Existence and nonexistence of solutions for the mean curvature equation with weights

Abstract

In this paper we study existence and nonexistence of positive radial solutions of a Dirichlet problem for the prescribed mean curvature operator with weights in a ball with a suitable radius. Because of the presence of different weights, possibly singular or degenerate, the problem under consideration appears rather delicate, it requires an accurate qualitative analysis of the solutions, as well as the use of Liouville type results based on an appropriate Pohozaev type identity. In addition, sufficient conditions for global solutions to be oscillatory are given.

1 Introduction

In this paper we study the existence and nonexistence of positive radial solutions for the Dirichlet boundary value problem

$$\begin{aligned} {\left\{ \begin{array}{ll} -\text {div}\left( \frac{|x|^\alpha \nabla u}{\sqrt{1+|\nabla u|^2}}\right) =|x|^\beta u^p&{}\text {in }B_R\backslash \{0\},\\ u=0&{}\text {on }\partial B_R, \end{array}\right. } \end{aligned}$$

(1.1)

where $p>1$, $\alpha ,\beta \in {\mathbb {R}}$, $R>0$, $B_R=\left\{ x\in {\mathbb {R}}^N:|x|<R\right\}$ and $N>2$.

In the model case $\alpha =\beta =0$, then (1.1) reduces to

$$\begin{aligned} {\left\{ \begin{array}{ll} -\text {div}\left( \frac{\nabla u}{\sqrt{1+|\nabla u|^2}}\right) = u^p&{}\text {in }B_R,\\ u=0&{}\text {on }\partial B_R, \end{array}\right. } \end{aligned}$$

(1.2)

in which the differential operator at the left hand side is known as the prescribed mean curvature operator. In particular, thanks to Gidas et al. [16, Corollary 1], all positive solutions of (1.2) have radial symmetry. Ni and Serrin in [25, Theorem 3.4] proved that if

$$\begin{aligned} p\ge (N+2)/(N-2)=2^*-1, \end{aligned}$$

with $2^*=2N/(N-2)$ Sobolev exponent, then (1.2) admits no positive radial solutions for any ball in ${\mathbb {R}}^N$. While, Serrin in [32, Theorem 3] proved that when $1<p<2^*-1$ there exists a positive number $R_1$ depending only on p and N such that problem (1.2) has no positive radial solutions for $0<R<R_1$. Moreover, positive radial solutions to (1.2) must satisfy $u(0)<\left( 4N^2p\right) ^{1/(p+1)}$, for any $p>1$, cfr. Corollary of Theorem 2 in [32].

Later, Cl$\acute{\textrm{e}}$ment, Man$\acute{\textrm{a}}$sevich and Mitidieri in [8, Theorem 3.6] proved that if $1<p<2^*-1$ there exists $R^*\ge 0$ such that (1.2) has at least one positive (radial) solution for every $R>R^*$.

On the other hand, if we consider the associated problem to (1.2), but in the entire space, precisely

$$\begin{aligned} -\text {div}\left( \frac{\nabla u}{\sqrt{1+|\nabla u|^2}}\right) = u^p\quad \text {in }{\mathbb {R}}^N, \end{aligned}$$

(1.3)

Ni and Serrin, in [24, 25], started the study of ground states for (1.3), namely of positive radial solutions u in ${\mathbb {R}}^N$ tending to zero at $\infty$. It was proved that if $1<p\le N/(N-2)=2_*-1$, where $2_*(<2^*)$ is Serrin exponent, no ground states of (1.3) can exist, cfr. [25, Theorem 2.2]. On the contrary, if $p\ge 2^*-1$ then there exists at least one ground state of (1.3), cfr. [25, Theorem 5.2]. In the range $2_*-1<p<2^*-1$ some contributions have been given by Cl$\acute{\textrm{e}}$ment, Man$\acute{\textrm{a}}$sevich and Mitidieri in [8] who proved nonexistence for problem (1.3), giving a Liouville type theorem for positive ground state solutions with $u(0)<C$, for some constant $0<C=C(N,p)$, cfr. [8, Theorem 3.5]. Another progress was achieved by Del Pino and Guerra in [12] obtaining existence of many ground states provided that $p<2^*-1$ but sufficiently close to $2^*-1$.

Radial solutions of (1.3) have been studied in the context of the analysis of capillary surfaces when the reaction has the form $f(u)=ku$, for $k>0$. Roughly, capillarity takes into account the effects of two opposing forces: adhesion, that is the attractive (or repulsive) force between the molecules of the liquid and those of the container; and cohesion, namely the attractive force between the molecules of the liquid. The study of capillary phenomena has attracted much attention even recently, to descrive phenomena such as motion of drops, bubbles and waves. Its importance is also known in applied fields ranging from industrial and biomedical and pharmaceutical to microfluidic systems.

Generalizations of problems (1.2) and (1.3), including reactions involving a competition between power type nonlinearities, have been deeply studied in recent years in the context of nonlinear equations on bounded domains with different types of boundary conditions (see [3,4,5, 9, 10, 13, 17, 18, 21, 23, 25,26,27, 32] and the references therein), as well as either in the entire ${\mathbb {R}}^N$ or in unbounded domains for various classes of nonlinearities (see [1, 10, 12, 15, 19, 20, 24, 25, 30]).

In particular, in [10], Conti and Gazzola considered existence and nonexistence of positive radial solutions for

$$\begin{aligned} -\text {div}\left( \frac{\nabla u}{\sqrt{1+|\nabla u|^2}}\right) =-au^q+u^p \end{aligned}$$

(1.4)

either in the whole space ${\mathbb {R}}^N$ with the condition $\lim _{|x|\rightarrow \infty }u(x)=0$, or in the ball $B_R$ subject to Dirichlet–Neumann free-boundary conditions $u=\frac{\partial u}{\partial n}=0$ on $\partial B_R$, for different ranges of parameters a, p and q.

A variational approach has been applied in [13, 17, 21, 22, 27] to prove several existence and multiplicity results concerning solutions of the Dirichlet problems with the prescribed mean curvature operator and more general nonlinear terms in non-symmetric domains

$$\begin{aligned} {\left\{ \begin{array}{ll} -\text {div}\left( \frac{\nabla u}{\sqrt{1+|\nabla u|^2}}\right) =\lambda f(x,u)+g(x,u)&{}\text {in }\Omega ,\\ u=0&{}\text {on }\partial \Omega , \end{array}\right. } \end{aligned}$$

(1.5)

where $\Omega \subset {\mathbb {R}}^N$ is a bounded smooth domain. Habets and Omari in [17] assumed that $f,g:\bar{\Omega }\times {\mathbb {R}}_+\rightarrow {\mathbb {R}}$ are continuous functions satisfying $\int _0^uf(x,s)ds$ is locally subquadratic at 0 and $\int _0^ug(x,s)ds$ is superquadratic at 0, and $\lambda >0$ is a sufficiently small real parameter. Later, Le [21, 22] and Obersnel and Omari [27] considered problem (1.5) with $g=0$ and $f: \mathcal {B}\times {\mathbb {R}}\rightarrow {\mathbb {R}}$ is a Carath$\acute{\textrm{e}}$odory function satisfies

$$\begin{aligned} |f(x,u)|\le c_1|u|^{q-1}+c_2(x)\quad \text {for }x\in \mathcal {B},u\in {\mathbb {R}}, \end{aligned}$$

with $q\in \left( 1,N/(N-1)\right)$, $c_1>0$ and $c_2\in L^{q'}(\mathcal {B})$, where $\mathcal {B}$ is an open ball in ${\mathbb {R}}^N$ containing $\overline{\Omega }$. In [13], Figueiredo and Pimenta studied the problem (1.5) in the case of $f=|u|^{q-2}u$ and $g=|u|^{2^*-2}u$ with $1<q<2$.

A more general class of equations associated with the mean curvature operators, is given by

$$\begin{aligned} {\left\{ \begin{array}{ll} -\text {div}\left( \frac{\nabla u}{\sqrt{1+|\nabla u|^2}}\right) =f\left( |x|,u,\nabla u\right) &{}\text {in }B_1,\\ u=0&{}\text {on }\partial B_1, \end{array}\right. } \end{aligned}$$

(1.6)

where $B_1\subset {\mathbb {R}}^N$ denotes the unit ball. In particular, when $f(|x|,u,v)=-au+b/\sqrt{1+|v|^2}$, $a,b>0$, then the radial version of (1.6) has been derived in [28] for describing the geometry of the human cornea. Indeed, the surface of the human cornea is modeled as a membrane, whose shape is described by the graph of the function u and is determined by balancing all forces acting over, that is, surface tension, elasticity, and intraocular pressure. The relevant physical parameters are incorporated into the coefficients a and b, which respectively measure the relative importance of the elasticity and of the intra-ocular pressure versus the surface tension. For further details we refer to [11].

Bereanu et al. in [3], investigated (1.6) both in Euclidean and Minkowski spaces, focusing also in an annular domain, obtaining existence results for a suitable f. Moreover, when f is independent of $u'$ and $f(r,\cdot )$ is nondecreasing for each fixed $r\in [0,1]$, they obtained uniqueness results for radial solutions of problem (1.6).

A first attempt in considering (1.3) with a potential $|x|^\beta$ in the nonlinearity, is due to Azzollini in [1] where he studied

$$\begin{aligned} -\text {div}\left( \frac{\nabla u}{\sqrt{1+|\nabla u|^2}}\right) = |x|^\beta u^p\quad \text {in }{\mathbb {R}}^N, \end{aligned}$$

(1.7)

with $\beta >0$. It was proved that if $1<p\le (N+\beta )/(N-2)$, then there exists no radial ground state solutions to (1.7). In particular, the exponent $(N+\beta )/(N-2)$ reduces to Serrin exponent for $\beta =0$.

As far as we know, differently for the p-Laplacian operator (cfr. [2, 6, 31]), the case of the mean curvature equation with different weights inside the divergence and in the nonlinearity, seems completely new. With this in mind, the aim of our paper is to prove results similar to those in [8, 25, 32], for problems (1.1) and (1.3), where different weights appear. Since we deal with radial solutions, we restrict our attention to the radial version of (1.1), namely

$$\begin{aligned} {\left\{ \begin{array}{ll} -\left( \frac{r^{N+\alpha -1}u'(r)}{\sqrt{1+\left( u'(r)\right) ^2}}\right) '= r^{N+\beta -1}u^p(r),\quad r\in (0,R),\\ u(R)=0, \end{array}\right. } \end{aligned}$$

(1.8)

where $r=|x|$, $x\in B_R$.

Define

$$\begin{aligned} 2^*_{\alpha ,\beta }=\frac{2(N+\beta )}{N+\alpha -2}, \end{aligned}$$

and assume

$(H_0)$:

$\alpha ,\beta \in {\mathbb {R}}$ such that $N+\alpha -2>0$ and $\beta -\alpha +1>0$.

In particular, $(H_0)$ implies $N+\beta >1$. The proof technique of the first two results we have obtained requires the use of many different arguments, such as Liouville type theorems, some properties of global oscillatory solutions, a deep qualitative analysis of solutions and some suitable estimates of the first derivative, in addition several auxiliary functions are involved.

Theorem 1.1

Assume $(H_0)$. If $p\ge 2^*_{\alpha ,\beta }-1$, then problem (1.1) has no positive radial solutions.

Theorem 1.2

Assume $(H_0)$. If $1<p<2^*_{\alpha ,\beta }-1$, then there exists $R^*\ge 0$ such that problem (1.1) has at least one nontrivial positive solution for every $R>R^*$.

Remark 1.3

Theorem 1.2, which extends Theorem 3.6 in [8], implies the existence of a large solution set for problem (1.1), and provides a lower bound on the radius of the ball $B_R$ within which a positive solution can exist.

Remark 1.4

When $\alpha =\beta$ and $N+\alpha -2>0$, we can use the results of [32] to conclude that any positive radial solution to (1.1) satisfies $u(0)<\left[ 4(N+\alpha )^2p\right] ^{1/(p+1)}$. Moreover, there exists a positive number $R_1$ depending only on N, p and $\alpha$ such that problem (1.1) has no positive radial solution for $0<R<R_1$. Unfortunately, when $\alpha \ne \beta$ the proof technique cannot be applied because the property that $u''(0)$ exists and it is not zero fails. For details we refer to Sect. 2.

The next theorem extends Theorem 3 in [32] to the case both $\alpha =\beta \ne 0$ and $\alpha \ne \beta$. The proof is very delicate since it is based on the use of some tricky and cumbersome estimates we need to produce because of the presence of different weights.

Theorem 1.5

Assume $(H_0)$ and $p>1$. Let u be a nonnegative radial solution of problem (1.1).

• (i) If $\beta \ge \alpha$, then there exists a positive number $R_1=R_1(N, p, \alpha , \beta )$ such that u is identically zero for every $0<R<R_1$.

• (ii) If $\beta <\alpha$, then there exists a positive number $R_2=R_2(N, p, \alpha , \beta ,u(0))$ such that u is identically zero for every $0<R<R_2$.

The fact that the upper bound for R depends also on u(0) when $\beta <\alpha$, differently from the case $\beta \ge \alpha$, is discussed in details at the end of the proof of Theorem 1.5.

Remark 1.6

The methods used in this paper to treat the case with different weights, but of power type, can be applied to deal with Dirichlet problems for the mean curvature operator with more general weights and nonlinearities, such as

$$\begin{aligned} {\left\{ \begin{array}{ll} -\text {div}\left( h(|x|)\frac{\nabla u}{\sqrt{1+|\nabla u|^2}}\right) =g(|x|)f(u)&{}\text {in }B_R\backslash \{0\},\\ u=0&{}\text {on }\partial B_R, \end{array}\right. } \end{aligned}$$

(1.9)

whose radial version is

$$\begin{aligned} {\left\{ \begin{array}{ll} -\left( \frac{a(r)u'(r)}{\sqrt{1+\left( u'(r)\right) ^2}}\right) '= b(r)f(u(r)),\quad r\in (0,R),\\ u(R)=0, \end{array}\right. } \end{aligned}$$

(1.10)

where $a(r)=r^{N-1}h(r)$ and $b(r)=r^{N-1}g(r)$ satisfy

$(G_0)$:

$a(r),b(r)\in C^1\left( (0,R]\right)$, $a(0)=0$, and there exist real exponents $\gamma ,\delta >-1$ such that $a(r)>r^{\delta +1}$ as $r\rightarrow 0$ and

$$\begin{aligned}{} & {} ra'(r)\ge \left( \frac{2(\delta +1)}{\gamma +1}+1\right) a(r),\\{} & {} rb'(r)\le \delta b(r), \end{aligned}$$

for all $r\in [0,R]$, and

$$\begin{aligned} (\gamma +1)F(u)\le uf(u) \end{aligned}$$

for all $u>0$, where $F(u)=\int _0^uf(t)dt$.

