Navier-Stokes equations

Transcription

1 Solomon I. Khmelnik Naier-Stokes equations On the eistence and the search method for global solutions Second edition Israel 11

2 Copright 11 b Solomon I. Khmelnik All right resered. No portion of this book ma be reproduced or transmitted in an form or b an means, electronic or mechanical, without written permission of the author. Published b MiC - Mathematics in Computer Comp. BOX 153, Bene-Aish, Israel, 686 Fa: solik@netision.net.il Printed in United States of America, Lulu Inc., First Edition 1, First Edition,.8.1 ID 93671, ISBN Second Edition, ID , ISBN

3 Annotation In this book we formulate and proe the ariational etremum principle for iscous incompressible and compressible fluid, from which principle follows that the Naiet-Stokes equations represent the etremum conditions of a certain functional. We describe the method of seeking solution for these equations, which consists in moing along the gradient to this functional etremum. We formulate the conditions of reaching this etremum, which are at the same time necessar and sufficient conditions of this functional global etremum eistence. Then we consider the so-called closed sstems. We proe that for them the necessar and sufficient conditions of global etremum for the named functional alwas eist. Accordingl, the search for global etremum is alwas successful, and so the unique solution of Naiet- Stokes is found. We contend that the sstems described b Naiet-Stokes equations with determined boundar solutions (pressure or speed) on all the boundaries, are closed sstems. We show that such tpe of sstems include sstems bounded b impermeable walls, b free space under a known pressure, b moable walls under known pressure, b the socalled generating surfaces, through which the fluid flow passes with a known speed. The book is supplemented b open code programs in the MATLAB sstem functions realiing the calculation method and test programs. Links on test programs are gien in the tet of the book when the eamples are described. The programs ma be obtained from the author b request at solik@netision.net.il 3

4 Contents Detailed contents \ 6 Introduction \ 1 Сhapter 1. Principle etremum of full action \ 1 Chapter. Principle etremum of full action for iscous incompressible fluid \ 18 Chapter 3. Computational Algorithm \ 33 Chapter 5. Stationar Problems \ 34 Chapter 6. Dnamic Problems \ 35 Chapter 7. An Eample: Computations for a Mier \ 37 Chapter 8. An Eample: Flow in a Pipe \ 49 Chapter 9. Principle etremum of full action for iscous compressible fluid \ 66 Discussion \ 7 Supplements \ 74 References \ 1 4

5 Detailed contents Introduction \ 7 Сhapter 1. Principle etremum of full action \ The Principle Formulation \ Energian in electrical engineering \ Energian in Mechanics \ Mathematical Ecursus \ Energian- in mechanics \ Energian- in Electrical Engineering \ Energian- in Electrodnamics \ Conclusion \ 17 Chapter. Principle etremum of full action for iscous incompressible fluid \ Hdrodnamic equations for iscous incompressible fluid \ 18.. The power balance \ Energian and quasietremal \ 1.4. The split energian \.5. About sufficient conditions of etremum \ 4.6. Boundar conditions \ Absolutel hard and impenetrable walls \ Sstems with a determined eternal pressure \ Sstems with generating surfaces \ A closed sstems \ 9.7. Modified Naier-Stokes equations \ 3.8. Conclusions \ 3 Chapter 3. Computational Algorithm \ 33 Chapter 5. Stationar Problems \ 34 Chapter 6. Dnamic Problems \ Absolutel closed sstems \ Closed sstems with ariable mass forces and eternal pressures \ 36 Chapter 7. An Eample: Computations for a Mier \ The problem formulation \ Polar coordinates \ Cartesian coordinates \ Mier with walls \ Ring Mier \ Mier with bottom and lid \ Acceleration of the mier \ 47

6 Chapter 8. An Eample: Flow in a Pipe \ Ring pipe \ Long pipe \ Variable mass forces in a pipe \ Long pipe with shutter \ Variable mass forces in a pipe with shutter \ Pressure in a long pipe with shutter \ 6 Chapter 9. Principle etremum of full action for iscous compressible fluid \ The equations of hdrodnamics \ Energian- and quasietremal \ The split energian- \ About sufficient conditions of etremum \ 69 Discussion \ 7 Supplement 1. Certain formulas \ 74 Supplement. Ecerpts from the book of Nicholas Umo \ 81 Supplement 3. Proof that Integral (.84) is of Constant Sign \ 88 Supplement 4. Soling Variational Problem with Gradient Descent Method \ 9 Supplement 5. The Surfaces of Constant Lagrangian \ 93 Supplement 6. Discrete Modified Naier-Stokes Equations \ Discrete modified Naier-Stokes equations for stationar flows \ 95. Discrete modified Naier-Stokes equations for dnamic flows \ 98 Supplement 7. An electrical model for soling the modified Naier- Stokes equations \ 99 References \ 1 6

7 Introduction In his preious works [6-5, 37, 38] the author presented the full action etremum principle, allowing to construct the functional for arious phsical sstems, and, which is most important, for dissipatie sstems. In [31, 34, 35, 36, 39] described this principle as applied to the hdrodnamics of iscous fluids. In this book (unlike the first edition of [34, 35]) used a more rigorous etension of this principle for power and also is considered hdrodnamics of compressible fluids. The first step in the construction of such functional consists in writing the equation of energ conseration or the equation of powers balance for a certain phsical sstem. Here we must take into account the energ losses (such as friction or heat losses), and also the energ flow into the sstem or from it. This principle has been used b the author in electrical engineering, electrodnamics, mechanics. In this book we make an attempt to etend the said principle to hdrodnamics. In Chapter 1 the full action etremum principle is stated and its applicabilit in electrical engineering theor, electrodnamics, mechanics is shown. In Chapter the full action etremum principle is applied to hdrodnamics for iscous incompressible fluid. It is shown that the Naiet-Stokes equations are the conditions of a certain functional's etremum. A method of searching for the solution of these equations, which consists in moing along the gradient towards this functional's etremum. The conditions for reaching this etremum are formulated, and the are proed to be the necessar and sufficient conditions of the eistence of this functional's global etremum. Then the closed sstems are considered. For them it is proed that the necessar and sufficient conditions of global etremum for the named functional are alwas alid. Accordingl, the search for global etremum is alwas successful, and thus the unique solution of Naiet- Stokes is found. It is stated that the sstems described b Naiet-Stokes and haing determined boundar conditions (pressures or speeds) on all the bounds, 7

8 belong to the tpe of closed sstems. It is shown that such tpe includes the sstems that are bounded b: o Impermeable walls, o Free surfaces, находящимися под известным давлением, o Moable walls being under a known pressure, o So-called generating surfaces through which the flow passes with a known speed. Thus, for closed sstems it is proed that there alwas eists a unique solution of Naiet-Stokes equations. In Chapter 3 the numerical algorithm is briefl described. In Chapter 5 the numerical algorithm for stationar problems is described in detail. In Chapter 6 the algorithm for dnamic problems solution is suggested, as a sequence of stationar problems solution, including problems with jump-like and impulse changes in eternal effects. Chapter 7 shows arious eamples of soling the problems in calculations of a mier b the suggested method. In chapter 8 we consider the fluid moement in pipe with arbitrar form of section. It is shown that regardless of the pipe section form the speed in the pipe length is constant along the pipe and is changing parabolicall along its section, if there is a constant pressure difference between the pipe's ends. Thus, the conclusion reached b the proposed method for arbitrar profile pipe is similar to the solution of a known Poiseille problem for round pipe. In Chapter 9 it is shown tat the suggested method ma be etended for iscous compressible fluids. Into Supplement 1 some formulas processing was placed in order not to oerload the main tet. For the analsis of energ processes in the fluid the author had used the book of Nikola Umo, some fragments of which are sited in Supplement for the reader's conenience. In Supplement 3 there is a deduction of a certain formula used for proing the necessar and sufficient condition for the eistence of the main functional's global etremum. In Supplement 4 the method of solution for a certain ariational problem b gradient descend is described. In Supplement 5 we are giing the deriation of some formulas for the surfaces whose Lagrangian has a constant alue and does not depend on the coordinates. In Supplement 6 dealt with a discrete ersion of modified Naier- Stokes equations and the corresponding functional. 8

9 In Supplement 7 we discuss an electrical model for soling modified Naier-Stokes equations and the solution method for these equations which follows this model. 9

10 Сhapter 1. Principle etremum of full action 1.1. The Principle Formulation The Lagrange formalism is widel known it is a uniersal method of deriing phsical equations from the principle of least action. The action here is determined as a definite integral - functional t S ( q) = ( K( q) P( q) ) dt (1) t1 from the difference of kinetic energ K (q) and potential energ P (q), which is called Lagrangian Λ ( q) = K( q) P( q). () Here the integral is taken on a definite time interal t1 t t, and q is a ector of generalied coordinates, dnamic ariables, which, in their turn, are depending on time. The principle of least action states that the etremals of this functional (i.e. the equations for which it assumes the minimal alue), on which it reaches its minimum, are equations of real dnamic ariables (i.e. eisting in realit). For eample, if the energ of sstem depends onl on functions q and their deriaties with respect to time q, then the etremal is determined b the Euler formula 1 ( K P) d ( K P) q dt q =. (3) As a result we get the Lagrange equations. The Lagrange formalism is applicable to those sstems where the full energ (the sum of kinetic and potential energies) is kept constant. The principle does not reflect the fact that in real sstems the full energ (the sum of kinetic and potential energies) decreases during motion, turning into other tpes of energ, for eample, into thermal energq, i. e. there occurs energ dissipation. The fact, that for dissipatie sstems (i.e., for sstem with energ dissipation) there is no formalism similar to

