then, by (5.1), with ℛ=M≥M~, and the invariance of the Leray–Schauder degree under compact homotopies, we obtain that
d
LS
(
I
-
𝒯
(
⋅
,
λ
)
,
ℬ
(
0
,
M
)
,
0
)
=
0
for all
λ
∈
[
0
,
1
]
,
and hence, in particular, that
d
LS
(
I
-
𝒯
(
⋅
,
0
)
,
ℬ
(
0
,
M
)
,
0
)
=
0
.
We next define the operator S:C[0,R]×C[0,R]×[0,1]↦C[0,R]×C[0,R] by
S
(
(
u
,
v
)
,
λ
)
:=
(
∫
r
R
ϕ
-
1
(
λ
s
α
∫
0
s
ξ
χ
f
11
(
|
u
(
ξ
)
|
)
f
12
(
|
v
(
ξ
)
|
d
ξ
)
d
s
,
∫
r
R
ψ
-
1
(
λ
s
α
~
∫
0
s
ξ
χ
~
f
21
(
|
u
(
ξ
)
|
)
f
22
(
|
v
(
ξ
)
|
d
ξ
)
d
s
)
.
Then S is a completely continuous operator that satisfies
𝒯
(
(
u
,
v
)
,
0
)
=
S
(
(
u
,
v
)
,
1
)
,
and hence
d
LS
(
I
-
S
(
⋅
,
1
)
,
ℬ
(
0
,
M
)
,
0
)
=
0
.
Claim.
Assume that (H2) and (H4) hold. Then there exists ρ0>0 such that for any ρ∈[0,ρ0], the equation
(
u
,
v
)
=
S
(
(
u
,
v
)
,
λ
)
does not have a solution in (B(0,ρ)¯∖{0})×[0,1]. Thus, the index i(S(⋅,1),0,0) is defined and we have i(S(⋅,1),0,0)=1. Then
i
(
𝒯
(
⋅
,
0
)
,
0
,
0
)
=
1
.
Proof of the claim. By contradiction, we assume there exist {(uk,vk)}, {λk}, {ρk}, with 0≤λk≤1, ∥uk∥+∥vk∥=ρk→0 as k→∞, such that
(
u
k
,
v
k
)
=
S
(
(
u
k
,
v
k
)
,
λ
k
)
.
This equation is equivalent to the system
(5.2)
{
u
k
(
r
)
=
∫
r
R
ϕ
-
1
(
λ
k
s
α
∫
0
s
ξ
χ
f
11
(
|
u
k
(
ξ
)
|
)
f
12
(
|
v
k
(
ξ
)
|
d
ξ
)
d
s
,
r
∈
[
0
,
R
]
,
v
k
(
r
)
=
∫
r
R
ψ
-
1
(
λ
k
s
α
~
∫
0
s
ξ
χ
~
f
21
(
|
u
k
(
ξ
)
|
)
f
22
(
|
v
k
(
ξ
)
|
d
ξ
)
d
s
,
r
∈
[
0
,
R
]
.
