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

Abstract

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

$$\begin{aligned} -{\textrm{div}}(|\nabla u |^{p-2}\nabla u+ a(x)|\nabla u |^{q-2}\nabla u)=0 \end{aligned}$$

and the normalized double phase parabolic equation

$$\begin{aligned} u_t=|\nabla u |^{2-p}{\textrm{div}}(|\nabla u |^{p-2}\nabla u+ a(x,t)|\nabla u |^{q-2}\nabla u), \quad 1<p\le q<\infty . \end{aligned}$$

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.

1 Introduction

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

$$\begin{aligned} -{\textrm{div}}(|\nabla u |^{p-2}\nabla u+ a(x)|\nabla u |^{q-2}\nabla u)=0 \quad \text {in } \Omega , \end{aligned}$$

(1.1)

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

$$\begin{aligned} W^{1,1}(\Omega )\ni w\mapsto \int _{\Omega }\left( \frac{1}{p}|\nabla w|^p+\frac{a(x)}{q}|\nabla w|^q\right) dx. \end{aligned}$$

Originally, this functional is connected to the Homogenization theory and Lavrentiev phenomenon [24, 30, 35], 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, 14,15,16,17,18,19, 21, 31] 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

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

suffices to characterize harmonic functions (see [6, 25, 34]). Moreover, a nonlinear mean value property was explored in [28] that continuous function u is a viscosity solution of the p-Laplace equation

$$\begin{aligned} -\Delta _pu=-{\textrm{div}}(|\nabla u |^{p-2}\nabla u)=0 \quad \text {in } \Omega \end{aligned}$$

if and only if the asymptotic expansion

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 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 $\ge$ ($\le$) 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 [22, 26, 27, 29, 32, 33]. For more related asymptotic mean value results, we refer to [23] for p-harmonic functions in the Heisenberg group, [3] for Monge-Ampère equation, [4, 29] for the nonlinear parabolic equations and 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 [28, 29], 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 $1<p\le q<\infty$, 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

(1.2)

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

$$\begin{aligned}&\alpha _p+\beta _p=1, \frac{\alpha _p}{\beta _p}=\frac{p-2}{N+2}, \quad \alpha _q+\beta _q=1, \frac{\alpha _q}{\beta _q}=\frac{q-2}{N+2}, \nonumber \\&M_u(x)=a(x)\dfrac{N+q}{N+p}|\nabla u(x) |^{q-p}. \end{aligned}$$

(1.3)

Remark 1.2

Note that here we only require that u is a continuous function. However, $\nabla u$ appears in the formula (1.2) 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 Sect. 2.

Remark 1.3

From the formula (1.2), 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

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 $a(x,t)\ge 0$ be a function that is $C^1$ in the space variable and continuous in the time variable, respectively. We consider the following parabolic equation

$$\begin{aligned} u_t=|\nabla u |^{2-p}{\textrm{div}}(|\nabla u |^{p-2}\nabla u+ a(x,t)|\nabla u |^{q-2}\nabla u)\quad {\textrm{in }}\ \Omega _T, \end{aligned}$$

(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 $1<p\le q<\infty$, the positive 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

(1.5)

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

$$\begin{aligned}&\alpha _p+\beta _p=1, \frac{\alpha _p}{\beta _p}=\dfrac{p-2}{N+2}, \quad \alpha _q+\beta _q=1, \frac{\alpha _q}{\beta _q}=\dfrac{q-2}{N+2},\nonumber \\&M_u(x,t)=a(x,t)\frac{N+q}{N+p}|\nabla u(x,t) |^{q-p}, \nonumber \\&\frac{N+p}{A_u(x,t)}+\frac{a(x,t)(N+q)|\nabla u(x,t) |^{q-p}}{B_u(x,t)}=1, \quad A_u(x,t), B_u(x,t)>0. \end{aligned}$$

(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

$$\begin{aligned} \frac{N+p}{A_u(x,t)}+\frac{a(x,t)(N+q)|\nabla u(x,t) |^{q-p}}{B_u(x,t)}=1 \end{aligned}$$

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

This manuscript is organized as follows. In Sect. 2, we introduce the basic definitions and give some necessary lemmas that will be used later. In Sect. 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 Sect. 4.

