Nonlinear Analysis 222 (2022) 113000
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Nonlinear Analysis
www.elsevier.com/locate/na
Singular solutions to k-Hessian equations with fast-growing
nonlinearities
João Marcos do Ó a ,∗, Evelina Shamarova a , Esteban da Silva b
a
b
Departamento de Matemática, Universidade Federal da Paraíba, João Pessoa, Brazil
Departamento de Matemática, Universidade Federal do Rio Grande do Norte, Natal, Brazil
article
info
Article history:
Received 3 February 2022
Accepted 13 May 2022
Communicated by Francesco Maggi
MSC:
34B15
34C23
35A24
35J15
35J62
35J92
abstract
We study a class of elliptic problems, involving a k-Hessian and a very fast-growing
nonlinearity, on a unit ball. We prove the existence of a radial singular solution and
obtain its exact asymptotic behavior in a neighborhood of the origin. Furthermore,
we study the multiplicity of regular solutions and bifurcation diagrams. An
essential ingredient of this study is analyzing the number of intersection points
between the singular and regular solutions for rescaled problems. In the particular
case of the exponential nonlinearity, we obtain the convergence of regular solutions
to the singular and analyze the intersection number depending on the parameter
k and the dimension d.
© 2022 Elsevier Ltd. All rights reserved.
Keywords:
k-Hessian equation
Singular solution
Multiplicity of regular solutions
Number of intersection points
Bifurcation diagrams
Liouville–Bratu–Gelfand problem
1. Introduction
In this paper, we are concerned with the following Dirichlet
⎧
2
⎪
⎨Sk (D w) = λ F (−w)
w < 0,
⎪
⎩
w = 0,
problem:
in Ω ,
in Ω ,
on ∂Ω ,
(1)
where Ω ⊂ Rd is a bounded domain, F is a positive function increasing to infinity, whose exact properties
will be specified later, and λ > 0 is a parameter. Above, 1 ⩽ k ⩽ d is an integer, where d > 2 is the
∗ Corresponding author.
E-mail addresses: jmbo@pq.cnpq.br (J.M. do Ó), evelina.shamarova@academico.ufpb.br (E. Shamarova),
esteban.silva@ufrn.br (E. da Silva).
https://doi.org/10.1016/j.na.2022.113000
0362-546X/© 2022 Elsevier Ltd. All rights reserved.
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
dimension, and Sk (D2 w) is the k-Hessian, i.e., the symmetric elementary function of the kth order of the
eigenvalues of D2 w.
We prove the existence of a specific radial singular solution to (1) in the case when Ω is a unit ball and
obtain its exact asymptotic behavior in a neighborhood of the origin. We furthermore demonstrate that
this specific singular solution allows to study properties of radial regular solutions to (1) such as bifurcation
diagrams and multiplicity, where an important tool is the number of intersection points between the singular
and regular solutions for rescaled problems.
1.1. Elliptic problems with nonlinearities increasing to infinity
In recent years, semilinear elliptic problems of the form
{
−∆u = λG(u),
u = 0,
in Ω ,
on ∂Ω ,
(2)
where G is smooth and increasing to infinity, have been of significant interest; see, e.g., [1–3,7,10,11,14,15,
20,22,26,30,31,33]. This list of references is by no means complete, so we refer the reader to [1,3,7,22,33],
and monograph [11] for a survey of results on this subject. The following important points are usually
addressed when dealing with this type problems: (1) the existence of a critical parameter λ# such that
for each λ ∈ (0, λ# ) there exists a regular solution (uλ , λ) to problem (2), and there are no solutions to (2)
for λ > λ# ; (2) the existence of a parameter λ∗ ∈ (0, λ# ] such that problem (2) possesses a singular solution
for λ = λ∗. Since a tremendous amount of papers (see, e.g., the above-cited ones and references therein) has
been dedicated to problems of type (2) and, in particular, to answering the above questions, this topic can
be regarded as fundamental.
In this work, we are concerned with a related problem; namely, with problem (1) involving a k-Hessian and
a function F under quite flexible imposed assumptions. k-Hessian equations constitute an important class
of fully nonlinear PDEs; so they have been studied by many authors [6,8,9,34,35,38–41]. When restricted to
so-called k-admissible solutions (for the definition see e.g. [39]), these equations are elliptic and their solutions
enjoy properties similar to those of solutions to semilinear equations involving the Laplacian. Furthermore,
k-Hessian equations are of interest in geometric PDEs [18] and differential geometry [36].
Problem (1) is in turn a larger generalization, compared to those that exist in the literature [15,20,26,30],
of the classical Gelfand problem. The latter deals with positive solutions to the system
{
∆u + λ eu = 0
in Ω ,
(3)
u=0
on ∂Ω .
Problem (3) is a fundamental problem that arises in many theories, such as non-linear diffusions generated
by non-linear sources [21,23,25], thermal ignition of a mixture of chemically active gases [14], unstable
membrane [4], and gravitational equilibrium of polytropic stars [5,13], among others. For problem (3), one
is mainly interested in the existence and multiplicity of regular solutions, as well as bifurcation diagrams.
Most of authors were concentrated on the case of a unit ball, e.g., [14,22], since in this case, it is known that
all solutions are radial [17] and problem (3) has the following simple representation:
{
′
u
u′′ + d−1
r u + λe = 0,
(4)
′
u (0) = u(1) = 0.
In [14], Gelfand proved the existence of the value λ for which (4) has an infinite number of solutions. In [22],
the authors completely described the relation between the multiplicity of solutions to (4) and the dimension
d of the ball.
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J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
Problem (4) was further generalized by various authors in the following two directions: replacing eu with
a more general function while keeping the radial Laplacian [15,26,30,31]; and replacing the Laplacian with
the radial operator
β
L(u) = r−γ (rα |u′ (r)| u′ (r))′
(5)
while preserving the term eu [20]. More specifically, in [15], the term eu was replaced with the n-times iterated
p
exponential (henceforth denoted by exp◦n (u)), in [26], it was replaced with eu , and in [30], this term was
replaced with eu + “lower-order term”.
1.2. Our setting: operator L and function F
In this work, we are concerned with negative radial solutions to problem (1) on a unit ball. At the same
time, our methods allow to study the convergence of positive regular solutions to the singular for the equation
L(u) + λeu = 0. To hit both of these goals, in (1), we make the substitution w = −u, and by this, we reduce
problem (1) to its radial form (see the Appendix for details) as follows:
⎧
f (u)
⎪
= 0, 0 ⩽ r < 1,
⎨L(u) + λ e
(Pλ )
u > 0, 0 ⩽ r < 1,
⎪
⎩
u(1) = 0,
where L is defined by (5) and f = ln F satisfies the assumptions below. For the sake of clarity, we formulate
these assumptions (in terms of f and the inverse function g = f −1 ) in a simplified manner. For the actual
and more general set of assumptions, we redirect the reader to Section 2.1.
(1)
(2)
(3)
(4)
f is smooth and increasing to infinity;
|g ′′ (t)| monotonically decreases to zero;
limt→+∞ f (g(t)+ε(t))
= 1, where ε(t) = O(g ′′ (s));
t ′
g (t)
g ′′ (t)
As t → +∞, g′ (t+o(t)) → 1 and g′′ (t+o(t))
= O(1);
′′
(u)
′
2
(5) limu→+∞ f ′ (u) = +∞, limu→+∞ ff′ (u)
2 (ln f (u)) = 0;
(6) α > β + 1 (i.e., d > 2k for the k-Hessian); β ⩾ 0; γ + 1 > α.
We emphasize that problem (Pλ ) is an extension of problem (4) in the two aforementioned directions
simultaneously, i.e., replacing the Laplacian with the operator (5), which can be roughly regarded as the
radial case of a k-Hessian, and replacing exp(u) with a rather general term exp{f (u)}, which encompasses
exponential-type nonlinearities previously considered in the literature [15,26,30], such as exp(u), exp(up ), the
iterated exponential exp◦n (u), exp(u) + “lower-order term” (see Section 2.2). In addition, due to flexibility
of our assumptions, exp{f (u)} provides a plenty of new examples of nonlinearities which have not been
considered in the literature, but can be treated within our framework.
We allow the parameters α, β, and γ to take a larger range of values than it is required for a k-Hessian
since it does not give us extra work in the proofs and allows to include the important case of a p-Laplacian.
The exact values of α, β, and γ for a k-Hessian and a p-Laplacian are summarized in the table below. The
table includes the parameter θ = γ + 2 + β − α whose role will become clear later.
k-Hessian
p-Laplacian
α
β
γ
θ
d−k
d−1
k−1
p−2
d−1
d−1
2k
p
Remark 1.1. Remark that since the solutions of interest to problem (1) are negative, according to [39],
the operator Sk (D2 w) (regarded as a d × d matrix) is elliptic on those solutions. On the other hand, if u is
a solution to problem (Pλ ), the operator Sk (D2 u(| · |)) may not be elliptic since Sk (D2 u) = (−1)k Sk (D2 w).
Its ellipticity depends on the parity of k.
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J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
1.3. Contributions of this work and comparison with other results
One of the main results of this work is a construction of a negative radial singular solution to problem (1)
on a unit ball, which is reduced to a construction of a positive radial singular solution to problem (Pλ ) (see
the Appendix), with an exact asymptotic behavior in a neighborhood of the origin. By a positive singular
solution to problem (1) in a ball, we understand a C2 -function which solves problem (1) at all points of the
ball except for the origin, and tends to the negative infinity as the argument goes to 0. By a singular solution
to problem (Pλ ), we understand a C2 -function which solves problem (Pλ ) at all points of (0, 1] and tends to
the positive infinity as the argument goes to 0+.
Radial singular solutions on a unit ball have attracted significant interest from many researchers
[2,15,16,19,26,27,29–32]. Most of these works deal however with semilinear equations involving the Laplacian.
Papers [16,19] study radial singular solutions to equations with the operator (5), but they do not discuss
asymptotics for these solutions, and furthermore, they study different types of nonlinearities. Among
applications of singular solutions are bifurcation diagrams for elliptic problems [15,26,30–32] and blow-up
behavior of solutions to parabolic problems [28,37]. Asymptotic representations for radial singular solutions
on a unit ball were previously obtained in [15,26,29,30]. However, these papers deal with the radial Laplacian
and very particular types of the function f (u). Moreover, the asymptotics of singular solutions obtained
in [15,26,30] are included in ours as particular cases. Our assumptions on f are quite flexible, so examples
of f include not only exp◦n (u), up , and u + “lower-order term” (cf. [15,26,30]), but also many others. As
such, we show that f (u) = exp◦n (up ), satisfies our assumptions, encompassing both exp◦n (u) from [15] and
up from [26]. We also remark that, compared to [27,31,32], our results are valid for the operator L, given
by (5) and therefore generalizing the Laplacian, and, on the whole, our work contains a sharp asymptotic
representation of a singular solution which is not present in the aforementioned works.
We state our main theorem on the asymptotic representation of a radial singular solution as follows:
Theorem 1.1. Assume (1)–(6). Then, there exists λ∗ > 0 such that problem (Pλ ) with λ = λ∗ possesses a
radial singular solution of the form
(
)
u∗λ∗ (r) = g(Z(r)) + O g ′′ (ln r−θ ) , where
{
}
Z(r) = ln θβ+1 (α − β − 1) − ln λ∗ − θ ln r
(6)
{β + 1}
{ ′
}
′′
′
.
+ (β + 1) ln g (τ ) + (β + 1)g (τ ) ln g (τ )
with τ = ln
λ∗ rθ
The important particular case is the exact form for the singular solution to problem (Pλ ) in the case
f (u) = u. Namely, the following corollary holds.
Corollary 1.1.
Assume (6) and consider problem (Pλ ) with f (u) = u. Then, (u∗λ∗ , λ∗ ), given by
u∗λ∗ (x) = −θ ln |x|,
λ∗ = θβ+1 (α − β − 1),
is a radial singular solution to problem (Pλ ).
We furthermore use the singular solution (u∗λ∗ , λ∗ ), constructed in Theorem 1.1, to study properties of
regular solutions such as bifurcation diagrams and multiplicity. Specifically, we obtain the following result.
4θ
(d < 2k + 8). Further, we let (uλ(ρ) , λ(ρ)) be
Theorem 1.2. Assume (1)–(6) and let α − β − 1 < β+1
the family of regular solutions to problem (Pλ ) parametrized by ρ = supr⩽1 |uλ (r)|, and let (u∗λ∗ , λ∗ ) be the
singular solution constructed in Theorem 1.1. Then, as ρ → +∞, λ(ρ) oscillates around λ∗ . In particular,
there are infinitely many regular solutions to problem (Pλ ) for λ(ρ) = λ∗ .
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J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
An important tool for obtaining the above result is the number of intersection points between the regular
and singular solutions to the equation
L(u) + ef (u) = 0.
(7)
1
Note that u∗ (r) = u∗λ∗ (r(λ∗ )− θ ) is a singular solution to (7). In general, the change of variable
1
u(r) = uλ (rλ− θ )
(8)
brings the equation in (Pλ ), containing λ, to Eq. (7). As such, the radial regular solution u( · , ρ) to (7) is
obtained from the radial regular solution (uλ(ρ) , λ(ρ)) to (Pλ ) by the change of variable (8). We have the
following result.
4θ
Theorem 1.3. Assume (1)–(6) and let α − β − 1 < β+1
. Then, for any δ > 0, it holds that
∗
∗
limρ→+∞ L(0,δ) (u( · , ρ) − u ( · )) = +∞, where u and u are regular and singular solutions to (7) and
L(0,δ) (u − u∗ ) is the number of zeros of u − u∗ on the interval (0, δ).
Furthermore, we turn our attention to a study of radial regular solutions. The result that we obtain in
this direction is the following.
Theorem 1.4. Assume (1), (5), and (6). Then, there exists a parameter λ# > 0, depending on α, β, γ, and
f , such that any regular solution (uλ , λ) to problem (Pλ ) satisfies λ ⩽ λ# . Moreover, for any λ < λ# , there
exists a regular solution to (Pλ ), and for λ = λ# , the solution can be either regular or singular.
The actual set of assumptions for Theorem 1.4 is, again, somewhat weaker, and the reader is redirected
to Section 3.3.
Finally, we study the equation
L(u) + eu = 0
(9)
In [20], the authors considered problem (Pλ ) with the nonlinearity eu and obtained the result of Theorem 1.2
along with the convergence λ(ρ) → λ∗ as ρ → +∞. In the current work, we complement the aforementioned
results by studying the number of intersection points between the regular and singular solutions to (9),
depending on α, β, and γ, and by proving the convergence of regular solutions to the singular as ρ → +∞.
Namely, we have the following results.
Theorem 1.5. Assume (6). Suppose 0 < α − β − 1 <
4θ
β+1
(2k < d < 2k + 8). Then, L(0,+∞) (u − u∗ ) = +∞.
Theorem 1.6. Assume (6). Suppose α − β − 1 ⩾ 4θ(β + 1) (d ⩾ 2k + 8k 2 ). Then, L(0,+∞) (u − u∗ ) = 0.
Theorem 1.7. Assume (6). Then,
lim u( · , ρ) = u∗
ρ→+∞
and
lim (uλ(ρ) , λ(ρ)) = (u∗λ∗ , λ∗ )
ρ→+∞
in Cloc (0, +∞).
(10)
Remark 1.2. Remark that the results on the intersection number, similar to those of Theorems 1.5 and
p−1
1.6, are known for the equation L(u) + |u| u = 0 (see e.g. [32]). Here they are obtained for the exponential
counterpart of this equation.
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J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
1.4. Methods
In this work, we use combinations of newly developed methods along with extensions, from the Laplacian
to the k-Hessian, of existing techniques. New methods are used for construction of a singular solution and
obtaining its sharp asymptotic (Theorem 1.1), for proving the convergence of regular solutions to the singular
(Theorem 1.7), and for showing that the regular and singular solutions to Eq. (9) do not intersect each other
if α − β − 1 ⩾ 4θ(β + 1) (Theorem 1.5). To be specific, in the proof of Theorem 1.1, we arrive at the equation
v ′′ − (α̂ − 1)v ′ = −
e−θt+f (v)
(11)
β
|v ′ |
which is equivalent to the equation in (Pλ ), but the variable t varies in a neighborhood of infinity. The
β
difficulty arises due to the fact that the denominator on the right-hand side contains |v ′ | , the term which
is not present in the Laplacian case (β = 0). This does not allow to apply methods known for semilinear
equations with the Laplacian (see, e.g., [15,26,30]). In particular, the singular solution v ∗ to (11) takes the
form
v ∗ (t) = g(θt + φ1 (t) + φ2 (t)) + η(t),
(12)
where
(
)β
φ1 = ln{θβ g ′ (θt) + (β + 1)g ′′ (θt) ln g ′ (θt) },
{ ( α
)}
φ2 = ln θ
− 1 + ln{g ′ (θt) + (β + 1)g ′′ (θt) ln g ′ (θt)}
β+1
(13)
and η(t) = O(g ′′ (θt)). Remark that the term φ1 does not appear in the Laplacian case; it was for the first
β
time introduced in this work. Its role is to make certain cancellations with the denominator |v ∗ ′ | to ensure
the possibility to reduce (11) to a pair of semilinear equations with respect to η and η ′ . The order of the
reminder η, in the general setting, was also first established in this work. Additional technical difficulties
arise due to the fact that the function f (unlike [15,26,30]) is arbitrary, so the asymptotic representation (6)
is expressed via the derivatives of f −1 .
