Spectral decomposition approach to macroscopic parameters of Fokker-Planck flows: Part 2 Igor A. Tanski Moscow, Russia
[email protected]
ABSTRACT In this paper we proceed with investigation of connections between Fokker Planck equation and continuum mechanics. We base upon expressions from our work [2], based upon the spectral decomposition of Fokker - Planck equation solution. In this decomposition we preserve only terms with the smallest degrees of damping. We find, that macroscopic parameters of Fokker-Planck flows, obtained in this way, satisfy the set of conservation laws of classic hydrodynamics. The expression for stresses (30) contains additional term - this term is negligible in big times limit. We proved also, that the velocities field alone satisfy Burgers equation without mass forces - but with some additional term. This term is also negligible in big times limit. For the zero degree theory, considered in [1], there are no additional terms. But this theory is valid only for the potential velocities field, fully deductible from density - the potential is proportional to density logarithm. In this theory we can not specify initial conditions for velocities independently from density. Taking in account of the next degree terms could partly solve this problem, but result in some loss of exactness.
Keywords Fokker-Planck equation, continuum mechanics
This paper is the second part of our work [1]. The aim of these papers is to investigate connections between Fokker - Planck equation and continuum mechanics. We follow the method, proposed in [4]. Namely, in spectral decomposition of solution we preserve only terms with the smallest degree of damping. In the first part we considered only zero degree terms. Here we discuss results of the next degree terms consideration. In the course of calculations we use results of our work [3]. We know from [1], that for zero degree terms satisfy the set of classic hydrodynamics equations for isothermal compressible fluid with friction mass force, proportional to velocity. Velocities alone satisfy in this case the Burgers equation without mass forces. But all this is true only for the potential velocities field, fully deductible from density - the potential is proportional to density logarithm. This is, of course, very special class of flows. In particular, we can not specify initial conditions for velocities independently from density.
-2-
We expect, that the taking in account of he next degree terms could solve this problem. All references to formulae [1] we shall write according to following format: (P1-##), where ## is the placeholder for corresponding formula number.
1. Density To get the first approximation of Fokker - Planck equation solution we keep terms with pi = 1 in (P1-1) - (P1-3). exp −t −∞ −∞ −∞ +∞ +∞ +∞
ρ1 =
× Aω 1ω 2ω 3 100
∫∫∫
k α + 2 α
3
Σ j=1
ω 2j ×
(1)
3/2 1/2 k ω j 2 k 2k iω 1 j=3 ω x ) exp dω 1 dω 2 dω 3. exp(i − j j 2π α α α Π j=1 2α α
It is useful to express ρ i as a derivative of some potential φ i to simplify handling terms with negative degrees of ω j in following expressions (see, for example (7)). Besides that we write e−α t factor explicitly to have clear insight of damping velocities of different terms in following expressions. We write ρ 1 = e−α t
∂φ 1 ; ∂x 1
(2)
where +∞ +∞ +∞
φ1 =
× Aω 1ω 2ω 3 100
k exp −t 2 α −∞ −∞ −∞
∫∫∫
3
ω 2j × Σ j=1
(3)
3/2 1/2 k ω j 2 k 2k 1 j=3 dω 1 dω 2 dω 3. exp(iω j x j ) exp − Π 2π α α α j=1 2α α
We see, that potential φ 1 satisfy diffusion equation ∂φ 1 k ∂2φ 1 ∂2φ 1 ∂2φ 1 = 2 + + , ∂t α ∂x 2 ∂y 2 ∂z 2
(4)
and density satisfy diffusion with damping equation k ∂2 ρ 1 ∂2 ρ 1 ∂2 ρ 1 ∂ρ1 + α ρ1 = 2 + + . ∂t α ∂x 2 ∂y 2 ∂z 2
(5)
Two another functions ρ 2 , ρ 3 are defined in the similar to (2) way by potentials φ 2 and φ 3 . These potentials satisfy the same equation (4), densities satisfy (5). Let us now denote the previously considered
-3density (P1-8) as ρ 0 . Then full density ρ for first degree approximation can be expressed as a sum ρ = ρ 0 + ρ 1 + ρ 2 + ρ 3 = ρ 0 + e−α t
∂φ k . ∂x k
(6)
Here and in the following repeated indices are understood to be summed (Einstein’s summation convention).
