Chapter 3 The Maximum Principle: Mixed Inequality Constraints

Mixed inequality constraints: Inequality constraints involving control and possibly state variables.

Examples:

g(u,t) 0 ,

g(x,u,t) 0 .

3.1 A Maximum Principle for Problems with Mixed Inequality Constraints

State equation:

where x(t) En, u(t) Em and f: En x Em xE1 En

is assumed to be continuously differentiable. Objective function:

where F: En x Em x E1 E1, and S: En x E1 E1 are

continuously differentiable and T is the terminal time.

u(t), t[0,T] is admissible if it is piecewise

continuous and satisfies the mixed constraints.

where g: En x Em xE1 Eq is continuously differentiable and terminal inequality and equality constraints:

where a: En x E1 Ela and b: En x E1 Elb are continuously differentiable.

Notes: (i) (3.6) does not depend explicitly on T.

(ii) Feasible set defined by (3.4) and (3.5) need

not be convex.

(iii) (3.6) may not be expressible by a simple set

of inequalities.

where Y is a convex set, X is the reachable set from the initial state x0, i.e.,

Interesting case of the terminal inequality constraint:

Full rank or constraint qualifications condition

holds for all arguments x(t), u(t), t, t [0,T], and

hold for all possible values of x(T) and T.

Hamiltonian function H: En x Em x En x E1 E1 is

where En (a row vector).

Lagrangian function L: En x Em x En x Eq x E1 E1 is

where Eq is a row vector, whose components are called Lagrange multipliers.

Lagrange multipliers satisfy the complimentary slackness conditions:

where Ela and Elb are constant vectors.

The necessary conditions for u* by the maximum principle are that there exist , , , such that (3.11) holds, i.e.,

The adjoint vector satisfies the differential equation

with the boundary conditions

Special Case: In the case of terminal constraint (3.6), the terminal conditions on the state and the adjoint variables in (3.11) will be, respectively,

Furthermore, if the terminal time T in (3.1)-(3.5) is

unspecified, there is an additional necessary

transversality condition for T* to be optimal

if T* (0,) .

Remark 3.1: We should have H=0F+f in (3.7) with 0 0. However, we can set 0=1 in most applications

Remark 3.2: If the set Y in (3.6) consists of a single point Y={k}, then as in (2.75), the transversality condition reduces to simply (T) equals to a constant to be determined, since x*(T)=k. In this case, salvage value function S can be disregarded.

Example 3.1: Consider the problem:

subject to

Note that constraints (3.16) are of the mixed type (3.3).

They can also be rewritten as 0 u x.

Solution: The Hamiltonian is

so that the optimal control has the form

To get the adjoint equation and the multipliers associated with constraints (3.16), we form the Lagrangian:

From this we get the adjoint equation

Also note that the optimal control must satisfy

and 1 and 2 must satisfy the complementary slackness conditions

It is obvious for this simple problem that u*(t)=x(t) should be the optimal control for all t[0,1]. We now show that this control satisfies all the conditions of the Lagrangian form of the maximum principle.

Since x(0)=1, the control u*= x gives x= et as the solution of (3.15). Because x=et >0, it follows that u*= x > 0; thus 1 =0 from (3.20).

From (3.19) we then have 2 =1+. Substituting this into (3.18) and solving gives

Since the right-hand side of (3.22) is always positive, u*= x satisfies (3.17). Note that 2 = e1-t 0 and x-u* = 0, so (3.21) holds.

3.2 Sufficiency Conditions

Let D En be a convex set. A function : D E1 is concave, if for all y,z D and for all p[0,1],

The function is quasiconcave if (3.23) is relaxed to

is strictly concave if y z and p (0,1), and (3.23) holds with a strict inequality.

is convex, quasiconvex, or strictly convex if - is concave, quasiconcave, or strictly concave,

respectively.

Theorem 3.1

Let (x*,u*,,μ,,) satisfy the necessary conditions in (3.11). If H(x,u,(t),t) 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), a in (3.4) is quasiconcave in x, and b in (3.5) is linear in x, then (x*,u*) is optimal.

The concavity of the Hamiltonian with respect to (x,u) is a crucial condition in Theorem 3.1. So we replace the concavity requirement on the Hamiltonian in Theorem 3.1 by a concavity requirement on H0, where

Theorem 3.2

and, if in addition, we drop the quasiconcavity requirement on g and replace the concavity requirement on H in Theorem 3.1 by the following assumption: For each t[0,T], if we define

A1(t) = {x| g(x,u,t) 0 for some u}, then H0(x,(t),t) is concave on A1(t), if A1(t) is convex.If A1(t) is not convex,we assume that H0 has a concave extension to co(A1(t)), the convex hull of A1(t).

