In this paper, we consider the following form of Hessian quotient type equation

$ \begin{align} \frac{\sigma _k(u_{ij}+u\delta_{ij} )}{\sigma _l(u_{ij}+u\delta_{ij} )} =u^{p-1}f(x), \; \; \; x \in \mathbb{S}^n, \end{align} $

(1.1)

where $ \sigma _k $ is the $ k $-th elementary symmetric function and $ u_{ij} $ is the second order covariant derivative of $ u $ with respect to an orthonormal frame on $ \mathbb{S}^n $, and a function $ u\in C^2(\mathbb{S}^n) $ is called convex if

$ \begin{align} (u_{ij}+u\delta _{ij})>0, \; \; \; on\; \; \; \mathbb{S}^n. \end{align} $

(1.2)

In fact, the equation (1.1) corresponds to a class of $ Lp $ Minkowski type problem. The $ Lp $ Minkowski problem introduced by Lutwak[3] is a generalisation of the classical Minkowski problem.

Given a Borel measure $ \mu $ on the unit sephere $ \mathbb{S}^n $, the $ Lp $ Minkowski problem concerns with the existence of a unique convex body $ \mathbb{K} $ in $ \mathbb{R}^{n+1} $ so that $ \mu $ is the $ Lp $ surface area measure of $ \mathbb{K} $,

$ \begin{align} d\mu=u^{1-p}dS_k, \end{align} $

(1.3)

where $ S_k $ is the ordinary surface area measure of $ \mathbb{K} $ and $ u:\mathbb{S}^n\mapsto \mathbb{R} $ is the support function of $ \mathbb{K} $. In the case of $ p=1 $, the $ Lp $ Minkowski problem reduces to the classical Minkowski problem. The classical Minkowski problem was considered by Minkowski in [4], which is to find the necessary and sufficient conditions on a given measure so that it is exactly the surface area measure of a convex body. The classical Minkowski problem corresponds to solve a Monge-Ampère type equation

$ \begin{align} \det (u_{ij}+u\delta_{ij}) =f(x), \quad x \in \mathbb{S}^n. \end{align} $

(1.4)

Many important contributions to Minkowski problems were done by Minkowski [5, 6], Alexandrov [7], Nirenberg [8], and Cheng-Yau [9], et al. Since the classical Minkowski problem, many Minkowski type problems have been introduced and extensively studied.

The $ L_p $ Minkowski problem ($ p \geq 1 $) is the problem of prescribing $ L_p $ surface area measure which was introduced by Lutwak [3], and is to solve a Hessian type geometric PDE

$ \begin{align} \det (u_{ij}+u\delta_{ij} ) =u^{p-1}f(x), \quad x \in \mathbb{S}^n, \end{align} $

(1.5)

and many important contributions to $ L_p $ Minkowski problems were done by Lutwak[3], Chou-Wang [10], Guan-Lin [11], Böröczky-Lutwak-Yang-Zhang[12], Lutwak-Oliker[13] and Lutwak-Yang-Zhang[14] et al.

The Christoffel-Minkowski problem concerns with the existence of convex bodies with prescribed $ k $-th surface area measure, which corresponds to finding convex solutions of the following geometric PDE

$ \begin{align} \sigma_k (u_{ij}+u\delta_{ij} ) = f(x), \quad x \in \mathbb{S}^n. \end{align} $

(1.6)

Important contributions to Christoffel-Minkowski problems were done by Guan-Ma [15] and Guan-Ma-Zhou [16] et al. The key tool is the constant rank theorem for fully nonlinear partial differential equations.

The $ L_p $-Christoffel-Minkowski problem corresponds to finding convex solutions of the following geometric PDE

$ \begin{align} \sigma_k (u_{ij}+u\delta_{ij} ) = u^{p-1} f(x), \quad x \in \mathbb{S}^n. \end{align} $

(1.7)

Equation (1.7) has been studied by Hu-Ma-Shen [17] in the case $ p-1 \geq k $, and Guan-Xia[1] for $ 1<p<k+1 $ and even prescribed data, by using the constant rank theorem.

This article is organized as follows. In Section 2, we present some properties of $ \sigma_k (\lambda) $ in Gårding's cone $ \Gamma_k $, which are important to the a priori estimates. In Section 3, we prove Theorem 3.1. At last we prove Theorem 4.2 in Section 4.

