Concentration profile, energy, and weak limits of radial solutions to semilinear elliptic equations with Trudinger–Moser critical nonlinearities

Abstract

We investigate the next Trudinger–Moser critical equations,

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta u=\lambda ue^{u^2+\alpha |u|^\beta }&{}\text { in }B,\\ u=0&{}\text { on }\partial B, \end{array}\right. } \end{aligned}$$

where $\alpha >0$, $(\lambda ,\beta )\in (0,\infty )\times (0,2)$ and $B\subset {\mathbb {R}}^2$ is the unit ball centered at the origin. We classify the asymptotic behavior of energy bounded sequences of radial solutions. Via the blow–up analysis and a scaling technique, we deduce the limit profile, energy, and several asymptotic formulas of concentrating solutions together with precise information of the weak limit. In particular, we obtain a new necessary condition on the amplitude of the weak limit at the concentration point. This gives a proof of the conjecture by Grossi et al. (Math Ann, to appear) in 2020 in the radial case. Moreover, in the case of $\beta \le 1$, we show that any sequence carries at most one bubble. This allows a new proof of the nonexistence of low energy nodal radial solutions for $(\lambda ,\beta )$ in a suitable range. Lastly, we discuss several counterparts of our classification result. Especially, we prove the existence of a sequence of solutions which carries multiple bubbles and weakly converges to a sign-changing solution.

1 Introduction

We study the following Trudinger–Moser critical equations,

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta u=\lambda ue^{u^2+\alpha |u|^\beta }&{}\text { in }B,\\ u=0&{}\text { on }\partial B, \end{array}\right. } \end{aligned}$$

(1.1)

where $B\subset {\mathbb {R}}^2$ is the unit ball centered at the origin, $\alpha >0$ and $(\lambda ,\beta )\in (0,\infty )\times (0,2)$. Our aim is to investigate the asymptotic behavior of any energy bounded radial solutions. Especially, we are interested in the concentrating behavior of them.

The study of (1.1) is motivated by the Trudinger–Moser inequality ([33, 36] and [32]). It gives the critical embedding of the Sobolev space in dimension two. Indeed, Moser [32] proves that for any bounded domain $\Omega \subset {\mathbb {R}}^2$, it holds that

$$\begin{aligned} \sup _{u\in H^1_0(\Omega ),\ \int _{\Omega }|\nabla u|^2dx\le 1}\int _{\Omega } e^{a u^2}dx {\left\{ \begin{array}{ll}<\infty \text { if }a\le 4\pi ,\\ =\infty \text { otherwise}. \end{array}\right. } \end{aligned}$$

(1.2)

A surprising fact is that the maximizer exists even in the critical case $\alpha =4\pi$. This is first proved by Carleson-Chang when $\Omega =B$ in their celebrated paper [10]. The generalization of the domain and the dimension are given by [16] and [28] respectively. More recently, the sharp form of the inequality (1.2) is discussed in [26, 35], and [23].

(1.2) suggests that critical nonlinearities of semilinear elliptic problems in dimension two have the exponential growth. This leads us to investigate the problem,

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta u=\lambda h(u)e^{u^2}&{}\text { in }\Omega ,\\ u=0&{}\text { on }\partial \Omega , \end{array}\right. } \end{aligned}$$

(1.3)

where $\lambda >0$, $\Omega \subset {\mathbb {R}}^2$ is a smooth bounded domain and $h:{\mathbb {R}}\rightarrow {\mathbb {R}}$ is a continuous function which has the subcritical growth at infinity, $\lim _{|t|\rightarrow \infty }h(t)/e^{at^2}=0$ for any $a>0$, and satisfies $h(0)=0$ and some appropriate conditions. Due to the lack of the compactness of the critical case of (1.2), solutions to (1.3) can be non-compact. From this point of view, we can regard (1.3) as the two dimensional counterpart of the Brezis-Nirenberg problem [9] in higher dimension. Due to the exponential nonlinearity, which is much stronger than the polynomial one in higher dimension, (1.3) seems to contain new phenomena and difficulties. For example, as discussed in [5] and [11], Palais-Smale sequences for the energy functional of (1.3) may admit more complicated bahavior.

Now, we start our discussion with the previous study of positive solutions of (1.3). For simplicity, we only consider the typical case $h(t)=t$. Certain generalization is given in the following results. First, Adimurthi ([1, 2]) proves that (1.1) admits at least one positive solution for all $\lambda \in (0,\Lambda _1)$ where $\Lambda _1>0$ is the first eigenvalue of $-\Delta$ on $\Omega$ with the Dirichlet boundary condition. After that, in [4, 6] and references therein, the authors investigate the asymptotic behavior of low energy positive solutions. By their results, we see that the least energy solution, obtained in [2], exhibits the single concentration and its full energy converges to $2\pi$ as $\lambda \rightarrow 0$. Moreover, we also observe in those results that the limit profile of the concentration is described by the Liouville equation via a suitable scaling. Furthermore, in [14] and [15], the classification of the asymptotic behavior of energy bounded sequences of positive solutions is accomplished. Especially, in [14], Druet proves that the limit full energy of any sequence is given by the sum of the energy of the weak limit and $2\pi N$ with a number $N\in {\mathbb {N}}\cup \{0\}$. Additionally, in [15], Druet-Thizy show that the concentration occurs if and only if $\lambda \rightarrow 0$. This implies that the weak limit of any concentrating solutions must be zero under their setting. Furthermore, they also give precise information of the location of concentration points and prove that they must be distinct. On the other hand, del Pino-Musso-Ruf [13] obtain a sufficient condition for the existence of sequences of solutions carrying multiple bubbles. It gives a counterpart of the classification result by [14] and [15].

More recently, the concentration behavior of sign-changing solutions is investigated. In this case, in view of the results in [7] and [8], it is reasonable to consider a stronger perturbation $h(t)=te^{|t|^{\beta }}$ with $\beta \in (0,2)$. (We are still considering only the typical case for simplicity.) In fact, Adimrthi-Yadava [8] show that (1.3) admits at least one pair of sign-changing solutions for any $\lambda \in (0,\Lambda _1)$ and $\beta \in (1,2)$. They also prove that if $\Omega =B$, there exist infinitely many radial sign-changing solutions under the same assumption for $\lambda$ and $\beta$. This stronger assumption is essential in the sense of the non-existence result by the same authors [8]. In fact, they prove that if $\Omega =B$ and $\beta \le 1$, there exists a constant $\lambda _{\text {AY}}(\beta )>0$ such that (1.3) permits no radial nodal solutions for all $\lambda \in (0,\lambda _{\text {AY}}(\beta ))$.

Motivated by this result, in [20], the authors attack the blow–up analysis of low energy nodal radial solutions. They investigate the behavior by fixing $\lambda \in (0,\lambda _{\text {AY}}(1))$ and taking the limit $\beta \downarrow 1$. Interestingly, the solutions sequence exhibits the multiple concentration at the origin and weakly converges to a nontrivial sign-definite solution of (1.1) with $\lambda \not =0$ and $\beta =1$. This behavior is very different from that in the positive case explained above. After that Grossi-Mancini-Naimen-Pistoia [19] construct a family of sign-changing solutions which concentrates at a $C^1$ stable critical point of a nontrivial residual mass. We remark that, in their construction, they choose the residual mass to be a sign-definite solution and assume that it is nondegenerate and its amplitude is larger than 1/2 at the concentration point. See (A1) and (A2) in their paper for more precise statements. Moreover, the authors conjecture that the largeness condition on the residual mass is essential for the existence of the concentrating solutions in the limit $\beta \downarrow 1$. See Remark 1.2 in their paper for the detail.

We lastly refer to some interesting results in the radial case. In [29] and [30], the precise asymptotic expansion of concentrating positive solutions is obtained. It allows the proof of the multiplicity and nonexistence of critical points of the Trudinegr-Moser functional in the super critical case. Under another setting, Manicini-Thizy [31] construct concentrating radial solutions which weakly converge to a radial eigenfunction. On the other hand, in two dimensional problem with the power type nonlinearity, Grossi-Grumiau-Pacella [18] find that the singular limit profile appears in the asymptotic behavior of the sign-changing solutions. In higher dimensional problems with nearly Sobolev critical growth, Grossi-Saldaña-Hugo [21] obtain sharp concentration estimates of nodal radial solutions for both of the Dirichlet and Neumann boundary value problems.

In this paper, we proceed with our blow–up analysis of (1.3) and answer some questions raised in the previous works. Moreover, we would like to find new concentration phenomena on (1.3) in the sign-changing case. To this end, inspired by the previous works [20] and [19], we focus on the strong perturbation problem (1.1). Motivated also by [29, 30] and [21], we focus on the radial case and establish explicit and sharp estimates.

More precisely, our first aim is to determine the limit profile and energy of every concentrating sequence. Especially, we would like to answer if the singular limit profile (which was actually observed in the power type problem by [18]) may appear or not in the sign changing case. We have already asked and negatively answered this question in the previous work [20]. But, in the crucial step to avoid the singular limit profile, we used the low energy characterization of the solutions. (See Case 2 in the proof of Lemma 4.3 in [20].) Hence there remains a question if such characterization is essential for the conclusion. In the present paper, we will complete the answer for any energy bounded sequence of radial solutions.

Furthermore, our second goal is to deduce precise information of the weak limit of each concentrating sequence. The first question in this direction arose in the positive case by [15]. They asked if concentrating solutions could weakly converge to a nontrivial solution. Their answer was “no” under their setting as noted above. On the other hand, the result in [20] implies that we may have a different answer in the sign-changing case. Moreover, a new question has arisen in Remark 2.1 of [19] about the relation between the concentration phenomenon and the amplitude of the weak limit at the concentration point. We will give an answer to this question in the radial case.

As a consequence of our new calculation, we arrive at the next classification result with precise concentration estimates. It answers all the questions above.

1.1 Main theorems

Let us give our main results. Throughout this paper, we fix $\alpha >0$ and regard $(\lambda ,\beta )$ as a parameter. Then, for any $(\lambda ,\beta )\in (0,\infty )\times (0,2)$, we define the energy functional associated to (1.1),

$$\begin{aligned} I_{\lambda ,\beta }(u)=\frac{1}{2}\int _B|\nabla u|^2dx-\lambda \int _BF_\beta (u)dx \end{aligned}$$

for all $u\in H_0^1(B)$ where $F_\beta (t)=\int _0^tse^{s^2+\alpha |s|^{\beta }}ds$. Then $I_{\lambda ,\beta }$ is well-defined and $C^1$ functional on $H_0^1(B)$ by (1.2). Furthermore, the standard argument shows that every critical point of $I_{\lambda ,\beta }$ corresponds to each solution of (1.1). Moreover, let ${\mathcal {N}}_{\lambda ,\beta }$ be the Nehari manifold defined by

$$\begin{aligned} {\mathcal {N}}_{\lambda ,\beta }:=\left\{ u\in H^1_0(B)\ | u\text { is radially symmetric and }\langle I_{\lambda ,\beta }'(u),u\rangle =0 \right\} . \end{aligned}$$

Using this, we put for any $k\in \{0\}\cup {\mathbb {N}}$,

$$\begin{aligned} \begin{aligned} c_{k,\lambda ,\beta }:=\inf \Big \{&I_{\lambda ,\beta }(u)\ |\ u\in {\mathcal {N}}_{\lambda ,\beta }, \exists r_0,\cdots , r_{k+1} \in [0,1] \text { such that}\ \\&0=r_0<r_1<\cdots<r_{k+1}=1, \ (0=r_0<r_1=1\text { if }k=0,)\\&u(x)=0\text { if }|x|=r_i,\ u_i:=u|_{\{r_{i-1}<|x|<r_{i}\}},\ (-1)^{i-1}u_i>0,\text { and } \\&u_i\in {\mathcal {N}}_{\lambda ,\beta }\text { (by zero extension), for any }\ 1\le i\le k+1\Big \}. \end{aligned} \end{aligned}$$

In addition, we define the set of radial solutions as follows.

$$\begin{aligned} \begin{aligned} S_{k,\lambda ,\beta }:=\Big \{&u\in C^2(B)\cap C^0({\overline{B}})\text {: a radial solution of }(1.1) | \exists r_0,\cdots , r_{k+1} \in [0,1]\\ {}&\text {such that }0=r_0<r_1<\cdots<r_{k+1}=1,\ (0=r_0<r_1=1\text { if }k=0,)\\&u(x)=0\text { if }|x|=r_i,\ u_i:=u|_{\{r_{i-1}<|x|<r_{i}\}},\ \text {sgn}{(u(0))}(-1)^{i-1}u_i>0,\\&\text {for all }1\le i\le k+1\Big \}. \end{aligned} \end{aligned}$$

We remark that for any $(k,\lambda ,\beta )\in \{0\}\times (0,\Lambda _1)\times (0,2)\cup {\mathbb {N}}\times (0,\Lambda _1)\times (1,2)$, there exists an element $u\in S_{k,\lambda ,\beta }$ such that $I_{\lambda ,\beta }(u)=c_{k,\lambda ,\beta }$ by [2] and [7]. We call any element $u\in S_{k,\lambda ,\beta }$ a nodal radial solution if $k\not =0$ and a positive (negative) solution or a sign-definite solution if $k=0$ and $u(0)>0$ ($<0$ respectively). Note that the result in [17] shows that any element $u \in S_{0,\lambda ,\beta }$ satisfies $\max _{x\in {\overline{B}}}|u(x)|=|u(0)|$. Lastly, for any $k\in {\mathbb {N}}$, let $\Lambda _k$ be the eigenvalue of $-\Delta$ on B with the Dirichlet boundary condition which corresponds to the radial eigenfunction $\varphi _k$ which has numbers $0=\tau _0<\tau _1<\cdots <\tau _k=1$ such that $\varphi _k(x)=0$ if $|x|=\tau _i$ and $(-1)^{i-1}\varphi _k>0$ on $(\tau _{i-1},\tau _i)$ for all $i=1,\cdots ,k$ and is normalized by $\varphi _k(0)=1$. Our main result is the following.

Theorem 1.1

(Classification result) Assume $(k,\lambda _*,\beta _*)\in \{0\}\times [0,\infty )\times (0,2)\cup {\mathbb {N}}\times [0,\infty )\times (0,3/2)$ and let $\{(\lambda _n,\beta _n)\}\subset (0,\infty )\times (0,2)$ be a sequence of values such that $(\lambda _n,\beta _n)\rightarrow (\lambda _*,\beta _*)$ as $n\rightarrow \infty$. Furthermore, we suppose $(u_n)$ is a sequence of solutions of (1.1) such that $u_n\in S_{k,\lambda _n,\beta _n}$ and $u_n(0)>0$ for all $n\in {\mathbb {N}}$. In addition, if $k\not =0$, we assume,

$$\begin{aligned} \int _B|\nabla u_n|^2dx\text { is uniformly bounded for all }n\in {\mathbb {N}}. \end{aligned}$$

Then, after extracting a suitable subsequence if necessary, we have a function $u_0$ and a number $N\in \{0,1,\cdots ,k+1\}$ such that $u_n \rightarrow u_0$ in $C^{2}_{\text {loc}}({\overline{B}}\setminus \{0\})$,

$$\begin{aligned} I_{\lambda _n,\beta _n}(u_n)\rightarrow 2\pi N+I_{\lambda _*,\beta _*}(u_0), \end{aligned}$$

and

$$\begin{aligned} \int _B|\nabla u_n|^2dx\rightarrow 4\pi N+\int _B|\nabla u_0|^2dx, \end{aligned}$$

as $n\rightarrow \infty$. Moreover, if $\max _{x\in {\overline{B}}}|u_n(x)|\rightarrow \infty$ as $n\rightarrow \infty$, then, up to a subsequence, either one of the next assertions holds,

• (i) $\lambda _*=0$, $\beta _n>1$ for all $n\in {\mathbb {N}}$, $N=k+1$, and $u_0=0$, or

• (ii) $\lambda _*=0$, $\beta _n\le 1$ for all $n\in {\mathbb {N}}$, $k=0$, $N=1$, and $u_0=0$, or

• (iii) $\lambda _*\not =0$, $\beta _n\downarrow 1$, $k\not =0$, $0<N<k+1$, and $u_0\in S_{k-N,\lambda _*,1}$ with $(-1)^Nu_0(0)\ge \alpha /2$, or

• (iv) $\lambda _*\not =0$, $\beta _n=1$ for all $n\in {\mathbb {N}}$, $k\not =0$, $N=1$, and $u_0\in S_{k-1,\lambda _*,1}$ with $-u_0(0)=\alpha /2$ or

• (v) $\lambda _*\not =0$, $\beta _n\uparrow 1$, $k\not =0$, $N=1$, and $u_0\in S_{k-1,\lambda _*,1}$ with $0< -u_0(0)\le \alpha /2$, or otherwise

• (vi) $\lambda _*=\Lambda _k$, $\beta _n<1$ for all $n\in {\mathbb {N}}$, $k\not =0$, $N=1$, and $u_0=0$.

On the other hand, if $\max _{x\in {\overline{B}}}|u_n(x)|$ is uniformly bounded for all $n\in {\mathbb {N}}$, then

• (vii) $\lambda _*\not =0$, $N=0$, $u_n \rightarrow u_0$ in $C^{2}({\overline{B}})$, $u_0\in S_{k,\lambda _*,\beta _*}\cup \{0\}$, and further, we get $u_0=0$ only if $\lambda _*= \Lambda _{k+1}$.

In particular, (i) ((ii)) happens if and only if $\lambda _*=0$ ($k=0$ and $\lambda _*=0$ respectively) and (vii) occurs if $k=0$ and $\lambda _*\not =0$, or $k\ge 1$, $\lambda _*\not =0$, and $\beta _*>1$, or $k\ge 1$, $0<\lambda _*\not =\Lambda _k$, and $\beta _*<1$.

Theorem 1.1 implies that the full (Dirichlet) energy of the sequence is decomposed by $2\pi N$ ($4\pi N$ respectively) for a number $N\in \{0,1,\cdots ,k+1\}$ and the energy of the weak limit $u_0$. This is the typical energy quantization phenomenon observed also in the previous works. In our theorem, (i)-(vi) describe the non-compact behavior and (vii) corresponds to the compact one. In the former case, there are three situations. The first one is found in (i) and (ii) which means that the $(k+1)$-concentration occurs with the zero weak limit. This phenomenon happens if and only if $\lambda _*=0$. The second one is observed in (iii), (iv), (v). It shows the $N(<k+1)$-concentration happens with the nontrivial weak limit. (Notice that the weak limit is possibly sign-changing.) This behavior yields $\lambda _*\not =0$ and $\beta _*=1$. Moreover, the sum of the number N of bubbles, and the number $k-N+1$ of nodal domains of the weak limit is always given by $N+(k-N+1)=k+1$. It comes from the fact that, in this case, if we focus on the behavior of the solution on each nodal domain, it weakly converges to zero if and only if it blows up if and only if it exhibits the single concentration. See the next theorem for the detail. The third one is observed in (vi). This means that the $N(=1<k+1)$-concentration occurs with the zero weak limit. This behavior requires $\lambda _*=\Lambda _k$ and $\beta _n<1$ for all $n\in {\mathbb {N}}$. As we will see in the next theorems, in this case, only the local maximum value $u_n(0)$ at the origin diverges to infinity and the other ones converge to zero.

Moreover, a remarkable result is found in the final assertions in (iii)-(vi). It gives the necessary condition on the amplitude $|u_0(0)|$ of the weak limit at the origin. Especially, (iii) gives a proof of the conjecture in Remark 1.2 of [19] in the radial case. It ensures that if the concentration occurs in the limit $\lambda _n\rightarrow \lambda _*\not =0$ and $\beta _n\downarrow 1$, then $|u_0(0)|$ needs to be greater than or equal to $\alpha /2$. Notice that our necessary condition is valid in any case of $\lambda _*>0$ as far as the concentration occurs as $\beta _n\downarrow 1$ while the previous results in [20] and [19] focus on the case of small $\lambda _*>0$. Moreover, in the cases $\beta _n=1$ for all $n\in {\mathbb {N}}$, $\beta _n\uparrow 1$, and $\beta _*<1$, we deduce new necessary conditions, $|u_0(0)|=\alpha /2$, $0\le |u_0(0)|\le \alpha /2$, and $u_0(0)=0$ respectively. (A related result is observed for a radial positive sequence in another setting by Theorem 0.3 in [31].) These conditions will be useful to detect new concentrating sequences of solutions. See Sect. 6 for more discussion. We will discuss several counterparts of assertions in the previous theorem later.

In addition, we remark that there is a striking difference between the cases $(\beta _n)\subset (1,2)$ ((i) and (iii)) and $(\beta _n)\subset (0,1]$ ((ii), (iv), (v), and (vi)). In the former case, N can be greater than 1 while in the latter case it has to be equal to one. This is a consequence of our blow-up analysis in Sects. 2 and 3. See Remark 3.4 for more explanation. Then, we notice that Theorem 1.1 contains the following nonexistence result.

Corollary 1.2

For any number $k\in {\mathbb {N}}$ and sequence $\{(\lambda _n,\beta _n)\}\subset (0,\infty )\times (0,1]$ such that $(\lambda _n,\beta _n)\rightarrow (\lambda _*,\beta _*)\in \{0\}\times (0,1]$, there exists no sequence of solutions $(u_n)$ such that $u_n\in S_{k,\lambda _n,\beta _n}$ and $\int _B|\nabla u_n|^2dx$ is uniformly bounded for all $n\in {\mathbb {N}}$.

This conclusion allows a partial proof of the nonexistence result by [8] via a different approach.

Corollary 1.3

Choose any $\beta \in (0,1]$. Then for each value $E>0$, there exists a constant ${\hat{\lambda }}={\hat{\lambda }}(\beta ,E)>0$ such that for all $\lambda \in (0,{\hat{\lambda }})$, (1.1) admits no radial nodal solution u with $\int _B|\nabla u|^2dx\le E$. In particular, for each number $k\in {\mathbb {N}}$, there is a value ${\hat{\lambda }}_0={\hat{\lambda }}_0(\beta ,k)>0$ such that for all $\lambda \in (0,{\hat{\lambda }}_0)$, there exists no solution $u\in S_{k,\lambda ,\beta }$ which attains $c_{k,\lambda ,\beta }$.

We finally remark on the additional condition on $\beta _*$ in the previous theorem.

Remark 1.4

In the case of $k\ge 1$, we additionally assumed $\beta _*<3/2$. This condition will first appear in (3.4) of Lemma 3.3 below. This does not seem simply a technical assumption. As discussed in Remark 3.5, in the case $k\ge 1$ and $\beta _*\ge 3/2$, we would have different formulas of the asymptotic energy expansion in Theorem 1.5 and also the concentration estimates in Theorem 1.6 below. Since the proof for the former case has already used many pages, we leave the latter case for our next works. We here restrict ourselves to only conjecture that, in the sign-changing case, the stronger perturbation ($\beta \ge 3/2$ in (1.1)) would delicately affect the precise asymptotic formulas of concentrating solutions.

Next, we shall check the detail of the behavior stated above. The previous theorem is a direct consequence of the next two theorems.

Theorem 1.5

(Limit profile and energy) Let values $k,\lambda _*,\beta _*$ and sequences $\{(\lambda _n,\beta _n)\}$, $(u_n)$ be chosen as in the assumption of Theorem 1.1 and assume $u_0\in H^1_0(B)$ is the weak limit of $(u_n)$ by extracting a subsequence if necessary. Furthermore, we write $u_n=u_n(|x|)$ and $u_0=u_0(|x|)$ for $x\in {\overline{B}}$ and then define values $0=r_{0,n}<r_{1,n}<\cdots <r_{k+1,n}=1$ ($0=r_{0,n}<r_{1,n}=1$ if $k=0$) so that $u_n(r_{i,n})=0$ and $(-1)^{i-1}u_n(r)>0$ if $r_{i-1,n}<r<r_{i,n}$ for all $1\le i\le k+1$. In addition, we set $u_{i,n}=u_n|_{[r_{i-1,n},r_{i,n}]}$ and define numbers $\rho _{i,n}\in [r_{i-1,n},r_{i,n})$ and $\mu _{i,n}>0$ by $\mu _{i,n}=|u_{i,n}(\rho _{i,n})|=\max _{r\in [r_{i-1,n},r_{i,n}]}|u_{i,n}(r)|$ for all $1\le i\le k+1$. Finally for each number $i=1,\cdots ,k+1$ such that $\mu _{i,n}\rightarrow \infty$, we set $\gamma _{i,n}>0$ so that

$$\begin{aligned} 1=2\lambda _n \mu _{i,n}f_n(\mu _{i,n})\gamma _{i,n}^2 \end{aligned}$$

where $f_n(t)=te^{t^2+|t|^{\beta _n}}$, and define

$$\begin{aligned} z_{i,n}(r):=2\mu _{i,n}(|u_{i,n}(\gamma _{i,n} r+\rho _{i,n})|-\mu _{i,n}) \text { for all } r\in \left[ \frac{r_{i-1,n}-\rho _{i,n}}{\gamma _{i,n}},\frac{r_{i,n}-\rho _{i,n}}{\gamma _{i,n}}\right] , \end{aligned}$$

and $z(r):=\log {(64/(8+r^2)^2)}$ which is a solution of the Liouville equation

$$\begin{aligned} -z''-\frac{1}{r}z'=e^z\text { in }(0,\infty ),\ z(0)=0=z'(0),\ \int _0^\infty e^zrdr<\infty . \end{aligned}$$

Then if $\max _{r\in [0,1]}|u_n(r)|\rightarrow \infty$, either one of the next assertions (i) and (ii) holds up to a subsequence.

• (i) For all $i=1,\cdots ,k+1$, we have $\mu _{i,n}\rightarrow \infty$, $\rho _{k+1,n}\rightarrow 0$, $\gamma _{i,n}\rightarrow 0$, $z_{i,n}\rightarrow z\text { in }C^2_{\text {loc}}((0,\infty ))\cap C^1_{\text {loc}}([0,\infty ))$,

  $$\begin{aligned}&\int _{r_{i-1,n}}^{r_{i,n}}u_{i,n}'(r)^2rdr= 2-\frac{\alpha \beta _*}{\mu _{i,n}^{2-\beta _n}}+o\left( \frac{1}{\mu _{i,n}^{2-\beta _n}}\right) , \end{aligned}$$

  (1.4)
  $$\begin{aligned}&\mu _{i,n}\int _{r_{i-1,n}}^{r_{i,n}}\lambda _n f_n(u_{i,n})rdr= 2-\frac{\alpha \beta _*}{\mu _{i,n}^{2-\beta _n}}+o\left( \frac{1}{\mu _{i,n}^{2-\beta _n}}\right) , \end{aligned}$$

  (1.5)

  and further, $\lambda _*=0$ and $u_n\rightarrow u_0=0$ in $C^2_{\text {loc}}((0,1])$.

• (ii) There exists a number $N\in \{1,\cdots ,k\}$ such that for all $i=1,\cdots ,N$, we have $\mu _{i,n}\rightarrow \infty$, $r_{i,n}\rightarrow 0$, $\gamma _{i,n}\rightarrow 0$, $z_{i,n}\rightarrow z\text { in }C^2_{\text {loc}}((0,\infty ))\cap C^1_{\text {loc}}([0,\infty ))$, (1.4), and (1.5), while for all $i=N+1,\cdots ,k+1$, there exist values $\mu _{i}\ge 0$, $r_i\in (0,1]$ and $\rho _i\in [0,1)$ such that $\mu _{i,n}\rightarrow \mu _{i}$, $r_{i,n}\rightarrow r_i$, $\rho _{i,n}\rightarrow \rho _i$, and $0=\rho _{N+1}<r_{N+1}<\cdots<\rho _{k+1}<r_{k+1}=1$ if $N<k$ and $0=\rho _{k+1}<r_{k+1}=1$ if $N=k$, and further, it holds that $\lambda _*\not =0$, $u_n|_{[r_{N,n},1]}\rightarrow u_0$ in $C^2_{\text {loc}}((0,1])$, $\lim _{r\rightarrow 0^+}(-1)^Nu_0(r)=\mu _{N+1}$, and

  $$\begin{aligned} \int _{r_{N,n}}^{1}u_{n}'(r)^2rdr\rightarrow \int _{0}^{1}u_0'(r)^2rdr. \end{aligned}$$

  Moreover, either one of the next assertions holds,

  • (a) $\mu _{N+1}>0$, $u_0(r_i)=0$, and $(-1)^{i-1}u_0>0$ on $(r_{i-1},r_i)$, for all $i=N+1,\cdots ,k+1$, or

  • (b) $\mu _{N+1}=0$, $u_0=0$, and further, $u_n|_{[r_{N,n},1]}/\mu _{N+1}\rightarrow (-1)^N\varphi _{k-N+1}$ in $C^2_{\text {loc}}((0,1])$ and $\lambda _*=\Lambda _{k-N+1}$.

On the other hand, if $\max _{r\in [0,1]}|u_n(r)|$ is uniformly bounded, choosing a subsequence if necessary, we get that,

• (iii) for any $i=1,\cdots ,k+1$, there exist values $\mu _i\ge 0$, $r_i\in (0,1]$ and $\rho _i\in [0,1)$ such that $\mu _{i,n}\rightarrow \mu _i$, $r_{i,n}\rightarrow r_{i}$, $\rho _{i,n}\rightarrow \rho _i$, and $0= \rho _1<r_1<\cdots<\rho _{k+1}<r_{k+1}=1$ if $k\ge 1$, and further, it holds that $\lambda _*\not =0$, $u_n\rightarrow u_0$ in $C^2([0,1])$, and

  $$\begin{aligned} \int _{0}^{1}u_{n}'(r)^2rdr\rightarrow \int _0^{1}u_0'(r)^2rdr. \end{aligned}$$

  In addition, either one of the next assertions is true,

  • (a) $u_0(0)>0$, $u_0(r_i)=0$, and $(-1)^{i-1}u_0>0$ on $(r_{i-1},r_i)$, for all $i=1,\cdots ,k+1$, or

  • (b) $u_0=0$, $u_n/u_{n}(0)\rightarrow \varphi _{k+1}$ in $C^2([0,1])$, and $\lambda _*=\Lambda _{k+1 }$.

