# A Numberable Set of Exact Solutions for the Macroscopic Approach to Extended Thermodynamics of Polyatomic Gases with Many Moments

Authors: Maria Cristina Carrisi; Rita Enoh Tchame; Marcel Obounou; Sebastiano Pennisi
Journal: Advances in Mathematical Physics
 (2016)

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## Abstract

A new model for Polyatomic Gases with an arbitrary but fixed number of moments has been recently proposed and investigated in the framework of Extended Thermodynamics; the arbitrariness of the number of moments is linked to a numberN and the resulting model is called an N-Model. This model has been elaborated in order to take into account the entropy principle, the Galilean relativity principle, and some symmetry conditions. It has been proved that the solution for all these conditions exists, but it has not been written explicitly because hard notation is necessary; it has only been shown how the theory is self-generating in the sense that if we know the closure of the N-Model, then we will be able to find that of (N+1)-Model. Up to now only a single particular solution has been found in this regard. Instead of this, we find here a numberable set of exact solutions which hold for every fixed number N.


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## Body

## 1. Introduction

The model for Polyatomic Gases considers for every numberN the following balance equations:(1)∂tFi1⋯in+∂kFki1⋯in=Pi1⋯inforn=0,…,N+2,∂tGi1⋯im+∂kGki1⋯im=Qi1⋯imform=0,…,Nand the resulting model is called an N-Model.HereFi1⋯in represents the momentum-like hierarchy, while Gi1⋯im represents the energy-like hierarchy; the quantities with an index k represent their fluxes, while Pi1⋯in and Qi1⋯im represent the production terms; for the conservation laws of mass and momentum we have P=0 and Pi=0, while for the conservation of energy we have Q=0.This model represents a novelty with respect to the earlier versions of Extended Thermodynamics (ET) which used only (1)1 obtaining the important result to have a symmetric hyperbolic system of partial differential equations, with finite speed of propagation of shock waves and other important properties. A very short list of significative articles in this regard is given in [1–7]. These results constituted an important improvement of the models of Ordinary Thermodynamics. A very significative article in this previous framework is given in [8]. But ET was still not completely satisfactory because it is too much restrictive also on the state function relating the pressure p to the mass density ρ and energy density ϵ. This problem has been overcome in Extended Thermodynamics of Polyatomic Gases by considering all the balance equations (1) of which the first one is called the “Mass-Block” of equations and the second is called the “Energy-Block” of equations. See, for example, the articles [9–16].It has to be noted that in the literature here cited the balance equations are not included in theN-Model, but they can be easily recovered with the theory of subsystems (see [17]) by simply putting the Lagrange multiplier of the last equation in the Mass-Block equal to zero. In [11, 12] it has been shown that the wave speeds of the N-Model are the same as those of monatomic gases; but this proof has been given with the kinetic equations which are only a subcase of those of the macroscopic approach (in effect, the whole closure here found is zero in the kinetic approach and only parts (5) and (6) remains). In any case there is no problem for what we have said at the beginning of this note. Moreover, a priori the Bhatnagar-Gross-Krook equation (see the equation before (14) in [10]) can be replaced by an infinite hierarchy of balance equations for moments and only for practical reasons some of them are eliminated; consequently, there is no inconsistence if another of these balance equations is eliminated in a second step with the theory of subsystems, as said above. The N-Model is here considered because it makes (in the macroscopic approach) the decomposition of tensors in nonconvective parts and in velocity dependent parts more linear. In the kinetic approach this is simply a consequence of the integrals, but in the macroscopic approach here followed this decomposition must be justified in a straightforward manner. In [11, 12] the (M,N)-system has been proposed with N≤M-1; but the macroscopic approach needs a much heavier notation and this is simplified by the present N-Model without losing generality because, in any case, the (M,N)-system can be recovered from the present one with the theory of subsystems.In (1) the independent variables are Fi1⋯in and Gi1⋯im; the quantities Fki1⋯in and Gki1⋯im are their corresponding fluxes. The problem will be closed when we know the expressions of Fki1⋯iN+2 and Gki1⋯iN as functions of the independent variables. Restrictions on their generality are obtained by imposing the entropy principle, the Galilean relativity principle, and the symmetry conditions. These last ones express the fact that each flux in (1) is equal to the independent variable of the subsequent equation, except for the flux in the last equation of the Mass-Block and for that in the last equation of the Energy-Block; for these last fluxes we know only that they are symmetric tensors.The Entropy Principle, exploited through Liu’s Theorem [18] and a bright idea conceived by Ruggeri and Strumia [19], becomes equivalent to assuming the existence of Lagrange Multipliers μA and λB which can be taken as independent variables and, after that, we have(2)Fi1⋯in=∂h′∂μi1⋯in,Gi1⋯im=∂h′∂λi1⋯im,Fki1⋯in=∂h′k∂μi1⋯in,Gki1⋯im=∂h′k∂λi1⋯im,which expresses all the moments in terms of only two unknown functions, the 4-potential h′ and h′k. Their relation with the entropy h and its flux hk is given by (3)h=-h′+Fi1⋯inμi1⋯in+Gi1⋯imλi1⋯im,hk=-h′k+Fki1⋯inμi1⋯in+Gki1⋯imλi1⋯im.A nice consequence of (2) is that the field equations assume the symmetric form. Another consequence of (2) is that the above-mentioned symmetry conditions and the Galilean relativity principle can be expressed as(4)∂h′k∂μi1⋯in=∂h′∂μki1⋯inforn=0,…,N+1;∂h′k∂μi1i2⋯iN+2=0,∂h′k∂λi1⋯im=∂h′∂λki1⋯imform=0,…,N-1;∂h′k∂λi1⋯iN=0,∂h′∂μkμi+∑n=1N+1∂h′∂μkj1⋯jnn+1μij1⋯jn+2λj1⋯jn-1δjni+∑s=0N-1∂h′∂λkh1⋯hss+1λih1⋯hs+h′δki=0.The equivalence of (4)5 to the Galilean relativity principle is proved in literature such as [13, 20, 21] and on Representation Theorems such as that of [22].In (5), (6), and (7) we report an already known particular solution (see [23]) for the conditions (4). But it is not the general solution; in fact, in the present article we will exhibit a significative set of other solutions depending on a numberable family of arbitrary single variable functions gi(λ) and we report it in the next section, (8) and (9). It is expressed through Taylor’s expansion around equilibrium which is defined as the state where μi1⋯in=0, for n=1,…,N+2, and λi1⋯im=0, for m=1,…,N, so that the only variables which are not zero at equilibrium are μ and λ. The first of these is the chemical potential, while λ=1/2T with T absolute temperature.Obviously, also the velocity is nonvanishing at equilibrium; but if we use a decomposition in nonconvective parts and in velocity dependent parts, we obtain the Lagrange multiplierμi of the momentum conservation law in terms of the other Lagrange multipliers and we see that its velocity independent part is zero at equilibrium. Moreover, from the fact that the velocity independent parts of μ⋯ and λ⋯ are zero at equilibrium, it follows that they are zero also as a whole, without decomposition; the only exceptions are for the Lagrange multipliers of the mass and of the energy conservation laws (see equation (25) of [21], e.g.).The proof of the fact that (8) and (9) give a solution for (4) is here split into two parts: the first one is reported in Section 3 and the second and final part is described in Section 4.In conclusion, the sum of the solution reported in (5), (6), and (7) and the present one reported in (8) and (9) is a more general solution for our conditions.


