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THESIS FOR THE DEGREE OF LICENTIATE OF PHILOSOPHY

Classication of Normal Discrete Kinetic Models

Mirela Cristina Vinerean Department of Mathematics Karlstad University Universitetsgatan 2 SE-65188 Karlstad, SWEDEN

Abstract In many interesting papers on discrete velocity models (DVMs), authors postulate from the beginning that the nite velocity space with ”good” properties is given and only after this step, study the Discrete Boltzmann Equation. Contrary to this approach, our aim is not to study the equation, but to discuss all possible choices of nite phase spaces (sets) satisfying this type of ”good” restrictions. Due to the velocity discretization it is well-known that it is possible to have DVMs with ”spurious” summational invariants (conservation laws which are not linear combinations of physical invariants). Our purpose is to give a method for constructing normal models (without spurious invariants) and to classify all normal plane models with small number of velocities (which usually appear in applications). On the rst step we describe DKMs as algebraic systems. We introduce an abstract discrete model (ADM) which is dened by the matrix of reactions (the same as for the concrete model). This matrix contains as rows all vector of reactions, which can be written as n-dimensional vectors (k1 , ., kn ) with ki Z, describing the ”jump” from a pre-reaction state to a new reaction state. The conservation laws corresponding to the many-particle system are uniquely determined by the ADM, or equivalently, by its matrix of reactions, and do not depend on the concrete realization. We nd the restrictions on ADM and then we give a general method of constructing concrete normal models (using the results on ADMs). Having the general algorithm, we consider in more detail, the particular cases of models with mass and momentum conservation (inelastic lattice gases with pair collisions) and models with mass, momentum and energy conservation (elastic lattice gases with pair collisions). Mathematics Subject Classications (2000): 82C40 (76P05) Keywords: Kinetic theory, discrete kinetic (velocity) models, conservation laws

ii Acknowledgement First of all, I want to thank my supervisor, Professor Alexander Bobylev, on whose ideas this paper is based, for his valuable guidance and support. His help and advises are of a great value to me. I thank Håkan Granath for his contribution in the process of generating matrices. This work was supported by the Graduate School in Mathematics and Computing (FMB), Sweden, and by the Department of Mathematics, Karlstad University Sweden. I especially want to thank Professor Christer Kiselman, the director of FMB and Maya Neytcheva, the deputy director of FMB, for their help and support. Special thanks go to Niclas Bernho for his careful proof-reading of this paper. Mirela Cristina Vinerean March 2004, Karlstad

Contents 1 Introduction and Preliminaries 1.1 Discrete velocity models of the Boltzmann equation and spurious conservation laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Preliminaries: How to nd an actual number of conservation laws for a given DVM? . . . . . . . . . . . . . . . . . . . 2 Discrete Kinetic Models as Algebraic Systems 2.1 Abstract model and its realization . . . . . . . 2.2 Conservation laws . . . . . . . . . . . . . . . . 2.3 Natural realization . . . . . . . . . . . . . . . . 2.4 Statement of the classication problem . . . . . 2.5 Inductive method (1-extension) . . . . . . . . . 2.6 Connection with functional equations . . . . . .

1 1 4

. . . . . .

5 5 10 12 16 18 20

3 Models with mass and momentum conservation 3.1 Statement of the problem and geometrical interpretation . . . . . . . . 3.2 Construction and classication of normal models . . . . . . . . . . . .

22 22 24

4 Models with mass, momentum and energy conservation 4.1 Statement of the problem and geometrical interpretation . . . . . . . . 4.2 General method (algorithm) . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Classication and construction of normal plane elastic models with small numbers of velocities . . . . . . . . . . . . . . . . . . . . . . . . 4.3.1 Starting point (n = 6) and 1-extension . . . . . . . . . . . . . . 4.3.2 Classication results for n {7, 8} . . . . . . . . . . . . . . . . 4.3.3 Algorithm for normal plane elastic models with small numbers of velocities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.4 Computer results for the cases n = 8 and n = 9 . . . . . . . . .

27 27 28

5 Conclusions

64

iii

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30 30 43 49 58

CONTENTS

iv

Chapter 1

Introduction and Preliminaries 1.1

Discrete velocity models of the Boltzmann equation and spurious conservation laws

The general discrete velocity model (DVM) of the Boltzmann equation reads ([6, 12]) μ ¶ (1.1) + vi fi (x, t) = Qi (f ) = Qi (f1 , ..., fn ) , i = 1, ..., n. t x where x Rd and t R+ denote the position and the time respectively, {v1 , ..., vn } Rd denote a set of velocities of the model. The functions fi (x, t) are understood as spatial densities of particles having velocities vi Rd , usually d = 2, 3 in applications. The collision operators Qi (f) in (1.1) are given by Qi (f ) =

n X

kl ij (fk fl fi fj ) for i = 1, ..., n,

(1.2)

j,k,l=1

such that the collision coecients kl ij , 1 i, j, k, l n, satisfy the relations ij kl kl ij = ji = kl 0,

(1.3)

with equality unless the conservation laws vi + vj = vk + vl , |vi |2 + |vj |2 = |vk |2 + |vl |2

(1.4)

are satised. DVM (1.1), (1.2) is called ”normal” if any solution of equations (vi ) + (vj ) = (vk ) + (vl )

(1.5)

where indices (i, j; k, l) take all possible values satisfying (1.4), is given by (v) = a + b · v + c |v|2 1

(1.6)

1. Introduction and Preliminaries

2

for some constants a, c R and b Rd (the terminology ”normal” was introduced in [8]). In such a case the equality n X

(vi )Qi (f1 , ..., fn ) = 0

(1.7)

i=1

has (for arbitrary f1 , ..., fn 0) the general solution (1.6), moreover D(f) =

n X

Qi (f) log fi 0 ,

(1.8)

i=1

D(f) = 0 if and only if fi = Mi = e(vi ) where (v) is given by Eq.(1.6). It is obvious that the ”normality” of DVM is a very important condition (similar to the condition of uniqueness of collision invariants for the Boltzmann equation [1, 2, 7]). Such a condition is usually assumed in most of the mathematical papers on DVMs (see for review [6], [13], [3]). On the other hand, it is not easy to construct a concrete normal d-dimensional DVM (to be more precise, its set of velocities {v1 , ..., vn } Rd , such that any conservation law (1.5) is given by Eq.(1.6)). This problem appeared already at the early stage of development of mathematical theory of DVMs [12] and still remains, generally speaking, unsolved. The same general problem ”How to construct DVMs without non-physical invariants?” remains open for more general discrete kinetic models related to gases with internal degrees of freedom, mixtures, chemically reacting gases, etc. To our knowledge, there is just one particular method to do it. The method proposed by Bobylev and Cercignani [5] in 1999 (see also [14, 15]), is based on the construction given in their previous paper [4]. The main idea is that, starting from an n-velocity normal DVM, we sometimes (under certain conditions described in detail in Chapter 2) can construct a normal DVM with (n + 1) velocities. We stress that all known normal DVMs are (or can be) constructed by this method. In particular, all regular ”quadratic” (d = 2) or ”cubic” (d = 3) models are normal since they can be obtained by this inductive approach starting from simple Broadwell-type models with a few velocities. Many new DVMs were obtained by the inductive method in the last years [10, 15]. On the other hand, this is just a particular method which does not answer many questions. For example it is still unclear whether or not all normal DVMs with a given number n of velocities can be obtained by the inductive approach (the answer is negative, as we shall see in Chapter 4). The main objective of this paper is to develop a very general approach to the problem of description of all non-isomorphic classes of normal discrete kinetic models with given conservation laws. The word ”normal”, in this case, means the absence of other (spurious) conservation laws.


3

What would such a theory mean for DVMs of the Boltzmann equation described at the beginning of this section? It would give the complete list of all non-isomorphic classes of normal d-dimensional DVMs with a given number n of velocities. This is what we call ”the complete classication of normal DVMs”. Our aim is to develop a general approach to the problem and to show how it works for small number of velocities in the plane case d = 2. We shall see below, that the cases n = 6, 7, 8 can be treated by analytical methods, however the computer becomes necessary already for the case n = 9. The use of the computer leads to quite new classes of normal DVMs with n = 9, which cannot be obtained by inductive approach. The paper is organized as follows: In Section 1.2 we present a known method for deciding the ”normality” of a given DVM. We stress that the way of constructing all normal DVM is still unknown (the inductive method of 1-extentions [5] does not cover, as we shall prove later, all cases). In Chapter 2 we consider the problem of the classication of normal discrete kinetic models (DKMs), when we know the order n and the dimension d of the model, the number p of linearly independent conservation laws, the function : Rd Rd of conservation laws and the set of reactions . We introduce an abstract version (ADM) of DKMs, dene the set (matrix) of reactions and show that all conservation laws are uniquely-dened by the ADM. By stating that a concrete discrete model (DM) is a d-dimensional realization of an ADM that has the same matrix of reactions, we prove in Theorem 1, that for any ADM fullling some special conditions, there exists at least one p-dimensional realization. In Lemma 4 we give necessary and sucient conditions such that a concrete DM is a realization of an ADM. This result is a tool for the complete classication of normal models with a given set {n, d, p, , } and we are able to give for this (in Section 2.4) a general algorithm. In section 2.6 we discuss the problem of the uniqueness of collision invariants for the Boltzmann equation, solved for the”continuous” case, and underline that the central problem of this thesis is to describe all nite sets for which the collision invariants are unique. In Chapter 3 and Chapter 4 we consider the particular cases of DVMs with mass and momentum conservation, and mass, momentum and energy conservation, respectively. We state the problem of classication and give a geometrical interpretation. In the rst case the general algorithm gives a complete solution of the classication problem. In the second case, we have a lot of admissible matrices of reactions which do not have normal DVMs as realizations. We give a modied algorithm in the case of elastic DVMs with pair-collisions, based on the algorithm given for DKMs. We study in more detail the case of plane elastic DVMs with small number of velocities. We give a complete classication of models up to n = 8 velocities (showing that they are all 1-extensions) and improve the algorithm for d = 2, by proving new properties of normal plane


4

DVMs. Finally, we use the algorithm to check the case of 8-velocity normal models and to obtain 9-velocity normal models. The result is that there exist DVMs which are not 1-extensions and we can construct them.

1.2

Preliminaries: How to nd an actual number of conservation laws for a given DVM?

We shall discuss a simple observation of Vedeniapin and Orlov [15] that represents (together with the paper [5]) another starting point for our present work. They have considered an arbitrary (not necessarily normal) DVM with the collision term (1.2) and introduced a set of vectors n kl ekl ij = (0, 0, ..., 1 , ..., 1 , ..., 1, ..., 1, ..., 0) R , ij > 0, (i)

(j)

(k)

(l)

(1.9)

for any combination of indecies (i, j|k, l) such that kl ij > 0. The conservation law (1.5) can be rewritten as a set of orthogonality conditions kl · ekl ij = 0, ij > 0,

(1.10)

= ((v1 ), ..., (vn )) Rn .

(1.11)

where The whole set of vectors (1.9) for all cases when kl ij > 0 can be written (one vector under another one, in an arbitrary order) in the form of a matrix with n columns. The rank p of this matrix cannot, by construction of the DVM, exceed (n (d + 2)), since the functions 0 (v) = 1; (v) = v , = 1, ..., d; d+1 (v) = |v|2

(1.12)

lead to (d + 2) linearly independent (in the general case) vectors (1.11) of conservation laws. Hence, the DVM is normal if and only if rank = n (d + 2) .

(1.13)

This gives a simple practical criterion of ”normality” for any given DVM: the number of linearly independent vectors (1.9) must be equal to (n d 2). By using this technique, the authors of [15] proved that some DVMs for mixtures proposed in [4] had spurious invariants. This idea helps to reject some ”bad” DVMs, however, it says nothing about the way to construct ”good” models. This question will be considered for general discrete kinetic models (DKMs) in the next chapter and vectors of type (1.9) will play an important role in our approach.

Chapter 2

Discrete Kinetic Models as Algebraic Systems 2.1

Abstract model and its realization

We consider an abstract version of a kinetic model with a nite set S of phase states. By using any numeration of the states we represent S as S = Zn = {z1 , ..., zn } .

