2
$\begingroup$

I am strugging on problem 3.2 in Gilbarg-Trudinger, which says that

if $L=a^{ij}(x)D_{ij}+b^{i}(x)D_{i}+c(x)$ is an elliptic operator in a bounded domain $\Omega \subset \mathbb{R}^{n}$ with $c<0$, and $u\in C^{2}(\Omega)\cap > C^(\overline\Omega)$ satisfies $Lu=f$ in $\Omega$, then we have $\sup_{\Omega}|u| \le\sup_{\partial \Omega}|u|+\sup_{\Omega}|\frac{f}{c}|$

In previous part of this chapter, we have actually shown that for the case $c\le0$, we have $\sup_{\Omega}|u| \le\sup_{\partial \Omega}|u|+C\sup_{\Omega}|\frac{f}{\lambda}|$, where $\lambda(x)$ is the minimum eigenvalue of $[a^{ij}(x)]$ and $C$ is a constant depending only on $diam(\Omega)$ and $\beta=\sup \frac{|\mathbf{b}|}{\lambda} (<\infty)$.

I have no idea why the bound can be independent of $diam(\Omega)$ and $\beta$, can anybody give me some idea?

$\endgroup$
1
  • $\begingroup$ Is $c$ a constant (as is indicated by its appearance in the estimate), and not a function of $x$ as at the start of the problem? $\endgroup$
    – Jeff
    Sep 3, 2016 at 0:42

1 Answer 1

1
$\begingroup$

There is a version of the maximum principle where you use the zeroth order term instead of uniform ellipticity to get the estimate. I'll sketch the proof below, and I'll take $c<0$ to be a constant (don't have the book in front of me right now). The argument should work just as well if $c(x)$ is negative and bounded away from zero.

If $u$ attains its max at a point $x\in \Omega$, then

$$f(x) = a^{ij}u_{x_ix_j} + b^iu_{x_i} + cu \leq cu(x),$$

due to the ellipticity of $a^{ij}$. Since $c$ is negative we have

$$\sup_\Omega u = u(x) \leq \frac{f(x)}{c} \leq \sup_{\Omega} \left|\frac{f}{c}\right|.$$

If the maximum is attained on the boundary, then $\sup_\Omega u \leq \sup_{\partial \Omega} |u|$, hence

$$\sup_\Omega u \leq \sup_{\partial \Omega} |u| + \sup_{\Omega} \left|\frac{f}{c}\right|.$$

Actually you can put the max of the two terms on the right hand side to get a slightly better estimate, if you like. You can obtain a similar estimate for $-u$ since $L(-u) = -f$.

Basically, when you use the zeroth order term to obtain estimates with the maximum principle, you do not need to play the perturbation trick (adding $\varepsilon e^{\alpha x_1}$ or something similar to $u$), so the size of the domain does not enter into the estimate, and you don't need uniform ellipticity.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .