Infinitely many radial solutions for a sub-super critical p-Laplacian problem

Abstract

We prove the existence of infinitely many sign-changing radial solutions for a p-sub-super critical p-Laplacian Dirichlet problem in a ball. We consider an equation defined by the p-Laplacian operator perturbed by a nonlinearity g(u) that is p-subcritical at $+\,\infty$ and p-supercritical at $-\,\infty$. Our results extend those in Castro et al. (Electron. J. Differ. Equ. 2007(111):1–10, 2007) for the corresponding semilinear case and those of El Hachimi and De Thelin (J. Differ. Equ. 128:78–102, 1996) where the subcritical case was studied.

1 Introduction

We study the quasilinear Dirichlet problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{aligned} \varDelta _p\, u + g(u)&{}=0 \ \quad \text { in }\, B_1(0)\subset \mathbb {R}^N, \\ u &{}=0 \quad \ \text { on }\, \partial B_1(0), \end{aligned} \end{array}\right. } \end{aligned}$$

(1.1)

where $\triangle _p u=\text {div}\,(|\nabla u|^{p-2}\nabla u)$, $1< p <N$, denotes the p-Laplacian operator. We assume that

$$\begin{aligned} g(s):={\left\{ \begin{array}{ll} |s|^{q_1} &{} \quad s\ge 0,\\ -|s|^{q_2} &{} \quad s< 0, \end{array}\right. } \end{aligned}$$

(1.2)

with

$$\begin{aligned} p-1<q_1<p^*-1<q_2<+\infty , \end{aligned}$$

(1.3)

where $p^*:=\frac{Np}{N-p}$ is the critical Sobolev exponent. That is, g has p-subcritical growth at $+\,\infty$ and p-supercritical growth at $-\,\infty$. Our main result is the following theorem.

Theorem 1.1

If g satisfies (1.2) and (1.3), then (1.1) has infinitely many radial solutions.

Remark 1.1

• (a) Theorem 1.1 remains valid when (1.2) is replaced by a locally Lipschitz function g satisfying $sg(s)>0$ for all $s\ne 0$, and either

  $$\begin{aligned} \lim _{s \rightarrow +\infty } \frac{g(s)}{s^{q_1}}=:L_1 \in (0,+\infty ) \ \ \hbox {and} \ \ \lim _{s \rightarrow -\infty } \frac{-g(s)}{|s|^{q_2}}=:L_2 \in (0,+\infty ), \end{aligned}$$

  (1.4)

  or

  $$\begin{aligned} \lim _{s \rightarrow +\infty } \frac{g(s)}{s^{q_2}}=:L_3 \in (0,+\infty ) \ \ \hbox {and} \ \ \lim _{s \rightarrow -\infty } \frac{-g(s)}{|s|^{q_1}}=:L_4 \in (0,+\infty ). \end{aligned}$$

  (1.5)

  In the appendix (see Sect. 5), we indicate the refinements needed to extend our result for this kind of nonlinearities.

• (b) Theorem 1.1 remains valid when the right-hand side in the partial differential equation in (1.1) is replaced by a continuous function of $(\Vert x\Vert , u)$ as long as such function is dominated at infinity by the nonlinearity g. This was done for the 2-Laplacian in [1] and for the p-Laplacian in [2] when g is a continuous function of $\Vert x\Vert$ for subcritical problems and some sub-super critical problems. The results in [1, 2] include cases where $1<p< N/(N-2)< (N+2)/(N-2) < q$ but not those where $N/(N-2)< p< (N+2)/(N-2) < q$. The discussion in the appendix includes the case in which the right-hand side in (1.1) is a function of u.

The radial solutions to (1.1) are the solutions to

$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{aligned} (p-1)|u'|^{p-2} u''+ \frac{N-1}{r} |u'|^{p-2}u' + g\big (u(r)\big )&{}=0, \qquad 0<r<1\\ u^\prime (0)=0, \quad u(1) &{}=0. \end{aligned} \end{array}\right. } \end{aligned}$$

(1.6)

Our proof of Theorem 1.1 is based on the analysis of the solutions to the initial value problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{aligned} (p-1)|u'|^{p-2} u''+ \frac{N-1}{r} |u'|^{p-2}u' + g\big (u(r)\big )&{}=0, \qquad 0<r<1\\ u(0)=d, \quad u^\prime (0)&{}=0. \end{aligned} \end{array}\right. } \end{aligned}$$

(1.7)

In fact, in Sect. 3, we show that the profile of the solutions of (1.7) is as shown in Fig. 1 (see below), with the number of oscillations increasing as $d\rightarrow +\infty .$

Fig. 1

Illustration of the oscillating behavior of the solutions to (1.7)

Following ideas that go back to [3], we simply prove the existence of a sequence $\{d_j\}_j$ converging to $+\,\infty$ such that the solution to (1.7) with $d = d_j$ satisfies $u(1) =0$.

Theorem 1.1 extends the results in [4] where the case $p=2$ is studied. It also extends the results in [2] where both $q_1, q_2<p^*$.

When $B_1(0)$ is replaced by an annulus, Eq. (1.6) has no singularity. This leads to the proof that (1.1) has infinitely many radial solutions, only assuming $\lim _{|s|\rightarrow \infty }g(s)/(s|s|^{p-2}) = +\infty$, see [5].

We obtain $L^\infty$ solutions that satisfy (1.1) in the classical sense in the unit ball for $r\ne 0,$ and weak solutions in the whole ball; those solutions are in $C^{1,\mu }[0,1]$ for $\mu \in (0,1)$. This is consistent with regularity theory for the p-Laplacian problems. Weak solutions in $W^{1,p}(\varOmega ) \cap L^{\infty } (\varOmega )$ are in $C^{1,\mu }(\overline{\varOmega })$, see [6, 7].

When $p=2$, $q_1=q_2=(N+2)/(N-2)$ and $\varOmega$ is star-shaped, $u=0$ is the only solution to (1.1), see [8]. However, if $\varOmega$ has nontrivial topology then (1.1) has nontrivial solutions and it is also proved that the critical problem (1.1) has a solution in annulus $\varOmega = \{x \in \mathbb {R}^N: \ 0< a< |x| < b\}$, see [9].

For the p-Laplacian problem, it is shown that the p-critical problem has no positive solution in star-shaped domains, see [10]. On the other hand, if we look for radial solutions for the p-critical problem in annulus, it is easy to prove that (1.1) has positive solutions, see [5].

For Liouville theorems in half spaces and $1<p<N$, a priori estimates and existence of positive solutions for a more general nonlinearity $g=g(x,u,Du)$, under subcritical growth of g with respect to u at infinity and zero, are studied, see [11]. In [12], a priori $L^\infty$ bounds and existence of solutions are proved for more general nonlinearities than pure powers.

The subcritical and critical cases, both semilinear and quasilinear, have been widely studied with different techniques. See [13] for an asymptotically linear problem, [14] for a p-superlinear problem, [15] for bifurcation techniques in the subcritical case and [16, 17] for minimax methods in critical point theory, the mountain pass theorem and variational methods in bifurcation theory. See also [18] for quasilinear equations and [19,20,21] and references therein for a priori bounds in general domains.

The supercritical case is not so extensively studied, see, for instance, [22,23,24,25,26,27,28] and references therein. For the existence and multiplicity of positive radial solutions, see [29]. See also [30, 31] where the existence of infinitely many (positive) solutions to a supercritical problem in a ball is proved ($p=2$ in [30], $1< p <2$ in [31]). The nonlinearities considered are linear for u small and supercritical for u large. Furthermore, the authors use some phase plane analysis after performing Fowler transformation.

To the best of our knowledge, the only study concerning problems with nonlinearities exhibiting a subcritical behavior at $+\,\infty$ and a supercritical behavior at $-\,\infty$ in the semilinear case is [4] (see also [1] for some related problems).

This paper is organized as follows. In Sect. 2, we introduce the energy and related functions, and their relationship. In Sect. 3, we study the oscillating qualitative properties of the solutions which will be used for proving Theorem 1.1 in Sect. 4. Finally, in the appendix, we indicate the refinements needed to extend our results to more general nonlinearities.

2 Preliminaries and known results

Our analysis of the solutions to the initial value problem (1.7) makes extensive use of the following quantities:

$$\begin{aligned} \mathcal {E}(r,d)&:=\mathcal {E}(r) = \dfrac{p-1}{p}|u^\prime (r)|^p+G\big (u(r)\big ), \nonumber \\ H (r,d)&:= r\mathcal {E}(r) + \frac{N-p}{p} \big |u'(r)\big |^{p-2}\, u'(r)\, u (r), \nonumber \\ P(r,d)&:= \int _0^rs^{N-1}\left[ N G\big (u(s)\big ) - \frac{N-p}{p} g\big (u(s)\big )u(s)\right] \, \mathrm{d}s. \end{aligned}$$

(2.1)

where $u(r)=u(r,d)$ is a solution to (1.7), and $G(s):=\int _0^s\, g(t)\, \hbox {d}t={\left\{ \begin{array}{ll} \frac{|s|^{q_1+1}}{q_1+1}, &{} \quad s\ge 0 \\ \frac{|s|^{q_2+1}}{q_2+1}, &{} \quad s< 0. \end{array}\right. }$

It is easy to check that for a given solution of (1.7),

$$\begin{aligned} \mathcal {E}^{\,\prime }(r)=\Big [(p-1)|u'|^{p-2}u''+g(u)\Big ]u'=-\dfrac{N-1}{r}|u^\prime |^p\le 0,\quad \text{ for } r>0, \end{aligned}$$

(2.2)

that is, the energy $\mathcal {E}$ is a decreasing function. The quantities in (2.1) are related by the Pohozaev-type identity

$$\begin{aligned} r^{N-1}H(r,d) - t^{N-1} H(t,d) =\int _t^r s^{N-1}\left[ N G (u(s)) - \frac{N-p}{p} g (u(s)) \, u(s) \right] \, \mathrm{d}s, \end{aligned}$$

(2.3)

see Proposition 4.1 of [32].

Taking $t=0$ in Eq. (2.3), we have the following Pohozaev identity

$$\begin{aligned} r^{N-1}H(r,d)= P(r,d), \end{aligned}$$

equivalently

$$\begin{aligned}&r^N\left[ \frac{p-1}{p} \big | u'(r) \big |^{p}\, + G\big (u(r)\big )\right] +\frac{N-p}{p}\, r^{N-1}\big |u'(r)\big |^{p-2}\, u'(r)\, u(r) \nonumber \\&\quad = \int _0^rs^{N-1}\left[ N G\big (u(s)\big ) - \frac{N-p}{p} g\big (u(s)\big )u(s)\right] \, \mathrm{d}s = P(r,d). \end{aligned}$$

(2.4)

Therefore,

$$\begin{aligned} \text {if}\quad u(r)=0,\qquad \text {then}\quad P(r,d)=\frac{p-1}{p} r^N \big | u'(r) \big |^{p}, \end{aligned}$$

(2.5)

and

$$\begin{aligned} \text {if}\quad u'(r)=0, \qquad \text {then}\quad P(r,d)= r^N G\big (u(r)\big ). \end{aligned}$$

(2.6)

If $u(s)\ge 0$, $u(s)\not \equiv 0$ for all $s\in (t_1,t_2)\subset [0,1]$, then

$$\begin{aligned} P(t_2,d)-P(t_1,d)= \frac{N-p}{p}\,\left[ \frac{p^*}{q_1+1} -1\right] \,\int _{t_1}^{t_2} s^{N-1}\, \big (u(s)\big )^{q_1+1}\, \mathrm{d}s> 0, \end{aligned}$$

(2.7)

and since the integrand is positive on $(t_1,t_2),$P(r) is an increasing function on $[t_1,t_2]$.

On the other hand, if $u(s)\le 0$, $u(s)\not \equiv 0$ for all $s\in (t_1,t_2)$, then

$$\begin{aligned} P(t_2,d)-P(t_1,d)= -\frac{N-p}{p}\left[ 1- \frac{p^*}{q_2+1} \right] \,\int _{t_1}^{t_2} s^{N-1} \big |u(s)\big |^{q_2+1}\, \mathrm{d}s< 0, \end{aligned}$$

(2.8)

and P(r) is a decreasing function on $[t_1,t_2]$.

Since (1.7) is equivalent to

$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{aligned} \left( r^{N-1}\big |u'\big |^{p-2}\, u'\right) ^\prime + r^{N-1}g\big (u(r)\big )&{}=0, \quad 0<r<1\\ u(0)=d, \quad u^\prime (0)&{}=0, \end{aligned} \end{array}\right. } \end{aligned}$$

(2.9)

integrating on $[t_1,t_2]$, we have

$$\begin{aligned} t_2^{N-1} \big |u'(t_2)\big |^{p-2}\, u'(t_2)-t_1^{N-1} \big |u'(t_1)\big |^{p-2}\, u'(t_1) = -\int _{t_1}^{t_2} s^{N-1} g\big (u(s)\big ) \, \mathrm{d}s. \end{aligned}$$

(2.10)

Thus, if $u(s)\ge 0$, $u(s)\not \equiv 0$ for all $s\in [t_1,t_2]\subset [0,1]$, then

$$\begin{aligned} t_2^{N-1} \big |u'(t_2)\big |^{p-2}\, u'(t_2)-t_1^{N-1} \big |u'(t_1)\big |^{p-2}\, u'(t_1) = -\int _{t_1}^{t_2} s^{N-1} \big (u(s)\big )^{q_1} \, \mathrm{d}s<0. \end{aligned}$$

(2.11)

On the other hand, if $u(s)\le 0$, $u(s)\not \equiv 0$ for all $s\in [t_1,t_2]$, then

$$\begin{aligned} t_2^{N-1} \big |u'(t_2)\big |^{p-2}\, u'(t_2)-t_1^{N-1} \big |u'(t_1)\big |^{p-2}\, u'(t_1) = \int _{t_1}^{t_2} s^{N-1} \big |u(s)\big |^{q_2} \, \mathrm{d}s>0. \end{aligned}$$

(2.12)

3 Oscillation properties of the solutions to (1.7)

Our arguments rely on the study of the oscillation behavior of solutions to (1.7). We estimate from below and from above the values of the first zero, the first point of local minimum, the second zero and the first positive point of local maximum value.

Let u be a solution of (1.7), and assume that $u(0) = d >0$. Let

$$\begin{aligned} \tau _1=\tau _1(d) := \sup \{r>0;\ u(s) \ge 0, \ \ \hbox {for all } \ s\in [0,r]\}. \end{aligned}$$

(3.1)

Thus, $\tau _1>0$ is the smallest zero of u. Let

$$\begin{aligned} \rho _1=\rho _1(d) := \sup \{r>0;\ u'(s) \le 0 \ \ \hbox {for all } \ s\in [0,r]\}. \end{aligned}$$

(3.2)

That is, $\rho _1$ is the smallest point of minimum of u. Let

$$\begin{aligned} \tau _2=\tau _2(d) := \sup \{r>\tau _1(d);\ u(s) \le 0,\ \ \hbox {for all } \ s\in [\tau _1(d),r]\}, \end{aligned}$$

(3.3)

i.e., $\tau _2$ is the second zero of u. Finally, let

$$\begin{aligned} \sigma _1=\sigma _1(d) := \sup \{r>0;\ u'(s) \ge 0,\ \ \hbox {for all } \ s \in [\tau _2(d),r]\}, \end{aligned}$$

(3.4)

i.e., $\sigma _1$ is the first point of maximum of u.