(2.1)

The number n is said to be the order of the model. We assume that there is a large number N of particles such that each particle can occupy at a given time instant t > 0 one of the discrete states of the phase set S. The instant state of the N-particle system is described by the n-vector of occupation numbers. ( = (N1 , ..., Nn ) , (2.2) N1 + ... + Nn = N such that Nk is the number of particles at the phase point zk (k = 1, ..., n). A stochastic dynamics of the system is described by the following process: at any time instant t the system may undergo, with certain probability dWs (t) = ps dt , s = 1, ..., m , one of the m elementary reactions (”jumps”). This can be written as Before reaction (0) Ni = Ni

ps dt

After reaction (s) Ni , i

= 1, ..., n and s = 1, ..., m.

The reaction is a collective interaction (collision) of several particles, which leads to a new state of the system (new occupation numbers). We do not assume that the number of reacting particles is preserved. 5

2. Discrete Kinetic Models as Algebraic Systems

6

If all possible reactions are numerated by numbers s = 1, 2, ..., m, then each reaction can be understood as a ”jump” from a pre-reaction state in (2.2) to a new state s , where ( (s) (s) s = (N1 , ..., Nn ) (2.3) (s) (s) N1 + ... + Nn = N (s) and N (s) might be dierent of N . We have ( (s) (s) Ni = Ni ki , i = 1, ..., n and s = 1, ..., m (s)

ki If we denote

{0, ±1, ±2, ...}

(s)

s = (k1 , ..., kn(s) ), s = 1, ..., m ,

then

s = s , s = 1, ..., m.

.

(2.4)

(2.5)

(2.6)

The vectors (2.5) (with integer components) are said to be vectors of reactions. (s) (s) The total number k+ of colliding (reacting) particles and k of products of reaction are given by P (s) (s) k = ki + (s) ki >0¯ (2.7) P ¯ (s) ¯¯ . (s) k = ¯ki ¯ (s) ki 0 .


11

Denition 5 A linear conservation law is dened by a linear functional l [] (l : Rn R) , such that l [] = const. (not depending on the the time t). A usual consideration leads to the following result. Lemma 1 There exists a unique vector u Rn (u = (u1 , ..., un )) such that l [] = (u, ) =

n X

ui i

(2.19)

i=1

with i = Ni from (2.2) , i = 1, ..., n. Proof. If = {1 , ..., n }, then l [] = l (1 , ..., n ) =

n P i=1

denote

i l (ei ) =

n P i=1

ui i .

Hence, u = (l (e1 ) , ..., l (en )), with { ek | k = 1, ..., n} the Euclidean basis of Rn . The next step is also straightforward. One can easily connect the vectors u of conservation laws with the matrix of reactions. Lemma 2 (a) A set of conservation laws (vectors u Rn ) is dened by (u, s ) = 0 , for all s in (2.10);

(2.20)

(b) If the number of linear independent vectors in is n p, then the model has p independent conservation laws {u1 , ..., up } Rn , forming a basis of a p-dimensional linear subspace U Rn . Furthermore, for L = Span

(2.21)

U = L and Rn = L U.

(2.22)

we have

Proof. (a) From (2.6) we have l [s ] = l [] l [ s ] , for all s . Using Eq.(2.19) and l [] = const., we obtain (u, s ) = 0, for all s . (b) Let L = Span . Then dim L = n p. Assuming that p 1, we obtain the natural decomposition Rn = L L and denote by U = L , dim U = p, the orthogonal complement of L in Rn . Any set of p linearly independent vectors (j)

uj = (u1 , ..., un(j) ) U, j = 1, ..., p

(2.23)


12

have the property (uj , s ) = 0 , for all s and hence, it forms a basis of conservation laws in U. The equality p = n dim L, L = Span (2.24) generalizes a similar result [15] (discussed in section 1.2) for classical DVMs with elastic pair collisions. In the sequel we will be mostly interested in conservation laws. This motivates the following. Denition 6 Two ADMs (or, equivalently, two sets of reactions 1 and 2 ) are said to be equivalent if they have the same space U of conservation laws (Span 1 = Span 2 ). Remark 2 10 . Any class of equivalent ADMs, given above, corresponds to a ((n p) × n) -matrix 1 1 · · · n1

(2.25) = ··· ··· ··· 1np · · · nnp whose rows are vectors of reactions ( rank = n p and ji = 0, ±1, ...). 20 . If the two numbers n and p 1 and the maximal nite set of reactions are given, then we obtain a nite number M (n, p; ) of pairwise disjoint classes of ADMs. 30 . The equality (2.24) helps to evaluate the exact number of conservation laws for any given model provided its set of reactions is known. It is not sucient, however, if we want to construct a concrete DM with some prescribed properties. We consider this problem in more details below. We stress that the conservation laws (vectors (2.23)) are dened completely by the ADM (matrix ) and do not depend on its concrete realization (the phase set Xn (2.17)).

2.3

Natural realization

We consider an ADM (, n), with the -matrix (2.25). By the above described method, we construct the orthogonal subspace U (2.22) and assume a basis of conservation laws (2.23) in U to be xed. We write the basis of conservation laws {u1 , u2 , ..., un } from (2.23) as a p×n matrix (denoted by U1 to distinguish it from the subspace U) in the following form (1) (1) ( u1 · · · un rank U1 = p

U1 = · · · · · · · · · , . (2.26) (k) (k) uk = (u1 , ..., un ) (p) (p) u1 · · · un


13

Since U = ker we have U1| = 0.

(2.27)

The n columns of the matrix U1 can be considered as n pairwise dierent p-dimensional vectors (1) (p) (2.28) wk = (uk , ..., uk ), k = 1, ..., n provided that wk = wl if and only if k = l. We denote the set of all vectors (2.28) by Wn = {w1 , ..., wn } Rp .

(2.29)

U1| = (w1 , ..., wn ) , wk Wn , k = 1, ..., n.

(2.30)

Note that

From Eqs.(2.19),(2.20),(2.27) and (2.30), we have that the most general (p-dimensional) linear conservation law of many-particle system reads n X

Nk (t) wk = const.

(2.31)

k=1

This remark gives a motivation for the next denition. Denition 7 The set Wn from (2.29) of vectors dened by (2.23) and (2.28) is said to be a set of linear invariants of the model. Remark 3 The set Wn is not unique (its denition depends on an arbitrary basis (2.23) of conservation laws). The vectors wk in (2.28) (k = 1, ..., n) are dened with

an accuracy up to any non-singular linear transformation (non-singular matrix) M : Rp Rp , such that 0

wk = Mwk , k = 1, ..., n.

(2.32) 0

0

We note that the transformation (2.32) cannot lead to equality wk = wl if wk 6= wl . We need to nd some conditions such that this restriction is fullled. The following denitions will help us to express such conditions. Denition 8 An ADM (, n) is said to be closed in if = {1 , ..., m } contains all linear combinations in of its elements, i.e. = Span .

(2.33)

Remark 4 There is one and only one closed ADM in the above dened class, since all models of the class have the same Span (see Denition 6).


14

Denition 9 An ADM is said to be well-dened if ek el / Span , 1 k < l n ,

(2.34)

for any pair of standard unit vectors in Rn , ei = (0, ...0, |{z} 1 , 0, ..., 0), i = 1, ..., n . i

The above two denitions are used in the process of obtaining a natural realization of an ADM with p linear conservation laws. In this section we have been describing the way one follows from the starting point were we choose an ADM (, n), until we obtain the corresponding set of invariants

Wn (2.29) (with accuracy up to any non-linear transformation M of Rp ). Having this set, one can construct a concrete p-dimensional DM of order n. Towards this goal, we x Wn and consider it as the phase set Xn (2.17) of the concrete DM. We dene by the set of all vectors s , such that (s , uj ) = 0, j = 1, ..., p , (j)

(j)

(2.35)

(j)

where uj = (u1 , ..., un ) and ui = wji , obtained from wk = (w1k , ..., wpk ), k = 1, ..., n. The total number of vectors s satisfying (2.35) is bounded because of the general restriction (2.8) ( is a nite set). The DM dened by the phase set (2.29) and Eqs.(2.35), which give the set of reactions, satises the two conditions in Denition 1. We note that the set of reactions of this model consists of all vectors such that and L from Lemma 2, i.e. = L , L = Span .

(2.36)

Generally speaking, 6= and the constructed model cannot always be considered as a realization of the initial ADM (, n). The necessary and sucient condition on , such that the above constructed DM (provided that it exists) is a realization of the corresponding ADM, reads (2.37) = or equivalently Eq.(2.33), so the ADM should be closed in . Remark 5 This condition is strong (in particular, all reactions must be invertible); however it is satised in many practically interesting cases. For a given , the model dened by a set of invariants (2.29), as the phase set, and conditions (2.35), as rules, is certainly non-unique. However, any invertible linear

transformation M : Rp Rp of its phase set does not change the set (2.36) of reactions and therefore, we consider such models as equivalent (isomorphic) ones.


15

Denition 10 The whole class of equivalent models described above is called the standard model for the set . It was implicitly assumed above that the standard model has exactly n distinct phase points, or, equivalently, that wk 6= wl if k 6= l. The following lemma yields corresponding necessary and sucient conditions on the matrix . Lemma 3 The condition wk 6= wl for k 6= l is fullled for all vectors in Wn (2.29) if and only if the ADM (, n) is well-dened. Proof. We note that all components of the vectors wk Wn are given by Eq.(2.28). Therefore (2.38) wk = (w1k , ..., wpk ), wjk = (uj , ek ) , k = 1, ..., n, j = 1, ..., p, where we used the notations (2.23) for uj . Hence, wk = wl if and only if (uj , ek el ) = 0, j = 1, ..., p . If ek el L, then wk = wl , since uj are orthogonal to L by construction. Conversely, if wk = wl then ek el L since {uj : j = 1, ..., p} is a basis in the orthogonal subspace U = L . Hence, the equality wk = wl is equivalent with ek el L and lemma is proved. Remark 6 In terms of the many-particle system, the condition (2.34) means that there is no nite sequence of reactions³ which leads to a transition ´ from the initial state 0 0 (N1 , ..., Nk , ..., Nl , ..., Nn ) to a state N1 , ..., Nk , ..., Nl , ..., Nn , where all Ni are the 0

0

same for i / {k, l} and Nk + Nl = Nk + Nl . Usually, we assume that conditions (2.34) are fullled. Hence, any satisfying condition (2.34) leads to a uniquely dened (as a whole class) standard model. ³ 0 0´ Remark 7 An ADM can be equivalently dened by pairs (Zn , ) and Zn , which 0

dier only by numeration. The set denes the same (with accuracy up to numera0 tion) standard model as the set . The conditions (2.34) are satised by if they are satised by . We are now able to formulate the main result of this section. All necessary steps of the proof are already done. Theorem 1 Let (, n) denote a well-dened abstract discrete model (ADM) (satisfying condition (2.34)) and such that p = n rank , p 1 . Then (a) there exists a uniquely dened standard p-dimensional model of order n


16

(a class of equivalent DMs such that (i) their phase states are the linear invariants (2.29) of ; (ii) their set of reactions is dened by the orthogonality conditions (2.35)); the standard model has, by construction, the same subspace U Rn of conservation laws as the corresponding ADM; (b) if, in addition, is closed in , then the standard model yields a class of equivalent p-dimensional realizations of the ADM (the standard representation is dened up

to any non-singular linear transformation (matrix) M : Rp Rp ). Thus, for any well-dened ADM there exists at least one (with accuracy up to nonsingular linear transformations of its phase points) p-dimensional realization, where p = n rank .

2.4

Statement of the classication problem

We can now formulate the main problem of this paper. In applications we usually need to construct a concrete DM with a phase set Xn = {x1 , ..., xn } Rd ,

(2.39)

a matrix of reactions and with p given conservation laws n X

Nk (t) (xk ) = const. , : Rd Rp .

