Some theoretical foundations of optimal control problem are expounded in the textbook: methods of variation calculus, ma

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Al-Farabi Kazakh National University

S.A. Aisagaliev, Zh.Kh. Zhunussova

OPTIMAL CONTROL Textbook

Almaty "Kɚzɚkh university" 2014

UDC 519. 7(075.8) LBC 22.311 ɹ73 A 28

Recommended by Educational and Methodical Section of the Humanities and Natural sciences specialties of the Republican Educational and Methodical Council of higher and postgraduate education of the Ministry of Education and Science of the Republic of Kazakhstan on the base of Al-Farabi Kazakh National University (Protocol ʋ 1, January 17, 2014)

Reviewers: Doctor of Physical and Mathematical Sciences, Professor M.Akhmet Doctor of Physical and Mathematical Sciences, Professor M.T.Dzhenaliev Doctor of Physical and Mathematical Sciences, Professor S. Ya. Serovaijsky

Aisagaliev S.Ⱥ., Zhunussova Zh.Kh. A 28

Optimal control: textbook/ S.A. Aisagaliev, Zh.Kh. Zhunussova. – Almaty: Kazakh university, 2014. – 200 p. ISBN 978-601-04-0588-2 Some theoretical foundations of optimal control problem are expounded in the textbook: methods of variation calculus, maximum principle, dynamical programming for solving of topical problems of economic macromodels. The tasks for independent work with solving of concrete examples, brief theory and solution algorithms of the problems, term tasks on sections of optimal control and variation calculus are presented in the appendix. It is intended as a textbook for students of the high schools training on specialties "mathematics", "mechanics", "economic cybernetics" and "informatics". It will be useful for the post-graduate students and scientific workers of the economic, mathematical and naturally-technical specialties. UDC 519. 7(075.8) LBC 22.311 ɹ73

ISBN 978-601-04-0588-2

© Aisagaliev S.Ⱥ., Zhunussova Zh.Kh., 2014 © KazNU al-Farabi, 2014

CONTENTS

FOREWORD ................................................................................................... 5 CHAPTER I NUMERICAL METHODS OF MINIMIZATION IN FINITE-DIMENSIONAL SPACE ............................................................ 6 LECTURE 1. MINIMIZATION METHODS OF ONE VARIABLE FUNCTIONS ............................................................. 6 LECTURE 2. GRADIENT METHOD. THE GRADIENT PROJECTION METHOD ............................................................................... 14 LECTURE 3. CONDITIONAL GRADIENT METHOD. CONJUGATE GRADIENT METHOD .......................................................... 22 LECTURE 4. NEWTON'S METHOD. THE PENALTY FUNCTION METHOD. LAGRANGE MULTIPLIER METHOD .............. 32 LECTURE 5. REVIEW ON OTHER NUMERICAL MINIMIZATION METHODS ........................................................................ 39 CHAPTER I I VARIATION CALCULUS............................................................................. 45 LECTURE 6. BRACHISTOCHRONE PROBLEM. THE SIMPLEST TASK. NECESSARY CONDITIONS FOR A WEAK MINIMUM. LAGRANGE'S LEMMA. EULER EQUATION ..... 45 LECTURE 7. DUBOIS - RAYMOND LEMMA. BOLZ PROBLEM. NECESSARY CONDITION OF WEIERSTRASS .................. 52 LECTURE 8. LEGENDRE CONDITION. JACOBI CONDITION. FUNCTIONALS DEPENDING ON N UNKNOWN FUNCTIONS. FUNCTIONALS DEPENDING ON HIGHER ORDER DERIVATIVES ................................................................ 61 LECTURE 9. ISOPERIMETRIC PROBLEM. CONDITIONAL EXTREMUM. LAGRANGE PROBLEM. GENERAL COMMENTS ....... 69 CHAPTER III OPTIMAL CONTROL. MAXIMUM PRINCIPLE ....................................... 77 LECTURE 10. PROBLEM STATEMENT OF OPTIMAL CONTROL. MAXIMUM PRINCIPLE FOR OPTIMAL CONTROL PROBLEM WITH FREE RIGHT END ............................................................................. 77 LECTURE 11. PROOF OF MAXIMUM PRINCIPLE FOR OPTIMAL CONTROL PROBLEM WITH FREE RIGHT END ..................................... 85 3

LECTURE 12. MAXIMUM PRINCIPLE FOR OPTIMAL CONTROL PROBLEM. CONNECTION BETWEEN MAXIMUM PRINCIPLE AND VARIATION CALCULUS ................................................................... 91 CHAPTER IV OPTIMAL CONTROL. DYNAMIC PROGRAMMING ............................. 100 LECTURE 13. OPTIMALITY PRINCIPLE. BELLMAN EQUATION........ 100 LECTURE 14. DISCRETE SYSTEMS. OPTIMALITY PRINCIPLE. BELLMAN EQUATION ................................................................................ 107 LECTURE 15. SUFFICIENT OPTIMALITY CONDITIONS ....................... 114 Appendix I. TASKS FOR INDEPENDENT WORK ................................... 123 Appendix II. TASKS ON OPTIMAL CONTROL ......................................... 154 Appendix III. KEYS ....................................................................................... 170 Appendix IV. TESTS ...................................................................................... 178 References....................................................................................................... 200

FOREWORD

The main sections of optimal control, the numerical methods of function minimization of the finite number variables are expounded in the textbook. It is written on the basis of the lectures on methods of optimization and variation calculus which have been delivered by the authors in Al-Farabi Kazakh National University. n connection with transition to credit technology of education the book is written in the manner of scholastic-methodical complex containing alongside with lectures the problems for independent work with solutions of the concrete examples, brief theory and algorithm on sections of this course, as well as term tasks for mastering of the main methods of the optimization problems solution. In the second half XX from necessity of practice appeared the new direction in mathematics as "Mathematical control theory" including the following sections: mathematical programming, optimal control with processes, theory of the extreme problems, differential and matrix games, controllability and observability theory, stochastic programming [1]-[14]. Mathematical control theory was formed at period of the tempestuous development and creating of the new technology and spacecrafts developing of the mathematical methods in economics and controlling by the different process in natural sciences. Emerged new problems could not be solved by classical methods of mathematics and required new approaches and theories. The different research and production problems were solved due to mathematical control theory, in particular: organizing of production to achieve maximum profit with provision for insufficiency resources, optimal control by nucleus and chemical reactors, electrical power and robotic systems, control by moving of the ballistic rockets, spacecrafts and satellites and others. Methods of the mathematical control theory were useful for developing mathematics. Classic boundary value problems of differential equations, problems of the best function approach, optimal choice of the parameters in the iterative processes, minimization of the difficulties with equations are reduced to studying of the extreme problems. Theory foundation and solution algorithms of the numerical methods of minimization in finite-dimensional space, variation calculus, optimal control, maximum principle, dynamical programming are expounded in the lectures 1-15. Execution of the term tasks for independent work of the students is provided for these sections. Solution methods of the extreme problems are related to one of the high developing areas of mathematics. That is why to make a textbook possessed by completion and without any shortage is very difficult. Authors will be grateful for critical notations concerning the textbook. 5

CHAPTER I NUMERICAL METHODS OF MINIMIZATION IN FINITE-DIMENSIONAL SPACE

The fundamentals to the theory of extreme problems in finite-dimensional space were supposed in the previous work [15]. Applied methods for solving of extreme problems based on the construction of a minimizing sequence for determining a minimum of the function on a given set the rapid development are obtained by appearance of the computers. Along with the applied nature of these methods the new mathematical problems related to convergence of iterative processes, rational choice of methods for the solution of certain extreme problems emerged. New methods for solving extreme problems focused on applying of computers have been developed. LECTURE 1. MINIMIZATION METHODS OF ONE VARIABLE FUNCTIONS The theory presented in the preceding work [15] for the function of several variables remains unconditionally true for functions of one variable. However, the numerical methods of the solution of extreme problems in finite-dimensional space at some stage require for solving the problem for functions of one variable on a given set. Therefore, in order to save computing time and computer memory, it is advisable to consider minimization methods of one variable functions. Bisection method of the interval. Let the function J(ɢ) be defined on the interval >ɫ, d @ E 1 . Necessary to find the minimum of J(ɢ) on the interval >ɫ, d @ E 1 . Definition 1. Function J(ɢ) defined on the interval >a, [email protected] is called unimodal, if it is continuous on >a, [email protected] and there exist the numbers D , E , a d D d E d b , such that 1) J (u) is strictly monotonously decreasing at a d u d D (a D ) ; 2) it is strictly monotonously increasing at E d u d b ( E b) ; J (u) at D d u d E , i.e. U * >D , E @ . In particular, if D E , 3) J (u ) J (u* ) uinf >a.b @ then

J (u ) is called strictly unimodal on the interval >a, [email protected] . We note, that if a continuous function J (u) is not unimodal on the prescribed interval >c, d @ , then the intervals >a, [email protected] >c, d @ can be found by partitioning of the interval >c, d @ , where function J (u) is unimodal on >a, [email protected] . We define the minimum of strictly unimodal function J (u) on the interval >a, [email protected] . According to the definition, unimodal (strictly) function is convex (strictly) function on the interval >a, [email protected] .

6

a) The points u1

a b G , u2 2

a b G are chosen, where G ! 0 is determined 2

by the prescribed accuracy of calculation, but it should not be less than the machine zero of computers. b) The values J (u1 ), J (u 2 ) are calculated. If J (u1 ) J (u2 ) , it is assumed a1 a, b1 u2 , if J (u1 ) ! J (u2 ) , then a1 u1 , b1 b . Since the function is unimodal, the point u* U * is found on the segment >a1 ,b1 @ and its length b1 a1 (b a G ) / 2 G . c) The points u3

a b G , u4 2

a b G 2

are selected and values J (u3 ) , J (u 4 ) are calculated, then they are compared and the (b1 a1 G ) / 2 segment >a2 ,b2 @ is determined. Its lengh is equal to b2 a2 G (b a G ) / 4 G etc. In general case the difference bk a k

(b a G ) / 2 k G , U * >a k , bk @ .

The method of the golden section. In the method of dividing a segment the point u1 (or the point u2 ) is on the interval >a1 ,b1 @ , however, in determining the segment >a2 ,b2 @ the value J (u1 ) [or J (u 2 ) ] is not used, since the point u1 (or u2 ) is far from the center of the segment >a2 ,b2 @ . Therefore, more efficient method of finding the minimum of a unimodal function on the interval >a, [email protected] , than the bisection method is the method of the golden section. The points u1 , u2 >a, [email protected] are selected as follows: 1) The points u1 , u 2 must be spaced equally from the ends of the segment >a, [email protected] , i.e. u1 a b u2 . 2) One of the points u1 , u 2 to be used on the next iteration, so the points u1 , u 2 are selected so that on the new segment of localization the remaining point locates the same position as on the original >a, [email protected] . Consequently, u1 a ba

u 2 u1 u2 a

b a 2u1 . b u1

Hence we get u1

u2

3 5 , 2 3 5 b (b a ) . 2

a (b a )

3) The values J (u1 ), J (u 2 ) are calculated. If J (u1 ) ! J (u2 ) , then a1 u1 , b1 b , and if J (u1 ) J (u2 ) , then a1 a, b1 u2 . At each iteration, except the first, the value J (u ) is calculated in a one point and the length of localization is reduced by 1.6 times. Note that bk ak

(( 5 1) / 2) k 1 (b a), k 1, 2, ... . 7

Optimal search. Optimal methods of searching for minimizing the unimodal function J (u ) on the segment >a, [email protected] are applied in cases, when the computation of the values J (u) requires a significant amount of computer time. In such cases it is advisable to calculate the function value as possible in a small number of points at the given accuracy, and possibly the quantity of calculations the values of function J (u ) on the interval >a, [email protected] are given. We present an optimal sequential search algorithm associated with the Fibonacci numbers. Frequency the method is called Fibonacci method. Sequence of numbers ^Fn ` are called by Fibonacci numbers, if Fn 2

Fn1 Fn , F1

F2 1, n 1, 2, 3, ...

(e.g., 1,1, 2, 3, 5, 8,13, ...). a) The length of the interval ' bk ak of uncertainty after k computing the values of function J (u ) are prescribed, i.e. u* >ak , bk @, the number k is not known, only ' is known. b) The number k

(b a ) / '

is found by the given ' . c) The item of Fibonacci number j is determined from the condition F j 1 k F j .

d) Step of searching G (b a ) / F j is defined. e) The value J (u ) in the point u a is calculated and the following point in which the value J (u ) is calculated we find by the formula u1

a G F j2 .

f) If J (a) ! J (u1 ),

the next point u2

u1 G F j 3 ,

u2

u1 G F j 3 .

otherwise

g) Further the process flows with decreasing value of step which for i -iteration is equal to G i rG F j i 2 . The point ui 1 ui G i . If J (ui 1 ) J (ui ) , then ui 2

ui 1 G F j ( i 1) 2 , and if 8

J (u i 1 ) ! J (u i ) ,

then ui2

u i 1 G F j ( i 1) 2 , i

0, 1, 2, ... , u 0

a.

More detailed exposition of the proof of Fibonacci method can be found in the book: Wild D.D. Metody poiska extremuma. M.: Nauka, 1967. Other methods for finding minimum of the function J (u ) on the segment >a, [email protected] such as broken method, the method of tangents, parabola method can be found in the book: Vasil'ev F.P. Chislenye metody reshenya extremalnih zadach. M.: Nauka, 1980. Auxiliary lemmas. The following two lemmas are often used in numerical methods of minimizing the function J (u ) on the set U of E n to prove the convergence of a sequence ^u n ` U , ^J (u k )` . Lemma 1. If the function J (u ) C 1 (U )

and J cu satisfies to Lipschitz condition on the convex set U, i.e. J c(u ) J c(Q ) d L u Q , u, Q U , L const ! 0,

(1)

then the inequality 1 2 J (Q ) J (u ) t J c(Q ), Q u LQ u , u, Q U . 2

(2)

holds. Proof. Let all conditions of the lemma be hold. We show, that the inequality (2) is true. If the vectors u , u h U , then according to the formula of finite increments we get 1

J (u h) J (u )

³

J c(u Dh), h dD .

0

Hence, in particular, under the vector h=v-u, we obtain 1

³

J (Q ) J (u)

J c(u D (Q u )), Q u dD

0

1

J c(Q ), Q u ³ J c(u D (Q u)) J c(Q ), Q u dD . 0

Since the scalar product J c(u D (Q u )) J c(Q ), Q u t J c(u D (Q u )) J c(Q ) Q u ,

then the integral 9

(3)

1

³

J c(u D (Q u )) J c(Q ), Q u dD t

0

1

t ³ J c(u D (Q u )) J c(Q ) Q u dD . 0

Hence, in view of (1) we get 1

³

J c(u D (Q u )) J c(Q ), Q u dD t

0

1

t ³ L u D (Q u ) Q dD Q u 0

1

³ (1 D )dD Q u 0

2

1 2 LQ u , u, Q U 2

Then from (3) follows, that 1 2 J (Q ) J (u ) t J c(Q ), Q u LQ u , u, Q U . 2

The inequality (2) is proved. The lemma is proved. Lemma 2. If the numeric sequence ^a n ` such that a n ! 0, a n a n 1 t Aa n2

for all n t n0 t 0,

then the inequality an

n0 1 for all n ! n0 An

holds. Proof. By assumption of the lemma the difference 1 ak 1

1 ak

ak ak 1 ak ak 1

ak § 1 ak 1 / a k ¨ ak 1 ¨© ak

ak § ak ak 1 · a ¨ ¸t A k ak 1 ¨© ak2 ¸¹ ak 1

for all k t n 0 . Then the sum

10

· ¸¸ ¹

(4)

n 1

§ 1

¦ ¨¨ a

k n0

©

k 1

1 ak

a k ! 0, a k a k 1

n 1 · a ¸¸ t A ¦ k , k n0 a k 1 ¹ 2 t Aa k , k t n0 t 0 .

Taking into account that a k / a k 1 t 1 Aa k2 / a k 1 ! 1

,

we obtain 1 1 ! A(n n0 ) . an an0

Consequently, 1 ! A( n n0 ), an

i.e. an

1 . A( n n 0 )

By condition of the lemma n ! n0 , therefore n t n0 1. Then nn 0 t n 02 n 0 , n 0 ! 0 , therefore nn 0 n 02 n0 ! 0 . It follows that n d nn 0 n02 n0 n

(n n0 )(n0 1) .

From this inequality we get n 1 1 , n ! n0 . d 0 n n0 n

Now inequality an

1 , n ! n0 , A( n n 0 )

can be written as (4). Lemma is proved. Gradient method. Let the function J (u ) C 1 ( E n ) .

We consider the optimization problem J (u ) o inf, 11

u U { E n .

Since the function J (u ) C 1 ( E n ) ,

then for any vector u E n the difference

J u h J u

and

J ' u , h oh, u ,

(5)

oh, u / h o 0 at h o 0 .

From the properties of the scalar product (the Cauchy - Bunyakovsky inequality), we get J ' u h d J ' u , h d J ' u h ,

moreover the left equality is possible only at h J ' u , and the right equality at h J ' u . Let the vector h DJ ' u , D const ! 0 . Then from (5) follows that J u DJ ' u J u D J ' u oD , u , 2

where oD , u / D o 0 at D o 0 . Thus, the direction of steepest decreasing function J (u ) in the point u coincides with the direction of the vector J ' u (anti-gradient) at J ' u z 0 . This property is the basis of a number of gradient methods for minimizing a differentiable function. The method algorithm: 10. Select the starting point u 0 E n . There is not any general rule for selecting the starting point. 20. Construct a sequence ^un ` by the formula un1

un D n J ' un ,

n

0, 1, 2, ... .

(6)

30. The number D n is called the step of gradient method and is chosen from the condition J u n1 J u n : a) In the method of steepest descent the number D n is determined from g n (D n )

min g n (D ) , D t0

where g n (D )

J (u n DJ ' (u n )) .

The function g n (D ), D t 0 depends on a variable D , and its minimum can be found by the methods of minimizing a function of one variable set out above. 12

b) If J ' (u), u E n satisfies to the Lipschitz condition, i.e., | J c(u ) J c(v) | d L u v , u , v E n , then the number D n can be selected by condition 0 H 0 d D n d 2 / L 2H ,

where H 0 ! 0, H ! 0 are parameters of the method. In particular, if H 0 1/ L, H

L/2,

then the number D n 1 / L at all n . c) The sequence ^D n ` can be prescribed by the conditions D n ! 0, n 0, 1,... ; f

¦D

f

n

n 0

(for example, D n that

c / 1 n , c D

f, ¦ D n2 f n 0

const ! 0, 1 / 2 d D d 1 ). Thus, it is necessary to ensure

J (u n1 ) J (un ), n

0,1,2,... ..

d) It can be accepted an

D, D ! 0 .

If the inequality J (u n1 ) J (u n )

is not fulfilled for some n , then the number D is broken up by up until this inequality is not executed (for example, to take D / 2, D / 4 , etc.). Depending on the choice of the number D n the various options for the gradient method which is more adapted for solving various problems of the form (4) can be obtained. Control questions 1. Give definition to the unimodal function. 2. Describe the bisection method of the interval. 3. Describe the method of the golden section. 4. Describe the gradient method. 5. What is the main idea of the minimization methods of one variable functions?

13

LECTURE 2. GRADIENT METHOD. THE GRADIENT PROJECTION METHOD Gradient method allows to find in the general case the stationary points of the function J (u ) C 1 ( E n ) on E n , and for convex and strongly convex functions the points to a global minimum J (u ) C 1 ( E n ) on E n . A decreasing sequence values of function J (u) C1 (U ) on a convex set U of n E can be constructed by the gradient projection method, and in the case when the function J (u) C1 (U ) is convex or strongly convex, the global minimum point J (u ) on U can be determined. Gradient method. We consider the optimization problem J (u ) o inf,

u U { E n ,

(1)

where the function J (u ) C 1 ( E n ) .

Theorem 1. If the function J (u ) C 1 ( E n ) ,

set U*

minn J (u) z uE

and sequence ^u n ` E is constructed by the rule u n 1

u n D n J c(u n ), n

g n (D n )

min g n (D ) D t0

0, 1, 2, ...,

min J (u n D J c(u n )),

(2)

D t0

gradient J cu satisfies to the Lipschitz condition J c(u ) J c(Q ) d L u Q , u , Q E n ,

then

(3)

lim J c(u n ) 0. n of

f, in addition, J (u ) is convex and set M (u 0 )

^u E

n

/ J (u) d J (u 0 )`

is bounded, then the sequence is minimizing for the problem (1), i.e. lim J (u* ) J * nof

14

inf J (u) minn J (u) ,

uE n

uE

(4)

and any limit point of the sequence ^u n ` belongs to the set U * . The estimation 0 d J (u n ) J * d 2 D 2 L / n , n

1, 2, 3, ... ,

(5)

where D sup u Q , u,Q M (u0 )

is held. If J (u ) is strongly convex on E n , then the estimation 0 d J (un ) J * d J (u0 ) J * q n

holds, where

q 1 P / 2 L, 0 q 1; P

const ! 0, P

2N ,

N is a constant of strong convexity of the function J (u ) on E n .

Proof. Let the function J (u ) C 1 ( E n ) , U * z

and the conditions (2) and (3) are fulfilled. We show, that lim J ' (u n )

nof

0.

If for some finite n the gradient J ' u n 0 , then it follows from (2), values un

u n1 ...

J ' u k 0, k

and such that

n, n 1,...,

lim J ' u n 0 . n of

We assume, that the gradient

J ' un z 0, n 0, 1, 2... ..

From the second relation of (2) follows that g n D n J u n D n J ' un J un1 d g n D J un DJ ' un

Then the following inequality holds: J (u n ) J (u n 1 ) t J (u n ) J (u n DJ ' (u n )),

D t 0, n 0, 1, 2 ... .

Since J (u ) C 1 (U ), U { E n 15

(7)

(6)

and the gradient J ' u satisfies the condition (3), then according to Lemma 1 (Lecture 18) the inequality holds v un , u un DJ ' un : J u n J u n DJ ' u n t J ' un , DJ ' u n u J ' un

2

§ ©

D ¨1

DL ·

¸ J ' u n , D t 0, n 2 ¹ 2

1 LD 2 u 2

(8)

0, 1, 2 ... .

Then inequality (7) with taking into account the relation (8) can be written as 2 § DL · J u n J u n 1 t D ¨1 ¸ J ' u n , D t 0, n 2 ¹ ©

0, 1, 2 ... .

Hence we get 2 § L · J u n J u n 1 t max D ¨1 D ¸ J ' u n D t0 2 ¹ ©

1 2 J ' u n ! 0, 2L

(9)

since maxD 1 DL / 2 at D t 0 is reached at D 1/ L . Since the values J ' u n ! 0, J ' u n z 0 , 2

then from (9) implies that the numerical sequence ^J u n ` is strictly decreasing

J un1 J un , n

0,1, 2... .

Since the set U * z , then the numerical sequence ^J u n ` is bounded from below, hence, there is a limit lim J (u n ) , and of the existence of the limit follows that n of

J (u n ) J (u n 1 ) o 0 at n o f .

By transition to the limit at n o f from (9) we get lim J ' (u n ) n of

0.

The first part of the theorem is proved. We suppose, in addition of the conditions above, the function J (u ) is convex and the set M (u 0 ) is bounded. We show that the sequence ^un ` of (2) is minimized. We note, that in force of the continuity J (u ) on E n the set M (u 0 ) is closed. In fact, if v E n is the limit point of the set M (u 0 ) , that there is a sequence ^wk ` M (u 0 ) , such that wk o v at k o f . 16

From inclusion ^wk ` M (u 0 ) follows, that J (wk ) d J (u 0 ), k

1,2... .

Hence, passing to the limit, taking into account the continuity J (u ) we obtain lim J ( wk ) k of

J ( v ) d J (u 0 ) .

Consequently, the point v M (u 0 ) . Finally, M (u 0 ) is bounded closed set, hence it is compact. We note, that as the numerical sequence ^J (u n )` is strictly decreasing, so J (u n ) d J (u 0 ), n 0, 1, 2... ,

it means the sequence ^u n ` M (u0 ) . On the other hand, the continuous function J (u ) reaches its lower boundary on the compact set M (u 0 ) . Therefore the set U * M (u0 ) , i.ɟ. J (u* ) J * , u* U * . It should be noted, that compact set M (u 0 ) is convex by Theorem 2 (Lecture 5). Since the convex function J (u ) ɋ 1 M (u 0 ) , M (u 0 )

is a convex set, then necessary and sufficiently fulfillment of inequality (Theorem 1 of Lecture 4) J (u ) J (v) t J ' (v), u v , u , v M (u 0 ).

Hence J (v) J (u ) d J ' (v), v u , u , v M (u 0 ).

(10)

From (10), in particular, at u

u* M (u 0 ),

v un M (u0 ) ,

we get 0 d J (u n ) J (u* ) d J ' (u n ), u n u* d d J ' (un ) un u* d D J ' (un ) ,

where D

sup u v

u ,vM u0

is the diameter of the set M (u 0 ) . Since J ' (u n ) o 0 at n o f , then from (10) we obtain 17

(11)

J (u* )

lim J (u n ) n of

J* .

This means that the sequence ^un ` is minimizing, moreover, any its limit point belongs to the set U * M (u 0 ) , M (u 0 ) is compact set. J (u ) is continuous on M (u 0 ) . The validity of (4) is proved. We prove the validity of (5). We denote as an

J (u n ) J * .

Now the inequality (9), (11) can be written as follows: a n a n1 t

1 2 J ' (u n ) , a n d D J ' (u n ) , n 2L

0, 1, 2, ... .

(12)

From (12) we get an an1 t

1 an2 , 2 LD 2

since J ' (u n ) t a n / D . Further, applying Lemma 2 (Lecture 18), we obtain the estimate (5), where A 1 / 2 LD 2 , n0

0.

We show faithfulness of estimation (6) in the case at J ( u ) ɋ 1 M ( u 0 )

is strongly convex function. In this case, the inequalities (7) and (8) of Lecture 4 (Theorems 4 and 5) are faithful. Since at the minimum point u* U * necessary and sufficiently performing of inequality J c(u* ), u u* t 0, u M (u 0 ) ,

then from (8) (Lecture 4) at u u* we get 2

J ' (u * ) J ' (v ), u * v t P u * v , v M (u0 ) .

It follows that 2

P u * v d J ' (v ), v u * J ' (u * ), v u* d d J ' (v), v u* d J ' (v) v u* , v M (u 0 ). 18

(13)

For values v u n M (u0 ) of the inequality (13) we obtain 2

P u n u* d J ' (u n ) u n u* , n

0, 1, 2, .... .

The inequality 0 d an d

1

P

J ' (u n )

2

follows from the inequality above and (11). Since a n a n 1 t

1 2 J ' (u n ) , 2L

then a n 1 d a n

1 2 P · J ' (u n ) d §¨ 1 ¸ u an , 2L 2L ¹ ©

2

as J ' (u n ) t Pa n . Hence we get a n 1 d qan , n

0, 1, ...,

where q 1 P / 2 L . Further, considering a n d an 1q d a n 2 q 2 d ... d a 0 q n ,

where a0

J (u 0 ) J * ,

we obtain (6). Theorem is proved. The gradient projection method. We consider the optimization problem J (u ) o inf, u U E n ,

(14)

where the function J (u) ɋ 1 (U ); U is a convex closed set. The method algorithm: 1°. Select the starting point u 0 U . 2°. Determine the point u1 u0 D 0 J c(u0 ) . 3°. Then we find the projection of the point u1 E n on the set U , as a result, we get u1

PU (u1 )

PU (u 0 D 0 J c(u 0 )) U .

In the general case, the sequence ^un ` U is constructed by the rule u n1

PU (u n D n J c(u n )), n

where the number D n is chosen from the condition 19

0,1,2,...,

(15)

J (u n1 ) J (u n ) .

There are different ways of the choice D n , some of which are considered above. Theorem 2. If the function J (u ) C1 U ,

U is a closed convex set, U * z , the gradient J ' (u ) satisfies the Lipschitz condition on the set U and the sequence ^un ` U is determined by the formula (15), where 0 H1 d D n d 2 / L 2H , H 1 , H ! 0

are prescribed numbers, then u n u n1 o 0 at n o f .

If, in addition, J (u) is convex on U , the set

^u U / J (u ) d J (u 0 )`

M (u 0 )

is bounded, then the sequence ^un ` is minimizing to the problem (14) and any its limit point belongs to the set U * . It is valid the estimation 0 d J (u n ) J * d

c2 1 , H n

n 1, 2, ... ,

(16)

wherɟ c

sup J ' (u )

u M u 0

1

H1

D,

D is the diameter of the set M (u 0 ) . Proof. Since the point un1 is the projection of the point u n D n J ' (u n ) E n

on the set U , then according to Theorem 5 (Lecture 5) we get [ w u n1 , v u n D n J ' (u n ), u U , (6)]: un 1 un D n J ' (un ), u un 1 t 0, u U .

Hence, we obtain J ' (u n ), u u n 1 t

1

Dn

u U , n

u n u n 1 , u u n 1 , 0, 1, ... .

According to Lemma 1 (Lecture 1, formula (2)) the inequality 20

(17)

1 2 J (un ) J (un 1 ) t J ' (un ), un un 1 L un un 1 2

(18)

holds. From relations (17) and (18) with taking into account 0 d H 1 d D n d 2 / L 2H

we get § 1 L· 2 2 J (u n ) J (u n1 ) t ¨¨ ¸¸ u n u n1 t H u n u n1 , H ! 0, ©Dn 2 ¹

(19)

From inequality (19) follows, that the numerical sequence ^J u n ` is strictly decreasing, and in view of the fact that U * z , it converges, therefore, J (u n ) J (u n 1 ) o 0 at n o f . Then from (19) we get u n u n1 o 0 , n o f .

The first part of the theorem is proved. Let, in addition of these conditions, the function J (u) be convex on U . Then M (u 0 ) is a limited and closed set, i.e. compact. Moreover, the sequence ^u n ` M (u 0 ) , set U * M (u0 ) and function J (u ) achieves the lower bound on the set M (u 0 ) (see proof of Theorem 1). We show, that the sequence

^u n ` M (u0 ) is minimizing. From (11) follows, that 0 d an

J (u n ) J * d J ' (u n ), u n u* J ' (u n ), u n u n 1 u n 1 u *

J ' (u n ), u n u n 1 J ' (u n ), u n 1 u * .

(20)

As follows from (17), the inequality u u* U : J ' (u n ), u* u n 1 t

holds. Then from (20), (21) follows, that 0 d an

J (u n ) J * d J ' (u n ), u n u n 1

21

1

Dn

u n u n 1 , u* u n 1

(21)

1

Dn

J ' (un )

u n u n 1 , u* u n 1

1

Dn

u* un1 , un un1 d J ' (u n )

1

Dn

d

u* u n1 u n u n .

Taking into account that J ' (u n )

1

Dn

u* u n1 d

J ' (u n )

1

Dn

u* u n1 d § D· d ¨¨ sup J ' (u) ¸¸, H1 ¹ © uM u0

1

Dn

d

1

H1

, u* u n 1 d D ,

we get J (u n ) J * d c u n u n 1 .

0 d an

Since by proved un un1 o 0 at n o f , consequently lim J (u ) nof

J* .

This means that the sequence ^un ` is minimizing and in force of the compactness of the set M (u 0 ) , the continuity J (u) all the limit points ^un ` belong to the set U * M (u 0 ) . From the inequalities 0 d an

J (u n ) J * d c un un 1 ,

J (u n ) J (u n 1 ) t H u n u n 1

follows, that an an1 t

H c

2

2

an2 , n 0, 1, ... .

Hence, in view of Lemma 2 (Lecture 1) we obtain the estimation (16). Theorem is proved. We note, that the gradient projection method requires the determination of the point projection v

u n D n J c( u n ) E n

of the set U. In general case its solution is quite difficult. Therefore, appropriate to apply this method in the cases when the point projection v of E n on the set U it is defined relatively simplify.

22

Control questions 1. Describe the gradient method. 2. Prove the Theorem 1. 3. Describe the gradient projection method. 4. Prove the Theorem 2. 5. What difference between the gradient and the gradient projection methods?

LECTURE 3. CONDITIONAL GRADIENT METHOD. CONJUGATE GRADIENT METHOD Minimum of a convex function J (u ) C 1 (U )

on a bounded closed convex set U of E n can be found by conditional gradient method. The conjugate gradient method is advisable to apply for the minimization of quadratic functions J (u ) C1 ( E n ) . In this case, if the quadratic function is convex, the conjugate gradient method converges to the minimum point of J (u ) no more than n steps. Conditional gradient method. We consider the optimization problem J (u ) o inf, u U ,

(1)

where J (u ) C 1 (U ) , U is bounded convex closed set of E n .

The method algorithm: 1°. Select the starting point u 0 U . We note, that the difference J (u ) J (u 0 )

J ' (u 0 ), u u 0 o u u 0 .

2°. Auxiliary point u0 is determined as the solution of optimization problem J 0 (u )

J ' (u 0 ), u u 0 o inf, u U .

Finally, J (u 0 ) 0

inf J 0 (u ) . uU

3 . The next approximation u1

u 0 D 0 u 0 u 0 , 0 d D 0 d 1 .

Since U is convex set, then the point u1 U . In the general case 23

u n D n un u n , 0 d D n d 1, n 0, 1, 2, ...,

u n1

(2)

where J n (u n )

inf J n (u ), uU

J ' (u n ), u u n .

J n (u )

Since the set U is compact, J n (u ) is a linear function, then there exists a point u n U . Theorem 1. If a function J (u ) C 1 (U ) , U is a convex closed bounded set, the gradient J ' u satisfies to Lipschitz condition on U and the sequence ^un `is determined by the formula (2), where the number D n is

determined by the condition g n (D n )

min g n (D ), g n (D )

0 dD d1

n

J u n D (u n u n ) ,

0, 1, 2, ... ,

(3)

then J n (u n )

J ' (u n ), u n u n o 0 at n o f .

(4)

If, in addition, J (u ) C 1 (U )

is convex on U , then the sequence ^un ` is minimizing to the problem (1) and any limit point belongs to the set U* , U* z . It is faithful the estimation 0 d an

c J (u n ) J * d , n

n 1, 2, ... ,

(5)

where J (u* )

J* , c

const ! 0 .

Proof. Since the set U is compact, J (u ) C 1 (U ) ,

then the set U * z . Since the set U is bounded, its diameter D f . From condition (3) follows that J un1 d J un D un un .

Then by Lemma (Lecture 1) the inequality 24

J (un ) J (un 1 ) t t J (un ) J un D (un un ) t

D2

Lu

2

u un un

2

t D J n (u n )

D2 2

LD 2 ,

0 d D d 1, n

0, 1, ... ,

(6)

holds, where J ' (u n ), u n u n , J n (u n ) 0 ,

u n u n d D, J n (u n )

since min J n (u ) d J n (u n )

J n (u n )

uU

0.

If J n (u n ) 0 , the first part of the theorem is proved. From expressions (6) we get 0 d J n (u n ) d

D

LD 2

2 0 d D d 1, n

We note, that since

J (u n ) J (u n 1 )

D

,

0, 1, 2, ....

g n (D n ) d g n (D ), 0 d D d 1,

then g n (D n ) d g n (0) .

Since g n (D n )

then the inequality

J ( u n 1 ), g n ( 0 )

J (u n ) ,

J (u n1 ) d J (u n )

holds. Hence, the sequence ^J (u n )` does not increase due to the fact that set U * z , J (u n ) t J * .

It follows that ^J (u n )` converges, i.e. J (u n ) J (u n1 ) o 0 at n o f .

Now, passing to the limit at n o f of (7), we get 0 d lim J n (u n ) d lim J n (u n ) d n of

n of

Hence, at D o 0 we obtain 25

D 2

LD 2 , 0 D d 1.

(7)

J n (u n ) o 0 at n o f .

The validity of (4) is proved. The first part of the theorem is proved. Let, in addition of these conditions, the function J (u ) is convex on U . We show that the sequence ^u n ` U is minimizing. In this case, as in the proof of the previous theorems, the inequality 0 d a n d J (u n ) J (u* ) d J ' (u n ), u n u*

J n (u* ) d J n (u n ),

n

0,1,2,...

(8)

holds. Hence, in force of proved above J n (u n ) o 0 , n o f ,

we get lim J (u n ) nof

J*

J (u * ) .

This means that sequence ^u n ` U is minimizing. We show, the fairness of estimation (5). From (6), (8), we get an an 1 t D J n (un )

D2

2 an d J n (un ), 0 d D d 1.

LD 2 ,

(9)

We note, that maximum of the function D J n (u n ) D 2 LD 2 / 2

on D is achieved at

D

and at

D

J n (u n ) / LD 2 ,

n t n0 , 0 d D d 1 ,

since J n (u n ) o 0 at n o f . Substituting the value D D of (9), we get 1 2 J n (un ) , 2 2 LD n t n0 , an d J n (un ).

a n a n 1 t

(10)

From (10) follows, that a n a n 1 t a n2 / 2 LD 2 , n t n 0 .

Then, according to Lemma 2 (Lecture 1), the estimation a n 2 LD 2 n 0 1 / n

is true for all n ! n0 . It follows that there exists a constant c ! 0 , such that the estimation (5) is true at all n 1, 2,... . Theorem is proved. 26

The conjugate gradient method. More efficient method for searching minimum of a quadratic function on E n , than the gradient, is the method of conjugate gradients. It is advisable to apply in the neighborhood of the minimum point of a smooth function, since the original function with a sufficient degree of accuracy can be approximated by a quadratic function in the neighborhood of the minimum point. We consider the following problem (11)

J (u ) o inf, u U { E n ,

where the function J (u ) C 1 ( E n ) .

The method algorithm: 10 . Select the starting point u 0 E n . 20 . Construct a sequence ^un ` by the rule u n D n pn , n

un1

where p0

J ' (u 0 ), p n

(12)

0, 1, ... ,

J ' (u n ) E n p n1 , n 1, 2, ... .

(13)

Numerical sequences ^D n `, ^E n ` are selected as follows: g n (D n ) g n (D )

En

min g n (D ), D t0

J (u n D p n ), n

0, 1, 2, ... ,

J ' (u n ), J ' (u n 1 ) J ' (u n ) , n I1 , ° 2 ® J ' (u n 1 ) ° 0 n I2, ¯

(14)

(15)

where the sets of indices I1 , I 2 are chosen that I1 I 2

In particular, if I2

^0,1,2,...`,

0 I2 .

^0, 1, 2,...`, I1

,

then E n 0 for all n , and we get the steepest descent method. In depending on the choice of indices sets I1 , I 2 the different variants of the conjugate gradient method can be obtained. Further, we consider the method of conjugate gradients for function J (u )

1 Au, u b, u 2 27

1 u ' Au b' u, 2

(16)

where A A* ! 0 is the symmetric positive definite matrix of n u n order, b E n is a prescribed vector. For the function (16) the sequences ^u n ` E n defined by (12) - (15), are written as follows: 1) Since g n (D )

J (u n D p n )

1 2 D Apn , pn D J ' (u n ), p n J (u n ) 2

then the number D n , where g n (D n )

min g n (D n ) D t0

is determined by the formula: g ' n (D n ) 0 : Dn

J ' (u n ), pn

J ' (u n ), J ' (u n ) E n pn1

Apn , pn

Apn , pn J ' (u n )

2

! 0,

Ap n , pn

(17)

In fact, for n 0 the scalar product 2

J ' (u 0 ) .

J ' (u 0 ), p 0

We note, that J ' (u n 1 ), p n

0, n

0, 1, 2, ... .

The proof is given below (see Lemma). 2) Since J ' (u n ) J ' (u n1 ) D n Apn , n

0, 1, ... ,

then J ' (u n ), J ' (u n 1 ) J ' (u n )

D n1 Ap n1 , p n1 , n 1, 2, ... ..

Hence J ' (u n ), Apn 1

En

Apn1 , p n1

, n 1, 2, ... .

(18)

Lemma. If the indices of the sets I1

^1, 2,...`,

I2

^0`,

then the sequence ^un ` defined by (12) - (14) for the function J (u ) of (16) satisfies to the conditions 28

0, i ! j;

J ' (u i ), p j

J ' (u i ), J ' (u j )

0, i z j ,

0, i z j , i, j

pi , Ap j

0, 1, 2,... .

