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16.2 Wave Motion on an Interval 573

1.5

1

0.5

0
0 0.5 1 1.5 2 2.5 3
x
–0.5

–1
FIGURE 16.5 Waves in Example 16.4.

This function satisfies the wave equation and boundary conditions, because both y f (x,t) and
y g (x,t) do. Furthermore,
y(x,0) = y f (x,0) + y g (x,0) = f (x) + 0 = f (x)
and
∂y ∂y f ∂y g
(x,0) = (x,0) + (x,0) = 0 + g(x) = g(x).
∂t ∂t ∂t

EXAMPLE 16.4
We will solve the wave equation on [0, L] with y(0,t) = y(L,t) = 0, subject to initial position

x for 0 ≤ x ≤ L/2
f (x) =
L − x for L/2 ≤ x ≤ L.
and initial velocity g(x) = x(1 + cos(πx/L)). The solution y(x,t) is the sum of the solutions of
the problems solved in Examples 16.1 and 16.3. If c = 1 and L = π, this solution is

∞

4
y(x,t) = sin(nπ/2)sin(nx)cos(nt)
2
n π
n=1
∞ n
3
2(−1)
+ sin(x)sin(t) + sin(nx)sin(nt).
2
2 n (n − 1)
2
n=2
In Figure 16.5, the string’s position at t = 0.2 is the fourth graph from the top. The wave
moves upward to the next graph at t = 0.6, then upward again at t = 1.2. The highest wave is at
t = 1.8. Following this, the wave is partly below the horizontal axis at t = 2.3, and completely
below this axis at t = 2.9.
16.2.4 Influence of Constants and Initial Conditions
It is interesting to observe how the initial conditions and the constant c influence the wave
motion.

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October 14, 2010 15:23 THM/NEIL Page-573 27410_16_ch16_p563-610