then 
f(t, z + dx, v + du) 
Now by assumption f is twice continuously differentiable with respect to 
x; hence 
ox = f(t, z+ ox, v + du) - E(t, z, v) 
= £, dx + £(t, 2, v + du) - £(t, z, v) + 0(|dx|*) (2.6) 
It is not necessary that du be uniformly small; indeed, in problems 
involving bang-bang controls, this is not at all true. However, there 
can be deviations du of order one only if their duration is short. It 
can be proved that if du satisfies the condition 
Ge) 
T 
{ |Su(o)| do <e 
0 
Since by assumption, f 
then the deviations 6x(t) are also of order €. 
is Lipschitz continuous with respect to u, 
= 0(du) 
l(E5 25 B) > Ges Bs w)) <ulo = a 
Moreover, it follows from Equation (1.6) that to the same order of 
approximation 
(1.8) 
Ses se Os ae (ES 5 Wy) = (65 5 WY) 
or in abbreviated form 
(1.9) 
Ox = £ Oxi E(WW)NS FW) 