This theorem shows the behavior on every part $u_{i,n}$ between neighboring two zero points $r_{i-1,n}<r_{i,n}$. From the behavior in (i) and (ii), we see that if $u_{i,n}$ blows up, it concentrates at the origin. Especially, the local maximum point $\rho _{i,n}$ converges to the origin. Actually, this is the reasonable and the only way for any blowing up solution to ensure the uniform boundedness of the energy. Furthermore, the limit profile is determined uniquely by the classical solution z of the Liouville equation and the limit energy is just equal to 2. This implies that neither the singular limit profile, observed in the power type problem in [18], nor the multiple concentration, occurs on any $u_{i,n}$.

Notice also that due to our strong perturbation, the second term of the right hand side of the energy expansion in (i) and (ii) is very different from that in the case of $\alpha =0$ in view of its sign and the exponent on $\mu _{i,n}$. (See Theorem 1 in [30].)

Finally, we obtain precise concentration estimates in terms of $(\lambda _n,\beta _n)$.

Theorem 1.6

(Concentration estimates) Under the same assumptions as in Theorem 1.5, let us assume that (i) or (ii) of the same theorem occurs. First suppose (i) happens. Then we have that $(\beta _n)\subset (0,1]$ yields $k=0$. Moreover, if $k\in \{0\}\cup {\mathbb {N}}$, we get that

$$\begin{aligned}&\lim _{n\rightarrow \infty }\frac{\log {\frac{1}{\lambda _n}}}{\mu _{k+1,n}^{\beta _n}}=\delta , \end{aligned}$$

(1.6)

where we defined the constant $\delta =\delta (\alpha ,\beta _*)$ by $\delta =\alpha (1-\beta _*/2)$ and if $k\ge 1$, we obtain for all $1\le i\le k$ that

$$\begin{aligned}&\lim _{n\rightarrow \infty }\frac{\log {\frac{1}{\lambda _n}}}{\mu _{i,n}^{\beta _n(\beta _n-1)^{k-i+1}}}=\delta ^{\frac{2-\beta _*(\beta _*-1)^{k-i+1}}{2-\beta _*}}, \end{aligned}$$

(1.7)

Furthermore, if $k\in \{0\}\cup {\mathbb {N}}$, we have that

$$\begin{aligned}&\lim _{n\rightarrow \infty }\left( \log {\frac{1}{\lambda _n}}\right) ^{\frac{1}{\beta _n}}|u_{k+1,n}'(1)|=2\delta ^{\frac{1}{\beta _*}}, \end{aligned}$$

(1.8)

and if $k\ge 1$, we get for all $1\le i\le k$ that

$$\begin{aligned} \begin{aligned} \lim _{n\rightarrow \infty }&\frac{\log {\frac{1}{\lambda _n}}}{\left( \log {\frac{1}{r_{i,n}}}\right) ^{(\beta _n-1)^{k-i+1}}}=2^{(\beta _*-1)^{k-i+1}}\delta ^{\frac{2-2(\beta _*-1)^{k-i+1}}{2-\beta _*}}, \end{aligned} \end{aligned}$$

(1.9)

and

$$\begin{aligned} \begin{aligned} \lim _{n\rightarrow \infty }&\frac{\log {\frac{1}{\lambda _n}}}{(\log {|u_{i,n}'(r_{i,n})|})^{(\beta _n-1)^{k-i+1}}}=2^{(\beta _*-1)^{k-i+1}}\delta ^{\frac{2-2(\beta _*-1)^{k-i+1}}{2-\beta _*}}. \end{aligned} \end{aligned}$$

(1.10)

In addition, if $k\ge 2$, we have for all $i=2,\cdots ,k$, that

$$\begin{aligned} \begin{aligned} \lim _{n\rightarrow \infty }&\frac{\log {\frac{1}{\lambda _n}}}{\left( \log {\frac{1}{\rho _{i,n}}}\right) ^{\frac{\beta _n(\beta _n-1)^{k-i+1}}{2}}}=2^{\frac{\beta _*(\beta _*-1)^{k-i+1}}{2}}\delta ^{\frac{2-\beta _*(\beta _*-1)^{k-i+1}}{2-\beta _*}}. \end{aligned} \end{aligned}$$

(1.11)

Lastly, suppose $k\ge 1$. Then we may assume that $\beta _n>1$ for all $n\in {\mathbb {N}}$ up to a subsequence (by the first conclusion of this theorem). Moreover, we define a number $L\in [0,\infty ]$ by

$$\begin{aligned} L:=\lim _{n\rightarrow \infty }\frac{\log {\log {\frac{1}{\lambda _n}}}}{(\beta _n-1)\left( \log {\frac{1}{\lambda _n}}\right) ^{\frac{2}{\beta _n}}}. \end{aligned}$$

(1.12)

Then if $L<\infty$, we get

$$\begin{aligned} \begin{aligned} \lim _{n\rightarrow \infty }\frac{\log {\frac{1}{\lambda _n}}}{\left( \log {\frac{1}{\rho _{k+1,n}}}\right) ^{\frac{\beta _n}{2}}}&=2^{\frac{\beta _*}{2}}\delta \left[ 1+L\delta ^{\frac{2}{\beta _*}}\right] ^{-\frac{\beta _*}{2}}, \end{aligned} \end{aligned}$$

(1.13)

and, on the other hand, if $L=\infty$, we necessarily have $\beta _*=1$ and obtain

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{\log {\log {\frac{1}{\lambda _n}}}}{(\beta _n-1)\log {\frac{1}{\rho _{k+1,n}}}}=2. \end{aligned}$$

(1.14)

Next, we suppose (ii) occurs. Then we get $\beta _*\le 1$ and that $(\beta _n)\subset (0,1]$ implies $N=1$. Moreover, if $1\le N\le k$, either one of the next assertions (a) and (b) is true.

• (a) $\mu _{N+1}>0$, $\beta _*=1$, and for all $1\le i\le N$, it holds that

  $$\begin{aligned}&\lim _{n\rightarrow \infty }\mu _{i,n}^{(\beta _n-1)^{N-i+1}}=\frac{2\mu _{N+1}}{\alpha }\left( =\frac{\mu _{N+1}}{\delta }\right) , \end{aligned}$$

  (1.15)
  $$\begin{aligned}&\lim _{n\rightarrow \infty }\left( \log {\frac{1}{r_{i,n}}}\right) ^{(\beta _n-1)^{N-i+1}}=\frac{2\mu _{N+1}}{\alpha }, \end{aligned}$$

  (1.16)
  $$\begin{aligned}&\lim _{n\rightarrow \infty }\left( \log {|u_{i,n}'(r_{i,n})|}\right) ^{(\beta _n-1)^{N-i+1}}=\frac{2\mu _{N+1}}{\alpha }, \end{aligned}$$

  (1.17)

  and

  $$\begin{aligned}&\lim _{n\rightarrow \infty }u_{N+1,n}'(r_{N+1,n})=-\frac{1}{r_{N+1}}\lambda _* \int _0^{r_{N+1}}f_*(u_0)rdr, \end{aligned}$$

  (1.18)
  $$\begin{aligned}&\lim _{n\rightarrow \infty }\rho _{N+1,n}^{\beta _n-1}= \sqrt{\frac{\alpha }{2\mu _{N+1}}}, \end{aligned}$$

  (1.19)

  where $f_*(t)=te^{t^2+\alpha |t|}$. Especially, $2\mu _{N+1}/\alpha >1$ ($\in (0,1)$) implies $\beta _n>1$ ($<1$ respectively) for all $n\in {\mathbb {N}}$. On the other hand, $\beta _n>1$ ($=1$, $<1$) for all $n\in {\mathbb {N}}$ requires $2\mu _{N+1}/\alpha \ge 1$ ($=1$, $\le 1$ respectively). Finally, if $1<N\le k$, (which yields $k\ge 2$, $\beta _n>1$ for all $n\in {\mathbb {N}}$, and $2\mu _{N+1}/\alpha \ge 1$), assuming $2\mu _{N+1}/\alpha >1$, we get for all $2\le i\le N$,

  $$\begin{aligned}&\lim _{n\rightarrow \infty }\left( \log {\frac{1}{\rho _{i,n}}}\right) ^{(\beta _n-1)^{N-i+1}}=\left( \frac{2\mu _{N+1}}{\alpha }\right) ^{2}. \end{aligned}$$

  (1.20)

• (b) $\mu _i=0$ for all $i=N+1,\cdots ,k+1$ and $\beta _n<1$ for all $n\in {\mathbb {N}}$.

Remark 1.7

We assumed $2\mu _{N+1}/\alpha >1$ for (1.20). This can be verified, for example, when $N=k$ and $\lambda _*\in (0,\Lambda _1)$ is small enough by Lemma 6.8 below.

This theorem describes the speed of the divergence or convergence of $\mu _{i,n}$, $r_{i,n}$, $\rho _{i,n}$ and $u_{i,n}'(r_{i,n})$ in terms of the parameter $(\lambda _n,\beta _n)$. Especially, in the case of (i), thanks to (1.6) and (1.7), we get that $\mu _{k+1,n}=(\alpha (1-\beta _*/2)+o(1))^{-1/\beta _*} (\log {(1/\lambda _n)})^{1/\beta _n}$ and, if $k\ge 1$, that $\mu _{k+1,n}/\mu _{i,n}\rightarrow 0$ as $n\rightarrow \infty$ for all $i=1,\cdots ,k$. Then, combining this together with the asymptotic energy formula in Theorem 1.5, we can also write the energy expansion in terms of $(\lambda _n,\beta _n)$ as follows.

Corollary 1.8

Assume as in Theorem 1.1 and suppose (i) or (ii) of the theorem occurs. Then we get

$$\begin{aligned} \int _B|\nabla u_n|^2dx=4\pi (k+1)-\frac{2\pi \alpha ^{\frac{2}{\beta _*}}\beta _*\left( 1-\frac{\beta _*}{2}\right) ^{\frac{2-\beta _*}{\beta _*}}}{\left( \log \frac{1}{\lambda _n}\right) ^{\frac{2-\beta _n}{\beta _n}}}+o\left( \frac{1}{{\left( \log \frac{1}{\lambda _n}\right) ^{\frac{2-\beta _n}{\beta _n}}}}\right) \end{aligned}$$

as $n\rightarrow \infty$.

Moreover, in the case of (i), we observe a delicate behavior when $(\lambda _*,\beta _*)=(0,1)$ by the formulas (1.13) and (1.14). They show that the asymptotic behavior $\rho _{k+1,n}\rightarrow 0$ of the local maximum point is described by either one of two different formulas (1.13) and (1.14). The choice is determined by the balance of the speed of two limits $\lambda _n\rightarrow 0$ and $\beta _n\rightarrow 1$. If former one is quicker than the latter one in the sense $L<\infty$ where L is the number defined by (1.12), we have (1.13) and otherwise we get (1.14). Actually, in the latter formula, the effect of the limit $\beta _n\rightarrow 1$ appears more clearly. This phenomenon comes from the combined effect of the two different behavior, the $(k+1)$-concentration with the zero weak limit in the case $\lambda _n\rightarrow 0$, and the k-concentration with the nontrivial weak limit in the case $0<\lambda _*\ll 1$ and $\beta _*=1$ (which is observed in [20] and (b) of Proposition 6.3 below). Notice also that for any $L\in [0,\infty ]$, there exist sequences $(\lambda _n)\subset (0,\Lambda _1)$ and $(\beta _n)\subset (1,2)$ which satisfy (1.12). Moreover, there exists a corresponding sequence $(u_n)$ of solutions by Theorem 1.3 in [7].

Finally, in the case of (ii), one of the most important results is (1.15). This proves the necessary condition, explained before, in the final assertions in (iii)-(v) of Theorem 1.1.

In the following sections, we give the proof of our main theorems.

1.2 Outline of the proof

We carry out the blow–up analysis based on a scaling technique. We begin with studying the limit profile of the concentration part as in [20]. The first difficulty arises here since we do not have the low energy characterization of solutions (Lemma 2.1 in [20]). In order to admit our wider setting, we change the proof and argue by induction. Our idea is to use useful assumptions ((2.4) and (2.5)) which are ensured by the previous step of the induction argument (Proposition 3.10). Using this idea, we succeed in avoiding the case of the singular limit profile. See the proof of Proposition 2.4.

Next, we will determine the energy of each concentrating part as in Theorem 1.5. An important step is to ensure that only the single concentration occurs in each nodal domain. The pointwise estimate in Lemma 3.7 will work for it. This is an extension of Lemma 5 in [29] (see also Lemma 13 in [30]) to our setting which allows the strong perturbation and also the sign-changing case. Using this and arguing as in Proof of Theorem 1 in [30], we obtain the energy expansion in Proposition 3.1. To accomplish these proofs, some careful remarks are needed. In particular, the estimate (3.4) for the error term is a key for the proof in the sign-changing case.

Finally, we will obtain the precise concentration estimate in Theorem 1.6. The proof is inspired by the argument in [21] for the power type problem in higher dimension. A new difficulty comes from the fact that, of course, the form of the nonlinearity is very different. In particular, the Pohozaev identity does not seem work well for our aim. In our proof, utilizing the useful identity in Lemma 2.2, instead of the Pohozaev identity, with the energy expansion in Proposition 3.1, we get the key assertions in Proposition 3.10. This is also crucial to proceed with our argument by induction.

We lastly remark that our approach mentioned above allows the proof without quoting the uniqueness of solutions which has not been completed for (1.1) yet except for large positive solutions ([3]).

1.3 Organization, definitions and notations

This paper consists of 6 sections. We begin with two sections, Sects.  2 and 3, which are mainly devoted to obtain the limit equation and the limit energy of concentrating solutions respectively. Next in Sect. 4, we analyze the behavior of non-concentrating parts of solutions. This is important to deduce the precise information of the weak limit of solutions. Next, in Sects. 5, we complete the proof of main theorems. Finally, in Sect. 6, we discuss some counterparts of our classification result. Additionally, the proof of Lemma 2.5 is given in Appendix 1 for the readers’ convenience.

Throughout these sections, we assume $\{(\lambda _n,\beta _n)\}\subset (0,\infty )\times (0,2)$, $(\lambda _*,\beta _*)\in [0,\infty )\times (0,2)$ and $(\lambda _n,\beta _n)\rightarrow (\lambda _*,\beta _*)$ as $n\rightarrow \infty$. We will impose more conditions on $\lambda _n,\beta _n,\lambda _*,\beta _*$ when needed. Moreover, we choose any $k\in \{0\}\cup {\mathbb {N}}$ and consider a sequence of solutions $(u_n)$ such that $u_n\in S_{k,\lambda _n,\beta _n}$ for all $n\in {\mathbb {N}}$. We set $f_n(t)=te^{t^2+\alpha |t|^{\beta _n}}$ and $f_*(t)=te^{t^2+\alpha |t|^{\beta _*}}$.

Furthermore, we define the norm in $H^1_0(B)$ by $\Vert \cdot \Vert _{H^1_0(B)}:=\left( \int _B|\nabla \cdot |^2dx\right) ^{1/2}$. Moreover, we denote the first kind Bessel function of order zero by $J_0$ which is defined by

$$\begin{aligned} J_0(r):=\sum _{j=0}^{\infty }\frac{(-1)^j}{(j!)^2}\left( \frac{r}{2}\right) ^{2j} \ (r\in {\mathbb {R}}). \end{aligned}$$

For any $k\in {\mathbb {N}}$, let $\Lambda _k$ and $\varphi _k$ be the eigenvalue and radial eigenfunction of $-\Delta$ on B with the Dirichlet boundary condition defined above. Then letting $0<t_1<t_2<\cdots$ be all the zeros of $J_0$ on $(0,\infty )$, we have that $\Lambda _k=t_k^2$ and $\varphi _k(x)=J_0(t_k |x|)$.

Finally, in the proofs, we often use the same character C to denote several constants when the explicit value is not very important.

2 Limit profile

Let us start the proof of main theorems. In the following, we refer to Radial Lemma in [34]. In our two dimensional setting, it is reduced to the following.

Lemma 2.1

([34]) There exists a constant $c>0$ such that every radial function $u\in H_0^1(B)$ is almost everywhere equal to a function ${\tilde{u}}(x)$, continuous for $x\not =0$, such that

$$\begin{aligned} |{\tilde{u}}(x)|\le c|x|^{-\frac{1}{2}}\Vert u\Vert _{H^1_0(B)}\ (x\in B\setminus \{0\}). \end{aligned}$$

Now, assume $k\in \{0\}\cup {\mathbb {N}}$. For any $1\le i\le k+1$, we define $u_{i,n}$, $r_{i-1,n}$, $r_{i,n}$, $\rho _{i,n}$, and $\mu _{i,n}$ as in Theorem 1.5. Then we have

$$\begin{aligned} {\left\{ \begin{array}{ll} -u_{i,n}''-\frac{1}{r}u_{i,n}'=\lambda _n f_n(u_{i,n}),\ (-1)^{i-1}u_{i,n}>0\text { in }(r_{i-1,n},r_{i,n}),\\ u_{i,n}(r_{i,n})=0=u_{i,n}'(\rho _{i,n}),\\ u_{i,n}(r_{i-1,n})=0\text { if }i\ge 2, \end{array}\right. } \end{aligned}$$

(2.1)

where $f_n(t):=te^{t^2+\alpha |t|^{\beta _n}}$. We often use the next identity.

Lemma 2.2

For any $i=1,\cdots ,k+1$, we have

$$\begin{aligned} u_{i,n}(\rho _{i,n})=\int _{\rho _{i,n}}^{r_{i,n}}\lambda _nf_n(u_{i,n})r\log {\frac{r_{i,n}}{r}}dr. \end{aligned}$$

Moreover, if $i\not =1$, we get

$$\begin{aligned} u_{i,n}(\rho _{i,n})=\int _{r_{i-1,n}}^{\rho _{i,n}}\lambda _nf_n(u_{i,n})r\log {\frac{r}{r_{i-1,n}}}dr \end{aligned}$$

Proof

Let us show the first formula. Fix any $i=1,\cdots ,k+1$. Multiplying the equation in (2.1) by $r\log {r}$, and integrating by parts from $\rho _{i,n}$ to $r_{i,n}$, we get

$$\begin{aligned} \begin{aligned} \int _{\rho _{i,n}}^{r_{i,n}}\lambda _n f_n(u_{i,n})r\log {r}dr&=\int _{\rho _{i,n}}^{r_{i,n}}(-ru_{i,n}'(r))'\log {r}dr\\&=\int _{\rho _{i,n}}^{r_{i,n}}\lambda _n f_n(u_{i,n})rdr\log {r_{i,n}}-u_{i,n}(\rho _{i,n}), \end{aligned} \end{aligned}$$

where we used $r_{i,n}u_{i,n}'(r_{i,n})=-\int _{\rho _{i,n}}^{r_{i,n}}\lambda _n f_n(u_{i,n})rdr$ for the last equality. This shows the first formula. Assuming $i\not =1$, the second assertion is similarly obtained by integrating by parts from $r_{i-1,n}$ to $\rho _{i,n}$. $\square$

We also use the next assertion.

Lemma 2.3

If $\int _{r_{i-1,n}}^{r_{i,n}}u_{i,n}'(r)^2rdr\rightarrow 0$, then $u_{i,n}\rightarrow 0$ in C([0, 1]). In particular, if $\limsup _{n\rightarrow \infty } \mu _{i,n}>0$, then by taking a subsequence if necessary, we get a constant $K_0>0$ such that $\int _{r_{i-1,n}}^{r_{i,n}}u_{i,n}'(r)^2rdr\ge K_0$ for all $n\in {\mathbb {N}}$.

Proof

We put $A_n^2:=\int _{r_{i-1,n}}^{r_{i,n}}u_{i,n}'(r)^2rdr$. Then, from Lemma 2.2 and the Hölder inequality, we get a constant $C>0$ such that

$$\begin{aligned} \begin{aligned} \mu _{i,n}&\le \left| \int _{\rho _{i,n}}^{r_{i,n}}\lambda _nf(u_n)r\log {\frac{r_{i,n}}{r}}dr\right| \\&\le C\left( \int _{\rho _{i,n}}^{r_{i,n}} u_{i,n}^4rdr\right) ^{\frac{1}{4}}\left( \int _{\rho _{i,n}}^{r_{i,n}} e^{4(1+\alpha )2\pi A_n^2\left( \frac{u_{i,n}}{\sqrt{2\pi }A_n}\right) ^2} rdr\right) ^{\frac{1}{4}} \\&\ \ \ \ \times \left( \int _{\rho _{i,n}}^{r_{i,n}}r\log ^2{\frac{1}{r}}dr\right) ^\frac{1}{2} \end{aligned} \end{aligned}$$

Then noting $A_n\rightarrow 0$, we use the Trudinger–Moser (1.2) and Sobolev inequalities to obtain that the right hand side converges to zero. This finishes the proof.$\square$

Let us begin our main discussion of this section. We study the limit equation of concentrating solutions. To this end, we fix a number $1\le i\le k+1$ and suppose

$$\begin{aligned} \mu _{i,n}\rightarrow \infty \text { as }n\rightarrow \infty . \end{aligned}$$

(2.2)

Moreover, if $i\not =1$, we also assume that

$$\begin{aligned}&\sup _{n\in {\mathbb {N}}}\int _{r_{i-1,n}}^{r_{i,n}}u_{i,n}'(r)^2rdr<\infty , \end{aligned}$$

(2.3)

$$\begin{aligned}&\lim _{n\rightarrow \infty }\frac{\log {\frac{1}{\lambda _nr_{i-1,n}^2}}}{\mu _{i-1,n}^{\beta _n}}=\delta , \end{aligned}$$

(2.4)

where $\delta =\delta (\alpha ,\beta _*)$ is a positive number defined after (1.6) of Theorem 1.6 and

$$\begin{aligned}&\lim _{n\rightarrow \infty }\mu _{i-1,n}r_{i-1,n} |u_{i-1,n}'(r_{i-1,n})|=2. \end{aligned}$$

(2.5)

Our goal is to prove the following.

Proposition 2.4

Assume (2.2)–(2.5), put $\gamma _{i,n}>0$ so that

$$\begin{aligned} 1=2\lambda _n \mu _{i,n} f_n(\mu _{i,n})\gamma _{i,n}^2, \end{aligned}$$

and define

$$\begin{aligned} z_{i,n}(r):=2\mu _{i,n}(|u_{i,n}(\gamma _{i,n} r+\rho _{i,n})|-\mu _{i,n})\ \left( r\in \left[ \frac{r_{i-1,n}-\rho _{i,n}}{\gamma _{i,n}},\frac{r_{i,n}-\rho _{i,n}}{\gamma _{i,n}}\right] \right) , \end{aligned}$$

for all $n\in {\mathbb {N}}$. Then we have $\gamma _{i,n}\rightarrow 0$, $(r_{i-1,n}-\rho _{i,n})/\gamma _{i,n}\rightarrow 0$, $(r_{i,n}-\rho _{i,n})/\gamma _{i,n}\rightarrow \infty$, and further, $z_{i,n}\rightarrow z$ in $C^2_{\text {loc}}((0,\infty ))\cap C^1_{\text {loc}}([0,\infty ))$ where

$$\begin{aligned} z(r)=\log {\frac{64}{(8+r^2)^2}} \end{aligned}$$

(2.6)

which satisfies

$$\begin{aligned} -z''-\frac{1}{r}z'=e^z\text { in }(0,\infty ),\ z(0)=0=z'(0) \text { and }\int _0^\infty e^zrdr=4. \end{aligned}$$

(2.7)

Before staring the proof, note that $z_{i,n}$ satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} -z_{i,n}''-\frac{1}{r+\frac{\rho _{i,n}}{\gamma _{i,n}}}z_{i,n}'=\left( \frac{z_{i,n}}{2\mu _{i,n}^2}+1\right) \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \times e^{z_{i,n}+\frac{z_{i,n}^2}{4\mu _{i,n}^2}+\alpha \mu _{i,n}^{\beta _{n}}\left\{ \left( \frac{z_{i,n}}{2\mu _{i,n}^2}+1\right) ^{\beta _n}-1\right\} },\\ z_{i,n}\le 0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \hbox { in }\left( \frac{r_{i-1,n}-\rho _{i,n}}{\gamma _{i,n}},\frac{r_{i,n}-\rho _{i,n}}{\gamma _{i,n}}\right) ,\\ z_{i,n}(0)=0=z_{i,n}'(0),\ z_{i,n}\left( \frac{r_{i,n}-\rho _{i,n}}{\gamma _{i,n}}\right) =-2\mu _{i,n}^2,\\ z_{i,n}\left( \frac{r_{i-1,n}-\rho _{i,n}}{\gamma _{i,n}}\right) =-2\mu _{i,n}^2\ (i\not =1). \end{array}\right. } \end{aligned}$$

(2.8)

Put $l:=\lim _{n\rightarrow \infty }(\rho _{i,n}-r_{i-1,n})/\gamma _{i,n}$. Then as in the proof of Lemma 4.3 in [20], the crucial step is to deduce $l=0$. Hence the case $i=1$ is easier. In the case $i>1$, we have to exclude the cases $l=\infty$ and $l\in (0,\infty )$. As a first step, we can prove by (2.3) that the case $l=\infty$ does not occur.

Lemma 2.5

Assume (2.2) and (2.3) and define $\gamma _{i,n}$ and $z_{i,n}$ as in the previous proposition. Moreover, put $l,m\in [0,\infty ]$ so that

$$\begin{aligned} l=\lim _{n\rightarrow \infty }\frac{\rho _{i,n}-r_{i-1,n}}{\gamma _{i,n}}\text { and }m=\lim _{n\rightarrow \infty }\frac{\rho _{i,n}}{\gamma _{i,n}}, \end{aligned}$$

by extracting a subsequence if necessary. Then we get that $\lim _{n\rightarrow \infty }\rho _{i,n}/r_{i,n}=0$, $\lim _{n\rightarrow \infty }(r_{i,n}-\rho _{i,n})/\gamma _{i,n}=\infty$, $0\le l=m<\infty$ and further, there exists a function z such that $z_{i,n} \rightarrow z$ in $C^2_{\text {loc}}((-l,\infty ))$ (in $C^2_{\text {loc}}((0,\infty ))\cap C^1_{\text {loc}}([0,\infty ))$ if $l=0$).

Proof

Using Lemmas 2.1, 2.3 and our assumptions (2.2) and (2.3), the proof is similar to that of Lemma 4.3 in [20]. (Especially, see the argument in “Case 1” there.) For the readers’ convenience we show the proof in Appendix 1. $\square$

Now our final aim becomes to prove $l=0$. In order to prove this, the variational characterization of solutions by [7] was useful in the previous work [20]. This allowed us to get the energy estimate in Lemma 2.1 (and also Lemma 2.5) in [20]. Using this, we could prove that the case $l\in (0,\infty )$ does not happen. (See the argument for “Case 2” in the proof of Lemma 4.3 in [20].) Since we only assume the boundedness of the energy in this paper, we need a new argument. We accomplish the proof with the aid of our new assumptions (2.4) and (2.5) as follows.

