Summary_ (Исследование некоторых типов дифференциальных уравнений с сильной нелинейностью), страница 3

A particular solution to the Navier-Stokes equations exponentially tending to zerofor large values of the radial coordinateLet us turn to the Navier-Stokes equations (6)-(9) in the considered axially symmetriccase and seek a particular solution to these equations in the form  A(t , r ) / r 2 ,   B(t , r ) / r 2 ,   C (t , r ) z / r 2 , q  q(t , r ) .(40)Then Eqs. (6)-(9) giveC  rBr ,At  (  B) Ae / r Arr  0 ,(41)Bt  B( Br  B / r ) / r  A2 / r 2  ( Brr  Br / r )  rqr ,(42)Brr  (  B)( Brr  Br / r ) / r  Br 2 / r  Btr  0 .(43)Let us now use the following dimensionless variables and functions:  (V* / l* )t ,   (r / l* ) 2 , P( , )  A(t , r ) / , Q( , )  B(t , r ) / ,S ( , )  (q r / r )(l* / V* ) 2 ,(44)where V* is the average absolute value of the fluid velocity and l* is its characteristicdimension.

Then Eqs. (41)-(43) give4P  2QP  Re P  0, Re  V*l* / , P  P( , ) ,(45)Re Q  4Q  Q(2Q  Q /  )  P 2 /   Re 2 S ( , ) , Q  Q( , ) ,(46)4(Q  Q )  2(QQ  Q 2 )  Re Q  0 ,(47)where Re is the Reynolds number.Consider Eq. (47) and seek its particular solution in the formQ  Ke f ( ) ,   Re  , K  const .(48)Then from (47) we find(4 f 2  f )(1  f )  0 .From (49) we come to the equationf  4 f 2 ,(49)f  f ( ) ,(50)1,4(51)which gives the following solution:f singular at the initial time t  0 .From (48) and (51) we find the following solution toEq. (47): Q  K exp   Re  .4 (52)12As to Eq. (46), it allows one to find the function S ( , ) which determines the distributionof the pressure in a fluid.Let us turn to Eq.

(45) and seek its solution in the formP  P( ),   Re.4(53)Substituting (52) and (53) into Eq. (45), we obtainP( )  (  12 Ke  ) P( )  0 .(54)From this equation we come to the solutionP( )  M exp    12 KEi(  ) , Ei( x)  x sesds ,(55)where M  const and Ei( x) is the exponential integral.From (55) we find the function P( ) of the formP  M  exp    12 KEi ( ) d ,(56)which exponentially tends to zero as r   .Let us determine the velocity components of the fluid. As follows from (44), (52) и (53),A  P( ), B  Ke  ,(57)where, taking into account the expression for the Reynolds number in (45),r2  Re .4 4t(58)From (41), (57) and (58) we also find the expression for the function С:r 2 e .2tApplying now formulas (57) and (59), from (5) and (40) we findC  rBr  Kv1 2 P( ) y  Ke  x,v2 2P( ) x  Ke  y ,rrwhere the function P( ) is determined by formula (56).(59)v3  Ke z,2t(60)Expressions (60) exponentially tend to zero for large values of the radial coordinate.The obtained solution satisfies the following initial and boundary conditions:1) vi  0 when t  0 and r  0 ;(61)2) rvi  0, r  , i  1, 2, 3 .(62)It describes a fluid which was initially at rest and which at t  0 was under the action of asource of pressure situated at the axis r  0 .1.5.

A particular solution to the Navier-Stokes equations for a liquid with cavitationLet us seek solution to the Navier-Stokes equations (6)-(9) in the form13A,  Fz,  Fr,r(63)r2r2where A  A(t , r, z ) and F  F (t , r , z ) are some differentiable functions.Then Eq. (6) is identically fulfilled..We will further assume that the potential   0 . Then the substitution of expressions(63) for  ,  ,  into Eqs.

(7)-(9) givesAt  ( Ar Fz  Az Fr ) / r  ( Arr  Ar / r  Azz )  0 ,Ftz  ( Fr Fzz  Fz Frz ) / r  ( A 2  Fz 2 ) / r 2   ( Frrz  Frz / r  Fzzz ) (64)1Ftr  ( Fr Frz  Fz Frr ) / r  Fr Fz / r 2   ( Frrr  Frr / r  Fr / r 2  Frzz )  rpr ,1(65)rp z . (66)Let us choose the variable   r 2 instead of r. Then the functions p, A and F can berepresented asp  p(t , , z ), A  A(t , , z ), F  F (t , , z),   r 2 .(67)Substituting expressions (67) into Eqs. (64)-(66), we come to the following equations:At  2( A Fz  Az F )  (4A  Azz )  0 ,(68)Ftz  2( F Fzz  Fz Fz )  ( A2  Fz 2 ) /   (4Fz  Fzzz )  (2 /  )p , (69)Ft  2( F Fz  Fz F )  (4F  4F  Fzz )  (1 /  ) p z / 2 .Let us seek particular solutions to Eqs.

