Contents

In the present work the classical problem of the kinetic theory of gases (the Smoluchowsky’ problem about temperature jump in rarefied gas) is considered. The rarefied gas fills half-space over a flat firm surface. logarithmic gradient of temperature is set far from surface. The kinetic equation with modelling integral of collisions in the form of BGK-model (Bhatnagar, Gross and Krook) is used.

The general mirror-diffuse boundary conditions of molecules reflexions of gas from a wall on border of half-space (Maxwell conditions) are considered. Expanding distribution function on two orthogonal directions in space of velocities, the Smoluchowsky’ problem to the solution of the homogeneous vector one-dimensional and one-velocity kinetic equation with a matrix kernel is reduced.

Then generalization of source-method is used and boundary conditions include in non-homogeneous vector kinetic equation. The solution in the form of Fourier integral is searched. The problem is reduced to the solution of vector Fredholm integral equation of the second sort with matrix kernel.

The solution of Fredholm equation in the form of Neumann’s polynoms with vector coefficients is searched. The system vector algebraic interengaged equations turns out. The solution of this system is under construction in the form of Neumann’s polynoms. Comparison with well-known Barichello—Siewert’ high-exact results is made. Zero and the first approach of jumps of temperature and numerical density are received. It is shown, that transition from the zero to the first approach raises 10 times accuracy in calculation coefficients of temperature and concentration jump.

Key words: the Smoluchowski’ problem, collisional gas, temperature and concentration jump, vector Fredholm equation of second sort.

PACS numbers: 05.20.Dd Kinetic theory, 47.45.-n Rarefied gas dynamics, 02.30.Rz Integral equations, 51. Physics of gases, 51.10.+y Kinetic and transport theory of gases.

A new method for solving of vector problems for kinetic equations with Maxwell boundary conditions

A. V. Latyshev111

Faculty of Physics and Mathematics,

Moscow State Regional University, 105005,

## 1 Introduction

The problem about temperature jump is known from the end of XIX century . M.Smoluhovsky has constructed in  the theory of temperature jump in the rarefied gas. Since then this problem the invariable attention already draws for a long time to itself (history of this question see in  and ).

This problem has been solved with use of approximate and numerical methods as for the modelling equations, and for full (non-linear) Boltzmann equation  - .

In 1972 in work  the problem about temperature jump with use of the modelling Boltzmann equation with collisional integral BGK (Bhatnagar, Gross, Krook) and with frequency of collisions of the molecules, proportional to the module of velocities of molecules has analytically been solved.

Attempts of the exact solution of this problem about temperature jump with diffusion boundary conditions and with use of the modelling Boltzmann equation with collisional integral BGK with constant frequency of collisions of molecules  -  begin with this moment.

The solution of this problem encounters considerable difficulties. This problem is formulated in the form of a vector boundary problem. The solution of last problem meets the solution of a vector boundary value Riemann—Hilbert problem with the matrix coefficient, having points of branchings. These difficulties have been overcome only in 1990 in our work  where the analytical solution of the Smoluchowsky problem has been received.

By the method developed in , further problems about temperature jump in molecular gases , and also in metal  have been analytically solved.

Along with the Smoluchowsky’ problem the big interest represents studying of behaviour of gas at weak evaporation (condensation) from a surface. These problems are called as the generalized Smoluchowsky’ problem in view of that boundary conditions in these problems differ slightly.

Let us notice, that in works [17, 18] the various kinetic models were used, in particular, the model of Shakhov (or, S-model, see ).

For the solution of boundary half-space problems with accommodation some methods  -  have been developed, allowing to receive the solution of this problem with any degree of accuracy.

In the present work the generalized source-method from  extends on a vector case to which the problem about temperature jump is reduced. Thus the effective method of the solution of boundary problems with mirror - diffusion boundary conditions (Maxwell conditions) develops. We will notice, that the method from  has already been applied in problems of electrodynamics of plasma  and in condensate problems of Bose—Einstein.

At the heart of an offered method the idea lays to include the boundary condition in the form of a source in the kinetic equation.

The method basis consists in the following. At first in half-space are formulated a vector problem about the temperature jump with boundary Maxwell conditions. Then unknown function continuations in conjugated half-space in the even method on spatial and on velocity variables. In half-space also are formulated the problem about temperature jump.

Now let us expand unknown function (which we will name also distribution function) on two composed: Chapman—Enskog’ distribution function and the second part of function distributions , correspoding to continuous spectrum (see  )

 h(x,μ)=has(x,μ)+hc(x,μ)

().

