Category Archives: PDE


Unit normal to a radial graph over sphere

Consider \Omega\subset \mathbb{S}^n is a domain in the sphere. S is a radial graph over \Omega.

\boldsymbol{F}(x)=\{v(x)x:x\in \Omega\}

What is the unit normal to this radial graph?

Suppose \{e_1,\cdots,e_n\} is a smooth local frame on \Omega. Let \nabla be the covariant derivative on \mathbb{S}^n. Tangent space of S consists of \{\nabla_{e_i}\boldsymbol{F}\}_{i=1}^n which are

\nabla_{e_i}\boldsymbol{F}=v(x)e_i+e_i(v)\cdot x

In order to get the unit normal, we need some simplification. Let us assume \{e_i\} are orthonormal basis of the tangent space of \Omega and \nabla v=e_1(v)e_1. Then

\nabla_{e_1}\boldsymbol{F}=v(x)e_1+e_1(v)x, \quad \nabla_{e_i}\boldsymbol{F}=v(x)e_i, \quad i\geq 2

Then we obtain an orthonormal basis of the tangent of S

\{\frac{1}{\sqrt{v^2+|\nabla v|^2}}\nabla_{e_1}\boldsymbol{F},\nabla_{e_2}\boldsymbol{F},\cdots,\nabla_{e_n}\boldsymbol{F}\}

We are able to get the normal by projecting x to this subspace

\nu=x-\frac{1}{v^2+|\nabla v|^2}\langle x,\nabla_{e_1}\boldsymbol{F}\rangle\nabla_{e_1}\boldsymbol{F}=\frac{v^2x-v\nabla v}{v^2+|\nabla v|^2}.

After normalization, the (outer)unit normal can be written

    \frac{vx-\nabla v}{\sqrt{v^2+|\nabla v|^2}}

Remark: Guan, Bo and  Spruck, Joel. Boundary-value Problems on \mathbb{S}^n for Surfaces of Constant Gauss Curvature.


Principle curvature of translator

Suppose {\Sigma\subset\mathbb{R}^{3}} is a translator in {e_{3}} direction. Denote {V=e_{3}^T} and {A} is the second fundamental form of {\Sigma}. Then it is well know that

\displaystyle \Delta A+\nabla_VA+|A|^2A=0 \ \ \ \ \ (1)

Choose a local orthonormal frame {\{\tau_1,\tau_2\}} such that {A(\tau_1,\tau_1)=\lambda}, {A(\tau_2,\tau_2)=\mu} and {A(\tau_1,\tau_2)=0} in the neighborhood of some point {p}. We want to change (1) to some expression on {\lambda} or {\mu}. To do that, we need to apply both sides of (1) to {\tau_1,\tau_1}, getting

\displaystyle (\nabla_VA)(\tau_1,\tau_1)=\nabla_V \lambda-2A(\nabla_V \tau_1,\tau_1)=\nabla_V\lambda \ \ \ \ \ (2)

where we have used {\nabla_V\tau_1\perp \tau_1}. Similarly

\displaystyle (\nabla_{\tau_k} A)(\tau_1,\tau_1)=\nabla_{\tau_k}\lambda

\displaystyle (\nabla_{\tau_k}A)(\tau_1,\tau_2)=(\lambda-\mu)\langle \nabla_{\tau_k}\tau_1,\tau_2\rangle

Now let us calculate the Laplacian of second fundamental form

\displaystyle (\nabla_{\tau_k}^2A)(\tau_k,\tau_1)=\tau_k(\nabla_{\tau_k}\lambda)-2(\nabla_{\tau_k}A)(\nabla_{\tau_k}\tau_1,\tau_1)-(\nabla_{\tau_k}\tau_k)\lambda

\displaystyle =\tau_1(\nabla_{\tau_1}\lambda)-2\frac{(\nabla_{\tau_k}A)(\tau_1,\tau_2)^2}{\lambda-\mu}.


\displaystyle (\Delta A)(\tau_1,\tau_1)=(\nabla_{\tau_k}^2A)(\tau_k,\tau_1)= \Delta \lambda-2\sum_{k=1}^2\frac{|A_{12,k}|^2}{\lambda-\mu}.

