Deep-sea Tides, and the Effect of Tidal Friction. 181 

 we have 



/= Lg ^, c = —p. — ft"} .... (K) 



the same formula and entailing the same conclusions as in the 

 case of the canal. The reasoning, however, fails if P and Q 



vanish when 0— ~ • 



But when we come to solve the equations We are met by a 

 difficulty. The values of u x and v x are to be found by the 

 equations 



(C — gc) (m -f n) sin cos = (n 2 cos 2 — m?)u ls 



(C-gc){ncos 2 + m) = (n*cos 2 0-m 2 )v 1 . 



Now, the moon's true motion being eastward, m < n. There is 



771 



therefore a latitude where cos 0— + — , for which both u, and v, 



are infinite, and our approximations fail in the neighbourhood. 

 If the moon moved westward, which is conceivable, the difficulty 

 would not occur : I proceed therefore with the general solution. 

 From the last-formed equations 



ncos 2 + m 

 1 ~~ (m + n) sin cos l> 



whence the equation of continuity, in its shorter form, may be 

 written 



_ . * /i d.Ksm0u x rt ncos 2 0-\-m . n 



—2mc sin 2 0= — = — a , Q —2-, — ■ — ? a - a.a • K sin ^ u i- 



sm d0 (m + n) cos snr J 



If we put cos 0=x and /csin 0u i =z J this becomes 



dz _ nx^ + m 



dx (m -f n) x(l — a? 2 ) 



uz tix -rm on 2\ 



the coefficient of £= — — - *.- + ^ =— — , and its integral 



m-\-n x \—x \-\-x • ° 



2rra 



ishyp.log -, whence the formula z=e f Fdx >\eJ Fdx Qdx gives 



om + n • s " 



making the constant of integration =0, because it would other- 

 wise give an infinite depth of sea at the equator. 



Replacing values of x, z } and u v and reducing, we get 



Q—gc 2mc 



' w 2 cos 2 0— m % "~ 3m + n 



