# Singular solutions to $\bm{k}$-Hessian equations
with fast-growing nonlinearities

João Marcos do Ó∗ Departamento de Matemática, Universidade Federal da Paraíba,
João Pessoa, Brazil<EMAIL_ADDRESS>,  Evelina Shamarova  Departamento de
Matemática, Universidade Federal da Paraíba, João Pessoa, Brazil
<EMAIL_ADDRESS>and  Esteban da Silva  Departamento de
Matemática, Universidade Federal do Rio Grande do Norte, Natal, Brazil
<EMAIL_ADDRESS>
###### 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 $d$.

∗Corresponding Author

Keywords: $k$-Hessian equation; singular solution; multiplicity of regular
solutions; number of intersection points; bifurcation diagrams; Liouville-
Bratu-Gelfand problem

2020 MSC: 34B15, 34C23, 35A24, 35J15, 35J62, 35J92

## 1\. Introduction

In this paper, we are concerned with the following Dirichlet problem:

(1) $\begin{cases}S_{k}(D^{2}w)=\lambda\,F(-w)&\quad\text{ in }\Omega,\\\
w<0,&\quad\text{ in }\Omega,\\\ w=0,&\quad\text{ on
}\partial\Omega,\end{cases}$

where $\Omega\subset{\mathbb{R}}^{d}$ is a bounded domain, $F$ is a positive
function increasing to infinity, whose exact properties will be specified
later, and $\lambda>0$ is a parameter. Above, $1\leqslant k\leqslant d$ is an
integer, where $d>2$ is the dimension, and $S_{k}(D^{2}w)$ is the $k$-Hessian,
i.e., the symmetric elementary function of the $k$-th order of the eigenvalues
of $D^{2}w$.

We prove the existence of a specific radial singular solution to (1) in the
case when $\Omega$ 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

(2) $\begin{cases}-\Delta u=\lambda G(u),&\quad\text{ in }\Omega,\\\
u=0,&\quad\text{ on }\partial\Omega,\end{cases}$

where $G$ is smooth and increasing to infinity, have been of significant
interest; see, e.g., [1, 2, 3, 7, 9, 10, 11, 12, 18, 19, 24, 28, 30, 32]. This
list of references is by no means complete, so we refer the reader to [1, 3,
7, 19, 32], and monograph [10] 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 $\lambda^{\\#}$ such that
for each $\lambda\in(0,\lambda^{\\#})$ there exists a regular solution
$(u_{{\scriptscriptstyle\lambda}},\lambda)$ to problem (2), and there are no
solutions to (2) for $\lambda>\lambda^{\\#}$; 2) the existence of a parameter
$\lambda^{*}\in(0,\lambda^{\\#}]$ such that problem (2) possesses a singular
solution for $\lambda=\lambda^{*}\\!.$ 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, 33, 34, 35,
38, 39, 40, 41]. When restricted to so-called $k$-admissible solutions (for
the definition see e.g. [38]), 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 [15] and differential geometry [36].

Problem (1) is in turn a larger generalization, compared to those that exist
in the literature [12, 18, 24, 28], of the classical Gelfand problem. The
latter deals with positive solutions to the system

(3) $\begin{cases}\Delta u+\lambda\,e^{u}=0&\quad\text{ in }\Omega,\\\
u=0&\quad\text{ on }\partial\Omega.\end{cases}$

Problem (3) is a fundamental problem that arises in many theories, such as
non-linear diffusions generated by non-linear sources [20, 21, 23], thermal
ignition of a mixture of chemically active gases [11], unstable membrane [4],
and gravitational equilibrium of polytropic stars [5, 17], 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., [11, 19], since in this case,
it is known that all solutions are radial [14] and problem (3) has the
following simple representation:

(4) $\begin{cases}u^{\prime\prime}+\frac{d-1}{r}u^{\prime}+\lambda e^{u}=0,\\\
u^{\prime}(0)=u(1)=0.\end{cases}$

In [11], Gelfand proved the existence of the value $\lambda$ for which (4) has
an infinite number of solutions. In [19], the authors completely described the
relation between the multiplicity of solutions to (4) and the dimension $d$ of
the ball.

Problem (4) was further generalized by various authors in the following two
directions: replacing $e^{u}$ with a more general function while keeping the
radial Laplacian [12, 24, 28, 30]; and replacing the Laplacian with the radial
operator

(5) $\displaystyle
L(u)=r^{-\gamma}(r^{\alpha}|u^{\prime}(r)|^{\beta}u^{\prime}(r))^{\prime}$

while preserving the term $e^{u}$ [18]. More specifically, in [12], the term
$e^{u}$ was replaced with the $n$-times iterated exponential (henceforth
denoted by $\exp^{\circ n}(u)$), in [24], it was replaced with $e^{u^{p}}\\!,$
and in [28], this term was replaced with $e^{u}+\text{``lower-order term''}$.

### 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 $L(u)+\lambda
e^{u}=0$. 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 Appendix for
details) as follows:

($P_{\lambda}$) $\begin{cases}L(u)+\lambda\,e^{f(u)}=0,\quad 0\leqslant
r<1,\\\ u>0,\quad 0\leqslant r<1,\\\ u(1)=0,\end{cases}$

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 Subsection 2.1.

  * (1)

$f$ is smooth and increasing to infinity;

  * (2)

$|g^{\prime\prime}(t)|$ monotonically decreases to zero;

  * (3)

$\lim_{t\to+\infty}\frac{f(g(t)+\varepsilon(t))}{t}=1$, where
$\varepsilon(t)=O(g^{\prime\prime}(s))$;

  * (4)

As $t\to+\infty$, $\frac{g^{\prime}(t)}{g^{\prime}(t+o(t))}\to 1$ and
$\frac{g^{\prime\prime}(t)}{g^{\prime\prime}(t+o(t))}=O(1)$;

  * (5)

$\lim_{u\to+\infty}f^{\prime}(u)=+\infty$,
$\lim_{u\to+\infty}\frac{f^{\prime\prime}(u)}{f^{\prime}(u)^{2}}(\ln
f^{\prime}(u))^{2}=0$;

  * (6)

$\alpha>\beta+1$ (i.e., $d>2k$ for the $k$-Hessian); $\beta\geqslant 0$;
$\gamma+1>\alpha$.

We emphasize that problem ($P_{\lambda}$) 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 [12, 24, 28], such as $\exp(u)$, $\exp(u^{p})$,
the iterated exponential $\exp^{\circ n}(u)$, $\exp(u)+\text{``lower-order
term''}$ (see Subsection 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 $\alpha$, $\beta$, and $\gamma$ 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 $\alpha$, $\beta$, and $\gamma$ for a
$k$-Hessian and a $p$-Laplacian are summarized in the table below. The table
includes the parameter $\theta=\gamma+2+\beta-\alpha$ whose role will become
clear later.

|  |  $\alpha$ |  $\beta$ |  $\gamma$ |  $\theta$
---|---|---|---|---|---
$k$-Hessian |  |  $d-k$ |  $k-1$ |  $d-1$ |  $2k$
$p$-Laplacian |  |  $d-1$ |  $p-2$ |  $d-1$ |  $p$

######  Remark 1.1.

Remark that since the solutions of interest to problem (1) are negative,
according to [38], the operator $S_{k}(D^{2}w)$ (regarded as a $d\times d$
matrix) is elliptic on those solutions. On the other hand, if $u$ is a
solution to problem ($P_{\lambda}$), the operator $S_{k}(D^{2}u(|\cdot|))$ may
not be elliptic since $S_{k}(D^{2}u)=(-1)^{k}S_{k}(D^{2}w)$. Its ellipticity
depends on the parity of $k$.

### 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 ($P_{\lambda}$)
(see 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 ${\rm C}^{2}$-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 ($P_{\lambda}$), we
understand a ${\rm C}^{2}$-function which solves problem ($P_{\lambda}$) 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, 12, 13, 24, 25, 27, 28, 29, 30, 31]. Most of these
works deal however with semilinear equations involving the Laplacian. Papers
[13] and [31] 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 [12, 24, 28, 29, 30]
and blow-up behavior of solutions to parabolic problems [26, 37]. Asymptotic
representations for radial singular solutions on a unit ball were previously
obtained in [12, 24, 27, 28]. 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 [12, 24, 28] are included in
ours as particular cases. Our assumptions on $f$ are quite flexible, so
examples of $f$ include not only $\exp^{\circ n}(u)$, $u^{p}$, and
$u+\text{``lower-order term''}$ (cf. [12, 24, 28]), but also many others. As
such, we show that $f(u)=\exp^{\circ n}(u^{p})$, satisfies our assumptions,
encompassing both $\exp^{\circ n}(u)$ from [12] and $u^{p}$ from [24]. We also
remark that, compared to [25, 29, 30], 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 $\lambda^{*}>0$ such that problem
($P_{\lambda}$) with $\lambda=\lambda^{*}$ possesses a radial singular
solution of the form

(6) $\displaystyle
u^{*}_{{\scriptscriptstyle\lambda}^{\\!*}}(r)=g(Z(r))+O\big{(}g^{\prime\prime}(\ln
r^{-\theta})\big{)},\quad\text{where}$ $\displaystyle
Z(r)=\ln\big{\\{}\theta^{\beta+1}(\alpha-\beta-1)\\}-\ln\lambda^{*}-\theta\ln
r$ $\displaystyle\hskip
19.91692pt+(\beta+1)\ln\\{g^{\prime}(\tau)+(\beta+1)g^{\prime\prime}(\tau)\ln
g^{\prime}(\tau)\big{\\}}\quad\text{with}\;\;\tau=\ln\Big{\\{}\frac{\beta+1}{\lambda^{*}r^{\theta}}\Big{\\}}.$

The important particular case is the exact form for the singular solution to
problem ($P_{\lambda}$) in the case $f(u)=u$. Namely, the following corollary
holds.

######  Corollary 1.1.

Assume (6) and consider problem ($P_{\lambda}$) with $f(u)=u$. Then,
$(u^{*}_{{\scriptscriptstyle\lambda}^{\\!*}},\lambda^{*})$, given by

$\displaystyle
u^{*}_{{\scriptscriptstyle\lambda}^{\\!*}}(x)=-\theta\ln|x|,\quad\lambda^{*}=\theta^{\beta+1}(\alpha-\beta-1),$

is a radial singular solution to problem ($P_{\lambda}$).

We furthermore use the singular solution
$(u^{*}_{{\scriptscriptstyle\lambda}^{\\!*}},\lambda^{*})$, constructed in
Theorem 1.1, to study properties of regular solutions such as bifurcation
diagrams and multiplicity. Specifically, we obtain the following result.

######  Theorem 1.2.

Assume (1)–(6) and let $\alpha-\beta-1<\frac{4\theta}{\beta+1}$ $(d<2k+8)$.
Further, we let
$(u_{\scriptscriptstyle{\scriptscriptstyle\lambda}(\rho)},\lambda(\rho))$ be
the family of regular solutions to problem ($P_{\lambda}$) parametrized by
$\rho=\sup_{r\leqslant 1}|u_{\scriptscriptstyle\lambda}(r)|$, and let
$(u^{*}_{{\scriptscriptstyle\lambda}^{\\!*}},\lambda^{*})$ be the singular
solution constructed in Theorem 1.1. Then, as $\rho\to+\infty$,
$\lambda(\rho)$ oscillates around $\lambda^{*}$. In particular, there are
infinitely many regular solutions to problem ($P_{\lambda}$) for
$\lambda(\rho)=\lambda^{*}$.

An important tool for obtaining the above result is the number of intersection
points between the regular and singular solutions to the equation

(7) $\displaystyle L(u)+e^{f(u)}=0.$

Note that
$u^{*}(r)=u^{*}_{{\scriptscriptstyle\lambda}^{\\!*}}(r(\lambda^{*})^{-\frac{1}{\theta}})$
is a singular solution to (7). In general, the change of variable

(8) $\displaystyle
u(r)=u_{\scriptscriptstyle\lambda}(r\lambda^{-\frac{1}{\theta}})$

brings the equation in ($P_{\lambda}$), containing $\lambda$, to equation (7).
As such, the radial regular solution $u(\,\cdot\,,\rho)$ to (7) is obtained
from the radial regular solution
$(u_{\scriptscriptstyle{\scriptscriptstyle\lambda}(\rho)},\lambda(\rho))$ to
($P_{\lambda}$) by the change of variable (8). We have the following result.

######  Theorem 1.3.

Assume (1)–(6) and let $\alpha-\beta-1<\frac{4\theta}{\beta+1}$. Then, for any
$\delta>0$, it holds that
$\lim_{\rho\to+\infty}\mathscr{L}_{(0,\delta)}(u(\,\cdot\,,\rho)-u^{*}(\,\cdot\,))=+\infty$,
where $u$ and $u^{*}$ are regular and singular solutions to (7) and
$\mathscr{L}_{(0,\delta)}(u-u^{*})$ is the number of zeros of $u-u^{*}$ on the
interval $(0,\delta)$.

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 $\lambda^{\\#}>0$,
depending on $\alpha$, $\beta$, $\gamma$, and $f$, such that any regular
solution $(u_{\scriptscriptstyle\lambda},\lambda)$ to problem ($P_{\lambda}$)
satisfies $\lambda\leqslant\lambda^{\\#}$. Moreover, for any
$\lambda<\lambda^{\\#}$, there exists a regular solution to ($P_{\lambda}$),
and for $\lambda=\lambda^{\\#}$, 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 Subsection 3.3.

Finally, we study the equation

(9) $\displaystyle L(u)+e^{u}=0$

In [18], the authors considered problem ($P_{\lambda}$) with the nonlinearity
$e^{u}$ and obtained the result of Theorem 1.2 along with the convergence
$\lambda(\rho)\to\lambda^{*}$ as $\rho\to+\infty$. 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
$\alpha$, $\beta$, and $\gamma$, and by proving the convergence of regular
solutions to the singular as $\rho\to+\infty$. Namely, we have the following
results.

######  Theorem 1.5.

Assume (6). Suppose $0<\alpha-\beta-1<\frac{4\theta}{\beta+1}$ $(2k<d<2k+8)$.
Then, $\mathscr{L}_{(0,+\infty)}(u-u^{*})=+\infty$.

######  Theorem 1.6.

Assume (6). Suppose $\alpha-\beta-1\geqslant 4\theta(\beta+1)$ $(d\geqslant
2k+8k^{2})$. Then, $\mathscr{L}_{(0,+\infty)}(u-u^{*})=0$.

######  Theorem 1.7.

Assume (6). Then,

(10)
$\displaystyle\lim_{\rho\to+\infty}u(\,\cdot\,,\rho)=u^{*}\quad\text{and}\quad\lim_{\rho\to+\infty}(u_{\scriptscriptstyle{\scriptscriptstyle\lambda}(\rho)},\lambda(\rho))=(u^{*}_{{\scriptscriptstyle\lambda}^{\\!*}},\lambda^{*})\quad\text{in}\;\;{\rm
C}_{loc}(0,+\infty).$

######  Remark 1.2.

Remark that the results on the intersection number, similar to those of
Theorems 1.5 and 1.6, are known for the equation $L(u)+|u|^{p-1}u=0$ (see e.g.
[29]). Here they are obtained for the exponential counterpart of this
equation.

### 1.4 Methods

In this work, we use combinations of newly developped 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 equation (9) do not intersect each other if
$\alpha-\beta-1\geqslant 4\theta(\beta+1)$ (Theorem 1.5). To be specific, in
the proof of Theorem 1.1, we arrive at the equation

(11) $\displaystyle
v^{\prime\prime}-(\hat{\alpha}-1)v^{\prime}=-\frac{e^{-\theta
t+f(v)}}{|v^{\prime}|^{\beta}}$

which is equivalent to the equation in ($P_{\lambda}$), but the variable $t$
varies in a neighbourhood of infinity. The difficulty arises due to the fact
that the denominator on the right-hand side contains $|v^{\prime}|^{\beta}$,
the term which is not present in the Laplacian case ($\beta=0$). This does not
allow to apply methods known for semilinear equations with the Laplacian (see,
e.g., [12, 24, 28]). In particular, the singular solution $v^{*}$ to (11)
takes the form

(12) $\displaystyle v^{*}(t)=g(\theta
t+\varphi_{1}(t)+\varphi_{2}(t))+\eta(t),$

where

(13) $\displaystyle\varphi_{1}=\ln\\{\theta^{\beta}\big{(}g^{\prime}(\theta
t)+(\beta+1)g^{\prime\prime}(\theta t)\ln g^{\prime}(\theta
t)\big{)}^{\beta}\\},$
$\displaystyle\varphi_{2}=\ln\Big{\\{}\theta\Big{(}\frac{\alpha}{\beta+1}-1\Big{)}\Big{\\}}+\ln\\{g^{\prime}(\theta
t)+(\beta+1)g^{\prime\prime}(\theta t)\ln g^{\prime}(\theta t)\\}$

and $\eta(t)=O(g^{\prime\prime}(\theta t))$. Remark that the term
$\varphi_{1}$ does not appear in the Laplacian case; it was for the first time
introduced in this work. Its role is to make certain cancelations with the
denominator $|{v^{*}}^{\prime}|^{\beta}$ to ensure the possibility to reduce
(11) to a pair of semilinear equations with respect to $\eta$ and
$\eta^{\prime}$. The order of the reminder $\eta$, in the general setting, was
also first established in this work. Additional technical difficulties arise
due to the fact that the function $f$ (unlike [12, 24, 28]) is arbitrary, so
the asymptotic representation (LABEL:as1) 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
$(0,+\infty)$ (along with the result of [18]). 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 equation (7):

(14)
$\displaystyle\tilde{u}(s,\rho)=\mathcal{F}^{-1}_{1}\varepsilon_{\rho}^{-\frac{\theta}{\beta+1}}\mathcal{F}(u(\varepsilon_{\rho}s,\rho)),\quad\varepsilon_{\rho}=\Big{(}\frac{\mathcal{F}(\rho)}{\mathcal{F}_{1}(1)}\Big{)}^{(\beta+1)/\theta},$

where
$\mathcal{F}(u)=\int_{u}^{+\infty}\exp\big{\\{}\\!\\!-\\!\frac{f(s)}{1+\beta}\big{\\}}ds$
and $\mathcal{F}_{1}(u)=(\beta+1)e^{-\frac{u}{\beta+1}}$. 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 [13]. The “Laplacian” analog of (14) can be found in,
e.g., [30]. The important difference with the Laplacian case is the following:
when we apply this tranformation, the first two terms
$L(\tilde{u})+e^{\tilde{u}}$ of the transformed equation are exactly as in the
canonical equation (9); however, the additional term
$s^{\alpha-\gamma}|\tilde{u}^{\prime}|^{\beta+2}\big{(}\frac{I(u(\varepsilon_{\rho}s))}{\beta+1}-1\big{)}$,
where $I(u)=\mathcal{F}(u)f^{\prime}(u)\exp\\{\frac{f(u)}{\beta+1}\\}$, has a
more complicated structure and its convergence to zero is rather difficult to
show. This convergence however is crucial to show the convergence of
$\tilde{u}$ to the regular solution of the canonical (limit) equation (9).
Remark that even in the Laplacian case, considered in [30], 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 (LABEL:as1) obtained in
Theorem 1.1. Thus, we reinforce that the proof of Theorem 1.3 on the
intersection number differs significanly from the Laplacian case and requires
new tools developed in this work.

Furthermore, the proof of Theorem 1.2 on bifurcation diagrams relies havily on
Theorem 1.3 discussed above. In addition, the aforementioned proof contains a
much more detailed analysis on the oscillation of $\lambda(\rho)$ arround
$\lambda^{*}$ compared to [30] and preceeding works.

Finally, we mention Proposition 3.1 that deals with upper and lower bounds for
$u(\,\cdot\,,\rho)^{-1}(B)$, $B>0$, whenever $\rho$ is sufficiently big. In
the Laplacian case, this result was obtained in [25] (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 equation (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 ($P_{\lambda}$) 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
($P_{\lambda}$). 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 between the regular and singular solutions for
the case $0<\alpha-\beta-1\leqslant\frac{4\theta}{\beta+1}$. The main results
of this section are Theorems 1.2 and 1.3. It is important to mention that we
employ here the tranformation (14) that reduces equation (7) to an equivalent
equation whose limit is equation (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 ($P_{\lambda}$)

In this section, we will construct a positive radial singular solution to
($P_{\lambda}$).

### 2.1 Standing assumptions and useful lemmas

We start by introducing the general set of assumptions implying assumptions
(1)–(6) from Subsection 1.2. Recall that $g$ denotes the inverse function for
$f$.

  * (A1)

$f\in{\rm C}^{4}([0,+\infty))$, $f^{\prime}>0$, and
$\lim_{u\to+\infty}f(u)=+\infty$.

  * (A2)

$\lim\limits_{t\to+\infty}g^{\prime\prime}(t)=0$ and
$\lim\limits_{t\to+\infty}\frac{f(g(t)+\varepsilon(t))}{t}=1$, where
$\varepsilon(t)=O(g^{\prime\prime}(t))$.

