电子科技大学：《数理方程与特殊函数》第四章 行波法（4.2）无界域上二维波动方程求解

l(x,y,二,D) 1 a rat r2r (x+rcos 0,y+rsin 0 rare 2Tal at Jo Jo r2n v(x+rcos 0,y+rsin 0) rare 2a √at-r 1arar2p(x+rcos, y+rsinerdrde 2al at Jo Jo 由初始条件得: (x+r(x+:9y+9i =x(x+y)+2x(x+)rcos 0+(x+y)r cos 8 +x r(sin 8+ cos 0)+ 2xr(sin 0+ cos 0)cos 0 +r(sin 0+cos 0)cos 8

0.8 1 0.6 0.4 0.2 0 x t 0 0.5 1 1.5 2 −1 −0.5 0 0.5 1 n 11 由初始条件得： 2 0 0 2 2 2 1 ( cos , sin ) ( , , , ) 2 at x r y r u x y z t rdrd a t a t r          + + =      −   2 0 0 2 2 2 1 ( cos , sin ) 2 at x r y r rdrd a a t r        + + +     −   2 0 0 2 2 2 1 ( cos , sin ) 2 at x r y r rdrd a t a t r          + + =      −   ( ) 2       x r y r x r x y r r + + = + + + + cos , sin ( cos ) ( cos sin ) 2 2 2 2 2 3 2 ( ) 2 ( ) cos ( ) cos (sin cos ) 2 (sin cos )cos (sin cos )cos x x y x x y r x y r x r xr r           = + + + + + + + + + + +

由三函数的周期性、正交性、倍角公式得,cose, Sinθ,cos0Sin,cos30,cos2Sin在0到2π上的积 分均为零,而 2丌 cos2a0=丌 0 所以 at r2r p(x+rose,y+rsin e drdo rdr r 2Tx(x+y) +(3x+y 2Tx(+y)at+I(3x+y)a't

0.8 1 0.6 0.4 0.2 0 x t 0 0.5 1 1.5 2 −1 −0.5 0 0.5 1 n 12 由三函数的周期性、正交性、倍角公式得，cosθ, Sinθ, cosθ Sinθ, cos3θ, cos2θ Sinθ在0到2π上的积 分均为零，而 2 2 0 cos d     =  所以： 2 0 0 2 2 2 at ( cos , sin ) x r y r rdrd a t r      + + −   3 2 0 0 2 2 2 2 2 2 2 3 3 2 ( ) (3 ) 2 2 ( ) (3 ) 3 at at rdr r dr x x y x y a t r a t r x x y at x y a t     = + + + − − = + + +  