Under condition $(G_0)$, it is possible to prove that problem (1.10) has no positive solutions in the spirit of Theorem 1.1. In particular, if we take $h(|x|)=|x|^{\alpha }$, $g(|x|)=|x|^{\beta }$ and $f(u)=u^p$ in (1.9), then $a(r)=r^\sigma$, $\sigma =N-1+\alpha$, $b(r)=r^\delta$, $\delta = N-1+\beta$, so that $(G_0)$ is satisfied with $\gamma =p$, $\sigma >1$ and $\delta>\sigma -1>0$, namely when ($H_0$) holds, and $p+1\ge \frac{2(\delta +1)}{\sigma -1}:=2^*_{\sigma ,\delta }$. When $\sigma =\delta =N-1$, we have $2^*_{\sigma ,\delta }=2^*$ with $2^*$ Sobolev exponent. For the existence result of problem (1.10), in the spirit of Theorem 1.2, we can only deal with the cases where $a(r)=r^\sigma a_0(r)$, $b(r)=r^\delta b_0(r)$, $f(u)=u^p$, with positive bounded continues functions $a_0(r)$ and $b_0(r)$ defined in $[0,\infty )$.

The paper is organized as follows. Section 2 presents preliminary results that include known results on the regularity of solutions. In Sect. 3, we give a new Pohozaev type identity for positive radial solutions of problem (1.1), and use it to prove the nonexistence result stated in Theorem 1.1. Section 4 is dedicated to proving some Liouville type results for the Cauchy problem associated with (1.1), while Sect. 5 deals with global oscillatory solutions for the Cauchy problem. In Sect. 6, we establish the existence of positive radial solutions of problem (1.1) by virtue of Banach Fixed Point Theorem together with some properties of oscillatory solutions obtained in Sect. 5. Finally, in Sect. 7, we give the proof of Theorem 1.5.

2 Regularity of solutions

In this section, we present some well-known results regarding the regularity and qualitative properties of positive radial solutions to problems of the form (1.1). A pioneering paper in this area is Ni and Serrin’s work in [24], where they studied positive radial solutions of $\text {div}\left( \mathcal {A}(|\nabla u|)\nabla u\right) +f(u)=0$, namely solutions of

$$\begin{aligned} {\left\{ \begin{array}{ll} -\left( r^{N-1}\mathcal {A}\left( |u'(r)|\right) u'(r)\right) '=r^{N-1}f(u(r)),\quad r>0,\\ u(0)=u_0>0, \end{array}\right. } \end{aligned}$$

(2.1)

where the operator $\mathcal {A}$ takes the form

$$\begin{aligned} \mathcal {A}:{\mathbb {R}}^+\rightarrow {\mathbb {R}},\quad \mathcal {A}\in C^1({\mathbb {R}}^+),\quad t\mathcal {A}(t)\text { strictly increasing},\quad \lim _{t\rightarrow 0^+}t\mathcal {A}(t)=0, \end{aligned}$$

and $f\in C({\mathbb {R}})$ with $f(0)=0$, $f>0$ in ${\mathbb {R}}^+$. Some notable examples of $\mathcal {A}$ in ${\mathbb {R}}^+$ include the m-Laplacian operator $\mathcal {A}(t)=t^{m-2}$ for $m>1$, the mean generalized curvature operator $\mathcal {A}(t)=(1+t^2)^{m/2-1}$ for $m\in (1,2]$, and the mean curvature operator $\mathcal {A}(t)=(1+t^2)^{-1/2}$.

Proposition 1 in [24] shows that every solution $u=u(r)$ of (2.1) is continuously differentiable on some interval $0\le r\le R$ with $u'(0)=0$. Their proof relies on Schauder’s fixed point theorem, applied to the compact operator

$$\begin{aligned} T[u](r)=u_0-\int _0^r\phi \left( \int _0^tf(u(s))\left( \frac{s}{t}\right) ^{N-1}ds\right) dt, \end{aligned}$$

(2.2)

where $\phi$ is the inverse function of $t\mathcal {A}(t)$ with $\phi (0)=0$, and R is chosen suitably small such that the value $\int _0^tf(v(s))(s/t)^{N-1}ds$ small for all $t\in (0,R]$, which ensures that $T[S]\subset S$.

Furthermore, they proved that positive solutions of (2.1) are of class $C^2$ as long as $u'(r)\ne 0$.

In the context of radial problems with different weights, such as

$$\begin{aligned} {\left\{ \begin{array}{ll} -\left( r^{N+\alpha -1}\mathcal {A}\left( |u'(r)|\right) u'(r)\right) '=r^{N+\beta -1}f(u(r)),\quad r>0,\\ u(0)=u_0>0, \end{array}\right. } \end{aligned}$$

(2.3)

the regularity of positive solutions can be analyzed by applying similar techniques as used in [24], cfr. for the m-Laplacian [2]. For example, when considering (2.3), the operator T defined in (2.2) needs to be modified to account for the different weights

$$\begin{aligned} T[u](r)=u_0-\int _0^r\phi \left( \frac{1}{t^{N+\alpha -1}}\int _0^tf(u(s))s^{N+\beta -1}ds\right) dt. \end{aligned}$$

The inclusion $T[S]\subset S$ is guaranteed if

$$\begin{aligned} 0\le \frac{1}{t^{N+\alpha -1}}\int _0^tf(u(s))s^{N+\beta -1}ds\le \frac{t^{\beta -\alpha +1}}{N+\beta }\max _{t\in [0,R]}f(u(t)) \end{aligned}$$

is sufficiently small for all $t\in [0,R]$, with $R>0$ suitable. As pointed out in [7, section 2], the condition $\beta -\alpha +1>0$ is necessary for obtaining regularity $C^1[0,R]$ of positive solutions of (2.3), and for ensuring that $u'(0)=0$.

Moreover, if u is a nonnegative solution of (2.3), we can obtain

$$\begin{aligned} -\mathcal {A}\left( |u'(r)|\right) u'(r)=\frac{1}{r^{N+\alpha -1}}\int _0^rs^{N+\beta -1}f(u(s))ds,\quad r\in (0,R]. \end{aligned}$$

From the positivity of $\mathcal {A}$ in ${\mathbb {R}}^+$ and the right hand side, we conclude that $u'(r)<0$ for all $0<r\le R$. Thus, we can obtain regularity $C^2(0,R]$ of u by utilizing the $C^1$ regularity of $\mathcal {A}$.

Let us reconsider problem (1.1) and discuss the regularity of positive solutions. We begin by defining weak solutions (distribution) of (1.1).

Definition 2.1

A weak distribution solution of (1.1) is a nonnegative function u of $C^1(B_R)\cap C(\overline{B_R})$ which verifies

$$\begin{aligned} \int _{B_R}\frac{|x|^\alpha \nabla u}{\sqrt{1+|\nabla u|^2}}\cdot \nabla \psi dx=\int _{B_R}|x|^\beta u^p\psi dx \end{aligned}$$

for all $C^1$ functions $\psi =\psi (x)$ with compact support in $B_R$.

Alternatively, in the radial case, a weak (distribution) solution of (1.8) is defined as a nonnegative function u that belongs to $C[0,R]\cap C^1(0,R)$ and satisfies

$$\begin{aligned} \int _0^R\frac{r^{N+\alpha -1}u'(r)}{\sqrt{1+\left( u'(r)\right) ^2}}\cdot \psi ' dr=\int _0^Rr^{N+\beta -1} u^p(r)\psi dr \end{aligned}$$

for all $C^1$ functions $\psi =\psi (r)$ with compact support in [0, R). Using distribution arguments, it can be shown that u satisfies

$$\begin{aligned} -\frac{r^{N+\alpha -1}u'(r)}{\sqrt{1+\left( u'(r)\right) ^2}}=\int _0^rs^{N+\beta -1} u^p(s)ds\quad \text {in }[0,R). \end{aligned}$$

(2.4)

As the right hand side of (2.4) is continuously differentiable in r, we conclude that

$$\begin{aligned} \frac{r^{N+\alpha -1}u'(r)}{\sqrt{1+\left( u'(r)\right) ^2}}\in C^1(0,R),\end{aligned}$$

(2.5)

implying that u is a classical solution of problem (1.1). Therefore, in our case, the definition of weak solution is consistent with that of classical solution. Indeed, if $u\in C[0,R]\cap C^1(0,R)$ is a weak positive solution of (1.8), then the right hand side of (2.4) is positive, yielding $u'(r)<0$ in (0, R]. Consequently,

$$\begin{aligned} \frac{|u'(r)|}{\sqrt{1+\left( u'(r)\right) ^2}}=\frac{1}{r^{N+\alpha -1}}\int _0^rs^{N+\beta -1} u^p(s)ds\quad \text {in }(0,R), \end{aligned}$$

so that by the continuity of u in [0, r], $r>0$ we reach

$$\begin{aligned} |u'(r)|=O\bigl (r^{\beta -\alpha +1}\bigr )\quad \text {as } r\rightarrow 0^+, \end{aligned}$$

and thus $u'(0)=0$ being $\beta -\alpha +1>0$. In conclusion, $u\in C^1[0,R]$ with

$$\begin{aligned} u'(0)=0\quad \text {and}\quad u'(r)<0\quad \text {for all }0<r\le R. \end{aligned}$$

(2.6)

In turn, $u\in C^2(0,R]$ due to the $C^1$ regularity of the function $\mathcal {A}(t)=(1+t^2)^{-1/2}$. So it remains to discuss the regularity of $u''$ at $r=0$. We will see that $u'$ is differentiable up to $r=0$ whenever $\alpha \le \beta$. Indeed, using (2.4) and $u'(0)=0$, by L’Hospital’s rule we have

$$\begin{aligned} \lim _{r\rightarrow 0}\frac{u'(r)}{r}=-\lim _{r\rightarrow 0}\frac{r^{N+\beta -1} u^p(r)}{(N+\alpha )r^{N+\alpha -1}}=-\frac{u_0^p}{N+\alpha }\lim _{r\rightarrow 0}r^{\beta -\alpha }, \end{aligned}$$

so that by Lagrange’s theorem

$$\begin{aligned} u''(0)= {\left\{ \begin{array}{ll} -\frac{u_0^p}{N+\alpha }&{}\text {if }\alpha =\beta ,\\ 0&{}\text {if }\alpha <\beta . \end{array}\right. } \end{aligned}$$

(2.7)

Summarizing, under the assumptions $(H_0)$, according to Lemma 2.1 in [7] devoted to m-Laplacian type operators, we obtain that every positive radial solution u of (1.1) is such that

$$\begin{aligned} {\left\{ \begin{array}{ll} u\in C^2[0,R]&{}\text {if }\alpha \le \beta ,\\ u\in C^1[0,R]\cap C^2(0,R]&{}\text {if }\beta<\alpha <\beta +1.\\ \end{array}\right. } \end{aligned}$$

(2.8)

3 Proof of Theorem 1.1

In this section, we utilize a Pohozaev type identity to prove Theorem 1.1, which establishes the nonexistence of positive radial solutions of problem (1.1). The Pohozaev type identity was initially proposed by Ni and Serrin in [25, section 2] to prove the nonexistence of solutions to problem (1.1) when $\alpha =\beta =0$. Our intention now is to present a more comprehensive and all-encompassing version of the Pohozaev type identity through an alternate construction that preserves the fundamental meaning, allowing for $\alpha$ and $\beta$ to take on values beyond zero.

Proposition 3.1

Let u(r) be a positive solution of (1.8). Then for any real constant a we have the identity

$$\begin{aligned}&\frac{d}{dr}\left[ r^{N+\alpha }\int _0^{\rho (r)} sE(s)ds+r^{N+\beta } F(u)-ar^{N+\alpha -1}A(\rho )\rho (r)u(r)\right] \nonumber \\&\quad =r^{N+\alpha -1}\left[ (N+\alpha )\int _0^{\rho (r)} sE(s)ds+\left[ a-(N+\alpha -1)\right] A(\rho )\rho ^2(r)\right] \nonumber \\&\qquad +r^{N+\beta -1}\left[ (N+\beta )F(u)-au(r)f(u)\right] , \end{aligned}$$

(3.1)

where

$$\begin{aligned} \rho (r)=|u'(r)|,\quad A(\rho )=\frac{1}{\sqrt{1+\rho ^2}},\quad f(u)=u^p,\quad F(u)=\int _0^uf(t)dt, \end{aligned}$$

and define

$$\begin{aligned} E(\rho )=A(\rho )+\rho (r)\frac{dA(\rho )}{d\rho }.\end{aligned}$$

(3.2)

Proof

Assuming that u is a positive solution of (1.8). By (2.6) it follows $u'(r)<0$ in (0, R], so that $\rho (r)=-u'(r)$. In particular, equation in (1.8) becomes

$$\begin{aligned} (N+\alpha -1)r^{N+\alpha -2}A(\rho )\rho (r)+r^{N+\alpha -1}E(\rho )\rho '(r)=r^{N+\beta -1}f(u),\quad r\in (0,R) \end{aligned}$$

(3.3)

where we have used the expression $E(\rho )$ in (3.2). Multiplying both sides of (3.3) by r, we obtain

$$\begin{aligned} r^{N+\beta }f(u)=(N+\alpha -1)r^{N+\alpha -1}A(\rho )\rho (r)+r^{N+\alpha }E(\rho )\rho '(r). \end{aligned}$$

(3.4)

To proceed, consider the left hand side of (3.1), by using (3.2), we arrive to

$$\begin{aligned} I:&=\frac{d}{dr}\left[ r^{N+\alpha }\int _0^{\rho (r)} sE(s)ds+r^{N+\beta } F(u)-ar^{N+\alpha -1}A(\rho )\rho (r) u(r)\right] \nonumber \\&=(N+\alpha )r^{N+\alpha -1}\int _0^{\rho (r)} sE(s)ds+r^{N+\alpha }\rho (r) E(\rho )\rho '(r)+(N+\beta )r^{N+\beta -1}F(u) \nonumber \\&\quad -r^{N+\beta }f(u)\rho (r)-a(N+\alpha -1)r^{N+\alpha -2}A(\rho )\rho (r) u(r) \nonumber \\&\quad -ar^{N+\alpha -1}\left[ E(\rho )\rho '(r)u(r)-A(\rho )\rho ^2(r)\right] . \end{aligned}$$

(3.5)

Then, replacing (3.4) into (3.5), we obtain

$$\begin{aligned} I&=(N+\alpha )r^{N+\alpha -1}\int _0^{\rho (r)}sE(s)ds+r^{N+\alpha }\rho (r) E(\rho )\rho '(r)+(N+\beta )r^{N+\beta -1}F(u) \nonumber \\&\quad -\left[ (N+\alpha -1)r^{N+\alpha -1}A(\rho )\rho (r)+r^{N+\alpha }E(\rho )\rho '(r)\right] \rho (r) \nonumber \\&\quad -a(N+\alpha -1)r^{N+\alpha -2}A(\rho )\rho (r) u(r) -ar^{N+\alpha -1}\left[ E(\rho )\rho '(r)u(r)-A(\rho )\rho ^2(r)\right] \nonumber \\&=(N+\alpha )r^{N+\alpha -1}\int _0^{\rho (r)}sE(s)ds+(N+\beta )r^{N+\beta -1}F(u) \nonumber \\&\quad +\left[ a-(N+\alpha -1)\right] r^{N+\alpha -1}A(\rho )\rho ^2(r) \nonumber \\&\quad -a(N+\alpha -1)r^{N+\alpha -2}A(\rho )\rho (r) u(r) -ar^{N+\alpha -1}E(\rho )\rho '(r)u(r). \end{aligned}$$

(3.6)

Using (3.3) in the last two terms on the right hand side of (3.6) we get

$$\begin{aligned} I&=(N+\alpha )r^{N+\alpha -1}\int _0^{\rho (r)}sE(s)ds+(N+\beta )r^{N+\beta -1}F(u)\\&\quad +\left[ a-(N+\alpha -1)\right] r^{N+\alpha -1}A(\rho )\rho ^2(r)-ar^{N+\beta -1}u(r)f(u), \end{aligned}$$

that is (3.1). $\square$

Now, to prove the nonexistence theorem for problem (1.1), our strategy is to combine Proposition 3.1 with the boundary condition $u(R)=0$ and the inequality

$$\begin{aligned} \frac{\sqrt{1+y}-1}{y}\le \frac{1}{2}\quad \text {for all }y\ge 0,\end{aligned}$$

(3.7)

following from the decreasing monotonicity of the function $\sqrt{1+y}-\frac{1}{2} y -1$ in ${\mathbb {R}}^+_0$.

