Parabolic Hessian Quotient Equation in Exterior Domain

Abstract

This study mainly focuses on the parabolic Hessian quotient equation in the exterior domain. The existence and uniqueness of generalized parabolically symmetric solutions with generalized asymptotic behavior are proven using Perron’s method.

1. Introduction

Let ${\mathbb{R}}^N$ be the N-dimensional Euclidean space, and ${\mathbb{R}}^{N+1}={\mathbb{R}}^N\times \mathbb{R}$ and ${\mathbb{R}}_{-}^{N+1}$ be the set of $(x,t)$ in ${\mathbb{R}}^{N+1}$ with $t\le 0.$ Denote in ${\mathbb{R}}_{-}^{N+1}$, $u=u(x,t)$. In this paper, we consider the parabolic Hessian quotient equation

$$-u_t\frac{{\sigma }_k\left(\lambda \left(D^{2}u\right)\right)}{{\sigma }_l\left(\lambda \left(D^{2}u\right)\right)}=1,$$

(1)

where $N\ge 3, N\ge k>l\ge 0,$

$${\sigma }_p\left(\lambda \right)={\sigma }_p\left(\lambda \left(D^{2}u\right)\right)=\sum _{1\le t_1<\cdots <t_p\le N}{\lambda }_{t_1}\dots {\lambda }_{t_p}(1\le p\le N),$$

and $\lambda =\lambda \left(D^{2}u\right)$ represents the eigenvalue vector $({\lambda }_1,{\lambda }_2,\cdots ,{\lambda }_N)$ of the Hessian matrix $D^{2}u$ of u. Additionally, we set ${\sigma }_0\left(\lambda \right)=1.$

(1) is the parabolic Hessian equation $-u_t{\sigma }_k\left(\lambda \left(D^{2}u\right)\right)=1$ if $l=0$, and in particular, (1) is the parabolic Monge–Ampère equation $-u_{t}det{D}^{2}u=1$ if $l=0,k=N$.

Regarding the parabolic Monge–Ampère equation, Gutiérrez-Huang [1] proved that if $u=u(x,t)\in C^{4,2}\left({\mathbb{R}}_{-}^{N+1}\right)$ is nonincreasing for t, convex for x and satisfies ${\mathbb{R}}_{-}^{N+1}$,

$$-u_{t}det{D}^{2}u=1,$$

(2)

and $-s_1\le u_t(x,t)\le -s_2$ with positive constants $s_1, s_2$, then $u(x,t)$ is the sum of $-ct$ and a convex quadratic polynomial, where c is a positive constant. Xiong-Bao [2] extended the result to $u_t=\mu \left(\log \det D^{2}u\right)$ with $\mu =\mu \left(z\right)\in C^2\left(\mathbb{R}\right)$. Wang-Bao [3] obtained that the solutions of $u_t-\log \det D^{2}u=f$ in ${\mathbb{R}}_{-}^{N+1}$ tend to $-\alpha t+\frac{1}{2}{x}^{T}Mx+\theta ·x+c$ at infinity for $\alpha >0, M$ being a matrix which is positive definite and $detM=1$, $\theta \in {\mathbb{R}}^N,c\in \mathbb{R}$, $x^T$ being the transpose of x. Zhang-Bao-Wang [4] found that the solutions of $-u_{t}det{D}^{2}u=f$ in ${\mathbb{R}}_{-}^{N+1}$ have the same asymptotic behavior as [3]. Dai [5] studied the exterior problem of $u_t-logdet{D}^{2}u=1$ and obtained the asymptotic behavior for the solutions; however, Gong-Zhou-Bao [6] studied the exterior problem of $-u_{t}det{D}^{2}u=f$. Other results can be seen in [7,8,9], etc.

Regarding the exterior problem of parabolic Hessian equation $-u_t{S}_k\left(D^{2}u\right)=1$, Zhou-Bao [10] obtained the generalized asymptotic behavior of the solutions

$${\displaystyle \underset{-t+{\left|x\right|}^2\to \infty }{lim\; sup}{(-t+|x|}^2{)}^{1-\frac{\chi (N-2)}{2}}|u(x,t)-(-t+\frac{1}{2}{x}^{T}Mx+\theta ·x+c)|<\infty }$$

as $-t+{\left|x\right|}^2\to \infty$, where $\chi \in [\frac{k-2}{N-2},1]$ is a constant and M is a matrix which is positive definite and satisfies ${\sigma }_k\left(\lambda \left(M\right)\right)=1$.

Li-Li [11] discussed the generalized asymptotic behavior of solutions for the elliptic Hessian quotient equation $\frac{{\sigma }_k\left(\lambda \left(D^{2}u\right)\right)}{{\sigma }_l\left(\lambda \left(D^{2}u\right)\right)}=f$ outside a bounded domain with $f=1$; however, Dai-Bao-Wang [12] discussed the generalized asymptotic behavior of solutions when f is a perturbation of a generalized symmetric function at infinity.

Herein, we study the generalized asymptotic behavior of solutions for the parabolic Hessian quotient Equation (1). Consider the exterior problem

$$\begin{array}{cc}\hfill & {\displaystyle -u_t\frac{{\sigma }_k\left(\lambda \left(D^{2}u\right)\right)}{{\sigma }_l\left(\lambda \left(D^{2}u\right)\right)}=1\mathrm{in}{\mathbb{R}}_{-}^{N+1}\setminus \overline{\mathsf{\Omega }},}\end{array}$$

(3)

$$\begin{array}{cc}\hfill & u=\varsigma \mathrm{on}{\partial }_p\mathsf{\Omega },\end{array}$$

(4)

where

$$\mathsf{\Omega }=\left\{\right(x,t\left)\right|0\ge t>V\left(x\right)\},$$

and $V\left(x\right)$ is second-order continuous differentiable and strictly convex such that $\mathsf{\Omega }$ is bounded and nonempty, ${\partial }_p\mathsf{\Omega }=\left\{(x,t)\right|V\left(x\right)=t\}$, and $\varsigma =\varsigma (x,t)\in C^0({\mathbb{R}}_{-}^{N+1}\setminus \overline{\mathsf{\Omega }})$ is a known function.

The following parts of the paper are divided into two sections. In the next section, several lemmas and a generalized radially symmetric function are provided. The main result and its proof are contained in the final section.

2. Preliminaries

The function $u\in C^{i,j}({\mathbb{R}}_{-}^{N+1}\setminus \overline{\mathsf{\Omega }})$ implies that u has the i-order continuous derivative in $x\in {\mathbb{R}}^N$ and j-order continuous derivative in $t\in \mathbb{R}$. Define a symmetric cone ${\mathsf{\Gamma }}_k$ as

$${\mathsf{\Gamma }}_k=\left\{\lambda |\lambda \mathrm{is} \mathrm{in}{\mathbb{R}}^N\mathrm{and}\mathrm{for}p=1,2,\cdots ,k,{\sigma }_p\left(\lambda \right)>0\right\}.$$

If $\lambda \left(D^{2}u(x,t)\right)\in {\mathsf{\Gamma }}_k$, then we call u k-convex. If u is non-increasing about t and k-convex, then we call u parabolically k-convex.

For ∀$(\widehat{y},\widehat{s})\in {\mathbb{R}}^{N+1}$ and $r>0$, let

$$O_r(\widehat{y},\widehat{s}):=\left\{(x,t)|\widehat{s}-r^2<t\le \widehat{s},|x-\widehat{y}|<r\right\}.$$

Definition 1. 

