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16.6 Characteristics and d’Alembert’s Solution 601

T
X – 5T = x – 5t
X + 5T = x + 5t

(x, t)

(x + 5t – 5T, T)

(x – 5t + 5T, T)
X
x – 5t x + 5t

FIGURE 16.21 Characteristic triangle in
Example 16.14

and, from Figure 16.21,

1 2 2 1 T x+5t−5T 2 2 1 2 4 5 6
X T dXdT = X T dX dT = x t + t .
10 10 0 x−5t+5T 12 36
The solution is
1
u(x,t) = [(x − 5t)cos(x − 5t) + (x + 5t)cos(x + 5t)]
2
1 5
1 −x+5t −x−5t 2 4 6
+ e − e + x t + t .
10 12 36

SECTION 16.6 PROBLEMS

In each of Problems 1 through 6, write the d’Alembert and
solution for the problem
u(x,0) = f (x),u t (x,0) = g(x) for −∞ < x < ∞.
2
u tt = c u xx for −∞ < x < ∞,t > 0
and 7. c = 4, f (x) = x, g(x) = e , F(x,t) = x + t
−x
u(x,0) = f (x),u t (x,0) = g(x) for −∞ < x < ∞. 8. c = 2, f (x) = sin(x), g(x) = 2x, F(x,t) = 2xt
2
9. c = 8, f (x) = x − x, g(x) = cos(2x), F(x,t) = xt 2
2
−x
2
1. c = 1, f (x) = x , g(x) =−x 10. c = 4, f (x) = x , g(x) = xe , F(x,t) = x sin(t)
2
2. c = 4, f (x) = x − 2x, g(x) = cos(x) 11. c = 3, f (x) = cosh(x), g(x) = 1, F(x,t) = 3xt 3
3. c = 7, f (x) = cos(πx), g(x) = 1 − x 2 12. c = 7, f (x) = 1 + x, g(x) = 0, F(x,t) = x − cos(t)
4. c = 5, f (x) = sin(2x), g(x) = x 3
In each of Problems 13 through 18, write the solution of
x
5. c = 14, f (x) = e , g(x) = x the problem
6. c = 12, f (x) =−5x + x , g(x) = 3
2
u tt = u xx for −∞ < x < ∞,t > 0
In each of Problems 7 through 12, solve the problem and
2
u tt = c u xx + F(x,t) for −∞ < x < ∞,t > 0 u(x,0) = f (x),u t (x,0) = 0for −∞ < x < ∞

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October 14, 2010 15:23 THM/NEIL Page-601 27410_16_ch16_p563-610