Proof of Theorem 1.1

Suppose u is a radial solution of problem (1.1). Using Proposition 3.1 with $a=\frac{N+\beta }{p+1}$, we get

$$\begin{aligned} (N+\beta )F(u)-au(r)f(u)=0. \end{aligned}$$

Hence

$$\begin{aligned}&\frac{d}{dr}\left[ r^{N+\alpha }\int _0^{\rho (r)} sE(s)ds+r^{N+\beta } F(u)-\frac{N+\beta }{p+1}r^{N+\alpha -1}A(\rho )\rho (r) u(r)\right] \nonumber \\&\quad =r^{N+\alpha -1}\left\{ (N+\alpha )\int _0^{\rho (r)} sE(s)ds+\left[ \frac{N+\beta }{p+1}-(N+\alpha -1)\right] A(\rho )\rho ^2(r)\right\} . \end{aligned}$$

(3.8)

Integrating (3.8) from 0 to R and using $u(R)=0$, $F(u(R))=0$, $N+\alpha -1>0$ and $N+\beta >0$, we obtain

$$\begin{aligned} R^{N+\alpha }\int _0^{\rho (R)} sE(s)ds =(N+\alpha )\int _0^Rr^{N+\alpha -1}\Psi (r)dr, \end{aligned}$$

(3.9)

with

$$\begin{aligned} \Psi (r):=\int _0^{\rho (r)} sE(s)ds+\frac{1}{N+\alpha }\left[ \frac{N+\beta }{p+1}-(N+\alpha -1)\right] A(\rho )\rho ^2(r). \end{aligned}$$

Now, since

$$\begin{aligned} E(\rho )=A(\rho )+\rho \frac{dA(\rho )}{d\rho }=\frac{1}{\sqrt{1+\rho ^2}}-\frac{\rho ^2}{\left( 1+\rho ^2\right) ^{3/2}}=\frac{1}{\left( 1+\rho ^2\right) ^{3/2}}, \end{aligned}$$

and

$$\begin{aligned} \int _0^\rho sE(s)ds=\int _0^\rho \frac{s}{\left( 1+s^2\right) ^{3/2}}ds =-\int _0^\rho \left( \frac{1}{\sqrt{1+s^2}}\right) 'ds=\frac{\sqrt{1+\rho ^2}-1}{\sqrt{1+\rho ^2}}, \end{aligned}$$

then

$$\begin{aligned} \Psi (r)=\frac{\rho ^2(r)}{\sqrt{1+\rho ^2(r)}}\left\{ \frac{\sqrt{1+\rho ^2(r)}-1}{\rho ^2(r)}+\frac{N+\beta }{(p+1)(N+\alpha )}-\frac{N+\alpha -1}{N+\alpha }\right\} . \end{aligned}$$

Then (3.9) becomes

$$\begin{aligned} R^{N+\alpha }\frac{\sqrt{1+\rho ^2(R)}-1}{\sqrt{1+\rho ^2(R)}} =(N+\alpha )\int _0^Rr^{N+\alpha -1}\Psi (r)dr. \end{aligned}$$

(3.10)

Note that the left hand side of (3.10) is positive due to $\rho (R)=|u'(R)|>0$ from (2.6). On the other hand, by $p\ge 2^*_{\alpha ,\beta }-1$ we have

$$\begin{aligned} p+1\ge 2^*_{\alpha ,\beta }=\frac{2(N+\beta )}{N+\alpha -2}. \end{aligned}$$

Thus

$$\begin{aligned} \frac{N+\beta }{(p+1)(N+\alpha )} -\frac{N+\alpha -1}{N+\alpha }&\le \frac{N+\alpha -2}{2(N+\alpha )}-\frac{N+\alpha -1}{N+\alpha }=-\frac{1}{2}. \end{aligned}$$

This, together with the fact that

$$\begin{aligned} \frac{\sqrt{1+\rho ^2(r)}-1}{\rho ^2(r)}\le \frac{1}{2}\quad \text {for all }0\le r\le R \end{aligned}$$

gives $\Psi (r)\le 0$, namely the right hand side of (3.10) is non-positive, which is a contradiction. $\square$

4 Liouville type theorems

In this section, we establish some Liouville type results for the Cauchy problem

$$\begin{aligned} {\left\{ \begin{array}{ll} -\left( \frac{r^{N+\alpha -1}u'(r)}{\sqrt{1+\left( u'(r)\right) ^2}}\right) '= r^{N+\beta -1}|u(r)|^{p-1}u(r),\quad r>0,\\ u(0)=u_0. \end{array}\right. } \end{aligned}$$

(4.1)

These results will be utilized in the final section to analyze the existence of positive solutions to the Dirichlet problem (1.1).

We introduce the notation

$$\begin{aligned} {2_*}_{\alpha ,\beta }=\frac{2(N-1)+\alpha +\beta }{N+\alpha -2}, \end{aligned}$$

and recall the definition of $2^*_{\alpha ,\beta }$ from the first section

$$\begin{aligned} 2^*_{\alpha ,\beta }=\frac{2(N+\beta )}{N+\alpha -2}. \end{aligned}$$

The main results of this section are presented as follows.

Theorem 4.1

Assume $(H_0)$. If $1<p\le {2_*}_{\alpha ,\beta }-1$, then there are no positive solutions $u\in C^2(0,\infty )$ of (4.1).

Theorem 4.2

Assume $(H_0)$. If ${2_*}_{\alpha ,\beta }-1<p<2^*_{\alpha ,\beta }-1$, then (4.1) admits no positive solutions $u\in C^2(0,\infty )$ for initial values

$$\begin{aligned} u_0\in (0,u_0^*],\qquad u_0^*=u_0^*(\alpha ,\beta ,p,N), \end{aligned}$$

(4.2)

precisely

$$\begin{aligned} u_0^*&=\left( \frac{N+\beta }{(N+\alpha -2)^{\beta -\alpha +1}}\right) ^{\frac{1}{p+\beta -\alpha +1}} \nonumber \\&\quad \cdot \left( \frac{(N+\alpha )(p+1)\left[ N-\alpha +2\beta +2-p(N+\alpha -2)\right] }{(N+\beta +p+1)^2}\right) ^{\frac{\beta -\alpha +2}{2(p+\beta -\alpha +1)}}. \end{aligned}$$

(4.3)

Remark 4.3

Theorem 4.2 above covers the result obtained by Cl$\acute{\textrm{e}}$ment, Man$\acute{\textrm{a}}$sevich and Mitidieri in [8, Theorem 3.5], since the value of $u_0^*$ we obtain in (4.3) when $\alpha =\beta =0$ is greater than that in [8]. Indeed, denoting

$$\begin{aligned} A:=u_0^*(0,0,p,N)=\left( \frac{N}{N-2}\right) ^{\frac{1}{p+1}}\left( \frac{N(p+1)[N+2-p(N-2)]}{(N+p+1)^2}\right) ^{\frac{1}{p+1}}, \end{aligned}$$

and B the value of $u_0^*$ in [8], that is

$$\begin{aligned} B:=\left( \frac{(p+1)[N+2-p(N-2)]}{N+p+1}\right) ^{\frac{1}{p+1}}. \end{aligned}$$

Then

$$\begin{aligned} \frac{A}{B}=\left( \frac{N^2}{(N-2)(N+p+1)}\right) ^{\frac{1}{p+1}}>\left( \frac{N^2}{(N-2)(N+2^*)}\right) ^{\frac{1}{p+1}}=1 \end{aligned}$$

being $p<2^*-1=\frac{N+2}{N-2}$. This implies that $A>B$. Therefore, Theorem 4.2 enlarges the range of initial data $u_0$ that guarantees any nonnegative solution of (4.1) is identically zero when $\alpha =\beta =0$ by employing a different technique. In addition, the result is completely new when either $\alpha =\beta \ne 0$ or $\alpha \ne \beta$.

In order to prove the two Liouville type theorems, we need a series of lemmas, the first of which will play a key role in obtaining the decay estimates of the solution to problem (4.1).

Lemma 4.4

Assume $(H_0)$. If $u\in C^2(0,\infty )$ is a positive solution of problem (4.1), then the function

$$\begin{aligned} W(r)=\frac{ru'(r)}{\sqrt{1+\left( u'(r)\right) ^2}}+(N+\alpha -2)\left( u(r)-u_\infty \right) \end{aligned}$$

is nonnegative and non-increasing in $(0,\infty )$, where $u_\infty =\inf _{r\ge 0}u(r)=\lim _{r\rightarrow \infty }u(r)$.

Proof

By (2.6) we have $u'(0)=0$ and $u'(r)<0$ for $r>0$. Using the equation in (4.1), we get

$$\begin{aligned} r^{N+\beta -1}u^p(r)&=-\left( \frac{r^{N+\alpha -2}ru'(r)}{\sqrt{1+\left( u'(r)\right) ^2}}\right) '\\&\le -r^{N+\alpha -2}\left( \frac{ru'(r)}{\sqrt{1+\left( u'(r)\right) ^2}}\right) ' -(N+\alpha -2)r^{N+\alpha -2}u'(r) , \end{aligned}$$

where in the last inequality we have used that $u'<0$, yielding

$$\begin{aligned} r^{\beta -\alpha +1}u^p(r)\le -\left( \frac{ru'(r)}{\sqrt{1+\left( u'(r)\right) ^2}}\right) '-(N+\alpha -2)u'(r)=-W'(r). \end{aligned}$$

This shows that W(r) is non-increasing since the solution u is positive. It remains to prove that W is nonnegative. Suppose by contradiction that there exist $r_1>0$ and $m<0$ such that $W(r_1)<m$. By the fact that W(r) is non-increasing, then $W(r)\le m$ for all $r>r_1$, which implies

$$\begin{aligned} \frac{ru'(r)}{\sqrt{1+\left( u'(r)\right) ^2}}\le m-(N+\alpha -2)\left( u(r)-u_\infty \right) \le m, \end{aligned}$$

since $N+\alpha -2>0$ and $u(r)\ge u_\infty$ by monotonicity. Thus, for all $r>r_1$, we obtain

$$\begin{aligned} u'(r)\le m\frac{\sqrt{1+\left( u'(r)\right) ^2}}{r}\le \frac{m}{r} \end{aligned}$$

and integrating from $r_1$ to r, we have

$$\begin{aligned} u(r)-u(r_1)\le m\ln \frac{r}{r_1}. \end{aligned}$$

Since $m<0$, this yields $\lim _{r\rightarrow \infty }u(r)=-\infty$, which contradicts the fact that u is a positive function. Therefore $W(r)\ge 0$ for all $r>0$. This completes the proof of Lemma 4.4. $\square$

A direct consequence of this lemma is that we can determine the rate at which the positive solution u(r) to problem (4.1) decays for all $r\in (0,\infty ]$. As a result, we can use this decay rate to prove our Liouville type result in Theorem 4.1.

Lemma 4.5

Assume $(H_0)$ and $p>1$, and let $u\in C^2(0,\infty )$ be a positive solution of problem (4.1). Then we have

$$\begin{aligned} u(r)\le Cr^{-\frac{\beta -\alpha +2}{p-1}}\quad \text {for }r>0, \end{aligned}$$

(4.4)

where $C=\left[ (N+\beta )(N+\alpha -2)\right] ^{1/(p-1)}$. In particular,

$$\begin{aligned} \lim _{r\rightarrow \infty }u(r)=0. \end{aligned}$$

(4.5)

Proof

Integrating the equation in (4.1) from 0 to r and using $N+\alpha -1>0$ and $N+\beta >0$, we obtain, thanks to the monotonicity of u,

$$\begin{aligned} -\frac{r^{N+\alpha -1}u'(r)}{\sqrt{1+\left( u'(r)\right) ^2}}=\int _0^rs^{N+\beta -1}u^p(s)ds\ge \frac{r^{N+\beta }}{N+\beta }u^p(r)\quad \text {for }r>0. \end{aligned}$$

This, together with Lemma 4.4, yields

$$\begin{aligned} (N+\alpha -2)u(r)\ge -\frac{ru'(r)}{\sqrt{1+\left( u'(r)\right) ^2}} \ge \frac{r^{\beta -\alpha +2}}{N+\beta }u^p(r)\quad \text {for }r>0, \end{aligned}$$

which allows us to derive (4.4) as required. $\square$

Based on the two lemmas above, we can deduce multiple decay estimates for the combination of the solution u(r), the variable r, and the first derivative $u'(r)$, which will be essential in proving our Liouville type result in Theorem 4.2.

