# Phorgy Phynance

## Network Theory and Discrete Calculus – Coordinates

This post is part of a series

When the binary tree was presented in the context of discrete calculus, the following small section of the tree was illustrated to establish the way the nodes are labelled

Two sets of coordinates were introduced on the binary tree:

1. Cartesian coordinates $(x,t)$
2. Graph coordinates $(u^+,u^-)$

The following illustrates the binary tree when we zoom out a bit:

Cartesian coordinates were defined such that

• $x(i\pm 1,j+1) = x(i,j) \pm \Delta x,$ and
• $t(i\pm 1,j+1) = t(i,j) + \Delta t$

resulting in

• $dx = \Delta x (\mathbf{e}^+ - \mathbf{e}^-),$ and
• $dt = \Delta t (\mathbf{e}^+ + \mathbf{e}^-).$

Although Cartesian coordinates often help to relate discrete calculus to continuum calculus, the expressions are often not the most natural to work with when performing computations. One reason for this can be understood by overlaying the Cartesian coordinate lines onto the binary tree.

On the other hand, graph coordinates are defined such that

• $u^+(i+1,j+1) = u^+(i,j) + \Delta u^+,$
• $u^+(i-1,j+1) = u^+(i,j),$
• $u^-(i+1,j+1) = u^-(i,j),$
• $u^-(i-1,j+1) = u^-(i,j) + \Delta u^-$

resulting in

• $du^+ = \Delta u^+ \mathbf{e}^+,$ and
• $du^- = \Delta u^- \mathbf{e}^-.$

Computations are often cleaner when using graph coordinates. One reason for this can be understood by overlaying the graph coordinate lines onto the binary tree.

For instance, the commutative relations in graph coordinates are given by

• $[du^\pm,u^\pm] = \Delta u^\pm du^\pm$
• $[du^\pm,u^\mp] = 0$

whereas the commutative relations for Cartesian coordinates are given by

• $[dx,x] = \frac{(\Delta x)^2}{\Delta t} dt$
• $[dt,t] = \Delta t dt.$
• $[dx,t] = [dt,x] = \Delta t dx.$

The cross commutative relations between the two sets of coordinates are given by

• $[du^\pm,x] = [dx,u^\pm] = \pm\Delta x du^\pm$
• $[du^\pm,t] = [dt,u^\pm] = \Delta t du^\pm$

As a final note, for any discrete 0-form $\phi$, the above indicates that

$d\phi = \frac{1}{\Delta t} [dt,\phi].$

Written by Eric

February 19, 2012 at 2:17 pm