(19)

Proof. We show, that J ' (u n 1 ), p n

0,

n 1, 2, ... .

We prove by mathematical induction method. For n 0 we get J ' (u1 ), p 0

Au1 b, Au 0 b

J ' (u1 ), J ' (u 0 )

Au 0 b, Au 0 b D 0 Ap 0 , Au 0 b

p0 , p0 D 0 Ap0 , p0 ,

where Au b, p0

J ' (u )

J ' (u 0 )

Au0 b .

According to the first relation of (17) the number D0

J ' (u 0 ), p 0 / Ap0 , p 0 p0 , p0 / Ap0 , p0 .

Substituting the value D 0 to the right-hand side of (20), we obtain J ' (u1 ), p0

0.

J ' (u n 1 ), p n

0

Let the equality

is true for some n ! 0 . We show, that it holds for n 1 . We note, that g ' n (D )

D Ap n , p n J ' (un ), pn ,

and g ' n (D n )

0.

Then g ' n (D n ) D n Ap n , p n J ' (u n ), p n J ' (u n ) D n Ap n , p n

D n Ap n J ' (u n ), p n

J ' (u n 1 ), p n

since J ' (u n1 )

J ' (u n ) Aun b, Au n1 b A(u n D n p n )

b

J ' (u n ) D n Ap n .

If D n 0 , then u n1

un , 0

g ' n (0) 29

J ' (un ), pn =

0,

(20)

2

J ' (u n ), J ' (u n ) E n p n 1

J ' (u n ) {

2

{ J ' (u n 1 ) d 0,

therefore, J ' (u n 1 ) { 0 .

Thus faithfulness of the equality J ' (un 1 ), pn

0, n

(21)

0, 1, 2,...

is proved. From (21) follows that 2

J ' (u n 1 ), J ' (u n 1 ) E n 1 p n n

0,1,2,...

J ' (u n 1 ) ,

J ' (u n 1 ), p n 1

(22)

On the basis of (21) and (22) it is easy to define pn

2

pn , pn 2

2

2

J ' (u n ) E n2 p n 1 t J ' (u n ) , n 1, 2, ...

J ' (u n ) E n p n 1 , J ' (u n ) E n p n 1

(23)

We show, that the equalities (19) are true. We prove by mathematical induction method. The equality J ' (u1 ), p0

0

is proved above, at i 1, j 0 , and Ap1 , p0

J ' (u1 ) E1 p0 , Ap0

p1 , Ap0

0

follows from (18). Let equations (19) be true for all i, j d n, n t 1 . We show, that they are true for i, j d n 1 . The equality J ' (u n 1 ), p n

0 at i

n 1, j

n

follows directly from (21) and (22). Since

J ' (u n ) J ' (u n1 ) D n Apn ,

then at j n J ' (u n 1 ), p j

J ' (u n ) D n Ap n , p j

J ' (u n ), p j D n Apn , p j

0

(24)

by induction assumption. Then J ' (u n 1 ), J ' (u j )

J ' (u n 1 ), p j E j p j 1

30

J ' (u n 1 ), p j E j J ' (u n 1 ), p j 1

0

by expression (24). The scalar product Ap n 1 , p j

p n 1 , Ap j

J ' (u n 1 ) E n 1 p n , Ap j

1

J ' (u n 1 ), Ap j

Dj

J ' (u n 1 ), J ' (u j ) J ' (u j 1 )

0,

since J ' (u n 1 ), J ' (u n )

J ' (u n 1 ), p n E n p n 1

0 (the case j

n ).

Finally, J ' (u n 1 ) E n 1 p n , Ap n

p n 1 , Ap n

0

by formula (18). Thus, it is shown, that J ' (u n 1 ), pi

J ' (u n 1 ), J ' (u i )

p n 1 , Api

0, i

0, 1, ..., n.

The lemma is proved. Theorem 2. If the indexes of the sets

^1, 2, 3, ...`,

I1

I2

^0`,

then the sequence ^un ` for the problem (11) with the function J (u ) of (16) converges to the minimum point no more than n steps. Proof. According to Lemma the vectors J ' (u i ), J ' ( u j ), i z j

are orthogonal. Such non-zero vectors in E n no more n . Consequently, there is a number m, 0 d m d n , such that J ' (u m ) 0 . This means that u*

u m , since the function

J (u ) C 1 E n

is convex. The theorem is proved. Control questions 1. Describe the conditional gradient method. 2. Prove the Theorem 1. 3. Describe the conjugate gradient method. 4. Prove the Lemma. 5. Prove the Theorem 2. 6. What difference between the conditional gradient and the conjugate gradient methods?

31

LECTURE 4. NEWTON'S METHOD. PENALTY FUNCTION METHOD. LAGRANGE MULTIPLIER METHOD Newton's method is applicable for finding the minimum J (u ) C 2 U

on the set U of En. It is advisable to adopt Newton's method in the final stage of search minimum J(ɢ) on U , when approximation is close enough to the point u* . The method of penalty functions is one of the common and therefore timeconsuming methods of solving extreme problems. It consists in replacing the original problem by auxiliary, which is solved by known numerical methods. Lagrange multiplier method is based on finding a saddle point of the Lagrange function. Newton's method. We consider the optimization problem (1)

J (u ) o inf, u U ,

where function

J u C 2 U ,

. For solving of the problem (1) the sequence ^un ` U is constructed by the following algorithm. The algorithm of the method: 1°. Select the starting point u0 U . Since the function J (u ) C 2 U , then n

U is given set of E

J (u ) J (u 0 )

J c(u0 ),u u0

1 2 J cc(u0 )(u u0 ), u u0 o u u0 . 2

2°. The point u1 U is determined by solving the following optimization problem: J 0 (u ) o min, u U ,

where J 0 (u )

J c(u 0 ), u u 0

1 J cc(u 0 )(u u 0 ), u u 0 . 2

In general, the point un1 is determined by the condition J n (u ) o min, u U ,

where the function 32

(2)

J c(u n ), u u n

J n (u)

1 J cc(u n )(u u n ), u u n . 2

a) If U { E n , the point u n1 can be determined by the solution of an algebraic equation J c(u n1 ) J c(u n ) J cc(un )(un 1 un ) 0 . Hence, in case when the matrix J cc(u n ) is nonsingular, we get u n >J cc(u n )@ J c(u n ), n 1

0,1,2,...

u n 1

(3)

b) In the case U { E n , the problem (2) can be solved by the conjugate gradient method. Theorem 1. If the function J u C 2 U , U is a closed convex set and 2

J cc(u )[ , [ t P [ , [ E n , u U ,

J cc(u ) J cc(v) d L u v , u, v U ,

then (2) is solvable. If, in addition, the value q

L u u

0

/ P 1,

the estimation un u* d

P

f

¦ q2 d

Lk

k

n

P

n

q2 n 0, 1,... n , L 1 q2

(4)

is valid, where J (u* )

Proof. As the function

min J (u ) . uU

J (u ) C 2 U

and 2

J cc(u )[ , [ t P [ , P ! 0, [ E n , u U ,

then by Theorem 6 (Lecture 4, formula (9)) [15] J (u ) is strongly convex function on U . It can be shown that a strongly convex function J (u ) achieves the lower bound on the closed convex set, moreover in a point u* U . Since the matrix J cc(u ) is positive definite at all u U and J cc(u n )

then the function

J ncc (u n ) ,

J n (u ), u U 33

is also strongly convex and gets the lower bound on U in a single point un1 U . The first part of the theorem is proved. Since the function J n (u ) is strongly convex on U and the global minimum is reached in the point un1 U , it is necessary and sufficiently to perform the inequality

0, 1, 2, ...

J ' n (u n 1 ), u u n 1 t 0,

u U , n

Taking into account that J ' (u n ) J cc(u n )(u u n ) ,

J ' n (u )

we obtain u U , n

0, 1, ...

J ' (u n ) J cc(u n )(u n1 u n ), u u n1 t 0,

(5)

If un1 un , then from (5) follows J ' (u n ), u u n 1 t 0, u U .

This means that un u* . We suppose that u n z u n 1 , n

0, 1, ... .

From inequality (5) at u u n we obtain J cc(u n )(u n 1 u n ), u n 1 u n d J ' (u n ), u n u n 1 ,

n

0, 1, ... .

Taking into account that P [ d J cc(u )u [ , [ , [ E n , u U 2

at [ un1 un we get 2

P u n 1 u n d J cc(u n )(u n 1 u n ), u n 1 u n d d J ' (un ), un un1 ,

From inequality

J 'n u n1 , u u n1 t 0, u U

after replacing n by n 1 (n t 1) we get J ' n 1 (u n ), u u n t 0, u U .

Hence, at u u n1 we get J ' n 1 (u n ), u n u n 1 d 0 . 34

n 0, 1, ... .

(6)

J ' (u n ), u n u n1 d J ' (u n ) J ' n 1 (u n ), u n u n 1 , n 1, 2,..

Then

(7)

Since derivative J ' n 1 (u n )

J ' (u n 1 ) J cc(u n 1 ) u u (un un 1 ) ,

then (7) is written as J ' (u n ), u n u n 1 d d J ' (un ) J ' (un 1 ) J cc(un 1 )(un un1 ), un un1 .

Taking into account that J cc(u n1 T (u n

J ' (u n ) J ' (u n1 )

u n 1 ))(u n u n 1 ), u n u n1 ,

where 0 d T d 1 , we obtain J ' (u n ), u n u n 1 d >J cc(u n1 T (u n u n 1 )) J cc(u n 1 )@u 2

u (u n u n 1 ), u n u n 1 d L u n u n 1 u n u n 1 , n 1, 2, ... ,

due to the fact that J cc(u ) J cc(v) d L u v , u, v U .

Now (6) can be written as: 2

P u n 1 u n d J cc(u n )(u n 1 u n ), u n 1 u n d 2

d L u n un 1 u n un 1 , n 1, 2,... .

Hence we get u n1 u n d

L

P

2

u n u n1 , n 1, 2,... .

(8)

From (8) at n 1 follows u 2 u1 d

L

P

u3 u 2 d

u1 u0 L

P

P

2

L

u 2 u1

In the general case, u n1 u n d

P L

q 2 at n 2 ,

P

2

L

q 4 , etc.

n

(9)

q 2 , n 1, 2,... .

We show, that the sequence ^uk ` is fundamental. In fact, for any m, n, m ! n , we have m

m

k n

k n

u m u n d ¦ u k 1 u k d ¦

P L

It follows, that 35

f

q2 d ¦ k

k n

P L

k

q2 d

P q2

n

L 1 q2

n

.

u m u n o 0 at n o f m ! n ,

consequently, ^u n ` U is fundamental, i.e. u n o u* at n o f ,

in addition u* U by the closure of the set U . From the inequality under fixed n and at m o f we get f

u n u* d P ¦ q 2 / L . k

k n

The validity of the estimation (4) is proved. It remains to show that u* U * . From the formula (5) at n o f we have J ' (u* ), u u* t 0, u U .

This means that u* U * . Theorem is proved. The method of penalty functions. We consider the optimization problem (10)

J (u ) o inf,

u U

^u E

n

u U 0 , g i (u ) d 0, i 1, m; g i (u ) 0, i

`

m 1, s ,

(11)

where U 0 is prescribed set of E n and the functions J (u ), g i (u ), i 1, s are defined on U0 . Definition. By penalty or penalty function of the set U on the set U 0 is called a sequence of functions ^Pk u ` defined on the set U 0 , if 0 at u U ; lim Pk u ® k of ¯f at u U 0 \ U .

For problem (1), (2) as a penalty function can take m

Pk (u )

Ak P (u ), P (u )

¦ ^maxg (u );0 ` i

i 1

u U 0 , Ak ! 0, k

1, 2, ..., lim Ak k of

p

s

¦ g (u )

i m 1

i

p

,

f,

where p t 1 is given number. We note, that if u U , that P (u ) 0 , consequently Pk (u )

Ak P(u ) 36

0,

(12)

and if u U 0 \ U , then P (u ) ! 0 and Ak P(u ) o f at k o f , since Ak o f at k o f .

Pk (u )

The values Ak , k 1, 2, ... are called by penalty coefficients. Penalty functions of sets U can be different, one of them is shown in (12). In the penalty function method the original problem (10), (11) is replaced by auxiliary problem ) k (u )

J (u ) Pk (u ) o inf, u U 0 , k

1, 2, ... .

(13)

Further, the problem (13) at the fixed k is solved by one of the numerical minimization methods represented above. Let ^u nk ` U 0 be such, that for each k 1, 2, ... the limit lim ) k u nk ) k * nof

inf ) k (u ) ! f, lim u nk

uU 0

nof

uk .

Then the sequence ^u k ` U 0 will be closer to the set U and lim J (u k ) k of

J ** , lim U (u k , U * ) k of

0,

since lim Pk (u )

0

k of

at u U 0 \ U (see formula (12)). Theorem 2. If the functions J (u ), g i (u ), i 1, s are defined on the set U 0 and the sequence ^u k ` U 0 satisfies to condition ) k (u k ) d ) k * H k , H k ! 0, lim H k k of

0,

then limJ (u k ) d lim ) k (u k ) k of

k of

lim ) k * d J ** . k of

If, moreover J*

inf J (u ) ! f , uU

then P (u k )

o( Ak1 ), k

1, 2, ...,

lim g i (u k ) d 0, i 1, m ; lim g i (u k ) k of

k of

0, i

m 1, s.

Proof of the theorem and other issues of penalty functions method can be found in the book: Vasil'ev F.P. Chislenye metody reshenia extremalnih zadach. M.: Nauka, 1980. The method of Lagrange multipliers. We consider the optimization problem (10), (11). Lagrange's function for the problem (10), (11) has the form 37

s

L (u , O )

J (u ) ¦ Oi g i (u ), u U 0 ,

O /0

^O E

(14)

i 1

s

/ O`1 t 0,..., O m t 0`.

We suppose, that the set U * z and Lagrange's function (14) has a saddle point u* , O U 0 u / 0 . By definition of saddle point u* , O* , it follows that *

L u * , O d L u * , O*

J * d L u , O* , u U 0 , O / 0 .

(15)

We note, that if a couple u * , O* is a saddle point, i.e., satisfies to the inequalities (15), then u* U * (see main theorem of Lecture 8 [15]) is the solution of (10), (11). Thus, the solution of (10), (11) is reduced to searching of the saddle point u* , O* satisfying (15). In particular, a saddle point

u , O U *

*

0

u /0

may be defined by the method of the gradient projection, through building the sequences ^u n `, ^On ` by the rule u n 1

O n 1

PU 0 u n D n L u u n , O n ,

P/ 0 O n D n L O u n , O n n

n

0, 1, 2,...,

P/ 0 O n D n g u n ,

(16) (17)

0, 1, 2, ...,

where L u (u 0 , On ) L O (u 0 , O n )

J ' (u n ) On , g ' (u n ) ,

g (u n ) g1 (u n ),..., g s (u n ) , n 0, 1, 2,...

Since the minimum L (u , O ) on u U 0 is determined, then in (16) the direction on anti-gradient is chosen and the maximum L (u, O ) is searched by O / 0 , whereas in (17) the search of the next point is carried out in the direction of the gradient function L (u , O ) by O . In the case of design points on the sets U 0 , / 0 , respectively, the search of the saddle point can be solved by penalty functions method (see the literature mentioned above.) Control questions 1. Describe Newton's method. 2. Prove the Theorem 1. 3. Describe the method of penalty functions. 4. Give definition to the penalty function. 5. Describe the method of Lagrange multipliers.

38

LECTURE 5. REVIEW ON OTHER NUMERICAL MINIMIZATION METHODS Algorithms of the methods of possible directions, coordinate descent, embedded functions and general comments to the numerical methods for solving extreme problems are presented. Literature on specific issues of numerical methods is provided. The method of feasible directions. We consider the optimization problem J (u ) o inf; u U

where

^u E

n

`

g i (u ) d 0, i 1, m ,

(1)

J (u ) C 1 U , g i (u ) C 1 U , i 1, m .

To solve the problem (1) is proposed to construct a sequence ^un ` U by the following algorithm. The algorithm of the method: 1°. Select the starting point u 0 U . We define a set of indices

^i

I0

1 d i d m, g i (u 0 )

0`.

2°. The possible direction e0 E n , e0 z 0 in the point u 0 U from the condition J (u 0 D 0 e0 ) J (u 0 ), 0 D 0 E 0 , E 0 ! 0, u 0 D 0 e0 U

is determined. It can be shown that the possible direction e0 E n is a solution of the linear programming problem: V o inf, ^(e,V ) E J c(u0 ), e d V ,

(e, V ) W0

n 1

i I0 , e

3°. The next point E0

g ic(u0 ), e d V ,

`

(e1 ,..., en ), e j d 1, j 1, n .

u1 u0 D 0 e0 , 0 D 0 d E 0 , sup^u0 te0 U , 0 d t d D ` ! 0. D

In the general case, u n 1

u n D n en , 0 D n E n , n

En

0,1, 2,...,

sup ^u n ten U , 0 d t d D `, D

where 39

(2)

en

e ,e ,...,e E n 1

n 2

n n

n

is the possible direction which is found of the solution of linear programming problem V o inf, ^(e,V ) E n1 J c(u n ), e d V ,

(e, V ) Wn

g ic(un ), e d V , i I n , e nj d 1, j 1, n, I n

^i

`.

1 d i d m, g i (u n ) 0`

(3)

Theorem 1. If the function J (u ) C 1 U , g i (u ) C 1 U , i 1, m ,

set U * z , that in the point u* U * problem (3) ( u n is replaced by u* , Wn , on W* , I n on the set of indices I * ^i 1 d i d m, g i (u* ) 0`) has a solution e* ,V * such that V * 0 inf V . W*

If, in addition, J (u ), g i (u ), i 1, m are convex on E n , and the set U is regular, that every point u* U * for which V * 0 is the solution of problem (1). Thus, if the vector en 0 for some n , that in the point u n1 u n the necessary conditions for minimum J (u ) on U hold, and in the case if J (u ) g i (u ), i 1, m

are convex functions and set U is regular, that minimum J (u ) on U , i.e. u n u* U * is reached in the point u n1 u n . The method of coordinate descent. We consider the problem J (u ) o inf, u U { E n ,

(4)

where J (u ) E n , in general case is not differentiable. The difference of the method of coordinate descent from all previous consists in that the minimum J (u) , u E n is determined by computing value of the function J (u ) . Sequence ^u k ` E n is constructed by the following algorithm for the problem (4). The algorithm of the method: 1°. Select the starting point u 0 E n and the number D 0 ! 0 which is the parameter of the method. We denote the unit vector as ek

0,...,0,1,0,... E n 40

with the k -coordinate equals to 1, and [k/n] is the integer part of the number k/n. Let the vector Pk e i , where ik k n>k [email protected] 1 . Hence we get k

e1 ,

P0

P1 e 2 , ..., Pn 1 e n , e1 ,

Pn

e 2 ,...,.

Pn 1 P2 n 1

n

e , P2 n

e1 ,...

2°. The next point is chosen as follows: u1

if J u 0 D 0 P0 J (u1 ) ; or u1

if J u 0 D 0 P0 J (u 0 ) . In the general case, if

u 0 D 0 P0 ,

u0 D 0 P0 ,

J u k D k Pk J (u k ) ,

then u k D k Pk , D k 1

u k 1

If the inequality

Dk ;

(5)

J u k D k Pk J (u k )

is not executed, then we calculate the value J u k D k Pk

and verify the inequality

J u k D k Pk J (u k ) .

If this inequality is satisfied, then Dk.

(6)

O D k at ik n, u k u k n 1 , ® ¯D k at ik z n, or u k z u k n 1 ; 0 d k d n 1,

(7)

u k 1

u k D k Pk , D k 1

If either condition is not satisfied (5) or (6), then we accept u k 1

D k 1

41

uk ,

where number O ! 0, 0 O 1 ; k is the number of iteration; n is dimension of ȿn . One cycle of ɩ iteration allows to carry out the descent by all components of the vector u E n . If at least one of the inequalities (5), (6) is satisfied, then D k D k 1 , in the opposite case the step D k ! 0 is fractioned. Theorem 2. If the function J (u ) C 1 E n

is convex on E , the set n

M (u 0 )

^u E

n

`

J (u ) d J ( u 0 )

is bounded, then the sequence ^u k ` E n of (5) - (7) converges to the set U * . We note, although the theorem is formulated for the function J (u ) C 1 E n ,

but numerical method to minimize (5) - (7) does not require the calculation of the derivative, which significantly reduces the amount of computing work. Method of embedded functions. We consider the problem

u U

^u E

J (u ) o inf; n

g i (u ) d 0, i 1, m, g i (u )

0, i

`

(8)

, u U 0 ,

(9)

m 1, s ,

where J (u ), g i (u ), i 1, s

are the functions defined on the set U 0 of ȿn . We denote by m

P (u )

¦ max^g (u );0`

pi

i

i 1

s

¦ g (u )

i m 1

i

pi

where pi t 1, i 1, s . We consider a family of functions of the parameter t f t f , defined by the formula p (10) Ɏ (u , t ) L max^J (u ) t ; 0` MP (u ), u U 0 , 0

where L ! 0, M !0 are the fixed parameters; p0 t 1 . We denote U (t )

inf Ɏ(u, t ),

uU 0

f t f .

If for the problem (1) the set U * z , then for the point u* U * , t * 42

J * ! f

(11)

value

Ɏu* , t* 0 ,

since g i (u* ) d 0, i 1, m,

m 1, s

g i (u* ) 0, i

consequently P(u* )

0,

and J*

J (u* )

t*

(see formulae (9) - (10)). As it follows from (9), for any u U 0 function P (u ) t 0 .

Then from (10) we get

Ɏu, t t 0, u U 0 , f t f.

Since for all t t*

max^J u t , 0`! 0

J* and u U 0 , then Ɏu , t ! 0, u U 0 , t J * .

Thus, in many cases the value J (u* ) ,

J*

u* U * is the smallest root of the equation

U (t )

0.

(12)

Thus, in the case, the set U * z , J (u* )

J * ! f , u* U * ,

search of the value J * reduces to determining of the smallest root of the equation (12). This problem can be solved by a method of minimizing the function of one variable (Lecture 1). Here we consider the most frequently used in practice numerical methods for solving of extreme problems. There are other methods, such as method with cubic rate of convergence, random search method, the method of barrier functions, method for finding the global minimum. It should be noted, that for application of a method of minimization for solving of a particular extreme problems, in addition to meeting its prerequisites and the speed of convergence, it is necessary to pay attention to the total amount of computation and computer time required to obtain the solution with reasonable accuracy. Under solving of the same optimization problem, at different stages of its solution is advisable to use a suitable numerical methods for minimization. This 43

feature is available under working with minimization program in interactive mode. Solving by numerical methods of optimization problems of large size (n is a sufficiently large number) is not possible, even with today's most powerful computers. In such cases it is advisable to break the original problem into subproblems of small dimension, loosely connected. In connection with the considered above, the development of common research methodology of convergence of the numerical methods represents an interest. There are excellent textbooks on numerical methods for solving optimization problems in finite-dimensional space in the native literature. Proofs of Theorems 1, 2 and other optimization issues can be found in the works [4, 5, 14]. Control questions 1. Describe the method of feasible directions. 2. Formulate the Theorem 1. 3. Describe the method of coordinate descent. 4. Formulate the Theorem 2. 5. Describe the method of embedded functions.

CHAPTER II VARIATION CALCULUS Basic theory and numerical methods for determination of the least value of a function J (u ) on a given set U of E n , where E n is the space of n vectors u u1 ,..., u n have been studied in the previous lectures. Let instead E n we take the space of continuously differentiable functions u (t ) on the interval >t 0 , t1 @ and denote C 1 >t0 , t1 @ , and as U

^u (t ) C >t , t @ u (t 1

0

1

0

)

u 0 , u (t 1 )

u1

`

which is a set of C 1 >t0 , t1 @ . To compare each element u (t ) U a number of Ǽ1, should take a definite integral t1

J (u )

³ F (t , u(t ), u (t ))dt 0

t0

instead of J (u ) . As a result, we obtain an optimization problem t1

J (u )

³ F (t, u(t ), u (t ))dt o inf, 0

u U C 1 >t 0 , t1 @

t0

again. In variation calculus the base of the theory of solutions of this optimization problem is studied. LECTURE 6. BRACHISTOCHRONE PROBLEM. THE SIMPLEST TASK. NECESSARY CONDITIONS FOR A WEAK MINIMUM. LAGRANGE LEMMA. EULER EQUATION Brachistochrone problem. I.Bernoulli proposed the following problem for all mathematicians of the world in 1696. There are two points A and B located on different levels are prescibed in the vertical plane. It is required to connect them such smooth line, moving along which the body by gravity traverses the path from A to B in the shortest possible time. It is said that a solution to this problem was given by I. Bernoulli, G.V. Leibniz, Ya. Bernoulli and another author without signature. It later emerged that the solution no sign gave I.Newton. Let the chosen coordinate system has its beginning in the point Ⱥ(0, 0), axis x is horizontal, axis ɭ is vertically downwards and the point Ⱥ is above the point B x0 , y0 45

. Let M x, y be a point on the origin curve y=y(x) passing through the points A and B. Since there is no friction, the sum of the kinetic T and potential energy P is constant in any point of the curve y y (x), 0 d x d x0 . Since in the point A the body is at rest, then T 3 A 0 , and in the point Ȃ the sum

T 3 M

0.

mv 2 / 2 mgy

Hence we determine the speed in the point M equals to v M instantaneous speed 1 y ( x) dx / dt ,

ds / dt

vM

2 gy . Since the

then 1 y 2 dx / dt .

2 gy( x)

vM

Hence (1 y 2 ( x)) / gy ( x) dx .

dt

Hence we get x0

T

³ 0

1 y 2 ( x) dx. 2 gy ( x)

(1)

Equation (1) defines the time of movement of the body for any curve y 0 d x d x0 passing through points A and B. We note, that the function y ( x ) C 1 >0, x 0 @ , set U ^y ( x ) C 1 >0, x 0 @ y (0) 0, y ( x 0 ) y 0 ` ,

y (x ),

and the task is written as follows: x0

T ( y)

³ 0

1 y 2 ( x) dx o inf, y 2 gy ( x)

(2)

y ( x) U .

The simplest problem. By generalization of the problem (2) is the simplest problem. Minimize the functional t1

J ( x, x )

³ F ( x, x, t )dt o inf

(3)

t0

on set U

^xt C >t , t @ xt 1

0

1

0

x 0 , xt1

We note, that if ɯ, t are replaced by ɭ, x respectively, and 46

`

x1 .

(4)

F ( x, x, t )

F ( y, y , x)

(1 y 2 ( x)) / 2 gy ( x) ,

then problem (2) follows from (3), (4). By introducing the notations x (t )

u (t ), t [t 0 , t1 ],

where u (t ), t [t 0 , t1 ] is continuous function, problem (3), (4) can be written as t1

³ F ( x, u, t )dt o inf,

J ( x, u )

x(t ) U .

(5)

t0

Definition 1. It is said, that function x 0 (t ) U delivers a strong local minimum to the functional J ( x, u ) in the problem (5) (or (3), (4)), if there exists the number H ! 0 , such that for any feasible function x(t ) U for which max | x(t ) x 0 (t ) | H , t0 dt dt1

the inequality J ( x, u ) t J ( x 0 , u 0 ) ,

where t 0 d t d t1 , u 0

u 0 (t )

x 0 (t ), t [t 0 , t1 ]

holds. Definition 2. It is said, that function x 0 (t ) U delivers a weak local minimum to the functional J ( x, u ) in the problem (5) (or (3), (4)), if there exists the number İ > 0 such that for any feasible function x(t ) U for which max | x(t ) x 0 (t ) | H , max | x (t ) x 0 (t ) | H , t0 dt dt1

t0 dt dt1

(6)

inequality J ( x, u ) t J ( x 0 , u 0 )

holds. We note, that in definition the value of functional J ( x, u ) is compared with the value J ( x 0 , u 0 ) on the set of feasible functions located in the İ-neighborhood x 0 for each t , t0 d t d t1 , and in the definition 2 such comparison is carried out on the set of feasible functions such that x(t) and x (t ) are in the İ-neighborhood for each t, t >t 0 , t1 @ , respectively x 0 (t ) and x 0 (t ) . Therefore, the necessary conditions for a weak local minimum in the point x 0 (t ) U will be necessary conditions for strong local minimum. On the other hand, if in the point x 0 (t ) U the strong local minimum is reached, then on it a weak local minimum will be reached, so as a condition J ( x, u ) t J ( x 0 , u 0 ) x (t ) U , max | x(t ) x 0 (t ) | H , t0 dt dt1

47

in particular, for those x(t ) U holds, for which max | x (t ) x 0 (t ) | H . t0 dt dt1

The converse is not true. Necessary conditions for a weak local minimum. As follows from the inclusion x (t ) U , x (t ) C 1 >t 0 , t1 @, x(t 0 ) x0 , x(t1 ) x1 . Therefore, it is advisable to choose feasible function x(t ) U in the form x(t )

x 0 (t ) J h(t ), t >t 0 , t1 @ ,

where h(t 0 )

h(t ) C 1 >t 0 , t1 @,

h(t1 )

0, J ! 0 are some number.

Thus, the function

^h(t ) C >t , t @

h(t ) U 0

1

0

1

h(t 0 )

h(t1 )

0`.

Since the difference x(t ) x 0 (t )

J h(t ) , h(t ) U 0 ,

then by selecting the number Ȗ the fullfilment of the inequalities given in the definitions 1, 2 can be provided. Let the function F(x,ɢ,t) is twice continuously differentiable in all arguments ( x, u, t ) E 1 u E 1 u >t 0 ,t1 @ . Then the difference

³ >F ( x

t1

J ( x 0 Jh, u 0 Jh) J ( x 0 , u 0 )

0

(t ) Jh(t ), x 0 (t ) Jh(t ), t )

t0 t1

ª wF ( x , x 0 , t ) wF ( x 0 , x 0 , t ) º F ( x 0 (t ), x 0 (t ), t ) dt J ³ « h(t ) h(t )»dt wx wx ¼ t0 ¬

@

J 2 ª w2 F ( x0 , x 0 , t ) 2 «¬

wx2

0

h2 (t ) 2

w2 F ( x0 , x 0 , t ) h(t )h(t ) wxwx

w 2 F ( x 0 , x 0 , t ) 2 º h (t )»dt o(J 2 ), wx 2 ¼

(7)

where x0 =x0(t)U. By introducing the notation GJ

t1

ª wF ( x 0 , x 0 , t ) wF ( x 0 , x 0 , t ) º h(t ) h(t )»dt , wx wx ¼ t0

³ «¬

48

(8)

t1

ª w 2 F ( x 0 , x 0 , t ) 2 w 2 F ( x 0 , x 0 , t ) h (t ) 2 h(t )h(t ) 2 wxwx wx t0

³ «¬

G 2J

w 2 F ( x 0 , x 0 , t ) 2 º h (t ) » dt , wx 2 ¼

(9)

relation (7) is written as 'J

J ( x 0 Jh, u 0 Jh) J ( x 0 , u 0 )

1 2

JGJ J 2G 2 J o(J 2 ) .

(10)

The value GJ determined by the formula (8) is called the first variation of the functional J ( x, u ) in the point x 0 U , and the value G 2 J of (9) is the second variation of the functional variation J ( x, u ) in the point x 0 U . Theorem 1. Let function x 0 (t ) U conducts a weak local minimum to the functional J ( x, u ) on the set U . Then the following conditions necessary hold: GJ =0, G 2 J t0.(11)

Proof. Let for some number Ȗ and function h(t ) U 0 the inequalities (6) hold, where x(t )

x 0 (t ) J h(t ) U

and J ( x, u ) t J ( x 0 , u 0 ) .

We show, that the relations of (11) are true. According to the expression (10) the difference 1 ª º J ( x, u) J ( x 0 , u 0 ) J «GJ JG 2 J » o(J 2 ), 2 ¬ ¼

where J ! 0 is sufficiently small number; o(J 2 ) / J 2 o 0 at J o 0 . It follows that if GJ z0, then we can always choose a number Ȗ the sign of which is opposite GJ , in addition J ( x, u ) J ( x 0 , u 0 ) 0 . This contradicts the condition J ( x, u ) t J ( x 0 , u 0 ) .

Consequently, GJ =0. Further, since GJ =0, then J ( x, u ) J ( x 0 , u 0 )

It follows that if G 2 J t0 , t1 @ is a solution of equation (13). In fact, according to the necessary condition of the first order in the point x 0 (t ) U the first variation t1

ª wF x 0 , x 0 , t wF x 0 , x 0 , t º h(t ) h(t )» dt wx wx ¼ t0

³ «¬

GJ

0,

(14)

h U 0 .

After integration by parts the second term can be written as t1

wF ( x 0 , x 0 , t ) ³ wx h(t )dt t0

U dV

wF d wF dU , dt wx dt wx h(t )dt , V h(t ) t1

t

1 § d wF ( x 0 , x 0 , t ) · wF ( x 0 , x 0 , t ) ¸¸h(t ) dt h(t ) ³ ¨¨ wx dt wx ¹ t0 © t0

t

1 § d wF ( x 0 , x 0 , t ) · ¸¸h(t )dt , ³ ¨¨ wx dt ¹ t0 ©

h(t ) U 0 .

Then, the relation (14) can be written as GJ

t1

ª wF ( x 0 , x 0 , t ) d wF ( x 0 , x 0 , t ) º »h(t )dt wx dt wx ¼ t0

³ «¬

0, h(t ) U 0 .

Hence by Lagrange's lemma, where the function a(t )

wF ( x 0 , x 0 , t ) d wF ( x 0 , x 0 , t ) , wx dt wx

we get the equation (13). The theorem is proved. The functions x 0 (t ) are called by extremals along which the Euler equation holds. 51

Control questions 1. What is the main idea of brachistochrone problem? 2. Formulate the simplest problem. 3. Give definition to the strong local minimum to the functional. 4. Give definition to the weak local minimum to the functional. 5. Prove the Theorem 2. 6. Prove the Lagrange's lemma. 7. What kind equation is called by Euler equation? 8. Prove the Theorem 2.

LECTURE 7. DUBOIS - RAYMOND LEMMA. BOLZ PROBLEM. WEIERSTRASS NECESSARY CONDITION We derive the Euler equation on the basis of lemma Dubois-Raymond and analysis of the Euler equation. Necessary conditions for a weak local minimum for Bolz problem are obtained. Necessary conditions for strong local minimum for the simplest problem are considered. Dubois - Raymond lemma. If the function b(t ) , t >t 0 ,t1 @ is continuous and for arbitrary continuous function v(t ), t >t 0 , t1 @ equals to zero on average, i.e. t1

³ v(t )dt

0,

(1)

t0

the definite integral t1

³ b(t )v(t )dt

0,

(2)

t0

then b(t ) b0 const. Proof. We suppose the contrary, i.e. that there exist the points T and T of >t 0 , t1 @ such that b(T ) z b(T ) . To certainty we assume, that T T , b(T ) ! b(T ) .

We choose a sufficiently small number İ0 so that the segments '1

>T H 0 ,T H 0 @,

'2

>T H

0

,T H 0

@

do not overlap with each other, moreover ' 1 >t 0 , t1 @, ' 2 >t 0 , t1 @

and E1

min b(t ) ! max b(t ) t'1

t' 2

E2 .

Since the function b(t ) , t >t 0 ,t1 @ is continuous, the last inequality holds. We 52

construct the function v(t ), t >t 0 , t1 @ which is equal to zero on average [see formula (1)] as follows:

v (t )

(t T H 0 ) 2 ( t T H 0 ) 2 , at t '1 , ° t1 ° 2 2 ° (t T H 0 ) (t T H 0 ) , at t ' 2 , ³ v(t ) dt 0, ® t0 ° at t >t0 , t1 @\ ('1 ' 2 ). °0, °¯

Then the integral t1

³ b(t )v(t )dt ³ b(t )v(t )dt ³ b(t )v(t )dt t (E

1

'1

t0

'2

E 2 ) ³ v(t )dt ! 0 . '1

This contradicts (2). The lemma is proved. The second conclusion of the Euler equation can be obtained from DuboisReymond lemma. In fact, from the necessary condition of the first order GJ

t1

ª wF ( x 0 , x 0 , t ) wF ( x 0 , x 0 , t ) º h(t ) h(t )» dt wx wx ¼ t0

³ «¬

0, h(t ) U 0

after integration by parts of the first term t1

ª wF ( x 0 , x 0 , t ) º ³t «¬ wx h(t )»¼ dt 0 U dV

dU h(t )dt , wF ( x 0 ([ ), x 0 ([ ), [ ) d[ ³t wx 0

h(t ), wF dt , V wx

t1

t1 § t wF ( x 0 ([ ), x 0 ([ ), [ ) · d[ ¸dt ³ h(t )¨ ³ ¨t ¸ wx t0 ©0 ¹

we get GJ

ª wF ( x 0 , x 0 , t ) t wF ( x 0 ([ ), x 0 ([ ), [ ) º ³ d[ » h(t ) dt ³t «« wx w x t0 0 ¬ ¼» t1

Therefore, taking into account the fact that t1

³ h(t )dt

t0

i.e. v (t ) h(t ) and denoting 53

0,

0,

wF ( x 0 , x 0 , t ) wF ( x 0 ([ ), x 0 ([ ), [ ) d[ , ³ wx wx t0 t

b(t )

by Dubois-Raymond lemma we get the Euler equation in the form of DuboisRaymond in the following form t wF ( x 0 , x 0 , t ) wF ( x 0 ([ ), x 0 ([ ), [ ) ³ d[ wx wx t0

b0 .

(3)

The second term in (3) is differentiable by t and the right side is also differentiable, hence the first term is differentiable by t. Then from (3) we have the well-known Euler equation Fx ( x 0 , x 0 , t )

d Fx ( x 0 , x 0 , t ) 0 . dt

(4)

Equation (4) can be written in the form which is convenient for solving of the applied problems: Fx Fxt Fxx x 0 Fxx x0

0, x 0 (t 0 )

x1 .

x0 , x 0 (t1 )

(5)

Since the function F ( x, u, t ) is twice continuously differentiable in all the variables ( x, u , t ) E n u E n u >t0 ,t1 @ .

the second order differential equation (5) has a solution x 0 (t , c1 , c 2 ), t >t 0 ,t1 @,

x 0 (t 0 )

constants c1 , c 2 are determined from the condition x 0 (t 0 )

x1 .

x 0 , x 0 ( t1 )

Example 1. In the problem of the brachistochrone the function (1 x 2 ) / 2 gx .

F ( x, x , t )

Consequently, (1 x 2 )

Fx Fxx

2 x 2 gx

,

x

Fx

x 2 x 2 gx(1 x ) 2

2 gx(1 x 2 )

,

Fxx

Equation (5) is written as 54

,

Fxt

0,

1 (1 x ) 2 gx(1 x 2 ) 2

.

1 x 0

2

2 x 0 2 gx 0

x 0

2

2 x 0 2 gx 0 1 x 0

2

x0

1 x 2gx 1 x 02

02

0

0.

It follows, that 2

2 x 0 x0 x 0 1 0, x 0 (0)

0, x 0 (t1 )

x0 ,

where the points A(0, 0), B(t , x 0 ) . The solution of this differential equation is the arc of the cycloid. Example 2 (example of Hilbert). Minimize the functional 1

J ( x, x )

³t

2/3

x 2 (t )dt

0

at conditions x ( 0)

0, x(1) 1 .

In this case Fx

0, Fx

2t 2 / 3 x , Fxt

3t 1/ 3 x , Fxx

0, Fxx

3t 1/ 3 .

Euler's equation has the form 3t 1 / 3 x 0 3t 1 / 3 x 0 0 , x 0 (0) 0, x 0 (1) 1 .

The solution of the differential equation can be written as: x 0 (t ) c1 t 1 / 3 c 2 .

By the condition x 0 (0) 0, x 0 (1) 1 follows that c 2 0, c1 1 . Then x 0 (t ) t 1 / 3 , t >0, [email protected] . However, the function x 0 (t ) t 1 / 3 is not continuously differentiable on the interval [0 , 1]. Example 3 (Example of Weierstrass). Let 1

J ( x, x )

³t

2

x 2 (t )dt ,

0

x ( 0)

0, x (1)

1.

