Research Article Exact Solution for Non-Self-Similar Wave

Research ArticleExact Solution for Non-Self-Similar Wave-Interaction Problemduring Two-Phase Four-Component Flow in Porous Media

S Borazjani1 P Bedrikovetsky1 and R Farajzadeh23

1 Australian School of Petroleum The University of Adelaide SA 5005 Australia2 Shell Global Solutions International Rijswijk The Netherlands3 Delft University of Technology The Netherlands

Correspondence should be addressed to S Borazjani saraborazjaniadelaideeduau

Received 6 September 2013 Revised 27 December 2013 Accepted 29 December 2013 Published 12 March 2014

Academic Editor Shuyu Sun

Copyright copy 2014 S Borazjani et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Analytical solutions for one-dimensional two-phase multicomponent flows in porous media describe processes of enhanced oilrecovery environmental flows of waste disposal and contaminant propagation in subterranean reservoirs and water managementin aquifers We derive the exact solution for 3 times 3 hyperbolic system of conservation laws that corresponds to two-phase four-component flow in porous media where sorption of the third component depends on its own concentration in water and also onthe fourth component concentration Using the potential function as an independent variable instead of time allows splitting theinitial system to 2 times 2 system for concentrations and one scalar hyperbolic equation for phase saturation which allows for fullintegration of non-self-similar problem with wave interactions

1 Introduction

Exact self-similar solutions of Riemann problems for hyper-bolic systems of conservation laws and non-self-similarsolutions of hyperbolic wave interactions have been derivedfor various flows in gas dynamics shallow waters andchromatography (see monographs [1ndash8]) For flow in porousmedia hyperbolic systems of conservation laws describe two-phase multicomponent displacement [9 10] Consider

120597119904

120597119905+120597119891 (119904 119888)

120597119909= 0 (1)

120597 (119888119904 + 119886 (119888))

120597119905+120597 (119888119891 (119904 119888))

120597119909= 0 (2)

where s is the saturation (volumetric fraction) of aqueousphase and f is the water flux Equation (1) is the massbalance for water and (2) is the mass balance for eachcomponent in the aqueous solution Under the conditionsof thermodynamic equilibrium the concentrations of the

components adsorbed on the solid phase (ai) and dissolved inthe aqueous phase (ci) are governed by adsorption isotherms

119886 = 119886 (119888) 119886 = (1198861 1198862 119886

119899) 119888 = (119888

1 1198882 119888

119899)

(3)

Exact and semianalytical solutions of one-dimensional flowproblems are widely used in stream-line simulation for flowprediction in three-dimensional natural reservoirs [10] Thesequence of concentration shocks in the one-dimensionalanalytical solution is important for interpretation of labo-ratory tests in two-phase multicomponent flow in naturalreservoir cores

The scalar hyperbolic equations (1) and (2) 119899 = 0correspond to displacement of oil by water [9 10] The(119899 + 1) times (119899 + 1) system (1) and (2) describes two-phaseflow of oleic and aqueous phases with 119899 components (suchas polymer and different salts) that may adsorb and bedissolved in both phases These flows are typical for so-calledchemical enhanced oil recovery displacements like injectionsof polymers or surfactants and for numerous environmentalflows [9 10] For polymer injection in oil reservoirs 119894 = 1

corresponds to polymer and 119894 = 2 3 119899 to different ions

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 731567 13 pageshttpdxdoiorg1011552014731567

2 Abstract and Applied Analysis

Therefore the system (1) and (2) is called themulticomponentpolymer-floodingmodel [11 12] Besides (119899minus1)times(119899minus1)hyper-bolic system (1) and (2) describes two-phase 119899-componentdisplacement which is typical for so-called gas methods ofenhanced oil recovery [9 10 13 14] The processes of hotwater injection with phase transitions secondary migrationof hydrocarbons with consequent formation of petroleumaccumulations enhanced geothermal energy projects andinjections into aquifers are described by the above systemsThe Riemann problems correspond to continuous injectionof chemical solutions or gases into oil reservoirs the solutionsare self-similar [3 9 14] The wave-interaction problems cor-respond to piece-wise-constant initial-boundary conditionsfor which the solutions are non-self-similar [1 10 15ndash17]Thewave-interaction solutions describe injection of limited slugs(banks) of chemical solutions or gaseous solvents driven inthe reservoirs by water or gas [9 10]

Riemann problem (1) and (2) with 119899 = 1 has been solvedwith applications to various injections of polymers [17 18]carbonized water and surfactants [19 20] and so forth Morecomplex self-similar solutions of (1) and (2) for 119899 = 2 3 wereobtained by Barenblatt et al [21] and Braginskaya and Entov[22] and later by Johansen et al andWinther et al [11 12 23ndash25] Analogous solutions for gas injection and 119899 = 3 4 have been obtained by Orr and others [9 13 26ndash31]

The system (1) and (2) describes two-phase multicom-ponent displacements in large scale approximation wherethe dissipative effects of capillary pressure diffusion andthermodynamic nonequilibrium are negligible if they arecompared with advective fluxes under the large length scaleof the natural subterranean reservoirs Travelling waves nearto shock discontinuities in dissipative systems have beenpresented in [10 32] A semianalytical global solutions havebeen obtained by Geiger et al [33] and Schmid et al [34] seealso [16]

The particular case of so-calledmulticomponent polymerflooding is the dependency of the component sorptionconcentration of its own concentration only 119886

119894(1198881 1198882 119888

119899) =

119886119894(119888119894) Exact solutions of the Riemann problem for this case

show that the concentration of each component performsthe jump without shocks of other components (see thecorresponding solution in the books [10 21]) Therefore inconcentration profiles the shocks are located in order ofdecrease of derivatives of the sorption functions In the case ofHenry isotherms 119886

119894(119888119894) = Γ119894119888119894 the shocks are located in order

of increase of Henryrsquos sorption coefficients Γ119894

The distinguished invariant feature of (119899 + 1) times (119899 +

1) conservation law systems for two-phase multicomponentflows in porous media with sorption and phase transitionsequations (1) and (2) is its splitting into an 119899 times 119899 auxiliarysystem for concentrations 119888

119894(119909 119905) and a scalar hyperbolic

equation for saturation 119904(119909 119905) [35 36]This splitting explainsthe simple form of Riemann problem solutions for system (1)and (2) as compared with gas dynamics or chromatography[1 2 37]

The non-self-similar solution of system (1) and (2) 119899 =

2 for slug injections has been considered by Fayers [17]where the qualitative behaviour of characteristic lines andshocks has been described The exact solutions of system

(1) and (2) for 119899 = 2 and 3 have been obtained in [15](see book [10] for detailed derivations in which the sorptionof component depends on its own concentration only 119886

119894=

119886119894(119888119894) 119894 = 1 2 ) Numerous interactions of different

saturation-concentration shocks occur after the injectionresulting in appearance of moving zones with differentcombinations of components However after all interactionsdifferent component slugs are separated from each otherAs in the case of continuous injection the slugs are finallypositioned in the order of decreasing sorption isothermderivatives (119889119886

119894119889119888119894) It seems that this simplified case draws

the line where the analytical solutions can be found from theanalysis of system (1) and (2) directly Consideration of cross-effects 119886

119894= 119886119894(1198881 1198882 119888

119899) in sorption functions equation

(3) introduces significant difficulties into wave analysis andeven the Riemann problem cannot be solved for any arbitrarycase 119899 = 2 (see [38] where the Riemann solutions have beenobtained for several particular cases)

The splitting technique reduces number of equations in(1) and (2) by one allowing for exact solutions in more com-plex multicomponent cases [35ndash40] The Riemann problemwith cross-effects for adsorption 119886

119894= 119886119894(1198881 1198882) has been

solved in [39 41 42] for continuous polymer injection withvarying salinity using the splitting method In the currentpaper the exact solution for non-self-similar problem ofinjection of polymer slug with varying salinity followed bywater drive is obtained

The structure of the text is as follows The particularcase of the general system (1) and (2) that is discussedin the current work is introduced in Section 2 along withphysics assumptions and initial-boundary conditions for sluginjection problem The detailed description of the splittingprocedure for the system is discussed along with formulationof initial and boundary conditions for the auxiliary systemwhich is presented in Section 3 Section 4 contains derivationof the Riemann solution that corresponds to the first stageof the slug injection The wave-interaction slug injectionproblem is solved in Section 5 Section 6 contains a simplifiedsolution for the particular case where the initial chemicalconcentration is zero which corresponds to the case ofpolymer slug injection The paper is concluded by physicalinterpretation of the solution obtained for chemical sluginjection with different water salinity into oilfield (Sections7 and 8)

2 Formulation of the Problem

Let us discuss the displacement of oil by aqueous chemicalsolution with water drive accounting for different salinitiesof formation and injected waters In the following text thecomponent 119899 = 1 is called the polymer and that 119899 = 2 iscalled the salt The assumptions of the mathematical modelare as follows both phases are incompressible dispersionand capillary forces are neglected there are two phases (oleicand aqueous phases) and two components dissolved in water(polymer and salt) water and oil phases are immisciblechemical and salt concentrations in water are negligiblysmall and do not affect the volume of the aqueous phase

Abstract and Applied Analysis 3

the fractional flow of the aqueous phase is affected byconcentration of the dissolved chemical the fractional flowis independent of salt concentration chemical and salt donot dissolve in oil linear sorption for the polymer 119886 = Γ119888Henryrsquos sorption coefficient Γ is salinity-dependent salt doesnot adsorb on the rock and temperature is constant

The system of governing equations consists of mass bal-ance equations for aqueous phase for dissolved and adsorbedchemical and for dissolved salt [8 9]

120597119904

120597119905+120597119891 (119904 119888)

120597119909= 0 (4)

120597 (119888119904 + 119886 (119888 120573))

120597119905+120597 (119888119891 (119904 119888))

120597119909= 0 (5)

120597 (120573119904)

120597119905+120597 (120573119891 (119904 119888))

120597119909= 0 (6)

where 119904 is the water saturation 119891 is the fractional flowfunction 119886 is the polymer sorption isotherm and 119888 and 120573 arechemical and salt concentrations respectively

The fractional flow function (water flux) depends onthe water saturation 119904 and on the chemical concentration119888 The typical S-shapes of fractional flow functions 119891 under119888 = const are shown in Figure 1 The fractional flow is amonotonically decreasing function of 119888 Sorption isothermsare linear for fixed salinity 119886(119888 120573) = Γ(120573)119888 The functions119891and 119886 are assumed to be bounded and smooth

The system (4)ndash(6) is a hyperbolic 3 times 3 system ofconservation laws with unknowns 119904 119888 and 120573

The displacement of oil by chemical slug corresponds tothe following initial-boundary problem

119909 = 0120573 = 0 119888 = 119888

1 119904 = 119904

119871 119905 lt 1

120573 = 0 119888 = 1198882 119904 = 119904

119871 119905 gt 1

(7)

119905 = 0 120573 = 1 119888 = 1198882 119904 = 119904

119877 (8)

For 119905 lt 1 during continuous injection of chemical solutionwith different salinity the solution of system (4)ndash(6) subjectto initial-boundary conditions equations (7) and (8) coin-cides with the solution of the Riemann problem

119909 = 0 120573 = 0 119888 = 1198881 119904 = 119904

119871 (9)

119905 = 0 120573 = 1 119888 = 1198882 119904 = 119904

119877 (10)

The initial condition is denoted by 119877 in Figure 1 and theboundary condition corresponding to injection of the slug isdenoted as 119871

Generally 119888(119909 0) = 1198882gt 0 is positive Further in the

text the component with concentration 119888 is called ldquochemicalrdquowhile for the case of the absence of this component initiallyin the reservoir 119888(119909 0) = 119888

2= 0 we use the term ldquopolymerrdquo

The solution of the Riemann problem is self-similar119904(119909 119905) = 119904(120585) 119888(119909 119905) = 119888(120585) 120573(119909 119905) = 120573(120585) 120585 = 119909119905 andit can be found in [37 39 40] The solution of the problem(7) and (8) in the neighbourhood of the point (0 1) in (119909 119905)-plane is also self-similar The global solution of the system(4)ndash(6) subject to the initial-boundary conditions equations

01

0 10

1L

s

f

R

D3

D2

D1

2

3

4

c = c1

c = clowast

c = c2

minusΓ(1)

Figure 1 Fractional flow curves and Riemann problem solutionwhere 119888

1is the slug concentration 119888

2is the initial concentration and

119888lowast is the intermediate concentration

(7) and (8) is non-self-similar it expresses the interactionsbetween hyperbolic waves occurring fromdecays of Riemanndiscontinuities in points (0 0) and (0 1) in (119909 119905)-plane

System of (4)ndash(6) subject to the initial and boundaryconditions equations (9) and (10) is solved in Section 4 usingthemethod so-called splitting procedure [35]This procedureis explained in the next section

3 Splitting Procedure

In the present section we briefly explain the splitting methodfor the solution of hyperbolic system of conservation lawsequations (4)ndash(6)

31 StreamlinePotential Function and Auxiliary System As itfollows from divergent (conservation law) form of equationfor mass balance for water (1) or (4) there does exist such apotential function 120593(119909 119905) that

119904 = minus120597120593

120597119909

119891 =120597120593

120597119905

(11)

that is

119889120593 = 119891119889119905 minus 119904119889119909 (12)

120593 (119909 119905) = int

119909119905

00

119891119889119905 minus 119904119889119909 (13)

Equation (4) is merely the condition of equality of secondderivatives of the potential 120593 as taken in different ordersIt also expresses that the differential of the first order formequation (12) equals zero The splitting procedure consistsof changing the independent variables from (119909 119905) to (119909 120593)

in system (4)ndash(6) Figures 2 and 3 show the correspondingmapping [43 44]

From fluid mechanics point of view 120593(119909 119905) is a potentialfunction which equals the volume of fluid flowing through


t

t1

0 0x xx1

x1

120593

120593 = 120593(x = 0 t1)

fL(x = 0 t1) cL(x = 0 t1)

fL(x = 0 t1) cL(x = 0 t1)

120593 = 120593(x1 t = 0)sR(x1 t = 0) cR(x1 t = 0)

sR(x1 t = 0) cR(x1 t = 0)

Figure 2 Introduction of potential function (Lagrangian coordinate) and mapping between independent variables

t

x x

120593 + Δ120593

120593 + Δ120593

120593

120593

x + Δx x + Δx

Figure 3 Derivation of mass balance equation in Eulerian and Lagrangian coordinate systems

a trajectory connecting points (0 0) and (119909 119905) As it followsfrom (12) two streamlines in Figure 3 correspond to constantvalues of potential that is there is no flux through stream-lines

120593 (119909 1199052) minus 120593 (119909 119905

1) = int

1199052

1199051

119891 (119909 119905) 119889119905 (14)

Equation (4) shows that the integral of (13) along the closedcontour is equal to zero that is the volume of fluid flowingthrough a trajectory connecting points (0 0) and (119909 119905) isindependent of trajectory and depends on end points onlyThe potential function equation (13) is determined in the waythat one end of trajectory is fixed at point (0 0)

Let us derive the relationship between the elementarywave speeds of the system in (120593 119909) coordinates and those ofthe large system in (119905 119909) Consider the trajectory 119909 = 119909

0(119905)

and its image 120593 = 1205930(119905) by the mapping equation (13)

1205930(119905) = 120593 (119909

0(119905) 119905) (15)

Define the trajectory speeds as

119863 =119889119909

119889119905 119881 =

119889119909

119889120593 (16)

Let us use 119909 as a parameter for both curves 119909 = 1199090(119905) and

120593 = 1205930(119905) Taking derivation of both parts of (13) in 119909 along

trajectories and using speed definitions in (16) we obtain

1

119881=119891

119863minus 119904 (17)

from which follows the relationship between elementarywave speed in planes (119909 119905) and (119909 120593)

119863 =119891

119904 + 1119881 (18)

For example the eigenvalues of the system of equation in(119905 119909) plane 120582

119894and in (120593 119909) Λ

119894 are related by (Figure 4

[43 44])

Λ119894(119904 119888) =

119891

119904 + 1120582119894

(19)

From now on the independent variables (119909 120593) are used in(4)ndash(6) instead of (119909 119905) Expressing the differential form 119889119905

from (12) as

119889119905 =119889120593

119891+119904119889119909

119891(20)


t

D

x

1 1

120593

V

x1120582i

+1V 1120582i

minus0

s

D

f

+

minus

Figure 4 Speeds of a particle in Eulerian and Lagrangian coordinates

and accounting for zero differential of the form 119889119905

1198892119905 = 0 = [

120597

120597119909(1

119891) minus

120597

120597120593(119904

119891)] 119889119909119889120593 (21)

we obtain the expression for (4) in coordinates (119909 120593)

120597 (119904119891)

120597120593minus120597 (1119891)

120597119909= 0 (22)

So (22) is themass balance for water that is it is (4) rewrittenin coordinates (119909 120593)

Let us derive (5) in (119909 120593) coordinates The conservationlaws for (5) in the integral form are

∮

120597Ω

(119888119891) 119889119905 minus (119888119904 + 119886) 119889119909 = 0 (23)

where Ω is a closed domain Ω sub 1198772 so the integral of (23) is

taken over the closed contourApplying the definition of the potential function equation

(13) into (23) yields

∮

120597Ω

119888 (119891119889119905 minus 119904119889119909) minus 119886119889119909 = ∮

120597Ω

119888119889120593 minus 119886119889119909 = 0 (24)

Tending the domain radius to zero and applying Greenrsquostheorem

120597119886 (119888 120573)

120597120593+120597119888

120597119909= 0 (25)