Let $u\in C^0({\mathbb{R}}_{-}^{N+1}\setminus \mathsf{\Omega })$. If for $\forall (\widehat{y},\widehat{s})\in {\mathbb{R}}_{-}^{N+1}\setminus \overline{\mathsf{\Omega }}$ and any parabolically k-convex function, $\varpi \in C^{2,1}\left(O_r(\widehat{y},\widehat{s})\right)$, satisfying

$$\varpi (\widehat{y},\widehat{s})=u(\widehat{y},\widehat{s}),\varpi (x,t)\le (\ge )u(x,t),for any(x,t)\in O_r(\widehat{y},\widehat{s})\subset {\mathbb{R}}_{-}^{N+1}\setminus \overline{\mathsf{\Omega }},$$

we can obtain $(\widehat{y},\widehat{s})$,

$$-{\varpi }_t\frac{{\sigma }_k\left(\lambda \left(D^2\varpi \right)\right)}{{\sigma }_l\left(\lambda \left(D^2\varpi \right)\right)}\le (\ge )1,$$

and then, we call u a viscosity supersolution (separately, subsolution) to (3).

For the subsolution, ϖ does not need to be parabolically k-convex.

If $u\in C^0({\mathbb{R}}_{-}^{N+1}\setminus \overline{\mathsf{\Omega }})$ is both a viscosity supersolution and a viscosity subsolution, then we call u the viscosity solution of (3).

Definition 2. 

If u is the viscosity supersolution (subsolution) of (3) and $u\ge (\le )\varsigma (x,t)$ on the boundary ${\partial }_p\mathsf{\Omega }$, then we call u the viscosity supersolution (subsolution) of problems (3) and (4).

If $u\in C^0\left(\overline{{\mathbb{R}}_{-}^{N+1}\setminus \mathsf{\Omega }}\right)$ is a viscosity solution of (3) and satisfies $u(x,t)=\varsigma (x,t)$ on the boundary ${\partial }_p\mathsf{\Omega }$, then we call u the viscosity solution of (3) and (4).

Let

$${\mathcal{M}}_{k,l}=\left\{M|M\mathrm{is}\mathrm{an}N-\mathrm{dimensional} \mathrm{positive} \mathrm{definite} \mathrm{square} \mathrm{matrix}, \mathrm{real} \mathrm{and} \mathrm{meets}\frac{{\sigma }_k\left(\lambda \left(M\right)\right)}{{\sigma }_l\left(\lambda \left(M\right)\right)}=1\right\}.$$

Definition 3. 

Suppose that the matrix $M=\mathrm{diag}\left(m_1,m_2,\cdots ,m_N\right)\in {\mathcal{M}}_{k,l}$. Let

$$\omega =\sqrt{-2t+x^{T}Mx}=\sqrt{-2t+\sum _{n=1}^N{m}_n{x}_n^2}.$$

(5)

If $u=u\left(\omega \right)$, then u is called a generalized parabolically symmetric function of M.

Lemma 1. 

Let $D_1\subset D_2\subset {\mathbb{R}}^{N+1}$ be two open subsets and $g\in C^0\left({\mathbb{R}}^{N+1}\right)$. Assume that $v\in C^0\left(D_2\right)$ and $w\in C^0\left({\overline{D}}_1\right)$ are, separately, the viscosity solutions of

$$-v_t\frac{{\sigma }_k\left(\lambda \left(D^{2}v\right)\right)}{{\sigma }_l\left(\lambda \left(D^{2}v\right)\right)}\ge g(x,t),(x,t)\in D_2,$$

(6)

and

$$-w_t\frac{{\sigma }_k\left(\lambda \left(D^{2}w\right)\right)}{{\sigma }_l\left(\lambda \left(D^{2}w\right)\right)}\ge g(x,t),(x,t)\in D_1.$$

(7)

Suppose

$$w=von\partial D_1\setminus \left(\partial D_1\cap \partial D_2\right),w\ge vin{D}_1.$$

Set

$$u=\left\{\begin{array}{cc}w\hfill & in{D}_1,\hfill \\ v\hfill & in{D}_2\setminus D_1.\hfill \end{array}\right.$$

Then, $u\in C^0\left(D_2\right)$ satisfies in the sense of viscosity

$$-u_t\frac{{\sigma }_k\left(\lambda \left(D^{2}u\right)\right)}{{\sigma }_l\left(\lambda \left(D^{2}u\right)\right)}\ge gin{D}_2.$$

Proof. 

For $\forall (\widehat{y},\widehat{s})\in D_2$, let any function $\varpi \in C^{2,1}\left(D_2\right)$ satisfy

$$u(\widehat{y},\widehat{s})-\varpi (\widehat{y},\widehat{s})\ge u(x,t)-\varpi (x,t),\forall (x,t)\in O_r(\widehat{y},\widehat{s})\subset D_2.$$

If $(\widehat{y},\widehat{s})\in D_1$, then

$$\begin{array}{cc}\hfill w(x,t)-\varpi (x,t)& =u(x,t)-\varpi (x,t)\le u(\widehat{y},\widehat{s})-\varpi (\widehat{y},\widehat{s})\hfill \\ \hfill & =w(\widehat{y},\widehat{s})-\varpi (\widehat{y},\widehat{s}),\forall (x,t)\in O_{r_1}(\widehat{y},\widehat{s})\subset O_r(\widehat{y},\widehat{s})\cap D_1.\hfill \end{array}$$

According to (7), we obtain

$$-{\varpi }_t(\widehat{y},\widehat{s})\frac{{\sigma }_k(\lambda \left(D^2\varpi (\widehat{y},\widehat{s})\right)}{{\sigma }_l(\lambda \left(D^2\varpi (\widehat{y},\widehat{s})\right)}\ge g(\widehat{y},\widehat{s}).$$

If $(\widehat{y},\widehat{s})\in D_2\setminus D_1$, then

$$\begin{array}{cc}\hfill v(x,t)-\varpi (x,t)& \le u(x,t)-\varpi (x,t)\le u(\widehat{y},\widehat{s})-\varpi (\widehat{y},\widehat{s})\hfill \\ \hfill & =v(\widehat{y},\widehat{s})-\varpi (\widehat{y},\widehat{s}),\forall (x,t)\in O_r(\widehat{y},\widehat{s}).\hfill \end{array}$$

According to (6), we obtain

$$\begin{array}{c}\hfill -{\varpi }_t(\widehat{y},\widehat{s})\frac{{\sigma }_k(\lambda \left(D^2\varpi (\widehat{y},\widehat{s})\right)}{{\sigma }_l(\lambda \left(D^2\varpi (\widehat{y},\widehat{s})\right)}\ge g(\widehat{y},\widehat{s}).\end{array}$$

According to Definition 1, we know that the Lemma is true. □

Lemma 2. 

(Comparison principle) Assume that Ω is a bounded open set on ${\mathbb{R}}^{N+1}$ and $u,v\in C^0(\overline{\mathsf{\Omega }})$, which are satisfied separately in the sense of viscosity.

$$-u_t\frac{{\sigma }_k\left(\lambda \left(D^{2}u\right)\right)}{{\sigma }_l\left(\lambda \left(D^{2}u\right)\right)}\ge gin\mathsf{\Omega }$$

and

$$-v_t\frac{{\sigma }_k\left(\lambda \left(D^{2}v\right)\right)}{{\sigma }_l\left(\lambda \left(D^{2}v\right)\right)}\le gin\mathsf{\Omega },$$

and then,

$${\displaystyle \underset{{\partial }_p\mathsf{\Omega }}{sup}(u-v)\ge \underset{\mathsf{\Omega }}{sup}(u-v).}$$

(8)

Proof. 

The proof of this lemma is almost identical to the proof of Lemma 2.5 in [10]. We will not provide detailed proof here. □

With regard to the comparison principle, we can also refer to references [13,14,15].