Furthermore, in the proof of Theorem 1.7, we use the result of Section 3 on the relative compactness
of regular solutions on each compact interval of (0, +∞) (along with the result of [20]). To the best of our
knowledge, the aforementioned technique is new. Also, it is worth to mention Theorem 1.6 whose proof
does not appear to have analogs in the literature; while the result itself, together with Theorem 1.5, can be
regarded as fundamental. Indeed, Theorems 1.5 and 1.6 study the number of intersection points between
the regular and singular solutions for the canonical equation (9), the result which did not appear in the
literature prior to this work.
Furthermore, many of our methods were previously unknown in the context of k-Hessian equations. They
represent non-trivial extensions, from the Laplacian to the k-Hessian, of existing techniques. For example, in
the proof of Theorem 1.3, dealing with the intersection number between the regular and singular solutions,
one obtains a certain equation via the following transformation of Eq. (7):
θ
− β+1
ũ(s, ρ) = F1−1 ερ
F(u(ερ s, ρ)),
ερ =
( F(ρ) )(β+1)/θ
F1 (1)
,
(14)
{ f (s) }
∫ +∞
− u
where F(u) = u exp − 1+β
ds and F1 (u) = (β + 1)e β+1 . We believe that in the k-Hessian case, this
transformation, in its exact form (14), was first used in this work. A version of this transformation, slightly
different from (14), was introduced in [16]. The “Laplacian” analog of (14) can be found in, e.g., [31]. The
important difference with the Laplacian case is the following: when we apply this transformation, the first
two terms L(ũ) + eũ of the transformed equation are exactly as in the canonical equation (9); however,
)
β+2 ( I(u(ερ s))
(u)
the additional term sα−γ |ũ′ |
− 1 , where I(u) = F(u)f ′ (u) exp{ fβ+1
}, has a more complicated
β+1
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J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
structure and its convergence to zero is rather difficult to show. This convergence however is crucial to show
the convergence of ũ to the regular solution of the canonical (limit) equation (9). Remark that even in the
Laplacian case, considered in [31], the aforementioned convergence was announced without a proof. It is
important to mention that in order to show the convergence of singular solutions, which is also crucial for
proving Theorem 1.3, we employ our novel asymptotic (6) obtained in Theorem 1.1. Thus, we reinforce that
the proof of Theorem 1.3 on the intersection number differs significantly from the Laplacian case and requires
new tools developed in this work.
Furthermore, the proof of Theorem 1.2 on bifurcation diagrams relies heavily on Theorem 1.3 discussed
above. In addition, the aforementioned proof contains a much more detailed analysis on the oscillation of
λ(ρ) around λ∗ compared to [31] and preceding works.
Finally, we mention Proposition 3.1 that deals with upper and lower bounds for u( · , ρ)−1 (B), B > 0,
whenever ρ is sufficiently big. For the Laplacian case, this result was obtained in [27] (Theorem 2.4) and
then was widely used in the same paper. However, an extension of Theorem 2.4 to the operator L is as
important for k-Hessian equations as the aforementioned theorem is important for semilinear equations
with the Laplacian. Besides of being a non-trivial extension from the Laplacian setting, Proposition 3.1
is an important tool in the absence of which many of the results of this work could not be obtained.
As such, Proposition 3.1 is used in the proof of the relative compactness of regular solutions to Eq. (7)
(Proposition 3.2) which is used to obtain the convergence (10) in Theorem 1.7. Next, Proposition 3.1 is used
to obtain Theorem 1.4. Finally, Proposition 3.1 is used in the following chain of results leading to one of
the main results of this work. Namely, it is directly applied in Lemma 4.2 implying Proposition 4.1 on the
convergence of the singular and regular solutions of the equation obtained via the transformation (14) to
the respective solutions of the canonical equation (9). This, in turn, implies Theorem 1.3 on the number of
intersection points while the latter is used in the proof of Theorem 1.2 on bifurcation diagrams.
1.5. Structure of this work
Our paper is structured as follows. In Section 2, we prove the existence of a singular solution to (Pλ )
and obtain its exact asymptotic behavior in a neighborhood of the origin. The main result of this section is
Theorem 1.1. In Section 3, we prove Theorem 1.4. In the same section, we obtain the existence and relative
compactness of regular solutions to problem (Pλ ). As a byproduct, we obtain a version of Pohozaev’s identity
for the operator (5). In Section 4, we study bifurcation diagrams and the number of intersection points
4θ
. The main results of this
between the regular and singular solutions for the case 0 < α − β − 1 ⩽ β+1
section are Theorems 1.2 and 1.3. It is important to mention that we employ here the transformation (14)
that reduces Eq. (7) to an equivalent equation whose limit is Eq. (9). The aforementioned equation is then
studied in Section 5, where we obtain Theorems 1.5, 1.6, 1.7.
2. Singular solution to problem (Pλ )
In this section, we will construct a positive radial singular solution to (Pλ ).
2.1. Standing assumptions and useful lemmas
We start by introducing the general set of assumptions implying assumptions (1)–(6) from Section 1.2.
Recall that g denotes the inverse function for f .
(A1) f ∈ C4 ([0, +∞)), f ′ > 0, and limu→+∞ f (u) = +∞.
(A2) limt→+∞ g ′′ (t) = 0 and limt→+∞ f (g(t)+ε(t))
= 1, where ε(t) = O(g ′′ (t)).
t
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J.M. do Ó, E. Shamarova and E. da Silva
g ′ (t)
g ′ (t+o(t)) = 1.
′′
(t)
limt→+∞ gg′ (t)
ln g ′ (t) =
Nonlinear Analysis 222 (2022) 113000
(A3) limt→+∞
(A4)
(A5) As t → +∞,
0.
(a) g ′′′ (t)(ln g ′ (t))n = O(g ′′ (t)), n = 0, 1, 2,
(b) sup |g ′′ (s)| = O(g ′′ (t)),
g (iv) (t) ln g ′ (t) = O(g ′′ (t)),
g ′′ (t + o(t)) = O(g ′′ (t)).
s⩾t
(A6) α > β + 1 (i.e., d > 2k for the k-Hessian); β ⩾ 0; θ > 0.
Remark 2.1. It will be rather convenient to have some alternative equivalent conditions at hand for (A2)
and (A4). Namely, (A2) and (A4) are equivalent to
f ′′ (u)
f ′ (u)3
′′
(u)
limu→+∞ ff′ (u)
2
(A2′ ) limu→+∞
= 0 and limt→+∞
(A4′ )
ln f ′ (u) = 0;
f (u+ε̃(u))
f (u)
′′
(u)
= 1, where ε̃(u) = O( ff′ (u)
3 ).
We start showing (A4′ ). A straightforward computation gives
g ′′ (t)
f ′′ (u)
′
ln
f
(u)
=
−
ln g ′ (t),
f ′ (u)2
g ′ (t)
where u = g(t).
′′
(u)
To show the equivalence of (A2) and (A2′ ), we note that g ′′ (f (u)) = ff′ (u)
3 , and hence, the first limits in
(A2) and (A2′ ) are equivalent. Substituting t = f (u), we represent η(t) as ϕ(u)g ′′ (f (u)), where ϕ(u) is a
′′
(u)
bounded function. Defining η̃(u) = ϕ(u) ff′ (u)
3 , we obtain the desired equivalence.
Remark 2.2.
For the p-Laplacian, the condition α > β + 1 is interpreted as p < d.
′′
(t)
Note that under (A4) and (A2), limt→+∞ gg′ (t)
= 0. Indeed, assume the opposite. Then,
⏐ g′′ (tn ) ⏐
there exists a sequence tn → +∞ such that ⏐ g′ (tn ) ⏐ > M for some constant M > 0. By (A4), it should hold
Remark 2.3.
that limn→+∞ g ′ (tn ) = 1, which, in turn, implies that limn→+∞
g ′′ (tn )
g ′ (tn )
= 0 by (A2).
Remark 2.4. Some of examples of the function ϕ(t) = g ′′ (t) satisfying the first expression in (A5)-(b)
are monotone functions and functions which can be represented as ϕ(t) = ϕ1 (t)ϕ2 (t), where ϕ1 is decreasing
to zero as t → +∞, and ϕ2 is bounded from above and below, with an infinite number of maximum and
minimum points, such as 2 + sin t or esin t .
Remark 2.5.
In what follows, we will need the following facts: as t → +∞,
t
t
t
− β+1
′
e β+1 g ′ (t) = O(t e β+1 ),
e
t
− β+1
g (t) = O(t e
).
The above expressions obviously hold by (A2). Indeed, g ′′ (t) is bounded, and therefore, by Taylor’s formula,
g ′ (t) has at most linear growth.
It turns out that some of the conditions (A5)-(a) are fulfilled if limu→+∞ f ′ (u) = +∞, or, which is the
same, limt→+∞ g ′ (t) = 0.
Lemma 2.1. Suppose limu→+∞ f ′ (u) = +∞. Then, conditions (A5)-(a), except for n = 2, are fulfilled. If,
′′
(u)
′
2
in addition, limu→+∞ ff′ (u)
2 (ln f (u)) = 0, then all conditions in (A5)-(a) are fulfilled.
8
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
Proof . Note that g ′′ (t) cannot be identically equal to zero in a neighborhood of +∞. Indeed, in this case
g ′ (t) = 0 in this neighborhood which is a contradiction.
Note that, by (A4), limt→∞ g ′′ (t) ln g ′ (t) = 0. By L’Hopital’s rule,
g ′′′ (t)
g ′′ (t)
g ′′ (t)
′
′
ln
g
(t)
=
lim
ln
g
(t)
−
lim
= 0.
t→+∞ g ′′ (t)
t→+∞ g ′ (t)
t→+∞ g ′ (t)
lim
The above inequality implies, in particular, that limt→+∞ g ′′′ (t) ln g ′ (t) = 0. Again, by L’Hopital’s rule,
g (4) (t)
g ′′′ (t)
g ′′ (t)
ln g ′ (t) = lim ′′
ln g ′ (t) − lim ′
= 0.
′′′
t→+∞ g (t)
t→+∞ g (t)
t→+∞ g (t)
lim
Note that limt→+∞
g ′′′ (t)
g ′′ (t)
= 0. Therefore,
g (4) (t)
g (4) (t)
g ′′′ (t)
′
′
ln
g
(t)
=
lim
ln
g
(t)
lim
= 0.
t→+∞ g ′′ (t)
t→+∞ g ′′′ (t)
t→+∞ g ′′ (t)
lim
Finally, a straightforward computation shows that
g ′′ (t)
f ′′ (u)
(ln f ′ (u))2 = ′ (ln g ′ (t))2 ,
′
2
f (u)
g (t)
where u = g(t).
The first identity in (A5)-(a) for n = 2 follows now by L’Hopital’s rule. □
Remark 2.6. Remark that Lemma 2.1 justifies condition (5) in Section 1.2.
Lemma 2.2. If ζ(t) is a differentiable function R+ → R such that limt→+∞ ζ ′ (t) = 0, then limt→+∞
0.
Proof . We have
∫
ζ(0) −
t
|ζ ′ (s)|ds ⩽ ζ(t) ⩽ ζ(0) +
0
∫
ζ(t)
t
=
t
|ζ ′ (s)|ds.
0
∫t
The integral 0 |ζ ′ (s)|ds is either bounded or tends to +∞ as t → +∞. In the first case the conclusion of
the lemma is obvious. In the second case it follows from L’Hopital’s rule. □
2.2. Examples of nonlinearities
Here, we give examples of functions f (u) in (Pλ ) satisfying (A1)–(A5).
Example 1 (f (u) = exp◦n (up ) (p > 0, n ⩾ 1)). First, introduce the notation for the iterated exponential
and the iterated logarithm:
exp◦n (t) = exp ◦ exp ◦ · · · ◦ exp(t);



n
ln◦n (t) = ln ◦ ln ◦· · · ◦ ln(t).
n
1
The inverse function to f is g(t) = [ln◦n (t)] p . We prove that (A1)–(A5) are fulfilled for the pair f, g.
Remark 2.7. Remark that in [15], f (u) = exp◦n (u), and in [26], f (u) = up , both are particular cases of
f (u) = exp◦n (up ) discussed in this example. In addition, we remark that [15,26] deal with the classical radial
Laplacian, while our partial differential operator takes the form (5).
9
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
First, we compute
1
1
,
p ln◦n (t)1− p1 ln◦(n−1) (t) ln◦(n−2) (t) · · · t
)
1
1
1 [( 1
−
−1
g ′′ (t) =
1
1
2− p
1− p
◦(n−1)
◦(n−1)
◦n
◦n
p p
2
2
ln
(t) · · · t
ln
(t)2 · · · t2
ln (t)
ln (t)
]
1
1
− ··· −
.
−
1
1
ln◦n (t)1− p ln◦(n−1) (t) ln◦(n−2) (t)2 · · · t2
ln◦n (t)1− p ln◦(n−1) (t) · · · t2
g ′ (t) =
Note that limt→+∞ g ′ (t) = 0. Moreover, limt→+∞
g ′′ (t)
g ′ (t)
g ′′ (t)
′
2
g ′ (t) (ln g (t))
= 0. To see the latter, we note that
=
and ln g (t) = O(ln(t)). Therefore, by Lemma 2.1, (A5)-(a) is fulfilled. Also, |g ′′ (t)| is
decreasing, so the first identity in (A5)-(b) is fulfilled. The above argument also implies (A4).
A verification of (A1) is straightforward. To verify (A3), it is sufficient to show that for the iterated
ln◦n (t)
logarithm, it holds that limt→+∞ ln◦n
(t+o(t)) = 1. We prove the following statement by induction on n:
O( 1t )
′
ln◦n (t + o(t)) = τ + o(τ ), where τ = ln◦n (t). For n = 1, it is obvious since ln(t + o(t)) = ln t + o(1).
Suppose we proved the statement for n − 1, i.e., ln◦(n−1) (t + o(t)) = τ + o(τ ) with τ = ln◦(n−1) (t). Then,
ln◦n (t) = ln(τ + o(τ )) = ln τ + o(ln(τ )) = ln◦n (t) + o(ln◦n (t)), which implies (A3). The same argument
implies the second expression in (A5)-(b).
It remains to verify (A2). Here, without loss of generality we can set p = 1. Indeed, define the function
g̃(t) = ln◦n (t) so that g(t)p = g̃(t). We have
(
(
)p
O(g ′′ (t)) )p
= g̃(t) + O(g ′′ (t))g(t)p−1 = g̃(t) + O(g̃ ′′ (t))
g(t) + O(g ′′ (t)) = g̃(t) 1 +
g(t)
(
′′
g ′ (t)2 )
(t)
since g̃ ′′ (t) = p g ′′ (t)g(t)p−1 1 + (p − 1) g(t)g
= g ′′ (t)g(t)p−1 O(1). Indeed, gg′ (t)
= 1t + o( 1t ) and
′′ (t)
g ′ (t)
g(t)
= O( 1t ). We further argue by induction on n. For n = 1, (A2) is clearly satisfied. Suppose (A2) is
true for n − 1. We have
exp(g̃(t) + O(g̃ ′′ (t))) = ln◦(n−1) (t) exp{O(g̃ ′′ (t))} = ln◦(n−1) (t)(1 + O(g̃ ′′ (t)))
= ḡ(t) + O(ḡ ′′ (t)),
where ḡ(t) = ln◦(n−1) (t).
Indeed, one immediately verifies that g̃ ′′ (t) ln◦(n−1) (t) = ḡ ′′ (t) + o(ḡ ′′ (t)). (A2) now follows now from the
induction hypothesis.
Example 2 (f (u) = up , p > 12 ). We restrict here the range of p to ( 12 , +∞) in order to guarantee that
1
limt→+∞ g ′′ (t) = 0. Indeed, g ′′ (t) = p1 ( p1 − 1) t p −2 . The verification of (A1)–(A5) is straightforward.
Example 3 (Perturbed Gelfand’s Problem). Take f (u) = u + ϑ(u), where ϑ(u) is a C4 -smooth function with
the following properties:
(i) limu→+∞ ϑ(n) (u) = 0 for n = 0, 1, 2;
(ii) u + ϑ(u) is increasing on [0, +∞);
(iii) ϑ′′′ (u) = O(ϑ′′ (u)), ϑ(iv) (u)ϑ′ (u) = O(ϑ′′ (u)), supv⩾u |ϑ′′ (v)| = O(ϑ′′ (u)), and ϑ′′ (u + o(u)) =
O(ϑ′′ (u)).