2. Velocities We have following expressions for velocities
u1k =
× Aω 1ω 2ω 3 100
exp −t ρ −∞ −∞ −∞ i
+∞ +∞ +∞
∫∫∫
k α + 2 α
3
Σ j=1
ω 2j ×
(7)
3/2 1/2 k ω j 2 k k ω k α δ 1k 2k iω 1 j=3 − − ω x ) exp dω 1 dω 2 dω 3. exp(i − j j 2π α ω1 α α Π j=1 α α 2α α
where ρ is defined by (6). Using this definition, we have u1k = e−α t
1 k ∂2φ 1 + α δ 1k φ 1 . − ρ α 2 ∂x 1 ∂x k
(8)
Recall expression (P1-11) for velocities u0k =
1 k ∂ρ0 − ρ α 2 ∂x k
(9)
and have u k = u0k + u1k + u2k + u3k ,
(10)
or uk =
1 k ∂ρ + α e−α t φ k . − 2 ρ α ∂x k
(11)
3. The continuity equation As we mentioned in [1], though we dropped all fast damping terms, (6) and (11) are still derived from exact solution of Fokker - Planck equation. Therefore they must satisfy certain conservation laws.
-4-
Let us check the continuity equation. ∂( ρ u k ) ∂( ρ u0k ) ∂( ρ u1k ) ∂( ρ u2k ) ∂( ρ u3k ) = + + + = ∂x k ∂x k ∂x k ∂x k ∂x k
(12)
2 k ∂2 ρ 1 k ∂2 ρ 2 k ∂2 ρ 3 ∂ρ ∂ρ ∂ρ ∂ρ ∂φ ∂ρ k ∂ ρ0 − − − + α e−α t k = − 0 − 1 − 2 − 3 = − . = − 2 2 2 2 2 2 2 2 α ∂x k α ∂x k α ∂x k α ∂x k ∂x k ∂t ∂t ∂t ∂t ∂t
and we get the continuity equation ∂ ρ ∂( ρ u k ) + = 0. ∂t ∂x k
(13)
We can eliminate velocities and write the continuity equation in another form ∂ρ k ∂2 ρ ∂φ − 2 + α e−α t k = 0. ∂t α ∂x k ∂x k ∂x k
(14)
4. Expressions for ρ derivatives We can express derivatives of ρ k in terms of velocities u k . For example, expression (P1-15) reads now ∂ρ0 α2 =− ρ u0 j , ∂x j k
(15)
∂ρ1 α2 =− ρ u1 j − α e−α t δ 1k φ 1 . ∂x j k
(16)
and similarly
Sum (15) and three expressions (16) and get ∂ρ α2 =− ρ u − α e−α t φ j . ∂x j k j
(17)
in full agreement with (11). k This means, that the field ρ ui − α e−α t φ i posses potential. This potential is equal to − 2 ρ . α We can also express second derivatives of ρ k in terms of velocities u k and their derivatives. For example, differentiation of (15) gives
-5∂2 ρ 0 α 2 ∂ρ α 2 ∂u0 j =− u0 j − ρ . ∂x i ∂x i ∂x j k ∂x i k
Pick expression for
(18)
∂ρ from (17) and insert it to (18) ∂x j 2
2 α 2 ∂u0 j ∂2 ρ 0 α = ρ ui − α e−α t φ i u0 j − ρ . ∂x i ∂x i ∂x j k k
(19)
Similarly differentiation of (16) gives ∂ρ1 α 2 ∂ρ α 2 ∂u1 j ∂φ =− u1 j − ρ − α e−α t δ 1k 1 = ∂x i ∂x j k ∂x i k ∂x i ∂x i
(20)
2
=
2 α 2 ∂u1 j ∂φ α ρ ui − α e−α t φ i u1 j − ρ − α e−α t δ 1k 1 . k k ∂x i ∂x i
Sum (19) and three expressions (20) and get 2
2 ∂φ j α 2 ∂u j ∂ρ α = ρ ui − α e−α t φ i u j − ρ − α e−α t . ∂x i ∂x i ∂x j k k ∂x i