Proposition 1 

([16]). Suppose $\mu \in {\mathsf{\Gamma }}_k$, $0\le l<k$ and $0\le \widehat{l}<\widehat{k},\widehat{k}\le k,\widehat{l}\le l$; then,

$${\left[\frac{{\sigma }_k\left(\mu \right)/C_N^k}{{\sigma }_l\left(\mu \right)/C_N^l}\right]}^{\frac{1}{k-l}}\le {\left[\frac{{\sigma }_{\widehat{k}}\left(\mu \right)/C_N^{\widehat{k}}}{{\sigma }_{\widehat{l}}\left(\mu \right)/C_N^{\widehat{l}}}\right]}^{\frac{1}{\widehat{k}-\widehat{l}}},$$

where $C_N^k=\frac{N!}{k!(N-k)!}.$

Proposition 2 

([17]). Let the matrix $M={(s_n{\delta }_{nj}+\overline{\gamma }{q}_n{t}_j)}_{N\times N},$ where $s=(s_1,\dots ,s_N)\in {\mathbb{R}}^N, q=(q_1,\dots ,q_N)\in {\mathbb{R}}^N$, $\overline{\gamma }\in \mathbb{R},$ and ${\left({\delta }_{nj}\right)}_{N\times N}$ is the identity matrix. Then,

$${\sigma }_k\left(\lambda \left(M\right)\right)=\overline{\gamma }\sum _{n=1}^N{t}_n^2{\sigma }_{k-1;n}\left(s\right)+{\sigma }_k\left(s\right),$$

where ${\sigma }_{k-1;n}\left(s\right):={\left.{\sigma }_{k-1}\left(s\right)\right|}_{s_n=0}$.

Let $m:=\left(m_1,m_2,\cdots ,m_N\right)\in {\mathsf{\Gamma }}_k$ satisfy ${\sigma }_k\left(m\right)={\sigma }_l\left(m\right)$. Define

$$\overline{{\eta }_k}=\overline{{\eta }_k}\left(m\right)=\underset{\genfrac{}{}{0pt}{}{x\in {\mathbb{R}}^N\setminus \left\{0\right\}}{t\in (-\infty ,0]}}{sup}\frac{{\sum }_{n=1}^N{\sigma }_{k-1;n}\left(m\right)m_n^2{x}_n^2}{{\sigma }_k\left(m\right)(-2t+{\sum }_{n=1}^N{m}_n{x}_n^2)},$$

(9)

and

$$\underline{{\eta }_l}=\underline{{\eta }_l}\left(m\right)=\underset{\genfrac{}{}{0pt}{}{x\in {\mathbb{R}}^N\setminus \left\{0\right\}}{t\in (-\infty ,0]}}{inf}\frac{{\sum }_{n=1}^N{\sigma }_{l-1;n}\left(m\right)m_n^2{x}_n^2}{{\sigma }_l\left(m\right)(-2t+{\sum }_{n=1}^N{m}_n{x}_n^2)}.$$

(10)

Lemma 3. 

Assume that $m:=\left(m_1,m_2,\cdots ,m_N\right)$ with $0<m_1\le m_2\le \cdots \le m_N$. Then,

$$\begin{array}{c}\hfill 0\le \underline{{\eta }_k}\left(m\right)\le \frac{m_1{\sigma }_{k-1;1}\left(m\right)}{{\sigma }_k\left(m\right)}\le \frac{k}{N}\le \overline{{\eta }_k}\left(m\right)=\frac{m_N{\sigma }_{k-1;N}\left(m\right)}{{\sigma }_k\left(m\right)}\le 1,1\le k\le N,\end{array}$$

(11)

$$\begin{array}{c}\hfill 0=\overline{{\eta }_0}\left(m\right)<\frac{1}{N}\le \frac{m_N}{{\sigma }_1\left(m\right)}=\overline{{\eta }_1}\left(m\right)\le \overline{{\eta }_2}\left(m\right)\le \cdots \le \overline{{\eta }_{N-1}}\left(m\right)<\overline{{\eta }_N}\left(m\right)=1,\end{array}$$

(12)

and

$$\begin{array}{c}\hfill 0=\underline{{\eta }_0}\left(m\right)\le \underline{{\eta }_1}\left(m\right)\le \underline{{\eta }_2}\left(m\right)\le \cdots \le \underline{{\eta }_{N-1}}\left(m\right)\le \underline{{\eta }_N}\left(m\right)<1.\end{array}$$

(13)

Proof. 

We know that if $m_n\le m_j,$ then

$$\begin{array}{c}\hfill m_n{\sigma }_{k-1;n}\left(m\right)\le m_j{\sigma }_{k-1;j}\left(m\right).\end{array}$$

(14)

The above inequality can be found in [11]. Then,

$$\begin{array}{cc}\hfill \overline{{\eta }_k}& =\underset{\genfrac{}{}{0pt}{}{x\in {\mathbb{R}}^N\setminus \left\{0\right\}}{t\in (-\infty ,0]}}{sup}\frac{{\sum }_{n=1}^N{\sigma }_{k-1;n}\left(m\right)m_n^2{x}_n^2}{{\sigma }_k\left(m\right)(-2t+{\sum }_{n=1}^N{m}_n{x}_n^2)}\hfill \\ \hfill & \ge \underset{\genfrac{}{}{0pt}{}{x_1=\cdots =x_{N-1}=t=0}{x_N\ne 0}}{sup}\frac{{\sum }_{n=1}^N{\sigma }_{k-1;n}\left(m\right)m_n^2{x}_n^2}{{\sigma }_k\left(m\right)(-2t+{\sum }_{n=1}^N{m}_n{x}_n^2)}\hfill \\ \hfill & =\underset{x_N\ne 0}{sup}\frac{{\sigma }_{k-1;N}\left(m\right)m_N^2{x}_N^2}{{\sigma }_k\left(m\right)m_N{x}_N^2}\hfill \\ \hfill & =\frac{m_N{\sigma }_{k-1;N}\left(m\right)}{{\sigma }_k\left(m\right)},\hfill \end{array}$$

and according to (14),

$$\begin{array}{cc}\hfill \overline{{\eta }_k}& =\underset{\genfrac{}{}{0pt}{}{x\in {\mathbb{R}}^N\setminus \left\{0\right\}}{t\in (-\infty ,0]}}{sup}\frac{{\sum }_{n=1}^N{\sigma }_{k-1;n}\left(m\right)m_n^2{x}_n^2}{{\sigma }_k\left(m\right)(-2t+{\sum }_{n=1}^N{m}_n{x}_n^2)}\hfill \\ \hfill & \le \underset{\genfrac{}{}{0pt}{}{x\ne 0}{t\le 0}}{sup}\frac{m_N{\sigma }_{k-1;N}\left(m\right){\sum }_{n=1}^N{m}_n{x}_n^2}{{\sigma }_k\left(m\right)(-2t+{\sum }_{n=1}^N{m}_n{x}_n^2)}\hfill \\ \hfill & \le \underset{\genfrac{}{}{0pt}{}{x\ne 0}{t\le 0}}{sup}\frac{m_N{\sigma }_{k-1;N}\left(m\right){\sum }_{n=1}^N{m}_n{x}_n^2}{{\sigma }_k\left(m\right){\sum }_{n=1}^N{m}_n{x}_n^2}\hfill \\ \hfill & =\frac{m_N{\sigma }_{k-1;N}\left(m\right)}{{\sigma }_k\left(m\right)}.\hfill \end{array}$$

