then: p1122(c2-1)+p11332=0 P1122=p1133 ·中umnm=9222c+p23c2+92323Sc2+p23322c2.… …+932232c2+p323252c2+0332252c2+033535 and: Φunnn=p2222 Then,taking into account the symmetries,we obtain the relation: p222(c-1)+93333+2s2c2(p23+202323)=0 (a) ·0mmmm=92222+92233c2+p23232c2+p23322c2.… …+9523252c2+9323252c2+93322c2+93333 and: Φmmmm=93333 then taking in account symmetry,we have: p2222s+p3333(c4-1)+2s2c2(p2233+2p2323)=0 (b) Examining member by member the difference of relations shown in (a)and (b) above,one obtains: P2222=03333 Replacing in (a): p2222(c4+s-1)+2s2c2(p2233+2p2323)=0 -2sc2p2222+2s2c2(p2233+202323)=0 2p2323=02222-92233 ·重1m1m=91212S2+91313C2=91313 2003 by CRC Press LLCthen: and: Then, taking into account the symmetries, we obtain the relation: (a) and: then taking in account symmetry, we have: (b) Examining member by member the difference of relations shown in (a) and (b) above, one obtains: Replacing in (a): j1122 c 2 ( ) – 1 j1133s 2 + = 0 j1122 = j1133 • FII II II II j2222c 4 j2233s 2 c 2 j2323s 2 c 2 j2332s 2 c = +++ 2º º j3223s 2 c 2 j3232s 2 c 2 j3322s 2 c 2 j3333s 4 ++++ FII II II II = j2222 j2222 c 4 ( ) – 1 j3333s 4 2s 2 c 2 j2233 + 2j2323 + + ( ) = 0 • FIII III III III j2222s 4 j2233s 2 c 2 j2323s 2 c 2 j2332s 2 c = +++ 2º º j3232s 2 c 2 j3232s 2 c 2 j3322s 2 c 2 j3333s 4 ++++ FIII III III III = j3333 j2222s 4 + j3333 c 4 ( ) – 1 2s 2 c 2 j2233 + 2j2323 + ( ) = 0 j2222 = j3333 j2222 c 4 s 4 ( ) + – 1 2s 2 c 2 j2233 + 2j2323 + ( ) = 0 –2s 2 c 2 j2222 2s 2 c 2 j2233 + 2j2323 + ( ) = 0 2j2323 = j2222 – j2233 • FI III I III j1212s 2 j1313c 2 = = + j1313 TX846_Frame_C13 Page 263 Monday, November 18, 2002 12:29 PM © 2003 by CRC Press LLC