Research ArticleBoltzmannrsquos Six-Moment One-Dimensional Nonlinear SystemEquations with the Maxwell-Auzhan Boundary Conditions

A Sakabekov1 and Y Auzhani2

1Research Laboratory ldquoApplied Modeling Oil and Gas Fieldsrdquo Kazakh-British Technical University 59 Tole Bi StreetAlmaty 050000 Kazakhstan2Al-Farabi Kazakh National University 71 Al-Farabi Avenue Almaty 050040 Kazakhstan

Correspondence should be addressed to A Sakabekov asakabekovkbtukz

Received 2 February 2016 Revised 1 June 2016 Accepted 5 June 2016

Academic Editor Peter G L Leach

Copyright copy 2016 A Sakabekov and Y AuzhaniThis is an open access article distributed under theCreative CommonsAttributionLicense which permits unrestricted use distribution and reproduction in anymedium provided the originalwork is properly cited

We prove existence and uniqueness of the solution of the problem with initial and Maxwell-Auzhan boundary conditions fornonstationary nonlinear one-dimensional Boltzmannrsquos six-moment system equations in space of functions continuous in timeand summable in square by a spatial variable In order to obtain a priori estimation of the initial and boundary value problem fornonstationary nonlinear one-dimensional Boltzmannrsquos six-moment system equations we get the integral equality and then use thespherical representation of vector Then we obtain the initial value problem for Riccati equation We have managed to obtain aparticular solution of this equation in an explicit form

1 Introduction

In case of one-atom gas any macroscopic system duringprocess of its evolution to an equilibrium state passes 3 stagesinitial transition period described in terms of a full functionof system distribution the kinetic period by means of a one-partial distribution function and the hydrodynamic periodbymeans of five firstmoments of the distribution function Inkinetic regime the behavior of a rarefied gas in space of timeand velocity is described by Boltzmannrsquos equation It is knownfrom gas dynamics that in most of the encountered problemsthere is no need to use detailed microscopic gas descriptionwith help of the distribution function Therefore it is naturalto look for a less detailed description using macroscopichydrodynamic variables (density hydrodynamic velocitytemperature etc) As these variables are defined in terms ofmoments of the distribution function we are faced with theproblem of analyzing the various moments of Boltzmannrsquosequation

Note that Boltzmannrsquos moment equations are intermedi-ate between Boltzmann (kinetic theory) and hydrodynamiclevels of description of state of the rarefied gas and formclass of nonlinear partial differential equations Existence ofsuch class of equations was noticed by Grad [1 2] in 1949

He obtained the moment system by expanding the particledistribution function in Hermite polynomials near the localMaxwell distribution Grad used Cartesian coordinates ofvelocities and Gradrsquos moment system contained as coeffi-cients such unknown hydrodynamic characteristics like den-sity temperature average speed and so forth Formulation ofboundary conditions for Gradrsquos system is almost impossibleas the characteristic equations for various approximations ofGradrsquos hyperbolic system contain unknown parameters likedensity temperature and average speed However 13- and20-moment Grad equations are widely used in solving manyproblems of the kinetic theory of gases and plasma

Boltzmannrsquos equation is equivalent to an infinite systemof differential equations relative to the moments of theparticle distribution function in the complete system ofeigenfunctions of linearized operator As a rule we limit studyby finitemoment system of equations because solving infinitesystem of equations does not seem to be possible

Finite system ofmoment equations for a specific task witha certain degree of accuracy replaces Boltzmannrsquos equationIt is necessary also roughly to replace the boundary condi-tions for the particle distribution function by a number ofmacroscopic conditions for the moments that is there arisesthe problem of boundary conditions for a finite system of

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2016 Article ID 5834620 8 pageshttpdxdoiorg10115520165834620

2 Journal of Applied Mathematics

equations that approximate the microscopic boundary con-ditions for Boltzmannrsquos equation The question of boundaryconditions for a finite system of moment equations canbe divided into two parts how many conditions must beimposed and how they should be prepared From micro-scopic boundary conditions for Boltzmannrsquos equation therecan be obtained an infinite set of boundary conditionsfor each type of decomposition However the number ofboundary conditions is determined not by the number ofmoment equations that is it is impossible for example totake as many boundary conditions as equations although thenumber of moment equations affects the number of bound-ary conditions In addition the boundary conditions mustbe consistent with the moment equations and the resultingproblemmust be correct Boundary condition problems arisein the following tasks (1) moment boundary conditions inrarefied gas slip-flow problems (2) definition of boundaryconditions on the surfaces of streamlined rarefied gas (3)prediction aerospace aerodynamic characteristics of aircraftat very high speeds and at high altitudes and so forth

In work [3] we have obtained the moment system whichdiffers from Gradrsquos system of equations We used sphericalvelocity coordinates and decomposed the distribution func-tion into a series of eigenfunctions of the linearized collisionoperator [4 5] which is the product of Sonine polynomialsand spherical functions The resulting system of equationswhich correspond to the partial sum of series and whichwe called Boltzmannrsquos moment system of equations is anonlinear hyperbolic system in relation to the moments ofthe particles distribution function

The structure of Boltzmannrsquosmoment systemof equationscorresponds to the structure of Boltzmannrsquos equation namelythe differential part of the resulting system is linear in relationto the moments of the distribution function and nonlinearityis included as moments of collision integral [6]

The linearity of differential part of Boltzmannrsquos momentsystem of equations simplifies the task of formulation of theboundary conditions In work [3] a homogeneous boundarycondition for particles distribution function was approxi-mated and proved the correctness of initial and bound-ary value problem for nonlinear nonstationary Boltzmannrsquosmoment system of equations in three-dimensional spaceIn work [7] the initial and boundary value problem forone-dimensional nonstationary Boltzmannrsquos equation withboundary conditions of Maxwell was approximated by acorresponding problem for Boltzmannrsquos moment system ofequationsThe boundary conditions for Boltzmannrsquosmomentsystem of equations were calledMaxwell-Auzhan conditions

In work [8] a systematic nonperturbative derivation ofa hierarchy of closed systems of moment equations corre-sponding to any classical theory has been presented Thispaper is a fundamental work where closed systems ofmoment equations describe a transition regime Moreoverhydrodynamical model is used to describe charge trans-port in a generic compound semiconductor Compoundsemiconductors have found wide use in the microelectronicindustryThe evolution equation for macroscopic variables isobtained by taking moments of the transport equation [9ndash13]

The study of various problems for Boltzmannrsquos momentsystem of equations is an important and actual task in thetheory of a rarefied gas and other applications of the momentsystem equations The correctness of initial and boundaryproblems for Boltzmannrsquos moment system of equations withMaxwell-Auzhan boundary conditions is being studied forthe first time

2 Existence and Uniqueness of the Solutionsof Initial and Boundary Value Problem forSix-Moment One-Dimensional BoltzmannrsquosSystem of Equations with Maxwell-AuzhanBoundary Conditions

In this section we prove the existence and uniqueness ofsolutions of the initial and boundary value problem forsix-moment one-dimensional Boltzmannrsquos system of equa-tions with Maxwell-Auzhan boundary conditions in space offunctions continuous in time and summable in square byspatial variable Note that the theorem of the existence ofa global solution in time of the initial and boundary valueproblem for 3-dimensional nonlinear Boltzmannrsquos equationwith boundary conditions of Maxwell is proved in work [14]

Wewrite in an expanded form systemof one-dimensionalBoltzmannrsquos moment equations in the kth approximationwhich corresponds to decomposition of the particle distri-bution function by eigenfunctions of the linearized collisionoperator

120597119891119899119897

120597119905+

1

120572

120597

120597119909[119897(radic

2 (119899 + 119897 + 12)

(2119897 minus 1) (2119897 + 1)119891119899119897minus1

minus radic2 (119899 + 1)

(2119897 minus 1) (2119897 + 1)119891119899+1119897minus1

) + (119897 + 1)

sdot (radic2 (119899 + 119897 + 12)

(2119897 + 1) (2119897 + 3)119891119899119897+1

minus radic2119899

(2119897 + 1) (2119897 + 3)119891119899minus1119897+1

)] = 119868119899119897

2119899 + 119897 = 0 1 119896

(1)

where the moments 119868119899119897of a nonlinear collision operator are

expressed through coefficients of Talmi and Klebsh-Gordon[15 16] and have the form

119868119899119897= sum⟨119873

3119871311989931198973 119897 | 11989911989700 119897⟩

sdot ⟨1198733119871311989931198973 119897 | 1198991119897111989921198972 119897⟩ (

1198971011989720

1198970) (1205901198973minus 1205900)

sdot 11989111989911198971

11989111989921198972

(2)

where ⟨1198733119871311989931198973 119897 | 1198991119897111989921198972 119897⟩ are generalized coefficients

of Talmi (11989710119897201198970) are Klebsh-Gordon coefficients 120572 =

1radic119877120579 is the constant 119877 is Boltzmannrsquos constant and 120579 is theideal gas temperature

Journal of Applied Mathematics 3

For generalized coefficients of Talmi there exists a table[15] for each value of quantum number 120585 = 2119899 + 119897 from 0to 6 Moreover there is a program on IBM for calculation ofgeneralized coefficients of Talmi

If in (1) 2119899+ 119897 takes values from 0 to 3 we get Boltzmannrsquosmoment system equations in the third approximation Wewrite it in an expanded form

12059711989100

120597119905+

1

120572

12059711989101

120597119909= 0

12059711989102

120597119905+

1

120572

120597

120597119909(

2

radic311989101

+3

radic511989103

minus2radic2

radic1511989111) = 11986902

12059711989110

120597119905+

1

120572

120597

120597119909(minusradic

2

311989101

+ radic5

311989111) = 0

12059711989101

120597119905+

1

120572

120597

120597119909(11989100

+2

radic311989102

minus radic2

311989110) = 0

12059711989103

120597119905+

1

120572

120597

120597119909

3

radic511989102

= 11986903

12059711989111

120597119905+

1

120572

120597

120597119909(minus

2radic2

radic1511989102

+ radic5

311989110) = 11986911

119909 isin (minus119886 119886) 119905 gt 0

(3)

where 11989100

= 11989100(119905 119909) and 119891

01= 11989101(119905 119909) 119891

11= 11989111(119905 119909)

are the moments of the particle distribution function 11986902

=

(1205902minus 1205900)(1198910011989102

minus 1198912

01radic3)2 119869

03= (14)(120590

3+ 31205901minus

41205900)1198910011989103

+ (14radic5)(21205901+ 1205900minus 31205903)1198910111989102 and 119869

11=

(1205901minus 1205900)(1198910011989111

+ (12)radic531198911011989101

minus (radic2radic15)1198910111989102) are

moments of the collision integral where 1205900 1205901 1205902 and 120590

3

are the Fourier coefficients of the cross section expansionby the Legendre polynomials The first third and fourthequations of system (3) correspond to mass conservation lawmomentum conservation law and energy conservation lawrespectively

We study the correctness of initial and boundary valueproblem for six-moment one-dimensional Boltzmannrsquos sys-tem equations

120597119906

120597119905+ 119860

120597119908

120597119909= 1198691(119906 119908)

120597119908

120597119905+ 1198601015840120597119906

120597119909= 1198692(119906 119908)

119905 isin (0 119879] 119909 isin (minus119886 119886)

(4)

119906|119905=0

= 1199060(119909)

119908|119905=0

= 1199080(119909)

119909 isin [minus119886 119886]

(5)

(119860119908minus

+ 119861119906minus

)1003816100381610038161003816119909=minus119886 =

1

120573(119860119908+

minus 119861119906+

)1003816100381610038161003816119909=minus119886

minus(1 minus 120573) 120578

120572120573radic120587119865 119905 isin [0 119879]

(6)

(119860119908minus

minus 119861119906minus

)1003816100381610038161003816119909=119886 =

1

120573(119860119908+

+ 119861119906+

)1003816100381610038161003816119909=119886

+(1 minus 120573) 120578

120572120573radic120587119865 119905 isin [0 119879]

(7)

where

119860 =1

120572

(

(

1 0 0

2

radic3

3

radic5minus2radic2

radic15

minusradic2

30 radic

5

3

)

)

119861 =1

120572radic120587

((

(

radic2 radic2

3minus

1

radic3

radic2

32radic2 minus1

minus1

radic3minus1 3radic2

))

)

1198691(119906 119908) = (0 119869

02 0)1015840

1198692(119906 119908) = (0 119869

03 11986911)1015840

119906 = (11989100 11989102 11989110)1015840

119908 = (11989101 11989103 11989111)1015840

119865 = (1

4radic2

1

8radic6

1

8radic3)

1015840

(8)

1198601015840 is the transpose matrix 119861 is the positive definite

matrix 1199060(119909) = (119891

0

00(119909) 119891

0

02(119909) 119891

0

10(119909))1015840 and 119908

0(119909) =

(1198910

01(119909) 119891

0

03(119909) 119891

0

11(119909))1015840 are given initial vector-functions

119908+

119906+ are vector moments of falling to the boundary particle

distribution function119908minus 119906minusare vectormoments of reflectingfrom the boundary particle distribution function Equation(4) is vector matrix form of system equations (3)

It is possible to check through direct calculations that

det1198601= det(

0 119860

1198601015840

0) = 0 (9)

and matrix 1198601has three positive and the same number of

negative nonzero eigenvalues From (6)-(7) it follows that thenumber of boundary conditions on the left and right ends ofinterval (minus119886 119886) is equal to the number of positive andnegativeeigenvalues of the matrix 119860

1

Thus system (4) is a symmetric hyperbolic nonlinearpartial differential equations system Let us show that 119869

02

4 Journal of Applied Mathematics

is a sign-nondefined square form It is easy to check that1198910011989102

minus (1198912

01radic3) = (119862119880119880) where 119880 = (119906 119908)

1015840 and

119862 =

((((((((

(

01

20 0 0 0

1

20 0 0 0 0

0 0 0 0 0 0

0 0 0 minus1

radic30 0

0 0 0 0 0 0

0 0 0 0 0 0

))))))))

)

(10)

Eigenvalues of matrix 119862 are minus12 minus1radic3 0 0 0 and 12Therefore 119869

02is a sign-nondefined square form Similarly we

can show that 11986903

and 11986911

are also sign-nondefined squareforms

For problem (4)ndash(7) the following theorem takes place

Theorem 1 If 1198800= (1199060(119909) 119908

0(119909)) isin 119871

2

[minus119886 119886] then problem(4)ndash(7) has a unique solution in domain [minus119886 119886] times [0 119879]

belonging to the space 119862([0 119879] 1198712[minus119886 119886]) moreover

10038171003817100381710038171198801198712[minus119886119886]

minus 1199031

1003817100381710038171003817119862[0119879]le 1198621(10038171003817100381710038171198800

10038171003817100381710038171198712[minus119886119886] minus 1199031(0)) (11)

where 1198621is a constant independent from 119880 and 119879 sim

M(11988001198712[minus119886119886]

minus 1199031(0))minus1

) 1199031(119905) is the particular solution of

Riccati equation (15) in an explicit form

Proof Let 1198800isin 1198712

[minus119886 119886] Let us prove estimation (11) Wemultiply the first equation in system (4) by 119906 and the secondequation by 119908 and perform integration from minus119886 to 119886

1

2

119889

119889119905int

119886

minus119886

[(119906 119906) + (119908 119908)] 119889119909

+ int

119886

minus119886

[(119860120597119908

120597119909 119906) + (119860

1015840120597119906

120597119909 119908)] 119889119909

= int

119886

minus119886

[(1198691 119906) + (119869

2 119908)] 119889119909

(12)

After integration by parts we get

1

2

119889

119889119905int

119886

minus119886

[(119906 119906) + (119908 119908)] 119889119909 + (119906minus

119860119908minus

)1003816100381610038161003816119909=119886

minus (119906minus

119860119908minus

)1003816100381610038161003816119909=minus119886 = int

119886

minus119886

[(1198691 119906) + (119869

2 119908)] 119889119909

(13)

Taking into account boundary conditions (6)-(7) we rewriteequality (13) in the following form

1

2

119889

119889119905int

119886

minus119886

[(119906 119906) + (119908 119908)] 119889119909 + (119861119906minus

119906minus

)1003816100381610038161003816119909=119886

+ (119861119906minus

119906minus

)1003816100381610038161003816119909=minus119886 minus

1

120573((119860119908

+

minus 119861119906+

) 119906minus

)10038161003816100381610038161003816119909=minus119886

+1

120573((119860119908+

+ 119861119906+

) 119906minus

)1003816100381610038161003816119909=119886 + (119865

1 119906minus

)1003816100381610038161003816119909=119886

+ (1198651 119906minus

)1003816100381610038161003816119909=minus119886

= int

119886

minus119886

[(1198691(119906 119908) 119906) + (119869

2(119906 119908) 119908)] 119889119909

(14)

where 1198651= ((1 minus 120573)120578120572120573radic120587)119865

Let us use the spherical representation [17] of vector119880(119905 119909) = 119903(119905)120596(119905 119909) where 120596(119905 119909) = (120596

1(119905 119909) 120596

2(119905 119909))

1015840119903(119905) = 119880(119905 sdot)

1198712[minus119886119886]

and 1205961198712[minus119886119886]

= 1 Furthermoreassume that120596(119905 119909) isin 119862

2

[0 119879] where the value of119879we definebelow

Substituting the values 119906 = 119903(119905)1205961(119905 119909) and 119908 =

119903(119905)1205962(119905 119909) into (14) we have that

119889119903

119889119905+ 119903119875 (119905) = 119903

2

119876 (119905) minus 119891 (119905) (15)

where

119875 (119905) = (119861120596minus

1 120596minus

1)1003816100381610038161003816119909=119886 + (119861120596

minus

1 120596minus

1)1003816100381610038161003816119909=minus119886

+1

120573[(119860120596+

2 120596minus

1)1003816100381610038161003816119909=119886 + (119861120596

+

1 120596minus

1)1003816100381610038161003816119909=119886

+ (119861120596+

1 120596minus

1)1003816100381610038161003816119909=minus119886 minus (119860120596

+

2 120596minus

1)1003816100381610038161003816119909=minus119886]

119876 (119905) = int

119886

minus119886

[(1198691(1205961 1205962) 1205961) + (1198692(1205961 1205962) 1205962)] 119889119909

119891 (119905) = (1198651 120596minus

1)1003816100381610038161003816119909=119886 + (119865

1 120596minus

1)1003816100381610038161003816119909=minus119886

(16)

Let us study (15) with an initial condition

119903 (0) =10038171003817100381710038171198800

1003817100381710038171003817 =10038171003817100381710038171198800

10038171003817100381710038171198712[minus119886119886] (17)

Let 1199031(119905) be a particular solution of Riccati equation (15)Then

the general solution of (15) has the form

119903 (119905) = 1199031(119905) + exp(int

119905

0

(2119876 (120591) 1199031(120591) minus 119875 (120591)) 119889120591) [119862

minus int

119905

0

119876 (120591)

sdot exp(int

120591

0

(2119876 (120585) 1199031(120585) minus 119875 (120585)) 119889120585) 119889120591]

minus1

(18)

Hence taking into consideration initial condition (17) weobtain

Journal of Applied Mathematics 5

119903 (119905) = 1199031(119905) + exp(int

119905

0

(2119876 (120591) 1199031(120591) minus 119875 (120591)) 119889120591)[

110038171003817100381710038171198800

1003817100381710038171003817 minus 1199031(0)

minus int

119905

0

119876 (120591) exp(int

120591

0

(2119876 (120585) 1199031(120585) minus 119875 (120585)) 119889120585) 119889120591]

minus1

(19)

If 119877(119905) equiv int119905

0

119876(120591) exp(int1205910

(2119876(120585)1199031(120585) minus 119875(120585))119889120585)119889120591 le 0 forall119905

then the second term of 119903(119905) is bounded forall119905 isin [0 +infin) underthe condition that difference 119880

0 minus 1199031(0) is positive Further

we suppose a difference 1198800 minus 1199031(0) is a positive number Let

119877(119905) gt 0 We denote by 1198791the moment of time at which

110038171003817100381710038171198800

1003817100381710038171003817 minus 1199031(0)

minus int

1198791

0

119876 (120591) exp(int

120591

0

(2119876 (120585) 1199031(120585) minus 119875 (120585)) 119889120585) 119889120591

= 0

(20)

Then 119903(119905) minus 1199031(119905) is bounded forall119905 isin [0 119879] where 119879 lt 119879

1and

1198791sim M(1(119880

0 minus 1199031(0))) Analyzing structure of 119903

1(119905) (see

(31)) from equality (20) we obtain 119876(119905) gt 0 for any 119905 isin [0 119879]Now we found one particular solution of (15) The

nonlinear Riccati equation can always be reduced to a second-order linear ordinary differential equation [18]

11991010158401015840

+ 1198911(119905) 1199101015840

+ 1198912(119905) 119910 = 0 (21)

where

119910 (119905) = exp(minusint

119905

0

119876 (120591) 119903 (120591) 119889120591)

1198911(119905) =

119876 (119905) 119875 (119905) minus 1198761015840

(119905)

119876 (119905)

1198912(119905) = 119876 (119905) 119891 (119905)

(22)

Solution of this equation will lead to a solution of 119903(119905) =

minus1199101015840

119910119876(119905) of the original Riccati equation By substitution

119910 (119905) = 119911 (119905) exp(minus1

2int

119905

0

1198911(120591) 119889120591) (23)

we write (21) in the following form [18]

11991110158401015840

(119905) + 119886 (119905) 119911 (119905) = 0 (24)

where 119886(119905) = 1198912(119905) minus (12)119891

1015840

1(119905) minus (14)119891

2

1(119905)

If instead of1198911(119905) and119891

2(119905)we substitute values from (22)

then we get

119886 (119905) = 119876119891 (119905)

minus2 (1198762

1198751015840

minus 11987610158401015840

119876 + 11987610158402

) + (119876119875 minus 1198761015840

)2

41198762

(25)

In work [19] it was proven that function

119911 (119905) = 1 +

infin

sum

119896=1

119887119896(119905) (26)

is the solution of (24) where

1198871(119905) = minusint

119905

0

int

120585

0

119886 (120591) 119889120591 119889120585

119887119896(119905) = minusint

119905

0

int

120585

0

119886 (120591) 119887119896minus1

(120591) 119889120591 119889120585

(27)

and the series on the right hand side of (26) uniformlyconverge

Then

119910 = (1 +

infin

sum

119896=1

119887119896(119905)) exp(minus

1

2int

119905

0

1198911(120591) 119889120591)

1199031(119905) = minus

1199101015840

119910119876= minus

1199111015840

(119905) minus (12) 119911 (119905) 1198911(119905)

119911 (119905) 119876 (119905)

(28)

is a particular solution of Riccati equation (15)Since 1199111015840(0) = 0 value of

1199031(0) = minus

1199111015840

(0) minus (12) 119911 (0) 1198911(0)

119911 (0) 119876 (0)=

1198911(0)

2119876 (0) (29)

Now instead of 1198911(0) we substitute its value 119875(0) minus

1198761015840

(0)119876(0) and get 1199031(0) = (119875(0)119876(0) minus 119876

1015840

(0))21198762

(0)The value of 119876(119905) with 119905 = 0 is

119876 (0)

= int

119886

minus119886

((1198691(1205961 1205962) 1205961) + (1198692(1205961 1205962) 1205962))1003816100381610038161003816119905=0 119889119909

gt 0

(30)

since 119876(119905) gt 0 for any 119905 isin [0 119879]

