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Technical Communique

On smooth optimal control determination

Ilya Ioslovich

, Per-Olof Gutman

Faculty of Civil and Environmental Engineering, Lowdermilk Division of Agricultural Engineering, Technion, Haifa 32000, Israel Received 15 January 2003; received in revised form 28 May 2004; accepted 15 July 2004

Abstract

When using the Pontryagin Maximum Principle in optimal control problems the most difficult part of the numerical solution is associated with the non-linear operation of the maximization of the Hamiltonian over the control variables. For a class of problems, the optimal control vector is a vector function with continuous time derivatives. A method is presented to find this smooth control without the maximization of Hamiltonian. Three illustrative examples are considered.

䉷2004 Elsevier Ltd. All rights reserved.

Keywords: Optimal control; Rigid body rotation; Crop density

1. Description of the method

Let us consider the classical optimal control problem (OCP) (Pontryagin, Boltyansky, Gamkrelidze, & Mis-chenko, 1962;Lee & Markus, 1967;Athans & Falb, 1966;

Pinch, 1993), in the form T

0

f0(x, u)dt →min, (1)

dx

dt =f (x, u), (2)

x(0)=x0, x(T )=xT, (3)

where the control variables u(t)Rm, the state variables

x(t)Rn, and f (x, u)Rn are column vectors, with

mn. The right-hand-side functions of the state equation (2), f (x, u) and the performance index function f0(x, u)

are smooth over all arguments. It is assumed that there are no other constraints except (3). The column vector

p(t)Rn is the vector of costate variables. According to

This paper was not presented at any IFAC meeting. This paper was recommended for publication in revised form by Assistant Editor Rick Middleton under the direction of Editor P. Van den Hof.

Corresponding author. Tel.: +972-4-8294254; fax: +972-4-8295696. E-mail address:[email protected](I. Ioslovich).

0005-1098/$ - see front matter䉷2004 Elsevier Ltd. All rights reserved. doi:10.1016/j.automatica.2004.07.011

Pontryagin’s Maximum Principle (PMP) (Pontryagin et al., 1962) the Hamiltonian H, is defined as

H=pTf (x, u)p

0f0(x, u). (4)

Let us consider the “normal” case when p0>0. Without

loss of generality we setp0=1, leading to

H=pTf (x, u)f

0(x, u). (5)

According to PMP the following system of differential equa-tions for the co-state variables p holds:

dp dt = −

jH

jx

T = −jf

jx

T

p+jf0

jx

T

. (6)

If there exists an optimal solution(x, u), then, according to PMP, there exists a costate vectorp∗such that the following conditions are satisfied:

(a) H (x, u, p)H (x, u, p), meaning that H has maximum over the control u,

(b) the variablesx, p∗satisfy Eqs. (2), (6), (c) the end conditions (3) must be hold.

(2)

time, u is determined from condition (a). There are as many unknown valuesp(0)as given end conditionsx(T )=xT. The maximization problem in (a) is often computationally costly. Since no other constraint than (3) is present, it follows from (a) that the control vector u∗ has to satisfy the necessary conditions

jH

ju =0. (7)

Often these equations are non-linear inu∗. In the following, we shall omit the symbol * for simplicity. From (5) and (7) one gets the system of equations

pTjf

ju

jf0

ju =0, (8)

wherejf0/juis a row vector, and the dimension of the

Ja-cobianjf/juisn×m.The Eq. (8) is linear in p. As men-tioned above, according to the usual procedure for solving an OCP with PMP, the non-linear system of m equations (8) must be solved in order to find the control vector u as a function of the costate vector p and the state vector x. In contrast, by linear operations on (8) it is very easy to com-pute m elements of the costate vector p as a vector function of the state vector x, the control vector u, and the nm remaining costates. Let us assume that the rank of the Ja-cobian matrix jf/ju is equal to m. This means that there exists a non-singularm×msub-matrix. We shall re-index the state variables in such a manner that the m variables cor-responding to this sub-matrix have indices 1 to m, and we shall denote the sub-vectors of the corresponding right-hand functions in (2) asfa, the costates aspa, the sub-vector of the remaining right-hand functions asfb, and the remaining costates aspb. Consequently,

x= [xa;xb]; p= [pa;pb];

f (x, u)= [fa(x, u);fb(x, u)].

