Hence 
lee 
QO < 4 WES = q' - g'f] Oxait [F - gine du} do 
+ ' f x i] 
q' (1) 6x(T) + G Ax, + GAT + u"(M Ax, + MpAT) 
T+AT . 
fs | F(o, x, u) + q'x - g'f(o, x, u) - F(T, & uD) 
a 
= GsK(GE) se GTEC, xx 
Xp» up) do 
ar (USE. Xp uy) + q'Zn aF @iYOxe(Gb) = @ VEC Xr up) | AT (1.33) 
The integral from T to T+AT is a second order contribution which goes to 
zero faster than the other terms as AT > 0. 
In order to determine AX» consider the solutions of the differ- 
ential Equation (1.1), which have the initial value x,. These solutions 
=0 
satisfy the integral equation 
t 
x(t) = X> + | £(0, x, u) do 
0 
Then he 
Noon (Ga) zi) | E@; = w) do 
ae bne a) T 
= 6x(T) + £ AT + 0(e*) 
where the geometry of the proof is illustrated in Figure l. 
A T+AT 
Figure 1 -- Geometry of the Proof 
17 