Euler's equation has the form tx 0 (t ) 2t 2 x0 (t )

0.

The solution of this equation x 0 (t ) c1t 1 c 2 , t >0, [email protected] . 55

However, any one curve of the family does not pass through the points x 0 (0) 0, x (1) 1 . We note, that Euler equation is obtained from the necessary condition of the first order for a weak local minimum, therefore for a number of tasks it gives positive solutions. However, the possibility of the following cases are not excluded: 1) the solution of Euler equation exists, is unique, but does not give a strong or a weak local minimum, 2) an infinite number of solutions, all of which deliver a global minimum J ( x, u ) on U; 3) an infinite number of solutions, but none one of them does not give any strong or weak local minimum; 4) there is no one solution of the Euler equation that passes through the given points (Example 3). Bolz problem. Minimize the functional 0

t1

J ( x, x ) ) 0 ( x (t 0 )) ) 1 ( x (t1 )) ³ F ( x, x , t )dt ,

(6)

t0

where t 0 , t1 are fixed numbers, and in contrast to the simplest problem values x(t 0 ), x (t1 ) are not fixed. Theorem 1. Let the functions ) 0 ( x ), ) 1 ( x ) C 1 ( E 1 )

and F ( x, u, t ) be twice continuously differentiable in all variables. In order the function x 0 ( t ) C 1 >t 0 , t 1 @

to conduct a weak local minimum to the functional (6) necessarily that it be a solution of the Euler equation Fx ( x 0 , x 0 , t )

d Fx ( x 0 , x 0 , t ) dt

0

(7)

and satisfies to the conditions Fx ( x 0 (t0 ), x 0 (t0 ), t ) ) 0 x ( x 0 (t0 )), Fx ( x 0 (t1 ), x 0 (t1 ), t ) )1x ( x 0 (t1 )).

Proof. Let for some number J functions x(t )

x 0 (t ) Jh(t ), h(t ) C 1 >t 0 , t1 @, h(t 0 ) z 0, h(t1 ) z 0

satisfy the conditions max x (t ) x 0 (t ) H ,

t 0 d t d t1

max x (t ) x 0 (t ) | H .

t0 dt dt1

56

(8)

Then the increment of the functional (6) can be written as 'J

J ( x Jh, x 0 Jh ) J ( x 0 , x 0 ) JGJ o(J 2 ) ,

where o(J 2 ) / J 2 o 0 at J 2 o 0

and the value GJ

) 0 x ( x 0 (t 0 )) h(t 0 ) ) 1x ( x 0 (t1 ))h(t1 ) t

1 ª wF ( x 0 (t ), x 0 (t ), t ) wF ( x 0 (t ), x 0 (t ), t ) º h(t ) h(t )» dt ³« wx wx ¼ t0 ¬

Integrating by parts, we obtain t1

t1

wF ( x 0 , x 0 , t ) wF ( x 0 , x 0 , t ) h ( t ) dt h(t ) ³ wx wx t0 t 0

t

1 § d wF ( x 0 , x 0 , t ) · ³ ¨¨ h(t ) ¸¸dt. wx dt ¹ t0 ©

Now, the first variation GJ has the form GJ

ª wF ( x 0 (t 0 ), x 0 (t 0 ), t 0 ) º 0 ( ( )) x t ) » h(t 0 ) « 0x 0 wx ¼ ¬

t2 ª ª wF ( x 0 (t ), x 0 (t ), t ) wF ( x 0 (t1 ), x 0 (t1 ), t1 ) º ( ) h t «) 1x ( x 0 (t1 )) » 1 ³« wx wx ¬ ¼ t1 ¬ d wF ( x 0 (t ), x 0 (t ), t ) º » h(t )dt GJ 1 GJ 2 GJ 3 . dt wx ¼

If x 0 ( t ) C 1 >t 0 , t 1 @ is the source solution of (6), it is necessary that GJ , h (t ) C 1 >t 0 , t1 @ .

From the independence of partial increments GJ 1 , GJ 2 , GJ 3 and from GJ follow that GJ 1

0, GJ 2

0, GJ 3

0.

It follow the formulas (7) and (8). The theorem is proved. We note, that the solution of equation (7), the function 57

x 0 (t , c1 , c2 ), t >t0 ,t1 @ ,

x 0 (t )

constants c1 , c2 are determined from the condition (8). Weierstrass necessary condition. We consider the simplest problem again. As a necessary condition of a weak local minimum is a necessary condition for strong local minimum, the function x 0 (t ) U that delivers a strong local minimum to the functional is a solution of Euler equation. However, in addition, additional necessary condition for a strong local minimum must exist, which is directly related to its definition. By such condition is a necessary condition of Weierstrass. Definition 1. By Weierstrass function for the simplest problem is called a function B( x, u, t, [ ) of four variables ( x, u, t , [ ) E 1 u E 1 u >t 0 , t1 @u E 1

defined by the formula B( x, u, t, [ ) F ( x, [ , t ) F ( x, u, t ) [ u, Fu ( x, u, t )

We note, that if the function F ( x, u, t ) is convex on the variable ɢ at fixed x E 1 , t >t 0 , t1 @ , then the function B( x, u, t, [ ) t 0 at all [ E 1 in view of Theorem 1 (Lecture 4 [15]), where u [ , v u . Theorem 2. In order the function x 0 (t ) U to deliver a strong local minimum to functional in the simplest problem, it is necessary that for any point t >t 0 , t1 @ along the solution of the Euler equation the inequality B ( x 0 (t ), x 0 (t ), t , [ ) t 0, [ E 1

(9)

holds. Proof. Let the functions x(t )

x 0 (t ) h(t , J ) U ,

such that max x (t ) x 0 (t )

t 0 d t d t1

max h (t , J ) H

t 0 d t d t1

and J ( x, u ) t J ( x 0 , u 0 ) .

We show, that the inequality (9) is executed. We note, that in the derivation of Euler's equation the function h(t , J ) Jh (t ) was taken, which is a particular case of the function h ( t , J ) . In the general case, the increment of the functional 'J

J ( x 0 h(t , J ), x 0 h(t , J )) J ( x 0 , u 0 )

1 2

JG J J 2G 2 J 2 o(J 2 ).

It follows that the value 58

wJ ( x 0 h(t , J ), x 0 h(t , J )) . wJ J 0

'J

(10)

It should be noted, that if J ! 0 , that for J ( x, u ) t J ( x 0 , u 0 ) ,

it is necessary the value GJ t 0 , since in this case the sign 'J coincides with the sign JGJ at sufficiently small Ȗ > 0. We choose the function h(t , J ), J ! 0, t >t 0 , t1 @, h(t 0 , J )

in the following form if 0, ° if °°J[ , if ®(t W )[ , ° ° J (H1 t W )[ , °¯ H1 J

h(t , J )

0, h(t1 , J )

0

t >W ,W H1 @, t W J , J ! 0, [ E1 , t >W ,W J @,

(11)

if t >W J ,W H1 @,

where H 1 ! 0, 0 J H 1 , W H 1 t1 , W (t 0 , t1 ) .

Thus, the function h ( t , J ) is linear on the segments >W ,W J @ and >W J ,W H 1 @ , therefore its derivative h(t , J )

°0, if t >W ,W H1 @, °° if t (W ,W J ), ®[ , ° J ° [ , if t (W J ,W H1 ), °¯ H1 J

(12)

reminds a needle, so h ( t , J ) it is often called the "needle" variations. Since the functions h(t , J ) h(t , J ) are defined by (11), (12), then J ( x 0 h(t , J ), x 0 h(t , J ))

t1

³ F (x

0

h(t , J ), x 0 h(t , J ), t )dt

t0

W J

³W F ( x

W H1

³ F (x W J

0

0

(t W )[ , x 0 [ , t )dt

J (H 1 J ) 1 (H 1 t W )[ , x 0 J (H 1 J ) 1[ , t )dt.

We calculate the first variation GJ according to formula (10): 59

GJ

wJ ( x 0 h(t , J ), x 0 h(t , J )) |J o0 wJ F ( x 0 (W ), x 0 (W ) [ ,W ) F ( x 0 (W ), x 0 (W ),W ) [

W H1

1 0 0 ³ Fx ( x (t ), x (t ), t )dt H1 [

W H1

W

³W F ( x (t ), x (t ), t )dt o(H ) 0

0

1

x

where o(H 1 )

W H 1

³ F ( x (t ), x (t ), t )H W 0

0

x

1 1

(t W )[dt .

(13)

Since the function x 0 (t ) U is solution of Euler equation, then the formula (3) is correct. From (3) we get W

Fx ( x 0 (W ), x 0 (W ),W ) ³ Fx ( x 0 (K ), x 0 (K ),K )dK

b0 ,

(14)

t0

Fx ( x 0 (W H 1 ), x 0 (W H 1 ),W H 1 )

W H1

³ F (x x

0

(K ), x 0 (K ),K )dK

b0 .

(15)

t0

By subtracting from (15) expression (14), we obtain W H1

³ F ( x (t ), x (t ), t )dt W 0

x

0

Fx ( x 0 (W H1 ), x 0 (W H1 ),W H1 ) Fx ( x 0 (W ), x 0 (W ),W ).

Then the expression (13) is written as GJ

F ( x 0 (W ), x 0 (W ) [ ,W ) F ( x 0 (W ), x 0 (W ),W )

[Fx ( x 0 (W H 1 ), x 0 (W H 1 ),W H 1 ) [Fx ( x 0 (W ), x 0 (W ),W ) [H 11

W H1

³W F ( x x

0

(t ), x 0 (t ), t )dt o(H 1 ).

Therefore with taking into account the fact that [Fx ( x (W H 1 ), x 0 (W H 1 ),W H 1 ) [H11 0

W H1

³W F ( x x

0

(t ), x 0 (t ), t )dt o 0,

at H 1 o 0, o(H 1 ) o 0 we get GJ

F ( x 0 (W ), x 0 (W ) [ ,W ) F ( x 0 (W ), x 0 (W ),W ) [Fx ( x 0 (W ), x 0 (W ),W ) t 0 .

By denoting [

(16)

x 0 [ for a fixed W the inequality (16) is written in the form

60

F ( x 0 , [ , t ) F ( x 0 , x 0 , t ) ( [ x 0 ) Fx ( x 0 , x 0 , t ) B( x 0 (t ), x 0 (t ), t , [ ) t 0

for arbitrary t (t 0 , t1 ) . Theorem is proved. Control questions 1. Prove the Dubois-Raymond lemma. 2. Formulate the Bolz problem. 3. Prove the Theorem 1. 4. Give definition to the Weierstrass function. 5. Prove the Theorem 2.

LECTURE 8. LEGENDRE CONDITION. JACOBI CONDITION. FUNCTIONALS DEPENDING ON N UNKNOWN FUNCTIONS. FUNCTIONALS DEPENDING ON HIGHER ORDER DERIVATIVES We consider necessary conditions of non-negativity of the second variation of the functional along the solution of Euler equation (the condition of Legendre, Jacobi's condition). Necessary conditions for a weak local minimum of the functionals depending on several functions and higher order derivatives are given. Legendre condition. We consider the simplest task. We have already proved that the necessary conditions for a weak local minimum J ( x, u ) on U are: 1) GJ 0 ; 2) G 2 J t 0 . The case at GJ 0 have been considered before. Now we consider, besides GJ 0 , non-negativity of the second variation of the functional to obtain stronest necessary conditions for a weak local minimum. Let the function x 0 (t ) U be solution of Euler equation that satisfies to the boundary conditions x 0 (t 0 ) x 0 , x 0 (t1 ) x1 . Introducing the notations w 2 F ( x 0 (t ), x 0 (t ), t ) wx 2 2 0 w F ( x (t ), x 0 (t ), t ) B(t ) wx 2 2 0 w F ( x (t ), x 0 (t ), t ) C (t ) wxwx

A(t )

Fxx ( x 0 (t ), x 0 (t ), t ), Fxx ( x 0 (t ), x 0 (t ), t ), Fxx ( x 0 (t ), x 0 (t ), t ),

the second variation G 2 J [see formula (9) of the lecture 6] is written in the form G 2J

³ >A(t )h t1

2

@

(t ) B(t )h 2 (t ) 2C (t )h(t )h(t ) dt .

t0

61

(1)

Theorem 1. In order the function x 0 (t ) U to conduct a weak local minimum in the simplest problem, it is necessary that along the solution of Euler equation the inequality Fxx ( x 0 (t ), x 0 (t ), t ) t 0, t >t 0 , t1 @

A(t )

(2)

holds (Legendre condition). Proof. Let along the solution of Euler equation the inequality G 2 J t 0 is satisfied. We show that the inequality (2) is true. We suppose the contrary, i.e., in the point W >t 0 , t1 @ the value A(W ) 0 . Let the function x (t )

h(t 0 , J )

x 0 (t ) h(t , J ), t >t 0 , t1 @,

h(t1 , J ) 0, h(t , J ) C 1 >t 0 ,t1 @ .

We represent the function h ( t , J ) h ( t1 , J ) o( t , J ) ,

where J ! 0 and h1 (t , J )

if t >W J / 2, W J / [email protected], 0, ° ® J / 2 (t W ) / J , if t >W J / 2, W @, ° ¯ J / 2 (t W ) / J , if t >W , W J / [email protected],

and the function o(t , J ) smooths the three angles of the function h1 (t, J ) so that h (t , J ) C 1 >t 0 , t1 @ .

Derivative

0, if t >W J / 2, W J / [email protected], ° h1 (t , J ) ® J / 2 (t W ) / J , if t >W J / 2, W @, ° ¯ J / 2 (t W ) / J , if t >W , W J / [email protected],

Since the value of the function o(t , J ), t >t 0 , t1 @

is quite small in comparison with the values h1 (t, J ) , as well as derivative o(t , J ) is small compared to h1 (t, J ) , the sign of the expression (1) is determined by the sign of the value

³ >A(t )h

t1

2 1

@

(t ) B(t )h12 (t ) 2C (t )h1 (t )h1 (t ) dt d max A(t )

t0

t W dJ / 2

o 0 JC1 max C (t ) max B(t ) o(J ) J o A(W ) 0, t W dJ / 2

t W dJ / 2

h1 (t , J ), h1 (t , J ) d C1. 62

Finally, G 2 J 0 . This contradicts the fact that G 2 J t 0 . Theorem is proved. Jacobi condition. Along the solution of Euler equation the second variation is a functional of the function h (t ) U 0 , i.e. G 2J

J 1 (h, h)

t1

³ F (h(t ), h(t ), t )dt , 1

h(t 0 )

0, h(t1 )

0,

(3)

t0

where F1 ( h(t ), h(t ), t )

A(t ) h 2 (t ) B (t ) h 2 (t ) 2C (t ) h(t ) h(t ) .

For the simplest problem (3) Euler equation has the form F1h

d F dt 1h

0.

Therefore, taking into account the fact that F1h F1h

2 B(t )h(t ) 2C (t )h(t ), 2 A(t )h(t ) 2C (t )h(t ),

we get A(t ) h0 (t ) A (t ) h 0 (t ) C (t ) h 0 (t ) B (t ) h 0 (t )

0.

(4)

We assume, that the strong Legendre condition A(t ) ! 0 is hold. Then equation (4) can be written as h0 P(t )h 0 (t ) Q(t )h 0 (t ) 0, h 0 (t 0 ) h 0 (t1 ) 0, (5) where P(t ) A (t ) / A(t ), Q(t ) ( B(t ) C (t )) / A(t ) . Let the function h 0 (t ), t >t 0 , t1 @

is solution of the differential equation (5). Definition. The zeros of the function h 0 (t ), t >t 0 , t1 @ ,

that are different from t 0 , called by points coupled with the point t 0 . Theorem 2. In order the function x 0 (t ) U to conduct a weak local minimum in the simplest problem, it is necessary that in the interval (t 0 , t1 ) there are not points, coupled with the point t 0 (Jacobi condition). Proof. Let J ( x, u ) t J ( x 0 , u 0 ) 63

for all x 0 (t ) h(t , J )

x(t )

for which max x(t ) x 0 (t ) H ,

t 0 d t d t1

max x (t ) x 0 (t ) H .

t 0 dt d t1

We show, that the function h 0 (t ) z 0, t , t (t 0 , t1 ) .

We suppose the contrary, i.e. that there is a point W (t 0 , t1 ) such that h 0 (W ) 0 . Note, that h(W ) z 0 , in the opposite case equation (5) has solution h 0 (t ) { 0, t >t 0 , t1 @ .

Let the function

h0 (t ), if t >t0 ,W @, h1 (t ) ® if t >W , t1 @. ¯0,

Since

d C (t )h 2 C (t )h 2 2C (t )h(t )h(t ) dt

and t1

d ³ dt C (t )h dt 2

0,

t0

in force of h(t 0 ) h(t1 ) 0 , then t1

t1

t0

t0

³ C (t )h 2 dt

³ 2C (t )h(t )h(t )dt .

Therefore, the functional J 1 ( h, h ) can be written as

³ >A(t )h t1

J1 (h, h)

2

@

(t ) ( B(t ) C (t ))h 2 (t ) dt .

(6)

t0

The value J1 (h1 , h1 )

³ >A(t )h t1

02

2

@

(t ) ( B(t ) C (t ))h 0 (t ) dt

0.

t0

In fact, the derivative

>

d A(t ) h 0 (t ) h 0 (t ) dt

@

2 A (t ) h 0 (t ) h 0 (t ) A(t ) h 0 (t ) A(t ) h 0 (t ) h0 (t )

>

2 A (t )h 0 (t )h 0 (t ) A(t )h 0 (t ) h 0 (t ) A (t )h 0 (t )

C (t )h 0 (t ) B(t )h 0 (t ) 64

@

2 2 A(t )h 0 (t ) B(t ) C (t ) h 0 (t ).

By integrating this expression with respect to t in limits from t 0 till W with taking into account h 0 (t 0 ) h 0 (W ) 0 , we obtain (6). Now we consider the simplest problem with a variation of the function h(t , J )

h1 (t ), if t >t0 , W J @, J ! 0, ° 1 ®h1 (W J ) (t W J )(H1 J ) h1 (W J ), if t >W J ,W H1 @, °0, if t t W H1 ¯

where

W H 1 t1 , H 1 ! 0, J ! 0

is the small number enough. Derivative h1 (t ) h 0 (t ), if t >t0 , W J @, ° 1 0 ® (H 1 J ) h (W J ), if t >W J ,W H1 @, °0, if t tW H 1. ¯

h(t , J )

We calculate the value of the functional J 1 (h(t , J ), h(t , J ))

t1

³ F (h(t , J ), h(t , J ), t )dt

G 2J

1

t0 t1

t1

t0

t0

³ F1 (h(t , J ), h(t , J ), t )dt J 1 (h, h)

W

W H 1

t0

³ F1 (h1 (t ), h1 (t ), t )dt

³ F (h(t , J ), h(t , J ), t )dt 1

F (h(t , J ), h(t , J ), t )dt ³ W J 1

W

³ F (h (t , J ), h (t , J ), t )dt , 1

1

1

W J

where J 1 ( h1 , h1 )

0

according to (6) and h(t , J ) h1 (t ) h 0 (t ), t >t 0 ,W J @, h(t , J ) { 0, h(t , J ) { 0, t >t H 1 , t1 @ .

We note, that by the average value theorem

>h h(t , J )

0

(W J )

@

J h 0 (T1 ) ,

(H 1 J ) h (W J ) 1

0

W J d T1 d W ; J (H 1 J ) 1 h 0 (T1 ) .

Then from (7) by the theorem about average value of the integral we get

65

(7)

J 1 (h(t , J ), h(t , J ))

W H1

³ >A(t )h

G 2J

2

@

(t , J ) ( B(t ) C (t ))h 2 (t , J ) dt

W J

³ >A(t )h W

02

@

2

(t ) ( B(t ) C (t ))h 0 (t ) dt

2

(J H 1 ) A(T 2 )

W J

J 2 h 0 (T1 ) (H 1 J ) 2

2

JA(T 3 )h 0 (T 3 ) o(J ) M (J ),

where W J d T 2 d W H 1 , W J d T 3 d W . We note, that G 2 J \ (J ) ,

moreover \ (0) 0 , and the derivative 2 A(W )h 0 (W ) 0 .

\ (0)

Consequently, there exist the numbers J 0 ! 0, H 0 ! 0 such that at 0 J J 0 , 0 H 1 H 0 , the value G 2 J 0 . This contradicts the condition G 2 J t 0 . For the correctness of the calculations is to smooth out the function h(t , J ) at the points t W J and t W H 1 , as in the case of proof of the Legendre theorem. The theorem is proved. Functionals depending on n unknown functions. We consider the simplest task, when x(t )

x1 (t ),, xn (t )

is n-vector function. Thus, to minimize the functional t1

J ( x, x )

³ F ( x(t ), x (t ), t )dt

t0

t1

³ F ( x (t ),, x 1

n

(t ), x1 (t ), , x n (t ), t )dt o inf

(8)

t0

at conditions

x1 (t ),, x n (t ) C 1 >t 0 , t1 @, x1 (t 0 ),, x n (t 0 ) x0 x10 ,, x n 0 E n , x1 (t1 ),, x n (t1 ) x1 x11 ,, x n1 E n . x(t )

x(t 0 ) x(t1 )

(9)

By introducing the notation U

^x(t ) x (t ), , x (t ) C >t , t @ / x(t ) 1

1

n

0

1

0

x0 , x (t1 )

`

x1 , x1 E n ,

problem (8), (9) can be written as t1

J ( x, x )

³ F ( x(t ), x (t ), t )dt o inf, x(t ) U .

t0

Theorem 3. In order the vector-function x 0 (t ) U to deliver a weak local 66

minimum to the functional J ( x, x ) on U, it is necessary that it be a solution of Euler equation d Fx ( x 0 , x 0 , t ) dt i

Fxi ( x 0 , x 0 , t )

0, i 1, n .

(10)

Proof. We select a feasible vector-function x 0 (t ) Jh U ,

x(t )

where h(t ) h1 (t ),, hn (t ) , and

hi (t 0 ) hi (t1 ) 0, i 1, n .

Then the first variation of the functional GJ is equal to t1 n

n

¦GJ i

GJ

ª

³ ¦ «¬ F

i 1

xi

t0 i 1

d º Fx hi (t )dt dt i »¼

0,

where GJ i

t1

ª

³ «¬ F

( x 0 , x 0 , t )

xi

t0

d º Fxi ( x 0 , x 0 , t )» hi (t )dt . dt ¼

(11)

Since the increments hi (t ), i 1, n are independent, then from GJ 0 follows that GJ i

0, i 1, n .

Then from (11) and in force of Lagrange lemma we obtain Euler equation (10). The theorem is proved. We note, that equation (11) is a differential equation of the 2n order and its solution x 0 (t ) x 0 (t , c1 , , c 2 n ), t >t 0 , t1 @ ; the constants c1 ,, c2 n are determined by the condition x 0 (t 0 )

x 0 , x 0 (t 1 )

x1 .

Functionals depending on derivatives of higher orders. We consider the following problem. To minimize the functional t1

³ F ( x(t ), x (t ), x(t ),, x

J ( x, x )

n

(12)

(t ), t )dt o inf,

t0

at conditions x(t ) U , U

^x(t ) C >t , t @ n

0

x (t1 )

1

x1 , x (t1 )

(13)

x (t 0 ) x 0 , x (t 0 ) x 01 , , x n 1 (t 0 ) x 0 n 1 ;

x11 , x(t1 )

67

x12 , , x n 1 (t1 )

`

x1n 1 .

Theorem 4. In order the function x 0 (t ) U to deliver a weak local minimum to the functional (12) at conditions (13) , it is necessary that it be solution of the EulerPoisson equation (n)

Fx ( x 0 , x 0 , , x 0 , t )

d Fx ( x 0 , , t ) dt dn (1) n n Fx n ( x 0 , , t ) dt

0.

(14)

Proof. For the function x 0 (t ) Jh U , h(t ) C n >t 0 , t1 @, h(t 0 ) h(t 0 ) h ( n 1) (t 0 ) 0,

x(t )

h(t1 ) h(t1 )

0,

h ( n 1) (t1 )

i.e. x(t ) U the first variation of the functional after integration has the form GJ

t1

n ª º d n d F F ( 1 ) Fx n » h(t )dt . n ³t «¬ x dt x dt ¼ 0

Then, from the condition GJ 0 in force of Lagrange lemma we obtain the equation (14). Theorem is proved. We note, that (14) is the differential equation of 2n order, and its solution x 0 (t )

x 0 (t , c1 , , c 2 n ),

t >t 0 ,t1 @;

the constants c1 , , c 2 n are determined from 2n conditions x 0 (t 0 )

x 0 n 1 , x 0 t1

x 0 , , x n 1 (t 0 ) x

0 n 1

(t 1 )

Control questions 1. Formulate the Legendre condition. 2. Prove the Theorem 1. 3. Formulate the Jacobi condition. 4. Give definition to the coupled points. 5. Prove the Theorem 2. 6. Prove the Theorem 3. 7. Prove the Theorem 4.

68

x1n 1 .

x1 , x t1

x11 ,....,

LECTURE 9. ISOPERIMETRIC PROBLEM. CONDITIONAL EXTREMUM. LAGRANGE PROBLEM. GENERAL COMMENTS The methods for solving the isoperimetric problem and problem on conditional extremum are presented. Necessary first-order conditions for the general Lagrange problem are formulated. The condition of Gilbert and necessary conditions of Weierstrass – E dmann in the break points of extremals are mentioned briefly. Isoperimetric problem. By isoperimetric is called the following problem: minimize the functional t1

³ F ( x(t ), x(t ), t )dt o inf

J ( x, x )

(1)

t0

at conditions x (t ) U

^

x (t ) C 1 >t 0 , t1 @ x (t 0 ) t1

³ G( x(t ), x(t ), t )dt

t0

x 0 , x (t 1 )

x1 ,

½° l ¾, °¿

(2)

where F ( x, u , t ), G ( x, u , t ) are twice continuously differentiable functions in the domain ( x, u , t ) E 1 u E 1 u >t 0 , t1 @; l is a prescribed number. Theorem 1. If the function x0 (t ) U delivers a weak local minimum to the functional (1) at conditions (2) and is not extremal of functional t1

K ( x, x )

³ G( x(t ), x (t ), t )dt ,

(3)

t0

then there exists a constant O such that the function x 0 (t ) U is a solution of the differential equation

>F ( x

0

@

, x 0 , t ) OG ( x 0 , x 0 , t ) x d F ( x 0 , x 0 , t ) OG ( x 0 , x 0 , t ) dt

>

@

x

0.

(4)

Proof. We choose the points T , W >t 0 , t1 @ . Let x(t )

x 0 (t ) h(t , H 1 ) U ,

where x 0 (t ) U is the weak point of local minimum of the functional (1) on U , and a function h(t , H 1 ) h1 (t , H 1 ) h2 (t , H 1 ), h1 (t , H 1 ) { 0, t >T H 1 , T H 1 @, h2 (t , H 1 ) { 0, t >W H 1 ,W H 1 @ . 69

Then the increment of the functional (1) is equal to 'J

T H1

ª

³ «F ( x T H ¬ x

0

, x 0 , t )

1

d Fx ( x 0 , x 0 , t )h1 (t , H 1 )dt dt W H 1

³ F (x W H

0

x

, x 0 , t )

1

d º Fx ( x 0 , x 0 , t )» h2 (t , H 1 )dt dt ¼

d º d º ª ª « Fx dt Fx » V 1 « Fx dt Fx » V 2 o(H 1 ) , ¬ ¼t T ¬ ¼t W

(5)

where V1

T H1

³ h (t, H 1

1

T H1

³ h (t , H

)dt , V 2

2

T H 1

1

)dt .

T H 1

We note, that V 1 , V 2 has the order H 1 , i.e., V 1 ~ H1 , V 2 ~ H1 .

Since x (t ) U , x 0 (t ) U ,

then K ( x, x ) l , K ( x 0 , x 0 ) l ,

consequently, 'K

K ( x, x ) K ( x 0 , x 0 )

0

d ª º 0 0 0 0 «¬G x ( x , x , t ) dt G x ( x , x , t )»¼ V 1 t T d ª º 0 0 «G x ( x , x , t ) G x ( x 0 , x 0 , t )» V 2 o1 (H 1 ). dt ¬ ¼t W

(6)

By condition of the theorem the function x 0 (t ) U is not an extremal of the functional (3), consequently, it is possible to choose a point T (t 0 , t1 ) such that d ª º «¬G x dt G x »¼ t

z0. T

Then from (6) follows, that V1

d ª º «¬G x dt G x »¼ t W V 2 o2 (H 1 ) . d º ª G G «¬ x dt x »¼ t T

(7)

By substituting the value V 1 of (7) to the right side of (5), we obtain 'J

ª d º d ª º ½ ®« Fx Fx » O «G x G x » ¾V 2 T 3 (H 1 ), dt dt ¼t W ¬ ¼t W ¿ ¯¬ 70

(8)

where the number d º ª « Fx Fx » dt ¼ t ¬

O

d ª º «G x dt G x » ¬ ¼t

T

. T

Thus, under the derivation of the increment of the functional in the form (8) it is taken into account the restriction K ( x, x ) l , i.e., it is an increment of functional as in the simplest case, therefore, GJ

Fx

d d ª º Fx O «G x G x » dx dt ¼ ¬

0, W >t 0 , t1 @, V 2 ~ H 1 .

Hence, it follows the statement (4). Theorem is proved. We note, condition (4) is the second order differential equation and its solution x0

that

the

x 0 (t , c1 , c 2 , O ), t >t 0 , t1 @ ,

constants c1 , c 2 , O are determined by x 0 (t 0 )

x0 , x 0 (t1 )

x1 , K ( x 0 , x 0 )

l.

In general case, the isoperimetric problem are formulated as follows: minimize the functional t1

J ( x, x )

J ( x1 , , x n , x1 , , x n )

³ F ( x(t ), x (t ), t )dt

t0

t1

³ F ( x (t ),, x 1

n

(t ) x1 (t ), , x n (t ), t )dt o inf

(9)

t0

at conditions x(t ) U

® x(t ) ¯

( x1 (t ), , x n (t )) C 1 >t 0 , t1 @ x(t 0 ) x0 , x(t1 ) x1 ; t1

Ki ( x )

³ Gi ( x( t ), x( t ),t )dt

t0

½° li , i 1, m¾ . °¿

(10)

Theorem 2. If the vector function x 0 (t ) U delivers a weak local minimum to the functional (9) at conditions (10) , that there exist the numbers O1 ,, Om , not all zero, such that the vector function x 0 (t ) U is a solution of the following equations: L ( x 0 , x 0 , t , O1 , , O m )

d L x ( x 0 , x 0 , t , O1 , , O m ) 0, i 1, m , dt i

where the function L ( x 0 , x 0 , t , O1 , , O m ) ,

called by Lagrangian is given by formula 71

(11)

L ( x 0 , x 0 , t , O1 , , O m )

m

F ( x, x, t ) ¦ Oi Gi ( x, x , t ) . i 1

In the case m 1 the proof of the theorem is presented above. In general case, the theorem can be proved by similar techniques. We note, that the solution of equation (11) , the vector function t >t 0 ,t1 @ , constants

x0

x 0 ( t , c1 , , c 2 n , O 1 , , O m ),

c1 , ,c2 n , O1 , ,O m

are determined from the condition x 0 ( t0 )

x0 ,

x 0 ( t1 ) x1 , K i ( x 0 , x 0 ) l , i 1,m .

Conditional extreme. We consider the following Lagrange problem: minimize the functional t1

³ F ( x(t ), x (t ), y(t ), y (t ), t )dt o inf

J ( x, y )

(12)

t0

at conditions

x(t ), y (t ) U

^x(t ), y(t ) C >t , t @ x(t 1

0

1

0

x(t1 ) x1 , y (t1 )

) x 0 , y (t 0 )

y0 ,

y1 , g x(t ), y (t ), t 0, t >t 0 , t1 @ ` .

(13)

In other words, to minimize the functional J ( x, y ) on the set of continuously differentiable functions x(t ), y (t ) C 1 >t 0 , t1 @

belonging to a prescribed surface

g x, y, t 0, t >t 0 , t1 @.

Theorem 3. If the function ( x 0 (t ), y 0 (t )) U delivers a weak local minimum to the functional (12) at conditions (13) and derivatives g y ( x 0 (t ), y 0 (t ), t )

and

g x ( x 0 (t ), y 0 (t ), t ) , t >t 0 ,t1 @

do not vanish simultenously, then there exists a function O (t ) C >t 0 , t1 @ such that the function ( x 0 (t ), y 0 (t )) U is a solution of differential equations Fx ( x 0 , x 0 , y 0 , y 0 , t ) O (t ) g x ( x 0 , y 0 t )

72

d Fx ( x 0 , x 0 , y 0 , y 0 , t ) dt

0,

(14)

Fy ( x 0 , x 0 , y 0 , y 0 , t ) O(t ) g y ( x 0 , y 0 t ) d Fy ( x 0 , x 0 , y 0 , y 0 , t ) dt

0.

(15)

Proof. Let the function

x where

( x (t ), y (t )) 0

(t ) h1 (t , H1 ), y 0 (t ) h2 (t , H1 )) U ,

h1 (t , H1 ) { 0, if t >W H1 ,W H1 @, h2 (t , H1 ) { 0, if t >W H1 , W H1 @ .

Then the increment of the functional (12) is equal to 'J

J ( x, y ) J ( x 0 , y 0 )

d ª º 0 0 0 0 0 0 0 0 « Fx ( x , x , y , y , t ) dt Fx ( x , x , y , y , t )» V 1 ¬ ¼t W d ª º « Fy ( x 0 , x 0 , y 0 , y 0 , t ) Fy ( x 0 , x 0 , y 0 , y 0 , t )» V 2 o(H 1 ) , dt ¬ ¼t W

(16)

where H 1 ! 0 is a sufficiently small number, W H1

³ h (t , H )dt ,

V1

1

1

W H 1

W H1

³ h (t , H

V2

2

1

)dt ,

W H 1

and the numbers V 1 , V 2 can be both positive and negative, both are of the order H 1 , i.e., V 1 k1H 1 , V 2 k 2 H 1 , V 1 ~ H 1 , V 2 ~ H 1 . Since ( x(t ), y (t )) U , ( x 0 (t ), y 0 (t )) U , then t1

0

³ >g ( x(t ), y(t ), t ) g ( x

0

@

(t ), y 0 (t ), t ) dt

g x ( x 0 (t ), y 0 (t ), t ) |t W V 1

t0

g y ( x 0 (t ), y 0 (t ), t ) |t W V 2 o(H 1 ).

Hence we get V2

g

x

/ g y |t W V 1 o2 (H 1 ) .

Substituting the value V 2 to the right-hand side of (16), the increment of the functional (12) with taking account the restriction as in the case of the simplest problem in the form 73

'J

° ª d º ® « Fx Fx » dt ¼ t °¯ ¬

W

gx gy

d º ª «¬ Fy dt Fy »¼ t

½° ¾V 1 o3 (H 1 ) W° ¿

is obtained. Hence, with this and the necessary condition for a weak local minimum GJ 0 follows that d d · § · g § ¨ Fx Fx ¸ x ¨ Fy Fy ¸ dt g dt © ¹ ¹ y ©

0, W >t 0 , t1 @.

(17)

by denoting d Fy dt gx

d Fx dt gy

Fx

Fy

O (t ), W

t

from (17), we obtain the differential equations (14), (15). Theorem is proved. Lagrange problem. By generalization of the problem of conditional extremum is the next Lagrange problem: minimize the functional t1

³ F ( x(t ), u(t ), t )dt o inf

(18)

f ( x, u, t ), t >t 0 , t1 @ ,

(19)

J ( x, u )

t0

at conditions x

) 0 ( x(t 0 ))

where x

u

)0

0, ) 1 ( x(t1 ))

(20)

0,

( x1 (t ), , x n (t )) C 1 >t 0 , t1 @;

(u1 (t ),, u r (t )) C>t0 , t1 @ , f ( f1 ,, f n ),

() 01 , , ) 0 s0 ), )1

()11 , , )1s1 )

are continuously differentiable functions in the variables at x E n , u E r , t >t 0 , t1 @. All considered problems are partial cases of (18) - (20). In fact, the simplest problem can be written as t1

J ( x, u )

³ F ( x(t ), u (t ), t )dt,

x u (t ),

t0

) 0 ( x(t 0 ))

x(t 0 ) x0

) 1 ( x(t1 ))

0,

x(t1 ) x1

0.

The isoperimetric problem can be written as t1

J ( x, u )

³ F ( x(t ), u(t ), t )dt o inf,

t0

x u, x1 G ( x(t ), u (t ), t ), ) 01 ( x(t 0 )) x(t 0 ) x0 0, ) 02 ( x1 (t 0 )) 0, ) 11 ( x(t1 )) x(t1 ) x0 0, ) 12 ( x1 (t1 )) x1 (t1 ) l 74

0.

For problem (18) - (20) the function L ( x, x , u ,\ , O0 , t )

O0 F ( x, u, t ) \ , x f ( x, u , t ) ,

O0 E 1 , x E n , x E n , t E 1 , u E r , \ E n

is called Lagrangian, and the functional K ( x, x , u ,\ , O0 , O1O2 )

t1

³ L ( x(t ), x (t ), u (t ),\ (t ), O , t )dt 0

t0

O1 , ) 0 ( x(t 0 )) O2 , ) 1 ( x(t1 )) ,

O1

(O11 ,, O1s0 ) E s0 ,

O2

(O21 ,, O2 s1 ) E s1

is called the Lagrange functional. Theorem 4. In order the pair ( x 0 (t ), u 0 (t )) C 1 >t0 , t1 @u C >t0 , t1 @

to deliver a weak local minimum in the problem (18) - (20), it is necessary that there exist Lagrange multipliers (O0 t 0, O11 , O2 ) , not all zero, such that: 1) Lx

d Lx dt

2) L x | x 0 (t ), u 0 (t ) 3) Lu

0 along the pair ( x 0 (t ), u 0 (t )) ;

O1 , )*0 x ( x 0 (t0 )) , L x | x

0

( t ), u 0 ( t )

O2 , )1x ( x 0 (t1 )) ;

0 along the pair ( x 0 (t ), u 0 (t )).

Partial cases of Theorem 4 are proved above, more general cases of the theorem will be proved in the next chapter. General comments to the simplest problem. 1. As it is shown above, Euler equation for the simplest problem has the form Fx ( x 0 , x 0 , t )

d Fx ( x 0 , x 0 , t ) dt

0.

This equation can be written as Fx ( x 0 , x 0 , t ) Fxt ( x 0 , x 0 , t ) Fxx ( x 0 , x 0 , t ) x 0 Fxx ( x 0 , x 0 , t ) x0

0.

(21)

It is shown, that despite of the fact that the function x 0 (t ) is sought in the class C 1 >t 0 ,t1 @ , Euler equation is reduced to defining of the function x 0 (t ) C 2 >t 0 , t1 @ .

Then the question arises: when the desired function x 0 (t ) C 2 >t 0 , t1 @? Since the first three terms along x 0 (t ), t >t 0 , t1 @ are continuous functions on t, then for the existence of x 0 (t ) C 2 >t 0 , t1 @ it is necessary that 75

Fxx ( x 0 (t ), x 0 (t ), t ) z 0, t >t 0 , t1 @ .

This condition is called the Hilbert condition. 2. It is seemed, that the derivative of the function x 0 (t ), t >t 0 ,t1 @ has discontinuities of the first order in the isolated points T1 , ,T k of the segment >t 0 ,t1 @ . In such cases, in the intervals between the points t 0 , T1 ,,T k , t1 the function x 0 (t ) satisfies to Euler equation, and in the angle points T1 , , T k are fulfilled the conditions Fx ( x 0 , x 0 , t ) |t

Ti 0

Fx ( x 0 , x 0 , t ) |t

Ti 0

, i 1, k

Conditions (22), obtained by GJ 0 , are called Weierstrass-Erdmann conditions. The continuity of the function Fx along x 0 (t ) is used for junction of solution of Euler equation in the points T1 , , T k , i.e., to determine the constants c1 , c2 in the general solutions of Euler equation [7]. Control questions 1. Formulate the isoperimetric problem. 2. Prove the Theorem 1. 3. Prove the Theorem 2. 4. Formulate the conditional extreme problem. 5. Prove the Theorem 3. 6. Formulate Lagrange problem.

