Alexandrov Maximum Principle and $W^{2}_p$ estimates

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I'm reading an article of N. V. Krylov: About an example of N. N. Ural'tseva and weak uniqueness for elliptic operators, Nonlinear partial differential equations and related topics, 131–144.

This article deals the solvability of following elliptic PDE: given $f\in L_p(\Omega)$, $$ a^{ij} u_{x^i x^j} +b^i u_{x^i} +c u -\lambda u=f$$ Here $a^{ij}$ is $d\times d$ symmetric matrix and there is $\kappa>0$ such that $$ \kappa^{-1} |\xi|^2 \geq a^{ij} \xi^i \xi^j \geq \kappa |\xi|^2$$for all $\xi\in\mathbb{R}^d$.

The article mainly concerns minimal regularity assumption on $a^{ij}$.

To solve the solvability of this equation in a smooth bounded domain $\Omega \subset \mathbb{R}^d$ in $\overset{\circ}{W^2_p}$(usual $W^2_p(\Omega)$ functions with vanishing at the boundary) we know that for any $f\in L_p(\Omega)$, it suffices to find a priori estimate: $$ \Vert u \Vert_{{W^2_p}(\Omega)}\leq C\Vert(t L +(1-t)\triangle)u-\lambda u\Vert_{L_p(\Omega)}.$$ Here $L=a^{ij} D_ij +b^i D_i +c$.

It is well known that in $d=2$, there is $W^{2}_2$-estimates with no additional regularity assumption on $a^{ij}$.

Also by Ural'tseva's example, in general, it is impossible to obtain $W^2_p$-estimates if $d\geq 3$ or $p>d$ if the coefficients are merely bounded measurable class.

Here is my question. According to the author, he says on $W^2_p$ estimates in $\Omega\subset \mathbb{R}^d$,

The emphasis is of course on large $p$ because the sharpness of the Alexandrov maximum principle shows that without additional restrctions on the coefficients such estimates do not exist for $p<d$.

But this sentence is not obvious to me. I search Fanghua Lin to see the Alexandrov maximum principle, but I cannot find the relevance on here.

Thank you for in advance.