114 Sir William Thomson on Gravitational 



travelling from east to west, and those of the southern part of 

 the Irish Channel by more powerful waves travelling from 

 south to north combined with less powerful waves travelling 

 from north to south. The problem of standing oscillations in 

 an endless rotating canal is solved by the following equations: — 



h = H | e~ l y cos (ma — at) — e^(cos ma + at) \ ; -v 



w = H^- 1 {€-^cos(m# — <re) + 6 l y cos (ma— at)}; > . (17) 



v = 0. * 



If we give ends to the canal, we fall upon the unsolved problem 

 referred to above of tesseral oscillations. If instead of being 

 rigorously straight we suppose the canal to be circular and 

 endless, provided the breadth of the canal be small in com- 

 parison with the radius of the circle, the solution (17) still 

 holds. In this case, if c denote the circumference of the canal, 



we must have m— , where i is an integer. 



II. Oscillations and Waves in Circular Basin (Polar 

 Coordinates). 



Let 



h=Y cos (iu -at) (18) 



be the solution for height, where P is a function of r. By 

 (8') P must satisfy the equation 



d 2 V 1 rJV ? 2 P a 2 — 4&> 2 



cLt 1 r dr r* gD v 7 



and by (V) we find 





(20) 



This is the solution for water in a circular basin, with or 

 without a central circular island. Let a be the radius of the 

 basin; and if there be a central island let a' be its radius. The 

 boundary conditions to be fulfilled are f=0 when r=a and 

 when r — a f . The ratio of one to the other of the two con- 

 stants of integration of (19), and the speed a of the oscillation, 

 are the two unknown quantities to be found by these two 

 equations. The ratio of the constants is immediately elimi- 

 nated; and the result is a transcendental equation for cr. There 

 is no difficulty, only a little labour, in thus finding as many as 