(21)
Alternate (21) on the indices i, j ∂φ j ∂ui ∂φ α 2 −α t ∂u j ρ − α e−α t − α e−α t i + α e (φ i u j − φ j ui ) = 0. − ρ ∂x i ∂x i ∂x j ∂x j k
(22)
We shall use this identity (see (41)). 2 2 2 α α ∂ρ α 1 α ∂u j ∂ui ∂φ j ∂φ i = ρ ui u j − e−α t φ i u j − e−α t φ j ui − ρ + + . (23) − α e−α t 2 k ∂x i ∂x j ∂x i ∂x j k 2 2 ∂x i ∂x j
Another interesting identity follows from (23) and (14) k
3
∂ρ
∂φ
Σ ∂x k ∂x k − α e−α t ∂x kk = α 2 k=1
∂ρ = ∂t
3
2 ∂u k α ρ u − α e−α t φ k u k − ρ . k ∂x k
Σ k=1 k
5. Current of momentum tensor and stresses The expression for current of momentum tensor is (see [2] and [3]):
(24)
-6 k exp −t α + 2 α −∞ −∞ −∞ +∞ +∞ +∞
J 1kl =
× Aω 1ω 2ω 3 100
∫∫∫
3
Σ j=1
ω 2j ×
(25)
3/2 k k ω k α δ 1k k ω l α δ 1l − − − − − δ kl 2π α α α ω1 α α ω1
k α 2δ 1l + × ω 12 α
1/2 k ω j 2 2k iω 1 j=3 ω x ) exp dω 1 dω 2 dω 3. × exp(i − j j α α Π j=1 2α α
We see after, that simplification 2 k ω k α δ 1k k ω l α δ 1l k α δ 1l − − + − − − δ kl = α α α ω1 α α ω1 ω 12
(26)
2
=
δ 1k k δ 1l k k k ω kω l + ωl + ω k − δ kl . 2 α ω1 α ω1 α α
cancel all negative degree terms. k 2 ∂3φ 1 k ∂φ 1 k ∂φ 1 k ∂φ 1 − δ 1k − δ 1l + δ kl J 1kl = e−α t 2 . α ∂x 1 ∂x k ∂x l α ∂x α ∂x α ∂x 1 l k
(27)
Recall (P1-14) 2 k k ∂ ρ0 + ρ 0δ ij . 2 α ∂x i ∂x j α
(28)
2 k k k ∂ ρ ∂φ k ∂φ l − e−α t + + δ kl ρ = α 2 ∂x k ∂x l α ∂x l ∂x k α
(29)
2
J 0ij =
Sum (28) and tree expressions (27) 2
J kl = J 0kl + J 1kl + J 2kl + J 3kl =
∂φ l k −α t ∂φ k ∂φ l k k ∂u = ρ u k − α e−α t φ k u l − 2 ρ l − α e−α t + − e + δ kl ρ = ∂x l ∂x k α ∂x k ∂x k α α α 1 k ∂u k ∂u l ∂φ l k −α t ∂φ k ∂φ l k −α t ∂φ k ρ + α e + − e + − + δ kl ρ = = ρ u k u l − e−α t (φ k u l + φ k u l ) − ∂x l ∂x k α ∂x l ∂x k 2 α 2 ∂x l ∂x k 2 α α 1 k ∂u k ∂u l k ∂φ k ∂φ l = ρ u k u l − e−α t (φ k u l + φ k u l ) − ρ + + + δ kl ρ . + α e−α t 2 ∂x l ∂x k 2 α ∂x l ∂x k 2 α
Tensor of stresses components are equal to
-7-
σ kl = ρ u k u l − J kl = α e−α t φ k u l +
=
α
2
e−α t (φ k u l + φ l u k ) +
k ∂u l k ∂φ ρ + α e−α t k − δ kl ρ = 2 α ∂x k ∂x l α
(30)
1 k ∂u k ∂u l k ∂φ k ∂φ l + + − δ kl ρ . + α e−α t ρ 2 ∂x l ∂x k 2 α ∂x l ∂x k α
Recall, that in [1] we get the state equation of compressible viscous fluid with kinematic viscosity k equal to ν = 2 . We see, that additional term is present in (30) - the first term. This term is proportional to α
e−α t and therefore it is relatively small for big t.
