Page 350 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 13] FOURIER SERIES 341

@y
With this equation in place the boundary condition ðx; 0Þ¼ 0, 0 < x < L can be considered.
@t
@y h n ih n n n n i
¼ c 1 sin x c 3 cos t c 4 sin t
@t L L L L L
At t ¼ 0
h n i n
0 ¼ c 1 sin x c 3
L L
n
Since c 1 6¼ 0and sin x is not identically zero, it follows that c 3 ¼ 0 and that
L
n n n
h ih i
y ¼ c 1 sin x c 4 cos t
L L L
The remaining initial condition is
2
yðx; 0Þ¼ mðLx x Þ; 0 < x < L
When it is imposed
n n
2
mðLx x Þ¼ c 1 c 4 sin x
L L
However, this relation cannot be satisfied for all x on the interval ð0; LÞ. Thus, the preceding
extensive analysis of the problem of the vibrating string has led us to an inadequate form
n n n
y ¼ c 1 c 4 sin x cos t
L L L
and an initial condition that is not satisfied. At this point the power of Fourier series is employed. In
particular, a theorem of differential equations states that any finite sum of a particular solution also is a
solution. Generalize this to infinite sum and consider
n n
1
X
b n sin x cos t
y ¼
L L
n¼1
with the initial condition expressed through a half range sine series, i.e.,
n
1
X 2
b n sin x ¼ mðLx x Þ; t ¼ 0
L
n¼1
According to the formula of Page 338 for coefficient of a half range sine series
L ð L 2 n x
ðLx x Þ sin dx
2m b n ¼ 0 L
That is
L ð L n x ð L 2 n x
Lx sin dx x sin dx
b n ¼
2m 0 L 0 L
Application of integration by parts to the second integral yields
L ð L n x L 3 ð L L n x
b n ¼ L x sin cos 2xdx
2m 0 L dx þ n cos n þ 0 n L
When integration by parts is applied to the two integrals of this expression and a little algebra is
employed the result is
4L 2
3
b n ¼ ð1 cos n Þ
ðn Þ