I Understanding the change of variables for PDEs

[Isaac0427]

TL;DR Summary

I have been trying to put together an understanding of change of variables for PDEs, and I am wondering if I am on the right track.

I've been trying to get change of variables in PDEs down (I don't particularly like my textbook or professor's approach to it), and I want to ask here if I am getting this right. Let $\vec{x}=(x_1,x_2,...,x_n)^T$ and $\partial_\vec{x}=(\partial_{x_1},\partial_{x_2},...,\partial_{x_n})^T$. I believe that a second-order linear PDE with constant coefficients (in n dimensions) can be written in the form
$$\left( \partial_\vec{x}^TA\partial_\vec{x}+B\partial_\vec{x}+C \right) u=f(\vec{x})$$
where $A$ is an nxn matrix, $B$ is an n-dimensional row vector, and $C$ is a scalar. Now, we can change coordinates to $\vec{y}=P\vec{x}$ where $P$ is an nxn matrix. From my understanding, this means $\partial_\vec{x}=P\partial_\vec{y}$. Now, though for a partial differential equation the choice of $A$ is not unique, there is a unique symmetric choice of $A$. Assuming $A$ is diagonalizable, we choose $P$ such that it diagonalizes $A$-- this also makes it an orthogonal matrix, and $P^{-1}=P^T$. So, changing variables gives us
$$\left( \partial_\vec{y}^TP^TAP\partial_\vec{y}+BP\partial_\vec{y}+C \right) u=f(P^T\vec{y}).$$

This gets rid of the cross terms for the second derivatives, which would then make life much easier when solving these equations. I did compare this for the $A=0$ case with problems done using a different method, and the answers were the same (this time though choosing $P$ such that $BP=(1,0)$).

If everything above is correct, this leaves me with two more questions:

1. I know that when $B=\vec{0}^T$ and $C=0$, in the 2-D case, we are left with either the diffusion equation, Laplace's equation, or the heat equation. How does one handle solving the equation when $B$ and/or $C$ are nonzero?

2. Is there a way to generalize this for non-constant coefficients?

Thank you very much in advance!

 

[wrobel]

changing variables in PDE is nothing more than the chain rule and other differentiation rules from calculus
if $y^i=f^i(x)$ then $\partial_{x^i}=\frac{\partial f^s}{\partial x^i}\Big|_{x\mapsto y}\partial_{y^s}$
In fact that is all

 

[Isaac0427]

changing variables in PDE is nothing more than the chain rule and other differentiation rules from calculus
if $y^i=f^i(x)$ then $\partial_{x^i}=\frac{\partial f^s}{\partial x^i}\Big|_{x\mapsto y}\partial_{y^s}$
In fact that is all

Right, but I'm trying to figure out this with matrices/rotating coordinates to diagonalize $A$. My understanding of what you wrote is that if you have $\vec{y}=P\vec{x}$, this means (assuming $P$ is a constant nxn matrix) $\partial_\vec{x}=P\partial_\vec{y}$. Is that correct? And if so, I believe that means the rest of the first part of my post is correct, and the other two questions still stand.