Combining all the above estimates to (1), we get

\displaystyle \Delta \lambda+\nabla_V\lambda+|A|^2\lambda-2\sum_{k=1}^2\frac{|A_{12,k}|^2}{\lambda-\mu}=0.

where we write {A_{12,k}=(\nabla_{\tau_k}A)(\tau_1,\tau_2)^2}. Using this equation and the other one on {\mu}, one can derive that if {\Sigma} is mean convex then it is actually convex.

Laplacian on graph

Suppose {\Omega\subset \mathbb{R}^n} is a domain and {\Sigma=graph(F)\subset \mathbb{R}^{n+1}} is a hypersurface, where {F=F(u_1,\cdots,u_n)} is a function on {\Omega}. Define {f=f(u_1,\cdots,u_n)} on {\Omega}. Then {f} also can be considered as a function on {\Sigma}. How do we understand {\Delta_\Sigma f}?

Denote {\partial_i=\frac{\partial}{\partial{u_i}}} for short. If we pull the metric of {\mathbb{R}^{n+1}} back to {\Omega}, denote as {g}, then

\displaystyle g_{ij}=g(\partial_i,\partial_j)=\delta_{ij}+F_{u_i}F_{u_j},\quad g^{ij}=\delta_{ij}-\frac{F_{u_i}F_{u_j}}{W^2},\quad \det g=W^2

where {W=\sqrt{1+|\nabla F|^2}} and {F_{u_i}=\frac{\partial F}{\partial u_i}}. Then one can use the local coordinate to calculate {\Delta_\Sigma f}

\displaystyle \Delta_\Sigma f=\frac{1}{\sqrt{\det g}}\partial_{u_i}\left(\sqrt{\det g}\,g^{ij}f_{u_j}\right)

\displaystyle =g^{ij}f_{u_iu_j}+f_{u_j}\frac{\partial_{u_i}(\sqrt{\det g}\,g^{ij})}{\sqrt{\det g}}

also one can see from another definition of Laplacian

\displaystyle \Delta_\Sigma f=g^{ij}Hess(f)(\partial_i,\partial_j)=g^{ij}[\partial_j(\partial_i(f))-(\nabla_{\partial_j}\partial_i)f]

\displaystyle =g^{ij}f_{u_iu_j}-g^{ij}(\nabla_{\partial_j}\partial_i)f

By using the expression of {g^{ij}} stated above, we can calculate

\displaystyle \frac{\partial_{u_i}(\sqrt{\det g}\,g^{ij})}{\sqrt{\det g}}=\frac{-F_{u_j}F_{u_{i}u_{k}}g^{ik}}{W^2}

It follows from the definition of tangential derivative on {\Sigma}, see, that

\displaystyle \langle\nabla^\Sigma f, e_{n+1}\rangle=g^{ij}f_{u_{i}}F_{u_{j}}=f_{u_{i}}F_{u_{i}}-\frac{f_{u_i}F_{u_i}|\nabla F|^2}{W^2}=\frac{f_{u_{i}}F_{u_{i}}}{W^2}


\displaystyle f_{u_j}\frac{\partial_{u_i}(\sqrt{\det g}\,g^{ij})}{\sqrt{\det g}}=-\langle \nabla^\Sigma f,e_{n+1}\rangle F_{u_iu_k}g^{ik}=-\langle \nabla^\Sigma f,e_{n+1}\rangle HW

where {H} is the mean curvature of the {\Sigma}. Combining all the above calculations,

\displaystyle \Delta_\Sigma f=g^{ij}f_{u_iu_j}-\langle \nabla^\Sigma f,e_{n+1}\rangle HW

One example of blowing up corner

Consider {u(x,y)=\sqrt{x^2+y^4}} on {\mathbb{R}^2_+=\{(x,y):x\geq 0, y\geq 0\}}. Notify {\mathbb{R}^2_+} has a corner at the origin and {u} is not smooth at the origin. {u\approx x} when {x\geq y^2} and {u\approx y^2} when {x\leq y^2}. We want to resolve {u} by blowing up the origin through a map {\beta}. After blowing up, {W} looks like the following picture.