Lemma 4.6

Assume $(H_0)$ and $p>1$. If $u\in C^2(0,\infty )$ is a positive solution of (4.1), then there exists a constant $C>0$ such that for all $r>0$, we have

$$\begin{aligned}&\frac{r^{N+\alpha }\left( u'(r)\right) ^2}{\sqrt{1+\left( u'(r)\right) ^2}} \le CY(r), \end{aligned}$$

(4.6)

$$\begin{aligned}&r^{N+\alpha }\left( \sqrt{1+\left( u'(r)\right) ^2}-1\right) \le CY(r), \end{aligned}$$

(4.7)

$$\begin{aligned}&\frac{r^{N+\alpha -1}u(r)|u'(r)|}{\sqrt{1+\left( u'(r)\right) ^2}}\le CY(r), \end{aligned}$$

(4.8)

$$\begin{aligned}&r^{N+\beta }u^{p+1}(r)\le CY(r), \end{aligned}$$

(4.9)

where

$$\begin{aligned} Y(r):=r^{N+\alpha -2-\frac{2(\beta -\alpha +2)}{p-1}}. \end{aligned}$$

In particular, if $p<2^*_{\alpha ,\beta }-1$ we have

$$\begin{aligned} \lim _{r\rightarrow \infty }Y(r)=0. \end{aligned}$$

(4.10)

Proof

We start by using Lemma 4.4 with $u_\infty =0$, so that the nonnegativity of W(r) yields

$$\begin{aligned} 0\le \frac{r|u'(r)|}{\sqrt{1+\left( u'(r)\right) ^2}}\le (N+\alpha -2)u(r), \quad r>0, \end{aligned}$$

(4.11)

which, by Lemma 4.5, implies that

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

(4.12)

From (4.11), we get

$$\begin{aligned} \frac{r^2\left( u'(r)\right) ^2}{\sqrt{1+\left( u'(r)\right) ^2}}\le (N+\alpha -2)^2u^2(r)\sqrt{1+\left( u'(r)\right) ^2} \end{aligned}$$

and by (4.12)$_2$, it follows that for r large

$$\begin{aligned} \frac{r^{N+\alpha }\left( u'(r)\right) ^2}{\sqrt{1+\left( u'(r)\right) ^2}} \le Cr^{N+\alpha -2}u^2(r) \end{aligned}$$

for some constant $C>0$. Applying Lemma 4.5 we have $u(r)\le Cr^{-\frac{\beta -\alpha +2}{p-1}}$, so that (4.6) follows immediately.

Next, we observe that

$$\begin{aligned} r^{N+\alpha }\left( \sqrt{1+\left( u'(r)\right) ^2}-1\right) =\frac{r^{N+\alpha }\left( u'(r)\right) ^2}{\sqrt{1+\left( u'(r)\right) ^2}+1} \le \frac{r^{N+\alpha }\left( u'(r)\right) ^2}{\sqrt{1+\left( u'(r)\right) ^2}}. \end{aligned}$$

Thus, (4.7) follows from (4.6).

We now prove (4.8) using again the nonnegativity of the function W(r) with $u_\infty =0$. From (4.11), we have

$$\begin{aligned} \frac{r^{N+\alpha -1}u(r)|u'(r)|}{\sqrt{1+\left( u'(r)\right) ^2}} \le (N+\alpha -2)r^{N+\alpha -2}u^2(r), \end{aligned}$$

thus, the decaying estimate for u in Lemma 4.5 gives immediately (4.8).

To prove (4.9), we use Lemma 4.5 to get

$$\begin{aligned} r^{N+\beta }u^{p+1}(r)\le Cr^{N+\beta -\frac{(p+1)(\beta -\alpha +2)}{p-1}}=Cr^{N+\alpha -2-\frac{2(\beta -\alpha +2)}{p-1}}. \end{aligned}$$

This proves (4.9).

Finally, to prove (4.10), we observe that when

$$\begin{aligned} p<2^*_{\alpha ,\beta }-1=\frac{N-\alpha +2\beta +2}{N+\alpha -2}, \end{aligned}$$

we have

$$\begin{aligned} p-1<\frac{2(\beta -\alpha +2)}{N+\alpha -2}. \end{aligned}$$

Then

$$\begin{aligned} N+\alpha -2-\frac{2(\beta -\alpha +2)}{p-1}<N+\alpha -2-2(\beta -\alpha +2)\frac{N+\alpha -2}{2(\beta -\alpha +2)}=0, \end{aligned}$$

which implies

$$\begin{aligned} \lim _{r\rightarrow \infty }Y(r)=\lim _{r\rightarrow \infty }r^{N+\alpha -2-\frac{2(\beta -\alpha +2)}{p-1}}=0. \end{aligned}$$

This completes the proof of Lemma 4.6. $\square$

The next lemma examines the characteristics of the first derivative $u'(r)$ of the solution to problem (4.1) for values of r that are suitably small.

Lemma 4.7

Assume $(H_0)$. If $u\in C^2(0,\infty )$ is a positive solution of problem (4.1), then

$$\begin{aligned} |u'(r)|\le \frac{u_0^p}{(N+\beta )\sqrt{\nu }}r^{\beta -\alpha +1}\qquad \text {if }0<r\le \left( \frac{(N+\beta )\sqrt{1-\nu }}{u_0^p}\right) ^{\frac{1}{\beta -\alpha +1}}, \end{aligned}$$

(4.13)

for any $\nu \in (0,1)$. In turn, $|u'(r)|\le \sqrt{(1-\nu )/\nu }$ for r satisfying (4.13)$_2$.

Proof

Integrating the equation in (4.1) from 0 to r and using $N+\alpha -1>0$, we obtain

$$\begin{aligned} -\frac{r^{N+\alpha -1}u'(r)}{\sqrt{1+\left( u'(r)\right) ^2}}=\int _0^rs^{N+\beta -1}u^p(s)ds, \end{aligned}$$

for all $r>0$. By the continuity and monotonicity of u along with the fact that $N+\beta >0$, we derive

$$\begin{aligned} \frac{|u'(r)|}{\sqrt{1+\left( u'(r)\right) ^2}}&\le \frac{u_0^p}{r^{N+\alpha -1}}\int _0^rs^{N+\beta -1}ds =\frac{u_0^p}{N+\beta }r^{\beta -\alpha +1}. \end{aligned}$$

This last inequality implies

$$\begin{aligned} \varphi (r)\left( u'(r)\right) ^2\le \left( \frac{u_0^p}{N+\beta }\right) ^2r^{2(\beta -\alpha +1)}, \end{aligned}$$

(4.14)

with

$$\begin{aligned} \varphi (r):=1-\left( \frac{u_0^p}{N+\beta }\right) ^2r^{2(\beta -\alpha +1)}. \end{aligned}$$

Since $\beta -\alpha +1>0$, we have that $\varphi (r)$ is a decreasing function with $\varphi (r)\le 1$, so that for any $\nu \in (0,1)$, inequality $\varphi (r)\ge \nu$ holds if

$$\begin{aligned} r\le \left( \frac{(N+\beta )\sqrt{1-\nu }}{u_0^p}\right) ^{\frac{1}{\beta -\alpha +1}}. \end{aligned}$$

(4.15)

Hence, when (4.15) is in force, by (4.14), it holds

$$\begin{aligned} \nu \left( u'(r)\right) ^2\le \left( \frac{u_0^p}{N+\beta }\right) ^2r^{2(\beta -\alpha +1)}. \end{aligned}$$

Thus, (4.13) follows. $\square$

Lemma 4.7 with $\nu =1/2$ can be rewritten as follows. This corollary will be used in the proofs of Theorems 1.2 and 1.5.

Corollary 4.8

Assume $(H_0)$. Let $u\in C^2(0,\infty )$ be a positive solution of problem (4.1). For any $M>0$, if

$$\begin{aligned} 0<u_0\le \left( \frac{N+\beta }{\sqrt{2}M^{\beta -\alpha +1}}\right) ^{1/p}:={u_0}_*(M), \end{aligned}$$

(4.16)

then for all $r\in (0,M]$

$$\begin{aligned} |u'(r)|\le \frac{\sqrt{2}u_0^p}{N+\beta }M^{\beta -\alpha +1}. \end{aligned}$$

(4.17)

In turn, $|u'(r)|\le 1$ for $r\in (0,M]$.

Proof

For $r\in (0,M]$, by (4.16), we have $r\le \left[ (N+\beta )/(\sqrt{2}u_0^p)\right] ^{\frac{1}{\beta -\alpha +1}}$. Using Lemma 4.7 with $\nu =1/2$, we obtain

$$\begin{aligned} |u'(r)|\le \frac{\sqrt{2}u_0^p}{N+\beta }r^{\beta -\alpha +1}\le \frac{\sqrt{2}u_0^p}{N+\beta }M^{\beta -\alpha +1}. \end{aligned}$$

$\square$

We also require the use of a Pohozaev type identity to prove Theorem 4.2.

Lemma 4.9

Assume $(H_0)$ and let $u\in C^2(0,\infty )$ be a positive solution of (4.1). Then for any $r>0$ we have

$$\begin{aligned}&\frac{N+\beta +p+1}{p+1}\int _0^r\frac{s^{N+\alpha -1}\left( u'(s)\right) ^2}{\sqrt{1+\left( u'(s)\right) ^2}}ds \nonumber \\&\qquad -(N+\alpha )\int _0^rs^{N+\alpha - 1}\left( \sqrt{1+\left( u'(s)\right) ^2}-1\right) ds \nonumber \\&\quad =\frac{r^{N+\alpha }\left( u'(r)\right) ^2}{\sqrt{1+\left( u'(r)\right) ^2}} -r^{N+\alpha }\left( \sqrt{1+\left( u'(r)\right) ^2}-1\right) +\frac{N+\beta }{p+1}\frac{r^{N+\alpha -1}u(r)u'(r)}{\sqrt{1+\left( u'(r)\right) ^2}} \nonumber \\&\qquad +\frac{1}{p+1}r^{N+\beta }u^{p+1}(r). \end{aligned}$$

(4.18)

Proof

We begin by multiplying the equation in (4.1) by u(r) and then integrating by parts from 0 to r, being $N+\alpha -1>0$, we obtain

$$\begin{aligned} -\frac{r^{N+\alpha -1}u(r)u'(r)}{\sqrt{1+\left( u'(r)\right) ^2}} +\int _0^r\frac{s^{N+\alpha -1}\left( u'(s)\right) ^2}{\sqrt{1+\left( u'(s)\right) ^2}}ds =\int _0^r s^{N+\beta -1} u^{p+1}(s)ds. \end{aligned}$$

(4.19)

Next, multiplying the equation in (4.1) by $ru'(r)$ and integrating by parts from 0 to r, using $N+\alpha >0$, we get

$$\begin{aligned} -\frac{r^{N+\alpha }\left( u'(r)\right) ^2}{\sqrt{1+\left( u'(r)\right) ^2}} +\int _0^r\frac{s^{N+\alpha -1}u'}{\sqrt{1+(u')^2}}\bigl [u'+su''\bigr ]ds=\int _0^r s^{N+\beta } u^pu'ds. \end{aligned}$$

(4.20)

Since

$$\begin{aligned}&\int _0^r\frac{s^{N+\alpha }u'(s)u''(s)}{\sqrt{1+\left( u'(s)\right) ^2}}ds =\int _0^rs^{N+\alpha }\left( \sqrt{1+\left( u'(s)\right) ^2}-1\right) 'ds \nonumber \\&\quad =r^{N+\alpha }\left( \sqrt{1+\left( u'(r)\right) ^2}-1\right) -(N+\alpha )\int _0^rs^{N+\alpha -1}\left( \sqrt{1+\left( u'(s)\right) ^2}-1\right) ds \end{aligned}$$

(4.21)

and

$$\begin{aligned} \int _0^r s^{N+\beta } u^p(s)u'(s)ds =\frac{1}{p+1}r^{N+\beta }u^{p+1}(r) -\frac{N+\beta }{p+1}\int _0^rs^{N+\beta -1}u^{p+1}(s)ds, \end{aligned}$$

(4.22)

by inserting (4.21) and (4.22) in (4.20), we arrive to

$$\begin{aligned}&-\frac{r^{N+\alpha }\left( u'(r)\right) ^2}{\sqrt{1+\left( u'(r)\right) ^2}} +\int _0^r\frac{s^{N+\alpha -1}\left( u'(s)\right) ^2}{\sqrt{1+\left( u'(s)\right) ^2}}ds +r^{N+\alpha }\left( \sqrt{1+\left( u'(r)\right) ^2}-1\right) \\&\qquad -(N+\alpha )\int _0^rs^{N+\alpha - 1}\left( \sqrt{1+\left( u'(s)\right) ^2}-1\right) ds \\&\quad =\frac{1}{p+1}r^{N+\beta }u^{p+1}(r) -\frac{N+\beta }{p+1}\int _0^rs^{N+\beta -1}u^{p+1}(s)ds. \end{aligned}$$

Replacing the last term with (4.19), we get

$$\begin{aligned}&-\frac{r^{N+\alpha }\left( u'(r)\right) ^2}{\sqrt{1+\left( u'(r)\right) ^2}} +\int _0^r\frac{s^{N+\alpha -1}\left( u'(s)\right) ^2}{\sqrt{1+\left( u'(s)\right) ^2}}ds +r^{N+\alpha }\left( \sqrt{1+\left( u'(r)\right) ^2}-1\right) \nonumber \\&\qquad -(N+\alpha )\int _0^rs^{N+\alpha - 1}\left( \sqrt{1+\left( u'(s)\right) ^2}-1\right) ds \nonumber \\&\quad =\frac{1}{p+1}r^{N+\beta }u^{p+1}(r) -\frac{N+\beta }{p+1}\left[ -\frac{r^{N+\alpha -1}u(r)u'(r)}{\sqrt{1+\left( u'(r)\right) ^2}} +\int _0^r\frac{s^{N+\alpha -1}\left( u'(s)\right) ^2}{\sqrt{1+\left( u'(s)\right) ^2}}ds\right] . \end{aligned}$$

By simplifying this expression, the required identity (4.18) is proved. $\square$

We are now ready to complete the proof of the Liouville type Theorems 4.1 and 4.2. We will begin by using Lemma 4.5 to prove Theorem 4.1.

Proof of Theorem 4.1

Assume that $u\in C^2(0,\infty )$ is a positive solution of (4.1), then we have

$$\begin{aligned} \left( {\mathcal {L}}(r):=\frac{r^{N+\alpha -1}u'(r)}{\sqrt{1+\left( u'(r)\right) ^2}}\right) '\le 0\quad \text {for }r\ge 0, \end{aligned}$$

hence ${\mathcal {L}}(r)$ is a decreasing non-positive function in ${\mathbb {R}}^+$. Consequently, for $r\rightarrow \infty$ it follows that ${\mathcal {L}}(r)\rightarrow \text { negative limit (possibly }-\infty )$, namely ${\mathcal {L}}(r)\le -C$ for large r, where C is some positive constant. In turn

$$\begin{aligned} r^{N+\alpha -1}u'(r)\le -C\sqrt{1+\left( u'(r)\right) ^2}\le -C, \quad r>>1. \end{aligned}$$

Integrating this relation from any r large to $\infty$, being $\lim _{r\rightarrow \infty }u(r)=0$ by (4.5), we get

$$\begin{aligned} u(r)\ge \frac{C}{N+\alpha -2}r^{-(N+\alpha -2)} \end{aligned}$$

(4.23)

for all sufficiently large r.

