## Tag Archives: harnack inequality

### 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 ${u}$

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

### Harnack inequality of simple case by Moser iteration

We are trying to give a brief proof of harnack inequality of the divergence form

$\displaystyle Lu=(a^{ij}u_{x_i})_{x_j}=0\quad (1)$

Harnack inequality says that on $B_r$ such that $B_{4r}\subset \Omega$, any weak solution $u>0$ of $(1)$ satisfies

$\displaystyle \sup_{B_r(x)}u\leq C\inf_{B_r(x)}u$

Define                  $\displaystyle I(p,r)=\left(\frac{1}{|B_r(x)|}\int_{B_r(x)}u^pdx\right)^{\frac{1}{p}}$ for $u>0$, then

$\displaystyle \sup_{B_r}u=\lim\limits_{p\to \infty} I(p,r)$

$\displaystyle \inf_{B_r}u=\lim\limits_{p\to -\infty} I(p,r)$

We will prove that

If $\displaystyle (a^{ij}u_{x_i})_{x_j}\geq 0$,  then $\displaystyle \sup_{B_{r/2}}u\leq CI(p,r)$, $\forall p>1$, $C$ only depends on $n,p, a^{ij}$

If $\displaystyle (a^{ij}u_{x_i})_{x_j}\leq 0$, then $\displaystyle C\inf_{B_{r/2}}u\geq I(p,r)$, $\forall 1\leq p< \frac{n}{n-2}$

$\textbf{Proposition:}$

If $p>1$, $Lu\geq 0$, $I\left(\frac{pn}{n-2}, r_1\right)\leq \left[\frac{Cr_2^n}{(r_2-r_1)^{2}r_1^{n-2}}\left(1+\frac{4p^2}{\lambda^4|p-1|^2}\right)\right]^{\frac{1}{p}}I(p,r_2)$

If $p<1$, $Lu\leq 0$, $I\left(\frac{pn}{n-2}, r_1\right)\leq \left[\frac{Cr_2^n}{(r_2-r_1)^{2}r_1^{n-2}}\left(1+\frac{4p^2}{\lambda^4|p-1|^2}\right)\right]^{\frac{1}{p}}I(p,r_2)$

If $p<0$, $Lu\leq 0$, $I\left(\frac{pn}{n-2}, r_1\right)\geq \left[\frac{Cr_2^n}{(r_2-r_1)^{2}r_1^{n-2}}\left(1+\frac{4p^2}{\lambda^4|p-1|^2}\right)\right]^{\frac{1}{p}}I(p,r_2)$

$\textbf{Proof: }$ Choose $v=\eta^2 u^{p-1}$, then

$\displaystyle \begin{cases} \displaystyle \int_{B_r} a^{ij}u_{x_i}v_{x_j}\leq 0 \text{ if } p>1\\\displaystyle \int_{B_r} a^{ij}u_{x_i}v_{x_j}\geq 0 \text{ if } p<1\end{cases}$

That is

$\displaystyle \int_{B_r} a^{ij}u_{x_i}u_{x_j}\eta^2 u^{p-2}\leq \frac{-2}{p-1}\int_{B_r} a^{ij}u_{x_i}\eta_{x_j}\eta u^{p-1}$

$\displaystyle \int_{B_r}|\nabla u|^2 \eta^2 u^{p-2}\leq \frac{2}{\lambda^2|p-1|}\int_{B_r}|u_{x_i}\eta_{x_j}\eta u^{p-1}|$

where $\lambda I\leq (a^{ij})\leq \lambda^{-1} I$. The left hand side is

$\displaystyle \int_{B_r}|\nabla u|^2 \eta^2 u^{p-2}\leq\frac{2}{\lambda^2|p-1|} \int_{B_r}\epsilon|\nabla u|^2 \eta^2 u^{p-2}+\frac{1}{\epsilon} |\nabla \eta|^2 u^p$

Let $\displaystyle \epsilon=\frac{\lambda^2|p-1|}{4}$,

$\displaystyle \int_{B_r}|\nabla u|^2 \eta^2 u^{p-2}\leq\frac{16}{\lambda^4|p-1|^2} \int_{B_r}|\nabla \eta|^2 u^p\quad (2)$

From the embedding $H^1(\mathbb{R}^n)\subset L^{\frac{2n}{n-2}}(\mathbb{R}^n)$, there is a constant $C=C(n)$

$\displaystyle \left(\int_{B_r}| \eta u^{\frac{p}{2}}|^{\frac{2n}{n-2}}\right)^{\frac{n-2}{n}}\leq C\int_{B_r}|\nabla(\eta u^{\frac{p}{2}})|^2$

Then we will have

$\displaystyle \left(\int_{B_r}| \eta u^{\frac{p}{2}}|^{\frac{2n}{n-2}}\right)^{\frac{n-2}{n}}\leq C\int_{B_r}|\nabla \eta|^2u^p+\frac{Cp^2}{4}\int_{B_r}\eta^2|\nabla u|^2u^{p-2}$

