Boundary problems for one-dimensional kinetic equation with constant collision frequency
aa r X i v : . [ m a t h - ph ] A p r Boundary problems for one-dimensional kinetic equationwith constant collision frequencyA. L. Bugrimov , A. V. Latyshev and A. A. Yushkanov Faculty of Physics and Mathematics,Moscow State Regional University, 105005,Moscow, Radio str., 10A
Abstract
For the one-dimensional linear kinetic equation analytical solutions of prob-lems about temperature jump and weak evaporation (condensation) over flatsurface are received. The equation has integral of collisions BGK (Bhatnagar,Gross and Krook) and constant frequency of collisions of molecules. Distributionof concentration, mass speed and temperature is received.
Key words: kinetic equation, frequency of collisions, preservation laws,separation of variables, characteristic equation, dispersion equation, eigenfunc-tions, analytical solution, boundary problems.PACS numbers: 05.60.-k Transport processes, 51.10.+y Kinetic and transporttheory of gases,
Introduction
In work [1] the linear one-dimensional kinetic equation with integralof collisions BGK (Bhatnagar, Gross and Krook) and frequency of colli-sions, affine depending on the module velocity of molecules has beenentered. Preservation laws of numerical density (concentration) of mole-cules, momentum of molecules and energy have been thus used.In [1] the theorem about structure of general solution of the enteredequation has been proved. f akul − f m @ mgou.ru avlatyshev @ mail.ru yushkanov @ inbox.ru In work [2], being continuation of [1], are received exact solutions ofthe problem on temperature jump and weak evaporation (condensation)in rarefied gas for kinetic equation with frequency of collisions, affinedepending on the module velocity of molecules.In the present work which is continuation of [1] and [2], exact solutionsof the problem about temperature jump and weak evaporation (conden-sation) in the rarefied gas are received. Here the one-dimensional equationwith constant frequency of collisions is used. This equation is a specialcase of the kinetic equation with frequency of collisions, affine dependingon the module velocity of molecules.These two problems following [3] we will name the generalized Smo-luchowsky’ problem , or simply the Smoluchowsky problem.Let us stop on history of exclusively analytical solutions of the genera-lized Smoluchowsky’ problem.For simple (one-nuclear) rarefied gas with a constant frequency ofcollisions of molecules the analytical solution of the generalized of Smo-luchowsky’ problems it is received in [4].In [5] the generalized of Smoluchowsky’ problem was analyticallysolved for simple rarefied gas with frequency of collisions the molecules,linearly depending on the module of molecular velocity. In [6] the problemabout strong evaporation (condensation) with constant frequency ofcollisions has been analytically solved.Let us notice, that for the first time the problem about temperaturejump with frequency of collisions of molecules, linearly depending onthe module molecular velocity, was analytically solved by Cassel andWilliams in work [7] in 1972.Then in works [8, 9, 10] the generalized Smoluchowsky’ problem alsoanalytical solution for case of multinuclear (molecular) gases has beenreceived.In works [11, 12, 13] the problem about behaviour of the quantumBoze-gas at low temperatures (similar to the temperature jump problem for electrons in metal) is considered. We used the kinetic equation withexcitation fonons agrees to N.N. Bogolyubov.In works [14, 15] the problem about temperature jump for electronsof degenerate plasmas in metal has been solved.In work [16] the analytical solution of the Smoluchowsky’ problemfor quantum gases it has been received.In work of Cercignani and Frezzotti [17] the Smoluchowsky’ problemit was considered with use of the one-dimensional kinetic equations. Thefull analytical solution of Smoluchowsky’ problem with use of Cercignani—Frezotti equation it has been received in work [18].At the same time there is an unresolved problem about temperaturejump and concentration with use of the BGK–equation with arbitrarydependence of frequency on velocity, in spite of on obvious importanceof the decision of a problem in similar statement.In the present work attempt to promote in this direction is made.Here the case of the affine dependence of collision frequency on molecularvelocity in models of one-dimensional gas is considered. Model of one-dimensional gas gave the good consent with the results devoted to thethree-dimensional gas [18].Let us start with statement problem. Then we will give the solutionof the Smoluchowsky’ problem for the one-dimensional kinetic equationwith frequency of collisions, affine depending on the module of molecularvelocity.