Suppose that $1<p<{2_*}_{\alpha ,\beta }-1=\frac{N+\beta }{N+\alpha -2}$, then

$$\begin{aligned} N+\alpha -2<\frac{\beta -\alpha +2}{p-1}, \end{aligned}$$

so that (4.23) contradicts Lemma 4.5, namely that $u(r)\le Cr^{-\frac{\beta -\alpha +2}{p-1}}$ for all $r>0$. Hence no solution of (4.1) can exist.

It remains to show the same result when $p={2_*}_{\alpha ,\beta }-1$. Using (4.1) and (4.23), we have

$$\begin{aligned} \left( \frac{r^{N+\alpha -1}u'(r)}{\sqrt{1+\left( u'(r)\right) ^2}}\right) '=-r^{N+\beta -1}u^{{2_*}_{\alpha ,\beta }-1}(r)\le -Cr^{-1} \end{aligned}$$

for all sufficiently large r. Integrating this inequality from a sufficiently large s to r, where $r>s$, leads to

$$\begin{aligned} \frac{r^{N+\alpha -1}u'(r)}{\sqrt{1+\left( u'(r)\right) ^2}}-\frac{s^{N+\alpha -1}u'(s)}{\sqrt{1+\left( u'(s)\right) ^2}}\le C\ln \frac{s}{r}. \end{aligned}$$

Therefore, we conclude that

$$\begin{aligned} \frac{r^{N+\alpha -1}u'(r)}{\sqrt{1+\left( u'(r)\right) ^2}}\rightarrow -\infty \quad \text {as }r\rightarrow \infty . \end{aligned}$$

Together with the fact that $\lim _{r\rightarrow \infty }u'(r)=0$ shown in (4.12)$_2$, it follows that

$$\begin{aligned} r^{N+\alpha -1}u'(r)\rightarrow -\infty \quad \text {as }r\rightarrow \infty . \end{aligned}$$

Hence, for any $M>0$, there exists $r_M=r_M(M)>0$ such that

$$\begin{aligned} r^{N+\alpha -1}u'(r)\le -M\quad \text {for any }r\ge r_M. \end{aligned}$$

Integrating this inequality from any fixed value $r\ge r_M$ to $\infty$, and using $\lim _{r\rightarrow \infty }u(r)=0$ by (4.5), we get

$$\begin{aligned} r^{N+\alpha -2}u(r)\ge \frac{M}{N+\alpha -2}. \end{aligned}$$

However, Lemma 4.5 shows that $r^{N+\alpha -2}u(r)\le C$ for all $r>0$, leading to a contradiction being the arbitrariness of M. Therefore, the proof is complete. $\square$

Finally we use Lemmas 4.6, 4.7 and 4.9 to finish the proof of Theorem 4.2.

Proof of Theorem 4.2

Suppose that $u\in C^2(0,\infty )$ is a positive solution of problem (4.1). From Lemma 4.9, the identity (4.18) holds. Taking $r\rightarrow \infty$, all the terms in the right hand side of (4.18) tend to zero by (4.6), (4.7), (4.8) and (4.9) in Lemma 4.6, respectively. Thus

$$\begin{aligned}&\frac{N+\beta +p+1}{p+1}\int _0^\infty \frac{s^{N+\alpha -1}\left( u'(s)\right) ^2}{\sqrt{1+\left( u'(s)\right) ^2}}ds \nonumber \\&\qquad \qquad -(N+\alpha )\int _0^\infty s^{N+\alpha - 1}\left( \sqrt{1+\left( u'(s)\right) ^2}-1\right) ds=0. \end{aligned}$$

(4.24)

Since

$$\begin{aligned} \sqrt{1+\left( u'(s)\right) ^2}-1=\frac{\left( u'(s)\right) ^2}{\sqrt{1+\left( u'(s)\right) ^2}+1}, \end{aligned}$$

we can rewrite (4.24) as

$$\begin{aligned} \int _0^\infty s^{N+\alpha -1}\left( u'(s)\right) ^2\left[ \frac{N+\beta +p+1}{(p+1)\sqrt{1+\left( u'(s)\right) ^2}}-\frac{N+\alpha }{\sqrt{1+\left( u'(s)\right) ^2}+1}\right] ds=0. \end{aligned}$$

(4.25)

If we can prove

$$\begin{aligned} \frac{N+\beta +p+1}{(p+1)\sqrt{1+\left( u'(r)\right) ^2}}\ge \frac{N+\alpha }{\sqrt{1+\left( u'(r)\right) ^2}+1}\quad \text {for all } r>0, \end{aligned}$$

(4.26)

namely that the integrand on the left hand side of (4.25) is nonnegative, in turn, by (4.25), necessarily

$$\begin{aligned} \frac{N+\beta +p+1}{(p+1)\sqrt{1+\left( u'(r)\right) ^2}}\equiv \frac{N+\alpha }{\sqrt{1+\left( u'(r)\right) ^2}+1}\quad \text {for all } r>0, \end{aligned}$$

which, thanks to the asymptotic estimates (4.12)$_2$ (or alternatively the continuity of u at zero), yields

$$\begin{aligned} \frac{N+\beta +p+1}{p+1}=\frac{N+\alpha }{2}. \end{aligned}$$

In particular, the above equality reads as

$$\begin{aligned} p+1=\frac{2(N+\beta )}{N+\alpha -2}=2^*_{\alpha ,\beta }, \end{aligned}$$

giving a contradiction since $p<2^*_{\alpha ,\beta }-1$, and thus, we can conclude that problem (4.1) does not admit any positive solution.

Hence, it remains to prove (4.26), or equivalently

$$\begin{aligned} \sqrt{1+\left( u'(r)\right) ^2}\le \Lambda :=\frac{N+\beta +p+1}{p(N+\alpha -1)+\alpha -\beta -1}\quad \text {for all } r>0. \end{aligned}$$

(4.27)

In particular, the positivity of $\Lambda$ follows from $p>2_{*\alpha ,\beta }-1$ and $(H_0)$, while $\Lambda >1$ by $p<2^*_{\alpha ,\beta }-1$.

To prove (4.27), we observe from (4.4) and (4.11) that

$$\begin{aligned} \frac{|u'(r)|}{\sqrt{1+\left( u'(r)\right) ^2}}\le (N+\alpha -2)^{\frac{p}{p-1}}(N+\beta )^{\frac{1}{p-1}}r^{-\frac{\beta -\alpha +2}{p-1}-1},\quad r>0, \end{aligned}$$

which implies

$$\begin{aligned} \left( u'(r)\right) ^2\le \frac{(N+\alpha -2)^{\frac{2p}{p-1}}(N+\beta )^{\frac{2}{p-1}}r^{-\frac{2(\beta -\alpha +2)}{p-1}-2}}{1-(N+\alpha -2)^{\frac{2p}{p-1}}(N+\beta )^{\frac{2}{p-1}}r^{-\frac{2(\beta -\alpha +2)}{p-1}-2}}\le \Lambda ^2-1, \end{aligned}$$

if

$$\begin{aligned} r\ge \left[ (N+\alpha -2)^{\frac{p}{p-1}}(N+\beta )^{\frac{1}{p-1}}\frac{\Lambda }{\sqrt{\Lambda ^2-1}}\right] ^{\frac{p-1}{p+\beta -\alpha +1}}:=K. \end{aligned}$$

Thus (4.27) holds for all $r\ge K$. To prove (4.27) for $r\in (0,K)$, we make use of Lemma 4.7 with $\nu =1/\Lambda ^2$, which gives immediately

$$\begin{aligned} |u'(r)|\le \frac{\Lambda K^{\beta -\alpha +1}}{N+\beta }u_0^{p}\le \sqrt{\Lambda ^2-1}\quad \text {for } r\in (0,K) \end{aligned}$$

provided that

$$\begin{aligned} K\le \left( \frac{(N+\beta )\sqrt{\Lambda ^2-1}}{\Lambda u_0^p}\right) ^{\frac{1}{\beta -\alpha +1}}. \end{aligned}$$

(4.28)

Note that (4.28) is equivalent to

$$\begin{aligned} 0<u_0\le u_0^*:=\left( \frac{(N+\beta )\sqrt{\Lambda ^2-1}}{\Lambda K^{\beta -\alpha +1}}\right) ^{\frac{1}{p}} . \end{aligned}$$

Replacing the values of K and $\Lambda$ in $u_0^*$ above, we immediately find (4.3). Thus, the proof of (4.27) is so completed. $\square$

5 Oscillatory solutions

In this section, our purpose is to determine the conditions that guarantee every solution of the Cauchy problem (4.1) to be oscillatory.

Let $\mathcal {S}$ denote the set of solutions of (4.1), with $u_0\ne 0$, which can be continued to the entire ${\mathbb {R}}_0^+$, namely,

$$\begin{aligned} \mathcal {S}=\left\{ u:{\mathbb {R}}_0^+\rightarrow {\mathbb {R}}\big |u\text { satisfies }(4.1) \text { with }u_0\ne 0 \right\} . \end{aligned}$$

Theorem 5.1

Assume $(H_0)$ and $1<p\le {2_*}_{\alpha ,\beta }-1$. Then every $u\in \mathcal {S}$ is oscillatory.

Theorem 5.2

Assume $(H_0)$ and ${2_*}_{\alpha ,\beta }-1<p<2^*_{\alpha ,\beta }-1$. Then every $u\in \mathcal {S}$ is oscillatory for $|u_0|\in (0,u_0^*]$, where $u_0^*$ is defined by (4.3).

Proof of Theorem 5.1

The proof is based on that of Theorem 3.1 in [14] adapted to case of different weights. Fix $u\in \mathcal {S}$ and first assume that $u_0>0$. We begin by showing that there exists $r_1>0$ such that

$$\begin{aligned} u(r)>0\text { in }[0,r_1),\quad u(r_1)=0,\quad u'(r)<0\text { in }(0,r_1]. \end{aligned}$$

(5.1)

It is obvious from Theorem 4.1 that conditions (5.1)$_{1,2}$ hold. Integrating the equation in (4.1) from 0 to $r\le r_1$, we obtain

$$\begin{aligned} -\frac{r^{N+\alpha -1}u'(r)}{\sqrt{1+\left( u'(r)\right) ^2}}=\int _0^rs^{N+\beta -1}|u(s)|^{p-1}u(s)ds, \end{aligned}$$

which implies that $u'(r)<0$ in $(0,r_1]$. This proves condition (5.1)$_3$ and completes the proof of (5.1).

Next we show that there exists $s_1>r_1$ such that

$$\begin{aligned} u'(r)<0\text { in }(0,s_1),\quad u'(s_1)=0,\quad u(s_1)<0. \end{aligned}$$

(5.2)

Assume for contradiction that (5.2) fails. Then necessarily $u'<0$ in ${\mathbb {R}}^+$ by (5.1) which forces $u,u'<0$ in a right neighbourhood of $r_1$. In turn, $u(r)<0$ in $(r_1,\infty )$ and

$$\begin{aligned} \lim _{r\rightarrow \infty }u(r)=\eta \in [-\infty ,0), \end{aligned}$$

(5.3)

$\frac{r^{N+\alpha -1}u'(r)}{\sqrt{1+\left( u'(r)\right) ^2}}$ is increasing in $(r_1,\infty )$ in view of (4.1), and

$$\begin{aligned} \lim _{r\rightarrow \infty }\frac{r^{N+\alpha -1}u'(r)}{\sqrt{1+\left( u'(r)\right) ^2}}= a_0\in (-\infty ,0]. \end{aligned}$$

(5.4)

Integrating the equation in (4.1) from $r_1$ to r, we obtain

$$\begin{aligned} \frac{r^{N+\alpha -1}u'(r)}{\sqrt{1+\left( u'(r)\right) ^2}}-\frac{r_1^{N+\alpha -1}u'(r_1)}{\sqrt{1+\left( u'(r_1)\right) ^2}}=\int _{r_1}^rs^{N+\beta -1}|u(s)|^pds. \end{aligned}$$

(5.5)

By (5.3), we get that the right hand side of (5.5) tends to $\infty$ as r tends to $\infty$ while the left hand side is bounded in view of (5.4). This yields a contraction, and thus (5.2) is proved.

Now consider the problem (4.1) in $(s_1,\infty )$ with 0 replaced by $s_1$ and $u(s_1)<0$. Repeating the argument above, we find two points $r_2,s_2$ with $s_2>r_2>s_1$ such that

$$\begin{aligned} u(r)<0\text { in }(s_1,r_2),\quad u(r_2)=0,\quad u'(r)>0\text { in }(s_1,s_2),\quad u'(s_2)=0. \end{aligned}$$

We can now iterate the argument yielding that u is oscillatory.

Analogously, we can argue when $u\in \mathcal {S}$ and $u_0<0$. $\square$

Proof of Theorem 5.2

We can repeat word by word the proof of Theorem 5.1 except for the use of Theorem 4.1 which needs to be replaced by Theorem 4.2. $\square$

6 Proof of Theorem 1.2

In this section, we establish the existence of positive solutions for problem (1.1), which is stated in Theorem 1.2. To achieve this, we utilize oscillatory solutions obtained in Sect. 5 along with the maximal existence interval and the features of solutions to problem (4.1).

To begin with, we need to analyze the existence as well as some properties of the local solution to the following auxiliary problem

$$\begin{aligned} {\left\{ \begin{array}{ll} -\left( \frac{r^{N+\alpha -1}w'}{\sqrt{1+\varepsilon w'^2}}\right) '= r^{N+\beta -1}|w|^{p-1}w,\quad r>0,\\ w(0)=\bar{\xi }, \quad w'(0)=0, \end{array}\right. } \end{aligned}$$

(6.1)

where $\varepsilon >0$.

The local existence of solutions to problem (6.1) follows in a standard way, see [24, Proposition 1]. For the sake of completeness, we present the proposition below.

Proposition 6.1

Assume $(H_0)$. Let $p>1$ and $\bar{\xi }>0$. Then, problem (6.1) has a unique continuously differentiable solution w(r) on some interval $0\le r\le r_0$.

Proof

Define the operator T by

$$\begin{aligned} T[w](r)=\bar{\xi }-\int _0^r\phi \left( \frac{1}{t^{N+\alpha -1}}\int _0^ts^{N+\beta -1}|w(s)|^{p-1}w(s)ds\right) dt,\quad r\in [0,r_0], \end{aligned}$$

(6.2)

where

$$\begin{aligned} \phi (\tau )=\frac{\tau }{\sqrt{1-\varepsilon \tau ^2}}\quad \text {for }\tau \in \left( 0,1/\sqrt{\varepsilon }\right) . \end{aligned}$$

We claim that T is a contraction map acting on the space

$$\begin{aligned} S=\left\{ w\in C\left( [0,r_0]\right) :\Vert w(r)-\bar{\xi }\Vert <\frac{1}{2}\bar{\xi }\right\} \end{aligned}$$

where $\Vert \cdot \Vert$ is the uniform norm, $\bar{\xi }>0$ and $r_0$ to be chosen suitably small.

