Parabolic Hessian Equations Outside a Cylinder

Abstract

In this article, we mainly review the parabolic Hessian equation on the exterior region. The existence and uniqueness of solutions with asymptotic properties to the exterior problem of the parabolic Hessian equation were obtained by using the Perron method.

1. Introduction

Let $\mathsf{\Omega }$ be a smooth bounded domain within ${\mathbb{R}}^n$, $\mathsf{\Psi }$ be a constant. Let ${\mathbb{R}}_{\mathsf{\Psi }}^{n+1}={\mathbb{R}}^n\times (0,\mathsf{\Psi }]$, the cylinder domain $U=\mathsf{\Omega }\times (0,\mathsf{\Psi }]$, $SU=\partial \mathsf{\Omega }\times (0,\mathsf{\Psi })$ represents the lateral boundary of U, $\overline{SU}=\partial \mathsf{\Omega }\times [0,\mathsf{\Psi }]$, $BU=\overline{\mathsf{\Omega }}\times \left\{0\right\}$ represents the bottom boundary of U, ${\partial }_{p}U=SU\cup BU$ represents the parabolic boundary of U. Let $\theta =\theta (x,\psi )$, for $0\le \psi \le \mathsf{\Psi }$, $\theta (.,\psi )\in C^2\left(\mathsf{\Omega }\right)$, $D^2\theta$ represents the Hessian matrix of the function $\theta$ about the variable x, $\lambda =\lambda \left(D^2\theta \right)=({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n)$ represents the characteristic root of the matrix $D^2\theta$.

$${\sigma }_k\left(\lambda \right)=\sum _{1\le j_1<j_2<\dots <j_k\le n}{\lambda }_{j_1}{\lambda }_{j_2}\dots {\lambda }_{j_k}.$$

${\sigma }_k\left(\lambda \right)$ represents the k-th elementary symmetric function of $\lambda$. Define $S_k\left(D^2\theta \right)$$={\sigma }_k\left(\lambda \left(D^2\theta \right)\right)$.

At present, we discuss the exterior problem of the parabolic Hessian equation

$$-{\theta }_{\psi }+\mu \left(S_k\left(D^2\theta \right)\right)=f,(x,\psi )\in {\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash \overline{U},$$

(1)

$$\theta (x,\psi )=\phi (x,\psi ),(x,\psi )\in SU,$$

(2)

$$\theta (x,0)={\theta }_0\left(x\right),x\in {\mathbb{R}}^n\backslash \mathsf{\Omega },$$

(3)

where the function ${\theta }_{\psi }=\partial \theta /\partial \psi$, $\mu \in C^2(0,+\infty )$, f is positive, $\phi$ and ${\theta }_0$ satisfy the compatibility condition—that is, for $\forall x\in \partial \mathsf{\Omega }$, then we have ${\theta }_0\left(x\right)=\phi (x,0)$.

The parabolic Hessian Equation (1) is fully nonlinear. In [1], Equation (1) plays a very important role in the existence of solutions of elliptic Hessian equations by the estimation of solutions of parabolic Hessian equations. For research on boundary value problems in the parabolic Hessian Equation (1) on the interior domain, please refer to [2,3,4], and other references can also verify it. For the study of the exterior problem of the elliptic Hessian equation, one can refer to [5,6,7], etc. Li-Dai [6] researched the existence of the solution to the exterior problem

$$\begin{array}{cc}\hfill S_k\left(D^2\theta \right)& =f,x\in {\mathbb{R}}^n\backslash \overline{\mathsf{\Omega }},\hfill \\ \hfill \theta & =\phi ,x\in \partial \mathsf{\Omega }\hfill \end{array}$$

for the elliptic Hessian equation with asymptotic property

$$\begin{array}{c}\hfill \underset{\left|x\right|\to \infty }{lim\; sup}\left({\left|x\right|}^{\alpha +min\{n,\beta \}-\frac{\alpha }{k}-2}|\theta \left(x\right)-g_0\left(\right|x\left|\right)-b_0·x-m|\right)<\infty \end{array}$$

at infinity and uniqueness, where $\mathsf{\Omega }$ is a bounded convex domain, $f={O\left(\right|x|}^{\alpha }{)+O(\left|x\right|}^{-\beta }),$$\left|x\right|\to \infty$, $g_0\left(\left|x\right|\right)$ is the radially symmetric solution of $S_k\left(D^2\theta \right)=f_0,g_0\left(0\right)=0,g_0^{\prime }\left(0\right)=0$, and $b_0,m$ are constants, $\left|x\right|={(x_1^2+x_2^2+\dots +x_n^2)}^{\frac{1}{2}}$. The recent study related to Hessian equations can be found in [8,9,10,11].

For the study of the parabolic equations on the exterior problem, please refer to [12,13,14,15,16], where Dai [14] studied the existence and uniqueness of the solution of the exterior problem

$$\begin{array}{cc}\hfill -{\theta }_{\psi }+\mu \left(S_k\left(D^2\theta \right)\right)=1,& (x,\psi )\in {\mathbb{R}}_{{\mathsf{\Psi }}_{1.2}}^{n+1}\backslash \overline{U_1},\hfill \\ \hfill \theta (x,\psi )=\phi (x,\psi ),& (x,\psi )\in S{U}_1,\hfill \\ \hfill \theta \left(x,-{\mathsf{\Psi }}_1\right)={\theta }_1\left(x\right),& x\in {\mathbb{R}}^n\backslash \mathsf{\Omega }\hfill \end{array}$$

of the parabolic Hessian equation with the asymptotic property at infinity,

$$\underset{\left|x\right|\to \infty }{lim\; sup}\left({\left|x\right|}^{\rho (n-2)}\left|\theta (x,\psi )-\left(-\psi +\frac{1}{2}{x}^{T}Ax+b·x+c\right)\right|\right)<\infty ,$$

where $U_1=\mathsf{\Omega }\times (-{\mathsf{\Psi }}_1,{\mathsf{\Psi }}_2],{\mathsf{\Psi }}_1,{\mathsf{\Psi }}_2$ are positive constants, $\rho$ depends only on $n,k,A\in \mathcal{A}$, $x^T$ represents the transpose of x, and

$$\mathcal{A}=\{A:A\mathrm{is}\mathrm{a}\mathrm{real}n\times n\mathrm{symmetric}\mathrm{positive}\mathrm{definite}\mathrm{matrix}\mathrm{and}\mathrm{satisfies}{S}_k\left(A\right)=1\}.$$

Dai-Cheng [17] studied the parabolic Monge–Ampère equations

$$-{\theta }_{\psi }det{D}^2\theta =g$$

outside of the bowl-shaped domain with g being the perturbation of $g_0\left(\right|x\left|\right)$ at infinity. For studies on the parabolic equations, refer to [18,19,20,21,22].

Assume that

$${\mathsf{\Gamma }}_k=\left\{\lambda \in {\mathbb{R}}^n\mid {\sigma }_j\left(\lambda \right)>0,j=1,2,\dots ,k\right\}.$$

The function $\theta$ is said to be k-admissible for $x\in \mathsf{\Omega }$ if $\lambda \left(D^2\theta (x,\psi )\right)\in {\mathsf{\Gamma }}_k$. The function $\theta \in C^{2k,k}({\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash \overline{U})$ indicates the continuity of the derivative $D_x^i{D}_{\psi }^j\theta (i+2j\le 2k)$ of $\theta$ in ${\mathbb{R}}_{\psi }^{n+1}\backslash \overline{U}$. If the function $\theta \in C_{x,\psi }^{2,1}\left(U\right)$ is convex-correlated with x and non-increased correlated with $\psi$, then we can say that $\theta$ is a parabolically convex function in U. Take into account the initial boundary value problem

$$-{\theta }_{\psi }+\mu \left(S_k\left(D^2\theta \right)\right)=f(x,\psi ),(x,\psi )\in U,$$

(4)

$$\theta =\phi (x,\psi ),(x,\psi )\in {\partial }_{p}U.$$

(5)

Definition 1.

A function $\theta \in C^0\left(\overline{U}\right)$ is known as a viscosity subsolution (respectively, supersolution) to (4), if for $\forall \varphi \in C^{2,1}\left(U\right)$, $(\overline{x},\overline{\psi })\in U$ satisfies

$$\varphi (\overline{x},\overline{\psi })=\theta (\overline{x},\overline{\psi }),\varphi (x,\psi )\ge (\le )\theta (x,\psi ),(x,\psi )\in U\cap \{\psi \le \overline{\psi }\},$$

we can reach

$$-{\varphi }_{\psi }(\overline{x},\overline{\psi })+\mu \left(S_k\left(D^2\varphi (\overline{x},\overline{\psi })\right)\right)\ge (\le )f(\overline{x},\overline{\psi }).$$

For the supersolution, ϕ is required to be k-admissible.

If the function $\theta \in C^0\left(\overline{U}\right)$ is both a viscosity supersolution and viscosity subsolution at the same time, then it is said to be the viscosity solution of (4).

Definition 2.

