Category Archives: Elliptic PDE


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.


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.

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

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.

Compensated compactness

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


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}

Subcriticality and supercriticality

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.

Approriate scalling in Yamabe equation

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


\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


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

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

Harnack inequality under scaling

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

\displaystyle \sup_{\Omega'}u\leq C(n,\Omega,\Omega')\inf_{\Omega'}u

To prove this Harnack inequality, there is an intermediate step

\displaystyle \sup_{B_R}u\leq 3^n\inf_{B_R}u

whenever {B_{4R}\subset\Omega}.

Here the constant is independent of {R}, actually the constant {3^n} is not so important. We can use the above Thm to give another proof.

Suppose {v(x)=u(Rx)} for {x\in B_1}. Let {\Omega=B_2} and {\Omega'=B_1}, applying the thm

\displaystyle \sup_{B_1}v\leq C(n)\inf_{B_1}v

convert back to {u}

\displaystyle \sup_{B_R}u\leq C(n)\inf_{B_R}u