In this section, we recall the definition and some basic properties of elementary symmetric functions, which could be found in [18].

Definition 2.1  For any $ k = 1, 2, \ldots, n, $ we set

$ \begin{align} \sigma_k(\lambda) = \sum _{1 \le i_1 < i_2 <\cdots<i_k\leq n}\lambda_{i_1}\lambda_{i_2}\cdots\lambda_{i_k}, \qquad \text {for any} \quad\lambda=(\lambda_1, \ldots, \lambda_n)\in {\Bbb R}^n. \end{align} $

(2.1)

For convenience, let $ \sigma_0=1 $ and $ \sigma_k =0 $ for $ k>n $.

Denote by $ \sigma _k (\lambda \left| i \right) $ the symmetric function with $ \lambda_i = 0 $ and $ \sigma _k (\lambda \left| ij \right) $ the symmetric function with $ \lambda_i =\lambda_j = 0 $.

The following standard formulas of elementary symmetric functions are needed.

Proposition 2.2  Let $ \lambda=(\lambda_1, \ldots, \lambda_n)\in\mathbb{R}^n $ and $ k =0, 1, \ldots, n $. Then

$ \begin{align*} &\sigma_k(\lambda)=\sigma_k(\lambda|i)+\lambda_i\sigma_{k-1}(\lambda|i), \quad \forall \, 1\leq i\leq n, \\ &\sum\limits_{i=1}^n \lambda_i\sigma_{k-1}(\lambda|i)=k\sigma_{k}(\lambda), \\ &\sum\limits_{i=1}^n \sigma_{k}(\lambda|i)=(n-k)\sigma_{k}(\lambda). \end{align*} $

Proposition 2.3  Let $ W=\left \{W_{ij} \right \} $ be an $ n \times n $ symmetric matrix and $ \lambda(W)= (\lambda_1, \lambda _2, \ldots , \lambda _{n}) $ be the eigenvalues of $ W $. If $ W=\left \{ W_{ij} \right \} $ is diagonal and $ \lambda_i= W_{ii} $, then

$ \begin{align} &\frac{{\partial \lambda _i }}{{\partial W_{ii} }} = 1, \quad \frac{{\partial \lambda _k }}{{\partial W_{ij} }} = 0, \text{otherwise} , \\ &\frac{{\partial ^2 \lambda _i }}{{\partial W_{ij} \partial W_{ji} }} = \frac{1}{{\lambda _i - \lambda _j }}, \quad i \ne j \text{ and } \lambda _i \ne \lambda _j, \\ &\frac{{\partial ^2 \lambda _i }}{{\partial W_{kl} \partial W_{pq} }} = 0 , \text{otherwise}. \end{align} $

Definition 2.1 can be extended to symmetric matrices by letting $ \sigma_k(W) = \sigma_k(\lambda(W)) $, where $ \lambda(W)= (\lambda _1(W), \lambda _2 (W), \cdots , \lambda _{n}(W)) $ are the eigenvalues of the symmetric matrix $ W $. We also denote by $ \sigma _k (W \left| i \right.) $ the symmetric function with $ W $ deleting the $ i $-row and $ i $-column and $ \sigma _k (W \left| ij \right.) $ the symmetric function with $ W $ deleting the $ i, j $-rows and $ i, j $-columns. Then we have the following properties.

Proposition 2.4  If $ W=\left \{ W_{ij} \right \} $ is diagonal and $ m $ is a positive integer, then

$ \begin{align} \frac{{\partial \sigma _m (W)}} {{\partial W_{ij} }} = \begin{cases} \sigma _{m - 1} (W\left| i \right.), &\text{if } i = j, \\ 0, &\text{if } i \ne j, \notag \end{cases} \end{align} $

and

$ \begin{align} \frac{{\partial ^2 \sigma _m (W)}} {{\partial W_{ij} \partial W_{kl} }} =\begin{cases} \sigma _{m - 2} (W\left| {ik} \right.), &\text{if } i = j, k = l, i \ne k, \\ - \sigma _{m - 2} (W\left| {ik} \right.), &\text{if } i = l, j = k, i \ne j, \\ 0, &\text{otherwise}.\notag \end{cases} \end{align} $

Recall that the classic Gårding's cone is defined as

$ \begin{equation} \label{1.2.7} \Gamma_k = \{ \lambda \in \mathbb{R}^n :\sigma _i (\lambda ) > 0, \forall 1 \le i \le k\}, \notag \end{equation} $

and the following properties are well known.