(2.40)

k=1

The function (map) (x) is usually known in advance. The model is said to be normal if it has no other conservation laws except (2.40) (we discussed already in the introduction, section 1.1, an example). The set (2.39) is assumed to be non-degenerate in the following sense: the equality n X

k (xk ) = 0, k = 1, ..., n

(2.41)

k=1

implies that all k = 0, k = 1, ..., n. By the classication of normal DMs we mean the description of all normal DMs (the phase sets Xn (2.39) and matrices of reactions), for given numbers n, p, d, given function : Rd Rp and given maximal set of reactions. It is clear in advance, that there exists just a nite number (perhaps zero) of distinct classes of such DMs. In order to explain how, in principle, this problem can be solved, we assume that there exists at least one normal DVM with the phase set Xn and the matrix of reactions. The matrix has, by construction, n columns and at


17

least (n p) linearly independent rows. Assuming that is well-dened (it is always so provided (x) = (y) if and only if x = y), we can construct its set of invariants (2.29) Wn = {w1 , ..., wn } Rp . and conclude that any conservation law can be written in the form (2.31) n X

Nk (t) wk = const., wk Rp , k = 1, ..., n.

k=1

The invariants are dened with accuracy up to a linear transformation 0

wk = Mwk , k = 1, ..., n,

(2.42)

where M is any non-singular matrix p × p. We note that all vectors w1 , ..., wn are distinct. Hence, we obtain the following set of equations (xk ) = Mwk , k = 1, ..., n,

(2.43)

or, in the scalar form j (x1k , ..., xdk ) =

p X

mjl wkl , j = 1, ..., p; k = 1, ..., n,

(2.44)

l=1

where {wk = (wk1 , ..., wkp ), k = 1, ..., n} is any basis of invariants of . The result can be formulated as follows. Lemma 4 If is a matrix of reactions of a normal DVM with the phase set (2.39) and given conservation laws (2.40), where = (1 , ..., p ), then there exists a matrix M = {mij ; i, j = 1, ..., p} , det M 6= 0, such that Eqs.(2.43), (2.44) are satised. The converse statement is also true. Lemma 4 gives us a practical tool for the complete classication of normal models with given numbers (n, d) and given function : Rd Rp. The procedure (algorithm) can be arranged in the following way. Step1. Consider the whole set of admissible matrices (well-dened, with (n p) linearly independent rows { 1 , ..., np } ). This set is nite under natural assumptions (called general restrictions in section 2.1) on . Therefore we can denote it by {1 , ..., N }. Step2. invariants.

Take the matrix 1 and construct an arbitrary set Wn (2.29) of its


18 (1)

Step3. Consider the Eqs.(2.43), (2.44) with unknowns Xn = {x1 , ..., xn } Rd and unknown (p × p) matrix M. If the equations are solvable, then the matrix 1 and (1) the set Xn (any solution of Eqs.(2.43), (2.44)) is accepted, otherwise the matrix 1 is rejected. Step4. Return to Step2 and take the next matrix 2 , etc. It is clear that the algorithm can be modied in some specic cases (see below) in order to simplify and accelerate the procedure. Its only non-trivial part is Step3. There are three possible cases: (a) p = d ; (b) p < d ; (c) p > d. The cases (a) and (b) are relatively simple since the number of equations (2.44) is equal to N1 = np and the number of unknowns is equal to N2 = nd + p2 . Hence N1 < N2 for any n 1 and p 1 provided that p d. This means that, under certain conditions on the map : Rd Rp , the equations (2.44) are solvable for any admissible matrix . Hence, all distinct classes of equivalent (in the sense of Denition 6) matrices lead to distinct classes of normal models. This solves the classication problem in the case p d. A practical application (DVMs with mass and momentum conservation) of the case p = d is discussed in detail in Chapter 3. Hence, the only time when we really need the above-described algorithm is in the case (c) p > d (the number of conservation laws is greater than the dimension of the phase space). Then the Eqs.(2.43), (2.44) are solvable if all n invariants {w1 , ..., wn } Rp belong to d-dimensional ”surface” w = M 1 (x)

(2.45)

and this, of course, is not the case for any admissible matrix . This explains the diculty of constructing normal DVMs for the classical Boltzmann equation with p = d + 2 (conservation of mass, momentum and energy).

2.5

Inductive method (1-extension)

In [5] the authors introduced a method of constructing DVMs without non-physical invariants (normal models) using an inductive scheme of 1-extension. The idea is that one can start with the simplest DM such as, for example, the modied Broadwell model for a simple gas. The Broadwell model with n = 2d velocities is not (formally) normal since it has only d + 1 conservation laws; to obtain a normal DVM, one add one more velocity as in the gure


19

Here (a) represents the Broadwell model , (b) the modied Broadwell model and (c) the extended Broadwell model. Given a normal DM (2.46) Xn = {x1 , ..., xn } Rd one constructs its extended version Xn+s = {x1 , ..., xn , xn+1 , ..., xn+s } , s 1,

(2.47)

where all new states {n + 1, ..., n + s} satisfy the conditions > 0, ( ) 1 r s () (i, j, k) , 1 i, j, k n, such that k,n+r ij

(2.48)

in other words, each new state (n + r) is a product of a certain reaction (i) + (j) (k) + (n + r)

(2.49)

which includes three states {i, j, k} present in the initial DM (2.46). The extended Broadwell model (c) corresponds to the case n = 5, s = 4. Lemma 5 (Bobylev-Cercignani,[5]) If the initial DM (2.46) is normal, then the same is true for the extended DM (2.47). The proof is quite obvious, it is enough to consider the case s = 1. This approach can be easily generalized to a wide class of discrete kinetic models. We shall use below this approach in the case of DVMs for the classical Boltzmann equation. By using 1-extensions one can prove that such normal DVMs exist for any given n 6 (we exclude DVMs with an isolated phase state (b) for n = 5). A straight forward continuation of the above described procedure, starting from the 9-velocity model (c), leads to an innite hierarchy of normal DVMs whose velocities belong to the regular lattice in R2 [5]. On the other hand, it is clear that (at least for small number of velocities n = 6, 7) there are normal DVMs of dierent structure, with ”rational”


20

and ”irrational” velocities. All normal DVMs obtained by 1-extensions are obviously ”reducible” (by taking o one phase state after another) to some ”irreducible” normal models. Can we reduce any normal model to the simplest 6-velocity one? If so, then the whole class of normal DVMs could be described in a simple way: just consider 6-velocity DVMs and make all possible 1-extensions. We shall see in the last chapter of this paper that the answer is negative: already in the case n = 9 there are some normal models which can not be obtained by 1extensions.

2.6

Connection with functional equations

The concept of normal DVMs is similar, to some extend, to the famous problem of uniqueness of collision invariants for Boltzmann equation. Therefore we briey discuss this problem here, in order to outline dierences between discrete (most important nite) and continuous cases. In the Boltzmann equation we have the following transformation which connects 0 0 pre-collisional velocities (v, w) with post-collisional velocities (v , w ): 0

0

v = v (v w, ) , w = w + (v w, ) ,

(2.50)

where v, w Rd (d 2) and S d1 is a parameter of the collision. Uniqueness of collision invariants means the following: we consider the functional equation 0 0 (2.51) (v) + (w) = (v ) + (w ) and assume that it holds for any v, w Rd and for any S d1 . Then any solution of the equation is a linear combination of (d + 2) collision invariants 1 (v) = 1 ; (v) = v , = 1, .., d ; +1 (v) = |v|2 .

(2.52)

How to prove this? Carleman [7] suggested the following way: choose the pair (v, w) such that v ·w = 0 v (the case v = w = 0 is trivial, therefore one can always assume and choose = |v| 0 0 that v 6= 0), then v = 0, w = v + w and (v) + (w) = (v + w) + (0) , v · w = 0.

(2.53)

By changing (v) to (v) (0) one obtains (in the same notation ) the equation (v) + (w) = (v + w) ,(0) = 0 , v · w = 0.

(2.54)


21

This is almost the Cauchy equation f (x) + f (y) = f(x + y) , f(0) = 0 , x, y Rd ,

(2.55)

except for the additional condition v · w = 0. The next step is to consider separately even and odd parts of (v) and nally to reduce the problem to the Cauchy equation in R or R+ [7, 9]. A more dicult case, when (v) is assumed to be locally integrable and the equality (2.53) holds almost everywhere in Rd , was solved by Arkeryd [1] and by Arkeryd and Cercignani [2]. A review on this problem can be found in the book [11]. It is very typical nowadays that many ”continuous” (in a broad sense) problems have to be stated and solved in the discrete (even nite) case. How to describe all nite sets (i.e. phase sets for normal DVMs) for which the collision invariants are unique? This is a central problem for this thesis. The above arguments obviously do not work. Even Cauchy equation (2.55) on a nite set A assumes implicitly that A has a structure of an Abelian group and this is not the case in many applied problems. On the other hand, it is possible to use similar methods for regular lattices in Rd (and nite sublattices), but this leads to well-known models which can be obtained by 1-extensions.

Chapter 3

Models with mass and momentum conservation 3.1

Statement of the problem and geometrical interpretation

We consider a set of particles with unit mass and n distinct velocities from the set Vn = {v1 , ..., vn } Rd .

(3.1)

We assume that the particles undergo inelastic (without conservation of energy) pair collisions (i) + (j) (k) + (l) (3.2) satisfying the momentum conservation law ()

vi

()

+ vj

()

()

= vk + vl

, = 1, ..., d.

(3.3)

Thus, we have a model with d-dimensional phase space (3.1) and (d + 1) scalar conservation laws of mass and momentum. The conservation of mass is equivalent in this case to the conservation of number of particles. We assume in addition that v 6= v , for 6= , , {i, j, k, l} .

(3.4)

Then the vector of the reaction (3.2) reads = (0, ..., |{z} 1 , ..., |{z} 1 , ..., |{z} 1 , ..., |{z} 1 , ..., 0). i

j

k

(3.5)

l

The set consists of vectors having four non-zero elements: two of them are equal to 1 and the other two are equal to (1). The phase set (3.1) has an obvious geometrical 22

3. Models with mass and momentum conservation

23

interpretation since any four vectors vi , vj , vk , vl , satisfying the conditions (3.3), (3.4), form a parallelogram in Rd

Remark 8 We do not exclude ”degenerate parallelograms” of the kind

It seems natural to introduce an alternative (geometrical) denition of these models. Denition 11 A DVM for identical particles with pair collisions, satisfying mass and momentum conservation laws, is a set of points (velocities) Vn (3.1) such that each point is a vertex of one or several parallelograms whose other vertices also belong to Vn . This denition is invariant under the numeration of points. On the other hand, if one chooses any xed numeration, then the matrix of the DVM can be easily reconstructed. In order to do this, we associate to any parallelogram a vector . This can be done in a unique way with accuracy up to the sign of . However, the sign of does not play any role in the problem of constructing normal models. In the case of inelastic collisions with energy dissipation, the sign of can be (”almost always”) dened uniquely by the condition |vi |2 + |vj |2 |vk |2 + |vl |2 .

(3.6)

Denition 12 The parallelograms are said to be independent if the corresponding vectors of reactions are linearly independent in Rn . Hence, the maximal number r of independent parallelograms of such a DVM is equal to rank , where is the matrix of reactions of the model. Therefore the number of conservation laws p reads p =nr .

(3.7)


24

Example 2 A DVM with 8 points and 2 independent parallelograms on the plane R2

has 6 conservation laws. The same example can be considered in Rd with any d 3 (we just assume that all 8 vertices are located somewhere in Rd , not in plane). Then, one can conclude that the model has at least (d + 1) conservation laws. This contradicts Eq.(3.7) if d 8. To avoid the contradiction, we introduce the following denition. Denition 13 The set (3.1) is said to be non-degenerate provided that the equalities a · vk = const., a Rd , k = 1, ..., n,

(3.8)

imply that a = 0. The same terminology is used for DVMs with the phase set (3.1). The meaning of degeneracy is obvious: a degenerate model has a real dimension d1 d 1 and should be considered after corresponding change of coordinates. It is clear that any set (3.1) with n < d is degenerate. This explains the above-mentioned contradiction (the example with two parallelograms is degenerate also in the case d = n = 8). From now on we consider only non-degenerate models. It is clear that any such model with n points (velocities) can not have more than (n d 1) independent parallelograms, otherwise the model would have less than (d + 1) conservation laws and this is a contradiction to the above-described construction. Therefore the normal models have a simple geometrical meaning: they maximize (for given n and d) the number of independent parallelograms. We briey describe in the next section how to construct all such models.

3.2

Construction and classication of normal models

We consider a normal DVM with given numbers n 4 (order) and d 2 (dimension) and with the set of velocities (3.1). The normal DVM has by denition exactly p = d+1 conservation laws n X k=1

Nk (t)vk = const. R , d

n X k=1

Nk (t) = const. R+

(3.9)


25

The set of all possible reactions consists of vectors (3.5) and the matrix of reactions must have (n (d + 1)) linearly independent rows { 1 , ..., nd1 } . The number of conservation laws p = d + 1 relates formally to the ”dicult” case (c) in Section 2.4 since p > d. Fortunately, this is not so because we have an ”universal” law: the vector u = (1, ..., 1) Rn

(3.10)

is orthogonal to all vectors (3.5) of the maximal set . This suggested the following procedure for all normal models. Step1. Take any well-dened (see section 2.3) (n d 1) × n matrix . Step2. Add the vector u (3.10) as an additional row to the matrix and consider the extended (n d) × n matrix . Step3. Construct d linearly independent vectors uk satisfying the equations uk = 0, k = 1, ..., d,

(3.11)

and represent them in the matrix form (1) (1) u1 ... un

(k) U = ... ... ... , uk = (u1 , ..., u(k) n ). (d) (d) u1 ... un Step4. Write down the set of linear invariants of by (2.28) (1)

(d)

vk = (uk , ..., uk ) Rd , k = 1, .., n.

