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International Journal of Modern Engineering Research (IJMER)www.ijmer.com Vol.2, Issue.4, July-Aug. 2012 pp-2242-2286 ISSN: 2249-6645

www.ijmer.com 2242 | Page

1Dr. K. N. Prasanna Kumar,

2Prof. B. S. Kiranagi,

3Prof. C. S. Bagewadi

.

ABSTRACT: We provide a series of Models for the problems that arise in Yang Mills Theory. No claim is made that the

problem is solved. We do factorize the Yang Mills Theory and give a Model for the values of LHS and RHS of the yang Mills

theory. We hope these forms the stepping stone for further factorizations and solutions to the subatomic denominations at

Plancks scale. Work also throws light on some important factors like mass acquisition by symmetry breaking, relationbetween strong interaction and weak interaction, Lagrangian Invariance despite transformations, Gauge field,

Noncommutative symmetry group of Gauge Theory and Yang Mills Theory itself..

We take in to consideration the following parameters, processes and concepts:

(1) Acquisition of mass(2) Symmetry Breaking(3) Strong interaction(4) Unified Electroweak interaction(5)

Continuous group of local transformations(6) Lagrangian Variance

(7) Group generator in Gauge Theory(8) Vector field or Gauge field(9) Non commutative symmetry group in Gauge Theory(10)Yang Mills Theory (We repeat the same Banks example. Individual debits and Credits are conservative so

also the holistic one. Generalized theories are applied to various systems which are parameterized. And we

live in measurement world. Classification is done on the parameters of various systems to which the Theory

is applied. ).

(11)First Term of the Lagrangian of the Yang Mills Theory(LHS)

(12)RHS of the Yang Mills Theory

SYMMETRY BREAKING AND ACQUISITION OF MASS:

MODULE NUMBERED ONENOTATION: : CATEGORY ONE OF SYMMETRY BREAKING : CATEGORY TWO OF SYMMETRY BREAKING

: CATEGORY THREE OF SYMMETRY BREAKING

: CATEGORY ONE OF ACQUISITION OF MASS

: CATEGORY TWO OF ACQUISITION OF MASS :CATEGORY THREE OF ACQUISITION OF MASSUNIFIED ELECTROWEAK INTERACTION AND STRONG INTERACTION:

MODULE NUMBERED TWO:

============================================================================= : CATEGORY ONE OF UNIFIED ELECTROWEAK INTERACTION : CATEGORY TWO OFUNIFIED ELECTROWEAK INTERACTION : CATEGORY THREE OFUNIFIED ELECTROWEAK IONTERACTION :CATEGORY ONE OF STRONG INTERACTION : CATEGORY TWO OF STRONG INTERACTION

: CATEGORY THREE OF STRONG INTERACTION

LAGRANGIAN INVARIANCE AND CONTINOUS GROUP OF LOCAL TRANSFORMATIONS:MODULE NUMBERED THREE:

============================================================================= : CATEGORY ONE OF CONTINUOUS GROUP OF LOCAL TRANSFORMATIONS

Some Contributions to Yang Mills Theory Fortification

Dissipation Models

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:CATEGORY TWO OFCONTINUOUS GROUP OF LOCAL TRANSFORMATIONS : CATEGORY THREE OF CONTINUOUS GROUP OF LOCAL TRANSFORMATION : CATEGORY ONE OF LAGRANGIAN INVARIANCE :CATEGORY TWO OF LAGRANGIAN INVARIANCE : CATEGORY THREE OF LAGRANGIAN INVARIANCEGROUP GENERATOR OF GAUGE THEORY AND VECTOR FIELD(GAUGE FIELD):: MODULE NUMBERED FOUR:

============================================================================

: CATEGORY ONE OF GROUP GENERATOR OF GAUGE THEORY : CATEGORY TWO OF GROUP GENERATOR OF GAUGE THEORY : CATEGORY THREE OF GROUP GENERATOR OF GAUGE THEORY :CATEGORY ONE OF VECTOR FIELD NAMELY GAUGE FIELD :CATEGORY TWO OF GAUGE FIELD : CATEGORY THREE OFGAUGE FIELDYANG MILLS THEORYAND NON COMMUTATIVE SYMMETRY GROUP IN GAUGE THEORY:

MODULE NUMBERED FIVE:

=============================================================================

: CATEGORY ONE OF NON COMMUTATIVE SYMMETRY GROUP OF GAUGE THEORY : CATEGORY TWO OF NON COMMUTATIVE SYMMETRY GROUP OPF GAUGE THEORY :CATEGORY THREE OFNON COMMUTATIVE SYMMETRY GROUP OF GAUGE THEORY : CATEGORY ONE OFYANG MILLS THEORY (Theory is applied to various subatomic particle systems andthe classification is done based on the parametricization of these systems. There is not a single system known which is

not characterized by some properties) :CATEGORY TWO OF YANG MILLS THEORY :CATEGORY THREE OF YANG MILLS THEORYLHS OF THE YANG MILLS THEORY AND RHS OF THE YANG MILLS THEORY.TAKEN TO THE OTHER

SIDE THE LHS WOULD DISSIPATE THE RHS WITH OR WITHOUT TIME LAG :

MODULE NUMBERED SIX:

============================================================================= : CATEGORY ONE OF LHS OF YANG MILLS THEORY : CATEGORY TWO OF LHS OF YANG MILLS THEORY : CATEGORY THREE OF LHS OF YANG MILLS THEORY : CATEGORY ONE OF RHS OF YANG MILLS THEORY : CATEGORY TWO OF RHS OF YANG MILLS THEORY : CATEGORY THREE OF RHS OF YANG MILLS THEORY (Theory applied to various characterized systemsand the systemic characterizations form the basis for the formulation of the classification).

===============================================================================

,

,

,

,

,

,

,

,

,

:

, , , , , , , , , , , , , , , , , , , , , , are Accentuation coefficients , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , are Dissipation coefficients.

SYMMETRY BREAKING AND ACQUISITION OF MASS:

MODULE NUMBERED ONE

The differential system of this model is now (Module Numbered one).113 = 13114 13 1 + 13 114 , 13 .214 = 14113 14 1 + 14 114 , 14 .3

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15 = 15114 15 1 + 15 114 , 15 .413 = 13114 13 1 13 1, 13 .514 = 14113 14 1 14 1, 14 .615

=

15

1

14

15

1

15

1

,

15 .7

+13

1

14 , = First augmentation factor .813 1, = First detritions factor.UNIFIED ELECTROWEAK INTERACTION AND STRONG INTERACTION:MODULE NUMBERED TWO

The differential system of this model is now ( Module numbered two).916 = 16217 16 2 + 16 217 , 16 .1017 = 17216 17 2 + 17 217 , 17.1118 = 18217 18 2 + 18 217 , 18.1216 = 16217 16 2 16 219, 16.1317

=

17

2

16

17

2

17

2

19

,

17 .14

18 = 18217 18 2 18 219, 18 .15+16 217 , = First augmentation factor .1616 219, = First detritions factor .17LAGRANGIAN INVARIANCE AND CONTINOUS GROUP OF LOCAL TRANSFORMATIONS:

MODULE NUMBERED THREE

The differential system of this model is now (Module numbered three).1820 = 20321 20 3 + 20 321 , 20 .1921 = 21320 21 3 + 21 321 , 21 .2022

=

22

3

21

22

3 +

22

3

21 ,

22 .21

20

= 203

21 20

3

20

3

23 , 20 .2221 = 21320 21 3 21 323 , 21 .2322 = 22321 22 3 22 323 , 22 .24+20 321 , = First augmentation factor.20 323 , = First detritions factor .25

GROUP GENERATOR OF GAUGE THEORY AND VECTOR FIELD(GAUGE FIELD):: MODULE NUMBERED FOUR:

============================================================================The differential system of this model is now (Module numbered Four).2624

=

24

4

25

24

4 +

24

4

25 ,

24 .27

25

= 25424 25 4

+ 25 4

25 , 25 .2826 = 26425 26 4 + 26 425 , 26 .2924 = 24425 24 4 24 427, 24 .3025 = 25424 25 4 25 427, 25 .3126 = 26425 26 4 26 427, 26 .32+24 425 , = First augmentation factor.3324 427, = First detritions factor .34

YANG MILLS THEORYAND NON COMMUTATIVE SYMMETRY GROUP IN GAUGE THEORY:

MODULE NUMBERED FIVE

The differential system of this model is now (Module number five).3528 = 28529 28 5 + 28 529 , 28 .3629 = 29528 29 5 + 29 529 , 29 .37

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30 = 30529 30 5 + 30 529 , 30 .3828 = 28529 28 5 28 531, 28 .3929 = 29528 29 5 29 531, 29 .4030 = 30529 30 5 30 531, 30 .41+

28

5

29 ,

= First augmentation factor .42

28 531, = First detritions factor .43LHS OF THE YANG MILLS THEORY AND RHS OF THE YANG MILLS THEORY.TAKEN TO THE OTHERSIDE THE LHS WOULD DISSIPATE THE RHS WITH OR WITHOUT TIME LAG :

MODULE NUMBERED SIX

:

The differential system of this model is now (Module numbered Six).444532 = 32633 32 6 + 32 633 , 32 .4633 = 33632 33 6 + 33 633 , 33 .4734

=

34

6

33

34

6 +

34

6

33 ,

34 .48

32

= 326

33 32

6

32

6

35, 32 .4933 = 33632 33 6 33 635, 33 .5034 = 34633 34 6 34 635, 34 .51+32 633 , = First augmentation factor.5232 635, = First detritions factor .53HOLISTIC CONCATENATE SYTEMAL EQUATIONS HENCEFORTH REFERRED TO AS GLOBALEQUATIONS

We take in to consideration the following parameters, processes and concepts:

(1) Acquisition of mass(2) Symmetry Breaking(3) Strong interaction(4) Unified Electroweak interaction(5) Continuous group of local transformations(6) Lagrangian Variance(7) Group generator in Gauge Theory(8) Vector field or Gauge field(9) Non commutative symmetry group in Gauge Theory(10)Yang Mills Theory (We repeat the same Banks example. Individual debits and Credits are conservative so

also the holistic one. Generalized theories are applied to various systems which are parameterized. And we

live in measurement world. Classification is done on the parameters of various systems to which the Theory

is applied. ).(11)First Term of the Lagrangian of the Yang Mills Theory(LHS)

(12)RHS of the Yang Mills Theory

54

13 = 13114 13 1 +13 114 , +16 2,2,17 , +20 3,3,21 , +24 4,4,4,4,25 , +28 5,5,5,5,29, +32 6,6,6,6,33 , 13 .55

14

=

14

1

13

14 1 +14 114 , +17 2,2,17 , +21 3,3,21 , +25

4,4,4,4,

25 , +29

5,5,5,5,

29, +33

6,6,6,6,

33 , 14 .56

15 = 15114 15 1 +15 114 , +18 2,2,17 , +22 3,3,21 , +26 4,4,4,4,25 , +30 5,5,5,5,29, +34 6,6,6,6,33 , 15 .57

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Where 13 114 , , 14 114 , , 15 114 , are first augmentation coefficients for category 1, 2 and 3+16 2,2,17 , , +17 2,2,17 , , +18 2,2,17 , are second augmentation coefficient for category 1, 2 and

3

+20 3,3,21 , , +21 3,3,21 , , +22 3,3,21 , are third augmentation coefficient for category 1, 2 and 3+24

4,4,4,4,

25 , , +25

4,4,4,4,

25 , , +26

4,4,4,4,

25 , are fourth augmentation coefficient for category 1,2 and 3+28 5,5,5,5,29 , , +29 5,5,5,5,29, , +30 5,5,5,5,29 , are fifth augmentation coefficient for category 1, 2and 3

+32 6,6,6,6,33 , , +33 6,6,6,6,33 , , +34 6,6,6,6,33 , are sixth augmentation coefficient for category 1, 2and 3.58 59

