Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider the sets: $$Q_0=\{x\in\mathbf{R}^n:0<x_1<2, \ |x_\alpha|<1, \ for \ 1<\alpha\leq n\},$$ $$Q_l=\{x\in\mathbf{R}^n:0<x_1<2l, \ |x_\alpha|<l \ for \ 1<\alpha\leq n\}.$$ Suppose $0\leq u\leq K$ in $Q_l$, and $$|Mu|\leq A(|\nabla u|+u+k), \ \ in \ Q_0,$$ where $$M=\sum_{i,j=1}^na_{ij}(x)\frac{\partial^2}{\partial x_i\partial x_j}.$$ The, by scaling $x=ly$, $$|Mu|\leq A\left(\frac{|\nabla u|}{l}+\frac{u}{l^2}+k\right) \ \ in \ Q_l.$$

share|cite|improve this question

Take $y=\frac{x}{l}$. Then $y(Q_l)=Q_0$. Now, with $v=u \circ y$, by the chain rule:

$|Mv|=\frac{1}{l^2}|Mu| \leq A(\frac{|\nabla u|}{l^2} + \frac{u}{l^2} + \frac{k}{l^2})= A(\frac{|\nabla v|}{l} + \frac{v}{l^2} + \frac{k}{l^2})$.

If $l\geq1$, the result follows. For $l<1$, it is actually wrong:

Take $n=1$, $a_{ij}=1$ and $u(y)=y^2$. Then $u$ satisfies the first inequality (for $A=1, k=2$) but not the second since

$|v''|=\frac{2}{l^2} > \frac{2x}{l^3} + \frac{x^2}{l^4} + 2 = A(\frac{|v'|}{l} + \frac{v}{l^2} + k)$ for $x$ small enough.

So one has to scale $k$, too (or require a lower bound for $u$).

share|cite|improve this answer
    
This change of variable is truth in $Q_l$ to $Q_0$? – José Carlos Nov 17 '12 at 1:31

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.