Now let us perform change of independent variables in (6) in(119909 120593) coordinates as follows

∬(120597 (120573119904)

120597119905+120597 (120573119891 (119904 119888))

120597119909)119889119909119889119905 = ∮

120597Ω

120573119904119889119909 minus 120573119891 (119904 119888) 119889119905

= ∮120573119889120593 = ∬120597120573

120597119909= 0

(26)

Finally the (119899 + 1) times (119899 + 1) system of conservation laws fortwo-phase n component chemical flooding in porous mediawith adsorption can be split into an 119899 times 119899 auxiliary systemequations (25) and (26) and one independent lifting equation

(22)The splitting is obtained from the change of independentvariables (119909 119905) to (119909 120593)This change of coordinates also trans-forms the water conservation law into the lifting equationThe solution of hyperbolic system (22) (25) and (26) consistsof three steps (1) solution of the auxiliary problem (25) and(26) subject to initial and boundary conditions (2) solutionof the lifting equation (22) and (3) determining time 119905 foreach point of the plane (119909 120593) from (13)

The auxiliary system contains only equilibrium thermo-dynamic variables while the initial system contains bothhydrodynamic functions (phasersquos relative permeabilities andviscosities) and equilibrium thermodynamic variables

The above splitting procedure is applied to the solution ofdisplacement of oil by polymer slug with alternated salinityin the next section

32 Formulation of the Splitting Problem for Two-Phase Flowwith Polymers and Salt Introducing new variables ldquodensityrdquo119865 and ldquofluxrdquo119880 and applying the splitting technique the 3 times 3system (4)ndash(6) is transformed to the following form

119865 = minus119904

119891 119880 =

1

119891(27)

120597 (119865 (119880 119888))

120597120593+120597 (119880)

120597119909= 0 (28)

120597119886 (119888 120573)

120597120593+120597119888

120597119909= 0 (29)

120597120573

120597119909= 0 (30)

The auxiliary system equations (29) and (30) are independentof (28)The auxiliary systemhas thermodynamic nature sinceit contains only sorption function 119886(119888 120573) and the unknownsare the component concentrations c and 120573 Equation (28) isthe volume conservation for two immiscible phases For theknown auxiliary solution of (29) and (30) equation (28) is ascalar hyperbolic equation Figure 5 shows the projection ofthe space of the large system into that of auxiliary system andthe lifting procedure [43 44]


+

+

+

minus

minus

minus

s

cn

c1

Figure 5 Projection of the space of the large system into that ofauxiliary system and the lifting procedure using the solution ofauxiliary system

The boundary conditions for slug problem equation (7)are reformulated for coordinates (119909 120593) as

119909 = 0120573 = 0 119888 = 119888

1 119880 = 1 120593 lt 1

120573 = 0 119888 = 1198882 119880 = 1 120593 gt 1

(31)

Figure 2 shows how the initial and boundary conditionsfor the large system (4) and (6) are mapped into those forauxiliary system and the lifting equations (28)ndash(30)

The initial conditions for slug problem equation (8) arereformulated for coordinates (119909 120593) as

120593 = minus 119904119877119909 120573 = 1 119888 = 119888

2 119880 = +infin (32)

The solution of the Riemann problem for 120593 lt 1 correspondsto the following initial and boundary conditions

119909 = 0 120573 = 0 119888 = 1198881 119880 = 1

120593 = minus119904119877119909 120573 = 1 119888 = 119888

2 119880 = +infin

(33)

4 Solution for the Riemann Problem

Let us discuss the solution of the problem equations (7) and(8) for 119905 lt 1 which is self-similar that is the boundary andinitial conditions become (9) and (10)

Themass balance conditions on shockswhich follow fromthe conservation law (Hugoniot-Rankine condition) form ofthe system (28)ndash(30) are

120590 [119880] = [119865] (34)

120590 [119888] = [119886] (35)

120590 [120573] = 0 (36)

where 120590 is reciprocal to the shock velocity of (28)ndash(30)As salt and polymer concentration are connected by thethermodynamic equilibrium relationship 119886(119888 120573) function119886 is discontinuous if 119888 is discontinuous so is 120573 Since 119865is a function of 119888 and 119880 discontinuity of 119888 and 119880 yieldsdiscontinuity of 119865

As it follows from equality (36) either 120590 = 0 or [120573] = 0From (34) and (35) it follows that if120590 = 0 [119886] = 0 and [119865] = 0If [120573] = 0 from (35) and (36) it follows that 120590 = [119886][119888] and120590 = [119865][119880] therefore it yields to 120590 = [119886][119888] = [119865][119880]Finally from (34) if [120573] = 0 and [119888] = 0 this leads to 120590 =

[119865][119880]The shock waves must obey the Lax evolutionary condi-

tions [1ndash4 9]

41 Solution for the Auxiliary System The solution of auxil-iary system is presented in Figure 6 by sequence of c-shockfrom point 119871 into intermediate point and (119888 120573)-shock intopoint 119877 The corresponding formulae are as follows

119888 (119909 120593)

120573 (119909 120593)=

1198881 120573 = 0 120593 gt Γ (0) 119909

119888lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198882 120573 = 1 minus119904

119877119909 lt 120593 lt 0

(37)

where the condition of continuity of function 119886(119888 120573) on theshock with 120590 = 0 and (35) allows finding the intermediateconcentration

119888lowast=Γ (1)

Γ (0)1198882 (38)

42 Solution for the Lifting Equation Figure 7 exhibits initialand boundary conditions for hydrodynamics lifting equation(28) Curves 119865 = 119865(119880 119888) are shown for constants 119888 = 119888

1 119888 =

1198882 and 119888 = 119888

lowast they are obtained from fractional flow curves119891 = 119891(119904 119888) for the same constant values of concentration 119888Point 119877 corresponds to119880 tending to infinity and 119865 tending tominus infinity where the fractional flow 119891 tends to zero Thetangent of the segment (0 0)ndash(119880 119865) tends to minus119904119877

The solution of lifting equation with known concentra-tions (37) is given by centred wave 119871ndash2 (119888-119880)-shock 2ndashgt3 (120573-119888-119880)-shock 3ndashgt4 and 119880-shock 4ndashgtR (Figure 7) Thecentred wave (119871ndash2) is given by (39)

120593

119909=

120597119865 (1198801 1198881)

120597119880 (39)

Points 2 and 3 are determined by the condition of equality of119880 and 119888 shock speeds

120597119865 (1198802 1198881)

120597119880=1198652(1198802 1198881) minus 1198653(1198803 119888lowast)

1198802minus 1198803

= Γ (0) (40)

Point 4 is determined by condition of equality of the shockvelocities 119888 120573 and 119880

1198653(1198803 119888lowast) = 1198654(1198804 1198882) = 0 (41)

Point 4 is connected to point 119877 by 119880-shock

1198654(1198804 1198882) minus 119865119894(119880119877 1198882)

1198804minus 119880119894

=minus1199044119891119877+ 1199041198771198914

119891119877 minus 1198914

= minus119904119877 (42)


c1c2 clowast c

R

a

a1

a2

L

120573 = 1

120573 = 0

(a)

c1

c2

clowast

c

R

L

1205731

(b)

Figure 6 Solution of the auxiliary problem (a) Adsorption isotherm for chemical for different water salinities and the Riemann problemsolution (b) Riemann problem solution on the plane of chemical concentration c and salinity 120573

F = minussf

U = 1f01

2

34

67

LminussL

minussR

R

c = c1c = clowast

c = c2

Figure 7 The image of the solution in (119865-119880) plane

The solution of the Riemann problem equations (28)ndash(30)with free variables (119909 120593) is given by the following formulae

119880(119909 120593)

119888 (119909 120593)

120573 (119909 120593)

=

1198801(120593

119909) 1198881 120573 = 0 120593 gt Γ (0) 119909

1198803 119888

lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198804 119888

2 120573 = 1 minus119904

119877119909 lt 120593 lt 0

infin 1198882 120573 = 1 120593 gt minus119904

119877119909

(43)

The expression 119904 = minus119880119865(119880 119888) allows calculating saturation119904(119909 120593)

119904 (119909 120593)

119888 (119909 120593)

120573 (119909 120593)

=

1199041(120593

119909) 1198881 120573 = 0 120593 gt Γ (0) 119909

1199043 119888

lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1199044 119888

2 120573 = 1 minus119904

119877119909 lt 120593 lt 0

119904119877 119888

2 120573 = 1 120593 lt minus119904

119877119909

(44)

Figure 8 shows the solution of the system (28)ndash(30) in (120593 119909)-plane For 120593 lt 1 the solution is self-similar the waveinteraction occurs at 120593 gt 1

I

II

III

IV

V120593

(f = 1 120573 = 0 c2)

(f = 0 120573 = 1 c2)

(f = 1 120573 = 0 c1)

1

0

U5(120593 x)

120573 = 0

120573 = 0

120573 = 0

c2

c2

U1(120593x)

c1

120593 = Γ(0)x + 1

120593 = Γ(0)x

clowastU3

U4 120573 = 1x

U = infin

120593 = minussRx

67

(x998400 120593998400)

Figure 8 Solution of the auxiliary and lifting system for slugproblem in (120593 119909)-plane

43 Inverse Mapping Change of Variables from (120593 119909) to (119905 119909)Time 119905 = 119905(119909 120593) for solution is calculated from (12) along anypath from point (119909 120593) to point (0 0) The expression for timet in zone II is

119905 =1

1198914

int

120593

0

119889120593 +1199044

1198914

int

119909

0

119889119909 = (minus119904119877+ 1199044

1198914

)119909 (45)

The expression for time 119905 in zone III is

119905 =1

1198913

int

120593

0

119889120593 +1199043

1198913

int

119909

0

119889119909 =1199043

1198913

119909 (46)


I

V

II

III

IV

t

(f = 1 120573 = 0 c2)

(f = 1 120573 = 0 c1)

1

0

s5(t x)

120573 = 0

120573 = 0 120573 = 0

c2

c2

c2

c1 clowasts3

s4120573 = 1

120573 = 1

x

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s4)xf4

s1(tx)

(x998400 t998400)

sR

(f = 0 120573 = 1 c2)

Figure 9 Non-self-similar solution of the problem for wave inter-actions in (119909 119905)-plane

In zone IV integral for calculation time 119905 = int119909120593

00(119889120593119891 +

119904119889119909119891) is calculated along the characteristic in centred 119880-wave

119905 =120593

119891 (1199041 (120593119909) 1198881)+

1199041(120593119909)

119891 (1199041 (120593119909) 1198881)119909 (47)

Figure 9 shows the solution for the Riemann problem at 119905 lt 1see Figure 10 for detailed description of the Riemann solutionand profiles of unknown functions Finally the solution of theRiemann problem for the system (4)ndash(6) is

119904 (119909 119905)

119888 (119909 119905)

120573 (119909 119905)

=

1199041(119905

119909) 1198881 120573 = 0 119905 gt

Γ (0) + 1199042

1198912

119909

1199043 119888

lowast 120573 = 0

1199043

1198913

119909 lt 119905 ltΓ (0) + 119904

2

1198912

119909

1199044 119888

2 120573 = 1

minus119904119877+ 1199044

1198914

119909 lt 119905 lt1199043

1198913

119909

119904119877 119888

2 120573 = 1 119905 lt

minus119904119877+ 1199044

1198914

119909

(48)

5 Solution of the Slug Problem

Now let us solve the slug problem equations (31) and (32)for auxiliary system (29) and (30) The solution of Riemannproblem at the point (0 1) is given by 119888-shock with 119888minus = 119888

2

and 119888+ = 1198881under constant 120573

119888 (119909 120593)

120573 (119909 120593)=

1198882 120573 = 0 120593 gt Γ (0) 119909 + 1

1198881 120573 = 0 Γ (0) 119909 lt 120593 lt Γ (0) 119909 + 1

119888lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198882 120573 = 1 minus119904

119877119909 lt 120593 lt 0

(49)

The solution of the auxiliary system is given by (49)So zone I in Figure 8 corresponds to initial conditions

the solution is given by point 4 in zone II and point 3 holds

I

II

III

IV

t

0

120573 = 0

120573 = 0

c2

c2

clowasts3

s3

s2

s4

s4

120573 = 1

120573 = 1

x

x

x

x

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s4)xf4

sR

sR

sL

c1

c2

clowast

c1

s

c

1

0

120573

D1t1 D2t1 D3t1

(a)

(b)

(c)

(d)

t1

s1(tx)

Figure 10 Solution of the Riemann problem (a) trajectories ofshock fronts and characteristic lines in (119909 119905)-plane (b) saturationprofile (c) chemical concentration profile (d) salinity profile

in zone III Centred waves equation (39) fills in zone IV Inzone V 119888 = 119888

2and 120573 = 0

Now let us solve the lifting equation (28) with given119888(119909 120593) and 120573(119909 120593)

TheHugoniot-Rankine condition for the rear slug front is

119865 (119880+ 1198881) minus 119865 (119880

minus 1198882)

119880+ minus 119880minus= Γ (0) (50)


119880 is constant along the characteristic lines behind the rearfront

1198805(119909 120593) = 119880

minus(1199091015840 1205931015840) (51)

where point (1199091015840 1205931015840) is located on the rear front and is locatedon the same characteristic line with point (119909 120593)

120593 minus 1205931015840

119909 minus 1199091015840=120597119865 (119880

5 1198881)

120597119880 (52)

The solution of lifting equation 119880(119909 120593) is given by differentformulae in zones IndashV

119880 (119909 120593)

119888 (119909 120593)

120573 (119909 120593)

=

1198805(119909 120593) 119888

2 120573 = 0 120593 gt Γ (0) 119909 + 1

1198801(120593

119909) 1198881 120573 = 0 Γ(0) 119909lt120593ltΓ(0) 119909+1

1198803 119888

lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198804 119888

2 120573 = 1 minus119904

119877119909 lt 120593 lt 0

infin 1198882 120573 = 1 120593 gt minus119904

119877119909

(53)

where the equation for rear front of the chemical slug in theauxiliary plane is

120593 = Γ (0) 119909 + 1 (54)

Finally the solution of auxiliary problem equation (53) allowscalculating 119905(119909 120593) for zones I II V Let us start withdetermining time 119905 along the rear front of the slug Thecentred 119904-wave propagates ahead of the rear front

120593

119909=119891 (119904+ 1198881) minus 119904+1198911015840(119904+ 1198881)

1198911015840 (119904+ 1198881)

(55)

From (54) (55) follow the expression for 119909119863(120593119863) in a

parametric form

119909119863(119904+) =

1198911015840(119904+ 1198881)

119891 (119904+ 1198881) minus 1198911015840 (119904+ 119888

1) (119904+ + Γ)

=1198911015840(119904+ 1198881)

Δ(56)

120593119863(119904+) =

119891 (119904+ 1198881) minus 1199041+

1198911015840(119904+ 1198881)

119891 (119904+ 1198881) minus 1198911015840 (119904+ 119888

1) (119904+ + Γ)

=119891 (119904+ 1198881) minus 1199041+

1198911015840(119904+ 1198881)

Δ

(57)

Integration of the form (41) along the rear front gives

119905119863=

120593

119891 (119904+ (120593 119909) 1198882)+

119904+(120593 119909)

119891 (119904+ (120593 119909) 1198882)119909

119905119863=

1

119891 (119904+ 1198881) minus 1198911015840 (119904+ 119888

1) (1199041++ Γ)

=1

Δ

(58)

Finally the solution of the slug problem for the system (4)ndash(6) subject to initial and boundary conditions equations (7)and (8) is (Figure 9)

119904 (119909 119905)

119888 (119909 119905)

120573 (119909 119905)

=

1199045 (119905 119909) 1198882 120573 = 0 119905 gt

Γ (0) + 1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909 + 1

1199041(119905

119909) 1198881 120573 = 0

Γ (0) + 1199042

1198912

119909 lt 119905 ltΓ (0) + 1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909 + 1

1199043 119888

lowast 120573 = 0

1199043

1198913

119909 lt 119905 ltΓ (0) + 1199042

1198912

119909

1199044 119888

2 120573 = 1

minus119904119877+ 1199044

1198914

119909 lt 119905 lt1199043

1198913

119909

119904119877 119888

2 120573 = 1 119905 lt

minus119904119877+ 1199044

1198914

119909

(59)Figure 11 presents trajectories of shock fronts in (119909 119905)-planealong with profiles of unknowns 119904 119888 and 120573 at typicalmoments

Here the trajectory of the rear slug front 119909119863= 119909119863(119905) is

given in a parametric form (Figure 12)1

119909119863

= 119904119861+ Γ (0)

1

119905119863

= 119891119860

(60)

where 119904119861is the abscissa of point 119861 and 119891

119860is the ordinate of

point119860 (Figure 12) Equations (60) can be solved graphicallyStraight line AB is a tangent to the fractional flow curve119888 = 119888

1 the tangent point in 119904+ The rear front position 119909

119863

is determined by the interval BC at the moment determinedby AC

6 Particular Case for the Polymer Absence inthe Reservoir before the Injection

In reality there is no chemical initially in the reservoir duringthe majority of chemical enhanced oil recovery applicationsthat is 119888(119909 0) = 0 For zero initial polymer concentrationthe intermediate polymer concentration is equal to zero sothe points ahead and behind the 120573-shock coincide in planes(119888 119886) and (119904 119891) The particular simplified solution is (Figures13 and 14)119904 (119909 119905)

119888 (119909 119905)

120573 (119909 119905)

=

1199045 (119905 119909) 119888 = 0 120573 = 0 119905 gt

Γ (0) + 1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909 + 1

1199041(119905

119909) 119888 = 119888

1 120573 = 0

Γ (0) + 1199042

1198912

119909lt119905ltΓ (0)+1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909+1