11 Lagrange formalism, seems to be strange: so the phsical world is found to be diided to a harmonious (with the principle of least action) part, and a chaotic ("unprincipled") part. The author puts forward the principle etremum of full action, applicable to dissipatie sstems. We propose calling full action a definite integral the functional t Φ( q ) = R( q) dt (4) t1 from the alue R ( q) = ( K( q) P( q) Q( q) ), (5) which we shall call energian (b analog with Lagrangian). In it Q (q) is the thermal energ. Further we shall consider a full action quasietremal, haing the form: ( K P) d ( K P) q dt q Q = q. (6) Functional (4) reaches its etremal alue (defined further) on quasietremals. The principle etremum of full action states that the quasietremals of this functional are equations of real dnamic processes. Right awa we must note that the etremals of functional (4) coincide with etremals of functional (1) - the component corresponding to Q (q), disappears Let us determine the etremal alue of functional (5). For this purpose we shall "split" (i.e. replace) the function q (t) into two independent functions (t) and (t), and the functional (4) will be associated with functional Φ ( t, ) = R(, ) dt, (7) t1 which we shall call "split" full action. The function R (, ) will be called "split" energian. This functional is minimied along function (t) with a fied function (t) and is maimied along function (t) with a fied function (t). The minimum and the maimum are sole ones. Thus, the etremum of functional (7) is a saddle line, where one group of functions o minimies the functional, and another - o, maimies it. The sum of the pair of optimal alues of the split functions gies us the sought function q = o o, satisfing the quasietremal 11

12 equation (6). In other words, the quasietremal of the functional (4) is a sum of etremals o, o of functional (7), determining the saddle point of this functional. It is important to note that this point is the sole etremal point there is no other saddle points and no other minimum or maimum points. Therein lines the essence of the epression "etremal alue on quasietremals". Our statement 1 is as follows: In eer area of phsics we ma find correspondence between full action and split full action, and b this we ma proe that full action takes global etremal alue on quasietremals. Let us consider the releance of statement 1 for seeral fields of phsics. 1.. Energian in electrical engineering Full action in electrical engineering takes the form (1.4, 1.5), where =, Q( q) = Rq q. (1) Lq Sq K( q) P( q) = Eq, Here stroke means deriatie, q - ector of functions-charges with respect to time, E - ector of functions-oltages with respect to time, L - matri of inductiities and mutual inductiities, R - matri of resistances, S - matri of inerse capacities, and functions K ( q), P( q), Q( q) present magnetic, electric and thermal energies correspondingl. Here and further ectors and matrices are considered in the sense of ector algebra, and the operation with them are written in short form. Thus, a product of ectors is a product of column-ector b row-ector, and a quadratic form, as, for eample, R q q is a product of row-ector q b quadratic matri R and b column-ector q. In [, 3] the author shown that such interpretation is true for an electrical circuit. The equation of quasietremal (1.6) in this case takes the form: Lq Rq E = Sq. () Substituting (1) to (1.5), we shall write the Energian (1.5) in epanded form: 1

13 Lq R( q) = Sq Eq Rq q. (3) Let us present the split energian in the form ( L S E R ) ( ) R (, ) =. (4) L S E R Here the etremals of integral (1.7) b functions (t) and (t), found b Euler equation, will assume accordingl the form: S L R E =, (5) S L R E =. (6) B smmetr of equations (5, 6) it follows that optimal functions and, satisfing these equations, satisf also the condition =. (7) Adding the equations (5) and (6), we get equation (), where q = o o. (8) It was shown in [, 3] that conditions (5, 6) are necessar for the eistence of a sole saddle line. It was also shown in [, 3] that sufficient condition for this is that the matri L has a fied sign, which is true for an electric circuit. Thus, the statement 1 for electrical engineering is proed. From it follows also statement : An phsical process described b an equation of the form (), satisfies the principle etremum of full action. Note that equation () is an equation of the circuit without knots. Howeer, in [, 3] has shown that to a similar form can be transformed into an equation of an electrical circuit (with an accurac) Energian in Mechanics Here we shall discuss onl one eample - line motion of a bod with mass m under the influence of a force f and drag force k q, where k - known coefficient, q - bod's coordinate. It is well known that f = mq kq. (1) 13

14 In this case the kinetic, potential and thermal energies are accordingl: K q mq ( ) =, P( q) = fq, Q( q) = kqq. () Let us write the energian (1.5) for this case: q mq R( ) = fq kqq. (3) The equation for energian in this case is (1) Let us present the split energian as: ( m f k ) R (, ) =. (4) 14 ( ) m f k It is eas to notice an analog between energians for electrical engineering and for this case, whence it follows that Statement 1 for this case is proed. Howeer, it also follows directl from Statement Mathematical Ecursus Let us introduce the following notations: t = d, ˆ = dt. (1) dt There is a known Euler s formula for the ariation of a functional of function f (,,,...) [1]. B analog we shall now write a similar formula for function f (..., ˆ,,,,...) : f (..., ˆ,,,,...) : () t ' ' d d ' ' ar =... f... ˆ dt f f f (3) dt dt In particular, if f () =, then ar = ; if f () = ˆ, then ar = ˆ. The equalit to ero of the ariation (1) is a necessar condition of the etremum of functional from function () Full Action for Powers In this case full action- is a definite integral - functional t Φˆ ( i ) = Rˆ ( i) dt (1) t1 from the alue R ˆ ( i ) = Kˆ( i) Pˆ( i) Qˆ( i), () ( )

15 which we shall call Energian-. In this case we shall define full action quasietremal- as Qˆ ˆ ˆ P K = i. (3) Functional (1) assumes etremal alue on these quasietremals. The principle etremal of full action- asserts that quasietremals of this functional are equations of real dnamic processes oer integral generalied coordinates i. Let us now determine the etremal alue of functional (1, ). For this purpose we, as before, will split the function i (t) to two independent functions (t) and (t), and put in accordance to functional (1) the functional t Φˆ (, ) = Rˆ (, ) dt, (4) t1 which we shall call "split full action-. We shall call the function R ˆ (, ) "split " Energian--. This functional is being minimied b the function (t) with fied function (t) and maimied b function (t) with fied function (t). As before, the quasietremal (3) of i = o o of etremals o, o of the functional (1) is a sum functional (4), determining the saddle point of this functional Energian- in mechanics As in Section 3 we shall consider an eample, for which the equation (3.1) is applicable, or m i k i f =. (1) In this case the kinetic, potential and thermal powers are accordingl: Kˆ ( i) = m i i, Pˆ( i) = f i, Qˆ( q) = k i. () Let us write the energian- (6.) for this case: R ˆ ( i) = m i i f i k i. (3) Уравнение квазиэкстремали в этом случае принимает вид (1). 15

16 1.7. Energian- in Electrical Engineering Let us consider an electrical circuit which equation has the form, (.) or iˆ L i R i E = S. (1) In this case the kinetic, potential and thermal powers are accordingl: K ˆ ( i ) = L i i, P ˆ( i ) = S i ˆ i E i, Q ˆ( i ) = R i. () Let us write the energian- (6.) for this case: R ˆ ( i ) = L i i S i ˆ i E i R i. (3) The equation of quasietremal in this case assumes the form (1). Let us now present the split Energian- as R S ˆ (, ) = R ( ) L( ) ˆ ( ) E( ) ˆ. (4) The etremals of integral (6.4) b the functions (t) and (t), found according to equation (4.3), will assume accordingl the form: S ˆ L R E =, (5) S ˆ L R E =. (6) From the smmetr of equations (5, 6) it follows that optimal functions and, satisfing these equations, satisf also the condition =. (7) Adding together the equations (5) and (6), we get the equation (1), where q = o o. (8) Therefore, the equation (1) is the necessar condition of the eistence of saddle line. In [, 3] it is shown that the sufficient condition for the eistence of a sole saddle line is matri L haing fied sign, which is true for eer electrical circuit Energian- in Electrodnamics In [, 3, 38], the proposed method is also applied to electrodnamics. 16

17 1.9. Conclusion The functionals (1.7) and (6.4) hae global saddle line and therefore the gradient descent to saddle point method ma be used for calculating phsical sstems with such functional. As the global etremum eists, then the solution also alwas eists. Further, the proposed method is applied to the hdrodnamics. 17

18 Chapter. Principle etremum of full action for iscous incompressible fluid.1. Hdrodnamic equations for iscous incompressible fluid The hdrodnamic equations for iscous incompressible liquid are as follows []: di ( ) =, (1) where p µ ρ t ( ) ρf = ρ, () ρ = const is constant densit, µ - coefficient of internal friction, p - unknown pressure, = [,, ] - unknown speed, ector, F = [ F, F, F ] - known mass force, ector,,,, t - space coordinates and time. p,, are repeatedl gien below The reminder notations ( ) in Supplement 1. Further the letter "p" will denote the formulas gien in this application... The power balance Umo [1] discussed for the liquids the condition of balance for specific (b olume) powers in a liquid flow. For a non-iscous and incompressible liquid this condition is of the form (see (56) in [1] and Supplement ) P 1 ( ) P5 ( ) P4 ( p, ) =, (3) and for iscous and incompressible liquid - another form (see (8) in [1] and Supplement ) P 1 ( ) P5 ( ) P ( p, ) =, (4) where 18

