Existence of Positive Solutions for a Singular Hessian Equation with a Negative Augmented Term

Abstract

In this paper, we focus on the existence of positive solutions for a singular Hessian equation with a negative augmented term. By finding more appropriate upper and lower solutions, we not only overcome the difficulty due to the negative augmented term but also remove a critical condition required in the existing work and establish new results for the existence of positive solutions of the equations under study. Our results improve and complement many existing works.

1 Introduction

In this paper, we focus on the existence of positive solutions for the following singular augmented Hessian equation

$$\begin{aligned} \left\{ \begin{aligned}&-\varrho ^{\frac{1}{k}}_k[\mu (D^2u-\varsigma (x) I)]=f(u),\ \ \hbox {in} \ \ {\Omega },\\&u=0 \ \ \ \hbox {on} \ \ \ \partial {\Omega }, \end{aligned} \right. \end{aligned}$$

(1.1)

where ${\Omega }$ is an open unit ball in $\mathbb {R}^n$, $k\le n<2k$, $D^2u-\varsigma (x) I$ is an augmented Hessian matrix, obtained from the standard Hessian matrix by subtracting a lower order symmetric matrix function, $\varrho _k(\mu (D^2u-\varsigma (x) I))$ is an augmented k-Hessian operator defined by

$$\begin{aligned} \varrho _k(\mu (D^2u-\varsigma (x) I))=\sum _{1\le i_1<i_2<\cdot \cdot \cdot <i_k\le n}\mu _{i_1}\cdot \mu _{i_2}\cdot \cdot \cdot \mu _{i_k},\ \ k=1,2,\cdot \cdot \cdot ,n, \end{aligned}$$

where $\mu _1,\mu _2,...,\mu _n$ are the eigenvalues of the augmented Hessian matrix $D^2u-\varsigma (x) I$, and

$$\begin{aligned} \mu (D^2u-\varsigma (x) I)=(\mu _1,\mu _2,\cdot \cdot \cdot ,\mu _n) \end{aligned}$$

is the vector of eigenvalues of $D^2u-\varsigma (x) I$.

The general augmented Hessian equation of the form

$$\begin{aligned} F[\mu (D^2u-A(x, u, Du))]=B(x,u,Du), \ \ \hbox {in} \ \ \Omega , \end{aligned}$$

(1.2)

is a class of fully nonlinear partial differential equations, where $\Omega$ is an open set in $\mathbb {R}^n$, the scalar function F is defined on an open cone in the linear space $S^{n\times n}$ of $n\times n$ real symmetric matrices, $A:\Omega \times \mathbb {R}\times \mathbb {R}^n\rightarrow S^{n\times n}$ is a symmetric matrix function and $A:\Omega \times \mathbb {R}\times \mathbb {R}^n\rightarrow \mathbb {R}$ is a scalar function. Such questions arise in the study of optimal transportation, geometric optics and conformal geometry [1,2,3,4]. They can also be applied to Neumann problems arising from prescribed mean curvature problems in conformal geometry as well as general oblique boundary value problems for augmented k-Hessian, Hessian quotient equations and certain degenerate equations. Recently, Amal et al. [5] enlarged the pluri-potentiel theory for complex Hessian equations on certain bounded domains. They proved the Hölder continuity for the Hessian equation on a m-hyperconvex domain of m-subharmonic type k, and under suitable hypotheses on the data, the existence of Hölder continuity solutions to the Dirichlet problem was established. In a recent work [2], by assuming a natural convexity condition for the domain and some appropriate convexity conditions for the matrix function in the augmented Hessian equation, Jiang and Trudinger established the global regularity of classical elliptic solutions for the augmented Hessian Eq. (1.2) subject to an oblique boundary condition. In [4], by using various derivative estimates and barrier constructions, Jiang and Trudinger extended their previous results for the Monge-Ampère and k-Hessian cases to general classes of augmented Hessian equations under Dirichlet boundary conditions in Euclidean space. Recently, Dai [6] considered the augmented Hessian equations

$$\begin{aligned} \varrho ^{\frac{1}{k}}_k[\mu (D^2u+\varsigma (x) I)]=f(u), \end{aligned}$$

(1.3)

in the whole space $\mathbb {R}^n$ and in the half space $\mathbb {R}^n_+:=\{x\in \mathbb {R}^n | x_n > 0\}$. The necessary and sufficient condition for the existence of classical subsolutions to the equations in $\mathbb {R}^n$ for $\varsigma (x) = \alpha ,\alpha \ge 0$, was established. The authors also obtained the nonexistence of positive viscosity subsolutions of the equations in $\mathbb {R}^n$ or $\mathbb {R}^n_+$ for $f (u) = u^p, p>1$. By employing the method of upper and lower solutions, Zhang et al. [7] established a new result on the existence of radial solutions for the following eigenvalue problem of a singular augmented Hessian equation

$$\begin{aligned} \left\{ \begin{aligned}&-S^{\frac{1}{k}}_k(\mu (D^2u+\lambda \alpha I))=\lambda f(|x|,u),\ \ \hbox {in} \ \ {\Omega }, \\&u=0 \ \ \ \hbox {on} \ \ \ \partial {\Omega }, \end{aligned}\right. \end{aligned}$$

(1.4)

where ${\Omega }$ is an open unit ball in $\mathbb {R}^n$, in which $n\in [k,2k]$, and the nonlinearity in f may be singular in some space variables. However there is a typo in (2.1) of [7], namely the $-\lambda \alpha I$ in (2.1) should be replaced by $+\lambda \alpha I$.

From [7] and the related literature, the case with the augmented term being $-\lambda \alpha I$ has not been studied in detail. Thus motivated by the aforementioned work, in this paper, by optimizing the method of upper and lower solutions, we consider the existence of positive solutions for the singular augmented Hessian Eq. (1.1) when $\varsigma (x)=\alpha , \alpha \ge 0$, and f satisfies the following basic assumption:

$(\textbf{G})$ $f:(0,+\infty )\rightarrow (\alpha _0,+\infty )$ is continuous and non-increasing, where $\alpha _0>(C_n^k)^{\frac{1}{k}}(\alpha +2)$.

Obviously, this case is more difficult to deal with than the case with $\lambda \alpha I$ because it will affect the sign of the corresponding operator. The main contributions of this paper include the following aspects:

• Derive a priori estimates for the first and second derivatives to overcome the difficulty due to the negative augmented term;

• Derive more appropriate upper and lower solutions which is usually difficult;

• Remove an essential condition $(\textbf{H}_{2})$ of [7] which is crucial in the proof of the work [7];

• Establish various new results for the existence of positive solutions for the singular augmented Hessian Eq. (1.1), filling the gap in the work [7].

