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596 CHAPTER 16 The Wave Equation

Add this to equation (16.13) to get
1 x
2G(x) = f (x) + g(w)dw − F(0) + G(0).
c 0
Then
1 1 x 1 1
G(x) = f (x) + g(w)dw − F(0) + G(0). (16.15)
2 2c 0 2 2
But then, from equation (16.13),
1 1 x 1 1
F(x) = f (x) − G(x) = f (x) − g(w)dw + F(0) − G(0). (16.16)
2 2c 0 2 2
Finally, use equations (16.15) and (16.16) to obtain the solution
u(x,t) = F(x − ct) + G(x + ct)
1 1 x−ct 1 1
= f (x − ct) − g(w)dw + F(0) − G(0)
2 2c 0 2 2
1 1 x+ct 1 1
+ f (x + ct) + g(w)dw − F(0) + G(0).
2 2c 0 2 2
After cancellations and combining the integrals, we have the solution
1 1 x+ct
u(x,t) = ( f (x − ct) + f (x + ct)) + g(w)dw. (16.17)
2 2c x−ct
This is d’Alembert’s solution of the Cauchy problem for the wave equation on the real line. It
gives an explicit solution in terms of the initial position and velocity functions.

EXAMPLE 16.12
We will solve the initial-boundary value problem
u tt = 4u xx for −∞ < x < ∞,t > 0
and
u(x,0) = e −|x| ,u t (x,0) = cos(4x) for −∞ < x < ∞.
Immediately,
1 x+2t
1
u(x,t) = e −|x−2t| + e −|x+2t| + cos(4w)dw
2 4 x−2t
1
1
= e −|x−2t| + e −|x+2t| + (sin(4(x + 2t)) − sin(4(x − 2t)))
2 16
1
1
= e −|x−2t| + e −|x+2t| + sin(4x)cos(8t).
2 8
16.6.1 Forward and Backward Waves
Write d’Alembert’s solution as
1 1 x−ct
u(x,t) = f (x − ct) − g(w)dw
2 c 0
1 1 x+ct
+ f (x + ct) + g(w)dw
2 c 0
=ϕ(x − ct) + β(x + ct),

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October 14, 2010 15:23 THM/NEIL Page-596 27410_16_ch16_p563-610