CHAPTER III OPTIMAL CONTROL. MAXIMUM PRINCIPLE Developing of the variation calculus and automatic optimal control theory led to the new theory as optimal control theory. We consider optimal control problems described by ordinary differential equations. The main method for solving of the optimal control problems is L.S.Pontryagin maximum principle which represented by itself necessary condition of optimality in such problems. The maximum principle is one of the famous achievements of modern mathematics which is applied in mechanics, physics, economics and technique. LECTURE 10. PROBLEM STATEMENT OF OPTIMAL CONTROL. MAXIMUM PRINCIPLE FOR OPTIMAL CONTROL PROBLEM WITH FREE RIGHT END Problem statement. We consider several examples of specific optimal control problems. Example 1. The body is moving along the line under influence of force from the point A to the point B. Let mass of the body be m. The point A has the coordinate x0 respects to the chosen coordinate system. The point B has the coordinate x1 respectively. We denote the force by u. Let the initial speed of the body in A be equal to v0. The body must have the speed v1 in the point B. We suppose, that absolute value of the force is less then the value a, i.e. |u| d a. It is necessary to find the law of changing the force by the time, i.e. to find a function u(t), that the body gets the way from A to B for the shot time. According to the second Newton’s law the equation of the body motion has the form mx u (t ), x (0) x0 , x (0) v0 and in the finite time moment T (T is unknown) x(T )

x1 , x (T ) v1 ,

absolute value |u| d a, 0 d t d T . By denoting x1=x, x2= x , the problem can be written in the form: to minimize a functional 77

T

J ( x1 , x2 , u )

³1 dt

T o inf

(1)

0

at conditions x1 x1 (0)

1 u (t ), | u (t ) |d a m

x2 , x 2

x1 , x1 (T ) v1 .

x0 , x1 (0) v0 , x1 (T )

(2) (3)

By the other words, to find the function u(t), |u| d a, 0 d t d T , which gets minimum to the functional (1) with taking into account differential equation (2) under restrictions (3). Example 2. As known, motion equation of the mathematical pendulum in relative units under external moment u(t) has the form M kM (t ) sin M

u (t )

By changing of external moment u(t) it is possible to influence on pendulum motion. We suppose, that equipment creating the external moment has limited resources, so absolute value u (t ) d ɫ, c const ! 0 . By denoting x 1 M , x2

M

Pendulum motion equation can be written in the form x1 (t )

x 2 (t ), x 2

kx2 (t ) sin x1 (t ) u (t ), u (t ) d c, t t 0.

Now we formulate the problems: 1) Let x1 (0)

x0 , x 2 ( 0)

v0 .

It is necessary for short time to calm the pendulum by choice u (t ) , u (t ) d ɫ

i.e. lead to the station x1 (T )

0 , x 2 (T ) 78

0.

This problem can be written in the form: to minimize the functional T

³1 dt

J ( x1 , x2 , u )

(4)

T o inf

0

at conditions x2 , x 2

x1 x1 (0)

(5)

kx2 sin x1 u (t ), u (t ) d c,

x0 , x2 (0) v0 ;

(6)

x1 (T ) 0, x2 (T ) 0.

2) Let x1 (0) x0 , x 2 (0) v0 . To find the law of changing the external moment u (t ) , u (t ) d ɫ , t >0, t1 @ ,

in order to the time moment t1 pendulum has deviation x1 ,

x1 (t1 )

speed x2 (t1 ) v1 and the generalized carried out work would be minimal. The problem can be formulated as following: to minimize the functional

³ >x

@

T

2 1

J ( x1 , x2 , u )

(7)

(t ) x22 (t ) dt o inf

0

at conditions x1

x2 , x 2

x0 , x2 (0) v0 ; x1 (t1 )

x1 (0)

(8)

kx2 sin x1 u (t ), x1 , x2 (t1 )

u (t ) d ɫ , t >t 0 , t1 @.

v1 .

(9) (10)

Thus by selecting of the external influence u (t ) (control function u (t ) , t >t0 , t1 @) on system motion can be affected, furthermore, to give to the motion of the system one or other properties. However, in the problems (1) - (3), (4) - (6), (7) - (10) remains a question: in which class of functions is the control function selected? Often in many applied researches it is assumed, that the function u (t ) KC >t 0 , t1 @ ,

where KC >t 0 ,t1 @ is the space of piecewise continuous functions on the interval >t 0 ,t1 @ . We note, that the function 79

u (t ) , t >t 0 , t1 @

is called piecewise continuous, if u (t ) is continuous at all points on t >t0 , t1 @ , in exception, perhaps, only a finite number of points W 1 ,..., W p >t 0 , t1 @ , where the function u (t ) may have discontinuities of the first order, i.e. lim u (t ) u (W i 0) z lim u (t ) u (W i 0).

t oW i 0

t oW i 0

Now we formulate the optimal control problem: minimize the functional t1

J ( x, u, x0 , t 0 , t1 )

³f

0

( x(t ), u (t ), t )dt

t0

) ( x(t1 ), t1 ) ) ( x(t 0 ), t 0 ) o inf

(11)

at conditions x (t )

f ( x(t ), u (t ), t ), t 0 d t d t1 ,

x(t ) G (t ) E n , t 0 d t d t1 ,

x(t 0 ) S 0 (t 0 ) E n ,

t0 T 0 E1;

x(t1 ) S1 (t1 ) E n ,

t1 T1 E 1 ,

u (t ) KC >t 0 , t1 @, u (t ) V (t ) E r , t >t 0 , t1 @,

(12) (13) (14) (15)

where x(t ) ( x1 (t ),..., xn (t ))

are the phase coordinates of the controlled object, u (t ) (u1 (t ),...,u r (t ))

is control, by choosing of which we can affect on the motion of object, f ( x(t ), u (t ), t ) ( f1 ( x(t ), u (t ), t ),..., f n ( x(t ), u (t ), t ))

is a function describing the device object of control and external factors acting on the object, G (t ) E n for each t , t >t 0 , t1 @ are restrictions on the phase coordinates of the object of control, often called phase constraints, T 0 , T1 are prescribed sets on the axis E 1 . Moments of time t 0 , t1 cannot always be fixed, but belong to the sets T1 , T 2 . In the general case, the initial and final states of the system are not fixed, and belong to the sets S 0 (t 0 ) E n , S1 (t1 ) E n respectively. We note, that 80

S 0 (t 0 ) G (t 0 ), S1 (t1 ) G(t1 ) .

Control u (t ) (u1 (t ),..., u r (t ))

is piecewise continuous vector function with values of set V E r for each t , t >t 0 , t1 @ (for instance, for the first example V

^u E

V

^u E

for the second example

1

1

/ a d u d a`, / c d u d c` ).

In theoretical investigations often it is chosen the control of a class L 2 >t 0 ,t1 @ with values from set V E r almost everywhere on >t 0 ,t1 @ . The set ( x(t ), u (t ), x0 , t 0 , t1 ) is called by feasible, if a prescribed set of variables satisfies to the conditions (12) - (15). The functional (11) for each set is assigned a number. The optimal control problem is to find a set of feasible collections ( x 0 (t ), u 0 (t ), x00 , t 00 , t10 )

whi ch minimizes the functional (11). Feasible collection ( x 0 (t ), u 0 (t ), x00 , t 00 , t10 ) is called by solution of the optimal control, if J ( x 0 , u 0 , x00 , t 00 , t10 ) inf J ( x, u , x0 , t 0 , t1 )

J*

on an ensemble defined by conditions (12) - (15). Finally, we note that among the constraints (12) - (15) can be functionals of the following types: g j ( x, u, x0 , t 0 , t1 ) d 0, j 1, m; g j ( x , u , x 0 , t 0 , t1 )

0, j

m 1, s,

(16)

where functionals t1

g j ( x, u, x0 , t 0 , t1 )

³ G ( x(t ), u(t ), t )dt F ( x(t j

j

0

), t 0 ; x(t1 ), t1 ), j 1, s.

t0

Constraints (16) are often called by integral constraints. Maximum principle for optimal control problem with free right end. Solution of the optimal control problem in general form (11) – (16) is complicated. We consider a particular case of the general problem (11) – (16). Optimal control problem with free right end: to minimize the functional 81

t1

³f

J ( x, u )

0

( x(t ), u (t ), t )dt ) ( x(t1 )) o inf

(17)

x 0 , t 0 d t d t1 ,

(18)

t0

at conditions x (t )

f ( x(t ), u (t ), t ), x(t 0 )

u (t ) KC >t 0 , t1 @, u (t ) V E r , t >t 0 , t1 @,

(19)

where x(t ) ( x1 (t ),..., xn (t )) are phase coordinates; u (t ) (u1 (t ),..., u r (t )) is control which is piecewise continuous function with values at the ensemble V E r , V does not depend on t ; inclusion u (t ) V occurs everywhere on t >t 0 ,t1 @ ; f ( x, u, t ) ( f1 ( x, u, t ),..., f n ( x, u, t )) ;

functions f 0 ( x, u , t ), f ( x, u , t ), ) (x ) have partial derivatives by values x and continuous with its derivatives f 0 x , f x , ) x by assembly its arguments in the domain x E n , u V E r , t >t 0 , t1 @ ;

the time moments t 0 , t1 are prescribed; the right end of the trajectory x(t1 ) E n is free. Theorem. If pair

( x 0 (t ), u 0 (t )), t >t 0 , t1 @

is solution of (17) – (19), then H ( x 0 (t ), u 0 (t ), t ,\ (t )), t >t 0 , t1 @,

sup H ( x 0 (t ), u , t ,\ (t )) uV

(20)

where function H ( x, u, t ,\ ) f 0 ( x, u, t ) \ , f ( x, u, t ) ,

and function

(21)

\ \ (t ) \ (t , u ), t 0 d t d t1

is solution of the next adjoint system: \ (t )

wH ( x 0 (t ), u 0 (t ), t ,\ (t )) wx

f 0 x ( x 0 (t ), u 0 (t ), t ) ( f x ( x 0 (t )), u 0 (t ), t ))*\ (t ), t0 d t d t1 ,

\ (t1 ) 82

w) ( x 0 (t1 )) . wx

(22) (23)

Proof of the theorem is represented in the next lecture. Solution algorithm of the problem (17) – (19) is followed from this theorem: 1. To construct the function by source data of the problem (17) – (19) H ( x, u, t ,\ ) f 0 ( x, u, t ) \ , f ( x, u, t ) , \ E n .

2. To solve the problem of nonlinear programming (in particular, it can be linear or convex programming problem): H ( x 0 , u, t ,\ ) o sup, u V ,

it is supposed, that ( x 0 ,\ , t ) are parameters. As result, we find u 0 u 0 ( x 0 ,\ , t ), t >t 0 , t1 @ . 3. To solve system of differential equations x 0 (t )

f ( x 0 (t ), u 0 ( x 0 (t ),\ (t ), t ), t ), x 0 (t 0 )

x0 , t 0 d t d t1 ,

\ (t ) H x ( x (t ), u ( x (t ),\ (t ), t ), t ,\ ), \ (t1 ) ) x ( x 0 (t1 )), 0

0

0

and find x 0 (t ), \ (t ), t >t 0 , t1 @ . 4. To define optimal control

u 0 (t ) u 0 ( x 0 (t ),\ (t ), t ), t >t 0 , t1 @

and optimal trajectory x 0 (t ), t >t 0 , t1 @ .

Example. As an example, we consider the linear system with a quadratic functional: to minimize the functional t

J ( x, u )

11 >xc(t )Q(t ) x(t ) u c(t ) R(t )u (t )@dt 1 xc(t1 ) Fx(t1 ) o inf ³ 2 t0 2

at conditions x

A(t ) x(t ) B(t )u (t ), x(t 0 )

x0 , t 0 d t d t1 ,

u (t ) E , t >t 0 , t1 @, r

u (t ) Ʉɋ >t 0 , t1 @,

where Q(t ) Q * (t ) t 0 ; R (t ) F

R * (t ) ! 0 ;

F * t 0 ; A(t ), B (t )

are given matrices with piecewise continuous elements of n u n, n u r orders respectively. 1. Function H ( x, u , t ,\ )

2. Since V

1 xcQ (t ) x u cR (t )u \ , A(t ) x B (t )u . 2

E r , the solution of the optimization problem

83

H ( x 0 , u , t ,\ ) o sup , u

E r is reduced to

wH ( x 0 , u 0 , t ,\ ) wu

w 2 H ( x 0 , u 0 , t ,\ ) 0. wu 2

0,

Hence we get wH ( x 0 , u 0 , t ,\ ) R (t )u 0 B * (t )\ wu w 2 H ( x 0 , u 0 , t ,\ ) R(t ) 0. wu 2

0,

From the first condition we get R 1 (t ) B * (t )\ ,

u0

and in view of the strict convexity H ( x 0 , u , t ,\ ) by u on the convex set E r , max H in this point is reached. 3. We solve systems of differential equations x 0

A(t ) x 0 B(t ) R 1 (t ) B * (t )\ (t ), x 0 (t 0 )

\ (t )

H x

\ (t1 )

We assume, that

) x

x0 ,

t 0 d t d t1 ,

Qx 0 A* (t )\ , Fx 0 (t1 ), t 0 d t d t1 .

(24)

(25)

\ (t ) K (t ) x 0 (t ), t >t 0 , t1 @ ,

where K (t ) K * (t ) are unknown symmetric matrix. We define a matrix K (t ) of conditions (24) and (25). Since \ (t ) K (t ) x 0 (t ) K (t ) x 0 (t ) Qx 0 (t ) A* (t )\ Qx 0 (t ) A* (t ) K (t ) x 0 (t ) ,

then substituting the value x 0 (t ) of the equation (25), we obtain K (t ) x 0 (t ) K (t ) A(t ) x 0 (t ) K (t ) B(t ) R 1 (t ) u u B * (t ) K (t ) x 0 (t )

Qx 0 (t ) A* (t ) K (t ) x 0 (t ), t >t 0 , t1 @.

It follows that the matrix K (t ) is solution of the following Riccati equation: K (t ) K (t ) A(t ) A* (t ) K (t ) K (t ) B (t ) R 1 (t ) B * (t ) K (t ) Q (t ), K (t1 ) F , t 0 d t d t1 .

4. We solve the Riccati equation (29), as a result we find the matrix K (t ), t >t 0 , t1 @. Then the initial optimal control u 0 (t ) R 1 (t ) B * (t ) K (t ) x 0 (t ) , t 84

>t0 ,t1 @ ,

and the optimal trajectory is determined by the solutions of the differential equation (24) at u 0 (t ) R 1 (t ) B * (t ) K (t ) x 0 (t ) . Control questions 1. Formulate the optimal control problem statement. 2. Formulate maximum principle for optimal control problem with free right end. 3. Formulate the theorem about solution of the adjoint system. 4. Represent solution algorithm for the optimal control problem with free right end. 5. Give an example to the optimal control problem.

LECTURE 11. PROOF OF MAXIMUM PRINCIPLE FOR OPTIMAL CONTROL PROBLEM WITH FREE RIGHT END

We prove the maximum principle for the simplest optimal control problem, since the proof of the maximum principle in general case requires knowledge of functional analysis. More complete proof of the maximum principle is given in elective courses. Proof. We prove the theorem under the assumption that the functions f , f x , f 0 x , ) x satisfy Lipschitz condition with respect to values ( x , u ) , i.e. f ( x 'x, u h, t ) f ( x, u, t )) d L 'x h ,

(26)

f x ( x 'x, u h, t ) f x ( x, u, t )) d L 'x h ,

(27)

f 0 x ( x 'x, u h, t ) f 0 x ( x, u, t )) d L 'x h ,

(28)

) x ( x 'x) ) x ( x) d L 'x ,

(29)

( x, u , t ), ( x 'x, u h, t ) E n u V u >t 0 , t1 @,

where is the Euclidean norm of the vector, is the norm of a rectangular matrix. We note, that the norm of the matrix A of n u m order is determined by the formula A

sup Az , z d 1, z E m .

Let u 0 (t ), u 0 (t ) h(t )

be piecewise continuous functions with values on the set V and x 0 (t ), x (t , u 0 h )

x 0 (t ) 'x (t ), t >t 0 , t1 @

85

are solutions of the differential equation (18) respectively at u 0 (t ), u 0 (t ) h(t ) . Since 'x ( t )

x(t , u 0 h) x 0 (t ) ,

then 'x (t )

f ( x 0 (t ) 'x(t ), u 0 (t ) h(t ), t ) f ( x 0 (t ), u 0 (t ), t ), 'x(t 0 ) 0, t 0 d t d t1 .

(30)

The solution of the differential equation (30) is written in the form

³ > f (x t

'x(t )

0

@

(W ) 'x(W ), u 0 (W ) h(W ),W ) f ( x 0 (W ), u 0 (W ),W ) dW .

t0

Hence, in view of (26) we obtain t

'x(t )

³

f ( x 0 (W ) 'x(W ), u 0 (W ) h(W ),W ) f ( x 0 (W ), u 0 (W ),W ) dW d

t0

t

t

t0

t0

d L ³ 'x(W ) dW L ³ h(t ) dW , t0 d t d t1.

(31)

According to Gronual lemma from inequality (31) we obtain t1

'x(t ) d c1 ³ h(t ) dt , c1 t0

We note, that (Gronuol's lemma) if continuous functions M (t ) t 0, b(t ) t 0

and t

M (t ) d a ³ M (W )dW b(t ), t0

t 0 d t d t1 , a

const ! 0,

then t

0 d M (t ) d a ³ b(W )e a ( t W ) dW b(t ), t0

t0 d t d t1

In particular, if b (t ) b

const ! 0 ,

then 0 d M (t ) d be a (t t0 ) , t 0 d t d t1 .

In the case above M (t )

'x(t ) , b

t1

L ³ h(t ) dt . t0

86

Le L ( t1 t0 ) , t 0 d t d t1 . (32)

We calculate the increment of the functional 'J

J ( x 0 'x, u 0 h) J ( x 0 , u 0 ) t1

³ >f

0

@

( x 0 (t ) 'x(t ), u 0 (t ) h(t ), t ) f 0 ( x 0 (t ), u 0 (t ), t ) dt

t0

) ( x 0 (t1 ) 'x(t1 )) ) ( x 0 (t1 )).

(33)

Since the difference )( x 0 (t1 ) 'x(t1 )) ) ( x 0 (t1 ))

)( x 0 (t1 ) T 'x(t1 )), 'x(t1 ) , 0 dT d1,

then equality (33) can be written as 'J

t1

³>f

0

@

( x 0 (t ) 'x(t ), u 0 (t ) h(t ), t ) f 0 ( x 0 (t ), u 0 (t ), t ) dt

t0

) x ( x 0 (t1 )), 'x(t1 ) R1 , R1

(34)

) x ( x0 (t1 ) T 'x(t1 )) ) x ( x0 (t1 )), 'x(t1 ) .

Since R1 d ) x ( x 0 (t1 ) T 'x(t1 )) ) x ( x 0 (t1 )) 'x(t1 ) ,

then according to (29) in view of (32) we get 2

§ t1 · R1 d L 'x(t1 ) d Lc ¨ ³ h(t ) dt ¸ . ¨t ¸ ©0 ¹ 2

2 1

(35)

We consider the second term of expression (34). By condition of the theorem (23) \ (t1 ) ) x ( x 0 (t1 )) , therefore ) x ( x 0 (t1 ), 'x(t1 )

\ (t1 ), 'x(t1 )

t1

d \ (t ), 'x(t ) dt , dt t0

³

since 'x(t 0 ) 0 . With taking into account the fact that \ (t ), t >t 0 , t1 @

is the solution of the differential equation (22), we obtain ) x ( x 0 (t1 ), 'x(t1 )

t1

t1

t0

t0

³ \ (t ), 'x(t ) dt ³ \ (t ), 'x (t ) dt 87

t1

³

H x ( x 0 (t ), u 0 (t ), t ,\ (t )), 'x(t ) dt

t0

t1

³ \ (t ), f ( x 0 (t ) 'x(t ), u 0 (t ) h(t ), t ) t0

f ( x 0 (t ), u 0 (t ), t ) dt.

By substituting this expression to the right hand side of equation (34), we obtain 'J

t1

>

³ H ( x 0 (t ) 'x(t ), u 0 (t ) h(t ), t ,\ (t )) t0

@

H ( x 0 (t ), u 0 (t ), t ,\ (t )) dt t1

³ H x ( x 0 (t ), u 0 (t ), t ,\ (t )), 'x(t ) dt R1 .

(36)

t0

Since the function

H ( x, u , t ,\ )

is continuously differentiable by x, then according to the formula of finite increments we get H ( x 0 'x, u 0 h, t ,\ ) H ( x 0 , u 0 h, t ,\ )

H x ( x 0 T 'x, u 0 h, t ,\ ), 'x ,

0 d T d 1.

Now, equation (36) can be written as 'J

t1

>

@

³ H ( x 0 (t ), u 0 (t ) h(t ), t ,\ (t )) H ( x 0 (t ), u 0 (t ), t ,\ (t )) dt t0

R1 R2 ,

(37)

where t1

R2

³ H x ( x 0 (t ) T 'x(t ), u 0 (t ) h(t ), t ,\ (t )) t0

H x ( x 0 (t ), u 0 (t ), t,\ (t )), 'x(t ) .

We estimate the term R2 of the expression (37). The difference H x ( x 0 T 'x, u 0 h, t ,\ ) H x ( x 0 , u 0 , t ,\ ) f 0 x ( x 0 T 'x, u 0 h, t ) ( f x ( x 0 T 'x, u 0 h, t )) *\ f 0 x ( x 0 , u 0 , t ) ( f x ( x 0 , u 0 , t ))*\ .

Then 88

t1

R2 d ³ f 0 x ( x 0 (t ) T 'x(t ), u 0 (t ) h(t ), t ) f 0 x ( x 0 (t ), u 0 (t ), t ) u t0

u 'x(t ) dt max \ (t , u 0 ) t 0 dt dt1

t1

³

f x ( x 0 T 'x, u 0 h, t )

t0

t1

f x ( x 0 (t ), u 0 (t ), t ) 'x(t ) dt d L ³ ( 'x(t ) h(t ) ) 'x(t ) dt t0

max\ t , u 0 L³ 'xt ht 'xt dt t0 dt dt1

t1

t0

t1

L§¨1 max \ (t , u 0 ) ·¸ ³ ( 'x(t ) h(t ) ) 'x(t ) dt. © t0 dt dt1 ¹ t0

(38)

Here the inequalities (27) and (28) are used. From inequality (32) follows that t1

³

t0

2

§ t1 · 'x(t ) dt d (t1 t 0 )c ¨ ³ h(t ) dt ¸ , ¨t ¸ ©0 ¹ 2

2 1

2

§ t1 · ¨ ¸ . x ( t ) h ( t ) dt c ' d 1 ³ h(t ) dt ³t ¨t ¸ 0 ©0 ¹ t1

Now inequality (38) can be written as 2

· § t1 R2 d c 2 ¨ ³ h(t ) dt ¸ , ¸ ¨t ¹ ©0 c2

>

(39)

@

L§¨1 max \ (t , u 0 ) ·¸ c12 (t1 t 0 ) c1 . © t0 dt dt1 ¹

From inequalities (35) and (39) we get R

· § t1 R1 R2 d R1 R2 d ( Lc c2 )¨ ³ h(t ) dt ¸ ¸ ¨t ¹ ©0 2 1

2

2

· § t1 c3 ¨ ³ h(t ) dt ¸ . ¸ ¨t ¹ ©0

where c3

We note, that since

Lc12 c2 , R

R1 R2 .

( x 0 (t ), u 0 (t )), t >t 0 , t1 @

is the optimal pair, then the increment of the functional defined by (37), 'J

t1

J ( x 0 'x, u 0 h) J ( x 0 , u 0 ) ³ g (t )dt R t 0, t0

where 89

(40)

g (t )

H ( x 0 (t ), u 0 (t ) h(t ), t ,\ (t )) H ( x 0 (t ), u 0 (t ), t ,\ (t )), t >t 0 ,t1 @ .

We choose the increment of control, function v u 0 (t ), if t >W ,W H @ >t 0 , t1 @, ® if W >t 0 , t1 @ \ (W ,W H ), ¯0,

h (t )

where

(41)

W ,W H >t 0 , t1 @, H ! 0

is sufficiently small number, v V is a point. In the case W

t1 the function

h(t ) v u 0 (t ) , v V , t >t1 H ,t1 @ ,

and in the other points of the segment >t 0 ,t1 @ is identically zero. With this choice of function h(t ), t >t 0 , t1 @ , u 0 (t ) h(t ) V and expression (40) can be written as 2

W H

'J

· § W H ³ g (t )dt R t 0, R d c3 ¨¨ ³ h(t ) dt ¸¸ . W ¹ ©W

(42)

According to the average value theorem the integral W H

³W g (t )dt

Hg (W TH ), 0 d T d 1,

(43)

and by the Cauchy - Bunyakovsky theorem 2

W H · § W H ¨ ³ h(t ) dt ¸ d H ³ h(t ) 2 dt. ¸ ¨ W ¹ © W

(44)

Since u 0 (t ) is piecewise continuous, x(t ), \ (t ) are continuous with respect to t, and the function H ( x, u , t ,\ ) is a non-continuous function in all of its arguments, then for sufficiently small H ! 0 , the function g (t ), t >W ,W H @ is continuous. Then inequality (42) with taking into account relations (43) and (44) can be written as: 0 d 'J

Hg (W TH ) R d Hg (W TH ) H c3

W H

³ h(t ) W

2

dt.

Hence, after dividing by H ! 0 , tending H o 0 , we get 0 d g (W ) , i.e. 90

g (W )

H ( x 0 (W ), v,W ,\ (W )) H ( x 0 (W ), u 0 (W ),W ,\ (W )) d 0, W >t 0 , t1 @.

Since W can be arbitrary point of the segment >t 0 ,t1 @ and the vector v V is any point, then from the last inequality we get H ( x 0 , v, t ,\ (t )) d H ( x 0 (t ), u 0 (t ), t ,\ (t )), t >t 0 , t1 @, v V .

This means that sup H ( x 0 , v, t ,\ ) vV

H ( x 0 , u 0 , t ,\ ), t , t >t 0 , t1 @.

Theorem is proved. It can be shown, that the theorem is true in the case, if instead of conditions (26) - (29) it is required only the continuity of the functions f , f x , f 0 , f 0 x , ), ) x

of the combination of their arguments. On these issues we recommend the following books: J. Varga, Optimalnoe upravlenie differentialnymi and functionalnymi uravneniami. M.; Nauka, 1977; Gabasov R., Kirillova F.M. Prinzip maxsimuma v teori optimalnogo upravlenia. Minsk: Nauka and Technika, 1974; Gamkrelidze R.V. Osnovi optimalnogo upravlenia. Tbilisi: TGU, 1977. Control questions 1. Prove the maximum principle for optimal control problem with free right end. 2. Formulate the Lipschitz condition. 3. Formulate the Gronuol lemma. 4. Formulate the Cauchy-Bunyakovsky theorem. 5. Formulate the average value theorem.

LECTURE 12. MAXIMUM PRINCIPLE FOR OPTIMAL CONTROL. CONNECTION BETWEEN MAXIMUM PRINCIPLE AND VARIATION CALCULUS

The rule for solving of the optimal control problem is presented. The connection between the maximum principle and variations calculus is established. Maximum principle for optimal control problem without phase restrictions. We consider optimal control problem without phase restrictions: to minimize the functional J ( x, u , t 0 , t1 ) t1

³f

0

( x(t ), u (t ), t )dt ) 0 ( x(t 0 ), x(t1 ), t 0 , t1 ) o inf

t0

at conditions 91

(1)

x (t )

f ( x (t ), u (t ), t ), t 0 d t d t1 , t 0 , t1 ',

g i ( x, u, t 0 , t1 ) d 0, i 1, m, g i ( x, u, t 0 , t1 ) 0, i

(2)

m 1, s,

u (t ) KC >t 0 , t1 @, u (t ) V E r , t >t 0 , t1 @,

(3) (4)

where KC >t 0 ,t1 @ is space of the piecewise continuous functions; t 0 , t1 ' are not fixed, but belong to the segment ' E1 ; u (t ) (u1 (t ),..., u r (t )) is control; ( x1 (t ),..., x n (t )) KC 1 >t0 , t1 @

x(t )

are phase coordinates; KC 1 >t 0 ,t1 @ is space of the piecewise continuous differentiable functions; f0 , )0 , f

( f 1 ,..., f n ), f x , f 0 x , ) 0 x

are continuous functions by assembly its arguments, and functionals t1

g i ( x , u , t 0 , t1 )

³ f ( x(t ), u(t ), t )dt ) ( x(t i

i

0

), x(t1 ), t 0 , t1 ),

(5)

t0

i 1, s , the vectors x(t 0 ) E n , x(t1 ) E n are not fixed; f i , ) i , f ix , ) ix

are continuous by assembly its arguments. Feasible controllable process is defined by four ( x(t ), u (t ), t 0 , t1 ) satisfied to differential connection (2), restrictions (3) and inclusions (4). Feasible controllable process ( x 0 (t ), u 0 (t ), t 00 , t10 )

is called by locally optimal (exactly, to delivery strong local minimum to the functional (1)), if there is a number H ! 0 , such, that J ( x, u , t 0 , t1 ) t J ( x 0 , u 0 , t 00 , t10 )

at all feasible ( x(t ), u (t ), t 0 , t1 ) satisfied to inequality max x(t ) x 0 (t ) max u (t ) u 0 (t ) t 0 t 00 t1 t10 H . t0 d t dt1

t 0 dt dt1

Optimal control problem (1) – (5) is solved by the next algorithm: 1. Lagrange’s functional is constructed 92

/( x, x , u,\ , x0 , x1 , t 0 , t1 , O0 , O1 ,..., O s ) t1

§

s

³ ¨© ¦ O

i

i 0

t0

· f i ( x, u , t ) \ , x f ( x, u , t ) ¸dt ¹ s

¦ Oi ) i ( x(t 0 ), x(t1 ), t 0 , t1 ),

(6)

i 0

where \

(\ 1 (t ),...,\ n (t )) KC 1 >t 0 , t1 @,

\ (t ) x0

O

x(t 0 ), x1

(O0 , O1 ,..., O s ) E

s 1

x(t1 ),

, O0 t 0, O1 t 0, Om t 0.

Integrand s

¦ O f ( x, u, t ) \ (t ), x f ( x, u, t )

L( x, x, u, t ,\ , O )

i

i

(7)

i 0

is called by Lagrangian for the problem (1)–(5). 2. Optimal pair ( x 0 (t ), u 0 (t ))

satisfies to Euler’s equation L x ( x 0 , x 0 , u 0 , t ,\ , O )

d L x ( x 0 , x 0 , u 0 , t ,\ , O ) dt x

0.

(8)

By substituting the value L from (7) into equation (8) we can write \ (t )

s

¦O f i

ix

( x 0 (t ), u 0 (t ), t ) ( f x ( x 0 (t ), u 0 (t ), t ))*\ (t ).

(9)

i 0

Differential equation (9) defines the adjoint system. 3. The boundary conditions for adjoint system (9) are defined by conditions \ (t 00 )

s

¦O ) i

i 0

\ (t ) 0 1

ix0

( x00 , x10 , t 00 , t10 ),

s

¦ Oi ) ix1 ( x , x , t , t ), 0 0

0 1

0 0

i 0

where x00

x 0 (t 00 ), x10

x 0 (t10 )

are initial and final stations of the optimal trajectory

>

@

x 0 (t ), t t 00 , t10 ; t , t ' 0 0

0 1

are optimal values of the initial and final time moments. 93

0 1

(10)

4. Optimal control u 0 ( x 0 ,\ , t )

u0

is defined by condition max H ( x 0 , u, t ,\ ) uV

H ( x 0 , u 0 , t ,\ ), t >t 0 , t1 @,

(11)

where ɇ is Pontryagin function which is equal to s

H ( x, u, t ,\ ) ¦ Oi f i ( x, u, t ) \ , f ( x, u, t ) . i 0

5. The optimal time moments t 00 , t10 are defined by equations s

s

i 0

i 0

¦ Oi f i ( x(t 00 ), u(t 00 ), t 00 ) ¦ Oi ) it0 ( x00 , x10 , t 00 , t10 )

x 0 (t 00 ), ) ix0 ( x00 , x10 , t 00 , t10 ) s

¦ O f ( x(t i

i

i 0

0 1

s

0; (12)

0.

), u(t10 ), t10 ) ¦ Oi ) it1 ( x00 , x10 , t 00 , t10 ) i 0

x 0 (t10 ), ) ix1 ( x00 , x10 , t 00 , t10 )

(13)

6. Lagrange multipliers O0 t 0, O1 t 0, ..., Om t 0, Om1 , ... , Os

are defined by condition Oi g i ( x 0 , u 0 , t 00 , t10 ) 0, i 1, s.

(14)

Usually it is considered two cases separately: 1) O0 0 (degenerated problems); 2) O0 z 0 . In this case possible to accept O0 1 . Theorem (Pontryagin's maximum principle). If ( x 0 (t ), u 0 (t ), t 00 , t10 )

is the optimal process for problem (1) - (5), then there exist Lagrange multipliers (O0 t 0, O1 t 0, .. .. , Om t 0, O m 1 , ... , O s ) E s 1

and function

\ (t ) KC 1 >t 0 , t1 @ ,

which are not equal to zero simultaneously, such the conditions (8) - ( 4) hold. 94

Example 1. Minimize the functional 4

³ (x u

J ( x, u )

2

)dt

0

at conditions u , u d 1 , x ( 0)

x

0.

For this example x u 2 , f i { 0 , i 1, s ,

f0

)0

0 , )1

x(0) ,

all the rest )i V

2, s , f

0, i

^u E

1

u,

/ 1 d u d 1` ;

the time moments t 0 0, t1 4 are prescribed. Lagrange functional (6) has the form

³ >O 4

/

0

@

( x u 2 ) \ ( x u ) dt O1 x(0) .

0

Lagrangian O0 (u 2 x) \ ( x u )

L

[see formula (7)]. By (9) we have \ O0 , from (10) follows that \ (0) O1 , \ (4) 0 .

Condition (11) is written as follows:

max O0u 2 O0 x \u

1du d 1

or

min O0u 2 \u

1du d 1

It is undenstandable, at

O0

2

O0u 0 O0 x 0 \u 0 2

O0 u 0 \ u 0 .

0, p

0, p

O

0

all the Lagrange multipliers are zero. This should not be held, therefore, O0 1 . By solution of the convex programming problem

min u 2 \u

1du d 1

we get

95

u0

\ °sign\ , 2 ! 1, ° ® \ °\ , d 1. °¯ 2 2

Then 0

x (t )

0 d t d 2, t , ° 2 ®t ° 2t 1, 2 d t d 4; ¯4

\ (t ) t 4.

Example 2. We consider the following problem in detail (see example 1, Lecture 10): minimize the functional T

J ( x1 , x2 , u )

³1 dt

T o inf

0

at conditions x2 , x1 x2 , x1 x2 , x 2 x1 (0) x0 , x2 (0) v0 ,

x1

u,

x1 (T ) x2 (T ) 0 , u (t ) KC>0,[email protected] , 1 d u d 1

For this problem f

x1 (0) x0 , ) 2

)1 )3

( x2 , u ) , f 0

x1 (T ) , ) 4

1, x2 (0) v0 ,

x2 (T ) , T '

f i { 0 , i 1,4 , ) 0

E1 ,

0.

10. Lagrange functional T

/

³ >\

1

( x1 x2 ) \ 2 ( x 2 u )@dt O0T O1 ( x1 (0) x0 )

0

O2 ( x2 (0) v0 ) O3 x1 (T ) O4 x2 (T ).

Lagrangian L \ 1 ( x1 x2 ) \ 2 O ( x 2 u ) [see formula (6), (7)]. 2 0. We define the adjoint system according to formula (9) \ 1

Hence we get

0, \ 2

\ 1 .

\ 2 (t ) ct c1 ,\ 1 (t ) c .

30. The boundary conditions of the adjoint system we calculate by the formula

(10):

\ 1 (0) O1 , \ 2 (0) O2 , \ 1 (T )

O3 , \ 2 (T )

4 0. Condition (11) is written as follows: function H

1 \ 1 x 2 \ 2 u ,

96

max[\ 2 u ]

1d u d1

O 4 .

Since the other terms do not dependent on u , at determining u 0 they are not involved. From this it follows that min(\ 2 u ), 1 d u d 1 .

We have the linear programming problem, hence u 0 sign\ 2 . 50. The optimal time T is determined from (13): O0 O3 x10 (T ) O 4 x 20 (T )

0, ) iT { 0 .

Since x10 (T )

x 20 (T )

0 , x 20 (T )

sign\ 2 ,

u0

consequently, x 20 (T ) 1 . Then O0

O4 x 20 (T ) \ 2 (T ) sign\ 2 (T )

\ 2 (T ) .

We consider the case O0 0 . Then from 50 we have \ 2 (T ) 0 , therefore O4 0 . From 2 0 it follows that \ 1 (t ) c , \ 2 (t ) c(t T ) .

Then from 4 0 we have u 0 signc(t T ) . It follows, that u 0 1 , either u 0 1 . By integrating the equation we get

x20 , x 20 1 at u 0

x10

x20 (t ) t c2 , x10

1 ,

(t c2 ) 2 . 2 c3

We define c3 , c2 from the conditions x0 , x 2 (0)

x1 (0)

v0 , x1 (T )

x2 (T ) 0 .

These conditions are satisfied, if there is a connection between the values x0 , v0 : v0

M ( x 0 ) , M ( x0 ) 2 x 0 ,

if x0 t 0, M ( x0 )

2x0 ,

if x0 d 0 . In the case u0

x10 (t )

1 , x 20 (t )

T t,

v0 (T t ) , 2 2 2 2

We consider the case O0 1 . In this case \ 2 (T ) 1 . Then from 2 0 we have 97

x0 .

\ 2 (T ) c(t T ) 1 at \ 2 (T ) 1

either \ 2 (T ) 1 c(t T ) at \ 2 (T ) 1 .

In this case 4 0 can be written as:

1, 0 d t d W , u 0 (t ) ® ¯ 1, W d t d T ,

or

1, 0 d t d W , u 0 (t ) ® ¯ 1, W d t d T .

By integrating the equations x20 , x 20

x10

with the conditions

u 0 (t )

x0 , x 10 (0) v0 , x10 (T )

x 10 (0)

x20 (T ) 0

for indicated values u0 (t ), 0 d t d T we define W and Ɍ. The connection between maximum principle and variation calculus. We consider the simplest problem. If the notation x u is introduced, then the simplest problem can be written in the following form: minimize the functional t1

³f

J ( x, y )

0

( x(t ), u (t ), t )dt o inf ( f 0

F)

(15)

t0

at conditions u (t ), t 0 d t d t1 ; x(t 0 )

x (t )

x 0 , x (t1 )

x1 , u E 1 ,

(16)

where t 0 , t1 , x0 , x1 are fixed numbers. Pontryagin function H

O0 f 0 ( x, u, t ) \ 1u .

Then the adjoint system \ 1

H x

O0 f 0 x ( x 0 , u 0 , t ) .

Optimal control u0 is determined by the condition max H ( x 0 , u, t ,\ ) , u E 1 . It follows that u0 satisfies to the equation wH wu

O0

wf 0 ( x 0 , u 0 , t ) \ 1 wu

The value O0 z 0 , in the opposite case \ 1 (t ) 0 , t >t 0 ,t1 @ . 98

0.

(17)

We accept O0 1 . Then the adjoint system (17) can be written as \ 1 (t )

wf 0 ( x 0 (t ), u 0 (t ), t ) . wu

f 0 x ( x 0 , u 0 , t ), \ 1 (t )

It follows that for arbitrary t , t0 d t d t1 t

f 0u ( x 0 (t ), u 0 (t ), t )

³f

0x

( x 0 (W ), u 0 (W ),W )dW \ (t 0 ).