(3.12)

We obtain the set (3.1) for a normal DVM with given matrix . This set, by construction, satises the condition n X vk = 0, (3.13) k=1

since uk · u = 0 for any k = 1, .., d and u given in (3.10). The most general class of the 0 phase sets Vn for given matrix is described by equalities 0

0

0

Vn = {v1 , ..., vn } Rd , 0

vk = a + Mvk , k = 1, .., n,

(3.14) (3.15)

with any a Rd and any non-singular (d × d) matrix M (a singular matrix M leads to a degenerate phase set). The models with sets (3.14) are isomorphic. Thus, it is proved that any well-dened matrix (note that is also well-dened) leads to a uniquely dened (with accuracy up to the isomorphism (3.14)) normal non-degenerate DVM. The latter property follows from linear independence of vectors


26

e uk , k = 1, .., d. On the other hand, any such DVM has a uniquely dened matrix , e = nd1. By taking away, if needed, linearly dependent rows (this means that rank we ”forbid” for a moment some reactions which do not inuence conservation laws) e to an (n d 1) × n matrix 1 . All such matrices form an equivalence we reduce class in the sense of Denition 6 (Section 2.2). By taking from the class any matrix and repeating the above procedure, one can reconstruct the phase set Vn . hence , the following result is proved. Proposition 1 There is a one-to-one correspondence (with accuracy up to isomorphism) between well-dened (n d) × n matrices and normal non-degenerate ddimensional DVMs (, Vn ) satisfying the mass (number of particles) and momentum conservation laws. This is the main result of this chapter. We shall see below that the situation becomes much more complicated if we add the energy conservation. Finally, we mention a simple geometrical property of normal DVMs ”with parallelograms” on the plane R2 : any such model admits 1-extension (a geometrical proof is obvious, since one can always nd 3 points which do not belong to any admissible parallelogram of the model and construct a new parallelogram with a new fourth vertex). The same is true for higher dimensions d 3. Thus, it is very easy to construct normal DVMs with mass and momentum conservation, by both algebraic (taking any well-dened (n d 1) × n matrix) and geometrical ways.

Chapter 4

Models with mass, momentum and energy conservation 4.1

Statement of the problem and geometrical interpretation

In this chapter, we consider the DVM introduced in the beginning of Section 3.1 with an additional restriction: any collision (reaction) (3.2) satises also the energy conservation (4.1) |vi |2 + |vj |2 = |vk |2 + |vl |2 . Thus, we have (d + 2) conservation laws. We describe the reactions by the vectors (3.5) and note that the parallelograms from Section 3.1 become now rectangles because of the conservation law (4.1). This is the main ”dierence” between the present models and those discussed in Section 3.2. The case mentioned in Remark 8 does not occur for rectangles (trivial collision vi À vj ). Denitions 11 and 12 are obviously modied for rectangles, however Denition 13 of non-degeneracy needs to be extended. Denition 14 The phase set Vn = {v1 , ..., vn } Rd of elastic (satisfying the energy conservation law (4.1)) DVM is said to be non-degenerate provided that the equalities a + b · vk + c |vk |2 = 0, b Rd , a, c R, k = 1, ..., n,

(4.2)

imply a = c = 0, b = 0. This means that we exclude two cases: (1) all velocities lying on the hyperplane b · v = const.; (2) all velocities lying on a sphere |v|2 = const. The case (1) can be reduced to the same problem, but in Rd1 . The case (2) is very simple: the only class

27

4. Models with mass, momentum and energy conservation

28

of normal elastic models on the sphere S d1 (its radius can always be made equal to one) is a class of the Broadwell-type models with 2d velocities. From now on we do not consider degenerate models. Similarly to the inelastic case (see the end of Section 3.1), we note that the normal models represent sets of n points in Rd which correspond to the maximal possible number (n d 2) of independent rectangles. One can easily guess that to construct the elastic normal models is much more dicult now, comparing to the inelastic case. In order to do this, we follow the procedure described in Section 2.4, for the general DKMs with p conservation laws.

4.2

General method (algorithm)

In Section 2.4 we described the steps one does in the process of the classication of normal discrete models with given order n, given dimension d and given function : Rd Rp of conservation laws. We have seen above that a discrete model with mass, momentum and energy conservation (elastic lattice gases with pair collisions) is normal if the number of p independent conservation laws is given by p = d + 2,

(4.3)

where d is the dimension of the phase set. We present below an equivalent form of the general algorithm (Step1-Step4 in Section 2.4), which we found to be a better option in the process of implementation. Step1. Generate all ((n d 2) × n) matrices having as rows (n d 2) linearly independent vectors of type 1 , ..., |{z} 1 , ...., |{z} 1 , ..., |{z} 1 , ..., 0) , = 1, ..., n d 2 . = (0, ..., |{z} i

j

k

(4.4)

l

In addition, these matrices should be well-dened (necessary condition in the general algorithm given in section 2.4): by adding to a new row, containing one element 1, one element (1) and the rest of elements zero (all possible combinations), we should obtain new matrices having the rank equal to (n d 1) . The set 1 = {1 , ..., N }, of all generated matrices , is nite, since the set is a nite set. Step2. For j = 1, ..., N do: Take = j ; Suppose that there exists a normal discrete model with the matrix of reactions and denote by (4.5) X = (x1 , ..., xn ), xk Rd , k = 1, ..., n

4. Models with mass, momentum and energy conservation the vector of phase points of the model; e = (|x1 |2 , |x2 |2 , ..., |xn |2 ); Denote X Check the solvability of the system ( X = 0 . e =0 X

29

(4.6)

First equation has as motivation the momentum conservation, second equation the energy conservation. Since the vector u = (1, ..., 1) is orthogonal on every matrix of reactions , we do not need an extra equation for the mass conservation. One way to check the solvability is the following. From the system X = 0, with rank = n d 2, we can express (n d 2) variables through (d + 2) parameters; under certain numeration we obtain xk =

n X

ak x , k = 1, ..., (n d 2) ,

(4.7)

=nd1

where ak are uniquely dened by the matrix , choice of ”independent” variables {xnd1 , ..., xn } and given numeration. Since we supposed that the model exists, we have that |xi |2 , i = 1, 2, ..., n satisfy Eqs.(4.6). Hence, |xk |2 =

n X

ak |x |2 , k = 1, ..., (n d 2) .

(4.8)

=nd1

We know that

n X

ak = 1 , k = 1, ..., (n d 2)

(4.9)

=nd1

since (1, ..., 1) is a solution to the system (4.6). We obtain the system of (n d 2) equations !2 Ã n n X X ak x = ak |x |2 , k = 1, ..., (n d 2) , (4.10) =nd1

=nd1

for (d + 2) d scalar variables (components of d-dimensional vectors {xnd1 , ..., xn } Rd ). If the Eqs.(4.10) have a non-trivial solution, then we compute the phase set (4.5) by Eqs.(4.7) and save the normal model (, Xn ). Otherwise, we reject the matrix j .


4.3

4.3.1

30

Classication and construction of normal plane elastic models with small numbers of velocities Starting point (n = 6) and 1-extension

We are going to study the particular case of plane elastic models with small numbers of velocities. It is already known that all ”quadratic” DVMs are normal and they are obtained by the 1-extension method (see Section 2.5). Thus, we are especially interested in the case of non-quadratic models obtained by the 1-extension scheme. Using geometrical arguments, we shall obtain all models up to 10 velocities, that are results of the 1-extension scheme. We include also the ”quadratic” cases and write the corresponding -matrices of reactions. Starting with n = 11, the normal models are ”reducible”. As we have seen before, from a geometrical point of view, we can represent a normal plane elastic model as a lattice of rectangles. The simplest model has six vertices (velocities) and two ”independent”rectangles; for normal models, the number m of independent rectangles is given by the equality m = n d 2, where n is the number of points and d is the dimension (for plane models d = 2). Lemma 6 For any given rectangle in a normal model with n = 6 velocities there exists only one ”neighbour” to it; in a model with n = 7 velocities there exist two ”neighbours” to the given rectangle; in a normal model with n 8 velocities there exist at least three ”neighbours” of a given rectangle, except the case of a ”quadratic” 8-velocity model of type (E) where there exist only two ”neighbours” (by a ”neighbour” of a rectangle in a normal plane DM we understand any other independent rectangle (reaction) in the model having at least a vertex in common with the initial rectangle). Proof. We denote by A = {x1 , x2 , x3 , x4 } Xn , the set of phase points that are vertices in some xed rectangle belonging to the model with phase set Xn ; B = Xn \A, the rest of s = (n 4) vertices of the model; R(B), the number of independent rectangles having as vertices only elements of B; R(AB), the number of ”neighbours” of the xed rectangle. Since the model is normal, the total number of rectangles is (n 4). Hence, it is equal to s. In the same time, this number is given by the sum (1 + R(B) + R(AB)).


31

We need to evaluate the number R(AB) to be able to prove the lemma. We have from the above relations R(AB) = s 1 R(B) = n 5 R(B)

(4.11)

We study the following possible cases: (i) s 3; (ii) s = 4; (iii) s 5. Case (i). To this case belong the models with n {6, 7} velocities, since s 3 implies that n 7. Since s 3 we have that R(B) = 0 (a rectangle has four vertices). From (4.11) we get R(AB) = s 1 = n 5.Therefore, R(AB) = 1 for n = 6 and R(AB) = 2 for n = 7. Case (ii). Since s = 4 we get the necessary condition R(B) 1. If R(B) = 0 then R(AB) = 3 and the lemma is proved. Suppose now that R(B) = 1. This means that the 8 vertices of the model can be seen as two groups belonging to two dierent circles (if the circle is the same, the energy conservation is automatically fullled and we have degeneracy). Since we have four linearly independent rectangles for all 8-velocity normal models, we should have R(AB) = 2. It is clear that both these independent rectangles contain two velocities from the rst rectangle (1) with vertices in A and two velocities in the second rectangle (2) with vertices in B. Both of these two rectangles can be constructed having as parallel sides: one side of (1) and one side of (2), one diagonal of (1) and one diagonal of (2) or one side of (1) and one diagonal of (2). Geometrically, all these cases can be rejected, except the last case side-diagonal which gives a normal ”quadratic” model of type (E) (see Section 4.3.1). Hence, R(AB) = 3 , except this particular case of ”quadratic” model. Case (iii). To this case belong the models with n 9 velocities, since s 5. We rewrite Eq.(4.11) as R(AB) = 3 + (s 4 R(B)). If R(B) 1, then R(AB) = 3 + (s 4 R(B)) 3 + (s 5) 3. Suppose now that R(B) 2. Then the model has at least two independent rectangles with all vertices in B. Since the set B contains s vertices, then the number of ”independent” rectangles that we can construct with vertices from B is less or equal (s 4), if B is not degenerate. For this case we obtain the necessary condition (s 4) 2 and this implies R(AB) 3. We study now the situation when R(B) 2 and B is degenerate (all points lying on a line or a circle). The case when the points are lying on a line is impossible, since we have two independent reactions on a straight line. Let us study the case when all s points of B are lying on a circle. Assume that there exist q free points (do not participate in any reaction) in B. It is clear that the number of the other points in B must be even (every rectangle has as diagonals


32

diameters of the circle). We x a rectangle on the circle. One can prove that to get N more independent reactions, one needs to add 2N particles. We x one reaction on the circle. If we want one more reaction we need to add two more particles (number of new reactions N = 1 , number of added particles 2N = 2); if we add now two more particles we get formally three rectangles, but only two of them are independent (number of new reactions N = 2 , number of added particles 2N = 4); by induction one can prove that this is true for any N. We have the total number of reactions in B given by R(B) = N + 1, with N 1 (because R(B) 2) . _The number of particles in B is s = 4 + 2N + q. Hence, (s 4 R(B)) = N 1 + q 0 and from Eq.(4.11), R(AB) 3. The lemma is now proved. Lemma 7 All 6-velocity normal models contain two independent reactions (rectangles) having two points (vertices) in common. Proof. Suppose that there are not two vertices in common. From Lemma 8 we know that for a given rectangle in a 6-velocity models exist exact one ”neighbour”. Then the -matrix corresponding to the model can be arranged in the following form Ã

¥

¥

¥

¥

0

0

¥

0

0

0

¥

¥

! (4.12)

where rst line represent rst reaction (¥-symbol is a non-zero element of the row, belonging to the set {±1}). In the second row we take in the rst position a nonzero element (so we get a vertex in common). Since there can not be two vertices in common, we are force to put on the next three columns zero elements. But this force us to have only two free places left for the three non-zero elements of the second reaction and we arrive in an impossible situation. Our supposition is false and lemma is proved. This result leads us to all three possibilities for plane 6-velocity normal models.