6. Equations of movement Now we consider another conservation law - equation of movement. The equation is: ∂( ρ ui ) ∂( ρ ui u j ) ∂σ ij + − + α ρ ui = 0. ∂x j ∂x j ∂t
(31)
We make following substitutions in (31) (all expressions are known from the previous text) : ρ uk = −
k ∂ρ + α e−α t φ k . α 2 ∂x k
(32)
ρ u k u l − σ kl = J kl .
(33)
∂ k ∂ρ + α e−α t φ i + − ∂t α 2 ∂x i
(34)
The result is:
+
∂ ∂x j
∂φ j α −α t k 1 k ∂ui ∂u j −α t ∂φ i ρ + α e + + δ ρ + + ρ ui u j − e (φ i u j + φ j ui ) − ij ∂x j ∂x i 2 α 2 α 2 ∂x j ∂x i +α ρ ui = 0.
We make further substitutions: 3 ∂( ρ u ) ∂ρ k =−Σ ; ∂t ∂x k=1 k
(35)
∂φ i k ∂2φ i ∂2φ i ∂2φ i = 2 + + 2 ; ∂t α ∂x 2 ∂y 2 ∂z
(36)
-8∂ k k ∂ρ δ ij ρ = = −α ρ ui − α e−α t φ i ∂x j α α ∂x i
(37)
2 k ∂2 ( ρ u j ) −α t ∂ φ i 2 −α t − α φ + α e + e i α 2 ∂x i ∂x j ∂x j x j
(38)
and get
+
∂ ∂x j
α −α t 1 k ∂ui ∂u j ∂φ i ∂φ j −α t ρ + + + α e−α t ρ ui u j − e (φ i u j + φ j ui ) − − α ρ ui − α e φ i + 2 ∂x ∂x ∂x 2 α ∂x 2 i i j j +α ρ ui = 0. We can place all the expression under the brackets: ∂ ∂x j
k ∂( ρ u j ) ∂φ α + α e−α t i + ρ ui u j − e−α t (φ i u j + φ j ui ) − 2 ∂x α ∂x 2 i j −
(39)
1 k ∂ui ∂u j ∂φ i ∂φ j + + =0 + α e−α t ρ ∂x j ∂x i 2 α 2 ∂x j ∂x i
and get ∂u j ∂ ∂φ α k −α t + α e−α t i + ρ ui u j − e−α t (φ i u j + φ j ui ) − − ρ ui − α e φ i u j + 2 ρ ∂x i ∂x j α ∂x j 2 −
(40)
∂φ j 1 k ∂ui ∂u j −α t ∂φ i ρ + α e + = 0. + ∂x j ∂x i 2 α 2 ∂x j ∂x i
We perform last simplifications and bring expression to the following form ∂ ∂x j
α −α t 1 k ∂ui ∂u j ∂φ i ∂φ j − − = 0. + α e−α t e (φ i u j − φ j ui ) + ρ 2 ∂x j ∂x i 2 α ∂x j ∂x i 2
(41)
This equality is true because of identity (22). In this way we checked, that variables ρ and v i satisfy the set of conservation laws of classic hydrodynamics. They are not quite the set of equations of classic hydrodynamics of isothermal compressible fluid, because expression for stresses (30) contains additional term. This term is negligible in big times limit.
