Positive Radially Symmetric Entire Solutions of p-k-Hessian Equations and Systems

Abstract

In this paper, we discuss the existence of positive radially symmetric entire solutions of the p-k-Hessian equation ${\sigma }_k^{\frac{1}{k}}\left(\lambda \left(D_i\left({\left|Du\right|}^{p-2}{D}_{j}u\right)\right)\right)={\alpha }^{\frac{1}{k}}\left(\right|x\left|\right)f\left(u\right),$ and the general p-k-Hessian system ${\sigma }_k^{\frac{1}{k}}\left(\lambda \left(D_i\left({\left|Du\right|}^{p-2}{D}_{j}u\right)\right)\right)={\alpha }^{\frac{1}{k}}\left(\right|x\left|\right)f_1\left(v\right)f_2\left(u\right)$, ${\sigma }_k^{\frac{1}{k}}\left(\lambda \left(D_i\left({\left|Dv\right|}^{p-2}{D}_{j}v\right)\right)\right)={\beta }^{\frac{1}{k}}\left(\right|x\left|\right)g_1\left(u\right)g_2\left(v\right)$.

1. Introduction

For a $C^2$ function u defined in a subset of the Euclidean space ${\mathbb{R}}^n(n\ge 2)$, $p\ge 2$ and $1\le k\le n$, the p-k-Hessian operator of u, firstly studied by Trudinger and Wang in [1], is defined as ${\sigma }_k\left(\lambda \left(D_i\left({\left|Du\right|}^{p-2}{D}_{j}u\right)\right)\right)$, where $\lambda \left(D_i\left({\left|Du\right|}^{p-2}{D}_{j}u\right)\right)$ denotes the eigenvalues of the $n\times n$ real symmetric matrix ${\left(D_i\left({\left|Du\right|}^{p-2}{D}_{j}u\right)\right)}_{i,j=1,2,\cdots ,n}$ and ${\sigma }_k\left(\lambda \right)={\sum }_{1\le i_1<\cdots <i_k\le n}{\lambda }_{i_1}\cdots {\lambda }_{i_k}$ denotes the k-th elementary symmetric function. The p-k-Hessian operator includes lots of well-known operators such as the Laplacian operator ($k=1$ and $p=2$), the p-Laplacian operator ($k=1$), the k-Hessian operator ($p=2$), and the Monge–Ampère operator ($k=n$ and $p=2$). We refer to [2,3,4,5,6,7] for more properties of the above operators.

In this paper, we firstly investigate the existence of positive radially symmetric entire solutions of the following p-k-Hessian equation:

$${\sigma }_k^{\frac{1}{k}}\left(\lambda \left(D_i\left({\left|Du\right|}^{p-2}{D}_{j}u\right)\right)\right)={\alpha }^{\frac{1}{k}}\left(\right|x\left|\right)f\left(u\right),x\in {\mathbb{R}}^n,$$

(1)

where $\alpha$ and f satisfy the following conditions:

($F_1$) $\alpha :[0,\infty )\to (0,\infty )$ is a continuous function;

($F_2$) $f:[0,\infty )\to (0,\infty )$ is a continuous, non-decreasing function.

Definition 1.

A function u is called p-k-convex in ${\mathbb{R}}^n$ if

$$\begin{array}{cc}\hfill u\in {\Phi }^{p,k}\left({\mathbb{R}}^n\right):=& \left\{u\in C^2\left({\mathbb{R}}^n\backslash \left\{0\right\}\right)\cap C^1\left({\mathbb{R}}^n\right):\right.\hfill \\ \hfill & \left.{\left|Du\right|}^{p-2}Du\in C^1\left({\mathbb{R}}^n\right),\lambda \left(D_i\left({\left|Du\right|}^{p-2}{D}_{j}u\right)\right)\in {\Gamma }_k\mathit{in}{\mathbb{R}}^n\right\},\hfill \end{array}$$

where ${\Gamma }_k=\left\{\lambda \in {\mathbb{R}}^n\mid {\sigma }_i\left(\lambda \right)>0,i=1,2,\cdots ,k\right\}$. In particular, a function u is called k-convex in ${\mathbb{R}}^n$ if $u\in {\Phi }^{2,k}\left({\mathbb{R}}^n\right)\cap C^2\left({\mathbb{R}}^n\right)$.

Definition 2.

A function u is called an entire solution of (1) if $u\in {\Phi }^{p,k}\left({\mathbb{R}}^n\right)$ satisfies (1) in ${\mathbb{R}}^n$.

For $p=2$, $k=1$, and $\alpha \equiv 1$, Keller [8] and Osserman [9] proved that Equation (1) admits an entire subsolution if and only if f satisfies the following Keller–Osserman condition:

$${\int }_0^{\infty }{\left({\int }_0^{t}f\left(s\right)ds\right)}^{-\frac{1}{2}}dt=\infty .$$

The above condition is widely applied in the study of boundary blow up problems, see [10,11,12].

For $p=2$, $1\le k\le n$, and $\alpha \equiv 1$, Ji and Bao [13] proved that (1) admits an entire subsolution if and only if f satisfies the generalized Keller–Osserman condition:

$${\int }_0^{\infty }{\left({\int }_0^t{f}^k\left(s\right)ds\right)}^{-\frac{1}{k+1}}dt=\infty .$$

Later on, Zhang and Zhou [14] proved that, for any fixed $a>0$, such that $A(\infty )\le D_{1a}(\infty )$, where

$$A\left(r\right):={\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}{\int }_0^r{\left(t^{k-n}{\int }_0^t{s}^{n-1}\alpha \left(s\right)ds\right)}^{\frac{1}{(p-1)k}}dt\forall r\ge 0,$$

(2)

$$A(\infty ):=\underset{r\to \infty }{lim}A\left(r\right)\in (0,\infty ],$$

(3)

and

$$D_{1a}\left(r\right):={\int }_a^r{\left(f\left(t\right)\right)}^{-\frac{1}{p-1}}dt\forall r\ge a,D_{1a}(\infty ):=\underset{r\to \infty }{lim}{D}_{1a}\left(r\right)\in (0,\infty ],$$

(4)

there exists a positive radially symmetric entire solution u of (1) satisfying $u\left(0\right)=a$ and

$$a+f\left(a\right)A\left(r\right)\le u\left(r\right)\le D_{1a}^{-1}\left(A\left(r\right)\right).$$

(5)

Note that A and $D_{1a}$ are strictly increasing functions. Thus, under the assumption that $A(\infty )\le D_{1a}(\infty )$, then $D_{1a}^{-1}\left(A\left(r\right)\right)$ in (5) is well-defined. For instance, when $p=2$, $1\le k\le n$, $\alpha \equiv 1$, $a=0$ and $f\left(u\right)=u^2$, $D_{10}(\infty )=\infty$; meanwhile, f does not satisfy the generalized Keller–Osserman condition.

For $p\ge 2$, $k=1$ and $\alpha \equiv 1$, Naito and Usami [15] proved that there exists a positive entire subsolution u of (1) if and only if

$${\int }_0^{\infty }{\left({\int }_0^{t}f\left(s\right)ds\right)}^{-\frac{1}{p}}dt=\infty .$$

Ni and Serrin [16] obtained some non-existence results. Later on, Filippucci [17,18] obtained the existence of radially symmetric entire solutions for some types of p-Laplace equations. Recently, the fractional p-Laplace differential operator has attracted a lot of attention and has been applied in many fields. For instance, Anthal, Giacomoni, and Sreenadh [19] obtained the existence, nonexistence, uniqueness, and regularity of weak solutions under suitable and general assumptions on the nonlinearity. As for applications, they treat cases of the subdiffusive type logistic Choquard problem and consider in the superdiffusive case the Brezis–Nirenberg type problem with logistic Choquard. Wang, Huang, and Li [20] obtained the existence of positive solutions for Riemann–Liouville fractional differential equations at resonance by using the theory of fixed point index and spectral theory of linear operators.

For $p\ge 2$, $1\le k\le n$, and $\alpha \equiv 1$, Bao and Feng [21] proved that there exists a positive entire subsolution u of (1) if and only if

$${\int }_0^{\infty }{\left({\int }_0^t{f}^k\left(s\right)ds\right)}^{-\frac{1}{(p-1)k+1}}dt=\infty .$$

We refer to [22,23,24,25,26,27,28,29,30] for more results concerning the k-Hessian type operator.