Owing to that Chapman—Enskog’ distribution function there is a linear combination of discrete solutions of the initial equation, function also is the solution of the kinetic equations. Function vanishes in zero far from wall. On a wall this function satisfies to boundary Maxwell condition.

Further we will transform the equation for function . We include in this equation boundary condition on wall for function in the form of a member of source-type laying in a plane .

We will underline, that function satisfies to the received equation in both conjugated half-spaces and .

We solve this equation in the second and the fourth quarters of a phase plane as the linear differential equation of the first order, considering known the right part of the equation . From the received solutions we found the boundary values of unknown function at , entering into the equation.

Now we expand by Fourier integrals unknown function , an unknown right part and Dirac delta-function. Boundary values of the unknown functions are thus expressed by the same integral on the spectral density functions .

Substitution of Fourier integrals in the kinetic equation and expression for the right part leads to the vector characteristic system of equations. If to exclude from this system the spectral density of function , we will receive vector Fredholm integral equation of the second sort.

Believing the gradient of the logarithm of temperature is setting, we will expand the unknown quantities of temperature and concentration jumps and also spectral density by polynoms on degrees of coefficient of diffusion (these are Neumann’s polynoms). On this way we receive system of the hooked equations on coefficients of polynoms for spectral density. Thus all equations on coefficients of spectral density have singularity (a pole of the second order in zero). Excepting these singularities consistently, we will construct all members of the polynoms for quantities of temperature and concentration jumps and for spectral density .

## 2 Statement problem

Let the rarefied one-nuclear gas occupies half-space over the flat firm surface laying in a plane . Far from a wall the logarithmic gradient of temperature is set

 gT=(dlnTdx)x=+∞.

We take the stationary kinetic equation of relaxation type with collisional integral BGK (Bhatnagar, Gross and Krook)

 vx∂f(x,v)∂x=feq(x,v)−f(x,v)τ,

where is the time between two consecutive collisions of molecules, is the collisional frequence of gaseous molecules, is the equilibrium distribution function,

 feq(x,v)=n(x)(m2πkT)3/2exp(−m2kT(x)v2),

where is the mass of molecule, is the Boltzmann constant, is the gas temperature,

 T(x)=2E(x)3kn(x),E(x)=∫m2v2f(x,v)d3v,

is the gas number density (concentration),

 n(x)=∫f(x,v)d3v.

Further we will be linearize the kinetic equation and search distribution function in the form

 f(x,v)=f0(v)(1+φ(x,v)),

where is the absolute Maxwellian,

 f0(v)=n0(m2πkT0)3/2exp(−mv22kT0),

where are number density and gas temperature in some point, for example, in origin of coordinates.

Let us be linerize distribution of numerical density and temperature concerning parametres and

 n(x)=n0+δn(x),T(x)=T0+δT(x).

According to (2.1) for distribution of numerical density we have

 n(x)=∫f0(v)[1+φ(x,v)]d3v=n0+δn(x),

where

 n0=∫f0(v)d3v,δn(x)=∫f0(v)φ(x,v)d3v.

For distribution of temperature we receive

 T(x)=23kn(x)∫mv22f0(v)[1+φ(x,v)]d3v.

We notice that

 n0n(x)=1−δnn0+o(h),h→0.

 T(x)=T023n0(1−δnn0)∫mv22kT0f0(v)[1+φ(x,v)]d3v=
 =2T03n0(1−δnn0)∫mv22kT0f0(v)d3v+2T03n0(1−δnn0)∫mv22kT0f0(v)φ(x,v)d3v.

We notice that

 23n0∫mv22kT0f0(v)d3v=1.

Hence, for relative change of temperature it is had

 δTT0=−δnn0+23n0∫mv22kT0f0(v)φ(x,v)d3v.

We will be linearize equilibrium function of distribution

 feq(v)=f0(v)[1+δn(x)n0+(mv22kT0−32)δT(x)T0].

We receive the following equation after linearizing the kinetic BGK–equation according to (2.1)

 vx∂φ(x,v)∂x=ν[δn(x)n0+(mv22kT0−32)δT(x)T0−φ(x,v)].

Let us enter dimensionless velocities and parametres — dimensionless velocity of molecules , where , dimensionless time , dimensionless coordinate

 x1=ν√2kT0mx=xvTτ=xl,

where is the mean free path of gaseous molecules, is the thermal velocity of the molecules movements, having an order of velocity of a sound.

Now the kinetic equation will be transformed to the form

 Cx∂φ∂x1+φ(x1,C)=δn(x1)n0+(C2−32)δT(x1)T0.