An expansion form of the function

1199031(119905) =

1

119876 (119905)[

1

119911 (119905)int

119905

0

119886 (120591) 119911 (120591) 119889120591

+119876 (119905) 119875 (119905) minus 119876

1015840

(119905)

2119876 (119905)]

(31)

It is not difficult to prove using immediate substitutionthat function 119903

1(119905) satisfies Riccati equation (15)

From equality (19) we obtain1003816100381610038161003816119903 (119905) minus 119903

1(119905)

1003816100381610038161003816 le 119862 (10038171003817100381710038171198800

1003817100381710038171003817 minus 1199031(0)) (32)

Hence1003816100381610038161003816119880 (119905 sdot)

1198712[minus119886119886]

minus 1199031(119905)

1003816100381610038161003816 le 119862 (10038171003817100381710038171198800

1003817100381710038171003817 minus 1199031(0))

10038171003817100381710038171198801198712[minus119886119886]

minus 1199031

1003817100381710038171003817119862[0119879]le 119862 (

100381710038171003817100381711988001003817100381710038171003817 minus 1199031(0))

(33)

6 Journal of Applied Mathematics

We take as interval of the solution of problem (15)ndash(17) segment [0 1(119880

0 minus 119903

1(0))] since the integrand

119876(120591) exp(int1205910

(2119876(120585)1199031(120585) minus 119875(120585))119889120585) is bounded Hence forall119905 isin

[0 119879] (where 119879 asymp 1198791) a priori estimation (11) takes place

Now we prove the existence of solution for (4)ndash(7) usingGalerkin method Let V

119897(119909)infin

119897=1be a basis in space 1198712[minus119886 119886]

where the dimension of vector V119897(119909) is equal to the dimension

of vector119880 For each119898we define an approximate solution119880119898

of (4)ndash(7) as follows

119880119898=

119898

sum

119895=1

119888119895119898

(119905) V119895(119909) (34)

int

119886

minus119886

((120597119880119898

120597119905+ 1198601

120597119880119898

120597119909) V119894(119909)) 119889119909

= int

119886

minus119886

(119869 (119880119898) V119894(119909)) 119889119909 119894 = 1119898 119905 isin (0 119879]

(35)

119880119898

1003816100381610038161003816119905=0 = 1198800119898

(119909) 119909 isin 119877 (36)

(119860119908minus

119898∓ 119861119906minus

119898)1003816100381610038161003816119909=∓119886

=1

120573(119860119908+

119898plusmn 119861119906+

119898)1003816100381610038161003816119909=plusmn119886 plusmn

(1 minus 120573) 120578

120572120573radic120587119865

(37)

where 1198800119898

is the orthogonal projection in 1198712 of the function

1198800on the subspace spanned by V

1 V

119898

119869 (119880119898) = (119869

1(119906119898 119908119898) 1198692(119906119898 119908119898))1015840

(38)

We represent V119895(119909) in form V

119895(119909) = (V(1)

119895 V(2)119895

)1015840 where

V(1)119895

= (V1198951 V1198952 V1198953)1015840

V(2)119895

= (V1198954 V1198955 V1198956)1015840

(39)

The coefficients 119888119895119898

(119905) are determined from the equations

119898

sum

119895=1

119889119888119895119898

119889119905int

119886

minus119886

(V119895 V119894

) 119889119909 + 119888119895119898

[(119861Vminus(1)119895

Vminus(1)119894

)10038161003816100381610038161003816119909=119886

+ (119861Vminus(1)119895

Vminus(1)119894

)10038161003816100381610038161003816119909=minus119886

+ (119861Vminus(1)119894

Vminus(1)119895

)10038161003816100381610038161003816119909=119886

+ (119861Vminus(1)119894

Vminus(1)119895

)10038161003816100381610038161003816119909=minus119886

+1

120573((119860V+(2)119894

+ 119861V+(1)119894

) Vminus(1)119895

)10038161003816100381610038161003816119909=119886

minus1

120573((119860V+(2)119894

minus 119861V+(1)119894

) Vminus(1)119895

)10038161003816100381610038161003816119909=minus119886

+1

120573((119860V+(2)119895

+ 119861V+(1)119895

) Vminus(1)119894

)10038161003816100381610038161003816119909=119886

minus1

120573(119860V+(2)119895

minus 119861V+(1)119895

Vminus(1)119894

)10038161003816100381610038161003816119909=minus119886

+ (1198651 Vminus(1)119894

)10038161003816100381610038161003816119909=119886

+ (1198651 Vminus(1)119894

)10038161003816100381610038161003816119909=minus119886

+ (1198651 Vminus(1)119895

)10038161003816100381610038161003816119909=119886

+ (1198651 Vminus(1)119895

)10038161003816100381610038161003816119909=minus119886

]

minus int

119886

minus119886

((119860120597V(2)119894

120597119909 V(1)119895) + (119860

1015840120597V(1)119894

120597119909 V(2)119895

))119889119909

= int

119886

minus119886

(119869(

119898

sum

119895=1

119888119895119898V119895) V119894)119889119909

119894 = 1119898 119905 isin (0 119879]

(40)

119888119894119898

(0) = 119889119894119898 119894 = 1119898 (41)

where 119889119894119898

is the 119894th component of 1198800119898

We multiply (35) by 119888119894119898(119905) and sum over 119894 from 1 to119898

int

119886

minus119886

((120597119880119898

120597119905+ 1198601

120597119880119898

120597119909) 119880119898)119889119909

= int

119886

minus119886

(119869 (119880119898) 119880119898) 119889119909

(42)

With help of the above shown arguments we now provethat 119903

119898(119905) is bounded in some time interval [0 119879

119898] where

119880119898(119905 119909) = 119903

119898(119905)120596119898(119905 119909) 119879

119898asymp M((119880

0119898 minus 1199031198981

(0))minus1

) 119879119898

ge

119879 forall119898 and

10038171003817100381710038171003817

100381710038171003817100381711988011989810038171003817100381710038171198712[minus119886119886] minus 119903

1198981

10038171003817100381710038171003817119862[0119879]

le 1198622(10038171003817100381710038171198800

10038171003817100381710038171198712[minus119886119886] minus 1199031(0))

(43)

where 1198622is constant and independent from119898 and 119903

1198981(119905) is a

particular solution of Riccati equation about 119903119898(119905)

Problem (40)-(41) represents a Cauchy problem for theordinary system of differential equations Existence of thesolution of problem (40)-(41) follows from theory of ordi-nary system of differential equations (Picardrsquos existence anduniqueness theorem) [20]Hence the existence of the solutionof problem (34)ndash(37) follows

Thus according to estimation (43) the sequence 119880119898

minus

1199031198981

of approximate solutions of problem (5)ndash(7) is uniformlybounded in function space119862([0 119879] 1198712[minus119886 119886])Moreover thehomogeneous system of equations 120591119864+ (1120572)119860120585with respectto 120591 120585 has only a trivial solution Then it follows from resultsin [21] that 119880

119898minus 1199031198981

rarr 119880minus 1199031is week in 119862([0 119879] 119871

2

[minus119886 119886])

and 119869(119880119898) rarr 119869(119880) is week in 119862([0 119879] 119871

2

[minus119886 119886]) as 119898 rarr

infin Further it can be shown with standard methods that limitelement is a weak solution of problem (5)ndash(7)

The uniqueness of solution of problem (5)ndash(7) is provedby contradiction Let problem (5)ndash(7) have two different solu-tions 119906

1 1199081and 119906

2 1199082 We denote them again by 119906 = 119906

1minus 1199062

Journal of Applied Mathematics 7

and119908 = 1199081minus1199082 Then with respect to new values of 119906 and119908

we obtain the following problem

120597119906

120597119905+ 119860

120597119908

120597119909= 1198691(1199061 1199081) minus 1198691(1199062 1199082)

120597119908

120597119905+ 1198601015840120597119906

120597119909= 1198692(1199061 1199081) minus 1198692(1199062 1199082)

119909 isin (minus119886 119886)

(44)

119906|119905=0

= 0

119908|119905=0

= 0

(45)

(119860119908minus

+ 119861119906minus

)1003816100381610038161003816119909=minus119886 =

1

120573(119860119908+

minus 119861119906+

)1003816100381610038161003816119909=minus119886

(119860119908minus

minus 119861119906minus

)1003816100381610038161003816119909=119886 =

1

120573(119860119908+

+ 119861119906+

)1003816100381610038161003816119909=119886

(46)

We prove that solution of problem (44)ndash(46) is trivialHence the uniqueness of the solution of problem (5)ndash(7)follows

Using the method which was mentioned before we get(see equality (14))

1

2

119889

119889119905int

119886

minus119886

[(119906 119906) + (119908 119908)] 119889119909 + (119861119906minus

119906minus

)1003816100381610038161003816119909=119886

+ (119861119906minus

119906minus

)1003816100381610038161003816119909=minus119886 minus

1

120573((119860119908

+

minus 119861119906+

) 119906minus

)10038161003816100381610038161003816119909=minus119886

+1

120573((119860119908

+

+ 119861119906+

) 119906minus

)10038161003816100381610038161003816119909=119886

= int

119886

minus119886

[((1198691(1199061 1199081) minus 1198691(1199062 1199082)) 119906)

+ ((1198692(1199061 1199081) minus 1198692(1199062 1199082)) 119908)] 119889119909

(47)

Let us transform the integrand on the right side of theequality as

((1198691(1199061 1199081) minus 1198691(1199062 1199082)) 119906)

=1205902minus 1205900

2[(119891001

119891021

minus1198912

011

radic3) minus (119891

002119891022

minus1198912

012

radic3)] (119891

021minus 119891022

) =1205902minus 1205900

2[(119891001

119891021

minus 119891002

119891022

) minus1

radic3(1198912

011minus 1198912

012)] (119891021

minus 119891022

)

=1205902minus 1205900

2[(119891001

minus 119891002

) 119891021

+ 119891002

(119891021

minus 119891022

) minus1

radic3(119891011

minus 119891012

) (119891011

+ 119891012

)] (119891021

minus 119891022

)

((1198692(1199061 1199081) minus 1198692(1199062 1199082)) 119906) = ((119869

03(1199061 1199081)

minus 11986903

(1199062 1199082)) (119891031

minus 119891032

)) + ((11986911

(1199061 1199081)

minus 11986911

(1199062 1199082)) (119891111

minus 119891112

)) = (1

4(1205903+ 31205901

minus 41205900) [(119891001

minus 119891002

) 119891031

+ 119891002

(119891031

minus 119891032

)]

+1

4radic5(21205901+ 1205900minus 31205903) [(119891011

minus 119891012

) 119891021

+ 119891012

(119891021

minus 119891022

)]) (119891031

minus 119891032

) + (1205901minus 1205900)

sdot ((119891001

minus 119891002

) 119891111

+ 119891002

(119891111

minus 119891112

)

+ 05radic5

2[(119891101

minus 119891102

) 119891011

+ 119891102

(119891011

minus 119891012

)] minus radic2

15[(119891011

minus 119891012

) 119891021

+ 119891012

(119891021

minus 119891022

)]) (119891111

minus 119891112

)

(48)

Once again we use spherical representation of 119906 = 119903(119905)1199081(119905

119909) and 119908 = 119903(119905)1199082(119905 119909) where 119903(119905) = 119880(119905 sdot)

1198712[minus119886119886]

Thenconcerning 119903(119905) we obtain new initial value problem

119889119903

119889119905+ 119903119875 (119905) = 119903

2

1198761(119905) (49)

119903 (0) = 0 (50)

where 119875(119905) has the same value as in (15) and

1198761(119905) =

1205902minus 1205900

2int

119886

minus119886

[(120596001

minus 120596002

) 120596021

+ 120596002

(120596021

minus 120596022

) minus1

radic3(120596011

minus 120596012

) (120596011

+ 120596012

)] (120596021

minus 120596022

) 119889119909 +1

4(1205903+ 31205901

minus 41205900) int

119886

minus119886

[(120596001

minus 120596002

) 120596031

+ +120596002

(120596031

minus 120596032

)] (120596031

minus 120596032

) 119889119909 +1

4radic5(21205901+ 1205900minus 31205903)

sdot int

119886

minus119886

[(120596011

minus 120596012

) 120596021

+ 120596012

(120596021

minus 120596022

)]

sdot (120596031

minus 120596032

) 119889119909 + (1205901minus 1205900) int

119886

minus119886

[(120596001

minus 120596002

) 120596111

+ 120596002

(120596111

minus 120596112

) +1

2

8 Journal of Applied Mathematics

sdot radic5

2[(120596101

minus 120596102

) 120596011

+ 120596102

(120596011

minus 120596012

)]

minus radic2

15[(120596011

minus 120596012

) 120596021

+ 120596012

(120596021

minus 120596022

)]

sdot (120596111

minus 120596112

)] 119889119909

(51)

The general solution of (49) is

119903 (119905) = exp(minusint

119905

0

119875 (120591) 119889120591)

sdot [119862 minus int

119905

0

1198761(120591) exp(minusint

120591

0

119875 (120585) 119889120585) 119889120591]

minus1

(52)

Solution of (49) which is satisfying homogeneous condition(50) is trivial that is 119903(119905) = 0 Hence 119880

119862([0119879]1198712[minus119886119886])

= 0

and 1199061= 1199062and 119908

1= 1199082

The theorem is proved

Competing Interests

The authors declare that they have no competing interests

References

[1] H Grad ldquoOn the kinetic theory of rarefied gasesrdquo Communica-tions on Pure andAppliedMathematics vol 2 no 4 pp 331ndash4071949

[2] H Grad ldquoPrinciple of the kinetic theory of gasesrdquo in Thermo-dynamics of Gases vol 12 ofHandbuch der Physik pp 205ndash294Springer Berlin Germany 1958

[3] A Sakabekov Initial-Boundary Value Problems for the Boltz-mannrsquos Moment System Equations Gylym Almaty 2002

[4] C Cercignani Theory and Application of the Boltzmann Equa-tion Instituto di Matematica Milano Italy 1975

[5] M N KoganDynamic of Rarefied Gas NaukaMoscow Russia1967

[6] K Kumar ldquoPolynomial expansions in kinetic theory of gasesrdquoAnnals of Physics vol 37 no 1 pp 113ndash141 1966

[7] A Sakabekov and Y Auzhani ldquoBoundary conditions for theonedimensional nonlinear nonstationary Boltzmannrsquos momentsystem equationsrdquo Journal of Mathematical Physics vol 55Article ID 123507 2014

[8] C D Levermore ldquoMoment closure hierarchies for kinetictheoriesrdquo Journal of Statistical Physics vol 83 no 5-6 pp 1021ndash1065 1996

[9] G Mascali and V Romano ldquoA hydrodynamical model for holesin silicon semiconductors the case of non-parabolic warpedbandsrdquo Mathematical and Computer Modelling vol 53 no 1-2pp 213ndash229 2011

[10] G Mascali and V Romano ldquoA non parabolic hydrodynamicalsubband model for semiconductors based on the maximumentropy principlerdquoMathematical and Computer Modelling vol55 no 3-4 pp 1003ndash1020 2012

[11] V D Camiola and V Romano ldquo2DEG-3DEG charge transportmodel for MOSFET based on the maximum entropy principlerdquo

SIAM Journal on Applied Mathematics vol 73 no 4 pp 1439ndash1459 2013

[12] G Alı G Mascali V Romano and R C Torcasio ldquoA hydrody-namical model for covalent semiconductors with a generalizedenergy dispersion relationrdquo European Journal of Applied Math-ematics vol 25 no 2 pp 255ndash276 2014

[13] V D Camiola and V Romano ldquoHydrodynamical model forcharge transport in graphenerdquo Journal of Statistical Physics vol157 no 6 pp 1114ndash1137 2014

[14] S Mischler ldquoKinetic equations with Maxwell boundary condi-tionsrdquo Annales Scientifiques de lrsquoEcole Normale Superieure vol43 no 5 pp 719ndash760 2010

[15] V G Neudachin and U F SmirnovNucleon Association of EasyKernel Nauka Moscow Russia 1969

[16] MMoshinskyTheHarmonic Oscillator inModern Physics fromAtoms to Quarks 1960

[17] S I Pokhozhaev ldquoOn an approach to nonlinear equationrdquoDoklady Akademii Nauk SSSR vol 247 pp 1327ndash1331 1979

[18] E Kamke Differentialgleichungen Losungsmethoden und Losu-ngen I Gewohuliche Differentialgleichungen BGTeubnerLeipzig Germany 1977

[19] A Tungatarov and D K Akhmed-Zaki ldquoCauchy problemfor one class of ordinary differential equationsrdquo InternationalJournal of Mathematical Analysis vol 6 no 14 pp 695ndash6992012

[20] M Tenenbaum andH PollardOrdinary Differential EquationsHarper and Row New York NY USA 1963

[21] L Tartar ldquoCompensated compactness and applications topartial differential equationsrdquo in Proceedings of the Non-LinearAnalysis and Mechanics Heriot-Watt Symposium R J KnopsEd vol 4 of Research Notes in Math pp 136ndash212 1979

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Journal of Applied Mathematics

equations that approximate the microscopic boundary con-ditions for Boltzmannrsquos equation The question of boundaryconditions for a finite system of moment equations canbe divided into two parts how many conditions must beimposed and how they should be prepared From micro-scopic boundary conditions for Boltzmannrsquos equation therecan be obtained an infinite set of boundary conditionsfor each type of decomposition However the number ofboundary conditions is determined not by the number ofmoment equations that is it is impossible for example totake as many boundary conditions as equations although thenumber of moment equations affects the number of bound-ary conditions In addition the boundary conditions mustbe consistent with the moment equations and the resultingproblemmust be correct Boundary condition problems arisein the following tasks (1) moment boundary conditions inrarefied gas slip-flow problems (2) definition of boundaryconditions on the surfaces of streamlined rarefied gas (3)prediction aerospace aerodynamic characteristics of aircraftat very high speeds and at high altitudes and so forth

In work [3] we have obtained the moment system whichdiffers from Gradrsquos system of equations We used sphericalvelocity coordinates and decomposed the distribution func-tion into a series of eigenfunctions of the linearized collisionoperator [4 5] which is the product of Sonine polynomialsand spherical functions The resulting system of equationswhich correspond to the partial sum of series and whichwe called Boltzmannrsquos moment system of equations is anonlinear hyperbolic system in relation to the moments ofthe particles distribution function

The structure of Boltzmannrsquosmoment systemof equationscorresponds to the structure of Boltzmannrsquos equation namelythe differential part of the resulting system is linear in relationto the moments of the distribution function and nonlinearityis included as moments of collision integral [6]

The linearity of differential part of Boltzmannrsquos momentsystem of equations simplifies the task of formulation of theboundary conditions In work [3] a homogeneous boundarycondition for particles distribution function was approxi-mated and proved the correctness of initial and bound-ary value problem for nonlinear nonstationary Boltzmannrsquosmoment system of equations in three-dimensional spaceIn work [7] the initial and boundary value problem forone-dimensional nonstationary Boltzmannrsquos equation withboundary conditions of Maxwell was approximated by acorresponding problem for Boltzmannrsquos moment system ofequationsThe boundary conditions for Boltzmannrsquosmomentsystem of equations were calledMaxwell-Auzhan conditions

In work [8] a systematic nonperturbative derivation ofa hierarchy of closed systems of moment equations corre-sponding to any classical theory has been presented Thispaper is a fundamental work where closed systems ofmoment equations describe a transition regime Moreoverhydrodynamical model is used to describe charge trans-port in a generic compound semiconductor Compoundsemiconductors have found wide use in the microelectronicindustryThe evolution equation for macroscopic variables isobtained by taking moments of the transport equation [9ndash13]

The study of various problems for Boltzmannrsquos momentsystem of equations is an important and actual task in thetheory of a rarefied gas and other applications of the momentsystem equations The correctness of initial and boundaryproblems for Boltzmannrsquos moment system of equations withMaxwell-Auzhan boundary conditions is being studied forthe first time

2 Existence and Uniqueness of the Solutionsof Initial and Boundary Value Problem forSix-Moment One-Dimensional BoltzmannrsquosSystem of Equations with Maxwell-AuzhanBoundary Conditions

In this section we prove the existence and uniqueness ofsolutions of the initial and boundary value problem forsix-moment one-dimensional Boltzmannrsquos system of equa-tions with Maxwell-Auzhan boundary conditions in space offunctions continuous in time and summable in square byspatial variable Note that the theorem of the existence ofa global solution in time of the initial and boundary valueproblem for 3-dimensional nonlinear Boltzmannrsquos equationwith boundary conditions of Maxwell is proved in work [14]

Wewrite in an expanded form systemof one-dimensionalBoltzmannrsquos moment equations in the kth approximationwhich corresponds to decomposition of the particle distri-bution function by eigenfunctions of the linearized collisionoperator

120597119891119899119897

120597119905+

1

120572

120597

120597119909[119897(radic

2 (119899 + 119897 + 12)

(2119897 minus 1) (2119897 + 1)119891119899119897minus1

minus radic2 (119899 + 1)

(2119897 minus 1) (2119897 + 1)119891119899+1119897minus1

) + (119897 + 1)

sdot (radic2 (119899 + 119897 + 12)

(2119897 + 1) (2119897 + 3)119891119899119897+1

minus radic2119899

(2119897 + 1) (2119897 + 3)119891119899minus1119897+1

)] = 119868119899119897

2119899 + 119897 = 0 1 119896

(1)

where the moments 119868119899119897of a nonlinear collision operator are

expressed through coefficients of Talmi and Klebsh-Gordon[15 16] and have the form

119868119899119897= sum⟨119873

3119871311989931198973 119897 | 11989911989700 119897⟩

sdot ⟨1198733119871311989931198973 119897 | 1198991119897111989921198972 119897⟩ (

1198971011989720

1198970) (1205901198973minus 1205900)

sdot 11989111989911198971

11989111989921198972

(2)

where ⟨1198733119871311989931198973 119897 | 1198991119897111989921198972 119897⟩ are generalized coefficients

of Talmi (11989710119897201198970) are Klebsh-Gordon coefficients 120572 =

1radic119877120579 is the constant 119877 is Boltzmannrsquos constant and 120579 is theideal gas temperature

Journal of Applied Mathematics 3

For generalized coefficients of Talmi there exists a table[15] for each value of quantum number 120585 = 2119899 + 119897 from 0to 6 Moreover there is a program on IBM for calculation ofgeneralized coefficients of Talmi

If in (1) 2119899+ 119897 takes values from 0 to 3 we get Boltzmannrsquosmoment system equations in the third approximation Wewrite it in an expanded form

12059711989100

120597119905+

1

120572

12059711989101

120597119909= 0

12059711989102

120597119905+

1

120572

120597

120597119909(

2

radic311989101

+3

radic511989103

minus2radic2

radic1511989111) = 11986902

12059711989110

120597119905+

1

120572

120597

120597119909(minusradic

2

311989101

+ radic5

311989111) = 0

12059711989101

120597119905+

1

120572

120597

120597119909(11989100

+2

radic311989102

minus radic2

311989110) = 0

12059711989103

120597119905+

1

120572

120597

120597119909

3

radic511989102

= 11986903

12059711989111

120597119905+

1

120572

120597

120597119909(minus

2radic2

radic1511989102

+ radic5

311989110) = 11986911

119909 isin (minus119886 119886) 119905 gt 0

(3)

where 11989100

= 11989100(119905 119909) and 119891

01= 11989101(119905 119909) 119891

11= 11989111(119905 119909)

are the moments of the particle distribution function 11986902

=

(1205902minus 1205900)(1198910011989102

minus 1198912

01radic3)2 119869

03= (14)(120590

3+ 31205901minus

41205900)1198910011989103

+ (14radic5)(21205901+ 1205900minus 31205903)1198910111989102 and 119869

11=

(1205901minus 1205900)(1198910011989111

+ (12)radic531198911011989101

minus (radic2radic15)1198910111989102) are

moments of the collision integral where 1205900 1205901 1205902 and 120590

3

are the Fourier coefficients of the cross section expansionby the Legendre polynomials The first third and fourthequations of system (3) correspond to mass conservation lawmomentum conservation law and energy conservation lawrespectively

