Variation calculus and methods of optimization: educational manual 9786010416949

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AL-FARABI KAZAKH NATIONAL UNIVERSITY

S.A. Aisagaliev Zh.Kh. Zhunussova

VARIATION CALCULUS AND METHODS OF OPTIMIZATION Educational manual

Almaty «Qɜzɜq university» 2016

UDɍ 517.97 (075.8) LBC 22.161.8ɻ73 A 28 Recommended for publication by the Academic Council of the Faculty of Mechanics and Mathematics, Editorial-Publishing Council and Educational and Methodical Union 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 ʍ 1, October 7, 2015)

Reviewers: Doctor of Physical and Mathematical Sciences, Professor M.I. Tleubergenov Candidate of Physical and Mathematical Sciences, Ass. professor A.U. Kalizhanova Doctor of Physical and Mathematical Sciences, Professor S.Ya. Serovaijsky

A 28

Aisagaliev S.ȼ., Zhunussova Zh.Kh. Variation calculus and methods of optimization: educational manual / S.ȼ. Aisagaliev, Zh.Kh. Zhunussova. – Almaty: Qɜzɜq university, 2016. – 322 p. ISBN 978-601-04-1694-9

Some theoretical foundations of optimal control problem are expounded in the educational manual: 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 an educational manual for students of the high schools training on specialties «mathematics», «mechanics», «mathematical and computer modeling» and «informatics». It will be useful for the post-graduate students and scientific workers of the economic, mathe-matical and naturally-technical specialties.

UDɍ 517.97 (075.8) LBC 22.161.8ɻ73

ISBN 978-601-04-1694-9

© Aisagaliev S.ȼ., Zhunussova Zh.Kh., 2016 © KazNU al-Farabi, 2016

CONTENTS FOREWORD

................................................................................................. 5

INTRODUCTION ................................................................................................. 6 LECTURE 1.

THE MAIN DETERMINATIONS. STATEMENT OF THE PROBLEM ...................................... 6 LECTURE 2. WEIERSTRASS’ THEOREM ............................................... 12 CHAPTER I CONVEX PROGRAMMING. ELEMENTS OF THE CONVEX ANALYSIS ...................... 16 LECTURE 3. CONVEX SETS ..................................................................... 16 LECTURE 4. CONVEX FUNCTIONS ........................................................ 22 LECTURE 5. THE PRESCRIPTION WAYS OF THE CONVEX SETS. THEOREM ABOUT GLOBAL MINIMUM. OPTIMALITY CRITERIA. THE POINT PROJECTION ON SET ................................................................................... 29 LECTURES 6, 7. SEPARABILITY OF THE CONVEX SETS ......................... 35 LECTURE 8. LAGRANGE’S FUNCTION. SADDLE POINT .................. 40 LECTURES 9, 10. KUHN-TUCKER’S THEOREM............................................ 46 CHAPTER II NONLINEAR PROGRAMMING.......................................... 57 LECTURES 11. PROBLEM STATEMENT. NECESSARY CONDITIONS OF THE OPTIMALITY ....... 57 LECTURE 12. SOLUTION ALGORITHM OF NONLINEAR PROGRAMMING PROBLEM .............................................. 66 LECTURE 13. DUALITY THEORY ............................................................. 73 CHAPTER III LINEAR PROGRAMMING .................................................. 80 LECTURES 14. PROBLEM STATEMENT. SIMPLEX-METHOD. .............. 80 LECTURE 15. DIRECTION CHOICE. NEW SIMPLEX-TABLE CONSRUCTION. THE INITIAL EXTREME POINT CONSTRUCTION ........ 90 CHAPTER IV NUMERICAL METHODS OF MINIMIZATION IN FINITE-DIMENSIONAL SPACE .................................... 97 LECTURE 16. MINIMIZATION METHODS OF ONE VARIABLE FUNCTIONS .................................... 97 LECTURE 17. GRADIENT METHOD. THE GRADIENT PROJECTION METHOD ........................ 106 LECTURE 18. CONDITIONAL GRADIENT METHOD. CONJUGATE GRADIENT METHOD ................................. 115 LECTURE 19. NEWTON'S METHOD. THE PENALTY FUNCTION METHOD. LAGRANGE MULTIPLIER METHOD ............................... 125 LECTURE 20. REVIEW OF THE NUMERICAL MINIMIZATION METHODS ............................................... 132 3

CHAPTER V LECTURE 21.

Appendix III.

VARIATION CALCULUS .................................................... 138 BRACHISTOCHRONE PROBLEM. THE SIMPLEST TASK. NECESSARY CONDITIONS FOR A WEAK MINIMUM. LAGRANGE'S LEMMA. EULER EQUATION .............................................................. 138 DUBOIS - RAYMOND LEMMA. BOLZ PROBLEM. WEIERSTRASS NECESSARY CONDITION .......................................................................... 145 LEGENDRE CONDITION. JACOBI CONDITION. ................................................................................................. FUNCTIONALS DEPENDING ON THE N UNKNOWN FUNCTIONS. FUNCTIONALS DEPENDING ON HIGHER ORDER DERIVATIVES ................................ 154 ISOPERIMETRIC PROBLEM. CONDITIONAL EXTREMUM. LAGRANGE PROBLEM. GENERAL COMMENTS ................................. 162 OPTIMAL CONTROL. MAXIMUM PRINCIPL ................. 170 PROBLEM STATEMENT OF OPTIMAL CONTROL. MAXIMUM PRINCIPLE FOR OPTIMAL CONTROL PROBLEM WITH FREE RIGHT END ................................. 170 PROOF OF MAXIMUM PRINCIPLE FOR OPTIMAL CONTROL PROBLEM WITH FREE RIGHT END ..................................................... 178 MAXIMUM PRINCIPLE FOR OPTIMAL CONTROL PROBLEM. CONNECTION BETWEEN MAXIMUM PRINCIPLE AND VARIATION CALCULUS ..................... 184 OPTIMAL CONTROL. DYNAMIC PROGRAMMING ...... 193 OPTIMALITY PRINCIPLE. BELLMAN EQUATION ....... 193 DISCRETE SYSTEMS. OPTIMALITY PRINCIPLE. BELLMAN EQUATION ....................................................... 200 SUFFICIENT OPTIMALITY CONDITIONS ....................... 207 TASKS FOR INDEPENDENT WORK ................................. 216 VARIANTS OF THE TASKS FOR INDEPENDENT WORK............................................... 274 KEYS ...................................................................................... 312

References

................................................................................................. 321

LECTURE 22. LECTURE 23.

LECTURE 24. CHAPTER VI LECTURE 25. LECTURE 26. LECTURE 27. CHAPTER VII LECTURE 28. LECTURE 29. LECTURE 30. Appendix I. Appendix II.

4

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. In 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

INTRODUCTION

L E C T U R E 1. THE MAIN DETERMINATIONS. PROBLEM STATEMENT Let E n be the euclidean space elements of which are determined by collection of n numbers, i.e. u  E n , u (u1 , u2 ,...,un ) and a scalar function J (u) J (u1 ,..., u n ) be determined on a set U of the space E n . It is necessary to find the maximum (or minimum) of the function J u on set U, where U  E n is the prescribed set. Production problem. The products of the five types are made on the enterprise. The cost of the unit product of each type accordingly c1 , c2 , c3 , c4 , c5 , in particular c1 3, c2 5, c 4 1, c5 8 . For fabrication specified products the enterprise has some determined resources expressed by the following normative data: Type of the product 1 2 3 4 5

Materials Energy

Labor expenses

a11

3

a 21

6

a31 1

a12

1

a 22

3

a32

a13

4

a23

1

a33 1

a14

2

a 24

4

a34

2

a15 1

a 25

5

a35

4

120

b3

20

b1

50

b2

3

Here a11 is an amount of the materials required for fabricating the unit of the 1-st type product; a12 is a consumption of the material for fabrication the unit of the 2-nd type product and so on; aij , i 1, 2, 3, j 1, 2, 3, 4, 5 are expenses of the material, energy, labor expenses for fabrication the unit of the j type products. It is required to find a such production plan output such that provides the maximal profit. So as the profit is proportional to total cost of marketed commodities, that hereinafter it is identified with total cost of marketed commodities. Let u1 , u2 , u3 , u4 , u5 be amount output products of five type. Then mathematical formalization of the problem maximizations function (profit) has the form J (u)

J (u1 , u 2 , u3 , u 4 , u5 )

c1u1  c2 u 2  c3u3  c4 u 4  c5 u5 3u1  5u 2  10u3  u 4  8u5 o max

6

(1)

at the conditions (restrictions of the resources) g1 (u ) a11u1  a12u2  a13u3  a14u4  a15u5 d b1 , ½ ° g 2 (u ) a21u1  a22u2  a23u3  a24u4  a25u5 d b2 ,¾ g 3 (u ) a31u1  a32u2  a33u3  a34u4  a35u5 d b3 , °¿

(2)

u1 t 0, u2 t 0, u3 t 0, u4 t 0, u5 t 0,

(3)

where aij , i 1,3, j 1,5 are normative coefficients which values presented above; b1 , b2 , b3 are recourses of the enterprise. Since amount of the products are nonnegative numbers that necessary the condition (3). If we introduce the notations § a11 ¨ A ¨ a21 ¨a © 31

c

§ c1 · ¨ ¸ ¨ c2 ¸ ¨ c ¸; ¨ 3¸ ¨ c4 ¸ ¨c ¸ © 5¹

u

a12 a22 a32

a13 a23 a33

a14 a24 a34

a15 · ¸ a25 ¸; a35 ¸¹

§ u1 · ¨ ¸ § b1 · ¨ u2 ¸ ¨ u ¸; b ¨ b ¸; ¨ 2¸ ¨ 3¸ ¨b ¸ ¨ u4 ¸ © 3¹ ¨u ¸ © 5¹

g

§ g1 · ¨ ¸ ¨ g 2 ¸, ¨g ¸ © 3¹

then optimization problem (1) – (3) can be written as J (u) c' u o max; g (u) Au d b; u t 0,

(1' ) ( 2' ) (3' )

where c' (c1 , c2 , c3 , c4 , c5 ) is a vector-line;  ' is a sign of transposition. Let vector-function be g (u) g (u)  b Au  b , but a set U 0 ^u  E 5 / u t 0`. Then problem (1) – (3) is written as: J (u) c' u o max; ^u  E 5 u  U 0 , g (u) d 0`  E 5 .

u U

(4) (5)

The problem (1) – (3) (or (1' ) – (3' ) , or (4), (5)) belongs to type so called problem of linear programming, since J (u ) is linear with respect to u function; the vectorfunction g (u ) also is linear with respect to u . 7

We considered the case above, when vector u has five components, vectorfunction g (u ) – three and matrix A has an order 3u 5 . In the case, vector u has dimensionality n , g (u ) is m - dimensional function, but matrix A has an order mu n and there are restrictions of the equality type (for instance g1 (u ) A1u  b1 , where A1 is a matrix of the order m1 u n , b1  E m1 ) problem (4), (5) is written:

U

^u  E

J u c ' u o max; u U , n

u  U 0 , g (u )

Au  b d 0, g1 (u )

(6) A1u  b1

(7)

`

0,

where set U 0 ^u  E n u t 0`, c  E n . Problem (6), (7) belongs to type of the general problem of linear programming. We suppose that J (u ) is convex function determined on the convex set U 0 (unnecessary linear); g (u ) is a convex function determined on the convex set U 0 . Now problem (6), (7) is written as

u U

J (u) o max; ^u  E u  U 0 , g (u) d 0, g1 (u) n

A1u  b1

`

0.

(8) (9)

Problem (8), (9) belongs to type so called problem of convex programming. Let J (u ) , g (u ) , g1 (u ) be arbitrary functions determined on the convex set U 0 . In this case, problem (8), (9) can be written as

u U

J (u) o max; ^u  E n u  U 0 , g (u) d 0, g1 (u)

`

0.

(10) (11)

Optimization problem (10), (11) belongs to type so called problem of nonlinear programming. In all presented problems of linear, convex and nonlinear programming the concrete ways of the prescription set U from E n are specified. If it is distracted from concrete way of the prescription set U from E n , so optimization problem in finite-dimensional space possible to write as J (u) o max; u  U , U  E n .

Finally, we note that problem of the function maximization J (u ) on set U tantamount to the problem of the function minimization – J (u ) on set U. So further it is possible to limit by consideration of the problem J (u) o min; u  U ,

8

U  En,

(12)

for instance, in the problem (1) - (3) instead of maximization J (u ) to minimize the function  3u1  5u2  10u3  u4  8u5 o min . Finally, the first part of the course is devoted to the solution methods of convex, nonlinear, linear programming. The question arises: whether the problem statement (12) correct in general case? It is necessary the following definitions from mathematical analysis for answer. Definition 1. The point u*  U is called the minimum point of the function J (u ) on a set U, if inequality J (u * ) d J (u ) holds at all u U . The value J (u* ) is called the least or minimum value of the function J (u ) on set U. The set U * u*  U J (u* ) min J (u ) contains all minimum points of the function

^

`

uU

on set U. It follows from the definition, that the global (or absolute) minimum of the function J (u ) on the set U is reached on the set U *  U . We remind, that the point u**  U is called the local minimum point of the function J (u ) , if inequality J (u ** ) d J (u ) is valid under all u  o (u ** , H ) ˆ U , where set J (u )

o(u** , H )

^u  E

`

| u  u** | H  E n

n

is an open sphere with the centre in u** and radius H ! 0 ; | a | is the Euclidean norm of the vector a a1 ,..., an  E n , i.e. | a |

n

¦a . 2 i

i 1

Example 1. Let function be J (u ) cos 2

S

,

u ^2 3` ;

and a) set U is determined as

following: U ^u  E 1 1 2 d u d 1`, then set U * b) U ^u  E 1 1 4 d u d 2`, then set U * ^2 / 7, 2 / 5, 2 / 3, 2` is countable; c) U ^u  E1 / 2  u  f`, then set U * ‡ , where ‡ is empty set. Example 2. The function J (u) ln u , set U ^u  E 1 0  u d 1`. The set U * ‡ . Example 3. The function J (u ) J 1 (u )  c, c const , where function

J1 (u )

­| u  a |, if u  a; ° ® u  b, if u ! b; °c, if a d u d b, ¯

and set U E1 . The set U * >a, b@ . Definition 2. It is spoken, that function J (u ) is bounded below on set U if there exists the number M such that J (u) t M under all u U . The function J (u ) is not bounded below on set U if there exists the sequence ^uk `  U such that lim J (u k ) f . k of

Function J (u ) is bounded below on set U in the examples 1, 3, but function J (u ) is not bounded on U in the example 2. 9

Definition 3. We suppose, that function J (u ) is bounded from below on set U . Then value J * inf J u is called the lower bound of the function J (u ) on set U, if: uU

1) J * d J (u ) under all u U ; 2) for arbitrary small number H ! 0 is found the point u(H )  U such that value J u H  J *  H . In the case, when the function J (u ) is not bounded from below on set U, lower bound J * f . We notice, that for example 1 value J * 0 , but for examples 2, 3 values J * f , J * c , accordingly. If set U * z ‡ , so J * min J u (see examples). Value J * uU

always exists, but min J u does not always exist. Since lower bound J * for function uU determined on set U always exists, independently of that, whether set U * is emptily or in emptily, problem (12) can be written in the manner of J (u )

J u o inf,

u U , U  E n .

(13)

Since value min J u does not exist, when set U * ‡ , that correct form of the uU optimization problem in finite-dimensional space has the form of (13). Definition 4. The sequence ^uk `  U is called minimizing for function J (u ) determined on set U, if limit lim J u k inf J u J * . k of

uU

As follows from definition of lower bound, in the case H k 1 k , k 1,2,3... we have the sequences ^u H k ` ^u 1 / k ` ^uk `  U k 1,2,..., for which J uk  J *  1/ k Thence follows that limit lim J u k J * . Finally, minimizing sequence ^uk `  U always k of

exists.

Definition 5. It is spoken, that sequence ^uk `  U converges to set U *  U , if | uk  u* | is a distance from point uk U till limit lim U u k , U * 0 , where U u k ,U * uinf U k of

*

*

set U * . It is necessary to note, if set U * z ‡ , that the minimizing sequence always exists which converges to set U * . However statement about any minimizing sequence converges to set U * , in general case, untrue. Example 4. Let function be J u u 4 (1  u 6 ) , U E 1 . For this example set U * ^0` , and the sequences ^uk 1/ k , k 1,2,...`  U , ^uk k , k 1,2,...`  U are minimizing. The first sequence converges to set U * , the second infinitely is distanced from it. Finally, source optimization problem has the form of (13). We consider the following its solutions: 1-st problem. Find the value J * inf J u . In this case independently of that uU

whether set U * is emptily or in emptily, problem (13) has a solution. 2-nd problem. Find the value J * inf J u and the point u*  U * . In order to the uU

problem (13) has a solution necessary the set U * z ‡ . 10

3-rd problem. Find the minimizing sequence ^uk `  U which converges to set U * . In this case necessary that set U * z ‡ . Most often in practice it is required solution of the 2-nd problem (see the production problem). CONTROL QUESTIONS 1. Give definition to the minimum point of the function on a set. 2. Give definition to the lower bound of the function on a set. 3. Formulate the matrix form of the production problem 4. Formulate convex programming problem. 5. Formulate mathematical formalization of the production problem.

11

LECTURE 2. WEIERSTRASS’ THEOREM We consider the optimization problem J u o inf,

u U , U  E n .

(1)

It is necessary to find the point u*  U * and value J * inf J u . uU

We notice, if set U * z ‡ , that J * J u*

min J u . uU

The question arises: what requirements are imposed on the function J (u ) and on set U that the set U * u*  U J (u* ) min J (u ) be in emptily? For answer it is uU

^

`

necessary to enter the notions of compact set and half-continuity from below of the function J (u ) on set U. The compact sets. Let ^uk `  E n be a certain sequence. We remind, that: a) the point v  E n is called the limiting point of the sequence ^uk `, if there exists a subsequence ^ukm ` for which limit lim uk v ; b) sequence ^uk `  E n is identified m of

m

bounded , if there exists a number M t 0 , such the norm | uk |d M for all k 1,2,3,... ; b) set U  E n is identified bounded, if there exists a number R t 0 , such the norm | u |d R at all u U ; c) the point v  E n is called the limiting point of the set U, if any its H is neighborhood set ov, H contains the points from U differenced from v ; d) for any limiting point v of the set U there exists the sequence ^uk `  U for which lim u k v ; e) set U  E n is called closed, if it contains all their own limiting points. k of

Definition 1. The set U  E n is called compact, if any sequence ^u k `  U has at least one limiting point v, moreover v U . It is easy make sure, the definition is equally to known from the course of the mathematical analysis statement about any bounded and closed set is compactly. In fact, according to Bolzano-Weierstrass theorem any bounded sequence has at least one limiting point (the set U is bounded ), but from inclusion v U follows the closure of the set U . Half-continuously from below. Let ^uk `  E n be a sequence. Then ^J k ` ^J uk ` is the numeric sequence. We notice, that: a) numeric sequence ^J k ` is bounded from below, if the number D exists, such that J k t D , k 1,2,3,... ; b) numeric sequence ^J k ` is not bounded from below, if ^J km ` subsequence exists, such the limit lim J k f . mof

m

Definition 2. By the lower limit of the bounded from below numeric sequence a lim J k if: 1) there exists subsequence ^J km `

^J k ` is identified the value ɜ denoted by

mof

12

for which lim J k m of

m

a ; 2) all other limiting points to sequences ^J k ` is not less the

value ɜ. If numeric sequence ^J k ` is not bounded from below, then the value a f. Example 1. Let J k 1   1 k , k 0,1,2,... . The value a 0 . Definition 3. It is said, the function J u determined on set U  E n , halfcontinuous from below in the point u U , if for any sequence ^uk `  U for which the limit lim u k u , the inequality lim J u k t J u holds. The function J (u ) is halfk of

k of

continuous from below on set U, if it is half-continuous from below in each point of the set U. Example 2. Let set be U ^u  E 1  1 d u d 1`, function J u u 2 at 0 | u | 1, J 0 1 . The function J (u ) is continuous on set 0 | u |d 1 , consequently, it is half-continuous from below on the set. We show, that function J (u ) is halfcontinuous from below in the point u 0 . In fact, the sequence ^1 / k `, k 1,2,... . belongs to the set U and the limit of the sequence is equal to zero. Numeric sequence ^J uk ` ^1/ k 2 `, moreover limit lim J uk lim J uk 0 . Consequently, lim J u k 0 ! 1 . k of

k of

k of

It means the function J (u ) is half-continuous from below on set U. Similarly possible to enter the notion of the half-continuity from above of the function J (u ) on set U . We notice, if the function J (u ) is continuous in the point u U , that it is half-continuous in it as from below, so and from above. The most suitable check way of the half-continuity from below of the function J (u ) on set U gives the following lemma. The Lemma. Let function J (u ) be determined on closed set U  E n . In order the function J (u ) to be half-continuous from below on set U , necessary and sufficiently that Lebesgue’s set M (ɭ) ^u  E n J (u) d c` be closed at all c  E 1 . Proof. Necessity. Let the function J (u ) be half-continuous from below on closed set U . We show, that set M (c) is closed at all c  E 1 . We notice, that empty set is considered as closed. Let v be any limiting point of the set M (c ) . From definition of the limiting point follows the existence of the sequence ^u k `  M (c) which converges to the point v . From inclusion ^u k `  M (c) follows that value J (u k ) d c, k 1,2,... . With consideration of ^u k `  U and half-continuity from below J (u ) on U follows J v d lim J u k d c . Consequently, the point v  M c . Closure of the set M (c ) k of

is proved. Sufficiency. Let U be closed set, but set M (c ) is closed at all c  E 1 . We show, that function J (u ) is half-continuous from below on U. Let ^uk `  U be a sequence which converges to the point u U . We consider the numeric sequence ^J uk `. Let the value a lim J uk . By definition of the lower limit there exists the k of

sequence ^J ukm ` for which lim J u k m of H ! 0 there exists the number N

m



N H ,

a . Consequently, for any sufficiently small

such that J u km d a  H under m ! N . Thence 13

with consideration of the limit lim u k m of

m

u , u km  M a  H , we have J u d a  H . Then

with consideration of arbitrarily H ! 0 it is possible to write the following inequality: J u d lim J u k a , i.e. the function J (u ) is half-continuous from below in the point k of

u U . Since the set U is closed, the function J (u ) is half-continuous from below in any point u U . Lemma is proved.

We notice, in particular, at c J * from lemma follows that set M ( J * ) ^u  E n u  U , J (u) d J * ` U * is closed. Theorem 1. Let function J (u ) be determined, finite and half-continuous from below on compact set U  E n . Then J * inf J u ! f , set uU

U*

^u  E *

n

u*  U , J (u* )

`

min J (u ) uU

is inempty, compactly and any minimizing sequence converges to set U * . Proof. Let ^uk `  U be any minimizing sequence, i.e. the limit of the numeric set lim J u k J * . We notice, that such minimizing sequence always exists. Let u* be k of

any limiting point of the minimizing sequence. Consequently, subsequence ^ukm `  U exists for which lim u k u* . With consideration of the compactness of the set U all m of

m

limiting points of the minimizing sequence belongs to the set U . As follows from definition of lower bound and half-continuity from below of the function J (u ) on set U the following inequalities



J * d J u* d lim J ukm mof

lim J uk k of

J*

(2)

are faithful. Since the sequence ^J uk ` converges to value J * , that any its subsequence also converges to value J * . It is followed, that value J * J u* from inequality (2). Consequently, set U * z ‡ and J * J u* ! f . Since the statement faithfully for any limiting point of the minimizing sequence, so it is possible to confirm that any minimizing sequence from U converges to set U * . We show that set U *  U is compact. Let ^wk ` be any sequence taking from set U * . From inclusion wk  U follows that ^wk `  U . Then with consideration of compactness of set U the subsequence ^wkm ` exists which converges to a point w*  U . Since ^wk `  U , that values J (wk ) J * , k 1,2, ... . Consequently, the sequence ^wk `  U is minimizing. Then, with consideration of proved above, the limiting point w*  U * . Finally, closeness of the set U * is proved. Restrictedness of the set U * follows from inclusion U *  U . Compactness of the set U * is proved. Theorem is proved. Set U is not often bounded in the applied problems. In such cases the following theorems are useful. 14

Theorem 2. Let function J (u ) be determined, finite and half-continuous from below on inempty closed set U  E n . Let for certain point v U Lebesgue’s set M (v )

^u  E

n

`

u  U , J (u ) d J (v)

is bounded. Then J * ! f , set U * is inempty, compactly and any minimizing sequence ^uk `  M (v) converges to set U * . Proof. Since all conditions of the lemma are held, that set M (v) is closed. From restrictedness and closeness of the set M (v) follows its compactness. The set U M (v )  M 1 (v ) , where set M 1 ( v ) ^u  E n u  U , J (u ) ! J (v )`, moreover on set M 1 (v ) the function J (u ) does not reach its lower bound J * . Hereinafter proof of the theorem 1 is repeated with change set U on compact set M (v) . Theorem is proved. Theorem 3. Let function J (u ) be determined, finite and half-continuous from below on inempty closed set U  E n . Let for any sequence ^vk `  U , v k o f , at k o f the equality lim J (v k ) f is held. Then J * ! f , set U * is inempty, compactly k of

and any minimizing sequence ^uk `  U converges to set U * . Proof. Since lim J (vk ) f , that the point v U , such that J (v ) ! J * exists. We k of enter Lebesgue’s set M (v)

^u  E

n

`

u  U , J (u ) d J (v) . With consideration of lemma

the set M (v) is closed. It is easy to show that set M (v) is bounded. In fact, if set M (v) is not bounded, then the sequence ^wk `  M (v) exists, such that wk o f at k o f . By condition of the theorem for such sequence the value J (wk ) o f at k o f . Since J (wk ) d J (v)  f it is not possible. Finally, set M (v) is bounded and closed, consequently, it is compact. Hereinafter proof of the theorem 1 is repeated for set M (v) . Theorem is proved. CONTROL QUESTIONS 1. Give definition to the compact set. 2. Give definition to the lower limit of the bound from below numeric sequence. 3. Give definition to the half-continuous from below function in the point. 4. Prove the Theorem 1. 5. Prove the Theorem 2. 6. Prove the Theorem 3.

15

Chapter I CONVEX PROGRAMMING. ELEMENTS OF THE CONVEX ANALYSIS

Amongst methods of solution of the optimization problems in finitedimensional space the most completed nature has a method of solution of the convex programming problem. Characteristics of the convex sets and functions which allow generalizing the known solution methods of the problems on conditional extreme are studied in the convex analysis developing intensive last years.

LECTURE 3. CONVEX SETS The problems of the convex programming are often met in the applied researches in the following type:

u U

^u  E

J (u ) o inf, n

u  U 0 , g i (u ) d 0, i 1, m,

g i (u )  a i , u ! bi

0, i

`

(1)

m  1, s ,

where J (u ) , g i (u) , i 1, s , are convex functions determined on convex set U 0  E n . Definition 1. Set U  E n is called by convex, if for any u U , v U and under all D , 0 d D d 1 , the point uD Du  (1  D )v v  D (u  v)  U . Example 1. We show that closed sphere S (u 0 , R )

^u  E

n

u  u0 d R

`

is a convex set. Let the points u  S (u 0 , R) , Q  S (u 0 , R) , i.e. the norms u  u 0 d R , v  u 0 d R . We take a number D , D  >0, 1@ , and define the point uD Du  (1  D )v . Norm 16

Du  (1  D )v  u 0

uD  u 0

D (u  u 0 )  (1  D )(v  u 0 ) d

d D u  u 0  (1  D ) v  u 0 d DR  (1  D ) R

R.

Consequently, the point uD  S (u 0 , R ) , and set S (u0 , R) is convex. Example 2. We show that hyperplane *

^u  E

n

 c, u ! J

`

is a convex set, where ɭ  E n is a vector, J is a number. Let the points u  * , v  * . Consequently, scalar products  c, u ! J ,  c, v ! J . Let uD Du  (1  D )v , D  >0, 1@ . Then scalar product  c, uD !  c, Du  (1  D )v ! D  c, u !   (1  D )  c, v ! DJ  (1  D )J

J.

Thence follows that the point uD  * and set * is convex. Example 3. We show that affine set Ɉ ^u  E n Au b`, where A is a constant matrix of the order mu n , b  E n is the vector which is convex. For the point uD Du  (1  D )v , u  Ɉ , v  Ɉ , D  >0, 1@ we have AuD

A(Du  (1  D )v) DAu  (1  D ) Av Db  (1  D )b b

Thence follows that uD  Ɉ and set Ɉ is convex. Let u1 , u 2 , …, u nr be linear independent solutions of the linear homogeneous system Au 0 . Then set M can be presented in the manner of M

^u  E

n

u

`

u 0  v, v  L ,

L

­ n ®u  E v ¯

nr

½

¦D u ¾¿ i

i

i 1

where vector u0  E n is a partial solution of the nonhomogeneous system Au b , L is a subspace to dimensionality n  r formed by the vectors u1 , u 2 , …, u nr , D i , i 1, n  r are the numbers and dimension of the affine set M is taken equal to the dimension of the space L . Definition 2. By affine cover of the arbitrary set U  E n called the intersection of all affine sets containing set U and it is denoted by aff U . Dimensionality of the set U is called the dimension its affine cover and it is denoted by dimU . Since the intersection of any number convex sets is convex set, so the set aff U is convex. Instead of source problem J (u) o inf , u U , where U is an arbitrary set it is considered the approximate problem J (u) o inf , u  aff U , since solution of the last problem in the many cases more simpler, than source. 17

Definition 3. The set A called by the sum of the sets A1 , A2 ,..., Am , i.e. A

A1  A2  ...Am if it contains that and only that points a

m

¦a , i

ai  Ai , i 1, m. Set

i 1

A is called by the set difference Ⱦ and ɍ i.e A B  C , if it contains that and only that points a b  c, b  B, c  C . The set A OD , where O is a real number, if it contains the points a O d , d  D. Theorem 1. If the sets A1 , A2 ,..., Am , B, C, D are convex, then the sets A A1  A2  ...  Am , A B  C , A OD are convex.

Proof. Let the points be a

m

¦a

i

 A, e

i 1

point uD

D a  1  D e

m

¦ e  A, i

m

¦ D a  1  D e , D  >0,1@ . i

a i , ei  Ai , i

1, m. Then the

i 1

i

Since the sets Ai , i 1, m are

i 1

convex, aD

that

m

¦a , i

ai

D a i  1  D ei  Ai , i 1, m .

Consequently,

the

point

D ai  1  D ei  Ai , i 1, m. Thence follows that point aD  A . Convexity

i 1

of the set A is proved. We show, that set A B  C is convex. Let the points be a b  c  A, aD D a  1  D e where The point e b1  c1  A , b, b1  B , c, c1  C . D b  c  1  D b1  c1 >D b   1  D b1 @  >D c  1  D c1 @, D  >0,1@ . Since the sets B and C are convex, the points b Db  1  D b1  B, c Dc  1  D c1  C . Then the point aD b  c  A, b  B, c  C . It is followed that set A is convex. From aD D a  1  D e follows that a Od1  A, e Od 2  A, d1 , d 2  D Od  A, d Dd1  1  D d 2  D . Theorem is D O d 1  1  D Od 2 , D  >0,1@ . Then aD proved. Let U be the prescribed set from E n , v  E n is a certain point. There is one and only one of the following possibilities: a) There exists a number H ! 0 , such that set o(X , H )  U . In this case point X is an internal point of the set U . We denote through int U the ensemble of all internal points of the set U. b) Set o(X , H ) does not contain nor one point of the set U . The point X is called by external with respect to U . c) Set o(X , H ) contains both points from set U , and points from set E n \ U . The point X is called by bound point of the set U. The set of all border points of the set U is denoted by *p U . d) The point X  U , but set o(X , H ) does not contain nor one point from set U , except the points X . The point X is called the isolated point of the set U . Convex set does not contain the isolated points. Definition 4. The point X  U called comparatively internal point of the set U if the intersection o(X , H ) ˆ aff U  U . We denote through riU the ensemble of all comparatively internal points of the set. Example 4. Let set 18

^u  E

U

1

0 d u d 1; 2 d u d 3`  E 1 .

For the example the sets int U

affU

^u  E

1

^u  E

`

1

`

0  u  1; 2  u  3 .

0 d u d 3 , riU

^u  E

1

`

0 d u  1; 2  u d 3 .

the following statements are faithful: 1) If ȼ is a convex set, so closing A is convex too. In fact, if a, b  A , so subsequences ^ak `  A, ^bk `  A, such that ak o a, bk o b under k o f exists. With consideration of convexity of the set A the point D ak  1  D bk  A, k 1,2,3,... . Then >D ak  1  D bk @ D a  1  D b  A under all D  >0,1@. Thence limiting point a klim of follows the convexity of the set A . 2) If U is a convex set and intU z ‡ , so the point v  D u0  v  int U , u0  inf U , v U , 0  D d 1 .

vD

To prove. 3) If U is a convex set, that int U is convex too. To prove. 4) If U is a convex nonempty set, that riU z ‡ and ri U is convex. To prove. 5) If U is a convex set and riU z ‡ , that the point vD

v  vD

v  D u0  v  riU , u0  riU , v U , 0  D d 1

To prove. Definition 5. The point u  E n called the convex combination of the points u1 , u 2 ,...,u m from E n , if it is represented in the manner of u

m

¦D u , i

where the

i

i 1

numbers D i t 0, i 1, m but their sum D1  D 2  ...  D m 1. Theorem 2. The set U is convex if and only if it contains all convex combinations of any finite number their own points. Proof. Necessity. Let U be a convex set. We show, that it contains all convex combinations of the finite number of their own points. We use the method of induction to proof of the theorem. From definition of the convex set follows that statement faithfully for any two points from U. We suppose, that set U contains the convex combinations m  1 of their own points, i.e. the point v

m1

¦E u i

i

U , E i t 0,

i 1

i

1, m  1, E1  ...  E m 1

1 . We prove, that it contains the convex combinations ɮ their

Di u i  D m u m . We i 1 1  Dm

m 1

own points. In fact, expression u D1u 1  ...  D m u m 1  D m ¦ 19

denote E i D i / 1  D m . We notice, that E i t 0 , i 1, m  1 , but sum E1  ...  E m1 1, D1  ...  D m1 1  D m , D i t 0, D1  D 2  ...  D m 1. since Then the point m m u 1  D m v  D mu , v U , u U . Thence with consideration of the set convexity U the point u U . Necessity is proved. Sufficiency. Let the set U contains all convex combinations of any finite number its own points. We show that U - a convex set. In particular, for m 2 we have uD Du1  1  D u 2  U under any u1 , u 2 U under all D  >0,1@. The inclusion means that U - a convex set. Theorem is proved. Definition 6. By convex cover of arbitrary set U  E n called the intersection of all convex sets containing set U and it is denoted by CoU . From given definition follows that CoU is the least (on remoteness from set U ) convex set containing the set U . We notice, that source problem J (u) o inf , u U , with arbitrary set U  E n can be replaced on the approximate problem J (u) o inf , u  CoU . We note, that if U is closed and bounded(compact) set, that CoU is also bounded and closed (compactly). Theorem 3. The set CoU contains that and only that points which are convex combination of the finite number points from U . Proof. To proof of theorem it is enough to show that CoU =W, where W is set of the points which are convex combination of any finite number points from set U . The inclusion W Ž CoU follows from theorem 2, since W Ž CoU and set CoU is convex.

On

the

other

hand,

if

the

u , v W ,

points

i.ɡ.

u

m

¦D u , i

i

i 1

p

m

¦D

D i t 0, i 1, m;

¦E v , E

1; v

i

point

1  D E i t 0,

t 0, i 1, p,

m

i

¦E

1, that under all D  >0,1@ the

i

i 1

p

¦D u  ¦ E v i

i 1

Ei

i

i 1

Du  1  D v

uD

i

i

i 1

p

i

i

,

where

D i DDi t 0,

i 1, m,

i 1

i 1, p moreover the sum

p

m

¦D  ¦ E i

i 1

i

1 . Thence follows that uD  W ,

i 1

consequently, the set W is convex. Since set U Ž W , so the inclusion Co U Ž W exists. From inclusion W Ž CoU , CoU Ž W follows that CoU W . Theorem is proved. Theorem 4. Any point u  CoU can be presented as convex combination no more than n  1 points from U . Proof. As follows from theorem 3, the point u  CoU is represented in the manner of

m

¦D u ,

m

¦D u ,1 , i

D i t 0,

i

i

i 1

i

1 . We suppose, that number m ! n  1 . Then n  1 

i 1

dimensional vectors u i

i

1, m, m ! n  1 are linearly dependent. Consequently,

there are the numbers J 1 ,...,J m , not all equal to zero, such that sum

m

¦ J i ui 0 . Thence

i 1

follows, that

m

¦J u i

i 1

i

0,

m

¦J

i

0 . Using first equality, the point u we present in the

i 1

20

manner

of

m

i

i

m

¦D

i

i 1

i

i

i 1

D i D i  tJ i t 0,

m

¦D u ¦D u

u

i 1

m

m

¦ D

 t¦J iu i i 1

i

 tJ i u i

i 1

m

¦D u , i

i

where

i 1

min J i , amongst J i ! 0 , where

1 under enough small t . Let J i*

index i, 1 d i d m . We choose the number t from the condition D i* / J i* t . Since m

¦J i 1

i

0 , that such J i* ! 0 always exists. Now the point

u

m

m

¦ D u ¦ D u ,D i

i

i

i

i 1

i

t 0,

i 1 i z i*

m

¦D

i

(2)

1.

i 1 i z i*

Specified technique are used for any m ! n  1 . Iterating the given process, eventually get m n  1 . Theorem is proved. Definition 7. The convex cover drawing on the points u 0 , u1 ,..., u m from E n is called m -dimensional simplex, if vectors ^u1  u 0 `, i 1, m are linear independent and it is denoted by S m . The points u 0 , u1 ,..., u m are called the tops of the simplex. Set S m is a convex polyhedron with dimension m and by theorem 3 it is represented in the manner of Sm

­ n ®u  E u ¯

m

¦D u , i

i

D i t 0, i

i 1

0, m,

m

¦D i 0

CONTROL QUESTIONS 1. Give definition to the convex set. 2. Give definition to the affine cover of the arbitrary set. 3. Give definition to the sum of the sets. 4. Prove the Theorem 1. 5. Prove the Theorem 2. 6. Prove the Theorem 3. 7. Prove the Theorem 4. 8. Give definition to the convex cover of the arbitrary set.

21

i

½ 1¾. ¿

LECTURE 4. CONVEX FUNCTIONS Convex

programming problem in general type is formulated: n J u o inf, u  U , U  E n , where U is convex set of E ; J (u ) is a convex function determined on convex set U . Definition 1. Let function J (u ) be determined on convex set U from E n . Function J (u ) is called convex on set U, if for any points u, v U and under all D , 0 d D d 1 the inequality J D u  1  D v d D J u  1  D J v

(1)

holds. If in the inequality (1) possible an equality only at D 0 and D 1 , then function J (u ) is called strongly convex on convex set U . Function J (u ) is concave (strongly concave), if function - J (u ) is concave (strongly concave) on set U . Definition 2. Let function J (u ) be determined on convex set U. Function J (u ) is called strictly convex on set U, if there exists a constant N ! 0 , such that for any points u, v U and under all D , 0 d D d 1 the inequality J Du  1  D v d DJ u  1  D J v  D 1  D N | u  v |2

(2)

holds.

Example 1. Let function J u c, u be determined on convex set U from E n . We show, that J u c, u is a convex function on U . In fact, u, v U are arbitrary points, number D  >0,1@, then the point uD Du  1  D v U on the strength of convexity of the set U . Then J uD

c , uD

D c, u  1  D c, v

DJ u  1  D J v .

In this case relationship (1) is executed with sign of the equality. The symbol is a scalar product. .,. Example 2. Let function J u | u |2 be determined on convex set U  E n . We show, that J (u ) is strictly convex function on set U . In fact, for any u, v U and under all D  >0,1@ the point uD Du  1  D v U . The value J uD Du  1  D v

2

Du  1  D v, Du  1  D v

D 2 | u |2 2D 1  D u, v  1  D 2 | v |2 . (3)

The scalar product 22

u  v, u  v

2 | u  v | 2 | u | 2 2 u , v  | v | .

Thence we have 2 u , v | u | 2  | v |2  | u  v | 2 . Having substituted the equality to the right part of expression (3), we get J uD D | u |2 1  D | v |2 D 1  D | u  v |2

DJ u  1  D J v  D 1  D | u  v |2 .

Finally, relationship (2) is executed with sign of the equality, moreover the number N 1. Example 3. The function J u | u | 2 determined on convex set U  E n is strongly convex. In fact, value J uD

Du  1  D v d D | u | 2 1  D | v | 2 DJ u  1  D J v , 2

moreover equality possible only at D 0 and D 1 . In general case check of convexity or strict convexity of the function J (u ) on convex set U by the definitions 1, 2 rather complex. In such events the following theorems are useful. The criteria to convexity of the smooth functions. Let C 1 (U ), C 2 (U ) be accordingly spaces of the continuously differentiated and twice continuously differentiated functions J (u ) , determined on set U . We notice, that gradient J ' u

wJ u / wu1 ,..., wJ u / wun  E n

under any fixed u U . Theorem 1. In order to the function J u  C 1 (U ) be convex on the convex set U  E n , necessary and sufficiently the inequality J u  J v t J ' v , u  v ,

u , v  U

holds.

(4)

Proof. Necessity. Let function J u  C 1 (U ) be convex on U. We show, that inequality (4) is executed. From inequality (1) we have J v  D u  v  J v d D >J u  J v @,

D  >0,1@,

u , v  U .

Thence on base of the formula of the finite increments we get D J ' v  TD u  v , u  v d D >J u  J v @,

23

0 d T d 1.

Both parts of the given inequality are divided into D ! 0 and turned to the limit under D o 0 , with consideration of J u  C 1 (U ) we get the inequality (4). Necessity is proved. Sufficiency. Let for function J u  C 1 (U ) , U be a convex set inequality (4) is executed. We show, that J (u ) is convex on U. Since the set U is convex, the point uD Du  1  D v U , u, v U , D  >0,1@ . Then from inequality (4) follows, that J u  J uD t J ' uD , u  uD , J v  J uD t J ' uD , v  u D , u, uD , v U .

We multiply the first inequality by D , but the second inequality by the number

1  D and add them. As a result we give DJ u  1  D J v t J uD . Thence follows the convexity to functions J (u ) on set U. Theorem is proved. Theorem 2. In order to the function J (u)  C 1 (U ) be convex on convex set U necessary and sufficiently the inequality J ' u  J ' v , u  v t 0, u ,

v U

(5)

holds. Proof. Necessity. Let the function J (u)  C 1 (U ) be convex on convex set U. We show, that inequality (5) is executed. Since inequality (4) faithfully for any u, v U , that, in particular, we have J v  J u t J ' u , v  u . We add the inequality with inequality (4), as a result we get the inequality (5). Necessity is proved. Sufficiency. Let for the function J (u )  C 1 (U ) , U be a convex set, inequality (5) is executed. We show, that J (u ) is convex on U . For proving it is enough to show, that difference DJ u  1  D J v  J uD

D J u  1  D J v  J Du  1  D v t 0, u , v  U .

Since J (u )  C 1 (U ) , that the following equality faithfully: J u  h  J u

J ' u  T1h , h

1

³ J ' u  th , h dt , 0

u ,

u  h U .

The first equality follows from formula of the finite increments, but the second from theorem about average value, 0 d T1 d 1 Then the difference DJ u  1  D J v  J uD D >J u  J uD @ 

24

1

 1  D >J v  J uD @ D ³ J ' uD  t u  uD , u  uD dt  0

1

 1  D ³ J ' uD  t v  uD , v  uD dt. 0

Let z1 uD  t u  uD uD  t 1  D u  v , z 2 uD   t v  u D uD  tD v  u . Then z1  z 2 t u  v , but the differences u  uD 1  D  z1  z2 / t , v  uD D  z1  z 2 / t . Now previous inequality is written: 1

DJ u  1  D J v  J uD D (1  D ) ³ J c( z1 )  J c( z2 ), z1  z2 0

1 dt. t

Since the points z1 , z 2  U , that according to inequality (5) we have DJ u

 1  D J v  J uD t 0 .

Theorem is proved. Theorem 3. In order to the function J (u)  C 2 (U ) be convex on convex set U , int U z ‡ , necessary and sufficiency the inequality J " (u )[ , [ t 0,

(6)

[  E n , u  U

holds. Proof. Necessity. Let function J (u )  C 2 (U ) be convex on U . We show, that inequality (6) is executed. If the point u  intU , then there exists a number H 0 ! 0 , such that points u  H[ U for any [  E n and under all H , | H |d H 0 . Since all conditions of the theorem 2 are executed, that inequality faithfully J ' (u  H[ )  J ' (u ), H[ t 0,

[  E n , | H |d H 0 .

Thence, considering that J ' (u  H[ )  J ' (u ), H[

J " (u  TH[ )[ , [ H ,

0 d T d 1,

and transferring to limit under H o 0 , we get the inequality (6). If u U is a border point, then there exists the following sequence ^uk `  intU , such that uk o 0 under k o 0 . Hereinafter, by proved J " (u k )[ , [ t 0, [  E n , uk  int U . Transferring to limit and taking into account that lim J " (u k ) J " (u ) with k of

considering of J (u )  C 2 (U ) , we get the inequality (6). Necessity is proved. Sufficiency. Let for function J (u )  C 2 (U ) , U be a convex set, intU z ‡ , inequality (6) is executed. We show, that J (u ) is convex on U . Since the equality faithfully 25

J ' u  J ' v , u  v

J " v  T u  v u  v, u  v ,

by denoting [ u  v and considering that uT J ' u  J ' v , u  v

0 d T d 1,

v  T u  v  U , we get

J " uT [ , [ t 0,

[  E n ,

uT  U .

Thence, with considering of the theorem 2, follows the convexity to function J (u ) on U . Theorem is proved. We note, that symmetrical matrix § w 2 J u / wu12 w 2 J u / wu1wu 2 ¨ 2 w 2 J u / wu 22 ¨ w J u / wu 2 wu1 J " u ¨ ¨. . . . . . . . . . . . . . . . ¨ w 2 J u / wu wu w 2 J u / wu n wu 2 n 1 © and a scalar product J " u [ , [ [ ' J " u [ .

... w 2 J u / wu1wu n · ¸ ... w 2 J u / wu 2 wu n ¸ . . . . . . . . ¸¸ ... w 2 J u / wu n2 ¸¹

The theorems 1 - 3 for strongly convex function J (u ) on U are formulated in the manner specified below and are proved by the similar way. Theorem 4. In order to the function J (u )  C 1 (U ) be strongly convex on the convex set U necessary and sufficiently executing of the inequality J u  J v t J ' v , u  v  k | u  v | 2 , u , v  U .

(7)

Theorem 5. In order to the function J (u )  C 1 (U ) be strongly convex on the convex set U, necessary and sufficiently the inequality J ' u  J ' v , u  v t P | u  v |2 ,

P k const ! 0, u , v  U

P

(8)

holds. Theorem 6. In order to the function J (u )  C 2 (U ) be strongly convex on the convex set U , int U z ‡ necessary and sufficiently the inequality J " u [ , [ t P | [ |2 , [  E n , u  U ,

P

P k const ! 0

(9)

holds. The formulas (4) - (9) can be applied for defining of the convexity and strong convexity of the smooth functions J (u ) determined on the convex set U  E n . For studying of convergence of the successive approximation methods are useful the following lemma. Definition 3. It is said, that gradient J ' u to function J (u)  C 1 (U ) satisfies to Lipschitz’s condition on the set U, if 26

| J ' u  J ' v |d L | u  v |, u , v  U , L

const t 0.

(10)

Space of such functions is denoted by C1,1 U . Lemma. If function J u  C1,1 U , U be convex set, then the inequality faithfully | J u  J v  J ' v , u  v |d L | u  v |2 / 2,

(11)

 u, v  U .

