Ou 
G. PLARR ON THE SOLUTION OF THE EQUATION Vpgp=0 : 5 
whole, we will put in evidence in all the terms those parts which depend on e 
also implicitly, and then only make the development. 
Let P’ and Q’ represent the values of P and Q which correspond to «=0 , 
and let us put: 
(32) P= Po ey 
Q=Q +d. 
First, as to y and ¢, they do not depend one. Because if we decompose ao 
and @ into their self-conjugate part @ and a vector depending on «, namely 
putting : 
(33) eae a} with S6%0'=S6's0, 
wo =o—Ve 
we get by (12) and (27) : 
n=2aSawa 
‘ld | (=) 2aS (Bay +756) =SaSBay, 
the terms in e disappear, being 
LaSaVea=0, LaS(BVey+yVeB)=0. 
For the partition of « and G we put 
a 
§ K =m—35P 6 
n=n+n;  K=K’+x" and then 
4 " u 2, uw 
( K eae ) . 
Now, by the expression (12) of y,, and because : 
wo =(6+ Ve =B'+ BVE+ Veo+ VeVe, 
we have 
(35) n, = YaSaB"a 
n, = 2aSal@Vea+ Vewa+t VeVea]. 
But SalsVea+ Vewa|=Se Vawsa+ Voa.a], which is zero. Then 
SaVeVea= — V’ca= —e?— Sea. 
Thus : 
n, = Lal —’ —S’ea]= —e°8 — SaS7ea . 
As to Pa +(—e’, the partition is made. We have 
/ 1 2 2 I 2 
P’=5n +0, PS=—e | 
Thus : (35 bis) K =3aSas'a—3P'S 
c= —23—SaSien + 28, 