2 Preliminaries

In this section, inspired by the ideas developed in [28], 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

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 $\nabla \phi (x)\ne 0$, we have

  

  (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 $\nabla \phi (x)\ne 0$, we have

  

  (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:

$$\begin{aligned}&{\textrm{div}}(|\nabla u |^{p-2}\nabla u+ a(x)|\nabla u |^{q-2}\nabla u)\\&\quad =|\nabla u |^{p-2}((p-2)\Delta _{\infty }^N u+\Delta u)+a(x)|\nabla u |^{q-2}((q-2)\Delta _{\infty }^N u+\Delta u)\\&\quad \quad +|\nabla u |^{q-2}\langle \nabla a, \nabla u\rangle , \end{aligned}$$

where $\Delta _{\infty }^N u=|\nabla u|^{-2}\langle D^2u\nabla u,\nabla u\rangle$.

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

$$\begin{aligned}&-(p-2)\Delta _{\infty }^N u-\Delta u-a(x)|\nabla u |^{q-p}((q-2)\Delta _{\infty }^N u+\Delta u)\nonumber \\&-|\nabla u |^{q-p}\langle \nabla a, \nabla u\rangle =0. \end{aligned}$$

(2.3)

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

Definition 2.2

([20], Definition 2.5) Let $1<p\le q<\infty$ and consider the equation

$$\begin{aligned} -{\textrm{div}}(|\nabla u |^{p-2}\nabla u+a(x)|\nabla u |^{q-2}\nabla u)=0. \end{aligned}$$

• (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 $\nabla \phi (x)\ne 0$ we have

  $$\begin{aligned}&\quad -\left( (p-2)\Delta _{\infty }^N \phi (x)+\Delta \phi (x)\right) -a(x)|\nabla \phi (x)\vert ^{q-p}\left( (q-2)\Delta _{\infty }^N \phi (x)+\Delta \phi (x)\right) \nonumber \\&\quad -|\nabla \phi (x)\vert ^{q-p}\langle \nabla a(x),\nabla \phi (x)\rangle \ge 0. \end{aligned}$$

  (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 $\nabla \phi (x)\ne 0$ we have

  $$\begin{aligned} \begin{aligned}&\quad -\left( (p-2)\Delta _{\infty }^N\phi (x)+\Delta \phi (x)\right) -a(x)|\nabla \phi (x)\vert ^{q-p}\left( (q-2)\Delta _{\infty }^N\phi (x)+\Delta \phi (x)\right) \\&\quad -|\nabla \phi (x)\vert ^{q-p}\langle \nabla a(x),\nabla \phi (x)\rangle \le 0. \end{aligned} \end{aligned}$$

  (2.5)

• (iii) Finally, u is a viscosity solution if and only if u is both a viscosity supersolution and a viscosity subsolution.

Remark 2.3

In the case that $2\le p\le q<\infty$, the equation is pointwise well-defined, so the requirement, $\nabla \phi (x)\ne 0$ in Definition 2.2, can be eliminated.

We next state the following useful results (Lemmas 2.4–2.6), which can be found in [28, Section 2].

Lemma 2.4

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

$$\begin{aligned} -\phi (x)+\frac{1}{2}\left\{ \mathop {\textrm{max}}\limits _{\overline{B_\varepsilon (x)}}\phi +\mathop {\textrm{min}}\limits _{\overline{B_\varepsilon (x)}}\phi \right\} \ge \frac{1}{2}\langle D^2\phi (x)(x^\varepsilon _1-x),(x^\varepsilon _1-x) \rangle +o(\varepsilon ^2)\nonumber \\ \end{aligned}$$

(2.6)

and

$$\begin{aligned} -\phi (x)+\frac{1}{2}\left\{ \mathop {\textrm{max}}\limits _{\overline{B_\varepsilon (x)}}\phi +\mathop {\textrm{min}}\limits _{\overline{B_\varepsilon (x)}}\phi \right\} \le \frac{1}{2}\langle D^2\phi (x)(x^\varepsilon _2-x),(x^\varepsilon _2-x) \rangle +o(\varepsilon ^2).\nonumber \\ \end{aligned}$$

(2.7)

Lemma 2.5

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

$$\begin{aligned} \lim \limits _{\varepsilon \rightarrow 0+}\dfrac{x^\varepsilon _1-x}{\varepsilon }=-\dfrac{\nabla \phi }{|\nabla \phi |}(x), \end{aligned}$$

(2.8)

where $x_1^{\varepsilon }$ is defined as in Lemma 2.4.