Proof. From equality J u  J v  J ' v , u  v

1

³ J ' v  t u  v  J ' v , u  v dt 0

follows that 1

J u  J v  J ' v , u  v d ³ J ' v  t u  v  J ' v u  v dt . 0

Thence with consideration of inequality (10) and after integration by t , we get the formula (11). Lemma is proved. The properties of the convex functions. The following statements are faithful. 1) If J i u , i 1, m are convex functions on convex set, then function J u

m

¦D J u , D i

i

i

t 0, i 1, m is convex on set U .

i 1

In fact, J uD

m

m

¦D i J i uD d ¦D i >DJ i u  1  D J i v @ i 1

i 1

m

m

i 1

i 1

D ¦D i J i u  1  D ¦D i J i v DJ u  1  D J v , u, v U .

2) If J i u , i  I is a certain family of the convex functions on convex set U, then function J u sup J i u , i  I is convex on set U . In fact, by determination of the upper boundary there exists an index i*  I and a number H ! 0 , such that J (uD )  H d J i* (uD ), i* i* H , D  I . Thence we have J (uD ) d H  D J i* (u )  1  D J i* (v), uD

D u  1  D v.

Consequently, J uD d D J u  1  D J v , at H o 0 . 3) Let J u be a convex function determined on convex set U  E n . Then the inequality is correct 27

§ m · m J ¨ ¦ D i ui ¸ d ¦ D i J u i , D i t 0, ©i1 ¹ i1

m

¦D

i

1.

i 1

To prove by the method to mathematical induction. 4) If function f t , t  >a, b @ is convex and it is not decrease, and function F u is convex on the convex set U  E n , moreover the values F u  >a, b @ , then function J u f F u is convex on U . In fact, the value J uD

f F uD d f D F u  1  D F v d

d D f F u  1  D f F v

D J u  1  D J v .

We notice, that t1 F u  >a , b @, t 2 F v  >a, b @ and f D F u  1  D F v

f D t1  1  D t 2 d

d D f t1  1  D f t 2 , t1 , t 2  [ a , b ] ,

with consideration of convexity of the function f t on segment >a, b @ .

CONTROL QUESTIONS 1. Give definition to the convex function on set. 2. Give definition to the strongly convex function on set. 3. Prove the Theorem 1. 5. Prove the Theorem 2. 6. Prove the Theorem 3. 7. Formulate the Theorem 4, the Theorem 5, the Theorem 6. 8. Formulate the properties of the convex functions.

28

LECTURE 5. THE PRESCRIPTION WAYS OF THE CONVEX SETS. THEOREM ABOUT GLOBAL MINIMUM. OPTIMALITY CRITERIA. THE POINT PROJECTION ON SET Some properties of the convex functions and convex set required for solution of convex programming problem are studied separately in the previous lectures. Now we consider the connection between them, necessary and sufficiently conditions of the optimality condition for convex programming problem and the other questions which are important for solution of the optimization problems in finite-dimensional space The prescription ways of the convex sets. Under studying of the properties to convex function J (u ) and convex set U any answer to the question has not been given how convex set U in space E n is prescribed. Definition 1. Let J (u ) be a certain function determined on set U from E n . The set epiJ

^u,J E

n1

/ u U  E n , J t J (u)` E n1

(1)

is called the above-graphic (or epigraph) to function J (u ) on set U. Theorem 1. In order to the function J u be convex on the convex set U necessary and sufficiently that set epi J be convex. Proof. Necessity. Let function J (u ) be convex on the convex set U . We show, that set epi J is convex. It is sufficiently to make sure in that for any z (u , J 1 )epi J , w (Q , J 2 ) epi J and under all D , 0 d D d 1, the point zD Dz  (1  D ) w epi J . In fact, z D Du  (1  D )Q ,DJ 1  (1  D )J 2 , moreover Du  (1  D )Q U with the point consideration of set convexity U, and the value DJ 1  (1  D )J 2 t DJ (u)  (1  D ) J (w) t J (uD ) .

Finally, the point zD uD , J D , where uD Du  (1  D )Q U , and value J D t J (uD ) . Consequently, the point zD epi J . Necessity is proved. Sufficiency. Let set epi J be convex, U be a convex set. We show that function J (u ) is convex on U . If the points u, v  U , so z u , J (u )  epi J , w v, J (v )  epi J . For any D ,D  >0,1@, the point zD uD ,DJ (u)  (1  D ) J (v)  epi J with consideration of convexity the set epi J . From the inclusion follows that value J D DJ (u)  (1  D ) J (v) t J (uD ) . It means that function J (u ) is convex on U . Theorem is proved. Theorem 2. If function J (u ) is convex on the convex set U  E n , then the set 29

^u  E

M (C )

n

u  U , J (u ) d C `

is convex under all C  E 1 . Proof. Let the points be ɤ, v  Ɉ(ɍ), i.e. J (u) d C, J (Q ) d C, u,Q U . The point uD Du  (1  D )Q U under all D , D  >0,1@ with consideration of convexity the set U. J (u ) Since function is convex on U, consequently value J (uD ) d DJ (u)  (1  D ) J (Q ), u,Q U . Thence with consideration of J (u) d C, J (Q ) d C , we get J (uD ) d DC  (1  D )C C . Finally, the point uD U , value J (uD ) d C . Consequently, for any u,Q U the point uD  M (C) under all C  E 1 . It means that set M C is convex. Theorem is proved. We consider the following optimization problem as appendix: J (u ) o inf ,

^u  E

u U

g i (u )

n

(2)

/ u  U 0 , g i (u ) d 0, i

¢ ai , u ²  bi

1, m;

`

m  1, s ,

0, i

(3)

where J (u ) , g i (u ), i 1, m are convex functions determined on convex set U 0 ; a i  E n , i m  1, s are the given vectors; bi , i m  1, s are prescribed numbers. We introduce the following sets: Ui

^u

U m 1

i

 E n /u  U 0 , g i (u ) d 0`, i

^u  E

n

/ ¢ ai , u ²  bi

0, i

1, m,

`

m  1, s .

The sets U i , i 1, m are convex, since g i (u ), i 1, m are convex functions determined on convex set U 0 (refer to theorem 2, C 0 ), and set U m1

^u  E

n

Au

` ^u  E

b

n

¢ a i , u ²  bi

m  1, s

0, i

`

where ȼ is a matrix of the order s  m u n ; the vectors ai , i m  1, s are the lines of the matrix ȼ; b bm1 ,, bs E ns is affine set. Now the problem (2), (3) possible write as J (u) o inf, u U

m1

U

i

 U0 .

(4)

i 0

Thereby, the problem (2), (3) is problem of the convex programming, since the intersection of any number convex sets is convex set. 30

Theorem about global minimum. We consider the problem of the convex programming (4), where J (u ) is convex function determined on convex set U from En . Theorem 3. If J (u ) is a convex function determined on convex set U and n J * inf J (u ) ! f , U * u*  E / u*  U , J (u* ) min J (u ) z ‡ , then any point of the local uU

^

uU

`

minimum to function J (u ) on U simultaneously is the point of its global minimum on U, moreover the set U * is convex. If J (u ) is strictly convex on U, then set U * contains not more one point. Proof. Let u*  U be a point of the local minimum of the function J (u) on set U, i.e. J (u* ) d J (u ) under all u o(u* , H ) ˆ U . Let Q U be an arbitrary point. Then the point w u*  D (Q  u* ) o(u* , H ) ˆ U with consideration of convexity the set U, if D Q  u*  H . Consequently, inequality J (u* ) d J ( w) is held. On the other hand, since J (u ) is a convex function on U, that the inequality J ( w)

J (u *  D (Q  u * ))

J (D Q  (1  D )u * ) d D J (Q )  (1  D ) J (u * )

holds and D ! 0 is sufficiently small number i.e. D  >0,1@ . Now inequality J (u* ) d J ( w) is written by J (u* ) d J ( w) d DJ (Q )  (1  D ) J (u* ) . This implies, that 0 d D >J (Q )  J (u* )@ . Consequently, J (u* ) d J (v ) , Q U . This means that in the point u*  U the global minimum to function J (u) on U is reached. We show, that set U * is convex. In fact, for any u ,Q  U * , i.e. and J (u ) J (Q ) J * under all D ,D  >0,1@ the value J (uD ) d J (Du  (1  D )Q ) d DJ (u )  (1  D ) J (Q )

J* .

Thence follows that J (uD ) J * , consequently, the point uD U * . Convexity of the set is proved. We notice, that from U * z Ø follows J * J (u* ) . J (u ) If is strictly convex function, then must be J (uD )  DJ (u)  (1  D ) J (Q ) J * , 0  D  1 . Contradiction is taken off, if u v , i.e. the set U * contains not more one of the point. Theorem is proved. Finally, in the convex programming problems any point of the local minimum to function J (u ) on U will be the point its global minimum on U, i.e. it is a solution of the problem. Optimization criterion. Once again we consider the convex programming problem (4) for case J (u)  C 1 (U ) . Theorem 4. If J (u)  C 1 (U ) is an arbitrary function, U is a convex set, the set U * z ‡ , then in any point u*  U * necessary the inequality ¢ J c(u* ), u  u* ² t 0, u  U

is held. 31

(5)

If J (u)  C1 (U ) is a convex function, U is a convex set, U * z ‡ , then in any point u*  U * necessary and sufficiently performing of the condition (5). Proof. Necessity. Let the point u*  U * . We show, that inequality (5) is held for any function J (u)  C 1 (U ) , U is a convex set (in particular, J (u ) is a convex function on U ). Let uU be an arbitrary point and number D  >0,1@ . Then the difference J (uD )  J (u* ) t 0, where uD Du  (1  D )u* U . Thence follows that 0 d J (D u  (1  D )u * )  J (u * )

J (u *  D (u  u * ))  J (u * )

D ¢ J c(u* ), u  u* ²  o(D ),

moreover o(D ) / D o 0 , at D o 0 . Both parts are divided into D ! 0 and by transferring to the limit under D o 0 we get the inequality (5). Necessity is proved. Sufficiency. Let J (u)  C 1 (U ) be a convex function, U be a convex set, U * z ‡ and inequality (5) is held. We show that the point u*  U * . Let uU be an arbitrary point. Then by theorem 2 (see the section "Convexity criteria of the smooth functions"), we have J (u )  J (u* ) t ¢ J c(u* ), u  u* ² t 0 under all uU . This implies that J (u * ) d J (u ), u  U . Consequently, the point u*  U * . Theorem is proved. Consequence. If J (u)  C 1 (U ) , U is a convex set, U * z ‡ and the point u*  U * , u*  int U , then the equality J c(u* ) 0 necessary is held. In fact, if u*  int U , then there exists the number H 0 ! 0 such that for any e  E n the point u u*  H e U under all H , H d H 0 . Then from (5) follows ¢ J c(u * ), H e² t 0, e E n , under all H , H d H 0 . Thence we have J c(u* ) 0 . Formula (5) as necessary condition of the optimality for nonlinear programming problem and as necessary and sufficiently condition of the optimality for convex programming problem will find applying in the following lectures. The point projection on set. As application to theorems 1, 2 we consider the point projection on convex closed set. Definition 2. Let U be a certain set from E n , and the point Q  E n . The point wU is called the projection of the point Q  E n on set U , if norm Q  w inf Q  u , uU and it is denoted by w PU (Q ) Example 1. Let the set be U

S (u 0 , R )

^u  E

n

/ u  u 0 d R`,

and the point Q  E n . The point w PU (Q ) u 0  R (Q  u 0 ) Q  u 0 , that follows from geometric interpretation. Example 2. If the set U

*

^u  E 32

n

`

/ ¢c, u² J ,

projection Q  E n on U is defined by the formula 2 w PU (Q ) Q  >J  ¢ c,Q ² @c c . Norm of the second composed is equal to the distance from the point v till U . Theorem 5. Any point Q  E n has an unique projection on convex closed set U  E n , moreover for point w PU (Q ) necessary and sufficiently the condition then

the

point

¢ w  v, u  w² t 0, u  U

(6)

holds. In particular, if U is affine set, that condition (6) is written by ¢ w  v, u  w² 0, u U .

(7)

Proof. The square of the distance from the point v till uU is equal to 2 n J (u ) u  v . Under fixed Q  E the function J (u ) is strictly convex function on convex closed set U . Any strictly convex function is strongly convex. Then by theorem 1 the function J (u) reaches the lower boundary in the unique point w , moreover wU with consideration of its completeness. Consequently, the point w PU (Q ) . Since the point w  U *  U and gradient J c( w) 2( w  v) , that necessary and sufficient optimality condition for the point w, ( w u* ) according to formula (5) is written in the manner of 2¢ w  v, u  w² t 0, u U . This implies the inequality (6). If U ^u  E n / Au b` is affine set, then from u0 , u  U , u0 z u follows that 2u0  u U . In fact, A(2u0  u) 2 Au0  Au 2b  b b . In particular, if u0 w PU (Q ) , then the point 2w  u U . Substituting to the formula (6) instead of the point uU we get ¢ w  v, w  u² t 0, u  U .

(8)

From inequalities (6) and (8) follow the expression (7). Theorem is proved. Example 3. Let the set be U ^u  E n / u t 0`, the point Q  E n . Then projection w PU (v ) (v1 , , v n ), where v i max( 0, v i ), i 1, n . To prove. Example 4. Let the set U

^u

(u 1 ,  , u n )  E n / D i d u i d E i , i

`

1, n

be n-dimensional parallelepiped, and the point Q  E n . Then components of the vector w  PU (Q ) are defined by formula wi

D i , if vi ! D i ­ ° Ei , if vi ! E i ® °v , if D d v d E , i i i i ¯ i 33

1, n.

In fact, from formula (6) follows that sum n

¦ (w  v )(u i

i

i

 wi ) t 0

i 1

under all D i d vi d E i , i 1, n . If vi  D i , that wi D i , u i t D i , consequently the product ( wi  vi )(u i  wi ) (  v i  D i )(u i  D i ) t 0 . Similarly, if vi ! E i , that wi

E i , u i d E i so ( wi  v i )(u i  wi )

(  vi  E i )(u i  E i ) t 0 .

Finally, if D i d vi d E i , then wi vi , so ( wi  vi )(u i  wi ) 0 . In the end, for the point w ( w1 , , wn )  PU (Q ) inequality (6) is held.

CONTROL QUESTIONS 1. Give definition to the epigraph to function on a set. 2. Prove the Theorem 1. 3. Prove the Theorem 2. 4. Prove the Theorem 3. 5. Prove the Theorem 4. 6. Give definition to the projection of the point on set. 7. Prove the Theorem 5.

34

LECTURES 6, 7. SEPARABILITY OF THE CONVEX SETS Problems on conditional extreme and method of the indefinite Lagrange multipliers as application to the theory of the implicit functions in course of the mathematical analysis were considered. In the last years this section of mathematics was essentially development. And the new theory called "Convex analysis" is appeared. The main idea in the theory is the separability theory of the convex sets. Definition 1. It is said, that hyperplane ¢c, u² J with normal vector c, c z 0 divides (separates) the sets ȼ and Ⱦ from E n if inequalities sup¢c, b² d J d inf ¢c, a²

(1)

aA

bB

are held. ¢c, a² , then sets ȼ and Ⱦ are strongly separated, and if If sup¢c, b²  inf aA bB

¢c, b²  ¢c, a², a  A, b  B , then they are strictly separated. We notice, that if hyperplane ¢c, u² J separates the sets ȼ and Ⱦ, then hyperplane ¢D c, u² D J , where D z 0 is any number also separates them, so under necessity possible to suppose the norm | c | 1 . Theorem 1. If U is a convex set of E n , and the point v  intU , then there exists hyperplane ¢c, u² J dividing set U and the point v. If U is a convex set, and the point v U , then set U and the point v are strongly separable. Proof. We consider the case, when the point v U . In this case by Theorem 5 the point Q  E n has an unique projection w PU (Q ) , moreover w  v, u  w t 0, u U . ¢ c , u  v²

Let ¢ w  v, u  v²

the

c wv . Then we have 2 ¢ w  v, w  v²  ¢ w  v , u  w² t c ! 0 . Thence

vector

¢ w  v , u  w  w  v²

it follows, that ¢c, u² ! ¢c, v², u U , i.e. set U and the point v  E n are strongly separable. Consequently, the set U and the point v are strongly separable. We notice, if v  intU , then v  *pU . Then by definition of the border point there exists the sequence {vk } Œ U , such that vk o v under k o f . Since the point vk U , then by proved the inequality ¢ c k , u ² ! ¢ c k , v k ² , u  U , c k 1 is held. By Bolzano–Weierstrass theorem from bounded sequence ^ck ` possible to select subsequence ^ckm ` , moreover ckm o c , under m o f and c 1 . For these elements of this subsequence the previous inequality is written as: ¢ c k , u ². ! ¢ ck , vk ² , u  U . By transferring to the limit under m o f with consideration of vkm o v, we get ¢ c , u ² t ¢ c , v ² ,  u  U . Theorem is proved. From proof of Theorem 1 follows that through any border point v of the convex set U possible to conduct hyperplane for which ¢c, u² t ¢c, v², u U . m

35

m

m

Hyperplane ¢c, u² J , where J ¢ c , v ² , v  U is called by supportive to set U in the point v , and vector c  E n is a supportive vector of the set U in the point v U . Theorem 2. If convex sets A and B of E n have a no points in common, then there exists hyperplane ¢c, u² J separating set ȼ and Ⱦ, as well as their closures A and B , in the case of point y  A ˆ B , the number J ¢c, y² . Proof. We denote U A  B . As it is proved earlier (see Theorem 1, Lecture 3), the set U is convex. Since the intersection A ˆ B ‡ , that 0 U . Then by Theorem 1 there exists hyperplane ¢c, u² J , J ¢c, 0² 0, separating set U and the point 0, i.e. ¢c, u² t 0, u  U . Thence with consideration of u a  b  U , a  A, b  B , we get ¢ c , a ² t ¢ c , b ² , a  A, b B . Finally, the hyperplane ¢c, u² J , where the number J ¢c, a² t J t sup¢c, b² separates the sets ȼ and B. Let the points satisfies to inequality inf aA bB

a  A , b  B . Then there exists the subsequences ^ak `  A, ^bk `  B such that ak o a, bk o b at k o f , moreover with consideration of proved above the inequalities ¢c, ak ² t J t ¢c, bk ². Thence, by transferring to the limit under k o f , we get ¢c, a² t J t ¢c, b² a  A, b  B . In particular, if the point y  A ˆ B , then J

¢c, y² .

Theorem is proved. Theorem 3. If the convex closed sets A and B have a no points in common and one of them is limited, then there exists hyperplane separating sets A and B strictly. Proof. Let the set U A  B . We notice, that the point 0 U , since the intersection A ˆ B Ø, U is a convex set. We show, that set U is closed. Let ɤ be a limiting point of the set U . Then there exists sequence ^uk `  U , such that u k o u under k o f , moreover uk ak  bk , ak  A, bk  B . If the set ȼ is bounded, then sequence ^ak `  A is bounded, consequently, there exists subsequence ^akm ` which converges to a certain point ɜ, moreover with consideration of completeness of the set ȼ the point a A . We consider the sequence ^bk `  B , where bk ak  uk . Since a km o a, u km o u at m o f that bkm o a  u b at m o f . With consideration of completeness of the set B the point b B . Finally, the point u a  b, a  A, b  B , consequently, u U . Completeness of the set U is proved. Since the point 0 U , U U , that with consideration of Theorem 1 there exists hyperplane ¢c, u² J strongly separated the set U and the point 0, i.e ¢c, u² ! 0 u U . This implies, that ¢ c, a ² ! ¢ c, b² , a  A, b  B . Theorem is proved. Theorem 4. If the intersection of the nonempty convex sets A0 , A1 ,, Am is empty set, i.e. A0 ˆ A1 ˆˆ Am Ø, then necessary there exists the vectors c0 , c1 ,, cm of E n , not all equal zero and the numbers J 0 , J 1 ,, J m such that ¢ ci , ai ² t J i , ai  Ai , i

c0  c1    cm 36

0, m;

0;

(2) (3)

J 0  J 1    J m 0;

(4)

Proof. We introduce the set A A0 u A1 uu Am which is a direct production of the sets Ai , i 0, m with elements a (a0 , a1 ,, am )  A , where a i  Ai , i 0, m . We notice, that set A  E , where E E n ( m 1) . It is easy to show, that set ȼ is convex. In fact, if a1 (a01 , a11 ,, a1m )  A, a 2 (a02 , a12 ,, am2 )  A ,

then under all D , D  >0,1@ , the point aD D a1  (1  D )a 2 (a0D , a1D ,, amD )  A , since a iD D a i1  (1  D ) a i2  Ai , i 0, m . We introduce the diagonal set B

^b

(b0 , b1 ,  , bm )  E b0

b1



bm

b, bi  E n , i

`

0, m .

The set B is convex. It is not difficult to make sure, the intersection A0 ˆ A1 ˆˆ Am ‡ , if and only if, when the intersection A ˆ B ‡ . Finally, under performing the condition of the theorem the sets ȼ and B are convex and A ˆB ‡ , i.e all conditions of Theorem 2 are held. Consequently, there exists the vector c (c0 , c1 ,, cm )  E, c z 0 such that ¢ c, a ² t ¢ c, b² , a  A, b  B . The inequality can be written in the manner of m

¦ i 0

m

m

ci , ai t ¦ ci , bi

¦c , b i

i 0

, ai  A, bi  E n .

(5)

i 0

As follows from inequality (5) linear function J (b)

m

¦ c ,b i

, b En

is

i 0

bounded, since in (5) ai  Ai , i 0, m . The linear function J (b) is bounded if and only if, when

m

¦c

i

0 . Thence it follows fairness of the formula (3). Now the inequality

i 0

(5) is written as

m

¦ c ,a i

i 0

then

m

ck , ak t ¦ ci , ai i 0

i

t 0, a i  Ai , i const,

0, m . The vectors ai

ai  Ai , i z k are fixed,

ak  Ak . Finally, the value

J k ! f , a k  Ak , k 1, m . We denote J 0 c0 , a0 t  inf

ck , ak t inf ck , ak ak Ak

(J 1  J 2    J m ) . Then m

ak Ak

¦ k 1

ck , ak

m

 ¦J k J 0 . k 1

Thereby, the values ci , ai t J i ai  Ai , i 0, m , J 0  J 1    J m Faithfulness of the expressions (2), (4) is proved. Theorem is proved. 37

0.

Convex cones. Theorems 1 - 5 are formulated and proved for convex sets of E n . Separability theorems are often applied in the extreme problems theory, when

convex sets are convex cones. Definition 2. Set K of E n is called by cone with vertex in zero, if it together with any its point u K contains the points Ou K at all O ! 0 . If the set K is convex, then it is called the convex cone, if K is closed, then it is called by closed cone, if K is opened, then it is called by open cone. Example 1. The set K ^u  E n / u t 0` is a convex closed cone. In fact if u K , then the point O u  K at all O ! 0, Ou ! 0 . Example 2. The set K ^u  E n / ¢a, u² 0` is a convex closed cone. Since for any O ! 0 , scalar product ¢ a, Ou² O ¢ a, u² 0, u K , consequently, vector Ou  K . Example 3. The set K ^u  E n / ¢a, u²  0` is an open convex cone, K ^u  E n / ¢ a, u² d 0` is closed convex cone, K ^u  E n ` E n is a convex cone. We introduce the set K*

^c  E

n

`

/ ¢c, u² t 0, u  K .

(6)

We notice, that set K  K * , since from c u  K follows that ¢ c, u ² c 2 t 0 . The set K * z ‡ , since it contains the element c 0 . The condition ¢c, u² t 0 , u  K means that vector c  K * forms an acute angle (including S / 2 ) with vectors u  K . To give a geometric interpretation of the set K * . Finally, the set K * of E n is the cone, if c  K * , i.e. ¢c, u² t 0 , u  K , then for vector O c, O ! 0 we get ¢O c, u² O ¢c, u² t 0 . Consequently, vector O c  K * . Definition 3. The set K * determined by formula (6) is called dual (or conjugate) cone to cone K . It is easy to make sure, that for cone K from example 1 the dual cone * K ^c  E n / c t 0`, for cone K from example 2 the conjugate cone * n K ^c  E c ED , E is a number `, for example 3 the dual cones K*

^c  E

n

`

/ c ED , E t 0 , K *

^c  E

n

`

/ c ED , E t 0 , K *

^0`

correspondingly. Theorem 5. If the intersection of the inempty convex with vertex in zero cones K 0 , K1 ,, K m is empty set, then necessary there exists the vectors ci  K i* , i 0, m not all are equal to zero and such that c0  c1    cm 0 . Proof. Since all conditions of Theorem 4 are held, that the expressions (2) - (4) are faithful. We notice, since K i , i 0, m are cones, that from ai  K i , i 0, m follows the vectors Oai  K i , i 0, m , at all O ! 0 . Then inequality (2) is written: ci , Oai t J i , ai  K i , i 0, m , for any O ! 0 . Thence we get ci , ai t J i / O , i 0, m . In particular, at ci , ai O o f , we get ci , ai t 0 , i 0, m, a i  Ai . On the other hand, J i inf a A i

38

0,

i 0, m .

From condition ci , ai t 0, a i  Ai , i 0, m, follows that vectors ci  K , i 0, m . Theorem is proved. Theorem 6. (Dubovicky–Milyutin theorem). In order to the intersection of the nonempty convex with vertex in zero cones K 0 , K1 ,, K m be empty set, when all these cones, except, one of them is open, necessary and sufficiently existence of the vectors ci  K i* , i 0, m , not all are equal to zero and such that c0  c1    cm 0 . Proof. Sufficiency. We suppose inverse, i.e. that vectors ci  K i* , i 0, m , c0  c1    cm 0 , but K 0 ˆ K1 ˆ  ˆ K m z ‡ . Then there exists the point w  K 0 ˆ K1 ˆ  ˆ K m . Since vectors ci  K i* , i 0, m , that inequalities c i , w t 0, * i

i 0, m are held. Since the sum c0  c1    cm

0 and c i , w t 0, i 0, m , that from

equality ¢c0  c1    cm , w² 0 follows ci , w 0, i 0, m . By condition of the Theorem not all ci , i 0, m are equal to zero, consequently, at least, there exists two vectors ci , c j , i z j differenced from zero. Since all cones are open, except one of them, that in particular, it is possible to suppose that cone K i is an open set. Then K i contains the set o( w, 2H ) ^u  K i / u  w  2H , H ! 0`. In particular, the point u w  H ci / ci  K i . Since ¢ci , u² t 0, u  K i , ci  K i* , that we get

¢c , w  H c / c ² i

where

i

i

¢ ci , w²  H ci

H ! 0, c i z 0 .

 H c t 0, i

It is impossible. Consequently, the intersection K 0 ˆ K1 ˆ  ˆ K m z ‡ . Necessity follows from Theorem 5 directly. Theorem is proved. CONTROL QUESTIONS 1. Give definition to the hyperplane which divives the sets. 2. Prove the Theorem 1. 3. Prove the Theorem 2. 4. Prove the Theorem 3. 5. Prove the Theorem 4. 6. Give definition to the convex cone. 7. Prove the Theorem 5. 8. Prove the Theorem 6.

39

LECTURE 8. LAGRANGE’S FUNCTION. SADDLE POINT Searching for the least value (the global minimum) to function J(u) determined on set U of E n is reduced to determination of the saddle point to Lagrange’s function. By such approach to solution of the extreme problems the necessity to proving of the saddle point existence of Lagrange’s function appears. We consider the next problem of nonlinear programming: J (u) o inf , u U

^u  E

n

(1)

u  U 0 , g i (u ) d 0, i

1, m;

g i (u )

m  1, s ,

`

0, i

(2)

where U 0 is the prescribed convex set of E n ; J (u ), g i (u ), i 1, m are the functions determined on set U 0 . In particular, J (u ), g i (u ), i 1, m are convex functions determined on convex set U 0 , and functions g i (u) ¢ ai , u²  bi , ai  E n , i m  1, s, bi , i m  1, s, are the given numbers. In this case problem (1), (2) is related to the convex programming problems. In the beginning we consider a particular type of the Lagrange’s function for problem (1), (2). Lagrange’s Function. Saddle point. Function s

J (u )  ¦ Oi g i (u ), u  U 0 ,

L (u, O )

^O  E

O  /0

i 1

s

`

(3)

/ O1 t 0,, Om t 0

is called by Lagrange function to problem (1), (2). Definition 1. Pair u* , O* U 0 u / 0 , i.e. u*  U 0 , O*  /0 is called by saddle point to Lagrange’s function (3), if the inequalities









L u* , O d L u* , O* d L u, O* , u U 0 , O  / 0

(4)

are held. We notice, that in the point u* U 0 minimum to function L u, O* is reached on set U 0 , but in the point O*  / 0 the maximum to function L u * , O is reached on set /0 . By range of definition of the Lagrange function is the set U 0 u /0 .

40

The main lemma. In order to the pair u* , O* U 0 u / 0 be a saddle point to Lagrange’s function (3) necessary and sufficiently performing of the following conditions:









L u* , O* d L u, O* , u  U 0 ; O g i (u * ) * i

0, i

1, s, u *  U .

(5) (6)

Proof. Necessity. Let the pair u* , O* U 0 u / 0 be a saddle point. We show, that the condition (5). (6) are held. Since pair u* , O* U 0 u / 0 is saddle point to Lagrange’s function (3), that inequalities (4) are held. Then from the right inequality follows the condition (5). It is enough to prove fairness of the equality (6). The left inequality of (4) is written so: s

s

i 1

i 1

J (u* )  ¦ Oi g i (u* ) d J (u* )  ¦ O*i g i (u* ) ,

O

(O1 ,, O s )  / 0  E s .

Consequently, the inequality holds s

¦ (O

* i

 Oi ) g i (u* ) t 0, O  / 0 .

(7)

i 1

At first, we show that u*  U , i.e. g i (u * ) d 0, i 1, m, g i (u* ) 0, i m  1, s . It is easy to make sure in that vector O

(O1 ,, O s )

­°Oi ® °¯O j

O*i ,

i 1, s, i z j ,

O  1, for j , 1 d j d m, * j

(8)

belongs to the set / ^O  E s / O1 t 0,, Om t 0`. Substituting value O  / 0 from (8) to the inequality (7) we get (1) gi (u* ) t 0 . Thence follows that g i (u* ) d 0, i 1, m (since it is faithful for any j , 1 d j d m ). Similarly the vector O

(O1 ,, O s )

­°Oi ® °¯O j

O*i ,

i 1, s, i z j ,

O  g j (u* ), for j, m  1 d j d s, * j

also belongs to the set / 0 . Then from inequality (7) we have  g i (u * ) 2 t 0, j m  1, s . Consequently, the values g j (u* ) 0, j m  1, s . From relationship g j (u* ) d 0, j 1, m , g j (u* ) 0, j m  1, s follows that the point u* U . 41

We choose the vector O  / 0 as follows: O

­°Oi ® °¯O j

(O1 , , Os )

O*i , i 1, s, i z j , 0, for j , 1 d j d m.

In this case inequality (7) is written so: O*j g j (u* ) 0, j 1, m . Since the value O*j t 0, j 1, m under proved above value g j (u* ) d 0, j 1, m , consequently, the equality O*j g j (u* ) 0, j 1, m g j (u* )

0, j

holds. The equalities O*j g j (u* ) 0, j m  1, s follow from equality

m  1, s . Necessity is proved.

Sufficiency. We suppose, that the conditions (5), (6) for a certain pair u* , O U 0 u / 0 are held. We show, that u* , O* is a saddle point to Lagrange’s function (3). It is easy to make sure in that product (O*i  Oi ) gi (u* ) t 0 for any i , 1 d i d m . In fact, from condition u*  U follows that g i (u* ) d 0, for any i , 1 d i d m . If gi (u* ) 0 , that (O*i  Oi ) g i (u* ) 0 . In the event of gi (u* )  0 the value O*i 0 , since the product O*i g i (u* ) 0 , consequently, (O*i  Oi ) g i (u* ) Oi g i (u* ) t 0, Oi t 0 . Then sum *

m

¦ (O*i  Oi ) g i (u* )  i 1

s

s

¦ (O*i  Oi ) g i (u* )

¦ (O

i m 1

* i

 Oi ) g i (u* ) t 0,

i 1

since g i (u * ) 0, i m  1, s, u*  U . Thence follows that s

s

¦ O g (u ) t ¦ O g (u ), * i

i

i

*

i 1

i

*

i 1

s

s

i 1

i 1

J (u* )  ¦ Oi g i (u* ) t J (u* )  ¦ Oi g i (u* ) .

The second inequality can be written in the manner of L u* , O d L u* , O* . The inequality in the ensemble with (5) defines the condition









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

This means that pair u* , O* U 0 u / 0 is the saddle point to Lagrange’s function (3). Lemma is proved. The main theorem. If pair u* , O* U 0 u / 0 is a saddle point to Lagrange’s function (3), then the vector u*  U is a solution of the problem (1), (2) i.e. u*  U *

^u  E *

n

u*  U , J (u* )

42

`

min J (u ) . uU

Proof. As follows from the main lemma for pair u* , O* U 0 u / 0 the conditions (5), (6) are held. Then value J (u* , O* ) J (u* ) 

s

¦ O g (u ) * i

i

J (u* ) ,

*

i 1

since the point u*  U . Now inequality (5) is written so: s

J (u* ) d J (u )  ¦ O*i g i (u), u U 0 .

(9)

i 1

Since the set U  U 0 , that inequality (9), in particular, for any u U  U 0 faithfully, i.e. s

J (u* ) d J (u)  ¦ O*i g i (u),  u  U , O*  / 0 .

(10)

i 1

As follows from condition (2), if uU , that g i (u ) d 0, i 1, m , and * g i (u ) 0, i m  1, s , consequently, the product Oi g i (u ) d 0 for any i, 1 d i d s . We notice, that O* (O1* ,, O*s )  / 0 , where O1* t 0,, O*m t 0 . Then from (10) follows J (u* ) d J (u ), u  U . It means that in the point u *  U the global (or absolute) minimum of the function J (u ) is reached on set U. Theorem is proved. We note the following: 1) The main lemma and the main theorem for the nonlinear programming problem (1), (2) are proved, in particular, its are true and for the convex programming problem. 2) Connection between solutions of the problem (1), (2) and saddle point to Lagrange’s function in type (3) is established. In general case, for problem (1), (2) Lagrange’s function is defined by the formula s

L(u,Ȟ ) O0 J (u )  ¦ Oi g i (u* ), u  U 0 ,

O  /0

^O

i 1

(O1 ,, Os )  E

s 1

O0 t 0,, Om t 0`.

(11)

If the value O0 t 0 , then Lagrange’s function (11) possible to present in the manner of L (u, O ) O0 L u , O , where Oi Oi / O0 , function L u , O is defined by formula (3). In this case the main lemma and the main theorem are true for Lagrange’s function in type (11). 3)We consider the following optimization problems: J (u ) o inf, u  U , J 1 (u ) o inf, u  U

, where function

s

J1 (u) d J (u)  ¦ O*i g i (u). i 1

43

As

follows

from

proof

of

the main theorem the conditions J1 (u ) d J (u ), are held. u  U ; J 1 (u * ) J (u * ), u *  U 4) For existence of the saddle point to Lagrange’s function necessary that set U * z ‡ . However this condition does not guarantee existence of the saddle point to Lagrange’s function for problem J (u) o inf, u  U . It is required to impose the additional requirements on function J (u ) and set U that reduces efficiency of the Lagrange’s multipliers method. We notice, that Lagrange’s multipliers method is an artificial technique solution of the optimization problems requiring spare additional condition to problem. Kuhn-Tucker’s theorem. The question arises: what additional requirements are imposed on function J (u) and on set U that Lagrange’s function (3) would have a saddle point? Theorems in which becomes firmly established under performing of which conditions Lagrange’s function has the saddle point are called Kuhn-Tucker theorems (Kuhn and Tucker are American mathematicians). It can turn out to be that source problem J (u) o inf, u U , U * z Ø , has a solution, i.e. the point u*  U * exists, however Lagrange’s function for the given problem has not saddle point. Example. Let function be J (u ) u  1 , and set U ^u  E1 / 0 d u d 1; (u  1) 2 d 0`. The function g (u) (u  1) 2 , set U 0 ^u  E1 / 0 d u d 1`, moreover functions J (u ) and g (u ) are convex on set U 0 . Since set U consists of unique element, i.e. U {1} , that set Consequently, the point Lagrange’s function U * {1} . u* 1 . 2 L u, O (u  1)  O (u  1) , O t 0, u  U 0 for the problem has not saddle point. In fact, from formula (4) follows



L u* , O 0 d L u* , O*





0 d L u , O*



(u  1)  O* (u  1) 2 ,

O t 0, 0 d u d 1.

Thence we have 0 d H  O*H 2 , where u 1 H , 0 d H d 1. Under sufficiently small H ! 0 does not exist the number O* t 0 such the given inequality is held. 1-st case. We consider the next problem of the convex programming: J (u) o inf; u U

^u  E

n

u  U 0 , g i (u ) d 0, i

(12)

`

1, m ,

(13)

where J (u ), g i (u ), i 1, m are convex functions determined on convex set U 0 . Definition 2. If there exists the point u i U 0 such that value g i (u i )  0 , then it is said, that restriction g i (u) d 0 on set is regularly. It is spoken that set U is regularly, if all restrictions g i (u )  0, i 1, m from the condition (13) are regular on set U 0 . We suppose, that point u U 0 exists such that 44

g i (u )  0, i

1, m .

(14)

The condition (14) is called by Sleyter condition. Let in the points u 1 , u 2 ,, u k U 0 all restrictions are regularly, i.e. g j (u i )  0, j 1, m, i 1, k . Then in the point k

¦D u

u

i

i 1

i

U 0 , D i t 0, i 1, k , D 1  D k

1

all restrictions g j (u ) d 0, j 1, m, are regular. In fact, g j (u )

§ k · k g j ¨ ¦ D i u i ¸ d ¦ D i g j (u i )  0, j 1, m. ©i1 ¹ i1

Theorem 1. If J (u ), g i (u ), i 1, m are convex functions determined on convex set U 0 , the set U is regularly and U * z Ø , then for each point u*  U * necessary there exist the Lagrange multipliers O* (O1* ,, O*m )  / 0

^O  E

m

`

/ O1 t 0,, Om t 0

such that pair u* , O* U 0 u / 0 forms a saddle point to Lagrange’s function Lu, O

m

J (u)  ¦ Oi g i (u), u  U 0 , O  / 0 . i 1

As follows from condition of the theorem, for the convex programming problem (12), (13) Lagrange’s function has a saddle point, if set U is regularly (under performing the condition that set U * z Ø ). We note, that in the example represented above the set U is not regular, since in the unique point u 1U 0 restriction g (1) 0 . We notice, that if set is regularly, then as follows from Sleyter’s condition (14), the point u  U  U 0 . Proof of Theorem 1 is provided in the following lectures. CONTROL QUESTIONS 1. Give definition to the saddle point. 2. Prove the main lemma. 3. Prove the main theorem. 4. Give definition to the regularly set. 5. Prove the Theorem 1.

45

LECTURES 9, 10. KUHN-TUCKER THEOREM For problem of convex programming the additional conditions imposed on convex set U under performing of which Lagrange’s function has a saddle point are obtained. We remind that for solving of the optimization problems in finitedimensional space by method of the Lagrangian multipliers necessary, except performing the condition U * z ‡ , existence of the saddle point to Lagrange’s function. In this case, applying of the main theorem is correct. Proof of the theorem 1. Proof is conducted on base of the separability theorem of the convex sets ȼ and Ⱦ in space E m1 . We define the sets ȼ and B as follows: A

^a

(a0 , a1 ,, am )  E m1 a0 t J (u ),

a1 t g1 (u), am t gm (u), u  U 0 `; B

^b

(15)

`

(b0 , b1 ,, bm )  E m1 b0  J * , b1  0, bm  0 , (16)

where J*

inf J (u ) uU

min J (u ) uU

since set U * z ‡ . a) We show, that sets ȼ and B have a no points in common. In fact, from the inclusion a A follows that a0 t J (u), a1 t g1 (u), am t g m (u), u  U 0 . If the point u U  U 0 , that inequality a0 t J (u ) t J * is held, consequently, the point a B . If the point u U 0 \ U , then for certain number i, 1 d i d m the inequality g i (u) ! 0 is held. Then ai t g i (u) ! 0 and once again the point a B . Finally, the intersection A ˆ B ‡ . b) We show, that sets ȼ and Ⱦ are convex. We take two arbitrary points a  A, d  A and the number D , D  >0,1@. From inclusion a  A follows that there exists the point u U 0 such that a0 t J (u), ai t g i (u), i 1, m, a (a0 , a1 ,, am ) . Similarly from d  A we have d 0 t J (v ), d i t g i (v ), i 1, m, where v  U 0 , d (d 0 , d1 ,, d m ) . Then the point aD

Da  (1  D )d

D a0  (1  D )d 0 ,D a1  (1  D )d1 ,,D am  (1  D )d m ,D  >0,1@,

46

moreover D a0  (1  D )d 0 a0D t DJ (u)  (1  D ) J (v) t J (uD ), i 1, m with consideration of convexity J (u ) on U 0 , where uD D u  (1  D )v . Similarly D ai  (1  D )di

aiD t D g i (u )  (1  D ) g i (v ) t gi (uD ), i

1, m ,

with consideration of function convexity g i (u), i 1, m on set U 0 . Finally, the point aD (a0D , a1D ,, amD ), a0D t t J (uD ), a iD t g i (uD ), i 1, m, moreover uD U 0 . This implies that point aD  A under all D , 0  D  1. Consequently, set ȼ is convex. By the similar way it is possible to prove the convexity of the set B. c) Since sets ȼ and Ⱦ are convex and A ˆ B Ø , that, by Theorem 2 (Lecture 6), there exists hyperplane c, w J , where normal vector c (O*0 , O1* , O*m )  E m1 , c z 0, w  E m 1 , separating sets ȼ and Ⱦ, as well as its closures A A, B ^b (b0 , b1 ,, bm )  E m1 b0 d J * , b1 d 0, bm d 0`. Consequently, inequality are held m

c, b

¦O b

* i i

d J d c, a

m

¦ O a , a  A, b  B . * i i

(17)

i 0

i 0

We notice, that if the point u*  U * , then J (u* ) J * , g i (u* ) d 0, i 1, m , consequently, the vector y ( J * ,0,,0)  A ˆ B and the value J c, y O*0 J * . Now inequalities (17) are written as m

O*0 b0  ¦ O*i bi d J i 0

m

O*0 J * d O*0 a0  ¦ O*i ai , a  A, b  B .

(18)

i 0

Thence, in particular, when vector b ( J *  1, 0,,0)  B , from the left inequality we have O*0 ( J *  1) d O*0 J * . Consequently, value O*0 t 0 . Similarly, choosing vector b ( J * ,0, ,0,1, 0 ,0) from the left inequality we get O*0 J *  O*i d O*0 J * . This implies that O*i t 0, i 1, m . d) We take any point u*  U * . It is easy to make sure in that, the point b ( J * ,0,,0, g i (u* ), 0,0)  A ˆ B , since value g i (u* ) d 0 . Substituting the point to the left and right inequalities (18), we get O*0 J *  O*i g i (u* ) d O*0 J * d O*0 J *   O*i gi (u* ) .

we have O*i g i (u* ) d 0 d O*i g i (u* ) . Consequently, the values O*i g i (u * ) 0, i 1, m . f) We show, that value O*0 ! 0 . In item c) it is shown that O*0 t 0 . By condition of the theorem set U is regular, i.e. Sleyter’s condition is executed (14). Consequently, Thence

47

the point u  U  U 0 such that g i (u )  0, i 1, m exists. We notice, that point a ( J (u ), g1 (u ),, g m (u ))  A . Then from the right inequality (18) we have O*0 J *  O*0 J * (u )  O1* g1 (u )   O*m g m (u) . We suppose opposite i.e. O*0 0 . Since vector c (O*0 , O1* ,, O*m ) z 0 , that at least one of the numbers O*i , 1 d i d m are different from zero. Then from the previous inequality we have 0 d O*i g i (u )  0 . It is impossible, since O*i z 0, O*i t 0 . Finally, the number O*0 is not equal to zero, consequently, O*0 ! 0 . Not derogating generalities, it is possible to put O*0 1 . g) Let u  U 0 be an arbitrary point. Then vector a ( J (u), g1 (u),, g m (u))  A ,

from the right inequality (18) we get m

J * d J (u)  ¦ O*i g i (u)





L u, O* ,  u  U 0 .

i 1

Since the product O*i g i (u* ) 0, i 1, m , that



m

J *  ¦ O*i g i (u)



L u* , O*



i 1



d L u, O* ,  u  U 0 . From the inequality and from equality O*i g i (u* ) 0, i 1, m, u*  U ,

where O*i t 0, i 1, m follows that pair u* , O* U 0 u / 0 is the saddle point to Lagrange’s function. Theorem is proved. The 2-nd case. Now we consider the next problem of the convex programming: J (u) o inf , u U

^u  E

n

u j ! 0, j  I , g i (u ) g i (u )

(19)

ai , u  bi d 0, i 1, m;

ai , u  bi

0, i

`

m  1, s ,

(20)

where I  ^1,2,, n` is a subset; the set U 0 ^u  E n / u j t 0, j  I `, J (u) is a convex function determined on convex set U 0 , a i  E n , i 1, s are prescribed vectors; bi , i 1, s are the numbers. For problem (19), (20) Lagrange’s function L u, O

m

J (u )  ¦ Oi g i (u ), u  U 0 ,

(21)

i 1

O

(O1 ,, O s )  / 0

^O  E

s

O1 t 0,, Om t 0`.

We notice that, if function J (u) ccu is linear, the task (19), (20) is called the general problem of linear programming. 48

We suppose, that set U * z ‡ . It is appeared, that for task of convex programming (19), (20) Lagrange’s function (21) always has saddle point without any additional requirements on convex set U . For proving of this fact it is necessary the following lemma. Lemma. If a1 ,, a p is a finite number of the vectors from E n and the numbers D i t 0, i 1, p , then set ­ n ®a  E a ¯

Q

p

¦D a , D i

i

t 0, i

i

i 1

½ 1, p ¾ ¿

(22)

is convex closed cone. Proof. We show, that set Q is a cone. In fact, if a  Q and O ! 0 is an arbitrary number, that O a

p

¦ O D i ai i 1

p

¦E a

i i

, where E i

OiD i t 0, i 1, p . This implies, that the

i 1

vector O a  Q . Consequently, set Q is the cone. We show, that Q is a convex cone. In fact, from a  Q, b  Q follows that D a  (1  D )b

¦ >DD a p

i

i

 (1  D ) E i ai

@

i 1

p

¦ (DD

i

p

¦J a ,J

 (1  D ) E i )ai

i

i 1

i 1, m , where D  >0,1@,

p

b

¦E a , i

i

i

i

DD i  (1  D ) E i t 0,

i 1

E i t 0, i 1, p .

i 1

Thence we have D a  (1  D )b  Q under all D , 0 d D d 1 . Consequently, set Q is the convex cone. We show, that Q is convex closed cone. We prove this by the method of mathematical induction. For values p 1 we have Q ^a  E n a Da1 , D t 0 ` is a halfline. Therefore, Q is closed set. We suppose, that set Q p 1

­ n ®a  E a ¯

p

¦D a , D i

i 1

i

i

½ t 0¾ ¿

is closed. We prove, that set Q Q p1  Ea p , E t 0 is closed. Let c  E n is limiting point of the set Q. Consequently, the sequence ^cm `  Q exists, moreover cm o c, m o f . The sequence ^cm ` is represented in the manner of cm bm  E m a p , where ^bm `  Q p1 , ^E m ` is a numeric sequences. As far as set Q p1 is closed, that bm o b, m o f , under b  Q p1 . It is remains to prove that numeric sequence ^E m ` is bounded. We suppose opposite i.e. E mk o f under k o f . Since bmk / E mk cmk / E mk  a p 49

and ^bmk / E mk ` Q p1 the sequence {cmk } is bounded, so far as cmk o c, k o f , then under k o f we have  a p  Q p1 . Since set Q p1 Ž Q (in the event of E 0, Q p 1 Q ), that vector  a p  Q . It is impossible. Therefore, the sequence ^E m ` is bounded. Consequently, vector c b  Ea p  Q , where E mk o E , E t 0 under k o f . Theorem is proved. We formulate and prove Farkas’s theorem having important significant in theories of the convex analysis previously than prove the theorem about existence of saddle point to Lagrange’s function for task (19), (20). Farkas theorem. If cone K with vertex in zero is determined by inequalities K

^e  E

n

ci , e d 0, i 1, m; ci , e  0, i ci , e

then the dual to it cone K *

^c  E

`

c, e t 0, e  K

n

(23)

p  1, s, ci  E n , i 1, s ,

0, i

­ n ®c  E ¯

K*

m  1, p;

`

has the type

s ½  ¦ O i c i , O1 t 0, , O p t 0 ¾ . i 1 ¿

c

(24)

Proof. Let cone K be defined by formula (23). We show, that set of the vectors c  K * for which c, e t 0, e  K is defined by formula (24), i.e. set Q

­ n ®c  E c ¯

s ½ ¦ Oi ci , O1 t 0, , O p t 0¾ i 1 ¿

with K * ^c  E n c, e t 0, e  K `. We E i t 0, i p  1, s . Then set Q from (25) is written as

coincides

Q

­ n ®ɭ  E c ¯

p

represent

s

Oi

(25) D i  E i , D i t 0,

s

¦ O (  c )  ¦ D (  c )  ¦ E ( c ), i

i

i 1

i p 1

Oi t 0, i 1, p, D i t 0, i

i

i

i p 1

i

p  1, s, E i t 0, i

i

½ p  1, s ¾ . ¿

(26)

As follows from expressions (26) and (22) the set Q is convex closed cone generated by the vectors  c1 , ,c p ,  c p 1 ,....,cs , c p 1 , , cs (see Lemma). s

We show, that Q Ž K * . In fact, if c  Q , i.e. c ¦ Oi ci , O1 t 0,, O p t 0 , then i 1

product ¢c, e²

s

¦ Oi ¢ci , e² t 0 with consideration of correlations (23). Consequently, i 1

vector c  Q belongs to set 50

K*

^c  E

n

c, e t 0, e  K ` .