Here

 δn(x1)n0=1π3/2∫e−C2φ(x1,C)d3C,
 δT(x1)T0=23π3/2∫e−C2(C2−32)φ(x1,C)d3C.

Further a variable we will designate again through .

Let us transform the linear kinetic equation to the form

 Cx∂φ∂x+φ(x,v)=1π3/2∫K(C,C′)φ(x,C)e−C′2d3C′

with kernel

It is easy to check up, that the equation (2.2) has the following partial solutions

 φ1(x,μ)=1,φ2(x,μ)=C2−32,φ3(x,μ)=(x−Cx)(C2−52).

Let us construct asymptotic Chapman—Enskog distribution in the form of the linear combination of partial solutions of the equation (2.2) with arbitrary constants

 φas(x,μ)=A0+A1(C2−32)+A2(x−μ)(C2−52),

where are arbitrary constants.

For finding of these constants we will take advantage of definition of the macroscopical parametres. From definition of numerical density (concentration)

 n(x)=∫f(x,v)d3v

follows, that the extrapolated concentration of gas on the wall is equal

 ne=nas(0)=∫fas(0,v)d3v=∫f0(v)(1+φas(0,v))d3v=
 =n0(β/π)3/2∫exp(−C2)(1+φas(0,C))d3v.

From here we have

 nen0=π−3/2∫exp(−C2)(1+φas(0,C))d3C,

or

 nen0=1+π−3/2∫exp(−C2)φas(0,C)d3C.

Hence, the quantity of jump of concentration is searched under the formula

 ne−n0n0=εn=π−3/2∫exp(−C2)φas(0,C)d3C.

Substituting expression (2.3) in this equality, we have

 εn=A0.

Setting gradient of temperature far from a wall means, that temperature distribution in half-space looks like

 T(x)=Te+(dTdx)x=+∞x=Te+GTx,% \T2A\cyrg\T2A\cyrd\T2A\cyreGT=(dTdx)x=+∞.

This distribution we will present in the form

 T(x)=T0(TeT0+gTx)=T0(1+Te−T0T0+gTx),x→+∞,

or

 T(x)=T0(1+εT+gTx),x→+∞,

where

 εT=Te−T0T0

is the required quantity of temperature jump.

From expression (2.4) it is visible, that relative change of temperature far from walls it is described by linear function

 δTas(x)T0=T(x)−T0T0=εT+gTx,x→+∞.

Relative change of temperature we will present in the form

 δT(x)T0=23π−3/2∫exp(−C2)(C2−32)φ(x,C)d3C.

Far from a wall relative change of temperature transforms as follows

 δTas(x)T0=23π−3/2∫exp(−C2)(C2−32)φas(x,C)d3C.

Substituting in this equality expression (2.3) for , we find, that

 δT(x)T0=A2+A3x(x→+∞).

Comparing expressions (2.5) and (2.6), we find, that

Thus, asymptotic part of function of distribution (at ) it is constructed and on the basis stated above transforms in the form

 φas(x,C)=εn+εT(C2−32)+gT(x−Cx)(C2−52).

Let us formulate down boundary conditions to the equation (2.2). At first let us formulate mirror–diffusion boundary condition on a wall for full function of distribution

 f(+0,v)=qf0(v)+(1−q)f(+0,−vx,vy,vz),vx>0.

Here is the accommodation coefficient, i.e. a part of the molecules flying after reflexion from a wall with Maxwell distribution on velocities, is the part of the molecules reflected from a wall purely mirror.

Using (2.1), from here we receive a boundary condition of a problem onto wall

 φ(0,C)=(1−q)φ(x,−Cx,Cy,Cz),Cx>0.

Let us demand, that far from a wall distribution function passed into Chapman—Enskog distribution with coordinate growth

 f(x,v)=f0(v)[1+εn+εT(mv22kT0−32)+gT(x−√m2kT0vx)(mv22kT0−52)],x→+∞.

From here according to (2.1) for function we receive the following boundary conditions

 φ(x,μ)=φas(x,μ)+o(1),x→+∞.

Here is the asymptotic Chapman—Enskog distribution, entered above.

So, the boundary problem about finding of jumps of temperature and concentration of gas (vapor) over a flat surface consists in finding of the such solution of the equation (2.2), which satisfies to boundary conditions (2.7) and (2.8).