Eq. (8) can be rewritten as

paTjfa

ju +pb

Tjfb

ju

jf0

ju =0. (9)

From here follows that

pa=

jfa ju

T−1 −jfb

ju

T

pb+jf0

ju

T .

=A(x, pb, u),

(10) where for convenience we have defined the vector function

A(x, pb, u). Taking the derivative of the costate pa from

(10) with respect to time t, one gets dpa

dt =

jA

jxf (x, u)+

jA

ju

du dt +

jA

jpb

dpb

dt . (11) The differential equations for the costates (6) can be rewrit-ten as

dpa dt = −

jH

jxa

T

= −jfa

jxa

T

pa−jfb

jxa

T

pb+jf0

jxa

T

, (12)

dpb dt = −

jH

jxb

T

= −jfa

jxb

T

pa−jfb

jxb

T

pb+jf0

jxb

T

. (13)

By substituting the expressions for pa from (10) into the right-hand sides of (12), (13), one gets

dpa dt = −

jfa jxa

T

A−jfb

jxa

T

pb+jf0

jxa

T

, (14)

dpb dt = −

jfa jxb

T

A−jfb

jxb

T

pb+jf0

jxb

T

. (15)

For convenience we shall denote the right-hand side of Eq. (15) asS(x, pb, u), namely,

dpb dt = −

jfa jxb

T

A−jfb

jxb

T

pb+jf0

jxb

T .

=S(x, pb, u).

(16) Exploiting the fact that the right-hand side of (14) is equal to the right-hand side of (11), one gets, with a simple matrix transformation, the equations for the time derivatives du/dt as functions of x,pb, and u only, as follows

B .=jA

ju.

We shall assume that B is non-singular. Thus,

jA

jxf (x, u)+

jA

jpb

dpb dt +B

du dt

= −jfa

jxa

T

A−jfb

jxa

T

pb+jf0

jxa

T

,

and hence, taking into account (16), one gets

du dt =B

−1

−jfa

jxa

T

A−jfb

jxa

T

pb+jf0

jxa

T

− jA

jxf (x, u)

jA

jpbS(x, pb, u)

.

=F (x, pb, u), (17)

where the notationF (x, pb, u)is defined. Thus, the problem is reduced to integrating the system of equations

du

dt =F (x, p

b, u),

dx

dt =f (x, u), dpb

dt =S(x, p

b, u), (18)

with the initial values x(0)=x0, and instead of the pair

of initial values pa(0), pb(0), the appropriate pair of ini-tial valuesu(0), pb(0)are to be found. The end conditions

(3)

have the differential equations for u. Note that no maximiza-tion of the Hamiltonian, nor any non-linear operamaximiza-tions are needed for the determination ofu(t). Instead, just a number of linear matrix operations have to be performed, which is a much easier task. The following theorem can be formulated based on the notations and derivations given above.

Theorem. If the OCP problem (1)–(3),mn,has the op-timal solutionx, usuch thatuis smooth, and if also the problem is normal in the sense that the costatep0can be set to 1, the Jacobianjfa/juis non-singular, the Jacobian B is non-singular, then the optimal states,x, costates,pb,

and controlusatisfy the Eqs. (18).

Note. In the case when n=m the set of variables and costatesxb, pbis empty and Eqs. (18) are changed accord-ingly. In this case the last set of these equations, namely those related topb,vanishes.

2. Rigid body rotation

As the first example, let us consider the following axisym-metric rigid body rotation problem, which is a modification of the problem considered inAthans and Falb (1966)

dx

dt =ay+u1, dy

dt = −ax+u2. (19) The variables x and y are the components of the angular velocity that need to be stopped, i.e., the final conditions are

x(T )=0, y(T )=0, (20)

and final time T is given. The performance index to be min-imized is chosen as

J=1

4

T

0

(u2 1+u

2 2)

2dt min.