(18)

t0

Expression (18) is Euler equation in the form of Dubois - Raymond (see Lecture 7). If (18) is differentiated by t, we obtain Euler equation in the form of Lagrange. Further, if in the point u 0 the maximum ɇ is reached on E 1 , it should be w 2 H ( x 0 , u 0 , t ,\ ) d 0. wu 2

It follows, that f 0uu ( x 0 (t ), u 0 (t ), t ) t 0 .

This is Legendre condition. From the maximum condition follows, that max1 H ( x 0 (t ), v, t ,\ ) vE

H ( x 0 (t ), u 0 (t ), t ,\ ), t >t 0 ,t1 @ .

Hence it follows that (\ 1 (t ) f 0u ) : 0 d H ( x 0 , u 0 , t ,\ ) H ( x 0 , v, t ,\ )

f 0 ( x 0 , v, t ) f 0 ( x 0 , u 0 , t )

(v u 0 ) f 0 u ( x 0 , u 0 , t )

B ( x 0 , u 0 , t , v). v E 1 ,

t >t 0 ,t1 @ .

This is condition of Weierstrass. Weierstrass - Erdman condition follows from continuity H ( x 0 (t ), u 0 (t ), t ,\ ) .

Thus, for the simplest problem (15) and (16) all the main results of the variation calculus follow from the maximum principle. For deep study of the mathematical theory of optimal processes the following works are recommended [1,2,3,8,10,12,13]. Control questions 1. Formulate the maximum principle for optimal control problem without phase restrictions. 2. Give definition to the feasible controllable process. 3. Give definition to the locally optimal process. 4. Give definition to the Lagrangian. 5. Formulate the Pontryagin's maximum principle. 6. Represent the Legendre condition. 7. Represent the Weierstrass condition.

99

CHAPTER IV OPTIMAL CONTROL. DYNAMIC PROGRAMMING

Along with the maximum principle there is another method for solving optimal control problem proposed by American mathematician R.Bellman and received the title of the dynamic programming method. This method is better suited to solve the problem of synthesis of control and obtain sufficient conditions of optimality. LECTURE 13. OPTIMALITY PRINCIPLE. BELLMAN EQUATION

The exposition of the differential Bellman equation for the problem of optimal control with a free right end is given on the basis of the optimality principle. Algorithms for solving the problem by dynamic programming method and solution of the example are presented. We consider the optimal control problem with a free right end: to minimize the functional t1

J ( x, u )

³f

0

( x(t ), u (t ), t )dt Ɏ( x(t1 )) o inf

(1)

t0

at conditions f ( x, u, t ), t 0 d t d t1 , x(t 0 )

x

x0 ,

u (t ) KC >t 0 , t1 @, u (t ) V E r , t 0 d t d t1 ,

(2) (3)

where moments t 0 , t1 are known, x0 E n is a prescribed vector and the functions f 0 ( x, u, t ), f ( x, u, t ), Ɏ( x)

are continuous in all their arguments. Together with problem (1) - (3) we consider the family of problems in the following form: minimize the functional t1

J t ( x, u )

³f

0

( x(0), u (W ),W )dW Ɏ( x(t1 )) o inf,

(4)

t

x (W )

f ( x(W ), u (W ),W ), t d W d t1 , x(t )

x,

(5)

u (W ) KC >t , t1 @, u (W ) V E r , t d W d t1 .

(6)

100

We note, that (4) - (6) are family of optimal control problems depending on the scalar t and n vector x E n . The transition from an optimal control problem (1) (3) to the family of optimal control problems (4) - (6) is called by principle of invariant imbedding. Function B ( x, t )

(7)

inf J t ( x, u )

x (t ) x u (W )V

is called by Bellman function for family of the problems (4) - (6). The principle of optimality. If u 0 (W ), W >t , t1 @ is optimal control for the state (t , x ) of the problem (4) - (6) , then the control u 0 (W ), W >s, t1 @, s ! t

will be optimal for the state s, x 0 ( s) of the problem (4) - (6), where x 0 ( s) is the value of the optimal trajectory of the system (4) - (6) generated by the control u 0 (W ), W >t , s @ .

Further we assume, that the formulated optimality principle is satisfied for the problem (4) - (6). This heuristic approach to the optimal control problem which is proved to the contrary, is the basis of the exposition of the differential equations of Bellman. From the optimality principle and introduced notation (7) follows that s

B ( x, t )

B( x 0 ( s), s ) ³ f 0 ( x 0 (W , u 0 ), u 0 (W ),W )dW

J t ( x 0 , u 0 ),

(8)

t

s

B( x, t ) d B( x( s ), s) ³ f 0 ( x(W , u ), u (W ),W )dW . t

Equation (8) follows directly from the principle of optimality, since

x

0

(W ), u 0 (W ) , W >t 0 , t1 @

is the optimal pair and the state x 0 ( s) is generated by the optimal control u 0 (W ), W >t , s @ ,

the value B( x 0 ( s), s )

inf

x ( s ) x0 ( s ) u (W )V , W >s ,t1 @

J t ( x, u )

°t1 ½° inf0 ®³ f 0 ( x(W ), u (W ),W )dW Ɏ( x(t1 ))¾. x( s) x ( s)° °¿ ¯s u (W )V 101

(9)

We consider the inequality (9). On the time interval W >t, [email protected] is valid control

u (W ), W >t , s @ differented from optimal control

u 0 (W ), W >t , s @ ,

and control u (W ), W >t , [email protected] creates a state x( s ) x( s, u ) , the value °t1 ½° ® f 0 ( x(W ), u (W ),W )dW Ɏ( x(t1 ))¾. x ( s ) x ( s ,u ) ° ³ °¿ ¯s u (W )V 0 x ( s) z x( s ).

B( x( s ), s )

inf

From (8), (9) at s t 't we get t 't

B ( x, t )

B( x 0 (t 't , u 0 ), t 't )

³f

0

( x 0 (W , u 0 ), u 0 (W ),W )dW ,

(10)

t

t 't

B ( x, t ) d B( x(t 't , u ), t 't )

³f

0

( x(W , u ), u (W ),W )dW .

(11)

t

Since both trajectories

x 0 (W , u 0 ), x(W , u ), W >t , t 't @

initiated from the point ɯ, then x 0 (t , u 0 )

x, W >t , t 't @ .

x(t , u )

From equality (10) we get t 't

B( x 0 (t 't , u 0 ), t 't ) B( x, t )

³f

0

( x 0 (W , u 0 ), u 0 (W ),W )dW .

t

We divide this equation by 't ! 0 and pass to the limit at 't o 0 . As a result, we get dB( x 0 (t 0, u 0 ), t ) f 0 ( x 0 (t 0, u 0 ), u 0 (t ), t ) 0, t >t 0 , t1 @. dt

Hence, with taking into account that x 0 (t 0, u 0 ) x dB( x 0 (t 0, u 0 ), t ) dt

we obtain

Bx ( x, t ), f ( x, u 0 , t ) Bt ( x, t ),

Bx ( x, t ), f ( x, u 0 , t ) Bt ( x, t ) f 0 ( x, u 0 , t ) 0, t >t 0 , t1 @.

Similarly, from (11) we get

102

(12)

t 't

0 d B( x(t 't , u ), t 't ) B( x, t )

³f

0

( x(W , u ), u (W ),W )dW .

t

Hence, after dividing by 't ! 0 and passing to the limit at 't o 0 we obtain 0d

dB ( x(t 0, u ), t ) f 0 ( x, u , t ), u V , t >t 0 , t1 @. dt

Since x(t 0, u ) x , then this inequality can be written as: 0 d Bx ( x, t ), f ( x, u, t ) Bt ( x, t ) f 0 ( x, u, t ), u V , t >t 0 , t1 @.

(13)

The relations (12), (13) can be combined and presented in the form of min>Bt ( x, t ) Bx ( x, t ), f ( x, u, t ) f 0 ( x, u, t )@ 0, t0 d t d t1. uV

(14)

We note, that the vector x

x(t ), t 0 d t d t1 ,

then in particular, at t t1 , x(t1 ) x and the value B( x, t1 ) Ɏ( x) .

(15)

The relations (14) and (15) are called by Bellman equation for the problem (1) (3). The algorithm for solving of the problem (1) - (3). We show the sequence for solving of the problem (1) - (3) by the method of dynamic programming: 10 . It is compiled the function K ( B x , x, u , t )

Bx ( x, t ), f ( x, u , t ) f 0 ( x, u, t ), u V .

20 . The problem of nonlinear programming is solved min K ( Bx , x, u, t ), u V ,

where ( Bx , x, t ) are parameters. As a result we find u0

u 0 ( B x ( x, t ), x, t ) .

Essentially u 0 u 0 ( B x ( x, t ), x, t ) is determined by the condition (14) and the fact that the partial derivative Bt ( x, t ) does not depend on u, hence, the term Bt ( x, t ) does not affect to the minimization by u. 30 . We substitute the value 103

u0

u 0 ( B x ( x, t ), x, t )

into equation (12) [or (14) ]. The result is wB ( x, t ) wB ( x, t ) , f ( x, u 0 ( B x ( x, t ), x, t ), t ) wt wx f 0 ( x, u 0 ( B x ( x, t ), x, t ), t )

0,

B x ( x, t )

Ɏ( x) .

We solve this differential equation in partial derivatives of the functions B ( x, t ) . As a result, we find the Bellman function B ( x, t ) as a function of x and t, i.e., B ( x, t ) M ( x, t ) , where M ( x, t ) is the known function. 40 . We find the optimal control u 0 (M x ( x 0 , t ), x 0 , t )

u0

u 0 (x 0 ,t)

u ( x 0 (t ), t ), t >t 0 , t1 @.

We note, that in the dynamic programming method the optimal control is determined as function from the optimal trajectory u of t. This optimal control u0

u 0 ( x 0 (t ), t ), t >t 0 , t1 @

is called synthesizing unlike program control u 0 maximum principle.

u 0 (t ), t >t 0 , t1 @ found by the

E x a m p l e. We consider a linear system with a quadratic functional: to minimize the functional J ( x, u ) t

11 1 [ x c(t )Q(t ) x(t ) u c(t ) R(t )u (t )]dt x c(t1 ) Fx(t1 ) o inf (16) ³ 2 t0 2

at conditions x

A(t ) x B(t )u, x(t 0 )

x0 , t 0 d t d t1 ,

u (t ) KC >t 0 , t1 @, u (t ) V { E r , t 0 d t d t1 ,

(17) (18)

where Q(t ) Q * (t ) t 0; R (t )

R * (t ) ! 0; F

F* t 0 .

We solve the problem (16) - (18) according to the algorithm described above. a) The function K

1 Bx ( x, t ), A(t ) x B (t )u [ xcQ (t ) x u cR(t )u ]. 2

b) Since the set V { E r , then the control u 0 u 0 ( B x ( x, t ), x, t ) is determined by the conditions 104

wK wu

0,

w2K t 0. wu 2

We calculate the derivatives wK wu

0,

B * (t ) Bx ( x, t ) R(t )u 0

w2K wu 2

R(t ) ! 0.

Hence we find u0

R 1 (t ) B * (t ) Bx ( x, t ) .

c) By substituting the value u 0 to Bellman equation (14) [or (12)] , we obtain wB ( x, t ) wB ( x, t ) , A(t ) x B (t ) R 1 (t ) B * (t ) B x ( x, t ) wt wx 1 [ xcQ (t ) x Bx* ( x, t ) B (t ) R 1 (t ) B * (t ) Bx ( x, t )] 0, B ( x, t1 ) 2

1 xcFx. 2

We seek a solution of the differential equation in the following form: B ( x, t )

1 xcK (t ) x , 2

where K (t ) K * (t ) is unknown symmetric matrix. Since wB( x, t ) wt

wB ( x, t ) 1 xcK (t ) x , 2 wx

K (t ) x,

then the Bellman equation can be written as 1 1 x cK (t ) x K (t ) x, A(t ) x B (t ) R 1 (t ) B * (t ) K (t ) x x cQ (t ) x 2 2 1 xcK (t ) B (t ) R 1 (t ) B * (t ) Kx 2 1 1 B ( x, t1 ) xcK (t1 ) x xcFx. 2 2

0,

Hence we get 1 1 1 ª1 xc« K (t ) K (t ) A(t ) A* (t ) K (t ) K (t ) B (t ) R 1 (t ) B * (t ) K (t ) 2 2 2 ¬2 1 º Q (t ) » xc 0, K (t1 ) 2 ¼

Hence, the source matrix K (t )

K * (t ) , t >t 0 ,t1 @ 105

F.

is solution of the following Riccati equation: K (t ) K (t ) A(t ) A* (t ) K (t ) K (t ) B(t ) R 1 (t ) B * (t ) K (t ) Q(t ), K (t1 ) F , t 0 d t d t1 .

We solve this equation and find the matrix K (t )

K * (t ) , t >t 0 ,t1 @ .

Then the Bellman function B ( x, t )

1 xcK (t ) x , t >t 0 ,t1 @ . 2

d) We find the optimal control u 0 (x0 ,t)

R 1 (t ) B * (t ) K (t ) x 0 (t ), t 0 d t d t1 .

The optimal trajectory

x 0 (t ) , t >t 0 ,t1 @

is determined from the solution of the differential equation x 0 (t )

A(t ) x 0 B(t )u 0 [ A(t ) B(t ) R 1 (t ) B * (t ) K (t )]x 0 (t ), x 0 (t 0 ) x0 , t >t 0 , t1 @.

The minimum value of the functional J (x0 ,u0 )

B ( x00 , t 0 )

1 x0c K (t 0 ) x0 . 2

We note, that the Bellman equation (14) above is obtained under the assumption that the function B ( x, t ) is continuously differentiable in all arguments ( x, t ) . However there are cases, when the function B ( x, t ) does not possess this property. Solving the Bellman equation (14) - (15) presents a difficulty. The following literature is supposed on the method of dynamic programming: R. Bellman, Dynamicheskoe progrmmirovanie. M.: IL, 1960; R. Bellman, S. Dreyfus. Prikladnye zadachi dynamicheskogo programmirovania. M.: Nauka, 1965; Bryson A., Ho Yu-Chi.Prikladnaya teoria optimalnogo kontrolya. M.: Mir, 1972. Control questions 1. Give definition to the Bellman function. 2. Formulate the principle of optimality. 3. Give definition to the Bellman equation. 4. Describe the method of dynamic programming. 5. What kind optimal control is called by synthesizing?

106

LECTURE 14. DISCRETE SYSTEMS. OPTIMALITY PRINCIPLE. BELLMAN EQUATION

The optimal control problems with phase constraints for discrete systems are considered. Solving of such problems by the maximum principle is extremely difficult. Dynamic programming method is often used in the numerical solving of optimal control problems. Discrete systems. We consider the problem of optimal control with phase constraints in the form: minimize functional t1

J ( x, u )

³f

0

( x(t ), u (t ), t )dt Ɏ( x(t1 )) o inf

(1)

t0

at conditions x

f ( x, u , t ), x(t 0 )

x0 , t 0 d t d t1 , x0 G (t 0 ),

(2) (3)

x(t ) G (t ) E n , t 0 d t d t1 , u (t ) KC >t 0 , t1 @, u (t ) V (t ) E r , t 0 d t d t1 ,

(4)

where G (t ) E n , t 0 d t d t1

is a given set . We assume the set V

V (t ) E r , t 0 d t d t1 ,

and all the rest of data the same as before. After dividing the interval >t 0 ,t1 @ into N parts by the points t 0 t 01 ... t 0 N

t1

and by approximating it the equations (1) - (4) functional I 0 >xi @0 , >ui @0

N 1

¦F

are written as: minimize the

( xi , ui ) Ɏx N o inf

(5)

0,..., N 1, x0 G0

(6)

0i

i 0

at conditions xi 1

Fi ( xi , ui ), i

xi Gi E n , i

>ui @0

0,1,..., N , Gi

(u0 , u1 ,..., u N 1 ), ui Vi

G (t 0 ),

Gi (t i ),

V (t i ), i 0,..., N 1,

where

>xi @0

( x0 , x1 ,..., x N ), xk E n , k 107

0, N , u k E r , k

0, N 1;

(7) (8)

functions F0i ( xi , u i )

f 0 ( xi , u i , t i )(t i 1 t i ), Fi ( xi , u i ) xi (ti 1 ti ) f ( xi , ui , ti ).

Together with (5) - (8) we consider the following family of problems: I k >xi @k , >ui @k

N 1

¦F

0i

( xi , u i ) Ɏ x N o inf

(9)

i k

at conditions xi 1

k , k 1,..., N 1, xk

Fi ( xi , ui ), i

xi Gi E n , i

>ui @k

x Gk ,

(11)

k , k 1,..., N ,

(u k , u k 1 ,..., u N 1 ), ui Vi , i

(10)

k , k 1,..., N 1,

(12)

dependent on k and vector x E n . The transition from the problem (5) - (8) to (9) (12) is called by principle of invariance embedding. The principle of optimality. Let

>ui @0k u k0 , u k01 ,..., u N0 1

be optimal control for the state ( k , x ) of the problem (9) - (12). Then the control

>ui @0s u s0 , u s01 ,..., u N0 1 is optimal for the state ( s, x s0 ), s ! k ,

>xi @0s xs0 , xs01 ,..., x N0 . We note, that trajectory

>xi @0k xk0 corresponds to the optimal control >ui @0k .

x, x k01 ,..., x N0

We introduce the function B x ( x)

where the set ' k ( x)

>xi @k

( xk

inf

>ui @k ' k ( x )

^>ui @k / ui Vi , i

I k >xi @k , >ui @k , xk

x,

(13)

k , k 1,..., N 1,

x, xk 1 ,..., x N ), xi Gi , i

k , k 1,..., N ` .

Function Bk (x) is called the Bellman equation for the problem (9) - (12). From Bellman's principle of optimality and introduced notation (13) at s k 1 , it follows 108

that Bk ( x )

F0 k ( x k0 , u k0 ) Bk 1 ( x k01 ), x k0

x, x k01

Bk ( x ) d F0 k ( x k0 , u k0 ) Bk 1 ( x k 1 ), x k 1 x k 1 Gk 1 , x k0

Fk ( x k0 , u k0 ),

Fk ( x k0 , u k ), u k Vk ,

(14) (15)

x Gk .

If we introduce the notation

^>ui @k

Dk ( x )

u V ,

uk

u, u k 1 ,..., u N 1 ) u Vk , >u i @k ' k ( x)` ,

(u k

then the relations (14), (15) can be written as Bk ( x )

inf >F0 k ( x, u ) Bk 1 ( Fk ( x, u ))@, k

uDk ( x )

B N ( x)

0,1,..., N 1,

(16) (17)

Ɏ( x).

Relation (17) follows directly from (13) and (9). Recurrence relations (16) and (17) are Bellman equations for problem (9) - (12). Now we consider the algorithm for solving the problem (5) - (8) based on the recurrence relations (16) and (17). stage 0. At k N the function BN ( x) Ɏ( x) , where x G N . stage 1. For the value k N 1 the relation (16) is written in he form [see formula (14), (15)]: BN 1 ( x)

inf

>F0 N 1 ( x, u ) BN ( FN 1 ( x, u ))@,

uDN 1 ( x )

^u E

D N 1 ( x )

r

`

/ u V N 1 , FN 1 ( x, u ) G N .

(18) (19)

By introducing the notations M1 ( x, u ) F0 N 1 ( x, u ) BN ( FN 1 ( x, u )) ,

the problem (18) , (19) are represented in the form M1 ( x, u ) o inf, u VN 1 , FN 1 ( x, u ) G N .

(20)

We define the function u 0 u 0 ( x) as solution of nonlinear programming (20). We denote by X N 1

^x E

n

/ FN 1 ( x, u 0 ( x)) G N , x G N 1 `.

109

Note, that BN ( FN 1 ( x, u )) Ɏ( FN 1 ( x, u )) .

Substituting the value u 0 u 0 ( x) to the right hand side of (18) we obtain BN 1 ( x) M1 ( x, u 0 ( x)) M 1( x), x X N 1 , u 0 ( x) u N0 1 ( x).

(21)

Stage 2. We consider the equation (16) for values k N 2 . Equation (16) can be written as BN 2 ( x)

>F0 N 2 ( x, u ) BN 1 ( FN 2 ( x, u ))@,

inf

(22)

uDN 2 ( x )

^u VN 2 /

DN 2 ( x)

FN 2 ( x, u ) X N 1 `,

(23)

where BN 1 ( FN 2 ( x, u )) M 1 ( FN 2 ( x, u )) .

By denoting M 2 ( x, u ) F0 N 2 ( x, u ) M 2 ( x, u ) F0 N 2 ( x, u ) B N 1 ( FN 2 ( x, u )) ,

nonlinear programming problem (22), (23) is written as follows: M 2 ( x, u ) o inf, u V N 2 , FN 2 ( x, u ) X N 1 , x G N 2 ,

(24)

where X N 1 is the known set defined in the previous iteration. Solving the problem of nonlinear programming (24) we find u0

and the set

^x E

X N 2

n

u 0 ( x)

u N0 2 ( x )

`

/ FN 2 ( x, u 0 ( x)) X N 1 , x G N 2 .

By substituting the value u0

u 0 ( x)

u N0 2 ( x )

to the right hand side of (22) we obtain BN 2 ( x) M 2 ( x, u 0 ( x)) M 2 ( x), x X N 2 , u 0 ( x) u N0 2 ( x).

(25)

Continuing this process, we determine [see formula (21), (25)]: B N k ( x) 0

u ( x)

u

0 N k

M k ( x), x X N k , ( x), k

1,2,..., N 1.

110

(26)

N-th stage. For the value k 0 equation (16) is written as: B0 ( x)

inf >F00 ( x, u ) B1 ( F0 ( x, u ))@,

(27)

^u V0 /

(28)

uD0 ( x )

D0 ( x)

F0 ( x, u ) X 1 `

where X 1 is the known set. By denoting M N ( x, u ) F00 ( x, u ) B1 ( F0 ( x, u )), B1 ( F0 ( x, u )) M N 1 ( F0 ( x, u )) ,

the problem (27) , (28) is written in the form M N ( x, u ) o inf, u V0 , F0 ( x, u ) X 1 .

Hence, we find u 0 ( x) u 00 ( x), B0 ( x) M N ( x, u 0 ( x)) M N ( x), x X0

^x E

n

/ F0 ( x, u ( x)) X 1 , x G0 `.

(29)

0

Stage N +1. As follows from formula (29), the known function B0 ( x) M N ( x) , x X 0 , where X 0 E n is also known set. We define the initial state x00 from the solution of the nonlinear programming problem B0 ( x) M N ( x) o inf, x X 0 .

Then the minimum value of the functional (5) is equal to B0 ( x00 ) . Stage N +2. By assuming x x00 we define the optimal control u 00 ( x00 ) at the first step of the equation (6) from the first condition of (29). Further, we find the value x10

F0 ( x00 , u 00 )

of equation (6). Then, as follows from the formula (26) optimal control u10 u10 ( x10 ) , etc.: u 00

u 0 ( x00 ), x10 ..., u N0 1

F0 ( x00 , u 00 ), u10

u10 ( x10 ), x 20

u N0 1 ( x N0 1 ), x N0

FN 1 ( x N0 1 , u N0 1 ).

Example 1. Minimize the functional I 0 >xi @0 , >ui @0

1

¦(x

2 i

i 0

at conditions 111

F1 ( x10 , u10 ),

ui2 ) x22 o inf

xi ui , i 0,1, [i, i 2], i 0,1, 2,

xi 1 xi Gi

>ui @0

(u 0 , u1 ), u i E 1 , i

0,1.

10. For k 2 B2 ( x) x 2 , x G2 [2;4], X 2 G2 [2;4]. 20. For the values k 1 the function B1 ( x)

> ^u E

inf x 2 u 2 ( x u ) 2

D1 ( x )

D1 ( x)

1

@

/ x u G2

>

@

inf 2 x 2 2u 2 2 xu ,

D1 ( x )

[2; 4]`, G1 [1; 3].

This nonlinear programming problem can be written as M1 ( x, u )

2 x 2 2u 2 2 xu o inf, 2 d x u d 4, ( g1

2 x u, g 2

x u 4)

Lagrange function L (u, O1 , O2 ) 2 x 2 2u 2 2 xu O1 (2 x u ) O2 ( x u 4), O1 t 0, O2 t 0.

We define a saddle point of Lagrange function of the conditions 4u 0 2 x O10 O02

0, O10 (2 x u 0 ) 0, O02 ( x u 0 4) 0,

O10 t 0, O02 t 0.

We can show, that u0

2 x, O10

8 2 x ! 0, O02

0.

It follows that u0

2 x, x d 4 .

u10

Then the set X 1 [1;3] G1 , and the function 2 x 2 4 x 8 M 1 ( x ).

B1 ( x )

30. For the values k 0 the function B0 ( x) D0 ( x )

>

@

inf x 2 u 2 2( x u ) 2 4( x u ) 8 ,

D0 ( x )

^u E

1

/ x u X1

[1; 3]`, G0

[0; 2].

Corresponding problem of nonlinear programming can be written as: M 2 ( x, u ) x 2 u 2 2( x u ) 2 4( x u ) 8 o inf, g1 (u ) 1 x u d 0, g 2 (u ) x u 3 d 0.

Lagrange function 112

L (u, O1 , O2 )

x 2 u 2 2( x u ) 2 4( x u ) 8 O1 (1 x u ) O2 ( x u 3), O1 t 0, O2 t 0.

The saddle point is determined by the conditions 6u 0 4 x 4 O10 O02

0, O10 (1 x u 0 ) 0, O2 ( x u 0 3) 0, O1 t 0, O2 t 0.

Hence we get u0

O10

u 0 (0) 1 x , 2 2 x t 0, O02 0 .

Since O10 t 0 at x d 1 ,

then set X 0 [0;1] , control u 0 (0) 1 x , function B0 ( x) 2 x 2 2 x 7 M 2 ( x) .

40. We define the initial condition x 0 from the solution of the problem 2 x 2 2 x 7 o inf , x X 0 [0,1] .

Hence we have x 0

1 , the minimum value of the functional 2 13 . B0 ( x 0 ) 2

50. Now consequently we define the optimal control and optimal trajectory u 00

1 x0

1 0 , x1 2

u 00 x 0

1, u10

2 x10

1, x20

u10 x10

2.

Example 2. We consider the problem that described in Example 1, in the absence of phase constraints at 1 . 2

x0

For values k 1 we get B1 ( x )

>

@

inf 2 x 2 2u 2 2 xu , u10

uE1

1 x, B1 ( x) 2

For the values k 0 the function B0 ( x )

3 ª º inf « x 2 u 2 ( x u ) 2 ». 2 ¬ ¼

uE1

113

3 2 x . 2

3x 1 . Hence, at x x0 we obtain u 00 5 2 2 minimum value of the functional B0 ( x0 ) . The values 5 1 , x10 x0 u 00 5 1 1 0 . u10 , x 20 x10 u1 10 10

Consequently u 00

3 10

and the

Recommended literature: Aris R. Discretnoe dynamicheskoe programmirovanie. M.: Mir, 1969; Boltyanskii V.G. Optimalnoe upravlenie discretnymi systemami. M.: Nauka, 1973. Control questions 1. Formulate optimal control problem with phase constraints for discrete systems. 2. What kind principle is called by invariance embedding? 3. Give definition to the Bellman equation. 4. Give definition to the Lagrange function. 5. Formulate the algorithm for solving optimal control problem with phase constraints for discrete systems.

LECTURE 15. SUFFICIENT OPTIMALITY CONDITIONS

As it is shown above, the maximum principle is necessary condition for strong local minimum, therefore the question arises: when a couple determined from the principle of maximum will be the solution of the initial problem of optimal control? A positive answer to the question the sufficient optimality conditions for V.F. Krotov problems with a fixed time give. We consider the problem of optimal control with fixed time under phase constraints: minimize the functional t1

J ( x, u )

³f

0

( x(t ), u (t ), t )dt Ɏ0 ( x(t 0 )) Ɏ1 ( x(t1 )) o inf

(1)

t0

at conditions x (t )

f ( x(t ), u (t ), t ), t 0 d t d t1 ,

x(t ) G (t ), t 0 d t d t1 , u (t ) KC >t 0 , t1 @, u (t ) V (t ) E r , t 0 d t d t1 ,

where t 0 , t1 are precsribed;

G (t ) E n , V (t ) E r , t >t 0 ,t1 @

are given sets; Ɏ0 ( x), Ɏ1 ( x)

are known functions, all other notations are the same as in the previous lectures. 114

(2) (3) (4)

Krotov function. The set of all pairs ( x(t ), u (t )) that satisfy the conditions (2) (4) is denoted by D>t 0 ,t1 @ . For each fixed t , t >t 0 , t1 @ the pair ( x, u ) E n u E r .

We denote by Dt the set of all ( x, u ) E n u E r

for which

( x (t ), u (t )) D>t 0 ,t1 @ , Dt E n u E r .

The projection of the set Dt on E n is denoted X t , i.e., Xt

PE n ( Dt )

for each t , t >t 0 , t1 @ , similarly we denote Vt

PE r ( Dt )

which is projection Dt on E r for each t , t >t 0 ,t1 @ . We note, that X t G (t ) , Vt V

for each t , t >t 0 ,t1 @ . Definition. Function K ( x, t ), x X t , t , t 0 d t d t1

is called by Krotov function, if it is defined and piecewise continuous for x X t , t 0 d t d t1 , and has a piecewise continuous partial derivative Kx

in the domain

( K x1 ( x, t ),... .., K xn ( x, t )), K t

( x, t ) X t u >t 0 , t1 @ .

Based on the function K ( x, t ) and the initial data of the problem (1) - (4) we define the following functions: R ( x, u , t )

t >t 0 , t1 @.

r0 ( x)

K x ( x, t ), f ( x, u, t ) K t ( x, t ) f 0 ( x, u, t ),

K ( x, t 0 ) Ɏ0 ( x), r1 ( x) K ( x, t1 ) Ɏ1 ( x), x X t ,

115

(5) (6)

We show the possibility of representing the functional (1) through introduced functions R( x, u, t ), r0 ( x), r1 ( x)

of the formulas (5) and (6) along the feasible pairs ( x(t ), u (t )) D>t 0 , t1 @ ,

satisfying the conditions (2) - (4). Lemma 1. If along the feasible pairs ( x(t ), u (t )) D>t 0 , t1 @

the function K ( x(t ), t ) of t is continuous and piecewise smooth, i.e. belongs to the class KC 1 >t 0 ,t1 @ , the functional (1) can be represented as t1

J ( x, u )

³ R( x(t ), u(t ), t )dt r ( x(t 0

0

)) r1 ( x(t1 )).

(7)

t0

Proof. In fact, since

K ( x (t ), t ) KC 1 >t 0 , t1 @

K (t )

along feasible pairs

( x (t ), u (t )) D>t 0 ,t1 @ ,

then the derivative dK ( x(t ), t ) dt

K x ( x(t ), t ), f ( x(t ), u (t ), t ) K t ( x(t ), t )

(8)

exists everywhere on >t 0 , t1 @ , in exception of finite number of the points. The relation (8) with regard to the expression (5) is written in the form dK ( x (t ), t ) dt

R ( x(t ), u (t ), t ) f 0 ( x(t ), u (t ), t ), t >t 0 , t1 @ .

(9)

Since the function K ( x(t ), t ) is continuous in t, then by integrating equality (9) in t, we get t1

dK ( x(t ), t ) dt dt t0

³

K ( x(t1 ), t1 ) K ( x(t 0 ), t 0 )

t1

t1

t0

t0

³ R( x(t ), u(t ), t )dt ³ f 0 ( x(t ), u(t ), t )dt.

Hence, in view of (6) we obtain the relation (7). The lemma is proved. Let the pair ( x 0 (t ), u 0 (t )) be the solution of (1) - (4), i.e. 116

J (x0 , u 0 )

Let

inf J ( x, u )

D[ t 0 ,t1 ]

J* .

( x(t ), u (t )) D>t 0 , t1 @ .

We show, that the following estimation holds: t1

0 d J ( x, u ) J ( x 0 , u 0 ) d ³ [ R( x(t ), u (t ), t ) Rmin (t )]dt t0

r1 ( x(t1 )) r1 min r0 ( x(t 0 )) r0 min

'( x, u ),

(10)

where Rmin (t )

inf R ( x, u , t ), r1 min

inf r1 ( x ), r0 min

x X t1

( x ,u )Dt

inf r0 ( x).

x X t 0

In fact, t1

0 d J ( x, u ) J ( x 0 , u 0 ) d ³ [ R( x(t ), u (t ), t ) R( x 0 (t ), u 0 (t ), t )]dt t0

r1 ( x (t1 )) r1 ( x 0 (t1 )) r0 ( x (t 0 )) r0 ( x 0 (t 0 )) d '( x, u )

by (7) and r1 ( x 0 (t1 )) t r1min , r0 ( x 0 (t 0 )) t r0 min .

We note, that the estimate (10) remains valid if Rmin (t ) , r1min , r0 min

are determined on the sets more widely than Dt , X t1 , X t0 ,

namely, Rmin (t )

inf

inf R( x, u, t ), r1 min

xG ( t ) uV ( t )

inf r1 ( x), r0 min

xG ( t1 )

inf r0 ( x) ,

xG ( t 0 )

(11)

since Dt G (t ) u V (t ), X t1 G (t1 ), X t0 G (t 0 ) .

Sufficient optimality conditions. We assume, that the pair ( x 0 (t ), u 0 (t )) D>t 0 , t1 @

is found on the base of the principle of maximum or some other method as a suspect optimal solution of (1) - (4). Theorem 1. Let condition of lemma 1 holds. In order the pair 117

( x 0 (t ), u 0 (t )) D>t 0 , t1 @

to be the optimal solution of the problem (1) - (4) , it is sufficient to satisfy the following equations: R( x 0 (t ), u 0 (t ), t ) 0

Rmin (t ), t 0 d t d t1 ,

r1 min , r0 ( x 0 (t 0 ))

r1 ( x (t1 ))

r0 min .

(12)

Proof. Since condition of lemma 1 holds, then there is a function K ( x (t ), t ) KC 1 >t 0 , t1 @

along arbitrary feasible pairs

( x(t ), u (t )) D>t 0 , t1 @

of the problem (1) - (4), and the function (1) is represented in the form (7). Then for every feasible pair ( x(t ), u (t )) D>t 0 , t1 @

the difference J ( x, u ) J ( x 0 , u 0 )

t1

³ [ R( x(t ), u(t ), t ) R( x

0

(t ), u 0 (t ), t )]dt

t0

r1 ( x(t1 )) r1 ( x 0 (t1 )) r0 ( x(t 0 )) r0 ( x 0 (t 0 ))

t1

³ >R( x(t ), u(t ), t )

t0

Rmin (t )@dt r1 ( x(t1 )) r1 min r0 ( x(t 0 )) r0 min t 0

by (12). It follows that

J ( x 0 , u 0 ) d J ( x, u ), ( x(t ), u (t )) D>t 0 , t1 @.

This means that J (x0 ,u 0 )

inf J ( x, u )

D >t0 ,t1 @

J* .

Theorem is proved. It should be noted, that the function Rmin (t ) , t >t 0 ,t1 @ ,

the values r1min , r0 min

can be defined as solutions of the optimization problem in a finite-dimensional space represented in the obtaining of estimation (10) or by (11). We note, that in the cases, when it is difficult to define the sets Dt , X t1 , X t0 ,

118

it is advisable to replace them by G (t ) u V (t ), G (t1 ), G (t 0 )

respectively. We assume, that by the method of consequence approximations the sequence

^xk (t ), u k (t )` D>t 0 , t1 @ for the problem (1) - (4) is defined. The question arises: whether the sequence

^xk (t ), u k (t )` D>t 0 , t1 @ is minimized, i.e. lim J ( xk , u k ) k of

inf J ( x, u )

D >t0 ,t1 @

J* .

Theorem 2. Let the function K ( x(t ), t ) in the variable t is continuous and belongs to the class KC 1 >t 0 ,t1 @ along arbitrary feasible pair ( x(t ), u (t )) D>t 0 , t1 @ .

In order to the sequence

^xk (t ), u k (t )` D>t 0 , t1 @

to be minimized sufficiently, that t1

t1

lim ³ R( xk (t ), u k (t ), t )dt k of

lim r1 ( xk (t1 ))

min

(13)

(t )dt ,

t0

r1 min , lim r0 ( xk (t 0 ))

k of

Proof. Let

³R

t0

k of

r0 min .

(14)

( x(t ), u (t )) D>t 0 , t1 @

be an arbitrary feasible pair for the problem (1) - (4). Then the difference t1

³ [ R( x(t ), u(t ), t ) R( x

J ( x, u ) J ( x k , u k )

k

(t ), u k (t ), t )]dt

t0

r1 ( x(t1 )) r1 ( xk (t1 )) r0 ( x(t 0 )) r0 ( xk (t 0 )),

due to the fact that the condition of lemma 1 holds. Hence, passing to the limit at k o f , with taking into account relations (13), (14) , we obtain J ( x, u ) lim J ( xk , u k ) k of

t1

³ [ R( x(t ), u(t ), t ) R

min

(t )]dt

t0

r1 ( x(t1 )) r1min r0 ( x(t 0 )) r0 min t 0.

Consequently, 119

J ( x, u ) t lim J ( xk , u k ), ( x(t ), u (t )) D>t 0 , t1 @. k of

This means that lim J ( xk , u k ) k of

J*

inf J ( x, u ) .

D >t0 ,t1 @

Theorem is proved. Existence of Krotov function. Let K ( x, t ) , x X t , t >t 0 ,t1 @

be Krotov function for the problem (1) - (4), i.e. along each feasible pair ( x(t ), u (t )) D>t 0 , t1 @

the function

K ( x (t ), t ) KC 1 >t 0 , t1 @ .

We show, that function K ( x, t )

where

K ( x , t ) J (t ) ,

J (t ) KC 1 >t 0 , t1 @

is also Krotov function. In fact, in this case the functions R ( x, u , t )

K x ( x, t ), f ( x, u, t ) K t ( x, t ) f 0 ( x, u, t )

R ( x, u , t ) J (t ), r1 ( x)

r0 ( x)

K ( x, t1 ) Ɏ1 ( x )

K ( x , t 0 ) Ɏ0 ( x )

r0 ( x) J (t 0 ),

r1 ( x) J (t1 ), x X t , t >t 0 , t1 @,

and functional (7) can be written as t1

J ( x, u )

³ R ( x(t ), u(t ), t )dt r ( x(t 0

0

)) r1 ( x(t1 ))

t0

t1

³ R( x(t ), u(t ), t )dt J (t ) J (t 1

0

)

t0

r0 ( x) J (t0 ) r1 ( x) J (t1 )

Thus, the function K ( x, t )

K ( x, t ) J (t )

is also a function of Krotov for problem (1) - (4), where J (t ) KC 1 >t 0 , t1 @

is an arbitrary function. In particular, when the function 120

J ( x, u ).

t

J (t ) ³ R min (t )dt r1min , t >t 0 , t1 @, t1

we get R ( x, u, t ) R( x, u, t ) Rmin (t ) , r 1( x) r1 ( x) r1min , r0 ( x) r0 ( x) J (t 0 ) ,

moreover Rmin (t ) r1 min

inf R ( x, u, t ) 0 ,

( x ,u )Dt

inf r 1( x) 0 ,

xX t1

inf r0 ( x )

xX t 0

r0 min J (t 0 ) .

Hence we get R ( x, u , t ) t 0 , ( x, u ) Dt ,

r1 ( x) t 0 , x X t1 .