Let us check (A1)–(A5). (A1) is clear. (A2′ ) and (A4′ ), which are equivalent to (A2) and, respectively, (A4)
by Remark 2.1, follow immediately from (i).
To check (A3), note that g(t) = t + ϑ̃(t), where ϑ̃(t) = −ϑ(u) with u = g(t). We further note that
g ′ (t) = 1 + ϑ̃′ (t) = f ′ 1(u) = 1+ϑ1′ (u) , which shows that limt→+∞ ϑ̃′ (t) = 0 and immediately implies (A3). It
10
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
remains to verify (A5). It is straightforward to compute
ϑ̃′′ (t) = −
ϑ′′ (u)
(1 + ϑ′ (u))4
and ϑ̃′′′ (t) =
4ϑ′′ (u)
ϑ′′′ (u)
−
,
(1 + ϑ′ (u))6
(1 + ϑ′ (u))5
which, by (iii), implies the first identity in (A5)-(a). Computing the fourth derivative of ϑ̃(t), it is
straightforward to obtain the second identity in (A5)-(a). The expression for ϑ̃′′ (t) implies (A5)-(b) if we
note that t + o(t) = f (u + o(u)).
2.3. Frequently used notation
For the reader’s convenience, we introduce a list of symbols that will be frequently used throughout the
paper:
θ
α
; θ̂ = β+1
.
θ = γ + 2 + β − α; α̂ = β+1
∗
∗
(uλ , λ) and (uλ∗ , λ ) denote regular and singular solutions to problem (Pλ ).
u and u∗ are rescaled, by (8), regular and singular solutions that solve (7).
g ′ , g ′′ , g ′′′ , g (iv) denote derivatives of the respective orders w.r.t. their arguments.
2.4. Singular solution in a neighborhood of the origin
One of the main results of this work is the representation of a singular solution announced in Theorem 1.1.
The statement of this theorem holds, however, under the more general assumptions (A1)–(A6).
Theorem 2.1. Assume (A1)–(A6). Then, the statement of Theorem 1.1 holds true.
Let Γ (t) = F −1 (t); namely, Γ (t) = g(ln t). The result of Theorem 2.1 can be rewritten as follows.
Corollary 2.1. Assume (A1)–(A6). Then, there exists λ∗ > 0 such that problem (Pλ ) possesses a singular
radial solution of the form
(
)
β+1
u∗λ∗ (r) = Γ (Ẑ(τ )) + O Γ ′ (τ )τ + Γ ′′ (τ )τ 2 , where τ = ∗ θ ,
λ r
[
]β+1
and Ẑ(τ ) = θβ+1 (α̂ − 1) τ β+2 Γ ′ (τ ) + (β + 1) ln{Γ ′ (τ )τ }(Γ ′ (τ ) + Γ ′′ (τ )τ )
.
Proof . Note that g ′ (ln τ ) = Γ ′ (τ )τ and g ′′ (ln τ ) = Γ ′ (τ )τ + Γ ′′ (τ )τ 2 . Therefore,
( {
}
)
g ln θβ+1 (α̂ − 1) + ln τ + (β + 1) ln{g ′ (ln τ ) + (β + 1)g ′′ (ln τ ) ln g ′ (ln τ )}
( {
})
= g ln θβ+1 (α̂ − 1)τ (g ′ (ln τ ) + (β + 1)g ′′ (ln τ ) ln g ′ (ln τ ))β+1
(
{
}β+1 )
= Γ θβ+1 (α̂ − 1) τ β+2 Γ ′ (τ ) + (β + 1) ln{Γ ′ (τ )τ }(Γ ′ (τ ) + Γ ′′ (τ )τ )
.
This completes the proof of the corollary. □
In Corollary 2.2, we obtain asymptotic representations for singular solutions when ef (u) is one of the
nonlinearities considered in [15,26,30]. The respective representations from [15,26,30] agree with ours when
L is the radial Laplacian.
Corollary 2.2. Assume (A6) and consider the following three particular cases of f : (a) f (u) = eu ;
(b) f (u) = up , (p > 21 ); (c) f (u) = u + ϑ(u), where ϑ(u) satisfies assumptions (i), (ii), (iii) from Example 3
in Section 2.2 and such that |ϑ(u)| + |ϑ′ (u)| ⩽ C0 e−δu . Then,
11
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
{
}
{
(
ln θ ln kr )}
1
α̂ − 1
= ln θ ln + ln
+ (β + 1) ln 1 + (β + 1)
(a)
1
r
ln kr
θ ln k
λ∗ /θ
(( 1r)−2 )
+ O ln
;
r
β+1
( r ) (
(1
)
1
k
(α − β − 1)θ p
=
θ
ln
+
(β
+
1)
−
1
ln ln + ln
(b) u∗λ∗
1
/
θ
r
p
r
pβ+1
λ∗
{
(1
)2 ln τ ( 1
) ln p }) p1
(( 1 ) p1 −2 )
+ (β + 1) ln 1 +
−1
−
−1
+ O ln
;
p
τ
p
τ
r
{
}
(c) u∗λ∗ (r) = ln θβ+1 (α − β − 1) − ln λ∗ − θ ln r + O(rθδ ).
u∗λ∗
( r )
1
Above, k = (1 + β) θ , τ = θ ln kr .
Proof . (a) and (b) follow immediately from representation (6). To show (c), we note that in a neigh′′′
(u)
= O(1). By L’Hopital’s rule and (i),
borhood of infinity, by (iii), ϑ′′ (u) either equals zero or ϑϑ′′ (u)
⏐ ϑ′′ (u) ⏐
′′
lim supu→+∞ ⏐ ln{1+ϑ′ (u)} ⏐ is finite, which means that |ϑ (u)| ⩽ C1 e−δu for some constant C1 > 0.
Furthermore, the representations for ϑ̃, ϑ̃′ , and ϑ̃′′ imply that in a neighborhood of infinity, |ϑ̃(t)| + |ϑ̃′ (t)| +
|ϑ̃′′ (t)| ⩽ C2 e−δt for some constant C2 > 0. Representation (c) follows now from (6). □
Remark 2.8. In example (b), we deal with the same nonlinear term as in [26]. We would like to emphasize
the following fact. In [26], the asymptotic representation for the singular solution (Theorem 1.1), in the case
0 < p < 1, is given up to a rest term (not computed explicitly) which does not tend to zero. Unlike [26],
our asymptotic representation includes additional explicit terms, tending to infinity, so our rest term is of a
higher order than in [26] and goes to zero.
The proof of Theorem 2.1 is divided into the steps outlined below.
2.4.1. Equivalent problem in a neighborhood of infinity
It is convenient to rewrite problem (Pλ ) by using the change of variable t = ln( κr ), where κ > 0 is a
constant to be specified later. First, we note that the equation in (Pλ ) can be rewritten as follows:
β
β
u′′ (r)|u′ (r)| + α̂
λ rγ−α ef (u)
u′ (r)|u′ (r)|
+
= 0.
r
(β + 1)
Let us show that for a singular solution to (Pλ ), it holds that (u∗λ∗ )′ (r) < 0 for all r > 0. Note that if β > 0,
from the above equation, it follows that a singular solution cannot have local maximum or minimum points,
since in those points u′ (r) = 0. Suppose β = 0. Then, every local extremum point, is necessarily a local
maximum point, which contradicts to the fact that limr→0+ u∗λ∗ (r) = +∞.
By doing the change of variable t = ln( κr ) and defining v(t) = u(r), we transform (Pλ ) to the following
problem:
{
θ −θt f (v)
v ′′ − (α̂ − 1)v ′ + λκ e e′ β = 0, t ∈ (ln κ, +∞),
(β+1)|v |
v(ln κ) = 0.
θ
κ
By choosing κ in such a way that λβ+1
= 1, we arrive at Eq. (11) with the boundary condition v(ln κ) = 0,
1
( β+1 )
where κ = λ θ . This problem is equivalent to (Pλ ), but given in a neighborhood of infinity.
Proposition 2.1. Assume (A1)–(A6). Then, there exists T > 0 such that on [T, +∞), Eq. (11) possesses
a singular solution of the form (12), where η(t) = O(g ′′ (θt)) is a C2 -function.
Remark 2.9. Remark that by Lemma 2.2 and Remark 2.3, limt→+∞
o(t).
12
ln g ′ (θt)
t
= 0. Therefore, (φ1 +φ2 )(t) =
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
Proof . Step 1. Transformation to a fixed-point problem. The idea is to obtain a semilinear equation with
respect to η by substituting v ∗ = g + η into (11) and showing that the difference of singular terms that occur
on the both sides of the resulting equation make a regular function. More specifically, we are going to show
that the original problem is equivalent to a couple semilinear equations with respect to η and ζ = η ′
{
η ′′ (t) − (α − β − 1) η ′ (t) + θ(α̂ − 1) η(t) = Φ(t, η, ζ),
(15)
ζ ′ (t) − (α − β − 1) ζ(t) = Φ(t, η, ζ) − θ(α̂ − 1) η(t),
where the function Φ is expected to be regular. For simplicity of notation, define
φ = φ1 + φ2 ,
where φ1 and φ2 are defined by (13). We start by transforming exp{−θt + f (v ∗ )} = exp{−θt + f (g(θt +
φ) + η)}, taking into account that g = f −1 . It holds that
f (g(θt + φ) + η) = θt + φ + f ′ (g(θt + φ) + ϑ(t, η)η)η,
where ϑ(t, η) takes values between 0 and 1. Let us transform the term f ′ (g(θt + φ) + ϑ(t, η)η). Define
( f (g(θt + φ) + ϑ(t, η)η )
)
ξ(t) = (θt + φ)
−1 .
θt + φ
Then, g(θt + φ) + ϑ(t, η)η = g(θt + φ + ξ). By (A2) and Remark 2.9, ξ = o(θt + φ) = o(θt + o(t)) = o(t).
Therefore,
η
η
f (g(θt + φ) + η) = θt + φ + ′
= θt + φ + ′
.
(16)
g (θt + φ + ξ)
g (θt + o(t))
Therefore,
exp{−θt + f (v ∗ )}
|v ∗ ′ (t)|
β
=
eφ1
|v ∗ ′ (t)|
η
β
e
φ2 + ′
g (θt+o(t))
.
(17)
Let us transform the first factor on the right-hand side. Defining ϑ1 (t) = θ−1 φ′ (t), we obtain
eφ1
β
|v ∗ ′ (t)|
=
(
)β
θβ g ′ (θt) + (β + 1)g ′′ (θt) ln g ′ (θt)
(18)
β
θβ |g ′ (θt + φ)(1 + ϑ1 ) + θ−1 η ′ |
1
1
1
=⏐
,
⏐β
η′
⏐1 + ′
⏐ (1 + ϑ1 )β (1 + ϑ2 )β
θg (θt+φ)(1+ϑ1 )
where ϑ2 is defined through the formula
g ′ (θt + φ) = (g ′ (θt) + (β + 1)g ′′ (θt) ln g ′ (θt))(1 + ϑ2 (t)).
Let us obtain expressions for ϑ1 and ϑ2 convenient for future computations. By (A4) and (A5),
′′
ϑ1 (t) = θ
−1
′
φ (t) = (β + 1)
=O
2
g ′′ (θt) + (β + 1)g ′′′ (θt) ln g ′ (θt) + (β + 1) gg′(θt)
(θt)
( g ′′ (θt) )
g ′ (θt)
g ′ (θt) + (β + 1)g ′′ (θt) ln g ′ (θt)
(19)
.
Since
φ(t) = (β + 1) ln g ′ (θt) + O(1),
13
(20)
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
by using the expansion of g ′ (θt + φ) by Taylor’s formula around θt, we obtain
g ′ (θt + φ) − g ′ (θt) − (β + 1)g ′′ (θt) ln g ′ (θt)
g ′ (θt) + (β + 1)g ′′ (θt) ln g ′ (θt)
∫1
( g ′′ (θt) )
g ′′ (θt)O(1) + 0 (1 − r)g ′′′ (θt + rφ)dr φ2
.
=
O
=
g ′ (θt) + (β + 1)g ′′ (θt) ln g ′ (θt)
g ′ (θt)
ϑ2 (t) =
(21)
The last identity holds by (A4), (A5), and the following argument. First of all, we note that by Taylor’s
formula around θt and (20),
(
( g ′′ (θt) )
( g ′′ (θt)
))
ln g ′ (θt + φ) = ln g ′ (θt) + ln 1 + O ′
+O ′
ln g ′ (θt) = ln g ′ (θt) + o(1),
g (θt)
g (θt)
which, by Remark 2.9, implies that
g ′′′ (θt + rφ)φ2 = O(g ′′ (θt + φ)) = O(g ′′ (θt)).
Note that since we are interested in solving Eq. (11) in a neighborhood of infinity, the expressions (1+ϑ1 )β
1
and (1+ϑ1 )β are well-defined.
2
Next, by the results of Section 2.4.1, for any singular solution to (Pλ ), it holds that (u∗λ∗ )′ (r) < 0 for all
r > 0. This implies that v ∗ ′ (t) > 0 for all t. Then, by Taylor’s formula, (18) can be transformed as follows:
2
)(
( g ′′ (θt) ))
(
η′
η′
eφ1
βη ′
′
+
o(1)
+
σ(t,
η
)
1
+
O
,
=
1
−
v ∗ ′ (t)β
θg ′ (θt)
g ′ (θt) g ′ (θt)2
g ′ (θt)
where
σ(t, η ′ ) = O(1)
∫
0
Recalling that
g ′′ (θt)
g ′ (θt)
1
(1 − r)
dr
(1 + rx)β+2
with x =
θg ′ (θt
η′
.
+ φ)(1 + ϑ1 )
(22)
→ 0 as t → +∞ (see Remark 2.3), the last expression can be rewritten as
2
( g ′′ (θt) )
η′
η′
eφ1
βη ′
′
+
o(1)
+
O(1)
σ(t,
η
)
+
O
.
=
1
−
′
v ∗ (t)β
θg ′ (θt)
g ′ (θt)
g ′ (θt)2
g ′ (θt)
(23)
Taking into account that eφ2 = θ(α̂ − 1)g ′ (θt)(1 + o(1)), by (17), (19), and (23), expression (17) can be
transformed as follows:
[
η
η
exp{−θt + f (v ∗ )}
eφ1
= e g′ (θt+o(t)) eφ2 ∗ ′ β = e g′ (θt+o(t)) eφ2 − β(α̂ − 1)η ′ + o(1)η ′
′
∗
β
v (t)
v (t)
′2
η
(
)]
η
φ2 + ′
g (θt+o(t)) − β(α̂ − 1)η ′ + F (t, η, η ′ ) + F (t, η)
+ O(1) ′
σ(t, η ′ ) + O g ′′ (θt) = e
1
2
g (θt)
(
)
φ2
η
η
e η
= eφ2 e g′ (θt+o(t)) − ′
−1 + ′
+ eφ2 − β(α̂ − 1)η ′
g (θt + o(t))
g (θt + o(t))
+ F1 (t, η, η ′ ) + F2 (t, η),
where
2
[
]
η
η′
F1 (t, η, η ′ ) = o(1)η ′ + O(1) ′
σ(t, η ′ ) e g′ (θt+o(t))
g (θt)
η
( ′ η
)
(
)
′
− β(α̂ − 1)η e g (θt+o(t)) − 1 + O g ′′ (θt) (e g′ (θt+o(t)) − 1);
(
)
F2 (t, η) = O g ′′ (θt) .
14
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
Now we are ready to evaluate differences of suitable singular terms that would result in regular functions.
More specifically, we aim to obtain (15) from (11) by substituting v ∗ (t) = g(θt + φ) + η(t) into (11). We
have
( g ′ (θt)(1 + o(1))
)
eφ2 η
−
θ(α̂
−
1)
η
=
θ(
α̂
−
1)
−
1
η.
(24)
g ′ (θt + o(t))
g ′ (θt + o(t))
Further, by the definition of φ2 , representation (20), and Taylor’s expansion for g ′ (θt + φ) around θt (in the
same way as we applied it in the numerator of (21)),
eφ2 − (α̂ − 1)
(
d
g(θt + φ) = θ(α̂ − 1) g ′ (θt) + (β + 1)g ′′ (θt) ln g ′ (θt)
dt
)
− g ′ (θt + φ)(1 + ϑ1 ) = O(g ′′ (θt)).
(25)
In (25), we used (A5) and representation (19) for ϑ1 . Next,
d2
g(θt + φ) = g ′′ (θt + φ)(θ + φ′ )2 + g ′ (θt + φ)φ′′ = O(g ′′ (θt)).
dt2
(26)
This holds by (A3), (A5), formula (19) for φ′ , and the following expression for φ′′ :
′′
3g ′′′ (θt)g ′′ (θt)
g ′ (θt)
g ′′ (θt) ln g ′ (θt)
g ′′′ (θt) + g (iv) (θt) ln g ′ (θt) +
2
φ (t) = θ (β + 1)
g ′ (θt) +
−
g ′′ (θt)3
g ′ (θt)2
− (φ′ )2 .