Since
|
u
k
(
ξ
)
|
≤
∥
u
k
∥
≤
ρ
k
,
|
v
k
(
ξ
)
|
≤
∥
u
k
∥
≤
ρ
k
,
and fij, i,j=1,2, is AH at zero with exponent μij, from (1.9), for ρk<s1, we obtain
f
11
(
|
u
k
(
ξ
)
|
)
≤
C
|
u
k
(
ξ
)
|
μ
11
-
ε
≤
C
∥
u
k
∥
μ
11
-
ε
,
f
12
(
|
v
k
(
ξ
)
|
)
≤
C
|
v
k
(
ξ
)
|
μ
12
-
ε
≤
C
∥
v
k
∥
μ
12
-
ε
,
f
21
(
|
u
k
(
ξ
)
|
)
≤
C
|
u
k
(
ξ
)
|
μ
21
-
ε
≤
C
∥
u
k
∥
μ
21
-
ε
,
f
22
(
|
v
k
(
ξ
)
|
)
≤
C
|
v
k
(
ξ
)
|
μ
22
-
ε
≤
C
∥
v
k
∥
μ
22
-
ε
,
where C is a positive constant. Therefore, from (5.2), for all r∈[0,R], we find
u
k
(
r
)
=
∫
r
R
ϕ
-
1
(
λ
k
s
α
∫
0
s
ξ
χ
f
11
(
|
u
k
(
ξ
)
|
)
f
12
(
|
v
k
(
ξ
)
|
)
𝑑
ξ
)
𝑑
s
≤
∫
r
R
ϕ
-
1
(
λ
k
s
α
s
χ
+
1
χ
+
1
C
2
∥
u
k
∥
μ
11
-
ε
∥
v
k
∥
μ
12
-
ε
)
𝑑
s
and
v
k
(
r
)
=
∫
r
R
ψ
-
1
(
λ
k
s
α
~
∫
0
s
ξ
χ
~
f
21
(
|
u
k
(
ξ
)
|
)
f
22
(
|
v
k
(
ξ
)
|
)
𝑑
ξ
)
𝑑
s
≤
∫
r
R
ψ
-
1
(
λ
k
s
α
~
s
χ
~
+
1
χ
~
+
1
C
2
∥
u
k
∥
μ
21
-
ε
∥
v
k
∥
μ
22
-
ε
)
𝑑
s
.
Hence,
(5.3)
ϕ
(
∥
u
k
∥
R
)
≤
C
1
λ
k
R
χ
-
α
+
1
∥
u
k
∥
μ
11
-
ε
∥
v
k
∥
μ
12
-
ε
and
(5.4)
ψ
(
∥
v
k
∥
R
)
≤
C
1
λ
k
R
χ
~
-
α
~
+
1
∥
u
k
∥
μ
21
-
ε
∥
v
k
∥
μ
22
-
ε
.
Then, from (5.3) and (1.10), we obtain
∥
u
k
∥
p
~
-
1
+
ε
≤
C
2
∥
u
k
∥
μ
11
-
ε
∥
v
k
∥
μ
12
-
ε
,
and thus
∥
u
k
∥
≤
C
3
∥
v
k
∥
μ
12
-
ε
p
~
-
1
-
μ
11
+
2
ε
.
Similarly, from (5.4) and (1.10), we find
∥
v
k
∥
≤
C
4
∥
u
k
∥
μ
21
-
ε
q
~
-
1
-
μ
22
+
2
ε
,
where all the constants Ci are positive and independent of k.
Combining these inequalities, we find that
∥
u
k
∥
≤
C
5
∥
u
k
∥
μ
12
-
ε
p
~
-
1
-
μ
11
+
2
ε
μ
21
-
ε
q
~
-
1
-
μ
22
+
2
ε
,
implying that
(5.5)
1
≤
C
5
∥
u
k
∥
Θ
(
ε
)
,
where
Θ
(
ε
)
=
μ
12
-
ε
p
~
-
1
-
μ
11
+
2
ε
μ
21
-
ε
q
~
-
1
-
μ
22
+
2
ε
-
1
.
Now
Θ
(
0
)
=
μ
12
μ
21
(
p
~
-
1
-
μ
11
)
(
q
~
-
1
-
μ
22
)
-
1
>
0
,
and then, by continuity, there exists ε0>0 small such that Θ(ε)>0 for all 0<ε<ε0. By choosing from the beginning such ε small enough, we obtain a contradiction to (5.5) because ∥uk∥≤ρk→0. Thus, the index i(S(⋅,1),0,0) is defined and clearly i(S(⋅,1),0,0)=1=i(𝒯(⋅,0),0,0). Thus, the claim is proved.
Now, with M>ρ0, and by the excision property of the Leray–Schauder degree, we have that there exists a nontrivial solution (u,v)∈ℬ(0,M)∖{ℬ(0,ε)¯} to the equation
(
u
,
v
)
=
𝒯
(
(
u
,
v
)
,
0
)
.