1199043 119888 = 0 120573 = 0

1199043

1198913

119909 lt 119905 ltΓ (0) + 1199042

1198912

119909

1199043 119888 = 0 120573 = 1

minus119904119877+ 1199043

1198913

119909 lt 119905 lt1199043

1198913

119909

119904119877 119888 = 0 120573 = 1 119905 lt

minus119904119877+ 1199043

1198913

119909

(61)


I

II

III

IV

t

0

120573 = 0

120573 = 0

120573 = 0

c2

c2

c2

clowasts3

s3

s2

s4

s4

120573 = 1

120573 = 1

x

x

x

x

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s4)xf4

sR

sR

sL

c1

c2

clowast

c1

s

c

1

1

0

120573

D1t1 D1t2D2t1 D2t2D3t1 D3t2

(a)

(b)

(c)

(d)

V

s5(t x) (x998400 t998400)

t1

t1

t2

t2

s1(tx)

Figure 11The solution of the slug injection problem (a) trajectoriesof shock fronts and characteristic lines in (119909 119905)-plane (b) saturationprofile (c) chemical concentration profile (d) salinity profile

7 Fluid Mechanics Interpretationof the Solution

Following exact solution equations (4)ndash(9) let us describestructure of two-phase flow with chemical and salt additivesduring chemical slug injection

During continuous injection 119905 lt 1 the solution ofchemical slug injection coincides with that of continuouschemical injection Initial conditions equation (10) is shownby point 119877 that corresponds to low initial saturation andinitial concentrations of chemical 119888

2and of salt 120573 = 1

The boundary condition at 119909 = 0 corresponds to point 119871of injection of chemical solution with concentration 119888

1and

salinity 120573 = 0 The path of Riemann problem solution inplane (119904 119891) consists of centred 119904-wave with injected chemicalconcentration and unity salinity 119888-119904-shock 2ndashgt3 119888 119904 120573-shock 3ndashgt4 and 119904-shock 4ndashgtR into initial point (Figure 1)Following nomenclature by Courant and Friedrichs [1] andLake [9] the Riemann solution is Lndash2ndashgt3ndashgt4ndashgtR Shock

01

0 10

1L

f

R

2

3

4

c = c1

c = clowast

c = c2

s

s+sminus

c

ab

Δ

Δf998400(s+)

Figure 12 Solution of the lifting equation in (119904 119891)-plane

01

0 10

1L

s

f

R

2

3

c = c1

minusΓ(0)

c = c2 = 0

Figure 13 Solution of the lifting system in (119891-119904) plane when 1198882= 0

2ndashgt3 in plane (119888 120573) is horizontal shock 3ndashgt4 is vertical(Figure 6(b)) Shock 2ndashgt3 in plane (119888 119886) occurs along thesorption isotherm shock 3ndashgt4 is a horizontal jump fromisotherm 119888 = 119888

lowastto that 119888 = 1198882(Figure 7)

The trajectories of shocks 2ndashgt3 3ndashgt4 and 4ndashgt 119877 areshown in Figure 7 Shock velocities are constant so thetrajectories are straight lines Let us fix the position 119909 = 1

of the raw of production wells Before arrival of the front 4ndashgt 119877 at themoment 119905 = 1119863

3 oil with fraction of water119891119877 and

initial concentrations of chemical and salt is produced Afterarrival of the front water-oil mixture with water fraction 119891

4

and initial concentrations of chemical and salt is produceduntil the arrival of the 3ndashgt4 front at the moment 119905 = 1119863

2

The corresponding profiles of saturation and concentra-tions are shown in Figure 10 The moment 119905

1for profiles is

fixed in Figure 10(a) allowing defining positions of all frontsin this moment Corresponding profiles at that moment forsaturation chemical concentration and salinity are shownin Figures 10(b) 10(c) and 10(d) respectively The saturationprofile consists of declining interval 119904119871ndash119904

2in 119904-wave two oil-

water banks 1199043and 1199044 and the initial undisturbed zone 119904119877

The chemical concentration profile consists of injected value119888 = 1198881in 119904-wave intermediate value 119888lowast in 119904

3-bank and initial


I

II

III

IVV

(f = 1 120573 = 0 c1)

1

0

120573 = 0

120573 = 1

120573 = 1

120573 = 0

120573 = 0

x

s5(t x)

s3

s3

(x998400 t998400)

sR

c1

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s3)xf3

c = 0

c = 0

c = 0

c = 0

(f = 0 120573 = 1 c = 0)

(f = 1 120573 = 0 c = 0)

t

s1(tx)

Figure 14 Non-self-similar solution of the problem for waveinteractions in (119909 119905)-plane when 119888

2= 0

concentration 1198882in 1199044-bank and in the initial zone Salinity 120573-

profile consists of injected value in zones IV and III and initialvalue in other zones

Injection of water without chemical and with salinity120573 = 0 starts at the moment 119905 = 1 The flow is not self-similar anymoreThe front trajectories and profiles are shownin Figure 11 The solution for slug problem coincides withthe solution for continuous injection ahead of the rear front119909119863(119905) The profiles taken at the moment 119905

1lt 1 during

continuous injection (Figure 11) coincide with those fromFigure 10

The propagation of the rear slug front from the begin-ning of water drive injection in the reservoir is shown inFigure 11(a) The rear front velocity decreases reaching thevalue of the forward front119863

3when time tends to infinityThe

slug thickness increases and stabilisesThe profiles are shown in Figures 11(b) 11(c) and 11(d)

for the moment after the beginning of slug injection 1199052gt 1

Saturation decreases in a simple wave behind the rear slugfront jumps down on the front decreases in centred 119904-wavein the slug and is constant in zones II II and I Chemicalslug dissolution during the water drive injection is shown inFigure 11(c) There does occur the full concentration shockfrom zero behind the read slug front to the injected value inthe slug Further in the reservoir there does appear a zone ofintermediate chemical concentration 119888lowast in the bank 119904

3 The

concentration is equal to its initial value in banks 1199044and in

the initial zone So dissolution of slug occurs in the initialwater by formation of oil-water bank 119904

3with lower chemical

concentration Salinity changes by full shock on the frontbetween zones III and II

8 Conclusions

Application of the splitting method to 3 times 3 conservation lawsystem describing two-phase four-component flow in porousmedia allows drawing the following conclusions

(1) The method of splitting between hydrodynamics andthermodynamics in system of two-phase multicom-ponent flow in porous media allows obtaining anexact solution for non-self-similar problem of dis-placement of oil by chemical slug with different watersalinity for the case of linear polymer adsorptionaffected by water salinity

(2) The solution consists of explicit formulae for watersaturation and polymer and salt concentrations in thecontinuity domains and of implicit formulae for fronttrajectories

(3) First integrals for front trajectories allow for graphicalinterpretation at the hodograph plane yielding agraphical method for finding the front trajectories

(4) For linear sorption isotherms the solution dependson three fractional flow curves that correspond toinitial reservoir state 119877 injected fluid 119871 and an inter-mediate curve for intermediate polymer concentra-tion and injected salinity the value for intermediatepolymer concentration is the part of exact solution

(5) For linear sorption isotherms the only continuouswave is 119904-wave with constants 119888 and 120573 concentrations119888 and 120573 change only across the fronts by jumpsthus the solution of any problem with piece-wiseconstant initial and boundary conditions is reducedto interactions between 119904-waves and shocks

(6) Introduction of salinity dependency for sorption ofthe chemical introduces the intermediate (119888 120573)-shockinto the solution of the Riemann problem this shockinteracts with 119904-wave and concentration shocks inthe solution of any problem with piece-wise constantinitial and boundary conditions

(7) The exact solution shows that the injected chemicalslug dissolves in the connate reservoir water ratherthan in the chemical-free water injected after the slug

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

Long-term cooperation in hyperbolic systems and fruitfuldiscussions with Professors A Shapiro (Technical Universityof Denmark) Y Yortsos (University of Southern California)A Roberts (University of Adelaide) A Polyanin (RussianAcademy of Sciences) M Lurie and V Maron (MoscowOil and Gas Gubkin University) are gratefully acknowledgedThe reviewers are gratefully acknowledged for their criticalcomments yielding to significant improvement of the text

References

[1] R Courant and K O Friedrichs Supersonic Flow and ShockWaves Springer New York NY USA 1976


[2] I M Gelrsquofand ldquoSome problems in the theory of quasi-linearequationsrdquo Uspekhi Matematicheskikh Nauk vol 14 no 2 pp87ndash158 1959

[3] A G Kulikovskii N V Pogorelov and A Yu SemenovMathe-matical Aspects of Numerical Solution of Hyperbolic Systems vol118 Chapman amp HallCRC Press Boca Raton Fla USA 2001

[4] B L Rozhdestvenski and N N Ianenko Systems of QuasilinearEquations and Their Application to the Dynamics of Gases vol55 American Mathematical Society 1983

[5] G BWhitham Linear andNonlinearWaves vol 42 JohnWileyamp Sons Hoboken NJ USA 2011

[6] A Kulikovskii and E Sveshnikova Nonlinear Waves in ElasticMedia CRC Press Boca Raton Fla USA 1995

[7] J D Logan An Introduction to Nonlinear Partial DifferentialEquations vol 93 JohnWiley amp Sons Hobokon NJ USA 2010

[8] H K Rhee R Aris and N R Amundson First-Order PartialDifferential Equations Theory and Application of HyperbolicSystems of Quasilinear Equations Dover New York NY USA2001

[9] L W Lake Enhanced Oil Recovery Prentice Hall EnglewoodCliffs NJ USA 1989

[10] P BedrikovetskyMathematicalTheory of Oil and Gas RecoveryWith Applications to Ex-USSR Oil and Gas Fields KluwerAcademic Publishers Boston Mass USA 1993

[11] T Johansen A Tveito and R Winther ldquoA Riemann solver for atwo-phasemulticomponent processrdquo SIAM Journal on Scientificand Statistical Computing vol 10 no 5 pp 846ndash879 1989

[12] T Johansen and R Winther ldquoThe Riemann problem for multi-component polymer floodingrdquo SIAM Journal on MathematicalAnalysis vol 20 no 4 pp 908ndash929 1989

[13] R Johns and F M Orr Jr ldquoMiscible gas displacement ofmulticomponent oilsrdquo SPE Journal vol 1 pp 39ndash50 1996

[14] F M Orr Jr Theory of Gas Injection Processes Tie-LinePublications Copenhagen Denmark 2007

[15] P Bedrikovetsky ldquoDisplacement of oil by a chemical slug withwater driverdquo Journal of Fluid Dynamics vol 3 pp 102ndash111 1982

[16] V G Danilov and D Mitrovic ldquoSmooth approximations ofglobal in time solutions to scalar conservation lawsrdquo Abstractand Applied Analysis vol 2009 Article ID 350762 26 pages2009

[17] F Fayers ldquoSome theoretical results concerning the displacementof a viscous oil by a hot fluid in a porous mediumrdquo Journal ofFluid Mechanics vol 13 pp 65ndash76 1962

[18] E L Claridge and P L Bondor ldquoA graphical method for calcu-lating linear displacement with mass transfer and continuouslychanging mobilitiesrdquo SPE Journal vol 14 no 6 pp 609ndash6181974

[19] G J Hirasaki ldquoApplication of the theory of multicomponentmultiphase displacement to three-component two-phase sur-factant floodingrdquo SPE Journal vol 21 no 2 pp 191ndash204 1981

[20] G A Pope L W Lake and F G Helfferich ldquoCation exchangein chemical floodingmdashpart 1 basic theory without dispersionrdquoSPE Journal vol 18 no 6 pp 418ndash434 1978

[21] G I Barenblatt V M Entov and V M RyzhikTheory of FluidFlows through Natural Rocks Kluwer Academic PublishersLondon UK 1989

[22] G S Braginskaya and V M Entov ldquoNonisothermal displace-ment of oil by a solution of an active additiverdquo Fluid Dynamicsvol 15 no 6 pp 873ndash880 1980

[23] O Dahl T Johansen A Tveito and R Winther ldquoMulicom-ponent chromatography in a two phase environmentrdquo SIAMJournal on Applied Mathematics vol 52 no 1 pp 65ndash104 1992

[24] T Johansen Y Wang F M Orr Jr and B Dindoruk ldquoFour-component gasoil displacements in one dimensionmdashpart Iglobal triangular structurerdquo Transport in Porous Media vol 61no 1 pp 59ndash76 2005

[25] T Johansen and R Winther ldquoThe solution of the Riemannproblem for a hyperbolic system of conservation laws modelingpolymer floodingrdquo SIAM Journal onMathematical Analysis vol19 no 3 pp 541ndash566 1988

[26] P Bedrikovetsky and M Chumak ldquoRiemann problem fortwo-phase four-and morecomponent displacement (Ideal Mix-tures)rdquo in Proceedings of the 3rd European Conference on theMathematics of Oil Recovery 1992

[27] V Entov F Turetskaya and D Voskov ldquoOn approximation ofphase equilibria of multicomponent hydrocarbonmixtures andprediction of oil displacement by gas injectionrdquo inProceedings ofthe 8th EuropeanConference on theMathematics of Oil Recovery2002

[28] V Entov and D Voskov ldquoOn oil displacement by gas injectionrdquoin Proceedings of the 7th European Conference on theMathemat-ics of Oil Recovery 2000

[29] R T Johns B Dindoruk and F M Orr Jr ldquoAnalytical theoryof combined condensingvaporizing gas drivesrdquo SPE AdvancedTechnology Series vol 1 no 2 pp 7ndash16 1993

[30] C Wachmann ldquoA mathematical theory for the displacement ofoil and water by alcoholrdquo Old SPE Journal vol 4 pp 250ndash2661964

[31] A Zick ldquoA combined condensingvaporizingmechanism in thedisplacement of oil by enriched gasesrdquo in Proceedings of the SPEAnnual Technical Conference and Exhibition 1986

[32] O M Alishaeva V M Entov and A F Zazovskii ldquoStructuresof the conjugate saturation and concentration discontinuitiesin the displacement of oil by a solution of an active materialrdquoJournal of Applied Mechanics and Technical Physics vol 23 no5 pp 675ndash682 1982

[33] S Geiger K S Schmid andY Zaretskiy ldquoMathematical analysisand numerical simulation of multi-phase multi-componentflow in heterogeneous porous mediardquo Current Opinion inColloid and Interface Science vol 17 no 3 pp 147ndash155 2012

[34] K S Schmid S Geiger and K S Sorbie ldquoAnalytical solutionsfor co- and counter-current imbibition of sorbing dispersivesolutes in immiscible two-phase flowrdquo in Proceedings of the 12thEuropean Conference on the Mathematics of Oil Recovery 2012

[35] A P Pires P G Bedrikovetsky and A A Shapiro ldquoA splittingtechnique for analytical modelling of two-phase multicompo-nent flow in porous mediardquo Journal of Petroleum Science andEngineering vol 51 no 1-2 pp 54ndash67 2006

[36] A P Pires Splitting between thermodynamics and hydrody-namics in the processes of enhanced oil recovery [PhD thesis]Laboratory of Petroleum Exploration and Production NorthFluminense State University UENF 2003

[37] H-K Rhee R Aris and N R Amundson ldquoOn the theory ofmulticomponent chromatographyrdquo Philosophical Transactionsof the Royal Society A vol 267 no 1182 pp 419ndash455 1970

[38] V M Entov and A F Zazovskii ldquoDisplacement of oil by asolution of an active and a passive additiverdquo Fluid Dynamicsvol 17 no 6 pp 876ndash884 1982

[39] C B Cardoso R C A Silva and A P Pires ldquoThe roleof adsorption isotherms on chemical-flooding oil recoveryrdquo


in Proceedings of the SPE Annual Technical Conference andExhibition (ATCE rsquo07) pp 773ndash780 November 2007

[40] B Vicente V Priimenko and A Pires ldquoStreamlines simula-tion of polymer slugs injection in petroleum reservoirsrdquo inProceedings of the SPE Latin America and Caribbean PetroleumEngineering Conference 2012

[41] S Oladyshkin and M Panfilov ldquoSplitting the thermodynamlicsand hydrodynamics in compositional gas-liquid flow throughporous reservoirsrdquo in Proceedings of the 10th European Confer-ence on the Mathematics of Oil Recovery 2006

[42] P M Ribeiro and A P Pires ldquoThe displacement of oil bypolymer slugs considering adsorption effectsrdquo in Proceedings ofthe SPEAnnual Technical Conference and Exhibition (ATCE rsquo08)pp 851ndash865 September 2008

[43] P Bedrikovetski A Pires and A Shapiro ldquoConservation lawsystem for two-phase n-component flow in porous mediasplitting between thermodynamics and hydrodynamicsrdquo inProceedings of the 10th International Congress on HyperbolicProblems Theory 2004

[44] A Pires P Bedrikovetsky and A Shapiro ldquoSplitting betweenthermodynamics and hydrodynamics in compositional mod-ellingrdquo in Proceedings of the 9th European Conference on theMathematics of Oil Recovery 2004

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


Therefore the system (1) and (2) is called themulticomponentpolymer-floodingmodel [11 12] Besides (119899minus1)times(119899minus1)hyper-bolic system (1) and (2) describes two-phase 119899-componentdisplacement which is typical for so-called gas methods ofenhanced oil recovery [9 10 13 14] The processes of hotwater injection with phase transitions secondary migrationof hydrocarbons with consequent formation of petroleumaccumulations enhanced geothermal energy projects andinjections into aquifers are described by the above systemsThe Riemann problems correspond to continuous injectionof chemical solutions or gases into oil reservoirs the solutionsare self-similar [3 9 14] The wave-interaction problems cor-respond to piece-wise-constant initial-boundary conditionsfor which the solutions are non-self-similar [1 10 15ndash17]Thewave-interaction solutions describe injection of limited slugs(banks) of chemical solutions or gaseous solvents driven inthe reservoirs by water or gas [9 10]

