singular Perturbation Problems in Ship Hydrodynamias 



at infinity has been included in ^q, and so we might expect that $| 

 will represent a flow with velocity vanishing at infinity, and there 

 appears to be no reason to expect a circulation in the 2-D problem. 



It would be tedious to go through the same arguments that 

 were used previously, and so I shall only summarize the results that 

 would be obtained after a careful matching process. In the near 

 field, $1 does indeed yield a velocity field which is bounded in 

 magnitude at infinity, and there is no circulation. Thus, it can be 

 represented by the series: 



:r. I \ ^ j_ Ain ■, , An COS 9 + Bii S in , / -, / ,« 



$, (x,y,z) = C, +"2^ log r +— IJ j;^ U + ... . (2-64) 



The "constants" are all functions of x. In the near field, all terms 

 must be the same order of magnitude, by definition, and so An 

 and Bii are O(eAio). (lam, as usual, ignoring quantities which 

 are 0(log e).) In the matching, the l/r terms are lost in the first 

 round, and the log r and constant terms are forced to match the 

 inner expansion of the outer expansion. 



In the outer expansion, (2-60), only a line of sources in the 

 <^l term of (2-60) can match the near-field expansion properly. 

 That is, in (2-59) and (2-60), we have the following: 



a. (k) = b. (k) = except for n = 0. 



The two-term outer expansion and its two -term inner expansion are: 



<|>(x,y,z) ~ Ux + -Ij^ \ dk e''*''KQ(|k|r)a*(k) (2-65a) 



~ Ux + -1^ 0-, (x) log r - ^ f,(x) , (2- 65b) 



where (2-61 a') has been used to express the latter. 



Matching between the near-field and far- field then shows that: 



A,o=(r,(x) (2- 66a) 



C, = -^f,(x). (2-66b) 



In obtaining an actual solution, one proceeds through the 

 following steps: 1) Matching shows that $i represents a flow with 

 bounded velocity at infinity. 2) Then the 4>| problem is completely 



719 