By using (2.9)–(2.12), it is straightforward to check that

$$\begin{aligned} 0<\tau _1<\rho _1<\tau _2<\sigma _1, \end{aligned}$$

(3.5)

and that

$$\begin{aligned}&u(r)>0,\qquad u'(r)<0,\qquad \text {for all}\quad r\in (0,\tau _1),\nonumber \\&u(r)<0,\qquad u'(r)<0,\qquad \text {for all}\quad r\in (\tau _1,\rho _1),\nonumber \\&u(r) <0,\qquad u'(r)>0,\qquad \text {for all}\quad r\in (\rho _1,\tau _2),\nonumber \\&u(r)>0,\qquad u'(r) >0,\qquad \text {for all}\quad r\in (\tau _2,\sigma _1). \end{aligned}$$

(3.6)

In Lemmas 3.2, 3.3, 3.5 and 3.6 we prove, in particular, that

$$\begin{aligned} \left( \frac{1}{d}\right) ^{(q_1+1-p)/p} = O(\tau _1) = O(\rho _1)=O(\tau _2) = O(\sigma _1) = O\left( \frac{1}{d}\right) ^{(q_1+1-p)/p}, \end{aligned}$$

(3.7)

for $d\gg 1$. The proof of Theorem 1.1 is based on estimating $\tau _1, \rho _1, \tau _2$, $\sigma _1$, and the value of u at those points. We achieve this in the following four steps.

Step 1.:

In this step, we prove that $\tau _1(d)=O\left( \left( \dfrac{1}{d} \right) ^\frac{q_1+1-p}{p}\right)$. Let

$$\begin{aligned} r_1=r_1(d) := \sup \{r>0;\ u(s) \ge d/2, \ \ \hbox {for all } \ s\in [0,r]\}. \end{aligned}$$

(3.8)

From now on, all throughout the paper, C and $C_0,\ C_1,\ C_2,\ldots$ will denote positive constants independent of d that may change from line to line.

Lemma 3.1

Let u be a solution to (1.7) with $d > 0$. If $r_1=r_1(d)$ is as defined by (3.8), then

• (i) There exist two constants $C_1,C_2>0$, independent of d, such that

  $$\begin{aligned} C_1 \bigg (\frac{1}{d}\bigg )^{\frac{q_1+1-p}{p}}\le r_1 \le C_2 \bigg (\frac{1}{d}\bigg )^{\frac{q_1+1-p}{p}}\ \ \hbox {for} \ \ d \gg 1. \end{aligned}$$

  (3.9)

• (ii) There exists a constant $C_3>0$, independent of d, such that

  $$\begin{aligned} P(r_1,d) \ge C_3\, d^{\, N\, \left[ 1-\frac{q_1+1}{p^*} \right] } \rightarrow +\infty \qquad \text {as } d\rightarrow +\infty . \end{aligned}$$

  (3.10)

Proof of Lemma 3.1

• (i) Let $u(0) = d>0$. Taking $(t_1,t_2)=(0,r)$ in (2.11),

  $$\begin{aligned} -r^{N-1} \big |u'(r)\big |^{p-2}\, u'(r) = \int _0^rs^{N-1} \big (u(s)\big )^{q_1} \, \mathrm{d}s \ge \frac{d^{\, q_1}}{2^{q_1}N}\ r^N,\quad \text {for } r\in [0,r_1]. \end{aligned}$$

  (3.11)

  Hence,

  $$\begin{aligned} -u'(r) \ge \left( \dfrac{d^{\, q_1}}{2^{q_1}N}\ \right) ^{\frac{1}{p-1}}\, r^{\frac{1}{p-1}},\qquad \text {for } r\in [0,r_1]. \end{aligned}$$

  (3.12)

  Integrating on $[0, r_1]$, we have

  $$\begin{aligned} \frac{d}{2} \ge \, \frac{p-1}{p}\, \left( \frac{d^{\, q_1}}{2^{q_1}N}\right) ^{\frac{1}{p-1}}r_1^{\, \frac{p}{p-1}}, \quad \hbox {or} \ \ \ r_1 \le C_2\bigg (\frac{1}{d}\bigg )^{\frac{q_1+1-p}{p}}, \end{aligned}$$

  (3.13)

  where $C_2>0$ is independent of d. Similarly, using that $u(r)\le d$ for all $r\in [0,r_1]$, we see that there exists $C_1>0$, independent of d, such that

  $$\begin{aligned} r_1 \ge C_1 \bigg (\frac{1}{d}\bigg )^{\frac{q_1+1-p}{p}} \ \ \hbox {for} \ \ d >0. \end{aligned}$$

  (3.14)

  Thus, the inequality (3.9) holds. By (1.3), $\frac{q_1+1}{p}>1$. This and (3.9) imply that $r_1(d)\rightarrow 0$ as $d\rightarrow +\infty$.

• (ii) From (2.7) with $(t_1,t_2)=(0,r_1)$, (3.14) and (1.3), there exists a constant $C>0$ such that

  $$\begin{aligned} P(r_1,d) \ge Cr_1^N d^{\, q_1+1} \ge C d^{\, N\, \left[ 1-\frac{q_1+1}{p^*} \right] } \rightarrow +\infty ,\qquad \text{ as } d\rightarrow +\infty , \end{aligned}$$

  (3.15)

  which proves this lemma.

$\square$

Next, we estimate the first zero $\tau _1(d)$ and we prove that $\tau _1(d)=O\left( \left( \dfrac{1}{d} \right) ^\frac{q_1+1-p}{p}\right) .$

Lemma 3.2

If u is a solution to (1.7) with $d > 0$, then

• (i) There exist two constants $C_1,C_2>0$, independent of d, such that

  $$\begin{aligned} C_1 \bigg (\frac{1}{d}\bigg )^{\frac{q_1+1-p}{p}}\le \tau _1 \le C_2 \bigg (\frac{1}{d}\bigg )^{\frac{q_1+1-p}{p}}\ \ \hbox {for} \ \ d \gg 1. \end{aligned}$$

  (3.16)

• (ii) There exist two constants $C_3,C_4>0$, independent of d, such that

  $$\begin{aligned} C_3\, d^{\, \frac{q_1+1}{p}}\le - \,u'(\tau _1) \le C_4\, d^{\, \frac{q_1+1}{p}},\qquad \hbox {for} \ \ d \gg 1. \end{aligned}$$

Remark 3.1

Observe that, from (2.4), the above lemma implies that there exist two constants $C_5,C_6>0$ such that

$$\begin{aligned} C_5\, d^{\, N\, \left[ 1-\frac{q_1+1}{p^*} \right] } \le P(\tau _1,d)= \frac{p-1}{p}\tau _1^N \big | u'(\tau _1) \big |^{p}\le C_6\, d^{\, N\, \left[ 1-\frac{q_1+1}{p^*} \right] },\quad \hbox {for} \ d \gg 1. \end{aligned}$$

(3.17)

Proof of Lemma 3.2

• (i) Since $\tau _1>r_1$, by Lemma 3.1(i), $\tau _1\ge C \ \left( \frac{1}{d}\right) ^{\, \frac{q_1+1-p}{p}}.$

From (2.11) for any $r\in [0,\tau _1]$,

$$\begin{aligned} r^{N-1}\big |u'(r)\big |^{p-1}\ge u(r)^{\, q_1}\, \frac{r^{N}}{N}. \end{aligned}$$

Hence,

$$\begin{aligned} -u'(r)\ge \left( \frac{1}{N}\right) ^{\frac{1}{p-1}}\, u(r)^{\, \frac{q_1}{p-1}}\ r^\frac{1}{p-1},\qquad \forall r\in [0,\tau _1]. \end{aligned}$$

Separating variables and integrating on $[r_1,r]$, we have

$$\begin{aligned} -\frac{u(r)^{\,1- \frac{q_1}{p-1}}- u(r_1)^{\,1- \frac{q_1}{p-1}}}{\frac{p-1-q_1}{p-1}}\ge \left( \frac{1}{N}\right) ^{\frac{1}{p-1}}\frac{r^\frac{p}{p-1}-r_1^\frac{p}{p-1}}{\frac{p}{p-1}}, \end{aligned}$$

or equivalently, for all $r\in [r_1,\tau _1]$

$$\begin{aligned} u(r)^{\,1- \frac{q_1}{p-1}}- u(r_1)^{\,1- \frac{q_1}{p-1}}\ge C\ \left[ r^\frac{p}{p-1}-r_1^\frac{p}{p-1}\right] , \end{aligned}$$

where $C=C(p,q_1,N)=\frac{q_1+1-p}{p}\left( \frac{1}{N}\right) ^{\frac{1}{p-1}}>0.$ This and (1.3) yield

$$\begin{aligned} u(r)^{\,\frac{q_1+1-p}{p-1}} \le \frac{1}{C\ r^\frac{p}{p-1}\, \left[ 1-\left( \dfrac{r_1}{r}\right) ^\frac{p}{p-1}\right] + \left( \dfrac{1}{u(r_1)}\right) ^{\,\frac{q_1+1-p}{p-1}}},\qquad \forall r\in [r_1,\tau _1]. \end{aligned}$$

If $\tau _1\le 2r_1$ we are done, so let us assume $\tau _1>2r_1$. From the preceding inequality,

$$\begin{aligned} u(r)^{\,\frac{q_1+1-p}{p-1}} \le \frac{1}{C\ r^\frac{p}{p-1}\, + \left( \dfrac{1}{u(r_1)}\right) ^{\,\frac{q_1+1-p}{p-1}}},\qquad \text{ for } \text{ all } r\in [2r_1,\tau _1], \end{aligned}$$

where $C=C(p,q_1,N)=\left[ 1-\left( \frac{1}{2}\right) ^{\frac{p}{p-1}}\right] \frac{q_1+1-p}{p}\left( \frac{1}{N}\right) ^{\frac{1}{p-1}}>0.$ Thus,

$$\begin{aligned} u(r) \le C_1\ \left( \frac{1}{ r}\right) ^{\,\frac{p}{q_1+1-p}},\quad \forall r\in [2r_1,\tau _1], \qquad \text{ where } \ C_1 = \left( \frac{1}{C}\right) ^{\,\frac{p-1}{q_1+1-p}}. \end{aligned}$$

(3.18)

Since P(r) is an increasing function on $[0,\tau _1]$ (see (2.7) for $(t_1,t_2)=(0,r)$ with $r\in [0,\tau _1]$), from (2.4) and Lemma 3.1(ii), it follows that

$$\begin{aligned} P(r)= & {} r^N\left[ \frac{p-1}{p} \big | u'(r) \big |^{p}\, + \frac{1}{q_1+1} \big | u(r) \big |^{q_1+1}\right] +\frac{N-p}{p}\, r^{N-1}\big |u'(r)\big |^{p-2}\, u'(r)\, u(r)\nonumber \\\ge & {} P(r_1) \ge C\ d^{\, N\, \left[ 1-\frac{q_1+1}{p^*} \right] },\qquad \text{ for } \text{ all } r\in [r_1,\tau _1]. \end{aligned}$$

(3.19)

Since $u\ge 0$ and $u'\le 0$ on $[0,\tau _1],$

$$\begin{aligned} r^N\left[ \frac{p-1}{p} \big | u'(r) \big |^{p}\, + \frac{1}{q_1+1} \big | u(r) \big |^{q_1+1}\right] \ge C\ d^{\, N\, \left[ 1-\frac{q_1+1}{p^*} \right] },\qquad \forall r\in [r_1,\tau _1]. \end{aligned}$$

(3.20)

If the inequality

$$\begin{aligned} \frac{1}{q_1+1} \big | u(\hat{r}) \big |^{q_1+1}\ge \frac{p-1}{p} \big | u'(\hat{r}) \big |^{p} \end{aligned}$$

(3.21)

holds true for some $\hat{r}\in [2r_1, \tau _1]$, then, from (3.20), there exists a constant $C>0$ such that

$$\begin{aligned} \hat{r}^N\, \big | u(\hat{r}) \big |^{q_1+1}\ge C\ d^{\, N\, \left[ 1-\frac{q_1+1}{p^*} \right] }. \end{aligned}$$

(3.22)

This and (3.18) imply

$$\begin{aligned} \left( \frac{1}{\hat{r}}\right) ^{\frac{p(q_1+1)}{q_1+1-p} -N}\ge C\ d^{\, N\, \left[ 1-\frac{q_1+1}{p^*} \right] }. \end{aligned}$$

(3.23)

Since $q_1$ is subcritical,

$$\begin{aligned} \frac{p(q_1+1)}{q_1+1-p}-N=\frac{Np}{q_1+1-p}\left[ 1-\frac{q_1+1}{p^*}\right] >0. \end{aligned}$$

(3.24)

Therefore, there exists a constant $C_0>0$ such that

$$\begin{aligned} \hat{r}\le C_0\ \left( \frac{1}{d}\right) ^\frac{q_1+1-p}{p}. \end{aligned}$$

(3.25)

Let $\tilde{r}:= C_0\, M\left( \frac{1}{d}\right) ^\frac{q_1+1-p}{p}$ for some constant $M>1$ such that

$$\begin{aligned} M>3C_2/C_0, \end{aligned}$$

(3.26)

where $C_2$ is the constant in (3.9). If $\tau _1 \le \tilde{r}$, then we are done. So, we can assume that $\tau _1 > \tilde{r}.$ From (3.26), $\tilde{r}\ge 2r_1$ (if $\tilde{r}< 2r_1$ then $M<3C_2/C_0$, which is a contradiction). Thus, $\tau _1>2r_1$ and (3.21) is false for all $r\in [\hat{r}, \tau _1]$, and then,

$$\begin{aligned} \frac{1}{q_1+1} \big | u({r}) \big |^{q_1+1}\le \frac{p-1}{p} \big | u'({r}) \big |^{p}\, \qquad \forall r\in \left( \tilde{r}, \tau _1\right) . \end{aligned}$$

(3.27)

This and (3.20) show that there exists $K >0$ such that

$$\begin{aligned} r^N \big | u'(r) \big |^{p}\ge K d^{\, N\, \left[ 1-\frac{q_1+1}{p^*} \right] },\qquad \forall r\in [\tilde{r},\tau _1]. \end{aligned}$$

(3.28)

Since $q_1$ is subcritical, see (3.24), $\frac{p}{q_1+1-p}>\frac{N}{q_1+1}>\frac{N-p}{p}$. Let $M\gg 1$ be a constant such that

$$\begin{aligned} C_1 \left( \frac{1}{M}\right) ^{\,\frac{p}{q_1+1-p}} - \frac{pK^{1/p}}{(N-p)C_2^{(N-p)/p}} \left[ 1-\left( \frac{1}{2}\right) ^{\,\frac{N-p}{p}}\right] \left( \frac{1}{M}\right) ^{\,\frac{N-p}{p}} < 0, \end{aligned}$$

(3.29)

where $C_1$ is as in (3.18), $C_2$ as in (3.25) and K as in (3.28).