1. Statement of the problem and the basic equations
Let us start with statement of a problem Smoluchowsky for the one-dimensional kinetic equation with frequency of collisions, affine dependingon the module velocity of molecules.Let us begin with the general statement. Let gas occupies half-space x > . The surface temperature T s and concentration of sated steam ofa surface n are set. Far from a surface gas moves with some velocity u , being velocity of evaporation (or condensation), also has the temperaturegradient g T = (cid:16) d ln Tdx (cid:17) x =+ ∞ . It is necessary to define jumps of temperature and concentrationdepending on velocity and temperature gradient.In a problem about weak evaporation it is required to define tempera-ture and concentration jumps depending on velocity, including a tempe-rature gradient equal to zero, and velocity of evaporation (condensation)is enough small. The last means, that u ≪ v T . Here v T is the heat velocity of molecules, having order of soundvelocity order, v T = 1 √ β s , β s = m k B T s ,m is the mass of molecule, k B is the Boltzmann constant.In the problem about temperature jump it is required to define tem-perature and concentration jumps depending on a temperature gradient,thus evaporation (condensation) velocity it is considered equal to zero,and the temperature gradient is considered as small. It means, that lg T ≪ , l = τ v T , τ = 1 ν , where l is the mean free path of gas molecules, τ is the mean relaxationtime, i.e. time between two consecutive collisions of molecules.Let us unite both problems (about weak evaporation (condensation)and temperature jump) in one. We will assume that the gradient oftemperature is small (i.e. relative difference of temperature on lengthof mean free path is small) and the velocity of gas in comparison withsound velocity is small. In this case the problem supposes linearizationand distribution function it is possible to search in the form f ( x, v ) = f ( v )(1 + h ( x, v )) , where f ( v ) = n s (cid:16) m πk B T s (cid:17) / exp h − mv k B T s i is the absolute Maxwellian.Let us pass in the equation (1.1) to dimensionless velocity C = p βv = vv T and dimensionless coordinate x ′ = ν r m k B T s x = xl The variable x ′ let us designate again through x .We take the linear kinetic equation [1] µ ∂h∂x + (1 + √ πa | µ | ) h ( x, µ ) == (1 + √ πa | µ | ) 1 √ π ∞ Z −∞ e − µ ′ (1 + √ πa | µ ′ | ) q ( µ, µ ′ , a ) h ( x, µ ′ ) dµ ′ . (1 . Here q ( µ, µ ′ , a ) is the kernel of equation, q ( µ, µ ′ , a ) = r ( a ) + r ( a ) µµ ′ + r ( a )( µ − β ( a ))( µ ′ − β ( a )) ,r ( a ) = 1 a + 1 , r ( a ) = 22 a + 1 , r ( a ) = 4( a + 1)4 a + 7 a + 2 ,β ( a ) = 2 a + 12( a + 1) ,a is the arbitrary positive paramater, a < + ∞ .Let us notice, that at a → the equation (1.1) passes in the equation µ ∂h∂x + h ( x, µ ) = 1 √ π ∞ Z −∞ e − µ ′ q ( µ, µ ′ ) h ( x, µ ) dµ (1 . with kernel q ( µ, µ ′ ) = 1 + 2 µµ ′ + 2 (cid:16) µ − (cid:17)(cid:16) µ ′ − (cid:17) . This equation is one-dimensional BGK-equation with constant frequ-ency of collisions.Let us consider the second limiting case of the equation (1.1). We willreturn to expression of frequency of collisions also we will copy it in theform ν ( µ ) = ν (1 + √ πa | µ | ) = ν + ν | µ | , where ν = √ πν a. Let us tend ν to zero. In this limit the quantity a tends to + ∞ ,because a = ν √ πν . It is easy to see, that in this limit lim a → + ∞ (1 + √ πa | C ′ | ) q ( µ, µ ′ , a ) = √ π | µ ′ | q ( µ, µ ′ ) , where q ( µ, µ ′ ) = 1 + µµ ′ + ( µ − µ ′ − . The equation (1.1) will thus be copied in the form µ | µ | ∂h∂x + h ( x , µ ) == ∞ Z −∞ e − µ ′ | µ ′ | [1 + µµ ′ + ( µ − µ ′ − dµ ′ . (1 . In this equation x = ν p β s x = xl , l = v T τ , τ = 1 ν . This equation is the one-dimensional kinetic equation with the frequ-ency of collisions proportional to the module of the molecular velocity.