  * (A3)

$\lim_{t\to+\infty}\frac{g^{\prime}(t)}{g^{\prime}(t+o(t))}=1$.

  * (A4)

$\lim_{t\to+\infty}\frac{g^{\prime\prime}(t)}{g^{\prime}(t)}\ln
g^{\prime}(t)=0$.

  * (A5)

As $t\to+\infty$,

$\displaystyle{\rm(a)}\;\;g^{\prime\prime\prime}(t)(\ln
g^{\prime}(t))^{n}=O(g^{\prime\prime}(t)),\;\;n=0,1,2,\quad
g^{(\rm{iv})}(t)\ln g^{\prime}(t)=O(g^{\prime\prime}(t)),$
$\displaystyle{\rm(b)}\;\;\sup\nolimits_{s\geqslant
t}|g^{\prime\prime}(s)|=O(g^{\prime\prime}(t)),\quad
g^{\prime\prime}(t+o(t))=O(g^{\prime\prime}(t)).$

  * (A6)

$\alpha>\beta+1$ (i.e., $d>2k$ for the $k$-Hessian); $\beta\geqslant 0$;
$\theta>0$.

######  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

  * (A2’)

$\lim_{u\to+\infty}\frac{f^{\prime\prime}(u)}{f^{\prime}(u)^{3}}=0$ and
$\lim_{t\to+\infty}\frac{f(u+\tilde{\varepsilon}(u))}{f(u)}=1$, where
$\tilde{\varepsilon}(u)=O(\frac{f^{\prime\prime}(u)}{f^{\prime}(u)^{3}})$.

  * (A4’)

$\lim_{u\to+\infty}\frac{f^{\prime\prime}(u)}{f^{\prime}(u)^{2}}\ln
f^{\prime}(u)=0$;

We start showing (A4’). A straightforward computation gives

$\displaystyle\frac{f^{\prime\prime}(u)}{f^{\prime}(u)^{2}}\ln
f^{\prime}(u)=-\frac{g^{\prime\prime}(t)}{g^{\prime}(t)}\ln
g^{\prime}(t),\quad\text{where}\;\;u=g(t).$

To show the equivalence of (A2) and (A2’), we note that
$g^{\prime\prime}(f(u))=\frac{f^{\prime\prime}(u)}{f^{\prime}(u)^{3}}$, and
hence, the first limits in (A2) and (A2’) are equivalent. Substituting
$t=f(u)$, we represent $\eta(t)$ as $\phi(u)g^{\prime\prime}(f(u))$, where
$\phi(u)$ is a bounded function. Defining
$\tilde{\eta}(u)=\phi(u)\frac{f^{\prime\prime}(u)}{f^{\prime}(u)^{3}}$, we
obtain the desired equivalence.

######  Remark 2.2.

For the $p$-Laplacian, the condition $\alpha>\beta+1$ is interpreted as $p<d$.

######  Remark 2.3.

Note that under (A4) and (A2),
$\lim_{t\to+\infty}\frac{g^{\prime\prime}(t)}{g^{\prime}(t)}=0$. Indeed,
assume the opposite. Then, there exists a sequence $t_{n}\to+\infty$ such that
$\big{|}\frac{g^{\prime\prime}(t_{n})}{g^{\prime}(t_{n})}\big{|}>M$ for some
constant $M>0$. By (A4), it should hold that
$\lim_{n\to+\infty}g^{\prime}(t_{n})=1$, which, in turn, implies that
$\lim_{n\to+\infty}\frac{g^{\prime\prime}(t_{n})}{g^{\prime}(t_{n})}=0$ by
(A2).

######  Remark 2.4.

Some of examples of the function $\phi(t)=g^{\prime\prime}(t)$ satisfying the
first expression in (A5)-(b) are monotone functions and functions which can be
represented as $\phi(t)=\phi_{1}(t)\phi_{2}(t)$, where $\phi_{1}$ is
decreasing to zero as $t\to+\infty$, and $\phi_{2}$ is bounded from above and
below, with an infinite number of maximum and minimum points, such as $2+\sin
t$ or $e^{\sin t}$.

######  Remark 2.5.

In what follows, we will need the following facts: as $t\to+\infty$,

$\displaystyle
e^{\frac{t}{\beta+1}}g^{\prime}(t)=O(t\,e^{\frac{t}{\beta+1}}),\qquad
e^{-\frac{t}{\beta+1}}g^{\prime}(t)=O(t\,e^{-\frac{t}{\beta+1}}).$

The above expressions obviously hold by (A2). Indeed, $g^{\prime\prime}(t)$ is
bounded, and therefore, by Taylor’s formula, $g^{\prime}(t)$ has at most
linear growth.

It turns out that some of the conditions (A5)-(a) are fulfilled if
$\lim_{u\to+\infty}f^{\prime}(u)=+\infty$, or, which is the same,
$\lim_{t\to+\infty}g^{\prime}(t)=0$.

######  Lemma 2.1.

Suppose $\lim_{u\to+\infty}f^{\prime}(u)=+\infty$. Then, conditions (A5)-(a),
except for $n=2$, are fulfilled. If, in addition,
$\lim_{u\to+\infty}\frac{f^{\prime\prime}(u)}{f^{\prime}(u)^{2}}(\ln
f^{\prime}(u))^{2}=0$, then all conditions in (A5)-(a) are fulfilled.

###### Proof.

Note that $g^{\prime\prime}(t)$ cannot be identically equal to zero in a
neighborhood of $+\infty$. Indeed, in this case $g^{\prime}(t)=0$ in this
neighborhood which is a contradiction.

Note that, by (A4), $\lim_{t\to\infty}g^{\prime\prime}(t)\ln g^{\prime}(t)=0$.
By L’Hopital’s rule,

$\displaystyle\lim_{t\to+\infty}\frac{g^{\prime\prime\prime}(t)}{g^{\prime\prime}(t)}\ln
g^{\prime}(t)=\lim_{t\to+\infty}\frac{g^{\prime\prime}(t)}{g^{\prime}(t)}\ln
g^{\prime}(t)-\lim_{t\to+\infty}\frac{g^{\prime\prime}(t)}{g^{\prime}(t)}=0.$

The above inequality implies, in particular, that
$\lim_{t\to+\infty}g^{\prime\prime\prime}(t)\ln g^{\prime}(t)=0$. Again, by
L’Hopital’s rule,

$\displaystyle\lim_{t\to+\infty}\frac{g^{(4)}(t)}{g^{\prime\prime\prime}(t)}\ln
g^{\prime}(t)=\lim_{t\to+\infty}\frac{g^{\prime\prime\prime}(t)}{g^{\prime\prime}(t)}\ln
g^{\prime}(t)-\lim_{t\to+\infty}\frac{g^{\prime\prime}(t)}{g^{\prime}(t)}=0.$

Note that
$\lim_{t\to+\infty}\frac{g^{\prime\prime\prime}(t)}{g^{\prime\prime}(t)}=0$.
Therefore,

$\displaystyle\lim_{t\to+\infty}\frac{g^{(4)}(t)}{g^{\prime\prime}(t)}\ln
g^{\prime}(t)=\lim_{t\to+\infty}\frac{g^{(4)}(t)}{g^{\prime\prime\prime}(t)}\ln
g^{\prime}(t)\lim_{t\to+\infty}\frac{g^{\prime\prime\prime}(t)}{g^{\prime\prime}(t)}=0.$

Finally, a straightforward computation shows that

$\displaystyle\frac{f^{\prime\prime}(u)}{f^{\prime}(u)^{2}}(\ln
f^{\prime}(u))^{2}=\frac{g^{\prime\prime}(t)}{g^{\prime}(t)}(\ln
g^{\prime}(t))^{2},\quad\text{where}\;\;u=g(t).$

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 Subsection 1.2.

######  Lemma 2.2.

If $\zeta(t)$ is a differentiable function ${\mathbb{R}}_{+}\to{\mathbb{R}}$
such that $\lim_{t\to+\infty}\zeta^{\prime}(t)=0$, then
$\lim_{t\to+\infty}\frac{\zeta(t)}{t}=0$.

###### Proof.

We have

$\displaystyle\zeta(0)-\int_{0}^{t}|\zeta^{\prime}(s)|ds\leqslant\zeta(t)\leqslant\zeta(0)+\int_{0}^{t}|\zeta^{\prime}(s)|ds.$

The integral $\int_{0}^{t}|\zeta^{\prime}(s)|ds$ is either bounded or tends to
$+\infty$ as $t\to+\infty$. 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 ($P_{\lambda}$) satisfying
(A1)–(A5).

Example 1. $f(u)=\exp^{\circ n}(u^{p})$ ($p>0,n\geqslant 1$). First, introduce
the notation for the iterated exponential and the iterated logarithm:

$\displaystyle\exp^{\circ
n}(t)=\underbrace{\exp\circ\exp\circ\dots\circ\exp}_{n}(t);\qquad\ln^{\circ
n}(t)=\underbrace{\ln\circ\ln\circ\dots\circ\ln}_{n}(t).$

The inverse function to $f$ is $g(t)=[\ln^{\circ n}(t)]^{\frac{1}{p}}$. We
prove that (A1)–(A5) are fulfilled for the pair $f,g$.

######  Remark 2.7.

Remark that in [12], $f(u)=\exp^{\circ n}(u)$, and in [24], $f(u)=u^{p}$, both
are particular cases of $f(u)=\exp^{\circ n}(u^{p})$ discussed in this
example. In addition, we remark that [12] and [24] deal with the classical
radial Laplacian, while our partial differential operator takes the form (5).

First, we compute

$\displaystyle g^{\prime}(t)$ $\displaystyle=\frac{1}{p}\,\frac{1}{\ln^{\circ
n}(t)^{1-\frac{1}{p}}\ln^{\circ(n-1)}(t)\ln^{\circ(n-2)}(t)\cdots t}\,,$
$\displaystyle g^{\prime\prime}(t)$
$\displaystyle=\frac{1}{p}\Big{[}\Big{(}\frac{1}{p}-1\Big{)}\frac{1}{\ln^{\circ
n}(t)^{2-\frac{1}{p}}\ln^{\circ(n-1)}(t)^{2}\cdots t^{2}}-\frac{1}{\ln^{\circ
n}(t)^{1-\frac{1}{p}}\ln^{\circ(n-1)}(t)^{2}\cdots t^{2}}$
$\displaystyle-\frac{1}{\ln^{\circ
n}(t)^{1-\frac{1}{p}}\ln^{\circ(n-1)}(t)\ln^{\circ(n-2)}(t)^{2}\cdots
t^{2}}-\cdots-\frac{1}{\ln^{\circ
n}(t)^{1-\frac{1}{p}}\ln^{\circ(n-1)}(t)\cdots t^{2}}\Big{]}.$

Note that $\lim_{t\to+\infty}g^{\prime}(t)=0$. Moreover,
$\lim_{t\to+\infty}\frac{g^{\prime\prime}(t)}{g^{\prime}(t)}(\ln
g^{\prime}(t))^{2}=0$. To see the latter, we note that
$\frac{g^{\prime\prime}(t)}{g^{\prime}(t)}=O(\frac{1}{t})$ and $\ln
g^{\prime}(t)=O(\ln(t))$. Therefore, by Lemma 2.1, (A5)-(a) is fulfilled.
Also, $|g^{\prime\prime}(t)|$ 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 logarithm, it holds that
$\lim_{t\to+\infty}\frac{\ln^{\circ n}(t)}{\ln^{\circ n}(t+o(t))}=1$. We prove
the following statement by induction on $n$: $\ln^{\circ
n}(t+o(t))=\tau+o(\tau)$, where $\tau=\ln^{\circ n}(t)$. 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., $\ln^{\circ(n-1)}(t+o(t))=\tau+o(\tau)$ with
$\tau=\ln^{\circ(n-1)}(t)$. Then, $\ln^{\circ
n}(t)=\ln(\tau+o(\tau))=\ln\tau+o(\ln(\tau))=\ln^{\circ n}(t)+o(\ln^{\circ
n}(t))$, 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 $\tilde{g}(t)=\ln^{\circ n}(t)$ so that
$g(t)^{p}=\tilde{g}(t)$. We have

$\displaystyle\big{(}g(t)+O(g^{\prime\prime}(t))\big{)}^{p}=\tilde{g}(t)\Big{(}1+\frac{O(g^{\prime\prime}(t))}{g(t)}\Big{)}^{p}=\tilde{g}(t)+O(g^{\prime\prime}(t))g(t)^{p-1}=\tilde{g}(t)+O(\tilde{g}^{\prime\prime}(t))$

since
$\tilde{g}^{\prime\prime}(t)=p\,g^{\prime\prime}(t)g(t)^{p-1}\big{(}1+(p-1)\frac{g^{\prime}(t)^{2}}{g(t)g^{\prime\prime}(t)}\big{)}=g^{\prime\prime}(t)g(t)^{p-1}O(1)$.
Indeed, $\frac{g^{\prime\prime}(t)}{g^{\prime}(t)}=\frac{1}{t}+o(\frac{1}{t})$
and $\frac{g^{\prime}(t)}{g(t)}=O(\frac{1}{t})$. We further argue by induction
on $n$. For $n=1$, (A2) is clearly satisfied. Suppose (A2) is true for $n-1$.
We have

$\exp(\tilde{g}(t)+O(\tilde{g}^{\prime\prime}(t)))=\ln^{\circ(n-1)}(t)\exp\\{O(\tilde{g}^{\prime\prime}(t))\\}=\ln^{\circ(n-1)}(t)(1+O(\tilde{g}^{\prime\prime}(t)))\\\
=\bar{g}(t)+O(\bar{g}^{\prime\prime}(t)),\quad\text{where}\;\bar{g}(t)=\ln^{\circ(n-1)}(t).$

Indeed, one immediately verifies that
$\tilde{g}^{\prime\prime}(t)\ln^{\circ(n-1)}(t)=\bar{g}^{\prime\prime}(t)+o(\bar{g}^{\prime\prime}(t))$.
(A2) now follows now from the induction hypothesis.

Example 2. $f(u)=u^{p}$, $p>\frac{1}{2}$. We restrict here the range of $p$ to
$(\frac{1}{2},+\infty)$ in order to guarantee that
$\lim_{t\to+\infty}g^{\prime\prime}(t)=0$. Indeed,
$g^{\prime\prime}(t)=\frac{1}{p}(\frac{1}{p}-1)\,t^{\frac{1}{p}-2}$. The
verification of (A1)–(A5) is straightforward.

Example 3. Perturbed Gelfand’s problem. Take $f(u)=u+\vartheta(u)$, where
$\vartheta(u)$ is a ${\rm C}^{4}$-smooth function with the following
properties:

  * (i)

$\lim_{u\to+\infty}\vartheta^{(n)}(u)=0$ for $n=0,1,2$;

  * (ii)

$u+\vartheta(u)$ is increasing on $[0,+\infty)$;

  * (iii)

$\vartheta^{\prime\prime\prime}(u)=O(\vartheta^{\prime\prime}(u))$,
$\vartheta^{(\rm{iv})}(u)\vartheta^{\prime}(u)=O(\vartheta^{\prime\prime}(u))$,
$\sup\nolimits_{v\geqslant
u}|\vartheta^{\prime\prime}(v)|=O(\vartheta^{\prime\prime}(u))$, and
$\vartheta^{\prime\prime}(u+o(u))=O(\vartheta^{\prime\prime}(u))$.

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 $g(t)=t+\tilde{\vartheta}(t)$, where
$\tilde{\vartheta}(t)=-\vartheta(u)$ with $u=g(t)$. We further note that
$g^{\prime}(t)=1+\tilde{\vartheta}^{\prime}(t)=\frac{1}{f^{\prime}(u)}=\frac{1}{1+\vartheta^{\prime}(u)}$,
which shows that $\lim_{t\to+\infty}\tilde{\vartheta}^{\prime}(t)=0$ and
immediately implies (A3). It remains to verify (A5). It is straightforward to
compute

$\displaystyle\tilde{\vartheta}^{\prime\prime}(t)=-\frac{\vartheta^{\prime\prime}(u)}{(1+\vartheta^{\prime}(u))^{4}}\quad\text{and}\quad\tilde{\vartheta}^{\prime\prime\prime}(t)=\frac{4\vartheta^{\prime\prime}(u)}{(1+\vartheta^{\prime}(u))^{6}}-\frac{\vartheta^{\prime\prime\prime}(u)}{(1+\vartheta^{\prime}(u))^{5}},$

which, by (iii), implies the first identity in (A5)-(a). Computing the fourth
derivative of $\tilde{\vartheta}(t)$, it is straightforward to obtain the
second identity in (A5)-(a). The expression for
$\tilde{\vartheta}^{\prime\prime}(t)$ 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:

$\theta=\gamma+2+\beta-\alpha$; $\hat{\alpha}=\frac{\alpha}{\beta+1}$;
$\hat{\theta}=\frac{\theta}{\beta+1}$.

$(u_{\scriptscriptstyle\lambda},\lambda)$ and
$(u^{*}_{{\scriptscriptstyle\lambda}^{\\!*}},\lambda^{*})$ denote regular and
singular solutions to problem ($P_{\lambda}$).

$u$ and $u^{*}$ are rescaled, by (8), regular and singular solutions that
solve (7).

$g^{\prime}$, $g^{\prime\prime}$, $g^{\prime\prime\prime}$, $g^{(\rm{iv})}$
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 $\Gamma(t)=F^{-1}(t)$; namely, $\Gamma(t)=g(\ln t)$. The result of Theorem
2.1 can be rewritten as follows.

######  Corollary 2.1.

Assume (A1)–(A6). Then, there exists $\lambda^{*}>0$ such that problem
($P_{\lambda}$) possesses a singular radial solution of the form

$\displaystyle u^{*}_{{\scriptscriptstyle\lambda}^{\\!*}}(r)=$
$\displaystyle\,\Gamma(\widehat{Z}(\tau))+O\big{(}\Gamma^{\prime}(\tau)\tau+\Gamma^{\prime\prime}(\tau)\tau^{2}\big{)},\quad\text{where}\;\;\tau=\frac{\beta+1}{\lambda^{*}r^{\theta}},$
and
$\displaystyle\widehat{Z}(\tau)=\theta^{\beta+1}(\hat{\alpha}-1)\,\tau^{\beta+2}\big{[}\Gamma^{\prime}(\tau)+(\beta+1)\ln\\{\Gamma^{\prime}(\tau)\tau\\}(\Gamma^{\prime}(\tau)+\Gamma^{\prime\prime}(\tau)\tau)\big{]}^{\beta+1}.$

###### Proof.

Note that $g^{\prime}(\ln\tau)=\Gamma^{\prime}(\tau)\tau$ and
$g^{\prime\prime}(\ln\tau)=\Gamma^{\prime}(\tau)\tau+\Gamma^{\prime\prime}(\tau)\tau^{2}$.
Therefore,

$g\big{(}\ln\big{\\{}\theta^{\beta+1}(\hat{\alpha}-1)\big{\\}}+\ln\tau+(\beta+1)\ln\\{g^{\prime}(\ln\tau)+(\beta+1)g^{\prime\prime}(\ln\tau)\ln
g^{\prime}(\ln\tau)\\}\big{)}\\\
=g\big{(}\ln\\{\theta^{\beta+1}(\hat{\alpha}-1)\tau(g^{\prime}(\ln\tau)+(\beta+1)g^{\prime\prime}(\ln\tau)\ln
g^{\prime}(\ln\tau))^{\beta+1}\big{\\}}\big{)}\\\
=\Gamma\big{(}\theta^{\beta+1}(\hat{\alpha}-1)\,\tau^{\beta+2}\big{\\{}\Gamma^{\prime}(\tau)+(\beta+1)\ln\\{\Gamma^{\prime}(\tau)\tau\\}(\Gamma^{\prime}(\tau)+\Gamma^{\prime\prime}(\tau)\tau)\big{\\}}^{\beta+1}\big{)}.$

This completes the proof of the corollary. ∎

In Corollary 2.2, we obtain asymptotic representations for singular solutions
when $e^{f(u)}$ is one of the nonlinearities considered in [12, 24, 28]. The
respective representations from [12, 24, 28] agree with ours when $L$ is the
radial Laplacian.