We study the correctness of initial and boundary valueproblem for six-moment one-dimensional Boltzmannrsquos sys-tem equations

120597119906

120597119905+ 119860

120597119908

120597119909= 1198691(119906 119908)

120597119908

120597119905+ 1198601015840120597119906

120597119909= 1198692(119906 119908)

119905 isin (0 119879] 119909 isin (minus119886 119886)

(4)

119906|119905=0

= 1199060(119909)

119908|119905=0

= 1199080(119909)

119909 isin [minus119886 119886]

(5)

(119860119908minus

+ 119861119906minus

)1003816100381610038161003816119909=minus119886 =

1

120573(119860119908+

minus 119861119906+

)1003816100381610038161003816119909=minus119886

minus(1 minus 120573) 120578

120572120573radic120587119865 119905 isin [0 119879]

(6)

(119860119908minus

minus 119861119906minus

)1003816100381610038161003816119909=119886 =

1

120573(119860119908+

+ 119861119906+

)1003816100381610038161003816119909=119886

+(1 minus 120573) 120578

120572120573radic120587119865 119905 isin [0 119879]

(7)

where

119860 =1

120572

(

(

1 0 0

2

radic3

3

radic5minus2radic2

radic15

minusradic2

30 radic

5

3

)

)

119861 =1

120572radic120587

((

(

radic2 radic2

3minus

1

radic3

radic2

32radic2 minus1

minus1

radic3minus1 3radic2

))

)

1198691(119906 119908) = (0 119869

02 0)1015840

1198692(119906 119908) = (0 119869

03 11986911)1015840

119906 = (11989100 11989102 11989110)1015840

119908 = (11989101 11989103 11989111)1015840

119865 = (1

4radic2

1

8radic6

1

8radic3)

1015840

(8)

1198601015840 is the transpose matrix 119861 is the positive definite

matrix 1199060(119909) = (119891

0

00(119909) 119891

0

02(119909) 119891

0

10(119909))1015840 and 119908

0(119909) =

(1198910

01(119909) 119891

0

03(119909) 119891

0

11(119909))1015840 are given initial vector-functions

119908+

119906+ are vector moments of falling to the boundary particle

distribution function119908minus 119906minusare vectormoments of reflectingfrom the boundary particle distribution function Equation(4) is vector matrix form of system equations (3)

It is possible to check through direct calculations that

det1198601= det(

0 119860

1198601015840

0) = 0 (9)

and matrix 1198601has three positive and the same number of

negative nonzero eigenvalues From (6)-(7) it follows that thenumber of boundary conditions on the left and right ends ofinterval (minus119886 119886) is equal to the number of positive andnegativeeigenvalues of the matrix 119860

1

Thus system (4) is a symmetric hyperbolic nonlinearpartial differential equations system Let us show that 119869

02

4 Journal of Applied Mathematics

is a sign-nondefined square form It is easy to check that1198910011989102

minus (1198912

01radic3) = (119862119880119880) where 119880 = (119906 119908)

1015840 and

119862 =

((((((((

(

01

20 0 0 0

1

20 0 0 0 0

0 0 0 0 0 0

0 0 0 minus1

radic30 0

0 0 0 0 0 0

0 0 0 0 0 0

))))))))

)

(10)

Eigenvalues of matrix 119862 are minus12 minus1radic3 0 0 0 and 12Therefore 119869

02is a sign-nondefined square form Similarly we

can show that 11986903

and 11986911

are also sign-nondefined squareforms

For problem (4)ndash(7) the following theorem takes place

Theorem 1 If 1198800= (1199060(119909) 119908

0(119909)) isin 119871

2

[minus119886 119886] then problem(4)ndash(7) has a unique solution in domain [minus119886 119886] times [0 119879]

belonging to the space 119862([0 119879] 1198712[minus119886 119886]) moreover

10038171003817100381710038171198801198712[minus119886119886]

minus 1199031

1003817100381710038171003817119862[0119879]le 1198621(10038171003817100381710038171198800

10038171003817100381710038171198712[minus119886119886] minus 1199031(0)) (11)

where 1198621is a constant independent from 119880 and 119879 sim

M(11988001198712[minus119886119886]

minus 1199031(0))minus1

) 1199031(119905) is the particular solution of

Riccati equation (15) in an explicit form

Proof Let 1198800isin 1198712

[minus119886 119886] Let us prove estimation (11) Wemultiply the first equation in system (4) by 119906 and the secondequation by 119908 and perform integration from minus119886 to 119886

1

2

119889

119889119905int

119886

minus119886

[(119906 119906) + (119908 119908)] 119889119909

+ int

119886

minus119886

[(119860120597119908

120597119909 119906) + (119860

1015840120597119906

120597119909 119908)] 119889119909

= int

119886

minus119886

[(1198691 119906) + (119869

2 119908)] 119889119909

(12)

After integration by parts we get

1

2

119889

119889119905int

119886

minus119886

[(119906 119906) + (119908 119908)] 119889119909 + (119906minus

119860119908minus

)1003816100381610038161003816119909=119886

minus (119906minus

119860119908minus

)1003816100381610038161003816119909=minus119886 = int

119886

minus119886

[(1198691 119906) + (119869

2 119908)] 119889119909

(13)

Taking into account boundary conditions (6)-(7) we rewriteequality (13) in the following form

1

2

119889

119889119905int

119886

minus119886

[(119906 119906) + (119908 119908)] 119889119909 + (119861119906minus

119906minus

)1003816100381610038161003816119909=119886

+ (119861119906minus

119906minus

)1003816100381610038161003816119909=minus119886 minus

1

120573((119860119908

+

minus 119861119906+

) 119906minus

)10038161003816100381610038161003816119909=minus119886

+1

120573((119860119908+

+ 119861119906+

) 119906minus

)1003816100381610038161003816119909=119886 + (119865

1 119906minus

)1003816100381610038161003816119909=119886

+ (1198651 119906minus

)1003816100381610038161003816119909=minus119886

= int

119886

minus119886

[(1198691(119906 119908) 119906) + (119869

2(119906 119908) 119908)] 119889119909

(14)

where 1198651= ((1 minus 120573)120578120572120573radic120587)119865

Let us use the spherical representation [17] of vector119880(119905 119909) = 119903(119905)120596(119905 119909) where 120596(119905 119909) = (120596

1(119905 119909) 120596

2(119905 119909))

1015840119903(119905) = 119880(119905 sdot)

1198712[minus119886119886]

and 1205961198712[minus119886119886]

= 1 Furthermoreassume that120596(119905 119909) isin 119862

2

[0 119879] where the value of119879we definebelow

Substituting the values 119906 = 119903(119905)1205961(119905 119909) and 119908 =

119903(119905)1205962(119905 119909) into (14) we have that

119889119903

119889119905+ 119903119875 (119905) = 119903

2

119876 (119905) minus 119891 (119905) (15)

where

119875 (119905) = (119861120596minus

1 120596minus

1)1003816100381610038161003816119909=119886 + (119861120596

minus

1 120596minus

1)1003816100381610038161003816119909=minus119886

+1

120573[(119860120596+

2 120596minus

1)1003816100381610038161003816119909=119886 + (119861120596

+

1 120596minus

1)1003816100381610038161003816119909=119886

+ (119861120596+

1 120596minus

1)1003816100381610038161003816119909=minus119886 minus (119860120596

+

2 120596minus

1)1003816100381610038161003816119909=minus119886]

119876 (119905) = int

119886

minus119886

[(1198691(1205961 1205962) 1205961) + (1198692(1205961 1205962) 1205962)] 119889119909

119891 (119905) = (1198651 120596minus

1)1003816100381610038161003816119909=119886 + (119865

1 120596minus

1)1003816100381610038161003816119909=minus119886

(16)

Let us study (15) with an initial condition

119903 (0) =10038171003817100381710038171198800

1003817100381710038171003817 =10038171003817100381710038171198800

10038171003817100381710038171198712[minus119886119886] (17)

Let 1199031(119905) be a particular solution of Riccati equation (15)Then

the general solution of (15) has the form

119903 (119905) = 1199031(119905) + exp(int

119905

0

(2119876 (120591) 1199031(120591) minus 119875 (120591)) 119889120591) [119862

minus int

119905

0

119876 (120591)

sdot exp(int

120591

0

(2119876 (120585) 1199031(120585) minus 119875 (120585)) 119889120585) 119889120591]

minus1

(18)

Hence taking into consideration initial condition (17) weobtain

Journal of Applied Mathematics 5

119903 (119905) = 1199031(119905) + exp(int

119905

0

(2119876 (120591) 1199031(120591) minus 119875 (120591)) 119889120591)[

110038171003817100381710038171198800

1003817100381710038171003817 minus 1199031(0)

minus int

119905

0

119876 (120591) exp(int

120591

0

(2119876 (120585) 1199031(120585) minus 119875 (120585)) 119889120585) 119889120591]

minus1

(19)

If 119877(119905) equiv int119905

0

119876(120591) exp(int1205910

(2119876(120585)1199031(120585) minus 119875(120585))119889120585)119889120591 le 0 forall119905

then the second term of 119903(119905) is bounded forall119905 isin [0 +infin) underthe condition that difference 119880

0 minus 1199031(0) is positive Further

we suppose a difference 1198800 minus 1199031(0) is a positive number Let

119877(119905) gt 0 We denote by 1198791the moment of time at which

110038171003817100381710038171198800

1003817100381710038171003817 minus 1199031(0)

minus int

1198791

0

119876 (120591) exp(int

120591

0

(2119876 (120585) 1199031(120585) minus 119875 (120585)) 119889120585) 119889120591

= 0

(20)

Then 119903(119905) minus 1199031(119905) is bounded forall119905 isin [0 119879] where 119879 lt 119879

1and

1198791sim M(1(119880

0 minus 1199031(0))) Analyzing structure of 119903

1(119905) (see

(31)) from equality (20) we obtain 119876(119905) gt 0 for any 119905 isin [0 119879]Now we found one particular solution of (15) The

nonlinear Riccati equation can always be reduced to a second-order linear ordinary differential equation [18]

11991010158401015840

+ 1198911(119905) 1199101015840

+ 1198912(119905) 119910 = 0 (21)

where

119910 (119905) = exp(minusint

119905

0

119876 (120591) 119903 (120591) 119889120591)

1198911(119905) =

119876 (119905) 119875 (119905) minus 1198761015840

(119905)

119876 (119905)

1198912(119905) = 119876 (119905) 119891 (119905)

(22)

Solution of this equation will lead to a solution of 119903(119905) =

minus1199101015840

119910119876(119905) of the original Riccati equation By substitution

119910 (119905) = 119911 (119905) exp(minus1

2int

119905

0

1198911(120591) 119889120591) (23)

we write (21) in the following form [18]

11991110158401015840

(119905) + 119886 (119905) 119911 (119905) = 0 (24)

where 119886(119905) = 1198912(119905) minus (12)119891

1015840

1(119905) minus (14)119891

2

1(119905)

If instead of1198911(119905) and119891

2(119905)we substitute values from (22)

then we get

119886 (119905) = 119876119891 (119905)

minus2 (1198762

1198751015840

minus 11987610158401015840

119876 + 11987610158402

) + (119876119875 minus 1198761015840

)2

41198762

(25)

In work [19] it was proven that function

119911 (119905) = 1 +

infin

sum

119896=1

119887119896(119905) (26)

is the solution of (24) where

1198871(119905) = minusint

119905

0

int

120585

0

119886 (120591) 119889120591 119889120585

119887119896(119905) = minusint

119905

0

int

120585

0

119886 (120591) 119887119896minus1

(120591) 119889120591 119889120585

(27)

and the series on the right hand side of (26) uniformlyconverge

Then

119910 = (1 +

infin

sum

119896=1

119887119896(119905)) exp(minus

1

2int

119905

0

1198911(120591) 119889120591)

1199031(119905) = minus

1199101015840

119910119876= minus

1199111015840

(119905) minus (12) 119911 (119905) 1198911(119905)

119911 (119905) 119876 (119905)

(28)

is a particular solution of Riccati equation (15)Since 1199111015840(0) = 0 value of

1199031(0) = minus

1199111015840

(0) minus (12) 119911 (0) 1198911(0)

119911 (0) 119876 (0)=

1198911(0)

2119876 (0) (29)

Now instead of 1198911(0) we substitute its value 119875(0) minus

1198761015840

(0)119876(0) and get 1199031(0) = (119875(0)119876(0) minus 119876

1015840

(0))21198762

(0)The value of 119876(119905) with 119905 = 0 is

119876 (0)

= int

119886

minus119886

((1198691(1205961 1205962) 1205961) + (1198692(1205961 1205962) 1205962))1003816100381610038161003816119905=0 119889119909

gt 0

(30)

since 119876(119905) gt 0 for any 119905 isin [0 119879]

An expansion form of the function

1199031(119905) =

1

119876 (119905)[

1

119911 (119905)int

119905

0

119886 (120591) 119911 (120591) 119889120591

+119876 (119905) 119875 (119905) minus 119876

1015840

(119905)

2119876 (119905)]

(31)

It is not difficult to prove using immediate substitutionthat function 119903

1(119905) satisfies Riccati equation (15)

From equality (19) we obtain1003816100381610038161003816119903 (119905) minus 119903

1(119905)

1003816100381610038161003816 le 119862 (10038171003817100381710038171198800

1003817100381710038171003817 minus 1199031(0)) (32)

Hence1003816100381610038161003816119880 (119905 sdot)

1198712[minus119886119886]

minus 1199031(119905)

1003816100381610038161003816 le 119862 (10038171003817100381710038171198800

1003817100381710038171003817 minus 1199031(0))

10038171003817100381710038171198801198712[minus119886119886]

minus 1199031

1003817100381710038171003817119862[0119879]le 119862 (

100381710038171003817100381711988001003817100381710038171003817 minus 1199031(0))

(33)

6 Journal of Applied Mathematics

We take as interval of the solution of problem (15)ndash(17) segment [0 1(119880

0 minus 119903

1(0))] since the integrand

119876(120591) exp(int1205910

(2119876(120585)1199031(120585) minus 119875(120585))119889120585) is bounded Hence forall119905 isin

[0 119879] (where 119879 asymp 1198791) a priori estimation (11) takes place

Now we prove the existence of solution for (4)ndash(7) usingGalerkin method Let V

119897(119909)infin

119897=1be a basis in space 1198712[minus119886 119886]

where the dimension of vector V119897(119909) is equal to the dimension

of vector119880 For each119898we define an approximate solution119880119898

of (4)ndash(7) as follows

119880119898=

119898

sum

119895=1

119888119895119898

(119905) V119895(119909) (34)

int

119886

minus119886

((120597119880119898

120597119905+ 1198601

120597119880119898

120597119909) V119894(119909)) 119889119909

= int

119886

minus119886

(119869 (119880119898) V119894(119909)) 119889119909 119894 = 1119898 119905 isin (0 119879]

(35)

119880119898

1003816100381610038161003816119905=0 = 1198800119898

(119909) 119909 isin 119877 (36)

(119860119908minus

119898∓ 119861119906minus

119898)1003816100381610038161003816119909=∓119886

=1

120573(119860119908+

119898plusmn 119861119906+

119898)1003816100381610038161003816119909=plusmn119886 plusmn

(1 minus 120573) 120578

120572120573radic120587119865

(37)

where 1198800119898

is the orthogonal projection in 1198712 of the function

1198800on the subspace spanned by V

1 V

119898

119869 (119880119898) = (119869

1(119906119898 119908119898) 1198692(119906119898 119908119898))1015840

(38)

We represent V119895(119909) in form V

119895(119909) = (V(1)

119895 V(2)119895

)1015840 where

V(1)119895

= (V1198951 V1198952 V1198953)1015840

V(2)119895

= (V1198954 V1198955 V1198956)1015840

(39)

The coefficients 119888119895119898

(119905) are determined from the equations

119898

sum

119895=1

119889119888119895119898

119889119905int

119886

minus119886

(V119895 V119894

) 119889119909 + 119888119895119898

[(119861Vminus(1)119895

Vminus(1)119894

)10038161003816100381610038161003816119909=119886

+ (119861Vminus(1)119895

Vminus(1)119894

)10038161003816100381610038161003816119909=minus119886

+ (119861Vminus(1)119894

Vminus(1)119895

)10038161003816100381610038161003816119909=119886

+ (119861Vminus(1)119894

Vminus(1)119895

)10038161003816100381610038161003816119909=minus119886

+1

120573((119860V+(2)119894

+ 119861V+(1)119894

) Vminus(1)119895

)10038161003816100381610038161003816119909=119886

minus1

120573((119860V+(2)119894

minus 119861V+(1)119894

) Vminus(1)119895

)10038161003816100381610038161003816119909=minus119886

+1

120573((119860V+(2)119895

+ 119861V+(1)119895

) Vminus(1)119894

)10038161003816100381610038161003816119909=119886

minus1

120573(119860V+(2)119895

minus 119861V+(1)119895

Vminus(1)119894

)10038161003816100381610038161003816119909=minus119886

+ (1198651 Vminus(1)119894

)10038161003816100381610038161003816119909=119886

+ (1198651 Vminus(1)119894

)10038161003816100381610038161003816119909=minus119886

+ (1198651 Vminus(1)119895

)10038161003816100381610038161003816119909=119886

+ (1198651 Vminus(1)119895

)10038161003816100381610038161003816119909=minus119886

]

minus int

119886

minus119886

((119860120597V(2)119894

120597119909 V(1)119895) + (119860

1015840120597V(1)119894

120597119909 V(2)119895

))119889119909

= int

119886

minus119886

(119869(

119898

sum

119895=1

119888119895119898V119895) V119894)119889119909

119894 = 1119898 119905 isin (0 119879]

(40)

119888119894119898

(0) = 119889119894119898 119894 = 1119898 (41)

where 119889119894119898

is the 119894th component of 1198800119898

We multiply (35) by 119888119894119898(119905) and sum over 119894 from 1 to119898

int

119886

minus119886

((120597119880119898

120597119905+ 1198601

120597119880119898

120597119909) 119880119898)119889119909

= int

119886

minus119886

(119869 (119880119898) 119880119898) 119889119909

(42)

With help of the above shown arguments we now provethat 119903

119898(119905) is bounded in some time interval [0 119879

119898] where

119880119898(119905 119909) = 119903

119898(119905)120596119898(119905 119909) 119879

119898asymp M((119880

0119898 minus 1199031198981

(0))minus1

) 119879119898

ge

119879 forall119898 and

10038171003817100381710038171003817

100381710038171003817100381711988011989810038171003817100381710038171198712[minus119886119886] minus 119903

1198981

10038171003817100381710038171003817119862[0119879]

le 1198622(10038171003817100381710038171198800

10038171003817100381710038171198712[minus119886119886] minus 1199031(0))

(43)

where 1198622is constant and independent from119898 and 119903

1198981(119905) is a

particular solution of Riccati equation about 119903119898(119905)

Problem (40)-(41) represents a Cauchy problem for theordinary system of differential equations Existence of thesolution of problem (40)-(41) follows from theory of ordi-nary system of differential equations (Picardrsquos existence anduniqueness theorem) [20]Hence the existence of the solutionof problem (34)ndash(37) follows

Thus according to estimation (43) the sequence 119880119898

minus

1199031198981

of approximate solutions of problem (5)ndash(7) is uniformlybounded in function space119862([0 119879] 1198712[minus119886 119886])Moreover thehomogeneous system of equations 120591119864+ (1120572)119860120585with respectto 120591 120585 has only a trivial solution Then it follows from resultsin [21] that 119880

119898minus 1199031198981

rarr 119880minus 1199031is week in 119862([0 119879] 119871

2

[minus119886 119886])

and 119869(119880119898) rarr 119869(119880) is week in 119862([0 119879] 119871

2

[minus119886 119886]) as 119898 rarr

infin Further it can be shown with standard methods that limitelement is a weak solution of problem (5)ndash(7)

The uniqueness of solution of problem (5)ndash(7) is provedby contradiction Let problem (5)ndash(7) have two different solu-tions 119906

1 1199081and 119906

2 1199082 We denote them again by 119906 = 119906

1minus 1199062

Journal of Applied Mathematics 7

and119908 = 1199081minus1199082 Then with respect to new values of 119906 and119908

we obtain the following problem

120597119906

120597119905+ 119860

120597119908

120597119909= 1198691(1199061 1199081) minus 1198691(1199062 1199082)

120597119908

120597119905+ 1198601015840120597119906

120597119909= 1198692(1199061 1199081) minus 1198692(1199062 1199082)

119909 isin (minus119886 119886)

(44)

119906|119905=0

= 0

119908|119905=0

= 0

(45)

(119860119908minus

+ 119861119906minus

)1003816100381610038161003816119909=minus119886 =

1

120573(119860119908+

minus 119861119906+

)1003816100381610038161003816119909=minus119886

(119860119908minus

minus 119861119906minus

)1003816100381610038161003816119909=119886 =

1

120573(119860119908+

+ 119861119906+

)1003816100381610038161003816119909=119886

(46)

We prove that solution of problem (44)ndash(46) is trivialHence the uniqueness of the solution of problem (5)ndash(7)follows

Using the method which was mentioned before we get(see equality (14))

1

2

119889

119889119905int

119886

minus119886

[(119906 119906) + (119908 119908)] 119889119909 + (119861119906minus

119906minus

)1003816100381610038161003816119909=119886

+ (119861119906minus

119906minus

)1003816100381610038161003816119909=minus119886 minus

1

120573((119860119908

+

minus 119861119906+

) 119906minus

)10038161003816100381610038161003816119909=minus119886

+1

120573((119860119908

+

+ 119861119906+

) 119906minus

)10038161003816100381610038161003816119909=119886

= int

119886

minus119886

[((1198691(1199061 1199081) minus 1198691(1199062 1199082)) 119906)

+ ((1198692(1199061 1199081) minus 1198692(1199062 1199082)) 119908)] 119889119909

(47)

Let us transform the integrand on the right side of theequality as

((1198691(1199061 1199081) minus 1198691(1199062 1199082)) 119906)

=1205902minus 1205900

2[(119891001

119891021

minus1198912

011

radic3) minus (119891

002119891022

minus1198912

012

radic3)] (119891

021minus 119891022

) =1205902minus 1205900

2[(119891001

119891021

minus 119891002

119891022

) minus1

radic3(1198912

011minus 1198912

012)] (119891021

minus 119891022

)

=1205902minus 1205900

2[(119891001

minus 119891002

) 119891021

+ 119891002

(119891021

minus 119891022

) minus1

radic3(119891011

minus 119891012

) (119891011

+ 119891012

)] (119891021

minus 119891022

)

