252 Mr G. H. Bryan, On the waves on a [June 4, 
Substituting these values of F, G and the expressions for y and 
ain in (22), (23) we get, omitting the constant terms, 
i (2m — va) Aa” + 2oBJ, (ha) 
_m . poe: eel lino — || 
=, ida + BJ, (ha)} ie 2a py = aeagieal 
ta {(n —1) Aa" + B [had (ha) — J,,(ha)]} ... (24), 
a, 
- =— - {2n(n — 1) Aa” 
+ Bih'a’ J,” (ha) — haJ,’ (ha) + J, (ha) ]}...(25). 
These two equations determine the two unknown constants A, 
B as linear functions of the given coefficients 7, g. 
6. Ifthe waves be free we must put f=0 and g=0 im these 
equations. By means of the well-known relations 
Ward,” (ha) + had,’ (ha) — (n® — ha?) J, (ha) = 0...... (26), 
hadi (ho) = nd, (ha) — Wade Aid). een. cack eee eres (27), 
(25) now gives 
Re 2n (n—1)Aa” 
~ 2had,’ (ha) — (2n?— h?a?) J, (ha) 
2n(n —1) Aa” 
te a’ —2n(n—1)} J, (ha) — 2haJ,, (ha) 28): 
To simplify (24) let us write 
1 Tr-1 
b= 7 ee (29) 
and multiply throughout by va. We thus obtain 
Aa” |a(a+ 21) + nk, — en (n — 1)a| 
+B |(2eu + nk, — = n(n—1) a) J (ha) + = nahaJ,., (ha)} = 
By eliminating the ratio of A to B from (28), (30), we obtain a 
relation which reduces to 
\2 (a+ 2v) + nk, — — ~ n(n —1) “— fhaJ,, (ha) — 2J,,,, (ha)} 
ste = n(n—1)?aJ,., (ha) =0 ...(81). 
