THE TWO GREAT SOLITARY WAVES OF RUSSELL. 339 



We must now integrate equation (8'). Proceeding on the same plan as before, we obtain 



w'- {d.cpy = {c -(p)di<p., 

 .: c -<p = Cs/rii - {d^\ 



And integrating this equation, and assuming a convenient epoch for the commencement of t, 

 we find 



c - u = Cn cos — — — (16') ; 



^ (^ f 



.-. d,u = n sin ■ (17'); 



e -"' + ■'- e"'-" . X - ct 



• '■ ; -, — = sin ; 



e-"' + '' + e"'-° C 



,^ X - ct 

 and .-. e"'-" + 6""' + ° = 2sec — — (is'). 



j'4 = ct^, and 



Let .r^, A\ be the values of x for a given particle at the times <,,, <^. Then (18') gives 



2 sec ' ^* ~ " * • = e"'" - "'' + e"'' " "'' 



2/t , 

 = — from (14) ; 



, k 

 .'. Xf^ - ct,^ = C cos - . 

 h 



Now while a particle is transferred by the wave through the space 2 (x^ - x ) (=3) the wave 

 itself has travelled its own length (= \) in addition to this space; and the time occupied is 



••■ t- (f, - t„) = - + a:^ - x^ ; 



•■• ^k - «'/, = -) because c^ = x^; 



.-. X = 2Ccos-' - . 

 /( 



But when x = x,^ and t = t,,, >i = 0, and consequently from (16') 



n cos ~ = Cn . ~\ 



C h 



2cA At 



V = 



Again, X + /3 = 2 c (<j - i^) 



= ?ilog.tan (^ + ^co8-'^); 



.-. nX = -cos"'- (20'). 



X X 2 



