# Asymptotic mean value properties for the elliptic and parabolic double phase
equations

Weili Meng and Chao Zhang∗ Weili Meng School of Mathematics, Harbin Institute
of Technology, Harbin 150001, P.R. China<EMAIL_ADDRESS>Chao Zhang
School of Mathematics, Harbin Institute of Technology, Harbin 150001, P.R.
China<EMAIL_ADDRESS>
###### Abstract.

We characterize an asymptotic mean value formula in the viscosity sense for
the double phase elliptic equation

$-\text{\rm{div}}(\lvert\nabla u\rvert^{p-2}\nabla u+a(x)\lvert\nabla
u\rvert^{q-2}\nabla u)=0$

and the normalized double phase parabolic equation

$u_{t}=\lvert\nabla u\rvert^{2-p}\text{\rm{div}}(\lvert\nabla
u\rvert^{p-2}\nabla u+a(x,t)\lvert\nabla u\rvert^{q-2}\nabla u),\quad 1<p\leq
q<\infty.$

This is the first mean value result for such kind of nonuniformly elliptic and
parabolic equations. In addition, the results obtained can also be applied to
the $p(x)$-Laplace equations and the variable coefficient $p$-Laplace type
equations.

###### Key words and phrases:

Mean value property; Viscosity solutions; Elliptic and parabolic double phase
equations

###### 2020 Mathematics Subject Classification:

35B05, 35D40, 35J92, 35K92

∗Corresponding author.

##  1\. Introduction

Let $\Omega$ be a bounded domain in $\mathbb{R}^{N}(N\geq 2)$. We consider the
following double phase elliptic equation

$\displaystyle-\text{\rm{div}}(\lvert\nabla u\rvert^{p-2}\nabla
u+a(x)\lvert\nabla u\rvert^{q-2}\nabla u)=0\quad\text{in }\Omega,$ (1.1)

where $1<p\leq q<\infty$ and $a(x)\geq 0$. It is the Euler-Lagrange equation
of the non-autonomous functional

$W^{1,1}(\Omega)\ni w\mapsto\int_{\Omega}\left(\frac{1}{p}\lvert\nabla
w\rvert^{p}+\frac{a(x)}{q}\lvert\nabla w\rvert^{q}\right)dx.$

Originally, this functional is connected to the Homogenization theory and
Lavrentiev phenomenon [22, 28, 33], which reflects the behavior of strongly
anisotropic materials, where the coefficient $a(\cdot)$ is used to regulate
two mixtures with $p$ and $q$ hardening, respectively.

During the last years, problems of the type considered in (1.1) have received
great attention from the variational point of view. The regularity of
minimizers and weak solutions is determined via a delicate interaction between
the growth conditions and the pointwise behaviour of $a(\cdot)$. Starting from
a series of remarkable works of Colombo and Mingione et. al. [2, 10, 11],
despite its relatively short history, double phase problems has already
achieved an elaborate theory with several connections to other branches. We
refer the readers to [1, 7, 8, 9, 12, 13, 14, 15, 16, 17, 19, 29] and the
references therein.

It is well-known that a continuous function $u$ is harmonic if and only if it
obeys the mean value formula discovered by Gauss. That is, $u$ solves the
Laplace equation $\Delta u=0$ in $\Omega$ if and only if

$u(x)=\dfrac{1}{|B_{\varepsilon}(x)|}\int_{B_{\varepsilon}(x)}u(y)dy=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}u(y)dy$

holds for all $x\in\Omega$ and $B_{\varepsilon}(x)\subset\Omega$. In fact, an
asymptotic version of the mean value property

$u(x)=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}u(y)dy+o(\varepsilon^{2})\quad\textmd{as
}\varepsilon\rightarrow 0$

suffices to characterize harmonic functions (see [6, 23, 32]). Moreover, a
nonlinear mean value property was explored in [26] that continuous function
$u$ is a viscosity solution of the $p$-Laplace equation

$-\Delta_{p}u=-\text{\rm{div}}(\lvert\nabla u\rvert^{p-2}\nabla
u)=0\quad\text{in }\Omega$

if and only if the asymptotic expansion

$u(x)=\frac{\alpha_{p}}{2}\left\\{\mathop{\rm{max}}\limits_{\overline{B_{\varepsilon}(x)}}u+\mathop{\rm{min}}\limits_{\overline{B_{\varepsilon}(x)}}u\right\\}+\beta_{p}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}u(y)dy+o(\varepsilon^{2})\quad\text{as
}\varepsilon\rightarrow 0$

holds for all $x\in\Omega$ in the viscosity sense, where
$\alpha_{p}+\beta_{p}=1$ and $\frac{\alpha_{p}}{\beta_{p}}=\frac{p-2}{N+2}$.
The second term is a linear one, while the first term counts for the
nonlinearity: the greater the $p$, the more nonlinear the formula. The
expression holds in the viscosity sense, which means that when the $C^{2}$
test function $\phi$ with non-vanishing gradient is close to $u$ from below
(above), the expression is satisfied with $\geq$ ($\leq$) for the test
function at $x$ respectively.

These mean value formulas originate from the study of dynamic programming for
tug-of-war games. The viscosity solution of the normalized parabolic
$p$-Laplace equation is characterized by an asymptotic mean value formula,
which is related to the tug-of-war game with noise, see [20, 24, 25, 27, 30,
31]. For more related asymptotic mean value results, we refer to [21] for
$p$-harmonic functions in the Heisenberg group, [3] for Monge-Ampère equation,
[4, 27] for the nonlinear parabolic equations and the recently published
monograph [5] for historical references and more general equations.

From the results mentioned above, we can see that there are few results
concerning the asymptotic mean value properties for the general nonuniformly
elliptic and parabolic equations. Motivated by the previous works [26, 27],
our intention in the present paper is to build a new bridge between the
viscosity solutions and the asymptotic mean value formula for the double phase
equations (1.1) and (1.4). In addition, the method developed here can also be
used to more equations, such as the $p(x)$-Laplace equations and the variable
coefficient $p$-Laplace type equations. The first result is stated as follows.

######  Theorem 1.1.

Let $1<p\leq q<\infty$, the non-negative function $a(x)$ be a $C^{1}$ function
in $\Omega$ and let $u(x)$ be a continuous function in $\Omega$. Then Eq.
(1.1) holds in the viscosity sense if and only if the asymptotic expansion

$\displaystyle u(x)$
$\displaystyle=\frac{\alpha_{p}+M_{u}(x)\alpha_{q}}{2(1+M_{u}(x))}\left\\{\max\limits_{\overline{B_{\varepsilon}(x)}}u+\min\limits_{\overline{B_{\varepsilon}(x)}}u\right\\}+\dfrac{\beta_{p}+M_{u}(x)\beta_{q}}{1+M_{u}(x)}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}u(y)dy$
$\displaystyle\quad+\frac{\varepsilon\lvert\nabla
u(x)\rvert^{q-p}}{4(N+p)(1+M_{u}(x))}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}u(y+\varepsilon\nabla
a(x))-u(y-\varepsilon\nabla a(x))dy+o(\varepsilon^{2})$ (1.2)

as $\varepsilon\rightarrow 0$, holds for all $x\in\Omega$ in the viscosity
sense. Here

$\displaystyle\alpha_{p}+\beta_{p}=1,\quad\frac{\alpha_{p}}{\beta_{p}}=\frac{p-2}{N+2},$
$\displaystyle\alpha_{q}+\beta_{q}=1,\quad\frac{\alpha_{q}}{\beta_{q}}=\frac{q-2}{N+2},$
(1.3) $\displaystyle M_{u}(x)=a(x)\dfrac{N+q}{N+p}\lvert\nabla
u(x)\rvert^{q-p}.$

######  Remark 1.2.

Note that here we only require that $u$ is a continuous function. However,
$\nabla u$ appears in the formula (1.1) due to the fact that it is an
expression in the viscosity sense. In other words, we focus on the $C^{2}$
test function $\phi$ that approaches $u$ from above and below. More details
will be given in Section 2.

