1 Introduction
This paper analyzes the quasilinear indefinite Neumann problem
(1.1)
{
-
(
u
′
1
+
(
u
′
)
2
)
′
=
λ
a
(
x
)
f
(
u
)
in
(
0
,
1
)
,
u
′
(
0
)
=
u
′
(
1
)
=
0
,
which is a one-dimensional prototype of
{
-
div
(
∇
u
1
+
|
∇
u
|
2
)
=
g
(
x
,
u
)
in
Ω
,
-
∇
u
⋅
ν
1
+
|
∇
u
|
2
=
σ
on
∂
Ω
,
where Ω is a bounded regular domain of ℝN with outward pointing normal ν, and g:Ω×ℝ→ℝ and σ:∂Ω→ℝ are given functions. Problems involving the mean curvature operator play a pivotal role in the mathematical analysis of a number of geometrical and physical issues, such as prescribed mean curvature problems for cartesian surfaces in the Euclidean space [29, 3, 22, 30, 9, 17, 15, 18, 16], capillarity phenomena for compressible or incompressible fluids [5, 11, 10, 19, 20, 12, 13, 14, 7, 6], and reaction-diffusion processes where the flux features saturation at high regimes [28, 21, 4].
When dealing with problem (1.1), λ∈ℝ is viewed as a parameter and the following assumptions will be considered eventually:
a
∈
L
∞
(
0
,
1
)
satisfies ∫01a(x)𝑑x<0 and there exists a point z∈(0,1) such that a(x)>0 a.e. in (0,z) and a(x)<0 a.e. in (z,1).
There holds
∫
0
z
(
∫
x
z
a
(
t
)
d
t
)
-
1
/
2
d
x
=
+
∞
or
∫
z
1
(
∫
x
z
a
(
t
)
d
t
)
-
1
/
2
d
x
=
+
∞
.
f
∈
𝒞
1
(
ℝ
)
is strictly increasing and, for some constant h>0, satisfies
lim
s
→
0
+
f
(
s
)
s
=
1
and
lim
s
→
+
∞
f
(
s
)
=
h
.
As the function (∫xza(t)𝑑t)-1/2 is continuous and positive in x∈[0,1]∖{z}, the integrals
∫
0
z
(
∫
x
z
a
(
t
)
d
t
)
-
1
/
2
d
x
and
∫
z
1
(
∫
x
z
a
(
t
)
d
t
)
-
1
/
2
d
x
are well-defined and each of them is either finite or equals +∞. Condition (A2) requires some of them, or both, to be divergent.
Condition (F) entails that f(0)=0 and f(s)>0 for all s>0. Moreover, the associated potential,
F
(
s
)
=
∫
0
s
f
(
t
)
𝑑
t
,
is quadratic at 0 and linear at +∞. Some paradigmatic examples of f satisfying (F), for a given h>0, are provided by
(1.2)
f
(
s
)
=
h
s
h
2
+
s
2
and
f
(
s
)
=
h
tanh
(
s
h
)
.
Throughout this paper, we are going to use the next notions of solution.
A couple (λ,u) is said to be a regular solution of (1.1) if u∈W2,1(0,1) and it satisfies the differential equation a.e. in (0,1), as well as the boundary conditions.
A couple (λ,u) is said to be a bounded variation solution, or shortly BV-solution, of (1.1) if u∈BV(0,1) and it satisfies
∫
0
1
D
a
u
D
a
ϕ
1
+
|
D
a
u
|
2
𝑑
x
+
∫
0
1
D
s
u
|
D
s
u
|
D
s
ϕ
=
∫
0
1
λ
a
f
(
u
)
ϕ
𝑑
x
for all ϕ∈BV(0,1) such that |Dsϕ| is absolutely continuous with respect to |Dsu| (cf. [2]).
A couple (λ,u) is said to be a singular solution of (1.1) if it is a bounded variation solution of (1.1) such that u∈BV(0,1)∖W2,1(0,1).
Once the pair (λ,u) solves (1.1) in any of the previous senses, it is said that (λ,u) is a positive solution if, in addition,
λ
≥
0
and
ess
inf
u
>
0
.
As usual, for any function v∈BV(0,1),
D
v
=
D
a
v
d
x
+
D
s
v
stands for the Lebesgue decomposition of the Radon measure Dv and Dsv|Dsv| denotes the density function of the measure Dsv with respect to its total variation |Dsv| (cf. [1]).
By (A1), any regular solution, (λ,u), of (1.1), with λ>0, satisfies u′(x)<0 for every x∈(0,1), u′(0)=u′(1)=0, and, since u is concave on [0,z) and convex on (z,1],
(1.3)
-
u
′
(
z
)
=
∥
u
′
∥
L
∞
(
0
,
1
)
.
Moreover, by [24, Proposition 3.6], any positive singular solution (λ,u) of (1.1), with λ>0, satisfies
u
|
[
0
,
z
)
∈
W
loc
2
,
1
[
0
,
z
)
∩
W
1
,
1
(
0
,
z
)
and is concave,
u
|
(
z
,
1
]
∈
W
loc
2
,
1
(
z
,
1
]
∩
W
1
,
1
(
z
,
1
)
and is convex,
u
′
(
x
)
<
0
for every x∈(0,1)∖{z}, u′(0)=u′(1)=0 and
(1.4)
u
′
(
z
-
)
=
u
′
(
z
+
)
=
-
∞
,
where u′(z-) and u′(z+) are the left and the right Dini derivatives of u at z.
The left and the right limits of u at z will be respectively denoted by u(z-) and u(z+). According to (1.4), throughout this paper we set
-
u
′
(
z
)
1
+
(
u
′
(
z
)
)
2
=
1
.
Any solution (λ,u) of (1.1) with λ≥0 and u(x)≥0 for a.e. x∈(0,1) satisfies either u=0 or essinfu>0. Indeed, if λ=0, then u is necessarily constant (see, e.g., [24, Section 3]), whereas in case λ>0 the conclusion easily follows by combining (1.3) with the uniqueness of solution to the Cauchy problem associated with the differential equation of (1.1). Moreover, it follows from [27, Proposition 1.1] that λ≥0 is necessary for the existence of a bounded variation solution, either regular or singular, satisfying essinfu>0. In particular, λ>0 is necessary for the existence of non-constant positive solutions.
Under the assumption (A1), the linear eigenvalue problem
(1.5)
{
-
φ
′′
=
λ
a
(
x
)
φ
,
x
∈
(
0
,
1
)
,
φ
′
(
0
)
=
φ
′
(
1
)
=
0
,
possesses two principal eigenvalues: λ=0 and λ=λ0>0. By a principal eigenvalue, it is meant a value of λ for which (1.5) admits a positive eigenfunction φ. Moreover, these eigenvalues are algebraically simple and the associated eigenfunctions, 1 and φ0, are positive and separated away from zero; a recent synthesis of the available results on this topic is contained in [23, Chapter 9].
Throughout this paper, we will denote by 𝒮+ the set of the positive bounded variation solutions of (1.1) closed by adding (0,0) and (λ0,0), which are the unique possible bifurcation points to positive solutions from the trivial solution branch (λ,u)=(λ,0), i.e.,
𝒮
+
=
{
(
λ
,
u
)
∣
(
λ
,
u
)
is a positive BV-solution of (1.1)
}
∪
{
(
0
,
0
)
,
(
λ
0
,
0
)
}
⊂
ℝ
×
BV
(
0
,
1
)
.
The set 𝒮+ is endowed with the topology of the strict convergence of ℝ×BV(0,1). Namely, if ((λn,un))n≥1 is a sequence in 𝒮+ and (λ,u)∈𝒮+, it is said that
lim
n
→
+
∞
(
λ
n
,
u
n
)
=
(
λ
,
u
)
whenever
lim
n
→
+
∞
(
|
λ
n
-
λ
|
+
∥
u
n
-
u
∥
L
1
+
|
∫
0
1
|
D
u
n
|
-
∫
0
1
|
D
u
|
|
)
=
0
.
The model (1.1) has been recently analyzed by the authors in [27, 26, 24, 25]. In [27] the existence of bounded variation solutions was investigated by means of the methods of the calculus of variations and in [26] the existence of regular solutions was dealt with by means of phase plane analysis and bifurcation techniques. The main result of [24] established the existence of a component of bounded variation solutions bifurcating from the trivial solution (λ,0) when the associated potential of f, F, is quadratic at 0. Finally, assuming that F is super-quadratic at 0 and superlinear at +∞, the main result of [25] characterized whether the solutions of (1.1) for sufficiently small λ>0 are regular, or singular, in terms of the integrability properties of the weight function a.