######  Corollary 2.2.

Assume (A6) and consider the following three particular cases of $f$:
$(a)\;f(u)=e^{u};$ $(b)\;f(u)=u^{p},$ $(p>\frac{1}{2});$
$(c)\;f(u)=u+\vartheta(u),$ where $\vartheta(u)$ satisfies assumptions
(i),(ii),(iii) from Example 3 in Subsection 2.2 and such that
$|\vartheta(u)|+|\vartheta^{\prime}(u)|\leqslant C_{0}e^{-\delta u}$. Then,

$\displaystyle(a)\;u^{*}_{{\scriptscriptstyle\lambda}^{\\!*}}\Big{(}\frac{r}{{\lambda^{*}}^{\nicefrac{{1}}{{\theta}}}}\Big{)}=\ln\Big{\\{}\theta\ln\frac{1}{r}+\ln\frac{\hat{\alpha}-1}{\ln\frac{k}{r}}+(\beta+1)\ln\Big{(}1+(\beta+1)\frac{\ln\big{\\{}\theta\ln\frac{k}{r}\big{\\}}}{\theta\ln\frac{k}{r}}\Big{)}\Big{\\}}$
$\displaystyle\hskip
256.0748pt+O\Big{(}\Big{(}\ln\frac{1}{r}\Big{)}^{-2}\Big{)};$
$\displaystyle(b)\;u^{*}_{{\scriptscriptstyle\lambda}^{\\!*}}\Big{(}\frac{r}{{\lambda^{*}}^{\nicefrac{{1}}{{\theta}}}}\Big{)}=\Big{(}\theta\ln\frac{1}{r}+(\beta+1)\Big{(}\frac{1}{p}-1\Big{)}\ln\ln\frac{k}{r}+\ln\frac{(\alpha-\beta-1)\theta^{\frac{\beta+1}{p}}}{p^{\beta+1}}$
$\displaystyle\phantom{(b)\;u^{*}_{{\scriptscriptstyle\lambda}^{\\!*}}\Big{(}\frac{r}{{\lambda^{*}}^{\nicefrac{{1}}{{\theta}}}}\Big{)}}+(\beta+1)\ln\Big{\\{}1+\Big{(}\frac{1}{p}-1\Big{)}^{2}\frac{\ln\tau}{\tau}-\Big{(}\frac{1}{p}-1\Big{)}\frac{\ln
p}{\tau}\Big{\\}}\Big{)}^{\frac{1}{p}}+O\Big{(}\Big{(}\ln\frac{1}{r}\Big{)}^{\frac{1}{p}-2}\Big{)};$
$\displaystyle(c)\;u^{*}_{{\scriptscriptstyle\lambda}^{\\!*}}(r)=\ln\big{\\{}\theta^{\beta+1}(\alpha-\beta-1)\\}-\ln\lambda^{*}-\theta\ln
r+O(r^{\theta\delta}).$

Above, $k=(1+\beta)^{\frac{1}{\theta}}$, $\tau=\theta\ln\frac{k}{r}$.

###### Proof.

(a) and (b) follow immediately from representation (LABEL:as1). To show (c),
we note that in a neighborhood of infinity, by (iii),
$\vartheta^{\prime\prime}(u)$ either equals zero or
$\frac{\vartheta^{\prime\prime\prime}(u)}{\vartheta^{\prime\prime}(u)}=O(1)$.
By L’Hopital’s rule and (i), $\lim{\rm
sup}_{u\to+\infty}\big{|}\frac{\vartheta^{\prime\prime}(u)}{\ln\\{1+\vartheta^{\prime}(u)\\}}\big{|}$
is finite, which means that $|\vartheta^{\prime\prime}(u)|\leqslant
C_{1}e^{-\delta u}$ for some constant $C_{1}>0$. Furthermore, the
representations for $\tilde{\vartheta}$, $\tilde{\vartheta}^{\prime}$, and
$\tilde{\vartheta}^{\prime\prime}$ imply that in a neighborhood of infinity,
$|\tilde{\vartheta}(t)|+|\tilde{\vartheta}^{\prime}(t)|+|\tilde{\vartheta}^{\prime\prime}(t)|\leqslant
C_{2}e^{-\delta t}$ for some constant $C_{2}>0$. Representation (c) follows
now from (LABEL:as1). ∎

######  Remark 2.8.

In example (b), we deal with the same nonlinear term as in [24]. We would like
to emphasize the following fact. In [24], 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 [24],
our asymptotic representation includes additional explicit terms, tending to
infinity, so our rest term is of a higher order than in [24] 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 ($P_{\lambda}$) by using the change of
variable $t=\ln(\frac{\kappa}{r})$, where $\kappa>0$ is a constant to be
specified later. First, we note that the equation in ($P_{\lambda}$) can be
rewritten as follows:

$\displaystyle
u^{\prime\prime}(r)|u^{\prime}(r)|^{\beta}+\hat{\alpha}\,\frac{u^{\prime}(r)|u^{\prime}(r)|^{\beta}}{r}+\frac{\lambda\,r^{\gamma-\alpha}\,e^{f(u)}}{(\beta+1)}=0.$

Let us show that for a singular solution to ($P_{\lambda}$), it holds that
$(u^{*}_{{\scriptscriptstyle\lambda}^{\\!*}})^{\prime}(r)<0$ for all $r>0$.
Note that if $\beta>0$, from the above equation, it follows that a singular
solution cannot have local maximum or minimum points, since in those points
$u^{\prime}(r)=0$. Suppose $\beta=0$. Then, every local extremum point, is
necessarily a local maximum point, which contradicts to the fact that
$\lim_{r\to 0+}u^{*}_{{\scriptscriptstyle\lambda}^{\\!*}}(r)=+\infty$.

By doing the change of variable $t=\ln(\frac{\kappa}{r})$ and defining
$v(t)=u(r)$, we transform ($P_{\lambda}$) to the following problem:

$\begin{cases}v^{\prime\prime}-(\hat{\alpha}-1)v^{\prime}+\frac{\lambda\kappa^{\theta}e^{-\theta
t}e^{f(v)}}{(\beta+1)|v^{\prime}|^{\beta}}=0,\quad t\in(\ln\kappa,+\infty),\\\
v(\ln\kappa)=0.\end{cases}$

By choosing $\kappa$ in such a way that
$\frac{\lambda\,\kappa^{\theta}}{\beta+1}=1$, we arrive at equation (11) with
the boundary condition $v(\ln\kappa)=0$, where
$\kappa=\big{(}\frac{\beta+1}{\lambda}\big{)}^{\frac{1}{\theta}}$. This
problem is equivalent to ($P_{\lambda}$), but given in a neighborhood of
infinity.

######  Proposition 2.1.

Assume (A1)–(A6). Then, there exists $T>0$ such that on $[T,+\infty)$,
equation (11) possesses a singular solution of the form (12), where
$\eta(t)=O(g^{\prime\prime}(\theta t))$ is a ${\rm C}^{2}$-function.

######  Remark 2.9.

Remark that by Lemma 2.2 and Remark 2.3, $\lim_{t\to+\infty}\frac{\ln
g^{\prime}(\theta t)}{t}=0$. Therefore, $(\varphi_{1}+\varphi_{2})(t)=o(t)$.

###### Proof.

Step 1. Transformation to a fixed-point problem. The idea is to obtain a
semilinear equation with respect to $\eta$ by substituting $v^{*}=g+\eta$ 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 $\eta$ and $\zeta=\eta^{\prime}$

(15)
$\begin{cases}\eta^{\prime\prime}(t)-(\alpha-\beta-1)\,\eta^{\prime}(t)+\theta(\hat{\alpha}-1)\,\eta(t)=\Phi(t,\eta,\zeta),\\\
\zeta^{\prime}(t)-(\alpha-\beta-1)\,\zeta(t)=\Phi(t,\eta,\zeta)-\theta(\hat{\alpha}-1)\,\eta(t),\end{cases}$

where the function $\Phi$ is expected to be regular. For simplicity of
notation, define

$\displaystyle\varphi=\varphi_{1}+\varphi_{2},$

where $\varphi_{1}$ and $\varphi_{2}$ are defined by (LABEL:ffi). We start by
transforming $\exp\\{-\theta t+f(v^{*})\\}=\exp\\{-\theta t+f(g(\theta
t+\varphi)+\eta)\\}$, taking into account that $g=f^{-1}$. It holds that

$\displaystyle f(g(\theta t+\varphi)+\eta)=\theta
t+\varphi+f^{\prime}(g(\theta t+\varphi)+\vartheta(t,\eta)\eta)\eta,$

where $\vartheta(t,\eta)$ takes values between $0$ and $1$. Let us transform
the term $f^{\prime}(g(\theta t+\varphi)+\vartheta(t,\eta)\eta)$. Define

$\displaystyle\xi(t)=(\theta t+\varphi)\Big{(}\frac{f\big{(}g(\theta
t+\varphi)+\vartheta(t,\eta)\eta\big{)}}{\theta t+\varphi}-1\Big{)}.$

Then, $g(\theta t+\varphi)+\vartheta(t,\eta)\eta=g(\theta t+\varphi+\xi)$. By
(A2) and Remark 2.9, $\xi=o(\theta t+\varphi)=o(\theta t+o(t))=o(t)$.
Therefore,

(16) $\displaystyle f(g(\theta t+\varphi)+\eta)=\theta
t+\varphi+\frac{\eta}{g^{\prime}(\theta t+\varphi+\xi)}=\theta
t+\varphi+\frac{\eta}{g^{\prime}(\theta t+o(t))}.$

Therefore,

(17) $\displaystyle\frac{\exp\\{-\theta
t+f(v^{*})\\}}{|{v^{*}}^{\prime}(t)|^{\beta}}=\frac{e^{\varphi_{1}}}{{|{v^{*}}^{\prime}(t)|^{\beta}}}\,e^{\varphi_{2}+\frac{\eta}{g^{\prime}(\theta
t+o(t))}}.$

Let us transform the first factor on the right-hand side. Defining
$\vartheta_{1}(t)=\theta^{-1}\varphi^{\prime}(t)$, we obtain

(18)
$\frac{e^{\varphi_{1}}}{|{v^{*}}^{\prime}(t)|^{\beta}}=\frac{\theta^{\beta}\big{(}g^{\prime}(\theta
t)+(\beta+1)g^{\prime\prime}(\theta t)\ln g^{\prime}(\theta
t)\big{)}^{\beta}}{\theta^{\beta}|g^{\prime}(\theta
t+\varphi)(1+\vartheta_{1})+\theta^{-1}\eta^{\prime}|^{\beta}}\\\
=\frac{1}{\big{|}1+\frac{\eta^{\prime}}{\theta g^{\prime}(\theta
t+\varphi)(1+\vartheta_{1})}\big{|}^{\beta}}\,\frac{1}{(1+\vartheta_{1})^{\beta}}\,\frac{1}{(1+\vartheta_{2})^{\beta}},$

where $\vartheta_{2}$ is defined through the formula

$\displaystyle g^{\prime}(\theta t+\varphi)=(g^{\prime}(\theta
t)+(\beta+1)g^{\prime\prime}(\theta t)\ln g^{\prime}(\theta
t))(1+\vartheta_{2}(t)).$

Let us obtain expressions for $\vartheta_{1}$ and $\vartheta_{2}$ convenient
for future computations. By (A4) and (A5),

(19)
$\vartheta_{1}(t)=\theta^{-1}\varphi^{\prime}(t)=(\beta+1)\,\frac{g^{\prime\prime}(\theta
t)+(\beta+1)g^{\prime\prime\prime}(\theta t)\ln g^{\prime}(\theta
t)+(\beta+1)\frac{g^{\prime\prime}(\theta t)^{2}}{g^{\prime}(\theta
t)}}{g^{\prime}(\theta t)+(\beta+1)g^{\prime\prime}(\theta t)\ln
g^{\prime}(\theta t)}\\\ =O\Big{(}\frac{g^{\prime\prime}(\theta
t)}{g^{\prime}(\theta t)}\Big{)}.$

Since

(20) $\displaystyle\varphi(t)=(\beta+1)\ln g^{\prime}(\theta t)+O(1),$

by using the expansion of $g^{\prime}(\theta t+\varphi)$ by Taylor’s formula
around $\theta t$, we obtain

(21) $\vartheta_{2}(t)=\frac{g^{\prime}(\theta t+\varphi)-g^{\prime}(\theta
t)-(\beta+1)g^{\prime\prime}(\theta t)\ln g^{\prime}(\theta
t)}{g^{\prime}(\theta t)+(\beta+1)g^{\prime\prime}(\theta t)\ln
g^{\prime}(\theta t)}\\\ =\frac{g^{\prime\prime}(\theta
t)O(1)+\int_{0}^{1}(1-r)g^{\prime\prime\prime}(\theta
t+r\varphi)dr\,\varphi^{2}}{g^{\prime}(\theta
t)+(\beta+1)g^{\prime\prime}(\theta t)\ln g^{\prime}(\theta
t)}=O\Big{(}\frac{g^{\prime\prime}(\theta t)}{g^{\prime}(\theta t)}\Big{)}.$

The last identity holds by (A4), (A5), and the following argument. First of
all, we note that by Taylor’s formula around $\theta t$ and (20),

$\displaystyle\ln g^{\prime}(\theta t+\varphi)=\ln g^{\prime}(\theta
t)+\ln\Big{(}1+O\Big{(}\frac{g^{\prime\prime}(\theta t)}{g^{\prime}(\theta
t)}\Big{)}+O\Big{(}\frac{g^{\prime\prime}(\theta t)}{g^{\prime}(\theta t)}\ln
g^{\prime}(\theta t)\Big{)}\Big{)}=\ln g^{\prime}(\theta t)+o(1),$

which, by Remark 2.9, implies that

$\displaystyle g^{\prime\prime\prime}(\theta
t+r\varphi)\varphi^{2}=O(g^{\prime\prime}(\theta
t+\varphi))=O(g^{\prime\prime}(\theta t)).$

Note that since we are interested in solving equation (11) in a neighborhood
of infinity, the expressions $\frac{1}{(1+\vartheta_{1})^{\beta}}$ and
$\frac{1}{(1+\vartheta_{2})^{\beta}}$ are well-defined.

Next, by the results of Subsection 2.4.1, for any singular solution to
($P_{\lambda}$), it holds that
$(u^{*}_{{\scriptscriptstyle\lambda}^{\\!*}})^{\prime}(r)<0$ for all $r>0$.
This implies that ${v^{*}}^{\prime}(t)>0$ for all $t$. Then, by Taylor’s
formula, (18) can be transformed as follows:

$\displaystyle\frac{e^{\varphi_{1}}}{{v^{*}}^{\prime}(t)^{\beta}}=\Big{(}1-\frac{\beta\eta^{\prime}}{\theta
g^{\prime}(\theta t)}+o(1)\,\frac{\eta^{\prime}}{g^{\prime}(\theta
t)}+\frac{{\eta^{\prime}}^{2}}{g^{\prime}(\theta
t)^{2}}\,\sigma(t,\eta^{\prime})\Big{)}\Big{(}1+O\Big{(}\frac{g^{\prime\prime}(\theta
t)}{g^{\prime}(\theta t)}\Big{)}\Big{)},$

where

(22)
$\displaystyle\sigma(t,\eta^{\prime})=O(1)\int_{0}^{1}\frac{(1-r)}{(1+rx)^{\beta+2}}\,dr\quad\text{with}\;\;x=\frac{\eta^{\prime}}{\theta
g^{\prime}(\theta t+\varphi)(1+\vartheta_{1})}.$

Recalling that $\frac{g^{\prime\prime}(\theta t)}{g^{\prime}(\theta t)}\to 0$
as $t\to+\infty$ (see Remark 2.3), the last expression can be rewritten as

(23)
$\displaystyle\frac{e^{\varphi_{1}}}{{v^{*}}^{\prime}(t)^{\beta}}=1-\frac{\beta\eta^{\prime}}{\theta
g^{\prime}(\theta t)}+o(1)\frac{\eta^{\prime}}{g^{\prime}(\theta
t)}+O(1)\,\frac{{\eta^{\prime}}^{2}}{g^{\prime}(\theta
t)^{2}}\,\sigma(t,\eta^{\prime})+O\Big{(}\frac{g^{\prime\prime}(\theta
t)}{g^{\prime}(\theta t)}\Big{)}.$

Taking into account that
$e^{\varphi_{2}}=\theta(\hat{\alpha}-1)g^{\prime}(\theta t)(1+o(1))$, by (17),
(19), and (23), expression (17) can be transformed as follows:

$\frac{\exp\\{-\theta
t+f(v^{*})\\}}{{v^{*}}^{\prime}(t)^{\beta}}=e^{\frac{\eta}{g^{\prime}(\theta
t+o(t))}}e^{\varphi_{2}}\frac{e^{\varphi_{1}}}{{v^{*}}^{\prime}(t)^{\beta}}=e^{\frac{\eta}{g^{\prime}(\theta
t+o(t))}}\Big{[}e^{\varphi_{2}}-\beta(\hat{\alpha}-1)\eta^{\prime}+o(1)\eta^{\prime}\\\
+O(1)\,\frac{{\eta^{\prime}}^{2}}{g^{\prime}(\theta
t)}\sigma(t,\eta^{\prime})+O\big{(}g^{\prime\prime}(\theta
t)\big{)}\Big{]}=e^{\varphi_{2}+\frac{\eta}{g^{\prime}(\theta
t+o(t))}}-\beta(\hat{\alpha}-1)\eta^{\prime}+F_{1}(t,\eta,\eta^{\prime})+F_{2}(t,\eta)\\\
=e^{\varphi_{2}}\Big{(}e^{\frac{\eta}{g^{\prime}(\theta
t+o(t))}}-\frac{\eta}{g^{\prime}(\theta
t+o(t))}-1\Big{)}+\frac{e^{\varphi_{2}}\eta}{g^{\prime}(\theta
t+o(t))}+e^{\varphi_{2}}-\beta(\hat{\alpha}-1)\eta^{\prime}\\\
+F_{1}(t,\eta,\eta^{\prime})+F_{2}(t,\eta),$

where

$\displaystyle
F_{1}(t,\eta,\eta^{\prime})=\Big{[}o(1)\eta^{\prime}+O(1)\,\frac{{\eta^{\prime}}^{2}}{g^{\prime}(\theta
t)}\sigma(t,\eta^{\prime})\Big{]}e^{\frac{\eta}{g^{\prime}(\theta t+o(t))}}$
$\displaystyle\hskip
45.52458pt-\beta(\hat{\alpha}-1)\eta^{\prime}\big{(}e^{\frac{\eta}{g^{\prime}(\theta
t+o(t))}}-1\big{)}+O\big{(}g^{\prime\prime}(\theta
t)\big{)}(e^{\frac{\eta}{g^{\prime}(\theta t+o(t))}}-1);$ $\displaystyle
F_{2}(t,\eta)=O\big{(}g^{\prime\prime}(\theta t)\big{)}.$

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 $v^{*}(t)=g(\theta t+\varphi)+\eta(t)$ into (11). We have

(24) $\displaystyle\frac{e^{\varphi_{2}}\eta}{g^{\prime}(\theta
t+o(t))}-\theta(\hat{\alpha}-1)\,\eta=\theta(\hat{\alpha}-1)\Big{(}\frac{g^{\prime}(\theta
t)(1+o(1))}{g^{\prime}(\theta t+o(t))}-1\Big{)}\eta.$

Further, by the definition of $\varphi_{2}$, representation (20), and Taylor’s
expansion for $g^{\prime}(\theta t+\varphi)$ around $\theta t$ (in the same
way as we applied it in the numerator of (21)),

(25) $e^{\varphi_{2}}-(\hat{\alpha}-1)\frac{d}{dt}\,g(\theta
t+\varphi)=\theta(\hat{\alpha}-1)\big{(}g^{\prime}(\theta
t)+(\beta+1)g^{\prime\prime}(\theta t)\ln g^{\prime}(\theta t)\\\
-g^{\prime}(\theta
t+\varphi)(1+\vartheta_{1})\big{)}=O(g^{\prime\prime}(\theta t)).$

In (25), we used (A5) and representation (19) for $\vartheta_{1}$. Next,

(26) $\displaystyle\frac{d^{2}}{dt^{2}}\,g(\theta
t+\varphi)=g^{\prime\prime}(\theta
t+\varphi)(\theta+\varphi^{\prime})^{2}+g^{\prime}(\theta
t+\varphi)\varphi^{\prime\prime}=O(g^{\prime\prime}(\theta t)).$

This holds by (A3), (A5), formula (19) for $\varphi^{\prime}$, and the
following expression for $\varphi^{\prime\prime}$:

$\displaystyle\varphi^{\prime\prime}(t)=\theta^{2}(\beta+1)\,\frac{g^{\prime\prime\prime}(\theta
t)+g^{(\rm{iv})}(\theta t)\ln g^{\prime}(\theta
t)+\frac{3g^{\prime\prime\prime}(\theta t)g^{\prime\prime}(\theta
t)}{g^{\prime}(\theta t)}-\frac{g^{\prime\prime}(\theta
t)^{3}}{g^{\prime}(\theta t)^{2}}}{g^{\prime}(\theta
t)+g^{\prime\prime}(\theta t)\ln g^{\prime}(\theta
t)}-(\varphi^{\prime})^{2}.$

Thus, we obtain (15) with

(27)
$\displaystyle\Phi(t,\eta,\zeta)=-\big{(}F_{1}(t,\eta,\zeta)+F_{2}(t,\eta)+F_{3}(t,\eta)+F_{4}(t)+F_{5}(t,\eta)\big{)},$

where $\eta^{\prime}$ is substituted by $\zeta$ on the right-hand side of the
first equation of (15) and in the second equation. Furthermore, $F_{3}$ is
defined by (24), $F_{4}(t)=O(g^{\prime\prime}(\theta t))$ comes due to (25)
and (26), and $F_{5}$ is defined as follows:

(28) $F_{5}(t,\eta)=e^{\varphi_{2}}\Big{(}e^{\frac{\eta}{g^{\prime}(\theta
t+o(t))}}-\frac{\eta}{g^{\prime}(\theta t+o(t))}-1\Big{)}\\\
=\theta(\hat{\alpha}-1)\int_{0}^{1}\\!(1-r)\,e^{\frac{r\eta}{g^{\prime}(\theta
t+o(t))}}dr\;\frac{g^{\prime}(\theta t)O(1)}{g^{\prime}(\theta
t+o(t))^{2}}\,\eta^{2}.$

We will be searching for pairs $(\eta,\zeta)$ that solve (15) in a certain
metric space which we denote by $X_{T}$. If $g^{\prime\prime}(t)$ is not an
identical zero in a neighborhood of infinity, we define

$\displaystyle X_{T}=\\{x\in{\rm C}[T,\infty):x(t)=O(g^{\prime\prime}(\theta
t))\;\text{as}\;t\to+\infty\\}.$

Note that under (A5)-(b), either $g^{\prime\prime}(\theta t)$ 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 $X_{T}$ is well-defined:

$\displaystyle\|x\|_{X_{T}}=\sup_{t\geqslant T}\\{|g^{\prime\prime}(\theta
t)|^{-1}|x(t)|\\}.$

In the case (ii), we define $X_{T}={\rm C}_{b}([T,+\infty))$ with the supremum
norm. Here, ${\rm C}_{b}$ stands for the space of bounded continuous
functions.