So,

$$\overline{{\eta }_k}=\frac{m_N{\sigma }_{k-1;N}\left(m\right)}{{\sigma }_k\left(m\right)}.$$

(15)

We can similarly prove that

$$0\le \underline{{\eta }_k}\le \frac{m_1{\sigma }_{k-1;1}\left(m\right)}{{\sigma }_k\left(m\right)}.$$

Since

$$m_1{\sigma }_{k-1;1}\left(m\right)\le m_2{\sigma }_{k-1;2}\left(m\right)\le \cdots \le m_N{\sigma }_{k-1;N}\left(m\right)$$

and

$$\sum _{n=1}^N\frac{m_n{\sigma }_{k-1;n}\left(m\right)}{{\sigma }_k\left(m\right)}=k,$$

then

$$\begin{array}{c}\hfill \underline{{\eta }_k}\left(m\right)\le \frac{k}{N}\le \overline{{\eta }_k}\left(m\right).\end{array}$$

So,

$$\begin{array}{c}\hfill 0\le \underline{{\eta }_k}\le \frac{m_1{\sigma }_{k-1;1}\left(m\right)}{{\sigma }_k\left(m\right)}\le \frac{k}{N}\le \overline{{\eta }_k}=\frac{m_N{\sigma }_{k-1;N}\left(m\right)}{{\sigma }_k\left(m\right)}\le 1,1\le k\le N.\end{array}$$

Then, (11) is proven.

Since

$$\begin{array}{c}\hfill {\sigma }_k\left(m\right)={\sigma }_{k;n}\left(m\right)+m_n{\sigma }_{k-1;n}\left(m\right),1\le n\le N,\end{array}$$

(16)

then

$$\begin{array}{c}\hfill m_n{\sigma }_{k-1;n}\left(m\right)<{\sigma }_k\left(m\right),1\le n\le N,1\le k\le N-1\end{array}$$

(17)

and

$$\begin{array}{c}\hfill m_n{\sigma }_{N-1;n}\left(m\right)={\sigma }_N\left(m\right),1\le n\le N.\end{array}$$

(18)

Thus, according to (17),

$$\underline{{\eta }_k}\le \overline{{\eta }_k}<1,1\le k\le N-1$$

and according to (18),

$$\overline{{\eta }_N}=1.$$

According to (15) and (16) and the inequality

$$\frac{{\sigma }_{k-1;N}\left(m\right)}{{\sigma }_{k;N}\left(m\right)}\le \frac{{\sigma }_{k;N}\left(m\right)}{{\sigma }_{k+1;N}\left(m\right)},$$

we know that

$$\begin{array}{cc}\hfill \overline{{\eta }_k}& =\frac{m_N{\sigma }_{k-1;N}\left(m\right)}{{\sigma }_k\left(m\right)}=\frac{m_N{\sigma }_{k-1;N}\left(m\right)}{{\sigma }_{k;N}\left(m\right)+m_N{\sigma }_{k-1;N}\left(m\right)}\hfill \\ \hfill & \le \frac{m_N{\sigma }_{k;N}\left(m\right)}{{\sigma }_{k+1;N}\left(m\right)+m_N{\sigma }_{k;N}\left(m\right)}=\frac{m_N{\sigma }_{k;N}\left(m\right)}{{\sigma }_{k+1}\left(m\right)}=\overline{{\eta }_{k+1}}.\hfill \end{array}$$

Then, we prove (12). We can similarly prove (13). □

Remark 1. 

Then, according to (11), we have

$$\frac{k-l}{N}\le \overline{{\eta }_k}-\underline{{\eta }_l}<\overline{{\eta }_k}\le 1.$$

For $M=diag\left(m_1,m_2,\cdots ,m_N\right)\in {\mathcal{M}}_{k,l}$, let $\omega =\sqrt{-2t+{\displaystyle \sum _{n=1}^N}{m}_n{x}_n^2}$; then,

$$\frac{\partial \omega }{\partial t}=-\frac{1}{\omega },$$

$$\frac{\partial \omega }{\partial x_n}=\frac{m_n{x}_n}{\omega },$$

$$\frac{{\partial }^2\omega }{\partial x_n\partial x_j}=\frac{m_n}{\omega }{\delta }_{nj}-\frac{m_n{x}_n}{{\omega }^2}\frac{m_j{x}_j}{\omega }.$$

Let $\psi (x,t)=\psi \left(\omega \right)\in C^{2,1}({\mathbb{R}}^{N+1}\setminus \left\{0\right\})$ be a generalized parabolically symmetric function. Then, for $n,j=1,\dots ,N,$

$${\psi }_t={\psi }^{\prime }\left(\omega \right)\frac{\partial \omega }{\partial t},$$

$${\psi }_n={\psi }^{\prime }\left(\omega \right)\frac{\partial \omega }{\partial x_n},$$

$$\begin{array}{c}\hfill {\psi }_{ij}={\psi }^{″}\left(\omega \right)\frac{\partial \omega }{\partial x_n}\frac{\partial \omega }{\partial x_j}+{\psi }^{\prime }\left(\omega \right)\frac{{\partial }^2\omega }{\partial x_n\partial x_j}.\end{array}$$

So,

$$\begin{array}{c}\hfill {\psi }_{ij}=\frac{{\psi }^{\prime }\left(\omega \right)}{\omega }{m}_n{\delta }_{nj}+\frac{{\psi }^{″}\left(\omega \right)-\frac{{\psi }^{\prime }\left(\omega \right)}{\omega }}{{\omega }^2}\left(m_n{x}_n\right)\left(m_j{x}_j\right).\end{array}$$

(19)

Theorem 1. 

Consider the problem

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\displaystyle \frac{\phi {\left(\omega \right)}^k+\overline{{\eta }_k}\omega \phi {\left(\omega \right)}^{k-1}{\phi }^{\prime }\left(\omega \right)}{\phi {\left(\omega \right)}^l+\underline{{\eta }_l}\omega \phi {\left(\omega \right)}^{l-1}{\phi }^{\prime }\left(\omega \right)}}=1,\omega >1,\hfill \\ \phi \left(1\right)=\alpha ,\hfill \end{array}\right.\end{array}$$

(20)

where $N\ge 3, N\ge k>l\ge 0, m=(m_1,\dots ,m_N), m_1\le \cdots \le m_N, \mathrm{diag}(m_1,\dots ,m_N)\in {\mathcal{M}}_{k,l}$ and $\alpha \ge 1.$ Then, in $[1,+\infty )$, (20) has a unique smooth solution $\phi =\phi (\omega ,\alpha )$, satisfying the following:

(i) $1\le \phi \left(\omega \right)=\phi (\omega ,\alpha )\le \alpha ,{\partial }_{\omega }\phi (\omega ,\alpha )\le 0$ for $\omega \ge 1,\alpha \ge 1.$ In particular, $\phi (\omega ,1)\equiv 1,\phi (1,\alpha )=\alpha$ and $1<\phi (\omega ,\alpha )<\alpha$ for $\omega >1,\alpha >1.$

(ii) $\phi (\omega ,\alpha )$ is strictly increasing and continuous in α and

$$\underset{\alpha \to +\infty }{lim}\phi (\omega ,\alpha )=+\infty ,\omega \ge 1.$$

(iii) $\phi (\omega ,\alpha )=1+O({\omega }^{-\frac{k-l}{\overline{{\eta }_k}-\underline{{\eta }_l}}})$ as $\omega \to +\infty$.

Proof. 

The proof can be seen in [11]. □

Theorem 2. 

