1209 
dad (2), == = ale) = Ak 
d(y), =d(y), =--.=d(yn= AK, | we oot (LON 
in whieh d indicates now that we have to differentiate according to 
all variables except P and 7. 
It must be possible to solve the ratios between then’ differentials 
Pa tt. © Ayre tye, «LC, ASKy. 0e from, then” equations (9) 
and (10); this is only then possible, when a relation exists between 
the coefficients. With the aid of (10) we write for (9): 
“e, AK, + y, BK, -- ..== AK | 
OPA K, grijpen = AK SP rs | 
so that we must be able to satisfy (10° and (11). Here this is the 
case when we are able to satisfy (11). 
When we add the n equations (11) after having multiplied the 
first one by À,, the second one by 2,, etc, then we obtain: 
SAE SPAR OPA ele 2 CE) 
Hence it is apparent that we are able to solve the ratios between 
Ax, Ar... from (9) and (10), when 
Ed aR eee. 
Arp et Aes Ae a = 0 
= (2) =A, EE AY. ag TAR aE AnYn = 0 
(15) 
can be satisfied. 
We might also satisfy (12) by putting equal to zero AK, AK, …; 
now, however, we leave this case out of consideration and we shall 
refer. to this later; then we shall see that the equilibrium is situated 
on the limit of its stability. 
(Mr. W. van per Woupr has drawn my attention to the fact that 
we can easily express the condition that (9) and (10) can be satis- 
fied in a determinant. It appears that this can be written like the 
product of different other determinants, so that we know at once 
all the conditions looked for. 
We have in (13) » equations between the n—1 ratios of 2, a, … An; 
consequently (13) can only be satisfied when a ratio exists between 
the variables. We may find it by eliminating from the equations 
(13) 2,...2,; we may also write this equation in the form of the 
following determinant ; 