Suppose further that $\tau _1 > 2M\tilde{r}$. From (3.28),

$$\begin{aligned} - \,u'(r) \ge K^{1/p}\, d^{\, \frac{N}{p}\ \left[ 1-\frac{q_1+1}{p^*} \right] }\, \left( \frac{1}{r}\right) ^{\frac{N}{p}},\qquad \forall r\in [M\tilde{r},\tau _1]. \end{aligned}$$

Integrating on $[M\tilde{r},r]$ for $r\le \tau _1$, we have

$$\begin{aligned} u(M\tilde{r})- u(r) \ge -\frac{pK^{1/p}}{N-p}\, d^{\, \frac{N}{p}\, \left[ 1-\frac{q_1+1}{p^*} \right] }\, \left[ \left( \frac{1}{r}\right) ^{\frac{N}{p}-1}- \left( \frac{1}{ M\tilde{r} }\right) ^{\frac{N}{p}-1}\right] ,\quad \forall r\in [M\tilde{r},\tau _1]. \end{aligned}$$

Therefore,

$$\begin{aligned} u(r) \le u(M\tilde{r})+ C\, d^{\, \frac{N}{p}\, \left[ 1-\frac{q_1+1}{p^*} \right] }\, \left( \frac{1}{ M\tilde{r} }\right) ^{\frac{N}{p}-1}\, \left[ \left( \frac{M\tilde{r}}{r}\right) ^{\frac{N}{p}-1}- 1\right] ,\quad \forall r\in [M\tilde{r},\tau _1], \end{aligned}$$

(3.30)

where

$$\begin{aligned} C:=\frac{pK^{1/p}}{N-p}. \end{aligned}$$

(3.31)

Moreover, from (3.18)

$$\begin{aligned} d^{\, \frac{N}{p}\, \left[ 1-\frac{q_1+1}{p^*} \right] }\, \left( \frac{1}{ M\tilde{r} }\right) ^{\frac{N}{p}-1}= \left( \frac{1}{C_2 M}\right) ^{\,\frac{N-p}{p}}\ d^{\, \frac{N}{p}\, \left[ 1-\frac{q_1+1}{p^*} \right] +\frac{(q_1+1-p)}{p}\frac{(N-p)}{p}}, \end{aligned}$$

(3.32)

and

$$\begin{aligned} \frac{N}{p}\, \left[ 1-\frac{q_1+1}{p^*} \right] +\frac{(q_1+1-p)}{p}\frac{(N-p)}{p}= 1. \end{aligned}$$

(3.33)

Substituting inequality (3.18) in the first term of (3.30), (3.32) and (3.33) in the second term, by (3.29) and (3.31), we conclude that for all $r \in [2M\tilde{r}, \tau _1]$

$$\begin{aligned} 0 \le u(r)\le & {} C_1\left( \frac{1}{M\tilde{r}}\right) ^{\,\frac{p}{q_1+1-p}} + C\left( \frac{1}{C_2 M}\right) ^{\,\frac{N-p}{p}} \left[ \left( \frac{M\tilde{r}}{r}\right) ^{\,\frac{N-p}{p}}-1\right] \ d \nonumber \\\le & {} \left[ C_1\, \left( \frac{1}{M}\right) ^{\,\frac{p}{q_1+1-p}} - \frac{C}{C_2^{\,\frac{N-p}{p}}} \left[ 1-\left( \frac{1}{2}\right) ^{\,\frac{N-p}{p}}\right] \left( \frac{1}{M}\right) ^{\,\frac{N-p}{p}} \right] \ d<0.\qquad \qquad \end{aligned}$$

(3.34)

This contradiction proves that $\tau _1 \le \max \{2r_1, 2M\tilde{r}\} \le C\left( \frac{1}{d}\right) ^{\frac{q_1+1-p}{p}}.$ Thus, part (i) of the lemma is proven.

• (ii) From (2.5) and (3.19) for $r=\tau _1$, we may write

  $$\begin{aligned} P(\tau _1,d)= \frac{p-1}{p}\tau _1^N \big | u'(\tau _1) \big |^{p}\ge P(r_1,d)\ge C d^{\, N\, \left[ 1-\frac{q_1+1}{p^*} \right] }. \end{aligned}$$

  This and (3.16) imply

  $$\begin{aligned} - u'(\tau _1) \ge C d^{\, \frac{N(q_1+1)}{p}\, \left[ \frac{1}{p}-\frac{1}{p^*} \right] }=C\, d^{\, \frac{q_1+1}{p}}. \end{aligned}$$

On the other hand, from (2.5) and (2.7) for $(t_1,t_2)=(0,\tau _1)$, we may write

$$\begin{aligned} P(\tau _1,d)= & {} \frac{p-1}{p}\tau _1^N \big | u'(\tau _1) \big |^{p} \le \left[ \frac{1}{q_1+1} -\frac{1}{p^*}\right] \, d^{\, q_1+1}\, \tau _1^N, \end{aligned}$$

(3.35)

and therefore,

$$\begin{aligned} \big | u'(\tau _1) \big |^{p}\le & {} \frac{p}{p-1}\left[ \frac{1}{q_1+1} -\frac{1}{p^*}\right] \, d^{\, q_1+1}. \end{aligned}$$

(3.36)

Consequently, $\big | u'(\tau _1) \big | \le C\, d^{\, \frac{q_1+1}{p}}$, with $C:=\left[ \frac{p}{p-1}\left( \frac{1}{q_1+1} -\frac{1}{p^*}\right) \right] ^{\frac{1}{p}}$, ending the proof. $\square$

Remark 3.2

If we restrict our attention to the case $g(s)=\vert s\vert ^{q_1}$ for $s\ge 0$ and $g(s)=-\vert s\vert ^{q_2}$ for $s< 0$, a much simpler proof of Lemma 3.2 can be made using scaling invariance: Indeed, let us denote by $u_d(r)$ the solution of the initial value problem (1.7), so in particular $u_1(r)$ denotes the solution of (1.7) for $d=1$, $r>0$. And let us denote by $\tau _1(d)$ the first zero of $u_d$, so in particular $\tau _1(1)$ denotes the first zero of $u_1.$ One can verify that for any $d > 0$, $u_d(r) = du_1(rd^{(q_1+1-p)/p})$, when $0< r <\tau _1(1)d^{-(q_1+1-p)/p}$. So, it is easy to see that the first zero of $u_d(r)$ satisfies $\tau _1(d) = \tau _1(1)d^{-(q_1+1-p)/p}$. This proves (i) in Lemma 3.2 with $C_1=C_2=\tau _1(1)$. Differentiating the expression for $u_d$ and evaluating at $r=\tau _1(d)$, we get (ii) in Lemma 3.2 with $C_3=C_4=-\,u_1'(\tau _1(1)).$ More in general, if u(r) is a positive solution when $0< r < \tau _1$, then $v(r) = du(rd^{(q_1+1-p)/p})$ is a solution for $d > 0$, $0< r < \tau _1d^{-(q_1+1-p)/p}$ and if u(r) is negative when $\tau _1< r < \tau _2$, then $v(r) = du_1(rd^{(q_2+1-p)/p})$ is a solution for $d > 0$ and $\tau _1d^{-(q_2+1-p)/p}< r < \tau _2 d^{-(q_2+1-p)/p}.$

However, this scale invariance argument cannot be applied to more general nonlinearities as described in Remark 1.1.

Step 2.:

Now, we estimate the first point of minimum $\rho _1(d)$.

In the following lemma, we prove that $\rho _1(d)=O\left( \left( \dfrac{1}{d} \right) ^\frac{q_1+1-p}{p}\right) .$ Notice that, again, this will imply $O(\tau _1(d))=O(\rho _1(d)).$

Lemma 3.3

Let u be a solution to (1.7) with $d > 0$, if $\rho _1=\rho _1(d)$ is as defined by (3.2), then

• (i) There exist two constants $C_1,C_2>0$, independent of d, such that

  $$\begin{aligned} C_1 \bigg (\frac{1}{d}\bigg )^{\frac{q_1+1-p}{p}}\le \rho _1 \le C_2 \bigg (\frac{1}{d}\bigg )^{\frac{q_1+1-p}{p}}\ \ \hbox {for} \ \ d \gg 1. \end{aligned}$$

  (3.37)

• (ii) There exist two constants $C_3,\, C_4>0$, independent of d, such that

  $$\begin{aligned} C_3\, d^{\, \frac{q_1+1}{q_2+1} }\le -u(\rho _1) \le C_4\, d^{\, \frac{q_1+1}{q_2+1} },\ \ \ \hbox {for} \ \ d \gg 1. \end{aligned}$$

  (3.38)

• (iii) There exists a constant $C_5>0$, independent of d, such that

  $$\begin{aligned} \rho _1-\tau _1\le C_5\left( \frac{1}{d}\right) ^{\, (q_1+1)\left[ \,\frac{1}{p}-\frac{1}{q_2+1}\right] },\qquad \text {for}\ d\gg 1. \end{aligned}$$

  (3.39)

Remark 3.3

Observe that, from definition of P, the above lemma implies that there exist two constants $C_6,C_7>0$ such that for $d\gg 1$,

$$\begin{aligned} C_6\, d^{\, N\, \left[ 1-\frac{q_1+1}{p^*} \right] }\le P(\rho _1,d)= \rho _1^N G\big ( u(\rho _1) \big ) = \rho _1^N \frac{\big | u(\rho _1) \big |^{q_2+1}}{q_2+1} \le C_7\, d^{\, N\, \left[ 1-\frac{q_1+1}{p^*} \right] } \end{aligned}$$

(3.40)

Proof of Lemma 3.3

• (i) From the definition of $\rho _1(d)$, we have

  $$\begin{aligned} u(r)\le 0\qquad \text {and}\quad u'(r)\le 0,\qquad \forall r\in [\tau _1(d),\rho _1(d)]. \end{aligned}$$

  (3.41)

  Integrating (2.9) on $[\tau _1,r]$ for any $r\in [\tau _1, \rho _1]$, we may write

  $$\begin{aligned} -r^{N-1}\big |u'(r)\big |^{p-2}\, u' (r) =- \,\tau _1^{\, N-1}\big |u'(\tau _1)\big |^{p-2}\, u' (\tau _1) + \int _{\tau _1}^r s^{N-1}|u(s)|^{\, q_2-1} u(s)\, \mathrm{d}s \end{aligned}$$

  (3.42)

  or, equivalently,

  $$\begin{aligned} r^{N-1}\big |u'(r)\big |^{p-1}= \tau _1^{\, N-1}\big |u'(\tau _1)\big |^{p-1} - \int _{\tau _1}^r s^{N-1}|u(s)|^{\, q_2} \, \mathrm{d}s. \end{aligned}$$

  (3.43)

  Assuming that $\rho _1(d)\ge 3\tau _1(d)$ and

  $$\begin{aligned} \int _{\tau _1}^{2\tau _1} s^{N-1}|u(s)|^{\, q_2} \, \mathrm{d}s > \frac{1}{2}\tau _1^{\, N-1}\big |u'(\tau _1)\big |^{p-1}, \end{aligned}$$

  (3.44)

  we have

  $$\begin{aligned} \int _{2\tau _1}^{3\tau _1} s^{N-1}|u(s)|^{\, q_2} \, \mathrm{d}s\ge & {} (2\tau _1)^{N-1}\int _{2\tau _1}^{3\tau _1} |u(s)|^{\, q_2} \, \mathrm{d}s \nonumber \\\ge & {} (2\tau _1)^{N-1}\tau _1\ |u(2\tau _1)|^{\, q_2} \nonumber \\\ge & {} \int _{\tau _1}^{2\tau _1} s^{N-1}|u(s)|^{\, q_2} \, \mathrm{d}s. \end{aligned}$$

  (3.45)

  Thus,

  $$\begin{aligned} \int _{\tau _1}^{3\tau _1} s^{N-1}|u(s)|^{\, q_2} \, \mathrm{d}s > \tau _1^{\, N-1}\big |u'(\tau _1)\big |^{p-1}. \end{aligned}$$

  This and (3.43) imply

  $$\begin{aligned} - (3\tau _1)^{N-1}\big |u'(3\tau _1)\big |^{p-2}\, u' (3\tau _1) <0, \end{aligned}$$

  which contradicts that $\rho _1(d)\ge 3\tau _1(d)$. Thus, if $\rho _1(d)\ge 3\tau _1(d)$ then

  $$\begin{aligned} \int _{\tau _1}^{2\tau _1} s^{N-1}|u(s)|^{\, q_2} \, \mathrm{d}s \le \frac{1}{2}\tau _1^{\, N-1}\big |u'(\tau _1)\big |^{p-1}. \end{aligned}$$

  (3.46)

  This and (3.43) give

  $$\begin{aligned} -u'(r)\ge \left( \frac{1}{2}\right) ^\frac{1}{p-1} \left( \frac{\tau _1}{r}\right) ^\frac{N-1}{p-1} \big |u'(\tau _1)\big |,\qquad \forall r\in [\tau _1, 2\tau _1]. \end{aligned}$$

  (3.47)

  Integrating on $[\tau _1,r]$, for $r\in [\tau _1, 2\tau _1]$,

  $$\begin{aligned} -u(r)\ge \frac{p-1}{N-p}\left( \frac{1}{2}\right) ^\frac{1}{p-1} \left( \tau _1\right) ^\frac{N-1}{p-1} \big |u'(\tau _1)\big | \left[ \left( \frac{1}{\tau _1}\right) ^\frac{N-p}{p-1}-\left( \frac{1}{r}\right) ^\frac{N-p}{p-1} \right] . \end{aligned}$$

  (3.48)

  From (2.7),

  $$\begin{aligned} P(r,{d})= & {} \frac{N}{q_1+1}\left[ 1 - \frac{q_1+1}{p^*} \right] \, \int _0^{\tau _1}s^{N-1}\, \big |u(s)\big |^{q_1+1}\, \mathrm{d}s \nonumber \\&- \frac{N}{q_2+1}\left[ \frac{q_2+1}{p^*} -1\right] \, \int _{\tau _1}^r s^{N-1}\, \big |u(s)\big |^{q_2+1}\, \mathrm{d}s. \end{aligned}$$

  (3.49)

  Since $q_2 + 1 > p^{*}$,

  $$\begin{aligned} \frac{1}{q_2+1}\ r^N \big | u(r) \big |^{q_2+1} \le P(r,{d})\le P(\tau _1,{d})= \frac{p-1}{p}\tau _1^N \big | u'(\tau _1) \big |^{p},\qquad \forall r\in [\tau _1,\rho _1], \end{aligned}$$

