2 32 3 2 32 3
G 0 pln(p   ln p p Gp[f;[f;g]](z) =6 76 q) 76 q q 76 7
4 54 5 4 54 5
2Gb 0 bp( + d(p3))q b 2 pd 33 3 pq  2dp Gbp
2 3 
2Gp  +  ln p
=6 q
+ Gbpq  Gp2 ln pq 7
:   4 5
2Gbp ln p
3q Gbp  + dp + Gp b + 2 p3d q3 p
Finally,
2 32 3 2 32 3
G 0 0 0 0 Gp[g;[f;g]z(t)] =6 76 76 76 7
4 54 5 4 54 5
Gb 0 Gq 0 G Gbp
2 3
0=6 7: (5.36)
4 5
G2bp
We can now check the Legendre-Clebsch condition (4.32) , which in our case yields the form:
@ d
2 @H
= @  = h;[g;[f;g]]z(t)i > 0 (5.37)@udt2 @u @u
which is equivalent to:
2 3T 2 3
h 1 0;[g;[f;g]]z(t)i =6 7 6 7= G2bp : (5.38)
4 5 4 5 2
2 G2bp
Considering the fact that (t) = 1Gq2 = 0;2 must be positive and thus
h;[g;[f;g]]z(t)i < 0: (5.39)
Therefore, the Legendre-Clebsch condition is satis…ed. I.e., our singular control is expected
to be locally optimal. Now that we have calculated all the Lie brackets, we can …nd the
formula of the singular control
31
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