Mathematical Methods for Physics and Engineering : A Comprehensive Guide

20.5 THE DIFFUSION EQUATION

written entirely in terms ofη,

4 η

d^2 f(η)

dη^2

+(2+η)

df(η)

dη

=0.

This is a straightforward ODE, which can be solved as follows. Writingf′(η)=

df(η)/dη, etc., we have

f′′(η)

f′(η)

=−

1

2 η

−

1

4

⇒ ln[η^1 /^2 f′(η)] =−

η

4

+c

⇒ f′(η)=

A

η^1 /^2

exp

(−η

4

)

⇒ f(η)=A

∫η

η 0

μ−^1 /^2 exp

(−μ

4

)

dμ.

If we now write this in terms of a slightly different variable

ζ=

η^1 /^2

2

=

x

2(κt)^1 /^2

,

thendζ=^14 η−^1 /^2 dη, and the solution to (20.34) is given by

u(x, t)=f(η)=g(ζ)=B

∫ζ

ζ 0

exp(−ν^2 )dν. (20.36)

HereBis a constant and it should be noticed thatxandtappear on the RHS

only in the indefinite upper limitζ, and then only in the combinationxt−^1 /^2 .If

ζ 0 is chosen as zero thenu(x, t) is, to within a constant factor,§the error function

erf[x/2(κt)^1 /^2 ], which is tabulated in many reference books. Only non-negative

values ofxandtare to be considered here, so thatζ≥ζ 0.

Let us try to determine what kind of (say) temperature distribution and flow

this represents. For definiteness we takeζ 0 = 0. Firstly, sinceu(x, t) in (20.36)

depends only upon the productxt−^1 /^2 , it is clear that all pointsxat timestsuch

thatxt−^1 /^2 has the same value have the same temperature. Put another way, at

any specific timetthe region having a particular temperature has moved along

the positivex-axis a distance proportional to the square root oft.Thisisatypical

diffusionprocess.

Notice that, on the one hand, att= 0 the variableζ→∞andubecomes

quite independent ofx(except perhaps atx= 0); the solution then represents a

uniform spatial temperature distribution. On the other hand, atx= 0 we have

thatu(x, t) is identically zero for allt.

§TakeB=2π− 1 / (^2) to give the usual error function normalised in such a way that erf(∞)=1.See
the Appendix.