2. Kinetic equation with constant collision frequency.Statement of boundary problem
Rectilinear substitution it is possible to check up, that the kineticequation (1.1) has following four private solutions h ( x, µ ) = 1 ,h ( x, µ ) = µ,h ( x, µ ) = µ ,h ( x, µ ) = (cid:16) µ − (cid:17)(cid:16) x − µ ) . Let us consider, that molecules are reflected from a wall purely dif-fusively, i.e. they are reflected from a wall with Maxwell distribution byvelocities, i.e. f ( x, v ) = f ( v ) , v x > . From here we receive for function h ( x, C ) condition h (0 , µ ) = 0 , µ > . (2 . Condition (2.1) is the first boundary condition to the equation (1.2).For asymptotic distribution of Chapmen—Enskog we will search inthe form of a linear combination of its partial solutions with unknowncoeffitients h as ( x, µ ) = A + A µ + A (cid:16) µ − (cid:17) + A (cid:16) µ − (cid:17) ( x − µ ) . (2 . We consider the distribution of number density n ( x ) = ∞ Z −∞ f ( x, v ) dv = ∞ Z −∞ f ( v )(1 + h ( x, v )) dv = n + δn ( x ) . Here n = ∞ Z −∞ f ( v ) dv, δn ( x ) = ∞ Z −∞ f ( v ) h ( x, v ) dv. From here we receive that δn ( x ) n = 1 √ π ∞ Z −∞ e − µ h ( x, µ ) dµ. We denote n e = n √ π ∞ Z −∞ e − µ (1 + h as ( x = 0 , µ )) dµ. From here we receive that ε n ≡ n e − n n = 1 √ π ∞ Z −∞ e − µ h as ( x = 0 , µ ) dµ. (2 . The quantity ε n is the unknown jump of concentration.Substituting (2.2) in (2.3), we find, that ε n = A . (2 . From definition of dimensional velocity of gas u ( x ) = 1 n ( x ) ∞ Z −∞ f ( x, v ) vdv we receive, that in linear approximation dimensional mass velocity isequal U ( x ) = 1 √ π ∞ Z −∞ e − µ h ( x, µ ) µdµ. Setting "far from a wall" velocity of evaporation (condensation), let uswrite U = 1 √ π ∞ Z −∞ e − µ h as ( x, µ ) µdµ. (3 . Substituting in (2.5) distribution (2.2), we receive, that A = 2 U. (2 . Let us consider temperature distribution T ( x ) = 2 kn ( x ) ∞ Z −∞ m v − u ( x )) f ( x, v ) dv. From here we find, that δT ( x ) T = − δnn + 2 √ π ∞ Z e − µ h ( x, µ ) µ dµ == 2 √ π ∞ Z e − µ h ( x, µ )( µ −
12 ) dµ.
From here follows, that at x → + ∞ asymptotic distribution is equal δT as ( x ) T = 2 √ π ∞ Z e − µ h as ( x, µ )( µ −
12 ) dµ. (2 . Setting of the gradient of temperature far from a wall means, thatdistribution of temperature looks like T ( x ) = T e + (cid:16) dTdx (cid:17) x =+ ∞ · x = T e + G T x, where G T = (cid:16) dTdx (cid:17) + ∞ . This distribution we will present in the form T ( x ) = T s (cid:16) T e T s + g T x (cid:17) = T s (cid:16) T e − T s T s + g T x (cid:17) , x → + ∞ , where g T = (cid:16) d ln Tdx (cid:17) x =+ ∞ , or T ( x ) = T s (1 + ε T + g T x ) , x → + ∞ , where ε T = T e − T s T s is the unknown temperature jump.From expression (2.7) is visible, that relative change of temperaturefar from a wall is described by linear function δT as ( x ) T s = T ( x ) − T s T s = ε T + g T x, x → + ∞ (2 . Substituting (2.2) in (2.7), we receive, that δT as ( x ) T s = A + A x. (2 . Comparing (2.7) and (2.10), we find A = ε T , A = g T . So, asymptotic function of Chapmen—Enskog’ distribution is const-ructed h as ( x, µ ) = ε n + ε T + 2 U µ + (cid:16) µ − (cid:17) [ ε T + g T ( x − µ )] . Now we will formulate the second boundary condition to the equation(1.2) h ( x, µ ) = h as ( x, µ ) + o (1) , x → + ∞ . (2 . Now we will formulate the basic boundary problem, which is gene-ralized Smoluchowsky’ problem. This problem consists in finding ofsuch solution of the kinetic equation (2.2) which satisfies to boundaryconditions (2.1) and (2.11).
3. Eigenvalues and eigenfunctions
Seperation of variables in the equation (1.2), taken in the form Раз-деление переменных в уравнении (1.2), взятое в виде h η ( x, µ ) = exp (cid:16) − xη (cid:17) Φ( η, µ ) , η ∈ C , (3 . reduces this equation to the characteristic ( η − µ )Φ( η, µ ) = η √ π n ( η ) + 2 η √ π µn ( η ) + 2 η √ π (cid:16) µ − (cid:17) n ( η ) , (3 . where η, µ ∈ ( − α, + α ) ,n ( η ) = ∞ Z −∞ e − µ ′ Φ( η, µ ) dµ, n ( η ) = ∞ Z −∞ e − µ ′ µ Φ( η, µ ) dµ, n ( η ) = ∞ Z −∞ e − µ ′ µ Φ( η, µ ) dµ are the zeroes, first and second moments of eigenfunction Φ( η, µ ) .Multiplying the characteristic equation (3.1) on e − µ ′ and integratingon all real axis, we receive, that n ( η ) ≡ . Multiplying the characteristic equation (3.1) on µ ′ e − µ ′ and integratingon all real axis, we receive, that n ( η ) ≡ . We obtain the characteristic equation ( η − µ )Φ( η, µ ) = η √ π (cid:16) − µ (cid:17) ∞ Z −∞ e − µ Φ( η, µ ) dµ. Let us accept further the