Category Archives: Elliptic PDE

32519989

Non-Holder continuous solution

Consider the following elliptic equation on \mathbb{R}^2

\displaystyle \partial_{x_i}\left(\frac{1}{r^2(\log r)^2}\partial_{x_j}u\right)=0,\quad r^2=x_1^2+x_2^2<1

It has a solution u=(\log r^{-1})^{-1} which is not holder continuous around origin. Note that the above equation is not uniformly elliptic.

See the paper, Trudinger, on the regularity of generalized solutions of linear, non-unformly elliptic equations, 1970.

Another example is w(x_1,\cdots, x_n)=|x_1|\log^2\frac{1}{|x_1|} and the function u(x_1,\cdots, x_n)=\frac{\text{sign } x_1}{\log (1/|x_1|)} satisfy the elliptic equation \partial_i(w(x)\partial_j u)=0 in \mathbb{R}^n. This equation is degenerate elliptic. But w is not an A_2 weight.

Check the paper Fabes Kenig and Serapioni, the local regularity of solutions of degenerate elliptic equations. 1982.

Upper half plane and disc

Suppose {\mathbb{R}^{n+1}_+=\{(y,t)|y\in \mathbb{R}^n,t>0\}} is the upper half plane. Define the map {\pi: \mathbb{R}^{n+1}_+=\{(y,t)\}\rightarrow \mathbb{R}^{n+1}=\{(x,s)\}} by the following 

\displaystyle x=\frac{y}{|y|^2+(t+1)^2}

\displaystyle s=\frac{t+1}{|y|^2+(t+1)^2}-1

One can see that {\pi} maps {\mathbb{R}_+^n} onto the open ball {B=B_{\frac12}((0,-\frac12))\subset \mathbb{R}^{n+1}}. It is easy to verify that {\pi^{-1}} looks like 

\displaystyle y=\frac{x}{|x|^2+(s+1)^2}

\displaystyle t=\frac{s+1}{|x|^2+(s+1)^2}-1

Upper half plane and disc

We want to pull the metric of {\mathbb{R}^{n+1}_+} to {B}, that is {(\pi^{-1})^*(dy^2+dt^2)}. Denote {A=|x|^2+(s+1)^2}. We have 

\displaystyle (\pi^{-1})^*dy_i=A^{-1}dx_i-A^{-2}x_idA

\displaystyle (\pi^{-1})^*dt=A^{-1}ds-A^{-2}(s+1)dA

\displaystyle (\pi^{-1})^*(dy^2+dt^2)=A^{-2}(dx^2+ds^2)

Therefore, {(\pi^{-1})^*(dy^2+dt^2)} is conformal to {dx^2+ds^2}

Next, for some {\alpha\geq 0}, we want to pull the solution {u} of 

\displaystyle -\text{div}(t^\alpha \nabla u)=\alpha(n+\alpha-1)t^{\alpha-1}u^{\frac{n+\alpha+1}{n+\alpha-1}}

to {B}. That is defining {\psi(x,s)=A^{-\frac{n+\alpha-1}{2}}u} on {B}. We shall derive the equation {\psi} satisfy on {B}. Note that the equation means 

\displaystyle \Delta u+\frac{\alpha}{t}\partial_t u=-\alpha(n+\alpha-1)\frac{1}{t}u^{\frac{n+\alpha+1}{n+\alpha-1}}.

Let us use the following notations 

\displaystyle \beta=n+\alpha-1.

It follows from the covariant property of conformal laplacian that for any {u=u(y,t)}

\displaystyle \Delta_{(y,t)}u=A^{\frac{n+3}{2}}\Delta_{(x,s)}(A^{-\frac{n-1}{2}}u).

Note that {A^{-\frac{n-1}{2}}u=A^{\frac{\alpha}{2}}\psi}. We turn to calculate 

\displaystyle \Delta_{(x,t)}(A^{\frac{\alpha}{2}}\psi)=(\Delta A^{\frac{\alpha}{2}})\psi+A^{\frac{\alpha}{2}}\Delta\psi+2\nabla A^{\frac{\alpha}{2}}\nabla \psi.

It is not hard to see 

\displaystyle \Delta A^{\frac{\alpha}{2}}=\alpha\beta A^{\frac{\alpha}{2}-1},\quad 2\nabla A^{\frac{\alpha}{2}}\nabla \psi=2\alpha A^{\frac{\alpha}{2}-1}\nabla\psi\cdot(x,s+1).

Therefore 

\displaystyle \Delta_{(y,t)}u=A^{\frac{n+3}{2}}\Delta_{(x,s)}(A^{\frac{\alpha}{2}}\psi)=A^{\frac{\beta+4}{2}}\Delta\psi+2\alpha A^{\frac{\beta+2}{2}}\nabla \psi\cdot(x,s+1)+\alpha\beta A^{\frac{\beta+2}{2}}\psi.