  (3.50)

  see also (3.41). Therefore,

  $$\begin{aligned} \big | u(r) \big | \le C \left( \frac{\tau _1}{r}\right) ^{\frac{N}{q_2+1}} \big | u'(\tau _1) \big |^{\frac{p}{q_2+1}},\qquad \forall r\in [\tau _1,\rho _1]. \end{aligned}$$

  (3.51)

  Integrating (3.47),

  $$\begin{aligned} |u(r)|\ge C \tau _1 \big |u'(\tau _1)\big |,\qquad \forall r\in \left[ \frac{3}{2}\tau _1, 2\tau _1\right] . \end{aligned}$$

  This and (3.51) yield

  $$\begin{aligned} \tau _1 \big |u'(\tau _1)\big |^{1-\frac{p}{q_2+1}}\le C \left( \frac{\tau _1}{r}\right) ^{\frac{N}{q_2+1}},\qquad \forall r\in \left[ \frac{3}{2}\tau _1, 2\tau _1\right] . \end{aligned}$$

  These and $\frac{p}{N}<\frac{q_2+1-p}{q_2+1}$ give

  $$\begin{aligned} \tau _1 \big |u'(\tau _1)\big |^\frac{p}{N}\le \tau _1 \big |u'(\tau _1)\big |^{1-\frac{p}{q_2+1}}\le C \left( \frac{\tau _1}{r}\right) ^{\frac{N}{q_2+1}}\le C,\quad \forall r\in \left[ \frac{3}{2}\tau _1, 2\tau _1\right] . \end{aligned}$$

  (3.52)

  From $P(\tau _1,d)\ge P(r_1,d)$, (3.52) and (3.50), we conclude that $P(r_1,d)$ is bounded. Since this contradicts (3.10), we have

  $$\begin{aligned} \tau _1\le \rho _1\le 3\tau _1,\qquad \text {for }\quad d\gg 1. \end{aligned}$$

• (ii) From (2.4), and (2.8) for $(t_1,t_2)=(\tau _1,\rho _1)$,

  $$\begin{aligned} \rho _1^N \frac{\big | u(\rho _1) \big |^{q_2+1}}{q_2+1}=P(\rho _1,d)\le & {} P(\tau _1,d)=\frac{p-1}{p}\tau _1^N \big | u'(\tau _1) \big |^{p}. \end{aligned}$$

  (3.53)

  This, (3.37), and Lemma 3.2(ii) imply

  $$\begin{aligned} \big | u(\rho _1) \big |\le & {} C\, d^{\, \frac{q_1+1}{q_2+1}}, \end{aligned}$$

  for some constant $C>0$ independent of d. On the other hand,

  $$\begin{aligned} P(\rho _1,d)= & {} \rho _1^N \frac{\big | u(\rho _1) \big |^{q_2+1}}{q_2+1} \nonumber \\\ge & {} P(\tau _1,d)-\left[ \frac{1}{p^*} -\frac{1}{q_2+1}\right] \, \big | u(\rho _1) \big |^{q_2+1}\, \left( \rho _1^N-\tau _1^N \right) . \end{aligned}$$

  (3.54)

  Rearranging terms and dividing by $\tau _1^N$

  $$\begin{aligned} \big | u(\rho _1) \big |^{q_2+1}\left[ \frac{1}{p^*} \left( \frac{\rho _1}{\tau _1}\right) ^N- \left( \frac{1}{p^*}-\frac{1}{q_2+1}\right) \right] \, \ge \frac{p-1}{p}\, \big | u'(\tau _1) \big |^{p}. \end{aligned}$$

  Since (3.16), (3.37), and Lemma 3.2(ii),

  $$\begin{aligned} \big | u(\rho _1) \big | \ge C\, d^{\, \frac{q_1+1}{q_2+1}}, \end{aligned}$$

  for some constant $C>0$ independent of d, ending this part of the proof.

• (iii) Let $r_2=r_2(d)$ be defined by

  $$\begin{aligned} r_2=r_2(d) := \inf \left\{ r>\tau _1;\ u(s) \le \frac{1}{2} u(\rho _1), \ \ \hbox {for all } \ s\in [r,\rho _1]\right\} . \end{aligned}$$

  (3.55)

Let $r\in [r_2,\rho _1]$. Integrating (2.9) on $[r,\rho _1]$,

$$\begin{aligned} \begin{aligned} -\big |u'(r)\big |^{p-2}u'(r) = -\int _r^{\rho _1}\left( \frac{s}{r}\right) ^{N-1} g\big (u(s)\big ) \, \mathrm{d}s \ge \frac{|u(\rho _1)|^{q_2}}{2^{q_2}}\ \left( \rho _1-r\right) , \end{aligned} \end{aligned}$$

(3.56)

and therefore, $u'(r)\le 0$ and

$$\begin{aligned} -u'(r)= \big |u'(r)\big |\, \ge C\, |u(\rho _1)|^{\frac{q_2}{p-1}}\ \left( \rho _1-r\right) ^{\frac{1}{p-1}}. \end{aligned}$$

(3.57)

Integrating on $[r_2,\rho _1]$,

$$\begin{aligned} \frac{|u(\rho _1)|}{2}= u(r_2)-u(\rho _1)\ge C\, |u(\rho _1)|^{\frac{q_2}{p-1}}\ \left( \rho _1-r_2\right) ^{\frac{p}{p-1}}, \end{aligned}$$

and so, from part (ii),

$$\begin{aligned} \rho _1-r_2\le C\, \left( \frac{1}{|u(\rho _1)|}\right) ^{\frac{q_2+1-p}{p}}\ \le C\, \left( \frac{1}{d}\right) ^{\, \frac{q_1+1}{q_2+1}\frac{q_2+1-p}{p}}. \end{aligned}$$

(3.58)

Since u is convex and decreases in $[\tau _1, \rho _1]$, $r_2 - \tau _1 \le (\rho _1 - \tau _1)/2$. This and (3.58) imply

$$\begin{aligned} \rho _1 - \tau _1 \le 2(\rho _1 - r_2) \le C\, \left( \frac{1}{d}\right) ^{\, \frac{q_1+1}{q_2+1}\frac{q_2+1-p}{p}}, \end{aligned}$$

(3.59)

which concludes the proof of Lemma. $\square$

Step 3.:

Estimating the second zero $\tau _2(d)$. Let $r'_2=r'_2(d)$ be defined by

$$\begin{aligned} r'_2=r'_2(d) := \sup \left\{ r>\rho _1;\ u(s) \le \frac{1}{2} u(\rho _1),\ \ \hbox {for all } \ s\in [\rho _1,r]\right\} . \end{aligned}$$

(3.60)

In the following lemma, we first estimate $r'_2=r'_2(d)$.

Lemma 3.4

Let u be a solution to (1.7) with $d \gg 1$. If $r'_2=r'_2(d)$ is as defined by (3.60), then there exists a constant $C_3>0$, independent of d, such that

$$\begin{aligned} r'_2-\rho _1 \le C_3\, \left( \frac{1}{d}\right) ^{\, (q_1+1)\left[ \,\frac{1}{p}-\frac{1}{q_2+1}\right] },\qquad \text {for } d\gg 1. \end{aligned}$$

(3.61)

Proof of Lemma 3.4

Let $r \in [ \rho _1,r'_2].$ Thus,

$$\begin{aligned} \begin{aligned} \big (u'(r)\big )^{p-1} = -\int _{\rho _1}^r \left( \frac{s}{r}\right) ^{N-1} g\big (u(s)\big ) \, \mathrm{d}s \ge \frac{|u(\rho _1)|^{q_2}}{2^{q_2}C_0^{N-1}}\ \left( r-\rho _1\right) . \end{aligned} \end{aligned}$$

(3.62)

Therefore,

$$\begin{aligned} \big (u'(r)\big )\, \ge C\, |u(\rho _1)|^{\frac{q_2}{p-1}}\ \left( r-\rho _1\right) ^{\frac{1}{p-1}}. \end{aligned}$$

Integrating on $[\rho _1,r'_2]$, we have

$$\begin{aligned} \frac{|u(\rho _1)|}{2}= u(r'_2)-u(\rho _1)\ge C\, |u(\rho _1)|^{\frac{q_2}{p-1}}\ \left[ r'_2-\rho _1\right] ^{\frac{p}{p-1}}. \end{aligned}$$

This and (3.38) give

$$\begin{aligned} r'_2-\rho _1\le C\, \left( \frac{1}{|u(\rho _1)|}\right) ^{\frac{q_2+1-p}{p}}\ \le C\, \left( \frac{1}{d}\right) ^{\, \frac{q_1+1}{q_2+1}\frac{q_2+1-p}{p}}, \end{aligned}$$

which proves the Lemma. $\square$

In the following lemma, we estimate the second zero $\tau _2(d)$ and its distance to $r'_2=r'_2(d)$, see (3.60).

Lemma 3.5

Let u be a solution to (1.7) with $d > 0$. Let $\tau _2=\tau _2(d)$ be the second zero defined by (3.3). Then,

• (i) There exists a constant $C_1>0$, independent of d, such that

  $$\begin{aligned} \tau _2-r'_2 \le C_1\, \left( \frac{1}{d}\right) ^{\, (q_1+1)\left[ \,\frac{1}{p}-\frac{1}{q_2+1}\right] },\qquad \text {for } d\gg 1. \end{aligned}$$

  (3.63)

• (ii) There exists a constant $C_2>0$, independent of d, such that

  $$\begin{aligned} P(\tau _2,d) \ge C_2\, d^{\, N\, \left[ 1-\frac{q_1+1}{p^*} \right] },\qquad \text {for } d\gg 1. \end{aligned}$$

  (3.64)

Remark 3.4

Let us observe that by (3.61), (3.63), and Lemma 3.3(i) there exists a constant $C_3>0$, independent of d, such that

$$\begin{aligned} \tau _2 \le C_3 \bigg (\frac{1}{d}\bigg )^{\frac{q_1+1-p}{p}}\ \ \hbox {for} \ \ d \gg 1. \end{aligned}$$

(3.65)

We note that (3.61) and (3.63) imply that there exists a constant $C_4>0$, independent of d, such that

$$\begin{aligned} \tau _2-\rho _1 \le C_4\, \left( \frac{1}{d}\right) ^{\, (q_1+1)\left[ \,\frac{1}{p}-\frac{1}{q_2+1}\right] },\qquad \text {for } d\gg 1. \end{aligned}$$

(3.66)

Also, (3.39) and (3.66) imply that there exists a constant $C_5>0$, independent of d, such that

$$\begin{aligned} \tau _2-\tau _1 \le C_5\, \left( \frac{1}{d}\right) ^{\, (q_1+1)\left[ \,\frac{1}{p}-\frac{1}{q_2+1}\right] },\qquad \text {for } d\gg 1. \end{aligned}$$

(3.67)

Proof of Lemma 3.5

(i).:

We argue as in the proof of part (iii) of Lemma 3.3. By the mean value theorem, (3.38) and (3.61), there exists $z\in (\rho _1, r'_2)$, such that

$$\begin{aligned} u'(z) = \frac{u(r'_2)-u(\rho _1)}{r'_2-\rho _1}=\frac{\dfrac{u(\rho _1)}{2}- u(\rho _1)}{r'_2-\rho _1}=\frac{1}{2}\, \frac{|u(\rho _1)|}{r'_2-\rho _1} \ge C\ d^{\, \frac{q_1+1}{p}}. \end{aligned}$$

(3.68)

Suppose that $u(s)\le 0$ for all $s\in [z,r]$ with $r\in [z,2z]$. Integrating Eq. (2.9) on [z, s] and using that $g(u)<0$ for all $u<0$, we have

$$\begin{aligned} \begin{aligned} s^{N-1}\, \big (u'(s)\big )^{p-1} \ge z^{N-1}\, \big (u'(z)\big )^{p-1}, \end{aligned} \end{aligned}$$

for all $s \in [z,r]$. Hence, by (3.68)

$$\begin{aligned} u'(r)\, \ge \left( \frac{z}{r}\right) ^\frac{N-1}{p-1}\, u'(z)\ge \left( \frac{1}{2}\right) ^\frac{N-1}{p-1}C\ d^{\, \frac{q_1+1}{p}} \ge C_1\ d^{\, \frac{q_1+1}{p}}. \end{aligned}$$

Integrating on [z, r],

$$\begin{aligned} 0\ge u(r) \ge u(z)\,+ C_0 \ d^{\, \frac{q_1+1}{p}}(r-z). \end{aligned}$$

Thus, $-u(z)\ge C_0 \ d^{\, \frac{q_1+1}{p}}(r-z).$ Since $z\in (\rho _1, r'_2), \ -u(z)\in (\vert u(\rho _1)\vert /2, \vert u(\rho _1)\vert ).$ This and (3.38) give

$$\begin{aligned} C_1d^{\frac{q_1 +1}{q_2+1}}\ge \vert u(\rho _1)\vert \ge \vert u(z)\vert \ge C_0 \ d^{\, \frac{q_1+1}{p}}(r-z). \end{aligned}$$

Hence,

$$\begin{aligned} (r-z) \le \frac{C_1}{C_0}d^{\frac{q_1 +1}{q_2+1}- \frac{q_1 + 1}{p}}=C\left( \frac{1}{d}\right) ^{\, (q_1+1)\left[ \,\frac{1}{p}-\frac{1}{q_2+1}\right] },\qquad \text {for } d\gg 1. \end{aligned}$$

(3.69)

By taking $r=2z$ in the previous inequality, and due to $z\in (\rho _1,r'_2)$,

$$\begin{aligned} \rho _1\le C\ \left( \frac{1}{d}\right) ^{\, (q_1+1)\left[ \,\frac{1}{p}-\frac{1}{q_2+1}\right] },\qquad \text {for } d\gg 1. \end{aligned}$$

Since $(q_1+1)\left[ \,\frac{1}{p}-\frac{1}{q_2+1}\right] >\frac{q_1+1-p}{p}$, this contradicts Lemma 3.3, and consequently, $u(2z)>0$, and $\tau _2<2z\le 2r'_2$.

Moreover, taking $r=\tau _2$ in (3.69),

$$\begin{aligned} \tau _2-r'_2<\tau _2-z\le C\ \left( \frac{1}{d}\right) ^{\, (q_1+1)\left[ \,\frac{1}{p}-\frac{1}{q_2+1}\right] },\qquad \text {for } d\gg 1. \end{aligned}$$

This proves part (i) of the lemma.

