## Tag Archives: maximum principle

### f-extremal disk

In the last nonlinear analysis seminar, Professor Espinar talked about the overdetermined elliptic problem(OEP) which looks like the following

$\Delta u+f(u)=0\quad\text{ in }\Omega$

$u>0\quad \text{ in }\Omega$

$u=0 \quad \text{on }\partial \Omega$

$\frac{\partial u}{\partial\eta}=cst\quad\text{on }\partial \Omega$

There is a BCN conjecture related to this

BCN: If $f$ is Lipschitz, $\Omega\subset \mathbb{R}^n$ is a smooth(in fact, Lipschitz) connected domain with $\mathbb{R}^n\backslash\Omega$ connected where OEP admits a bounded solution, then $\Omega$ must be either a ball, a half space, a generalized cylinder or the complement of one of them.

BCN is false in $n\geq 3$. Epsinar wih Mazet proved BCN when $n=2$. This implies the Shiffer conjecture in dimension 2. In higher dimension of Shiffer conjecture, if we know the domain is contained in one hemisphere of $\mathbb{S}^n$, then one can use the equator or the great circle to perform the moving plane.

### Strong maximum principle

Strong maximum principle: Suppose ${Lu\geq 0}$ with ${c=0}$, ${u\in C^2(\Omega)\cap C^1(\overline{\Omega})}$ where ${\Omega}$ is connected and open in ${\mathbb{R}^n}$. If ${u}$ achieves maximum over ${\overline{\Omega}}$ at an interior point, then ${u}$ must be a constant.

Proof: Suppose ${M=\sup\limits_{\overline{\Omega}}u}$. Set ${V=\{x\in \Omega|u(x). If ${u}$ achieves maximum at an interior pt and not constant over all ${\Omega}$, then ${\partial V\cap \Omega\neq \emptyset}$. choose a point ${x_0}$ such that ${\text{dist}(x_0,\partial V)< \text{dist}(x_0,\partial \Omega)}$. Consider the largest ball ${B_r(x_0)\subset V}$ with radius ${r}$ centering at ${x_0}$. Suppose ${y\in \partial B_r\cap \partial V}$, then ${u(y)=M}$ and ${D_u(y)=0}$, because ${M}$ is the maximum. However, applying the hopf lemma to ${u}$ in ${V}$, we know ${\displaystyle \frac{\partial u}{\partial n}>0}$ where ${\nu}$ is the outer normal vector of ${B_r(x_0)}$. Contradiction.

Remark: If ${c\leq 0}$ and ${u}$ achieves nonngative maximum in the interior of ${\Omega}$. Then ${u}$ must be a constant.

### General approach for fully nonlinear elliptic equation

Consider the Dirichlet problem in a bounded domian $\Omega\subset \mathbb{R}^n$ with smooth boundary $\partial \Omega$

$\displaystyle \begin{cases} F(D^2u)=\psi(x) \text{ in } \Omega\\\quad \quad u=\phi\quad\text{ on }\Omega\end{cases}\quad(1)$

The function $F$ are represented by a smooth symmetric function

$\displaystyle F(D^2u)=f(\lambda_1,\lambda_2,\cdots,\lambda_n)$

here $\lambda_1,\lambda_2,\cdots,\lambda_n$ are the eigenvalues of $D^2u$. In order to be elliptic, we require

$\displaystyle \frac{\partial f}{\partial \lambda_i}>0$, $\forall\, i>0\quad (2)$

$f$ is defined in an open convex cone $\Gamma\subset \mathbb{R}^n$ with vertex at origin, and

$\displaystyle \bigcap\limits_{i=1}^n \{\lambda_i>0\}\subset \Gamma\subset \left\{\sum\lambda_i>0\right\}$

Since $f$ is symmetric, we also require $\Gamma$ to be symmetric.

The following are the assumptions for  (1) to be solvable,

$\psi\in C^\infty(\overline{\Omega})$, $\phi\in C^\infty(\partial \Omega)$

$\displaystyle \psi_0= \min_{\overline{\Omega}}\psi\leq \max_{\overline{\Omega}}\psi=\psi_1\quad (4)$

$f$ is a concave function  $(5)$

$\displaystyle \overline{\lim\limits_{\lambda\to\partial \Gamma}}f(\lambda)\leq \tilde{\psi_0}<\psi_0\quad (6)$

For every compact set $K$ in $\Gamma$ and every constant $C>0$, there is a number $R=R(K,C)$ such that

$\displaystyle f(\lambda_1,\lambda_2,\cdots,\lambda_n+R)\geq C$ for all $\lambda\in K\quad(7)$

$\displaystyle f(R\lambda)\geq C$ for all $\lambda\in K\quad (8)$

Also we need restrict $\partial \Omega$. There exists sufficiently large constant $R$ such that for every point on $\partial \Omega$, if $\kappa_1,\kappa_2,\cdots,\kappa_{n-1}$ are the principle curvatures of $\partial \Omega$

$\displaystyle (\kappa_1,\kappa_2,\cdots,\kappa_{n-1},R)\in\Gamma\quad (9)$

$\mathbf{Thm(CNS):}$ If $(2-9)$ are satisfied, then $(1)$ has a unique solution $u\in C^\infty(\overline{\Omega})$ with $\lambda(D^2u)\in \Gamma$.

