Consider the sets: $Q_0=\{x\in\mathbf{R}^n:0
How to prove this change of variables?
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pde
1 Answers
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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 $u$).
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0This change of variable is truth in $Q_l$ to $Q_0$? – 2012-11-17