(ii):

From definition of P, of $\rho _1$, Lemma 3.2–3.3, and (3.67)

$$\begin{aligned} P(\tau _2,d)\ge & {} P(\tau _1,d)-\left[ \frac{1}{p^*} -\frac{1}{q_2+1}\right] \, \big | u(\rho _1) \big |^{q_2+1}\, \left( \tau _2^N-\tau _1^N \right) \nonumber \\\ge & {} P(\tau _1,d)-\left[ \frac{1}{p^*} -\frac{1}{q_2+1}\right] \, \big | u(\rho _1) \big |^{q_2+1}\, \tau _2^{N-1}\left( \tau _2-\tau _1 \right) \nonumber \\\ge & {} C_1\,d^{\, N\, \left[ 1-\frac{q_1+1}{p^*} \right] }-C_2\,d^{\, N\, \left[ 1-\frac{q_1+1}{p^*} \right] -\left( 1-\frac{q_1+1}{q_2+1}\right) }, \end{aligned}$$

(3.70)

where the latter inequality comes from (3.17), (3.38), (3.65), and (3.67). Due to $q_2>q_1$, part (ii) is proven and hence the lemma.

$\square$

Step 4.:

Estimating the first positive point of maximum, $\sigma _1(d)$. In the following lemma, we estimate $\sigma _1(d)$, defined by (3.4), and also $u(\sigma _1)$.

Lemma 3.6

Let u be a solution to (1.7) with $d > 0$. Let $\sigma _1=\sigma _1(d)$ be the first point of maximum strictly positive defined by (3.4). Then,

• (i) There exist two constants $C_1,C_2>0$, independent of d, such that

  $$\begin{aligned} C_1 \bigg (\frac{1}{d}\bigg )^{\frac{q_1+1-p}{p}}\le \sigma _1 \le C_2 \bigg (\frac{1}{d}\bigg )^{\frac{q_1+1-p}{p}}\ \ \hbox {for} \ \ d \gg 1. \end{aligned}$$

  (3.71)

• (ii) There exists a constant $C_3>0$, independent of d, such that

  $$\begin{aligned} u(\sigma _1) \ge C_3\, \, d,\qquad \text {for } d\gg 1. \end{aligned}$$

• (iii) There exists a constant $C_4>0$, independent of d, such that

  $$\begin{aligned} P(\sigma _1,d) \ge P(r_1,d) \ge C_4\,d^{\, N\, \left[ 1-\frac{q_1+1}{p^*} \right] },\qquad \text {for } d\gg 1. \end{aligned}$$

Proof of Lemma 3.6

• (i) By (3.5) and Lemma 3.3(i), $\sigma _1>\rho _1\ge C_1 \big (\frac{1}{d}\big )^{\frac{q_1+1-p}{p}}$. Assuming that $p^{1/p}u'(r) > u'(\tau _2)$ for all $r \in (\tau _2, e \tau _2)$, from (2.2), we have

  $$\begin{aligned} \mathcal {E}(e\tau _2)\le & {} \mathcal {E}(\tau _2) - \int _{\tau _2}^{e\tau _2} \frac{N-1}{p}|u'(\tau _2)|^p \mathrm{d}s \nonumber \\= & {} \frac{p-1}{p} |u'(\tau _2)|^p - \frac{N-1}{p} |u'(\tau _2)|^p \nonumber \\= & {} \frac{p-N}{p} |u'(\tau _2)|^p < 0. \end{aligned}$$

  (3.72)

Since this contradicts that $\mathcal {E}(r) \ge 0$ for all $r \in [0,1]$, we conclude that there exist $\hat{r} \in [\tau _2, e\tau _2]$ such that $u'(\hat{r}) = p^{-1/p} u'(\tau _2)$ and

$$\begin{aligned} u'(s) \ge p^{-1/p} u'(\tau _2)\ \text { for all}\ s \in [\tau _2, \hat{r}]. \end{aligned}$$

(3.73)

Integrating (2.9) on $[\tau _2, \hat{r}]$,

$$\begin{aligned} \begin{aligned} \int _{\tau _2}^{\hat{r}} s^{N-1} u^{q_1}(s) \mathrm{d}s = \tau _2^{\, N-1}\big |u'(\tau _2)\big |^{p-1} -\hat{r}^{N-1}\big |u'(\hat{r})\big |^{p-1}. \end{aligned} \end{aligned}$$

(3.74)

If $\hat{r} \le p^{(p-1)/(2p(N-1))}\tau _2$, then

$$\begin{aligned} \int _{\tau _2}^{\hat{r}} s^{N-1} u^{q_1}(s) \mathrm{d}s\le & {} u^{q_1}(\hat{r})\frac{1}{N}\Big (p^{N(p-1)/(2p(N-1))}\tau _2^{N} -\tau _2^{N}\Big ) \nonumber \\= & {} u^{q_1}(\hat{r})\frac{1}{N}\tau _2^{N}\Big (p^{N(p-1)/(2p(N-1))} -1\Big ). \end{aligned}$$

(3.75)

Now,

$$\begin{aligned} \tau _2^{\, N-1}\big |u'(\tau _2)\big |^{p-1} -\hat{r}^{N-1}\big |u'(\hat{r})\big |^{p-1}\ge & {} \tau _2^{\, N-1}\big |u'(\tau _2)\big |^{p-1}\nonumber \\&-p^{(p-1)/(2p)}\tau _2^{\, N-1}p^{-(p-1)/p}|u'(\tau _2)\big |^{p-1}\nonumber \\\ge & {} \tau _2^{\, N-1}\big |u'(\tau _2)\big |^{p-1}\Big (1-p^{(1-p)/(2p)} \Big ). \end{aligned}$$

(3.76)

From (3.74), (3.75), and (3.76),

$$\begin{aligned} \frac{1}{N} \tau _2 \left( p^{N(p-1)/(2p(N-1))} - 1 \right) u^{q_1}(\hat{r}) \ge \big |u'(\tau _2)\big |^{p-1}\left( 1 - p^{(1-p)/(2p)}\right) . \end{aligned}$$

(3.77)

From (1.7) with $g(u)=u^{q_1}$, it follows that if $r \ge \hat{r}$ is such that $u'(s) > 0$ for all $s\in [\hat{r}, r]$ then u is concave on $[\hat{r}, r]$. We claim that $u''(s) \le -K |u'(\tau _2)|/\tau _2$ where K is a positive constant depending only on p and N. In fact, due to (1.7) and (3.77),

$$\begin{aligned} (p-1)\left( u'(s)\right) ^{p-2}u''(s)<-u^{q_1}(s)\le -u^{q_1}(\hat{r})\le -C(N,p)\left( u'(\tau _2)\right) ^{p-1}/\tau _2 \end{aligned}$$

(3.78)

and since $u''(s)<0$, we obtain $(p-1)\left( u'(s)\right) ^{p-2}u''(s)\ge (p-1)\left( u'(\hat{r})\right) ^{p-2}u''(s)$, provided $p\ge 2$. Combining these last two inequalities and taking into account $u'(\hat{r}) = p^{-1/p} u'(\tau _2)$ the claim follows. If $p<2$, (1.7) gives us

$$\begin{aligned} (p-1)u''(s)\le -u^{q_1}(s) \left( u'(s)\right) ^{2-p}\le -u^{q_1}(\hat{r})\left( u'(s)\right) ^{2-p}. \end{aligned}$$

(3.79)

Combining (3.73), (3.78), and (3.79), the claim is obtained. Thus, integrating $u''(s) \le -K |u'(\tau _2)|/\tau _2$ on $[\hat{r}, r]$, we have

$$\begin{aligned} 0 \le u'(r) \le p^{-1/p} u'(\tau _2) - K \frac{u'(\tau _2)}{\tau _2}(r - \hat{r}). \end{aligned}$$

(3.80)

Assume $u'(r)>0$ for all $r>\hat{r}$. By taking r such that $r-\hat{r}>\frac{2\tau _2}{K}p^{-1/p}$, from (3.80) we have $u'(r)<0$, which is a contradiction. This proves that there exists $\sigma _1>\hat{r}>\tau _2$ such that $u'(\sigma _1) =0$ and $u'(s)>0$ on $[\tau _2, \sigma _1]$. Moreover, from (3.80),

$$\begin{aligned} \sigma _1\le \hat{r}+ \frac{\tau _2}{K}p^{-1/p}\le \left( e+\frac{p^{-1/p}}{K}\right) \tau _2. \end{aligned}$$

On the other hand, if $\hat{r} \ge p^{(p-1)/(2p(N-1))}\tau _2$ and since $u'(s) \ge p^{-1/p} u'(\tau _2)$ for all $s \in [\tau _2, \hat{r}]$, then integrating the last inequality on $[\tau _2, \hat{r}]$,

$$\begin{aligned} u(\hat{r})\ge & {} p^{-1/p}u'(\tau _2)(\hat{r}-\tau _2)\ge p^{-1/p}u'(\tau _2)\tau _2\left( p^{(p-1)/(2p(N-1))}-1\right) \nonumber \\= & {} K_1u'(\tau _2)\tau _2, \end{aligned}$$

(3.81)

where $K_1=p^{-1/p}\left( p^{(p-1)/(2p(N-1))}-1\right) >0.$ If we assume $u'(r)>0$ for all $r>\hat{r}$, because of (2.9), since $\hat{r}\le e\tau _2$ and $u'(\hat{r}) = p^{-1/p} u'(\tau _2)$, we obtain

$$\begin{aligned} r^{N-1}u'(r)^{p-1}= & {} \hat{r}^{N-1}u'(\hat{r})^{p-1}-\int _{\hat{r}}^{r}s^{N-1}u(s)^{q_1}\,\mathrm{d}s\nonumber \\\le & {} \hbox {e}^{N-1}\tau _2^{N-1}p^{-(p-1)/p}u'(\tau _2)^{p-1}-\int _{\hat{r}}^{r}s^{N-1}u(s)^{q_1}\,\mathrm{d}s. \end{aligned}$$

(3.82)

From (3.81), it follows

$$\begin{aligned} \int _{\hat{r}}^{r}s^{N-1}u(s)^{q_1}\,\mathrm{d}s\ge & {} u(\hat{r})^{q_1}\int _{\hat{r}}^{r}s^{N-1}\ge u(\hat{r})^{q_1}\tau _2^{N-1}(r-\hat{r})\nonumber \\\ge & {} K_1^{q_1}u'(\tau _2)^{q_1}\tau _2^{q_1}\tau _2^{N-1}(r-\hat{r}). \end{aligned}$$

(3.83)

From (3.82) and (3.83),

$$\begin{aligned} 0\le r^{N-1}u'(r)^{p-1}\le \tau _2^{N-1}u'(\tau _2)^{p-1}\left[ \hbox {e}^{N-1}p^{-(p-1)/p}- K_1^{q_1}u'(\tau _2)^{q_1-p+1}\tau _2^{q_1}(r-\hat{r}) \right] . \end{aligned}$$

(3.84)

Taking r such that

$$\begin{aligned} r-\hat{r}>\frac{2\hbox {e}^{N-1}p^{-(p-1)/p}}{K_1^{q_1}u'(\tau _2)^{q_1-p+1}\tau _2^{q_1}}, \end{aligned}$$

from (3.84), we have $u'(r)<0$, which is a contradiction. Thus, there exists $\sigma _1>\hat{r}>\tau _2$ such that $u'(\sigma _1) =0$ and $u'(s)>0$ on $[\tau _2, \sigma _1]$. Moreover, from (3.84)

$$\begin{aligned} \sigma _1-\hat{r}\le C(N,p,q_1)\frac{1}{u'(\tau _2)^{q_1-p+1}}\frac{1}{\tau _2^{q_1}}. \end{aligned}$$

(3.85)

We claim that

$$\begin{aligned} u'(\tau _2)\ge Cd^{(q_1+1)/p}. \end{aligned}$$

(3.86)

In fact, due to (2.5) and (3.64), $\frac{p-1}{p}\tau _2^N\vert u'(\tau _2)\vert ^p=P(\tau _2,d)\ge C_2d^{N[1-\frac{q_1+1}{p^*}]}$. Thus, by using (3.65),

$$\begin{aligned} \vert u'(\tau _2)\vert ^p\ge & {} C\frac{d^{N[1-\frac{q_1+1}{p^*}]}}{\tau _2^N}\ge C d^{N[1-\frac{q_1+1}{p^*}]} d^{\frac{N}{p}(q_1+1-p)}\\= & {} Cd^{q_1+1}. \end{aligned}$$

The claim follows. Now, from (3.85), (3.86), and (3.65),

$$\begin{aligned} \sigma _1-\hat{r}\le C\left( \frac{1}{d}\right) ^{\frac{q_1+1}{p}(q_1+1-p)}\left( \frac{1}{d}\right) ^{\frac{-q_1}{p}(q_1+1-p)}, \end{aligned}$$

and then,

$$\begin{aligned} \sigma _1\le \hat{r}+C\left( \frac{1}{d}\right) ^{\frac{q_1+1-p}{p}}. \end{aligned}$$

Since $\hat{r}\le e\tau _2$ and (3.65),

$$\begin{aligned} \sigma _1\le C\left( \frac{1}{d}\right) ^{\frac{q_1+1-p}{p}}. \end{aligned}$$

This proves part (i) of the lemma.

• (ii) From the definition of P, see (2.6) and (2.7) for $(t_1,t_2)=(\tau _2,\sigma _1)$, and (3.64), we have

  $$\begin{aligned} \begin{aligned} \frac{\sigma _1^N\vert u(\sigma _1)\vert ^{q_1+1}}{q_1+1}=P(\sigma _1,d) >P(\tau _2,d) \ge C\, d^{\, N\, \left[ 1-\frac{q_1+1}{p^*} \right] },\qquad \text {for } d\gg 1. \end{aligned} \end{aligned}$$

  Hence,

  $$\begin{aligned} u(\sigma _1)\ge C\, d^{\, N\, \left[ \frac{1}{q_1+1}-\frac{1}{p^*} \right] } \left( \frac{1}{\sigma _1}\right) ^{\frac{N}{q_1+1}}\ge C\, d. \end{aligned}$$

• (iii) The definition of P and the nonlinearity g give us

  $$\begin{aligned} P(\sigma _1,d)= & {} P(r_1,d)+N\left( \frac{1}{q_1+1}-\frac{1}{p^*}\right) \int _{r_1}^{\tau _1}s^{N-1} \big ( u(s)\big )^{q_1+1}\mathrm{d}s\nonumber \\&-N\left( \frac{1}{p^*}-\frac{1}{q_2+1}\right) \int _{\tau _1}^{\tau _2}s^{N-1} \vert u(s)\vert ^{q_2+1}\mathrm{d}s\nonumber \\&+N\left( \frac{1}{q_1+1}-\frac{1}{p^*}\right) \int _{\tau _2}^{\sigma _1}s^{N-1} \big ( u(s)\big )^{q_1+1}\mathrm{d}s. \end{aligned}$$

  (3.87)

Let $r'_1=r'_1(d) := \sup \{r>0;\ u(s) \ge d/4 \ \hbox {on} \ [0,r]\}.$ Since ${\mathcal {E}}$ is a nonincreasing function, see (2.2), $|u'(s)| \le \frac{p}{(p-1)(q_1+1)}d^{(q_1+1)/p}$ for all $s \in [r_1, r'_1]$. Integrating $-u'$ in the latter interval, we have $d \le 4 \frac{p}{(p-1)(q_1+1)}d^{(q_1+1)/p}(r'_1 - r_1)$, or $r'_1 - r_1 \ge Cd^{(p - q_1-1)/p}$. Therefore,

$$\begin{aligned} \int _{r_1}^{\tau _1}s^{N-1} \big ( u(s)\big )^{q_1+1}\mathrm{d}s\ge & {} \int _{r_1}^{r_1^\prime }s^{N-1} \big ( u(s)\big )^{q_1+1}\mathrm{d}s \nonumber \\\ge & {} \frac{1}{N}\left( \frac{d}{4}\right) ^{q_1+1}r_1^{N-1}\left[ r_1^\prime -r_1\right] \nonumber \\\ge & {} \hat{C} d^{q_1+1-\frac{q_1+1-p}{p}\left[ (N-1)+1\right] }\equiv \hat{C}d^\xi , \end{aligned}$$

(3.88)

where $\xi :=N\left[ 1-\frac{q_1+1}{p^*}\right]$.

