432 Chapter 15Orthogonal expansions. Fourier analysis
and, after integration by parts,
(15.54)
Therefore
(15.55)
0 Exercise 12, 13
15.5 The vibrating string
It was shown in Section 14.7 that the solution of the wave equation for the vibrating
string of length lis (equation (14.103))
(15.56)
in which the coefficientsA
nandB
nare determined by the initial conditions (equations
(14.104))
(15.57)
where f(x) and g(x) are given displacement and
velocity functions that describe the state of the system
at timet 1 = 10.
We suppose that, att 1 = 10 , we pull the string side-
ways at its centre, and then let go. The initial shape of
the string is shown in Figure 15.10, and the displace-
ment function is therefore identical to the function
discussed in Example 15.7, except that it is defined
only within the half-interval 01 ≤ 1 x 1 ≤ 1 l. The Fourier representation of the function is
then given by equation (15.55):
yx f x (15.58)
Ax
l
x
l
( ) ( ),= 0 = sin −sin +sin
81
1
1
3
31
5
22 2 2π
ππ 55 πx
l
−
tnnny
t
gx B
nx
l
==∂
∂
==
∑
01() sin
∞ω
π
yx f x A
nx
l
nn() (),= = sin
=∑
0
1∞π
yxt
nx
l
AtBt
nnnnn( ),=sin cos +sin
=∑
1∞π
ωω
fx
Ax
l
x
l
x
l
( )=sin −+−sin sin
81
1
1
3
31
5
5
22 2 2π
πππ
b
n
A
n
n
A
n
n
n=+ =,,,,
−=
0
8
15913
8
2222if even
if
if
π
π
...
3 3 7 11 15,, , ,
...
A
0 l
x
f(x)
- •
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Figure 15.10