Proposition 2.5  For $ \lambda \in \Gamma_k $ and $ k > l \geq 0 $, $ r > s \geq 0 $, $ k \geq r $, $ l \geq s $, we have

$ \begin{align} \Bigg[\frac{{\sigma _k (\lambda )}/{C_n^k }}{{\sigma _l (\lambda )}/{C_n^l }}\Bigg]^{\frac{1}{k-l}} \le \Bigg[\frac{{\sigma _r (\lambda )}/{C_n^r }}{{\sigma _s (\lambda )}/{C_n^s }}\Bigg]^{\frac{1}{r-s}}, \end{align} $

with equality if and only if $ \lambda_1 = \lambda_2 = \cdots =\lambda_n >0 $.

Proposition 2.6  (1) $ \Gamma_k $ are convex cones, and $ \Gamma_1 \supset \Gamma_2 \supset \cdots \supset \Gamma_n $.

(2) If $ \lambda=(\lambda_1, \ldots, \lambda_n) \in \Gamma_k $ with $ \lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_n $, then $ \lambda _{k} > 0 $,

$ \begin{align} \label{1.2.9} \sigma_{k-1} (\lambda|n) \geq\sigma_{k-1} (\lambda|n-1) \geq \cdots \geq \sigma_{k-1} (\lambda|1) >0. \end{align} $

(3) If $ \lambda=(\lambda_1, \ldots, \lambda_n) \in \Gamma_k $, then $ \sigma_k(\lambda)^{\frac{1}{k}} $ and $ \Big[\frac{\sigma_k(\lambda)}{\sigma_l(\lambda)}\Big]^{\frac{1}{k-l}} $ $ (0 \leq l <k \leq n) $ are concave with respect to $ \lambda $. Equivalently, for any $ (\xi_1, \ldots, \xi_n) $, we have

$ \begin{align} \sum\limits_{i, j=1}^n\frac{\partial ^2 \Big[\frac{\sigma_k(\lambda)}{\sigma_l(\lambda)}\Big]}{\partial\lambda_i \partial \lambda_j} \xi_i \xi_j \leq& (1- \frac{1}{k-l}) \frac{[\sum\limits_{i=1}^n \frac{\partial \Big[\frac{\sigma_k(\lambda)}{\sigma_l(\lambda)}\Big]}{\partial\lambda_i} \xi_i ]^2}{\frac{\sigma_k(\lambda)}{\sigma_l(\lambda)}}. \end{align} $

Theorem 3.1  Suppose $ u \in C^{3}(\mathbb{S}^{n}) $ is a positive convex solution of the equation (1.1), where $ 0 \leq l < k \leq n $, $ p-1>0 $, and $ f\in C^1(\mathbb{S}^n) $ is a smooth, positive function. Then we have the weighted gradient estimate

$ \begin{align} \frac{|\nabla u|^2}{u^\gamma} \leq A (\max\limits_{\mathbb{S}^{n}} u) ^{2-\gamma}, \quad \forall\; x \in \mathbb{S}^n, \end{align} $

(3.1)

where $ \gamma =\min \{1, \frac{p-1}{k-l}\} $ and $ A $ is a positive constant depending on $ n $, $ k $, $ l $, $ p $, $ min_{\mathbb{S}^{n}} f $ and $ \left \| f \right \| _{C^{1} } $.

Proof  Following the idea of [1, 2], we prove Theorem 3.1 by a contradiction argument.