Thus, we obtain (15) with
(
)
Φ(t, η, ζ) = − F1 (t, η, ζ) + F2 (t, η) + F3 (t, η) + F4 (t) + F5 (t, η) ,
(27)
where η ′ is substituted by ζ on the right-hand side of the first equation of (15) and in the second equation.
Furthermore, F3 is defined by (24), F4 (t) = O(g ′′ (θt)) comes due to (25) and (26), and F5 is defined as
follows:
(
)
η
η
F5 (t, η) = eφ2 e g′ (θt+o(t)) − ′
−1
(28)
g (θt + o(t))
∫ 1
rη
g ′ (θt)O(1) 2
= θ(α̂ − 1) (1 − r) e g′ (θt+o(t)) dr ′
η .
g (θt + o(t))2
0
We will be searching for pairs (η, ζ) that solve (15) in a certain metric space which we denote by XT . If
g ′′ (t) is not an identical zero in a neighborhood of infinity, we define
XT = {x ∈ C[T, ∞) : x(t) = O(g ′′ (θt)) as t → +∞}.
Note that under (A5)-(b), either g ′′ (θt) does not take zero values in a neighborhood of infinity (case (i)), or
identically equals zero in a neighborhood of infinity (case (ii)). In the case (i), for sufficiently large T , the
following norm in XT is well-defined:
∥x∥XT = sup{|g ′′ (θt)|
−1
|x(t)|}.
t⩾T
In the case (ii), we define XT = Cb ([T, +∞)) with the supremum norm. Here, Cb stands for the space of
bounded continuous functions.
In XT , (15) uniquely transforms to the following couple of fixed-point equations:
η = Ψ (η, ζ),
ζ = Ψ̂ (η, ζ).
Since α̂ > 1, the explicit formula for Ψ̂ is the following:
∫ +∞
(
)
Ψ̂ (η, ζ)(t) =
e2δ(t−s) Φ(s, η, ζ) − θ(α̂ − 1) η(s) ds,
t
15
(29)
(30)
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
where δ = α−β−1
. The explicit formula for Ψ depends on the sign of the discriminant D = 4(δ 2 − θ(α̂ − 1))
2
of the corresponding characteristic equation. In the case D < 0, we have
√
∫ +∞
(
)
|D|
1
δ(t−s)
e
sin µ(s − t) Φ(s, η, ζ) ds, where µ =
Ψ (η, ζ)(t) =
.
(31)
µ t
2
In the case D > 0, we have
Ψ (η, ζ)(t) =
∫
1
2µ
+∞
(
)
e(δ−µ)(t−s) 1 − e2µ(t−s) Φ(s, η, ζ) ds.
(32)
t
Finally, if D = 0, we have
∫
+∞
eδ(t−s) (s − t)Φ(s, η, ζ) ds.
Ψ (η, ζ)(t) =
(33)
t
Step 2. Existence of a fixed point. Let us show that we can find constants M, M̂ > 0 and a number T > 0
such that the map (Ψ , Ψ̂ ) acts from Σ to Σ , where the latter is defined as a complete metric space of the
following form:
Σ = Σ (M, M̂ , T ) = {(η, ζ) ∈ XT × XT : ∥η∥XT ⩽ M and ∥ζ∥XT ⩽ M̂ }.
We define the metric in Σ as follows:
ρ{(η1 , ζ1 ), (η2 , ζ2 )} = ∥η1 − η2 ∥XT + ∥ζ1 − ζ2 ∥XT .
Moreover, we show that T can be chosen in such a way that (Ψ , Ψ̂ ) is a contraction map Σ → Σ .
The integral operator applied to Φ and defined by (31), (32), or (33) splits into five terms (which we
denote by I1 , I2 , I3 , I4 , and I5 ), corresponding to the decomposition of Φ by formula (27). We evaluate each
of these terms. First, we compute the integrals
∫ +∞
∫ +∞
∫ +∞
δ(t−s)
(δ−µ)(t−s)
1
1
1
1
;
;
e
ds
=
e
ds
=
eδ(t−s) (s − t)ds = δ12 ;
µ
µδ
2µ
2µ(δ−µ)
t
t
t
and define
κ = max
{
1
1
1
1
2δ , 2µ(δ−µ) , µδ , δ 2
}
.
1
which appears from the
The expression for κ involves the values of the above integrals and the term 2δ
2δ(t−s)
′′
integration of e
in (30). We make analysis separately in the case (i), when g (θt) ̸= 0 in a neighborhood
of infinity, and in the case (ii), when g ′′ (θt) identically equals zero in a neighborhood of infinity. We start
from the case (i). Introduce constants M1 , M2 , and a number T1 > 0 such that
sup |g ′′ (θs)| ⩽ M1 |g ′′ (θt)|,
s⩾t
|Fi (t)| ⩽ M2 |g ′′ (θt)|,
t ⩾ T1 ;
i = 2, 4,
(34)
t ⩾ T1 .
M1
Let us fix the metric space for the fixed point argument. Take M > 5κM1 M2 and M̂ = θM
1+β + M . Choose
T > T1 in such a way that if the a priori estimates |η(t)| ⩽ M |g ′′ (θt)| and |ζ(t)| < M̂ |g ′′ (θt)| hold, then,
|Fi (t)| ⩽ M2 |g ′′ (θt)|,
i = 1, 3, 5,
t ⩾ T.
We can always do this since for η and ζ with the above property, F1 , F3 , and F5 are o(g ′′ (t)). Now take
(η, ζ) ∈ Σ (M, M̂ , T ) and note that each of the terms Ii , i = 1, 2, 3, 4, 5, becomes smaller than κM1 M2 ,
16
J.M. do Ó, E. Shamarova and E. da Silva
which is smaller than
Ψ1 + Ψ2 , where
M
5 .
Nonlinear Analysis 222 (2022) 113000
Thus, we have proved that |Ψ (η, ζ)(t)| ⩽ M |g ′′ (t)|. Next, we represent Ψ̂ as
∫
+∞
e2δ(t−s) Φ(s, η(s), ζ(s))ds,
∫ +∞
Ψ2 (η)(t) = −θ(α̂ − 1)
e2δ(t−s) η(s)ds.
Ψ1 (η, ζ)(t) =
t
(35)
t
M1 ′′
From the above argument, we obtain that |Ψ1 (η, ζ)(t)| ⩽ M |g ′′ (t)|. Furthermore, |Ψ2 (η)(t)| ⩽ θM
1+β |g (t)|.
Therefore, by the choice of M̂ , we obtain that |Ψ̂ (η, η ′ )(t)| ⩽ M̂ |g ′′ (t)|. Thus, we proved that Ψ (η, ζ) is a
map Σ → Σ .
Let us show that if the number T is sufficiently large, then Ψ : Σ → Σ is a contraction map. Let χ(s − t)
denote one of the factors sin{µ(s − t)}, 1 − e2µ(t−s) , or s − t which occur in the integrals (31), (32), and (33).
By (A5)-(b) and the integrability of the exponential factor, multiplied by χ, it suffices to observe that for
two elements (η1 , ζ1 ) and (η2 , ζ2 ) from Σ , and for each t ∈ (T, +∞),
|F1 (t, η1 , ζ1 ) − F1 (t, η2 , ζ2 )| ⩽ ϵ(t)(|η1 − η2 | + |ζ1 − ζ2 |),
|Fi (t, η1 ) − Fi (t, η2 )| ⩽ ϵ(t)|η1 − η2 |,
i = 3, 5,
(36)
where ϵ it is a positive function tending to 0 as t → +∞. Inequalities (36) follow immediately from the
expressions for Fi , i = 1, 3, 5, if we take into account the estimates |ηi (t)| ⩽ M |g ′′ (t)|, |ζi (t)| ⩽ M̂ |g ′′ (t)|,
i = 1, 2, Remark 2.3, and assumption (A3). We also observe that, by (22), σ1 (t, ζ) is Lipschitz in its second
argument for sufficiently large t. By (34) and (36),
sup |g ′′ (θt)|
−1
|Ψ (η1 , ζ1 )(t) − Ψ (η2 , ζ2 )(t)|
t⩾T
+∞
∫
{
−1
′′
⩽ 5 sup |g (θt)|
t⩾T
∫
t
+∞
⩽ 5M1 sup ϵ(t)
t⩾T
(
) }
eδ(t−s) χ(t − s)ϵ(s) |η1 (s) − η2 (s)| + |ζ1 (s) − ζ2 (s)| ds
−1 (
eδ(t−s) χ(t − s)|g ′′ (θs)|
)
|η1 (s) − η2 (s)| + |ζ1 (s) − ζ2 (s)| ds.
T
Thus, by choosing T sufficiently large, we can make the factor in front of the integral small, so that there
exists r > 0 sufficiently small (whose exact bound will be specified later) such that
∥Ψ (η1 , ζ1 ) − Ψ (η2 , ζ2 )∥XT ⩽ r(∥η1 − η2 ∥XT + ∥ζ1 − ζ2 ∥XT ).
(37)
By (30),
∥Ψ̂ (η1 , ζ1 ) − Ψ̂ (η2 , ζ2 )∥XT ⩽ ∥Ψ1 (η1 , ζ1 ) − Ψ1 (η2 , ζ2 )∥XT + ∥Ψ2 (η1 ) − Ψ2 (η2 )∥XT ,
where Ψ1 and Ψ2 are given by (35). The first term in the above identity is evaluated exactly as in (37). For
the second term, we have
∥Ψ2 (η1 ) − Ψ2 (η2 )∥XT ⩽ R∥η1 − η2 ∥XT
with R =
θM1
1+β
(recall that 2δ = α − β − 1). Therefore,
∥Ψ̂ (η1 , ζ1 ) − Ψ̂ (η2 , ζ2 )∥XT ⩽ (R + r)∥η1 − η2 ∥XT + r∥ζ1 − ζ2 ∥XT .
Now let (Ψ (2) , Ψ̂ (2) ) = (Ψ , Ψ̂ )◦(Ψ , Ψ̂ ). We show that one can find a number r ∈ (0, 1) such that (Ψ (2) , Ψ̂ (2) )
is a contraction map. We have
∥Ψ (2) (η1 , ζ1 ) − Ψ (2) (η2 , ζ2 )∥XT ⩽ r(2r + R)∥η1 − η2 ∥XT + 2r2 ∥ζ1 − ζ2 ∥XT ,
∥Ψ̂ (2) (η1 , ζ1 ) − Ψ̂ (2) (η2 , ζ2 )∥XT ⩽ 2r(r + R)∥η1 − η2 ∥XT + r(2r + R)∥ζ1 − ζ2 ∥XT .
17
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
Therefore,
∥Ψ (2) (η1 , ζ1 ) − Ψ (2) (η2 , ζ2 )∥XT + ∥Ψ̂ (2) (η1 , ζ1 ) − Ψ̂ (2) (η2 , ζ2 )∥XT
⩽ (4r2 + 3Rr)(∥η1 − η2 ∥XT + ∥ζ1 − ζ2 ∥XT ).
Picking r > 0 in such a way that 4r2 + 3Rr < 1, we obtain that (Ψ (2) , Ψ̂ (2) ) is a contraction map Σ → Σ .
Let (η, ζ) be its unique fixed point. Then, (η, ζ) is also a fixed point of (Ψ , Ψ̂ ), while η determines solution
(12) with the prescribed singular part given by the first term. To see this, it suffices to note that the solution
(η, ζ), that we just found, satisfies the property η ′ = ζ. Indeed, ζ̄ = η ′ also satisfies the second equation in
(15) which implies that
η ′ (t) = Ψ̂ (η, ζ)(t) + Ce(α−β−1)t ,
where C is a constant. On the other hand, (31), (32), or (33) imply that
∫
}
1 +∞ d { δ(t−s)
e
χ(t − s) Φ(s, η, ζ) ds,
η′ =
γ t
dt
which implies that η ′ (t) = o(1). Therefore, C = 0 and ζ = η ′ .
Thus, we obtained a singular solution v ∗ (t) = g(θt + φ(t)) + η(t) to (11).
Consider now the case (ii), i.e., g ′′ (θt) = 0 on [T̂ , +∞) for some T̂ > 0. In this case, the terms F2 and
F4 are equal to zero. The non-zero terms are F1 , F3 , and F5 . Since F1 (t, 0, 0) = F3 (t, 0, 0) = F5 (t, 0, 0) = 0,
η = 0 is a solution to
η ′′ (t) − (α − β − 1) η ′ (t) + θ(α̂ − 1) η(t) = Φ(t, η, η ′ ).
Hence, in the case (ii), v ∗ (t) = g(θt + φ(t)) is an exact singular solution to (11). □
Now we extend the singular solution to (11), that we constructed in Proposition 2.1, to a solution to the
same equation on the entire real line.
Proposition 2.2. In the assumptions of Proposition 2.1, the solution v ∗ can be extended to a solution
of (11) on (−∞, +∞). Moreover, the extended solution v ∗ is strictly increasing and there exists a number
T ∗ ∈ (−∞, T ) such that v ∗ (T ∗ ) = 0.
Proof . By Proposition 2.1, there exists a singular solution v ∗ to (11) on some interval [T, +∞). Eq. (11)
can be rewritten as follows:
( (β+1−α)t ′ β ′ )′
e
|v (t)| v (t) = −(β + 1)e−(γ+1)t ef (v) .
(38)
Substituting v ∗ into (38) and integrating it from t to +∞, we obtain
∫ +∞
∗
β
e(β+1−α)t |v ∗ ′ (t)| v ∗ ′ (t) = (β + 1)
e−(γ+1)s ef (v ) ds.
(39)
t
Indeed, since limt→+∞ g ′′ (t) = 0, we conclude that g ′ (t) has at most linear growth. Furthermore, since
( ′′ (θt) )
(see formula (19)) and η ′ = O(g ′′ (θt)),
v ∗ ′ (t) = g ′ (θt + φ)(θ + φ′ ) + η ′ , where φ = o(t), φ′ = O gg′ (θt)
∗′
we obtain that v (t) also has at most linear growth on [T, +∞). Since β + 1 − α < 0, the expression on
left-hand side of (39) goes to zero as t → +∞, and hence, (39) holds. Recall that v ∗ ′ (t) > 0 (as we showed
in the proof of Proposition 2.1) for any extension of v ∗ . Let w(t) = e(β+1−α)t v ∗ ′ (t)β+1 . We then obtain the
following system with respect to w and v ∗ :
{
∗
w′ (t) = −(β + 1)e−(γ+1)t ef (v ) ,
(40)
1
v ∗ ′ (t) = e(α̂−1)t w β+1 .
18
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
Pick S > T and define c0 = v ∗ (S), c1 = w(S). Furthermore, we note that on (−∞, S], c1 ⩽ w(t) ⩽
c1 + c2 e−(γ+1)t = ct for some constant c2 > 0. Define ν = 1(−∞,c0 +ε] ∗ ρε and ςt = 1[c1 −ε,ct +ε] ∗ ρε , where
ρε , ε < c1 , is a standard mollifier supported on the ball of radius ε. Note that ν( · ) = 1 on (−∞, c0 ] and
ςt ( · ) = 1 on [c1 , ct ]. Instead of (40), consider the system of first-order ODEs on (−∞, S]
{
∗
∗
w′ (t) = −(β + 1)e−(γ+1)t eν(v )f (v ) ,
(41)
1
v ∗ ′ (t) = e(α̂−1)t (ςt (w)w) β+1 .
On [T, S], (v ∗ (t), w(t)) is also a solution to (41), since over this interval ν(v ∗ (t)) = 1 and ςt (w(t)) = 1.
1
1
Next, since eν(v)f (v) ⩽ ef (c0 +2ε) and 0 < (ςt (w)w) β+1 ⩽ (ct + 2ε) β+1 , one can extend (v ∗ , w) to (−∞, S]
(see, e.g., [12], Chapter 2, §6), obtaining by this a solution (ṽ, w̃) to (41) which coincides with (v ∗ , w) on
[T, S]. Since on (−∞, S], ν(ṽ) = 1 and ςt (w̃) = 1, (ṽ, w̃) is also a solution to (38), and therefore, to (11). We
then use the same notation for this extended solution, i.e., we write (v ∗ , w) instead of (ṽ, w̃). There are two
possible situations: either v ∗ is strictly positive over (−∞, S], or there exists T ∗ ∈ R such that v ∗ > 0 on
(T ∗ , S] and v ∗ (T ∗ ) = 0. Let us show that the first situation cannot be realized. If v ∗ is strictly positive over
(−∞, S], then there exists a finite limit L = limt→−∞ v ∗ (t). Integrating (38) from −R to 0 (where R > 0 is
∗
sufficiently large) and taking into account that ef (v ) ⩾ ef (L) , we obtain that
)
β + 1 f (L) ( θR
e
e − e−(α−β−1)R
γ+1
∫0
′
which shows that v ∗ (−R) → +∞ as R → +∞. The latter implies that v ∗ (−R) = v ∗ (0)− −R v ∗ ′ (t)dt → −∞
as R → +∞. This contradicts to the fact that limt→−∞ v ∗ (t) = L < +∞.
