## Tag Archives: poincare ineq

### Holder continuity of weak solution and inverse holder inequality

$\textbf{Thm:}$  Any weak solution of

$\displaystyle (a^{ij}\partial_iu)_j=0$

is automatically holder continuous.

This is a corollary of the celebrated thm of De Giorgi, John Nash and J, Moser. It is obtained from the Harnack inequality of weak solution. Actually in dimension 2, this theorem is easily known by mathematicians before the technique of Moser iteration. In the following, we will give a short expository of the proof

From the Poincare inequality

$\displaystyle |u-u_B|\leq C\int_{B_r(x_0)}\frac{|\nabla u(y)|}{|x-y|^{n-1}}dy=I_{1}(\nabla u)$

$||I_1(\nabla u)||_2\leq C||\nabla u||_p$, where $\displaystyle \frac{1}{p}-\frac{1}{n}=\frac{1}{2}$, $p>1$ when $n=2$, $p=\frac{2n}{n+2}<2$

So this means that $\displaystyle \left(\frac{1}{|B_r|}\int_{B_r}|u-u_B|^2dx\right)^{1/2}\leq Cr\left(\frac{1}{|B_r|}\int_{B_r}|\nabla f|^p\right)^{1/p}$

Combining with Caccippoli’s inequality, we can get

$\displaystyle \left(\frac{1}{|B_\frac{r}{2}|}\int_{B_{\frac{r}{2}(x_0)}}|\nabla u|^2dx\right)^{1/2}\leq C\left(\frac{1}{|B_r|} \int_{B_r(x_0)}|\nabla u|^pdx\right)^{1/p}$

with $p<2$. This is the reverse holder inequality

So $\displaystyle |\nabla u|^p\in RH_{\frac{2}{p}}$, $\frac{2}{p}>1$. Thus by Gehring, $w\in RH_{\frac{2}{p}+\epsilon}$. That is $\nabla u\in L^{2+\epsilon}(B^r)$

The by the Morrey’s embedding thm, in dimension 2, $\displaystyle u\in C^{0,1-\frac{2}{2+\epsilon}}(B_r)$.

### Caccioppoli inequality

A Caccioppoli inequality is of the form

$\displaystyle r^p\int_{B_{\frac{r}{2}}(x_0)}|\nabla u|^pdx\leq C\int_{B_r(x_0)}\bigg|u(x)-(u)_{B_{r}(x_0)}\bigg|^pdx\quad\quad (1)$

where $C$ is a constant independent of $r$ and $u$, $1, and $\displaystyle (u)_{B_{r}(x_0)}=\frac{1}{|B_r(x_0)|}\int_{B_r(x_0)}udx$.

Caccioppoli inequality can be considered as the reverse Poincare inequality

$\displaystyle \int_{B_r(x_0)}\bigg|u(x)-(u)_{B_{r}(x_0)}\bigg|^pdx\leq Cr^p\int_{B_{r}(x_0)}|\nabla u|^pdx$

holds for $u\in W^{1,p}(B_r(x_0))$.

Of course $(1)$ does not hold for any $u\in W^{1,p}(B_r(x_0))$, but if $u$ satisfies some second order elliptic equation or certain variational problems, $u$ can satisfies $(1)$.

For example, if $u$ is the weak solution of the divergence form

$\displaystyle \text{div}(A\nabla u)=0$ or $(a^{ij}\partial_ju)_i=0$ in $B_r(x_0)$

here $a^{ij}$ are just bounded measurable and strictly elliptic. $\nu I\leq A\leq \nu^{-1}I$.

Choose a cut off function $\eta\in C^\infty_c(B_r(x_0))$ and $\eta=1$ in $B_{\frac{r}{2}}(x_0)$ and $\displaystyle |\nabla\eta|\leq \frac{4}{r}$. Let $v=\eta^2(u-\overline{u})\in H^1_0(B_r(x_0))$, here we use $\overline{u}=(u)_{B_{r}(x_0)}$ for abbreviation. Then

$\displaystyle \int_{B_r(x_0)}a^{ij}\partial_ju\partial_iv=0$

$\displaystyle \int_{B_r(x_0)} a^{ij}\partial_ju\eta^2\partial_iu=-\int_{B_r(x_0)} 2a^{ij}\partial_ju\eta\nabla\eta(u-\overline{u})$