Thus, in general, we can assume Rmin (t ) 0 , r1min

0,

at the conditions of Theorems 1 and 2 . Then the basic relations for determining of Krotov function are K x ( x, t ), f ( x, u, t ) K t ( x, t ) f 0 ( x, u, t ) t 0,

R ( x, u , t )

( x, u ) Dt , t >t 0 , t1 @,

inf

( x , u )Dt

>K

x

( x, t ), f ( x, u, t ) K t ( x, t ) f 0 ( x, u, t )@ 0, K ( x, t1 ) Ɏ1 ( x ) t 0, x X t1 ,

r1 ( x )

where

(15) (16) (17)

r1 ( x 0 (t1 )) K ( x 0 (t1 )) Ɏ1 ( x 0 (t1 )) 0,

(18)

r0 ( x ) { K ( x, t 0 ) Ɏ0 ( x ) t r0 min , x X t0 ,

(19)

K ( x 0 (t 0 ), t 0 ) Ɏ0 ( x 0 (t1 ))

(20)

r0 min ,

( x 0 (t ), u 0 (t )) D>t 0 , t1 @

is the optimal pair for the problem (1) - (4). Thus, Krotov function is determined from the differential inequality (15) and relations (17), (19), and the optimal pair ( x 0 (t ), u 0 (t )) D>t 0 , t1 @

from correlations (16), (18), (20). 121

Unfortunately, the general methods for solving the Cauchy-Krotov problem (15), (17), (19) is currently not available. Since the constructive description of the sets Dt , X t , X t are often absent, it is advisable to consider the Cauchy-Krotov problem under simplifying assumptions by replacing 1

0

Dt , X t1 , X t0

on G (t ) u V (t ), G (t1 ), G (t 0 )

respectively. In particular, as a result of simplifications can be obtained Bellman equations inf

uV ( t )

>K

x

( x, t ), f ( x, u, t ) K t ( x, t ) f 0 ( x, u , t )@ 0, t >t 0 , t1 @, x X t , K ( x 0 (t1 ), t1 ) K ( x 0 (t 0 ), t 0 )

Ɏ1 ( x 0 (t1 )), x 0 (t1 ) X t1 , Ɏ0 ( x 0 (t 0 )), x 0 (t 0 ) X t0 .

(21)

(22)

We note, that Krotov function K ( x, t ) is defined of the more common conditions (15), (17), (19) and it can exist even when Bellman function defined by (21), (22) does not exist. Recommended literature: V.F. Krotov, V.I. Gurman. Methody i zadachi optimalnogo upravlenia. M.: Nauka, 1973. Control questions 1. Give definition to the Krotov function. 2. Prove the Lamma 1. 3. Prove the Theorem 1. 4. Prove the Theorem 2. 5. Represent the Cauchy-Krotov problem.

APPENDIX 1 TASKS FOR INDEPENDENT WORK

P. 1.1. NUMERICAL METHODS OF MINIMIZATION 1. Let

J u

Au, u b, u ,

where A is prescribed symmetric matrix of n u n, b E n order. a) Describe the method of steepest descent to minimize of function J (u ) . Specify an explicit expression for D n . b) Describe Newton method for minimizing of function J (u ) . 2. Let 2 J u Au b , where A is prescribed matrix of m u n, b E m order. a) Describe the method of steepest descent for minimization of function J (u ) . Specify an explicit expression for D n . b) Describe the method of conjugate gradients and Newton's method for minimization of function J (u ) . 3. Solve the equation e u 2 0 by Newton's method, u 0 1 . 4. Minimize the following functions using method of steepest descent, conjugate gradients method, Newton's method: ɚ) J u 5u12 2u1u 2 2u1u 3 4u 22 4u 2 u 3 u 32 2u1 u 2 u3 6, u 0

ɛ) J u 4u12 2u1u 2 2u 2 u 3 3u 22 2u 32 ɜ) J u 6u 4u1u 2 2u1u 3 5u 6u 2 u 3 4u 2 1

2 2

1,1,1 .

1,1,1 .

u1 u 2 u3 1, u 0

2 3

3u1 2u 2 u3 10, u 0

2,1,1

. ɝ) J (u ) (2u1 u 2 u 3 ) 2 (u1 u 2 u 3 ) 2 (u1 u 2 3u3 ) 2 ,

123

u0

1,1,1 .

P. 1.2. VARIATION CALCULUS The simplest problem. Minimize the functional t1

³ F ( x(t ), x (t ), t )dt o inf

J ( x, x )

(1)

t0

at conditions

x(t ) C 1 >t 0 , t1 @, x(t 0 )

x 0 , x (t 0 )

x1 .

The function F ( x, y , t ) has continuous partial derivatives with respect to all arguments including the second order. It is said that the feasible function x 0 (t ) C 1 >t 0 , t1 @, x 0 (t 0 )

x0 , x 0 (t1 )

x1

yields a weak local minimum to functional J ( x, x ) , if there exists number H ! 0 such that for any feasible function x (t ) C 1 >t 0 , t1 @ , x(t 0 )

x0 , x(t1 )

x1

for which 0 max x(t ) x (t ) H ,

t 0 dt %t1

0 max x (t ) x (t ) H

t 0 dt %t1

the inequality J ( x, x ) t J ( x 0 , x 0 )

holds.

Theorem 1. In order the function x 0 (t ) C 1 >t 0 , t1 @ x 0 (t 0 )

x 0 , x 0 (t1 )

x1

conducts a weak local minimum to the functional (1), it is necessary that it satisfies Euler equation Fx ( x 0 , x 0 , t )

d Fx ( x 0 , x 0 , t ) dt

0, x 0 (t 0 )

x0 , x 0 (t1 )

x1 .

(2)

Euler equation (2) can be written as Fx Fxt Fxx x 0 Fxx x0

0, x(t 0 )

x 0 , x ( t1 )

Example 1. J x, x

³ x t dt o inf, x0 1

2

0

0, x1 1 .

Function F x , therefore 2

Fx

0, Fx

2 x , Fxt 124

0, Fxx

0, Fxx

2.

x1 .

(3)

Then according to (3) we get 2 x0 t 0 . It follows that x0 t 0 . The solution of this differential equation x 0 t c1t c2 . We define the constants c1 , c2 by the conditions x 0 0 0, x 0 1 1 .

Constants c1 1, c 2

0 . The initial solution x 0 t t , 0 d t d 1 .

In the case, the function

x1 t ,..., xn t C 1 >t0 , t1 @, x t 0 x0 , x t1 x1 , x0 , x1 E n , xt

Euler equation (2) can be written as d Fxi ( x 0 , x 0 , t ) 0, dt x0 , xt1 x1 .

Fxi ( x 0 , x 0 , t ) xt 0

Example 2. J x1 , x 2 , x1 , x 2

1

³ x t x t dt o inf, 1

2

0

x1 0 0, x1 1 0, x2 0 0, x2 1 1.

Function F

x 1 x 2 , therefore Fx1

0, Fx2

0, Fx1

x 2 , Fx2

0, x10

0.

x1 .

Then from (4) follows, that It follows that

x20

x10 t c1t c2 , x20

c3 t c 4 .

We define the constants c1 , c2 , c3 , c4 from the conditions x10 0 0, x10 1 0, x 20 0 0, x20 1 1 .

Constants c1

0, c2

0, c4

125

0, c3

1.

i 1, n,

(4)

The source solutions of the problem x10 t { 0, t >0,[email protected], x 20 t t , t >0,[email protected] .

We consider the functionals, that depend on higher order derivatives: minimize the functional t1

J x

³ F ( x(t ), x (t ),..., x

n

(t ), t ) dt o inf

(5)

t0

at conditions

xt C n >t 0 , t1 @, xt0 x0 ,

x t 0

x01 ,..., x n 1 t0 x0 n 1 ,

xt1

x t1 x11 ,..., x

x1 ,

n 1

t1

x1n 1 .

Functions x ( t ) C n [t 0 , t 1 ]

that satisfy the conditions x k t j x jk , k

0, n 1, j

0,1, x00

x0 , x10

x1

are called by feasible. We say that the feasible function x 0 t conducts a weak local minimum to the functional (5), if there exists a number H ! 0 such that for arbitrary feasible function xt for which max xt x 0 t H ,...,

t0 dt dt1

max x n t x 0

t 0 d t dt1

n

t H

the equality J x t J x 0 holds. Necessary condition for a weak local minimum to the functional (5) on the set of feasible functions is determined by Euler - Puasson equation Fx

along

n d2 d n d Fx 2 Fx ... 1 F n dt dt dt n x

x 0 t C n >t 0 , t1 @, x k t j

x jk , k

0, n 1, j

0. 0,1 .

Example 3. 1

³ x t dt o inf, x0 2

0, x 0 0, x1 0, x 1 1 .

0

126

(6)

Function F x2 , therefore the equation of Euler-Puasson takes the form x0 t 0 .

The solution of this equation x 0 t c1t 3 c2 t 2 c3t c4 , t >0,[email protected] .

Constants ci , i 1,4 are determined by the conditions Hence we get

x 0 0 0, x 0 0 0, x1 0, x 1 1 . c1

The source solution

1, c2

1, c3

0, c4

0.

x 0 t t 3 t 2 , t >0,[email protected] .

Hilbert condition. Let the function x 0 (t ), t >t 0 , t1 @ be solution of the Euler equation (2). If the function Fxx x 0 t , x 0 t , t z 0, t >t0 ,t1 @ , the function x0 t , t >t 0 , t1 @

has a continuous second derivative. Necessary conditions of higher order. It is said that on x 0 (t ), t >t 0 , t1 @

(the solution of Euler equation (2)) satisfies the Legendre (strong Legendre condition), if (7) Fxx x 0 t , x 0 t , t t 0; Fxx x 0 t , x 0 t , t ! 0 , t , t >t 0 , t1 @ . Let At Fxx x 0 t , x 0 t , t , B t Fxx x 0 t , x 0 t , t , C t Fxx x 0 t , x 0 t , t . We assume that the strongest Legendre condition At ! 0 is performed. Let the function h 0 t , t >t 0 ,t1 @

be solution of the following Jacobi equation h0 t Pt h 0 t Qt h 0 t 0, h 0 t 0 h 0 t1 0 ,

where 127

(8)

Pt

A t / At , Qt

Bt C t / At .

The point W is called the conjugate to the point t0 , if there is a nontrivial solution h 0 t of the Jacobi equation (8), for which h 0 t 0 h 0 W 0 .

It is said that on x 0 t the condition of Jacobi holds, if in the segment t0 , t1 there are not any points conjugated to t 0 . Theorem 2. Let the function x 0 t , t >t 0 , t1 @

be solution of the Euler equation (2). If x 0 t C 2 >t 0 , t1 @

and it delivers a weak local minimum to the functional (1), the conditions Legendre and Jacobi must be held. Strong local minimum. Weierstrass necessary condition. It is said, that function x 0 (t ) C 1 >t 0 , t1 @,

x 0 (t 0 )

x 0 , x 0 (t1 )

x1

delivers a strong local minimum to the functional (1), if there exists such number H ! 0 , that for arbitrary feasible function x(t ) C 1 >t 0 , t1 @,

x(t 0 )

for which

x0 , x(t1 )

x1

max x t x 0 t H

t0 dt dt1

the inequality

J x, x t J x 0 , x 0

is performed. Since the necessary conditions for a weak local minimum are necessary conditions for strong local minimum, then the function x 0 (t ) C 1 >t 0 , t1 @,

is a solution of the Euler equation (2). 128

Theorem 3. In order the function x 0 (t ) C 1 >t 0 , t1 @, x 0 , x 0 (t1 )

x 0 (t 0 )

x1

to deliver a strong local minimum to functional in the simplest problem, it is necessary that for arbitrary point t t 0 ,t1 along the solution of the Euler equation the inequality Bx 0 t , x 0 t , t , [ F x 0 t , [ , t F x 0 t , x 0 t , t –

[ x 0 t , Fx x 0 t , x 0 t , t t 0, [ , [ E 1

holds (necessary condition of Weierstrass) . Example 4. S

J x, x

³ ( x

(t ) x 2 (t ))dt o inf,

2

0

x0 0, xS 0 .

1. Function x 2 x 2 .

F

Then Fx

2 x, Fx

2 x ,

consequently, the Euler equation can be written as x0 t x 0 t 0 .

The solution of this equation is x 0 t c1 sin t c2 cos t .

We define the constants c1 , c 2 by conditions x 0 0 0, x 0 S 0 .

Hence we get

x 0 t c1 sin t , t >0, S @ t 0

2 . Since Fxx

2

0, t1

S .

At ! 0 ,

then the strength Legendre condition (7) holds. Functions Bt

Fxx

2, C t Fxx 129

0,

therefore Qt

Pt

A t / At 0, 1 .

( B (t ) C (t )) / A(t )

Then the Jacobi equation (8) can be written as: h0 t h 0 t 0, h 0 0 0, h 0 S 0 .

Solution

h 0 t c3 sin t , t >0, S @ .

If c3 z 0 , then there is no point in the interval (0, S ) conjugated with t 0 Thus, for the function

0.

x 0 t c1 sin t , t >0, S @

the necessary conditions of the second order are held. Bolz problem. Minimize the functional J x, x

³

t1

t0

F ( xt , x t , t ) dt ) xt 0 , xt1 o inf ,

(9)

where t 0 ,t1 are the fixed points, the values xt 0 , xt1 are not fixed. It is said that the function x 0 (t ) C 1 >t 0 , t1 @,

delivers weak local minimum to the functional (9), if there exists number H ! 0 such that for arbitrary function xt C 1 >t 0 ,t1 @ ,

for which

max xt x 0 t H , t0 dt dt1

max x t x 0 t H

t0 dt dt1

the inequality

J x, x t x 0 , x 0

is fulfilled. Theorem 4. In order the function

x 0 (t ) C 1 >t 0 , t1 @

to deliver a weak local minimum to the functional (9) , it is necessary that it is a solution of the Euler equation

Fx x 0 , x 0 , t

and satisfies to the conditions 130

d Fx x 0 , x 0 , t dt

0

(10)

0 0 0 0 Fx x0 , x 0 , t 0 ) x x t 0 , x t1 , Fx x t1 , x t1 , t1 0

Further, if

) x1 x 0 t 0 , x 0 t1 .

xt

x1 t ,..., xn t , F F x1 ,..., xn , x1 ,..., x n , t , )x1 t 0 ,...xn t0 , xt1 ,..., xn t1 ,

then relations (10) and (11) can be written as

Fxi t k

1 k ) x

d Fx x 0 , x 0 , t dt i

Fxi x 0 , x 0 , t

ki

0, i 1, n;

0,1 .

, i 1, n, k

Example 5. 1

³ x t xt dt x 1 o inf .

J x, x

2

2

0

Functions

x 2 1 .

x 2 x, )

F

Since Fx

1, Fx

2 x ,

then the Euler equation (10) has the form 2 x0 t 1 0 .

Solution

x 0 t t 2 / 4 c1t c2 .

Constants c1 , c2 are determined by the transversality conditions (11). Since

Fx x 0 t 0 , x t 0 , t 0

2 x 0 t0 ) x0

0,

Fx x t1 , x t1 , t1 0

0

2 x 0 1

then c1

The source solution

0, c2

3/ 4 .

x 0 t t 2 / 4 3 / 4, t >0,[email protected] .

Isoperimetric problem. Minimize the functional 131

) x1

2 x 0 1 ,

(11)

t1

³ F ( x(t ), x (t ), t )dt o inf

J ( x, x )

(12)

t0

at conditions t1

K i ( x)

³G

.i

( x(t ), x (t ), t )dt

li , i

1, m,

(13)

t0

x(t 0 )

x 0 , x(t1 )

x1 ,

where x(t )

( x1 (t ),..., x n (t )), t 0 , t1

are fixed and x0 , x1 E n are prescribed vectors. Theorem 5. If the vector function

x t ,..., x t ,

x 0 t

0 1

0 n

x 0 (t ) C 1 >t 0 , t1 @

delivers a weak local minimum of the functional (12) at conditions (13), then there are numbers O1 ,..., Om , not all zero, such that the vector function x 0 t is solution of the following equations:

Lxi x 0 , x 0 , t , O1 ,..., Om

d Lx x 0 , x 0 , t , O1 ,..., Om dt i

0, i 1, n

where the function (Lagrangian) m

Lx, x , t , O1 ,..., Om F x, x , t ¦ Oi Gi x, x , t . i 1

1

³ x t dt o inf ,

Example 6.

2

0

1

³ xdt

0, x0 0 , x1 1 .

0

Functions F

x 2 , G

x 2 Ox .

x, L

Equation (14) for this problem can be written as: Hence we have

2 x0 t O

0.

O / 4, t >0,[email protected] .

x 0 t c1t 2 c2 t c3 , c1

Constants c1 , c2 , c3 are determined by the conditions 1

x 0 0 0, x 0 1 1, ³ x 0 t dt 0

132

0.

Consequently

x 0 0 c3

0, x 0 1 c1 c2

1

³ x t dt

0.

c3 / 3 c 2 / 2

0

1,

0

Hence we get c1

The source solution

0.

2, c3

3, c2

x 0 t 3t 2 2t , t >0,[email protected], O 12 .

Lagrange problem. By Lagrange problem is called the following optimization problem: minimize the functional J x, u, t 0 , t1

t1

³ f xt , ut , t dt ) xt , xt , t , t o inf

(15)

f xt , u t , t , t 0 d t d t1 ,

(16) (17)

0

0

0

1

0

1

t0

at conditions

x t

g j x, u , t 0 , t1 d 0, i 1, m; g j x, u, t 0 , t1 0, j

m 1, s ,

where the functional g j x, u , t 0 , t1

t1

³ f xt , u t , t dt ) xt , xt , t , t ; j

j

0

1

0

1

t0

values t 0 , t1 , in general case are not fixed, furthermore t 0 , t1 ', '

is given finite interval. Constraint (16) is called by differential connection, the vector function xt

x1 t ,..., xn t

u t

u t ,..., u t

is phase variable, vector function 1

p

is control. We note, that x (t ) C 1 >t 0 , t1 @, u t C >t0 ,t1 @ .

It should be noted, that all prescribed problems above: the simplest, the problem of Bolz, isoperimetric and the so-called problem with the mobile end-points are partial cases of the Lagrange problem (15) - (17) at x t u t , g j x, u, t 0 , t1 0, j 1, s . Four 133

xt , ut , t0 , t1 is called by feasible controlled process, if xt C 1 >t 0 , t1 @, u t C >t 0 , t1 @, t 0 , t1 int ', t 0 t1

and everywhere on segment >t0 , t1 @ differential connection (16) and constraints (17) hold. It is said, that the feasible controlled process

x t , u t , t 0

0

0 0

, t10

delivers a weak local minimum to the functional (15) at conditions (16) , (17), if there exists such number H ! 0 , that for arbitrary feasible controlled process

xt , ut , t0 , t1 satisfying to condition max xt x 0 t max x t x 0 t max u t u 0 t

t 0 dt dt1

t0 dt dt1

t 0 d t dt1

t 0 t 00 t1 t10 H

inequality

J x, u , t 0 , t1 t J x 0 , u 0 , t 00 , t10

holds. Algorithm for solving Lagrange problem (15) - (17) 1. Construct the Lagrange functional / x, u, t 0 , t1 ,\ , O

t1

³ Ldt l

(18)

t0

where Lagrangian s

L

¦ O f x, u, t \ t , x f x, u, t i

i

,

i 0

s

l

¦ O ) xt , xt , t , t , \ t \ t ,...\ t C >t , t @ . 1

i

i

0

1

0

1

1

0

n

1

i 0

2. From the Euler equation

Lx x 0 , u 0 , t ,\ , O

find the conjugate variable

d Lx x 0 , u 0 , t ,\ , O dt

\ t \ 1 t ,...\ n t . 134

0

(19)

By substituting the value of Lagrangian L of equation (19), we obtain

¦ O f x t , u t , t f x t , u t , t \ t , s

\ t

0

i

0

0

ix

0

x

i 0

(20)

t >t 0 , t1 @.

3. The boundary conditions for the adjoint system (20) are determined by the conditions k (21) Lx x 0 t k0 , u 0 t k0 , t k0 ,\ t k0 , O 1 l x t , k 0,1 . k

Substituting the values Lx , Lx t from (21), we get k

\ t k0

1 k ¦ Oi ) ix t x 0 t00 , x 0 t10 , t00 , t10 , k s

k

i 0

4. Optimal control

>

u 0 t , t t 00 , t10

(22)

0,1.

@

is determined by the condition

Lu x 0 t , u 0 t , t 00 , t10 ,\ , O

0.

It follows that

¦ O f x t , u t , t f x t , u t , t \ t s

0

i

0

0

iu

0

u

>

@

0, t t 00 , t10 .

(23)

i 0

5. The optimal values t 00 , t10 ' are determined by the following conditions:

/ tk x 0 t k0 , u 0 t k0 , t00 , t10 ,\ t k0 , O

0,1 .

0, k

It follows that

1 k 1 ¦ Oi f i x00 t k0 , t k0 ¦ Oi ) it s

s

i 0

i 0

k

) ix tk x 0 t k0

0.

(24)

6. The values Oi , i 1, s are determined by the conditions Oi t 0, i

0, m, Oi g i x 0 , u 0 , t 00 , t10

0

0

0 0

gi x , u , t ,t

Theorem 6. If

x t , u t , t 0

0

0 0

, t10

0 1

0, i 1, m; 0, i

m 1, s.

(25)

is an optimal (in a weak sense) process, then there are the Lagrange multipliers O O0 , O1 ,..., Os and the function 135

\ t C 1 >t 00 ,t10 @

which is not equal to zero simultaneously, such that the conditions (20), (22), (23), (24), (25) hold. Example 7. Minimize the functional S

J x, u

2

³ u t dt o inf 2

0

at conditions x1

u x1 , x1 0 0, x 2 0 0, x1 S / 2 1.

x 2 , x 2

For this problem u2, )0

f0

g3

the vector

0, g 1

g 2 ) 2 x 2 (0) 0 , ) 3 x1 (S / 2) 1 0 ,

x2 , u x1 ,

f

x1 0 0,

)1

t0

0, t1

S /2

is fixed. 1. Lagrange functional S /2

/

³ Ldt l , 0

l

L

O0 u 2 \ 1 x1 x 2 \ 2 x 2 u x1 ,

O1 x1 0 O2 x2 0 O3 x1 S / 2 1 .

2. Euler equation (19) takes the form d Lx dt 1

0, Lx2

\ 2 , Lx1

\ 1 , L x2

Lx1

d Lx dt 2

0.

Since Lx1

\ 1 , Lx2

\2,

then the adjoint system \ t \ 1 t ,\ 2 t

is the solution of differential equations \ 1 t \ 2 , \ 2 t \ 1 t , t >0,S / [email protected]

3. Since

l x1 0

O1 , l x2 0

O 2 , l x1 S / 2

then 136

O3 , l x2 S / 2

0,

Lx1 L x1

t 0 t S /2

\ 1 0 l x1 0

O1 ,

\ 1 S / 2

l x1 S / 2

O3 , Lx2

t 0

\ 2 0 l x2 0 L x2

O2 ,

t S /2

\ 2 S / 2 l x2 S / 2

0 .

The same results can be obtained from (22). 4. By the condition Lu 0 [see formula (23)] it follows 2O0 u 0 \ 2

0.

5. Since t 0 , t1 are fixed t0 0, t1 S / 2 , then there is no neccesity to determine them. 6. The values O0 t 0, but O1 , O2 , O3 are arbitrary numbers, x10 0 0, x20 0 0, x10 S / 2 1 ,

where

x 0 t , t >0, S / [email protected]

is the solution of the differential equation x10

x20 , x 20

u 0 x10 .

Further, we analyze the obtained results: a) Let O0 0 . Then from 4 item follows that \ 2 t { 0, t >t0 ,t1 @ .

Since \ 2 0 O2 , then O2

0 . Since \ 2

\ 1 , then

\ 1 t { 0, t >0, S / [email protected] .

We note, that \ 1 0 O1 , \ 1 S / 2 O3 , consequently O1 0, O3 0 . Thus, in the case O0 0 we have \ 1 t { 0, \ 2 t { 0 , O1 0, O2 0, O3 0 . It can not be (see Theorem 6). b) We assume O0 1 / 2 . Then we have u 0 t \ 0 t , t >0, S / [email protected] of 4 item. The optimal process is defined by the equations: x10

\ 1 0

x 20 ,

x10 0 0, \ 1

x 20

u 0 x10

\ 2 t x10 ;

x 20 0 0, x10 S / 2 1; \ 2, \ 2 \ 1 , O1 , \ 1 S / 2 O3 , \ 2 0 O 2 , \ 2 S / 2 0,

137

where O1 , O2 , O3 are arbitrary numbers. Since \2 \ 2

0 , then

\ 2 t c1 sin t c2 cos t .

Taking into account that \ 2 S / 2 0 we have \ 2 t c2 cos t . Therefore, the optimal control u 0 t c2 cos t . The solution of differential equations x10

can be written as

x 20 , x 20

x10 t

c 2 cos t x10

c3 c 4 t sin t c5 cos t .

Constants c3 , c 4 , c5 are determined by the given conditions at the ends x10 0 0, x10 0

x 20 0 0, x10 S / 2 1 .

It follows that c3

0, c 4

x10 t

2

2 / S , c5

0.

Source solution S

t sin t , t >0, S / [email protected] .

4 / S from the equation

We define the constant c2

x10 t x10 t c 2 cos t ,

where Then optimal control Solution of the problem

x10 t

2t sin t / S .

u 0 t

4 cos t / S .

x t , u t 2t sin t / S , 4 cos t / S ,

x 20 t

0 1

0

x10 t 2 sin t / S 2t cos t / S , t >0, S / [email protected] .

Solve the simplest problems 1-10: 1.

1

³ x t dt o inf, x0 2

1, x1 0 .

0

2.

T0

³ x t dt o inf, x0 2

0, xT0 [ .

0

3.

3/ 2

³ x

2

2 x dt o inf, x0

0, x3 / 2 1 .

0

138

4.

e

³ tx dt o inf, x1

e, xe 1 .

2

1

5.

3

³ t

1 x 2 dt o inf, x2

2

0, x3 1 .

2

6.

1

³ x

2

xx 12tx dt o inf, x0 x1

0.

0

7.

1

³ 4 x sin t x

x1

x 2 4 xsht dt o inf, x0

1, x1

2

x 2 dt o inf, x0

0.

0

8.

1

³ x

2

0.

0

9.

S /2

³ 2 x x

2

x 2 dt o inf, x0 xS / 2

0.

0

10.

T0

³ sin xdt o inf, x0

0, xT0 [ .

0

Solve the problems of Bolz 11 - 17: 11.

1

³ x

2

dt 4 x 2 0 5 x 2 1 o inf .

2

x 2 )dt 2 x1 sh1 o inf .

0

12.

1

³ ( x 0

S

³ ( x

13.

2

x 2 4 x sin t )dt 2 x 2 0 2 xS x 2 S o inf .

0

14.

S /2

³ ( x

2

x 2 )dt x 2 0 x 2 S / 2 4 xS / 2 o inf .

2

x 2 )dt Dx 2 T0 o inf .

0

T0

³ ( x

15.

0

16.

3

³ 4 x x dt x 0 8x3 o inf . 2 2

2

0

139

17.

1

³ e x dt 4e 2

x

x 0

32e x 1 o inf .

0

Solve tasks with several functions and with higher derivatives 18 - 27: 18.

1

³ ( x

x 22 2 x1 x 2 )dt o inf ,

2 1

0

x1 0 x2 0 0, x1 1 sh1, x2 1 sh1.

19.

1

³ ( x

x 22 x1 x 2 )dt o inf ,

2 1

0

x1 0 x2 0 1, x1 1 e, x2 1 1 / e.

20.

1

³ ( x x

1 2

6 x1t 12 x 2 t 2 )dt o inf ,

0

x2 0 x1 0 0, x1 1 x2 1 1.

21.

S /2

³ ( x x

1 2

x1 x 2 )dt o inf ,

0

x1 0 x2 0 0, x1 S / 2 1, x2 S / 2 1. S /2

22. ³ ( x12 x 22 2 x1 x2 2 x2 x3 )dt o inf , x1 0 x3 0 1, 0

x2 0 1, x1 S / 2 S / 2, x2 S / 2

23.

1

³ x dt o inf , x0 2

x 0 x 1

0, x3 S / 2

S / 2 .

0, x1 1.

0

24.

1

2

48 x)dt o inf , x1

2

24tx)dt o inf , x0

³ ( x

x 1

0, x0 1, x 0

4.

0

25.

1

³ ( x

x 0 0, x1 1 / 5, x 1 1.

0

140

1

³ ( x

26.

x2 )dt o inf ,

2

0

x0 x0 0, x 0 1, x 1 ch1, x1 x1 sh1. S

³ (x

27.

2

x2 )dt o inf ,

0

x0

x 0

x0

xS 0,

xS S ,

xS 2.

Solve the isoperimetric problems 28-33: 1

1

0

0

1

1

0

0

3, x0 1, x1

2 2 ³ x dt o inf , ³ x dt

28.

0, x0

2 ³ x dt o inf , ³ txdt

29.

1

³ x

30.

0

dt o inf ,

2,

0

14

S

S

0

0

1

1

0

0

3S / 2, x0

2 2 t ³ ( x x )dt o inf , ³ xe dt

32.

³ txdt

3 / 2,

0

x0 2, x1

4.

1

³ xdt

³ x sin tdt o inf , ³ xdt

31.

4, x1

1

2

6.

e

2

0, xS S .

1 / 4, x0

0, x1

e.

1

³ x x dt o inf ,

33.

1 2

0

1

³ x dt 1

1, x1 0 x1 1 0,

0

1

³ x dt 2

0, x2 0 0, x2 1 1 .

0

Solve the problems of Lagrange 34-40: 34.

1

³ u dt o inf ,x x 2

u, x0 1.

0

141

35.

S /2

u , x 0 1.

³ u dt o inf ,x x 2

0

36.

S /2

³u

2

dt x 2 0 o inf , x x

u, xS / 2 1.

0

37.

1

³ (x

2

x u, x1 1.

u 2 )dt o inf , x

0

38.

1

³ (x

2

u 2 )dt o inf , x 2 x

u, x0 1.

0

39.

S /2

³u

2

dt x 2 0 o inf , x x

u, x0 0, xS / 2 1.

0

40.

1

³ ( x

2 1

x 22 )dt o inf ,

0

x1 x2 x2 x1 1, x1 0 0, x1 1 sin 1, x2 0 1, x2 1 cos1.

P 1.3. OPTIMAL CONTROL. MAXIMUM PRINCIPLE We consider the problem of optimal control: minimize the functional t1

J x , u , t 0 , t1

³ F xt , u t , t dt ) xt , xt , t 0

0

0

1

0

, t1 o inf

(1)

t0

at conditions x

f x, u, t , t 0 d t d t1 ,

u t KC >t 0 , t1 @,

g j x, u, t 0 , t1 d 0, j 1, m,

u t U E , t >t 0 , t1 @, r

g j x, u , t 0 , t1 0, j

m 1, s ,

where functionals g j x , u , t 0 , t1

t1

³ F xt , ut , t dt ) xt , xt , t j

j

0

1

0

, t1 , j

1, s .

t0

Here

u t

u1 t ,..., u r t KC >t 0 , t1 @

is control, the values of vector function u t belong to the given set U of E r ; xt

x1 t ,..., x n t KC 1 >t 0 , t1 @ 142

(2) (3) (4)

are phase variables, equation (2) is called by differential connection; relations (4) are called by constraints, values t 0 , t1 ', ' is prescribed interval. Four

xt , u t , t 0 , t1

is called by feasible controlled process, if xt KC 1 >t 0 , t1 @, u t KC >t 0 , t1 @

and differential connection (2), explosion (3) and constraints (4) are satisfied. It is said, that the feasible controlled process

x t , u t , t 0

0

0 00

, t 101

delivers a strong local minimum to the functional (1) at conditions (2) - (4), if there is a number H ! 0 , such that for every feasible process

xt , u t , t 0 , t1 for which

max xt x 0 t max u t u 0 t t0 t00 t1 t10 H

t 0 d t d t1

t 0 d t d t1

inequality

J x, u , t0 , t1 t J x 0 , u 0 , t00 , t10

holds. Four

x t , u t , t 0

0

0 00

, t 101

is called by optimal in the strong sense of the process, or, in short, the optimal process. Solution algorithm: 1. Construct the Lagrange functional /

t1

§

·

s

³ ¨© ¦ O F xt , u t , t \ t , x f xt , u t , t ¸¹dt i

t0

i

i 0

t1

s

¦ O ) xt , xt , t , t ³ Ldt l , i

i

i 0

s

¦ O F x, u, t \ t , x f x, u, t , i

i

i 0

l

s

¦ O ) xt , xt , t , t , \ t KC >t , t @ . 1

i

i

0

1

0

i 0

2. Function 143

1

1

0

1

t0

where L

0

0

1

\ t

\ 1 t ,...,\ n t KC 1 >t 00 , t10 @

is a solution of Euler equation Lx x 0 t , u 0 t , t ,\

>

d Lx x 0 t , u 0 t , t ,\ 0, t t10 , t20 dt

@

which can be written as \ t

¦ O F x t , u t , t f x t , u t , t \ t , t >t s

0

i

0

0

ix

0

0 1

x

@

, t20 .

(5)

i 0

3. The boundary conditions for adjoint system (5) are determined by the conditions Lx x 0 tk , u .0 tk , tk0 ,\ tk , O

1 k lx t

,

k

k

0,1.

These conditions are written as \ tk0

s

1 k ¦ Oi ) ix t i 0

4. Optimal control

>

u 0 t , t t00 , t10

u0

is determined by the condition of maximum

max H x 0 t , v, t ,\ , O vU

k

xt , xt , t , t , k 0 0

0 1

0 0

0 1

0,1.

(6)

@

H x 0 t , u 0 t , t ,\ , O ,

(7)

where Pontryagin function H x, v, t ,\ , O

s

¦ Oi Fi x, v, t \ , f x, v, t . i 0

5. The optimum values t 00 , t10 are determined by the relations / tk

which can be written as

0, k

0,1,

1 k 1 ¦ Oi Fi x 0 tk , u 0 tk , tk0 ¦ Oi >) it s

s

i 0

where ) it k ) ix ( t k )

i 0

k

@

) ix t k x 0 tk0

0,

(8)

) x t , x t , t , t ,

) it k x 0 t00 , x 0 t10 , t00 , t10 , 0

ix ( t k )

0 0

0

0 1

0 0

0 1

6. The conditions of complementary slackness Oi g i x 0 , u 0 , t 00 , t10 0, i 1, s, O0 t 0, O1 t 0, ..., Om t 0 . (9)

are held. 144

Theorem. If

x t , u t , t 0

0

0 00

, t 101

is an optimal process in the problem (1) - (4), then there exist Lagrange multipliers O

(O0 t 0, O1 t 0,...Om t 0, Om 1 ,..., Os ) ,

\ t KC 1 >t0 ,t1 @,

which are not equal to zero such that the conditions (5) - (9) are performed. Example. 4

³ u

J ( x, u )

2

u, u d 1, x0 0.

x dt o inf; x

0

For this example, u 2 x, Fi { 0, f

F0

^u E

) i { 0, U

u, gi 1

x0

`

0,

/ 1 d u d 1 .

1. Lagrange functional

³ >O u 4

/

0

2

@

x \ x u dt O1 x0 ,

0

L

O0 u 2 x \ x u , l

O1 x0 , O0 t 0.

2. Since

O0 , Lx \ then the adjoint system can be written as: \ t O0 . Lx

3. Since l x 0

then

O1 , l x 1

0,

\ 0 O1 , \ 4 0.

4. Optimal control u 0 t is determined by the condition max H x 0 , u , t ,\ , O uU1

where H

Hence we get

H x 0 , u 0 , t ,\ , O ,

\u O0 u 2 x .

>

min O0 u 2 x 0 \ u

1du d 1

@

2

O0 u 0 x 0 \ u 0 .

(10)

The problem of convex programming is solved by method of Lagrange multipliers. Lagrange function 145

O0 u 2 x 0 \ u O2 1 u O3 u 1 , O2 t 0, O3 t 0.

L

(11)

We note, that the optimization problem (10) can be written as O 0 u 2 x 0 \u o inf, g 1 u 1 u d 0, g 2 u u 1 d 0 .

Lagrange function for the problem (12) has the form (11). If u g1 (u )

1 0, g 2 (u )

(12)

0 we get

1 0 ,

so that the condition of Slater holds. We define a saddle point of the Lagrange function L u , O2 , O3 , u E 1 , O2 t 0, O3 t 0

by the condition Lu

0, O 2 1 u 0 0, O3 u 0 1 0.

2O 0 u 0 O 2 O 3 \

Hence, in the case a) O 2 0, O3 0, u 0 b) O2 0, O3 ! 0, u 0 c) O2 ! 0, O3 0, u 0

O0

1 we get:

\ t / 2, t >0,[email protected];

1, 2 O3 \

0, O3

1, 2 O2 \

\ 2 ! 0;

0,

O2 2 \ ! 0 ; d) O2 ! 0, O3 ! 0 is impossible.

Thus, the optimal control u

0

\ t / 2, if \ t / 2 d 1, ° ® 1, if \ t ! 2, ° 1, if \ t 2. ¯

(13)

5 . The values t0 0, t1 4 are fixed, therefore in this example the relations (8) are not used. 6. Condition (9) can be written in the form O0 t 0, g1

x0

0.

Further we present analysis of the obtained results in items 1 - 6. We consider two cases: O0 0 and O0 1 . Let O0 0 . In this case \ 0 0, \ 0 O1 , \ 4 0

(see [15]) . Consequently

\ t { 0, t >0,[email protected], O1 146

0.

This is impossible, since all Lagrange multipliers are equal to zero. Let O0 1 Then \ t 1, \ 0 O1 , \ 4 0 . It follows \ t t c1 , \ 4 4 c1

Hence

4

0, c1

\ t t 4, t >0,[email protected] .

From (13) we get u 0 t

1, 0 d t 2, ® ¯t 4 / 2, 2 d t d 4.

By integrating the equations x 0 t

u 0 t , x 0 0

0, t >0,[email protected] ,

we get x 0 t

t , 0 d t 2, ®2 / 4 2t 1, 2 d t d 4. t ¯

Solve the following optimal control problems: 1. J x, u

S

³ x sin tdt o inf, x

u, u d 1, xr S

0.

S

2. J x, u

4

³ u

2

x dt o inf, x

u , u d 1, x4

0.

0

3. J x, u

1

³ xdt o inf, x

u, u d 2, x0

x 0

0.

0

4. J x, u

4

³ xdt o inf, 0

x u, u d 2, x0 x4 0, x0 x 4 0. T

5. J x, u, T

³ 1 dt

T o inf,

1

x u, u d 2, x 1 1, xT 1, x 1

x T 0.

T

6. J x, u, T

³1 dt

T o inf,

0

x u, u d 2, x 0 x T 0, x0 1, xT 3. T

7. J x, u, T

³ 1 dt

T o inf,

0

x u, 3 d u d 1, x0 3, x 0 x T 0, xT 5.

8. J x, u

2

³ u dt o inf, 0

147

x u, u t 2, x0 x 0 0, x2 3. 1

9. J x, u

³u

2

dt o inf,

0

x u, u d 1, x0 x 0 0, x1 11 / 24. 2

10. J x, u x2 o inf, x u, u d 2, ³ u 2 dt 2, x0 0. 0

P.1.4 . OPTIMAL CONTROL. DYNAMIC PROGRAMMING Continuous systems. We consider the problem of optimal control with free right end: minimize the functional t1

J x, u

³f

0

( x(t ), u (t ), t )dt )( xt1 ) o inf

(1)

x0 , t >t0 , t1 @ ,

(2)

t0

at conditions x

f xt , u t , t , xt0

u t KC >t0 , t1 @, u t V t E r , t >t0 , t1 @ ,

where t 0 , t1 are fixed, x0 E n is given vector, V t , t >t0 ,t1 @

is prescribed set of E , r

xt

Solution algorithm 1. Function

K Bx , x, u, t

x1 t ,..., xn t KC 1 >t 0 , t1 @ .

Bx x, t , f x, u, t f 0 x, u, t , u V t

is compiled, where Bx, t is the Bellman function. 2. The problem of nonlinear programming K Bx , x, u, t o inf, u V t

is solved with respect to the variable u . As a result, we find the optimal control u0

u 0 Bx , x, t , t >t0 , t1 @.

3. Bellman equation

w x, t wt

Bx x, t , f x, u 0 Bx , x, t , t f 0 x, u 0 Bx , x, t , t , B x, t1 ) x , t0 d t d t1

is solved. Function Bx, t M ( x, t ) is defined. 148

(3)

4. We find the optimal control u 0 x 0 , t

u0

by the known function M x, t , where x 0 the equation

u 0 M x x0 ,t , x0 ,t

x 0 t is the optimal trajectory satisfying to

x0 , t >t0 , t1 @.

f x 0 , u 0 x 0 , t , t , x 0 t0

x 0

Example 1. T

J x, u

³ u t dt Dx T o inf, D 2

0

2

x

u , x0

f0

u2,)

For this example

x0 , u KC >0, T @, u E 1 .