following normalization condition for theeigenfunctions Φ( η, µ ) : n ( η ) ≡ ∞ Z −∞ e − µ ′ Φ( η, µ ) dµ = 1 . (3 . Now the characteristic equation becomes ( η − µ )Φ( η, µ ) = η √ π (cid:16) − µ (cid:17) . (3 . Eigenfunctions of the continuous spectrum filling by the continuousfashion the interval ( −∞ , ∞ ) , We find [19] in space of the generalizedfunctions Φ( η, µ ) = η √ π (cid:16) − µ (cid:17) P η − µ + e η λ ( η ) δ ( η − µ ) , η ∈ ( −∞ , ∞ ) . (3 . Here λ ( η ) is the dispersion fuction, defined by equation (3.3), P x − isthe distribution, meaning principal value of integral at intrgration x − , δ ( x ) is the Dirac function, λ ( z ) = 1 + z √ π ∞ Z −∞ e − τ / − τ τ − z dτ = −
12 + (cid:16) − z (cid:17) λ ( z ) ,λ ( z ) = 1 + z √ π ∞ Z −∞ e − τ dττ − z = 1 √ π ∞ Z −∞ e − τ τ dττ − z . Apparently from the solution of the characteristic equation, continuousspectrum of the characteristic equation is the set σ µ = { η : −∞ < η < + ∞} . By definition by the discrete spectrum of the characteristic equationis set of zero of dispersion function.Expanding dispersion function in Laurent series in a vicinity infinitelyremote point, we are convinced, that it in this point has zero of the fourthorder. Applying an argument principle from the theory of functionscomplex variable, it is possible to show, that other zero, except z i = ∞ , dispersion function not has. Thus, the discrete spectrum of thecharacteristic equations consists of one point z i = ∞ , which multiplicityis equal four, σ d = { z i = ∞} . To point z i = ∞ , as to a 4-fold point of the discrete spectrum,corresponds the following four discrete (partial) solutions the kineticdecision (1.2): h ( x, µ ) , h ( x, µ ) , h ( x, µ ) and h ( x, µ ) .Let us result Sokhotsky formulas for the difference and the sum ofthe boundary values of dispersion function from above and from belowon the cut ( −∞ , + ∞ ) : λ + ( µ ) − λ − ( µ ) = 2 √ πiµe − µ (cid:16) − µ (cid:17) , µ ∈ ( −∞ , + ∞ ) , and λ + ( µ ) + λ − ( µ )2 = −
12 + (cid:16) − µ (cid:17) λ ( µ ) , µ ∈ ( −∞ , + ∞ ) . On the real axis function λ ( µ ) is calculated on to the formula λ ( µ ) = 1 − µ Z e − µ (1 − t ) dt.
4. Homogeneous boundary Riemann problem
Here we will consider homogeneous boundary Riemann problem fromtheories of functions complex variable which is required further. Thisproblem consists in the finding of such function X ( z ) , which is analyticalin a complex plane, cut along the real positive half-axis C ′ = C \ R + .Boundary values of this function from above and from below on thereal half-axis satisfy to the boundary condition X + ( µ ) X − ( µ ) = λ + ( µ ) λ − ( µ ) , µ > . (4 . We note that | λ + ( µ ) | = | λ − ( µ ) | , λ + ( µ ) = λ − ( µ ) , µ ∈ ( −∞ , + ∞ ) . Let us enter the principal value of argument θ ( µ ) = arg λ + ( µ ) ,defined in the cut plane C ′ and fixed in zero by the condition θ (0) = 0 .Then λ + ( µ ) = | λ + ( µ ) | e iθ ( µ ) , λ − ( µ ) = | λ − ( µ ) | e − iθ ( µ ) . Noe the problem (4.1) will be rewritten in the form X + ( µ ) X − ( µ ) = e iθ ( µ ) , µ > . (4 . Taking the logarithm of the problem (4.2), we receive following nume-rable family of problems of the finding of analytical function on its zerojump on the positive real half-axis R + = { µ : µ > } : ln X + ( µ ) − ln X − ( µ ) = 2 iθ ( µ ) + 2 πik, k ∈ Z , µ > . (4 . Рис. 1. The angle θ = θ ( µ ) monotonously increases from to π . The solution of problems (4.3) is expressed by integral of Cauchy type ln X ( z ) = 1 π ∞ Z θ ( µ ) + kπµ − z dµ, k ∈ Z . Let us notice, that the angle θ ( µ ) is on a semiaxis R + monotonouslyincreasing function from to 2 π . It means, that index of coefficient G ( µ ) = λ + ( µ ) λ − ( µ ) of homogeneous Riemann problem (5.1) on the positivereal half-axis is equal to unit κ = κ ( G ) = 12 π h arg G ( µ ) i(cid:12)(cid:12)(cid:12) ∞ = 1 . From here follows, that among family of solutions (4.3) only one (at k = − ) is expressed by the converging integral of Cauchy type ln X ( z ) = 1 π ∞ Z θ ( µ ) − πµ − z dµ. (4 . We denote further V ( z ) = ln X ( z ) , whence X ( z ) = e V ( z ) . Let us redefine the received solution as follows X ( z ) = 1 z e V ( z ) . (4 . Let us notice, that the solution (4.5) isbounded function in a vicinitythe point z = 0 . Really, at z → it is had V ( z ) = − θ (0) − ππ ln z + O ( z ) , z → , where O ( z ) is the bounded function in a vicinity the point z = 0 . Hence,in a vicinity of the point z = 0 function X ( z ) = e O ( z ) is the boundedfunction.