Let us first show that

$$\begin{aligned} T[S]\subset S. \end{aligned}$$

Consider $w\in S$, thus $0<\frac{1}{2}\bar{\xi }<w(r)<\frac{3}{2}\bar{\xi }$ for all $r\in [0,r_0]$. Now, take $t\le r_0$, using $N+\beta >0$ and $\beta -\alpha +1>0$, we have

$$\begin{aligned} \frac{1}{t^{N+\alpha -1}}\int _0^ts^{N+\beta -1}w^p(s)ds <\left( \frac{3\bar{\xi }}{2}\right) ^p\frac{r_0^{\beta -\alpha +1}}{N+\beta }:=b. \end{aligned}$$

(6.3)

For $r_0$ suitably small, we have $b<1/\sqrt{\varepsilon }$ by $\beta -\alpha +1>0$, which implies that $\phi (b)$ is finite and positive. Since

$$\begin{aligned} \phi '(\tau )=\frac{1}{\left( 1-\varepsilon \tau ^2\right) ^{3/2}}>0\quad \text {for }\tau \in \left( 0,1/\sqrt{\varepsilon }\right) , \end{aligned}$$

then $\phi$ is increasing in $\left( 0,1/\sqrt{\varepsilon }\right)$. We can use this to bound the integral

$$\begin{aligned} \big |T[w](r)-\bar{\xi }\big |&=\int _0^r\phi \left( \frac{1}{t^{N+\alpha -1}}\int _0^ts^{N+\beta - 1}w^p(s)ds\right) dt\le \int _0^r\phi (b)dt \le r_0\phi (b). \end{aligned}$$

(6.4)

Using the expression for $\phi (b)$, we find

$$\begin{aligned} \big |T[w](r)-\bar{\xi }\big |\le \frac{r_0b}{\sqrt{1-\varepsilon b^2}} =\frac{\left( \frac{3}{2}\right) ^pr_0^{\beta -\alpha +2}\bar{\xi }^{p-1}}{(N+\beta )\sqrt{1-\varepsilon b^2}}\bar{\xi }. \end{aligned}$$

(6.5)

For $r_0$ suitably small and hence b small, we have

$$\begin{aligned} \frac{\left( \frac{3}{2}\right) ^pr_0^{\beta -\alpha +2}\bar{\xi }^{p-1}}{(N+\beta )\sqrt{1-\varepsilon b^2}}<\frac{1}{2}. \end{aligned}$$

This inequality, together with (6.5), gives

$$\begin{aligned} \big |T[w](r)-\bar{\xi }\big |<\frac{1}{2}\bar{\xi }, \end{aligned}$$

which implies $T[w]\in S$ and hence $T[S]\subset S$.

Then, let us show that T is a contraction map. For $w_1,w_2\in S$, we have

$$\begin{aligned} \big |T[w_1]-T[w_2]\big |=\left| \int _0^r\left( \phi (t_2)-\phi (t_1)\right) dt\right| , \end{aligned}$$

(6.6)

where we defined

$$\begin{aligned} t_i:=\frac{1}{t^{N+\alpha -1}}\int _0^ts^{N+\beta -1}w_i^p(s)ds,\quad i=1,2. \end{aligned}$$

By (6.3), we have

$$\begin{aligned} t_i<b<1/\sqrt{\varepsilon },\quad i=1,2 \end{aligned}$$

(6.7)

for $r_0$ suitably small. We estimate $|t_2-t_1|$ as follows

$$\begin{aligned} |t_2-t_1|&\le \frac{1}{t^{N+\alpha -1}}\int _0^ts^{N+\beta -1}\left| w_1^p(s)-w_2^p(s)\right| ds \nonumber \\&\le p\left( \frac{3\bar{\xi }}{2}\right) ^{p-1}\Vert w_1-w_2\Vert \frac{1}{t^{N+\alpha -1}}\int _0^ts^{N+\beta -1}ds \nonumber \\&\le p\left( \frac{3\bar{\xi }}{2}\right) ^{p-1}\frac{r_0^{\beta -\alpha +1}}{N+\beta }\Vert w_1-w_2\Vert =\frac{2bp}{3\bar{\xi }}\Vert w_1-w_2\Vert , \end{aligned}$$

(6.8)

where we have used the fact that

$$\begin{aligned} |w_1^p-w_2^p|\le p\sup \left\{ w_1^{p-1},w_2^{p-1}\right\} |w_1-w_2|. \end{aligned}$$

Now we compute

$$\begin{aligned} \phi (t_2)-\phi (t_1)&=\frac{(t_2-t_1)\sqrt{1-\varepsilon t_1^2}+t_1\left( \sqrt{1-\varepsilon t_1^2}-\sqrt{1-\varepsilon t_2^2}\right) }{\sqrt{\left( 1-\varepsilon t_1^2\right) \left( 1-\varepsilon t_2^2\right) }} \nonumber \\&=(t_2-t_1)\frac{\sqrt{1-\varepsilon t_1^2}+\frac{\varepsilon t_1(t_2+t_1)}{\sqrt{1-\varepsilon t_1^2}+\sqrt{1-\varepsilon t_2^2}}}{\sqrt{\left( 1-\varepsilon t_1^2\right) \left( 1-\varepsilon t_2^2\right) }}. \end{aligned}$$

(6.9)

Using (6.7), (6.8) and (6.9), we obtain

$$\begin{aligned} \left| \phi (t_2)-\phi (t_1)\right| \le \frac{2bp}{3\bar{\xi }}\frac{1+\frac{\varepsilon b^2}{\sqrt{1-\varepsilon b^2}}}{1-\varepsilon b^2}\Vert w_1-w_2\Vert . \end{aligned}$$

This, together with (6.6), yields

$$\begin{aligned} \big |T[w_1]-T[w_2]\big |\le \int _0^r\left| \phi (t_2)-\phi (t_1)\right| dt\le \frac{2bpr_0}{3\bar{\xi }}\frac{1+\frac{\varepsilon b^2}{\sqrt{1-\varepsilon b^2}}}{1-\varepsilon b^2}\Vert w_1-w_2\Vert . \end{aligned}$$

For $r_0$ suitably small and hence b small, we have

$$\begin{aligned} \frac{2bpr_0}{3\bar{\xi }}\frac{1+\frac{\varepsilon b^2}{\sqrt{1-\varepsilon b^2}}}{1-\varepsilon b^2}<1. \end{aligned}$$

Thus T is a contraction map. By the Banach Fixed Point Theorem, T has a unique fixed point w in S which is continuously differentiable for $r\ge 0$ by the representation (6.2).

In the rest of the proof, we verify that the fixed point w satisfies (4.1) for $0\le r\le r_0$. From (6.2), we obtain

$$\begin{aligned} w(r)=\bar{\xi }-\int _0^r\phi \left( \frac{1}{t^{N+\alpha -1}}\int _0^ts^{N+\beta -1}|w(s)|^{p-1}w(s)ds\right) dt,\quad r\in [0,r_0].\nonumber \\ \end{aligned}$$

(6.10)

Taking the derivative of both sides of (6.10) with respect to r, we get

$$\begin{aligned} w'(r)=-\phi \left( \frac{1}{r^{N+\alpha -1}}\int _0^rs^{N+\beta -1}|w(s)|^{p-1}w(s)ds\right) . \end{aligned}$$

(6.11)

Since $\phi (\tau )=\frac{\tau }{\sqrt{1-\varepsilon \tau ^2}}$ and $\phi '>0$, we know that $\phi$ is an odd invertible function and that its inverse is given by $\psi (\sigma )=\frac{\sigma }{\sqrt{1+\varepsilon \sigma ^2}}$. Thus, we can use (6.11) to obtain

$$\begin{aligned} \psi (w'(r))=-\frac{1}{r^{N+\alpha -1}}\int _0^rs^{N+\beta -1}|w(s)|^{p-1}w(s)ds, \end{aligned}$$

which implies that

$$\begin{aligned} -\frac{r^{N+\alpha -1}w'}{\sqrt{1+\varepsilon w'^2}}=\int _0^rs^{N+\beta -1}|w|^{p-1}wds. \end{aligned}$$

(6.12)

Since the right hand side is continuously differentiable in r, it follows that

$$\begin{aligned} \frac{r^{N+\alpha -1}w'}{\sqrt{1+\varepsilon w'^2}}\in C^1(0,r_0), \end{aligned}$$

so that w is a classical solution of problem (6.1). This proves Proposition 6.1. $\square$

Next, we shall prove that for given initial data $u_0$, the local solution u of problem (4.1), namely of problem (6.1) with $\varepsilon =1$, is sign-changing in its maximal domain of existence. To this end, we give the following two lemmas, cfr. [1, Lemmas 2.2 and 2.3].

We will denote, if it exists, by $R_0(u_0)>0$ the point such that

$$\begin{aligned} u\left( R_0(u_0)\right) =0\quad \text {and}\quad u(r)>0\text { for }r\in \left[ 0,R_0(u_0)\right) . \end{aligned}$$

Lemma 6.2

Assume $(H_0)$. Let $1<p<2^*_{\alpha ,\beta }-1$. Then, there exists $\bar{\xi }>0$ such that for all $\delta >0$ sufficiently small, there exists $\bar{\varepsilon }>0$ for which the unique local solution w(r) of problem (6.1) corresponding to $\varepsilon \in (0,\bar{\varepsilon }]$, defined in $[0,\mathcal R_\varepsilon )$, $\mathcal R_\varepsilon <\infty$, is sign-changing and $R_0(\bar{\xi })\in [ \overline{\mathcal R} -\delta ,\overline{\mathcal R}+\delta ],$ for some $\overline{\mathcal R}<\mathcal R_\varepsilon$.

Proof

Assume that w(r) is the local solution of problem (6.1), then it satisfies

$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} \left( \frac{w'}{\sqrt{1+\varepsilon w'^2}}\right) '+\frac{N+\alpha -1}{r}\frac{w'}{\sqrt{1+\varepsilon w'^2}}+r^{\beta -\alpha }|w|^{p-1}w=0,\\ w'(0)=0,\\ w(0)=\bar{\xi }, \end{array}\right. } \end{aligned} \end{aligned}$$

(6.13)

with $\bar{\xi }$ to be chosen. Let $[0,{\mathcal {R}}_\varepsilon )$ its maximal domain of existence and assume that $\mathcal R_\varepsilon <\infty$ (if $\mathcal R_\varepsilon =\infty$ we have done).

Take any $\overline{\mathcal R}<\mathcal R_\varepsilon$ and consider the following Dirichlet boundary value problem

$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} \text {div}\left( |x|^\alpha \nabla v\right) +|x|^\beta v^p=0&{}\text {in }B_{\overline{\mathcal R}},\\ v>0&{}\text {in }B_{\overline{\mathcal R}},\\ v=0&{}\text {on }\partial B_{\overline{\mathcal R}}. \end{array}\right. } \end{aligned}\end{aligned}$$

(6.14)

In particular, $v(\overline{{\mathcal {R}}})=0$ and for $\delta$ small enough w is well defined in $(\overline{{\mathcal {R}}}-\delta , \overline{{\mathcal {R}}}+\delta )\subset (0,{\mathcal {R}}_\varepsilon )$. By [2, Theorem 1.1], it is established that under condition $(H_0)$, problem (6.14) has at least one positive radial solution v satisfying

$$\begin{aligned} {\left\{ \begin{array}{ll} v''+\frac{N+\alpha -1}{r}v'+r^{\beta -\alpha }|v|^{p-1}v=0,\quad 0<r<\overline{\mathcal R},\\ v'(0)=0,\\ v(0)=\bar{\xi }, \quad v(\overline{\mathcal R})=0. \end{array}\right. } \end{aligned}$$

(6.15)

Now, consider problem (6.13) for $\bar{\xi }=v(0)$, and noting that the equation in (6.13) is a regular perturbation of the equation in (6.15), we can deduce for $\delta >0$ sufficiently small, the existence of a sufficiently small $\bar{\varepsilon }$ such that, for any $\varepsilon \in (0,\bar{\varepsilon }]$, the solution w to (6.13) intersects the axis at least once within the interval $[\overline{\mathcal R}-\delta ,\overline{\mathcal R}+\delta ]$, hence w is sign-changing in $[0,\mathcal R_\varepsilon )$. $\square$

Lemma 6.3

Assume $(H_0)$. Let $1<p<2^*_{\alpha ,\beta }-1$, $\bar{\delta }>0$ sufficiently small, and let $\bar{\xi }$ and $\bar{\varepsilon }$ be given in Lemma 6.2. If w is the sign-changing solution of (6.13), then the function

$$\begin{aligned} u(r)=\varepsilon ^{\frac{\beta -\alpha +2}{2(p+\beta -\alpha +1)}}w\left( \varepsilon ^{\frac{p-1}{2(p+\beta -\alpha +1)}}r\right) \end{aligned}$$

(6.16)

for any $\varepsilon \in (0,\bar{\varepsilon }]$ is the unique local sign-changing solution of problem (4.1) with $\varepsilon$ such that

$$\begin{aligned} u_0=\varepsilon ^{\frac{\beta -\alpha +2}{2(p+\beta -\alpha +1)}}\bar{\xi } \quad \text {with}\quad R_0(u_0)\in \bigl (\varepsilon ^{-\frac{p-1}{2(p+\beta -\alpha +1)}}(\overline{\mathcal R}-\bar{\delta }) , \varepsilon ^{-\frac{p-1}{2(p+\beta -\alpha +1)}}(\overline{\mathcal R}+\bar{\delta })\bigr ). \end{aligned}$$

(6.17)

In particular, $R_0(u_0)=\varepsilon ^{-\frac{p-1}{2(p+\beta -\alpha +1)}}R_0(\bar{\xi })$.

Proof

Set $t=\varepsilon ^{\frac{p-1}{2(p+\beta -\alpha +1)}}r$, we have $w'(t)=\varepsilon ^{-\frac{1}{2}}u'(r)$ and $1+\varepsilon w'^2(t)=1+u'^2(r)$. Moreover, we compute

$$\begin{aligned}{} & {} \left( \frac{w'(t)}{\sqrt{1+\varepsilon w'^2(t)}}\right) '=\varepsilon ^{-\frac{1}{2}-\frac{p-1}{2(p+\beta -\alpha +1)}}\left( \frac{u'(r)}{\sqrt{1+u'^2(r)}}\right) ',\\{} & {} \qquad \frac{N+\alpha -1}{t}\frac{w'(t)}{\sqrt{1+\varepsilon w'^2(t)}}=\varepsilon ^{-\frac{1}{2}-\frac{p-1}{2(p+\beta -\alpha +1)}}\frac{N+\alpha -1}{r}\frac{u'(r)}{\sqrt{1+u'^2(r)}}, \end{aligned}$$

and

$$\begin{aligned} t^{\beta -\alpha }|w(t)|^{p-1}w(t)=\varepsilon ^{\frac{(p-1)(\beta -\alpha )-p(\beta -\alpha +2)}{2(p+\beta -\alpha +1)}}r^{\beta -\alpha }|u(r)|^{p-1}u(r). \end{aligned}$$

Note that

$$\begin{aligned} \frac{(p-1)(\beta -\alpha )-p(\beta -\alpha +2)}{2(p+\beta -\alpha +1)}+\frac{1}{2}+\frac{p-1}{2(p+\beta -\alpha +1)}=0. \end{aligned}$$

Since w(t) satisfies (6.13), we get that u(r) satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} \left( \frac{u'}{\sqrt{1+ u'^2}}\right) '+\frac{N+\alpha -1}{r}\frac{u'}{\sqrt{1+ u'^2}}+r^{\beta -\alpha }|u|^{p-1}u=0,\\ u'(0)=0,\\ u(0)=\varepsilon ^{\frac{\beta -\alpha +2}{2(p+\beta -\alpha +1)}}\bar{\xi }. \end{array}\right. } \end{aligned}$$

Thus, u is a local sign-changing solution of problem (4.1). $\square$

Now, inspired by [8, Proposition 2.1], but facing the further difficulty due to the presence of different weights, we use Theorems 5.1 and 5.2 and Lemmas 6.2 and 6.3 to prove our existence result for the Dirichlet problem (1.1).

