Singular solutions for semilinear elliptic equations with general supercritical growth

Abstract

A positive radial singular solution for $\Delta u+f(u)=0$ with a general supercritical growth is constructed. An exact asymptotic expansion as well as its uniqueness in the space of radial functions are also established. These results can be applied to the bifurcation problem $\Delta u+\lambda f(u)=0$ on a ball. Our method can treat a wide class of nonlinearities in a unified way, e.g., $u^p\log u$, $\exp (u^p)$ and $\exp (\cdots \exp (u)\cdots )$ as well as $u^p$ and $e^u$. Main technical tools are intrinsic transformations for semilinear elliptic equations and ODE techniques.

1 Introduction and main results

We are concerned with singular radial solutions of the semilinear elliptic equation

$$\begin{aligned} \Delta u+f(u)=0 \quad \text{ in } \ \Omega \setminus \{0\}, \end{aligned}$$

where $\Omega = \{x\in \mathbf{R}^N: |x|<R\}$ is a finite ball with $N \ge 3$ and $f \in C^2[0, \infty )$. In this study we consider solutions to the ordinary differential equation

$$\begin{aligned} u''+ \frac{N-1}{r}u'+ f(u)=0 \quad \text{ for } \ r>0. \end{aligned}$$

(1.1)

By a singular solution $u^*(r)$ of (1.1) we mean that $u^*(r)$ is a classical solution of (1.1) for $0 < r \le r_0$ with some $r _0 > 0$ and it satisfies $u^*(r) \rightarrow \infty$ as $r \rightarrow 0$.

It is well known that the singular solution plays an important role in the study of the solution structure of (1.1) in the supercritical case (see, e.g., [6, 11, 20, 21, 23]), and various properties of the singular solutions have been studied extensively. Let us recall some known results for the existence and uniqueness of the singular solution of (1.1). When $f(u)=u^p$ with $p > N/(N-2)$, (1.1) has the exact singular solution

$$\begin{aligned} u^*(r)= Ar^{-2/(p-1)}, \quad \text{ where } \ A=\left\{ \frac{2}{p-1}\left( N-2-\frac{2}{p-1}\right) \right\} ^{1/(p-1)}. \end{aligned}$$

(1.2)

It was shown by Serrin–Zou [25, Proposition 3.1] that, if $p>p_S = (N+2)/(N-2)$, the singular solution of (1.1) is unique. In the case $f(u) = e^u$, it was shown by Mignot-Puel [19] that

$$\begin{aligned} u^*(r) = -2\log r + \log 2(N-2) \end{aligned}$$

is the unique singular solution of (1.1) when $N \ge 3$. Later, the singular radial solutions have been studied for various nonlinearities. For $f(u)=u^p+g(u)$ with lower order term g, we refer to [2, 4, 11, 12, 18, 20, 23, 24]. For $f(u)=e^u+g(u)$, see [21, 24]. In [24] the authors of this paper showed the existence and uniqueness of the singular solution if f(u) satisfies either $f(u) = u^p + o(u^p)$ or $f(u) = e^u + o(e^u)$ as $u \rightarrow \infty$, where $p > p_S$. (For precise statements, see [24, Theorem 1.1].)

Recently, the singular radial solutions have been constructed by [14] for $f(u)=\exp (u^p)$ with $p>0$, and by [9] for the cases $f(u) = \exp (e^u)$ and, more generally,

$$\begin{aligned} f(u)=\underbrace{\exp (\cdots \exp (u)\cdots )}_{n+1 \ \mathrm{times}}. \end{aligned}$$

(1.3)

Furthermore, the existence of the singular solution was obtained by [16, 22] for a certain general class of supercritical nonlinearities. On the other hand, the uniqueness of the singular radial solution seems hard to obtain, and it was left open in [9, 14, 16, 22].

In this paper, we will investigate qualitative properties of the singular solution to (1.1) with the following general hypotheses (f1) and (f2) on f.

• (f1) $f \in C^2[0, \infty )$, $f(u) > 0$ and $F(u) < \infty$ for $u \ge u_0$ with some $u_0 \ge 0$, where

  $$\begin{aligned} F(u) = \int ^{\infty }_u \frac{ds}{f(s)}. \end{aligned}$$

  (1.4)

• (f2) There exists a finite limit

  $$\begin{aligned} q = \lim _{u \rightarrow \infty }\frac{f'(u)^2}{f(u)f''(u)}. \end{aligned}$$

  (1.5)

If (f1) and (f2) are satisfied, then $q \ge 1$ in (1.5) and the exponent q is also given by

$$\begin{aligned} q = \lim _{u \rightarrow \infty }F(u)f'(u). \end{aligned}$$

(1.6)

(See Lemma 2.1 below.) Define the growth rate of f by $p = \lim _{u\rightarrow \infty }uf'(u)/f(u)$. Then, applying L’Hospital’s rule, we obtain

$$\begin{aligned} \frac{1}{p} = \lim _{u\rightarrow \infty }\frac{f(u)/f'(u)}{u} = \lim _{u \rightarrow \infty }\left( 1-\frac{f(u)f''(u)}{f'(u)^2}\right) = 1 - \frac{1}{q}, \end{aligned}$$

and hence $1/p + 1/q = 1$. The exponent q, defined by (1.5), was first introduced by Dupaigne and Farina [7] in the study of the stable solutions in $\mathbf{R}^N$. Fujishima and Ioku [8] investigated the solvability of semilinear parabolic equations with the nonlinear term f satisfying (f1) and (1.6). It is easy to see that various examples of f satisfy (f1) and (f2) with some exponent $q \ge 1$. One can easily check that $q = p/(p-1) > 1$ if $f(u) = u^p$ with $p > 1$, and that $q = 1$ if $f(u) = e^{u}$. For example, if $f(u) = u^p(\log (u+e))^{\gamma }$ with $p > 1$ and $\gamma \in \mathbf{R}$, then $q = p/(p-1)$. Hence, the principal term of f is not necessarily $u^p$ if $q = p/(p-1) > 1$. The case $q=1$ corresponds to a super-power case, and various examples of f were provided in [22, Sect. 2]. In particular, the case $q = 1$ includes the super-exponential nonlinearities, e.g., $f(u)=\exp (u^p)$ with $p > 0$ and (1.3) with $n \ge 1$. (See also Lemma 5.1 below).

Let $q_S$ denote the Hölder conjugate of the critical Sobolev exponent $p_S = (N+2)/(N-2)$, i.e.,

$$\begin{aligned} q_S = \frac{N+2}{4}. \end{aligned}$$

Then the supercritical case $p > p_S$ corresponds to the case $q < q_S$. In this paper, we study the properties of singular solutions to (1.1) with a nonlinearity f satisfying (f1) and (f2) with $q < q_S$. For $\alpha > 0$, we denote by $u(r, \alpha )$ a regular solution of (1.1) satisfying $u(0) = \alpha$ and $u'(0) = 0$. We also show the convergence property of $u(r, \alpha )$ to the singular solution.

Theorem 1.1

Suppose that $N \ge 3$ and $\mathrm{(f1)}$ and $\mathrm{(f2)}$ with $q < q_S$ hold. Then, there exists a unique singular solution $u^*(r)$ of (1.1) for $0 < r \le r_0$ with some $r_0 > 0$, and the regular solution $u(r,\alpha )$ satisfies

$$\begin{aligned} u(r,\alpha ) \rightarrow u^*(r) \quad \text{ in } \ C^2_{\mathrm{loc}}(0,r_0] \quad \text{ as } \ \alpha \rightarrow \infty . \end{aligned}$$

(1.7)

Furthermore, the singular solution $u^*$ satisfies

$$\begin{aligned} u^*(r) = F^{-1}\left[ \frac{r^2}{2N-4q}(1+o(1))\right] \quad \text{ as } \ r \rightarrow 0. \end{aligned}$$

(1.8)

Remark 1.1

• (i) In [22, Theorem 1.1] the singular solution of (1.1) was constructed by a contraction mapping theorem. On the other hand, the uniqueness of the singular solution and the convergence property of the regular solution (1.7) were left open in [22]. In the proof of Theorem 1.1 we construct the singular solution in a completely different way, which will enable us to obtain the convergence property simultaneously.

• (ii) When $f(u)=u^p$, it was shown in [3, 5] that (1.1) has a continuum of singular solutions if $p \le p_S$. When $f(u) = e^u$, (1.1) has no singular solution if $N = 1$, and (1.1) has a continuum of singular solutions if $N = 2$. (See [26].) Therefore, we cannot expect the uniqueness of the singular solution in the critical and subcritical cases $q \ge q_S$ with $N \ge 3$ and in the super-power case $q = 1$ with $N = 1, 2$.

• (iii) By Theorem 1.1, we can easily calculate an exact asymptotic expansion of the singular solution near the origin in the typical cases. We use here the notation $f(u) \sim g(u)$ means that $f(u)/g(u) \rightarrow 1$ as $u \rightarrow \infty$. In the case $f(u) \sim u^p$ and $f'(u) \sim pu^{p-1}$, we have $F(u) \sim u^{1-p}/(p-1)$ by L’Hospital’s rule. Then $q = p/(p-1)$ and $F^{-1}(u) \sim ((p-1)u)^{-1/(p-1)}$, and hence (1.8) can be written as

  $$\begin{aligned} u^*(r) = Ar^{-2/(p-1)}(1+o(1)) \quad \text{ as } \ r \rightarrow 0, \end{aligned}$$

  where A is the constant defined in (1.2). Thus a principal term of the asymptotic expansion of the singular solution $u^*$ at $r = 0$ is given by (1.2). In the case $f(u) \sim e^u$ and $f'(u) \sim e^u$, we obtain

  $$\begin{aligned} u^*(r) = -2\log r + \log 2(N-2) + o(1) \quad \text{ as } \ r \rightarrow 0 \end{aligned}$$

  by a similar argument.

Let us consider the asymptotic expansion of the singular solution near the origin in the super-exponential case $f(u) = \exp (g(u))$. Define $g_n(u)$ with $n \ge 1$ by

$$\begin{aligned} g_1(u) = \exp u \quad \text{ and } \quad g_{n}(u) = \exp (g_{n-1}(u)) \quad \text{ for } \ n \ge 2. \end{aligned}$$

(1.9)

Then f(u) in (1.3) is represented by $\exp (g_n(u))$. In the cases $f(u) = \exp (g(u))$, where $g(u) = u^p$ with $p > 0$ and $g(u) = g_n(u)$ with $n \ge 1$, it was shown by [22] that (f1) and (f2) hold with $q = 1$. (See also Lemma 5.1 below.) Then (1.1) has a unique singular solution $u^*$, which satisfies (1.8) by Theorem 1.1. In particular, we obtain the following asymptotic expansion near the origin.

Corollary 1.1

Let $N \ge 3$, and let $f(u) = \exp (g(u))$ in (1.1).