Lemma 2.6

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

Although Lemmas 2.4 and 2.6 provide the bridge between the viscosity solution of p-Laplace equation $-\Delta _pu=0$ and the asymptotic mean value formula in [28], it is not enough for the double phase elliptic equation due to the presence of the term $\langle \nabla a,\nabla u \rangle$ in Eq. (2.3). Therefore, we need the following lemma.

Lemma 2.7

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

Proof

Observe that

and

Thus, we obtain

This finishes the proof. $\square$

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 (2.4) from (2.1).

For the case that $p>2$, we know from (1.3) 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

From $\alpha _p+\beta _p=1, \alpha _q+\beta _q=1$, we write

$$\begin{aligned} 0&\ge \dfrac{\alpha _p+M_{\phi }(x)\alpha _q}{1+M_{\phi }(x)}I+\dfrac{\beta _p+M_{\phi }(x)\beta _q}{1+M_{\phi }(x)}{} \textit{II}+\dfrac{\varepsilon |\nabla \phi (x) |^{q-p}}{4(N+p)(1+M_{\phi }(x))}{} \textit{III}+o(\varepsilon ^2), \end{aligned}$$

where

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

$$\begin{aligned} 0&\ge \alpha _pI+\beta _p\textit{II}+M_{\phi }(x)\left( \alpha _qI+\beta _q\textit{II}\right) +\dfrac{\varepsilon |\nabla \phi (x) |^{q-p}}{4(N+p)}{} \textit{III}+o(\varepsilon ^2). \end{aligned}$$

It follows from Lemmas 2.4, 2.6 and 2.7 that

$$\begin{aligned} 0&\ge \frac{\alpha _p}{2}\langle D^2\phi (x)(x^\varepsilon _1-x),(x^\varepsilon _1-x) \rangle +\dfrac{\varepsilon ^2\beta _p}{2(N+2)}\Delta \phi (x) \\&\quad +M_{\phi }(x)\Bigg (\frac{\alpha _q}{2}\langle D^2\phi (x)(x^\varepsilon _1-x),(x^\varepsilon _1-x) \rangle +\left. \dfrac{\varepsilon ^2\beta _q}{2(N+2)}\Delta \phi (x)\right) \\&\quad +\dfrac{\varepsilon ^2|\nabla \phi (x) |^{q-p}}{2(N+p)}\langle \nabla \phi (x),\nabla a(x)\rangle +o(\varepsilon ^2). \end{aligned}$$

Dividing by $\frac{\varepsilon ^2}{2}$, taking the limit as $\varepsilon \rightarrow 0$ and by Lemma 2.5, we have

$$\begin{aligned} 0&\ge \alpha _p\Delta _{\infty }^N\phi (x)+\dfrac{\beta _p}{N+2}\Delta \phi (x)+M_{\phi }(x)\left( \alpha _q\Delta _{\infty }^N\phi (x)+\dfrac{\beta _q}{N+2}\Delta \phi (x)\right) \\&\quad +\dfrac{|\nabla \phi (x) |^{q-p}}{N+p}\langle \nabla \phi (x),\nabla a(x)\rangle . \end{aligned}$$

Multipling by $N+p$, we get

$$\begin{aligned} 0&\ge (p-2)\Delta _{\infty }^N\phi (x)+\Delta \phi (x)+a(x)|\nabla \phi (x) |^{q-p}((q-2)\Delta _{\infty }^N\phi (x)+\Delta \phi (x))\\&\quad +|\nabla \phi (x) |^{q-p}\langle \nabla a(x), \nabla \phi (x)\rangle . \end{aligned}$$

Therefore, u is a viscosity supersolution according to (2.4). 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

$$\begin{aligned} 0&\le (p-2)\Delta _{\infty }^N\phi (x)+\Delta \phi (x)+a(x)|\nabla \phi (x) |^{q-p}((q-2)\Delta _{\infty }^N\phi (x)+\Delta \phi (x))\\&\quad +|\nabla \phi (x) |^{q-p}\langle \nabla a(x), \nabla \phi (x)\rangle . \end{aligned}$$