######  Remark 1.3.

From the formula (1.1), we can see that all the terms on the right-hand side
are nonlinear, which is different from the standard $p$-Laplace equation. The
exponents $p,q$ and the non-negative coefficient $a(x)$ coupling together
influence on the nonlinearity in a delicate way. In particular, when $a(x)$ is
a positive constant, Eq. (1.1) is nothing but the $(p,q)$-Laplace equation.
Then the third term

$\frac{\varepsilon\lvert\nabla
u(x)\rvert^{q-p}}{4(N+p)(1+M_{u}(x))}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}u(y+\varepsilon\nabla
a(x))-u(y-\varepsilon\nabla a(x))dy$

will vanish.

Next, we turn to the parabolic case. Let $T>0$, $\Omega_{T}=\Omega\times(0,T)$
be a space-time cylinder, and let $a(x,t)\geq 0$ be a function that is $C^{1}$
in the space variable and continuous in the time variable, respectively. We
consider the following parabolic equation

$u_{t}=\lvert\nabla u\rvert^{2-p}\text{\rm{div}}(\lvert\nabla
u\rvert^{p-2}\nabla u+a(x,t)\lvert\nabla u\rvert^{q-2}\nabla
u)\quad\text{\rm{in }}\Omega_{T},$ (1.4)

which is called the normalized double phase parabolic equation. The difference
between elliptic and the normalized parabolic case is that we have to consider
the influence of time variable $t$ in parabolic setting. To this end, we try
to separate the estimates according to $p$ and $q$, and consider the integrals
in different time intervals. Finally, we find that when the two time lags
satisfy certain viscosity condition, $u$ satisfies the asymptotic mean value
formula in the viscosity sense is equivalent to $u$ is the viscosity solution
to Eq. (1.4). The second result is stated as follows.

######  Theorem 1.4.

Let $1<p\leq q<\infty$ , the non-negative function $a(x,t)$ be a function that
is $C^{1}$ in the space variable, and continuous in the time variable and let
$u(x,t)$ be a continuous function in $\Omega_{T}$. Then Eq. (1.4) holds in the
viscosity sense if and only if the asymptotic expansion

$\displaystyle u(x,t)$
$\displaystyle=\dfrac{1}{1+M_{u}(x,t)}\left(\dfrac{\alpha_{p}}{2}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{t-\frac{\varepsilon^{2}}{A_{u}(x,t)}}^{t}\left\\{\max\limits_{y\in\overline{B_{\varepsilon}(x)}}u(y,s)+\min\limits_{y\in\overline{B_{\varepsilon}(x)}}u(y,s)\right\\}ds\right.$
$\displaystyle\quad+\left.\beta_{p}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{t-\frac{\varepsilon^{2}}{A_{u}(x,t)}}^{t}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}u(y,s)dyds\right)$
$\displaystyle\quad+\dfrac{M_{u}(x,t)}{1+M_{u}(x,t)}\left(\dfrac{\alpha_{q}}{2}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{t-\frac{\varepsilon^{2}}{B_{u}(x,t)}}^{t}\left\\{\max\limits_{y\in\overline{B_{\varepsilon}(x)}}u(y,s)+\min\limits_{y\in\overline{B_{\varepsilon}(x)}}u(y,s)\right\\}ds\right.$
$\displaystyle\quad\left.+\beta_{q}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{t-\frac{\varepsilon^{2}}{B_{u}(x,t)}}^{t}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}u(y,s)dyds\right)$
$\displaystyle\quad+\dfrac{\varepsilon\lvert\nabla
u(x,t)\rvert^{q-p}}{4(N+p)(1+M_{u}(x,t))}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}u(y+\varepsilon\nabla
a(x,t),t)-u(y-\varepsilon\nabla a(x,t),t)dy$
$\displaystyle\quad+o(\varepsilon^{2})$ (1.5)

as $\varepsilon\rightarrow 0$, holds for all $(x,t)\in\Omega_{T}$ in the
viscosity sense. Here

$\displaystyle\begin{split}&\alpha_{p}+\beta_{p}=1,\quad\frac{\alpha_{p}}{\beta_{p}}=\dfrac{p-2}{N+2},\\\
&\alpha_{q}+\beta_{q}=1,\quad\frac{\alpha_{q}}{\beta_{q}}=\dfrac{q-2}{N+2},\\\
&M_{u}(x,t)=a(x,t)\frac{N+q}{N+p}\lvert\nabla u(x,t)\rvert^{q-p},\\\
&\frac{N+p}{A_{u}(x,t)}+\frac{a(x,t)(N+q)\lvert\nabla
u(x,t)\rvert^{q-p}}{B_{u}(x,t)}=1,\quad A_{u}(x,t),B_{u}(x,t)>0.\end{split}$
(1.6)

######  Remark 1.5.

It is worth mentioning that the positive functions $A(x,t)$ and $B(x,t)$
depend on the test function $\phi$. It means that

$\frac{N+p}{A_{u}(x,t)}+\frac{a(x,t)(N+q)\lvert\nabla
u(x,t)\rvert^{q-p}}{B_{u}(x,t)}=1$

holds in the viscosity sense, which we called the viscosity condition.

This manuscript is organized as follows. In Section 2, we introduce the basic
definitions and give some necessary lemmas that will be used later. In Section
3, we give the proof of Theorem 1.1 and present some corollaries, including
the $p(x)$-Laplace type equation. Finally, we prove Theorem 1.4 in Section 4.

##  2\. Preliminaries

In this section, inspired by the ideas developed in [26], we first give the
definition of the asymptotic mean value formula for $u$ at $x\in\Omega$.

######  Definition 2.1.

A continuous function u satisfies

$\displaystyle u(x)$
$\displaystyle=\dfrac{\alpha_{p}+M_{u}(x)\alpha_{q}}{2(1+M_{u}(x))}\left\\{\mathop{\rm{max}}\limits_{\overline{B_{\varepsilon}(x)}}u+\mathop{\rm{min}}\limits_{\overline{B_{\varepsilon}(x)}}u\right\\}+\dfrac{\beta_{p}+M_{u}(x)\beta_{q}}{1+M_{u}(x)}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}u(y)dy$
$\displaystyle\quad+\dfrac{\varepsilon\lvert\nabla
u(x)\rvert^{q-p}}{4(N+p)(1+M_{u}(x))}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}u(y+\varepsilon\nabla
a(x))-u(y-\varepsilon\nabla a(x))dy+o(\varepsilon^{2})$

as $\varepsilon\rightarrow 0$, in the viscosiy sense if

  * (i)

for every $\phi\in C^{2}$ such that $u-\phi$ has a strict minimum at the point
$x\in\Omega$ with $u(x)=\phi(x)$ and $\nabla\phi(x)\neq 0$, we have

$\displaystyle\begin{split}0&\geq-\phi(x)+\dfrac{\alpha_{p}+M_{\phi}(x)\alpha_{q}}{2(1+M_{\phi}(x))}\left\\{\mathop{\rm{max}}\limits_{\overline{B_{\varepsilon}(x)}}\phi+\mathop{\rm{min}}\limits_{\overline{B_{\varepsilon}(x)}}\phi\right\\}+\dfrac{\beta_{p}+M_{\phi}(x)\beta_{q}}{1+M_{\phi}(x)}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}\phi(y)dy\\\
&\quad+\dfrac{\varepsilon\lvert\nabla\phi(x)\rvert^{q-p}}{4(N+p)(1+M_{\phi}(x))}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}\phi(y+\varepsilon\nabla
a(x))-\phi(y-\varepsilon\nabla a(x))dy+o(\varepsilon^{2}).\end{split}$ (2.1)

  * (ii)

for every $\phi\in C^{2}$ such that $u-\phi$ has a strict maximum at the point
$x\in\Omega$ with $u(x)=\phi(x)$ and $\nabla\phi(x)\neq 0$, we have

$\displaystyle\begin{split}0&\leq-\phi(x)+\dfrac{\alpha_{p}+M_{\phi}(x)\alpha_{q}}{2(1+M_{\phi}(x))}\left\\{\mathop{\rm{max}}\limits_{\overline{B_{\varepsilon}(x)}}\phi+\mathop{\rm{min}}\limits_{\overline{B_{\varepsilon}(x)}}\phi\right\\}+\dfrac{\beta_{p}+M_{\phi}(x)\beta_{q}}{1+M_{\phi}(x)}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}\phi(y)dy\\\
&\quad+\dfrac{\varepsilon\lvert\nabla\phi(x)\rvert^{q-p}}{4(N+p)(1+M_{\phi}(x))}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}\phi(y+\varepsilon\nabla
a(x))-\phi(y-\varepsilon\nabla a(x))dy+o(\varepsilon^{2}).\end{split}$ (2.2)

Next, we consider the viscosity solution of the double phase elliptic
equations. Let us expand the left-hand side of Eq. (1.1) as follows:

$\displaystyle\quad\text{\rm{div}}(\lvert\nabla u\rvert^{p-2}\nabla
u+a(x)\lvert\nabla u\rvert^{q-2}\nabla u)$ $\displaystyle=\lvert\nabla
u\rvert^{p-2}((p-2)\Delta_{\infty}u+\Delta u)+a(x)\lvert\nabla
u\rvert^{q-2}((q-2)\Delta_{\infty}u+\Delta u)$
$\displaystyle\quad+\lvert\nabla u\rvert^{q-2}\langle\nabla a,\nabla
u\rangle,$

where $\Delta_{\infty}u=|\nabla u|^{-2}\left<D^{2}u\nabla u,\nabla u\right>$.

