Generalized Boltzmann Equation

Abstract

In what follows, we intend to construct the generalized Boltzmann physical kinetics using the different methods of the kinetic equation derivations from the Bogolubov hierarchy.

Keywords

Generalized Boltzmann equation

Bogolubov hierarchy

Method of many scales

1.1 Mathematical Introduction-Method of Many Scales

In the sequel asymptotic methods will be used, but first of all, we will look at the method of many scales. The method of many scales is so popular that Nayfeh in his book [68] written more than 40 years ago said that method of many scales (MMS) is discovered by different authors every half a year. As a result, there exist many different variants of MMS. As a minimum four variants of MMS are considered in [68]. We are interested only in the main ideas of MMS, which are used further in theory of kinetic equations.

From this standpoint we demonstrate MMS possibilities using typical example of solution linear differential equation which also has the exact solution for comparing the results [68]. But in contrast with usual consideration, which can be found in literature, we intend to bring this example up to table and a graph.

Therefore let us consider the linear differential equation

x¨+?x?+x=0.

  (1.1.1)

We begin with the special case when d = 1 and is a small parameter. Equation (1.1.1) has the exact solution

=ae-?t/2cost1-14?2+?,

  (1.1.2)

where a and ? are arbitrary constants of integrating. In typical case of small parameter d in front of senior derivative-in this case it would be ¨-the effects of boundary layer can be observed. Using the derivatives

x?=-12?x-ae-?t/21-14?2sint1-14?2+?,x¨=-12?x?-x1-14?2+12a?e-?t/21-14?2sint1-14?2+?,

for substitution in Eq. (1.1.1) we find the identical satisfaction of Eq. (1.1.1).

We begin with a direct expansion in small , using series

=x0+?x1+?2x2+?,

  (1.1.3)

and after differentiating

x?=x?0+?x?1+?2x?2+?,x¨=x¨0+?x¨1+?2x¨2+?.

Substitute series (1.1.3) into (1.1.1) and equalize coefficients in front of equal powers of , having

¨0+x0=0,

  (1.1.4)

¨1+x1=-x?0,

  (1.1.5)

¨2+x2=-x?1,

  (1.1.6)

¨3+x3=-x?2,

  (1.1.7)

and so on. The general solution of homogeneous Eq. (1.1.4) has the form

0=acost+?.

  (1.1.8)

Substitute (1.1.8) in (1.1.5):

¨1+x1=asint+?.

  (1.1.9)

General solution (1.1.1) should contain only two arbitrary constants. In this case, both constants a and ? are contained in the main term of expansion defined by relation (1.1.8). Then we need find only particular solution of Eq. (1.1.9); which can be found as follows

1=-12atcost+?.

  (1.1.10)

Really,

x?1=-12acost+?+12atsin(t+?),x¨1=12asint+?-x1+12asin(t+?).

After substitution in (1.1.9), we find identity. Equation (1.1.6) can be rewritten as

¨2+x2=12acost+?-12atsint+?,

  (1.1.11)

and its solution

2=18at2cost+?+18atsint+?.

  (1.1.12)

Really,

?2=14atcost+?-18at2sint+?+18asint+?+18atcost+?,

¨2=12acost+?-58atsint+?-18at2cost+?.

Substitution into left-hand side of Eq. (1.1.11), lead to result

2acost+?-58atsin(t+?)-18at2cos(t+?)+18at2cost+?+18atsin(t+?)=12acost+?-12atsin(t+?).

Then we state the identical satisfaction of Eq. (1.1.11) by solution (1.1.12). In analogous way the solution of Eq. (1.1.7) is written as cubic polynomial in t. For the first three terms of Eq. (1.1.3) series the solution is

=acost+?-12?atcos(t+?)+18?2at2cost+?+tsint+?+O?3.

  (1.1.13)

At our desire the variable t can be considered as dimensionless time. Suppose, of course, that we wish to have a solution for arbitrary time moments. But it is not possible in the developed procedure, because the series construction regards the successive terms of the series to be smaller than the forgoing terms; in other case, it is impossible to speak about series convergence. But for fixed , the time moment can be found when successive term of expansion is no smaller than the forgoing term. Figure 1.1 contains comparison of the exact solution (1.1.2) for concrete parameters of calculations a = 1, ? = 0,  = 0.2 with approximate solutions

0=costx1=cost-0.1tcostx2=cost-0.1tcost+0.005t2cost+tsintxex=e-0.1tcost0.99.

As we could expect, the divergence of solution 2x and exact solution exx appears when t is of order 10; or, in the common case, if t ~ - 1. But as it follows from Fig. 1.1, the situation is much worse, because, for example, for t = 4p approximate solution 1x gives wrong sign in comparison with the exact solution exx. For the mathematical model of an oscillator with damping-which is reflecting by Eq. (1.1.1), it means that approximate solution 1x forecasts a deviation in the opposite direction for the mentioned oscillator. By the way, solution 1x is also worse in comparison with solution 0x, in this sense the minor approximation is better than senior ones.

This poses the question how to improve the situation remaining in the frame of asymptotic methods. To answer this question, let us consider the exact solution (1.1.2). Exponential and cosine terms containing in this solution, can be expand in the following series for small and fixed t:

-?t/2=1-12?t+18?t2-164?t3+?

  (1.1.14)

cost1-14?2+?=cost1-18?2-1128?4-11024?6-?+?=cost+?-18t?2-1128t?4-11024t?6-??cost+?+18t?2+1128t?4+11024t?6+?sin(t+?)=cost+?+18t?2sin(t+?)+1128t?4sin(t+?)+?.

  (1.1.15)

Obviously, product of the first terms in expansions (1.1.14) and (1.1.15) gives 0x, and retaining of terms of O(3) lead to result

?a1-12?t+18?2t2cost+?+18t?2sint+??acost+?-12a?tcos(t+?)+18at?2sin(t+?)+18a?2t2cos(t+?).

Then we state that the used construction of asymptotic solution is based in deed on the assumption that combination ?t is small. If it is not so (for t having the order - 1), then expansions (1.1.14) and (1.1.15) are wrong or need to take into account all terms of expansions. But asymptotic expansion can be organized by another way, using additional variables:

1=?t

  (1.1.16)

and

2=?2t.

  (1.1.17)

In this case

-?t/2=e-T1/2,

  (1.1.18)

and expansion (1.1.15) is replaced by other...