Page 152 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

142 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
and applying the boundary conditions and noting that

d 2
2
[sin(λx)] =−λ sin(λx)
dx 2
we have
1

−λ 2 sin(λx)u(x, t)dx + [u x sin(λx) − λu cos(λx)] 1 0
x=0
2
=−λ U(λ, t) − u(1)[λ cos λ + H sin λ]
Defining
1

S λ {u(x, t)}= u(x, t)sin(λx)dx = U(λ, t)
x=0
as the Fourier sine transform of u(x, t) and setting

λ cos λ + H sin λ = 0

we find
2
U t (λ, t) =−λ U(λ, t)

whose solution is
2
U(λ, t) = Ae −λ t
The initial condition of the transformed function is
1
1

U(λ, 0) = sin(λx)dx = [1 − cos(λ)]
λ
x=0
Applying the initial condition we find

1 2
U(λ, t) = [1 − cos(λ)]e −λ t
λ
It now remains to find from this the value of u(x, t).
Recall from the general theory of Fourier series that any odd function of x defined on
0 ≤ x ≤ 1 can be expanded in a Fourier sine series in the form

1
∞
sin(λ n x)

u(x, t) = u(ξ, t)sin(λ n ξ)dξ
# # 2
sin(λ n ) #
n=1
#
ξ=0