Our work is related to some recent works on other Hessian equation [8,9,10,11,12,13], and improve and generalize some results in [14, 15, 25,26,27]. In addition, this work is related to some recent works on theories and methods of analysis and numerical techniques such as fixed-point theory [18,19,20,21], method of reduction of dimension [22], function spaces theories [23, 28,29,30,31], regularity theories [24, 32,33,34,35]. For example, based on fixed point theorems for monotone and mixed monotone operators in a normal cone, Duan, Liao and Tang [19] prove that the nonlinear matrix equation

$$\begin{aligned} X-\sum ^m_{i=1} A^*_i X^{\delta _i}A_i=Q, \ 0< |\delta _i| < 1 \end{aligned}$$

always has a unique positive definite solution, and a multi-step stationary iterative method is proposed to compute the unique positive definite solution. Recently, Ran and Reurings [20] showed an analogue of Banach’s fixed point theorem in partially ordered sets for solving the general nonlinear matrix equations

$$\begin{aligned} X=Q\pm \sum ^m_{j=1} A^*_j\mathcal {F}(X)A_j, \end{aligned}$$

where Q is a positive definite matrix and $A_1,...,A_m$ are arbitrary $n \times n$ matrices, and F is a continuous and monotone map from the set of positive definite matrices to itself. In [18], the authors studied a parametric nonlinear elliptic problem driven by the (p, q)-Laplacian operator and with singular, concave and convex terms. By employing a topological approach based on the Schauder-Tychonov fixed-point theorem, positive solutions and a bifurcation-type theorem describing the changes in the set of positive solutions were derived.

This paper is organised as follows. In Sect. 2, we firstly calculate the eigenvalues of the Hessian matrix with an augmented term, and then give the definition of lower and upper solutions and a lemma which will be used in the rest of this paper. In Sect. 3, we find more appropriate upper and lower solutions of the equation and give the proof of our main results. In Sect. 4, two examples are given to illustrate our main results.

2 Basic Definitions and Preliminaries

Let

$$\begin{aligned} r = |x| ={\left( \sum _{i=1}^n x^2_i\right) ^\frac{1}{2}}. \end{aligned}$$

Then we have the following Lemma.

Lemma 2.1

Assume that $v:[0,+\infty )\rightarrow \mathbb {R}$ is a $C^2$ radially symmetric function with $v'(0)=0,$ and $\varsigma :[0,+\infty )\rightarrow \mathbb {R}^+$ is a $C^1$ radially symmetric function. Then for $u(x)=v(r)$ where $r=|x|$, we have $u(x) \in C^2(\mathbb {R}^n)$. Moreover the vector of eigenvalues of $D^2u-\varsigma (x) I$ and the augmented k-Hessian operator $\varrho _k(D^2u(x)-\varsigma (|x|)I$ are as follows

$$\begin{aligned}{} & {} \mu \left( D^2u(x)-\varsigma (|x|) I\right) \nonumber \\{} & {} \quad =\left\{ \left( v''(r)-\varsigma (r), \frac{v'(r)}{r}-\varsigma (r),..., \frac{v'(r)}{r}-\varsigma (r)\right) ,\ \ r\in (0,\infty ),\right. \nonumber \\{} & {} \quad \left. \left( v''(0)-\varsigma (0), v''(0)-\varsigma (0),..., v''(0)-\varsigma (0) \right) \ \ \ r=0, \right. \end{aligned}$$

(2.1)

$$\begin{aligned}{} & {} \varrho _k(D^2u(x)-\varsigma (|x|) I)\nonumber \\{} & {} \quad =\left\{ C_{n-1}^{k-1}(v''(r){-}\varsigma (r) )\left( \frac{v'(r){-}\varsigma (r) r}{r}\right) ^{k-1}{+}C_{n-1}^{k}\left( \frac{v'(r){-}\varsigma (r) r}{r}\right) ^{k}, r\in (0,\infty ), \right. \nonumber \\{} & {} \left. \quad C_{n}^{k}\left( v''(0)-\varsigma (0) \right) ^{k}, \ \ \ r=0, \right. \end{aligned}$$

(2.2)

where I denotes the unit matrix.

Proof

Firstly, for $u\not =0, 1\le i,j\le n,$ it follows from $u(x)=v(r)$ that

$$\begin{aligned} \frac{\partial u(x)}{\partial x_i}=\frac{v'(r)}{r}x_i,\ \ \ \ \frac{\partial ^2u(x)}{\partial x_i\partial x_j}=\frac{v''(r)}{r^2}x_ix_j-\frac{v'(r)}{r^3}x_ix_j+\frac{v'(r)}{r}\delta _{ij} \end{aligned}$$

(2.3)

where

$$\begin{aligned} \delta _{ij}=\left\{ \begin{aligned}&1,\ \ i=j,\ \\&0, \ \ \ i\not = j. \end{aligned} \right. \end{aligned}$$

(2.4)

Thus by using (2.3) and $v'(0)=0$, one has

$$\begin{aligned} u'(0)=\lim _{x\rightarrow 0} \frac{\partial u(x)}{\partial x_i}=\lim _{x\rightarrow 0}\left( \frac{v'(r)-v'(0)}{r-0} x_i \right) =v''(0)\cdot 0=0, \end{aligned}$$

(2.5)

and

$$\begin{aligned} u''(0)=\lim _{x\rightarrow 0} \frac{\partial ^2u(x)}{\partial x_i\partial x_j}=\lim _{x\rightarrow 0}\left[ \left( {v''(r)-\frac{v'(r)}{r}}\right) \frac{x_ix_j}{r^2} +\frac{v'(r)}{r}\delta _{ij}\right] =v''(0)\delta _{ij}. \nonumber \\ \end{aligned}$$

(2.6)

So (2.3) and (2.6) imply that $u(x) \in C^2(\mathbb {R}^n)$.

In what follows, we calculate the eigenvalues of $D^2u(x)-\varsigma (|x|) I$. To do this, let

$$\begin{aligned} c=\left\{ \begin{aligned}&\frac{v''(r)}{r^2}-\frac{v'(r)}{r^3},\ \ r\in (0,\infty ),\ \\&0, \ \ \ r=0. \end{aligned} \right. \ \ \ \ d=\left\{ \begin{aligned}&\frac{v'(r)}{r},\ \ r\in (0,\infty ),\ \\&0, \ \ \ r=0. \end{aligned} \right. \end{aligned}$$

(2.7)

and $x=(x_1,x_2,\cdot \cdot \cdot ,x_n)$, then

$$\begin{aligned} A=D^2u(x)-\varsigma (|x|) I=c x^Tx+(d-\varsigma (r))I. \end{aligned}$$

