Nonlinear coordinate transformation

[shawnstrausser]

Homework Statement

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/

Homework Equations

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..

 

[Ray Vickson]

Homework Statement

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/

Homework Equations

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..

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 $-\arctan(m) \leq \theta \leq \arctan(m)$.