The function $\theta \in C^0\left(\overline{U}\right)$ is called the viscosity subsolution (supersolution) of the initial boundary value problem (4) and (5), if θ is the viscosity subsolution (supersolution) of (4), and there is $\theta \le (\ge )\phi (x,\psi )$ on the boundary ${\partial }_{p}U$.

The function $\theta \in C^0\left(\overline{U}\right)$ is said to be a viscosity solution of (4) and (5), if θ is a viscosity solution of (4) and satisfies $\theta (x,\psi )=\phi (x,\psi )$ on the boundary ${\partial }_{p}U$.

Suppose $\mathsf{\Omega }$ is a bounded strictly convex region in ${\mathbb{R}}^n$ and the following conditions hold:

$\left(H_1\right)f=f\left(x\right)\in C^0\left({\mathbb{R}}^n\right)$ satisfies

$$\underset{x\in {\mathbb{R}}^n}{inf}f\left(x\right)=f_1>0,$$

(6)

and for constants $\tau >0$ ($\tau$ also satisfies the following (12)) and $\beta >2$,

$${\mu }^{-1}(f\left(x\right)-\tau )={\mu }^{-1}(f_0{\left(\right|x\left|\right)-\tau )+O(\left|x\right|}^{-\beta }),|x|\to +\infty ,$$

(7)

where $f_0\in C^0\left([0,+\infty )\right)$ is positive and radially symmetric in x,

$${\mu }^{-1}\left(f_0\left(\right|x\left|\right)-\tau \right)=O\left({\left|x\right|}^{\alpha }\right),\left|x\right|\to +\infty ,$$

(8)

and for the constant $\alpha$,

$$\frac{k(2-min\{n,\beta \left\}\right)}{k-1}<\alpha <\infty ,n+\alpha >0,\alpha +\beta >0.$$

(9)

We remark that $F\left(x\right)=O\left(G\right(x\left)\right),\left|x\right|\to +\infty$ means that there exists some constant L, such that $\left|F\right(x)/G(x\left)\right|\le L,$ as $\left|x\right|\to +\infty$.

$\left(H_2\right)$ The function $\mu \in C^2(0,+\infty )$ satisfies

$${\mu }^{\prime }>0,{\mu }^{″}<0,$$

(10)

and for $y>0,$

$$\mu \left(y^k\right)<f_1+y.$$

(11)

A function that satisfies (10) and (11) is $\mu \left(y\right)=f_1+\frac{1}{k}lny$. An example of f satisfying (6)–(8) may be chosen as $f\left(x\right)=\tau +f_1+\frac{1}{k}{ln\left(\right|x|}^{\alpha }+{\left|x\right|}^{-\beta })$ and $f_0$ satisfying (8) may be chosen as $f_0=\tau +f_1+\frac{\alpha }{k}ln\left|x\right|.$

$\left(H_3\right)\phi \in C^{2,1}\left(\overline{U}\right)$ is k-admissible, and

$$\underset{\begin{array}{c}x\in \overline{\mathsf{\Omega }}\\ 0\le \psi \le \mathsf{\Psi }\end{array}}{sup}{\phi }_{\psi }(x,\psi )\le -\tau .$$

(12)

An example of $\phi$ may be $\phi (x,\psi )=-\psi +\frac{1}{2}{\left|x\right|}^2.$

$\left(H_4\right)$ The function ${\theta }_0\in C^0\left(\overline{{\mathbb{R}}^n\backslash \mathsf{\Omega }}\right)$ (see [6]) satisfies

$$S_k\left(D^2{\theta }_0\right)={\mu }^{-1}(f\left(x\right)-\tau ),x\in {\mathbb{R}}^n\backslash \overline{\mathsf{\Omega }},$$

(13)

$${\theta }_0=\phi (x,0),x\in \partial \mathsf{\Omega },$$

(14)

and for some constant vector $d\in {\mathbb{R}}^n$ and a sufficiently large constant c,

$$\left\{\begin{array}{c}\underset{\left|x\right|\to \infty }{lim\; sup}\left({\left|x\right|}^{min\{n,\beta \}+\alpha -\frac{\alpha }{k}-2}\left|{\theta }_0\left(x\right)-\left(g_0\left(\right|x\left|\right)+d·x+c\right)\right|\right)<\infty ,\beta \ne n,\hfill \\ \underset{\left|x\right|\to \infty }{lim\; sup}\left({\left|x\right|}^{n+\alpha -\frac{\alpha }{k}-2}{(ln|x\left|\right)}^{-1}\left|{\theta }_0\left(x\right)-\left(g_0\left(\right|x\left|\right)+d·x+c\right)\right|\right)<\infty ,\beta =n,\hfill \end{array}\right.$$

(15)

where

$$g_0\left(\right|x\left|\right)={\int }_0^{z}b{s}^{1-\frac{n}{k}}{\left[{\int }_0^{s}n{y}^{n-1}{\mu }^{-1}\left(f_0\left(y\right)-\tau \right)dy\right]}^{\frac{1}{k}}ds,x\in {\mathbb{R}}^n\backslash \left\{0\right\}$$

(16)

satisfies $S_k\left(D^2{g}_0\right)={{\mu }^{-1}(f}_0-\tau )$, $g_0\left(0\right)=0,g_0^{\prime }\left(0\right)=0$ with $b={(1/c_n^k)}^{\frac{1}{k}}$.

This paper mainly uses the Perron method to obtain the following results:

Theorem 1.

Let $n\ge 3$, the conditions $\left(H_1\right)-\left(H_4\right)$ hold, then for the $d\in {\mathbb{R}}^n$ and sufficiently large constant c in (15), the problem (1)–(3) possesses a unique viscosity solution $\theta \in C^0\left(\overline{{\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash U}\right)$ satisfying

$$\left\{\begin{array}{c}{lim\; sup}_{\left|x\right|\to \infty }\left({\left|x\right|}^{min\{n,\beta \}+\alpha -\frac{\alpha }{k}-2}\left|\theta (x,\psi )-\left(-\tau \psi +g_0\left(\right|x\left|\right)+d·x+c\right)\right|\right)<\infty ,\beta \ne n,\hfill \\ {lim\; sup}_{\left|x\right|\to \infty }\left({\left|x\right|}^{n+\alpha -\frac{\alpha }{k}-2}{(ln|x\left|\right)}^{-1}\left|\theta (x,\psi )-\left(-\tau \psi +g_0\left(\right|x\left|\right)+d·x+c\right)\right|\right)<\infty ,\beta =n.\hfill \end{array}\right.$$

(17)

This article is organized as follows. In the next section, we provide some useful lemmas and a radially symmetric solution. In the last section, we will prove Theorem 1 by the Perron method.

2. Several Lemmas

Lemma 1.

Set $\mathsf{\Phi }\left(x,\psi \right)\in C^{2,1}\left(\overline{U}\right)$, then there is a constant C that depends only on $n,\mathsf{\Phi },U$, so that for $\forall \xi \in \partial \mathsf{\Omega },0\le \psi \le \mathsf{\Psi }$, there are $\overline{x}(\xi ,\psi )\in {\mathbb{R}}^n$, satisfying $C\ge |\overline{x}(\xi ,\psi )|$, and

$$\mathsf{\Phi }(x,\psi )>w_{\xi }(x,\psi ),(x,\psi )\in \overline{U}\backslash \left\{(\xi ,\psi )\right\},$$

(18)

where $w_{\xi }(x,\psi )=\mathsf{\Phi }(\xi ,\psi )+\frac{\overline{c}}{2}\left[|x-\overline{x}{(\xi ,\psi )|}^2-{|\xi -\overline{x}(\xi ,\psi )|}^2\right],(x,\psi )\in {\mathbb{R}}^n\times [0,\mathsf{\Psi }],$ $\overline{c}$ is any bounded constant.

Proof. 

This lemma has been proved in reference [12]. □

Set $V\subset {\mathbb{R}}^n\backslash \overline{\mathsf{\Omega }}$ is a convex region, $D=V\times (0,\mathsf{\Psi }]$, then $D\subset {\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash \overline{U}$. Let $\overline{S}D=\partial V\times [0,\mathsf{\Psi }]$.

Lemma 2.

Let $g\in C^0\left(\overline{D}\right)$. Assume that the function $v\in C^0\left(\overline{D}\right)$, $\theta \in C^0({\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash \overline{U})$ satisfies, respectively,

$$-v_{\psi }+\mu \left(S_k\left(D^{2}v\right)\right)\ge g(x,\psi ),(x,\psi )\in D,$$

$$-{\theta }_{\psi }+\mu \left(S_k\left(D^2\theta \right)\right)\ge g(x,\psi ),(x,\psi )\in {\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash \overline{U},$$

in the viscosity sense, and

$$\theta \le v,(x,\psi )\in D,\theta =v,(x,\psi )\in \overline{S}D.$$

(19)

Define

$$\begin{array}{cc}\hfill w(x,\psi )& =\left\{\begin{array}{cc}v(x,\psi ),\hfill & (x,\psi )\in D,\hfill \\ \theta (x,\psi ),\hfill & (x,\psi )\in ({\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash \overline{U})\backslash D.\hfill \end{array}\right.\hfill \end{array}$$

Then $w\in C^0({\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash \overline{U})$ satisfies in the viscosity sense

$$-w_{\psi }+\mu \left(S_k\left(D^{2}w\right)\right)\ge g(x,\psi ),(x,\psi )\in {\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash \overline{U}.$$

Proof. 