5. Analytical splution of the boundary problem for kineticequation with constant collision frequency
Here we will prove the theorem about the analytical solution of thebasic boundary problem (1.2), (2.1) and (2.11).
Theorem.
Boundary problem (1.2), (2.1) and (2.11) has the uniquedecision, representable in the form of the sum linear combinations ofdiscrete (partial) solutions of this equation and integral on the continuousspectrum from eigenfunctions correponding to the continuous spectrum h ( x, µ ) = h as ( x, µ ) + ∞ Z exp (cid:16) − xη (cid:17) Φ( η, µ ) A ( η ) dη. (5 . In equality (5.1) ε n and ε T are unknown coefficient (discrete spectrum), U and g T are the given qualities, A ( η ) is the unknown function (coefficientof the continuous spectrum). Coefficients of discrete and continuous spectra are subject to findingfrom boundary conditions.Expansion (5.1) it is possible to present in the explicit form in classicalsense h ( x, µ ) = ε n + ε T + 2 U µ + (cid:16) µ − (cid:17) [ ε T + g T ( x − µ )]++ e µ − x/µ λ ( µ ) A ( µ ) + (cid:16) − µ (cid:17) √ π ∞ Z e − x/η ηA ( η ) η − µ dη. (5 . ′ ) Proof.
Let us substitute expansion (5.1) in the boundary condition(2.1). We receive the integral equation h as (0 , µ ) + ∞ Z Φ( η, µ ) A ( η ) dη = 0 , < µ < ∞ . In the explicit form this equation looks like h as (0 , µ ) + e µ λ ( µ ) A ( µ )++ (cid:16) − µ (cid:17) √ π ∞ Z ηA ( η ) η − µ dη = 0 , < µ < ∞ . (5 . Here h as (0 , µ ) = ε n + ε T + 2 U µ + (cid:16) µ − (cid:17) ( ε T − g T µ ) . Let us enter auxiliary function N ( z ) = 1 √ π ∞ Z ηA ( η ) η − z dη, (5 . for which according to formulas Sokhotsky it is had N + ( µ ) − N − ( µ ) = 2 √ πiµA ( µ ) , < µ < ∞ , (5 . N + ( µ ) + N − ( µ )2 = 1 √ π ∞ Z ηA ( η ) η − µ dη, < µ < ∞ . (5 . Let us transform the equation (5.2), considering formulas Sokhotskyfor dispersion function and according to equalities (5.4) and (5.5). Wereceive non-uniform the boundary Riemann conditionConsidering formulas Sokhotsky for dispersive function, Let’s transformthe equation (5.6) to a non-uniform regional problem Римана: λ + ( µ ) h(cid:16) − µ (cid:17) N + ( µ ) + h as (0 , µ ) i −− λ − ( µ ) h(cid:16) − µ (cid:17) N − ( µ ) + h as (0 , µ ) i = 0 , < µ < ∞ . (5 . Let us consider the corresponding homogeneous boundary Riemannproblem X + ( µ ) X − ( µ ) = λ + ( µ ) λ − ( µ ) , < µ < ∞ . (5 . The solution of this problem which is bounded and not disappearingin points z = 0 and z = α it is resulted in the previous item X ( z ) = 1 z exp V ( z ) , (5 . where V ( z ) = 1 π ∞ Z θ ( µ ) − πµ − z dµ, (6 . θ ( µ ) = arg λ + ( µ ) is the principal value of argument, fixed by condition θ (0) = 0 .Let us transform the problem (5.6) by means of the homogeneousproblem (5.7) to the problem of finding of analytical function on itsjump on the cut X + ( µ ) h(cid:16) − µ (cid:17) N + ( µ ) + h as (0 , µ ) i == X − ( µ ) h(cid:16) − µ (cid:17) N − ( µ ) + h as (0 , µ ) i , < µ < ∞ . (5 . Let us find singularities of the boundary condition (5.10). Consideringbehaviour of the functions entering into boundary condition (5.10), we receive the common solution corresponding to boundary problem (cid:16) z − (cid:17) N ( z ) = h as (0 , z ) + C + C zX ( z ) , (5 . where C and C are arbitrary constans, and h as (0 , z ) = ε n + ε T + 2 U µ + (cid:16) z − (cid:17) ( ε T − g T z ) . Let us notice, that the solution (5.11) has in infinitely removed point z = ∞ a pole of the third order, while function N ( z ) , defined by equality(5.3), has in this point a pole the first order.That the solution (5.11) could be accepted in quality of function N ( z ) , defined by equality (5.3), we will lower order of a pole at thesolution (5.11) from three