Denote {W=[\mathbb{R}^2_+,(0,0)]} and {\beta:W\rightarrow \mathbb{R}^2_+} is the blow down map. On {W\backslash lb}(near A), {\beta} takes the form

\displaystyle \beta_1(\xi_1,\eta_1)=(\xi_1^2,{\xi_1\eta_1})

where {\xi_1} is the boundary defining function for ff and {\eta_1} is boundary defining function for rb. Similarly on {W\backslash rb}(near B), {\beta} takes the form

\displaystyle \beta_2(\xi_2,\eta_2)=(\xi_2\eta^2_2,\eta_2)

where {\xi_2} is a bdf for lb and {\eta_2} is a bdf for ff. One can verify that {\beta} is a diffeomorphism {\mathring{W}\rightarrow \mathring{\mathbb{R}}^2_+}. Let {w=\beta^* u}. Then {w} is a polyhomogeneous conormal function on {W}. Its index can be denoted {(E,F,H)} correspond to lb, ff and rb.

\displaystyle E=\{(n,0)\}, F=\{(2n,0)\}, H=\{(2n,0)\}

Suppose {\pi_1} is the projection to {x} coordinate. Consider {f=\pi_1\circ \beta:W\rightarrow \mathbb{R}_+}, {f} is actually a {b-}fibration. Then the push forward map {f_*} maps {w} to a polyhomogeneous function on {\mathbb{R}^+}.

\displaystyle f_*w=\pi_* u=\int_0^\infty u(x,y)dy

In order to make {u} is integrable, let us assume {u} support {x\leq 1} and {y\leq 1}. What is the index for {f_* w} on {\mathbb{R}^+}?

\displaystyle \int_0^1\sqrt{x^2+y^4}dy=\int_0^{\sqrt{x}}\sqrt{x^2+y^4}dy+\int_{\sqrt{x}}^1\sqrt{x^2+y^4}dy

For the first integral, letting {y^2/x=t}

\displaystyle \int_0^{\sqrt{x}}\sqrt{x^2+y^4}dy=x\sqrt{x}\int_0^1\sqrt{1+t^4}dt=c_0x\sqrt{x}

For the second integral, letting {x/y^2=t}

\displaystyle \int_{\sqrt{x}}^1\sqrt{x^2+y^4}dy=\frac{1}{2}x\sqrt{x}\int_x^1t^{-\frac{5}{2}}\sqrt{t^2+1}dt

Since the Taylor series

\displaystyle \sqrt{1+t^2}=1+\frac{1}{2}t^2-\frac{1}{8}t^4+\cdots,\quad \text{for }|t|<1


\displaystyle \int_{\sqrt{x}}^1\sqrt{x^2+y^4}dy=a_0+a_1x+a_2x\sqrt{x}+\cdots

Combining all the above analysis, the index for {f_*w} is {\{(n,0)\}\cup \{\frac{n}{2},0\}}

From another point of view, the vanishing order of {f} on each boundary hypersurface of {W} are {e_f(lb)=1}, {e_f(\text{ff})=2} and {e_f(rb)=0}. {f} maps lb and ff to the boundary of {\mathbb{R}^+}. Therefore the index of {f_*w} is contained in

\displaystyle \frac{1}{e_f(lb)}E\overline{\cup}\frac{1}{e_f(\text{ff})}F=E\overline{\cup}\frac{1}{2}F

Remark: Daniel Grieser, Basics of {b-}Calculus.

Frist and second variation of translating soliton

Suppose {F(x,t):\Sigma\times(-\epsilon,\epsilon)\rightarrow \mathbb{R}^{n+1}} be a variation of {\Sigma} with compact support and fixed boundary. Consider the weighted area functional

\displaystyle \mathcal{A}(\Sigma)=\int_\Sigma e^{x_{n+1}}dv

What is the critical point of this area functional? To see that,

\displaystyle \frac{\partial}{\partial t}\mathcal{A}(\Sigma)=\int_\Sigma F_t(x_{n+1})e^{x_{n+1}}dv+e^{x_{n+1}}\partial_t dv

From some basic calculation in minimal surface(see colding minicozzi’s book)

\displaystyle \partial_t dv=-\langle F_t,\mathbf{H}\rangle+div_{\Sigma}F_t^T

Stokes’ theorem implies

\displaystyle \int_\Sigma div_\Sigma F_t^Te^{x_{n+1}}dv=-\int_{\Sigma}\langle F_t^T,\nabla_\Sigma x_{n+1}\rangle e^{x_{n+1}}dv=-\int_\Sigma F_t^T(x_{n+1})e^{x_{n+1}}dv

Combining these above fact, we get

\displaystyle \partial_t\mathcal{A}(\Sigma)=\int_\Sigma F_t^{\perp}(x_{n+1})-\langle F_t,\mathbf{H}\rangle dv=\int_\Sigma \langle F_t^{\perp},N\rangle N(x_{n+1})-\langle F_t,\mathbf{H}\rangle dv

