68 G. PLARR ON THE SOLUTION OF THE EQUATION Vp¢p=0 “ 
which is real, the determinations of the fractionary powers are all supposed 
only numerical. 
For E = E,= = , the equation gives : 
0=(¢+ Py + 4 L (Se e Q’) 
and the second term can be put under the form : 
27 Q? (2 +P). 
4 pz 
Therefore we get 

0=(+PY[e +P 42h), 
or ; 
eed dts Sara 8 
We call ¢, the single root, and e, the double root for E = E,, and we have 
thus 
a <a > = J — 
It is clear that the double root corresponds to P = 0, @ = 0, because 
P= 2p PS Q; 
Q= -—E¢@, + QY = eS FiO. = 0. 
We remark also that 
é, >é;, because IY = — 4P? — 97 Q” 
gives 
lig 27 Q? ia 
e, == P’ = 4p? + Zp2 > ¢@ = Tp: 
In fact ¢,, outpasses all other values of ¢ given by '=0. 
§ 10. We establish now the expression of the vector v normal to the surface. 
For the differentiation we take up the originary expression (42) of T, and put 
it under the form : 
fy =9= (+ @) 
P=P—2, Q=@ + Sewe. 
Differentiating we get 
0=()aP + 24Q, 
dP = — 28ede, dQ = 2 Sede. 