Here, our attention focuses on the case where F is quadratic at 0 and linear at +∞. Our main aim is to discuss the fine structure of the set 𝒮+ of positive solutions of (1.1), in particular, establishing the existence of a component bifurcating from (λ,0) at some λ0>0 and from (λ,∞) at some λ∞>0, as well as characterizing the formation of singularities and describing the precise asymptotic profile of the solutions with large norm. Figure 1 provides us with a picture of such profiles. Precisely, the main findings of this paper can be summarized as follows.
Theorem 1.1.
Assume (A1) and (F). Then the set S+ contains at least, two connected components, C0+ and Cλ0+, which satisfy the following conditions: Bifurcation from 0 and from ∞.
𝒞
0
+
=
{
0
}
×
[
0
,
+
∞
)
.
𝒞
λ
0
+
∩
(
ℝ
×
{
0
}
)
=
(
λ
0
,
0
)
.
There exist
λ
*
,
λ
*
∈
(
0
,
+
∞
)
such that
λ
*
<
λ
0
<
λ
*
and
proj
ℝ
(
𝒮
+
∖
𝒞
0
+
)
⊆
[
λ
*
,
λ
*
]
;
thus, in particular, the component
𝒞
λ
0
+
is bounded in the parameter
λ
∈
ℝ
.
proj
L
∞
(
0
,
1
)
𝒞
λ
0
+
is unbounded.
For every sequence in
𝒮
+
∖
𝒞
0
+
, ((λn,un))n≥1, such that
lim
n
→
+
∞
∥
u
n
∥
L
∞
=
+
∞
,
necessarily
lim
n
→
+
∞
λ
n
=
λ
∞
=
(
h
∫
0
z
a
(
x
)
𝑑
x
)
-
1
;
thus,
λ
=
0
and
λ
=
λ
∞
are the unique values of λ for which bifurcation from infinity of positive bounded variation solutions occurs.
Asymptotic profile.
For every sequence in
𝒮
+
∖
𝒞
0
+
, ((λn,un))n≥1, such that
lim
n
→
+
∞
∥
u
n
∥
L
∞
=
+
∞
,
necessarily,
lim
n
→
+
∞
u
n
(
x
)
=
+
∞
and
lim
n
→
+
∞
u
n
′
(
x
)
=
-
∫
0
x
a
(
t
)
𝑑
t
(
∫
0
z
a
(
t
)
𝑑
t
)
2
-
(
∫
0
x
a
(
t
)
𝑑
t
)
2
for all
x
∈
[
0
,
z
)
, while, for every
x
∈
(
z
,
1
]
,
lim
n
→
+
∞
u
n
(
x
)
=
u
∞
(
x
)
,
where
u
∞
∈
W
loc
2
,
∞
(
z
,
1
]
solves the problem
{
-
(
u
′
1
+
(
u
′
)
2
)
′
=
λ
∞
a
(
x
)
f
(
u
)
in
(
z
,
1
)
,
u
′
(
z
+
)
=
-
∞
,
u
′
(
1
)
=
0
,
with
u
∞
(
x
)
>
0
for all
x
∈
(
z
,
1
]
and either
u
∞
(
z
+
)
<
+
∞
if
∫
z
1
(
∫
x
z
a
(
t
)
d
t
)
-
1
/
2
d
x
<
+
∞
or
u
∞
(
z
+
)
=
+
∞
if
∫
z
1
(
∫
x
z
a
(
t
)
𝑑
t
)
-
1
/
2
𝑑
x
=
+
∞
.
Regularity versus development of singularities.
There exists a neighborhood,
𝒰
, of
(
λ
0
,
0
)
in
ℝ
×
L
∞
(
0
,
1
)
such that
𝒞
λ
0
+
∩
𝒰
consists of regular solutions.
The set
𝒮
+
and, in particular, the component
𝒞
λ
0
+
consist only of regular solutions if and only if
(A2)
holds.
If
(A2)
fails, then all solutions
(
λ
,
u
)
in
𝒮
+
and, in particular, in
𝒞
λ
0
+
, with
λ
>
0
and sufficiently large
∥
u
∥
L
∞
, are singular.
In particular, under conditions (A1) and (F), the failure of (A2) characterizes the development of singularities by the positive solutions of (1.1) along the component 𝒞λ0+. A simple example for which (A2) fails occurs when a is assumed to be a positive constant, A>0, in [z-η,z), and a negative constant, -B<0, in (z,z+η], for some small η>0. In such case, owing to Theorem 1.1, all solutions in 𝒞λ0+ develop a singularity when they become sufficiently large. On the contrary, all solutions are regular if, for instance, the weight a is differentiable at the nodal point z.
According to [26, Theorem 1.4], if in addition to (F) we assume that f∈𝒞2(ℝ), then, thanks to the local bifurcation theorem of [8], in a neighborhood of (λ0,0) the component 𝒞λ0+ consists of a 𝒞1-curve, (λ(s),s(φ0+v(s))), for s close to 0, such that v(0)=0, λ(0)=λ0 and
λ
′
(
0
)
=
-
λ
0
f
′′
(
0
)
2
∫
0
1
a
(
x
)
φ
0
3
(
x
)
𝑑
x
∫
0
1
a
(
x
)
φ
0
2
(
x
)
𝑑
x
,
with
∫
0
1
a
(
x
)
φ
0
2
(
x
)
𝑑
x
>
0
and
∫
0
1
a
(
x
)
φ
0
3
(
x
)
𝑑
x
>
0
.
Therefore, 𝒞λ0+ bifurcates transcritically from (λ0,0) if f′′(0)≠0. Precisely, this bifurcation is supercritical if f′′(0)<0, while it is subcritical if f′′(0)>0. Should it occurs that f′′(0)=0 and f∈𝒞3(ℝ), like for the special choices made in (1.2), then, by [26, equation (4.2)], λ′′(0)<0. Therefore, the bifurcation of 𝒞λ0+ in this case also is subcritical.
By modulating the constant h in (F), the value of λ where 𝒞λ0+ bifurcates from infinity, λ∞, can be at any side of λ0; hence, any of the components 𝒞λ0+ plotted in Figure 2 are admissible. Indeed, the bifurcation value from infinity
λ
∞
=
λ
∞
(
h
)
=
1
h
∫
0
z
a
(
x
)
𝑑
x
satisfies
lim
h
→
+
∞
λ
∞
(
h
)
=
0
,
lim
h
→
0
+
λ
∞
(
h
)
=
+
∞
.
This clearly produces the existence of multiple solutions for λ>λ0, if λ∞≤λ0 and the bifurcation from λ0 is supercritical, as it is displayed by the first two plots of Figure 2, as well as for λ<λ0, if λ∞≥λ0 and the bifurcation from λ0 is subcritical. Figure 2 illustrates the special case when 𝒞λ0+ bifurcates supercritically from λ0. Similar bifurcation diagrams occur if the bifurcation from zero is subcritical.
The structure of this paper is the following. Section 2 contains the main non-existence result, Section 3 analyzes the bifurcation of the large solutions from infinity, Section 4 ascertains the exact limiting profiles of the large positive solutions of (1.1) and, finally, Section 5 delivers the proof of Theorem 1.1.
2 Non-existence of Positive Solutions for Small and Large λ>0
The main result of this section can be stated as follows.
Theorem 2.1.
Assume (A1) and (F). Then problem (1.1) has no positive solutions both for sufficiently small and for sufficiently large λ>0.
Proof.
In order to prove that (1.1) has no positive solutions for all small λ>0, we will argue by contradiction, assuming that (1.1) admits a sequence of positive solutions, ((λn,un))n≥1, such that λn>0 for all n≥1 and
(2.1)
lim
n
→
+
∞
λ
n
=
0
.
Integrating the differential equation of (1.1) in (0,z), we get, for all n≥1,
(2.2)
-
u
n
′
(
z
)
1
+
(
u
n
′
(
z
)
)
2
=
λ
n
∫
0
z
a
(
x
)
f
(
u
n
(
x
)
)
𝑑
x
.
