Murthy 



It will be noted that the contour L,, and L c have in common the 

 curves L and L at the bow and stern which are taken in different 

 directions. Choosing the clockwise direction for the integration, the 

 combined line integral may be written 



v 2 



47Tg 



<f[I>] G { - G [* { ] +-^f|>] g1 dr, 



L B +L S 

 + (line integrals along the hull intersection L , L , 



L and L ) 



2 + 2- ; 



where [^] and [ty t] denote the "jumps" in these functions across 

 the cushion boundary in crossing from a point on the EFS outside to 

 a point on the IFS just within. These jumps will exist due to the 

 singularity of the potential at the boundary of the cushion indicated 

 earlier. We are, of course, assuming that G is continuous across 

 the boundary. 



The line integrals along the hull surfaces need not be discussed 

 in detail as their order will be 0( 5 ) higher than that of the potential 

 under investigation. This is because the total width of each hull is of 

 0( 5 ) and the line integral is taken with respect to 7? on the hull 

 surfaces. On the other hand, the line integrals along L R and L. 

 will be of the same order as the potential as v can now take a value 

 up to the semi-width of the ACV on either side. However, the line 

 integrals around the hulls will have to be taken into account when 

 evaluating ^ 110 by including the contribution of 0(5/8 ) arising from 

 the integration of the potential ¥ Q10 of 0(/3 ). 



It may be added that the line integral along the cushion 

 boundary will vanish if there is no discontinuity of ^ at the bounda- 

 ry. As the discontinuity arises mainly because of a pressure distri- 

 bution within the IFS higher than atmospheric, a suitable distribution 

 of pressure will remove the discontinuity and the need for evaluating 

 the line integral. This will be discussed presently. 



The surface of integration S_ is that part of the plane z = 

 contained within the instantaneous position of the IFS, which is a 

 fluctuating region oscillating about the steady state position S Q say. As 

 the instantaneous position S Q is unknown and has to be solved as 

 part of the problem, we will reduce the surface of integration to the 

 known region S Q . 



226 