## 3 Reduction to vector boundary problem

If to use substitution

 φ(x,C)=h1(x,μ)+(C2−32)h2(x,μ),μ=Cx,

that equation (2.2) is reduced breaks up to two equations

 μ∂h1∂x+h1(x,μ)=
 =1√π∞∫−∞e−μ′2[h1(x,μ′)+(μ′2−12)h2(x,μ′)]dμ′

and

 μ∂h2∂x+h2(x,μ)=

This equation we will present in the vector form

 μ∂h∂x+h(x,μ)=1√π∞∫−∞e−μ′2K(μ′)h(x,μ′)dμ′.

Here is the vector-column

 h(x,μ)=(h1(x,μ)h2(x,μ)),

and matrix kernel have the following form

 K(μ)=⎛⎜ ⎜⎝1(μ2−12)23(μ2−12)23[(μ2−12)2+1]⎞⎟ ⎟⎠.

The right part of the equation (3.2)

 U(x)=1√π∞∫−∞e−μ′2K(μ′)h(x,μ′)dμ′

has clear physical sense. The vector-column looks like

 U(x)=⎛⎜ ⎜ ⎜ ⎜⎝δn(x)n0δT(x)T0⎞⎟ ⎟ ⎟ ⎟⎠,

i.e., components of this vector consist of the relative changes of numerical density of gas and relative change temperatures (concerning equilibrium values ). It is possible to present this vector in the form

 U(x)=Uas(x)+Uc(x),

where

 Uas(x)=⎛⎜ ⎜ ⎜ ⎜⎝δnas(x)n0δTas(x)T0⎞⎟ ⎟ ⎟ ⎟⎠,Uc(x)=⎛⎜ ⎜ ⎜ ⎜⎝δnc(x)n0δTc(x)T0⎞⎟ ⎟ ⎟ ⎟⎠.

According to (3.1) from (2.7) and (2.8) for the vector-functions we receive the following vector boundary conditions

 h(+0,μ)=(1−q)h(+0,−μ),μ>0,

and

 h(x,μ)=has(x,μ)+o(1),x→+∞,

where

 has(x,μ)=(εnεT)+gT(x−μ)(−11).

Function is the solution of the equation (3.2). Hence, if to search for the solution of the equation (3.2) in the form

 h(x,μ)=has(x,μ)+hc(x,μ),

then function is still searched from the equation (3.2)

 μ∂hc∂x+hc(x,μ)=1√π∞∫−∞e−μ′2K(μ′)hc(x,μ′)dμ′.

and boundary conditions (3.3) and (3.4) will be transformed thanking (3.5) to the following form

 hc(+0,μ)=h+0(μ)+(1−q)hc(+0,−μ),μ>0,
 hc(+∞,μ)=0,0=(00).

Here

 h+0(μ)=−q(εnεT)+(2−q)g+Tμ(−11).

Let us solve further the problem consisting of the solution of the equation (3.6) with boundary conditions (3.7) – (3.10).

## 4 Kinetic equation with source

For solution of this problem the auxiliary problem is required to us in "negative"  half-space . That it to formulate, we will continue function as follows

 h(x,μ)=h(−x,−μ).

Let us notice, that at continuation (4.1) logarithmic temperature gradient , which for "positive"  half-spaces we will designate through , changes the sign

 g−T=(dlnTdx)x=−∞=−(dlnTdx)x=+∞=−g+T.

Besides, we will notice, that function automatically satisfies to equality (4.1): . This equality means, that the equality (3.1) is carried out for function : .

Hence, boundary conditions in "negative"  space are formulated as follows

 hc(−0,μ)=h−0(μ)+(1−q)hc(−0,−μ),μ<0,
 hc(−∞,μ)=0.

Here

 h−0(μ)=−q(εnεT)+(2−q)g−Tμ(−11).

Let us unite both problems — in "positive"  and "negative"  half-spaces — in one, having included boundary conditions in the kinetic equation by means of member of type of the source

Here

 Uc(x)=1√π∞∫−∞e−μ2K(μ)hc(x,μ)dμ,
 h±0(μ)=−q(εnεT)+(2−q)g+T|μ|(−11),
 hc(∓0,μ)=limx→∓0,±x<0hc(x,μ),±μ>0.

These function are finding from equalities

 h+c(x,μ)=−1μ+∞∫xet/μUc(t)dt,h−c(x,μ)=1μx∫−∞et/μUc(t)dt.

## 5 Vector Fredholm equation of second sort

The solution of the equations (4.2) and (4.3) we search in the form of Fourier integrals

 Uc(x)=12π∞∫−∞eikxE(k)dk,δ(x)=12π∞∫−∞eikxdk,
 hc(x,μ)=12π∞∫−∞eikxΦ(k,μ)dk.