.60

13 = 13114 13 1 13 1, 16 2,2,19, 20 3,3,23 , 24 4,4,4,4,27 , 28 5,5,5,5,31 , 32 6,6,6,6,35 , 13.61

14 = 14113 14

1

14

1

,

17

2,2,

19,

21

3,3,

23 ,

25 4,4,4,4,27 , 29 5,5,5,5,31 , 33 6,6,6,6,35, 14 .6215 = 15114 15 1 15 1, 18 2,2,19, 22 3,3,23 , 26 4,4,4,4,27 , 30 5,5,5,5,31 , 34 6,6,6,6,35, 15 .63Where 13 1, , 14 1, , 15 1, are first detrition coefficients for category 1, 2 and 316 2,2,19, , 17 2,2,19, , 18 2,2,19, are second detritions coefficients for category 1, 2 and 320 3,3,23 , , 21 3,3,23 , , 22 3,3,23 , are third detritions coefficients for category 1, 2 and 3

24

4,4,4,4,

27 ,

,

25

4,4,4,4,

27 ,

,

26

4,4,4,4,

27 ,

are fourth detritions coefficients for category 1, 2

and 328 5,5,5,5,31 , , 29 5,5,5,5,31 , , 30 5,5,5,5,31 , are fifth detritions coefficients for category 1, 2and 332 6,6,6,6,35 , , 33 6,6,6,6,35, , 34 6,6,6,6,35 , are sixth detritions coefficients for category 1, 2and 3 .64.65

16 = 16217 16 2 +16 217 , +13 1,1,14 , +20 3,3,321 , +24 4,4,4,4,425 , +28 5,5,5,5,529, +32 6,6,6,6,633 , 16 .66

17 = 17216 17 2 +17 217 , +14 1,1,14 , +21 3,3,321 , +

25

4,4,4,4,4

25 ,

+

29

5,5,5,5,5

29,

+

33

6,6,6,6,6

33 ,

17 .67

18 = 18217 18 2 +18 217 , +15 1,1,

14 , +22 3,3,321 , +26 4,4,4,4,425 , +30 5,5,5,5,529, +34 6,6,6,6,633 , 18 .68Where +16 217 , , +17 217 , , +18 217 , are first augmentation coefficients for category 1, 2 and 3+13 1,1,14 , , +14 1,1,14 , , +15 1,1,14 , are second augmentation coefficient for category 1, 2 and 3+20 3,3,321 , , +21 3,3,321 , , +22 3,3,321 , are third augmentation coefficient for category 1, 2 and3

+24 4,4,4,4,425 , , +25 4,4,4,4,425 , , +26 4,4,4,4,425 , are fourth augmentation coefficient for category1, 2 and 3

+

28

5,5,5,5,5

29 ,

, +

29

5,5,5,5,5

29,

, +

30

5,5,5,5,5

29 ,

are fifth augmentation coefficient for category

1, 2 and 3

+32 6,6,6,6,633 , , +33 6,6,6,6,633 , , +34 6,6,6,6,633 , are sixth augmentation coefficient for category1, 2 and 3 .69 70.71

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16 = 16217 16 2 16 219, 13 1,1,, 20 3,3,3,23 , 24 4,4,4,4,427 , 28 5,5,5,5,531 , 32 6,6,6,6,635 , 16 .7217 = 17216 17 2 17 219, 14 1,1,, 21 3,3,3,23 ,

25

4,4,4,4,4

27 ,

29

5,5,5,5,5

31 ,

33

6,6,6,6,6

35 ,

17 .73

18 = 18217 18 2 18 219, 15 1,1,, 22 3,3,3,23 , 26 4,4,4,4,427 , 30 5,5,5,5,531 , 34 6,6,6,6,635 , 18 .74where b16 2G19, t , b17 2G19, t , b18 2G19, t are first detrition coefficients for category 1, 2 and 313 1,1,, , 14 1,1,, , 15 1,1,, are second detrition coefficients for category 1,2 and 320 3,3,3,23 , , 21 3,3,3,23 , , 22 3,3,3,23 , are third detrition coefficients for category 1,2 and 324 4,4,4,4,427 , , 25 4,4,4,4,427 , , 26 4,4,4,4,427 , are fourth detritions coefficients for category 1,2and 3

28

5,5,5,5,5

31 ,

,

29

5,5,5,5,5

31 ,

,

30

5,5,5,5,5

31 ,

are fifth detritions coefficients for category 1,2

and 332 6,6,6,6,635 , , 33 6,6,6,6,635 , , 34 6,6,6,6,635 , are sixth detritions coefficients for category 1,2and 3.75

20 = 20321 20 3 +20 321 , +16 2,2,217 , +13 1,1,1,14 , +24 4,4,4,4,4,425 , +28 5,5,5,5,5,529, +32 6,6,6,6,6,633 , 20 .7621 = 21320 21 3 +21 321 , +17 2,2,217 , +14 1,1,1,14 ,

+25 4,4,4,4,4,425 , +29 5,5,5,5,5,529 , +33 6,6,6,6,6,633 , 21 .7722 = 22321

22

3

+

22

3

21 ,

+

18

2,2,2

17 ,

+

15

1,1,1,

14 ,

+26 4,4,4,4,4,425 , +30 5,5,5,5,5,529 , +34 6,6,6,6,6,633 , 22 .78+20 321 , , +21 321 , , +22 321 , are first augmentation coefficients for category 1, 2 and 3+16 2,2,217 , , +17 2,2,217 , , +18 2,2,217 , are second augmentation coefficients for category 1, 2and 3

+13 1,1,1,14 , , +14 1,1,1,14 , , +15 1,1,1,14 , are third augmentation coefficients for category 1, 2and 3

+24 4,4,4,4,4,425 , , +25 4,4,4,4,4,425 , , +26 4,4,4,4,4,425 , are fourth augmentation coefficients forcategory 1, 2 and 3

+28 5,5,5,5,5,529 , , +29 5,5,5,5,5,529 , , +30 5,5,5,5,5,529, are fifth augmentation coefficients forcategory 1, 2 and 3+32 6,6,6,6,6,633 , , +33 6,6,6,6,6,633 , , +34 6,6,6,6,6,633 , are sixth augmentation coefficients for

category 1, 2 and 3 .79 80.81

20 = 20321 20 3 20 323 , 16 2,2,219, 13 1,1,1,, 24 4,4,4,4,4,427 , 28 5,5,5,5,5,531 , 32 6,6,6,6,6,635, 20 .8221 = 21320 21 3 21 323 , 17 2,2,219, 14 1,1,1,,

25

4,4,4,4,4,4

27 ,

29

5,5,5,5,5,5

31 ,

33

6,6,6,6,6,6

35,

21 .83

22 = 22321 22 3 22 323 , 18 2,2,219, 15 1,1,1,, 26 4,4,4,4,4,427 , 30 5,5,5,5,5,531 , 34 6,6,6,6,6,635, 22 .84

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20 323 , , 21 323 , , 22 323 , are first detritions coefficients for category 1, 2 and 316 2,2,219, , 17 2,2,219, , 18 2,2,219, are second detritions coefficients for category 1, 2 and313 1,1,1,, , 14 1,1,1,, , 15 1,1,1,, are third detrition coefficients for category 1,2 and 324

4,4,4,4,4,4

27 , , 25

4,4,4,4,4,4

27 , , 26

4,4,4,4,4,4

27 , are fourth detritions coefficients forcategory 1, 2 and 328 5,5,5,5,5,531 , , 29 5,5,5,5,5,531 , , 30 5,5,5,5,5,531 , are fifth detritions coefficients forcategory 1, 2 and 332 6,6,6,6,6,635 , , 33 6,6,6,6,6,635, , 34 6,6,6,6,6,635 , are sixth detritions coefficients for category1, 2 and 3 .85.86

24 = 24425 24 4 +24 425 , +28 5,5,29 , +32 6,6,33 , +13 1,1,1,114 , +16 2,2,2,217 , +20 3,3,3,321 , 24 .87

25

= 254

24 25

4

+

25

4

25 ,

+

29

5,5,

29 ,

+

33

6,6

33 ,

+14 1,1,1,114 , +17 2,2,2,217 , +21 3,3,3,321 , 25 .8826 = 26425 26 4 +26 425 , +30 5,5,29, +34 6,6,33 , +15 1,1,1,114 , +18 2,2,2,217 , +22 3,3,3,321 , 26 .89

24 425 , , 25 425 , , 26 425 , 1, 2 3+28 5,5,29, , +29 5,5,29, , +30 5,5,29 , 1, 2 3+32 6,6,33 , , +33 6,6,33 , , +34 6,6,33 , 1, 2 3+

13

1,1,1,1

14 ,

, +

14

1,1,1,1

14 ,

, +

15

1,1,1,1

14 ,

are fourth augmentation coefficients for category 1,

2,and 3

+16 2,2,2,217 , , +17 2,2,2,217 , , +18 2,2,2,217 , are fifth augmentation coefficients for category 1,2,and 3

+20 3,3,3,321 , , +21 3,3,3,321 , , +22 3,3,3,321 , are sixth augmentation coefficients for category 1,2,and 3 .90 91.92

24 = 24425 24 4 24 427 , 28 5,5,31 , 32 6,6,35 , 13 1,1,1,1, 16 2,2,2,219, 20 3,3,3,323 , 24 .9325

=

25

4

24

25 4 25 427 , 29 5,5,31 , 33 6,6,35 , 14

1,1,1,1

, 17

2,2,2,2

19, 21

3,3,3,3

23 , 25 .94

26 = 26425 26 4 26 427 , 30 5,5,31 , 34 6,6,35 , 15 1,1,1,1, 18 2,2,2,219, 22 3,3,3,323 , 26 .95 24 427 , , 25 427 , , 26 427 , 1, 2 328 5,5,31 , , 29 5,5,31 , , 30 5,5,31 , 1, 2 332 6,6,35 , , 33 6,6,35 , , 34 6,6,35 , 1, 2 313 1,1,1,1, , 14 1,1,1,1, , 15 1,1,1,1,

1, 2

3

16

2,2,2,2

19, , 17

2,2,2,2

19, , 18

2,2,2,2

19, 1, 2 3 20 3,3,3,323 , , 21 3,3,3,323 , , 22 3,3,3,323 , 1, 2 3 .96, 9798

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28 = 28529 28 5 +28 529, +24 4,4,25 , +32 6,6,633 , +13 1,1,1,1,114 , +16 2,2,2,2,217 , +20 3,3,3,3,321 , 28 .99

29 = 29528 29 5 +29 529, +25 4,4,25 , +33 6,6,633 , +

14

1,1,1,1,1

14 ,

+

17

2,2,2,2,2

17 ,

+

21

3,3,3,3,3

21 ,

29 .100

30 = 30529 30 5 +30 529, +26 4,4,25 , +34 6,6,633 , +15 1,1,1,1,114 , +18 2,2,2,2,217 , +22 3,3,3,3,321 , 30 .101 +28 529, , +29 529, , +30 529 , 1, 2 3 +24 4,4,25 , , +25 4,4,25 , , +26 4,4,25 , 1, 2 +32 6,6,633 , , +33 6,6,633 , , +34 6,6,633 , 1, 2 3+13 1,1,1,1,114 , , +14 1,1,1,1,114 , , +15 1,1,1,1,114 , are fourth augmentation coefficients for category1,2, and 3

+

16

2,2,2,2,2

17 ,

, +

17

2,2,2,2,2

17 ,

, +

18

2,2,2,2,2

17 ,

are fifth augmentation coefficients for category

1,2,and 3

+20 3,3,3,3,321 , , +21 3,3,3,3,321 , , +22 3,3,3,3,321 , are sixth augmentation coefficients for category1,2, 3 .102.103

28 = 28529 28 5 28 531 , 24 4,4,27 , 32 6,6,635 , 13 1,1,1,1,1, 16 2,2,2,2,219, 20 3,3,3,3,323 , 28 .10429 = 29528 29 5 29 531 , 25 4,4,27 , 33 6,6,635, 14 1,1,1,1,1, 17 2,2,2,2,219, 21 3,3,3,3,323 , 29 .10530 = 30529