((1198692(1199061 1199081) minus 1198692(1199062 1199082)) 119906) = ((119869

03(1199061 1199081)

minus 11986903

(1199062 1199082)) (119891031

minus 119891032

)) + ((11986911

(1199061 1199081)

minus 11986911

(1199062 1199082)) (119891111

minus 119891112

)) = (1

4(1205903+ 31205901

minus 41205900) [(119891001

minus 119891002

) 119891031

+ 119891002

(119891031

minus 119891032

)]

+1

4radic5(21205901+ 1205900minus 31205903) [(119891011

minus 119891012

) 119891021

+ 119891012

(119891021

minus 119891022

)]) (119891031

minus 119891032

) + (1205901minus 1205900)

sdot ((119891001

minus 119891002

) 119891111

+ 119891002

(119891111

minus 119891112

)

+ 05radic5

2[(119891101

minus 119891102

) 119891011

+ 119891102

(119891011

minus 119891012

)] minus radic2

15[(119891011

minus 119891012

) 119891021

+ 119891012

(119891021

minus 119891022

)]) (119891111

minus 119891112

)

(48)

Once again we use spherical representation of 119906 = 119903(119905)1199081(119905

119909) and 119908 = 119903(119905)1199082(119905 119909) where 119903(119905) = 119880(119905 sdot)

1198712[minus119886119886]

Thenconcerning 119903(119905) we obtain new initial value problem

119889119903

119889119905+ 119903119875 (119905) = 119903

2

1198761(119905) (49)

119903 (0) = 0 (50)

where 119875(119905) has the same value as in (15) and

1198761(119905) =

1205902minus 1205900

2int

119886

minus119886

[(120596001

minus 120596002

) 120596021

+ 120596002

(120596021

minus 120596022

) minus1

radic3(120596011

minus 120596012

) (120596011

+ 120596012

)] (120596021

minus 120596022

) 119889119909 +1

4(1205903+ 31205901

minus 41205900) int

119886

minus119886

[(120596001

minus 120596002

) 120596031

+ +120596002

(120596031

minus 120596032

)] (120596031

minus 120596032

) 119889119909 +1

4radic5(21205901+ 1205900minus 31205903)

sdot int

119886

minus119886

[(120596011

minus 120596012

) 120596021

+ 120596012

(120596021

minus 120596022

)]

sdot (120596031

minus 120596032

) 119889119909 + (1205901minus 1205900) int

119886

minus119886

[(120596001

minus 120596002

) 120596111

+ 120596002

(120596111

minus 120596112

) +1

2

8 Journal of Applied Mathematics

sdot radic5

2[(120596101

minus 120596102

) 120596011

+ 120596102

(120596011

minus 120596012

)]

minus radic2

15[(120596011

minus 120596012

) 120596021

+ 120596012

(120596021

minus 120596022

)]

sdot (120596111

minus 120596112

)] 119889119909

(51)

The general solution of (49) is

119903 (119905) = exp(minusint

119905

0

119875 (120591) 119889120591)

sdot [119862 minus int

119905

0

1198761(120591) exp(minusint

120591

0

119875 (120585) 119889120585) 119889120591]

minus1

(52)

Solution of (49) which is satisfying homogeneous condition(50) is trivial that is 119903(119905) = 0 Hence 119880

119862([0119879]1198712[minus119886119886])

= 0

and 1199061= 1199062and 119908

1= 1199082

The theorem is proved

Competing Interests

The authors declare that they have no competing interests

References

[1] H Grad ldquoOn the kinetic theory of rarefied gasesrdquo Communica-tions on Pure andAppliedMathematics vol 2 no 4 pp 331ndash4071949

[2] H Grad ldquoPrinciple of the kinetic theory of gasesrdquo in Thermo-dynamics of Gases vol 12 ofHandbuch der Physik pp 205ndash294Springer Berlin Germany 1958

[3] A Sakabekov Initial-Boundary Value Problems for the Boltz-mannrsquos Moment System Equations Gylym Almaty 2002

[4] C Cercignani Theory and Application of the Boltzmann Equa-tion Instituto di Matematica Milano Italy 1975

[5] M N KoganDynamic of Rarefied Gas NaukaMoscow Russia1967

[6] K Kumar ldquoPolynomial expansions in kinetic theory of gasesrdquoAnnals of Physics vol 37 no 1 pp 113ndash141 1966

[7] A Sakabekov and Y Auzhani ldquoBoundary conditions for theonedimensional nonlinear nonstationary Boltzmannrsquos momentsystem equationsrdquo Journal of Mathematical Physics vol 55Article ID 123507 2014

[8] C D Levermore ldquoMoment closure hierarchies for kinetictheoriesrdquo Journal of Statistical Physics vol 83 no 5-6 pp 1021ndash1065 1996

[9] G Mascali and V Romano ldquoA hydrodynamical model for holesin silicon semiconductors the case of non-parabolic warpedbandsrdquo Mathematical and Computer Modelling vol 53 no 1-2pp 213ndash229 2011

[10] G Mascali and V Romano ldquoA non parabolic hydrodynamicalsubband model for semiconductors based on the maximumentropy principlerdquoMathematical and Computer Modelling vol55 no 3-4 pp 1003ndash1020 2012

[11] V D Camiola and V Romano ldquo2DEG-3DEG charge transportmodel for MOSFET based on the maximum entropy principlerdquo

SIAM Journal on Applied Mathematics vol 73 no 4 pp 1439ndash1459 2013

[12] G Alı G Mascali V Romano and R C Torcasio ldquoA hydrody-namical model for covalent semiconductors with a generalizedenergy dispersion relationrdquo European Journal of Applied Math-ematics vol 25 no 2 pp 255ndash276 2014

[13] V D Camiola and V Romano ldquoHydrodynamical model forcharge transport in graphenerdquo Journal of Statistical Physics vol157 no 6 pp 1114ndash1137 2014

[14] S Mischler ldquoKinetic equations with Maxwell boundary condi-tionsrdquo Annales Scientifiques de lrsquoEcole Normale Superieure vol43 no 5 pp 719ndash760 2010

[15] V G Neudachin and U F SmirnovNucleon Association of EasyKernel Nauka Moscow Russia 1969

[16] MMoshinskyTheHarmonic Oscillator inModern Physics fromAtoms to Quarks 1960

[17] S I Pokhozhaev ldquoOn an approach to nonlinear equationrdquoDoklady Akademii Nauk SSSR vol 247 pp 1327ndash1331 1979

[18] E Kamke Differentialgleichungen Losungsmethoden und Losu-ngen I Gewohuliche Differentialgleichungen BGTeubnerLeipzig Germany 1977

[19] A Tungatarov and D K Akhmed-Zaki ldquoCauchy problemfor one class of ordinary differential equationsrdquo InternationalJournal of Mathematical Analysis vol 6 no 14 pp 695ndash6992012

[20] M Tenenbaum andH PollardOrdinary Differential EquationsHarper and Row New York NY USA 1963

[21] L Tartar ldquoCompensated compactness and applications topartial differential equationsrdquo in Proceedings of the Non-LinearAnalysis and Mechanics Heriot-Watt Symposium R J KnopsEd vol 4 of Research Notes in Math pp 136ndash212 1979

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Applied Mathematics 3

For generalized coefficients of Talmi there exists a table[15] for each value of quantum number 120585 = 2119899 + 119897 from 0to 6 Moreover there is a program on IBM for calculation ofgeneralized coefficients of Talmi

If in (1) 2119899+ 119897 takes values from 0 to 3 we get Boltzmannrsquosmoment system equations in the third approximation Wewrite it in an expanded form

12059711989100

120597119905+

1

120572

12059711989101

120597119909= 0

12059711989102

120597119905+

1

120572

120597

120597119909(

2

radic311989101

+3

radic511989103

minus2radic2

radic1511989111) = 11986902

12059711989110

120597119905+

1

120572

120597

120597119909(minusradic

2

311989101

+ radic5

311989111) = 0

12059711989101

120597119905+

1

120572

120597

120597119909(11989100

+2

radic311989102

minus radic2

311989110) = 0

12059711989103

120597119905+

1

120572

120597

120597119909

3

radic511989102

= 11986903

12059711989111

120597119905+

1

120572

120597

120597119909(minus

2radic2

radic1511989102

+ radic5

311989110) = 11986911

119909 isin (minus119886 119886) 119905 gt 0

(3)

where 11989100

= 11989100(119905 119909) and 119891

01= 11989101(119905 119909) 119891

11= 11989111(119905 119909)

are the moments of the particle distribution function 11986902

=

(1205902minus 1205900)(1198910011989102

minus 1198912

01radic3)2 119869

03= (14)(120590

3+ 31205901minus

41205900)1198910011989103

+ (14radic5)(21205901+ 1205900minus 31205903)1198910111989102 and 119869

11=

(1205901minus 1205900)(1198910011989111

+ (12)radic531198911011989101

minus (radic2radic15)1198910111989102) are

moments of the collision integral where 1205900 1205901 1205902 and 120590

3

are the Fourier coefficients of the cross section expansionby the Legendre polynomials The first third and fourthequations of system (3) correspond to mass conservation lawmomentum conservation law and energy conservation lawrespectively

We study the correctness of initial and boundary valueproblem for six-moment one-dimensional Boltzmannrsquos sys-tem equations

120597119906

120597119905+ 119860

120597119908

120597119909= 1198691(119906 119908)

120597119908

120597119905+ 1198601015840120597119906

120597119909= 1198692(119906 119908)

119905 isin (0 119879] 119909 isin (minus119886 119886)

(4)

119906|119905=0

= 1199060(119909)

119908|119905=0

= 1199080(119909)

119909 isin [minus119886 119886]

(5)

(119860119908minus

+ 119861119906minus

)1003816100381610038161003816119909=minus119886 =

1

120573(119860119908+

minus 119861119906+

)1003816100381610038161003816119909=minus119886

minus(1 minus 120573) 120578

120572120573radic120587119865 119905 isin [0 119879]

(6)

(119860119908minus

minus 119861119906minus

)1003816100381610038161003816119909=119886 =

1

120573(119860119908+

+ 119861119906+

)1003816100381610038161003816119909=119886

+(1 minus 120573) 120578

120572120573radic120587119865 119905 isin [0 119879]

(7)

where

119860 =1

120572

(

(

1 0 0

2

radic3

3

radic5minus2radic2

radic15

minusradic2

30 radic

5

3

)

)

119861 =1

120572radic120587

((

(

radic2 radic2

3minus

1

radic3

radic2

32radic2 minus1

minus1

radic3minus1 3radic2

))

)

1198691(119906 119908) = (0 119869

02 0)1015840

1198692(119906 119908) = (0 119869

03 11986911)1015840

119906 = (11989100 11989102 11989110)1015840

119908 = (11989101 11989103 11989111)1015840

119865 = (1

4radic2

1

8radic6

1

8radic3)

1015840

(8)

1198601015840 is the transpose matrix 119861 is the positive definite

matrix 1199060(119909) = (119891

0

00(119909) 119891

0

02(119909) 119891

0

10(119909))1015840 and 119908

0(119909) =

(1198910

01(119909) 119891

0

03(119909) 119891

0

11(119909))1015840 are given initial vector-functions

119908+

119906+ are vector moments of falling to the boundary particle

distribution function119908minus 119906minusare vectormoments of reflectingfrom the boundary particle distribution function Equation(4) is vector matrix form of system equations (3)

It is possible to check through direct calculations that

det1198601= det(

0 119860

1198601015840

0) = 0 (9)

and matrix 1198601has three positive and the same number of

negative nonzero eigenvalues From (6)-(7) it follows that thenumber of boundary conditions on the left and right ends ofinterval (minus119886 119886) is equal to the number of positive andnegativeeigenvalues of the matrix 119860

1

Thus system (4) is a symmetric hyperbolic nonlinearpartial differential equations system Let us show that 119869

02

4 Journal of Applied Mathematics

is a sign-nondefined square form It is easy to check that1198910011989102

minus (1198912

01radic3) = (119862119880119880) where 119880 = (119906 119908)

1015840 and

119862 =

((((((((

(

01

20 0 0 0

1

20 0 0 0 0

0 0 0 0 0 0

0 0 0 minus1

radic30 0

0 0 0 0 0 0

0 0 0 0 0 0

))))))))

)

(10)

Eigenvalues of matrix 119862 are minus12 minus1radic3 0 0 0 and 12Therefore 119869

02is a sign-nondefined square form Similarly we

can show that 11986903

and 11986911

are also sign-nondefined squareforms

For problem (4)ndash(7) the following theorem takes place

Theorem 1 If 1198800= (1199060(119909) 119908

0(119909)) isin 119871

2

[minus119886 119886] then problem(4)ndash(7) has a unique solution in domain [minus119886 119886] times [0 119879]

belonging to the space 119862([0 119879] 1198712[minus119886 119886]) moreover

10038171003817100381710038171198801198712[minus119886119886]

minus 1199031

1003817100381710038171003817119862[0119879]le 1198621(10038171003817100381710038171198800

10038171003817100381710038171198712[minus119886119886] minus 1199031(0)) (11)

where 1198621is a constant independent from 119880 and 119879 sim

M(11988001198712[minus119886119886]

minus 1199031(0))minus1

) 1199031(119905) is the particular solution of

Riccati equation (15) in an explicit form

Proof Let 1198800isin 1198712

[minus119886 119886] Let us prove estimation (11) Wemultiply the first equation in system (4) by 119906 and the secondequation by 119908 and perform integration from minus119886 to 119886

1

2

119889

119889119905int

119886

minus119886

[(119906 119906) + (119908 119908)] 119889119909

+ int

119886

minus119886

[(119860120597119908

120597119909 119906) + (119860

1015840120597119906

120597119909 119908)] 119889119909

= int

119886

minus119886

[(1198691 119906) + (119869

2 119908)] 119889119909

(12)

After integration by parts we get

1

2

119889

119889119905int

119886

minus119886

[(119906 119906) + (119908 119908)] 119889119909 + (119906minus

119860119908minus

)1003816100381610038161003816119909=119886

minus (119906minus

119860119908minus

)1003816100381610038161003816119909=minus119886 = int

119886

minus119886

[(1198691 119906) + (119869

2 119908)] 119889119909

(13)

Taking into account boundary conditions (6)-(7) we rewriteequality (13) in the following form

1

2

119889

119889119905int

119886

minus119886

[(119906 119906) + (119908 119908)] 119889119909 + (119861119906minus

119906minus

)1003816100381610038161003816119909=119886

+ (119861119906minus

119906minus

)1003816100381610038161003816119909=minus119886 minus

1

120573((119860119908

+

minus 119861119906+

) 119906minus

)10038161003816100381610038161003816119909=minus119886

+1

120573((119860119908+

+ 119861119906+

) 119906minus

)1003816100381610038161003816119909=119886 + (119865

1 119906minus

)1003816100381610038161003816119909=119886

+ (1198651 119906minus

)1003816100381610038161003816119909=minus119886

= int

119886

minus119886

[(1198691(119906 119908) 119906) + (119869

2(119906 119908) 119908)] 119889119909

(14)

where 1198651= ((1 minus 120573)120578120572120573radic120587)119865

Let us use the spherical representation [17] of vector119880(119905 119909) = 119903(119905)120596(119905 119909) where 120596(119905 119909) = (120596

1(119905 119909) 120596

2(119905 119909))

1015840119903(119905) = 119880(119905 sdot)

1198712[minus119886119886]

and 1205961198712[minus119886119886]

= 1 Furthermoreassume that120596(119905 119909) isin 119862

2

[0 119879] where the value of119879we definebelow

Substituting the values 119906 = 119903(119905)1205961(119905 119909) and 119908 =

119903(119905)1205962(119905 119909) into (14) we have that

119889119903

119889119905+ 119903119875 (119905) = 119903

2

119876 (119905) minus 119891 (119905) (15)

where

119875 (119905) = (119861120596minus

1 120596minus

1)1003816100381610038161003816119909=119886 + (119861120596

minus

1 120596minus

1)1003816100381610038161003816119909=minus119886

+1

120573[(119860120596+

2 120596minus

1)1003816100381610038161003816119909=119886 + (119861120596

+

1 120596minus

1)1003816100381610038161003816119909=119886

+ (119861120596+

1 120596minus

1)1003816100381610038161003816119909=minus119886 minus (119860120596

+

2 120596minus

1)1003816100381610038161003816119909=minus119886]

119876 (119905) = int

119886

minus119886

[(1198691(1205961 1205962) 1205961) + (1198692(1205961 1205962) 1205962)] 119889119909

119891 (119905) = (1198651 120596minus

1)1003816100381610038161003816119909=119886 + (119865

1 120596minus

1)1003816100381610038161003816119909=minus119886

(16)

Let us study (15) with an initial condition

119903 (0) =10038171003817100381710038171198800

1003817100381710038171003817 =10038171003817100381710038171198800

10038171003817100381710038171198712[minus119886119886] (17)

Let 1199031(119905) be a particular solution of Riccati equation (15)Then

the general solution of (15) has the form

119903 (119905) = 1199031(119905) + exp(int

119905

0

(2119876 (120591) 1199031(120591) minus 119875 (120591)) 119889120591) [119862

minus int

119905

0

119876 (120591)

sdot exp(int

120591

0

(2119876 (120585) 1199031(120585) minus 119875 (120585)) 119889120585) 119889120591]

minus1

(18)

Hence taking into consideration initial condition (17) weobtain

Journal of Applied Mathematics 5

119903 (119905) = 1199031(119905) + exp(int

119905

0

(2119876 (120591) 1199031(120591) minus 119875 (120591)) 119889120591)[

110038171003817100381710038171198800

1003817100381710038171003817 minus 1199031(0)

minus int

119905

0

119876 (120591) exp(int

120591

0

(2119876 (120585) 1199031(120585) minus 119875 (120585)) 119889120585) 119889120591]

minus1

(19)

If 119877(119905) equiv int119905

0

119876(120591) exp(int1205910

(2119876(120585)1199031(120585) minus 119875(120585))119889120585)119889120591 le 0 forall119905

then the second term of 119903(119905) is bounded forall119905 isin [0 +infin) underthe condition that difference 119880

0 minus 1199031(0) is positive Further

we suppose a difference 1198800 minus 1199031(0) is a positive number Let

119877(119905) gt 0 We denote by 1198791the moment of time at which

110038171003817100381710038171198800

1003817100381710038171003817 minus 1199031(0)

minus int

1198791

0

119876 (120591) exp(int

120591

0

(2119876 (120585) 1199031(120585) minus 119875 (120585)) 119889120585) 119889120591

= 0

(20)

Then 119903(119905) minus 1199031(119905) is bounded forall119905 isin [0 119879] where 119879 lt 119879

1and

1198791sim M(1(119880

0 minus 1199031(0))) Analyzing structure of 119903

1(119905) (see

(31)) from equality (20) we obtain 119876(119905) gt 0 for any 119905 isin [0 119879]Now we found one particular solution of (15) The

nonlinear Riccati equation can always be reduced to a second-order linear ordinary differential equation [18]

11991010158401015840

+ 1198911(119905) 1199101015840

+ 1198912(119905) 119910 = 0 (21)

where

119910 (119905) = exp(minusint

119905

0

119876 (120591) 119903 (120591) 119889120591)

1198911(119905) =

119876 (119905) 119875 (119905) minus 1198761015840

(119905)

119876 (119905)

1198912(119905) = 119876 (119905) 119891 (119905)

(22)

Solution of this equation will lead to a solution of 119903(119905) =

minus1199101015840

119910119876(119905) of the original Riccati equation By substitution

119910 (119905) = 119911 (119905) exp(minus1

2int

119905

0

1198911(120591) 119889120591) (23)

we write (21) in the following form [18]

11991110158401015840

(119905) + 119886 (119905) 119911 (119905) = 0 (24)

where 119886(119905) = 1198912(119905) minus (12)119891

1015840

1(119905) minus (14)119891

2

1(119905)

If instead of1198911(119905) and119891

2(119905)we substitute values from (22)

then we get

119886 (119905) = 119876119891 (119905)

minus2 (1198762

1198751015840

minus 11987610158401015840

119876 + 11987610158402

) + (119876119875 minus 1198761015840

)2

41198762

(25)

In work [19] it was proven that function

119911 (119905) = 1 +

infin

sum

119896=1

119887119896(119905) (26)

is the solution of (24) where

1198871(119905) = minusint

119905

0

int

120585

0

119886 (120591) 119889120591 119889120585

119887119896(119905) = minusint

119905

0

int

120585

0

119886 (120591) 119887119896minus1

(120591) 119889120591 119889120585

(27)

and the series on the right hand side of (26) uniformlyconverge

Then

119910 = (1 +

infin

sum

119896=1

119887119896(119905)) exp(minus

1

2int

119905

0

1198911(120591) 119889120591)

1199031(119905) = minus

1199101015840

119910119876= minus

1199111015840

(119905) minus (12) 119911 (119905) 1198911(119905)

119911 (119905) 119876 (119905)

(28)

is a particular solution of Riccati equation (15)Since 1199111015840(0) = 0 value of

1199031(0) = minus

1199111015840

(0) minus (12) 119911 (0) 1198911(0)

119911 (0) 119876 (0)=

1198911(0)

2119876 (0) (29)

Now instead of 1198911(0) we substitute its value 119875(0) minus

1198761015840

(0)119876(0) and get 1199031(0) = (119875(0)119876(0) minus 119876

1015840

(0))21198762

(0)The value of 119876(119905) with 119905 = 0 is

119876 (0)

= int

119886

minus119886

((1198691(1205961 1205962) 1205961) + (1198692(1205961 1205962) 1205962))1003816100381610038161003816119905=0 119889119909

gt 0

(30)

since 119876(119905) gt 0 for any 119905 isin [0 119879]

An expansion form of the function

1199031(119905) =

1

119876 (119905)[

1

119911 (119905)int

119905

0

119886 (120591) 119911 (120591) 119889120591

+119876 (119905) 119875 (119905) minus 119876

1015840

(119905)

2119876 (119905)]

(31)

It is not difficult to prove using immediate substitutionthat function 119903

1(119905) satisfies Riccati equation (15)

From equality (19) we obtain1003816100381610038161003816119903 (119905) minus 119903

1(119905)

1003816100381610038161003816 le 119862 (10038171003817100381710038171198800

1003817100381710038171003817 minus 1199031(0)) (32)

Hence1003816100381610038161003816119880 (119905 sdot)

1198712[minus119886119886]

minus 1199031(119905)

1003816100381610038161003816 le 119862 (10038171003817100381710038171198800

1003817100381710038171003817 minus 1199031(0))

10038171003817100381710038171198801198712[minus119886119886]

minus 1199031

1003817100381710038171003817119862[0119879]le 119862 (

100381710038171003817100381711988001003817100381710038171003817 minus 1199031(0))

(33)

6 Journal of Applied Mathematics

We take as interval of the solution of problem (15)ndash(17) segment [0 1(119880

0 minus 119903

1(0))] since the integrand

119876(120591) exp(int1205910

(2119876(120585)1199031(120585) minus 119875(120585))119889120585) is bounded Hence forall119905 isin

[0 119879] (where 119879 asymp 1198791) a priori estimation (11) takes place

Now we prove the existence of solution for (4)ndash(7) usingGalerkin method Let V