Let φ satisfy (20) and define

$$\begin{array}{c}\hfill {\displaystyle \mathsf{\Psi }(x,t)=\mathsf{\Psi }\left(\omega \right)={\mathsf{\Psi }}_{\beta ,\gamma ,\alpha }\left(\omega \right)=\beta +{\int }_{\gamma }^{\omega }\varrho \phi (\varrho ,\alpha )d\varrho ,\omega \ge \gamma .}\end{array}$$

(21)

Then, Ψ is parabolically k-convex and satisfies

$${\displaystyle -{\mathsf{\Psi }}_t\frac{{\sigma }_k\left(\lambda \left(D^2\mathsf{\Psi }\right)\right)}{{\sigma }_l\left(\lambda \left(D^2\mathsf{\Psi }\right)\right)}=1}$$

for $\omega \ge \gamma .$

Proof. 

Clearly,

$${\mathsf{\Psi }}^{\prime }\left(\omega \right)=\omega \phi \left(\omega \right),$$

$${\mathsf{\Psi }}^{″}\left(\omega \right)=\phi \left(\omega \right)+\omega {\phi }^{\prime }\left(\omega \right).$$

So,

$$\frac{{\mathsf{\Psi }}^{\prime }\left(\omega \right)}{\omega }=\phi \left(\omega \right),$$

$$\frac{{\mathsf{\Psi }}^{″}\left(\omega \right)-\frac{{\mathsf{\Psi }}^{\prime }\left(\omega \right)}{\omega }}{\omega }={\phi }^{\prime }\left(\omega \right).$$

On the other hand,

$${\mathsf{\Psi }}_t=-\phi \left(\omega \right).$$

Therefore, according to (19), we know that

$$\begin{array}{c}\hfill D^2\mathsf{\Psi }={\left(\phi \left(\omega \right)m_n{\delta }_{nj}+\frac{{\phi }^{\prime }\left(\omega \right)}{\omega }\left(m_n{x}_n\right)\left(m_j{x}_j\right)\right)}_{N\times N}.\end{array}$$

According to Proposition 2, we can know that

$${\sigma }_k\left(\lambda \left(D^2\mathsf{\Psi }\right)\right)={\sigma }_k\left(m\right)\phi {\left(\omega \right)}^k+\frac{{\phi }^{\prime }\left(\omega \right)}{\omega }\phi {\left(\omega \right)}^{k-1}\sum _{n=1}^N{\sigma }_{k-1;n}\left(m\right){\left(m_n{x}_n\right)}^2.$$

Since ${\phi }^{\prime }\left(\omega \right)\le 0$, then for $1\le p\le k,$

$$\begin{array}{cc}\hfill {\sigma }_p\left(\lambda \left(D^2\mathsf{\Psi }\right)\right)& ={\sigma }_p\left(m\right)\phi {\left(\omega \right)}^p+\frac{{\phi }^{\prime }\left(\omega \right)}{\omega }\phi {\left(\omega \right)}^{p-1}\sum _{n=1}^N{\sigma }_{p-1;n}\left(m\right){\left(m_n{x}_n\right)}^2\hfill \\ \hfill & ={\sigma }_p\left(m\right)\left[\phi {\left(\omega \right)}^p+\omega {\phi }^{\prime }\left(\omega \right)\phi {\left(\omega \right)}^{p-1}\frac{{\displaystyle \sum _{n=1}^N{\sigma }_{p-1;n}\left(m\right)m_n^2{x}_n^2}}{{\sigma }_p\left(m\right)\left({\displaystyle -2t+\sum _{n=1}^N{m}_n^2{x}_n^2}\right)}\right]\hfill \\ \hfill & \ge {\sigma }_p\left(m\right)\phi {\left(\omega \right)}^{p-1}\left[\phi \left(\omega \right)+\overline{{\eta }_p}\left(m\right)\omega {\phi }^{\prime }\left(\omega \right)\right].\hfill \end{array}$$

(22)

From the equation in (20), we have

$$\begin{array}{c}\hfill {\phi }^{\prime }=-\frac{1}{\omega }·\frac{\phi }{\overline{{\eta }_k}}·\frac{{\phi }^{k-l}-1}{{\phi }^{k-l}-\frac{\underline{{\eta }_l}}{\overline{{\eta }_k}}}.\end{array}$$

On the other hand, according to Lemma 3 and Theorem 1, we know that

$$0\le \frac{{\phi }^{k-l}-1}{{\phi }^{k-l}-\frac{\underline{{\eta }_l}}{\overline{{\eta }_k}}}<1\le \frac{\overline{{\eta }_k}}{\overline{{\eta }_p}},$$

which implies that

$$\phi +\overline{{\eta }_p}\omega {\phi }^{\prime }>0.$$

So, according to (22), $\mathsf{\Psi }$ is parabolically k-convex.

According to Theorem 1, we have $\phi \left(\omega \right)\ge 1$ and ${\phi }^{\prime }\left(\omega \right)\le 0.$ So,

$$\begin{array}{cc}\hfill -{\mathsf{\Psi }}_t\frac{{\sigma }_k\left(\lambda \left(D^2\mathsf{\Psi }\right)\right)}{{\sigma }_l\left(\lambda \left(D^2\mathsf{\Psi }\right)\right)}& =\phi \left(\omega \right)\frac{{\displaystyle {\sigma }_k\left(m\right)\phi {\left(\omega \right)}^k+\frac{{\phi }^{\prime }\left(\omega \right)}{\omega }\phi {\left(\omega \right)}^{k-1}\sum _{n=1}^N{\sigma }_{k-1;n}\left(m\right){\left(m_n{x}_n\right)}^2}}{{\displaystyle {\sigma }_l\left(m\right)\phi {\left(\omega \right)}^l+\frac{{\phi }^{\prime }\left(\omega \right)}{\omega }\phi {\left(\omega \right)}^{l-1}\sum _{n=1}^N{\sigma }_{l-1;n}\left(m\right){\left(m_n{x}_n\right)}^2}}\hfill \\ \hfill & =\phi \left(\omega \right)\frac{{\sigma }_k\left(m\right)\left[{\displaystyle \phi {\left(\omega \right)}^k+\frac{{\phi }^{\prime }\left(\omega \right)}{\omega }\phi {\left(\omega \right)}^{k-1}\sum _{n=1}^N\frac{{\sigma }_{k-1;n}\left(m\right)}{{\sigma }_k\left(m\right)}{m}_n^2{x}_n^2}\right]}{{\sigma }_l\left(m\right)\left[{\displaystyle \phi {\left(\omega \right)}^l+\frac{{\phi }^{\prime }\left(\omega \right)}{\omega }\phi {\left(\omega \right)}^{l-1}\sum _{n=1}^N\frac{{\sigma }_{l-1;n}\left(m\right)}{{\sigma }_l\left(m\right)}{m}_n^2{x}_n^2}\right]}\hfill \end{array}$$