By Lemma 2.5,

$$\begin{aligned} 0&\le (p-2)\left\langle D^2\phi (x)\left( \dfrac{x^\varepsilon _1-x}{\varepsilon }\right) ,\left( \dfrac{x^\varepsilon _1-x}{\varepsilon }\right) \right\rangle +\Delta \phi (x)\\&\quad +a(x)|\nabla \phi (x)\vert ^{q-p}\left( (q-2)\left\langle D^2\phi (x)\left( \dfrac{x^\varepsilon _1-x}{\varepsilon }\right) ,\left( \dfrac{x^\varepsilon _1-x}{\varepsilon }\right) \right\rangle +\Delta \phi (x)\right) \\&\quad +|\nabla \phi (x)\vert ^{q-p}\left\langle \nabla a(x),\nabla \phi (x)\right\rangle +o(1). \end{aligned}$$

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

$$\begin{aligned} 0&\le (p-2)\langle D^2\phi (x)\left( x^\varepsilon _1-x\right) ,\left( x^\varepsilon _1-x\right) \rangle +\varepsilon ^2\Delta \phi (x)\\&\quad +a(x)|\nabla \phi (x)\vert ^{q-p}\left( (q-2)\langle D^2\phi (x)\left( x^\varepsilon _1-x\right) ,\left( x^\varepsilon _1-x\right) \rangle +\varepsilon ^2\Delta \phi (x) \right) \\&\quad +\varepsilon ^2|\nabla \phi (x)\vert ^{q-p}\langle \nabla a(x),\nabla \phi (x)\rangle +o(\varepsilon ^2). \end{aligned}$$

By Lemmas 2.4, 2.6 and 2.7, we have

$$\begin{aligned} 0&\le 2(p-2)I+2(N+2)\textit{II}+a(x)|\nabla \phi (x)\vert ^{q-p}(2(q-2)I+2(N+2)\textit{II})\\&\quad +\frac{\varepsilon }{2}|\nabla \phi (x)\vert ^{q-p}{} \textit{III}+o(\varepsilon ^2). \end{aligned}$$

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

$$\begin{aligned} 0&\le \left( \alpha _pI+\beta _p\textit{II}\right) +a(x)\dfrac{N+q}{N+p}|\nabla \phi (x)\vert ^{q-p}\left( \alpha _qI+\beta _q\textit{II}\right) \\&+\dfrac{\varepsilon |\nabla \phi (x) |^{q-p}}{4(N+p)}{} \textit{III}+o(\varepsilon ^2). \end{aligned}$$

Then separating $\phi (x)$ from I and II, we get

Thus

Similarly, we can also prove that u satisfies (2.1) if u is the viscosity supersolution.

When $1<p\le 2$, 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 2.4 if necessary.

Combining the arguments above, we complete the proof. $\square$

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

$$\begin{aligned} -{\textrm{div}}(|\nabla u |^{p-2}\nabla u)=0 \quad \text {in } \Omega \end{aligned}$$

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

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 [28]. 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 ${\tilde{a}}(x)\ge 1$ and let u(x) be a continuous function in $\Omega$. The equation

$$\begin{aligned} -{\textrm{div}}({\tilde{a}}(x)|\nabla u |^{p-2}\nabla u)=0 \quad \text {in } \Omega \end{aligned}$$

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

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

$$\begin{aligned} -{\textrm{div}}((a(x)+1)|\nabla u |^{p-2}\nabla u)=0. \end{aligned}$$

In this situation, we have

$$\begin{aligned} \alpha _p=\alpha _q, \quad \beta _p=\beta _q, \quad M_u(x)=a(x). \end{aligned}$$

Thus, the asymptotic mean value formula (1.2) reads as

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

Remark 3.4

In fact, the condition ${\tilde{a}}(x)\ge 1$ 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

$$\begin{aligned} -{\textrm{div}}(|\nabla u|^{p(x)-2}\nabla u)=0 \quad \text {in } \Omega . \end{aligned}$$

(3.1)

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

$$\begin{aligned}&{\textrm{div}}(|\nabla u|^{p(x)-2}\nabla u)\\&\quad =\sum _{i=1}^{N}|\nabla u|^{p(x)-2}(\ln |\nabla u|\partial _ip+(p(x)-2)\partial _i\ln |\nabla u|)u_i+|\nabla u|^{p(x)-2}\Delta u\\&\quad =\sum _{i=1}^{N}|\nabla u|^{p(x)-2}\left( \ln |\nabla u|p_i+(p(x)-2)|\nabla u|^{-2} \sum _{j=1}^{N}u_{ij}u_j\right) u_i\!+\!|\nabla u|^{p(x)-2}\Delta u\\&\quad =|\nabla u|^{p(x)-2}\left( \ln |\nabla u|\langle \nabla p,\nabla u\rangle +(p(x)-2)\Delta _{\infty }^N u+\Delta u) \right) . \end{aligned}$$

