of Oceanic Tides. 269 



the given circumstances of the surface of the fluid ; and if it be 

 found capable of satisfying all the other conditions, it must give 

 the true solution of the problem, because, from the nature of the 

 case, the solution must be of a unique and definite character. 

 Putting /(r) for rP(r), we have 



r<f>=f(r) cos 2 X sin 2(0— pt) . 



The substitution of this value in the left-hand side of the equa- 

 tion (0) gives the result 



(rf'( r ) - y^ty cos 2 X sin 2(0- fit), 



which, for the present instance, is the quantity I have called <I>. 

 We have seen that = does not give an appropriate solution. 



d& d$> 



Also jq =0 and — =0 are inapplicable to the inquiry, because 



and X are contained in an explicit manner in the above expres- 

 sion. But the equation -j- =0 becomes 



which is proper for finding the form of f(r) by integration. By 

 assuming that f(r) = cr m , there results for finding the values of 

 m the equation 



m(m— 1) (m- 2) + m(m— 1) — 6m + 6 = 0. 

 Hence m has the three values 3, 1, and — 2, and consequently 

 f{r)=c 1 r 3 + c 2 r + c s r-2, 

 *&- ) =F(r) = c 1 r 2 + r 2 + c 3 r-3. 



The form of F(r) being thus found, the following equations 

 immediately result : — 



= ( c ,r 8 + c 2 + c 3 r~ 3 ) cos 2 X sin 2 (6 —fit) , 

 ■jfc = —2fi(c/ z + c 2 + c 3 r~ 3 ) cos 2 X cos 2 {0—fit) > 



jjt=-4fjL*(c 1 r* + c q + c 3 r- 3 ) cos 2 X sin 2(0-fit), 



u= -Z= (2c 1 r-Sc 3 r- 4 ) cos 2 X sin 2 {0- fit), 



V=Z FcosXat^ 2{c i r + c< l r-* + c 3 r-*) C os\cos2(0-fit) > 



w=z 1 d £ - _( c r + c 2 r- ] +<y- 4 ) sin2X sm2{0-fii). 

 r dX 