(68)-(70) in the formp  p(t , ), A(t , , z)  F (t , , z),   const  0 .(70)(71)Then substituting the expression for A in (71) into Eq. (68), we obtainFt  (4F  Fzz )  0 .(72)After differentiation of this equation with regard to z and  , we findFtz  (4Fz  Fzzz )  0, Ft  (4F  4F  Fzz )  0(73)The substitution of equalities (71) and (73) into Eqs. (69) and (70) results in theequations2( F Fzz  Fz Fz )  (2 F 2  Fz 2 ) /   (2 /  )p ,(74)F Fz  Fz F  0 .(75)Thus, we come to the three equations (72), (74) and (75).Let us seek their particular solutions in the formF  U (t , ) sin(z   ),   const ,(76)where U (t , ) is some differentiable function.Substituting expression (76) into Eqs.

(72), (74) and (75), we obtainU t  (4U  2U )  0 ,(77)14(2 /  )p  2U (2U  U /  ) ,(78)U 2  UU  0 .(79)Consider Eq. (78). From it we find22  (U ) U 2 2 2p   2 U /  .2  2(80)Taking into account (71) and relation   r 2 in (67), from Eq. (80) we come to thefollowing expression for the pressure p:p  p* (t ) 2U 22r 2,(81)where p* (t ) is some continuous and positive function.Consider now Eq.

(79). It can be rewritten asU / U  U / U ,U  U (t , ) .(82)Integrating this equation with respect to the variable  , we findU  h(t ) exp  f (t )  ,(83)where f (t ) and h(t ) are arbitrary functions.Let us substitute expression (83) for the function U (t , ) into Eq. (77). Then we obtainh  hf h(4f 2  2 )  0 .(84)h  2h  0,(85)From this equation we findf  4f 2 .Equations (85) give1, h0  const ,4twhere t  0 corresponds to the singularity of the function f (t ) .h  h0 exp( 2t ),f (86)As a result we obtainr 2 ,U  h0 exp   2t 4tr 2 F  h0 exp   2t sin(z   ),4t(87)A  F .(88)Using formulas (5), (63) and (88), we come to the following formulas for the componentsvi of the vector of velocity:v1  h0r2r 2 ,[sin(z   ) y  cos(z   ) x] exp   2t 4t(89)15v2 h0r2r 2 ,[sin(z   ) x  cos(z   ) y ] exp   2t 4t (90)h0r 2 .sin(z   ) exp   2t 2t4t (91)v3  Formulas (81) and (87) givep  p* (t ) 2 h022r 2r 2 ,exp   22t 2t(92)where p* (t ) can be arbitrary positive function.The obtain solution has the following properties for t  0 :1) The functions vi exponentially tend to zero as r   and as t  and   0 , they have sinusoidal dependence on the coordinate z.2) When t  0  and r  0 , the functions vi tend to zero.3) As follows from (92), the function p   as r  0 .

This implies that the radialcoordinate r should be positive in order to have the pressure p positive in the consideredliquid. Moreover, the coordinate r for its points should satisfy the inequality r  r0 (t )  0 .Here the values r  r0 (t ) correspond to a sufficiently small value of the pressure p in a liquidbelow which the phenomenon of cavitation occurs.

In this case, the process of rupturing aliquid and forming a vapor-filled cavity in it takes place.Therefore, the obtained solution (89)-(92) is applicable to the region occupied by a liquidand satisfying the inequality r  r0 (t )  0 . As to the region r  r0 (t ) , it corresponds to thecavity filled by the vapor of this liquid.From (92) we find that the function r0 (t ) should satisfy the equalityp* (t ) 2  22t  r0 (t )   pcav ,exp2t 2r02 (t )2 h02(93)where pcav is the small positive pressure below which cavitation occurs in a liquid. Thevalue pcav is equal to the saturated vapor pressure in a liquid.Consider t  0 and assume that p* (t )  pcav .

Then Eq. (93) has a solution r0 (t )  0since the function p in (92) becomes larger than pcav for sufficiently large r and smaller thanpcav for sufficiently small r. Moreover, the positive function r0 (t ) satisfying Eq. (93) isunique since expression (92) is an increasing function of r for any t  0 .As an analysis shows, the function r0 (t ) has the following asymptotic behavior:r0 (t )  O t ln(1 / t ) , t  0 and hence r0 (0)  0 and(94)16r0 (t ) 2( p* ()  pcav )h0 exp( 2t ), t   .(95)2. New classes of solutions to the classical Yang-Mills equations2.1. Stationary spherically symmetric solution to the Yang-Mills equationsLet us turn to the Yang-Mills equations with SU(2) symmetry which play an importantroleс in the quantum theory of physical fields. They can be represented in the form3  F k ,  g klm F l , Am  (4 / c) J k , ,(96)F k ,    Ak ,   Ak ,  g klm Al , Am, ,(97)where  ,  0,1, 2,3 , k , l , m  1, 2,3 , Ak , , F k , are potentials and strengths of a YangMills field,  klm is the antisymmetric tensor, 123  1 , g is the constant of electroweakinteractions and J k , are three 4-vectors of current densities.Consider Eqs.