V cr 所以,根据 Fourier变换的折积公式,就有 (2x)3/2 v()sin kct p{ik·r}dk (2)/2V2c 6(r-rl-ct)v(r)dr 1 其中∑是以r为球心、ct为半径的球面|r-rl=ct (k:)cos kct explikr)dk (2丌)3/2Ot 中(k)nk expiry 10 (2)3 k如F(k一:r)4k {a∥/P (sinke F(k, t-rlexplik r)dk dr {=// r-rI 6(r-rT-cr)f(r, t-T)dr's dr 显然 二f(-p-r1/,|r-r1<ct 8(r-rl-cr)f(r, t-r)dr 所以, 公人如 kcT F(k, t-)p读, f(r, t-lr-r/c)dr 把上面的结果集中起来,就最后求得 (T,t) u(r)d∑+ p(r)d∑Wu Chong-shi §27.2 Fourier ➘➴ Ð 8 Ñ = 1 √ 2πc 2 r Z ∞ 0 sin kct sin kr dk = r π 2 1 cr δ(r − ct). ➯➲✛ÒÓ Fourier ✤✥✕❞✣ÔÕ✛■✬ 1 (2π) 3/2 ZZZ Ψ(k) sin kct kc exp{ik · r} dk = 1 (2π) 3/2 r π 2 1 c Z ZZ 1 |r − r 0 | δ(|r − r 0 | − ct) ψ(r 0 ) dr 0 = 1 4πc ZZ Σ0 1 |r − r 0 | ψ(r 0 ) dΣ0 , ❧ ❍ Σ 0 ❍➲ r ▲❤♠❇ ct ▲♥♦✕❤ ❭ |r − r 0 | = ct ✪ F ✦❨æ 1 (2π) 3/2 ZZ Z Φ(k) cos kct exp{ik · r} dk = 1 (2π) 3/2 ∂ ∂t ZZ Z Φ(k) sin kct kc exp{ik · r} dk = 1 4πc ∂ ∂t ZZ Σ0 1 |r − r 0 | φ(r 0 ) dΣ0 . F ✦❵æ 1 (2π) 3/2 ZZZ  1 kc Z t 0 sin kcτ F(k, t − τ) dτ  exp{ik · r} dk = Z t 0  1 (2π) 3/2 ZZZ  sin kcτ kc F(k, t − τ)  exp{ik · r} dk  dτ = Z t 0  1 4πc ZZZ 1 |r − r 0 | δ(|r − r 0 | − cτ) f(r 0 , t − τ) dr 0  dτ = 1 4πc Z ZZ 1 |r − r 0 | Z t 0 δ(|r − r 0 | − cτ) f(r 0 , t − τ) dτ  dr 0 , ♣❴✛ Z t 0 δ(|r − r 0 | − cτ) f(r 0 , t − τ) dτ =    1 c f(r 0 , t − |r − r 0 |/c), |r − r 0 | < ct; 0, |r − r 0 | > ct. ➯➲✛ 1 (2π) 3/2 ZZZ  1 kc Z t 0 sin kcτ F(k, t − τ) dτ  exp{ik · r} dk = 1 4πc2 ZZZ |r−r0 |<ct 1 |r − r 0 | f(r 0 , t − |r − r 0 |/c) dr 0 . ➨ ü❭ ✕❪✛q ❍r❈✛■Ú❁☛❱ u(r, t) = 1 4πc   Z Z Σ0 1 |r − r 0 | ψ(r 0 ) dΣ0 + ∂ ∂t ZZ Σ0 1 |r − r 0 | φ(r 0 ) dΣ0   + 1 4πc2 ZZ Z |r−r0 |<ct 1 |r − r 0 | f(r 0 , t − |r − r 0 |/c) dr 0