usin = h;[f;[f;g]z(t)]ih;[g;[f;g]z(t)]i; (5.40)
2 3T 2 3 
1 Gp  +  ln p + Gbp26 7 6 q q  Gp2 ln pq 7   
4 5 4 52
2 Gbp ln pq Gbp  + dp3 + Gp b + 23d pq3 p
= (5.41)2 3T 2 3
1 06 7 6 7
4 5 4 5
2 G2bp
       
2
1Gp  + bpq + 2Gp b ln pq b  + dp3 +  b + 2 p3d q3 p= (5.42)
G2bp2       
2
1  + bpq +  p2 b ln q b  + dp3 +  b + 2 p3d q3 p= (5.43)
Gb2
1 p 1 p   1=  ln 1  1  +  1 2+ dqp + 21 1 2dp
3: (5.44)G q G G Gq 
2 3Gb 3 p Gb 2 G
Knowing that h;[f;g]z(t)i = 0;
2 3T 2 3
1 Gp6 7 6 7= Gbp
4 5 4 5 2 + Gp1 = 0 () 1 b= : (5.45)
 2 2 Gbp
Hence  
1 p p 2 2 du
sin = b  +  ln dp3 + p q : (5.46)G q q 3b 3 p
In summary, we can state the following:
Proposition 14 Let (z;u) be an extremal pair for our OC problem, if the control u is
singular on a open interval, then u is of order 1 and the Legendre-Clebsch condition is
satis…ed. The singular control can be expressed as
!
1 p(t) p(t) du
sin(t) = b  +  ln
2 2dp(t)
3 + p q(t) : (5.47)G q(t) q(t) 3b
3 p(t)
32
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