Suppose that $u$ is a smooth function with $\nabla u\neq 0$, we can see that
$u$ is a solution to Eq. (1.1) if and only if

$\displaystyle-(p-2)\Delta_{\infty}u-\Delta u-a(x)\lvert\nabla
u\rvert^{q-p}((q-2)\Delta_{\infty}u+\Delta u)$
$\displaystyle\quad-\lvert\nabla u\rvert^{q-p}\langle\nabla a,\nabla
u\rangle=0.$ (2.3)

Then we give the definition of viscosity solutions to Eq. (1.1).

######  Definition 2.2 ([18], Definition 2.5).

Let $1<p\leq q<\infty$ and consider the equation

$-\text{\rm{div}}(\lvert\nabla u\rvert^{p-2}\nabla u+a(x)\lvert\nabla
u\rvert^{q-2}\nabla u)=0.$

  * (i)

A lower semi-continuous function $u$ is a viscosity supersolution if for every
$\phi\in C^{2}$ such that $u-\phi$ has a strict minimum at the point
$x\in\Omega$ with $\nabla\phi(x)\neq 0$ we have

$\displaystyle-\left((p-2)\Delta_{\infty}\phi(x)+\Delta\phi(x)\right)-a(x)\lvert\nabla\phi(x)|^{q-p}\left((q-2)\Delta_{\infty}\phi(x)+\Delta\phi(x)\right)$
$\displaystyle\quad-\lvert\nabla\phi(x)|^{q-p}\langle\nabla
a(x),\nabla\phi(x)\rangle\geq 0.$ (2.4)

  * (ii)

An upper semi-continuous function $u$ is a viscosity subsolution if for every
$\phi\in C^{2}$ such that $u-\phi$ has a strict maximum at the point
$x\in\Omega$ with $\nabla\phi(x)\neq 0$ we have

$\displaystyle\begin{split}&-\left((p-2)\Delta_{\infty}\phi(x)+\Delta\phi(x)\right)-a(x)\lvert\nabla\phi(x)|^{q-p}\left((q-2)\Delta_{\infty}\phi(x)+\Delta\phi(x)\right)\\\
&\quad-\lvert\nabla\phi(x)|^{q-p}\langle\nabla a(x),\nabla\phi(x)\rangle\leq
0.\end{split}$ (2.5)

  * (iii)

Finally, $u$ is a viscosity solution if and only if $u$ is both a viscosity
supersolution and a viscosity subsolution.

We next state the following useful results (Lemmas 2.3–2.5), which can be
found in [26, Section 2].

######  Lemma 2.3.

Let $\phi$ be a $C^{2}$ function in a neighborhood of $x$ and let
$x_{1}^{\varepsilon}$ and $x_{2}^{\varepsilon}$ be the points at which $\phi$
attains its minimum and maximum in $\overline{B_{\varepsilon}(x)}$
respectively. We have

$\displaystyle-\phi(x)+\frac{1}{2}\left\\{\mathop{\rm{max}}\limits_{\overline{B_{\varepsilon}(x)}}\phi+\mathop{\rm{min}}\limits_{\overline{B_{\varepsilon}(x)}}\phi\right\\}\geq\frac{1}{2}\langle
D^{2}\phi(x)(x^{\varepsilon}_{1}-x),(x^{\varepsilon}_{1}-x)\rangle+o(\varepsilon^{2})$
(2.6)

and

$\displaystyle-\phi(x)+\frac{1}{2}\left\\{\mathop{\rm{max}}\limits_{\overline{B_{\varepsilon}(x)}}\phi+\mathop{\rm{min}}\limits_{\overline{B_{\varepsilon}(x)}}\phi\right\\}\leq\frac{1}{2}\langle
D^{2}\phi(x)(x^{\varepsilon}_{2}-x),(x^{\varepsilon}_{2}-x)\rangle+o(\varepsilon^{2}).$
(2.7)

######  Lemma 2.4.

Let $\phi$ be a $C^{2}$ function in a neighborhood of $x$ with
$\nabla\phi(x)\neq 0$. We have

$\displaystyle\lim\limits_{\varepsilon\rightarrow
0+}\dfrac{x^{\varepsilon}_{1}-x}{\varepsilon}=-\dfrac{\nabla\phi}{\lvert\nabla\phi\rvert}(x),$
(2.8)

where $x_{1}^{\varepsilon}$ is defined as in Lemma 2.3.

######  Lemma 2.5.

Let $\phi$ be a $C^{2}$ function in a neighborhood of $x$. We have

$\displaystyle-\phi(x)+\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}\phi(y)dy=\dfrac{\varepsilon^{2}}{2(N+2)}\Delta\phi(x)+o(\varepsilon^{2})\quad\text{as
}\varepsilon\rightarrow 0.$

Although Lemma 2.3 and Lemma 2.5 provide the bridge between the viscosity
solution of $p$-Laplace equation $-\Delta_{p}u=0$ and the asymptotic mean
value formula in [26], it is not enough for the double phase elliptic equation
due to the presence of the term $\left<\nabla a,\nabla u\right>$ in Eq. (2).
Therefore, we need the following lemma.

######  Lemma 2.6.

Let $\phi$ be a $C^{2}$ function in a neighborhood of $x$. We have

$\displaystyle\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}\phi(y+\varepsilon\nabla
a(x))-\phi(y-\varepsilon\nabla a(x))dy=2\varepsilon\langle\nabla\phi(x),\nabla
a(x)\rangle+o(\varepsilon^{2})$

as $\varepsilon\rightarrow 0$.

###### Proof.

Observe that

$\displaystyle\quad\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}\phi(y+\varepsilon\nabla
a(x))dy$
$\displaystyle=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B(0,1)}\phi(x+(\nabla
a(x)+z)\varepsilon)dz$
$\displaystyle=\phi(x)+\varepsilon\langle\nabla\phi(x),\nabla
a(x)\rangle+\frac{\varepsilon^{2}}{2}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B(0,1)}\langle
D^{2}\phi(x)\nabla a(x),\nabla a(x)\rangle+\langle D^{2}\phi(x)z,z\rangle dz$
$\displaystyle\quad+o(\varepsilon^{2})$

and

$\displaystyle\quad\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}\phi(y-\varepsilon\nabla
a(x))dy$
$\displaystyle=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B(0,1)}\phi(x+(z-\nabla
a(x))\varepsilon)dz$
$\displaystyle=\phi(x)-\varepsilon\langle\nabla\phi(x),\nabla
a(x)\rangle+\frac{\varepsilon^{2}}{2}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B(0,1)}\langle
D^{2}\phi(x)\nabla a(x),\nabla a(x)\rangle+\langle D^{2}\phi(x)z,z\rangle dz$
$\displaystyle\quad+o(\varepsilon^{2}).$

Thus, we obtain

$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}\phi(y+\varepsilon\nabla
a(x))-\phi(y-\varepsilon\nabla a(x))dy=2\varepsilon\langle\nabla\phi(x),\nabla
a(x)\rangle+o(\varepsilon^{2})\quad\text{as }\varepsilon\rightarrow 0.$

This finishes the proof. ∎

##  3\. Elliptic case

In this section, we will prove Theorem 1.1 and consider several special cases
as corollaries. Then we apply the ideas to the $p(x)$-Laplace type equations
and give the corresponding conclusions.