In $X_{T}$, (15) uniquely transforms to the following couple of fixed-point
equations:

(29)
$\displaystyle\eta=\Psi(\eta,\zeta),\qquad\zeta={\hat{\Psi}}(\eta,\zeta).$

Since $\hat{\alpha}>1$, the explicit formula for $\hat{\Psi}$ is the
following:

(30)
$\displaystyle{\hat{\Psi}}(\eta,\zeta)(t)=\int_{t}^{+\infty}e^{2\delta(t-s)}\big{(}\Phi(s,\eta,\zeta)-\theta(\hat{\alpha}-1)\,\eta(s)\big{)}ds,$

where $\delta=\frac{\alpha-\beta-1}{2}$. The explicit formula for $\Psi$
depends on the sign of the discriminant
$D=4(\delta^{2}-\theta(\hat{\alpha}-1))$ of the corresponding characteristic
equation. In the case $D<0$, we have

(31)
$\displaystyle\Psi(\eta,\zeta)(t)=\frac{1}{\mu}\,\int_{t}^{+\infty}e^{\delta(t-s)}\sin\big{(}\mu(s-t)\big{)}\Phi(s,\eta,\zeta)\,ds,\quad\text{where}\;\;\mu=\frac{\sqrt{|D|}}{2}.$

In the case $D>0$, we have

(32)
$\displaystyle\Psi(\eta,\zeta)(t)=\frac{1}{2\mu}\,\int_{t}^{+\infty}e^{(\delta-\mu)(t-s)}\big{(}1-e^{2\mu(t-s)}\big{)}\Phi(s,\eta,\zeta)\,ds.$

Finally, if $D=0$, we have

(33)
$\displaystyle\Psi(\eta,\zeta)(t)=\int_{t}^{+\infty}e^{\delta(t-s)}(s-t)\Phi(s,\eta,\zeta)\,ds.$

Step 2. Existence of a fixed point. Let us show that we can find constants
$M,\hat{M}>0$ and a number $T>0$ such that the map $(\Psi,{\hat{\Psi}})$ acts
from $\Sigma$ to $\Sigma$, where the latter is defined as a complete metric
space of the following form:

$\displaystyle\Sigma=\Sigma(M,\hat{M},T)=\\{(\eta,\zeta)\in X_{T}\times
X_{T}:\|\eta\|_{X_{T}}\leqslant
M\;\text{and}\;\|\zeta\|_{X_{T}}\leqslant\hat{M}\\}.$

We define the metric in $\Sigma$ as follows:

$\displaystyle\rho\\{(\eta_{1},\zeta_{1}),(\eta_{2},\zeta_{2})\\}=\|\eta_{1}-\eta_{2}\|_{X_{T}}+\|\zeta_{1}-\zeta_{2}\|_{X_{T}}.$

Moreover, we show that $T$ can be chosen in such a way that
$(\Psi,{\hat{\Psi}})$ is a contraction map $\Sigma\to\Sigma$.

The integral operator applied to $\Phi$ and defined by (31), (32), or (33)
splits into five terms (which we denote by $I_{1}$, $I_{2}$, $I_{3}$, $I_{4}$,
and $I_{5}$), corresponding to the decomposition of $\Phi$ by formula (27). We
evaluate each of these terms. First, we compute the integrals

$\displaystyle{\frac{1}{\mu}}\int_{t}^{+\infty}e^{\delta(t-s)}ds={\frac{1}{\mu\delta}};\;\;{\frac{1}{2\mu}}\int_{t}^{+\infty}e^{(\delta-\mu)(t-s)}ds={\frac{1}{2\mu(\delta-\mu)}};\;\;\int_{t}^{+\infty}e^{\delta(t-s)}(s-t)ds={\frac{1}{\delta^{2}}};$

and define

$\displaystyle\varkappa=\max\big{\\{}{\frac{1}{2\delta},\frac{1}{2\mu(\delta-\mu)},\frac{1}{\mu\delta},\frac{1}{\delta^{2}}}\big{\\}}.$

The expression for $\varkappa$ involves the values of the above integrals and
the term $\frac{1}{2\delta}$ which appears from the integration of
$e^{2\delta(t-s)}$ in (30). We make analysis separately in the case (i), when
$g^{\prime\prime}(\theta t)\neq 0$ in a neighborhood of infinity, and in the
case (ii), when $g^{\prime\prime}(\theta t)$ identically equals zero in a
neighborhood of infinity. We start from the case (i). Introduce constants
$M_{1}$, $M_{2}$, and a number $T_{1}>0$ such that

(34) $\displaystyle\sup_{s\geqslant t}|g^{\prime\prime}(\theta s)|\leqslant
M_{1}|g^{\prime\prime}(\theta t)|,\quad t\geqslant T_{1};$
$\displaystyle|F_{i}(t)|\leqslant M_{2}|g^{\prime\prime}(\theta t)|,\quad
i=2,4,\quad t\geqslant T_{1}.$

Let us fix the metric space for the fixed point argument. Take $M>5\varkappa
M_{1}M_{2}$ and $\hat{M}=\frac{\theta MM_{1}}{1+\beta}+M$. Choose $T>T_{1}$ in
such a way that if the a priori estimates $|\eta(t)|\leqslant
M|g^{\prime\prime}(\theta t)|$ and $|\zeta(t)|<\hat{M}|g^{\prime\prime}(\theta
t)|$ hold, then,

$\displaystyle|F_{i}(t)|\leqslant M_{2}|g^{\prime\prime}(\theta t)|,\quad
i=1,3,5,\quad t\geqslant T.$

We can always do this since for $\eta$ and $\zeta$ with the above property,
$F_{1}$, $F_{3}$, and $F_{5}$ are $o(g^{\prime\prime}(t))$. Now take
$(\eta,\zeta)\in\Sigma(M,\hat{M},T)$ and note that each of the terms $I_{i}$,
$i=1,2,3,4,5$, becomes smaller than $\varkappa M_{1}M_{2}$, which is smaller
than $\frac{M}{5}$. Thus, we have proved that $|\Psi(\eta,\zeta)(t)|\leqslant
M|g^{\prime\prime}(t)|$. Next, we represent ${\hat{\Psi}}$ as
$\Psi_{1}+\Psi_{2}$, where

(35)
$\displaystyle\Psi_{1}(\eta,\zeta)(t)=\int_{t}^{+\infty}e^{2\delta(t-s)}\Phi(s,\eta(s),\zeta(s))ds,$
$\displaystyle\Psi_{2}(\eta)(t)=-\theta(\hat{\alpha}-1)\int_{t}^{+\infty}e^{2\delta(t-s)}\eta(s)ds.$

From the above argument, we obtain that $|\Psi_{1}(\eta,\zeta)(t)|\leqslant
M|g^{\prime\prime}(t)|$. Furthermore,
$|\Psi_{2}(\eta)(t)|\leqslant\frac{\theta
MM_{1}}{1+\beta}|g^{\prime\prime}(t)|$. Therefore, by the choice of $\hat{M}$,
we obtain that
$|{\hat{\Psi}}(\eta,\eta^{\prime})(t)|\leqslant\hat{M}|g^{\prime\prime}(t)|$.
Thus, we proved that $\Psi(\eta,\zeta)$ is a map $\Sigma\to\Sigma$.

Let us show that if the number $T$ is sufficiently large, then
$\Psi:\Sigma\to\Sigma$ is a contraction map. Let $\chi(s-t)$ denote one of the
factors $\sin\\{\mu(s-t)\\}$, $1-e^{2\mu(t-s)}$, or $s-t$ which occur in the
integrals (31), (32), and (33). By (A5)-(b) and the integrability of the
exponential factor, multiplied by $\chi$, it suffices to observe that for two
elements $(\eta_{1},\zeta_{1})$ and $(\eta_{2},\zeta_{2})$ from $\Sigma$, and
for each $t\in(T,+\infty)$,

(36)
$\displaystyle|F_{1}(t,\eta_{1},\zeta_{1})-F_{1}(t,\eta_{2},\zeta_{2})|\leqslant\epsilon(t)(|\eta_{1}-\eta_{2}|+|\zeta_{1}-\zeta_{2}|),$
$\displaystyle|F_{i}(t,\eta_{1})-F_{i}(t,\eta_{2})|\leqslant\epsilon(t)|\eta_{1}-\eta_{2}|,\quad
i=3,5,$

where $\epsilon$ it is a positive function tending to $0$ as $t\to+\infty$.
Inequalities (LABEL:Fi) follow immediately from the expressions for $F_{i}$,
$i=1,3,5$, if we take into account the estimates $|\eta_{i}(t)|\leqslant
M|g^{\prime\prime}(t)|$,
$|\zeta_{i}(t)|\leqslant\hat{M}|g^{\prime\prime}(t)|$, $i=1,2$, Remark 2.3,
and assumption (A3). We also observe that, by (22), $\sigma_{1}(t,\zeta)$ is
Lipschitz in its second argument for sufficiently large $t$. By (34) and
(LABEL:Fi),

$\sup_{t\geqslant T}|g^{\prime\prime}(\theta
t)|^{-1}|\Psi(\eta_{1},\zeta_{1})(t)-\Psi(\eta_{2},\zeta_{2})(t)|\\\ \leqslant
5\sup_{t\geqslant T}\Big{\\{}|g^{\prime\prime}(\theta
t)|^{-1}\int_{t}^{+\infty}e^{\delta(t-s)}\chi(t-s)\epsilon(s)\big{(}|\eta_{1}(s)-\eta_{2}(s)|+|\zeta_{1}(s)-\zeta_{2}(s)|\big{)}ds\Big{\\}}\\\
\leqslant 5M_{1}\sup_{t\geqslant
T}\epsilon(t)\int_{T}^{+\infty}e^{\delta(t-s)}\chi(t-s)|g^{\prime\prime}(\theta
s)|^{-1}\big{(}|\eta_{1}(s)-\eta_{2}(s)|+|\zeta_{1}(s)-\zeta_{2}(s)|\big{)}ds.$

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

(37)
$\displaystyle\|\Psi(\eta_{1},\zeta_{1})-\Psi(\eta_{2},\zeta_{2})\|_{X_{T}}\leqslant
r(\|\eta_{1}-\eta_{2}\|_{X_{T}}+\|\zeta_{1}-\zeta_{2}\|_{X_{T}}).$

By (30),

$\|{\hat{\Psi}}(\eta_{1},\zeta_{1})-{\hat{\Psi}}(\eta_{2},\zeta_{2})\|_{X_{T}}\leqslant\|\Psi_{1}(\eta_{1},\zeta_{1})-\Psi_{1}(\eta_{2},\zeta_{2})\|_{X_{T}}+\|\Psi_{2}(\eta_{1})-\Psi_{2}(\eta_{2})\|_{X_{T}},$

where $\Psi_{1}$ and $\Psi_{2}$ are given by (LABEL:ps23). The first term in
the above identity is evaluated exactly as in (37). For the second term, we
have

$\displaystyle\|\Psi_{2}(\eta_{1})-\Psi_{2}(\eta_{2})\|_{X_{T}}\leqslant
R\|\eta_{1}-\eta_{2}\|_{X_{T}}\quad\text{with}\;\;R=\frac{\theta
M_{1}}{1+\beta}$

(recall that $2\delta=\alpha-\beta-1$). Therefore,

$\displaystyle\|{\hat{\Psi}}(\eta_{1},\zeta_{1})-{\hat{\Psi}}(\eta_{2},\zeta_{2})\|_{X_{T}}\leqslant(R+r)\|\eta_{1}-\eta_{2}\|_{X_{T}}+r\|\zeta_{1}-\zeta_{2}\|_{X_{T}}.$

Now let
$(\Psi^{(2)},{\hat{\Psi}^{(2)}})=(\Psi,{\hat{\Psi}})\circ(\Psi,{\hat{\Psi}})$.
We show that one can find a number $r\in(0,1)$ such that
$(\Psi^{(2)},{\hat{\Psi}^{(2)}})$ is a contraction map. We have

$\displaystyle\|\Psi^{(2)}(\eta_{1},\zeta_{1})-\Psi^{(2)}(\eta_{2},\zeta_{2})\|_{X_{T}}\leqslant
r(2r+R)\|\eta_{1}-\eta_{2}\|_{X_{T}}+2r^{2}\|\zeta_{1}-\zeta_{2}\|_{X_{T}},$
$\displaystyle\|{\hat{\Psi}^{(2)}}(\eta_{1},\zeta_{1})-{\hat{\Psi}^{(2)}}(\eta_{2},\zeta_{2})\|_{X_{T}}\leqslant
2r(r+R)\|\eta_{1}-\eta_{2}\|_{X_{T}}+r(2r+R)\|\zeta_{1}-\zeta_{2}\|_{X_{T}}.$

Therefore,

$\|\Psi^{(2)}(\eta_{1},\zeta_{1})-\Psi^{(2)}(\eta_{2},\zeta_{2})\|_{X_{T}}+\|{\hat{\Psi}^{(2)}}(\eta_{1},\zeta_{1})-{\hat{\Psi}^{(2)}}(\eta_{2},\zeta_{2})\|_{X_{T}}\\\
\leqslant(4r^{2}+3Rr)(\|\eta_{1}-\eta_{2}\|_{X_{T}}+\|\zeta_{1}-\zeta_{2}\|_{X_{T}}).$

Picking $r>0$ in such a way that $4r^{2}+3Rr<1$, we obtain that
$(\Psi^{(2)},{\hat{\Psi}^{(2)}})$ is a contraction map $\Sigma\to\Sigma$. Let
$(\eta,\zeta)$ be its unique fixed point. Then, $(\eta,\zeta)$ is also a fixed
point of $(\Psi,{\hat{\Psi}})$, while $\eta$ determines solution (12) with the
prescribed singular part given by the first term. To see this, it suffices to
note that the solution $(\eta,\zeta)$, that we just found, satisfies the
property $\eta^{\prime}=\zeta$. Indeed, $\bar{\zeta}=\eta^{\prime}$ also
satisfies the second equation in (15) which implies that

$\displaystyle\eta^{\prime}(t)=\hat{\Psi}(\eta,\zeta)(t)+Ce^{(\alpha-\beta-1)t},$

where $C$ is a constant. On the other hand, (31), (32), or (33) imply that

$\displaystyle\eta^{\prime}=\frac{1}{\gamma}\,\int_{t}^{+\infty}\frac{d}{dt}\big{\\{}e^{\delta(t-s)}\chi(t-s)\big{\\}}\Phi(s,\eta,\zeta)\,ds,$

which implies that $\eta^{\prime}(t)=o(1)$. Therefore, $C=0$ and
$\zeta=\eta^{\prime}$.

Thus, we obtained a singular solution $v^{*}(t)=g(\theta
t+\varphi(t))+\eta(t)$ to (11).

Consider now the case (ii), i.e., $g^{\prime\prime}(\theta t)=0$ on
$[\hat{T},+\infty)$ for some $\hat{T}>0$. In this case, the terms $F_{2}$ and
$F_{4}$ are equal to zero. The non-zero terms are $F_{1}$, $F_{3}$, and
$F_{5}$. Since $F_{1}(t,0,0)=F_{3}(t,0,0)=F_{5}(t,0,0)=0$, $\eta=0$ is a
solution to

$\displaystyle\eta^{\prime\prime}(t)-(\alpha-\beta-1)\,\eta^{\prime}(t)+\theta(\hat{\alpha}-1)\,\eta(t)=\Phi(t,\eta,\eta^{\prime}).$

Hence, in the case (ii), $v^{*}(t)=g(\theta t+\varphi(t))$ 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 $(-\infty,+\infty)$. Moreover, the extended solution
$v^{*}$ is strictly increasing and there exists a number $T^{*}\in(-\infty,T)$
such that $v^{*}(T^{*})=0$.

###### Proof.

By Proposition 2.1, there exists a singular solution $v^{*}$ to (11) on some
interval $[T,+\infty)$. Equation (11) can be rewritten as follows:

(38)
$\displaystyle\big{(}e^{(\beta+1-\alpha)t}|{v}^{\prime}(t)|^{\beta}{v}^{\prime}(t)\big{)}^{\prime}=-(\beta+1)e^{-(\gamma+1)t}e^{f(v)}.$

Substituting $v^{*}$ into (38) and integrating it from $t$ to $+\infty$, we
obtain

(39) $\displaystyle
e^{(\beta+1-\alpha)t}|{v^{*}}^{\prime}(t)|^{\beta}{v^{*}}^{\prime}(t)=(\beta+1)\int_{t}^{+\infty}e^{-(\gamma+1)s}e^{f(v^{*})}ds.$

Indeed, since $\lim_{t\to+\infty}g^{\prime\prime}(t)=0$, we conclude that
$g^{\prime}(t)$ has at most linear growth. Furthermore, since
${v^{*}}^{\prime}(t)=g^{\prime}(\theta
t+\varphi)(\theta+\varphi^{\prime})+\eta^{\prime}$, where $\varphi=o(t)$,
$\varphi^{\prime}=O\big{(}\frac{g^{\prime\prime}(\theta t)}{g^{\prime}(\theta
t)}\big{)}$ (see formula (19)) and $\eta^{\prime}=O(g^{\prime\prime}(\theta
t))$, we obtain that ${v^{*}}^{\prime}(t)$ also has at most linear growth on
$[T,+\infty)$. Since $\beta+1-\alpha<0$, the expression on left-hand side of
(39) goes to zero as $t\to+\infty$, and hence, (39) holds. Recall that
${v^{*}}^{\prime}(t)>0$ (as we showed in the proof of Proposition 2.1) for any
extension of $v^{*}$. Let
$w(t)=e^{(\beta+1-\alpha)t}{v^{*}}^{\prime}(t)^{\beta+1}$. We then obtain the
following system with respect to $w$ and $v^{*}$:

(40) $\begin{cases}w^{\prime}(t)=-(\beta+1)e^{-(\gamma+1)t}e^{f(v^{*})},\\\
{v^{*}}^{\prime}(t)=e^{(\hat{\alpha}-1)t}w^{\frac{1}{\beta+1}}.\end{cases}$

Pick $S>T$ and define $c_{0}=v^{*}(S)$, $c_{1}=w(S)$. Furthermore, we note
that on $(-\infty,S]$, $c_{1}\leqslant w(t)\leqslant
c_{1}+c_{2}\,e^{-(\gamma+1)t}=\mathfrak{c}_{t}$ for some constant $c_{2}>0$.
Define
$\nu=\operatorname{\mathbbm{1}}_{(-\infty,c_{0}+\varepsilon]}\ast\rho_{\varepsilon}$
and
$\varsigma_{t}=\operatorname{\mathbbm{1}}_{[c_{1}-\varepsilon,\mathfrak{c}_{t}+\varepsilon]}\ast\rho_{\varepsilon}$,
where $\rho_{\varepsilon}$, $\varepsilon<c_{1}$, is a standard mollifier
supported on the ball of radius $\varepsilon$. Note that $\nu(\,\cdot\,)=1$ on
$(-\infty,c_{0}]$ and $\varsigma_{t}(\,\cdot\,)=1$ on
$[c_{1},\mathfrak{c}_{t}]$. Instead of (40), consider the system of first-
order ODEs on $(-\infty,S]$

(41)
$\begin{cases}w^{\prime}(t)=-(\beta+1)e^{-(\gamma+1)t}e^{\nu(v^{*})f(v^{*})},\\\
{v^{*}}^{\prime}(t)=e^{(\hat{\alpha}-1)t}(\varsigma_{t}(w)w)^{\frac{1}{\beta+1}}.\end{cases}$

On $[T,S]$, $(v^{*}(t),w(t))$ is also a solution to (41), since over this
interval $\nu(v^{*}(t))=1$ and $\varsigma_{t}(w(t))=1$. Next, since
$e^{\nu(v)f(v)}\leqslant e^{f(c_{0}+2\varepsilon)}$ and
$0<(\varsigma_{t}(w)w)^{\frac{1}{\beta+1}}\leqslant(\mathfrak{c}_{t}+2\varepsilon)^{\frac{1}{\beta+1}}$,
one can extend $(v^{*},w)$ to $(-\infty,S]$ (see, e.g., [16], Chapter 2, §6),
obtaining by this a solution $(\tilde{v},\tilde{w})$ to (41) which coincides
with $(v^{*},w)$ on $[T,S]$. Since on $(-\infty,S]$, $\nu(\tilde{v})=1$ and
$\varsigma_{t}(\tilde{w})=1$, $(\tilde{v},\tilde{w})$ 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 $(\tilde{v},\tilde{w})$. There
are two possible situations: either $v^{*}$ is strictly positive over
$(-\infty,S]$, or there exists $T^{*}\in{\mathbb{R}}$ 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 $(-\infty,S]$, then there
exists a finite limit $L=\lim_{t\to-\infty}{v^{*}}(t)$. Integrating (38) from
$-R$ to $0$ (where $R>0$ is sufficiently large) and taking into account that
$e^{f({v^{*}})}\geqslant e^{f(L)}$, we obtain that

$\displaystyle{v^{*}}^{\prime}(-R)^{\beta+1}\geqslant
e^{-(\alpha-\beta-1)R}{v^{*}}^{\prime}(0)^{\beta+1}+\frac{\beta+1}{\gamma+1}e^{f(L)}\big{(}e^{\theta
R}-e^{-(\alpha-\beta-1)R}\big{)}$

which shows that ${v^{*}}^{\prime}(-R)\to+\infty$ as $R\to+\infty$. The latter
implies that
${v^{*}}(-R)={v^{*}}(0)-\int_{-R}^{0}{v^{*}}^{\prime}(t)dt\to-\infty$ as
$R\to+\infty$. This contradicts to the fact that
$\lim_{t\to-\infty}{v^{*}}(t)=L<+\infty$.