Proof of Theorem 1.2

Denote by $\left[ 0, \mathcal {R}\right)$ the maximal interval of existence of the corresponding solution u of problem (4.1). We claim that u admits a zero in $(0,\mathcal {R})$. Indeed, if $\mathcal {R}=\infty$, we divided the proof in two cases: $p\in ({2_*}_{\alpha ,\beta }-1,2^*_{\alpha ,\beta }-1)$ and $p\in (1,{2_*}_{\alpha ,\beta }-1]$. In the first case, we let $u_0$ be fixed first smaller than $u_0^*$ given in (4.3), then the claim follows from Theorem 5.2. In the second case, the claim follows directly from Theorem 5.1.

If $\mathcal {R}<\infty$, let assume that u has no zeros in $[0,{\mathcal {R}})$, consequently $0<u(r)\le u_0$ in $[0,{\mathcal {R}})$. Observe that, because of the boundedness of the solution it follows $\limsup _{r\rightarrow \mathcal R^+}|u'(r)|=\lim _{r\rightarrow \mathcal R^+} |u'(r)|=\infty$ and

$$\begin{aligned} 1=\lim _{r\rightarrow \mathcal R^+}\frac{u'(r)}{\sqrt{1+\left( u'(r)\right) ^2}}=\mathcal R^{-(N+\alpha -1)}\int _{0}^{\mathcal R}s^{N+\beta -1}u^p(s)ds, \end{aligned}$$

in turn by monotonicity of u it follows ${\mathcal {R}}^{\beta -\alpha +1}\ge u_0^{-p}(N+\beta )\ge (u_0^*)^{-p}(N+\beta )$.

On the other hand u can be written as in (6.16), where w is the solution given in Lemma 6.2 with $\overline{\xi }= v(0)$ and v is the positive radial solution of problem (6.14) for a suitable $\overline{\mathcal R}$ (thus $\overline{\xi }$ is fixed) and $\varepsilon =\varepsilon (u_0,\overline{\xi })$. By Lemma 6.3, we know that u has a zero in an interval contained in $[0,{\mathcal {R}})$, thus we have reached a contradiction.

In turn, we conclude that for all $u_0$, say $u_0>0$, there exists $r_1=r_1(u_0)>0$ such that

$$\begin{aligned} {\left\{ \begin{array}{ll} -\left( \frac{r^{N+\alpha -1}u'(r)}{\sqrt{1+\left( u'(r)\right) ^2}}\right) '= r^{N+\beta -1}u^p(r),\quad r\in (0,r_1),\\ u(r_1)=0, \quad u(r)>0 \text { in } [0,r_1). \end{array}\right. } \end{aligned}$$

(6.18)

Next, we claim that the function $u_0\rightarrow r_1(u_0)$ is continuously differentiable. We can prove this by using the Implicit Function Theorem. Since the $C^1$ function $u(r,u_0)$ satisfies $u(r_1(u_0),u_0)=0$ and $u'(r_1(u_0),u_0)\ne 0$, the Implicit Function Theorem guarantees the existence of $\varepsilon \in (0,u_0)$ such that, for every ${\widetilde{u}}_0\in (u_0-\varepsilon ,u_0+\varepsilon )$, the function $r_1({\widetilde{u}}_0)$ is continuously differentiable and satisfies $u(r_1({\widetilde{u}}_0),{\widetilde{u}}_0)=0$.

To complete the proof of the theorem, it is sufficient to prove that

$$\begin{aligned} \lim _{u_0\rightarrow 0^+}r_1=\infty . \end{aligned}$$

(6.19)

Assume by contradiction that there exists $C>0$ such that for all $\sigma >0$ there exists $u_{0,\sigma }$ with $0<u_{0,\sigma }<\sigma$ such that $r_\sigma =r_1(u_{0,\sigma })< C$. In particular, it is possible to choose $\sigma$ so small that

$$\begin{aligned} u_{0,\sigma }\le \min \left\{ \left( \frac{N+\beta }{\sqrt{2}\,C^{\beta -\alpha +1}}\right) ^{\frac{1}{p}},\left( \frac{N+\beta }{\sqrt{2}\,C^{\beta -\alpha +2}}\right) ^{\frac{1}{p-1}}\right\} . \end{aligned}$$

In turn

$$\begin{aligned} u_{0,\sigma }< \min \left\{ \left( \frac{N+\beta }{\sqrt{2}\,r_\sigma ^{\beta -\alpha +1}}\right) ^{\frac{1}{p}},\left( \frac{N+\beta }{\sqrt{2}\,r_\sigma ^{\beta -\alpha +2}}\right) ^{\frac{1}{p-1}}\right\} . \end{aligned}$$

(6.20)

Applying Corollary 4.8 with $M=r_\sigma$ we obtain

$$\begin{aligned} -u'(r)\le \frac{\sqrt{2}u_{0,\sigma }^p}{N+\beta }r_\sigma ^{\beta -\alpha +1} \end{aligned}$$

(6.21)

being $u'(r)<0$ for $r\in (0,r_\sigma )$. Integrating (6.21) with respect to r from 0 to $r_\sigma$ we find

$$\begin{aligned} -u(r_\sigma )+u_{0,\sigma }\le \frac{\sqrt{2}u_{0,\sigma }^p}{N+\beta }r_\sigma ^{\beta -\alpha +2}. \end{aligned}$$

Using $u(r_\sigma )=0$, we get

$$\begin{aligned} u_{0,\sigma }^{p-1}\ge \frac{N+\beta }{\sqrt{2}r_\sigma ^{\beta -\alpha +2}} \end{aligned}$$

which contradicts (6.20), yielding (6.19). In conclusion we obtain that there exists $R^*\ge 0$ such that the Dirichlet problem (1.1) has at least one positive solution in $B_R$ with $R>R^*$. $\square$

7 Proof of Theorem 1.5

Throughout this section, for simplicity in the notation, for the initial value of the solution we use $\xi$ instead of $u_0$.

Theorem 7.1

Assume $(H_0)$ and $\beta \ge \alpha$. Consider the initial value problem

$$\begin{aligned} {\left\{ \begin{array}{ll} -\left( \frac{r^{N+\alpha -1}u'(r)}{\sqrt{1+\left( u'(r)\right) ^2}}\right) '= r^{N+\beta -1}g(u(r)),\quad r>0,\\ u(0)=\xi ,\quad u'(0)=0, \end{array}\right. } \end{aligned}$$

(7.1)

where g(u) is a given function on $[0,\infty )$ with $g(0)=0$, $g'(u)>0$ and $g''(u)\ge 0$ for $u>0$. If $\xi >0$ and

$$\begin{aligned} \frac{1}{\xi ^{\beta -\alpha +2}}\le g'(\xi )\le \left( \frac{g(\xi )}{\mathcal {C}}\right) ^{\frac{\beta -\alpha +2}{\beta -\alpha +1}}, \end{aligned}$$

(7.2)

where $\mathcal {C}=(N+\beta )2^{\frac{3(\beta -\alpha )}{2}+1}$, then the solution cannot be continued to the entire ${\mathbb {R}}_0^+$.

Remark 7.2

Theorem 7.1 above covers the result obtained by Serrin [32, Theorem 2], since $\mathcal {C}=2N$ when $\beta =\alpha =0$, that yields Serrin’s condition. In addition, the result is completely new when $\beta >\alpha$.

Proof of Theorem 7.1

Let us assume by contradiction that u can be continued to the entire ${\mathbb {R}}_0^+$. Since $u'(0)=0$ and $u(0)>0$, then, being g positive in ${\mathbb {R}}^+$, we have $u'(r)<0$ for all $r>0$. Thus, we get from (7.1) that

$$\begin{aligned} -u''(r)=\left( 1+\left( u'(r)\right) ^2\right) ^{\frac{3}{2}}\left[ r^{\beta -\alpha }g(u(r))-\frac{N+\alpha -1}{r}\frac{|u'(r)|}{\sqrt{1+\left( u'(r)\right) ^2}}\right] ,\quad r>0. \end{aligned}$$

(7.3)

Note that $g(u)\le g(\xi )$ since $u(r)<\xi$ for $r>0$ and $g'\ge 0$. We have

$$\begin{aligned} \frac{|u'(r)|}{\sqrt{1+\left( u'(r)\right) ^2}}\le \frac{g(\xi )}{r^{N+\alpha -1}}\int _0^rs^{N+\beta -1}ds =\frac{g(\xi )}{N+\beta }r^{\beta -\alpha +1},\quad r>0. \end{aligned}$$

(7.4)

Thus

$$\begin{aligned} -u''(r)\ge \left( 1+\left( u'(r)\right) ^2\right) ^{\frac{3}{2}}\left[ g(u(r))-\frac{N+\alpha -1}{N+\beta }g(\xi )\right] r^{\beta -\alpha },\quad r>0. \end{aligned}$$

(7.5)

By $(H_0)$ we have $\frac{N+\alpha -1}{N+\beta }<1$.

Next, we consider the auxiliary function

$$\begin{aligned} w(r)=\xi -d+\left( d^{\beta -\alpha +2}-r^{\beta -\alpha +2}\right) ^{\frac{1}{\beta -\alpha +2}},\quad 0\le r\le d, \end{aligned}$$

where d is chosen such that

$$\begin{aligned} \left( \frac{N+\beta }{g(\xi )}\right) ^{\frac{1}{\beta -\alpha +1}}<d\le \xi . \end{aligned}$$

(7.6)

This is possible by virtue of (7.2). By $(H_0)$ we have $\beta -\alpha +2>1$. Note that $w(0)=\xi$, $w(d)=\xi -d$ and $w(r)\ge \xi -d$ in [0, d]. Furthermore, $w'(0)=0$, indeed in [0, d) we have

$$\begin{aligned} w'(r)=-\left( d^{\beta -\alpha +2}-r^{\beta -\alpha +2}\right) ^{-\frac{\beta -\alpha +1}{\beta -\alpha +2}}r^{\beta -\alpha +1}, \end{aligned}$$

and

$$\begin{aligned} w''(r) =-(\beta -\alpha +1)d^{\beta -\alpha +2}\left( d^{\beta -\alpha +2}-r^{\beta -\alpha +2}\right) ^{-\frac{2(\beta -\alpha )+3}{\beta -\alpha +2}}r^{\beta -\alpha }. \end{aligned}$$

(7.7)

From (7.5), (7.6) and (7.7) we can infer that

$$\begin{aligned} -u''(r)>-w''(r) \end{aligned}$$

for $r>0$ sufficiently close to zero being

$$\begin{aligned} \left( 1+\left( u'(r)\right) ^2\right) ^{\frac{3}{2}}\biggl [g(u(r))-\frac{N+\alpha -1}{N+\beta }g(\xi )\biggr ]r^{\beta -\alpha }\sim \frac{\beta -\alpha +1}{N+\beta }g(\xi ) r^{\beta -\alpha }\quad \text {as } r\rightarrow 0^+, \end{aligned}$$

and

$$\begin{aligned} -w''(r)\sim (\beta -\alpha +1)d^{-(\beta -\alpha +1)}r^{\beta -\alpha }\quad \text {as } r\rightarrow 0^+. \end{aligned}$$

Hence, the graph of u lies below the graph of w, at least for sufficiently small values of r.

Since by contradiction u is defined on the entire ${\mathbb {R}}_0^+$, then there exists a value $r_0\in (0,d]$ such that

$$\begin{aligned} 0<u(r)<w(r)\quad \text {for}\quad 0<r<r_0, \end{aligned}$$

with either

$$\begin{aligned} \text {CASE I: }u(r_0)=w(r_0)\quad \text {or}\quad \text { CASE II: }u(r_0)=\xi -d, \end{aligned}$$

see Fig. 1.

Fig. 1

(r, u, w)

In particular, $r_0\in (0,d]$ in CASE I, while $r_0\in (0,d)$ in CASE II. By analyzing the different possible cases, it can be shown that there exist values t and s, with $0<t<r_0$ and $t<s<d$ such that

$$\begin{aligned} u(t)=w(s),\quad u'(t)=w'(s),\quad u''(t)\ge w''(s). \end{aligned}$$

(7.8)

Indeed, following [29], consider the function $h(\tau )=w^{-1}(u(\tau ))$ on the interval $[0,r_0]$. Then h is well-defined due to the strictly decreasing nature of w. Now define the function

$$\begin{aligned} H(\tau )=h(\tau )-\tau \quad \text {on }[0,r_0]. \end{aligned}$$

Note that $H(0)=0$. Furthermore, since w lies above u in $(0,r_0)$, it follows that $h(\tau )>\tau$ in $(0,r_0)$, i.e., $H(\tau )>0$ in $(0,r_0)$. Note also that in $(0,r_0)$

$$\begin{aligned} H'(\tau )&=h'(\tau )-1=\frac{u'(\tau )}{w'(h(\tau ))}-1,\\ H''(\tau )&=\frac{u''(\tau )w'(h(\tau ))-u'(\tau )w''(h(\tau ))h'(\tau )}{w'^2(h(\tau ))}. \end{aligned}$$

In CASE I, we have $H(r_0)=0$, while in CASE II, since $h(r_0)= w^{-1}(\xi -d)=d$, and by ${\displaystyle {\lim \nolimits _{\tau \rightarrow r_0^-}H'(\tau )=u'(r_0)\lim \nolimits _{\tau \rightarrow d^-}[w'(\tau )]^{-1}-1=-1,}}$ then H is locally strictly decreasing in $r_0$. Hence there exists $t\in (0,r_0)$ such that t is a maximal point of H on $[0,r_0]$. This implies that $H'(t)=0$ and $H''(t)\le 0$. Since $w'<0$, it follows that

$$\begin{aligned} u'(t)=w'\left( h(t)\right) \quad \text {and}\quad u''(t)\ge w''\left( h(t)\right) . \end{aligned}$$

Thus, we take $s=h(t)$. This proves (7.8).

