inside a Hollow Unlimited Boundary. 227 



roughly described as hyperbolic (see figs. 2 and 3). In (11) if 

 c = we cannot make vanish unless a, 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 



ixa cos a x a cos „ *\ 

 u x = c + — 1 1- «&c. 



aa sin a±a sin . 



Vy = " h h &C 



(12). 



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 (//, + a x ) : fx. The strength of the original source is 

 27r/ia. The strength of the additional fictitious source which re- 

 presents the effect of the boundary is 27ra 1 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 DABA'D' (Fig. 1). The modification is that equation 

 (5) instead of holding over the arc AGA' must now hold over the 

 arc AC A'. This condition is more easily satisfied if A is in the 

 quadrant yOC] 



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 Diff. and Int. Calculus, that if 

 lie between and tt/2 (but excluding the actual limit ir/2) 



sin - sin 30 + sin 50 - &c. = (13), 



provided that (13) may be regarded as the limit of the series 



p sin 6 — p 3 sin 30 + p 5 sin 50 — &c, 



and in that way we shall use it, so that in (5) we can take 



am = (-l)* (M x6, 



where in has the successive odd values 1, 3, 5, &c. to oo , 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 (2n -1)0 and give n all integer- 

 values from 1 to oo . Another way of satisfying equation (5) is to 



VOL. XI. PT. III. 17 



