and
( t) = h;[f + ug;[f;g]]z(t)i = 0: (5.28)
Using a simple property of Lie Bracket,
h;[f + ug;[f;g]]z(t)i = h;[f;[f;g]z(t)]i+ uh;[g;[f;g]z(t)]i = 0: (5.29)
I.e, provided h;[g;[f;g]z(t)]i =6 0, we have a singular control or order 1: Finally, solving for
the singular control u, we obtain the following expression
usin = h;[f;[f;g]z(t)]ih;[g;[f;g]z(t)]i: (5.30)
Now, we need to evalulate the necessary Lie Brackets. Using (5.17) and (5.20), we get
[f;g] = Dg(z)f(z)Df(z)g(z) (5.31)
2 32 3 2 32 3
0 0 pln(p 0=6 76 q)   ln p p76 q q 76 7 (5.32)
4 54 5 4 54 52
0 G bp( + d(p3))q b 2 pd3 3 pq  2dp3 Gq
2 3
Gp=6 7 (5.33)    
4 5
2 2Gq  + dp3 G bpq  + dp3
2 3
Gp= 6 7 (5.34)
4 5
Gbp
2 3
= Gp6 7: (5.35)
4 5
b
Also calculating the expression of [f;[f;g]z(t)] leads to:
30
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