Riemann problem (1) and (2) with 119899 = 1 has been solvedwith applications to various injections of polymers [17 18]carbonized water and surfactants [19 20] and so forth Morecomplex self-similar solutions of (1) and (2) for 119899 = 2 3 wereobtained by Barenblatt et al [21] and Braginskaya and Entov[22] and later by Johansen et al andWinther et al [11 12 23ndash25] Analogous solutions for gas injection and 119899 = 3 4 have been obtained by Orr and others [9 13 26ndash31]

The system (1) and (2) describes two-phase multicom-ponent displacements in large scale approximation wherethe dissipative effects of capillary pressure diffusion andthermodynamic nonequilibrium are negligible if they arecompared with advective fluxes under the large length scaleof the natural subterranean reservoirs Travelling waves nearto shock discontinuities in dissipative systems have beenpresented in [10 32] A semianalytical global solutions havebeen obtained by Geiger et al [33] and Schmid et al [34] seealso [16]

The particular case of so-calledmulticomponent polymerflooding is the dependency of the component sorptionconcentration of its own concentration only 119886

119894(1198881 1198882 119888

119899) =

119886119894(119888119894) Exact solutions of the Riemann problem for this case

show that the concentration of each component performsthe jump without shocks of other components (see thecorresponding solution in the books [10 21]) Therefore inconcentration profiles the shocks are located in order ofdecrease of derivatives of the sorption functions In the case ofHenry isotherms 119886

119894(119888119894) = Γ119894119888119894 the shocks are located in order

of increase of Henryrsquos sorption coefficients Γ119894

The distinguished invariant feature of (119899 + 1) times (119899 +

1) conservation law systems for two-phase multicomponentflows in porous media with sorption and phase transitionsequations (1) and (2) is its splitting into an 119899 times 119899 auxiliarysystem for concentrations 119888

119894(119909 119905) and a scalar hyperbolic

equation for saturation 119904(119909 119905) [35 36]This splitting explainsthe simple form of Riemann problem solutions for system (1)and (2) as compared with gas dynamics or chromatography[1 2 37]

The non-self-similar solution of system (1) and (2) 119899 =

2 for slug injections has been considered by Fayers [17]where the qualitative behaviour of characteristic lines andshocks has been described The exact solutions of system

(1) and (2) for 119899 = 2 and 3 have been obtained in [15](see book [10] for detailed derivations in which the sorptionof component depends on its own concentration only 119886

119894=

119886119894(119888119894) 119894 = 1 2 ) Numerous interactions of different

saturation-concentration shocks occur after the injectionresulting in appearance of moving zones with differentcombinations of components However after all interactionsdifferent component slugs are separated from each otherAs in the case of continuous injection the slugs are finallypositioned in the order of decreasing sorption isothermderivatives (119889119886

119894119889119888119894) It seems that this simplified case draws

the line where the analytical solutions can be found from theanalysis of system (1) and (2) directly Consideration of cross-effects 119886

119894= 119886119894(1198881 1198882 119888

119899) in sorption functions equation

(3) introduces significant difficulties into wave analysis andeven the Riemann problem cannot be solved for any arbitrarycase 119899 = 2 (see [38] where the Riemann solutions have beenobtained for several particular cases)

The splitting technique reduces number of equations in(1) and (2) by one allowing for exact solutions in more com-plex multicomponent cases [35ndash40] The Riemann problemwith cross-effects for adsorption 119886

119894= 119886119894(1198881 1198882) has been

solved in [39 41 42] for continuous polymer injection withvarying salinity using the splitting method In the currentpaper the exact solution for non-self-similar problem ofinjection of polymer slug with varying salinity followed bywater drive is obtained

The structure of the text is as follows The particularcase of the general system (1) and (2) that is discussedin the current work is introduced in Section 2 along withphysics assumptions and initial-boundary conditions for sluginjection problem The detailed description of the splittingprocedure for the system is discussed along with formulationof initial and boundary conditions for the auxiliary systemwhich is presented in Section 3 Section 4 contains derivationof the Riemann solution that corresponds to the first stageof the slug injection The wave-interaction slug injectionproblem is solved in Section 5 Section 6 contains a simplifiedsolution for the particular case where the initial chemicalconcentration is zero which corresponds to the case ofpolymer slug injection The paper is concluded by physicalinterpretation of the solution obtained for chemical sluginjection with different water salinity into oilfield (Sections7 and 8)

2 Formulation of the Problem

Let us discuss the displacement of oil by aqueous chemicalsolution with water drive accounting for different salinitiesof formation and injected waters In the following text thecomponent 119899 = 1 is called the polymer and that 119899 = 2 iscalled the salt The assumptions of the mathematical modelare as follows both phases are incompressible dispersionand capillary forces are neglected there are two phases (oleicand aqueous phases) and two components dissolved in water(polymer and salt) water and oil phases are immisciblechemical and salt concentrations in water are negligiblysmall and do not affect the volume of the aqueous phase




120597119904

120597119905+120597119891 (119904 119888)

120597119909= 0 (4)

120597 (119888119904 + 119886 (119888 120573))

120597119905+120597 (119888119891 (119904 119888))

120597119909= 0 (5)

120597 (120573119904)

120597119905+120597 (120573119891 (119904 119888))

120597119909= 0 (6)





119909 = 0120573 = 0 119888 = 119888

1 119904 = 119904

119871 119905 lt 1

120573 = 0 119888 = 1198882 119904 = 119904

119871 119905 gt 1

(7)

119905 = 0 120573 = 1 119888 = 1198882 119904 = 119904

119877 (8)


119909 = 0 120573 = 0 119888 = 1198881 119904 = 119904

119871 (9)

119905 = 0 120573 = 1 119888 = 1198882 119904 = 119904

119877 (10)






01

0 10

1L

s

f

R

D3

D2

D1

2

3

4

c = c1

c = clowast

c = c2

minusΓ(1)










119904 = minus120597120593

120597119909

119891 =120597120593

120597119905

(11)

that is

119889120593 = 119891119889119905 minus 119904119889119909 (12)

120593 (119909 119905) = int

119909119905

00

119891119889119905 minus 119904119889119909 (13)





t

t1

0 0x xx1

x1

120593

120593 = 120593(x = 0 t1)

fL(x = 0 t1) cL(x = 0 t1)

fL(x = 0 t1) cL(x = 0 t1)

120593 = 120593(x1 t = 0)sR(x1 t = 0) cR(x1 t = 0)

sR(x1 t = 0) cR(x1 t = 0)


t

x x

120593 + Δ120593

120593 + Δ120593

120593

120593

x + Δx x + Δx



120593 (119909 1199052) minus 120593 (119909 119905

1) = int

1199052

1199051

119891 (119909 119905) 119889119905 (14)



0(119905)


1205930(119905) = 120593 (119909

0(119905) 119905) (15)


119863 =119889119909

119889119905 119881 =

119889119909

119889120593 (16)




1

119881=119891

119863minus 119904 (17)


119863 =119891

119904 + 1119881 (18)


119894and in (120593 119909) Λ


[43 44])

Λ119894(119904 119888) =

119891

119904 + 1120582119894

(19)


from (12) as

119889119905 =119889120593

119891+119904119889119909

119891(20)


t

D

x

1 1

120593

V

x1120582i

+1V 1120582i

minus0

s

D

f

+

minus



1198892119905 = 0 = [

120597

120597119909(1

119891) minus

120597

120597120593(119904

119891)] 119889119909119889120593 (21)


120597 (119904119891)

120597120593minus120597 (1119891)

120597119909= 0 (22)



∮

120597Ω

(119888119891) 119889119905 minus (119888119904 + 119886) 119889119909 = 0 (23)




∮

120597Ω

119888 (119891119889119905 minus 119904119889119909) minus 119886119889119909 = ∮

120597Ω

119888119889120593 minus 119886119889119909 = 0 (24)


120597119886 (119888 120573)

120597120593+120597119888

120597119909= 0 (25)


∬(120597 (120573119904)

120597119905+120597 (120573119891 (119904 119888))

120597119909)119889119909119889119905 = ∮

120597Ω

120573119904119889119909 minus 120573119891 (119904 119888) 119889119905

= ∮120573119889120593 = ∬120597120573

120597119909= 0

(26)






119865 = minus119904

119891 119880 =

1

119891(27)

120597 (119865 (119880 119888))

120597120593+120597 (119880)

120597119909= 0 (28)

120597119886 (119888 120573)

120597120593+120597119888

120597119909= 0 (29)

120597120573

120597119909= 0 (30)



+

+

+

minus

minus

minus

s

cn

c1



119909 = 0120573 = 0 119888 = 119888

1 119880 = 1 120593 lt 1

120573 = 0 119888 = 1198882 119880 = 1 120593 gt 1

(31)



120593 = minus 119904119877119909 120573 = 1 119888 = 119888

2 119880 = +infin (32)


119909 = 0 120573 = 0 119888 = 1198881 119880 = 1

120593 = minus119904119877119909 120573 = 1 119888 = 119888

2 119880 = +infin

(33)




120590 [119880] = [119865] (34)

120590 [119888] = [119886] (35)

120590 [120573] = 0 (36)




tions [1ndash4 9]


119888 (119909 120593)

120573 (119909 120593)=

1198881 120573 = 0 120593 gt Γ (0) 119909

119888lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198882 120573 = 1 minus119904

119877119909 lt 120593 lt 0

(37)


119888lowast=Γ (1)

Γ (0)1198882 (38)


1 119888 =

1198882 and 119888 = 119888



120593

119909=

120597119865 (1198801 1198881)

120597119880 (39)


120597119865 (1198802 1198881)

120597119880=1198652(1198802 1198881) minus 1198653(1198803 119888lowast)

1198802minus 1198803

= Γ (0) (40)


1198653(1198803 119888lowast) = 1198654(1198804 1198882) = 0 (41)


1198654(1198804 1198882) minus 119865119894(119880119877 1198882)

1198804minus 119880119894

=minus1199044119891119877+ 1199041198771198914

119891119877 minus 1198914

= minus119904119877 (42)


c1c2 clowast c

R

a

a1

a2

L

120573 = 1

120573 = 0

(a)

c1

c2

clowast

c

R

L

1205731

(b)


F = minussf

U = 1f01

2

34

67

LminussL

minussR

R

c = c1c = clowast

c = c2



119880(119909 120593)

119888 (119909 120593)

120573 (119909 120593)

=

1198801(120593

119909) 1198881 120573 = 0 120593 gt Γ (0) 119909

1198803 119888

lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198804 119888

2 120573 = 1 minus119904

119877119909 lt 120593 lt 0

infin 1198882 120573 = 1 120593 gt minus119904

119877119909

(43)


119904 (119909 120593)

119888 (119909 120593)

120573 (119909 120593)

=

1199041(120593

119909) 1198881 120573 = 0 120593 gt Γ (0) 119909

1199043 119888

lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1199044 119888

2 120573 = 1 minus119904

119877119909 lt 120593 lt 0

119904119877 119888

2 120573 = 1 120593 lt minus119904

119877119909

(44)


I

II

III

IV

V120593

(f = 1 120573 = 0 c2)

(f = 0 120573 = 1 c2)

(f = 1 120573 = 0 c1)

1

0

U5(120593 x)

120573 = 0

120573 = 0

120573 = 0

c2

c2

U1(120593x)

c1

120593 = Γ(0)x + 1

120593 = Γ(0)x

clowastU3

U4 120573 = 1x

U = infin

120593 = minussRx

67

(x998400 120593998400)



119905 =1

1198914

int

120593

0

119889120593 +1199044

1198914

int

119909

0

119889119909 = (minus119904119877+ 1199044

1198914

)119909 (45)


119905 =1

1198913

int

120593

0

119889120593 +1199043

1198913

int

119909

0

119889119909 =1199043

1198913

119909 (46)


I

V

II

III

IV

t

(f = 1 120573 = 0 c2)

(f = 1 120573 = 0 c1)

1

0

s5(t x)

120573 = 0

120573 = 0 120573 = 0

c2

c2

c2

c1 clowasts3

s4120573 = 1

120573 = 1

x

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s4)xf4

s1(tx)

(x998400 t998400)

sR

(f = 0 120573 = 1 c2)



00(119889120593119891 +


119905 =120593

119891 (1199041 (120593119909) 1198881)+

1199041(120593119909)

119891 (1199041 (120593119909) 1198881)119909 (47)


119904 (119909 119905)

119888 (119909 119905)

120573 (119909 119905)

=

1199041(119905

119909) 1198881 120573 = 0 119905 gt

Γ (0) + 1199042

1198912

119909

1199043 119888

lowast 120573 = 0

1199043

1198913

119909 lt 119905 ltΓ (0) + 119904

2

1198912

119909

1199044 119888

2 120573 = 1

minus119904119877+ 1199044

1198914

119909 lt 119905 lt1199043

1198913

119909

119904119877 119888

2 120573 = 1 119905 lt

minus119904119877+ 1199044

1198914

119909

(48)



2


119888 (119909 120593)

120573 (119909 120593)=

1198882 120573 = 0 120593 gt Γ (0) 119909 + 1

1198881 120573 = 0 Γ (0) 119909 lt 120593 lt Γ (0) 119909 + 1

119888lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198882 120573 = 1 minus119904

119877119909 lt 120593 lt 0

(49)



I

II

III

IV

t

0

120573 = 0

120573 = 0

c2

c2

clowasts3

s3

s2

s4

s4

120573 = 1

120573 = 1

x

x

x

x

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s4)xf4

sR

sR

sL

c1

c2

clowast

c1

s

c

1

0

120573

D1t1 D2t1 D3t1

(a)

(b)

(c)

(d)

t1

s1(tx)



2and 120573 = 0



119865 (119880+ 1198881) minus 119865 (119880

minus 1198882)

119880+ minus 119880minus= Γ (0) (50)



1198805(119909 120593) = 119880

minus(1199091015840 1205931015840) (51)


120593 minus 1205931015840

119909 minus 1199091015840=120597119865 (119880

5 1198881)

120597119880 (52)


119880 (119909 120593)

119888 (119909 120593)

120573 (119909 120593)

=

1198805(119909 120593) 119888

2 120573 = 0 120593 gt Γ (0) 119909 + 1

1198801(120593

119909) 1198881 120573 = 0 Γ(0) 119909lt120593ltΓ(0) 119909+1

1198803 119888

lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198804 119888

2 120573 = 1 minus119904

119877119909 lt 120593 lt 0

infin 1198882 120573 = 1 120593 gt minus119904

119877119909

(53)


120593 = Γ (0) 119909 + 1 (54)


120593

119909=119891 (119904+ 1198881) minus 119904+1198911015840(119904+ 1198881)

1198911015840 (119904+ 1198881)

(55)


parametric form

119909119863(119904+) =

1198911015840(119904+ 1198881)

119891 (119904+ 1198881) minus 1198911015840 (119904+ 119888

1) (119904+ + Γ)

=1198911015840(119904+ 1198881)

Δ(56)

120593119863(119904+) =

119891 (119904+ 1198881) minus 1199041+

1198911015840(119904+ 1198881)

119891 (119904+ 1198881) minus 1198911015840 (119904+ 119888

1) (119904+ + Γ)

=119891 (119904+ 1198881) minus 1199041+

1198911015840(119904+ 1198881)

Δ

(57)


119905119863=

120593

119891 (119904+ (120593 119909) 1198882)+

119904+(120593 119909)

119891 (119904+ (120593 119909) 1198882)119909

119905119863=

1

119891 (119904+ 1198881) minus 1198911015840 (119904+ 119888

1) (1199041++ Γ)

=1

Δ

(58)


119904 (119909 119905)

119888 (119909 119905)

120573 (119909 119905)

=

1199045 (119905 119909) 1198882 120573 = 0 119905 gt

Γ (0) + 1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909 + 1

1199041(119905

119909) 1198881 120573 = 0

Γ (0) + 1199042

1198912

119909 lt 119905 ltΓ (0) + 1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909 + 1

1199043 119888

lowast 120573 = 0

1199043

1198913

119909 lt 119905 ltΓ (0) + 1199042

1198912

119909

1199044 119888

2 120573 = 1

minus119904119877+ 1199044

1198914

119909 lt 119905 lt1199043

1198913

119909

119904119877 119888

2 120573 = 1 119905 lt

minus119904119877+ 1199044

1198914

119909




119909119863

= 119904119861+ Γ (0)

1

119905119863

= 119891119860

(60)





119863




119888 (119909 119905)

120573 (119909 119905)

=

1199045 (119905 119909) 119888 = 0 120573 = 0 119905 gt

Γ (0) + 1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909 + 1

1199041(119905

119909) 119888 = 119888

1 120573 = 0

Γ (0) + 1199042

1198912

119909lt119905ltΓ (0)+1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909+1

1199043 119888 = 0 120573 = 0

1199043

1198913

119909 lt 119905 ltΓ (0) + 1199042

1198912

119909

1199043 119888 = 0 120573 = 1

minus119904119877+ 1199043

1198913

119909 lt 119905 lt1199043

1198913

119909

119904119877 119888 = 0 120573 = 1 119905 lt

minus119904119877+ 1199043

1198913

119909

(61)