19 ρ W P1 =, t (5) dp dp dp d d d dp dp dp P = d d d dp dp dp d d d (6) P = p 4, (7) 1 dw dw dw P = 5 ρ, (8) d d d W = ( ) (9) p and so on tensions (see (р4) in Supplement 1). Here P 1 is the power of energ ariation, P 4 is the power of work of pressure ariation, P 5 - the power of ariation of energ ariation for direction change, and the alue P 7( p, ) = P5 ( ) P4 ( p, ) (1) is, as it was shown b Umo, the ariation of energ flow power through a gien liquid olume see (56) и (58) в [1]. In [] it was shown, that for incompressible liquid the following equalit is alid dp dp dp d d d dp dp dp = p d d d µ (11) dp dp dp d d d This follows from (р4). From this it follows that 19

20 P ( p ) = µ. (1) or subject to (6) P = P4 P3 (13) where P = µ 3 (14) - power of change of energ loss for internal friction during the motion. Therefore, we rewrite (4) in the form P 1 ( ) P5 ( ) P4 ( p, ) P3 ( ) =, (15) We shall supplement the condition (15) b mass forces power P6 = ρf. (16) Then for eer iscous incompressible liquid this balance condition is of the form P1 ( ) P5 ( ) P4 ( p, ) P3 ( ) P6 ( ) =. (17) Taking into condition(1) and formula (p1a) let us rewrite (7) in the form P = di ( p) 4, (18) Taking into account (p9a), condition(1) and formula (p1a) let us rewrite (8) in the form P5 = di( W ). (19) From (18, 19) and Ostrogradsk formula (p8) we find: P4 dv = di p dv = ps n ds, () V V ( ) S ( W ) dv = P5 dv = di W n ds (а) V V S or, subject to (р15), P dv = G( dv = W ds. (1) V ( ) 5 ) V Returning again to the definitions of powers (7, 8), we will get ( p) dv = ps n ds, (1а) V S ( ( W ) dv = W n ds (1в) V S or S n

21 ( G ) ) dv = V ( W ds. (1с) S.3. Energian and quasietremal For further discussion we shall assemble the unknown functions into a ector q = [ p, ] = [ p,,, ]. () This ector and all its components are functions of (,,, t). We are considering a liquid flow in olume V. The full action- in hdrodnamics takes a form T Φ = R ( q(,,, t) dv dt, (3) V Haing in mind (17) and the definition of energian -, let us write the energian- in the following form 1 R( q) = P1 ( ) P3 ( ) P4 ( q) P5 ( ) P6 ( ) n. (4) Below in Supplement 1 will be shown see (р8, р15, р18): d P1 = ρ, dt (5) P5 = ρ G( ), (6) where G( ) = ( ). (7) Taking this into account let us rewrite the energian (4) in a detailed form d 1 R( q) = ρ µ di( p) ρ G( ) ρf. (8) dt Further we shall denote the deriatie computed according to Ostrogradsk formula (р3), b the smbol, as distinct from o ordinar deriatie. Taking this into account (р19), we get 1

22 d d P1, = ρ ; o dt dt di( ) q 4 ( p) 5 o ( P () ) ( P ( q) ) = ; ( P (, G( ) )) = ρ( ) ( P () ) 6 = ρf. 3 = µ ; ; (9) In accordance with Chapter 1 we write the quasietremal in the following form: d o P1, dt 5 1 ( P () ) ( P ( q) ) 3 q 4 =. (3) ( P (, G( ) )) o ( P6 ( ) ) From (9) it follows that the quasietremal (3) after differentiation coincides with equations (1, )..4. The split energian- Let us consider the split functions () in the form q = [ p, ] = [ p,,, ], (31) q = [ p, ] = [ p,,, ]. (3) Let us present the split energian taking into account the formula (р18) in the form d d ρ µ ( ) dt dt R( q, q ) = ( di( p ) di( p )). (33) ρ ( G( ) G( )) ρ F( ) Let us associate with the functional (3) functional of split full action T Φ = R( q, q ) dv dt, (34) V With the aid of Ostrogradsk formula (р3) we ma find the ariations of functional (34) with respect to functions q. In this we shall take into

23 account the formulas (р), obtained in the Supplement 1. Then we hae: or = b p, (35) p or = b, (36) b p = di( ), (37) d ( ) ρ µ p dt b =. (38) ρ G, G, ρ F X X So, the ector b = [ b p, b ] (39) is a ariation of functional (34), and the condition b = [ b p, b ] = (4) is the necessar condition for the eistence of the etremal line. Similarl, b = [ b p, b ] = (41) The equations (4, 41) are necessar condition for the eistence of a saddle line. B smmetr of these equations we conclude that the optimal functions q and q, satisfing these equations, satisf also the condition q = q. (4) Subtracting in couples the equations (4, 41) taking into consideration (37, 38), we get di ( ) =, (43) d( ) ( ) ( ) ρ µ p p ρ F dt =. (44) ρ,,,, G G G G X X X X For = according to (р14а), we hae G( ) G, G( ) G, = G( ) X X (45) 3

24 Taking into account (7, 45) and reducing (43, 44) b, получаем we get the equations (1, ), where q = q o q o. (46) - see (, 31, 3), i.e. the equations of etremal line are Naiet-Stokes equations..5. About sufficient conditions of etremum Let us rewrite the functional (34) in the form T Φ = R q, q ) d d d dt q q X,,, t (, (47), are determined b (31, 3), ( ) where ectors = ector of independent ariables. Further onl the functions q ( X ) = [ p ( X ), ( X )] will be aried. Vector b, defined b (39), is a ariation of functional Φ b the function q and depends on function q, i.e. b = b( q ). Here the function q here is fied. Let S be an etremal, and subsequentl, the gradient in it is b s =. To find out which tpe of etremum we hae, let us look at the sign of functional's increment δ Φ = Φ( S) Φ( C), (48) where С is the line of comparison, where b = b с. Let the alues ector q on lines S и С differ b qc qs = q qs = δ q = a b, (49) where b is the ariation on the line С, а a known number. Thus, p S bp q = q S a b = a. (5) b S where b p, b are determined b (35, 36) accordingl, and do not depend on q. If δ Φ = a A, (51) 4

25 where A has a constant sign in the icinit of etremal b s =, then this etremal is sufficient condition of etremum. If, furthermore, A is of constant sign in all definitional domain of the function q, then this etremal determines a global etremum. From (48) we find δr = R( S) R( C) = R( qs ) R( q ), (5) or, taking into account (33, 5), ( ) ( ) d d ab ρ ab s s dt dt µ ( ( s ab ) ( s ab ) ( )) δr = ( s ab ) ( p ) ( ps abp ) (53) ρ ( ( s ab ) G( ) G( s ab )) ρ F( ( s ab ) ) Taking into account (р1), we get: [ G (, b ) G (, b )] a G( b ) G ( s ab ) = G( s ) a 1 s s. (54) Here (53) is transformed into δ R = R R1a Ra, (55) where R, R1, R are functions not dependent on а, of the form ( ) d d ρ s s dt dt R = µ ( s ( s ) ( )) ( s ( p ) ( p s )), ρ ( G G( )) F( ) s ( ) s ρ s (56) d db ρ b µ ( b s s ( b )) dt dt R = ( b ( p ) ( b ), (57) 1 p ρ( bg( ) ( G1 ( s, b ) G ( s, b ))) ρ F b 5

26 R = µ b ( b ) ρ G( b ). (58) Now we must find ( δr ) = R (59) a This function depends on q. To proe that the necessar condition (4) is also a sufficient condition of global etremum of the functional (47) with respect to function q, we must proe that the integral T Φ = R( q, q ) dv dt (6) a V or, which is the same, the integral T Φ = R dv dt (61) a V is of constant sign. Similarl, to proe that the necessar condition (41) is also a sufficient condition of a global etremum of the functional (47) with respect to function q, we hae to proe that the integral similar to (6) is also of the same sign. Specifing the concepts, we will sa that the Naier-Stokes equations hae a global solution, if for them there eists a unique nonero solution in a gien domain of the fluid eistence. In the aboe-cited integrals the energ flow through the domain's boundaries was not taken into account. Hence the aboe-stated ma be formulated as the following lemma Lemma 1. The Naier-Stokes equations for incompressible fluid hae a global solution in an unlimited domain, if the integral (61, 58) has constant sign for an speed of the flow..6. Boundar conditions The boundar conditions determine the power flow through the boundaries, and, generall speaking, the ma alter the power balance equation. Let us iew some specific cases of boundaries Absolutel hard and impenetrable walls If the speed has a component normal to the wall, then the wall gets energ from the fluid, and full returns it to the fluid. (changing the speed direction). The tangential component of speed is equal to ero (adhesion effect). Therefore such walls do not change the sstem's 6

27 energ. Howeer, the energ reflected from walls creates an internal energ flow, circulating between the walls. So in this case all the aboestated formulas remain unchanged, but the conditions on the walls (impenetrabilit, adhesion) should not be formulated eplicitl the appear as a result of soling the problem with integrating in a domain bounded b walls. Then the second lemma is alid: Lemma. The Naier-Stokes equations for incompressible fluid hae a global solution in a domain bonded b absolutel hard and impenetrable walls, if the integral (61, 58) is of the same sign for an flow speed..6.. Sstems with a determined eternal pressure In the presence of eternal pressure the power balance condition (17) is supplemented b one more component the power of pressure forces work P8 = p s n, (6) where p s - eternal pressure, S - surfaces where the pressure determined, n - normal component of flow incoming into aboe surface, In this case the full action is presented as follows: T Φ = R( q(,,, t) dv P8 ( q(,,, t) dv dt. (63) V S For conenience sake let us consider the functions Q, determined on the domain of the flow eistence and taking ero alue in all the points of this domain, ecept the points belonging to the surface S. Then the restraint (63) ma be written in the form T Φ = R ˆ ( q(,,, t) dv dt, (64) V where energian R ˆ ( q) = R( q) Q P8 ( n). (65) One ma note that here the last component is identical to the power of bod forces in the sense that both of them depend onl on the speed. So all the preious formulas ma be etended on this case also, b performing substitution in them. 7