• (i) In the case $g(u) = u^p$ with $p > 0$, the singular solution $u^*$ of (1.1) satisfies

  $$\begin{aligned} u^*(r) = \left( -2\log r -\frac{p-1}{p}\log (-2\log r) + \log \frac{2N-4}{p} + o(1)\right) ^{1/p}\quad \text{ as }\ r\to 0. \end{aligned}$$

  (1.10)

• (ii) In the case $g(u) = g_n(u)$, defined by (1.9), with $n \ge 1$, the singular solution $u^*$ of (1.1) satisfies

  $$\begin{aligned} u^*(r) = \log ^{n}\Bigl (-2\log r + \log (2N-4) -\sum _{k = 1}^{n}\log ^{k}(-2\log r)\Bigr ) + o(1)\quad\text{ as }\ r\to 0, \end{aligned}$$

  (1.11)

  where $\log ^1u= \log u$ and $\log ^n u = \log (\log ^{n-1} u)$ for $n \ge 2$. In particular, if $g(u) = \exp u$, then the singular solution $u^*$ of (1.1) satisfies

  $$\begin{aligned} u^*(r) = \log \Bigl (-2\log r + \log (2N-4) - \log (-2\log r)\Bigr ) + o(1)\quad\text{ as }\ r\to 0. \end{aligned}$$

  (1.12)

Remark 1.2

In the case $g(u) = u^p$, Kikuchi and Wei [14] constructed the singular solution $u^*$ satisfying (1.10) by using a contraction mapping principle. Also, in the case $g(u) = g_n(u)$ with $n \ge 1$, Ghergu and Goubet [9] constructed the singular solution which has a prescribed behavior around the origin. In our proof, we make use of the formulas (1.8) to derive the asymptotic expansions (1.10) and (1.11) in a unified way.

As an application, we consider the nonlinear eigenvalue problem

$$\begin{aligned} \left\{ \begin{array}{rl} \Delta v + \lambda f(v) = 0 \quad &{} \text{ in } \ B, \\ v = 0 \quad &{} \text{ on } \ \partial B, \\ v > 0 \quad &{} \text{ in } \ B, \end{array} \right. \end{aligned}$$

(1.13)

where $B = \{x \in \mathbf{R}^N: |x| < 1\}$ with $N \ge 3$ and $\lambda >0$ is a parameter. We assume that $f(u) > 0$ for $u \ge 0$. By the symmetry result of Gidas, Ni and Nirenberg [10], every classical solution of (1.13) is radially symmetric about the origin. Hence, the problem (1.13) can be reduced to an ODE problem. Denote by $\{(\lambda (\alpha ), v(r, \alpha ))\}$ a solution of (1.13) with $v(0, \alpha ) = \alpha > 0$. It is well known that the solution set of (1.13) can be described as the curve $\{(\lambda (\alpha ), v(r, \alpha )): 0< \alpha < \infty \}$. (see, e.g., [15, 20, 21]). In the cases $f(u) = e^u$ and $f(u) = (u+1)^p$, the bifurcation diagram of the problem (1.13) was completely characterized by Joseph-Lundgren [13]. In these cases, there is a special change of variables such that (1.13) can be transformed into an autonomous first order system. For a general nonlinearity, we cannot expect to find such a change of variables, but we obtain the following result as a corollary of Theorem 1.1.

Corollary 1.2

Suppose that $N \ge 3$ and $\mathrm{(f1)}$ and $\mathrm{(f2)}$ with $q < q_{S}$ hold. Assume that $f(u) > 0$ for $u \ge 0$. Then the following (i) and (ii) hold.

• (i) The problem (1.13) has a unique radial singular solution $(\lambda ^*, v^*)$, that is, there exists a unique $\lambda ^* > 0$ such that the problem (1.13) with $\lambda = \lambda ^*$ has a singular solution $v^*$, and the singular solution $v^*$ is a unique singular radial solution of (1.13) with $\lambda = \lambda ^*$.

• (ii) As $\alpha \rightarrow \infty$, the solution $(\lambda (\alpha ), v(r, \alpha ))$ of (1.13) described above satisfies

  $$\begin{aligned} \lambda (\alpha ) \rightarrow \lambda ^* \quad \text{ and } \quad v(r, \alpha ) \rightarrow v^*(r) \quad \text{ in } \ C^2_{\mathrm{loc}}(0, 1], \end{aligned}$$

  where $(\lambda ^*, v^*)$ is the singular radial solution in (i).

Remark 1.3

One of the open problems raised by [1] is whether $\lambda$ for which (1.13) has a singular solution is unique or not. Corollary 1.2 (i) gives an affirmative answer to this question in the radially symmetric case.

Under the assumptions (f1) and (f2) the following quasi-scaling, which was introduced in [8], works well:

$$\begin{aligned} v(r):=F^{-1}\left[ \lambda ^{-2}F(u(\lambda r))\right] . \end{aligned}$$

(1.14)

It was found by [22] that the limit equation of (1.1) with respect to (1.14) as $\lambda \rightarrow 0$ is

$$\begin{aligned} v''+\frac{N-1}{r}v'+f(v)+\frac{q-F(v)f'(v)}{F(v)f(v)}(v')^2=0 \quad \text {for} \ r>0 \end{aligned}$$

(1.15)

and that (1.15) has the exact singular solution

$$\begin{aligned} v^*(r)=F^{-1}\left[ \frac{r^2}{2N-4q}\right] \quad \text {for}\ r>0. \end{aligned}$$

In this paper we will find a singular solution to (1.1) of the form

$$\begin{aligned} u^*(r)=F^{-1}\left[ \frac{r^2e^{-x(t)}}{2N-4q}\right] \quad \text {with} \ t=-\log r. \end{aligned}$$

(1.16)

Then $u^*$ given by (1.16) solves (1.1) for $0 < r \le r_0$ if and only if x satisfies

$$\begin{aligned} x''-ax'+b(e^x-1)+(q-1)(x')^2+(f'(u)F(u)-q)(x'+2)^2=0 \quad \text {for} \ t \ge t_0, \end{aligned}$$

(1.17)

where $a=N+2-4q$, $b=2N-4q$ and $t_0=-\log r_0$. In the proof of Theorem 1.1, we will show that $x(t) \rightarrow 0$ as $t \rightarrow \infty$ if $u^*$ given by (1.16) is a singular solution of (1.1). We also show that (1.17) has a unique solution x satisfying $x(t) \rightarrow 0$ as $t \rightarrow \infty$. The transformation (1.16) enables us to deal with a quite wide class of nonlinear terms.

The paper is organized as follows. In Sect. 2, we give some preliminary results and in Sect. 3, we investigate the asymptotic behavior of the singular solution. In Sect. 4, we show the uniqueness and the existence of singular solution, and give the proof of Theorem 1.1. In Sect. 5, we consider the super-exponential cases, and give the proof of Corollaries 1.1 and 1.2.

2 Preliminaries

First we show the following lemma.

Lemma 2.1

If (f1) and (f2) are satisfied, then $f'(u) \rightarrow \infty$ as $u \rightarrow \infty$. Furthermore, the exponent q in (1.5) satisfies $q \ge 1$ and q is also given by (1.6).

Proof

If we assume that $f'(u)$ is bounded as $u \rightarrow \infty$, then $f(u) \le Cu$ with some constant $C > 0$. This contradicts $F(u) < \infty$. Then we obtain $\limsup _{u \rightarrow \infty }f'(u) = \infty$. If we assume that $\liminf _{u \rightarrow \infty }f'(u) < \infty$, then there exists a sequence $u_n \rightarrow \infty$ of local maximum of $f'(u)$ such that $f'(u_n) > 0$ and $f''(u_n) = 0$. This contradicts (1.5). Thus we obtain $f'(u) \rightarrow \infty$ as $u \rightarrow \infty$. By L’Hospital’s rule, we have

$$\begin{aligned} \lim _{u \rightarrow \infty }F(u)f'(u) = \lim _{u \rightarrow \infty }\frac{(F(u))'}{(1/f'(u))'} = \lim _{u \rightarrow \infty }\frac{f'(u)^2}{f(u)f''(u)} = q. \end{aligned}$$

By [22, Lemma 2.1] (see also [8, Remark 1.1]), we obtain $q \ge 1$. $\square$

In the remaining part of this paper except the last section, we assume that $\mathrm{(f1)}$ and $\mathrm{(f2)}$ with $q < q_S$ hold. By Lemma 2.1 we may assume that

$$\begin{aligned} f'(u) > 0 \quad \text{ for } \ u \ge u_0 \end{aligned}$$

(2.1)

by replacing $u_0$ in (f1).

For a solution u of (1.1), define x(t) by

$$\begin{aligned} \frac{F(u(r))}{r^2}=\frac{e^{-x(t)}}{2N-4q} \quad \text{ and } \quad t=-\log r. \end{aligned}$$

(2.2)

Since $q_{S} = (N+2)/4 < N/2$ for $N>2$, the condition $q < N/2$ is always satisfied if $1 \le q < q_S$.

Lemma 2.2

Let u be a solution of (1.1), and define x(t) by (2.2) with $1 \le q < q_S$. Then x(t) satisfies

$$\begin{aligned} x''-ax'+ b(e^x-1) +(q-1)(x')^2 +(f'(u)F(u)-q)(x'+2)^2 = 0, \end{aligned}$$

(2.3)

where

$$\begin{aligned} a=N+2-4q> 0 \quad \text{ and } \quad b=2N-4q > 0. \end{aligned}$$

(2.4)

In the case $q > 1$, put $z(t) = e^{(q-1)x(t)}$. Then z(t) satisfies

$$\begin{aligned} \frac{F(u(r))}{r^2}=\frac{z(t)^{-1/(q-1)}}{2N-4q} \end{aligned}$$

(2.5)

and

$$\begin{aligned} z''-az'+(q-1)b(z^p-z)+(q-1)(f'(u)F(u)-q)\left( \frac{z'}{(q-1)z}+2\right) ^2 z=0, \end{aligned}$$

(2.6)

where $p=q/(q-1)$.

Proof

From (2.2) we have

$$\begin{aligned} F(u(r)) = \frac{e^{-x(t)-2t}}{2N-4q}. \end{aligned}$$

(2.7)

Differentiating the above twice, we obtain

$$\begin{aligned} -\frac{u'(r)}{f(u)} = \frac{d}{dt}\left( \frac{e^{-x(t)-2t}}{2N-4q}\right) \frac{dt}{dr} = \frac{x'(t)+2}{2N-4q}e^{-x(t)-t} \end{aligned}$$

(2.8)

and

$$\begin{aligned} -\frac{u''(r)}{f(u)} + \frac{f'(u)}{f(u)^2}u'(r)^2 = -\frac{e^{-x(t)}}{2N-4q}(x''(t) -3x'(t) - 2 -x'(t)^2). \end{aligned}$$

(2.9)

From (2.7) and (2.8) it follows that

$$\begin{aligned} \frac{f'(u)}{f(u)^2}u'(r)^2 = f'(u)\frac{(x'(t)+2)^2}{(2N-4q)^2}e^{-2x(t)-2t} =\frac{e^{-x(t)}}{2N-4q}f'(u)F(u)(x'(t)+2)^2. \end{aligned}$$

Then, from (2.9), we obtain

$$\begin{aligned} \frac{u''(r)}{f(u)} = \frac{e^{-x(t)}}{2N-4q}\bigl (x''(t) -3x'(t) - 2 -x'(t)^2 + f'(u)F(u)(x'(t) + 2)^2\bigr ). \end{aligned}$$

From (2.8) and $1/r = e^t$, we have

$$\begin{aligned} \frac{u'(r)}{rf(u)} = -\frac{e^{-x(t)}}{2N-4q}(x'(t)+2). \end{aligned}$$

Thus we obtain

$$\begin{aligned} \begin{array}{rcl} 0 &{} = &{} \displaystyle (2N-4q)e^{x(t)}\left( \frac{u''}{f(u)}+\frac{N-1}{r}\frac{u'}{f(u)}+1\right) \\ &{} = &{} \displaystyle x''(t) -x'(t)^2 -(N+2)x'(t) + f'(u)F(u)(x'+2)^2 +(2N-4q)e^{x(t)} -2N. \end{array} \end{aligned}$$

Note that $q(2+x'(t))^2 = qx'(t)^2 + 4qx'(t) + 4q$. Then we obtain (2.3). In the case $q > 1$, by a direct calculation, we obtain (2.5) and (2.6). $\square$

Lemma 2.3

Let u be a positive solution of (1.1) satisfying $u(r) \ge u_0$ for $0 < r \le r_0$. Then the following (i) and (ii) hold.