119897(119909)infin

119897=1be a basis in space 1198712[minus119886 119886]

where the dimension of vector V119897(119909) is equal to the dimension

of vector119880 For each119898we define an approximate solution119880119898

of (4)ndash(7) as follows

119880119898=

119898

sum

119895=1

119888119895119898

(119905) V119895(119909) (34)

int

119886

minus119886

((120597119880119898

120597119905+ 1198601

120597119880119898

120597119909) V119894(119909)) 119889119909

= int

119886

minus119886

(119869 (119880119898) V119894(119909)) 119889119909 119894 = 1119898 119905 isin (0 119879]

(35)

119880119898

1003816100381610038161003816119905=0 = 1198800119898

(119909) 119909 isin 119877 (36)

(119860119908minus

119898∓ 119861119906minus

119898)1003816100381610038161003816119909=∓119886

=1

120573(119860119908+

119898plusmn 119861119906+

119898)1003816100381610038161003816119909=plusmn119886 plusmn

(1 minus 120573) 120578

120572120573radic120587119865

(37)

where 1198800119898

is the orthogonal projection in 1198712 of the function

1198800on the subspace spanned by V

1 V

119898

119869 (119880119898) = (119869

1(119906119898 119908119898) 1198692(119906119898 119908119898))1015840

(38)

We represent V119895(119909) in form V

119895(119909) = (V(1)

119895 V(2)119895

)1015840 where

V(1)119895

= (V1198951 V1198952 V1198953)1015840

V(2)119895

= (V1198954 V1198955 V1198956)1015840

(39)

The coefficients 119888119895119898

(119905) are determined from the equations

119898

sum

119895=1

119889119888119895119898

119889119905int

119886

minus119886

(V119895 V119894

) 119889119909 + 119888119895119898

[(119861Vminus(1)119895

Vminus(1)119894

)10038161003816100381610038161003816119909=119886

+ (119861Vminus(1)119895

Vminus(1)119894

)10038161003816100381610038161003816119909=minus119886

+ (119861Vminus(1)119894

Vminus(1)119895

)10038161003816100381610038161003816119909=119886

+ (119861Vminus(1)119894

Vminus(1)119895

)10038161003816100381610038161003816119909=minus119886

+1

120573((119860V+(2)119894

+ 119861V+(1)119894

) Vminus(1)119895

)10038161003816100381610038161003816119909=119886

minus1

120573((119860V+(2)119894

minus 119861V+(1)119894

) Vminus(1)119895

)10038161003816100381610038161003816119909=minus119886

+1

120573((119860V+(2)119895

+ 119861V+(1)119895

) Vminus(1)119894

)10038161003816100381610038161003816119909=119886

minus1

120573(119860V+(2)119895

minus 119861V+(1)119895

Vminus(1)119894

)10038161003816100381610038161003816119909=minus119886

+ (1198651 Vminus(1)119894

)10038161003816100381610038161003816119909=119886

+ (1198651 Vminus(1)119894

)10038161003816100381610038161003816119909=minus119886

+ (1198651 Vminus(1)119895

)10038161003816100381610038161003816119909=119886

+ (1198651 Vminus(1)119895

)10038161003816100381610038161003816119909=minus119886

]

minus int

119886

minus119886

((119860120597V(2)119894

120597119909 V(1)119895) + (119860

1015840120597V(1)119894

120597119909 V(2)119895

))119889119909

= int

119886

minus119886

(119869(

119898

sum

119895=1

119888119895119898V119895) V119894)119889119909

119894 = 1119898 119905 isin (0 119879]

(40)

119888119894119898

(0) = 119889119894119898 119894 = 1119898 (41)

where 119889119894119898

is the 119894th component of 1198800119898

We multiply (35) by 119888119894119898(119905) and sum over 119894 from 1 to119898

int

119886

minus119886

((120597119880119898

120597119905+ 1198601

120597119880119898

120597119909) 119880119898)119889119909

= int

119886

minus119886

(119869 (119880119898) 119880119898) 119889119909

(42)

With help of the above shown arguments we now provethat 119903

119898(119905) is bounded in some time interval [0 119879

119898] where

119880119898(119905 119909) = 119903

119898(119905)120596119898(119905 119909) 119879

119898asymp M((119880

0119898 minus 1199031198981

(0))minus1

) 119879119898

ge

119879 forall119898 and

10038171003817100381710038171003817

100381710038171003817100381711988011989810038171003817100381710038171198712[minus119886119886] minus 119903

1198981

10038171003817100381710038171003817119862[0119879]

le 1198622(10038171003817100381710038171198800

10038171003817100381710038171198712[minus119886119886] minus 1199031(0))

(43)

where 1198622is constant and independent from119898 and 119903

1198981(119905) is a

particular solution of Riccati equation about 119903119898(119905)

Problem (40)-(41) represents a Cauchy problem for theordinary system of differential equations Existence of thesolution of problem (40)-(41) follows from theory of ordi-nary system of differential equations (Picardrsquos existence anduniqueness theorem) [20]Hence the existence of the solutionof problem (34)ndash(37) follows

Thus according to estimation (43) the sequence 119880119898

minus

1199031198981

of approximate solutions of problem (5)ndash(7) is uniformlybounded in function space119862([0 119879] 1198712[minus119886 119886])Moreover thehomogeneous system of equations 120591119864+ (1120572)119860120585with respectto 120591 120585 has only a trivial solution Then it follows from resultsin [21] that 119880

119898minus 1199031198981

rarr 119880minus 1199031is week in 119862([0 119879] 119871

2

[minus119886 119886])

and 119869(119880119898) rarr 119869(119880) is week in 119862([0 119879] 119871

2

[minus119886 119886]) as 119898 rarr

infin Further it can be shown with standard methods that limitelement is a weak solution of problem (5)ndash(7)

The uniqueness of solution of problem (5)ndash(7) is provedby contradiction Let problem (5)ndash(7) have two different solu-tions 119906

1 1199081and 119906

2 1199082 We denote them again by 119906 = 119906

1minus 1199062

Journal of Applied Mathematics 7

and119908 = 1199081minus1199082 Then with respect to new values of 119906 and119908

we obtain the following problem

120597119906

120597119905+ 119860

120597119908

120597119909= 1198691(1199061 1199081) minus 1198691(1199062 1199082)

120597119908

120597119905+ 1198601015840120597119906

120597119909= 1198692(1199061 1199081) minus 1198692(1199062 1199082)

119909 isin (minus119886 119886)

(44)

119906|119905=0

= 0

119908|119905=0

= 0

(45)

(119860119908minus

+ 119861119906minus

)1003816100381610038161003816119909=minus119886 =

1

120573(119860119908+

minus 119861119906+

)1003816100381610038161003816119909=minus119886

(119860119908minus

minus 119861119906minus

)1003816100381610038161003816119909=119886 =

1

120573(119860119908+

+ 119861119906+

)1003816100381610038161003816119909=119886

(46)

We prove that solution of problem (44)ndash(46) is trivialHence the uniqueness of the solution of problem (5)ndash(7)follows

Using the method which was mentioned before we get(see equality (14))

1

2

119889

119889119905int

119886

minus119886

[(119906 119906) + (119908 119908)] 119889119909 + (119861119906minus

119906minus

)1003816100381610038161003816119909=119886

+ (119861119906minus

119906minus

)1003816100381610038161003816119909=minus119886 minus

1

120573((119860119908

+

minus 119861119906+

) 119906minus

)10038161003816100381610038161003816119909=minus119886

+1

120573((119860119908

+

+ 119861119906+

) 119906minus

)10038161003816100381610038161003816119909=119886

= int

119886

minus119886

[((1198691(1199061 1199081) minus 1198691(1199062 1199082)) 119906)

+ ((1198692(1199061 1199081) minus 1198692(1199062 1199082)) 119908)] 119889119909

(47)

Let us transform the integrand on the right side of theequality as

((1198691(1199061 1199081) minus 1198691(1199062 1199082)) 119906)

=1205902minus 1205900

2[(119891001

119891021

minus1198912

011

radic3) minus (119891

002119891022

minus1198912

012

radic3)] (119891

021minus 119891022

) =1205902minus 1205900

2[(119891001

119891021

minus 119891002

119891022

) minus1

radic3(1198912

011minus 1198912

012)] (119891021

minus 119891022

)

=1205902minus 1205900

2[(119891001

minus 119891002

) 119891021

+ 119891002

(119891021

minus 119891022

) minus1

radic3(119891011

minus 119891012

) (119891011

+ 119891012

)] (119891021

minus 119891022

)

((1198692(1199061 1199081) minus 1198692(1199062 1199082)) 119906) = ((119869

03(1199061 1199081)

minus 11986903

(1199062 1199082)) (119891031

minus 119891032

)) + ((11986911

(1199061 1199081)

minus 11986911

(1199062 1199082)) (119891111

minus 119891112

)) = (1

4(1205903+ 31205901

minus 41205900) [(119891001

minus 119891002

) 119891031

+ 119891002

(119891031

minus 119891032

)]

+1

4radic5(21205901+ 1205900minus 31205903) [(119891011

minus 119891012

) 119891021

+ 119891012

(119891021

minus 119891022

)]) (119891031

minus 119891032

) + (1205901minus 1205900)

sdot ((119891001

minus 119891002

) 119891111

+ 119891002

(119891111

minus 119891112

)

+ 05radic5

2[(119891101

minus 119891102

) 119891011

+ 119891102

(119891011

minus 119891012

)] minus radic2

15[(119891011

minus 119891012

) 119891021

+ 119891012

(119891021

minus 119891022

)]) (119891111

minus 119891112

)

(48)

Once again we use spherical representation of 119906 = 119903(119905)1199081(119905

119909) and 119908 = 119903(119905)1199082(119905 119909) where 119903(119905) = 119880(119905 sdot)

1198712[minus119886119886]

Thenconcerning 119903(119905) we obtain new initial value problem

119889119903

119889119905+ 119903119875 (119905) = 119903

2

1198761(119905) (49)

119903 (0) = 0 (50)

where 119875(119905) has the same value as in (15) and

1198761(119905) =

1205902minus 1205900

2int

119886

minus119886

[(120596001

minus 120596002

) 120596021

+ 120596002

(120596021

minus 120596022

) minus1

radic3(120596011

minus 120596012

) (120596011

+ 120596012

)] (120596021

minus 120596022

) 119889119909 +1

4(1205903+ 31205901

minus 41205900) int

119886

minus119886

[(120596001

minus 120596002

) 120596031

+ +120596002

(120596031

minus 120596032

)] (120596031

minus 120596032

) 119889119909 +1

4radic5(21205901+ 1205900minus 31205903)

sdot int

119886

minus119886

[(120596011

minus 120596012

) 120596021

+ 120596012

(120596021

minus 120596022

)]

sdot (120596031

minus 120596032

) 119889119909 + (1205901minus 1205900) int

119886

minus119886

[(120596001

minus 120596002

) 120596111

+ 120596002

(120596111

minus 120596112

) +1

2

8 Journal of Applied Mathematics

sdot radic5

2[(120596101

minus 120596102

) 120596011

+ 120596102

(120596011

minus 120596012

)]

minus radic2

15[(120596011

minus 120596012

) 120596021

+ 120596012

(120596021

minus 120596022

)]

sdot (120596111

minus 120596112

)] 119889119909

(51)

The general solution of (49) is

119903 (119905) = exp(minusint

119905

0

119875 (120591) 119889120591)

sdot [119862 minus int

119905

0

1198761(120591) exp(minusint

120591

0

119875 (120585) 119889120585) 119889120591]

minus1

(52)

Solution of (49) which is satisfying homogeneous condition(50) is trivial that is 119903(119905) = 0 Hence 119880

119862([0119879]1198712[minus119886119886])

= 0

and 1199061= 1199062and 119908

1= 1199082

The theorem is proved

Competing Interests

The authors declare that they have no competing interests

References

[1] H Grad ldquoOn the kinetic theory of rarefied gasesrdquo Communica-tions on Pure andAppliedMathematics vol 2 no 4 pp 331ndash4071949

[2] H Grad ldquoPrinciple of the kinetic theory of gasesrdquo in Thermo-dynamics of Gases vol 12 ofHandbuch der Physik pp 205ndash294Springer Berlin Germany 1958

[3] A Sakabekov Initial-Boundary Value Problems for the Boltz-mannrsquos Moment System Equations Gylym Almaty 2002

[4] C Cercignani Theory and Application of the Boltzmann Equa-tion Instituto di Matematica Milano Italy 1975

[5] M N KoganDynamic of Rarefied Gas NaukaMoscow Russia1967

[6] K Kumar ldquoPolynomial expansions in kinetic theory of gasesrdquoAnnals of Physics vol 37 no 1 pp 113ndash141 1966

[7] A Sakabekov and Y Auzhani ldquoBoundary conditions for theonedimensional nonlinear nonstationary Boltzmannrsquos momentsystem equationsrdquo Journal of Mathematical Physics vol 55Article ID 123507 2014

[8] C D Levermore ldquoMoment closure hierarchies for kinetictheoriesrdquo Journal of Statistical Physics vol 83 no 5-6 pp 1021ndash1065 1996

[9] G Mascali and V Romano ldquoA hydrodynamical model for holesin silicon semiconductors the case of non-parabolic warpedbandsrdquo Mathematical and Computer Modelling vol 53 no 1-2pp 213ndash229 2011

[10] G Mascali and V Romano ldquoA non parabolic hydrodynamicalsubband model for semiconductors based on the maximumentropy principlerdquoMathematical and Computer Modelling vol55 no 3-4 pp 1003ndash1020 2012

[11] V D Camiola and V Romano ldquo2DEG-3DEG charge transportmodel for MOSFET based on the maximum entropy principlerdquo

SIAM Journal on Applied Mathematics vol 73 no 4 pp 1439ndash1459 2013

[12] G Alı G Mascali V Romano and R C Torcasio ldquoA hydrody-namical model for covalent semiconductors with a generalizedenergy dispersion relationrdquo European Journal of Applied Math-ematics vol 25 no 2 pp 255ndash276 2014

[13] V D Camiola and V Romano ldquoHydrodynamical model forcharge transport in graphenerdquo Journal of Statistical Physics vol157 no 6 pp 1114ndash1137 2014

[14] S Mischler ldquoKinetic equations with Maxwell boundary condi-tionsrdquo Annales Scientifiques de lrsquoEcole Normale Superieure vol43 no 5 pp 719ndash760 2010

[15] V G Neudachin and U F SmirnovNucleon Association of EasyKernel Nauka Moscow Russia 1969

[16] MMoshinskyTheHarmonic Oscillator inModern Physics fromAtoms to Quarks 1960

[17] S I Pokhozhaev ldquoOn an approach to nonlinear equationrdquoDoklady Akademii Nauk SSSR vol 247 pp 1327ndash1331 1979

[18] E Kamke Differentialgleichungen Losungsmethoden und Losu-ngen I Gewohuliche Differentialgleichungen BGTeubnerLeipzig Germany 1977

[19] A Tungatarov and D K Akhmed-Zaki ldquoCauchy problemfor one class of ordinary differential equationsrdquo InternationalJournal of Mathematical Analysis vol 6 no 14 pp 695ndash6992012

[20] M Tenenbaum andH PollardOrdinary Differential EquationsHarper and Row New York NY USA 1963

[21] L Tartar ldquoCompensated compactness and applications topartial differential equationsrdquo in Proceedings of the Non-LinearAnalysis and Mechanics Heriot-Watt Symposium R J KnopsEd vol 4 of Research Notes in Math pp 136ndash212 1979

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Stochastic AnalysisInternational Journal of

4 Journal of Applied Mathematics

is a sign-nondefined square form It is easy to check that1198910011989102

minus (1198912

01radic3) = (119862119880119880) where 119880 = (119906 119908)

1015840 and

119862 =

((((((((

(

01

20 0 0 0

1

20 0 0 0 0

0 0 0 0 0 0

0 0 0 minus1

radic30 0

0 0 0 0 0 0

0 0 0 0 0 0

))))))))

)

(10)

Eigenvalues of matrix 119862 are minus12 minus1radic3 0 0 0 and 12Therefore 119869

02is a sign-nondefined square form Similarly we

can show that 11986903

and 11986911

are also sign-nondefined squareforms

For problem (4)ndash(7) the following theorem takes place

Theorem 1 If 1198800= (1199060(119909) 119908

0(119909)) isin 119871

2

[minus119886 119886] then problem(4)ndash(7) has a unique solution in domain [minus119886 119886] times [0 119879]

belonging to the space 119862([0 119879] 1198712[minus119886 119886]) moreover

10038171003817100381710038171198801198712[minus119886119886]

minus 1199031

1003817100381710038171003817119862[0119879]le 1198621(10038171003817100381710038171198800

10038171003817100381710038171198712[minus119886119886] minus 1199031(0)) (11)

where 1198621is a constant independent from 119880 and 119879 sim

M(11988001198712[minus119886119886]

minus 1199031(0))minus1

) 1199031(119905) is the particular solution of

Riccati equation (15) in an explicit form

Proof Let 1198800isin 1198712

[minus119886 119886] Let us prove estimation (11) Wemultiply the first equation in system (4) by 119906 and the secondequation by 119908 and perform integration from minus119886 to 119886

1

2

119889

119889119905int

119886

minus119886

[(119906 119906) + (119908 119908)] 119889119909

+ int

119886

minus119886

[(119860120597119908

120597119909 119906) + (119860

1015840120597119906

120597119909 119908)] 119889119909

= int

119886

minus119886

[(1198691 119906) + (119869

2 119908)] 119889119909

(12)

After integration by parts we get

1

2

119889

119889119905int

119886

minus119886

[(119906 119906) + (119908 119908)] 119889119909 + (119906minus

119860119908minus

)1003816100381610038161003816119909=119886

minus (119906minus

119860119908minus

)1003816100381610038161003816119909=minus119886 = int

119886

minus119886

[(1198691 119906) + (119869

2 119908)] 119889119909

(13)

Taking into account boundary conditions (6)-(7) we rewriteequality (13) in the following form

1

2

119889

119889119905int

119886

minus119886

[(119906 119906) + (119908 119908)] 119889119909 + (119861119906minus

119906minus

)1003816100381610038161003816119909=119886

+ (119861119906minus

119906minus

)1003816100381610038161003816119909=minus119886 minus

1

120573((119860119908

+

minus 119861119906+

) 119906minus

)10038161003816100381610038161003816119909=minus119886

+1

120573((119860119908+

+ 119861119906+

) 119906minus

)1003816100381610038161003816119909=119886 + (119865

1 119906minus

)1003816100381610038161003816119909=119886

+ (1198651 119906minus

)1003816100381610038161003816119909=minus119886

= int

119886

minus119886

[(1198691(119906 119908) 119906) + (119869

2(119906 119908) 119908)] 119889119909

(14)

where 1198651= ((1 minus 120573)120578120572120573radic120587)119865

Let us use the spherical representation [17] of vector119880(119905 119909) = 119903(119905)120596(119905 119909) where 120596(119905 119909) = (120596

1(119905 119909) 120596

2(119905 119909))

1015840119903(119905) = 119880(119905 sdot)

1198712[minus119886119886]

and 1205961198712[minus119886119886]

= 1 Furthermoreassume that120596(119905 119909) isin 119862

2

[0 119879] where the value of119879we definebelow

Substituting the values 119906 = 119903(119905)1205961(119905 119909) and 119908 =

119903(119905)1205962(119905 119909) into (14) we have that

119889119903

119889119905+ 119903119875 (119905) = 119903

2

119876 (119905) minus 119891 (119905) (15)

where

119875 (119905) = (119861120596minus

1 120596minus

1)1003816100381610038161003816119909=119886 + (119861120596

minus

1 120596minus

1)1003816100381610038161003816119909=minus119886

+1

120573[(119860120596+

2 120596minus

1)1003816100381610038161003816119909=119886 + (119861120596

+

1 120596minus

1)1003816100381610038161003816119909=119886

+ (119861120596+

1 120596minus

1)1003816100381610038161003816119909=minus119886 minus (119860120596

+

2 120596minus

1)1003816100381610038161003816119909=minus119886]

119876 (119905) = int

119886

minus119886

[(1198691(1205961 1205962) 1205961) + (1198692(1205961 1205962) 1205962)] 119889119909

119891 (119905) = (1198651 120596minus

1)1003816100381610038161003816119909=119886 + (119865

1 120596minus

1)1003816100381610038161003816119909=minus119886

(16)

Let us study (15) with an initial condition

119903 (0) =10038171003817100381710038171198800

1003817100381710038171003817 =10038171003817100381710038171198800

10038171003817100381710038171198712[minus119886119886] (17)

Let 1199031(119905) be a particular solution of Riccati equation (15)Then

the general solution of (15) has the form

119903 (119905) = 1199031(119905) + exp(int

119905

0

(2119876 (120591) 1199031(120591) minus 119875 (120591)) 119889120591) [119862

minus int

119905

0

119876 (120591)

sdot exp(int

120591

0

(2119876 (120585) 1199031(120585) minus 119875 (120585)) 119889120585) 119889120591]

minus1

(18)

Hence taking into consideration initial condition (17) weobtain

Journal of Applied Mathematics 5

119903 (119905) = 1199031(119905) + exp(int

119905

0

(2119876 (120591) 1199031(120591) minus 119875 (120591)) 119889120591)[

110038171003817100381710038171198800

1003817100381710038171003817 minus 1199031(0)

minus int

119905

0

119876 (120591) exp(int

120591

0

(2119876 (120585) 1199031(120585) minus 119875 (120585)) 119889120585) 119889120591]

minus1

(19)

If 119877(119905) equiv int119905

0

119876(120591) exp(int1205910

(2119876(120585)1199031(120585) minus 119875(120585))119889120585)119889120591 le 0 forall119905

then the second term of 119903(119905) is bounded forall119905 isin [0 +infin) underthe condition that difference 119880

0 minus 1199031(0) is positive Further

we suppose a difference 1198800 minus 1199031(0) is a positive number Let

119877(119905) gt 0 We denote by 1198791the moment of time at which

110038171003817100381710038171198800

1003817100381710038171003817 minus 1199031(0)

minus int

1198791

0

119876 (120591) exp(int

120591

0

(2119876 (120585) 1199031(120585) minus 119875 (120585)) 119889120585) 119889120591

= 0

(20)

Then 119903(119905) minus 1199031(119905) is bounded forall119905 isin [0 119879] where 119879 lt 119879

1and

1198791sim M(1(119880

0 minus 1199031(0))) Analyzing structure of 119903

1(119905) (see

(31)) from equality (20) we obtain 119876(119905) gt 0 for any 119905 isin [0 119879]Now we found one particular solution of (15) The

nonlinear Riccati equation can always be reduced to a second-order linear ordinary differential equation [18]

11991010158401015840

+ 1198911(119905) 1199101015840

+ 1198912(119905) 119910 = 0 (21)

where

119910 (119905) = exp(minusint

119905

0

119876 (120591) 119903 (120591) 119889120591)

1198911(119905) =

119876 (119905) 119875 (119905) minus 1198761015840

(119905)

119876 (119905)

1198912(119905) = 119876 (119905) 119891 (119905)

(22)

Solution of this equation will lead to a solution of 119903(119905) =

minus1199101015840

119910119876(119905) of the original Riccati equation By substitution

119910 (119905) = 119911 (119905) exp(minus1

2int

119905

0

1198911(120591) 119889120591) (23)

we write (21) in the following form [18]

11991110158401015840

(119905) + 119886 (119905) 119911 (119905) = 0 (24)

where 119886(119905) = 1198912(119905) minus (12)119891

1015840

1(119905) minus (14)119891

2

1(119905)