Proof of Proposition 2.4

Without losing the generality, we may suppose $u_{i.n}\ge 0$. Let l, m and z as in Lemma 2.5. Then we get $l=m<\infty$ by the lemma. Let us prove $l=0$. If $i=1$, this is trivial. Hence, we suppose $i\ge 2$ and $l>0$ on the contrary. Then, by (2.8) and Lemma 2.5, the limit function z satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} -z''-\frac{1}{l+r}z'= e^{z},\ z\le 0\ \ \hbox { in }\left( -l,+\infty \right) \\ z(0)=0=z'(0). \end{array}\right. } \end{aligned}$$

It follows that

$$\begin{aligned} z(r)=\log \frac{4A^2l^{A+2}(r+l)^{A-2}}{\left( (A+2)l^A+(A-2)(r+l)^A\right) ^2}, \end{aligned}$$

where $A=\sqrt{2l^2+4}$. (See Proof of Proposition 3.1 in [18] or the proof of Lemma 4.3 in [20]). Then, we use Lemma 2.2 to get

$$\begin{aligned} \begin{aligned}&2\mu _{i,n}^2\\&\ =2\mu _{i,n}\int _{r_{i-1,n}}^{\rho _{i,n}}\lambda _n f(u_{i,n})r\log {\frac{r}{r_{i-1,n}}}dr\\&=\log {\frac{\gamma _{i,n}}{r_{i-1,n}}}\int _{\frac{r_{i-1,n}-\rho _{i,n}}{\gamma _{i,n}}}^{0}\left( \frac{z_{i,n}(r)}{2\mu _{i,n}^2}+1\right) \\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \times e^{z_{i,n}(r)+\frac{z_{i,n}^2(r)}{4\mu _{i,n}^2}+\alpha \mu _{i,n}^{\beta _n}\left\{ \left( \frac{z_{i,n}(r)}{2\mu _{i,n}^2}+1\right) ^{\beta _n}-1\right\} }\left( r+\frac{\rho _{i,n}}{\gamma _{i,n}}\right) dr\\&\ \ +\int _{\frac{r_{i-1,n}-\rho _{i,n}}{\gamma _{i,n}}}^{0}\left( \frac{z_{i,n}(r)}{2\mu _{i,n}^2}+1\right) e^{z_{i,n}(r)+\frac{z_{i,n}^2(r)}{4\mu _{i,n}^2}+\alpha \mu _{i,n}^{\beta _n}\left\{ \left( \frac{z_{i,n}(r)}{2\mu _{i,n}^2}+1\right) ^{\beta _n}-1\right\} }\\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \times \left( r+\frac{\rho _{i,n}}{\gamma _{i,n}}\right) \log {\left( r+\frac{\rho _{i,n}}{\gamma _{i,n}}\right) }dr. \end{aligned} \end{aligned}$$

(2.9)

Noting that $l=m$ implies $r_{i-1,n}/\gamma _{i,n}\rightarrow 0$, we apply the Lebesgue convergence theorem to obtain

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{2\mu _{i,n}^2 }{\log {\frac{\gamma _{i,n}}{r_{i-1,n}}}}=\int _{-l}^{0}e^z(r+l)dr =\sqrt{2l^2+4}-2. \end{aligned}$$

(2.10)

On the other hand, we have by the definition of $\gamma _{i,n}$,

$$\begin{aligned} \begin{aligned} \frac{2\mu _{i,n}^2}{\log {\frac{\gamma _{i,n}}{r_{i-1,n}}}}&=\frac{4\mu _{i,n}^2}{2\log {\frac{1}{r_{i-1,n}}}-\log {2\lambda _n}-2\log {\mu _{i,n}}-\mu _{i,n}^2-\alpha \mu _{i,n}^{\beta _n}} \\&=\frac{4}{\frac{\log {\frac{1}{\lambda _n r_{i-1,n}^2}}}{\mu _{i,n}^2}-1+o(1)}. \end{aligned} \end{aligned}$$

(2.11)

Combining (2.10) with (2.11), we get

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{\log {\frac{1}{\lambda _n r_{i-1,n}^2}}}{\mu _{i,n}^2}=\frac{2\sqrt{2l^2+4}+4+l^2}{l^2}\in (0,\infty ). \end{aligned}$$

Then since

$$\begin{aligned} \frac{\log {\frac{1}{\lambda r_{i-1,n}^2}}}{\mu _{i,n}^2}=\frac{\log {\frac{1}{\lambda r_{i-1,n}^2}}}{\mu _{i-1,n}^{\beta _n}}\left( \frac{\mu _{i-1,n}}{\mu _{i,n}}\right) ^{\beta _n}\frac{1}{\mu _{i,n}^{2-\beta _n}}, \end{aligned}$$

using our assumptions (2.2) and (2.4), we deduce that

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{\mu _{i,n}}{\mu _{i-1,n}}=0. \end{aligned}$$

(2.12)

On the other hand, since $u_n=u_n(r)$ satisfies $-u_n''-u_n'/r=\lambda _n f_n(u_n)$ on $(\rho _{i-1,n},\rho _{i,n})$, multiplying this equation by r and integrating over $(\rho _{i-1,n},\rho _{i,n})$, we get

$$\begin{aligned} \int _{\rho _{i-1,n}}^{r_{i-1,n}}\lambda f(u_{i-1,n})rdr=-\int _{r_{i-1,n}}^{\rho _{i,n}}\lambda f(u_{i,n})rdr. \end{aligned}$$

Then, it follows from (2.5) and the similar scaling argument as in (2.9) that

$$\begin{aligned} \frac{\mu _{i,n}}{\mu _{i-1,n}}=-\frac{\mu _{i,n}\int _{r_{i-1,n}}^{\rho _{i,n}}\lambda f(u_{i,n})rdr}{\mu _{i-1,n}\int _{\rho _{i-1,n}}^{r_{i-1,n}}\lambda f(u_{i-1,n})rdr}\rightarrow \frac{\sqrt{2l^2+4}-2}{4}>0 \end{aligned}$$

(2.13)

as $n\rightarrow \infty$. This contradicts (2.12). Hence we get $l=0$. Then (2.8) and Lemma 2.5 prove that z satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} -z''-\frac{1}{r}z'= e^{z},\ z\le 0\ \ \hbox { in }\left( 0,+\infty \right) \\ z(0)=0=z'(0). \end{array}\right. } \end{aligned}$$

After integration (see Proof of Proposition 3.1 in [18] or the proof of Lemma 4.3 in [20]), we conclude that z satisfies (2.6) and (2.7). We complete the proof. $\square$

In the proof above, we get the following.

Lemma 2.6

Assume (2.2)–(2.5). Then we have

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{\rho _{i,n}}{\gamma _{i,n}}=0, \end{aligned}$$

(2.14)

and

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{\mu _{i,n}}{\mu _{i-1,n}}=0. \end{aligned}$$

(2.15)

Proof

In the previous proof, we get $l=m=0$. This proves (2.14). Using $l=0$ in (2.13), we obtain (2.15). This completes the proof. $\square$

3 Limit energy

In this section, we study the limit energy of concentrating solutions. As in the previous section, we fix $i=1,\cdots ,k+1$ and suppose (2.2)–(2.5). Without loss of the generality we assume $u_{i,n}\ge 0$. Moreover, we define $\gamma _{i,n}$ and $z_{i,n}$ as in the previous section. Our main goal is to prove the next asymptotic energy expansion.

Proposition 3.1

Assume (2.2)–(2.5). Moreover, if $i\not =1$, we suppose $\beta _*<3/2$. Then we have (1.4) and (1.5).

For the proof, we begin with the next lemma.

Lemma 3.2

Let $i\ge 2$ and suppose (2.2)–(2.5). Then we get $r_{i-1,n}/\rho _{i,n}\rightarrow 0$ and

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{\lambda _n \rho _{i,n}^2 f_n(\mu _{i,n})\log {\frac{\rho _{i,n}}{r_{i-1,n}}}}{\mu _{i,n}}=2=\lim _{n\rightarrow \infty }\frac{\lambda _n \rho _{i,n}^2 f_n(\mu _{i,n})}{r_{i-1,n} u_{i,n}'(r_{i-1,n})}. \end{aligned}$$

(3.1)

Proof

We put ${\tilde{r}}_n:=r_{i-1,n}/\rho _{i,n}$. Moreover, we write $\mu _n=\mu _{i,n}$ and $\rho _n=\rho _{i,n}$ for simplicity. We first claim that $\lim _{n\rightarrow \infty }{\tilde{r}}_n=0$. Actually, we get

$$\begin{aligned} \mu _n=\int _{r_{i-1,n}}^{\rho _n} u_n'(r)dr\le \left( \int _{r_{i-1,n}}^{\rho _n}|u_n'(r)|^2rdr\right) ^{\frac{1}{2}}\left( \log {\frac{1}{{\tilde{r}}_n}}\right) ^{\frac{1}{2}}. \end{aligned}$$

Then the claim follows by (2.2) and (2.3). Next we define a scaled function ${\tilde{z}}_n(r):=2\mu _n(u_n(\rho _nr)-\mu _n)$ for $r\in ({\tilde{r}}_n,1)$. Then it satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} -{\tilde{z}}_n''-\frac{1}{r}{\tilde{z}}_n'=2\lambda _n \rho _n^2\mu _nf_n(\mu _n)\frac{f_n\left( \frac{{\tilde{z}}_n}{2\mu _n}+\mu _n\right) }{f_n(\mu _n)} \text { in }({\tilde{r}}_n,1),\\ {\tilde{z}}_n(1)=0={\tilde{z}}_n'(1). \end{array}\right. } \end{aligned}$$

Thanks to (2.14), we get that

$$\begin{aligned} -{\tilde{z}}_n''-\frac{1}{r}{\tilde{z}}_n'=o(1) \end{aligned}$$

where $o(1)\rightarrow 0$ uniformly in $({\tilde{r}}_{n},1)$. Integrating this formula and using ${\tilde{z}}_n'(1)=0={\tilde{z}}_n(1)$, we see ${\tilde{z}}_n\rightarrow 0$ in $C^1_{\text {loc}}((0,1])$. Then similarly to (2.9), we use Lemma 2.2 to derive

$$\begin{aligned} \begin{aligned} \mu _n&= \lambda _n \rho _n^2f_n(\mu _n)\int _{{\tilde{r}}_n}^{1}\frac{f_n\left( \frac{{\tilde{z}}_n}{2\mu _n}+\mu _n\right) }{f_n(\mu _n)}r\log {\frac{r}{{\tilde{r}}_n}}dr\\&=\lambda _n \rho _n^2f_n(\mu _n)\log {\frac{1}{{\tilde{r}}_n}}\int _{{\tilde{r}}_n}^{1}\left( \frac{{\tilde{z}}_n}{2\mu _n^2}+1\right) e^{{\tilde{z}}_n+\frac{{\tilde{z}}_n^2}{4 \mu _n^2}+\alpha \mu _n^{\beta _n}\left\{ \left( \frac{{\tilde{z}}_n}{2\mu _n^2}+1\right) ^{\beta _n}-1\right\} } rdr\\&\ \ \ \ +\lambda _n \rho _n^2f_n(\mu _n)\int _{{\tilde{r}}_n}^{1}\left( \frac{{\tilde{z}}_n}{2\mu _n^2}+1\right) e^{{\tilde{z}}_n+\frac{{\tilde{z}}_n^2}{4 \mu _n^2}+\alpha \mu _n^\beta \left\{ \left( \frac{{\tilde{z}}_n}{2\mu _n^2}+1\right) ^{\beta _n}-1\right\} } r\log {r}dr. \end{aligned} \end{aligned}$$

Therefore, it follows from the Lebesgue convergence theorem and our first claim that

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{\mu _n}{\lambda _n \rho _n^2f_n(\mu _n)\log {\frac{1}{{\tilde{r}}_n}}}=\frac{1}{2}. \end{aligned}$$

This shows the first equality of (3.1). Finally, since

$$\begin{aligned} \begin{aligned} r_{i-1,n}&u_n'(r_{i-1,n})\\&=\int _{r_{i-1,n}}^{\rho _n} \lambda _n f_n(u_n)rdr\\&=\lambda _n \rho _n^2f_n(\mu _n)\int _{{\tilde{r}}_n}^{1}\left( \frac{{\tilde{z}}_n}{2\mu _n^2}+1\right) e^{{\tilde{z}}_n+\frac{{\tilde{z}}_n^2}{4 \mu _n^2}+\alpha \mu _n^{\beta _n}\left\{ \left( \frac{{\tilde{z}}_n}{2\mu _n^2}+1\right) ^{\beta _n}-1\right\} }rdr, \end{aligned} \end{aligned}$$

we similarly get the second one. This completes the proof. $\square$

By the previous lemma, (2.4) and (2.5), we get the following.

Lemma 3.3

Let $i\ge 2$ and assume (2.2)–(2.5). Then we have that

$$\begin{aligned} \limsup _{n\rightarrow \infty }\mu _{i,n}^{2-\beta _n}\left( \frac{\rho _{i,n}}{\gamma _{i,n}}\right) ^{2(\beta _n-1)}= 8^{\beta _*-1}\delta , \end{aligned}$$

(3.2)

and

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{\mu _{i,n}}{\mu _{i-1,n}^{\beta _n-1}}=\delta . \end{aligned}$$

(3.3)

In particular, if $\beta _*<3/2$, we get

$$\begin{aligned} \lim _{n\rightarrow \infty }\mu _{i,n}^{2-\beta _n}\frac{\rho _{i,n}}{\gamma _{i,n}}=0. \end{aligned}$$

(3.4)

Proof

From the first equality in (3.1), we get

$$\begin{aligned} \frac{\lambda _n \rho _{i,n}^2f_n(\mu _{i,n})}{\mu _{i,n}}\left( \log {(\lambda _n\rho _{i,n}^2)}+\log {\frac{1}{\lambda _n r_{i-1,n}^2}}\right) =4+o(1). \end{aligned}$$

(3.5)

On the other hand, using (2.5) and (2.4) for the second equality in (3.1) implies

$$\begin{aligned} \begin{aligned} 2+o(1)&=\frac{\lambda _n \rho _{i,n}^2f_n(\mu _{i,n})\mu _{i-1,n}}{2+o(1)}\\&=\frac{\lambda _n \rho _{i,n}^2f_n(\mu _{i,n})}{2+o(1)}\left\{ \frac{\log {\frac{1}{\lambda _n r_{i-1,n}^2}}}{\delta +o(1)}\right\} ^{\frac{1}{\beta _n}}. \end{aligned} \end{aligned}$$

It follows from the first equality of this formula that

$$\begin{aligned} \lambda _n \rho _{i,n}^2=\frac{4+o(1)}{f_n(\mu _{i,n})\mu _{i-1,n}}, \end{aligned}$$

(3.6)

and from the second one that

$$\begin{aligned} \log {\frac{1}{\lambda _n r_{i-1,n}^2}}=\frac{4^{\beta _*}\delta +o(1)}{\left\{ \lambda _n \rho _{i,n}^2 f_n(\mu _{i,n})\right\} ^{\beta _n}}. \end{aligned}$$

(3.7)

We substitute (3.7) into (3.5) and get

$$\begin{aligned} \begin{aligned} 4+o(1)&=\frac{\lambda _n \rho _{i,n}^2f(\mu _{i,n})}{\mu _{i,n}}\left[ \log {(\lambda _n\rho _{i,n}^2)}+\frac{4^{\beta _*}\delta +o(1)}{\left\{ \lambda _n \rho _{i,n}^2 f_n(\mu _{i,n})\right\} ^{\beta _n}}\right] \\&=\frac{4+o(1)}{\mu _{i,n}\mu _{i-1,n}}\log {\frac{4+o(1)}{f_n(\mu _{i,n})\mu _{i-1,n}}}+\frac{4^{\beta _*}\delta +o(1)}{\mu _{i,n}\left\{ \lambda _n \rho _{i,n}^2 f_n(\mu _{i,n})\right\} ^{\beta _n-1}} \end{aligned} \end{aligned}$$

where we used (3.6) for the second equality. Notice that (2.15) implies that the first term on the right hand side converges to zero as $n\rightarrow \infty$. Consequently, by the definition of $\gamma _{i,n}$, we obtain

$$\begin{aligned} \mu _{i,n}^{2-\beta _n}\left( \frac{\rho _{i,n}}{\gamma _{i,n}}\right) ^{2(\beta _n-1)}= 8^{\beta _*-1}\delta +o(1). \end{aligned}$$

This proves (3.2). Furthermore, substituting the definition of $\gamma _{i,n}$ and (3.6) into this formula, we see

$$\begin{aligned} \mu _{i,n}^{2-\beta _n}\left( \frac{(8+o(1))\mu _{i,n}}{\mu _{i-1,n}}\right) ^{\beta _n-1}= 8^{\beta _*-1}\delta +o(1). \end{aligned}$$

This ensures (3.3). Finally, if $\beta _*\in (0,3/2)$, (3.2) and (2.14) show that

$$\begin{aligned} \mu _{i,n}^{2-\beta _n}\frac{\rho _{i,n}}{\gamma _{i,n}}= \left( 8^{\beta _*-1}\delta +o(1)\right) \left( \frac{\rho _{i,n}}{\gamma _{i,n}}\right) ^{3-2\beta _n}\rightarrow 0 \end{aligned}$$

as $n\rightarrow \infty$. We finish the proof. $\square$

Remark 3.4

By (3.2) and (2.14), we see that if $(\beta _n)\subset (0,1]$, then $\mu _{i,n}$ is uniformly bounded for all $n\in {\mathbb {N}}$. This contradicts our basic assumption (2.2). This will prove that, interestingly, if $\int _0^1u_n'(r)^2rdr$ is uniformly bounded and $\beta _n\le 1$ for all $n\in {\mathbb {N}}$, then $u_{i,n}$ does never blow up for any $i\ge 2$. For the detail, see Proof of Theorem 1.6 in Sect. 5. This remark suggests that, in the rest of the argument in this section, we may restrict our attention only on the case $i=1$ if $(\beta _n)\subset (0,1]$.

Remark 3.5

As in the statement, the assumption $\beta _*<3/2$ ensures (3.4). We will see that the next lemmas strongly depend on this fact. For example, it allows the assertion $\varepsilon _n=o(\mu _{i,n}^{2-\beta _n})$ in Lemma 3.7 below. On the other hand, if $i\not =1$ and $(\beta _n)\subset [3/2,2)$, this assertion fails by (3.2) above. This implies that the effect of the error term $\rho _{i,n}/\gamma _{i,n}$ would appear in the strong pointwise estimate like (3.17). More precisely, the term $-\alpha \beta _*/(2\mu _{i,n}^{2-\beta _*})$ in (3.17) would be modified to the one with $\mu _{i,n}^{-(2-\beta _n)/(2(\beta _n-1))}$ in view of (3.2). This change would affect all the results, for example, the energy expansion in Theorem 1.5 and the asymptotic formulas in Theorem 1.6, based on (3.17).

Notice that in the following lemmas, we additionally assume $\beta _*<3/2$ if $i\not =1$. We next prove the following.

Lemma 3.6

Assume $i\ge 1$ and (2.2)–(2.5). Moreover, we suppose $\beta _*<3/2$ if $i\not =1$. Let $z_{i,n},z$ be functions defined in Proposition 2.4 and put $\phi _n:=\mu _{i,n}^{2-\beta _n}(z_{i,n}-z)$. Then we get $\phi _n\rightarrow \phi$ in $C^2_{\text {loc}}(0,\infty )\cap C^0_{\text {loc}}([0,\infty ))$ where $\phi$ satisfies

$$\begin{aligned} -\phi ''-\frac{1}{r}\phi '=e^{z}\left( \phi +\frac{\alpha \beta _*}{2}z\right) \text { in }(0,\infty ),\ \phi (0)=0. \end{aligned}$$

(3.8)

In particular, we obtain

$$\begin{aligned} \begin{aligned} \phi (r) =\alpha \beta _*\left( \log {(8+r^2)}+\frac{8}{8+r^2}-1-\log {8}\right) \ (r\in [0,\infty )). \end{aligned} \end{aligned}$$

(3.9)

Proof

We write $\rho _n=\rho _{i,n}$, $\mu _n=\mu _{i,n}$, $\gamma _n=\gamma _{i,n}$ and $z_n=z_{i,n}$ for simplicity. Then using the definition of $\phi _n$ and the equations in (2.8) and (2.7), we get

$$\begin{aligned} \begin{aligned}&\displaystyle -\phi _n''-\frac{1}{r+\frac{\rho _{n}}{\gamma _{n}}}\phi _n'\\&=\mu _{n}^{2-\beta _n}\Bigg \{\left( \frac{z_n}{2\mu _n^2}+1\right) e^{z_n+\frac{z_n^2}{4\mu _n^2}+\alpha \mu _n^{\beta _n}\left( \left( \frac{z_n}{2\mu _n^2}+1\right) ^{\beta _n}-1\right) }\\&\quad \quad -e^z+\left( \frac{1}{r+\frac{\rho _{n}}{\gamma _{n}}}-\frac{1}{r}\right) z'\Bigg \}, \end{aligned} \end{aligned}$$

(3.10)

for all $r\in \left( 0,\frac{r_{i,n}-\rho _{n}}{\gamma _{n}}\right)$. Here recalling that $z_n$ is locally uniformly bounded in $[0,\infty )$, we use the Taylor theorem to see

$$\begin{aligned} \left( \frac{z_n}{2\mu _n^2}+1\right) e^{z_n+\frac{z_n^2}{4\mu _n^2}+\alpha \mu _n^{\beta _n}\left\{ \left( \frac{z_n}{2\mu _n^2}+1\right) ^{\beta _n}-1\right\} }=e^{z_n}+\frac{\alpha \beta _n }{2\mu _n^{2-\beta _n}}z_ne^{z_n}+o\left( \frac{1}{\mu _n^{2-\beta _n}}\right) \end{aligned}$$

where $\mu _n^{2-\beta _n}\cdot o(1/\mu _n^{2-\beta _n})\rightarrow 0$ locally uniformly in $[0,\infty )$. Then after substituting this into (3.10), for all $r\in \left( 0,\frac{r_{i,n}-\rho _{n}}{\gamma _{n}}\right)$ and $n\in {\mathbb {N}}$, we apply the mean value theorem to obtain a constant $\theta =\theta (n,r)\in (0,1)$ such that

$$\begin{aligned} {\left\{ \begin{array}{ll}\displaystyle -\phi _n''(r)-\frac{1}{r+\frac{\rho _{n}}{\gamma _{n}}}\phi _n'(r)=e^{z(r)+\theta (z_n(r)-z(r))}\phi _n(r)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\frac{\alpha \beta _n}{2}z_n(r) e^{z_n(r)}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\mu _{n}^{2-\beta _n}\left( \frac{1}{r+\frac{\rho _{n}}{\gamma _{n}}}-\frac{1}{r}\right) z'(r)+o(1),\\ \phi _n(0)=0=\phi _n'(0), \end{array}\right. } \end{aligned}$$

(3.11)

where $o(1)\rightarrow 0$ uniformly locally in $[0,\infty )$. Here, notice that third term of the right hand side of the equation is nontrivial if $i\ge 2$. But thanks to (3.4), we have

$$\begin{aligned} \mu _{n}^{2-\beta _n}\left( \frac{1}{r+\frac{\rho _{n}}{\gamma _{n}}}-\frac{1}{r}\right) z'(r)=\mu _{n}^{2-\beta _n}\frac{\rho _{n}}{\gamma _{n}}\frac{4}{(8+r^2)\left( r+\frac{\rho _{n}}{\gamma _{n}}\right) }\rightarrow 0 \end{aligned}$$

(3.12)

uniformly locally in $(0,\infty )$ as $n\rightarrow \infty$. Now we claim that $\phi _n$ is locally uniformly bounded in $[0,\infty ).$ If not, there exist a constant $R>0$ and a sequence $(\xi _n)\subset [0,R]$ such that $\phi _n(\xi _n)=\max _{r\in [0,R]}|\phi _n(r)|$ and $\lim _{n\rightarrow \infty }|\phi _n(\xi _n)|=\infty$. Then putting ${\tilde{\phi }}_n:=\phi _n/\phi _n(\xi _n)$ and multiplying the equation in (3.11) by $\phi (\xi _n)^{-1}$, we obtain that for all $r\in (0,R]$

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle -{\tilde{\phi }}_n''(r)-\frac{1}{r+\frac{\rho _{n}}{\gamma _{n}}}{\tilde{\phi }}_n'(r)=e^{z(r)+\theta (z_n(r)-z(r))}{\tilde{\phi }}_n(r)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \displaystyle +\frac{\mu _{n}^{2-\beta _n}}{\phi (\xi _n)}\frac{\rho _{n}}{\gamma _{n}}\frac{4}{(8+r^2)\left( r+\frac{\rho _{n}}{\gamma _{n}}\right) } +o(1),\\ {\tilde{\phi }}_n(0)=0={\tilde{\phi }}_n'(0), \end{array}\right. } \end{aligned}$$

(3.13)

where $o(1)\rightarrow 0$ uniformly in [0, R]. It follows that

$$\begin{aligned} -{\tilde{\phi }}_n''(r)-\frac{1}{r+\frac{\rho _{n}}{\gamma _{n}}}{\tilde{\phi }}_n'(r)=O(1) +\frac{\mu _{n}^{2-\beta _n}}{\phi (\xi _n)}\frac{\rho _{n}}{\gamma _{n}}\frac{4}{(8+r^2)\left( r+\frac{\rho _{n}}{\gamma _{n}}\right) }\ (r\in (0,R]), \end{aligned}$$

where O(1) is uniformly bounded in [0, R]. Then, for any $r\in (0,R]$, multiplying this formula by $(r+\rho _n/\gamma _n)$ and integrating over (0, r) give

$$\begin{aligned} -{\tilde{\phi }}_n'(r)=O\left( \frac{\frac{1}{2}r^2+\frac{\rho _{n}}{\gamma _{n}}r}{r+\frac{\rho _{n}}{\gamma _{n}}}\right) +\frac{\mu _{n}^{2-\beta _n}}{\phi _n(\xi _n)}\frac{\rho _{n}}{\gamma _{n}}\left( r+\frac{\rho _n}{\gamma _n}\right) ^{-1}\int _0^r\frac{4}{(8+r^2)}dr. \end{aligned}$$

(3.14)

Consequently, with the aid of (3.4), we have that ${\tilde{\phi }}_n$ is uniformly bounded in $C^1([0,R])$. Therefore, it follows from the Ascoli-Arzelà theorem, the equation in (3.13) and (3.12) that there exists a function ${\tilde{\phi }}$ such that ${\tilde{\phi }}_n\rightarrow {\tilde{\phi }}$ in $C^2_{\text {loc}}((0,R])\cap C([0,R])$. Then ${\tilde{\phi }}$ satisfies

$$\begin{aligned} -{\tilde{\phi }}''-\frac{1}{r}{\tilde{\phi }}'=e^z{\tilde{\phi }} \text { in }(0,R],\ {\tilde{\phi }}(0)=0. \end{aligned}$$

This implies

$$\begin{aligned} {\tilde{\phi }}(r)={\tilde{c}}_1\frac{8-r^2}{8+r^2}+{\tilde{c}}_2\frac{(8-r^2)\log {r}+16}{8+r^2}\ (r\in (0,R]) \end{aligned}$$

for some constants ${\tilde{c}}_1,{\tilde{c}}_2\in {\mathbb {R}}$. Since $\lim _{r\rightarrow 0+0}{\tilde{\phi }}(r)=0$, we get ${\tilde{c}}_1={\tilde{c}}_2=0$ and thus, obtain ${\tilde{\phi }}=0$ in [0, R]. This contradicts the fact that ${\tilde{\phi }}_n(\xi _n)=1$ for all $n\in {\mathbb {N}}$. This proves the claim. Then, for any $r\in (0,\infty )$, multiplying the equation in (3.11) by $(r+\rho _n/\gamma _n)$ and integrating over (0, r), a similar calculation as above gives that $\phi _n$ is uniformly bounded in $C_{\text {loc}}^1([0,\infty ))$ thanks to (3.4). Then again by the Ascoli–Arzelà theorem, (3.11) and (3.12), we obtain a function $\phi$ such that $\phi _n\rightarrow \phi$ in $C^2_{\text {loc}}((0,\infty ))\cap C^0_{\text {loc}}([0,\infty ))$. It follows from (3.11) that

$$\begin{aligned} -\phi ''-\frac{1}{r}\phi '=e^{z} \phi +\frac{\alpha \beta _*}{2}ze^{z}\text { in }(0,\infty ),\ \phi (0)=0. \end{aligned}$$

(3.15)

Then we compute that for all $r>0$,

$$\begin{aligned} \begin{aligned} \frac{1}{\alpha \beta _*}\phi (r)&=c_1\frac{8-r^2}{8+r^2}+c_2\frac{(8-r^2)\log {r}+16}{8+r^2}\\ {}&\ \ +\log {(8+r^2)}+\frac{2(8-r^2)\log {r}+8(3-2\log {8})}{8+r^2},\\ \end{aligned} \end{aligned}$$

(3.16)

where $c_1,c_2\in {\mathbb {R}}$ are some constants. By $\lim _{r\rightarrow 0+0}\phi (r)=0$, we get $c_2=-2$ and then conclude $c_1=1+\log {8}$. This completes the proof. $\square$

The next estimate is important.