• (i) $u'(r) \le 0$ for $0 < r \le r_0$.

• (ii) $F(u(r)) \ge r^2/(2N)$ for $0 < r \le r_0$.

Proof

• (i) Assume to the contrary that there exists $r_1 \in (0, r_0]$ such that $u'(r_1) > 0$. Since $(r^{N-1}u'(r))'=-r^{N-1}f(u(r)) \le 0$ for $0 < r \le r_0$, the function $r^{N-1}u'(r)$ is nonincreasing. Then it follows that $r^{N-1}u'(r) \ge r_1^{N-1}u'(r_1) > 0$ for $0 < r \le r_1$. This implies that

  $$\begin{aligned} u'(r) \ge C r^{1-N}, \quad 0 < r \le r_1, \end{aligned}$$

  with $C = r_1^{N-1}u'(r_1) > 0$. Integrating the above on $(r, r_1]$, we obtain

  $$\begin{aligned} u(r_1) -u(r) \ge C\int ^{r_1}_r s^{1-N}ds. \end{aligned}$$

  Letting $r \rightarrow 0$, we obtain $u(r) \rightarrow -\infty$. This is a contradiction. Thus we obtain $u'(r) \le 0$ for $0 < r \le r_1$.

• (ii) Integrating (1.1) on $[\rho , r]$, by (i), we have

  $$\begin{aligned} -r^{N-1}u'(r)=-\rho ^{N-1}u'(\rho )+\int _{\rho }^rs^{N-1}f(u(s))ds \ge \int _{\rho }^rs^{N-1}f(u(s))ds. \end{aligned}$$

  Letting $\rho \rightarrow 0$, we have

  $$\begin{aligned} -r^{N-1}u'(r) \ge \int _{0}^rs^{N-1}f(u(s))ds \ge f(u(r))\int _0^rs^{N-1}ds =\frac{f(u(r))}{N}r^N, \end{aligned}$$

  where we used the fact that f(u) is increasing for $u \ge u_0$ which follows from (2.1). Then it follows that

  $$\begin{aligned} \frac{d}{dr}F(u(r))=-\frac{u'(r)}{f(u)}\ge \frac{r}{N}. \end{aligned}$$

  Integrating the above on $[\rho ,r]$ with $0<\rho <r$, and letting $\rho \rightarrow 0$, we have $F(u(r)) \ge r^2/(2N)$.

$\square$

Lemma 2.4

For any $\delta > 0$, there exists a constant $C > 0$ such that

$$\begin{aligned} f(u) \ge Cu^{\frac{q+\delta }{q+\delta -1}} \end{aligned}$$

(2.10)

for sufficiently large u.

Proof

From (1.5) there exists $u_1 \ge u_0$ such that

$$\begin{aligned} \frac{f'(u)^2}{f(u)f''(u)} \le q + \delta \quad \text{ for } \ u \ge u_1. \end{aligned}$$

Then it follows that

$$\begin{aligned} \frac{d}{dr}\left( f(u)^{-\frac{1}{q+\delta }}f'(u)\right) = f(u)^{-\frac{1}{q+\delta }-1}\left( f''(u)f(u)-\frac{1}{q+\delta }f'(u)^2\right) \ge 0 \quad \text{ for } \ u \ge u_1. \end{aligned}$$

Then $f(u)^{-\frac{1}{q+\delta }}f'(u)$ is nondecreasing for $u \ge u_1$, and hence we obtain

$$\begin{aligned} f(u)^{-\frac{1}{q+\delta }}f'(u) \ge C \quad \text{ for } \ u \ge u_1 \end{aligned}$$

with $C = {f(u_1)^{-\frac{1}{q+\delta }}}f'(u_1) > 0$. Integrating the above on $[u_1, u]$, we obtain

$$\begin{aligned} \frac{q+\delta }{q+\delta -1} \left( f(u)^{-\frac{1}{q+\delta }+1}- f(u_1)^{-\frac{1}{q+\delta }+1}\right) = \int ^u_{u_1} f(t)^{-\frac{1}{q+\delta }}f'(t)dt \ge C(u-u_1). \end{aligned}$$

Then we obtain (2.10) for sufficiently large u. $\square$

Lemma 2.5

Let u be a singular solution of (1.1) for $0 < r \le r_0$. Then

$$\begin{aligned} -r^{N-1}u'(r) = \int ^r_0 s^{N-1}f(u(s))ds \quad \text{ for } \ 0 < r \le r_0. \end{aligned}$$

(2.11)

Proof

Since $q < q_S = (N+2)/4$ and $N \ge 3$, we have

$$\begin{aligned} -N + 2q< -N + \frac{N+2}{2} < 0. \end{aligned}$$

Then there exists $\delta > 0$ such that

$$\begin{aligned} -N + 2q + 2\delta < 0. \end{aligned}$$

(2.12)

First we will show that

$$\begin{aligned} f(u(r)) = O(r^{-2q -2\delta }) \quad \text{ as } \ r \rightarrow 0. \end{aligned}$$

(2.13)

In fact, from (1.6) there exists $u_1 \ge u_0$ such that

$$\begin{aligned} F(u)f'(u) \le q + \delta \quad \text{ for } \ u \ge u_1. \end{aligned}$$

Observe that

$$\begin{aligned} \frac{d}{du}\left( f(u)F(u)^{q+\delta }\right) = F(u)^{q+\delta -1}\left( f'(u)F(u)-(q+\delta )\right) \le 0 \quad \text{ for } \ u \ge u_1. \end{aligned}$$

Then $f(u)F(u)^{q+\delta }$ is nonincreasing in $u \ge u_1$, and hence

$$\begin{aligned} f(u)F(u)^{q+\delta } \le C \quad \text{ for } \ u \ge u_1, \end{aligned}$$

where $C = f(u_1)F(u_1)^{q+\delta } > 0$. From Lemma 2.3 (ii), we obtain

$$\begin{aligned} f(u(r)) \le CF(u(r))^{-q-\delta } \le Cr^{-2q-2\delta }, \end{aligned}$$

which implies (2.13).

From Lemma 2.3 (i) we have $-r^{N-1}u'(r) \ge 0$ for $0 < r \le r_0$. We will show that

$$\begin{aligned} \liminf _{r \rightarrow 0}\left( -r^{N-1}u'(r)\right) = 0. \end{aligned}$$

(2.14)

Assume to the contrary that $\liminf _{r \rightarrow 0}\left( -r^{N-1}u'(r)\right) > 0$. Then there exist $L > 0$ and $r_1 \le r_0$ such that $-r^{N-1}u'(r) \ge L$ for $0 < r \le r_1$. This implies that there exists a constant $C > 0$ such that $u(r) \ge Cr^{2-N}$ for sufficiently small $r > 0$. By Lemma 2.4 we obtain

$$\begin{aligned} f(u(r)) \ge Cr^{-\frac{(q+\delta )(N-2)}{q+\delta -1}} \end{aligned}$$

for sufficiently small $r > 0$. From (2.13) it follows that

$$\begin{aligned} 2q + 2\delta \ge \frac{(q+\delta )(N-2)}{q+\delta -1}. \end{aligned}$$

This implies that $-N+2q +2\delta \ge 0$. This contradicts (2.12). Thus we obtain (2.14). Then there exists $r_n \rightarrow 0$ such that $-r_n^{N-1}u'(r_n) \rightarrow 0$ as $n \rightarrow \infty$. Integrating (1.1) on $[r_n, r]$, and letting $n \rightarrow \infty$, we obtain (2.11). $\square$

Lemma 2.6

Let u be a singular solution of (1.1). Then

$$\begin{aligned} \limsup _{r \rightarrow 0}\frac{r^2}{F(u(r))} > 0. \end{aligned}$$

Proof

Assume by contradiction that

$$\begin{aligned} \lim _{r \rightarrow 0}\frac{r^2}{F(u(r))} = 0. \end{aligned}$$

(2.15)

Take $q_0 \in (q, q_{\mathrm{S}})$, and define $z_0(t)$ by

$$\begin{aligned} \frac{F(u(r))}{r^2}=\frac{z_0(t)^{-{1}/{(q_0-1)}}}{2N-4q_0}. \end{aligned}$$

(2.16)

Applying Lemma 2.2 with $q = q_0$, we see that $z_0(t)$ satisfies

$$\begin{aligned} z_0'' - a_0 z_0' + (q_0-1)b_0 (z_0^{p_0}-z_0) + (q_0-1)(f'(u)F(u)-q_0)\left( \frac{z_0'}{(q_0-1)z_0} +2\right) ^2 z_0 = 0, \end{aligned}$$

(2.17)

where $a_0 = N+2-4q_0 > 0$, $b_0 = 2N-4q_0 > 0$ and $p_0 = q_0/(q_0-1) > 1$. From (2.15) and (2.16), we have $z_0(t) \rightarrow 0$ as $t \rightarrow \infty$. From $q_0 > q$, there exists $t_1 \in \mathbf{R}$ such that

$$\begin{aligned} z_0(t) < 1\quad \text{ for } \ t \ge t_1 \end{aligned}$$

(2.18)

and

$$\begin{aligned} f'(u(r))F(u(r)) \le q_0 \quad \text{ for } \ 0 < r \le r_1 \end{aligned}$$

(2.19)

with $r_1 = e^{-t_1}$. Note that $q_0 > 1$ and $p_0 > 1$. From (2.18) and (2.19) we have

$$\begin{aligned} (q_0-1)b_0 (z_0^{p_0}-z_0) + (q_0-1)\left( f'(u)F(u)-q_0\right) \left( \frac{z_0'}{(q_0-1)z_0} +2\right) ^2 z_0 < 0 \quad \text{ for } \ t \ge t_1. \end{aligned}$$

From (2.17) we have $z_0'' - a_0 z_0' > 0$, and hence $(e^{-a_0t}z_0')' > 0$ for $t \ge t_1$. Then $e^{-a_0 t}z_0'(t)$ is increasing for $t \ge t_1$. We show that

$$\begin{aligned} z_0'(t) \le 0 \quad \text{ for } \ t \ge t_1. \end{aligned}$$

(2.20)