\displaystyle =\int_\Sigma \langle F_t,N\rangle \langle N,e_{n+1}\rangle-\langle F_t,\mathbf{H}\rangle dv=\int_\Sigma\langle F_t,\langle N,e_{n+1}\rangle N-\mathbf{H}\rangle dv

where {N} is the unit normal of {\Sigma}. Therefore the critical point of {\mathcal{A}} will satisfy

\displaystyle \mathbf{H}=\langle N, e_{n+1}\rangle N=e_{n+1}^{\perp}

This is so called translating soliton.

Now consider the second variation at a translation soliton,

\displaystyle \partial_t^2\mathcal{A}(\Sigma)=\int_\Sigma e^{x_{n+1}}\left[F_{tt}(x_{n+1})dv+2F_t(x_{n+1})^2dv+2F_t(x_{n+1})\partial_tdv+\partial_{tt}dv\right]

\displaystyle =\int_\Sigma e^{x_{n+1}}\left[F_{tt}(x_{n+1})dv+\partial_{tt}dv\right]


\displaystyle \partial_{tt}dv=-|\langle A(\cdot,\cdot),F_t\rangle|^2+|\nabla_\Sigma^NF_t|^2+div_\Sigma(F_{tt})

Suppose {F_t=\eta N}, where {\eta\in C_c^\infty(\Sigma)}, then {|\nabla_\Sigma^NF_t|^2=|\nabla \eta|^2}

\displaystyle \partial_t^2\mathcal{A}(\Sigma)=\int_\Sigma [|\nabla\eta|^2-|A|^2\eta^2]e^{x_{n+1}}dv

Bach flat four dimensional manifold and sigma2 functional

We want to find the necessary condition of being the critical points of {\int\sigma_2} on four dimensional manifold.

1. Preliminary

Suppose {(M^n,g)} is a Riemannian manifold with {n=4}. {P_g} is the Schouten tensor

\displaystyle P_g=\frac{1}{n-2}\left(Ric-\frac{R}{2(n-1)}g\right)

and denote {J=\text{\,Tr\,} P_g}. Define

\displaystyle \sigma_2(g)=\frac{1}{2}[(\text{\,Tr\,} P_g)^2-|P_g|_g^2]

\displaystyle I_2(g)=\int_M \sigma_2(g)d\mu_g

where {|P|_g^2=\langle P,P\rangle_g}. It is well known that {I_2(g)} is conformally invariant.

Suppose {g(t)=g+th} where {h} is a symmetric 2-tensor. We want to calculate the first derivative of {I_2(g(t))} at {t=0}. To that end, let us list some basic facts (see the book of Toppings). Firstly denote {(\delta h)_j=-\nabla^i{h_{ij}}} the divergence operator and

\displaystyle G(h)=h-\frac{1}{2}(\text{\,Tr\,} h) g

\displaystyle (\Delta_L h)_{ij}=(\Delta h)_{ij}-h_{ik}Ric_{jl}g^{kl}-h_{jk}Ric_{il}g^{kl}+2R_{ikjl}h^{kl}

where {\Delta_L} is the Lichnerowicz Laplacian. Then the first variation of Ricci curvature and scalar curvature are

\displaystyle \dot{R}=\delta^2h-\Delta(\text{\,Tr\,} h)-\langle h,Ric\rangle \ \ \ \ \ (1)

\displaystyle \dot{Ric}=-\frac{1}{2}\Delta_Lh-\frac{1}{2}L_{(\delta G(h))^\sharp}g=-\frac{1}{2}\Delta_Lh-\frac{1}{2}L_{(\delta h)^\sharp}g-\frac{1}{2}Hess(\text{\,Tr\,} h)

\displaystyle =-\frac{1}{2}\Delta_Lh-d(\delta h)-\frac{1}{2}Hess(\text{\,Tr\,} h)

where we were using upper dot to denote the derivative with respect to {t}.

2. First variation of the sigma2 functional

Lemma 1 {(M^4,g)} is a critical point of {I_2(g)} if and only if

\displaystyle \Delta P-Hess(J)+2\mathring{Rm}(P)-2JP-|P|_g^2g=0 \ \ \ \ \ (2)

where {(\mathring{Rm}(P))_{ij}=R_{ikjl}P^{kl}}.


