MINIMA SOLUTIONS IN THE CALCULUS OF VARIATIONS. 
125 
integral change sign rapidly. In fact, if the order of dAz/dx i is unity, and its 
fluctuations recur at intervals of order /3“, then the terms in the integral 
' Xl dAz 
( O ’ ] 
dx x 
will cancel, provided x 1 — x{ = m(3", m being any integer; and so the integration up 
to any point can only contain terms of the order /3". Moreover, since, if any function 
is periodic with respect to any variable, all its fluxions with respect to that variable 
are of the same period, it follows that, if one differentiation introduces a coefficient of 
the order two will bring in /3 -3 q and so on. 
It thus appears that, if we assume Ay x = fi p , multiplied by a function whose fluc¬ 
tuation-periods with regard to x x , x 2 , &c., are p x , p 2 , &c., the order of 
^^ — will be that of (3 (p pi) ; 
(A/X-y 
d A :p 
dx 2 
and, in general, that of d ai+a2+&c -Ayjdx^ 1 dx 2 2 . . . will be Now, by the 
conditions in Art. 18, none of the quantities A 2 , &c., can be infinite ; hence we must 
reject any values of Ay v A y 2 , &c., which do not fulfil this condition. Thus the term 
just considered must be rejected if it makes any of the expressions 
p — app l — a. 2 p 2 — &c. —a n p M < 0, 
where cq, a 2 , . . . ci m have the values corresponding to any fluxion occurring in U. And 
it may be rejected as useless unless it makes some one, at least, of these expressions 
zero. An example will render this more intelligible. Suppose we consider a 
function in which d a y 1 /dx 1 8 , ddyjdxp dx 2 , ddy jdx^ dx 2 , d 5 yjdx. 2 b , are the highest 
fluxions. Here the equations to be satisfied are 
p — Spi = >0 ‘ • • (a); p-^3p 1 —p 2 —>0 . . . ( b); 
p - 2 Pl - 2 \p 2 = >0 ...(c); p - 5p, = >0 . • • (d), 
together with those obtained from the lower fluxions. But, as the lower fluxions must 
be smaller than the higher in the ratio 1//3 at least, it is unnecessary to consider the 
conditions obtained from them. If we put p Y = 1, we get, from (a), p = 8, and it is 
obvious that p 2 = 1 will make (b), (c), and (d) each > 0. If we put ( b ) = 0 and 
eliminate p from (l), (c), and (d), we get, as our conditions from ( a),p 2 — 5p 1 > 0; 
therefore yq>5, as p x = 1 ; from (3), p x — p 2 >0, and therefore p 2 < 1. Hence no 
value for p x , p 2 , or p can satisfy (b) = 0 and the other inequalities. Hence we 
may leave d^Ay/dx 8 dx 2 out of the final condition, as it could only give terms 