By contradiction, assume that there exists $t_2 \ge t_1$ such that $z_0'(t_2) > 0$. Then $e^{-a_0 t}z_0'(t) \ge e^{-a_0 t_2}z_0'(t_2) > 0$ for $t \ge t_2$. This implies that $z_0'(t) \ge e^{a_0 (t-t_2)}z_0'(t_2) > 0$ for $t \ge t_2$, and hence $z_0'(t) \rightarrow \infty$ as $t \rightarrow \infty$. This contradicts (2.18). Thus (2.20) holds. Then, from (2.16) we obtain

$$\begin{aligned} \frac{d}{dr}\left( \frac{r^2}{F(u(r))}\right) = -\frac{2N-4q_0}{q_0-1}e^t z_0(t)^{-(q_0-2)/(q_0-1)}z_0'(t) \ge 0 \end{aligned}$$

(2.21)

for $0 < r \le r_1$. From (2.19) and Lemma 2.3 (i) we see that

$$\begin{aligned} \frac{d}{dr}\left( F(u(r))^{q_0}f(u(r))\right) = F(u(r))^{q_0-1}\left( f'(u(r))F(u(r))-q_0\right) u'(r) \ge 0 \end{aligned}$$

(2.22)

for $0 < r \le r_1$. Take $\varepsilon > 0$ such that $\varepsilon q_0 < 1$. From (2.15) we have

$$\begin{aligned} \frac{r^2 f(u)}{F(u)f(u)} = \frac{r^2}{F(u)} \rightarrow 0 \quad \text{ for } \ r \rightarrow 0. \end{aligned}$$

Then there exists $r_2 \le r_1$ such that

$$\begin{aligned} r^2f(u) \le (N-2q_0)\varepsilon F(u)f(u) \quad \text{ for } \ 0 < r \le r_2. \end{aligned}$$

(2.23)

From Lemma 2.5, (2.21), (2.22) and (2.23) we have

$$\begin{aligned} \begin{array}{rcl} \displaystyle -r^{N-1}u'(r) &{} = &{} \displaystyle \int ^r_0 s^2 f(u)s^{N-3}ds \le (N-2q_0)\varepsilon \int ^r_0 F(u(s))f(u(s))s^{N-3}ds \\ &{} = &{} \displaystyle (N-2q_0)\varepsilon \int ^r_0 \left( \frac{s^2}{F(u(s))}\right) ^{q_0-1} F(u(s))^{q_0}f(u(s))s^{N-1-2q_0}ds \\ &{} \le &{} \displaystyle (N-2q_0)\varepsilon \left( \frac{r^2}{F(u(r))}\right) ^{q_0-1}F(u(r))^{q_0}f(u(r)) \int ^r_0 s^{N-1-2q_0}ds \\ &{} = &{} \varepsilon r^{N-2}F(u(r))f(u(r)) \end{array} \end{aligned}$$

for $0 < r \le r_2$. Hence $ru'(r) + \varepsilon F(u)f(u) \ge 0$ for $0 < r \le r_2$. Observe that

$$\begin{aligned} \frac{d}{dr}\left( \frac{r^{\varepsilon }}{F(u(r))}\right) = \frac{r^{\varepsilon -1}}{F(u(r))^2f(u)}\left( \varepsilon F(u(r))f(u(r))+ ru'(r)\right) \ge 0 \end{aligned}$$

for $0 < r \le r_2$. Then $r^{\varepsilon }/F(u(r))$ is nondecreasing and bounded for $0 < r \le r_2$. From (2.22) we see that $F(u)^{q_0}f(u)$ is also bounded for $0 < r \le r_2$. Thus we have

$$\begin{aligned} -r^{N-1}u'(r) = \int ^r_0 F(u)^{q_0}f(u)\left( \frac{s^{\varepsilon }}{F(u)}\right) ^{q_0}s^{N-1-\varepsilon q_0}ds \le C\int ^r_0 s^{N-1-\varepsilon q_0}ds = \frac{Cr^{N-\varepsilon q_0}}{N-\varepsilon q_0}, \end{aligned}$$

which implies that $u'(r) = O(r^{1-\varepsilon q_0})$ as $r \rightarrow 0$. Since $\varepsilon q_0 < 1$, we have $u'(r) \rightarrow 0$ as $r \rightarrow 0$, and hence $\lim _{r \rightarrow 0}u(r) < \infty$. This is a contradiction. Thus we obtain the conclusion. $\square$

3 Asymptotic behavior of the singular solution

Let u be a singular solution of (1.1) satisfying $u(r) \ge u_0$ for $0 < r \le r_0$. Define x(t) by (2.2) for $t \ge t_0$, where $t_0 = -\log r_0$. In this section, we will show the following proposition.

Proposition 3.1

One has $\lim _{t \rightarrow \infty }x(t) = 0$ and $\lim _{t \rightarrow \infty }x'(t) = 0$.

To prove Proposition 3.1, we first show the following lemma.

Lemma 3.1

• (i) $x(t) \le \log (N/(N-2q))$ for $t \ge t_0$.

• (ii) $x'(t) \ge -2$ for $t \ge t_0$.

Proof

• (i) By Lemma 2.3 (ii) and (2.2) we have

  $$\begin{aligned} \frac{r^2e^{-x(t)}}{2N-4q}=F(u(r))\ge \frac{r^2}{2N}. \end{aligned}$$

  This implies that $e^{x(t)} \le N/(N-2q)$, and hence $x(t) \le \log (N/(N-2q))$ for $t \ge t_0$.

• (ii) From (2.7) we have $x(t) = -2t - \log (2N-4q) - \log F(u(r))$. By differentiating the both sides, since $dr = -rdt$, we obtain

  $$\begin{aligned} x'(t) = -2 - \frac{ru'(r)}{f(u)F(u)}. \end{aligned}$$

  By Lemma 2.3 (i), we obtain $x'(t) \ge -2$.

$\square$

We will show that $x(t) \rightarrow 0$ as $t \rightarrow \infty$ by dividing into the following two cases:

• (i) $x'(t)$ is nonoscillatory at $t=\infty$, that is, $x'(t)\ge 0$ or $x'(t)\le 0$ for sufficiently large t.

• (ii) $x'(t)$ is oscillatory at $t=\infty$, that is, the sign of $x'(t)$ changes infinitely many times as $t\rightarrow \infty$.

First we consider the case (i).

Lemma 3.2

Assume that $x'(t)$ is nonoscillatory at $t=\infty$. Then $x(t)\rightarrow 0$ as $t\rightarrow \infty$.

Proof

By Lemma 2.6 and (2.2) we obtain $\limsup _{t\rightarrow \infty }x(t)>-\infty$. Since x(t) is monotone increasing or decreasing for sufficiently large t, we have $\lim _{t \rightarrow \infty }x(t) > -\infty$. Then, by Lemma 3.1 (i), x(t) is bounded, and hence there exists a constant $c\in \mathbf{R}$ such that $x(t)\rightarrow c$ as $t\rightarrow \infty$. We show that $c=0$. On the contrary, assume that $c \ne 0$. We first show that

$$\begin{aligned} \lim _{t\rightarrow \infty }x'(t)=0. \end{aligned}$$

(3.1)

We consider the case $x'(t) \ge 0$ for all t large. Since x(t) is bounded, we have

$$\begin{aligned} \liminf _{t\rightarrow \infty }x'(t) = 0. \end{aligned}$$

(3.2)

Assume by contradiction that $\limsup _{t\rightarrow \infty }x'(t) > 0$. Let $t_n\rightarrow \infty$ be a sequence of local minimum points of $x'(t)$. Then, from (3.2), we have

$$\begin{aligned} x''(t_n)=0 \quad \text{ and } \quad x'(t_n)\rightarrow 0 \quad \text{ as } \ n \rightarrow \infty . \end{aligned}$$

By (2.3) and the fact that $f'(u)F(u)\rightarrow \ q$ as $t\rightarrow \infty$, we obtain

$$\begin{aligned} \begin{array}{rcl} b(e^{x(t_n)}-1) &{} = &{} -x''(t_n)+ax'(t_n)-(q-1)x'(t_n)^2-(f'(u)F(u)-q)(x'(t_n)+2)^2 \\ &{} \rightarrow &{} 0 \quad \text{ as } \ n \rightarrow \infty . \end{array} \end{aligned}$$

On the other hand,

$$\begin{aligned} b(e^{x(t_n)}-1) \rightarrow b(e^c-1) \ne 0 \quad \text{ as } \ n \rightarrow \infty . \end{aligned}$$

This is a contradiction. In the case where $x'(t)\le 0$ for all t large, we get the contradiction by a similar argument. Thus (3.1) holds.

Letting $t \rightarrow \infty$ in (2.3), from (3.1) we obtain $x''(t)\rightarrow -b(e^c-1)\ne 0$ as $t\rightarrow \infty$. This implies that $|x'(t)|\rightarrow \infty$ as $t\rightarrow \infty$, which is a contradiction. Thus, we obtain $x(t)\rightarrow 0$ as $t\rightarrow \infty$. $\square$

Next, we consider the case (ii). First we show that $x'(t)$ is bounded for $t\ge t_0$.

Lemma 3.3

Assume that the sign of $x'(t)$ changes infinitely many times as $t\rightarrow \infty$. Then $x'(t)$ is bounded for $t\ge t_0$.

Proof

Assume by contradiction that $\limsup _{t\rightarrow \infty }|x'(t)|=\infty$. Since $x'(t)$ is oscillatory and $x'(t) \ge -2$ by Lemma 3.1 (ii), we have

$$\begin{aligned} -2 \le \liminf _{t\rightarrow \infty }x'(t) \le 0 < \limsup _{t\rightarrow \infty }x'(t) = \infty . \end{aligned}$$

(3.3)

First we consider the case $q>1$. From (3.3) there exists a sequence $t_n\rightarrow \infty$ satisfying

$$\begin{aligned} x''(t_n)=0 \quad \text{ and } \quad x'(t_n) \rightarrow \infty \quad \text{ as } \ n \rightarrow \infty . \end{aligned}$$

By (2.3) and the fact that $f'(u)F(u)\rightarrow \ q$ as $t\rightarrow \infty$, we have

$$\begin{aligned} \begin{array}{rcl} 0 &{}= &{}x''(t_n)-ax'(t_n)+b(e^{x(t_n)}-1)+(q-1)x'(t_n)^2+(f'(u)F(u)-q)(x'(t_n)+2)^2 \\ &{} \rightarrow &{} \infty \quad \text{ as } \ n \rightarrow \infty , \end{array} \end{aligned}$$

which is a contradiction. Next we consider the case $q=1$. From (3.3), the function $x'(t)$ oscillates between 0 and an arbitrary large fixed constant. Hence, for any $M > 0$, there is a sequence $t_n\rightarrow \infty$ such that $x'(t_n) = M$ and $x''(t_n) \le 0$. By (2.3) we have

$$\begin{aligned} 0 \le -x''(t_n)=-aM + b(e^{x(t_n)}-1)+(f'(u)F(u)-1)(M+2)^2. \end{aligned}$$

Since $f'(u(r))F(u(r)) \rightarrow 1$ as $t \rightarrow \infty$, we have

$$\begin{aligned} \liminf _{n \rightarrow \infty }e^{x(t_n)} \ge 1+\frac{aM}{b}. \end{aligned}$$

This contradicts Lemma 3.1 (i), since $M > 0$ is arbitrary. Thus, $x'(t)$ is bounded in both cases $q > 1$ and $q = 1$. $\square$

In order to show that $x(t) \rightarrow 0$ as $t \rightarrow \infty$ in the case (ii), we consider the following ordinary differential equation

$$\begin{aligned} w'' - c(t)w' + \gamma (w^p-w) + G(t) = 0 \quad \text{ for } \ t \ge t_0, \end{aligned}$$

(3.4)

where $\gamma > 0$ and $p > 1$ are constants, $c \in C[t_0, \infty )$ and $G \in C[t_0, \infty )$. In (3.4), we assume that

$$\begin{aligned} c(t) \ge c_* > 0 \quad \text{ for } \ t \ge t_0 \end{aligned}$$

(3.5)

with some constant $c_* > 0$, and that $G(t) \rightarrow 0$ as $t \rightarrow \infty$. We can state the following lemma.