Let

$ \Phi =\frac{|\nabla u|^2}{u ^\gamma}, $

where $ \gamma =\min \{1, \frac{p-1}{k-l}\}\in (0, 1] $. Denote $ M_u = \max_{\mathbb{S}^n} u $. Assume $ \Phi $ attains maximum at $ x_{0} $. We may choose an orthonormal frame on $ \mathbb{S}^n $ such that

$ \begin{eqnarray*} u_{1}=|\nabla u|, \quad \text{and}\; \{u_{ij}\}_{2\leq i, j\leq n}\; \text{is diagonal}. \end{eqnarray*} $

In the following, we compute at $ x_0 $. Then we have at $ x_0 $,

$ \begin{eqnarray} 0=(\log \Phi)_{i}= \frac{2u_{k}u_{ki}}{|\nabla u|^{2}} - \gamma \frac{u_{i}}{u}, \end{eqnarray} $

(3.2)

hence we have

$ \begin{align} &u_{11} = \frac{\gamma}{2}\frac{u_1^2}{u}, \\ &u_{1i} = 0, \quad i = 2, \cdots, n. \end{align} $

(3.3)

Hence $ \{ u_{ij} \}_{1 \leq i, j \leq n} $ is diagonal, $ \{ b_{ij} \}_{1 \leq i, j \leq n} $ is diagonal with $ b_{ij}:=u_{ij}+u \delta_{ij} $, and $ \{ F^{ij} \}_{1 \leq i, j \leq n} $ is diagonal, where

$ \begin{align} F^{ij}=\frac{\partial \left [ \frac{\sigma_k(\lambda)}{\sigma_l(\lambda)} \right ] }{\partial b_{ij}}. \end{align} $

(3.4)

Also, at $ x_0 $, we have

$ \begin{align} 0 &\ge F^{ii}(\log{\Phi } )_{ii}\\ &= F^{ii}\frac{2u_{ii}^2+2u_lu_{lii}}{\left | \nabla u \right |^2 } -\gamma \frac{F^{ii}u_{ii}}{u}+\gamma (1-\gamma ) \frac{F^{ii}u_i^2}{u^2} \\ &=\frac{2F^{ii}u_{ii}^2}{u_1^2}+\frac{2F^{ii}u_1(b_{ii, 1}-u_i\delta _{1i})}{u_1^2}-\gamma \frac{F^{ii}u_{ii}}{u}+\gamma (1-\gamma )\frac{F^{ii}u_i^2}{u^2} \\ &\ge \frac{2F^{ii}u_{ii}^2}{u_1^2}+2(p-1)u^{p-2}f+\frac{2u^{p-1}f_1}{u_1}-2F^{11}-\gamma \frac{F^{ii}b_{ii}}{u}+2(1-\gamma )\frac{F^{11}u_{11}}{u} \\ &\ge 2F^{11}(\frac{u_{11}^2}{u_1^2}-1 )+2(p-1)\frac{F}{u}+\frac{2u^{p-1}f_1}{u_1}-\gamma (k-l)\frac{F}{u}. \end{align} $

(3.5)

By (3.3), if $ A\ge\frac{4}{\gamma^2} $,

$ \begin{align} \frac{u_{11}^2}{u_1^2}-1 \ge \frac{\gamma ^2}{4} A \frac{M_u}{u}-1 \ge 0. \end{align} $

(3.6)

By the definition of $ \Phi $, we have

$ \begin{align} \frac{2u^{p-1}f_1}{u_1}&\ge -Cu^{p-1}\Phi^{-\frac{1}{2}}u^{-\frac{\gamma}{2} } \\ &\ge -\frac{C}{\sqrt{A}}M_u^{-1+\frac{\gamma}{2}}u^{p-1}u^{-\frac{\gamma}{2}}\\ &\ge -\frac{C}{\sqrt{A} }u^{p-1}u^{-1}\\ &\ge -\frac{C}{\sqrt{A} }\frac{F}{u}. \end{align} $

(3.7)

Combining (3.5), (3.6) and (3.7), we have

$ \begin{align} 0 &\ge F^{ii}(\log\Phi)_{ii} \\ & \ge 2F^{11}(\frac{u_{11}^2}{u_1^2}-1 )+2(p-1)\frac{F}{u}-\frac{C}{\sqrt{A}}\frac{F}{u}-\gamma (k-l)\frac{F}{u}\\ & \ge [2(p-1)-\frac{C}{\sqrt{A}}-\gamma (k-l) ]\frac{F}{u} \\ & > 0, \end{align} $

(3.8)

if we choose $ A > \frac{C^2}{p-1} $. This is a contradiction. Hence Theorem 3.1 holds.