Thus

$$\begin{aligned} \begin{aligned} |\mu I-A|&= \left| \begin{array}{ccccc} \mu -c x^2_1-d+\varsigma (r) &{}-cx_1x_2 &{}-cx_1x_3 &{} \cdot \cdot \cdot &{} -cx_1x_n \\ -cx_2x_1 &{} \mu -c x^2_2-d+\varsigma (r) &{} -cx_2x_3 &{} \cdot \cdot \cdot &{} -cx_2x_n \\ \cdot \cdot \cdot &{} \cdot \cdot \cdot &{} \cdot \cdot \cdot &{} \cdot \cdot \cdot &{} \cdot \cdot \cdot \\ -cx_nx_1 &{} -cx_nx_2 &{} -cx_nx_3&{} \cdot \cdot \cdot &{} \mu -c x^2_n-d+\varsigma (r) \end{array} \right| \\&\overset{r_i-\frac{x_i}{x_1}r_1}{\underset{i=2,3,\cdot \cdot \cdot ,n}{=}} \left| \begin{array}{ccccc} \mu -c x^2_1-d+\varsigma (r) &{}-cx_1x_2 &{}-cx_1x_3 &{} \cdot \cdot \cdot &{} -cx_1x_n \\ (d+\varsigma (r)-\mu )\frac{x_2}{x_1} &{} \mu -d+\varsigma (r) &{} 0 &{} \cdot \cdot \cdot &{} 0 \\ \cdot \cdot \cdot &{} \cdot \cdot \cdot &{} \cdot \cdot \cdot &{} \cdot \cdot \cdot &{} \cdot \cdot \cdot \\ (d+\varsigma (r)-\mu )\frac{x_n}{x_1} &{} 0 &{} 0&{} \cdot \cdot \cdot &{} \mu -d+\varsigma (r) \end{array} \right| \\ {}&\overset{c_1+\frac{x_i}{x_1}c_i}{\underset{i=2,3,\cdot \cdot \cdot ,n}{=}} \left| \begin{array}{ccccc} \mu -c r^2-d+\varsigma (r) &{}-cx_1x_2 &{}-cx_1x_3 &{} \cdot \cdot \cdot &{} -cx_1x_n \\ 0 &{} \mu -d+\varsigma (r) &{} 0 &{} \cdot \cdot \cdot &{} 0 \\ \cdot \cdot \cdot &{} \cdot \cdot \cdot &{} \cdot \cdot \cdot &{} \cdot \cdot \cdot &{} \cdot \cdot \cdot \\ 0 &{} 0 &{} 0&{} \cdot \cdot \cdot &{} \mu -d+\varsigma (r) \end{array} \right| \\ {}&=( \mu -d+\varsigma (r))^{n-1}(\mu -c r^2-d+\varsigma (r)). \end{aligned} \end{aligned}$$

(2.8)

Therefore, the eigenvalues of the matrix $D^2u(x)-\varsigma (r) I$ are

$$\begin{aligned}{} & {} \left( c r^2+d-\varsigma (r), d-\varsigma (r),d-\varsigma (r),\cdot \cdot \cdot ,d-\varsigma (r)\right) \nonumber \\{} & {} \quad =\left( v''(r)-\varsigma (r),\frac{v'(r)}{r}-\varsigma (r),...,\frac{v'(r)}{r}-\varsigma (r) \right) . \end{aligned}$$

(2.9)

Consequently, it follows from (2.6) to (2.9) that (2.1) holds. From the definition of $\varrho _k$, it is easy to obtain the Eq. (2.2). $\square$

It follows from Lemma 2.1 that in radial coordinates, the equation (1.1) is equivalent to the following equation (also see [16])

$$\begin{aligned} \left\{ \begin{aligned}&(-1)^k\left[ C_{n-1}^{k-1}(v''(r)-\varsigma (r) )\left( \frac{v'(r)-\varsigma (r) r}{r}\right) ^{k-1}+C_{n-1}^{k}\left( \frac{v'(r)-\varsigma (r) r}{r}\right) ^{k}\right] \\&=f^k(v(r)),\ \ r\in (0,1), \\&v'(0)=0,\ \ v(1)=0, \end{aligned} \right. \nonumber \\ \end{aligned}$$

(2.10)

which can be written in the following equivalent form

$$\begin{aligned} \left\{ \begin{aligned}&(-1)^kC_{n-1}^{k-1}\left[ {r^{n-k}}(v'(r)-\varsigma (r) r)^k\right] '\\&=k{r^{n-1}}\left[ f(v(r))\right] ^k-kC_{n-1}^{k-1}\varsigma '(r) r^k(v'(r)-\varsigma (r) r)^{k-1},\ \ r\in (0,1), \\&v'(0)=0,\ \ v(1)=0. \end{aligned} \right. \nonumber \\ \end{aligned}$$

(2.11)

Remark 2.1

How to eliminate the effects of $kC_{n-1}^{k-1}\varsigma '(r) r^k(v'(r)-\varsigma (r) r)^{k-1}$, in order to transform (2.11) into an effective integral equation, is still an open problem. In this paper, we only focus on the Hessian Eq. (1.1) under the case where the augmented term is $-\alpha I$, namely $\varsigma (r)=\alpha$.

Lemma 2.2

Assume that f(r) is a continuous function on $(0,+\infty )$ satisfying $f>\alpha _0$ and $\varsigma (r)=\alpha$. Let $v(r)\in C[0,1]\ \cap C^1(0,1)$ satisfies

$$\begin{aligned} v(r)=\int _r^1\left[ {t^{\frac{k-n}{k}}}\left( \int _0^t\frac{ks^{n-1}}{C_{n-1}^{k-1}} f^k(v(s))ds\right) ^{\frac{1}{k}}-\alpha t\right] dt,\ \ r\in [0,1], \end{aligned}$$

(2.12)

then $v(r)\in C^2[0,1]$ and satisfies (2.10).

Proof

Firstly, it is obvious that $v(1)=0$ from (2.12). Moreover,

$$\begin{aligned} v(0)=\int _0^1\left[ {t^{\frac{k-n}{k}}}\left( \int _0^t\frac{ks^{n-1}}{C_{n-1}^{k-1}} f^k(v(s))ds\right) ^{\frac{1}{k}}-\alpha t\right] dt\ge \frac{1}{2}((C^k_n)^{-\frac{1}{k}}\alpha _0-\alpha )>0,\nonumber \\ \end{aligned}$$

(2.13)

and

$$\begin{aligned} v'(r)=-{r^{\frac{k-n}{k}}}\left( \int _0^r\frac{ks^{n-1}}{C_{n-1}^{k-1}} f^k(v(s))ds\right) ^{\frac{1}{k}}+\alpha r, \end{aligned}$$

(2.14)

that is

$$\begin{aligned} \ -v'(r)+\alpha r={r^{\frac{k-n}{k}}}\left( \int _0^r\frac{ks^{n-1}}{C_{n-1}^{k-1}} f^k(v(s))ds\right) ^{\frac{1}{k}}\ge 0. \end{aligned}$$

(2.15)

Hence it follows from (2.15) and L’Hôpital’s rule that

$$\begin{aligned} \begin{aligned}&v'(0)=\lim _{r\rightarrow 0}v'(r)=\lim _{r\rightarrow 0}\left[ -{r^{\frac{k-n}{k}}}\left( \int _0^r\frac{ks^{n-1}}{C_{n-1}^{k-1}} f^k(v(s))ds\right) ^{\frac{1}{k}}+\alpha r\right] =0, \end{aligned} \nonumber \\ \end{aligned}$$

(2.16)

which implies that $v(r)\in C^1[0,1].$

On the other hand, we have

$$\begin{aligned} \begin{aligned} v''(0)&=\lim _{r\rightarrow 0}\frac{v'(r)-v'(0)}{r-0}\\&=\lim _{r\rightarrow 0}\frac{\left[ -{r^{\frac{k-n}{k}}}\left( \int _0^r\frac{ks^{n-1}}{C_{n-1}^{k-1}} f^k(v(s))ds\right) ^{\frac{1}{k}}+\alpha r\right] }{r}\\&=-\left( \frac{k}{nC_{n-1}^{k-1}}\right) ^{\frac{1}{k}}f(v(0))+\alpha , \end{aligned} \end{aligned}$$