This implies, that Q Ž K * . We show, that K * Ž Q . We suppose opposite, i.e. the vector a  K * , however a  Q . Since set Q is convex closed cone and the point a  Q , that by Theorem 1 (Lecture 6) the point a  E n is strongly separable from set Q i.e. d , c ! d , a , c  Q where d , d z 0 is a normal vector to the hyperplane d , u J d , a . Thence we get s

¦ Oi d , ci ! d , a , Oi , i 1, s, O1 t 0,, O p t 0 .

d,c

(27)

i 1

We choose the vector O ( O1 t 0,, O p t 0, O p1 ,, Os ) : O

­°Oi ® °¯O j

( O1 , , O s )

i z j , i 1, s

0,

t , t ! 0, j , 1 d j d p.

For the vector O inequality (27) is written in the form  t d , ci ! d , a . . We divide into t ! 0 and t o f , as a result we get c j , d d 0, j 1, p. Hereinafter, we take the vector O

( O1 ,, O s )

­°Oi ® °¯O j

From inequality (27) we get  t c j , d we get c j , d

0, j

p  1, s

2

0,

i z j, i 1, s,

t c j , d , t ! 0,

j , p  1 d j d s.

! d,a

. We divide into t ! 0 and tending t o f ,

. Finally, for the vector d  E n inequalities

c j , d d 0, j 1, p, c j , d

0, j

p  1, s

are held. Then from (23) follows, that d  K , where K is a closure of the set K . Since the vector a K * , that the inequality a, e t 0, e  K holds. Thence, in particular, for e d  K follows a, d t 0 . However from (27) under Oi 0, i 1, s , we get a, d  0 . We have got the contradiction. Consequently, vector a K * belongs to the set Q. From inclusions Q Ž K * , K * Ž Q follows that K * Q . Theorem is proved. Now we formulate theorem about existence of the saddle point to Lagrange’s function (21) for problem of convex programming (19), (20).

51

Theorem 2. If J (u ) is a convex function on convex set U 0 , J (u)  C 1 (U 0 ) and set U * z ‡ for problems (19), (20), then for each point u*  U * necessary there exists the Lagrange multipliers O*

^O  E

(O1* ,, O*s )  / 0

s

`

/ O1 t 0,, Om t 0

such that pair u* , O* U 0 u / 0 forms a saddle point to Lagrange’s function (21) on set U0 u /0 . Proof. Let the condition of the theorem is held. We show, the pair u* , O* U 0 u / 0 is saddle point to Lagrange’s function (21). Let u*  U * be an arbitrary point. We define some feasible directions coming from point u* for convex set U . We notice, that the vector e (e1 ,, en )  E n is called by the feasible direction in the point u* , if there exists a number H 0 ! 0 such that u u *  H e  U under all H , 0 d H d H 0 . From inclusion u u *  H e  U with consideration of the expressions (20) we get u *j  H e j t 0, j  I ; ai , u*  He  bi

ai , u*  He  bi d 0, i 1, m ;

0, i

(28)

m  1, p, H , 0 d H d H 0 .

If the sets of the indexes

^j

I1

u *j

`

0, j  I ,

I2

^j

ai , u*

bi , 1 d i d m`

are introduced, then conditions (28) are written: e j t 0, j  I 1 ; a i , e d 0, i  I 2 ;

ai , e

0, i

m  1, s.

(29)

Finally, the set of the feasible directions in the point u* according to correlation (29) is cone K

^e  E

n

 ej

 e j , e d 0, j  I 1 ;

ai , e

0, i

a i , e d 0, i  I 2 ,

`

m  1, s ,

(30)

where e j 0, ,0,1,0, ,0  E n is a unique vector. And inverse statement faithfully, i.e. if e K , then ɡ is feasible direction. Since J (u ) is a convex function on convex set U  U 0 and J (u )  C 1 (U ) , that by Theorem 4 (Lecture 5) in the point u*  U * , necessary and sufficiently executing of the inequality J c(u* ), u  u* t 0, u  U . Thence with consideration of that u  u * H e, 0 d H d H 0 , e  K , we get J c(u* ), e t 0, e  K . Consequently, the vector 52

J c(u* )  K * .

By Farkas theorem dual cone K * to cone (30) is defined by the formula (24), so the numbers P *j t 0, j  I 1 ; O*i t 0, i  I 2 ; O*m 1 ,  , O*s such that

exist.

s

¦ P e  ¦O a  ¦O a

J c(u* )

* j

jI1

* i i

j

iI 2

(31)

* i i

i m 1

Let O*i 0 for values i  ^1,2,  , m` \ I 2 . Then expression (31) is written as s

J c(u* )  ¦ O*i ai

¦P e jI1

i 1

* j

.

j

(32)

We notice, that O*i g i (u* ) 0, i 1, s , since g i (u * ) 0, i  I 2 and i m  1, s . As follows from the expression (21) u  U 0 the difference







L u, O*  L u* , O*



s

J (u )  J (u* )  ¦ O*i ai , u  u* .

(33)

i 1

Since convex function J (u )  C 1 (U ) , that according to the Theorem 1 (Lecture 4) the difference J (u )  J (u* ) t J c(u* ), u  u* , u  U 0 . Now inequality (33) with consideration of correlations (32) can be written in the manner of









s

L u, O*  L u* , O* t J (u* )  ¦ O*i ai , u  u* i 1

¦P jI1

since e j , u  u *

* j

u j  u *j , u *j

e , u  u* j

0, j  I 1 .

¦P jI1

* j

(u j  u *j )

¦P u jI1

* j

j

t 0,

This implies, that L u* , O* d L u, O* , u  U 0 .

From the inequality and equality O*i g i (u* ) 0, i 1, s follows that pair u* , O* U 0 u / 0 is the saddle point to Lagrange’s function (21). Theorem is proved. The 3-rd case. We consider more general problem of the convex programming: J (u ) o inf , u U

^u  E

n

u  U 0 , g i (u ) d 0, i

1, m; g i (u )

 bi d 0 , i

ai , u 

m  1, p,

53

(34)

g i (u )

ai , u  bi

0, i

`

p  1, s , (35)

where

are convex functions determined on convex set U 0 ; a i  E , i m  1, s are the vectors; bi , i m  1, s are the numbers. Lagrange’s function for tasks (34), (35) J (u ), g i (u ), i

1, m

n

s

L u , O

J (u )  ¦ Oi g i (u ), u  U 0 ,

O  /0

^O  E

i 1

s

O1 t 0,, Om t 0`.

(36)

Theorem 3. If J (u ), g i (u ), i 1, m are convex functions determined on convex set U 0 , the set U * z Ø for tasks (34), (35) and there exists the point u  riU 0 ˆ U , such that g i (u )  0, i 1, m , then for each point u*  U * necessary there exist the Lagrange multipliers O*

(O1* ,  , O*s )  / 0

^O  E

s

O1 t 0,  , O p t 0`

such that the pair u* , O* U 0 u / 0 forms a saddle point to Lagrange’s function (36) on set U 0 u / 0 . Proof of theorem requires of using more fine separability theorems of the convex set, than Theorems 1 - 6 (Lectures 6, 7). For studying more general separability theorems of the convex set and other Kuhn-Tucker theorems is recommended the book: Rocafellar R. Convex analysis. Moskow, 1973. We note the following: 10. The theorems 1-3 give the sufficient conditions of existence of the saddle point for convex programming problem. Example 1. Let function be J (u) 1  u , but set U ^u  E 1 0 d u d 1; (1  u ) 2 d 0`. For this example the functions J (u ), g (u ) (1  u ) 2 are convex on convex set U 0 ^u  E 1 0 d u d 1`. The set U ^1` , consequently U * ^1`, set U * ^1`. The point u* 1 and the value J (u* ) 0 are solution of the problem J (u ) o inf, u  U . Lagrange’s function L u, O (1  u)  O (1  u) 2 , u  U 0 , O t 0 . The pair u* 1, O* t 0 is saddle point to Lagrange’s function, since L u * , O* 0 d L u , O* (1  u )  O* (1  u ) 2 , 0 d u d 1, O* g (u * ) 0 . We notice, that for the example neither Sleyter’s theorem from Theorem 1, nor condition of Theorem 3 is not held. 2°. In general case Lagrange multipliers for the point u*  U * are defined ambiguous. In specified example the pair u* 1, O* t 0 is the saddle point to Lagrange’s function at any O* t 0 . 3°. Performing of the condition of Theorems 1 - 3 guarantees the existence of the saddle point to Lagrange’s function. Probably, this circumstance for problem as convex, so and nonlinear programming is a slight learned area in theories of the extreme problems. 54

Solution algorithm of convex programming problem. The problems of convex programming written in the manner of (34), (35) often are occurred in the applied researches. On base of the theories stated above we briefly give a solution sequence of convex programming problem. 10. To make sure in that for task (34), (35) set U * z ‡ . To use Theorems 1-3 from Lecture 2 (Weierstrass’ theorem). 20. To check performing the conditions of Theorems 1-3 in depending on type of convex programming problem to be a warranty of existence saddle points to Lagrange’s function. For instance, if task has the form of (12), (13), then to show that set U is regularly; if task has the form of (19), (20), that necessary J (u)  C1 (U 0 ) , but for task (34), (35) necessary existence of the point u  riU 0 ˆ U for which g i (u )  0, i 1, m .

30. To form Lagrange’s function L u , O J (u ) 

s

¦ O g (u) i

i

with definitional

i 1

domain U 0 u / 0 , where

^O  E

/0

s

O1 t 0,, Om t 0` .

40. To find a saddle point u* , O* U 0 u / 0 to Lagrange’s function from the conditions (The main lemma, Lecture 8):







0, i

1, s,



L u * , O* d L u , O* , u  U 0 ,

O g i (u * ) * i

(37)

O1* t 0,  , O*m t 0.

a) As follows from the first condition, function L u , O* reaches the minimum on set U 0 in the point u*  U *  U 0 . Since functions J (u ), g i (u ), i 1, m are convex on convex set U 0 , that function L u , O* is convex on U 0 . If functions J (u )  C 1 (U 0 ), i 1, m , then according to the optimality criterion (Lecture 5) and theorems about global minimum (Lecture 5) the first condition from (37) can be replaced on L u u * , O* , u  u * t 0, u  U 0 , where L u u* , O* J c(u* )  ¦ O*i g ic (u* ) . Now s

i 1

condition (37) is written as





L u u* , O* , u  u* t 0 , u  U 0 ,

O g i (u* ) 0, i 1, s, O1* t 0,, O*m t 0. * i

(38)

b) If we except J (u)  C 1 (U 0 ), g i (u)  C 1 (U 0 ), i 1, m , set U 0 E n , then according to optimality criteria (Lecture 5) the conditions (38) possible present in the manner of 55



L u u * , O*



0, O*i g i (u * )

0, i

1, s,

O1* t 0,  , O*m t 0 .

(39)

The conditions (39) are represented by the system n  s of the algebraic equations comparatively n  s unknowns u* u1* ,, un* , O* O1* ,, O*n . We notice, that if Lagrange’s function has the saddle point, then system of the algebraic equations (39) has a solution, moreover must be O1* t 0,, O*m t 0. Conditions (39) are used and in that events, when U 0  E n , however in this case necessary to make sure the point u*  U 0 . 5°. We suppose, that pair u* , O* U 0 u / 0 is determined. Then the point u*  U and value J * J (u* ) is the solution of the problem (34), (35) (see The main theorem, Lecture 8). CONTROL QUESTIONS

1. Prove the Theorem 1. 2. Prove the lemma. 3. Prove the Farkas' theorem. 4. Prove the Theorem 2. 5. Prove the Theorem 3. 6. Formulate solution algorithm of the convex programming problem.

56

Chapter II NONLINEAR PROGRAMMING

There are not similar theorems for problem of nonlinear programming as for problems of convex programming guaranteed existence of the saddle point to Lagrange’s function. It is necessary to note that if some way is installed that pair (u * , O* )  U 0 u / 0 is the saddle point to Lagrange’s function, then according to the main theorem the point u*  U * is the point of the global minimum in the problems of nonlinear programming. We formulate the sufficient conditions of the optimality for nonlinear programming problem by means of generalized Lagrange’s function. We notice, that point u*  U determined from sufficient optimality conditions, in general case, is not a solution of the problem, but only "suspected" point. It is required to conduct some additional researches; at least, to answer on the question: will the point u*  U be by point of the local minimum to function J (u ) on ensemble U ?

LECTURES 11. PROBLEM STATEMENT. NECESSARY CONDITIONS OF THE OPTIMALITY Problem statement. The following problem often is occurred in practice: J (u) o inf , u U

^u  E

n

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

(1) 0, i

`

m  1, s ,

(2)

where J ( u ), g i ( u ), i 1 , s are the functions determined on convex ensemble U 0 of E n . By introducing the notations Ui U m 1

^u  E

^u  E 57

n

n

`

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

0, i

1, m,

`

m  1, s ,

(3)

the ensemble U possible to present in the manner of Ui U m 1

^u  E

^u  E

n

n

`

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

0, i

1, m,

`

m  1, s ,

(3)

Now problem (1), (2) can be written in such form: J (u ) o inf, u  U

We suppose, that J * inf J (u ) ! f , ensemble

^u  E

U*

*

n

u *  U , J (u* )

`

min J (u ) z ‡ . uU

We notice, that if ensemble U* z ‡ ,

then J (u* ) min J (u ) . uU

It is necessary to find the point u*  U * and value J * J (u* ) . The generalized Lagrange’s function for problem (1), (2) has the form L (u, O )

s

O0 J (u )  ¦ Oi g i (u ), u  U 0 , O i 1

O  /0

^O  E

(O0 , O1 ,, Os ),

O0 t 0, O1 t 0,, Om t 0`.

s 1

(5)

Let U 01 be an open ensemble containing set U 0  U 01 . Theorem 1 (the necessary optimality conditions). If functions be J (u)  1 C (U 01 ), g i (u)  C 1 (U 01 ), int U 0 z 0, U 0 is a convex ensemble, and ensemble U * z ‡ , then for each point u*  U * the Lagrange’s multipliers O * ( O*0 , O1* , , O*s )  / 0 necessary exist, such that the following condition is held: O * z 0, O*0 t 0, O1* t 0, , O*m t 0 , L u (u* , O * ), u  u*

s

O*0 J c(u* )  ¦ O*i g ic(u* ), u  u* t 0,  u  U 0 ,

(6) (7)

i 1

O*i g i (u* )

0, i 1, s.

(8)

Proof of the theorem is released by theorems 5, 6 (Lecture 7) about condition of the emptiness intersection of the convex cones (Dubovicky-Milutin theorem) and it is represented below. We comment the conditions of Theorem 1. 58

We note the following: a) In contrast to similar theorems in the convex programming problems it is not become firmly established, that pair (u * , O * ) U 0 u / 0 is a saddle point to Lagrange’s function, i.e. the conditions of the main theorem are not held. b) Since pair (u* , O * ) in general case is not a saddle point, that from the condition (6) - (8) does not follow that point u*  U is solution of the problem (1), (2). c) If the value O*0 ! 0 , then problem (1), (2) is called nondegenerate. In this case it is possible to take O*0 1 , since Lagrange’s function is linear function comparatively O . g) If O*0 0 , then problem (1), (2) is called degenerate. The number of the unknown Lagrangian multipliers, independently the problem (1), (2) is degenerate or nondegenerate, possible to reduce on unit by entering the normalization condition, i.e. condition (6) can be changed by O*

D , O*0 t 0, O1* t 0,  , O*m t 0 ,

(9)

where D ! 0 is any given number, in particular, D 1 . Example 1. Let J (u ) u  1,

^u  E

U

Generalized

1

Lagrange’s

L (u, O ) O0 J (u)  O (u  1) 2 ,

function 1

for

(u  1) 2 d 0`.

problem has the form O t 0 . Ensemble U ^1` , ^1` moreover J (u * ) 0, g (u * ) 0 . The condition (7) is

^u  E

u U 0

consequently, ensemble U * written as

0 d u d 1; g (u )

`

0 d u d1 ,

the

O 0 t 0,

(O*0 J c(u* )  O* g c (u* )) (u  u * ) t 0, u  U 0 .

Thence we get ( O*0 )( u  1) t 0,  u  U 0 , or (  O*0 )(1  u ) t 0, 0 d u d 1. Since expression (1  u) t 0 under 0 d u d 1 , then for executing the inequality necessary that value O*0 0 , so as O*0 t 0 by the condition (6). Finally, the source problem is degenerate. The condition (8) is held for any O* ! 0 . Thereby, all conditions of Theorem 1 are held in the point (u * 1, O*0 0, O* ! 0 ) . We notice, the ordinary Lagrange’s function L (u , O ) (u  1)  O (u  1) 2 for the problem has not saddle point. Example 2. Let J (u )

J (u 1 , u 2 )

u1  cos u 2 ,

U

^(u , u )  E 1

2

2

g (u1 )

59

u1 d 0 ` .

Ensemble

U0  E2.

Generalized Lagrange’s function

L (u 1 , u 2 , O 0 , O )

O 0 (u1  cos u 2 )  O u1 , O 0 t 0, O t 0,

. Since ensemble U 0  E 2 , that condition (7) is written as L u (u * , O * ) 0 . Thence follows that g ic (u* ), e 0, i m  1, s . Consequently, O*0 O ! 0, u 2* Sk , k 0,r1,r2,, . From condition (9) follows that it is possible to take O*0 O* 1 , where D 2 . From condition (8) we get u1* 0 . Finally, necessary conditions of the optimality (6)-(8) are held in the point u* (u1* , u 2* ) (0, kS ), O*0 1, O* 1 . To define in which points from ^(0, Sk )` is reached minimum J (u ) on U the additional researches is required. It is easy to make sure in the points u* 0, S (2m  1)  ^0, Sk ` minimum J (u ) on U is reached, where m 0,r1, . It is required to construct the cones to prove Theorem 1: K y are decrease directions of the function J (u ) in the point u* ; K b are internal directions of the ensembles U i , i 0, m in the point u * ; K m 1 K k are tangent directions for ensemble U m 1 in the point u * . Cone construction. We define the cones K y , K bi , i 1, m, K k in the point u*  U . Definition 1. By decrease direction of the function J (u ) in the point u*  U is called the vector e  E n , e z 0 if there exist the numbers H 0 ! 0, G ! 0 such that under all e o(G , e) ^e  E n e  e  G ` the following inequality u

(u1 , u 2 )  E

2

i

J (u *  He )  J (u * ), H , 0  H  H 0

(10)

is held. We denote through K y K y (u* ) the ensemble of all decrease directions of the function J (u ) in the point u * . Finally, the ensemble Ky

^e  E

n

e  e  G , J (u *  He )  J (u* ), 0  H  H 0 `.

(11)

As follows from the expression (11), ensemble K y contains together with the point ɡ and its G -neighborhood, consequently, K y is open ensemble. The point e  K y . Since function J (u )  C 1 (U 01 ) , that difference [see the formula (10), e  K y ]: J (u *  H e )  J (u * )

J c(u*  THe), e H  0, H , 0  H  H 0 .

60

We notice, that open ensemble U 01 contains the neighborhood of the point u*  U . Thence, dividing by H ! 0 and tending H o 0 , we get J c(u* ), e  0 . Consequently, ensemble

^e  E

Ky

`

(12)

J c(u * ), e  0

n

is open convex cone. By Farkas’ theorem the dual cone to cone (12) is defined by formula

^c

K y*

`

O*0 J c(u* ), O*0 t 0 .

 E n cy

y

(13)

Definition 2. By internal direction of the ensemble U i , in the point u*  U is called the vector e  E n , e z 0 , if there exist the numbers H 0 ! 0, G ! 0 , such that under all e  o(G , e) inclusion u *  He  U i , H , 0  H  H 0 holds. We denote through K b K b (u * ) the ensemble of all internal ensemble U i directions in the point u*  U . Finally, the ensemble i

K bi

i

^e  E

n

`

We notice, that K b is the open ensemble, moreover the point problem (1), (2) ensemble U i , i 1, m is defined by the formula (3), i.e. i

Ui

Then ensemble

K bi

(14)

e  e  G , u *  He  U i , H , 0  H  H 0 .

^u  E

n

`

g i (u ) d 0 , i

e  K bi .

For

1, m

defined by formula (14) can be written as K bi

^e  E

n

g i ( u *  He )  0, H , 0  H  H 0 `.

(15)

Since the point u*  U , that it is possible two cases: 1) g i (u* )  0 ; 2) g i (u* ) 0 . We consider the case, when g i (u* )  0 . In this case, on the strength of function continuity g i (u ) on ensemble U 0 there exists a number H 0 ! 0 such that value g i ( u*  He)  0 under all H , 0  H  H 0 and for any vector e  E n . Then ensemble K b E n is open cone, but dual to it cone K b ^0` . In the case g i (u* ) 0 we get g i ( u*  He)  g i ( u* )  0 [see the formula (15)]. Thence with consideration of that function g i (u )  C 1 (U 01 ) , we get g ic (u*  THe), e H  0, H , 0  H  H 0 . We divide into H ! 0 and tend H o 0 , as a result the given inequality is written in the manner of g ic (u* ), e  0 . Then open ensemble K b is defined by formula i

i

i

61

^e  E

K bi

g ic (u* ), e  0`, i 1, m.

n

(16)

By Farkas’ theorem the dual cone to cone (16) is written as:

^c  E

K b*i

n

i

ci

`

O*i g ic (u* ), O*i t 0 , i 1, m.

(17)

We notice, that Kbi , i 1, m is opened convex cones. Definition 3. By tangent direction to ensemble U m 1 in the point u*  U is called the vector e  E n , e z 0 , if there exist a number H 0 ! 0 and function r (H ) ( r1 (H ),  , rn (H )), 0 d H d H 0 such that r (0) 0; r (H ) / H o 0 under H o 0 and u*  He  r (H )  U m 1 , H , 0  H  H 0 . We denote through K m 1 K m 1 (u* ) the ensemble of all tangent directions of the ensemble U m 1 in the point u*  U . From given determinations follows that ensemble K m1 {e  E n / u*  He  r (H )  U m1 , H , 0  H  H 0 ; r (0)

0, r (H ) / H o 0, H o 0}.

According to formula (3) the ensemble U m 1

^u  E

n

g i (u )

0, i

`

m  1, s ,

consequently, K m 1

^e  E

n

g i (u *  He  r (H ))

0, i

m  1, s, H , 0  H  H 0 `,

(18)

where vector-function r (H ) possesses by the properties r (0) 0; r (H ) / H o 0 under H o 0 . Since functions g i (u)  C 1 (U 01 ), i m  1, s and g i (u* ) 0, i m  1, s, u*  U that differences g i (u*  He  r (H ))  g i (u* ) g ic (u* ), He  r (H )   oi (H , u * ) 0, i m  1, s . Thence, dividing in H ! 0 and tending H o 0 , with consideration of that under H o 0 , we get r (H ) / H o 0, oi (H , u* ) / H o 0 g ic (u* ), e 0, i m  1, s . Consequently, ensemble (18) K m1

^e  E

n

g ic (u* ), e

0, i

`

m  1, s .

(19)

is closed convex cone. Dual cone to cone (19) by Farkas’ theorem is defined by formula

62

­ n ®c  E c ¯

K m* 1

s ½  ¦ O*i g ic (u* )¾ . i m 1 ¿

(20)

Finally, we define K 0 which is an ensemble internal directions of the convex ensemble U 0 in the point u*  U  U 0 . It is easy to make sure in that, if u* intU 0 , then * ^0`. If u*  *pU 0 , then K 0 E n , consequently, K K0

^e  E

n

e

O (u  u * ),  u  int U 0 , O ! 0`

(21)

is an open convex cone, but dual to it cone K0

{c0  E n / c0 , u t c0 , u* , at all u  U 0 }.

(22)

Cone’s constructions are stated without strict proofs. Full details of these facts with proofs reader can find in the book: Vasiliev F. P. Numerical solution methods of the extreme problems. M.: Nauka, 1980. Hereinafter, We denote by K i , i 1, m the cones Kbi , i 1, m . Lemma. If u*  U is a point of the function minimum J (u ) on ensemble U , then necessary the intersection of the convex cones be K y ˆ K 0 ˆ K 1 ˆ  ˆ K m ˆ K m 1

‡.

(23)

Proof. Let the point be u*  U *  U . We show, that correlation (23) is held. We suppose opposite, i.e. existence of the vector e  K y ˆ K 0 ˆ K 1 ˆ  ˆ K m ˆ K m 1 .

From inclusion e  K y follows that J (u*  He )  J (u* ) under all H , 0  H  H y , e  e  G y , but from e  K i , i 0, m follows that u*  He  U i under all H , 0  H  H i , e  e  G i , i 0, m . Let numbers be G min( G y , G 0 , , G m ), D min( H y , H 0 , , H m ) . Then m

inequality J (u*  He )  J (u* ) is true and inclusion u*  He  U i exists under all i 0

H , 0  H  D and e  e  G .

Since vector e  K m 1 , the point u (H ) u *  He   r (H )  U m 1 under all H , 0  H  H m 1 , r (0) 0; r (H ) / H o 0 at H o 0 . We choose the vector e e  r (H ) / H . If r (H ) / H  G . H m 1 d D and H m 1 ! 0 is sufficiently small number, then the norm e  e Then the point u *  He

u *  He  r (H )  U 0 ˆ U 1 ˆ  ˆ U m ˆ U m 1

63

U

.

Finally, the point u *  He  U and J (u*  He )  J (u* ) . It is impossible, since the point u*  U  U 0 is solution of the problem (1), (2). The lemma is proved. We note the following before transfer to proof of the theorem: 1) If J c(u* ) 0, then under O*0 1, O1* O*2 ... O*s 0 all conditions of the Theorem 1 are fulfilled. In fact, the norm O * 1 z 0 is a scalar product O*0 J c(u* ), u  u*

L u (u* , O * ), u  u*

 u  U 0 , O g i (u * ) * i

2) O

* i

1, O

* j

O*

If

0, i

0,

1, s.

under a certain i, 1 d i d m , then for 0, j 1, s, j z i , all conditions of Theorem 1 are also held. In fact, g ic (u* )

0

1 , L u (u* , O * ), u  u*

O*i g ic (u * ), u  u*

values

O*i g i (u* ) 0, i 1, s .

0,  u  U ,

3) Finally, if the vectors ^g ic (u* ), i m  1, s` are linearly dependent, also the conditions (6) - (8) of Theorem 1 are held. In fact, in this case there exist the numbers * * O*m 1 , Om 2 ,., Os not all equal to zero, such that * * O*m 1 g mc 1 (u * )  Om 2 g mc  2 (u* )    Os g cs (u* )

We take O*0 O1*



O*m

0.

0.

Then the norm, O * z 0 ,

L u ( u * , O * ), u  u *

s

¦O

i m 1

 u  U 0 , O *i g i ( u * )

* i

g ic ( u * ) , u  u * 0, i

0,

1, s ,

since g i (u* ), i m  1, s . From items 1 - 3 follow, that Theorem 1 should prove for case, when J c(u* ) z 0, g ic (u* ) z 0 under all i 1, m , and vectors ^g ic (u* ), i m  1, s` are linearly independent. Proof of Theorem 1. By condition of the theorem the ensemble u*  U . Then, as follows from proved lemma, the intersection of the convex cones K y ˆ K 0 ˆ K 1 ˆ  ˆ K m ˆ K m 1

‡

(24)

in the point u*  U * , moreover all cones, except K m1 , are open. We notice, that in the case of

^

J c(u* ) z 0, g ic (u* ) z 0 , i 1, m ; g ic (u* ), i

64

m  1, s

`

are linearly independent, all cones K y , K 0 ,  , K m 1 are nonempty [see the formulas (12), (16), (19), (21)]. Then by Dubovicky-Milutin theorem, for performing of the expression (24) necessary and sufficiently existence of the vectors c y  K *y , c0  K 0* , c1  K 1* , , c m 1  K m* 1 , not all equal to zero and such that c y  c 0  c1    c m 1

0.

(25)

As follows from formulas (13), (17), (20) the vectors cy

O*0 J c(u* ), O*0 t 0, ci cm1

We have cone

K 0*

O*i g ic (u* ), O*i t 0, i 1, m, s

 ¦ O*i g ic (u* ) . i m1

s

c0

 c y  c1    c m  c m 1

O*0 J c(u* )  ¦ O*i g ic (u* ) from equality (25). Since i 1

is defined by formula (22), that c0 , u  u 0

s

O*0 J c(u* )  ¦ O*i g ic (u* ), u  u 0 i 1

L u (u* , O* ), u  u 0 t 0, u  U 0 .

Thence follows fairness of the expression (7). The condition (6) follows from that not all c y , c 0 , c1 ,  , c m , are equal to zero. We notice, that if g i (u* )  0 for a certain i, 1 d i d m , then cone K i E n , consequently, K i* ^0`. It means that c i  O*i g ic (u * ) 0, g ic (u * ) z 0 . Thence follows, that O*i 0 . Thereby, the products O*i g ic (u* ) 0, i 1, s , i.e. the condition (8) is held. Theorem is proved. CONTROL QUESTIONS

1. Formulate problem statement of nonlinear programming problem. 2. Formulate the necessary optimality conditins. 3. Prove the Theorem 1. 4. Give definition to the internal direction of the ensemble in the point. 5. Give definition to the tangent direction of the ensemble in the point. 6. Prove the Lemma.

65

LECTURE 12. SOLUTION ALGORITHM OF NONLINEAR PROGRAMMING PROBLEM We show a sequence of the solution of nonlinear programming problem in the following type: J (u) o inf ,

^u  E

u U

n

(1)

u  U 0 , g i (u ) d 0, i

1, m;

g i (u )

m  1, s ,

0, i

`

(2)

where functions J (u)  C1 (U 01 ), gi (u)  C1 (U 01 ), i 1, s, U 01 is an open ensemble containing convex

ensemble U 0 of E n , in particular, U 01 E n on base of the correlations (6) - (8) from Theorem 1 of the previous lecture. We notice, that these correlations are true not only for the point u*  U * , but also for points of the local function minimum J (u ) on ensemble U. The question arises: under performing which conditions of the problem (1), (2) will be nondegenerate and the point u*  U will be a point of the local minimum J (u ) on U? 10. It is necessary to make sure in that ensemble

^u  E

U*

*

n

u *  U , J (u* )

`

min J (u ) z ‡ uU

for which to use the theorems 1-3 (Lecture 2). 20. To compose the generalized Lagrange’s function for problems (1), (2): s

L (u, O ) O 0 J (u )  ¦ Oi g i (u ), u U 0 ; i 1

O

(O 0 , O1 ,  , O s )  / 0

^O  E

s 1

`

/ O 0 t 0, O1 t 0,  , O m t 0 .

30. To find the points u*

(u 1* ,  , u n* )  U , O

from the following conditions: 66

*

(O*0 ,, O*s )  / 0

O*

(3)

D , O*0 t 0, O1* t 0,  , O*m t 0 ,

L u (u * , O * ), u  u * s

O*0 J c(u* )  ¦ O*i g ic(u* ), u  u* t 0,  u U 0 ,

(4)

i 1

O*i g ic (u * )

0, i

(5)

1, s,

where D ! 0 is the number, in particular D 1 . a) If the point u*  int U 0 or U 0 E n , that condition (4) can be replaced on s

L u (u* , O * ) O*0 J c(u* )  ¦ O*i g ic (u* ) 0 .

(6)

i 1

In this case we get a system n 1  s of the algebraic equations (3), (5), (6) to determinate n 1  s of the unknowns u1* , , u n* ; O*0 , , O*s . b) If after solution of the algebraic equations system (3), (5), (6) [or formulas (3) - (5)] it is turned out that value O*0 ! 0 , that problem (1), (2) is identified by nondegenerate. The conditions (3) in it can be changed by more simpler condition If in the nondegenerate problem a pair O*0 1, O1* t 0, O*2 t 0, , O*m t 0 . * * * * (u * , O ), O s ( O1 ,  , O s ) forms saddle point of the Lagrange’s function L (u, O )

s

J (u)  ¦ Oi g i (u ), u U 0 , O  / 0 i 1

^O  E

s

`

/ O1 t 0,, Om t 0 ,

then u*  U is the point of the global minimum. 4°. We consider the necessary problem of nonlinear programming: (7)

J (u) o inf u U

^u  E

n

g i (u )

0, g i (u )  C 1 ( E n ), i

`

1, m .

(8)

The problem (7), (8) is a particular case of the problem (1), (2). The ensemble U 0 E n . The point in the problem (7), (8) is called by normal point of the minimum, if the vectors ^g ic (u* ), i 1, m` are linearly independent. We notice, that if u*  U is normal point, that problem (7), (8) is nondegenerate. In fact, for problem (7), (8) the equality

67

m

O*0 J c(u* )  ¦ O*i g ic (u* ) 0

(9)

i 1

holds. If O*0 0 , then on the strength of linearly independence of the vectors ^g ic (u* ), i 1, m` we would get O*i 0, i 1, m . Then the vector O * 0 that contradicts to (3). Let u*  U be a normal point for problem (7), (8). Then it is possible to take * , O 0 1 and Lagrange’s function has the form m

L (u, O ) J (u )  ¦ Oi g i (u ), u  E n , O (O1 , , Om )  E m i 1

Theorem. Let functions J (u ), g i (u ), i 1, m be determined and twice continuously differentiable in neighborhoods of the point u*  U . In order to the point u*  U be a point of the local minimum J (u ) on ensemble U , i.e. J (u * ) d J (u ), u , u  o(u * , H ) ˆ U

sufficiently that quadratic form yc

w 2 L (u* , O* ) y to be wu 2

positively determined on hyperplane § wg u* · ¨ ¸ y © wu ¹ *

§ wg 1 u * / wu1 ¨ ¨ wg 2 u * / wu1 ¨  ¨ ¨ wg u / wu 1 © m *

wg1 u * / wu n ·§ y1 · ¸¨ ¸  wg 2 u* / wu n ¸¨ y 2 ¸ ¸¨  ¸   ¸¨ ¸  wg m u * / wu n ¸¹¨© y n ¸¹ 

0.

Proof. Let quadratic form be yc

w 2 L (u* , O* ) y wy 2 § w 2 L (u* , O* ) / wu12 ¨ 2 ¨ w L (u* , O* ) / wu 2 wu1 ( y1 ,  , yn )¨  ¨ ¨ w 2 L (u , O* ) / wu wu n * 1 ©

 w 2 L (u* , O* ) / wu1wu n · ¸  w 2 L (u* , O* ) / wu 2 wu n ¸ ¸u   ¸ w 2 L (u* , O* ) / wu n2 ¸¹  § y1 · ¨ ¸ ¨y ¸ u ¨ 2 ¸ ! 0, y z 0,  ¨ ¸ ¨y ¸ © n¹

68

(10)

on hyperplane (10). We show, that u*  U is a point of the local function minimum J (u ) on ensemble U. We notice, that u * is a normal point and it is determined from the conditions L u (u * , O* )

0, O*i g i (u * )

0, i

1, m , i.e. g i (u* )

0, i

1, m .

Finally, pair (u* , O* ) is known. Since functions J (u )  C 2 (o(u * , G )), g i (u )  C 2 (o(u * , G )) ,

that continuous function y c(w 2 L (u* , O* ) / wu 2 ) y relatively to variable ɯ reaches the lower bound on compact ensemble V ^y  E n y 1, (wg u* / wu ) * y 0`. Let the number be J

min y c

w 2 L (u* , O* ) y, y  V wu 2

We introducw the ensemble AH

^u  E

n

u

u*  Hy, y

1, (wg u* / wu )* y

0`,

(11)

where * is a transposition sign for matrixes; H ! 0 is sufficiently small number. If the point u  AH , then the quadratic form (u  u * ) cL uu (u * , O* )( u  u * )

where u  AH

L uu (u * , O * )

w 2 L (u * , O* ) / w u 2

H 2 y cL uu (u * , O* ) y t H 2 J

,

(12)

on the strength of correlation (11). For the points

the difference

L (u , O* )  L (u* , O* )

L u (u* , O* ), u  u* 

1 (u  u* )c u 2



u L uu (u* , O* )(u  u* )  o u  u*

2

t JH2

2

 o(H 2 )

§ J o(H 2 ) · J H 2 ¨¨  2 ¸¸ t H 2 , 0  H  H 1 , H ¹ 4 ©2

and since L u (u * , O * ) 0 inequality (12) faithfully and o(H 2 ) / H 2 o 0 under H o 0 . We introduce the ensemble 69

(13)

BH

^u  E

n

`

0, i 1, m  U .

H , g i (u )

u  u*

(14)

Since hyperplane (10) is tangent to manifold g i (u ) 0, i 1, m in the point u * , that for each point u~  BH there exists the point u  AH such that norm u~  u d KH 2 , K const ! 0 . In fact, if u  AH : g i (u )  g i (u* )

g i (u )

g ic (u* ), u  u* 



1 2  (u  u* )c g iuu (u* ) (u  u* )  oi u  u* 2



1 2 H y cg iuu (u * ) y  o i (H ), 2

u~  BH : g i (u~ )  g i (u* )

0

1; i 1, m;

(15)

1 g ic(u* ), u~  u*  (u~  u* )c u 2



u g iuu (u* )(u~  u* )  oi u~  u* 

y

2



g ic (u* ), u~  u  u  u* 

1 ~ (u  u  u  u* )c g iuu (u* )(u~  u  u  u* )  oi (H 2 ) 2 1 1 g ic (u* ), u~  u  (u~  u )c g iuu (u* )(u~  u* )  (u  u* )c u 2 2 u g iuu (u* )(u  u * )  (u~  u ) c g iuu (u* )(u  u* )  oi (H 2 ), i

1, m; (16)

From (11), (14) and (15), (16) follow that the norm u~  u d KH 2 , u~  BH , u  AH . Since the function L u (u , O* ) is continuously differentiable by u neighborhoods of the point u* , and derivative L u (u * , O* ) 0 , that the difference L u (u, O* )  L u (u* , O* )

in

L u (u, O* ) L uu (u * , O* )(u  u * )  o u  u *

in neighborhoods of the point u * . This implies that, in particular, u  AH , the norm L (u , O* ) d K 1H , if 0  H  H 1 , H 1 ! 0 is sufficiently small number. For the points the difference L (u~ , O* )  L (u , O* ) u  AH , u~  BH



1 L u (u , O* ), u~  u  (u~  u )cL uu (u * , O* )(u~  u )  o u~  u* 2 L (u~ , O* )  L (u , O* ) d KK 1H 3  o1 H 3 d

2

,

consequently,

the

norm



d K 2H

3

under sufficiently small H 1 ! 0, 0  H  H 1 . Then the difference (u  AH , u~  BH )

70

L (u~, O* )  L (u* , O* )

L (u, O* )  L (u* , O* )  L (u~, O* )   L (u, O* ) t L (u, O* )  L (u* , O* )  L (u~, O* )  L (u, O* ) t t

J 4

H 2  K 2H 3 t 0, 0  H  H 1 , (17)

on the strength of correlations (13). So as values L (u~ , O* ) J (u~ ), L (u* , O* ) J (u* ) , since g i (u~ ) g i (u* ) 0, i 1, m then from (17) we get J (u* ) d J (u~ ) . Consequently, u*  U is the point of the local function minimum J (u ) on ensemble U . The theorem is proved. J u u12  5u1u2  u22 , Example. Let function be ensemble 2 2 2 U ^u (u1 , u 2 )  E u1  u 2 1`. To find the function minimum J (u ) on ensemble U. For this example the ensemble U 0 E 2 , functions J (u )  C 2 ( E 2 ), g i (u ) u12  u 22  1  C 2 ( E 2 ) , ensemble U * z ‡ , since U  E 2 is compact ensemble. Necessary optimality conditions L u (u * , O* )

­° 2u1*  5u 2*  2O*u1* 0, 0: ® °¯2u 2*  5u1*  2O*u 2* 0,

g (u * )

where function

  u

0 : u1*

2

* 2 2

J (u )  O g (u ), u  E 2 , O  E 1 .

L (u , O )

1,

Thence we find the points

u ,u , O : * 1

* 2

*

1) O*

3 / 2, u1* * 1

 5 / 6 , u 2*

*

2) O

3 / 2, u

*

3) O

3 / 2, u

4) O

*

5 / 6, u * 1

5 / 6, u

3 / 2, u

1/ 6;  1/ 6;

* 2 * 2

1/ 6;

 5 / 6, u

* 1

* 2

 1 / 6.

It is necessary to find, in which point (O* , u1* , u2* ) from 1-4 the minimum J (u ) on U is reached. First of all we select the points where local minimum J (u ) on U is reached. We note, that problem is nondegenerate and matrix L uu (u * , O* ) and vector g u (u* ) are equal to §  2  2O* ¨ ¨ 5 ©

L uu (u* , O* )

For

the

first

y cL uu (u* , O ) y *

Thence we get

point

O*

· ¸, g u (u * ) *¸ 2  2O ¹ 5

 5 / 6 , u 2*

3 / 2, u 1*

g c(u* )

is  2 5 / 6 y1  2 1 / 6 y 2 0 . 5 y1 to the quadric form we

1/ 6

y  2 5 y1 y 2  5 y , a hyperplane equation 2 1

y2

2 2

5 y1 .

By substituting the values 71

y2

§ 2u1* · ¨ * ¸. ¨ 2u ¸ © 2¹

quadric

form

get y cL uu (u * , O* ) y 36 y12 ! 0, y1 z 0 . Consequently, ( O * 3 / 2, u 1*  5 / 6 , u 2* 1 / 6 ) is the point of the local minimum J (u ) on U . By the similar way it is possible to make sure in that O * 3 / 2, u 1* 5 / 6 , u 2*  1 / 6 is a point of the local minimum, but the * 1 / 6 and O *  3 / 2, u 1*  5 / 6 , u 2*  1 / 6 are not points O*  3 / 2, u 1* 5 / 6 , u2 the points of the local minimum J (u ) on U . In order to find minimum J (u ) on U necessary to calculate the values of the function J (u ) in the points 1) and 2). It can be shown, that J (  5 / 6 , 1 / 6 ) J ( 5 / 6 , 1 / 6 ) 3 / 2 . Consequently, in the points 1) and 2) the global minimum J (u ) on U is reached. 50. Now we consider the problem (1), (2) in the case U 0 E n , J (u )  C 2 ( E n ), g i (u )  C 2 ( E n ), i 1, s . We suppose, that problem (1), (2) is nondegenerate and the points u * , O*0 ! 0, O* are determined by algorithm 10 - 30. We select amongst constraints for which g i (u ) d 0, i 1, m, g i (u ) 0, i m  1, s g i (u* ) 0, i  I , where ensemble of indexes I

^i

i

m  1, s, g i (u* )

`

0, 1 d i d m .

If the problem (1), (2) is nondegenerate, then the vectors ^gic(u* ), i  I ` are linearly independent. According to specified above theorem from item 40 the point u*  U is the point of the local minimum to functions J (u ) on U , if quadratic form y cL uu (u* , O* ) y t 0, y z 0 on hyperplane wg i u * / wu * u y=0, i  I .

CONTROL QUESTIONS

1. Formulate the Theorem 1. 2. Prove the Theorem 1. 3. Formulate solution algorithm of nonlinear programming problem. 4. Formulate the necessary problem of nonlinear programming. 5. Give definition to the point of the local minimun on ensemble.

72

LECTURE 13. DUALITY THEORY The main and dual problem are formulated on the base of the Lagrange’s function and relationship between its solutions is established. The dual problems for the main, general and canonical forms of linear programming problem are determined. Since dual problem is a problem of convex programming regardless of that whether the main problem is the problem of convex programming or not, then in many cases it is reasonable to study the dual problem and use the relationship between its solutions, to return to the source problem. Such approach often is used under solving of linear programming problem. We consider the nondegenerate nonlinear programming problem in the following type: J (u) o inf , (1) u U

^u  E

n

u  U 0 , g i (u ) d 0, i

1, m;

g i (u )

m  1, s ,

0, i

`

(2)

For problem (1), (2) Lagrange’s function s

L (u, O) J (u)  ¦Oi gi (u), u U0 ; i 1

O (O0 , O1 ,, Os ) /0

^O  E

s

O1 t 0,, Om t 0`.

(3)

The main task. We introduce the function [see the formula (3)]: sup L (u, O ), u  U 0 .

X (u )

O/ 0

(4)

We show, that function X (u )

­ J (u ), u  U , ® ¯ f, u  U 0 \ U .

In fact, if u  U , then g i (u) d 0, i 1, m, g i (u ) 0, i m  1, s , consequently, X (u )

s ­ ½ sup ® J (u )  ¦ Oi g i (u )¾ O/ 0 ¯ i 1 ¿

m ­ ½ sup ® J (u )  ¦ Oi g i (u )¾ O/ 0 ¯ i 1 ¿

73

J (u ) ,

(5)

m

since O1 t 0,, Om t 0, ¦ Oi g i (u) d 0 under all u  U , moreover O 0  / 0 . If u  U 0 \ U , i 1

that possible for a certain number i, 1 d i d m, g i (u) ! 0 and for a certain j , m  1 d j d s , g j (u ) z 0 . In the both cases by choice a sufficient large Oi ! 0 or O j kg j (u ), k ! 0 is sufficiently large number, the value X (u ) can be done by arbitrary large number. Now the source problem (1), (2) can be written in the equal type: X (u ) o inf, u  U 0 ,

(6)

on the strength of correlations (4), (5). We notice, that inf X (u ) uU 0

J* ,

inf J (u ) uU

U*

consequently, if the ensemble U * z ‡ , then

^u  U min J (u) *

uU

J (u* )

J*

^u U *

0

` X (u* )

J (u* )

J*

`

min X (u ) .

uU

The source problem (1), (2) or tantamount its problem (6) is called by the main task. Dual problem. On base of the Lagrange’s function (3) we introduce function \ (O ) inf L (u, O ), O  / 0 . uU

(7)

Optimization problem of the following type: \ (O ) o sup, O  / 0 ,

(8)

is called by dual problem to problem (1), (2) or tantamount its problem (6), but Lagrange’s multipliers O (O1 ,, Os )  / 0 are dual variables with respect to variables u (u1 ,, un ) U 0 . We denote through /*

­O*  E s O*  / , \ (O* ) ® 0 ¯

max\ (O )½¾ O/ 0 ¿

We notice, that if /* z Ø , that \ (O* ) sup\ (O ) \ * . O/0

Lemma. The values J *

inf X (u), \ *

uU 0

sup\ (O ) for main (6) and dual (8)

O/ 0

problems accordingly satisfy to the inequalities \ O d \ * d J * d X (u ),  u  U 0 ,  O  / 0 .