Thus, since a and f are bounded, letting n→+∞ in (2.2), it follows from (2.1) that
lim
n
→
+
∞
-
u
n
′
(
z
)
1
+
(
u
n
′
(
z
)
)
2
=
0
.
Hence, by (1.3), we find that
(2.3)
0
=
lim
n
→
+
∞
u
n
′
(
z
)
=
lim
n
→
+
∞
∥
u
′
∥
L
∞
(
0
,
1
)
and, consequently, (λn,un) is a regular solution of (1.1) for sufficiently large n. Further, possibly passing to a subsequence, relabeled by n, there exists L∈[0,+∞] such that
lim
n
→
+
∞
u
n
(
0
)
=
L
.
According to (2.3), from the identity
u
n
(
x
)
=
u
n
(
0
)
+
∫
0
x
u
n
′
(
t
)
𝑑
t
for
x
∈
[
0
,
1
]
,
it is easily seen that
lim
n
→
+
∞
u
n
=
L
uniformly on
[
0
,
1
]
.
Next, we will distinguish between three different cases according to the value of L.
Suppose that L=0. Then setting
v
n
=
u
n
∥
u
n
∥
L
∞
(
0
,
1
)
and
A
n
=
a
f
(
u
n
)
u
n
[
1
+
(
u
n
′
)
2
]
3
/
2
,
we find that
(2.4)
{
-
v
n
′′
=
λ
n
A
n
(
x
)
v
n
in
(
0
,
1
)
,
v
n
′
(
0
)
=
v
n
′
(
0
)
=
0
.
Since
∥
v
n
∥
L
∞
(
0
,
1
)
=
1
for all
n
≥
1
,
and due to (A1), (F) and (2.3), we have that
lim
n
→
+
∞
∥
A
n
-
a
∥
L
∞
(
0
,
1
)
=
0
,
letting n→+∞ in (2.4), it follows from (2.1) that
lim
n
→
+
∞
∥
v
n
′′
∥
L
∞
(
0
,
1
)
=
0
.
Thus, since
v
n
′
(
x
)
=
v
n
′
(
0
)
+
∫
0
x
v
n
′′
(
t
)
𝑑
t
=
∫
0
x
v
n
′′
(
t
)
𝑑
t
for
x
∈
[
0
,
1
]
,
we also have that
lim
n
→
+
∞
∥
v
n
′
∥
L
∞
(
0
,
1
)
=
0
.
Therefore, possibly passing to a further subsequence, still labeled by n, we find that
lim
n
→
+
∞
v
n
(
x
)
=
1
uniformly in
x
∈
[
0
,
1
]
.
Finally, dividing (2.4) by λn and integrating on [0,1] yields
0
=
∫
0
1
A
n
(
x
)
v
n
(
x
)
𝑑
x
for all
n
≥
1
.
Thus, we conclude that
0
=
lim
n
→
+
∞
∫
0
1
A
n
(
x
)
v
n
(
x
)
𝑑
x
=
∫
0
1
a
(
x
)
𝑑
x
<
0
,
which is a contradiction.
Now, suppose that L∈(0,+∞). Then, integrating (1.1) on [0,1], we see that
0
=
-
λ
n
-
1
∫
0
1
(
u
n
′
(
x
)
1
+
(
u
n
′
(
x
)
)
2
)
′
𝑑
x
=
∫
0
1
a
(
x
)
f
(
u
n
(
x
)
)
𝑑
x
for all n≥1. Thus, letting n→+∞, we find that
0
=
f
(
L
)
∫
0
1
a
(
x
)
𝑑
x
<
0
,
which is a contradiction, because f(L)>0 and ∫01a(x)𝑑x<0.
Finally, suppose that L=+∞. Then the same argument yields
0
=
h
∫
0
1
a
(
x
)
𝑑
x
<
0
,
which is as well impossible. This concludes the proof of the non-existence of positive solutions for sufficiently small λ>0.
To prove the non-existence for sufficiently large λ>0, we will also argue by contradiction. So, assume that there is a sequence of positive solutions of (1.1), ((λn,un))n≥1, with
(2.5)
lim
n
→
+
∞
λ
n
=
+
∞
.
Then it follows from (2.2) and (2.5) that
lim
n
→
+
∞
∫
0
z
a
(
x
)
f
(
u
n
(
x
)
)
𝑑
x
=
lim
n
→
+
∞
(
1
λ
n
-
u
n
′
(
z
)
1
+
(
u
n
′
(
z
)
)
2
)
=
0
,
and hence, along some subsequence, relabeled by n, it becomes apparent that
lim
n
→
+
∞
(
a
(
x
)
f
(
u
n
(
x
)
)
)
=
0
for a.e.
x
∈
[
0
,
z
]
.
Thus, owing to (A1) and (F), we deduce that
lim
n
→
+
∞
u
n
(
x
)
=
0
for a.e.
x
∈
[
0
,
z
]
.
Therefore, since un is decreasing,
(2.6)
lim
n
→
+
∞
u
n
(
x
)
=
0
for all
x
∈
(
0
,
1
]
.
Moreover, possibly passing to a further subsequence, there exists L∈[0,+∞] such that
(2.7)
lim
n
→
+
∞
u
n
(
0
)
=
L
.
Subsequently, we will distinguish between two different cases. First, suppose that L∈(0,+∞]. Then the concavity of un on [0,z) implies that, for any given y∈(0,z),
u
n
′
(
y
)
≤
u
n
(
y
)
-
u
n
(
0
)
y
<
0
,
and hence (2.6) and (2.7) provide us with
(2.8)
lim
n
→
+
∞
u
n
′
(
y
)
≤
lim
n
→
+
∞
u
n
(
y
)
-
u
n
(
0
)
y
=
{
-
L
y
if
L
∈
(
0
,
+
∞
)
,
-
∞
if
L
=
+
∞
.
By the concavity, we also have that
(2.9)
u
n
(
x
)
≤
u
n
(
y
)
+
u
n
′
(
y
)
(
x
-
y
)
for all
x
∈
(
y
,
z
)
.
Note that the right-hand side of (2.9) vanishes at
x
n
=
y
-
u
n
(
y
)
u
n
′
(
y
)
.
Moreover, thanks to (2.6) and (2.8), it becomes apparent that
lim
n
→
+
∞
x
n
=
y
<
z
.
Thus, for sufficiently large n, we find that xn<z. This forces un to vanish in (0,z), which is impossible. The proof is completed in this case.
Lastly, suppose that L=0. Then, since un(0)=∥u∥L∞(0,1),
lim
n
→
+
∞
u
n
(
x
)
=
0
uniformly in
x
∈
[
0
,
1
]
.
Thus, by (F), we can infer that, for sufficiently large n,
f
(
u
n
(
x
)
)
≥
1
2
u
n
(
x
)
for all
x
∈
[
0
,
1
]
.
Hence, it follows from the differential equation of (1.1) that
(2.10)
-
u
n
′′
(
x
)
=
λ
n
a
(
x
)
f
(
u
n
(
x
)
)
(
1
+
(
u
n
′
(
x
)
)
2
)
3
/
2
≥
1
2
λ
n
a
(
x
)
u
n
(
x
)
for a.e.
x
∈
[
0
,
z
]
.
Subsequently, we denote by μ1>0 the unique principal eigenvalue of the weighted eigenvalue problem
(2.11)
{
-
φ
′′
=
μ
a
(
x
)
φ
in
(
0
,
z
)
,
φ
(
0
)
=
φ
(
z
)
=
0
,
and consider any positive eigenfunction, φ1>0, associated to μ1. We have that μ1>0, because a(x)>0 for all x∈(0,z) (cf. [23, Chapter 9]), and, moreover, φ1(x)>0 for all x∈(0,z), φ1′(0)>0 and φ1′(z)<0. Thus, multiplying (2.10) by φ1 and integrating by parts twice in (0,z) yields
1
2
λ
n
∫
0
z
a
(
x
)
u
n
(
x
)
φ
1
(
x
)
𝑑
x
≤
-
∫
0
z
u
n
′′
(
x
)
φ
1
(
x
)
𝑑
x
=
u
n
′
(
0
)
φ
1
(
0
)
-
u
n
′
(
z
)
φ
1
(
z
)
+
∫
0
z
u
n
′
(
x
)
φ
1
′
(
x
)
𝑑
x
=
∫
0
z
u
n
′
(
x
)
φ
1
′
(
x
)
𝑑
x
=
u
n
(
z
)
φ
1
′
(
z
)
-
u
n
(
0
)
φ
1
′
(
0
)
-
∫
0
z
u
n
(
x
)
φ
1
′′
(
x
)
𝑑
x
<
-
∫
0
z
φ
1
′′
(
x
)
u
n
(
x
)
𝑑
x
.
