PROFESSOR KELLAND ON THE THEORY OF WAVES. 535 



/V = -F'^ (5) 



(pt=.--^'t (6). 



34. Let us next solve these six equations. To find C, we must combine 



(1), (2), and (3). 



From (1) and (2), 



C((pt4>t+ftFt) = E.((ptF(-^t/f) (a) 



and ft-om (3), 



-GBB. = G(ftFt + (pt^t) 



-G'D = {<ptYt-^tft) (b) 

 To find H, we eliminate C by (1) and (2), and obtain 



H{(Fiy+(^(y}+j)(ftFt+(pt^t)=o (c) 



Putting for H its value from (3), we get 



(Fty+i-^tf^B^ (d) 



Also, by (5) and (6), 



ftF t + cpt'^t^-i^ tF' t + -^t-^' t) (e) 



= -D^=DH 



dt 



But by (3), 



ftFt + (pt-^t =- DH 



DH = 0, /7F^+</)^-v].^= 0. 



35. We may satisfy the equation DH=0 in two ways: 1. by making D=0, 

 in which case, by virtue of equation {d), both F^ and 4'^ equal zero, or z is itself 

 constant. This case corresponds to equilibrium, and we have nothing to do with 

 it. 2. By making H=0 ; which hypothesis requires that D should be a constant, 

 independent of the time, by equation (4). To prevent error, in consequence of a 

 quantity equal to zero appearing in our equations, it wiU be desirable to wiite 

 them down again, omitting H. We shall also omit t for the sake of brevity, and 

 write /instead otft, (p instead of (pt, &c. 



The equations are 



D/=C4. (1) 



D0 = -CF (2) 



/F^-0^i/ = O (3) 



/= -F' (5) 



<P = -^' (6) 



The equation (3), combined with (4) and (5), gives 