Our first main result of this paper can be stated as follows.

Theorem 1.

Let $n\ge 2$, $1\le k\le n$, $p\ge 2$, and $a>0$. Assume that ($F_1$) and ($F_2$) hold. If $0\le A(\infty )\le D_{1a}(\infty )\le \infty$, there exists a positive radially symmetric entire solution u of (1) satisfying $u\left(0\right)=a$ and

$$a+{\left(f\left(a\right)\right)}^{\frac{1}{p-1}}A\left(r\right)\le u\left(r\right)\le D_{1a}^{-1}\left(A\left(r\right)\right)\forall r\ge 0,$$

where A and $D_{1a}$ are defined as in (2) and (4), respectively. Moreover, if $A(\infty )=D_{1a}(\infty )=\infty$, we have that $\underset{r\to \infty }{lim}u\left(r\right)=\infty$; if $A(\infty )<D_{1a}(\infty )=\infty$ or $A(\infty )\le D_{1a}(\infty )<\infty$, we have that u is bounded.

Remark 1.

Since the domain of $D_{1a}^{-1}$ is $\left[0,D_{1a}(\infty )\right)$, we could not discuss the condition of $D_{1a}(\infty )<A(\infty )\le \infty$ occurring. It would be interesting to see whether this condition is optimal in obtaining the existence result.

The second part of this paper is to investigate the existence of radially symmetric entire solutions of the general p-k-Hessian system

$$\left\{\begin{array}{c}{\sigma }_k^{\frac{1}{k}}\left(\lambda \left(D_i\left({\left|Du\right|}^{p-2}{D}_{j}u\right)\right)\right)={\alpha }^{\frac{1}{k}}\left(\right|x\left|\right)f_1\left(v\right)f_2\left(u\right),\\ {\sigma }_k^{\frac{1}{k}}\left(\lambda \left(D_i\left({\left|Dv\right|}^{p-2}{D}_{j}v\right)\right)\right)={\beta }^{\frac{1}{k}}\left(\right|x\left|\right)g_1\left(u\right)g_2\left(v\right),\end{array}\right.$$

(6)

where $\alpha ,\beta$, and $f_1,f_2,g_1,g_2$ satisfy the following conditions:

($G_1$) $\alpha ,\beta :[0,\infty )\to (0,\infty )$ are two continuous functions;

($G_2$) $f_1,f_2,g_1,g_2:[0,\infty )\to (0,\infty )$ are continuous, non-decreasing functions.

Definition 3.

A pair of functions $(u,v)$ is called an entire solution of (6) if $(u,v)\in {\Phi }^{p,k}\left({\mathbb{R}}^n\right)\times {\Phi }^{p,k}\left({\mathbb{R}}^n\right)$ satisfies (6) in ${\mathbb{R}}^n$.

For $p=2$, $k=1$, $\alpha ,\beta ,f_2,g_2\equiv 1$, $f_1\left(v\right)=v^s$ and $g_1\left(u\right)=u^t$ with $0<s\le t$, Lair and Wood [31] obtained the existence and nonexistence of entire subsolutions of (6). See [32,33] and the references therein for more results.

For $p=2$, $1\le k\le n$, and $f_2,g_2\equiv 1$, Zhang and Zhou [14] proved that, for any fixed $a>0$ such that $A(\infty )+B(\infty )\le D_{2a}(\infty )$, there exists a positive radially symmetric entire solution $(u,v)$ of (6) satisfying $u\left(0\right)=v\left(0\right)=\frac{a}{2}$,

$$\frac{a}{2}+f_1\left(\frac{a}{2}\right)A\left(r\right)\le u\left(r\right)\le D_{2a}^{-1}(A\left(r\right)+B\left(r\right)),$$

(7)

and

$$\frac{a}{2}+g_1\left(\frac{a}{2}\right)B\left(r\right)\le v\left(r\right)\le D_{2a}^{-1}(A\left(r\right)+B\left(r\right)),$$

(8)

where

$$B\left(r\right):={\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}{\int }_0^r{\left(t^{k-n}{\int }_0^t{s}^{n-1}\beta \left(s\right)ds\right)}^{\frac{1}{(p-1)k}}dt\forall r\ge 0,$$

(9)

$$B(\infty ):=\underset{r\to \infty }{lim}B\left(r\right)\in (0,\infty ],$$

(10)

and

$$D_{2a}\left(r\right):={\int }_a^r{(f_1\left(t\right)f_2\left(t\right)+g_1\left(t\right)g_2\left(t\right))}^{-\frac{1}{p-1}}dt\forall r\ge a,D_{2a}(\infty ):=\underset{r\to \infty }{lim}{D}_{2a}\left(r\right)\in (0,\infty ].$$

(11)

Note that A, B, and $D_{2a}$ are strictly increasing functions. Thus, under the assumption that $A(\infty )+B(\infty )\le D_{2a}(\infty )$, $D_{2a}^{-1}(A\left(r\right)+B\left(r\right))$ in (7) and (8) are well-defined. For more results, we refer to [34,35,36].

The second main result of this paper is as follows.

Theorem 2.

Let $n\ge 2$, $1\le k\le n$, $p\ge 2$, and $a>0$. Assume that ($G_1$) and ($G_2$) hold. If $0\le A(\infty )+B(\infty )\le D_{2a}(\infty )\le \infty$, there exists a positive radially symmetric entire solution $(u,v)$ of (6) satisfying $u\left(0\right)=v\left(0\right)=\frac{a}{2}$,

$$\frac{a}{2}+{\left(f_1\left(\frac{a}{2}\right)f_2\left(\frac{a}{2}\right)\right)}^{\frac{1}{p-1}}A\left(r\right)\le u\left(r\right)\le D_{2a}^{-1}(A\left(r\right)+B\left(r\right))\forall r\ge 0,$$

(12)

and

$$\frac{a}{2}+{\left(g_1\left(\frac{a}{2}\right)g_2\left(\frac{a}{2}\right)\right)}^{\frac{1}{p-1}}B\left(r\right)\le v\left(r\right)\le D_{2a}^{-1}(A\left(r\right)+B\left(r\right))\forall r\ge 0,$$

(13)

where A, B and $D_{2a}$ are defined as in (2), (9) and (11), respectively. Moreover, if $A(\infty )=B(\infty )=D_{2a}(\infty )=\infty$, we have that $\underset{r\to \infty }{lim}u\left(r\right)=\underset{r\to \infty }{lim}v\left(r\right)=\infty$; if $A(\infty )+B(\infty )<D_{2a}(\infty )=\infty$ or $A(\infty )+B(\infty )\le D_{2a}(\infty )<\infty$, we have that u and v are bounded.

Remark 2.

Since the domain of $D_{2a}^{-1}$ is $\left[0,D_{2a}(\infty )\right)$, we could not discuss the condition $D_{2a}(\infty )<A(\infty )+B(\infty )\le \infty$ occurring. It would be interesting to see whether this condition is optimal in obtaining the existence result.

The rest of this paper is organized as follows: in Section 2, we give some preliminaries. In Section 3, we prove Theorem 1. In Section 4, we prove Theorem 2.

2. Preliminaries

In this section, we present some lemmas which will be needed in the proof of our main results.

Let $n\ge 2$, $1\le k\le n$ and $p\ge 2$. For $R\in (0,\infty ]$, let $B_R:=\left\{x\in {\mathbb{R}}^n:\left|x\right|<R\right\}$.

Lemma 1.

