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566 CHAPTER 16 The Wave Equation

θ + Δθ
T(x + Δx, t)
y
T(x, t) θ

y = y(x, t)
(fixed t)

x
x x + Δx

FIGURE 16.1 Deriving the wave equation.

As x → 0, x → x and we obtain
2
∂v ∂ y
= ρ .
∂x ∂t 2
The horizontal component of the tension is h(x,t) = T (x,t)cos(θ),so
∂y
v(x,t) = h(x,t)tan(θ) = h(x,t) .
∂x
Then
2
∂ ∂y ∂ y
h = ρ .
∂x ∂x ∂t 2
To compute the left side of this equation, use the fact that the horizontal component of the tension
of the segment is zero, so

h(x + x,t) = h(x,t).
This means that h is independent of x,so
∂ y ∂ y
2
2
h = ρ .
∂x 2 ∂t 2
2
Let c = h/ρ to get the one-dimensional wave equation
2
2
∂ y ∂ y
= c 2 . (16.1)
∂t 2 ∂x 2
The fact that the ends are held fixed is reflected in the boundary conditions
y(0,t) = y(L,t) = 0for t ≥ 0.
The initial displacement of the string, and the velocity with which it is released at time 0, are the
initial conditions, given by
y(x,0) = f (x) for 0 ≤ x ≤ L.
and
∂y
(x,0) = g(x).
∂t
The wave equation, together with these initial and boundary conditions, is called an initial-
boundary value problem for y(x,t).

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October 14, 2010 15:23 THM/NEIL Page-566 27410_16_ch16_p563-610