I am reading Qing Han & Fanghua Lin’s book Elliptic partial differential equations. In the section 2.4 Gradient estimate, I come across a trouble understanding the following step:
Here is some of my manuscript:
use the ellipticity assumption we can get $$\textcircled{1}\geqslant 2\lambda |D^{2}u|^{2}$$ And $\textcircled{4}$ is easily: $$\textcircled{4}\geqslant -|Du|^{2}-|Df|^{2}$$ For $\textcircled{2}$, I estimate it as follows: since $$|2D_{k}a_{ij}D_{ij}uD_{k}u|\leq \frac{1}{2}[2(D_{k}a_{ij})^{2}+2(D_{ij}u)^{2}(D_{k}u)^{2} ]=(D_{k}a_{ij})^{2}+(D_{ij}u)^{2}(D_{k}u)^{2} $$ So $$\sum_{i}\sum_{j}\sum_{k}2D_{k}a_{ij}D_{ij}uD_{k}u\leq\sum_{i}\sum_{j}\sum_{k} (D_{k}a_{ij})^{2} +\sum_{i}\sum_{j}\sum_{k} (D_{ij}u)^{2}(D_{k}u)^{2} =\sum_{i}\sum_{j}\sum_{k} (D_{k}a_{ij})^{2} +|D^{2}u|^{2}|Du|^{2}$$ Thus we get $$\textcircled{2}\geq-\sum_{i}\sum_{j}\sum_{k} (D_{k}a_{ij})^{2} -|D^{2}u|^{2}|Du|^{2} $$ Similarly I get estimation for $\textcircled{3}$ $$\textcircled{3}\geq -\sum_{i}\sum_{k}(D_{k}b_{i})^{2}-|Du|^{2}|Du|^{2}=-\sum_{i}\sum_{k}(D_{k}b_{i})^{2}-|Du|^{4}$$
I am not sure whether I am on the right way. My confusion is: I will get a $|Du|^{4}$ term which vanished on the book. Can anyone help me figure out that estimation?
The point is that you have $L^{\infty}$ control on the coefficients $a_{ij}, b_j$ and its first derivatives, so you can ignore them. Then each term is essentially quadratic (or of lower order) in $Du$ and $D^2u,$ which you can split appropriately using Young's inequality.
More precisely if we let $K$ be the sum of $C^1(\overline\Omega)$ norms of $a_{ij}, b_j$ and $f,$ then we can estimate
\begin{align} 2a_{ij}D_{ki}uD_{kj}u &\geq 2\lambda |D^2u|^2, \\ -2D_ka_{ij}D_kuD_{ij}u &\geq -2K |Du||D^2u| \geq -\lambda|D^2u|^2 - \frac{K^2}{\lambda}|Du|,\\ -2D_kb_iD_kuD_iu &\geq - 2K|Du|^2, \\ 2D_zf|Du|^2 &\geq -K^2 - |Du|^2, \\ 2D_kfD_ku &\geq - K^2 - |Du|^2. \end{align} Hence putting it all together, $$ L(|Du|^2) \geq \lambda |D^2u|^2 - \left( \frac{K^2}{\lambda} + 2K + K\right)|Du|^2 - 2K^2.$$