Suppose that $\alpha ,\beta :[0,R)\to (0,\infty )$ are continuous functions for some $R\in (0,\infty ]$, and ($G_2$) holds. Let $a>0$. Assume that $\zeta ,\eta \in C^0\left([0,R)\right)\cap C^1\left((0,R)\right)$ are positive solutions of the following system:

$$\left\{\begin{array}{cc}\hfill {\zeta }^{\prime }& ={\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}{\left(r^{k-n}{\int }_0^r{s}^{n-1}\alpha \left(s\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right)ds\right)}^{\frac{1}{(p-1)k}},r\in (0,R),\hfill \\ \hfill {\eta }^{\prime }& ={\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}{\left(r^{k-n}{\int }_0^r{s}^{n-1}\beta \left(s\right)g_1^k\left(\zeta \right)g_2^k\left(\eta \right)ds\right)}^{\frac{1}{(p-1)k}},r\in (0,R),\hfill \\ \hfill \zeta \left(0\right)& =\eta \left(0\right)=\frac{a}{2}.\hfill \end{array}\right.$$

(14)

Then, $\zeta ,\eta \in C^2\left((0,R)\right)\cap C^1\left([0,R)\right)$ satisfy ${\zeta }^{\prime }\left(0\right)={\eta }^{\prime }\left(0\right)=0$, ${\zeta }^{\prime },{\eta }^{\prime }>0$ in $(0,R)$ and

$$\left\{\begin{array}{c}{C}_{n-1}^{k-1}{\left({\left({\zeta }^{\prime }\right)}^{p-1}\right)}^{\prime }{\left(\frac{{\left({\zeta }^{\prime }\right)}^{p-1}}{r}\right)}^{k-1}+C_{n-1}^k{\left(\frac{{\left({\zeta }^{\prime }\right)}^{p-1}}{r}\right)}^k=\alpha \left(r\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right),\\ C_{n-1}^{k-1}{\left({\left({\eta }^{\prime }\right)}^{p-1}\right)}^{\prime }{\left(\frac{{\left({\eta }^{\prime }\right)}^{p-1}}{r}\right)}^{k-1}+C_{n-1}^k{\left(\frac{{\left({\eta }^{\prime }\right)}^{p-1}}{r}\right)}^k=\beta \left(r\right)g_1^k\left(\zeta \right)g_2^k\left(\eta \right).\end{array}\right.$$

(15)

Proof of Lemma 1.

It is easy to see that $\zeta ,\eta \in C^2\left((0,R)\right)$ since we can differentiate the first two lines in (14) directly. By the positivity of $\alpha$, $\beta$, $f_1,f_2,g_1,g_2$ and the first two lines in (14), we have that ${\zeta }^{\prime },{\eta }^{\prime }>0$ in $(0,R)$. Integrating the first two lines in (14) on $[0,r]$ for any $r\in (0,R)$, we have that

$$\zeta \left(r\right)=\zeta \left(0\right)+{\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}{\int }_0^r{\left(t^{k-n}{\int }_0^t{s}^{n-1}\alpha \left(s\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right)ds\right)}^{\frac{1}{(p-1)k}}dt,$$

(16)

and

$$\eta \left(r\right)=\eta \left(0\right)+{\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}{\int }_0^r{\left(t^{k-n}{\int }_0^t{s}^{n-1}\beta \left(s\right)g_1^k\left(\zeta \right)g_2^k\left(\eta \right)ds\right)}^{\frac{1}{(p-1)k}}dt.$$

It follows from (16) and the L’Hôpital’s rule that

$$\begin{array}{cc}\hfill {\zeta }^{\prime }\left(0\right)& =\underset{r\to 0^{+}}{lim}\frac{\zeta \left(r\right)-\zeta \left(0\right)}{r}\hfill \\ \hfill & ={\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}\underset{r\to 0^{+}}{lim}\frac{{\int }_0^r{\left(t^{k-n}{\int }_0^t{s}^{n-1}\alpha \left(s\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right)ds\right)}^{\frac{1}{(p-1)k}}dt}{r}\hfill \\ \hfill & ={\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}\underset{r\to 0^{+}}{lim}{\left(r^{k-n}{\int }_0^r{s}^{n-1}\alpha \left(s\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right)ds\right)}^{\frac{1}{(p-1)k}}\hfill \\ \hfill & ={\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}\underset{r\to 0^{+}}{lim}{\left(\frac{{\int }_0^r{s}^{n-1}\alpha \left(s\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right)ds}{r^{n-k}}\right)}^{\frac{1}{(p-1)k}}\hfill \\ \hfill & ={\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}{\left(\underset{r\to 0^{+}}{lim}\frac{{\int }_0^r{s}^{n-1}\alpha \left(s\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right)ds}{r^{n-k}}\right)}^{\frac{1}{(p-1)k}}\hfill \\ \hfill & ={\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}{\left(\underset{r\to 0^{+}}{lim}\frac{r^{n-1}\alpha \left(r\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right)}{(n-k)r^{n-k-1}}\right)}^{\frac{1}{(p-1)k}}\hfill \\ \hfill & ={\left(\frac{k}{(n-k)C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}{\left(\underset{r\to 0^{+}}{lim}{r}^k\alpha \left(r\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right)\right)}^{\frac{1}{(p-1)k}}\hfill \\ \hfill & ={\left(\frac{k\alpha \left(0\right)f_1^k\left(\frac{a}{2}\right)f_2^k\left(\frac{a}{2}\right)}{(n-k)C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}\underset{r\to 0^{+}}{lim}{r}^{\frac{1}{(p-1)}}\hfill \\ \hfill & =0.\hfill \end{array}$$

In addition, by the first line in (14), we also have that

$$\underset{r\to 0^{+}}{lim}{\zeta }^{\prime }\left(r\right)={\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}\underset{r\to 0^{+}}{lim}{\left(r^{k-n}{\int }_0^r{s}^{n-1}\alpha \left(s\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right)ds\right)}^{\frac{1}{(p-1)k}}=0.$$

Therefore, we have that $\zeta \in C^1\left([0,R)\right)$. Similarly, we can prove that $\eta \in C^1\left([0,R)\right)$.

By taking $(p-1)k$ power of the first line in (14), we have

$${\left({\zeta }^{\prime }\right)}^{(p-1)k}=\frac{k}{C_{n-1}^{k-1}}\left(r^{k-n}{\int }_0^r{s}^{n-1}\alpha \left(s\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right)ds\right).$$

(17)

Multiplying (17) with $\frac{C_{n-1}^{k-1}}{k}{r}^{n-k}$, we have

$$\frac{C_{n-1}^{k-1}}{k}{r}^{n-k}{\left({\zeta }^{\prime }\right)}^{(p-1)k}={\int }_0^r{s}^{n-1}\alpha \left(s\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right)ds,$$

(18)

Differentiating (18) with respect to r directly, we have

$$\frac{(n-k)C_{n-1}^{k-1}}{k}{r}^{n-k-1}{\left({\zeta }^{\prime }\right)}^{(p-1)k}+(p-1)C_{n-1}^{k-1}{r}^{n-k}{\left({\zeta }^{\prime }\right)}^{\left[\right(p-1)k-1]}{\zeta }^{″}=r^{n-1}\alpha \left(s\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right),$$

(19)

Multiplying (19) with $r^{1-n}$, we have

$$\begin{array}{cc}\hfill & C_{n-1}^k{r}^{-k}{\left({\zeta }^{\prime }\right)}^{(p-1)k}+(p-1)C_{n-1}^{k-1}{r}^{1-k}{\left({\zeta }^{\prime }\right)}^{\left[\right(p-1)k-1]}{\zeta }^{″}\hfill \\ \hfill =& C_{n-1}^k{\left(\frac{{\left({\zeta }^{\prime }\right)}^{(p-1)}}{r}\right)}^k+(p-1)C_{n-1}^{k-1}{\left(\frac{{\left({\zeta }^{\prime }\right)}^{p-1}}{r}\right)}^{k-1}{\left({\zeta }^{\prime }\right)}^{p-2}{\zeta }^{″}\hfill \\ \hfill =& C_{n-1}^k{\left(\frac{{\left({\zeta }^{\prime }\right)}^{p-1}}{r}\right)}^k+C_{n-1}^{k-1}{\left({\left({\zeta }^{\prime }\right)}^{p-1}\right)}^{\prime }{\left(\frac{{\left({\zeta }^{\prime }\right)}^{p-1}}{r}\right)}^{k-1}\hfill \\ \hfill =& \alpha \left(r\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right),\hfill \end{array}$$

that is the first line of (15). Similarly, we can prove that $\eta$ satisfies the second line of (15). □

Lemma 2.