Lemma 3.4

Let $w \in C^2[t_0, \infty )$ be a bounded positive solution of (3.4). Assume that the sign of $w'(t)$ changes infinitely many times as $t \rightarrow \infty$. Then $w(t) \rightarrow 1$ as $t \rightarrow \infty$.

Since the proof of Lemma 3.4 is rather complicated, we will give the proof after completing the proof of Proposition 3.1. By using Lemma 3.4, we will show the following lemma.

Lemma 3.5

Assume that the sign of $x'(t)$ changes infinitely many times as $t\rightarrow \infty$. Then $x(t) \rightarrow 0$ as $t \rightarrow \infty$.

Proof

First we consider the case $q=1$. By Lemma 3.3, $x'(t)$ is bounded. Then we can take $q_0>1$ such that

$$\begin{aligned} (q_0-1)|x'(t)| < a \quad \text{ for } \ t \ge t_0, \end{aligned}$$

(3.6)

where a is the constant in (2.4) with $q = 1$. Define $z(t)= e^{(q_0-1)x(t)}$. Since x(t) satisfies (2.3) with $q = 1$, we see that z(t) satisfies

$$\begin{aligned} z''-az' - \frac{(z')^2}{z} +(q_0-1)b(z^{p_0}-z) +(q_0-1)H(t, z(t)) = 0, \end{aligned}$$

where a and b are constants given by (2.4), $p_0 = q_0/(q_0-1)$ and

$$\begin{aligned} H(t,z(t)) = \left( f'(u)F(u)-q\right) \left( \frac{z'(t)}{(q_0-1)z(t)}+2\right) ^2z(t) \end{aligned}$$

(3.7)

with $q = 1$. Since $z'/z = (q_0-1)x'$, we have

$$\begin{aligned} z''-\alpha (t)z'+(q_0-1)b(z^{p_0}-z)+(q_0-1)H(t, z(t)) = 0, \end{aligned}$$

(3.8)

where $\alpha (t) =a+(q_0-1)x'(t)$. From (3.6) we have

$$\begin{aligned} \inf _{t> t_0}\alpha (t) = \alpha _* > 0 \end{aligned}$$

with some constant $\alpha _* > 0$. In the case $q>1$, define $z(t) = e^{(q_0-1)x(t)}$ with $q_0 = q$. Then, by Lemma 2.2, z(t) satisfies (3.8) with $\alpha (t) \equiv a$ and $p_0 = q/(q-1)$. Therefore, z(t) satisfies (3.8) in both cases $q = 1$ and $q > 1$.

We show that H(t, z(t)), defined by (3.7), satisfies

$$\begin{aligned} H(t, z(t)) \rightarrow 0 \quad \text{ as } \ t \rightarrow \infty . \end{aligned}$$

(3.9)

In fact, we observe that

$$\begin{aligned} H(t, z(t)) = e^{(q_0-1)x(t)}\left( f'(u)F(u)-q\right) (x'(t)+2)^2. \end{aligned}$$

Lemmas 3.1 (i) and 3.3 imply that x(t) is bounded above and $x'(t)$ is bounded. Since $f'(u)F(u)-q\rightarrow 0$ as $t\rightarrow \infty$, we see that (3.9) holds. Then, applying Lemma 3.4 with $c(t) = \alpha (t)$, $\gamma = (q_0-1)b$ and $G(t) = (q_0-1)H(t, z(t))$, we obtain $z(t) \rightarrow 1$ as $t \rightarrow \infty$, which implies that $x(t) \rightarrow 0$ as $t \rightarrow \infty$. $\square$

Proof of Proposition 3.1

Combining Lemmas 3.2 and 3.5, we obtain $\lim _{t\rightarrow \infty }x(t)=0$. We will show that $x'(t) \rightarrow 0$ as $t \rightarrow \infty$. Define $\alpha$ and $\beta$, respectively, by

$$\begin{aligned} \limsup _{t\rightarrow \infty }x'(t)=\alpha \ \ \text {and}\ \ \liminf _{t\rightarrow \infty }x'(t)=\beta . \end{aligned}$$

(3.10)

We first show that $\alpha = \beta$. On the contrary, assume that $\alpha \ne \beta$. Since we have either $\alpha \ne 0$ or $\beta \ne 0$, we may assume here that $\alpha \ne 0$. In the case $q = 1$, x satisfies (2.3), which is reduced into

$$\begin{aligned} x''-ax'+b\left( e^x-1\right) +\left( f'(u)F(u)-1\right) (x'+2)^2=0. \end{aligned}$$

(3.11)

Since $\alpha \ne \beta$ in (3.10), there exists a sequence $t_n \rightarrow \infty$ such that $x''(t_n) = 0$ and $x'(t_n) \rightarrow \alpha \ne 0$ as $n \rightarrow \infty$. From (3.11) and the fact that $f'(u)F(u) \rightarrow 1$ as $t \rightarrow \infty$, we have

$$\begin{aligned} ax'(t_n) = b\left( e^{x(t_n)}-1\right) + \left( f'(u)F(u)-1\right) (x'(t_n) +2)^2 \rightarrow 0 \quad \text{ as } \ n \rightarrow \infty , \end{aligned}$$

which is a contradiction. In the case $q > 1$, let $z(t) = e^{(q-1)x(t)}$. Then, by Lemma 2.2, z satisfies (2.6). Since $z'(t) = x'(t)e^{(q-1)x(t)}$ and $x(t) \rightarrow 0$ as $t \rightarrow \infty$, we have

$$\begin{aligned} \lim _{t \rightarrow \infty }z(t) = 1, \quad \limsup _{t \rightarrow \infty }z'(t) = \alpha \quad \text{ and } \quad \liminf _{t \rightarrow \infty }z'(t) = \beta . \end{aligned}$$

Since $\alpha \ne \beta$, there exists a sequence $t_n \rightarrow \infty$ such that $z''(t_n) = 0$ and $z'(t_n) \rightarrow \alpha \ne 0$ as $n \rightarrow \infty$. Define H(t, z(t)) by (3.7) with $q_0 = q$. Then, by the fact that $f'(u)F(u) \rightarrow q$ as $t \rightarrow \infty$, we have $H(t_n, z(t_n)) \rightarrow 0$ as $n \rightarrow \infty$. From (2.6) we obtain

$$\begin{aligned} az'(t_n) = (q-1)(z(t_n)^p-z(t_n)) + (q-1)H(t_n, z(t_n)) \rightarrow 0 \quad \text{ as } \ n \rightarrow \infty , \end{aligned}$$

which is a contradiction. Thus $\alpha = \beta$ holds in (3.10) in both cases $q = 1$ and $q > 1$, and hence we have $x'(t) \rightarrow \alpha$ as $t \rightarrow \infty$. Since $x(t) \rightarrow 0$ as $t \rightarrow \infty$, we obtain $\alpha = 0$. Thus we obtain $x'(t) \rightarrow 0$ as $t \rightarrow \infty$. $\square$

Proof of Lemma 3.4

Define $\alpha$ and $\beta$, respectively, by

$$\begin{aligned} \limsup _{t\rightarrow \infty }w(t)=\alpha \quad \text{ and } \quad \liminf _{t\rightarrow \infty }w(t)=\beta . \end{aligned}$$

(3.12)

Since w(t) is bounded above, we have $0 \le \beta \le \alpha <\infty$. We will show that $\alpha = \beta = 1$ by dividing the proof into three steps.

Step 1 We have

$$\begin{aligned} \int _{\beta }^{\alpha }(w^p - w)dw \le 0. \end{aligned}$$

(3.13)

In fact, by (3.12), there exist $\{t_i\}$ and $\{\tau _i\}$ with $t_i<\tau _i$ such that, for $i=1,2,\ldots$,

$$\begin{aligned} \lim _{i\rightarrow \infty }w(t_i)=\alpha , \quad w'(t)< 0 \quad \text{ for } \ t_i< t < \tau _i, \quad \text{ and } \quad w'(t_i) = w'(\tau _i)=0. \end{aligned}$$

Multiplying (3.4) by $w'(t)$ and integrating it over $[t_i,\tau _i]$, we obtain

$$\begin{aligned} \int _{t_i}^{\tau _i}\left( w''w'- c(t)(w')^2\right) dt + \int _{t_i}^{\tau _i}\left( \gamma (w^p-w)+G(t)\right) w'(t)dt=0 \end{aligned}$$

for $i=1,2,\ldots$. Integrating by parts, from $w'(t_i)=w'(\tau _i)=0$ and (3.5), we have

$$\begin{aligned} \int _{t_i}^{\tau _i}\left( \gamma (w^p- w) + G(t)\right) w'(t)dt = \int _{t_i}^{\tau _i} c(t)(w')^2dt \ge 0 \quad \text{ for } \ i=1,2,\ldots . \end{aligned}$$

Since $w'(t) < 0$ on $(t_i,\tau _i)$, there exists an inverse function $t=t_i(w)$ of $w = w(t)$ for $t_i< t < \tau _i$. Then we obtain

$$\begin{aligned} \int _{w(\tau _i)}^{w(t_i)}\left( \gamma (w^p-w)+ G(t_i(w))\right) dw \le 0 \quad \text{ for } \ i=1,2,\ldots . \end{aligned}$$

(3.14)

Since $G(t) \rightarrow 0$ as $t \rightarrow \infty$, we have $G(t_i(w))\rightarrow 0$ as $i \rightarrow \infty$. Put ${\tilde{\beta }} = \liminf _{i\rightarrow \infty }w(\tau _i)$. Then ${\tilde{\beta }} \ge \beta$ and there exists a subsequence, which we denote by $\{\tau _i\}$ again, such that $w(\tau _i)\rightarrow {\tilde{\beta }}$ as $i\rightarrow \infty$. Letting $i \rightarrow \infty$ in (3.14), we obtain

$$\begin{aligned} \gamma \int _{{\tilde{\beta }}}^{\alpha }(w^p-w)dw \le 0. \end{aligned}$$

This implies ${\tilde{\beta }} \le 1$. From $\beta \le {\tilde{\beta }}$, we have (3.13). Thus, Step 1 is proved.