Set $(\overline{x},\overline{\psi })\in {\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash \overline{U},$ and $\varphi \in C^{2,1}({\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash \overline{U})$, which satisfy

$$\varphi (\overline{x},\overline{\psi })=w(\overline{x},\overline{\psi }),w(x,\psi )\le \varphi (x,\psi ),(x,\psi )\in ({\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash \overline{U})\cap \{\psi \le \overline{\psi }\}.$$

If $(\overline{x},\overline{\psi })\in D$, then

$$\varphi (\overline{x},\overline{\psi })=v(\overline{x},\overline{\psi }),\varphi (x,\psi )\ge v(x,\psi ),(x,\psi )\in D.$$

So

$$-{\varphi }_{\psi }(\overline{x},\overline{\psi })+\mu \left(S_k\left(D^2\varphi (\overline{x},\overline{\psi })\right)\right)\ge g(\overline{x},\overline{\psi }).$$

If $(\overline{x},\overline{\psi })\in ({\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash \overline{U})\backslash D$, then

$$\varphi (\overline{x},\overline{\psi })=\theta (\overline{x},\overline{\psi }),\varphi (x,\psi )\ge u(x,\psi ),(x,\psi )\in ({\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash \overline{U})\backslash D.$$

Due to (19), we have

$$\varphi (\overline{x},\overline{\psi })=\theta (\overline{x},\overline{\psi }),\theta (x,\psi )\le \varphi (x,\psi ),(x,\psi )\in {\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash \overline{U}.$$

Since $\theta$ is a viscosity subsolution, thus,

$$-{\varphi }_{\psi }(\overline{x},\overline{\psi })+\mu \left(S_k\left(D^2\varphi (\overline{x},\overline{\psi })\right)\right)\ge g(\overline{x},\overline{\psi }).$$

Thus, w is a viscosity subsolution. The lemma proof is completed. □

Lemma 3.

If $\theta \in C^0\left(U\right)$ satisfies

$$-{\theta }_{\psi }+\mu \left(S_k\left(D^2\theta \right)\right)\ge f(x,\psi )$$

(20)

in the viscosity sense. Then,

$$-{\theta }_{\psi }+\mathsf{\Delta }\theta >0$$

(21)

in the viscosity sense.

Proof. 

For any function $h\in C^{2,1}\left(U\right)$, any point $(\overline{x},\overline{\psi })\in U$ satisfies

$$h(\overline{x},\overline{\psi })=\theta (\overline{x},\overline{\psi }),h(x,\psi )\ge \theta (x,\psi ),(x,\psi )\in U\cap \{\psi \le \overline{\psi }\}.$$

Since $\theta$ is a viscosity subsolution, then

$$-h_{\psi }(\overline{x},\overline{\psi })+\mu \left(S_k\left(D^{2}h(\overline{x},\overline{\psi })\right)\right)\ge f(\overline{x},\overline{\psi }).$$

Thus, by (11),

$$\begin{array}{cc}\hfill S_k\left(D^{2}h(\overline{x},\overline{\psi })\right)& \ge {\mu }^{-1}(f_1+h_{\psi }(\overline{x},\overline{\psi }))\hfill \\ \hfill & \ge h_{\psi }^k(\overline{x},\overline{\psi }).\hfill \end{array}$$

Then according to the Newton–Maclaurin inequality, at $(\overline{x},\overline{\psi })$, there are

$$\begin{array}{cc}\hfill \mathsf{\Delta }h(\overline{x},\overline{\psi })& \ge n{\left(\frac{S_k\left(D^{2}h(\overline{x},\overline{\psi })\right)}{C_n^k}\right)}^{\frac{1}{k}}\hfill \\ \hfill & \ge {\left(\frac{n^k}{C_n^k}\right)}^{\frac{1}{k}}{h}_{\psi }(\overline{x},\overline{\psi })\hfill \\ \hfill & >h_{\psi }(\overline{x},\overline{\psi }).\hfill \end{array}$$

Thus,

$$-h_{\psi }(\overline{x},\overline{\psi })+\mathsf{\Delta }h(\overline{x},\overline{\psi })>0,$$

and then (21) is established. □

The proof of Lemma 4 below can be directly obtained from Proposition 3 in Appendix A of Reference [23].

Lemma 4.

Suppose $\underline{\theta },\overline{\theta }\in C^0\left(\overline{D}\right)$ are the viscosity subsolution and the viscosity supersolution of $-{\theta }_{\psi }+\mu \left(S_k\left(D^2\theta \right)\right)=f$, respectively, and ${\left.\underline{\theta }\right|}_{{\partial }_{p}D}={\left.\overline{\theta }\right|}_{{\partial }_{p}D}=\varphi \in C^0\left({\partial }_{p}D\right)$, then the problem

$$\left\{\begin{array}{cc}-{\theta }_{\psi }+\mu \left(S_k\left(D^2\theta \right)\right)=f(x,\psi ),& (x,\psi )\in D,\hfill \\ \theta =\varphi (x,\psi ),& (x,\psi )\in {\partial }_{p}D\hfill \end{array}\right.$$

possesses a unique viscosity solution $\theta \in C^0\left(\overline{D}\right)$.

Lemma 5.

Set $\underline{\theta }\in C^0\left(\overline{D}\right)$, satisfying

$$-{\underline{\theta }}_{\psi }+\mu \left(S_k\left(D^2\underline{\theta }\right)\right)\ge f(x,\psi ),(x,\psi )\in D$$

in the viscosity sense, then the problem

$$\left\{\begin{array}{cc}-{\theta }_{\psi }+\mu \left(S_k\left(D^2\underline{\theta }\right)\right)=f(x,\psi ),& (x,\psi )\in D,\hfill \\ \theta =\underline{\theta }(x,\psi ),& (x,\psi )\in {\partial }_{p}D\hfill \end{array}\right.$$

(22)

possesses a unique viscosity solution $\theta \in C^0\left(\overline{D}\right)$.

Proof. 

Apparently, $\underline{\theta }$ is a viscosity subsolution of (22). According to Lemma 4, we just have to prove that (22) has a viscosity supersolution $\overline{\theta }$, which satisfies $\overline{\theta }=\underline{\theta }$ on ${\partial }_{p}D$.

Suppose $v\in C^{2,1}\left(D\right)\cap C^0\left(\overline{D}\right)$ satisfy

$$\left\{\begin{array}{cc}-v_{\psi }+\mathsf{\Delta }v=0,& (x,\psi )\in D,\hfill \\ v=\underline{\theta },& (x,\psi )\in {\partial }_{p}D.\hfill \end{array}\right.$$

Then we claim that v is a viscosity supersolution of (22).

In fact, if we assume that v is not a viscosity solution to (22), then there exists some point $(\overline{x},\overline{\psi })\in D$ and x—convex function $\varphi \in C^{2,1}\left(D\right)$, such that there are

$$v(\overline{x},\overline{\psi })=\varphi (\overline{x},\overline{\psi }),\varphi (x,\psi )\le v(x,\psi ),(x,\psi )\in D\cap \{\psi \le \overline{\psi }\},$$

(23)

but

$$-{\varphi }_{\psi }(\overline{x},\overline{\psi })+\mu (S_k\left(D^2\varphi (\overline{x},\overline{\psi })\right)>f(\overline{x},\overline{\psi }).$$

Thus, in light of Lemma 3, we know

$$-{\varphi }_{\psi }(\overline{x},\overline{\psi })+\mathsf{\Delta }\varphi (\overline{x},\overline{\psi })>0.$$

However, by (23), we have

$$v_{\psi }(\overline{x},\overline{\psi })={\varphi }_{\psi }(\overline{x},\overline{\psi }),S_k\left(D^{2}v(\overline{x},\overline{\psi })\right)\ge S_k\left(D^2\varphi (\overline{x},\overline{\psi })\right),k=1,\dots ,n.$$

Thus, we have

$$-{\varphi }_{\psi }(\overline{x},\overline{\psi })+\mathsf{\Delta }\varphi (\overline{x},\overline{\psi })\le -v_{\psi }(\overline{x},\overline{\psi })+\mathsf{\Delta }v(\overline{x},\overline{\psi })=0,$$

which is a contradiction.

Lemma 5 is proved. □

The following is the process of finding the radial symmetric solution of $S_k\left(D^2{g}_0\right)={{\mu }^{-1}(f}_0-\tau )$.

Suppose $g_0\left(x\right)=g_0\left(p\right)=g_0\left(\right|x\left|\right)$ is radial symmetric, where $p=\left|x\right|={(x_1^2+x_2^2+\dots +x_n^2)}^{\frac{1}{2}}$, then

$$\frac{\partial p}{\partial x_i}=\frac{x_i}{p},\frac{\partial g_0}{\partial x_i}=\frac{\partial g_0}{\partial p}·\frac{x_i}{p}.$$

After calculating,

$$D_{ij}{g}_0\left(x\right)=\left(p{g}_0^{″}\left(p\right)-g_0^{\prime }\left(p\right)\right)\frac{x_i{x}_j}{p^3}+g_0^{\prime }\left(p\right)\frac{{\gamma }_{ij}}{p},i,j=1,\dots ,n,$$

among them

$${\gamma }_{ij}=\left\{\begin{array}{cc}1,\hfill & i=j,\hfill \\ 0,\hfill & i\ne j.\hfill \end{array}\right.$$

Let $x={(p,0,\dots ,0)}^T$, then

$$\begin{array}{cc}\hfill D^2{g}_0& =\left(\begin{array}{ccccc}{g}_0^{″}\left(p\right)& 0& \dots & 0& 0\\ 0& \frac{g_0^{\prime }\left(p\right)}{p}& \dots & 0& 0\\ \dots & \dots & \dots & \dots & \dots \\ 0& 0& \dots & \frac{g_0^{\prime }\left(p\right)}{p}& 0\\ 0& 0& \dots & 0& \frac{g_0^{\prime }\left(p\right)}{p}\end{array}\right),\hfill \end{array}$$

so we have,

$$S_k\left(\lambda \left(D^2{g}_0\right)\right)=C_{n-1}^{k-1}{g}_0^{″}\left(p\right)g_0^{\prime }{\left(p\right)}^{k-1}{p}^{1-k}+C_{n-1}^k{g}_0^{\prime }{\left(p\right)}^k{p}^{-k}={\mu }^{-1}\left(f_0-\tau \right).$$

(24)

According to (24), we can know

$${\left(p^{n-k}{g}_0^{\prime }{\left(p\right)}^k\right)}^{\prime }=\frac{n{p}^{n-1}}{C_n^k}{\mu }^{-1}\left(f_0-\tau \right).$$

By the integration from 1 to p,

$$p^{n-k}{g}_0^{\prime }{\left(p\right)}^k={\int }_1^p\frac{n{y}^{n-1}}{C_n^k}{\mu }^{-1}\left(f_0-\tau \right)dy+g_0^{\prime }{\left(1\right)}^k.$$

Thus,

$$g_0^{\prime }\left(p\right)={\left(\frac{1}{p^{n-k}}\left({\int }_1^p\frac{n{y}^{n-1}}{C_n^k}{\mu }^{-1}\left(f_0-\tau \right)dy+g_0^{\prime }{\left(1\right)}^k\right)\right)}^{\frac{1}{k}}.$$

For some $R_1>0$, by the integration from $2R_1$ to p again,

$$g_0\left(p\right)={\int }_{2R_1}^p{\left(\frac{1}{C_n^k}\right)}^{\frac{1}{k}}{s}^{1-\frac{n}{k}}{\left({\int }_1^{s}n{y}^{n-1}{\mu }^{-1}\left(f_0-\tau \right)dy+C_n^k{g}_0^{\prime }{\left(1\right)}^k\right)}^{\frac{1}{k}}ds+g_0\left(2R_1\right).$$

Assume that $\tilde{d}=g_0\left(2R_1\right),b={(1/C_n^k)}^{\frac{1}{k}},\tilde{a}=C_n^k{g}_0^{\prime }{\left(1\right)}^k$, then

$$g_0\left(p\right)={\int }_{2R_1}^{p}b{s}^{1-\frac{n}{k}}{\left({\int }_1^{s}n{y}^{n-1}{\mu }^{-1}\left(f_0-\tau \right)dy+\tilde{a}\right)}^{\frac{1}{k}}ds+\tilde{d}.$$

When $g_0\left(0\right)=0,g_0^{\prime }\left(0\right)=0$, the radial symmetry solution of $S_k\left(D^2{g}_0\right)={{\mu }^{-1}(f}_0-\tau )$ is

$$g_0\left(\right|x\left|\right)={\int }_0^{p}b{s}^{1-\frac{n}{k}}{\left({\int }_0^{s}n{y}^{n-1}{\mu }^{-1}\left(f_0-\tau \right)dy\right)}^{\frac{1}{k}}ds.$$

3. Proof of Theorem 1

Proof of Theorem 1.

We might as well fix $z>2$, such that

$$B_2\left(0\right)\subset \subset \mathsf{\Omega }\subset \subset B_z\left(0\right):=\{x\in {\mathbb{R}}^n:\left|x\right|<z\}.$$

(25)

By subtracting a linear function from $\theta$, we just have to prove that the theorem holds for d equals 0. The following proof can be divided into six parts.

Step 1. Construct a viscosity subsolution of (1)–(3).

According to Lemma 1, for any $\xi \in \partial \mathsf{\Omega }$, there is $\overline{x}(\xi ,\psi )\in {\mathbb{R}}^n,0\le \psi \le \mathsf{\Psi },\left|\overline{x}(\xi ,\psi )\right|<\infty$, such that

$$\phi (x,\psi )>w_{\xi }(x,\psi ),(x,\psi )\in \overline{U}\backslash \left\{(\xi ,\psi )\right\},$$

where

$$w_{\xi }(x,\psi )=\phi (\xi ,\psi )+\frac{c_{*}}{2}\left[|x-\overline{x}{(\xi ,\psi )|}^2-{|\xi -\overline{x}(\xi ,\psi )|}^2\right],(x,\psi )\in {\mathbb{R}}^n\times [0,\mathsf{\Psi }],$$

$${\displaystyle \mu \left(C_n^k{c}_{*}{}^k\right)\ge \underset{B_{3z}\left(0\right)}{sup}f-\tau .}$$

Let $B_{3z}=B_{3z}\left(0\right).$ Then $\phi (\xi ,\psi )=w_{\xi }(\xi ,\psi )$, and according to condition $\left(H_2\right)$,

$$-{\left(w_{\xi }\right)}_{\psi }+\mu \left(S_k\left(D^2{w}_{\xi }\right)\right)\ge f\left(x\right),(x,\psi )\in B_{3z}\times (0,\mathsf{\Psi }],$$

and

$$S_k\left(D^2{w}_{\xi }(x,0)\right)\ge {\mu }^{-1}(f\left(x\right)-\tau ),x\in B_{3z}.$$

Let

$$w(x,\psi )=\underset{\xi \in \partial \mathsf{\Omega }}{sup}{w}_{\xi }(x,\psi ),(x,\psi )\in {\mathbb{R}}^n\times [0,\mathsf{\Psi }].$$

Then

$$\phi (x,\psi )\ge w(x,\psi ),(x,\psi )\in \overline{U},$$

$$\phi (x,\psi )=w(x,\psi ),(x,\psi )\in SU,$$

(26)

and

$$-w_{\psi }+\mu \left(S_k\left(D^{2}w\right)\right)\ge f\left(x\right),(x,\psi )\in B_{3z}\times (0,\mathsf{\Psi }].$$

Set

$$W(x,\psi )=\left\{\begin{array}{cc}\phi (x,\psi ),\hfill & (x,\psi )\in U,\\ w(x,\psi ),\hfill & (x,\psi )\in \left({\mathbb{R}}^n\times [0,\mathsf{\Psi }]\right)\backslash U.\end{array}\right.$$

According to Lemma 2 and (26), there are $W\in C^0({\mathbb{R}}^n\times [0,\mathsf{\Psi }])$, and

$$\phi (x,\psi )=W(x,\psi ),(x,\psi )\in \partial \mathsf{\Omega }\times [0,\mathsf{\Psi }],$$

(27)

$$-W_{\psi }+\mu \left(S_k\left(D^{2}W\right)\right)\ge f\left(x\right),(x,\psi )\in (B_{3z}\backslash \overline{\mathsf{\Omega }})\times (0,\mathsf{\Psi }],$$

(28)

$$S_k\left(D^{2}W(x,0)\right)\ge {\mu }^{-1}(f\left(x\right)-\tau ),x\in B_{3z}\backslash \overline{\mathsf{\Omega }}.$$

(29)

Let $\overline{f}\left(\right|x\left|\right),\underline{f}\left(\right|x\left|\right)$ be smooth functions satisfying the conditions

$$\overline{f}={\mu }^{-1}\left(f_0\left(\right|x\left|\right)-\tau \right)+C_1{\left|x\right|}^{-\beta },\left|x\right|\to +\infty ,$$

$$\underline{f}={\mu }^{-1}\left(f_0\left(\right|x\left|\right)-\tau \right)-C_2{\left|x\right|}^{-\beta },\left|x\right|\to +\infty ,$$

where $C_1$ and $C_2$ are positive constants, and

$$\overline{f}\left(\right|x\left|\right)\ge {\mu }^{-1}(f\left(x\right)-\tau )\ge \underline{f}\left(\right|x\left|\right)>0.$$