The case (c), diagonal-diagonal, is forbidden since all velocities lie on a circle (1,v,|v|2 are linearly dependent) and the conservation of energy is automatically ful-


33

lled. Hence, the only possible cases for n = 6 are given in the above gures (a) and (b). All other possibilities are obtained by translation, rotation or scaling. The corresponding -matrices for ! the above two cases are Ã 1 1 1 1 0 0 ; (a) = 1 0 0 1 1 1 Ã ! 1 1 1 1 0 0 (b) = . 1 0 1 0 1 1 All matrices equivalent with (a) and (b), in the sense of Denition 6, have as a concrete realization a model of type (a) and (b), respectively (translated or scaled or rotated). Using the 1-extension method for constructing normal models (Bobylev-Cercignani, 1999), we can start with the simplest normal model (in our case for n = 6) and add one more velocity such that the new point is a vertex of some new independent rectangle. By Lemma 5 we obtain a new normal model. If we approach the problem in this way, using only geometrical arguments and checking the rule for the number of independent rectangles, we get the complete result of 1-extensions. We start from a 6-velocity model, given in gures (a) or (b), with the angles t and s as free parameters, 0 < t, s < . We add a new velocity such that the new model is 2 normal (a 7-velocity model with 3 ”independent” rectangles). Let us consider the case of the 6-velocity model (a). The only new possible reactions (independent rectangles) are (1367) and (4257), and they are possible only for s = t. Other combinations are eliminated since we can take at most two vertices from 2 d > , 126 d = , 435 d = . d > , 436 an already existent rectangle and the angles 125 2 2 Both possible cases to extend the 6-velocity model (a) lead to the same type of 7velocity model, given below in gure (1). Let us consider now the case of the 6-velocity model (b). We can take at most two vertices from one already existent rectangle. New quadrangles (1257), (1467), (2367), (3457), (4567) are not rectangles (angles dierent of ). Since the vertices 1 2 and 3 belong to both initial rectangles, we cannot construct with them a new reaction. The quadrangles (1267), (1475), (2357), (3467) are rectangles if the vertices 3, 2, 6, the vertices 1, 2, 5, the vertices 2, 3, 5 and the vertices 3, 2, 6, respectively, are collinear (the corresponding models are of the type given in gure (1)). A reaction (4267) is possible d = ; the obtained model is of type (2). New reactions (2567) and (2457) only if 426 2 25 = , respectively, and we obtain a models of are possible only if 6d 25 = and if 4d 2 2 type (3) . After checking all possible new reactions, we obtained three dierent 7-velocity models with 1 free parameter t (in the case (1) s = t, where we mean s from gure 2


34

(a); in the case (2) s = t and in the case (3) s = 2t, where we mean s from gure 2 (b)).

The corresponding matrices of reactions are 1 1 1 1 0 0 0

(1) = 1 0 0 1 1 1 0 for models of type (1); 1 0 1 0 0 1 1 1 1 1 1 0 0 0

(2) = 1 0 1 0 1 1 0 for models of type (2); 0 1 0 1 0 1 1 1 1 1 1 0 0 0

(3) = 1 0 1 0 1 1 0 for models of type (3); 0 1 0 0 1 1 1 and all matrices equivalent to them in the sense of Denition 6. Knowing all 7-velocity models obtained by 1-extension method, we try to add a new velocity in order to get 8-velocity normal models (with 4 ”independent” rectangles). We start with a 7-velocity model of type (1). For arbitrary t we can construct a model of type (A), by drawing a new rectangle (4258).


35

The only possibilities to add a new velocity would be in the quadrangles (1278), (1478), (1578), (2378), (2478), (2578), (2678), (3478), (3578), (4578), (4678), (5678). After analyzing all these possibilities we obtain that (1278) gives a normal 8-velocity model only if t = . In this case, we have a model of type (B). 4

The new quadrangles (2678), (3478), (3578) form rectangles only if t = ; we get 4 models of type (B). By construction of model (1), new quadrangles (1478), (1578), (2378), (2578), (4578), (4678) cannot form rectangles (angles dierent of ). 2 1 It is possible to construct a new rectangle (2478) only if t = arctan (one proves 2 rst that the vertices 8, 1, 2, 6 are collinear, second the segments equality (12) = 2(26) and then computes tan t in two dierent ways). The model is of type (C)

We now try to do the same type of analyses on the model of type (2). The only possibilities to add a new velocity would be in the quadrangles (1258), (1268), (1278), (1378), (1458), (1468), (1478), (1578), (1678), (2358), (2368), (2378), (2458), (2568), (2578), (2578), (3458), (3468), (3478), (3578), (3678), (4568), (4578), (5678). The following new quadrangles are not rectangles (angles dierent of ): (1258), (1268), 2 (1278), (1468), (2368), (3458), (3468), (3478), (3678), (4568). It is possible to construct new rectangles (1458), (1478), (2358), (2378) only if t = 1 arctan (one proves rst that the vertices 1, 2, 5, 7 are collinear and then computes 2 tan t in two dierent ways). The model is of type (C).


36

New rectangles (1678), (2568), (3578), (4578) are possible to construct only if t = 6 (one uses that 6d 25 = and prove that 5d 62 = 2d 60 using that (6820) is a parallelogram; 2 see gure (D)). The model is of type (D) .

We can construct new rectangles (1378), (2458), (5678) only if t = (one uses 4 d = ). The model is of type (E) . that 765 2

11 3 We can construct new rectangles (2578) , (1578) only if t = arctan( ) (one 4 15 = respectively, shows that 1d 27 = and computes tan t). uses that 7d 25 = , or 7d 2 2 6 The models are of type (F ) .


37

We study now the model of type (3). The only possibilities of adding a new velocity would be in the quadrangles (1258), (1268), (1278), (1378), (1458), (1468), (1478), (1578), (1678), (2358), (2368), (2378), (2458), (2468), (2478), (3458), (3468), (3478), (3578), (3678), (4568), (4578), (4678). Because of angles dierent of , the 2 quadrangles: (1258), (1378), (1468), (1478), (1678), (2368), (2378), (2458), (3458), (3578), (4568) , (4678) cannot be rectangles. New rectangles (1278), (2468), (3478) are possible to construct only if t = . The 6 model is of type (D). We can construct new rectangles (1268), (1458), (1578), (2358), (3468), (3678), (4578) only if t = . The model is of type (B). 4 (one uses that It is possible to construct a new rectangle (2478) only if t = 8 d 128 d 826 d 627 d = , so 427 d = ). The models are of type (G). 421 8 2

We have now the complete description of 1-extension 8-velocity models.

4. Models with mass, momentum and energy conservation The corresponding matrices 1 1 1 1 1 0 0 1 (A) = 1 0 1 0 0

1 1 (B) = 1 1 1 1 (C) = 1 0 1 1 (D) = 0 0 1 1 (E) = 0 1 1 1 (F ) = 0 0 1 1 (G) = 0 0

1

0

1

1 1 1 0 0 1 0 1 0 1 0 0 1 1 1 0 0 1 0 1 0 1 0 1 1 1 1 0 1 0 1 0 1 1 0 0 1 1 1 0 1 0 1 0 1 0 1 0 1 1 1 0 1 0 1 0 1 1 0 0 1 1 1 0 1 0 1 0 0 1 0 1

38

are: 0 0 0

1 0 0

1 1 0

1 0 0 1 0 0 0 0

1 1 0 0

0 1 1 0

0 0 1 1 0 0 0 0

1 1 0 0

0 1 1 0

0 0 1 1 0 0 0 0

1 1 0 0

0 1 1 0

1 1 0 1 0 0 0 0

1 1 0 0

0 1 1 0

0 0 1 1 0 0 0 0

1 1 0 0

0 1 1 0

1 0 1 1 0 0 0 0

1 1 0 0

1 1 1 0

0 0 1 1 0 1 0

and all matrices equivalent to them, in the sense of Denition 6. Knowing all 8-velocity models obtain by 1-extension method, we try to add a new velocity in order to get 9-velocity normal models. We construct this time models with 5 1 ”independent” rectangles. We can extend a model (A) only for t = or t = arctan 4 2 and we obtained models of type (i) or type (iii), respectively.


39

(computing all angles, we arrive It is not possible to extend model (G), t = 8 to contradictions). Hence, we found a maximal 8-velocity model (the model does not admit 1-extension). 1-extensions of models (B) and (E), t = , lead to the same 4 model of type (ii).

1 1-extension of model (C), t = arctan , leads to models of type (iii) and (iv) . 2


40

11 3 1-extension of model (F ), t = arctan , leads to a model of type (v) . 4

1-extension of model (D), t =

, leads to a model of type (vi). 6

All other attempts to make 1-extension lead us to geometrical contradictions. Besides the ”quadratic” models (i) and (ii), only the models (iii) and (iv) admit 1-


41

extension (to the same 10-velocity maximal model of type (I)). We obtained two 9-velocity maximal models (v) and (vi) and a 10-velocity maximal model (I).

After the maximal case n = 10, the only models which admit 1-extension are ”quadratic” models. 1-extensions of models (i) and (ii) give us models of type (II) , (III) , (IV ) , (V ).

All these models admit 1-extension on innitely many steps. The corresponding matrices for 9-velocity and 10-velocity models, obtained by 1-extension are


(i)

= =

(ii)

1 1 1 1 0 0 0 0 1 0 0 1 1 1 0 0 1 0 1 0 0 1 1 0 1 1 0 0 0 0 1 1 1

(iii)

=

(iv)

=

(v)

=

1 1 1 1 0 0 0 0 0 1 0 0 1 1 1 0 0 0 1 0 1 0 0 1 1 0 0 0 1 0 1 1 0 0 1 0 0 1 0 0 0 1 1 0 1

0

0

0

0 1

0

0 0 0 0

1 1

1 1 1 1 0 0 0 0 0 1 0 0 1 1 1 0 0 0 1 0 1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 1 0 1 1 0 0 0 1 1 1 1 1 0 0 0 0 1 0 0 1 1 1 0 0 1 0 1 0 0 1 1 0 0 1 0 1 0 0 1 1 1 0 0 1 0 0 0 1 1 1 1 1 0 0 0 0 1 0 1 0 1 1 0 0 0 1 0 1 0 1 1 0 0 1 0 0 1 0 1 1 1 1 0 0 0 0 0 1

1 1 (vi) = 0 0 1 1 1 1 (I) = 0 0 1

0 0 0 0 1

0

0

0

0 1

1 1 1 0 0 0 0 0 0 1 0 1 1 0 0 0 1 0 1 0 1 1 0 0 1 0 0 1 1 0 1 0 1 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 1 1 1 0 0 0 1 0 0 1 1 0 1 0 1 0 0 1 1 1 0 1 1 0 0 0 0 0 1 0 0 0 1

0 0 0 0 1 0

0 0 0 0 0 1

42


1 1 1 0 1 0 (II) = 0 1 0 1 0 0 1 1 1 0 1 0 (III) = 1 1 1 0 0

(IV )

=

(V )

=

1

1 0

1 1

1 0 0 1 0 0 1 0 1 1 0 1 1 0 0 0 0 0 0

1

1 1 1 1 1 0 0 1 1 0 1 0 0 1 0 1 0 1 0 0 0 0 1 1 1 1 1 1 1 0 0 1 1 0 1 0 0 1 0 1 0 1 0 0 0 1 0 0

43

0 0

0 0

0 1 1 0 0 0

1 0 0 1 0 0

0 1 1 0 1 0

1 0 0 1 0 1 0 0 0 0 0 0

1 1 0 0 0 0

0 1 1 0 0 0

0 0 1 1 0 0

0 1 0 1 1 0

0 0 0 1 0 1 0 0 0 0 0 0

1 1 0 0 0 0

0 1 1 0 0 0

1 0 0 1 0 0

0 1 1 0 1 0

0 0 0 1 0 1 0 0 0 0 0 0

1 1 0 0 0 0

0 1 1 0 0 0

1 0 0 1 0 0

0 1 1 0 1 0

1 0 0 0 1 1 0 1

0 1

0 0

0 0

and all matrices equivalent to them in the sense of Denition 6.