Theorem 3.1 remains valid if

3.3 Current-Value Formulation

Assume a constant continuous discount rate 0. The time dependence in (3.2) comes only through the discount factor.

The objective is to

subject to (3.1) and (3.3)-(3.5).

and the standard Lagrangian is

with s and s and s satisfying

The standard Hamiltonian is

and the current-value Lagrangian

We define

we can rewrite (3.27) and (3.28) as

From (3.35), we have

and then from (3.29)

The current-value Hamiltonian is defined as

The complementary slackness conditions satisfied by the current-value Lagrange multipliers and are

on account of (3.31), (3.32), (3.35), and (3.39).

From (3.14), the necessary transversality condition for T* to be optimal is:

where (T) follows immediately from terminal conditions for s(T) in (3.30) and (3.36)

The current-value maximum principle

Special Case: When the terminal constraints is given by (3.6) instead of (3.4) and (3.5), we need to replace the terminal condition on the state and the adjoint variables, respectively, by (3.12) and

subject to the wealth dynamics

where W0>0. Note that the condition W(T) = 0 is sufficient to make W(t) 0 for all t. We can interpret lnC(t) as the utility of consuming at the rate C(t) per unit time at time t.

Example 3.2: Use the current-value maximum principle to solve the following consumption problem for = r:

Solution: The current-value Hamiltonian formulation:

where the adjoint equation is

since we assume = r, and where is some constant to be determined. The solution of (3.44) is simply (t)= for 0t T.

To find the optimal control, we maximize H bydifferentiating (3.43) with respect to C and setting theresult to zero:

which implies C=1/=1/. Using this consumption level in the wealth dynamics gives

which can be solved as

Setting W(T)=0 gives

Therefore, the optimal consumption

since = r.

3.4 Terminal Conditions/Transversality Conditions

Case 1: Free-end point. In this case x(T) X.

From (3.11), it is obvious that for the free-end-point problem

if (x)0, then (T)=0.

Case 2: Fixed-end point. In this case, the terminal condition is

and the transversality condition in (3.11) does not

provide any information for (T). *(T) will be some constant .

Case 3: One-sided constraints. In this case, the ending value of the state variable is in a one-sided interval

where k X. In this case it is possible to show that

and

Case 4: A general case. A general ending condition is

Table 3.1 Summary of the Transversality Conditions

Example 3.3 Consider the problem

subject to

Solution: The Hamiltonian is

The optimal control has the form

The adjoint equation is

with the transversality conditions

Since (t) is monotonically increasing, the control (3.51) can switch at most once, and it can only switch from u* = -1 to u* = 1. Let the switching time be t* 2. The optimal control is

Since the control switches at t*, (t*) must be 0. Solving (3.52) we get

There are two cases t* < 2 and t* = 2. We analyze the first case first. Here (2)=2-t* > 0; therefore from (3.53), x(2) = 0. Solving for x with u* given in (3.54), we obtain

which makes t*=3/2. Since this satisfies t* < 2, we do not have to deal with the case t*= 2.

Figure 3.1 State and Adjoint Trajectories in Example 3.3

Isoperimetric or budget constraint: It is of the form

where l: En x Em x E1 E1 is assumed nonnegative,

bounded,and continuously differentiable and K is a positive constant representing the amount of the budget. It can be converted into a one-sided constraint by the state equation

3.4.1 Examples Illustrating Terminal Conditions

Example 3.4: The problem is:

where B is a positive constant.

Solution: The Hamiltonian for the problem is given in (3.43) and the adjoint equation is given in (3.44) except that the transversality conditions are from Row 3 of Table 3.1:

In Example 3.2 the value of , the terminal value of (T), was

We now have two cases: (i) B and (ii) < B.

In case (i), the solution of the problem is the same as that of Example 3.2, because by setting (T)= and recalling that W(T) = 0 in that example, it follows that (3.59) holds.

In case (ii), we set (T)= B and use (3.44) which is = 0. Hence (t)= B for all t. The Hamiltonian

maximizing condition remains unchanged. Therefore, the optimal consumption is

C=1/ =1/B

Solving (3.58) with this C gives

It is easy to show that

is nonnegative since < B. Note that (3.59) holds for case (ii).

Example 3.5: A Time-Optimal Control Problem.