Therefore, according to (2.11), we see that
(2.12)
1
2
λ
n
∫
0
z
a
(
x
)
u
n
(
x
)
φ
1
(
x
)
𝑑
x
<
μ
1
∫
0
z
a
(
x
)
φ
1
(
x
)
u
n
(
x
)
𝑑
x
for all n≥1. Since
∫
0
z
a
(
x
)
φ
1
(
x
)
u
n
(
x
)
𝑑
x
>
0
,
it follows from (2.12) that λn<2μ1, which contradicts (2.5) and ends the proof. ∎
According to Theorem 2.1, the values of the parameter λ, defined by
λ
*
=
inf
{
λ
>
0
∣
(
λ
,
u
)
∈
𝒮
+
}
,
λ
*
=
sup
{
λ
>
0
∣
(
λ
,
u
)
∈
𝒮
+
}
,
satisfy
0
<
λ
*
≤
λ
*
<
+
∞
.
Therefore, by Theorem 2.1,
(2.13)
λ
∈
[
λ
*
,
λ
*
]
⊂
(
0
,
+
∞
)
if
(
λ
,
u
)
∈
𝒮
+
∖
(
{
0
}
×
[
0
,
+
∞
)
)
.
3 Bifurcation of Positive Solutions from Zero and from Infinity
According to [24, Theorem 5.13], the component of 𝒮+ bifurcating from (λ,u)=(λ0,0), denoted in this paper by 𝒞λ0+, is unbounded in ℝ×L∞(0,1). Moreover, Theorem 2.1 implies that λ∈[λ*,λ*] if (λ,u)∈𝒞λ0+. Therefore, we conclude that projL∞(0,1)𝒞λ0+ and hence projL∞(0,1)(𝒮+∖({0}×[0,+∞))) must be unbounded. Accordingly, from any unbounded sequence of positive solutions in 𝒮+∖({0}×[0,+∞)), one can extract a subsequence ((λn,un))n≥1 and find a number λ∞ such that
(3.1)
lim
n
→
+
∞
λ
n
=
λ
∞
∈
[
λ
*
,
λ
*
]
and
lim
n
→
∞
∥
u
n
∥
L
∞
(
0
,
1
)
=
+
∞
.
In particular, since un(0)=∥un∥L∞(0,1), we have that
(3.2)
lim
n
→
+
∞
u
n
(
0
)
=
+
∞
.
Throughout this section we will assume that (3.2) holds and the precise value of λ∞ will be ascertained by showing that
λ
∞
=
(
h
∫
0
z
a
(
x
)
𝑑
x
)
-
1
.
The independence of λ∞ from the particular sequence ((λn,un))n≥1, satisfying (3.1), proves that λ∞ is the only bifurcation point from infinity of 𝒮+∖({0}×[0,+∞)). In particular, 𝒞λ0+ is a connected component of 𝒮+ bifurcating from zero and from infinity.
In the sequel ((λn,un))n≥1 stands for any sequence satisfying (3.1), for some number λ∞. We begin with the next result of technical nature.
Lemma 3.1.
Assume (A1), (F) and (3.2). Then the following hold:
(3.3)
lim sup
n
→
+
∞
u
n
(
1
)
<
+
∞
𝑎𝑛𝑑
lim
n
→
+
∞
u
n
′
(
z
)
=
-
∞
.
Proof.
To prove the first conclusion, we will argue by contradiction assuming that, along some subsequence, also labeled by n, one has that
lim
n
→
+
∞
u
n
(
1
)
=
+
∞
and hence
lim
n
→
+
∞
u
n
(
x
)
=
+
∞
uniformly in
x
∈
[
0
,
1
]
.
Thus, taking ϕ=1 as the test function in the definition of bounded variation solution and letting n→+∞, assumption (F) yields
h
∫
0
z
a
(
x
)
d
x
=
lim
n
→
+
∞
∫
0
z
a
(
x
)
f
(
u
n
(
x
)
)
)
d
x
=
-
lim
n
→
+
∞
∫
z
1
a
(
x
)
f
(
u
n
(
x
)
)
d
x
=
-
h
∫
1
z
a
(
x
)
d
x
and hence ∫01a(x)𝑑x=0, contradicting (A1). Therefore, the first assertion holds. The second assertion, which requires to be proven only for regular solutions, follows readily from (3.2) and the first assertion of (3.3) by taking into account that
-
u
n
′
(
z
)
=
∥
u
n
′
∥
L
∞
(
0
,
1
)
.
This ends the proof. ∎
The next result provides us with uniform a priori bounds for the first and second derivatives of the solutions un on any interval of the form [0,z-η], η>0.
Proposition 3.2.
Assume (A1), (F) and (3.2). Then, for every η∈(0,z), there exists a constant C>0 such that
(3.4)
max
{
∥
u
n
′
∥
L
∞
(
0
,
z
-
η
)
,
∥
u
n
′′
∥
L
∞
(
0
,
z
-
η
)
}
≤
C
.
Proof.
Fix η∈(0,z) and, arguing by contradiction, suppose that there is a subsequence of ((λn,un))n≥1, still labeled by n, such that
lim
n
→
+
∞
u
n
′
(
z
-
η
)
=
-
∞
.
The monotonicity of un′ on [0,z) implies that
lim
n
→
+
∞
u
n
′
(
x
)
=
-
∞
for all
x
∈
[
z
-
η
,
z
]
.
Thus, integrating the differential equation on (z-η,z) and letting n→+∞ yields
0
=
lim
n
→
+
∞
(
-
u
n
′
(
z
)
1
+
(
u
n
′
(
z
)
)
2
+
u
n
′
(
z
-
η
)
1
+
(
u
n
′
(
z
-
η
)
)
2
)
=
lim
n
→
+
∞
(
λ
n
∫
z
-
η
z
a
(
x
)
f
(
u
n
(
x
)
)
𝑑
x
)
.
Hence, thanks to (2.13), we infer that
lim
n
→
+
∞
∫
z
-
η
z
a
(
x
)
f
(
u
n
(
x
)
)
𝑑
x
=
0
.
Since a(x)>0 for a.e. x∈(0,z), this implies that, for a subsequence relabeled by n,
lim
n
→
+
∞
f
(
u
n
(
x
)
)
=
0
for a.e.
x
∈
(
z
-
η
,
z
)
.
Then, using (F) and taking into account that un is decreasing, we find that
lim
n
→
+
∞
u
n
(
x
)
=
0
for all
x
∈
(
z
-
η
,
1
]
.
Fix y∈(z-η,z). Then, since
lim
n
→
+
∞
u
n
(
y
)
=
0
and
lim
n
→
+
∞
u
n
′
(
y
)
=
-
∞
,
the concavity of un on [0,z) entails that
lim
n
→
+
∞
u
n
(
z
-
)
≤
lim
n
→
+
∞
(
u
n
(
y
)
+
u
n
′
(
y
)
(
z
-
y
)
)
=
-
∞
,
which is impossible, because un(x)≥0 for all x∈[0,1]. Therefore, there exists a constant C>0 such that
∥
u
n
′
∥
L
∞
(
0
,
z
-
η
)
≤
C
for all n≥1. By (F), the estimate for the second derivatives follows readily by rewriting the differential equation in the form
-
u
n
′′
=
λ
n
a
f
(
u
n
)
(
1
+
(
u
n
′
)
2
)
3
/
2
in
[
0
,
z
)
,
and taking into account that λn∈[λ*,λ*] for all n≥1, because of Theorem 2.1. This ends the proof. ∎
The next results provides us with the pointwise behavior of the solutions un, as n→+∞, on the interval [0,z).
Theorem 3.3.
Assume (A1), (F) and (3.2). Then, for every η∈(0,z), one has
(3.5)
lim
n
→
+
∞
u
n
(
x
)
=
+
∞
uniformly in
x
∈
[
0
,
z
-
η
]
,
and, in particular,
(3.6)
lim
n
→
+
∞
u
n
(
x
)
=
+
∞
for all
x
∈
[
0
,
z
)
.
Proof.