7. Burgers equation
-9-
We write equations of movement in slightly modified form (compare with [1]): ∂u 1 ∂σ ij ∂ui + uj i − + α ui = 0. ∂x j ρ ∂x j ∂t
(42)
We take expression for stresses in asymmetric form (see 30)) σ ij = α e−α t φ j ui +
k α2
ρ
∂ui k −α t ∂φ j k + e − δ ij ρ . ∂x i ∂x j α α
(43)
Perform substitution ∂ui ∂u 1 ∂ −α t k ∂ui k −α t ∂φ j k + uj i − α e φ j ui + 2 ρ + e − δ ij ρ + α ui = 0 ∂t ∂x i ∂x j ρ ∂x j α ∂x j α α
(44)
∂u 1 ∂ k ∂φ j ∂ui + u j i − e−α t α φ j ui + − ∂x j ρ ∂x j α ∂x i ∂t
(45)
and get
−
1 k ∂ ρ ∂ui k ∂2 ui 1 k ∂ρ − + + α ui = 0. 2 2 ρ α ∂x j ∂x j α ∂x j ∂x j ρ α ∂x i
We use expression for derivatives α2 ∂ρ =− ρ u − α e−α t φ j ∂x j k j
(46)
∂ui ∂u 1 ∂ k ∂φ j + u j i − e−α t αφ u + + ∂t ∂x j ρ ∂x j j i α ∂x i
(47)
once again:
+
1 k ∂2 ui 1 ∂ui ρ u j − α e−α t φ j − 2 − α ρ u − α e−α t φ i + α ui = 0. ∂x j α ∂x j ∂x j ρ i ρ
It follows ∂u k ∂2 ui 1 ∂ k ∂φ j ∂ui + 2u j i − 2 − e−α t α φ j ui + − ∂x j α ∂x j ∂x j ρ ∂x j α ∂x i ∂t −
1 ρ
α e−α t φ j
∂ui 1 2 −α t + α e φ i = 0. ∂x j ρ
(48)
-10-
Simple rearrangement of terms bring expression to its final form: ∂ui ∂u k ∂2 ui + 2u j i − 2 + ∂t ∂x j α ∂x j ∂x j +
(49)
∂ k ∂φ j ∂ui e−α t α 2φ i − α φ j ui + − αφ j = 0. ρ ∂x j α ∂x i ∂x j 1
As a result we get following equation for ui only. Similarly to [1], this equation strongly resembles Burgers equation. To get more usual form of Burgers equation we could perform substitution t′ = 2t and k ν′ = , but we omit this calculation. 2α 2 We see from (49), that velocities v i satisfy the Burgers equation with some additional term. This term (the last term in (49)) is negligible in big times limit. Moreover, only this last term depend on ρ variable. It is inversely proportional to ρ and directly proportional to φ and its derivatives.
DISCUSSION In this paper we proceed with investigation of connections between Fokker - Planck equation and continuum mechanics. We base upon expressions from our work [2], based upon the spectral decomposition of Fokker - Planck equation solution. In this decomposition we preserve only terms with the smallest degrees of damping - zero and first degree. We get expressions for density, velocities and stresses from potentials ρ 0 , φ i . Potentials all satisfy the same diffusion equation. Starting from initial density and velocities, we could solve diffusion equations and find potentials. We find, that macroscopic parameters of Fokker-Planck flows, obtained in this way, satisfy the set of conservation laws of classic hydrodynamics. They are not quite the set of equations of classic hydrodynamics of isothermal compressible fluid, because expression for stresses (30) contains additional term. This term is negligible in big times limit. We proved also, that the velocities field alone, with potential, which satisfy diffusion equation, satisfy Burgers equation without mass forces - with some additional term. This term (the last term in (49)) is negligible in big times limit. Another approach to considered problem see in [5].
REFERENCES [1]
Igor A. Tanski. Spectral decomposition approach to macroscopic parameters of Fokker-Planck flows: Part 1 arXiv:0707.3306v2 [nlin.CD]
[2]
Igor A. Tanski. Macroscopic parameters of Fokker-Planck flows. arXiv:nlin.SI/0606019 v1 06 Jun 2006
[3]
Igor A. Tanski. Spectral decomposition of 3D Fokker - Planck differential operator. arXiv:nlin.CD/0607050 v3 25 Jun 2007
[4]
P. Re´sibois, M. De Leener. Classical kinetic theory of fluids, John Wiley and sons, NY, 1977
[5]
Alex A. Samoletov. A Remark on the Kramers Problem. arXiv:physics/0311042 v1 10 Nov 2003