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

$$\begin{aligned} -(p(x)-2)\Delta _{\infty }^N u-\Delta u-\ln |\nabla u|\langle \nabla p,\nabla u\rangle =0. \end{aligned}$$

We find that the term $\langle \nabla p,\nabla u\rangle$ appears in the equation above. We can still use Lemma 2.7 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

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 [29] 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 [29, 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 $(x_0,t_0)\in \Omega _T$ and $\phi \in C^2(\Omega _T)$ is such that

• (i) $u(x_0,t_0)=\phi (x_0,t_0)$.

• (ii) $u(x,t)>\phi (x,t)$ for all $(x,t)\in \Omega _T,(x,t)\ne (x_0,t_0)$, then we have at the point $(x_0,t_0)$

  $$\begin{aligned} {\left\{ \begin{array}{ll} \phi _t \ge (p-2)\Delta _{\infty }^N\phi +\Delta \phi +a|\nabla \phi \vert ^{q-p}\left( (q-2)\Delta _{\infty }^N\phi +\Delta \phi \right) \\ \qquad +|\nabla \phi \vert ^{q-p}\langle \nabla a,\nabla \phi \rangle &{}\text { if } \nabla \phi (x_0,t_0)\ne 0,\\ \phi _t \ge \lambda _{\min }((p-2)D^2\phi )+\Delta \phi &{}\text { if } \nabla \phi (x_0,t_0)=0. \end{array}\right. } \end{aligned}$$

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 $D^2\phi \ne 0$. We state it as the following lemma. The proof follows the ideas in [29, Lemma 2]. However, we need to require the additional assumption that $a(x,t)>0$ due to the technical reason, which has been used in [19, 20, 22] for the double phase equations. For the convenience of the reader, we give a sketched proof.

Lemma 4.2

Let $1<p\le q<\infty$, the positive function a(x, t) be a function that is $C^1$ in the space variable, and continuous in the time variable. A function $u:\Omega _T\rightarrow \mathbb {R}$ is a viscosity solution to Eq. (1.4) if u is continuous and whenever $(x_0, t_0)\in \Omega _T$ and $\phi \in C^2(\Omega _T)$ is such that

• (i) $u(x_0,t_0)=\phi (x_0,t_0)$.

• (ii) $u(x,t)>\phi (x,t)$ for all $(x,t)\in \Omega _T, (x,t)\ne (x_0,t_0)$, then at the point $(x_0,t_0)$, if $\nabla \phi (x_0,t_0)\ne 0$, we have

  $$\begin{aligned} \begin{aligned} \phi _t&\ge (p-2)\Delta _{\infty }^N\phi +\Delta \phi +a|\nabla \phi \vert ^{q-p}\left( (q-2)\Delta _{\infty }^N\phi +\Delta \phi \right) \\&\quad +|\nabla \phi \vert ^{q-p}\langle \nabla a,\nabla \phi \rangle ; \end{aligned} \end{aligned}$$

  (4.1)

  if $\nabla \phi (x_0,t_0)=0$ and $D^2\phi (x_0,t_0)=0$, we have

  $$\begin{aligned} \phi _t(x_0,t_0)\ge 0. \end{aligned}$$

  (4.2)

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

Proof

We prove it by contradiction. If the claim is not true, we will have $\phi \in C^2(\Omega _T)$ and $(x_0,t_0)\in \Omega _T$ such that

$$\begin{aligned} {\left\{ \begin{array}{ll} u(x_0,t_0)=\phi (x_0,t_0)\\ u(x,t)>\phi (x,t) \quad \text {for } (x,t)\in \Omega _T,(x,t)\ne (x_0,t_0), \end{array}\right. } \end{aligned}$$

for which $\nabla \phi (x_0,t_0)=0$, $D^2\phi (x_0,t_0)\ne 0$ and

$$\begin{aligned} \phi _t(x_0,t_0)<\lambda _{\min }\left( (p-2)D^2\phi (x_0,t_0)\right) +\Delta \phi (x_0,t_0), \end{aligned}$$

(4.3)

or the similar inequality when testing from above. Denote

$$\begin{aligned} v_j(x,t,y,s)=u(x,t)-\phi (y,s)+\dfrac{j}{4}|x-y|^4+\dfrac{j}{2}|t-s|^2. \end{aligned}$$

Then let $(x_j,t_j,y_j,s_j)$ mean the minimum point of $v_j$ in $\overline{\Omega }_T\times \overline{\Omega }_T$. We know that $(x_j,t_j,y_j,s_j)\in \Omega _T\times \Omega _T$ for j large enough and $(x_j,t_j,y_j,s_j)\rightarrow (x_0,t_0,x_0,t_0)$ as $j\rightarrow \infty$ (see [13]).

