Fourier Transform pair as (note the sign difference in the definition from that of
Haberman, § 10.3.2)
ƒ(k) =
f(x)e ikzdr f(x)
(a) Show that the general solution of (§ 10.6.1 of Haberman)
Pu
Ət²
f(x),
is
1
ikr
2/7 f(k) e³kx dk,
2π
2 Pu
əx²
u(x,0)
(-∞0 Ju(x, 0) Ət u(x,t) = ½ [ƒ(x − ct) + f(x + ct)] 0 (1) (2) (3)
Fig: 1