(3.5) 
Since , v) is zero on this interval, either z, v, or both are 
on the boundary of their allowable range. From previous arguments, it 
is known that one cannot freely choose du. Only those values of du are 
allowed which satisfy 
o, (2 + Sx, v + du) < 0 
or by (3.4) 
go, (2 + Ox, v + du) - $ (2, vy) < 0 
On the other hand, for a neighboring trajectory to z 
ott OS War OH a Ws Ob = O 
on [t,> thayl- Simca.) 
WE a OX Wa Ot) = Os, W) = Ol = © (3.6) 
In order that the above inequality and (3.6) hold, 
sy, 2 0 (3.7) 
Moreover, provided the variations are sufficiently small, I) is free. 
If » is twice continuously differentiable, then it follows from 
(3.6) that 
>, Sx +o, Sut Sp =O (3.8) 
29 
