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I'm trying to understand the following estimation: $ - \int_U u \left( \Delta u |Du|^{p-2} + (p-2) (\nabla u^T D^2 u \nabla u) |Du|^{p-4})\right) \leq C \int_U u |Du|^{p-2} |D^2 u| dx$

taken from Question $5.9$ - Evans PDE $2$nd edition.

I don't see how the other terms can be estimated in an appropriate way.

Maybe someone can help me to understand it?

AplusB
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1 Answers1

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The estimate should have $|u|$ instead of $u$. You have $|\Delta u|\le C|\nabla^2u|$ and $|\nabla u^t D^2u \nabla u|\le C |\nabla u|^2 | D^2u |$. Why does this bother you?

Gio67
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  • Thanks for your answer! I'm quite new to the topic and a bit confused about all those estimations. I have $ |\Delta u| = |\sum \frac{\partial ^2 u}{\partial x_i^2}|$ and $ |\nabla^2u| = |\nabla \cdot \nabla u|=|\nabla \cdot (\frac{\partial u}{\partial x_1},\ldots,\frac{\partial u}{\partial x_n})|$ which is the same as $ |\Delta u|$. Am i right so far? Why do I need the constant $C$. Or is it just for construction? – AplusB Jun 30 '17 at 08:24
  • $|\nabla^2u|=|\nabla(\nabla u)|$ so $\nabla^2u$ is an $N\times N$ matrix and its norm is not the same as $|\Delta u|$. The constant $C$ is just to play safe. It depends on which norm you take for $\nabla^2u$. – Gio67 Jun 30 '17 at 10:26
  • Thanks for the answer! That helps a bit. But still I'm not sure about the inequality. Is there a theorem stating this fact? Or is it obvious bit I don't see it? – AplusB Jul 03 '17 at 07:27
  • No theorem. You just have to do the calculations. You have an $N\times N$ matrix $A$ with entries $a_{i,j}$ and you are saying that $\left\vert\sum_i a_{i,i} \right\vert\le c\Vert A\Vert$ and that for every vector $b$, $\left\vert b^tAb \right\vert\le c\Vert A\Vert$ – Gio67 Jul 03 '17 at 10:38