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572 CHAPTER 16 The Wave Equation

1.2

0.8

0.4

0
0 0.5 1 1.5 2 2.5 3
x

FIGURE 16.4 Wave moving downward for
increasing times in Example 16.3.

EXAMPLE 16.3
Suppose the string is released from its horizontal position with an initial velocity given by g(x)=
x(1 + cos(πx/L)). Equation (16.6) is the solution. First compute the integral
L L
g(ξ)sin(nπξ/L)dξ = ξ (1 + cos(πξ/L))sin(nπξ/L)dξ
0 0
3L
⎧ 2
for n = 1
⎪
⎪
4π
⎨
=
2
⎪ L (−1) n
for n = 2,3,···.
⎪
⎩
2
nπ(n − 1)
The solution is
2 3L 2 πx πct
y(x,t) = sin sin
πc 4π L L
∞
2
2
L (−1) n nπx nπct
+ sin sin .
2
πc n π(n − 1) L L
2
n=2
Then with c = 1 and L = π, this solution is
∞ n
3
2(−1)
y(x,t) = sin(x)sin(t) + sin(nx)sin(nt).
2
2
2 n (n − 1)
n=2
Figure 16.4 shows the wave moving downward at times t = 0.2,0.5, 0.7,1.3, and 3.4.
16.2.3 Nonzero Initial Displacement and Velocity
Suppose the string has an initial displacement f (x) and an initial velocity g(x). Write the solu-
tion y f (x,t) for the problem with initial displacement f and zero initial velocity and the solution
y g (x,t) for the problem with zero initial displacement and initial velocity g(x). Then the solution
for the current problem is
y(x,t) = y f (x,t) + y g (x,t).
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October 14, 2010 15:23 THM/NEIL Page-572 27410_16_ch16_p563-610