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16.6 Characteristics and d’Alembert’s Solution 599

Notice that by time t = 3 the wave has split into two disjoint copies of the original (t = 0
wave, one continuing to move to the right, the other to the left.) This separation is because f (x)
in this example is nonzero on only a bounded interval.

16.6.2 Forced Wave Motion
Using the characteristics we will write a solution for the Cauchy problem on the line with a
forcing term:
2
u tt = c u xx + F(x,t) for −∞ < x < ∞,t > 0
and
u(x,0) = f (x),u t (x,0) = g(x) for −∞ < x < ∞.
Suppose we want the solution at P 0 : (x 0 ,t 0 ). There are two characteristics through P 0 , namely
the straight lines
x − ct = x 0 − ct 0 and x + ct = x 0 + ct 0 .
Use parts of these to form the characteristic triangle of Figure 16.20, with vertices (x 0 −
ct 0 ,0), (x 0 + ct 0 ,0) and (x 0 ,t 0 ) and sides L, M, and I.Let denote this solid triangle and
compute the double integral of −F over .

∂ ∂

2 2
− F(x,t)dA = (c u xx − u tt )dA = (c u x ) − (u t ) dA.
∂x ∂t

Apply Green’s theorem to the last integral to obtain

2
− F(x,t)dA = u t dx + c u x dt,
C
where C is the boundary of , oriented counterclockwise. C consists of the line segments L, M,
and I. Compute the line integral over each side.
On I, T = 0, and x varies from x 0 − ct 0 to x 0 + ct 0 ,so
x 0 +ct 0 x 0 +ct 0

2
u t dx + c u x dt = u t (x,0)dx = g(w)dw.
I x 0 −ct 0 x 0 −ct 0
On L, x + ct = x 0 + ct 0 ,so dx =−cdt and

1
2 2
u t dx + c u x dt = u t (−c)dt + c u x − dx =−c du
c
L L L
=−c[u(x 0 ,t 0 ) − u(x 0 + ct 0 ,0)].
t P :(x , t )
0
0
0
M L

x
(x – ct , 0) I (x + ct , 0)
0
0
0
0
Characteristic triangle
FIGURE 16.20 The characteristic triangle.

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October 14, 2010 15:23 THM/NEIL Page-599 27410_16_ch16_p563-610