28 F F Q p s ρ. (66) Therefore the following lemma is true: Lemma 3. The Naier-Stokes equations for incompressible fluid hae a global solution in a domain bounded b surfaces with a certain pressures, if the integral (61, 58) has constant sign for an flow speed. Such surface ma be a free surface or a surface where the pressure is determined b the problem's conditions (for eample, b a gien pressure in the pipe section). p s ma be included in the full action Note also that the pressure functional formall, without bringing in phsical considerations. Indeed, in the presence of eternal pressure there appears a new constraint - (1а). In [4] it is shown that such problem of a search for a certain functional with integral constraints (certain integrals of fied alues) is equialent to the search for the etremum of the of the sum of our functional and integral constraint. More precisel, in our case we must seek for the etremum of the following functional: T Φ = R ( q(,,, t)) dv dt V R( q(,,, t)) Rˆ ( q(,,, t) = λ - p Q p s n ˆ, (67) ( ), (68) where λ an unknown scalar multiplier. It is determined or known initial conditions [4]. For λ = 1 after collecting similar terms the Energian (68) again assumes the form (65), which was to be proed Sstems with generating surfaces There is a conception often used in hdrodnamics of a certain surface through which a flow enters into a gien fluid olume with a certain constant speed, i.e., NOT dependent on the processes going on in this olume. The energ entering into this olume with this flow, eidentl will be proportional to squared speed module and is constant. We shall call such surface a generating surface (note that this is to some etent similar to a source of stabilied direct current whose magnitude does not depend on the electric circuit resistance). If there is a generating surface, the power balance condition (17) is supplemented b another component the power of flow with constant squared speed module. 8

29 P9 = W s n, (69) где W s - squared module of input flow speed, S - surfaces where the pressure determined, n - normal component of flow incoming into aboe surface, One ma notice a formal analog between W s and p s. So here we also ma consider the functional (64), where the energian is R ˆ ( q) = R( q) Q P9 ( n), (7) and then perform the substitution F F Q W s ρ. (71) Consequentl, the following lemma is true: Lemma 4. The Naier-Stokes equations for incompressible fluid hae a global solution in a domain bounded b generating surface with a certain pressure, if the integral (61, 58) has constant sign for an flow speed. Note also that W s the pressure p s ma be included in the full action functional formall, without bringing in phsical considerations.(similar with pressure p s ). Indeed, in the presence of eternal pressure there appears a new constraint - (1c). Including this integral constraint into the problem of the search for functional's etremum, we again get Energian (7) Closed sstems We will call the sstem closed if it is bounded b o absolutel hard and impenetrable walls, o surfaces with certain eternal pressure,, o generating surfaces, or o not bounded b anthing. In the last case the sstem will be called absolutel closed. Such case is possible. For eample, local bod forces in a bondless ocean create such a sstem, and we shall discuss this case later. There is a possible case when the sstem is bounded b walls, but there is no energ echange between fluid and walls. An eample a flow in endless pipe under the action of ais bod forces. Such eample will also be considered below. In consequence of Lemmas 1-4, the following theorem is true: 9

30 Theorem 1. The Naier-Stokes equations for incompressible fluid hae a global solution in a gien domain, if o the domain of fluid eistence is closed, o the integral (61, 58) has constant sign for an flow speed. The free surface, which is under certain pressure, ma also be the boundar of a closed sstem. But the boundaries of this sstem are changeable, and the integration must be performed within the fluid olume. It is well known that the fluid flow through a certain surface S is determined as ws = ρ di( ) dθ. (7) S Thus, the boundar conditions in the form of free surface are full considered, b the fact that the integration must be performed within the changeable boundaries of the free surface. We hae indicated aboe, that the power of energ flow change is determined b (1). In a closed sstem this power is equal to ero. Therefore for such sstem the Energian (4) or (8) turns into Energian (accordingl) R q) = P( ) P ( ) P ( ), (73) 3 ( d R ( q) = ρ µ ρf. (74) dt For such sstems the Naier-Stokes equations take the form (1) and µ ρf = t ρ, (75) Some eamples of such sstem will be cited below..7. Modified Naier-Stokes equations From (p19a) we find that ( ) = ( W ). (76) Substituting (76) in (), we get ( ) = ( W ). (77) Let us consider the alue D = p ρ W, (78) which we shall call quasipressure. Then (77) will take the form

31 ρ µ D ρ F =. (79) t The equations sstem (1, 79) will be called modified Naier-Stokes equations. The solution of this sstem are functions, D, and the pressure ma be determined from (9, 78). It is eas to see that the equation (79) is much simpler than (). The aboe said ma be formulated as the following lemma. Lemma 5. If a gien domain of incompressible fluid is described b Naier-Stokes equations, then it is also described b modified Naier- Stokes equations, and their solutions are similar. Phsics aside, we ma note that from mathematical point of iew the equation (79) is a particular case of equation (), and so all the preious reasoning ma be repeated for modified Naier-Stokes equations. Let us do it. The functional of split full action (34) contains modified split Energian d d ρ µ ( ) R( q, q ) = dt dt ( ( ) ( )) ( ). (8) di D di D ρ F - see (33). Gradient of this functional with respect to function q is (37) and d b = ρ µ ( D ) ρ F. (81) dt - see (38). The components of equation (55) take the form d db ρ b ( µ b s s ( b )) R = dt dt, (8) 1 ( ( ) ( )) b D bp ρ F b R = µ b ( b ). (83) Thus, for modified Naier-Stokes equations b analog with Theorem 1 we ma formulate the following theorem Theorem. Modified Naier-Stokes equations for incompressible fluid hae a global solution in the gien domain, if o the fluid domain of eistence is a closed sstem o the integral (61, 83) has the same sign for an fluid flow speed. Lemma 6. Integral (61, 83) alwas has positie alue. Proof. Consider the integral 31

32 T J = µ ( ) dv dt (84) V This integral epresses the thermal energ, eoled b the liquid due to internal friction. This energ is positie not depending on what function connects the ector of speeds with the coordinates. A stricter proof of this statement is gien in Supplement 3. Hence, integral (84) is positie for an speed. Substituting in (84) = b, we shall get integral (61, 83), which is alwas positie, as was to be proed. From Lemmas 5, 6 and Theorem there follows a following. Theorem 3. The equations of Naier-Stokes for incompressible fluid alwas hae a solution in a closed domain. The solution of equation (1, 79) permits to find the speeds. Calculation of pressures inside the closed domain with known speeds is performed with the aid of equation (78) or ( ) = p ρ. (85).8. Conclusions 1. Among the computed olumes of fluid flow the closed olumes of fluid flow ma be marked, which do not echange flow with adjacent olumes the so-called closed sstems.. The closed sstems are bounded b: o Impermeable walls, o Surfaces, located under the known pressure, o Moable walls being under a known pressure, o So-called generating surfaces through which the flow passes with a known speed. 3. It ma be contended that the sstems described b Naiet-Stokes equations, and haing certain boundar conditions (pressures or speeds) on all boundaries, are closed sstems. 4. For closed sstems the global solution of modified Naier-Stokes equations alwas eists. 5. The solution of Naier-Stokes equations ma alwas be found from the solution of modified Naier-Stokes equations. Therefore, for closed sstems there alwas eists a global solution of modified Naier- Stokes equations. 3

33 Chapter 3. Computational Algorithm The method of solution for hdrodnamics equations with a known functional, haing a global saddle point, is based on the following outlines [, 3]. For the gien functional from two functions q 1, q two more secondar functionals are formed from those functions q 1, q. Each of these functionals has its own global saddle line. Seeking for the etremum of the main functional is substituted b seeking for etremums of two secondar functionals, and we are moing simultaneousl along the gradients of these functionals. In general operational calculus should be used for this purpose. Howeer, in some particular cases the algorithm ma be considerabl simplified. Another complication is caused b the fact that in the computations we hae to integrate oer all the flow area. But the area ma be infinite, and full integration is impossible. Neertheless, the solution is possible also for an infinite area, if the flow speed is damping. Here we shall discuss onl these particular cases. 33

34 Chapter 5. Stationar Problems Note that in stationar mode the equations (.1,.) assumes the form di( ) =, p µ ρ ( ). ρf =. (1) The modified equations (1, 79) in stationar mode take the form: di( ) =, µ D ρ F =.. () In Appendi 6 we considered the discrete ersion of modified Naier- Stokes equations for stationar sstems (). It was shown that for stationar closed sstems the solution of modified Naier-Stokes equations is reduced to a search for quadratic functional minimum (and not a saddle points, as in general case). After soling these equations the pressure is calculated b the equation (.78), i.e. or p ρ W D =. (3) ( ) = p = D ρ (4) The equation (75) for absolutel closed sstems in stationar mode takes the form µ ρf =. (5) The solution of equation () has been discussed in detail in Supplement 4. After soling it the pressures are calculated b the equation (4). 34

35 Chapter 6. Dnamic Problems 6.1. Absolutel closed sstems Let us consider the equation (.75) for absolutel closed sstems and rewrite is as η F = t (1) where µ η =. () ρ Assuming that time is a discrete ariable with step dt, we shall rewrite (1) as n n 1 η n F n =, (3) dt = 1,,3 n dt where n,... the number of a time point. Let us write (3) as where F η n Fn 1 =. (4) = F n n dt 1 n1. (5) For a known speed n 1 the alue n is determined b (4). Soling this equation is similar to soling a stationar problem see Supplement 4. On the whole the algorithm of soling a dnamic problem for a closed sstem is as follows Algorithm 1 1. n 1 and F n are known. Computing n b (4, 5). 3. Checking the deiation norm 35