74

(9)

Proof. As follows from formula (7), function \ (O )

inf L (u, O ) d L (u, O ), u  U 0 , O  / 0 .

uU 0

Thence we get \*

sup\ (O ) d sup L (u, O )

O/0

O/0

X (u), u U 0 ,

(10)

on the strength of correlations (4). From correlations (10) transferring to lower bound on u we get \ * d inf X (u ) J * . Thence and from determinations of the lower bound uU 0

the inequalities (9) are followed. Lemma is proved. Theorem 1. In order to execute the correlations U * z ‡ , /* z ‡ , X (u* )

J * \ * \ O* ,

(11)

necessary and sufficiently that Lagrange’s function (3) has saddle point on ensemble U 0 u / 0 . The ensemble of the saddle points to function L (u, O ) on U 0 u / 0 coincides with ensemble U* u /* . Proof. Necessity. Let for the points u*  U * , O*  /* correlations (11) are complied. We show, that pair u* , O* is a saddle point to Lagrange’s function (3) on ensemble U 0 u / 0 . Since \ * \ O*

* inf L (u , O* ) d d L (u* , O ) d sup L (u* , O )

uU 0

O/ 0

X (u* )

J* ,

that on the strength of correlations (11) we get L (u* , O* )

inf L (u, O* )

sup L (u* , O ) , u*  U * , O*  / 0 .

(12)

L (u* , O ) d L (u* , O* ) d L (u, O* ), u U 0 , O  / 0 .

(13)

uU 0

O/ 0

From inequality (12) follows that

It means that pair u* , O* U * u /* is the saddle point. Moreover, ensemble U* u /* belongs to the ensemble of the saddle points to Lagrange’s function, since u* U * , O*  /* is an arbitrary taken points from ensemble U * , /* accordingly. Necessity is proved. Sufficiency. Let pair u * , O*  U 0 u / 0 be a saddle point to Lagrange’s function (3). We show that correlations (11) are fulfilled. 75

As follows from determination of the saddle point which has the form (13) inequality L (u* , O ) d L (u* , O* ), O  / 0 faithfully. Consequently, X (u* )

sup L (u* , O )

O/ 0

L (u* , O* ) .

(14)

Similarly from the right inequality (13) we get \ (O* )

inf L (u, O* )

uU 0

L (u* , O* ).

(15)

From inequalities (14), (15) with consideration of correlation (9) we get L (u* , O* ) \ (O* ) d \ * d J * d X (u* )

L (u* , O* ) .

Thence we have \ (O* ) \ * J * X (u* ) . Consequently, the ensemble U * z ‡ , /* z ‡ and moreover ensemble of the saddle points to function (3) belongs to the ensemble U* u /* . The theorem is proved. The following conclusions can be made on base of the lemma and Theorem 1: 1°. The following four statements are equivalent: a) or u * , O*  U 0 u / 0 is the saddle point to Lagrange’s function (3) on ensemble U 0 u / 0 ; b) or correlations (11) are held; c) or there exists the points u *  U 0 , O*  / 0 such that X (u* ) \ (O* ) ; d) or equality max inf L (u, O ) O/ 0 uU 0

min sup L (u, O ) uU 0 O/ 0

is true.

2°. If u * , O* , a* , b *  U 0 u / 0 are saddle points of the Lagrange’s function (3) on U 0 u / 0 , then u* , b* , a* , O* are also saddle points to function (3) on U 0 u / 0 , moreover L (u* , b* )

X (u* )

L (a* , O* )

L (u* , O* )

L (a* , b* ) \ (O* )

J* \ * .

However the inverse statement, i.e. that from L (u* , O* ) L (a, b) follows a, b U 0 u / 0 is a saddle point in general case untrue. 3°. Dual problem (8) it is possible to write as  \ (O ) o inf, O  / 0 .

(16)

Since function L (u, O ) is linear by O on convex ensemble / 0 , that optimization problem (16) is the convex programming problem, \ (O ) is convex on 76

/ 0 regardless of that, the main task (1), (2) would be convex or no. We notice, that in

general case, the dual problem to dual problem does not comply with source, i.e. with the main problem. There is such coincidence only for tasks of linear programming. We consider the problem of linear programming as applications to duality theories. The main task of linear programming has the form c, u o inf,

J (u ) u U

^u  E

n

(17)

`

u t 0, Au  b d 0 ,

where c  E n , b  E m are the vectors; ȼ is the matrix of the order mu n ; the ensemble

^u  E

U0

n

u

u1 t 0,  , u n

t 0 t 0

`.

Lagrange’s function for task (17) is written as L (u, O ) u U 0 , O

c, u  O , Au  b

c  A*O , u  b, O ,

^O  E

(O1 ,, Om )  / 0

m

O1 t 0,, Om t 0`.

(18)

As follows from formulas (17), (18), function \ (O )

inf L(u, O )

uU 0

­  b, O , c  A*O t 0 ® * ¯ f, (c  A O )i  0, O  / 0 .

We notice, that under c  A*O t 0 lower bound which is equal to  b, O is got under u 0 U 0 . If ( c  A*O ) i  0 , then it is possible to choose ui o f , but all u j 0, j 1, n, i z j , herewith \ O o f, O*  / 0 . Finally, the dual task to task (18) has the form  \ O

O/

^O  E

b, O o inf,

m

The dual task to task (19) complies with source task (17). By introducing of the additional variables u ni t 0, problem (17) can be written in the following type:

>Au @i  bi  u n i

J (u ) o inf 0, u t 0, u n  i t 0, i

General task of linear programming has the form 77

(19)

O t 0, c  A * O t 0`.

i 1, m , optimization

½ ¾. 1, m ¿

(20)

^u  E

u U

c, u o inf,

J (u ) n

`

u j t 0, j  I , Au  b d 0, A u  b d 0 ,

(21)

where c  E n , b  E m , b  E s are the vectors; A, A are the matrixes accordingly to the orders m u n, s u n ; index ensemble I Ž ^1,2,, n`. The ensemble U0

^u u , , u  E 1

n

n

u j t 0, j  I `.

Lagrange’s function for task (21) is written as L (u , O )

c, u  P , Au  b  P , A u  b c  A* P  A * P , u  b, P  b , P ,

u U 0 , O

(P , P )  / 0

^O

`

(P , P )  E m u E s P t 0 .

Function \ O inf L(u , O ) uU 0





­ b, P  b , P , if c  A* P  A * P i t 0, i  I ; °° c  A* P  A * P j 0, j  I ; ® ° under rest. °¯ f





The dual task to task (21) is written as  \ O

c  A P  A P *

ui





b, P  b , P o inf; c  A* P  A * P i t 0, i  I ; *

j

0, j  I ; O

(22)

( P , P )  E n u E s , P t 0.

It is possible to show, that dual task to task (22) complies with (21). By introducing of the additional variables u ni t 0, i 1, m and representations qi  vi , qi t 0, vi t 0, i  I the problem (21) possible write as J (u )

>Au @i  bi  u n i

c, u o inf,

u j t 0, j  I , u i

0, A u  b

q i  vi , q i t 0, vi t 0, i  I .

0,

(23)

Canonical task of linear programming has the form c, u o inf,

J (u ) u U

^u  E

n

78

u t 0, Au  b

`

0,

(24)

where c  E n , b  E s are the vectors; ȼ is the matrix of the order s u n ; the ensemble U0

^u  E

n

u

u 1 ,  , u n t 0 `.

Lagrange’s function for task (24) is written so: L (u, O )

c, u  O , Au  b u U 0 , O  / 0

c  A* O , u  b, O ,

^O  E `. s

Function \ (O )

inf L(u, O )

uU 0

­ b, O , c  A*O t 0 ® * ¯  f, c  A O  0.

Then the dual problem to the problem (24) has the form  \ (O )

b, O o inf;

c  A * O t 0, O  E s .

(25)

It is easy to make sure in the dual task to task (25) complies with task (24). Finally, we note that main and the general task of linear programming by the way of introducing some additional variables are reduced to the canonical tasks of linear programming [see the formulas (20), (23)].

CONTROL QUESTIONS

1. Formulate nondegenerate nonlinear programming problem 2. Formulate the main task. 3. Formulate the dual problem. 4. Prove the Lemma. 5. Formulate the Theorem 1. 6. Prove the Theorem 1. 7. Formulate the canonical task of linear programming.

79

Chapter III LINEAR PROGRAMMING

As it is shown above, the main and the general problem of linear programming are reduced to the canonical problems of linear programming. So it is reasonable to develop the general solution method for the canonical linear programming problem. Such general method is a simplex-method. Simplex-method for nondegenerate problems of linear programming in canonical form is stated below.

LECTURES 14. PROBLEM STATEMENT. SIMPLEX-METHOD We consider the linear programming problem in the canonical form c, u o inf,

J (u )

u U

^u  E

n

u t 0, Au  b

`

0,

(1)

where c  E n , b  E m are the vectors; ȼ is the matrix of the order m u n . Matrix A a ij , i 1, m , j 1, n can be represented in the manner of

A

a , a ,..., a , 1

2

n

aj

§ a1 j · ¨ ¸ ¨ a2 j ¸ ¨ ... ¸, j ¨ ¸ ¨a ¸ © mj ¹

1, n .

The vectors a j , j 1, n are called by the condition vectors, and the vector b  E m is a vector of the restrictions. Now the equation Au b can be written in the manner of a1u1  a 2u 2  ...  a n u n b . Since ensemble U 0 ^u  E n u t 0` and U ^u  E n Au b` is affine ensemble which are convex, that (1) is a problem of convex programming. We notice, that if ensemble 80

U*

^u  E

n

*

u*  U , J (u* )

`

min c, u z ‡ , uU

that Lagrange’s function for problem (1) always has saddle point, any point of the local minimum simultaneously is the point of the global minimum and necessary and sufficiently condition of the optimality is written as J ' (u * ), u  u * c, u  u * t 0, u  U 0 . We suppose, that ensemble U * z ‡ . It is necessary to find the point u*  U * and value J * inf J (u ) J (u * ) min J (u ) . uU

uU

Simplex-method. For the first time solution of the problem (1) was considered on simplex U

n ­ n ®u  E u t 0, ¦ ui i 1 ¯

½ 1¾ ¿

so solution method of such linear programming problem was called by simplexmethod. Then method was generalized for case of the ensemble U specified in the problem (1), although initial name of the method is kept. Definition 1. The point u U is called by extreme (or angular), if it is not represented in the manner of u Du1  1  D u 2 , 0  D  1, u1 , u 2 U . From given definition follows that extreme point is not an internal point of any segment belonging to ensemble U. Lemma 1. Extreme point uU has not more m positive coordinates. Proof. Not derogating generalities, hereinafter we consider that first components k , k d m of the extreme point are positive, since by the way of recalling the variables always possible to provide the given condition. We suppose opposite, i.e. that extreme point uU has m 1 positive coordinates

u u1 ! 0, u2 ! 0,...,um1 ! 0,

0,...,0 .

We compose the matrix A1 (a1 , a 2 ,..., a m1 ) of the order m u  m  1 from condition vectors corresponding to positive coordinates of the extreme point. We consider the homogeneous linear equation A1 z 0 for vector z  E m1 . The equation has a nonzero solution ~z , ~z z 0 . We define n-vector u~ ~z ,0 and consider two vectors: u1 u  H u~, u 2 u  H u~ , where H ! 0 is sufficiently small number. We notice, that vectors u1 , u 2 U under 0  H  H 1 , where H 1 ! 0 is sufficiently small number. In fact, Au1

Au  H A u~

Au  A1 ~ z 81

Au

b, u1 u  H u~ t 0

under sufficiently small H 1 ! 0 similarly Au 2 b, u 2 t 0 . Then the extreme point u 1 / 2 u 1  1 / 2 u 2 , D 1/ 2 . It opposites to the definition of the extreme point. The lemma is proved. Lemma 2. Condition vectors corresponding to positive coordinates of the extreme point are linear independent. Proof. Let u

u1 ! 0, u2 ! 0,...,uk ! 0,

0,...,0 U , k d m

be an extreme point. We show that the vectors a1 , a 2 ,..., a k , k d m are linear independent. We suppose opposite, i.e. that there are the numbers O1 ,..., Ok not all equal to zero such that O1a1  O 2 a 2  ...  Ok a k 0 (the vectors a1 ,..., a k are linear dependent). From inclusion uU follows that a1u1  a 2u 2  ...  a k u k

b,

u i ! 0, i 1, k .

We multiply the first equality on H ! 0 and add (subtract) from the second equality as a result we get a1 u1 r H O1   a 2 u2 r H O2  ...  a k uk r H Ok b .

We denote by u1

u1  H O1 ,..., u k  H O k ,0,...,0  E n ,

u2

u1  H O1 ,..., uk  H O k ,0,...,0  E n .

There is a number H 1 ! 0 such that u1 U , u 2 U under all H , 0  H  H 1 . Then vector u

1 / 2 u1  1 / 2 u 2 U ,

u 1  U , u 2 U .

We have obtained the contradiction in that uU is extreme point. Lemma is proved. From lemmas 1, 2 follow, that: a) The number of the extreme points of the ensemble U is finite and it does not exceed the sum

m

¦C

k n

, where Cnk is a combinations number from n-elements on k. In

k 1

fact, the number of the positive coordinates of the extreme points is equal to k , k d m (by Lemma 1), but the number of the linear independent vectors corresponding to positive coordinates of the extreme point is equal to Cnk (by Lemma 2). Adding on k within from 1 till m we get the maximum possible number of the extreme points. b) The ensemble U

^u  E

n

u t 0, Au  b 0` 82

is convex polyhedron with final number of the extreme points under any matrix ȼ of the order m u n . Definition 2. The problem of the linear programming in canonical form (1) is called nondegenerate, if the number of the positive coordinates of the feasible vectors not less than rank of the matrix ȼ, i.e. in the equation a1u1  a 2u2  ...  a nun b under u i ! 0, i 1, n the number differenced from zero summands not less, than rank of the matrix A. Lemma 3. Let rangA m, m  n . If in the nondegenerate problem the feasible vector has exactly m positive coordinates, then u is an extreme point of the ensemble U. Proof. Let the feasible vector u u1 ! 0, um ! 0, 0,...,0 U has exactly m positive coordinates. We show, that u is an extreme point of the ensemble U. We suppose opposite i.e. there are the points u1 , u 2 U , u1 z u 2 , and a number D , 0  D  1 such that u Du1  1  D u 2 (the point uU is not an extreme point). From given presentations follows, that u 1 u11 ,..., u 1m ,0,...,0 , u 2 u12 ,..., um2 ,0,...,0 . Let the point be u(H ) u  H (u1  u 2 ), u1 z u 2 . Consequently, u(H ) (u1  H (u11  u12 ), u 2  H (u12  u22 ),..., um  H (u1m  um2 ),0,...,0 . We notice, that Au(H ) Au  H Au1  Au2 b at any H . We assume, there is negative amongst the first m coordinates of the vector u1  u 2 . Then by increasing H ! 0 from 0 to f we find the number H 1 ! 0 such that one from m first coordinates of the vector u H 1 becomes equal to zero, but all rest will be nonnegative. But it isn’t possible in the nondegenerate problem. Similarly if u1  u 2 ! 0 , then reducing H from 0 till  f we get the contradiction. Lemma is proved. We notice, that from rangA m does not follow that problem of linear programming (1) will be nondegenerate. Example 1. Let the ensemble U

^u

(u1 , u 2 , u 3 , u 4 )  E 4 u j t 0, j 1,4;

3u1  u 2  u3  u 4

3, u1  u 2  2u3  u 4

1`.

In this case the matrix § 3 1 1 1· ¸¸ A ¨¨ © 1  1 2 1¹

a , a 1

2

, a 3 , a 4 ,

rangA 2.

points of the ensemble U are u1 1,0,0,0 , u 2 0,5 / 3, 4 / 3, 0 , u 3 0,1,0,2 . Here u 2 , u 3 are nondegenerate extreme points, u1 is degenerate extreme point. Since there is a feasible vector u1 , number of the positive coordinates less than rangA , the problem of linear programming (1) is not nondegenerate. The number of the extreme points is equal to 3, that does not exceed By

the

extreme

83

the amount c14  c42 10 ; the vectors corresponding to the positive coordinates of the extreme point u 2  a 2 , a 3 , u 3  a 2 , a 4 are linear independent. Example 2. Let ensemble be U

^u

(u1 , u 2 , u 3 , u 4 )  E 4 u j t 0, j 1,4;

3u1  u 2  u3  u 4

3,  u1  u 2  2u3  u 4

1`.

extreme points u1 1/ 2, 0, 0, 3 / 2 , u 2 5 / 7, 0, 6 / 7, 0 , u 3 0, 5 / 3, 0, 4 / 3 , u 4 0, 1, 0, 2 . The problem (1) is nondegenerate. We notice that in nondegenerate problem the number of the extreme points no more than C nm . Lemma 4. Any point uU can be represented as convex linear combination

Here

rangA 2 ,

the

points of the ensemble U , i.e. u

s

¦D k u k , D k t 0, k 1

s

¦D

k

1, u1 , u 2 ,..., u s are extreme

k 1

points of the ensemble U . Proof. We prove the lemma for case, when U is convex bounded closed ensemble. In fact, for order the ensemble U * z ‡ , necessary and sufficiency the ensemble U is compactly. It means, that U is convex closed bounded ensemble. We prove the lemma by method of mathematical induction for cases, when u  *p U and u  intU . Let u  *p U , uU . If n 1, then ensemble U is a segment; consequently, statement of the lemma faithfully, since any point of the segment ( u  *p U , u  intU ) can be represented in the manner of the convex combination of the extreme points (the end of the segment). Let the lemma be true for ensemble U  E n1 n t 2 . We conduct supportive hyperplane to ensemble U  E n through the point u  *pU , i.e. c, u t c, u , u  U . We denote by U 1 U ˆ * , where * ^u  E n c , u c , u ` are ensemble points of the supportive hyperplane. We notice that ensemble U 1 is convex, bounded and closed, moreover U 1  E n 1 . Let u1 , u 2 ,...,u s1 be extreme points of the ensemble U 1 , then by hypothesis the point u

s1

¦ E k u k  E n1 , E k t 0, k 1, s1 , k 1

s1

¦E

k

1.

k 1

It is remains to show, the points u1 , u 2 ,...,u s1 are extreme points and ensembles U. Let u i D w  1  D v, w, v  U , 0  D  1 . We show that u i w v for any extreme point u i U1 . In fact, since c, w t c, u , c, v t c, u and c, u i c , u , that c, u i D c, w   1  D c, v t c , u

c, u i

. Thence follows, that c, w

c, v

c, u , i.e. w, v  U 1 .

However the point u i is an extreme point U 1 , consequently, u i w v . Therefore the points u1 , u 2 ,...,u s1 are extreme points of the ensemble U . For border point u  *pU . Lemma is proved. Let the point u  intU . We conduct through u U a line l which crosses the borders of the ensemble U in the points a U , b U . And the point u  intU is 84

represented in the manner of u D a  1  D b, 0  D  1 . The points a  *pU , b  *pU on the strength of proved a

s1

¦P v k

k

, Pk t 0,

s1

¦P

k 1, s1 ,

k 1

s2

¦E w ,

1, b

k

E k t 0, k 1, s 2 ,

k

k

k 1

k 1

s2

¦E

k

1,

k 1

where v1 ,..., v s1 ; w1 ,..., ws2 are extreme points of the ensemble U . Then the point s1

u D a  1  D b

s1

¦ D P v  ¦ 1  D E w , k

k

k

k 1

k

k 1

moreover Dk

D P k t 0, k 1, s1 , D k

1  D E k

s1

s2

¦D  ¦D

t 0, k 1, s2 ,

k

k 1

k

D  1  D 1 .

k 1

Lemma is proved. Lemma 5. Let U be convex bounded closed ensemble from E n , i.e. ensemble U * z ‡ . Then minimum to function J (u ) on ensemble U is reached in the extreme point of the ensemble U. If minimum J (u ) on U is reached in several extreme points u1 ,...,u k of the ensemble U, then J (u ) has same minimum value in any point k

u

¦D i u i , D i t 0, i 1

k

¦D

i

1.

i 1

Proof. Let minimum J (u ) on U is reached in the point u*  U * . If u*  U is an extreme point, then the lemma is proved. Let u*  U be certain border or internal point of the ensemble U. Then on the s

s

¦D u ,

strength of Lemma 4 we get u*

D i t 0, ¦D i 1 , where u1 ,...,u s are extreme

i

i

i 1

i 1

points of the ensemble U . The value s

J (u* )

c, u*

¦D

i

c, u i

i 1

where J i Let J 0

J (u i ) min J i 1di d s

s

¦D J , i

i

i 1

c , u i , i 1, s.

J (u i0 ) . Then s

J (u* ) t J 0 ¦D i

J0

J (u i0 ) .

i 1

Thence we get J (u* ) J (u i0 ) , consequently, minimum J (u ) on U is reached in the extreme point u i0 . 85

Let the value J (u 2 ) ... J (u k ) ,

J (u 1 )

J (u* )

where u1 ,...,u k are extreme points of the ensemble U. We show that J u

§ k · J ¨ ¦ Di u i ¸ ©i1 ¹

J  u* ,

k

where Di t 0, ¦ Di 1 . In fact, the value i 1

J u

k

k

¦ Di J  u i

¦ D J u J u . i

i 1

*

*

i 1

Lemma is proved. We notice, that Lemmas 4, 5 are true for any problem of linear programming in the canonical form of the type (1) with restrict closed convex ensemble U. From lemmas 1 - 5 follows that solution algorithm of the linear programming problem in canonical form must be based on transition from one extreme point of the ensemble U to another, moreover under such transition the value of function J (u ) in the following extreme point less, than previous. Such algorithm converges to solution of the problem (1) through finite number steps, since number of the extreme points of the ensemble U doesn’t exceed the number

m

¦C

k n

(in the case of the nondegenerate

k 1

problem Cnm ). Under each transition from extreme point u i U to the extreme point u i 1  U necessary to make sure in that, whether the given extreme point u i U be solution of the problem (1). For this a general optimality criterion which easy checked in each extreme point must be existed. Optimality criterion for nondegenerate problem of linear programming in canonical form is brought below. Let the problem (1) be nondegenerate and the point u*  U be a solution of the problem (1). According to Lemma 5 the point u*  U is extreme. Since the problem (1) is nondegenerate, that extreme point u*  U has exactly m positive coordinates. Not derogating generalities, we consider the first m components of the vector u*  U are positive, i.e. u*

u , u ,..., u * 1

* 2

* m



,0,...,0 , u1* ! 0, u 2* ! 0,..., u m* ! 0 .

We present the vector c  E n and matrix ȼ in the manner of where c Ƚ (c1 , c2 ,..., cm )  E m , c H (cm1 , cm2 ,..., cn )  E nm , c c Ƚ , c ɉ , A  AȽ , AH , AȽ (a1 , a 2 ,..., a m ) , AH (a m1 , a m2 ,..., a n ). We notice, that according to Lemma 2 condition vectors a1 , a 2 ,..., a m corresponding to the positive coordinates of the extreme 86

point u*  U are linear independent, i.e. matrix AȽ is nonsingular, consequently, there is the inverse matrix AȽ1 . Lemma 6 (optimality criterion). In order to the extreme point u*  U be a solution of the nondegenerate problem of linear programming in canonical form (1), necessary and sufficiency the inequality holds cH'  cȽ' AȽ1 AH t 0.

(2)

Proof. Necessary. Let the extreme point be u* u Ƚ , u H* U , where u Ƚ* u1* , u 2* ,..., u m* , uH* 0,...,0 is a solution of the nondegenerate problem (1). We show, that inequality (2) is held. Let where u uȽ , uɉ  U , u Ƚ  u1 ,..., um , u H  um 1 ,..., un is an arbitrary point. We define the ensemble of the feasible directions ^l` to the point u*  U . We notice that vector l  E n , l z 0 is called by feasible direction in the point u*  U , if there is a number H 0 ! 0 such that vector u u*  H l  U under all 0 d H d H0 . We present the vector l  E n in the manner of l  lȽ , lH , where lȽ  l1 ,..., lm  E m , lH  lm 1 ,..., ln  E n  m . From inclusion u*  Hl  U follows that uȽ*  H lȽ t 0, u H*  H lH H l H t 0, Au*  H l AȽ uȽ*  H lȽ  AH H lH b . Since u*  U , that Au* Au Ƚ* b consequently from the last equality we get AȽ lȽ  AH lH 0 . Thence follows the vector lȽ  ȼȽ1 ȼɉ lH . Since under sufficiency small H ! 0 the inequality u Ƚ*  HlȽ t 0 is held, the feasible directions in the point u* are defined by the correlations l H t 0,

lȽ

 AȽ1 AH l H .

(3)

Finally, having chosen the arbitrary vectors lH  E n  m , lH t 0 possible to find lȽ  ȼȽ1 ȼɉ lH and construct the ensemble of the feasible directions L, which each element has the form l   ȼȽ1 ȼɉ l H , lH t 0  E n , i.e. L ^l  E n l l Ƚ , l H , l Ƚ  AȽ1 AH l H , l H t 0` E n . Now any point u U can be represented in the manner of u u *  H l , l  L , H ! 0, 0 d H d H 0 , H 0 H 0 l . Consequently, u  u* H l , l  L . Since the

function

J (u )

c, u  C 1 U ,

J c  u* , u  u* t 0,  u  U

J ' u *

c, u  u *

then in the point u*  U necessary inequality (Lecture 5) is held. Thence with provision for that

H l , l  L we get

H ! 0 , that we get cȽ , lȽ  ɭɉ , lH

¬ª c Ƚ , l Ƚ  ɭ ɉ , l H ¼º H t 0 .

Since the number t 0 . By substituting the value lȽ  ȼȽ1 ȼɉ lH from c, Hl

formula (3), we get c Ƚ ,  ȼȽ1 ȼɉ l ɉ  ɭ ɉ , l H

 cc

H

 c Ƚc ȼȽ1 ȼɉ l H t 0

under all lH t 0 . Thence follows the inequality (2). Necessity is proved. 87

Sufficiency. Let the inequality (2) be executed. We show, that u*  U *  U is an extreme point of the ensemble U . Since the function J  u  C 1 U is convex on convex ensemble U, that necessary and sufficiency the execution of the inequality J (u )  J (v ) t J c(v ), u  v , u , v  U

(Lecture 4). Hence in particular, v u*  U , then we get J (u )  J (u * ) t J ' (u * ), u  u *

H  cȽ , lȽ  cH , lH

H c

' H



c, u  u *

c, H l

 c A AH l H t 0, l  L, u U . ' Ƚ

1 Ƚ

Then J (u * ) d J (u ), u  U . Consequently, minimum is reached in the point u*  U on U . According to Lemma 5 u*  U is extreme point. Lemma is proved. It is easy to check the optimality criterion (2) by simplex table formed for the extreme point u* U 0 . Ba sis

ɭ

AȽ

ɍɝ

b

ɭ1

… ɭj

… ɭ j0

… ɭn

ɜ1

aj … u1 j

… … … ui j

… aj … u1 j … … … ui j

0 0

… … … …

… … … uij … … … umj … zj …z c

… … … uij … … … umj … zj … zj cj

… … … … … …

ɭ1

u

…

…

…

a i0

ci0

u

* i0

…

…

…

…

ai

ci

ui*

u i1

…

…

…

…

cm

* m

ɜ

1

a

m

z z c

* 1

u

u11

… ui0 1

um1 z1

0

0

0

0

j

0

0

0

j

0

0

an

T

u1n

…

…

ui0n

T0

…

…

uin

Ti

…

…

umn zn zn  cn

The condition vectors a1 , a 2 ,..., a m corresponding to positive coordinates of the extreme point u*  u1* ,..., u m* , 0,..., 0 are leaded in the first column of the table. Matrix Coordinates of the vector c c1 ,...,cn corresponding to positive coordinates of the extreme point are given in the second column; in the third corresponding to a i positive coordinates of the extreme point. In the following columns the decomposition coefficients of the vector a j , j 1, n on base vector ai , i 1, n are reduced. Finally, in the last column the values Ti which will be ȼȽ

 a ,..., a . 1

m

88

explained in the following lecture are shown. We consider the values specified in the last two lines more detail. Since vectors a1 , a 2 ,..., a m are linear independent (Lemma 2), that they form the base in the Euclidean space E m , i.e. any vector a j , j 1, n can be singularity decomposed by this base. Consequently, m

¦a u

aj

i

ij

ȼȽ u j , u j

i 1

Thence, we get u j

u

1j

, u2 j ,..., umj  E m , j 1, n . m

¦c u

ȼȽ1ɜ j , j 1, n . We denote through z j

i ij

, j 1, n . We notice,

i 1

that z j c j , j 1, m , since u j (0,...,0,1,0,...,0), j 1, m .. Then vector z c

where z  c Ƨ

 z  c

Ƚ

,  z  c H

 0,  z  c , H

z1  c1 , z 2  c2 ,..., z m  cm 0,...,0 , z  c H z m1  cm1 ,..., z n  cn .

Since the values z j

m

¦c u

i ij

cȽc u j

ccȽ ȼȽ1ɜ j , j 1, n , that z j  c j

c cȽ ȼȽ1 a j  c j , j

1, n

i 1

Consequently, vector ( z  c)c (0, ccȽ ȼȽ1 AH  ccH ) . Comparing the correlation with optimality criterion (2), we make sure in that the extreme point u*  U to be a solution of the problem necessary and sufficiency that the values z j  c j d 0, j 1, n . Finally, by sign of the values in the last row of the simplex-table it is possible to define whether the point u*  U a solution of the nondegenerate problem (1). CONTROL QUESTIONS

1. Formulate linear programming problem in the canonical form. 2. Describe a simplex-method. 3. Give definition to the extreme point. 4. Prove the Lemma 1. 5. Prove the Lemma 2. 6. Prove the Lemma 3. 7. Give definition to the nondegenerate linear programming problem. 8. Prove the Lemma 4. 9. Prove the Lemma 5. 10. Prove the Lemma 6.

89

LECTURE 15. DIRECTION CHOICE. NEW SIMPLEX-TABLE CONSRUCTION. THE INITIAL EXTREME POINT CONSRUCTION In the linear programming problem of the canonical form minimum to linear function is reached in the extreme point of the convex polyhedron U. A source extreme point in the simplex-method is defined by transition from one extreme point to the following, moreover value of the linear function in the next extreme point less, than in previous. It is necessary to choose the search direction from the extreme point and find the following, in the last to check the optimality criterion etc. Necessity in determination of the initial extreme point to use the simplex-method for problem solution is appeared. We consider the nondegenerate problem of linear programming in canonical form c, u o inf, u  U

J (u )

^u  E

n

u t 0, Au  b

`

0.

(1)

Let u U be an extreme point and in the point u U the minimum J (u ) on U isn’t reached, i.e. optimality criterion z j  c j d 0, j 1, n is not held. Then necessary to transfer from given extreme point to other extreme point u  U , where value J  u  J  u . It is necessary to choose the direction of the motion from given extreme point u . Direction choice. Since u U is an extreme point, not derogating generalities, it is possible to consider, that the first m components of the vector u are positive, i.e. u (u1 , u 2 ,..., u m , 0,..., 0 ), ui ! 0, i 1, m . According to the formula (2) (Lecture 16), feasible directions in the point u are defined from correlations lH t 0, lȽ  ȼȽ1 ȼɉ lH . Derivative of function J (u ) on feasible direction l in the point u is equal to wJ u / wl

cȽ , lȽ  cH , lH

c, l

¦ z n



j m1

j

 c j l j , l j t 0, j

c

' H



 cȽ' AȽ1 AH lH

m  1, n .

Necessary to choose the feasible direction l 0 in the point u from minimum condition wJ  u wl on ensemble of the feasible directions L . The source direction l 0  L is chosen by the following in the simplex-method: a) Index j0  I1 is defined, where I1

^j

m  1 d j d n, z j  c j ! 0` 90

from condition z j  c j 0

0

max ( z j  c j ) . Since in the point u U minimum J (u ) on U is jI1

not reached, i.e. the inequality z j  c j d 0 under all j , 1 d j d n aren’t held, that ensemble I1 z ‡ . b) Vector l H0 t 0 , l H0  E nm is chosen, so that l H0 (0,...,0,1,0,...,0) , i.e. j0 is a component of the vector lH0 is equal to 1, but all rest components are equal to zero. Finally, the motion direction l 0 in the point u U is defined by the correlations: l 0  L, l 0 lȽ0

 AȽ1 AH lH0

l

0 Ƚ

, l H0 , l H0

 AȽ1a j0

0,...,0, 1, 0,...0 ,

u j0

(u1 j0 ,  u2 j0 ,...,um j0 ) .

(2)

We notice, that derivative of the function J (u ) in the point u in the row of l 0 is equal to wJ  u wl 0





 z j0  c j0  0 .

c, l 0

It is follows to note that in general case wJ  u wl 0 isn’t the least value wJ  u wl on ensemble L . However such direction choice l 0 allows to construct a transition algorithm from one extreme point to other. We consider the points ensemble from U along chosen direction l 0 . These points uT u  T l 0 U , T t 0. Since u  u1 ! 0,..., um ! 0, 0,..., 0  U is an extreme point, T l 0 T lȽ0 , lH0 T  u j0 , that the points u T u 1  T u 1 j , u 2  T u 2 j ,... We introduce the indexes ensemble .., u m  T u m j ,0 ,..., 0 , T ,0 ,..., 0 , T t 0 . I 2 ^i 1 d i d m , uij ! 0` . We define the values T i u i u ij , i  I 2 . Let 0

0

0

0

0

T0

Then

for

min Ti iI 2

values



min ui uij0 iI 2

T

T0



Ti0 ! 0, i0  I 2 .

vector

u (T 0 )

( u 1  T 0 u 1 j0 ,...

We notice, that u(T 0 ) t 0, 0 Au (T 0 ) A u  T 0 Al 0 A u  A Ƚ l Ƚ0  AH lH Au b, consequently, the point u T 0 U . On the other hand, vector u  T0 has exactly m positive coordinates. It means that u  T0 u is the extreme point of the ensemble U. We calculate value .., u i0 1  T 0 u i0 1 j0 ,0 , u i0 1  T 0 u i0 1 j0 ,..., u m  T 0 u mj 0 ,0 ,..., 0, T 0 ,0 ,..., 0 ).

J ( u (T 0 ))

J (u  T 0 l 0 )

c, u  T 0 c, l 0

c, u  T 0l 0

Thence we get 91

J (u )  T 0 ( z j0  c j0 ).

J (u (T 0 ))  J (u )

 T 0 ( z j0  c j0 )  0 .

Then J (u(T 0 )) J (u )   J (u ) , i.e. in the extreme point u  T0 u value J  u less, than in extreme point u . We note the following: 1) If ensemble of the indexes I 2 ‡ , i.e. uij  0, i 1, m , then for any T ! 0 the point u  T  U , moreover 0

J (u (T ))

J (u )  T ( z j 0  c j 0 ) o f

at T o f .

In this case source problem has a no solution. However, if ensemble U is compact, that it isn’t possible. 2) It can turn out so that T 0 Ti Ti , i0  I 2 , i0  I 2 . In this case the vector u  T 0 has m 1 positive coordinates. It means that source problem is degenerate. In such cases possible appearance "thread", i.e. through determined number steps once again render in the extreme point u  T0 . There are the different methods from "recirculations". We recommend the following books on this questions: Gabasov R., Kirillova F. M. Methods of optimization. Minsk: BGU, 1975, Karmanov V.G. Mathematical programming. M.: Science, 1975; Moiseev N.N., Ivanilov U.P., Stolyarova E.M. Methods of optimization. M.: Science, 1978. To make sure in the extreme point u  T0 u  U is a solution of the problem (1), need to build the simplex-table for the extreme point u with aim to check the optimality criterion (2) (Lecture 16) in the point u . Construction of the new simplex-table. The simplex-table built for the extreme point u* U in the previous lecture, in particular for the extreme point u U is true. In the simplex-table column j0 and row i0 and values Ti , i  I 2 are indicated. We build the simplex-table for the extreme point u  T0 u  U on base of simplex-table for the point u* u . We notice, that for point u U by the base vectors were condition vectors a1 ,..., a m corresponding to positive coordinates of the extreme point u . For extreme point u(T0 ) u by the base vectors will be 0

0

a1 , a 2 ,..., a i0 1 , a j0 , a i0 1 , a m ,

(2*)

as condition vectors corresponding to positive coordinates. Thereby, the first column of the new simplex-table differs from previous, that instead of vector ai is written the vector a j . In the second column instead ci is written c j , in the third column nonnegative components of the vector u are written, since 0

0

0

92

0

a1u1  a 2u2  ...  a i0 1ui0 1  a j0 T 0  a i0 1ui0 1  ...  a mum

b,

where u i  T 0 u ij 0 , i

ui

1, m , i z i0 .

In the rest columns of the new simplex-table must be decompositions coefficients of the vector a j , j 1, n , on the new basis (2*). There were the decompositions m

¦a u , j

aj

i

ij

1, n .

i 1

in the previous base a1 ,..., a m . Thence follows that m

¦a u

aj

i

ij

i 1 i z i0

then under j

j0

 a i0 ui0 j , j 1, n .

we get a j0

m

¦a u i

ij0

i 1 i z i0

 a i0 ui0 j0 , ai0

uij0

m

¦ a i

ui0 j0

i 1 i z i0



1 ui0 j0

a j0 .

(3)

By substituting value ai from formula (3) we get 0

m

¦a u

aj

i

ij

i 1 i z i0

m

§

¦ a ¨¨ u i

i 1 i z i0

©

ij



 a i0 ui0 j

ui j0 ui0 j · ui0 j j0 ¸ a , ui 0 j 0 ¸¹ ui 0 j 0

(3*)

j 1, n.

The formula (3*) presents by the decompositions of the vectors a j , j 1, n by new base (2*). From expression (3*) follows that in the new simplex-table coefficients  uij ɩɪɞ are determined by formula

u

ij ɩɪɞ

uij ui0 j0  uij0 ui0 j ui0 j0

, i z i0 ,

but in row i0 of the new simplex-tables must be (u i j ) ɩɪɞ u i j u i j , vector a j in base, that in column j0 of the new simplex-table all (u i j 0

0

0

j

0

0

93

0

) ɩɪɞ

1, n. 0,

Since i z i0 ,

Finally, coefficients (u i j ) ɩɪɞ , i 1, m , j 1, n , are calculated by the known coefficients u i j , i 1, m , j 1, n of the previous simplex-table. Hereinafter, the last two rows of the new simplex-table are calculated by the known ( u i0 j 0 ) ɩɪɞ

1.

 cȽ ɩɪɞ

 ɭ ,..., ɭ 1

i0 1

, c j0 , ci0 1 ,..., cm



and (u i j ) ɩɪɞ , i 1, m , j 1, n . If it turns out ( z j  c j ) ɩɪɞ d 0, j 1, n , that u  U is a problem solution. Otherwise, it is fulfilled the transition to the following extreme point of the ensemble U and so on. Construction of the initial extreme point. As follows from Lemmas 2, 3, the extreme point can be determined from system of the algebraic equations AȽ uȽ b, uȽ ! 0 , where AÁ (a j , a j ,..., a j ), a j , i 1, m , are linear independent vectors, column of the matrix A u Ƚ (u j , u j ,..., u j ) . However such determination way of the extreme point u ( 0,..., 0, u j ,0,..., 0, u j ,0 ,..., 0 ) is complete enough, when matrix A has the great dimensionality. 1. Let the main problem of linear programming is given 1

m

2

1

2

k

m

m

1

J (u)

c, u o inf, u U

^u  E

n

`

u t 0, Au  b d 0 ,

(4)

where c  E n , b  E n are the vectors; A is the matrix of the order m u n . We suppose, that vector b ! 0 . By introducing of the additional variables un 1 t 0, i 1, m problem (4) can be represented to the canonical form J (u)

c, u o inf, [ Au]i  u ni

bi , i 1, m,

(5)

u (u1 ,...,un ) t 0, uni t 0, i 1, m.

u, u n1 ,..., u nm 

 A, I m A, a , a ,..., a , a (0,..., 0,1, 0,..., 0), k 1, m , where I m is an unit matrix of the order m u m , problem (5) is written in the manner of

By introducing some notions ɭ (c,0)  E n m , u n 1

n 2

n m

 E n m , A

n k

J (u )

c , u o inf, u U

^u  E

n m

u t 0, Au

`

b.

(6)

For problem (6) initial extreme point u  0,..., 0, b1 ,..., bm  E n  m , since condition vectors a n 1 ,..., a n  m (the columns of the unit matrix I m ) corresponding to positive coordinates of the extreme point u~ , b ! 0 are linear independent. 2. Danzig’s method (two-phase method). We consider canonical problem of linear programming

94

J (u)

c, u o inf, u U

^u  E

n

`

u t 0, Au  b 0 ,

(7)

where c  E n , b  E n ; ȼ is a matrix of the order m u n . We suppose that vector b ! 0 (if bi z 0, i 1, m , then being multiplied by the first row, where bi  0 , always possible to provide to bi ! 0, i 1, m ). The following canonical problem on the first stage (phase) is solved: m

¦u >Au@i  u ni

ni

o inf,

i 1

(8)

bi , i 1, m, u t 0, u n i t 0, i 1, m.

By initial extreme point for problem (8) is vector u  0,..., 0, b1 ,..., bm t 0 . Hereinafter problem (8) is solved by the simplex-method and its solution u*  u1* , u 2* ,..., u n* , 0,..., 0 is found. We notice, that if the source problem (7) has a

solution and it is nondegenerate, then the vector u*  u1* , u2* ,..., un*  E n has exactly m positive coordinates and it is the initial extreme point for the problem (7), since lower bound in (8) is reached, under un i 0, i 1, m . On the second stage the problem (7) with initial extreme point u*  U is solved by the simplex-method. 3. Charnes’ method (Ɉ-method). Charnes’ method is a generalization of Danzig’s method by associations of two stages of the problem solution. So called Ɉ-problem of the following type instead of source problem (7) is considered: m

c, u  M ¦ uni o inf ,

(9)

i 1

> Au @i  uni

bi , u t 0, un i t 0, i 1, m ,

where M ! 0 is a feasible large number. For M-problem (9) the initial extreme point u  0,..., 0, b1 , b2 ,..., bm  E n  m , b ! 0 , and it is solved by the simplex-method. If the source problem (7) has a solution, then Ɉ-problem has such solution: u~* u~1* ,..., u~n* ,0,..., 0 , where components un*i 0, i 1, m . Vector u * u~1* , u~2* ,..., u~n*  E n is the solution of the problem (7). We notice, that for Ɉ-problem the values z j  c j D j M   E j , j 1, n . So in the simplex-table for rows z j , z j  c j are conducted two rows: one for coefficients D j other for E j . Index j0 , where z j  c j max ( z j  c j ), z j  c j ! 0 is defined by the 0

value of the coefficients D j , D j

0

!0.

95

j

CONTROL QUESTIONS

1. Formulate a nondegenerate problem of linear programming. 2. Formulate the direction choice. 3. Present construction of the new simplex table. 4. Present construction of the initial extreme point. 5. Describe Danzig's method. 6. Describe the M-method.

96

Chapter IV 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 16. 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(ɤ) be defined on the interval >ɭ, d @  E 1 . Necessary to find the minimum of J(ɤ) on the interval >ɭ, d @  E 1 . Definition 1. Function J(ɤ) defined on the interval >a, b@ is called unimodal, if it is continuous on >a, b@ 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) ;

97

>D , E @. In particular, if D is called strictly unimodal on the interval >a, b@ .

3) J (u ) J (u* )

inf J (u ) at D d u d E , i.e. U *

u>a .b @

E,

then J (u) We note, that if a continuous function J (u) is not unimodal on the prescribed interval >c, d @ , then the intervals >a, b@ >c, d @ can be found by partitioning of the interval >c, d @ , where function J (u) is unimodal on >a, b@ . We define the minimum of strictly unimodal function J (u) on the interval >a, b@ . According to the definition, unimodal (strictly) function is convex (strictly) function on the interval >a, b@ . a) The points u1

a b G , u2 2

a b G 2

are chosen, where G ! 0 is

determined 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 (u 2 ) , it is assumed a1 a, b1 u 2 , if J (u1 ) ! J (u 2 ) , 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 (u4 ) 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 u 1 (or the point u 2 ) is on the interval >a1 ,b1 @ , however, in determining the segment >a 2 ,b2 @ the value J (u1 ) [or J (u2 ) ] is not used, since the point u1 (or u 2 ) is far from the center of the segment >a 2 ,b2 @ . Therefore, more efficient method of finding the minimum of a unimodal function on the interval >a, b@ , than the bisection method is the method of the golden section. The points u1 , u 2  >a, b@ are selected as follows: 1) The points u1 , u 2 must be spaced equally from the ends of the segment >a, b@ , i.e. u1  a b  u 2 . 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, b@ . Consequently, u1  a ba

u2  u1 u2  a 98

b  a  2u1 . b  u1

Hence we get 3 5 , 2 3 5 . u2 b  (b  a) 2 3) The values J (u1 ), J (u 2 ) are calculated. If J (u1 ) ! J (u 2 ) , then a1 u1 , b1 b , and if J (u1 )  J (u 2 ) , then a1 a, b1 u 2 . At each iteration, except the first, the value J (u ) is u1

a  (b  a)

calculated in a one point and the length of localization is reduced by 1.6 times. Note that bk  a k

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

Optimal search. Optimal methods of searching for minimizing the unimodal function J (u ) on the segment >a, b@ 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, b@ 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

Fn 1  Fn , F1

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

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

k

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 99

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 u i 1 u i  G i . If J (ui 1 )  J (ui ) , then u i  2

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

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

then u i2

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. Methods for extremum search. M.: Nauka, 1967. Other methods for finding minimum of the function J (u ) on the segment >a, b@ such as broken method, the method of tangents, parabola method can be found in the book: Vasil'ev F.P. Numerical methods for solving the extremum problems. 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 J (Q )  J (u ) t J c(Q ), Q  u 

1 2 LQ  u , u , Q  U . 2

(2)

holds. Proof. Let all conditions of the lemma be held. 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

100

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 J (Q )  J (u )

1

³

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 . (3) 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 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 n 0 t 0,

then the inequality 101

an 

n0  1 for all n ! n0 An

holds. Proof. By assumption of the lemma the difference 1 a k 1



a k § 1  a k 1 / a k ¨ a k 1 ¨© ak

a k  a k 1 a k a k 1

1 ak

a ak § ak  ak 1 · ¨¨ ¸¸ t A k 2 ak 1 ak 1 © ak ¹

for all k t n 0 . Then the sum § 1

n 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  n0 )

By condition of the lemma n ! n0 , therefore n t n 0  1. Then nn 0 t n 02  n 0 , n 0 ! 0

therefore nn 0  n 02 It follows that

,

 n0 ! 0 . 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 102

· ¸¸ ¹

(4)

an 

1 , n ! n0 , A(n  n0 )

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

We consider the optimization problem u U { E n .

J (u ) o inf,

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

that 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  D J ' u  J u D J ' u  o D , 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 u n 1

u n  D n J ' u n ,

103

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 n 1  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  D J ' (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. 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 H0

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 , D

c

f, ¦D n2  f n 0

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

0,1,2,... ..

d) It can be accepted an

D, D ! 0 .

If the inequality J (u n 1 )  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. 104

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?