I

II

III

IV

t

0

120573 = 0

120573 = 0

120573 = 0

c2

c2

c2

clowasts3

s3

s2

s4

s4

120573 = 1

120573 = 1

x

x

x

x

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s4)xf4

sR

sR

sL

c1

c2

clowast

c1

s

c

1

1

0

120573


(a)

(b)

(c)

(d)

V

s5(t x) (x998400 t998400)

t1

t1

t2

t2

s1(tx)





2and of salt 120573 = 1


1and


01

0 10

1L

f

R

2

3

4

c = c1

c = clowast

c = c2

s

s+sminus

c

ab

Δ

Δf998400(s+)


01

0 10

1L

s

f

R

2

3

c = c1

minusΓ(0)

c = c2 = 0








4


2


1for profiles is





3-bank and initial


I

II

III

IVV

(f = 1 120573 = 0 c1)

1

0

120573 = 0

120573 = 1

120573 = 1

120573 = 0

120573 = 0

x

s5(t x)

s3

s3

(x998400 t998400)

sR

c1

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s3)xf3

c = 0

c = 0

c = 0

c = 0

(f = 0 120573 = 1 c = 0)

(f = 1 120573 = 0 c = 0)

t

s1(tx)


2= 0




1lt 1 during







3 The





8 Conclusions











Acknowledgments


References























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra












120597119904

120597119905+120597119891 (119904 119888)

120597119909= 0 (4)

120597 (119888119904 + 119886 (119888 120573))

120597119905+120597 (119888119891 (119904 119888))

120597119909= 0 (5)

120597 (120573119904)

120597119905+120597 (120573119891 (119904 119888))

120597119909= 0 (6)





119909 = 0120573 = 0 119888 = 119888

1 119904 = 119904

119871 119905 lt 1

120573 = 0 119888 = 1198882 119904 = 119904

119871 119905 gt 1

(7)

119905 = 0 120573 = 1 119888 = 1198882 119904 = 119904

119877 (8)


119909 = 0 120573 = 0 119888 = 1198881 119904 = 119904

119871 (9)

119905 = 0 120573 = 1 119888 = 1198882 119904 = 119904

119877 (10)






01

0 10

1L

s

f

R

D3

D2

D1

2

3

4

c = c1

c = clowast

c = c2

minusΓ(1)










119904 = minus120597120593

120597119909

119891 =120597120593

120597119905

(11)

that is

119889120593 = 119891119889119905 minus 119904119889119909 (12)

120593 (119909 119905) = int

119909119905

00

119891119889119905 minus 119904119889119909 (13)





t

t1

0 0x xx1

x1

120593

120593 = 120593(x = 0 t1)

fL(x = 0 t1) cL(x = 0 t1)

fL(x = 0 t1) cL(x = 0 t1)

120593 = 120593(x1 t = 0)sR(x1 t = 0) cR(x1 t = 0)

sR(x1 t = 0) cR(x1 t = 0)


t

x x

120593 + Δ120593

120593 + Δ120593

120593

120593

x + Δx x + Δx



120593 (119909 1199052) minus 120593 (119909 119905

1) = int

1199052

1199051

119891 (119909 119905) 119889119905 (14)



0(119905)


1205930(119905) = 120593 (119909

0(119905) 119905) (15)


119863 =119889119909

119889119905 119881 =

119889119909

119889120593 (16)




1

119881=119891

119863minus 119904 (17)


119863 =119891

119904 + 1119881 (18)


119894and in (120593 119909) Λ


[43 44])

Λ119894(119904 119888) =

119891

119904 + 1120582119894

(19)


from (12) as

119889119905 =119889120593

119891+119904119889119909

119891(20)


t

D

x

1 1

120593

V

x1120582i

+1V 1120582i

minus0

s

D

f

+

minus



1198892119905 = 0 = [

120597

120597119909(1

119891) minus

120597

120597120593(119904

119891)] 119889119909119889120593 (21)


120597 (119904119891)

120597120593minus120597 (1119891)

120597119909= 0 (22)



∮

120597Ω

(119888119891) 119889119905 minus (119888119904 + 119886) 119889119909 = 0 (23)




∮

120597Ω

119888 (119891119889119905 minus 119904119889119909) minus 119886119889119909 = ∮

120597Ω

119888119889120593 minus 119886119889119909 = 0 (24)


120597119886 (119888 120573)

120597120593+120597119888

120597119909= 0 (25)


∬(120597 (120573119904)

120597119905+120597 (120573119891 (119904 119888))

120597119909)119889119909119889119905 = ∮

120597Ω

120573119904119889119909 minus 120573119891 (119904 119888) 119889119905

= ∮120573119889120593 = ∬120597120573

120597119909= 0

(26)






119865 = minus119904

119891 119880 =

1

119891(27)

120597 (119865 (119880 119888))

120597120593+120597 (119880)

120597119909= 0 (28)

120597119886 (119888 120573)

120597120593+120597119888

120597119909= 0 (29)

120597120573

120597119909= 0 (30)



+

+

+

minus

minus

minus

s

cn

c1



119909 = 0120573 = 0 119888 = 119888

1 119880 = 1 120593 lt 1

120573 = 0 119888 = 1198882 119880 = 1 120593 gt 1

(31)



120593 = minus 119904119877119909 120573 = 1 119888 = 119888

2 119880 = +infin (32)


119909 = 0 120573 = 0 119888 = 1198881 119880 = 1

120593 = minus119904119877119909 120573 = 1 119888 = 119888

2 119880 = +infin

(33)




120590 [119880] = [119865] (34)

120590 [119888] = [119886] (35)

120590 [120573] = 0 (36)




tions [1ndash4 9]


119888 (119909 120593)

120573 (119909 120593)=

1198881 120573 = 0 120593 gt Γ (0) 119909

119888lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198882 120573 = 1 minus119904

119877119909 lt 120593 lt 0

(37)


119888lowast=Γ (1)

Γ (0)1198882 (38)


1 119888 =

1198882 and 119888 = 119888



120593

119909=

120597119865 (1198801 1198881)

120597119880 (39)


120597119865 (1198802 1198881)

120597119880=1198652(1198802 1198881) minus 1198653(1198803 119888lowast)

1198802minus 1198803

= Γ (0) (40)


1198653(1198803 119888lowast) = 1198654(1198804 1198882) = 0 (41)


1198654(1198804 1198882) minus 119865119894(119880119877 1198882)

1198804minus 119880119894

=minus1199044119891119877+ 1199041198771198914

119891119877 minus 1198914

= minus119904119877 (42)


c1c2 clowast c

R

a

a1

a2

L

120573 = 1

120573 = 0

(a)

c1

c2

clowast

c

R

L

1205731

(b)


F = minussf

U = 1f01

2

34

67

LminussL

minussR

R

c = c1c = clowast

c = c2



119880(119909 120593)

119888 (119909 120593)

120573 (119909 120593)

=

1198801(120593

119909) 1198881 120573 = 0 120593 gt Γ (0) 119909

1198803 119888

lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198804 119888

2 120573 = 1 minus119904

119877119909 lt 120593 lt 0

infin 1198882 120573 = 1 120593 gt minus119904

119877119909

(43)


119904 (119909 120593)

119888 (119909 120593)

120573 (119909 120593)

=

1199041(120593

119909) 1198881 120573 = 0 120593 gt Γ (0) 119909

1199043 119888

lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1199044 119888

2 120573 = 1 minus119904

119877119909 lt 120593 lt 0

119904119877 119888

2 120573 = 1 120593 lt minus119904

119877119909

(44)


I

II

III

IV

V120593

(f = 1 120573 = 0 c2)

(f = 0 120573 = 1 c2)

(f = 1 120573 = 0 c1)

1

0

U5(120593 x)

120573 = 0

120573 = 0

120573 = 0

c2

c2

U1(120593x)

c1

120593 = Γ(0)x + 1

120593 = Γ(0)x

clowastU3

U4 120573 = 1x

U = infin

120593 = minussRx

67

(x998400 120593998400)



119905 =1

1198914

int

120593

0

119889120593 +1199044

1198914

int

119909

0

119889119909 = (minus119904119877+ 1199044

1198914

)119909 (45)


119905 =1

1198913

int

120593

0

119889120593 +1199043

1198913

int

119909

0

119889119909 =1199043

1198913

119909 (46)


I

V

II

III

IV

t

(f = 1 120573 = 0 c2)

(f = 1 120573 = 0 c1)

1

0

s5(t x)

120573 = 0

120573 = 0 120573 = 0

c2

c2

c2

c1 clowasts3

s4120573 = 1

120573 = 1

x

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s4)xf4

s1(tx)

(x998400 t998400)

sR

(f = 0 120573 = 1 c2)



00(119889120593119891 +


119905 =120593

119891 (1199041 (120593119909) 1198881)+

1199041(120593119909)

119891 (1199041 (120593119909) 1198881)119909 (47)


119904 (119909 119905)

119888 (119909 119905)

120573 (119909 119905)

=

1199041(119905

119909) 1198881 120573 = 0 119905 gt

Γ (0) + 1199042

1198912

119909

1199043 119888

lowast 120573 = 0

1199043

1198913

119909 lt 119905 ltΓ (0) + 119904

2

1198912

119909

1199044 119888

2 120573 = 1

minus119904119877+ 1199044

1198914

119909 lt 119905 lt1199043

1198913

119909

119904119877 119888

2 120573 = 1 119905 lt

minus119904119877+ 1199044

1198914

119909

(48)



2


119888 (119909 120593)

120573 (119909 120593)=

1198882 120573 = 0 120593 gt Γ (0) 119909 + 1

1198881 120573 = 0 Γ (0) 119909 lt 120593 lt Γ (0) 119909 + 1

119888lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198882 120573 = 1 minus119904

119877119909 lt 120593 lt 0

(49)



I

II

III

IV

t

0

120573 = 0

120573 = 0

c2

c2

clowasts3

s3

s2

s4

s4

120573 = 1

120573 = 1

x

x

x

x

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s4)xf4

sR

sR

sL

c1

c2

clowast

c1

s

c

1

0

120573

D1t1 D2t1 D3t1

(a)

(b)

(c)

(d)

t1

s1(tx)



2and 120573 = 0



119865 (119880+ 1198881) minus 119865 (119880

minus 1198882)

119880+ minus 119880minus= Γ (0) (50)



1198805(119909 120593) = 119880

minus(1199091015840 1205931015840) (51)


120593 minus 1205931015840

119909 minus 1199091015840=120597119865 (119880

5 1198881)

120597119880 (52)


119880 (119909 120593)

119888 (119909 120593)

120573 (119909 120593)

=

1198805(119909 120593) 119888

2 120573 = 0 120593 gt Γ (0) 119909 + 1

1198801(120593

119909) 1198881 120573 = 0 Γ(0) 119909lt120593ltΓ(0) 119909+1

1198803 119888

lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198804 119888

2 120573 = 1 minus119904

119877119909 lt 120593 lt 0

infin 1198882 120573 = 1 120593 gt minus119904

119877119909

(53)


120593 = Γ (0) 119909 + 1 (54)


120593

119909=119891 (119904+ 1198881) minus 119904+1198911015840(119904+ 1198881)

1198911015840 (119904+ 1198881)

(55)


parametric form

119909119863(119904+) =

1198911015840(119904+ 1198881)

119891 (119904+ 1198881) minus 1198911015840 (119904+ 119888

1) (119904+ + Γ)

=1198911015840(119904+ 1198881)

Δ(56)

120593119863(119904+) =

119891 (119904+ 1198881) minus 1199041+

1198911015840(119904+ 1198881)

119891 (119904+ 1198881) minus 1198911015840 (119904+ 119888

1) (119904+ + Γ)

=119891 (119904+ 1198881) minus 1199041+

1198911015840(119904+ 1198881)

Δ

(57)


119905119863=

120593

119891 (119904+ (120593 119909) 1198882)+

119904+(120593 119909)

119891 (119904+ (120593 119909) 1198882)119909

119905119863=

1

119891 (119904+ 1198881) minus 1198911015840 (119904+ 119888

1) (1199041++ Γ)

=1

Δ

(58)


119904 (119909 119905)

119888 (119909 119905)

120573 (119909 119905)

=

1199045 (119905 119909) 1198882 120573 = 0 119905 gt

Γ (0) + 1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909 + 1

1199041(119905

119909) 1198881 120573 = 0

Γ (0) + 1199042

1198912

119909 lt 119905 ltΓ (0) + 1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909 + 1

1199043 119888

lowast 120573 = 0

1199043

1198913

119909 lt 119905 ltΓ (0) + 1199042

1198912

119909

1199044 119888

2 120573 = 1

minus119904119877+ 1199044

1198914

119909 lt 119905 lt1199043

1198913

119909

119904119877 119888

2 120573 = 1 119905 lt

minus119904119877+ 1199044

1198914

119909




119909119863

= 119904119861+ Γ (0)

1

119905119863

= 119891119860

(60)





119863




119888 (119909 119905)

120573 (119909 119905)

=

1199045 (119905 119909) 119888 = 0 120573 = 0 119905 gt

Γ (0) + 1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909 + 1

1199041(119905

119909) 119888 = 119888

1 120573 = 0

Γ (0) + 1199042

1198912

119909lt119905ltΓ (0)+1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909+1

1199043 119888 = 0 120573 = 0

1199043

1198913

119909 lt 119905 ltΓ (0) + 1199042

1198912

119909

1199043 119888 = 0 120573 = 1

minus119904119877+ 1199043

1198913

119909 lt 119905 lt1199043

1198913

119909

119904119877 119888 = 0 120573 = 1 119905 lt

minus119904119877+ 1199043

1198913

119909

(61)


I

II

III

IV

t

0

120573 = 0

120573 = 0

120573 = 0

c2

c2

c2

clowasts3

s3

s2

s4

s4

120573 = 1

120573 = 1

x

x

x

x

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s4)xf4

sR

sR

sL

c1

c2

clowast

c1

s

c

1

1

0

120573


(a)

(b)

(c)

(d)

V

s5(t x) (x998400 t998400)

t1

t1

t2

t2

s1(tx)





2and of salt 120573 = 1


1and


01

0 10

1L

f

R

2

3

4

c = c1

c = clowast

c = c2

s

s+sminus

c

ab

Δ

Δf998400(s+)


01

0 10

1L

s

f

R

2

3

c = c1

minusΓ(0)

c = c2 = 0








4


2


1for profiles is





3-bank and initial


I

II

III

IVV

(f = 1 120573 = 0 c1)

1

0

120573 = 0

120573 = 1

120573 = 1

120573 = 0

120573 = 0

x

s5(t x)

s3

s3

(x998400 t998400)

sR

c1

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s3)xf3

c = 0

c = 0

c = 0

c = 0

(f = 0 120573 = 1 c = 0)

(f = 1 120573 = 0 c = 0)

t

s1(tx)


2= 0




1lt 1 during







3 The





8 Conclusions











Acknowledgments


References























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










t

t1

0 0x xx1

x1

120593

120593 = 120593(x = 0 t1)

fL(x = 0 t1) cL(x = 0 t1)

fL(x = 0 t1) cL(x = 0 t1)

120593 = 120593(x1 t = 0)sR(x1 t = 0) cR(x1 t = 0)

sR(x1 t = 0) cR(x1 t = 0)


t

x x

120593 + Δ120593

120593 + Δ120593

120593

120593

x + Δx x + Δx



120593 (119909 1199052) minus 120593 (119909 119905

1) = int

1199052

1199051

119891 (119909 119905) 119889119905 (14)



0(119905)


1205930(119905) = 120593 (119909

0(119905) 119905) (15)


119863 =119889119909

119889119905 119881 =

119889119909

119889120593 (16)




1

119881=119891

119863minus 119904 (17)


119863 =119891

119904 + 1119881 (18)


119894and in (120593 119909) Λ


[43 44])

Λ119894(119904 119888) =

119891

119904 + 1120582119894

(19)


from (12) as

119889119905 =119889120593

119891+119904119889119909

119891(20)


t

D

x

1 1

120593

V

x1120582i

+1V 1120582i

minus0

s

D

f

+

minus



1198892119905 = 0 = [

120597

120597119909(1

119891) minus

120597

120597120593(119904

119891)] 119889119909119889120593 (21)


120597 (119904119891)

120597120593minus120597 (1119891)

120597119909= 0 (22)



∮

120597Ω

(119888119891) 119889119905 minus (119888119904 + 119886) 119889119909 = 0 (23)




∮

120597Ω

119888 (119891119889119905 minus 119904119889119909) minus 119886119889119909 = ∮

120597Ω

119888119889120593 minus 119886119889119909 = 0 (24)


120597119886 (119888 120573)

120597120593+120597119888

120597119909= 0 (25)


∬(120597 (120573119904)

120597119905+120597 (120573119891 (119904 119888))

120597119909)119889119909119889119905 = ∮

120597Ω

120573119904119889119909 minus 120573119891 (119904 119888) 119889119905

= ∮120573119889120593 = ∬120597120573

120597119909= 0

(26)






119865 = minus119904

119891 119880 =

1

119891(27)

120597 (119865 (119880 119888))

120597120593+120597 (119880)

120597119909= 0 (28)

120597119886 (119888 120573)

120597120593+120597119888

120597119909= 0 (29)

120597120573

120597119909= 0 (30)



+

+

+

minus

minus

minus

s

cn

c1



119909 = 0120573 = 0 119888 = 119888

1 119880 = 1 120593 lt 1

120573 = 0 119888 = 1198882 119880 = 1 120593 gt 1

(31)



120593 = minus 119904119877119909 120573 = 1 119888 = 119888

2 119880 = +infin (32)