36 t n t ε = n 1 (6) and, if it doesn't eceed a gien alue, the calculation is oer. расчет заканчивается. Otherwise we assign n 1 n (7) and go to p. 1. Eample 1. Let the bod forces on a certain time point assume instantl a certain alue there is a jump of bod forces. Then in the initial moment the speed o =, and on the first iteration we assign n 1 =. Further we perform the computation according to Algorithm Closed sstems with ariable mass forces and eternal pressures Consider the modified equation (1, 79) in the case when the mass forces are sinusoidal functions of time with circular frequenc ω. In this case equations (1, 79) take the form of equations with comple ariables: di( ) =, j ω ρ µ D ρ F =,. (8) where j - the imaginar unit. In Supplement 6 the discrete ersion of these equations is considered. There it is shown that their solution is reduced to the search of saddle point of a certain function of comple ariables. After soling these equations the pressure is calculated b equation(4). 36

37 Chapter 7. An Eample: Computations for a Mier 7.1. The problem formulation Let us consider a mier, whose lades are made of fine-mesh material and are located close enough to one another. Then the pressure forces of the blades on the fluid ma be equated to bod forces. The bod forces might hae a limited area of action Θ (less than the fluid olume) It mean onl that outside this area the bod forces are equal to ero. In addition, these forces ma be a function of speed, coordinates and time. Let us discuss some cases. For eample, the blades of a mier work in a closed fluid olume Θ, and the force F m, applied to the blades, is passed to the fluid elements. The bod force F ma be determined as F m = ( µ ρf ) dθ. Θ Let us assume also that the mier is long enough, and so in its middle the problem of calculation of the field of speeds ma be considered as a two-dimensional problem. Let us first consider a structure without walls. In such a problem there is no restraints, and so the sstem is a closed one (in the sense that was defined aboe). Let us use for our calculations the method described in Chapter 5. Let us assume that the bod forces created b mier's blades and acting along a circle with its center in the coordinate origin, are described as follows σ ( ) ( R a F R = e ), (1) where R is the distance from the current point to the rotation ais, σ, a are certain constants. Function (1) is shown on Fig. 1, and gradient of forces (1) is shown on Fig.. 37

38 Fig Fig.. 38

39 7.. Polar coordinates If bod forces are plane and do not depend on the angle, then the Naiet-Stokes equations assume the form []: r 1 p ρ r = 1 F µ r r r r, () ρ. (3) = Interestingl enough in this sstem the equation for the calculation of pressure using speed is etracted from the main equation. Phsicall it ma be eplained b the fact that our sstem is absolutel closed (in the determined aboe sense). It confirms our assertion that speed calculation and pressure calculation in a absolutel closed sstem ma be parted. The condition of continuit in this sstem is also absent, which also corresponds with our statement for absolutel closed sstem. Thus, as the pressure in this case is not included into equation (3), the latter cannot be soled independentl, and the pressure ma be found afterwards b direct integration of the equation (). But the equation (3) ma not be soled b direct integration. Indeed, depending on the direction of integration (from infinit to ero or ice ersa) the results will be quite different. When integrating "from the ero:" the result depends on initial alues of speed and on its deriatie, which are not determined b the problem's conditions. Neertheless, the unique solution should eist, and it ma be obtained b the proposed method. To achiee it, we must better return to Cartesian coordinates Cartesian coordinates Projections of forces (1) on coordinate aes are F F R = e R σ (, ) ( R a = e ) σ (, ) ( R a ), (4a). (4b) The equation for this absolutel closed stationar sstem is as follows: µ ρf =, (5) 39

40 To sole the equation (5) use the method described aboe in Chapter 5. This method is realied in the program testpostokpuas (mode=1), which builds the following graphs 1. Logarithm of relatie mistake function ( ε 1 = µ ρf ) dd ( ρf ) dd (7),, of the residual of equation (.7) in dependence of iteration number see the first window on Fig 3;. Logarithm of relatie mistake function d d ( ) dd ε = di( ) dd d d (8),, - of the residual in the continuit condition in dependence of iteration number see the second window on Fig. 3; note that this mistake is a methodic one it is caused b boundedness of the surface of integration plane and decreases with the surface etension; 3. speed function R (on the last iteration) in dependence of radius see the third window on Fig 3; thus, this Figure shows the problem solution; 4. force function ρ F and Lagrangian function µ in dependence of radius see the fourth window on Fig 3, where these functions a denoted b dot line and full line accordingl. The calculation was performed for σ =.1, a = 5, µ = 1, ρ = 1, n = 35, where n n - the dimensions of the integration domain. The dimensions are chosen large enough, so that the speed on a large distance from center would be close to ero, and thus the sstem ma be considered absolutel closed. Here ε 1 =.1, ε =.7, k = 86, where k is the number of iterations. 4

41 testpostokpuas -3.5 Error Neprer Er-g,L-r Fig Mier with walls Contrar to the preious case (in Cartesian coordinates we shall now consider a mier with clindrical walls, located on the circle with radius R s. We hae shown aboe that the walls create a closed sstem and do not change the power balance in the sstem. In essence, the calculation is done in the same wa, b (5.) and the program testmiermodif, mode=, as in the preious case. The integration area is restricted b the circle with radius R s. Calculation results are shown on Fig. 4. In this case 11 ε1 = 5 1, ε =.6, k = 7, R s =. It is important to note that on the circle of radius R s the speed is =. This answers the known fact that due to icious friction the speed of fluid on the surface of a bod surrounding it, is equal to ero. It is also important to note that to get this result we had not hae to add more equations in the main equation - it was enough to restrict the integration domain. 41

42 Error testmiermodif Neprer Er-g,L-r Fig Ring Mier Let us consider now a mier with internal and eternal clindrical walls, located on circles correspondingl with radius R 1 and R. Fig. 4а shows the result of computation b (5.), b the program testkolomodif, ariant=,, which has built the following graphs: 1. function (.7) see the first window;. function (.8) see the second window; 3. the speed function R depending on radius see the third window; 4. the speed module function depending on Cartesian coordinates see the fourth window. The calculations hae been made for σ =.1, a = 5, µ = 1, ρ = 1, r = 33 and R 1 = 3, R = 7. 4 We got ε 1 = 4 1, ε =.8, k = 5. 4

43 -4 testkolomodif Neprer Error Iterations Iterations r Fig. 4а. testkolomodif Neprer Error Iterations Iterations r Fig. 4в. 43

44 Fig. 4с. Fig. 4d. 44

45 In the similar wa we shall consider a mier where interior part has the form of a square with half-side of R 1. Fig. 4в shows the result of calculation b (5.), b the program testkolomodif, ariant=1. We got ε 1 =.45, ε =.43, k = 5. Fig. 4c and 4d show the speed gradient distribution for a round and square interior parts accordingl Mier with bottom and lid O Hi 3 6 Hs Hm O 7 Rs Ri 1 Fig. 5. Let us consider now a mier with bottom and lid see Fig. 5, where (9,1,11,1) unlimited integration domain,, 45

46 (,3,6,7) the area of mier's blades, (1,8) the mier's bottom, (4,5) the mier's lid, (1,4; 5,8) the mier's clindrical wall, o - the ais passing along the diameter through the mier's center, o - the ais passing along the rotation ais of mier's blades, R s - the radius of mier's can, R i - the radius of initial integration domain, H s - half-height of the mier's can, bounded b bottom and lid, H m half-height of mier's blades, H i - half-height of initial integration domain. Bottom, lid and walls of the can create a closed sstem and do not change the power balance in the sstem. The calculations are performs eactl as in the preious case. The calculations results are shown on Fig. 4. It is important to note that on the circle of radius R s, along the bottom and along the lid the speed is = - see further. This answers the alread mentioned fact that due to iscous friction the fluid's speed on the surface of a bod surrounded b the fluid, is alwas equal to ero. It is significant that to get this result it was no need to add an more conditions to the main equations it was enough to restrict the integration domain in the course of calculations. The calculations were performed b the program testmiermodif3 (mode=1), which has built the following graphs: 1. the function (.7) see the first window on the first ertical line on Fig 6;. the function (.8) see the second window on the first ertical line on Fig 6; 3. the function of speed R depending on radius see the first window on the second ertical line on Fig 6; 4. the function of speed R depending on the distance along the height up to the mier's center for constant alue of radius equal to a see the third window on the second ertical line on Fig 6; the rectangle in this window depicts the force action area; 5. the function of force ρ F and function of Lagrangian µ depending on radius see the fourth window on Fig. 6, where these functions are depicted b dot line and full line accordingl. 46

47 The calculations were performed for: σ =.1, a = 5, µ = 1, R R = 15, H = 15, H = 3, = 35, s i m s We got ε 1 =.4, ε =.4, k = 133. i H = 7, r = testmiermodif3 Error1 Error iterrations (r) =f(),f=f() F, Lap Fig r 7.7. Acceleration of the mier In Section we hae discussed the case of stead-state moement of the fluid in the mier. Now we shall consider the period of acceleration, assuming (as in Eample 1 in Section 6.1), that the bod forces in a certain moment instantl assume a certain alue there occurs a jump of bod forces.. Then in the first moment ( 1) = and on the first iteration we assume ( 1) =, and then we calculate the transient process according to algorithm 1 from Section 6.1. This algorithm is realied in the program testragonmier, which builds the following graphs (see Fig. 8): 1. the speed function with radius 5.. the relatie residual function (6.4); 3. the relatie diergence from ero function. 47