4.3.2

Classication results for n {7, 8}

We have described in the previous section all ”reducible” normal models with n = 7 and n = 8 (results of 1-extension scheme). We wonder now, if there exist other normal models with n = 7 and n = 8 which are ”irreducible”. We can arrange the corresponding of any normal model in the ”special” equivalent table-form, described below.


¥

¥

¥

¥

0

0

0

0

¥

?

?

?

?

.

.

.

0

¥

?

?

?

.

.

.

0 0

0 0

¥ 0

? ¥

? ?

. .

. .

. .

0

0

0

0

?

.

.

.

0

44

?

?

?

?

?

(4.13)

-rst we rearrange the matrix (if needed) in order to obtain a row which has all its four non-zeros in the rst four columns: denote the new matrix by (we can always interchange columns and get an equivalent matrix); -every line of the table contains at least one row of the matrix ; -every column of the table contains exact one column of the matrix ; -rst line contains one row of the matrix having in the rst four columns all its non-zeros {1, 1, 1, 1} (because of the rst step, this row exists); -the second line contains all k1 0 rows of , left after we wrote the rst line, having in the rst column a non-zero; -the third line contains all k2 0 rows of , left after we wrote the two rst lines of the table, having in the second column a non-zero; -the fourth line contains all k3 0 rows of , left after we wrote the rst three lines of the table, having in the third column a non-zero; -the fth line contains all k4 0 rows of , left after we wrote the rst four lines of the table, having in the fourth column a non-zero; -the last line of the table contains all left m1 0 (these rows will have in the rst four columns only zeros); -we have 1 + (k1 + k2 + k3 + k4 ) + m1 = n 4, if the model is of order n and dimension d = 2. Remark 9 The rectangles given by the rows k1 , k2 , k3 , k4 represent all the ”neighbours” of the rectangle given in the rst line of the table and because of Lemma 6 we have that for n 8 k1 + k2 + k3 + k4 3 except for the particular case of the ”quadratic” 8-velocity normal model (E). Lemma 8 All normal models with n = 7 or n = 8 contain a 6-velocity normal model (in other words, two independent rectangles having two vertices in common). Proof. (a) n = 7 : We suppose the contrary. We can arrange the corresponding -matrix of the model in the ”special” form (4.13)


¥

¥

¥

¥

0

0

¥

0

0

0

¥

¥

0

¥

0

0

¥

¥

0 0

0 0

¥ 0

0 ¥

¥ ¥

¥ ¥

0

0

0

0

¥

¥

0

45

¥

¥

. ¥

¥

¥

(4.14)

We are forced, because of our supposition (no two vertices in common), to ll the rst four columns with zero elements. Since we have four non-zero elements in each row, we need to ll the rest of columns, with non-zero-elements. Since the matrix has only three rows we need to fulll the condition k1 + k2 + k3 + k4 + m1 = 2.

(4.15)

From the table (4.14) we notice that m1 = 0 is a necessary condition (we have only three non-zero elements in a reaction). Then k1 + k2 + k3 + k4 = 2, so there are two rows having three common non-zero elements (they describe the same reaction) and this is a contradiction (all rows describe linear independent reactions). Hence, our supposition for n = 7 was false. (b) n = 8 : For the particular case of ”quadratic” models it is know that all 8velocity normal DM contain a 6-velocity normal DM. So we only need to prove the lemma for the non-quadratic models. We suppose the contrary. Then we can write the corresponding -matrix of the model in the ”special” form (4.13) ¥ ¥ ¥ ¥ 0 0 0 0

¥ 0 0 0 ? ? ? ?

0 ¥ 0 0 ? ? ? ?

(4.16)

. 0 0 ¥ 0 ? ? ? ?

0 0 0 ¥ ? ? ? ?

0 0 0 0 ¥ ¥ ¥ ¥ We arranged -matrix, using the scheme described above. In this case k1 + k2 + k3 + k4 + m1 = 3.

(4.17)

From Remark 9 we have that k1 + k2 + k3 + k4 3. Hence, m1 = 0. Since we cannot arrange in the three last rows (having each of them three non-zeros in the last four columns) the elements such that two dierent rows do not have two vertices in common, we have that our assumption is false and the lemma is proved. With the help of Lemma 8 we can now prove the following theorem.


46

Theorem 2 All normal models with n = 7 and n = 8 are 1-extensions of models with n = 6 and n = 7, respectively. Proof. We treat the two cases separately . (a) n = 7: We proved that all normal models with n = 7 velocities contain a 6-velocity normal model. Hence they are all 1-extensions. (b) n = 8: We proved that all normal models with n = 8 velocities contain a 6velocity normal model. If they all contain a 7-velocity normal model, then the theorem is true. Suppose that there exist normal models with n = 8 velocities, obtained as 2extensions from a 6-velocity normal model. Then the corresponding -matrix of the model has in all columns at least two non-zeros (otherwise the model is reducible to a 7-velocity normal model). Case 1. The model contains a 6-velocity model of type (a). The corresponding -matrix is given by Ã ! 1 1 1 1 0 0 , 1 0 0 1 1 1

(4.18)

(all other models of this type lead to matrices which are equivalent with , in the sense of the Denition 6). Then the matrix corresponding to the 8-velocity model has the form (we use now the normal way to write a matrix) 1 1 1 1 0 0 0 0

1 0 0 1 1 1 0 0

. (4.19)

0 ? ? 0 ? ? ¥ ¥ 0 ? ? 0 ? ? ¥ ¥ Since we cannot have (by our supposition) a column with only one non-zero, we are forced to put non-zeros in the last two columns (last two rows) and we also need to have at least one more non-zero in columns 2,3,5 and 6. Since for each of the last two rows we have only two more available non-zeros, we need to put zero in columns 1 and 4, in the last two rows. These arguments explain the form (4.19). Noting the fact that by interchanging two rows or multiplying them with (1) or changing two columns in the matrix of reactions, the model is the same, we obtain the following

4. Models with mass, momentum and energy conservation possible matrices for the 8-velocity normal 1 1 1 1 1 0 0 1 0 0 ¥ 0 0 ¥ 0 0 or

1

1 0 0 or

1

1 0 0

47

model in case 1. 0

0

0

0

1

1

0

¥ 0

0 ¥

¥ ¥

0

, ¥

¥ 0

(4.20)

1

1

1

0

0

0

0

0

1

1

1

0

¥

¥

0

0

0

¥

0

0

0

¥

¥

¥

0

, ¥

¥

1

1

1

0

0

0

0

0

0

1

1

1

0

¥

0

0

¥

0

¥

0

¥

0

0

¥

¥

(4.21)

0

. ¥

¥

(4.22)

Using geometrical arguments, one can prove that all these three cases are impossible. Case (4.20): If we denote by (i, j, k, l) a reaction of the model (where i, j, k, l are the positions in the vector of reaction containing non-zeros), then the last two reactions are (3, 5, 7, 8) and (2, 6, 7, 8) and the rst two reactions are given in the 6velocity model of type (a). Suppose that the last two reactions are possible. Then (3, 5) and (2, 6) can be both of them sides or diagonals or one diagonal and the other one side of their corresponding rectangles. In the rst situation, we can construct such two rectangles, but the 8-velocity models will not be normal since they will contain only three independent rectangles, instead of four. In the second situation, if both are diagonals they intersect the diagonal (7, 8) in its middle point and hence, they have a point in common, which is false, since they are parallel (see g.(a) for 6-velocity models). The last possibility is not working either (we have (3, 5) and (2, 6) parallel and (7, 8) is parallel with one of them and intersect the other one). Case (4.21): The last two reactions are (2, 3, 7, 8) and (5, 6, 7, 8). We can easily prove, using similar arguments as in case (4.20), that they are impossible. Case (4.22): The last two reactions are (2, 5, 7, 8) and (3, 6, 7, 8). We have (2, 5) and (3, 6) diagonals in the rectangles (2, 3, 5, 6). This means that they cannot be both sides in the rectangles (2, 5, 7, 8) and (3, 6, 7, 8) since this leads to (2, 5) parallel with (3, 6). They cannot be both diagonals in these two rectangles since then the model will have only three independent rectangles instead of four. If one of (2, 5) and (3, 6) is a diagonal in the new rectangles, then it intersects (7, 8) in the middle point and


48

hence, (7, 8) intersects the other one also in the middle point, so the case when one is diagonal and the other one is side is excluded. Case 2. The model contains a 6-velocity model of type (b). We repeat the same steps as for the case 1. The corresponding -matrix of the 6-velocity model of type (b) is given by Ã ! 1 1 1 1 0 0 , (4.23) 1 0 1 0 1 1 (all other models of this type lead to matrices which are equivalent with , in the sense of the Denition 6). The matrix corresponding to the 8-velocity model has the form 1 1 1 1 0 0 0 0

1 0 1 0 1 1 0 0

. (4.24) 0 ? 0 ? ? ? ¥ ¥

0 ? 0 ? ? ? ¥ ¥ Using the same reasoning as for the case 1, we get that the only possibilities (nonequivalent) are

1

1 0 0 or

1

1 0 0 or

1

1 0 0

1

1

1

0

0

0

0

1

0

1

1

0

0

0

¥

¥

0

¥

¥

0

0

0

¥

¥

0

, ¥

¥

1

1

1

0

0

0

0

0

1

0

1

1

0

¥

0

¥

0

0

¥

0

0

0

¥

¥

¥

0

, ¥

¥

1

1

1

0

0

0

0

0

1

0

1

1

0

¥

0

0

¥

0

¥

0

0

¥

0

¥

¥

0

(4.25)

(4.26)

0

. ¥

¥

(4.27)

We shall prove that the above matrices do not lead to normal models. All our arguments relate now to the gure (b) for 6-velocity models. Case (4.26): The new added reactions are (2, 4, 7, 8) and (5, 6, 7, 8). If we suppose (2, 4) and (5, 6) to be both diagonals in the new rectangles, we have that they both intersect the diagonal (7, 8) in their middle point. Then (1, 3) and (5, 6) have also,


49

as common point, the middle point. But this leads to a contradiction, since they are parallel (see g.(b) for the 6-velocity model). Suppose that (2, 4) and (5, 6) are both sides in the rectangles. Then they are parallel (they are both parallel with (7, 8)). But this result is also false since (5, 6) is parallel with (1, 3), which intersects (2, 4) in the middle point. The only left possibility is that one of (2, 4) and (5, 6) is diagonal and the other one is a side in the new rectangles. If for example (2, 4) is a side, then the segment which reunites the middle point of (2, 4) and the middle point of (7, 8) is perpendicular on (2, 4) and (7, 8); let us denote this segment (AB). But the middle point A of (2, 4) coincides with the middle point of (1, 3) and by considering (5, 6) diagonal we have that its middle point coincides with the middle point B of (7, 8), so (AB) is also perpendicular on (1, 3). In this way we obtain a contradiction since (1, 3) and (2, 4) cannot be parallel. If we have (2, 4) as a diagonal and (5, 6) as side in the new rectangles, then (7, 8) has to coincide with (1, 3) (they both have to be congruent with (2, 4), parallel with (5, 6) and to intersect (2, 4) in its middle point). Hence, this case is impossible. Cases (4.25) and (4.27): These two cases admit the same type of proof. It is enough to prove, for example for case (4.25). The new added reactions are (4, 5, 7, 8) and (2, 6, 7, 8). We construct a segment (2, 9) parallel and congruent with (1, 6). Assuming that the new reactions are possible, we need to have the segments (4, 5) , (2, 6) and (7, 8) congruent (as opposite sides or diagonals in the same rectangle). From these assumptions we obtain that (1, 4, 5, 9) and (2, 9, 5, 3) are parallelograms and hence, (4, 5) is congruent with (1, 9). This leads to (2, 6) congruent with (1, 9). Thus, (1, 6, 9, 2) should be a rectangle. This implies (1, 6) perpendicular to (1, 2) and this is a contradiction (by construction of the 6-velocity model (b)). Remark 10 We checked these results using computer and the algorithm given in the next section. We shall describe the method and present the conclusions in Section 4.3.4. Since we proved that all normal models with n = 7 and n = 8 are 1-extensions and in the previous section we described all these models, we have now a complete image over the case n = 7 and n = 8 for normal plane elastic models.

4.3.3

Algorithm for normal plane elastic models with small numbers of velocities

We start by proving a result that will enable us to give a more explicit algorithm for classifying elastic plane normal discrete models, based on the general algorithm, given in Section 4.2.


