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Bach flat four dimensional manifold and sigma2 functional

October 5, 2017 – 12:25 am

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

Proof:

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

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 ${\sigma_2}$ ,

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

where

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

$\Box$

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.

Unique continuation property on the boundary

April 26, 2017 – 3:13 pm

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 $D^\beta u(0)=0$ .

By Sun's World | Also posted in PDE, Uncategorized | Tagged uniqueness continuation | Comments (0)

Interior estimate for Monge Ampere equation

April 24, 2017 – 6:16 pm

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.

By Sun's World | Also posted in PDE | Comments (0)

Newton tensor

April 3, 2017 – 2:56 pm

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.

By Sun's World | Also posted in PDE, Uncategorized | Tagged fully nonlinear | Comments (1)

Some calculations of sigma_2

March 20, 2017 – 2:12 pm

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

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

By Sun's World | Also posted in PDE, Uncategorized | Tagged sigma2 | Comments (0)

f-extremal disk

March 2, 2017 – 8:06 pm

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.

By Sun's World | Also posted in Geometry, PDE, Riemannian Geo | Tagged maximum principle | Comments (0)

Compensated compactness

February 25, 2017 – 5:02 pm

Suppose $T$ is a vector field and $\nabla\cdot T = 0$ . $E= \nabla \psi$ and $\psi$ is a scalar function. We have following theorem(Coifman-Lions-Meyers-Semmes)

Theorem: If $T\in L^2(\mathbb{R}^n)$ and $T\in L^2(\mathbb{R}^n)$ , then $E\cdot T\in \mathcal{H}^1(\mathbb{R}^n)$ , which is the hardy space.
Given $f(x)\in L^1(\mathbb{R}^n)$ , it has harmonic extension $\mathbb{R}^{n+1}_+=\{(x,t)|x\in\mathbb{R}^n, t>0\}$

$\tilde{f}(x,t)=c_n\int_{\mathbb{R}^n}\frac{ f(x-y)t}{(t^2+|x|^2)^{\frac{n+1}{2}}}dy$

Definition: the non-tangential maximal function

$N(f)=\sup_{(\xi,t)\in \Gamma(x)}|\tilde f(\xi, t)|$

It is easy to prove that $N(f)\leq c_n f^*(x)$ the Hardy-Littlewood maximal function. From this we can Hardy norm as

$||f||_{\mathcal{H}^1}=||f||_{L^1}+||N(f)||_{L^1}$

Hardy space consists of all $f$ having finite hardy norm. There is well know fact that the dual space of $\mathcal{H}^1$ is BMO, which is defined as the following.

Define $f\in L^1_{loc}(\mathbb{R}^n)$ , if for any cube $Q$ ,

$\sup_Q\frac{1}{|Q|}\int_Q|f-f_Q|<\infty,\quad \text{where }f_Q=\frac{1}{|Q|}\int_Qf$

then $f\in BMO$ . $L^\infty \subset BMO$ and $\log|x|\in BMO$ but not in $L^\infty$ .

Let us see how do we use the main theorem. Suppose on $\mathbb{R}^2$ , $u$ is the solution of the following elliptic equation

$\displaystyle\frac{\partial}{\partial x_i}\left(a_{ij}(x)\frac{\partial u}{\partial x_j}\right)=\frac{\partial f}{\partial x_1}\frac{\partial g}{\partial x_2}-\frac{\partial f}{\partial x_2}\frac{\partial f}{\partial x_2}$

where $||\nabla f||_{L^2}<\infty$ , $||\nabla g||_{L^2}<\infty$ and $(a_{ij})$ is uniform elliptic. YanYan Li and Sagun Chanillo proved that the green function of this elliptic operator belongs to BMO. The right hand side of this equation can be rewritten as $T\cdot E$ , where

$T=\left(\frac{\partial f}{\partial x_2}, -\frac{\partial f}{\partial x_1}\right),\quad E=\left(\frac{\partial g}{\partial x_1},\frac{\partial g}{\partial x_a}\right)$

therefore the right hand side belong to $\mathcal{H}^1$ . Since

$u(x)=\int G_x(y)T\cdot E(y)dy$

therefore from the theorem we stated at the beginning, we get

$||u||_\infty\leq C||\nabla f||_{L^2}||\nabla g||_{L^2}$

By Sun's World | Also posted in Analysis, Harmonic Analysis, PDE | Tagged hardy space | Comments (0)

