60 G. PLARR ON THE SOLUTION OF THE EQUATION V pop =O : 
which comprises the points for which the vector e (as primitively defined) 
satisfies to 
I>0. 
By the expressions (39) of P, Q, the equation of the surface will become : 
(42 bis) 0=4(P’—e)? + 27(Q’ + Seas)”. 
We call ¢ the tensor of «, which satisfies to the equation, and ¢’ its unit 
vector, putting 
(43) e=Te, e=VUe- 
Then we have Se@e=eSe’we’ and we put 
(44) E=—Sewe , 
and generally we put 
1 rh oe 
(45) F?=—7T=P'+7Q', 
so that we have 
a ae 
(46) F(e)=(? + P+ {(He?—-Q)’. 
Developing we get 
Fle 6 1, 21a 04 2 arp e 3. 2a, 
(?)=e + (8P' +7 Ee +(8P?—- 5 QE )e’+ (P* +7 Q”). 
The equation F(e’)=0 , being of the 3d degree in ¢’, assigns to é at least 
one real value, and this value is positive because the last term is 
a SO 
P?4+7Q?=—7, 
and we know by the preceding that —I” is negative. Therefore there will be. 
at least one real value for me 
Te=+ J 
satisfying the equation, whatever be the value of E= —Se@e’ ; and therefore 
the surface will have at /east one point in every direction of ¢’, in fact in all 
possible directions. 
Moreover, when e=0 then I takes the value I” which is positive. It follows 
that the origin O of ¢ is one of the points for which the condition I> 0 is satis- 
fied ; and if we consider, for the present, the directions ¢ for which the 
equation F(e?)=0 gives only one positive root, e, then we may conclude that 
all the points on those directions, from the origin O, up to the surface belong 
to the space for which I> 0 is satisfied. 
The directions of ¢«’ for which Fe’=0 assigns three positive values to e must 
now be determined in order to draw the corresponding conclusion in respect 
to the fulfilment of the condition, [> 0, in those directions. 
