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Rotation of coordinates (context of solving simple PDE)

  1. Jul 21, 2014 #1
    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


    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. jcsd
  3. Jul 21, 2014 #2

    George Jones

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    This is the equation of the x' axis, i.e, it is the equation for y' = 0.
  4. Jul 25, 2014 #3
    So I need an arbitrary constant?
  5. Jul 25, 2014 #4
    My lecturer did the change of coordinates for a more general constant coefficient PDE [itex]\sum_{j=1}^n a_j\frac{\partial f}{\partial x_j}=b(x,u)[/itex] in n-variables by defining the new variables as:


    How do you get this?
  6. Jul 26, 2014 #5


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

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

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