F(x) = -| Sgn x(£n(2K|x|)-iTT+Y) + it f G-(£) d? 



(44) 



Here, Sgn x is defined as 1 when x < 0, -1 when x > 0, and undefined at x = 0, and y , 

 Euler's constant, is 0.57721.... The integrated term in Equation (43) can be 

 expressed as 



£im 

 e+0 



-q..(DF(x-5) 



£=X-£ 



-q,(£)F( X -0 

 S=-L/2 J 



?=W2 



£=x+£ 



(45) 



Since q.(-L/2) and q.(L/2) are assumed to be zero, Equation (45) is further reduced 



ilim {-q. (x-e)F(£)+q. (x+e)F(-£)} 



(46) 



£+0 



As q.(x) is a continuous function along the ship's length, Equation (46) can be 

 written as 



- £im {q.(x)[F(£)-F(-e)]} 



(47) 



e+0 



F(e) has a singularity as £ -* 0. This singularity occurs because of the limits of 



the integral and can be omitted with the interpretation that the "finite part" 



11 12 

 defined by Hadamard obeys many of the ordinary rules of integration. ' By dis- 

 carding the integrated term, the integral equation for q . (x) becomes 



qjOO " i 



mi- 



(?) F(x-C) d? = o. + a (j=3,5) 



(48). 



The numerical solution of Equation (48) is accomplished by iterating. One dis- 

 advantage of this method is that the numerical method does not converge when the 



13 