From equalities (4.3) and (5.1) follows, that

 E(k)=1√π∞∫−∞e−μ2K(μ)Φ(k,μ)dk.

Two following equalities follow from equalities (4.4)

 h±c(0,μ)=12π∞∫−∞E(k1)dk11+ik1μ=1π∞∫0E(k1)dk11+k21μ2.

From the kinetic equation (4.2) by means of (5.4) it is found

 Φ(k,μ)=E(k)1+ikμ−q(εnεT)|μ|1+ikμ+
 +(2−q)gT(−11)μ21+ikμ−|μ|1+ikμq√π∞∫0E(k1)dk11+k21μ2.

Substituting (5.5) in (5.3), we come to the vector integral Fredholm equation of the second sort

 L(k)E(k)=−q^T1(k)(εnεT)+(2−q)gT^T2(k)(−11)−qπ∞∫0^J(k,k1)E(k1)dk1.

Here is the dispersion matrix-function

 L(k)=E2−1√π∞∫−∞e−μ2K(μ)dμ1+ikμ=
 =E2−2√π∞∫0e−μ2K(μ)dμ1+k2μ2=E2−^T0(k),

where is the unit matrix of the second order,

 ^Tn(k)=2√π∞∫0e−μ2K(μ)μndμ1+k2μ2,n=1,2,⋯,

the matrix kernel of integral Fredholm equation is defined by integral expression

 ^J(k,k1)=2√π∞∫0e−μ2K(μ)μdμ(1+k2μ2)(1+k21μ2).

It is obvious, that

 ^J(k,0)=^T1(k),^J(0,k1)=^T1(k1).

## 6 Solution of vector Fredholm equation

For the solution of the equation (5.6) we will search in the form of Neumann’s polynoms

 E(k)=(2−q)gT[E0(k)+E1(k)q+E2(k)q2+⋯+Em(k)qm]
 (εnεT)=2−qqgT(ε∘n+ε1nq+ε2nq2+⋯+εmnqmε∘T+ε1Tq+ε2Tq2+⋯+εmTqm).

Let us substitute (6.1) and (6.2) in the equation (5.6). We receive system of the hooked equations

 L(k)E0(k)=−^T1(k)(ε∘nε∘T)+^T2(k)(−11),
 L(k)E1(k)=−^T1(k)(ε1nε1T)−1π∞∫0^J(k,k1)E0(k1)dk1,
 ........................
 L(k)Em(k)=−^T1(k)(εmnεmT)−1π∞∫0^J(k,k1)Em−1(k1)dk1,m=1,2,⋯

Let us calculate in an explicit form the matrixes entering into the equation (5.6)

 L(k)=(1001)−2√π∞∫−∞e−μ2⎛⎜ ⎜⎝1μ2−1223(μ2−12)23(μ4−μ2+54)⎞⎟ ⎟⎠dμ1+k2μ2=
 =⎛⎜ ⎜⎝1−T0(k)−T2(k)+12T0(k)−23(T2(k)−12T0(k))1−23(T4(k)−T2(k)+54T0(k))⎞⎟ ⎟⎠.

Here integrals are entered

 Tm(k)=2√π∞∫0e−μ2μmdμ1+k2μ2,m=0,1,2,⋯.

Let us notice, that the dispersion matrix is proportional to . Let us notice, that

 2√π∞∫0e−t2K(t)dt=1√π∞∫−∞e−t2K(t)dt=E2.

Therefore the dispersion matrix–function is equal

 L(k)=E2−^T0(k)=2√π∞∫0e−t2K(t)dt−2√π∞∫0e−t2K(t)1+k2t2dt=
 =k22√π∞∫0e−t2K(t)t2dt1+k2t2=k2^T2(k).

Let us write out matrix elements

 T211(k)=T2(k),T212(k)=2√π∞∫0e−t2(t2−1/2)t2dt1+k2t2=T4(k)−12T2(k),
 T221(k)=23T212(k)=23(T4(k)−12T2(k)),T222(k)=23(T6(k)−T4(k)+54T2(k)).

Therefore

 ^T2(k)=⎛⎜ ⎜⎝T2(k)T4(k)−12T2(k)23(T4(k)−12T2(k))23(T6(k)−T4(k)+54T2(k))⎞⎟ ⎟⎠.

Further we find a matrix

 ^T1(k)=⎛⎜ ⎜⎝T1(k)T3(k)−12T1(k)23(T3(k)−12T1(k))23(T5(k)−T