$$\begin{array}{cc}\hfill & =\phi \left(\omega \right)\frac{{\sigma }_k\left(m\right)\left[{\displaystyle \phi {\left(\omega \right)}^k+\omega \frac{{\phi }^{\prime }\left(\omega \right)}{{\omega }^2}\phi {\left(\omega \right)}^{k-1}\sum _{n=1}^N\frac{{\sigma }_{k-1;n}\left(m\right)}{{\sigma }_k\left(m\right)}{m}_n^2{x}_n^2}\right]}{{\sigma }_l\left(m\right)\left[{\displaystyle \phi {\left(\omega \right)}^l+\omega \frac{{\phi }^{\prime }\left(\omega \right)}{{\omega }^2}\phi {\left(\omega \right)}^{l-1}\sum _{n=1}^N\frac{{\sigma }_{l-1;n}\left(m\right)}{{\sigma }_l\left(m\right)}{m}_n^2{x}_n^2}\right]}\hfill \\ \hfill & =\phi \left(\omega \right)\frac{{\sigma }_k\left(m\right)\left[\phi {\left(\omega \right)}^k+\omega {\phi }^{\prime }\left(\omega \right)\phi {\left(\omega \right)}^{k-1}\frac{{\displaystyle \sum _{n=1}^N{\sigma }_{k-1;n}\left(m\right)m_n^2{x}_n^2}}{{\sigma }_k\left(m\right)\left({\displaystyle -2t+\sum _{n=1}^N{m}_n^2{x}_n^2}\right)}\right]}{{\sigma }_l\left(m\right)\left[\phi {\left(\omega \right)}^l+\omega {\phi }^{\prime }\left(\omega \right)\phi {\left(\omega \right)}^{l-1}\frac{{\displaystyle \sum _{n=1}^N{\sigma }_{l-1;n}\left(m\right)m_n^2{x}_n^2}}{{\sigma }_l\left(m\right)\left({\displaystyle -2t+\sum _{n=1}^N{m}_n^2{x}_n^2}\right)}\right]}\hfill \\ \hfill & \ge \frac{\phi {\left(\omega \right)}^k+\omega {\phi }^{\prime }\left(\omega \right)\phi {\left(\omega \right)}^{k-1}\overline{{\eta }_k}}{\phi {\left(\omega \right)}^l+\omega {\phi }^{\prime }\left(\omega \right)\phi {\left(\omega \right)}^{l-1}\underline{{\eta }_l}}=1.\hfill \end{array}$$

(23)

□

3. Main Results

The following are our main results.

Theorem 3. 

Set $N\ge 3$, $N\ge k>l\ge 0$, $\frac{k-l}{\overline{{\eta }_k}-\underline{{\eta }_l}}>2$ and $\varsigma \in C^{2,2}\left(\overline{\mathsf{\Omega }}\right)$. Then for $\forall M\in {\mathcal{M}}_{k,l}, \theta \in {\mathbb{R}}^N, c\in \mathbb{R}$, there exists a constant $\widehat{c}=\widehat{c}{(N,M,\theta ,\mathsf{\Omega },|\left|\varsigma \right||}_{C^{2,2}\left(\overline{\mathsf{\Omega }}\right)})$ such that for $\forall c>\widehat{c}$, problems (3) and (4) possesses only one viscosity solution $u\in C^0\left(\overline{{\mathbb{R}}_{-}^{N+1}\setminus \mathsf{\Omega }}\right)$, satisfying

$$\underset{-t+{\left|x\right|}^2\to \infty }{lim\; sup}\left({(-t+|x|}^2{)}^{\frac{1}{2}\frac{k-l}{\overline{{\eta }_k}-\underline{{\eta }_l}}-1}\left|u(x,t)-\left(-t+\frac{1}{2}{x}^{T}Mx+\theta ·x+c\right)\right|\right)<\infty .$$

(24)

Remark 2. 

We have just learned that Zhou [18] obtained the asymptotic behavior of the radially symmetric solution of (3) and (4). But the asymptotic behavior in [18] is different from the generalized asymptotic behavior in (24), and the result in [18] is a particular case of Theorem 3 when $M=I$, with I being the identity matrix.

The proof of Theorem 3 consists of the following three subsections.

3.1. Construction of Subsolutions

Lemma 4. 

If $N\ge 3$, $\varsigma \in C^{2,2}\left(\overline{\mathsf{\Omega }}\right)$, $M\in {\mathcal{M}}_{k,l}$, then there is a constant $c_0=c_0(N,|\left|\varsigma \right|{|}_{C^{2,2}\left(\overline{\mathsf{\Omega }}\right)},\mathsf{\Omega },M)>0$ such that for any $\overline{c}>c_0$ and $(\overline{\zeta },\overline{\lambda })\in {\partial }_p\mathsf{\Omega }$, we can find $C_0=C_0({N,\left|\right|\varsigma \left|\right|}_{C^2\left(\overline{\mathsf{\Omega }}\right)},\mathsf{\Omega },M,\overline{c})$ and $\rho (\overline{\zeta },\overline{\lambda })\in {\mathbb{R}}^N$, satisfying

$$|\rho (\overline{\zeta },\overline{\lambda })|\le C_0$$

and

$$\varsigma >{\mathsf{\Theta }}_{\overline{\zeta },\overline{\lambda }}on{\partial }_p\mathsf{\Omega }\setminus \left\{(\overline{\zeta },\overline{\lambda })\right\},$$

(25)

where

$${\mathsf{\Theta }}_{\overline{\zeta },\overline{\lambda }}={\mathsf{\Theta }}_{\overline{\zeta },\overline{\lambda }}(x,t)=\varsigma (\overline{\zeta },\overline{\lambda })-\overline{c}(t-\lambda )+\frac{1}{2}{(x-\rho )}^{T}M(x-\rho )-\frac{1}{2}{(\zeta -\rho )}^{T}M(\zeta -\rho ),(x,t)\in {\mathbb{R}}_{-}^{N+1}.$$

Proof. 

See [10]. □

According to Lemma 4, for $M\in {\mathcal{M}}_{k,l}$, $(\overline{\zeta },\overline{\lambda })\in {\partial }_p\mathsf{\Omega }$, there exists $C_0>0$, $\rho (\overline{\zeta },\overline{\lambda })\in {\mathbb{R}}^N$, $|\rho (\overline{\zeta },\overline{\lambda })|\le C_0$, satisfying (25). Choose $\overline{c}=max\left\{C_0,1\right\}$; then,

$$-{\left({\mathsf{\Theta }}_{\overline{\zeta },\overline{\lambda }}\right)}_t\frac{{\sigma }_k\left(\lambda \left(D^2{\mathsf{\Theta }}_{\overline{\zeta },\overline{\lambda }}\right)\right)}{{\sigma }_l\left(\lambda \left(D^2{\mathsf{\Theta }}_{\overline{\zeta },\overline{\lambda }}\right)\right)}=\overline{c}\ge 1,\forall (x,t)\in {\mathbb{R}}_{-}^{N+1}.$$

For $(x,t)\in {\mathbb{R}}_{-}^{N+1}$, let

$$\mathsf{\Theta }(x,t)=\underset{(\overline{\zeta },\overline{\lambda })\in {\partial }_p\mathsf{\Omega }}{sup}{\mathsf{\Theta }}_{\overline{\zeta },\overline{\lambda }}(x,t),$$

and then,

$$\mathsf{\Theta }=\varsigma \mathrm{on}{\partial }_p\mathsf{\Omega },$$

(26)

and in the sense of viscosity, we have

$$-{\mathsf{\Theta }}_t\frac{{\sigma }_k\left(\lambda \left(D^2\mathsf{\Theta }\right)\right)}{{\sigma }_l\left(\lambda \left(D^2\mathsf{\Theta }\right)\right)}\ge 1,(x,t)\in {\mathbb{R}}_{-}^{N+1}.$$

(27)

Without losing generality, we can suppose that M is diagonal and $\theta =0$. Set ${\mathsf{\Omega }}_R=\left\{(x,t)|{\sum }_{n=1}^N{m}_n{x}_n^2-R^2<2t\le 0\right\}.$ Let ${\mathsf{\Omega }}_{R_1}\subset \subset \mathsf{\Omega }\subset \subset {\mathsf{\Omega }}_{R_2}$, where $R_2>max\{R_1,2\}$. Let $\omega$ be defined in (5).

