Lemma from PDE book This is Lemma 6.1 from Gilbarg - Trudinger.  It states "Let $\textbf{P}$ be a constant matrix which defines a nonsingular linear transformation $y=x\textbf{P}$ from $\mathbb{R}^n \rightarrow \mathbb{R}^n$.  Letting $u(x) \rightarrow \tilde{u}(y)$ under this transformation one verifies easily that $A^{ij}D_{ij}u(x) = \tilde{A}^{ij}D_{ij}\tilde{u}(y)$, where $\tilde{\textbf{A}} = \textbf{P}^t\textbf{A}\textbf{P}$."  
Here, we are using the summation convention, and $A^{ij}$ is a constant matrix with $A^{ij} = A^{ji}$.
I have no idea where this is coming from.  It says it's an easy verification... but even trying to do this with 2 by 2 matrices gives a huge mess.  Second, what exactly does $u(x) \rightarrow \tilde{u}(y)$ mean?  Is $\tilde{u}$ the same function but just a different variable?  Why not call it $u(y)$ then?  Any help is appreciated.
 A: The idea is to perform a "rotation and stretching" $(P)$ of coordinates which transforms $u$ (defined on $\Omega$) into a function $\tilde{u}$ defined on $P(\Omega)$ so that $\tilde{u}$ satisfies a nice equation. 
Computationally, we have $u(x) = \tilde{u}(Px)$. The general formula (by the Chain Rule) for $D^2u$ is 
$$D^2u(x) = P^T D^2\tilde{u}(Px) P.$$
Thus, we have
$$0 = tr(AD^2u(x)) = tr(AP^T D^2\tilde{u}(Px) P) = tr(PAP^T D^2\tilde{u}(Px)).$$
For a simple example try $u_{xx} + u_{xy} + u_{yy} = 0$, say defined on $B_1$. By rotating coordinates to the $(1,1)$ and $(1,-1)$ directions we can write the equation without mixed derivatives, and by stretching in one direction and squeezing in the other we obtain harmonic $\tilde{u}$ defined on some rotated ellipse.
The idea of changing coordinates is also very useful in scaling arguments for PDE, where we have some estimate in $B_1$ which we would like to apply at all scales.
A: The definition is
$$
\tilde{u}(xP)=u(x),
\qquad\textrm{or equivalently,}\qquad
\tilde{u}(y)=u(xP^{-1}).
$$
Then use the beloved rule.
