# Geometric intuition of mean value property of nonlinear elliptic equation

I wish to gain some understanding of the MVP of nonlinear elliptic equation by geometric intuition.

## Linear elliptic equation case

First of all, I have a very good geometric explain of the MVP of Laplace equation, i.e.

MVP of laplace equation

$\Delta u=0$ in $\Omega$ , $\forall B(x_0,r)\subset \Omega$ is a Ball, we have following identity:

$\frac{1}{\mu(\partial(B))}\int_{\partial B}u(x)dx=u(x_0)$

I need to point out first, this property is not difficult to proof by standard integral by part method, but the following method have more geometric intuition. And in some sense explain why this property holds.

The proof is not very difficult to explain by mathematic formula, but I wish to divide the proof into two part, explain one part by graph and literal interpretation.

part 1 of the proof:

we consider a 1-parameter group of foliation, and consider the integral identity with this foliation.

$\int_{v\in S_{n-1}}\int_{\gamma_v} \frac{\partial \partial_{v}u}{\partial t}dt=\int_{B(x_0,r)}\partial_{n}u-\int_{S_{n-1}}\partial_{v}udv=\int_{B(x_0,r)}\partial_{n}u$

Part 2 of the proof:

and we have:

$\int_{v\in S_{n-1}}\int_{\gamma_v} \frac{\partial \partial_{v}u}{\partial t}dt=0$

by the pointwise equation $\Delta u=0$, one key point is $\partial_{-v-v}u=\partial_{vv}u, \forall v\in S_{n-1}$.

This approach cloud easily to transform to the general elliptic equation case and it seems a little difficult to transform to Possion equation, the non-hemomorphism case.

## Nonlinear elliptic equation case

A-B-P estimate for general nonlinear uniformly elliptic equation

ABP estimate is the most basic estimate in fully nonliear elliptic equation.
The ABP maximum principle states (roughly) that, if

$a^{ij} \partial _i \partial _j u \geq f, in \ \Omega \subset \mathbb{R}^n (a^{ij} \geq C Id >0),$

Then (assuming sufficient regularity of the coefficients),

\sup _{\Omega} u \leq \sup _{\partial \Omega} u + C (\int _{\Omega} \vert f \vert^n )^{1/n} ………. (*)

I will give an intuitive explanation of the proof of (*) .
Usually, in order to prove maximum principles, the key idea is to use that at a local max the second derivative is negative-definite, then choose a good basis and get some identity of 1-order drivative and inequality for 2-order’s. This process is used in such like the proof of the Hopf lemma, and some inter gradient estimate, consider some flexiable function like $e^{Au}$ or sometihng else anyway.

But in the proof of ABP we need more geometric intution and more trick.

First we do a rescaling:
if $a^{ij} \partial _i \partial _j u \geq 1, in B_1 \subset \mathbb{R}^n (a^{ij} \geq C Id >0),u|_{\partial B_1}\geq 0.$
then:

$|inf_{B_1}u| \leq C |A|^{1/n} ....(**)$

And then We explain what is the contact set. It is the subset $\Gamma^{+} of \Omega$ such that u agree with it convex envelop. i.e. $\Gamma^{+}=\{x|u=convex \ evolap \ of \ u\ at \ x\}$ . The geometric meaning is it has at least one lower support plane. So what is $\Gamma^{+}$ it is just the set that u is very low on it. Or in another way of view you consider $-u$ as a lot of mountains then $\Gamma^+$ is the place near the tops of which mountain can see every thing (locally).
Then we look at every point in $\Gamma^{+}$ , then determination of the hessian matrix $det(u_{ij})$ at this point have a control due to the PDE $a^{ij} \partial _i \partial _j u \geq 1$, and the uniformly elliptic property.

The determination of hessian matrix could be view as a determination of Jacobe matrix of the map $(u_1,...,u_n)\to (e_1,...,e_n)$ .and by Area formula we have:

$\int_{\Phi(\Omega)} f( \Phi^{-1}(y)) dy =\int_{\Omega}f(x)|J({\Phi(x)})| dx,$

It is easy to see for a constant $c$ ,$B_{c|sup_{\Omega}|u||}(0)\subset \Phi(\Omega)$ (Base on the PDE on every point, the geometric intution is just the function u could not be very narrow cone at every point). So we have , take $f=\chi_{\Gamma^{+}}$ ,

$|B_{c\sup_{\Omega}|u|}(0)|^{n}\leq \int_{\Gamma^{+}}\chi_{\Gamma_{+}}(x)|J_{\Phi}(x)|dx$

so we have:

$|B_{c\sup_{\Omega}|u|}(0)|\lesssim |{\Gamma_{+}}|^{\frac{1}{n}}...(***)$

and the classical matrix inequality for every positive definite matrix A we have
:

$det(AB)\leq (\frac{tr(AB)}{n})^n...(****) .$

combine (***),(****) ,we have:

$sup_{\Omega}|u|\lesssim ||\frac{a^{ij}u_{ij}}{D^*}||_{L^n({\Gamma^+})} .$

graph

General approach to get MVP for elliptic equation which is come from geometry

MVP for K-hessian equation, with geometric explanation

MVP for K-curvature equation, with geometric explanation

MVP for p-Laplace equation, with geometric explanation

# Schauder estimate and Sobelov inequality

In this note we discuss the Schauder theory for uniformly elliptic linear equations and Sobelov inequality.

the three main topics ars a priori estimate in Holder norms,regularity of arbitrary solutions and the solvability of the Dirichlet problem.Among these topics,a priori estimates are the most fundamental and the basis of the follows two.we will discuss both the interior Schauder estimate and global Schauder estimate.

## -Schauder Theory-

1. Interior Schauder Theory

${\Omega}$ be a domain in ${R^n}$,bounded most of the time.
${a_{ij},b_i,c}$ be defined in ${\Omega}$,with ${a_{ij}=a_{ji}}$.where ${1\leq i,j\leq n}$.
we consider the operator ${L}$ given by,

$\displaystyle Lu=a_{ij}\partial_{ij}u+b_i\partial_iu+c,in \ \Omega.$

easy to see ${Lu}$ is defined for any ${u\in C^2(\Omega)}$.
the operator ${L}$ is always be assumed to be strictly elliptic in ${\Omega}$;namely,
$\displaystyle a{ij}\xi_i\xi_j \geq \lambda|\xi|^2$

for any ${\xi\in R^n,x\in \Omega}$,where ${\lambda}$ is a positive constant.
1.1. Interior Schauder Estimate

define the weighted ${C^{k,\alpha}}$ norm,

$\displaystyle |u|^*_{C^{k,\alpha}(B_R)}=\sum_{i=0}^k R^i|D^iu|_{L^{\infty}(B_R)}+R^{k+\alpha}[D^ku]_{C^{\alpha}(B_R)}$

easy to see ${R}$ come from a scaling.
consider the PDE.
$\displaystyle Lu=a_{ij}\partial_{ij}u+b_i\partial_iu+c=f,in \ \Omega.$

we want to proof this type estimate,
$\displaystyle |u|_{C^{2,\alpha}(A)} \leq C(|u|_{L^{\infty}(\Omega)}+|f|_{C^{\alpha}(\Omega)})$

where ${A\subset \Omega }$
we first deal with a easy case,${a_{ij}}$ is constant. in this case we proof the estimate:

Lemma 1 ${f \in C^{\alpha}(B_R)}$,for some ${\alpha \in (0,1)}$,and ${(a_{ij})}$ be a constant symmetric ${n\times n}$ matrix satisfying
$\displaystyle \lambda |\xi|^2 \leq a_{ij}\xi_i\xi_j \leq \Lambda|\xi|^2$

${\exists \lambda,\Lambda >0,\forall \xi \in R^n}$. suppose ${u\in C^2(B_R)}$ satisfies:
$\displaystyle a_{ij}\partial_{ij}u=f, in \ B_R$

then ,${u \in C^{2,\alpha}(B_{\frac{R}{2}})}$,moreover,
$\displaystyle |u|^*_{C^{2,\alpha}(B_{\frac{R}{2}})}\leq C[|u|_{L^{\infty}(B_R)}+R^2|f|^*_{C^{\alpha}(B_R)}]$

Proof: $\Box$ to continue,we prove an interpolation inequality for Holder continuous functions.

Lemma 2 Let ${\alpha,\mu \in (0,1)}$ and ${B_R}$ be a ball of radius ${R}$ in ${R^n}$,then, (1)for any ${u\in C^{1,\alpha}(\overline B_R)}$,
$\displaystyle \mu^{\alpha}R^{\alpha}[u]_{C^{\alpha}(B_R)}\leq C[\mu R |\nabla u|_{L^{\infty}(B_R)}+|u|_{L^{\infty}(B_1)}]$

(2)for any ${u\in C^{1,\alpha}(\overline B_R)}$,
$\displaystyle \mu R |\nabla u|_{L^{\infty}(B_R)} \leq C[\mu^{1+\alpha}R^{1+\alpha}|\nabla u|_{C^{\alpha}(B_R)}+|u|_{L^{\infty}(B_1)}]$