Thus, we conclude that solution (12) can be extended to $(-\infty,+\infty)$ in
such a way that there exists a finite number $T^{*}$ such that
$v^{*}(T^{*})=0$. ∎

#### 2.4.2 Proof of Theorem 2.1

The proof of Theorem 2.1 now follows from the inverse change of variable.

###### Proof of Theorem 2.1.

Let $T^{*}$ be as in Proposition 2.2. Define $\lambda^{*}=(\beta+1)e^{-\theta
T^{*}}$ and
$\kappa=\big{(}\frac{\beta+1}{\lambda^{*}}\big{)}^{\frac{1}{\theta}}$. Further
define $u^{*}_{{\scriptscriptstyle\lambda}^{\\!*}}(r)=v^{*}(\ln(\kappa/r))$.
By the argument in paragraph 2.4.1,
$(u^{*}_{{\scriptscriptstyle\lambda}^{\\!*}},\lambda^{*})$ is a singular
solution to problem ($P_{\lambda}$). ∎

### 2.5 Exact singular solution to the Gelfand problem for a $k$-Hessian

Consider the problem

(42) $\begin{cases}-L(u)=\lambda\,e^{u}&\text{ in }B,\\\ u=0&\text{ on
}\partial B,\end{cases}$

where $L$ can be a $k$-Hessian with $k<\frac{d}{2}$, a $p$-Laplacian with
$p<d$, or, in general, the operator (5) with $\alpha,\beta,\gamma$ 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 $k<\frac{d}{2}$. Then, the singular solution constructed in Theorem 2.1
takes the form

(43) $\displaystyle
u^{*}_{{\scriptscriptstyle\lambda}^{\\!*}}(x)=-2k\ln|x|,\quad\lambda^{*}=(2k)^{k}(d-2k).$

If $L$ is a $p$-Laplacian with $p<d$, then, the aforementioned singular
solution is

(44) $\displaystyle
u^{*}_{{\scriptscriptstyle\lambda}^{\\!*}}(x)=-p\ln|x|,\quad\lambda^{*}=p^{p-1}(d-p).$

###### Proof of Corollaries 1.1 and 2.3.

Note that for problem (42), $f(t)=g(t)=t$. Therefore, $g^{\prime}(t)=1$,
$g^{\prime\prime}(t)=0$. By Proposition 2.1, in a neighborhood of $+\infty$,
the solution (12) takes the form

(45) $\displaystyle v^{*}(t)=\theta
t+\ln\\{\theta^{\beta+1}(\hat{\alpha}-1)\\}.$

It is straightforward to check that (45) verifies (11) with $f(v)=v$ on the
entire real line. Furthermore, $v(t)=0$ if
$t=T^{*}=-\frac{\ln\\{\theta^{\beta+1}(\hat{\alpha}-1)\\}}{\theta}$.
Therefore,

$\displaystyle\lambda^{*}=(\beta+1)e^{-\theta
T^{*}}=(\beta+1)\theta^{\beta+1}(\hat{\alpha}-1)=\theta^{\beta+1}(\alpha-\beta-1).$

Recall that $u^{*}_{{\scriptscriptstyle\lambda}^{\\!*}}(x)=v^{*}(t)$ with
$t=\ln\big{\\{}\big{(}\frac{\beta+1}{\lambda^{*}}\big{)}^{\frac{1}{\theta}}\frac{1}{|x|}\big{\\}}=-\frac{1}{\theta}\ln\\{\theta^{\beta+1}(\hat{\alpha}-1)\\}-\ln|x|$.
Comparing it with (45), we obtain that
$u^{*}_{{\scriptscriptstyle\lambda}^{\\!*}}(x)=-\theta\ln|x|$.

In the case of a $k$-Hessian with $k<\frac{d}{2}$, we have $\theta=2k$,
$\beta+1=k$, $\alpha-\beta-1=d-2k$, which implies (43).

In the case of a $p$-Laplacian with $p<d$, we have $\theta=p$, $\beta+1=p-1$,
$\alpha-\beta-1=d-p$, which implies (44). ∎

## 3\. Regular solutions to problem ($P_{\lambda}$)

Here, we will be interested in regular solutions to problem ($P_{\lambda}$).
For regular solutions, problem ($P_{\lambda}$) should be stated as follows:

($P^{\prime}_{\lambda}$)
$\begin{cases}-L(u_{\scriptscriptstyle\lambda}(r))=\lambda\,e^{f(u_{\scriptscriptstyle\lambda}(r))},\\\
u^{\prime}_{\scriptscriptstyle\lambda}(0)=u_{\scriptscriptstyle\lambda}(1)=0.\end{cases}$

Indeed, from Lemma 3.1 below, it follows that any solution to ($P_{\lambda}$)
is decreasing in $r\in[0,+\infty)$ and achieves the maximum at $r=0$.
Moreover, since the actual solution $u_{\scriptscriptstyle\lambda}$ should be
regular in a ball, we add the condition
$u^{\prime}_{\scriptscriptstyle\lambda}(0)=0$.

######  Lemma 3.1.

Let $(u_{\scriptscriptstyle\lambda},\lambda)$ be a regular radial solution to
($P^{\prime}_{\lambda}$). Then, $u^{\prime}_{\scriptscriptstyle\lambda}(r)<0$
on $(0,+\infty)$.

###### Proof.

The equation in ($P^{\prime}_{\lambda}$) implies

$\displaystyle
r^{\alpha}|u^{\prime}_{\scriptscriptstyle\lambda}(r)|^{\beta}u^{\prime}_{\scriptscriptstyle\lambda}(r)=-\int_{0}^{r}s^{\gamma}e^{f(u_{\scriptscriptstyle\lambda}(s))}ds+C.$

Since $u^{\prime}_{\scriptscriptstyle\lambda}(0)=0$, we obtain that $C=0$.
This implies that $u^{\prime}(r)<0$ for all $r>0$. ∎

######  Remark 3.1.

If $\gamma+1>\alpha$ (as in the case of the $k$-Hessian), then in
($P^{\prime}_{\lambda}$), we do not need to state that $u^{\prime}(0)=0$ if we
are interested in a regular solution. This assumption is fulfilled
automatically. Indeed, by L’Hopital’s rule,

$\displaystyle(-u^{\prime}_{\scriptscriptstyle\lambda}(0))^{\beta+1}=\lim_{r\to
0}\alpha^{-1}r^{\gamma+1-\alpha}e^{f(u_{\scriptscriptstyle\lambda}(r))}=0.$

### 3.1 Standing assumptions: regular solutions

  * (B1)

$f\in{\rm C}^{1}({\mathbb{R}},{\mathbb{R}})$, $f^{\prime}>0$, and
$\lim_{u\to+\infty}f(u)=+\infty$.

  * (B2)

$\lim_{u\to+\infty}u^{-1}\exp\big{\\{}\frac{f(u)}{\beta+1}\big{\\}}=\lim_{u\to+\infty}uf^{\prime}(u)=+\infty$.

  * (B3)

$\alpha>\beta+1>0$; $\theta>0$.

######  Remark 3.2.

Remark that the second limit in (B2) is implied by (A4) and the first
condition of (A2) ($\lim_{t\to+\infty}g^{\prime\prime}(t)=0$). We have

(46)
$\displaystyle\lim_{u\to+\infty}uf^{\prime}(u)=\lim_{t\to+\infty}\frac{g(t)}{g^{\prime}(t)}\geqslant\lim_{t\to+\infty}\frac{g(t)}{\int_{0}^{t}|g^{\prime\prime}(s)|ds+g^{\prime}(0)}=+\infty.$

Indeed, we know that $\lim_{t\to+\infty}g(t)=+\infty$. If
$\lim_{t\to+\infty}\int_{0}^{t}|g^{\prime\prime}(s)|ds<+\infty$, then (46) is
straightforward. Otherwise, if
$\lim_{t\to+\infty}\int_{0}^{t}|g^{\prime\prime}(s)|ds=+\infty$, then (46)
follows from L’Hopital’s rule and Remark 2.3. The first expression in (B2)
tends to $+\infty$ by L’Hopital’s rule and Remark 2.5.

### 3.2 Existence and uniqueness of a regular solution to
($P^{\prime}_{\lambda}$)

First of all, we note that by the change of variable (8), problem
($P^{\prime}_{\lambda}$) can be transformed to

($P_{\rho}$) $\begin{cases}-L(u)=e^{f(u)},\\\ u(0)=\rho,\quad
u^{\prime}(0)=0.\end{cases}$

Above, $\rho$ and $\lambda$ are connected through the identity

$\displaystyle\rho=\lambda^{\frac{1}{\beta+1}}\int_{0}^{1}t^{-\frac{\alpha}{\beta+1}}\Big{(}\int_{0}^{t}s^{\gamma}e^{f(u_{\scriptscriptstyle\lambda}(s))}ds\Big{)}^{\frac{1}{\beta+1}}dt,$

where $(u_{\scriptscriptstyle\lambda},\lambda)$ is the solution to
($P^{\prime}_{\lambda}$). Note that problem ($P_{\rho}$) is equivalent to the
integral equation

(47) $\displaystyle
u(r)=\rho-\int_{0}^{r}t^{-\frac{\alpha}{\beta+1}}\Big{(}\int_{0}^{t}s^{\gamma}e^{f(u(s))}ds\Big{)}^{\frac{1}{\beta+1}}dt.$

We have the following result on the existence and uniqueness of solution to
(47).

######  Lemma 3.2.

Assume (B1) and (B3). Then, equation (47) has a unique solution.

###### Proof.

Let the map $\Gamma(u)$ 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

$\displaystyle u\leqslant\rho\quad\text{and}\quad
u\geqslant\rho-e^{\frac{f(\rho)}{1+\beta}}(\gamma+1)^{-\frac{1}{\beta+1}}\hat{\theta}^{-1}T^{\hat{\theta}}.$

Let $A_{\rho,T}$ denote the constant on the right-hand side of the second
estimate. Introduce the complete metric space

$\displaystyle\Sigma=\\{u\in{\rm C}[0,T]:\;A_{\rho,T}\leqslant
u\leqslant\rho\\}$

with the supremum norm as the metric, and take two functions $u_{1}$ and
$u_{2}$ from $\Sigma$. It is easy to see that $\Gamma(u_{1})$ and
$\Gamma(u_{2})$ are functions $\Sigma\to\Sigma$. Furthermore,

$\Gamma(u_{1})(r)-\Gamma(u_{2})(r)=\frac{1}{\beta+1}\int_{0}^{r}t^{-\frac{\alpha}{\beta+1}}\Big{(}\int_{0}^{1}\Lambda(t,\rho,\lambda)^{-\frac{\beta}{\beta+1}}d\lambda\Big{)}\\\
\int_{0}^{t}s^{\gamma}(e^{f(u_{2}(s))}-e^{f(u_{1}(s))})ds\,dt,$

where $\Lambda(t,\rho,\lambda)=\int_{0}^{t}s^{\gamma}(\lambda
e^{f(u_{1}(s))}+(1-\lambda)e^{f(u_{2}(s))})ds$, and we have that

$\displaystyle
e^{f(A_{\rho,T})}(\gamma+1)^{-1}t^{\gamma+1}\leqslant\Lambda(t,\rho,\lambda)\leqslant
e^{f(\rho)}(\gamma+1)^{-1}t^{\gamma+1}.$

Taking into account that $|u_{1}|$ and $|u_{2}|$ are bounded, we obtain that
there exists a constant $K=K(\rho,T,\gamma,\beta,f)$ such that

$\displaystyle\sup_{[0,r]}|\Gamma(u_{1})-\Gamma(u_{2})|\leqslant
K\int_{0}^{r}t^{\frac{-\alpha+\gamma+1}{\beta+1}}\sup_{[0,t]}|u_{1}-u_{2}|\,ds.$

Since $\frac{-\alpha+\gamma+1}{\beta+1}=\hat{\theta}-1$, we obtain that for
the map
$\Gamma^{(n)}=\underbrace{\Gamma\circ\Gamma\circ\dots\circ\Gamma}_{n}$, it
holds that

$\displaystyle\sup_{[0,r]}|\Gamma^{(n)}(u_{1})-\Gamma^{(n)}(u_{2})|\leqslant\frac{(Kr^{\hat{\theta}}\hat{\theta}^{-1})^{n}}{n!}\sup_{[0,r]}|u_{1}-u_{2}|\,ds.$

This implies that $\Gamma^{(n)}$ is a contraction for some $n$. Clearly, the
unique fixed point of $\Gamma^{(n)}$ is also the unique fixed point of
$\Gamma$, 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 $f\in{\rm C}({\mathbb{R}},{\mathbb{R}})$, $a\in{\mathbb{R}}$ be an
arbitrary constant, and $u\in{\rm C}^{2}[0,+\infty)$ be a solution to problem
($P_{\rho}$). Then, for all $r>0$,

(48)
$r^{\alpha}|u^{\prime}|^{\beta+2}\Big{(}a-\frac{\alpha-\beta-1}{\beta+2}\Big{)}+r^{\gamma}\Big{\\{}(\gamma+1)\int_{0}^{u}e^{f(t)}dt-a\,u\,e^{f(u)}\Big{\\}}\\\
=\frac{d}{dr}\Big{\\{}r^{\alpha+1}\Big{(}\frac{\beta+2}{\beta+1}\,|u^{\prime}|^{\beta+2}+r^{\gamma-\alpha}\int_{0}^{u}e^{f(t)}dt+ar^{-1}|u^{\prime}|^{\beta}u^{\prime}u\Big{)}\Big{\\}}.$

###### Proof.

In [22], a version of Pohozaev’s identity was obtained for solutions to
equations whose particular type is

(49) $\displaystyle
r^{-(d-1)}(r^{(d-1)}|u^{\prime}(r)|^{\beta}u^{\prime}(r))^{\prime}+h(r,u)=0,$

where $d$ is the space dimension and $h\in{\rm
C}({\mathbb{R}}_{+}\times{\mathbb{R}})$. As it follows from the proof of
Theorem 2.1, exposed in paragraph 2 of [22], instead of $d-1$, one can use any
number $\alpha>0$. 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 [22]. Multiplying the both parts of the equation in
($P_{\rho}$) by $r^{\alpha-\gamma}$, we bring this equation to a form similar
to (49):

$\displaystyle
r^{-\alpha}(r^{\alpha}|u^{\prime}(r)|^{\beta}u^{\prime}(r))^{\prime}+r^{\gamma-\alpha}e^{f(u)}=0.$

We then apply Theorem 2.1 from [22] to the above equation, substituting $d$ by
$\alpha+1$ and taking into account that $u^{\prime}(0)=0$. Referring to the
notation of the aforementioned theorem, we have $A(\rho)=\rho^{\beta}$ and
$E(\rho)=(\beta+1)\rho^{\beta}$. It is then straightforward to verify the
assumptions of Theorem 2.1 from [22] as well as to make necessary computations
leading to equation (48). ∎

Let $u(\,\cdot\,,\rho)$ be the solution to problem ($P_{\rho}$). Our next
result, given by Proposition 3.1, is about upper and lower bounds for the
number

(50) $\displaystyle R(B,\rho)=u(\,\cdot\,,\rho)^{-1}(B),\quad 0\leqslant
B\leqslant\rho.$

######  Proposition 3.1.

Assume (B1)–(B3). Then, for each $B\geqslant 0$, the number $R(B,\rho)$ is
well-defined. Moreover, there exist two positive constants $\underline{R}(B)$
and $\bar{R}(B)$, and a number $\rho_{0}(B)>B$, all depending only on $B$,
such that for all $\rho\geqslant\rho_{0}(B)$,

$\displaystyle\underline{R}(B)\leqslant R(B,\rho)\leqslant\bar{R}(B).$

Moreover, the second inequality holds for all $\rho\geqslant B$.

###### Proof.

Fix a number $B\geqslant 0$. We start by showing that $R(B,\rho)$ is well-
defined for $\rho>B$. This is equivalent to the fact that a zero of
$u(r,\rho)-B$ exists. Suppose this is not the case. Since $u(r,\rho)$ is
decreasing, then $u(r,\rho)>B$ for all $r\geqslant 0$. From ($P_{\rho}$), we
obtain that
$u(r,\rho)\leqslant\rho-e^{\frac{f(B)}{1+\beta}}(\gamma+1)^{-\frac{1}{\beta+1}}\hat{\theta}^{-1}r^{\hat{\theta}}\to-\infty$,
as $r\to+\infty$. This is a contradiction. Hence, there exists a unique $r$
such that $u(r,\rho)=B$.