Suppose that $\alpha ,\beta :[0,R)\to (0,\infty )$ are continuous functions for some $R\in (0,\infty ]$, and ($G_2$) holds. Let $a>0$. Assume that $\zeta ,\eta \in C^0\left([0,R)\right)\cap C^1\left((0,R)\right)$ are positive solutions of (14). Let $u\left(x\right)=\zeta \left(r\right)$ and $v\left(x\right)=\eta \left(r\right)$, where $r=\left|x\right|$. Then, we have $u,v\in C^2(B_R\backslash \left\{0\right\})\cap C^1\left(B_R\right)$ with ${\left|Du\right|}^{p-2}Du$, ${\left|Dv\right|}^{p-2}Dv\in C^1\left(B_R\right)$ satisfying $\lambda \left(D_i\left({\left|Du\right|}^{p-2}{D}_{j}u\right)\right),\lambda \left(D_i\left({\left|Dv\right|}^{p-2}{D}_{j}v\right)\right)$$\in {\Gamma }_k$ in $B_R\backslash \left\{0\right\}$. Moreover, $(u,v)$ is a radially symmetric solution of

$$\left\{\begin{array}{c}{\sigma }_k\left(\lambda \left(D_i\left({\left|Du\right|}^{p-2}{D}_{j}u\right)\right)\right)=\alpha \left(\right|x\left|\right)f_1^k\left(v\right)f_2^k\left(u\right),x\in B_R,\\ {\sigma }_k\left(\lambda \left(D_i\left({\left|Dv\right|}^{p-2}{D}_{j}v\right)\right)\right)=\beta \left(\right|x\left|\right)g_1^k\left(u\right)g_2^k\left(v\right),x\in B_R.\end{array}\right.$$

(20)

Proof of Lemma 2.

We firstly prove that $u\in C^2(B_R\backslash \left\{0\right\})\cap C^1\left(B_R\right)$. Indeed, by Lemma 1, we know that $\zeta \in C^2\left((0,R)\right)\cap C^1\left([0,R)\right)$ satisfies ${\zeta }^{\prime }\left(0\right)=0$ and ${\zeta }^{\prime }>0$ in $(0,R)$. Then, by the definition of u, we have that $u\in C^2(B_R\backslash \left\{0\right\}).$ Moreover, for any $x\ne 0,$ we have

$$\frac{\partial u}{\partial x_i}\left(x\right)={\zeta }^{\prime }\left(r\right)\frac{x_i}{r},\left|Du\left(x\right)\right|=|{\zeta }^{\prime }\left(r\right)|={\zeta }^{\prime }\left(r\right),i=1,2,\cdots ,n,$$

(21)

and

$$\frac{{\partial }^{2}u}{\partial x_i\partial x_j}\left(x\right)=\left({\zeta }^{″}\left(r\right)-\frac{{\zeta }^{\prime }\left(r\right)}{r}\right)\frac{x_i{x}_j}{r^2}+\frac{{\zeta }^{\prime }\left(r\right)}{r}{\delta }_{ij},i,j=1,2,\cdots ,n,$$

(22)

It follows from

$$\begin{array}{cc}\hfill \frac{\partial u}{\partial x_i}\left(0\right)& =\underset{h\to 0}{lim}\frac{u(0,\cdots ,h,0,\cdots ,0)-u\left(0\right)}{h}\hfill \\ \hfill & =\underset{h\to 0}{lim}\frac{\zeta \left(\right|h\left|\right)-\zeta \left(0\right)}{h}\hfill \\ \hfill & =\underset{h\to 0}{lim}\frac{\zeta \left(\right|h\left|\right)-\zeta \left(0\right)}{\left|h\right|}·\frac{\left|h\right|}{h}\hfill \\ \hfill & ={\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}\underset{h\to 0}{lim}\frac{{\int }_0^{\left|h\right|}{\left(t^{k-n}{\int }_0^t{s}^{n-1}\alpha \left(s\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right)ds\right)}^{\frac{1}{(p-1)k}}dt}{\left|h\right|}·\frac{\left|h\right|}{h}\hfill \\ \hfill & ={\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}\underset{h\to 0}{lim}{\left(\frac{{\int }_0^{\left|h\right|}{s}^{n-1}\alpha \left(s\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right)ds}{{\left|h\right|}^{n-k}}\right)}^{\frac{1}{(p-1)k}}·\frac{\left|h\right|}{h}\hfill \\ \hfill & =0.\hfill \end{array}$$

In addition, by (21) and the fact that ${\zeta }^{\prime }\left(0\right)=0$, we also have that $\underset{x\to \infty }{lim}Du\left(x\right)=0$. Therefore, we can conclude that $u\in C^1\left(B_R\right)$.

We secondly prove that ${\left|Du\right|}^{p-2}Du\in C^1\left(B_R\right)$. Indeed, by the definition of u, we have that ${\left|Du\right|}^{p-2}{D}_{j}u\in C^1\left(B_R\backslash \left\{0\right\}\right)$. Moreover, for any $x\ne 0$, by (21) and (22), we have

$$\begin{array}{cc}\hfill D_i\left({\left|Du\right|}^{p-2}{D}_{j}u\right)=D_i\left({\left({\zeta }^{\prime }\right)}^{p-1}\frac{x_j}{r}\right)=\left({\left({\left({\zeta }^{\prime }\right)}^{p-1}\right)}^{\prime }-\frac{{\left({\zeta }^{\prime }\right)}^{p-1}}{r}\right)& \frac{x_i{x}_j}{r^2}+\frac{{\left({\zeta }^{\prime }\right)}^{p-1}}{r}{\delta }_{ij},\hfill \\ \hfill & i,j=1,2,\cdots ,n.\hfill \end{array}$$

It follows that, for $i,j=1,2,\cdots ,n$,

$$\begin{array}{cc}\hfill & D_i\left({\left|Du\left(0\right)\right|}^{p-2}{D}_{j}u\left(0\right)\right)\hfill \\ \hfill =& \underset{h\to 0}{lim}\frac{{\left|Du\left(0,\cdots ,h,\cdots ,0\right)\right|}^{p-2}{D}_{j}u\left(0,\cdots ,h,\cdots ,0\right)-{\left|Du\left(0\right)\right|}^{p-2}{D}_{j}u\left(0\right)}{h}\hfill \\ \hfill =& \underset{h\to 0}{lim}\frac{({\zeta }^{\prime }{\left(\right|h\left|\right))}^{p-1}}{\left|h\right|}{\delta }_{ij}\hfill \\ \hfill =& {\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{k}}\underset{h\to 0}{lim}\frac{{\left({\left|h\right|}^{k-n}{\int }_0^{\left|h\right|}{s}^{n-1}\alpha \left(s\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right)ds\right)}^{\frac{1}{k}}}{\left|h\right|}{\delta }_{ij}\hfill \\ \hfill =& {\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{k}}\underset{h\to 0}{lim}{\left(\frac{{\int }_0^{\left|h\right|}{s}^{n-1}\alpha \left(s\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right)ds}{{\left|h\right|}^n}\right)}^{\frac{1}{k}}{\delta }_{ij}\hfill \\ \hfill =& {\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{k}}{\left(\underset{h\to 0}{lim}\frac{{\left|h\right|}^{n-1}\alpha \left(\right|h\left|\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right)}{{n\left|h\right|}^{n-1}}\right)}^{\frac{1}{k}}{\delta }_{ij}\hfill \\ \hfill =& {\left(\frac{\alpha \left(0\right)f_1^k\left(\frac{a}{2}\right)f_2^k\left(\frac{a}{2}\right)}{C_n^k}\right)}^{\frac{1}{k}}{\delta }_{ij}.\hfill \end{array}$$

In addition, we also have that, for $i,j=1,2,\cdots ,n,$

$$\begin{array}{cc}\hfill & \underset{x\to 0}{lim}{D}_i\left({\left|Du\right|}^{p-2}{D}_{j}u\right)\hfill \\ \hfill =& \underset{x\to 0}{lim}\left[\left({\left({\left({\zeta }^{\prime }\right)}^{p-1}\right)}^{\prime }-\frac{{\left({\zeta }^{\prime }\right)}^{p-1}}{r}\right)\frac{x_i{x}_j}{r^2}+\frac{{\left({\zeta }^{\prime }\right)}^{p-1}}{r}\right]{\delta }_{ij}\hfill \\ \hfill =& {\left(\frac{\alpha \left(0\right)f_1^k\left(\frac{a}{2}\right)f_2^k\left(\frac{a}{2}\right)}{C_n^k}\right)}^{\frac{1}{k}}{\delta }_{ij}.\hfill \end{array}$$