105

LECTURE 17. 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 )  C 1 (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 )  C 1 (U ) is convex or strongly convex, the global minimum point J (u ) on U can be determined. Gradient method. We consider the optimization problem u U { E n ,

J (u ) o inf,

(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 ‡ uE

and sequence ^u n `  E is constructed by the rule u n  D n J c(u n ), n

u n 1 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

lim J c(u n ) nof

0.

If, 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. 106

(3)

lim J (u* )

inf J (u ) minn J (u ) ,

J*

nof

uE n

uE

(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 (u 0 )

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

(6)

holds, where q 1  P / 2L, 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 u n u n1 ... J ' u k 0, k n, n  1,..., and such that lim J ' u n 0 . nof

We assume, that the gradient

J ' u n z 0, n

0, 1, 2 ... ..

From the second relation of (2) follows that g n D n

J u n  D n J ' u n

J u n 1 d g n D

Then the following inequality holds:

107

J u n  D J ' u n

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

D t 0, n

0, 1, 2 ... .

(7)

Since J (u )  C 1 (U ), U { E n

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

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 un  J un1 t D ¨1  ¸ J ' un , D t 0, n 0, 1, 2... . 2 ¹ ©

Hence we get 2 § L · J un  J un1 t maxD ¨1  D ¸ J ' un D t0 © 2 ¹

1 2 J ' un ! 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 u n 1  J u n , 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 nof

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 ) nof

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 108

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

1,2 ... .

Hence, passing to the limit, taking into account the continuity J (u ) we obtain lim J ( w k ) 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 (u 0 ) . 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 (u 0 ) , i.ɡ. 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 )  ɍ 1 M (u 0 ) , M (u 0 )

is a convex set, that 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

u n  M (u 0 ) ,

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

where 109

(11)

D

sup u  v

u ,vM u0

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

lim J (u n ) nof

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 a n J (u n )  J * . Now the inequality (9), (11) can be written as follows: a n  a n 1 t

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

0, 1, 2, ... .

(12)

From (12) we get a n  a n 1 t

1 a n2 , 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 )  ɍ 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 (u 0 ) . 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 ).

For values v u n  M (u 0 ) 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 110

(13)

a n 1 d a n 

P · § 1 2 J ' ( u n ) d ¨1  ¸u a , 2 L¹ n 2L ©

as J ' (u n ) 2 t Pa n . Hence we get a n 1 d qa n , n

0, 1, ...,

where q 1  P / 2L . Further, considering a n d a n 1 q 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 )  ɍ (U ); U is a convex closed set. The method algorithm: 1°. Select the starting point u 0  U . 2°. Determine the point u1 u 0  D 0 J c(u 0 ) . 3°. Then we find the projection of the point u1  E n on the set U , as a result, we get 1

u1

PU (u 0  D 0 J c(u 0 ))  U .

PU (u1 )

In the general case, the sequence ^u n `  U is constructed by the rule u n 1

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

0,1,2,...,

(15)

where the number D n is chosen from the condition J (u n 1 )  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 )  C 1 U ,

is a closed convex set, U * z ‡ , the gradient J ' (u) satisfies the Lipschitz condition on the set U and the sequence ^u n `  U is determined by the formula (15), where 0  H 1 d D n d 2 / L  2H , H 1 , H ! 0 are prescribed numbers, then u n  u n 1 o 0 at n o f . If, in addition, J (u) is convex on U , the set U

M (u 0 )

^u  U / J (u ) d J (u 0 )`

is bounded, then the sequence ^u n ` 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

wherɡ 111

c2 1 ˜ , H n

n 1, 2, ... ,

(16)

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 u n1 is the projection of the point u n  D n J ' (u n )  E n

on the set U , that according to Theorem 5 (Lecture 5) we get [ w u n 1 , v u n  D n J ' (u n ), u U , (6)]: u n 1  u n  D n J ' (u n ), u  u n 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 ,

(17)

0, 1, ... .

According to Lemma 1 (Lecture 1, formula (2)) the inequality J (un )  J (un 1 ) t J ' (un ), un  un 1 

1 2 L u n  u n 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 (19)  ¸¸ u n  u n 1 t H u n  u n 1 , H ! 0, J (u n )  J (u n 1 ) t ¨¨ D 2 © n ¹ 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 n 1 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 (u 0 ) 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 (u 0 )

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 * .

112

(20)

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

1

Dn

un  un 1, u*  un 1

(21)

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



1

Dn

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

un  un1 , u*  un1

J ' (u n ) 

1

Dn

u*  un1 , un  un 1 d J ' (u n ) 

d 1

Dn

u*  u n 1 u n  u n .

Taking into account that J ' (u n ) 

1

Dn

u*  u n 1 d

J ' (u n ) 

1

Dn

u*  u n 1 d

§ D· d ¨¨ sup J ' (u )  ¸¸, H u  M  u 0 1¹ ©

1

Dn

d

1

H1

, u*  un1 d D,

we get 0 d an

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

Since by proved u n  u n1 o 0 at n o f , consequently lim J (u ) J * . nof

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 u n  u n 1 ,

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

follows, that an  an 1 t

H c2

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.

113

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?

114

LECTURE 18. 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 )  C 1 ( 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 ,

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 u 0 is determined as the solution of optimization problem J ' (u 0 ), u  u 0 o inf, u  U .

J 0 (u )

Finally, J (u 0 )

inf J 0 (u ) . uU

0

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 115

(1)

u n 1

u n  D n u n  u n , 0 d D n d 1, n

0, 1, 2, ... ,

(2)

where J n (u n )

inf J n (u ), uU

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 ^u n ` 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

J (u n )  J * d

c , n

n 1, 2, ... ,

where J (u* )

J* , c

const ! 0 .

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

that the set U * z ‡ . Since the set U is bounded, that its diameter D  f . From condition (3) follows that J u n 1 d J u n  D u n  u n . 116

(5)

Then by Lemma (Lecture 1) the inequality J (un )  J (un 1 ) t

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

D2 2

Lu 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 J n (u n )

min J n (u ) d J n (u n ) uU

0.

If J n (u n ) 0 , then 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

J (u n )  J (u n 1 )

D

,

0, 1, 2, ....

We note, that since 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 )

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

J (u n ) ,

that the inequality J (u n 1 ) 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 n 1 ) 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

117

D 2

LD 2 , 0  D d 1.

(7)

Hence, at D o 0 we obtain 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 (u n ) , 2 LD 2 n t n0 , an d  J n (un ).

a n  a n 1 t

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

118

(10)

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. 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 ^u n ` by the rule un  D n pn ,

u n 1

0, 1, ... ,

n

(12)

where p0

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

J ' (u 0 ), p n

1, 2, ... .

(13)

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

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, ¯

En

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

In particular, if I2

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

^0, 1, 2,...`, I1 119

0  I2 .

‡,

(14)

(15)

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

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 )

1 2 D Ap n , p n  D J ' (u n ), p n  J (u n ) 2

J (u n  D p n )

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 ), p n

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

Ap n , p n

Ap n , p n J ' (u n )

2

! 0,

Ap n , p n

(17)

In fact, for n 0 the scalar product J ' (u 0 ), p 0

J ' (u 0 )

2

.

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 n 1 )

D n Ap n , n

0, 1, ... ,

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

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

Hence En

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

Lemma. If the indices of the sets 120

, n 1, 2, ... .

(18)

I1

^1, 2,...`,

^0`,

I2

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

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

Au1  b, Au 0  b

Au 0  b, Au 0  b  D 0 Ap 0 , Au 0  b p0 , p0  D 0 Ap0 , p0 , (20)

where J ' (u )

Au  b, p 0

J ' (u 0 )

Au 0  b .

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

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

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

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 ' (u n ), 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

121

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

 J ' (u n 1 ), p n

0,

since Au n  b,

J ' (u n )

A(u n  D n p n ) 

Au n 1  b

J ' (u n 1 )

J ' (u n ) D n Ap n .

b

If D n 0 , then u n 1

un , 0

 J ' (u n ), pn =

g ' n ( 0)

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

J ' (u n )  E n2 p n 1

2

2

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 ), p 0

0

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

p1 , Ap 0

J ' (u1 )  E 1 p 0 , Ap 0

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

122

n  1, j

n

follows directly from (21) and (22). Since D n Ap n ,

J ' (u n )  J ' (u n 1 )

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

J ' (u n 1 ), p j

0

(24)

J ' (u n1 ), J ' (u j )  J ' (u j 1 )

0,

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

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

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

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

0 (the case j

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

n ).

Finally, p n 1 , Ap n

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

0

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

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

p n 1 , Ap i

0, i

0, 1, ..., n.

The lemma is proved. Theorem 2. If the indexes of the sets I1

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

I2

^0`,

then the sequence ^u n ` 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 123

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?

124

LECTURE 19. NEWTON'S METHOD. PENALTY FUNCTION METHOD. LAGRANGE MULTIPLIERS 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(ɤ) 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 J (u) o inf, u  U ,

where function

(1)

J u  C 2 U ,

n U is given set of E . For solving of the problem (1) the sequence ^u n `  U is

constructed by the following algorithm. The algorithm of the method: 1°. Select the starting point u 0  U . Since the function J (u )  C 2 U ,

that 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 u n1 is determined by the condition J n (u ) o min, u  U ,

125

(2)

where the function J n (u )

J c(u n ), u  u n 

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 n 1 ) J c(u n )  J cc(u n )(u n 1  u n ) 0 . u n  >J cc(u n ) @ J c(u n ), n 1

u n 1

0,1,2,...



Hence, in case when the matrix J cc(u n ) is nonsingular, we get

(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 n J cc(u )[ , [ t P [ , [  E , u U , 2

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

¦q

Lk

n

2k

P

n

q2 d ˜ n n , L 1  q2

0, 1,...

(4)

is valid, where J (u * )

Proof. As the function

min J (u ) . uU

J (u )  C 2 U

and J cc(u )[ , [ t P [ , P ! 0, [  E n , u  U , 2

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 )

J ncc (u n ) ,

then the function J n (u ), u  U

126

is also strongly convex and gets the lower bound on U in a single point u n 1  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 u n1  U , it is necessary and sufficiently to perform the inequality J ' n (u n 1 ), u  u n 1 t 0,



u U , n 0, 1, 2, ...

Taking into account that J ' n (u )

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

we obtain J ' (u n )  J cc(u n )(u n 1  u n ), u  u n 1 t 0, 0, 1, ...

(5)



u U , n

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

This means that u n 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[

at [

u n 1  u n

2

n d J cc(u )u [ , [ , [  E , u U

we get P u n 1  u n

2

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

d J ' (un ), un  un 1 ,

From inequality J ' n u n 1 , u  u n 1 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 127

n 0, 1, ... .

(6)

J ' n 1 (u n ), u n  u n 1 d 0 .

Then 

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

(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(un1 )(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  u n 1 u n  u n 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 n

q 4 , etc.

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

(9)

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

128

m

m

k n

k n

u m  u n d ¦ u k 1  u k d ¦

It follows, that

P L

q

2k

f

d¦ k n

P L

q

2k

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

u U

^u  E

(10)

J (u ) o inf, 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 ® ¯f at u U 0 \ U .

k of

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

Ak P (u ), P (u )

m

¦ ^maxg (u );0 `

p

i

i 1

u  U 0 , Ak ! 0, k

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



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 129

(12)

Pk (u )

Ak P(u )

0,

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

uU 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 , uU

then P (u k )

o( Ak1 ), k

1, 2, ...,

lim g i (u ) d 0, i 1, m ; lim g i (u k ) 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. Numerical methods for solving the extremum problems. 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 130

s

L (u, O )

J (u )  ¦ Oi g i (u ), u  U 0 ,

O  /0

^O  E

i 1

s

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

(14)

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

u n 1

O n 1

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 , On )

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. 131

LECTURE 20. REVIEW OF 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 I0

^i

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: (e, V ) W0

V o inf, ^(e,V )  E J c(u0 ), e d V , n 1

i  I0 , e

`

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

3°. The next point E0

g ic(u0 ), e d V ,

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,

132

u n 1

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

sup ^u n  ten  U , 0 d t d D `,

En

where en

0,1, 2,...,

(2)

D

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 ,

^i

e nj d 1, j 1, n, I n

`.

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). 133

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

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 n@  1 . Hence we get k

e1 ,

P0

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

Pn Pn 1 P2 n 1

e 2 ,...,.

e n , P2 n

e1 ,...

2°. The next point is chosen as follows: if J u 0  D 0 P0  J (u1 ) ; or if J u 0  D 0 P0  J (u 0 ) . In the general case, if

u1

u0  D 0 P0 ,

u1

u0  D 0 P0 ,

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

then u k 1

If the inequality

u k  D k Pk , D k 1

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 u k 1 134

u k  D k Pk , D k 1

Dk.

(6)

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

D k 1

uk ,

­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)

where number O ! 0, 0  O  1 ; k is the number of iteration; n is dimension of Ɂn . One cycle of ɫ 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

^u  E

M (u 0 )

`

J (u ) d J (u 0 )

n

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

`

m  1, s ,

(8)

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

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

P (u )

¦ max^g (u );0`

pi

i



s

¦ g (u )

i m 1

i 1

i

pi

, u U 0 ,

(9)

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) ɐ (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 ɐ(u, t ),

uU 0

135

 f  t  f .

(11)

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

value

J * ! f

ɐu* , t* 0 ,

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

g i (u* ) 0, i

m  1, s

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 ɐu, 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 ɐu , 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 136

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 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.

137

Chapter V 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 >t0 , t1 @ . To compare each element u (t )  U a number of Ǿ1, necessary take a definite integral 1

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 21. 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. 138

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 ȼ(0, 0), axis x is horizontal, axis ɯ is vertically downwards and the point ȼ is above the point B x0 , y0 . 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, that T  3 A 0 , and in the point Ȅ the sum T  3 M mv 2 / 2  mgy 0 . 2 gy . Since the Hence we determine the speed in the point M equals to v M instantaneous speed vM

1  y ( x) dx / dt ,

ds / dt

that vM

1  y 2 dx / dt .

2 gy ( x)

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

y (x ),

0 d x d x0 passing through points A and B.

We note, that the function set U

y ( x )  C 1 >0, x 0 @ ,

^y( x)  C >0, x @ y(0) 1

0

0, y ( x 0 )

y0 ` ,

and the task is written as follows: x0

T ( y)

³ 0

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

y ( x)  U .

(2)

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

t0

on set 139

(3)

^xt  C >t , t @ xt 1

U

0

1

0

x 0 , xt1

`

x1 .

(4)

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

J ( x, u )

³ F ( x, u, t )dt o inf,

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 , t 1 ]

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 IJ > 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 IJ-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 IJ-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 140

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

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 0 (t )  J h(t ), t  >t 0 , t1 @ ,

x(t )

where h(t 0 )

h(t )  C 1 >t 0 , t1 @,

h(t1 )

0, J ! 0 are some number.

Thus, the function h(t )  U 0

^h(t )  C >t , t @ 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 Ș the fullfilment of the inequalities given in the definitions 1, 2 can be provided. Let the function F(x,ɤ,t) be 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 w 2 F ( x0 , x 0 , t ) 2 «¬

wx 2

0

h2 (t )  2

w 2 F ( x0 , x 0 , t ) h(t )h(t )  wxwx



where x0 =x0(t)U. 141

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

(7)

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 , ³ « wx wx ¼ t0 ¬

(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 )  ³t «¬ wx 2  w w x x 0

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 Ș 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 Ș 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 142

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



1 2 2 J G J oJ2 . 2

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  « ³ wx wx ¼ t0 ¬ h  U 0 .

GJ

0,

(14)

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

wF ( x 0 , x 0 , t )  h(t )dt ³t wx 0

U dV t1

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

t

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

0

0

t

1 § d wF ( x 0 , x 0 , t ) · ¸¸h(t )dt ,  ³ ¨¨ dt wx 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 ³ « wx dt wx ¼ t0 ¬

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. 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.

144

LECTURE 22. 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 IJ0 so that the segments

>T  H 0 ,T  H 0 @,

'1

>T  H

'2

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 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 ) , at t , T H T H ° 0 0 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

t0

'1

'2

This contradicts (2). The lemma is proved. 145

 E 2 ) ³ v(t )dt ! 0 . '1

The second conclusion of the Euler equation can be obtained from DuboisReymond lemma. In fact, from the necessary condition of the first order t1

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

³ «¬

GJ

0, h(t )  U 0

after integration by parts of the first term t1

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

³ «¬

U dV

h(t ), wF dt , V wx

dU h(t )dt , wF ( x 0 ([ ), x 0 ([ ), [ ) d[ ³ wx t0 t1

t1 § t wF ( x 0 ([ ), x 0 ([ ), [ ) ·  ³ h(t )¨ ³ d[ ¸dt ¨t ¸ w x 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 ³ « wx w x »¼ t0 « t 0 ¬ t1

0,

Therefore, taking into account the fact that t1

³ h(t )dt

0,

t0

i.e. v (t ) h(t ) and denoting wF ( x 0 , x 0 , t ) wF ( x 0 ([ ), x 0 ([ ), [ ) ³ d[ , wx wx t0 t

b(t )

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

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: 146

Fx  Fxt  Fxx x 0  Fxx x0

0, x 0 (t 0 )

x0 , x 0 (t1 )

x1 .

(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 0 )

t  >t 0 ,t1 @,

x 0 (t , c1 , c 2 ),

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 gx(1  x 2 )

,

Fxx





2 x 2 gx(1  x ) 2

,

Fxt

0,

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

.

Equation (5) is written as  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 2 gx 1  x 02

0

02

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 147

Fx

0, Fx

2t 2 / 3 x , Fxt

3t 1/ 3 x , Fxx

3t 1/ 3 .

0, Fxx

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, 1@ . 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 The solution of this equation

tx 0 (t )  2t 2 x0 (t )

0.

x 0 (t ) c1t 1  c 2 , t  >0, 1@ .

However, any one curve of the family does not pass through the points x 0 (0) 0, x 0 (1) 1 . We note, that Euler equation is obtained from 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 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. 148

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 to the function x 0 ( t )  C 1 >t 0 , t 1 @

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 .

t 0 dt dt1

Then the increment of the functional (6) can be written as J ( x  Jh, x 0  Jh )  J ( x 0 , x 0 ) JGJ  o(J 2 ) ,

'J

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 )  ³t wx wx t 0 0

t1

§ d wF ( x 0 , x 0 , t ) ·  ³ ¨¨ h(t ) ¸¸dt.  w dt x ¹ t0 © 149

(8)

Now, the first variation GJ has the form GJ

ª wF ( x 0 (t 0 ), x 0 (t 0 ), t 0 ) º 0 » h(t 0 )  «) 0 x ( x (t 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, then 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 x 0 (t )

x 0 (t , c1 , c2 ), t  >t0 ,t1 @ ,

constants c1 , c 2 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 ɤ at fixed x  E , 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 to the function x 0 (t ) U 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 1

150

B( x 0 (t ), x 0 (t ), t , [ ) t 0, [  E 1

holds. Proof. Let the functions x(t )

(9)

x 0 (t )  h(t , J )  U ,

such that max h (t , J )  H

max x (t )  x 0 (t )

t 0 d t d t1

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

It follows that the value 'J

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

(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 Ș > 0. We choose the function h(t , J ), J ! 0, t  >t 0 , t1 @, h(t 0 , J )

in the following form

h(t , J )

if ­0, ° if °°J[ , if ®(t  W )[ , ° ° J (H1  t  W )[ , °¯ H1  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 151

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

h(t , 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 t1

³ F (x

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

0

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

t0

W J

³W F ( x



W H1

³ F (x

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.

W J

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

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

W H1

³W F ( x x

W

0

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

where o(H 1 )

W H1

³ F (x W

0

x

(t ), x 0 (t ), t )H 11 (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 .

t0

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

³W F ( x (t ), x (t ), t )dt 0

0

x

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 152

(15)

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 0 (W  H 1 ), x 0 (W  H 1 ),W  H1 )  [H11

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 [

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

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.

153

(16)

LECTURE 23. LEGENDRE CONDITION. JACOBI CONDITION. FUNCTIONALS DEPENDING ON THE 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

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

A(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 .

(1)

t0

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

Fxx ( x 0 (t ), x 0 (t ), t ) t 0, t  >t0 , t1 @

(2)

holds (Legendre condition). Proof. Let along the solution of Euler equation the inequality G 2 J t 0 is satisfied. 154

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 )

x 0 (t )  h(t , J ), t  >t 0 , t1 @,

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

h(t 0 , J )

We represent the function h ( t , J ) h ( t 1 , J )  o( t , J ) ,

where J ! 0 and

h1 (t , J )

­0, if t  >W  J / 2, W  J / 2@, ° ® J / 2  (t  W ) / J , if t  >W  J / 2, W @, ° ¯ J / 2  (t  W ) / J , if t  >W , W  J / 2@,

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 / 2@, ° h1 (t , J ) ® J / 2  (t  W ) / J , if t  >W  J / 2, W @, ° ¯ J / 2  (t  W ) / J , if t  >W , W  J / 2@,

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.

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 155

h (t ) U 0 ,

i.e. J 1 (h, h)

G 2J

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 )

(4)

0.

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 )

Let the function

A (t ) / A(t ), Q(t )

( B(t )  C (t )) / A(t ) .

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 ) 156

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 dt1

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 h1 (t )

Since

­h 0 (t ), if t  >t0 ,W @, ® if t  >W , t1 @. ¯0,

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 , that t1

t1

 ³ C (t )h 2 dt

³ 2C (t )h(t )h(t )dt .

t0

t0

Therefore, the functional J 1 ( h, h ) can be written as J1 (h, h)

³ >A(t )h

t1

2

@

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

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 ) 157

@





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

(6)

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

h(t , J )

­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. ¯

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

G 2J

t1

³ F (h(t , J ), h(t , J ), t )dt 1

t0 t1

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

t0

W

W H 1

t0

W J

 ³ F1 (h1 (t ), h1 (t ), t )dt

t1

³ F (h(t , J ), h(t , J ), t )dt  1

t0

³ F (h(t , J ), h(t , J ), t )dt  1

W



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

1

1



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 ) 1 h 0 (W  J )

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 158

(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 ),, xn (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 )

By introducing the notation U

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

1

n

0

problem (8), (9) can be written as 159

1

0

x0 , x (t1 )

`

x1 , x1  E n ,

(9)

t1

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

J ( x, x )

t0

Theorem 3. In order to the vector-function x 0 (t )  U deliver a weak local minimum to the functional J ( x, x ) on U, it is necessary that it be a solution of Euler equation Fxi ( x 0 , x 0 , t ) 

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

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 GJ

t1 n

n

¦GJ i

ª

³ ¦ «¬ F

i 1

t0 i 1

xi



d º Fx hi (t )dt dt i »¼

0,

where GJ i

t1

ª

³ «¬ F

xi

( x 0 , x 0 , t ) 

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, that 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 (t1 )

x1 .

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

J ( x, x )

³ F ( x(t ), x (t ), x(t ),, x

t0

160

n

(t ), t )dt o inf,

(12)

at conditions U

^x(t )  C >t , t @

x (t1 )

n

0

x1 , x (t1 )

1

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

x12 ,  , x n 1 (t1 )

x11 , x(t1 )

(13)

`

x1n 1 .

Theorem 4. In order to the function x 0 (t )  U 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

 (1) n

dn F n ( x 0 ,, t ) dt n x

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 )  h ( n 1) (t1 )

0,

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 .  x x « n ³t ¬ dt 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 )

t  >t 0 ,t1 @;

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

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

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

x 0 n 1 , x 0 t1

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.

161

x1n 1 .

x1 , x t1

x11 ,....,

LECTURE 24. 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 - Erdmann in the break points of extremals are mentioned briefly. Isoperimetric problem. By isoperimetric is called the following problem: minimize the functional t1

J ( x, x )

at conditions x (t )  U

^

³ F ( x(t ), x (t ), t )dt o inf

(1)

t0

x (t )  C 1 >t 0 , t1 @ x (t 0 )

x 0 , x (t 1 )

x1 ,

t1

³ G( x(t ), x(t ), t )dt

t0

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 @, 162

h2 (t , H 1 ) { 0, t  >W  H 1 ,W  H 1 @ .

Then the increment of the functional (1) is equal to T H1

0

x





ª

³ «F (x T H ¬

'J

1

W H 1

³ F (x W H x



, x 0 , t ) 

0

, x 0 , t ) 

1

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

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 W ¬ ¼t T ¬

(5)

where V1

T H1

T H1

T H 1

T H 1

³ h1 (t , H 1 )dt , V 2

³ h (t , H 2

1

)dt .

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 ,

that 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 º ª  «G x ( x 0 , x 0 , t )  G x ( x 0 , x 0 , t )» V 2  o1 (H 1 ). (6) dt ¼t W ¬

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

163

(7)

By substituting the value V 1 of (7) to the right side of (5), we obtain ­ª d d º ª º ½ ®« Fx  Fx »  O «G x  G x » ¾V 2  T 3 (H 1 ), dt ¼ t W ¿ dt ¼ t W ¬ ¯¬

'J

(8)

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

O

d º ª «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 º ª 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, that the condition (4) is the second order differential equation and its solution x 0 (t , c1 , c 2 , O ), t  >t 0 , t1 @ ,

x0

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 G ( x( t ), x( t ),t )dt i

t0

½° li , i 1, m¾ . (10) °¿

Theorem 2. If the vector function x (t )  U delivers a weak local minimum to the functional (9) at conditions (10) , then 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: 0

164

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

(11)

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

called by Lagrangian is given by formula m

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

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

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

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

x0

c1 , ,c2 n , O1 , ,O m

are determined from the condition x 0 ( t0 )

x0 ,

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

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

J ( x, y )

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

(12)

t0

at conditions

x(t ), y (t )  U

^x(t ), y(t )  C >t , t @ x(t 1

0

x(t1 ) x1 , y (t1 )

1

0

) 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 165

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

0.

(14)

(15)

Proof. Let the function

x where

0

( x (t ), y (t )) (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 º ª 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 º ª  « 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, V1

W  H1

³ h (t , H )dt , 1

1

W H 1

V2

W H1

³ h (t , H 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., V1

k1H 1 , V 2

k 2H 1 , V 1 ~ H 1 , V 2 ~ H 1 .

Since ( x(t ), y (t ))  U , ( x 0 (t ), y 0 (t ))  U , 166

that 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 '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

J ( x, u )

(18)

t0

at conditions x

f ( x, u, t ), t  >t 0 , t1 @ ,

) 0 ( x(t 0 ))

where x

0, ) 1 ( x(t1 ))

0,

(19) (20)

( 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 )

u )0

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 167

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

³ F ( x(t ), u(t ), t )dt o inf,

J ( x, u )

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

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 to the pair ( x 0 (t ), u 0 (t ))  C 1 >t0 , t1 @u C >t0 , t1 @

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

168

0.

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

(21) It is shown, that despite of the fact that the function x (t ) is sought in the class C 1 >t 0 ,t1 @ , Euler equation is reduced to defining of the function  Fxx ( x 0 , x 0 , t ) x 0  Fxx ( x 0 , x 0 , t ) x0

0.

0

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 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.

169

Chapter VI 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 25. 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) 170

x1 , x (T ) v1 ,

x (T )

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 T

J ( x1 , x2 , u )

³1˜ dt

T o inf

(1)

0

at conditions 1 u (t ), | u (t ) |d a m x0 , x1 (0) v0 , x1 (T ) x1 , x1 (T ) v1 .

x1 x1 (0)

x2 , x 2

(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 ɭ, 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) 171

v0 .

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

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

0 , x 2 (T )

0.

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

J ( x1 , x2 , u )

³1˜ dt

(4)

T o inf

0

at conditions x1 x1 (0)

x2 , x 2

(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 ɭ , 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

J ( x1 , x2 , u )

2 1

@

(7)

(t )  x22 (t ) dt o inf

0

at conditions x1 (0)

x1 x2 , x 2 kx2  sin x1  u (t ), x0 , x2 (0) v0 ; x1 (t1 ) x1 , x2 (t1 ) u (t ) d ɭ , t  >t 0 , t1 @.

v1 .

(8) (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 @ ,

172

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

x(t 0 )  S 0 (t 0 )  E n ,

t0 T 0  E1;

x(t1 )  S1 (t1 )  E n ,

t1  T 1  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 173

S 0 (t 0 ) Ž G (t 0 ), S1 (t1 ) Ž 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 t1

J ( x, u )

³f

0

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

t0

at conditions 174

(17)

x (t )

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

x 0 , t 0 d t d t1 ,

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

(18) (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 sup H ( x 0 (t ), u , t ,\ (t )) uV

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

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

\ (t1 )

 w) ( x (t1 )) . wx 0

(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): 175

H ( x 0 , u , t ,\ ) o sup, u  V

it is supposed, that u

u ( x ,\ , t ), t  >t 0 , t1 @ .

0

0

( x 0 ,\ , t )

,

are parameters. As result, we find

0

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

x

x0 , t 0 d t d t1 ,

u (t )  E , t  >t 0 , t1 @, r

u (t )  Ɇɍ >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 ,\ )



1 xcQ (t ) x  u cR (t )u  \ , A(t ) x  B (t )u . 2

2. Since V

E r , the solution of the optimization problem H ( x , u , t ,\ ) o sup , u E r is reduced to 0

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

0,

w 2 H ( x 0 , u 0 , t ,\ )  0. wu 2

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

From the first condition we get 176

0,

u0

R 1 (t ) B * (t )\ ,

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 ) \ (t1 )

We assume, that

H x ) x

x0 ,

(24)

t 0 d t d t1 ,

Qx 0  A* (t )\ ,

(25)

 Fx 0 (t1 ), t 0 d t d t1 .

\ (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 >t 0 ,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. 177

LECTURE 26. 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 @

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 .

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

178

(30)

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

Le L (t1 t0 ) , t 0 d t d t1 .

(32)

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

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 , (34)

R1

) x ( x0 (t1 )  T 'x(t1 ))  ) x ( x0 (t1 )), 'x(t1 ) . 179

Since R1 d ) x ( x 0 (t1 )  T 'x(t1 ))  ) x ( x 0 (t1 )) 'x(t1 ) ,

that 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

(35)

2 1

We consider the second term of expression (34). By condition of the theorem (23) \ (t1 ) ) x ( x 0 (t1 )) , therefore t1

d \ (t ), 'x(t ) dt , dt t0

³

 \ (t1 ), 'x(t1 )

) x ( x 0 (t1 ), 'x(t1 )

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

 ³ \ (t ), 'x(t ) dt  ³ \ (t ), 'x (t ) dt t0

t0

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, that 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 , 180

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 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

t1

u 'x(t ) dt  max \ (t , u 0 ) ³ f x ( x 0  T 'x, u 0  h, t )  t 0 dt dt1

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. (38) © t0 dt dt1 ¹ t0

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

t1

³

t0

2 1

2

§ t1 · 'x (t ) h(t ) dt d c1 ¨ ³ h(t ) dt ¸ . ¨t ¸ ©0 ¹

Now inequality (38) can be written as 2

§ t1 · R2 d c 2 ¨ ³ h(t ) dt ¸ , ¨t ¸ ©0 ¹ c2

>

@

L§¨1  max \ (t , u 0 ) ·¸ c12 (t1  t 0 )  c1 . © t0 dt dt1 ¹

181

(39)

From inequalities (35) and (39) we get § t1 · R1  R2 d R1  R2 d ( Lc  c2 )¨ ³ h(t ) dt ¸ ¨t ¸ ©0 ¹

2

2

§ t1 · c3 ¨ ³ h(t ) dt ¸ . ¨t ¸ ©0 ¹

2 1

R

where We note, that since

R1  R2 .

Lc12  c2 , R

c3

( 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,

(40)

t0

where 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 ·  ³ g (t )dt  R t 0, R d c3 ¨¨ ³ h(t ) dt ¸¸ . W ©W ¹ W H

'J

(42)

According to the average value theorem the integral W H

³ g (t )dt

W

182

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. (44) ¨ ¸ W © W ¹ 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, that 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. 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, that 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 ,\ ) vV

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, Optimal control by differential and functional equations. M.; Nauka, 1977; Gabasov R., Kirillova F.M. Maxsimum principle in the theory of optimal control. Minsk: Nauka and Technika, 1974; Gamkrelidze R.V. Basics of optimal control. 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.

183

LECTURE 27. 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

(1)

f ( x (t ), u (t ), t ), t 0 d t d t1 , t 0 , t1  ',

(2) (3) (4)

t0

at conditions x (t )

g i ( x, u, t 0 , t1 ) d 0, i 1, m, g i ( x, u, t 0 , t1 ) 0, i

m  1, s,

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

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; x(t )

( x1 (t ),..., x n (t ))  KC 1 >t 0 , t1 @

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 , f0x , )0x

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 )

184

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 / ( 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 L( x, x, u, t ,\ , O )

s

¦ O f ( x, u, t )  \ (t ), x  f ( x, u, t ) 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 ).

i 0

Differential equation (9) defines the adjoint system. 185

(9)

3. The boundary conditions for adjoint system (9) are defined by conditions s

¦O )

\ (t 00 )

i

i 0

ix0

( x00 , x10 , t 00 , t10 ),

s

(10)

¦ Oi ) ix1 ( x , x , t , t ),

\ (t ) 0 1

0 0

0 1

0 0

0 1

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. 4. Optimal control u 0 ( x 0 ,\ , t )

u0

is defined by condition

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

max H ( x 0 , u, t ,\ ) uV

(11)

where ɉ 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 ) 

s

¦ O f ( x(t i

i 0

i

0 1

s

 x 0 (t 00 ), ) ix0 ( x00 , x10 , t 00 , t10 )



0; (12)

 x 0 (t10 ), ) ix1 ( x00 , x10 , t 00 , t10 )



0. (13)



), u(t10 ), t10 )  ¦ Oi ) it1 ( x00 , x10 , t 00 , t10 )  i 0

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.

Usually it is considered two cases separately: 1) O0 0 (degenerated problems); 186

(14)

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. Example 1. Minimize the functional 4

J ( x, u )

³ (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:





O0 u 0  O0 x 0 \u 0



 O0 u 0 \ u 0 .

max  O0 u 2  O0 x \u

1du d 1

or



min O0 u 2 \u

1du d 1

187

2

2

It is undenstandable, at

O0

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 u0

­ \ °sign\ , 2 ! 1, ° ® \ °\ , d 1. °¯ 2 2

Then 0 d t d 2, ­ t , ° 2 ®t °  2t  1, 2 d t d 4; ¯4

0

x (t )

\ (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 >0,1@ ,  1 d u d 1

For this problem f )1 )3

( x2 , u ) , f 0

x1 (0)  x0 , ) 2 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 188

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 )

O 4 .

4 0. Condition (11) is written as follows: function H

1  \ 1 x 2  \ 2 u ,

max[\ 2 u ]

1d u d1

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 . 5 0. 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 )

u0

sign\ 2 ,

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 x10 x20 , x 20 1 at u 0 1 , we get x20 (t ) t  c2 , x10

(t  c2 ) 2 . 2  c3

We define c3 , c2 from the conditions x1 (0)

x0 , x 2 (0)

v0 , x1 (T )

x2 (T ) 0 .

These conditions are satisfied, if there is a connection between the values x0 , v0 : 189

M ( x 0 ) , M ( x0 )  2 x 0 ,

v0

if x0 t 0, M ( x0 )

 2x0 ,

if x0 d 0 . In the case 1 , x 20 (t )

u0

T  t,

v0  (T  t ) , 2 2 2 2

x10 (t )

x0 .

We consider the case O0 1 . In this case \ 2 (T ) 1 . Then from 2 0 we have \ 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:

or

­  1, 0 d t d W , u 0 (t ) ® ¯ 1, W d t d T , ­ 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 x 10 (0)

u 0 (t )

x0 , x 10 (0) v0 , x10 (T )

x20 (T ) 0

for indicated values u0 (t ), 0 d t d T we define W and Ɏ. 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

J ( x, y )

³f

0

( x(t ), u (t ), t )dt o inf ( f 0

F)

(15)

t0

at conditions x (t )

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

where t 0 , t1 , x0 , x1 are fixed numbers. 190

x 0 , x (t1 )

x1 , u  E 1 ,

(16)

Pontryagin function O0 f 0 ( x, u, t ) \ 1u .

H

Then the adjoint system \ 1

H x

O0 f 0 x ( x 0 , u 0 , t ) .

(17)

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

0.

The value O0 z 0 , in the opposite case \ 1 (t ) 0 , t  >t 0 ,t1 @ .

We accept O0 1 . Then the adjoint system (17) can be written as \ 1 (t )

f 0 x ( x 0 , u 0 , t ), \ 1 (t )

wf 0 ( x 0 (t ), u 0 (t ), t ) . wu

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 ɉ 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 ,\ ) vE

H ( x 0 (t ), u 0 (t ), t ,\ ), t  >t 0 ,t1 @ . 191

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'a maximum principle. 6. Represent the Legendre condition. 7. Represent the Weierstrass condition.

192

Chapter VII 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 28. 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

³f

J ( x, u )

0

( x(t ), u (t ), t )dt  ɐ( 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 , t 0 d t d t1 , r

(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 ), ɐ( 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  ɐ( x(t1 )) o inf,

(4)

t

x (W ) f ( x(W ), u (W ),W ), t d W d t1 , x(t ) x, u (W )  KC >t , t1 @, u (W )  V  E r , t d W d t1 . 193

(5) (6)

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 (7) B( x, t ) 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)

B( x, t ) d B( x( s ), s)  ³ f 0 ( x(W , u ), u (W ),W )dW .

(9)

t

s

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 )

½° ­° 1 inf0 ®³ f 0 ( x(W ), u (W ),W )dW  ɐ( x(t1 ))¾. x( s) x ( s)° °¿ ¯s u (W )V t

194

We consider the inequality (9). On the time interval W  >t, s@ is valid control

u (W ), W  >t , s @ differented from optimal control

u 0 (W ), W  >t , s @ ,

and control u (W ), W  >t , s @ creates a state x( s ) x( s, u ) , the value ½° ­°t1 ® f 0 ( x(W ), u (W ),W )dW  ɐ( 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

³f

B ( x, t ) d B( x(t  't , u ), t  't ) 

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 ɱ, then x 0 (t , u 0 )

x(t , u )

x, W  >t , t  't @ .

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 ),

B x ( 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

195

(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. uV

(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 ) ɐ( 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. 196

30 . We substitute the value 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 )

ɐ( 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 u0

u 0 (M x ( x 0 , t ), x 0 , t )

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 u 0 (t ), t  >t 0 , t1 @ found by the maximum principle. Example. We consider a linear system with a quadratic functional: to minimize the functional J ( x, u ) t

1 11 [ x c(t )Q(t ) x(t )  u c(t ) R(t )u (t )]dt  x c(t1 ) Fx(t1 ) o inf ³ 2 t0 2

(16)

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 , t 0 d t d t1 , r

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

197

(17) (18)

b) Since the set V { E r , that the control u 0 u 0 ( B x ( x, t ), x, t ) is determined by the conditions wK wu

We calculate the derivatives

wK wu

0,

w2K t 0. wu 2

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 )  wx wt 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 198

F.

K (t )

K * (t ) , t  >t 0 ,t1 @

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 1 xcK (t ) x , t  >t 0 ,t1 @ . 2

B ( x, t )

d) We find the optimal control u 0 (x0 ,t)

The optimal trajectory

 R 1 (t ) B * (t ) K (t ) x 0 (t ), t 0 d t d t1 .

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 ,u 0 )

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 conreol is called by synthesizing? 199

LECTURE 29. 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  ɐ( x(t1 )) o inf

(1)

t0

at conditions x

x0 , t 0 d t d t1 , x0  G (t 0 ),

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

x(t )  G (t )  E , t 0 d t d t1 , n

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

(2) (3) (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

0i

are written as: minimize the

( xi , ui )  ɐ x N o inf

(5)

i 0

at conditions xi 1

Fi ( xi , ui ), i x i  Gi  E , i n

>ui @0

0,..., N  1, x0  G0 0,1,..., N , Gi

(u0 , u1 ,..., u N 1 ), ui Vi

where 200

G (t 0 ),

Gi (t i ),

V (t i ), i 0,..., N  1,

(6) (7) (8)

>xi @0

( x0 , x1 ,..., x N ), xk  E n , k

0, N , u k  E r , k

0, N  1;

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 , >u i @k

N 1

¦F

0i

( xi , ui )  ɐ x N o inf

(9)

i k

at conditions xi 1

Fi ( xi , ui ), i

k , k  1,..., N  1, xk

xi  Gi  E n , i

>ui @k

x  Gk ,

k , k  1,..., N ,

(u k , u k 1 ,..., u N 1 ), ui Vi , i

k , k  1,..., N  1,

(10) (11) (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

x, x k01 ,..., x N0

corresponds to the optimal control >ui @0k . 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

k , k  1,..., N  1,

x, xk 1 ,..., x N ), xi  Gi , i 201

k , k  1,..., N ` .

x,

(13)

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 that (14) Bk ( x ) F0 k ( x k0 , u k0 )  Bk 1 ( x k01 ), x k0 x, x k01 Fk ( x k0 , u k0 ), 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 k ), u k  Vk ,

(15)

x  Gk .

If we introduce the notation Dk ( x )

^>ui @k

uk

(u k

u V ,

u, u k 1 ,..., u N 1 ) u  Vk , >u i @k  ' k ( x)` ,

then the relations (14), (15) can be written as Bk ( x)

inf >F0 k ( x, u )  Bk 1 ( Fk ( x, u ))@, k

uDk ( x )

B N ( x)

0,1,..., N  1,

(16) (17)

ɐ( 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) ɐ( 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 ))@, (18) uDN 1 ( x )

D N 1 ( x)

^u  E

r

`

/ u  V N 1 , FN 1 ( x, u )  G N .

(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

Note, that BN ( FN 1 ( x, u )) ɐ( FN 1 ( x, u )) . 202

`

/ FN 1 ( x, u 0 ( x))  G N , x  G N 1 .

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) inf >F0 N 2 ( x, u )  BN 1 ( FN 2 ( x, u ))@, (22) uDN  2 ( x )

D N  2 ( x)

^u VN 2 /

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

u 0 ( x)

u N0  2 ( x )

and the set X N 2

^x  E

n

/ 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.

N-th stage. For the value k 0 equation (16) is written as:

203

(26)

inf >F00 ( x, u )  B1 ( F0 ( x, u ))@,

B0 ( x)

uD0 ( x )

^u V0 /

D0 ( x)

F0 ( x, u )  X 1 `

(27) (28)

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 x X0

^x  E

u 0 ( x) u00 ( x), B0 ( x) M N ( x, u 0 ( x)) M N ( x),

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

F0 ( x00 , u 00 ), u10

..., u N0 1

u N0 1 ( x N0 1 ), x N0

u10 ( x10 ), x 20

F1 ( x10 , u10 ),

FN 1 ( x N0 1 , u N0 1 ).

Example 1. Minimize the functional I 0 >xi @0 , >ui @0

1

¦(x

2 i

 ui2 )  x22 o inf

i 0

at conditions xi 1 xi  Gi

>ui @0

0

xi  ui , i 0,1, [i, i  2], i 0,1, 2,

(u 0 , u1 ), u i  E 1 , i

0,1.

1 . For k 2 B2 ( x) x , x  G2 [2;4], X 2 G2 [2;4]. 20. For the values k 1 the function 2

B1 ( x)

>

inf x 2  u 2  ( x  u ) 2

D1 ( x )

204

@

>

@

inf 2 x 2  2u 2  2 xu ,

D1 ( x )

^u  E

D1 ( x)

1

/ x  u  G2

[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, 2  x  u, g 2

( g1

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

u10

2  x, x d 4 .

Then the set X 1 [1;3] G1 , and the function B1 ( x )

2 x 2  4 x  8 M 1 ( 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 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 . 205

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)

>

@

1  x, B1 ( x) 2

inf 2 x 2  2u 2  2 xu , u10

uE1

3 2 x . 2

For the values k 0 the function 3 ª º B0 ( x) inf1 « x 2  u 2  ( x  u ) 2 ». uE ¬ 2 ¼ 3x 1 0 Consequently u 0  . Hence, at x x0 we obtain u 00 5 2 2 mum 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



3 and the mini10

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. 206

LECTURE 30. 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  ɐ0 ( x(t 0 ))  ɐ1 ( x(t1 )) o inf

(1)

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

(2) (3) (4)

t0

at conditions x (t )

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; ɐ0 ( x), ɐ1 ( x)

are known functions, all other notations are the same as in the previous lectures. 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 )

207

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 Ž G (t ) , Vt Ž 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 )  ɐ0 ( x), r1 ( x)  K ( x, t1 )  ɐ1 ( x), x  X t ,

(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 208

t1

J ( x, u )

³ R( x(t ), u(t ), t )dt  r ( x(t

))  r1 ( x(t1 )).

(7)

K x ( x(t ), t ), f ( x(t ), u (t ), t )  K t ( x(t ), t )

(8)

0

0

t0

Proof. In fact, since K (t )

along feasible pairs

K ( x(t ), t )  KC 1 >t 0 , t1 @

( x (t ), u (t ))  D>t 0 ,t1 @ ,

then the derivative dK ( x(t ), t ) dt

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. 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

where Rmin (t )

inf R ( x, u , t ), r1 min

( x ,u )Dt

In fact, 209

inf r1 ( x), r0 min

x X t1

inf r0 ( x).

x X t 0

'( x, u ), (10)

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

xG ( t ) uV ( t )

inf r1 ( x), r0 min

xG ( t1 )

inf r0 ( x) ,

xG ( t 0 )

(11)

since Dt Ž G (t ) u V (t ), X t1 Ž G (t1 ), X t0 Ž 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 to the pair ( x 0 (t ), u 0 (t ))  D>t 0 , t1 @

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

r1 ( x (t1 ))

Rmin (t ), t 0 d t d t1 ,

r1 min , r0 ( x 0 (t 0 ))

r0 min .

Proof. Since condition of lemma 1 holds, that there is a function K ( x (t ), t )  KC 1 >t 0 , t1 @

along arbitrary feasible pairs 210

(12)

( 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 ,

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. 211

lim J ( xk , u k )

J* .

inf J ( x, u )

D >t0 ,t1 @

k of

Theorem 2. Let the function K ( x(t ), t ) in the variable t be continuous and belong 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 @

be minimized sufficiently, that t1

lim ³ R( xk (t ), u k (t ), t )dt k of

t0

lim r1 ( xk (t1 )) k of

Proof. Let

t1

³R

min

(13)

(t )dt ,

t0

r1 min , lim r0 ( xk (t 0 )) 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 J ( x, u )  J ( x k , u k )

t1

³ [ R( x(t ), u(t ), t )  R( x

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 t1

³ [ R( x(t ), u(t ), t )  R

J ( x, u )  lim J ( xk , u k ) k of

min

(t )]dt 

t0

 r1 ( x(t1 ))  r1min  r0 ( x(t 0 ))  r0 min t 0.

Consequently,

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 @ 212

the function

K ( x (t ), t )  KC 1 >t 0 , t1 @ .

We show, that function K ( x , t )  J (t ) ,

K ( x, t )

where

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 )  ɐ1 ( x )

r0 ( x)  J (t 0 ),

K ( x , t 0 )  ɐ0 ( x )

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 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 )

inf R ( x, u, t ) 0 ,

( x ,u )Dt

213

J ( x, u ).

r1 min

inf r 1( x) 0 ,

xX t1

inf r0 ( x) r0 min  J (t 0 ) .

x X 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, r1 ( x )

 K ( x, t1 )  ɐ1 ( x ) t 0, x  X t1 ,

r1 ( x 0 (t1 ))  K ( x 0 (t1 ))  ɐ1 ( x 0 (t1 )) 0, r0 ( x ) { K ( x, t 0 )  ɐ0 ( x ) t r0 min , x  X t0 , K ( x 0 (t 0 ), t 0 )  ɐ0 ( x 0 (t1 ))

where

r0 min ,

(15) (16) (17) (18) (19) (20)

( 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). 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

214

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

uV ( 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 , ɐ1 ( x 0 (t1 )), x 0 (t1 )  X t1 ,

K ( x 0 (t1 ), t1 ) K ( x 0 (t 0 ), t 0 )

ɐ0 ( 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.

