## Category Archives: Elliptic PDE

32519989

### 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

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

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

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

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

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

### 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 ${u}$

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

### Nonuniqueness of strong solution

If we assume $u\in W^{2,p}_{loc}(\Omega)\cap C^0(\bar{\Omega})$, where $p, then the uniqueness of Dirichlet problem will fails. See the following example

$\begin{cases}\displaystyle \Delta u+\left(-1+\frac{n-1}{1-\lambda}\right)D_{ij}u=0\\ u(x)=1\quad \text{ on }\partial B_1(0)\end{cases}$

where $n>2(2-\lambda)>2$. One can verify that $u(x)=1$ and $u(x)=|x|^\lambda\in W^{2,2}(B_1)$ are two solutions.