Page 91 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

SOLUTIONS USING FOURIER SERIES AND INTEGRALS 81
2
If we choose the length scale as r 1 then we define η = r/r 1 ,ζ = z/L,and τ = αt/r .The
1
normalized temperature can be chosen as U = (u − u 1 )(u 0 − u 1 ). With these we find that
1 r 1 2
U τ = (ηU η ) η + U ςς
η L
U ς (ς =±1) = 0
(5.35)
U η (η = 1) + BU(η = 1) = 0
U(τ = 0) = 1

where B = hr 1 /k.
Let U = T(τ)R(η)Z(ζ). Insert into the differential equation and divide by U.

T 1 r 1 2 Z

= (ηR ) + (5.36)
T ηR L Z

Z ς (ς =±1) = 0

R η (η = 1) + BR(η = 1) = 0
U(τ = 0) = 1

Again, the dance is the same. The left-hand side of Eq. (5.36) cannot be a function of η or ζ so
each side must be a constant. The constant must be negative for the time term to be bounded.

Experience tells us that Z /Z must be a negative constant because otherwise Z would
be exponential functions and we could not simultaneously satisfy the boundary conditions at
ζ =±1. Thus, we have

2

T =−λ T
2
2 2

η R + ηR + β η R = 0
(5.37)
L
2

Z =−γ 2 Z
r 1
with solutions
2
T = Ae −λ t
Z = C 1 cos(γ Lς/r 1 ) + C 2 sin(γ Lς/r 1 ) (5.38)
R = C 3 J 0 (βη) + C 4 Y o (βη)

It is clear that C 4 must be zero always when the cylinder is not hollow because Y 0 is unbounded
when η = 0. The boundary conditions at ς =±1 imply that Z is an even function, so that C 2