Page 78 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

68 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
After applying the initial condition

2q 0 2
A j (τ) = 1 − exp(−λ τ) (4.127)
j
3
λ J 1 (λ j )
j
The solution is therefore
∞
2q 0 2
u(τ,r) = 1 − exp(−λ τ) J 0 (λ j r) (4.128)
j
3
λ J 1 (λ j )
j=1 j
Example 4.12 (Time dependent heat generation). Suppose that instead of constant heat
generation, the generation is time dependent, q(τ). The differential equation for A(τ)then
becomes
2q(τ)
2
A (τ) + λ A(τ) = (4.129)
j
λ j J 1 (λ j )
2
An integrating factor for this equation is exp(λ τ) so that the equation can be written as
j
2q(τ)
2 2
d
A j exp(λ τ) = exp(λ τ) (4.130)
j
j
dτ λ j J 1 (λ j )
Integrating and introducing as a dummy variable t
τ
2 2
A j (τ) = q(t)exp(−λ (τ − t))dt (4.131)
j
λ j J 1 (λ j )
t=0
Problems
1. By differentiating the series form of J 0 (x) term by term show that

J (x) =−J 1 (x)

0
2. Show that

xJ 0 (x)dx = xJ 1 (x) + constant

x n
3. Using the expression for s J 0 (s )ds show that
s =0
x

2
2
5
s J 0 (s )ds = x(x − 8)[4xJ 0 (x) + (x − 8)J 1 (x)]
s =0
4. Express 1 − x as a Fourier–Bessel series.