Let us prove that the function $[B,+\infty)\to{\mathbb{R}}$, $\rho\mapsto
R(B,\rho)$ is bounded from above. Rescale problem ($P_{\rho}$) by setting
$\lambda^{\frac{1}{\theta}}=R(B,\rho)$ as follows:

$\begin{cases}-L(u(\lambda^{\frac{1}{\theta}}r))=\lambda\,e^{f(u(\lambda^{\frac{1}{\theta}}r))},\\\
u(0)=\rho,\quad u^{\prime}(0)=0\end{cases}$

which is equivalent to

$\displaystyle
u(\lambda^{\frac{1}{\theta}}r)=B+\int_{r}^{1}\lambda^{\frac{1}{\beta+1}}t^{-\frac{\alpha}{\beta+1}}\Big{(}\int_{0}^{t}s^{\gamma}e^{f(u(\lambda^{\frac{1}{\theta}}s))}ds\Big{)}^{\frac{1}{\beta+1}}dt.$

For $r\in(0,1)$, it holds that

$\displaystyle
u(\lambda^{\frac{1}{\theta}}r)-B\geqslant\lambda^{\frac{1}{\beta+1}}\int_{r}^{1}t^{-\frac{\alpha}{\beta+1}}dt\,\Big{(}\int_{0}^{r}s^{\gamma}e^{f(u(\lambda^{\frac{1}{\theta}}r))}ds\Big{)}^{\frac{1}{\beta+1}}.$

This implies that for $r\in(0,1)$, we have the estimate

(51) $\displaystyle\sup_{u\geqslant
B}\big{\\{}(u-B)\,e^{-\frac{f(u)}{\beta+1}}\big{\\}}\geqslant(u(\lambda^{\frac{1}{\theta}}r)-B)e^{-\frac{f(u(\lambda^{\frac{1}{\theta}}r))}{\beta+1}}\geqslant
K_{1}\lambda^{\frac{1}{\beta+1}}(r^{\frac{\theta}{\beta+1}}-r^{\frac{\gamma+1}{\beta+1}}),$

where $K_{1}>0$ is a constant. By (B2), $0<\sup_{u\geqslant
B}(u-B)\,e^{-\frac{f(u)}{\beta+1}}<+\infty$. Hence, the right-hand side of
(51) is bounded uniformly in $r\in(0,1)$. Evaluating it (say) at
$r=\frac{1}{2}$, we obtain that for all $B\geqslant 0$ and $\rho\geqslant B$,

(52) $\displaystyle R(B,\rho)=\lambda^{\frac{1}{\theta}}\leqslant
K_{2}\sup_{u\geqslant
B}\big{\\{}(u-B)\,e^{-\frac{f(u)}{\beta+1}}\big{\\}}^{\frac{\beta+1}{\theta}}=\bar{R}(B),$

where $K_{2}$ is a constant. Now we prove the existence of a lower bound
$\underline{R}(B)$ by using identity (48). By L’Hopital’s rule and (B2),

$\displaystyle\lim_{u\to+\infty}\frac{u\,e^{f(u)}}{\int_{0}^{u}e^{f(t)}dt}=\lim_{u\to+\infty}\frac{e^{f(u)}+u\,e^{f(u)}f^{\prime}(u)}{e^{f(u)}}=1+\lim_{u\to+\infty}uf^{\prime}(u)=+\infty.$

As we show below, it suffices to prove the existence of $\underline{R}(B)$ for
those $B>0$ which make the following inequality hold:

(53) $\displaystyle(\gamma+1)\int_{0}^{u}e^{f(t)}dt<au\,e^{f(u)}\quad\text{for
all}\;\;u\geqslant B,$

where $a\in{\mathbb{R}}$ is to be fixed later. Indeed, suppose the existence
of $\underline{R}(B)$ is obtained for those $B$ which make inequality (53)
fulfilled for $u\geqslant B$. If $B\geqslant 0$ is arbitrary, we find a number
$\tilde{B}>B$ such that (53) holds for $u\geqslant\tilde{B}$. Then,
$\underline{R}(\tilde{B})\leqslant R(\tilde{B},\rho)\leqslant R(B,\rho)$ for
$\rho\geqslant\rho_{0}(\tilde{B})$, and we can define
$\underline{R}(B)=\underline{R}(\tilde{B})$ and
$\rho_{0}(B)=\rho_{0}(\tilde{B})$. 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 $(x+y)^{\frac{1}{1+\beta}}\leqslant
x^{\frac{1}{1+\beta}}+y^{\frac{1}{1+\beta}}$ if $\beta\geqslant 0$ and
$(x+y)^{\frac{1}{1+\beta}}\leqslant
2^{-\frac{\beta}{\beta+1}}(x^{\frac{1}{1+\beta}}+y^{\frac{1}{1+\beta}})$ if
$-1<\beta<0$. We set $A_{\beta}=\max\\{1,2^{-\frac{\beta}{\beta+1}}\\}$ and
fix
$\hat{a}\in(0,A_{\beta}^{-1}-\frac{a(\beta+1)^{2}}{(\beta+2)(\alpha-\beta-1)})$.
The number $a>0$ is then has to be chosen in such a way that
$\frac{a(\beta+1)^{2}}{(\beta+2)(\alpha-\beta-1)}<A_{\beta}^{-1}$. In
addition, we choose $a$ to satisfy $a<\frac{\alpha-\beta-1}{\beta+2}$ so that
the first term on the left-hand side of (48) is negative. By (48) and (53), on
$(0,\bar{R}(B)]$, it holds that

$\displaystyle\frac{\beta+2}{\beta+1}\,|u^{\prime}|^{\beta+2}+ar^{-1}|u^{\prime}|^{\beta}u^{\prime}u<0.$

Since $u^{\prime}<0$, for $r\in(0,\bar{R}(B)]$,

(54) $\displaystyle r\,|u^{\prime}(r)|<a\,\frac{\beta+1}{\beta+2}\,u(r).$

Next, we define

$\displaystyle
D=B\Big{(}1-\hat{a}A_{\beta}-\frac{A_{\beta}\,a(\beta+1)^{2}}{(\beta+2)(\alpha-\beta-1)}\Big{)}^{-1}\;\text{and}\quad\underline{R}(B)=(\hat{a}\,D\,\hat{\theta}(\gamma+1)^{\frac{1}{\beta+1}}e^{-\frac{f(D)}{\beta+1}})^{\frac{1}{\hat{\theta}}}.$

We aim to prove that $R(B,\rho)\geqslant\underline{R}(B)$ for all
$\rho\geqslant D$. If, for some $\rho\geqslant D$,
$R(D,\rho)\geqslant\underline{R}(B)$, then (since $D>B$)
$R(B,\rho)>\underline{R}(B)$. Otherwise, if $R(D,\rho)<\underline{R}(B)$,
consider the following problem on $[R(D,\rho),\underline{R}(B)]$:

(55) $\displaystyle L(v)=-e^{f(D)},\quad v(R(D,\rho))=D,\quad
v^{\prime}(R(D,\rho))=u^{\prime}(R(D,\rho),\rho).$

To simplify notation, define $\tilde{D}=e^{f(D)}$ and $\tilde{R}=R(D,\rho)$.
Since

(56) $\displaystyle
r^{\alpha}|v^{\prime}|^{\beta}v^{\prime}=-\tilde{D}\int_{\tilde{R}}^{r}s^{\gamma}ds-\tilde{R}^{\alpha}|u^{\prime}(\tilde{R})|^{\beta+1},$

we obtain that on $[\tilde{R},\underline{R}(B)]$, $v^{\prime}(r)<0$. Taking
into account this observation, by (54) and (56), for
$r\in[\tilde{R},\underline{R}(B)]$, we obtain

$v(r)=D-\int_{\tilde{R}}^{r}\Big{(}\frac{\tilde{D}}{\gamma+1}(s^{\gamma+1}-\tilde{R}^{\gamma+1})+\tilde{R}^{\alpha}|u^{\prime}(\tilde{R})|^{\beta+1}\Big{)}^{\frac{1}{\beta+1}}s^{-\hat{\alpha}}ds\\\
\geqslant
D-A_{\beta}\Big{[}\Big{(}\frac{\tilde{D}}{\gamma+1}\Big{)}^{\frac{1}{\beta+1}}\int_{\tilde{R}}^{r}s^{\frac{\gamma+1}{\beta+1}}s^{-\frac{\alpha}{\beta+1}}ds+\frac{\tilde{R}^{\hat{\alpha}}|u^{\prime}(\tilde{R})|r^{-\hat{\alpha}+1}}{\hat{\alpha}-1}-\frac{\tilde{R}\,|u^{\prime}(\tilde{R})|}{\hat{\alpha}-1}\Big{]}\\\
>D-A_{\beta}\Big{(}\frac{\tilde{D}}{\gamma+1}\Big{)}^{\frac{1}{\beta+1}}\frac{\underline{R}(B)^{\hat{\theta}}}{\hat{\theta}}-\frac{A_{\beta}\,a(\beta+1)D}{(\beta+2)(\hat{\alpha}-1)}=B,$

where the last identity follows from the definition of $\underline{R}(B)$ and
$D$.

It remains to show that on $[\tilde{R},\underline{R}(B)]$, $u(r,\rho)\geqslant
v(r)$. Indeed, by (55), for $r\in[\tilde{R},\underline{R}(B)]$, we have

$\displaystyle(r^{\alpha}|v^{\prime}|^{\beta+1})^{\prime}=r^{\gamma}e^{f(D)}\geqslant
r^{\gamma}e^{f(u)}=(r^{\alpha}|u^{\prime}|^{\beta+1})^{\prime}\text{ and}$
$\displaystyle r^{\alpha}|v^{\prime}|^{\beta+1}\geqslant
r^{\alpha}|u^{\prime}|^{\beta+1}.$

Since the derivatives $u^{\prime}$ and $v^{\prime}$ are negative, again by
(55), we obtain that on $[\tilde{R},\underline{R}(B)]$, $u(r,\rho)\geqslant
v(r)\geqslant B$. Therefore, $R(B,\rho)\geqslant\underline{R}(B)$, and we can
take $\rho_{0}(B)=D$. The proof is now complete. ∎

######  Proposition 3.2.

Assume (B1)–(B3). Then, for any interval $[r_{1},r_{2}]\subset(0,\infty)$, the
family $\\{u(\,\cdot\,,\rho)\\}_{\rho\geqslant 0}$ is relatively compact in
${\rm C}([r_{1},r_{2}])$. Furthermore, each limit point of this family is a
solution to (7), regular or singular.

###### Proof.

Fix an interval $[r_{1},r_{2}]\subset(0,+\infty)$. Let $B>0$ and let
$\underline{R}(B)$, $\bar{R}(B)$, and $\rho_{0}(B)$ be as they were defined in
Proposition 3.1 , so we have that $R(B,\rho)\in[\underline{R}(B),\bar{R}(B)]$
for all $\rho>\rho_{0}(B)$, where $R(B,\rho)$ is defined by (50). By (B2),
$\lim_{B\to+\infty}\bar{R}(B)=0$. Choose a number $B>0$ such that
$r_{1}>\bar{R}(B)$, and note that on $[r_{1},r_{2}]$, $u(r,\rho)<B$ for all
$\rho\geqslant B$. By (47), it holds that for $r\geqslant R(B,\rho)$,

(57)
$u(r,\rho)=B-\int_{R(B,\rho)}^{r}t^{-\frac{\alpha}{\beta+1}}\Big{(}\int_{0}^{t}s^{\gamma}e^{f(u(s,\rho))}ds\Big{)}^{\frac{1}{\beta+1}}dt\\\
\geqslant
B-e^{\frac{f(B)}{\beta+1}}\int_{R(B,\rho)}^{r}t^{-\frac{\alpha}{\beta+1}}\Big{(}\int_{0}^{t}s^{\gamma}ds\Big{)}^{\frac{1}{\beta+1}}dt\geqslant
B\,-\,e^{\frac{f(B)}{\beta+1}}(\gamma+1)^{-\frac{1}{\beta+1}}\hat{\theta}^{-1}r^{\hat{\theta}}.$

Therefore, the family $\\{u(r,\rho)\\}_{\rho\geqslant B}$ is uniformly bounded
on $[r_{1},r_{2}]$.

Let us show that the family $\\{u(r,\rho)\\}_{\rho\leqslant B}$ is uniformly
bounded on $[0,r_{2}]$. Note that this family is bounded from above by $B$. It
is also bounded from below by the same argument as (57). Namely,

$\displaystyle
u(r,\rho)=\rho-\int_{0}^{r}t^{-\frac{\alpha}{\beta+1}}\Big{(}\int_{0}^{t}s^{\gamma}e^{f(u(s,\rho))}ds\Big{)}^{\frac{1}{\beta+1}}dt\geqslant-e^{\frac{f(B)}{\beta+1}}(\gamma+1)^{-\frac{1}{\beta+1}}\hat{\theta}^{-1}r_{2}^{\hat{\theta}}.$

Furthermore, (47) implies the uniform continuity of the family
$\\{u(r,\rho)\\}_{\rho\geqslant 0}$ on $[r_{1},+\infty)$. Indeed, for all
$r^{\prime},r^{\prime\prime}$, such that $r_{1}<r^{\prime}<r^{\prime\prime}$,
and $\rho\geqslant 0$,

(58)
$0<u(r^{\prime},\rho)-u(r^{\prime\prime},\rho)=\int_{r^{\prime}}^{r^{\prime\prime}}t^{-\frac{\alpha}{\beta+1}}\Big{(}\int_{0}^{t}s^{\gamma}e^{f(u(s,\rho))}ds\Big{)}^{\frac{1}{\beta+1}}dt\\\
\leqslant
e^{\frac{f(B)}{\beta+1}}(\gamma+1)^{-\frac{1}{\beta+1}}\hat{\theta}^{-1}({r^{\prime\prime}}^{\hat{\theta}}-{r^{\prime}}^{\hat{\theta}}).$

By the Arzelà-Ascolli theorem, the family $\\{u(r,\rho)\\}$ is relatively
compact on $[r_{1},r_{2}]$.

Let us prove now that each limit point of the family
$\\{u(r,\rho)\\}_{\rho\geqslant 0}$ is either regular or singular solution.
Let $\\{u(r,\rho_{n})\\}$ converge to a function $u^{**}(r)$ in ${\rm
C}([r_{1},r_{2}])$. We claim that the sequence $\partial_{r}u(r_{1},\rho_{n})$
is bounded. Suppose it is not the case, that is, we can find a subsequence
$\partial_{r}u(r_{1},\rho_{n_{k}})\to-\infty$ as $k\to\infty$. Without loss of
generality, we assume that
$\lim_{n\to\infty}\partial_{r}u(r_{1},\rho_{n})=-\infty$. For
$r\in[r_{1},r_{2}]$, it holds that,

$\displaystyle
u(r,\rho_{n})=u(r_{1},\rho_{n})-\int_{r_{1}}^{r}t^{-\frac{\alpha}{\beta+1}}\Big{(}r_{1}^{\alpha}|\partial_{r}u(r_{1},\rho_{n})|^{\beta+1}+\int_{r_{1}}^{t}s^{\gamma}e^{f(u(s,\rho_{n}))}ds\Big{)}^{\frac{1}{\beta+1}}dt.$

The last term in this identity converges to $\infty$ which contradicts to the
convergence of $\\{u(r,\rho_{n})\\}$ in ${\rm C}([r_{1},r_{2}])$. Therefore,
$\partial_{r}u(r_{1},\rho_{n})$ is bounded. Let a subsequence
$r_{1}^{\alpha}|\partial_{r}u(r_{1},\rho_{n_{k}})|^{\beta+1}$ converge to
$k(r_{1})$. In the previous expression, passing to the limit as $k\to\infty$,
we obtain

$\displaystyle
u^{**}(r)=u^{**}(r_{1})-\int_{r_{1}}^{r}t^{-\frac{\alpha}{\beta+1}}\Big{(}k(r_{1})+\int_{r_{1}}^{t}s^{\gamma}e^{f(u^{**}(s))}ds\Big{)}^{\frac{1}{\beta+1}}ds.$

From the above identity, we obtain that $u^{**}(r)$ is differentiable in $r$,
$(u^{**})^{\prime}(r)<0$,
$k(r_{1})=r_{1}^{\alpha}|(u^{**})^{\prime}(r_{1})|^{\beta+1}$, and $u^{**}(r)$
solves (7) on $[r_{1},r_{2}]$. By extracting further subsequences from
$\\{u(r,\rho_{n})\\}$ and repeating the argument on compact subintervals of
$(0,+\infty)$, we can extend $u^{**}(r)$ to a solution of (7) on
$(0,+\infty)$. Note that $\\{u(r,\rho_{n})\\}$ contains a subsequence that
converges pointwise to $u^{**}(r)$ on $(0,+\infty)$.

Let us show now the following fact. Suppose $u(r,\rho_{n})$ converges to
$u^{**}(r)$ pointwise on $(0,+\infty)$. If $\rho_{n}$ is unbounded, then
$u^{**}(r)$ is a singular solution, if $\rho_{n}$ is bounded, then it is a
regular solution. Let us show that, in the first case,
$\lim_{r\to+0}u^{**}(r)=+\infty$. Suppose this is not the case, so we can find
a sequence $r_{n}\to 0$ as $n\to+\infty$ such that
$\lim_{n\to\infty}u^{**}(r_{n})=B<+\infty$. Since $(u^{**})^{\prime}(r)<0$,
$\sup_{r\geqslant r_{n}}u^{**}(r)\leqslant u^{**}(r_{n})$, which implies that
$\sup_{r\geqslant 0}u^{**}(r)\leqslant B$. Let $\rho_{n_{k}}\to+\infty$ be a
subsequence of $\\{\rho_{n}\\}$. By Proposition 3.1, we can find
$\underline{R}(2B)$ and $\rho_{0}(2B)$ such that whenever
$\rho_{n_{k}}>\rho_{0}(2B)$, $u(r,\rho_{n_{k}})>2B$ on
$(0,\underline{R}(2B))$. Therefore, $u^{**}(r)\geqslant 2B$ on
$(0,\underline{R}(2B))$. This contradicts to the previously obtained
inequality $\sup_{r\geqslant 0}u^{**}(r)\leqslant B$. Hence,
$\lim_{r\to+0}u^{**}(r)=+\infty$.

Suppose now $\rho_{n}$, introduced above, is bounded. Then, we can extract a
subsequence $\lim_{k\to+\infty}\rho_{n_{k}}=\bar{\rho}<+\infty$. Let us prove
that $u^{**}(r)$ is a regular solution with $u^{**}(0)=\bar{\rho}$, i.e.,
$u^{**}(r)=u(r,\bar{\rho})$. It follows from (47), that $u(r,\rho)$ is
continuous in $\rho$ uniformly in $r$ varying over compact intervals. Indeed,

$\sup_{[0,r]}|u(\,\cdot\,,\rho_{1})-u(\,\cdot\,,\rho_{2})|\leqslant|\rho_{1}-\rho_{2}|\\\
+\frac{1}{\beta+1}\int_{0}^{r}s^{-\frac{\alpha}{\beta+1}}\Big{(}\int_{0}^{1}\Lambda(s,\lambda)^{-\frac{\beta}{\beta+1}}d\lambda\Big{)}\int_{0}^{s}t^{\gamma}(e^{f(u(t,\rho_{2}))}-e^{f(u(t,\rho_{1}))})dt\,ds\\\
\leqslant|\rho_{1}-\rho_{2}|+K_{r,\rho_{1},\rho_{2}}\int_{0}^{r}s^{\hat{\theta}-1}\sup_{[0,s]}|u(\,\cdot\,,\rho_{1})-u(\,\cdot\,,\rho_{2})|ds,$

where $K_{r,\rho_{1},\rho_{2}}>0$ is a constant that only depends on
$r,\rho_{1},\rho_{2}$. Furthermore, above,

$\displaystyle\Lambda(s,\lambda)=\int_{0}^{s}t^{\gamma}(\lambda
e^{f(u(t,\rho_{1}))}+(1-\lambda)e^{f(u(t,\rho_{2}))})dt,$ $\displaystyle
K^{(1)}_{r,\rho_{1},\rho_{2}}\,s^{\gamma+1}\leqslant\Lambda(s,\lambda)\leqslant
K^{(2)}_{r,\rho_{1},\rho_{2}}\,s^{\gamma+1},$

where $K^{(1)}_{r,\rho_{1},\rho_{2}}$ and $K^{(2)}_{r,\rho_{1},\rho_{2}}$ are
positive constants. Now by Gronwalls’s inequality, one can find another
constant $K^{(3)}_{r,\rho_{1},\rho_{2}}$, such that

$\displaystyle\sup_{[0,r]}|u(\,\cdot\,,\rho_{1})-u(\,\cdot\,,\rho_{2})|\leqslant
K^{(3)}_{r,\rho_{1},\rho_{2}}|\rho_{1}-\rho_{2}|.$

This implies that $\lim_{k\to+\infty}u(r,\rho_{n_{k}})=u(r,\bar{\rho})$ on
compact intervals of $[0,+\infty)$, and hence, $u^{**}(r)=u(r,\bar{\rho})$. ∎

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
$\\{u(\,\cdot\,,\rho)\\}_{\rho\geqslant 0}$ that can be obtained as a limit of
a subsequence with $\rho_{n}\to+\infty$, 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 $u(r,\rho)$ to problem
($P_{\rho}$), there is also a unique solution
$(u_{\scriptscriptstyle\lambda},\lambda)$ with
$u_{\scriptscriptstyle\lambda}(r)=u(\lambda^{\frac{1}{\theta}}r,\rho)$ to
problem ($P^{\prime}_{\lambda}$). For $u$ and $u_{\scriptscriptstyle\lambda}$
related through this identity, it holds that the zero $R(0,\rho)$ of $u$ is
expressed via $\lambda$ as follows:

(59) $\displaystyle R(0,\rho)=\lambda^{\frac{1}{\theta}}.$

By Proposition 3.1, $R(0,\rho)$ has an upper bound. Therefore, the set of
$\lambda$ with the property that $(u_{\scriptscriptstyle\lambda},\lambda)$ is
a solution to ($P^{\prime}_{\lambda}$) is bounded. Let $\lambda^{\\#}$ be the
exact upper bound for this set. Let us show that for all
$\lambda<\lambda^{\\#}$, there exists a regular solution to problem
($P^{\prime}_{\lambda}$). Indeed, $R(0,\rho)$ is the unique solution to the
equation $u(r,\rho)=0$ with respect to $r$ for each fixed $\rho>0$. By the
implicit function theorem, the function
$R(0,\,\cdot\,):(0,+\infty)\to{\mathbb{R}}$ is continuous. Moreover,
$\lim_{\rho\to 0}R(0,\rho)=0$ which is implied by the inequality

$\rho=u(0,\rho)-u(R(0,\rho),\rho)=\int_{0}^{R(0,\rho)}t^{-\frac{\alpha}{\beta+1}}\Big{(}\int_{0}^{t}s^{\gamma}e^{f(u(s,\rho))}ds\Big{)}^{\frac{1}{\beta+1}}dt\\\
\geqslant
e^{\frac{f(0)}{\beta+1}}(\gamma+1)^{-\frac{1}{\beta+1}}\hat{\theta}^{-1}R(0,\rho)^{\hat{\theta}}.$

As a continuous function of $\rho$, $\lambda=R(0,\rho)^{\theta}$ takes all the
values on the interval $[0,\lambda^{\\#})$. Consider the case
$\lambda=\lambda^{\\#}$. We prove that there exists a solution to (7),
singular or regular, such that $(\lambda^{\\#})^{\frac{1}{\theta}}$ is its
zero. Find a sequence $\\{\rho_{n}\\}$ such that
$\lambda_{n}=R(0,\rho_{n})^{\theta}$ converges to $\lambda^{\\#}$. Let
$u^{**}$ be a limit point of $u(\,\cdot\,,\rho_{n})$ on an interval
$[a,b]\subset(0,+\infty)$ such that
$(\lambda^{\\#})^{\frac{1}{\theta}}\in(a,b]$, i.e., there exists a subsequence
$u(\,\cdot\,,\rho_{n_{k}})\to u^{**}$, as $k\to\infty$, uniformly on $[a,b]$.
By Corollary 3.1, if $\rho_{n_{k}}$ is unbounded, then $u^{**}$ is a singular
solution. Otherwise, it is a regular solution. Without loss of generality, we
write $\rho_{n}$ for $\rho_{n_{k}}$. Let
$R^{*}(0)=(\lambda^{\\#})^{\frac{1}{\theta}}$. We have

(60) $|u^{**}(R^{*}(0))|=|u(R(0,\rho_{n}),\rho_{n})-u^{**}(R^{*}(0))|\\\
\leqslant|u(R(0,\rho_{n}),\rho_{n})-u(R^{*}(0),\rho_{n})|+|u(R^{*}(0),\rho_{n})-u^{**}(R^{*}(0))|.$

By the uniform continuity of the family $u(\,\cdot\,,\rho)_{\rho\geqslant 0}$,
the first term on the right-hand side of (60) goes to zero. The second term
goes to zero by the uniform convergence $u(\,\cdot\,,\rho_{n})\to u^{**}$ on
$[a,b]$. This implies that $u^{**}((\lambda^{\\#})^{\frac{1}{\theta}})=0$,
which completes the proof. ∎

## 4\. Study of regular solutions by means of the singular solution

Here we demonstrate how the asymptotic representation (LABEL:as1) can be used
to study regular solutions to problem ($P_{\lambda}$).