\displaystyle \frac{d}{dt}\big|_{t=0} I_2(g(t))=\int_M J\dot J-\langle\dot P,P\rangle+\langle h,P\wedge P\rangle+\frac{1}{2}\sigma_2\text{\,Tr\,} h \,d\mu_g

where {(P\wedge P)_{ij}=P_{ik}P_{jl}g^{kl}}. Since we have

\displaystyle \int_M\langle P,\dot P\rangle\\ =\frac{1}{n-2}\int_M\langle P, \dot Ric-\dot Jg-Jh\rangle =\frac{1}{n-2}[\langle P, \dot Ric\rangle-\dot J J-J\langle h,P\rangle]

\displaystyle =\frac{1}{n-2}[-\frac{1}{2}\langle h,\Delta_L P\rangle+\langle h,Hess(J)\rangle-\frac{1}{2}\Delta J \text{\,Tr\,} h-\dot J J-J\langle h,P\rangle]

Plugging this into the derivative of {I_2} to get

\displaystyle (n-2)\frac{d}{dt}\big|_{t=0} I_2(g(t))\\ =\int_M\frac{1}{2}\langle h,\Delta_L P\rangle-\langle h,Hess(J)\rangle+\frac{1}{2}\Delta J \text{\,Tr\,} h\\

\displaystyle \quad +(n-1)\dot J J+J\langle h,P\rangle+(n-2)\langle h,P\wedge P\rangle+\frac{n-2}{2}\sigma_2\text{\,Tr\,} h d\mu_g

In order to simplify the above equation, we recall the definition of Lichnerowicz Laplacian {\Delta_L}

\displaystyle (\Delta_LP)_{ij}=(\Delta P)_{ij}-2P_{ik}Ric_{jl}g^{kl}+2R_{ikjl}P^{kl}

\displaystyle =(\Delta P)_{ij}-2(n-2)P_{ik}P_{jl}g^{kl}-2JP_{ij}+2R_{ikjl}P^{kl}

Apply (1) to get

(n-1)\dot J J=\frac12J\dot R=\frac12J[\delta^2h-\Delta(\text{\,Tr\,} h)-\langle h,Ric\rangle]
=\frac{1}{2}[\langle h, Hess(J)\rangle-\text{\,Tr\,} h\Delta J-(n-2)J\langle h, P\rangle-J^2\text{\,Tr\,} h]

Therefore we can simplify it to be

\displaystyle (n-2)\frac{d}{dt}\big|_{t=0} I_2(g(t)) =\int_M\frac{1}{2}\langle h,\Delta P\rangle-\frac{1}{2}\langle h, Hess(J)\rangle+h^{ij}R_{ikjl}P^{kl}

\displaystyle -\frac{n-2}{2}J\langle h,P\rangle-\frac{1}{2} J^2 \text{\,Tr\,} h+\frac{n-2}{2}\sigma_2\text{\,Tr\,} h d\mu_g

Let us denote {(\mathring{Rm}(P))_{ij}=R_{ikjl}P^{kl}}. Using the fact {n=4} and the definition of {\sigma_2},

\displaystyle \frac{d}{dt}\big|_{t=0} I_2(g(t))=\frac{1}{4}\int_M\langle h,Q\rangle d\mu_g


\displaystyle Q=\Delta P-Hess(J)+2\mathring{Rm}(P)-2JP-|P|_g^2g


Remark 1 It is easy to verify {\text{\,Tr\,} Q=0}, this is equivalent to say {I_2} is invariant under conformal change. More precisely, letting {h=2ug}, then

\displaystyle \frac{d}{dt}\big|_{t=0} I_2(g(t))=\frac{1}{4}\int_M\langle h,Q\rangle d\mu_g=\frac{1}{2}\int_M u\text{\,Tr\,} Q d\mu_g=0.

Remark 2 If {g} is an Einstein metric with {Ric=2(n-1)\lambda g}, then {P=\lambda g}, {J=n\lambda} and

\displaystyle \mathring{Rm}(P)=\lambda Ric=2(n-1)\lambda^2 g

It is easy to verify that {Q=0}. In other words, Einstein metrics are critical points of {I_2}.

Are there any non Einstein metric which are critical points of {I_2}?