Thus, we conclude that solution (12) can be extended to (−∞, +∞) in such a way that there exists a
finite number T ∗ such that v ∗ (T ∗ ) = 0. □
v ∗ ′ (−R)β+1 ⩾ e−(α−β−1)R v ∗ ′ (0)β+1 +
2.4.2. Proof of Theorem 2.1
The proof of Theorem 2.1 now follows from the inverse change of variable.
)1
(
∗
θ
.
Proof of Theorem 2.1. Let T ∗ be as in Proposition 2.2. Define λ∗ = (β + 1)e−θT and κ = β+1
λ∗
∗
∗
∗
∗
Further define uλ∗ (r) = v (ln(κ/r)). By the argument in paragraph 2.4.1, (uλ∗ , λ ) is a singular solution to
problem (Pλ ). □
2.5. Exact singular solution to the Gelfand problem for a k-Hessian
Consider the problem
{
−L(u) = λ eu
u=0
in B,
on ∂B,
(42)
where L can be a k-Hessian with k < d2 , a p-Laplacian with p < d, or, in general, the operator (5) with α, β, γ
satisfying (A6). Here, we prove Corollary 1.1, stating the exact form of the singular solution constructed in
Theorem 2.1. Furthermore, we “decode” the formula for a singular solution announced in Corollary 1.1 for
the cases of a k-Hessian and a p-Laplacian. Namely, we have the following corollary.
Corollary 2.3. Assume (A6) and consider problem (42) for the case when L is a k-Hessian with k <
Then, the singular solution constructed in Theorem 2.1 takes the form
u∗λ∗ (x) = −2k ln |x|,
λ∗ = (2k)k (d − 2k).
d
2.
(43)
If L is a p-Laplacian with p < d, then, the aforementioned singular solution is
u∗λ∗ (x) = −p ln |x|,
λ∗ = pp−1 (d − p).
19
(44)
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
Proof of Corollaries 1.1 and 2.3. Note that for problem (42), f (t) = g(t) = t. Therefore, g ′ (t) = 1,
g ′′ (t) = 0. By Proposition 2.1, in a neighborhood of +∞, the solution (12) takes the form
v ∗ (t) = θt + ln{θβ+1 (α̂ − 1)}.
(45)
It is straightforward to check that (45) verifies (11) with f (v) = v on the entire real line. Furthermore,
β+1
v(t) = 0 if t = T ∗ = − ln{θ θ(α̂−1)} . Therefore,
∗
λ∗ = (β + 1)e−θT = (β + 1)θβ+1 (α̂ − 1) = θβ+1 (α − β − 1).
)1 1 }
{(
1
β+1
θ
Recall that u∗λ∗ (x) = v ∗ (t) with t = ln β+1
(α̂ − 1)} − ln |x|. Comparing it with (45),
λ∗
|x| = − θ ln{θ
∗
we obtain that uλ∗ (x) = −θ ln |x|.
In the case of a k-Hessian with k < d2 , we have θ = 2k, β + 1 = k, α − β − 1 = d − 2k, which implies (43).
In the case of a p-Laplacian with p < d, we have θ = p, β + 1 = p − 1, α − β − 1 = d − p, which implies
(44). □
3. Regular solutions to problem (Pλ )
Here, we will be interested in regular solutions to problem (Pλ ). For regular solutions, problem (Pλ ) should
be stated as follows:
{
−L(uλ (r)) = λ ef (uλ (r)) ,
(Pλ′ )
u′λ (0) = uλ (1) = 0.
Indeed, from Lemma 3.1, it follows that any solution to (Pλ ) is decreasing in r ∈ [0, +∞) and achieves the
maximum at r = 0. Moreover, since the actual solution uλ should be regular in a ball, we add the condition
u′λ (0) = 0.
Lemma 3.1.
Let (uλ , λ) be a regular radial solution to (Pλ′ ). Then, u′λ (r) < 0 on (0, +∞).
Proof . The equation in (Pλ′ ) implies
r
α
β
|u′λ (r)| u′λ (r)
∫
r
=−
sγ ef (uλ (s)) ds + C.
0
Since u′λ (0) = 0, we obtain that C = 0. This implies that u′ (r) < 0 for all r > 0. □
Remark 3.1.
If γ + 1 > α (as in the case of the k-Hessian), then in (Pλ′ ), we do not need to state that u′ (0) = 0 if we
are interested in a regular solution. This assumption is fulfilled automatically. Indeed, by L’Hopital’s rule,
(−u′λ (0))β+1 = lim α−1 rγ+1−α ef (uλ (r)) = 0.
r→0
3.1. Standing assumptions: regular solutions
(B1) f ∈ C1 (R, R), f ′ > 0, and limu→+∞ f (u) = +∞.
{ (u) }
(B2) limu→+∞ u−1 exp fβ+1
= limu→+∞ uf ′ (u) = +∞.
(B3) α > β + 1 > 0; θ > 0.
Remark 3.2. Remark that the second limit in (B2) is implied by (A4) and the first condition of (A2)
(limt→+∞ g ′′ (t) = 0). We have
lim uf ′ (u) = lim
u→+∞
t→+∞
g(t)
g(t)
⩾ lim ∫
= +∞.
g ′ (t) t→+∞ t |g ′′ (s)|ds + g ′ (0)
0
20
(46)
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
∫t
Indeed, we know that limt→+∞ g(t) = +∞. If limt→+∞ 0 |g ′′ (s)|ds < +∞, then (46) is straightforward.
∫t
Otherwise, if limt→+∞ 0 |g ′′ (s)|ds = +∞, then (46) follows from L’Hopital’s rule and Remark 2.3. The first
expression in (B2) tends to +∞ by L’Hopital’s rule and Remark 2.5.
3.2. Existence and uniqueness of a regular solution to (Pλ′ )
First of all, we note that by the change of variable (8), problem (Pλ′ ) can be transformed to
{
−L(u) = ef (u) ,
u(0) = ρ, u′ (0) = 0.
(Pρ )
Above, ρ and λ are connected through the identity
∫ 1
(∫ t
) 1
1
α
β+1
− β+1
β+1
dt,
ρ=λ
t
sγ ef (uλ (s)) ds
0
0
where (uλ , λ) is the solution to (Pλ′ ). Note that problem (Pρ ) is equivalent to the integral equation
∫
r
u(r) = ρ −
α
− β+1
t
(∫
0
t
) 1
β+1
sγ ef (u(s)) ds
dt.
(47)
0
We have the following result on the existence and uniqueness of solution to (47).
Lemma 3.2.
Assume (B1) and (B3). Then, Eq. (47) has a unique solution.
Proof . Let the map Γ (u) be given by the right-hand side of (47). Fix an arbitrary interval [0, T ]. Note that
for any solution u of (47), on [0, T ], it holds that
f (ρ)
u ⩽ ρ and u ⩾ ρ − e 1+β (γ + 1)
1
− β+1
−1
θ̂
T θ̂ .
Let Aρ,T denote the constant on the right-hand side of the second estimate. Introduce the complete metric
space
Σ = {u ∈ C[0, T ] : Aρ,T ⩽ u ⩽ ρ}
with the supremum norm as the metric, and take two functions u1 and u2 from Σ . It is easy to see that
Γ (u1 ) and Γ (u2 ) are functions Σ → Σ . Furthermore,
Γ (u1 )(r) − Γ (u2 )(r) =
∫ r
(∫ 1
)
β
1
− α
−
t β+1
Λ(t, ρ, λ) β+1 dλ
β+1 0
0
∫ t
sγ (ef (u2 (s)) − ef (u1 (s)) )ds dt,
0
where Λ(t, ρ, λ) =
∫t
0
sγ (λef (u1 (s)) + (1 − λ)ef (u2 (s)) )ds, and we have that
ef (Aρ,T ) (γ + 1)−1 tγ+1 ⩽ Λ(t, ρ, λ) ⩽ ef (ρ) (γ + 1)−1 tγ+1 .
Taking into account that |u1 | and |u2 | are bounded, we obtain that there exists a constant K =
K(ρ, T, γ, β, f ) such that
∫ r −α+γ+1
sup |Γ (u1 ) − Γ (u2 )| ⩽ K
t β+1 sup |u1 − u2 | ds.
[0,r]
0
21
[0,t]
J.M. do Ó, E. Shamarova and E. da Silva
Since
−α+γ+1
β+1
Nonlinear Analysis 222 (2022) 113000
= θ̂ − 1, we obtain that for the map Γ (n) = Γ
 ◦ Γ ◦· · · ◦ Γ, it holds that
n
sup |Γ (n) (u1 ) − Γ (n) (u2 )| ⩽
[0,r]
(Krθ̂ θ̂−1 )n
sup |u1 − u2 | ds.
n!
[0,r]
This implies that Γ (n) is a contraction for some n. Clearly, the unique fixed point of Γ (n) is also the unique
fixed point of Γ , the solution to (47). □
3.3. Properties of regular solutions
We start by deriving a version of Pohozaev’s identity suitable for our applications.
Lemma 3.3. Let f ∈ C(R, R), a ∈ R be an arbitrary constant, and u ∈ C2 [0, +∞) be a solution to problem
(Pρ ). Then, for all r > 0,
∫ u
{
}
(
α − β − 1)
β+2
+ rγ (γ + 1)
ef (t) dt − a u ef (u)
(48)
rα |u′ |
a−
β+2
0
∫ u
)}
d { α+1 ( β + 2 ′ β+2
β
=
r
|u |
+ rγ−α
ef (t) dt + ar−1 |u′ | u′ u .
dr
β+1
0
Proof . In [24], a version of Pohozaev’s identity was obtained for solutions to equations whose particular
type is
β
r−(d−1) (r(d−1) |u′ (r)| u′ (r))′ + h(r, u) = 0,
(49)
where d is the space dimension and h ∈ C(R+ × R). As it follows from the proof of Theorem 2.1, exposed in
paragraph 2 of [24], instead of d − 1, one can use any number α > 0. More specifically, the fact that d − 1
is an integer which differs from the space dimension by one, does not play any role in the computation on
p. 545 of [24]. Multiplying the both parts of the equation in (Pρ ) by rα−γ , we bring this equation to a form
similar to (49):
β
r−α (rα |u′ (r)| u′ (r))′ + rγ−α ef (u) = 0.
We then apply Theorem 2.1 from [24] to the above equation, substituting d by α + 1 and taking into
account that u′ (0) = 0. Referring to the notation of the aforementioned theorem, we have A(ρ) = ρβ and
E(ρ) = (β + 1)ρβ . It is then straightforward to verify the assumptions of Theorem 2.1 from [24] as well as
to make necessary computations leading to Eq. (48). □
Let u( · , ρ) be the solution to problem (Pρ ). Our next result, given by Proposition 3.1, is about upper
and lower bounds for the number
R(B, ρ) = u( · , ρ)−1 (B),
0 ⩽ B ⩽ ρ.
(50)
Proposition 3.1. Assume (B1)–(B3). Then, for each B ⩾ 0, the number R(B, ρ) is well-defined. Moreover,
there exist two positive constants R(B) and R̄(B), and a number ρ0 (B) > B, all depending only on B, such
that for all ρ ⩾ ρ0 (B),
R(B) ⩽ R(B, ρ) ⩽ R̄(B).
Moreover, the second inequality holds for all ρ ⩾ B.
22
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
Proof . Fix a number B ⩾ 0. We start by showing that R(B, ρ) is well-defined for ρ > B. This is equivalent
to the fact that a zero of u(r, ρ) − B exists. Suppose this is not the case. Since u(r, ρ) is decreasing, then
f (B)
− 1
u(r, ρ) > B for all r ⩾ 0. From (Pρ ), we obtain that u(r, ρ) ⩽ ρ−e 1+β (γ +1) β+1 θ̂−1 rθ̂ → −∞, as r → +∞.
This is a contradiction. Hence, there exists a unique r such that u(r, ρ) = B.
Let us prove that the function [B, +∞) → R, ρ ↦→ R(B, ρ) is bounded from above. Rescale problem (Pρ )
1
by setting λ θ = R(B, ρ) as follows:
{
1
1
−L(u(λ θ r)) = λ ef (u(λ θ r)) ,
u(0) = ρ, u′ (0) = 0
which is equivalent to
∫
1
1
u(λ θ r) = B +
1
λ β+1 t
(∫
α
− β+1
r
t
) 1
1
β+1
dt.
sγ ef (u(λ θ s)) ds
0
For r ∈ (0, 1), it holds that
1
1
1
∫
u(λ θ r) − B ⩾ λ β+1
t
α
− β+1
dt
r
(∫
r
) 1
1
β+1
sγ ef (u(λ θ r)) ds
.
0
This implies that for r ∈ (0, 1), we have the estimate
f (u)
{
sup (u − B) e
− β+1 }
1
θ
⩾ (u(λ r) − B)e
−
1
f (u(λ θ r))
β+1
1
θ
γ+1
⩾ K1 λ β+1 (r β+1 − r β+1 ),
(51)
u⩾B
−
f (u)
where K1 > 0 is a constant. By (B2), 0 < supu⩾B (u − B) e β+1 < +∞. Hence, the right-hand side of (51)
is bounded uniformly in r ∈ (0, 1). Evaluating it (say) at r = 21 , we obtain that for all B ⩾ 0 and ρ ⩾ B,
f (u) } β+1
{
1
−
R(B, ρ) = λ θ ⩽ K2 sup (u − B) e β+1 θ = R̄(B),
(52)
u⩾B
where K2 is a constant. Now we prove the existence of a lower bound R(B) by using identity (48). By
L’Hopital’s rule and (B2),
u ef (u)
ef (u) + u ef (u) f ′ (u)
lim ∫ u f (t) = lim
= 1 + lim uf ′ (u) = +∞.
f (u)
u→+∞
u→+∞
u→+∞
e
e
dt
0
As we show below, it suffices to prove the existence of R(B) for those B > 0 which make the following
inequality hold:
∫ u
(γ + 1)
ef (t) dt < au ef (u) for all u ⩾ B,
(53)
0
where a ∈ R is to be fixed later. Indeed, suppose the existence of R(B) is obtained for those B which
make inequality (53) fulfilled for u ⩾ B. If B ⩾ 0 is arbitrary, we find a number B̃ > B such that (53)
holds for u ⩾ B̃. Then, R(B̃) ⩽ R(B̃, ρ) ⩽ R(B, ρ) for ρ ⩾ ρ0 (B̃), and we can define R(B) = R(B̃) and
ρ0 (B) = ρ0 (B̃). Thus, without loss of generality, we assume that B is sufficiently large so that (53) holds.
The number a is chosen as follows. Note that for any two positive numbers x and y, it holds that
β
1
1
1
1
1
1
−
(x + y) 1+β ⩽ x 1+β + y 1+β if β ⩾ 0 and (x + y) 1+β ⩽ 2 β+1 (x 1+β + y 1+β ) if −1 < β < 0. We set
β
− β+1
Aβ = max{1, 2
} and fix â ∈ (0, A−1
β −
a(β+1)2
(β+2)(α−β−1)
a(β+1)2
(β+2)(α−β−1) ).
The number a > 0 is then has to be chosen in
A−1
β .
such a way that
<
In addition, we choose a to satisfy a < α−β−1
β+2 so that the first term
on the left-hand side of (48) is negative. By (48) and (53), on (0, R̄(B)], it holds that
β + 2 ′ β+2
β
|u |
+ ar−1 |u′ | u′ u < 0.
β+1
23
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
Since u′ < 0, for r ∈ (0, R̄(B)],
r |u′ (r)| < a
β+1
u(r).
β+2
(54)
Next, we define
(
D = B 1 − âAβ −
f (D) 1
1
Aβ a(β + 1)2 )−1
−
and R(B) = (â D θ̂(γ + 1) β+1 e β+1 ) θ̂ .
(β + 2)(α − β − 1)
We aim to prove that R(B, ρ) ⩾ R(B) for all ρ ⩾ D. If, for some ρ ⩾ D, R(D, ρ) ⩾ R(B), then (since D > B)
R(B, ρ) > R(B). Otherwise, if R(D, ρ) < R(B), consider the following problem on [R(D, ρ), R(B)]:
L(v) = −ef (D) ,
v(R(D, ρ)) = D,
v ′ (R(D, ρ)) = u′ (R(D, ρ), ρ).
