# Rotation of coordinates (context of solving simple PDE)

1. Jul 21, 2014

### chipotleaway

If you rotate your rectangular coordinate system (x,y) so that the rotated x'-axis is parallel to a vector (a,b), in terms of the (x,y) why is it given by

x'=ax+by
y'=bx-ay

I got x'=ay-bx, y'=by+ax from y=(b/a)x.

By the way this is from solving the PDE aux+buy=0 by making one of the partial derivatives disappear. The general solution I got from this change of variables is u(ay-bx) rather than u(bx-ay) - is u(ay-bx) wrong?

2. Jul 21, 2014

### George Jones

Staff Emeritus
This is the equation of the x' axis, i.e, it is the equation for y' = 0.

3. Jul 25, 2014

### chipotleaway

So I need an arbitrary constant?

4. Jul 25, 2014

### chipotleaway

My lecturer did the change of coordinates for a more general constant coefficient PDE $\sum_{j=1}^n a_j\frac{\partial f}{\partial x_j}=b(x,u)$ in n-variables by defining the new variables as:

$y_1=\frac{x_1}{a_1}$
and
$y_j=x_j-\frac{a_j}{a_1}x_1$

How do you get this?

5. Jul 26, 2014

### HallsofIvy

Strictly speaking, x'= ax+ by, y'= bx- ay does NOT give a rotation.

If $a^2+ b^2> 1$ there will be both a rotation and an expansion. If $a^2+ b^2< 1$ there will be both a rotation and a shrinking It is only when $a^2+ b^2= 1$ that x'= ax+ by, y'= bx- ay gives a pure rotation.

One can show that a rotation, counterclockwise, about the origin, through angle $\theta$ is given by $x'= cos(\theta)x+ sin(\theta)y$, $y'= sin(\theta)x- cos(\theta) y$. In your formulation, "a" must be $cos(\theta)$ and "b" must be $sin(\theta)$ where $\theta$ is the angle rotated through.