Ogilvie 



Every term in (2-51) must be of the same order of magnitude 

 at a point in the near field, that is for r = 0(e). If a term were 

 of some other order of magnitude with respect to c, the definition 

 of "consistency" would eliminate it from this series. The orders 

 of magnitude of most of the unknown constants can then be written 

 down. Since the first term, Ux, is 0(c), we can make the follow- 

 ing statements: 



tiq, Aoo= 0(c); Aqt, Bo,= 0(c ); Aon, Bon = 0(e ). 



The term containing the logarithm does not fit the pattern quite so 

 well -- unless we follow my arbitrary practice of saying that 

 log € = 0(1). (See the discussion of "consistency" in Section 1.2.) 

 Then we can say that: 



60= 0(€). 



The Laurent series expression for the near-field expansion 

 is very convenient when It comes to finding the outer expansion of 

 the inner expansion. All we need to do is to interpret r differently 

 and rearrange the terms according to their dependence on €. Thus, 

 if we consider that r = 0(1), the outer expansion of the one-term 

 inner expansion is: 



«(>(x,y,z) ~ $o(x,y;z) ~ Ux + 60 log r + tiq tan' ^ + A 



^00 



0(1) 0(€) 0(€) 0(c) 



^ Aq, cos e ^ Bo, sine +0(^3^^ ^2-51') 

 r r 



0(c2) 0(€^) 



This obviously matches the one-term outer expansion, with an 

 asymptotically small error which is 0(e). 



We could keep two terms in (2-51'); that would be the two- 

 term outer expansion of the one-term inner expansion, which would 

 have to match the one-term inner expansion of the two-term outer 

 expansion. From (2-51') and (2-45b), we thus construct the equality: 



Ux + 60 log r + Tiotan' ^ + A^= Ux + -^ y, (z) [l - i tan ' X] . 



This can be true only if the following are separately true: 



6o=0; Tio= - — Y,(z); Aoo = -2V|(z)- (2-52) 



700 