Subcriticality and supercriticality

July 27, 2014 – 7:26 pm

Consider the equation

$\displaystyle \Delta u=u^p\text{ on }\mathbb{R}^n$

usually we call the equation is subcritical when ${p<\frac{n+2}{n-2}}$ , supercritical when ${p>\frac{n+2}{n-2}}$ . The reason comes from the scalling the solution. Suppose ${u(x)}$ is a solution of the equation, then ${u^\lambda(x)=\lambda^{\frac{2}{p-1}}u(\lambda x)}$ is another solution. Consider the energy possessed by ${u}$ around any point ${x_0}$ of radius ${{\lambda}}$ can be bounded

$\displaystyle \int_{B_{\lambda}(x_0)}|\nabla u(x)|^2dx\leq E$

when ${\lambda\rightarrow 0}$ , we scale ${B_\lambda(x_0)}$ to ${B_1(x_0)}$ , then ${u}$ will become ${u^\lambda}$ in order to be a solution and ${u^\lambda}$ lives on ${B_1(x_0)}$ . While the energy will be

$\displaystyle \int_{B_{1}(x_0)}|\nabla u^\lambda(x)|^2dx=\lambda^{\frac{4}{p-1}+2-n}\int_{B_{\lambda}(x_0)}|\nabla u(y)|^2dy$

If the ${\delta=\frac{4}{p-1}+2-n<0}$ , which is ${p> \frac{n+2}{n-2}}$ , the energy bound of ${u^\lambda}$ will become ${\lambda^\delta E}$ . Since ${\lambda\rightarrow 0}$ , the bound deteriorates by ‘zooming in’. In this case, we call the equation is supercritical. The solution looks more singular at this time.

Remark: The energy should include ${\int_{B_{\lambda}(x_0)}u^2dx}$ , but somehow this term scale differently with ${\int_{B_{\lambda}(x_0)}|\nabla u(x)|^2dx}$ and can not give one the critical exponent exactly.

By Sun's World | Also posted in PDE | Comments (0)

Approriate scalling in Yamabe equation

May 22, 2014 – 1:32 am

Suppose ${(M,g)}$ is a Riemannian manifold, and ${L_g=\Delta_g -\frac{n-2}{4(n-1)}R_g}$ is the conformal Laplacian. Assume ${u>0}$ satisfies

$\displaystyle L_gu+Ku^p=0$

where ${K}$ is some fixed constant, ${1<p\leq \frac{n+2}{n-2}}$ . Suppose near a point ${x_0\in M}$ , there is a coordinates ${x^1,x^2,\cdots, x^n}$ . We want to scale the coordinates to ${x^i=\lambda y^i}$ ,

$\displaystyle g(x)=g_{ij}(x)dx^idx^j=\lambda^2 g_{ij}(\lambda y)dy^idy^j=\lambda^2 \hat{g}(y)$

By the conformal invariance of ${L}$ , for any ${\phi}$ , we get

$\displaystyle L_{g}(\lambda^{-\frac{n-2}{2}}\phi)=\lambda^{-\frac{n+2}{2}}L_{\hat{g}}(\phi)$

We want to choose ${\phi(y)=\lambda^{\alpha}u(\lambda y)}$ such that

$\displaystyle L_{\hat{g}}(\phi)+K\phi^p=0$

which means

$\displaystyle L_{\hat{g}}(\lambda^{\alpha}u(\lambda y))=\lambda^{\frac{n+2}{2}}L_g (\lambda^{\alpha-\frac{n-2}{2}}u(\lambda y))=-K\lambda^{\alpha+2} (u(\lambda y))^p$

Letting

$\displaystyle \alpha+2=\alpha p$

we get ${\alpha=\frac{2}{p-1}}$ .

The above proof may not be right.

Or we should look it more directly

$\displaystyle L_g(u(x))=\lambda^2 L_{\hat{g}}(u(\lambda y))=\lambda^2 Ku(\lambda y)^p$

then

$\displaystyle L_{\hat{g}}(\lambda^\alpha u(\lambda y))=K(\lambda^\alpha u(\lambda y))^p$

with ${\alpha=\frac{2}{p-1}}$ .

By Sun's World | Also posted in PDE | Tagged yamabe | Comments (0)

Harnack inequality under scaling

November 7, 2013 – 4:46 pm

Thm: Suppose ${\Omega}$ is domain in ${\mathbb{R}^n}$ . ${u}$ is a harmonic function in ${\Omega}$ , ${u\geq 0}$ . For any subdomain ${\Omega'\subset\subset \Omega}$ , there exists a constant ${C}$ such that