30

5

30

5

31 ,

26

4,4,

27 ,

34

6,6,6

35 ,

15 1,1,1,1,1,, 18 2,2,2,2,219, 22 3,3,3,3,323 , 30 .106 28 531 , , 29 531 , , 30 531 , 1, 2 324 4,4,27 , , 25 4,4,27 , , 26 4,4,27 , 1,2 332 6,6,635, , 33 6,6,635, , 34 6,6,635, 1,2 313 1,1,1,1,1, , 14 1,1,1,1,1, , 15 1,1,1,1,1,, are fourth detrition coefficients for category 1,2, and3

16

2,2,2,2,2

19,

,

17

2,2,2,2,2

19,

,

18

2,2,2,2,2

19,

are fifth detrition coefficients for category 1,2,

and 3 20 3,3,3,3,323 , , 21 3,3,3,3,323 , , 22 3,3,3,3,323 , are sixth detrition coefficients for category 1,2,and 3.107.108

32 = 32633 32 6 +32 633 , +28 5,5,529 , +24 4,4,4,25 , +13 1,1,1,1,1,114 , +16 2,2,2,2,2,217 , +20 3,3,3,3,3,321 , 32 .109

33 = 33632 33 6 +33 633 , +29 5,5,529 , +25 4,4,4,25 , +14 1,1,1,1,1,114 , +17 2,2,2,2,2,217 , +21 3,3,3,3,3,321 , 33 .110

34

=

34

6

33

34

6 +

34

6

33 ,

+

30

5,5,5

29,

+

26

4,4,4,

25 ,

+15 1,1,1,1,1,114 , +18 2,2,2,2,2,217 , +22 3,3,3,3,3,321 , 34 .111

+32 633 , , +33 633 , , +34 633 , 1, 2 3+28 5,5,529 , , +29 5,5,529 , , +30 5,5,529, 1, 2 3

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+24 4,4,4,25 , , +25 4,4,4,25 , , +26 4,4,4,25 , 1, 2 +13 1,1,1,1,1,114 , , +14 1,1,1,1,1,114 , , +15 1,1,1,1,1,114 , - are fourth augmentation coefficients+16 2,2,2,2,2,217 , , +17 2,2,2,2,2,217 , , +18 2,2,2,2,2,217 , - fifth augmentation coefficients+

20

3,3,3,3,3,3

21 ,

, +

21

3,3,3,3,3,3

21 ,

, +

22

3,3,3,3,3,3

21 ,

sixth augmentation coefficients .112

.11332 = 32633 32 6 32 635 , 28 5,5,531 , 24 4,4,4,27 , 13 1,1,1,1,1,1, 16 2,2,2,2,2,219, 20 3,3,3,3,3,323 , 32 .11433 = 33632 33 6 33 635 , 29 5,5,531 , 25 4,4,4,27 , 14 1,1,1,1,1,1, 17 2,2,2,2,2,219, 21 3,3,3,3,3,323 , 33 .11534 = 34633 34 6 34 635 , 30 5,5,531 , 26 4,4,4,27 , 15 1,1,1,1,1,1, 18 2,2,2,2,2,219, 22 3,3,3,3,3,323 , 34 .116

32

6

35 ,

,

33

6

35 ,

,

34

6

35,

1, 2

3

28 5,5,531 , , 29 5,5,531 , , 30 5,5,531 , 1, 2 324 4,4,4,27 , , 25 4,4,4,27 , , 26 4,4,4,27 , 1,2 313 1,1,1,1,1,1, , 14 1,1,1,1,1,1, , 15 1,1,1,1,1,1, are fourth detrition coefficients for category 1, 2,and 316 2,2,2,2,2,219, , 17 2,2,2,2,2,219, , 18 2,2,2,2,2,219, are fifth detrition coefficients for category 1,2, and 3 20 3,3,3,3,3,323 , , 21 3,3,3,3,3,323 , , 22 3,3,3,3,3,323 , are sixth detrition coefficients for category 1,2, and 3

.117118

Where we suppose.119

(A) 1, 1, 1, 1,1,1 > 0,, = 13,14,15(B) The functions 1, 1 are positive continuous increasing and bounded.Definition of()1, ()1:1(14 , ) ()1 (13 )(1)1(, ) ()1 ()1 (13 )(1).120121

(C) 21 14 , = ()1limG

1

,

= (

)

1

Definition of(13 )(1), ( 13 )(1) :Where (13 )(1), ( 13 )(1), ()1, ()1 are positive constants and = 13,14,15 .122They satisfy Lipschitz condition:

|()114 , ()114 , | ( 13 )(1)|14 14 |( 13 )(1) |()1 , ()1, | < ( 13 )(1)|| ||( 13 )(1) .123

124

125With the Lipschitz condition, we place a restriction on the behavior of functions

()114 , and()114 , . 14 , and 14 , are points belonging to the interval ( 13 )(1), ( 13 )(1) . It is to benoted that ()114 , is uniformly continuous. In the eventuality of the fact, that if( 13 )(1) = 1 then the function(

)1

14 ,

, the first augmentation coefficient WOULD be absolutely continuous. .126

Definition of(13 )(1)

, ( 13 )(1)

:(D) (13 )(1), ( 13 )(1), are positive constants()1

( 13 )(1) , ()1

( 13 )(1) < 1.127

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Definition of( 13 )(1), ( 13 )(1) :(E) There exists two constants ( 13 )(1) and (13 )(1) which together with ( 13 )(1), ( 13 )(1), (13)(1) and

(13 )(1) and the constants ()1, ( )1, ()1, ( )1, ()1, ()1, = 13,14,15,satisfy the inequalities

1

( 13 )(1) [ ()1 + ()1 + (13 )(1) + ( 13 )(1) ( 13 )(1)] < 11

( 13 )(1) [ (

)

1

+ (

)

1

+ (

13 )

(1) + (

13 )

(1) (

13 )

(1)] < 1.128

129

130

131

132Where we suppose.1342, 2, 2, 2,2,2 > 0, , = 16,17,18.135The functions 2, 2 are positive continuous increasing and bounded..136Definition of(pi)

2, (ri)2:.137 217 , ()2 162 .138 2(19 , ) ()2 ()2 ( 16 )(2) .139lim

2

2

17 ,

= (

)

2

.140

lim 2 19, = ()2 .141Definition of(16 )(2), ( 16 )(2) :Where (16 )(2), ( 16 )(2), ()2, ()2 are positive constants and = 16,17,18 .142They satisfy Lipschitz condition:.143

|( )217 , ( )217 , | ( 16 )(2)|17 17 |( 16 )(2) .144|( )219 , ( )219, | < ( 16 )(2)||19 19 ||( 16 )(2) .145With the Lipschitz condition, we place a restriction on the behavior of functions ( )217 , and( )217 , .17 , And 17 , are points belonging to the interval ( 16 )(2), ( 16 )(2) . It is to be noted that ( )217 , is uniformlycontinuous. In the eventuality of the fact, that if( 16 )(2) = 1 then the function ( )217 , , the SECONDaugmentation coefficient would be absolutely continuous. .146

Definition of(

16 )

(2), (

16 )

(2) :.147

(F) (16 )(2), ( 16 )(2), are positive constants()2( 16 )(2) , ()2( 16 )(2) < 1.148Definition of( 13 )(2), ( 13 )(2) :There exists two constants ( 16 )(2) and (16 )(2) which together with (16 )(2), ( 16 )(2), (16)(2) ( 16 )(2) and theconstants ()2, ()2 , ()2, ()2, ()2, ()2 , = 16,17,18,

satisfy the inequalities .1491

( M 16 )(2) [ (ai)2 + (ai)2 + ( A16 )(2) + ( P16 )(2) ( k16 )(2)] < 1 .1501

( 16 )(2) [ ()2 + ()2 + ( 16 )(2) + ( 16 )(2) ( 16 )(2)] < 1 .151Where we suppose.152

(G)

3,

3,

3,

3,

3,

3 > 0,

,

= 20,21,22

The functions 3

, 3

are positive continuous increasing and bounded.Definition of()3, (ri)3: 3(21 , ) ()3 (20 )(3) 3(23 , ) ()3 ()3 ( 20 )(3).1532 3 21 , = ()3limG 3 23 , = ()3Definition of(20 )(3), ( 20 )(3) :

Where (20 )(3), ( 20 )(3), ()3, ()3 are positive constants and = 20,21,22 .154155

156

They satisfy Lipschitz condition:

|(

)

3

21

,

(

)

3

21 ,

|

(

20 )

(3)|

21

21

|

(

20 )

(3)

|( )323 , ( )323 , | < ( 20 )(3)||23 23 ||( 20 )(3) .157158159

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With the Lipschitz condition, we place a restriction on the behavior of functions ( )321 , and( )321 , .21 , And 21 , are points belonging to the interval ( 20 )(3), ( 20 )(3) . It is to be noted that ( )321 , isuniformly continuous. In the eventuality of the fact, that if( 20 )(3) = 1 then the function ( )321 , , the THIRDaugmentation coefficient, would be absolutely continuous. .160

Definition of(20 )(3), ( 20 )(3) :(H) (

20 )

(3), (

20 )

(3), are positive constants

()3

( 20 )(3) , ()3

( 20 )(3) < 1.161There exists two constants There exists two constants ( 20 )(3) and ( 20 )(3) which together with(20 )(3), ( 20 )(3), (20)(3) ( 20 )(3) and the constants ()3, ()3, ()3, ()3, ()3, ()3, = 20,21,22,satisfy the inequalities

1

( 20 )(3) [ ()3 + ( )3 + (20 )(3) + ( 20 )(3) ( 20 )(3)] < 11

( 20 )(3) [ ()3 + ()3 + ( 20 )(3) + ( 20 )(3) ( 20 )(3)] < 1.162163

164

165

166

167

Where we suppose.168(I) 4, 4, 4, 4, 4 , 4 > 0, , = 24,25,26

(J) The functions 4, 4 are positive continuous increasing and bounded.Definition of()4, ()4: 4(25 , ) ()4 (24 )(4) 427, ()4 ()4 (24 )(4).169(K) 2 4 25 , = ()4

limG 4 27, = ()4Definition of(

24 )

(4), (

24 )

(4) :

Where (24 )(4), ( 24 )(4), ()4, ()4 are positive constants and = 24,25,26 .170They satisfy Lipschitz condition:|( )425 , ( )425 , | ( 24 )(4)|25 25 |( 24 )(4) |( )427 , ( )427, | < ( 24 )(4)||27 27 ||( 24 )(4) .171With the Lipschitz condition, we place a restriction on the behavior of functions ( )425 , and( )425 , .25 , And 25 , are points belonging to the interval ( 24 )(4), ( 24 )(4) . It is to be noted that ( )425 , isuniformly continuous. In the eventuality of the fact, that if( 24 )(4) = 4 then the function ( )425 , , the FOURTHaugmentation coefficient WOULD be absolutely continuous. .172

173

Defi174nition of(

24 )

(4), (

24 )

(4) :

(L) ( 24 )176175(4), ( 24 )(4), are positive constants(M) ()4( 24 )(4) , ()

4( 24 )(4) < 1 .174

Definition of( 24 )(4), ( 24 )(4) :(N) There exists two constants ( 24 )(4) and (24 )(4) which together with ( 24 )(4), ( 24 )(4), (24)(4) (24 )(4)

and the constants ()4, ()4, ()4, ()4, ()4, ()4, = 24,25,26,satisfy the inequalities

1

( 24 )(4) [ ()4 + ( )4 + (24 )(4) + ( 24 )(4) ( 24 )(4)] < 11

( 24 )(4) [ ()4 + ()4 + ( 24 )(4) + ( 24 )(4) ( 24 )(4)] < 1.175Where we suppose.176

(O)

5

,

5

,

5

,

5

,

5

,

5

> 0,

,

= 28,29,30

(P) The functions 5, 5 are positive continuous increasing and bounded.Definition of()5, ()5: 5(29 , ) ()5 (28 )(5)