## 2. A Set of Exact Solutions for Conditions (4)

Before writing our solution we need an apparently complicatednotation.To do the derivatives with respect toμi1⋯in a number pn of times, we use the compact form An,1 to indicate a set of n indexes, An,2 to indicate another set of n indexes, and so on up to An,pn, so that the first index of A⋯,⋯ indicates how many indexes have the corresponding μ⋯, while the second index of A⋯,⋯ indicates how many derivatives we are taking with respect to it. Similarly, to do the derivatives with respect to λi1⋯im a number rm of times, we use the compact form Bm,1 to indicate a set of m indexes, Bm,2 to indicate another set of m indexes, and so on up to Bm,rm. After describing this notation, we can now report from [23] the already known solution for our conditions and it is(5)h′=∑p1,…,pN+2,r1,…,rNp1+2p2+⋯+N+2pN+2+r1+2r2+⋯+NrNeven0⋯∞1p1!1p2!⋯1pN+2!1r1!1r2!⋯1rN!·1+p1+2p2+⋯+N+2pN+2+r1+2r2+⋯+NrN!!1+p1+2p2+⋯+N+2pN+2+r1+2r2+⋯+NrN·∂1+p1+p2+⋯+pN+2+r1+r2+⋯+rN∂λr1+r2+⋯+rN∂μ1+p1+p2+⋯+pN+2-12λ1/2p1+2p2+⋯+N+2pN+2+r1+2r2+⋯+NrN·ψ1/2p1+2p2+⋯+N+2pN+2+r1+2r2+⋯+NrN-12λ1/2p1+2p2+⋯+N+2pN+2+r1+2r2+⋯+NrN·δA1,1⋯A1,p1⋯AN+2,1⋯AN+2,pN+2B1,1⋯B1,r1⋯BN,1⋯BN,rN·μA1,1⋯μA1,p1⋯μAN+2,1⋯μAN+2,pN+2λB1,1⋯λB1,r1⋯λBN,1⋯λBN,rN,(6)h′k=∑p1,…,pN+2,r1,…,rNp1+2p2+⋯+N+2pN+2+r1+2r2+⋯+NrNodd0⋯∞1p1!1p2!⋯1pN+2!1r1!1r2!⋯1rN!·p1+2p2+⋯+N+2pN+2+r1+2r2+⋯+NrN!!·∂1+p1+p2+⋯+pN+2+r1+r2+⋯+rN∂λr1+r2+⋯+rN∂μ1+p1+p2+⋯+pN+2-12λ1/21+p1+2p2+⋯+N+2pN+2+r1+2r2+⋯+NrN·ψ1/21+p1+2p2+⋯+N+2pN+2+r1+2r2+⋯+NrN-12λ1/21+p1+2p2+⋯+N+2pN+2+r1+2r2+⋯+NrN·δkA1,1⋯A1,p1⋯AN+2,1⋯AN+2,pN+2B1,1⋯B1,r1⋯BN,1⋯BN,rN·μA1,1⋯μA1,p1⋯μAN+2,1⋯μAN+2,pN+2λB1,1⋯λB1,r1⋯λBN,1⋯λBN,rN,where ψn is a family of functions depending only on μ and λ and constrained only by(7)∂ψn+1∂μ=ψn.But the above one is not the unique solution; in fact, by using the same notation, we can now say that a more general solution can be obtained by adding to (5) and (6) the following expressions (8) and (9); that is,(8)h′=∑p0,…,pN+2,r1,…,rN∑i=1N+2ipi+∑i=1Nirieven0⋯∞1p0!1p1!⋯1pN+2!1r1!1r2!⋯1rN!·1+∑i=1N+2ipi+∑i=1Niri!!1+∑i=1N+2ipi+∑i=1Niri·∂r1+r2+⋯+rN∂λr1+r2+⋯+rN-12λ∑i=0N+23-ipi-∑i=1Nirig∑i=0N+2i-2pi+∑i=1Niri-1·δA1,1⋯A1,p1⋯AN+2,1⋯AN+2,pN+2B1,1⋯B1,r1⋯BN,1⋯BN,rNμp0·μA1,1⋯μA1,p1⋯μAN+2,1⋯μAN+2,pN+2λB1,1⋯λB1,r1⋯λBN,1⋯λBN,rN,(9)h′k=∑p0,…,pN+2,r1,…,rN∑i=1N+2ipi+∑i=1Niriodd0⋯∞1p0!1p1!⋯1pN+2!1r1!1r2!⋯1rN!∑i=1N+2ipi+∑i=1Niri!!