Next, we consider two scenarios that $x_j=y_j$ infinitely often and $x_j\ne y_j$ for all j large enough, respectively. Since the case $x_j=y_j$ can be treated in a very similar way to that in [29, Lemma 2], we focus on the case $x_j\ne y_j$. By applying the parabolic theorem of sums for $v_j$, there are two symmetric matrices $X_j$, $Y_j$ such that $X_j-Y_j$ is postive semidefinite and

$$\begin{aligned} \left( j(t_j-s_j),j|x_j-y_j|^2(x_j-y_j),Y_j\right) \in \overline{\mathcal {P}}^{2,+}\phi (y_j,s_j), \\ \left( j(t_j-s_j),j|x_j-y_j|^2(x_j-y_j),X_j\right) \in \overline{\mathcal {P}}^{2,-}u(x_j,t_j). \end{aligned}$$

Noting that $\nabla \phi (x_0,t_0)=0$. By (4.3) and the assumptions on u and a, after some arrangements we obtain

$$\begin{aligned} \begin{aligned} 0&=j(t_j-s_j)-j(t_j-s_j)\\&<(p-2)\left\langle Y_j\dfrac{x_j-y_j}{|x_j-y_j|},\dfrac{x_j-y_j}{|x_j-y_j|}\right\rangle +{\textrm{tr}}(Y_j)\\&\quad -(p-2)\left\langle X_j\dfrac{x_j-y_j}{|x_j-y_j|},\dfrac{x_j-y_j}{|x_j-y_j|}\right\rangle -{\textrm{tr}}(X_j)\\&\quad +a(y_j,s_j)(j|x_j-y_j|^3)^{q-p}\left[ (q-2)\left\langle Y_j\dfrac{x_j-y_j}{|x_j-y_j|},\dfrac{x_j-y_j}{|x_j-y_j|}\right\rangle +\textrm{tr}(Y_j)\right] \\&\quad -a(x_j,t_j)(j|x_j-y_j|^3)^{q-p}\left[ (q-2)\left\langle X_j\dfrac{x_j-y_j}{|x_j-y_j|},\dfrac{x_j-y_j}{|x_j-y_j|}\right\rangle +\textrm{tr}(X_j)\right] \\&\quad +(j|x_j-y_j|^3)^{q-p+1}\left\langle \nabla a(y_j,s_j)-\nabla a(x_j,t_j),\dfrac{x_j-y_j}{|x_j-y_j|}\right\rangle \\&:=I_1+I_2+I_3. \end{aligned} \end{aligned}$$

(4.4)

We first get $I_1\le 0$, which is identical to that in [29, Lemma 2]. Then we can follow the estimates on $J_3$ in the proof of [22, Proposition 5.1], and deduce that $I_2\rightarrow 0$ as $j\rightarrow \infty$. It is worth mentioning that we need to require $a(x,t)>0$ to get this limit for the technical reason. The details can be found in [22, pp. 28–30], where the number l should be replaced with 4 here. As for the third term $I_3$,

$$\begin{aligned} I_3 \le (j|x_j-y_j|^3)^{q-p+1}| \nabla a(y_j,s_j)-\nabla a(x_j,t_j)|\rightarrow 0 \end{aligned}$$

as $j\rightarrow \infty$, provided that $\nabla a$ is continuous. Indeed, from the minimum point we can see that

$$\begin{aligned}&u(x_j,t_j)-\phi (y_j,s_j)+\dfrac{j}{4}|x_j-y_j|^4+\dfrac{j}{2}|t_j-s_j|^2\\&\quad \le u(x_j,t_j)-\phi (x_j,s_j)+\dfrac{j}{2}|t_j-s_j|^2. \end{aligned}$$

