528 Mr. A. 0. Allen on 



Initially he = /, z = w = 0, 



L(R+B') . .._ ■■ M(R + B ') 



.'_ M2/' ■" ~" > T.!.'M2 /j 



# = 



LL'-M 3y ' ' LL'-M 5 



m 



and the third derivative o£ . is =2^2+20/; 



general this is not zero. The determinant 



L'D+E+E' -MD -(MD+E) 



ED #D 2 +GD -(ED+l/K) 



MD _( L +^)D-(G+r) -(LD+r) 



is a quartic in D, and in general does not factorize even 

 when M=KRr; calling its roots a, /3, 7, 8, we may write 



as = a ie nt + aj* 1 + a 5 e^ + a J* + ~ R , - / ; 



in m and £ we shall have b's and c 9 s for the a's, and the 

 coDstants will be — KR/'and zero. To find the c's, we have 



Ci + c 2 + c 3 + c A = 0, c x a + c 2 /3 + c 3 y + ^8 = 0, 



C!« H-6 2 p +637 -1-C4O — LI7— M 2 ' 



so that the c's all vanish and the balance is a continuous 

 one when L = M. In this special case the quartic yields 

 two factors, viz. i%D 2 -f(MG + MR + ?R)D + R(G + r) and 

 M(L'-M)D 2 + (LV + MR')D + r(R + R'). If a and /3 are 

 the roots of the first equation, and 7 and 8 those of the 

 second equation corresponding to these factors, [ have no 

 doubt that just as in § 2 the coefficients a lt a 2 , b u b 2 should 

 be zero, because Gr and g can play no real part in deter- 

 mining the flow, since no current crosses by Gr ; but I have 

 not succeeded in verifying this. It is interesting to notice 

 that if, with L = M, we adjust the balance M = KRr in such 

 a way that R = r= A /M/K, the two quadratics break up into 

 linear factors, viz. MD + R and gD + G + r in the first case, 

 MD + r and (L'-M)D + R + R/ in the second. The time- 



. . , . M L'-M , g 

 constants involved, viz. — . -g-— - l7/ , and ^-— , are respec- 



XV XV + It UT'<' 



tively those of L (or K) discharging through r (or R), L' 

 through R + R', and G through G + r; the actual value of 