to unit.Then let us equate values of the left and right parts of equality (5.11)in points of the real axis µ , = ± p / .Decomposition is required to us V ( z ) = V z + V z + · · · , z → ∞ . Here V n = − π ∞ Z τ n − [ θ ( τ ) − π ] dτ, n = 1 , , · · · . Lowering an order of pole on two units in infinitely removed point atthe soltution (5.11), we find, that C = V g T − ε T ,C = g T . The pole of function in the point µ = p / is eliminated by twolimiting conditions from above and from below the real axis, for thispoint lays on a cut (the real axis) C + C µ + X + ( µ )( ε n + ε T + 2 U µ ) = 0 , (5 . and C + C µ + X − ( µ )( ε n + ε T + 2 U µ ) = 0 , (5 . The point µ = − µ does not belong to the cut, therefore we receive C − C µ + X ( − µ )( ε n + ε T − U µ ) = 0 . (5 . We take a half-sum of conditions (5.12) and (5.13) C + C µ + X + ( µ ) + X − ( µ )2 ( ε n + ε T + 2 U µ ) = 0 (5 . We note that X ± ( µ ) = 1 µ e V ± ( µ ) , where V ± ( µ ) = V ( µ ) ± i [ θ ( µ ) − π ] . Hence X + ( µ ) + X − ( µ )2 = X ( µ ) . Taking into account this equality from the equations (5.14) and (5.15)it is received expressions of required qualities of jump of temperatureand jump concentration ε T = g T h V − µ X ( µ ) + X ( − µ ) X ( µ ) − X ( − µ ) i − U µ X ( µ ) X ( − µ ) X ( µ ) − X ( − µ ) (5 . and ε n = g T h V + µ X ( µ ) + X ( − µ ) − X ( µ ) − X ( − µ ) i −− U µ X ( µ ) + X ( − µ ) − X ( µ ) X ( − µ ) X ( µ ) − X ( − µ ) . (5 . Coefficient of continuous spectrum A ( η can be found on the basis ofthe formula Sokhotsky (5.4) and formulas of the difference of boundaryvalues N ( z ) , received with the help solutions (5.11) N + ( µ ) − N − ( µ ) = C + C µµ − / h X + ( µ ) − X − ( µ ) i . (5 . From equalities (5.4) and (5.18) we find coefficient of the continuousspectrum √ πiηA ( η ) = g T ( V + η ) − ε T η − / h X + ( η ) − X − ( η ) i . (5 . We note that X + ( η ) − X − ( η ) = − iX ( η ) sin θ ( η ) . By means of this equality coefficient of the continuous spectrum(5.19) it is definitively equal ηA ( η ) = − ( V + η ) g T − ε T √ πX ( η )( η − /
2) sin θ ( η ) . So, all coefficients of expansion (5.1) are established. On to construction,expansion (5.1) satisfies to boundary conditions (2.1) and (2.11). Thatfact, that expansion (5.1) satisfies to the equation (1.2), it is checkeddirectly.Uniqueness of decomposition (5.1) is proved by a method from theopposite. The theorem is proved.
6. Temperature jump and weak evaporation (condensation).Numerical calculations
Numerical calculations of coefficients V n lead to the following results V = 2 . · · · , V = 2 . , V = 3 . · · · , and also µ = 1 . · · · , X ( µ ) = 3 . · · · , X ( − µ ) = 0 . · · · . Now it is required to us following Theorem.
For dispersion function λ ( z ) takes place the followingfactorization formula λ ( z ) = − X ( z ) X ( − z ) , z ∈ C ′ ,λ + ( µ ) = − X + ( µ ) X ( − µ ) , µ > ,λ − ( µ ) = − X ( µ ) X + ( − µ ) , µ < . Proof.
This theorem is proved in the same way, as well as the proofof similar theorems in our works [3].By means of this theorem it is found exact value X ( µ ) X ( − µ ) = 23 . We rewrite thiese formulas (5.16) and (5.17) in the form ε T = K T T g T + K T U (2 U ) , ε n = K nT g T + K nU (2 U ) . Here K T T = V − µ X ( µ ) + X ( − µ ) X ( µ ) − X ( − µ ) ,K T U = − µ X ( µ ) X ( − µ ) X ( µ ) − X ( − µ ) ,K nT = V + µ X ( µ ) + X ( − µ ) − X ( µ ) − X ( − µ ) ,K nU = − µ X ( µ ) + X ( − µ ) − X ( µ ) X ( − µ ) X ( µ ) − X ( − µ ) . Now it is easy to find that K T T = 1 . , K T U = − . ,K nT = 3 . , K nU = − . . Hence, coefficient of jump of temperature and jump of concentrationare calculated under formulas ε T = 1 . g T − . U ) , (6 . and ε n = − . g T − . U ) . (6 .