Dx 2 T , f

1. Function K

const ! 0,

u, U { E1 .

Bx x, t u u 2 .

2. We define the optimal control u 0 of the solution Bx x, t u u 2 o inf, u E 1 .

Hence we get

2u 0 Bx x, t

therefore u0

0,

Bx x , t / 2 .

3. The Bellman equation can be written as: Bt x , t

B x x, t u 0 u 0

2

B x2 x, t / 4,

B x, T Dx 2 , x E 1 .

The function Bx, t is sought in the form B x, t \ 0 t \ 1 t x \ 2 t x 2 .

Substituting this expression to the Bellman equation, we get \ 0 t \1 t x \ 2 t x 2 >\ 1 t 2\ 2 t x @2 / 4 0 d t d T ,\ 0 T \ 1 T x \ 2 T x 2

0, x E1 ,

Dx 2 , x E1.

Equating the coefficients of the same degree x , we get \ 0 t \ 1 t / 4 0, \ 1 t \ 1 t \ 2 t 0, \ 2 t \ 22 t 0, 0 d t d T , \ 0 T 0, \ 1 T 0, \ 2 T D . 149

Hence, we find

\ 0 t \ 1 t { 0, \ 2 t D />1 D t T @, 0 d t d T .

Now the Bellman function B x, t M x, t Dx 2 />1 D t T @.

4. We find the optimal control u0

Bx x , t / 2

M x x, t / 2

The optimal trajectory

Dx 0 />1 D t T @.

x 0 t is

x0

determined by solution of the differential equation x 0 t Dx 0 t />1 D t T @, x 0 0

x 0 , t >0, T @.

Control Dx 0 />1 D t T @, t >0, T @

u0

is called by synthesizing control. Discrete systems. The method of dynamic programming for discrete systems we illustrate with example.

Example 2. Mminimize the functional I >xi @0 , >ui @0

N 1

¦ F x , u )x o inf 0i

i

i

N

(4)

i 0

at conditions xi 1

f i xi , ui , xi i

0

x0 , i

>ui @0 u0 , u1 ,..., uN 1 , ui Vi E r , i where

>x1 @0 x0 , x1,..., xN , xi E n , i

0, N 1,

0, N 1,

(5) (6)

0, N , ui E r , i 1, N 1.

The problem (4) - (6) is a discrete analogue of the optimal control problem with a free right end without phase constraints. 0 step. At i N the Bellman function B N x N ) x N ,

where xN E . 1 step. According to the optimality control principle on the last interval u N0 1 should be chosen that u N 1 V N 1 and minimize the partial sum n

150

F0 N 1 xN 1 , u N 1 )xN

F0 N 1 xN 1 , u N 1 ) f N 1 xN 1 , u N 1 .

We note, that in the absence of a phase constraint the set DN 1 VN 1 . Thus, the optimal control u N0 1 is determined by solution of the following nonlinear programming problem: F0 N 1 xN 1 , u N 1 ) f N 1 xN 1 , u N 1 o inf, u N 1 VN 1 .

Solving this optimization problem in the finite-dimensional space, we find u N0 1 xN 1

u N0 1

We denote

u N0 1 xN 1 , N 1 .

F0 N 1 xN 1 , u N0 1 ) f N 1 xN 1 , u N0 1 .

BN 1 xN 1

2 step. For values i N 2 the optimal control u N0 2 is determined by solving of the following optimization problem: F0 N 2 x N 2 , u N 2 B N 1 x N 1

F0 N 2 x N 2 , u N 2

B N 1 f N 2 x N 2 , u N 2 o inf, u N 2 V N 2

where

f N 2 xN 2 , u N 2 .

xN 1

Therefore we determine

u N0 2

and the value

BN 2 x N 2

u N0 2 xN 2

u N0 2 xN 2 , N 1, N 2

F0 N 2 x N 2 , u N0 2 BN 1 f N 2 xN 2 , u N0 2 .

3 step. Optimal control u N0 3 we find of solution of the problem F0 N 3 x N 3 , u N 3 B N 2 x N 2

F0 N 3 x N 3 , u N 3

B N 2 f N 3 x N 3 , u N 3 o inf, u N 3 V N 3 .

It follows that u N0 3

and

u N0 3 xN 3

B N 3 x N 3

u N0 3 xN 3 , N 3, N 2, N 1

F0 N 3 x N 3 , u N0 3 B N 2 f N 3 x N 3 , u N0 3

etc. N step. For values i 0 the optimal control u00 is defined by solution of the optimization problem F00 x0 , u0 B1 x1

F00 x0 , u0 B1 f 0 x0 , u0 o inf, u0 V0 . 151

Hence we find the optimal control u00 x0

u00

and value

u00 x0 ,0,1,..., N 1

B0 x 0

F00 x 0 , u 00

B1 f 0 x 0 , u 00 inf I >xi @0 , >u i @0 .

Since according to the condition of problem the vector x0 E n is prescribed, then the value u00 is known. Next, we define x10 f 0 x0 ,u00 . By the known vector x10 E n we find u10 x10

u10

and so on. By the known vector xN0 1 we define

u N0 1 x N0 1 .

u N0 1

Solve the following problems:

³ >x t u t @dt o inf, x t

T

1. J x, u

2

x0 x0 ,

xt u t ,

2

0

u t KC >0, T @, u t E1 , 0 d t d T .

2. J x, u x 2 1 o inf, x t u t , x0 x0 , 0 d t d 1, u t KC >0,[email protected], u t V

^u E

1

/ 0 d u d 1`.

3. J x, u x 2 T o inf, x t u t , x0 x0 , 0 d t d T , u t KC >0,[email protected], u t E1 , 0 d u d T .

4. J x, u, T T o inf, x t xT

f xt , u t , t , 0 d t d T , x0

x1 , u t KC >0, T @, u t V E , 0 d t d T . r

Write the Bellman equation for this problem. 5. Find Bellman function for the problem J x, u x t

T

³ > at , xt

f 0 u t , t @dt c, xT o inf,

t0

At xt C t u t f t , t0 d t d T , xt0

u t KC >t0 , T @, u

u t V t E , t0 d t d T .

x0 ,

r

(to find function Bx, t in the form Bx, t

152

\ t , x ) .

x0 ,

T

6. J x, u D 1 ³ x 2 (t )dt D 2 x 2 (T ) o inf, 0

x t a(t ) x(t ) c(t )u t f (t ), x0

x0 ,

u t KC >0, T @, 0 d t d T ; a) u t E 1 , b) u t V (t ) u t E 1 u d 1 .

^

`

5

¦[x

7. I ([ xi ]0 , [u i ]0 )

2 i

u i2 ] o inf, xi 1

xi u i ,

i 0

xi

0,5 .

x0 , u i E 1 , i

i 0

3

¦ [( x )

8. I ([ xi ]0 , [u i ]0 )

1 2 i

( xi2 ) 2 u i2 ] ( x 14 ) 2 ( x 42 ) 2 o inf,

i 0

xi1 xi2 u i ,

xi11

xi1

x01 , xi2

i 0

xi21

x02 , i

i 0

xi1 xi2 ,

0,3, u i E 1 , i

9. I ([ xi ]0 , [u i ]0 ) x o inf, 2 6

xi

i 0

x0 , i

10. I ([ xi ]0 , [u i ]0 )

xi 1

0,5, u i E , i 1

4

¦[x

2 i

ui ,

0,5 .

u i2 ] x52 o inf,

i 0

xi 1

xi

i 0

xi u i ,

x0 , i

0,4, u i E 1 , i

0,4 .

0,3 .

APPENDIX II TASKS ON "OPTIMAL CONTROL"

Term tasks for the following sections are worked out for students of the 2 and 3 courses studied of the specialties "Mathematics", "Mechanics", "Informatics", "Economic cybernetics". variation calculus (1st task) optimal control (2nd task). Further, we present options for tasks of the course "Optimal control". 1st task To solve the simplest or isoperimetric problems (variation calculus)

1. J x

1

³ x t dt o extr; x0 2

1, x1

0.

0

2. J x

T0

³ x t dt o extr; x0 2

0, xT0 [ .

0

3. J x

1

³ ( xt x t )dt o extr; x0 2

x1

0.

0

4. J x

T0

³ x t xt dt o extr; x0 2

0, xT0 [ .

0

5. J x

1

³ x t txt dt o extr; x0 2

x1

0.

0

6. J x

1

³ t xt x t dt o extr; x0 2

2

x1

0

154

0.

7. J x

T0

³ x t dt o extr; x0 3

0, xT0 [ .

0

8. J x

3/2

³ x t 2 xt dt o extr; x0

0, x3 / 2 1.

3

0

9. J x

T0

³ x t x t dt o extr; x0 3

0, xT0 [ .

2

0

10. J x

T0

³ x t x t dt o extr; x0 3

0, xT0 [ .

2

0

11. J x

e

³ tx t dt o extr; x1 2

0, xe 1.

1

12. J x

1

³ 1 t x t dt o extr; x0

0, x1 1.

2

0

13. J x

e

³ (tx t 2 xt )dt o extr; x1

1, xe 0.

2

1

14. J x

e

³ ( xt tx t )dt o extr; x1

1, xe

2

2.

1

15. J x

2

³t

2

x 2 (t ) dt o extr ;

x(1)

3, x( 2) 1.

1

16. J x

3

³ t

2

1 x 2 t dt o extr ; x2

0, x3 1.

2

17. J x

e

³ 2 xt t x t dt o extr; x1 2

2

e, xe

0.

1

18. J x

1

³ x t x t dt o extr; x0 2

2

1, x1

2.

0

4/3

19. J x P ³ 0

xt dt o extr; x0 1, x4 / 3 x 2 t

155

1 . 9

20. J x

1

³ e x t dt o extr; x0 2

x

0, x1

ln 4.

0

21. J x

1

³ x t xt xt 12txt dt o extr; x0 2

x1

0.

0

22. J x

e

³ (tx

2

(t ) x(t ) x (t ))dt o extr ; x1 0, xe 1.

1

23. J x

1

³ t x t 12 x t dt o extr; x0 2 2

2

0, x1 1.

0

24. J x

1

³ x t x t dt o extr; x 1 2

2

x1 1.

1

25. J x

1

³ x t 4 x t dt o extr; x 1 2

2

1, x1 1.

1

26. J x

1

³ x t x t 2 xt dt o extr; x0 2

2

x1

0.

x1

0.

0

27. J x

1

³ x t x t txt dt o extr; x0 2

2

0

28. J x

1

³ 4 xt sin t x t x t dt o extr; x0 2

2

x1

0.

x1

0.

0

29. J x

1

³ x t x t 6 xt sh2t dt o extr; x0 2

2

0

30. J x

T0

³ x t x t 4 xt sin t dt o extr; 2

2

0

x0 0, xT0 [ .

31. J x

T0

³ x t x t 6 xt sh2t dt o extr; 2

2

0

x0 0, xT0 [ . 156

32. J x

1

³ x t x t 4 xt sht dt o extr; 2

2

0

x0 1, x1 0.

33. J x

T0

³ x t x t 4 xt sht dt o extr; 2

2

0

x0 0, xT0 [ .

34. J x

1

³ x t x t 4 xt cht dt o extr; x0 2

x1

2

0.

0

35. J x

T0

³ x t x t 4 xt cht dt o extr; 2

2

0

x(0)

36. J x

0, xT0 [ . S /2

³ x t x t dt o extr; x0 2

2

0

37. J x

S /4

³ 4 x t x t dt o extr; x0 2

2

0

38. J x

S /4

³ x t 4 x t dt o extr; x0 2

2

0

39. J x

3S / 4

³ x t 4 x t dt o extr; x0 2

2

0

40. J x

S /2

2

2

3S / 2

³ x t x t 2 xt dt o extr; 2

2

0

§ 3S · x0 0, x¨ ¸ 0. © 2 ¹

157

0.

§S · 1, x¨ ¸ ©4¹

0.

§S · 0, x¨ ¸ 1. ©4¹ § 3S · 0, x¨ ¸ © 4 ¹

³ 2 xt x t x t dt o extr; x0 0

41. J x

§S · 1, x¨ ¸ ©2¹

§S · x¨ ¸ ©2¹

1.

0.

S /2

³ x t x t txt dt o extr; x0

42. J x

2

2

0

S /2

³ x t x t 4 xt sht dt o extr;

43. J x

2

2

0

x0

§S · x¨ ¸ ©2¹

44. J x

0.

T0

³ x t x t 4 xt sht dt o extr; 2

2

0

x0

§S · x¨ ¸ ©2¹

45. J x

0. S /2

³ 6 xt sin 2t x t x t dt o extr; 2

2

0

§S · x0 x¨ ¸ ©2¹

46. J x

0.

T0

³ x t x t 6 xt sin 2t dt o extr; 2

2

0

x0 0, xT0 [ .

47. J x

S /2

³ 4 xt sin t x t x t dt o extr; 2

2

0

§S · x0 x¨ ¸ ©2¹

48. J x

0.

3S / 2

³ x t x t 4 xt sin t dt o extr; 2

2

0

§ 3S · x0 x¨ ¸ 0. © 2 ¹

49. J x

S /2

³ x t x t 4 xt cos t dt o extr; 2

2

0

§S · x0 x¨ ¸ ©2¹

0.

158

§S · 0, x¨ ¸ ©2¹

0.

50. J x

S /2

³ x t x t 4 xt cos t dt o extr; 2

2

0

§S · x0 0, x¨ ¸ ©2¹

51. J x

S 2

.

3S / 2

³ x t 4 xt cos t x t dt o extr; 2

2

0

§ 3S · x0 0, x¨ ¸ © 2 ¹

52. J x

3S . 2

T0

³ x t x t 4 xt cos t dt o extr; 2

2

0

x0 0, xT0 [ .

53. J x

1

³ x t 3x t e 2

2

2t

dt o extr; x0 1, x1

e.

0

54. J x

1

³ x t x t e 2

2

2t

dt o extr; x0

0, x1

2t

dt o extr; x0

0, xT0 [ .

e.

0

55. J x

T0

³ x t x t e 2

2

0

56. J x

1

³ sin xt dt o extr; x0

0, x1

0

57. J x

1

³ cos x t dt o extr; x0

S 2

.

0, x1 S .

0

58. J x

T0

³ sin x t dt o extr; x0

0, xT0 [ .

0

59. J x

T0

³ cos x t dt o extr; x0

0, xT0 [ .

0

159

T0

60. J x

³ xt e

dt o extr; x0

x t

0, xT0 [ .

0

T0

61. J x

³ ( x t 5xt )dt o extr; x0 5

0, xT0 [ .

0

1

62. J x P ³ (1 x 2 t )2 dt o extr; x0 0, x1 [ . 0

63. J x

1

1

0

0

1

1

0

0

2 ³ x t dt o extr; ³ xt dt

64. J x

2 ³ x t dt o extr; ³ xt dt

1

65. J x

1

³ x t dt o extr; ³ txt dt 2

0

66. J x

0, x0 1, x1

0.

3, x0 1, x1

6.

0, x0

0, x1 1.

0, x0

4, x1

0

1

1

0

0

2 ³ x t dt o extr; ³ txt dt

1

67. J x

³ x t dt o extr; 2

0

1

³ xt dt 0

1

1, ³ txt dt

0, x0 x1 0.

0

1

68. J x

³ x t dt o extr; 2

0

1

1

3 , ³ txt dt 2 0

³ xt dt 0

69. J x

2, x0 2, x1 14.

S

³ x t dt o extr; 2

0

S

³ xt cos tdt 0

S 2

, x0 1, xS 1.

160

4.

70. J x

S

S

0

0

2 ³ x t dt o extr; ³ xt sin tdt

0, x0 0, xS 1.

S

71. J x

³ xt sin tdt o extr; 0

S

3S , x0 0, x(S ) S . 2

³ x t dt 2

0

S

72. J x

³ x t dt o extr; 2

0

S

S

³ xt cos tdt

2

0

S

³ xt sin tdt

,

S 2, x0 2, xS 0.

0

1

73. J x

³ x t dt o extr; 2

0

74. J x

1

1

³ x t dt o extr; ³ xt e dt 2

t

0

1

³ xt e

t

0, x0 0, x1 1.

0

e, x1 2, x0 2e 1.

dt

0

1

75. J x

³ ( x t x t )dt o extr; 2

2

0

76. J x

2

³t

2

3

1

1

³ xt e

t

x 2 t dt o extr ; ³ xt dt

1 3e 2 1 , x0 0, x1 . 4 e

dt

0

77. J x

2, x1 4, x2 1.

1

S

³ ( x t x t )dt o extr; 2

2

0

1

t ³ xt e dt 0

1 3e 2 1 , x0 0, x1 . 4 e

161

S /2

78. J x

³ ( x t x t )dt o extr; 2

2

0

S /2

§S · 1, x0 x¨ ¸ 0. ©2¹

³ xt sin tdt 0

T0

79. J x

³ xt

1 x 2 t dt o extr;

T0

T0

³

1 x 2 t dt

l , x T0 xT0 0.

T0

1

80. J x

³ x t dt o extr; 2

x0

x 0

x 1 0,

x1 1.

x1

x 1

x0 0,

x 0 1.

0

1

81. J x

³ x t dt o extr; 2

0

1

82. J x

³ ( x

48 x) dt o extr;

2

0

x (1)

x (1)

0, x (0) 1, x (0)

4.

1

83. J x

³ (24tx x ) dt o extr; 2

0

x (0)

x (0)

x(1)

0, x (1)

1 . 10

1

84. J x

³ ( x

2

24tx) dt o extr;

0

x ( 0)

x (0)

85. J x

0, x(1)

1 , x (1) 1. 5

S

³ ( x

2

x 2 ) dt o extr;

0

x ( 0)

0,

x (0) 1, x (S )

shS , x (S )

162

chS .

S

86. J x

³ ( x

2

x 2 ) dt o extr;

0

x ( 0)

0, x (S )

x (0) T0

87. J x

³ ( x

2

chS 1, x (S )

shS .

x 2 ) dt o extr;

0

x ( 0)

x (0) x(T0 ) x (T0 )

0.

2nd task

To solve the following tasks by the maximum principle and dynamic programming: 1.

S

³ x sin tdt o extr;

x u, u d 1, x S 0, xS 0.

S

2.

7S / 4

³ x sin tdt o extr;

u, u d 1, x0 0.

x

0

3.

T0

³ x dt o extr;

u, u t A, x0 0, xT0 [ , A 0 .

x

0

4. 5. 6. 7. 8. 9.

4

³ x

2

x dt o extr; x

³ x

2

x dt o extr; x

³ x

2

x dt o extr; x

³ x

2

³ x

2

x dt o extr; x

³ x

2

x dt o extr ; x

0 T0

0 T0

0 T0

0 T0

0 T0

u , u d 1, x4 0.

u, u d 1, x0 0.

u, u d 1, x0 [ .

x dt o extr; x u, u d 1, x0 [ .

u , u d 1, xT0 [ .

u, u d 1, x0 0, xT0 0.

0

10. 11.

T0

³ x

2

x dt o extr ; x

³ x

2

x 2 dt o extr; x

0 T0

u, u d 1, x0 0, xT0 [ .

u, u d 1, x0 [ .

0

12.

1

³ x dt o extr; 1

x1

x 2 , x 2

u , u d 2, x1 0

x 2 0 0.

x1

x 2 , x 2

u, u d 2, x1 1

x 2 1 0.

0

13.

1

³ x dt o extr; 1

0

163

1

³ x dt o extr;

14.

1

x1

x 2 , x 2

u , u d 2, x1 0

x 2 1 0.

0

2

³ x dt o extr;

15.

1

0

x1

x2 , x2

u, u d 2, x1 0 x2 0 0, x2 2 0.

2

³ x dt o extr;

16.

1

0

x1

u, x1 0 x2 2 0, x2 0 0, u d 2.

x2 , x2

1

17. ³ x1 dt o extr; 0

x1

x 2 , x 2

u,

x1 0 x2 2 0, x 2 0 0, u d 2.

2

³ x dt o extr;

18.

1

0

x1

u, u d 2, x1 0 x2 0 x1 2 0.

x2 , x2

2

³ x dt o extr;

19.

1

0

x1

u, u d 2, x2 0 x2 2 x1 2 0.

x2 , x2

2

³ x dt o inf ;

20.

1

0

x1

x2 , x2

u, u d 2, x1 0 x2 0 x2 2 0.

4

21. ³ x1 dt o extr; 0

x1

u, u d 2, x1 0 x1 4 0, x2 0 x2 4 0.

x2 , x2

22. T o inf, x1 x2 , x2 u, u d 2, x1 1 1, x1 T 1, x 2 1

23. T o inf, x1

x 2 , x 2

x 2 T 0.

u,

u d 2, x1 1 1, x1 T 1, x2 1 x2 T 0.

24. T o inf, x1

x 2 , x 2 u, u d 2, x2 0 x2 T 0, x1 0 1, x1 T 3.

25. T o inf, x1

x 2 , x 2 u, 1 d u d 3, x1 0 1, x2 0 x2 T 0, x1 T 1.

26. T o inf, x1

x 2 , x 2

u,

3 d u d 1, x1 0 3, x2 0 x2 T 0, x1 T 5.

27. T o inf, x1

x 2 , x 2 u, 0 d u d 1, x1 0 [1 , x2 0 [ 2 , x1 T x2 T 0.

28. T o inf, x1

x 2 , x 2

u, 164

u d 1, x1 0 [1 , x2 0 [ 2 , x2 T 0.

29. T o inf, x1

x 2 , x 2

u,

u d 1, x1 0 [1 , x2 0 [ 2 , x1 T 0.

30.

2

³ u dt o inf, 0

31.

³ u dt o inf,

x1

x 2 , x 2

u,

u t 2, x1 0 x2 0 0, x1 2 3.

2

³ u dt o inf,

x1

x 2 , x 2

u,

u t 2, x1 0 x2 2 0, x1 2 3.

2

³ u dt o inf, x

x 2 , x 2

1

0

34.

u,

u t 2, x1 0 0, x1 2 1, x2 2 2.

0

33.

x 2 , x 2

2

0

32.

x1

u,

u d 2, x1 0 0, x1 2 1, x2 2 2.

2

³ u dt o inf, 0

x1

x 2 , x 2

u,

u d 2, x1 0 1, x2 0 2, x1 2 0.

1

35. ³ u 2 dt o inf, x1

x 2 , x 2

u,

0

u d 24, x1 0 11, x1 1 x2 1 0. 2

36. ³ u 2 dt o inf, x1

x 2 , x 2

u,

0

u t 6, x1 0 x2 0 0, x1 2 17. 1

37. ³ u 2 dt o inf,

x1

x 2 , x 2

u,

0

u d 1, x1 0 x1 0 0, x1 1

11 . 24

2

38. x(2) o extr;

x

u, u d 2,

³u

2

dt

2, x(0)

0.

0

39. x(T0 ) o extr;

x

u, u d 1,

T0

³u

2

dt

2, x(0)

0

§ x2 u2 · u ¸¸dt o extr ; 2 ¹ 0©

1

40. ³ ¨¨ 41.

T0

³ ( xu

2

x

u, x(1) [ .

yu1 )dt o sup;

0

x

u1 , y

y (0)

u 2 , u12 u 22 d 1, x (0)

y (T0 ). 165

x (T0 ),

0.

42.

T0

³ ( xu

2

yu1 )dt o sup;

0

(u1 [ ) 2 u22 d 1, x u1 , y u 2 , y (0)

43.

T0

³ ( xu

2

y (T0 ), x(0)

x(T0 ).

y (T0 ), x(0)

x(T0 ).

y (T0 ), x(0)

x(T0 ).

y (T0 ), x(0)

x(T0 ).

yu1 )dt o sup;

0

2

2

§ u1 · § u1 · ¨ ¸ ¨ ¸ d 1, ©a¹ ©b¹ x u1 , y u 2 , y (0)

44.

T0

³ ( xu

2

yu1 )dt o sup;

0

u1 d 1, u 2 d 1, x

45.

u1 , y

T0

³ ( xu

2

u 2 , y (0)

yu1 )dt o sup;

0

u1 u 2 d 1, x

46.

u1 , y

T0

tdt

³ 1 u

2

u 2 , y (0)

dt o inf ;

u, u t 0, x(0)

x

0, x(T0 )

[.

0

47.

1

³x

dt o sup;

2

u , u d 1, x(0)

x

0.

0

1

48. ³ u 2 dt o extr; x x u, x(0) 1. 0

1

49. ³ u 2 dt o extr; x x u, x(0) 1,

x (0)

0.

0

1

50. ³ u 2 dt o extr; 0

x x

u,

x ( 0)

0,

x(1)

sh1, x (1)

ch1 sh1.

1

51. ³ u 2 dt o extr; 0

x x

52.

u,

x ( 0)

x (0)

0,

x(1)

S /2

³u

2

dt o extr;

0

x x

53.

u , x (0) 1.

S /2

³u

2

dt o extr;

0

x x

u , x ( 0)

§S · 0, x¨ ¸ 1. ©2¹

166

sh1, x (1)

ch1 sh1.

54.

S /2

³u

2

dt o extr;

0

x x

55.

§S · x¨ ¸ ©2¹

u , x (0)

§S · 0, x ¨ ¸ ©2¹

S 2

.

S /2

³u

2

dt o extr;

0

x x

56.

u , x ( 0)

§S · x¨ ¸ ©2¹

§S · 0, x ¨ ¸ 1, x 0 ©2¹

S /2

³u

2

dt x 2 (0) o extr;

0

x x

57.

u.

S /2

³u

2

dt x 2 (0) o extr;

0

§S · u , x ¨ ¸ 1. ©2¹

x x

58.

S /2

³u

2

dt x 2 (0) o extr;

0

x x

59.

§S · u , x¨ ¸ ©2¹

§S · 0, x ¨ ¸ 1. ©2¹

S /2

³u

2

dt x 2 (0) o extr;

0

x x

§S · u , x¨ ¸ ©2¹

x (0)

§S · 0, x ¨ ¸ 1. ©2¹

1

60. ³ u 2 dt x 2 (0) o extr; 0

1

x x

u.

61. ³ u 2 dt o extr; 0

u, u d 1, x(0)

x

62.

T0

³ (u

2

0,

x(1) [ .

x)dt o extr;

0

x

63.

T0

³ (u

2

u, u d 1, x(0)

0.

x)dt o extr;

0

x

u, u d 1, x(0)

x(T0 )

0.

T

64. ³ (u 2 x)dt o extr; 0

x

u, u d 1, x(0)

0. 167

S 2

.

T

65. ³ (u 2 x)dt o extr; 0

x

66.

u, u d 1, x(0)

T0

³ (u

2

x(T )

0.

x 2 )dt o extr;

0

x

u, u d 1, x(T0 ) [ .

T

67. ³ (u 2 ɯ 2 )dt o extr; 0

x

68.

u, u d 1, x(T ) [ .

T0

³ (u

2

x 2 )dt o extr;

0

x

u, u d 1, x(0)

0, x(T0 ) [ .

T

69. ³ (u 2 x 2 )dt o extr; 0

x

u, u d 1, x(0)

0, x(T ) [ .

S

70. ³ (u 2 x 2 )dt o extr; 0

x

71.

u, u d 1, x(0)

T0

³ (u

2

x

u, u d 1, x(0)

0.

x 2 )dt o extr;

0

72.

T0

³ (u

2

x

u, u d 1, x(0)

0.

x 2 )dt o extr;

0

x(T0 )

0.

4

73. ³ x1dt o extr; 0

x1

x2 ,

x 2

u, u d 2, x1 (0)

x1 (4)

0.

u, u d 2, x1 (0)

x 2 (4)

x1 (4)

0.

u, u d 2, x1 (0) x 2 (0)

x 2 (4)

x1 (4)

4

74. ³ x1dt o extr; 0

x1

x2 ,

x 2

4

75. ³ x1dt o extr; 0

x1

x2 ,

x 2

168

0.

APPENDIX III KEYS

to the 1st task 1. 2. 3. 4. 5. 6. 7.

1 t abs min,

f .

J max

[ t / T0 abs min, J max

t t / 4 abs max, 2

f .

J min

f .

t / 4 [ / T0 T0 / 4 t abs min, J max 2

t t /12 abs min, J t t / 24 abs max, J 3

max

4

min

f .

f .

f .

xˆ t [ t / T0 is unit extreme, [ ! 0 xˆ loc min, [ 0 xˆ loc max, [

locextr , J min

8. xˆ 2t / 3

f, J max

0 xˆ locextr , [ xˆ is non-strong

f .

is unit extreme, xˆ loc min, xˆ is non-strong f, J max f . 9. xˆ t [ t / T0 is unit extreme, [ ! T0 / 3 3/ 2

locextr , J min

xˆ loc min, [ T0 / 3 xˆ loc max, [

T0 / 3 xˆ locextr ,

(ln(3(t 1) /(t 1))) / ln(3 / 2) abs min, J max locextr , J min

f, J max

f is non-strong

f .

10. xˆ t [ t / T0 is unit extreme, [ ! T0 / 3 xˆ loc min, [ T0 / 3 xˆ loc max, [

T0 / 3

xˆ locextr , [ xˆ is non-strong extreme, J min

J max

f .

11. ln t abs min, J max f . 12. (ln(t 1)) / ln 2 abs min, J max f . 13. t e ln t abs min, J max f . 1 e 3t ln t abs max, J min 2 2 15. 4 / t 1 abs min, J max f .

14.

f .

16. (ln(3(t 1) /(t 1))) / ln(3 / 2) abs min, J max f . 17. e / t ln t abs max, J min f . 18. xˆ t 1 abs min, J max f . 169

f,

19. xˆ 2 t 2 2 / 4 conducts strong local minimum, J min f, J max f . 20. 2 lnt 1 abs min, J max f . 21. t 3 t abs min, J max f . 22. ln t abs min, J max f . 23. t 3 abs min, J max f . 24. cht / ch1 abs min, J max f . 25. sh2t / sh2 abs min, J max f . 26. e t e1t / 1 e 1 abs min, J max f . 27. sht / 2 sh1 t / 2 abs min, J max f . 28. sin t sin1sht / sh1 abs max, J min f . 29. sh2t sh2sht / sh1 abs min, J max f . 30. sin t [ sin T0 sht / shT0 abs min, J max f . 31. sh2t [ sh2T0 sht / shT0 abs min, J max f . 32. (t 1)cht abs min, J max f . 33. tcht [ T0 chT0 sht / shT0 abs min, J max f . 34. (t 1) sht abs min, J max f . 35. (([ / shT0 ) T0 t ) sht abs min, J max f . 36. cos t abs min, J max f . 37. cos 2t abs max, J min f . 38. sin 2t abs min, J max f . 39. J min f, J max f ; 40. cos t sin t 1 abs max, J min f . 41. J min f, J max f ; 42. (S sin t 2t ) / 4 abs min, J max f . 43. sht ( shS / 2) sin t abs min, J max f . 44. 0 T0 S sht sin t ([ shT0 ) / sin T0 abs min ; T0 S at [ shS , xˆ sht C sin t abs min C R , J min f, J max f . 45. sin 2t abs max, J min f; 46. 0 T S xˆ ( 47. 48. 49. 50. 51.

[

sin T0

2 cos T0 ) sin t sin 2t abs min ;

t cos t abs max ;

J min

f, J max

f ;

t sin t (S / 2) sin t abs min ; t sin t abs min ;

t sin t loc extr, J min

f ;

170

§ [

·

T0 ¸¸ sin t t sin t abs min ; 52. 0 T0 S ¨¨ © sin T0 ¹ t 53. e abs min ; 54. te 2t abs max ; 55. t[e t T T0 abs min ; 0

56. 57. 58. 59. 60.

tS abs max ; 2 tS abs min ; t[ loc max ; T0 t[ loc max ; T0 t[ loc min ; T0

61. [ t 4T05 / 4 ,

4(t c) 5 / 4 c 5 / 4 loc min ; 5

62. t[ abs min ; 63. 3t 2 4t 1 abs min ; 64. 3t 2 2t 1 abs min 5t 3 3t abs min ; 2 66. 5t 3 3t 4 abs min ; 67. 60t 3 96t 2 36t abs min ;

65.

68. 10t 3 12t 2 6t 2 abs min ; 69. cos t abs min ; 70.

t 2 sin t

S

abs min ;

71. t sin t abs max; 72. 2 sin t cos t 1 abs min; 73. 2e1t 1 t abs min ; 74. 75. 76. 77. 78. 79.

2(1 e t ) (e 1)t abs min ; 2 e3 e 4e 3 te t abs min ; 4 abs min ; t2 4 t sin t C sin t , C R 1 ; S 8 t cos t ; S l 2T0 there is not extreme, l xˆ

2T0 , x { 0, l ! 2T0

T · § t rC ¨ ch ch 0 ¸, C ! 0 ; C¹ © C 171

80. 2t 3 3t 2 abs min ; 81. t (t 1) 2 abs min ; 82. t 4 4t 3 6t 2 4t 1 abs min; 83. 84. 85. 86. 87.

t 2 (t 3 2t 1) abs max ; 10 t 5 3t 3 2t 2 abs min ; 10 sht abs min ; cht cos t abs min ; C1 sin t C 2 cos t C 3 sht C 4 cht .

to the 2nd task S t , S d t d S / 2, ° t S / 2, xˆ abs min, xˆ abs max . 1. xˆ ® t , ° t S, S /2 d t d S ¯ t , t d S / 4 2. xˆ ® xˆ abs min, xˆ abs max . ¯ t S / 2, t ! S / 4 3 . T here is not any feasible functions fo r [ AT0 ; there is a feasible functionextreme xˆ t At for [ AT0 ; at AT0 [ 0 feasible extreme is an arbitrary monotony

decreasing feasible function; there is a feasible extreme xˆ { 0 at [ 0 ; finally, if [ ! 0 , then arbitrary monotony increasing function is feasible extreme. An arbitrary extreme conducts abs min . t 2 / 4 3, 0 d t d 2

4. xˆ ®

xˆ abs min, 4 t abs max . ¯t 4, 2 d t d 4 0 d t d T0 2, t, t 2 tT 5. T0 d 2 0 abs min ; T0 ! 2 xˆ ® 2 / 4 1 , t T T 4 2 0 0 T0 2 d t d T0 , ¯ xˆ abs min, t abs max . 0 d t d T0 2, t [, 6. T0 ! 2 xˆ ® xˆ abs min, 2 ¯t T0 / 4 1 [ T0 , T0 2 d t d T0 ,

xˆ

t T0 2 / 4 [ T02 / 4 abs min, 7. S min f xn t [ t , Tˆn

any Tˆ

t [ abs max .

f x n t [ t , Tˆn

n , S max

n , [ d 0 there are not

is feasible extreme; t /4 [ t [ 0 d t d [ 1, t [ , 2 [ , [ ! 1 is feasible extreme xˆ min ® Tˆ 1 [ . 2 ˆ ¯t [ 1 / 4, [ 1 d t d T , 8. T0 d 2 xˆ t 2 / 4 [ T02 / 4 abs min ; feasible

extreme;

0 [ d 1 xˆ

T0 d 2

t 2 / 4 1 [ T0 , 0 d t d 2, ® 2 d t d T0 , ¯t [ T0 , xˆ abs min, t T0 [ abs max .

T0 ! 2 xˆ

172

2

9. T0 d 4 xˆ t t T0 / 4 abs min, T0 ! 4 0 d t d T0 / 2 2, t, ° 2 / 2 / 4 1 / 2 , / 2 2 d t d T0 / 2 2, t T T T ® 0 0 0 ° t T , T0 / 2 2 d t d T0 , 0 ¯

xˆ

xˆ abs min; xˆ

0 d t d T0 / 2, t, xˆ abs max; ® ¯T0 t , T0 / 2 d t d T0 ,

10. S min f , S max f , [ d 0 there are not any feasible extremes; 0 [ d 1 is feasible extreme; xˆ t 2 / 4, Tˆ 2 [ , [ ! 1 is feasible extreme t 2 / 4, 0 d t d 2, Tˆ 1 [ . ® ˆ ¯t 1, 2 t d T , [ 11. [ d cthT0 xˆ cht T0 abs min, [ ! cthT0 chT0 xˆ min

° [ tsign[ , 0 d t d [ 1 C2 , ® 2 °¯Csign[cht T0 , [ 1 C t d T0 , xˆ abs min , where C ! 0 is root of equation

xˆ

Csh [ 1 C 2 T0

12. 13. 14. 15. 16. 17.

1; [ t abs max .

t abs min, t abs max . 2

2

t 1 abs min, t 1 abs max . t 2 2t abs min, 2t t 2 abs max . 2

2

t 2 2 2 abs min,

2 t 2 abs max . 2 2 t 2 abs min, 2 t abs max . t 2 t abs min, t t 2 abs max . 2

t2,

0 d t d 2 2, 2 t t ( 8 4 2 ) 12 8 2 , 2 2 t d 2, ¯ xˆ abs min, xˆ abs max .

18. xˆ ®

t 2 2, 2 ¯ (t 2) ,

19. xˆ ®

t 2 ,

0 d t d 1, xˆ abs min, xˆ abs max . 1 t d 2,

0 d t d 1, ¯t 2 2, 1 d t d 2,

20. xˆ ®

2

t2,

xˆ abs min .

0 d t d 1, 1 t d 3,

xˆ abs min, xˆ abs max . ¯(t 2) 2, § t 2 2t , 1 d t d 0, ˆ · 22. ¨¨ xˆ ® 2 T 1¸ abs min . ¸ 0 t d 1, ¯ t 2t , © ¹ 2 § t 2t , 1 d t d 0, ˆ · 23. ¨¨ xˆ ® 2 T 1¸ abs min . ¸ 2 , 0 d 1 , t t t ¯ © ¹ § · t 2 1, 0 d t d 1, ˆ 24. ¨¨ xˆ ® 2 T 2 ¸ abs min . ¸ 4 1 , 1 d 2 , t t t ¯ © ¹

21. xˆ ®

2

173

§ 25. ¨ xˆ °®

0 d t d 3, 4 ·¸ Tˆ abs min . 2 ¨ °¯ 3t 4 / 2 1, 3 ¹¸ 3 t d Tˆ , © § 3t 2 / 2 3, 0 d t d 3 , ˆ 8 ·¸ 26. ¨ xˆ °® T abs min . 2 ¨ °¯ t 8 / 3 / 2 5, 2 / 3 t d 8 3 , 3 ¹¸ © 28. [ 2 t 0 xˆ t 2 / 2 [ 2 t [1 , Tˆ [ 2 abs min, [ 2 0 xˆ t 2 / 2 [ 2 t [1 , Tˆ [ 2 abs min . 29. [ t 0 xˆ t 2 / 2 [ t [ , Tˆ [ [ 2 2[ abs min, t 2 / 2 1,

1

2

1

[1 0 xˆ t / 2 [ 2 t [1 , Tˆ [1 0 Tˆ 0 abs min . 2

2

2

[ 2 [

2 2

2[ abs min, 1

1

0 d t d 1, 0, xˆ abs min . 2 ¯ (t 1) , 1 d t d 2,

30. xˆ ®

t 2 ,

0 d t d 1, xˆ abs min . t d t d 2, 2 1 , 1 ¯ 0 d t d 1, 2t , 32. xˆ ® xˆ abs min . 2 ¯3 (t 2) , 1 d t d 2, 0 d t d 1, 0, 33. xˆ ® xˆ abs min . 2 ¯ (t 1) , 1 d t d 2,

31. xˆ ®

t 1 2 , 0 d t d 1,

xˆ abs min . 1 d t d 2, ¯0, 8t 3 18t 11, 0 d t d 1 / 2, xˆ abs min . ® 2 1 / 2 d t d 1, ¯ 12(t 1) , t 3 6t 2 , 0 d t d 1, xˆ abs min . ® 2 ¯3t 3t 1, 1 d t d 2, t 2 / 2, 0 d t d 1 / 2, xˆ abs min . ® 3 2 ¯t / 3 t t / 4 1 / 24, 1 / 2 d t d 1,

34. xˆ ® 35. xˆ 36. xˆ 37. xˆ

38. x t abs max, 39. T0 t 2, x t

x

t abs min ;

2 abs max ; T0

40. [ d 1 x [ , [ ! 1 1 0dt d , °c, c ° ; x ® °c ch§¨ t 1 ·¸, 1 d t d 1, c ch§¨1 1 ·¸ [ . ¨ ° ¨ c ¸¹ c c ¸¹ © ¯ © T 41. Sphere with radius 0 is optimal trajectory. 2S

42. Optimal trajectory is ellipse x 2 y 2 2 [y const . 1

2

x y 43. Optimal trajectory is ellipse §¨ ·¸ §¨ ·¸ ©b¹

©a¹ 174

2

R2 .