(2.17)

and for any $r\in (0,1)$, by (2.15), we also have

$$\begin{aligned} \begin{aligned} v''(r)&=-r^{-\frac{n}{k}}\left( \int _0^r\frac{ks^{n-1}}{C_{n-1}^{k-1}} f^k(v(s))ds\right) ^{\frac{1}{k}-1}\left[ \frac{k-n}{k}\left( \int _0^r\frac{ks^{n-1}}{C_{n-1}^{k-1}} f^k(v(s))ds\right) \right. \\&\left. \quad +\frac{r^n f^k(v(r))}{C_{n}^{k}}\right] +\alpha . \end{aligned}\nonumber \\ \end{aligned}$$

(2.18)

Thus (2.17) and (2.18) imply that $v(r)\in C^2[0,1].$

Finally, by using $v'(0)=0, v(1)=0$ and integrating (2.18), we get that v satisfies (2.10). $\square$

In order to establish the existence of solutions of the Eq. (2.10), we choose $X=C[0,1]$ as our work space, which is a Banach space with the norm $||v(r)||=\max _{r\in [0,1]}|v(r)|$. Then define an integral operator T as follows

$$\begin{aligned} (T v)(r)=\int _r^1\left[ {t^{\frac{k-n}{k}}}\left( \int _0^t\frac{ks^{n-1}}{C_{n-1}^{k-1}} f^k(v(s))ds\right) ^{\frac{1}{k}}-\alpha t\right] dt,\ \ r\in [0,1].\nonumber \\ \end{aligned}$$

(2.19)

In the following, we give the definitions of the upper and lower solutions for the Eq. (2.10).

Definition 2.1

A function $v\in X$ is called a lower solution (resp. upper solution) of (2.10) if

$$\begin{aligned} \left\{ \begin{aligned}&(-1)^kC_{n-1}^{k-1}\left[ {r^{n-k}}(v'(r)-\alpha r)^k\right] '\ge 0\ \ \ (resp. \le ),\ \ r\in (0,1), \\&v'(0)\le 0,\ \ v(1)\le 0 \ \ \ (resp. v'(0)\ge 0,\ \ v(1)\ge 0 ). \end{aligned} \right. \end{aligned}$$

(2.20)

It follows from Lemma 2.2 that we have the following comparison principle.

Lemma 2.3

If $v \in C([0,1], \mathbb {R})$ satisfies

$$\begin{aligned} \begin{aligned}&\left\{ (-1)^kC_{n-1}^{k-1}\left[ {r^{n-k}}(v'(r)-\alpha r)^k\right] '\ge 0,\ \ r\in (0,1),\right. \\&\left. v'(0)=0,\ \ v(1)=0. \right. \end{aligned} \end{aligned}$$

(2.21)

Then $(-1)^k(v'(r)-\alpha r)^k\ge 0,\ r\in [0,1].$

Proof

Let $h(r)\ge 0$ and

$$\begin{aligned} (-1)^kC_{n-1}^{k-1}\left[ {r^{n-k}}(v'(r)-\alpha r)^k\right] '=h(r). \end{aligned}$$

Then

$$\begin{aligned} (-1)^k(v'(r)-\alpha r)^k={r^{{k-n}}}\int _0^r\frac{h(s)}{C_{n-1}^{k-1}} ds \ge 0,\ r\in [0,1]. \end{aligned}$$

$\square$

3 Main Results

In this section, we first list the hypotheses to be used in this paper.

(F) For any constant $\varpi >0$,

$$\begin{aligned} 0<\int _0^1{\xi ^{n-1}}f^k( \varpi \left( 1-{\xi ^2}\right) )d\xi <+\infty . \end{aligned}$$

(3.1)

Denote a constant

$$\begin{aligned} \vartheta =\left[ \frac{ k^{1+\frac{1}{k}}}{(2k-n)(C_{n-1}^{k-1})^{\frac{1}{k}} }\left( \int _0^1{\xi ^{n-1}}f^k\left( 1-{\xi ^2}\right) d\xi \right) ^{\frac{1}{k}}-\frac{\alpha }{2}\right] . \end{aligned}$$

Let $X=C[0,1]$, and define a cone $K=\{v\in X: v(r)\ge 0\}$ and a subset $K^*$ of X

$$\begin{aligned} \begin{aligned} K^*=\Big \{ v\in X:&\hbox { there exists a number}\ \ 0<\lambda _{v}<1 \ \ \hbox {such that} \\&\ \lambda _v \left( 1-{r^2}\right) \le v(r)\le \lambda ^{-1}_v \left( 1-{r^2}\right) , \ \ r\in [0,1]\Big \}. \end{aligned} \end{aligned}$$

Theorem 3.1

Suppose $(\textbf{G})$ and $(\textbf{F})$ hold. Then the singular augmented Hessian Eq. (1.1) has at least one positive solution u(x) satisfying the following asymptotic properties

$$\begin{aligned} \begin{aligned} 1-|x|^2\le u(|x|)\le \vartheta ( 1-|x|^2). \end{aligned} \end{aligned}$$

Proof

We firstly show that $T: K^* \rightarrow K^*$ is completely continuous.

Indeed, for any $v\in K^*$, by using the definition of $K^*$, for any $r\in [0, 1]$, there exists $0<\lambda _{v}<1$ such that

$$\begin{aligned} \lambda _{v} \left( 1-{r^2}\right) \le v(r)\le \lambda ^{-1}_{v} \left( 1-{r^2}\right) . \end{aligned}$$

(3.2)

Since T is non-increasing on v, it follows from (3.1) to (3.2) that

$$\begin{aligned}{} & {} (Tv)(r)=\int _r^1\left[ {t^{\frac{k-n}{k}}}\left( \int _0^t\frac{k\xi ^{n-1}}{C_{n-1}^{k-1}} f^k( v(\xi ))d\xi \right) ^{\frac{1}{k}}-\alpha t\right] dt \nonumber \\{} & {} \quad \le \int _r^1\left[ {t^{\frac{k-n}{k}}}\left( \int _0^t\frac{k\xi ^{n-1}}{C_{n-1}^{k-1}} f^k( \lambda _{v} \left( 1-{\xi ^2}\right) )d\xi \right) ^{\frac{1}{k}}-\alpha t\right] dt \nonumber \\{} & {} \quad \le \frac{ k^{1+\frac{1}{k}}}{(2k-n)(C_{n-1}^{k-1})^{\frac{1}{k}} }\left( \int _0^1{\xi ^{n-1}}f^k( \lambda _{v} \left( 1-{\xi ^2}\right) )d\xi \right) ^{\frac{1}{k}}(1-r^{\frac{2k-n}{k}})-\frac{\alpha }{2}(1-r^2)\nonumber \\{} & {} \quad \le \left[ \frac{ k^{1+\frac{1}{k}}}{(2k-n)(C_{n-1}^{k-1})^{\frac{1}{k}} }\left( \int _0^1{\xi ^{n-1}}f^k( \lambda _{v} \left( 1-{\xi ^2}\right) )d\xi \right) ^{\frac{1}{k}}-\frac{\alpha }{2}\right] (1-r^2)\nonumber \\{} & {} \quad <+\infty . \end{aligned}$$

(3.3)

Thus T is uniformly bounded.