Combing with $(2)$, we can get

$\displaystyle \left(\int_{B_r}| \eta u^{\frac{p}{2}}|^{\frac{2n}{n-2}}\right)^{\frac{n-2}{n}}\leq C\left(1+\frac{4p^2}{\lambda^4|p-1|^2}\right)\int_{B_r}|\nabla \eta|^2u^p (3)$

Let us suppose $r=r_2$, $\eta$ is cut off function on $B_{r_2}$, $\eta=1$ on $B_{r_1}$ for some $r_1 and $\eta=0$ outside $B_{r_2}$, $\nabla\eta\leq \frac{2}{r_2-r_1}$, then $(3)$ means

$\displaystyle \left(\int_{B_{r_1}}u^{\frac{pn}{n-2}}\right)^{\frac{n-2}{n}}\leq \frac{C}{(r_2-r_1)^2}\left(1+\frac{4p^2}{\lambda^4|p-1|^2}\right)\int_{B_{r_2}}u^p$

$\begin{cases}\displaystyle I\left(\frac{pn}{n-2}, r_1\right)\leq \left[\frac{Cr_2^n}{(r_2-r_1)^{2}r_1^{n-2}}\left(1+\frac{4p^2}{\lambda^4|p-1|^2}\right)\right]^{\frac{1}{p}}I(p,r_2)\text{ if }p>0\hfill (4)\\ \displaystyle I\left(\frac{pn}{n-2}, r_1\right)\geq \left[\frac{Cr_2^n}{(r_2-r_1)^{2}r_1^{n-2}}\left(1+\frac{4p^2}{\lambda^4|p-1|^2}\right)\right]^{\frac{1}{p}}I(p,r_2)\text{ if }p<0\hfill (5)\end{cases}$

$\hfill \square$

Denote $\gamma=\frac{n}{n-2}$.Suppose $p>1$, and $Lu\geq 0$ choose $p_k=\gamma^kp$ and $r_k=\frac{r}{2}+\frac{r}{2^k}$, from $(4)$,

$\displaystyle I(p_{k+1}, r_{k+1})\leq \left[C(n,\lambda)2^k\right]^{\frac{1}{p_k}}I(p_k,r_k)$, for all $k\geq 0$

$\displaystyle \sup_{B_{r/2}}u=\lim_{k\to \infty}I(p_k,r_k)\leq C(n,\lambda)^{\sum_0^\infty \frac{k}{p_{k}}}I(p,\frac{3}{2}r)\leq C(n,\lambda,p)I(p,\frac{3}{2}r)$

because we know $\displaystyle \sum_0^\infty \frac{k}{p_{k}}<\infty$.

Let us consider the case $\boldsymbol{Lu\leq 0}$ to get the other inequality involves $\inf_{B_{r/2}}u$.

Choose $p_k=p\gamma^{-k}$, $r_k=2r-\frac{r}{2^{k}}$, apply $(4)$, we can get

$\displaystyle I(p_{k}, r_k)\leq \left[C(n,\lambda)2^k\right]^{\frac{1}{p_{k+1}}}I(p_{k+1},r_{k+1})$, for $0\leq k\leq N$

Remember from the proposition, the case $k=0$ needs $p\gamma^{-1}<1$, that is $p<\gamma$.

So by iteration                       $\displaystyle I(p,r)\leq C(n,\lambda)^{Q_1} I(p\gamma^{-N}, r_{N})\quad (6)$

where $\displaystyle Q_1=\sum_0^{N-1} \frac{k}{p_{k+1}}$, because $p<\gamma$.

By similar iteration, from $(5)$, we get

$\displaystyle I(-\infty, \frac{r}{2})\geq C^{Q_2}I(-p_N, r_N)\quad (7)$

where $\displaystyle Q_2=\sum_0^{\infty} \frac{-k}{p_N\gamma^k}<\infty$.

If we can prove that $I(p_{N}, r_N)\leq C I(-p_N, r_N)$, for some $N$, then combine $(7)$ and $(6)$,

$\displaystyle I(p,r)\leq CI(-\infty, \frac{r}{2})=C\inf_{B_{r/2}}u$

We need the following theorem, which is actually the consequence of John-Nirenberg thm and poincare inequality

$\textbf{Thm: }$ Let $u\in W^{1,1}(\Omega)$, where $\Omega$ is convex, and suppose there exists a constant $K$ such that

$\displaystyle \int_{\Omega\,\cap B_r}|Du|\leq Kr^{n-1}$ for any balls $B_r=B_r(x_0)$

Then there exist positive constants $\sigma_0$ and $C$ depending only on $n$ such that

$\displaystyle \int_\Omega e^{\frac{\sigma}{K}|u-u_\Omega|}\leq C(\text{diam }\Omega)^n$

$\textbf{Lemma:}$ There exists two numbers $p_0>0, C$, such that

$\displaystyle I(p_0,r)\leq CI(-p_0,r)$

for any $u>0$ and $Lu\leq 0$.