To simplify notation, define D̃ = ef (D) and R̃ = R(D, ρ). Since
∫ r
β+1
β
rα |v ′ | v ′ = −D̃
sγ ds − R̃α |u′ (R̃)|
,
(55)
(56)
R̃
we obtain that on [R̃, R(B)], v ′ (r) < 0. Taking into account this observation, by (54) and (56), for
r ∈ [R̃, R(B)], we obtain
∫ r(
) 1
D̃
β+1 β+1 −α̂
s ds
v(r) = D −
(sγ+1 − R̃γ+1 ) + R̃α |u′ (R̃)|
R̃ γ + 1
[( D̃ ) 1 ∫ r γ+1
R̃α̂ |u′ (R̃)|r−α̂+1
R̃ |u′ (R̃)| ]
β+1
− α
⩾ D − Aβ
s β+1 s β+1 ds +
−
γ+1
α̂ − 1
α̂ − 1
R̃
( D̃ ) 1 R(B)θ̂
Aβ a(β + 1)D
β+1
= B,
> D − Aβ
−
γ+1
(β + 2)(α̂ − 1)
θ̂
where the last identity follows from the definition of R(B) and D.
It remains to show that on [R̃, R(B)], u(r, ρ) ⩾ v(r). Indeed, by (55), for r ∈ [R̃, R(B)], we have
(rα |v ′ |
α
β+1 ′
) = rγ ef (D) ⩾ rγ ef (u) = (rα |u′ |
′ β+1
r |v |
′ β+1
α
⩾ r |u |
β+1 ′
) and
.
Since the derivatives u′ and v ′ are negative, again by (55), we obtain that on [R̃, R(B)], u(r, ρ) ⩾ v(r) ⩾ B.
Therefore, R(B, ρ) ⩾ R(B), and we can take ρ0 (B) = D. The proof is now complete. □
Proposition 3.2. Assume (B1)–(B3). Then, for any interval [r1 , r2 ] ⊂ (0, ∞), the family {u( · , ρ)}ρ⩾0 is
relatively compact in C([r1 , r2 ]). Furthermore, each limit point of this family is a solution to (7), regular or
singular.
Proof . Fix an interval [r1 , r2 ] ⊂ (0, +∞). Let B > 0 and let R(B), R̄(B), and ρ0 (B) be as they were defined
in Proposition 3.1, so we have that R(B, ρ) ∈ [R(B), R̄(B)] for all ρ > ρ0 (B), where R(B, ρ) is defined by
(50). By (B2), limB→+∞ R̄(B) = 0. Choose a number B > 0 such that r1 > R̄(B), and note that on [r1 , r2 ],
u(r, ρ) < B for all ρ ⩾ B. By (47), it holds that for r ⩾ R(B, ρ),
∫ r
(∫ t
) 1
β+1
− α
u(r, ρ) = B −
t β+1
sγ ef (u(s,ρ)) ds
dt
(57)
R(B,ρ)
⩾B−e
f (B)
β+1
∫
0
r
t
R(B,ρ)
α
− β+1
(∫
t
) 1
f (B)
β+1
− 1
sγ ds
dt ⩾ B − e β+1 (γ + 1) β+1 θ̂−1 rθ̂ .
0
Therefore, the family {u(r, ρ)}ρ⩾B is uniformly bounded on [r1 , r2 ].
24
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
Let us show that the family {u(r, ρ)}ρ⩽B is uniformly bounded on [0, r2 ]. Note that this family is bounded
from above by B. It is also bounded from below by the same argument as (57). Namely,
∫ r
(∫ t
) 1
f (B)
β+1
− α
− 1
dt ⩾ −e β+1 (γ + 1) β+1 θ̂−1 r2θ̂ .
u(r, ρ) = ρ −
t β+1
sγ ef (u(s,ρ)) ds
0
0
Furthermore, (47) implies the uniform continuity of the family {u(r, ρ)}ρ⩾0 on [r1 , +∞). Indeed, for all r′ , r′′ ,
such that r1 < r′ < r′′ , and ρ ⩾ 0,
∫ r′′
(∫ t
) 1
β+1
− α
sγ ef (u(s,ρ)) ds
dt
(58)
0 < u(r′ , ρ) − u(r′′ , ρ) =
t β+1
r′
⩽e
0
f (B)
β+1
(γ + 1)
1
− β+1
−1
θ̂
θ̂
θ̂
(r′′ − r′ ).
By the Arzelà-Ascolli theorem, the family {u(r, ρ)} is relatively compact on [r1 , r2 ].
Let us prove now that each limit point of the family {u(r, ρ)}ρ⩾0 is either regular or singular solution. Let
{u(r, ρn )} converge to a function u∗∗ (r) in C([r1 , r2 ]). We claim that the sequence ∂r u(r1 , ρn ) is bounded.
Suppose it is not the case, that is, we can find a subsequence ∂r u(r1 , ρnk ) → −∞ as k → ∞. Without loss
of generality, we assume that limn→∞ ∂r u(r1 , ρn ) = −∞. For r ∈ [r1 , r2 ], it holds that,
∫ r
∫ t
(
) 1
β+1
− α
β+1
u(r, ρn ) = u(r1 , ρn ) −
t β+1 r1α |∂r u(r1 , ρn )|
dt.
+
sγ ef (u(s,ρn )) ds
r1
r1
The last term in this identity converges to ∞ which contradicts to the convergence of {u(r, ρn )} in C([r1 , r2 ]).
β+1
Therefore, ∂r u(r1 , ρn ) is bounded. Let a subsequence r1α |∂r u(r1 , ρnk )|
converge to k(r1 ). In the previous
expression, passing to the limit as k → ∞, we obtain
∗∗
∫
∗∗
r
u (r) = u (r1 ) −
α
− β+1
t
(
∫
t
k(r1 ) +
γ f (u∗∗ (s))
s e
r1
1
) β+1
ds
ds.
r1
β+1
From the above identity, we obtain that u∗∗ (r) is differentiable in r, (u∗∗ )′ (r) < 0, k(r1 ) = r1α |(u∗∗ )′ (r1 )|
,
∗∗
and u (r) solves (7) on [r1 , r2 ]. By extracting further subsequences from {u(r, ρn )} and repeating the
argument on compact subintervals of (0, +∞), we can extend u∗∗ (r) to a solution of (7) on (0, +∞). Note
that {u(r, ρn )} contains a subsequence that converges pointwise to u∗∗ (r) on (0, +∞).
Let us show now the following fact. Suppose u(r, ρn ) converges to u∗∗ (r) pointwise on (0, +∞). If ρn
is unbounded, then u∗∗ (r) is a singular solution, if ρn is bounded, then it is a regular solution. Let us
show that, in the first case, limr→+0 u∗∗ (r) = +∞. Suppose this is not the case, so we can find a sequence
rn → 0 as n → +∞ such that limn→∞ u∗∗ (rn ) = B < +∞. Since (u∗∗ )′ (r) < 0, supr⩾rn u∗∗ (r) ⩽ u∗∗ (rn ),
which implies that supr⩾0 u∗∗ (r) ⩽ B. Let ρnk → +∞ be a subsequence of {ρn }. By Proposition 3.1, we
can find R(2B) and ρ0 (2B) such that whenever ρnk > ρ0 (2B), u(r, ρnk ) > 2B on (0, R(2B)). Therefore,
u∗∗ (r) ⩾ 2B on (0, R(2B)). This contradicts to the previously obtained inequality supr⩾0 u∗∗ (r) ⩽ B. Hence,
limr→+0 u∗∗ (r) = +∞.
Suppose now ρn , introduced above, is bounded. Then, we can extract a subsequence limk→+∞ ρnk = ρ̄ <
+∞. Let us prove that u∗∗ (r) is a regular solution with u∗∗ (0) = ρ̄, i.e., u∗∗ (r) = u(r, ρ̄). It follows from
(47), that u(r, ρ) is continuous in ρ uniformly in r varying over compact intervals. Indeed,
sup |u( · , ρ1 ) − u( · , ρ2 )| ⩽ |ρ1 − ρ2 |
[0,r]
+
1
β+1
∫
r
α
− β+1
s
(∫
0
1
∫ 0r
⩽ |ρ1 − ρ2 | + Kr,ρ1 ,ρ2
0
β
− β+1
Λ(s, λ)
dλ
)∫
s
tγ (ef (u(t,ρ2 )) − ef (u(t,ρ1 )) )dt ds
0
sθ̂−1 sup |u( · , ρ1 ) − u( · , ρ2 )|ds,
[0,s]
25
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
where Kr,ρ1 ,ρ2 > 0 is a constant that only depends on r, ρ1 , ρ2 . Furthermore, above,
∫ s
Λ(s, λ) =
tγ (λef (u(t,ρ1 )) + (1 − λ)ef (u(t,ρ2 )) )dt,
0
(1)
(2)
Kr,ρ
sγ+1 ⩽ Λ(s, λ) ⩽ Kr,ρ
sγ+1 ,
1 ,ρ2
1 ,ρ2
(1)
(2)
where Kr,ρ1 ,ρ2 and Kr,ρ1 ,ρ2 are positive constants. Now by Gronwall’s inequality, one can find another
(3)
constant Kr,ρ1 ,ρ2 , such that
(3)
sup |u( · , ρ1 ) − u( · , ρ2 )| ⩽ Kr,ρ
|ρ − ρ2 |.
1 ,ρ2 1
[0,r]
This implies that limk→+∞ u(r, ρnk ) = u(r, ρ̄) on compact intervals of [0, +∞), and hence, u∗∗ (r) =
u(r, ρ̄). □
From the proof of Proposition 3.2, we obtain the following corollary.
Corollary 3.1. Under assumptions of Proposition 3.2, any limit point of the family {u( · , ρ)}ρ⩾0 that can be
obtained as a limit of a subsequence with ρn → +∞, is a singular solution. Otherwise, it is a regular solution.
The theorem below is the main result of this subsection.
Theorem 3.1. Assume (B1)–(B3). Then, the result of Theorem 1.4 holds true.
Proof . Since we know that there exists a unique solution u(r, ρ) to problem (Pρ ), there is also a unique
1
solution (uλ , λ) with uλ (r) = u(λ θ r, ρ) to problem (Pλ′ ). For u and uλ related through this identity, it holds
that the zero R(0, ρ) of u is expressed via λ as follows:
1
R(0, ρ) = λ θ .
(59)
By Proposition 3.1, R(0, ρ) has an upper bound. Therefore, the set of λ with the property that (uλ , λ)
is a solution to (Pλ′ ) is bounded. Let λ# be the exact upper bound for this set. Let us show that for
all λ < λ# , there exists a regular solution to problem (Pλ′ ). Indeed, R(0, ρ) is the unique solution to the
equation u(r, ρ) = 0 with respect to r for each fixed ρ > 0. By the implicit function theorem, the function
R(0, · ) : (0, +∞) → R is continuous. Moreover, limρ→0 R(0, ρ) = 0 which is implied by the inequality
∫ R(0,ρ)
(∫ t
) 1
α
β+1
− β+1
sγ ef (u(s,ρ)) ds
dt
ρ = u(0, ρ) − u(R(0, ρ), ρ) =
t
0
0
⩾e
f (0)
β+1
(γ + 1)
1
− β+1
−1
θ̂
R(0, ρ)θ̂ .
As a continuous function of ρ, λ = R(0, ρ)θ takes all the values on the interval [0, λ# ). Consider the case
1
λ = λ# . We prove that there exists a solution to (7), singular or regular, such that (λ# ) θ is its zero. Find a
sequence {ρn } such that λn = R(0, ρn )θ converges to λ# . Let u∗∗ be a limit point of u( · , ρn ) on an interval
1
[a, b] ⊂ (0, +∞) such that (λ# ) θ ∈ (a, b], i.e., there exists a subsequence u( · , ρnk ) → u∗∗ , as k → ∞,
uniformly on [a, b]. By Corollary 3.1, if ρnk is unbounded, then u∗∗ is a singular solution. Otherwise, it is a
1
regular solution. Without loss of generality, we write ρn for ρnk . Let R∗ (0) = (λ# ) θ . We have
|u∗∗ (R∗ (0))| = |u(R(0, ρn ), ρn ) − u∗∗ (R∗ (0))|
∗
(60)
∗
∗∗
∗
⩽ |u(R(0, ρn ), ρn ) − u(R (0), ρn )| + |u(R (0), ρn ) − u (R (0))|.
By the uniform continuity of the family u( · , ρ)ρ⩾0 , the first term on the right-hand side of (60) goes to
zero. The second term goes to zero by the uniform convergence u( · , ρn ) → u∗∗ on [a, b]. This implies that
1
u∗∗ ((λ# ) θ ) = 0, which completes the proof. □
26
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
4. Study of regular solutions by means of the singular solution
Here we demonstrate how the asymptotic representation (6) can be used to study regular solutions to
problem (Pλ ).
The change of variable (8) transforms problem (Pλ ) to the following:
{
1
−L(u) = ef (u) , 0 ⩽ r ⩽ λ θ ,
1
u(λ θ ) = 0.
1
In the above problem, one can actually exclude λ out of consideration keeping in mind that λ θ is the zero of
the solution (by Proposition 3.1, we know that a zero of the solution exists). In Sections 4.1 and 4.2, we deal
1
1
with the singular solution u∗ (r) = u∗λ∗ ((λ∗ )− θ r) and regular solutions u(r, ρ) = uλ (λ− θ r, ρ) to Eq. (7). In
the case of regular solutions, Eq. (7) is complemented with the initial conditions u(0, ρ) = ρ, ∂r u(0, ρ) = 0.
4.1. Transformation of problem (Pρ ) and its limit equation
For simplicity of notation, in this section, we define the function
ψ = ef.
Note that if limt→+∞ g ′′ (t) = 0, then, by Remark 2.5, F(u), involved in (14), is finite for all u. Indeed, by
the change of variable s = g(t), we obtain
∫ +∞
∫ +∞ f (s)
− t
− β+1
ds =
(61)
e β+1 g ′ (t)dt < +∞.
e
f (u)
u
Remark 4.1. Sometimes we write ρ as the second parameter of ũ. However, we do not mean that ρ is its
maximal value, unlike the notation u(r, ρ). Note that
{
F(u(r, ρ)) }
ũ(s, ρ) = F1−1 F1 (1)
⩽ F1−1 (F1 (1)) = 1,
F(ρ)
and ũ(0, ρ) = 1.
s ⩾ 0,
We just want to emphasize that ũ actually depends on ρ. Moreover, we will study the limit of ũ as ρ goes
to infinity.
Remark 4.2. Sometimes we skip the dependence on ρ in both u and ũ to simplify notation.
Lemma 4.1. Assume (61) is fulfilled. Let u(r) be a C2 -solution to problem (Pρ ). Then, the function ũ(s),
defined by (14), is a solution to the problem
β+2
L(ũ) + eũ + sα−γ |ũ′ |
ũ(0) = 1,
( I(u(ε s))
)
ρ
− 1 = 0,
β+1
(62)
′
ũ (0) = 0,
1
where ũ is everywhere evaluated at s and I(u) = F(u)f ′ (u)ψ(u) β+1 .
Proof . First, we note that the ũ(s), defined by (14), satisfies the initial conditions in (62). Define G(u) =
θ
θ̂
−1
F1−1 ε−
(ũ(s)), where r = ερ s. Computing L(u) by using the
ρ F(u), where θ̂ = β+1 . Then, u(r) = G
27
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
previous expression, we obtain
(
)′
β
−γ α ′
L(u(r)) = ε−θ
s |ũ (s)| ũ′ (s)G′ (u(r))−β−1 s
ρ s
′
−β−1
α−γ ′
= ε−θ
− ε−θ
|ũ (s)|
ρ L(ũ(s))G (u(r))
ρ (β + 1)s
β+2
G′ (u(r))−β−3 G′′ (u(r)).
Problem (Pρ ) is then transformed to the following:
β+2
L(ũ(s)) + εθρ ψ(u(r))G′ (u(r))β+1 − (β + 1)sα−γ |ũ′ (s)|
G′′ (u(r))
= 0.
G′ (u(r))2
By the definition of F(u) and F1 (u) in (14),
ũ(s)
θ̂ β+1
G′ (u(r)) = ε−
ψ(u(r))
ρ e
1
− β+1
,
(63)
which transforms the second term into eũ(s) . Furthermore,
G′′ (u)
(ln G′ (u))′
1 G′ (u) − f ′ (u)
=
=
.
′
2
′
G (u)
G (u)
β+1
G′ (u)
(64)
In (64), u is everywhere evaluated at r. Recall that ũ(s) = G(u(r)). For simplicity of notation, we skip the
dependence on s in ũ and, with a slight abuse of notation, simply write ũ = G(u). By (63),
−
β
ψ ′ (u)ψ(u) β+1 F(u)
I(u)
f ′ (u)
=
=
ũ
−θ̂
G′ (u)
β
+1
ερ F(u) e β+1
θ̂
since ε−
ρ F(u) = F1 (ũ) and F1 (ũ) = (β + 1)e
Lemma 4.2.
ũ
− β+1
. The above computations imply (62).
