Rotation of coordinates (context of solving simple PDE)

[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 au_x+bu_y=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?

 

[George Jones]

y=(b/a)x

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

 

[chipotleaway]

So I need an arbitrary constant?

 

[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?

 

[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.