Step 2 We have

$$\begin{aligned} \int _{\beta }^{\alpha }(w^p -w)dw = 0. \end{aligned}$$

(3.15)

In fact, by (3.12), there exist $\{s_i\}$ and $\{\sigma _i\}$ with $s_i<\sigma _i$ such that

$$\begin{aligned} \lim _{i\rightarrow \infty }w(s_i)=\beta , \quad w'(t)>0 \quad \text{ for } \ s_i<s<\sigma _i \quad \text{ and } \quad w'(s_i) = w'(\sigma _i) = 0 \end{aligned}$$

for each $i=1,2,\ldots$. Multiplying (3.4) by $w'(t)$, and integrating it over $[s_i,\sigma _i]$, we obtain

$$\begin{aligned} \int _{s_i}^{\sigma _i}\left( w''w'-c(t)(w')^2\right) dt + \int _{s_i}^{\sigma _i}\left( \gamma (w^p-w)+G(t)\right) w'(t)dt=0 \end{aligned}$$

for $i=1,2,\ldots$. From $w'(s_i) = w'(\sigma _i)=0$ and (3.5), we have

$$\begin{aligned} -c_*\int _{s_i}^{\sigma _i}(w')^2dt +\int _{s_i}^{\sigma _i}\left( \gamma (w^p-w) + G(t)\right) w'(t)dt \ge 0. \end{aligned}$$

Since $w'(t) > 0$ on $(s_i,\sigma _i)$, there exists an inverse function $t= t_i(w)$ of $w = w(t)$ for $s_i< t < \sigma _i$. Putting $w'(t) = w'(t_i(w))= q_i(w)$, we obtain

$$\begin{aligned} \int _{w(s_i)}^{w(\sigma _i)}\left( \gamma (w^p-w) + G(t_i(w)\right) dw \ge c_*\int _{w(s_i)}^{w(\sigma _i)}q_i(w)dw > 0. \end{aligned}$$

(3.16)

Put ${\tilde{\alpha }} = \limsup _{i\rightarrow \infty }w(\sigma _i)$. Then ${\tilde{\alpha }} \le \alpha$ and there exists a subsequence, which we denote by $\{\sigma _i\}$ again, such that

$$\begin{aligned} w(\sigma _i) \rightarrow {\tilde{\alpha }} \quad \text{ as } \ i \rightarrow \infty . \end{aligned}$$

(3.17)

Note that $G(t_i(w))\rightarrow 0$ as $i \rightarrow \infty$. Then, letting $i\rightarrow \infty$ in (3.16), we obtain

$$\begin{aligned} \gamma \int _{\beta }^{{\tilde{\alpha }}}(w^p-w)dw \ge \limsup _{i \rightarrow \infty }c_* \int _{w(s_i)}^{w(\sigma _i)}q_i(w)dw \ge 0. \end{aligned}$$

This implies that ${\tilde{\alpha }} > 1$. From $\alpha \ge {\tilde{\alpha }}$, we have

$$\begin{aligned} \gamma \int _{\beta }^{\alpha }(w^p-w)dw \ge \gamma \int _{\beta }^{{\tilde{\alpha }}}(w^p-w)dw \ge \limsup _{i \rightarrow \infty }c_*\int _{w(s_i)}^{w(\sigma _i)}q_i(w)dw \ge 0. \end{aligned}$$

(3.18)

From (3.13) and (3.18), we obtain (3.15) and

$$\begin{aligned} \gamma \int _{\beta }^{\alpha }(w^p-w)dw = \gamma \int _{\beta }^{{\tilde{\alpha }}}(w^p-w)dw = 0, \end{aligned}$$

which implies that ${\tilde{\alpha }}=\alpha$. Then, from (3.17), we find that

$$\begin{aligned} \left\{ \begin{array}{l} \lim _{i\rightarrow \infty }w(s_i)=\beta ,\quad \lim _{i\rightarrow \infty }w(\sigma _i)=\alpha , \\ w'(t)>0 \quad \text{ for } \ s_i< t <\sigma _i \quad \text{ and } \quad w'(s_i) = w'(\sigma _i)=0 \quad \text{ for } \text{ each } \ i=1,2,\ldots . \end{array} \right. \end{aligned}$$

(3.19)

Step 3 We show that $\alpha =\beta =1$.

From (3.15), we have $\beta \le 1 \le \alpha$. Assume by contradiction that $\beta < 1$. Then $\alpha > 1$. Put

$$\begin{aligned} A=-\int _{\beta }^{(\beta +1)/2}(w^p - w)dw > 0. \end{aligned}$$

Let $\{s_i\}$ and $\{\sigma _i\}$ be the sequences appearing in (3.19), and let $t = t_i(w)$ is the inverse function of $w = w(t)$ for $s_i< t < \sigma _i$. Put $w'(t) = w'(t_i(w)) = q_i(w)$. We will show that

$$\begin{aligned} \liminf _{i\rightarrow \infty }q_i(w) \ge \sqrt{2\gamma A} \quad \text{ for } \text{ all } \ w \in \left[ \frac{\beta +1}{2},1\right] . \end{aligned}$$

(3.20)

Take any ${\hat{w}} \in \left[ \frac{\beta +1}{2},1\right]$. From (3.19), there exists ${\hat{s}}_i$ satisfying $s_i< {\hat{s}}_i < \sigma _i$ and $w({\hat{s}}_i)={\hat{w}}$ for sufficiently large i. Multiplying (3.4) by $w'(t)$, and integrating it over $[s_i,{\hat{s}}_i]$, we obtain

$$\begin{aligned} \int _{s_i}^{{\hat{s}}_i}\left( w''w'-c(t)(w')^2\right) dt+ \int _{s_i}^{{\hat{s}}_i}\left( \gamma (w^p-w) +G(t)\right) w'(t)dt=0. \end{aligned}$$

From $w'(s_i)=0$ and (3.5), we obtain

$$\begin{aligned} \frac{w'({\hat{s}}_i)^2}{2} -c_*\int _{s_i}^{{\hat{s}}_i}(w')^2dt + \int _{s_i}^{{\hat{s}}_i}\left( \gamma (w^p-w)+G(t)\right) w'(t)dt \ge 0. \end{aligned}$$

Then it follows that

$$\begin{aligned} \frac{q_i({\hat{w}})^2}{2} -c_*\int _{w(s_i)}^{{\hat{w}}}q_i(w)dw + \int _{w(s_i)}^{{\hat{w}}}\left( \gamma (w^p-w)+G(t_i(w))\right) dw \ge 0. \end{aligned}$$

(3.21)

Note here that, from (3.15) and (3.18), we have

$$\begin{aligned} \lim _{i \rightarrow \infty }\int _{w(s_i)}^{w(\sigma _i)}q_i(w)dw = 0. \end{aligned}$$

(3.22)

Since $q_i(w) > 0$ and $w(s_i)< {\hat{w}} < w(\sigma _i)$, if follows that

$$\begin{aligned} 0 \le \int _{w(s_i)}^{{\hat{w}}}q_i(w)dw \le \int _{w(s_i)}^{w(\sigma _i)}q_i(w)dw \rightarrow 0 \quad \text{ as } \ i \rightarrow \infty . \end{aligned}$$

Note that $G(t_i(w))\rightarrow 0$ as $i \rightarrow \infty$. Then, letting $i \rightarrow \infty$ in (3.21) we have

$$\begin{aligned} \liminf _{i\rightarrow \infty }\frac{q_i({\hat{w}})^2}{2} \ge \gamma \int _{\beta }^{{\hat{w}}}(w^p-w)dw \ge \gamma \int _{\beta }^{(\beta +1)/2}(w^p-w)dw =\gamma A. \end{aligned}$$

Since $q_i(w)>0$, we obtain (3.20).

From (3.19) and $\beta< 1 < \alpha$, there exist ${\tilde{s}}_i$ and $\tilde{\sigma _i}$ satisfying

$$\begin{aligned} s_i< {\tilde{s}}_i< {\tilde{\sigma }}_i < \sigma _i, \quad w({\tilde{s}}_i)=\frac{\beta +1}{2} \quad \text{ and } \quad w({\tilde{\sigma }}_i)=1 \end{aligned}$$

for sufficiently large i. Then, from $q_i(w)>0$, it follows that

$$\begin{aligned} \int _{w(s_i)}^{w(\sigma _i)}q_i(w)dw \ge \int _{w({\tilde{s}}_i)}^{w({\tilde{\sigma }}_i)}q_i(w)dw = \int _{(\beta +1)/2}^1 q_i(w)dw. \end{aligned}$$

Letting $i \rightarrow \infty$ in the above, from (3.20), we obtain

$$\begin{aligned} \liminf _{i \rightarrow \infty }\int _{w(s_i)}^{w(\sigma _i)}q_i(w)dw \ge \frac{1-\beta }{2}\sqrt{2\gamma A} > 0. \end{aligned}$$

This contradicts (3.22). Thus, we obtain $\beta = 1$. By (3.15) we obtain $\alpha =1$. $\square$

4 Proof of Theorem 1.1

We first show the uniqueness of the singular solution.

Theorem 4.1

There exists at most one singular solution of (1.1).

To show Theorem 4.1, we need the following lemma by [17, Lemma 4.2].

Lemma 4.1

Let y(t) be a solution of

$$\begin{aligned} y''-A(t)y'+B(t)y=0, \end{aligned}$$

(4.1)

where A(t) and B(t) are continuous functions satisfying

$$\begin{aligned} \lim _{t \rightarrow \infty }A(t) = \alpha>0 \quad \text{ and } \quad \lim _{t \rightarrow \infty }B(t) = \beta > 0. \end{aligned}$$

If y(t) is bounded as $t \rightarrow \infty$, then $y(t) \equiv 0$.