(3)for any ${u\in C^2(\overline B_R)}$,
$\displaystyle \mu R|\nabla u|_{L^{\infty}(B_R)}\leq C[\mu^2R^2|\nabla^2 u|_{L^{\infty}(B_R)}+|u|_{L^{\infty}(B_R)}]$

where ${C}$ is a positive constant depending on n and ${\alpha}$.
Proof: $\Box$

Corollary 3 Let ${\alpha,\mu\in (0,1)}$ and ${B_R}$ be a ball of radius ${R}$ in ${R^n}$.Then,for any ${u\in C^{2,\alpha}(\overline B_R)}$,
$\displaystyle \sum_{i=0}^2(\mu R)^i|\nabla^i u|_{L^{\infty}(B_R)}+\sum_{i=0}^1(\mu R)^{i+\alpha}[\nabla^i u]_{C^{\alpha}(B_R)}\leq C[(\mu R)^{2+\alpha}[\nabla^2 u]_{C^{\alpha}(B_R)}+|u|_{L^{\infty}(B_R)}]$

Proof: $\Box$

Now we are ready to prove an interior estimate for ${C^{2,\alpha}}$-norms of solutions of uniformly elliptic equations.The trick is to freeze coefficients.

Lemma 4
2. Global Schauder Theory

## -Sobelov inequality-

Theorem 5
$\displaystyle W_0^{1,p}(\Omega)\longrightarrow L^{\frac{np}{n-p}}(\Omega),1\leq p

moreover,we have: ${\exists C=C(n,p)}$, ${\forall u\in W^{1,p}_0(\Omega)}$,
$\displaystyle ||u||_{\frac{np}{n-p}} \leq C||Du||_p,1\leq p

Proof:

$\displaystyle p=1$

suffice to proof:
$\displaystyle ||u||_{\frac{n}{n-1}}\leq C||Du||_1$

obvious we have:
$\displaystyle |u(x)|\leq \int_{-\infty}^{\infty}|Du(x)|dx$

so ${\int_{\Omega} |u|^{\frac{n}{n-1}}\leq \int_{\Omega} \Pi_{i=1}^n(\int_{-\infty}^{\infty}|D_iu(x)|dx)^{\frac{1}{n-1}}}$.
so ${||u||_{\frac{n}{n-1}}\leq (\int_{\Omega}\Pi_{i=1}^n(\int_{-\infty}^{\infty}|D_iu|)^{\frac{1}{n-1}})^{\frac{n-1}{n}}\leq \int_{\Omega} \Pi_{i=1}^n(\int_{-\infty}^{\infty}|D_iu|)^{\frac{1}{n}} \leq \int_{\Omega} \frac{1}{n} \sum_{i=1}^n(\int_{-\infty}^{\infty}|D_iu|)\leq C||Du||_1}$
$\displaystyle 1

use the similar argument as ${p=1}$ to prove the situation ${1.
suffice to prove ${||u||_{\frac{np}{n-p}}\leq C||Du||_p}$.
obvious we have:${|u(x)|^p\leq \int_{-\infty}^{\infty}p|u|^{p-1}|Du|}$.
${(\int_{\Omega}|u(x)|^{\frac{np}{n-p}})^{\frac{n-p}{np}}}$
${\leq (\int_{\Omega} \Pi_{i=1}(\int_{-\infty}^{\infty} p|u|^{p-1}|D_iu| )^{\frac{1}{n-p}})^{\frac{n-p}{np}} }$
${\leq C\int_{\Omega} \Pi_{i=1}^n(\int_{-\infty}^{\infty}p|u|^{p-1}|D_iu|)^{\frac{1}{np}}}$
${\leq\frac{c}{n}\sum_{i=1}^n\int_{\Omega}(\int_{-\infty}^{\infty}p|u|^{p-1}|D_iu|)^{\frac{1}{p}}}$
${\leq \frac{c}{n}\sum_{i=1}^n\tilde C p[(\int_{\Omega} (|u|^{p-1})^{\frac{p}{p-1}})^{\frac{p-1}{p}}+(\int_{\Omega} |D_iu|^p)^{\frac{1}{p}}]^{\frac{1}{p}} }$
${\leq C||Du||_p}$. Q.E.D. $\Box$
4.

$\displaystyle W_0^{1,p}(\Omega)\longrightarrow C(\bar\Omega),n

moreover,we have: ${\exists C=C(n,p)}$, ${\forall u\in W^{1,p}_0(\Omega)}$,
$\displaystyle sup_{\Omega}|u| \leq C|\Omega|^{\frac{1}{n}-\frac{1}{p}}||Du||_p,p>n$