Taking into account that, for every n≥1 and x∈[0,z),
u
n
(
x
)
=
u
n
(
0
)
+
∫
0
x
u
n
′
(
t
)
𝑑
t
,
it is easily seen that (3.5) follows from (3.2) and (3.4). As (3.4) holds for arbitrarily small η>0, (3.6) holds too. This ends the proof. ∎
The next result establishes the exact value of λ∞ in (3.1).
Theorem 3.4.
Assume (A1), (F) and (3.2). Then one has
(3.7)
lim
n
→
+
∞
λ
n
=
λ
∞
=
(
h
∫
0
z
a
(
x
)
𝑑
x
)
-
1
.
Proof.
By (2.2), we have that
λ
n
=
-
u
n
′
(
z
)
1
+
(
u
n
′
(
z
)
)
2
(
∫
0
z
a
(
x
)
f
(
u
n
(
x
)
)
𝑑
x
)
-
1
for all n≥1. Thus, letting n→+∞ in the previous identity, (3.7) follows from (3.3) and (3.6) through the dominated convergence theorem, because f is bounded. ∎
4 Sharp Limiting Behavior of the Large Positive Solutions
Thanks to the analysis done in Section 3, we already know that for any sequence ((λn,un))n≥1 in 𝒮+∖({0}×[0,+∞)) satisfying (3.2), condition (3.7) holds, and, moreover,
lim
n
→
+
∞
u
n
(
x
)
=
+
∞
for all
x
∈
[
0
,
z
)
.
This section aims to establish the precise asymptotic behavior of the solutions un, for large n.
The next result provides us with the exact profile of the derivatives, un′, of the solutions on the interval [0,z).
Theorem 4.1.
Assume (A1), (F) and (3.2). Then, for every η∈(0,z), one has
lim
n
→
+
∞
u
n
′
(
x
)
=
-
∫
0
x
a
(
t
)
𝑑
t
(
∫
0
z
a
(
t
)
𝑑
t
)
2
-
(
∫
0
x
a
(
t
)
𝑑
t
)
2
uniformly in
x
∈
[
0
,
z
-
η
]
,
and, in particular,
lim
n
→
+
∞
u
n
′
(
x
)
=
-
∫
0
x
a
(
t
)
𝑑
t
(
∫
0
z
a
(
t
)
𝑑
t
)
2
-
(
∫
0
x
a
(
t
)
𝑑
t
)
2
for all
x
∈
[
0
,
z
)
.
Proof.
Let ψ:ℝ→(-1,1) be the map defined by
(4.1)
ψ
(
s
)
=
s
1
+
s
2
.
The function ψ is invertible with inverse ψ-1:(-1,1)→ℝ defined by
ψ
-
1
(
t
)
=
t
1
-
t
2
.
Fix η∈(0,z) and pick any x∈[0,z-η]. Then integrating the differential equation on (0,x) yields, for all n≥1,
ψ
(
-
u
n
′
(
x
)
)
=
λ
n
∫
0
x
a
(
t
)
f
(
u
n
(
t
)
)
𝑑
t
or, equivalently,
u
n
′
(
x
)
=
ψ
-
1
(
-
λ
n
∫
0
x
a
(
t
)
f
(
u
n
(
t
)
)
𝑑
t
)
.
Therefore, letting n→+∞ in this identity and taking into account Theorems 3.3 and 3.4, the dominated convergence theorem and the continuity of ψ-1 imply that, for every x∈[0,z-η],
lim
n
→
+
∞
u
n
′
(
x
)
=
ψ
-
1
(
-
λ
∞
h
∫
0
x
a
(
t
)
𝑑
t
)
=
-
∫
0
x
a
(
t
)
𝑑
t
(
∫
0
z
a
(
t
)
𝑑
t
)
2
-
(
∫
0
x
a
(
t
)
𝑑
t
)
2
.
The uniform convergence on [0,z-η] is a direct consequence of Proposition 3.2 and the Ascoli–Arzelà theorem. The proof is complete. ∎
The next result complements Proposition 3.2 on the interval [z+η,1].
Proposition 4.2.
Assume (A1), (F) and (3.2). Then, for every η∈(0,1-z), there exists a constant C>0 such that
max
{
∥
u
n
′
∥
L
∞
(
z
+
η
,
1
)
,
∥
u
n
′′
∥
L
∞
(
z
+
η
,
1
)
}
≤
C
.
Proof.
The proof basically follows the same patterns as the one of Proposition 3.2. Fix η∈(0,1-z) and, arguing by contradiction, suppose that there is some subsequence of ((λn,un))n≥1, relabeled by n, such that
lim
n
→
+
∞
u
n
′
(
z
+
η
)
=
-
∞
.
Then, by the monotonicity of un on [z,1], this actually implies that
lim
n
→
+
∞
u
n
′
(
x
)
=
-
∞
for all
x
∈
[
z
,
z
+
η
]
.
Thus, integrating the differential equation in (1.1) on (z,z+η) and letting n→+∞, we find that
lim
n
→
+
∞
(
λ
n
∫
z
z
+
η
a
(
x
)
f
(
u
n
(
x
)
)
𝑑
x
)
=
0
.
Consequently, since, by (2.13), we already know that λn∈[λ*,λ*], it follows that
lim
n
→
+
∞
∫
z
z
+
η
a
(
x
)
f
(
u
n
(
x
)
)
𝑑
x
=
0
.
Hence, by (A1), along some subsequence, still labeled by n, we find that
lim
n
→
+
∞
f
(
u
n
(
x
)
)
=
0
for a.e.
x
∈
[
z
,
z
+
η
]
.
Therefore, by the monotonicity of f(un), this implies that
lim
n
→
+
∞
f
(
u
n
(
x
)
)
=
0
for all
x
∈
(
z
,
1
]
.
Lastly, using ϕ=1 as a test function in the definition of bounded variation solution, it follows from Theorem 3.3 and the dominated convergence theorem that
0
=
-
lim
n
→
+
∞
∫
z
1
a
(
x
)
f
(
u
n
(
x
)
)
𝑑
x
=
lim
n
→
+
∞
∫
0
z
a
(
x
)
f
(
u
n
(
x
)
)
𝑑
x
=
h
∫
0
z
a
(
x
)
𝑑
x
>
0
,
which is impossible. This contradiction provides us with the uniform a priori bounds for un′ on [z+η,1]. The uniform bounds on un′′ can be obtained as in the proof of Proposition 3.2; therefore, the technical details are omitted here. ∎
The next result complements Theorem 3.3 by providing us with the limiting behavior of the solutions un, as n→+∞, on the interval (z,1].
Proposition 4.3.
Assume (A1), (F) and (3.2). Then there exists a subsequence, ((λnk,unk))k≥1 of ((λn,un))n≥1 for which the pointwise limit
u
∞
(
x
)
=
lim
k
→
+
∞
u
n
k
(
x
)
for all
x
∈
(
z
,
1
]
is well defined, satisfies u∞∈Wloc2,∞(z,1] and solves the problem
(4.2)
{
-
(
u
′
1
+
(
u
′
)
2
)
′
=
λ
∞
a
(
x
)
f
(
u
)
in
(
z
,
1
)
,
u
′
(
z
)
=
-
∞
,
u
′
(
1
)
=
0
.
Moreover, the following hold:
u
∞
(
x
)
>
0
for all
x
∈
(
z
,
1
]
,
lim
k
→
+
∞
u
n
k
=
u
∞
in
W
loc
2
,
∞
(
z
,
1
]
,
and either
u
∞
(
z
+
)
<
+
∞
if
∫
z
1
(
∫
x
z
a
(
t
)
d
t
)
-
1
/
2
d
x
<
+
∞
or
u
∞
(
z
+
)
=
+
∞
if
∫
z
1
(
∫
x
z
a
(
t
)
d
t
)
-
1
/
2
d
x
=
+
∞
.
Proof.
Fix any η∈(0,1-z). By Proposition 4.2, there exists a subsequence, ((λnk,unk))k≥1, of ((λn,un))n≥1 and a function u∞,η∈𝒞1[z+η,1] such that
(4.3)
lim
k
→
+
∞
u
n
k
=
u
∞
,
η
in
𝒞
1
[
z
+
η
,
1
]
.
