G. PLARR ON THE SOLUTION OF THE EQUATION Vpgp=0 : 
Thus we get 
S(nk’ + 266") = ZS[ aSae+ BSBe + ySyc |waSae 
= —ScGBaSace= —SeBLaSace 
= —Sea(—e)= + SeBe. 
On the other hand, remembering ¢,= —aw:, we have 
—2 See = + Wee, 
and as we and @e are identical, because 
we=—Bet+ ViP=—Be, 
we have finally owing to (30) : 
38Q” =SeBe + 2ewe = 3S8eBe 
Q” =Sewe .* 
Then also if we partition—;P* we have 
2 seid 
—3P= =(5P —¢) . 
D7 
The first term of the square will be — : P®, namely what =: P’ be- 
comes when we replace x’ by x’, @ by G? in the expression (31) of — =P? Thus: 
7 fA 7 
SP + 2°, 
and we put 
2 2 lA vss 
—sPt= RP? +P”, 
the explicit expression of P’”” is not wanted in the following. 
We have now, by recapitulation : 
P=Pp’-—é 2P’ =7? +20 
a estate 
Q=Q+SeHe { 8Q) =S(ye' + 20C/). 
- §6. We now call IY what I becomes when we suppose init e=0. Then, 
introducing this supposition into the expression (20) and dividing by 3 we get : 
1 ld 7 2 i Lf 
(40) I” = (2P’)(—3P*)—8Q') 
and substituting the preceding values we get : : 
(40 bis) BT’ = (+ 20)(? + 262) —S*( qu’ +200) 
* A direct treatment of the expression (8) of Q will give Q”=SeWe by a somewhat shorter process. 
VOL. XXVIII. PART I. 