On the other hand, from (3.38), (3.67), and Lemma 3.5 we have

$$\begin{aligned} \begin{aligned} \int _{\tau _1}^{\tau _2}s^{N-1} \vert u(s)\vert ^{q_2+1}\mathrm{d}s&\le \bar{C} d^{q_1+1-\frac{q_1+1-p}{p}(N-1)-(q_1+1)\left[ \,\frac{1}{p}-\frac{1}{q_2+1}\right] } \equiv \bar{C} d^\eta , \end{aligned} \end{aligned}$$

(3.89)

where $\eta :=N\left[ 1-\frac{q_1+1}{p^*}\right] -\left( 1-\frac{q_1+1}{q_2+1}\right) .$ Since $0<\eta <\xi$, for d large, the first integral in (3.87) is larger than the second. Since, in addition, the third integral in (3.87) is positive we have $P(\sigma _1,d) \ge P(r_1,d)$, which proves the lemma. $\square$

4 Proof of Theorem 1.1

Proof

Let $d_2>0$ be such that if $d>d_2$ then

$$\begin{aligned} \left( \frac{1}{q_1+1}-\frac{1}{p^*}\right) \hat{C} d^\xi -\left( \frac{1}{p^*}-\frac{1}{q_2+1}\right) \bar{C}d^\eta > 0, \end{aligned}$$

(4.1)

where $\hat{C}, \xi , \bar{C}, \eta$ are as in (3.88) and (3.89). Repeating the arguments developed in Sect. 3, we see that if $d > d_2$, then there exists $\sigma _2(d) := \sigma _2$ such that u has a local maximum at $\sigma _2>\sigma _1$, there exists $\rho _2\in (\sigma _{1},\sigma _2)$ such that $u'<0$ on $(\sigma _{1},\rho _2)$, $u'>0$ in $(\rho _2,\sigma _2)$, and there exist $\tau _{3} \in (\sigma _{1},\rho _2)$ and $\tau _{4} \in (\rho _2, \sigma _2)$ with $u(\tau _{3}) =u(\tau _{4}) =0, \sigma _2 =O(\sigma _1)$ and for $r \in [\sigma _{1}, \tau _{3}]$

$$\begin{aligned} \begin{aligned} P(r,d)&=P(\sigma _{1},d)+N\left( \frac{1}{q_1+1}-\frac{1}{p^*}\right) \int _{\sigma _{1}}^{r}s^{N-1} \big ( u(s)\big )^{q_1+1}\mathrm{d}s\\&\ge P(\sigma _{1},d). \end{aligned} \end{aligned}$$

(4.2)

Also, for $r \in [\tau _{3}, \sigma _2]$

$$\begin{aligned} P(r,d)\ge & {} P(\sigma _{1},d)+N\left( \frac{1}{q_1+1}-\frac{1}{p^*}\right) \int _{\sigma _{1}}^{\tau _{3}}s^{N-1} \big ( u(s)\big )^{q_1+1}\mathrm{d}s\nonumber \\&-N\left( \frac{1}{p^*}-\frac{1}{q_2+1}\right) \int _{\tau _{3}}^{\tau _{4}}s^{N-1} \vert u(s)\vert ^{q_2+1}\mathrm{d}s\nonumber \\\ge & {} P(\sigma _{1},d)+ N\left[ \left( \frac{1}{q_1+1}-\frac{1}{p^*}\right) \hat{C} d^\xi -\left( \frac{1}{p^*}-\frac{1}{q_2+1}\right) \bar{C}d^\eta \right] \nonumber \\\ge & {} P(\sigma _{1},d). \end{aligned}$$

(4.3)

Hence, for all $r \in [\sigma _1, \sigma _2]$,

$$\begin{aligned} P(r,d) \ge P(\sigma _1, d) \rightarrow \infty \qquad \text {as}\ \ d\rightarrow \infty . \end{aligned}$$

(4.4)

Iterating this argument, we find $\sigma _3(d):=\sigma _3< \cdots< \sigma _k(d):=\sigma _k \le 1 < \sigma _{k+1}$ such that for each $i=2, \ldots , k+1, u$ has a maximum at $\sigma _i$ (recall $\sigma _0=0$), there exists $\rho _i \in (\sigma _{i-1}, \sigma _i)$ with $u'<0$ on $(\sigma _{i-1},\rho _{i})$, $u'>0$ in $(\rho _{i},\sigma _i)$, and there exist $\tau _{2i-1} \in (\sigma _{i-1},\rho _i)$ and $\tau _{2i} \in (\rho _i, \sigma _i)$ with $u(\tau _{2i-1}) =u(\tau _{2i}) =0, \sigma _i=O(\sigma _{i-1})$ (see (3.7)) and

$$\begin{aligned} P(r,d) \ge P(\sigma _1, d) \rightarrow \infty \qquad \text {as}\ \ d\rightarrow \infty \end{aligned}$$

(4.5)

for all $r \in [\sigma _1, 1]$, see Lemma 3.6(iii). For $r\in [0, \sigma _1],\ \mathcal {E}(r)\ge \mathcal {E}(\sigma _1)=G(u(\sigma _1))\ge Cd^{q_1+1}.$ From this inequality, (4.5), and (2.4), we may assume that $\mathcal {E}(r,d) \ge 1$ for all $r \in [0, 1]$. Hence, $(u(r,d),u'(r,d))\ne (0,0)$ for all $r\in (0,1]$ which implies that there exists a differentiable function $\phi (r,d)$ such that $\phi (0, d) =0$,

$$\begin{aligned} \begin{aligned} u(r,d)&=\sqrt{\big (u(r,d)\big )^2+\big (u'(r,d)\big )^2}\, \, \cos \phi (r,d),\\ u'(r,d)&= - \sqrt{\big (u(r,d))^2+\big (u'(r,d)\big )^2}\, \, \sin \phi (r,d). \end{aligned} \end{aligned}$$

(4.6)

Suppose now that $u(T,d) =0$, i.e., $\phi (T,d) = j\pi + \pi /2$ for some nonnegative integer j. Differentiating the first equation in (4.6) with respect to r, we have

$$\begin{aligned} \begin{aligned} u'(T,d)&= -\sqrt{\big (u(T,d))^2+\big (u'(T,d)\big )^2}\sin (\phi (T,d)) \phi '(T,d). \end{aligned} \end{aligned}$$

(4.7)

Because $u(T,d) =0$, $u'(T,d)\not = 0$. Thus, the second equation in (4.6) and (4.7) give $\phi '(T,d) =1$. Hence, $\phi (t,d)> \phi (T,d)$ for all $t \in (T,1]$. In fact, assuming to the contrary that $\phi (T_1, d) =\phi (T,d)$ for some $T_1 \in (T, 1]$, then by the continuity of $\phi (\cdot , d)$ we may assume $\phi (t,d) >\phi (T,d)$ for all $t\in (T, T_1)$ (simply take $T_1 = \sup \{t \in (T, 1]; \phi (t,d) > \phi (T,d)\}$). Since $\phi (t,d) > \phi (T_1,d)$ for $t \in (T, T_1)$, $\phi '(T_1, d) \le 0$ which contradicts that $\phi '(T_1, d) = 1$ since $\phi (T_1,d) = \phi (T,d) = j\pi + \pi /2$.

Since u decreases in $[0, \rho _1]$ and $u'(\rho _1) =0$, $\phi (\rho _1) = \pi$. Similarly, $\phi (\sigma _1) = 2\pi$. Iterating this, we see that $\phi (\sigma _k) = 2k\pi$. By the definition of $\tau _{2k}$, $\phi (\tau _{2k},d)=2k\pi - \pi /2.$ Hence, $\lim _{d \rightarrow +\infty }\phi (1,d)=+\infty .$ This and the intermediate value theorem imply the existence of a sequence $\big \{\delta _j\big \}\rightarrow +\infty$ such that $\ \ \phi (1,\delta _j)=j\pi +\frac{\pi }{2}$, in other words $u(1,\delta _j)=0.$ This proves the theorem. $\square$

Appendix: On a more general nonlinearity

In this appendix, we indicate the refinements needed to extend our results to a more general nonlinearity g satisfying hypothesis (1.4) of Remark 1.1(a).

Note that under hypothesis (1.4), for every $\varepsilon >0$, there exist two constants $M_\varepsilon>0,\ M'_\varepsilon >0$ such that

$$\begin{aligned} (L_1-\varepsilon ) s^{q_1}-M_\varepsilon\le & {} g(s)\le (L_1+\varepsilon ) s^{q_1}+M_\varepsilon ,\qquad \forall s>0, \end{aligned}$$

(5.1)

$$\begin{aligned} (L_2-\varepsilon ) |s|^{q_2}-M_\varepsilon\le & {} -g(s)=|g(s)|\le (L_2+\varepsilon ) |s|^{q_2}+M_\varepsilon , \qquad \forall s<0, \end{aligned}$$

(5.2)

and

$$\begin{aligned} \frac{(L_1-\varepsilon )}{q_1+1} s^{q_1+1}-M'_\varepsilon\le & {} G(s)\le \frac{(L_1+\varepsilon )}{q_1+1} s^{q_1+1}+M'_\varepsilon ,\qquad \forall s>0, \end{aligned}$$

(5.3)

$$\begin{aligned} \frac{(L_2-\varepsilon )}{q_1+1} |s|^{q_2+1}-M'_\varepsilon\le & {} G(s)\le \frac{(L_2+\varepsilon )}{q_2+1} |s|^{q_2+1}+M'_\varepsilon , \qquad \forall s<0, \end{aligned}$$

(5.4)

where $G(s):=\int _0^s\, g(t)\, dt$.

Moreover, for all $s>0$,

$$\begin{aligned} \left( \frac{p^*(L_1-\varepsilon )}{q_1+1}-(L_1+\varepsilon )\right) s^{q_1+1}-M_\varepsilon ''\le & {} p^* G(s) - s g(s)\nonumber \\\le & {} \left( \frac{p^*(L_1+\varepsilon )}{q_1+1}-(L_1-\varepsilon )\right) s^{q_1+1}+M_\varepsilon '',\nonumber \\ \end{aligned}$$

(5.5)

and for all $s<0$,

$$\begin{aligned} -\left( (L_2+\varepsilon )-\frac{p^*(L_2-\varepsilon )}{q_2+1}\right) |s|^{q_2+1}-M_\varepsilon ''\le & {} p^* G(s) - s g(s)\nonumber \\\le & {} - \left( (L_2-\varepsilon )-\frac{p^*(L_2+\varepsilon )}{q_2+1}\right) |s|^{q_2+1}+M_\varepsilon ''.\nonumber \\ \end{aligned}$$

(5.6)

Hence, there exist constants $C_i>0$, $i=1,\ldots ,4$ and $M>0$ such that

$$\begin{aligned} C_1 s^{q_1+1}-M\le & {} p^* G(s) - s g(s)\le C_2 s^{q_1+1}+M,\qquad \forall s>0, \end{aligned}$$

(5.7)

$$\begin{aligned} -C_3 |s|^{q_2+1}-M\le & {} p^* G(s) - s g(s)\le -C_4 |s|^{q_2+1}+M,\qquad \forall s<0. \end{aligned}$$

(5.8)

$$\begin{aligned} C_5\, |s|^{q_1+1}-M\le & {} G(s)\le C_6\, |s|^{q_1+1}+M,\qquad \forall s\in {\mathbb {R}}, \end{aligned}$$

(5.9)

and

$$\begin{aligned} \begin{aligned} C_7\, s^{q_1}-M&\le g(s)\le C_8\, s^{q_1}+M,\qquad \forall s>0,\\ C_9\, |s|^{q_2}-M&\le -g(s)=|g(s)|\le C_{10}\, |s|^{q_2}+M, \qquad \forall s<0. \end{aligned} \end{aligned}$$

(5.10)

Furthermore, Eqs. (2.7) and (2.8) should both be substituted by

$$\begin{aligned} P(t_2,d)-P(t_1,d)= \frac{N-p}{p}\,\int _{t_1}^{t_2} s^{N-1}\,\Big [ p^* G \big (u(s)\big ) - g \big (u(s)\big ) \, u(s) \Big ]\, \mathrm{d}s, \end{aligned}$$

(5.11)

and Eqs. (2.11) and (2.12) should be substituted by

$$\begin{aligned} t_2^{N-1} \big |u'(t_2)\big |^{p-2}\, u'(t_2)-t_1^{N-1} \big |u'(t_1)\big |^{p-2}\, u'(t_1)= & {} -\int _{t_1}^{t_2} s^{N-1} g \big (u(s)\big ) \, \mathrm{d}s<0, \end{aligned}$$

(5.12)

$$\begin{aligned} t_2^{N-1} \big |u'(t_2)\big |^{p-2}\, u'(t_2)-t_1^{N-1} \big |u'(t_1)\big |^{p-2}\, u'(t_1)= & {} \int _{t_1}^{t_2} s^{N-1} \Big |g \big (u(s)\big )\Big | \, \mathrm{d}s>0,\qquad \end{aligned}$$

(5.13)

respectively.

Remark 5.1

The proof of Lemma 3.1 should be modified in the following way.

Let $u(0) = d\gg 1$. Taking $(t_1,t_2)=(0,r)$ in (2.11), Eq. (3.11) in part (i) should be substituted by

$$\begin{aligned} \begin{aligned} -r^{N-1} \big |u'(r)\big |^{p-2}\, u'(r)&= \int _0^rs^{N-1} g \big (u(s)\big ) \, \mathrm{d}s\ge (C\, d^{\, q_1}-M)\ r^N \\&\ge \frac{C}{2}\, d^{\, q_1}\, r^N,\quad \text {for } r\in [0,r_1], \end{aligned} \end{aligned}$$

(5.14)

and Eq. (3.15) in part (ii) should be substituted by

$$\begin{aligned} P(r_1,d) \ge Cr_1^N (d^{\, q_1+1} -M) \ge \frac{C}{2} d^{\, N\, \left[ 1-\frac{q_1+1}{p^*} \right] } \rightarrow +\infty ,\qquad \text{ as } d\rightarrow +\infty , \end{aligned}$$

(5.15)

respectively. All the others arguments remain unchanged, except possibly the constants involved.