If instead of1198911(119905) and119891

2(119905)we substitute values from (22)

then we get

119886 (119905) = 119876119891 (119905)

minus2 (1198762

1198751015840

minus 11987610158401015840

119876 + 11987610158402

) + (119876119875 minus 1198761015840

)2

41198762

(25)

In work [19] it was proven that function

119911 (119905) = 1 +

infin

sum

119896=1

119887119896(119905) (26)

is the solution of (24) where

1198871(119905) = minusint

119905

0

int

120585

0

119886 (120591) 119889120591 119889120585

119887119896(119905) = minusint

119905

0

int

120585

0

119886 (120591) 119887119896minus1

(120591) 119889120591 119889120585

(27)

and the series on the right hand side of (26) uniformlyconverge

Then

119910 = (1 +

infin

sum

119896=1

119887119896(119905)) exp(minus

1

2int

119905

0

1198911(120591) 119889120591)

1199031(119905) = minus

1199101015840

119910119876= minus

1199111015840

(119905) minus (12) 119911 (119905) 1198911(119905)

119911 (119905) 119876 (119905)

(28)

is a particular solution of Riccati equation (15)Since 1199111015840(0) = 0 value of

1199031(0) = minus

1199111015840

(0) minus (12) 119911 (0) 1198911(0)

119911 (0) 119876 (0)=

1198911(0)

2119876 (0) (29)

Now instead of 1198911(0) we substitute its value 119875(0) minus

1198761015840

(0)119876(0) and get 1199031(0) = (119875(0)119876(0) minus 119876

1015840

(0))21198762

(0)The value of 119876(119905) with 119905 = 0 is

119876 (0)

= int

119886

minus119886

((1198691(1205961 1205962) 1205961) + (1198692(1205961 1205962) 1205962))1003816100381610038161003816119905=0 119889119909

gt 0

(30)

since 119876(119905) gt 0 for any 119905 isin [0 119879]

An expansion form of the function

1199031(119905) =

1

119876 (119905)[

1

119911 (119905)int

119905

0

119886 (120591) 119911 (120591) 119889120591

+119876 (119905) 119875 (119905) minus 119876

1015840

(119905)

2119876 (119905)]

(31)

It is not difficult to prove using immediate substitutionthat function 119903

1(119905) satisfies Riccati equation (15)

From equality (19) we obtain1003816100381610038161003816119903 (119905) minus 119903

1(119905)

1003816100381610038161003816 le 119862 (10038171003817100381710038171198800

1003817100381710038171003817 minus 1199031(0)) (32)

Hence1003816100381610038161003816119880 (119905 sdot)

1198712[minus119886119886]

minus 1199031(119905)

1003816100381610038161003816 le 119862 (10038171003817100381710038171198800

1003817100381710038171003817 minus 1199031(0))

10038171003817100381710038171198801198712[minus119886119886]

minus 1199031

1003817100381710038171003817119862[0119879]le 119862 (

100381710038171003817100381711988001003817100381710038171003817 minus 1199031(0))

(33)

6 Journal of Applied Mathematics

We take as interval of the solution of problem (15)ndash(17) segment [0 1(119880

0 minus 119903

1(0))] since the integrand

119876(120591) exp(int1205910

(2119876(120585)1199031(120585) minus 119875(120585))119889120585) is bounded Hence forall119905 isin

[0 119879] (where 119879 asymp 1198791) a priori estimation (11) takes place

Now we prove the existence of solution for (4)ndash(7) usingGalerkin method Let V

119897(119909)infin

119897=1be a basis in space 1198712[minus119886 119886]

where the dimension of vector V119897(119909) is equal to the dimension

of vector119880 For each119898we define an approximate solution119880119898

of (4)ndash(7) as follows

119880119898=

119898

sum

119895=1

119888119895119898

(119905) V119895(119909) (34)

int

119886

minus119886

((120597119880119898

120597119905+ 1198601

120597119880119898

120597119909) V119894(119909)) 119889119909

= int

119886

minus119886

(119869 (119880119898) V119894(119909)) 119889119909 119894 = 1119898 119905 isin (0 119879]

(35)

119880119898

1003816100381610038161003816119905=0 = 1198800119898

(119909) 119909 isin 119877 (36)

(119860119908minus

119898∓ 119861119906minus

119898)1003816100381610038161003816119909=∓119886

=1

120573(119860119908+

119898plusmn 119861119906+

119898)1003816100381610038161003816119909=plusmn119886 plusmn

(1 minus 120573) 120578

120572120573radic120587119865

(37)

where 1198800119898

is the orthogonal projection in 1198712 of the function

1198800on the subspace spanned by V

1 V

119898

119869 (119880119898) = (119869

1(119906119898 119908119898) 1198692(119906119898 119908119898))1015840

(38)

We represent V119895(119909) in form V

119895(119909) = (V(1)

119895 V(2)119895

)1015840 where

V(1)119895

= (V1198951 V1198952 V1198953)1015840

V(2)119895

= (V1198954 V1198955 V1198956)1015840

(39)

The coefficients 119888119895119898

(119905) are determined from the equations

119898

sum

119895=1

119889119888119895119898

119889119905int

119886

minus119886

(V119895 V119894

) 119889119909 + 119888119895119898

[(119861Vminus(1)119895

Vminus(1)119894

)10038161003816100381610038161003816119909=119886

+ (119861Vminus(1)119895

Vminus(1)119894

)10038161003816100381610038161003816119909=minus119886

+ (119861Vminus(1)119894

Vminus(1)119895

)10038161003816100381610038161003816119909=119886

+ (119861Vminus(1)119894

Vminus(1)119895

)10038161003816100381610038161003816119909=minus119886

+1

120573((119860V+(2)119894

+ 119861V+(1)119894

) Vminus(1)119895

)10038161003816100381610038161003816119909=119886

minus1

120573((119860V+(2)119894

minus 119861V+(1)119894

) Vminus(1)119895

)10038161003816100381610038161003816119909=minus119886

+1

120573((119860V+(2)119895

+ 119861V+(1)119895

) Vminus(1)119894

)10038161003816100381610038161003816119909=119886

minus1

120573(119860V+(2)119895

minus 119861V+(1)119895

Vminus(1)119894

)10038161003816100381610038161003816119909=minus119886

+ (1198651 Vminus(1)119894

)10038161003816100381610038161003816119909=119886

+ (1198651 Vminus(1)119894

)10038161003816100381610038161003816119909=minus119886

+ (1198651 Vminus(1)119895

)10038161003816100381610038161003816119909=119886

+ (1198651 Vminus(1)119895

)10038161003816100381610038161003816119909=minus119886

]

minus int

119886

minus119886

((119860120597V(2)119894

120597119909 V(1)119895) + (119860

1015840120597V(1)119894

120597119909 V(2)119895

))119889119909

= int

119886

minus119886

(119869(

119898

sum

119895=1

119888119895119898V119895) V119894)119889119909

119894 = 1119898 119905 isin (0 119879]

(40)

119888119894119898

(0) = 119889119894119898 119894 = 1119898 (41)

where 119889119894119898

is the 119894th component of 1198800119898

We multiply (35) by 119888119894119898(119905) and sum over 119894 from 1 to119898

int

119886

minus119886

((120597119880119898

120597119905+ 1198601

120597119880119898

120597119909) 119880119898)119889119909

= int

119886

minus119886

(119869 (119880119898) 119880119898) 119889119909

(42)

With help of the above shown arguments we now provethat 119903

119898(119905) is bounded in some time interval [0 119879

119898] where

119880119898(119905 119909) = 119903

119898(119905)120596119898(119905 119909) 119879

119898asymp M((119880

0119898 minus 1199031198981

(0))minus1

) 119879119898

ge

119879 forall119898 and

10038171003817100381710038171003817

100381710038171003817100381711988011989810038171003817100381710038171198712[minus119886119886] minus 119903

1198981

10038171003817100381710038171003817119862[0119879]

le 1198622(10038171003817100381710038171198800

10038171003817100381710038171198712[minus119886119886] minus 1199031(0))

(43)

where 1198622is constant and independent from119898 and 119903

1198981(119905) is a

particular solution of Riccati equation about 119903119898(119905)

Problem (40)-(41) represents a Cauchy problem for theordinary system of differential equations Existence of thesolution of problem (40)-(41) follows from theory of ordi-nary system of differential equations (Picardrsquos existence anduniqueness theorem) [20]Hence the existence of the solutionof problem (34)ndash(37) follows

Thus according to estimation (43) the sequence 119880119898

minus

1199031198981

of approximate solutions of problem (5)ndash(7) is uniformlybounded in function space119862([0 119879] 1198712[minus119886 119886])Moreover thehomogeneous system of equations 120591119864+ (1120572)119860120585with respectto 120591 120585 has only a trivial solution Then it follows from resultsin [21] that 119880

119898minus 1199031198981

rarr 119880minus 1199031is week in 119862([0 119879] 119871

2

[minus119886 119886])

and 119869(119880119898) rarr 119869(119880) is week in 119862([0 119879] 119871

2

[minus119886 119886]) as 119898 rarr

infin Further it can be shown with standard methods that limitelement is a weak solution of problem (5)ndash(7)

The uniqueness of solution of problem (5)ndash(7) is provedby contradiction Let problem (5)ndash(7) have two different solu-tions 119906

1 1199081and 119906

2 1199082 We denote them again by 119906 = 119906

1minus 1199062

Journal of Applied Mathematics 7

and119908 = 1199081minus1199082 Then with respect to new values of 119906 and119908

we obtain the following problem

120597119906

120597119905+ 119860

120597119908

120597119909= 1198691(1199061 1199081) minus 1198691(1199062 1199082)

120597119908

120597119905+ 1198601015840120597119906

120597119909= 1198692(1199061 1199081) minus 1198692(1199062 1199082)

119909 isin (minus119886 119886)

(44)

119906|119905=0

= 0

119908|119905=0

= 0

(45)

(119860119908minus

+ 119861119906minus

)1003816100381610038161003816119909=minus119886 =

1

120573(119860119908+

minus 119861119906+

)1003816100381610038161003816119909=minus119886

(119860119908minus

minus 119861119906minus

)1003816100381610038161003816119909=119886 =

1

120573(119860119908+

+ 119861119906+

)1003816100381610038161003816119909=119886

(46)

We prove that solution of problem (44)ndash(46) is trivialHence the uniqueness of the solution of problem (5)ndash(7)follows

Using the method which was mentioned before we get(see equality (14))

1

2

119889

119889119905int

119886

minus119886

[(119906 119906) + (119908 119908)] 119889119909 + (119861119906minus

119906minus

)1003816100381610038161003816119909=119886

+ (119861119906minus

119906minus

)1003816100381610038161003816119909=minus119886 minus

1

120573((119860119908

+

minus 119861119906+

) 119906minus

)10038161003816100381610038161003816119909=minus119886

+1

120573((119860119908

+

+ 119861119906+

) 119906minus

)10038161003816100381610038161003816119909=119886

= int

119886

minus119886

[((1198691(1199061 1199081) minus 1198691(1199062 1199082)) 119906)

+ ((1198692(1199061 1199081) minus 1198692(1199062 1199082)) 119908)] 119889119909

(47)

Let us transform the integrand on the right side of theequality as

((1198691(1199061 1199081) minus 1198691(1199062 1199082)) 119906)

=1205902minus 1205900

2[(119891001

119891021

minus1198912

011

radic3) minus (119891

002119891022

minus1198912

012

radic3)] (119891

021minus 119891022

) =1205902minus 1205900

2[(119891001

119891021

minus 119891002

119891022

) minus1

radic3(1198912

011minus 1198912

012)] (119891021

minus 119891022

)

=1205902minus 1205900

2[(119891001

minus 119891002

) 119891021

+ 119891002

(119891021

minus 119891022

) minus1

radic3(119891011

minus 119891012

) (119891011

+ 119891012

)] (119891021

minus 119891022

)

((1198692(1199061 1199081) minus 1198692(1199062 1199082)) 119906) = ((119869

03(1199061 1199081)

minus 11986903

(1199062 1199082)) (119891031

minus 119891032

)) + ((11986911

(1199061 1199081)

minus 11986911

(1199062 1199082)) (119891111

minus 119891112

)) = (1

4(1205903+ 31205901

minus 41205900) [(119891001

minus 119891002

) 119891031

+ 119891002

(119891031

minus 119891032

)]

+1

4radic5(21205901+ 1205900minus 31205903) [(119891011

minus 119891012

) 119891021

+ 119891012

(119891021

minus 119891022

)]) (119891031

minus 119891032

) + (1205901minus 1205900)

sdot ((119891001

minus 119891002

) 119891111

+ 119891002

(119891111

minus 119891112

)

+ 05radic5

2[(119891101

minus 119891102

) 119891011

+ 119891102

(119891011

minus 119891012

)] minus radic2

15[(119891011

minus 119891012

) 119891021

+ 119891012

(119891021

minus 119891022

)]) (119891111

minus 119891112

)

(48)

Once again we use spherical representation of 119906 = 119903(119905)1199081(119905

119909) and 119908 = 119903(119905)1199082(119905 119909) where 119903(119905) = 119880(119905 sdot)

1198712[minus119886119886]

Thenconcerning 119903(119905) we obtain new initial value problem

119889119903

119889119905+ 119903119875 (119905) = 119903

2

1198761(119905) (49)

119903 (0) = 0 (50)

where 119875(119905) has the same value as in (15) and

1198761(119905) =

1205902minus 1205900

2int

119886

minus119886

[(120596001

minus 120596002

) 120596021

+ 120596002

(120596021

minus 120596022

) minus1

radic3(120596011

minus 120596012

) (120596011

+ 120596012

)] (120596021

minus 120596022

) 119889119909 +1

4(1205903+ 31205901

minus 41205900) int

119886

minus119886

[(120596001

minus 120596002

) 120596031

+ +120596002

(120596031

minus 120596032

)] (120596031

minus 120596032

) 119889119909 +1

4radic5(21205901+ 1205900minus 31205903)

sdot int

119886

minus119886

[(120596011

minus 120596012

) 120596021

+ 120596012

(120596021

minus 120596022

)]

sdot (120596031

minus 120596032

) 119889119909 + (1205901minus 1205900) int

119886

minus119886

[(120596001

minus 120596002

) 120596111

+ 120596002

(120596111

minus 120596112

) +1

2

8 Journal of Applied Mathematics

sdot radic5

2[(120596101

minus 120596102

) 120596011

+ 120596102

(120596011

minus 120596012

)]

minus radic2

15[(120596011

minus 120596012

) 120596021

+ 120596012

(120596021

minus 120596022

)]

sdot (120596111

minus 120596112

)] 119889119909

(51)

The general solution of (49) is

119903 (119905) = exp(minusint

119905

0

119875 (120591) 119889120591)

sdot [119862 minus int

119905

0

1198761(120591) exp(minusint

120591

0

119875 (120585) 119889120585) 119889120591]

minus1

(52)

Solution of (49) which is satisfying homogeneous condition(50) is trivial that is 119903(119905) = 0 Hence 119880

119862([0119879]1198712[minus119886119886])

= 0

and 1199061= 1199062and 119908

1= 1199082

The theorem is proved

Competing Interests

The authors declare that they have no competing interests

References

[1] H Grad ldquoOn the kinetic theory of rarefied gasesrdquo Communica-tions on Pure andAppliedMathematics vol 2 no 4 pp 331ndash4071949

[2] H Grad ldquoPrinciple of the kinetic theory of gasesrdquo in Thermo-dynamics of Gases vol 12 ofHandbuch der Physik pp 205ndash294Springer Berlin Germany 1958

[3] A Sakabekov Initial-Boundary Value Problems for the Boltz-mannrsquos Moment System Equations Gylym Almaty 2002

[4] C Cercignani Theory and Application of the Boltzmann Equa-tion Instituto di Matematica Milano Italy 1975

[5] M N KoganDynamic of Rarefied Gas NaukaMoscow Russia1967

[6] K Kumar ldquoPolynomial expansions in kinetic theory of gasesrdquoAnnals of Physics vol 37 no 1 pp 113ndash141 1966

[7] A Sakabekov and Y Auzhani ldquoBoundary conditions for theonedimensional nonlinear nonstationary Boltzmannrsquos momentsystem equationsrdquo Journal of Mathematical Physics vol 55Article ID 123507 2014

[8] C D Levermore ldquoMoment closure hierarchies for kinetictheoriesrdquo Journal of Statistical Physics vol 83 no 5-6 pp 1021ndash1065 1996

[9] G Mascali and V Romano ldquoA hydrodynamical model for holesin silicon semiconductors the case of non-parabolic warpedbandsrdquo Mathematical and Computer Modelling vol 53 no 1-2pp 213ndash229 2011

[10] G Mascali and V Romano ldquoA non parabolic hydrodynamicalsubband model for semiconductors based on the maximumentropy principlerdquoMathematical and Computer Modelling vol55 no 3-4 pp 1003ndash1020 2012

[11] V D Camiola and V Romano ldquo2DEG-3DEG charge transportmodel for MOSFET based on the maximum entropy principlerdquo

SIAM Journal on Applied Mathematics vol 73 no 4 pp 1439ndash1459 2013

[12] G Alı G Mascali V Romano and R C Torcasio ldquoA hydrody-namical model for covalent semiconductors with a generalizedenergy dispersion relationrdquo European Journal of Applied Math-ematics vol 25 no 2 pp 255ndash276 2014

[13] V D Camiola and V Romano ldquoHydrodynamical model forcharge transport in graphenerdquo Journal of Statistical Physics vol157 no 6 pp 1114ndash1137 2014

[14] S Mischler ldquoKinetic equations with Maxwell boundary condi-tionsrdquo Annales Scientifiques de lrsquoEcole Normale Superieure vol43 no 5 pp 719ndash760 2010

[15] V G Neudachin and U F SmirnovNucleon Association of EasyKernel Nauka Moscow Russia 1969

[16] MMoshinskyTheHarmonic Oscillator inModern Physics fromAtoms to Quarks 1960

[17] S I Pokhozhaev ldquoOn an approach to nonlinear equationrdquoDoklady Akademii Nauk SSSR vol 247 pp 1327ndash1331 1979

[18] E Kamke Differentialgleichungen Losungsmethoden und Losu-ngen I Gewohuliche Differentialgleichungen BGTeubnerLeipzig Germany 1977

[19] A Tungatarov and D K Akhmed-Zaki ldquoCauchy problemfor one class of ordinary differential equationsrdquo InternationalJournal of Mathematical Analysis vol 6 no 14 pp 695ndash6992012

[20] M Tenenbaum andH PollardOrdinary Differential EquationsHarper and Row New York NY USA 1963

[21] L Tartar ldquoCompensated compactness and applications topartial differential equationsrdquo in Proceedings of the Non-LinearAnalysis and Mechanics Heriot-Watt Symposium R J KnopsEd vol 4 of Research Notes in Math pp 136ndash212 1979

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Applied Mathematics 5

119903 (119905) = 1199031(119905) + exp(int

119905

0

(2119876 (120591) 1199031(120591) minus 119875 (120591)) 119889120591)[

110038171003817100381710038171198800

1003817100381710038171003817 minus 1199031(0)

minus int

119905

0

119876 (120591) exp(int

120591

0

(2119876 (120585) 1199031(120585) minus 119875 (120585)) 119889120585) 119889120591]

minus1

(19)

If 119877(119905) equiv int119905

0

119876(120591) exp(int1205910

(2119876(120585)1199031(120585) minus 119875(120585))119889120585)119889120591 le 0 forall119905

then the second term of 119903(119905) is bounded forall119905 isin [0 +infin) underthe condition that difference 119880

0 minus 1199031(0) is positive Further

we suppose a difference 1198800 minus 1199031(0) is a positive number Let

119877(119905) gt 0 We denote by 1198791the moment of time at which

110038171003817100381710038171198800

1003817100381710038171003817 minus 1199031(0)

minus int

1198791

0

119876 (120591) exp(int

120591

0

(2119876 (120585) 1199031(120585) minus 119875 (120585)) 119889120585) 119889120591

= 0

(20)

Then 119903(119905) minus 1199031(119905) is bounded forall119905 isin [0 119879] where 119879 lt 119879

1and

1198791sim M(1(119880

0 minus 1199031(0))) Analyzing structure of 119903

1(119905) (see

(31)) from equality (20) we obtain 119876(119905) gt 0 for any 119905 isin [0 119879]Now we found one particular solution of (15) The

nonlinear Riccati equation can always be reduced to a second-order linear ordinary differential equation [18]

11991010158401015840

+ 1198911(119905) 1199101015840

+ 1198912(119905) 119910 = 0 (21)

where

119910 (119905) = exp(minusint

119905

0

119876 (120591) 119903 (120591) 119889120591)

1198911(119905) =

119876 (119905) 119875 (119905) minus 1198761015840

(119905)

119876 (119905)

1198912(119905) = 119876 (119905) 119891 (119905)

(22)

Solution of this equation will lead to a solution of 119903(119905) =

minus1199101015840

119910119876(119905) of the original Riccati equation By substitution

119910 (119905) = 119911 (119905) exp(minus1

2int

119905

0

1198911(120591) 119889120591) (23)

we write (21) in the following form [18]

11991110158401015840

(119905) + 119886 (119905) 119911 (119905) = 0 (24)

where 119886(119905) = 1198912(119905) minus (12)119891

1015840

1(119905) minus (14)119891

2

1(119905)

If instead of1198911(119905) and119891

2(119905)we substitute values from (22)

then we get

119886 (119905) = 119876119891 (119905)

minus2 (1198762

1198751015840

minus 11987610158401015840

119876 + 11987610158402

) + (119876119875 minus 1198761015840

)2

41198762

(25)

In work [19] it was proven that function

119911 (119905) = 1 +

infin

sum

119896=1

119887119896(119905) (26)

is the solution of (24) where

1198871(119905) = minusint

119905

0

int

120585

0

119886 (120591) 119889120591 119889120585

119887119896(119905) = minusint

119905

0

int

120585

0

119886 (120591) 119887119896minus1

(120591) 119889120591 119889120585

(27)

and the series on the right hand side of (26) uniformlyconverge

Then

119910 = (1 +

infin

sum

119896=1

119887119896(119905)) exp(minus

1

2int

119905

0

1198911(120591) 119889120591)

1199031(119905) = minus

1199101015840

119910119876= minus

1199111015840

(119905) minus (12) 119911 (119905) 1198911(119905)

119911 (119905) 119876 (119905)

(28)

is a particular solution of Riccati equation (15)Since 1199111015840(0) = 0 value of

1199031(0) = minus

1199111015840

(0) minus (12) 119911 (0) 1198911(0)

119911 (0) 119876 (0)=

1198911(0)

2119876 (0) (29)

Now instead of 1198911(0) we substitute its value 119875(0) minus

1198761015840

(0)119876(0) and get 1199031(0) = (119875(0)119876(0) minus 119876

1015840

(0))21198762

(0)The value of 119876(119905) with 119905 = 0 is

119876 (0)

= int

119886

minus119886

((1198691(1205961 1205962) 1205961) + (1198692(1205961 1205962) 1205962))1003816100381610038161003816119905=0 119889119909

gt 0

(30)

since 119876(119905) gt 0 for any 119905 isin [0 119879]

An expansion form of the function

1199031(119905) =

1

119876 (119905)[

1

119911 (119905)int

119905

0

119886 (120591) 119911 (120591) 119889120591

+119876 (119905) 119875 (119905) minus 119876

1015840

(119905)