Here, according to (10), the vector A has the form

p=A= u

1(u21+u22)

u2(u21+u22)

. (21)

The costate equations are dpx

dt =apy, dpy

dt = −apx. (22)

One may notice that the Hamiltonian is strictly concave. HerejA/jx=0. The matrix B is symmetric and has the form

B=

3u21+u22 2u1u2

2u1u2 u21+3u22

(23)

with

B−1= 1

(3[u21+u22]2)

u2

1+3u22 −2u1u2

−2u1u2 3u21+u22 . (24)

Extracting the time derivatives foru1, u2one gets after some

simplifications

F = du1

dt du2

dt

=

au2 −au1

, (25)

implying du1

dt =au2, du2

dt = −au1. (26)

From (26) it is easily seen that

u2

1+u22=C2=const.

Hence the control vector (u1, u2)T has constant length

C >0, and it rotates around the origin of the coordinate system in the u-plane with the constant angular velocity a. Introducing the variablesr,by the coordinate transforma-tion

x=r sin

y=r cos (27)

one realizes that the system of Eqs. (19), (26) has the solution

u1= −C x

x2+y2,

u2= −C y

x2+y2. (28)

Thus, the initial value of the vectoru= [u1;u2]is collinear and has opposite direction to the initial vector of the state coordinates [x;y]. To check the admissibility of (28), one may substitute (27) into (19) and find the equations for the time derivatives of r,. To find the derivative of r, the re-sulting equations must be multiplied by sin,cos, respec-tively, and then summed. To find the derivative ofthe same equations must be multiplied by cosand−sin, respec-tively, and then summed. Substituting (28) into the obtained equations one gets

dr

dt = −C, d

dt =a. (29)

Thus one can see that both vectors(u1, u2)Tand(x, y)T

ro-tate with the same angular velocity a and their collinearity is maintained in steady state. Let us denote

x(0)2+y(0)2=

(4)

3. Optimal spacing for greenhouse lettuce growth

Let us consider the problem of the optimal variable spac-ing policy for greenhouse lettuce growth (seeIoslovich & Gutman, 1999, 2000; Seginer & Ioslovich, 1999). The dynamical model of the lettuce growth in constant climate conditions has the form

dv

dt =aG(W), (30)

wherevis the dry mass of a single plant (kg/plant), a is the spacing(m2/plant), G is the net photosynthesis(kg/m2), W is the plant density(kg/m2). The net photosynthesisG(W) is a strictly concave function. The spacing is variable in time and must be adjusted in an optimal way during the growth process to minimize the cost of the process, which is set as

J= T

0

acRdt, (31)

wherecR($/m2s)is the price of rent, including operational costs. The spacing and plant density are connected via

v=aW. (32)

Thus plant density W can be considered as a control variable instead of the spacing a. The state equation and the expres-sion for the performance index can be rewritten in the form

dv dt =

v WG(W), J =

T

0

v

WcRdt. (33)

The final time T is free and the final moment T is determined through the condition

v(T )=vT, (34)

where vT is a given final marketable weight of a single lettuce plant. The Hamiltonian, H, has the form

H= v

W(pG(W)cR), (35)

where p is the scalar costate variable. The maximization of the Hamiltonian over the control W gives

jH

jW = − v

W2(pG(W)cR)+

vp W

jG

jW =0. (36)

From here the value of p is found as

p= cR

G(W)WjjWG. (37)

The differential equation for the co-state p has form dp

dt = −

jH

jv = −

pG(W)cR

W . (38)

Substituting an equation for p from (37) to (38) one gets after some transformations

dp dt = −

cRjjWG

(GWjjWG). (39)

Differentiating (37) with respect to t one gets

dp dt =

dW dt

cRWjG 2

jW2

(GWjjWG)2. (40)

From (39) and (40) one gets the differential equation for the control variable W as follows

dW dt = −

jG

jW(GWjjWG)

WjjWG22

. (41)

For the free final time we have the transversality condition

H (T )=0. (42)

Substituting the expression for p (37) and the expression for

H (35) into the Eq. (42) one gets for the final moment T the

following equality

v cR jG

jW

(GWjjWG)=0. (43)

From (43) and (41) it follows that there exists an optimal steady-state solution with the constant control value W∗ which satisfies the condition

jG(W)

jW =0. (44)

Thus, the optimal lettuce density is constant and corresponds to the maximal net photosynthesis. The problem is solved without maximization of the Hamiltonian, which in this case seems rather simple, but unexpected in advance. However, if the final time is not free, and is fixed a priori at some non-optimal value tf, Eq. (41) makes is possible to avoid the maximization of the Hamiltonian, which constitutes a difficult numerical problem for each time step. For details, see Ioslovich and Gutman, 1999.