Remark 5.2

We now write the proof of Lemma 3.2 for a generalized nonlinearity.

• (i) Since $\tau _1>r_1$, by Lemma 3.1(i), $\tau _1\ge C \ \left( \frac{1}{d}\right) ^{\, \frac{q_1+1-p}{p}}.$

From (5.12) and (5.10), for any $r\in [0,\tau _1]$,

$$\begin{aligned} r^{N-1}\big |u'(r)\big |^{p-1}\ge \left( C_7 u(r)^{\, q_1}-M\right) \, \frac{r^{N}}{N}. \end{aligned}$$

Thanks to (5.1) and (5.7), there exist positive constants $C_7, C_1, M$ such that $g(u(r))\ge C_7 u(r)^{\, q_1}-M$ and $p^*G(u(r))-u(r)g(u(r))\ge C_1 \big (u(s)\big )^{q_1+1} - M$, $\forall r\in [0, \tau _1]$. Let us define $r^*$ such that

$$\begin{aligned} r^*=r^*(d) := \sup \left\{ r>0;\ u(s) \ge \max \left\{ \left( \frac{M}{C_7}\right) ^{\frac{1}{q_1}}, \left( \frac{M}{C_1}\right) ^{\frac{1}{q_1+1}} \right\} , \ \forall s\in [0,r]\right\} .\nonumber \\ \end{aligned}$$

(5.16)

By definition, $r^*\le \tau _1.$ Moreover, we observe that $g(u(r))\ge 0$ and

$$\begin{aligned} p^*G(u(r))-u(r)g(u(r))\ge 0,\ \forall r\in [0, r^*). \end{aligned}$$

We claim that $r^* \le C\left( \frac{1}{d}\right) ^{\frac{q_1+1-p}{p}}$ for some $C>0$: since

$$\begin{aligned} x^q-a\ge \big (x-a^{1/q}\big )^q,\qquad \text {for all}\ \ x\ge a^{1/q},\ \ q>1,\ \ a>0, \end{aligned}$$

we have

$$\begin{aligned} \big |u'(r)\big |^{p-1}\ge C_7\left( u(r)- \left( \frac{M}{C_7}\right) ^{\frac{1}{q_1}}\right) ^{q_1}\, \frac{r}{N},\qquad \forall r\in [0,r^*). \end{aligned}$$

Hence,

$$\begin{aligned} -u'(r)\ge \left( \frac{C_7}{N}\right) ^{\frac{1}{p-1}}\, \left( u(r)- \left( \frac{M}{C_7}\right) ^{\frac{1}{q_1}}\right) ^{\, \frac{q_1}{p-1}}\ r^\frac{1}{p-1},\qquad \forall r\in [0,r^*). \end{aligned}$$

Separating variables and integrating on [0, r] we have

$$\begin{aligned} -\frac{\left( u(r)- \left( \frac{M}{C_7}\right) ^{\frac{1}{q_1}}\right) ^{\,1- \frac{q_1}{p-1}}- \left( d- \left( \frac{M}{C_7}\right) ^{\frac{1}{q_1}}\right) ^{\,1- \frac{q_1}{p-1}}}{\frac{p-1-q_1}{p-1}}\ge \left( \frac{C_7}{N}\right) ^{\frac{1}{p-1}}\frac{r^\frac{p}{p-1}}{\frac{p}{p-1}}, \end{aligned}$$

or equivalently, for all $r\in [0,r^*)$

$$\begin{aligned} \left( u(r)- \left( \frac{M}{C_7}\right) ^{\frac{1}{q_1}}\right) ^{\,1- \frac{q_1}{p-1}}- \left( d- \left( \frac{M}{C_7}\right) ^{\frac{1}{q_1}}\right) ^{\,1- \frac{q_1}{p-1}}\ge C\ r^\frac{p}{p-1}, \end{aligned}$$

where $C=C(p,q_1,N)=\frac{q_1+1-p}{p}\left( \frac{C_7}{N}\right) ^{\frac{1}{p-1}}>0.$ This and (1.3) yield

$$\begin{aligned} \left( u(r)- \left( \frac{M}{C_7}\right) ^{\frac{1}{q_1}}\right) ^{\,-\frac{q_1+1-p}{p-1}} \ge C\ r^\frac{p}{p-1},\quad \forall r\in [0,r^*). \end{aligned}$$

Thus,

$$\begin{aligned} u(r)-\left( \frac{M}{C_7}\right) ^{\frac{1}{q_1}} \le C_1\ \left( \frac{1}{ r}\right) ^{\,\frac{p}{q_1+1-p}},\quad \forall r\in (0,r^*), \quad \text{ where }\ C_1 = \left( \frac{1}{C}\right) ^{\,\frac{p-1}{q_1+1-p}}. \end{aligned}$$

(5.17)

Let $r \in [r_1, r^*)$. Replacing $(t_1,t_2)$ by $(r_1,r)$ in (5.11), from (5.7), (5.16), and (5.15),

$$\begin{aligned} \begin{aligned} P(r)&= r^N\left[ \frac{p-1}{p} \big | u'(r) \big |^{p}\, + G\big ( u(r) \big )\right] +\frac{N-p}{p}\, r^{N-1}\big |u'(r)\big |^{p-2}\, u'(r)\, u(r)\\&\ge P(r_1) +\frac{N-p}{p}\int _{r_1}^{r} s^{N-1}\,\Big [ \tilde{C}_1 \big (u(s)\big )^{q_1+1} - M \Big ]\, \mathrm{d}s \ge P(r_1)\ge C\ d^{\, N\, \left[ 1-\frac{q_1+1}{p^*} \right] }. \end{aligned} \end{aligned}$$

(5.18)

Since $u\ge 0$ and $u'\le 0$ on $[0,r^*),$

$$\begin{aligned} r^N\left[ \frac{p-1}{p} \big | u'(r) \big |^{p}\, + G\big ( u(r) \big )\right] \ge C\ d^{\, N\, \left[ 1-\frac{q_1+1}{p^*} \right] },\qquad \forall r\in [r_1,r^*). \end{aligned}$$

(5.19)

If the inequality

$$\begin{aligned} G\big ( u(\hat{r}) \big ) \ge \frac{p-1}{p} \big | u'(\hat{r}) \big |^{p}, \end{aligned}$$

(5.20)

holds true for some $\hat{r}\in [r_1, r^*)$, then, from (5.9), and (5.19) we have

$$\begin{aligned} \hat{r}^N\, \Big [C_6\big | u(\hat{r}) \big |^{q_1+1}+M\Big ]\ge \hat{r}^N\, G\big ( u(\hat{r}) \big )\ge \frac{C}{2}\ d^{\, N\, \left[ 1-\frac{q_1+1}{p^*} \right] }. \end{aligned}$$

This, and (5.17) imply

$$\begin{aligned} C_6 \hat{r}^N \left[ C_1\left( \frac{1}{ \hat{r}}\right) ^{ \frac{p}{q_1+1-p}}+\left( \frac{M}{C_7}\right) ^{\frac{1}{q_1}} \right] ^{q_1+1} \ge \frac{C}{2} d^{N\left[ 1-\frac{q_1+1}{p^*} \right] }-M \hat{r}^N\ge \frac{C}{4} d^{N\left[ 1-\frac{q_1+1}{p^*} \right] }, \end{aligned}$$

(5.21)

and hence,

$$\begin{aligned} C_6^\frac{1}{q_1+1} \hat{r}^\frac{N}{q_1+1} \left[ C_1 \left( \frac{1}{ \hat{r}}\right) ^{ \frac{p}{q_1+1-p}}+\left( \frac{M}{C_7}\right) ^{\frac{1}{q_1}} \right] \ge \left( \frac{C}{4}\right) ^\frac{1}{q_1+1} d^{\frac{N}{q_1+1}\left[ 1-\frac{q_1+1}{p^*} \right] }, \end{aligned}$$

(5.22)

and thus,

$$\begin{aligned} \left( \frac{1}{\hat{r}}\right) ^{\frac{p}{q_1+1-p} -\frac{N}{q_1+1}}\ge C\ d^{\frac{N}{q_1+1}\left[ 1-\frac{q_1+1}{p^*} \right] }. \end{aligned}$$

(5.23)

Since $q_1$ is subcritical,

$$\begin{aligned} \frac{p}{q_1+1-p} -\frac{N}{q_1+1}=\frac{1}{q_1+1}\,\frac{Np}{q_1+1-p}\left[ 1-\frac{q_1+1}{p^*}\right] >0. \end{aligned}$$

(5.24)

Therefore, there exists a constant $C_4>0$ such that

$$\begin{aligned} \hat{r}\le C_4\ \left( \frac{1}{d}\right) ^\frac{q_1+1-p}{p}, \quad \text{ where }\ C_4 = \left( \frac{1}{C}\right) ^{\frac{1}{\frac{p}{q_1+1-p} -\frac{N}{q_1+1}}}. \end{aligned}$$

(5.25)

Let

$$\begin{aligned} \tilde{r}:= C_5\, \left( \frac{1}{d}\right) ^\frac{q_1+1-p}{p}, \quad \text {where}\ C_5:=\max \{C_4, 3C_2\}, \end{aligned}$$

(5.26)

and where $C_2$ is the constant in (3.9). If $r^* \le \tilde{r}$, then we are done. So, we can assume that $r^* > \tilde{r}.$ From (3.9), $\tilde{r}>3C_2\,\left( \frac{1}{d}\right) ^\frac{q_1+1-p}{p}\ge 2r_1$. Thus, $r^*>2r_1$ and (5.20) is false for all $r\in [ \tilde{r},r^*)$, then

$$\begin{aligned} G\big ( u(r) \big )\le \frac{p-1}{p} \big | u'({r}) \big |^{p}\, \qquad \forall r\in \left( \tilde{r}, r^*\right) . \end{aligned}$$

(5.27)

This and (5.19) show that there exists $K >0$ such that

$$\begin{aligned} r^N \big | u'(r) \big |^{p}\ge K d^{\, N\, \left[ 1-\frac{q_1+1}{p^*} \right] },\qquad \forall r\in [\tilde{r}, r^*). \end{aligned}$$

(5.28)

Since $q_1$ is subcritical, see (3.24), $\frac{p}{q_1+1-p}>\frac{N}{q_1+1}>\frac{N-p}{p}$. Let $\tilde{M}\gg 1$ be a constant such that

$$\begin{aligned} C_1 \left( \frac{1}{\tilde{M}}\right) ^{\,\frac{p}{q_1+1-p}}\left( \frac{1}{C_5}\right) ^{\,\frac{p}{q_1+1-p}} - \frac{pK^{1/p}}{(N-p)C_5^{(N-p)/p}} \left[ 1-\left( \frac{1}{2}\right) ^{\,\frac{N-p}{p}}\right] \left( \frac{1}{\tilde{M}}\right) ^{\,\frac{N-p}{p}} < 0, \end{aligned}$$

(5.29)

where $C_1$ is as in (5.17), $C_5$ as in (5.26), and K as in (5.28).

Suppose further that $r^*> 2\tilde{M}\tilde{r}$. From (5.28)

$$\begin{aligned} - u'(r) \ge K^{1/p}\, d^{\, \frac{N}{p}\ \left[ 1-\frac{q_1+1}{p^*} \right] }\, \left( \frac{1}{r}\right) ^{\frac{N}{p}},\qquad \forall r\in [\tilde{M}\tilde{r},r^*). \end{aligned}$$

Integrating on $[\tilde{M}\tilde{r},r]$  for $r< r^*$, we have

$$\begin{aligned} u(\tilde{M}\tilde{r})- u(r) \ge -\frac{pK^{1/p}}{N-p}\, d^{\, \frac{N}{p}\, \left[ 1-\frac{q_1+1}{p^*} \right] }\, \left[ \left( \frac{1}{r}\right) ^{\frac{N}{p}-1}- \left( \frac{1}{ \tilde{M}\tilde{r} }\right) ^{\frac{N}{p}-1}\right] ,\quad \forall r\in [\tilde{M}\tilde{r},r^*). \end{aligned}$$

Therefore,

$$\begin{aligned} u(r) \le u(\tilde{M}\tilde{r})+ C\, d^{\, \frac{N}{p}\, \left[ 1-\frac{q_1+1}{p^*} \right] }\, \left( \frac{1}{ \tilde{M}\tilde{r} }\right) ^{\frac{N}{p}-1}\, \left[ \left( \frac{\tilde{M}\tilde{r}}{r}\right) ^{\frac{N}{p}-1}- 1\right] ,\quad \forall r\in [\tilde{M}\tilde{r},r^*), \end{aligned}$$

(5.30)

where

$$\begin{aligned} C:=\frac{pK^{1/p}}{N-p}. \end{aligned}$$

(5.31)

Moreover, from (5.26)

$$\begin{aligned} d^{\, \frac{N}{p}\, \left[ 1-\frac{q_1+1}{p^*} \right] }\, \left( \frac{1}{ \tilde{M}\tilde{r} }\right) ^{\frac{N}{p}-1}= \left( \frac{1}{C_5 \tilde{M}}\right) ^{\,\frac{N-p}{p}}\ d^{\, \frac{N}{p}\, \left[ 1-\frac{q_1+1}{p^*} \right] +\frac{(q_1+1-p)}{p}\frac{(N-p)}{p}}, \end{aligned}$$

(5.32)

and

$$\begin{aligned} \frac{N}{p}\, \left[ 1-\frac{q_1+1}{p^*} \right] +\frac{(q_1+1-p)}{p}\frac{(N-p)}{p}= 1. \end{aligned}$$

(5.33)

Substituting inequality (5.17) in the first term of (5.30), (5.32) and (5.33) in the second term, by (5.29) and (5.31), we conclude that for all $r \in [2\tilde{M}\tilde{r}, r^*)$

$$\begin{aligned} \begin{aligned} 0&\le u(r) \le \left( \frac{M}{C_7}\right) ^{\frac{1}{q_1}} +C_1\left( \frac{1}{\tilde{M}\tilde{r}}\right) ^{\,\frac{p}{q_1+1-p}} + C\left( \frac{1}{C_5 \tilde{M}}\right) ^{\,\frac{N-p}{p}} \left[ \left( \frac{\tilde{M}\tilde{r}}{r}\right) ^{\,\frac{N-p}{p}}-1\right] d \\&\le \left( \frac{M}{C_7}\right) ^{\frac{1}{q_1}}{+} \left[ C_1\, \left( \frac{1}{\tilde{M}}\right) ^{\,\frac{p}{q_1+1-p}}\left( \frac{1}{C_5}\right) ^{\,\frac{p}{q_1+1-p}} - \frac{C}{C_5^{\,\frac{N-p}{p}}} \left[ 1-\left( \frac{1}{2}\right) ^{\,\frac{N-p}{p}}\right] \left( \frac{1}{\tilde{M}}\right) ^{\,\frac{N-p}{p}} \right] d\\&<0, \end{aligned} \end{aligned}$$

for d big enough. This contradiction proves that for all $r<r^*$, $r \le \max \{2r_1, 2\tilde{M}\tilde{r}\} \le C\left( \frac{1}{d}\right) ^{\frac{q_1+1-p}{p}}.$ Hence, $r^*\le C\left( \frac{1}{d}\right) ^{\frac{q_1+1-p}{p}}$. This implies $r^*<\infty$ and $r^*<\tau _1.$ The proof of the claim is completed.