###### Proof of Theorem 1.1.

Considering the sufficiency, we need to show that $u$ is a viscosity solution
to Eq. (1.1) by $u$ satisfying the asymptotic mean value formula. We first
prove that $u$ is a viscosity supersolution. To be precise, we intend to prove
((i)) from (2.1).

For the case that $p>2$, we know from (1.1) that $\alpha_{p}>0$ and
$\alpha_{q}>0$. Suppose that the function $u$ satisfies the asymptotic mean
value formula in the viscosity sense. Recalling (2.1), we have

$\displaystyle 0$
$\displaystyle\geq-\phi(x)+\dfrac{\alpha_{p}+M_{\phi}(x)\alpha_{q}}{2(1+M_{\phi}(x))}\left\\{\mathop{\rm{max}}\limits_{\overline{B_{\varepsilon}(x)}}\phi+\mathop{\rm{min}}\limits_{\overline{B_{\varepsilon}(x)}}\phi\right\\}+\dfrac{\beta_{p}+M_{\phi}(x)\beta_{q}}{1+M_{\phi}(x)}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}\phi(y)dy$
$\displaystyle\quad+\dfrac{\varepsilon\lvert\nabla\phi(x)\rvert^{q-p}}{4(N+p)(1+M_{\phi}(x))}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}\phi(y+\varepsilon\nabla
a(x))-\phi(y-\varepsilon\nabla a(x))dy+o(\varepsilon^{2}).$

From $\alpha_{p}+\beta_{p}=1,\alpha_{q}+\beta_{q}=1$, we write

$\displaystyle 0$
$\displaystyle\geq\dfrac{\alpha_{p}+M_{\phi}(x)\alpha_{q}}{1+M_{\phi}(x)}\@slowromancap
i@+\dfrac{\beta_{p}+M_{\phi}(x)\beta_{q}}{1+M_{\phi}(x)}\@slowromancap
ii@+\dfrac{\varepsilon\lvert\nabla\phi(x)\rvert^{q-p}}{4(N+p)(1+M_{\phi}(x))}\@slowromancap
iii@+o(\varepsilon^{2}),$

where

$\displaystyle\@slowromancap
i@=-\phi(x)+\frac{1}{2}\left\\{\mathop{\rm{max}}\limits_{\overline{B_{\varepsilon}(x)}}\phi+\mathop{\rm{min}}\limits_{\overline{B_{\varepsilon}(x)}}\phi\right\\},$
$\displaystyle\@slowromancap
ii@=-\phi(x)+\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}\phi(y)dy,$
$\displaystyle\@slowromancap
iii@=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}\phi(y+\varepsilon\nabla
a(x))-\phi(y-\varepsilon\nabla a(x))dy.$

The non-negativity of $M_{\phi}(x)$ implies that

$\displaystyle 0$ $\displaystyle\geq\alpha_{p}\@slowromancap
i@+\beta_{p}\@slowromancap ii@+M_{\phi}(x)\left(\alpha_{q}\@slowromancap
i@+\beta_{q}\@slowromancap
ii@\right)+\dfrac{\varepsilon\lvert\nabla\phi(x)\rvert^{q-p}}{4(N+p)}\@slowromancap
iii@+o(\varepsilon^{2}).$

It follows from Lemmas 2.3, 2.5 and 2.6 that

$\displaystyle 0$
$\displaystyle\geq\frac{\alpha_{p}}{2}\left<D^{2}\phi(x)(x^{\varepsilon}_{1}-x),(x^{\varepsilon}_{1}-x)\right>+\dfrac{\varepsilon^{2}\beta_{p}}{2(N+2)}\Delta\phi(x)$
$\displaystyle\quad+M_{\phi}(x)\Bigg{(}\frac{\alpha_{q}}{2}\langle
D^{2}\phi(x)(x^{\varepsilon}_{1}-x),(x^{\varepsilon}_{1}-x)\rangle+\left.\dfrac{\varepsilon^{2}\beta_{q}}{2(N+2)}\Delta\phi(x)\right)$
$\displaystyle\quad+\dfrac{\varepsilon^{2}\lvert\nabla\phi(x)\rvert^{q-p}}{2(N+p)}\left<\nabla\phi(x),\nabla
a(x)\right>+o(\varepsilon^{2}).$

Dividing by $\frac{\varepsilon^{2}}{2}$, taking the limit as
$\varepsilon\rightarrow 0$ and by Lemma 2.4, we have

$\displaystyle 0$
$\displaystyle\geq\alpha_{p}\Delta_{\infty}\phi(x)+\dfrac{\beta_{p}}{N+2}\Delta\phi(x)+M_{\phi}(x)\left(\alpha_{q}\Delta_{\infty}\phi(x)+\dfrac{\beta_{q}}{N+2}\Delta\phi(x)\right)$
$\displaystyle\quad+\dfrac{\lvert\nabla\phi(x)\rvert^{q-p}}{N+p}\langle\nabla\phi(x),\nabla
a(x)\rangle.$

Multipling by $N+p$, we get

$\displaystyle 0$
$\displaystyle\geq(p-2)\Delta_{\infty}\phi(x)+\Delta\phi(x)+a(x)\lvert\nabla\phi(x)\rvert^{q-p}((q-2)\Delta_{\infty}\phi(x)+\Delta\phi(x))$
$\displaystyle\quad+\lvert\nabla\phi(x)\rvert^{q-p}\langle\nabla
a(x),\nabla\phi(x)\rangle.$

Therefore, $u$ is a viscosity supersolution according to ((i)). We can use
(2.7) instead of (2.6) to prove that $u$ is a viscosity subsolution and we
omit the proof.

For the necessity of the theorem, we need to prove that $u$ satisfies the
asymptotic mean value formula in the viscosity sense if $u$ is a viscosity
solution to Eq. (1.1). Assume that $u$ is a viscosity solution to Eq. (1.1).
In particular, $u$ is a viscosity subsolution. From (2.5), we have

$\displaystyle 0$
$\displaystyle\leq(p-2)\Delta_{\infty}\phi(x)+\Delta\phi(x)+a(x)\lvert\nabla\phi(x)\rvert^{q-p}((q-2)\Delta_{\infty}\phi(x)+\Delta\phi(x))$
$\displaystyle\quad+\lvert\nabla\phi(x)\rvert^{q-p}\left<\nabla
a(x),\nabla\phi(x)\right>.$

By Lemma 2.4,

$\displaystyle 0$
$\displaystyle\leq(p-2)\left<D^{2}\phi(x)\left(\dfrac{x^{\varepsilon}_{1}-x}{\varepsilon}\right),\left(\dfrac{x^{\varepsilon}_{1}-x}{\varepsilon}\right)\right>+\Delta\phi(x)$
$\displaystyle\quad+a(x)\lvert\nabla\phi(x)|^{q-p}\left((q-2)\left<D^{2}\phi(x)\left(\dfrac{x^{\varepsilon}_{1}-x}{\varepsilon}\right)\right.\right.,\left(\dfrac{x^{\varepsilon}_{1}-x}{\varepsilon}\right)\bigg{>}+\Delta\phi(x)\bigg{)}$
$\displaystyle\quad+\lvert\nabla\phi(x)|^{q-p}\langle\nabla
a(x),\nabla\phi(x)\rangle+o(1).$

Multipling by $\varepsilon^{2}$ on the inequality above, we get

$\displaystyle 0$ $\displaystyle\leq(p-2)\langle
D^{2}\phi(x)\left(x^{\varepsilon}_{1}-x\right),\left(x^{\varepsilon}_{1}-x\right)\rangle+\varepsilon^{2}\Delta\phi(x)$
$\displaystyle\quad+a(x)\lvert\nabla\phi(x)|^{q-p}\left((q-2)\langle
D^{2}\phi(x)\left(x^{\varepsilon}_{1}-x\right),\left(x^{\varepsilon}_{1}-x\right)\rangle+\varepsilon^{2}\Delta\phi(x)\right)$
$\displaystyle\quad+\varepsilon^{2}\lvert\nabla\phi(x)|^{q-p}\langle\nabla
a(x),\nabla\phi(x)\rangle+o(\varepsilon^{2}).$