119909 = 0 120573 = 0 119888 = 1198881 119880 = 1

120593 = minus119904119877119909 120573 = 1 119888 = 119888

2 119880 = +infin

(33)




120590 [119880] = [119865] (34)

120590 [119888] = [119886] (35)

120590 [120573] = 0 (36)




tions [1ndash4 9]


119888 (119909 120593)

120573 (119909 120593)=

1198881 120573 = 0 120593 gt Γ (0) 119909

119888lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198882 120573 = 1 minus119904

119877119909 lt 120593 lt 0

(37)


119888lowast=Γ (1)

Γ (0)1198882 (38)


1 119888 =

1198882 and 119888 = 119888



120593

119909=

120597119865 (1198801 1198881)

120597119880 (39)


120597119865 (1198802 1198881)

120597119880=1198652(1198802 1198881) minus 1198653(1198803 119888lowast)

1198802minus 1198803

= Γ (0) (40)


1198653(1198803 119888lowast) = 1198654(1198804 1198882) = 0 (41)


1198654(1198804 1198882) minus 119865119894(119880119877 1198882)

1198804minus 119880119894

=minus1199044119891119877+ 1199041198771198914

119891119877 minus 1198914

= minus119904119877 (42)


c1c2 clowast c

R

a

a1

a2

L

120573 = 1

120573 = 0

(a)

c1

c2

clowast

c

R

L

1205731

(b)


F = minussf

U = 1f01

2

34

67

LminussL

minussR

R

c = c1c = clowast

c = c2



119880(119909 120593)

119888 (119909 120593)

120573 (119909 120593)

=

1198801(120593

119909) 1198881 120573 = 0 120593 gt Γ (0) 119909

1198803 119888

lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198804 119888

2 120573 = 1 minus119904

119877119909 lt 120593 lt 0

infin 1198882 120573 = 1 120593 gt minus119904

119877119909

(43)


119904 (119909 120593)

119888 (119909 120593)

120573 (119909 120593)

=

1199041(120593

119909) 1198881 120573 = 0 120593 gt Γ (0) 119909

1199043 119888

lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1199044 119888

2 120573 = 1 minus119904

119877119909 lt 120593 lt 0

119904119877 119888

2 120573 = 1 120593 lt minus119904

119877119909

(44)


I

II

III

IV

V120593

(f = 1 120573 = 0 c2)

(f = 0 120573 = 1 c2)

(f = 1 120573 = 0 c1)

1

0

U5(120593 x)

120573 = 0

120573 = 0

120573 = 0

c2

c2

U1(120593x)

c1

120593 = Γ(0)x + 1

120593 = Γ(0)x

clowastU3

U4 120573 = 1x

U = infin

120593 = minussRx

67

(x998400 120593998400)



119905 =1

1198914

int

120593

0

119889120593 +1199044

1198914

int

119909

0

119889119909 = (minus119904119877+ 1199044

1198914

)119909 (45)


119905 =1

1198913

int

120593

0

119889120593 +1199043

1198913

int

119909

0

119889119909 =1199043

1198913

119909 (46)


I

V

II

III

IV

t

(f = 1 120573 = 0 c2)

(f = 1 120573 = 0 c1)

1

0

s5(t x)

120573 = 0

120573 = 0 120573 = 0

c2

c2

c2

c1 clowasts3

s4120573 = 1

120573 = 1

x

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s4)xf4

s1(tx)

(x998400 t998400)

sR

(f = 0 120573 = 1 c2)



00(119889120593119891 +


119905 =120593

119891 (1199041 (120593119909) 1198881)+

1199041(120593119909)

119891 (1199041 (120593119909) 1198881)119909 (47)


119904 (119909 119905)

119888 (119909 119905)

120573 (119909 119905)

=

1199041(119905

119909) 1198881 120573 = 0 119905 gt

Γ (0) + 1199042

1198912

119909

1199043 119888

lowast 120573 = 0

1199043

1198913

119909 lt 119905 ltΓ (0) + 119904

2

1198912

119909

1199044 119888

2 120573 = 1

minus119904119877+ 1199044

1198914

119909 lt 119905 lt1199043

1198913

119909

119904119877 119888

2 120573 = 1 119905 lt

minus119904119877+ 1199044

1198914

119909

(48)



2


119888 (119909 120593)

120573 (119909 120593)=

1198882 120573 = 0 120593 gt Γ (0) 119909 + 1

1198881 120573 = 0 Γ (0) 119909 lt 120593 lt Γ (0) 119909 + 1

119888lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198882 120573 = 1 minus119904

119877119909 lt 120593 lt 0

(49)



I

II

III

IV

t

0

120573 = 0

120573 = 0

c2

c2

clowasts3

s3

s2

s4

s4

120573 = 1

120573 = 1

x

x

x

x

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s4)xf4

sR

sR

sL

c1

c2

clowast

c1

s

c

1

0

120573

D1t1 D2t1 D3t1

(a)

(b)

(c)

(d)

t1

s1(tx)



2and 120573 = 0



119865 (119880+ 1198881) minus 119865 (119880

minus 1198882)

119880+ minus 119880minus= Γ (0) (50)



1198805(119909 120593) = 119880

minus(1199091015840 1205931015840) (51)


120593 minus 1205931015840

119909 minus 1199091015840=120597119865 (119880

5 1198881)

120597119880 (52)


119880 (119909 120593)

119888 (119909 120593)

120573 (119909 120593)

=

1198805(119909 120593) 119888

2 120573 = 0 120593 gt Γ (0) 119909 + 1

1198801(120593

119909) 1198881 120573 = 0 Γ(0) 119909lt120593ltΓ(0) 119909+1

1198803 119888

lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198804 119888

2 120573 = 1 minus119904

119877119909 lt 120593 lt 0

infin 1198882 120573 = 1 120593 gt minus119904

119877119909

(53)


120593 = Γ (0) 119909 + 1 (54)


120593

119909=119891 (119904+ 1198881) minus 119904+1198911015840(119904+ 1198881)

1198911015840 (119904+ 1198881)

(55)


parametric form

119909119863(119904+) =

1198911015840(119904+ 1198881)

119891 (119904+ 1198881) minus 1198911015840 (119904+ 119888

1) (119904+ + Γ)

=1198911015840(119904+ 1198881)

Δ(56)

120593119863(119904+) =

119891 (119904+ 1198881) minus 1199041+

1198911015840(119904+ 1198881)

119891 (119904+ 1198881) minus 1198911015840 (119904+ 119888

1) (119904+ + Γ)

=119891 (119904+ 1198881) minus 1199041+

1198911015840(119904+ 1198881)

Δ

(57)


119905119863=

120593

119891 (119904+ (120593 119909) 1198882)+

119904+(120593 119909)

119891 (119904+ (120593 119909) 1198882)119909

119905119863=

1

119891 (119904+ 1198881) minus 1198911015840 (119904+ 119888

1) (1199041++ Γ)

=1

Δ

(58)


119904 (119909 119905)

119888 (119909 119905)

120573 (119909 119905)

=

1199045 (119905 119909) 1198882 120573 = 0 119905 gt

Γ (0) + 1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909 + 1

1199041(119905

119909) 1198881 120573 = 0

Γ (0) + 1199042

1198912

119909 lt 119905 ltΓ (0) + 1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909 + 1

1199043 119888

lowast 120573 = 0

1199043

1198913

119909 lt 119905 ltΓ (0) + 1199042

1198912

119909

1199044 119888

2 120573 = 1

minus119904119877+ 1199044

1198914

119909 lt 119905 lt1199043

1198913

119909

119904119877 119888

2 120573 = 1 119905 lt

minus119904119877+ 1199044

1198914

119909




119909119863

= 119904119861+ Γ (0)

1

119905119863

= 119891119860

(60)





119863




119888 (119909 119905)

120573 (119909 119905)

=

1199045 (119905 119909) 119888 = 0 120573 = 0 119905 gt

Γ (0) + 1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909 + 1

1199041(119905

119909) 119888 = 119888

1 120573 = 0

Γ (0) + 1199042

1198912

119909lt119905ltΓ (0)+1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909+1

1199043 119888 = 0 120573 = 0

1199043

1198913

119909 lt 119905 ltΓ (0) + 1199042

1198912

119909

1199043 119888 = 0 120573 = 1

minus119904119877+ 1199043

1198913

119909 lt 119905 lt1199043

1198913

119909

119904119877 119888 = 0 120573 = 1 119905 lt

minus119904119877+ 1199043

1198913

119909

(61)


I

II

III

IV

t

0

120573 = 0

120573 = 0

120573 = 0

c2

c2

c2

clowasts3

s3

s2

s4

s4

120573 = 1

120573 = 1

x

x

x

x

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s4)xf4

sR

sR

sL

c1

c2

clowast

c1

s

c

1

1

0

120573


(a)

(b)

(c)

(d)

V

s5(t x) (x998400 t998400)

t1

t1

t2

t2

s1(tx)





2and of salt 120573 = 1


1and


01

0 10

1L

f

R

2

3

4

c = c1

c = clowast

c = c2

s

s+sminus

c

ab

Δ

Δf998400(s+)


01

0 10

1L

s

f

R

2

3

c = c1

minusΓ(0)

c = c2 = 0








4


2


1for profiles is





3-bank and initial


I

II

III

IVV

(f = 1 120573 = 0 c1)

1

0

120573 = 0

120573 = 1

120573 = 1

120573 = 0

120573 = 0

x

s5(t x)

s3

s3

(x998400 t998400)

sR

c1

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s3)xf3

c = 0

c = 0

c = 0

c = 0

(f = 0 120573 = 1 c = 0)

(f = 1 120573 = 0 c = 0)

t

s1(tx)


2= 0




1lt 1 during







3 The





8 Conclusions











Acknowledgments


References























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










t

D

x

1 1

120593

V

x1120582i

+1V 1120582i

minus0

s

D

f

+

minus



1198892119905 = 0 = [

120597

120597119909(1

119891) minus

120597

120597120593(119904

119891)] 119889119909119889120593 (21)


120597 (119904119891)

120597120593minus120597 (1119891)

120597119909= 0 (22)



∮

120597Ω

(119888119891) 119889119905 minus (119888119904 + 119886) 119889119909 = 0 (23)




∮

120597Ω

119888 (119891119889119905 minus 119904119889119909) minus 119886119889119909 = ∮

120597Ω

119888119889120593 minus 119886119889119909 = 0 (24)


120597119886 (119888 120573)

120597120593+120597119888

120597119909= 0 (25)


∬(120597 (120573119904)

120597119905+120597 (120573119891 (119904 119888))

120597119909)119889119909119889119905 = ∮

120597Ω

120573119904119889119909 minus 120573119891 (119904 119888) 119889119905

= ∮120573119889120593 = ∬120597120573

120597119909= 0

(26)






119865 = minus119904

119891 119880 =

1

119891(27)

120597 (119865 (119880 119888))

120597120593+120597 (119880)

120597119909= 0 (28)

120597119886 (119888 120573)

120597120593+120597119888

120597119909= 0 (29)

120597120573

120597119909= 0 (30)



+

+

+

minus

minus

minus

s

cn

c1



119909 = 0120573 = 0 119888 = 119888

1 119880 = 1 120593 lt 1

120573 = 0 119888 = 1198882 119880 = 1 120593 gt 1

(31)



120593 = minus 119904119877119909 120573 = 1 119888 = 119888

2 119880 = +infin (32)


119909 = 0 120573 = 0 119888 = 1198881 119880 = 1

120593 = minus119904119877119909 120573 = 1 119888 = 119888

2 119880 = +infin

(33)




120590 [119880] = [119865] (34)

120590 [119888] = [119886] (35)

120590 [120573] = 0 (36)




tions [1ndash4 9]


119888 (119909 120593)

120573 (119909 120593)=

1198881 120573 = 0 120593 gt Γ (0) 119909

119888lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198882 120573 = 1 minus119904

119877119909 lt 120593 lt 0

(37)


119888lowast=Γ (1)

Γ (0)1198882 (38)


1 119888 =

1198882 and 119888 = 119888



120593

119909=

120597119865 (1198801 1198881)

120597119880 (39)


120597119865 (1198802 1198881)

120597119880=1198652(1198802 1198881) minus 1198653(1198803 119888lowast)

1198802minus 1198803

= Γ (0) (40)


1198653(1198803 119888lowast) = 1198654(1198804 1198882) = 0 (41)


1198654(1198804 1198882) minus 119865119894(119880119877 1198882)

1198804minus 119880119894

=minus1199044119891119877+ 1199041198771198914

119891119877 minus 1198914

= minus119904119877 (42)


c1c2 clowast c

R

a

a1

a2

L

120573 = 1

120573 = 0

(a)

c1

c2

clowast

c

R

L

1205731

(b)


F = minussf

U = 1f01

2

34

67

LminussL

minussR

R

c = c1c = clowast

c = c2



119880(119909 120593)

119888 (119909 120593)

120573 (119909 120593)

=

1198801(120593

119909) 1198881 120573 = 0 120593 gt Γ (0) 119909

1198803 119888

lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198804 119888

2 120573 = 1 minus119904

119877119909 lt 120593 lt 0

infin 1198882 120573 = 1 120593 gt minus119904

119877119909

(43)


119904 (119909 120593)

119888 (119909 120593)

120573 (119909 120593)

=

1199041(120593

119909) 1198881 120573 = 0 120593 gt Γ (0) 119909

1199043 119888

lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1199044 119888

2 120573 = 1 minus119904

119877119909 lt 120593 lt 0

119904119877 119888

2 120573 = 1 120593 lt minus119904

119877119909

(44)


I

II

III

IV

V120593

(f = 1 120573 = 0 c2)

(f = 0 120573 = 1 c2)

(f = 1 120573 = 0 c1)

1

0

U5(120593 x)

120573 = 0

120573 = 0

120573 = 0

c2

c2

U1(120593x)

c1

120593 = Γ(0)x + 1

120593 = Γ(0)x

clowastU3

U4 120573 = 1x

U = infin

120593 = minussRx

67

(x998400 120593998400)



119905 =1

1198914

int

120593

0

119889120593 +1199044

1198914

int

119909

0

119889119909 = (minus119904119877+ 1199044

1198914

)119909 (45)


119905 =1

1198913

int

120593

0

119889120593 +1199043

1198913

int

119909

0

119889119909 =1199043

1198913

119909 (46)


I

V

II

III

IV

t

(f = 1 120573 = 0 c2)

(f = 1 120573 = 0 c1)

1

0

s5(t x)

120573 = 0

120573 = 0 120573 = 0

c2

c2

c2

c1 clowasts3

s4120573 = 1

120573 = 1

x

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s4)xf4

s1(tx)

(x998400 t998400)

sR

(f = 0 120573 = 1 c2)



00(119889120593119891 +


119905 =120593

119891 (1199041 (120593119909) 1198881)+

1199041(120593119909)

119891 (1199041 (120593119909) 1198881)119909 (47)


119904 (119909 119905)

119888 (119909 119905)

120573 (119909 119905)

=

1199041(119905

119909) 1198881 120573 = 0 119905 gt

Γ (0) + 1199042

1198912

119909

1199043 119888

lowast 120573 = 0

1199043

1198913

119909 lt 119905 ltΓ (0) + 119904

2

1198912

119909

1199044 119888

2 120573 = 1

minus119904119877+ 1199044

1198914

119909 lt 119905 lt1199043

1198913

119909

119904119877 119888

2 120573 = 1 119905 lt

minus119904119877+ 1199044

1198914

119909

(48)



2


119888 (119909 120593)

120573 (119909 120593)=

1198882 120573 = 0 120593 gt Γ (0) 119909 + 1

1198881 120573 = 0 Γ (0) 119909 lt 120593 lt Γ (0) 119909 + 1

119888lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198882 120573 = 1 minus119904

119877119909 lt 120593 lt 0

(49)



I

II

III

IV

t

0

120573 = 0

120573 = 0

c2

c2

clowasts3

s3

s2

s4

s4

120573 = 1

120573 = 1

x

x

x

x

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s4)xf4

sR

sR

sL

c1

c2

clowast

c1

s

c

1

0

120573

D1t1 D2t1 D3t1

(a)

(b)

(c)

(d)

t1

s1(tx)



2and 120573 = 0



119865 (119880+ 1198881) minus 119865 (119880

minus 1198882)

119880+ minus 119880minus= Γ (0) (50)



1198805(119909 120593) = 119880

minus(1199091015840 1205931015840) (51)


120593 minus 1205931015840

119909 minus 1199091015840=120597119865 (119880

5 1198881)

120597119880 (52)


119880 (119909 120593)

119888 (119909 120593)

120573 (119909 120593)

=

1198805(119909 120593) 119888

2 120573 = 0 120593 gt Γ (0) 119909 + 1

1198801(120593

119909) 1198881 120573 = 0 Γ(0) 119909lt120593ltΓ(0) 119909+1

1198803 119888

lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198804 119888

2 120573 = 1 minus119904

119877119909 lt 120593 lt 0

infin 1198882 120573 = 1 120593 gt minus119904

119877119909

(53)


120593 = Γ (0) 119909 + 1 (54)


120593

119909=119891 (119904+ 1198881) minus 119904+1198911015840(119904+ 1198881)

1198911015840 (119904+ 1198881)

(55)


parametric form

119909119863(119904+) =

1198911015840(119904+ 1198881)

119891 (119904+ 1198881) minus 1198911015840 (119904+ 119888

1) (119904+ + Γ)

=1198911015840(119904+ 1198881)

Δ(56)

120593119863(119904+) =

119891 (119904+ 1198881) minus 1199041+

1198911015840(119904+ 1198881)

119891 (119904+ 1198881) minus 1198911015840 (119904+ 119888

1) (119904+ + Γ)