$\mathbf{Proof:}$ The existence get from continuity method.

Krylov has shown how from a priori estimates

$|u|_{C^2(\overline{\Omega})}\leq C\quad (10)$

and uniform ellipticity of the linearized opeartor $L=\sum{F_{ij}}\partial_{ij}$ to derive

$\displaystyle |u|_{C^{2,\nu}({\overline{\Omega}})}\leq C$

So we only need to derive $(10)$ for any solution of $(1)$.

Using a maximum principle of fully nonlinear equation and $\psi\leq \psi_1$ and $(9)$, it is possible to construct a subsolution $\underline{u}$of $(1)$. So $u\geq \underline{u}$

If $\lambda(D^2u)\in \Gamma$ then $u$ can be bounded above by a harmonic function $v$ in $\Omega$.

$\underline{u}\leq u\leq v$

Additionally,                                      $|\nabla u|\leq C$ on $\partial \Omega$

Then differentiate $F(D^2u)=\psi$ to get a linear elliptic function. Using maximum principle to get $|u|_{C^1}\leq C$.

For the second derivative estimate, it is also estimate $u_{ij}$ on the boundary first and then differentiate $F(D^2u)=\psi$ another time to get a linear elliptic function of $u_{ij}$. And then apply maximum principle to get interior gradient estimate.

The bound of $u_{\alpha n}$, $\alpha is achieved from  constructing a barrier function.

The estimate of $u_{nn}$ is complicate. Still constructing a barrier function.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$Cafferelli, Nirenberg, Spruck: The Dirichlet problem for nonlinear second order elliptic equations III.

### Monge-Ampere equation and bounday behavior of domain

$\mathbf{Problem:}$ Suppose $\Omega\subset \mathbb{R}^n$ is a bounded domain and $\partial \Omega$ is $C^2$. If there exists a convex function $u\in C^2(\overline{\Omega})$ satisfies

$\displaystyle \begin{cases}\det D^2u=1 \text{ in }\Omega\\ u=0\quad \quad\text{ on }\partial \Omega\end{cases}$

Then $\Omega$ is uniformly convex. In other words, the principle curvature  of every point on $\partial \Omega$, namely $\kappa_1,\kappa_2,\cdots,\kappa_{n-1}$, are positive. Moreover, $\partial \Omega$ is connect.

$\mathbf{Proof:}$ For any boundary point of $\Omega$, we may suppose the point is origin. Since $\partial \Omega\in C^2$, there exists a neighborhood of 0 such that $\partial \Omega$ is represented by $x_n=\rho(x_1,x_2,\cdots,x_{n-1})$ with $\rho\in C^2$. Choosing the principle coordinate system, poositive $x_n$ axis is interior normal at origin and

$\displaystyle \rho(x')=\frac{1}{2}\sum\limits_{i=1}^{n-1}\kappa_ix_i^2+O(|x'|^3)$ here $x'=x_1,x_2,\cdots, x_{n-1}$.

Since $u(x',\rho(x'))=0$ near the origin, it following, on differentiation,

$\displaystyle u_{ij}=-u_n\rho_{ij}=-u_n\kappa_i\delta_{ij}$    for $i,j.

So at origin, we get

$\displaystyle D^2u=\left( \begin{array}{cccc} \kappa_1 & 0 &\cdots & u_{1n} \\ 0 & \kappa_2 & \ldots & u_{2n} \\ \vdots &\vdots & \ddots &\vdots\\ u_{1n}&u_{2n}& \cdots & u_{nn} \end{array} \right)$

which means

$\displaystyle 1=\det D^2u=|u_n|^{n-2}\prod\limits_{i=1}^{n-1}\kappa_i\left\{|u_n|u_{nn}-\sum\limits_{i=1}^{n-1}\frac{(u_{in})^2}{\kappa_i}\right\}.$

So $\kappa_i\neq 0$ on $\partial \Omega$ and for all $\, i=1,2,\cdots,n$. However, $\kappa_i$ is continuous function on $\partial \Omega$, which means $\kappa_i$ does not change sign on $\partial \Omega$.