44. Optimal trajectory is quadric x y const . 45. Optimal trajectory is quadric x const, y const . p 1 3 7 46. x ¨§ ln u 2 u 4 ·¸ p, 2©

47. x(t )

t

u

4

¹

S

³ sign cos(2n 1) 2 W dW ,

t

8

p§1 2· ¨ 2u u ¸, p 0. 2 ©u ¹

0,r1,.... ;

n

0

48. 49. 50. 51. 52. 53. 54.

x x x x x x x

ch t Csh t , u 0 ; ch t , u 0 ; tch t , u 2 sh t ; tsh t , u 2ch t ; sin t C cos t , u 0 ; sin t , u 0 ; t cos t , u 2 sin t ;

S 55. x §¨ t ·¸ sin t , u 2 cos t ;

56. x 57. x 58. x 59. x 60. x 61. [

2¹ © C sin t , u 0 ; sin t , u 0 ; 2(t 2) cos t 4 sin t ; , u 4S 4S ( 2tS 4S ) cos t ( 4t 2S ) sin t , u 4 4S S 2 Cch t , u 0 ; d 1, J max 1 ;

62. T0 d 2, x

8 cos t 4S sin t ; 4 4S S 2

2tT0 t 2 , T0 ! 2, 4 0 d t d T0 2,

t , ° ; x ® (t T0 ) 2 T , T0 2 d t d T0 . 1 ° 0 4 ¯ tT0 t 2 63. T0 d 4, , x 4

T0 ! 4,

64. x(t ) t , J

°t , ° ° § T0 ¨t ° T 2 ° x ® 0 1 © 4 °2 ° °T0 t , ° °¯ T2 T o f at T 2

0dtd

T0 2, 2

2

· ¸ ¹ ,

T0 T 2 d t d 0 2, ; 2 2 T0 2 d t d T0 . 2

o f , x (t )

T2 o f at T o f ; 2 175

t , J

T

65. x

°°t , ® °T t , °¯

66. [ d ɫthT0 , [ ! ɫthT0 ,

67. [

69. [ x (t )

T , 2

T d t d T. 2 [chT , x chT0 x

J

T

T2 . 4

ch t , °sign [ sh W ® °[ (t T ) sign [ , 0 ¯

0 d t dW,

;

W d t d T0 .

0, x { 0, [ z 0 there is no any feasible extreme;

68. [ d thT0 , x x

0dtd

[sht , thT0 [ d T0 , shT0

sh t sign [ , ° ® chW °[ (t T ) sign [ , 0 ¯

0 d t dW,

thW T0 W

W d t d T0 ,

0, x { 0, T ! 0 is arbitrary, [ 1,

[t

x

[ sht shT

,

[ . [ t 1,

;

T 0 d t dW, t , 70. x r °® cos t W tgW 1 ; °¯ sin W , W d t d S , sign p, p ! 1, 71. u °® p x , p d 1, °¯ p, t , t dWn, ° x r® T0 · T0 § 2 ° 1 W n sign t cos¨ t 2n 1 ¸, W n d t d 2n 1 , ¹ © ¯ § 1· (2n 1)¨¨W n arctg ¸¸ T0 , n 1,..., k ; W n ¹ © 72. T0 S , x { 0, T0 ! S

x

0 d t dW, t , ° § T · §T · ° r ® 1 W 2 cos¨ t 0 ¸, W d t d T0 W , W tg ¨ 0 W ¸ 1 ; 2¹ © ©2 ¹ ° °¯T0 t , T0 W d t d T0 ,

73. x t 2 4t abs min ; t 2 (8 8 2 )t , 0 d t d 2 2 , 74. x °® ; 2

75. x

°¯ (t 4) , 2 2 d t d 4, 2 t , 0 d t d 1, ° 2 ®(t 2) 2, 1 d t d 3, ° (t 4) 2 , 3 d t d 4. ¯

176

APPENDIX IV TESTS

1) The equation Fx x 0 , x 0 , t

d Fx x 0 , x 0 , t dt

0 is

A) Legendre equation B) Euler-Puasson equation C) Euler equation D) Euler-Ostrogradsky equation E) Jacobi equation 2) I x, x

³ x t dt o min, K x, x ³ xt dt

3) I x, x

³

4) I x, x

2 ³0 x t dt o min,

1

1

2

0

0

A) xt 6t , t [0,1] B) xt 3t 2 2t 1, t [0,1] C) xt t 2 / 4 3 / 4, t [0,1] D) xt t 3 t 2 , t [0,1] E) there is not any extreme S

0

x 2 t dt o min,

A) xt t , t [0,1] B) xt 3t 2 2t 1, t [0,1] C) x t sin t , t [0, S ] D) x t cos t , t [0, S ] E) there is not any extreme S

A) x t t , t [0,1] B) x t t 2 sin t / S C) xt t 2 / 4 3 / 4 D) xt t 3 t 2 E) there is not any extreme

K x, x

K x, x

177

3, x0 0, x1 6

S

³ xt cos t dt 0

S

³0 xt sin t dt

S / 2, x 0 1, x S 1

S / 2, x0 0, xS 1

5) I x, x

S

³0 xt sin t dt o min,

A) xt t sin t B) xt t sin t C) xt t 2 / 4 3 / 4 D) xt t 3 t 2 E) there is not any extreme I x, x

S

K x, x

S

2 ³0 x t dt

S

³0 x t dt o min, K x, x ³0 xt sin t dt 6) S G x, x ³ xt cos t dt S / 2, x0 2, xS 0 0 2

3S / 2, x0 0, xS S

S 2,

A) x t t cos t B) xt 2 sin t cost 1 C) xt t 2 / 4 3 / 4 D) xt t 3 t 2 E) there is not any extreme

7) I x, x

2 ³0 x t dt o min, 1

A) xt te t 5 B) xt 2e1t 1 t C) xt 2e1t 1 t D) xt t 3 t 2 E) there is not any extreme

³0 ( x 48 x)dt o extr , x0 A) xt t cost B) xt 2 sin t cost 1

8)

1

2

K x, x

³0 xt e 1

t

dt

e, x0 2e 1, x1 2

0, x1 0, x 0 1, x 1 1

C) x t t 2 / 4 3 / 4 D) xt t 4 2t 2 t E) there is not any extreme

³0 ( x 48 x ) dt o extr , x0 A) xt t cost B) xt 2 sin t cost 1

9)

1

2

0, x 1 0, x 0 0, x 1 3 / 2

C) x t t 4 5t 3 / 2 3t 2 / 2 D) x t t 3 t 2 E) there is not any extreme

178

10) ³ ( x2 48 x)dt o extr , x0 0, x1 0, x 0 0, x 1 0 0 1

A) xt t cost B) xt 2 sin t cost 1 C) xt t 2 / 4 3 / 4 D) xt t 4 2t 3 t 2 E) there is not any extreme 1

³0 ( x A) xt B) xt

11)

2

48 x)dt o extr, x0 0, x1 1, x 0 0, x 1 3

t cost 2 sin t cost 1

C) x t t 2 / 4 3 / 4 D) x t t 3 t 2 E) x t 2t 3 t 4 S /2

³0 ( x x )dt o extr , x0 A) xt t cost B) xt 2 sin t cost 1 C) xt sin t / 2 sht / 2shS / 2

12)

2

2

0, xS / 2 1, x 0 0, x S / 2 1

D) xt 2t 3 t 4 E) there is not any extreme 13)

S /2

³

0

( x2 x 2 ) dt o extr , x0 0, xS / 2 0, x 0 0, x S / 2 1

A) xt t cost B) xt 2 sin t cost 1 C) xt t 2 / 4 3 / 4 D) xt t 3 t 2 E) there is not any extreme S

³0 ( x A) xt B) xt

14)

2

x 2 )dt o extr , x0 0, x S 0, x 0 1, x S 1

t cost 2 sin t cost 1

C) xt t 2 / 4 3 / 4 D) xt t 3 t 2 E) xt sin t 15) I x, x

1 2

³t 0

x 2 t dt o min, x0 0, x1 1

A) xt t 1 / 3 , t [0,1]

179

B) xt 2t 1 / 3 , t [0,1] C) xt t 1 / 2 , t [0,1] D) xt t , t [0,1] E) there is not any extreme 16) I x, x

2 ³0 x t dt o min, x0 1

0, x1 1

A) xt t , t [0,1] B) xt 2t 1 / 3 , t [0,1] C) xt t 1 / 2 , t [0,1] D) xt t , t [0,1] E) there is not any extreme 1/ 3

17) I x , x

³ x t dt o min, x 0

18) I x, x

2 ³0 x t dt o min, K x, x ³0 txt dt

0, x 0 0, x1 1

19) I x, x

2 ³0 x t dt o min, K x, x ³0 txt dt

0, x0 4, x1 4

I x, x

³ x t dt o min, K x, x ³ xt dt

1

2

0

A) xt t, t [0,1] B) xt 0, t [0,1] C) x t t 1, t [0,1] D) xt t 3 2t 2 , t [0,1] E) there is not any extreme

x 0 0, x 1

1

1

x 1 1

A) xt t , t [0,1] B) xt 3t 2 2t 1, t [0,1] C) xt t 2 / 4 3 / 4, t [0,1] D) xt 5t 3 / 2 3t / 2, t [0,1] E) there is not any extreme 1

1

A) xt t , t [0,1] B) xt 3t 2 2t 1, t [0,1] C) xt t 2 / 4 3 / 4, t [0,1] D) xt 5t 3 3t 4, t [0,1] E) there is not any extreme

20)

1

1

2

0

0

x0 x1 0 A) xt t , t [0,1]

180

1, G x, x

³ txt dt 1

0

0,

B) x t 3t 2 2t 1, t [0,1] C) xt t 2 / 4 3 / 4, t [0,1] D) xt 60t 3 96t 2 36t , t [0,1] E) there is not any extreme 21) If for every function y ( x) M the number J is stated by some law , then it is said that in class Ɇ defined the following A) Factorial J B) vector J C) functional J D) multiplier J E) division J 22) Gy y( x) y0 ( x) ( y ( x), y0 ( x) M ) is called A) derivative B) argument increment C) division D) integral E) sum of argument 23) The most of maximums f1 ( x) f ( x) , f1c( x) f c( x) , …, f ( k )1 ( x) f ( k ) ( x) for curves f , f1 is A) functional B) sufficient condition of maximum C) sufficient condition of minimum D) the length of k-th order E) the length of 0-th order 24) The equation Fx x 0 , x 0 , t

d Fx x 0 , x 0 , t dt

0 is

A) Legendre equation B) Euler-Puasson equation C) Euler equation D) Euler-Ostrogradsky equation E) Jacobi equation 25) I x, x

2 ³0 x t dt o min, K x, x ³0 xt dt 1

1

A) xt 6t , t [0,1] B) xt 3t 2 2t 1, t [0,1] C) xt t 2 / 4 3 / 4, t [0,1] D) xt t 3 t 2 , t [0,1] E) there is not any extreme

181

3, x0 0, x1 6

S

S

26) I x , x

³

x 2 t dt o min,

K x , x

³ x t cos t dt

27) I x, x

2 ³0 x t dt o min,

K x, x

³0 xt sin t dt

28) I x, x

³0 xt sin t dt o min,

0

A) xt t , t [0,1] B) xt 3t 2 2t 1, t [0,1] C) x t sin t , t [0, S ] D) x t cos t , t [0, S ] E) there is not any extreme S

A) x t t , t [0,1] B) x t t 2 sin t / S C) xt t 2 / 4 3 / 4 D) xt t 3 t 2 E) there is not any extreme S

A) xt t sin t B) xt t sin t C) xt t 2 / 4 3 / 4 D) xt t 3 t 2 E) there is not any extreme S

I x, x

S

K x, x

S

S

2

S / 2, x0 0, xS 1

2 ³0 x t dt

³ x t dt o min, K x, x ³ xt sin t dt 29) S G x, x ³ xt cos t dt S / 2, x0 2, xS 0 0

S / 2, x 0 1, x S 1

0

0

3S / 2, x0 0, xS S

S 2,

0

A) x t t cos t B) xt 2 sin t cost 1 C) xt t 2 / 4 3 / 4 D) xt t 3 t 2 E) there is not any extreme

30) I x, x

2 ³0 x t dt o min, 1

K x, x

A) xt te 5 B) xt 2e1t 1 t C) xt 2e1t 1 t D) xt t 3 t 2 E) there is not any extreme t

182

³0 xt e 1

t

dt

e, x0 2e 1, x1 2

1

³ ( x A) xt B) xt

31)

2

0

48 x)dt o extr , x0 0, x1 0, x 0 1, x 1 1

t cost 2 sin t cost 1

C) x t t 2 / 4 3 / 4 D) xt t 4 2t 2 t E) there is not any extreme

³ ( x 48 x )dt o extr , x 0 A) xt t cost B) xt 2 sin t cost 1

32)

1

2

0

0, x 1 0, x 0 0, x 1 3 / 2

C) x t t 4 5t 3 / 2 3t 2 / 2 D) x t t 3 t 2 E) there is not any extreme 1

³0 ( x A) xt B) xt

33)

2

48 x) dt o extr , x0 0, x1 0, x 0 0, x 1 0

t cost 2 sin t cost 1

C) xt t 2 / 4 3 / 4 D) xt t 4 2t 3 t 2 E) there is not any extreme 34) I x, x

A) xt B) xt C) xt D) x t E) xt

1 2/3

³0 t

x 2 t dt o min, x0 0, x1 1

t 1 / 3 , t [0,1] 2t 1 / 3 , t [0,1] t 1 / 2 , t [0,1] t , t [0,1] t 4 / 3 , t [0,1]

35) I x, x

³ x t xt dt x 1 o min

36) I x, x

2 ³0 x t dt o min, K x, x ³0 xt dt

1

2

2

0

A) xt t , t [0,1] B) x t 0, t [0,1] C) xt t 2 / 4 3 / 4, t [0,1] D) xt t 3 t 2 , t [0,1] E) there is not any extreme 1

1

A) x t t , t [0,1]

183

0, x0 0, x1 1

B) xt 3t 2 2t , t [0,1]

C) xt t / 4 3 / 4, t [0,1] D) x t t 3 t 2 , t [0,1] E) there is not any extreme 2

37) Necessary condition of a weak local minimum A) G I 0, G 2 I 0 B) G I ! 0, G 2 I 0 C) G I 0, G 2 I ! 0 D) G I 0, G 2 I 0 E) G I 0, G 2 I t 0 1

³0 ( x A) xt B) xt

38)

2

48 x)dt o extr, x0 0, x1 1, x 0 0, x 1 3

t cost 2 sin t cost 1

C) x t t 2 / 4 3 / 4 D) x t t 3 t 2 E) x t 2t 3 t 4 S /2

³0 ( x x )dt o extr , x0 A) xt t cost B) xt 2 sin t cost 1 C) xt sin t / 2 sht / 2shS / 2

39)

2

2

0, xS / 2 1, x 0 0, x S / 2 1

D) xt 2t 3 t 4 E) there is not any extreme S /2

³0 ( x x )dt o extr , x0 A) xt t cost B) xt 2 sin t cost 1

40)

2

2

0, xS / 2 0, x 0 0, x S / 2 1

C) xt t 2 / 4 3 / 4 D) xt t 3 t 2 E) there is not any extreme S

³0 ( x A) xt B) xt

41)

2

x 2 )dt o extr , x0 0, x S 0, x 0 1, x S 1

t cost 2 sin t cost 1

C) xt t 2 / 4 3 / 4

184

D) xt t 3 t 2 E) xt sin t 42) I x, x

1 2

³0 t

x 2 t dt o min, x0 0, x1 1

A) xt t 1 / 3 , t [0,1] B) xt 2t 1 / 3 , t [0,1] C) xt t 1 / 2 , t [0,1] D) xt t , t [0,1] E) there is not any extreme 43) I x, x

2 ³0 x t dt o min, x0 1

0, x1 1

A) xt t , t [0,1] B) xt 2t 1 / 3 , t [0,1] C) xt t 1 / 2 , t [0,1] D) xt t , t [0,1] E) there is not any extreme 1/ 3

44) I x , x

2 ³0 x t dt o min, x 0

45) I x, x

2 ³0 x t dt o min, K x, x ³0 txt dt

I x, x

³ x t dt o min, K x, x ³ xt dt

1

A) xt t, t [0,1] B) xt 0, t [0,1] C) x t t 1, t [0,1] D) xt t 3 2t 2 , t [0,1] E) there is not any extreme

x 0 0, x 1

1

1

x 1 1

0, x0 4, x1 4

A) xt t , t [0,1] B) xt 3t 2 2t 1, t [0,1] C) xt t 2 / 4 3 / 4, t [0,1] D) xt 5t 3 3t 4, t [0,1] E) there is not any extreme

46)

x0 A) xt B) x t C) xt

1

1

2

0

0

x1 0 t , t [0,1] 3t 2 2t 1, t [0,1] t 2 / 4 3 / 4, t [0,1]

185

1, G x, x

³ txt dt 1

0

0,

D) xt 60t 3 96t 2 36t , t [0,1] E) there is not any extreme 47) One of the linearity conditions of functional J[y(x)] is written as A) J[y1(x)+y2(x)]= J[y1(x)]+ J[y2(x)] A) J[y1(x)]= J[y2(x)] B) J[y1(x)]= J[y2(x)]=0 C) J[y1(x)]> J[y2(x)] D) J[y1(x)]< J[y2(x)] 48) The value 'J J > y( x) Gy( x)@ J > y( x)@ is called A) an integral equality B) Euler equation C) extreme D) Jacobean E) increment of the functional J 49) If y ( x) y1 ( x) , yc( x) y1c( x) are small on [a, b], then curves y (x) , y1 ( x) close in sense of closeness of A) the first order B) the second order C) the third order D) the k-th order E) the 0-th order S /2

³0 ( x x )dt o extr , x0 A) xt t cost B) xt 2 sin t cost 1 C) xt sin t / 2 sht / 2shS / 2

50)

2

2

0, xS / 2 1, x 0 0, x S / 2 1

D) xt 2t 3 t 4 E) there is no any extreme S /2

³0 ( x x )dt o extr , x0 A) xt t cost B) xt 2 sin t cost 1

51)

2

2

0, xS / 2 0, x 0 0, x S / 2 1

C) xt t 2 / 4 3 / 4 D) xt t 3 t 2 E) there is no any extreme 52)

A)

S

³0 ( x xt

2

x 2 )dt o extr , x0 0, x S 0, x 0 1, x S 1

t cost 186

B) C) D) E)

xt 2 sin t cost 1 xt t 2 / 4 3 / 4 xt t 3 t 2

xt sin t

53) I x, x

³0 t

54) I x, x

2 ³0 x t dt o min, x0

1 2

x 2 t dt o min, x0 0, x1 1

A) xt t 1 / 3 , t [0,1] B) xt 2t 1 / 3 , t [0,1] C) xt t 1 / 2 , t [0,1] D) xt t , t [0,1] E) there is no any extreme 1

0, x1 1

A) xt t , t [0,1] B) xt 2t 1 / 3 , t [0,1] C) xt t 1 / 2 , t [0,1] D) xt t , t [0,1] E) there is no any extreme 1/ 3

55) Weierstrass’ theorems define A) Sufficient conditions that ensemble U {u U / I u min I u } z * * * u U B) Necessary and sufficient conditions that ensemble U {u U / I §¨¨u ·¸¸ min I u } z * * © * ¹ uU C) Necessary conditions that ensemble U {u U / I u min I u } z * * * u U D) Necessary and sufficient conditions to ensemble convexity

U

*

{u U / I u * *

min I u } uU

E) Necessary conditions that the point u* U is a point of the minimum to function Iu on ensemble U 56) Sufficient conditions that the ensemble U {u U / I u min I u } z give

*

*

*

uU

A) Weierstrass’ theorems B) Farkas’s theorems C) Kuhn-Tucker theorems D) Lagrange theorems E) There isn’t any correct answer amongst A)-D). 187

57) Ensemble U E n is identified convex if A) u U , v U , D >0,[email protected] the point u D

D u 1 D v U B) u U , v U , D >0,[email protected], uD d Du 1 D v C) u U , v U , D >0,[email protected], u t Du 1 D v D D) uU,vU, number D>0,[email protected] for which u D d D u 1 D v U E) u U , v U , D E the point u D d D u 1 D v U 1

58) Ensemble U E n is convex iff A) it contains all convex linear combinations of any final number own points B) uU, vU possible construct the point u Du 1 D v which is a

D

convex linear combination of the points u , v C) It is limited and closed D) v U , H ! 0 vicinity S H v U E) From any sequence {u } U it is possible to select converging subsequence k

59) Function I u determined on convex ensemble U E n is identified convex if: A) u U , v U , D >0,[email protected], I D u 1 D v d D I u 1 D I v B) {u }U : lim u u, I u d lim I §¨ u ·¸

k k of k k of © k ¹ C) u U, v U, D>0,[email protected], IDu 1 D v t DIu 1 D Iv D) {u }U : lim u u, I u t lim I §¨ u ·¸ k k k of k of © k ¹ E) There isn’t any correct answer amongst A)-D).

I u

60) Necessary and sufficient condition that continuously differentiated function is convex on convex ensemble U E n : A) I u I v t I cv , u v !, u U , v U B) I u I v d I D u 1 D v , u U , v U , D >0,[email protected] C) I cu* , u u* !t 0, u U D) I cu* , u u* !d 0, u U E) I u I v t I D u 1 D v , u U , v U , D >0,[email protected] 61) Necessary and sufficient condition that continuously differentiated function

I u is convex on convex ensemble U E n A) I cu I cv , u v !t 0, u U , v U B) I cu I cv , u v !d 0, u U , v U 188

C) I cu I cv t I u I v , u U , v U D) I D u 1 D v t I cu I cv , u U , v U E) I D u 1 D v d I cu I cv , u U , v U 62) Function I u determined on convex ensemble strongly convex on U , if

U E n is identified

2 A) N ! 0, I D u 1 D v d D I u 1 D I v D1 D N u v ,

u U ,v U ,D>0,[email protected] B) I D u 1 D v d D I u 1 D I v , u U , v U , D >0,[email protected]

C) I D u 1 D v t D I u 1 D I v , u U , v U , D >0,[email protected]

D) N ! 0, I D u 1 D v t D I u 1 D I v D1 D N u v 2 , u U , vU ,

D>0,[email protected]

E) There isn’t any correct answer amongst A)-D). 63) In order to twice continuously differentiated on convex ensemble U function I (u) is convex on U necessary and sufficiently performing the condition

A) I ccu [ , [ !t 0, u U , [ E n B) det I ccu z 0 u U C) I ccu [ , [ !d 0, u U , [ E n D) I ccu u , v !t 0, u U , v U E) I ccu u , v !d 0, u U , v U

64) Choose correct statement A) If I u , G u are convex functions determined on convex ensemble U ,

D t 0, E t 0 that function B) If

D I u E G u is also convex on ensemble U

I u , G u are convex functions determined on convex ensemble

U E n , that function I u Gu is also convex on U C) If I u , G u are convex functions determined on convex ensemble U E n , that function I u Gu is also convex on U D) If I u , G u are convex functions determined on convex ensemble U E n , Gu z 0, u U , that I u / Gu is convex function on U E) There isn’t any correct answer amongst A)-D). 65) Choose correct statement A) Intersection of two convex ensembles is convex ensemble B) Union of two convex ensembles is convex ensemble 189

C) If U E n is convex ensemble, that ensemble En \U is also convex D) If U E n , U E n is convex ensemble, that ensemble U \ U is also 1 2 1 2 convex E) There isn’t any correct answer amongst A)-D). 66) Define type of the following problem I u

c, u !o inf n u U {u E / u t 0, j I , g u a i , u ! b d 0, i 1, m, j i i g u a i , u ! b 0, i m 1, s } i i

A) General problem of linear programming B) Canonical problem of linear programming C) Nondegenerate problem of linear programming D) Problem of nonlinear programming E) The simplest variational problem

67) Define type of the following problem I u c, u !oinf

u U

{u E n / u

j

0, i 1, n, Au

b}

A) General problem of linear programming B) Canonical problem of linear programming C) Nondegenerate problem of linear programming D) Problem of nonlinear programming E) The simplest variational problem 68) Function I u u 2 2u u u 2 on ensemble

1

1 2

2

U En is

A) convex B) concave C) neither convex, nor concave D) convex under u t 0, u t 0 and concave under u d 0, u d 0

1 2 1 2 E) convex under u d 0, u d 0 and concave under u t 0, u t 0 1 2 1 2

69) Define type of the problem I u

2u 2 u 2 o inf 1 2

u U {u E n / 2u u d 3, u 4u 5} 1 2 1 2

A) Problem of nonlinear programming B) Canonical problem of linear programming C) Convex programming problem D) General problem of linear programming 190

E) The simplest variational problem 70) Define type of the problem I u

2u u o inf 1 2 n 2 {u E / u t 0, u u d 4, 2u u d 2 2 1 2 1 2

u U

A) Convex programming problem B) Canonical problem of linear programming C) General problem of linear programming D) Problem of nonlinear programming E) The simplest variational problem 71) Define type of the problem

2u u o inf 1 2 u U {u E n / u t 0, u 4u d 2, u 3u 1 1 2 1 2 I (u )

4}

A) General problem of linear programming B) Canonical problem of linear programming C) Convex programming problem D) Problem of nonlinear programming E) The simplest variational problem 72) Sequence

{u } U k

I u

is identified minimizing to function

determined on ensemble U , if A) lim I (u ) I , where I * k

inf I u * uU B) I u d I u , k 1,2,... k 1 k

k of

C) lim u

u , moreover u U

k1 t I uk , k k of k

D) I u

1,2,...

E) There isn’t any correct answer amongst A)-D). 73) Choose correct statement for problem I u o inf, u U E n A) For any function I (u) and ensemble U E n always exists a minimizing sequence {u } U for function I (u)

k

B) If function

I (u)

reaches minimum value on

is continuously differentiable on ensemble

U,

i.e.

u U *

*

such that I u

C) If minimizing sequence {u } U exists for function

* umin I u U

k

such that I u

191

U,

min I u uU

that it

I (u), that u U *

D) If the point

u U *

* umin I u U

exists such that I u

sequence {u } U exists for function

k

I (u)

, that minimizing

E) There isn’t any correct answer amongst A)-D).

74) Choose correct statement for problem

I u o inf, u U E n , U

*

{u U / I u * *

min I u } uU

U,

A) If function I u is semicontinuous from below on compact ensemble that ensemble U z

*

U

B) If ensemble ensemble U z

is convex, but function

*

C) If ensemble U is limited, but function ensemble U z

*

I u

is continuous on

U,

but

I u is continuously differentiable on closed ensemble U , that U

U z *

that

I u is convex on ensemble U , that

D) If function I u is semicontinuous from below on ensemble ensemble U is limited, that E) If function

U,

*

z

75) Simplex method is used for solution A) linear programming problem in canonical form B) linear programming problem in general form C) convex programming problem D) nonlinear programming problem E) optimal programming problem 76) Kuhn-Tucker theorems A) define necessary and sufficient conditions that in the convex programming {u U / I u min I u } exist Lagrange’s problem for each point U

multipliers

O /

*

0

*

such that pair

* u*, O*

uU

forms saddle point to Lagrange’

function B) define necessary and sufficient existence conditions of function minimizing sequences {u } U k C) define necessary and sufficient convexity conditions to function I u

I(u)

n convex ensemble U E

192

on

D) define necessary conditions that in the convex programming problem ensemble U {u U / I u min I u } consists of single point

* uU {u U / I u min I u } z * * uU

*

*

E) define necessary and sufficient conditions that ensemble

U

*

For

77)

the

I u o inf, u U

convex

programming

problem

{u E n / u U , g u d 0, i 1, m} 0 i

of

the

Lagrange’s function has the form

m Lu , O I u ¦ O g u , u U , 0 A) i 1 i i O / {O E m / O t 0,..., Om t 0} 0 1 m m B) Lu, O ¦ O g u , O t 0, i 1, m, ¦ O 1, u U i i i i 0 i 1 i 1 C) L u , O O I (u ), u U O E1 0 m D) Lu , O I (u ¦ O g u ), u U , O E m 0 i 1 i i m L u , O I u ¦ O g u , u U , 0 E) i 1 i i O / {O E m / O t 0,..., Om t 0} 0 1 78) For the convex programming problem of the type

u U g u i

{u E n / u U , g u d 0, i 1, m, 0 i a i , u ! b 0, i m 1, s} i

Lagrange’s function has the form

I u o inf

s I u ¦ O g u , u U , 0 A) i 1 i i O / {u E s / u t 0, u t 0, ..., u m t 0} 0 1 2 m m B) L u , O ¦ O g u , O t 0, i 1, m, ¦ O 1, u U i 0 i 1 i i i 1 i Lu, O

193

type

m I u ¦ O g u , u U , O E m 0 i 1 i i

C) L u , O

s I u ¦ O g u , u U , 0 D) i 1 i i O / {u E s / u t 0, u t 0, ..., u s t 0} 0 1 2 E) Lu , O OI u , u U , O E1 Lu, O

*

79) For the convex programming problem necessary and sufficient conditions that for any point u U {u U / I u min I u exist Lagrange’s

multipliers O* /

*

0

*

uU

(u , O ) forms saddle point to Lagrange’s * *

such that pair

function are defined by A) Kuhn-Tucker theorems B) Lagrange’s theorems C) Weierstrass’ theorem D) Farkas’ theorems E) Bellman’s theorems

I u be a convex function determined and continuously differentiated {u U / I u min I u z . In order on convex ensemble U , ensemble U 80) Let

*

*

*

the point u * U to be a point of the function minimum necessary and sufficient performing the condition A) I c u , u u !t 0, u U B) C) D) E)

* * I cu , u u !d 0, u U * * I u I u t I cu , u u !, * * I ccu [ , [ !t 0, [ E * I cu 0 *

uU

I u

on ensemble

U

u U

n

*

81) Pair u , O* U u /

0

0

is identified saddle point to Lagrange’s function

s Lu , O I u ¦ O g u , if i 1 i i A) L u , O d L u , O* d L u , O* , u U , O / 0 0 * *

194

B) L u , O d L u , O* , u U , O / 0 0 * * C) L u , O 0

*

*

D) Lu , O d L u , O* d Lu , O , u U , O / E) u U

*

*

{u U / I u * *

*

0

0 min I u }, O* t 0, i 1, s j uU

82) If pair u , O* U u / is a saddle point to Lagrange’s function

0

0

*

in the convex programming problem, then A) the point u U {u U / I u

*

*

*

B) Lebesgue ensemble M u

min I u } uU {u U / I u d I u } is compact *

Lu, O

C) there is a minimizing sequence {u } U for function I u , such that

*

lim u f k of k D) the convex programming problem is nondegenerate {u U / I u min I u } contains single point E) ensemble U

*

*

*

uU

83) For the nonlinear programming problem

I u o inf, u U

{u E n / u U , g u d 0, i 1, m, g u 0, i 1, m i 0 i

generalized Lagrange’s function has the form A)

L u , O

s

O I u ¦ O g u , u U , O /

0 0 i 1 i i {O E s 1 / O t 0 , O t 0 ,..., O m t 0} 0 1

B) Lu , O C) Lu , O D) Lu , O E) Lu , O

0

m I u ¦ g u , u U , O E m 0 i 1 i m O I u ¦ O g u , u U , O t E m 0 0 i 1 i i s O I u ¦ O g u , u U , O t E s 1 0 0 i 1 i i s O I u ¦ g u , u U , O t 0 0 i 1 i 195

84) Let U E n be convex ensemble, function I u C1 U the condition I c u , u u !t 0, u U is

*

*

umin I u } U {u U / I u min I u } * * uU {u U / I u min I u } * * uU

A) necessary condition that the point

u U * *

{u U / I u * *

B) necessary and sufficient condition that the point

u U * *

C) sufficient condition that the point

u U * *

D) necessary and sufficient condition that the function I u is convex in the point u U E)

U

*

*

necessary

{u U / I u * *

and

sufficient

min I u } z uU

condition

that

ensemble

85) Indicate faithful statement A) convex programming problem is a partial case of the nonlinear programming problem B) nondegenerate problem of nonlinear programming can be reduced to the convex programming problem C) convex programming problem is a partial case of the linear programming problem D) any nonlinear programming problem can be reduced to the convex programming problem E) nonlinear programming problem is a partial case of the convex programming problem 86) For solving of the nonlinear programming problem is used A) Lagrange multipliers method B) Simplex-method C) Method of the least squares D) Pontryagin maximum principle E) Bellman’s dynamic programming method 87) What of enumerated methods can be used for solving of the convex programming problem A) Lagrange multipliers method B) Simplex-method C) Method of the least squares D) Pontryagin maximum principle E) Bellman’s dynamic programming method 196

88) What of enumerated methods can be used for solving of the linear programming problem A) Simplex-method B) Method of the least squares C) Pontryagin maximum principle D) Bellman’s dynamic programming method E) any method from A)- D) 89) In the convex programming problem the minimum to function ensemble U can be reached A) in internal or border points ensemble U B) only in the border points ensemble U C) only in the isolated points ensemble U D) only in the internal points ensemble U E) in internal, border, isolated points ensemble U

I u on

90) In the nonlinear programming problem minimum to function I u on ensemble U can be reached A) in internal, border, isolated points ensemble U B) only in the border points ensemble U C) only in the isolated points ensemble U D) in internal or border points ensemble U E) only in the internal points ensemble U 91)

If

in

the

linear

programming

problem

in

canonical

form

ensemble I u o inf, u U {u E n / u t 0, j 1, n, Au b} j U {u U / I u min I u } contains single point u , that the point is * * * uU

*

A) an extreme point ensemble U B) an isolated points ensemble U C) an internal point ensemble U D) an internal or extreme point ensemble U E) extreme or isolated point ensemble U

92) The linear programming problem in canonical form has the form A) I u c, u !o inf, u U {u E n / u t 0, j 1, n, Au

j

B)

b}

I u c, u !o inf, u U {u E n / u t 0, j I , g u a i , u ! b d 0, i 1, m, j i i g u a i , u ! b 0, i m 1, s} i i 197

C) I u c , u !o inf, u U D) E) 93)

{u E / Au n

b}

I u c, u !o inf, u U {u E n / u t 0, j I , g u a i , u ! b d 0, i 1, m} j i i

I u c, u !o inf, u U The

linear

I u c, u !o inf, u U

{u E n / 0 d g u d b , i 1, m} i i

programming

problem

in

{u E n / u t 0, j 1, n, Au j

canonical

form

b}

is identified nondegenerate, if: A) any point u U has not less than rang A positive coordinates B) rangA m, where A is a constant matrix of dimensionality m u n, m n C) ensemble U {u U / I u min I u } z

rangA m u n, m n D)

E) any point

*

*

m, where

u U

A

*

uU

is a constant matrix of dimensionality

has no more than

94) By extreme point ensemble U

rang A positive coordinates

{u E n / u t 0, j 1, n, Au b} is j

called A) Point

u

u U which can not be presented in the manner of D v 1 D w , where D 0,1 , v U , w U B) Isolated point ensemble U C) Border point ensemble U D) Point u U presented in the manner of

D 0,1 , v U , w U

E) Internal point ensemble U

u

D v 1 D w ,

where

REFERENCES the main references: 1. Alekseev B.M., Tychomirov B.M., F o m i n C .B. Optimalnoe upravlenie. – M.: Nauka, 1979. 2. Boltyunsky B.G.Matematicheskie metody optimalnogo upravlenia. – M.: Nauka, 1969. 3. Brison A., Ho Yu-shi. Prikladnaia teoria optimalnogo upravlenia. – M.: Mir, 1972. 4. Vasilev F.P.Lekzii po metodam reshenia extremalnih zadach. – M.: MSU, 1974. 5. Vasilev F.P. Chislenie metody reshenia extremalnih zadach. – M.: Nauka, 1980. 6. Gabasov R., Kirillova F.M. Metody optimizazii. – Minsk: BSU, 1980.

the additional references: 7. Gelfand I.M., Fomin S.B. Variozionnoe ischislenie. – M.: PhisMath, 1961. 8. Zubov B.I. Lekzii po teorii upravlenia. – M.: Nauka, 1975. 9. Karmanov B.G. Matematicheskoe programmirovanie. – M.: Nauka, 1975. 10. Krasovsky N.N. Teoria upravlenia dvizheniem. – M.: Nauka, 1968. 11. Krotov B.F., Gurman B.I. Metody i zadachi optimalnogo upravlenia. – M.: Nauka, 1973. 12. Lie E.B., Markus L. Osnovi teorii optimalnogo upravlenia. – M.: Nauka, 1972. 13. Pontryagin L.S., Boltyansky B.G., Gamkleridze R.B., Mishenko E.F. Matematicheskaya teoria optimalnih prozessov. – M.: Nauka, 1976. 14. Pshenichny B.N., Danilin Yu.M. Chislenie metody v extremalnih zadachah. – M.: Nauka, 1975. 15. Aisagaliev S.Ⱥ., Zhunussova Zh.Kh. Mathematical programming: textbook. – Almaty: Kɚzɚkh university, - 208 p. 2011. 16. Ⱥɣɫɚɝɚɥɢɟɜ ɋ.Ⱥ., Ʉɚɛɢɞɨɥɞɚɧɨɜɚ Ⱥ.Ⱥ. Optimalnoe upravlenie dinamicheskih sistem. Verlag, Palmarium academic publishing, - 288ɫ. 2012.

ɍɱɟɛɧɨɟ ɢɡɞɚɧɢɟ

Aisagaliev Serikbay Abdigalievich Zhunussova Zhanat Khaphizovna

OPTIMAL CONTROL Textbook ȼɵɩɭɫɤɚɸɳɢɣ ɪɟɞɚɤɬɨɪ Ɂ. ɍɫɟɧɨɜɚ Ʉɨɦɩɶɸɬɟɪɧɚɹ ɜɟɪɫɬɤɚ Ɍ.ȿ. ɋɚɩɚɪɨɜɚ Ⱦɢɡɚɣɧ ɨɛɥɨɠɤɢ: Ɋ.ȿ. ɋɤɚɤɨɜ ɂȻ ʋ 7393 ɉɨɞɩɢɫɚɧɨ ɜ ɩɟɱɚɬɶ 01.07.14. Ɏɨɪɦɚɬ 100ɯ80 1/12. Ȼɭɦɚɝɚ ɨɮɫɟɬɧɚɹ. ɉɟɱɚɬɶ ɰɢɮɪɨɜɚɹ. Ɉɛɴɟɦ 16,75 ɩ.ɥ. Ɍɢɪɚɠ 200 ɷɤɡ. Ɂɚɤɚɡ ʋ 1329. ɂɡɞɚɬɟɥɶɫɬɜɨ «Ԕɚɡɚԕ ɭɧɢɜɟɪɫɢɬɟɬi» Ʉɚɡɚɯɫɤɨɝɨ ɧɚɰɢɨɧɚɥɶɧɨɝɨ ɭɧɢɜɟɪɫɢɬɟɬɚ ɢɦ. ɚɥɶ-Ɏɚɪɚɛɢ. 050040, ɝ. Ⱥɥɦɚɬɵ, ɩɪ. ɚɥɶ-Ɏɚɪɚɛɢ, 71. Ʉɚɡɇɍ. Ɉɬɩɟɱɚɬɚɧɨ ɜ ɬɢɩɨɝɪɚɮɢɢ ɢɡɞɚɬɟɥɶɫɬɜɚ «Ԕɚɡɚԕ ɭɧɢɜɟɪɫɢɬɟɬi».