By Lemmas 2.3, 2.5 and 2.6, we have

$\displaystyle 0$ $\displaystyle\leq 2(p-2)\@slowromancap
i@+2(N+2)\@slowromancap
ii@+a(x)\lvert\nabla\phi(x)|^{q-p}(2(q-2)\@slowromancap
i@+2(N+2)\@slowromancap ii@)$
$\displaystyle\quad+\frac{\varepsilon}{2}\lvert\nabla\phi(x)|^{q-p}\@slowromancap
iii@+o(\varepsilon^{2}).$

Furthermore, dividing by $2(N+p)$, we obtain

$\displaystyle 0$ $\displaystyle\leq\left(\alpha_{p}\@slowromancap
i@+\beta_{p}\@slowromancap
ii@\right)+a(x)\dfrac{N+q}{N+p}\lvert\nabla\phi(x)|^{q-p}\left(\alpha_{q}\@slowromancap
i@+\beta_{q}\@slowromancap
ii@\right)+\dfrac{\varepsilon\lvert\nabla\phi(x)\rvert^{q-p}}{4(N+p)}\@slowromancap
iii@+o(\varepsilon^{2}).$

Then separating $\phi(x)$ from $\@slowromancap i@$ and $\@slowromancap ii@$,
we get

$\displaystyle(1+M_{\phi}(x))\phi(x)$
$\displaystyle\leq\dfrac{\alpha_{p}+M_{\phi}(x)\alpha_{q}}{2}\left\\{\mathop{\rm{max}}\limits_{\overline{B_{\varepsilon}(x)}}\phi+\mathop{\rm{min}}\limits_{\overline{B_{\varepsilon}(x)}}\phi\right\\}$
$\displaystyle\quad+\left(\beta_{p}+M_{\phi}(x)\beta_{q}\right)\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}\phi(y)dy$
$\displaystyle\quad+\dfrac{\varepsilon\lvert\nabla\phi(x)\rvert^{q-p}}{4(N+p)}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}\phi(y+\varepsilon\nabla
a(x))-\phi(y-\varepsilon\nabla a(x))dy+o(\varepsilon^{2}).$

Thus

$\displaystyle 0$
$\displaystyle\leq-\phi(x)+\dfrac{\alpha_{p}+M_{\phi}(x)\alpha_{q}}{2(1+M_{\phi}(x))}\left\\{\mathop{\rm{max}}\limits_{\overline{B_{\varepsilon}(x)}}\phi+\mathop{\rm{min}}\limits_{\overline{B_{\varepsilon}(x)}}\phi\right\\}+\dfrac{\beta_{p}+M_{\phi}(x)\beta_{q}}{1+M_{\phi}(x)}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}\phi(y)dy$
$\displaystyle\quad+\dfrac{\varepsilon\lvert\nabla\phi(x)\rvert^{q-p}}{4(N+p)(1+M_{\phi}(x))}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}\phi(y+\varepsilon\nabla
a(x))-\phi(y-\varepsilon\nabla a(x))dy+o(\varepsilon^{2}).$

Similarly, we can also prove that $u$ satisfies (2.2) if $u$ is the viscosity
supersolution.

When $1<p\leq 2$, we can divide it into the following cases: $p=2,q>2$;
$p=2,q=2$; $1<p<2,q=2$; $1<p<2,q>2$; $1<q<2$. The proofs of these cases are
similar to the case that $p>2$, by using (2.7) instead of (2.6) in Lemma 2.3
if necessary.

Combining the arguments above, we complete the proof. ∎

From the proof of Theorem 1.1, the following corollaries will follow.

######  Corollary 3.1 ($p$-Laplace equation).

Let $1<p<\infty$ and $u(x)$ be a continuous function in a domain
$\Omega\subset\mathbb{R}^{N}$. The equation

$-\text{\rm{div}}(\lvert\nabla u\rvert^{p-2}\nabla u)=0\quad\text{in }\Omega$

holds in the viscosity sense if and only if the asymptotic expansion

$\displaystyle u(x)$
$\displaystyle=\dfrac{\alpha_{p}}{2}\left\\{\mathop{\rm{max}}\limits_{\overline{B_{\varepsilon}(x)}}u+\mathop{\rm{min}}\limits_{\overline{B_{\varepsilon}(x)}}u\right\\}+\beta_{p}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}u(y)dy+o(\varepsilon^{2})\quad\text{as
}\varepsilon\rightarrow 0$

holds for all $x\in\Omega$ in the viscosity sense. Here
$\alpha_{p}+\beta_{p}=1,\frac{\alpha_{p}}{\beta_{p}}=\frac{p-2}{N+2}.$

######  Remark 3.2.

Corollary 3.1 is the main result in [26]. In fact, Corollary 3.1 also holds
for $p=\infty$ with $\alpha_{p}=1$, $\beta_{p}=0$.

######  Corollary 3.3 (Variable coefficient $p$-Laplace equation).

Let $1<p<\infty$, $\tilde{a}(x)$ be a $C^{1}$ function in a domain
$\Omega\subset\mathbb{R}^{N}$ with $\tilde{a}(x)\geq 1$ and let $u(x)$ be a
continuous function in $\Omega$. The equation

$\displaystyle-\text{\rm{div}}(\tilde{a}(x)\lvert\nabla u\rvert^{p-2}\nabla
u)=0\quad\text{in }\Omega$

holds in the viscosity sense if and only if the asymptotic expansion

$\displaystyle u(x)$
$\displaystyle=\dfrac{\alpha_{p}}{2}\left\\{\mathop{\rm{max}}\limits_{\overline{B_{\varepsilon}(x)}}u+\mathop{\rm{min}}\limits_{\overline{B_{\varepsilon}(x)}}u\right\\}+\beta_{p}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}u(y)dy$
$\displaystyle\quad+\dfrac{\varepsilon}{4(N+p)\tilde{a}(x)}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}u(y+\varepsilon\nabla\tilde{a}(x))-u(y-\varepsilon\nabla\tilde{a}(x))dy+o(\varepsilon^{2})$

as $\varepsilon\rightarrow 0$, holds for all $x\in\Omega$ in the viscosity
sense. Here
$\alpha_{p}+\beta_{p}=1,\frac{\alpha_{p}}{\beta_{p}}=\frac{p-2}{N+2}$.

###### Proof.

When $q=p$ in Eq. (1.1), we have

$-\text{\rm{div}}((a(x)+1)\lvert\nabla u\rvert^{p-2}\nabla u)=0.$

In this situation, we have

$\alpha_{p}=\alpha_{q},\quad\beta_{p}=\beta_{q},\quad M_{u}(x)=a(x).$

Thus, the asymptotic mean value formula (1.1) reads as

$\displaystyle u(x)$
$\displaystyle=\dfrac{\alpha_{p}+M_{u}(x)\alpha_{q}}{2(1+M_{u}(x))}\left\\{\mathop{\rm{max}}\limits_{\overline{B_{\varepsilon}(x)}}u+\mathop{\rm{min}}\limits_{\overline{B_{\varepsilon}(x)}}u\right\\}+\dfrac{\beta_{p}+M_{u}(x)\beta_{q}}{1+M_{u}(x)}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}u(y)dy$
$\displaystyle\quad+\dfrac{\varepsilon\lvert\nabla
u(x)\rvert^{q-p}}{4(N+p)(1+M_{u}(x))}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}u(y+\varepsilon\nabla
a(x))-u(y-\varepsilon\nabla a(x))dy+o(\varepsilon^{2})$
$\displaystyle=\dfrac{\alpha_{p}}{2}\left\\{\mathop{\rm{max}}\limits_{\overline{B_{\varepsilon}(x)}}u+\mathop{\rm{min}}\limits_{\overline{B_{\varepsilon}(x)}}u\right\\}+\beta_{p}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}u(y)dy$
$\displaystyle\quad+\dfrac{\varepsilon}{4(N+p)(1+a(x))}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}u(y+\varepsilon\nabla
a(x))-u(y-\varepsilon\nabla a(x))dy+o(\varepsilon^{2}).$

Let $\tilde{a}(x)=a(x)+1$. We finish the proof. ∎

######  Remark 3.4.

In fact, the condition $\tilde{a}(x)\geq 1$ can be replaced by
$\tilde{a}(x)>0$ and we can prove the conclusion by the same method as in
Theorem 1.1.