Since $u$ is convex and $u=0$ on $\partial \Omega$, then $u\leq 0$ in $\Omega$. Also we know that $D^2u$ is positive definite matrix. Thus $\Delta u>0$ in $\Omega$, by the hopf lemma,

$u_n>0$ at $\partial \Omega$

So near origin $\kappa_iu_n$ is almost $u_{ii}$, which is positive by the property of $D^2u$. So $\kappa_i$ is positive.

For the argument $\partial \Omega$ is connect, see the paper in the following remark.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ Caffarellli, Nirenberg and Spruck. The Dirichlet problem for nonlinear second order elliptic equations,III:Functions of the eigenvalues of the Hessian.

Also refer to the formula [GT, p471].

### Improvement of Aleksandrov maximum principle

$\mathbf{Problem:}$ For $u\in C^2(\Omega)\cap C^0(\overline{\Omega})$ we have

$\displaystyle \sup\limits_{\Omega}u\leq \sup\limits_{\partial \Omega}u+\frac{d}{2w_n^{\frac{1}{n}}}\left(\int_{\Gamma^+}|det D^2u|\right)^{\frac{1}{n}}$

where $d=diam \Omega$.

There is an easy proof when the RHS is $\displaystyle \frac{d}{w_n^{\frac{1}{n}}}$. We will adjust the original proof.

$\mathbf{Proof:}$ By replacing $u$ with $\displaystyle u-\sup\limits_{\partial \Omega}u$, it suffices to assume $u\leq 0$ on $\partial \Omega$. Suppose $0\in\Omega$ and $u$ achieves maximum $M$ at 0.

Firstly, $\displaystyle |\chi(\Gamma^+)|\leq \int_{\Gamma^+}|det D^2u|$. Let $v$ be the function whose graph is the cone $K$ with vertex $(0,M)$ and base $\partial \Omega$. Since $diam(\Omega)=d$, there exists a ball $B_\frac{d}{2}(x_0)$ such that $\Omega\subset B_\frac{d}{2}(x_0)$. Let $\bar{v}$ be the function whose graph is cone $\bar{K}$ with vertex $(0,M)$ and base $B_\frac{d}{2}(x_0)$. Then

$\chi_{\bar{v}}(B_\frac{d}{2}(x_0))\subset \chi_{v}(\Omega)\subset \chi(\Omega)$

Define $d(y)$ is the length of the ray in direction $y$ tha lies in $B_\frac{d}{2}(x_0)$,

$\displaystyle \chi_{\bar{v}}(y)=\begin{cases}-\frac{M}{d(y)}\frac{y}{|y|}\quad y\neq 0, y\in B_\frac{d}{2}(x_0)\\ E\quad\quad\quad y=0\end{cases}$

$E$ is defined as the region bounded by $-\frac{M}{d(y)}\frac{y}{|y|}$, $y\in \partial B_\frac{d}{2}(x_0)$. This is because $p\in\mathbb{R}^n$ such that $\bar{v}(y)\leq \bar{v}(0)+p\cdot y$ means that

$\displaystyle M-\frac{M|y|}{d(y)}\leq M+p\cdot y\quad \forall\, y\in B_\frac{d}{2}(x_0)$

$\displaystyle -\frac{M|y|}{d(y)}\leq p\cdot y\quad \forall\, y\in B_\frac{d}{2}(x_0).$

Suppose a bounded region $G$ in $\mathbb{R}^n$ with boundary $(r(\omega), \omega)$ , $\omega\in(\mathbb{R}^{n-1})$,  then the volume of $\displaystyle G=\frac{1}{n}\int_{\mathbb{S}^{n-1}}r^n(\omega)d\sigma(\omega)$. With this fact in hand,

$\displaystyle |\chi_{\bar{v}}(B_\frac{d}{2}(x_0))|=\frac{1}{n}\int_{\mathbb{S}^{n-1}}\left(\frac{M}{d(\omega)}\right)^nd\sigma(\omega)$ and $\displaystyle |B_\frac{d}{2}(x_0)|=\frac{1}{n}\int_{\mathbb{S}^{n-1}}d(\omega)^nd\sigma(\omega)$

By the holder inequality

$\displaystyle |\chi(B_\frac{d}{2}(x_0))||B_\frac{d}{2}(x_0)|\geq \left(\frac{1}{n}\int_{\mathbb{S}^{n-1}}\sqrt{M^n}d\sigma(\omega)\right)^2=M|B_1(0)|^2$.