48 The computation was performed for the conditions taken in Section, i.e. σ =.1, a = 5, µ = 1, ρ = 1, n = testragonmier; dnamic t Error.15 Neprer Fig

49 Chapter 8. An Eample: Flow in a Pipe 8.1. Ring pipe We shall begin with an eample. Let there be a ring pipe with rectangular section see Fig. 1, where o is center of construction, s center of rectangular pipe section, R the distance from o ais of the ring to a certain point of pipe section measured along the ais o; also the Figure shows the main dimensions of the construction and the directions of Cartesian coordinates aes. O O ro Ro Zo r R O s o Fig. 1 Such ring pipe is a closed sstem. Let us assume that in this sstem the bod forces directed perpendicularl to the section plane of the pipe are in effect. Such forces do not depend on the coordinate and are defined b formulas 49

50 5 F (, ) = Fo R F (, ) = Fo R,, =,, (1),. () ( ) F. (3) The definitional domain of bod forces is the interior of the pipe. At this FR ( R, ) = ( F (,, ) ) ( F (,, ) ) (5) or F R ( R, ) = 1. (6) The calculation is performed b program testmiermodif3 (mode=) and, in accordance with Chapter 5, in two stages: the speed was calculated b the equation (5.), and the pressure deriaties b equation (5.3) for gien speed. The following initial data was used: F o =, ρ = 1.7, µ =.7, ro = 1, o = 11, R o = 17. The calculations were performed for ( ) ( R, ) ( (,, ) ) (,, ) R =, (7) ( R, ) dp(,, ) dp(,, ) dp dr = d d. (8) Let us further denote the distance from a point in the section to the center of the section along ох ais as r = R R o. (9) The calculations results are shown on Fig.. as follows: 1. function (7..7) see first window on the first ertical;. function (7..8) see the second window on the first ertical 3. the speed function R depending on radius and on the coordinate for constant =, = see the first window on the second ertical; 4. the speed function R the center of the pipe section with constant radius depending on the distance b height to R o see the second window on the first ertical;

51 5. the pressure deriatie function dp dr depending on the radius see the second window on the second ertical. 5 testmiermodif3, mode= Error1 Error iterrations (r) r 6 =f(),f=f() p Fig r The mentioned calculation (see the first window) shows that this speed satisfies equation (5.). It is important to note that the solution obtained b the proposed method without indicating the initial conditions, knowing onl the domain of the flow eistence. Distribution of speeds ( R, ) along the pipe section drawn b the plane = is shown on Fig.. 3. The same function depending on the coordinates of one pipe section will be denoted as ( r, ) or Π ( r, ). From (5.) it follows that this function has a constant alue of Lagrangian on its definition domain the pipe section. We shall call such functions functions of constant Lagrangian. Since for each form of section these functions hae different form, we shall denote the function ( r, ) for a rectangular section as п( r, ). 51

52 testmiermodif s Fig. 3. Важно отметить, что это решение получено предложенным методом без указания начальных условий, а только с указанием области существования течения. Zero alues of speed on the pipe walls appeared as the result of computations. Распределение скоростей ( R, ) по сечению трубы, проведенному по плоскости =, показано на фиг Long pipe Here we shall discuss flow in a infinitel long pipe of arbitrar profile in which bod forces are in action. Let us mark a certain segment of this pipe and assume that the section forms and speeds on both ends of the segment are similar. Then instead of this segment we ma consider an equialent sstem of such segment where the ends are connected in such wa that the fluid flow from, sa, the left end flows directl into the right end. Such sstem is a closed one and we can use the proposed method for its calculation. Eidentl, the flow in eer part of an infinitel long pipe coincide with the flow in the built sstem. For eample, let us look at a "connected" in the described wa segment of pipe of the length o, where constant bod forces 5 F o are

53 acting, directed along the pipe's ais o. Let also the pipe's section is determined in coordinates (, ) and is a square with half-side n, and the following alues are known: F o = 1, ρ = 1, µ = 1, n = 13, o = testdawlemodif (mode=) Error iterrations -1 - L(,) (,) - - F; -L 1.5 -, =, =1 Fig. 3. This sstem is absolutel closed, because the fluid does not interact with the walls. The computation is performed according to (5.5). The result of calculation using the program testdawlemodif (mode=) are depicted on Fig 3, where the following functions are shown: 1. function (.7) see the first window on the first ertical,. function of speed (, ) for constant see second window on the first ertical, 3. Lagrangian function µ in dependence of coordinates (, ) of the section for constant see the first window on the second ertical, 4. functions ρ F and Lagrangian µ depending on with = and with constant see the second window on the 53

54 54 second ertical where these function are depicted b straight and broken lines accordingl. The speed diergence and the pressure gradient are equal to ero. Thus, for constant bod force the pressure in a linear pipe is constant. From (5.5) it follows that for constant bod force the Lagrangian also has a constant alue on all pipe section, ecluding the boundaries, where the force and the Lagrangian eperience a jump see Fig. 3. The function of speed distribution on the pipe section, which corresponds to the constant Lagrangian, is shown on Fig. 3. We shall call such functions the functions of constant Lagrangian. As for each form of pipe section the functions are different, we shall denote the function (, ) for a rectangular section as п(, ). So, on a rectangular section of a pipe the speeds are distributed according to the function п( r, ) with a constant Lagrangian. In Supplement 5 it is shown that elliptic paraboloid is also a function with constant Lagrangian. Therefore, in a similar wa we ma proe that on an elliptic section of ring pipe the speeds are distributed according to a function э( r, ) of elliptic paraboloid. In particular, the speeds on a circular section of ring pipe are distributed according to paraboloid of reolution function. Let us consider now another mode of flow in pipe; we shall call this mode a conjugated mode (with regard to the aboe considered mode). In this mode the bod forces are absent, but beside the pressure p there eists a certain additional pressure p f. If p f = ρ F, (1) then the equation (5.5) ma be substituted b equation p f µ =. (13) From (1) there also follows that the gradient has a constant alue in the direction perpendicular to the pipe section, i.e. dp p f = (14) d and dp = µ (15) d or

55 dp = ρ F o (16) d Thus, in a pipe the speed along the pipe is distributed according to the function п( r, ) of a constant Lagrangian, if onl the pressure is constant on all the points of the pipe section, and is changing uniforml along the pipe. The difference of pressures between two pipe sections spaced at a distance L, is equal to dp p 1 p = L (17) d and, taking into account (15), p1 p = µ. (18) L Eidentl, the same conclusion ma be reached regarding an part of a pipe. Therefore, The speed in a part of the pipe with rectangular section is constant along the pipe and is changing on the section according to function п( r, ), if there eists a constant difference of potentials on the ends of the pipe. If the analtical dependence is known: п(, ) = п f (, ), (19) then, as it follows from (18), п p (, ) 1 p = L µ f (, ). () In a similar wa we ma get the function э( r, ) of speed distribution in a pipe with elliptic section, and, particularl with a circular section. In this case there eists an analtical dependence of the form (19), namel dependence (c16) see Supplement 5. Specificall, for circular section it has the form (с), and then the formula () becomes: p () 1 p r = ( r ( r ) k o 4L µ. (1) where o r is the radius of circular pipe section. The latter formula coincides with the known Poiseille formula []. This ma sere as an additional confirmation of the proposed method applicabilit. 55

56 In the same wa we ma calculate the flow in a pipe of arbitrar section and/or in a pipe bent in arbitrar wa (if onl the form of sections and speeds on both ends of the segment are the same). Thus, an infinite sstem is formall transformed into a closed sstem Variable mass forces in a pipe Here we shall assume that in a long pipe there are mass forces aring sinusoidall with time. Then for speeds calculation we ma use the equations (6.8) and their solution method, gien in Supplement 6. Fig. 3а and Fig. 1 show the results of calculations b the program testdawlemodiftime (mode=) for Fo = 1, ρ = 1, n = 13, o = 7 and for seeral alues of µ, ω. Fig. 3а presents the speeds distribution for =, and the table shows the alues of speeds amplitudes and cosine of phase-shifts of speed sinusoid from mass forces sinusoid in the point = 1, = 1. We ma note that for high frequenc the distribution function of the speed b pipe section tends to a constant, with the eception of section contour, where it is alwas equal to ero. Howeer in this process the speed amplitude decreases significantl. Table 1. Variant µ ω Amplitude Сosine

57 om=,mu=1 - - om=1,mu= om=1,mu=1 - - om=1,mu=1 - - Fig. 3а Long pipe with shutter Here we shall discuss the flow in an infinitel long pipe with square section with square side n, in which an absolutel hard cube with halfside R o is placed. As in the preious case, we shall consider a "connected" pipe segment of length o, where constant bod forces F, are acting, directed along ais o see Fig. 4. Let also the pipe o section be defined in coordinates (, ) and be a square with half side n, and also the following alues are known F o = 1, ρ = 1, µ = 1, µ = 1, r = 39, n = 7, o = 57, Ro = 4. 57

58 Fo n Ro o Fig. 4. This sstem is closed, and in it the fluid interacts with the cube's walls. The calculation is performed according to (5.). The ruslts of calculation with the aid of program testdawlemodif (mode=8) are presented on Fig. 5, 6, 7. The alues obtained are: ε 1 =.35, ε =.6, k = 9. On Fig. 4 the ertical lines (-6,-5,-4,-3) are drawn, passing through the centers of sections distant b (-6,-5,-4,-3) from the cube's center.. Fig. 5 shows distribution of speeds b these sections, and Fig. 6 shows distribution of speeds b the same sections. Fig. 7 shows distribution of speeds and b the ais of these sections for fied alue of =. These figures permit to gie a picture of speeds distribution when flowing around the cube under the influence of bod forces in an infinitel long pipe. 58

59 4 4 (,), = (,), = (,), = (,), = Fig (,), = (,), = (,), = (,), = Fig

60 (), =na, =-3,4,5, (), =na, =-3,4,5,6 Fig Variable mass forces in a pipe with shutter Here we, as in Section 8.3, shall assume that in a long pipe with shutter the bod forces, aring sinusoidall with time, are acting. Then for speeds calculation we ma use equations (6.8) and methods of their solution gien in Supplement 6. Fig. 7а, 7b and Table. show the results of calculation b the program testdawlemodiftime (mode=5) for Fo = 1, ρ = 1, n = 13, o = 3 and for seeral alues of µ, ω. Fig. 7а and 7b present the speeds distribution and accordingl b the pipe section for =. Table shows the alues of speeds amplitudes for ( 1, 1,) and ( 8, 8, 6) cosine of phase-shifts of speed sinusoid from bod forces sinusoid in the point. We ma note that for high frequenc the distribution function of the speed b pipe section tends to a constant, with the eception of section contour, where it is alwas equal to ero. Howeer in this process the speed amplitude decreases significantl. The amplitude of speed also decreases with frequenc growth.