2119876 (119905)]

(31)

It is not difficult to prove using immediate substitutionthat function 119903

1(119905) satisfies Riccati equation (15)

From equality (19) we obtain1003816100381610038161003816119903 (119905) minus 119903

1(119905)

1003816100381610038161003816 le 119862 (10038171003817100381710038171198800

1003817100381710038171003817 minus 1199031(0)) (32)

Hence1003816100381610038161003816119880 (119905 sdot)

1198712[minus119886119886]

minus 1199031(119905)

1003816100381610038161003816 le 119862 (10038171003817100381710038171198800

1003817100381710038171003817 minus 1199031(0))

10038171003817100381710038171198801198712[minus119886119886]

minus 1199031

1003817100381710038171003817119862[0119879]le 119862 (

100381710038171003817100381711988001003817100381710038171003817 minus 1199031(0))

(33)

6 Journal of Applied Mathematics

We take as interval of the solution of problem (15)ndash(17) segment [0 1(119880

0 minus 119903

1(0))] since the integrand

119876(120591) exp(int1205910

(2119876(120585)1199031(120585) minus 119875(120585))119889120585) is bounded Hence forall119905 isin

[0 119879] (where 119879 asymp 1198791) a priori estimation (11) takes place

Now we prove the existence of solution for (4)ndash(7) usingGalerkin method Let V

119897(119909)infin

119897=1be a basis in space 1198712[minus119886 119886]

where the dimension of vector V119897(119909) is equal to the dimension

of vector119880 For each119898we define an approximate solution119880119898

of (4)ndash(7) as follows

119880119898=

119898

sum

119895=1

119888119895119898

(119905) V119895(119909) (34)

int

119886

minus119886

((120597119880119898

120597119905+ 1198601

120597119880119898

120597119909) V119894(119909)) 119889119909

= int

119886

minus119886

(119869 (119880119898) V119894(119909)) 119889119909 119894 = 1119898 119905 isin (0 119879]

(35)

119880119898

1003816100381610038161003816119905=0 = 1198800119898

(119909) 119909 isin 119877 (36)

(119860119908minus

119898∓ 119861119906minus

119898)1003816100381610038161003816119909=∓119886

=1

120573(119860119908+

119898plusmn 119861119906+

119898)1003816100381610038161003816119909=plusmn119886 plusmn

(1 minus 120573) 120578

120572120573radic120587119865

(37)

where 1198800119898

is the orthogonal projection in 1198712 of the function

1198800on the subspace spanned by V

1 V

119898

119869 (119880119898) = (119869

1(119906119898 119908119898) 1198692(119906119898 119908119898))1015840

(38)

We represent V119895(119909) in form V

119895(119909) = (V(1)

119895 V(2)119895

)1015840 where

V(1)119895

= (V1198951 V1198952 V1198953)1015840

V(2)119895

= (V1198954 V1198955 V1198956)1015840

(39)

The coefficients 119888119895119898

(119905) are determined from the equations

119898

sum

119895=1

119889119888119895119898

119889119905int

119886

minus119886

(V119895 V119894

) 119889119909 + 119888119895119898

[(119861Vminus(1)119895

Vminus(1)119894

)10038161003816100381610038161003816119909=119886

+ (119861Vminus(1)119895

Vminus(1)119894

)10038161003816100381610038161003816119909=minus119886

+ (119861Vminus(1)119894

Vminus(1)119895

)10038161003816100381610038161003816119909=119886

+ (119861Vminus(1)119894

Vminus(1)119895

)10038161003816100381610038161003816119909=minus119886

+1

120573((119860V+(2)119894

+ 119861V+(1)119894

) Vminus(1)119895

)10038161003816100381610038161003816119909=119886

minus1

120573((119860V+(2)119894

minus 119861V+(1)119894

) Vminus(1)119895

)10038161003816100381610038161003816119909=minus119886

+1

120573((119860V+(2)119895

+ 119861V+(1)119895

) Vminus(1)119894

)10038161003816100381610038161003816119909=119886

minus1

120573(119860V+(2)119895

minus 119861V+(1)119895

Vminus(1)119894

)10038161003816100381610038161003816119909=minus119886

+ (1198651 Vminus(1)119894

)10038161003816100381610038161003816119909=119886

+ (1198651 Vminus(1)119894

)10038161003816100381610038161003816119909=minus119886

+ (1198651 Vminus(1)119895

)10038161003816100381610038161003816119909=119886

+ (1198651 Vminus(1)119895

)10038161003816100381610038161003816119909=minus119886

]

minus int

119886

minus119886

((119860120597V(2)119894

120597119909 V(1)119895) + (119860

1015840120597V(1)119894

120597119909 V(2)119895

))119889119909

= int

119886

minus119886

(119869(

119898

sum

119895=1

119888119895119898V119895) V119894)119889119909

119894 = 1119898 119905 isin (0 119879]

(40)

119888119894119898

(0) = 119889119894119898 119894 = 1119898 (41)

where 119889119894119898

is the 119894th component of 1198800119898

We multiply (35) by 119888119894119898(119905) and sum over 119894 from 1 to119898

int

119886

minus119886

((120597119880119898

120597119905+ 1198601

120597119880119898

120597119909) 119880119898)119889119909

= int

119886

minus119886

(119869 (119880119898) 119880119898) 119889119909

(42)

With help of the above shown arguments we now provethat 119903

119898(119905) is bounded in some time interval [0 119879

119898] where

119880119898(119905 119909) = 119903

119898(119905)120596119898(119905 119909) 119879

119898asymp M((119880

0119898 minus 1199031198981

(0))minus1

) 119879119898

ge

119879 forall119898 and

10038171003817100381710038171003817

100381710038171003817100381711988011989810038171003817100381710038171198712[minus119886119886] minus 119903

1198981

10038171003817100381710038171003817119862[0119879]

le 1198622(10038171003817100381710038171198800

10038171003817100381710038171198712[minus119886119886] minus 1199031(0))

(43)

where 1198622is constant and independent from119898 and 119903

1198981(119905) is a

particular solution of Riccati equation about 119903119898(119905)

Problem (40)-(41) represents a Cauchy problem for theordinary system of differential equations Existence of thesolution of problem (40)-(41) follows from theory of ordi-nary system of differential equations (Picardrsquos existence anduniqueness theorem) [20]Hence the existence of the solutionof problem (34)ndash(37) follows

Thus according to estimation (43) the sequence 119880119898

minus

1199031198981

of approximate solutions of problem (5)ndash(7) is uniformlybounded in function space119862([0 119879] 1198712[minus119886 119886])Moreover thehomogeneous system of equations 120591119864+ (1120572)119860120585with respectto 120591 120585 has only a trivial solution Then it follows from resultsin [21] that 119880

119898minus 1199031198981

rarr 119880minus 1199031is week in 119862([0 119879] 119871

2

[minus119886 119886])

and 119869(119880119898) rarr 119869(119880) is week in 119862([0 119879] 119871

2

[minus119886 119886]) as 119898 rarr

infin Further it can be shown with standard methods that limitelement is a weak solution of problem (5)ndash(7)

The uniqueness of solution of problem (5)ndash(7) is provedby contradiction Let problem (5)ndash(7) have two different solu-tions 119906

1 1199081and 119906

2 1199082 We denote them again by 119906 = 119906

1minus 1199062

Journal of Applied Mathematics 7

and119908 = 1199081minus1199082 Then with respect to new values of 119906 and119908

we obtain the following problem

120597119906

120597119905+ 119860

120597119908

120597119909= 1198691(1199061 1199081) minus 1198691(1199062 1199082)

120597119908

120597119905+ 1198601015840120597119906

120597119909= 1198692(1199061 1199081) minus 1198692(1199062 1199082)

119909 isin (minus119886 119886)

(44)

119906|119905=0

= 0

119908|119905=0

= 0

(45)

(119860119908minus

+ 119861119906minus

)1003816100381610038161003816119909=minus119886 =

1

120573(119860119908+

minus 119861119906+

)1003816100381610038161003816119909=minus119886

(119860119908minus

minus 119861119906minus

)1003816100381610038161003816119909=119886 =

1

120573(119860119908+

+ 119861119906+

)1003816100381610038161003816119909=119886

(46)

We prove that solution of problem (44)ndash(46) is trivialHence the uniqueness of the solution of problem (5)ndash(7)follows

Using the method which was mentioned before we get(see equality (14))

1

2

119889

119889119905int

119886

minus119886

[(119906 119906) + (119908 119908)] 119889119909 + (119861119906minus

119906minus

)1003816100381610038161003816119909=119886

+ (119861119906minus

119906minus

)1003816100381610038161003816119909=minus119886 minus

1

120573((119860119908

+

minus 119861119906+

) 119906minus

)10038161003816100381610038161003816119909=minus119886

+1

120573((119860119908

+

+ 119861119906+

) 119906minus

)10038161003816100381610038161003816119909=119886

= int

119886

minus119886

[((1198691(1199061 1199081) minus 1198691(1199062 1199082)) 119906)

+ ((1198692(1199061 1199081) minus 1198692(1199062 1199082)) 119908)] 119889119909

(47)

Let us transform the integrand on the right side of theequality as

((1198691(1199061 1199081) minus 1198691(1199062 1199082)) 119906)

=1205902minus 1205900

2[(119891001

119891021

minus1198912

011

radic3) minus (119891

002119891022

minus1198912

012

radic3)] (119891

021minus 119891022

) =1205902minus 1205900

2[(119891001

119891021

minus 119891002

119891022

) minus1

radic3(1198912

011minus 1198912

012)] (119891021

minus 119891022

)

=1205902minus 1205900

2[(119891001

minus 119891002

) 119891021

+ 119891002

(119891021

minus 119891022

) minus1

radic3(119891011

minus 119891012

) (119891011

+ 119891012

)] (119891021

minus 119891022

)

((1198692(1199061 1199081) minus 1198692(1199062 1199082)) 119906) = ((119869

03(1199061 1199081)

minus 11986903

(1199062 1199082)) (119891031

minus 119891032

)) + ((11986911

(1199061 1199081)

minus 11986911

(1199062 1199082)) (119891111

minus 119891112

)) = (1

4(1205903+ 31205901

minus 41205900) [(119891001

minus 119891002

) 119891031

+ 119891002

(119891031

minus 119891032

)]

+1

4radic5(21205901+ 1205900minus 31205903) [(119891011

minus 119891012

) 119891021

+ 119891012

(119891021

minus 119891022

)]) (119891031

minus 119891032

) + (1205901minus 1205900)

sdot ((119891001

minus 119891002

) 119891111

+ 119891002

(119891111

minus 119891112

)

+ 05radic5

2[(119891101

minus 119891102

) 119891011

+ 119891102

(119891011

minus 119891012

)] minus radic2

15[(119891011

minus 119891012

) 119891021

+ 119891012

(119891021

minus 119891022

)]) (119891111

minus 119891112

)

(48)

Once again we use spherical representation of 119906 = 119903(119905)1199081(119905

119909) and 119908 = 119903(119905)1199082(119905 119909) where 119903(119905) = 119880(119905 sdot)

1198712[minus119886119886]

Thenconcerning 119903(119905) we obtain new initial value problem

119889119903

119889119905+ 119903119875 (119905) = 119903

2

1198761(119905) (49)

119903 (0) = 0 (50)

where 119875(119905) has the same value as in (15) and

1198761(119905) =

1205902minus 1205900

2int

119886

minus119886

[(120596001

minus 120596002

) 120596021

+ 120596002

(120596021

minus 120596022

) minus1

radic3(120596011

minus 120596012

) (120596011

+ 120596012

)] (120596021

minus 120596022

) 119889119909 +1

4(1205903+ 31205901

minus 41205900) int

119886

minus119886

[(120596001

minus 120596002

) 120596031

+ +120596002

(120596031

minus 120596032

)] (120596031

minus 120596032

) 119889119909 +1

4radic5(21205901+ 1205900minus 31205903)

sdot int

119886

minus119886

[(120596011

minus 120596012

) 120596021

+ 120596012

(120596021

minus 120596022

)]

sdot (120596031

minus 120596032

) 119889119909 + (1205901minus 1205900) int

119886

minus119886

[(120596001

minus 120596002

) 120596111

+ 120596002

(120596111

minus 120596112

) +1

2

8 Journal of Applied Mathematics

sdot radic5

2[(120596101

minus 120596102

) 120596011

+ 120596102

(120596011

minus 120596012

)]

minus radic2

15[(120596011

minus 120596012

) 120596021

+ 120596012

(120596021

minus 120596022

)]

sdot (120596111

minus 120596112

)] 119889119909

(51)

The general solution of (49) is

119903 (119905) = exp(minusint

119905

0

119875 (120591) 119889120591)

sdot [119862 minus int

119905

0

1198761(120591) exp(minusint

120591

0

119875 (120585) 119889120585) 119889120591]

minus1

(52)

Solution of (49) which is satisfying homogeneous condition(50) is trivial that is 119903(119905) = 0 Hence 119880

119862([0119879]1198712[minus119886119886])

= 0

and 1199061= 1199062and 119908

1= 1199082

The theorem is proved

Competing Interests

The authors declare that they have no competing interests

References

[1] H Grad ldquoOn the kinetic theory of rarefied gasesrdquo Communica-tions on Pure andAppliedMathematics vol 2 no 4 pp 331ndash4071949

[2] H Grad ldquoPrinciple of the kinetic theory of gasesrdquo in Thermo-dynamics of Gases vol 12 ofHandbuch der Physik pp 205ndash294Springer Berlin Germany 1958

[3] A Sakabekov Initial-Boundary Value Problems for the Boltz-mannrsquos Moment System Equations Gylym Almaty 2002

[4] C Cercignani Theory and Application of the Boltzmann Equa-tion Instituto di Matematica Milano Italy 1975

[5] M N KoganDynamic of Rarefied Gas NaukaMoscow Russia1967

[6] K Kumar ldquoPolynomial expansions in kinetic theory of gasesrdquoAnnals of Physics vol 37 no 1 pp 113ndash141 1966

[7] A Sakabekov and Y Auzhani ldquoBoundary conditions for theonedimensional nonlinear nonstationary Boltzmannrsquos momentsystem equationsrdquo Journal of Mathematical Physics vol 55Article ID 123507 2014

[8] C D Levermore ldquoMoment closure hierarchies for kinetictheoriesrdquo Journal of Statistical Physics vol 83 no 5-6 pp 1021ndash1065 1996

[9] G Mascali and V Romano ldquoA hydrodynamical model for holesin silicon semiconductors the case of non-parabolic warpedbandsrdquo Mathematical and Computer Modelling vol 53 no 1-2pp 213ndash229 2011

[10] G Mascali and V Romano ldquoA non parabolic hydrodynamicalsubband model for semiconductors based on the maximumentropy principlerdquoMathematical and Computer Modelling vol55 no 3-4 pp 1003ndash1020 2012

[11] V D Camiola and V Romano ldquo2DEG-3DEG charge transportmodel for MOSFET based on the maximum entropy principlerdquo

SIAM Journal on Applied Mathematics vol 73 no 4 pp 1439ndash1459 2013

[12] G Alı G Mascali V Romano and R C Torcasio ldquoA hydrody-namical model for covalent semiconductors with a generalizedenergy dispersion relationrdquo European Journal of Applied Math-ematics vol 25 no 2 pp 255ndash276 2014

[13] V D Camiola and V Romano ldquoHydrodynamical model forcharge transport in graphenerdquo Journal of Statistical Physics vol157 no 6 pp 1114ndash1137 2014

[14] S Mischler ldquoKinetic equations with Maxwell boundary condi-tionsrdquo Annales Scientifiques de lrsquoEcole Normale Superieure vol43 no 5 pp 719ndash760 2010

[15] V G Neudachin and U F SmirnovNucleon Association of EasyKernel Nauka Moscow Russia 1969

[16] MMoshinskyTheHarmonic Oscillator inModern Physics fromAtoms to Quarks 1960

[17] S I Pokhozhaev ldquoOn an approach to nonlinear equationrdquoDoklady Akademii Nauk SSSR vol 247 pp 1327ndash1331 1979

[18] E Kamke Differentialgleichungen Losungsmethoden und Losu-ngen I Gewohuliche Differentialgleichungen BGTeubnerLeipzig Germany 1977

[19] A Tungatarov and D K Akhmed-Zaki ldquoCauchy problemfor one class of ordinary differential equationsrdquo InternationalJournal of Mathematical Analysis vol 6 no 14 pp 695ndash6992012

[20] M Tenenbaum andH PollardOrdinary Differential EquationsHarper and Row New York NY USA 1963

[21] L Tartar ldquoCompensated compactness and applications topartial differential equationsrdquo in Proceedings of the Non-LinearAnalysis and Mechanics Heriot-Watt Symposium R J KnopsEd vol 4 of Research Notes in Math pp 136ndash212 1979

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Journal of Applied Mathematics

We take as interval of the solution of problem (15)ndash(17) segment [0 1(119880

0 minus 119903

1(0))] since the integrand

119876(120591) exp(int1205910

(2119876(120585)1199031(120585) minus 119875(120585))119889120585) is bounded Hence forall119905 isin

[0 119879] (where 119879 asymp 1198791) a priori estimation (11) takes place

Now we prove the existence of solution for (4)ndash(7) usingGalerkin method Let V

119897(119909)infin

119897=1be a basis in space 1198712[minus119886 119886]

where the dimension of vector V119897(119909) is equal to the dimension

of vector119880 For each119898we define an approximate solution119880119898

of (4)ndash(7) as follows

119880119898=

119898

sum

119895=1

119888119895119898

(119905) V119895(119909) (34)

int

119886

minus119886

((120597119880119898

120597119905+ 1198601

120597119880119898

120597119909) V119894(119909)) 119889119909

= int

119886

minus119886

(119869 (119880119898) V119894(119909)) 119889119909 119894 = 1119898 119905 isin (0 119879]

(35)

119880119898

1003816100381610038161003816119905=0 = 1198800119898

(119909) 119909 isin 119877 (36)

(119860119908minus

119898∓ 119861119906minus

119898)1003816100381610038161003816119909=∓119886

=1

120573(119860119908+

119898plusmn 119861119906+

119898)1003816100381610038161003816119909=plusmn119886 plusmn

(1 minus 120573) 120578

120572120573radic120587119865

(37)

where 1198800119898

is the orthogonal projection in 1198712 of the function

1198800on the subspace spanned by V

1 V

119898

119869 (119880119898) = (119869

1(119906119898 119908119898) 1198692(119906119898 119908119898))1015840

(38)

We represent V119895(119909) in form V

119895(119909) = (V(1)

119895 V(2)119895

)1015840 where

V(1)119895

= (V1198951 V1198952 V1198953)1015840

V(2)119895

= (V1198954 V1198955 V1198956)1015840

(39)

The coefficients 119888119895119898

(119905) are determined from the equations

119898

sum

119895=1

119889119888119895119898

119889119905int

119886

minus119886

(V119895 V119894

) 119889119909 + 119888119895119898

[(119861Vminus(1)119895

Vminus(1)119894

)10038161003816100381610038161003816119909=119886

+ (119861Vminus(1)119895

Vminus(1)119894

)10038161003816100381610038161003816119909=minus119886

+ (119861Vminus(1)119894

Vminus(1)119895

)10038161003816100381610038161003816119909=119886

+ (119861Vminus(1)119894

Vminus(1)119895

)10038161003816100381610038161003816119909=minus119886

+1

120573((119860V+(2)119894

+ 119861V+(1)119894

) Vminus(1)119895

)10038161003816100381610038161003816119909=119886

minus1

120573((119860V+(2)119894

minus 119861V+(1)119894

) Vminus(1)119895

)10038161003816100381610038161003816119909=minus119886

+1

120573((119860V+(2)119895

+ 119861V+(1)119895

) Vminus(1)119894

)10038161003816100381610038161003816119909=119886

minus1

120573(119860V+(2)119895

minus 119861V+(1)119895

Vminus(1)119894

)10038161003816100381610038161003816119909=minus119886

+ (1198651 Vminus(1)119894

)10038161003816100381610038161003816119909=119886

+ (1198651 Vminus(1)119894

)10038161003816100381610038161003816119909=minus119886

+ (1198651 Vminus(1)119895

)10038161003816100381610038161003816119909=119886

+ (1198651 Vminus(1)119895

)10038161003816100381610038161003816119909=minus119886

]

minus int

119886

minus119886

((119860120597V(2)119894

120597119909 V(1)119895) + (119860

1015840120597V(1)119894

120597119909 V(2)119895

))119889119909

= int

119886

minus119886

(119869(

119898

sum

119895=1

119888119895119898V119895) V119894)119889119909

119894 = 1119898 119905 isin (0 119879]

(40)

119888119894119898

(0) = 119889119894119898 119894 = 1119898 (41)

where 119889119894119898

is the 119894th component of 1198800119898

We multiply (35) by 119888119894119898(119905) and sum over 119894 from 1 to119898

int

119886

minus119886

((120597119880119898

120597119905+ 1198601

120597119880119898

120597119909) 119880119898)119889119909

= int

119886

minus119886

(119869 (119880119898) 119880119898) 119889119909

(42)

With help of the above shown arguments we now provethat 119903

119898(119905) is bounded in some time interval [0 119879

119898] where

119880119898(119905 119909) = 119903

119898(119905)120596119898(119905 119909) 119879

119898asymp M((119880

0119898 minus 1199031198981

(0))minus1

) 119879119898

ge

119879 forall119898 and

10038171003817100381710038171003817

100381710038171003817100381711988011989810038171003817100381710038171198712[minus119886119886] minus 119903

1198981

10038171003817100381710038171003817119862[0119879]

le 1198622(10038171003817100381710038171198800

10038171003817100381710038171198712[minus119886119886] minus 1199031(0))

(43)

where 1198622is constant and independent from119898 and 119903

1198981(119905) is a

particular solution of Riccati equation about 119903119898(119905)

Problem (40)-(41) represents a Cauchy problem for theordinary system of differential equations Existence of thesolution of problem (40)-(41) follows from theory of ordi-nary system of differential equations (Picardrsquos existence anduniqueness theorem) [20]Hence the existence of the solutionof problem (34)ndash(37) follows

Thus according to estimation (43) the sequence 119880119898

minus

1199031198981

of approximate solutions of problem (5)ndash(7) is uniformlybounded in function space119862([0 119879] 1198712[minus119886 119886])Moreover thehomogeneous system of equations 120591119864+ (1120572)119860120585with respectto 120591 120585 has only a trivial solution Then it follows from resultsin [21] that 119880

119898minus 1199031198981

rarr 119880minus 1199031is week in 119862([0 119879] 119871

2

[minus119886 119886])

and 119869(119880119898) rarr 119869(119880) is week in 119862([0 119879] 119871

2

[minus119886 119886]) as 119898 rarr

infin Further it can be shown with standard methods that limitelement is a weak solution of problem (5)ndash(7)

The uniqueness of solution of problem (5)ndash(7) is provedby contradiction Let problem (5)ndash(7) have two different solu-tions 119906

1 1199081and 119906

2 1199082 We denote them again by 119906 = 119906

1minus 1199062

Journal of Applied Mathematics 7

and119908 = 1199081minus1199082 Then with respect to new values of 119906 and119908

we obtain the following problem

120597119906

120597119905+ 119860

120597119908

120597119909= 1198691(1199061 1199081) minus 1198691(1199062 1199082)

120597119908

120597119905+ 1198601015840120597119906

120597119909= 1198692(1199061 1199081) minus 1198692(1199062 1199082)

