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}


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<r_2 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}


\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<p<p_0, we also have

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

So we can choose N large enough such that 0<p_N<p_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)<c+\epsilon.

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)<Kv(x_m), \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.