$\textbf{Proof:}$ Let $w=\eta^2\log u$, $\eta\in C^\infty_c(B_{2\rho}(x_0))$ and $\eta=1$ on $B_\rho(x_0)$ and $|\nabla \eta|<\frac{2}{\rho}$

$\displaystyle \int_{B_{2\rho}}\eta^2|\nabla w|^2=\int_{B_{2\rho}} \eta^2u^{-2}|\nabla u|^2\leq\frac{1}{\lambda}\int_{B_{2\rho}}\eta^2 u^{-2}a^{ij}u_{x_i}u_{x_j}$

while

$\displaystyle 0\leq \int_{B_r}a^{ij}u_{x_i}\left(\frac{\eta^2}{u}\right)_{x_j}=-\int_{B_r} a^{ij}u_{x_i}\eta^2 u_{x_j}u^{-2}+\int_{B_r}a^{ij}u_{x_i}2\eta \eta_{x_j}u^{-1}$

$\displaystyle \int_{B_r} a^{ij}u_{x_i}\eta^2 u_{x_j}u^{-2}\leq\int_{B_r}|a^{ij}u_{x_i}2\eta \eta_{x_j}u^{-1}|\leq \lambda^{-1} \int_{B_r}\epsilon$$\eta^2 u^{-2}|\nabla u|^2+\frac{1}{\epsilon}|\nabla \eta|^2$

So letting $\epsilon=\frac{1}{2\lambda^2}$, then

$\displaystyle \int_{B_{2\rho}}\eta^2|\nabla w|^2\leq 4\lambda\int_{B_{2\rho}}|\nabla \eta|^2\leq C\rho^{n-2}$

By holder inequality

$\displaystyle \int_{B_{\rho}}|\nabla w|\leq C\rho^{\frac{n}{2}}\left(\int_{B_\rho}|\nabla w|^2\right)^{1/2}\leq C\rho^{n-1}$

By the theorem we have stated, there exists $p_0$ such that

$\displaystyle \int_{B_r}e^{p_0(w-w_{B_r})}\leq Cr^n$

$\displaystyle \int_{B_r}e^{p_0(w_{B_r}-w)}\leq Cr^n$

Multiply them together

$\displaystyle \int_{B_r}e^{p_0w}\int_{B_r}e^{-p_0w}\leq C r^{2n}$

This is equivalent to $I(p_0.r)\leq CI(-p_0,r)\quad \qquad \square$

Carefully examine the proof of this lemma, not only for some $p_0$, but also for $0, we also have

$\displaystyle I(p.r)\leq CI(-p,r)$

So we can choose $N$ large enough such that $0, then

$I(p_N.r_N)\leq CI(-p_N,r_N)$

Hence we complete the proof of Harnack inequality.

$\textbf{Remark: }$ consult much to the GT’s book and MA600A: Theory of Partial Differential Equation, Roger Moser.

### Isolated singularity of elliptic function

$\mathbf{Problem: }$ Let $Lu=au_{xx}+2bu_{xy}+cu_{yy}=0$ in an exterior domain $r>r_0$, $L$ being uniformly elliptic. Prove that if $u$ is bounded on one side then $u$ has a limit(possibly infinite) as $r\to \infty$.

$\mathbf{Proof: }$ WLOG, assume $u(x)\geq 0$, when $|x|\geq r_0$, here $x=(x',x'')\in\mathbb{R}^2$
We have $\displaystyle \lim\limits_{\overline{|x|\to \infty}}u(x)=c\geq 0$.
If $c=+\infty$, then $u$ has a limit which is  infinity.
Otherwise, $\forall\, \epsilon>0$, $\exists\, x_m\to \infty$, $m=1,2,\cdots$, such that $u(x_m).

Define $v(x)=u(x)-c+\epsilon$,  note that $v(x_m)<2\epsilon$, $m=1,2,\cdots$ and $Lv=0$. By the Harnack inequality(see Thm 3.10, need a little bit of work)
$v(x), $\forall\, |x|=r_m$, $m=1,2,\cdots$
By maximum principle
$v(x)<2K\epsilon$, $r_m\leq|x|\leq r_{m+1}$,$m=1,2,\cdots$,
So $\overline{\lim\limits_{x\to\infty}}u(x)\leq 2K\epsilon$, since $\epsilon$ is arbitrary,, we know $\overline{\lim\limits_{x\to\infty}}v(x)\leq 0$ which means $\overline{\lim\limits_{x\to \infty}}u(x)\leq c$. So $u$ has a limit as $r\to\infty$.
$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ Gilbarg Trudinger’s book. Chapter 3, exercise 3.3.