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531, ()5 ()5 (28 )(5).177(Q) 2 5 29 , = ()5

limG 5 31 , = ()5Definition of(28 )(5), ( 28 )(5) :Where (

28 )

(5), (

28 )

(5), (

)

5

, (

)

5

are positive constants and

= 28,29,30 .178


|( )529 , ( )529 , | ( 28 )(5)|29 29 |( 28 )(5) |( )531 , ( )531, | < ( 28 )(5)||31 31 ||( 28 )(5) .179With the Lipschitz condition, we place a restriction on the behavior of functions ( )529 , and( )529 , .29 , and 29 , are points belonging to the interval ( 28 )(5), (28 )(5) . It is to be noted that ( )529, is uniformlycontinuous. In the eventuality of the fact, that if( 28 )(5) = 5 then the function ( )529 , , theFIFTH augmentationcoefficient attributable would be absolutely continuous. .180

Definition of(28 )(5), ( 28 )(5) :(R) ( 28 )(5), ( 28 )(5), are positive constants

()5( 28 )(5) , ()

5( 28 )(5) < 1.181

Definition of(

28)(5), (

28)(5) :

(S) There exists two constants ( 28 )(5) and ( 28 )(5) which together with ( 28 )(5), ( 28 )(5), (28)(5) (28 )(5) andthe constants ()5, ()5, ()5, ()5, ()5, ()5, = 28,29,30, satisfy the inequalities1

( 28 )(5) [ ()5 + ( )5 + (28 )(5) + ( 28 )(5) ( 28 )(5)] < 11

( 28 )(5) [ ()5 + ()5 + ( 28 )(5) + ( 28 )(5) ( 28 )(5)] < 1.182Where we suppose.1836, 6, 6, 6, 6 , 6 > 0, , = 32,33,34(T) The functions 6, 6 are positive continuous increasing and bounded.

Definition of()6, ()6:

6(

33 ,

)

(

)6

(

32 )

(6)

6

(

35,

)

(

)

6

(

)

6

(

32)(6).184

(U) 2 6 33 , = ()6limG 6 35, = ()6

Definition of(32 )(6), ( 32 )(6) :Where (32 )(6), ( 32 )(6), ()6, ()6 are positive constants and = 32,33,34 .185


|( )633 , ( )633 , | ( 32 )(6)|33 33 |( 32 )(6) |( )635 , ( )635, | < ( 32 )(6)||35 35 ||( 32 )(6) .186With the Lipschitz condition, we place a restriction on the behavior of functions ( )633 , and( )633 , .

33 ,

and

33 ,

are points belonging to the interval

(

32 )(6), (

32 )(6)

. It is to be noted that (

)6

33 ,

is

uniformly continuous. In the eventuality of the fact, that if(

32)(6) = 6 then the function (

)

6

33,

, the SIXTH

augmentation coefficient would be absolutely continuous. .187

Definition of(32 )(6), ( 32 )(6) :(32 )(6), ( 32 )(6), are positive constants

()6( 32 )(6) , ()

6( 32 )(6) < 1.188

Definition of( 32 )(6), ( 32 )(6) :There exists two constants ( 32 )(6) and (32 )(6) which together with ( 32 )(6), ( 32 )(6), (32)(6) ( 32 )(6) and theconstants ()6, ()6 , ()6, ()6, ()6, ()6 , = 32,33,34,satisfy the inequalities

1

( 32 )(6) [ ()6 + ( )6 + (32 )(6) + ( 32 )(6) ( 32 )(6)] < 11

(

32 )

(6) [ (

)6 + (

)6 + (

32 )

(6) + (

32 )

(6) (

32 )

(6)] < 1.189

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.190Theorem 1: if the conditions IN THE FOREGOING above are fulfilled, there exists a solution satisfying the conditions

Definition of 0 ,0 : 13 113 1 , 0 = 0 > 0 () (13 )(1)( 13 )(1) , 0 = 0 > 0 .191.192

Definition of 0 ,0 ( 16 )(2)(16 )(2) , 0 = 0 > 0() (16 )(2)( 16 )(2) , 0 = 0 > 0.193.194

( 20 )(3)(20 )(3) , 0 = 0 > 0() (20 )(3)( 20 )(3) , 0 = 0 > 0.195Definition of 0 ,0 : 24 424 4 , 0 = 0 > 0

(

)

(

24 )

(4)

(

24 )

(4)

,

0

=

0 > 0 .196

Definition of 0 ,0 : 28 528 5 , 0 = 0 > 0 () (28 )(5)( 28 )(5) , 0 = 0 > 0 .197198

Definition of 0 ,0 : 32 632 6 , 0 = 0 > 0 () (32 )(6)( 32 )(6) , 0 = 0 > 0 .199Proof: Consider operator (1) defined on the space of sextuples of continuous functions , :+ + which satisfy.200

0 = 0

, 0 = 0

, 0

( 13 )(1)

,0

(13 )(1)

, .2010 0 ( 13 )(1)(13 )(1) .2020 0 ( 13 )(1)(13 )(1) .203By13 = 130 + (13)11413 (13 )1 + 13 )11413, 131313130 .20414 = 140 + (14)11313 (14 )1 + (14 )11413, 13 1413 130 .20515 = 150 + (15)11413 (15 )1 + (15 )11413, 13 1513130 .20613 = 130 + (13)11413 (13 )1 (13 )113, 131313 130 .20714 = 140 + (14)11313 (14 )1 (14 )113, 131413 130 .208T

15t

= T

15

0 +

(

15)

1

1413

(

15)

1

(

15)

1

13,

13 1513130

Where 13 is the integrand that is integrated over an interval 0, .209.210Proof:

Consider operator (2) defined on the space of sextuples of continuous functions , :+ + which satisfy.2110 = 0 , 0 = 0 , 0 ( 16 )(2) ,0 (16 )(2), .2120 0 ( 16 )(2)(16 )(2) .2130 0 ( 16 )(2)(16 )(2) .214By16 = 160 + (16)21716 (16 )2 + 16 )21716, 161616160 .21517

=

170 +

(

17)

2

1616 (

17)

2

+ (

17)

2

1716,

17 171616

0.216

18 = 180 + (18)21716 (18 )2 + (18 )21716, 16 1816160 .21716 = 160 + (16)21716 (16 )2 (16 )216, 161616 160 .218

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17 = 170 + (17)21616 (17 )2 (17 )216, 161716 160 .21918 = 180 + (18)21716 (18 )2 (18 )216, 161816 160 Where 16 is the integrand that is integrated over an interval 0, .220Proof:

Consider operator

(3) defined on the space of sextuples of continuous functions

,

:

+

+ which satisfy .221

0 = 0

, 0 = 0

, 0

( 20 )(3)

,0

(20 )(3)

, .2220 0 ( 20 )(3)( 20 )(3) .2230 0 ( 20 )(3)(20 )(3) .224By20 = 200 + (20)32120 (20 )3 + 20 )32120, 20 2020 200 .22521 = 210 + (21)32020 (21 )3 + (21 )32120, 20 2120 200 .22622 = 220 + (22)32120 (22 )3 + (22 )32120, 20 2220 200 .22720 = 200 + (20)32120 (20 )3 (20 )320, 202020200 .22821 = 210 + (21)32020 (21 )3 (21 )320, 202120200 .229T

22

t

= T22

0 +

(

22)

3

21

20

(

22

)

3

(

22

)

3

20

,

20

22

20

20

0

Where 20 is the integrand that is integrated over an interval 0, .230Consider operator (4) defined on the space of sextuples of continuous functions , :+ + which satisfy.2310 = 0 , 0 = 0 , 0 ( 24 )(4) ,0 (24 )(4), .2320 0 ( 24 )(4)( 24 )(4) .2330 0 ( 24 )(4)(24 )(4) .234By24 = 240 + (24)42524 (24 )4 + 24 )42524, 24 2424 240 .23525 = 250 + (25)42424 (25 )4 + (25 )42524, 24 2524 240 .236

26

=

260 +

(

26)

4

25

24

(

26 )4 + (

26 )4

25

24

,

24

26

24

24

0

.237

24 = 240 + (24)42524 (24 )4 (24 )424, 242424240 .23825 = 250 + (25)42424 (25 )4 (25 )424, 242524240 .239T26t = T260 + (26)42524 (26 )4 (26 )424, 242624 240 Where 24 is the integrand that is integrated over an interval 0, .240

Consider operator (5) defined on the space of sextuples of continuous functions , :+ + which satisfy.241

2420 = 0 , 0 = 0 , 0 ( 28 )(5) ,0 (28 )(5), .2430 0 ( 28 )(5)( 28 )(5) .2440

0

(

28 )

(5)

(28 )(5) .245

By28 = 280 + (28)52928 (28 )5 + 28 )52928, 282828 280 .24629 = 290 + (29)52828 (29 )5 + (29 )52928, 28 2928280 .24730 = 300 + (30)52928 (30 )5 + (30 )52928, 28 3028 280 .24828 = 280 + (28)52928 (28 )5 (28 )528, 282828280 .24929 = 290 + (29)52828 (29 )5 (29 )528, 28 2928280 .250T30t = T300 + (30)52928 (30 )5 (30 )528, 28 3028280 Where 28 is the integrand that is integrated over an interval 0, .251Consider operator

(6) defined on the space of sextuples of continuous functions

,

:

+

+ which satisfy .252

0 = 0 , 0 = 0 , 0 ( 32 )(6) ,0 (32 )(6), .2530 0 ( 32 )(6)( 32 )(6) .2540 0 ( 32 )(6)(32 )(6) .255

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By32 = 320 + (32)63332 (32 )6 + 32 )63332, 32 3232 320 .25633 = 330 + (33)63232 (33 )6 + (33 )63332, 32 3332 320 .25734 = 340 + (34)63332 (34 )6 + (34 )63332,32 3432 320 .258

32

=

320 +

(

32)

6

33

32

(

32

)

6

(

32

)

6

32

,

32

32

32

32

0

.259

33 = 330 + (33)63232 (33 )6 (33 )632, 323332320 .260T34t = T340 + (34)63332 (34 )6 (34 )632, 323432 320 Where 32 is the integrand that is integrated over an interval 0, .261262

(a) The operator (1) maps the space of functions satisfying GLOBAL EQUATIONS into itself .Indeed it is obvious that13 130 + (13)1 140 +(13 )(1)( 13 )(1)130 13 =1 + (13)1140 + (13)1( 13 )(1)( 13 )(1) ( 13 )(1) 1.263From which it follows that

13

130

(

13 )

(1)

(

13)

1( 13 )

(1)

(

13 )

(1) +

140

( 13 )(1)+140

140

+ (

13 )

(1)

0 is as defined in the statement of theorem 1.264Analogous inequalities hold also for 14 ,15 ,13 ,14 ,15.265The operator (2) maps the space of functions satisfying GLOBAL EQUATIONS into itself .Indeed it is obvious that.26616 160 + (16)2 170 +(16 )(6)( 16 )(2)160 16 = 1 + (16)2170 + (16 )2(16 )(2)( 16 )(2) ( 16 )(2) 1.267From which it follows that

16 160 (16 )(2) (16)2( 16 )(2) ( 16 )(2) + 170 ( 16 )(2)+170170 + ( 16 )(2).268

Analogous inequalities hold also for 17 ,18 ,16 ,17 ,18 .269(a) The operator

(3) maps the space of functions satisfying GLOBAL EQUATIONS into itself .Indeed it is obvious that

20 200 + (20)3 210 +(20 )(3)( 20 )(3)200 20 =1 + (20)3210 + (20)3( 20 )(3)(20 )(3) ( 20 )(3) 1.270From which it follows that

20 200 ( 20 )(3) (20)3( 20 )(3) ( 20 )(3) + 210 ( 20 )(3)+210210 + ( 20 )(3).271

Analogous inequalities hold also for 21 ,22 ,20 ,21 , 22 .272(b) The operator (4) maps the space of functions satisfying GLOBAL EQUATIONS into itself .Indeed it is obvious that24 240 + (24)4 250 +(24 )(4)( 24 )(4)240 24 =