For the positive constants $c,\alpha$ to be determined, let

$$\overline{\psi }(x,t)=\frac{1}{2}\sum _{i=1}^N{m}_n{x}_n^2+c-t\mathrm{in}{\mathbb{R}}_{-}^{N+1},$$

and

$$\underline{\psi }(x,t)=\mathsf{\Psi }\left(\omega \right)={\int }_{R_2}^{\omega }\varrho \phi (\varrho ,\alpha )d\varrho +\underset{{\mathsf{\Omega }}_{R_2}}{inf}\mathsf{\Theta }\mathrm{in}{\mathbb{R}}_{-}^{N+1}.$$

(28)

Then, clearly,

$$\underline{\psi }(x,t)\le {\int }_{R_2}^{R_2}\varrho \phi (\varrho ,\alpha )d\varrho +\underset{{\mathsf{\Omega }}_{R_2}}{inf}\mathsf{\Theta }\le \mathsf{\Theta }(x,t)\mathrm{on}{\partial }_p\mathsf{\Omega }.$$

(29)

Let $R_3=R_2+1$. Since $\phi$ is increasing in $\alpha$, then we can select sufficiently large ${\alpha }_1,c_1$ to make the next three inequalities correct for $\alpha >{\alpha }_1, c>c_1$:

$$\underline{\psi }(x,t)={\int }_{R_2}^{R_3}\varrho \phi (\varrho ,\alpha )d\varrho +\underset{{\mathsf{\Omega }}_{R_2}}{inf}\mathsf{\Theta }\ge \mathsf{\Theta }(x,t)\mathrm{on}{\partial }_p{\mathsf{\Omega }}_{R_3},$$

(30)

$$\overline{\psi }=\frac{1}{2}\sum _{n=1}^N{m}_n{x}_n^2+c-t\ge \mathsf{\Theta }\mathrm{on}{\partial }_p{\mathsf{\Omega }}_{R_3},$$

(31)

$$\overline{\psi }=\frac{1}{2}\sum _{n=1}^N{m}_n{x}_n^2+c-t\ge \mathsf{\Theta }\ge \underline{\psi }\mathrm{on}{\partial }_p{\mathsf{\Omega }}_{R_1}.$$

(32)

According to Theorem 2, $\underline{\psi }$ is parabolically k-convex. In addition, according to (23), we can determine that

$$\begin{array}{c}\hfill -{\underline{\psi }}_t\frac{{\sigma }_k\left(\lambda \left(D^2\underline{\psi }\right)\right)}{{\sigma }_l\left(\lambda \left(D^2\underline{\psi }\right)\right)}\ge 1.\end{array}$$

So, $\underline{\psi }$ is a sub-solution of (3).

By performing further calculations, we obtain the following from Theorem 1-(iii):

$$\begin{array}{cc}\hfill \underline{\psi }=& \underset{{\mathsf{\Omega }}_{R_2}}{inf}\mathsf{\Theta }+{\int }_{R_2}^{\omega }\varrho \left[\phi (\varrho ,\alpha )-1\right]d\varrho +\frac{1}{2}{\omega }^2-\frac{1}{2}{R}_2^2\hfill \\ \hfill =& {\displaystyle \frac{1}{2}{\omega }^2-{\int }_{\omega }^{\infty }\varrho \left[\phi (\varrho ,\alpha )-1\right]d\varrho +{\int }_{R_2}^{\infty }\varrho \left[\phi (\varrho ,\alpha )-1\right]d\varrho -\frac{1}{2}{R}_2^2+\underset{{\mathsf{\Omega }}_{R_2}}{inf}\mathsf{\Theta }}\hfill \\ \hfill =& {\displaystyle -t+\frac{1}{2}\sum _{n=1}^N{m}_n{x}_n^2+\mu \left(\alpha \right)-{\int }_{\omega }^{\infty }\varrho \left[\phi (\varrho ,\alpha )-1\right]d\varrho }\hfill \\ \hfill =& {\displaystyle -t+\frac{1}{2}\sum _{n=1}^N{m}_n{x}_n^2+\mu \left(\alpha \right)+O\left({(-t+|x|}^2{)}^{1-\frac{1}{2}\frac{k-l}{\overline{{\eta }_k}-\underline{{\eta }_l}}}\right),\mathrm{as}-t+|x{|}^2\to +\infty ,}\hfill \end{array}$$

(33)

where

$$\mu \left(\alpha \right)={\int }_{R_2}^{\infty }\varrho \left[\phi (\varrho ,\alpha )-1\right]d\varrho -\frac{1}{2}{R}_2^2+\underset{{\mathsf{\Omega }}_{R_2}}{inf}\mathsf{\Theta }.$$

Obviously, $\mu \left(\alpha \right)$ is strictly increasing about $\alpha$ on $(0,+\infty )$, and we have

$$\underset{\alpha \to \infty }{lim}\mu \left(\alpha \right)=+\infty .$$

Let $\widehat{c}=max\{c_1,\mu \left({\alpha }_1\right)\}.$ Then for all $c>\widehat{c}$, we can deduce that $\alpha =\alpha \left(c\right)$ exists and satisfies $\mu \left(\alpha \right(c\left)\right)=c$. Consequently, when $-t+|x{|}^2\to +\infty$, we can obtain

$$\underline{\psi }(x,t)=-t+\frac{1}{2}\sum _{n=1}^N{m}_n{x}_n^2+c+O\left({{(-t+|x|}^2)}^{1-\frac{1}{2}\frac{k-l}{\overline{{\eta }_k}-\underline{{\eta }_l}}}\right).$$

(34)

Proposition 3. 

Let $N\ge 3$, $N\ge k>l\ge 0$, $\frac{k-l}{\overline{{\eta }_k}-\underline{{\eta }_l}}>2$, $M=diag\left(m_1,m_2,\cdots ,m_N\right)\in {\mathcal{M}}_{k,l}$ and $\underline{\psi }$ be defined as (28). Then, we can determine that $\underline{\psi }$ is a smooth parabolically k-convex subsolution of (3) in ${\mathbb{R}}_{-}^{N+1}\setminus \left\{0\right\}$, and (34) is true.

Define, for all $c>\widehat{c}$,

$$\vartheta =\left\{\begin{array}{c}max\left\{\mathsf{\Theta }(x,t),\underline{\psi }(x,t)\right\},(x,t)\in {\mathsf{\Omega }}_{R_3}\setminus {\mathsf{\Omega }}_{R_1},\hfill \\ \underline{\psi }(x,t),(x,t)\in {\mathbb{R}}_{-}^{N+1}\setminus {\mathsf{\Omega }}_{R_3}.\hfill \end{array}\right.$$

3.2. Proof of $\overline{\psi }\ge \vartheta$

From Lemma 2 and (32)–(34),

$$\overline{\psi }\ge \underline{\psi }\mathrm{in}{\mathbb{R}}_{-}^{N+1}\setminus {\mathsf{\Omega }}_{R_1}.$$

(35)

Through (30), we know that $\vartheta \in C^0\left({\mathbb{R}}_{-}^{N+1}\setminus \mathsf{\Omega }\right)$. According to (27) and Lemma 1, we determine that $\vartheta$ satisfies in the sense of viscosity.

$$-{\vartheta }_t\frac{{\sigma }_k\left(\lambda \left(D^2\vartheta \right)\right)}{{\sigma }_l\left(\lambda \left(D^2\vartheta \right)\right)}\ge 1,(x,t)\in {\mathbb{R}}_{-}^{N+1}\setminus \overline{\mathsf{\Omega }}.$$

(36)

According to (29) and (26),

$$\vartheta =\mathsf{\Theta }=\varsigma \mathrm{on}{\partial }_p\mathsf{\Omega }.$$

(37)

Thus, $\vartheta$ satisfies (3) and (4) in the sense of the viscosity sub-solution. According to (34), if $-t+|x{|}^2\to +\infty$, we can obtain

$$\vartheta (x,t)=c-t+\frac{1}{2}\sum _{n=1}^N{m}_n{x}_n^2+O\left({(-t+|x|}^2{)}^{1-\frac{1}{2}\frac{k-l}{\overline{{\eta }_k}-\underline{{\eta }_l}}}\right).$$

(38)

In addition, according to (27), (31) and (32), and Lemma 2,

$$\overline{\psi }\ge \mathsf{\Theta }\mathrm{in}{\mathsf{\Omega }}_{R_3}\setminus {\mathsf{\Omega }}_{R_1}.$$

And then, combining this with (35), we deduce that

$$\overline{\psi }\ge \vartheta \mathrm{in}{\mathbb{R}}_{-}^{N+1}\setminus \mathsf{\Omega }.$$

(39)

3.3. Proving Theorem 3

Proof of Theorem 3. 

