a Horizontal Layer of Fluid. 543 



o£ this circle, k is given by J '(ka) = 0, J being the Bessel's 

 function of zero order, or &a = 3 - 832. If b be the side of 

 the hexagon, a 2 = 3^3 . b 2 /27r. 



Appendix. 



On the nearly symmetrical solution for a nearly circular 

 area, when w satisfies (d 2 /dx 2 + d 2 /dy 2 -t-& 2 )io = and makes 

 dw/dn = on the boundary. 



Starting with the true circle of radius a, we have w a 

 function of r (the radius vector) only, and the solution is 

 w = J (kr) with the condition J '(ka) =0, yielding &a = 3'832, 

 which determines k if a be given, or a if k be given. In 

 the problem proposed the boundary is only approximately 

 circular, so that we write r = a-\-p, where a is the mean 

 value and 



p = a 1 cos# + /3 1 sin#+ . . . + « H cos nd + /3 n smn0. . (57) 



In (57) 6 is the vectorial angle and a Y &c. are quantities 

 small relatively to a. The general solution of the differ- 

 ential equation being 



K; = A J (A;r , ) + J 1 (/;r){A 1 cos^ + B J sin 0} 



+ ... + J n (&r){A n cosn0 + B H sin0}, . . (58) 



we are to suppose now that A ]5 &c, are small relatively 

 to A . It remains to consider the boundary condition. 



If (f> denote the small angle between r and the normal dn 

 measured outwards, 



dw dw , die . 



dn 



and 



dr 



tan + = ;,ie 



dr 

 dp 



(59) 



= ( — ot n sin n6+ ft n cos n0) (GO) 



n 

 add a 



with sufficient approximation, only the general term being- 

 written. In formulating the boundary condition dw/dn = 

 correct to the second order of small quantities, we require 

 div/dr to the second order, but div/dO to the first order only. 

 We have 



| ^ = A { J ' (ka) + kp J "(*a) + lPp % Jo'"(ha) } 

 + {JJ(ka) + kpJ n "(ka)}{A n cos n6 + B n sin nd}, 



7/i = J,i(kd)\ — A„ sin7i^ + B„ cos uO} 

 ado a 