61 1 5 1 om=,mu=1 - - om=1,mu= om=1,mu=1 - - om=1,mu=1 - - Fig. 7а om=,mu=1 - - om=1,mu= om=1,mu=1 - - om=1,mu=1 - - Fig. 7в. 61

62 Table. Variant µ ω Amplitude Сosine Amplitude Сosine Pressure in a long pipe with shutter Let us return to the eample in section 8.4 and anale the distribution of pressures in a pipe with shutter. For this purpose we shall anale the following alues: - quasipressure see (18) in Appendi 6 or D = r di( ) ; (1) - gradient of quasipressure, as deriaties of (1) or b (.77), i.e. D = µ ρf. () - gradient of dnamic pressure see (р19d) or ( P d ) = ρ G (3) or, taking into account (р19a, p19c, p19d), ρ ( P d ) = ( W ) = ρ G ; (4) - gradient of pressure see (.78) or ρ p = D or, taking into account (4), ( W ), (5) p = D ρ G. (6) Furtherfore, we shall calculate aerage alues b pipe's section dp p mid = (, ), dg d G mid = (, ), mid d mid dd D mid = (, ) for a fied alue of, and also aerage alue d mid of pressure P = p d mid. min

63 dp,g,d,p() dp(),g(),d() dp() dp() ; == , ==-11(k), ==9(g) Fig , =1, =1(b); 19(r) Fig. 8 shows the results of the calculation program testdawlemodif (mode = 8): 1. functions p mid, Gmid, Dmid, P of see the first window on the first ertical ;. functions p, G, D of for fied alues of = = 9 see the first window on the second ertical ; 3. functions p of for fied alues of = = 9 (the upper cure) and = = 11 (the lower cure) see the second window on the first ertical ; 4. functions p of for fied alues = 1 and = 1 (the upper cure) and = 19 (the lower cure) see the second window on the second ertical. 63

64 P(,), =1 - - P(,), = P(,), =5 - - P(,), = = 6 64 Fig. 9. dp d Fig. 9 shows distribution functions (, ) 4 for fied alues of One ma notice the following:. 1) Quasipressure is equal to ero (a closed sstem!). ) Aerage pressure gradient b eer section is equal to ero. 3) Difference of pressures, as an integral of pressures gradient on the ends of the pipe are equal to ero, i.e. ma pd =. (7) min 4) The distribution of pressure gradient b the pipe's section is irregular. 5) The proposed method permits to calculate the pressure distribution in the pipe with shutter for gien bod forces. We must note that the precision of calculation increases with the etension of the pipe's segment length, due to the fact that as the distance between the segments ends and the shutter grows, the dependence of speeds distribution on the ends decreases, and the distributions themseles

65 become equal this same assumption is made when we "connect" the ends of infinite pipe. Let us now consider the case when the bod forces are absent, but there is a difference between pressures on the ends of the segment. In the aboe treated problem the equation of the tpe (5.1) has been soled. We shall now rewrite the last of equations as p µ ρg ρf =. (8) Let us perform a substitution ρ F p, (9) and call the alue p a force pressure.. Then the equation (8) will take the form ( p ) µ ρg =. (1) Here p = p p. (11) We hae: ma p d = L F, (1) min where L - length of the pipe. From this and from (7) it follows that the solution of equation (1) satisfies the constraint ma p d = δp, (13) min where δ P = L F (14) - the known pressures difference on the pipe ends. Consequentl, the solution of equation (8) is also solution of equation (1) with constraint (13). But it was shown aboe that the solution of modified equations (1, 77) is unique. Therefore, the solution of equation (8) alwas is the solution of equation (1) with constraint (13). So, the solution of equation (1) with constraint (13), i.e. calculation of speeds in a pipe with shutter and pressures difference of the pipe's ends, ma be substituted b solution of equation (8), where F = δp L. (15) For breit sake we hae omitted here to mention that the equations (8) and (1) should be soled together with the equation (.1). 65

66 Chapter 9. Principle etremum of full action for iscous compressible fluid In this section we shall use this principle for the Naier-Stokes equations describing compressible fluid. Naier-stokes equation for iscous compressible fluid are considered. It is shown that these equations are the conditions of a certain functional s etremum. The method of finding the solution of these equations is described. It consists of moing along the gradient towards the etremum of his functional. The conditions of reaching this etremum are formulated the are simultaneousl necessar and sufficient conditions of the eistence of this functional s global etremum where 9.1. The equations of hdrodnamics Recall the equation for a iscous incompressible fluid (.1.1,.1.): di ( ) =, (1) ρ p µ ρ G( ) ρ F = t G ) = ( ), () ( (3) In contrast with the equations for iscous incompressible fluid, the equations for iscous compressible fluid hae the following form []: ρ di( ρ ) = t ρ p µ ρ G ρ F µ Ω t 3 where () = ( ), (4) () () =, (5) Ω. (6) The Supplement 1 functions (3) and (6) are presented in epanded form - see (p14, p9, p3). For a compressible fluid densit is a known function of pressure: = f ( p) ρ. (7) 66

67 Further the reasoning will be b analog with the preious. In this case we hae to consider also the power of energ loss ariation in the course of epansion/compression due to the friction. P8 ( ) = µ Ω( ). (9) 3 We hae also: ( P8 ( ) ) = µ Ω( ). (1) 3 We ma note that the function Ω () in the present contet behaed in the same wa as the function (). This allows to appl the proposed method also for compressible fluids. 9.. Energian- and quasietremal B analog with preious reasoning we shall write the formula for quasietremal for compressible fluid in the following form: d 1 1 ρ µ o ( ) di( ρ p ) dt q ρ ( ρ G( ) ) o ( ρ F ) =. (11) p ρ p ρ t 1 µ o 3 ( Ω( ) ) The split energian- B analog with preious reasoning we shall write the formula for split energian- for compressible fluid in the following form: d d ρ µ ( ) dt dt (( ( ) ( ))) di p di p R ( ρ ρ q, q ) = ρ. (1) ρ ( G( ) G( )) ρ F( ) dρ dρ µ p p ( Ω( ) Ω ( )) ρ dt dt 3 With the aid of Ostrogradsk formula (р3) we ma find the ariations of functional of spilt full action- with respect to functions q : 67

68 or = b p, p (13) or = b, (14) These ariations are determined b aring the functions p and, whereas the functions ρ, p, do not change. Then we shall get: d d d 1), ρ = ρ dt dt dt ) [ µ ( )] = µ, 3) [ ρ( G( ) G( ))] = ρ G, G,, X X 4) [ ρ F( )] = ρ F, µ µ 5) ( ( ) ( )) = Ω( ), Ω Ω 3 3 6) ρ ( di( ρ p ) di( ρ p )) = grad( p ) 7) ( di( p ) di( p )) di( ), p ρ ρ = ρ ρ ρ (15) dρ dρ dρ 8) p p =. p ρ dt dt ρ dt Remarks for these formulas: 1,, 3, 4) the deriation is gien below, 5) is similar to formula ), 6, 7) the deriation is gien in the Supplement 1 see (p34, p35) accordingl Then we hae: dρ b p = di( ρ ), (16) dt, 68

69 d µ ( ) ( ) ( ) ρ µ Ω p dt 3 b =. (17) ρ G, G, ρ F X X As was shown aboe, the condition b = [ b p, b ] = (18) and the similar condition b = [ b p, b ] = (19) Are necessar conditions for the eistence of a saddle line. From the smmetr of these equations it follows that the optimal functions q and q, satisfing the equations (18, 19), must satisf also the condition q = q. () Subtracting in pairs the equations (18, 19) taking into account (16, 17), we get dρ - - di( ) =, (1) dt d( ) µ ρ µ ( ) Ω( ) dt 3,, G G =.() ( ) X X p p ρ F ρ G, G, X X Taking into account (1.45) and cancelling (1, ) b, we get the equations (4, 5), where q = q o q o, (3) i.e. equations etreme lines are the Naier-Stokes equations for compressible fluids About sufficient conditions of etremum Aboe we hae proed for incompressible fluid, that the necessar conditions (18, 19) of the eistence of etremum for the full action- functional are also sufficient conditions, if the integral 69