Next, to handle the term {\partial_t u}, applying {\partial_t A=-2(1+t)A^{2}=-2(1+s)A^{3}}

\displaystyle \partial_t u=\partial_t(A^{\frac{\beta}{2}}\psi)=-\beta(1+t)A^{\frac{\beta+2}{2}}\psi+A^{\frac{\beta}{2}}\partial_x \psi[-2y(1+t)A^2]+

\displaystyle A^{\frac{\beta}{2}}\partial_s\psi[A-2(1+t)^2A^2].

That is 

\displaystyle \partial_t u=A^{\frac{\beta}{2}}(1+s)[-\beta\psi-2x\partial_x\psi-2(1+s)\partial_s\psi]+A^{\frac{\beta+2}{2}}\partial_s\psi

\displaystyle =-A^{\frac{\beta}{2}}(1+s)[\beta\psi+2\nabla\psi\cdot(x,s+\frac12)]+A^{\frac{\beta}{2}}[A-(1+s)]\partial_s\psi.

It follows from the transformation formula that 

\displaystyle \frac{\alpha}{t}\partial_t u=\frac{\alpha A}{s+1-A}\partial_t u

To summarize the above calculation, one the one hand, 

\displaystyle \Delta u+\frac{\alpha}{t}\partial_t u=A^{\frac{\beta+4}{2}}\Delta\psi-\frac{2\alpha}{s+1-A}A^{\frac{\beta+4}{2}}\nabla\psi\cdot(x,s+\frac12)-\frac{\alpha\beta}{s+1-A}A^\frac{\beta+4}{2}\psi

on the other hand 

\displaystyle -\alpha(n+\alpha-1)\frac{1}{t}u^{\frac{n+\alpha+1}{n+\alpha-1}}=-\alpha\beta\frac{A}{s+1-A}A^{\frac{n+\alpha+1}{2}}\psi^{\frac{n+\alpha+1}{n+\alpha-1}}.

Then we get the equation of {\psi} 

\displaystyle \Delta \psi-\frac{2\alpha \nabla\psi\cdot(x,s+\frac12)}{s+1-A}=\alpha\beta\frac{\psi-\psi^{\frac{n+\alpha+1}{n+\alpha-1}}}{s+1-A}.

Notice that {s+1-A=\frac14-r^2} where {r=\sqrt{|x|^2+(s+\frac12)^2}} is the distance of {(x,s)} to the center {(0,-\frac12)} of the the ball {B}. The equation that {\psi} satisfies is rotationally symmetric with respect to the center of {B}

Green’s function and parametrix

Suppose {(M^n,g)} is a Riemannian manifold without boundary with {n\geq 2}. Consider the Beltrami-Laplacian on {M}

\displaystyle \Delta_g f= |g|^{-\frac{1}{2}}\partial_i[|g|^{\frac12}g^{ij}\partial_jf]

We want to find the Green’s function {G} for {\Delta}, namely 

\displaystyle \Delta^{dist}_{g,y} G(x,y)=\delta_x(y)

in distribution sense, or equivalently for any {\psi,\phi\in C^2(M)}

\displaystyle \phi(x)=\int_{M}G(x,y)\Delta_g \phi(y)dv_y.

\displaystyle \psi(y)=\Delta_{g,y}\int_M G(x,y)\psi(x)dv_x

Moreover {\psi\rightarrow \int_M G\psi dv} looks like the inverse operator of {\Delta_g}. Of course such {G} would only be unique modulo some constants.

To do that, we know on {(\mathbb{R}^n,|dx|^2)}, the Green function is 

\displaystyle K(x,y)=\begin{cases}\frac{1}{n(2-n)\omega_n}|x-y|^{2-n}& n>2\\\frac{1}{2\pi}\log|x-y|,&n=2.\end{cases}

In a general Riemannian manifold, {G} should look like {K} more or less. 

Suppose {M} has injectivity radius {\delta} and {\eta} is a smooth radial cut off function on {\mathbb{R}^n} whose support lies in {B_\delta(0)} and {\eta=1} in {B_{\delta/2}(0)} . Then {K(x,y)\eta(|x-y|)} is well defined on {M} through local coordinate patches, which we still deonte as {K}. Easy computation shows 

\displaystyle |\Delta_{g,y}K(x,y)|\leq C|x-y|^{2-n}\in L^1(M).

Green’s formula reads 

\displaystyle \psi(y)=\int_M K(x,y)\Delta \psi(y)dv_y-\int_M\Delta_yK(x,y)\psi(y)dv_y

for all {\psi\in C^2(M)}. This actually means 

\displaystyle \Delta_{g,y}^{dist} K(x,y)=\Delta_{g,y}K(x,y)+\delta_x(y)

and 

\displaystyle \int_M \Delta_{g,y}K(x,y)dv_y=-1.

