Nonlinear Optimal Control Theory

Examples 269

dλ

5

dt

= −H

rν

= λ

3

(t)

Z

Ω

cz

2

ν

2

cos z

1

dµ

t

+ λ

4

(t)

Z

Ω

cz

2

ν

2

sin z

1

dµ

t

.

Hence

λ

1

(t) = a

1

λ

2

(t) = a

2

for all t in [0, t

1

] (8.7.6)

λ

3

(t) = −a

1

t + a

3

λ

4

(t) = −a

2

t + a

4

for consta nts a

1

, a

2

, a

3

, a

4

.

Since the initial point is fixe d and the terminal set J

∞

is given by (8.7.3),

the transversality condition (6.3.24) gives that the n + 1 vector (−H

r

(π(t

1

)),

λ(t

1

)) is orthogonal to J

∞

at (t

1

, x(t

1

), y(t

1

), p(t

1

), q(t

1

), m(t

1

)). From (8.7.3)

we get that the vectors (dσ

0

, dσ

3

, dσ

4

, dσ

5

) of the form (1, 0, 0, 0), (0, 1, 0, 0),

(0, 0, 1, 0), and (0, 0, 0, 1) are a basis to the tangent space to J

1

at (t

1

, x(t

1

),

y(t

1

), p(t

1

), q(t

1

), m(t

1

)). Thus,

H(π(t

1

)) = 0 λ

3

(t

1

) = λ