Remark 3.2  If we choose

$ \begin{align} \Phi = \frac{|\nabla u|^2}{(u - \min u)^\gamma}, \end{align} $

(3.9)

following the idea of [1, 2], we can prove

$ \begin{align} \frac{|\nabla u(x)|^2}{(u(x)- \min u)^\gamma} \leq A \max\limits_{\mathbb{S}^n} u^{2- \gamma}, \quad \forall\; x \in \mathbb{S}^n. \end{align} $

(3.10)

Following Lemma 3.1 in [2], we can get the positive lower bound and upper bound of $ u $. In fact, we can prove the following lemma (see [19]).

Lemma 4.1  Assume $ u $ is a positive even convex function on $ \mathbb{S}^n $ satisfying condition

$ \begin{align} \frac{|\nabla u|^2}{u^\gamma}(x) \leq A \max\limits_{\mathbb{S}^n} u^{2-\gamma}, \quad \forall\; x \in \mathbb{S}^n, \end{align} $

(4.1)

for some $ \gamma \in (0, 1) $ and $ A>0 $. Then the following non-collapsing estimate holds,

$ \begin{align} \frac{\max u}{\min u} \leq C, \end{align} $

(4.2)

where $ C $ depends only on $ n $, $ \gamma $ and $ A $.

Proof  The proof is similar with [2]. For complete, we give the proof here.

Let $ \Omega $ be the convex body with support function $ u $. Since $ u $ is even, the center of mass of $ \Omega $ is the origin. From the John Lemma, there is an ellipsoid $ E $ centered at the origin, such that

$ \begin{align} E \subset \Omega \subset (n+1)^{\frac{3}{2}} E. \end{align} $

Write $ E $

$ \begin{align} \frac{x_1^2}{b^2_1}+ \cdots + \frac{x_{n+1}^2}{b^2_{n+1}} \leq 1, \end{align} $

with longest axis $ b_1 $, and the shortest axis $ b_{n+1} $. We have

$ \begin{align} b_1 \leq \max u \leq (n+1)^{\frac{3}{2}} b_1, \quad b_{n+1} \leq \min u \leq (n+1)^{\frac{3}{2}} b_{n+1}. \end{align} $

Recall that the support function of $ E $ is

$ \begin{align} u_E(x) = \sqrt{b^2_1 x_1^2 + \cdots + b^2_{n+1} x_{n+1}^2 }, \quad x \in \mathbb{S}^n, \end{align} $

and then

$ \begin{align} u_E(x) \leq u(x) \leq (n+1)^{\frac{3}{2}} u_E(x) , \quad x \in \mathbb{S}^n. \end{align} $

Restrict the support function $ u_E $ to the slice $ S := \{x \in \mathbb{S}^n | x = (x_1, 0, \cdots, 0, x_{n+1})\} $. Set

$ \begin{align} v(s) = u_E(s, 0, \cdots, 0, \sqrt{1-s^2})= \sqrt{b^2_1 s^2 + b^2_{n+1} (1-s^2) } \geq b_1 s, \quad s \in [0, 1]. \end{align} $

Hence

$ \begin{align} v(t (\frac{b_{n+1}}{b_1})^{\frac{2-\gamma}{2}}) \geq t b_1^{\frac{\gamma}{2}} b_{n+1}^{\frac{2-\gamma}{2}}, \end{align} $

for $ t \in [0, (\frac{b_1}{b_{n+1}})^{\frac{2-\gamma}{2}}] $. On the other hand, set $ q(s) = u(s, 0, \cdots, 0, \sqrt{1-s^2})^{\frac{2-\gamma}{2}} $. By the weighted gradient estimate

$ \begin{align} |\frac{d}{ds} q(s)| \leq \frac{|\nabla u(s, 0, \cdots, 0, \sqrt{1-s^2})|}{u ^{\frac{\gamma}{2}}} \leq A^{\frac{1}{2}} (\max u)^{\frac{2-\gamma}{2}} \leq A^{\frac{1}{2}}(n+1)^{\frac{3(2-\gamma)}{4}} b_1 ^{\frac{2-\gamma}{2}}. \end{align} $

