Page 159 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

STURM–LIOUVILLE TRANSFORMS 149
2
and according to the boundary conditions we must choose = sin(nx)and λ =−n .The sine
transform of q(x)is Q(λ).

2
U t =−n U + Q(λ)
U = U(λ, t)

The homogeneous and particular solutions give

2
U n = Ce −n t + Q n
n 2
when t = 0, U = 0sothat

Q n
C =−
n 2
where Q n is given by

π

Q n = q(x)sin(nx)dx
x=0

2
Q n −n t
Since U n = 2 [1 − e ] the solution is
n
∞
2
Q n −n t
sin(nx)
u(x, t) = [1 − e
n 2 ] # sin(nx) # 2
n=1 # #
Note that Q n is just the nth term of the Fourier sine series of q(x). For example, if
q(x) = x,
π n+1
Q n = (−1)
n
Example 9.5 (A mixed transform). Consider steady temperatures in a half cylinder of infi-
nite length with internal heat generation, q(r) that is a function of the radial position. The
appropriate differential equation is

1 1
u rr + u r + u θθ + u zz + q(r) = 0 0 ≤ r ≤ 10 ≤ z ≤∞ 0 ≤ θ ≤ π
r r 2
with boundary conditions

u(1,θ, z) = 1

u(r, 0, z) = u(r,π, z) = u(r,θ, 0) = 0