·∂r1+r2+⋯+rN∂λr1+r2+⋯+rN-12λ∑i=0N+23-ipi-∑i=1Niri-1g∑i=0N+2i-2pi+∑i=1Niri·δkA1,1⋯A1,p1⋯AN+2,1⋯AN+2,pN+2B1,1⋯B1,r1⋯BN,1⋯BN,rNμp0·μA1,1⋯μA1,p1⋯μAN+2,1⋯μAN+2,pN+2λB1,1⋯λB1,r1⋯λBN,1⋯λBN,rN,where gi is a family of functions depending only on λ such that gi=0 for i<0. Here we use only a letter δ⋯ as a shortened symbol denoting the product of some δ⋯ each one with 2 indexes and with a final symmetrization over all these indexes.We note the following:(i)
IfN=0, the function g in the expression of h′k has index -2p0-p1 and consequently this g may be different from zero only if p0=0 and p1=0; but in this case p1+2p2 is not odd. This fact implies that h′k=0. Similarly, the function g in the expression of h′ has index -2p0-p1-1 so that g=0 and h′=0. This confirms the results of [24] for this case N=0, because in the sum of (8) and (9) and the corresponding results (5) and (6) of [23] only these last ones remain in the case N=0.(ii)
IfN=1, the function g in the expression of h′k has index -2p0-p1+p3+r1 and this is an odd number because p1+2p2+3p3+r1 must be odd. Consequently, the index p0 belongs to the interval 0,-p1+p3+r1-1/2 if this is not the empty set. By taking into account this fact we see that the sum of the solution reported in (6) and the present expression (9) for the case N=1 is the same as that in [24]. Similarly, the function g in the expression of h′ has index -2p0-p1+p3+r1-1 and this is an odd number because p1+2p2+3p3+r1 must be even. Consequently, the index p0 belongs to the interval 0,-p1+p3+r1-2/2 if this is not the empty set. By taking into account this fact we see that the sum of our expression (8) and that in (5) for the case N=1 is the same as that in [24].(iii)
IfN≥2 our solution is completely new.


## 3. The New Solution (8) and (9) Satisfies (4)1–4

In this section we prove that (8) and (9) are a solution for (4)1–4, putting off (4)5 to Section 4:(i)
Let us substitute firstly (8) and (9) in (4)1.We obtain for both sides the following expression:(10)∑p0,…,pN+2,r1,…,rNn+∑i=1N+2ipi+∑i=1Niriodd0⋯∞1p0!1p1!⋯1pN+2!1r1!1r2!⋯1rN!n+∑i=1N+2ipi+∑i=1Niri!!