1 + (

24)

4

250 +

(24)4( 24 )(4)(

24 )

(4)

( 24 )(4)

1

.273

From which it follows that24 240 ( 24 )(4) (24)4( 24 )(4) ( 24 )(4) + 250 ( 24 )(4)+250250 + ( 24 )(4)0 is as defined in the statement of theorem 1.274

(c) The operator (5) maps the space of functions satisfying GLOBAL EQUATIONS into itself .Indeed it is obvious that28 280 + (28)5 290 +(28 )(5)( 28 )(5)280 28 =1 + (28)5290 + (28)5( 28 )(5)( 28 )(5) ( 28 )(5) 1.275From which it follows that

28

280

( 28 )(5)

(28)5

(

28 )

(5)

(

28 )

(5) +

290

( 28 )(5)+290290 + (

28 )

(5)

0 is as defined in the statement of theorem 1.276(d) The operator (6) maps the space of functions satisfying GLOBAL EQUATIONS into itself .Indeed it is obvious that

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32 320 + (32)6 330 +(32 )(6)( 32 )(6)320 32 =1 + (32)6330 + (32)6( 32 )(6)(32 )(6) ( 32 )(6) 1.277From which it follows that

32

320

( 32 )(6)

(32)6

(

32 )

(6)

(

32 )

(6) +

330

( 32 )(6)+330330 + (

32 )

(6)

0 is as defined in the statement of theorem 6Analogous inequalities hold also for 25 ,26 ,24 ,25 , 26 .278279

.280

It is now sufficient to take()1

( 13 )(1) , ( )1

( 13 )(1) < 1 and to choose( P13 )(1) and ( Q13 )(1) large to have.281

282

()1(13)1 ( 13)1 + ( 13 )(1) + 0

( 13 )(1)+00 ( 13 )(1).283()1

(13)1 ( 13 )(1) + 0 ( 13 )(1)+0

0 + (13 )(1) (13 )(1).284In order that the operator (1) transforms the space of sextuples of functions , satisfying GLOBAL EQUATIONS intoitself.285

The operator (1) is a contraction with respect to the metric 1,1, 2,2 = {+

1 2(13)1 ,+ 1 2(13)1}.286Indeed if we denote

Definition of

,

:

, = (1)

(,)It results131 2 (13)10 141 142( 13)113( 13)113 13 + {(13 )1131 132( 13)113(13)1130 +(13 )1141, 13131 132( 13)113( 13)113 +132|(13 )1141, 13 (13 )1142, 13| (13)113( 13)113}13Where 13 represents integrand that is integrated over the interval 0, tFrom the hypotheses it follows.2871 2( 13)1 1

( 13)1 (13)1 + (13 )1 + (13)1 + ( 13)1( 13)1 1,1; 2,2And analogous inequalities for

. Taking into account the hypothesis the result follows.288

Remark 1: The fact that we supposed (

13)

1

and (

13)

1

depending also on tcan be considered as not conformal with the

reality, however we have put this hypothesis ,in order that we can postulate condition necessary to prove the uniqueness of

the solution bounded by ( 13)1( 13)1 ( 13)1(13)1 respectively of+.If instead of proving the existence of the solution on +, we have to prove it only on a compact then it suffices to considerthat ( )1 and ( )1, = 13,14,15 depend only on T14 and respectively on ( ) and hypothesis canreplaced by a usual Lipschitz condition..289Remark 2: There does not exist any where = 0 = 0From 19 to 24 it results 0 ( )1( )11413,13130 0 0( )1 > 0 for t > 0.290291

Definition of ( 13)1

1,(13)1

3 :Remark 3: if13 is bounded, the same property have also 14 15 . indeed if13 < ( 13)1 it follows 14 (13)11 (14 )114 and by integrating

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14 (13)12 = 140 + 2(14)1( 13)11/(14 )1In the same way , one can obtain15 (13)13 = 150 + 2(15)1( 13)12/(15 )1If14 15 is bounded, the same property follows for 13 , 15 and 13 , 14 respectively..292

Remark 4: If13 bounded, from below, the same property holds for 14 15 . The proof is analogous with thepreceding one. An analogous property is true if

14 is bounded from below..293

Remark 5: If T13 is bounded from below and lim (( )1 (, ) ) = (14 )1 then 14 .Definition of 1 and 1 :Indeed let 1 be so that for > 1(14)1 ( )1(, ) < 1,13 () > 1.294Then

14 (14)11 114 which leads to14 (14)111 1 1+ 140 1 If we take t such that 1 = 12 it results14 (14)112 , = 21 By taking now 1 sufficiently small one sees that T14 is unbounded. The same propertyholds for 15 iflim (15 )1 , = (15 )1We now state a more precise theorem about the behaviors at infinity of the solutions . 295.296

It is now sufficient to take(

)

2

( 16 )(2) ,(

)

2

( 16 )(2) < 1 and to choose( 16 )(2) (16 )(2) large to have.297()2

(16)2 ( 16)2 + ( 16 )(2) + 0( 16 )(2)+00 ( 16 )(2).298

()2(16)2 ( 16 )(2) + 0

( 16 )(2)+00 + (16 )(2) (16 )(2).299In order that the operator (2) transforms the space of sextuples of functions , satisfying .300The operator

(2) is a contraction with respect to the metric

191

, 191

, 192

, 192

= {+


Definition of19 ,19 : 19 ,19 = (2)(19 ,19).302It results161 2 (16)20 171 172( 16)216( 16)216 16 + {(16 )2161 162( 16)216(16)2160 +(16 )2171, 16161 162( 16)216( 16)216 +

16

2|(

16 )2

17

1,

16

(

16 )2

17

2,

16

|

(16)216

( 16)216}

16 .303

Where

16represents integrand that is integrated over the interval

0,

From the hypotheses it follows.304191 192e( M16)2t1

( M16)2 (16)2 + (16 )2 + ( A 16)2 + ( P16)2( 16)2d191, 191; 192, 192 .305And analogous inequalities for G and T. Taking into account the hypothesis the result follows.306Remark 1: The fact that we supposed (16 )2 and (16 )2 depending also on tcan be considered as not conformal with thereality, however we have put this hypothesis ,in order that we can postulate condition necessary to prove the uniqueness of

the solution bounded by ( P16)2e( M16)2t and ( Q 16)2e( M16)2t respectively of+.If instead of proving the existence of the solution on +, we have to prove it only on a compact then it suffices to considerthat ( )2 and ( )2, = 16,17,18 depend only on T17 and respectively on 19(and not on t) and hypothesis canreplaced by a usual Lipschitz condition..307

Remark 2: There does not exist any t where G

t

= 0 and T

t

= 0

From 19 to 24 it results

Gt G0e ( )2( )2T1716,16d16t0 0Tt T0e( )2t > 0 for t > 0.308

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Definition of ( M16)21 ,( M16)22 and ( M16)23 :Remark 3: ifG16 is bounded, the same property have also G17 and G18 . indeed if

G16 < ( M16)2 it follows dG 17dt ( M16)21 (17 )2G17 and by integratingG17 ( M16)22 = G170 + 2(17)2( M16)21/(17 )2In the same way , one can obtain

G18

( M

16)

2

3= G

18

0 + 2(

18)

2

( M

16)

2

2/(

18)

2

IfG17 or G18 is bounded, the same property follows for G16 , G18 and G16 , G17 respectively..309

310

Remark 4: IfG16 is bounded, from below, the same property holds for G17 and G18 . The proof is analogous with thepreceding one. An analogous property is true ifG17 is bounded from below..311

Remark 5: If T16 is bounded from below and limt(( )2 (19t, t)) = (17 )2 then T17 .Definition of 2 and 2 :Indeed let t2 be so that for t > t2(17)2 ( )2(19t, t) < 2, T16 (t) > 2 .312Then

dT 17

dt (17)22 2T17 which leads to

T17

(

17)

22

2 1

e

2t

+ T17

0 e

2t If we take t such that e

2t =

1

2

it results .313

T17 (17)222 , = log 22 By taking now 2 sufficiently small one sees that T17 is unbounded. The same propertyholds for T18 iflim (18 )2 19t, t = (18 )2We now state a more precise theorem about the behaviors at infinity of the solutions . 314.315


( 20 )(3) , ( )3


()3(20)3 ( 20)3 + ( 20 )(3) + 0

( 20 )(3)+00 ( 20 )(3).317()

3

(20)3 ( 20 )(3) + 0

(

20 )(3)+

0

0

+ (20 )(3) ( 20 )(3).318In order that the operator (3) transforms the space of sextuples of functions , into itself.319The operator (3) is a contraction with respect to the metric 231, 231, 232, 232 = {+


Definition of23 ,23 :23 , 23 = (3)23, 23.321It results

20

1

2

(

20)

3

0

21

1

21

2

(

20)3

20

(

20)3

20

20

+

{(20 )3201 202( 20)320( 20)3200 +(20 )3211, 20201 202( 20)320( 20)320 +202|(20 )3211, 20 (20 )3212,20| ( 20)320(20)320}20Where 20 represents integrand that is integrated over the interval 0, t

From the hypotheses it follows.322

3231 2( 20)3 1

( 20)3 (20)3 + (20 )3 + ( 20)3 + ( 20)3( 20)3 231, 231; 232 , 232And analogous inequalities for

. Taking into account the hypothesis the result follows.324

Remark 1: The fact that we supposed (20 )3

and (20 )3

depending also on tcan be considered as not conformal with thereality, however we have put this hypothesis ,in order that we can postulate condition necessary to prove the uniqueness ofthe solution bounded by ( 20)3( 20)3 ( 20)3( 20)3 respectively of+.

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If instead of proving the existence of the solution on +, we have to prove it only on a compact then it suffices to considerthat ( )3 and ( )3, = 20,21,22 depend only on T21 and respectively on 23( ) and hypothesis canreplaced by a usual Lipschitz condition..325

Remark 2: There does not exist any where = 0 = 0From 19 to 24 it results

0

( )3( )32120,20200

0

0( )3

> 0 for t > 0.326Definition of ( 20)31,( 20)32( 20)33 :Remark 3: if20 is bounded, the same property have also 21 22 . indeed if20 < (20)3 it follows 21 (20)31 (21 )321 and by integrating21 (20)32 = 210 + 2(21)3(20)31/(21 )3In the same way , one can obtain22 (20)33 = 220 + 2(22)3(20)32/(22 )3If21 22 is bounded, the same property follows for 20 , 22 and 20 , 21 respectively..327

Remark 4: If20 bounded, from below, the same property holds for 21 22 . The proof is analogous with thepreceding one. An analogous property is true if21 is bounded from below..328Remark 5: If T20 is bounded from below and lim

((

)3

23

,

)

= (

21 )3 then

21

.