Finally, we consider the $p(x)$-Laplace equation

$\displaystyle-\text{\rm{div}}(\lvert\nabla u\rvert^{p(x)-2}\nabla
u)=0\quad\text{in }\Omega.$ (3.1)

Let us formally expand the left-hand side of Eq. (3.1) as follows:

$\displaystyle\quad\text{\rm{div}}(\lvert\nabla u\rvert^{p(x)-2}\nabla u)$
$\displaystyle=\sum_{i=1}^{N}\partial_{i}(\lvert\nabla u\rvert^{p(x)-2}u_{i})$
$\displaystyle=\sum_{i=1}^{N}\partial_{i}(\lvert\nabla
u\rvert^{p(x)-2})u_{i}+\lvert\nabla u\rvert^{p(x)-2}\Delta u$
$\displaystyle=\sum_{i=1}^{N}\partial_{i}(e^{(p(x)-2)\ln\lvert\nabla
u\rvert})u_{i}+\lvert\nabla u\rvert^{p(x)-2}\Delta u$
$\displaystyle=\sum_{i=1}^{N}\lvert\nabla u\rvert^{p(x)-2}(\ln\lvert\nabla
u\rvert\partial_{i}p+(p(x)-2)\partial_{i}\ln\lvert\nabla
u\rvert)u_{i}+\lvert\nabla u\rvert^{p(x)-2}\Delta u$
$\displaystyle=\sum_{i=1}^{N}\lvert\nabla u\rvert^{p(x)-2}(\ln\lvert\nabla
u\rvert p_{i}+(p(x)-2)\lvert\nabla
u\rvert^{-2}\sum_{j=1}^{N}u_{ij}u_{j})u_{i}+\lvert\nabla
u\rvert^{p(x)-2}\Delta u$ $\displaystyle=\lvert\nabla
u\rvert^{p(x)-2}\left(\ln\lvert\nabla u\rvert\langle\nabla p,\nabla
u\rangle+(p(x)-2)\Delta_{\infty}u+\Delta u)\right).$

Suppose that $u$ is a smooth function with $\nabla u\neq 0$, we can see that
$u$ is a solution to Eq. (3.1) if and only if

$-(p(x)-2)\Delta_{\infty}u-\Delta u-\ln\lvert\nabla u\rvert\langle\nabla
p,\nabla u\rangle=0.$

We find that the term $\langle\nabla p,\nabla u\rangle$ appears in the
equation above. We can still use Lemma 2.6 to obtain the asymptotic mean value
formula by the same method as in Theorem 1.1, which is given as the following
theorem without proof.

######  Theorem 3.5 ($p(x)$-Laplace equation).

Let $p(x)$ be a $C^{1}$ function in a domain $\Omega\subset\mathbb{R}^{N}$
with $1<p(x)<\infty$ and $u(x)$ be a continuous function in $\Omega$. Then Eq.
(3.1) holds in the viscosity sense if and only if the asymptotic expansion

$\displaystyle u(x)$
$\displaystyle=\dfrac{\alpha_{p}(x)}{2}\left\\{\mathop{\rm{max}}\limits_{\overline{B_{\varepsilon}(x)}}u+\mathop{\rm{min}}\limits_{\overline{B_{\varepsilon}(x)}}u\right\\}+\beta_{p}(x)\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}u(y)dy$
$\displaystyle\quad+\dfrac{\varepsilon\ln\lvert\nabla
u(x)\rvert}{4(N+p(x))}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}u(y+\varepsilon\nabla
p(x))-u(y-\varepsilon\nabla p(x))dy+o(\varepsilon^{2})$

as $\varepsilon\rightarrow 0$, holds for all $x\in\Omega$ in the viscosity
sense. Here
$\alpha_{p}(x)+\beta_{p}(x)=1,\frac{\alpha_{p}(x)}{\beta_{p}(x)}=\frac{p(x)-2}{N+2}$.

##  4\. Parabolic case

In this section, we start from the definition of viscosity solutions to the
normalized double phase parabolic equation, and combine the ideas in [27] to
investigate the possible form of the mean value formula. We integrate the
terms with $p$ and $q$ over different time intervals, and find that when these
two time lags satisfy the viscosity condition, the mean value formula holds.

We first give the definition of viscosity solutions to Eq. (1.4). The similar
definition can be found in [27, Definition 1].

######  Definition 4.1.

A function $u:\Omega_{T}\rightarrow\mathbb{R}$ is a viscosity solution to
(1.4) if $u$ is continuous and whenever $(x,t)\in\Omega_{T}$ and $\phi\in
C^{2}(\Omega_{T})$ is such that

  * (i)

$u(x,t)=\phi(x,t)$.

  * (ii)

$u(y,s)>\phi(y,s)$ for all $(y,s)\in\Omega_{T},(y,s)\neq(x,t)$, then we have
at the point $(x,t)$

$\begin{cases}\phi_{t}\geq(p-2)\Delta_{\infty}\phi+\Delta\phi+a\lvert\nabla\phi|^{q-p}\left((q-2)\Delta_{\infty}\phi+\Delta\phi\right)\\\
\qquad+\lvert\nabla\phi|^{q-p}\langle\nabla a,\nabla\phi\rangle&\text{ if
}\nabla\phi(x,t)\neq 0,\\\
\phi_{t}\geq\lambda_{\min}((p-2)D^{2}\phi)+\Delta\phi&\text{ if
}\nabla\phi(x,t)=0.\end{cases}$

In addition, when the test function $\phi$ touches $u$ from above, all
inequalities are reversed and $\lambda_{\min}((p-2)D^{2}\phi)$ is replaced by
$\lambda_{\max}((p-2)D^{2}\phi)$.

In fact, the number of test functions $\phi$ can be reduced, if the gradient
of a test function $\phi$ vanishes, we can suppose $D^{2}\phi=0$. Nothing is
required if $\nabla\phi=0$ and $D^{2}\phi\neq 0$. We state the following lemma
without proof (see [27, Lemma 2] for details).

######  Lemma 4.2.

A function $u:\Omega_{T}\rightarrow\mathbb{R}$ is a viscosity solution to Eq.
(1.4) if $u$ is continuous and whenever $(x,t)\in\Omega_{T}$ and $\phi\in
C^{2}(\Omega_{T})$ is such that

  * (i)

$u(x,t)=\phi(x,t)$.

  * (ii)

$u(y,s)>\phi(y,s)$ for all $(y,s)\in\Omega_{T},(y,s)\neq(x,t)$, then at the
point $(x,t)$, if $\nabla\phi(x,t)\neq 0$, we have

$\displaystyle\begin{split}\phi_{t}&\geq(p-2)\Delta_{\infty}\phi+\Delta\phi+a\lvert\nabla\phi|^{q-p}\left((q-2)\Delta_{\infty}\phi+\Delta\phi\right)\\\
&\quad+\lvert\nabla\phi|^{q-p}\langle\nabla a,\nabla\phi\rangle;\end{split}$
(4.1)

if $\nabla\phi(x,t)=0$ and $D^{2}\phi(x,t)=0$, we have

$\displaystyle\phi_{t}(x,t)\geq 0.$ (4.2)

In addition, when the test function $\phi$ touches $u$ from above, all
inequalities are reversed.

The definition of $u$ satisfying the asymptotic mean value formula (1.4) at
the point $(x,t)$ in the viscosity sense is similar to Definition 2.1, so we
omit it. But it is worth to mentioning that $\nabla\phi(x,t)=0$ is allowed in
the parabolic case, which is consistent with Definition 4.1.

Similar to the elliptic case, we also need the following lemmas. The ideas of
Lemmas 4.3–4.5 come from [27, Section 3] and the proofs are similar.