□
Assume (B1)–(B3). Then, limρ→+∞ u(ερ s, ρ) = +∞ in Cloc [0, +∞).
Proof . Fix an arbitrary interval [0, T ]. From Proposition 3.1 it follows that for any B > 0, there exist a
number R(B) > 0 and a constant ρ0 > 0 such that for all ρ > ρ0 and r ∈ (0, R(B)), u(r, ρ) > B. Note that
limρ→+∞ ερ = 0. This proves that limρ→+∞ sups∈[0,T ] u(ερ s, ρ) = +∞. □
Lemma 4.3.
Cloc [0, +∞).
Assume (61), (B1)–(B3), and that limt→+∞
Proof . We have
∫ +∞
I(u) =
u
f (s)
e
− β+1
e
ds f ′ (u)
f (u)
− β+1
g ′′ (t)
g ′ (t)
∫ +∞
=
f (u)
e
e
= 0. Then, limρ→∞ I(u(ερ s)) = β + 1 in
t
− β+1
′
f (u)
− β+1
g (t) dt
.
g ′ (f (u))
Let v = f (u). By L’Hopital’s rule,
∫ +∞ − t ′
− v
e β+1 g (t) dt
e β+1 g ′ (v)
= lim
lim v − v
v
v
v→+∞
v→+∞ (β + 1)−1 e− β+1 g ′ (v) − e− β+1 g ′′ (v)
e β+1 g ′ (v)
(β + 1)
= lim
= β + 1.
′′
v→+∞ 1 − (β + 1) g (v)
g ′ (v)
The statement now follows from Lemma 4.2.
□
Lemma 4.4. Assume (A1)–(A6). Then, the following representation holds for the singular solution (12) of
Eq. (11):
v ∗ (t) = F −1 (A−1 e−θ̂t (1 + ε(t))),
(65)
1
where A = θ̂(α̂ − 1) β+1 and limt→+∞ ε(t) = 0.
28
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
Proof . Identity (65) is equivalent to ε(t) = A eθ̂t F(v ∗ (t))−1. Let us show that limt→+∞ eθ̂t F(v ∗ (t)) = A−1 .
Recall that v ∗ (t) = g(θt + φ(t)) + η(t), where φ = φ1 + φ2 is defined by (13), η = O(g ′′ (θt)), and
η ′ = O(g ′′ (θt)). By L’Hopital’s rule,
lim
F(v ∗ (t))
e−θ̂t
t→+∞
−
= lim
e
f (v ∗ (t))
β+1
(g ′ (θt + φ)(θ + φ′ ) + η ′ )
θ̂e−θ̂t
t→+∞
.
Transforming f (v ∗ (t)) by formula (16), we obtain that the limit on the right-hand side equals to
lim θ̂−1 e
φ
− β+1 ′
t→+∞
g (θt + φ)(θ + φ′ ) = lim
t→+∞
g ′ (θt + φ)(θ + φ′ )
θ̂θ(α̂ − 1)
1
β+1
We used representation (13) for φ = φ1 + φ2 and the fact that limt→+∞
zero by (19). This concludes the proof. □
= A−1 .
g ′ (θt)
η
g ′ (θt+o(t))
= 0. Finally, φ′ goes to
Lemma 4.5. Assume (A1)–(A6). Then, the singular solution u∗ to (7) has the representation
(
)
u∗ (r) = F −1 K rθ̂ (1 + δ(r)) ,
where K =
β+1
θ (α
− β − 1)
1
− β+1
1
, δ(r) = ε(ln κ − ln r), κ = (β + 1) θ .
Proof . The proof follows immediately from Lemma 4.4 by the substitution t = ln κ − ln r. The constant
1
− 1
− β+1
K can be computed by the definition of A as follows: K = A−1 (β + 1) β+1 = β+1
. □
θ (α − β − 1)
Lemma 4.5 allows to obtain a representation for the singular solution ũ∗ to (62).
θ̂
∗
Lemma 4.6. Assume (A1)–(A6). Then, the singular to Eq. (62) ũ∗ (s) = F1−1ε−
ρ F(u (ερ s)) can be
represented as follows:
ũ∗ (s) = ln{θβ+1 (α − β − 1)} − θ ln s − (β + 1) ln{1 + δ(ερ s)}.
(66)
Proof . Formula (66) is the result of application of transformation (14) to u∗ (ερ s). Indeed, ũ∗ (s) =
t
F1−1 (K sθ̂ (1 + δ(ερ s))), where F1−1 (t) = −(β + 1) ln β+1
. This immediately implies (66). □
Proposition 4.1.
Assume (A1)–(A6). Then, for the singular and regular solutions to (62), it holds that
lim ũ(s) = w(s)
ρ→+∞
and
lim ũ∗ (s) = w∗ (s)
ρ→+∞
(67)
in Cloc [0, +∞) and Cloc (0, +∞), respectively. Here, w(s) is the regular solution to the problem
L(w) + ew = 0,
w(0) = 1,
w′ (0) = 0,
(68)
and w∗ (s) is the singular solution to the equation in (68) given explicitly as follows:
κ
w∗ (s) = ln{θβ+1 (α − β − 1)} − θ ln s = ln{θβ+1 (α̂ − 1)} + θ ln .
s
(69)
Remark 4.3. The last expression in (69) agrees with (45) with t = ln κs .
For the proof of Proposition 4.1, we need the following lemma.
Lemma 4.7. Under assumptions of Lemma 4.3, problem (68) is the limit problem for (62) as ρ → +∞.
More specifically, the last term on the right-hand side of (62) goes to zero in Cloc [0, +∞) as ρ → +∞.
29
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
β+1
Proof . Let ω̃(s) = |ũ′ (s)|
. Recall that by Lemma 3.1, u′ (r) < 0. Therefore, by (14), |ũ′ (s)| = −ũ′ (s).
Fix an arbitrary interval [0, T ] and introduce the notation,
Iρ (s) =
I(u(ερ s))
− 1.
β+1
(70)
Then, ω̃(s) satisfies the following first-order differential inequality:
β+2
ω̃ ′ (s) = sγ−α eũ − αs−1 ω̃ + ω̃ β+1 Iρ (s) ⩽ C + εω̃ + εω̃ 2 .
Here, C = T γ−α e is the bound for sγ−α eũ . Furthermore, ε is an arbitrary sufficiently small number. Note
β+2
that |Iρ (s)| < ε for sufficiently large ρ. Finally, it holds that ω̃ β+1 ⩽ ω̃ + ω̃ 2 .
We aim to estimate ω̃(s) via the solution ω(s) to the problem
ω ′ = C + εω + εω 2 ,
ω(0) = 0.
(71)
′
It is known that ω = − 1ε yy is a solution to the ODE in (71) if y is the solution to
y ′′ − εy ′ + Cεy = 0.
For sufficiently small ε, the both roots of the characteristic equation are complex, and therefore, the solution
takes the form
√
ε
4Cε − ε2
y(s) = K̄e 2 s (cos(Ds) + K sin(Ds)), where D =
.
2
It is straightforward to obtain now the solution ω to problem (71), computing K from the condition
ω(0) = 0:
ε2 + 4D2
1
ω(s) =
ε .
4Dε ctg(Ds) − 2D
[ 2 arcctg √ ε 2 )
√ 4Cε−ε . When ε is sufficiently small,
Note that the solution ω(s) is well-defined on the interval 0,
2
4Cε−ε
the numerator of the right endpoint is close to π; however, the denominator is small. Thus, we can choose
ε sufficiently small (which makes ρ sufficiently large) such that ω(s) is well-defined on [0, T ]. Therefore, on
[0, T ], it holds that
β+1
|ũ′ (s, ρ)|
⩽ ω(s)
(72)
for sufficiently large ρ. This, along with Lemma 4.3, proves the lemma.
□
Proof of Proposition 4.1. Everywhere in the proof, Ki , i = 1, 2, 3, are positive constants. Fix an arbitrary
interval [0, T ]. We have
∫ r
(∫ 1
)
β
1
− α
−
w(r) − ũ(r) =
t β+1
Λ(t, ρ, λ) β+1 dλ
β+1 0
0
∫ t
β+2
(sγ (eũ − ew ) + sα |ũ′ (s)|
Iρ (s))ds dt,
0
where Λ(t, ρ, λ) is defined by the first identity below and can be estimated by the expression on the right-hand
side:
∫ t
∫ t
∫ t
( γ ũ
)
β+2
α ′
γ w
w(T )
s e ds ⩾ (1 − λ)e
sγ ds.
Λ(t, ρ, λ) = λ
s e + s |ũ (s)|
Iρ ds + (1 − λ)
0
0
Indeed, it holds that
∫
t
β+2
(sγ eũ + sα |ũ′ (s)|
Iρ (s))ds = tα |ũ′ (t)|
0
30
0
β+1
⩾ 0.
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
By Gronwall’s inequality,
tα |ũ′ (t)|
β+1
⩽e
∫t
0
|ũ′ (s)Iρ (s)|ds
t
∫
sγ eũ ds ⩽ K1 tγ+1 ,
(73)
0
where the last estimate holds by (72) for sufficiently large ρ. Furthermore, estimate (73) implies
∫ r γ+1−α
{
}
sup |ũ − w| ⩽ K2
t β+1 sup |ũ − w| + sup |Iρ | dt,
[0,r]
[0,t]
0
where, in particular, it is taken into account that
∫1
0
[0,r]
β
− β+1
(1 − λ)
dλ = β + 1. By Gronwall’s inequality,
θ̂
| rθ̂ eK3 r .
sup[0,r] |ũ − w| ⩽ K3 sup[0,r] |Iρ
This implies that sup[0,T ] |ũ − w| → 0 as ρ → +∞.
We prove now the second convergence in (67). It follows from the proof of Corollary 1.1 that the singular
solution to (68) takes the form (69). If, in (66), we pass to the limit as ρ goes to +∞, we obtain exactly (69).
The convergence holds in Cloc (0, +∞) since the convergence limρ→+∞ δ(ερ s) = 0 holds in this space. □
4.2. Number of intersection points. Case α − β − 1 <
4θ
β+1
(d < 2k + 8)
In this subsection, we first study the number of intersection points between the regular and singular
solutions w and w∗ to the equation in (68), and then use this result to study the limit, as ρ → +∞, of
the number of intersection points between u( · , ρ) and u∗ . Recall that the number of zeros of w − w∗ on an
interval (R, S) is denoted by L(R,S) (w − w∗ ). Although Theorem 4.1 has its own interest, we prove it in this
subsection as an auxiliary result.
Theorem 4.1. Assume (A6). Suppose α − β − 1 < 4θ̂. Then, for any R > 0, w and w∗ have an infinite
number of intersection points on (R, +∞), i.e., L(R,∞) (w − w∗ ) = +∞. Furthermore, L(0,R] (w − w∗ ) is
finite.
1
Proof . Define v(t) = w(s) and ω(t) = v ′ (t)β+1 , where t = ln{ κs } with κ = (β + 1) θ . By Lemma 3.1,
v ′ (t) > 0, and hence, by (38),
( (β+1−α)t
)′
e
ω(t) = −(β + 1)e−(γ+1)t ev .
(74)
Now let w∗ (s) be given by (69) and v ∗ (t) = w∗ (s), where t = ln{ κs }. Then, v ∗ (t) = θt + ln{θβ+1 (α̂ − 1)}.
Define x(t) = v(t) − v ∗ (t) and y(t) = ω(t) − ω ∗ (t), where ω ∗ (t) = v ∗ ′ (t)β+1 = θβ+1 . Eq. (74) implies that x
and y satisfy the following system of ODEs
{
y ′ = (α − β − 1)(y − θβ+1 (ex − 1)),
(75)
1
x′ = (y + θβ+1 ) β+1 − θ,
1
where the last equation holds due to the fact that v ′ = ω β+1 . Note that system (75) has a unique steady
state (0, 0). We are going to determine its type. The linearization of (75) has the form
{
y ′ = (α − β − 1)(y − θβ+1 x),
(76)
y
x′ = (β+1)θ
β.
Since 0 < α − β − 1 < 4θ̂, the roots of the characteristic equation for the linearized system are complex
numbers with the positive real part α−β−1
, and hence, (0, 0) is an unstable focus. This means that as t goes
2
to −∞, x(t) crosses zero infinitely many times at any neighborhood of −∞. This implies that for any R > 0,
L(−∞,−R) (v − v ∗ ) = L(κeR ,+∞) (w − w∗ ) = +∞.
31
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
Suppose L(0,R] (w − w∗ ) = +∞ for some R > 0. Then, there exists an accumulation point, say s0 , of
zeros of w − w∗ belonging to [0, R]. Note that (w − w∗ )(s0 ) = 0 which implies that s0 ∈ (0, R]. Furthermore,
(w − w∗ )′ (s0 ) = 0. This implies that v(t0 ) = v ∗ (t0 ) and ω(t0 ) = ω ∗ (t0 ) for t0 = ln κ − ln s0 , which is a
contradiction. □
Corollary 4.1. Under assumptions of Theorem 4.1, there exists R > 0 such that w − w∗ changes the sign
at each zero point on (R, +∞).
Proof . The proof follows immediately from the fact that (0, 0) is a focus for (75).
□
Theorem 4.2. Assume (A1)–(A6) and let α − β − 1 < 4θ̂. Let u( · , ρ) be the regular solution to problem
(Pρ ) and u∗ be the singular solution to (7) defined by (8) via (u∗λ∗ , λ∗ ). Then, for any δ > 0, it holds that
limρ→+∞ L(0,δ) (u( · , ρ) − u∗ ( · )) = +∞.
Proof . Recall that ũ − ũ∗ → w − w∗ as ρ → +∞ uniformly on any interval [R, S] ⊂ (0, +∞). Note that for
a sufficiently large ρ, L[R,S] (ũ − ũ∗ ) is not smaller than L[R,S] (w − w∗ ) − 2, where the latter number is finite
by Theorem 4.1. Recall that u(ερ s, ρ) = F1−1 εθ̂ρ F(ũ(s, ρ)) with limρ→+∞ ερ = 0. Fix N ∈ N. There exist
R, S > 0 and a number ρR,S > 0 such that L[R,S] (ũ − ũ∗ ) ⩾ N for ρ > ρR,S . Note that for each zero point
s ∈ [R, S] of ũ − ũ∗ , there exists exactly one zero point r ∈ [ερ R, ερ S] of u( · , ρ) − u∗ ( · ). Choose ρδ > ρR,S
such that for all ρ > ρδ , [ερ R, ερ S] ⊂ (0, δ). Therefore, for any N ∈ N, one can find ρδ such that for all
ρ ⩾ ρδ , L(0,δ) (u( · , ρ) − u∗ ( · )) ⩾ N . The theorem is proved. □
4.3. Bifurcation diagram. Case α − β − 1 <
4θ
β+1
(d < 2k + 8)
Theorem 4.3. Assume (A1)–(A6) and α − β − 1 < 4θ̂. Let (uλ(ρ) , λ(ρ)) be the family of regular solutions
to problem (Pλ ) and let (u∗λ∗ , λ∗ ) be the singular solution constructed in Theorem 2.1. Then, as ρ → ∞, λ(ρ)
oscillates around λ∗ . In particular, there are infinitely many regular solutions to problem (Pλ ) for λ(ρ) = λ∗ .
Proof . Let u( · , ρ) and u∗ be the regular and singular solutions to (7), as in Theorem 4.2. Define v̂(r, ρ) =
u(r, ρ) − u∗ and show that
∂r v̂(r0 , ρ) ̸= 0 whenever v̂(r0 , ρ) = 0.
(77)
Suppose v̂(r0 , ρ) = ∂r v̂(r0 , ρ) = 0. For r ∈ (0, r0 ), we have
∫ r0
∫ r0
[(
) 1
β+1
β+1
− α
s β+1 r0α |u′ (r0 )|
ξ γ ef (u(ξ)) dξ
u(r) − u∗ (r) =
−
r
s
∫ r0
(
) 1 ]
∗
β+1
β+1
− r0α |u′ (r0 )|
−
ξ γ ef (u (ξ)) dξ
ds
s
∫ r0
∫
∫
( 1
) r0
β
∗
1
− α
−
=
s β+1
Λ(s, λ) β+1 dλ
ξ γ (ef (u (ξ)) − ef (u(ξ)) )dξ ds,
β+1 r
0
s
where
Λ(s, λ) = r0α |u′ (r0 )|
β+1
∫
−
r0
∗
ξ γ (λef (u(ξ)) + (1 − λ)ef (u
(ξ))
)ds
s
β+1
= sα (λ|u′ (s)|
+ (1 − λ)|u∗ ′ (s)|
β+1
).
Take ϵ ∈ (0, r0 ). For s ∈ [ϵ, r0 ], we have
(
β+1 )
β+1
β+1
inf sα min{|u′ (s)|
, |u∗ ′ (s)|
} ⩽ Λ(s, λ) ⩽ r0α |u′ (r0 )|
.
s∈[ϵ,r0 ]
32
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
Therefore, there exists a constant K > 0 such that for r ∈ [ϵ, r0 ],
sup |u − u∗ | ⩽ K
[r,r0 ]
∫
r
r0
sup |u(s) − u∗ (s)|ds.