$\displaystyle \nu\int_{B_r(x_0)}\eta^2|\nabla u|^2dx\leq C\int_{B_r(x_0)} |\partial_ju\eta\nabla\eta(u-\overline{u})|$

$\displaystyle \int_{B_r(x_0)}\eta^2|\nabla u|^2dx\leq C(\nu)\int_{B_r(x_0)}|\nabla\eta(u-\overline{u})|^2dx\leq \frac{C}{r^2}\int_{B_r(x_0)} |u-\overline{u}|^2dx$

Noticing that $\eta=1$ on $B_{\frac{r}{2}(x_0)}$, we can get $(1)$

$\textbf{Remark:}$ If $u$ satisfies $(a^{ij}\partial_ju)_i+b^i\partial_iu=0$, we can also get the same conclusion.

MA6000A: Theory of Partial Differential Equations. Roger Moser

### Fredholm alternative and second order elliptic equation

$\textbf{Thm:}$ (Fredholm alternative) Suppose $T:H\to H$ is a compact mapping of Hilbert space $H$ into itself. Then there exists a countable set $\Lambda\subset\mathbb{R}$ having limit points except possible $\lambda=0$ such that

(1)If $\lambda\not\in \Lambda$, then

$\displaystyle \lambda x-Tx=y$ and $\displaystyle \lambda x-T^\ast x=y$

is uniquely solvable for any $y\in H$. And the inverse mappings $(\lambda I-T)^{-1}$, $(\lambda I-T^\ast)^{-1}$ are bounded.

(2)If $\lambda\in \Lambda$, the the $\text{Ker}(\lambda I-T)$ has positive finite dimension and $\lambda x-Tx=y$ is solvable if and only if $y$ is orthogonal to the $\text{Ker}(\lambda I-T^\ast)$. And vice versa.

For a generalized second order elliptic equation

$\displaystyle Lu=(a^{ij}u_{i})_j+b^iu_i+cu=f$

Usually we require $(a^{ij})$ is strictly elliptic, $b^i, c\in L^\infty(\Omega)$. Since the

$\displaystyle \theta||u||_{H^1_0}\leq (-Lu,u)+\gamma ||u||_{L^2}$

then $L_\sigma=L-\sigma$ is invertible for $\sigma$ big enough. So $Lu=f$ is equivalent to

$\displaystyle L_\sigma u+\sigma u=f$

$\displaystyle u+\sigma L_\sigma^{-1}u=L_{\sigma}^{-1}f$

Let $T=-\sigma IL_\sigma^{-1}: L^2\to L^2$, since $I(u):H_0^1\to L^2$ is compact, then $T$ is compact. Then any $u$ satisfies the above equation is a weak solution in $H^1_0(\Omega)$

$\displaystyle T^\ast=-\sigma I(L^\ast)_\sigma^{-1}$ with $L^\ast v=(a^{ij}v_j)_i-b^iv_i+(c-\partial_ib^i )v$

In the general case to solve $Lu=\mu u+f$, this is equivalent to

$\displaystyle u+(\sigma-\mu )L_\sigma^{-1}u=L_\sigma^{-1}f$

$\displaystyle (Tu,v)_{L^2}=(u,T^*v)_{L^2}$

So there exists a series of $\mu$ such that $\displaystyle Lu=\mu u$ has nontrivial solutions. And the above equation is solvable if and only if

$\displaystyle 0=(L_\sigma^{-1} f,v)_{L^2}=(f,(L^\ast)^{-1}v)_{L^2}=(f, \frac{1}{\mu-\sigma}v)_{L^2}=\frac{1}{\mu-\sigma}(f,v)_{L^2}$.

for any $v\in \text{Ker}(id+(\sigma-\mu)(L^\ast)_\sigma^{-1})$

For instance, consider the following equation

$\displaystyle (1)\quad \begin{cases}\Delta u+u=\sin x\text{ on }\Omega\\u=0\text{ on }\partial \Omega\end{cases}$

If

$(2)\quad \displaystyle \begin{cases}\Delta u+u=0\text{ on }\Omega\\u=0\text{ on }\partial \Omega\end{cases}$

has no nontrivial solution, then $(1)$ is uniquely solvable.

Otherwise $(1)$ is solvable if and only if $\displaystyle \int_\Omega \sin x v=0$ for any $v$ solves $(2)$. The question is when $(2)$ has only the trivial solution.