From (7.8), it follows that

$$\begin{aligned} \frac{t^{\alpha -\beta }u''(t)}{\left( 1+\left( u'(t)\right) ^2\right) ^{3/2}}\ge \frac{t^{\alpha -\beta }w''(s)}{\left( 1+\left( w'(s)\right) ^2\right) ^{3/2}}=-(\beta -\alpha +1)d^{\beta -\alpha +2}\frac{F(t,s)}{[z(s)]^{3/2}}, \end{aligned}$$

(7.9)

where

$$\begin{aligned} F(t,s):=\left( d^{\beta -\alpha +2}-s^{\beta -\alpha +2}\right) ^{\frac{\beta -\alpha }{\beta -\alpha +2}}\left( \frac{s}{t}\right) ^{\beta -\alpha },\quad t<s<d, \end{aligned}$$

and

$$\begin{aligned} z(s):=s^{2(\beta -\alpha +1)}+\left( d^{\beta -\alpha +2}-s^{\beta -\alpha +2}\right) ^{\frac{2(\beta -\alpha +1)}{\beta -\alpha +2}}. \end{aligned}$$

It is obvious that

$$\begin{aligned} \min _{s\in (0,d)}z(s)=z\left( 2^{-\frac{1}{\beta -\alpha +2}}d\right) =2^{-\frac{\beta -\alpha }{\beta -\alpha +2}}d^{2(\beta -\alpha +1)}. \end{aligned}$$

(7.10)

On the other hand, when $\beta >\alpha$ otherwise is trivial, to estimate the upper bound of F(t, s), we use

$$\begin{aligned} w^{-1}(\tau )=\left[ d^{\beta -\alpha +2}-(\tau -\xi +d)^{\beta -\alpha +2}\right] ^{\frac{1}{\beta -\alpha +2}},\quad \xi -d<\tau <\xi , \end{aligned}$$

so that, by the definition of h,

$$\begin{aligned} s=h(t)=w^{-1}\left( u(t)\right) =\left[ d^{\beta -\alpha +2}-(u(t)-\xi +d)^{\beta -\alpha +2}\right] ^{\frac{1}{\beta -\alpha +2}}. \end{aligned}$$

In turn

$$\begin{aligned} \frac{s}{t}=\frac{\left[ d^{\beta -\alpha +2}-(u(t)-\xi +d)^{\beta -\alpha +2}\right] ^{\frac{1}{\beta -\alpha +2}}}{t}. \end{aligned}$$

This gives

$$\begin{aligned} F(t,s)=\left( d^{\beta -\alpha +2}-s^{\beta -\alpha +2}\right) ^{\frac{\beta -\alpha }{\beta -\alpha +2}}[f(t)]^{\frac{\beta -\alpha }{\beta -\alpha +2}},\quad 0<t<\min (s,r_0), \end{aligned}$$

where

$$\begin{aligned} f(t):=\frac{d^{\beta -\alpha +2}-(u(t)-\xi +d)^{\beta -\alpha +2}}{t^{\beta -\alpha +2}}. \end{aligned}$$

We will prove that

$$\begin{aligned} f(t)\le \frac{\sqrt{2}d^{\beta -\alpha +1}g(\xi )}{N+\beta }. \end{aligned}$$

(7.11)

Indeed, for all $0<t<\min (s,r_0)$

$$\begin{aligned} \frac{t^{\beta -\alpha +3}}{\beta -\alpha +2} f'(t)&= -t(u(t)-\xi +d)^{\beta -\alpha +1}u'(t)+(u(t)-\xi +d)^{\beta -\alpha +2}-d^{\beta -\alpha +2} \\&=-t(w(s)-\xi +d)^{\beta -\alpha +1}w'(s)+(w(s)-\xi +d)^{\beta -\alpha +2}-d^{\beta -\alpha +2}\\&=ts^{\beta -\alpha +1}+d^{\beta -\alpha +2}-s^{\beta -\alpha +2}-d^{\beta -\alpha +2}\\&=s^{\beta -\alpha +1}(t-s)<0, \end{aligned}$$

being $t<s$, where in the second equality we used (7.8), and in the third equality we used the definition of the function w. Thus, f is decreasing in $\left( 0,\min (s,r_0)\right)$ and hence

$$\begin{aligned} f(t)\le f(0^+):=\lim _{t\rightarrow 0}f(t), \end{aligned}$$

so that, by definition of f and by Lagrange’s theorem,

$$\begin{aligned} f(0^+)=\lim _{t\rightarrow 0}\frac{-(u(t)-\xi +d)^{\beta -\alpha +1}u'(t)}{t^{\beta -\alpha +1}}\le \frac{\sqrt{2}d^{\beta -\alpha +1}g(\xi )}{N+\beta }, \end{aligned}$$

where in the last inequality we have used (7.4) and that $|u'(t)|=|w'(s)|\le 1$. Thus (7.11) follows. Consequently

$$\begin{aligned} F(t,s)\le d^{\beta -\alpha }\left( \frac{\sqrt{2}d^{\beta -\alpha +1}g(\xi )}{N+\beta }\right) ^{\frac{\beta -\alpha }{\beta -\alpha +2}},\quad 0<t<\min (s,r_0). \end{aligned}$$

(7.12)

Combining (7.9), (7.10) and (7.12), we obtain

$$\begin{aligned} \frac{t^{\alpha -\beta }u''(t)}{\left( 1+\left( u'(t)\right) ^2\right) ^{3/2}}\ge -c_1d^{-\frac{2(\beta -\alpha +1)}{\beta -\alpha +2}}g(\xi )^{\frac{\beta -\alpha }{\beta -\alpha +2}}, \end{aligned}$$

(7.13)

where $c_1:=(\beta -\alpha +1)\left( \frac{4}{N+\beta }\right) ^{\frac{\beta -\alpha }{\beta -\alpha +2}}$.

From (7.5), evaluated in $r=t$, and by (7.13), we get

$$\begin{aligned} -\frac{N+\alpha -1}{N+\beta }g(\xi )+g(u(t))\le c_1d^{-\frac{2(\beta -\alpha +1)}{\beta -\alpha +2}}g(\xi )^{\frac{\beta -\alpha }{\beta -\alpha +2}} \end{aligned}$$

or in turn

$$\begin{aligned} \frac{\beta -\alpha +1}{N+\beta }g(\xi )&\le c_1d^{-\frac{2(\beta -\alpha +1)}{\beta -\alpha +2}}g(\xi )^{\frac{\beta -\alpha }{\beta -\alpha +2}}+g(\xi )-g(u(t)) \nonumber \\&<c_1d^{-\frac{2(\beta -\alpha +1)}{\beta -\alpha +2}}g(\xi )^{\frac{\beta -\alpha }{\beta -\alpha +2}}+dg'(\xi ), \end{aligned}$$

(7.14)

using the fact that, by Lagrange theorem,

$$\begin{aligned} g(\xi )-g(u(t))=\bigl (\xi -u(t)\bigr )g'(\eta )\le \bigl (\xi -w(s)\bigr )g'(\xi )<dg'(\xi ) \end{aligned}$$

for $\eta \in (u(t),\xi )$, thanks to $g''(u)\ge 0$, $g'(\xi )>0$ and $u',w'<0$.

We now explicitly select d as $g'(\xi )^{-\frac{1}{\beta -\alpha +2}}$. This choice in (7.6) is possible by condition (7.2) being ${\mathcal {C}}>N+\beta$. Subsequently, (7.14) can be rewritten as

$$\begin{aligned} \frac{\beta -\alpha +1}{N+\beta }g(\xi )&<c_1g(\xi )^{\frac{\beta -\alpha }{\beta -\alpha +2}}g'(\xi )^{\frac{2(\beta -\alpha +1)}{(\beta -\alpha +2)^2}}+g'(\xi )^{\frac{\beta -\alpha +1}{\beta -\alpha +2}}\nonumber \\&<2c_1g(\xi )^{\frac{\beta -\alpha }{\beta -\alpha +2}}g'(\xi )^{\frac{2(\beta -\alpha +1)}{(\beta -\alpha +2)^2}}, \end{aligned}$$

(7.15)

where in the last inequality we used that

$$\begin{aligned} \frac{\beta -\alpha +1}{\beta -\alpha +2}=\frac{2(\beta -\alpha +1)}{(\beta -\alpha +2)^2}+\frac{(\beta -\alpha )(\beta -\alpha +1)}{(\beta -\alpha +2)^2} \end{aligned}$$

and (7.6) which gives

$$\begin{aligned} \frac{N+\beta }{g(\xi )}<d^{\beta -\alpha +1}=g'(\xi )^{-\frac{\beta -\alpha +1}{\beta -\alpha +2}} \end{aligned}$$

that is, since $\beta >\alpha$,

$$\begin{aligned} g'(\xi )^{\frac{\beta -\alpha +1}{\beta -\alpha +2}}<\frac{g(\xi )}{N+\beta }<c_1^{\frac{\beta -\alpha +2}{\beta -\alpha }}g(\xi ), \end{aligned}$$

being

$$\begin{aligned} c_1^{\frac{\beta -\alpha +2}{\beta -\alpha }}=(\beta -\alpha +1)^{\frac{\beta -\alpha +2}{\beta -\alpha }}\frac{4}{N+\beta }>\frac{1}{N+\beta }. \end{aligned}$$

It follows from (7.15) that

$$\begin{aligned} g(\xi )<\mathcal {C}g'(\xi )^{\frac{\beta -\alpha +1}{\beta -\alpha +2}}, \end{aligned}$$

(7.16)

where

$$\begin{aligned} \mathcal {C}=\left( 2c_1\frac{N+\beta }{\beta -\alpha +1}\right) ^{\frac{\beta -\alpha +2}{2}}=2^{\frac{3(\beta -\alpha )}{2}+1}(N+\beta ). \end{aligned}$$

Since (7.16) contradicts (7.2), the proof is complete. $\square$

Remark 7.3

In the case $\beta <\alpha$ the proof of the above theorem cannot be performed because of the possible lack of boundedness of the function F(t, s), indeed condition (7.12) fails.

Corollary 7.4

Assume $(H_0)$, $\beta \ge \alpha$ and $p>1$. No positive radial solution of problem (1.1) can exist unless the initial value $\xi =u(0)$ satisfies

$$\begin{aligned} \xi <\left( p\mathcal {C}^{\frac{\beta -\alpha +2}{\beta -\alpha +1}}\right) ^{\frac{\beta -\alpha +1}{p+\beta -\alpha +1}}:=\xi _*, \end{aligned}$$

(7.17)

where $\mathcal {C}$ is given in Theorem 7.1.

Proof

If (7.17) does not true, we can verify that condition (7.2) holds for $g(\xi )=\xi ^p$, indeed (7.2) becomes

$$\begin{aligned} 1\le p \xi ^{p+\beta -\alpha +1}\le p \biggl (\frac{\xi ^{\beta -\alpha +2}}{\xi _*}\biggr )^{\frac{p+\beta -\alpha +1}{\beta -\alpha +1}}, \end{aligned}$$

thus the second inequality immediately follows from $\xi \ge \xi _*$, while the first is a consequence of $\xi \ge \xi _*>1$ being $p>1$ and $\mathcal C>N+\beta >1$. Hence the solution u(r) cannot be continued to the line $u=0$ as it is necessary to satisfy the boundary condition $u=0$ when $r=R$ for any $R>0$. $\square$

Lemma 7.5

Let u(r) be a positive radial solution of problem (1.1) with initial value $u(0)=\xi >0$. Then there exists $r_1\in (0,R]$ such that

$$\begin{aligned} |u'(r_1)|\ge \frac{\xi }{R}. \end{aligned}$$

(7.18)

Proof

Consider the auxiliary function

$$\begin{aligned} z(r)=-\frac{\xi }{R}r+\xi ,\quad 0\le r\le R, \end{aligned}$$

whose graph is a line segment with $z(0)=\xi$, $z(R)=0$ and $z'(r)=-\frac{\xi }{R}$. By a contradiction argument (if $u'(r)>z'(r)$ in (0, R), then $u(R)-u(0)>z(R)-z(0)$ yielding the absurd inequality $u(0)<z(0)$), it immediately follows that the slope of the graph of z is greater than or equal to the slope of the graph of u for some point $r_1$, i.e.,

$$\begin{aligned} u'(r_1)\le z'(r_1). \end{aligned}$$

This proves (7.18) in view of $u'(r)<0$ in (0, R]. $\square$

Now we use Corollary 4.8, Corollary 7.4 and Lemma 7.5 to prove our nonexistence result for Dirichlet problem (1.1) with R small.

Proof of Theorem 1.5

Assume that u is a positive radial solution of (1.1) with $u(0)=\xi >0$.

(i) When $\beta \ge \alpha$, according to Corollary 7.4, we know that $\xi$ must satisfy the condition (7.17). Let

$$\begin{aligned} 0<R<\min \left\{ \left( \frac{N+\beta }{\sqrt{2}\xi _*^p}\right) ^{\frac{1}{\beta -\alpha +1}},\left( \frac{N+\beta }{\sqrt{2}\xi _*^{p-1}}\right) ^{\frac{1}{\beta -\alpha +2}}\right\} :=R_1=R_1(N,p,\alpha ,\beta ). \end{aligned}$$

Then, by (7.17) we have, for all $\xi <\xi _*$,

$$\begin{aligned} 0<R<\min \left\{ \left( \frac{N+\beta }{\sqrt{2}\xi ^p}\right) ^{\frac{1}{\beta -\alpha +1}},\left( \frac{N+\beta }{\sqrt{2}\xi ^{p-1}}\right) ^{\frac{1}{\beta -\alpha +2}}\right\} :=R_2=R_2(N,p,\alpha ,\beta ,\xi ). \end{aligned}$$

Applying also Corollary 4.8 with $M=R<R_1<R_2$, we obtain, for all $0<r\le R$,

$$\begin{aligned} |u'(r)|\le \frac{\sqrt{2}\xi ^p}{N+\beta }R^{\beta -\alpha +1}<\frac{\xi }{R} \end{aligned}$$

(7.19)

being $R<R_2$, which contradicts Lemma 7.5. Thus, problem (1.1) has no positive radial solution in $B_R$ with $0<R<R_1$.

(ii) When $\beta <\alpha$, we don’t have a bound from above for the initial value of u of the type in (7.17), thus $R_2$ depends also on the initial value $\xi$. In particular, let $0<R<R_2$, by Corollary 4.8 with $M=R$, as above (7.19) is in force, leading to a contradiction with Lemma 7.5. Therefore, problem (1.1) does not have any positive radial solutions in $B_R$ with $0<R<R_2$. $\square$

Remark 7.6

The main difference between the two cases above is that $R_1<\infty$ in case (i), roughly because of the necessary condition for existence, while in case (ii) $R_2$ doesn’t have an upper bound and could be as large as we want depending on the smallness of u(0).