Lemma 3.7

Let $i\ge 1$ and assume (2.2)–(2.5). In addition, if $i\not =1$, we suppose $\beta _*<3/2$. Then for any $R>0$, there exists a constant $C>0$ such that

$$\begin{aligned} z_{i,n}(r)\le \left( 1-\frac{\alpha \beta _*}{2\mu _{i,n}^{2-\beta _n}}\right) z(r)+C \varepsilon _n\log {r} \end{aligned}$$

(3.17)

for all $r\in [R,(r_{i,n}-\rho _{i,n})/\gamma _{i,n}]$ and all large $n\in {\mathbb {N}}$ where we put

$$\begin{aligned} \varepsilon _n:=\max \left\{ \frac{1}{\mu _{i,n}^{4-2\beta _n}},\ \frac{1}{\mu _n^2},\ \frac{|\beta _n-\beta _*|}{\mu _n^{2-\beta _n}},\ \frac{\rho _{i,n}}{\gamma _{i,n}}\right\} =o\left( \frac{1}{\mu _{i,n}^{2-\beta _n}}\right) . \end{aligned}$$

Proof

We put $\rho _n=\rho _{i,n}$, $\mu _n=\mu _{i,n}$, $\gamma _{n}=\gamma _{i,n}$, and $z_n=z_{i,n}$ for simplicity. We apply the contraction mapping argument in the proof of Lemma 5 in [29] (see also [30]). We define a function $\psi _n$ on $[0,(r_{i,n}-\rho _n)/\gamma _n]$ by

$$\begin{aligned} \psi _n:=z_n-z-\frac{\phi }{\mu _n^{2-\beta _n}}, \end{aligned}$$

where $\phi$ is taken from the previous lemma. Then from (2.8), (2.7), and (3.8), we get

$$\begin{aligned} \begin{aligned}&-\psi _n''-\frac{1}{r+\frac{\rho _n}{\gamma _n}}\psi _n'\\&=\left( \frac{z_n}{2\mu _n^2}+1\right) e^{z_n+\frac{z_n^2}{4\mu _n^2}+\mu _n^{\beta _n}\left\{ \left( \frac{z_n}{2\mu _n^2}+1\right) ^{\beta _n}-1\right\} }\\&\ \ \ \ -e^z-\frac{1}{\mu _n^{2-\beta _n}}e^z\left( \phi +\frac{\alpha \beta _*}{2}z\right) +\left( \frac{1}{r+\frac{\rho _n}{\gamma _n}}-\frac{1}{r}\right) \left( z'+\frac{\phi '}{\mu _n^{2-\beta _n}}\right) \\&=\Phi _n(\psi _n) \end{aligned} \end{aligned}$$

(3.18)

where we defined

$$\begin{aligned} \begin{aligned} \Phi _n(\psi ):=&e^z\Bigg [\left\{ 1+\frac{1}{2\mu _n^2}\left( z+\frac{\phi }{\mu _n^{2-\beta _n}}+\psi \right) \right\} e^{h_n(\psi )}\\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -1-\frac{1}{\mu _n^{2-\beta _n}}\left( \phi +\frac{\alpha \beta _*}{2}z\right) \Bigg ]\\&+\frac{\rho _n}{\gamma _n}\frac{4}{(r+\frac{\rho _n}{\gamma _n})(r^2+8)}\left( 1-\frac{\alpha \beta _*}{2\mu _n^{2-\beta _n}}\frac{r^2}{r^2+8}\right) \end{aligned} \end{aligned}$$

(3.19)

with

$$\begin{aligned} \begin{aligned} h_n(\psi )&:=\psi +\frac{\phi }{\mu _n^{2-\beta _n}}+\frac{1}{4\mu _n^{2}}\left( z+\frac{\phi }{\mu _n^{2-\beta _n}}+\psi \right) ^2\\ {}&+\alpha \mu _n^{\beta _n}\left[ \left\{ \frac{1}{2\mu _n^{2}}\left( z+\frac{\phi }{\mu _n^{2-\beta _n}}+\psi \right) +1\right\} ^{\beta _n}-1\right] . \end{aligned} \end{aligned}$$

(3.20)

We first claim that for any $T>0$, there exists a constant $C(T)>0$ such that

$$\begin{aligned} |\psi _n(r)|\le C(T)\varepsilon _n \text { and }|\psi _n'(r)|\le C(T)\varepsilon _n\ (r\in [0,T]), \end{aligned}$$

(3.21)

for all large $n\in {\mathbb {N}}$ where $\varepsilon _n$ is defined as in the statement of this lemma and $\varepsilon _n=o\left( \mu _n^{-(2-\beta _n)}\right)$ by (3.4). To show the claim, fix any $T>0$. Then since $\psi _n\rightarrow 0$ uniformly in [0, T] by Lemma 3.6, we have $|h_n(\psi _n)|\le 1$ for large $n\in {\mathbb {N}}$. It follows that $e^{h_n(\psi _n)}=1+h_n(\psi _n)+O(h_n(\psi _n)^2)$ on [0, T] by the Taylor theorem. Using this for (3.19) with $\psi =\psi _n$, we compute that

$$\begin{aligned} \begin{aligned} \Phi _n(\psi _n)&=e^{z}\left[ \left( 1+O\left( \frac{1}{\mu _n^{2-\beta _n}}\right) \right) \psi _n+O\left( \psi _n^2\right) +O(\varepsilon _n)\right] \\&+O\left( \frac{\rho _n}{\gamma _n}\frac{1}{\left( r+\frac{\rho _n}{\gamma _n}\right) \left( 8+r^2\right) }\right) \text { on } [0,T]. \end{aligned} \end{aligned}$$

(3.22)

Then, putting $\psi _n=\varepsilon _n{\bar{\psi }}_n$ in (3.18), we get

$$\begin{aligned} \begin{aligned} -{\bar{\psi }}_n''-\frac{1}{r+\frac{\rho _n}{\gamma _n}}{\bar{\psi }}_n'&= e^z\left( 1+O\left( \frac{1}{\mu _n^{2-\beta _n}}\right) \right) {\bar{\psi }}_n+O\left( \varepsilon _n{\bar{\psi }}_n^2\right) +O\left( 1\right) \\&\ \ \ \ +O\left( \frac{1}{\varepsilon _n}\frac{\rho _n}{\gamma _n}\frac{1}{\left( r+\frac{\rho _n}{\gamma _n}\right) \left( 8+r^2\right) }\right) \text { on } [0,T], \end{aligned} \end{aligned}$$

and ${\bar{\psi }}_n(0)=0={\bar{\psi }}'(0)$. Using this equation and noting $\varepsilon _n^{-1}\rho _n/\gamma _n\le 1$, we get that ${\bar{\psi }}_n$ is uniformly bounded in $C^1([0,T])$. (The detail of the proof is similar to the argument in confirming the locally uniformly boundedness of $\phi _n$ and $\phi _n'$ in the proof of Lemma 3.6.) This ensures the claim.

Next let us extend the estimate (3.21) to a suitable expanding interval. To this end, we choose a sufficiently large number $T>0$ and a sequence $(s_n)\subset (T,(r_{i,n}-\rho _n)/\gamma _n)$ so that $s_n\rightarrow \infty$. (More precise choice of T and $(s_n)$ is given later.) Then we consider an initial value problem

$$\begin{aligned} -\psi ''-\frac{1}{r+\frac{\rho _n}{\gamma _n}}\psi '=\Phi _n(\psi )\text { on }(T,s_n],\ \psi (T)=\psi _n(T),\ \psi '(T)=\psi _n'(T). \end{aligned}$$

Putting $\omega =\left( r+\frac{\rho _n}{\gamma _n}\right) \psi '$, we get the equivalent system,

$$\begin{aligned} {\left\{ \begin{array}{ll} \psi '=\frac{\omega }{\left( r+\frac{\rho _n}{\gamma _n}\right) }\text { on }(T,s_n],\\ \omega '=-\left( r+\frac{\rho _n}{\gamma _n}\right) \Phi _n(\psi )\text { on }(T,s_n],\\ \psi (T)=\psi _n(T),\ \omega (T)=\left( T+\frac{\rho _n}{\gamma _n}\right) \psi _n'(T). \end{array}\right. } \end{aligned}$$

(3.23)

Notice that, by the uniqueness, the solution $(\psi ,\omega )$ satisfies $(\psi ,w)=(\psi _n,(r+\rho _n/\gamma _n)\psi _n')$ on $[T,s_n]$. In order to construct the solution with an appropriate estimate, we reduce (3.23) into an integral equation on a suitable function space. To do this, we define the norms $\Vert \cdot \Vert _1$ and $\Vert \cdot \Vert _2$ by

$$\begin{aligned} \Vert f\Vert _1=\sup _{r\in (T,s_n]}\left| \frac{f(r)}{\log {\left( r+\frac{\rho _n}{\gamma _n}\right) }-\log {\left( T+\frac{\rho _n}{\gamma _n}\right) }}\right| ,\ \ \Vert f\Vert _2=2\sup _{r\in [T,s_n]}|f(r)|. \end{aligned}$$

We fix a constant ${\tilde{C}}>0$ such that

$$\begin{aligned} {\tilde{C}}\ge 4(C(T)(T+2)+1), \end{aligned}$$

(3.24)

where $C(T)>0$ is taken from (3.21). Then we consider a set of functions

$$\begin{aligned} \begin{aligned} {\mathcal {B}}_{{\tilde{C}}}:=&\Big \{(\psi ,\omega )\in C^0([T,s_n])\times C^0([T,s_n])\ \big |\ \Vert \psi -\psi _n(T)\Vert _1\le {\tilde{C}} \varepsilon _n,\\ {}&\ \ \ \ \Vert \omega \Vert _2\le {\tilde{C}}\varepsilon _n,\ \psi (T)=\psi _n(T),\ \omega (T)=\left( T+\frac{\rho _n}{\gamma _n}\right) \psi _n'(T)\Big \}. \end{aligned} \end{aligned}$$

Moreover, we define a map ${\mathcal {F}}:{\mathcal {B}}_{{\tilde{C}}}\rightarrow C^0([T,s_n])\times C^0([T,s_n])$ so that ${\mathcal {F}}(\psi ,\omega )=(F_{1}(\psi ,\omega ),F_{2}(\psi ,\omega ))$ and

$$\begin{aligned} \begin{aligned}&F_{1}(\psi ,\omega )(r):=\psi (T)+\int _T^{r}\frac{\omega }{\left( s+\frac{\rho _n}{\gamma _n}\right) }ds,\\&F_{2}(\psi ,\omega )(r):=\omega (T)-\int _T^{r}\left( s+\frac{\rho _n}{\gamma _n}\right) \Phi _n(\psi )ds, \end{aligned} \end{aligned}$$

for $r\in [T,s_n]$. We shall find a fixed point of ${\mathcal {F}}$ in ${\mathcal {B}}_{{\tilde{C}}}$.

To this end, we fix a small number $0<d\ll 1$ (independently of T) and choose the sequence $(s_n)$ so that $s_n\le \sqrt{e^{d \mu _n^{\min \{1,2-\beta _n\}}}-8}$ for all $n\in {\mathbb {N}}$ and $\liminf _{n\rightarrow \infty }\varepsilon _n^{1/2}s_n>0$. It follows that

$$\begin{aligned} \frac{1}{\mu _n^{\min \{1,2-\beta _n\}}}\log {(8+r^2)}\le d \end{aligned}$$

(3.25)

for all $r\in [T,s_n]$ and $n\in {\mathbb {N}}$. Moreover, by (3.21) and (3.25), there exists a number $n_0=n_0(T,d)$ such that if $n\ge n_0$, it holds that

$$\begin{aligned} \sup _{s\in [T,s_n]}|\psi (s)|\le d \end{aligned}$$

(3.26)

for any $\psi \in C^0([T,S])$ with $\Vert \psi -\psi _n(T)\Vert _1\le {\tilde{C}}\varepsilon _n$. Then similarly to (3.22), we calculate by (2.6), (3.9), (3.25) and (3.26) that

$$\begin{aligned} \begin{aligned} \Phi _n(\psi )&=e^z\left[ \left( 1+O\left( \delta \right) \right) \psi +O\left( \varepsilon _n\log ^2{(8+r^2)}\right) \right] \\&+O\left( \frac{\rho _n}{\gamma _n}\frac{1}{(r+\frac{\rho _n}{\gamma _n})(8+r^2)}\right) \text { on }[T,s_n] \end{aligned} \end{aligned}$$

(3.27)

for any $\psi \in C^0([T,S])$ such that $\Vert \psi -\psi _n(T)\Vert _1\le {\tilde{C}}\varepsilon _n$ and all $n\ge n_0$. Analogously, we compute that

$$\begin{aligned} |\Phi _n(\psi )-\Phi _n({\bar{\psi }})|\le \left( 1+O\left( \delta \right) \right) e^z|\psi -{\bar{\psi }}|\text { on }[T,s_n], \end{aligned}$$

(3.28)

for all $\psi , {\bar{\psi }} \in C^0([T,S])$ verifying $\Vert \psi -\psi _n(T)\Vert _1\le {\tilde{C}}\varepsilon _n$ and $\Vert {\bar{\psi }}-\psi _n(T)\Vert _1\le {\tilde{C}}\varepsilon _n$ and all $n\ge n_0$. After this we always assume $n\ge n_0$.

Now, we first claim that ${\mathcal {F}}: {\mathcal {B}}_{{\tilde{C}}}\rightarrow {\mathcal {B}}_{{\tilde{C}}}$. In fact, for any $(\psi ,\omega )\in {\mathcal {B}}_{{\tilde{C}}}$, we get

$$\begin{aligned} \begin{aligned} |F_{1}(\psi ,\omega )(r)-\psi _n(T)|&\le \frac{1}{2}\Vert \omega \Vert _2\left| \int _T^r\frac{1}{s+\frac{\rho _n}{\gamma _n}}ds\right| . \end{aligned} \end{aligned}$$

It follows that

$$\begin{aligned} \Vert F_{1}(\psi ,\omega )-\psi _n(T)\Vert _1\le \frac{1}{2}{\tilde{C}}\varepsilon _n. \end{aligned}$$

On the other hand, from our choice of $(\psi ,\omega )$, (3.21) and (3.27), we have for any $r\in [T,s_n]$ that

$$\begin{aligned} \begin{aligned}&|F_{2}(\psi ,\omega )(r)|\\&\le \varepsilon _n\Bigg [C(T)\left( T+\frac{\rho _n}{\gamma _n}\right) +65\int _T^r\frac{\left( s+\frac{\rho _n}{\gamma _n}\right) \left( C(T)+{\tilde{C}}\log {\frac{r+\frac{\rho _n}{\gamma _n}}{T+\frac{\rho _n}{\gamma _n}}}\right) }{(8+s^2)^2}ds\\&\ \ \ \ \ \ \ \ \ +O\left( \int _T^r\frac{\left( s+\frac{\rho _n}{\gamma _n}\right) \log ^2(8+s^2)}{(8+s^2)^2}ds\right) +O\left( \varepsilon _n^{-1}\frac{\rho _n}{\gamma _n}\int _T^r\frac{1}{8+s^2}ds\right) \Bigg ]\\&\le \varepsilon _n\Bigg [C(T)\left( T+1+65\int _T^r\frac{\left( s+\frac{\rho _n}{\gamma _n}\right) }{(8+s^2)^2}ds\right) +65{\tilde{C}}\int _T^r\frac{\left( s+\frac{\rho _n}{\gamma _n}\right) \log {\frac{r+\frac{\rho _n}{\gamma _n}}{T+\frac{\rho _n}{\gamma _n}}}}{(8+s^2)^2}ds\\&\ \ \ \ \ \ \ \ \ +O\left( \int _T^r\frac{\left( s+\frac{\rho _n}{\gamma _n}\right) \log ^2(8+s^2)}{(8+s^2)^2}ds\right) +O\left( \varepsilon _n^{-1}\frac{\rho _n}{\gamma _n}\int _T^r\frac{1}{8+s^2}ds\right) \Bigg ]. \end{aligned} \end{aligned}$$

Since $\varepsilon _n^{-1}\rho _n/\gamma _n\le 1$, taking $T>0$ large enough, we get

$$\begin{aligned} |F_{2}(\psi ,\omega )(r)|\le \varepsilon _n\left[ C(T)(T+2)+\frac{{\tilde{C}}}{4}+1\right] . \end{aligned}$$

We fix this T. Then it follows from (3.24) that

$$\begin{aligned} \Vert F_{2}(\psi ,\omega )\Vert _2\le {\tilde{C}}\varepsilon _n. \end{aligned}$$

This proves the claim. Next we shall show that ${\mathcal {F}}$ is a contraction mapping. Indeed, for any $(\psi ,\omega ),({\bar{\psi }},{\bar{\omega }})\in {\mathcal {B}}_{{\tilde{C}}}$, we obtain

$$\begin{aligned} \begin{aligned} |F_{1}(\psi ,\omega )(r)-F_{1}({\bar{\psi }},{\bar{\omega }})(r)|&\le \frac{1}{2} \Vert \omega -{\bar{\omega }}\Vert _2\log {\frac{r+\frac{\rho _n}{\gamma _n}}{T+\frac{\rho _n}{\gamma _n}}}. \end{aligned} \end{aligned}$$

This implies that

$$\begin{aligned} \Vert F_{1}(\psi ,\omega )-F_{1}({\bar{\psi }},{\bar{\omega }})\Vert _1 \le \frac{1}{2}\Vert \omega -{\bar{\omega }}\Vert _2. \end{aligned}$$

Moreover, we get by (3.28) that

$$\begin{aligned} \begin{aligned} |F_{2}(\psi ,\omega )(r)-&F_{2}({\bar{\psi }},{\bar{\omega }})(r)|\\&\le 65\int _T^r\left( s+\frac{\rho _n}{\gamma _n}\right) \frac{|\psi -{\bar{\psi }}|}{(8+s^2)^2}ds\\&\le 65 \sup _{s\in (T,s_n]}\left| \frac{\psi -{\bar{\psi }}}{\log {\frac{s+\frac{\rho _n}{\gamma _n}}{T+\frac{\rho _n}{\gamma _n}}}}\right| \int _T^r\left( s+\frac{\rho _n}{\gamma _n}\right) \frac{\log {\frac{s+\frac{\rho _n}{\gamma _n}}{T+\frac{\rho _n}{\gamma _n}}}}{(8+s^2)^2}ds. \end{aligned} \end{aligned}$$

Choosing $T>0$ larger if necessary, we see

$$\begin{aligned} \Vert F_{1}(\psi ,\omega )-F_1({\bar{\psi }},{\bar{\omega }})\Vert _2 \le \frac{1}{2}\Vert \psi -{\bar{\psi }}\Vert _1. \end{aligned}$$

Consequently, ${\mathcal {F}}$ is a contraction mapping from ${\mathcal {B}}_{{\tilde{C}}}$ to itself. This suggests that there exists a fixed point $(\psi ,\omega )\in {\mathcal {B}}_{{\tilde{C}}}$ of ${\mathcal {F}}$. Then, as noted above, we get $(\psi ,\omega )=(\psi _n, (r+\rho _n/\gamma _n)\psi _n')$ on $[T,s_n]$. Since $(\psi ,\omega )\in {\mathcal {B}}_{{\tilde{C}}}$, we have by (3.21),

$$\begin{aligned} |\psi _n(r)|\le \varepsilon _n\left( C(T)+{\tilde{C}} \log {\frac{r+\frac{\rho _n}{\gamma _n}}{T+\frac{\rho _n}{\gamma _n}}}\right) , \end{aligned}$$

(3.29)

and

$$\begin{aligned} \left( r+\frac{\rho _n}{\gamma _n}\right) |\psi _n'(r)|\le \frac{{\tilde{C}}}{2}\varepsilon _n \end{aligned}$$

(3.30)

for all $r\in [T,s_n]$.

Let us finish the proof. Fix $T>0$ as above and choose any $R>0$. If $R< T$, we get by the definition of $\psi _n$, (2.6), (3.9) and (3.21) that

$$\begin{aligned} z_n(r)\le \left( 1-\frac{\alpha \beta _*}{2\mu _n^{2-\beta _n}}\right) z(r)-\frac{\alpha \beta _n}{\mu _n^{2-\beta _n}}\frac{r^2}{r^2+8}+O(\varepsilon _n)<\left( 1-\frac{\alpha \beta _*}{2\mu _n^{2-\beta _n}}\right) z(r) \end{aligned}$$

for all $r\in [R,T]$ if n is large enough. Hence we may assume $T\le R$. Then, similarly, it follows from (3.29) that

$$\begin{aligned} \begin{aligned}&z_n(r)\\&\le \left( 1-\frac{\alpha \beta _*}{2\mu _n^{2-\beta _n}}\right) z(r)-\frac{\alpha \beta _n}{\mu _n^{2-\beta _n}}\frac{r^2}{r^2+8}+\left( C(T)+{\tilde{C}} \log {\frac{r+\frac{\rho _n}{\gamma _n}}{T+\frac{\rho _n}{\gamma _n}}}\right) \varepsilon _n\\&\le \left( 1-\frac{\alpha \beta _*}{2\mu _n^{2-\beta _n}}\right) z(r)+O(\varepsilon _n\log {r}) \end{aligned} \end{aligned}$$

(3.31)

for all $r\in [R,s_n]$. Moreover, for any $r\in [s_n,(r_{i,n}-\rho _n)/\gamma _n]$, we have

$$\begin{aligned} \begin{aligned} \left( r+\frac{\rho _n}{\gamma _n}\right) z_n'(r)&\le \int _0^{s_n}\left\{ \left( r+\frac{\rho _n}{\gamma _n}\right) z_n'(r)\right\} 'dr\\&=\left( s_n+\frac{\rho _n}{\gamma _n}\right) \left( z'(s_n)+\frac{\phi '(s_n)}{\mu _n^{2-\beta _n}}+\psi _n'(s_n)\right) . \end{aligned} \end{aligned}$$

(3.32)

Here we use (2.6), (3.9) and (3.30) to see

$$\begin{aligned} \begin{aligned}&\left( s_n+\frac{\rho _n}{\gamma _n}\right) z'(s_n)=-4+O\left( \frac{1}{s_n^2}\right) +o\left( \frac{\rho _n}{\gamma _n}\right) ,\\&\left( s_n+\frac{\rho _n}{\gamma _n}\right) \frac{\phi '(s_n)}{\mu _n^{2-\beta _n}}=\frac{2\alpha \beta _*}{\mu _n^{2-\beta _n}}+o\left( \frac{1}{s_n^2}\right) +o\left( \frac{\rho _n}{\gamma _n}\right) , \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \left| \left( s_n+\frac{\rho _n}{\gamma _n}\right) \psi _n'(s_n)\right| \le \frac{{\tilde{C}}}{2}\varepsilon _n. \end{aligned}$$

Substituting these formulas into (3.32) and using $\liminf _{n\rightarrow \infty }s_n\varepsilon _n^{1/2}>0$, we get

$$\begin{aligned} \begin{aligned} \left( r+\frac{\rho _n}{\gamma _n}\right) z_n'(r)&\le -4+\frac{2\alpha \beta _*}{\mu _n^{2-\beta _n}}+O(\varepsilon _n) \end{aligned} \end{aligned}$$

for all $r\in [s_n,(r_{i,n}-\rho _n)/\gamma _n]$. Lastly, for any $r\in [s_n,(r_{i,n}-\rho _n)/\gamma _n]$, dividing this inequality by $(r+\rho _n/\gamma _n)$ and integrating over $[s_n,r]$, we compute that

$$\begin{aligned} \begin{aligned} z_n(r)&\le -\left( 4-\frac{2\alpha \beta _*}{\mu _n^{2-\beta _n}}\right) \log {\left( r+\frac{\rho _n}{\gamma _n}\right) }+\left( 4-\frac{2\alpha \beta _*}{\mu _n^{2-\beta _n}}\right) \log {\left( s_n+\frac{\rho _n}{\gamma _n}\right) }\\ {}&\ \ \ +O(\varepsilon _n\log {r})+O(\varepsilon _n\log {s_n})+z_n(s_n)\\&\le \left( 1-\frac{\alpha \beta _*}{2\mu _n^{2-\beta _n}}\right) z(r)+O(\varepsilon _n\log {r}) \end{aligned} \end{aligned}$$

if $n\in {\mathbb {N}}$ is large enough. This completes the proof. $\square$

Using the previous lemma, we deduce the following asymptotic expansion of the energy.

Proposition 3.8

Assume $i\ge 1$ and (2.2)–(2.5). Moreover, suppose $\beta _*<3/2$ if $i\not =1$. Then we get

$$\begin{aligned} \mu _{i,n}\int _{\rho _{i,n}}^{r_{i,n}}\lambda _nf(u_{i,n})rdr=2-\frac{\alpha \beta _*}{\mu _{i,n}^{2-\beta _n}}+o\left( \frac{1}{\mu _{i,n}^{2-\beta _n}}\right) , \end{aligned}$$

(3.33)

and

$$\begin{aligned} \int _{\rho _{i,n}}^{r_{i,n}}\lambda _nu_{i,n}f(u_{i,n})rdr=2-\frac{\alpha \beta _*}{\mu _{i,n}^{2-\beta _n}}+o\left( \frac{1}{\mu _{i,n}^{2-\beta _n}}\right) . \end{aligned}$$

(3.34)

Proof

We write $\mu _n=\mu _{i,n}$, $\rho _n=\rho _{i,n}$, $r_n=r_{i,n}$, $\gamma _n=\gamma _{i,n}$, and $z_n=z_{i,n}$. We refer to the proof of Theorem 1 in [30]. We first note that

$$\begin{aligned} \begin{aligned}&\mu _n\int _{\rho _{n}}^{r_{n}}\lambda _nf(u_{n})rdr \\ {}&=\frac{1}{2}\int _{0}^{\frac{r_n-\rho _{n}}{\gamma _{n}}}\left( \frac{z_n}{2\mu _n^2}+1\right) e^{z_n+\frac{z_n^2}{4\mu _n^2}+\alpha \mu _n^{\beta _n}\left\{ \left( \frac{z_n}{2\mu _n^2}+1\right) ^{\beta _n}-1\right\} }\left( r+\frac{\rho _{n}}{\gamma _{n}}\right) dr \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&\int _{\rho _{n}}^{r_{n}}\lambda _nu_{n}f(u_{n})rdr \\ {}&=\frac{1}{2}\int _{0}^{\frac{r_n-\rho _{n}}{\gamma _{n}}}\left( \frac{z_n}{2\mu _n^2}+1\right) ^2 e^{z_n+\frac{z_n^2}{4\mu _n^2}+\alpha \mu _n^{\beta _n}\left\{ \left( \frac{z_n}{2\mu _n^2}+1\right) ^{\beta _n}-1\right\} }\left( r+\frac{\rho _{n}}{\gamma _{n}}\right) dr. \end{aligned} \end{aligned}$$

So it suffices to show that

$$\begin{aligned} \begin{aligned}&\frac{1}{2}\int _{0}^{\frac{r_{n}-\rho _{n}}{\gamma _{n}}}\left( \frac{z_n}{2\mu _n^2}+1\right) ^m e^{z_n+\frac{z_n^2}{4\mu _n^2}+\alpha \mu _n^{\beta _n}\left\{ \left( \frac{z_n}{2\mu _n^2}+1\right) ^{\beta _n}-1\right\} }\left( r+\frac{\rho _{n}}{\gamma _{n}}\right) dr\\&=2-\frac{\alpha \beta _*}{\mu _n^{2-\beta _n}}+o\left( \frac{1}{\mu _n^{2-\beta _n}}\right) \end{aligned} \end{aligned}$$

(3.35)

for $m=1,2$. To prove (3.35), we first claim

$$\begin{aligned} I_n&:=\frac{1}{2}\int _{0}^{s_n}\left( \frac{z_n}{2\mu _n^2}+1\right) ^m e^{z_n+\frac{z_n^2}{4\mu _n^2}+\alpha \mu _n^{\beta _n}\left\{ \left( \frac{z_n}{2\mu _n^2}+1\right) ^{\beta _n}-1\right\} }\left( r+\frac{\rho _{n}}{\gamma _{n}}\right) dr\nonumber \\&=2-\frac{\alpha \beta _*}{\mu _n^{2-\beta _n}}+o\left( \frac{1}{\mu _n^{2-\beta _n}}\right) \end{aligned}$$

(3.36)

for $m=1,2$ where $s_n>0$ is chosen as in the proof of Lemma 3.7. In fact, using the equation in (2.8) and noting $z_n=O(\log {(8+r^2)})$ on $[0,s_n]$ by (2.6), (3.9), (3.21) and (3.29), we get

$$\begin{aligned} \begin{aligned} I_n&=-\frac{1}{2}\int _0^{s_n}\left\{ 1+O\left( \frac{\log {(8+r^2)}}{\mu _n^{2}}\right) \right\} ^{m-1}\\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \times \left\{ \left( r+\frac{\rho _{n}}{\gamma _{n}}\right) \left( z'+\frac{\phi '}{\mu _n^{2-\beta _n}}+\psi _n'\right) \right\} 'dr, \end{aligned} \end{aligned}$$

for $m=1,2$. Here, we estimate by (2.6), our choice of $(s_n)$ and (3.4) that

$$\begin{aligned} \begin{aligned} I_{1,n}:=\int _0^{s_n}\left\{ \left( r+\frac{\rho _{n}}{\gamma _{n}}\right) z'\right\} 'dr=-4+o\left( \frac{1}{\mu _n^{2-\beta _n}}\right) , \end{aligned} \end{aligned}$$

and by (3.9) that

$$\begin{aligned} \begin{aligned} I_{2,n}:=\frac{1}{\mu _n^{2-\beta _n}}\int _0^{s_n}\left\{ \left( r+\frac{\rho _{n}}{\gamma _{n}}\right) \phi '\right\} 'dr =\frac{2\alpha \beta _*}{\mu _n^{2-\beta _n}}+o\left( \frac{1}{\mu _n^{2-\beta _n}}\right) , \end{aligned} \end{aligned}$$

and further, from (3.30), that

$$\begin{aligned} \begin{aligned} I_{3,n}:=\int _0^{s_n}\left\{ \left( \frac{\rho _{n}}{\gamma _{n}}+r\right) \psi _n'\right\} 'dr=o\left( \frac{1}{\mu _n^{2-\beta _n}}\right) . \end{aligned} \end{aligned}$$

Moreover we assert

$$\begin{aligned} \begin{aligned} \int _0^{s_n}&\frac{\log {(8+r^2)}}{\mu _n^2} \left\{ \left( r+\frac{\rho _{n}}{\gamma _{n}}\right) \left( z'+\frac{\phi '}{\mu _n^{2-\beta _n}}+\psi _n'\right) \right\} 'dr=o\left( \frac{1}{\mu _n^{2-\beta _n}}\right) . \end{aligned} \end{aligned}$$

Indeed, if $\beta _*>1$, we have $\mu _n^{2-\beta _n}\cdot \mu _n^{-2}\log {(8+r^2)}\rightarrow 0$ uniformly on $[0,s_n]$ by (3.25) and then, using the previous three formulas for $I_{1,n},I_{2,n}$ and $I_{3,n}$ above we easily get the assertion. On the other hand, if $\beta _*\le 1$, we get by integrating by parts and (2.6), (3.9), (3.21) and (3.30) that

$$\begin{aligned} \begin{aligned} \int _0^{s_n}&\frac{\log {(8+r^2)}}{\mu _n^2} \left\{ \left( r+\frac{\rho _{n}}{\gamma _{n}}\right) \left( z'+\frac{\phi '}{\mu _n^{2-\beta _n}}+\psi _n'\right) \right\} 'dr\\&=\frac{1}{\mu _n^2}\bigg \{\log {(8+s_n^2)}\left( I_{1,n}+I_{2,n}+I_{3,n}\right) \\&\ \ \ \ \ \ \ \ -\int _0^{s_n}\frac{2r}{8+r^2}\left( r+\frac{\rho _n}{\gamma _n}\right) \left( \frac{-4r}{(8+r^2)^2}+O\left( \frac{1}{\mu _n^{2-\beta _n}(1+r)}\right) \right) dr\bigg \}\\&=o\left( \frac{1}{\mu _n^{2-\beta _n}}\right) \end{aligned} \end{aligned}$$

by the three formulas for $I_{1,n},I_{2,n}$ and $I_{3,n}$ above and a direct calculation. This proves the assertion. As a consequence, we get

$$\begin{aligned} I_n=-\frac{1}{2}(I_{1,n}+I_{2,n}+I_{3,n})+o\left( \frac{1}{\mu _n^{2-\beta _n}}\right) . \end{aligned}$$