In addition, we also have

$$\begin{aligned} (Tv)(r)= & {} \int _r^1\left[ {t^{\frac{k-n}{k}}}\left( \int _0^t\frac{k\xi ^{n-1}}{C_{n-1}^{k-1}} f^k( v(\xi ))d\xi \right) ^{\frac{1}{k}}-\alpha t\right] dt\nonumber \\\ge & {} \int _r^1\left[ {t^{\frac{k-n}{k}}}\left( \int _0^t\frac{k\xi ^{n-1}}{C_{n-1}^{k-1}} \alpha _0^kd\xi \right) ^{\frac{1}{k}}-\alpha t\right] dt\nonumber \\\ge & {} \frac{\alpha _0 k^{\frac{1}{k}}}{2(n C_{n-1}^{k-1})^{\frac{1}{k}} }(1-r^2)-\frac{\alpha }{2}(1-r^2)\nonumber \\\ge & {} \frac{1}{2}\left[ \frac{\alpha _0}{(C_{n}^{k})^{\frac{1}{k}} }-{\alpha } \right] (1-r^2). \end{aligned}$$

(3.4)

Take

$$\begin{aligned} \begin{aligned}&\widetilde{\lambda }_{v}=\min \left\{ \frac{1}{4},\frac{1}{2}\left[ \frac{\alpha _0}{(C_{n}^{k})^{\frac{1}{k}} }-{\alpha } \right] ,\right. \\&\qquad \quad \left. \left[ \frac{ k^{1+\frac{1}{k}}}{(2k-n)(C_{n-1}^{k-1})^{\frac{1}{k}} }\left( \int _0^1{\xi ^{n-1}}f^k( \lambda _{v} \left( 1-{\xi ^2}\right) )d\xi \right) ^{\frac{1}{k}}-\frac{\alpha }{2}\right] ^{-1} \right\} , \end{aligned} \end{aligned}$$

then by (3.3) and (3.4), one has

$$\begin{aligned} \begin{aligned}&\widetilde{\lambda }_{v}(1-r^2) \le (Tv)(r)\le \widetilde{\lambda }^{-1}_{v}(1-r^2), \end{aligned} \end{aligned}$$

which implies that $T(K^*)\subset K^*.$

On the other hand, it is obvious that T is continuous in X and also equicontinuous on any bounded set of $K^*$. It follows from the Arezela-Ascoli theorem that $T:K^* \rightarrow K^*$ is completely continuous.

Next by Lemma 2.2 and (2.19), we get

$$\begin{aligned} \left\{ \begin{aligned}&(-1)^{k}C_{n-1}^{k-1}\left[ {r^{n-k}}((Tv)'(r)-\alpha r)^k\right] '=k{r^{n-1}}\left[ f(v(r))\right] ^k,\ \ r\in (0,1), \\&(Tv)'(0)=0,\ \ (Tv)(1)=0. \end{aligned} \right. \end{aligned}$$

(3.5)

In the following, we shall find a pair of lower and upper solutions for the augmented Hessian Eq. (2.10). By (3.3), (3.4) and $(\textbf{G})$, one has

$$\begin{aligned} \begin{aligned}&1-r^2\le \frac{1}{2}\left[ \frac{\alpha _0}{(C_{n}^{k})^{\frac{1}{k}} }-{\alpha } \right] (1-r^2) \le T(1-r^2)\\&\quad \le \left[ \frac{ k^{1+\frac{1}{k}}}{(2k-n)(C_{n-1}^{k-1})^{\frac{1}{k}} }\left( \int _0^1{\xi ^{n-1}}f^k\left( 1-{\xi ^2}\right) d\xi \right) ^{\frac{1}{k}}-\frac{\alpha }{2}\right] (1-r^2). \end{aligned} \end{aligned}$$

(3.6)

Let

$$\begin{aligned} \beta (r)= T(1-r^2), \end{aligned}$$

then by (3.6), we have

$$\begin{aligned} \begin{aligned}&1-r^2 \le \beta (r)\le \vartheta (1-r^2). \end{aligned} \end{aligned}$$

(3.7)

Moreover, from (3.7) and noting that T is a decreasing operator, it follows that

$$\begin{aligned} T(\vartheta (1-r^2))\le (T\beta )(r)\le T(1-r^2)=\beta (r), \end{aligned}$$

and

$$\begin{aligned} T(\vartheta (1-r^2))= & {} \int _r^1\left[ {t^{\frac{k-n}{k}}}\left( \int _0^t\frac{k\xi ^{n-1}}{C_{n-1}^{k-1}} f^k( \vartheta (1-\xi ^2))d\xi \right) ^{\frac{1}{k}}-\alpha t\right] dt\nonumber \\\ge & {} \int _r^1\left[ {t^{\frac{k-n}{k}}}\left( \int _0^t\frac{k\xi ^{n-1}}{C_{n-1}^{k-1}} \alpha _0^kd\xi \right) ^{\frac{1}{k}}-\alpha t\right] dt\nonumber \\\ge & {} \frac{\alpha _0 k^{\frac{1}{k}}}{2(n C_{n-1}^{k-1})^{\frac{1}{k}} }(1-r^2)-\frac{\alpha }{2}(1-r^2)\nonumber \\= & {} \frac{1}{2}\left[ \frac{\alpha _0}{(C_{n}^{k})^{\frac{1}{k}} }-{\alpha } \right] (1-r^2)\ge 1-r^2, \end{aligned}$$

(3.8)

i.e.,

$$\begin{aligned} 1-r^2\le (T\beta )(r)\le \beta (r)\le \vartheta (1-r^2). \end{aligned}$$

(3.9)

Let

$$\begin{aligned} \zeta (r) =T\beta (r), \ \ \beta (r)= T(1-r^2). \end{aligned}$$

(3.10)

In the following, we show that $\zeta (r),\beta (r)$ are a pair of lower and upper solutions of the Eq. (2.10). In fact, it follows from $(\textbf{G})$ that T is a decreasing operator with respect to v, thus by (3.6)–(3.10), we have $\zeta (r), \beta (r)\in K^*$ and

$$\begin{aligned} \begin{aligned}&1-r^2 \le \zeta (r)= T\beta (r)\le \beta (r)\le \vartheta (1-r^2). \end{aligned} \end{aligned}$$

(3.11)

Next, from (3.5) to (3.11), one gets

$$\begin{aligned} \begin{aligned}&(-1)^kC_{n-1}^{k-1}\left[ {r^{n-k}}(\zeta '(r)-\alpha r)^k\right] '-k{r^{n-1}}f^k(\zeta (r)) \\&\quad =(-1)^kC_{n-1}^{k-1}\left[ {r^{n-k}}((T\beta )'(r)-\alpha r)^k\right] '-k{r^{n-1}}f^k((T\beta )(r)) \\&\quad \le (-1)^kC_{n-1}^{k-1}\left[ {r^{n-k}}((T\beta )'(r)-\alpha r)^k\right] '-k{r^{n-1}}f^k(\beta (r))=0, \end{aligned} \end{aligned}$$

(3.12)

and

$$\begin{aligned} \begin{aligned}&(-1)^kC_{n-1}^{k-1}\left[ {r^{n-k}}(\beta '(r)-\alpha r)^k\right] '-k{r^{n-1}}f^k(\beta (r)) \\&\quad \ge (-1)^kC_{n-1}^{k-1}\left[ {r^{n-k}}(\beta '(r)-\alpha r)^k\right] '-k{r^{n-1}}f^k(1-r^2)=0. \end{aligned} \end{aligned}$$

(3.13)

Moreover, by (3.5) and (3.10), we get that $\beta$ and $\zeta$ satisfy

$$\begin{aligned} \begin{aligned}&\zeta '(0)=0,\ \ \zeta (1)=0,\\&\beta '(0)=0, \ \ \beta '(1)=0. \end{aligned} \end{aligned}$$

(3.14)

Thus (3.6)–(3.14) imply that $\beta (t)$ and $\zeta (r)$ are a pair of upper and lower solutions of the eq. (2.10) with $\zeta (r), \beta (r)\in K^*$.

