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Unique continuation property on the boundary

I am writing a theorem proved by Jin Zhiren in his thesis.

 

Suppose \Omega is a smooth domain in \mathbb{R}^n, x_0\in \partial \Omega and u is a harmonic function in \Omega. If there exists A, b>0 such that

\displaystyle |u(x)|\leq Ae^{-\frac{b}{|x-x_0|}}\quad x\in \Omega

for |x-x_0| small, then u=0. If n=2, the same conclusion holds for the solutions of a general second order linear elliptic equation.

A borderline example for this theorem is u be the real part of e^{-1/z^\alpha}, \alpha\in (0,1). u is harmonic in the right half plane and u\leq Ae^{-1/|x|^\alpha} and consequently D^\beta u(0)=0.

 

 

 

 

Newton tensor

Suppose {A:V\rightarrow V} is a symmetric endomorphism of vector space {V}, {\sigma_k} is the {k-}th elementary symmetric function of the eigenvalue of {A}. Then

\displaystyle \det(A+tI)=\sum_{k=0}^n \sigma_k t^{n-k}

One can define the {k-}th Newton transformation as the following

\displaystyle \det(A+tI)(A+tI)^{-1}=\sum_{k=0}^{n-1}T_k(A)t^{n-k-1}

This means

\displaystyle \det(A+tI)=\sum_{k=0}^{n-1}T_k(A+tI)t^{n-k-1}

\displaystyle =T_0 t^n+\sum_{k=0}^{n-2}(A\cdot T_k(A)+T_{k+1}(A))t^{n-k-1}+T_{n-1}(A)

By comparing coefficients of {t}, we get the relations of {T_k}

\displaystyle T_0=1,\quad A\cdot T_k(A)+T_{k+1}(A)=\sigma_{k+1}I,\, 0\leq k\leq n-2\quad T_{n-1}(A)=\sigma_n

Induction shows

\displaystyle T_{k}(A)=\sigma_kI-\sigma_{k-1}A+\cdots+(-1)^kA^k

For example

\displaystyle T_1(A)=\sigma_1I-A

\displaystyle T_2(A)=\sigma_2-\sigma_1A+A^2

One of the important property of Newton transformation is that: Suppose {F(A)=\sigma_k(A)}, then

\displaystyle F^{ij}=\frac{\partial F}{\partial A_{ij}}=T_{k-1}^{ij}(A)

The is because

\displaystyle \frac{\partial }{\partial A_{ij}}\det(A+tI)=\det(A+tI)((A+tI)^{-1})_{ij}.

If {A\in \Gamma_k}, then {T_{k-1}(A)} is positive definite and therefore {F} is elliptic.

Remark: Hu, Z., Li, H. and Simon, U. . Schouten curvature functions on locally conformally flat Riemannian manifolds. Journal of Geometry, 88(1{-}2), (2008), 75{-}100.

Self-shrinker and polynomial volume growth

Proposition: If M is an entire graph of at most polynomial volume growth and H=\langle X,\nu\rangle, namely M is a self-shrinker. Then M is a plane.

Proof: Suppose

\displaystyle v=\frac{1}{\langle \nu, w\rangle}

Then one can derive the following equation

\displaystyle \Delta v=\langle\nabla v,X\rangle+|A|^2v+2v^{-1}|\nabla v|^2

Multiplying both sides by e^{-|X|^2/2} and integration on M, which makes sense because of the polynomial volume growth, we get

\int_M (\Delta v-\langle\nabla v,X\rangle)e^{-\frac{|X|^2}{2}}d\mu=\int_M (|A|^2v+2v^{-1}|\nabla v|^2)e^{-\frac{|X|^2}{2}}d\mu

However, integration by parts shows the LHS is zero. Thus A\equiv v\equiv 0, M must be a plane.

