inside a Hollow Unlimited Boundary. 
227 
roughly described as hyperbolic (see figs. 2 and 3). In (11) if 
c = 0 we cannot make 6 vanish unless ol x vanishes, which we do 
not suppose, so in the two-dimensional part of the problem there 
is no parabolic form in those cases where the least value of m in 
(5) is unity. 
Next, from (3) and (4) the liquid velocities at distant points 
are given by 
iia cos 0 oqa cos 6 D ) 
U x = c + - 4- — + &c. 
l (12). 
iia sin 6 a-u sin 6 0 
v y = - +— + &c. 
y r r 
We see thus that in all cases where the least value of m in (5) 
is unity the effect of the rigid boundary on the velocity at distant 
points is as it were to multiply the effectiveness of the source in 
the ratio of (y + Oj) : y. The strength of the original source is 
27 T/jia. The strength of the additional fictitious source which re- 
presents the effect of the boundary is 27 ra^a. May it not be that 
in some cases, at any rate, of the Reflection of Sound at curved 
surfaces a similar explanation would apply, if only the exact 
solution of the problem could be found ? 
[It may be remarked, but the remark will not be pursued into 
detail, that equations (1), (2), (3), (4) are, with a certain modifica- 
tion, applicable to the motion through liquid of a solid whose 
boundary is BABA'B' (Fig. 1). The modification is that equation 
(5) instead of holding over the arc AG A' must now hold over the 
arc AC' A'. This condition is more easily satisfied if A is in the 
quadrant yOG'.] 
5. We have now to shew how the condition (5) can be 
satisfied. Probably the simplest way is the following. It is 
shewn on page 607 of De Morgan’s Biff, and Int. Calculus , that if 
6 lie between 0 and tt/ 2 (but excluding the actual limit tt/2) 
sin 0 — sin 3 6 + sin 50 — &c. = 0 (13), 
provided that (13) may be regarded as the limit of the series 
p sin 6 — p 3 sin S0 + p 5 sin 50 — &c., 
and in that way we shall use it, so that in (5) we can take 
a m =(~ l) 4(m_1) x b, 
where m has the successive odd values 1 , 3, 5, &c. to go , and b is a 
constant, or what is the same thing and more convenient, we can 
take S ( a m sin m0) = bS (— l) n+1 sin (2 n — 1)0 and give n all integer 
values from 1 to oo . Another way of satisfying equation (5) is to 
17 
VOL. XI. PT. III. 
