Page 94 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

84 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
saw one way of dealing with this. Note that it wouldn’t have worked had q been a function of
time. We approach this problem somewhat differently. From experience, we expect a solution
of the form
∞

y(ξ, τ) = B n (τ)sin(nπξ) (5.54)
n=1
where the coefficients B n (τ) are to be determined. Note that the equation above satisfies the
end conditions. Inserting this series into the differential equation and using the Fourier sine
series of C
∞ n
2C[1 − (−1) ]

C = sin(nπξ) (5.55)
nπ
n=1
∞ ∞
2

B (τ)sin(nπξ) = [−(nπ) B n (τ)] sin(nπξ)
n
n=1 n=1
∞ n
2[1 − (−1) ]
+ C sin(nπξ)sin( τ) (5.56)
nπ
n=1
Thus
n
2[1 − (−1) ]
2
B =−(nπ) B n + C sin( τ) (5.57)

n
nπ
subject to initial conditions y = 0and y τ = 0at τ = 0. When n is even the solution is zero.
That is, since the right-hand side is zero when n is even,
B n = C 1 cos(nπτ) + C 2 sin(nπτ) (5.58)
But since both B n (0) and B (0) are zero, C 1 = C 2 = 0. When n is odd we can write

n
4C
2
B + [(2n − 1)π] B 2n−1 = sin(ωτ) (5.59)
2n−1
(2n − 1)π
(2n − 1)π is the natural frequency of the system, ω n . The homogeneous solution of the above
equation is
B 2n−1 = D 1 cos(ω n τ) + D 2 sin(ω n τ) . (5.60)

To obtain the particular solution we assume a solution in the form of sines and cosines.

B P = E 1 cos(ωτ) + E 2 sin(ωτ) (5.61)

Differentiating and inserting into the differential equation we find
4C
2 2 2
−E 1 ω cos(ωτ) − E 2 ω sin(ωτ) + ω [E 1 cos(ωτ) + E 2 sin(ωτ)] = sin(ωτ) (5.62)
n
ω n