=119891 (119904+ 1198881) minus 1199041+

1198911015840(119904+ 1198881)

Δ

(57)


119905119863=

120593

119891 (119904+ (120593 119909) 1198882)+

119904+(120593 119909)

119891 (119904+ (120593 119909) 1198882)119909

119905119863=

1

119891 (119904+ 1198881) minus 1198911015840 (119904+ 119888

1) (1199041++ Γ)

=1

Δ

(58)


119904 (119909 119905)

119888 (119909 119905)

120573 (119909 119905)

=

1199045 (119905 119909) 1198882 120573 = 0 119905 gt

Γ (0) + 1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909 + 1

1199041(119905

119909) 1198881 120573 = 0

Γ (0) + 1199042

1198912

119909 lt 119905 ltΓ (0) + 1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909 + 1

1199043 119888

lowast 120573 = 0

1199043

1198913

119909 lt 119905 ltΓ (0) + 1199042

1198912

119909

1199044 119888

2 120573 = 1

minus119904119877+ 1199044

1198914

119909 lt 119905 lt1199043

1198913

119909

119904119877 119888

2 120573 = 1 119905 lt

minus119904119877+ 1199044

1198914

119909




119909119863

= 119904119861+ Γ (0)

1

119905119863

= 119891119860

(60)





119863




119888 (119909 119905)

120573 (119909 119905)

=

1199045 (119905 119909) 119888 = 0 120573 = 0 119905 gt

Γ (0) + 1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909 + 1

1199041(119905

119909) 119888 = 119888

1 120573 = 0

Γ (0) + 1199042

1198912

119909lt119905ltΓ (0)+1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909+1

1199043 119888 = 0 120573 = 0

1199043

1198913

119909 lt 119905 ltΓ (0) + 1199042

1198912

119909

1199043 119888 = 0 120573 = 1

minus119904119877+ 1199043

1198913

119909 lt 119905 lt1199043

1198913

119909

119904119877 119888 = 0 120573 = 1 119905 lt

minus119904119877+ 1199043

1198913

119909

(61)


I

II

III

IV

t

0

120573 = 0

120573 = 0

120573 = 0

c2

c2

c2

clowasts3

s3

s2

s4

s4

120573 = 1

120573 = 1

x

x

x

x

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s4)xf4

sR

sR

sL

c1

c2

clowast

c1

s

c

1

1

0

120573


(a)

(b)

(c)

(d)

V

s5(t x) (x998400 t998400)

t1

t1

t2

t2

s1(tx)





2and of salt 120573 = 1


1and


01

0 10

1L

f

R

2

3

4

c = c1

c = clowast

c = c2

s

s+sminus

c

ab

Δ

Δf998400(s+)


01

0 10

1L

s

f

R

2

3

c = c1

minusΓ(0)

c = c2 = 0








4


2


1for profiles is





3-bank and initial


I

II

III

IVV

(f = 1 120573 = 0 c1)

1

0

120573 = 0

120573 = 1

120573 = 1

120573 = 0

120573 = 0

x

s5(t x)

s3

s3

(x998400 t998400)

sR

c1

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s3)xf3

c = 0

c = 0

c = 0

c = 0

(f = 0 120573 = 1 c = 0)

(f = 1 120573 = 0 c = 0)

t

s1(tx)


2= 0




1lt 1 during







3 The





8 Conclusions











Acknowledgments


References























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










+

+

+

minus

minus

minus

s

cn

c1



119909 = 0120573 = 0 119888 = 119888

1 119880 = 1 120593 lt 1

120573 = 0 119888 = 1198882 119880 = 1 120593 gt 1

(31)



120593 = minus 119904119877119909 120573 = 1 119888 = 119888

2 119880 = +infin (32)


119909 = 0 120573 = 0 119888 = 1198881 119880 = 1

120593 = minus119904119877119909 120573 = 1 119888 = 119888

2 119880 = +infin

(33)




120590 [119880] = [119865] (34)

120590 [119888] = [119886] (35)

120590 [120573] = 0 (36)




tions [1ndash4 9]


119888 (119909 120593)

120573 (119909 120593)=

1198881 120573 = 0 120593 gt Γ (0) 119909

119888lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198882 120573 = 1 minus119904

119877119909 lt 120593 lt 0

(37)


119888lowast=Γ (1)

Γ (0)1198882 (38)


1 119888 =

1198882 and 119888 = 119888



120593

119909=

120597119865 (1198801 1198881)

120597119880 (39)


120597119865 (1198802 1198881)

120597119880=1198652(1198802 1198881) minus 1198653(1198803 119888lowast)

1198802minus 1198803

= Γ (0) (40)


1198653(1198803 119888lowast) = 1198654(1198804 1198882) = 0 (41)


1198654(1198804 1198882) minus 119865119894(119880119877 1198882)

1198804minus 119880119894

=minus1199044119891119877+ 1199041198771198914

119891119877 minus 1198914

= minus119904119877 (42)


c1c2 clowast c

R

a

a1

a2

L

120573 = 1

120573 = 0

(a)

c1

c2

clowast

c

R

L

1205731

(b)


F = minussf

U = 1f01

2

34

67

LminussL

minussR

R

c = c1c = clowast

c = c2



119880(119909 120593)

119888 (119909 120593)

120573 (119909 120593)

=

1198801(120593

119909) 1198881 120573 = 0 120593 gt Γ (0) 119909

1198803 119888

lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198804 119888

2 120573 = 1 minus119904

119877119909 lt 120593 lt 0

infin 1198882 120573 = 1 120593 gt minus119904

119877119909

(43)


119904 (119909 120593)

119888 (119909 120593)

120573 (119909 120593)

=

1199041(120593

119909) 1198881 120573 = 0 120593 gt Γ (0) 119909

1199043 119888

lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1199044 119888

2 120573 = 1 minus119904

119877119909 lt 120593 lt 0

119904119877 119888

2 120573 = 1 120593 lt minus119904

119877119909

(44)


I

II

III

IV

V120593

(f = 1 120573 = 0 c2)

(f = 0 120573 = 1 c2)

(f = 1 120573 = 0 c1)

1

0

U5(120593 x)

120573 = 0

120573 = 0

120573 = 0

c2

c2

U1(120593x)

c1

120593 = Γ(0)x + 1

120593 = Γ(0)x

clowastU3

U4 120573 = 1x

U = infin

120593 = minussRx

67

(x998400 120593998400)



119905 =1

1198914

int

120593

0

119889120593 +1199044

1198914

int

119909

0

119889119909 = (minus119904119877+ 1199044

1198914

)119909 (45)


119905 =1

1198913

int

120593

0

119889120593 +1199043

1198913

int

119909

0

119889119909 =1199043

1198913

119909 (46)


I

V

II

III

IV

t

(f = 1 120573 = 0 c2)

(f = 1 120573 = 0 c1)

1

0

s5(t x)

120573 = 0

120573 = 0 120573 = 0

c2

c2

c2

c1 clowasts3

s4120573 = 1

120573 = 1

x

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s4)xf4

s1(tx)

(x998400 t998400)

sR

(f = 0 120573 = 1 c2)



00(119889120593119891 +


119905 =120593

119891 (1199041 (120593119909) 1198881)+

1199041(120593119909)

119891 (1199041 (120593119909) 1198881)119909 (47)


119904 (119909 119905)

119888 (119909 119905)

120573 (119909 119905)

=

1199041(119905

119909) 1198881 120573 = 0 119905 gt

Γ (0) + 1199042

1198912

119909

1199043 119888

lowast 120573 = 0

1199043

1198913

119909 lt 119905 ltΓ (0) + 119904

2

1198912

119909

1199044 119888

2 120573 = 1

minus119904119877+ 1199044

1198914

119909 lt 119905 lt1199043

1198913

119909

119904119877 119888

2 120573 = 1 119905 lt

minus119904119877+ 1199044

1198914

119909

(48)



2


119888 (119909 120593)

120573 (119909 120593)=

1198882 120573 = 0 120593 gt Γ (0) 119909 + 1

1198881 120573 = 0 Γ (0) 119909 lt 120593 lt Γ (0) 119909 + 1

119888lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198882 120573 = 1 minus119904

119877119909 lt 120593 lt 0

(49)



I

II

III

IV

t

0

120573 = 0

120573 = 0

c2

c2

clowasts3

s3

s2

s4

s4

120573 = 1

120573 = 1

x

x

x

x

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s4)xf4

sR

sR

sL

c1

c2

clowast

c1

s

c

1

0

120573

D1t1 D2t1 D3t1

(a)

(b)

(c)

(d)

t1

s1(tx)



2and 120573 = 0



119865 (119880+ 1198881) minus 119865 (119880

minus 1198882)

119880+ minus 119880minus= Γ (0) (50)



1198805(119909 120593) = 119880

minus(1199091015840 1205931015840) (51)


120593 minus 1205931015840

119909 minus 1199091015840=120597119865 (119880

5 1198881)

120597119880 (52)


119880 (119909 120593)

119888 (119909 120593)

120573 (119909 120593)

=

1198805(119909 120593) 119888

2 120573 = 0 120593 gt Γ (0) 119909 + 1

1198801(120593

119909) 1198881 120573 = 0 Γ(0) 119909lt120593ltΓ(0) 119909+1

1198803 119888

lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198804 119888

2 120573 = 1 minus119904

119877119909 lt 120593 lt 0

infin 1198882 120573 = 1 120593 gt minus119904

119877119909

(53)


120593 = Γ (0) 119909 + 1 (54)


120593

119909=119891 (119904+ 1198881) minus 119904+1198911015840(119904+ 1198881)

1198911015840 (119904+ 1198881)

(55)


parametric form

119909119863(119904+) =

1198911015840(119904+ 1198881)

119891 (119904+ 1198881) minus 1198911015840 (119904+ 119888

1) (119904+ + Γ)

=1198911015840(119904+ 1198881)

Δ(56)

120593119863(119904+) =

119891 (119904+ 1198881) minus 1199041+

1198911015840(119904+ 1198881)

119891 (119904+ 1198881) minus 1198911015840 (119904+ 119888

1) (119904+ + Γ)

=119891 (119904+ 1198881) minus 1199041+

1198911015840(119904+ 1198881)

Δ

(57)


119905119863=

120593

119891 (119904+ (120593 119909) 1198882)+

119904+(120593 119909)

119891 (119904+ (120593 119909) 1198882)119909

119905119863=

1

119891 (119904+ 1198881) minus 1198911015840 (119904+ 119888

1) (1199041++ Γ)

=1

Δ

(58)


119904 (119909 119905)

119888 (119909 119905)

120573 (119909 119905)

=

1199045 (119905 119909) 1198882 120573 = 0 119905 gt

Γ (0) + 1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909 + 1

1199041(119905

119909) 1198881 120573 = 0

Γ (0) + 1199042

1198912

119909 lt 119905 ltΓ (0) + 1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909 + 1

1199043 119888

lowast 120573 = 0

1199043

1198913

119909 lt 119905 ltΓ (0) + 1199042

1198912

119909

1199044 119888

2 120573 = 1

minus119904119877+ 1199044

1198914

119909 lt 119905 lt1199043

1198913

119909

119904119877 119888

2 120573 = 1 119905 lt

minus119904119877+ 1199044

1198914

119909




119909119863

= 119904119861+ Γ (0)

1

119905119863

= 119891119860

(60)





119863




119888 (119909 119905)

120573 (119909 119905)

=

1199045 (119905 119909) 119888 = 0 120573 = 0 119905 gt

Γ (0) + 1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909 + 1

1199041(119905

119909) 119888 = 119888

1 120573 = 0

Γ (0) + 1199042

1198912

119909lt119905ltΓ (0)+1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909+1

1199043 119888 = 0 120573 = 0

1199043

1198913

119909 lt 119905 ltΓ (0) + 1199042

1198912

119909

1199043 119888 = 0 120573 = 1

minus119904119877+ 1199043

1198913

119909 lt 119905 lt1199043

1198913

119909

119904119877 119888 = 0 120573 = 1 119905 lt

minus119904119877+ 1199043

1198913

119909

(61)


I

II

III

IV

t

0

120573 = 0

120573 = 0

120573 = 0

c2

c2

c2

clowasts3

s3

s2

s4

s4

120573 = 1

120573 = 1

x

x

x

x

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s4)xf4

sR

sR

sL

c1

c2

clowast

c1

s

c

1

1

0

120573


(a)

(b)

(c)

(d)

V

s5(t x) (x998400 t998400)

t1

t1

t2

t2

s1(tx)





2and of salt 120573 = 1


1and


01

0 10

1L

f

R

2

3

4

c = c1

c = clowast

c = c2

s

s+sminus

c

ab

Δ

Δf998400(s+)


01

0 10

1L

s

f

R

2

3

c = c1

minusΓ(0)

c = c2 = 0








4


2


1for profiles is





3-bank and initial


I

II

III

IVV

(f = 1 120573 = 0 c1)

1

0

120573 = 0

120573 = 1

120573 = 1

120573 = 0

120573 = 0

x

s5(t x)

s3

s3

(x998400 t998400)

sR

c1

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s3)xf3

c = 0

c = 0

c = 0

c = 0

(f = 0 120573 = 1 c = 0)

(f = 1 120573 = 0 c = 0)

t

s1(tx)


2= 0




1lt 1 during







3 The





8 Conclusions











Acknowledgments


References























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










c1c2 clowast c

R

a

a1

a2

L

120573 = 1

120573 = 0

(a)

c1

c2

clowast

c

R

L

1205731

(b)


F = minussf

U = 1f01

2

34

67

LminussL

minussR

R

c = c1c = clowast

c = c2



119880(119909 120593)

119888 (119909 120593)

120573 (119909 120593)

=

1198801(120593

119909) 1198881 120573 = 0 120593 gt Γ (0) 119909

1198803 119888

lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198804 119888

2 120573 = 1 minus119904

119877119909 lt 120593 lt 0

infin 1198882 120573 = 1 120593 gt minus119904

119877119909

(43)


119904 (119909 120593)

119888 (119909 120593)

120573 (119909 120593)

=

1199041(120593

119909) 1198881 120573 = 0 120593 gt Γ (0) 119909

1199043 119888

lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1199044 119888

2 120573 = 1 minus119904

119877119909 lt 120593 lt 0

119904119877 119888

2 120573 = 1 120593 lt minus119904

119877119909

(44)


I

II

III

IV

V120593

(f = 1 120573 = 0 c2)

(f = 0 120573 = 1 c2)

(f = 1 120573 = 0 c1)

1

0

U5(120593 x)

120573 = 0

120573 = 0

120573 = 0

c2

c2

U1(120593x)

c1

120593 = Γ(0)x + 1

120593 = Γ(0)x

clowastU3

U4 120573 = 1x

U = infin

120593 = minussRx

67

(x998400 120593998400)



119905 =1

1198914

int

120593

0

119889120593 +1199044

1198914

int

119909

0

119889119909 = (minus119904119877+ 1199044

1198914

)119909 (45)


119905 =1

1198913

int

120593

0

119889120593 +1199043

1198913

int

119909

0

119889119909 =1199043

1198913

119909 (46)


I

V

II

III

IV

t

(f = 1 120573 = 0 c2)

(f = 1 120573 = 0 c1)

1

0

s5(t x)

120573 = 0

120573 = 0 120573 = 0

c2

c2

c2

c1 clowasts3

s4120573 = 1

120573 = 1

x

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s4)xf4

s1(tx)

(x998400 t998400)

sR

(f = 0 120573 = 1 c2)



00(119889120593119891 +


119905 =120593

119891 (1199041 (120593119909) 1198881)+

1199041(120593119909)

119891 (1199041 (120593119909) 1198881)119909 (47)


119904 (119909 119905)

119888 (119909 119905)

120573 (119909 119905)

=

1199041(119905

119909) 1198881 120573 = 0 119905 gt

Γ (0) + 1199042

1198912

119909

1199043 119888

lowast 120573 = 0

1199043

1198913

119909 lt 119905 ltΓ (0) + 119904

2

1198912

119909

1199044 119888

2 120573 = 1

minus119904119877+ 1199044

1198914

119909 lt 119905 lt1199043

1198913

119909

119904119877 119888

2 120573 = 1 119905 lt

minus119904119877+ 1199044

1198914

119909

(48)



2


119888 (119909 120593)

120573 (119909 120593)=

1198882 120573 = 0 120593 gt Γ (0) 119909 + 1

1198881 120573 = 0 Γ (0) 119909 lt 120593 lt Γ (0) 119909 + 1

119888lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198882 120573 = 1 minus119904

119877119909 lt 120593 lt 0

(49)



I

II

III

IV

t

0

120573 = 0

120573 = 0

c2

c2

clowasts3

s3

s2

s4

s4

120573 = 1

120573 = 1

x

x

x

x

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s4)xf4

sR

sR

sL

c1

c2

clowast

c1

s

c

1

0

120573

D1t1 D2t1 D3t1

(a)

(b)

(c)

(d)

t1

s1(tx)



2and 120573 = 0



119865 (119880+ 1198881) minus 119865 (119880

minus 1198882)

119880+ minus 119880minus= Γ (0) (50)



1198805(119909 120593) = 119880

minus(1199091015840 1205931015840) (51)


120593 minus 1205931015840

119909 minus 1199091015840=120597119865 (119880

5 1198881)

120597119880 (52)


119880 (119909 120593)

119888 (119909 120593)

120573 (119909 120593)