For $a>0$, let

$${\theta }_1(x,\psi )=-\tau \psi +{inf}_{B_z\times [0,\mathsf{\Psi }]}W+{\int }_{2z}^{\left|x\right|}b{s}^{1-\frac{n}{k}}{\left({\int }_1^{s}n{y}^{n-1}\overline{f}\left(y\right)dy+a\right)}^{\frac{1}{k}}ds,(x,\psi )\in {\mathbb{R}}^n\times [0,\mathsf{\Psi }],$$

$${\theta }_2(x,\psi )=-\tau \psi +{sup}_{B_z\times [0,\mathsf{\Psi }]}W+{\int }_2^{\left|x\right|}b{s}^{1-\frac{n}{k}}{\left({\int }_1^{s}n{y}^{n-1}\underline{f}\left(y\right)dy+a\right)}^{\frac{1}{k}}ds,(x,\psi )\in {\mathbb{R}}^n\times [0,\mathsf{\Psi }].$$

Therefore, ${\theta }_1,{\theta }_2$ are parabolically convex, and

$${\theta }_2(x,\psi )\ge {\theta }_1(x,\psi ),(x,\psi )\in \partial B_2\times [0,\mathsf{\Psi }],$$

(30)

$$-{\left({\theta }_1\right)}_{\psi }+\mu \left(S_k\left(D^2{\theta }_1\right)\right)=\tau +\mu \left(\overline{f}\right)\ge f\left(x\right),(x,\psi )\in ({\mathbb{R}}^n\backslash B_2)\times (0,\mathsf{\Psi }],$$

(31)

$$-{\left({\theta }_2\right)}_{\psi }+\mu \left(S_k\left(D^2{\theta }_2\right)\right)=\tau +\mu \left(\underline{f}\right)\le f\left(x\right),(x,\psi )\in ({\mathbb{R}}^n\backslash B_2)\times (0,\mathsf{\Psi }],$$

(32)

$$S_k\left(D^2{\theta }_1(x,\psi )\right)=\overline{f}\ge {\mu }^{-1}(f\left(x\right)-\tau ),(x,\psi )\in ({\mathbb{R}}^n\backslash B_2)\times [0,\mathsf{\Psi }],$$

(33)

$$S_k\left(D^2{\theta }_2(x,\psi )\right)=\underline{f}\le {\mu }^{-1}(f\left(x\right)-\tau ),(x,\psi )\in ({\mathbb{R}}^n\backslash B_2)\times [0,\mathsf{\Psi }].$$

(34)

Take $a_0\ge 0$, so that for any $a\ge a_0$, the following three inequalities hold simultaneously

$$\begin{array}{cc}\hfill {\theta }_1(x,\psi )& =-\tau \psi +\underset{B_z\times [0,\mathsf{\Psi }]}{inf}W+{\int }_{2z}^{3z}b{s}^{1-\frac{n}{k}}{\left({\int }_1^{s}n{y}^{n-1}\overline{f}\left(y\right)dy+a\right)}^{\frac{1}{k}}ds\hfill \\ \hfill & \ge W(x,\psi ),\left|x\right|=3z,0\le \psi \le \mathsf{\Psi },\hfill \end{array}$$

$$\begin{array}{cc}\hfill {\theta }_2(x,\psi )& =-\tau \psi +\underset{B_z\times [0,\mathsf{\Psi }]}{sup}W+{\int }_2^{3z}b{s}^{1-\frac{n}{k}}{\left({\int }_1^{s}n{y}^{n-1}\underline{f}\left(y\right)dy+a\right)}^{\frac{1}{k}}ds\hfill \\ \hfill & \ge W(x,\psi ),\left|x\right|=3z,0\le \psi \le \mathsf{\Psi },\hfill \end{array}$$

(35)

$${\theta }_2(x,\psi )\ge \phi (x,\psi ),x\in \partial \mathsf{\Omega },0\le \psi \le \mathsf{\Psi }.$$

(36)

Further, it can be found that for $(x,\psi )\in \left({\mathbb{R}}^n\backslash B_2\right)\times [0,\mathsf{\Psi }]$,

$$\begin{array}{cc}\hfill {\theta }_1(x,\psi )=& -\tau \psi +\underset{B_z\times [0,\mathsf{\Psi }]}{inf}W+{\int }_{2z}^{\infty }b{s}^{1-\frac{n}{k}}{\left({\int }_1^{s}n{y}^{n-1}\overline{f}\left(y\right)dy+a\right)}^{\frac{1}{k}}ds\hfill \\ \hfill & -{\int }_{\left|x\right|}^{\infty }b{s}^{1-\frac{n}{k}}{\left({\int }_1^{s}n{y}^{n-1}\overline{f}\left(y\right)dy+a\right)}^{\frac{1}{k}}ds\hfill \\ \hfill =& -\tau \psi +\underset{B_z\times [0,\mathsf{\Psi }]}{inf}W+{\int }_{2z}^{\infty }b{s}^{1-\frac{n}{k}}{\left({\int }_1^{s}n{y}^{n-1}\overline{f}\left(y\right)dy+a\right)}^{\frac{1}{k}}ds\hfill \\ \hfill & -{\int }_{2z}^{\infty }b{s}^{1-\frac{n}{k}}{\left({\int }_0^{s}n{y}^{n-1}{\mu }^{-1}\left(f_0\left(y\right)-\tau \right)dy\right)}^{\frac{1}{k}}ds\hfill \\ \hfill & +{\int }_{2z}^{\infty }b{s}^{1-\frac{n}{k}}{\left({\int }_0^{s}n{y}^{n-1}{\mu }^{-1}\left(f_0\left(y\right)-\tau \right)dy\right)}^{\frac{1}{k}}ds\hfill \\ \hfill & -{\int }_{\left|x\right|}^{\infty }b{s}^{1-\frac{n}{k}}{\left({\int }_1^{s}n{y}^{n-1}\overline{f}\left(y\right)dy+a\right)}^{\frac{1}{k}}ds\hfill \\ \hfill =& -\tau \psi +\underset{B_z\times [0,\mathsf{\Psi }]}{inf}W\hfill \end{array}$$

$$\begin{array}{c}\\ & +{\int }_{2z}^{\infty }b{s}^{1-\frac{n}{k}}\left[{\left({\int }_1^{s}n{y}^{n-1}\overline{f}\left(y\right)dy+a\right)}^{\frac{1}{k}}-{\left({\int }_0^{s}n{y}^{n-1}{\mu }^{-1}\left(f_0\left(y\right)-\tau \right)dy\right)}^{\frac{1}{k}}\right]ds\hfill \\ \hfill & -{\int }_0^{2z}b{s}^{1-\frac{n}{k}}{\left({\int }_0^{s}n{y}^{n-1}{\mu }^{-1}\left(f_0\left(y\right)-\tau \right)dy\right)}^{\frac{1}{k}}ds\hfill \\ \hfill & +{\int }_0^{\left|x\right|}b{s}^{1-\frac{n}{k}}{\left({\int }_0^{s}n{y}^{n-1}{\mu }^{-1}\left(f_0\left(y\right)-\tau \right)dy\right)}^{\frac{1}{k}}ds\hfill \\ \hfill & +{\int }_{\left|x\right|}^{\infty }b{s}^{1-\frac{n}{k}}{\left({\int }_0^{s}n{y}^{n-1}{\mu }^{-1}\left(f_0\left(y\right)-\tau \right)dy\right)}^{\frac{1}{k}}ds\hfill \\ \hfill & -{\int }_{\left|x\right|}^{\infty }b{s}^{1-\frac{n}{k}}{\left({\int }_1^{s}n{y}^{n-1}\overline{f}\left(y\right)dy+a\right)}^{\frac{1}{k}}ds\hfill \\ \hfill =& -\tau \psi +\underset{B_z\times [0,\mathsf{\Psi }]}{inf}W\hfill \\ \hfill & +{\int }_{2z}^{\infty }b{s}^{1-\frac{n}{k}}\left[{\left({\int }_1^{s}n{y}^{n-1}\overline{f}\left(y\right)dy+a\right)}^{\frac{1}{k}}-{\left({\int }_0^{s}n{y}^{n-1}{\mu }^{-1}\left(f_0\left(y\right)-\tau \right)dy\right)}^{\frac{1}{k}}\right]ds\hfill \\ \hfill & -{\int }_{\left|x\right|}^{\infty }b{s}^{1-\frac{n}{k}}\left[{\left({\int }_1^{s}n{y}^{n-1}\overline{f}\left(y\right)dy+a\right)}^{\frac{1}{k}}-{\left({\int }_0^{s}n{y}^{n-1}{\mu }^{-1}\left(f_0\left(y\right)-\tau \right)dy\right)}^{\frac{1}{k}}\right]ds\hfill \\ \hfill & -{\int }_0^{2z}b{s}^{1-\frac{n}{k}}{\left({\int }_0^{s}n{y}^{n-1}{\mu }^{-1}\left(f_0\left(y\right)-\tau \right)dy\right)}^{\frac{1}{k}}ds\hfill \\ \hfill & +{\int }_0^{\left|x\right|}b{s}^{1-\frac{n}{k}}{\left({\int }_0^{s}n{y}^{n-1}{\mu }^{-1}\left(f_0\left(y\right)-\tau \right)dy\right)}^{\frac{1}{k}}ds.\hfill \end{array}$$