Define an auxiliary function

$$\begin{aligned} F(v)=\left\{ \begin{aligned}&f( \zeta (r)), \ \ \ \ v<\zeta (r),\\&f(v(r)),\ \ \ \zeta (r)\le v\le \beta (r),\\&f(\beta (r)),\ \ \ \ v>\beta (r),\end{aligned}\right. \end{aligned}$$

(3.15)

and consider the following modified equation

$$\begin{aligned} \left\{ \begin{aligned}&(-1)^kC_{n-1}^{k-1}\left[ {r^{n-k}}(v'(r)-\alpha r)^k\right] '=k{r^{n-1}}\left[ F( v(r))\right] ^k,\ \ r\in (0,1), \\&v'(0)=0,\ \ v(1)=0. \end{aligned} \right. \end{aligned}$$

(3.16)

Let us define an operator S in X

$$\begin{aligned} (S v)(r) =\int _r^1\left[ {t^{\frac{k-n}{k}}}\left( \int _0^t\frac{k\xi ^{n-1}}{C_{n-1}^{k-1}} F^k( v(\xi ))d\xi \right) ^{\frac{1}{k}}-\alpha t\right] dt,\ \ r\in [0,1], \ \ \ \forall v\in X. \end{aligned}$$

By Lemma 2.2, the fixed point of S is the solution of the boundary value problem (3.16). So we firstly seek for the fixed point of the operator S.

It follows from $\zeta \in K^*$ that there exists a constant $0<\lambda _\zeta <1$ such that

$$\begin{aligned} \lambda _\zeta (1-r^2)\le \zeta (r)\le \lambda ^{-1}_\zeta (1-r^2),\ r\in [0,1]. \end{aligned}$$

(3.17)

Thus from (3.15) to (3.17), for all $v\in X$, one has

$$\begin{aligned} \begin{aligned} (S_{ }v)(r)&=\int _r^1\left[ {t^{\frac{k-n}{k}}}\left( \int _0^t\frac{k\xi ^{n-1}}{C_{n-1}^{k-1}} F^k( v(\xi ))d\xi \right) ^{\frac{1}{k}}-\alpha t\right] dt \\&\le \int _r^1\left[ {t^{\frac{k-n}{k}}}\left( \int _0^t\frac{k\xi ^{n-1}}{C_{n-1}^{k-1}} f^k( \zeta (\xi ))d\xi \right) ^{\frac{1}{k}}-\alpha t\right] dt \\&\le \int _r^1\left[ {t^{\frac{k-n}{k}}}\left( \int _0^t\frac{k\xi ^{n-1}}{C_{n-1}^{k-1}} f^k( \lambda _{\zeta } \left( 1-{\xi ^2}\right) )d\xi \right) ^{\frac{1}{k}}-\alpha t\right] dt\\&\le \frac{ k^{1+\frac{1}{k}}}{(2k-n)(C_{n-1}^{k-1})^{\frac{1}{k}} }\left( \int _0^1{\xi ^{n-1}}f^k( \lambda _{\zeta } \left( 1-{\xi ^2}\right) )d\xi \right) ^{\frac{1}{k}}(1-r^{\frac{2k-n}{k}})-\frac{\alpha }{2}(1-r^2)\\&\le \left[ \frac{ k^{1+\frac{1}{k}}}{(2k-n)(C_{n-1}^{k-1})^{\frac{1}{k}} }\left( \int _0^1{\xi ^{n-1}}f^k( \lambda _{\zeta } \left( 1-{\xi ^2}\right) )d\xi \right) ^{\frac{1}{k}}-\frac{\alpha }{2}\right] (1-r^2)\\&<+\infty .\end{aligned}\nonumber \\ \end{aligned}$$

(3.18)

Hence S is bounded. As F is continuous, $S: X \rightarrow X$ is also continuous.

Let $B \subset X$ be a bounded set, then for any $v\in B$, there exists a positive constant $M>0$ such that $||v||\le M$. Now let

$$\begin{aligned} L =\max _{ 0\le v\le M}|F(v)| +C_{n}^{k} \alpha ^k, \end{aligned}$$

and for any $\epsilon > 0$, take

$$\begin{aligned} \sigma =\left[ \left( \frac{ L}{C_{n}^{k} }\right) ^{\frac{1}{k}}- \alpha \right] ^{-1}\epsilon , \end{aligned}$$

then for any $r_{1},r_{2} \in [0,1]$ with $\mid r_{1}-r_{2}\mid <\sigma$, we have

$$\begin{aligned} |Sv(r_{1})-Sv(r_{2})|\le & {} \left| \int _{r_1}^{r_2}{t^{\frac{k-n}{k}}}\left( \int _0^t\frac{k\xi ^{n-1}}{C_{n-1}^{k-1}} F^k(v(\xi ))d\xi \right) ^{\frac{1}{k}}-\alpha tdt\right| \nonumber \\\le & {} \frac{1}{2} \left[ \left( \frac{ L}{C_{n}^{k} }\right) ^{\frac{1}{k}}- \alpha \right] |(r_2+r_1)(r_2-r_1)|\nonumber \\\le & {} \left[ \left( \frac{ L}{C_{n}^{k} }\right) ^{\frac{1}{k}}- \alpha \right] |r_2-r_1| <\epsilon , \end{aligned}$$

(3.19)

which implies that S(B) is equicontinuous.

From the Arzela-Ascoli theorem, $S:X\rightarrow X$ is a completely continuous operator. Consequently, by the Schauder fixed point theorem, S has a fixed point u such that $u=Su$.