Definition of 3 and 3 :Indeed let 3 be so that for > 3(21)3 ( )323, < 3 ,20 () > 3.329330

Then21 (21)33 321 which leads to21 (21)333 1 3+ 210 3 If we take t such that 3 = 12 it results21 (21)332 , = 23 By taking now 3 sufficiently small one sees that T21 is unbounded. The same property

holds for 22 iflim (22 )3 23, = (22 )3We now state a more precise theorem about the behaviors at infinity of the solutions . 331

.332

It is now sufficient to take()4( 24 )(4) , ( )

4( 24 )(4) < 1 and to choose

( P24 )(4) and ( Q24 )(4) large to have.333()4

(24)4 ( 24)4 + ( 24 )(4) + 0( 24 )(4)+00 ( 24 )(4).334

()4(24)4 ( 24 )(4) + 0

( 24 )(4)+00 + (24 )(4) ( 24 )(4).335In order that the operator (4) transforms the space of sextuples of functions , satisfying IN to itself.336The operator

(4) is a contraction with respect to the metric

271

, 271

, 272

, 272

= {+

1 2(24)4 ,+ 1 2(24)4}Indeed if we denote

Definition of27 , 27 : 27 , 27 = (4)(27, 27)It results241 2 (24)40 251 252( 24)424( 24)424 24 + {(24 )4241 242( 24)424(24)4240 +

(24 )4251, 24241 242( 24)424( 24)424 +

24

2|(

24 )4

25

1,

24

(

24 )4

25

2,

24

|

( 24)424

( 24)424}

24

Where

24represents integrand that is integrated over the interval

0, t


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338271 272( 24 )4 1

( 24)4 (24)4 + (24 )4 + ( 24)4 + ( 24)4( 24)4 271, 271; 272 , 272And analogous inequalities for . Taking into account the hypothesis the result follows.339Remark 1: The fact that we supposed (

24 )4 and (

24 )4 depending also on tcan be considered as not conformal with the

reality, however we have put this hypothesis ,in order that we can postulate condition necessary to prove the uniqueness of

the solution bounded by ( 24)4( 24)4 ( 24)4( 24)4 respectively of+.If instead of proving the existence of the solution on +, we have to prove it only on a compact then it suffices to considerthat ( )4 and ( )4, = 24,25,26 depend only on T25 and respectively on 27( ) and hypothesis canreplaced by a usual Lipschitz condition..340

Remark 2: There does not exist any where = 0 = 0From 19 to 24 it results 0 ( )4( )42524,24240 0 0( )4 > 0 for t > 0.341Definition of ( 24)41,( 24)42( 24)43 :Remark 3: if24 is bounded, the same property have also 25 26 . indeed if

24 < (

24)

4

it follows

25

(

24)

4

1

(

25

)

4

25 and by integrating

25 (24)42 = 250 + 2(25)4(24)41/(25 )4In the same way , one can obtain26 (24)43 = 260 + 2(26)4(24)42/(26 )4If25 26 is bounded, the same property follows for 24 , 26 and 24 , 25 respectively..342

Remark 4: If24 bounded, from below, the same property holds for 25 26 . The proof is analogous with thepreceding one. An analogous property is true if25 is bounded from below..343Remark 5: If T24 is bounded from below and lim (( )4 (27, ) ) = (25 )4 then 25 .Definition of 4 and 4 :Indeed let 4 be so that for > 4(25)4 ( )4(27, ) < 4 ,24 () > 4.344Then

25

(

25)

4

4

4

25 which leads to

25 (25)444 1 4+ 250 4 If we take t such that 4 = 12 it results25 (25)442 , = 24 By taking now 4 sufficiently small one sees that T25 is unbounded. The same propertyholds for 26 iflim (26 )4 27, = (26 )4We now state a more precise theorem about the behaviors at infinity of the solutions ANALOGOUS inequalities hold also

for 29 ,30 ,28 ,29 , 30 .345.346


( 28 )(5) , ()5


(

)

5

(28)5 ( 28)5 + ( 28 )(5) + 0( 28 )(5)+

0

0

( 28 )(5).348()5

(28)5 ( 28 )(5) + 0( 28 )(5)+00 + (28 )(5) ( 28 )(5).349

In order that the operator (5) transforms the space of sextuples of functions , into itself.350The operator (5) is a contraction with respect to the metric 311, 311, 312, 312 = {+


Definition of31 , 31 : 31 , 31 = (5)31, 31It results281 2 (28)50 291 292( 28)528( 28)528 28 +

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{(28 )5281 282( 28)528( 28)5280 +(28 )5291, 28281 282( 28)528( 28)528 +282|(28 )5291, 28 (28 )5292,28| ( 28)528(28)528}28Where 28 represents integrand that is integrated over the interval 0, t


352311 312( 28 )5 1

( 28)5 (28)5 + (28 )5 + ( 28)5 + ( 28)5( 28)5 311, 311; 312 , 312And analogous inequalities for . Taking into account the hypothesis (35,35,36) the result follows.353Remark 1: The fact that we supposed (28 )5 and (28 )5 depending also on tcan be considered as not conformal with thereality, however we have put this hypothesis ,in order that we can postulate condition necessary to prove the uniqueness of


Remark 2: There does not exist any

where

= 0

= 0

From GLOBAL EQUATIONS it results 0 ( )5( )52928,28280 0 0( )5 > 0 for t > 0.355Definition of ( 28)51,( 28)52( 28)53 :Remark 3: if28 is bounded, the same property have also 2930 . indeed if28 < (28)5 it follows 29 (28)51 (29 )529 and by integrating29 (28)52 = 290 + 2(29)5( 28)51/(29 )5In the same way , one can obtain30 (28)53 = 300 + 2(30)5(28)52/(30 )5If

29

30 is bounded, the same property follows for

28 ,

30 and

28 ,

29 respectively..356

Remark 4: If

28

bounded, from below, the same property holds for

29

30. The proof is analogous with the

preceding one. An analogous property is true if29 is bounded from below..357Remark 5: If T28 is bounded from below and lim (( )5 (31, ) ) = (29 )5 then 29 .Definition of 5 and 5 :Indeed let 5 be so that for > 5

(29)5 ( )5(31, ) < 5,28 () > 5.358359

Then29 (29)55 529 which leads to29 (29)555 1 5+ 290 5 If we take t such that 5 = 12 it results

29 (

29)

5

5

2 , = 2

5 By taking now 5 sufficiently small one sees that T29 is unbounded. The same propertyholds for 30 iflim (30 )5 31, = (30 )5We now state a more precise theorem about the behaviors at infinity of the solutionsAnalogous inequalities hold also for 33 ,34 ,32 ,33 , 34 .360.361


( 32 )(6) , ( )6


()6(32)6 ( 32)6 + ( 32 )(6) + 0

( 32 )(6)+00 ( 32 )(6).363(

)

6

(32)6 ( 32 )(6) + 0

( 32 )(6)+0

0

+ (32 )(6) ( 32 )(6).364In order that the operator (6) transforms the space of sextuples of functions , into itself.365The operator (6) is a contraction with respect to the metric

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351, 351, 352, 352 = {+


Definition of

35

,

35

:

35

,

35

=

(6)

35

,

35

It results

321 2 (32)60 331 332( 32)632( 32)632 32 + {(32 )6321 322( 32)632( 32)6320 +(32 )6331, 32321 322( 32)632( 32)632 +322|(32 )6331, 32 (32 )6332,32| ( 32)632(32)632}32Where 32 represents integrand that is integrated over the interval 0, t


367351 352( 32 )6 1

(

32)

6

(

32)

6 + (

32 )6 + (

32)

6 + (

32)6(

32)

6

35

1,

35

1;

35

2,

35

2

And analogous inequalities for . Taking into account the hypothesis the result follows.368Remark 1: The fact that we supposed (32 )6 and (32 )6 depending also on tcan be considered as not conformal with thereality, however we have put this hypothesis ,in order that we can postulate condition necessary to prove the uniqueness of


Remark 2: There does not exist any where = 0 = 0From 69 to 32 it results 0 ( )6( )63332,32320 0

0

( )6 > 0 for t > 0.370

Definition of ( 32)6

1,

( 32)6

2( 32)6

3 :Remark 3: if32 is bounded, the same property have also 33 34 . indeed if32 < (32)6 it follows 33 (32)61 (33 )633 and by integrating33 (32)62 = 330 + 2(33)6(32)61/(33 )6In the same way , one can obtain34 (32)63 = 340 + 2(34)6(32)62/(34 )6If33 34 is bounded, the same property follows for 32 , 34 and 32 , 33 respectively..371

Remark 4: If32 bounded, from below, the same property holds for 33 34 . The proof is analogous with thepreceding one. An analogous property is true if33 is bounded from below..372Remark 5: If T32 is bounded from below and lim (( )6 (35, ) ) = (33 )6 then 33 .Definition of

6 and

6 :

Indeed let

6 be so that for

>

6

(33)6 ( )635, < 6,32 () > 6.373374

Then33 (33)66 633 which leads to33 (33)666 1 6+ 330 6 If we take t such that 6 = 12 it results33 (33)662 , = 26 By taking now 6 sufficiently small one sees that T33 is unbounded. The same property

holds for 34 iflim (34 )6 35, , = (34 )6We now state a more precise theorem about the behaviors at infinity of the solutions .375376

Behavior of the solutionsIf we denote and define

Definition of (1)1 , (2)1 , (1)1 , (2)1 :(a) 1)1 , (2)1 , (1)1 , (2)1 four constants satisfying

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(2)1 (13 )1 + (14 )1 (13 )114 , + (14 )114 , (1)1(2)1 (13 )1 + (14 )1 (13 )1, (14 )1, (1)1.377Definition of (1 )1, (2)1, (1)1, (2)1,1,1 :By (1)1 > 0 , (2)1 < 0 and respectively (1)1 > 0 , (2)1 < 0 the roots of the equations (14)112 +(

1)

1

1

(

13)

1 = 0 and (

14)

1

1

2

+ (

1)

1

1

(

13)

1 = 0.378Definition of (

1 )

1

, , (

2)

1

, (

1)

1

, (

2)

1

:

By (1)1 > 0 , (2)1 < 0 and respectively (1)1 > 0 , (2)1 < 0 the roots of the equations (14)112 +(2)11 (13)1 = 0 and (14)112 + (2)11 (13)1 = 0 .379Definition of (1)1 , (2)1 , (1)1, (2)1, (0 )1 :-(b) If we define (1)1 , (2)1 , (1)1, (2)1 by

(2)1 = (0)1, (1)1 = (1)1, (0)1 < (1)1(2)1 = (1)1, (1)1 = (1)1 , (1)1 < (0)1 < (1)1,and (0 )1 = 130140

( 2)1 = (1)1, (1)1 = (0 )1, (1)1 < (0)1 .380and analogously

(

2)

1 = (

0)1, (

1)

1 = (

1)1,

(

0)

1 < (

1)1

(2)1 = (1)1, (1)1 = (1)1 , (1)1 < (0)1 < (1)1,and (0)1 = 130140 ( 2)1 = (1)1, (1)1 = (0)1, (1)1 < (0)1 where (1)1, (1)1

are defined respectively.381382Then the solution satisfies the inequalities130 (1)1(13 )1 13 () 130 (1)1 where ()1 is defined

1

(1)1 130 (1)1(13)1 14 () 1(2)1 130 (1)1 .383(

(15 )1130(

1)

1

(

1)

1

(

13 )

1

(

2)

1

(1)1(13 )1

(2)1

+

150

(2)1

15(

)

(15 )1130

(

2)

1

(

1)

1

(

15

)

1

[

(1)1

(15 )1 ] + 150 (15 )1) .384130 (1)1 13 () 130 (1)1+(13 )1 .3851

(1)1 130 (1)1 13() 1(2)1130 (1)1+(13 )1 .386(15)1130

(1)1(1)1(15 )1 (1)1 (15 )1+ 150 (15 )1 15() (15)1130

(2)1(1)1+(13 )1+(2)1 (1)1+(13)1 (2)1 + 150 (2)1 .387

Definition of(1)1, (2)1, (1)1, (2)1:-Where (1)1 = (13)1(2)1 (13 )1

(

2)

1

= (

15)

1

(

15)

1

(1)1 = (13)1(2)1 (13 )1(2)1 = (15 )1 (15)1.388Behavior of the solutionsIf we denote and define.389

Definition of (1)2 , (2)2 , (1)2 , (2)2 :1)2 , (2)2 , (1)2 , (2)2 four constants satisfying.390(2)2 (16 )2 + (17 )2 (16 )2T17 , + (17 )2T17 , (1)2 .391(2)2 (16 )2 + (17 )2 (16 )219, (17 )219, (1)2 .392Definition of (1 )2, (2)2 , (1)2, (2)2 :.393By (1)2 > 0 , (2)2 < 0 and respectively (1)2 > 0 , (2)2 < 0 the roots.394of the equations (

17)

2

2

2

+ (

1)

2

2

(

16)

2 = 0 .395and (14)

