Math578 - Alexiades
                HW 1 ( ON PAPER)
      Similarity solution of 1D diffusion on half line

Derivation of exact similarity solution of the diffusion problem:
	 ut = D uxx ,   0<x<∞ , t>0
	 u(x,0) = uinit ,   0<x<∞
	 u(0,t) = uo,  (and limx→∞ u(x,t) = uinit),    t>0
with D, uinit, uo constants.

1. Seek u(x,t) as a function of a "similarity variable", i.e. as 
	u(x,t) = w(ξ)   with  ξ = x / √t 
   Plug into the heat equation to obtain the ODE for w(ξ):
	w ′′ + (ξ/(2D)) w ′ = 0

2. Solve the ODE explicitly to find its general solution:
	w(ξ) = (A√π /2) erf( ξ / (2√D) ) + B,
    with A, B arbitrary constants.  Therefore
	u(x,t) = (A√π /2) erf( x / (2√(D t) ) ) + B .

3. Apply the BC and IC to find A, B, thus obtaining the explicit similarity solution
	u(x,t) = uo + (uinit − uo) erf( x / (2√(Dt) ) ),
   valid for 0≤x<∞ , t≥0.