Next we show that the fixed point u of the operator S is also the fixed point of the operator T. Indeed, by the definition of F, we only need to prove

$$\begin{aligned} \zeta (r)\le u(r)\le \beta (r), \ \ \ r\in [0,1]. \end{aligned}$$

(3.20)

Firstly we prove $u(r)\le \beta (r)$, $r\in [0,1]$. To do this, let us construct a function as follows:

$$\begin{aligned} w(r)=-\int _r^1 \left[ (\beta '(\xi )-\alpha r)^k-(u'(\xi )-\alpha r)^k\right] ^{\frac{1}{k}}d\xi +\frac{1}{2}\alpha r^2-\frac{1}{2}\alpha ,\ \ r\in [0,1], \end{aligned}$$

then

$$\begin{aligned} w(1)=0,\ \ (w'(r)-\alpha r)^k =(\beta '(r)-\alpha r)^k-(u'(r)-\alpha r)^k. \end{aligned}$$

(3.21)

Since $\beta (r)$ is the upper solution of the problem (2.10) and u is a fixed point of S, one has

$$\begin{aligned} (w'(0))^k =(\beta '(0))^k-(u'(0))^k=0, \ \hbox {i.e.,} \ w'(0)=0. \end{aligned}$$

(3.22)

Moreover, it follows from $(\textbf{G})$, (3.11) and the definition of F that

$$\begin{aligned} \begin{aligned}&f( \beta (r))\le F(u(r))\le f(\zeta (r))\le f( 1-{r^2}), \ \ \hbox {for any}\ u\in K^*\ \hbox {and}\ r\in [0,1]. \end{aligned} \end{aligned}$$

(3.23)

By using (3.23) and (3.21), one has

$$\begin{aligned} \begin{aligned}&(-1)^kC_{n-1}^{k-1}\left[ {r^{n-k}}(w'(r)-\alpha r)^k\right] '=(-1)^kC_{n-1}^{k-1}\left[ {r^{n-k}}(\beta '(r)-\alpha r)^k\right] '\\&\quad -(-1)^kC_{n-1}^{k-1}\left[ {r^{n-k}}(u'(r)-\alpha r)^k\right] '\\&\quad = k{r^{n-1}}\left[ f(1-r^2)\right] ^k-k{r^{n-1}}\left[ F( v(r))\right] ^k\ge 0. \end{aligned} \end{aligned}$$

(3.24)

Thus it follows from (3.24), (3.21), (3.22) and Lemma 2.3 that

$$\begin{aligned} (-1)^k(w'(r)-\alpha r)^k\ge 0. \end{aligned}$$

Consequently, by (3.21), one has

$$\begin{aligned} (-1)^k(\beta '(r)-\alpha r)^k\ge (-1)^k(u'(r)-\alpha r)^k, \ r\in [0,1]. \end{aligned}$$

(3.25)

From (2.15), we get

$$\begin{aligned} -\beta '(r)+\alpha r\ge 0, -u'(r)+\alpha r\ge 0. \end{aligned}$$

(3.26)

Hence (3.25) and (3.26) yield $\beta '(r) \le u'(r)$. Integrating the above inequality from r to 1, we have $u(r) \le \beta (r)$ on [0, 1]. In the same way, one also has $u(r) \ge \zeta (r)$ on [0, 1]. Hence we obtain

$$\begin{aligned} \zeta (r)\le u(r)\le \beta (r), \ \ \ r\in [0,1]. \end{aligned}$$

(3.27)

Thus we have $F(u(r))=f( u(r)),\ r\in [0,1]$, that is, u(r) is also a fixed point of T, and so it is a positive solution of the singular augmented Hessian Eq. (1.1).

Finally, it follows from (3.27) to (3.11) that

$$\begin{aligned} 1-{r^2}\le \zeta (r)\le u(r)\le \beta (r) \le \vartheta (1-r^2), \end{aligned}$$

i.e.,

$$\begin{aligned} \begin{aligned} 1-|x|^2\le u(|x|)\le \vartheta ( 1-|x|^2). \end{aligned} \end{aligned}$$

$\square$

Theorem 3.2

Suppose f satisfies the following condition

$(\mathbf {G^*})$ $f:[0,+\infty )\rightarrow (\alpha _0,+\infty )$ is continuous and non-increasing, where $\alpha _0>(C_n^k)^{\frac{1}{k}}(\alpha +2)$.

Then the nonsingular augmented Hessian eq. (1.1) has at least one positive solution u(x) satisfying

$$\begin{aligned} 1-{|x|^2}\le u(|x|)\le \frac{1}{2}\left[ \frac{f(0)}{(C_{n}^{k})^{\frac{1}{k}} }-{\alpha } \right] (1-|x|^2). \end{aligned}$$

(3.28)

Proof

In this case, we choose

$$\begin{aligned} K =\{v\in X: v(r)\ge 0, r \in [0, 1]\}, \end{aligned}$$

as work space. Let

$$\begin{aligned} \zeta (r) = T(T0)(r), \ \ \beta (r) = (T0)(r)=\frac{1}{2}\left[ \frac{f(0)}{(C_{n}^{k})^{\frac{1}{k}} }-{\alpha } \right] (1-r^2), \end{aligned}$$

then we have $\beta (r), \zeta (r)\in K$ and

$$\begin{aligned} 0\le \beta (r) =T0,\ \ 0\le \zeta (r)=T(T0)=(T\beta )(r)\le T0=\beta (t). \end{aligned}$$

(3.29)

Consequently, one has

$$\begin{aligned} \begin{aligned}&(-1)^kC_{n-1}^{k-1}\left[ {r^{n-k}}(\zeta '(r)-\alpha r)^k\right] '-k{r^{n-1}}f^k(\zeta (r)) \\&=(-1)^kC_{n-1}^{k-1}\left[ {r^{n-k}}((T\beta )'(r)-\alpha r)^k\right] '-k{r^{n-1}}f^k((T\beta )(r))\\&\le (-1)^kC_{n-1}^{k-1}\left[ {r^{n-k}}((T\beta )'(r)-\alpha r)^k\right] '-k{r^{n-1}}f^k(\beta (r))=0, \end{aligned} \end{aligned}$$

(3.30)

and

$$\begin{aligned} \begin{aligned}&(-1)^kC_{n-1}^{k-1}\left[ {r^{n-k}}(\beta '(r)-\alpha r)^k\right] '-k{r^{n-1}}f^k(\beta (r)) \\&=(-1)^kC_{n-1}^{k-1}\left[ {r^{n-k}}((T0)'(r)-\alpha r)^k\right] '-k{r^{n-1}}f^k((T0)(r)) \\&\ge (-1)^kC_{n-1}^{k-1}\left[ {r^{n-k}}((T0)'(r)-\alpha r)^k\right] '-k{r^{n-1}}f^k(0)=0. \end{aligned} \end{aligned}$$

(3.31)

Hence from (3.29) to (3.31), $\zeta (r)$ and $\beta (r)$ are still a pair of upper and lower solutions of the equation (2.10), respectively. Proceeding as in the proof of Theorem 3.1, we get that T has a fixed point $u\in K$ satisfying (3.28). So we complete the proof of Theorem 3.2. $\square$

4 Examples

In this section, we give two examples to illustrate our main results.

Example 4.1

Consider the following singular augmented Hessian equation

$$\begin{aligned} \left\{ \begin{aligned}&-\varrho ^{\frac{1}{3}}_3[D^2u-5 I]=\frac{1}{u^{\frac{1}{4}}(x)}+14,\ \ \hbox {in} \ \ {\Omega }\subset \mathbb {R}^4, \\&u=0 \ \ \ \hbox {on} \ \ \ \partial {\Omega }, \end{aligned} \right. \end{aligned}$$

(4.1)

where $\Omega$ is a unit open ball. Then the singular augmented Hessian Eq. (4.1) has at least one positive radial solution u satisfying the asymptotic properties

$$\begin{aligned} 1-{|x|^2}\le u(|x|)\le 12.1917(1-|x|^{2}). \end{aligned}$$

Proof

Take $k=3,\ \alpha =5, \ n=4, \ \alpha _0=14, \ f(u)=u^{-\frac{1}{4}}+14,$ then $f:(0,+\infty )\rightarrow (14,+\infty )$ is continuous and non-increasing, with $\alpha _0=14 >(C_n^k)^{\frac{1}{k}}(\alpha +2)=7\root 3 \of {4}$. Thus $(\textbf{G})$ holds.