2

2

2

+ (1)2

2

(16)2

= 0 and.396Definition of (1 )2, , (2)2, (1)2, (2)2 :.397By (1)2 > 0 , (2)2 < 0 and respectively (1)2 > 0 , (2)2 < 0 the.398roots of the equations (17)222 + (2)22 (16)2 = 0.399

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and (17)222 + (2)22 (16)2 = 0 .400Definition of (1)2 , (2)2 , (1)2, (2)2 :-.401If we define (1)2 , (2)2 , (1)2, (2)2 by.402(2)2 = (0)2, (1)2 = (1)2, (0)2 < (1)2 .403(2)2 = (1)2, (1)2 = (1)2 , (1)2 < (0)2 < (1)2,and (0)

2

=G16

0

G170 .404( 2)2 = (1)2, (1)2 = (0 )2, (1)2 < (0 )2 .405and analogously

(2)2 = (0)2, (1)2 = (1)2, (0)2 < (1)2(2)2 = (1)2, (1)2 = (1)2 , (1)2 < (0)2 < (1)2,and (0)2 = T160T170 .406( 2)2 = (1)2, (1)2 = (0)2, (1)2 < (0)2 .407Then the solution satisfies the inequalities

G160 e(S1)2(16)2t 16 G160 e(S1)2t.408

(

)2 is defined.409

1

(1)2G

16

0 e

(S1)

2

(

16 )

2

t

17(

)

1

(2)2G

16

0 e(S1)

2

t .410

((18)2G160

(1)2(S1)2(16)2(S2)2 e(S1)2(16 )2t e(S2)2t + G180 e(S2)2t G18() (18)2G160(2)2(S1)2(18 )2 [e(S1)2te(18 )2t] + G180 e(18 )2t) .411T16

0 e(R1)2 16() T160 e(R1)2+(16)2 .412

1

(1)2 T160 e(R1)2 16() 1(2)2 T160 e(R1)2+(16 )2 .413(18 )2T160

(1)2(R1)2(18 )2 e(R1)2 e(18 )2 + T180 e(18 )2 18() (18)2T160

(2)2(R1)2+(16)2+(R2)2 e(R1)2+(16)2 e(R2)2 + T180 e(R2)2 .414Definition of(S1)

2, (S2)2, (R1)2, (R2)2:-.415Where (S1)

2

= (

16)

2

(

2)

2

(

16

)

2

(S2)2 = (18)2 (18)2 .416(1)2 = (16)2(2)1 (16 )2(R2)

2 = (18 )2 (18)2.417.418Behavior of the solutionsIf we denote and define

Definition of (1)3 , (2)3 , (1)3 , (2)3 :(a) 1)3 , (2)3 , (1)3 , (2)3 four constants satisfying(2)3 (20 )3 + (21 )3 (20 )321 , + (21 )321 , (1)3(2)3 (20 )3 + (21 )3 (20 )3, (21 )323, (1)3.419Definition of (

1 )

3, (

2)

3, (

1)

3, (

2)

3 :(b)

By (1)

3

> 0 , (2)3

< 0 and respectively (1)3

> 0 , (2)3

< 0 the roots of the equations (21)3

3

2

+(1)33 (20)3 = 0and (21)332 + (1)33 (20)3 = 0 andBy (1)3 > 0 , (2)3 < 0 and respectively (1)3 > 0 , (2)3 < 0 the

roots of the equations (21)332 + (2)33 (20)3 = 0and (21)332 + (2)33 (20)3 = 0.420

Definition of (1)3 , (2)3 , (1)3, (2)3 :-(c) If we define (1)3 , (2)3 , (1)3, (2)3 by

(2)3 = (0)3, (1)3 = (1)3, (0)3 < (1)3(2)3 = (1)3, (1)3 = (1)3 , (1)3 < (0)3 < (1)3,and (

0 )

3

=

200

210

( 2)3 = (1)3, (1)3 = (0 )3, (1 )3 < (0)3 .421and analogously(2)3 = (0)3, (1)3 = (1)3, (0)3 < (1)3

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(2)3 = (1)3, (1)3 = (1)3 , (1)3 < (0)3 < (1)3, and (0)3 = 200210 ( 2)3 = (1)3, (1)3 = (0)3, (1)3 < (0)3Then the solution satisfies the inequalities200 (1)3(20)3 20 () 200 (1)3

(

)3 is defined .422

4231

(1)3 200 (1)3(20)3 21() 1(2)3 200 (1)3 .424(

(22 )3200(1)3(1)3(20 )3(2)3 (1)3(20 )3 (2)3 + 220 (2)3 22() (22 )3200(2)3(1)3(22 )3 [(1)3 (22 )3 ] + 220 (22 )3) .425200 (1)3 20() 200 (1)3+(20)3 .426

1

(1)3 200 (1)3 20 () 1(2)3 200 (1)3+(20 )3 .427(22 )3200

(1)3(1)3(22 )3 (1)3 (22 )3+ 220 (22 )3 22() (22)3200

(

2)3

(

1)

3+(

20 )3+(

2)3

(1)3+(20)3

(2)3

+

220

(2)3 .428

Definition of(1)3, (2)3, (1)3, (2)3:-Where (1)3 = (20)3(2)3 (20 )3(2)3 = (22)3 (22)3(1)3 = (20)3(2)3 (20 )3(2)3 = (22 )3 (22 )3.429

.430

.431


Definition of (1)4 , (2)4 , (1)4 , (2)4 :(d)

(1)

4

, (2)4

, (1)4

, (2)4

four constants satisfying

(2)4 (24 )4 + (25 )4 (24 )425 , + (25 )425 , (1)4(2)4 (24 )4 + (25 )4 (24 )427, (25 )427, (1)4

432

Definition of (1 )4, (2)4, (1)4, (2)4,4,4 :(e) By (1)4 > 0 , (2)4 < 0 and respectively (1)4 > 0 , (2)4 < 0 the roots of the equations

(25)442 + (1)44 (24)4 = 0and (25)442 + (1)44 (24)4 = 0 and

433

Definition of (1 )4, , (2)4, (1)4, (2)4 :By (1)4 > 0 , (2)4 < 0 and respectively (1)4 > 0 , (2)4 < 0 theroots of the equations (25)442 + (2)44 (24)4 = 0and (25)442 + (2)44 (24)4 = 0

Definition of (1)4 , (2)4 , (1)4, (2)4, (0 )4 :-(f) If we define (1)4 , (2)4 , (1)4, (2)4 by

(2)4 = (0)4, (1)4 = (1)4, (0)4 < (1)4(

2)

4

= (

1)

4

, (

1)

4

= (

1)

4

,

(

4)

4

< (

0)

4

< (

1)

4

,

and (0 )4 = 240250 ( 2)4 = (4)4, (1)4 = (0 )4, (4)4 < (0)4

434435

436

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and analogously

(2)4 = (0)4, (1)4 = (1)4, (0)4 < (1)4(

2)

4 = (

1)

4, (

1)

4 = (

1)

4 ,

(

1)

4 < (

0)

4 < (

1)

4,and (0)4 = 24

0

250 ( 2)4 = (1)4, (1)4 = (0)4, (1)4 < (0)4 where (1)4, (1)4

are defined by 59 and 64 respectively

437

438

Then the solution satisfies the inequalities

240 (1)4(24 )4 24 240 (1)4 where ()4 is defined

439440

441442443

444445

1

(1)4 240 (1)4(24)4 25 1(2)4 240 (1)4 446447 (26)4240

(1)4(1)4(24 )4(2)4 (1)4(24 )4 (2)4 + 260 (2)4 26 (26)4240(2)4(1)4(26)4(1)4(26)4+260(26)4

448

240 (1)4 24 240 (1)4+(24)4 4491

(

1)4

240

(1)4

24 (

)

1

(

2)4

240

(1)4+(24 )4 450

(26 )4240(1)4(1)4(26 )4 (1)4 (26 )4+ 260 (26 )4 26()

(26)4240(2)4(1)4+(24 )4+(2)4 (1)4+(24)4 (2)4 + 260 (2)4

451


(2)4 = (26)4 (26)4(1)

4

= (24)4

(2)4

(24 )4

(2)4 = (26 )4 (26 )4

452

453


Definition of (1)5 , (2)5 , (1)5 , (2)5 :(g) (1)5 , (2)5 , (1)5 , (2)5 four constants satisfying

(

2)

5

(

28 )5 + (

29 )5

(

28 )5

29 ,

+ (

29 )5

29 ,

(

1)

5

(2)5 (28 )5 + (29 )5 (28 )531, (29 )531, (1)5

454

Definition of (1 )5, (2)5, (1)5, (2)5,5,5 : 455

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(h) By (1)5 > 0 , (2)5 < 0 and respectively (1)5 > 0 , (2)5 < 0 the roots of the equations(29)552 + (1)55 (28)5 = 0and (29)552 + (1)55 (28)5 = 0 and

Definition of (

1 )

5, , (

2)

5, (

1)

5, (

2)

5 :By (1)5 > 0 , (2)5 < 0 and respectively (1)5 > 0 , (2)5 < 0 theroots of the equations (29)552 + (2)55 (28)5 = 0

and (29)552 + (2)55 (28)5 = 0Definition of (1)5 , (2)5 , (1)5, (2)5, (0 )5 :-(i) If we define (1)5 , (2)5 , (1)5, (2)5 by

(2)5 = (0)5, (1)5 = (1)5, (0)5 < (1)5(2)5 = (1)5, (1)5 = (1)5 , (1)5 < (0)5 < (1)5,and (

0 )

5

=

28029

0

( 2)5 = (1)5, (1)5 = (0 )5, (1 )5 < (0)5

456

and analogously

(2)5 = (0)5, (1)5 = (1)5, (0)5 < (1)5(2)5 = (1)5, (1)5 = (1)5 , (1)5 < (0)5 < (1)5,

and (0)5 = 280290 (

2)

5

= (

1)

5

, (

1)

5

= (

0)

5

,

(

1)

5

< (

0)

5

where (

1)

5

, (

1)

5

are defined respectively

457


280 (1)5(28)5 28 () 280 (1)5 where ()5 is defined

458

1

(5)5 280 (1)5(28)5 29() 1(2)5280 (1)5 459460

(30)5280

(

1)5

(

1)5

(

28 )

5

(

2)5

(1)5(28 )5

(2)5

+

300

(2)5

30

(30)5

280(

2)5(

1)5

(30

)5

(1)5

(30

)5

+300

(30

)5

461

280 (1)5 28 () 280 (1)5+(28)5 4621

(1)5 280 (1)5 28 () 1(2)5 280 (1)5+(28 )5 463(30 )5280

(1)5(1)5(30 )5 (1)5 (30 )5+ 300 (30 )5 30() (30)5280

(

2)5

(

1)

5+(

28 )5+(

2)5

(1)5+(28)5

(2)5

+

300

(2)5

464


465

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(2)5 = (30)5 (30)5(1)5 = (28)5(2)5 (28 )5

(

2)

5 = (

30 )5

(

30 )

5Behavior of the solutionsIf we denote and define

Definition of (1)6 , (2)6 , (1)6 , (2)6 :(j) (1)6 , (2)6 , (1)6 , (2)6 four constants satisfying(2)6 (32 )6 + (33 )6 (32 )633 , + (33 )633 , (1)6(2)6 (32 )6 + (33 )6 (32 )635, (33 )635, (1)6

466

Definition of (

1 )

6

, (

2)

6

, (

1)

6

, (

2)

6

,

6

,

6

:

(k) By (1)6 > 0 , (2)6 < 0 and respectively (1)6 > 0 , (2)6 < 0 the roots of the equations(33)662 + (1)66 (32)6 = 0and (33)662 + (1)66 (32)6 = 0 and

467

Definition of (1 )6, , (2)6, (1)6, (2)6 :By (1)6 > 0 , (2)6 < 0 and respectively (1)6 > 0 , (2)6 < 0 the

roots of the equations (33)662 + (2)66 (32)6 = 0and (33)662 + (2)66 (32)6 = 0

Definition of (

1)

6

, (

2)

6

, (

1)

6

, (

2)