·∂r1+r2+⋯+rN∂λr1+r2+⋯+rN-12λ∑i=0N+23-ipi-∑i=1Niri-1+3-ng∑i=0N+2i-2pi+∑i=1Niri+n-2·δki1⋯inA1,1⋯A1,p1⋯AN+2,1⋯AN+2,pN+2B1,1⋯B1,r1⋯BN,1⋯BN,rNμp0·μA1,1⋯μA1,p1⋯μAN+2,1⋯μAN+2,pN+2λB1,1⋯λB1,r1⋯λBN,1⋯λBN,rN,where we have taken into account thatμi1⋯in here is denoted with μAn,1 or with μAn,2 and so on up to μAn,pn. Similarly, μki1⋯in is denoted with μAn+1,1 or with μAn+1,2 and so on up to μAn+1,pn+1. Consequently, the left hand side of (4)1 causes, with respect to expression (9), a rise of one unity of the index pn; similarly, the right hand side of (4)1 causes, with respect to expression (8), a rise of one unity of the index pn+1.(ii)
It is easy to verify that (9) satisfies (4)2,4.(iii)
By substituting (8) and (9) in (4)3 in the case m≥1, we obtain for both sides the following expression: (11)∑p0,…,pN+2,r1,…,rNm+∑i=1N+2ipi+∑i=1Niriodd0⋯∞1p0!1p1!⋯1pN+2!1r1!1r2!⋯1rN!m+∑i=1N+2ipi+∑i=1Niri!!·∂1+r1+r2+⋯+rN∂λ1+r1+r2+⋯+rN-12λ∑i=0N+23-ipi-∑i=1Niri-1-mg∑i=0N+2i-2pi+∑i=1Niri+m·δki1⋯imA1,1⋯A1,p1⋯AN+2,1⋯AN+2,pN+2B1,1⋯B1,r1⋯BN,1⋯BN,rNμp0·μA1,1⋯μA1,p1⋯μAN+2,1⋯μAN+2,pN+2λB1,1⋯λB1,r1⋯λBN,1⋯λBN,rN,where we have taken into account thatλi1⋯im here is denoted with λBm,1 or with λBm,2 and so on up to λBm,rm. Similarly, λki1⋯im is denoted with λBm+1,1 or with λBm+1,2 and so on up to λBm+1,rm+1. Consequently, the left hand side of (4)3 with m≥1 causes, with respect to expression (9), a rise of one unity of the index rm; similarly, the right hand side of (4)3 with m≥1 causes, with respect to expression (8), a rise of one unity of the index rm+1.(iv)
Let us conclude substituting (8) and (9) into (4)3 in the case m=0.We have the same situation of the above case for its right hand side, while for its left hand side we have to take simply the derivative of (9) with respect to λ. More precisely, we obtain for both sides the following expression:(12)
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## 4. The New Solution (8) and (9) Satisfies (4)5

First of all we note that (4)5 can be rewritten as(13)∑n=0N+1∂h′∂μkj1⋯jnn+1μij1⋯jn+∂h′∂μki2λ+∑s=0N-12λj1⋯js+1∂h′∂μkj1⋯js+1i+∑s=0N-1∂h′∂λkh1⋯hss+1λih1⋯hs+h′δki=0.(i) After that, we note that the first term of this relation thanks to (8) becomes (14)∑n=0N+1n+1μij1⋯jn∂h′∂μkj1⋯jn=∑n=0N+1n+1pn+1·∑p0,…,pN+2,r1,…,rN∑i=1N+2ipi+∑i=1Nirieven0⋯∞1p0!1p1!⋯1pN+2!1r1!1r2!⋯1rN!