100 THE TIDAL PROBLEM. 



VII. CASE i = i\ = t2 = a^dizO e = S = 0* 



This case differs from V only in that the rotational momentum and 

 energy of m^ are not supposed to be zero. In this case equations (9) and 

 (10) become, supposing that C2 = Ci, 



In these equations only P, D^, D^, and E can vary, the last decreasing 

 through loss of energy by friction. If by means of (55) we eliminate one 

 of the variables from (56) we have a relation among the other three. This 

 equation may be considered as defining a surface. Let the £'-axis be pointed 

 upward. Then starting from any point on the surface the variables other 

 than E may change in any way, so far as these considerations show, so 

 that the point descends. 



We shall now find the maximum and minimum values of E. In order 

 to simplify the algebra let us put 



u = 



^1 „ = /M' ^2 _'^i (57) 



Using this substitution and eliminating P between (55) and (56) we obtain 



E -1 u^ v^ 



7z [M—u—vY nil '^^1 

 The necessary conditions for a maximum or minimum of E are 



^ — ^ — -- =— w, +u [M—u—vf=0 



27: du 



— — — ^r ~—-=—Km,+v[M—u—vf==0 



ZlZ 01) 



(58) 



(59) 



Multiplying the first equation by v, the second by u, and taking their 

 difference, we have ku = v; whence, by (57), 



D, = Di (60) 



Then the first equation of (59) becomes 



-^ + [M-{l+K)uY=0 (61) 



* See 5, pp. 178-181. 