=

1198805(119909 120593) 119888

2 120573 = 0 120593 gt Γ (0) 119909 + 1

1198801(120593

119909) 1198881 120573 = 0 Γ(0) 119909lt120593ltΓ(0) 119909+1

1198803 119888

lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198804 119888

2 120573 = 1 minus119904

119877119909 lt 120593 lt 0

infin 1198882 120573 = 1 120593 gt minus119904

119877119909

(53)


120593 = Γ (0) 119909 + 1 (54)


120593

119909=119891 (119904+ 1198881) minus 119904+1198911015840(119904+ 1198881)

1198911015840 (119904+ 1198881)

(55)


parametric form

119909119863(119904+) =

1198911015840(119904+ 1198881)

119891 (119904+ 1198881) minus 1198911015840 (119904+ 119888

1) (119904+ + Γ)

=1198911015840(119904+ 1198881)

Δ(56)

120593119863(119904+) =

119891 (119904+ 1198881) minus 1199041+

1198911015840(119904+ 1198881)

119891 (119904+ 1198881) minus 1198911015840 (119904+ 119888

1) (119904+ + Γ)

=119891 (119904+ 1198881) minus 1199041+

1198911015840(119904+ 1198881)

Δ

(57)


119905119863=

120593

119891 (119904+ (120593 119909) 1198882)+

119904+(120593 119909)

119891 (119904+ (120593 119909) 1198882)119909

119905119863=

1

119891 (119904+ 1198881) minus 1198911015840 (119904+ 119888

1) (1199041++ Γ)

=1

Δ

(58)


119904 (119909 119905)

119888 (119909 119905)

120573 (119909 119905)

=

1199045 (119905 119909) 1198882 120573 = 0 119905 gt

Γ (0) + 1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909 + 1

1199041(119905

119909) 1198881 120573 = 0

Γ (0) + 1199042

1198912

119909 lt 119905 ltΓ (0) + 1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909 + 1

1199043 119888

lowast 120573 = 0

1199043

1198913

119909 lt 119905 ltΓ (0) + 1199042

1198912

119909

1199044 119888

2 120573 = 1

minus119904119877+ 1199044

1198914

119909 lt 119905 lt1199043

1198913

119909

119904119877 119888

2 120573 = 1 119905 lt

minus119904119877+ 1199044

1198914

119909




119909119863

= 119904119861+ Γ (0)

1

119905119863

= 119891119860

(60)





119863




119888 (119909 119905)

120573 (119909 119905)

=

1199045 (119905 119909) 119888 = 0 120573 = 0 119905 gt

Γ (0) + 1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909 + 1

1199041(119905

119909) 119888 = 119888

1 120573 = 0

Γ (0) + 1199042

1198912

119909lt119905ltΓ (0)+1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909+1

1199043 119888 = 0 120573 = 0

1199043

1198913

119909 lt 119905 ltΓ (0) + 1199042

1198912

119909

1199043 119888 = 0 120573 = 1

minus119904119877+ 1199043

1198913

119909 lt 119905 lt1199043

1198913

119909

119904119877 119888 = 0 120573 = 1 119905 lt

minus119904119877+ 1199043

1198913

119909

(61)


I

II

III

IV

t

0

120573 = 0

120573 = 0

120573 = 0

c2

c2

c2

clowasts3

s3

s2

s4

s4

120573 = 1

120573 = 1

x

x

x

x

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s4)xf4

sR

sR

sL

c1

c2

clowast

c1

s

c

1

1

0

120573


(a)

(b)

(c)

(d)

V

s5(t x) (x998400 t998400)

t1

t1

t2

t2

s1(tx)





2and of salt 120573 = 1


1and


01

0 10

1L

f

R

2

3

4

c = c1

c = clowast

c = c2

s

s+sminus

c

ab

Δ

Δf998400(s+)


01

0 10

1L

s

f

R

2

3

c = c1

minusΓ(0)

c = c2 = 0








4


2


1for profiles is





3-bank and initial


I

II

III

IVV

(f = 1 120573 = 0 c1)

1

0

120573 = 0

120573 = 1

120573 = 1

120573 = 0

120573 = 0

x

s5(t x)

s3

s3

(x998400 t998400)

sR

c1

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s3)xf3

c = 0

c = 0

c = 0

c = 0

(f = 0 120573 = 1 c = 0)

(f = 1 120573 = 0 c = 0)

t

s1(tx)


2= 0




1lt 1 during







3 The





8 Conclusions











Acknowledgments


References























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










I

V

II

III

IV

t

(f = 1 120573 = 0 c2)

(f = 1 120573 = 0 c1)

1

0

s5(t x)

120573 = 0

120573 = 0 120573 = 0

c2

c2

c2

c1 clowasts3

s4120573 = 1

120573 = 1

x

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s4)xf4

s1(tx)

(x998400 t998400)

sR

(f = 0 120573 = 1 c2)



00(119889120593119891 +


119905 =120593

119891 (1199041 (120593119909) 1198881)+

1199041(120593119909)

119891 (1199041 (120593119909) 1198881)119909 (47)


119904 (119909 119905)

119888 (119909 119905)

120573 (119909 119905)

=

1199041(119905

119909) 1198881 120573 = 0 119905 gt

Γ (0) + 1199042

1198912

119909

1199043 119888

lowast 120573 = 0

1199043

1198913

119909 lt 119905 ltΓ (0) + 119904

2

1198912

119909

1199044 119888

2 120573 = 1

minus119904119877+ 1199044

1198914

119909 lt 119905 lt1199043

1198913

119909

119904119877 119888

2 120573 = 1 119905 lt

minus119904119877+ 1199044

1198914

119909

(48)



2


119888 (119909 120593)

120573 (119909 120593)=

1198882 120573 = 0 120593 gt Γ (0) 119909 + 1

1198881 120573 = 0 Γ (0) 119909 lt 120593 lt Γ (0) 119909 + 1

119888lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198882 120573 = 1 minus119904

119877119909 lt 120593 lt 0

(49)



I

II

III

IV

t

0

120573 = 0

120573 = 0

c2

c2

clowasts3

s3

s2

s4

s4

120573 = 1

120573 = 1

x

x

x

x

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s4)xf4

sR

sR

sL

c1

c2

clowast

c1

s

c

1

0

120573

D1t1 D2t1 D3t1

(a)

(b)

(c)

(d)

t1

s1(tx)



2and 120573 = 0



119865 (119880+ 1198881) minus 119865 (119880

minus 1198882)

119880+ minus 119880minus= Γ (0) (50)



1198805(119909 120593) = 119880

minus(1199091015840 1205931015840) (51)


120593 minus 1205931015840

119909 minus 1199091015840=120597119865 (119880

5 1198881)

120597119880 (52)


119880 (119909 120593)

119888 (119909 120593)

120573 (119909 120593)

=

1198805(119909 120593) 119888

2 120573 = 0 120593 gt Γ (0) 119909 + 1

1198801(120593

119909) 1198881 120573 = 0 Γ(0) 119909lt120593ltΓ(0) 119909+1

1198803 119888

lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198804 119888

2 120573 = 1 minus119904

119877119909 lt 120593 lt 0

infin 1198882 120573 = 1 120593 gt minus119904

119877119909

(53)


120593 = Γ (0) 119909 + 1 (54)


120593

119909=119891 (119904+ 1198881) minus 119904+1198911015840(119904+ 1198881)

1198911015840 (119904+ 1198881)

(55)


parametric form

119909119863(119904+) =

1198911015840(119904+ 1198881)

119891 (119904+ 1198881) minus 1198911015840 (119904+ 119888

1) (119904+ + Γ)

=1198911015840(119904+ 1198881)

Δ(56)

120593119863(119904+) =

119891 (119904+ 1198881) minus 1199041+

1198911015840(119904+ 1198881)

119891 (119904+ 1198881) minus 1198911015840 (119904+ 119888

1) (119904+ + Γ)

=119891 (119904+ 1198881) minus 1199041+

1198911015840(119904+ 1198881)

Δ

(57)


119905119863=

120593

119891 (119904+ (120593 119909) 1198882)+

119904+(120593 119909)

119891 (119904+ (120593 119909) 1198882)119909

119905119863=

1

119891 (119904+ 1198881) minus 1198911015840 (119904+ 119888

1) (1199041++ Γ)

=1

Δ

(58)


119904 (119909 119905)

119888 (119909 119905)

120573 (119909 119905)

=

1199045 (119905 119909) 1198882 120573 = 0 119905 gt

Γ (0) + 1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909 + 1

1199041(119905

119909) 1198881 120573 = 0

Γ (0) + 1199042

1198912

119909 lt 119905 ltΓ (0) + 1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909 + 1

1199043 119888

lowast 120573 = 0

1199043

1198913

119909 lt 119905 ltΓ (0) + 1199042

1198912

119909

1199044 119888

2 120573 = 1

minus119904119877+ 1199044

1198914

119909 lt 119905 lt1199043

1198913

119909

119904119877 119888

2 120573 = 1 119905 lt

minus119904119877+ 1199044

1198914

119909




119909119863

= 119904119861+ Γ (0)

1

119905119863

= 119891119860

(60)





119863




119888 (119909 119905)

120573 (119909 119905)

=

1199045 (119905 119909) 119888 = 0 120573 = 0 119905 gt

Γ (0) + 1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909 + 1

1199041(119905

119909) 119888 = 119888

1 120573 = 0

Γ (0) + 1199042

1198912

119909lt119905ltΓ (0)+1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909+1

1199043 119888 = 0 120573 = 0

1199043

1198913

119909 lt 119905 ltΓ (0) + 1199042

1198912

119909

1199043 119888 = 0 120573 = 1

minus119904119877+ 1199043

1198913

119909 lt 119905 lt1199043

1198913

119909

119904119877 119888 = 0 120573 = 1 119905 lt

minus119904119877+ 1199043

1198913

119909

(61)


I

II

III

IV

t

0

120573 = 0

120573 = 0

120573 = 0

c2

c2

c2

clowasts3

s3

s2

s4

s4

120573 = 1

120573 = 1

x

x

x

x

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s4)xf4

sR

sR

sL

c1

c2

clowast

c1

s

c

1

1

0

120573


(a)

(b)

(c)

(d)

V

s5(t x) (x998400 t998400)

t1

t1

t2

t2

s1(tx)





2and of salt 120573 = 1


1and


01

0 10

1L

f

R

2

3

4

c = c1

c = clowast

c = c2

s

s+sminus

c

ab

Δ

Δf998400(s+)


01

0 10

1L

s

f

R

2

3

c = c1

minusΓ(0)

c = c2 = 0








4


2


1for profiles is





3-bank and initial


I

II

III

IVV

(f = 1 120573 = 0 c1)

1

0

120573 = 0

120573 = 1

120573 = 1

120573 = 0

120573 = 0

x

s5(t x)

s3

s3

(x998400 t998400)

sR

c1

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s3)xf3

c = 0

c = 0

c = 0

c = 0

(f = 0 120573 = 1 c = 0)

(f = 1 120573 = 0 c = 0)

t

s1(tx)


2= 0




1lt 1 during







3 The





8 Conclusions











Acknowledgments


References























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











1198805(119909 120593) = 119880

minus(1199091015840 1205931015840) (51)


120593 minus 1205931015840

119909 minus 1199091015840=120597119865 (119880

5 1198881)

120597119880 (52)


119880 (119909 120593)

119888 (119909 120593)

120573 (119909 120593)

=

1198805(119909 120593) 119888

2 120573 = 0 120593 gt Γ (0) 119909 + 1

1198801(120593

119909) 1198881 120573 = 0 Γ(0) 119909lt120593ltΓ(0) 119909+1

1198803 119888

lowast 120573 = 0 0 lt 120593 lt Γ (0) 119909

1198804 119888

2 120573 = 1 minus119904

119877119909 lt 120593 lt 0

infin 1198882 120573 = 1 120593 gt minus119904

119877119909

(53)


120593 = Γ (0) 119909 + 1 (54)


120593

119909=119891 (119904+ 1198881) minus 119904+1198911015840(119904+ 1198881)

1198911015840 (119904+ 1198881)

(55)


parametric form

119909119863(119904+) =

1198911015840(119904+ 1198881)

119891 (119904+ 1198881) minus 1198911015840 (119904+ 119888

1) (119904+ + Γ)

=1198911015840(119904+ 1198881)

Δ(56)

120593119863(119904+) =

119891 (119904+ 1198881) minus 1199041+

1198911015840(119904+ 1198881)

119891 (119904+ 1198881) minus 1198911015840 (119904+ 119888

1) (119904+ + Γ)

=119891 (119904+ 1198881) minus 1199041+

1198911015840(119904+ 1198881)

Δ

(57)


119905119863=

120593

119891 (119904+ (120593 119909) 1198882)+

119904+(120593 119909)

119891 (119904+ (120593 119909) 1198882)119909

119905119863=

1

119891 (119904+ 1198881) minus 1198911015840 (119904+ 119888

1) (1199041++ Γ)

=1

Δ

(58)


119904 (119909 119905)

119888 (119909 119905)

120573 (119909 119905)

=

1199045 (119905 119909) 1198882 120573 = 0 119905 gt

Γ (0) + 1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909 + 1

1199041(119905

119909) 1198881 120573 = 0

Γ (0) + 1199042

1198912

119909 lt 119905 ltΓ (0) + 1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909 + 1

1199043 119888

lowast 120573 = 0

1199043

1198913

119909 lt 119905 ltΓ (0) + 1199042

1198912

119909

1199044 119888

2 120573 = 1

minus119904119877+ 1199044

1198914

119909 lt 119905 lt1199043

1198913

119909

119904119877 119888

2 120573 = 1 119905 lt

minus119904119877+ 1199044

1198914

119909




119909119863

= 119904119861+ Γ (0)

1

119905119863

= 119891119860

(60)





119863




119888 (119909 119905)

120573 (119909 119905)

=

1199045 (119905 119909) 119888 = 0 120573 = 0 119905 gt

Γ (0) + 1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909 + 1

1199041(119905

119909) 119888 = 119888

1 120573 = 0

Γ (0) + 1199042

1198912

119909lt119905ltΓ (0)+1199045 (119905 119909)

1198915(1199045 (119905 119909) 1198882)

119909+1

1199043 119888 = 0 120573 = 0

1199043

1198913

119909 lt 119905 ltΓ (0) + 1199042

1198912

119909

1199043 119888 = 0 120573 = 1

minus119904119877+ 1199043

1198913

119909 lt 119905 lt1199043

1198913

119909

119904119877 119888 = 0 120573 = 1 119905 lt

minus119904119877+ 1199043

1198913

119909

(61)


I

II

III

IV

t

0

120573 = 0

120573 = 0

120573 = 0

c2

c2

c2

clowasts3

s3

s2

s4

s4

120573 = 1

120573 = 1

x

x

x

x

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s4)xf4

sR

sR

sL

c1

c2

clowast

c1

s

c

1

1

0

120573


(a)

(b)

(c)

(d)

V

s5(t x) (x998400 t998400)

t1

t1

t2

t2

s1(tx)





2and of salt 120573 = 1


1and


01

0 10

1L

f

R

2

3

4

c = c1

c = clowast

c = c2

s

s+sminus

c

ab

Δ

Δf998400(s+)


01

0 10

1L

s

f

R

2

3

c = c1

minusΓ(0)

c = c2 = 0








4


2


1for profiles is





3-bank and initial


I

II

III

IVV

(f = 1 120573 = 0 c1)

1

0

120573 = 0

120573 = 1

120573 = 1

120573 = 0

120573 = 0

x

s5(t x)

s3

s3

(x998400 t998400)

sR

c1

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s3)xf3

c = 0

c = 0

c = 0

c = 0

(f = 0 120573 = 1 c = 0)

(f = 1 120573 = 0 c = 0)

t

s1(tx)


2= 0




1lt 1 during







3 The





8 Conclusions











Acknowledgments


References























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










I

II

III

IV

t

0

120573 = 0

120573 = 0

120573 = 0

c2

c2

c2

clowasts3

s3

s2

s4

s4

120573 = 1

120573 = 1

x

x

x

x

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s4)xf4

sR

sR

sL

c1

c2

clowast

c1

s

c

1

1

0

120573


(a)

(b)

(c)

(d)

V

s5(t x) (x998400 t998400)

t1

t1

t2

t2

s1(tx)





2and of salt 120573 = 1


1and


01

0 10

1L

f

R

2

3

4

c = c1

c = clowast

c = c2

s

s+sminus

c

ab

Δ

Δf998400(s+)


01

0 10

1L

s

f

R

2

3

c = c1

minusΓ(0)

c = c2 = 0








4


2


1for profiles is





3-bank and initial


I

II

III

IVV

(f = 1 120573 = 0 c1)

1

0

120573 = 0

120573 = 1

120573 = 1

120573 = 0

120573 = 0

x

s5(t x)

s3

s3

(x998400 t998400)

sR

c1

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s3)xf3

c = 0

c = 0

c = 0

c = 0

(f = 0 120573 = 1 c = 0)

(f = 1 120573 = 0 c = 0)

t

s1(tx)


2= 0




1lt 1 during







3 The





8 Conclusions











Acknowledgments


References























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










I

II

III

IVV

(f = 1 120573 = 0 c1)

1

0

120573 = 0

120573 = 1

120573 = 1

120573 = 0

120573 = 0

x

s5(t x)

s3

s3

(x998400 t998400)

sR

c1

t =(Γ(

0)+ s 2

)xf2

t =(s 3

f3)x

t =(minuss

R + s3)xf3

c = 0

c = 0

c = 0

c = 0

(f = 0 120573 = 1 c = 0)

(f = 1 120573 = 0 c = 0)

t

s1(tx)


2= 0




1lt 1 during







3 The





8 Conclusions











Acknowledgments


References























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra






























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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