singular Perturbation Problems in Ship Hydrodynamics 



(Note the comparison between (2-28) and the relation between a„ 

 and (Tn in Table 2-1. In fact, (2-29) gives the inversion of the 

 formula in Table 2-1.) The inner expansion of f(x,y,z) can now be 

 written in terms of these two functions: 



f(x,y,z) = ^{p.(x,z)(sgn y) - Y(x,z)y - jy y^(sgn Y){[l,^^ + fijz) 



+ 37 y'(Yxx + \zz) + 4j y^sgn y)(fixxxx + ^\^%xiz + M-zzzz) + • • • } 



(2-30) 

 This may be compared with (2-12). 



Now let us assume that the two-term outer expansion is: 



♦(x.y,z;c)~Ux+if T ^i^^iifdiiL^ 



Furthermore, assume that |j.| and "Y| are both 0(c). (If these 

 assumptions are too restrictive, that fact will become clear in the 

 subsequent steps of the method of matched asymptotic expansions.) 

 Then the inner expansion of the two-term outer e3qpansion is: 



1 2 

 4Kx,y,z;e) ~ Ux ± |j,,(x,z;c) - y Y, (x,zje) t "2 Y (m-Ixx ■•■ f^izz^' 



0(1) 0(€) 0(€^) 0(€^) 



I have kept four terms, as indicated by the order-of-magnitude 

 notes under the terms. (Recall that y = 0(e) in the inner expan- 

 sion.) 



Matching with the appropriate forms of the outer expansion 

 of the inner expansion, we find that: 



A, (x,z;c) = ± fx,(x,z;€); (2-31) 



B2(x,z;c) = - e\, (x,z;€). (2-32) 



From (2-25), we find that: 



V,(x,z;e) = - €UG^(x,z) = - Ug^(x,z;e). (2-33) 



It appears now that we could use this knowledge of "Vj in (2-28) for 

 determining jx, . But this is wrong. Note from (2-30) that Yi(x,z) 



693 