50

Lemma 9 Let (1) + (2) ¿ (3) + (4) be one reaction of a normal model (it describes a rectangle geometrically). Then there exists 5 n such that xk = (ak1 x1 + ak2 x2 + ak3 x3 + ak x ) R2 , for all k = 1, ..., n, where

(

ak = k , = 1, 2, 3,

, a4 = 0 a41 = a42 = a43 = 1

(4.28)

(4.29)

1 if k = ; we have used for the last coecients the 0 , otherwise fact that x1 + x2 = x3 + x4 ); i.e. all velocities of a model can be expressed by a linear (we remind that k =

combination of three velocities belonging to a xed reaction, and a fourth velocity from a dierent reaction. Proof. From the system X = 0 , = 1, ..., n 4

(4.30)

with linearly independent from (4.4) and X in (4.5) , there exist 1 , , , n all dierent such that xk = Ak x + Ak x + Ak x + Ak x , k = 1, ..., n.

(4.31)

(a1 ) If 1 {, , , }, for example = 1, then xk = Ak1 x1 + Ak x + Ak x + Ak x , k = 1, ..., n.

(4.32)

x1 = A1 x + A1 x + A1 x + A1 x ,

(4.33)

(a2 ) Else with at least two coecients from A1 , A1 , A1 , A1 dierent of zero, since A1 + A1 + A1 + A1 = 1 from Eq.(4.9) and since, if we consider for example only A1 6= 0 then x1 = A1 x = A1 = 1 and x1 = x . This contradicts our supposition that all numbers , , , are not equal to 1. Without loss of generality, we can assume that A1 6= 0. Then x can be expressed as 0

0

0

0

x = A1 x1 + A x + A x + A x .

(4.34)

Hence, 00

00

00

00

xk = Ak1 x1 + Ak x + Ak x + Ak x , k = 1, ..., n.

(4.35)


51

In both cases (a1 ) and (a2 ) we have xk span{x1 , x , x , x } for all k = 1, ..., n. We 00 denote by Bkj , the coecients Akj or Akj , j {1, , , }, from the cases (a1 ) or (a2 ), respectively. (b1 ) If 2 {, , }, for example = 2, then xk = Bk1 x1 + Bk2 x2 + Bk x + Bk x , k = 1, ..., n.

(4.36)

/ {, , }), then (b2 ) If 2 x2 = B21 x1 + B2 x + B2 x + B2 x . Suppose that B2 = B2 = B2 = 0. From B21 + B2 + B2 + B2 = 1 (from Eq.(4.9)), we get B21 = 1 and hence, x1 = x2 (contradiction to the supposition that x1 , x2 belong to a reaction). This means that at least one of B2 , B2 , B2 is non-zero, for example B2 6= 0. We can now express x as a linear combination of x1 , x2 , x , x . Hence, we can write 0 0 0 0 (4.37) xk = Bk1 x1 + Bk2 x2 + Bk x + Bk x , k = 1, ..., n. In both cases (b1 ) and (b2 ) we have xk span{x1 , x2 , x , x } for all k = 1, ..., n. We 0 denote by Ckj , j {1, 2, , } the coecients Bkj or Bkj , from the cases (b1 ) or (b2 ), respectively. (c1 ) If 3 {, }, for example = 3, then xk = Ck1 x1 + Ck2 x2 + Ck3 x3 + Ck x , k = 1, ..., n.

(4.38)

(c2 ) Else x3 = C31 x1 + C32 x2 + C3 x3 + C3 x . If C3 = C3 = 0 we obtain using (4.9) C =

31 C32 = 1 C31 + C32 = 1 =

(4.39)

x3 = x1 + (1 ) x2

(4.40)

and in this case

and hence, we get x1 , x2 , x3 collinear. But these points belong to a reaction of the model (which, from geometrical point of view is a rectangle) and thus, this case is impossible. Hence at least one of the coecients C3 , C3 is dierent of zero. We can take, for example, C3 6= 0. We express x as a linear combination of x1 , x2 , x3 , x and obtain 0 0 0 0 (4.41) xk = Ck1 x1 + Ck2 x2 + Ck3 x3 + Ck x , k = 1, ..., n


52

In both cases (c1 ) and (c2 ) we have xk span{x1 , x2 , x3 , x } for all k = 1, ..., n. We 0 denote the coecients Ckj (or Ckj ), j {1, 2, 3, } by ak , bk , ck , dk , respectively. Then xk = ak x1 + bk x2 + ck x3 + dk x , k = 1, ..., n,

(4.42)

ak + bk + ck + dk = 1

(4.43)

with

and x1 , x2 , x3 belonging to the same reaction and x to another one (if x = x4 , since x4 span {x1 , x2 , x3 } we get xk = ak x1 + bk x2 + ck x3 , k = 1, ..., n,

(4.44)

which contradicts the fact that the system (4.30) has four free parameters) and the lemma is proved. From Lemma 9 we have that the solution of the system X = 0 has the form xk = ak x1 + bk x2 + ck x3 + dk x , k = 1, ..., n, where x1 , x2 , x3 belong to the same reaction and x to another reaction of the model and where the coecients fulll (4.43). Because of the invariance under translation, rotation and scaling transforms, we can take x1 = (0, 0) x2 = (1, 0) = e1 . (4.45) x3 = (1, ) x4 = (0, ) = e2 , R x = (x, y) / {x1 , x2 , x3 } Hence, the solution will have the general form xk = ak e1 + bk e2 + ck x , k = 1, ..., n,

(4.46)

where ak , bk , ck are real constants. One can prove that x = (x, y) fullls an additional restriction 1 + 2 1 (x )2 + (y )2 6= , (4.47) 2 2 4 i.e. it does not lie on the circle circumscribing the triangle having as vertices the points x1 , x2 , x3 from (4.45). Since we will not use this property of x , we do not prove it here, for brevity.


53

The system (4.6) given in the general algorithm in Section 4.2, will be replaced, using the Eq.(4.10) by x1 = (0, 0) x2 = (1, 0) = e1 x3 = (0, ) = e2 , R x4 = (1, ) . (4.48) x = (x, y) / {x1 , x2 , x3 } ; x fullls (4.47) xk = ak e1 + bk e2 + ck x , k {5, ..., n} \ { } X = 0 (a2k ak ) + (b2k bk ) 2 + (c2k ck )(x2 + y 2 ) + 2ak ck x + 2bk ck y = 0 for all k {5, ..., n} \ { } If this system admits a non-trivial solution (, x, y, ak , bk , ck : k 6) , then we can nd the normal model given by the phase states set Xn = {x1 , ..., xn } R2 and we can construct it. We try to improve the form of the system (4.48).We prove the following lemma. Lemma 10 All normal models with n {9, 10, 11} velocities contain a 6-velocity normal model (in other words, two independent rectangles having two vertices in common). Proof. (a)We suppose that the normal models with n = 9 do not contain two independent rectangles having two vertices in common. In this case, using the ”special” representation (4.14), we can write the corresponding matrix of reactions of the model as ¥ ¥ ¥ ¥ 0 0 0 0 0

¥ 0 0 0 ? ? ? ? ?

0 ¥ 0 0 ? ? ? ? ?

(4.49)

. 0 0 ¥ 0 ? ? ? ? ?

0 0 0 ¥ ? ? ? ? ?

0 0 0 0 ? ? ? ? ? We know that for a normal model with n = 9 velocities we have 1 + (k1 + k2 + k3 + k4 ) + m1 = 5 and we proved in Lemma 6 (see Remark 9) that (k1 + k2 + k3 + k4 ) 3. Hence a necessary condition is that m1 1.


54

(a1 ) m1 = 1 = k1 + k2 + k3 + k4 = 3. In this case, the last row has on the last ve free positions four non-zero elements and its above three rows have on the last ve free positions three non-zero elements. We cannot arrange the non-zero elements such that in two dierent rows there are not two columns with non-zeros (in both rows). Hence, we have two vertices in common in two dierent independent rectangles. (a2 ) m1 = 0 = k1 + k2 + k3 + k4 = 4. All last four rows have three non-zeros in the last ve free positions. We cannot arrange the non-zero elements such that in two dierent rows we have not in two dierent columns non-zeros, in both rows, so again, we have two vertices in common. Hence, the lemma is true for n = 9. (b) n = 10: We suppose that the normal models with n = 10 do not contain two independent rectangles having two vertices in common. Using the same arguments as for the above case, we can arrange the -matrix of the model as ¥

¥

¥

¥

0

0

0

0

0

0

¥

0

0

0

?

?

?

?

?

?

0

¥

0

0

?

?

?

?

?

?

0

0

¥

0

?

?

?

?

?

?

0

0

0

¥

?

?

?

?

?

?

0

0

0

0

?

?

?

?

?

?

.

(4.50)

For a normal model with n = 10 we have 1 + (k1 + k2 + k3 + k4 ) + m1 = 6 and we proved in Lemma 6 (see also Remark 9) that (k1 + k2 + k3 + k4 ) 3. Hence a necessary condition is that m1 2. We have the following cases. (b1 ) m1 = 2 = k1 + k2 + k3 + k4 = 3 : (b2 ) m1 = 1 = k1 + k2 + k3 + k4 = 4 : (b3 ) m1 = 0 = k1 + k2 + k3 + k4 = 5 : In all these cases, all the combinations one tries in order to arrange the non-zero elements, lead to the existence of two reactions with at least two vertices in common. Hence, the lemma is true for the case n = 10. (c) n = 11: Suppose that the normal models with n = 11 do not contain two independent rectangles having two corners in common. The -matrix of the model can be arranged as


¥

¥

¥

¥

0

0

0

0

0

0

0

¥

0

0

0

?

?

?

?

?

?

?

0

¥

0

0

?

?

?

?

?

?

?

0 0

0 0

¥ 0

0 ¥

? ?

? ?

? ?

? ?

? ?

? ?

? ?

0

0

0

0

?

?

?

?

?

?

?

55

(4.51)

For a normal model with n = 11 we have 1 + (k1 + k2 + k3 + k4 ) + m1 = 7 and we proved in Lemma 6 (see Remark 9) that (k1 + k2 + k3 + k4 ) 3. Hence, a necessary condition is that m1 3. We have the following possible cases. (c1 ) m1 = 3 = k1 + k2 + k3 + k4 = 3 : (c2 ) m1 = 2 = k1 + k2 + k3 + k4 = 4 : (c3 ) m1 = 1 = k1 + k2 + k3 + k4 = 5 : (c4 ) m1 = 0 = k1 + k2 + k3 + k4 = 6 : Similar with the cases for n = 9 and n = 10, we nd for every possible arrangement of the non-zero elements, two independent reactions (rows of the matrix) with at least two common vertices (two columns with non-zeros in both rows). The lemma is also true for the case n = 11. Lemma 10 and Lemma 8 imply that, for the cases n {8, 9, 10, 11}, in the system (4.48) the vector x = (x, y) can be taken as a vertex in a rectangle, ”neighbour” with the initial rectangle (with vertices (0, 0) , (1, 0) , (1, ) , (0, )), such that it has as rst common vertex (0, 0) and as a second common vertex (i) (0, ): in this case x = (y, 0) = (μ, 0) (the model contains a 6-velocity model of type (a)); (ii) (1, ): in this case x = (y, y) = (μ, μ) (the model contains a 6-velocity model of type (b)). If we take as second common vertex (1, 0), then x = (0, y), and we obtain an equivalent model with the one in case (i), due to the invariance under rotation. The point x fullls in both cases the condition (4.47). The system (4.48) will have now the following forms


56

Case (i) :

x1 = (0, 0) x2 = (1, 0) = e1 x3 = (1, ) x4 = (0, ) = e2 , R x5 = (μ, ) x6 = (μ, 0) , μ R \ {1} xk = ak e1 + bk e2 + ck (μ, 0) , k = 7, ..., n X = 0, X = (x1 , ..., xn ) , xk R2 , k = 1, ..., n (a2k ak ) + (b2k bk ) 2 + (c2k ck )μ2 + 2ak ck μ = 0, for all k = 7, ..., n. Case (ii) : x1 = (0, 0) x2 = (1, 0) = e1 x3 = (1, ) x4 = (0, ) = e2 , R x = (1 μ, + μ) / {xi , i < 6} 5 x6 = (μ, μ) / {xi , i < 5} , μ R xk = ak e1 + bk e2 + ck (μ, μ) , k = 7, ..., n X = 0, X = (x1 , ..., xn ) , xk R2 , k = 1, ..., n (a2k ak ) + (b2k bk )2 + (c2k ck )μ2 (1 + 2 ) + 2ck (bk ak )μ = 0, for all k = 7, ..., n.

