§19.2 Fourier变换 第10页 所以,根据 Fourier变换的折积公式,就有 sin kct 2丌)3/2 业(k) dk rd 其中∑是以r为球心、ct为半径的球面r-r|=c ★第二 (2)3/2 重(k) cos kct exp{ir}dk (2)3/2ot (r)d∑ ★第三项 sin kcr F(k, t-T)dr exp(ik.r)dk {a2z/ F(k,t {ik·r}dk 4丌c 6( 5(r-rl-cr)f(r t-Tdrdr 显然 8(r-rl-cr)f(r, t-T)dr f(r, t-r-r/c), r-rI 0 所以 sin kcT F(k, t-r)dr explik r]dk f(r,t-Ir-r'l/c)d 把上面的结果集中起来,就最后求得 v(r)d∑+ f(r, t-r-rl/c)dn§19.2 Fourier C† 1 10  ¤±§ŠâFourierC†òÈúª§Òk 1 (2π) 3/2 ZZZ Ψ(k) sin kct kc exp{ik · r} dk = 1 (2π) 3/2 r π 2 1 c ZZZ 1 |r − r0 | δ(|r − r 0 | − ct) ψ(r 0 ) dr 0 = 1 4πc ZZ Σ0 1 |r − r0 | ψ(r 0 ) dΣ0 , Ù¥Σ 0´±r¥%!ctŒ»¥¡|r − r 0 | = ct© F 1‘ 1 (2π) 3/2 ZZZ Φ(k) cos kct exp{ik · r} dk = 1 (2π) 3/2 ∂ ∂t ZZZ Φ(k) sin kct kc exp{ik · r} dk = 1 4πc ∂ ∂t ZZ Σ0 1 |r − r0 | φ(r 0 ) dΣ0 . F 1n‘ 1 (2π) 3/2 ZZZ · 1 kc Z t 0 sin kcτ F(k, t − τ ) dτ ¸ exp{ik · r} dk = Z t 0 ½ 1 (2π) 3/2 ZZZ · sin kcτ kc F(k, t − τ ) ¸ exp{ik · r} dk ¾ dτ = Z t 0 ½ 1 4πc ZZZ 1 |r − r0 | δ(|r − r 0 | − cτ ) f(r 0 , t − τ ) dr 0 ¾ dτ = 1 4πc ZZZ 1 |r − r0 | ·Z t 0 δ(|r − r 0 | − cτ ) f(r 0 , t − τ ) dτ ¸ dr 0 , w,§ 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 − r0 | f(r 0 , t − |r − r 0 |/c) dr 0 . rþ¡(J8¥å5§Ò￾¦ u(r, t) = 1 4πc   ZZ Σ0 1 |r − r0 | ψ(r 0 ) dΣ0 + ∂ ∂t ZZ Σ0 1 |r − r0 | φ(r 0 ) dΣ0   + 1 4πc2 ZZZ |r−r0 |<ct 1 |r − r0 | f(r 0 , t − |r − r 0 |/c) dr 0