2 V Ox ay ar ror r 80 2 极坐标相容方程VΦ=(2+2)①=(2+0+ ar- rar 72b2)2c=0 V4Φ + )(2+ ar 00- ar- r ar r-a0 (D 2 ag (D 20020 r4a4×32a3 ar3 r3 a02ar 1a2Φ402Φ1aΦ 0 2 4 3 ar 4=如,203①1a2Φ,1a,1o4Φ Or 24d 2 ag 4aC 0 r2a-206 02r400   +   +     +   +     = ) 1 1 )( 1 1 ( 2 2 2 2 2 2 2 2 2 2 4 r r r r  r r r r  r r r r r r r  r   −    +    +     +    = 2 3 3 3 3 4 4 2 2 4 4 4 2 4 2 1 2 2    0 1 4 1 2 3 2 2 4 2 2 =   +    +    − r r r  r r 4 4 2 3 4 2 3 2 3 4 4 4 2 1 1 1    +   +    −    +      = r r r r r r r r 0 2 2 4 2 2 2 4 3 2 2 3 4 2 =    +     −     + r r  r r  r  2 2 2 2 2 2 2 2 2 2   +   +   =   +    = x y r r r r 极坐标相容方程 ( ) ( ) 0 2 2 2 2 2 2 2 2 2 2 2 4  =   +   +    =   +     = x y r r r r 