最优控制理论第四章资料

When x=0, we impose
, in order to insure
that (4.10) holds. Therefore, the optimal control on the
state constraห้องสมุดไป่ตู้nt boundary is
Now we form the Lagrangian
where 1,2, and satisfy the complementary
Chapter 4
The Maximum Principle: General Inequality Constraints
4.1 Pure State Variable Inequality Constraints: Indirect Method
It is common to require state variable to remain nonnegative, i.e.,
We associate multipliers ηi with (4.3) whenever (4.3) must be imposed, i.e., whenever xi(t)=0. A convenient way to do this is to impose an “either or” condition
same as in Example 2.2, namely,
Since (t)-1 on [0,1] and x(0)=1>0, the original
optimal control given by (4.11) is u*(t)=-1. Substituting this into (4.8) we get x(t)=1-t, which is positive for t<1. Thus
i.e., xi(t) 0, i=1,2,…,n. Constraints exhibiting (4.1) are called pure state variable inequality constraints. The general form is
With (4.1): at any point where a component xi(t)>0, the corresponding constraint xi(t) 0 is not binding and can be ignored.
where is a multiplier associated with (4.1), then the
method is referred to as the direct adjoining approach.
Remark 4.1 The first two conditions in (4.5) are
ηi xi=0. This will make ηi=0 whenever xi >0.
We can now form the Lagrangian where H is as defined in (3.7) and η=(η1, η2 ,…,ηn ). We apply the maximum principle in (3.11) with additional necessary conditions satisfied by η
The complementary slackness conditions for
the direct multiplier are 0 and x*=0. Since 0,
it follows that
Example 4.1 Consider the problem:
Solution. The Hamiltonian is which implies the optimal control to be
and the modified transversality condition
where is a constant vector satisfying
Since the constraints are adjoined indirectly (in this case via their first time derivative) to form the Lagrangian, the method is called the indirect adjoining approach. If on the other hand the Lagrangian L is formed by adjoining directly the constraints (4.1), i.e.,
complementary slackness conditions on the multiplier
η. The last condition
is difficult to motivate. The
direct maximum principle multiplier is related to η as
slackness conditions
Furthermore, the optimal trajectory must satisfy From the Lagrangian we also get
Let us first try (2)= =0. Then the solution for is the
In the time interval [0,1) by (4.14), 2=0 since *<1, and by (4.15) =0 because x>0. Therefore, 1(t)=-(t)
In any interval where xi(t)=0,we must have
so
that xi does not become negative.
The control must be constrained to satisfy
making fi 0, as a constraint of the mixed type (3.3) over the interval. We can add the constraint