This implies that there exists a nontrivial solution to problem (1.4) such that u(r)>0,v(r)>0 for all r∈[0,R). In this form Theorem 5.1 is proved. ∎
6 Some Examples
In this section we give two simple examples of problems illustrating our results. Here B(0,R) denotes the ball with center at the origin and radius R in ℝn, n>2.
Example 1.
We consider the question of the existence of positive radially symmetric solutions for the following problem, which generalizes the model problem considered in [1]:
(6.1)
{
div
(
(
|
∇
u
|
p
1
-
2
+
|
∇
u
|
p
2
-
2
)
∇
u
)
+
|
x
|
ν
u
(
|
x
|
)
δ
11
v
(
|
x
|
)
δ
12
=
0
in
B
(
0
,
R
)
,
div
(
(
|
∇
u
|
q
1
-
2
+
|
∇
v
|
q
2
-
2
)
∇
v
)
+
|
x
|
ν
v
(
|
x
|
)
δ
21
v
(
|
x
|
)
δ
22
=
0
in
B
(
0
,
R
)
,
u
|
∂
Ω
=
0
,
v
|
∂
Ω
=
0
.
Here p2>p1>1, q2>q1>1, δij>0,i,j=1,2, and ν>0.
The AH function that generate the differential operators in (1.4) have the form
ϕ
(
s
)
=
s
(
|
s
|
p
1
-
2
+
|
s
|
p
2
-
2
)
,
ψ
(
s
)
=
s
(
|
s
|
q
1
-
2
+
|
s
|
q
2
-
2
)
,
and hence in radial coordinates system (6.1) takes the form
(6.2)
{
(
r
n
-
1
ϕ
(
u
′
)
)
′
+
r
χ
u
(
r
)
δ
11
v
(
r
)
δ
12
=
0
,
r
∈
(
0
,
R
)
,
(
r
n
-
1
ψ
(
v
′
)
)
′
+
r
χ
~
u
(
r
)
δ
21
v
(
r
)
δ
22
=
0
,
r
∈
(
0
,
R
)
,
u
′
(
0
)
=
u
(
R
)
=
0
,
v
′
(
0
)
=
v
(
R
)
=
0
,
where χ=n-1+ν and χ~=n-1+ν~. We notice that χ>0 and χ~>0. We assume here that n>max{p2,q2}, p1-δ11-1>0, q1-δ22-1>0, and that
δ
12
δ
21
>
(
p
2
-
δ
11
-
1
)
(
q
2
-
δ
22
-
1
)
.
By direct computation, it can be proved that
lim
s
→
∞
s
ϕ
(
s
)
Φ
(
s
)
=
p
2
and
lim
s
→
0
s
ϕ
(
s
)
Φ
(
s
)
=
p
1
,
lim
s
→
∞
s
ψ
(
s
)
Ψ
(
s
)
=
q
2
and
lim
s
→
0
s
ψ
(
s
)
Ψ
(
s
)
=
q
1
,
and that sϕ(s)Φ(s), sψ(s)Ψ(s) are increasing functions of s. Hence,
inf
s
∈
[
0
,
∞
)
s
ϕ
(
s
)
Φ
(
s
)
=
p
1
,
sup
s
∈
[
0
,
∞
)
s
ϕ
(
s
)
Φ
(
s
)
=
p
2
,
inf
s
∈
[
0
,
∞
)
s
ψ
(
s
)
Ψ
(
s
)
=
q
1
,
sup
s
∈
[
0
,
∞
)
s
ψ
(
s
)
Ψ
(
s
)
=
q
2
.
If
max
{
E
1
-
n
-
p
2
p
2
-
1
,
E
2
-
n
-
q
2
q
2
-
1
}
>
0
,
where
E
1
=
(
χ
+
p
2
-
n
+
1
)
(
q
2
-
δ
22
-
1
)
+
(
χ
~
+
q
2
-
n
+
1
)
δ
12
δ
12
δ
21
-
(
p
2
-
δ
11
-
1
)
(
q
2
-
δ
22
-
1
)
and
E
2
=
(
χ
~
+
q
2
-
n
+
1
)
(
p
2
-
δ
11
-
1
)
+
(
χ
+
p
2
-
n
+
1
)
δ
21
δ
12
δ
21
-
(
p
2
-
δ
11
-
1
)
(
q
2
-
δ
22
-
1
)
,
then it is not difficult to see that all the conditions of Theorem 5.1 are met, and hence from that theorem it follows that problem (6.2) has a solution (u,v) such that u(r)>0, v(r)>0 for all r∈[0,R), which gives the existence of positive radially symmetric solutions for problem (6.1).