215

APPENDIX 1 TASKS FOR INDEPENDENT WORK P.1.1. MULTIVARIABLE FUNCTION MINIMIZATION IN THE ABSENCE OF RESTRICTIONS Statement of the problem. Let scalar function J u J u1 ,...,un

be determined in all space E n . To solve the following optimization problem: J u o inf, u  E n

The point u  E n is identified by the point of the minimum J (u ) on E n , if J u* d J u ,  u  E n . The variable J u* is identified by least or minimum value of the function J (u ) on E n . We notice that absolute (global) minimum J (u ) on E n is reached in the point u*  E n . The point u0  E n is identified by the point of the local minimum J (u ) on E n , if J u0 d J u under all u  u 0  H , H ! 0 is sufficiently small number. Usually first define the points of the local minimum and then amongst of them find the points of the global minimum. Following theorems known from the course of the mathematical analysis: Theorem 1. If function J (u )  C1 E n , then in the point u0  E n the equality J cu0 0 (necessary first-order condition) is executed. Theorem 2. If function J (u )  C 2 E n , then in the point u0  E n the following conditions: J cu 0 0, J ccu0 t 0

(necessary condition of the second order) are executed. Theorem 3. In order to the point u0  E n be a point of the local minimum to function J (u )  C 2 E n , necessary and sufficiency J cu0 0, J ccu0 ! 0 .

216

We notice, that problem J(u) o sup, u  E n is tantamount to the problem - J(u) o inf , u  E n . To find points of the local and global minimum function: 1 . J (u )

J (u1 ,u 2 ,u 3 )

u12  u 22  u 32  u1u 2  u1  2u 3 , u

(u1 , u 2 , u 3 )  E 3 .

2. J (u1 ,u2 ) u13u 22 (6  u1  u 2 ), u 3. J (u1 ,u2 ) (u1  1) 2  2u 22 , u

(u1 , u 2 )  E 2 . (u1 , u 2 )  E 2 .

4. J (u1 ,u2 ) u14  u 24  2u12  4u1u 2  2u 22 , u 5 . J (u 1 , u 2 )

(u12  u 22 )e  u1 u2 , u 2

1  u1  u 2

6. J (u1 ,u 2 )

1  u12  u12

7. J (u1 ,u 2 ,u 3 )

u1 

2

, u

(u 1 , u 2 )  E 2 .

(u1 , u 2 )  E 2 .

u 22 u 32 2   , u1 ! 0, u 2 ! 0, u 3 ! 0. 4u1 u 2 u 3

8. J (u1 ,u2 ) u12  u1u 2  u 22  2u1  u 2 , u 9. J (u1 ,u 2 )

(u1 , u 2 )  E 2 .

(u1 ,u 2 )  E 2 .

sin u1 sin u 2 sin( u1  u 2 ) , 0 d u1 , u 2 d Þ .

10 . J (u1 ,...,u n )

u1u 22 ...u nn (1  u1  2u 2  ...  nu n ) , u i t 0 , i

1,n.

P.1.2. CONVEX ENSEMBLES AND CONVEX FUNCTIONS

Let U be a certain ensemble E n , but function J u J u1 ,...,un is determined on ensemble U . The ensemble U is identified by convex, if the point İu  ( 1  İ)Ƞ  U , u,Ƞ  U and under all D , D [0,1] . Function J (u ) is convex on convex ensemble U , if J (İu  ( 1  İ)Ƞ) d İJ (u )  (1  İ) J (Ƞ), u,Ƞ  U, İ  [0,1].

(1)

Function J (u ) is strictly convex on U , if in the equality (1) possible under only D 0, D 1 . Function J (u ) is concave (strictly concave) on U , if function  J u is concave (strictly concave) on U . Function J (u ) is strictly concave on U , if 217

J (İu  ( 1  İ)Ƞ) d İJ (u )  (1  İ ) J (Ƞ)  İ (1  İ )ȝ|u  Ƞ|2 , ȝ ! 0, u,Ƞ  U, İ  [0 ,1]

(2)

n Theorem 1. Let U be a convex ensemble of E . Then in order to the function J u  C1 U be a convex on U necessary and sufficiency to execute one of the following inequalities:

J (u )  J (Q ) t J cv , u  v , u ,Q  U ,

(3)

or J'(u)-J'(Ƞ),u-Ƞ t 0 ,u,Ƞ  U.

(4)

If int U z ‡ ɤ J(u) C 2 (U ) , then for convexity of J (u ) on U necessary and sufficiency that J''(u) [ , [ t 0, [  E n ,

u  U .

Theorem 2. In order to the function J(u) C1(U) be strictly convex on convex ensemble U necessary and sufficiency execution one of the following two conditions: 1) J (u )  J v t J' v , u-Ƞ  ȝ|u  Ƞ|2 , ȝ ! 0 , u,Ƞ  U; 2)

J cu  J cv , u  v t P u  v , P 2

2N ! 0, u , v  U .

(5) (6)

If int U z ‡ and J(u) C 2 (U ) , then for strictly convexity of J (u ) on U necessary and sufficiency that J''(u) [ , [ t P [ 2 , P

2N ! 0, u ,Q  U .

To solve the following problems on base of the determination (1), (2) and correlations (3) - (6): 1. To prove that intersection of any number convex ensembles are convex. Is this statement for unions of ensembles faithfully? To prove that closing of the convex ensemble is convex. 2. a) Is ensemble U  E n convex, if for any points u,Q U the point (u  Q ) / 2  U ? b) Is closed ensemble U convex (under performing of the previous condition)? 3. Let u0  E n , but number r  E1 , r ! 0 . Is ensemble V ^u  E n / u  u 0 d r ` \ ^u 0 ` (the ball without the centre) convex? 218

4. To show the convexity of the following function on E 1 : a) J (u )

e u ; ɝ ) J (u )

ɟ) J (u )

u;

­a(u-c) , a ! 0, u d c, ® ¯b(u-c) , b ! 0, u d c;

ɞ) J(u)

u d c, ­0, ® ¯a(u-c) , u t c, a ! 0.

5. To prove that function J u 1 / u is strictly convex under u ! 0 and strictly concave under u  0 [using only determinations to strict convexity (the concavity)]. 6. Let function M v is continuous and M v t 0,  f  v  f . Then function f

³ (Q  u )M (Q )dQ

J (u )

u

is convex on E 1 . To prove. 7. Let J (u1 , u 2 )

f

1 ([  u1 )M ([ )d[ , u 2 u³1

^u

convex function on ensemble U 8. To prove that J (u)

u 2 ! 0, M ([ ) t 0 ,  f  [  f .

Is J u1 , u 2

(u1 ,u2 )/u1  E 1 , u 2 ! 0 )`?

n

¦D J (u) – a convex function on convex ensemble U i

i

of

i 1

E n , if negative coefficients D i correspond to the concave functions J i u , but positive D i – convex functions J i u . 9. To show, if J (u ) is convex and ensemble of values u satisfying to condition J (u ) b, b, b  E 1 is convex that J (u ) certainly is a linear function. 10. Function J (u ) is convex on U , if and only if function g (D ) J (u  D (Q  u)) by one variable D , 0 d D d 1 is convex under any u,Q U . If J (u ) is strictly convex on U , that g (D ), 0 d D d 1 is strictly convex. To prove. 11. Function J (u ) determined on convex ensemble U from E n is identified quasiconvex, if J (Du  (1  D )Q ) d max ^J (u ), J (Q )`, u ,Q  U

and under all D , 0 d D d 1. Is any convex function quasiconvex, conversely? To prove that J (u ) is quasiconvex on U if and only if ensemble M (Q ) ^u  U J(u) d J (Ƞ)` is convex under all Q  U . 12. To check in each of the following exercises whether the function J (u ) convex (concave) on given ensemble U, or indicate such points from U in the neighborhood of which J (u ) is not neither convex, nor concave: a ) J (u )

u16  u 22  u 32  u 42  10u1  5u 2  3u 4  20, U { E 4 ;

219

ɝ ) J (u ) e

ɞ) J (u )

2u1u2

u15  0,5u32  7u1  u3  6, U

ɟ ) J (u )

, u  E2 ;

^u

`

(u1 , u 2 , u3 )  E 3 ui d 0, i 1,2,3 ;

6u12  u 23  6u32  12u1  8u 2  7, U

^u  E 3 / u t 0`.

13. Let J (u )

Au  b

2

Au  b, Au  b ,

where A is the matrix of the order m u n, b  Ɂ m - the vector. To prove that J (u ) is convex on E n . If A*A is nondegenerate, that J (u ) is strictly convex on E n . To find J cu , J ccu . * 14. Let J(u) 0 ,5 Au,u  b, u , where A A is the matrix of the order A* t 0 , that J u is convex on convex ensemble U from E n ; 2) if ȼ=ȼ*>0, that J (u ) is strictly convex on U, moreover N O1 / 2, where O1 ! 0 is the least proper number of the matrix ȼ . 15. To prove that ensemble ȼ is convex iff, when for any numbers O t 0; P t 0 the equality O  P A OA  PA is executed. 16. To prove that if A1 ,..., Am is a convex ensemble, then

n u n, b  E n is the vector. To prove that: 1) if A

§m · Co¨¨  Ai ¸¸ ©i 1 ¹

­ n ®u  E u ¯

n

n

D i u i , D i t 0, ¦ D i ¦ i 1 i 1

½ 1¾. ¿

17. Let J (u ) be a continuous function on convex ensemble U, moreover for any u,Q U the inequality § u Q · J¨ ¸ d >J (u)  J (Q )@ / 2. © 2 ¹

is executed. To prove that J (u ) is convex on U. 18. To realize, under what values of the parameter D the following functions are convex: a) J (u1 ,u2 ) D u12 u 22  (u1  u 2 ) 4 ; b) J (u) D u12 u 22  (u12  u 22 ) 2 . 220

19. To find on plane of the parameters D , E areas, where function J (u1 ,u2 ) u1İ u 2ȗ , u1 ! 0, u 2 ! 0

is convex (strictly convex) and concave (strictly concave). 20. To find on plane E2 areas, where function J (u1 , u 2 ) e u1u2 is convex, and areas in which it is concave. 21. Let J i (u ), i 1,m be convex, nonnegative, monotonous increasing on E1 functions. To prove that function J (u)

m

J i (u ). possesses by these characteristics. – i 1

22. Let J (u ) be a function determined and convex on convex ensemble U. To prove that: a) ensemble ȿ

­Q  E n / ½ sup > Q , u  J (u)@¾ ® uU ¯ ¿

is inempty and convex; b) function J * (u )

sup > Q , u  J (u )@

uU

is identified by conjugate to J (u ) . Will J (u ) convex on U? 23. Let function J (u ) be determined and convex on convex ensemble U from E n . To prove that for any (including for border) points Q U the inequality lim J (u ) d J (Q ) (the semicontinuity property from below) is executed. u oQ

24. Let function J (u ) be determined and convex on convex closed bounded ensemble U. Is it possible to confirm that: a) J (u ) is upper lower and bounded on U; b) J u reaches upper and lower border on ensemble U? 25. If J (u ) is convex (strictly convex) function in E n and matrix A z 0 of the order n u n, b  E n , that J  Au  b is also convex (strictly convex). To prove. P.1.3. CONVEX PROGRAMMING. KUNA-TUCKER’S THEOREM

We consider the next problem of the convex programming: J (u) o inf

at condition 221

(1)

u U

^u  E n

g i (u )

u  U 0 ; g i (u ) d 0, i

ai , u  bi d 0, i g i (u )

1, m;

m  1, p;

ai , u  bi

0, i

`

p  1, s ,

(2)

where J (u ), g i (u ), i 1, m are convex functions determined on convex ensemble U 0 from E n ; i m  1, s, ai  E n are the vectors; bi , i m  1, s are the numbers. Theorem 1 (the sufficient existence conditions of the saddle point). Let J (u ), g i (u ), i 1, m be convex functions determined on convex ensemble U 0 the ensemble U* z ‡ and let the point u  riU 0 ˆ U exists such that g i (u )  0, i 1, m. Then u* U* for each point Lagrange coefficients O* (O1* ,...O*s )  / 0 ^O  E s / O1 t 0,...O p t 0`, exist such that pair u* , O* U 0 u / 0 forms saddle point to Lagrange’s function L (u, O )

s

J (u)  ¦ Oi g i (u), u  U 0 , O  / 0 ,

(3)

i 1

on ensemble U 0 u / 0 , i.e. the inequality L (u* , O ) d L (u* , O* ) d L (u , O* ), u  U 0 , O  / 0

(4)

is executed. Lemma. In order to pair u* , O* U 0 u / 0 be saddle point of the Lagrange’s function (3), necessary and sufficiency performing the following conditions: L (u* ,O* ) d L (u,Ȟ* ), u  U 0 ,

O*i g i (u* ) 0, i 1, s, u*  U *  U , O*  / 0 ,

(5)

i.e. inequalities (4) is tantamount to the correlations (5). Theorem 2 (the sufficient optimality conditions). If pair u* , O* U 0 u / 0 is saddle point to Lagrange’s function (3), then vector u* U* is solution of the problem (1), (2). Problem algorithm of the convex programming based on the theorems 1,2 and lemma are provided in the lectures 10. We illustrate solution rule of the convex programming problem in the following example. Example. To maximize the function  8u12  10u22  12u1u2  50u1  80u2 o sup 222

(6)

at condition u1  u 2

1, 8u12  u 22 d 2, u1 t 0, u 2 t 0.

(7)

Solution. Problem (6), (7) is tantamount to the problem J (u) 8u12  10u22  12u1u2  50u1  80u2 o inf

(8)

at condition u1  u 2

1, 8u12  u 22 d 2, u1 t 0, u 2 t 0.

(9)

J (u) 8u12  10u 22  12u1u 2  50u1  80u 2 o inf,

(10)

Problem (8), (9) possible to write as

u U

^u u , u  E 1

2

2

/u U 0 ,g1 (u) 8u12  u 22  2 d 0,

^u u , u  E

g 2 (u) u1  u 2  1 0`, U 0

1

2

2

/ u1 t 0, u2 t 0`.

(11)

We notice that ensemble U 0  E 2 is convex, function J (u ) is convex on ensemble U 0 , since symmetrical matrix J ' ' (u )

§16 - 12 · ¸¸ ! 0, ¨¨ © - 12 20 ¹

J ' ' (u )[ , [ ! 0, [  E 2 , u  U 0 .

It is easy to check that functions g1 (u ), g 2 (u ) are convex on ensemble U 0 . Finally, problem (10), (11) is a problem of the convex programming. Entering ensembles U1

^u  E

2

`

/ g1 (u) d 0 , U 2

^u  E

2

`

/ g 2 (u) 0 ,

problem (10), (11) possible to write as J (u) o inf, u  U

U 0 ˆ U1 ˆ U 2 .

We notice that ensembles U 0 , U1 , U 2 are convex, consequently, ensemble U is convex. Further we solve problem (10), (11) on algorithm provided in the lecture 10. 10. We are convinced of ensemble U*

^u U J (u ) *

*

223

`

min J (u ) z ‡. uU

In fact, ensembles U 0 , U 1 , U 2 are convex and closed, moreover ensemble U is bounded, consequently, ensemble U is bounded and closed, i.e. U – a compact ensemble in En. As follows from correlation (10) function J (u ) is continuous (semicontinuous from below) on compact ensemble U. Then according to the theorem 1 (lecture 2) ensemble U * z ‡. Now problem (10), (11) possible to write as J (u) o min,u U  E 2 .

20. We show that Sleytera’s condition is executed. In fact, the point u

(0,1)  riU 0 ˆ U , (aff U 0

U 0 , aff U 0 ˆ U 0

U0 ) ,

moreover g1 (u ) 1  0 . Consequently, by theorem 1 Lagrange’s function for problem (10), (11) L (u , O )

8u12  10u 22  12u1u 2  50u1  80u 2   O1 (8u12  u 22  2)  O 2 (u1  u 2  1),

u

(u1 , u 2 )  U 0 , Ȟ

( Ȟ1 ,Ȟ2 )  / 0

^O  E

2

/ O1 t 0`

(12)

has saddle point. 30. Lagrange’s function has the form (12), area of its determination U 0 u / 0, , moreover O1 t 0, but O 2 can be as positive as negative. 40. We define saddle point u* , O* U 0 u / 0 to Lagrange’s function (12) on base of (5). From inequality









L u* , O* d L u, O* , u  U 0

follows that convex function





L u* , O* , u  U 0 ( O* is a fixed vector) reaches the least value on convex ensemble U 0

in the point u* . Then according to optimality criterion (lecture 5, theorem 4) necessary and sufficiently executing of the inequality





(13)

L u u * , O * , u  u * t 0,  u  U 0 ,

where L (u* , O* )

O*

§16u1*  12u 2*  50  16O1*u1*  O*2* · ¸, u ¨ ¨ 20u *  12u *  80  2O*u *  O* ¸ * 2 1 1 2 2 ¹ ©

(O1* , O*2 )

224

(u1* , u 2* ),

Conditions O*i g i (u* ) 0, i 1,2 (the conditions of the complementing elasticity) are written:



2

2

O1* 8u1*  u 2*  2



0, O*2 (u1*  u 2*  1)

0, O1* t 0.

(14)

a) We suppose that u*  intU 0 . In this case from inequality (13) we have L u (u* , O* ) 0. Saddle point is defined from conditions 16u1*  12u 2*  50  16O1*u1*  O*2 20u  12u  80  2O u  O * 2

* 1

2

* * 1 2

2

O1* (8u1*  u 2*  2)

* 2

0, 0,

0, u1*  u 2*  1

0, O*2 z 0, O1* t 0.

(15)

We notice, that from the second equation (14) and u *  U follows that u1*  u2*  1 0, O*2 z 0 . We solve the system of the algebraic equations (15). It is

possible several events: 1) O1* ! 0, O*2 z 0. In this case the point u* is defined from solution of the system * * * 8u1*  u 2* 2, u1*  u 2* 1 . Thence we have u1 0,47, u2 0,53. Then value O1 is solution of the equation 126,2  6,46O1* 0 . Thence we have 2

2

O1*



126,2  0. 6,46

It is impossible, since O1* ! 0. 2) O1* 0, O*2 z 0. The point u* is defined from equations u1*  u2* 1 , 20u2*  12u1*  80  O*2 0 . 16u1*  12u2*  50  O*2 0 , Thence we have * * u2 158 / 60, u1 98 / 60  0. The point u*  U 0 . So the case is excluded. Thereby, conditions (13), (14) are not executed in the internal points of the ensemble U 0 . It remains to consider the border points of the ensemble U 0 . b) Since the point u *  U , that remains to check the conditions (13), (14) in the border points (1,0), (0,1) of the ensemble U 0 . We notice that border point (1,0)  U , since in the point restriction 8u12  u22  2 d 0. is not executed. Then the conditions (13), (14) follows to check in the singular point u* (0,1)  U 0 . In the point g1 (u* ) 1  0, consequently, value O1* 0. and equalities (14) are executed. Since derivative L u (u* , O* ) (38  O*2 ,60  O*2 ), that inequality (13) is written as * * * (38  O2 )u1  (60  O2 )(u2  1) t 0 under all u1 t 0, u 2 t 0 . We choose O2 60 , then the inequality to write so: 98u1 t 0, u1 t 0, u 2 t 0 . Finally, the point u* (u1* 0, u2* 1) is the saddle point to Lagrange’s function (12). 50. Problem (10), (11) has the solutions u1* 0, u 2* 1, J (u * ) 70 . 225

To solve the following problems: 1. J (u) u12  u 22  6 o sup; u1  u 2 d 5, u1 t 0. 2. J (u) 2u12  2u1  4u2  3u3  8 o sup; 8u1  3u2  3u3

40,  2u1  u2  u3

3, u2 t 0.

3. J (u ) 5u12  u 22  4u1u 2  5u1  4u 2  3 o sup, u12  u22  2u1  2u2 d 4, u1 t 0, u2 t 0 . 4.

J (u )

 3u 22  11u1  3u 2  u 3  27 o sup,

u1  7u2  3u3 d 7, 5u  2u2  u3 d 2.

5. J (u) u1  u2  u3  30u4  8u5  56u6 o sup, 2u1  3u 2  u 3  u 4  u 5  10u 6 d 20,

u1  2u 2  3u 3  u 4  u 5  7u 6

11,

x1 t 0, x 2 t 0, x3 t 0.

6. The following problem: J (u )

c, u  0,5 Du , u o sup

at conditions Au d b, u t 0 , where D D* t 0, is identified by the problem of the quadratic programming. To define the solutions of the problem. 7. To define the least distance from origin of coordinates till ensemble u1  u 2 t 4, 2u1  u 2 t 5 .

8. To formulate Kuna-Tucker’s conditions and dual problem for the following problem: J (u)

4u12  4u1u2  2u22  3u1  eu12u2 o inf

at condition u1  2u 2

0, u12  u 22 d 10,

u 2 d 1 / 2, u 2 t 0

9. To find J (u

pu12  qu1u 2 o sup)

226

at condition u1  ru22 d 1, u1 t 0, u 2 t 0 , where - p, q, r - parameters. Under what values of the parameters p, q, r solution exists? 10. To solve the geometric problem: J (u) (u1  3) 2  u 22 o sup

at condition  u1  (u 2  1) 2 t 0, u1 t 0, u 2 t 0 . Are Kuna–Tucker’s conditions satisfied in the optimum point? P.1.4. NONLINEAR PROGRAMMING

We consider the next problem of the nonlinear programming:

U

^u  E

n

/ u  U 0 , g i (u ) d 0, i

1, m; g i (u )

J (u) o inf,

(1)

u U ,

(2)

`

m  1, s ,

0, i

where function J (u )  C 1 (U 01 ), g i (u )  C 1 (U 01 ), i

1, s , U 01

– open ensemble contained the convex ensembles U 0 from E n , in particular, U 01 E n . Theorem 1 (the necessary optimality condition). If the functions J (u )  C 1 ( E n ), g i (u )  C 1 ( E n ), i

1, s, int U 0 z ‡,

is a convex ensemble, and ensemble U * z ‡ , then for each point u*  U necessary exist the Lagrange’s coefficients U0

O

*

^O  E

(O*0 , O1* ,..., O*s )  / 0

s 1

`,

/ O 0 t 0, O1 t 0,..., O m t 0

such the following conditions are executed: *

O z 0, O*0 t 0, O1* t 0, ..., O*m t 0,

(3)

*

L u (u * , O )u  u * S

O*0 J c(u* )  ¦ O*i g ic (u* ), u  u* t 0, u  U 0 , i 1

227

(4)

O*i g i (u * )

0, i

1, s, u *  U .

(5)

Lagrange’s function for problem (1), (2) has the form L (u, O )

S

O0 J (u )  ¦ Oi g i (u ), u  U 0 , O

(O0 , O1 ,..., Os )  / 0

i 1

^O  E

S 1

`

/ O0 t 0, O1 t 0,..., Om t 0 .

At solution of the problem (1), (2) necessary to consider separately two events: 1) O*0 0 (degenerate problem); 2) O*0 ! 0 (nondegenerate problem). In this case possible to take O*0 1 . We suppose, that the points

u , O *

* 0

t 0, O*

(O1* t 0,..., O*m t 0, O*m1 ,..., O*s )

from conditions (3), (4), (5) are found. The point u*  U is identified by normal, if the vectors g ic (u* ), i  I , g cm1 (u* ),..., g cs (u* ) are linear independent, where ensemble I ^i  ^1,..., m`/ g i (u* ) 0`. If u*  U – a normal point, that problem (1), (2) is nondegenerate, i.e. O*0 1. In the event U 0 E n is executed the following theorem. Theorem 2. Let functions J (u ), g i (u ), i 1, s, be definite, continuous and twice continuously differentiable in the neighborhood of the normal point u*  U . In order to the normal point u*  U be a point of the local minimum J (u) on ensemble U, i.e. 2 * 2 J (u* ) d J (u ), u , u  D (u* , H ) ˆ U sufficiently that quadric form y cw L (u* , O ) / wu y, y z 0 be positive definite on the hyperplane *

§ wg i (u* ) · ¨ ¸ y © wu ¹

*

§ wg (u ) · 0, i  I , ¨ i * ¸ y © wu ¹

0, i

m  1, s.

We consider the solutions of the following example on base of the theorems 1, 2 and solution algorithm of the nonlinear programming (the lectures 13). Example 1. To find the problem solution 3  6u1  2u 2  2u1 u 2  2u 22 o sup

(6)

at conditions 3u1  4u2 d 8,  u1  4u22 d 2, u1 t 0, u2 t 0.

Solution. Problem (6), (7) is tantamount to the problem 228

(7)

J (u) 2u22  2u1 u2  6u1  2u2  3 o inf, u

^u  E

(u1 , u 2 )  U

2

(8)

`

/ u  U 0 , g1 (u ) d 0, g 2 (u ) d 0 ,

(9)

where U0

^u  E

2

`

/ u1 t 0, u 2 t 0 ; g1 (u ) 3u1  4u 2  8;

g 2 (u) u1  4u22  2 .

Function J (u) is not a convex function on ensemble U 0 , consequently, we have the nonlinear programming problem. 10. We show that ensemble U* z ‡ . Let

^u  E

U1

2

`

/ g 1 (u ) d 0 , U 2

^u  E

2

`

/ g 2 (u ) d 0 .

Then U U 0 ˆ U1 ˆ U 2 – closed bounded ensemble, consequently, it is convex. Function J (u) is continuous on ensemble U . Thence in effect Weierstrass theorems we have U* z ‡ . 20. Generalized Lagrange’s function for problem (8), (9) has the form L (u1 , u 2 , O 0 , O1O 2 )

O 0 ( 2u 22  2u1u 2  6u1  2u 2  3) 

 O1 (3u1  4u 2  8)  O 2 ( u1  4u 22  2),

O

^O  E

(O0 , O1 , O2 )  / 0

3

u  (u 1 , u 2 )  U 0 ,

`

/ O0 t 0, O1 t 0, O 2 t 0 .

0

3 . According to the conditions (3) - (5) quadruple

u , O O , O , O  U *

*

* 0

* 1

* 2

0

u /0

is defined from correlations | O * |z 0, O*0 t 0, O1* t 0, O*2 t 0,





L u u * , O * , u  u * t 0,  u  U 0 ,

O1* g1 (u* ) 0, O*2 g 2 (u* ) 0, u* U ,

where derivative L u (u * , O * )

· § O*0 ( 2u 2*  6)  3O1*  O*2 ¸ ¨ * ¨ O ( 4u *  2u *  2)  4O*  8O* u * ¸ . 2 1 1 2 2 ¹ © 0

229

(10) (11) (12)

a) We expect that u*  intU 0 . Then from (11) follows that L u (u * , O * ) 0 , consequently, pair (u* , O * ) is defined from solution of the following algebraic equations: O*0 (2u2*  6)  3O1*  O*2 0, O*0 (4u2*  2u1*  2)  4O1*  8O*2 u2* 0 .

O1* (3u1*  4u2*  8) 0, O*2 (u1*  4u2*2  2) 0 ,

(13)

where u1* ! 0, u 2* ! 0, O*0 t 0, O1* t 0, O*2 t 0 . We consider the event, when O*0 0 . Then O*2 3O1* , 4O1* (1  6u2* ) 0 . If O1* 0, that O*2 0, consequently, condition (10) is broken, since O*0 0, O1* 0, O*2 0 . So it is possible only O1* ! 0 . Then u 2* 1/ 6. It is impossible, since u2* ! 0. Thence follows that O*0 ! 0, i.e. problem (8), (9) is nondegenerate, so it is possible to take O10 1 and conditions (13) are written so: 2u 2*  6  3O1*  O*2

0, 4u 22  2u1*  2  4O1*  8O*2 u 2*

0,

O (3u  4u  8) 0, O (u  4u  2) 0 . * 1

* 1

* 2

* 2

* 1

* 2

(14)

Now we consider the different possibilities: 1) O1* ! 0, O*2 ! 0. In this case the point u* (u1* , u2* ) is defined from system 3u1*  4u2*  8 0,  u1*  4u2*2  2 0, u1* ! 0, u2* ! 0 [refer to formula (14)]. Thence we have u1* 1,43, u2* 0,926, J (u1* , u2* ) 9,07 . However O1* 1,216, O*2 0,5  0 , so the given point u* (1,43; 0,926 ) can not be solution of the problem (8), (9). 2) O1* 0, O*2 ! 0. In this case from (14) we have 2u 2*  6  O*2

0,  u1*  4u 2*2  2 0.

have u1* 117, u 2* 5,454, O*2 4,908 ! 0 . However * * (117 ; 5,454 )  U , since the inequality 3u1  4u2 d 8. aren’t executed. 3) O1* ! 0, O*2 0. Equations (14) are written so:

Thence u*

0, 4u 2*  2u1*  2  8O*2 u 2*

we

2u 2*  6  3O1*

0, 4u 2*  2u1*  2  4O1*

the

point

0, 3u1*  4u 2*  8 0.

Thence we have u1* 28 / 9, u 2* 1 / 3  0, O1* 16 / 9 ! 0 . The point u* ( 28 / 9,  1 / 3)  U . 4) O1* 0, O*2 0. In this case from equation (14) we have 2u2*  6 0, 4u2*  2u1*  2 0.

Thence we have u2* 3, u1* 5. However the point u* (5; 3)  U , since inequality  u1*  4u2*  2 d 0. are not executed. Thereby, the point u*  intU 0 . b) We expect, that the point u*  ȿɬU0 . Here possible the following events: 1) u (0, u2 )  ȿɬU0 , u2 t 0; 2) u u1 ,0  ȿɬU0 , u1 t 0. For the first type point of the g 2 4u22  2 d 0 , consequently, 0 d u2 d 1/ 2 . Then. restriction g 1 4u 2  8 d 0, 230

u*

u

* 1

0, u2



1 / 2 , J u* 3,4 . For the second type of the border point of the

restriction g 1 3u1  8 d 0, g 2 u1  2 d 0 . Then u* 8 / 3, 0 , but value J u* 19 . Finally, problem solution (8), (9): u* u1* 8 / 3, u2* 0 , J u* 19. Example 2. It is required from wire of the given length l to do the equilateral triangle and square which total area is maximum. Solution. Let u1 , u 2 be a sum of the lengths of the triangle sides of the square accordingly. Then sum u1  u 2 l , but side of the triangle has a length u1 / 3 , side of the square - u2 / 4 and total area is S u1 , u 2

3 2 1 2 u1  u 2 . 36 16

Now optimization problem can be formulated so: to minimize the function J u1 , u 2 

3 2 1 2 u1  u 2 o inf 36 16

(15)

at conditions u1  u 2

(16)

l , u1 t 0, u 2 t 0.

Introducing indications g 1 u1 , u 2 u1  u 2  l , g 2 u1 , u 2

u1 , g 3 u1 , u 2

u 2

problem (15), (16) is written as J u1 , u2 

3 2 1 2 u1  u2 o inf 36 16

(17)

at conditions u U

^u  E

2

g1 (u) u1  u 2  l

0, g 2 (u1 , u 2 )

g 3 (u1 , u 2 )

u1 d 0,

u 2 d 0`,

(18)

where possible U 0 E n . Unlike previous example the conditions u1 t 0, u 2 t 0 are enclosed in restrictions g 2 u1 , u 2 , g 3 u1 ,u 2 . Such approach allows using the theorem 2 for study properties of the functions J (u1 , u 2 ) in the neighborhood of the point u* u1* , u2*  U . Problem (17), (18) is the nonlinear programming problem since function J (u1 , u 2 ) is not convex on ensemble U 0 E n . 231

10. Ensemble U is bounded and closed, consequently, it is compact. Function J u1 , u2  ɍ 2 E 2 , so ensemble U * z ‡ 20. Generalized Lagrange’s function for problem (17), (18) to write as 3 2 1 2 u1  u 2 )  36 16  O1 (u1  u 2  l )  O2  u1  O3  u 2 , u u1 , u 2  E 2 ,

L (u, O )

O

L (u1 , u 2 , O0 , O1 , O2 , O3 )

O0 , O1 , O2 , O3  / 0

30. Since ensemble U 0

^O  E

O0 (  4

`

/ O0 t 0, O2 t 0, O3 t 0 .

E 2 , that conditions (3) – (5) are written: § · 3 * * ¨ u1 O 0  O1*  O*2 ¸ ¨ 18 ¸ ¨ 1 * * * * ¸ ¨  u 2 O 0  O1  O3 ¸ © 8 ¹

*

L (u * , O )

(19)

0,

O1* u1*  u2*  l 0, O*2  u1* 0, O*3  u2* 0 ,

(20)

*

O z 0, O*0 t 0, O*2 t 0, O*3 t 0 .

(21)

a) We consider the event O*0 0 . In this case, as follows from expression (19), O1* O*2 O*3 . If O*2 ! 0 , that O3 ! 0 , consequently, u1* 0, u2* 0 [refer to formula (20)]. It is impossible, since u2*  u2* l . It means, O1* O*2 O*3 0 . The equalities opposites to the condition (21). Then the source problem (17), (18) is nondegenerate. Consequently, value O*0 1 . b) Since problem (17), (18) is nondegenerate, the conditions (19) – (21) are written: 

3 * u1  O1*  O*2 18

O1*

1 0,  u 2*  O1*  O*3 8 0, O*2 u1*

0, O*3u 2*

0, u1*  u 2*  l

0,

0, O*2 t 0, O*3 t 0.

We consider the different cases: 1. O*2 0, O*3 0 . Then u1*

2. O*2 ! 0, O*3

9l , u2* 94 3

4l 3 , O1* 94 3

0 . In this case we get 232

l 3 18  8 3

.

(22)

u1*

3. O*2 l , u 2*

u1*

0, u2* 0

l / 8, O*2

l /8 .

0, O*3 ! 0 . Then from systems of the equations (22) we get

0, O1*

u1*

l , O1*

0, u2*

* * 3 ˜ l / 18 . The event O2 ! 0, O3 ! 0 is excluded, since

3 ˜ l / 18, O*3

0 and condition u1*  u2*

l. is not executed.

4 . Quadric form y cL uu (u* , O* ) y



3 2 1 2 y1  y 2 . 18 8

Hyperplane equations for events 1 - 3 are accordingly written: 1)

wg u wg1 u* y1  1 * y2 wu2 wu1

y1  y2

0 . Thence we have y1



 y 2 . Then



 4 3  9 y12  0 ,

y cL uu (u * , O* ) y

i.e. the point ( u1* 9l (9  4 3 ) , u 2*

4l 3 (9  4 3 ) )

is not the point of the local minimum. wg ( u ) wg 1 ( u * ) y1  1 * y 2 wu 2 wu 1

wg 2 (u* ) wg (u ) y1  2 * y 2 wu1 wu 2

y1  y 2

 y1

0,

0.

Thence we get y1

0, y 2

0; y cL uu (u * , O* ) y

0, y

0.

The sufficient conditions of the local minimum are degenerated. 3)

wg1 (u* ) wg (u ) y1  1 * y 2 wu1 wu 2

y1  y 2

0,

wg 3 (u* ) wg (u ) y1  3 * y 2 wu1 wu 2

 y2

0.

The data of the equation have the solutions y1 0, y 2 0 . Again the sufficient optimality conditions do not give the onedigit answer. Solution of the problem is found by comparison of function values J (u1 , u 2 ) in the last two events. In the second 233

event value J (u1* , u2* ) l 2 / 16 , but in the third event J (u1* , u 2* )  3l 2 / 36. Then the problem solution is the point ( u1* 0, u2* l ), i.e. from the whole wire is made only square. To solve the following problems: 1. To find J u1 , u2 , u3 u1u2u3 o inf at conditions: a) u1  u2  u3  3 0 , b) u1  u2  u3  8 0 ; c) u1u 2  u1u 3  u 2 u 3

a,

u i t 0, i

1,3.

2. To prove the equality n

u1n  u 2n § u1  u 2 · t¨ ¸ , n t 1, u1 t 0, u 2 t 0. 2 © 2 ¹

3. To find the sides of the maximum rectangle area inserted in circle u12  u22 R 2 . 4. To find the most short distance from the point (1,0) till ellipse 4u12  9u22 36 . 5. To find the distance between parabola u2 u12 and line u1  u 2 5 . 6. To find the most short distance from ellipse 2u12  3u22 12 till line u1  u 2 6 . 7. To find J u ucAu o sup, A A* at condition u cu 1 . To show, if u*  E n – the problem solution, that J (u * ) is equal to the most characteristic root of the matrix A . 8. To find the parameters of the cylindrical tank which under given area of the surface S has the maximum volume. 9. a) J (u) u1  u 2  u3 o inf,

1 1 1   u1 u 2 u3

b) J (u) u1u 2 u3 o inf, u1  u2  u3

c) J (u)

6, u1u2  u1u3  u2u3 12;

1 1  o inf, u1 u 2

1 1  2 2 u1 u 2

1;

d) J u 2u1  3u 22  u32 o inf, u1  u 2  u 3

8, u i t 0, i

1,3;

e) J u u12  u 22  u32 o inf, u1  u 2  u 3 d 12, u i t 0, i

1,3 ;

234

1;

f) J (u) u1u 2  u1u3  u 2 u3 o inf, u1  u2  u3 d 4; g) J (u) u12 u 2  u 22 u1  u1u 2 u3 o inf, u1  u 2  u3 d 15, ui t 0, i 1,3;

l) J (u) u12  2u1u 2  u32 o inf, u1  2u 2  u 3

1, 2u1  u 2  u 3

5, u i t 0, i

1,3.

10. To find the conditional extreme to function (u1  2) 2  (u2  3) 2

J (u)

at condition u12  u22 d 52 .

P.1.5. LINEAR PROGRAMMING. SIMPLEX-METHOD

General problem of the linear programming (in particular, basic task) is reduced to the linear programming problem in canonical form of the following type: J (u) ccu o inf,

(1)

^u  E

(2)

u U

n

/ u t 0, Au b`,

where A

a , a ,..., a 1

2

m

, a m1 ,..., a n

– matrix of the order m u n; a i  E m , i 1, n ; vectors are identified by the condition vectors, but b  E m – by the restriction vector. Simplex-method is a general method of the problem solution of the linear programming in canonical form (1), (2). Since the general and the basic problems of the linear programming are reduced to type (1), (2), it is possible to consider that simplex-method is the general method of the problem solution of the linear programming. The base of the linear programming theory is stated in the lectures 15 - 17. We remind briefly a rule of the nondegenerate problem solution of the linear programming in canonical form. 235

10. To build the initial extreme point u 0  U ensemble U . We notice, if U * z ‡ , that lower border to linear function (1) on U is reached in the extreme point ensemble U , moreover in the nondegenerate problem the extreme point has exactly m positive coordinates i.e. u 0 u10 ! 0,..., um0 ! 0,0,...,0 , and vectors a1 ,..., a m , corresponding to the positive coordinates of the extreme point are linear independent. The extreme point u 0  U is defined in the general event by Ɉ-method (Charnes’ method), but on a number of events – on source data ensemble U directly. 20. Simplex-table is built for extreme the point u 0  U . Vectors a i , i 1, m are presented in the first column, in the second – elements of the vector c with corresponding to lower indexes, in the third – positive coordinates of the extreme point u 0 U and in the rest column – decomposition coefficients of the vectors a i , i 1, n on base

a ,..., a i.e. a ¦ a u 1

m

j

m

i

ij

AȽ u j ,

i 1

where zj

AȽ

m

¦c u

i ij

a ,..., a 1

m

is

nonsingular

matrix;

uj

u

1j

,..., u mj , j 1, n .

Values

, j 1, n are brought in penultimate line, but in the last line – values

i 1

z j  c j , j 1, n . The main purpose of the simplex-table is to check the optimality

criterion for the point u 0  U . If it is turn out to be that values z j  c j d 0, j 1, n , the extreme point u 0  U – a solution of the problem (1), (2), but otherwise transition to the following extreme point u1  U is realized, moreover value J u1  J u 0 . 30. The extreme point u1  U and corresponding to it simplex-table is built on base of the simplex-table of the point u 0 U . The index j0 is defined from condition z j 0  c j 0 max(z j  c j ) amongst z j  c j ! 0 . Column j0 of the simplex-table of the point u 0 U is identified pivotal column, and vector is entered a j0 to the number of base in simplex-table of the point u1  U instead of vector a i0 . The index i0 is defined from condition min u1

ui0 uij

T i0 amongst uij0 ! 0 . The extreme point

(u10  T 0 u1 j0 ,..., u i00 1  T 0 u i0 1 j0 , 0, u i00 1  Tu i0 1 j0 ,..., u m0   T0umj0 ,0,...,0,T0 ,0,...,0), T 0

Ti0 .

The base consists of the vectors a1 ,..., a i0 1 , a j0 , a i0 1 ,..., a m . Decomposition coefficients of the vectors a j , j 1, n on the base are defined by the formula

236

(u ij ) ɩɪɞ

u ij0 u i0 j

u ij 

(u i0 j ) ɩɪɞ

u i0 j0 u i0 j

u i0 j0

, i z i0 , j z j 0 ;

(3)

, j 1, n.

Hereinafter values ( z j ) ɩɪɞ , ( z j  c j ) ɩɪɞ are calculated by the known coefficients (u ij ) ɩɪɞ , i 1, m, j 1, n . Thereby, new simplex-table is built for point u1  U . Further

optimality criterion is checked. If it is turn out to be that ( z j  ɭ j ) ɩɪɞ d 0, j 1, n , so the extreme point u1  U - solution of the problem (1), (2), but otherwise transition to the new extreme point u 2 U realized and etc. Example 1. To solve the linear programming problem J u1 , u2 , u3 , u4 , u5 , u6 6u1  3u2  5u3  2u4  4u5  2u6 o inf, u1  2u 2  u 4  3u5 4u 2  u3  u5

17,

12,

u2  8u 4  u5  u6

6,

u1 t 0, u 2 t 0, u3 t 0, u 4 t 0, u5 t 0, u6 t 0 .

Matrix A , vectors b and c are equal to

A

a

b

§17 · ¨ ¸ ¨12 ¸, c c ¨6 ¸ © ¹

1

§1 2 0 1 3 0· ¨ ¸ ¨0 4 1 0 1 0¸ ¨0 1 0 8 1 1¸ © ¹

§1· ¨ ¸ 2 ¨ 0 ¸, a ¨0¸ © ¹

§ 2· ¨ ¸ 3 ¨ 4 ¸, a ¨1¸ © ¹

a , a 1

§0· ¨ ¸ 4 ¨ 1 ¸, a ¨0¸ © ¹

2



, a3 , a 4 , a5 , a6 ,

§1· ¨ ¸ 5 ¨ 0 ¸, a ¨8¸ © ¹

§ 3· ¨ ¸ 6 ¨ 1 ¸, a ¨  1¸ © ¹

§ 0· ¨ ¸ ¨ 0 ¸, ¨1¸ © ¹

(6,  3,5,  2,  4,2) .

Solution. The initial extreme point u 0 17, 0,12, 0, 0, 6 easy is defined for the example. The condition vectors corresponding to positive coordinates of the extreme point - a1 , a 3 , a 6 .

Base

ɭ

b

6 a1

-3 a2 237

5 a3

-2 a4

-4 a5

2 a6

0

u0=(17, 0, 12, 0, 0, 6)

I A1

6

17

1

2

0

1

3

0

A3

5

12

0

(4) 1

0

1

0

A6

2

6

0

1

0

8

-1

1

0

37

0

2 4

25

0

zj cj z j0  c j0 j0

17 2 12 4 6 1

Ÿ i0

3

0

J (u )

174

37 , 2

u1=(11, 3, 0, 0, 0, 3)

II a1 a2

6 -3

11 3

1 0

0 1

-1/2 1 1/4 0

a6

2

3

0

0

-1/4 (8) -5/4 1

0

0

- 374

zj cj z j0  c j0

24, j0

5/2 1/4

24

63 4

0 0

11 3/ Ÿ i 0 8

0

1

J (u )

6 63

4

u2=( 858 ,3,0, 83 ,0,0)

III

a1 a2 4

a

zj cj

85 8

6 -3

3

-2

3 8

1 0 0 0

0 1

- 15 32

0

-

1 32

0

-

17 2

z j0  c j0

1 4

39 2

0 0

( 85 ) - 18 32

1

-

0 , j0

1 4 5 32

39 2

0

4 -

1 8

-

-3

i0 1

5

u3=(0, 2, 0, 1, 4, 0)

IV

238

2

J (u )

54

a5

-4

2

a

-3

4

a

-2

4

32 85

2

- 858

1

1 17

-

zj cj

624 85

0

- 15 85

0 1

- 854

1

25 85

0 0

0

-

5 85

1 0

1 85 10

0

-

430 85

0 0

zj cj d 0

j 1,6

85

- 177 85

3

J (u )

 24

Example 2. J (u) 2u1  3u2  5u3  6u4  4u5 o inf; 2u1  u2  u3  u4

5;

u1  3u2  u3  u4  2u5  u1  4u 2  u 4

8;

1, u j t 0,

j 1,5.

Corresponding Ɉ-problem has the form J (u )

2u1  3u 2  5u3  6u 4  4u5  Mu6  Mu7 o inf;

2u1  u2  u3  u4  u6

5;

u1  3u2  u3  u4  2u5  u1  4u 2  u 4  u 7

8;

1, u j t 0,

j 1,7.

For M-problem matrix ȼ, the vectors b and c are equal

A

§ 2 1 1 1 0 1 0 · ¨ ¸ ¨1 3 1  1 2 0 0 ¸ ¨  1 4 0 1 0 0 1¸ © ¹

(a 1 , a 2 , a 3 , a 4 , a 5 , a 6 , a 7 ) ,

§5· ¨ ¸ b ¨ 8 ¸, cc (2,3,5,6,4, M , M ) . ¨1 ¸ © ¹

Solution. The initial extreme point u0=(0, 0, 0, 0, 4, 5, 1). We present value z j  c j in the manner of z j  c j D j M  E j , j 1, n , i.e. instead of one line for z j  c j are entered two: in the first values E j are written, in the second – D j . Since M – sufficiently great positive number, then z j  c j ! zk  ck , if D j ! D k . But if D j D k , that E j ! Ek . Base

ɭ

B

-2 a1

-3 a2

5 a3 239

-6 a4

4 a5

Ɉ ɜ6

Ɉ ɜ7

u0

I J (u ) 16  6M A6 A5 A7

M 4 M

zj cj

5 4 1

2 1/2 -1

1 3/2 (4)

-1 1/2 0

1 -1/2 1

0 1 0

1 0 0

Ej

4

9

-3

4

0

0

0 0 1 Ÿ 0

Dj

1

5

-1

2

0

0

0

II J (u 1 ) (55 / 4)  (19 / 4)M a6 a5 a2

M 4 M

zj cj

u1

19/4 29/8 1/4 Ej

Dj

III J (u 2 ) 5 / 9 ɜ1 ɜ5 ɜ2 zj cj

19/9 16/9 7/9 Ej Dj

(0,14,0,0,29 / 8,19 / 4,0)

(9/4) 7/8 -1/4 25/4 9/4

u2

-2 4 -3

(0,0,0,0,4,5,1)

1 0 0 0 0

0 0 1 0 0

-1 1/2 0 -3 -1

3/4 -7/8 1/4 7/4 3/4

1Ÿ 0 0

0 1 0 0 0

0

(19 / 9,7 / 9,0,0,16 / 9,0,0)

0 0 1 0 0

-4/9 8/9 -1/9 -2/9 0

3/9 -21/18 3/9 -1/3 0

0 1 0 0 0

As follows from the last simplex-table by solution of the Ɉ-problem is a vector u 2 (19 / 9,7 / 9,0,0,16 / 9,0,0) . Then solution of the source problem is vector u2

(19 / 9,7 / 9,0,0,16 / 9) , J (u 2 ) 5 / 9 .

To solve the following linear programming problems: 1. J (u) 2u1  u2  2u3  3u4 o sup; 3u1  u3  u4 d 6; u2  u3  u4 d 2; 240

 u1  u 2  u3 d 5; u j t 0,

j 1,4 .