Indeed, differentiating the first line in (14) with respect to r directly, we have

$$\begin{array}{cc}\hfill {\zeta }^{″}=& \frac{1}{(p-1)C_{n-1}^{k-1}}{\left({\zeta }^{\prime }\right)}^{1-(p-1)k}\hfill \\ \hfill & ·\left[(k-n)r^{k-n-1}{\int }_0^r{s}^{n-1}\alpha \left(s\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right)ds+r^{k-1}\alpha \left(r\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right)\right].\hfill \end{array}$$

(23)

By the first line of (14), (23), and the L’Hôpital’s rule, we have

$$\underset{r\to 0^{+}}{lim}\frac{{\left({\zeta }^{\prime }\left(r\right)\right)}^{p-1}}{r}={\left(\frac{\alpha \left(0\right)f_1^k\left(\frac{a}{2}\right)f_2^k\left(\frac{a}{2}\right)}{C_n^k}\right)}^{\frac{1}{k}},$$

and

$$\begin{array}{cc}\hfill & \underset{r\to 0^{+}}{lim}{\left({\left({\zeta }^{\prime }\right)}^{p-1}\right)}^{\prime }=\underset{r\to 0^{+}}{lim}(p-1){\left({\zeta }^{\prime }\right)}^{p-2}{\zeta }^{″}\hfill \\ \hfill =& \frac{1}{C_{n-1}^{k-1}}\underset{r\to 0^{+}}{lim}{\left({\zeta }^{\prime }\right)}^{(p-1)(1-k)}\left[(k-n)r^{k-n-1}{\int }_0^r{s}^{n-1}\alpha \left(s\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right)ds+r^{k-1}\alpha \left(r\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right)\right]\hfill \\ \hfill =& \frac{1}{C_{n-1}^{k-1}}{\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1-k}{k}}\left[(k-n){\left(\underset{r\to 0^{+}}{lim}\frac{{\int }_0^r{s}^{n-1}\alpha \left(s\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right)ds}{r^n}\right)}^{\frac{1}{k}}\right.\hfill \\ \hfill & \left.+\alpha \left(0\right)f_1^k\left(\frac{a}{2}\right)f_2^k\left(\frac{a}{2}\right)\underset{r\to 0^{+}}{lim}{\left(\frac{r^n}{{\int }_0^r{s}^{n-1}\alpha \left(s\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right)ds}\right)}^{\frac{k-1}{k}}\right]\hfill \\ \hfill =& \frac{1}{k}{\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{k}}\left[(k-n){\left(\underset{r\to 0^{+}}{lim}\frac{r^{n-1}\alpha \left(r\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right)}{n{r}^{n-1}}\right)}^{\frac{1}{k}}\right.\hfill \\ \hfill & \left.+\alpha \left(0\right)f_1^k\left(\frac{a}{2}\right)f_2^k\left(\frac{a}{2}\right)\underset{r\to 0^{+}}{lim}{\left(\frac{n{r}^{n-1}}{r^{n-1}\alpha \left(r\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right)}\right)}^{\frac{k-1}{k}}\right]\hfill \\ \hfill =& \frac{1}{k}{\left(\frac{k\alpha \left(0\right)f_1^k\left(\frac{a}{2}\right)f_2^k\left(\frac{a}{2}\right)}{C_{n-1}^{k-1}}\right)}^{\frac{1}{k}}\left[(k-n){\left(\frac{1}{n}\right)}^{\frac{1}{k}}+n{\left(\frac{1}{n}\right)}^{\frac{1}{k}}\right]\hfill \\ \hfill =& {\left(\frac{\alpha \left(0\right)f_1^k\left(\frac{a}{2}\right)f_2^k\left(\frac{a}{2}\right)}{C_n^k}\right)}^{\frac{1}{k}}.\hfill \end{array}$$

Therefore, we can conclude that ${\left|Du\right|}^{p-2}Du\in C^1\left(B_R\right)$.

We thirdly prove that $\lambda \left(D_i\left({\left|Du\right|}^{p-2}{D}_{j}u\right)\right)\in {\Gamma }_k$ in $B_R\backslash \left\{0\right\}$. Indeed, the eigenvalues of $D_i\left({\left|Du\right|}^{p-2}{D}_{j}u\right)$ are

$$\lambda \left(D_i\left({\left|Du\right|}^{p-2}{D}_{j}u\right)\right)=\left({\left({\left({\zeta }^{\prime }\right)}^{p-1}\right)}^{\prime },\frac{{\left({\zeta }^{\prime }\right)}^{p-1}}{r},\cdots ,\frac{{\left({\zeta }^{\prime }\right)}^{p-1}}{r}\right),$$

It follows from the definition of ${\sigma }_l$, the first line of (15) in Lemma 1 and the positivity of $\alpha ,f_1,f_2$ that for $1\le l\le k$, $x\in B_R\backslash \left\{0\right\}$

$$\begin{array}{cc}\hfill {\sigma }_l\left(\lambda \left(D_i\left({\left|Du\right|}^{p-2}{D}_{j}u\right)\right)\right)& =C_{n-1}^{l-1}{\left(\frac{{\left({\zeta }^{\prime }\right)}^{p-1}}{r}\right)}^{l-1}\left[{\left({\left({\zeta }^{\prime }\right)}^{p-1}\right)}^{\prime }+\frac{n-l}{l}\frac{{\left({\zeta }^{\prime }\right)}^{p-1}}{r}\right]\hfill \\ \hfill & \ge C_{n-1}^{l-1}{\left(\frac{{\left({\zeta }^{\prime }\right)}^{p-1}}{r}\right)}^{l-1}\left[{\left({\left({\zeta }^{\prime }\right)}^{p-1}\right)}^{\prime }+\frac{n-k}{k}\frac{{\left({\zeta }^{\prime }\right)}^{p-1}}{r}\right]\hfill \\ \hfill & =\frac{C_{n-1}^{l-1}}{C_{n-1}^{k-1}}{\left(\frac{{\left({\zeta }^{\prime }\right)}^{p-1}}{r}\right)}^{l-k}\alpha \left(r\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right)\hfill \\ \hfill & >0.\hfill \end{array}$$

(24)

We fourthly prove that u satisfies the first line in (20). It follows from the definition of u, the first line of (15) in Lemma 1 and (24) that, for $x\in B_R$,

$$\begin{array}{cc}\hfill {\sigma }_k\left(\lambda \left(D_i\left({\left|Du\right|}^{p-2}{D}_{j}u\right)\right)\right)& =C_{n-1}^{k-1}{\left(\frac{{\left({\zeta }^{\prime }\right)}^{p-1}}{r}\right)}^{k-1}\left[{\left({\left({\zeta }^{\prime }\right)}^{p-1}\right)}^{\prime }+\frac{n-k}{k}\frac{{\left({\zeta }^{\prime }\right)}^{p-1}}{r}\right]\hfill \\ \hfill & =\alpha \left(r\right)f_1^k\left(\eta \right)f_2^k\left(\zeta \right)=\alpha \left(\right|x\left|\right)f_1^k\left(v\right)f_2^k\left(u\right).\hfill \end{array}$$

Analogous to the above argument, we can prove that $v\in C^2(B_R\backslash \left\{0\right\})\cap C^1\left(B_R\right)$ with ${\left|Dv\right|}^{p-2}Dv\in C^1\left(B_R\right)$, $\lambda \left(D_i\left({\left|Dv\right|}^{p-2}{D}_{j}v\right)\right)\in {\Gamma }_k$ in $B_R\backslash \left\{0\right\}$, and the second line of (20) holds. □

Applying Lemma 1 and Lemma 2 with $R=\infty$, $f_1=f$, $f_2\equiv 1$ and $\zeta \left(0\right)=a$, we have

Corollary 1.

Suppose that ($F_1$) and ($F_2$) holds. Let $a>0$. Assume that $\zeta \in C^0\left([0,\infty )\right)\cap C^1\left((0,\infty )\right)$ is a positive solution of

$$\left\{\begin{array}{cc}\hfill {\zeta }^{\prime }& ={\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}{\left(r^{k-n}{\int }_0^r{s}^{n-1}\alpha \left(s\right)f^k\left(\zeta \right)ds\right)}^{\frac{1}{(p-1)k}},r\in (0,\infty ),\hfill \\ \hfill \phi \left(0\right)& =a.\hfill \end{array}\right.$$

(25)

Let $u\left(x\right)=\zeta \left(\right|x\left|\right)$. Then, we have that u is a positive radially symmetric entire solution of (1).