Here is one example. Suppose {M=\mathbb{S}^2\times N}, where {\mathbb{S}^2} is the sphere with standard round metric and {(N,g_N)} is a two dimensional compact manifolds with sectional curvature {-1}. {M} is endowed with the product metric. We can prove {Ric=g_{S^2}-g_N}, {P=\frac{1}{2}g_{S^2}-\frac{1}{2}g_N}, {J=0}, {\mathring{Rm}(P)=g_{prod}} and consequently {Q=0}.

Note that the above example is a locally conformally flat manifold. For this type of manifold, we have the following lemma which can say

Lemma 2 Suppose {g} is locally conformally flat and {Q=0}, then

Proof: When {g} is locally conformally flat,

\displaystyle \mathring{Rm}(g)=JP+|P|_g^2g-2P\wedge P

{Q=0} is equivalent to

\displaystyle \Delta P-Hess(J)+|P|_g^2g-4P\wedge P=0

Actually this is equivalent to the Bach tensor {B} is zero. \Box

3. Another point of view

We have the Euler Characteristic formula for four dimensional manifolds

\displaystyle 8\pi^2\chi(M)=\int_M (|W|_g^2+\sigma_2) d\mu_g

therefore the critical points for {\int_M \sigma_2d\mu_g} will be the same as the critical points of {\int_M |W|_g^2d\mu_g}. However, the functional

\displaystyle g\rightarrow \int_M |W|_g^2d\mu_g

is well studied by Bach. The critical points of this functional satisfy Bach tensor equal to 0.

\displaystyle B_{ij}=\nabla^k\nabla^l W_{likj}+\frac{1}{2}Ric^{kl}W_{likj}

Obviously, {B=0} for Einstein metric, but not all Bach flat metrics are Einstein. For example {B=0} for any locally conformally flat manifolds.

Bubble functions under different setting

Bubble function can be defined either on \mathbb{R}^n, \mathbb{S}^n or \mathbb{B}^n. For the following notations, c_n will denote suitable constants which may be different from line to line.

  • For every \epsilon>0 and  \xi\in\mathbb{R}^n, define

\displaystyle u_{\epsilon,\xi}=c_n\left(\frac{\epsilon}{\epsilon^2+|x-\xi|^2}\right)^{\frac{n-2}{2}}

It is well know that -\Delta u= c_nu^{\frac{n+2}{n-2}}. Moreover (\mathbb{R}^n,u^{\frac{4}{n-2}}_{\epsilon,\xi}g_E) is isometric to the standard sphere minus one point.

  • For any a\in \mathbb{S}^n and \lambda>0 define

\displaystyle\delta(a,\lambda)=c_n\left(\frac{\lambda}{\lambda^2+1+(\lambda^2-1)\cos d(a,x)}\right)^{\frac{n-2}{2}}

where d(a,x) is the geodesic distance of a and x on \mathbb{S}^n. Actually \cos d(a,x)=a\cdot x

  • For each p\in \mathbb{B}^{n+1}, define \delta_p(x):\mathbb{S}^n\to \mathbb{R} by


Both the second and third one satisfy

\displaystyle \frac{4(n-1)}{n-2}\Delta_{\mathbb{S}^n}\delta-n(n-1)\delta+c_n\delta^{\frac{n-2}{n+2}}=0

If we make p=\frac{\lambda-1}{\lambda+1}a, the third one will be changed to the second one.

To get the second one from the first one, let us deonte \Phi_a:\mathbb{S}^n\to \mathbb{R}^n be the stereographic projection from point a. Then

\displaystyle \delta(a,\lambda)\circ \Phi^{-1}_a=c_n\left(\frac{\lambda(1+|y|^2)}{\lambda^2|y|^2+1}\right)^{\frac{n-2}{2}}=c_nu_{\lambda,0}u_{1,0}^{-1}

It should be able to see the third one from hyperbolic translation directly.

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.





Interior estimate for Monge Ampere equation

Suppose we have u is a generalized solution of the Monge-Ampere equation

\det(\nabla^2 u)=1 \text{ in } B_1\subset\mathbb{R}^n

when n=2, Heinz proved

    |\nabla^2 u|_{B_{1/2}}\leq \sup_{B_1}u

when n\geq 3, Pogorelov has a counter example. One can have a solution u\in C^1(B_1), but u\in C^{1,\beta}(B_1) for some \beta\in (0,1). See his book The Minkowski Multidimensional Problem, on page 83.

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.