This shows the claim. Next we claim that

$$\begin{aligned} \begin{aligned} J_n^m&:=\int _{s_n}^{\frac{r_n-\rho _{n}}{\gamma _{n}}}\left( \frac{z_n(r)}{2\mu _n^2}+1\right) ^m e^{z_n+\frac{z_n^2}{4\mu _n^2}+\alpha \mu _n^{\beta _n}\left\{ \left( \frac{z_n}{2\mu _n^2}+1\right) ^{\beta _n}-1\right\} }rdr\\&=o\left( \frac{1}{\mu _n^{2-\beta _n}}\right) \end{aligned} \end{aligned}$$

(3.37)

for $m=1,2$. For the proof, it suffices to consider the case $m=1$ since $J_n^2\le J_n^1$. Hence let us show (3.37) for $m=1$. To this end, we first set a sequence $(c_n)\subset (0,1]$ so that $c_n:=1$ if $n\in {\mathbb {N}}$ satisfies $\beta _n\ge 1$ and $c_n:=\beta _n$ otherwise. Then, we define a value $c_*>0$ by $\lim _{n\rightarrow \infty }c_n=c_*$. Furthermore, we fix a small constant $\eta >0$ so that $c_*\alpha -(\alpha \beta _*+\eta )/2>0$. Noting (3.17) we may assume

$$\begin{aligned} z_n(r)\le \left( 1-\frac{\alpha \beta _*+\eta }{2\mu _n^{2-\beta _n}}\right) z(r) \end{aligned}$$

(3.38)

for all $r\in [s_n,(r_n-\rho _n)/\gamma _n]$. After this, we put $a_n:=1-(\alpha \beta _*+\eta )/(2\mu _n^{2-\beta _n})$. In addition, we have that

$$\begin{aligned} \left( \frac{z_n(r)}{2\mu _n^2}+1\right) ^{\beta _n}-1\le \frac{c_n z_n(r)}{2\mu _n^2} \end{aligned}$$

for any $r\in [s_n,(r_n-\rho _n)/\gamma _n]$. This is clearly obtained by noting $z_n/(2\mu _n^2)+1\in [0,1]$ in the case $\beta _n\ge 1$ and by using the mean value theorem if $\beta _n\in (0,1)$. It follows that

$$\begin{aligned} \begin{aligned} J_n^1&=\int _{s_n}^{\frac{r_n-\rho _{n}}{\gamma _{n}}}\left( \frac{z_n}{2\mu _n^2}+1\right) e^{z_n+\frac{z_n^2}{4\mu _n^2}+\alpha \mu _n^{\beta _n}\left\{ \left( \frac{z_n}{2\mu _n^2}+1\right) ^{\beta _n}-1\right\} }rdr\\&\le \int _{s_n}^{\frac{r_n-\rho _{n}}{\gamma _{n}}}\left( \frac{z_n}{2\mu _n^2}+1\right) e^{\left( 1+\frac{\alpha c_n}{2\mu _n^{2-\beta _n}}\right) z_n+\frac{z_n^2}{4\mu _n^2}}rdr\\&\le \int _{s_n}^{\frac{r_n-\rho _{n}}{\gamma _{n}}}\left( \frac{a_n z}{2\mu _n^2}+1\right) e^{\left( 1+\frac{\alpha c_n}{2\mu _n^{2-\beta _n}}\right) a_nz+\frac{a_n^2z^2}{4\mu _n^2}}rdr \end{aligned} \end{aligned}$$

(3.39)

by (3.38). Here we note that $a_n z(r)/(2\mu _n^2)+1\ge 0$ if and only if

$$\begin{aligned} r\le \sqrt{8\left( e^{\frac{\mu _n^2}{a_n}}-1\right) }=:R_n. \end{aligned}$$

Then it is clear by (3.38) that $R_n\ge (r_n-\rho _n)/\gamma _n$. On the other hand since $\liminf _{n\rightarrow \infty }s_n\varepsilon _n^{\frac{1}{2}}>0$, it follows from (3.4) that there exists a sequence $(M_n)\subset (0,\infty )$ such that $M_n\rightarrow \infty$ and $s_n\ge \sqrt{ M_n\mu _n^{2-\beta _n}}$ for all $n\in {\mathbb {N}}$. Then from (3.39) we compute with changing the variable by $\tau =-a_n z(r)$ and putting ${\tilde{s}}_n:=-a_n z(s_n)$ and ${\tilde{R}}_n:=-a_n z(R_n)$ that

$$\begin{aligned} \begin{aligned} J_n^1&\le \int _{s_n}^{R_n}\left( \frac{a_n z}{2\mu _n^2}+1\right) e^{\left( 1+\frac{\alpha c_n}{2\mu _n^{2-\beta _n}}\right) a_n z+\frac{a_n^2z^2}{4\mu _n^2}}rdr\\&= \frac{2}{a_n}\int _{{\tilde{s}}_n}^{{\tilde{R}}_n}\left( 1-\frac{\tau }{2\mu _n^2}\right) e^{\frac{\tau ^2}{4\mu _n^2}+\left( \frac{1}{2a_n}-1-\frac{\alpha c_n}{2\mu _n^{2-\beta _n}}\right) \tau }d\tau \\ \end{aligned} \end{aligned}$$

Here notice that ${\tilde{s}}_n=4\log {s_n}+O(1)$ by (3.25) and ${\tilde{R}}_n=2\mu _n^2$ from the definition. Again changing the variable by

$$\begin{aligned} t=\frac{\tau }{2\mu _n}+\mu _n\left( \frac{1}{2a_n}-1-\frac{\alpha c_n}{2\mu _n^{2-\beta _n}}\right) \end{aligned}$$

and setting

$$\begin{aligned} \begin{aligned} {\bar{R}}_n&=\frac{{\tilde{R}}_n}{2\mu _n}+\mu _n\left( \frac{1}{2a_n}-1-\frac{\alpha c_n}{2\mu _n^{2-\beta _n}}\right) \end{aligned} \end{aligned}$$

and

$$\begin{aligned} {\bar{s}}_n=\frac{{\tilde{s}}_n}{2\mu _n}+\mu _n\left( \frac{1}{2a_n}-1-\frac{\alpha c_n}{2\mu _n^{2-\beta _n}}\right) , \end{aligned}$$

we get

$$\begin{aligned} \begin{aligned} J_n^1&\le \frac{4\mu _n}{a_n \exp {\left[ \mu _n^2\left( \frac{1}{2a_n}-1-\frac{\alpha c_n}{2\mu _n^{2-\beta _n}}\right) ^2\right] }}\\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \times \int _{{\bar{s}}_n}^{{\bar{R}}_n}\left( -\frac{t}{\mu _n}+\frac{1}{2a_n}-\frac{\alpha c_n}{2\mu _n^{2-\beta _n}}\right) e^{t^2}dt. \end{aligned} \end{aligned}$$

(3.40)

Now we calculate by (3.25) and our choices of $(M_n)$ and $\eta$ that

$$\begin{aligned} \begin{aligned}&\mu _n^{2-\beta _n}\frac{4\mu _n}{a_n\exp {\left[ \mu _n^2\left( \frac{1}{2a_n}-1-\frac{\alpha c_n}{2\mu _n^{2-\beta _n}}\right) ^2\right] }}\int _{{\bar{s}}_n}^{{\bar{R}}_n}\frac{t}{\mu _n}e^{t^2}dt\\&=O\left( \mu _n^{2-\beta _n}\exp {\left\{ -\left( c_*\alpha -\frac{\alpha \beta _*+\eta }{2}+o(1)\right) \mu _n^{\beta _n}\right\} }\right) +O\left( \frac{1}{M_n}\right) \\&\rightarrow 0, \end{aligned} \end{aligned}$$

(3.41)

as $n\rightarrow \infty$. Similarly, we get

$$\begin{aligned} \begin{aligned} A_n:&=\mu _n^{2-\beta _n}\frac{4\mu _n}{a_n\exp {\left[ \mu _n^2\left( \frac{1}{2a_n}-1-\frac{\alpha c_n}{2\mu _n^{2-\beta _n}}\right) ^2\right] }}\int _{{\bar{s}}_n}^{{\bar{R}}_n}e^{t^2}dt\\&\rightarrow 0 \end{aligned} \end{aligned}$$

(3.42)

as $n\rightarrow \infty$. In fact, noting ${\bar{s}}_n=-(1+o(1))\mu _n/2+O(1)$ by (3.25) and ${\bar{R}}_n=(1+o(1))\mu _n/2$ we decompose

$$\begin{aligned} A_n=A_{1,n}+A_{2,n} \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} A_{1,n}&:=\mu _n^{2-\beta _n}\frac{4\mu _n}{a_n\exp {\left[ \mu _n^2\left( \frac{1}{2a_n}-1-\frac{\alpha c_n}{2\mu _n^{2-\beta _n}}\right) ^2\right] }}\int _{\{|t|\le \mu _n/4\}}e^{t^2}dt. \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} A_{2,n}&:=\mu _n^{2-\beta _n}\frac{4\mu _n}{a_n\exp {\left[ \mu _n^2\left( \frac{1}{2a_n}-1-\frac{\alpha c_n}{2\mu _n^{2-\beta _n}}\right) ^2\right] }}\int _{[{\bar{s}}_n,{\bar{R}}_n]\cap \{|t|\ge \mu _n/4\}}e^{t^2}dt. \end{aligned} \end{aligned}$$

Since

$$\begin{aligned} A_{1,n}=O\left( \frac{\mu _n^{3-\beta _n}}{e^{(1+o(1))\mu _n^2/4}}\cdot \frac{\mu _n}{2} e^{\mu _n^2/16}\right) , \end{aligned}$$

we easily get $A_{1,n}\rightarrow 0$. On the other hand, since

$$\begin{aligned} \begin{aligned} A_{2,n}&\le \mu _n^{2-\beta _n}\frac{4\mu _n}{a_n\exp {\left[ \mu _n^2\left( \frac{1}{2a_n}-1-\frac{\alpha c_n}{2\mu _n^{2-\beta _n}}\right) ^2\right] }}\cdot 4\int _{[{\bar{s}}_n,{\bar{R}}_n]\cap \{|t|\ge \mu _n/4\}}\frac{|t|}{\mu _n}e^{t^2}dt, \end{aligned} \end{aligned}$$

we get $A_{2,n}\rightarrow 0$ similarly to the calculation above. Consequently, using (3.41) and (3.42) for (3.40), we get $J_n^1=o(\mu _n^{-(2-\beta _n)})$. This proves our claim (3.37). Then by (3.36) and (3.37), we readily obtain (3.35). This finishes the proof. $\square$

We also get the following.

Lemma 3.9

Suppose $i\ge 1$ and (2.2)–(2.5) and further, if $i\not =1$, let $\beta _*<3/2$. Then we have

$$\begin{aligned} \mu _{i,n}\int ^{\rho _{i,n}}_{r_{i-1,n}}\lambda _nf(u_{i,n})rdr=o\left( \frac{1}{\mu _{i,n}^{2-\beta _n}}\right) , \end{aligned}$$

(3.43)

and

$$\begin{aligned} \int ^{\rho _{i,n}}_{r_{i-1,n}}\lambda _nu_{i,n}f(u_{i,n})rdr=o\left( \frac{1}{\mu _{i,n}^{2-\beta _n}}\right) . \end{aligned}$$

(3.44)

Proof

By the first assertion in Lemma 3.2 and (3.4), we get $(\rho _{i,n}-r_{i-1,n})/\gamma _{i,n}=o(\mu _{i,n}^{-(2-\beta _n)})$. It folows that

$$\begin{aligned} \begin{aligned}&\frac{1}{2}\int _{\frac{r_{i-1,n}-\rho _{i.n}}{\gamma _{i,n}}}^0\left( \frac{z_{i,n}}{2\mu _n^2}+1\right) ^m e^{z_{i,n}+\frac{z_{i,n}^2}{4\mu _n^2}+\alpha \mu _n^{\beta _n}\left\{ \left( \frac{z_{i,n}}{2\mu _n^2}+1\right) ^{\beta _n}-1\right\} }\left( r+\frac{\rho _{n}}{\gamma _{n}}\right) dr\\&=o\left( \frac{1}{\mu _{i,n}^{2-\beta _n}}\right) , \end{aligned} \end{aligned}$$

for $m=1,2$. This proves (3.43) and (3.44). We finish the proof. $\square$

We get the proof of Proposition 3.1.

Proof of Proposition 3.1

(1.4) follows from (2.1), (3.34) and (3.44). (1.5) is proved by (3.33) and (3.43). This finishes the proof. $\square$

We end this section by proving the next key lemma.

Proposition 3.10

We assume $i\ge 1$ and (2.2)–(2.5). Moreover, suppose $\beta _*<3/2$ if $i>1$. Then we get

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{\log {\frac{1}{\lambda _nr_{i,n}^2}}}{\mu _{i,n}^{\beta _n}}=\delta , \end{aligned}$$

(3.45)

and

$$\begin{aligned} \lim _{n\rightarrow \infty }\mu _{i,n}r_{i,n} |u_{i,n}'(r_{i,n})|=2. \end{aligned}$$

(3.46)

Proof

We denote $\mu _n=\mu _{i,n}$, $\rho _n=\rho _{i,n}$, $r_n=r_{i,n}$, $\gamma _n=\gamma _{i,n}$, and $z_n=z_{i,n}$ as usual. To deduce (3.45), we first claim that there exists a constant $C>0$ such that

$$\begin{aligned} \lambda _n r_n^2 \mu _n^2e^{(\delta +o(1))\mu _n^{\beta _n}}\le C \end{aligned}$$

for all $n\in {\mathbb {N}}$. In fact, (3.17) implies that for any $r\in [1,(r_{n}-\rho _n)/\gamma _n]$,

$$\begin{aligned} z_n(r)\le \left( 1-\frac{\alpha \beta _*+o(1)}{2\mu _n^{2-\beta _n}}\right) z(r), \end{aligned}$$

if $n\in {\mathbb {N}}$ is large enough. Choosing $r=(r_n-\rho _n)/\gamma _n$, we get by the first and second assertions in Lemma 2.5 that

$$\begin{aligned} \begin{aligned} 0&\le \mu _n^2-\left( 1-\frac{\alpha \beta _*+o(1)}{2\mu _n^{2-\beta _n}}\right) \log {2\lambda _nr_{n}^2\mu _n^2e^{\mu _n^2+\alpha \mu _n^{\beta _n}}}+O(1).\end{aligned} \end{aligned}$$

This implies that there exists a constant ${\bar{C}}>0$ such that

$$\begin{aligned} \log {\lambda _nr_{n}^2\mu _n^2e^{(\delta +o(1))\mu _n^{\beta _n}}}\le {\bar{C}} \end{aligned}$$

for all large $n\in {\mathbb {N}}$. This proves the claim. It follows that

$$\begin{aligned} \liminf _{n\rightarrow \infty }\frac{\log {\frac{1}{\lambda _nr_{n}^2}}}{\mu _n^{\beta _n}}\ge \delta . \end{aligned}$$

(3.47)

Next, we shall show

$$\begin{aligned} \limsup _{n\rightarrow \infty }\frac{\log {\frac{1}{\lambda _nr_{n}^2}}}{\mu _n^{\beta _n}}\le \delta . \end{aligned}$$

(3.48)

To do this, we use the first formula in Lemma 2.2 to obtain

$$\begin{aligned} \begin{aligned}&2\mu _n^2\\&=\log {\frac{r_{n}}{\gamma _n}}\\&\times \int _0^{\frac{r_{n}-\rho _n}{\gamma _n}}\left( \frac{z_n}{2\mu _n^2}+1\right) e^{z_n+\frac{z_n^2}{4\mu _n^2}+\alpha \mu _n^{\beta _n}\left\{ \left( \frac{z_n}{2\mu _n^2}+1\right) ^{\beta _n}-1\right\} }\left( r+\frac{\rho _{n}}{\gamma _{n}}\right) dr\\&+\int _0^{\frac{r_{n}-\rho _n}{\gamma _n}}\left( \frac{z_n}{2\mu _n^2}+1\right) e^{z_n+\frac{z_n^2}{4\mu _n^2}+\alpha \mu _n^{\beta _n}\left\{ \left( \frac{z_n}{2\mu _n^2}+1\right) ^{\beta _n}-1\right\} }\\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \times \left( r+\frac{\rho _{n}}{\gamma _{n}}\right) \log {\frac{1}{\left( r+\frac{\rho _{n}}{\gamma _{n}}\right) }}dr. \end{aligned} \end{aligned}$$

(3.49)

Here, we observe that for any value $R_0>1$,

$$\begin{aligned} \begin{aligned}&\int _0^{\frac{r_{n}-\rho _n}{\gamma _n}}\left( \frac{z_n}{2\mu _n^2}+1\right) e^{z_n+\frac{z_n^2}{4\mu _n^2}+\alpha \mu _n^{\beta _n}\left\{ \left( \frac{z_n}{2\mu _n^2}+1\right) ^{\beta _n}-1\right\} }\\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \times \left( r+\frac{\rho _{n}}{\gamma _{n}}\right) \log {\frac{1}{\left( r+\frac{\rho _{n}}{\gamma _{n}}\right) }}dr\\&\le \int _0^{R_0}e^zr\log {\frac{1}{r}}dr+o(1) \end{aligned} \end{aligned}$$

where $o(1)\rightarrow 0$ as $n\rightarrow \infty$. Then, since we can choose $R_0>1$ so large that $\int _0^{R_0}e^zr\log {\frac{1}{r}}dr<0$, the second term of the right hand side of (3.49) is negative value for all large $n\in {\mathbb {N}}$. Hence using this and (3.33) for (3.49), we get

$$\begin{aligned} 2\mu _n^2\le \left( 2-\frac{\alpha \beta _*}{\mu _n^{2-\beta _n}}+o\left( \frac{1}{\mu _n^{2-\beta _n}}\right) \right) \log {2\lambda _n r_{n}^2\mu _nf_n(\mu _n)}. \end{aligned}$$

It follows that

$$\begin{aligned} \frac{\log {\frac{1}{\lambda _nr_{n}^2}}}{\mu _n^{\beta _n}}\le \delta +o(1). \end{aligned}$$

This proves (3.48). Then (3.47) and (3.48) show (3.45). Finally we shall ensure (3.46). For any $r\in (\rho _n,r_n)$, multiplying the equation in (2.1) by r and integrating by parts over $(\rho _n,r_n)$, we get

$$\begin{aligned} -r_{n}u_n'(r_{n})=\lambda _n\int _{\rho _{n}}^{r_{n}}f(u_n)rdr. \end{aligned}$$

Hence we obtain from (3.33) that

$$\begin{aligned} \lim _{n\rightarrow \infty }\mu _nr_{n}|u_n'(r_{n})|=2. \end{aligned}$$

This gives (3.46). We complete the proof. $\square$

4 Behavior of non-concentrating parts

In this section we mainly discuss the behavior of a sequence $(u_{i,n})$ which does not blow up. This is useful to deduce precise informations on the weak limit. Especially, Lemmas 4.6 and 4.7 will be important for the proof in the case of (ii) of Theorems 1.5 and 1.6. We begin with the next basic lemma. Let ${\mathcal {N}}_{\lambda ,\beta }$ be the Nehari manifold defined in Sect. 1.1.

Lemma 4.1

For any $\Lambda _0\in (0,\Lambda _1)$ and $\beta _0\in (0,2)$, we have a constant $K>0$ such that

$$\begin{aligned} \int _B|\nabla u|^2dx\ge K \end{aligned}$$

for all $u\in {\mathcal {N}}_{\lambda ,\beta }$ and all $(\lambda ,\beta )\in (0,\Lambda _0]\times [\beta _0,2]$.

Proof

The proof is standard. For the readers’ convenience we show the proof. First fix $\Lambda _0\in (0,\Lambda _1)$ and $\beta _0\in (0,2)$ and assume $\lambda \in (0,\Lambda _0]$ and $\beta \in [\beta _0,2]$. Next choose $\varepsilon >0$ so that $(1+\varepsilon )\Lambda _0<\Lambda _1$. Then for any $p>2$, we can find a constant $C_1>0$ independently of $\beta \in [\beta _0,2]$ such that $|t^2e^{t^2+\alpha |t|^\beta }|\le (1+\varepsilon )t^2+C_1t^pe^{(1+\alpha )t^2}$ for all $t\in {\mathbb {R}}$. Then the Hölder, Poincare, Sobolev, and Trudinger–Moser (1.2) inequalities suggest that there exists a constant $C_2>0$, which is independent of $\beta \in [\beta _0,2]$, such that

$$\begin{aligned} \begin{aligned} \int _Bu^2e^{u^2+\alpha |u|^\beta }dx&\le (1+\varepsilon )\int _Bu^2dx+C_1\left( \int _B|u|^{2p}dx\right) ^{\frac{1}{2}}\left( \int _Be^{2(1+\alpha )u^2}dx\right) ^{\frac{1}{2}}\\&\le \frac{1+\varepsilon }{\Lambda _1}\Vert u\Vert ^2_{H^1_0(B)}+C_2\Vert u\Vert _{H^1_0(B)}^p, \end{aligned} \end{aligned}$$

for all $u\in H^1_0(B)$ with $\Vert u\Vert _{H^1_0(B)}^2\le 4\pi /(2(1+\alpha ))$. Hence, it follows that

$$\begin{aligned} \langle I_{\lambda ,\beta }'(u),u\rangle \ge \left( 1-\frac{(1+\varepsilon )\Lambda _0}{\Lambda _1}\right) \Vert u\Vert _{H^1_0(B)}^2-\Lambda _0C_2\Vert u\Vert _{H^1_0(B)}^p \end{aligned}$$

for any $u\in H^1_0(B)$ satisfying $\Vert u\Vert _{H^1_0(B)}^2\le 4\pi /(2(1+\alpha ))$. Since $2<p$, we get a constant $C_3>0$, which is independent of $\lambda \in (0,\Lambda _0]$ and $\beta \in [\beta _0,2]$, such that $\langle I_{\lambda ,\beta }(u),u\rangle >0$ for all $u\in H^1_0(B)$ with $\Vert u\Vert _{H^1_0(B)}\le C_3$. Therefore $u\in {\mathcal {N}}_{\lambda ,\beta }$ implies $\Vert u\Vert _{H^1_0(B)}\ge C_3$. This finishes the proof. $\square$

From now on, as usual, let $k\in \{0\}\cup {\mathbb {N}}$ and $\{(\lambda _n,\beta _n)\}\subset (0,\infty )\times (0,2)$ be a sequence such that $(\lambda _n,\beta _n)\rightarrow (\lambda _*,\beta _*)$ as $n\rightarrow \infty$ for some value $(\lambda _*,\beta _*)\in [0,\infty )\times (0,2)$. Moreover, assume that $(u_n)$ is a sequence of solutions satisfying $u_n\in S_{k,\lambda _n,\beta _n}$ for all $n\in {\mathbb {N}}$. In the following lemmas, we always suppose

$$\begin{aligned} \int _{B}|\nabla u_n|^2dx\text { uniformly bounded for all }n\in {\mathbb {N}}, \end{aligned}$$

(4.1)

if $k\not =0$. All the other notations below are defined as in the main theorems. We get the following.

Lemma 4.2

Assume (4.1). If $\mu _{i,n}\rightarrow \infty$ as $n\rightarrow \infty$ for some $i\in \{1,\cdots ,k+1\}$, then we have $\lim _{n\rightarrow \infty }\rho _{i,n}=0.$ On the other hand, if $\lim _{n\rightarrow \infty }r_{i,n}=0$ for some $i\in \{1,\cdots ,k\}$, then we get $\mu _{j,n}\rightarrow \infty$ for all $j=1,\cdots ,i$. Finally, if $\lambda _*=0$, then we obtain $\mu _{k+1,n}\rightarrow \infty$.

Proof

First assume $\mu _{i,n}\rightarrow \infty$ as $n\rightarrow \infty$ for some $i\in \{1,\cdots ,k+1\}$. Then Lemma 2.1 implies that there exists a constant $c>0$ such that

$$\begin{aligned} \rho _{i,n}\mu _{i,n}^2\le c^2 \int _B|\nabla u_n|^2dx, \end{aligned}$$

for all $n\in {\mathbb {N}}$. Hence, by our assumptions, we get $\rho _{i,n}\rightarrow 0$ as $n\rightarrow \infty$. This shows the first assertion. Next, we suppose $r_{i,n}\rightarrow 0$ for some $i\in \{1,\cdots ,k\}$. Then, assume there exists a number $j\in \{1,\cdots ,i\}$ such that $\mu _{j,n}$ is uniformly bounded up to a subsequence on the contrary. Then for any $x\in B$, we put ${\tilde{u}}_{j,n}(x):=u_n(r_{j,n}x)$ if $r_{j-1,n}/r_{j,n}<|x|<1$ and ${\tilde{u}}_{j,n}(x):=0$ otherwise. It follows that ${\tilde{u}}_{j,n}\in {\mathcal {N}}_{\lambda _n r_{j,n}^2,\beta _n}$. Then since $\lambda _n r_{i,n}^2\rightarrow 0$, we get by Lemma 4.1 that there exists a constant $K>0$ such that

$$\begin{aligned} K\le \int _B\lambda _nr_{i,n}^2 {\tilde{u}}_{j,n}f_n({\tilde{u}}_{j,n})dx\rightarrow 0, \end{aligned}$$

since ${\tilde{u}}_{j,n}$ is uniformly bounded. This is a contradiction. Hence we prove the second assertion. Finally, assume $\lambda _*=0$ and $\mu _{k+1,n}$ is uniformly bounded up to a subsequence on the contrary. Then for all $x\in B$, we put ${\bar{u}}_{k+1,n}(x):=u_n(x)$ if $r_{k,n}<|x|<1$ and ${\bar{u}}_{k+1,n}(x):=0$ otherwise. Since ${\bar{u}}_{k+1,n}\in {\mathcal {N}}_{\lambda _n,\beta _n}$ and $\lambda _*=0$, we can apply Lemma 4.1 again and get a constant $K>0$ such that

$$\begin{aligned} K\le \lambda _n\int _B {\bar{u}}_{k+1,n}f_n({\bar{u}}_{k+1,n})dx\rightarrow 0 \end{aligned}$$

since ${\bar{u}}_{k+1,n}$ is uniformly bounded. This is a contradiction. This ensures the last assertion. We complete the proof. $\square$

After this, we regard $u_n=u_n(|x|)$ and study the behavior of the function $u_n(r)$ ($r\in [0,1]$). Let us give the next three standard lemmas.

Lemma 4.3

Suppose (4.1). Assume that there exists a number $i\in \{1,\cdots ,k\}$ such that $\mu _{i,n}$ is uniformly bounded for all $n\in {\mathbb {N}}$. Then $\mu _{i+1,n}$ is also uniformly bounded. Furthermore, there exist constants $r_{i-1},,r_i,\rho _i$, and $\rho _{i+1}$ such that $0\le r_{i-1}\le \rho _i<r_i<\rho _{i+1}<1$, $r_{i-1,n}\rightarrow r_{i-1}$, $r_{i,n}\rightarrow r_i$, $\rho _{i,n}\rightarrow \rho _i$ and $\rho _{i+1,n}\rightarrow \rho _{i+1}$ by extracting a subsequence if necessary. Moreover, $r_{i-1}=\rho _i$ if and only if $\rho _i=0$.

Proof

First, assume that $\mu _{i+1,n}\rightarrow \infty$ up to a subsequence on the contrary. Then the first assertion in Lemma 4.2 suggests that $r_{i,n}<\rho _{i+1,n}\rightarrow 0$. Then the second assertion in the same lemma implies $\mu _{i,n}\rightarrow \infty$ which is a contradiction. This proves the first assertion in the present lemma. Next, we choose constants $0\le r_{i-1}\le \rho _i\le r_{i}\le \rho _{i+1}\le 1$ so that $r_{i-1,n}\rightarrow r_{i-1}$, $r_{i,n}\rightarrow r_i$, $\rho _{i,n}\rightarrow \rho _i$, and $\rho _{i+1,n}\rightarrow \rho _{i+1}$ by taking a subsequence if necessary. We claim $\rho _i<r_{i}$. In fact, if $\rho _i=r_i$ on the contrary, Lemma 2.2 shows that

$$\begin{aligned} \begin{aligned} 1&=\int _{\rho _{i,n}}^{r_{i,n}}\lambda _n \left| \frac{f_n(u_n)}{\mu _{i,n}}\right| r\log {\frac{r_{i,n}}{r}}dr\le \lambda _n \left| \frac{f_n(\mu _{i,n})}{\mu _{i,n}}\right| \max _{r\in (0,1]}\left| r\log {\frac{1}{r}}\right| (r_{i,n}-\rho _{i,n})\\&\rightarrow 0. \end{aligned} \end{aligned}$$

This is a contradiction. Next we show $r_{i}<\rho _{i+1}$. Otherwise, we get $0<r_i=\rho _{i+1}$. Then again Lemma 2.2 implies

$$\begin{aligned} \begin{aligned} 1&=\int _{r_{i,n}}^{\rho _{i+1,n}}\lambda _n \left| \frac{f_n(u_n)}{\mu _{i+1,n}}\right| r\log {\frac{r}{r_{i,n}}}dr\le \lambda _n\left| \frac{f_n(\mu _{i+1,n})}{\mu _{i+1,n}}\right| \log {\frac{\rho _{i+1,n}}{r_{i,n}}}\frac{\rho _{i+1,n}^2-r_{i,n}^2}{2}\\&\rightarrow 0, \end{aligned} \end{aligned}$$

which is a contradiction. Next we ensure $\rho _{i+1}<1$. If not, we have $1\ge r_{i+1,n}>\rho _{i+1,n}\rightarrow 1$ and then analogously, we get

$$\begin{aligned} \begin{aligned} 1&=\int _{\rho _{i+1,n}}^{r_{i+1,n}}\lambda _n \left| \frac{f_n(u_n)}{\mu _{i+1,n}}\right| r\log {\frac{r_{i+1,n}}{r}}dr=O\left( \log {\frac{r_{i+1,n}}{\rho _{i+1,n}}}\frac{r_{i+1,n}^2-\rho _{i+1,n}^2}{2}\right) \rightarrow 0. \end{aligned} \end{aligned}$$

This is a contradiction. Finally, we suppose $r_{i-1}=\rho _i>0$ on the contrary. Then similarly, we see that

$$\begin{aligned} \begin{aligned} 1&=\int ^{\rho _{i,n}}_{r_{i-1,n}}\lambda _n \left| \frac{f_n(u_n)}{\mu _{i,n}}\right| r\log {\frac{r}{r_{i-1,n}}}dr=O\left( \log {\frac{\rho _{i,n}}{r_{i-1,n}}}\frac{\rho _{i,n}^2-r_{i-1,n}^2}{2}\right) \rightarrow 0. \end{aligned} \end{aligned}$$

This is a contradiction. This completes the proof. $\square$

Lemma 4.4

Assume (4.1). Suppose that for some $i\in \{1,\cdots ,k+1\}$, there exist constants $\mu _i\ge 0$, $r_{i-1}\le \rho _i<r_i\le 1$ such that $\mu _{i,n}\rightarrow \mu _i$, $\rho _{i,n}\rightarrow \rho _i$, and $r_{j,n}\rightarrow r_j$ for $j=i-1,i$. Then we have a nontrivial function $w_i$ in $(r_{i-1},r_i)$ such that $u_{i,n}/\mu _{i,n}\rightarrow w_i$ in $C^2_{\text {loc}}((r_{i-1},r_i))$. Furthermore, if $r_{i-1}<\rho _i$ (which yields $i\not =1$), then we get $r_{i-1}>0$. Finally, $\rho _i=0$ implies $\lim _{r\rightarrow 0+0}w_i(r)=1$.

