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)<M\}}. 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<n 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<n.

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:}