${\mu\in (0,1]}$,

$\displaystyle (V_{\mu}f)(x)=\int_{\Omega}|x-y|^{n(\mu-1)}f(y)dy$

then ${V_{\mu}: L^1(\Omega) \longrightarrow L^1(\Omega) }$ is well-defined by the following lemma:
Lemma 6 ${V_{\mu}:L^p \longrightarrow L^q}$ continously for any q,${1\leq q \leq \infty}$ satisfy ${0\leq \delta=\delta(p,q)=\frac{1}{p}-\frac{1}{q} \leq \mu}$.
furthermore,for any ${f\in L^p(\Omega)}$
$\displaystyle ||V_{\mu}f||_q \leq (\frac{1-\delta}{\mu -\delta})^{1-\delta}w_n^{1-\mu}|\Omega|^{\mu-\delta}||f||_p$

Proof: ${h(x-y)=|x-y|^n(\mu-1)}$ directly calculate follows that :

$\displaystyle ||h||_r \leq (\frac{1-\delta}{\mu -\delta})^{1-\delta} w_n^{1-\mu}|\Omega|^{\mu-\delta}$

now follows young inequality and this priori estimate we have:
${||V_{\mu}f||_q=(\int_{\Omega}(\int_{\Omega}|x-y|^{n(\mu-1)}f(y)dy)dx)^{\frac{1}{q}}}$
${\leq (\int_{\Omega}(\int_{\Omega}h^{\frac{r}{q}}h^{r(1-\frac{1}{p})}|f|^{\frac{p}{q}}|f|^{p\delta})^qdx)^{\frac{1}{q}}}$
${\leq (\int_{\Omega}(\int (h^r|f|^p)^{\frac{1}{q}}(\int h^r)^{1-\frac{1}{p}}(\int f^p)^{\delta})^{q})^{\frac{1}{q}}}$
${\Longrightarrow}$
$\displaystyle ||V_{\mu}f||_q \leq sup_{x \in \Omega} \{\int h^r(x-y)dy\}^{\frac{1}{r}}||f||_p$

and by the priori estimate,we have:
$\displaystyle ||V_{\mu}f||_q \leq (\frac{1-\delta}{\mu -\delta})^{1-\delta}w_n^{1-\mu}|\Omega|^{\mu-\delta}||f||_p$

Q.E.D. $\Box$
Lemma 7 ${f\in L^p(\Omega)}$,${g=V_{\mu}f}$.
${\Longrightarrow}$ ${\exists c_1,c_2}$ constant depend only on ${n,p}$,such that
$\displaystyle \int_{\Omega} exp[\frac{g}{c_1||f||_p}]^{p^}dx\leq c_2|\Omega|,p^=\frac{p}{p-1}$

Proof: we have

$\displaystyle ||g||_q \leq q^{1-\frac{1}{p}+\frac{1}{q}}w_n^{1-\frac{1}{p}}|\Omega|^{\frac{1}{q}}||f||_p$

${\Longrightarrow}$
$\displaystyle \int_{\Omega} |g|^{p^q}dx \leq p^q(w_np^q||f||_p^{p^})^q|\Omega|$

${\Longrightarrow}$
$\displaystyle \int_{\Omega}\sum_{N_0}^{N}\frac{1}{k!}(\frac{|g|}{c_1||f||_p})^{p^k}\leq p^|\Omega|\sum(\frac{p^`w_n}{c_1^p})^k\frac{k^k}{(k-1)!}$

then take ${c_1,c_2}$ suffice large. Q.E.D. $\Box$
Lemma 8 let ${u\in W^{1,1}_0(\Omega)}$
$\displaystyle u(x)=\frac{1}{nw_n} \int_{\Omega} \frac{(x_i-y_i)D_iu(y)}{|x-y|^n}$

a.e. in ${\Omega}$.
Proof: frist zero extended ${u}$ to whole space.and we have ${u(x)=\int_{-\infty}^xD_iu(x)}$.

$\displaystyle u(x)=\int_0^{\infty}D_ru(x+rw)dr$

forall ${w\in \partial B_1(0)}$,so
$\displaystyle u(x)=-\frac{1}{nw_n}\int_0^{\infty}\int_{|w|=1}D_ru(x+rw)drdw=\frac{1}{nw_n}\int_{\Omega}\frac{(x_i-y_i)D_iu(y))}{|x-y|^ndy}$

Q.E.D. $\Box$
Theorem 9 let ${u\in W^{1,n}_0(\Omega)}$,then there exists constant ${c_1,c_2}$ such that
$\displaystyle \int_{\Omega}exp[\frac{|u|}{c_1||Du||_n}]^{\frac{n}{n-1}}dx\leq c_2|\Omega|$

Proof: a $\Box$

Theorem 10 ${u\in W_0^{1,p}(\Omega),p>n}$,then ${u\in C^{\gamma}(\Omega)}$,${\gamma=1-\frac{n}{p}}$.
moreover ${\forall ball B=B_R}$
$\displaystyle osc_{\Omega \cap B_R}u \leq C R^{\gamma} ||Du||_p$

Proof: a $\Box$