Page 77 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS 67
Thus
∞
2
2
−λ τ
U(τ, η) = e n J 0 (λ n η) (4.118)
λ n J 1 (λ n )
n=0
Example 4.11 (Heat generation in a cylinder). Reconsider the problem of heat transfer in a
long cylinder but with heat generation and at a normalized initial temperature of zero.
1
u τ = (ru r ) r + q 0 (4.119)
r
u(τ, 1) = u(0,r) = 0, u bounded (4.120)

Our experience with the above example hints that the solution maybe of the form

∞

u = A j (τ)J 0 (λ j r) (4.121)
j=1
This equation satisfies the boundary condition at r = 1and A j (τ) is to be determined. Substi-
tuting into the partial differential equation gives
∞ ∞
1 d dJ 0

A (τ)J 0 (λ j ) = A j (τ) r + q 0 (4.122)
j
j=1 j=1 r dr dr
In view of Bessel’s differential equation, the first term on the right can be written as
∞
2
−λ J 0 (λ j r)A j (τ) (4.123)
j
j=1
The second term can be represented as a Fourier–Bessel series as follows:

∞
2J 0 (λ j r)

q 0 = q 0 (4.124)
λ j J 1 (λ j )
j=1
as shown in Example 4.9 above.
Equating coefficients of J 0 (λ j r)wefind that A j (τ) must satisfy the ordinary differential
equation
2
2
A (τ) + λ A(τ) = q 0 (4.125)
j
λ j J 1 (λ j )
with the initial condition A(0) = 0.
Solution of this simple first-order linear differential equations yields

2q 0 2
A j (τ) = + C exp(−λ τ) (4.126)
j
3
λ J 1 (λ j )
j