2. J (u) u1  u2  u3  3u4  u5  u6  3u7 o inf; 3u3  u5  u6

6;

u2  2u3  u4 10; u1  u6

0;

u3  u 6  u7

j 1,7 .

6; u j t 0,

3. J (u) u1  2u2  u3  2u4  u5 o inf; u1  2u2  u3 d 2; 2u1  u 2  u 4

0;

u1  3u2  u5

6;

u1 t 0; u2 t 0, u3 t 0, u4 t 0.

4. J (u) 2u1  3u2  2u3  u4 o sup; 2u1  2u2  3u3  u4 d 6;  u2  u3  u4 t 2;

u1  u2  2u3

5;

u1 t 0; u2 t 0, u3 t 0.

5. J (u) u1  2u2  u3 o sup; 3u1  u2  u3 t 4; 2u1  3u 2 t 6;

 u1  2 0 u 2 d 1, u j t 0, j 1,3.

6. J (u) u1  2u2  u3 o inf; u1  u 2 t 1;

u1  2u2  u3

8;

 u1  3u 2 t 3, u j t 0, j

1,3.

7. J (u) 2u1  u2  u3 o sup; u1  2u2  u3

4;

u1  u 2 d 2;

u1  u3 t 1, u j t 0, j 1,3.

8. The steel twig by length 111 cm have gone into blanking shop. It is necessary to cut them on stocking up on 19, 23 and 30 cm in amount 311, 215 and 190 pieces accordingly. To build the model on base of which it is possible to formulate the 241

extreme problem of the variant choice performing the work under which the number cut twig are minimum. 9. There are two products which must pass processing on four machines (I, II, III, IV) in process production. The time of the processing of each product on each of these machines is specified in the following form: Machine I II III IV

ȼ 2 4 3 1

The machines I, II, III and IV can be used accordingly during 45, 100, 300 and 50 hours. Price of the product ȼ – 6 tenge per unit, product Ⱦ – 4 tenge. In what correlation do follows to produce the product ȼ and Ⱦ to get maximum profit? To solve the problem in expectation that product ȼ is required in amount not less 22 pieces. 10. Refinery disposes by two oil grade – ȼ and Ⱦ, moreover petrol and fuel oil are got under processing. Three possible production processes are characterized by the following scheme: a) 1 unit of the sort A + 2 unit of the sort B o 2 unit of the fuel oil + 3 unit of petrol; b) 2 unit of the sort A + 1 unit of the sort B o 5 unit of the fuel oil + 1 unit of petrol; c) 2 unit of the sort A + 2 unit of the sort B o 2 unit of the fuel oil + 1 unit of petrol; We assume the price of the fuel oil 1 tenge per unit, but the price of the petrol 10 tenge per unit. To find the most profitable production plan if there are 10 units to oil of the sort ȼ and 15 units oil of the sort B.

P. 1.6. 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 242

J u

2

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: ɜ) J u 5u12  2u1u 2  2u1u 3  4u 22  4u 2 u 3  u 32   2u1  u 2  u3  6, u 0

1,1,1 .

ɝ) J u 4u12  2u1u 2  2u 2 u 3  3u 22  2u 32 

1,1,1 .

 u1  u 2  u3  1, u 0

ɞ) J u 6u12  4u1u 2  2u1u 3  5u 22  6u 2 u 3  4u 32   3u1  2u 2  u3  10, u 0

2,1,1 .

ɟ) J (u ) (2u1  u 2  u 3 )  (u1  u 2  u 3 )  2

2

 (u1  u 2  3u3 ) 2 ,

u0

 1,1,1 .

P. 1.7. VARIATION CALCULUS The simplest problem. Minimize the functional t1

J ( x, x )

³ F ( x(t ), x (t ), t )dt o inf

(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 )

243

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 to the function x 0 (t )  C 1 >t 0 , t1 @ x 0 (t 0 )

x 0 , x 0 ( t1 )

x1

conduct 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 )

(2)

x1 .

Euler equation (2) can be written as Fx  Fxt  Fxx x 0  Fxx x0

0, x(t 0 )

x 0 , x ( t1 )

x1 .

(3)

Example 1. J x, x

³ x t dt o inf, x0 1

2

0

0, x1 1 .

Function F x 2 , therefore Fx

0, Fx

2 x , Fxt

0, Fxx

0, Fxx

2.

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 244

x 0 0 0, x 0 1 1 .

Constants c1 1, c2

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 Fx ( x 0 , x 0 , t ) 0, dt i x0 , xt1 x1 .

Fxi ( x 0 , x 0 , t )  xt 0

Example 2. 1

J  x1 , x 2 , x1 , x 2

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

x1 .

Then from (4) follows, that x20

0, x10

0.

It follows that 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,

Constants 245

x 20 0 0, x20 1 1 .

i 1, n,

(4)

c1

0, c2

0, c4

0, c3

1.

The source solutions of the problem x10 t { 0, t  >0,1@, x 20 t t , t  >0,1@ .

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 t 0 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 to the conditions x k t j x jk , k

0, n  1, j

0,1,

x00

x0 , x10

x1

are called 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 

n d d2 n d Fx  2 Fx  ...   1 F n dt dt dt n x

246

0.

(6)

along

x 0 t  C n >t 0 , t1 @, x k t j

x jk , k

0, n  1, j

0,1 .

Example 3. 1

³ x t dt o inf, x0 2

0, x 0 0, x1 0, x 1 1 .

0

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  >0,1@ .

Constants ci , i 1,4 are determined by the conditions x 0 0 0, x 0 0 0, x1 0, x 1 1 .

Hence we get

c1

The source solution

1, c2

1, c3

0, c4

0.

x 0 t t 3  t 2 , t  >0,1@ .

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 ,

247

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

Pt

A t / At , Qt

(8)

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 248

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, that the function x 0 (t )  C 1 >t 0 , t1 @,

is a solution of the Euler equation (2). Theorem 3. In order to the function x 0 (t )  C 1 >t 0 , t1 @, x 0 , x 0 (t1 )

x 0 (t 0 )

x1

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 B x 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. J x, x

S

³ ( x

2

(t )  x 2 (t ))dt o inf,

0

x0 0, xS 0 .

1. Function F

x 2  x 2 .

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 . 249

Hence we get x 0 t c1 sin t , t  >0, S @ t 0

2 . Since Fxx

2

S .

0, t1

At ! 0 ,

that the strength Legendre condition (7) holds. Functions Bt

2, C t Fxx

Fxx

therefore

Pt

Qt

0,

A t / At 0,

( B (t )  C (t )) / A(t )

1 .

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 for the function

0 . Thus,

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 250

J  x, x t x 0 , x 0

is fulfilled. Theorem 4. In order to the function x 0 (t )  C 1 >t 0 , t1 @

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 



d Fx x 0 , x 0 , t dt



(10)

0

and satisfies to the conditions



Fx  x 0 , x 0 , t 0



) x0 x 0 t 0 , x 0 t1 , Fx x 0 t1 , x 0 t1 , t1

x1 t ,..., xn t ,

F  x1 ,..., xn , x1 ,..., x n , t ,

F

)x1 t 0 ,...xn t 0 , xt1 ,..., xn t1 ,

then relations (10) and (11) can be written as





Fxi x 0 , x 0 , t 

Fxi t k



d Fx x 0 , x 0 , t dt i

 1 k ) x

ki



0, i 1, n;

, i 1, n, k

0,1 .

Example 5. J  x, x

1

³ x t  xt dt  x 1 o inf . 2

2

0

Functions x 2  x, )

F

Since Fx

1, Fx

x 2 1 . 2 x ,

that the Euler equation (10) has the form Solution

2 x0 t  1 0 . x 0 t t 2 / 4  c1t  c2 . 251



 ) x1 x 0 t 0 , x 0 t1 . (11)

Further, if xt





Constants c1 , c2 are determined by the transversality conditions (11). Since



Fx x 0 t 0 , x t0 , t0





2 x 0 t 0 ) x0

Fx x t1 , x t1 , t1 0

0

0,



2 x 1 0

2 x 0 1 ,

) x1

then c1

The source solution

3/ 4 .

0, c2

x 0 t t 2 / 4  3 / 4, t  >0,1@ .

Isoperimetric problem. Minimize the functional 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,

t0

x(t 0 )

x 0 , x(t1 )

(13)

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 0 t

x t ,..., x t , 0 1

x 0 (t )  C 1 >t 0 , t1 @

0 n

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

Example 6. ³ x 2 t dt o inf , 0

252

1

0, x0 0 , x1 1 .

³ xdt 0

Functions x 2 , G

F

x 2  Ox .

x, L

Equation (14) for this problem can be written as:  2 x0 t  O

Hence we get

0.

O / 4, t  >0,1@ .

x 0 t c1t 2  c2 t  c3 , c1

Constants c1 , c2 , c3 are determined by the conditions x 0 0 0, x 0 1 1,

1

³ x t dt 0

0.

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 The source solution

2, c3

3, c2

c1

0.

x 0 t 3t 2  2t , t  >0,1@, O 12 .

Lagrange problem. By Lagrange problem is called the following optimization problem: minimize the functional t1

J  x, u, t 0 , t1

³ f xt , ut , t dt  ) xt , xt , t , t o inf 0

0

0

1

0

1

(15)

t0

at conditions

x t

f  xt , u t , t , t 0 d t d t1 ,

g j x, u, t 0 , t1 d 0, i 1, m; g j  x, u, t 0 , t1 0, j

(16) m  1, s ,

where the functional g j  x, u , t 0 , t1

t1

³ f xt , ut , t dt  ) xt , xt , t , t ; j

j

t0

values t 0 , t1 , in general case are not fixed, furthermore

253

0

1

0

1

(17)

t 0 , t1  ', '

is given finite interval. Constraint (16) is called by differential connection, the vector function is phase variable, vector function is control. We note, that

xt

x1 t ,..., xn t

u t

u t ,..., u t 1

p

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

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 254

t1

/  x, u, t 0 , t1 ,\ , O

³ Ldt  l

(18)

t0

where Lagrangian s

¦ O f x, u, t  \ t , x  f x, u, t

L

i

,

i

i 0

s

¦ O ) xt , xt , t , t , \ t \ t ,...\ t  C >t , t @ .

l

1

i

0

i

1

0

1

1

0

n

1

i 0

2. From the Euler equation





Lx x 0 , u 0 , t ,\ , O 



d Lx x 0 , u 0 , t ,\ , O dt



(19)

0

find the conjugate variable \ t \ 1 t ,...\ n t .

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 Lx x 0 t k0 , u 0 t k0 , t k0 ,\ t k0 , O  1 l x t , k 0,1 . (21) 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

It follows that 255

0, k

0,1

 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 \ 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 J  x, u

S

2

³ u t dt o inf 2

0

at conditions x1

For this problem

u  x1 , x1 0 0, x2 0 0, x1 S / 2 1.

x 2 , x 2

u2, )0

f0

0, g 1

)1

x1 0 0,

g 2 ) 2 x 2 (0) 0 , g 3 ) 3 x1 (S / 2)  1 0 ,

the vector f

x2 , u  x1 ,

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 Lx1 

d Lx dt 1

0, Lx2 

d Lx dt 2

0.

Since Lx1

\ 2 , Lx1

\ 1 , L x2 256

\ 1 , Lx2

\2,

that the adjoint system

\ t \ 1 t ,\ 2 t

is the solution of differential equations \ 1 t \ 2 , \ 2 t \ 1 t , t  >0,S / 2@

3. Since l x1 0

O1 , l x2 0

O 2 , l x1 S / 2

that Lx1

Lx1

t 0

t S /2

\ 1 0 l x1 0

\ 1 S / 2

O1 ,

 l x1 S / 2

O3 , Lx2

Lx2

0,

O3 , l x2 S / 2

\ 2 0 l x2 0

t 0

O2 ,

\ 2 S / 2

t 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

5. Since t 0 , t1 are fixed t0

0.

S / 2 , that there is no neccesity to determine

0, t1

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 / 2@

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 , that O2

0 . Since \ 2

\ 1 , that

\ 1 t { 0, t  >0, S / 2@ . 257

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 / 2@ of 4 item. The optimal process is defined by the equations: x10

\ 1 0

x 20 ,

u 0  x10

x 20

x10 0 0, \ 1

\ 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,

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

c 2 cos t  x10

x 20 , x 20

x10 t

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

It follows that

c3

x 20 0 0,

x10 S / 2 1 .

2 / S , c5

0, c 4

0.

Source solution x10 t

We define the constant c2

2

S

t sin t , t  >0, S / 2@ .

4 / S from the equation x10 t  x10 t c 2 cos t ,

where

x10 t

Then optimal control

2t sin t / S .

u 0 t

4 cos t / S .

Solution of the problem x 20 t

x t , u t 2t sin t / S , 4 cos t / S , 0 1 0 1

0

x t 2 sin t / S  2t cos t / S , t  >0, S / 2@ . 258

Solve the simplest problems 1-10: 1

³ x t dt o inf, x0

1.

2

1, x1 0 .

0

T0

³ x t dt o inf, x0

2.

2

0, xT0 [ .

0

3/ 2

³ x

3.



 2 x dt o inf, x0

2

0, x3 / 2 1 .

0

e

³ tx dt o inf, x1

4.

e, xe 1 .

2

1

3

³ t

5.

2



 1 x 2 dt o inf, x2

0, x3 1 .

2

1

³ x

6.

2



 xx  12tx dt o inf, x0 x1

0.

0

1

³ 4 x sin t  x

7.



x1

 x 2  4 xsht dt o inf, x0

1, x1

2

 x 2 dt o inf, x0

0.

0

1

³ x

8.

2



0

S /2

³ 2 x  x

9.

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

259

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

17.

1

³ e x dt  4e x

2

x 0

 32e  x 1 o inf .

0

Solve tasks with several functions and with higher derivatives 18 – 27: 18.

1

³ ( x

2 1

 x 22  2 x1 x 2 )dt o inf ,

0

x1 0 x2 0 0, x1 1 sh1, x2 1  sh1.

19.

1

³ ( x

2 1

x 22  x1 x 2 )dt o inf ,

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 x 2  2 x2 x3 )dt o inf , x1 0 x3 0 1, 0

260

x2 0 1, x1 S / 2 S / 2, x2 S / 2 0, x3 S / 2

23.

1

³ x dt o inf , x0

x 0 x 1

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

26.

1

³ ( x

2

 x2 )dt o inf ,

0

x0 x0 0, x 0 1, x 1 ch1, x1 x1 sh1.

27.

S

³ (x

2

 x2 )dt o inf ,

0

x0

x 0

x0

xS 0,

xS S ,

xS 2.

Solve the isoperimetric problems 28-33: 28.

29.

30.

1

1

2

2

0

0

1

1

0

0

1

2 ³ x dt o inf ,

x0 2, x1

32.

0, x0

2 ³ x dt o inf , ³ txdt

0

31.

3, x0 1, x1

³ x dt o inf , ³ x dt

1

3 / 2,

0

0

0

0

³ txdt

2,

 14 S

1

2

4.

0

³ x sin tdt o inf , ³ xdt 1

4, x1

1

³ xdt

S

³ ( x

6.

3S / 2, x0

 x 2 )dt o inf , ³ xet dt

e

2

0, xS S .



 1 / 4, x0

0

261

0, x1

e.

33.

1

³ x x dt o inf , 1 2

0

1

³ x1dt 1, x1 0 x1 1 0, 0

1

0, x2 0 0, x2 1 1 .

³ x dt 2

0

Solve the problems of Lagrange 34-40: 34.

1

³ u dt o inf ,x  x 2

u, x0 1.

0

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

 u 2 )dt o inf , x

2

x  u, x1 1.

0

38.

1

³ (x

 u 2 )dt o inf , x  2 x

2

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 1.8. OPTIMAL CONTROL. MAXIMUM PRINCIPLE

We consider the problem of optimal control: minimize the functional J  x , u , t 0 , t1

t1

³ F xt , u t , t dt  ) xt , xt , t 0

0

0

1

0

, t1 o inf

(1)

t0

at conditions x 262

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

(2)

u t  KC >t 0 , t1 @,

g j  x, u, t 0 , t1 d 0, j 1, m,

u t  U  E r , t  >t 0 , t1 @,

g j  x, u , t 0 , t1 0, j

m  1, s ,

(3) (4)

where functionals g j  x , u , t 0 , t1

t1

³ F xt , ut , t dt  ) xt , xt , t j

0

j

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 @

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 x t  x 0 t  max u t  u 0 t  t0  t00  t1  t10  H

t 0 d t d t1

inequality

t 0 d t d t1

J  x, u , t0 , t1 t J x 0 , u 0 , t00 , t10

holds. Four

x t , u t , t 0

0

263

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

0

i

1

0

1

i 0

t0

where s

¦ O F x, u, t  \ t , x  f x, u, t ,

L

i

i

i 0

s

l

¦ O ) xt , xt , t , t , \ t  KC >t , t @ . 1

i

i

0

1

0

1

0

1

i 0

2. Function

\ 1 t ,...,\ n t  KC 1 >t 00 , t10 @

\ t

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

¦ O F x t , u t , t  f x t , u t , t \ t , t  >t s

\ t

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 u0

>

u 0 t , t  t00 , t10

k



xt , xt , t , t , k 0 0

0 1

0 0

0 1

0,1.

(6)

@

is determined by the condition of maximum



max H x 0 t , v, t ,\ , O vU

264







H x 0 t , u 0 t , t ,\ , O ,

(7)

where Pontryagin function s

H x, v, t ,\ , O

¦ 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

i 0

where

k

 @

 ) ix t k x 0 tk0

0,

(8)

   ) x t , x t , t , t ,

) it k

) it k x 0 t00 , x 0 t10 , t00 , t10 ,

) ix (t k )

0

0 0

ix ( t k )

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 .

are held. 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 >t0 ,t1 @ , 1

which are not equal to zero such that the conditions (5) - (9) are performed. Example. 4

³ u

J ( x, u )

2



 x dt o inf; x

u, u d 1, x0 0.

0

For this example, u 2  x, Fi { 0, f

F0

) i { 0, U

^u  E

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

2. Since Lx

O0 , Lx 265

\

O1 x0 , O0 t 0.

(9)

that the adjoint system can be written as: \ t O0 . 3. Since l x 0 O1 , l x 1 0 , that

\ 0 O1 , \ 4 0.

4. Optimal control u 0 t is determined by the condition max H x 0 , u , t ,\ , O uU1

where H

Hence we get

H x 0 , u 0 , t ,\ , O ,

\u  O0 u 2  x .

>

min O0 u 2  x 0  \ u

1d u 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 L

O0 u 2  x 0  \ u  O2  1  u  O3 u  1 , O2 t 0, O3 t 0.

(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

2O 0 u 0  O 2  O 3  \

0, O 2  1  u 0 0, O3 u 0  1 0.

Hence, in the case O0 1 we get: a) O 2 0, O3 0, u 0 \ t / 2, t  >0,4@; b) O2 0, O3 ! 0, u 0 1, 2  O3  \ 0, O3 \  2 ! 0 ; c) O2 ! 0, O3 0, u 0 1,  2  O2  \ 0, O2 2  \ ! 0 ; 266

d) O2 ! 0, O3 ! 0 is impossible. Thus, the optimal control u0

­\ 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  >0,4@, O1

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

0, c1

\ t t  4, t  >0,2@ .

From (13) we get u 0 t

­  1, 0 d t  2, ® ¯t  4 / 2, 2 d t d 4.

By integrating the equations we get

0, t  >0,4@ ,

x 0 t

u 0 t , x 0 0

x 0 t

 t , 0 d t  2, ­ ®2 t  2t  1, 2 d t d 4. / 4 ¯

Solve the following optimal control problems:

267

4

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

u, u d 2, x0

³ xdt o inf, x

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

³ 1 ˜ 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

x u, u t 2, x0 x 0 0, x2 3.

9. J x, u

1

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.1.9 . 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)

t0

at conditions x

f xt , u t , t , xt0

x0 , t  >t0 , t1 @ ,

u t  KC >t0 , t1 @, u t  V t  E , t  >t0 , t1 @ , r

where t 0 , t1 are fixed, x0  E is given vector, n

V t , t  >t0 ,t1 @ 268

(2) (3)

is prescribed set of E r ,

xt

x1 t ,..., xn t 

Solution algorithm 1. Function

K Bx , x, u, 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 u 0 Bx , x, t , t  >t0 , t1 @.

u0

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. 4. We find the optimal control u 0 x 0 , t

u0

by the known function M x, t , where x the equation

0

 



x t is the optimal trajectory satisfying to 0

f x 0 , u 0 x 0 , t , t , x 0 t0

x 0



u 0 M x x0 ,t , x0 ,t

x0 , t  >t0 , t1 @.

Example 1. J  x, u

T

³ u t dt  Dx T o inf, D 2

0

x

2

u , x0

For this example f0

u2,)

1. Function K

const ! 0,

x0 , u  KC >0, T @, u  E 1 .

Dx 2 T , f

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 0 , therefore u 0  Bx  x , t / 2 . 3. The Bellman equation can be written as:  Bt  x , t

B x  x, t u 0  u 0

269

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 .

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

The optimal trajectory x0

M x  x, t / 2

Dx 0 />1  D t  T @.

x 0 t is

determined by solution of the differential equation

x 0 t Dx 0 t / >1  D t  T @, x 0 0

Control u0

x 0 , t  >0, T @.

Dx 0 />1  D t  T @, t  >0, T @

is called by synthesizing control. Discrete systems. The method of dynamic programming for discrete systems we illustrate with example.

Example 2. Minimize 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 where

>x1 @0 x0 , x1,..., xN , xi  E n , i

270

(5)

0, N  1, r

,i

0, N  1,

0, N , ui  E r , i 1, N  1.

(6)

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

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 u N0 1  xN 1 u N0 1  xN 1 , N  1 . We denote BN 1  xN 1 F0 N 1 xN 1 , u N0 1  )  f N 1 xN 1 , u N0 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 xN 1

f N  2  xN  2 , u N  2 .

Therefore we determine u N0  2

and the value

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 .

BN  2  x N  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

u N0  3 xN  3 u N0  3  xN  3 , N  3, N  2, N  1

and B N  3 x N  3









F0 N  3 x N  3 , u N0  3  B N  2 f N  3 x N  3 , u N0  3

etc. 271



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 .

Hence we find the optimal control u00 x0

u00

u00  x0 ,0,1,..., N  1

and value



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

u N0 1 x N0 1 .

Solve the following problems: 1. J x, u

³ >x t  u t @dt o inf, xt

T

2

 xt  u t ,

2

0

x0 x0 , 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  E

u t  KC >0,1@, u t  V

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 >0,1@, 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

x0 ,

x1 , u t  KC >0, T @, u t  V Ž E , 0 d t d T . r

Write the Bellman equation for this problem. 5. Find Bellman function for the problem J  x, u

T

³ > at , xt

 f 0 u t , t @dt  c, xT o inf,

t0

x t

At xt  C t u t  f t , t0 d t d T , xt0

u t  KC >t0 , T @, u

x0 ,

u t  V t  E r , t0 d t d T .

272

(to find function Bx, t in the form Bx, t

\ t , x ) .

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 )

7. I ([ xi ]0 , [u i ]0 )

^ut  E 5

¦[x

`

u d1.

1

 u i2 ] o inf, xi 1

2 i

 xi  u i ,

i 0

xi

i 0

0,5 .

x0 , u i  E 1 , i

8. I ([ xi ]0 , [u i ]0 )

3

¦ [( x )

1 2 i

 ( xi2 ) 2  u i2 ]  ( x 14 ) 2  ( x 42 ) 2 o inf,

i 0

xi11

xi1

i 0

xi1  xi2  u i ,

x10 , xi2

xi21

x02 , i

i 0

9. I ([ xi ]0 , [u i ]0 ) x62 o inf, xi

i 0

x0 , i

10. I ([ xi ]0 , [u i ]0 )

 xi1  xi2 ,

0,3, u i  E 1 , i xi 1

0,5, u i  E 1 , i 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 .

273

0,3 .

APPENDIX II VARIANTS OF THE TASKS FOR INDEPENDENT WORK

Term tasks for students of the 2-nd, 3-rd courses educating on the specialties «Mathematics», «Mechanics», «Informatics», «Mathematical and computer modeling» are worked out by the following parts: convex sets and convex function (1-th task), convex and nonlinear functions (2-nd task), linear programming (3-rd task), variation calculus (4-th task) optimal control (5-th task). Task 1

To check the function J (u ) is convex (concave) on the set U , or indicate such points from U in neighborhood of which J (u ) isn’t neither convex or concave (1 – 89-variants). 1. J u u16  u22  u32  u42  10u1  5u2  3u4  20; U 2. J u e2u u ; U 1

2

E 4.

E 2.

3. J u u13  u23  u32  10u1  u2  15u3  10; U 1 2

4. J u u12  u22  u32  u1u2  u3  10; U

E 3.

E 3.

5. J u u12  u22  2u32  u1u2  u1u3  u2u3  5u2  25; U 1 2

6. J u u25  u32  7u1  u3  6; U

^u  E

3

`

:u d 0.

7. J u 3u12  u22  2u32  u1u2  3u1u3  u2u3  3u2  6; U 1 2

8. J u 5u12  u22  4u32  u1u2  2u1u3  2u2u3  u3  1; U 1 2

E 3.

E 3. E 3.

1 2

9. J u 2u12  u 22  5u 32  u1u 2  2u1u 3  u 2 u 3  3u1  2u 2  6; U 10. J u u13  2u32  10u1  u 2  5u3  6; U {u  E 3 : u d 0}. 274

E 3.

11. J u 5u14  u26  u32  13u1  7u3  8; U

E 3.

12. J u 3u12  2u22  u32  3u1u2  u1u3  2u2u3  17; U

E 3.

1 2

13. J u 4u13  u24  u34  3u1  8u2  11; U {u  E 3 : u d 0}. 14. J u 8u13  12u32  3u1u3  6u2  17; U {u  E 3 : u t 0}. 15. J u 2u12  2u22  4u32  2u1u2  2u1u3  2u2u3  16; U 1 2

16. J u 2u12  u22  u32  2u1u2  8u3  12; U 1 2

E 3.

E 3.

1 2

17. J u  u27  u34  2u2u3  11u1  6; U {u  E 3 : u d 0}. 18. J u U

5 2 3 1 u1  u 22  4u 32  u1u 2  2u1u 3  u 2 u 3  8u3  13; 2 2 2

E 3. 1 2

19. J u 3u12  u 23  2u1u 2  5u1u 3  7u1  16; U {u  E 3 : u d 0}. 3 2

1 2

20. J u 2u12  u22  u32  u1u2  u1u3  2u2u3  10; U 3 2

5 2

21. J u 2u12  u32  u1u3  12u2  18; U

E 3.

E 3.

22. J u 6u12  u23  6u32  12u1  8u2  7; U {u  E 3 : u t 0}. 3 2

23. J u u12  u22  2u32  u1u2  u1u3  2u2u3  8u2 ; U 5 2

24. J u 4u12  u22  u32  4u1u2  11u3  14; U

275

E 3.

E 3.

25. J u

7 2 4 3 1 3 u1  u 2  u 3  13u1  7u 3  9; 2 2 2

U {u  E 3 : u d 0}. 5 6

1 4

3 2

26. J u  u13  u25  u33  22u2  17; U {u  E 3 : u d 0}. 3 2

5 6

27. J u  u13  2u 22  u 32  2u1u 2  3u1u 3  u 2 u 3 ; U {u  E 3 : u d 0}.

28. J u 2u12  4u22  u32  u1u2  9u1u3  u2u3  9; U 29. J u

3 2 5 2 9 2 u1  u2  u3  3u1u3  7u2u3 ; U 2 2 2 7 6

5 2

30. J u  u13  u 22 

E 3.

E 3.

5 4 1 3 u 3  u 4  3u 2 ; 12 2

U {u  E 3 : u t 0}. 1 2

31. J u 3u12  u 23  2u1u 2  5u1u 3  7u1  16; U {u  E 3 : u d 0}.

32. J u U

5 2 3 1 u1  u 22  4u 32  u1u 2  2u1u 3  u 2 u 3  8u 3  13; 2 2 2

E3. 1 2

1 2

33. J u  u27  u34  2u2u3  11u1  6; U {u  E 3 : u d 0}. 1 2

34. J u 2u12  u22  u32  2u1u2  8u3  12; U

E 3.

35. J u 2u12  2u22  4u32  2u1u2  2u1u3  2u2u3  16; U 7 6

5 2

36. J u  u13  u 22 

5 4 1 3 u 3  u 4  3u 2 ; 12 2

U {u  E 4 : u t 0}. 276

E 3.

37. J u

3 2 5 2 9 2 u1  u2  u3  3u1u3  7u2u3 ; U 2 2 2

E 3.

38. J u 2u12  2u22  u32  u1u2  9u1u3  u2u3  9; U

E 3.

3 2

5 6

39. J u  u13  2u 22  u 32  2u1u 2  3u1u 3  u 2 u 3 ; U {u  E 3 : u d 0}. 5 6

1 4

3 2

40. J u  u13  u25  u33  22u2  17; U {u  E 3 : u d 0}. 41. J u

7 2 4 3 1 3 u1  u 2  u 3  13u1  7u 3  9; 2 3 2

U {u  E 3 : u d 0}. 5 2

42. J u 4u12  u22  u32  4u1u2  11u3  14; U

E 3.

3 2

43. J u u12  u22  2u32  u1u2  u1u3  2u2u3  8u2 ; U

E 3.

44. J u 6u12  u32  6u32  12u1  8u2  7; U {u  E 3 : u t 0}. 3 2

5 2

45. J u 2u12  u32  u1u3  12u2  18; U 3 2

E 3.

1 2

46. J u 2u12  u22  u32  u1u2  u1u3  2u2u3  10; U 47. J u u16  u22  u32  u42  10u1  3u4  20; U 48. J u e2u  u ; U 1

2

E 4.

E 2.

49. J u u13  u23  u33  10u1  u2  15u3  10; U 1 2

E 3.

50. J u u12  u22  u32  u1u2  u3  10; U

E 3.

E 3.

51. J u u12  u22  2u32  u1u2  u1u3  u2u3  5u2  25; U 277

E 3.

1 2

52. J u u25  u32  7u1  u3  6; U {u  E 3 : u d 0}. 53. J u 3u12  u22  2u32  u1u2  3u1u3  u2u3  3u2  6; U

E 3.

1 2

54. J u 5u12  u 22  4u 32  u1u 2  2u1u 3  2u 2 u 3  u 3  1; U

E 3.

55. J u u13  2u33  10u1  u2  5u3  6; U {u  E 3 : u d 0}. 56. J u 5u14  u26  u32  13u1  7u3  8; U

E 3.

57. J u 3u12  2u22  u32  3u1u2  u1u3  2u3u2  17; U

E 3.

1 2

58. J u 4u13  u24  u34  3u1  8u2  11; U {u  E 3 : u d 0}. 59. J u 3u13  12u32  3u1u3  6u2  17; U {u  E 3 : u t 0}. 60. J u 2u12  2u22  4u32  2u1u2  2u1u3  2u2u3  16; U 61. J u u16  u22  u32  u42  10u1  3u4  20; U 62. J u e2u u ; U 1

E 3.

E 4.

E 2.

2

63. J u u13  u23  u33  10u1  u2  15u3  10; U 1 2

64. J u u12  u22  u32  u1u2  u3  10; U

E 3.

E 3.

65. J u u12  u22  2u32  u1u2  u1u3  u2u3  5u2  25; U

E 3.

1 2

66. J u u25  u32  7u1  u3  6; U {u  E 3 : u d 0}. 67. J u 3u12  u22  2u32  u1u2  3u1u3  u2u3  3u2  6; U 1 2

68. J u 5u12  u22  u32  u1u2  2u1u3  2u2u3  u3  1; U

278

E 3. E 3.

1 2

1 2

69. J u 2u12  u 22  5u 32  u1u 2  2u1u 3  u 2 u 3   3u1  2u 2  6;

U

E3.

70. J u u13  u33  10u1  u2  5u3  6; U {u  E 3 : u d 0}. 71. J u 5u14  u26  u32  13u1  7u3  8; U

E 3.

72. J u 3u12  2u22  2u32  3u1u2  u1u3  2u2u3  17; U

E 3.

1 2

73. J u 4u3  u24  u34  3u1  8u2  11; U {u  E 3 : u d 0}. 74. J u 8u13  12u32  3u1u3  6u2  17; U {u  E 3 : u t 0}. 75. J u 2u12  2u22  4u32  u1u2  2u2u3  16; U 1 2

76. J u 2u12  u22  u32  2u1u2  8u3  12; U 1 2

E 3. E 3.

1 2

77. J u  u27  u34  2u2u3  11u1  6; U {u  E 3 : u d 0}. 5 2

3 2

1 2

78. J u  u12  u 22  4u 32  u1u 2  2u1u 3  u 2 u 3  8u 3  13; U

E 3.

1 2

79. J u u12  u 23  2u1u 2  5u1u 3  7u1  16; U

{u  E 3 : u d 0}.

3 2

1 2

80. J u 2u12  u22  u32  u1u2  u1u3  2u2u3  10; U 3 2

5 2

81. J u u12  u32  u1u3  12u2  18; U

E 3.

E 3.

82. J u 6u12  u23  6u32  12u1  8u2  17; U {u  E 3 : u t 0}.

279

3 2

83. J u u12  u22  2u22  u1u2  u1u3  2u2u3  8u2 ; U 84. J u U

E 3.

7 2 4 3 1 3 u1  u 2  u 3  13u1  7u 3  9; 2 3 2

{u  E 3 : u d 0}.

85. J u

5 3 1 5 3 3 u1  u2  u3  22u2  10; U 6 4 2

{u  E 3 : u d 0}.

3 2

5 6

86. J u  u13  2u 22  u 32  2u1u 2  3u1u 3  u 2 u 3 ; U

{u  E 3 : u d 0}.

87. J u 2u12  4u22  u32  u1u2  9u1u3  u2u3  8; U 88. J u

3 2 5 2 9 2 u1  u2  u3  3u1u3  7u2u3 ; U 2 2 2

89. J u

7 3 5 2 5 4 1 3 u1  u2  u3  u4  3u2 ; U 6 2 12 2

E 3.

E 3.

{u  E 3 : u t 0}.

Task 2

To evaluate the task of the convex or nonlinear programming (1 – 87–variants): 1 2 2u1  u 2 d 2, u1 t 0,

1. J u 3u1  2u2  u12  u22  u1u2 o max, u1  2u 2 d 2, u 2 t 0.

1 2

2. J u 3u1  2u2  u12  u22  u1u2 o max,

u1 d 3, u2 d 6,

u1 t 0, u 2 t 0.

3 2 u1  u 2 d 1, u 2 t 0.

3. J u 4u1  8u 2  u12  u 22  2u1u 2 o max, u1  u 2 d 3, u1 t 0,

3 2 u1 d 4, u 2 t 0.

4. J u 4u1  8u 2  u12  u 22  2u1u 2 o max,  u1  u 2 d 1, u1 t 0,

280

3 2

5. J u 4u1  8u 2  u12  u 22  2u1u 2 o max, 3u1  5u 2 d 15, u1  u 2 d 1,

u1 t 0, u 2 t 0.

1 2  u1  2u 2 d 2, u1 t 0,

6. J u 3u1  2u 2  u12  u 22  u1u 2 o max, 2u1  u 2 d 2, u 2 t 0.

7. J u u1  6u 2  u12  3u 22  3u1u 2 o max, 4u1  3u 2 d 12, u1 t 0,

 u1  u2 d 1, u 2 t 0.

8. J u u1  6u 2  u12  3u 22  3u1u 2 o max, u1  u 2 d 3, u1 t 0,

 2u1  u 2 d 2, u 2 t 0.

9. J u u1  6u 2  u12  3u 22  3u1u 2 o max, u1  u 2 d 0, u1 t 0,

u2 d 5, u2 t 0.

3 2 3u1  4u 2 d 12, u1 t 0,

 u1  u 2 d 2, u 2 t 0.

3 2  u1  2u 2 d 2, u1 t 0,

u1 d 2, u 2 t 0.

3 2 3u1  4u 2 d 12, u1 t 0,

 u1  2u 2 d 2, u 2 t 0.

10. J u 6u 2  u12  u 22  2u1u 2 o max,

11. J u 6u 2  u12  u 22  2u1u 2 o max,

12. J u 6u 2  u12  u 22  2u1u 2 o max,

3 2

13. J u 8u1  12u 2  u12  u 22 o max,  2u1  u 2 d 4, u1 t 0,

2u1  5u2 d 10, u2 t 0.

3 2

14. J u 8u1  12u 2  u12  u 22 o max,  u1  2u 2 d 2, u1 t 0,

u1 d 6, u 2 t 0.

3 2

15. J u 8u1  12u 2  u12  u 22 o max,  3u1  2u 2 d 0, u1 t 0,

4u1  3u2 d 12, u 2 t 0.

1 2  2u1  u 2 d 2, u1 t 0,

2u1  3u2 d 6, u2 t 0.

16. J u 3u1  2u 2  u12  u 22  u1u 2 o max, 281

1 2

17. J u 6u1  4u 2  u12  u 22  u1u 2 o max, u1  2u 2 d 2, u1 t 0,

 2u1  u2 d 0, u 2 t 0.

1 2

18. J u 6u1  4u 2  u12  u 22  u1u 2 o max, 2u1  u 2 d 2, u1 t 0,

u 2 d 1, u 2 t 0.

1 2

19. J u 6u1  4u 2  u12  u 22  u1u 2 o max, 3u1  2u 2 d 6, u1 t 0,

 3u1  u2 d 3, u2 t 0.

20. J u 8u1  6u 2  2u12  u 22 o max,

 u1  u 2 d 1, u1 t 0,

3u1  2u 2 d 6, u2 t 0.

21. J u 8u1  6u2  2u12  u 22 o max,

 u1  u 2 d 1, u1 t 0, u1 d 3, u 2 t 0.

22. J u 8u1  6u 2  2u12  u 22 o max,

 u1  u 2 d 2, u1 t 0,

3u1  4u2 d 12, u 2 t 0.

23. J u 2u1  2u 2  u12  2u 22  2u1u 2 o max, 4u1  3u 2 d 12, u1 t 0,

u2 d 3, u2 t 0.

24. J u 2u1  2u 2  u12  2u 22  2u1u 2 o max, 2u1  u 2 d 4, u1 t 0,

 u1  u 2 d 2, u 2 t 0.

25. J u 2u1  2u 2  u12  2u 22  2u1u 2 o max, u 2 d 4, u 2 t 0. 2u1  u 2 d 2, u1 t 0, 26. J u 4u1  4u 2  3u12  u 22  2u1u 2 o max, 4u1  5u 2 d 20, u1 t 0,

u1 d 4, u 2 t 0.

27. J u 4u1  4u 2  3u12  u 22  2u1u 2 o max, 3u1  6u 2 d 18, u1 t 0,

u1  4u 2 d 4, u 2 t 0.

28. J u 4u1  4u 2  3u12  u 22  2u1u 2 o max, 3u1  4u 2 d 12, u1 t 0,

u1  2u 2 d 2, u 2 t 0.

29. J u 12u1  4u 2  3u12  u 22 o max,

u1  u 2 d 6, u1 t 0,

1 1  u1  u2 d 1, u2 t 0. 2 2

282

11 1 2 1 u1  u 2  u12  u 22  u1u 2 o max, 2 6 3 2 2u1  u 2 d 2, u1 t 0,  u1  2u2 d 2, u 2 t 0.

30. J u

31. J u 18u1  12u 2  2u12  u 22  2u1u 2 o max, 1 u1  u2 t 1, u2 t 0. 2

u1  u 2 d 4, u1 t 0,

1 2 5u1  2u 2 d 10, u1 t 0,

5 2 3u1  2u 2 t 6, u 2 t 0.

32. J u 6u1  16u 2  u12  u 22  2u1u 2 o max,

33. J u 11u1  8u 2  2u12  u 22  u1u 2 o max, u1  u 2 d 0, u1 t 0,

3u1  4u2 d 12, u 2 t 0.

34. J u 8u 2  4u12  2u 22  4u1u 2 o max, u1 d 4, u 2 d 3,

u1 t 0, u 2 t 0.

35. J u 18u1  20u 2  u12  2u 22  2u1u 2 o max,

u1  u 2 d 5,

u1 t 2, u 2 t 0.

3 1 36. J u 12u1  2u 2  u12  u 22  u1u 2 o max, 2 2 u1 d 4, u 2 d 3, u1  3u 2 d 6, u 2 t 0.

37. J u 26u1  20u 2  3u12  2u 22  4u1u 2 o max, 2u1  u 2 d 4, u1 t 0,

u2 d 2, u2 t 0.

38. J u 10u1  8u 2  u12  2u 22  2u1u 2 o max,  u1  u 2 d 2, u1 t 0,

u1 d 5, u2 t 0.

1 13 u1  5u 2  2u12  u 22  u1u 2 o max, 2 2 u1  u 2 d 3, u1 t 0, u1  u 2 d 1, u 2 t 0.

39. J u

1 9 27 1 u1  u 2  u12  2u 22  u1u 2 o max, 2 2 2 2 u1  u 2 t 2, u1 t 0, u1  u2 d 4, u2 t 0.

40. J u

41. J u 2u1  8u 2  u12  5u 22  4u1u 2 o max, u1  u 2 d 3, u1 t 0,

 2u1  3u2 d 6, u2 t 0.

42. J u 8u1  18u 2  u12  2u 22  2u1u 2 o max,  u1  u 2 d 3, u1 t 0,

u1 d 2, u 2 t 0.

283

3 23 u1  u12  2u 22  u1u 2 o max, 2 2 5u1  4u 2 d 20, u1 t 0, u1  u2 d 2, u2 t 0.

43. J u

44. J u 48u1  28u 2  4u12  2u 22  4u1u 2 o max, 2u1  u 2 d 6, u1 t 0,

 2u1  u 2 d 4, u 2 t 0.

45. J u u1  10u 2  u12  2u 22  u1u 2 o max, 3u1  5u 2 d 15, u1 t 0,

u1  2u 2 d 4, u 2 t 0.

46. J u 6u1  18u 2  2u12  2u 22  2u1u 2 o max, 2u1  u 2 d 2, u1 t 0,

5u1  3u 2 d 15, u 2 t 0.

47. J u u1  5u 2  u12  u 22  u1u 2 o max, u1  u 2 d 3, u1 t 0,

u1 d 2, u 2 t 0.

48. J u 14u1  10u 2  u12  u 22  u1u 2 o max,  3u1  2u 2 d 6, u1 t 0,

u1  u2 d 4, u2 t 0.

49. J u u12  u 22  6 o max,

u1  u 2 d 5,

u 2 t 0.

50. J u 5u12  u 22  4u1u 2 o max, u1 t 0, u2 t 0. u12  u 22  2u1  4u 2 d 4,

51. J u u12  u1u2  u2u3  6 o max, u1  2u2  u3 d 3, u j t 0, j  1,3.

52. J u u 2u3  u1  u2  10 o min,

2u1  5u 2  u3 d 10, u 2 t 0, u 3 t 0 .

53. J u u1u2  u1  u2  5 o min,

2u1  u 2 3, u1 t 0, 5u1  u 2 4, u 2 t 0.

54. J u 10u12  5u 22  u1  2u 2  10 o min, 2u12  u 2 d 4, u 2 t 0,

u1  u2 d 8.

55. J u u32  u1u2  6 o min,

u1  2u 2  u3 d 3, u j t 0, u1  u 2  u3 d 2, j

56. J u 2u12  3u 22  u1  6 o max,

u12  u 2 d 3, u 2 t 0, 2u1  u 2 d 5.

284

1,2,3 .

57. J u u12  3u1  5 o min,

u12  u 22  2u1  8u 2 d 16,  u12  6u1  u 2 d 7.

5 2

58. J u u12  u22  u1u2  7 o min,

u12  4u1  u2 d 5,  u12  6u1  u 2 d 7.

59. J u u15  7 o max, 7 2 2u12  9u 22 d 18, u 2 t 0,

u12  u 22  4u 2 d 0, u1u2  1 t 0, u1 t 0.

60. J u 4u12  u 22  u1u 2  u 2  5 o min,  u1  u2 d 1.

61. J u 2u15  3u 23  11 o min,

u12  u 22  6u1  16u 2 d 72, u 2  8 d 0.

62. J u 3u12  u 22  2u1u 2  5 o min, 25u12  4u 22 d 100 , u1 t 0,

u12  1 t 0.

63. J u 5u12  2u 25  3u1  4u 2  18 o max, 3u12  6  u 2 d 0,

u12  u 12 d 9,

u 1 t 0,

5 2

64. J u 3u12  u 22  3u1u 2  7 o min,

u 2 t 0.

3u1  u 2 d 1,

u 22 d 2.

65. J u u16  2u1u 2  u1  6 o max,

3u12 d 15, u1 t 0

 u1  5u 2 t 10, u 2 t 0.

66. J u 4u12  3u 22  4u1u 2  u1  6u 2  5 o max,  u12  u 22 t 3,

 3u12  u 2 t 4.

67. J u u12  2u 22  u1u 2  u1  26 o max, u12 d 25,

u1  2u 2 d 5, u 2 t 0.

68. J u u12  u 22 o max,

2u1  3u 2 d 6,

 u1  5u 2 t 10, u1 t 0.

69. J u 2u12  u 22  4u1u 2  u1  6 o max,  u1  u 2 t 1, u1 t 0 , 2u1  u 2 d 5, u 2 t 0. 285

70. J u 2u12  3u 22  u1u 2  6 o max, u1  u2 d 3.  u12  u 2 t 5, u1 t 0

71. J u u12  u 22  u1  5u 2  5 o max, , u1  u2 d 5. 72. J u u12  u 22  u1u 3 o min, u1  2u 2  2u3

3, j

3u1  u 2  u 3 d 4, u j t 0,

1, 2,3.

73. J u 5u12  6u 22  u32  8u1u 2  u1 o max, u12  u2  u3 d 5,

u1  5u 2 d 8, u1 t 0, u 2 t 0.

74. J u u12  u22  u32 o min,

u1  u2  u3

3,

2u1  u 2  u3 d 5.

75. J u 3u12  u22  u32  u1u2  6 o min,

2u1  u2  u3 d 5,

u 2  3u3 d 8, u 2 t 0 .

76. J u u1u 2  u1u 3  2u 2 u 3  u1  5 o max, u1  u 2  u3 3, 2u1  u 2 d 5, u1 t 0.

1 3 2 2 u1  5u 2  4u3 16 ,

77. J u u12  u 22  u 32  12u1  13u 2  5u 3 o min, 2u1  7u 2  3u 3 d 2, u1 t 0.

78. J u u12  2u 22  30u1  16u3  10 o min, 5u1  3u 2  4u 3

20 ,

u1  6u 2  3u3 d 0, u3 t 0.

1 2 u1  u22  5u1  u3  16 o min, u1  u2  2u3 d 3, 2 2u1  u 2  3u3 t 11, u j t 0, j 1,2,3.

79. J u

80. J u 2u12  2u1  4u 2  3u3  8 o max, 8u1  3u 2  3u3 d 40,

 2u 1  u 2  u 3

81. J u u1u3  u1  10 o min,

3,

u12  u3 d 3, u j t 0

u22  u3 d 3, j 1,2,3.

82. J u 3u12  u 22  u 2  7u3 o max, 4u1  u 2  2u3

5  1, u1 t 0,

2u2  u3 d 4.

286

u 2 t 0.

83. J u u12  u 22  u1u 2  6u1u3  u 2 u3  u1  25 o min, u12  u22  u1u2  u3 d 10, u1u 2  u1  2u 2  u3 4, u j t 0, j 1,2,3.

84. J u e u1 u2 u3 o max,

 u12  u 22  u32 d 10, u 2 t 0

u13 d 5, u3 t 0.

85. J u 3u 22  11u1  3u 2  u3  27 o max, u1  7u 2  3u3 d 7,

5u1  2u 2  u 3 d 2, u 3 t 0.

86. J u 4u12  8u1  u2  4u3  12 o min,

3u1  u2  u3 d 5 ,

u1  2u 2  u3 t 0, u j t 0, j 1,2,3. 1 2 3 2 u 2  u 3  2u 2  9u 3 o min, 2 2 3u1  5u 2  u3 t 19, 2u 2  u3 t 0, u j t 0, j 1,2,3.