Thus, letting n→+∞ in the differential equations
-
u
n
k
′′
(
x
)
=
λ
n
k
a
(
x
)
f
(
u
n
k
(
x
)
)
(
1
+
(
u
n
k
′
(
x
)
)
2
)
3
/
2
for a.e. in
x
∈
(
z
+
η
,
1
)
,
it becomes apparent, by Theorem 3.4, that actually u∞,η∈W2,∞(z+η,1) and it satisfies
-
u
∞
,
η
′′
(
x
)
=
λ
∞
a
(
x
)
f
(
u
∞
,
η
(
x
)
)
(
1
+
(
u
∞
,
η
′
(
x
)
)
2
)
3
/
2
for a.e. in
x
∈
(
z
+
η
,
1
)
.
Now, consider any sequence (ηj)j≥1 such that ηj∈(0,1-z), for all j≥1, with
lim
j
→
+
∞
η
j
=
0
,
and consider, for every j≥1, the associated u∞,ηj. By a diagonal argument, we can extract a further subsequence of ((λnk,unk))k≥1, still labeled by k, and a function u∞∈Wloc2,∞(z,1] such that
lim
k
→
+
∞
u
n
k
=
u
∞
in
W
loc
2
,
∞
(
z
,
1
]
and
-
u
∞
′′
(
x
)
=
λ
∞
a
(
x
)
f
(
u
∞
(
x
)
)
(
1
+
(
u
∞
′
(
x
)
)
2
)
3
/
2
for a.e. in
x
∈
(
z
,
1
)
,
or, in other words,
-
(
u
∞
′
(
x
)
1
+
(
u
∞
′
(
x
)
)
2
)
′
=
λ
∞
a
(
x
)
f
(
u
∞
(
x
)
)
for a.e.
x
∈
(
z
,
1
)
.
By the construction, it follows easily from (4.3) that
u
∞
(
x
)
≥
0
for all
x
∈
(
z
,
1
]
and
u
∞
′
(
1
)
=
0
.
Thus, u∞ is convex and decreasing in (z,1]. Moreover, since f(0)=0, the uniqueness of the solution of the associated Cauchy problem entails that u∞(1)>0 and hence
u
∞
(
x
)
>
0
for all
x
∈
(
z
,
1
]
.
Let us show that
u
∞
′
(
z
+
)
=
-
∞
if
u
∞
(
z
+
)
<
+
∞
.
Indeed, if u∞′(z+)∈ℝ, then, by continuous dependence of the solutions of the associated Cauchy problems on the initial conditions and on the parameters,
lim
k
→
+
∞
u
n
k
(
z
)
=
u
∞
′
(
z
)
∈
ℝ
,
which contradicts Lemma 3.1.
Moreover, by the convexity of u∞, we also must have that
u
∞
′
(
z
+
)
=
-
∞
if
u
∞
(
z
+
)
=
+
∞
.
Consequently, in any case we can conclude that
(4.4)
u
∞
′
(
z
+
)
=
-
∞
.
To complete the proof, it remains to characterize the finiteness of u∞(z+) in terms of the integrability properties of the weight function a. Integrating (4.2) in (z,x), with x∈(z,1), and using (4.4) yields
0
<
-
u
∞
′
(
x
)
1
+
(
u
∞
′
(
x
)
)
2
=
1
+
λ
∞
∫
z
x
a
(
t
)
f
(
u
∞
(
t
)
)
𝑑
t
<
1
and hence
-
u
∞
′
(
x
)
=
1
+
λ
∞
∫
z
x
a
(
t
)
f
(
u
∞
(
t
)
)
𝑑
t
2
+
λ
∞
∫
z
x
a
(
t
)
f
(
u
∞
(
t
)
)
𝑑
t
⋅
1
λ
∞
∫
x
z
a
(
t
)
f
(
u
∞
(
t
)
)
𝑑
t
.
Therefore, for sufficiently small η>0, the following estimates hold:
(4.5)
1
2
1
λ
∞
h
(
∫
x
z
a
(
t
)
𝑑
t
)
-
1
/
2
≤
-
u
∞
′
(
x
)
≤
1
λ
∞
f
(
u
∞
(
1
)
)
(
∫
x
z
a
(
t
)
𝑑
t
)
-
1
/
2
for every x∈(z,z+η). Pick x∈(z,z+η). Then integrating (4.5) on (x,z+η) yields
1
2
1
λ
∞
h
∫
x
z
+
η
(
∫
t
z
a
(
s
)
d
s
)
-
1
/
2
d
t
≤
u
∞
(
x
)
-
u
∞
(
z
+
η
)
≤
1
λ
∞
f
(
u
∞
(
1
)
)
∫
x
z
+
η
(
∫
t
z
a
(
s
)
d
s
)
-
1
/
2
d
t
.
Suppose
∫
z
1
(
∫
x
z
a
(
t
)
d
t
)
-
1
/
2
d
x
<
+
∞
.
Then letting x→z+ in the previous inequalities provides us with
u
∞
(
z
+
)
≤
u
∞
(
z
+
η
)
+
1
λ
∞
f
(
u
∞
(
1
)
)
∫
z
z
+
η
(
∫
t
z
a
(
s
)
d
s
)
-
1
/
2
d
t
<
+
∞
.
Similarly, in case
∫
z
1
(
∫
x
z
a
(
t
)
d
t
)
-
1
/
2
d
x
=
+
∞
,
we find that
u
∞
(
z
+
)
≥
u
∞
(
z
+
η
)
+
1
2
1
λ
∞
h
∫
z
z
+
η
(
∫
t
z
a
(
s
)
d
s
)
-
1
/
2
d
t
=
+
∞
.
The proof is complete. ∎
The next result describes the ultimate behavior of the solutions un at the endpoints of the interval [z,1].
Proposition 4.4.
Assume (A1), (F) and (3.2). Then one has
(4.6)
lim inf
n
→
+
∞
u
n
(
1
)
>
0
and
(4.7)
lim sup
n
→
+
∞
u
n
(
z
+
)
<
+
∞
if
∫
z
1
(
∫
x
z
a
(
t
)
d
t
)
-
1
/
2
d
x
<
+
∞
,
whereas
(4.8)
lim sup
n
→
+
∞
u
n
(
z
+
)
=
+
∞
if
∫
z
1
(
∫
x
z
a
(
t
)
d
t
)
-
1
/
2
d
x
=
+
∞
.
Proof.
To prove (4.6), we proceed by contradiction assuming that there exists a subsequence, ((λnk,unk))k≥1, of ((λn,un))n≥1 such that
(4.9)
lim
k
→
+
∞
u
n
k
(
1
)
=
0
.
Applying Proposition 4.3 to this subsequence yields
lim sup
k
→
+
∞
u
n
k
(
1
)
>
0
,
which contradicts (4.9) and ends the proof of (4.6).
Next, we will prove (4.7). Integrating the differential equation of (1.1), we obtain, for every n≥1 and x∈(z,1),
(4.10)
0
<
-
u
n
′
(
x
)
1
+
(
u
n
′
(
x
)
)
2
=
-
u
n
′
(
z
)
1
+
(
u
n
′
(
z
)
)
2
-
λ
n
∫
x
z
a
(
t
)
f
(
u
n
(
t
)
)
𝑑
t
<
1
;
thus, setting
φ
n
(
x
,
z
)
=
-
u
n
′
(
z
)
1
+
(
u
n
′
(
z
)
)
2
-
λ
n
∫
x
z
a
(
t
)
f
(
u
n
(
t
)
)
𝑑
t
,
we have that
(4.11)
-
u
n
′
(
x
)
=
φ
n
(
x
,
z
)
1
+
φ
n
(
x
,
z
)
⋅
1
1
-
φ
n
(
x
,
z
)
.
Thanks to Proposition 4.4, Lemma 3.1 and Theorem 3.4, respectively, we can find M>0 and n0∈ℕ such that, for every n≥n0,
(4.12)
M
2
≤
u
n
(
1
)
≤
M
and
λ
∞
2
≤
λ
n
.
By (4.10), for every n≥1 and x∈(z,1), we have that
φ
n
(
x
,
z
)
1
+
φ
n
(
x
,
z
)
<
φ
n
(
x
,
z
)
<
1
.
Thus, due to (4.11) and (4.10), we find that
-
u
n
′
(
x
)
≤
1
1
-
φ
n
(
x
,
z
)
=
1
1
+
u
n
′
(
z
)
1
+
(
u
n
′
(
z
)
)
2
+
λ
n
∫
x
z
a
(
t
)
f
(
u
n
(
t
)
)
𝑑
t
≤
1
λ
n
∫
x
z
a
(
t
)
f
(
u
n
(
t
)
)
𝑑
t
≤
1
λ
n
f
(
u
n
(
1
)
)
(
∫
x
z
a
(
t
)
𝑑
t
)
-
1
/
2
.
