# Coordinate method for PDE

Solving the PDE $au_x+bu_y+cu=0$ The PDE is transformed by the coordinate method via,
$\begin{cases}x'=ax+by\\y'=bx-ay\\\end{cases}$.

What I don't understand is how should I know I have to pick this as the transformation ?
How do I know that this is the transformation to be used?
Is it something to do with linear transformations in matrix?

You should recognize $au_x+bu_y$ as the dot product of $\nabla u$ with the vector $\left<a,b\right>$. Also known as the directional derivative. This makes the direction of vector $\left<a,b\right>$ seem special for this equation, so we take a coordinate system in which it is one of two coordinate axes. This is best done by a rotation matrix, or a multiple of it, so that the axes remain nicely orthogonal (although this isn't strictly necessary).
• So one transformation is selected using the directional derivative<a,b>.The other is the orthogonal of $<a,b>$ that is $<b,-a>$.Since $<-b,a>$ is also orthogonal to $<a,b>$ I could have selected $y=-bx+ay$ as well. Couldn't I? – clarkson Mar 21 '14 at 3:35
It would be useful for first order PDE of this kind to consider it as being extracted from the direction derivatives. For instance, the directional derivative of a function u(x,y) is given by $$\begin{equation} du=\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy=u_xdx+u_ydy\end{equation}$$ Rewriting your equation as $$\begin{equation} au_x+bu_y=-cu \end{equation}$$ What your directional derivative of u tells is that $$\begin{equation} du=-cu \end{equation}$$ One simply gets $$\begin{equation} \frac{dx}{a}=\frac{dy}{b}=\frac{du}{-cu}.\end{equation}$$ Solving the first two equations, you will just get $$\begin{equation} bx-ay=constant,\end{equation}$$ which is one of your transformed coordinates. In fact this kind of transformation is called methods of characteristics. It roughly suggests that you assume the solution depends on both x and y and then try to go to the space where the solution is constant, just like the previous expression has shown.