42 ROYAL SOCIETY OF CANADA 
an integral taken along a curve in the (u, x, y) space and defined by the 
differential equation, (6). 
(d). Now it happens that in the present problem, x can be 
eliminated from (7) by (6), giving 
soak [edie eee A GAL. den aeinsioie toate ceaeueee ease ote (8) 
0 y 
From (6) and (f’), we have : 
1 7 4 du 
x af du Tu Nos tres RS (9 
A 1 sd 
whence £ transforms into £’’, 
L's y=y (u), 
satisfying the conditions : 
(a y (u) is of class C’ in (0, a); 
(8) y (0) = Yo, and y (a) = y,, where: 
: du 
i)a,=a,+ f” — , and y, are fixed, or 
{> 
li) @ |. fee V} = 0, or 
tee a du . 
ili) æ + 1 5 and y, are arbitrary ; 
(v") y (4) > Oin (0 a). 
(e). Cases in which the problems of (b) and (d) are equivalent. 
For each £" and the corresponding integral, J, there is an £” and a 
corresponding J”, such that J = J'’. Now in order that the problems 
shall be equivalent, the converse must be true; i.e., for each £’’ and in- 
tegral J’, there is an £ and the corresponding integral J, such that 
I= 1”. To see under what circumstances this is the case, we apply the 
above transformations inversely ; i.e, we determine the function, x (u), 
by the equation. 
2 (tu), = ey JT Ca LE AE ETES (10) 
We thus transform the integral, Z'', given by (8) into J’ given by (7), 
and the transformed path of integration, £”, satisfies the conditions, 
(a)... (8), except those to which the end-point, M (x, y,) is sub- 
jected. In case (8 : i), in order that £” shall pass through (x, y, a), 
we must have, = t+ dE du 
y 
oO 
ee 