6

, (

0 )

6

:-

(l) If we define (1)6 , (2)6 , (1)6, (2)6 by(2)6 = (0)6, (1)6 = (1)6, (0)6 < (1)6(2)6 = (1)6, (1)6 = (6 )6 , (1 )6 < (0 )6 < (1)6,and (0 )6 = 320330

( 2)6 = (1)6, (1)6 = (0 )6, (1 )6 < (0)6

468

470

and analogously

(2)6 = (0)6, (1)6 = (1)6, (0)6 < (1)6(2)6 = (1)6, (1)6 = (1)6 , (1)6 < (0)6 < (1)6,

and (0)6 = 320330 ( 2)6 = (1)6, (1)6 = (0)6, (1)6 < (0)6 where (1)6, (1)6

are defined respectively

471


320

(

1)

6

(

32)

6

32 () 320

(

1)

6

where ()6 is defined

472

1

(1)6 320 (1)6(32)6 33() 1(2)6 320 (1)6 473

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(34)6320(1)6(1)6(32 )6(2)6 (1)6(32 )6 (2)6 + 340 (2)6 34 (34)6320(2)6(1)6(34)6(1)6(34)6+340(34)6

474

320

(

1)

6

32(

)

320

(

1)

6

+(

32)

6

475

1

(1)6 320 (1)6 32 () 1(2)6 320 (1)6+(32 )6 476(34 )6320

(1)6(1)6(34 )6 (1)6 (34 )6+ 340 (34 )6 34() (34)6320

(2)6(1)6+(32 )6+(2)6 (1)6+(32)6 (2)6 + 340 (2)6 477

Definition of(1)6, (2)6, (1)6, (2)6:-Where (

1)

6

= (

32)

6

(

2)

6

(

32

)

6

(2)6 = (34)6 (34)6(1)6 = (32)6(2)6 (32 )6(2)6 = (34 )6 (34 )6

478

479Proof : From GLOBAL EQUATIONS we obtain1 = (13)1 (13 )1 (14 )1 + (13 )114 , (14 )114 , 1 (14)11Definition of1 :- 1 = 1314

It follows

(14)112 + (2)11 (13)1 1 (14)112 + (1)11 (13)1From which one obtains

Definition of(1)1 , (0 )1 :-(a) For 0 < (0 )1 = 130140 < (1)1 < (1 )1 1() (1)1+()1(2)1 141(1)1(0)1

1+()1 141(1)1(0)1 , ()1 = (1)1(0)1(0)1(2)1 it follows (0)1 1() (1)1

480

481

In the same manner , we get

1() (1)1+()1(2)1 141(1)1(2)11+()1 141(1)1(2)1 , ()1 = (1)1(0)1(0)1(2)1 From which we deduce (0)1 1() (1)1

482

(b) If 0 < (1)1 < (0)1 = 130140 < (1 )1 we find like in the previous case,(1 )1 (1)1+1(2)1 141(1)1(2)1

1+1 141(1)1(2)1 1 (1)1+1(2)1 141(1)1(2)1

1+1 141(1)1(2)1 (1)1

483

(c) If 0 < (1)1 (1)1 (0)1 = 130

140 , we obtain(1)1 1 (1)1+1(2)1 141(1)1(2)1

1+1 141(1)1(2)1 (0)1484

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And so with the notation of the first part of condition (c) , we have

Definition of 1 :-(2)1 1 (1)1, 1 = 1314

In a completely analogous way, we obtain

Definition of

1

:-

(2)1 1 (1)1, 1 = 13

14 Now, using this result and replacing it in GLOBAL E486QUATIONS we get easily the result stated in thetheorem.Particular case :

If(13 )1 = (14 )1, (1)1 = (2)1 and in this case (1)1 = (1)1 if in addition (0)1 = (1 )1then 1 = (0)1 and as a consequence 13 () = (0)114() this also defines (0)1 for the special caseAnalogously if (13 )1 = (14 )1, (1)1 = (2)1 and then(1)1 =(1)1if in addition (0)1 = (1)1 then 13 () = (0)114 () This is an important consequenceof the relation between (1 )1 and (1)1 , and definition of(0)1.

485

486we obtain

d

2

dt = (16)2

(16 )2

(17 )2

+ (16 )2

T17 , t (17 )2

T17 , t2

(17)2

2

487

Definition of2 :- 2 = G16G17 488It follows(17)222 + (2)22 (16)2 d2dt (17)222 + (1)22 (16)2

489

From which one obtains

Definition of(1)2 , (0 )2 :-(d) For 0 < (0)2 = G160G170 < (1)2 < (1)2

2(

)

(1)2+(C)2(2)2 172(1)2(0)2

1+(C)

2

172(1)2(0)2 , (C)2 = (1)

2(0)2(

0)2

(

2)2

it follows (0)2 2() (1)2

490

In the same manner , we get2() (1)2+(C)2(2)2 172(1)2(2)21+(C)2 172(1)2(2)2 , (C)2 = (1)2(0)2(0)2(2)2

491

From which we deduce (0)2 2() (1)2 492(e) If 0 < (1)2 < (0)2 = G160G170 < (1)2 we find like in the previous case,(1)2 (1)2+C2(2)2 172(1)2(2)2

1+C2 172(1)2(2)2 2 (1)2+C2(2)2 172(1)2(2)2

1+C2 172(1)2(2)2 (1)2

493

(f) If 0 < (1)2

(1)2

(0 )2

=G16

0

G170 , we obtain

(1)2 2 (1)2+C2(2)2 172(1)2(2)21+C2 172(1)2(2)2 (0 )2

And so with the notation of the first part of condition (c) , we have

494

Definition of 2 :-(2)2 2 (1)2, 2 = 1617

495

In a completely analogous way, we obtain

Definition of 2 :-(

2)

2

2

(

1)

2,

2

=

16

17

496

. 497Particular case :

If(16 )2 = (17 )2, (1)2 = (2)2 and in this case (1)2 = (1)2 if in addition (0)2 = (1)2then 2 = (0)2 and as a consequence 16 () = (0)217()

498

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Analogously if (16 )2 = (17 )2, (1)2 = (2)2 and then(1)2 =(1)2if in addition (0)2 = (1)2 then 16 () = (0)217 () This is an important consequenceof the relation between (1 )2 and (1)2

499

From GLOBAL EQUATIONS we obtain3

= (

20)

3

(

20

)

3

(

21

)

3

+ (

20

)

3

21 ,

(

21

)

3

21 ,

3

(

21)

3

3

500

Definition of3 :- 3 = 2021 It follows(21)332 + (2)33 (20)3 3 (21)332 + (1)33 (20)3501


(a) For 0 < (0)3 = 200210 < (1)3 < (1)33() (1)3+()3(2)3 213(1)3(0)3

1+()3 213(1)3(0)3 , ()3 = (1)3(0)3(0)3(2)3 it follows (

0)

3

3

(

)

(

1)

3

502

In the same manner , we get3() (1)3+()3(2)3 213(1)3(2)31+()3 213(1)3(2)3 , ()3 = (1)3(0)3(0)3(2)3

Definition of(1)3 :-From which we deduce (0)3 3() (1)3

503

(b) If 0 < (1)3 < (0)3 = 200210 < (1 )3 we find like in the previous case,(1)3 (1)3+3(2)3 213(1)3(2)3

1+3 213(1)3(2)3 3 (

1)3+

3(

2)3

213(1)3(2)3

1+3 213

(1)3

(2)3

(

1)

3

504

(c) If 0 < (1)3 (1)3 (0)3 = 200210 , we obtain(1)3 3 (1)3+3(2)3 213(1)3(2)3

1+3 213(1)3(2)3 (0)3And so with the notation of the first part of condition (c) , we have

Definition of 3 :-(2)3 3 (1)3, 3 = 2021 In a completely analogous way, we obtain

Definition of

3

:-

(2)3 3 (1)3, 3 = 20

21 Now, using this result and replacing it in GLOBAL EQUATIONS we get easily the result stated in the theorem.Particular case :

If(20 )3 = (21 )3, (1)3 = (2)3 and in this case (1 )3 = (1)3 if in addition (0)3 = (1)3then 3 = (0)3 and as a consequence 20 () = (0)321()Analogously if (20 )3 = (21 )3, (1)3 = (2)3 and then(1)3 =(1)3if in addition (0)3 = (1)3 then 20 () = (0)321 () This is an important consequenceof the relation between (1 )3 and (1)3

505

506

: From GLOBAL EQUATIONS we obtain

4

= (24)4

(24 )4

(25 )4

+ (24 )4

25 , (25 )4

25 , 4

(25)4

4

507

508

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Definition of4 :- 4 = 2425 It follows(25)442 + (2)44 (24)4 4 (25)442 + (4)44 (24)4


Definition of(1)4 , (0 )4 :-(d) For 0 < (0 )4 = 240250 < (1)4 < (1 )4 4 (1)4+4(2)4 254(1)4(0)4

4+4 254(1)4(0)4 , 4 = (1)4(0)4(0)4(2)4 it follows (0)4 4() (1)4In the same manner , we get

4 (1)4+4(2)4 254(1)4(2)44+4 254(1)4(2)4 , ()4 = (1)4(0)4(0)4(2)4

From which we deduce (0 )4 4() (1 )4

509

(e) If 0 < (1)4 < (0)4 = 240250 < (1 )4 we find like in the previous case,(1 )4 (1)4+4(2)4 254(1)4(2)4

1+4 254(1)4(2)4 4 (1)4+4(2)4 25

4

(1)4

(2)4

1+4 254(1)4(2)4 (1)4

510

511

(f) If 0 < (1)4 (1)4 (0)4 = 240250 , we obtain(1)4 4 (1)4+4(2)4 254(1)4(2)4


Definition of 4 :-(2)4 4 (1)4, 4 = 2425 In a completely analogous way, we obtainDefinition of 4 :-(2)4 4 (1)4, 4 = 2425 Now, using this result and replacing it in GLOBAL EQUATIONS we get easily the result stated in the theorem.

Particular case :

If(

24)

4

= (

25)

4

,

(

1)

4

= (

2)

4

and in this case (

1)

4

= (

1)

4

if in addition (

0)

4

= (

1)

4

then 4 = (0)4 and as a consequence 24 () = (0)425()this also defines (0)4 for the specialcase .Analogously if (24 )4 = (25 )4, (1)4 = (2)4 and then

512

513

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(1)4 =(4)4if in addition (0)4 = (1)4 then 24 () = (0)425() This is an important consequenceof the relation between (1 )4 and (1)4 ,and definition of(0)4.

514From GLOBAL EQUATIONS we obtain

5

= (

28)

5

(

28 )5

(

29 )5 + (

28 )5

29 ,

(

29 )5

29,

5

(

29)

5

5

Definition of5 :- 5 = 2829 It follows(29)552 + (2)55 (28)5 5 (29)552 + (1)55 (28)5From which one obtains

Definition of(

1)

5

, (

0 )

5

:-

(g) For 0 < (0 )5 = 280290 < (1)5 < (1 )5 5() (1)5+()5(2)5 295(1)5(0)5

5+()5 295(1)5(0)5 , ()5 = (1)5(0)5(0)5(2)5 it follows (0)5 5() (1)5

515

In the same manner , we get

5

() (

1)

5

+(

)

5

(

2)

5

29

5

(

1)5

(

2)5

5+()5 295(1)5(2)5 , ()5

=(

1)

5

(

0)

5

(0)5(2)5 From which we deduce (0 )5 5() (5)5

516

(h) If 0 < (1)5 < (0)5 = 280290 < (1 )5 we find like in the previous case,(1 )5 (1)5+5(2)5 295(1)5(2)5

1+5 295(1)5(2)5 5 (1)5+5(2)5 295(1)5(2)5

1+

5

29

5

(

1)

5

(

2)

5

(

1)

5

517

(i) If 0 < (1)5 (1)5 (0)5 = 280290 , we obtain(1)5 5 (1)5+5(2)5 295(1)5(2)5


Definition of 5 :-(2

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