That is,

$$\begin{array}{cc}\hfill {\theta }_1(x,\psi )=& -\tau \psi +{\mu }_1\left(a\right)+g_0\left(\right|x\left|\right)-{\int }_{\left|x\right|}^{\infty }b{s}^{1-\frac{n}{k}}\left[{\left({\int }_1^{s}n{y}^{n-1}\overline{f}\left(y\right)dy+a\right)}^{\frac{1}{k}}\right.\hfill \\ \hfill & \left.-{\left({\int }_0^{s}n{y}^{n-1}{\mu }^{-1}\left(f_0\left(y\right)-\tau \right)dy\right)}^{\frac{1}{k}}\right]ds,\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill {\mu }_1\left(a\right)=& \underset{B_z\times [0,\mathsf{\Psi }]}{inf}W\hfill \\ \hfill & +{\int }_{2z}^{\infty }b{s}^{1-\frac{n}{k}}\left[{\left({\int }_1^{s}n{y}^{n-1}\overline{f}\left(y\right)dy+a\right)}^{\frac{1}{k}}-{\left({\int }_0^{s}n{y}^{n-1}{\mu }^{-1}\left(f_0\left(y\right)-\tau \right)dy\right)}^{\frac{1}{k}}\right]ds-g_0\left(2z\right).\hfill \end{array}$$

Likewise,

$$\begin{array}{cc}\hfill {\theta }_2(x,\psi )=& -\tau \psi +{\mu }_2\left(a\right)+g_0\left(\right|x\left|\right)-{\int }_{\left|x\right|}^{\infty }b{s}^{1-\frac{n}{k}}\left[{\left({\int }_1^{s}n{y}^{n-1}\underline{f}\left(y\right)dy+a\right)}^{\frac{1}{k}}\right.\hfill \\ \hfill & \left.-{\left({\int }_0^{s}n{y}^{n-1}{\mu }^{-1}\left(f_0\left(y\right)-\tau \right)dy\right)}^{\frac{1}{k}}\right]ds,\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill {\mu }_2\left(a\right)=& \underset{B_z\times [0,\mathsf{\Psi }]}{sup}W\hfill \\ \hfill & +{\int }_2^{\infty }b{s}^{1-\frac{n}{k}}\left[{\left({\int }_1^{s}n{y}^{n-1}\underline{f}\left(y\right)dy+a\right)}^{\frac{1}{k}}-{\left({\int }_0^{s}n{y}^{n-1}{\mu }^{-1}\left(f_0\left(y\right)-\tau \right)dy\right)}^{\frac{1}{k}}\right]ds-g_0\left(2\right).\hfill \end{array}$$

Then ${\mu }_1\left(a\right),{\mu }_2\left(a\right)$ are strictly monotone increasing on $(0,+\infty )$, and

$$\underset{a\to +\infty }{lim}{\mu }_1\left(a\right)=+\infty ,\underset{a\to +\infty }{lim}{\mu }_2\left(a\right)=+\infty .$$

In light of (7) and (8), we can assume that for $C_3>0$ and the sufficiently large constant $s_0$,

$${\mu }^{-1}(f_0\left(y\right)-\tau )=C_3{\left|y\right|}^{\alpha },y>s_0,$$

and

$$\overline{f}\left(y\right)=C_3{\left|y\right|}^{\alpha }+C_1{\left|y\right|}^{-\beta },y>s_0.$$

If $\beta \ne n$, then according to (9), we have

$$\begin{array}{cc}\hfill & \left.\left.{\int }_{\left|x\right|}^{\infty }b{s}^{1-\frac{n}{k}}\right[{\left({\int }_1^{s}n{y}^{n-1}\overline{f}\left(y\right)dy+a\right)}^{\frac{1}{k}}-{\left({\int }_0^{s}n{y}^{n-1}{\mu }^{-1}\left(f_0\left(y\right)-\tau \right)dy\right)}^{\frac{1}{k}}\right]ds\hfill \\ \hfill =& {\int }_{\left|x\right|}^{\infty }b{s}^{1-\frac{n}{k}}\left[{\left({\int }_1^{s_0}n{y}^{n-1}\overline{f}\left(y\right)dy+{\int }_{s_0}^{s}n{y}^{n-1}\overline{f}\left(y\right)dy+a\right)}^{\frac{1}{k}}\right.\hfill \\ \hfill & \left.-{\left({\int }_0^{s_0}n{y}^{n-1}{\mu }^{-1}\left(f_0\left(y\right)-\tau \right)dy+{\int }_{s_0}^{s}n{y}^{n-1}{\mu }^{-1}\left(f_0\left(y\right)-\tau \right)dy\right)}^{\frac{1}{k}}\right]ds\hfill \\ \hfill =& {\int }_{\left|x\right|}^{\infty }b{s}^{1-\frac{n}{k}}\left[{\left({\int }_{s_0}^{s}n{y}^{n-1}\overline{f}\left(y\right)dy+d_1\right)}^{\frac{1}{k}}-{\left({\int }_{s_0}^{s}n{y}^{n-1}{\mu }^{-1}\left(f_0\left(y\right)-\tau \right)dy+d_2\right)}^{\frac{1}{k}}\right]ds\hfill \end{array}$$

$$\begin{array}{cc}\hfill =& {\int }_{\left|x\right|}^{\infty }b{s}^{1-\frac{n}{k}}\left[{\left({\int }_{s_0}^{s}n{y}^{n-1}\left(C_3{y}^{\alpha }+C_1{y}^{-\beta }\right)dy+d_1\right)}^{\frac{1}{k}}-{\left({\int }_{s_0}^{s}n{y}^{n-1}{C}_3{y}^{\alpha }dy+d_2\right)}^{\frac{1}{k}}\right]ds\hfill \\ \hfill =& {\int }_{\left|x\right|}^{\infty }b{\left(d_3\right)}^{\frac{1}{k}}{s}^{1+\frac{\alpha }{k}}\left[{\left(1+\frac{d_4}{d_3}{s}^{-\alpha -\beta }+\frac{d_5}{d_3}{s}^{-n-\alpha }\right)}^{\frac{1}{k}}-{\left(1+\frac{d_6}{d_3}{s}^{-n-\alpha }\right)}^{\frac{1}{k}}\right]ds\hfill \\ \hfill =& {\int }_{\left|x\right|}^{\infty }\left[O\left(s^{1-n-\alpha +\frac{\alpha }{k}}\right)+O\left(s^{1-\beta -\alpha +\frac{\alpha }{k}}\right)\right]ds\hfill \\ \hfill =& O\left({\left|x\right|}^{-min\{n,\beta \}+\frac{\alpha }{k}+2-\alpha }\right),\left|x\right|\to +\infty ,\hfill \end{array}$$

(37)

where $d_i,i=1,\dots ,6$ are constants.

If $\beta =n$, by (37),

$$\begin{array}{cc}\hfill & {\int }_{\left|x\right|}^{\infty }b{s}^{1-\frac{n}{k}}\left[{\left({\int }_1^{s}n{y}^{n-1}\overline{f}\left(y\right)dy+a\right)}^{\frac{1}{k}}-{\left({\int }_0^{s}n{y}^{n-1}{\mu }^{-1}\left(f_0\left(y\right)-\tau \right)dy\right)}^{\frac{1}{k}}\right]ds\hfill \\ \hfill =& {\int }_{\left|x\right|}^{\infty }b{\left(\widetilde{d_3}\right)}^{\frac{1}{k}}{s}^{1+\frac{\alpha }{k}}\left[{\left(1+\frac{\widetilde{d_4}}{\widetilde{d_3}}{s}^{-n-\alpha }lns+\frac{\widetilde{d_5}}{\widetilde{d_3}}{s}^{-n-\alpha }\right)}^{\frac{1}{k}}-{\left(1+\frac{\widetilde{d_6}}{\widetilde{d_3}}{s}^{-n-\alpha }\right)}^{\frac{1}{k}}\right]ds\hfill \\ \hfill =& O\left({\left|x\right|}^{\frac{\alpha }{k}+2-\alpha -n}ln\left|x\right|\right),\left|x\right|\to +\infty ,\hfill \end{array}$$

where $\widetilde{d_i},i=1,\dots ,6$ are constants.