Proof

We may suppose $u_{i,n}\ge 0$. Put $w_{i,n}:=u_{i,n}/\mu _{i,n}$. Then it satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} -w_{i,n}''-\frac{1}{r}w_{i,n}'=\lambda _n w_{i,n}\frac{f_n(\mu _{i,n}w_{i,n})}{\mu _{i,n}w_{i,n}},\ 0< w_{i,n}\le 1\text { in }(r_{i-1,n},r_{i,n}),\\ w_{i,n}(r_{i,n})=0=w_{i,n}'(\rho _{i,n}),\ w_{i,n}(\rho _{i,n})=1,\\ w_{i,n}(r_{i-1,n})=0\text { if }i\not =1. \end{array}\right. } \end{aligned}$$

(4.2)

Notice that $\lambda _n w_{i,n}f_n(\mu _{i,n}w_{i,n})/(\mu _{i,n}w_{i,n})$ is uniformly bounded in $(r_{i-1,n},r_{i,n})$. Then using the equation and conditions in (4.2), we clearly get a function $w_i \ge 0$ such that $w_{i,n}\rightarrow w_i$ in $C^2_{\text {loc}}((r_{i-1},r_i))$. Now, let us assume $r_{i-1}<\rho _i$. We may suppose $i\not =1$. Then we have $\rho _i\in (r_{i-1},r_i)$ and thus, we obviously see $w_i(\rho _i)=1$. It follows that $r_{i-1}>0$. Otherwise, Lemma 2.2 shows that

$$\begin{aligned} \begin{aligned} 1&=\log {\frac{1}{r_{i-1,n}}}\left( \int _{0}^{\rho _{i}}\lambda _* w_i \frac{f_*(\mu _i w_i)}{\mu _i w_i}rdr+o(1)\right) +\int _{0}^{\rho _{i}}\lambda _* w_i\frac{f_*(\mu _i w_{i})}{\mu _i w_i}r\log {r}dr\\&\ \ +o(1) \end{aligned} \end{aligned}$$

where we defined $f_*(t)/t=1$ if $t=0$. Since $\lambda _*\not =0$ by Lemma 4.2, $f(t)/t\ge 1$ for any $t\ge 0$ and $w_i(\ge 0)$ is nontrivial on $(0,\rho _i)$, we get that the right hand side of the formula above diverges to infinity. This is a contradiction. This proves the second assertion of the lemma. Finally, let us suppose $0=\rho _i=r_{i-1}$. Then we claim that there exists a constant $C>0$ such that

$$\begin{aligned} |w_{i,n}'(r)|\le C \end{aligned}$$

(4.3)

for all $r\in [\rho _{i,n},r_{i,n}]$ and all $n\in {\mathbb {N}}$. To see this, for any $r\in [\rho _{i,n},r_{i,n}]$, we multiply the equation in (4.2) by r and integrate over $[\rho _{i,n},r]$ and get

$$\begin{aligned} -rw_{i,n}'(r)=O\left( \frac{r^2-\rho _{i,n}^2}{2}\right) , \end{aligned}$$

(4.4)

for all $r\in [\rho _{i,n},r_{i,n}]$. This readily proves the claim. Then we confirm that $\lim _{r\rightarrow 0+0}w_i(r)=1$. If not, we have a sequence $(\sigma _n)\subset (0,r_i)$ and a constant $\varepsilon _0\in (0,1]$ such that $\sigma _n\rightarrow 0$ and $w_i(\sigma _n)\rightarrow 1-\varepsilon _0$ as $n\rightarrow \infty$. Then we can choose a sequence $({\tilde{\sigma }}_n)\subset (\rho _{i,n},r_{i,n})$ so that ${\tilde{\sigma }}_n\rightarrow 0$ and $w_{i,n}({\tilde{\sigma }}_n)\rightarrow 1-\varepsilon _0$ by selecting a suitable subsequence. Consequently, it follows from the mean value theorem that there exists a sequence $({\bar{\sigma }}_n)\subset (\rho _{i,n},{\tilde{\sigma }}_{n})$ such that ${\bar{\sigma }}_n\rightarrow 0$ and

$$\begin{aligned} w_{i,n}'({\bar{\sigma }}_n)=\frac{w_{i,n}({\tilde{\sigma }}_n)-w_{i,n}(\rho _{i,n})}{{\tilde{\sigma }}_n-\rho _{i,n}}\rightarrow -\infty . \end{aligned}$$

This contradicts (4.3). This finishes the proof. $\square$

Lemma 4.5

We suppose (4.1). Assume that for some $i\in \{1,\cdots ,k\}$, there exists a value $\mu _i\ge 0$ such that $\mu _{i,n}\rightarrow \mu _i$. Then by extracting a subsequence, we have a constant $\mu _{i+1}\ge 0$ such that $\mu _{i+1,n}\rightarrow \mu _{i+1}$ and $\lim _{n\rightarrow \infty }(\mu _{i+1,n}/\mu _{i,n})\in (0,\infty )$. Especially, $\mu _i>0$ ($=0$) yields $\mu _{i+1}>0$ ($=0$ respectively).

Proof

We assume $u_i\ge 0$. Put $w_{j,n}:=u_{j,n}/\mu _{j,n}$ for $j=i,i+1$. Note that, Lemmas 4.2 and 4.3 imply $\lambda _*\not =0$ and there exist values $\mu _{i+1}\ge 0$ and $r_{i-1}\le \rho _i<r_i<\rho _{i+1}<r_{i+1}$ such that $\mu _{i+1,n}\rightarrow \mu _{i+1}$, $r_{j,n}\rightarrow r_j$ for $j=i-1,i,i+1$ and $\rho _{j,n}\rightarrow \rho _j$ for $j=i,i+1$ up to a subsequence. Moreover, by Lemma 4.4, there exist continuous functions $w_i\ge 0$ in $[\rho _i,r_i)$ and $w_{i+1}\le 0$ in $(r_{i},r_{i+1})$ such that $w_{j,n}\rightarrow w_j$ in $C^2_{\text {loc}}((r_{j-1},r_j))$ and $|w_j(\rho _j)|=1$ for $j=i,i+1$. Then as usual, multiplying the equation for $u_n$ by r and integrating over $(\rho _{i,n},\rho _{i+1,n})$, we get

$$\begin{aligned} \frac{\mu _{i+1,n}}{\mu _{i,n}}=-\frac{\int _{\rho _{i,n}}^{r_{i,n}}\lambda _n\frac{f_n(u_n)}{\mu _{i,n}}rdr}{\int _{r_{i,n}}^{\rho _{i+1,n}}\lambda _n\frac{f_n(u_n)}{\mu _{i+1,n}}rdr}\rightarrow -\frac{\int _{\rho _{i}}^{r_{i}}\lambda _*w_i\frac{f_*(\mu _i w_i)}{\mu _i w_i}rdr}{\int _{r_{i}}^{\rho _{i+1}}\lambda _*w_{i+1}\frac{f_*(\mu _{i+1} w_{i+1})}{\mu _{i+1} w_{i+1}}rdr}\in (0,\infty ), \end{aligned}$$

since $f_*(t)/t\ge 1$ for any $t\ge 0$ where we again defined $f_*(t)/t=1$ for $t=0$. This completes the former assertion. Then the latter one is clearly confirmed. This ends the proof. $\square$

Finally, we prove the next two important lemmas.

Lemma 4.6

Suppose (4.1). Let $k\ge 1$ and choose $N\in \{1,\cdots ,k\}$. Assume that $\mu _{N,n}\rightarrow \infty$ and the formula (3.45) holds for $i=N$. Moreover, suppose there exists a constant $\mu _{N+1}\ge 0$ such that $\lim _{n\rightarrow \infty }\mu _{N+1,n}= \mu _{N+1}$. Then we get $\lambda _*\not =0$ and, taking a subsequence if necessary, we have $\lim _{n\rightarrow \infty }\rho _{N+1,n}=0$, $\lim _{n\rightarrow \infty }(r_{N,n}/\rho _{N+1,n})=0$, and further,

$$\begin{aligned} \lim _{n\rightarrow \infty }\rho _{N+1,n}^2\frac{f_n(\mu _{N+1,n})}{\mu _{N+1,n}}\log {\frac{1}{r_{N,n}}}=\frac{2}{\lambda _* }, \end{aligned}$$

(4.5)

and

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{\rho _{N+1,n}^2f_n(\mu _{N+1,n})}{r_{N,n}u_{N,n}'(r_{N,n})}=\frac{2}{\lambda _* }. \end{aligned}$$

(4.6)

Proof

Without losing the generality we may assume $u_{N+1,n}\ge 0$. From Lemma 4.2, we get $\lambda _*\not =0$. Then (3.45) with $i=N$ and our assumption $\mu _{N,n}\rightarrow \infty$ imply $r_{N,n}\rightarrow 0$. Then Lemmas 4.3 and 4.4 yield $\rho _{N+1,n}\rightarrow 0$. Moreover, we claim $r_{N,n}/\rho _{N+1,n}\rightarrow 0$. In fact, using Lemma 2.2, we get

$$\begin{aligned} 1= \int _{r_{N,n}}^{\rho _{N+1,n}}\lambda _n\frac{f_n(u_n)}{\mu _{N+1,n}}r\log {\frac{r}{r_{N,n}}}dr\le \lambda _n \frac{f_n(\mu _{N+1,n})}{\mu _{N+1,n}}\log {\frac{\rho _{N+1,n}}{r_{N,n}}}\frac{\rho _{N+1,n}^2-r_{N,n}^2}{2}. \end{aligned}$$

This formula implies $\log {(\rho _{N+1,n}/r_{N,n})}\rightarrow \infty$. This shows the claim. Now, let us deduce (4.5) and (4.6). To this end, we put ${\hat{r}}_n:=r_{N,n}/\rho _{N+1,n}$ and ${\hat{w}}_n(r):=u_{N+1,n}(\rho _{N+1,n}r)/\mu _{N+1,n}$ for all $r\in [{\hat{r}}_n,1]$. Then again using the equation in (2.1) with the conditions ${\hat{w}}_n(1)=1$, ${\hat{w}}_n'(1)=0$ and previous claims, we find a function ${\hat{w}}_0$ such that ${\hat{w}}_n\rightarrow {\hat{w}}_0$ in $C^2_{\text {loc}}((0,1])$ and get

$$\begin{aligned} {\left\{ \begin{array}{ll} -{\hat{w}}_0''(r)-\frac{1}{r} {\hat{w}}_0'(r)=0,\ 0\le {\hat{w}}_0\le 1,\ {\hat{w}}_0'\ge 0\text { in }(0,1),\\ {\hat{w}}_0(1)=1,\ {\hat{w}}_0'(1)=0. \end{array}\right. } \end{aligned}$$

(4.7)

We readily compute that ${\hat{w}}_0=1$. Finally, we use Lemma 2.2 to see

$$\begin{aligned} \begin{aligned} \mu _{N+1,n}&=\int _{r_{N,n}}^{\rho _{N+1,n}}f_n(u_n)r\log {\frac{r}{r_{N,n}}}dr\\&=\lambda _n \rho _{N+1,n}^2f_n(\mu _{N+1,n})\log {\frac{\rho _{N+1,n}}{r_{N,n}}}\int _{{\hat{r}}_n}^{1}\frac{f_n(\mu _{N+1,n}{\hat{w}}_n)}{f_n(\mu _{N+1,n})}rdr\\&\ \ \ +\lambda _n \rho _{N+1,n}^2f_n(\mu _{N+1,n})\int _{{\hat{r}}_n}^{1}\frac{f_n(\mu _{N+1,n}{\hat{w}}_n)}{f_n(\mu _{N+1,n})}r\log {r}dr \end{aligned} \end{aligned}$$

Since $f_n(\mu _{N+1,n}{\hat{w}}_n)/f_n(\mu _{N+1,n})\rightarrow 1$ on (0, 1), the Lebesque convergence theorem and previous claims give

$$\begin{aligned} \frac{\mu _{N+1,n}}{\rho _{N+1,n}^2f_n(\mu _{N+1,n})\log {\frac{\rho _{N+1,n}}{r_{N,n}}}}=\frac{\lambda _*}{2}+o(1). \end{aligned}$$

This proves (4.5). On the other hand, multiplying the equation in (2.1) with $i=N+1$ by r and integrating over $(r_{N,n},\rho _{N+1,n})$, we see

$$\begin{aligned} \begin{aligned} r_{N,n}u_n'(r_{N,n})&=\int _{r_{N,n}}^{\rho _{N+1,n}}\lambda _nf_n(u_n)rdr \\&=\lambda _n \rho _{N+1,n}^2f_n(\mu _{N+1,n})\int _{{\hat{r}}_n}^{1}\frac{f_n(\mu _{N+1,n}{\hat{w}}_n)}{f_n(\mu _{N+1,n})}rdr. \end{aligned} \end{aligned}$$

Hence similarly we obtain (4.6). This finishes the proof. $\square$

By the previous lemma, we deduce the following.

Lemma 4.7

Suppose as in the previous lemma. In addition, we assume that (3.46) with $i=N$ is true. Then we get

$$\begin{aligned} \lim _{n\rightarrow \infty }\mu _{N+1,n}\left( \log {\frac{1}{r_{N,n}}}\right) ^{\frac{1-\beta _n}{\beta _n}}=2^{\frac{\beta _*-1}{\beta _*}} \delta ^{\frac{1}{\beta _*}}, \end{aligned}$$

(4.8)

and

$$\begin{aligned} \lim _{n\rightarrow \infty }\mu _{N+1,n}f_n(\mu _{N+1,n})^{\beta _n-1}\rho _{N+1,n}^{2(\beta _n-1)}=\frac{4^{\beta _*-1}\delta }{\lambda _*^{\beta _*-1}}. \end{aligned}$$

(4.9)

Proof

Noting $\lambda _*\not =0$ by the previous lemma, we combine (4.6) together with (3.46) and (3.45) for $i=N$ and get

$$\begin{aligned} \begin{aligned} \frac{2}{\lambda _*}+o(1)&=\frac{\rho _{N+1,n}^2f_n(\mu _{N+1,n}) \mu _{N,n}}{2+o(1)}\\&=\frac{\rho _{N+1,n}^2f_n(\mu _{N+1,n})}{2+o(1)}\left( \frac{2\log {\frac{1}{r_{N,n}}}(1+o(1))}{\delta +o(1)}\right) ^{\frac{1}{\beta _n}}. \end{aligned} \end{aligned}$$

Then it holds that

$$\begin{aligned} \begin{aligned} \rho _{N+1,n}^2&f_n(\mu _{N+1,n})\\&=\left\{ 2^{2-\frac{1}{\beta _*}}\delta ^{\frac{1}{\beta _*}}\lambda _*^{-1} +o(1)\right\} \left( \log {\frac{1}{r_{N,n}}}\right) ^{-\frac{1}{\beta _n}}. \end{aligned} \end{aligned}$$

(4.10)

Substituting this into (4.5), we obtain

$$\begin{aligned} \begin{aligned} \frac{2}{\lambda _* }+o(1)&=\frac{1}{\mu _{N+1,n}}\left\{ 2^{2-\frac{1}{\beta _*}}\delta ^{\frac{1}{\beta _*}}\lambda _*^{-1} +o(1)\right\} \left( \log {\frac{1}{r_{N,n}}}\right) ^{1-\frac{1}{\beta _n}}. \end{aligned} \end{aligned}$$

It follows that

$$\begin{aligned} \mu _{N+1,n}\left( \log {\frac{1}{r_{N,n}}}\right) ^{\frac{1-\beta _n}{\beta _n}}=2^{\frac{\beta _*-1}{\beta _*}} \delta ^{\frac{1}{\beta _*}}+o(1). \end{aligned}$$

This proves (4.8). Using this and (4.10), we get

$$\begin{aligned} f_n(\mu _{N+1,n})^{\beta _n-1}\rho _{N+1,n}^{2(\beta _n-1)}=\frac{1}{\mu _{N+1,n}}\left( \frac{4^{\beta _*-1}\delta }{\lambda _*^{\beta _*-1}}+o(1)\right) . \end{aligned}$$

This shows (4.9). We finish the proof. $\square$

5 Proof of main theorems

Let us complete the proof of main theorems. We shall first show Theorems 1.5 and 1.6. Then Theorem 1.1 will readily follow from them. We begin with the proof of Theorem 1.5.

Proof of Theorem 1.5

We first note that the standard argument shows that the weak limit $u_0$ of $(u_n)$ is a radially symmetric smooth solution of (1.1) with $(\lambda ,\beta )=(\lambda _*,\beta _*)$. In particular, writing $u_0=u_0(|x|)$, we have that $u_0(r)$ ($r\in [0,1]$) satisfies

$$\begin{aligned} -u_0''-\frac{1}{r}u_0'=\lambda _* f_*(u_0)\text { in }(0,1)\text { and }u_0'(0)=0=u_0(1). \end{aligned}$$

(5.1)

We begin with the case $\max _{r\in [0,1]}|u_n(r)|\rightarrow \infty$. Set $N:=\max \{i=1,\cdots ,k+1\ |\ (\mu _{i,n})\text { is unbounded}\}$. Suppose $N=k+1$. Then we have $\mu _{k+1}\rightarrow \infty$ as $n\rightarrow \infty$ by extracting a subsequence if necessary. We shall confirm all the assertions in the case of (i). By Lemma 4.2, we get $\rho _{k+1,n}\rightarrow 0$ and $\mu _{i,n}\rightarrow \infty$ for all $i=1,\cdots ,k+1$. Especially, the assumption (2.2) is satisfied for $i=1$. As a consequence, Propositions  2.4, 3.1 and 3.10 hold true for $i=1$. It follows that all of the assumptions (2.2)–(2.5) are verified for $i=2$. Consequently, the assertions in Propositions  2.4, 3.1 and 3.10 are true for $i=2$. Repeating the same argument, we ensure all the assertions in Propositions  2.4, 3.1 and 3.10 for any $i=1,\cdots ,k+1$. This completes the former assertions in (i). Finally, by (3.45) with $i=k+1$ and the facts that $r_{k+1,n}=1$ and $\mu _{k+1,n}\rightarrow \infty$, we get $\lambda _*=0$. This yields that $u_0=0$ by (5.1). Moreover, Lemma 2.1 implies that $u_n$ is locally uniformly bounded in (0, 1]. Then using the equation for $u_n$ as usual, it is easy to show that $u_n\rightarrow 0\text { in }C^2_{\text {loc}}((0,1])$ up to a subsequence. This finishes the case of (i).

Next we assume $N<k+1$. Then similarly to the previous argument, we get $\rho _{N,n}\rightarrow 0$ and $\mu _{i,n}\rightarrow \infty$ for all $i=1,\cdots ,N$ up to a subsequence. Then, analogously, we have that all the assertions in Propositions 2.4, 3.1 and 3.10 are true for all $i=1,\cdots ,N$. On the other hand, by the definition of N, for each $i=N+1,\cdots ,k+1$, there exists a value $\mu _i\ge 0$ such that $\mu _{i,n}\rightarrow \mu _i$ up to a subsequence. Then from Lemma 4.3, we get numbers $0\le r_N\le \rho _{N+1}<r_{N+1}<\cdots<\rho _{k+1}< r_{k+1}=1$ if $N<k$ and $0\le r_N\le \rho _{N+1}<r_{N+1}=1$ if $N=k$ such that $r_{i,n}\rightarrow r_i$ for all $i=N,\cdots ,k+1$ and $\rho _{i,n}\rightarrow \rho _i$ for all $i=N+1,\cdots ,k+1$ by taking a subsequence again if necessary. Moreover, from Lemmas 4.6, we get $\lambda _*\not =0$, and $r_N=\rho _{N+1}=0$. Furthermore, a usual argument shows that $u_n|_{[r_{N,n},1]}\rightarrow u_0$ in $C^2_{\text {loc}}((0,1])$ and $\lim _{r\rightarrow 0+0}(-1)^Nu_0(r)=\mu _{N+1}$ by Lemma 4.4. It follows that

$$\begin{aligned} \begin{aligned} \int _{r_{N,n}}^1u_n'(r)^2rdr&=\int _{r_{N,n}}^1\lambda _nf_n(u_n)u_nrdr\\&\rightarrow \int _{0}^1\lambda _*f_*(u_0)u_0rdr=\int _0^1u_0'(r)^2rdr \end{aligned} \end{aligned}$$

(5.2)

by (5.1). This proves the former part of (ii).

Now, we assume $\mu _{N+1}>0$. Then noting (5.1) and Lemmas 4.4 and 4.5, we get that $u_0(r_i)=0$, $(-1)^{i-1}u_0\ge 0$ on $[r_{i-1},r_i]$, $u_0'(\rho _i)=0$ and $(-1)^{i-1}u_0(\rho _i)=\mu _i>0$ for all $i=N+1,\cdots ,k+1$. Moreover, by (5.1), we readily see $(-1)^{i-1}u_i>0$ on $(r_{i-1},r_i)$ for all $i=N+1,\cdots ,k+1$. This completes the case of (a). Next we suppose $\mu _{N+1}=0$. By Lemma 4.5, it is obvious that $u_0=0$. Put $w_n:=u_n|_{[r_{N,n},1]}/\mu _{N+1,n}$ on $[r_{N,n},1]$. By Lemma 4.5 again, for every $i=N+1,\dots ,k+1$, we have a constant $\mu _i^*>0$ such that $\max _{r\in [r_{i-1,n},r_{i,n}]}w_n(r)=\mu _{i,n}/\mu _{N+1,n}\rightarrow \mu _i^*$ up to a subsequence. In particular, $w_n$ is uniformly bounded in $[r_{N,n},1]$. Then, by the standard argument and Lemma 4.4, we get a continuous function $w_0$ in [0, 1] such that $w_n\rightarrow w_0$ in $C^2_{\text {loc}}((0,1])$ and

$$\begin{aligned} {\left\{ \begin{array}{ll} -w_0''-\frac{1}{r} w_0'=\lambda _* w_0\text { in }(0,1),\\ (-1)^Nw_0(0)=1,\ w_0(r_i)=0,\\ (-1)^{i-1}w_0>0 \text { on }(r_{i-1},r_i)\ (i=N+1,\cdots ,k+1). \end{array}\right. } \end{aligned}$$

Using the equation and the condition $(-1)^Nw_0(0)=1$, we obtain $w_0(r)=(-1)^N J_0(\sqrt{\lambda _*}r)$ in [0, 1] where $J_0$ is the first kind Bessel function of order zero defined in Sect. 1.3. Moreover, since $w_0$ has just $(k-N)$ interior zero points in (0, 1) and $w_0(1)=0$, we get that $\sqrt{\lambda _*}$ coincides with the $(k-N+1)$–th zero point of $J_0$ on $(0,\infty )$, i.e., $\sqrt{\lambda _*}=t_{N-k+1}$. It follows that $w_0=(-1)^N\varphi _{k-N+1}$ and $\lambda _*=\Lambda _{k-N+1}$. This completes the case of (ii).

Finally, if $u_n$ is uniformly bounded in [0, 1], repeating the similar (and simpler) argument based on Lemmas 4.2-4.5 as above, we can confirm all the assertions in (iii). This finishes the proof. $\square$

Next we prove Theorem 1.6.

Proof of Theorem 1.6

We first assume that $k>0$, $(\beta _n)\subset (0,1]$, and $\mu _{1,n}\rightarrow \infty$. Then we claim that for every $i=2,\cdots ,k+1$, $\mu _{i,n}$ is bounded uniformly for all $n\in {\mathbb {N}}$. To see this, we shall show that $\mu _{2,n}$ is uniformly bounded. Otherwise, we get $\mu _{2,n}\rightarrow \infty$ up to a subsequence. Then arguing as in the previous proof, we ensure that all the assumptions (2.2)–(2.5) are satisfied for $i=2$. Then we get (2.14) and (3.2) for $i=2$ by Lemmas 2.6 and  3.3 respectively. But if $\beta _n\le 1$ for all $n\in {\mathbb {N}}$, (2.14) and (3.2) with $i=2$ yield that $\mu _{2,n}$ is uniformly bounded. This contradicts (2.2). Consequently, Lemma 4.3 proves the claim.

