Page 38 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

28 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
and

2
Q ξξ + λ Q = 0 (2.100)
Q(0) = 0
Q(1) = 0

The solutions are

P = A sin(λτ) + B cos(λτ) (2.101)
Q = C sin(λξ) + D cos(λξ) (2.102)

The first boundary condition of Eq. (2.100) requires that D = 0. The second requires that
C sin(λ) be zero. Our eigenvalues are again λ n = nπ. The boundary condition at τ = 0, that
P τ = 0 requires that A = 0. Thus

PQ n = K n sin(nπξ)cos(nπτ) (2.103)
The final form of the solution is then

∞

y(τ, ξ) = K n sin(nπξ)cos(nπτ) (2.104)
n=0

2.2.3 Orthogonality
Applying the final (nonhomogeneous) boundary condition (the initial position).
∞

f (ξ) = K n sin(nπξ) (2.105)
n=0
In particular, if f (x) = hx, 0 < x < 1/2

= h(1 − x), 1/2 < x < 1 (2.106)

1 1/2 1

f (x)sin(nπx)dx = hx sin(nπx)dx + h(1 − x)sin(nπx)dx
0 0 1/2
2h nπ 2h
= sin = (−1) n+1 (2.107)
2
2
n π 2 2 n π 2
and
1

2
K n sin (nπx)dx = K n /2 (2.108)
0