(4.52)

(4.53)

Trying to implement the algorithm, using the above systems, we realize that it is easier to use the systems (4.52) and (4.53) in the equivalent forms given below. Case (i) : x = (0, 0) 1 x = (1, 0) = e1 2 x3 = (1, ) x4 = (0, ) = e2 , R (4.54) x5 = (μ, ) x6 = (μ, 0) , μ R \ {1} xk = (ak , bk ) , k = 7, ..., n X = 0, X = (x1 , ..., xn ) , xk R2 , k = 1, ..., n e e = (|x1 |2 , ..., |xn |2 ) X = 0, X Case (ii) : system (4.54) with modied x5 = (1 μ, + μ) , x6 = (μ, μ) , μ

R .


57

Since for n {7, 8, 9, 10, 11} all normal models have in correspondence an ((n 4) × n) matrix which contains a 6-velocity model (of type (a) or type (b), see section 5.1), and since by interchanging rows or columns, we get an equivalent matrix to the initial one, we can state that all matrices corresponding to normal models of order n have one the following two forms

1 1 1 1 0

1 = ? ? or

=

0

.

. . 0

0 1 1 1 0 . . 0

. . . . . . . ?

. . . . . . . ?

0 ? ?

1 1 1 1 0 0 . . . 1 0 1 0 1 1 0 . . ? ? . . . . . . . ? ? . . . . . . .

0 0 ? ?

(4.55)

,

(4.56)

where the rows with ”?” contain vectors of type i,j;k,l = (0, ..., 1, ..., 1, ..., 1, ..., 1, ..., 0) Rn , i

k

j

l

such that rank = n 4. Note also the fact that a matrix (4.55) corresponds to the above case (i) and a matrix (4.56) corresponds to the above case (ii). Lemma 10 enables us to consider only matrices of the forms (4.55) and (4.56), having the rst two rows xed. This fact is a big advantage in the process of generating the matrices (the number decrease considerable) and also in the implementation of the algorithm. The algorithm follows the next steps Step 1. Generate all ((n 4) × n) matrices (of type (4.55), respective (4.56)), with rank = n 4, fullling the ”3 corners rule” (in two dierent rows there are not three columns having non-zero elements in both rows) since three vertices dene completely a rectangle. More than this, by adding to a new row, containing one element 1, one element (1) and the rest of elements zero (all possible combinations), we should obtain new matrices having rank equal to (n 3) ( then is well-dened). The 0 0 sets 1 = {1 , ..., N } of all generated matrices of type (4.55) and 2 = {1 , ..., M } of all generated matrices of type (4.56), will be nite, as we discussed for the general situation, in section 2.4. Step 2. Consider rst the set 1 .


58

For j = 1, ..., N , make = j in the system (4.54) and check if the system is solvable. In the armative case, save the matrix and use the non-trivial solution (, μ, , ak , bk , : k 7) to compute the phase set Xn = {x1 , ..., xn } R2 and construct the normal model by plotting the phase points. In the negative case, simply reject the matrix j . Step 3. Repeat Step 2, for the set 2 , by checking, this time, the solvability of the system from case (ii).

4.3.4

Computer results for the cases n = 8 and n = 9

For the particular cases n = 8 and n = 9, we obtained in advance all 1-extension normal models (see Section 4.3.1). In section 4.3.2 we proved analytically that all normal models with n = 8 are ”reducible”, but it is still not answered what happens in the case n = 9. Are all 9-velocity normal models ”reducible”, like all normal models with n < 9? An analytical approach is almost impossible in this case and the only solution is to use the algorithm above and computer. A necessary condition such that a matrix of reactions is ”irreducible” is that all its columns contain at least two non-zeros (otherwise, one can remove the column and the row corresponding to the non-zero element and obtain a normal model). We denote this condition with ( ), so we can refer to it later. Condition ( ) is not sucient, in general. We modify Step 1 in the algorithm above and generate instead all matrices having in all columns at least two non-zero elements and fullling all initial conditions of Step1. Then we follow Step 2 and Step 3 and check all this matrices. In the case n = 8, the result is the expected one: all generated matrices (in modied Step1) were rejected by the algorithm. Hence, the property that all 8-velocity normal models are 1-extentions, is veried. As we proved analytically in section 4.3.2, in this case, the condition ( ) is not only necessary, but also sucient. We are going to describe how we implemented the algorithm for the case n = 9 and present the results. We split the problem in two cases (by (i : j, k : l) we mean the matrix containing the rows from i to j and the columns from k to l of matrix ): (1) det (3 : 5, 7 : 9) = 3 (2) det (3 : 5, 7 : 9) = 2. In this paper we are going to present the method and the results for the case (1); for the case (2) we have done the needed changes in the algorithm, but we are not able to present the nal results yet. By generating the matrices, we get |1 | = 4292 and |2 | = 6757, where by |A| we denote the number of elements in the set A.


59

Denote by ij the set of all matrices from i belonging to the case (j), i = 1, 2 and j = 1, 2. We have |11 | = 3486, |12 | = 1806, |21 | = 4293, |22 | = 2464. Hence in the case (1) we check in total |11 | + |21 | = 7779 matrices. To check the matrices in 11 we use the system (4.54) in the equivalent form

x1 = (0, 0) ; x2 = (1, 0) ; x3 = (1, ) x4 = (0, ) , (0, 1] R x5 = (μ, ) ; x6 = (μ, 0) , μ R \ {1} xk = (ak , bk ) , k = 7, ..., 9 X = (x1 , ..., x9 ) ; X1 = X(1 : 6, 1 : 2) A = (3 : 5, 7 : 9); N = (3 : 5, 1 : 6); B = A1 ; Y = B N X1 ; , x7 = (Y (1, 1) , Y (1, 2)) ; x8 = (Y (2, 1) , Y (2, 2)) ; x9 = (Y (3, 1) , Y (3, 2)) ; ¡ ¢ e = x1 xt , ..., x9 xt ; e = X; e X 1 9 e (3) =0 e (4) = 0 e (5) = 0

(4.57)

where xt is the transposed vector x and Y (i, j) is the element in row i and column e R5 , but because of the form of x1 , ..., x6 , the rst j of matrix Y . The vector two components are zero and just the last three components are interesting for us. The initial system (4.54) has solution if the new formed system of equations (perhaps e (3) = 0 e non-linear) (4) = 0 with unknowns , μ has solution. e (5) = 0 We have restriction on and μ: (0, 1] , μ / {0, 1} , because = tan t, t (0, ], 4 / {x1 , x2 }. where t is the free parameter in 6-velocity models (see section 5.1) and x6 We checked all 3486 matrices in 11 and we obtained three dierent models (all known from before, as 1-extensions, from Section 4.3.1: models of type (iv) , (v) and (vi)). We found, in this way, our rst examples of matrices for which the condition ( ) is not sucient. To check the matrices in 21 we use the system (4.57), with the corresponding changes for x5 , x6 : x5 = (1 μ, + μ) and x6 = (μ, μ) (see case (ii) above). We have the on (0, 1] as before and μ / {0}, from x6 6= x1 . We checked all 4293 matrices in 21 and we found 7 dierent models (one already known, representing the 1-extension normal model (iv)). The other six normal models are new and more important they are ”irreducible”.

4. Models with mass, momentum and energy conservation d for the Model i). We present the models below (we denote by ti the angle 213

1 t1 = arctan 2 2

1 t2 = arctan 7

60


t3 =

6

t4 = arctan( 3 2)

61


1 t5 = arctan 7

1 t6 = arctan 2 2

62


63

Model 1 and Model 6 are 3-extensions of a 6-velocity normal model of type (b); Model 2 and Model 3 are 2-extensions of a 7-velocity normal model of type (2); Model 4 and Model 5 are 2-extensions of a 7-velocity normal model of type (3). All 9-velocity normal models, containing a 6-velocity model of type (a), that we obtained, were ”reducible”. It seems, by now, that only the 9-velocity normal models containing a 6-velocity model of type (b) can be ”irreducible”. All normal models that we know seem to lead by extensions to axial symmetric normal models. We observe that Model 1 and Model 6 accept 1-extention and the result is the same 10-velocity normal model given below.

This is the rst ”irreducible” 10-velocity model we found. A next step would be to classify all normal plane discrete models with 10 velocities.

Chapter 5

Conclusions In this thesis we consider dierent questions like ”How to construct normal DMs?”, ”Is it possible to construct all normal DVMs by the inductive method given by Bobylev and Cercignani [5]?”, ”How to describe all possible non-isomorphic normal DVMs?”. What answers are given in the present paper? In Section 2.4, we present a general algorithm for the complete classication of normal DMs with a given set {n, d, p, , } (order, dimension, number of linearly independent conservation laws, vector of conservation laws and maximal set of reactions). This algorithm gives also a method to construct normal models. In the particular case of models with mass and momentum conservation, the general algorithm gives us the complete classication. It is not so simple in the particular case of models with mass, momentum and energy conservation. We nd, in this case, an algorithm which allows us to give a full description of all possible non-isomorphic normal DVMs. Unfortunately, since the number of admissible matrices that we have to check with our algorithm increases very fast with the order of the model, we are able to give results only for the cases of small number of velocities, where we can nd restrictions that decrease the number of admissible matrices. By analytical methods (and by computer checking) we classify all normal models up to n = 8 velocities and show that they are all results of the inductive (1-extension) method. We obtain all ”reducible” models up to n = 10 velocities by analytical methods and nd for the rst time, in our knowledge, ”irreducible” normal models (n {9, 10}). Hence, the inductive method is certainly not enough to construct all normal models. We believe that one can proceed forward using similar methods and obtain a complete classication for normal DVMs with small numbers of velocities. The methods we develop can be applied to a very general class of DKMs. We restricted ourself, in this paper, to the case of classical DVMs for simple gases. The natural next step would be a generalization to the case DVMs for mixtures. 64

Bibliography [1] L. Arkeryd, On the Boltzmann equation. Part II: The full initial value problem, Arch. Rat. Mech. Anal., 45 (1972), pp. 17—34. [2] L. Arkeryd and C. Cercignani, On a functional equation arising in the kinetic theory of gases, Rend. Mat. Acc. Lincei, 1 (1990), pp. 139—149. [3] A. V. Bobylev and N. Bernho, Discrete velocity models and dynamical systems, in Lecture Notes on the Discretization of the Boltzmann Equation, N. Bellomo and R. Gatignol, eds., World Scientic, 2003, pp. 203—222. [4] A. V. Bobylev and C. Cercignani, Discrete velocity models for mixtures, J. Statist. Phys., 91 (1998), pp. 327—342. [5]

, Discrete velocity models without non-physical invariants, J. Statist. Phys., 97 (1999), pp. 677—686.

[6] H. Cabannes, The discrete velocity model. http://research.nianet.org/ luo/nonjournalpubs.html, 1980 (2003). Lecture notes given at the University of California at Berkeley, 1980, revised with R. Gatignol and L-S. Luo, 2003. [7] T. Carleman, Problème Mathématiques dans la Théorie Cinétique des Gaz, Almqvist-Wiksell, Uppsala, 1957. [8] C. Cercignani, Sur des critère d’existence globale en théorie cinétique discrète, C. R. Acad. Sc. Paris, 301 (1985), pp. 89—92. [9]

, Are there more than ve linearly independent collision invariants for the Boltzmann equation?, J. Statist. Phys., 58 (1990), pp. 817—823.

[10] C. Cercignani and H. Cornille, Shock waves for discrete velocity gas mixture, J. Statist. Phys., 99 (2000), pp. 115—140. [11] C. Cercignani, R. Illner, and M. Pulvirenti, The Mathematical Theory of Dilute Gases, Springer-Verlag, New York, 1994. 65

BIBLIOGRAPHY

66

[12] R. Gatignol, Théorie Cinétique des Gaz à Répartition Discrète de Vitesses, Springer-Verlag, New York, 1975. [13] T. Platkowski and R. Illner, Discrete velocity models of the Boltzmann equation: A survey on the mathematical aspects of the theory, SIAM Rev., 30 (1988), pp. 213— 255. [14] V. V. Vedenyapin, Velocity inductive construction for mixtures, Transport Theory and Statist. Phys., 28 (1999), pp. 727—742. [15] V. V. Vedenyapin and Y. N. Orlov, Conservation laws for polynomial Hamiltonians and for discrete models for Boltzmann equation, Teoret. and Math. Phys., 121 (1999), pp. 1516—1523.


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