Let $r \in [r^*, \tau _1)$. From (2.10) and (5.1) we may write

$$\begin{aligned} r^{N-1}\big |u'(r)\big |^{p-1}= & {} \int _{0}^{r_1} s^{N-1} g\big (u(s)\big ) \, \mathrm{d}s +\int _{r_1}^{r} s^{N-1} g\big (u(s)\big ) \, \mathrm{d}s\nonumber \\\ge & {} \left( C_7 \left( \frac{d}{2}\right) ^{\, q_1}-M\right) \, \frac{r_1^{N}}{N} \nonumber \\\ge & {} C d^{\, q_1}r_1^{N}. \end{aligned}$$

(5.34)

Hence, by using the estimate for $r_1$, we get

$$\begin{aligned} -u'(r) \ge Cd^{\frac{pq_1-N(q_1+1-p)}{p(p-1)}} r^{(1-N)/(p-1)}. \end{aligned}$$

Letting $r = mr^*$ with $m>1$ and such that $mr{^*}<\tau _1$. Integration on $[r^*,r]$ yields

$$\begin{aligned} \begin{aligned} u(r^*)&\ge u(mr^*)+ Cd^{\frac{pq_1-N(q_1+1-p)}{p(p-1)}}\frac{p-1}{N-p}\left( (r^*)^{(p-N)/(p-1)}(1 - m^{(p-N)/(p-1)})\right) \\&\ge Cd(1 - m^{(p-N)/(p-1)}). \end{aligned} \end{aligned}$$

(5.35)

From (5.16) and (5.35), it follows that, for $d > 0$ large we have $m \le 2$. This proves that $r < 2r^*$ which in turn proves that $\tau _1 < 2r^* \le Cd^{(p-q_1 -1)/p}$. Replacing this in (5.34) we have

$$\begin{aligned} \big | u'(\tau _1) \big |^{p-1 } \ge Cd^{q_1}\tau _1 \ge C d^{(q_1+1)(p-1)/p}, \end{aligned}$$

(5.36)

which implies $|u'(\tau _1)| \ge C d^{(q_1+1)/p}$.

(ii):

From (2.5) and (5.11) for $(t_1,t_2)=(0,\tau _1)$, we may write

$$\begin{aligned} P(\tau _1,d)= & {} \frac{p-1}{p}\tau _1^N \big | u'(\tau _1) \big |^{p} \nonumber \\= & {} \frac{N-p}{p}\,\int _{0}^{\tau _1} s^{N-1}\,\Big [ p^* G \big (u(s)\big ) - g \big (u(s)\big ) \, u(s) \Big ]\, \mathrm{d}s \nonumber \\\le & {} \frac{N-p}{p}\, \int _{0}^{\tau _1} s^{N-1}\,\Big [ C_1 \big (u(s)\big )^{q_1+1} +M \Big ]\, \mathrm{d}s \nonumber \\\le & {} C\, d^{\, q_1+1}\, \tau _1^N + \frac{M}{N}\, \tau _1^N \le 2C\, d^{\, q_1+1}\, \tau _1^N, \end{aligned}$$

(5.37)

and therefore,

$$\begin{aligned} \big | u'(\tau _1) \big |^{p}\le & {} C\, d^{\, q_1+1}. \end{aligned}$$

(5.38)

The proof of the lemma is completed.

Remark 5.3

We show the changes in the proof of Lemma 3.3.

• (i) Equations (3.42)–(3.45) and the arguments mentioned there, should be substituted by the following.

  $$\begin{aligned} -r^{N-1}\big |u'(r)\big |^{p-2}\, u' (r) =- \,\tau _1^{\, N-1}\big |u'(\tau _1)\big |^{p-2}\, u' (\tau _1) + \int _{\tau _1}^r s^{N-1}g\big (u(s)\big )\, \mathrm{d}s \end{aligned}$$

or, equivalently,

$$\begin{aligned} r^{N-1}\big |u'(r)\big |^{p-1}= \tau _1^{\, N-1}\big |u'(\tau _1)\big |^{p-1} - \int _{\tau _1}^r s^{N-1}\big |g\big (u(s)\big )\big | \, \mathrm{d}s. \end{aligned}$$

(5.39)

Assume that $\rho _1(d)\ge 3\tau _1(d)$ and

$$\begin{aligned} \int _{\tau _1}^{2\tau _1} s^{N-1}\big |g\big (u(s)\big )\big | \, \mathrm{d}s > \frac{1}{2}\tau _1^{\, N-1}\big |u'(\tau _1)\big |^{p-1}. \end{aligned}$$

(5.40)

For any $\varepsilon >0$ small enough, there exists an $M_\varepsilon >0$ satisfying the following

$$\begin{aligned} \int _{2\tau _1}^{3\tau _1} s^{N-1}\big |g\big (u(s)\big )\big | \, \mathrm{d}s\ge & {} \int _{2\tau _1}^{3\tau _1} s^{N-1} \big ((L_2-\varepsilon )|u(s)|^{\, q_2} -M_\varepsilon \big )\, \mathrm{d}s \nonumber \\\ge & {} \big ((L_2-\varepsilon )|u(2\tau _1)|^{\, q_2} -M_\varepsilon \big ) \frac{\left( 3^N-2^N\right) }{N}\tau _1^{N} \nonumber \\= & {} \big ((L_2-\varepsilon )|u(2\tau _1)|^{\, q_2} -M_\varepsilon \big ) \frac{\left( 3^N-2^N\right) }{\left( 2^N-1\right) }\int _{\tau _1}^{2\tau _1} s^{N-1} \, \mathrm{d}s \nonumber \\\ge & {} \frac{\left( 3^N-2^N\right) }{\left( 2^N-1\right) }\int _{\tau _1}^{2\tau _1} s^{N-1} \big ((L_2-\varepsilon )|u(s)|^{\, q_2} -M_\varepsilon \big )\, \mathrm{d}s\nonumber \\\ge & {} \frac{\left( 3^N-2^N\right) }{\left( 2^N-1\right) }\int _{\tau _1}^{2\tau _1} s^{N-1} \left( \frac{(L_2-\varepsilon )}{(L_2+\varepsilon )}\big |g\big (u(s)\big )\big | -M' \right) \, \mathrm{d}s\nonumber \\\ge & {} \frac{9}{8} \int _{\tau _1}^{2\tau _1} s^{N-1}\big |g\big (u(s)\big )\big | \, \mathrm{d}s-CM'\tau _1^{N}\nonumber \\\ge & {} \int _{\tau _1}^{2\tau _1} s^{N-1}\big |g\big (u(s)\big )\big | \, \mathrm{d}s, \end{aligned}$$

(5.41)

where we have used that $\frac{\left( 3^N-2^N\right) }{\left( 2^N-1\right) }\ge \left( \left( \frac{3}{2}\right) ^N-1\right) \ge \frac{5}{4} \$ for any $N\ge 2$, and $\varepsilon >0$ is fixed so that $\frac{5}{4}\frac{(L_2-\varepsilon )}{(L_2+\varepsilon )}\ge \frac{9}{8}$.

Furthermore, Eqs. (3.49)–(3.52) and the arguments involved, should be rewritten in the following way.

From (5.11),

$$\begin{aligned} \begin{aligned} P(r,{d}) = P(\tau _1,d) + \frac{N-p}{p}\,\int _{\tau _1}^r s^{N-1}\,\Big [ p^* G \big (u(s)\big ) - g \big (u(s)\big ) \, u(s) \Big ]\, \mathrm{d}s. \end{aligned} \end{aligned}$$

(5.42)

From (2.5) and (5.8),

$$\begin{aligned} P(r,{d})\le & {} \frac{p-1}{p}\tau _1^N \big | u'(\tau _1) \big |^{p} + \frac{N-p}{p}\int _{\tau _1}^r s^{N-1}\,\Big [- C_4 |u(s)|^{q_2+1}+M\Big ]\, \mathrm{d}s\nonumber \\\le & {} \frac{p-1}{p}\tau _1^N \big | u'(\tau _1) \big |^{p} +\frac{M}{p^*}\,r^N,\qquad \forall r\in [\tau _1,\rho _1]. \end{aligned}$$

(5.43)

From this, (2.4), and (3.41),

$$\begin{aligned} r^N G\big ( u(r) \big ) \le P(r,{d})\le \frac{p-1}{p}\tau _1^N \big | u'(\tau _1) \big |^{p}+\frac{M}{p^*}\,r^N,\quad \forall r\in [\tau _1,\rho _1]. \end{aligned}$$

(5.44)

And since (5.4), $\ C_1|u(r)|^{\, q_2+1} \le C_2 \left( \frac{\tau _1}{r}\right) ^N \big | u'(\tau _1) \big |^{p}+C_3.$   Therefore,

$$\begin{aligned} |u(r)|^{\, q_2+1} \le C \left[ 1+\big | u'(\tau _1) \big |^{p}\right] ,\quad \forall r\in \left[ \tau _1,\rho _1\right] , \end{aligned}$$

and using Lemma 3.2(ii),

$$\begin{aligned} |u(r)|^{\, q_2+1} \le C \big | u'(\tau _1) \big |^{p},\quad \forall r\in \left[ \tau _1,\rho _1\right] . \end{aligned}$$

(5.45)

Integrating (3.47),

$$\begin{aligned} |u(r)|\ge C \tau _1 \big |u'(\tau _1)\big |,\qquad \forall r\in \left[ \frac{3}{2}\tau _1, 2\tau _1\right] . \end{aligned}$$

This, (5.45), and Lemma 3.2(i)–(ii) yield

$$\begin{aligned} d^{\, q_2+1}\le C\left( \tau _1 \big |u'(\tau _1)\big |\right) ^{\, q_2+1}\le C \big |u'(\tau _1)\big |^p \le C d^{\, q_1+1}. \end{aligned}$$

Now the facts $q_2>q_1$ and $d\gg 1$, provides the contradiction.

• (ii) Equations (3.53)–(3.55) and the arguments involved contained in part (ii) of the proof, should be rewritten in the following way.

From (5.43),

$$\begin{aligned} \rho _1^N G\big ( u(\rho _1) \big )=P(\rho _1,d)\le & {} \frac{p-1}{p}\tau _1^N \big | u'(\tau _1) \big |^{p}+\frac{M}{p^*}\rho _1^N. \end{aligned}$$

(5.46)

Since (5.4), $C\big | u(\rho _1) \big |^{q_2+1}-M\le \left( \frac{\tau _1}{\rho _1}\right) ^N\big | u'(\tau _1) \big |^{p}$. This, (3.37), and Lemma 3.2(ii) imply

$$\begin{aligned} \big | u(\rho _1) \big |\le & {} C\, d^{\, \frac{q_1+1}{q_2+1}}, \end{aligned}$$

for some constant $C>0$ independent of d.

Since (5.6)

$$\begin{aligned} p^* G(u(s)) - u(s) g(u(s))\ge & {} -\left( (L_2+\varepsilon )-\frac{p^*(L_2-\varepsilon )}{q_2+1}\right) |u(s)|^{q_2+1}-M_\varepsilon ''.\nonumber \\ \end{aligned}$$

(5.47)

On the other hand, since (5.4), (5.42), and (5.8),

$$\begin{aligned} \bigg [\frac{(L_2+\varepsilon )}{q_2+1} |u(\rho _1)|^{q_2+1}+M'_\varepsilon \bigg ]\rho _1^N\ge & {} \rho _1^N G\big ( u(\rho _1) \big )=P(\rho _1,d)\nonumber \\\ge & {} P(\tau _1,d) -[C_3 \big | u(\rho _1) \big |^{q_2+1}+M]\frac{\rho _1^N-\tau _1^N}{N}.\nonumber \\ \end{aligned}$$

(5.48)

From (5.48),

$$\begin{aligned} \left[ C_1 \big | u(\rho _1) \big |^{q_2+1}+ C_2\right] \rho _1^N \ge \frac{p-1}{p}\tau _1^N\big | u'(\tau _1) \big |^{p}+\left[ C_3 \big | u(\rho _1) \big |^{q_2+1}+ M\right] \frac{\tau _1^N}{N}. \end{aligned}$$

Dividing by $\tau _1^N$, $\left[ C_1 \big | u(\rho _1) \big |^{q_2+1}+ C_2\right] (\rho _1/\tau _1)^N \ge \frac{p-1}{p}\big | u'(\tau _1) \big |^{p}.$

From (3.16), (3.37), and Lemma 3.2(ii),

$$\begin{aligned} |u(\rho _1)|\ge Cd^{^{\frac{q_1+1}{q_2+1}}},\quad \text {for}\ d\gg 1, \end{aligned}$$

for some constant $C>0$ independent of d, ending this part of the proof.

(iii) Equations (3.56) and (3.57) should be rewritten in the following way. Let $r\in [r_2,\rho _1]$ [see (3.55)]. Integrating (2.9) on $[r,\rho _1]$,

$$\begin{aligned} -\big |u'(r)\big |^{p-2}u'(r)= & {} -\int _r^{\rho _1}\left( \frac{s}{r}\right) ^{N-1} g\big (u(s)\big ) \, \mathrm{d}s \nonumber \\\ge & {} \int _r^{\rho _1}\left( \frac{s}{r}\right) ^{N-1} \left[ C\big |u(s)\big |^{q_2}-M\right] \, \mathrm{d}s \nonumber \\\ge & {} \left[ C\frac{|u(\rho _1)|^{q_2}}{2^{q_2}}-M\right] \, \left( \rho _1-r\right) , \end{aligned}$$

(5.49)

and since Lemma 3.3(ii), $-\big |u'(r)\big |^{p-2}u'(r)\ge \frac{C}{2}\frac{|u(\rho _1)|^{q_2}}{2^{q_2}}\, \left( \rho _1-r\right) ,$ for $d\gg 1$, and therefore, $u'(r)\le 0$ and

$$\begin{aligned} -u'(r)= \big |u'(r)\big |\, \ge C\, |u(\rho _1)|^{\frac{q_2}{p-1}}\ \left( \rho _1-r\right) ^{\frac{1}{p-1}}. \end{aligned}$$

(5.50)

Remark 5.4

In a similar way, following the ideas of the previous remarks, it is possible to prove the extensions of Lemmas 3.4–3.6 to the more general nonlinearity g satisfying hypothesis (1.4) of Remark 1.1(a).