7. Limiting transition in the general formulas
Here we will show, that if we will make limiting transition in thegeneral formulas (7.8) and (7.9) from [2] at a → , we in accuracy let usreceive formulas (5.16) and (5.17). We will remind, that formulas (7.8)and (7.9) for temperature and concentration jump are received for thecase frequencies of collisions, affine depending on the module of velocity.We will do this transition for temperature jump in the case of weakevaporations (i.e. for the case g T = 0 ).Let us transform the formula (5.16) to the form ε T = − U µ X ( − µ ) − µ X ( µ ) . (7 . From formula (7.8) from [2] we receive ε T = − U V + lim a → K , (7 . where K = 12 πi /a Z (cid:16) X + ( µ ) − X − ( µ ) (cid:17) dµQ ( µ, µ ) . For coincidence of equalities (7.1) and (7.2) it is required to proveequality lim a → K = − V + 12 µ X ( − µ ) − µ X ( µ ) (7 . We note that in considering case at a → : r ( a ) → , β ( a ) → , ω ( a ) → , r ( a ) → , r ( a ) → . Hence, Λ ( µ ) ≡ , Λ ( µ ) ≡ , Q ( µ, µ ) = 32 − µ . Therefore lim a → = 12 πi ∞ Z (cid:16) X + ( µ ) − X − ( µ ) (cid:17) dµQ ( µ, µ ) . We form the difficult contour Γ R , consisting of the external circles | z | = R with radius R = 1 /a , and two internal circles | z ± µ | = a withradiuses r = a .Under Cauchy theorem I Γ R dµX ( µ ) Q ( µ, µ ) = 0 . We denote lim a → K = I .
Then from the previous equality it is received I = I + I − I , where I = lim a → πi I | µ + µ | = a dµX ( µ ) Q ( µ, µ ) ,I = lim a → πi I | µ − µ | = a dµX ( µ ) Q ( µ, µ ) ,I = lim a → πi I | µ | =1 /a dµX ( µ ) Q ( µ, µ ) . It is easy to see, that I = 1 Q ′ ( µ, µ ) X ( µ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) µ = − µ = 12 µ X ( µ ) , in the same way I = − µ X ( µ ) . For calculation of integral I we will spread out its subintegral functionby Laurent series in a vicinity of infinitely remote point and let us presentit in the form Q ( z, z ) X ( z ) = − V z + ϕ ( z ) , µ → ∞ , where ϕ ( z ) = O ( z − ) , z → ∞ . Hence, this integral equals πi I | µ | =1 /a dµX ( µ ) Q ( µ, µ ) = V + 12 πi I | µ | =1 /a ϕ ( µ ) dµ. The limit of last integral is equal in this equality to zero at a → owing to previous asmptotic, therefore I = V .So, equality (7.3) is established.
8. Distribution of macroparameters of gas in "half-space"
Let us consider distribution of concentration, mass velocity and tem-perature depending on coordinate x .Let us begin with concentration distribution (numerical density) δn ( x ) n = 1 √ π ∞ Z −∞ e − µ h ( x, µ ) dµ == 1 √ π ∞ Z −∞ e − µ h h as ( x, µ ) + ∞ Z e − x/η Φ( η, µ ) A ( η ) dη i dµ == ε T − g T x + 1 √ π ∞ Z e − x/η dη ∞ Z −∞ e − µ Φ( η, µ ) A ( η ) dη. Having taken advantage of the normalizing equality (3.3), we receive δn ( x ) n = ε T − g T x + 1 √ π ∞ Z e − x/η A ( η ) dη. Let us transform coefficient of the continuous spectrum. Noticing,that sin θ ( η ) = √ πηe − η (3 / − η ) | λ + ( η ) | . Hence, A ( η ) = ( V + η ) g T − ε T X ( η ) | λ + ( η ) | e − η . Thus, we come to following distribution of concentration δn ( x ) n = [ K T T (1 − m ( x )) − x + V m ( x ) + m ( x )] g T ++ K T U (1 − m ( x ))(2 U ) . Here m k ( x ) = 1 √ π ∞ Z e − x/η − η η k dηX ( η ) | λ + ( η ) | , k = 0 , . Mass velocity U ( x ) is equal everywhere at x > to given on infinityquantity of velocity, i.e. U ( x ) ≡ U . Really, we have U ( x ) = 1 √ π ∞ Z −∞ e − µ h ( x, µ ) µdµ == 1 √ π ∞ Z −∞ e − µ " h as ( x, µ ) + ∞ Z e − x/η Φ( η, µ ) A ( η ) dη µdµ. From here we obtain that U ( x ) = U + ∞ Z e − x/η A ( η ) dη ∞ Z −∞ e − µ µ Φ( η, µ ) dµ ≡ U, because the first moment of eigenfunction Φ( η, µ ) is equal to zero as ithas been shown above.We consider the distribution of temperature δT ( x ) T = ε T + g T x + 2 √ π ∞ Z −∞ e − µ (cid:16) µ − (cid:17) h ( x, µ ) dµ = = ε T + g T x + 2 √ π ∞ Z e − x/η A ( η ) dη ∞ Z −∞ e − µ (cid:16) µ − (cid:17) Φ( η, µ ) dµ. Considering, that the second moment of eigenfunction Φ( η, µ ) is equalto zero, from here we receive the temperature distribution δT ( x ) T = ε T + g T x − √ π ∞ Z e − x/η A ( η ) dη == [ x − m ( x ) + K T T (1 + m ( x ))] g T + K T U (1 + m ( x ))(2 U ) .