This has much relation with the poincare inequality, since if $u$ satisfies $(2)$, we will have $\displaystyle \int_\Omega |\nabla u|^2= \int_\Omega u^2$($\partial \Omega$ must have some regularity to justify this).

$\textbf{Lemma1:}$ For $u\in H^1_0(\Omega)$, we have

$\displaystyle ||u||_{L^2}\leq \left(\frac{|\Omega|}{w_n}\right)^{1/n}||\nabla u||_{L^2}$

where $w_n$ is the volume of unit ball in $\mathbb{R}^n$.

$\textbf{Lemma2:}$ If $\Omega$ can be bounded by an n-dimensional strip with width $d$, we have

$\displaystyle ||u||_{L^2}\leq \left(\frac{1}{d}\right)^{2}||\nabla u||_{L^2}$

So if $|\Omega|$ is small enough, or $\Omega$ is thin enough, $(2)$ will only have the trivial solution.

Remark: Why the $T$ is an operator on $L^2(\Omega)$ instead of the $H^1_0(\Omega)$? Pay attention to the “normal to the $Ker(\lambda I- T^*)$” in fredholm alternative.

### Completeness of H(\Omega)

$\displaystyle D(\Omega)=\{\overset{\rightharpoonup}{u}\in C^\infty_c(\Omega)| \nabla\cdot \overset{\rightharpoonup}{u}=0 \}$

$\displaystyle H(\Omega)$ is  the closure of $D(\Omega)$ under the norm of $H^1_0(\Omega)$

$\langle\displaystyle \overset{\rightharpoonup}{u}, \overset{\rightharpoonup}{v}\rangle_{H(\Omega)}=\int_{\Omega}\nabla\overset{\rightharpoonup}{u}\cdot \nabla\overset{\rightharpoonup}{v}dx$

$H(\Omega)$ is defined for arbitrary domain $\Omega\subset \mathbb{R}^3$ or $\mathbb{R}^2$, unless $\Omega=\mathbb{R}^2$

Let us prove $H(\Omega)$ is complete.

Simply take $u$ as $\overset{\rightharpoonup}{u}$, $u$ is understood as a vector function.

$\textbf{Proof:}$ $n=3$. Recall the Hardy inequality, for all $u\in C^\infty_c(\Omega)$

$\displaystyle \int_{\mathbb{R}^3}\frac{u^2(x)}{|x-y|^2}dx\leq 4\int_{\mathbb{R}^3}|\nabla u|^2dx$, $\forall y\in \mathbb{R}^3$, $\quad (1)$

Suppose $u_n$ is cauchy sequence in $H(\Omega)$, namely $\int_{\Omega}|\nabla u_n-\nabla u_m|^2dx\to 0$ as $n,m\to \infty$.

Since $D(\Omega)$ is dense in $H(\Omega)$, WLOG assume $u_n\in D(\Omega)$, $\forall n$.

Then apply $(1)$, we get,

$\displaystyle \int_{\Omega}\frac{(u_n(x)-u_m(x))^2}{|x-y|^2}dx\leq 4\int_{\Omega}|\nabla u_n-\nabla u_m|^2dx\to \infty$, $y$ is fixed.

This means $\displaystyle \frac{u_n(x)}{|x-y|}\to \frac{u(x)}{|x-y|}$ in $L^2(\Omega)$.

Apparently, $\nabla u_n$ should converge to some $v$ in $L^2(\Omega)$. Let us prove $\nabla u=v$.

For any $w\in D(\Omega)$

$\displaystyle \int_{\Omega}u_n\cdot\nabla wdx=-\int_{\Omega}\frac{u_n}{|x-y|}\cdot w\nabla|x-y| dx+\int_{\Omega}\frac{u_n}{|x-y|}\cdot\nabla(|x-y|w)$  $(2)$

Since $w\nabla|x-y|$ and $\nabla(|x-y|w)$ are both $C^\infty_c(\Omega)$ functions except a singular point $y$,

$\displaystyle \int_{\Omega}\frac{u_n}{|x-y|}\cdot w\nabla|x-y| dx\to \int_{\Omega}\frac{u}{|x-y|}\cdot w\nabla|x-y| dx$ as $n\to \infty$