[s,r0 ]
By Gronwall’s inequality, sup[ϵ,r0 ] |u − u∗ | = 0. Since ϵ ∈ (0, r0 ) is arbitrary, we obtain that u = u∗ on (0, r0 )
which is a contradiction. Thus, (77) is true.
1
1
Recall that λ(ρ) θ and λ∗ θ are zeros of u( · , ρ) and u∗ , respectively. Since u and u∗ are decreasing in
1
1
r, they do not have other zeros. Define r(ρ) = min{λ(ρ) θ , λ∗ θ } and z(ρ) = L(0,r(ρ)) v̂( · , ρ). Note that
if ρ is sufficiently large, then, by Proposition 3.1, we can find two positive constants τ1 and τ2 such that
τ1 < r(ρ) < τ2 . Furthermore, we note that z(ρ) is positive and finite. Indeed, z(ρ) > 0 since, by Theorem 4.2,
limρ→+∞ L(0,τ1 ) (v̂( · , ρ)) = +∞. Assuming that z(ρ) is infinite, we can find a sequence {rn } of zeros of v̂
converging to a point r̃ ∈ [0, r(ρ)]. One has v̂(r̃, ρ) = ∂r v̂(r̃, ρ) = 0 which contradicts to (77). Thus, we
conclude that for each fixed ρ, the equation v̂(r, ρ) = 0 has a finite number of solutions on (0, r(ρ)). By the
same argument, for each fixed ρ > 0, the equation v̂(r, ρ) = 0 has a finite number of solutions on any finite
interval [0, R], and therefore, there is a countable number of zeros of v̂(r, ρ) on (0, +∞).
Fix ρ0 > 0 and take an arbitrary zero r0 of v̂( · , ρ0 ). By (77), ∂r v̂(r0 , ρ0 ) ̸= 0. By the implicit function
theorem, there exist an interval (ρ1 , ρ2 ), containing ρ0 , and a unique continuous function r(ρ) solving the
equation v̂(r, ρ) = 0 on (ρ1 , ρ2 ) and such that r(ρ0 ) = r0 . Without loss of generality, (ρ1 , ρ2 ) can be regarded
as the maximal interval to which r(ρ) can be extended. This extension will be unique by (77) and the
implicit function theorem. The curve r(ρ) may cross r(ρ) at some point from (ρ1 , ρ2 ) or may not. When
it happens a zero enters or exits the interval (0, r(ρ)). Remark that ρ1 can be either zero or positive; ρ2 can
be either finite or +∞. Let us show that if ρ2 is finite, then limρ→ρ2 − r(ρ) = +∞. Suppose ρ2 is finite, but
limρ→ρ2 − r(ρ) ̸= +∞. Then, we can find a sequence ρn → ρ2 as n → +∞ such that r(ρn ) converges to a
number r2 ∈ (0, +∞). By continuity, v̂(r2 , ρ2 ) = 0. If r2 is the limit of r(ρ) as ρ → ρ2 , then, by the implicit
function theorem, we can find an extension of r(ρ) beyond ρ2 which is a contradiction. Therefore, the limit at
ρ2 does not exist. Then, we can find a neighborhood U = U (r2 , ρ2 ) = (r2 −δ, r2 +δ)×(ρ2 −δ, ρ2 +δ), where (1)
the equation v̂(r, ρ) = 0 is uniquely solvable for each fixed ρ ∈ (ρ2 − δ, ρ2 + δ) and, furthermore, determines
r as a continuous function of ρ living inside U ; and (2) r(ρ) exits and enters U infinitely many times. This
contradicts to the implicit function theorem. Hence, if ρ2 is finite, we have that limρ→ρ2 − r(ρ) = +∞. By
the similar argument, one can show that if ρ1 > 0, then limρ→ρ1 + r(ρ) = +∞. Also remark that by the
implicit function theorem and (77), two functions r1 (ρ) and r2 (ρ) representing two different solutions of
v̂(r, ρ) = 0 cannot exit or enter the interval (0, r(ρ)] through its right end at the same point ρ. One can
extend this observation stating that two functions r1 (ρ) and r2 (ρ) representing two different solutions of
v̂(r, ρ) = 0 never intersect each other, which means that two zeros cannot merge. The above arguments
allow us to conclude that z(ρ) is positive, finite, and changes only by 1 when a zero of v̂( · , ρ) enters or
exists the interval (0, r(ρ)) through its right end.
Let us prove now that λ(ρ) oscillates infinitely around λ∗ as ρ → +∞. First, we note that v̂(0, ρ) = −∞
for all ρ > 0, i.e., does not change the sign as ρ varies. Then, z(ρ) changes only if v̂(r(ρ), ρ) changes the
1
sign. Suppose there exists σ1 > 0 such that for all ρ > σ1 , λ(ρ) ⩾ λ∗ . Then, r(ρ) = λ∗ θ and we have that
1
v̂(r(ρ)) = u(λ∗ θ , ρ) ⩾ 0 for all ρ > σ1 . Likewise, if we assume that there exists σ2 > 0 such that for all
1
1
ρ > σ2 , λ(ρ) ⩽ λ∗ , then r(ρ) = λ(ρ) θ and v̂(r(ρ)) = −u∗ (λ(ρ) θ , ρ) ⩽ 0 for all ρ > σ2 . In both situations,
i.e., λ(ρ) ⩾ λ∗ for all ρ > σ1 or λ(ρ) ⩽ λ∗ for all ρ > σ2 , a zero of v̂ cannot enter the open interval
(0, r(ρ)) through its right end (however it can reach r(ρ) and then turn back) when ρ > max{σ1 , σ2 }, which
contradicts to Theorem 4.2 stating that limρ→+∞ L(0,τ1 ) (v̂( · , ρ)) = +∞. □
33
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
5. Liouville–Bratu-Gelfand Problem
In this section, we consider the equation
L(u) + eu = 0.
(78)
As before, when we deal with regular solutions, we complete Eq. (78) with the initial conditions u(0) = ρ
and u′ (0) = 0; the regular solution will be also denoted by u(r, ρ). Furthermore, we let u∗ be the singular
solution to (78) given by (69).
In this section, we aim to complement the results of [20], where the authors studied bifurcation diagrams
for problem (42), with a study of the number of intersection points between the regular and singular solutions
to Eq. (78) and the convergence of regular solutions to the singular. Recall that Eqs. (42) and (78) are
equivalent by the change of variable (8).
5.1. Number of intersection points between the regular and singular solutions
We already studied the number of intersection points between the regular and singular solutions u(ρ, · )
and u∗ to (78) in Section 4.2. Namely, we obtained Theorem 4.1 whose statement we repeat here for
completeness of the presentation.
Theorem 4.1.
Assume (A6). Suppose α − β − 1 < 4θ̂ (2k < d < 2k + 8). Then, L(0,∞) (u − u∗ ) = +∞.
Theorem 5.1. Assume (A6). Suppose α − β − 1 ⩾ 4θ(β + 1) (d ⩾ 2k + 8k 2 ). Then, L(0,+∞) (u − u∗ ) = 0.
Proof . First of all, we note that α − β − 1 ⩾ 4θ(β + 1) ⩾ 4θ̂. We keep the same notation as in the proof of
Theorem 4.1 in Section 4.2, except for the notation for the regular and singular solutions. Namely, we define
the same objects with w = u and w∗ = u∗ . The unique steady state (0, 0) of (76), and therefore of (75),
is then an unstable node. We further note that the eigenvectors for the linearized system (76) point to the
first quadrant of the phase plane. Therefore, in a small neighborhood of the origin, the integral curves are
located at the first and the third quadrants. As t increases, the motion along the curves occurs clockwise.
Note that the solution to (75) satisfies the following conditions:
lim x(t) = −∞ and
t→+∞
From (75), we will evaluate the derivative
1
β+1 β+1
dy
dx
lim y(t) = −θβ+1 .
t→+∞
(79)
at the points of the curve y = c(x) = (ℓx + θ)β+1 − θβ+1 , which
is the same as (y + θ
)
− θ = ℓx, where the number ℓ is to be fixed later. Introducing the notation
δ = α − β − 1, we obtain that in the area {x < 0, −θβ+1 < y < 0}, the following estimate holds:
(
(
dy ⏐⏐
y
θβ+1 ex − 1 )
θ)
β
−
>
δθ
1
−
⩾ (β + 1)θβ ℓ,
⏐ =δ
1
dx c
ℓ
x
ℓ
β+1
β+1
(y + θ
)
−θ
(80)
where the number ℓ is to be chosen in such a way that the last inequality is satisfied. To see why the
y
second inequality is true, we note that the function y ↦→
is non-decreasing on (−θβ+1 , 0) (a
1
(y+θ β+1 ) β+1 −θ
straightforward computation shows that it has a positive derivative if β ⩾ 0), and therefore, on (−θβ+1 , 0),
it is greater than θβ . Now solving the inequality
(
θ)
δ 1−
⩾ (β + 1)ℓ
ℓ
34
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
with respect to ℓ, we obtain that under the assumption δ ⩾ 4θ(β + 1), it has at least one solution. Next, we
remark that c(x) = (ℓx + θ)β+1 − θβ+1 is a convex increasing function on [− θℓ , 0] whose graph connects the
points (− θℓ , −θβ+1 ) and (0, 0). We further note that the integral curves of (75) cannot intersect the curve c
dc
when approaching it from the right. Indeed, dx
= (β + 1)(ℓx + θ)β ℓ < (β + 1)θβ ℓ when x < 0. Assuming
⏐
dy ⏐
dc
> dx
that such an intersection occurs, by (80), we obtain that dx
. Therefore, the only possibility for (79)
c
to hold is when the integral curve, solving (75), never leaves the third quadrant. This means that this curve
never intersects the y-axis, and therefore v(t) ̸= v ∗ (t) for all t. This implies that u(ρ, r) ̸= u∗ (r) for all
r > 0. □
5.2. Convergence of regular solutions
Here, we prove Theorem 1.7.
Proof of Theorem 1.7. Fix an arbitrary interval [r1 , r2 ] ⊂ (0, +∞) and find a sequence u( · , ρn ),
limn→∞ ρn = +∞, converging uniformly on [r1 , r2 ] to a singular solution u∗∗ to (78). Let us prove that
u∗∗ = u∗ , where u∗ is the singular solution to (78) given by (69).
1
For any B ⩾ 0, we define R∗ (B) = (u∗ )−1 (B). Recall that R∗ (0) = (λ∗ ) θ . From [20] (Theorem 3.1), we
know that R(0, ρn ) → R∗ (0).
Note that uB ( · , ρn ) = u( · , ρn ) − B solves the equation L(u) + eB eu = 0. This implies that ũB (r, ρn ) =
B
uB (e− θ r, ρn ) satisfies Eq. (78) and the initial conditions ũB (0, ρn ) = ρn − B, ∂r ũB (0, ρn ) = 0. Therefore,
B
B
B
uB (e− θ r, ρn ) = u(r, ρn − B). Hence, uB (e− θ R(0, ρn − B), ρn ) = 0 which is the same as u(e− θ R(0, ρn −
B), ρn ) = B. Thus, as n → ∞,
B
B
R(B, ρn ) = e− θ R(0, ρn − B) → e− θ R∗ (0) = R∗ (B),
B
where the convergence holds by Theorem 3.1 in [20]. To see why R∗ (B) = e− θ R∗ (0), we observe that
B
B
u∗B (e− θ r) = λ∗ − θ ln r = u∗ (r) (a verification is straightforward). This implies that u∗B (e− θ R∗ (0)) = 0,
B
which is the same as u∗ (e− θ R∗ (0)) = B. Hence,
u−1 ( · , ρn ) → (u∗ )−1
in Cloc (0, +∞).
For simplicity, we write un for u( · , ρn ). Since un and u∗∗ are decreasing in r and un ⇒ u∗∗ on [r1 , r2 ], for
a small ε > 0, there exists N ∈ N such that [un (r2 ), un (r1 )] ⊂ [u∗∗ (r2 ) − ε, u∗∗ (r1 ) + ε] for n ⩾ N . Since
∗ −1
u−1
on [u∗∗ (r2 ) − ε, u∗∗ (r1 ) + ε], we obtain that uniformly on [r1 , r2 ],
n ⇒ (u )
∗ −1
|r − (u∗ )−1 (un (r))| = |u−1
(un (r))| → 0
n (un (r)) − (u )
as n → ∞.
By continuity of u∗ ,
|u∗ (r) − un (r)| → 0
as n → ∞
uniformly on [r1 , r2 ]. Therefore, u∗ = u∗∗ .
Let us obtain the first limit in (10). Suppose it is not true. Then, one can find an interval [r1 , r2 ] ⊂
(0, +∞), a sequence ρn → +∞, and a number ε > 0 such that |u(r, ρn ) − u∗ (r)| > ε on [r1 , r2 ] for all n.
On the other hand, we can find a subsequence u( · , ρnk ) which converges to a solution of (78) uniformly on
[r1 , r2 ]. By Corollary 3.1, this solution is singular. By what was proved, this singular solution is u∗ . This is
a contradiction. Thus, the first limit in (10) is obtained.
To show the second equality in (10), we note that by Theorem 3.1 from [20], λ(ρ) → λ∗ as ρ → +∞.
1
1
Since uλ(ρ) (r) = u(λ(ρ) θ r, ρ) and u∗λ∗ (r) = u∗ (λ∗ θ r), the result follows. □
35
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
Corollary 5.1. Let assumptions of Theorem 1.7 be fulfilled and let λ# be as in Theorem 1.4. Then, if
0 < α − β − 1 < 4θ̂ (2k < d < 2k + 8), (uλ# , λ# ) is a regular solution; if α − β − 1 ⩾ 4θ̂ (d ⩾ 2k + 8), then
λ# = λ∗ .
Proof . If 0 < α − β − 1 < 4θ̂, the statement follows from Theorems 4.3 and 1.7. If α − β − 1 ⩾ 4θ̂, the
statement follows from Theorem 3.1 (case III) in [20]. □
Open problems
In this work, some of the results were obtained for the case f (u) = u. It would be interesting to obtain
those results, along with the convergence λ(ρ) → λ∗ , for problem (Pλ ). More specifically, the results of
interest are
(i) For problem (Pλ ), to show that λ(ρ) → λ∗ as ρ → +∞.
(ii) For the same problem, to obtain the result of Theorem 1.7 about the convergence of regular solutions
to the singular.
(iii) For Eq. (7), to find a condition on α, β, γ so that the singular and regular solutions do not intersect
each other. Such a result would be a generalization of Theorem 1.6.
These problems are the subject of our future research.
Acknowledgments
The authors thank the anonymous referee for valuable suggestions on improvements of the readability
of the paper. J.M. do Ó acknowledges partial support from CNPq through the grants 312340/2021-4 and
429285/2016-7 and from Paraı́ba State Research Foundation (FAPESQ), grant no 3034/2021. E. Shamarova
acknowledges partial support from Universidade Federal da Paraı́ba (PROPESQ/PRPG/UFPB) through
the grant PIA13631-2020 (public calls no 03/2020 and 06/2021).
Appendix
Remark that the radial form of problem (1) is the following:
⎧ k
C
⎪
⎨ dd r1−d (rd−k w′ (r)k )′ = λef (−w) ,
w < 0, 0 ⩽ r < 1,
⎪
⎩
w(1) = 0.
0 ⩽ r < 1,
(81)
We show that for regular radial solutions, (81) can be reduced to (Pλ ), and vice versa. Since the solutions
of interests are radial in the ball, we have that w′ (0) = 0. Therefore, the integral form of the above equation
is
∫
λd r d−1 f (−w(s))
d−k ′
k
r
w (r) = k
s e
ds.
Cd 0
This shows that w′ (r) cannot take zero values inside the ball except for the origin. Since w(1) = 0 and
w′ (0) = 0, w(r) is strictly increasing, and therefore, w′ (r) > 0 for 0 < r ⩽ 1. By doing the substitution
u = −w, we rewrite problem (81) as follows:
⎧
Cd
k−1 ′
⎪
u (r))′ = λef (u) , 0 ⩽ r < 1,
⎨− dk r1−d (rd−k |u′ (r)|
u > 0, 0 ⩽ r < 1,
⎪
⎩
u(1) = 0.
36
J.M. do Ó, E. Shamarova and E. da Silva
Nonlinear Analysis 222 (2022) 113000
By this, in the context of regular solutions, problem (1) is equivalent to problem (Pλ ) with the operator L
defined by (5) and α = d − k, β = k − 1, γ = d − 1.
On the other hand, as we showed in Section 2.4.1, for any singular solution to (Pλ ), it holds that (u∗λ∗ )′ < 0.
This implies that wλ∗ = −u∗λ∗ is a singular solution to (81).
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