·1+∑i=1N+2ipi+∑i=1Niri!!1+∑i=1N+2ipi+∑i=1Niri·∂r1+r2+⋯+rN∂λr1+r2+⋯+rN-12λ∑i=0N+23-ipi-∑i=1Nirig∑i=0N+2i-2pi+∑i=1Niri-1μp0·δjn+1iδkj1⋯jnA1,1⋯A1,p1⋯An+1,1⋯An+1,pn+1-1⋯AN+2,1⋯AN+2,pN+2B1,1⋯B1,r1⋯BN,1⋯BN,rN·μA1,1⋯μA1,p1⋯μAn+1,1⋯μAn+1,pn+1-1μj1⋯jn+1⋯μAN+2,1⋯μAN+2,pN+2·λB1,1⋯λB1,r1⋯λBN,1⋯λBN,rN,because the derivation causes a presence of the factor pn+1 and the substitution of μAn+1,1⋯μAn+1,pn+1 with μAn+1,1⋯μAn+1,pn+1-1 and the new free indexes kj1⋯jn in the expression of δ⋯; moreover, we have also substituted μij1⋯jn with μj1⋯jn+1δjn+1i. Now in the resulting expression we can insert a symmetrization over all the indexes of the set j1⋯jn+1An+1,1⋯An+1,pn+1-1 because that expression remains the same if we exchange two of these indexes. This fact is evident if the two indexes are taken between j1⋯jn+1; for the proof in the other cases, let us consider the shortened expression δjn+1iδ(kj1⋯jnk1⋯knkn+1⋯)μj1⋯jn+1μk1⋯kn+1; here we can exchange the names of the indexes j· with those of k·, so that the above shortened expression becomes δkn+1iδ(kk1⋯knj1⋯jnjn+1⋯)μk1⋯kn+1μj1⋯jn+1.Now we can exchange the indexesk1⋯kn with the indexes j1⋯jn in the expression of δ(kk1⋯knj1⋯jnjn+1⋯) because this is a symmetric tensor.We obtainδkn+1iδ(kj1⋯jnk1⋯knjn+1⋯)μk1⋯kn+1μj1⋯jn+1. By comparing this result with the expression which we started from, it is the same if we had exchanged the indexes jn+1 and kn+1. This completes the proof of the fact that expression (14) remains the same if we exchange two indexes of the set j1⋯jn+1An+1,1⋯An+1,pn+1-1; so we can insert there a symmetrization over those indexes and (14) becomes(15)∑n=0N+1n+1μij1⋯jn∂h′∂μkj1⋯jn=∑n=0N+1n+1pn+1·∑p0,…,pN+2,r1,…,rN∑i=1N+2ipi+∑i=1Nirieven0⋯∞1p0!1p1!⋯1pN+2!1r1!1r2!⋯1rN!·1+∑i=1N+2ipi+∑i=1Niri!!1+∑i=1N+2ipi+∑i=1Niri·∂r1+r2+⋯+rN∂λr1+r2+⋯+rN-12λ∑i=0N+23-ipi-∑i=1Nirig∑i=0N+2i-2pi+∑i=1Niri-1μp0·δjn+1_iδkj1⋯jn_A1,1⋯A1,p1⋯An+1,1⋯An+1,pn+1-1_⋯AN+2,1⋯AN+2,pN+2B1,1⋯B1,r1⋯BN,1⋯BN,rN·μA1,1⋯μA1,p1⋯μAn+1,1⋯μAn+1,pn+1-1μj1⋯jn+1⋯μAN+2,1⋯μAN+2,pN+2·λB1,1⋯λB1,r1⋯λBN,1⋯λBN,rN,where underlined indexes denote symmetrization over these indexes. Now we observe that (n+1)pn+1 is exactly the number of the indexes of the set j1⋯jn+1An+1,1⋯An+1,pn+1-1 and that, thanks to the summation ∑n=0N+1, the index near i in δjn+1i can be every index of the set A1,1⋯A1,p1⋯An+1,1⋯An+1,pn+1⋯AN+2,1⋯AN+2,pN+2. These facts allow rewriting (15) as(16)∑n=0N+1n+1μij1⋯jn∂h′∂μkj1⋯jn=∑n=0N+1n+1pn+1·∑p0,…,pN+2,r1,…,rN∑i=1N+2ipi+∑i=1Nirieven0⋯∞1p0!1p1!⋯1pN+2!1r1!1r2!⋯1rN!