$\displaystyle \int_{\Omega}\frac{u_n}{|x-y|}\cdot\nabla(|x-y|w)\to \int_{\Omega}\frac{u}{|x-y|}\cdot\nabla(|x-y|w)$ as $n\to \infty$

$(2)$ becomes

$\displaystyle \int_{\Omega}u_n\cdot\nabla wdx\to \int_{\Omega}u\cdot\nabla wdx$

However $\displaystyle \int_{\Omega}u_n\cdot\nabla wdx=-\int_{\Omega}\nabla u_n\cdot wdx \to -\int_{\Omega}v\cdot wdx$, which means

$\displaystyle \int_{\Omega}u\cdot\nabla wdx\to -\int_{\Omega}v\cdot wdx$

Then $\nabla u=v$ and it is easy to prove $v\in H(\Omega)$. And $u_n\to v$ in $H(\Omega)$

When $n=2$, use the same method by

$\displaystyle \int_{|x-y|>1}\frac{u^2(x)}{|x-y|^2\ln^2|x-y|}dx\leq 4\int_{|x-y|>1}|\nabla u|^2dx$, $\forall y\in \mathbb{R}^3$

### Poincare inequality from different approach

$\mathbf{Problem:}$ Let $\Omega$ be convex and $u\in C^1(\Omega)$. Suppose $S\subset\Omega$ with $\displaystyle \frac{|S|}{|\Omega|}>0$,

$\displaystyle |u(x)-u_S|\leq \frac{d^n}{n|S|}\int_{\Omega}\frac{|\nabla u(y)|}{|x-y|^{n-1}}dy$

where $d$ is the diameter of $\Omega$ and $\displaystyle u_S=\frac{1}{|S|}\int_S u(y)dy$.

$\mathbf{Proof\,1: }$

$\displaystyle u(y)-u(x)=\int_0^1\frac{d}{dt}\left(u(ty+(1-t)x)\right)dt=\int_0^1\nabla u(ty+(1-t)x)\cdot(x-y)dt$

$\displaystyle |u(x)-u_S|\leq\frac{1}{|S|}\int_S|u(x)-u(y)|dy\leq \frac{1}{|S|}\int_S\int_0^1|\nabla u(ty+(1-t)x)||x-y|dtdy$

Set $z=ty+(1-t)x$, then $t|x-y|=|z-x|$. Since $|x-y|\leq d$, $\displaystyle t\geq \frac{|z-x|}{d}$. We also have $t^ndy=dz$

Then

$\displaystyle |u(x)-u_S|\leq \frac{d}{|S|}\int_\Omega\int_\frac{|z-x|}{d}^\infty \frac{|\nabla u(z)|}{t^n}\,dtdz=\frac{d^n}{(n-1)|S|}\int_\Omega\frac{|\nabla u(z)|}{|z-x|^{n-1}}dz$.

$\mathbf{Proof\,2: }$ Using

$\displaystyle u(x)-u(y)=\int_{0}^{|x-y|}\frac{d}{dt}u(x+tw)dt=\int_0^{|x-y|}\nabla u(x+tw)\cdot wdt$

where $\displaystyle w=\frac{x-y}{|x-y|}$. Then

$\displaystyle |u(x)-u_S|\leq\frac{1}{|S|}\int_S |u(x)-u(y)|dy\leq \frac{1}{|S|}\int_S\int_0^{|x-y|}|\nabla u(x+tw)|dtdy$

$\displaystyle \leq \frac{1}{|S|}\int_\Omega\int_0^{\infty}|\nabla u(x+tw)|\chi_{\{t<|x-y|\}}dtdy$

$\displaystyle \leq \frac{1}{|S|} \int_0^\infty\int_{|w|=1}\int_0^d|\nabla u(x+tw)|\chi_{\{t<|x-rw|\}}r^{n-1}dwdrdt$

$\displaystyle \leq \frac{d^n}{n|S|}\int_0^\infty\int_{|w|=1}|\nabla u(x+tw)|\chi_{\{t<|x-\xi(w)|\}}dwdt$

$\displaystyle =\frac{d^n}{n|S|}\int_\Omega \frac{|\nabla u(z)|}{|z-x|^{n-1}}dz$

where $\xi(w)$ is the intersection of ray $x+tw$ with $\partial \Omega$ (it is unique because $\Omega$ is convex).

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$