This section concerns the proof of uniqueness. Suppose that u and v are both viscosity solutions for (3), (4) and (24). Then, $\forall \epsilon >0$, and we can find $R_4>0$ to make $\mathsf{\Omega }\subset {\mathsf{\Omega }}_{R_4}$, and

$$v\le \epsilon +u,(x,t)\in {\mathbb{R}}_{-}^{N+1}\setminus {\mathsf{\Omega }}_{R_4}.$$

According to Lemma 2, in ${\mathsf{\Omega }}_{R_4}\setminus \mathsf{\Omega }$, $v\le \epsilon +u$ holds. Therefore, in ${\mathbb{R}}_{-}^{N+1}\setminus \mathsf{\Omega }$, we have $v\le \epsilon +u$. Let $\epsilon$ tend to 0; then, $v\le u$ on ${\mathbb{R}}_{-}^{N+1}\setminus \mathsf{\Omega }$ holds. Likewise, $v\ge u$ on ${\mathbb{R}}_{-}^{N+1}\setminus \mathsf{\Omega }$ holds. Therefore $v=u$ on ${\mathbb{R}}_{-}^{N+1}\setminus \mathsf{\Omega }$.

Now let us prove existence. We have already constructed the subsolution $\vartheta$ and the supersolution $\overline{\psi }$ of (3) in the sense of viscosity such that $\overline{\psi }\ge \vartheta$.

Let the set $\mathcal{S}$ be composed of the function $\mathrm{Y}$; these are subsolutions of (3), (4) in the sense of viscosity and satisfy

$$\overline{\psi }\ge \mathrm{Y}\mathrm{in}{\mathbb{R}}_{-}^{N+1}\setminus \mathsf{\Omega }.$$

Because $\vartheta \in \mathcal{S}$, $\mathcal{S}\ne \varnothing$. Define in ${\mathbb{R}}_{-}^{N+1}\setminus \mathsf{\Omega }$:

$$u_c=sup\left\{\mathrm{Y}|\mathrm{Y}\in \mathcal{S}\right\}.$$

Then, we have

$$\overline{\psi }\ge u_c\mathrm{in}{\mathbb{R}}_{-}^{N+1}\setminus \mathsf{\Omega }.$$

(40)

Because $\vartheta \in \mathcal{S}$, according to (38), when $-t+|x{|}^2\to +\infty$,

$$u_c=c-t+\frac{1}{2}{\sum }_{n=1}^N{m}_n{x}_n^2+O\left({(-t+|x|}^2{)}^{1-\frac{1}{2}\frac{k-l}{\overline{{\eta }_k}-\underline{{\eta }_l}}}\right).$$

(41)

For any $(\kappa ,\iota )\in {\partial }_p\mathsf{\Omega }$, on the one hand, according to (37), we have

$$\varsigma (\kappa ,\iota )=\underset{(x,t)\to (\kappa ,\iota )}{lim}\vartheta (x,t)\le \underset{(x,t)\to (\kappa ,\iota )}{liminf}{u}_c(x,t).$$

Additionally, we can prove that

$$\varsigma (\kappa ,\iota )\ge \underset{(x,t)\to (\kappa ,\iota )}{limsup}{u}_c(x,t).$$

Indeed, for each $\mathrm{Y}\in \mathcal{S}$, according to Proposition 1, in the sense of viscosity, we have

$$\left\{\begin{array}{c}-{\mathrm{Y}}_t+\Delta \mathrm{Y}\ge 0,(x,t)\in {\mathsf{\Omega }}_{R_2}\setminus \overline{\mathsf{\Omega }},\hfill \\ \mathrm{Y}\le \varsigma ,(x,t)\in {\partial }_p\mathsf{\Omega },\hfill \\ \mathrm{Y}\le {\displaystyle \underset{{\partial }_p{\mathsf{\Omega }}_{R_2}}{sup}}\overline{\psi }=:\mathsf{\Xi },(x,t)\in {\partial }_p{\mathsf{\Omega }}_{R_2}.\hfill \end{array}\right.$$

Choose $\mathsf{\Lambda }\in C^{2,1}\left({\mathsf{\Omega }}_{R_2}\setminus \overline{\mathsf{\Omega }}\right)\cap C^0\left(\overline{{\mathsf{\Omega }}_{R_2}}\setminus \mathsf{\Omega }\right)$ to satisfy

$$\left\{\begin{array}{c}-{\mathsf{\Lambda }}_t+\Delta \mathsf{\Lambda }=0,(x,t)\in {\mathsf{\Omega }}_{R_2}\setminus \overline{\mathsf{\Omega }},\hfill \\ \mathsf{\Lambda }=\varsigma ,(x,t)\in {\partial }_p\mathsf{\Omega },\hfill \\ \mathsf{\Lambda }=\mathsf{\Xi },(x,t)\in {\partial }_p{\mathsf{\Omega }}_{R_2}.\hfill \end{array}\right.$$

From the comparison principle, it can be directly proven that $\mathrm{Y}\le \mathsf{\Lambda }$, $(x,t)\in \overline{{\mathsf{\Omega }}_{R_2}}\setminus \mathsf{\Omega }$. So, we have $u_c\le \mathsf{\Lambda }$, $(x,t)\in \overline{{\mathsf{\Omega }}_{R_2}}\setminus \mathsf{\Omega }$, and accordingly,

$$\underset{(x,t)\to (\kappa ,\iota )}{limsup}{u}_c(x,t)\le \underset{(x,t)\to (\kappa ,\iota )}{lim}\mathsf{\Lambda }(x,t)=\varsigma (\kappa ,\iota ).$$

Finally, to demonstrate that $u_c$ satisfies (3) in the sense of viscosity, we can follow the techniques in [5].

Theorem 3 is proven. □

4. Conclusions and Future Directions

This paper mainly studies the generalized asymptotic behavior of generalized parabolically symmetric solutions for the parabolic Hessian quotient equation ${\displaystyle -u_t\frac{{\sigma }_k\left(\lambda \left(D^{2}u\right)\right)}{{\sigma }_l\left(\lambda \left(D^{2}u\right)\right)}=1}$ in the exterior domain. Employing Perron’s method, this paper proves the existence and uniqueness of generalized parabolically symmetric solutions with generalized asymptotic behavior. The proof of the main result contains three sections: Section 3.1, Section 3.2 and Section 3.3. Lemma 4 plays an important role in proving the main result.

It is interesting to study the generalized asymptotic behavior of the parabolic Hessian quotient equation ${\displaystyle -u_t\frac{{\sigma }_k\left(\lambda \left(D^{2}u\right)\right)}{{\sigma }_l\left(\lambda \left(D^{2}u\right)\right)}=f}$ with f being a perturbation of 1.