Example 2.
We consider the problem
(6.3)
{
div
(
|
x
|
ζ
a
(
|
∇
u
|
)
∇
u
)
+
|
x
|
ν
u
(
|
x
|
)
δ
11
v
(
|
x
|
)
δ
12
=
0
in
B
(
0
,
R
)
,
div
(
|
x
|
ζ
~
b
(
|
∇
v
|
)
∇
v
)
+
|
x
|
ν
~
u
(
|
x
|
)
δ
21
v
(
|
x
|
)
δ
22
=
0
in
B
(
0
,
R
)
,
u
|
∂
B
(
0
,
R
)
=
0
,
v
|
∂
B
(
0
,
R
)
=
0
,
where a,b↦(0,∞)↦[0,∞) are given by a(s)=|s|p-2log(1+|s|),b(s)=|s|q-2log(1+|s|), with p>1, q>1 and p>q. The exponents satisfy δij>0, i,j=1,2, 0<ζ≤ν,0<ζ~≤ν~, and we assume that n+ζ>p and n+ζ~>q. Furthermore, we suppose that p>δ11+1,q>δ22+1 and δ12δ21>(p-δ11)(q-δ22). In radial coordinates, the system takes the form
{
(
r
α
ϕ
(
u
′
)
)
′
+
r
χ
u
(
r
)
δ
11
v
(
r
)
δ
12
=
0
,
r
∈
(
0
,
R
)
(
r
α
~
ψ
(
v
′
)
)
′
+
r
χ
~
u
(
r
)
δ
21
v
(
r
)
δ
22
=
0
,
r
∈
(
0
,
R
)
,
u
′
(
0
)
=
0
=
v
′
(
0
)
,
u
(
R
)
=
0
=
v
(
R
)
,
where the functions ϕ and ψ have the form:
ϕ
(
s
)
=
|
s
|
p
-
2
s
log
(
1
+
|
s
|
)
and
ψ
(
s
)
=
|
s
|
q
-
2
s
log
(
1
+
|
s
|
)
,
and the exponents α,α~,χ,χ~ are given by (1.3). It can be proved that
lim
s
→
∞
s
ϕ
(
s
)
Φ
(
s
)
=
p
and
lim
s
→
0
s
ϕ
(
s
)
Φ
(
s
)
=
p
+
1
,
lim
s
→
∞
s
ψ
(
s
)
Ψ
(
s
)
=
q
and
lim
s
→
0
s
ψ
(
s
)
Ψ
(
s
)
=
q
+
1
,
and that sϕ(s)Φ(s), sψ(s)Ψ(s) are decreasing functions of s. Then if
max
{
E
1
-
n
+
ζ
-
p
p
-
1
,
E
2
-
n
+
ζ
~
-
q
q
-
1
}
>
0
,
where now
E
1
=
(
ν
+
p
-
ζ
)
(
q
-
δ
22
-
1
)
+
(
ν
~
+
q
-
ζ
~
)
δ
12
δ
12
δ
21
-
(
p
-
δ
11
-
1
)
(
q
-
δ
22
-
1
)
and
E
2
=
(
ν
~
+
q
-
ζ
~
)
(
p
-
δ
11
-
1
)
+
(
ν
+
p
-
ζ
)
δ
21
δ
12
δ
21
-
(
p
-
δ
11
-
1
)
(
q
-
δ
22
-
1
)
,
it can be verified that all the conditions of Theorem 5.1 are met, and hence from that theorem it follows that problem (6.3) has a radially symmetric solution (u,v) such that u(|x|)>0, v(|x|)>0 for all x∈B(0,R).
and
Satoshi Tanaka