So, by (4.12), we get
(4.13)
-
u
n
′
(
x
)
≤
2
λ
∞
f
(
M
/
2
)
(
∫
x
z
a
(
t
)
𝑑
t
)
-
1
/
2
and therefore
u
n
(
z
+
)
=
u
n
(
1
)
-
∫
z
1
u
n
′
(
x
)
d
x
≤
M
+
2
λ
∞
f
(
M
/
2
)
∫
z
1
(
∫
x
z
a
(
t
)
d
t
)
-
1
/
2
d
x
,
which implies (4.7).
To prove (4.8), we argue by contradiction assuming that there is a subsequence, ((λnk,unk))k≥1, of ((λn,un))n≥1 such that
lim sup
k
→
+
∞
u
n
k
(
z
+
)
<
+
∞
.
This implies the existence of a constant C>0 such that
sup
(
z
,
1
)
u
n
k
=
u
n
k
(
z
+
)
≤
C
for all k≥1. Applying Proposition 4.3 to this subsequence yields
(4.14)
sup
(
z
,
1
)
u
∞
≤
C
,
as u∞ is the pointwise limit in (z,1) of (unk)k≥1. Since we are assuming
∫
z
1
(
∫
x
z
a
(
t
)
d
t
)
-
1
/
2
d
x
=
+
∞
,
(4.14) contradicts the last assertion of Proposition 4.3, because u∞(z+)=+∞ in such case. The proof is complete. ∎
The next result establishes the uniqueness of the positive solution of problem (4.2).
Proposition 4.5.
For every λ>0, problem (4.2) has at most one solution u∈Wloc2,∞(z,1] with u(x)>0 for all x∈(z,1].
Proof.
First, suppose that u and v are two solutions of (4.2) satisfying u(x)>0, v(x)>0 for all x∈(z,1] and
u
(
z
+
)
<
+
∞
,
v
(
z
+
)
<
+
∞
.
Set w=u-v. As
(4.15)
u
′
(
z
+
)
=
v
′
(
z
+
)
=
-
∞
and hence
u
′
(
z
+
)
1
+
(
u
′
(
z
+
)
)
2
=
-
1
=
v
′
(
z
+
)
1
+
(
v
′
(
z
+
)
)
2
,
we easily get from (4.2)
λ
∫
z
1
a
(
x
)
f
(
u
(
x
)
)
w
(
x
)
𝑑
x
=
-
∫
z
1
(
u
′
(
x
)
1
+
(
u
′
(
x
)
)
2
)
′
w
(
x
)
𝑑
x
=
u
′
(
z
+
)
1
+
(
u
′
(
z
+
)
)
2
w
(
z
+
)
+
∫
z
1
u
′
(
x
)
w
′
(
x
)
1
+
(
u
′
(
x
)
)
2
𝑑
x
=
-
w
(
z
+
)
+
∫
z
1
u
′
(
x
)
w
′
(
x
)
1
+
(
u
′
(
x
)
)
2
𝑑
x
and
λ
∫
z
1
a
(
x
)
f
(
v
(
x
)
)
w
(
x
)
𝑑
x
=
-
∫
z
1
(
v
′
(
x
)
1
+
(
v
′
(
x
)
)
2
)
′
w
(
x
)
𝑑
x
=
v
′
(
z
+
)
1
+
(
v
′
(
z
+
)
)
2
w
(
z
+
)
+
∫
z
1
v
′
(
x
)
w
′
(
x
)
1
+
(
v
′
(
x
)
)
2
𝑑
x
=
-
w
(
z
+
)
+
∫
z
1
v
′
(
x
)
w
′
(
x
)
1
+
(
v
′
)
(
x
)
2
𝑑
x
.
Subtracting the above identities yields
∫
z
1
(
u
′
(
x
)
1
+
(
u
′
(
x
)
)
2
-
v
′
(
x
)
1
+
(
v
′
(
x
)
)
2
)
(
u
′
(
x
)
-
v
′
(
x
)
)
𝑑
x
=
λ
∫
z
1
a
(
x
)
(
f
(
u
(
x
)
)
-
f
(
v
(
x
)
)
)
(
u
(
x
)
-
v
(
x
)
)
𝑑
x
.
By using the monotonicity of the functions f and ψ, with ψ defined by (4.1), and taking also into account that λ>0 and a(x)<0 for a.e. x∈(z,1), we find hat
∫
z
1
a
(
x
)
(
f
(
u
(
x
)
)
-
f
(
v
(
x
)
)
)
(
u
(
x
)
-
v
(
x
)
)
𝑑
x
=
0
,
and therefore u=v.
Now, suppose that u and v are two solutions of (4.2) satisfying u(x)>0, v(x)>0 for all x∈(z,1] and
u
(
z
+
)
=
v
(
z
+
)
=
+
∞
.
The proof that u=v in this case is divided into two steps. Step 1.The solutions u and v are ordered, i.e., either u(x)≤v(x) for all x∈(z,1), or v(x)≤u(x) for all x∈(z,1). Since u′(1)=v′(1)=0, if u(1)=v(1), by the uniqueness of solution for the associated Cauchy problem, we conclude that u=v. So, suppose that, e.g., u(1)<v(1) and that there exists y∈(z,1) such that
u
(
y
)
=
v
(
y
)
and
u
(
x
)
<
v
(
x
)
for all
x
∈
(
y
,
1
]
.
Set w=u-v. As w(y)=0, we get
λ
∫
y
1
a
(
x
)
f
(
u
(
x
)
)
w
(
x
)
𝑑
x
=
-
∫
y
1
(
u
′
(
x
)
1
+
(
u
′
(
x
)
)
2
)
′
w
(
x
)
𝑑
x
=
∫
y
1
u
′
(
x
)
w
′
(
x
)
1
+
(
u
′
(
x
)
)
2
𝑑
x
,
λ
∫
y
1
a
(
x
)
f
(
v
(
x
)
)
w
(
x
)
𝑑
x
=
-
∫
y
1
(
v
′
(
x
)
1
+
(
v
′
(
x
)
)
2
)
′
w
(
x
)
𝑑
x
=
∫
y
1
v
′
(
x
)
w
′
(
x
)
1
+
(
v
′
(
x
)
)
2
𝑑
x
.
Therefore, subtracting these identities and arguing as above shows that
∫
y
1
(
u
′
(
x
)
1
+
(
u
′
(
x
)
)
2
-
v
′
(
x
)
1
+
(
v
′
(
x
)
)
2
)
(
u
′
(
x
)
-
v
′
(
x
)
)
𝑑
x
=
λ
∫
y
1
a
(
x
)
(
f
(
u
(
x
)
)
-
f
(
v
(
x
)
)
)
(
u
(
x
)
-
v
(
x
)
)
𝑑
x
=
0
.
Consequently, we have that u(x)=v(x) for all x∈(y,1], which is impossible. This ends the proof of Step 1. Step 2.The solutions u and v satisfy v=u. Assume that, e.g., v(x)≥u(x) for all x∈(z,1] and define, for every k>0, the function
w
k
=
min
{
v
-
u
,
k
}
.
Then we get, for a.e. x∈(z,1),
(
u
′
(
x
)
1
+
(
u
′
(
x
)
)
2
-
v
′
(
x
)
1
+
(
v
′
(
x
)
)
2
)
w
k
′
(
x
)
=
{
(
u
′
(
x
)
1
+
(
u
′
(
x
)
)
2
-
v
′
(
x
)
1
+
(
v
′
(
x
)
)
2
)
(
v
′
(
x
)
-
u
′
(
x
)
)
≤
0
if
v
(
x
)
-
u
(
x
)
≤
k
,
0
if
v
(
x
)
-
u
(
x
)
>
k
,
and thus
(4.16)
(
u
′
(
x
)
1
+
(
u
′
(
x
)
)
2
-
v
′
(
x
)
1
+
(
v
′
(
x
)
)
2
)
w
k
′
(
x
)
≤
0
.