Thus, when $\left|x\right|\to \infty$,

$$\left\{\begin{array}{cc}{\theta }_1(x,\psi )=g_0\left(\right|x\left|\right)-\tau \psi +{\mu }_1\left(a\right)+O\left({\left|x\right|}^{-min\{n,\beta \}+\frac{\alpha }{k}+2-\alpha }\right),\hfill & \beta \ne n,\hfill \\ {\theta }_1(x,\psi )=g_0\left(\right|x\left|\right)-\tau \psi +{\mu }_1\left(a\right)+O\left({\left|x\right|}^{-n+\frac{\alpha }{k}+2-\alpha }ln\left|x\right|\right),\hfill & \beta =n.\hfill \end{array}\right.$$

In a similar way, when $\left|x\right|\to \infty$,

$$\left\{\begin{array}{cc}{\theta }_2(x,\psi )=g_0\left(\right|x\left|\right)-\tau \psi +{\mu }_2\left(a\right)+O\left({\left|x\right|}^{-min\{n,\beta \}+\frac{\alpha }{k}+2-\alpha }\right),\hfill & \beta \ne n\hfill \\ {\theta }_2(x,\psi )=g_0\left(\right|x\left|\right)-\tau \psi +{\mu }_2\left(a\right)+O\left({\left|x\right|}^{-n+\frac{\alpha }{k}+2-\alpha }ln\left|x\right|\right),\hfill & \beta =n.\hfill \end{array}\right.$$

In addition, it can be seen that for a sufficiently large c in (15), there are $a_1\left(c\right)$, $a_2\left(c\right)>a_0$, such that ${\mu }_1\left(a_1\left(c\right)\right)={\mu }_2\left(a_2\left(c\right)\right)=c$. Then when $\left|x\right|\to \infty$,

$$\left\{\begin{array}{cc}{\theta }_1(x,\psi )=g_0\left(\right|x\left|\right)-\tau \psi +c+O\left({\left|x\right|}^{-min\{n,\beta \}+\frac{\alpha }{k}+2-\alpha }\right),\hfill & \beta \ne n\hfill \\ {\theta }_1(x,\psi )=g_0\left(\right|x\left|\right)-\tau \psi +c+O\left({\left|x\right|}^{-n+\frac{\alpha }{k}+2-\alpha }ln\left|x\right|\right),\hfill & \beta =n,\hfill \end{array}\right.$$

and

$$\left\{\begin{array}{cc}{\theta }_2(x,\psi )=g_0\left(\right|x\left|\right)-\tau \psi +c+O\left({\left|x\right|}^{-min\{n,\beta \}+\frac{\alpha }{k}+2-\alpha }\right),\hfill & \beta \ne n\hfill \\ {\theta }_2(x,\psi )=g_0\left(\right|x\left|\right)-\tau \psi +c+O\left({\left|x\right|}^{-n+\frac{\alpha }{k}+2-\alpha }ln\left|x\right|\right),\hfill & \beta =n.\hfill \end{array}\right.$$

Thus, we know

$$\underset{\left|x\right|\to \infty }{lim}\left({\theta }_1(x,\psi )-{\theta }_2(x,\psi )\right)=0,0\le \psi \le \mathsf{\Psi }.$$

(38)

According to (30), (33), (34), and (38), and the comparison principle, we have

$${\theta }_2(x,0)\ge {\theta }_1(x,0),x\in {\mathbb{R}}^n\backslash B_2,$$

(39)

$${\theta }_2(x,\mathsf{\Psi })\ge {\theta }_1(x,\mathsf{\Psi }),x\in {\mathbb{R}}^n\backslash B_2.$$

(40)

Likewise, by (31), (32), and (38)–(40), and the comparison principle, we can have

$${\theta }_2(x,\psi )\ge {\theta }_1(x,\psi ),(x,\psi )\in \left({\mathbb{R}}^n\backslash B_2\right)\times [0,\mathsf{\Psi }],$$

(41)

and by (29) and (34)–(36), and the comparison principle,

$${\theta }_2(x,0)\ge W(x,0),x\in B_{3z}\backslash \mathsf{\Omega }.$$

Moreover, for $U_{3z}=B_{3z}\times (0,\mathsf{\Psi }]$,

$$-{\left({\theta }_2\right)}_{\psi }+\mu \left(S_k\left(D^2{\theta }_2\right)\right)\le f,(x,\psi )\in U_{3z}\backslash \overline{U},$$

$${\theta }_2(x,\psi )\ge W(x,\psi ),(x,\psi )\in {\partial }_p{U}_{3z}.$$

Then according to (35) and (36) and the principle of comparison,

$${\theta }_2(x,\psi )\ge W(x,\psi ),(x,\psi )\in {\overline{U}}_{3z}\backslash U.$$

In addition, when $\left|x\right|\le z,0\le \psi \le \mathsf{\Psi }$, obviously there are

$${\theta }_1(x,\psi )\le \underset{B_z\times [0,\mathsf{\Psi }]}{inf}W\le W(x,\psi ).$$

(42)

Define for $a\ge a_0$,

$${\underline{\theta }}_a(x,\psi )=\left\{\begin{array}{cc}max\{W(x,\psi ),{\theta }_1(x,\psi )\},\hfill & x\in B_{3z}\backslash \mathsf{\Omega },0\le \psi \le \mathsf{\Psi }\hfill \\ {\theta }_1(x,\psi ),\hfill & x\in {\mathbb{R}}^n\backslash B_{3z},0\le \psi \le \mathsf{\Psi }.\hfill \end{array}\right.$$

Then ${\theta }_2(x,\psi )\ge {\underline{\theta }}_a(x,\psi ),x\in {\mathbb{R}}^n\backslash \mathsf{\Omega },0\le \psi \le \mathsf{\Psi }$, and according to Lemma 2, we know that

$$-{\left({\underline{\theta }}_a\right)}_{\psi }+\mu \left(S_k\left(D^2{\underline{\theta }}_a\right)\right)\ge f,(x,\psi )\in {\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash \overline{U}.$$

According to (26), (25), and (42), and the definition of W, we can know

$${\underline{\theta }}_a(x,\psi )=\phi (x,\psi ),(x,\psi )\in SU.$$

(43)

Moreover, when $\left|x\right|\to \infty$,

$$\left\{\begin{array}{cc}{\underline{\theta }}_a(x,\psi )=g_0\left(\right|x\left|\right)-\tau \psi +c+O\left({\left|x\right|}^{-min\{n,\beta \}+\frac{\alpha }{k}+2-\alpha }\right),\hfill & \beta \ne n,\hfill \\ {\underline{\theta }}_a(x,\psi )=g_0\left(\right|x\left|\right)-\tau \psi +c+O\left({\left|x\right|}^{-n+\frac{\alpha }{k}+2-\alpha }ln\left|x\right|\right),\hfill & \beta =n.\hfill \end{array}\right.$$

(44)

So according to the condition $\left(H_3\right)$,

$$\underset{\left|x\right|\to \infty }{lim}\left({\underline{\theta }}_a(x,0)-{\theta }_0\left(x\right)\right)=0.$$

In addition, ${\underline{\theta }}_a(x,0)=\phi (x,0)=W(x,0)={\theta }_0\left(x\right),x\in \partial \mathsf{\Omega }$,

$$S_k\left(D^2{\underline{\theta }}_a(x,0)\right)\ge {\mu }^{-1}(f\left(x\right)-\tau ),x\in {\mathbb{R}}^n\backslash \overline{\mathsf{\Omega }}.$$

Therefore, according to the comparison principle, for $x\in {\mathbb{R}}^n\backslash \mathsf{\Omega }$,

$${\underline{\theta }}_a(x,0)\le {\theta }_0\left(x\right).$$

Thus, ${\underline{\theta }}_a$ is a viscosity subsolution of (1)–(3).

Step 2. Define the Perron solution of (1)–(3).

For $X=(x,\psi )\in {\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash U$, let ${\mathcal{S}}_{c,X}$ denote the set of viscosity subsolutions $v\in C^0\left(\overline{{\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash U}\right)$ of (1)–(3) that satisfy $v(x,\psi )\le {\theta }_2(x,\psi )$. Then ${\underline{\theta }}_a\in {\mathcal{S}}_{c,X}$. So ${\mathcal{S}}_{c,X}$ is not empty, let

$${\theta }_c(x,\psi )=sup\left\{v(x,\psi ):v\in {\mathcal{S}}_{c,X}\right\},(x,\psi )\in {\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash U.$$

Step 3. Show that ${\theta }_c$ at infinity satisfies asymptotic behavior.

First, according to the definition of ${\theta }_c$,

$${\theta }_c(x,\psi )\le {\theta }_2(x,\psi ).$$

(45)

So, as $\left|x\right|\to +\infty$,

$${\theta }_c(x,\psi )+\tau \psi -g_0\left(\right|x\left|\right)-c\le O\left({\left|x\right|}^{-min\{n,\beta \}+\frac{\alpha }{k}+2-\alpha }\right),\beta \ne n,$$

and

$${\theta }_c(x,\psi )+\tau \psi -g_0\left(\right|x\left|\right)-c\le O\left({\left|x\right|}^{-n+\frac{\alpha }{k}+2-\alpha }ln\left|x\right|\right),\beta =n.$$

Because ${\underline{\theta }}_a\in {\mathcal{S}}_{c,X}$, by (44), when $\left|x\right|\to \infty$, we have

$${\theta }_c(x,\psi )+\tau \psi -g_0\left(\right|x\left|\right)-c\ge O\left({\left|x\right|}^{-min\{n,\beta \}+\frac{\alpha }{k}+2-\alpha }\right),\beta \ne n,$$

and

$${\theta }_c(x,\psi )+\tau \psi -g_0\left(\right|x\left|\right)-c\ge O\left({\left|x\right|}^{-n+\frac{\alpha }{k}+2-\alpha }ln\left|x\right|\right),\beta =n.$$

As a result,

$$\underset{\left|x\right|\to \infty }{lim\; sup}\left({\left|x\right|}^{min\{n,\beta \}+\alpha -\frac{\alpha }{k}-2}\left|{\theta }_c(x,\psi )-\left(-\tau \psi +g_0\left(\right|x\left|\right)+c\right)\right|\right)<\infty ,\beta \ne n,$$

and

$$\underset{\left|x\right|\to \infty }{lim\; sup}\left({\left|x\right|}^{n+\alpha -\frac{\alpha }{k}-2}{(ln|x\left|\right)}^{-1}\left|{\theta }_c(x,\psi )-\left(-\tau \psi +g_0\left(\right|x\left|\right)+c\right)\right|\right)<\infty ,\beta =n.$$

Step 4. Prove that ${\theta }_c=\phi ,(x,\psi )\in SU$, and ${\theta }_c(x,0)={\theta }_0\left(x\right),x\in {\mathbb{R}}^n\backslash \mathsf{\Omega }$.