·1+∑i=1N+2ipi+∑i=1Niri!!1+∑i=1N+2ipi+∑i=1Niri·∂r1+r2+⋯+rN∂λr1+r2+⋯+rN-12λ∑i=0N+23-ipi-∑i=1Nirig∑i=0N+2i-2pi+∑i=1Niri-1μp0·δjn+1_iδkj1⋯jnA1,1⋯A1,p1⋯An+1,1⋯An+1,pn+1-1⋯AN+2,1⋯AN+2,pN+2_B1,1⋯B1,r1⋯BN,1⋯BN,rN·μA1,1⋯μA1,p1⋯μAn+1,1⋯μAn+1,pn+1-1μj1⋯jn+1⋯μAN+2,1⋯μAN+2,pN+2·λB1,1⋯λB1,r1⋯λBN,1⋯λBN,rN.(ii) For the fourth term of (13) we can do similar passages (the difference is that we have λ⋯ instead of μ⋯, N-2 instead of N, and s instead of n); in this way that term becomes (17)∑s=0N-1s+1λih1⋯hs∂h′∂λkh1⋯hs=∑s=0N-1s+1rs+1·∑p0,…,pN+2,r1,…,rN∑i=1N+2ipi+∑i=1Nirieven0⋯∞1p0!1p1!⋯1pN+2!1r1!1r2!⋯1rN!·1+∑i=1N+2ipi+∑i=1Niri!!1+∑i=1N+2ipi+∑i=1Niri·∂r1+r2+⋯+rN∂λr1+r2+⋯+rN-12λ∑i=0N+23-ipi-∑i=1Nirig∑i=0N+2i-2pi+∑i=1Niri-1μp0·δhs+1_iδkh1⋯hs_A1,1⋯A1,p1⋯AN+2,1⋯AN+2,pN+2B1,1⋯B1,r1⋯Bs+1,1⋯Bs+1,rs+1-1⋯BN,1⋯BN,rN_·μA1,1⋯μA1,p1⋯μAN+2,1⋯μAN+2,pN+2·λB1,1⋯λB1,r1⋯λBs+1,1⋯λBs+1,rs+1-1⋯λBN,1⋯λBN,rN.(iii) If we look at the last term of (13), we see that it can be written together with (16) and (17) and they become (18)∑n=0N+1∂h′∂μkj1⋯jnn+1μij1⋯jn+∑s=0N-1∂h′∂λkh1⋯hss+1λih1⋯hs+h′δki=1+∑n=0N+1n+1pn+1+∑s=0N-1s+1rs+1·∑p0,…,pN+2,r1,…,rN∑i=1N+2ipi+∑i=1Nirieven0⋯∞1p0!1p1!⋯1pN+2!1r1!1r2!⋯1rN!·1+∑i=1N+2ipi+∑i=1Niri!!1+∑i=1N+2ipi+∑i=1Niri·∂r1+r2+⋯+rN∂λr1+r2+⋯+rN-12λ∑i=0N+23-ipi-∑i=1Nirig∑i=0N+2i-2pi+∑i=1Niri-1μp0·δk_iδA1,1⋯A1,p1⋯AN+2,1⋯AN+2,pN+2B1,1⋯B1,r1⋯BN,1⋯BN,rN_·μA1,1⋯μA1,p1⋯μAN+2,1⋯μAN+2,pN+2λB1,1⋯λB1,r1⋯λBN,1⋯λBN,rN,where also the index k has been put under the symmetrization and we recall that a property of symmetrization is δik_δ(⋯_)=δik_δ⋯_=δ(ik⋯). We note that the coefficient in square bracket at the beginning of the right hand side of (18) has become equal to the denominator of the half factorial of that right hand side.(iv) For the second term of (13), we can use (8) and the derivation causes someway the rising of one unity of the index p2; the result is (19)∂h′∂μki2λ=2λ∑p0,…,pN+2,r1,…,rN∑i=1N+2ipi+∑i=1Nirieven0⋯∞1p0!1p1!⋯1pN+2!1r1!1r2!⋯1rN!·1+∑i=1N+2ipi+∑i=1Niri!!·∂r1+r2+⋯+rN∂λr1+r2+⋯+rN-12λ∑i=0N+23-ipi-∑i=1Niri+1g∑i=0N+2i-2pi+∑i=1Niri-1μp0·δkiA1,1⋯A1,p1⋯AN+2,1⋯AN+2,pN+2B1,1⋯B1,r1⋯BN,1⋯BN,rN·μA1,1⋯μA1,p1⋯μAN+2,1⋯μAN+2,pN+2λB1,1⋯λB1,r1⋯λBN,1⋯λBN,rN.(v) There remains considering the third term of (13); for it we can use (8) and the derivation causes someway the rising of one unity of the index ps+3; in this way it becomes (20)∑s=0N-12λj1⋯js+1∂h′∂μkj1⋯js+1i=∑s=0N-12λj1⋯js+1·∑p0,…,pN+2,r1,…,rNs+3+∑i=1N+2ipi+∑i=1Nirieven0⋯∞1p0!1p1!⋯1pN+2!1r1!1r2!⋯1rN!