The change of variable (8) transforms problem ($P_{\lambda}$) to the
following:

$\begin{cases}-L(u)=e^{f(u)},&0\leqslant
r\leqslant\lambda^{\frac{1}{\theta}},\\\
u(\lambda^{\frac{1}{\theta}})=0.\end{cases}$

In the above problem, one can actually exclude $\lambda$ out of consideration
keeping in mind that $\lambda^{\frac{1}{\theta}}$ is the zero of the solution
(by Proposition 3.1, we know that a zero of the solution exists). In
Subsections 4.1 and 4.2, we deal with the singular solution
${u}^{*}(r)=u^{*}_{{\scriptscriptstyle\lambda}^{\\!*}}((\lambda^{*})^{-\frac{1}{\theta}}r)$
and regular solutions
$u(r,\rho)=u_{\scriptscriptstyle\lambda}(\lambda^{-\frac{1}{\theta}}r,\rho)$
to equation (7). In the case of regular solutions, equation (7) is
complemented with the initial conditions $u(0,\rho)=\rho$,
$\partial_{r}u(0,\rho)=0$.

### 4.1 Transformation of problem ($P_{\rho}$) and its limit equation

For simplicity of notation, in this section, we define the function

$\displaystyle\psi=e^{f}\\!.$

Note that if $\lim_{t\to+\infty}g^{\prime\prime}(t)=0$, then, by Remark 2.5,
$\mathcal{F}(u)$, involved in (14), is finite for all $u$. Indeed, by the
change of variable $s=g(t)$, we obtain

(61)
$\displaystyle\int_{u}^{+\infty}e^{-\frac{f(s)}{\beta+1}}ds=\int_{f(u)}^{+\infty}e^{-\frac{t}{\beta+1}}g^{\prime}(t)dt<+\infty.$

######  Remark 4.1.

Sometimes we write $\rho$ as the second parameter of $\tilde{u}$. However, we
do not mean that $\rho$ is its maximal value, unlike the notation $u(r,\rho)$.
Note that

$\displaystyle\tilde{u}(s,\rho)=\mathcal{F}^{-1}_{1}\Big{\\{}\mathcal{F}_{1}(1)\,\frac{\mathcal{F}(u(r,\rho))}{\mathcal{F}(\rho)}\Big{\\}}\leqslant\mathcal{F}^{-1}_{1}(\mathcal{F}_{1}(1))=1,\quad
s\geqslant 0,$ $\displaystyle\text{and}\quad\tilde{u}(0,\rho)=1.$

We just want to emphasize that $\tilde{u}$ actually depends on $\rho$.
Moreover, we will study the limit of $\tilde{u}$ as $\rho$ goes to infinity.

######  Remark 4.2.

Sometimes we skip the dependence on $\rho$ in both $u$ and $\tilde{u}$ to
simplify notation.

######  Lemma 4.1.

Assume (61) is fulfilled. Let $u(r)$ be a ${\rm C}^{2}$-solution to problem
($P_{\rho}$). Then, the function $\tilde{u}(s)$, defined by (14), is a
solution to the problem

(62)
$\displaystyle\,L(\tilde{u})+e^{\tilde{u}}+s^{\alpha-\gamma}|\tilde{u}^{\prime}|^{\beta+2}\Big{(}\frac{I(u(\varepsilon_{\rho}s))}{\beta+1}-1\Big{)}=0,$
$\displaystyle\,\tilde{u}(0)=1,\quad\tilde{u}^{\prime}(0)=0,$

where $\tilde{u}$ is everywhere evaluated at $s$ and
$I(u)=\mathcal{F}(u)f^{\prime}(u)\psi(u)^{\frac{1}{\beta+1}}$.

###### Proof.

First, we note that the $\tilde{u}(s)$, defined by (14), satisfies the initial
conditions in (LABEL:eq-tr). Define
$G(u)=\mathcal{F}^{-1}_{1}\varepsilon_{\rho}^{-\hat{\theta}}\mathcal{F}(u)$,
where $\hat{\theta}=\frac{\theta}{\beta+1}$. Then,
$u(r)=G^{-1}(\tilde{u}(s))$, where $r=\varepsilon_{\rho}s$. Computing $L(u)$
by using the previous expression, we obtain

$L(u(r))=\varepsilon_{\rho}^{-\theta}\,s^{-\gamma}\big{(}s^{\alpha}|\tilde{u}^{\prime}(s)|^{\beta}\tilde{u}^{\prime}(s)G^{\prime}(u(r))^{-\beta-1}\big{)}^{\prime}_{s}\\\
=\varepsilon_{\rho}^{-\theta}\,L(\tilde{u}(s))G^{\prime}(u(r))^{-\beta-1}-\varepsilon_{\rho}^{-\theta}\,(\beta+1)s^{\alpha-\gamma}|\tilde{u}^{\prime}(s)|^{\beta+2}G^{\prime}(u(r))^{-\beta-3}G^{\prime\prime}(u(r)).$

Problem ($P_{\rho}$) is then transformed to the following:

$\displaystyle
L(\tilde{u}(s))+\varepsilon_{\rho}^{\theta}\,\psi(u(r))G^{\prime}(u(r))^{\beta+1}-(\beta+1)s^{\alpha-\gamma}|\tilde{u}^{\prime}(s)|^{\beta+2}\frac{G^{\prime\prime}(u(r))}{G^{\prime}(u(r))^{2}}=0.$

By the definition of $\mathcal{F}(u)$ and $\mathcal{F}_{1}(u)$ in (14),

(63) $\displaystyle
G^{\prime}(u(r))=\varepsilon_{\rho}^{-\hat{\theta}}e^{\frac{\tilde{u}(s)}{\beta+1}}\psi(u(r))^{-\frac{1}{\beta+1}},$

which transforms the second term into $e^{\tilde{u}(s)}$. Furthermore,

(64) $\displaystyle\frac{G^{\prime\prime}(u)}{G^{\prime}(u)^{2}}=\frac{(\ln
G^{\prime}(u))^{\prime}}{G^{\prime}(u)}=\frac{1}{\beta+1}\frac{G^{\prime}(u)-f^{\prime}(u)}{G^{\prime}(u)}.$

In (64), $u$ is everywhere evaluated at $r$. Recall that
$\tilde{u}(s)=G(u(r))$. For simplicity of notation, we skip the dependence on
$s$ in $\tilde{u}$ and, with a slight abuse of notation, simply write
$\tilde{u}=G(u)$. By (63),

$\displaystyle\frac{f^{\prime}(u)}{G^{\prime}(u)}=\frac{\psi^{\prime}(u)\psi(u)^{-\frac{\beta}{\beta+1}}\mathcal{F}(u)}{\varepsilon_{\rho}^{-\hat{\theta}}\mathcal{F}(u)\,e^{\frac{\tilde{u}}{\beta+1}}}=\frac{I(u)}{\beta+1}$

since
$\varepsilon_{\rho}^{-\hat{\theta}}\mathcal{F}(u)=\mathcal{F}_{1}(\tilde{u})$
and $\mathcal{F}_{1}(\tilde{u})=(\beta+1)e^{-\frac{\tilde{u}}{\beta+1}}$. The
above computations imply (LABEL:eq-tr). ∎

######  Lemma 4.2.

Assume (B1)–(B3). Then,
$\lim_{\rho\to+\infty}u(\varepsilon_{\rho}s,\rho)=+\infty$ in ${\rm
C}_{loc}[0,+\infty)$.

###### Proof.

Fix an arbitrary interval $[0,T]$. From Proposition 3.1 it follows that for
any $B>0$, there exists a number $\underline{R}(B)>0$ and a constant
$\rho_{0}>0$ such that for all $\rho>\rho_{0}$ and $r\in(0,\underline{R}(B))$,
$u(r,\rho)>B$. Note that $\lim_{\rho\to+\infty}\varepsilon_{\rho}=0$. This
proves that
$\lim_{\rho\to+\infty}\sup_{s\in[0,T]}u(\varepsilon_{\rho}s,\rho)=+\infty$. ∎

######  Lemma 4.3.

Assume (61), (B1)–(B3), and that
$\lim_{t\to+\infty}\frac{g^{\prime\prime}(t)}{g^{\prime}(t)}=0$. Then,
$\lim_{\rho\to\infty}I(u(\varepsilon_{\rho}s))=\beta+1$ in ${\rm
C}_{loc}[0,+\infty)$.

###### Proof.

We have

$\displaystyle
I(u)=\frac{\int_{u}^{+\infty}e^{-\frac{f(s)}{\beta+1}}ds\,f^{\prime}(u)}{e^{-\frac{f(u)}{\beta+1}}}=\frac{\int_{f(u)}^{+\infty}e^{-\frac{t}{\beta+1}}g^{\prime}(t)\,dt\,}{e^{-\frac{f(u)}{\beta+1}}g^{\prime}(f(u))}.$

Let $v=f(u)$. By L’Hopital’s rule,

$\displaystyle\lim_{v\to+\infty}\frac{\int_{v}^{+\infty}e^{-\frac{t}{\beta+1}}g^{\prime}(t)\,dt}{e^{-\frac{v}{\beta+1}}g^{\prime}(v)}=$
$\displaystyle\lim_{v\to+\infty}\frac{e^{-\frac{v}{\beta+1}}g^{\prime}(v)}{(\beta+1)^{-1}e^{-\frac{v}{\beta+1}}g^{\prime}(v)-e^{-\frac{v}{\beta+1}}g^{\prime\prime}(v)}$
$\displaystyle=$
$\displaystyle\lim_{v\to+\infty}\frac{(\beta+1)}{1-(\beta+1)\frac{g^{\prime\prime}(v)}{g^{\prime}(v)}}=\beta+1.$

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 equation (11):

(65) $\displaystyle
v^{*}(t)=\mathcal{F}^{-1}(A^{-1}\,e^{-\hat{\theta}t}(1+\varepsilon(t))),$

where $A=\hat{\theta}(\hat{\alpha}-1)^{\frac{1}{\beta+1}}$ and
$\lim_{t\to+\infty}\varepsilon(t)=0$.

###### Proof.

Identity (65) is equivalent to
$\varepsilon(t)=A\,e^{\hat{\theta}t}\mathcal{F}(v^{*}(t))-1$. Let us show that
$\lim_{t\to+\infty}e^{\hat{\theta}t}\mathcal{F}(v^{*}(t))=A^{-1}$. Recall that
$v^{*}(t)=g(\theta t+\varphi(t))+\eta(t)$, where
$\varphi=\varphi_{1}+\varphi_{2}$ is defined by (LABEL:ffi),
$\eta=O(g^{\prime\prime}(\theta t))$, and
$\eta^{\prime}=O(g^{\prime\prime}(\theta t))$. By L’Hopital’s rule,

$\displaystyle\lim_{t\to+\infty}\frac{\mathcal{F}(v^{*}(t))}{e^{-\hat{\theta}t}}=\lim_{t\to+\infty}\frac{e^{-\frac{f(v^{*}(t))}{\beta+1}}(g^{\prime}(\theta
t+\varphi)(\theta+\varphi^{\prime})+\eta^{\prime})}{\hat{\theta}e^{-\hat{\theta}t}}.$

Transforming $f(v^{*}(t))$ by formula (16), we obtain that the limit on the
right-hand side equals to

$\displaystyle\lim_{t\to+\infty}\hat{\theta}^{-1}e^{-\frac{\varphi}{\beta+1}}g^{\prime}(\theta
t+\varphi)(\theta+\varphi^{\prime})=\lim_{t\to+\infty}\frac{g^{\prime}(\theta
t+\varphi)(\theta+\varphi^{\prime})}{\hat{\theta}\theta(\hat{\alpha}-1)^{\frac{1}{\beta+1}}g^{\prime}(\theta
t)}=A^{-1}.$

We used representation (LABEL:ffi) for $\varphi=\varphi_{1}+\varphi_{2}$ and
the fact that $\lim_{t\to+\infty}\frac{\eta}{g^{\prime}(\theta t+o(t))}=0$.
Finally, $\varphi^{\prime}$ goes to zero by (19). This concludes the proof. ∎

######  Lemma 4.5.

Assume (A1)–(A6). Then, the singular solution $u^{*}$ to (7) has the
representation

$\displaystyle
u^{*}(r)=\mathcal{F}^{-1}\big{(}K\,r^{\hat{\theta}}(1+\delta(r))\big{)},$

where $K=\frac{\beta+1}{\theta}(\alpha-\beta-1)^{-\frac{1}{\beta+1}}$,
$\delta(r)=\varepsilon(\ln{\kappa}-\ln{r})$,
$\kappa=(\beta+1)^{\frac{1}{\theta}}$.

###### Proof.

The proof follows immediately from Lemma 4.4 by the substitution
$t=\ln\kappa-\ln r$. The constant $K$ can be computed by the definition of $A$
as follows:
$K=A^{-1}(\beta+1)^{-\frac{1}{\beta+1}}=\frac{\beta+1}{\theta}(\alpha-\beta-1)^{-\frac{1}{\beta+1}}$.
∎

Lemma 4.5 allows to obtain a representation for the singular solution
$\tilde{u}^{*}$ to (LABEL:eq-tr).

######  Lemma 4.6.

Assume (A1)–(A6). Then, the singular to equation (LABEL:eq-tr)
$\tilde{u}^{*}(s)=\mathcal{F}^{-1}_{1}\\!\varepsilon_{\rho}^{-\hat{\theta}}\mathcal{F}(u^{*}(\varepsilon_{\rho}s))$
can be represented as follows:

(66)
$\displaystyle\tilde{u}^{*}(s)=\ln\\{\theta^{\beta+1}(\alpha-\beta-1)\\}-\theta\ln
s-(\beta+1)\ln\\{1+\delta(\varepsilon_{\rho}s)\\}.$

###### Proof.

Formula (66) is the result of application of transformation (14) to
$u^{*}(\varepsilon_{\rho}s)$. Indeed,
$\tilde{u}^{*}(s)=\mathcal{F}^{-1}_{1}(K\,s^{\hat{\theta}}(1+\delta(\varepsilon_{\rho}s)))$,
where $\mathcal{F}^{-1}_{1}(t)=-(\beta+1)\ln\frac{t}{\beta+1}$. This
immediately implies (66). ∎

######  Proposition 4.1.

Assume (A1)–(A6). Then, for the singular and regular solutions to (LABEL:eq-
tr), it holds that

(67)
$\displaystyle\lim_{\rho\to+\infty}\tilde{u}(s)=w(s)\quad\text{and}\quad\lim_{\rho\to+\infty}\tilde{u}^{*}(s)=w^{*}(s)$

in ${\rm C}_{loc}[0,+\infty)$ and ${\rm C}_{loc}(0,+\infty)$, respectively.
Here, $w(s)$ is the regular solution to the problem

(68) $\displaystyle L(w)+e^{w}=0,\quad w(0)=1,\quad w^{\prime}(0)=0,$

and $w^{*}(s)$ is the singular solution to the equation in (68) given
explicitly as follows:

(69) $\displaystyle
w^{*}(s)=\ln\\{\theta^{\beta+1}(\alpha-\beta-1)\\}-\theta\ln
s=\ln\\{\theta^{\beta+1}(\hat{\alpha}-1)\\}+\theta\ln\frac{\kappa}{s}.$

######  Remark 4.3.

The last expression in (69) agrees with (45) with $t=\ln\frac{\kappa}{s}$.

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
(LABEL:eq-tr) as $\rho\to+\infty$. More specifically, the last term on the
right-hand side of (LABEL:eq-tr) goes to zero in ${\rm C}_{loc}[0,+\infty)$ as
$\rho\to+\infty$.

###### Proof.

Let $\tilde{\omega}(s)=|\tilde{u}^{\prime}(s)|^{\beta+1}$. Recall that by
Lemma 3.1, $u^{\prime}(r)<0$. Therefore, by (14),
$|\tilde{u}^{\prime}(s)|=-\tilde{u}^{\prime}(s)$. Fix an arbitrary interval
$[0,T]$ and introduce the notation,

(70)
$\displaystyle\mathcal{I}_{\rho}(s)=\frac{I(u(\varepsilon_{\rho}s))}{\beta+1}-1.$

Then, $\tilde{\omega}(s)$ satisfies the following first-order differential
inequality:

$\displaystyle\tilde{\omega}^{\prime}(s)=s^{\gamma-\alpha}e^{\tilde{u}}-\alpha
s^{-1}\tilde{\omega}+\tilde{\omega}^{\frac{\beta+2}{\beta+1}}\mathcal{I}_{\rho}(s)\leqslant
C+\varepsilon\tilde{\omega}+\varepsilon\tilde{\omega}^{2}.$

Here, $C=T^{\gamma-\alpha}e$ is the bound for
$s^{\gamma-\alpha}e^{\tilde{u}}$. Furthermore, $\varepsilon$ is an arbitrary
sufficiently small number. Note that $|\mathcal{I}_{\rho}(s)|<\varepsilon$ for
sufficiently large $\rho$. Finally, it holds that
$\tilde{\omega}^{\frac{\beta+2}{\beta+1}}\leqslant\tilde{\omega}+\tilde{\omega}^{2}$.