Some calculations of sigma_2

On four-manifold {(M^4,g_0)}, we define Shouten tensor

\displaystyle A = Ric-\frac 16 Rg

and Einstein tensor and gravitational tensor

\displaystyle E=Ric - \frac 14 Rg\quad S=-Ric+\frac{1}{2}Rg

Suppose {\sigma_2} is the elemantary symmetric function

\displaystyle \sigma_2(\lambda)=\sum_{i<j}\lambda_i\lambda_j

Thinking of {A} as a tensor of type {(1,1)}. {\sigma_2(A)} is defined as {\sigma_2} applied to eigenvalues of {A}. Then

\displaystyle \sigma_2(A)= \frac{1}{2}[(tr_g A)^2-\langle A, A\rangle_g] \ \ \ \ \ (1)

Notice {A=E+\frac{1}{12}Rg}. Easy calculation reveals that

\displaystyle \sigma_2(A)=-\frac{1}{2}|E|^2+\frac{1}{24}R^2 \ \ \ \ \ (2)

Under conformal change of metric {g=e^{2w}g_0}, we have

\displaystyle R= e^{-2w}(R_0-6\Delta_0 w-6|\nabla_0 w|^2) \ \ \ \ \ (3)

\displaystyle A=A_0-2\nabla^2_0 w+2dw\otimes dw-|\nabla_0w|^2g_0 \ \ \ \ \ (4)

\displaystyle S=S_0+2\nabla_0^2w-2\Delta_0wg_0-2dw\otimes dw-|\nabla_0 w|^2g_0 \ \ \ \ \ (5)

We want to solve the equation {\sigma_2(A)=f>0}, which is equivalent to solve

\displaystyle \sigma_2(A_0-2\nabla^2_0w+2dw\otimes dw-|\nabla_0w|^2g_0)=f

This is an fully nonlinear equation of Monge-Ampere type. Under local coordinates, the above equation can be treated as

\displaystyle F(\partial_i\partial_j w,\partial_kw,w,x)=f

where {F(p_{ij},v_k,s,x):\mathbb{R}^{n\times n}\times\mathbb{R}^n\times\mathbb{R}\times\mathbb{R}^n\rightarrow \mathbb{R}}. This equation is elliptic if the matrix {\left(\frac{\partial F}{\partial p_{ij}}\right)} is positive definite. In order to find that matrix, we need the linearized operator

\displaystyle L[\phi]=\frac{\partial F}{\partial p_{ij}}(\nabla_0^2\phi)_{ij}=\frac{d}{dt}|_{t=0}F(\partial_i\partial_j w+t\partial_i\partial_j\phi,\partial_kw,w,x) \ \ \ \ \ (6)

Using the elementary identity

\displaystyle \frac{d}{dt}\rvert_{t=0}\sigma_2(H+tG)=tr_gH\cdot tr_gG-\langle H, G\rangle_g. \ \ \ \ \ (7)

for any fixed matrix {H} and {G}. Now plug in {H=A} is Schouten tensor and {B=-2\nabla_0^2\phi}. One can calculate them as

\displaystyle tr_g H\cdot tr_g G=\langle \frac{1}{3}Rg, G\rangle_g \ \ \ \ \ (8)

Then we get

\displaystyle L[\phi]=\langle S,G\rangle_g=-2\langle S,\nabla^2_0\phi\rangle_g

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.

When the norm map is surjective

\mathbf{Problem:} Suppose F is finite field with q=p^m elements, E an Galois extension of F such that [E:F]=n. Prove that N_{E/F}:E^\ast\to F^\ast is surjective.

\mathbf{Proof:} From theorem 4.26 we know that \text{Gal }E/F is cyclic over F generated by \eta:a\to a^q.

\displaystyle N_{E/F}(a)=aa^qa^{q^2}\cdots a^{q^{n-1}}=a^{\frac{q^n-1}{q-1}}.

N_{E/F}:E^\ast\to F^\ast is a homomorphism. It is well known that  any finite subgroup of the multiplicative group of field is cyclic, which means E^\ast and F\ast is cyclic.

Since |E\ast|=q^n-1 and |F^\ast|=q-1, and N_{E/F}(a)=a^m, where m={\frac{q^n-1}{q-1}}, then |\text{Ker } N|=(q^n-1,m)=m and |\text{Im }N|=\frac{q^n-1}{(q^n-1,m)}=q-1. Thus N is surjective.

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

\mathbf{Remark:} Jacobson p300.

Welcome, bro.

Hi, everyone.

This is my formula: f(x) = \int_0^\infty g(s)\, ds