3. Proof of Theorem 1

Proof. 

Let $a>0$. By Corollary 1, we only need to prove that there exists $u\in C^0\left([0,\infty )\right)\cap C^1\left((0,\infty )\right)$ satisfying (25). We define a sequence of positive functions ${\left\{u_m\right\}}_{m=0}^{\infty }\subset C^0\left([0,\infty )\right)$$\cap C^1\left((0,\infty )\right)$ by $u_0\left(r\right):=a$ and, for $m\in {\mathbb{N}}^{+}$,

$$u_m\left(r\right):=a+{\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}{\int }_0^r{\left(t^{k-n}{\int }_0^t{s}^{n-1}\alpha \left(s\right)f^k\left(u_{m-1}\right)ds\right)}^{\frac{1}{(p-1)k}}dt.$$

(26)

We firstly prove that ${\left\{u_m\right\}}_{m=0}^{\infty }$ is non-decreasing by the induction argument. Indeed, by (26) and the positivity of $\alpha$, f, it is easy to see that, for any $r\ge 0$,

$$u_1\left(r\right)-u_0\left(r\right)={\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}{\int }_0^r{\left(t^{k-n}{\int }_0^t{s}^{n-1}\alpha \left(s\right)f^k\left(a\right)ds\right)}^{\frac{1}{(p-1)k}}dt>0.$$

Suppose that, for some $m\in {\mathbb{N}}^{+}$, we have $u_m\left(r\right)\ge u_{m-1}\left(r\right),\forall r\ge 0$. Then, by the monotonicity of f, we have

$$\begin{array}{cc}\hfill u_{m+1}\left(r\right)& =a+{\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}{\int }_0^r{\left(t^{k-n}{\int }_0^t{s}^{n-1}\alpha \left(s\right)f^k\left(u_m\right)ds\right)}^{\frac{1}{(p-1)k}}dt\hfill \\ \hfill & \ge a+{\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}{\int }_0^r{\left(t^{k-n}{\int }_0^t{s}^{n-1}\alpha \left(s\right)f^k\left(u_{m-1}\right)ds\right)}^{\frac{1}{(p-1)k}}dt\hfill \\ \hfill & =u_m\left(r\right),\forall r\ge 0.\hfill \end{array}$$

We secondly prove that, for any $R\in (0,\infty )$, ${\left\{u_m\right\}}_{m=0}^{\infty }$ is uniformly bounded on $[0,R]$.

By (26), the monotonicity of f, $u_m$, the fact that $u_0\equiv a$ and the definition of A in (2), we have

$$\begin{array}{cc}\hfill u_m\left(r\right)& \ge a+{\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}{\int }_0^r{\left(t^{k-n}{\int }_0^t{s}^{n-1}\alpha \left(s\right)f^k\left(u_0\right)ds\right)}^{\frac{1}{(p-1)k}}dt\hfill \\ \hfill & \ge a+{\left(f\left(a\right)\right)}^{\frac{1}{p-1}}A\left(r\right),\forall r\ge 0.\hfill \end{array}$$

(27)

Differentiating (26) with respect to r directly, using the monotonicity of f, $u_m$ and the definition of A in (2), we have

$$\begin{array}{cc}\hfill u_m^{\prime }\left(r\right)& ={\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}{\left(r^{k-n}{\int }_0^r{s}^{n-1}\alpha \left(s\right)f^k\left(u_{m-1}\right)ds\right)}^{\frac{1}{(p-1)k}}\hfill \\ \hfill & \le {\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}{\left(r^{k-n}{\int }_0^r{s}^{n-1}\alpha \left(s\right)f^k\left(u_m\right)ds\right)}^{\frac{1}{(p-1)k}}\le f^{\frac{1}{(p-1)}}\left(u_m\right)A^{\prime }\left(r\right).\hfill \end{array}$$

(28)

Multiplying (28) with $f^{-\frac{1}{(p-1)}}\left(u_m\right)$ and integrating from 0 to r on both sides, using the facts that $u_m\left(0\right)=a$ and $A\left(0\right)=0$, we have

$${\int }_a^{u_m\left(r\right)}{f}^{-\frac{1}{(p-1)}}\left(\tau \right)d\tau \le A\left(r\right).$$

For any $r\ge 0$, by the fact that $u_m\left(r\right)\ge a$ and the definition of $D_{1a}$ in (4), we have that $D_{1a}\left(u_m\left(r\right)\right)\le A\left(r\right)$ and $u_m\left(r\right)\le D_{1a}^{-1}\left(A\left(r\right)\right)$. Therefore, by (27) and the monotonicity of A, we have

$$a+{\left(f\left(a\right)\right)}^{\frac{1}{p-1}}A\left(r\right)\le u_m\left(r\right)\le D_{1a}^{-1}\left(A\left(R\right)\right),\forall r\in [0,R].$$

We thirdly prove that, for any $R\in (0,\infty )$, ${\left\{u_m\right\}}_{m=0}^{\infty }$ is uniformly equicontinuous on $[0,R]$. For every $\epsilon >0$, there exists a

$$\delta ={\left(c_n^k\right)}^{\frac{1}{kp}}{\left(f^k\left(D_{1a}^{-1}\left(A\left(R\right)\right)\right)\underset{r\in [0,R]}{max}\alpha \left(r\right)\right)}^{-\frac{1}{kp}}{\left(\frac{p}{p-1}\epsilon \right)}^{\frac{p-1}{p}}>0,$$

such that, for all $m\ge 1$, $x,y\in [0,R]$ satisfy $0\le x-y<\delta$, it follows from (26), the continuity of $\alpha$, the monotonicity of f, and the boundedness of $u_m$ that

$$\begin{array}{cc}\hfill 0& \le u_m\left(x\right)-u_m\left(y\right)={\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}\left[{\int }_0^x{\left(t^{k-n}{\int }_0^t{s}^{n-1}\alpha \left(s\right)f^k\left(u_{m-1}\right)ds\right)}^{\frac{1}{(p-1)k}}dt\right.\hfill \\ \hfill & \left.-{\int }_0^y{\left(t^{k-n}{\int }_0^t{s}^{n-1}\alpha \left(s\right)f^k\left(u_{m-1}\right)ds\right)}^{\frac{1}{(p-1)k}}dt\right]\hfill \\ \hfill & \le {\left(\frac{k\underset{r\in [0,R]}{max}\alpha \left(r\right)f^k\left(u_{m-1}\left(R\right)\right)}{n{C}_{n-1}^{k-1}}\right)}^{\frac{1}{(P-1)k}}{\int }_y^x{t}^{\frac{1}{p-1}}dt\hfill \\ \hfill & \le {\left(\frac{\underset{r\in [0,R]}{max}\alpha \left(r\right)f^k\left(D_{1a}^{-1}\left(A\left(R\right)\right)\right)}{C_n^k}\right)}^{\frac{1}{(P-1)k}}\frac{p-1}{p}\left(x^{\frac{p}{p-1}}-y^{\frac{p}{p-1}}\right)\hfill \\ \hfill & \le {\left(\frac{\underset{r\in [0,R]}{max}\alpha \left(r\right)f^k\left(D_{1a}^{-1}\left(A\left(R\right)\right)\right)}{C_n^k}\right)}^{\frac{1}{(P-1)k}}\frac{p-1}{p}{(x-y)}^{\frac{p}{p-1}}\hfill \\ \hfill & <\epsilon .\hfill \end{array}$$

We fourthly apply the Arzelà–Ascoli theorem to conclude that ${\left\{u_m\right\}}_{m=0}^{\infty }$ has a uniformly convergent subsequence on $[0,R]$ for any $R\in (0,\infty )$.