######  Lemma 4.3.

Let $\phi$ be a $C^{2}$ function in a neighborhood of $(x,t)$,
$\varepsilon>0,A(x,t)>0,s\in(t-\frac{\varepsilon^{2}}{A(x,t)},t)$. Denote by
$x^{\varepsilon,s}_{1},x^{\varepsilon,s}_{2}$ points in which $\phi$ attains
its minimum and maximum over a ball $\overline{B_{\varepsilon}(x)}$ at time
$s$ respectively. We have

$\displaystyle\begin{split}&\quad\dfrac{1}{2}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{t-\frac{\varepsilon^{2}}{A(x,t)}}^{t}\left\\{\mathop{\rm{max}}\limits_{y\in\overline{B_{\varepsilon}(x)}}\phi(y,s)+\mathop{\rm{min}}\limits_{y\in\overline{B_{\varepsilon}(x)}}\phi(y,s)\right\\}ds-\phi(x,t)\\\
&\geq\frac{1}{2}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{t-\frac{\varepsilon^{2}}{A(x,t)}}^{t}\left<D^{2}\phi(x,t)(x^{\varepsilon,s}_{1}-x),(x^{\varepsilon,s}_{1}-x)\right>ds-\frac{\varepsilon^{2}}{2A(x,t)}\phi_{t}(x,t)+o(\varepsilon^{2})\end{split}$
(4.3)

and

$\displaystyle\begin{split}&\quad\dfrac{1}{2}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{t-\frac{\varepsilon^{2}}{A(x,t)}}^{t}\left\\{\max\limits_{y\in\overline{B_{\varepsilon}(x)}}\phi(y,s)+\min\limits_{y\in\overline{B_{\varepsilon}(x)}}\phi(y,s)\right\\}ds-\phi(x,t)\\\
&\leq\frac{1}{2}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{t-\frac{\varepsilon^{2}}{A(x,t)}}^{t}\left<D^{2}\phi(x,t)(x^{\varepsilon,s}_{2}-x),(x^{\varepsilon,s}_{2}-x)\right>ds-\frac{\varepsilon^{2}}{2A(x,t)}\phi_{t}(x,t)+o(\varepsilon^{2}).\end{split}$
(4.4)

######  Lemma 4.4.

Let $\phi$ be a $C^{2}$ function in a neighborhood of $(x,t)$ with
$\nabla\phi(x,t)\neq 0$. We have

$\displaystyle\lim\limits_{\varepsilon\rightarrow
0+}\frac{x^{\varepsilon,s}_{1}-x}{\varepsilon}=-\dfrac{\nabla\phi}{\lvert\nabla\phi\rvert}(x,t),$
(4.5)

where $x^{\varepsilon,s}_{1}$ is defined as in Lemma 4.3.

######  Lemma 4.5.

Let $\phi$ be a $C^{2}$ function in a neighborhood of $(x,t)$, $s$ and
$A(x,t)$ are defined as in Lemma 4.3. Then

$\displaystyle\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{t-\frac{\varepsilon^{2}}{A(x,t)}}^{t}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}\phi(y,s)dyds-\phi(x,t)=\dfrac{\varepsilon^{2}}{2(N+2)}\Delta\phi(x,t)-\dfrac{\varepsilon^{2}}{2A(x,t)}\phi_{t}(x,t)+o(\varepsilon^{2}).$

######  Lemma 4.6.

Let $\phi$ be a $C^{2}$ function in a neighborhood of $(x,t)$. We have

$\displaystyle\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}\phi(y+\varepsilon\nabla
a(x,t),t)-\phi(y-\varepsilon\nabla
a(x,t),t)dy=2\varepsilon\left<\nabla\phi(x,t),\nabla
a(x,t)\right>+o(\varepsilon^{2}).$

Now we are ready to prove the second main result.

###### Proof of Theorem 1.4.

We first prove the sufficiency. If $u$ satisfies the asymptotic mean value
formula 1.4 in the viscosity sense, we need to prove that $u$ is a viscosity
solution. If so, for a test function $\phi$, we have

$\displaystyle 0$
$\displaystyle\geq-\phi(x,t)+\dfrac{1}{1+M_{\phi}(x,t)}\left(\dfrac{\alpha_{p}}{2}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{t-\frac{\varepsilon^{2}}{A_{\phi}(x,t)}}^{t}\left\\{\max\limits_{y\in\overline{B_{\varepsilon}(x)}}\phi(y,s)+\min\limits_{y\in\overline{B_{\varepsilon}(x)}}\phi(y,s)\right\\}ds\right.$
$\displaystyle\quad\left.+\beta_{p}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{t-\frac{\varepsilon^{2}}{A_{\phi}(x,t)}}^{t}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}\phi(y,s)dyds\right)$
$\displaystyle\quad+\dfrac{M_{\phi}(x,t)}{1+M_{\phi}(x,t)}\left(\dfrac{\alpha_{q}}{2}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{t-\frac{\varepsilon^{2}}{B_{\phi}(x,t)}}^{t}\left\\{\max\limits_{y\in\overline{B_{\varepsilon}(x)}}\phi(y,s)+\min\limits_{y\in\overline{B_{\varepsilon}(x)}}\phi(y,s)\right\\}ds\right.$
$\displaystyle\quad\left.+\beta_{q}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{t-\frac{\varepsilon^{2}}{B_{\phi}(x,t)}}^{t}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}\phi(y,s)dyds\right)$
$\displaystyle\quad+\frac{\varepsilon\lvert\nabla\phi(x,t)\rvert^{q-p}}{4(N+p)(1+M_{\phi}(x,t))}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}\phi(y+\varepsilon\nabla
a(x,t),t)-\phi(y-\varepsilon\nabla a(x,t),t)dy$
$\displaystyle\quad+o(\varepsilon^{2}).$

By the non-negativity of $M_{\phi}(x,t)$ and splitting $\phi(x,t)$, we get

$\displaystyle 0$
$\displaystyle\geq\alpha_{p}\left(\dfrac{1}{2}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{t-\frac{\varepsilon^{2}}{A_{\phi}(x,t)}}^{t}\left\\{\max\limits_{y\in\overline{B_{\varepsilon}(x)}}\phi(y,s)+\min\limits_{y\in\overline{B_{\varepsilon}(x)}}\phi(y,s)\right\\}ds-\phi(x,t)\right)$
$\displaystyle\quad+\beta_{p}\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{t-\frac{\varepsilon^{2}}{A_{\phi}(x,t)}}^{t}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}\phi(y,s)dyds-\phi(x,t)\right)$
$\displaystyle\quad+M_{\phi}(x,t)\left[\alpha_{q}\left(\dfrac{1}{2}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{t-\frac{\varepsilon^{2}}{B_{\phi}(x,t)}}^{t}\left\\{\max\limits_{y\in\overline{B_{\varepsilon}(x)}}\phi(y,s)+\min\limits_{y\in\overline{B_{\varepsilon}(x)}}\phi(y,s)\right\\}ds-\phi(x,t)\right)\right.$
$\displaystyle\quad\left.+\beta_{q}\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{t-\frac{\varepsilon^{2}}{B_{\phi}(x,t)}}^{t}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}\phi(y,s)dyds-\phi(x,t)\right)\right]$
$\displaystyle\quad+\frac{\varepsilon\lvert\nabla\phi(x,t)\rvert^{q-p}}{4(N+p)}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}\phi(y+\varepsilon\nabla
a(x,t),t)-\phi(y-\varepsilon\nabla a(x,t),t)dy+o(\varepsilon^{2}),$ (4.6)

where
$\alpha_{p},\alpha_{q},\beta_{p},\beta_{q},M_{\phi}(x,t),A_{\phi}(x,t),B_{\phi}(x,t)$
are determined by (1.6).