Consider a subway train of mass m (assume m=1), which moves along a smooth horizontal track with negligible friction. The position x of the train along the track at time t is determined by Newton’s Law of Motion

Solution: The standard Hamiltonian function is:

where the adjoint variables 1 and 2 satisfy

Thus, 1 =c1 , 2 = c2 + c1 (T- t).

which together with the bang-bang control policy (3.64) implies either

The transversality condition (3.14) with y(T) = 0 and S 0 yields

The Hamiltonian maximizing condition yields the form of the optimal control to be

Table 3.2 State Trajectories and Switching Curve

We can put - and + into a single switching curve as

If the initial state (x0,y0) lies on the switching curve,

then we have u*= +1(resp., u*= -1) if x0 0 (resp., x0<0); i.e, if (x0,y0) lies on + (resp.,- ). If the initial state (x0,y0) is not on the switching curve, then we choose, between u*= 1 and u*= -1, that which moves the system toward the switching curve. By inspection, it is obvious that above the switching curve we must

choose u*= -1 and below we must choose u*= +1.

Figure 3.2: Minimum Time Optimal Response for Problem (3.63)

The other curves in Figure 3.2 are solutions of the differential equations starting from initial points (x0,y0). If (x0,y0) lies above the switching curve as shown in Figure 3.2, we use u* = -1 to compute the curve as follows:

Integrating these equations gives

Elimination of t between these two gives

This is the equation of the parabola in Figure 3.2 through (x0,y0) . The point of intersection of the curve (3.66) with the switching curve + is obtained by solving (3.66) and the equation for +, namely 2x = y2, simultaneously, which gives

where the minus sign in the expression for y* in (3.67) was chosen since the intersection occurs when y* is negative. The time t* to reach the switching curve, called the switching time, given that we start above it, is

To find the minimum total time to go from the starting point (x0,y0) to the origin (0,0), we substitute t* into the equation for + in Column (b) of Table 3.2; this gives

As a numerical example, start at the point (x0,y0) = (1 , 1 ). Then the equation of the parabola (3.66) is 2x = 3- y2 . The switching point (3.67) is . Finally, the switching time is t*= from (3.68). Substituting into (3.69), we find the minimum time to stop is T=

To complete the solution of this numerical example let us evaluate c1 and c2, which are needed to obtain 1

and 2. Since (1,1) is above the switching curve,

u*(T)= 1 and therefore, c2 = 1. To complete c1, we observe that c2 + c1(T-t*) = 0 so that

In exercises 3.14-3.17, you are asked to work other

examples with different starting points above, below,

and on the switching curve. Note that t*= 0 by

definition, if the starting point is on the switching curve.

3.5 Infinite Horizon and Stationarity

Transversality Conditions

Free-end problem:

one-side constraint:

Stationarity Assumption:

Long-run stationary equilibrium It is defined by the quadruple satisfying

Clearly, if the initial condition the optimal control is

for all t. If the constraint involving g is not imposed, may be dropped from the quadruple. In this case, the equilibrium is defined by the triple satisfying

Example 3.6: Consider the problem:

subject to

Solution: By (3.73) we set

where is a constant to be determined. This gives the

optimal control , and setting ,

we see all the conditions of (3.73) hold, including the

Hamiltonian maximizing condition.

Furthermore andW = W0 satisfy the transversality conditions (3.71). Therefore, by the sufficiency theorem, the control obtained is optimal. Note that the interpretation of the solution is that the trust spends only the interest from its endowment W0. Note further that the triple

is an optimal long-run stationary equilibrium for the problem.

3.6 Model Types

Table 3.3: Objective, State, and Adjoint Equations for Various Model Types

In Model Type (a) of Table 3.3, and f are linear. It is called the linear-linear case. The Hamiltonian is

Model Type (b) of Table 3.3 is the same as Model Type (a) except that the function C(x) is nonlinear. Model Type (c) has linear function in state equation and quadratic functions in the objective function. Model Type (d) is a more general version of Model Type (b) in which the state equation is nonlinear in x. In Model Type (e) and (f), the functions are scaler functions, and there is only one state equation so that is also a scaler function.

Remark 3.3 In order to use the absolute valuefunction |u| of a control variable u in forming the functions or f. We define the following equations:

We write

We need not impose (3.79) explicitly.

Remark 3.4 Table 3.1 and 3.3 are constructed for

continuous-time models.

Then the problem of maximizing the Hamiltonian function becomes an LP problem:

Remark 3.5 Consider Model Types (a) and (b) when the control variable constraints are defined by linear inequalities of the form:

Remark 3.7 One important model type that we did not include in Table 3.3 is the impulse control model of Bensoussan and Lions. In this model, an infinite control is instantaneously exerted on a state variable in order to cause a finite jump in its value.

Remark 3.6 The salvage value part of the objective function, S[x(T),T], makes sense in two cases:

(a) When T is free, and part of the problem is to determine the optimal terminal time.

(b) When T is fixed and we want to maximize the salvage value of the ending state x(T), which in this case can be written simply as S[x(T)].