By (26), the arbitrariness of R, and the convergence of $u_m$, we define that

$$u\left(r\right):=\underset{m\to \infty }{lim}{u}_m\left(r\right)=a+{\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}{\int }_0^r{\left(t^{k-n}{\int }_0^t{s}^{n-1}\alpha \left(s\right)f^k\left(u\right)ds\right)}^{\frac{1}{(p-1)k}}dt,r\ge 0.$$

(29)

Differentiating (29) with respect to r directly, we can derive (25). Applying Lemma 1 with $f_1=f,f_2\equiv 1$ and $R=\infty$, we can see that u is non-decreasing on $[0,\infty )$. Moreover, applying Corollary 1, we could infer that function u is a positive radially symmetric entire solution of (1) that satisfying $u\left(0\right)=a$ and

$$a+{\left(f\left(a\right)\right)}^{\frac{1}{p-1}}A\left(r\right)\le u\left(r\right)\le D_{1a}^{-1}\left(A\left(r\right)\right).$$

(30)

Letting $r\to \infty$ in (30), using the definitions of A and $D_{1a}$ in (3) and (4), respectively, we have that

$$a+{\left(f\left(a\right)\right)}^{\frac{1}{p-1}}A(\infty )\le \underset{r\to \infty }{lim}u\left(r\right)\le D_{1a}^{-1}\left(A(\infty )\right).$$

(31)

By (31), the monotonicity of u and the definition of $D_{1a}$ in (4), it is easy to see that, if $A(\infty )=D_{1a}(\infty )=\infty$, the solution u satisfies $\underset{r\to \infty }{lim}u\left(r\right)=\infty$; if $A(\infty )<D_{1a}(\infty )=\infty$ or $A(\infty )\le D_{1a}(\infty )<\infty$, the solution u satisfies $\underset{r\to \infty }{lim}u\left(r\right)\le D_{1a}^{-1}\left(A(\infty )\right)<\infty$, which implies u is bounded. □

4. Proof of Theorem 2

Proof. 

Let $a>0$. By Lemma 2, we only need to prove that there exist $u,v\in C^0\left([0,\infty )\right)\cap C^1\left((0,\infty )\right)$ satisfying (14). We define two sequences of positive functions ${\left\{u_m\right\}}_{m=0}^{\infty }$, ${\left\{v_m\right\}}_{m=0}^{\infty }$$\subset C^0\left([0,\infty )\right)\cap C^1\left((0,\infty )\right)$ by $u_0\left(r\right)=v_0\left(r\right):=\frac{a}{2}$ and, for $m\in {\mathbb{N}}^{+}$,

$$\left\{\begin{array}{c}{u}_m\left(r\right):=\frac{a}{2}+{\left(\frac{k}{C_{n-1}^{k-1}}{\int }_0^r\left(t^{k-n}{\int }_0^t{s}^{n-1}\alpha \left(s\right)f_1^k\left(v_{m-1}\right)f_2^k\left(u_{m-1}\right)ds\right)\right)}^{\frac{1}{(p-1)k}}dt,\hfill \\ v_m\left(r\right):=\frac{a}{2}+{\left(\frac{k}{C_{n-1}^{k-1}}{\int }_0^r\left(t^{k-n}{\int }_0^t{s}^{n-1}\beta \left(s\right)g_1^k\left(u_{m-1}\right)g_2^k\left(v_{m-1}\right)ds\right)\right)}^{\frac{1}{(p-1)k}}dt.\hfill \end{array}\right.$$

(32)

We firstly prove that ${\left\{u_m\right\}}_{m=0}^{\infty }$ is non-decreasing by the induction argument. Indeed, by the first line of (32) and the positivity of $\alpha$, $f_1$, $f_2$, it is easy to see that, for any $r\ge 0$,

$$u_1\left(r\right)-u_0\left(r\right)={\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}{\int }_0^r{\left(t^{k-n}{\int }_0^t{s}^{n-1}\alpha \left(s\right)f_1^k\left(a\right)f_2^k\left(a\right)ds\right)}^{\frac{1}{(p-1)k}}dt>0.$$

Suppose that, for some $m\in {\mathbb{N}}^{+}$, we have $u_m\left(r\right)\ge u_{m-1}\left(r\right),\forall r\ge 0$. Then, by the monotonicity of $f_1$ and $f_2$, we have

$$\begin{array}{cc}\hfill u_{m+1}\left(r\right)& =\frac{a}{2}+{\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}{\int }_0^r{\left(t^{k-n}{\int }_0^t{s}^{n-1}\alpha \left(s\right)f_1^k\left(v_m\right)f_2^k\left(u_m\right)\right)}^{\frac{1}{(p-1)k}}dt\hfill \\ \hfill & \ge \frac{a}{2}+{\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}{\int }_0^r{\left(t^{k-n}{\int }_0^t{s}^{n-1}\alpha \left(s\right)f_1^k\left(v_{m-1}\right)f_2^k\left(u_{m-1}\right)ds\right)}^{\frac{1}{(p-1)k}}dt\hfill \\ \hfill & =u_m\left(r\right),\forall r\ge 0.\hfill \end{array}$$

We secondly prove that, for any $R\in (0,\infty )$, ${\left\{u_m\right\}}_{m=0}^{\infty }$ is uniformly bounded on $[0,R]$.

By the first line of (32), the monotonicity of $f_1,f_2,u_m$, the fact that $u_0=v_0\equiv \frac{a}{2}$ and the definition of A in (2), we have

$$\begin{array}{cc}\hfill u_m\left(r\right)& \ge \frac{a}{2}+{\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}{\int }_0^r{\left(t^{k-n}{\int }_0^t{s}^{n-1}\alpha \left(s\right)f_1^k\left(v_0\right)f_2^k\left(u_0\right)ds\right)}^{\frac{1}{(p-1)k}}dt\hfill \\ \hfill & \ge \frac{a}{2}+{\left(f_1\left(\frac{a}{2}\right)f_2\left(\frac{a}{2}\right)\right)}^{\frac{1}{p-1}}A\left(r\right),\forall r\ge 0.\hfill \end{array}$$

(33)

Differentiating the first line of (32) with respect to r directly, using the monotonicity of $f_1$, $f_2$, $u_m$, the definition of A in (2) and the positivity of $g_1,g_2$, we have

$$\begin{array}{cc}\hfill u_m^{\prime }\left(r\right)& ={\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{p-1}}{\left(r^{k-n}{\int }_0^r{s}^{n-1}\alpha \left(s\right)f_1^k\left(v_{m-1}\right)f_2^k\left(u_{m-1}\right)ds\right)}^{\frac{1}{(p-1)k}}\hfill \\ \hfill & \le {\left(f_1\left(v_m\left(r\right)\right)f_2\left(u_m\left(r\right)\right)\right)}^{\frac{1}{p-1}}{A}^{\prime }\left(r\right)\hfill \\ \hfill & \le \left(f_1(v_m\left(r\right)+u_m\left(r\right))f_2(v_m\left(r\right)+u_m\left(r\right))\right.\hfill \\ \hfill & {\left.+g_1(u_m\left(r\right)+v_m\left(r\right))g_2(u_m\left(r\right)+v_m\left(r\right))\right)}^{\frac{1}{p-1}}{A}^{\prime }\left(r\right).\hfill \end{array}$$

(34)

Similarly, we can prove that

$$\begin{array}{cc}\hfill v_m^{\prime }\left(r\right)& \le \left(f_1(v_m\left(r\right)+u_m\left(r\right))f_2(v_m\left(r\right)+u_m\left(r\right))\right.\hfill \\ \hfill & {\left.+g_1(u_m\left(r\right)+v_m\left(r\right))g_2(u_m\left(r\right)+v_m\left(r\right))\right)}^{\frac{1}{p-1}}{B}^{\prime }\left(r\right).\hfill \end{array}$$

(35)

Adding both sides of (34) and (35), respectively, we have

$$\begin{array}{cc}\hfill u_m^{\prime }\left(r\right)+v_m^{\prime }\left(r\right)\le & \left(f_1(v_m\left(r\right)+u_m\left(r\right))f_2(v_m\left(r\right)+u_m\left(r\right))\right.\hfill \\ \hfill & {\left.+g_1(u_m\left(r\right)+v_m\left(r\right))g_2(u_m\left(r\right)+v_m\left(r\right))\right)}^{\frac{1}{p-1}}\left(A^{\prime }\left(r\right)+B^{\prime }\left(r\right)\right).\hfill \end{array}$$

(36)