Assume that $p>2$, where $\alpha_{p},\alpha_{q}>0$. For inequality (4), we
apply Lemmas 4.3, 4.5 and Lemma 4.6 to have

$\displaystyle 0$
$\displaystyle\geq\frac{\alpha_{p}}{2}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{t-\frac{\varepsilon^{2}}{A_{\phi}(x,t)}}^{t}\left<D^{2}\phi(x,t)(x^{\varepsilon,s}_{1}-x),(x^{\varepsilon,s}_{1}-x)\right>ds+\dfrac{\varepsilon^{2}\beta_{p}}{2(N+2)}\Delta\phi(x,t)$
$\displaystyle\quad-\dfrac{\varepsilon^{2}}{2A_{\phi}(x,t)}\phi_{t}(x,t)+M_{\phi}(x,t)\Bigg{(}\frac{\alpha_{q}}{2}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{t-\frac{\varepsilon^{2}}{B_{\phi}(x,t)}}^{t}\left<D^{2}\phi(x,t)(x^{\varepsilon,s}_{1}-x),(x^{\varepsilon,s}_{1}-x)\right>ds$
$\displaystyle\quad+\dfrac{\varepsilon^{2}\beta_{q}}{2(N+2)}\Delta\phi(x,t)-\dfrac{\varepsilon^{2}}{2B_{\phi}(x,t)}\phi_{t}(x,t)\Bigg{)}+\dfrac{\varepsilon^{2}\lvert\nabla\phi(x,t)\rvert^{q-p}}{2(N+p)}\left<\nabla\phi(x,t),\nabla
a(x,t)\right>$ $\displaystyle\quad+o(\varepsilon^{2}).$

When $\nabla\phi(x,t)\neq 0$, multiplying by $\dfrac{2}{\varepsilon^{2}}$ and
taking the limit as $\varepsilon\rightarrow 0$ on the inequality above, by
Lemma 4.4, we have

$\displaystyle 0$
$\displaystyle\geq\alpha_{p}\Delta_{\infty}\phi(x,t)+\dfrac{\beta_{p}}{N+2}\Delta\phi(x,t)-\frac{1}{A_{\phi}(x,t)}\phi_{t}(x,t)$
$\displaystyle\quad+M_{\phi}(x,t)\left(\alpha_{q}\Delta_{\infty}\phi(x,t)+\dfrac{\beta_{q}}{N+2}\Delta\phi(x,t)-\frac{1}{B_{\phi}(x,t)}\phi_{t}(x,t)\right)$
$\displaystyle\quad+\dfrac{\lvert\nabla\phi(x,t)\rvert^{q-p}}{N+p}\left<\nabla\phi(x,t),\nabla
a(x,t)\right>.$

Multipling by $N+p$ again, we get

$\displaystyle 0$
$\displaystyle\geq(p-2)\Delta_{\infty}\phi(x,t)+\Delta\phi(x,t)+a(x,t)\lvert\nabla\phi(x,t)\rvert^{q-p}((q-2)\Delta_{\infty}\phi(x,t)$
$\displaystyle\quad+\Delta\phi(x,t))-\left(\dfrac{N+p}{A_{\phi}(x,t)}+\dfrac{a(x,t)(N+q)\lvert\nabla\phi(x,t)\rvert^{q-p}}{B_{\phi}(x,t)}\right)\phi_{t}(x,t)$
$\displaystyle\quad+\lvert\nabla\phi(x,t)\rvert^{q-p}\left<\nabla\phi(x,t),\nabla
a(x,t)\right>.$

Recalling (1.6), we have

$\displaystyle\dfrac{N+p}{A_{\phi}(x,t)}+\dfrac{a(x,t)(N+q)\lvert\nabla\phi(x,t)\rvert^{q-p}}{B_{\phi}(x,t)}=1.$
(4.7)

Therefore, we obtain

$\phi_{t}\geq(p-2)\Delta_{\infty}\phi+\Delta\phi+a\lvert\nabla\phi|^{q-p}\left((q-2)\Delta_{\infty}\phi+\Delta\phi\right)+\lvert\nabla\phi|^{q-p}\langle\nabla
a,\nabla\phi\rangle.$

It follows that (4.1) holds when $\nabla\phi(x,t)\neq 0$. When
$\nabla\phi(x,t)=0$ and $D^{2}\phi(x,t)=0$, by (4.7), we get
$A_{\phi}(x,t)=N+p$ and $M_{\phi}(x,t)=0$. According to the asymptotic mean
value formula, we have

$\displaystyle 0$
$\displaystyle\geq-\phi(x,t)+\frac{\alpha_{p}}{2}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{t-\frac{\varepsilon^{2}}{N+p}}^{t}\left\\{\max\limits_{y\in\overline{B_{\varepsilon}(x)}}\phi(y,s)+\min\limits_{y\in\overline{B_{\varepsilon}(x)}}\phi(y,s)\right\\}ds$
$\displaystyle\quad+\beta_{p}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{t-\frac{\varepsilon^{2}}{N+p}}^{t}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}\phi(y,s)dyds+o(\varepsilon^{2}).$

By Lemma 4.5 and the expansion

$\phi(y,s)-\phi(x,t)=\phi_{t}(x,t)(s-t)+o(|s-t|+|y-x|^{2}),$

we have

$\displaystyle 0$
$\displaystyle\geq\alpha_{p}\left(\frac{1}{2}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{t-\frac{\varepsilon^{2}}{N+p}}^{t}\left\\{\max\limits_{y\in\overline{B_{\varepsilon}(x)}}\phi(y,s)+\min\limits_{y\in\overline{B_{\varepsilon}(x)}}\phi(y,s)\right\\}ds-\phi(x,t)\right)$
$\displaystyle\quad-\frac{\varepsilon^{2}\beta_{p}}{2(N+p)}\phi_{t}(x,t)+o(\varepsilon^{2})$
$\displaystyle=\frac{\alpha_{p}}{2}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{t-\frac{\varepsilon^{2}}{N+p}}^{t}\left\\{\max\limits_{y\in\overline{B_{\varepsilon}(x)}}(\phi(y,s)-\phi(x,t))+\min\limits_{y\in\overline{B_{\varepsilon}(x)}}(\phi(y,s)-\phi(x,t))\right\\}ds$
$\displaystyle\quad-\dfrac{\varepsilon^{2}\beta_{p}}{2(N+p)}\phi_{t}(x,t)+o(\varepsilon^{2})$
$\displaystyle=\alpha_{p}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{t-\frac{\varepsilon^{2}}{N+p}}^{t}\phi_{t}(x,t)(s-t)ds-\frac{\varepsilon^{2}\beta_{p}}{2(N+p)}\phi_{t}(x,t)+o(\varepsilon^{2})$
$\displaystyle=-\frac{\varepsilon^{2}\alpha_{p}}{2(N+p)}\phi_{t}(x,t)-\frac{\varepsilon^{2}\beta_{p}}{2(N+p)}\phi_{t}(x,t)+o(\varepsilon^{2})$
$\displaystyle=-\frac{\varepsilon^{2}}{2(N+p)}\phi_{t}(x,t)+o(\varepsilon^{2}).$

Dividing by $\varepsilon^{2}$ and taking the limit as $\varepsilon\rightarrow
0$, we have

$\phi_{t}(x,t)\geq 0.$

Thus, we prove that $u$ is a viscosity supersolution. We can use the same
method to prove that $u$ is a viscosity subsolution.

For the necessity and other cases, since it is similar to the proof of
elliptic case, so we omit it. The proof is complete. ∎

In particular, we consider the case that $a\equiv 0$. For this case, it
follows from (1.6) that $A_{u}(x,t)=N+p$. Then the following corollary holds.

######  Corollary 4.7 (Normalized parabolic $p$-Laplace equation).

Let $1<p<\infty$ and $u(x,t)$ be a continuous function in a domain
$\Omega_{T}$. The equation

$u_{t}=\lvert\nabla u\rvert^{2-p}\text{\rm{div}}(\lvert\nabla
u\rvert^{p-2}\nabla u)\quad\text{in }\Omega_{T}$

holds in the viscosity sense if and only if the asymptotic expansion

$\displaystyle u(x,t)$
$\displaystyle=\dfrac{\alpha_{p}}{2}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{t-\frac{\varepsilon^{2}}{N+p}}^{t}\left\\{\mathop{\rm{max}}\limits_{y\in\overline{B_{\varepsilon}(x)}}u(y,s)+\mathop{\rm{min}}\limits_{y\in\overline{B_{\varepsilon}(x)}}u(y,s)\right\\}ds$
$\displaystyle\quad+\beta_{p}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{t-\frac{\varepsilon^{2}}{N+p}}^{t}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B_{\varepsilon}(x)}u(y,s)dyds+o(\varepsilon^{2})\quad\text{as
}\varepsilon\rightarrow 0$

holds for all $(x,t)\in\Omega_{T}$ in the viscosity sense. Here
$\alpha_{p}+\beta_{p}=1,\frac{\alpha_{p}}{\beta_{p}}=\frac{p-2}{N+2}.$

######  Remark 4.8.

Corollary 4.7 is the main result in [27]. In fact, Corollary 4.7 also holds
for $p=\infty$ with $\alpha_{p}=1$, $\beta_{p}=0$.

## Acknowledgment

This work was supported by the National Natural Science Foundation of China
(No. 12071098) and the Fundamental Research Funds for the Central Universities
(No. 2022FRFK060022).

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