So $\displaystyle |\chi(B_\frac{d}{2}(x_0))|\geq w_n\left(\frac{2M}{d}\right)^n$

$\displaystyle M\leq \frac{d}{2w_n^{\frac{1}{n}}}\left(\int_{\Gamma^+}|det D^2u|\right)^{\frac{1}{n}}$

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$Gilbarg Trudinger’s book chapter 9. 9.1

### Strong barrier function and generalized linear elliptic equation

Suppose $\Omega$ is a domain in $\mathbb{R}^n$. $x_0\in \partial \Omega$ satisfies the exterior sphere condition. That is there exists a ball $B=B_R(y)$ such that $\bar{B}\cap \overline{\Omega}=x_0$. Then the function
$\displaystyle w(x)=\begin{cases}R^{2-n}-|x-y|^{2-n} \quad n\geq 3\\log\frac{|x-y|}{R} \qquad n=2\end{cases}$
is a barrier function for $\Delta$ at $x_0$.
For a strong barrier function $w$ at $x_0$, we mean $\Delta u\leq -1$ or $Lu\leq -1$, at such $x_0$ Define

$w=K(R^{-\alpha}-|x-y|^{-\alpha})$

Then $\displaystyle Lw=K\alpha|x-y|^{-\alpha-4}\left[-(\alpha+2)a^{ij}(x_i-y_i)(x_j-y_j)+|x-y|^2( a^{ii}+b^i(x_i-y_i))\right]+cw$
Assume $c\leq 0$ and let $|x-y|=r$
$Lw\leq K\alpha r^{-\alpha-4}\left[-(\alpha+2)a^{ij}(x_i-y_i)(x_j-y_j)+r^2(a^{ii}+b^i(x_i-y_i))\right]\leq -1$
when $\alpha=\alpha(\lambda, \Lambda,diam(\Omega),R)$ and $K$ large enough.
$\mathbf{Problem: }$ Let $u\in C^2(\Omega)\cap C^0(\overline{\Omega})$ be a solution of $Lu=f$ in a bounded $C^1$ domain $\Omega$ $n\geq 3$. Suppose $x_0\in\partial \Omega$ satisfy the exterior ball condition with $B_R(y)$. Moreover, assume

$a^{ij}\xi_i\xi_j\geq \lambda |\xi|^2$
$|a^{ij}|, |b^i|, |c|\leq \Lambda$

Suppose $u|_{\partial \Omega}=\phi$, $\phi\in C^2(\overline{\Omega})$. Show that $u$ satisfies a Lipschitz condition at $x_0$

$|u(x)-u(x_0)|\leq K(x-x_0)$, $x\in \Omega$.

where $K=K(\lambda,\Lambda,R,diam(\Omega), \sup|f|)$. If the sign $c$ is unrestricted, show that the same result holds provided $K$ also depends on $\sup|u|$.

$\mathbf{Proof:}$ When $c\leq 0$ suppose $w_0$ is the strong barrier function of $L$ at $x_0$, then there exists $M=M(\sup\limits_{\Omega}|f|, |\phi|_{2;\Omega})$ such that $\displaystyle w=Mw_0$ satisfies
$w+u(x_0)-u\geq 0$ on $\partial \Omega$ and $L(w+u(x_0)- u)\leq 0$

$-w+u(x_0)-u\leq 0$ on $\partial \Omega$ and $L(-w+u(x_0)-u)\geq 0$
By the maximum principle, we know that

$-w(x)\leq u(x)-u(x_0)\leq w(x)$ $|u(x)-u(x_0)|\leq w(x)=w(x)-w(x_0)\leq K|x-x_0|$.

When the sign of $c$ is undefined, let $\tilde{L}u=Lu-c^+u=f-c^+u$, apply the above procedure to $\tilde{L}$, only need to adjust that $M=M(\sup\limits_{\Omega}|f|, |\phi|_{2;\Omega},\Lambda,\sup|u|)$.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$Gilbarg Trudinger’s book. Chapter, exercise 3.6.

### One simple maximum principle

$\mathbf{Problem:}$If $L$ is elliptic with $c<0$ in a bounded domain $\Omega$, and $u\in C^2(\Omega)\cap C^0(\overline{\Omega})$ satisfies $Lu=f$ in $\Omega$, then
$\displaystyle \sup\limits_{\Omega}\leq \sup\limits_{\partial \Omega}|u|+\sup\limits_{\Omega}|f/c|$

$\mathbf{Proof:}$ Let $v=u-\sup\limits_{\Omega}|u|-\sup\limits_{\Omega}|f/c|$

Then $Lv=Lu-c(\sup\limits_{\Omega}|u|-\sup\limits_{\Omega}|f/c|)\geq 0$ and $v(x)\leq 0$ on $\partial \Omega$. By the maximum principle, we conclude, $v\leq 0$
So $u\leq \sup\limits_{\Omega}|u|-\sup\limits_{\Omega}|f/c|$
We also have $-u$ satisfies the above inequality, so the conclusion holds.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$