70 T I = R dv dt (4) V has constant sign, where R = µ b ( b ) ρ G( b ). (5) For compressible fluid the necessar conditions (18, 19) of the eistence of etremum for the full action- functional are also sufficient conditions, if the integral (4) has constant sign, where, contrar to (5), R µ = µ b ( b ) bω( b ) ρ G( b ). (6) 3 For closed sstems with a flow of систем incompressible fluid we hae shown aboe that the alue (5) assumes the form R = µ b ( b ). (7) Similarl, for closed sstems with a flow of compressible fluid the alue (6) assumes the form R µ = µ b ( b ) bω( b ). (8) 3 Let us consider now, similarl to (4), the integral T J = R dv dt V where R (9) µ = µ ( ) Ω( ). (3) 3 (i.e. in this formula instead of the function b there is the function of speed). As the proof of the integral s constanc of sign must be alid for an function, it is enough to proe the constanc of sign of integral (9) with speeds. For this we must note that: o the first term in (3) epresses the heat energ euded b the fluid as the result of internal friction, o the second tem in (3) is the heat energ euded/absorbed b the fluid as the result of epansion\compression. The first energ is positie regardless to the alue of ector-function of speed with respect to the coordinates (A more eact proof of this fact for the first term is gien in [4, 5]). The second term is equal to ero (as in our statement the temperature is not taken into account, i.e. assumed 7

71 to be constant). Therefore, integral (4, 3) is positie on an iteration, which was required to show. Thus, the Naier-Stokes equations for incompressible fluid hae a global solution. 71

72 Discussion Phsical assumptions are often built on mathematical corollar facts. So it ma be legitimate to build mathematical assumption on the base of phsical facts. In this book there are seeral such places 1. The equations are deried on the base of the presented principle of general action etremum.. The main equation is diided into two independent equations based on a phsical fact the absence of energ flow through a closed sstem. 3. The eclusion of continuit conditions for closed sstems is based in the phsical fact the continuit of fluid flow in a closed sstem 4. Usuall in the problem formulation we indicate the boundaries of solution search and the boundar conditions for speed, acceleration pressure on the boundaries These conditions usuall are formed on the base of phsical facts, for eample the fluid "adhesion" to the walls, the walls hardness, etc. In the presented method we do not include the boundar conditions into the problem formulation the are found in the process of solution. The solution method consists in moing along the gradient towards saddle point of the functional generated from the power balance equation. The obtained solutions: a. ma be interpreted as eperimentall found phsical effects (for instance, the walls impermeabilit, "sticking" of fluid to the walls, absence of energ flow through a closed sstem), b. coincide with solutions obtained earlier with the aid of other methods (for instance, the solution of Poiseille problem), c. ma иe seen as generaliation of known solutions (for instance, a generaliation of Poiseille problem solution for pipes with arbitrar form of section and/or with arbitrar form of ais line), d. belong to unsoled (as far as the author knows) problems (for instance, problems with bod as the functions of speed, coordinates and time). 7

73 We ma point also some possible directions of this approach deelopment, for eample o for compressible fluids, o for problems of electro- and magneto-hdrodnamics o for free surfaces dnamics (in changing boundaries for constant fluid olume). The proof of global solution eistence belongs to closed sstems Practicall, we must anale the bounded and closed sstems. Therefore aboe we hae discussed some methods of formal transformation of non-closed sstems into closed ones, such as: 1. long pipe as the limit of ring pipe,. transformation of a limited pipe segment into closed sstem At the same time it must be noted that the solution method has not been treated here on a full scale we considered onl special cases of stationar flows and changing with time flows. 73

74 74 Supplement 1. Certain formulas Here we shall consider the proof of some formulas that were used in the main tet. First of all we must remind that () = di, (р1) ( ) ( ) Q Q Q di di =, (р1а) = p p p p,,, (р) =, (р3) =, (р4) ( ) =, (р5) ( ) =. (р6) From (.5,.7а) it follows that ( ) 1 dt d P = ρ, (р7)

75 75 i.e. dt d P ρ = 1 (р8) Let us consider the function (.7) or ( ) ( ) ( ) = 5 1 d d d d d d P ρ (р9) or ( ) 5 W P = ρ. (р9а) Differentiating, we shall get: = d d d d d d d d d d d d d d d d d d P ρ 5. (р1) After rearranging the items, we get = d d d d d d d d d d d d d d d d d d P ρ 5. (р11) Let us denote:

76 76.,, = = = d d d d d d g d d d d d d g d d d d d d g (р1) Let us consider the ector = g g g G (р13) or. = G (р14) Note that ) ( ) ( 1 G G = (р14а) From (р11-р14) we get 5 G, P = ρ (р15) ( ) ) ( ) (, 5 G G P = ρ, (р16) Comparing (р6) and (р14), we find that ( ) G = ) (. (р18) Thus, ( ) ( ) G P = ρ, 5, (р19) Comparing (p9a, p15, p18), we find that

77 77 ( ) ( ) W =. (р19а) As dnamic pressure is determined [] b P d ρw =, (р19с) then from (p18, р19a) it follows that the gradient of dnamic pressure is ( ) G P d = ρ. (р19д) Let us consider also ( ) ( ) b G b G b G G b G,, ) ( ) ( ) ( 1 =, (р) where ( ),, 1 = b b b b b b b b b b G (ра) ( )., = b b b b b b b b b b G (рв) If b a b =, then ( ) ( ) b ag b ag b G a G b a G,, ) ( ) ( ) ( 1 =. (р1) We hae dt d dt d o =, dt d dt d o =, ( ) o =, ( ) ( ) = X G G o,, ( ) ( ) ( ) G G o =,

78 o p di p ( ( p )) = ( p ), o ( ( p )) = ( ) di. o di ( p ) = ( p ), o ( p ) = di ( ) -see (р1а). (р) The necessar conditions for etremum of functional from the functions with seeral independent ariables the Ostrogradsk equations [4] hae for each of the functions the form o f f f = =, (р3) a=,,, t a ( d da) where f the integration element, (,,,t) the ariable function, a independent ariable. The tensions (in hdrodnamics) are determined in the following wa []: p p, p p, p p = µ = µ = µ, p = p =, µ p = p = µ, p = p = µ. (р4) Let us consider formulas d = p p p, d = p p p, d = p p p. (р5) From (р4, р5) we find d = p µ, 78

79 79 = p d µ, = p d µ. (р6) From this it follows that the double integral in formula (81) in [1] and in Supplement ma be presented in the following form ( ) ( ) () ( ) () ( ) = J p n J p n J p n d J cos cos cos σ. (р7) The Ostrogradsk formula: integral of diergence of the ector field F, distributed in a certain olume V, is equal to ector flow F through the surface S, bounding this olume: ( ) = S V ds n F dv F di. (р8) () ( ) ( ) ( ) = Ω ) di(, ) di(, ) di(, (р9) = Ω ) (, (р3) If p, ρ are scalar fields, and is a ector field, then

80 ( ρ) ρ di( ) grad( p) p di( ρ ) di( ρ ) = grad, (р31) di( ρ p ) = ρ, (р3) i.е. di( ρ p ) = ρ grad. (р33) ( p) p grad( ρ) p ρ di( ) Consider di( ρ p ) and suppose that the etremum of a certain functional is determined or b aring the function p, or b aring the function. Then, differentiating the last epression b Ostrogradsk formula (p3), we shall find: o p o or [ di( p )] = grad( ρ) ρ di( ) ρ, [ di( ρ p )] = ρ grad( p ) p grad( ρ) p grad( ρ) o p di( p ) = ρ grad p [ di( p )] = di( ρ ) ρ, (р34) [ ] ( ) o ρ. (р35) 8

81 Supplement. Ecerpts from the book of Nicholas Umo 81

82 8

83 83

84 84

85 85

86 86

87 87

88 Supplement 3. Proof that Integral ( ) dv is of Constant Sign V Here we shall consider in detail the substantiation of the fact that integral (.84) alwas has positie alue. In other words we shall proe that the integral is of constant sign. 88 J = ( ) dv 1. (1) V Let us first consider the two-dimension case. Let us substitute the Laplacian b its discrete analog. To do this we shall take a twodimensional speeds network k, m, where m = 1, n - the number of point on the ais ОХ, k = 1, n - number of point on tee ais ОУ. The alue of discrete Laplacian in each point is determined b formula (see, for eample, the function DEL in MATLAB): 1 Lk, m = ( k, m 1 k, m 1 k 1, m k 1, m ) k, m. () 4 According to this the discrete Laplacian ma be found b the formula L = A, (3) where row ector 1,1,..., 1, m,..., 1, n,,1,...,, m,...,, n,... =, (4) k,1,..., k, m,..., k, n,... n,1,..., n, m,..., n, n, and А is a matri built according to formula (). For illustration Fig 1 shows matri А for n = 5, built according to formula () see for eample, [7]. This Figure shows also the numbering of ector k, m

89 elements. According to formula (3) the Laplacian also is presented in the form similar to (4). The discrete analog of integral (1) is T J1 = A. (5) To erif that the matri А is of constant sign, let us find for it the Kholetsk epansion T A = U U, (6) where U is the upper triangular matri. It is known [8], that if matri а A is smmetrical and positiel defined, then it has a unique Kholetsk epansion. The program testmatri.m computes epansion (6) and shows that matri A is smmetrical and positiel defined. It means that for an ector T A >. (7) Thus, it is proed that the alue (5) in two-dimensional case is positie. Decreasing the network spacing, in the limit we get that the integral (1) in two-dimensional case has positie alue. In the same wa it ma be shown that in three-dimensional case integral (1) is positie, which was to be proed. Fig