87. J u

Task 3

To evaluate the task of the linear programming (1 – 89 variants): J u

ccu o min; U

^u  E

§3 ¨

1

1 1 1· ¸

¨0 ©

5

6 1 0 ¸¹

n

/ Au

1. cc (5,1,1,0,0), bc 5,4,11 , A ¨ 2  1 3 0 0 ¸.

2. cc (6,1,1,2,0), bc 4,1,9 , A

§1 2 1 6 1· ¨ ¸ ¨ 3  1  1 1 0 ¸. ¨1 3 5 0 0 ¸¹ ©

3. cc (0,6,1,1,0), bc 6,6,6 , A

§3 1 1 6 1 · ¸ ¨ ¨ 1 0 5 1  7 ¸. ¨1 2 3 1 1 ¸ ¹ ©

4. cc (7,1,1,1,0), bc 5,3,2 , A

§5 1 1 3 1· ¨ ¸ ¨ 0  2 4 1 1 ¸. ¨1  3 5 0 0¸ © ¹

287

`

b, u t 0 .

2 1· §1 1 1 ¨ ¸ 0 1  3 5 ¸. ¨ 3 0 1 6 1¸ © ¹

5. c c (8,1,3,0,0), b c 4,3,6 , A ¨ 2

§  2 1 2 0 0· ¸ ¨ 1 4 1 3 ¸. ¨ 3 1  1 0 6 ¸¹ ©

6. cc (0,1,3,1,1), bc 2,8,5 , A ¨ 1

0 1 1 1· ¸ 1 3 1 2 ¸. ¨1 0 1 2 1¸ © ¹ § 2 ¨

7. cc (1,2,1,1,0), bc 2,7,2 , A ¨ 4

8. cc (0,1,6,1,3), bc 9,14,3 , A

§ 6 1 1 2 1· ¸ ¨ ¨  1 0  1 7 8 ¸. ¨ 1 0 2 1 1¸ ¹ ©

9. cc (8,1,1,1,0), bc 5,9,3 , A

§ 2 0 3 1 1· ¸ ¨ ¨ 3 1 1 6 2 ¸. ¨ 1 0 2 1 2¸ ¹ ©

10. cc (1,3,1,1,0), bc 4,4,15 , A

§ 2 0 3 1 0· ¨ ¸ ¨ 1 0  1 2 3 ¸. ¨ 3 3 6 3 6¸ © ¹

11. cc (0,2,0,1,3), bc 6,1,24 , A

§ 4 1 1 0 1· ¨ ¸ ¨  1 3  1 0 3 ¸. ¨ 8 4 12 4 12 ¸ © ¹

12. c c (10,5,25,5,0), b c

§ 8 16 8 8 24 · ¨ ¸ 32,1,15 , A ¨ 0 2  1 1 1 ¸. ¨ 0 3 2 1 1 ¸ © ¹

13. cc (6,0,1,1,2), bc 8,2,2 , A

§ 4 1 1 2 1· ¨ ¸ ¨ 2  1 0 1 0 ¸. ¨1 1 0 0 1¸ © ¹

14. cc (5,1,3,1,0), bc 7,7,12 , A

§1 2 3 4 1· ¸ ¨ ¨ 0 3  1 4 0 ¸. ¨0 4 0 8 1¸ ¹ ©

288

§3 ¨

4 1 0 0· ¸ 2 1 1 1 ¸. ¨1  3 0 0 1¸ © ¹

15. cc (5,3,2,1,1), bc 12,16,3 , A ¨ 3

§1 1 1 0 0· ¸ ¨ 2 1 1 2 ¸. ¨2 1 0 0 1¸ ¹ ©

16. cc (7,0,1,1,1), bc 1,12,4 , A ¨ 2

§ 1 1 1 0 0· ¨ ¸ 2 1 1 1 ¸. ¨ 3 2 0 0 1¸ © ¹

17. cc (6,1,2,1,1), bc 2,11,6 , A ¨ 5

18. cc (0,0,3,2,1), bc 5,7,2 , A

1 3· §2 1 1 ¸ ¨ ¨ 3 0 2  1 6 ¸. ¨1 0 1 2 1¸ ¹ ©

19. cc (1,7,2,1,1), bc 20,12,6 , A

§6 3 1 1 1· ¸ ¨ ¨ 4 3 0 1 0 ¸. ¨ 3  2 0 0 1¸ ¹ ©

20. cc (2,0,1,1,1), bc 2,14,1 , A

§ 1 2 1 0 0· ¨ ¸ 5 1 1 2 ¸. ¨ 3 ¨ 1 1 0 0 1¸ © ¹

21. cc (6,1,0,1,2), bc 2,18,2 , A

§ 1 2 1 0 0· ¨ ¸ 6 2 1 1 ¸. ¨ 2 ¨ 1  2 0 0 1¸ © ¹

22. cc (0,3,1,1,1), bc 2,2,6 , A

§ 1 2 1 0 0· ¨ ¸ ¨ 1 1 0 1 0 ¸. ¨ 2 1 1 1 2¸ © ¹

23. cc (3,0,1,2,1), bc 6,2,2 , A

§ 2 2 1 1 1· ¨ ¸ ¨ 2  1 0 1 0 ¸. ¨1 1 0 0 1¸ © ¹

24. cc (0,5,1,1,1), bc 2,2,10 , A

§ 1 1 1 0 0· ¸ ¨ ¨ 1  2 0 1 0 ¸. ¨ 2 1 1 1 2 ¸¹ ©

289

§ 3 ¨

4 1 0 0· ¸

¨ 3 ©

2 1 1 1 ¸¹

§ 1 ¨

1 1 0 0· ¸ 1 0 1 0 ¸. 2 1 1 2 ¸¹

25. cc (1,5,2,1,1), bc 12,1,3 , A ¨  1 1 0 1 0 ¸.

26. cc (5,0,1,1,1), bc 1,3,12 , A ¨  3 ¨ 2 ©

§1 ¨

27. cc (7,0,2,1,1), bc 2,3,11 , A ¨ 3 ¨5 ©

1 1 0 0· ¸  1 0 1 0 ¸. 2 1 1 1 ¸¹

28. cc (1,4,1,1,1), bc 28,2,12 , A

§ 5 5 1 2 1· ¸ ¨ ¨  1 2 0 1 0 ¸. ¨ 3 4 0 0 1¸ ¹ ©

29. c c (0,8,2,1,1), bc 2,20,6 , A

§ 1 2 1 0 0· ¸ ¨ 3 1 1 1 ¸. ¨6 ¨ 3  2 0 0 1¸ ¹ ©

30. cc (0,2,1,1,1), bc 14,10,1 , A

§ 3 5 1 1 2· ¨ ¸ ¨ 2 5 0 1 0 ¸. ¨1 1 0 0 1¸ © ¹

31. cc (7,2,0,1,2), bc 2,12,18 , A

§ 1 2 1 0 0· ¨ ¸ ¨ 3 4 0 1 0 ¸. ¨  2 6 2 1 1¸ © ¹

32. cc (1,3,1,1,1), bc 2,6,1 , A

§ 1 2 1 0 0· ¨ ¸ ¨ 2 1 1 1 2 ¸. ¨ 1 1 0 0 1¸ © ¹

33. cc (5,1,1,1,2), bc 2,8,2 , A

§ 1 2 1 0 0· ¨ ¸ ¨ 4 1 1 2 1 ¸. ¨ 1 1 0 0 1¸ © ¹

34. c c (1,2,1,1,1), bc 11,2,3 , A

§1 1 2 2 1 · ¸ ¨ ¨1  2 0 1 0 ¸. ¨1 1 0 0 1 ¸ ¹ ©

290

§ 2 ¨

35. cc (10,5,2,1,1), bc 17,1,3 , A ¨  1 ¨1 ©

3 1 2 1· ¸ 1 0 1 0 ¸.  3 0 0 1 ¸¹

2 1 3· ¸ 4 8 4 ¸. ¨1 0 1 7 1¸ ¹ © §1 0 ¨

36. cc (2,1,3,1,1), bc 6,16,7 , A ¨ 2 2

§ 1 1 1 0 0· ¨ ¸ 3 2 1 0 ¸. ¨ 3 2 0 0 1¸ © ¹

37. cc (4,1,1,2,1), bc 2,13,16 , A ¨ 4

38. cc (2,2,1,2,1), bc 12,2,26 , A

§ 4  3 1 0 0· ¸ ¨ ¨  1 2 0 1 0 ¸. ¨ 6 3 1 1 1 ¸¹ ©

39. cc (5,2,1,1,1), bc 26,12,6 , A

§9 1 1 1 2· ¸ ¨ ¨ 4 3 0 1 0 ¸. ¨3  2 0 0 1¸ ¹ ©

40. cc (1,11,1,2,1), bc 13,10,1 , A

§2 6 1 1 1· ¨ ¸ ¨ 2 5 0 1 0 ¸. ¨1 1 0 0 1¸ © ¹

41. cc (5,1,1,2,0), bc 1,6,2 , A

§ 3 1  3 1 0· ¨ ¸ 2 1 ¸. ¨2 3 1 ¨ 3 1  2 1 0¸ © ¹

42. cc (0,3,1,1,1), bc 6,2,1 , A

§ 2 1 1 1 2· ¨ ¸ ¨ 1 1 0 1 0 ¸. ¨1 1 0 0 1¸ © ¹

43. cc (8,1,3,0,0), bc 4,3,6 , A

2 1· §1 1 1 ¨ ¸ ¨ 2 0 1  3 5 ¸. ¨ 3 0 1 6 1¸ © ¹

44. cc (2,1,1,1,1), bc 2,11,3 , A

§ 1 1 1 0 0· ¸ ¨ ¨ 1 1 2 2 1 ¸. ¨ 1 1 0 0 1¸ ¹ ©

291

§ 4 3 2 1 1· ¨ ¸

45. cc (5,0,1,2,1), bc 13,3,6 , A ¨ 3 1 0 1 0 ¸. ¨ 3 2 0 0 1¸ © ¹

§1 1 1 0 0· ¸ ¨ 2 2 1 3 ¸. ¨2 1 0 0 1¸ ¹ ©

46. cc (1,3,1,1,1), bc 1,17,4 , A ¨ 5

§3 ¨

4 1 0 0· ¸ 3 1 2 1 ¸. ¨1  3 0 0 1¸ © ¹

47. cc (9,5,2,1,1), bc 12,17,3 , A ¨ 2

48. cc (1,1,1,2,1), bc 12,26,12 , A

§ 4  3 1 0 0· ¸ ¨ ¨ 6 3 1 1 1 ¸. ¨ 3 4 0 0 1¸ ¹ ©

49. cc (0,7,1,1,1), bc 2,26,6 , A

§1 2 1 0 0· ¸ ¨ 1 1 1 2 ¸. ¨9 ¨ 3  2 0 0 1¸ ¹ ©

50. cc (4,8,1,2,1), bc 2,13,1 , A

§ 1 2 1 0 0· ¨ ¸ 6 1 1 1 ¸. ¨ 2 ¨ 1 1 0 0 1¸ © ¹

51. cc (3,1,1,1,0), bc 3,6,5 , A

§ 1 1 0 1 2· ¨ ¸ ¨ 2 1 1 2 3 ¸. ¨3 2 0  3 8¸ © ¹

52. cc (1,3,1,2,1), bc 2,2,5 , A

§ 1 2 1 0 0· ¨ ¸ ¨ 1 1 0 1 0 ¸. ¨ 1 2 1 1 1¸ © ¹

53. c c (0,1,1,2,1), bc 2,2,6 , A

§ 1 2 1 0 0· ¨ ¸ ¨ 2  1 0 1 0 ¸. ¨ 2 2 1 1 1 ¸¹ ©

54. cc (0,5,1,1,1), bc 2,2,11 , A

§ 1 1 1 0 0· ¸ ¨ ¨ 1  2 0 1 0 ¸. ¨1 1 2 2 1 ¸¹ ©

292

§ 3 ¨

4 1 0 0· ¸

¨ 2 ©

3 1 2 1 ¸¹

55. cc (9,2,1,0,1), bc 12,1,17 , A ¨  1 1 0 1 0 ¸.

§1 1 1 0 0· ¸ ¨ 1 0 1 0 ¸. ¨5 2 2 1 3¸ ¹ ©

56. cc (1,0,1,1,1), bc 1,3,17 , A ¨ 3

§1 ¨

57. cc (3,2,1,2,1), bc 2,3,13 , A ¨ 3 ¨ 4 ©

1 1 0 0· ¸  1 0 1 0 ¸. 3 2 1 1 ¸¹

58. cc (9,0,1,1,1), bc 2,12,26 , A

§ 1 2 1 0 0· ¸ ¨ ¨ 4 3 0 1 0 ¸. ¨ 9 1 1 1 2¸ ¹ ©

59. cc (5,5,1,2,1), bc 26,2,12 , A

§ 6 3 1 1 1· ¸ ¨ ¨  1 2 0 1 0 ¸. ¨ 3 4 0 0 1¸ ¹ ©

60. cc (0,10,1,2,1), bc 2,10,1 , A

§ 1 2 1 0 0· ¨ ¸ ¨ 2 5 0 1 0 ¸. ¨ 2 6 1 1 1¸ © ¹

61. cc (3,2,1,1,0), bc 5,5,5 , A

62. cc (1,2,1,1,0), bc 25,3,5 , A

§3 1 3 1 2 · ¨ ¸ ¨ 3 2 1 1 1 ¸. ¨ 7  2 2 0  1¸ © ¹ § 5 10 5 15 10 · ¨ ¸ 2 ¸. ¨0 1 1 6 ¨ 0 6 1  1  1¸ © ¹

63. cc (1,1,2,1,0), bc 4,7,9 , A

§ 2 1 0 1 1· ¨ ¸ ¨ 3 2 0 1 1 ¸. ¨ 1 1 1 2 6¸ © ¹

64. cc (1,3,1,0,0), bc 4,3,6 , A

1 · §2 1 1 1 ¸ ¨ ¨ 1 0 2  1  3 ¸. ¨3 0 3 1 2 ¸¹ ©

293

§3 1 ¨

1 2 3· ¸ 3 2  1¸. ¨3 0 1 1 6 ¸ © ¹

65. cc (2,1,1,5,0), bc 7,1,9 , A ¨ 2 0

§4 ¨

1

1 2 1· ¸

¨1 ©

1

0 0 1 ¸¹

66. cc (6,0,1,1,2), bc 8,2,2 , A ¨ 2  1 0 1 0 ¸.

§1 2 ¨

3

4 1· ¸

¨0 4 ©

0

8 1 ¸¹

67. cc (5,1,3,1,0), bc 7,7,12 , A ¨ 0 3  1 4 0 ¸.

§3 4 1 0 0· ¸ ¨ ¨ 3 2 1 1 1 ¸. ¨1  3 0 0 1¸ ¹ ©

68. cc (5,3,2,1,1), bc 12,16,3 , A

§1 1 1 0 0· ¸ ¨ ¨ 2 2 1 1 2 ¸. ¨2 1 0 0 1¸ ¹ ©

69. cc (7,0,1,1,1), bc 1,12,4 , A

70. cc (6,1,2,1,1), bc 2,11,6 , A

§ 1 1 1 0 0· ¨ ¸ ¨ 5 2 1 1 1 ¸. ¨ 3 2 0 0 1¸ © ¹

71. cc (0,0,3,2,1), bc 5,7,2 , A

1 3· §2 1 1 ¨ ¸ ¨ 3 0 2  1 6 ¸. ¨1 0 1 2 1¸ © ¹

72. cc (1,7,2,1,1), bc 20,12,6 , A

§6 3 1 1 1· ¨ ¸ ¨ 4 3 0 1 0 ¸. ¨ 3  2 0 0 1¸ © ¹

73. cc (2,0,1,1,1), bc 2,14,1 , A

§ 1 2 1 0 0· ¨ ¸ 5 1 1 2 ¸. ¨ 3 ¨ 1 1 0 0 1¸ © ¹

74. cc (6,1,0,1,2), bc 2,18,2 , A

§ 1 2 1 0 0· ¸ ¨ 6 2 1 1 ¸. ¨ 2 ¨ 1  2 0 0 1¸ ¹ ©

294

§ 1 2 1 0 0· ¨ ¸ 1 0 1 0 ¸. ¨ 2 1 1 1 2¸ © ¹

75. cc (0,3,1,1,1), bc 2,2,6 , A ¨ 1

§2 ¨

2

1 1 1· ¸

¨1 ©

1

0 0 1 ¸¹

76. cc (3,0,1,2,1), bc 6,2,2 , A ¨ 2  1 0 1 0 ¸.

§1 ¨ ¨ 2 ©

1 1 0 0· ¸  2 0 1 0 ¸. 2 1 1 2 ¸¹

§ 3 ¨

4 1 0 0· ¸

¨ 3 ©

2 1 1 1 ¸¹

§ 1 ¨

1 1 0 0· ¸ 1 0 1 0 ¸. 2 1 1 2 ¸¹

77. cc (0,5,1,1,1), bc 2,2,10 , A ¨ 1

78. cc (1,5,2,1,1), bc 12,1,3 , A ¨  1 1 0 1 0 ¸.

79. cc (5,0,1,1,1), bc 1,3,12 , A ¨  3 ¨ 2 ©

80. cc (7,0,2,1,1), bc 2,3,11 , A

§ 1 1 1 0 0· ¸ ¨ ¨ 3  1 0 1 0 ¸. ¨5 2 1 1 1 ¸¹ ©

81. cc (1,4,1,1,1), bc 28,2,12 , A

§ 5 5 1 2 1· ¸ ¨ ¨  1 2 0 1 0 ¸. ¨ 3 4 0 0 1¸ ¹ ©

§1 ¨

82. cc (0,8,2,1,1), bc 2,20,6 , A ¨ 6

¨ 3 ©

2 1 0 0· ¸ 3 1 1 1 ¸.  2 0 0 1 ¸¹

§ 1 2 1 0 0· ¸ ¨ 4 0 1 0 ¸. ¨ 2 6 2 1 1¸ ¹ ©

83. cc (7,2,0,1,2), bc 2,12,18 , A ¨ 3

295

84. cc (1,3,1,1,1), bc 2,6,1 , A

§ 1 2 1 0 0· ¨ ¸ ¨ 2 1 1 1 2 ¸. ¨ 1 1 0 0 1¸ © ¹

§1 ¨

1

2 2 1· ¸

¨1 ©

1

0 0 1 ¸¹

85. cc (1,2,1,1,1), bc 11,2,3 , A ¨1  2 0 1 0 ¸.

86. cc (10,5,2,1,1), bc 17,1,3 , A

3 1 2 1· § 2 ¨ ¸ ¨  1 1 0 1 0 ¸. ¨ 1  3 0 0 1¸ © ¹

87. cc (2,1,3,1,1), bc 6,16,7 , A

§ 1 0 2 1 3· ¸ ¨ ¨ 2 2 4 8 4 ¸. ¨1 0 1 7 1¸ ¹ ©

88. cc (1,11,1,2,1), bc 13,10,1 , A

§ 2 6 1 1 1· ¨ ¸ ¨ 2 5 0 1 0 ¸. ¨7  4 0 0 1¸ © ¹

89. cc (2,1,1,1,1), bc 2,11,3 , A

§ 1 1 1 0 0· ¨ ¸ ¨ 1 1 2 2 1 ¸. ¨ 1 1 0 0 1¸ © ¹

Task 4

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

296

3. J x

1

³ ( xt  x t )dt o extr; x0

x1

2

0.

0

4. J x

T0

³ x t  xt dt o extr; x0

0, xT0 [ .

2

0

5. J x

1

³ x t  txt dt o extr; x0

x1

2

0.

0

6. J x

1

³ t xt  x t dt o extr; x0 2

x1

2

0.

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 3

0, x3 / 2 1.

0

9. J x

T0

³ x t  x t dt o extr; x0 3

2

0, xT0 [ .

0

10. J x

T0

³ x t  x t dt o extr; x0 3

2

0, xT0 [ .

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 2

1, xe 0.

1

14. J x

e

³ ( xt  tx t )dt o extr; x1 2

1, xe

1

15. J x

2

³t

2

x 2 (t )dt o extr ;

x(1)

3, x(2) 1.

1

297

2.

3

³ t

16. J x



 1 x 2 t dt o extr ; x2

2

0, x3 1.

2

e

³ 2 xt  t x t dt o extr; x1

17. J x

2

e, xe

2

0.

1

1

18. J x

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

1

20. J x

³ e x t dt o extr; x0 2

x

0, x1

1 . 9

ln 4.

0

1

³ x t  xt xt  12txt dt o extr; x0

21. J x

2

x1

2

(t )  x(t ) x (t ))dt o extr ; x1 0, xe 1.

0.

0

22. J x

e

³ (tx 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

0

298

x1

0.

29. J x

1

³ x t  x t  6 xt sh2t dt o extr; x0 2

2

x1

0.

x1

0.

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 [ .

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

2

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

§S · 1, x¨ ¸ ©2¹

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

0

2

§S · 1, x¨ ¸ ©4¹

0.

0.

§S · 0, x¨ ¸ 1. ©4¹

299

3S / 4

³ x t  4 x t dt o extr; x0

39. J x

2

2

0

40. J x

§ 3S · 0, x¨ ¸ © 4 ¹

S /2

³ 2 xt  x t  x t dt o extr; x0 2

2

0

41. J x

1.

§S · x¨ ¸ ©2¹

0.

3S / 2

³ x t  x t  2 xt dt o extr; 2

2

0

§ 3S · x0 0, x¨ ¸ 0. © 2 ¹

42. J x

S /2

³ x t  x t  txt dt o extr; x0 2

2

0

43. J x

S /2

³ x t  x t  4 xt sht dt o extr; 2

2

0

x0

44. J x

§S · x¨ ¸ ©2¹

0.

T0

³ x t  x t  4 xt sht dt o extr; 2

2

0

§S · x0 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

x0

§S · x¨ ¸ ©2¹

0.

300

§S · 0, x¨ ¸ ©2¹

0.

48. J x

3S / 2

³ x t  x t  4 xt sin t dt o extr; 2

2

0

x0

49. J x

§ 3S · x ¨ ¸ 0. © 2 ¹ S /2

³ x t  x t  4 xt cos t dt o extr; 2

2

0

x0

50. J x

§S · x¨ ¸ ©2¹

0.

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

57. J x

1

³ cos xt dt o extr; x0

0, x1

S 2

.

0, x1 S .

0

301

T0

58. J x

³ sin x t dt o extr; x0

0, xT0 [ .

0

T0

59. J x

³ cos x t dt o extr; x0

0, xT0 [ .

0

T0

60. J x

³ x t 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 67. 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

³ x t dt o extr; 2

0

1

³ xt dt 0

68. J x

1

1, ³ txt dt

0, x0 x1 0.

0

1

³ x t dt o extr; 2

0

1

³ xt dt 0

1

3  , ³ txt dt 2 0

2, x0 2, x1 14.

302

4.

S

69. J x

³ x t dt o extr; 2

0

S

S

³ xt cos tdt

2

0

70. J x

, x0 1, xS 1.

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

³ xt e

t

1

1

0

0

2 t ³ x t dt o extr; ³ xt e dt

0, x0 0, x1 1.

e, x1 2, x0 2e  1.

dt

0

75. J x

1

³ ( x t  x t )dt o extr; 2

2

0

76. J x 1

2

2

1

1

3 2 ³ t x t dt o extr; ³ xt dt

t ³ xt e dt

0

2, x1 4, x2 1.

1  3e 2 1 , x0 0, x1 . 4 e

303

77. J x

S

³ ( x t  x t )dt o extr; 2

2

0

1  3e 2 1 , x0 0, x1 . 4 e

1

t ³ xt e dt

0

78. J x

S /2

³ ( x t  x t )dt o extr; 2

2

0

S /2

§S · 1, x0 x¨ ¸ 0. ©2¹

³ xt sin tdt 0

79. J x

T0

³ xt

1  x 2 t dt o extr;

T0

T0

³

1  x 2 t dt

l , x T0 xT0 0.

T0

80. J x

1

³ x t dt o extr; 2

x0

x 0

x 1 0,

x1 1.

x1

x 1

x0 0,

x 0 1.

0

81. J x

1

³ x t dt o extr; 2

0

82. J x

1

³ ( x

 48 x) dt o extr;

2

0

x(1)

83. J x

x (1)

0, x (0) 1, x (0)

 4.

1

³ (24tx  x ) dt o extr; 2

0

x (0)

84. J x

x (0)

x(1)

0, x (1)

1 . 10

1

³ ( x

2

 24tx) dt o extr;

0

x ( 0)

x (0)

0, x(1)

1 , x (1) 1. 5

304

S

85. J x

³ ( x

2

 x 2 ) dt o extr;

0

x ( 0)

x (0) 1, x (S )

0,

shS , x (S )

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.

Task 5

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;

x

u , u d 1, x0 0.

0

3.

T0

³ x dt o extr;

x

u, u t A, x0 0, xT0 [ ,  A  0 .

0

4.

4

³ x



u , u d 1, x4 0.



u, u d 1, x0 0.



u, u d 1, x0 [ .

2

 x dt o extr; x

2

 x dt o extr; x

2

 x dt o extr; x

0

5.

T0

³ x 0

6.

T0

³ x 0

7.

T0

³ x

2



 x dt o extr; x u, u d 1, x0 [ .

0

8.

T0

³ x

2



 x dt o extr; x

u , u d 1, xT0 [ .

0

305

9.

T0

³ x



 x dt o extr; x

2

u, u d 1, x0 0, xT0 0.

0

10.

T0

³ x



2

 x dt o extr ; x

2

 x 2 dt o extr; x

u, u d 1, x0 0, xT0 [ .

0

11.

T0

³ x



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.

x1

x 2 , x 2

u , u d 2, x1 0

x 2 1 0.

0

13.

1

³ x dt o extr; 1

0

14.

1

³ x dt o extr; 1

0

15.

2

³ x dt o extr; 1

0

x1

16.

x2 , x2

u, u d 2, x1 0  x2 0 0, x2 2 0.

2

³ x dt o extr; 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

18.

x 2 , x 2

³ x dt o extr; 1

x1

x2 , x2

u, u d 2, x1 0 x2 0 x1 2 0.

2

³ x dt o extr; 1

0

x1

20.

x1 0  x 2 2 0, x 2 0 0, u d 2.

2

0

19.

u,

x2 , x2

u, u d 2, x2 0 x2 2 x1 2 0.

2

³ x dt o inf ; 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

x2 , x2

u, u d 2, x1 0  x1 4 0, x2 0 x2 4 0. 306

22. T o inf, x1

x2 , x2

u,

u d 2, x1  1 1, x1 T 1, x 2  1

x 2 T 0.

23. T o inf, x1

x 2 , x 2 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, u d 1, x1 0 [1 , x2 0 [ 2 , x2 T 0.

29. T o inf, x1

x2 , x 2 u, [1 , x2 0 [ 2 , x1 T 0.

u d 1, x1 0

30.

2

³ u dt o inf, 0

31.

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.

³ u dt o inf,

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 0

x 2 , x 2

u,

u d 24, x1 0 11, x1 1 x2 1 0.

2

36. ³ u 2 dt o inf, x1 0

x 2 , x 2

u,

u t 6, x1 0 x2 0 0, x1 2 17. 307

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 

38. x(2) o extr;

11 . 24

2

u, u d 2,

x

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.

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)

42.

T0

³ ( xu

2

u 2 , u12  u 22 d 1, x (0)

x (T0 ),

y (T0 ).

 yu1 )dt o sup;

0

(u1  [ ) 2  u22 d 1, x

43.

u1 , y

T0

³ ( xu

2

u 2 , y (0)

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

44.

u1 , y

T0

³ ( xu

2

u 2 , y (0)

 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

³1 u

2

u 2 , y (0)

dt o inf ;

x

u, u t 0, x(0)

0

308

0, x(T0 )

[.

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)

x ( 0)

x (0)

sh1, x (1)

ch1  sh1.

1

51. ³ u 2 dt o extr; 0

x  x

52.

u,

0,

x (1)

ch1  sh1.

sh1, 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

54.

u , x ( 0)

§S · 0, x¨ ¸ 1. ©2¹

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

x  x

§S · u , x ¨ ¸ 1. ©2¹

309



S 2

.

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

x  x

u.

1

61. ³ u 2 dt o extr; 0

x

62.

T0

³ (u

2

u, u d 1, x(0)

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.

T

65. ³ (u 2  x)dt o extr; 0

x

66.

T0

³ (u

2

u, u d 1, x(0)

x(T )

0.

 x 2 )dt o extr;

0

x

u, u d 1, x(T0 ) [ .

T

67. ³ (u 2  ɱ 2 )dt o extr; 0

x

u, u d 1, x(T ) [ .

310

68.

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.

T0

³ (u

2

u, u d 1, x(0)

0.

 x 2 )dt o extr;

0

x

72.

T0

³ (u

2

u, u d 1, x(0)

0.

 x 2 )dt o extr;

0

x

u, u d 1, x(0)

x(T0 )

0.

4

73. ³ x1dt o extr; 0

x1

x2 ,

x 2

u, u d 2, x1 (0)

x1 (4)

0.

4

74. ³ x1dt o extr; 0

x1

x2 ,

x 2

u, u d 2, x1 (0)

x 2 (4)

x1 (4)

0.

x 2

u, u d 2, x1 (0) x 2 (0)

x 2 (4)

x1 (4)

4

75. ³ x1dt o extr; 0

x1

x2 ,

311

0.

APPENDIX III KEYS To the 1-st task It is possible to construct the matrixes J cc(u) and check its determination on U according to the theorem 3 (lecture 4). Similarly tasks 2-89 are executed. To the 2-nd task 1. J (u* ) 2,50 ; 2. J (u* ) 4,75 ; 3. J (u * ) 11,00 ; 4. J (u * ) 11,00 ; 5. J (u * ) 11,12 ; 6. J (u* ) 2,50 ; 7. J (u* ) 5,67 ; 8. J (u* ) 5,29 ; 9. J (u* ) 6,25 ; 10. J (u * ) 9,59 ; 11. J (u * ) 12,50 ; 12. J (u * ) 9,59 ; 13. J (u * ) 24,34 ; 14. J (u * ) 38,91 ; 15. J (u * ) 27 ,99 ; 16. J (u* ) 4,57 ;

17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32.

J (u * )

8,00 ;

J (u * )

5,75 ;

J (u * ) J (u * )

8,40 ; 12,76;

J (u * )

17,00;

J (u * )

15,84

J (u * )

4,45 ;

J (u * )

3,77 ;

J (u* ) J (u * )

4,20 ; 11,02

J (u * )

10,12

J ( u* ) J (u * )

9,53 ; 13,00

J (u * ) J (u * )

6,10 ; 41,00;

J (u * )

25,81

33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48.

J (u * )

20,82;

J (u * ) 15,00; J (u * )

66,0;

J (u * )

25,12;

J (u * )

47 ,0;

J (u * )

23,50;

J (u * )

12,07;

J (u * )

25,13

J (u * ) J (u * )

4,60 ; 40,0;

J (u * )

23,45;

J (u * )

112,0;

J (u * ) 103,5; J (u * )

41,23;

J (u* ) J (u * )

6,75 ; 40,0.

To the 3-rd task 1.

J (u * )

5;

17. J (u * ) 15 ;

2.

J (u * )

6;

18. J (u* )

34. J (u* )

3.

J (u * )

6;

19.

35. J (u* )

4.

J (u* )

5.

J (u* )

6.

J (u* )

7.

J (u * )

8.

J (u * )

14 ; 3 134 ; 7 137  ; 33 0; 11 ; 3

4 ; 3 J (u * ) 14 ;

20. J (u* ) 21. J (u* ) 22. J (u* ) 23. J (u* )

43 ; 7 99 ; 5 17 ; 3 7;

24. J (u* ) 6 ; 312

33. J (u* )

36. J (u* )

20 ; 3 22 ; 3 437 ; 13 31 ; 3

37. U ‡ ; 89 ; 5 J (u * ) 19 ;

38. J (u* ) 39.

40. J (u * ) 21 ;

9.

J (u * )

0;

10. J (u* ) 7 ; 54 ; 13

41. J (u * ) 3 ;

25. U ‡ ; 26. J (u* )

27 ; 5

27. J (u * ) 16 ;

43.

12. J (u * ) 10 ;

28. J (u * ) 26 ;

44.

13. J (u* ) 6 ;

29. J (u * ) 12 ;

45.

11. J (u* )

7 6

3 7 118 ; 47. 5 19 ; 48. 3 438 ; 61. 13

14. J (u* )  ;

30. J (u* )  ; 46.

15. J (u * ) 26 ;

31. J (u* )

29 ; 5

32. J (u* )

16. J (u* ) 49.

J (u * )

16 ;

55. J (u* )

50.

J (u * )

23 ;

56.

51.

J (u * )

52.

23 ; 3 J (u * ) 6 ;

53.

J (u * )

54.

J (u * )

J (u* )

8;

62.

57.

J (u * )

58.

J (u * )

24 ; 5 22 ;

4;

59.

J (u * )

28 ;

26 ; 5

60.

U

17 ; 3 110 J (u* ) ; 7 19 J (u* ) ; 2 J (u * ) 5 ; 77 J (u * ) ; 5 389 J (u* ) ; 13 72 J (u * ) ; 5

42. J (u* )

63. 64. 65.

J (u * )

5;

133 ; 37 21 J (u* ) ; 10 J (u * )  1 ; 5 J (u* )  . 17

J (u* )

‡;

To the 4th 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

f .

f .

f . min xˆ t [ t / T0 is unitary extreme, [ ! 0 Ÿ 4

xˆ  loc min, [  0 Ÿ xˆ  loc max,

0 Ÿ xˆ  locextr , [ xˆ is non-strong locextr, J min

8. xˆ 2t / 3

f, J max

f .

is unitary extreme, xˆ  loc min, xˆ is non-strong locextr, J min f, J max f . 9. xˆ t [ t / T0 is unitary extreme, [ ! T0 / 3 Ÿ xˆ  loc min, [  T0 / 3 Ÿ xˆ  loc max, [ T0 / 3 Ÿ xˆ  locextr , (ln(3(t  1) /(t  1))) / ln(3 / 2)  abs min, J max f is non-strong locextr, J min f, J max f . 10. xˆ t [ t / T0 is unitary extreme, [ ! T0 / 3 Ÿ xˆ  loc min, [  T0 / 3 Ÿ xˆ  loc max, [ T0 / 3 Ÿ xˆ  locextr , [ xˆ is non-strong extreme, J min f, J max f . 3/ 2

313

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.

16. 17. 18. 19.

f .

(ln(3(t  1) /(t  1))) / ln(3 / 2)  abs min, J max

e / t  ln t  abs max, J min xˆ xˆ 2

J min

t  1  abs min, J max

t  2

2

f .

f . f .

/ 4 conducts strong local minimum,

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 Ÿ sht  sin t ([  shT0 ) / sin T0  abs min . T0 S Ÿ at [ shS , xˆ sht  C sin t  abs min C  R , J min f, J max f . 314

45. sin 2t  abs max, J min

f;

46. 0  T  S Ÿ xˆ (

 2 cos T0 ) sin t  sin 2t  abs min .

47. 48. 49. 50. 51.

[

sin T0

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 . ·

52. 0  T0  S Ÿ ¨¨  T0 ¸¸ sin t  t sin t  abs min . ¹ © sin T0 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;

2(1  e t ) (e  1)t   abs min . 75. te  t  abs min . e3 e 2  4e  3 4 4 8 76. 2  abs min . 77. t sin t  C sin t , C  R 1 ; 78. t cos t . S S t T · t 79. l  2T0 there is not extreme, l 2T0 , x { 0, l ! 2T0 Ÿ xˆ rC §¨ ch  ch 0 ¸, C ! 0 . C¹ © C

73. 2e1t  1  t  abs min . 74.

315

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; t 2 (t 3  2t  1) t 5  3t 3  2t 2  abs max . 84.  abs min . 85. sht  abs min . 10 10 86. cht  cos t  abs min . 87. C1 sin t  C 2 cos t  C3 sht  C 4 cht .

83.

To the 5th task 1. xˆ

­S  t ,  S d t d S / 2, ° t  S / 2, ®  t, ° t S, S / 2 d t d S ¯ ­ t , t d S / 4 ¯ t  S / 2, t ! S / 4

2. xˆ ®

xˆ  abs min,  xˆ  abs max .

xˆ  abs min,  xˆ  abs max .

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 Ÿ  0  abs min ; T0 ! 2 Ÿ xˆ ® 2 4 2 ¯t  T0 / 4  1  T0 , T0  2 d t d T0 ,

xˆ  abs min, t  abs max . 0 d t d T0  2, ­ t [, 6. T0 ! 2 Ÿ xˆ ® xˆ  abs min, 2 ¯t  T0 / 4  1  [  T0 , T0  2 d t d T0 ,

t  T0 2 / 4  [  T02 / 4  abs min, 7. S min f xn t [  t , Tˆn n , S max xˆ

feasible extreme; 0  [ d 1 Ÿ xˆ

t  [  abs max .



f x n t [  t , Tˆn



n , [ d 0 Ÿ there are not any t / 4  [ t  [ is feasible extreme; Tˆ 2 [ , [ ! 1  2

0 d t d [  1, ­ t  [ , 2 ˆ ¯t  [  1 / 4, [  1 d t d T ,

is feasible extreme xˆ min ®

8. T0 d 2 Ÿ xˆ t 2 / 4  [  T02 / 4  abs min . ­ 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 Ÿ xˆ

9. T0 d 4 Ÿ xˆ t t  T0 / 4  abs min, T0 ! 4 Ÿ Ÿ xˆ

T0 d 2 Ÿ

0 d t d T0 / 2  2, ­  t, ° 2 ®t  T0 / 2 / 4  1  T0 / 2, T0 / 2  2 d t d T0 / 2  2, ° t T , T0 / 2  2 d t d T0 , 0 ¯ 316

Tˆ 1  [ .

0 d t d T0 / 2, ­ t, xˆ  abs max; ® ¯T0  t , T0 / 2 d t d T0 ,

xˆ  abs min; xˆ

10. S min f , S max f , [ d 0 Ÿ there are not any feasible extremes; 0  [ d 1 Ÿ is feasible extreme; xˆ t 2 / 4, Tˆ 2 [ , [ ! 1 Ÿ is feasible extreme ­t 2 / 4, 0 d t d 2, ® ˆ ¯t  1, 2  t d T ,

xˆ min

[

11. [ d cthT0 Ÿ xˆ

Tˆ 1  [ .

cht  T0  abs min, [ ! cthT0 Ÿ

chT0

­° [  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

Ÿ 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 . 2

2

t 2  2t  abs min, 2t  t 2  abs max .

t  2 2  2  abs min,

2  t  2  abs max . 2

t 2  2  abs min, 2  t 2  abs max . t 2  t  abs min, t  t 2  abs max . ­  t2,

18. xˆ ®

0 d t d 2  2,

¯t  (8  4 2 ) t  12  8 2 , 2  2  t d 2, xˆ  abs min,  xˆ  abs max . 2

­ t 2  2,

0 d t d 1,

xˆ  abs min,  xˆ  abs max . 1  t d 2, ¯ (t  2) , ­ t 2 , 0 d t d 1, 20. xˆ ® xˆ  abs min . 2 ¯t  2  2, 1 d t d 2, ­  t2, 0 d t d 1, 21. xˆ ® xˆ  abs min,  xˆ  abs max . 2 1  t d 3, ¯(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 § ­ t 2  2t ,  1 d t d 0, ˆ · 23. ¨¨ xˆ ® 2 T 1¸  abs min . ¸ 0  t d 1, ¯ t  2t , © ¹ 2 § · ­ t  1, 0 d t d 1, ˆ 24. ¨¨ xˆ ® 2 T 2 ¸  abs min . ¸ ¯ t  4t  1, 1  t d 2, © ¹

19. xˆ ®

2

§ ­ 25. ¨ xˆ °® ¨ ©



 t 2 / 2  1,



2

°¯ 3t  4 / 2  1,

0 d t d 3, 3  t d Tˆ ,

Tˆ

4 ·¸  abs min . 3 ¸¹ 317

§ ­  3t 2 / 2  3, 26. ¨ xˆ °® 2

0 d t d 3 , ˆ 8 ·¸ T  abs min . ¨ °¯ t  8 / 3 / 2  5, 2 / 3  t d 8 3 , 3 ¸¹ © 28. [ 2 t 0 Ÿ xˆ t 2 / 2  [ 2 t  [1 , Tˆ [ 2  abs min, [ 2  0 Ÿ Ÿ xˆ t 2 / 2  [ t  [ , Tˆ [  abs min .











2

1

2





 2[  abs min,

29. [1 t 0 Ÿ xˆ t / 2  [ 2t  [1 , Tˆ [ 2  [ 22  2[1  abs min,



2

[1  0 Ÿ xˆ t / 2  [ 2 t  [ 1 , Tˆ 2

[ 2  [

2 2

1

[1 0 Ÿ Tˆ 0  abs min . 0 d t d 1, ­0, 30. xˆ ® xˆ  abs min . 2 ¯  (t  1) , 1 d t d 2, ­ t 2 ,

0 d t d 1,

xˆ  abs min . ¯ 2t  1, 1 d t d 2, 0 d t d 1, ­ 2t , 32. xˆ ® xˆ  abs min . 2   t 3 ( 2 ) , 1 d t d 2, ¯ 0 d t d 1, ­0, xˆ  abs min . 33. xˆ ® 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, t  abs max, x t  abs min .

34. xˆ ® 35. xˆ 36. xˆ 37. xˆ 38. x

39. T0 t 2, x t

2  abs max . T0

40. [ d 1 Ÿ x [ , [ ! 1 Ÿ

41. Sphere with radius

x

1 ­ 0dtd , °c, c ° . ® °c ch§¨ t  1 ·¸, 1 d t d 1, c ch§¨1  1 ·¸ [ . ¨ ° ¨ c ¸¹ c c ¸¹ © ¯ ©

T0 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 §¨ ·¸  §¨ ·¸

2

R2 . ©b¹ ©a¹ 44. Optimal trajectory is quadric x  y const .

45. Optimal trajectory is quadric x

const, y 318

const .

p 1 3 7 46. x  §¨ ln  u 2  u 4 ·¸  p, 2©

47. x(t )

u

4

S

t

³ sign cos(2n  1) 2 W dW ,



t

8

¹

n

p§1 2· ¨  2u  u ¸, p  0. 2 ©u ¹

0,r1,.... .

0

48. 49. 50. 51. 52. 53. 54.

x

0.

ch t  Csh t , u

x ch t , u 0 . x

tch t , u

2 sh t .

x

tsh t , u

2ch t .

x

sin t  C cos t , u

x

sin t , u

x

t cos t , u

0.

0.

2 sin t .

S 55. x §¨ t  ·¸ sin t , u 2 cos t . 2¹ © C sin t , u

56. x 0. 57. x sin t , u 0 . 58. x 60. x

 2(t  2) cos t , u 4S Cch t , u 0 .

[ d 1, J max

62. T0 d 2, x

63. T0 d 4,

4 sin t ; 59. x 4S

(2tS  4S ) cos t  (4t  2S ) sin t , u 4  4S  S 2

8 cos t  4S sin t 4  4S  S 2

1. 2tT0  t 2 , T0 ! 2, 4

tT0  t 2 , 4

x

64. x(t ) t , J T 

x

T0 ! 4,

0 d t d T0  2, ­t , ° . ® (t  T0 ) 2 , T0  2 d t d T0 . °T0  1  4 ¯ T ­ 0 d t d 0  2, °t , 2 ° 2 ° § T0 · ¨t  ¸ ° T T0 T 2¹ ° 0 ,  2 d t d 0  2, . x ® 1 © 4 2 2 °2 T0 °  2 d t d T0 . °T0  t , 2 ° °¯

T2 o f at T o f , x(t ) 2

T2 o f at T o f . 2 T ­ 0dtd , °°t , 2 65. x ® T °T  t , d t d T. °¯ 2

t , J

T



66. [ d ɭthT0 ,

x

[chT chT0

J

,

T

T2 . 4

[ ! ɭthT0 ,

x 319

ch t ­ , °sign [ sh W ® °[  (t  T ) sign [ , 0 ¯

0 d t dW,

W d t d T0 .

.

67. [

0, x { 0, [ z 0 there is no any feasible extreme;

68. [ d thT0 , x thW  T0  W

69. [ x(t )

70. x

x

[sht shT0

,

thT0  [ d T0 ,

x

[ sht shT

W d t d T0 ,

,

[ t 1,

;

T 0 d t dW, ­t , ° r ® cos t W tgW 1 ; 71. u °¯ sin W , W d t d S , ­t , t dWn, ° 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 ¸¸ W n ¹ © 72. T0  S , x { 0, T0 ! S

­°sign p, p ! 1, ® p d 1, °¯ p,

T0 , n 1,..., k ;

0 d t dW, ­t , ° T § · ° §T · x r ® 1  W 2 cos¨ t  0 ¸, W d t d T0  W , W tg ¨ 0  W ¸ 1 . 2 2 © ¹ © 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

0 d t dW,

[ .

0, x { 0, T ! 0 is arbitrary, [  1,

[t

x

­ sh t sign [ , ° ® chW °[  (t  T ) sign [ , 0 ¯

°¯ (t  4) , 2 2 d t d 4, ­ t 2 , 0 d t d 1, ° 2 ®(t  2)  2, 1 d t d 3, ° (t  4) 2 , 3 d t d 4. ¯

320

p

x ,

REFERENCES

The main references:

1. 2. 3. 4. 5. 6.

Alekseev B.M., Tychomirov B.M., Fomin C.B. Optimal control. – M.: Nauka, 1979. Boltyunsky B.G.Mathematical methods of optimal control. – M.: Nauka, 1969. Brison A., Ho Yu-shi. Applied theory of optimal control. – M.: Mir, 1972. Vasilev F.P.Lectures on methods for solving of extreme problems. – M.: MSU, 1974. Vasilev F.P.Numerical methods for solving of extreme problems. – M.: Nauka, 1980. Gabasov R., Kirillova F.M. Methods of optimizations. – Minsk: BSU, 1980. The additional references:

7. 8. 9. 10. 11. 12. 13.

Gelfand I.M., Fomin S.B. Variation calculus. – M.: PhisMath, 1961. Zubov B.I. Lectures on control theory. – M.: Nauka, 1975. Karmanov B.G. Matematical programming. – M.: Nauka, 1975. Krasovsky N.N. Motion control theory. – M.: Nauka, 1968. Krotov B.F., Gurman B.I. Methods and tasks of optimal control. – M.: Nauka, 1973. Lie E.B., Markus L. Basics of the theory of optimal control. – M.: Nauka, 1972. Pontryagin L.S., Boltyansky B.G., Gamkleridze R.B., Mishenko E.F. Mathematical theory of optimal processes. – M.: Nauka, 1976. 14. Pshenichny B.N., Danilin Yu.M. Numerical methods in extreme problems. – M.: Nauka, 1975. 15. Aisagaliev S.ȼ., Zhunussova Zh.Kh. Mathematical programming: textbook. – Almaty: Kɜzɜkh university, 2016. – 208 p. 16. Aisagaliev S.ȼ.,Kabidoldanova ȼ.ȼ. Optimal control of dynamic systems. Verlag, Palmarium academic publishing, 2012. – 288 ɭ.

321

Educational issue

Aisagaliev Serikbay Abdigalievich Zhunussova Zhanat Khaphizovna

VARIATION CALCULUS AND METHODS OF OPTIMIZATION Educational manual Computer page makeup: K. Umirbekova Cover designer: K. Umirbekova _www.maths.york.ac.uk

IB No9155 Signed for publishing 04.02.16. Format 60x84 1/16. Off set paper. Digital printing. Volume 26,83 printer’s sheet. Edition: 100. Order No43. Publishing house «Qazaq university» Al-Farabi Kazakh National University KazNU, 71 Al-Farabi, 050040, Almaty Printed in the printing offi ce of the «Kazakh Universitety» publishing house