x 
G. PLARR ON THE SOLUTION OF THE EQUATION Vpdp=0. 85 
we get: 
b =(G—H¢ +¢") 
w= (G—H¢’ is g”) 
and from these expressions there follows at once 
Saba =Sarb’a, &c., because 
Sag’a=Saga, Sap’a=Saga, &e., &e. 
Therefore we have 
t'Spp’ =tSp’p 
and therefore the values of ¢ and ¢ are one and the same. 
We now have: 
Aat+tBB+C y=tp’ 
A’a+BB+C’y=tp. 

Squaring we get: 
—SA?=t%p” , —SA?= tp? : 
On the other hand, if we sum the squares of Ja=Aa, &., Wa=A’a, &e., we 
form . 
2(Wal=p'2A’, Zr'a)?=p2ZA”, 
and replacing 2A’, &c., 
=(ya)’= —Up'p? =2(W'a)?. 
For simplyfying we put : 
pt =T = —D(ya) 
/G=iTpTp , 
in taking for ¢ and the radical a positive value. 
Then our previous equations become 
S (ha. a) = SEUpUp’ 
zWa. a)=./@Up Up. ae 
But now we will calculate scalar and vector of the first members by the 
explicit expressions of J, w’. 
Saya = ZSa(Ga—H¢a + ¢?a) 
= G2a’—H 2Saga+ 2Sag’a 
= —3G—H(—m,) + (2m,—m*) . 
VOL. XXVIII. PART I. 04 

