MINIMA SOLUTIONS IN THE CALCULUS OF VARIATIONS. 
113 
Next suppose that, in fig. 4, the point S represents the first intersection of a 
consecutive synclastic curve AB'S with ABS, and that the dotted line, through T, is 
a line whose equation is x = x, x being the value of x for which Y nn = 0 (if there is 
more than one value, take the first one you pass through in going from A vid the 
curve). Then, in that figure, S 3 U A first becomes capable of the value zero at C, where 
the curve meets the ordinate through T. For the second variation vanishes in passing 
from ABC to a curve ABC'C coincident with ABC up to O', a point infinitely close 
to C, and having at C and O' contact of the (n — 2) th order with ABC, since from A 
to C' By is zero, and from C to O' Y nn is zero, 
c 
Hence S 3 U A first becomes capable of a zero value when C coincides with the inter¬ 
section (S) of a consecutive synclastic curve having contact of the (n — l) th order with 
ABC, or with the point (T), given by Y nn = 0, whichever is first reached in passing 
from A along the curve. 
It follows, rigorously, that the synclastic property cannot cease until the first of 
these points is reached. It does not follow that, if the integration is extended 
beyond this limit, the integral is anticlastic. 
Three cases arise :—1st, S is nearer A than T is; or, 2nd, T is nearer to A than S ; 
or, 3rd, T and S coincide. 
Case 1.—In this case we can easily show (see fig. 3) that, C being the intersection 
of a consecutive synclastic curve ADC with ABCG, the synclastic property does 
cease at C, that is, we can join A and G by curves the integral along which 
is either greater or less than that along ABCG. Let ADC be a consecutive 
synclastic curve for which S 3 U A = 0, then evidently, representing by I (ABC . . . ) 
the integral along the curve ABC . . . , we have 
add 
and get 
I (ABC) = I (ADC); 
I (CG) = I (CG), 
I (ABCG) = I (ADCG) 
Q 
MDCCOLXXX VII.—A. 