·s+4+∑i=1N+2ipi+∑i=1Niri!!s+4+∑i=1N+2ipi+∑i=1Niri·∂r1+r2+⋯+rN∂λr1+r2+⋯+rN-12λ-s+∑i=0N+23-ipi-∑i=1Nirigs+1+∑i=0N+2i-2pi+∑i=1Niri-1μp0·δkj1⋯js+1iA1,1⋯A1,p1⋯AN+2,1⋯AN+2,pN+2B1,1⋯B1,r1⋯BN,1⋯BN,rN·μA1,1⋯μA1,p1⋯μAN+2,1⋯μAN+2,pN+2λB1,1⋯λB1,r1⋯λBN,1⋯λBN,rN.In this expression, we can substitute λj1⋯js+1 with λBs+1,rs+1+1 and j1⋯js+1 with Bs+1,rs+1+1; moreover, we substitute 1/rs+1! with rs+1+1/(rs+1+1)! and, after that, decrease rs+1 of one unity (which is equivalent to a change of index). So this expression becomes(21)∑s=0N-12λj1⋯js+1∂h′∂μkj1⋯js+1i=∑s=0N-12rs+1·∑p0,…,pN+2,r1,…,rN2+∑i=1N+2ipi+∑i=1Nirieven0⋯∞1p0!1p1!⋯1pN+2!1r1!1r2!⋯1rN!·3+∑i=1N+2ipi+∑i=1Niri!!3+∑i=1N+2ipi+∑i=1Niri·∂-1+r1+r2+⋯+rN∂λ-1+r1+r2+⋯+rN12λ1+∑i=0N+23-ipi-∑i=1Nirig∑i=0N+2i-2pi+∑i=1Niri-1μp0·δkiA1,1⋯A1,p1⋯AN+2,1⋯AN+2,pN+2B1,1⋯B1,r1⋯BN,1⋯BN,rN·μA1,1⋯μA1,p1⋯μAN+2,1⋯μAN+2,pN+2λB1,1⋯λB1,r1⋯λBN,1⋯λBN,rN.It is true that with change of index we have in the summation the extra term with rs+1=0; but it does not effect the result for the presence of the coefficient rs+1.If we look at the expressions (18), (19), and (21) we conclude that to prove (13) it is sufficient that the following relation holds: (22)∂r1+r2+⋯+rN∂λr1+r2+⋯+rN-12λ∑i=0N+23-ipi-∑i=1Nirig∑i=0N+2i-2pi+∑i=1Niri-1+2λ∂r1+r2+⋯+rN∂λr1+r2+⋯+rN-12λ1+∑i=0N+23-ipi-∑i=1Nirig∑i=0N+2i-2pi+∑i=1Niri-1+∑s=0N-12rs+1∂-1+r1+r2+⋯+rN∂λ-1+r1+r2+⋯+rN-12λ1+∑i=0N+23-ipi-∑i=1Nirig∑i=0N+2i-2pi+∑i=1Niri-1=0,and to prove this relation it is sufficient to prove that(23)∂r∂λrϑ+2λ∂r∂λr-12λϑ+2r∂-1+r∂λ-1+r-12λϑ=0,where we have put r=r1+r2+⋯+rN and ϑ=-1/2λ∑i=0N+2(3-i)pi-∑i=1Nirig∑i=0N+2(i-2)pi+∑i=1Niri-1.Obviously, (23) is an identity because we have (24)∂r∂λrϑ=∂r∂λr-2λ-12λϑ=-2λ∂r∂λr-12λϑ-2r∂-1+r∂λ-1+r-12λϑ.This concludes all the arguments which we had to prove.


## 5. Conclusions

We conclude that, in the particular caseN=0, we obtain the 11 moments’ model and the solution here indicated is exactly that already obtained in literature with the macroscopic approach. In the case N=1 we have that the present one is a particular solution for that already known in literature. So we have seen how it transfers itself also to higher values of N. In this way we are approaching the goal to obtain the most general solution and this will be the object of a future research.



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*Source: 1307813-2016-08-25.xml*