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Those two problems bothers me for a while. I think I got most of it but I do want to have a nice and clean solution, so I post it here for discussion.

All below I will use Einstein summation.

The first one is Problem 8 on page 367. It asks us to prove the gradient estimation for elliptic equation $$Lu:=-a_{ij}\partial_i\partial_ju=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$$

It suggest we start with $v:=|Du|^2+\lambda u^2$ and prove that $Lv\leq 0$ for $\lambda$ large enough. As usual, I did computation and end up with

$$Lv= -a_{ij}\,\partial_j\partial_ku\,\partial_i\partial_k u + \partial_ka_{ij}\,\partial_k u\,\partial_i\partial_ju-2\lambda a_{ij}\,\partial_ju\,\partial_i u-2\lambda a_{ij}\,u\, \partial_i\partial_ju,$$ in which I plug in function (1) with derivative with respect $\partial_k$ and multiply with $\partial_k u$. Then, apply ellipticity and use Cauchy inequality, I end up with $$Lv\leq -\frac{\theta}{2}|D^2u|^2+(-2\theta\lambda +A)|Du|^2-2\lambda a_{ij}\,u\, \partial_i\partial_ju $$ where $\theta>0$ is the ellipticity constant and $A$ only depends on $a_{ij}$. I got stuck on the last part of the right hand side.

I was planning to estimate it with $$ -2\lambda a_{ij}\,u\, \partial_i\partial_ju \leq B\lambda |u|^2+ \frac{\theta}{4}|D^2u|^2$$ and I would have $$Lv\leq -\frac{\theta}{4}|D^2u|^2+(-2\theta\lambda +A)|Du|^2+B\lambda |u|^2$$ Clearly as $\lambda$ large enough, I have $-2\theta\lambda+A<0$, but how may I get ride of the last part? Here I think $\lambda$ should not depends on $u$.

The next question regrading to Problem 9 on page 367. I could do all the way until $$ \left|\frac{\partial}{\partial \nu}u(x^0)\right|\leq C\left|\frac{\partial}{\partial \nu}w(x^0)\right| $$ but the actually question asks me to prove $$ |Du(x_0)|\leq\left|\frac{\partial}{\partial \nu}w(x^0)\right| $$ I don't have a clue how the last part come from...

Pleas help. Thank you!

spatially
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1 Answers1

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Question 1: the "last part on the right hand side" $-2\lambda a_{ij} u \partial_i \partial_j u = 2 \lambda uLu$ vanishes by assumption. So you don't have to estimate it.

Question 2: I don't have the book on front of me and therefore don't know the details of this barrier estimate. What else do you know about the behavior of $u$ at $x_0$? Is this perhaps a boundary point? Is $u = 0$ on this part of the boundary? In that case $Du(x_0) = \pm \partial_\nu u(x_0) \cdot \nu$, with the sign depending on the direction of the normal.

Hans Engler
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  • For question 1: the only comment I can say is I am a idiot... For Q2: yes $x_0$ at $\partial\Omega$ and $u=0$ as well. but how you get $Du(x_0)=\partial_\nu(x_0)\cdot\nu$? – spatially Dec 04 '14 at 16:35
  • Explain this to yourself when $\Omega = { x_n > 0}$ (half space) and $x_0 = 0$. What are the partial derivatives $\partial_i u(x_0)$ for $i = 1, \dots, n-1$ in this case? How is $\partial_nu(x_0)$ related to $\partial_\nu u(x_0)$? Then go to the general case via the Implicit Function Theorem. – Hans Engler Dec 04 '14 at 16:55
  • Sure. I'll definitely work that out. – spatially Dec 04 '14 at 17:24
  • Yes, I find that $\nabla u(x_0)$ must parallel to $\nu(x_0)$ because $u\equiv 0$ on $\partial \Omega$. Hence $|\nabla u(x_0)|=|\partial_\nu u(x_0)$. Am I correct sir? – spatially Dec 05 '14 at 01:35
  • That argument makes sense. But I am not your prof :). – Hans Engler Dec 05 '14 at 03:06