9. Conclusion
In the present work the analytical solution of boundary problemsfor the one-dimensional kinetic equation with constant frequency ofcollisions of molecules is considered. We consider the solution of thegeneralized Smoluchowsky problem (problems about temperature jumpand weak evaporation (condensation)). Numerical calculations are done.Distribution of concentration, mass speed and temperature is received.
REFERENCES [1]
Latyshev A.V., Yushkanov A.A.
The kinetic one-dimensionalequation with frequency of collisions, affine depending on themodule molecular velocity// arXiv:1403.2068v1 [math-ph] 9 Mar2014, 20pp.[2]
Latyshev A.V., Yushkanov A.A.
Boundary problems for the one-dimensional kinetic equation with frequency of collisions, affinedepending on the module velocity// ArXiv:1403. 5854, [math-ph]23 Mar 2014, 30 pp. [3] Latyshev A.V., Yushkanov A.A.
Kinetic equatios type Williamsand their exact solutions. Monograph. M.: MGOU (Moscow StateRegional University), 2004, 271 p.[4]
Latyshev A.V.
Application of case’ method to the solution of linearkinetic BGK equation in a problem about temperature jump//Appl. math. and mechanics. 1990. V. 54. Issue 4. P. 581–586.[russian][5]
Latyshev A.V., Yushkanov A.A.
Boundary problems for modelBoltzmann equation with frequency proportional to velocity ofmoleculs// Izvestiya Russian Academy of Science. Ser. Mechanika,Fluid and Gas (Russian "Fluids Dynamics"). 1993. №6. 143-155 pp.[russian][6]
Latyshev A.V., Yushkanov A.A.
Analytical solution of the problemabout strong evaporation (condensation)// Izvestiya RussianAcademy of Science. Ser. Mechanika, Fluid and Gas (Russian"Fluids Dynamics"). 1993. №6. 143-155 pp. [russian][7] cassell J.S., Williams M.M.R.
An exact solution of the temperatureslip problem in rarefied gases// Transport Theory and Statist.Physics, 2(1), 81–90 (1972).[8]
Latyshev A.V., Yushkanov A.A.
Temperature jump and weakevaporatuion in molecular gases// J. of experim. and theor. physics.1998. V. 114. Issue. 3(9). P. 956–971. [russian][9]
Latyshev A.V., Yushkanov A.A.
The Smoluchowski problem inpolyatomic gases// Letters in J. of Tech. Phys. 1998. V. 24. №17.P. 85–90. [russian][10]
Latyshev A.V., Yushkanov A.A.
Analytic solutions of boundaryvalue problem for model kinetic equatins// Math. Models of Non-Linear Excitations, Transfer, Dynamics, and control in condensed Systems and Other Media. Edited by L.A. Uvarova and A.V.Latyshev. Kluwer Academic. New York–Moscow. 2001. P. 17–24.[11]
Latyshev A.V., Yushkanov A.A.
Smolukhowski problem fordegenerate Bose gases// Theoretical and Mathematical Physics.Springer New York. Vol. 155, №3,June, 2008, pp. 936 – 948.[12]
Latyshev A.V., Yushkanov A.A.
Temperature jump in degeneratequantum gases in the presence of the Bose–Einstein condensate //Theor. and Mathem. Phys. 2010. V. 162(1). P. 95–105 [russian][13]
Latyshev A.V., Yushkanov A.A.
Temperature jump in degeneratequantum gases with the Bogoliubov excitation energy and in thepresence of the Bose–Einstein condensate, Theoret. and Math.Phys., 165:1 (2010), 1358–1370.[14]
Latyshev A.V., Yushkanov A.A.
Smoluchowski problem forelectrons in metal// Theor. and Mathem. Phys. 2005, январь, Т.142. №1. С. 93–111 [russian][15]
Latyshev A.V., Yushkanov A.A.
Smoluchowski problem for metalswith mirror-diffusive boundary conditions //Theoretical andMathematical Physics October 2009, Volume 161, Issue 1, pp. 1403-1414.[16]
Latyshev A.V., Yushkanov A.A.
Boundary value problems forquantum gases. Monograph M.: MGOU, 2012, 266 p.[russian][17]
Сercignani С., Frezzoti A.
Linearized analysis of a one-speedB.G.K. model in the case of strong condensation// BulgarianAcademy of sci. theor. appl. mech. Sofia. 1988. V.XIX. №3. 19-23P.[18]
Latyshev A.V., Yushkanov A.A.
Analytical solution of one-dimensional problem about moderate strong evaporation (and condensation) in half-space// Appl. mech. and tech. physics. 1993.№1. 102-109 p. [russian][19] Vladimirov V.S., Zharinov V.V.
Equations of mathematicalphysics. M.: Fizmatlit. 2000. 399 с.[russian][20]