Nonlinear Optimal Control Theory

Examples 259

If s > 1, we re w rite (8.4.8) a s

τ =

[sξ ± (ξ

2

− (s

2

− 1)η

2

)

1/2

]

(s

2

− 1)

. (8.4.9)

From (8.4.5), since s > 1, |α| ≤ 1, and τ > 0, we get that ξ > 0. For τ to be

real we require that ξ

2

− (s

2

− 1)η

2

≥ 0, or equivalently

−(s

2

− 1)

−1/2

≤

η

ξ

≤ (s

2

− 1)

−1/2

. (8.4.10)

Thus if, s > 1 any points (ξ, η) that lie in the region subtended by the angle

determined by the line segments from the origin with slope −(s

2

−1)

−1/2

and

(s

2

− 1)

1/2

can be reached.

We a ssert that in (8.4.9) we take the minus sign. We have

(ξ

2

− (s

2

− 1)η

2

)

1/2

=



ξ

2



1 −

(s

2

− 1)η

2

ξ

2



1/2

≤ (ξ

2

)

1/2

< sξ,

where the ne xt to the last inequality follows from (8.4.10). Therefore since we

want the smallest