We aim to estimate $\tilde{\omega}(s)$ via the solution $\omega(s)$ to the
problem

(71)
$\displaystyle\omega^{\prime}=C+\varepsilon\omega+\varepsilon\omega^{2},\qquad\omega(0)=0.$

It is known that $\omega=-\frac{1}{\varepsilon}\frac{y^{\prime}}{y}$ is a
solution to the ODE in (71) if $y$ is the solution to

$\displaystyle y^{\prime\prime}-\varepsilon y^{\prime}+C\varepsilon y=0.$

For sufficiently small $\varepsilon$, the both roots of the characteristic
equation are complex, and therefore, the solution takes the form

$\displaystyle
y(s)=\bar{K}e^{\frac{\varepsilon}{2}s}(\cos(Ds)+K\sin(Ds)),\quad\text{where}\quad
D=\frac{\sqrt{4C\varepsilon-\varepsilon^{2}}}{2}.$

It is straightforward to obtain now the solution $\omega$ to problem (71),
computing $K$ from the condition $\omega(0)=0$:

$\displaystyle\omega(s)=\frac{\varepsilon^{2}+4D^{2}}{4D\varepsilon}\frac{1}{\operatorname{\text{ctg}}(Ds)-\frac{\varepsilon}{2D}}.$

Note that the solution $\omega(s)$ is well-defined on the interval
$\Big{[}0,\frac{2\operatorname{\text{arcctg}}\frac{\varepsilon}{\sqrt{4C\varepsilon-\varepsilon^{2}}}}{\sqrt{4C\varepsilon-\varepsilon^{2}}}\Big{)}$.
When $\varepsilon$ is sufficiently small, the numerator of the right endpoint
is close to $\pi$; however, the denominator is small. Thus, we can choose
$\varepsilon$ sufficiently small (which makes $\rho$ sufficiently large) such
that $\omega(s)$ is well-defined on $[0,T]$. Therefore, on $[0,T]$, it holds
that

(72) $\displaystyle|\tilde{u}^{\prime}(s,\rho)|^{\beta+1}\leqslant\omega(s)$

for sufficiently large $\rho$. This, along with Lemma 4.3, proves the lemma. ∎

###### Proof of Proposition 4.1.

Everywhere in the proof, $K_{i}$, $i=1,2,3$, are positive constants. Fix an
arbitrary interval $[0,T]$. We have

$w(r)-\tilde{u}(r)=\frac{1}{\beta+1}\int_{0}^{r}t^{-\frac{\alpha}{\beta+1}}\Big{(}\int_{0}^{1}\Lambda(t,\rho,\lambda)^{-\frac{\beta}{\beta+1}}d\lambda\Big{)}\\\
\int_{0}^{t}(s^{\gamma}(e^{\tilde{u}}-e^{w})+s^{\alpha}|\tilde{u}^{\prime}(s)|^{\beta+2}\mathcal{I}_{\rho}(s))ds\,dt,$

where $\Lambda(t,\rho,\lambda)$ is defined by the first identity below and can
be estimated by the expression on the right-hand side:

$\Lambda(t,\rho,\lambda)=\lambda\int_{0}^{t}\big{(}s^{\gamma}e^{\tilde{u}}+s^{\alpha}|\tilde{u}^{\prime}(s)|^{\beta+2}\mathcal{I}_{\rho}\big{)}ds+(1-\lambda)\int_{0}^{t}s^{\gamma}e^{w}ds\geqslant(1-\lambda)e^{w(T)}\int_{0}^{t}s^{\gamma}ds.$

Indeed, it holds that

$\displaystyle\int_{0}^{t}(s^{\gamma}e^{\tilde{u}}+s^{\alpha}|\tilde{u}^{\prime}(s)|^{\beta+2}\mathcal{I}_{\rho}(s))ds=t^{\alpha}|\tilde{u}^{\prime}(t)|^{\beta+1}\geqslant
0.$

By Gronwall’s inequality,

(73) $\displaystyle t^{\alpha}|\tilde{u}^{\prime}(t)|^{\beta+1}\leqslant
e^{\int_{0}^{t}|\tilde{u}^{\prime}(s)\mathcal{I}_{\rho}(s)|ds}\int_{0}^{t}s^{\gamma}e^{\tilde{u}}ds\leqslant
K_{1}t^{\gamma+1},$

where the last estimate holds by (72) for sufficiently large $\rho$.
Furthermore, estimate (73) implies

$\displaystyle\sup_{[0,r]}|\tilde{u}-w|\leqslant
K_{2}\int_{0}^{r}t^{\frac{\gamma+1-\alpha}{\beta+1}}\big{\\{}\sup_{[0,t]}|\tilde{u}-w|+\sup_{[0,r]}|\mathcal{I}_{\rho}|\big{\\}}\,dt,$

where, in particular, it is taken into account that
$\int_{0}^{1}(1-\lambda)^{-\frac{\beta}{\beta+1}}d\lambda=\beta+1$. By
Gronwall’s inequality, $\sup_{[0,r]}|\tilde{u}-w|\leqslant
K_{3}\sup_{[0,r]}|\mathcal{I}_{\rho}|\,r^{\hat{\theta}}e^{K_{3}r^{\hat{\theta}}}$.
This implies that $\sup_{[0,T]}|\tilde{u}-w|\to 0$ as $\rho\to+\infty$.

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 $\rho$ goes to $+\infty$, we obtain exactly
(69). The convergence holds in ${\rm C}_{loc}(0,+\infty)$ since the
convergence $\lim_{\rho\to+\infty}\delta(\varepsilon_{\rho}s)=0$ holds in this
space. ∎

### 4.2 Number of intersection points. Case
$\alpha-\beta-1<\frac{4\theta}{\beta+1}$ ($d<2k+8$).

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 $\rho\to+\infty$, of the
number of intersection points between $u(\,\cdot\,,\rho)$ and $u^{*}$. Recall
that the number of zeros of $w-w^{*}$ on an interval $(R,S)$ is denoted by
$\mathscr{L}_{(R,S)}(w-w^{*})$. Although Theorem 4.1 has its own interest, we
prove it in this subsection as an auxiliary result.

######  Theorem 4.1.

Assume (A6). Suppose $\alpha-\beta-1<4\hat{\theta}$. Then, for any $R>0$, $w$
and $w^{*}$ have an infinite number of intersection points on $(R,+\infty)$,
i.e., $\mathscr{L}_{(R,\infty)}(w-w^{*})=+\infty$. Furthermore,
$\mathscr{L}_{(0,R]}(w-w^{*})$ is finite.

###### Proof.

Define $v(t)=w(s)$ and $\omega(t)=v^{\prime}(t)^{\beta+1}$, where
$t=\ln\\{\frac{\kappa}{s}\\}$ with $\kappa=(\beta+1)^{\frac{1}{\theta}}$. By
Lemma 3.1, $v^{\prime}(t)>0$, and hence, by (38),

(74)
$\displaystyle\big{(}e^{(\beta+1-\alpha)t}\omega(t)\big{)}^{\prime}=-(\beta+1)e^{-(\gamma+1)t}e^{v}.$

Now let $w^{*}(s)$ be given by (69) and $v^{*}(t)=w^{*}(s)$, where
$t=\ln\\{\frac{\kappa}{s}\\}$. Then, $v^{*}(t)=\theta
t+\ln\\{\theta^{\beta+1}(\hat{\alpha}-1)\\}$. Define $x(t)=v(t)-v^{*}(t)$ and
$y(t)=\omega(t)-\omega^{*}(t)$, where
$\omega^{*}(t)={v^{*}}^{\prime}(t)^{\beta+1}=\theta^{\beta+1}$. Equation (74)
implies that $x$ and $y$ satisfy the following system of ODEs

(75)
$\begin{cases}y^{\prime}=(\alpha-\beta-1)(y-\theta^{\beta+1}(e^{x}-1)),\\\
x^{\prime}=(y+\theta^{\beta+1})^{\frac{1}{\beta+1}}-\theta,\end{cases}$

where the last equation holds due to the fact that
$v^{\prime}=\omega^{\frac{1}{\beta+1}}$. 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

(76) $\begin{cases}y^{\prime}=(\alpha-\beta-1)(y-\theta^{\beta+1}x),\\\
x^{\prime}=\frac{y}{(\beta+1)\theta^{\beta}}.\end{cases}$

Since $0<\alpha-\beta-1<4\hat{\theta}$, the roots of the characteristic
equation for the linearized system are complex numbers with the positive real
part $\frac{\alpha-\beta-1}{2}$, and hence, $(0,0)$ is an unstable focus. This
means that as $t$ goes to $-\infty$, $x(t)$ crosses zero infinitely many times
at any neighborhood of $-\infty$. This implies that for any $R>0$,
$\mathscr{L}_{(-\infty,-R)}(v-v^{*})=\mathscr{L}_{(\kappa
e^{R},+\infty)}(w-w^{*})=+\infty$.

Suppose $\mathscr{L}_{(0,R]}(w-w^{*})=+\infty$ for some $R>0$. Then, there
exists an accumulation point, say $s_{0}$, of zeros of $w-w^{*}$ belonging to
$[0,R]$. Note that $(w-w^{*})(s_{0})=0$ which implies that $s_{0}\in(0,R]$.
Furthermore, $(w-w^{*})^{\prime}(s_{0})=0$. This implies that
$v(t_{0})=v^{*}(t_{0})$ and $\omega(t_{0})=\omega^{*}(t_{0})$ for
$t_{0}=\ln\kappa-\ln s_{0}$, 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 $(R,+\infty)$.

###### Proof.

The proof follows immediately from the fact that $(0,0)$ is a focus for (75).
∎

######  Theorem 4.2.

Assume (A1)–(A6) and let $\alpha-\beta-1<4\hat{\theta}$. Let
$u(\,\cdot\,,\rho)$ be the regular solution to problem ($P_{\rho}$) and
$u^{*}$ be the singular solution to (7) defined by (8) via
$(u^{*}_{{\scriptscriptstyle\lambda}^{\\!*}},\lambda^{*})$. Then, for any
$\delta>0$, it holds that
$\lim_{\rho\to+\infty}\mathscr{L}_{(0,\delta)}(u(\,\cdot\,,\rho)-u^{*}(\,\cdot\,))=+\infty$.

###### Proof.

Recall that $\tilde{u}-\tilde{u}^{*}\to w-w^{*}$ as $\rho\to+\infty$ uniformly
on any interval $[R,S]\subset(0,+\infty)$. Note that for a sufficiently large
$\rho$, $\mathscr{L}_{[R,S]}(\tilde{u}-\tilde{u}^{*})$ is not smaller than
$\mathscr{L}_{[R,S]}(w-w^{*})-2$, where the latter number is finite by Theorem
4.1. Recall that
$u(\varepsilon_{\rho}s,\rho)=\mathcal{F}^{-1}_{1}\varepsilon_{\rho}^{\hat{\theta}}\mathcal{F}(\tilde{u}(s,\rho))$
with $\lim_{\rho\to+\infty}\varepsilon_{\rho}=0$. Fix $N\in{\mathbb{N}}$.
There exist $R,S>0$ and a number $\rho_{R,S}>0$ such that
$\mathscr{L}_{[R,S]}(\tilde{u}-\tilde{u}^{*})\geqslant N$ for
$\rho>\rho_{R,S}$. Note that for each zero point $s\in[R,S]$ of
$\tilde{u}-\tilde{u}^{*}$, there exists exactly one zero point
$r\in[\varepsilon_{\rho}R,\varepsilon_{\rho}S]$ of
$u(\,\cdot\,,\rho)-u^{*}(\,\cdot\,)$. Choose $\rho_{\delta}>\rho_{R,S}$ such
that for all $\rho>\rho_{\delta}$,
$[\varepsilon_{\rho}R,\varepsilon_{\rho}S]\subset(0,\delta)$. Therefore, for
any $N\in{\mathbb{N}}$, one can find $\rho_{\delta}$ such that for all
$\rho\geqslant\rho_{\delta}$,
$\mathscr{L}_{(0,\delta)}(u(\,\cdot\,,\rho)-u^{*}(\,\cdot\,))\geqslant N$. The
theorem is proved. ∎

### 4.3 Bifurcation diagram. Case $\alpha-\beta-1<\frac{4\theta}{\beta+1}$
($d<2k+8$).

######  Theorem 4.3.

Assume (A1)–(A6) and $\alpha-\beta-1<4\hat{\theta}$. Let
$(u_{\scriptscriptstyle{\scriptscriptstyle\lambda}(\rho)},\lambda(\rho))$ be
the family of regular solutions to problem ($P_{\lambda}$) and let
$(u^{*}_{{\scriptscriptstyle\lambda}^{\\!*}},\lambda^{*})$ be the singular
solution constructed in Theorem 2.1. Then, as $\rho\to\infty$, $\lambda(\rho)$
oscillates around $\lambda^{*}$. In particular, there are infinitely many
regular solutions to problem ($P_{\lambda}$) for $\lambda(\rho)=\lambda^{*}$.

###### Proof.

Let $u(\,\cdot\,,\rho)$ and $u^{*}$ be the regular and singular solutions to
(7), as in Theorem 4.2. Define $\hat{v}(r,\rho)=u(r,\rho)-u^{*}$ and show that

(77) $\displaystyle\partial_{r}\hat{v}(r_{0},\rho)\neq 0\;\text{ whenever
}\;\hat{v}(r_{0},\rho)=0.$

Suppose $\hat{v}(r_{0},\rho)=\partial_{r}\hat{v}(r_{0},\rho)=0$. For
$r\in(0,r_{0})$, we have

$u(r)-u^{*}(r)=\int_{r}^{r_{0}}s^{-\frac{\alpha}{\beta+1}}\Big{[}\Big{(}r_{0}^{\alpha}|u^{\prime}(r_{0})|^{\beta+1}-\int_{s}^{r_{0}}\xi^{\gamma}e^{f(u(\xi))}d\xi\Big{)}^{\frac{1}{\beta+1}}\\\
-\Big{(}r_{0}^{\alpha}|u^{\prime}(r_{0})|^{\beta+1}-\int_{s}^{r_{0}}\xi^{\gamma}e^{f(u^{*}(\xi))}d\xi\Big{)}^{\frac{1}{\beta+1}}\Big{]}ds\\\
=\frac{1}{\beta+1}\int_{r}^{r_{0}}s^{-\frac{\alpha}{\beta+1}}\Big{(}\int_{0}^{1}\Lambda(s,\lambda)^{-\frac{\beta}{\beta+1}}d\lambda\Big{)}\int_{s}^{r_{0}}\xi^{\gamma}(e^{f(u^{*}(\xi))}-e^{f(u(\xi))})d\xi\,ds,$

where

$\displaystyle\Lambda(s,\lambda)=$
$\displaystyle\,r_{0}^{\alpha}|u^{\prime}(r_{0})|^{\beta+1}-\int_{s}^{r_{0}}\xi^{\gamma}(\lambda
e^{f(u(\xi))}+(1-\lambda)e^{f(u^{*}(\xi))})ds$ $\displaystyle=$
$\displaystyle\,s^{\alpha}(\lambda|u^{\prime}(s)|^{\beta+1}+(1-\lambda)|{u^{*}}^{\prime}(s)|^{\beta+1}).$

Take $\epsilon\in(0,r_{0})$. For $s\in[\epsilon,r_{0}]$, we have

$\displaystyle\inf_{s\in[\epsilon,r_{0}]}\big{(}s^{\alpha}\min\\{|u^{\prime}(s)|^{\beta+1},|{u^{*}}^{\prime}(s)|^{\beta+1}\\}\big{)}\leqslant\Lambda(s,\lambda)\leqslant
r_{0}^{\alpha}|u^{\prime}(r_{0})|^{\beta+1}.$

Therefore, there exists a constant $K>0$ such that for $r\in[\epsilon,r_{0}]$,

$\displaystyle\sup_{[r,r_{0}]}|u-u^{*}|\leqslant
K\int_{r}^{r_{0}}\sup_{[s,r_{0}]}|u(s)-u^{*}(s)|ds.$

By Gronwall’s inequality, $\sup_{[\epsilon,r_{0}]}|u-u^{*}|=0$. Since
$\epsilon\in(0,r_{0})$ is arbitrary, we obtain that $u=u^{*}$ on $(0,r_{0})$
which is a contradiction. Thus, (77) is true.

Recall that $\lambda(\rho)^{\frac{1}{\theta}}$ and
${\lambda^{*}}^{\frac{1}{\theta}}$ are zeros of $u(\,\cdot\,,\rho)$ and
$u^{*}$, respectively. Since $u$ and $u^{*}$ are decreasing in $r$, they do
not have other zeros. Define
$r(\rho)=\min\\{\lambda(\rho)^{\frac{1}{\theta}},{\lambda^{*}}^{\frac{1}{\theta}}\\}$
and $z(\rho)=\mathscr{L}_{(0,r(\rho))}\hat{v}(\,\cdot\,,\rho)$. Note that if
$\rho$ is sufficiently large, then, by Proposition 3.1 , we can find two
positive constants $\tau_{1}$ and $\tau_{2}$ such that
$\tau_{1}<r(\rho)<\tau_{2}$. Furthermore, we note that $z(\rho)$ is positive
and finite. Indeed, $z(\rho)>0$ since, by Theorem 4.2,
$\lim_{\rho\to+\infty}\mathscr{L}_{(0,\tau_{1})}(\hat{v}(\,\cdot\,,\rho))=+\infty$.
Assuming that $z(\rho)$ is infinite, we can find a sequence $\\{r_{n}\\}$ of
zeros of $\hat{v}$ converging to a point $\tilde{r}\in[0,r(\rho)]$. One has
$\hat{v}(\tilde{r},\rho)=\partial_{r}\hat{v}(\tilde{r},\rho)=0$ which
contradicts to (77). Thus, we conclude that for each fixed $\rho$, the
equation $\hat{v}(r,\rho)=0$ has a finite number of solutions on
$(0,r(\rho))$. By the same argument, for each fixed $\rho>0$, the equation
$\hat{v}(r,\rho)=0$ has a finite number of solutions on any finite interval
$[0,R]$, and therefore, there is a countable number of zeros of
$\hat{v}(r,\rho)$ on $(0,+\infty)$.

Fix $\rho_{0}>0$ and take an arbitrary zero $r_{0}$ of
$\hat{v}(\,\cdot\,,\rho_{0})$. By (77),
$\partial_{r}\hat{v}(r_{0},\rho_{0})\neq 0$. By the implicit function theorem,
there exists an interval $(\rho_{1},\rho_{2})$, containing $\rho_{0}$, and a
unique continuous function $\mathfrak{r}(\rho)$ solving the equation
$\hat{v}(r,\rho)=0$ on $(\rho_{1},\rho_{2})$ and such that
$\mathfrak{r}(\rho_{0})=r_{0}$. Without loss of generality,
$(\rho_{1},\rho_{2})$ can be regarded as the maximal interval to which
$\mathfrak{r}(\rho)$ can be extended. This extension will be unique by (77)
and the implicit function theorem. The curve $\mathfrak{r}(\rho)$ may cross
$r(\rho)$ at some point from $(\rho_{1},\rho_{2})$ or may not. When it happens
a zero enters or exits the interval $(0,r(\rho))$. Remark that $\rho_{1}$ can
be either zero or positive; $\rho_{2}$ can be either finite or $+\infty$. Let
us show that if $\rho_{2}$ is finite, then
$\lim_{\rho\to\rho_{2}-}\mathfrak{r}(\rho)=+\infty$. Suppose $\rho_{2}$ is
finite, but $\lim_{\rho\to\rho_{2}-}\mathfrak{r}(\rho)\neq+\infty$. Then, we
can find a sequence $\rho^{n}\to\rho_{2}$ as $n\to+\infty$ such that
$\mathfrak{r}(\rho^{n})$ converges to a number $r_{2}\in(0,+\infty)$. By
continuity, $\hat{v}(r_{2},\rho_{2})=0$. If $r_{2}$ is the limit of
$\mathfrak{r}(\rho)$ as $\rho\to\rho_{2}$, then, by the implicit function
theorem, we can find an extension of $\mathfrak{r}(\rho)$ beyond $\rho_{2}$
which is a contradiction. Therefore, the limit at $\rho_{2}$ does not exist.
Then, we can find a neighborhood
$U=U(r_{2},\rho_{2})=(r_{2}-\delta,r_{2}+\delta)\times(\rho_{2}-\delta,\rho_{2}+\delta)$,
where 1) the equation $\hat{v}(r,\rho)=0$ is uniquely solvable for each fixed
$\rho\in(\rho_{2}-\delta,\rho_{2}+\delta)$ and, furthermore, determines $r$ as
a continuous function of $\rho$ living inside $U$; and 2) $\mathfrak{r}(\rho)$
exits and enters $U$ infinitely many times. This contradicts to the implicit
function theorem. Hence, if $\rho_{2}$ is finite, we have that
$\lim_{\rho\to\rho_{2}-}\mathfrak{r}(\rho)=+\infty$. By the similar argument,
one can show that if $\rho_{1}>0$, then
$\lim_{\rho\to\rho_{1}+}\mathfrak{r}(\rho)=+\infty$. Also remark that by the
implicit function theorem and (77), two functions $\mathfrak{r}_{1}(\rho)$ and
$\mathfrak{r}_{2}(\rho)$ representing two different solutions of
$\hat{v}(r,\rho)=0$ cannot exit or enter the interval $(0,r(\rho)]$ through
its right end at the same point $\rho$. One can extend this observation
stating that two functions $\mathfrak{r}_{1}(\rho)$ and
$\mathfrak{r}_{2}(\rho)$ representing two different solutions of
$\hat{v}(r,\rho)=0$ never intersect each other, which means that two zeros
cannot merge. The above arguments allow us to conclude that $z(\rho)$ is
positive, finite, and changes only by $1$ when a zero of
$\hat{v}(\,\cdot\,,\rho)$ enters or exists the interval $(0,r(\rho))$ through
its right end.