196 



Here we have dropped the superscripts n, i and have 

 defined (r]^)j, (?k)j, (Bk'j' and (a)^)j by 



(Se) ■ = - ^ (mt, + a )u. , - m3W. , 

 j 2 n j-1 ^ j-1 



r=u, -u. +h.v. ,,„ 

 1 D-1 3 3 3-1/2 



r = w. , - w. + h.t. , ,. 

 2 3-1 3 3 3-1/2 



(60a) 

 {60b) 



(67). 



1 n-1 



2"n"j-l/2 (62f) 



1 (^4 + a„)w. - mgu, + ^ a^ w^lj^^ '^^g) 





'^-'j-1/2" <«^'3-l/2 



- (nij + a„) (u2) 



n j-l/2 



(60c) 



(Bs)^ = - J (mtt + a )w._ - m9U._j 



1 n-1 



^ 2 "n "j-l/2 



3 



(62h) 



{63a) 



'^"'j = Vl/2 - ™12 



(bt); , ,^ + (et)^ , ,^ - (m^ + ci„) Cuw) 



j-1/2 j-l/2 



n j-l/2 



(°2)j = -^ 



(03) j = - -J (mi + 2a^ 



(63b) 



{63c) 



m, {w2) . - mq (u2) . 



^ 3-1/2 ^ 3-1/2 



n-1 n-1 



+ fY(W, ,u — u w 



"nl j-l/2 j-l/2 j-l/2 j-l/2 



{60d) 



(04). 



(05) ■ 



- {mi + 2a^) 



me 



{63d) 



{63e) 



3 3-1/2 



3-1/2 



<^i'j=ir^i9j 



(m, + 2(Y ) u . , - m w . 

 1 ™n' 3-1/2 6 ]-l/2 



b. 



'3 = -^., 



,i,,. .- J^.i 



2 "3-1 



(tJe) j = - -^ me 



(60e) The boundary conditions become 



(61a) 



(61b) 



6uo = fiwg = 69q = 0, 6uj = 6w = 



{63f) 



(64) 



The solution of the linear system given by (59) 

 and (64) is obtained by using the block elimination 

 method. According to this method, the system is 

 written as 



(C3)j = 2-, 



{61c) 



Here 



/A 6 = r 



(65) 



'^'''j = 2 -j-1 



{C5) . = - (mo + a ) u . 

 3 "3 



{<;6)j = - (m2 + a^)u 



(Bl)j = {Cl)j 



(e2)j = {C2)j 



(e3)j=jt. 



(B14) . = ^ t 



J 2 j-1 



(61d) 



(61e) 



{61f) 



(62a) 



{62b) 



(62c) 



(62d) 



/A 



(Bs). = -| (m, +a^)u. -maw. - i a^ u^lj/^ 



{62e) 