Proof of Theorem 4.1

Let $u_j(r)$, $j=1,2$, be singular solutions of (1.1) for $0<r<r_0$. For $j = 1, 2$, define $x_j(t)$ by

$$\begin{aligned} \frac{F(u_j(r))}{r^2}=\frac{e^{-x_j(t)}}{2N-4q} \quad \text{ and } \quad t=-\log r. \end{aligned}$$

Proposition 3.1 implies that $x_j(t) \rightarrow 0$ as $t \rightarrow \infty$ for $j = 1, 2$. Define $y(t) = x_1(t)-x_2(t)$. We will show that $y(t) \equiv 0$. By Lemma 2.2, $x_j(t)$, $j = 1, 2$, satisfies

$$\begin{aligned} x_j''-ax_j'+b\left( e^{x_j}-1\right) + (q-1)\left( x_j'\right) ^2 + \left( f'(u_j)F(u_j)-q\right) \left( x_j'+2\right) ^2 = 0, \end{aligned}$$

where a and b are constants in (2.4). Then y(t) satisfies

$$\begin{aligned} \begin{array}{r} y''-ay'+bE\left( x_1, x_2\right) y +(q-1)\left( x_1'+x_2'\right) y' +\left( f'(u_1)F(u_1)-q\right) \left( x_1'+x_2'+4\right) y' \\ +\left( x_2'+2\right) ^2\left( f'(u_1)F(u_1)-f'(u_2)F(u_2)\right) =0, \end{array} \end{aligned}$$

where

$$\begin{aligned} E(x_1, x_2) = \left\{ \begin{array}{cl} \displaystyle \frac{e^{x_1}-e^{x_2}}{x_1-x_2}, \quad &{} \text{ if } \ x_1 \ne x_2, \\ e^{x_1}, \quad &{} \text{ if } \ x_1 = x_2. \end{array} \right. \end{aligned}$$

(4.2)

Let $w_j =F(u_j)$ for $j=1,2$. Then it follows that

$$\begin{aligned} f'(u_1)F(u_1)-f'(u_2)F(u_2) = w_1 f'\left( F^{-1}(w_1)\right) - w_2 f'\left( F^{-1}(w_2)\right) . \end{aligned}$$

Observe that

$$\begin{aligned} \frac{d}{dw}\left( wf'\left( F^{-1}(w)\right) \right) = f'\left( F^{-1}(w)\right) -wf''\left( F^{-1}(w)\right) f\left( F^{-1}(w)\right) . \end{aligned}$$

Then, by the mean value theorem, there exists a ${\bar{w}}$ which lies between $w_1$ and $w_2$ such that

$$\begin{aligned} w_1f'\left( F^{-1}(w_1)\right) -w_2f'\left( F^{-1}(w_2)\right) = \left( f'\left( F^{-1}({\bar{w}})\right) -{\bar{w}}f''\left( F^{-1}({\bar{w}})\right) f\left( F^{-1}({\bar{w}})\right) \right) (w_1-w_2). \end{aligned}$$

Since F is monotone, there exists a ${\bar{u}}$ which lies between $u_1$ and $u_2$ such that $F({\bar{u}}) = {\bar{w}}$. Then we have

$$\begin{aligned} \begin{array}{rcl} f'(u_1)F(u_1)-f'(u_2)F(u_2) &{} = &{} \left( f'({\bar{u}})-F({\bar{u}})f({\bar{u}})f''({\bar{u}})\right) \left( F(u_1)-F(u_2)\right) \\ &{} = &{} \displaystyle \left( 1-f'({\bar{u}})F({\bar{u}}) \frac{f({\bar{u}})f''({\bar{u}})}{f'({\bar{u}})^2}\right) f'({\bar{u}})\left( F(u_1)-F(u_2)\right) . \end{array} \end{aligned}$$

Define ${\bar{x}}$ as $F({\bar{u}}(t)) = r^2 e^{-{\bar{x}}(t)}/(2N-4q)$. Then it follows that

$$\begin{aligned} F(u_1(t)) - F(u_2(t)) = \frac{r^2}{2N-4q}\left( e^{-x_1(t)}-e^{-x_2(t)}\right) = F({\bar{u}}(t))e^{{\bar{x}}(t)}E(x_1, x_2)y, \end{aligned}$$

where $E(x_1, x_2)$ is defined by (4.2). As a consequence, y satisfies (4.1), where

$$\begin{aligned} \begin{array}{l} A(t) = a - (q-1)\left( x_1'+x_2'\right) - \left( f'(u_1)F(u_1)-q\right) \left( x_1'+x_2'+4\right) , \\ \displaystyle B(t) = bE(x_1, x_2) + \left( x_2'+2\right) ^2 \left( 1-f'({\bar{u}})F({\bar{u}}) \frac{f({\bar{u}})f''({\bar{u}})}{f'({\bar{u}})^2}\right) f'({\bar{u}})F({\bar{u}})e^{{\bar{x}}}E(x_1, x_2). \end{array} \end{aligned}$$

Note that (1.6) holds by Lemma 2.1. Since $u_1(r)\rightarrow \infty$ and $x_j'(t) \rightarrow 0$ as $t \rightarrow \infty$ by Proposition 3.1, we see that

$$\begin{aligned} A(t) \rightarrow a \quad \text{ as } \ t \rightarrow \infty . \end{aligned}$$

Since ${\bar{u}}$ lies between $u_1$ and $u_2$, ${\bar{x}}$ lies between $x_1$ and $x_2$. Then ${\bar{u}} \rightarrow \infty$ as $t \rightarrow \infty$ and ${\bar{x}} \rightarrow 0$ as $t \rightarrow \infty$. From (1.5) and (1.6) we obtain

$$\begin{aligned} \lim _{t \rightarrow \infty }f'({\bar{u}})F({\bar{u}})=q, \quad \lim _{t \rightarrow \infty }\frac{f({\bar{u}})f''({\bar{u}})}{f'({\bar{u}})^2} = \frac{1}{q} \quad \text{ and } \quad \lim _{t \rightarrow \infty }e^{{\bar{x}}} = 1. \end{aligned}$$

Since $x_j(t) \rightarrow 0$ as $t \rightarrow \infty$ and $E(x_1, x_2)$ is continuous at $x_1 = x_2 = 0$, we see that $E(x_1(t), x_2(t)) \rightarrow 1$ as $t \rightarrow \infty$. Thus we obtain

$$\begin{aligned} B(t) \rightarrow b \quad \text{ as } \ t \rightarrow \infty . \end{aligned}$$

Proposition 3.1 implies that $y(t) \rightarrow 0$ as $t\rightarrow \infty$. Then, by Lemma 4.1, we obtain $y(t)\equiv 0$, and hence (1.1) has at most one singular solution. $\square$

Recall that, for $\alpha > 0$, we denote by $u(r, \alpha )$ a regular solution of (1.1) satisfying $u(0) = \alpha$ and $u'(0) = 0$. In the proof of Theorem 1.1, we need the following lemma [24, Lemma 5.1].

Lemma 4.2

Assume that there exist constants $p_0 > 2N/(N-2)$ and ${\hat{u}}_0$ such that

$$\begin{aligned} 0< p_0 \int _0^uf(t)dt < uf(u) \quad \text{ for } \ u > {\hat{u}}_0. \end{aligned}$$

(4.3)

• (i) Let $\alpha > {\hat{u}}_0$. Assume that $u(r, \alpha ) \ge {\hat{u}}_0$ for $0 \le r \le {\hat{r}}_0$ with some ${\hat{r}}_0 > 0$. Then

  $$\begin{aligned} 0< -ru'(r, \alpha )< \frac{2N}{p_0}u(r, \alpha ) \quad \text{ for } \ 0 < r \le {\hat{r}}_0. \end{aligned}$$

• (ii) Put

  $$\begin{aligned} \eta =\frac{1}{2}\left( 1-\frac{2N}{p_0(N-2)}\right) . \end{aligned}$$

  (4.4)

  Take any $\beta > {\hat{u}}_0$, and define $r_{\beta }$ by

  $$\begin{aligned} r_{\beta }=\left( \frac{2N\beta }{f_M(\beta /\eta )}\right) ^{1/2}, \end{aligned}$$

  (4.5)

  where $f_M(r)=\max _{0\le s\le r}f(s)$. If $\alpha > \beta /\eta$, then $u(r,\alpha ) > \beta$ for $0 \le r \le r_{\beta }$.

Proof of Theorem 1.1

By (1.5) and L’Hospital’s rule, we have

$$\begin{aligned} \lim _{u \rightarrow \infty }\frac{f(u)}{uf'(u)} = \lim _{u \rightarrow \infty }\frac{f(u)/f'(u)}{u} = \lim _{u \rightarrow \infty }\left( 1-\frac{f(u)f''(u)}{f'(u)^2}\right) = \frac{q-1}{q}. \end{aligned}$$

Since $f'(u) > 0$ for sufficiently large u by Lemma 2.1, we obtain $uf'(u)/f(u) \rightarrow \infty$ as $u \rightarrow \infty$ if $q = 1$, Then, using L’Hospital’s rule again, we obtain

$$\begin{aligned} \lim _{u \rightarrow \infty }\frac{uf(u)}{\int _0^uf(t)dt} = \lim _{u \rightarrow \infty }\left( 1+\frac{uf'(u)}{f(u)}\right) = \left\{ \begin{array}{cl} (2q-1)/(q-1) \quad &{} \text{ if } \ q > 1, \\ \infty \quad &{} \text{ if } \ q = 1. \end{array} \right. \end{aligned}$$

Since $q < q_S$, we have $(2q-1)/(q-1) > 2N/(N-2)$. Take $p_0 \in (2N/(N-2), (2q-1)/(q-1))$. Then there exists ${\hat{u}}_0 \ge u_0$ such that (4.3) holds. Take $\beta > {\hat{u}}_0$, and define $\eta$ and $r_{\beta }$ by (4.4) and (4.5), respectively. Let $\alpha > \beta /\eta$. Then, by Lemma 4.2 (ii), we have $u(r, \alpha )> \beta > {\hat{u}}_0 \ge u_0$ for $0 \le r \le r_{\beta }$. By Lemma 2.3 (ii) we have

$$\begin{aligned} u(r, \alpha ) \le F^{-1}\left[ \frac{r^2}{2N}\right] \quad \text{ for } \ 0 < r \le r_{\beta }. \end{aligned}$$

(4.6)

By Lemma 4.2 (i), we obtain

$$\begin{aligned} 0< -ru'(r, \alpha )< \frac{2N}{p_0}F^{-1}\left[ \frac{r^2}{2N}\right] \quad \text{ for } \ 0 < r \le r_{\beta }. \end{aligned}$$

(4.7)

Let $\{\alpha _k\}$ be a sequence such that $\alpha _k\rightarrow \infty$ as $k\rightarrow \infty$. From (4.6) and (4.7), $u(r,\alpha _k)$ and $u_r(r,\alpha _k)$ are uniformly bounded in $k \in \mathbf{N}$ on any compact subset of $(0,r_{\beta }]$. Since $f\in C^2[0,\infty )$ in (1.1), $u_{rr}(r,\alpha _k)$ and $u_{rrr}(r,\alpha _k)$ are also uniformly bounded on the subset. Then, by the Ascoli-Arzelá theorem with the diagonal argument, there exist $u^*\in C^2(0,r_{\beta }]$ and a subsequence, which is denoted by $\{u(r,\alpha _k)\}$, such that

$$\begin{aligned} u(r,\alpha _k)\rightarrow \ u^*(r) \quad \text{ in } \ C^2_{\mathrm{loc}}(0,r_{\beta }] \quad \text{ as } \ k \rightarrow \infty . \end{aligned}$$

It is clear that $u^*$ satisfies (1.1) for $0< r \le r_{\beta }$. Take any ${\tilde{\beta }} > \beta$. Lemma 4.2 (ii) implies that $u(r_{{\tilde{\beta }}},\alpha _k) > {\tilde{\beta }}$ if $\alpha _k > {\tilde{\beta }}/\eta$. Letting $k \rightarrow \infty$, we obtain $u^*(r_{{\tilde{\beta }}}) \ge {\tilde{\beta }}$. Since $u^*(r)$ is nonincreasing for $0 < r \le r_{\beta }$ by Lemma 2.3 (i), we obtain $u^*(r) \rightarrow \infty$ as $r \rightarrow 0$, and hence $u^*$ is a singular solution. We can define $u^*(r)$ on $(0, r_0]$ as a positive solution of (1.1) with some $r_0 > 0$. Since the singular solution $u^*$ of (1.1) is unique by Theorem 4.1, we obtain $u(r,\alpha ) \rightarrow u^*(r)$ in $C^2_{\mathrm{loc}}(0,r_0]$ as $\alpha \rightarrow \infty$, and hence (1.7) holds. By Proposition 3.1 and (2.2), we obtain (1.8). Thus the proof is complete. $\square$

5 Super-exponential nonlinearity

In this section we consider the case where f(u) has the form $f(u) = \exp (g(u))$.

