ON THE HYDRODYNAMICAL 'THEORY OF SEICHES. 621 
We have therefore 
éh(1 — w?) =u={AC(c, wv) +BS(c, w)} sinn(t-7), 
where A and B are arbitrary constants. 
Also 
= = { AC(c,w)+BS(c, w) } sin n(t =) ; 
where the dashes denote differentiation with respect to w. 
Since € is finite, we have w=0, when w=+1. Therefore, since C(c,-—1)=C(c, 1), 
and S(c ,—1)= —S(c, 1), the following boundary conditions must be satisfied : 
A O(c, 1)+B S(c, 1)=0, 
A O(c, 1)-BS(c,1)=0. 
These are equivalent to 
ALO(en I) =0nae BS(e.1) 08 
Now we see from the relation 
C(c, w)S'(c, w) -—C'(c, w)S(c, w)=1, 
that C(c,1) and S(c, 1) cannot vanish simultaneously. Therefore either 
BO nC(c al) =O); 
or 
MeO) Sie) On 
mae we have seen, the roots of C(c,1)=0 are -c,=1.2,¢,=3.4,.. 
Muee=(2s—1)2s,....; and the roots of S(c,1)=0,c,=2.3,¢4=45,...., 
€,=2s(2s+1),... . Hence we have the two sets of solutions 
E= A C(Co._1 ,w) sin Noo (t 7 T) . 
=e See ee 
h 1 - 0? : G3): 
c= _— 2 CG: wv) sin Mog (t aa: t) 
Here O(c,,,,w) is a polynomial of degree 25; and O'(cy_1,w) a polynomial of 
degree 2s—1. 
Also 
= B S(@s ,&) sin N2,(t — 7), 
h 1 — w? 5 | ee 
C= B S’(¢o5) W) sin 2»,(¢ — 7) 
aa 
where S(c,,, w) and S’(c,,, w) are polynomials of degrees 25+ 1 and 2s respectively. 
*Since the abstract of this paper was published, I have discovered that the solution for the particular case of a 
symmetric parabolic basin was given by Lams in the new edition of his Hydrodynamics (1895), § 182. He arrives 
at his result by means of LeGenpDR#’s function, which is closely allied to the seiche functions. 
