Page 44 - 6637

Supplementary Problems

1 Z t 0
+cu (x + cτ, 0)) dτ = u(x , 0) + (−cu (x − cτ, 0) + cu (x + cτ, 0)) dτ
0
x
0
0
0
x
x
2 0
1 Z t 0
+ (u (x − cτ, 0) + u (x + cτ, 0)) dτ
t
0
t
0
2 0
1
= u(x , 0) + (u(x − ct , 0) − u(x , 0) + u(x + ct , 0) − u(x , 0))
0
0
0
0
0
0
0
2
Z Z
1 x 0 −ct 0 1 x 0 +ct 0
+ −u (τ, 0) dτ + u (τ, 0) dτ
t
t
2c 2c
x 0 x 0
Z
1 1 x 0 +ct 0
= (u(x − ct , 0) + u(x + ct , 0)) + u (τ, 0) dτ
0
0
0
t
0
2 2c
x 0 −ct 0
We have D’Alembert’s solution.
Z x+ct
1 1
u(x, t) = (u(x − ct, 0) + u(x + ct, 0)) + u (τ, 0) dτ
t
2 2c x−ct

6.2 Supplementary Problems
Exercise 120. Solve initial value problems for a wave equation:
2
a) u = u , u(x, 0) = x , u (x, 0) = 0; b) u = 4u , u(x, 0) = 0, u (x, 0) =
tt
tt
xx
xx
t
t
x;
c) u = u , u(x, 0) = x, u (x, 0) = −x.
t
xx
tt
2
2
Solution: a) u = x + t ; b) u = xt; c) u = x(1 − t).
Exercise 121. Find the shape of the string at t = π, defined by the initial value
problem: u = u , u(x, 0) = sin x, u (x, 0) = cos x.
tt
xx
t
Solution: u = − sin x.
2
x
Exercise 122. Solve u = c u , u(x, 0) = e , u (x, 0) = sin x.
xx
tt
t
2
2
Exercise 123. Solve u = c u , u(x, 0) = log(1 + x ), u (x, 0) = 4 + x.
t
xx
tt
Exercise 124. The midpoint of a piano string of tension T, density ρ, and length
l is hit by a hammer whose head diameter is 2a. A flea is sitting at a distance l/4
from one end. (Assume that a < l/4; otherwise, poor flea!) How long does it take
for the disturbance to reach the flea?
Exercise 125. (The hammer blow) Let φ(x) ≡ 0 and ψ(x) = 1 for |x| < a
and ψ(x) = 0 for lxl ≥ a. Sketch the string profile (u versus x) at each of
the successive instants t = a/2c, a/c, 3a/2c, 2a/c, and 5a/c. [Hint: Calcu-
late u(x, t) = 1 R x+ct ψ(s)ds = 1 {length of (x − ct, x + ct) ∩ (−a, a)}.Then
2c x−ct 2c
40