4. Maximal area surrounded by a curve of given length with fixed end points on a horizontal axis

This is a very well-known problem described in e.g.

(5)

problem. We shall solve it in the case of given parameters as an OCP

1 0

x1dt→max, (45)

dx1

dt =u, (46)

dx2

dt =

1+u2. (47)

The considered end conditions are

x1(0)=x1(1)=0, x2(0)=0, x2(1)=/3. (48)

The Hamiltonian has the form

H=p1u+p2

1+u2+x

1. (49)

The differential equations for the costate variables are dp1

dt = −1, dp2

dt =0. (50)

From (49) one gets

jH

ju =p1+p2√ u

1+u2=0. (51)

In fact (51) is possible to solve with respect to u. However this is not always the case in general, and we want to demon-strate our new method. Here we usep1aspaandp2aspb.

From (51) the value ofp1is found in the form

p1= −p2√ u

1+u2. (52)

Taking the derivative with respect to t of (52) and using (50) one gets

du dt =

(1+u2)3/2

p2 .

(53)

Now, taking into account thatp2is a constant according to

(50), we have to integrate numerically Eqs. (46), (47), and (53) in order to find the unknown constantsp2andu(0)such

that conditions (48) are satisfied. The answer isp2= −1,

andu(0)=tan/6=√1

3.The resulting curvex1(t)is a part

of a circle. As in the other examples, we did not have to solve the non-linear equation (51).

This problem also can be solved analytically as follows. One can divide (53) over (47) and get

du dx2 =

1+u2

p2 ,

(54)

thus changing independent variable from t tox2. Eq. (54)

can be solved and be given as

arctanu= x2

p2 +C,

(55)

where C is an unknown integration constant. From (55) one gets

u=tan(x2/p2+C). (56)

From here we can get

1+u2=1/cos(x

2/p2+C).

Now one can integrate (47) and get

sin(x2/p2+C)=(t+C1)/p2, (57)

where C1 is another integration constant. Doing the same

substitution of x2 instead of t in (46), (47) as above and

using (56), one gets dx1

dx2=

sin(x2/p2+C). (58)

From here it follows

(C2−x1)/p2=cos(x2/p2+C), (59)

whereC2again is an additional integration constant. From

(57) and (59) it follows

(x1−C2)2+(t+C1)2=p22, (60)

which means that the optimal curve is a part of a circle. Unknown constantsp2, C, C1, C2are easily found from the

end conditions fort, x1, x2and we get

p2= −1, C=/6, C1= −1/2, C2= − √

3/2, from which it follows from (56) thatu(0)=tan/6=1/√3.

5. Conclusion

A method for extracting smooth optimal control for a class of optimal control problems was presented and three illus-trative examples were shown. The method does not require the maximization of the Hamiltonian over the control, and instead substitute the ODE for the part of costates by the ODE for the smooth control. One could note that the method has advantage only if the equations jH/ju=0 cannot be solved analytically.

Acknowledgements

The authors are grateful to Professors V.F. Krotov, A. Ioffe, and B.T. Polyak for fruitful discussions. The authors also want to thank the Associate Editor and anonymous reviewers for useful notes.

References

(6)

Gelfand, I. M., & Fomin, S. V. (1969). Calculus of variations (p. 228). Englwood Cliffs, NJ: Prentice-Hall.

Ioslovich, I., & Gutman, P. -O. (1999). Plant factory optimization. Third international conference on industrial automation Montreal, Canada; international Association for Industrial Automation (IAIA), Conference Proceedings (pp. 15.5–15.8).

Ioslovich, I., & Gutman, P. -O. (2000). Optimal control of crop spacing in a plant factory. Automatica, 36, 1665–1668.

Lee, E. B., & Markus, L. (1967). Foundations of optimal control theory (p. 576). NY: Wiley.

Pinch, E. R. (1993). Optimal control and the calculus of variations (p. 234). UK: Oxford Science Publication.

Pontryagin, L., Boltyansky, V., Gamkrelidze, R., & Mischenko, E. (1962). Mathematical theory of optimal processes (p. 383). NY: Wiley. Seginer, I., & Ioslovich, I. (1999). Optimal spacing and cultivation

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