First prove ${\theta }_c(x,0)={\theta }_0\left(x\right)$, $x\in {\mathbb{R}}^n\backslash \mathsf{\Omega }.$ Since $\phi \in C^{2,1}\left(\overline{U}\right)$, then according to (12), select the constants $m_1,m_2$, which satisfy $m_2\le \tau \le -{\phi }_{\psi }(x,\psi )\le m_1$. Let

$$A(x,\psi )=-m_1\psi +{\theta }_0\left(x\right),(x,\psi )\in {\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash U,$$

$$B(x,\psi )=-m_2\psi +{\theta }_0\left(x\right),(x,\psi )\in {\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash U.$$

Then

$$-A_{\psi }+\mu \left(S_k\left(D^{2}A\right)\right)=f\left(x\right)+m_1-\tau \ge f\left(x\right),(x,\psi )\in {\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash \overline{U},$$

$$-B_{\psi }+\mu \left(S_k\left(D^{2}B\right)\right)=f\left(x\right)+m_2-\tau \le f\left(x\right),(x,\psi )\in {\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash \overline{U}.$$

Moreover, on $SU$,

$$A(x,\psi )=-m_1\psi +{\theta }_0\left(x\right)=\phi (x,0)-m_1\psi \le \phi (x,\psi ),$$

$$B(x,\psi )=-m_2\psi +{\theta }_0\left(x\right)=\phi (x,0)-m_2\psi \ge \phi (x,\psi ).$$

When $\left|x\right|\to \infty$,

$$\underset{\left|x\right|\to \infty }{lim}\left(A(x,\psi )-{\theta }_c(x,\psi )\right)\le 0,$$

$$\underset{\left|x\right|\to \infty }{lim}\left(B(x,\psi )-{\theta }_c(x,\psi )\right)\ge 0.$$

Obviously, when $x\in {\mathbb{R}}^n\backslash \mathsf{\Omega }$, ${\theta }_0\left(x\right)=A(x,0)=B(x,0)$, then $A(x,\psi )$, $B(x,\psi )$ are the viscosity subsolution and viscosity supersolution of (1)–(3), respectively, so there are

$$B(x,\psi )\ge {\theta }_c\ge A(x,\psi ),(x,\psi )\in \overline{{\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash U}.$$

Therefore,

$${\theta }_c(x,0)={\theta }_0\left(x\right),x\in {\mathbb{R}}^n\backslash \mathsf{\Omega }.$$

The following is to prove that ${\theta }_c=\phi ,(x,\psi )\in SU$. For any $\overline{\xi }\in \partial \mathsf{\Omega },0<\overline{\tau }\le \mathsf{\Psi }$, for one thing, due to ${\underline{\theta }}_a\in S_{c,X}$, according to (43), we have

$$\underset{(x,\psi )\to (\overline{\xi },\overline{\tau })}{lim\; inf}{\theta }_c(x,\psi )\ge \underset{(x,\psi )\to (\overline{\xi },\overline{\tau })}{lim}{\underline{\theta }}_a(x,\psi )=\phi (\overline{\xi },\overline{\tau }).$$

For another, we claim that

$$\underset{(x,\psi )\to (\overline{\xi },\overline{\tau })}{lim\; sup}{\theta }_c(x,\psi )\le \phi (\overline{\xi },\overline{\tau }).$$

In fact, we take $B_R=\{x:|x|\le R\}$, so that $\mathsf{\Omega }\subset \subset B_R\subset \subset {\mathbb{R}}^n$, and let $B_R^{\mathsf{\Psi }}=\left(B_R\backslash \overline{\mathsf{\Omega }}\right)\times (0,\mathsf{\Psi }]$.

Then ${\partial }_p{B}_R^{\mathsf{\Psi }}\supset SU$, and for every $v\in {\mathcal{S}}_{c,X}$, there are

$$\left\{\begin{array}{cc}-v_{\psi }+\mathsf{\Delta }v\ge 0,\hfill & (x,\psi )\in B_R^{\mathsf{\Psi }},\hfill \\ v\le \phi ,\hfill & (x,\psi )\in SU,\hfill \\ v\le B,\hfill & (x,\psi )\in {\partial }_p{B}_R^{\mathsf{\Psi }}\backslash SU.\hfill \end{array}\right.$$

Select the barrier function $w^{+}$ satisfying

$$\left\{\begin{array}{cc}-w_{\psi }^{+}+\mathsf{\Delta }{w}^{+}=0,\hfill & (x,\psi )\in B_R^{\mathsf{\Psi }},\hfill \\ w^{+}=\phi ,\hfill & (x,\psi )\in SU,\hfill \\ w^{+}=B,\hfill & (x,\psi )\in {\partial }_p{B}_R^{\mathsf{\Psi }}\backslash SU.\hfill \end{array}\right.$$

According to the comparison principle, there are $v\le w^{+},(x,\psi )\in \overline{B_R^{\mathsf{\Psi }}}$; therefore, ${\theta }_c\le w^{+},(x,\psi )\in \overline{B_R^{\mathsf{\Psi }}}$, and so

$$\underset{(x,\psi )\to (\overline{\xi },\overline{\tau })}{lim\; sup}{\theta }_c(x,\psi )\le \underset{(x,\psi )\to (\overline{\xi },\overline{\tau })}{lim}{w}^{+}(x,\psi )=\phi (\overline{\xi },\overline{\tau }).$$

Step 5. Show that ${\theta }_c$ is a viscosity solution of (1).

According to the definition of ${\theta }_c$, it is obvious that ${\theta }_c$ is the viscosity subsolution of (1), so we have to prove that ${\theta }_c$ is the viscosity supersolution of (1). For any $(\overline{x},\overline{\psi })\in {\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash \overline{U}$, fix $ϵ$ to make $B_{ϵ}\left(\overline{x}\right)\subset {\mathbb{R}}^n\backslash \overline{\mathsf{\Omega }}$, then for $0\le \tau <\overline{\psi },U_{ϵ}=B_{\epsilon }\left(\overline{x}\right)\times (\tau ,\mathsf{\Psi }]\subset {\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash \overline{U}$. According to the Lemma 5,

$$\left\{\begin{array}{cc}-{\tilde{\theta }}_{\psi }+\mu \left(S_k\left(D^2\tilde{\theta }\right)\right)=f,\hfill & \hfill (x,\psi )\in U_{ϵ},\\ \tilde{\theta }={\theta }_c,\hfill & \hfill (x,\psi )\in {\partial }_p{U}_{ϵ}\end{array}\right.$$

has a unique viscosity solution $\tilde{\theta }\in C^0\left(\overline{U_{\epsilon }}\right)$. According to the comparison principle, ${\theta }_c\le \tilde{\theta },(x,\psi )\in U_{ϵ}$. Let

$$\tilde{w}(x,\psi )=\left\{\begin{array}{cc}\tilde{\theta }(x,\psi ),\hfill & (x,\psi )\in U_{ϵ},\hfill \\ {\theta }_c(x,\psi ),\hfill & (x,\psi )\in {\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash \left(U\cup U_{ϵ}\right).\hfill \end{array}\right.$$

Since ${\theta }_2(x,\psi )\ge \tilde{\theta }(x,\psi ),(x,\psi )\in {\overline{U}}_{ϵ}$, so $\tilde{w}\in {\mathcal{S}}_{c,X}$. According to the definition of ${\theta }_c$,

$${\theta }_c\ge \tilde{w},(x,\psi )\in {\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash U,$$

so ${\theta }_c\equiv \tilde{\theta },(x,\psi )\in U_{ϵ}$, and then ${\theta }_c$ is the viscosity solution of (1).

Step 6. Prove the uniqueness.

Let $\theta$ and v satisfy (1)–(3) and (17), then

$$\underset{\left|x\right|\to \infty }{lim}(\theta (x,\psi )-v(x,\psi ))=0.$$

According to the comparison principle, there is $\theta \equiv v,(x,\psi )\in {\mathbb{R}}_{\mathsf{\Psi }}^{n+1}\backslash U$.

Theorem 1 has been proven. □

4. Conclusions

This paper employed the Perron method to prove the existence and uniqueness of viscosity solutions with asymptotic behavior for the first initial boundary value problem of a parabolic Hessian equation outside a cylindrical domain. The proof is divided into six steps and the main ingredient is the first step in which the viscosity subsolutions are constructed. Lemma 1 plays a crucial role in constructing the viscosity subsolutions. Our method is also suitable for the exterior problem of other parabolic equations.