Now, we assume that (i) of Theorem 1.5 occurs. Then the first conclusion follows by the previous claim. Moreover, if $k\in {\mathbb {N}}\cup \{0\}$, arguing as in the previous proof again, we get that all the assertions in Lemma 3.3 and Proposition 3.10 hold true for any $i=1,\cdots ,k+1$. It follows from (3.45) with $i=k+1$ that (1.6) holds true. Then if $k\ge 1$, it follows from (3.3) that

$$\begin{aligned} \frac{(\delta +o(1))^{-\frac{1}{\beta _n}}\left( \log {\frac{1}{\lambda _n}}\right) ^{\frac{1}{\beta _n}}}{\mu _{k,n}^{(\beta _n-1)}}=\delta +o(1). \end{aligned}$$

This gives (1.7) with $i=k$ after an easy calculation. Then we get (1.7) for all $i=1,\cdots ,k$ by induction. In fact, we assume (1.7) is true for some $i=j\in \{2,\cdots ,k\}$. Then using (3.3) with $i=j$, we similarly get

$$\begin{aligned} \begin{aligned}&\frac{\left( \delta ^{\frac{2-\beta _*(\beta _*-1)^{k-j+1}}{2-\beta _*}}+o(1)\right) ^{-\frac{1}{\beta _n(\beta _n-1)^{k-j+1}}}\left( \log {\frac{1}{\lambda _n}}\right) ^{\frac{1}{\beta _n(\beta _n-1)^{k-j+1}}}}{\mu _{j-1,n}^{\beta _n-1}}\\&=\delta +o(1). \end{aligned} \end{aligned}$$

It follows that

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{\log {\frac{1}{\lambda _n}}}{\mu _{j-1,n}^{\beta _n(\beta _n-1)^{k-j+2}}}=\delta ^{\frac{2-\beta _*(\beta _*-1)^{k-j+2}}{2-\beta _*}}. \end{aligned}$$

This is (1.7) with $i=j-1$. This shows the desired conclusion. Moreover, since (3.45) and (3.46) with $i=k+1$ imply

$$\begin{aligned} \mu _{k+1,n}=\left( \frac{\log {\frac{1}{\lambda _n}}}{\delta +o(1)}\right) ^{\frac{1}{\beta _n}}, \end{aligned}$$

and

$$\begin{aligned} \mu _{k+1,n}|u_{k+1,n}'(1)|=2+o(1) \end{aligned}$$

respectively, combining these two formulas, we get (1.8). Next assume $k\ge 1$ and $i=1,\cdots ,k$. Noting the first conclusion, we may assume $\beta _n>1$ for all $n\in {\mathbb {N}}$. Then by (1.7), we have that

$$\begin{aligned} \frac{\log {\frac{1}{\lambda _n}}}{\mu _{i,n}^{\beta _n}}=\frac{\log {\frac{1}{\lambda _n}}}{\mu _{i,n}^{\beta _n(\beta _n-1)^{k-i+1}}}\mu _{i,n}^{\beta _n\{(\beta _n-1)^{k-i+1}-1\}}\rightarrow 0 \end{aligned}$$

since $1\le k-i+1\le k$ and $1\le \beta _*<2$. Using this and (3.45), we obtain

$$\begin{aligned} \frac{\delta }{2}+o(1)=\frac{\log {\frac{1}{r_{i,n}}}}{\mu _{i,n}^{\beta _n}}. \end{aligned}$$

Therefore, it follows from (1.7) that

$$\begin{aligned} \begin{aligned}&\left( \frac{\delta }{2}+o(1)\right) ^{(\beta _n-1)^{k-i+1}}=\frac{\left( \log {\frac{1}{r_{i,n}}}\right) ^{(\beta _n-1)^{k-i+1}}}{\log {\frac{1}{\lambda _n}}}\left( \delta ^{\frac{2-\beta _*(\beta _*-1)^{k-i+1}}{2-\beta _*}}+o(1)\right) . \end{aligned} \end{aligned}$$

Using this formula, we readily get (1.9). It follows that

$$\begin{aligned} \frac{\log { \frac{1}{r_{i,n}}}}{\log {\frac{1}{\lambda _n}}}=\frac{\left( \log {\frac{1}{r_{i,n}}}\right) ^{(\beta _n-1)^{k-i+1}}}{\log {\frac{1}{\lambda _n}}}\left( \log {\frac{1}{r_{i,n}}}\right) ^{1-(\beta _n-1)^{k-i+1}}\rightarrow \infty \end{aligned}$$

(5.3)

as $n\rightarrow \infty$. Then we get by (3.46), (3.45) and (5.3) that,

$$\begin{aligned} \begin{aligned} \log {|u_{i,n}'(r_{i,n})|}&=\log {\frac{1}{r_{i,n}}}-\log {\mu _{i,n}}+O(1)\\&=\log {\frac{1}{r_{i,n}}}-\frac{1}{\beta _n}\log {\log {\frac{1}{r_{i,n}}}}-\frac{1}{\beta _n}\log {\left( 1+\frac{\log {\frac{1}{\lambda _n}}}{\log {\frac{1}{r_{i,n}^2}}}\right) }+O(1)\\&=\log {\frac{1}{r_{i,n}}}(1+o(1))\\&=\left( \frac{\log {\frac{1}{\lambda _n}}}{2^{(\beta _*-1)^{k-i+1}}\delta ^{\frac{2-2(\beta _*-1)^{k-i+1}}{2-\beta _*}}+o(1)}\right) ^{\frac{1}{(\beta _n-1)^{k-i+1}}}(1+o(1)) \end{aligned} \end{aligned}$$

by (1.9). This proves (1.10). Next, for any $i=2,\cdots ,k+1$, from (3.2) and the definition of $\gamma _{i,n}$, we get

$$\begin{aligned} \begin{aligned} \log {\left( 8^{\beta _*-1}\delta +o(1)\right) }&=(\beta _n-1)\mu _{i,n}^2\left( 1+o(1)\right) +\beta _n \log {\mu _{i,n}}\\&\ \ \ -2(\beta _n-1)\log {\frac{1}{\rho _{i,n}}} \end{aligned} \end{aligned}$$

(5.4)

where we noted

$$\begin{aligned} \frac{\log {\frac{1}{\lambda _n}}}{\mu _{i,n}^2}\le \frac{\log {\frac{1}{\lambda _nr_{i,n}^2}}}{\mu _{i,n}^{\beta _n}}\frac{1}{\mu _{i,n}^{2-\beta _n}}\rightarrow 0 \end{aligned}$$

as $n\rightarrow \infty$ by (3.45). Now, we suppose $k\ge 2$ and $i=2,\cdots ,k$. Then we have by (1.7) that ,

$$\begin{aligned} \begin{aligned}&\frac{\log {\mu _{i,n}}}{(\beta _n-1)\mu _{i,n}^2}=\frac{(1+o(1))\log {\log {\frac{1}{\lambda _n}}}}{\beta _n(\beta _n-1)^{k-i+2}\left( \frac{\log {\frac{1}{\lambda _n}}}{\delta ^{(2-\beta _*(\beta _*-1)^{k-i+1})/(2-\beta _*)}+o(1)}\right) ^{\frac{2}{\beta _n(\beta _n-1)^{k-i+1}}}}. \end{aligned} \end{aligned}$$

Since $i<k+1$, we get $(\log {\mu _{i,n}})/((\beta _n-1)\mu _{i,n}^2)\rightarrow 0$. Therefore, we obtain from (5.4) and (1.7) that

$$\begin{aligned} \begin{aligned} \frac{1}{2}+o(1)&=\frac{\log {\frac{1}{\rho _{i,n}}}}{\mu _{i,n}^2}\\&=\left( \frac{\delta ^{(2-\beta _*(\beta _*-1)^{k-i+1})/(2-\beta _*)}+o(1)}{\log {\frac{1}{\lambda _n}}}\right) ^{\frac{2}{\beta _n(\beta _n-1)^{k-i+1}}}\log {\frac{1}{\rho _{i,n}}}. \end{aligned} \end{aligned}$$

This proves (1.11). On the other hand, if $k\ge 1$ and $i=k+1$, we use (1.6) and see

$$\begin{aligned} \frac{\log {\mu _{k+1,n}}}{(\beta _n-1)\mu _{k+1,n}^2}= \frac{ \left( \delta +o(1)\right) ^{\frac{2}{\beta _n}}\log {\log {\frac{1}{\lambda _n}}}}{\beta _n(\beta _n-1)\left( \log {\frac{1}{\lambda _n}}\right) ^{\frac{2}{\beta _n}}}. \end{aligned}$$

Hence if there exists a constant $L\ge 0$ such that

$$\begin{aligned} \frac{\log {\log {\frac{1}{\lambda _n}}}}{(\beta _n-1)\left( \log {\frac{1}{\lambda _n}}\right) ^{\frac{2}{\beta _n}}}=L, \end{aligned}$$

we get from (5.4) and (1.6) that

$$\begin{aligned} \begin{aligned} 1+\delta ^{\frac{2}{\beta _*}} L+o(1)&=\frac{2\log {\frac{1}{\rho _{k+1,n}}}}{\mu _{k+1,n}^2}=2\left( \frac{\delta +o(1)}{\log {\frac{1}{\lambda _n}}}\right) ^{\frac{2}{\beta _n}}\log {\frac{1}{\rho _{k+1,n}}}. \end{aligned} \end{aligned}$$

This ensures (1.13). On the other hand, if

$$\begin{aligned} \frac{\log {\log {\frac{1}{\lambda _n}}}}{(\beta _n-1)\left( \log {\frac{1}{\lambda _n}}\right) ^{\frac{2}{\beta _n}}}\rightarrow \infty , \end{aligned}$$

we necessarily have $\beta _*=1$ and then, by (5.4) and (1.6), we get

$$\begin{aligned} 1+o(1)=\frac{2(\beta _n-1)\log {\frac{1}{\rho _{k+1,n}}}}{\beta _n \log {\mu _{k+1,n}}}=\frac{2(\beta _n-1)\log {\frac{1}{\rho _{k+1,n}}}}{ (1+o(1))\log {\log {\frac{1}{\lambda _n}}}}. \end{aligned}$$

This proves (1.14). This completes the case of (i).

Next we assume that (ii) of Theorem 1.5 happens. Then, since $\mu _{N+1,n}$ is uniformly bounded and $r_{N,n}\rightarrow 0$, we have $\beta _*\le 1$ by (4.8) in Lemma 4.7. Then, the first claim above completes the first assertion in the case of (ii). Let us suppose $\mu _{N+1}>0$ and complete the case of (a). Again by (), we see $\beta _*=1$. We shall prove (1.15)–(1.20). Multiplying the equation in (2.1) with $i=N+1$ by r and integrating over $(\rho _{N+1,n},r_{N+1,n})$, we get

$$\begin{aligned} r_{N+1,n}u_{N+1,n}'(1)=-\int _{\rho _{N+1,n}}^{r_{N+1,n}}\lambda _n f_n(u_{N+1,n})rdr. \end{aligned}$$

Then recalling the assertions in (ii) of Theorem 1.5, we ensure (1.18). Moreover, noting $\beta _*=1$, we get by (4.9) that

$$\begin{aligned} \rho _{N+1,n}^{\beta _n-1}=\sqrt{\frac{\delta }{\mu _{N+1}}}+o(1)=\sqrt{\frac{\alpha }{2\mu _{N+1}}}+o(1). \end{aligned}$$

This gives (1.19). Next we see by (4.8) that

$$\begin{aligned} \left( \log {\frac{1}{r_{N,n}}}\right) ^{\beta _n-1}=\frac{2\mu _{N+1}}{\alpha }+o(1). \end{aligned}$$

(5.5)

Then we use (3.45) with $i=N$ and (5.5) to obtain

$$\begin{aligned} \mu _{N,n}^{\beta _n-1}=(1+o(1))\left( \log {\frac{1}{r_{N,n}}}\right) ^{\frac{\beta _n-1}{\beta _n}}=\frac{2\mu _{N+1}}{\alpha }+o(1). \end{aligned}$$

This gives (1.15) with $i=N$. Then if $N\ge 2$, (1.15) is true for all $i=1,\cdots ,N$ by induction. Indeed, assuming (1.15) is true for some $i=j\in \{2,\cdots ,N\}$, we use (3.3) with $i=j$ to see

$$\begin{aligned} \mu _{j-1,n}^{\beta _n-1}=\left( \frac{2}{\alpha }+o(1)\right) \mu _{j,n}. \end{aligned}$$

This and the assumption suggest

$$\begin{aligned} \mu _{j-1,n}^{(\beta _n-1)^{N-j+2}}=(1+o(1))\mu _{j,n}^{(\beta _n-1)^{N-j+1}}=\frac{2\mu _{N+1}}{\alpha }+o(1). \end{aligned}$$

This shows (1.15) with $i=j-1$. This finishes (1.15). This and (3.45) show

$$\begin{aligned} \left( \log {\frac{1}{r_{i,n}}}\right) ^{(\beta _n-1)^{N-i+1}}=(1+o(1))\mu _{i,n}^{\beta _n(\beta _n-1)^{N-i+1}}=\frac{2\mu _{N+1}}{\alpha }+o(1) \end{aligned}$$

for all $i=1,\cdots ,N$. This proves (1.16). Moreover, for any $i=1,\cdots ,N$, we get by (3.46) and (3.45) that

$$\begin{aligned} \begin{aligned} \log {|u_{i,n}'(r_{i,n})|}&=\log {\frac{1}{r_{i,n}}}\left( 1-\frac{\log {\mu _{i,n}}}{\log {\frac{1}{r_{i,n}}}}+o(1)\right) =\log {\frac{1}{r_{i,n}}}\left( 1+o(1)\right) , \end{aligned} \end{aligned}$$

where we noted

$$\begin{aligned} \frac{\log {\mu _{i,n}}}{\log {\frac{1}{r_{i,n}}}}=\frac{\mu _{i,n}^{\beta _n}}{\frac{1}{2}(1+o(1))\log {\frac{1}{\lambda _n r_{i,n}^2}}}\frac{\log {\mu _{i,n}}}{\mu _{i,n}^{\beta _n}}\rightarrow 0 \end{aligned}$$

by (3.45). Then it follows from (1.16) that

$$\begin{aligned} \left( \log {|u_{i,n}'(r_{i,n})|}\right) ^{(\beta _n-1)^{N-i+1}}=\frac{2\mu _{N+1}}{\alpha }(1+o(1)). \end{aligned}$$

This proves (1.17). In particular, (1.15) clearly shows that $2\mu _{N+1}/\alpha >1$ ($\in (0,1)$) yields $\beta _n>1$ ($<1$ respectively) for all $n\in {\mathbb {N}}$. On the other hand, $\beta _n>1$ ($=1$, $<1$) for all $n\in {\mathbb {N}}$ suggests $2\mu _{N+1}/\alpha \ge 1$ ($=1$, $\le 1$ respectively). Finally, suppose $1<N\le k$ and $2\mu _{N+1}/\alpha >1$. Then we have $\beta _n>1$ for all $n\in {\mathbb {N}}$ by the first claim above. Then, for any $i=2,\dots ,N$, we use (3.2) and the definition of $\gamma _{i,n}$ to deduce

$$\begin{aligned} \mu _{i,n}f_n(\mu _{i,n})^{\beta _n-1}\rho _{i,n}^{2(\beta _n-1)}=\frac{\alpha }{2}+o(1). \end{aligned}$$

It follows that

$$\begin{aligned} \log {\frac{1}{\rho _{i,n}}}=\frac{\mu _{i,n}^2}{2}\left( 1+\frac{\log {\mu _{i,n}}}{\mu _{i,n}^2(\beta _n-1)}(1+o(1))+o(1)\right) . \end{aligned}$$

(5.6)

Here note that (1.15) implies

$$\begin{aligned} \frac{\log {\mu _{i,n}}}{\mu _{i,n}^2(\beta _n-1)}=\frac{\log {(2\mu _{N+1}/\alpha +o(1))}}{(\beta _n-1)^{N-i+2}(2\mu _{N+1}/\alpha +o(1))^{2/(\beta _n-1)^{N-i+1}}}\rightarrow 0 \end{aligned}$$

since $2\mu _{N+1}/\alpha >1$. Consequently, (5.6) and (1.15) ensure

$$\begin{aligned} \left( \log {\frac{1}{\rho _{i,n}}}\right) ^{(\beta _n-1)^{N-i+1}}=\mu _{i,n}^{2(\beta _n-1)^{N-i+1}}\left( 1+o(1)\right) =\left( \frac{2\mu _{N+1}}{\alpha }\right) ^2+o(1). \end{aligned}$$

This gives (1.20). This completes the case of (a).

Lastly, if $\mu _{N+1}=0$, Lemma 4.5 shows $\mu _i=0$ for all $i=N+1,\cdots ,k+1$. Moreover, (4.8) confirms that $\beta _n<1$ for all $n\in {\mathbb {N}}$ since $r_{N,n}\rightarrow 0$. This completes the case of (b). This finishes the proof. $\square$

Let us complete the proof of Theorem 1.1.

Proof of Theorem 1.1

Assume $\max _{n\rightarrow \infty }|u_n(x)|\rightarrow \infty$. Then writing $u_n=u_n(|x|)$, the function $u_n(r)$ $(r\in [0,1])$ verifies (i) or (ii) of Theorem 1.5. If (i) occurs and $\beta _n>1$ for all $n\in {\mathbb {N}}$, the assertions in (i) of Theorem 1.5 completes (i) of Theorem 1.1. On the other hand, if (i) of Theorem 1.5 happens and $\beta _n\le 1$ for all $n\in {\mathbb {N}}$, by the first conclusion in Theorems 1.6, we get $k=0$. This shows (ii) of Theorem 1.1. Next, we suppose (ii) of Theorem 1.5 occurs. Then if $\beta _n>1$ for all $n\in {\mathbb {N}}$, we have that (a) of Theorem 1.6 occurs and thus, $\beta _*=1$. Moreover, the second assertion below (1.19) ensures $(-1)^Nu_0(0)\ge \alpha /2$. This implies (iii) of Theorem 1.1 occurs. If $\beta _n=1$ for all $n\in {\mathbb {N}}$, then the first conclusion in the case of (ii) of Theorem 1.6 implies $N=1$. Furthermore, again the second assertion below (1.19) proves $-u_0(0)=\alpha /2$. This is (iv) of Theorem 1.1. Lastly, if $\beta _n < 1$ for all $n\in {\mathbb {N}}$, similarly we get $N=1$. Moreover, we have two cases. The first case is $u_0\not =0$. Since this case corresponds to (a) of Theorems 1.5 and  1.6, we have $u_0(0)\not =0$ and $\beta _n\uparrow 1$. In addition, we get $-u_0(0)\in (0,\alpha /2]$ by the second assertion below (1.19). This confirms (v) of Theorem 1.1. The second case is $u_0=0$. This implies that (b) of Theorems 1.5 and 1.6 occurs. Hence we have $\lambda _*=\Lambda _{k}$ from the final assertion of (b) of Theorem 1.5 since $N=1$. This shows that (vi) occurs. Finally, if $u_n$ is uniformly bounded, we get (iii) of Theorem 1.5. This completes (vii) of Theorem 1.1. This finishes the proof. $\square$

Corollary 1.2 immediately follows from Theorem 1.1.

Proof of Corollary 1.2

We assume that there exist such sequences of values $\{(\lambda _n,\beta _n)\}\subset (0,\infty )\times (0,1]$ and nodal radial solutions $(u_n)$ on the contrary. Then, in view of the fact that $\lambda _n\rightarrow 0$ and $\beta _n\le 1$ for all $n\in {\mathbb {N}}$, we have that (ii) of Theorem 1.1 occurs. But then we get $k=0$ which is a contradiction. This finishes the proof. $\square$

Next, we shall prove Corollary 1.3. We recall Lemma 2.1 in [20] with slight generalization.

Lemma 5.1

Assume $k\in {\mathbb {N}}\cup \{0\}$, $\{(\lambda _n,\beta _n)\}\subset (0,\Lambda _1)\times (0,2)$, and $(\lambda _n,\beta _n)\rightarrow (\lambda _*,\beta _*)\in (0,\Lambda _1)\times (0,2)$. Then we have

$$\begin{aligned} \limsup _{n\rightarrow \infty }c_{k,\lambda _n,\beta _n}\le 2\pi k+c_{0,\lambda _*,\beta _*}, \end{aligned}$$

where the number $c_{k,\lambda ,\beta }$ is defined as in Sect. 1.1.

Proof

Noting $\lambda _*\not =0$ and $\beta _*>0$, we can repeat the completely same argument with the proof of Lemma 2.1 in [20]. This ensures the proof. $\square$

Using this, we give the proof.

Proof of Corollary 1.3

We first assume that the first conclusion does not hold. Then we have sequences of positive values $(\lambda _n)$, natural numbers $(k_n)$ and nodal radial solutions $(u_n)$ such that $\lambda _n\rightarrow 0$ as $n\rightarrow \infty$, $u_n\in S_{k_n,\lambda _n,\beta _n}$, $u_n(0)>0$, and $\int _B|\nabla u_n|^2dx$ is uniformly bounded for all $n\in {\mathbb {N}}$. Then, we claim that, up to a suitable subsequence, there exists a number $k\in {\mathbb {N}}$ such that $u_n\in S_{k,\lambda _n,\beta }$ for all $n\in {\mathbb {N}}$. Otherwise, we get $k_n\rightarrow \infty$ as $n\rightarrow \infty$. Then choose numbers $0=r_{0,n}<r_{1,n}<\cdots <r_{k+1,n}=1$ so that $u_n(x)=0$ if $|x|=r_{i,n}$ and $(-1)^{i-1}u_n(x)>0$ if $r_{i-1,n}<|x|<r_{i,n}$ for all $i=1,\cdots ,k+1$. Moreover, for all $i=1,\cdots ,k+1$, define a function $u_{i,n}\in {\mathcal {N}}_{\lambda _n,\beta }$ by $u_{i,n}:=u_n|_{\{r_{i-1,n}<|x|<r_{i,n}\}}$ with zero extension to whole B. Then since $\lambda _n\rightarrow 0$, Lemma 4.1 implies that there exists a constant $K>0$ such that

$$\begin{aligned} \int _{B}|\nabla u_n|^2dx=\sum _{i=1}^{k+1}\int _{B}|\nabla u_{i,n}|^2dx\ge (k_n+1)K \end{aligned}$$

for all $n \in {\mathbb {N}}$. Since the right hand side diverges to infinity, we get a contradiction. This proves the claim. But, then the existence of such sequence $(u_n)$ contradicts Corollary 1.2 since $\beta \le 1$. This proves the first assertion. Next we suppose the latter conclusion fails on the contrary. Then there exists a number $k\in {\mathbb {N}}$ and sequences of positive values $(\lambda _n)$ and solutions $(u_n)$ such that $\lambda _n\rightarrow 0$, $u_n\in S_{k,\lambda _n,\beta }$, and $I_{\lambda _n,\beta }(u_n)=c_{k,\lambda _n,\beta }$ for all $n\in {\mathbb {N}}$. In addition, for any $k\in {\mathbb {N}}$ and $\lambda \in (0,\Lambda _1)$, it holds that $c_{k,\lambda ,\beta }\le 2\pi k+c_{0,\lambda ,\beta }<2\pi (k+1)$. In fact, the first inequality is obtained by just choosing $\lambda _n=\lambda$ and $\beta _n=\beta$ for all $n\in {\mathbb {N}}$ in Lemma 5.1 and the second one comes from the fact that $c_{0,\lambda ,\beta }<2\pi$ by [2]. In particular, we get $I_{\lambda _n,\beta }(u_n)<2\pi (k+1)$ for all $n\in {\mathbb {N}}$. Consequently, the standard argument shows that $(u_n)$ is bounded in $H^1_0(B)$. But this is again impossible in view of Corollary 1.2 since $\beta \le 1$. This completes the proof. $\square$

Finally we prove Corollary 1.8.

Proof of Corollary 1.8

Assume as in the corollary. We write $u_n=u_n(|x|)$ $(x\in {\overline{B}})$ and consider the function $u_n(r)$ $(r\in [0,1])$. Then we get all the assertions in (i) of Theorems 1.5 and 1.6. It follows from (1.6) that

$$\begin{aligned} \mu _{k+1,n}=\left( \frac{\log {\frac{1}{\lambda _n}}}{\alpha \left( 1-\frac{\beta _*}{2}\right) +o(1)}\right) ^{\frac{1}{\beta _n}}. \end{aligned}$$

Moreover, if $k\ge 1$, we also have by (1.7) (or (2.15)) that $\mu _{k+1,n}/\mu _{i,n}\rightarrow 0$ as $n\rightarrow \infty$ for all $i=1,\cdots ,k$. Then from Theorem 1.5, we derive

$$\begin{aligned} \begin{aligned} \int _0^1u_n'(r)^2rdr&=\sum _{i=1}^{k+1}\int _{r_{i-1,n}}^{r_{i,n}}u_{i,n}'(r)^2rdr=2(k+1)-\frac{\alpha \beta _*}{\mu _{k+1,n}^{2-\beta _n}}+o\left( \frac{1}{\mu _{k+1,n}^{2-\beta _n}}\right) \\&=2(k+1)-\frac{\alpha ^{\frac{2}{\beta _*}}\beta _*\left( 1-\frac{\beta _*}{2}\right) ^{\frac{2-\beta _*}{\beta _*}}}{\left( \log \frac{1}{\lambda _n}\right) ^{\frac{2-\beta _n}{\beta _n}}}+o\left( \frac{1}{{\left( \log \frac{1}{\lambda _n}\right) ^{\frac{2-\beta _n}{\beta _n}}}}\right) . \end{aligned} \end{aligned}$$

This finishes the proof.

6 Counterparts

In this final section, we discuss the counterparts of our classification result, Theorem 1.1. In the following, we suppose $k\in \{0\}\cup {\mathbb {N}}$. Then we first remark that for any sequences $(\lambda _n)\subset (0,\Lambda _1)$ and $(\beta _n)\subset (1,2)$ ((0, 2) if $k=0$) of values, there exists a sequence $(u_n)$ of radial solutions which satisfies the assumptions in the theorem. To see this, for any such sequences $(\lambda _n)$ and $(\beta _n)$, we define the sequence $(u_n)$ of solutions so that $u_n\in S_{k,\lambda _n,\beta _n}$ and $I_{\lambda _n,\beta _n}(u_n)=c_{k,\lambda _n,\beta _n}$ for all $n\in {\mathbb {N}}$ where $c_{k,\lambda ,\beta }$ is the number defined in Sect. 1.1. This choice is possible by [2] and [7]. Consequently, since $I_{\lambda _n,\beta _n}(u_n)<2\pi k+c_{0,\lambda _n,\beta _n}<2\pi (k+1)$ for any $n\in {\mathbb {N}}$, which is proved in the same papers, a standard argument shows that $(u_n)$ is bounded in $H^1_0(B)$. Hence $(u_n)$ satisfies all the assumptions in the theorem.

Then we can immediately show some easy examples with this sequence $(u_n)$. Indeed, let us suppose $(\lambda _n,\beta _n)\rightarrow (\lambda _*,\beta _*)$ for $(\lambda _*,\beta _*)\in [0,\Lambda _1)\times (0,2)$ if $k=0$ and $(\lambda _*,\beta _*)\in [0,\Lambda _1)\times [1,3/2)$ if $k\ge 1$. Then, it follows from Theorem 1.1 that if $\lambda _*=0$ and $\beta _n>1$ ($\beta _n\le 1$) for all $n\in {\mathbb {N}}$, then $(u_n)$ behaves as in (i) ((ii) respectively) of the theorem. On the other hand, if $k=0$ and $\lambda _*\not =0$, or $k\ge 1$, $\lambda _*\not =0$, and $\beta _*>1$, then $(u_n)$ behaves as in (vii) with $u_0\not =0$.

We shall find more examples for (iii) and (vii). To this end, recalling $S_{0,\lambda ,1}\not =\emptyset$ if and only if $\lambda \in (0,\Lambda _1)$ ([2]), we define

$$\begin{aligned} \begin{aligned} \Lambda ^*:=\inf \{\Lambda \in (0,\Lambda _1)\ |\ u(0)<\alpha /2\text { for any }u\in S_{0,\lambda ,1}\text { with }&I_{\lambda ,1}(u)=c_{0,\lambda ,1}\\&\text { if } \lambda \in (\Lambda ,\Lambda _1)\}. \end{aligned} \end{aligned}$$

It follows that $\Lambda ^*\in (0,\Lambda _1)$. (See Lemma 6.8 below.) On the other hand, noting the nonexistence result by [8], we define

$$\begin{aligned} \begin{aligned} \Lambda _*:=\inf \{\Lambda>0\ |\ S_{1,\Lambda ,1}\not =\emptyset \}(=\sup \{\Lambda >0\ |\ S_{k,\Lambda ,1}=\emptyset \text { for any }k\in {\mathbb {N}} \}), \end{aligned} \end{aligned}$$

and get $\Lambda _*>0$. Moreover, our necessary condition on the weak limit in (iii) of Theorem 1.1 allows to see $\Lambda _*\le \Lambda ^*$ as follows.

Corollary 6.1

Assume $\{(\lambda _n,\beta _n)\}\subset (0,\Lambda _1)\times (1,2)$ and let $(u_n)$ be a sequence of solutions such that $u_n\in S_{1,\lambda _n,\beta _n}$ and $I_{\lambda _n,\beta _n}(u_n)=c_{1,\lambda _n,\beta _n}$ for all $n\in {\mathbb {N}}$. Moreover, suppose $(\lambda _n,\beta _n)\rightarrow (\lambda _*,\beta _*)\in (0,\Lambda _1)\times \{1\}$ and $|u(0)|<\alpha /2$ for all $u\in S_{0,\lambda _*,1}$ with $I_{\lambda _*,1}(u)=c_{0,\lambda _*,1}$ (which is verified if $\lambda _*\in (\Lambda ^*,\Lambda _1)$). Then $(u_n)$ behaves as in (vii) of Theorem 1.1 with $u_0\not =0$. In particular, there exists at least one pair of solutions $u^{\pm }\in S_{1,\lambda ,1}$ such that $u^-(0)<0<u^+(0)$, $u^+=-u^-$, and $I_{\lambda ,1}(u^{\pm })\le 2\pi +c_{0,\lambda ,\beta }$, for all $\lambda \in (\Lambda ^*,\Lambda _1)$ and thus, it holds that $0<\Lambda _*\le \Lambda ^*<\Lambda _1$.

Remark 6.2

The latter assertion is not covered by Theorem 1.3 in [7] since the nonlinearity $f(t)=ts^{t^2+\alpha |t|}$ does not satisfy (2) of Theorem 1.2 there.

Then we can give the next result. Notice that by the previous corollary and the argument in the first paragraph of this section, we ensure the existence of a sequence satisfying each assumption of (a)-(d) below.

Proposition 6.3

Let $k\in \{0\}\cup {\mathbb {N}}$ and $\{(\lambda _n,\beta _n)\}\subset (0,\Lambda _1)\times (0,2)$ and suppose $(\lambda _n,\beta _n)\rightarrow (\lambda _*,\beta _*)\in (0,\Lambda _1]\times (0,2)$. Moreover, we assume $(u_n)$ is a sequence of solutions such that $u_n\in S_{k,\lambda _n,\beta _n}$ and $\int _B|\nabla u_n|^2dx$ is uniformly bounded for all $n\in {\mathbb {N}}$. Then we have the next assertion.

• (a) Let $\lambda _*=\Lambda _1$. Then, if $k=0$, (vii) of Theorem 1.1 occurs with $u_0=0$. Moreover, assume $\beta _n\ge 1$ for all $n\in {\mathbb {N}}$. Then if $k\ge 1$ and $\beta _*\in (1,3/2)$, or $k=1$ and $\beta _*=1$, we have that $(u_n)$ behaves as in (vii) with $u_0\not =0$.

Moreover, we assume $k\ge 1$, $\beta _n>1$ for all $n\in {\mathbb {N}}$, $\beta _*=1$ and $\lambda _*\in (0,\Lambda _1)$. Then we get the following.

• (b) If $S_{\kappa ,\lambda _*,1}=\emptyset$ for all $1\le \kappa \le k$, (which is satisfied if $\lambda _*\in (0,\Lambda _*)$,) then $(u_n)$ behaves as in (iii) of Theorem 1.1 with $N=k$.

In addition, we suppose $I_{\lambda _n,\beta _n}(u_n)=c_{k,\lambda _n,\beta _n}$ for any $n\in {\mathbb {N}}$. Then we obtain the following.

• (c) For any $\lambda _*\in [\Lambda _*,\Lambda _1)$, there exists a natural number $k_{\lambda _*}$ such that if $k\ge k_{\lambda _*}$, $(u_n)$ behaves as in (iii) of Theorem 1.1 with some natural number $k-k_{\lambda _*}< N\le k$. Moreover, we have $k_{\lambda _*}\ge 2$ if $\lambda _*> \Lambda ^*$.

• (d) Assume that any solution $u\in S_{0,\lambda _*,1}$ with $I_{\lambda _*,1}(u)=c_{0,\lambda _*,1}$ satisfies $|u(0)|<\alpha /2$ (which is verified if $\lambda _*\in (\Lambda ^*,\Lambda _1)$). Then if $k\ge 2$ and (iii) of Theorem 1.1 holds true, then $N\not =k$. In particular, for any $\lambda _*\in (\Lambda ^*,\Lambda _1)$, chossing the number $k_{\lambda _*}\ge 2$ from (c) above, we get that for all $k\ge k_{\lambda _*}$, $(u_n)$ behaves as in (iii) of Theorem 1.1 with some number $k-k_{\lambda _*}< N<k$.

From the conclusion in (a), we get an additional existence result for $\lambda =\Lambda _1$ as follows.