Lemma 5.1

Let $f(u) = \exp (g(u))$, where $g \in C^2(0, \infty )$.

• (i) If g(u) satisfies

  $$\begin{aligned} \liminf _{u \rightarrow \infty }\frac{g(u)}{\log u} > 1, \end{aligned}$$

  (5.1)

  then (f1) holds.

• (ii) The hypothesis (f2) with $q = 1$ holds if and only if

  $$\begin{aligned} \lim _{u \rightarrow \infty }\frac{g''(u)}{g'(u)^2} = 0. \end{aligned}$$

  (5.2)

• (iii) If (f1) and (f2) with $q = 1$ hold, then

  $$\begin{aligned} \lim _{u \rightarrow \infty }g(u) = \infty \quad \text{ and } \quad \lim _{u \rightarrow \infty }\frac{\log g'(u)}{g(u)} = 0. \end{aligned}$$

  (5.3)

Proof

• (i) If (5.1) holds, then there exist $\kappa > 1$ and $u_1 \ge 0$ such that $g(u) \ge \kappa \log u$ for $u \ge u_1$. This implies that $f(u) \ge u^{\kappa }$ for $u \ge u_1$, and hence (f1) holds.

• (ii) Observe that

  $$\begin{aligned} \frac{f'(u)^2}{f(u)f''(u)} = \frac{1}{1+\frac{g''(u)}{g'(u)^2}}. \end{aligned}$$

  Then we have

  $$\begin{aligned} \lim _{u \rightarrow \infty }\frac{g''(u)}{g'(u)^2} = 0 \quad \text{ if } \text{ and } \text{ only } \text{ if } \quad \lim _{u \rightarrow \infty }\frac{f'(u)^2}{f(u)f''(u)} =1. \end{aligned}$$

  Thus (ii) holds.

• (iii) By Lemma 2.1 we have $f(u) \rightarrow \infty$ as $u \rightarrow \infty$, and hence $g(u) \rightarrow \infty$ as $u \rightarrow \infty$. By (ii) we obtain (5.2). By L’Hospital’s rule, we have

  $$\begin{aligned} \lim _{u \rightarrow \infty }\frac{\log g'(u)}{g(u)} = \lim _{u\rightarrow \infty }\frac{g''(u)}{g'(u)^2} = 0. \end{aligned}$$

  Thus (5.3) holds.

$\square$

To prove Corollary 1.1, we need the following lemma.

Lemma 5.2

Assume that f has the form $f(u) = \exp (g(u))$, and that (f1) and (f2) with $q = 1$ hold. Let $u^*$ be a singular solution of (1.1). Then

$$\begin{aligned} \log g(u^*(r))= & {} \log (-2\log r) + o(1) \quad \text{ as } \ r \rightarrow 0, \end{aligned}$$

(5.4)

$$\begin{aligned} g(u^*(r)) + \log g'(u^*(r))= & {} -2\log r +\log (2N-4) + o(1) \quad \text{ as } \ r \rightarrow 0. \end{aligned}$$

(5.5)

Proof

From (1.6) we have $F(u)f'(u) = F(u)g'(u)\exp (g(u)) \rightarrow 1$ as $u \rightarrow \infty$. Then

$$\begin{aligned} \log F(u) + \log g'(u) + g(u) = o(1) \quad \text{ as } \ u \rightarrow \infty . \end{aligned}$$

(5.6)

By Lemma 5.1 (iii) we obtain (5.3). Then we have

$$\begin{aligned} \frac{\log F(u)}{g(u)} \rightarrow -1 \quad \text{ as } \ u \rightarrow \infty . \end{aligned}$$

(5.7)

Since $u^*$ satisfies (1.8) by Theorem 1.1, we obtain

$$\begin{aligned} \frac{F(u^*(r))}{r^2} = \frac{1}{2N-4} + o(1) \quad \text{ as } \ r \rightarrow 0. \end{aligned}$$

Then it follows that

$$\begin{aligned} \log F(u^*(r)) - 2\log r = -\log (2N-4) + o(1) \quad \text{ as } \ r \rightarrow 0. \end{aligned}$$

(5.8)

Substituting this equality into (5.6) with $u = u^*(r)$, we obtain (5.5). From (5.8) we have

$$\begin{aligned} \frac{\log F(u^*(r))}{-2\log r} \rightarrow -1 \quad \text{ as } \ r \rightarrow 0. \end{aligned}$$

Then, from (5.7) with $u = u^*(r)$, we obtain

$$\begin{aligned} \frac{g(u^*(r))}{-2\log r} = \frac{g(u^*(r))}{\log F(u^*(r))} \frac{\log F(u^*(r))}{-2\log r} \rightarrow 1 \quad \text{ as } \ r \rightarrow 0, \end{aligned}$$

which implies (5.4). $\square$

Proof of Corollary 1.1

• (i) In the case $g(u) = u^p$ with $p > 0$, we have $\log g'(u) = \log p +(p-1)\log u$. From (5.5) it follows that

  $$\begin{aligned} u^*(r)^p + (p-1)\log u^*(r) = -2\log r + \log \frac{2N-4}{p} + o(1) \quad \text{ as } \ r \rightarrow 0. \end{aligned}$$

  From (5.4) we have

  $$\begin{aligned} \log u^*(r) = \frac{1}{p}\log (-2\log r) + o(1) \quad \text{ as } \ r \rightarrow 0. \end{aligned}$$

  Then it follows that

  $$\begin{aligned} u^*(r) ^p = -2\log r + \log \frac{2N-4}{p} -\frac{p-1}{p}\log (-2\log r) + o(1) \quad \text{ as } \ r \rightarrow 0. \end{aligned}$$

  Thus we obtain (1.10).

• (ii) First we consider the case $n = 1$. Since $g_1'(u) = \exp u$, from (5.5) we have

  $$\begin{aligned} e^{u^*(r)} + u^*(r) = -2\log r +\log (2N-4) + o(1) \quad \text{ as } \ r \rightarrow 0. \end{aligned}$$

  From (5.4) we have $u^*(r) = \log (-2\log r) + o(1)$ as $r \rightarrow 0$. Then it follows that

  $$\begin{aligned} e^{u^*(r)} = -2\log r + \log (2N-4) - \log (-2\log r) + o(1) \quad \text{ as } \ r \rightarrow 0. \end{aligned}$$

  Thus we obtain

  $$\begin{aligned} u^*(r) = \log \Bigl (-2\log r + \log (2N-4) - \log (-2\log r) + o(1)\Bigr ) \quad \text{ as } \ r \rightarrow 0, \end{aligned}$$

  and hence (1.12) holds. Next we consider the case $n \ge 2$. Since $g_n'(u)=g_1(u)g_2(u)\cdots g_n(u)$, we have

  $$\begin{aligned} \log g_n'(u)=u+g_1(u)+\cdots +g_{n-1}(u). \end{aligned}$$

  From (5.5) we have

  $$\begin{aligned} u^*(r) + g_1(u^*(r)) + \cdots + g_{n-1}(u^*(r)) + g_n(u^*(r)) = -2\log r +\log (2N-4) + o(1) \end{aligned}$$

  (5.9)

  as $r \rightarrow 0$. From (5.4) it follows that $g_{n-1}(u^*(r)) = \log (-2\log r)+o(1)$ as $r \rightarrow 0$. Then, as $r \rightarrow 0$, we have

  $$\begin{aligned} \begin{array}{rcl} g_{n-2}(u^*(r)) &{} = &{} \log \Bigl (\log (-2\log r) + o(1)\Bigr ) = \log ^2(-2\log r) + o(1), \\ &{} &{} \cdots \\ g_{1}(u^*(r)) &{} = &{} \log ^{n-1}(-2\log r) + o(1), \\ u^*(r) &{} = &{} \log ^{n}(-2\log r) + o(1). \end{array} \end{aligned}$$

  Substituting these equalities into (5.9), we have

  $$\begin{aligned} g_n(u^*(r)) = -2\log r + \log (2N-4) -\sum _{k = 1}^{n}\log ^{k}(-2\log r) + o(1) \quad \text{ as } \ r \rightarrow 0. \end{aligned}$$

  Thus we obtain (1.11).

$\square$

Corollary 1.2 can be proved by the similar argument as in Corollary 1.1 in [24], but we give a sketch of the proof for the convenience of the reader.

Proof of Corollary 1.2

• (i) Since $f(u) > 0$ for $u \ge 0$, $u^*(r)$ has the first zero $r_0^*$. Put

  $$\begin{aligned} \lambda ^* = (r_0^*)^2 \quad \text{ and } \quad v^*(r) = u^*(r_0^* r). \end{aligned}$$

  (5.10)

  Then $v^*(r)$ with $r = |x|$ solves (1.13) with $\lambda = \lambda ^*$. It is clear that $v^*(r) \rightarrow \infty$ as $r \rightarrow 0$. Since the singular solution $u^*$ of (1.1) is unique by Theorem 1.1, $(\lambda ^*, v^*)$ is the unique singular solution of (1.13).

• (ii) Since $f(u) > 0$ for all $u \ge 0$, $u(r, \alpha )$ has the first zero, which we denote by $r_0(\alpha )$. Define $f(u) = f(0)$ for all $u \le 0$, and we extend the domains of $u^*(r)$ and $u(r, \alpha )$ to $(0, r_0^* + \delta ]$ with some $\delta > 0$. Then, from (1.7), we obtain

  $$\begin{aligned} u(r, \alpha ) \rightarrow u^*(r) \quad \text{ in } \ C^2_{\mathrm{loc}}(0, r_0^* + \delta ] \quad \text{ and } \quad r_0(\alpha ) \rightarrow r_0^* \quad \text{ as } \ \alpha \rightarrow \infty . \end{aligned}$$

  (5.11)

  Put

  $$\begin{aligned} \lambda (\alpha ) = r_0(\alpha )^2 \quad \text{ and } \quad v(r, \alpha ) = u(r_0(\alpha )r, \alpha ). \end{aligned}$$

  (5.12)

  Then $(\lambda (\alpha ), v(r, \alpha ))$ solves (1.13) with $\Vert v\Vert _{L^{\infty }(B)} = \alpha$. From (5.10) and (5.12) we have

  $$\begin{aligned} \begin{array}{rcl} |v(r, \alpha )-v^*(r)| &{} = &{} |u(r_0(\alpha )r, \alpha ) -u^*(r_0^* r)| \\ &{} \le &{} |u(r_0(\alpha )r, \alpha )-u^*(r_0(\alpha )r)| + |u^*(r_0(\alpha )r)-u^*(r_0^* r)|. \end{array} \end{aligned}$$

  From (5.11) we obtain $\lambda (\alpha ) \rightarrow \lambda ^*$ and $v(\cdot , \alpha ) \rightarrow v^*$ in $C_{\mathrm{loc}}^2(0, 1]$ as $\alpha \rightarrow \infty$. Thus (ii) holds.

$\square$