Page 620 - Advanced_Engineering_Mathematics o'neil

600 CHAPTER 16 The Wave Equation

Finally, on M, x − ct = x 0 − ct 0 so dx = cdt and

1
2 2
u t dx + c u x dt = u t cdt + c dx = c du
c
M M M
= c[u(x 0 − ct 0 ,0) − u(x 0 ,t 0 )].
M has initial point (x 0 ,t 0 ) and terminal point (x 0 − ct 0 ,0) because of the counterclockwise
orientation on C. Upon summing these integrals, we obtain

x 0 +ct 0
− F(x, y)dA = g(w)dw
x 0 −ct 0
− c[u(x 0 ,t 0 ) − u(x 0 + ct 0 ,0)]+ c[u(x 0 − ct 0 ,0) − u(x 0 ,t 0 )].
Then
x 0 +ct 0

− F(x, y)dA = g(w)dw − 2cu(x 0 ,t 0 ) + c[ f (x 0 + ct 0 ) + f (x 0 − ct 0 )].
x 0 −ct 0
Solve this equation for u(x 0 ,t 0 ) to get
1
u(x 0 ,t 0 ) = ( f (x 0 − ct 0 ) + f (x 0 + ct 0 ))
2
1 x 0 +ct 0 1
+ g(w)dw + F(x,t)dA.
2c x 0 −ct 0 2c
Since x 0 is any real number and t 0 any positive number, we can drop the subscripts and write the
solution u(x,t) as
1 1 x+ct
u(x,t) = ( f (x − ct) + f (x + ct)) + g(w)dw
2 2c x−ct
1
+ F(X, T )dX dT
2c
in whichwehaveused X and T as the variables of integration to avoid confusion with the point
(x,t) at which the solution is given.

EXAMPLE 16.14
We will solve the problem
2 2
u tt = 25u xx + x t for −∞ < x < ∞,t > 0
and
u(x,0) = x cos(x),u t (x,0) = e −x for −∞ < x < ∞.
The solution at any x and time t is
1
u(x,t) = [(x − 5t)cos(x − 5t) + (x + 5t)cos(x + 5t)]
2
1 x+5t 1
2
2
w
+ e dw + X T dX dT.
10 x−5t 10
Compute
1 x+5t 1 −x+5t −x−5t
e −w dw = e − e
10 x−5t 10

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October 14, 2010 15:23 THM/NEIL Page-600 27410_16_ch16_p563-610