Corollary 6.4

Let $\beta \in [1,3/2)$. Then for any $k\in {\mathbb {N}}$ if $\beta >1$ and for $k=1$ if $\beta =1$, there exists at least one pair of solutions $u_{k,\beta }^{\pm }\in S_{k,\Lambda _1,\beta }$ such that $u_{k,\beta }^-(0)<0<u_{k,\beta }^+(0)$, $u_{k,\beta }^+=-u_{k,\beta }^-$ and $I_{\Lambda _1,\beta }(u_{k,\beta }^\pm )\le 2\pi k$. Moreover, choosing a suitable sequence $(\beta _n)\subset (1,3/2)$ such that $\beta _n\rightarrow 1$, we have that $(u_{1,\beta _n}^+)$ behaves as in (vii) of Theorem 1.1 with $u_0\not =0$ and there exists a natural number $k_{\Lambda _1}\ge 2$, such that if $k\ge k_{\Lambda _1}$, $(u_{k,\beta _n}^+)$ behaves as in (iii) of Theorem 1.1 with $k-k_{\Lambda _1}<N<k$.

The behavior in (b) of Proposition 6.3 has already been observed in [20] for low energy nodal radial solutions. Our present work gives new information by Theorem 1.6 without imposing the low energy characterization. Moreover, notice that our necessary condition in (iii) of Theorem 1.1 suggests that such behavior (i.e., (iii) with $k=N$) can happen only if $0<\lambda _*<\Lambda _1$ and $\lambda _*$ is not too closed to $\Lambda _1$ (by Lemma 6.8 below). On the other hand, in view of (v) of Theorem 1.1, a similar phenomenon seems possibly to occur also in the case of $\beta _n\uparrow 1$. Interestingly, the corresponding necessary condition has the inequality opposite to that in the case of $\beta _n\downarrow 1$. It leads us to expect the following.

Conjecture 6.5

Let $(\lambda _*,\beta _*)\in (\Lambda ^*,\Lambda _1)\times \{1\}$. Then there exist sequences of values $(\lambda _n)\subset (0,\Lambda _1)$ and $(\beta _n)\subset (0,1)$ and solutions $(u_n)$ such that $(\lambda _n,\beta _n)\rightarrow (\lambda _*,\beta _*)$, $u_n\in S_{1,\lambda _n,\beta _n}$ for all $n\in {\mathbb {N}}$, and $(u_n)$ behaves as in (v) of Theorem 1.1.

Remark 6.6

Similar behavior would also occur in the general bounded domain case.

Remark 6.7

We also expect that there exist sequences of concentrating solutions which behave as in (iv) with $k=1$ and $0<\lambda _*<\Lambda _1$ and (vi) with $k=1$, $\lambda _*=\Lambda _1$, and $\beta _*\in (0,1]$ respectively. The corresponding phenomena on the Brezis-Nirenberg problem are observed in [24] and [25].

We also remark on the final assertions in (d) of Proposition 6.3 and in Corollary 6.4. Since $0<N<k$, these assertions prove the existence of a concentrating sequence of solutions which weakly converges to a sign-changing solution of (1.1). We emphasize that this conclusion holds true when $\lambda _*\le \Lambda _1$ is sufficiently closed to $\Lambda _1$. This phenomenon is new in view of the previous works, [20] and [19], where the authors observed a concentrating sequence of solutions which weakly converges to a sign-definite solution of (1.1) with sufficiently small $\lambda >0$. We naturally expect that we can choose $k_{\lambda _*}=2$ for any $\lambda _*\in (\Lambda ^*,\Lambda _1]$.

We finally conjecture that more counterparts of (iii)-(vi) would exist in the case $\lambda _*>\Lambda _1$ and $\beta _*\le 1$. Our necessary condition on the weak limit will be useful to detect such sequences of solutions.

6.1 Proofs

Let us prove the results above. We first put a basic lemma. Recall that $\max _{x\in {\overline{B}}}|u(x)|=|u(0)|$ for any $u\in S_{0,\lambda ,\beta }$ and all $(\lambda ,\beta )\in (0,\Lambda _1)\times (0,2)$ by [17].

Lemma 6.8

Fix any $\beta \in (0,2)$. Then, for any constants $M>\varepsilon >0$, there exist values $0<{\bar{\lambda }}\le {\tilde{\lambda }}<\Lambda _1$ such that if $\lambda \in (0,{\bar{\lambda }})$, then $|u(0)|\ge M$ for any $u\in S_{0,\lambda ,\beta }$, and if $\lambda \in ({\tilde{\lambda }},\Lambda _1)$, then $|u(0)|<\varepsilon$ for all $u\in S_{0,\lambda ,\beta }$.

Proof

If the former assertion fails, there exists a constant $M>0$ and sequences $(\lambda _n)\subset (0,\Lambda _1)$ and $(u_n)$ such that $\lambda _n\rightarrow 0$, $u_n\in S_{0,\lambda _n,\beta }$ and $0<u_n(0)\le M$ for all $n\in {\mathbb {N}}$. This is impossible since by Theorem 1.1, we get that $\lambda _n\rightarrow 0$ yields $u_n(0)\rightarrow \infty$. On the other hand, if the latter conclusion does not hold, there exists a constant $\varepsilon _0>0$ and sequences $(\lambda _n)\subset (0,\Lambda _1)$ and $(u_n)$ such that $\lambda _n\rightarrow \Lambda _1$, $u_n\in S_{0,\lambda _n,\beta }$ and $u_n(0)\ge \varepsilon _0$ for any $n\in {\mathbb {N}}$. This is again impossible since by Theorem 1.1 and the fact that $S_{0,\Lambda _1,\beta }=\emptyset$, we get that $\lambda _n\rightarrow \Lambda _1$ implies $u_n\rightarrow 0$ in $C^2({\overline{B}})$. This finishes the proof. $\square$

We show Corollary 6.1.

Proof of Corollary 6.1

Assume as in the corollary. Then we claim that $(u_n)$ behaves as in (vii) of Theorem 1.1. If not, since $\lambda _*\not =0$, (iii) would happen for $N=1$ and then, by Lemma 5.1, the weak limit $u_0$ of $(u_n)$ would verify $u_0\in S_{0,\lambda _*,1}$, $I_{\lambda _*,1}(u_0)=c_{0,\lambda _*,1}$ and $|u_0(0)|\ge \alpha /2$. This contradicts our choice of $\lambda _*$. This proves the claim. Then, since $\lambda _*<\Lambda _1$, the weak limit $u_0$ of $(u_n)$ is nontrivial. This ensures the former conclusion. Moreover notice that by Lemma 5.1, we get $I_{\lambda _*,1}(u_0)\le 2\pi +c_{0,\lambda _*,1}$. Then, since there exists a sequence of solutions $(u_n)$ verifying the assumption of this corollary for any $\lambda _*\in (\Lambda ^*,\Lambda _1)$ by the argument in the first paragraph of this section, the latter conclusion clearly follows. This finishes the proof. $\square$

In order to prove (d) of Proposition 6.3, we use Lemma 6.11 below which ensures the nonexistence of low energy solutions with many nods. To show the lemma, we refer to the argument in [27] and obtain the a priori lower estimate of the energy of elements in $S_{k,\lambda ,1}$. We apply the next lemma. (See Corollary 5.2 on p346 in [23] or Lemma 3 in [27].) In the following we let $a,b\in {\mathbb {R}}$ be constants such that $a<b$.

Lemma 6.9

Let q(t) be a continuous function on [a, b]. Let $v(t)\not =0$ be a solution of the equation:

$$\begin{aligned} v''+q(t)v=0,\ \ t\in [a,b]. \end{aligned}$$

Assume that v(t) has exactly k zeros in (a, b]. Then we have

$$\begin{aligned} k<\frac{1}{2}\left( (b-a)\int _a^bq^+(t)dt\right) ^{\frac{1}{2}}+1 \end{aligned}$$

where $q^+(t)\equiv \max \{q(t),0\}$.

Using this, we prove the next lemma.

Lemma 6.10

Assume $k\in {\mathbb {N}}$, $k\ge 2$ and $\lambda \ge \Lambda _*$. Then there exists a number $L=L(\lambda )>0$ which depends on $\lambda$ and is independent of k such that

$$\begin{aligned} \int _B|\nabla u|^2dx\ge L(k-2)^2-1 \end{aligned}$$

for any $u\in S_{k,\lambda ,1}$.

Proof

Fix $\lambda \ge \Lambda _*$ and suppose $k\ge 2$. For any $u\in S_{k,\lambda ,1}$ with $u(0)>0$, we write $u=u(|x|)$ and define a number $r_{u,2}\in (0,1)$ so that there exists a value $r_{u,1}\in (0,r_{u,2})$ such that $u(r_{u,i})=0$ for $i=1,2$ and $u(r)>0$ for all $r\in [0,r_{u,1})$ and $u(r)<0$ for all $r\in (r_{u,1},r_{u,2})$. Then we claim that

$$\begin{aligned} r_\lambda :=\inf _{k\ge 2}\inf _{u\in S_{k,\lambda ,1},u(0)>0}r_{u,2}>0. \end{aligned}$$

Assume that $r_\lambda =0$ on the contrary. Then for any $\varepsilon \in (0,1)$, we can choose $k_\varepsilon \ge 2$ and a nodal radial solution $u_\varepsilon \in S_{k_\varepsilon ,\lambda ,1}$ with $u_\varepsilon (0)>0$ such that $r_{u_{\varepsilon },2}\in (0,\varepsilon )$. Here we fix $\varepsilon \in (0,1)$ so small that $\varepsilon ^2\lambda \in (0,\Lambda _*)$ and define $v_\varepsilon (x):=u_\varepsilon (r_{u_{\varepsilon },2} |x|)$ ($x\in B$). Then we get $v_\varepsilon \in S_{1,r_{u_{\varepsilon },2}^2\lambda ,1}$. This is impossible by the definition of $\Lambda _*$. Hence we ensure the claim. Choose any $u\in S_{k,\lambda ,1}$ with $u(0)>0$ and regard $u=u(|x|)$. Then we perform the Liouville transformation $r=1/(1-\log {t})$, $v(r)=ru(t)$ for $t\in (r_{u,2},1]$. It follows that

$$\begin{aligned} v''+q(r)v=0,\ \ \ \ r\in \left[ \frac{1}{1-\log {r_{u,2}}},1\right] \end{aligned}$$

where

$$\begin{aligned} q(r)=t^2\left( \log {\frac{e}{t}}\right) ^4\frac{\lambda f(u(t))}{u(t)}\ \ (t=e^{1-1/r}), \end{aligned}$$

and $f(u)=ue^{u^2+\alpha |u|}$. Notice that v has exactly $k-1$ zeros in $\left( 1/(1-\log {r_{u,2}}),1\right]$. Then noting $f(t)/t\le e^{1+\alpha }+t^2e^{t^2+\alpha |t|}$ for any $t\in {\mathbb {R}}$, we apply Lemma 6.9 and get

$$\begin{aligned} \begin{aligned}&k-1\\&<\frac{1}{2}\left( \left( 1-\frac{1}{1-\log {r_{u,2}}}\right) \int _{\frac{1}{1-\log {r_{u,2}}}}^1 q(r)dr\right) ^{\frac{1}{2}}+1\\&=\frac{1}{2}\left( \left( 1-\frac{1}{1-\log {r_{u,2}}}\right) \int _{r_{u,2}}^1 \frac{\left( \log {\frac{e}{t}}\right) ^4}{(1-\log {t})^2}\frac{\lambda f(u(t))}{u(t)}tdt\right) ^{\frac{1}{2}}+1\\&\le \frac{1}{2}\left( 1-\frac{1}{1-\log {r_\lambda }}\right) ^{\frac{1}{2}}\left( \log {\frac{e}{r_\lambda }}\right) ^2\left( \int _{r_{u,2}}^1 \lambda (e^{1+\alpha }+u(t)f(u(t)))tdt\right) ^\frac{1}{2}+1\\&\le \frac{1}{\sqrt{L}}\left( 1+\int _B|\nabla u|^2dx\right) ^{\frac{1}{2}}+1 \end{aligned} \end{aligned}$$

for some constant $L=L(\lambda )>0$. This completes the proof. $\square$

Then we can prove the next key lemma.

Lemma 6.11

For any $\lambda \in [\Lambda _*,\Lambda _1]$, there exists a natural number $k_\lambda$ such that for any $k\ge k_\lambda$, there exists no radial nodal solution $u\in S_{k,\lambda ,1}$ satisfying $I_{\lambda ,1}(u)\le 2\pi k+c_{0,\lambda ,1}$.

Proof

Assume $\lambda \in [\Lambda _*,\Lambda _1]$, $k\in {\mathbb {N}}$ and $u\in S_{k,\lambda ,1}$ satisfies $I_{\lambda ,1}(u)\le 2\pi k+c_{0,\lambda ,1}$. We claim that there exists a constant $M=M(\lambda )>0$ which depends only on $\lambda$ and independent of k and u such that

$$\begin{aligned} \int _B|\nabla u|^2dx\le 4\pi (k+1)+M. \end{aligned}$$

(6.1)

To see this, choose u as above and put $f(t)=te^{t^2+|t|}$ and $F(t)=\int _0^t f(s)ds$. Then an elementary calculation shows that for any $\varepsilon >0$, there exists a constant $t_\varepsilon >0$ such that $\varepsilon f(t)t-F(t)\ge 0$ for any $|t|\ge t_\varepsilon$. (See formula (2.3) in [12]). We choose $\varepsilon \in (0,1/2)$ so small that $(2\pi k+c_{0,\lambda ,1})/(1/2-\varepsilon )\le 4\pi (k+1)$. This choice is possible since $c_{0,\lambda ,1}<2\pi$ ([2]). Then there exists a constant $C_\varepsilon >0$ which is independent of k and u such that

$$\begin{aligned} \begin{aligned} 2\pi k+c_{0,\lambda ,1}&\ge I_{\lambda ,\beta }(u)-\varepsilon \langle I_{\lambda ,\beta }'(u),u\rangle \\&\ge \left( \frac{1}{2}-\varepsilon \right) \int _B|\nabla u|^2dx+\lambda \int _{B\cap \{u\le t_\varepsilon \}}(\varepsilon f(u)u-F(u))dx\\&\ge \left( \frac{1}{2}-\varepsilon \right) \int _B|\nabla u|^2dx-\lambda C_\varepsilon . \end{aligned} \end{aligned}$$

This proves (6.1) by putting $M=\lambda C_\varepsilon /(1/2-\varepsilon )$. Then take a constant $L=L(\lambda )>0$ from the previous lemma. Choose a natural number ${\tilde{k}}_\lambda \ge 2$ so that $L(k-2)^2-1> 4\pi (k+1)+M$ for all $k\ge {\tilde{k}}_\lambda$. As a consequence, it follows from the previous lemma and (6.1) that $k\ge {\tilde{k}}_\lambda$ and $u\in S_{k,\lambda ,1}$ yield $I_{\lambda ,1}(u)> 2\pi k+c_{0,\lambda ,1}$. This finishes the proof. $\square$

Let us complete the proof of Proposition 6.3.

Proof of Proposition 6.3

Assume as in the proposition. First suppose $\lambda _*=\Lambda _1$. Then if $k=0$, since $\lambda _*\not =0$ and $S_{0,\Lambda _1,\beta _*}=\emptyset$, we clearly have that (vii) of Theorem 1.1 occurs with $u_0=0$. In addition, assume $\beta _n\ge 1$ for all $n\in {\mathbb {N}}$ and suppose $k\ge 1$ and $\beta _*\in (1,3/2)$ or $k=1$ and $\beta _*=1$. Then in the former case, noting $\beta _*>1$ and on the other hand, in the latter case, using $S_{0,\Lambda _1,1}=\emptyset$, we see that $(u_n)$ does not blow up and thus, $(u_n)$ behaves as in (vii) with $u_0\not =0$ since $0<\lambda _*=\Lambda _1<\Lambda _{k+1}$ in both cases. This completes (a). Next suppose $k\ge 1$, $\lambda _*\in (0,\Lambda _1)$, $\beta _n>1$ for all $n\in {\mathbb {N}}$ and $\beta _*=1$. Then, since $\lambda _*\not =0$, $(u_n)$ behaves as in (iii) or (vii) of Theorem 1.1. Hence, if we additionally suppose $S_{\kappa ,\lambda _*,1}=\emptyset$ for all $0<\kappa \le k$, the only possibility is (iii) with $N=k$ to happen. This completes (b). Finally, suppose $I_{\lambda _n,\beta _n}(u_n)=c_{k,\lambda _n,\beta _n}$ for all $n\in {\mathbb {N}}$. Set $\lambda _*\in [\Lambda _*,\Lambda _1)$ and $k\ge k_{\lambda _*}$ where $k_\lambda$ is chosen from Lemma 6.11. It follows that $(u_n)$ behaves as in (iii) with a natural number $k-k_{\lambda _*}<N\le k$. Otherwise, there would be integers $0\le N\le k-k_{\lambda _*}$, $\kappa =k-N$ and an element $u_0\in S_{\kappa ,\lambda _*,1}$ such that

$$\begin{aligned} I_{\lambda _n,\beta _n}(u_n)\rightarrow 2\pi N+I_{\lambda _*,1}(u_0)\le 2\pi k+c_{0,\lambda _*,1} \end{aligned}$$

by Lemma 5.1. This implies $I_{\lambda _*,1}(u_0)\le 2\pi \kappa +c_{0,\lambda _*,1}$. Since $\kappa \ge k_{\lambda _*}$, this is impossible by Lemma 6.11. This proves the former conclusion of (c). Then noting the first conclusion of Corollary 6.1, we get the latter one. Lastly let us show (d). Set $\lambda _*$ as in the assumption. Suppose $k\ge 2$ and $(u_n)$ behaves as in (iii). If $N=k$ on the contrary, then by Lemma 5.1, the weak limit $u_0$ of $(u_n)$ satisfies $u_0\in S_{0,\lambda _*,1}$, $I_{\lambda _*,1}(u_0)=c_{0,\lambda _*,1}$ and $|u_0(0)|\ge \alpha /2$. This is a contradiction. Hence we get $N\not =k$. This proves the first assertion. Especially, combining (c) with the previous conclusion, we obviously get the final conclusion. This completes the proof. $\square$

Finally, we complete the proof of Corollary 6.4.

Proof of Corollary 6.4

Fix $\beta \in [1,3/2)$. Note that the first assertion in (a) of the previous proposition implies $c_{0,\lambda _n,\beta }\rightarrow 0$ if $\lambda _n\uparrow \Lambda _1$. Then, we set $k=1$ if $\beta =1$ and $k\ge 1$ if $\beta >1$ and choose sequences $(\lambda _n)\subset (0,\Lambda _1)$ and $(u_n)$ so that $\lambda _n\uparrow \Lambda _1$, $u_n\in S_{k,\lambda _n,\beta }$ and $I_{\lambda _n,\beta }(u_n)\le 2\pi k+c_{0,\lambda _n,\beta }$ for all $n\in {\mathbb {N}}$. This choice is valid by the argument in the first paragraph of this section and Corollary 6.1. Then from the latter assertion in (a) of the previous proposition, we have a solution $u_0\in S_{k,\Lambda _1,\beta }$ such that $I_{\lambda _n,\beta }(u_n)\rightarrow I_{\Lambda _1,\beta }(u_0)\le 2\pi k$. This proves the former conclusion. The latter assertion is clear by noting $S_{0,\Lambda _1,1}=\emptyset$, choosing $k_{\Lambda _1}$ from Lemma 6.11 and arguing as in the proof of Corollary 6.1 and (d) of Proposition 6.3. This finishes the proof. $\square$

Proof of Lemma 2.5

In this appendix we show the proof of Lemma 2.5.

Proof of Lemma 2.5

Without loss of the generality we may assume $u_{i,n}\ge 0$. First we claim that $r_{i,n}/\gamma _{i,n}\rightarrow \infty$. If not, we have a constant $C>0$ such that $r_{i,n}/\gamma _{i,n}\le C$ for all $n\in {\mathbb {N}}$. Then putting $v_{n}(r)=u_{i,n}(r_{i,n}r)$ for $r\in [r_{i-1,n}/r_{i,n},1]$, we get from (2.1) that

$$\begin{aligned} {\left\{ \begin{array}{ll} -v_n''-\frac{1}{r}v_n'=\lambda _n r_{i,n}^2 f_n(v_n),\ (-1)^{i-1}v_n>0\text { in }(r_{i-1,n},r_{i,n}),\\ v_n(1)=0=v_n'(\rho _{i,n}/r_{i,n}),\\ v_n(r_{i-1,n}/r_{i,n})=0\text { if }i\ge 2. \end{array}\right. } \end{aligned}$$

Then the above equation implies

$$\begin{aligned} -v_n''-\frac{1}{r}v_n'\le \lambda _n r_{i,n}^2f_n(\mu _{i,n})=\frac{1}{2\mu _{i,n}}\left( \frac{r_{i,n}}{\gamma _{i,n}}\right) ^2\le \frac{C^2}{2\mu _{i,n}}\rightarrow 0, \end{aligned}$$

uniformly on $[r_{i-1,n}/r_{i,n},1]$. It follows that $v_n\rightarrow 0$ uniformly in $[r_{i-1,n}/r_{i,n},1]$. This contradicts our assumption (2.2). This proves the claim. In particular, we get $\gamma _{i,n}\rightarrow 0$. Next, we claim $\rho _{i,n}/r_{i,n}\rightarrow 0$. This is trivial for $i=1$. Hence we assume $i\ge 2$. Define $v_n$ as above. It follows from our assumption (2.3) and Lemma 2.1 that there exist constants $c,C>0$ such that

$$\begin{aligned} C\ge \int _{r_{i-1,n}/r_{i,n}}^{1} v_n'(r)^2rdr\ge \frac{\mu _{i,n}^2}{2\pi c^2}\left( \frac{\rho _{i,n}}{r_{i,n}}\right) . \end{aligned}$$

Then (2.2) shows the claim. In particular, we get $\rho _{i,n}\rightarrow 0$ and $(r_{i,n}-\rho _{i,n})/\gamma _{i,n}\rightarrow \infty$ by the first claim. Next, by the definition of $z_{i,n}$ and (2.8), we get that

$$\begin{aligned} 0\le -z_{i,n}''-\frac{1}{r+\frac{\rho _{i,n}}{\gamma _{i,n}}}z_{i,n}'\le 1 \text { on }\left[ \frac{r_{i-1,n}-\rho _{i,n}}{\gamma _{i,n}},\frac{r_{i,n}-\rho _{i,n}}{\gamma _{i,n}}\right] . \end{aligned}$$

Then, for any $r\in \left[ \frac{r_{i-1,n}-\rho _{i,n}}{\gamma _{i,n}},\frac{r_{i,n}-\rho _{i,n}}{\gamma _{i,n}}\right]$, multiplying the equation by $r+\frac{\rho _{i,n}}{\gamma _{i,n}}$ and integrating over (0, r) if $r\ge 0$ and over (r, 0) if $r<0$ give

$$\begin{aligned} {\left\{ \begin{array}{ll} 0\le -z_{i,n}'(r)\le \ \frac{\frac{r^2}{2}+\frac{\rho _{i,n}}{\gamma _{i,n}}r}{r+\frac{\rho _{i,n}}{\gamma _{i,n}}} \text { if }r\ge 0, \text { and }\\ 0\le z_{i,n}'(r)\le -\frac{\frac{r^2}{2}+\frac{\rho _{i,n}}{\gamma _{i,n}}r}{r+\frac{\rho _{i,n}}{\gamma _{i,n}}} \text { if }r<0. \end{array}\right. } \end{aligned}$$

(1.1)

Integrating this again, we get

$$\begin{aligned} \begin{aligned} 0\le -z_{i,n}(r)&\le \int _0^r\frac{\frac{r^2}{2}+\frac{\rho _{i,n}}{\gamma _{i,n}}r}{r+\frac{\rho _{i,n}}{\gamma _{i,n}}}dr\\&= {\left\{ \begin{array}{ll} \frac{r^2}{4}\text { if }i=1,\\ \frac{r^2}{4}+\left( \frac{\rho _{i,n}}{\gamma _{i,n}}\right) \frac{r}{2}-\frac{1}{2}\left( \frac{\rho _{i,n}}{\gamma _{i,n}}\right) ^2\log {\frac{\frac{\rho _{i,n}}{\gamma _{i,n}}+r}{\frac{\rho _{i,n}}{\gamma _{i,n}}}},\text { if }i\ge 2, \end{array}\right. } \end{aligned} \end{aligned}$$

(1.2)

for any $r\in \left[ \frac{r_{i-1,n}-\rho _{i,n}}{\gamma _{i,n}},\frac{r_{i,n}-\rho _{i,n}}{\gamma _{i,n}}\right]$. Then from (1.1) and (1.2), we get that $z_{i,n}$ is uniformly bounded in $C^1_{\text {loc}}((-l,\infty ))$ ($C^1_{\text {loc}}([0,\infty ))$ if $l=0$). Furthermore, since (1.1) implies $|z_{i,n}'(r)/(r+\rho _{i,n}/\gamma _{i,n})|$ is locally uniformly bounded in $(-l,\infty )$ ($[0,\infty )$ if $l=0$), using the equation in (2.8), we get that $z_{i,n}$ is uniformly bounded in $C^2_{\text {loc}}((-l,\infty ))$ ($C^2_{\text {loc}}([0,\infty ))$ if $l=0$). Then it follows from the Ascoli-Arzelà theorem and the equation in (2.8) that there exists a function z such that $z_{i,n}\rightarrow z$ in $C^2_{\text {loc}}((-l,\infty ))$ ($C^2_{\text {loc}}((0,\infty ))\cap C^1_{\text {loc}}([0,\infty ))$ if $l=0$).

Now our final aim is to show $l=m<\infty$. This is clear if $i=1$. Hence after this we assume $i\ge 2$. We first claim $l<\infty$. We suppose $l=\infty$ on the contrary. Then $m=\infty$. It follows from (2.8) that z satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} -z''=e^z\hbox { in }{\mathbb {R}},\\ z(0)=0=z'(0). \end{array}\right. } \end{aligned}$$

This implies $z(r)=\log \frac{4e^{\sqrt{2}r}}{\left( 1+e^{\sqrt{2}r}\right) ^2}$ $(r\in {\mathbb {R}})$. Then by (2.3), there exits a constant $C>0$ such that

$$\begin{aligned} C&\ge \lambda _n\int _{\rho _{i,n}}^{r_{i,n}}u_nf_n(u_n)rdr\nonumber \\&\ge \frac{1}{2} \int _{0}^{{\frac{r_{i,n}-\rho _{i,n}}{\gamma _{i,n}}}}\left( \frac{z_{i,n}}{2\mu _{i,n}^2}+1\right) e^{z_{i,n}+\frac{z_{i,n}^2}{4\mu _{i,n}^2}+\alpha \mu _{i,n}^{\beta _n}\left\{ \left( \frac{z_{i,n}}{2\mu _{i,n}^2}+1\right) ^{\beta }-1\right\} }\left( r+\frac{\rho _{i,n}}{\gamma _{i,n}}\right) dr\nonumber \\&\ge \frac{1}{2} \frac{\rho _{i,n}}{\gamma _{i,n}}\int _{0}^{{\frac{r_{i,n}-\rho _{i,n}}{\gamma _{i,n}}}}\left( \frac{z_{i,n}}{2\mu _{i,n}^2}+1\right) e^{z_{i,n}+\frac{z_{i,n}^2}{4\mu _{i,n}^2}+\alpha \mu _{i,n}^{\beta _n}\left\{ \left( \frac{z_{i,n}}{2\mu _{i,n}^2}+1\right) ^{\beta }-1\right\} }dr. \end{aligned}$$

(1.3)

Here the Fatou lemma implies

$$\begin{aligned} \liminf _{n\rightarrow \infty }\int _{0}^{{\frac{r_{i,n}-\rho _{i,n}}{\gamma _{i,n}}}}\left( \frac{z_{i,n}}{2\mu _{i,n}^2}+1\right) e^{z_{i,n}+\frac{z_{i,n}^2}{4\mu _{i,n}^2}+\alpha \mu _{i,n}^{\beta _n}\left\{ \left( \frac{z_{i,n}}{2\mu _{i,n}^2}+1\right) ^{\beta }-1\right\} }dr\ge \int _0^\infty e^zdr. \end{aligned}$$

Since the right hand side of the inequality above is positive value and $m= \infty$, we get that the right hand side of (1.3) diverges to infinity which is a contradiction. This proves the claim. Finally we show $l=m$. Let us assume $l<m$ on the contrary. Then we claim that there exists a constant $C>0$ such that

$$\begin{aligned} |z_{i,n}'(r)|\le C \text { for all }n\in {\mathbb {N}}\text { and }r\in \left[ \frac{r_{i-1,n}-\rho _{i,n}}{\gamma _{i,n}},0\right] . \end{aligned}$$

In fact, if $m=\infty$, the claim follows easily by the latter formula in (1.1). If $m<\infty$, using (1.1) again we get a constant $C>0$ such that

$$\begin{aligned} 0\le z_{i,n}'(r)\le \frac{C}{m-l+o(1)} \end{aligned}$$

uniformly for all $r\in \left[ \frac{r_{i-1,n}-\rho _{i,n}}{\gamma _{i,n}},0\right]$. This proves the claim. On the other hand, by the mean value theorem, we have a sequence $(\xi _n)\subset \left( \frac{r_{i-1,n}-\rho _{i,n}}{\gamma _{i,n}},0\right)$ such that

$$\begin{aligned} z_{i,n}'(\xi _n)=\frac{-z_{i,n}\left( \frac{r_{i-1,n}-\rho _{i,n}}{\gamma _{i,n}}\right) }{\frac{\rho _{i,n}-r_{i-1,n}}{\gamma _{i,n}}}=\frac{2\mu _{i,n}^2}{l+o(1)}\rightarrow \infty \end{aligned}$$

since $l\in [0,\infty )$. This is a contradiction. We finish the proof.