Then we have

$$\begin{aligned} \dfrac{j}{4}|x_j-y_j|^4\le \phi (y_j,s_j)-\phi (x_j,s_j)\le C|x_j-y_j|, \end{aligned}$$

so

$$\begin{aligned} \dfrac{j}{4}|x_j-y_j|^3\le C. \end{aligned}$$

That is, the term $j|x_j-y_j|^3$ is bounded. At this moment, merging the estimates on $I_1$–$I_3$ together with (4.4), we arrive at $0<0$, which is a contradiction. This leads to the desired conclusion. $\square$

The definition of u satisfying the asymptotic mean value formula (1.5) at the point (x, t) in the viscosity sense is similar to Definition 2.1, so we omit it. But $\nabla \phi (x,t)=0$ 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 [29, 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

(4.5)

and

(4.6)

Lemma 4.4

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

$$\begin{aligned} \lim \limits _{\varepsilon \rightarrow 0+}\frac{x^{\varepsilon ,s}_1-x}{\varepsilon }=-\dfrac{\nabla \phi }{|\nabla \phi |}(x,t), \end{aligned}$$

(4.7)

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

Lemma 4.6

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

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 (1.5) in the viscosity sense, we need to prove that u is a viscosity solution. If so, for a test function $\phi$, we have

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

(4.8)

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 (4.8), we apply Lemmas 4.3, 4.5 and Lemma 4.6 to have

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

$$\begin{aligned} 0&\ge \alpha _p\Delta _{\infty }^N\phi (x,t)+\dfrac{\beta _p}{N+2}\Delta \phi (x,t)-\frac{1}{A_{\phi }(x,t)}\phi _t(x,t)\\&\quad +M_\phi (x,t)\left( \alpha _q\Delta _{\infty }^N\phi (x,t)+\dfrac{\beta _q}{N+2}\Delta \phi (x,t)-\frac{1}{B_{\phi }(x,t)}\phi _t(x,t)\right) \\&\quad +\dfrac{|\nabla \phi (x,t) |^{q-p}}{N+p}\langle \nabla \phi (x,t),\nabla a(x,t)\rangle . \end{aligned}$$

Multipling by $N+p$ again, we get

$$\begin{aligned} 0&\ge (p-2)\Delta _{\infty }^N\phi (x,t)+\Delta \phi (x,t)+a(x,t)|\nabla \phi (x,t) |^{q-p}((q-2)\Delta _{\infty }^N\phi (x,t)\\&\quad +\Delta \phi (x,t))-\left( \dfrac{N+p}{A_{\phi }(x,t)}+\dfrac{a(x,t)(N+q)|\nabla \phi (x,t) |^{q-p}}{B_{\phi }(x,t)}\right) \phi _t(x,t)\\&\quad +|\nabla \phi (x,t) |^{q-p}\langle \nabla \phi (x,t),\nabla a(x,t)\rangle . \end{aligned}$$

Recalling (1.6), we have

$$\begin{aligned} \dfrac{N+p}{A_{\phi }(x,t)}+\dfrac{a(x,t)(N+q)|\nabla \phi (x,t) |^{q-p}}{B_{\phi }(x,t)}=1. \end{aligned}$$

(4.9)

Therefore, we obtain

$$\begin{aligned} \phi _t \ge (p-2)\Delta _{\infty }^N\phi +\Delta \phi +a|\nabla \phi \vert ^{q-p}\left( (q-2)\Delta _{\infty }^N\phi +\Delta \phi \right) +|\nabla \phi \vert ^{q-p}\langle \nabla a,\nabla \phi \rangle . \end{aligned}$$

It follows that (4.1) holds when $\nabla \phi (x,t)\ne 0$. When $\nabla \phi (x,t)=0$ and $D^2\phi (x,t)=0$, by (4.9), we get $A_{\phi }(x,t)=N+p$ and $M_{\phi }(x,t)=0$. According to the asymptotic mean value formula, we have

By Lemma 4.5 and the expansion

$$\begin{aligned} \phi (y,s)-\phi (x,t)=\phi _t(x,t)(s-t)+o(|s-t|+|y-x|^2), \end{aligned}$$

we have

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

$$\begin{aligned} \phi _t(x,t)\ge 0. \end{aligned}$$

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, it is similar to the proof of elliptic case, so we omit it. The proof is complete. $\square$

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

$$\begin{aligned} u_t=|\nabla u |^{2-p}{\textrm{div}}(|\nabla u |^{p-2}\nabla u) \quad \text {in }\Omega _T \end{aligned}$$

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

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 [29]. In fact, Corollary 4.7 also holds for $p=\infty$ with $\alpha _p=1$, $\beta _p=0$.