M578 - Alexiades
HW 1 (may do by hand ON PAPER if you prefer)
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.