The Forces on a Circular Cylinder 528 
where z = x + iy, and the coefficients are complex. Now we have 
—n (= i)” % m—1 pixz ie ‘ 
% ome Cue y>0. (6) 
Hence, in order to satisfy (4), we write ¢, in the form 
bo = i F (x) e*-*¥ die + i G (rc) e@+« U-29) dic, (7) 
0 0 
where the real part is to be taken, and 
F() = (2° A, (n= 1) 1. (8) 
Putting (7) in (4), we obtain 
p\eees Ban Goar Ue Tai, 
G(x) =~ SESE FW, (9) 
With this value in (7) the surface condition is satisfied. Further, we may 
change the sign of 7 throughout the second term of (7), and we obtain 
feo) foe} . 
! sbi . im 
| F(x) eik% an} OTE TS (x) eaixx te (y— 2h) de (10) 
h b K—Ky—t 
where the real part is to be taken, and the asterisk denotes the conjugate 
complex quantity. It may be noted that this method of satisfying the 
condition at the free surface is quite general, and independent of the 
form of the submerged body. 
It is convenient for the present problem to alter the notation slightly 
from (8), and we write 
F (x) = — ica®f (Kk) 
5 ‘ . Il 
F (6) = By + by (a) + 82 (xa? + 23 (Ka)? + ine | (11) 
Further, the expression (10) is a function of the complex variable z; 
hence we have for the complex potential function w, or ¢ + iv, 
{oo} foo) . 
EK Sites! 
ace ica*| f(x) e%? aue—ica® | ena Omar (2) GUE ES Gh. (12) 
Fs KK tp 
this being in a form valid in the liquid in the region y > 0, it also being 
noted that ultimately wu is to be made zero. 
3—We have now to determine the function f(«) so as to satisfy the 
condition 0¢/ér = 0 for r=a. For this we turn the second term in 
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