On the other hand, for every y∈(z,1), we have that
λ
∫
y
1
a
(
x
)
(
f
(
u
(
x
)
)
-
f
(
v
(
x
)
)
)
w
k
(
x
)
𝑑
x
=
-
∫
y
1
(
u
′
(
x
)
1
+
(
u
′
(
x
)
)
2
-
v
′
1
+
(
v
′
(
x
)
)
2
)
′
w
k
(
x
)
𝑑
x
=
(
u
′
(
y
)
1
+
(
u
′
(
y
)
)
2
-
v
′
(
y
)
1
+
(
v
′
(
y
)
)
2
)
w
k
(
y
)
+
∫
y
1
(
u
′
(
x
)
1
+
(
u
′
(
x
)
)
2
-
v
′
(
x
)
1
+
(
v
′
(
x
)
)
2
)
w
k
′
(
x
)
𝑑
x
.
Thus, by (4.16), we find that
λ
∫
y
1
a
(
x
)
(
f
(
u
(
x
)
)
-
f
(
v
(
x
)
)
)
w
k
(
x
)
𝑑
x
≤
(
u
′
(
y
)
1
+
(
u
′
(
y
)
)
2
-
v
′
(
y
)
1
+
(
v
′
(
y
)
)
2
)
w
k
(
y
)
for all y∈(z,1). So, letting y→z+ in this inequality, condition (4.15) and the boundedness of wk imply that
λ
∫
z
1
a
(
x
)
(
f
(
u
(
x
)
)
-
f
(
v
(
x
)
)
)
w
k
(
x
)
𝑑
x
≤
0
.
But, since λ>0 and, for a.e. x∈(z,1), a(x)<0, f(u(x))≤f(v(x)), wk(x)≥0, we conclude that
∫
z
1
a
(
x
)
(
f
(
u
(
x
)
)
-
f
(
v
(
x
)
)
)
w
k
(
x
)
𝑑
x
=
0
.
Hence, we infer that
a
(
x
)
(
f
(
u
(
x
)
)
-
f
(
v
(
x
)
)
)
w
k
(
x
)
=
0
for a.e.
x
∈
(
z
,
1
)
,
and then
(
f
(
v
(
x
)
)
-
f
(
u
(
x
)
)
)
w
k
(
x
)
=
0
for all
x
∈
(
z
,
1
]
.
Letting k→+∞ yields
(
f
(
v
(
x
)
)
-
f
(
u
(
x
)
)
)
(
v
(
x
)
-
u
(
x
)
)
=
0
for all
x
∈
(
z
,
1
]
.
The monotonicity of f entails u=v. The proof is complete. ∎
Finally, thanks to Propositions 4.3, 4.4 and 4.5, we can describe the exact asymptotic behavior of the solutions un in the interval (z,1].
Theorem 4.6.
Assume (A1), (F) and (3.2). Then the whole sequence, ((λn,un))n≥1 converges to (λ∞,u∞) in R×Wloc2,∞(z,1] as n→+∞, where λ∞ is given by (3.7) and u∞ is the unique solution of (4.2) satisfying u∞(x)>0 for all x∈(z,1].
5 Proof of Theorem 2.1
From [24, Theorem 5.1.3] it is easily inferred the existence of two connected components, 𝒞0+ and 𝒞λ0+, satisfying the following conditions:
𝒞
0
+
and 𝒞λ0+ are unbounded in ℝ×L∞(0,1);
𝒞
0
+
and 𝒞λ0+ are closed and connected subsets of ℝ×BV(0,1) with BV(0,1) endowed with the topology of the strict convergence;
(
0
,
0
)
∈
𝒞
0
+
and (λ0,0)∈𝒞λ0+;
{
(
0
,
r
)
:
r
≥
0
}
⊆
𝒞
0
+
;
ess
inf
u
>
0
if (λ,u)∈𝒞0+∪𝒞λ0+ with u≠0;
λ
=
λ
0
if (λ,0)∈𝒞0+∪𝒞λ0+ with λ>0;
either 𝒞0+∩𝒞λ0+=∅, or (0,0)∈𝒞λ0+ and (λ0,0)∈𝒞0+, and, in such case, 𝒞0+=𝒞λ0+;
there is a neighborhood, 𝒱, of (0,0) in ℝ×L∞(0,1) such that 𝒞0+∩𝒱 consists of regular solutions;
there is a neighborhood, 𝒰, of (λ0,0) in ℝ×L∞(0,1) such that 𝒞λ0+∩𝒰 consists of regular solutions.
Combining these properties with the theory developed in the previous sections allows us to verify all the conclusions of Theorem 1.1.
Indeed, from conditions (3), (4), (6), (7) and Theorem 2.1, it readily follows that
𝒞
0
+
=
{
(
0
,
r
)
:
r
≥
0
}
and
𝒞
0
+
∩
𝒞
λ
0
+
=
∅
.
Thus, (a) and (b) hold. The conclusions (c) and (d) are consequences of (1) and Theorem 2.1. Theorem 3.4 yields (e). The conclusion (f) follows from Theorems 3.3 and 4.1, for what concerns the behavior on the interval [0,z), and from Theorem 4.6 and Proposition 4.3, for what concerns the behavior on the interval (z,1]. While assertion (g) is precisely (9).
In order to prove (h), we proceed by contradiction. So, assume that (A2) holds and that (λ,u)∈𝒮+, with λ>0, is a singular solution. Then, thanks to (1.4), for any x∈(0,1)∖{z}, integrating the differential equation in (x,z) yields
0
<
-
u
′
(
x
)
1
+
(
u
′
(
x
)
)
2
=
1
-
λ
∫
x
z
a
(
t
)
f
(
u
(
t
)
)
𝑑
t
<
1
,
and hence
-
u
′
(
x
)
=
1
-
λ
∫
x
z
a
(
t
)
f
(
u
(
t
)
)
𝑑
t
2
-
λ
∫
x
z
a
(
t
)
f
(
u
(
t
)
)
𝑑
t
⋅
1
λ
∫
x
z
a
(
t
)
f
(
u
(
t
)
)
𝑑
t
.
So, for sufficiently small η>0 and every x∈[z-η,z+η]∖{z}, we have that
-
u
′
(
x
)
≥
1
2
λ
h
(
∫
x
z
a
(
t
)
𝑑
t
)
-
1
/
2
.
Under condition (A2), this estimate entails that either u′∉L1(0,z), or u′∉L1(z,1), which contradicts the fact that
u
∈
W
1
,
1
(
0
,
z
)
∩
W
1
,
1
(
z
,
1
)
.
Therefore, assuming (A2), the set 𝒮+ consists of regular solutions, as claimed by assertion (h).
Lastly, we will prove the conclusion (h). More generally, we will show that all the solutions of 𝒮+ with sufficiently large L∞-norm must develop singularities if (A2) fails. We will argue by contradiction assuming the existence of a sequence of regular solutions, ((λn,un))n≥1, in 𝒮+ such that
(5.1)
lim
n
→
+
∞
∥
u
n
∥
L
∞
=
+
∞
.
Then, by Theorem 3.4, we also have that
lim
n
→
+
∞
λ
n
=
λ
∞
.
As we are assuming that
∫
z
1
(
∫
x
z
a
(
t
)
d
t
)
-
1
/
2
d
x
<
+
∞
,
it follows from Proposition 4.4 that
0
<
lim inf
n
→
∞
u
n
(
z
)
≤
lim sup
n
→
+
∞
u
n
(
z
)
<
+
∞
.
Thus, there exist n0∈ℕ and M>0 such that, for every n≥n0,
M
2
≤
u
n
(
z
)
≤
M
and
λ
n
≥
λ
∞
2
.
Since these estimates coincide with (4.12), reasoning as in the proof of Proposition 4.4, it becomes apparent that (4.13) holds, i.e.,
-
u
n
′
(
x
)
≤
2
λ
∞
f
(
M
/
2
)
(
∫
x
z
a
(
t
)
𝑑
t
)
-
1
/
2
,
and hence
u
n
(
0
)
=
u
n
(
z
)
-
∫
0
z
u
n
′
(
x
)
d
x
≤
M
+
2
λ
∞
f
(
M
/
2
)
∫
0
z
(
∫
x
z
a
(
t
)
d
t
)
-
1
/
2
d
x
=
C
<
+
∞
,
as we are assuming that (A2) fails. This implies that, for all n≥n0,
∥
u
n
∥
L
∞
=
u
n
(
0
)
≤
C
,
which contradicts (5.1) and ends the proof of Theorem 1.1.