Multiplying (36) with

$${\left(f_1(v_m\left(r\right)+u_m\left(r\right))f_2(v_m\left(r\right)+u_m\left(r\right))+g_1(u_m\left(r\right)+v_m\left(r\right))g_2(u_m\left(r\right)+v_m\left(r\right))\right)}^{-\frac{1}{(p-1)}}$$

and integrating from 0 to r on both sides, using the facts that $u_m\left(0\right)+v_m\left(0\right)=a$ and $A\left(0\right)=B\left(0\right)=0$, we have

$${\int }_a^{u_m\left(r\right)+v_m\left(r\right)}{(f_1\left(\tau \right)f_2\left(\tau \right)+g_1\left(\tau \right)g_2\left(\tau \right))}^{-\frac{1}{p-1}}d\tau \le A\left(r\right)+B\left(r\right),$$

for any $r\ge 0$, by the fact that $u_m\left(r\right)+v_m\left(r\right)\ge a$ and the definition of $D_{2a}$ in (11), we have $D_{2a}\left(u_m\left(r\right)+v_m\left(r\right)\right)\le A\left(r\right)+B\left(r\right)$ and $u_m\left(r\right)+v_m\left(r\right)\le D_{2a}^{-1}(A\left(r\right)+B\left(r\right).$ Therefore, by (33) and the monotonicity of A, B, we have

$$\frac{a}{2}+{\left(f_1\left(\frac{a}{2}\right)f_2\left(\frac{a}{2}\right)\right)}^{\frac{1}{p-1}}A\left(r\right)\le u_m\left(r\right)\le D_{2a}^{-1}(A\left(R\right)+B\left(R\right)),\forall r\in [0,R].$$

We thirdly prove that, for any $R\in (0,\infty )$, ${\left\{u_m\right\}}_{m=0}^{\infty }$ is uniformly equicontinuous on $[0,R]$. For every $\epsilon >0$, there exists a

$$\begin{array}{cc}\hfill \delta =& {\left(c_n^k\right)}^{\frac{1}{kp}}{\left(f_1^k\left(D_{2a}^{-1}(A\left(R\right)+B\left(R\right)\right)f_2^k\left(D_{2a}^{-1}(A\left(R\right)+B\left(R\right))\right)\underset{r\in [0,R]}{max}\alpha \left(r\right)\right)}^{-\frac{1}{kp}}{\left(\frac{p}{p-1}\epsilon \right)}^{\frac{p-1}{p}}\hfill \\ \hfill >& 0,\hfill \end{array}$$

such that, for all $m\ge 1$, $x,y\in [0,R]$ satisfy $0\le x-y<\delta$, it follows from (26), the continuity of $\alpha$, the monotonicity of $f_1,f_2$ and the boundedness of $u_m$ that

$$\begin{array}{cc}\hfill & 0\le u_m\left(x\right)-u_m\left(y\right)={\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}\left[{\int }_0^x{\left(t^{k-n}{\int }_0^t{s}^{n-1}\alpha \left(s\right)f_1^k\left(u_{m-1}\right)f_2^k\left(u_{m-1}\right)ds\right)}^{\frac{1}{(p-1)k}}dt\right.\hfill \\ \hfill & \left.-{\int }_0^y{\left(t^{k-n}{\int }_0^t{s}^{n-1}\alpha \left(s\right)f_1^k\left(u_{m-1}\right)f_2^k\left(u_{m-1}\right)ds\right)}^{\frac{1}{(p-1)k}}dt\right]\hfill \\ \hfill & \le {\left(\frac{k\underset{r\in [0,R]}{max}\alpha \left(r\right)f_1^k\left(u_{m-1}\left(R\right)\right)f_2^k\left(u_{m-1}\left(R\right)\right)}{n{C}_{n-1}^{k-1}}\right)}^{\frac{1}{(P-1)k}}{\int }_y^x{t}^{\frac{1}{p-1}}dt\hfill \\ \hfill & \le {\left(\frac{\underset{r\in [0,R]}{max}\alpha \left(r\right)f_1^k\left(D_{2a}^{-1}(A\left(R\right)+B\left(R\right)\right)f_2^k\left(D_{2a}^{-1}(A\left(R\right)+B\left(R\right))\right)}{C_n^k}\right)}^{\frac{1}{(P-1)k}}\frac{p-1}{p}\left(x^{\frac{p}{p-1}}-y^{\frac{p}{p-1}}\right)\hfill \\ \hfill & \le {\left(\frac{\underset{r\in [0,R]}{max}\alpha \left(r\right)f_1^k\left(D_{2a}^{-1}(A\left(R\right)+B\left(R\right)\right)f_2^k\left(D_{2a}^{-1}(A\left(R\right)+B\left(R\right))\right)}{C_n^k}\right)}^{\frac{1}{(P-1)k}}\frac{p-1}{p}{(x-y)}^{\frac{p}{p-1}}\hfill \\ \hfill & <\epsilon .\hfill \end{array}$$

We fourthly apply the Arzelà–Ascoli theorem to conclude that ${\left\{u_m\right\}}_{m=0}^{\infty }$ has a uniformly convergent subsequence on $[0,R]$ for any $R\in (0,\infty )$.

Analogous to the above argument, we can prove that ${\left\{v_m\right\}}_{m=0}^{\infty }$ is non-decreasing. Moreover, for any $R\in (0,\infty )$, on $[0,R]$, ${\left\{v_m\right\}}_{m=0}^{\infty }$ is uniformly bounded, uniformly equicontinuous and has a uniformly convergent subsequence.

By (32), the arbitrariness of R and the convergence of $u_m,v_m$, we define that

$$u\left(r\right):=\underset{m\to \infty }{lim}{u}_m\left(r\right)=\frac{a}{2}+{\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}{\int }_0^r{\left(t^{k-n}{\int }_0^t{s}^{n-1}\alpha \left(s\right)f_1^k\left(v\right)f_2^k\left(u\right)ds\right)}^{\frac{1}{(p-1)k}}dt,r\ge 0,$$

(37)

and

$$v\left(r\right):=\underset{m\to \infty }{lim}{v}_m\left(r\right)=\frac{a}{2}+{\left(\frac{k}{C_{n-1}^{k-1}}\right)}^{\frac{1}{(p-1)k}}{\int }_0^r{\left(t^{k-n}{\int }_0^t{s}^{n-1}\beta \left(s\right)g_1^k\left(u\right)g_1^k\left(v\right)ds\right)}^{\frac{1}{(p-1)k}}dt,r\ge 0.$$

(38)

Differentiating (37) and (38) with respect to r directly, we can derive the first two lines of (14). Applying Lemma 1 with $R=\infty$, we can see that u, v are non-decreasing on $[0,\infty )$. Moreover, applying Lemma 2, we could infer that ($u,v$) is a positive radially symmetric entire solution of (6) satisfying $u\left(0\right)=v\left(0\right)=\frac{a}{2}$, (12) and (13).

Letting $r\to \infty$ in (12) and (13), respectively, using the definitions of $A,B$ and $D_{2A}$ in (3), (10) and (11), respectively, we have that

$$\frac{a}{2}+{\left(f_1\left(\frac{a}{2}\right)f_2\left(\frac{a}{2}\right)\right)}^{\frac{1}{p-1}}A(\infty )\le \underset{r\to \infty }{lim}u\left(r\right)\le D_{2a}^{-1}(A(\infty )+B(\infty )),$$

(39)

and

$$\frac{a}{2}+{\left(g_1\left(\frac{a}{2}\right)g_2\left(\frac{a}{2}\right)\right)}^{\frac{1}{p-1}}B(\infty )\le \underset{r\to \infty }{lim}v\left(r\right)\le D_{2a}^{-1}(A(\infty )+B(\infty )).$$

(40)

By (39), (40), the monotonicity of $u,v$, and the definition of $D_{2A}$ in (11), it is easy to see that if $A(\infty )=B(\infty )=D_{2a}(\infty )=\infty$, the solution $(u,v)$ satisfies $\underset{r\to \infty }{lim}u\left(r\right)=\underset{r\to \infty }{lim}v\left(r\right)=\infty$; if $A(\infty )+B(\infty )<D_{2a}(\infty )=\infty$ or $A(\infty )+B(\infty )\le D_{2a}(\infty )<\infty$, the solution $(u,v)$ satisfies $\underset{r\to \infty }{lim}u\left(r\right)\le D_{2a}^{-1}(A(\infty )+B(\infty ))<\infty$, $\underset{r\to \infty }{lim}v\left(r\right)\le D_{2a}^{-1}(A(\infty )+B(\infty ))<\infty$, which implies that u and v are bounded. □