Singula!' Perturbation Problems in Ship Hydrodynamics 



solution has a satisfactory behavior at infinity, viz. , it represents 

 outgoing waves. 



We need the inner expansion of this potential function, that 

 is, we must find its behavior as r = (y^ + zr)''^ — 0. The basic idea 

 here in finding the inner expansion is to use the second form of 

 solution, convert the contour integral into an integral along a closed 

 contour, and use the calculus of residues. The integrand of the 

 inner integral has four singularities, located at i = ± io and at 



i = ± i|k| , where Iq - {v^ - k*^) 



2\J/2 



The first two are simple poles. 



but the second two are branch points. We "connect" the latter via 

 the point at infinity; see Fig. (3-3). It is drawn for the case that 

 |k| < v; if |k| > V, aJLl four singularities are purely imaginary. 

 The contour is closed as shown if y > 0. (Otherwise, the contour 

 is closed below.) The integrals along the large circular arcs 

 approach zero as the radius of the arcs approaches infinity. Then 

 the inner integral in (^(x,y,z) is equal to Ziri times the residue 

 at i = - io> less the value of the contour integral down and back up 

 the imaginary axis. The latter can be shown to be 0(e), and so the 

 inner integral in ^(x,y,z) is: 



I 



ZttIv 



-iyVv^ 



-CD (k^ + i2)'/2 



(v' 



k^)'/2 



+ 0(e) 



Next, we assume that the source distribution is smooth 

 enough that (J'(x) does not vary rapidly on a length scale comparable 

 with ship beam. This assumption implies that cr (k) decreases 

 rapidly with increasing values of k, and so the value of the above 

 inner integral --a function of k -- does not really matter except 

 when k is small in magnitude. Accordingly, we expand the above 

 expression in a manner appropriate for small |k|. We obtain 



Fig. (3-3). Contour of Integration Defining the 

 Velocity Potential of a Line of Pul- 

 sating Sources: Zero-Speed Case 



751 



