56 G. PLARR ON THE SOLUTION OF THE EQUATION Vp¢p=0 : 
namely : 
(36) k= — 528 —DaS’ae. 
Then putting ¢=¢ +, we have by (12) and by (34) : 
heal 
(37) (,=52aS(Ba*y + ya"B)= ZaSBo’y , 
1 
2 
aa S(BaVey+yBVep)) 1 
—sda oe € € 
Gg { avai } +52aS(BVeVey +yVeVeB). 
Now the first sum > of the 2d member is zero, because in another order the 
terms contain : 
Se[V(yoB+GB . y)+ V(Bay+ By), 
which is identically zero. 
The second sum = contains SBVeVey=SyVeVeB=—SVcBVey= —SeBSey . 
Thus we get: 
(38) C’=—ZaSPSye. 
Now we have 
3Q = 8Q'+3Q"=S [nlm +x!) +200 + 0) —2ee] 
thus : 
{208i +2) 
3Q" =S(nk" + 2¢¢' —2ee,) . 
By (36) we have 
1 
Sy" = —3€°Snd— WyaS*ae. 
We have already shown that Sy is zero, being= — 2Sama=0, by (8). 
Thus, as Sna=Sawa, &c., we have: 
Sk = + SaaS’ ae. 
Then, by (38): 2S¢¢’ = =22SalSPeSye ; and because by definition (27) 
and (12): 
2Sal = Sao +Sar= —S(BBy+ ya), we have 
2800. = + DPBySPSye+ WywsBSPSye. 
Now, in the general terms of the two sums, we may operate a circular 
permutation of a, 8, y, in the first sum one step forward, in the second sum 
two steps forward (or one backward), so that 
QSL’ = WyBaSyeSae + PSBBaSacSPe . 