Hence

$ \begin{align} q(t (\frac{b_{n+1}}{b_1})^{\frac{2-\gamma}{2}}) \leq& q(0) + A^{\frac{1}{2}}(n+1)^{\frac{3(2-\gamma)}{4}} b_1 ^{\frac{2-\gamma}{2}} \cdot t (\frac{b_{n+1}}{b_1})^{\frac{2-\gamma}{2}} \\ \leq& [ (n+1)^{\frac{3}{2}} b_{n+1} ]^{\frac{2-\gamma}{2}} + t A^{\frac{1}{2}}(n+1)^{\frac{3(2-\gamma)}{4}} b_{n+1}^{\frac{2-\gamma}{2}} \\ =& (n+1)^{\frac{3(2-\gamma)}{4}}[ 1 + t A^{\frac{1}{2}}] b_{n+1}^{\frac{2-\gamma}{2}}. \end{align} $

Thus

$ \begin{align} u(t (\frac{b_{n+1}}{b_1})^{\frac{2-\gamma}{2}}, 0, \cdots, 0, \sqrt{1-t^2 (\frac{b_{n+1}}{b_1})^{2-\gamma}}) \leq(n+1)^{\frac{3}{2}} [1 + t A^{\frac{1}{2}}]^{\frac{2}{2-\gamma}} b_{n+1}. \end{align} $

Since $ u_E(x) \leq u(x) $, we obtain

$ \begin{align} t b_1^{\frac{\gamma}{2}} b_{n+1}^{\frac{2-\gamma}{2}}\leq (n+1)^{\frac{3}{2}} [ 1+ t A^{\frac{1}{2}}]^{\frac{2}{2-\gamma}} b_{n+1}. \end{align} $

Let $ t= A^{-\frac{1}{2}} $, then we have

$ \begin{align} \frac{b_1}{b_{n+1}}\leq (n+1)^{\frac{3}{\gamma}}2^{\frac{4}{(2-\gamma)\gamma}} A^{\frac{1}{\gamma}}. \end{align} $

Hence

$ \begin{align} \frac{\max u}{\min u} \leq (n+1)^{\frac{3}{2}} \frac{b_1}{b_{n+1}}\leq (n+1)^{(\frac{3}{2}+\frac{3}{\gamma})}2^{\frac{4}{(2-\gamma)\gamma}} A^{\frac{1}{\gamma}}. \end{align} $

Now we start to prove Theorem 4.2.

Theorem 4.2  Suppose $ u \in C^{3}(\mathbb{S}^{n}) $ is a positive convex even solution of the equation (1.1), where $ 0 \leq l < k \leq n $, $ p-1>0 $, and $ f\in C^1(\mathbb{S}^n) $ is a positive even function. Then we have the following uniform $ C^0 $ estimate

$ \begin{align} &0 < c_0 \leq u \leq C_0 , \quad \text{if } \; p-1\ne k-l, \end{align} $

(4.3)

$ \begin{align} &0 < 1 \leq \frac{u}{\min u} \leq C_0 , \quad \text{if } \; p-1 = k-l, \end{align} $

(4.4)

where $ c_0 $ and $ C_0 $ are two positive constants depending only on $ n $, $ k $, $ l $, $ p $, $ \min_{\mathbb{S}^{n}} f $ and $ ||f||_{C^1} $.

Proof  When $ p-1> k-l $, we can get directly from the equation (1.1)

$ \begin{align} \min u \geq \big[ \frac{C_{n}^k}{C_{n}^l} \frac{1}{\min f}\big]^{\frac{1}{p-1-(k-l)}}, \quad \max u \geq \big[ \frac{C_{n}^k}{C_{n}^l} \frac{1}{\max f}\big]^{\frac{1}{p-1-(k-l)}}. \end{align} $

When $ p-1 = k-l $, Theorem 4.2 holds from Theorem 3.1 and Lemma 4.1. When $ 0<p-1<k-l $, we know from the equation (1.1)

$ \begin{align} \min u \leq \big[ \frac{C_{n}^k}{C_{n}^l} \frac{1}{\max f}\big]^{\frac{1}{p-1-(k-l)}}, \quad \max u \geq \big[ \frac{C_{n}^k}{C_{n}^l} \frac{1}{\min f}\big]^{\frac{1}{p-1-(k-l)}}. \end{align} $

Remark 4.3  From [20], the constant rank theorem holds if $ f^{-\frac{1}{p-1 +k-l}} $ is spherical convex and $ p-1 \geq 0 $. From [16], the existence theorem of the positive convex even solutions of (1.1) holds by the method of degree theory.