After changing the order of integration in the Green’s formula, we could establish that 

\displaystyle \phi(y)=\Delta_{g,y}\int_M K(x,y)\phi(x)dv_x-\int_M \Delta_{g,y}K(x,y)\phi(x)dv_x

for any {\phi\in C^2(M)}. One can compare this with the equation {G} satisfy. Denote {\Gamma_1=\Delta_{g,y}K(x,y)\in L^1}. Define two operators 

\displaystyle Q\phi(y)=\int_M K(x,y)\phi(x)dv_x

\displaystyle R\phi(y)=\int_{M}\Gamma_1(x,y)\phi(x)dv_x

Also denote {\Gamma_{k+1}=R\Gamma_k}. Then Green’s formula means 

\displaystyle Q\Delta_g=\Delta_g Q=I-R

To find the Green’s function, it is equivalent to find the inverse operator of {\Delta_g}. One can apply the idea of parametrix, which says the inverse operator can be constructed by 

\displaystyle (I-R)^{-1}Q=Q+RQ+R^2Q+\cdots

Therefore if we define {\Gamma_k=} 

\displaystyle G(x,y)=K(x,y)+\sum \int_{M}\Gamma_k(x,z) K(z,y)dv_z+remainder

Some calculation of sigma2 II

We want find the critical metric of the following functional restricted to the space of conformal metrics of unit volume 

\displaystyle I(g)=\frac{n-4}{2}t\int_M J^2_gd\mu_g+4\int_M\sigma_2(A_g)d\mu_g

Here {(M^n,g)} is a Riemannian metric with {n\geq 5}, {A} is the Schouten tensor and {J} is {tr_gA}.

Now let us differentiate the two terms respectively, suppose {g(s)=(1+s\psi)^{\frac{4}{n-4}}g=u(s)^{\frac{4}{n-2}}g}. From conformal change property of scalar curvature, we get 

\displaystyle J_{g(s)}=u(s)^{-\frac{n+2}{n-2}}\left(-\frac{2}{n-2}\Delta_{g}u(s)+Ju(s)\right)

\displaystyle \dot J=-\frac{2}{n-2}\Delta_g\dot u-\frac{4}{n-2}J\dot u

Then 

\displaystyle \frac{d}{ds}|_{t=0}\int_{M}J_{g(s)}^2d\mu_{g(s)}=\int_{M}2J\dot{J}+\frac{2n}{n-2}J^2\dot ud\mu

\displaystyle =\int_M -\frac{4}{n-2}J\Delta \dot u+\frac{2(n-4)}{n-2}J^2\dot ud\mu

\displaystyle =\int_M\left(-\frac{4}{n-4}\Delta J+2J^2\right)\psi d\mu

where we have used the fact {\dot u(0)=\frac{n-2}{n-4}\psi}. Next, it is easy to know 

\displaystyle \frac{d}{ds}|_{t=0}\int_M \sigma_2(A_{g(s)})d\mu_{g(s)}=2\int_{M}\sigma_2(A)\psi d\muSo 

\displaystyle I'(g)\psi=2t\int_{M}\left(-\Delta J+\frac{n-4}{2}J\right)\psi d\mu_g+8\int_{M}\sigma_2(A)\psi d\mu_g

Now assume {g(s)} keep the unit volume infinitesimally, that is {\int_M \psi d\mu=0}, then {I'(g)=0} under this constraint means 

\displaystyle t\left(-\Delta J+\frac{n-4}{2}J\right)+4\sigma_2(A)=const.

Remark: M.J. Gursky, F. Hang, Y.-J. Lin, Riemannian Manifolds with Positive Yamabe Invariant and Paneitz Operator, Int. Math. Res. Not.

Concavity maximum principle

Let u\in C^2(\Omega)\cap C(\bar{\Omega}) satisfy the elliptic equation

\displaystyle Lu=a^{ij}(Du)u_{ij}-b(x,u,Du)=0\quad \text{in }\Omega

Assume b is jointly concave with respect to (x,u) and

\displaystyle \frac{\partial b}{\partial u}\geq 0

Then

\mathcal{C}(y_1,y_3,\lambda)=u(\lambda y_3+(1-\lambda)y_1)-\lambda u(y_3)-(1-\lambda)u(y_1)

can not achieve positive maximum in the interior of \Omega\times \Omega.

