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Nonlinear coordinate transformation

  1. Jan 9, 2016 #1
    1. The problem statement, all variables and given/known data
    Evening all, I'm trying to solve the 2-D diffusion equation in a region bounded by y = m x + b, and y = -m x -b. The boundary condition makes it complicated to work with numerically, and I recall a trick that involves a coordinate transformation so that y = m x + b, and y = -m x -b are mapped to y = 1, and y = 0, which I already have a program to solve numerically. This will transform the PDE, but that is easier (I think) to work with..

    Reference https://www.physicsforums.com/threads/nonlinear-coordinate-transformation.851414/

    2. Relevant equations


    3. The attempt at a solution
    My attempt: I simplified the problem to first consider y = m x, and y = - m x. The curve y = m x can be written as (x, m x), and y = - m x as (x, -m x). The transformation, T, should satisfy T(x, m x) = (x, 1) and T(x, - m x) = (x,0). As far as I can tell, (x, m x), and (x, - m x) form a basis(?) and so this should completely determine the transformation, from what I recall in Linear Algebra (or maybe that only works for a linear transformation?). It looks like the transformation is non linear as well. I'm not quite sure where to go, I want to map from the coordinates (x,y) to (x',y') that transform the curves as described above. If I could find the relationships x = f(x',y') and y = g(x',y'), then I could transform the derivatives in the PDE. Any help would be appreciated! Thank you in advance..
     
  2. jcsd
  3. Jan 9, 2016 #2

    Ray Vickson

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    The two lines ##y = mx+b## and ##y = -mx-b## are of the form ##y = s(x \:+ \: b/m)## for ##s = -m## and ##s = +m##. So, if you change variables to ##x## and ##s##, where ##s## is the slope of the line through ##(-b/m,0)## you will have ##y = s (x\:+\:b/m)##. This gives a PDE in the variables ##(x,s)## on the region ##-m \leq s \leq m##. However, the form of the DE will become more complicated, due to the nonlinear coordinate transformation (which now involves a product ##sx## when determining ##y##.

    Alternatively, you can use polar coordinates centered at ##(-b/m,0)##, so that ##x+b/m = r \cos(\theta), y = r \sin(\theta)##. This gives a PDE in ##(r,\theta)## over the region ##-\arctan(m) \leq \theta \leq \arctan(m)##.
     
    Last edited: Jan 9, 2016
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