119909 isin (minus119886 119886)

(44)

119906|119905=0

= 0

119908|119905=0

= 0

(45)

(119860119908minus

+ 119861119906minus

)1003816100381610038161003816119909=minus119886 =

1

120573(119860119908+

minus 119861119906+

)1003816100381610038161003816119909=minus119886

(119860119908minus

minus 119861119906minus

)1003816100381610038161003816119909=119886 =

1

120573(119860119908+

+ 119861119906+

)1003816100381610038161003816119909=119886

(46)

We prove that solution of problem (44)ndash(46) is trivialHence the uniqueness of the solution of problem (5)ndash(7)follows

Using the method which was mentioned before we get(see equality (14))

1

2

119889

119889119905int

119886

minus119886

[(119906 119906) + (119908 119908)] 119889119909 + (119861119906minus

119906minus

)1003816100381610038161003816119909=119886

+ (119861119906minus

119906minus

)1003816100381610038161003816119909=minus119886 minus

1

120573((119860119908

+

minus 119861119906+

) 119906minus

)10038161003816100381610038161003816119909=minus119886

+1

120573((119860119908

+

+ 119861119906+

) 119906minus

)10038161003816100381610038161003816119909=119886

= int

119886

minus119886

[((1198691(1199061 1199081) minus 1198691(1199062 1199082)) 119906)

+ ((1198692(1199061 1199081) minus 1198692(1199062 1199082)) 119908)] 119889119909

(47)

Let us transform the integrand on the right side of theequality as

((1198691(1199061 1199081) minus 1198691(1199062 1199082)) 119906)

=1205902minus 1205900

2[(119891001

119891021

minus1198912

011

radic3) minus (119891

002119891022

minus1198912

012

radic3)] (119891

021minus 119891022

) =1205902minus 1205900

2[(119891001

119891021

minus 119891002

119891022

) minus1

radic3(1198912

011minus 1198912

012)] (119891021

minus 119891022

)

=1205902minus 1205900

2[(119891001

minus 119891002

) 119891021

+ 119891002

(119891021

minus 119891022

) minus1

radic3(119891011

minus 119891012

) (119891011

+ 119891012

)] (119891021

minus 119891022

)

((1198692(1199061 1199081) minus 1198692(1199062 1199082)) 119906) = ((119869

03(1199061 1199081)

minus 11986903

(1199062 1199082)) (119891031

minus 119891032

)) + ((11986911

(1199061 1199081)

minus 11986911

(1199062 1199082)) (119891111

minus 119891112

)) = (1

4(1205903+ 31205901

minus 41205900) [(119891001

minus 119891002

) 119891031

+ 119891002

(119891031

minus 119891032

)]

+1

4radic5(21205901+ 1205900minus 31205903) [(119891011

minus 119891012

) 119891021

+ 119891012

(119891021

minus 119891022

)]) (119891031

minus 119891032

) + (1205901minus 1205900)

sdot ((119891001

minus 119891002

) 119891111

+ 119891002

(119891111

minus 119891112

)

+ 05radic5

2[(119891101

minus 119891102

) 119891011

+ 119891102

(119891011

minus 119891012

)] minus radic2

15[(119891011

minus 119891012

) 119891021

+ 119891012

(119891021

minus 119891022

)]) (119891111

minus 119891112

)

(48)

Once again we use spherical representation of 119906 = 119903(119905)1199081(119905

119909) and 119908 = 119903(119905)1199082(119905 119909) where 119903(119905) = 119880(119905 sdot)

1198712[minus119886119886]

Thenconcerning 119903(119905) we obtain new initial value problem

119889119903

119889119905+ 119903119875 (119905) = 119903

2

1198761(119905) (49)

119903 (0) = 0 (50)

where 119875(119905) has the same value as in (15) and

1198761(119905) =

1205902minus 1205900

2int

119886

minus119886

[(120596001

minus 120596002

) 120596021

+ 120596002

(120596021

minus 120596022

) minus1

radic3(120596011

minus 120596012

) (120596011

+ 120596012

)] (120596021

minus 120596022

) 119889119909 +1

4(1205903+ 31205901

minus 41205900) int

119886

minus119886

[(120596001

minus 120596002

) 120596031

+ +120596002

(120596031

minus 120596032

)] (120596031

minus 120596032

) 119889119909 +1

4radic5(21205901+ 1205900minus 31205903)

sdot int

119886

minus119886

[(120596011

minus 120596012

) 120596021

+ 120596012

(120596021

minus 120596022

)]

sdot (120596031

minus 120596032

) 119889119909 + (1205901minus 1205900) int

119886

minus119886

[(120596001

minus 120596002

) 120596111

+ 120596002

(120596111

minus 120596112

) +1

2

8 Journal of Applied Mathematics

sdot radic5

2[(120596101

minus 120596102

) 120596011

+ 120596102

(120596011

minus 120596012

)]

minus radic2

15[(120596011

minus 120596012

) 120596021

+ 120596012

(120596021

minus 120596022

)]

sdot (120596111

minus 120596112

)] 119889119909

(51)

The general solution of (49) is

119903 (119905) = exp(minusint

119905

0

119875 (120591) 119889120591)

sdot [119862 minus int

119905

0

1198761(120591) exp(minusint

120591

0

119875 (120585) 119889120585) 119889120591]

minus1

(52)

Solution of (49) which is satisfying homogeneous condition(50) is trivial that is 119903(119905) = 0 Hence 119880

119862([0119879]1198712[minus119886119886])

= 0

and 1199061= 1199062and 119908

1= 1199082

The theorem is proved

Competing Interests

The authors declare that they have no competing interests

References

[1] H Grad ldquoOn the kinetic theory of rarefied gasesrdquo Communica-tions on Pure andAppliedMathematics vol 2 no 4 pp 331ndash4071949

[2] H Grad ldquoPrinciple of the kinetic theory of gasesrdquo in Thermo-dynamics of Gases vol 12 ofHandbuch der Physik pp 205ndash294Springer Berlin Germany 1958

[3] A Sakabekov Initial-Boundary Value Problems for the Boltz-mannrsquos Moment System Equations Gylym Almaty 2002

[4] C Cercignani Theory and Application of the Boltzmann Equa-tion Instituto di Matematica Milano Italy 1975

[5] M N KoganDynamic of Rarefied Gas NaukaMoscow Russia1967

[6] K Kumar ldquoPolynomial expansions in kinetic theory of gasesrdquoAnnals of Physics vol 37 no 1 pp 113ndash141 1966

[7] A Sakabekov and Y Auzhani ldquoBoundary conditions for theonedimensional nonlinear nonstationary Boltzmannrsquos momentsystem equationsrdquo Journal of Mathematical Physics vol 55Article ID 123507 2014

[8] C D Levermore ldquoMoment closure hierarchies for kinetictheoriesrdquo Journal of Statistical Physics vol 83 no 5-6 pp 1021ndash1065 1996

[9] G Mascali and V Romano ldquoA hydrodynamical model for holesin silicon semiconductors the case of non-parabolic warpedbandsrdquo Mathematical and Computer Modelling vol 53 no 1-2pp 213ndash229 2011

[10] G Mascali and V Romano ldquoA non parabolic hydrodynamicalsubband model for semiconductors based on the maximumentropy principlerdquoMathematical and Computer Modelling vol55 no 3-4 pp 1003ndash1020 2012

[11] V D Camiola and V Romano ldquo2DEG-3DEG charge transportmodel for MOSFET based on the maximum entropy principlerdquo

SIAM Journal on Applied Mathematics vol 73 no 4 pp 1439ndash1459 2013

[12] G Alı G Mascali V Romano and R C Torcasio ldquoA hydrody-namical model for covalent semiconductors with a generalizedenergy dispersion relationrdquo European Journal of Applied Math-ematics vol 25 no 2 pp 255ndash276 2014

[13] V D Camiola and V Romano ldquoHydrodynamical model forcharge transport in graphenerdquo Journal of Statistical Physics vol157 no 6 pp 1114ndash1137 2014

[14] S Mischler ldquoKinetic equations with Maxwell boundary condi-tionsrdquo Annales Scientifiques de lrsquoEcole Normale Superieure vol43 no 5 pp 719ndash760 2010

[15] V G Neudachin and U F SmirnovNucleon Association of EasyKernel Nauka Moscow Russia 1969

[16] MMoshinskyTheHarmonic Oscillator inModern Physics fromAtoms to Quarks 1960

[17] S I Pokhozhaev ldquoOn an approach to nonlinear equationrdquoDoklady Akademii Nauk SSSR vol 247 pp 1327ndash1331 1979

[18] E Kamke Differentialgleichungen Losungsmethoden und Losu-ngen I Gewohuliche Differentialgleichungen BGTeubnerLeipzig Germany 1977

[19] A Tungatarov and D K Akhmed-Zaki ldquoCauchy problemfor one class of ordinary differential equationsrdquo InternationalJournal of Mathematical Analysis vol 6 no 14 pp 695ndash6992012

[20] M Tenenbaum andH PollardOrdinary Differential EquationsHarper and Row New York NY USA 1963

[21] L Tartar ldquoCompensated compactness and applications topartial differential equationsrdquo in Proceedings of the Non-LinearAnalysis and Mechanics Heriot-Watt Symposium R J KnopsEd vol 4 of Research Notes in Math pp 136ndash212 1979

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Applied Mathematics 7

and119908 = 1199081minus1199082 Then with respect to new values of 119906 and119908

we obtain the following problem

120597119906

120597119905+ 119860

120597119908

120597119909= 1198691(1199061 1199081) minus 1198691(1199062 1199082)

120597119908

120597119905+ 1198601015840120597119906

120597119909= 1198692(1199061 1199081) minus 1198692(1199062 1199082)

119909 isin (minus119886 119886)

(44)

119906|119905=0

= 0

119908|119905=0

= 0

(45)

(119860119908minus

+ 119861119906minus

)1003816100381610038161003816119909=minus119886 =

1

120573(119860119908+

minus 119861119906+

)1003816100381610038161003816119909=minus119886

(119860119908minus

minus 119861119906minus

)1003816100381610038161003816119909=119886 =

1

120573(119860119908+

+ 119861119906+

)1003816100381610038161003816119909=119886

(46)

We prove that solution of problem (44)ndash(46) is trivialHence the uniqueness of the solution of problem (5)ndash(7)follows

Using the method which was mentioned before we get(see equality (14))

1

2

119889

119889119905int

119886

minus119886

[(119906 119906) + (119908 119908)] 119889119909 + (119861119906minus

119906minus

)1003816100381610038161003816119909=119886

+ (119861119906minus

119906minus

)1003816100381610038161003816119909=minus119886 minus

1

120573((119860119908

+

minus 119861119906+

) 119906minus

)10038161003816100381610038161003816119909=minus119886

+1

120573((119860119908

+

+ 119861119906+

) 119906minus

)10038161003816100381610038161003816119909=119886

= int

119886

minus119886

[((1198691(1199061 1199081) minus 1198691(1199062 1199082)) 119906)

+ ((1198692(1199061 1199081) minus 1198692(1199062 1199082)) 119908)] 119889119909

(47)

Let us transform the integrand on the right side of theequality as

((1198691(1199061 1199081) minus 1198691(1199062 1199082)) 119906)

=1205902minus 1205900

2[(119891001

119891021

minus1198912

011

radic3) minus (119891

002119891022

minus1198912

012

radic3)] (119891

021minus 119891022

) =1205902minus 1205900

2[(119891001

119891021

minus 119891002

119891022

) minus1

radic3(1198912

011minus 1198912

012)] (119891021

minus 119891022

)

=1205902minus 1205900

2[(119891001

minus 119891002

) 119891021

+ 119891002

(119891021

minus 119891022

) minus1

radic3(119891011

minus 119891012

) (119891011

+ 119891012

)] (119891021

minus 119891022

)

((1198692(1199061 1199081) minus 1198692(1199062 1199082)) 119906) = ((119869

03(1199061 1199081)

minus 11986903

(1199062 1199082)) (119891031

minus 119891032

)) + ((11986911

(1199061 1199081)

minus 11986911

(1199062 1199082)) (119891111

minus 119891112

)) = (1

4(1205903+ 31205901

minus 41205900) [(119891001

minus 119891002

) 119891031

+ 119891002

(119891031

minus 119891032

)]

+1

4radic5(21205901+ 1205900minus 31205903) [(119891011

minus 119891012

) 119891021

+ 119891012

(119891021

minus 119891022

)]) (119891031

minus 119891032

) + (1205901minus 1205900)

sdot ((119891001

minus 119891002

) 119891111

+ 119891002

(119891111

minus 119891112

)

+ 05radic5

2[(119891101

minus 119891102

) 119891011

+ 119891102

(119891011

minus 119891012

)] minus radic2

15[(119891011

minus 119891012

) 119891021

+ 119891012

(119891021

minus 119891022

)]) (119891111

minus 119891112

)

(48)

Once again we use spherical representation of 119906 = 119903(119905)1199081(119905

119909) and 119908 = 119903(119905)1199082(119905 119909) where 119903(119905) = 119880(119905 sdot)

1198712[minus119886119886]

Thenconcerning 119903(119905) we obtain new initial value problem

119889119903

119889119905+ 119903119875 (119905) = 119903

2

1198761(119905) (49)

119903 (0) = 0 (50)

where 119875(119905) has the same value as in (15) and

1198761(119905) =

1205902minus 1205900

2int

119886

minus119886

[(120596001

minus 120596002

) 120596021

+ 120596002

(120596021

minus 120596022

) minus1

radic3(120596011

minus 120596012

) (120596011

+ 120596012

)] (120596021

minus 120596022

) 119889119909 +1

4(1205903+ 31205901

minus 41205900) int

119886

minus119886

[(120596001

minus 120596002

) 120596031

+ +120596002

(120596031

minus 120596032

)] (120596031

minus 120596032

) 119889119909 +1

4radic5(21205901+ 1205900minus 31205903)

sdot int

119886

minus119886

[(120596011

minus 120596012

) 120596021

+ 120596012

(120596021

minus 120596022

)]

sdot (120596031

minus 120596032

) 119889119909 + (1205901minus 1205900) int

119886

minus119886

[(120596001

minus 120596002

) 120596111

+ 120596002

(120596111

minus 120596112

) +1

2

8 Journal of Applied Mathematics

sdot radic5

2[(120596101

minus 120596102

) 120596011

+ 120596102

(120596011

minus 120596012

)]

minus radic2

15[(120596011

minus 120596012

) 120596021

+ 120596012

(120596021

minus 120596022

)]

sdot (120596111

minus 120596112

)] 119889119909

(51)

The general solution of (49) is

119903 (119905) = exp(minusint

119905

0

119875 (120591) 119889120591)

sdot [119862 minus int

119905

0

1198761(120591) exp(minusint

120591

0

119875 (120585) 119889120585) 119889120591]

minus1

(52)

Solution of (49) which is satisfying homogeneous condition(50) is trivial that is 119903(119905) = 0 Hence 119880

119862([0119879]1198712[minus119886119886])

= 0

and 1199061= 1199062and 119908

1= 1199082

The theorem is proved

Competing Interests

The authors declare that they have no competing interests

References

[1] H Grad ldquoOn the kinetic theory of rarefied gasesrdquo Communica-tions on Pure andAppliedMathematics vol 2 no 4 pp 331ndash4071949

[2] H Grad ldquoPrinciple of the kinetic theory of gasesrdquo in Thermo-dynamics of Gases vol 12 ofHandbuch der Physik pp 205ndash294Springer Berlin Germany 1958

[3] A Sakabekov Initial-Boundary Value Problems for the Boltz-mannrsquos Moment System Equations Gylym Almaty 2002

[4] C Cercignani Theory and Application of the Boltzmann Equa-tion Instituto di Matematica Milano Italy 1975

[5] M N KoganDynamic of Rarefied Gas NaukaMoscow Russia1967

[6] K Kumar ldquoPolynomial expansions in kinetic theory of gasesrdquoAnnals of Physics vol 37 no 1 pp 113ndash141 1966

[7] A Sakabekov and Y Auzhani ldquoBoundary conditions for theonedimensional nonlinear nonstationary Boltzmannrsquos momentsystem equationsrdquo Journal of Mathematical Physics vol 55Article ID 123507 2014

[8] C D Levermore ldquoMoment closure hierarchies for kinetictheoriesrdquo Journal of Statistical Physics vol 83 no 5-6 pp 1021ndash1065 1996

[9] G Mascali and V Romano ldquoA hydrodynamical model for holesin silicon semiconductors the case of non-parabolic warpedbandsrdquo Mathematical and Computer Modelling vol 53 no 1-2pp 213ndash229 2011

[10] G Mascali and V Romano ldquoA non parabolic hydrodynamicalsubband model for semiconductors based on the maximumentropy principlerdquoMathematical and Computer Modelling vol55 no 3-4 pp 1003ndash1020 2012

[11] V D Camiola and V Romano ldquo2DEG-3DEG charge transportmodel for MOSFET based on the maximum entropy principlerdquo

SIAM Journal on Applied Mathematics vol 73 no 4 pp 1439ndash1459 2013

[12] G Alı G Mascali V Romano and R C Torcasio ldquoA hydrody-namical model for covalent semiconductors with a generalizedenergy dispersion relationrdquo European Journal of Applied Math-ematics vol 25 no 2 pp 255ndash276 2014

[13] V D Camiola and V Romano ldquoHydrodynamical model forcharge transport in graphenerdquo Journal of Statistical Physics vol157 no 6 pp 1114ndash1137 2014

[14] S Mischler ldquoKinetic equations with Maxwell boundary condi-tionsrdquo Annales Scientifiques de lrsquoEcole Normale Superieure vol43 no 5 pp 719ndash760 2010

[15] V G Neudachin and U F SmirnovNucleon Association of EasyKernel Nauka Moscow Russia 1969

[16] MMoshinskyTheHarmonic Oscillator inModern Physics fromAtoms to Quarks 1960

[17] S I Pokhozhaev ldquoOn an approach to nonlinear equationrdquoDoklady Akademii Nauk SSSR vol 247 pp 1327ndash1331 1979

[18] E Kamke Differentialgleichungen Losungsmethoden und Losu-ngen I Gewohuliche Differentialgleichungen BGTeubnerLeipzig Germany 1977

[19] A Tungatarov and D K Akhmed-Zaki ldquoCauchy problemfor one class of ordinary differential equationsrdquo InternationalJournal of Mathematical Analysis vol 6 no 14 pp 695ndash6992012

[20] M Tenenbaum andH PollardOrdinary Differential EquationsHarper and Row New York NY USA 1963

[21] L Tartar ldquoCompensated compactness and applications topartial differential equationsrdquo in Proceedings of the Non-LinearAnalysis and Mechanics Heriot-Watt Symposium R J KnopsEd vol 4 of Research Notes in Math pp 136ndash212 1979

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Journal of Applied Mathematics

sdot radic5

2[(120596101

minus 120596102

) 120596011

+ 120596102

(120596011

minus 120596012

)]

minus radic2

15[(120596011

minus 120596012

) 120596021

+ 120596012

(120596021

minus 120596022

)]

sdot (120596111

minus 120596112

)] 119889119909

(51)

The general solution of (49) is

119903 (119905) = exp(minusint

119905

0

119875 (120591) 119889120591)

sdot [119862 minus int

119905

0

1198761(120591) exp(minusint

120591

0

119875 (120585) 119889120585) 119889120591]

minus1

(52)

Solution of (49) which is satisfying homogeneous condition(50) is trivial that is 119903(119905) = 0 Hence 119880

119862([0119879]1198712[minus119886119886])

= 0

and 1199061= 1199062and 119908

1= 1199082

The theorem is proved

Competing Interests

The authors declare that they have no competing interests

References

[1] H Grad ldquoOn the kinetic theory of rarefied gasesrdquo Communica-tions on Pure andAppliedMathematics vol 2 no 4 pp 331ndash4071949

[2] H Grad ldquoPrinciple of the kinetic theory of gasesrdquo in Thermo-dynamics of Gases vol 12 ofHandbuch der Physik pp 205ndash294Springer Berlin Germany 1958

[3] A Sakabekov Initial-Boundary Value Problems for the Boltz-mannrsquos Moment System Equations Gylym Almaty 2002

[4] C Cercignani Theory and Application of the Boltzmann Equa-tion Instituto di Matematica Milano Italy 1975

[5] M N KoganDynamic of Rarefied Gas NaukaMoscow Russia1967

[6] K Kumar ldquoPolynomial expansions in kinetic theory of gasesrdquoAnnals of Physics vol 37 no 1 pp 113ndash141 1966

[7] A Sakabekov and Y Auzhani ldquoBoundary conditions for theonedimensional nonlinear nonstationary Boltzmannrsquos momentsystem equationsrdquo Journal of Mathematical Physics vol 55Article ID 123507 2014

[8] C D Levermore ldquoMoment closure hierarchies for kinetictheoriesrdquo Journal of Statistical Physics vol 83 no 5-6 pp 1021ndash1065 1996

[9] G Mascali and V Romano ldquoA hydrodynamical model for holesin silicon semiconductors the case of non-parabolic warpedbandsrdquo Mathematical and Computer Modelling vol 53 no 1-2pp 213ndash229 2011

[10] G Mascali and V Romano ldquoA non parabolic hydrodynamicalsubband model for semiconductors based on the maximumentropy principlerdquoMathematical and Computer Modelling vol55 no 3-4 pp 1003ndash1020 2012

[11] V D Camiola and V Romano ldquo2DEG-3DEG charge transportmodel for MOSFET based on the maximum entropy principlerdquo

SIAM Journal on Applied Mathematics vol 73 no 4 pp 1439ndash1459 2013

[12] G Alı G Mascali V Romano and R C Torcasio ldquoA hydrody-namical model for covalent semiconductors with a generalizedenergy dispersion relationrdquo European Journal of Applied Math-ematics vol 25 no 2 pp 255ndash276 2014

[13] V D Camiola and V Romano ldquoHydrodynamical model forcharge transport in graphenerdquo Journal of Statistical Physics vol157 no 6 pp 1114ndash1137 2014

[14] S Mischler ldquoKinetic equations with Maxwell boundary condi-tionsrdquo Annales Scientifiques de lrsquoEcole Normale Superieure vol43 no 5 pp 719ndash760 2010

[15] V G Neudachin and U F SmirnovNucleon Association of EasyKernel Nauka Moscow Russia 1969

[16] MMoshinskyTheHarmonic Oscillator inModern Physics fromAtoms to Quarks 1960

[17] S I Pokhozhaev ldquoOn an approach to nonlinear equationrdquoDoklady Akademii Nauk SSSR vol 247 pp 1327ndash1331 1979

[18] E Kamke Differentialgleichungen Losungsmethoden und Losu-ngen I Gewohuliche Differentialgleichungen BGTeubnerLeipzig Germany 1977

[19] A Tungatarov and D K Akhmed-Zaki ldquoCauchy problemfor one class of ordinary differential equationsrdquo InternationalJournal of Mathematical Analysis vol 6 no 14 pp 695ndash6992012

[20] M Tenenbaum andH PollardOrdinary Differential EquationsHarper and Row New York NY USA 1963

[21] L Tartar ldquoCompensated compactness and applications topartial differential equationsrdquo in Proceedings of the Non-LinearAnalysis and Mechanics Heriot-Watt Symposium R J KnopsEd vol 4 of Research Notes in Math pp 136ndash212 1979

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of