See the paper Korevaar 1983

Some identities related to mean curvature of order m

For any {n\times n} (not necessarily symmetric) matrix {\mathcal{A}}, we let {[\mathcal{A}]_m} denote the sum of its {m\times m} principle minors. For any hypersurface which is a graph of {u\in C^2(\Omega)}, where {\Omega\subset \mathbb{R}^n}. We have its downward unit normal is

\displaystyle (\nu,\nu_{n+1})=\left(\frac{Du}{\sqrt{1+|Du|^2}},\frac{-1}{\sqrt{1+|Du|^2}}\right).

The principle curvatures are taken from the eigenvalues of the Jacobian matrix {[D\nu]}. One can define its {m} mean curvature using the notation of above

\displaystyle H_m=\sum_{i_1<\cdots<i_m}\kappa_{i_1}\cdots\kappa_{i_m}=[D\nu]_m

Now let us consider general matrix {\mathcal{A}},

\displaystyle A_m=[\mathcal{A}]_m=\frac{1}{m!}\sum \delta\binom{i_1,\cdots,i_m}{j_1,\cdots,j_m}a_{i_1j_1}\cdots a_{i_mj_m}

where {\delta} is the generalized Kronecker delta.

\displaystyle \delta\binom{i_1 \dots i_p }{j_1 \dots j_p} = \begin{cases} +1 & \quad \text{if } i_1 \dots i_p \text{ are distinct and an even permutation of } j_1 \dots j_p \\ -1 & \quad \text{if } i_1 \dots i_p \text{ are distinct and an odd permutation of } j_1 \dots j_p \\ \;\;0 & \quad \text{in all other cases}.\end{cases}

Then we define the Newton tensor

\displaystyle T^{ij}_m=\frac{\partial A_m}{\partial a_{ij}}=\frac{1}{(m-1)!}\sum\delta\binom{i,i_2 \dots i_m }{j,j_2 \dots j_m}a_{i_2j_2}\cdots a_{i_m j_m}.

For any vector field {X} on {\mathbb{R}^n}, {DX} is a matrix, where {D=(D_{1},\cdots,D_n)} and {|X|\neq 0}, denote {\tilde X=X/|X|}, we have

\displaystyle (m-1)!T^{ij}_m(DX)X_iX_j=\sum\delta\binom{i,i_2 \dots i_m }{j,j_2 \dots j_m}X_iX_j[D_{i_2}X_{j_2}]\cdots [D_{i_m}X_{j_m}].

Since for any {1\leq p,k,l\leq n}

\displaystyle D_k\tilde X_l=\frac{D_kX_{l}}{|X|}-\frac{\sum_{p=1}^nX_pD_kX_pX_l}{|X|^3}

\displaystyle \sum_{i,j,i_2,j_2}\delta\binom{i,i_2 \dots i_m }{j,j_2 \dots j_m}X_iX_jX_pX_{j_2}D_{i_2}X_p=0

because \delta is skew-symmetric in j,j_2. Then

\displaystyle (m-1)!T^{ij}_m(DX)X_iX_j=|X|^{m-1}\sum\delta\binom{i,i_2 \dots i_m }{j,j_2 \dots j_m}X_iX_j[D_{i_2}\tilde X_{j_2}]\cdots [D_{i_m}\tilde X_{j_m}].

=(m-1)!|X|^{m-1}T_m^{ij}(D\tilde X)X_iX_j=(m-1)!|X|^{m+1}T_m^{ij}(D\tilde X)\tilde X_i\tilde X_j.

Applying the formula (T^{ij}_m(\mathcal{A}))=[\mathcal{A}]_{m-1}I-T_{m-1}(\mathcal{A})\cdot \mathcal{A}(Check [1] Propsition 1.2) and (D\tilde X)\tilde X=0, we get

T^{ij}_m(DX)X_iX_j=|X|^{m+1}[D\tilde X]_{m-1}

It follows from the result of Reilly, Remark 2.3(a), that

\displaystyle mA_m[DX]=D_i[T^{ij}_mX_j]

Suppose {X} is a vector field normal to {\partial \Omega}, then

\displaystyle m\int_\Omega [DX]_m=\int_{\partial \Omega} T^{ij}_m X_j\gamma_i=\int_{\partial \Omega}(X\cdot \gamma)^m[D\gamma]_{m-1}=\int_{\partial \Omega} (X\cdot \gamma)^m H_{m-1}(\partial \Omega)

where {\gamma} is the outer ward unit normal to {\partial \Omega}.

Remark: [1] R.C. Reilly, On the Hessian of a function and the curvatures of its graph., Michigan Math. J. 20 (1974) 373–383. doi:10.1307/mmj/1029001155.

[2] N. Trudinger, Apriori bounds for graphs with prescribed curvature.

 

We probably have to assume DX is a symmetric matrix in order to use the formula of Reilly. Not sure about this.

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}}.

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

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.