Chapter 4

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The Maximum Principle: General Inequality Constraints. Chapter 4. 4.1 Pure State Variable Inequality Constraints: Indirect Method. It is common to require state variable to remain nonnegative, i.e., i.e., x i ( t )  0, i =1,2,…, n . Constraints exhibiting (4.1) are

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4.1 Pure State Variable Inequality Constraints: Indirect Method

It is common to require state variable to remain

nonnegative, i.e.,

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)  0is not binding and can

be ignored.

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

We associate multipliers ηiwith (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

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

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.,

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

complementary slackness conditions on the multiplier

η. The last condition is difficult to motivate. The

direct maximum principle multiplier  is related to η as

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

When x=0, we impose , in order to insure

that (4.10) holds. Therefore, the optimal control on the

state constraint boundary is

Now we form the Lagrangian

where 1,2, and  satisfy the complementary

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

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

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)

=2-t >0 for 0t<1, and this with u=-1 satisfies (4.13).

At t=1 we have x(1)=0 so the optimal control is given

by (4.12), which is u*(1)=0. Now assume that we

continue to use the control u*(t)=0 in the interval

1 t  2.

With this control, we can solve (4.8) beginning with

x(1)=0, and obtain x(t)=0 for 1t2. Since (t)  0 in the

same interval, we see that u*(t)=0 satisfies (4.12)

throughout this interval.To complete the solution, we

calculate the Lagrange multipliers.since u*=0, we have

1=2=0 throughout 1t2. Then from (4.16) we obtain

=-=2-t0 which, with x=0, satisfies (4.15). This

completes the solution.

Remark 4.2 In instances where the initial state or the

final state or both are on the constraint boundary, the

maximum principle may degenerate I.e., there is no nontrivial solution of the necessary conditions,i.e., (t)0, t[0,T], where T is the terminal time.

4.1.1 Jump Conditions

There may be a piecewise continuous (t) satisfying a

jump condition

at a time at which the state trajectory hits its

boundary value zero.

Example 4.2 Consider Example 4.1 with T=3 and the

terminal state constraint

Clearly, the optimal control u* will be the one that

keeps x as small as possible, subject to the state

constraint (4.1) and the boundary condition

x(0)=x(3)=1. Thus,

We only compute the adjoint function and multipliers

that satisfy the optimality conditions. These are

4.2 A Maximum Principle: Indirect Method

Mixed constraints as in Chapter 3 and the pure state

inequality constraints

where h: En x E1  Ep. By the definition of function h,

(4.26) represents a set of p constraints hi(x,t)0,

i=1,2,…,p.It is noted that the constraint hi0 is called

a constraint of rth order if the rth time derivative of

hiis the first time a term in control u appears in the

expression by putting f(x,u,t) for after each

differentiation.

In case of first order constraints, h1(x,u,t) is as follows:

With respect to the ith constraint hi(x,t)0, a sub

Interval (1,2)  [0,T] with 1< 2is called an interior

interval if hi(x(t),t)>0 for all t(1,2). If the optimal

trajectory satisfies hi(x(t),t)=0 for for

some i, then is called a boundary interval. An

instant is called an entry time if there is an interior

interval ending at t= and a boundary interval

starting at . Correspondingly, is called an exit

time if a boundary interval ends and an interior interval

starts at .

If the trajectory just touches the boundary at time ,

i.e., and if the trajectory is in

the interior just before and just after ,then is called

a contact time. Entry, exit and contact times are called

junction times.

1

1

1

Entry

Exit

Contact

To formulate the maximum principle for the problem

with mixed constraints as well as first-order pure state

constraints, we form the Lagrangian as

H is defined in (3.7), u satisfies the complementary

slackness conditions stated in (3.9), and Eqsatisfies

the conditions

The maximum principle sates that the necessary

conditions for u* to be an optimal control are that there

exists multipliers ,,,,,and , and the jump

parameters , which satisfy (4.29) that follows.

Note that the jump conditions on the adjoint variables in (4.29) generalize the jump condition on H in (4.29)requires that the Hamiltonian should be continuous at if ht = 0.Example 4.3 Consider the following problem with the discount rate  0:

Solution. As one can see from Figure 4.2 and 4.3, the optimal solution is:

Figure 4.2: Feasible State Space and Optimal State Trajectory

for Example 4.3

Note that at the entry time t=1 to the state constraint (4.34), the control u* and, therefore, h1*=u*+2(t-2) is discontinuous, i.e., the entry is non-tangential. on the other hand, u* and h1* are continuous at t=2 so that the exit is tangential. With u* and x* thus obtained, we must obtain , 1 ,2, , and  so that the necessary optimality conditions (4.29) holds, i.e.,

and

which, along with u* and x*, satisfy (4.29).

Note, furthermore, that  is continuous at the exit

timet=2. At the entry time so that (4.30) also holds.

4.3 Current-Value Maximum Principle: Indirect method

With the Hamiltonian H as defined in (3.33), we can

write The Lagrangian

We can now state the current-value form of the

maximum principle as given in (4.42)

4.4 Sufficiency Conditions

The sufficiency results can be stated in the indirect

adjoining framework. In order to do so, let us define the

Hamiltonian H and the Lagrangian Ld in the direct

method as

where d, d, and dare multipliers in the direct

formulation, corresponding to , , and  in the direct

formulation.

It can be shown that:

Theorem 4.1Let satisfy the necessary

conditions in (4.29) and let

If is concave in (x,u) at each t [0,T], S in

(3.2) is concave in x,g in (3.3) is quasiconcave in (x,u)

h in (4.26) and a in (3.4) are quasiconcave in x, and b

in (3.5) is linear in x, then (x*,u*) is optimal.

Theorem 4.2 Theorem 4.1 remains valid if the concavity of in (x,u) at each t is replaced by the concavity of the maximized Hamiltonian

in x, where

Theorem 4.1 and Theorem 4.2 are written for finitehorizon problems and remains valid if the transversality conditions on the adjoint variables (4.29) is replaced by the following limiting transversality condition

Example 4.1 (Continued) First we obtain the direct

It is easy to see that

is linear and hence concave in (x,u) at each t[0,2].

Functions

and

are linear and hence quasiconcave in (x,u) and x,

respectively. Functions S  0, a 0 and b 0 satisfy

The conditions of Theorem 4.1 trivially.