For any $\varpi >0,$ through some calculation, we have

$$\begin{aligned} \begin{aligned}&0<\int _0^1{\xi ^{n-1}}f^k(\varpi (1-{\xi ^2}))d\xi =\int _0^1{\xi ^{3}}\left[ \varpi ^{-\frac{1}{4}} (1-\xi ^2)^{-\frac{1}{4}}+14\right] ^{3}d\xi \\&=\varpi ^{-\frac{3}{4}}\int _0^1{\xi ^{3}} (1-\xi ^2)^{-\frac{3}{4}}d\xi +42\varpi ^{-\frac{1}{2}}\int _0^1{\xi ^{3}} (1-\xi ^2)^{-\frac{1}{2}}d\xi \\&\ \ \ +588\varpi ^{-\frac{1}{4}}\int _0^1{\xi ^{3}} (1-\xi ^2)^{-\frac{1}{4}}d\xi +14^3\int _0^1{\xi ^{3}}d\xi \\&=\frac{8}{5}\varpi ^{-\frac{3}{4}}+42\varpi ^{-\frac{1}{2}}\times \frac{2}{3} +588\varpi ^{-\frac{1}{4}}\times \frac{8}{21}+2744\times \frac{1}{4}\\&=\frac{8}{5}\varpi ^{-\frac{3}{4}}+28\varpi ^{-\frac{1}{2}}+224\varpi ^{-\frac{1}{4}}+686 <+\infty . \end{aligned} \end{aligned}$$

(4.2)

Hence $(\textbf{F})$ holds.

Now we calculate $\vartheta$ step by step as follows

$$\begin{aligned} \begin{aligned}&0<\int _0^1{\xi ^{n-1}}f^k(1-{\xi ^2})d\xi =\int _0^1{\xi ^{3}}\left[ (1-\xi ^2)^{-\frac{1}{4}}+14\right] ^{3}d\xi \\&=\int _0^1{\xi ^{3}} (1-\xi ^2)^{-\frac{3}{4}}d\xi +42\int _0^1{\xi ^{3}} (1-\xi ^2)^{-\frac{1}{2}}d\xi \\&\ \ \ +588\int _0^1{\xi ^{3}} (1-\xi ^2)^{-\frac{1}{4}}d\xi +14^3\int _0^1{\xi ^{3}}d\xi \\&=\frac{8}{5}+42\times \frac{2}{3} +588\times \frac{8}{21}+2744\times \frac{1}{4}\\&=\frac{8}{5}+28+224+686=939.6. \end{aligned} \end{aligned}$$

(4.3)

Since $k=3,\ \alpha =5, \ n=4,$ we have

$$\begin{aligned} \begin{aligned} \vartheta&=\frac{ k^{1+\frac{1}{k}}}{(2k-n)(C_{n-1}^{k-1})^{\frac{1}{k}} }\left( \int _0^1{\xi ^{n-1}}f^k\left( 1-{\xi ^2}\right) d\xi \right) ^{\frac{1}{k}}-\frac{\alpha }{2}\\&=\frac{ 3^{\frac{4}{3}}}{2(C_{3}^{2})^{\frac{1}{3}} }\times 939.6^{\frac{1}{3}}-\frac{5}{2}=1.5\times 9.7945-2.5=12.1917. \end{aligned} \end{aligned}$$

Thus it follows from Theorem 3.1 that the conclusion is valid. $\square$

Example 4.2

Consider the following nonsingular augmented Hessian equation

$$\begin{aligned} \left\{ \begin{aligned}&-\varrho ^{\frac{1}{3}}_3[D^2u-5 I]=\frac{1}{u^4(x)+1}+14,\ \ \hbox {in} \ \ {\Omega }\subset \mathbb {R}^4, \\&u=0 \ \ \ \hbox {on} \ \ \ \partial {\Omega }, \end{aligned} \right. \end{aligned}$$

(4.4)

where $\Omega$ is a unit open ball. Then the nonsingular augmented Hessian Eq. (4.4) has at least one positive radial solution u satisfying the asymptotic properties

$$\begin{aligned} 1-{|x|^2}\le u(|x|)\le 11.6741(1-|x|^{2}). \end{aligned}$$

Proof

Take $k=3,\ \alpha =5, \ n=4, \ \alpha _0=14, \ f(u)=\frac{1}{u^4+1}+14,$ then $f:(0,+\infty )\rightarrow (14,15]$ is continuous and non-increasing, with $\alpha _0=14 >(C_n^k)^{\frac{1}{k}}(\alpha +2)=7\root 3 \of {4}$. Thus $(\textbf{G})$ holds.

For any $\varpi >0,$ noticing

$$\begin{aligned} 0<\frac{1}{\varpi ^{4} (1-\xi ^2)^{{4}}+1}<1, \end{aligned}$$

we have the following estimate

$$\begin{aligned}{} & {} 0<\int _0^1{\xi ^{n-1}}f^k(\varpi (1-{\xi ^2}))d\xi \nonumber \\= & {} \int _0^1{\xi ^{3}}\left[ \frac{1}{\varpi ^{4} (1-\xi ^2)^{{4}}+1}+14\right] ^{3}d\xi<15^{3}\int _0^1{\xi ^{3}}d\xi \nonumber \\= & {} 843.75<+\infty . \end{aligned}$$

(4.5)

Hence $(\textbf{F})$ holds.

In the following, we caculate the value of $\vartheta$ by (4.5) and obtain

$$\begin{aligned} \vartheta= & {} \frac{ k^{1+\frac{1}{k}}}{(2k-n)(C_{n-1}^{k-1})^{\frac{1}{k}} }\left( \int _0^1{\xi ^{n-1}}f^k\left( 1-{\xi ^2}\right) d\xi \right) ^{\frac{1}{k}}-\frac{\alpha }{2}<\frac{ 3^{\frac{4}{3}}}{2(C_{3}^{2})^{\frac{1}{3}} }\\\times & {} 843.75^{\frac{1}{3}}-\frac{5}{2}=11.6741. \end{aligned}$$

Hence it follows from Theorem 3.2 that the conclusion holds. $\square$

5 Conclusion

The equation with form (1.1) arises in the theory of geometric optics [17]. Following the work of Caffarelli, Nirenberg and Spruck [3] on fully nonlinear Hessian equations (corresponding to $\varsigma (x) \equiv 0$ in (1.1)), a lot of work has been devoted to equations of the form (1.1). In this work, by finding more appropriate upper and lower solutions, we establish the existence of positive solutions for a singular Hessian equation with a negative augmented term. We also remove a critical condition required in the existing work [7] and fill the gap in the work [7]. Thus our main results can be applied to handle some geometric optics models with the negative augmented term. However the problem for the case where the augmented term is a general matrix function A(x, u, Du) reminds an open problem for further research. In future work, we will continue to focus on the existence of solutions for the Hessian equation with the general augmented term.