Open access peer-reviewed chapter - ONLINE FIRST

Singularly Perturbed Parabolic Problems

By Asan Omuraliev and Ella Abylaeva

Submitted: September 19th 2018Reviewed: January 11th 2019Published: September 27th 2019

DOI: 10.5772/intechopen.84339

Downloaded: 46

Abstract

The aim of this work is to construct regularized asymptotic of the solution of a singularly perturbed parabolic problems. Namely, in the first paragraph, we consider the case when the scalar equation contains a free term consisting of a finite sum of the rapidly oscillating functions. In the first paragraph, it is shown that the asymptotic solution of the problem contains parabolic, power, rapidly oscillating, and angular boundary layer functions. Angular boundary layer functions have two components: the first one is described by the product of a parabolic boundary layer function and a boundary layer function, which has a rapidly oscillating change. The second section is devoted to a two-dimensional equation of parabolic type. Asymptotic of the scalar equation contains a rapidly oscillating power, parabolic boundary layer functions, and their product; then, the multidimensional equation additionally contains a multidimensional composite layer function.

Keywords

  • singularly perturbed parabolic problem
  • asymptotic
  • stationary phase
  • power boundary layer
  • parabolic boundary layer
  • angular boundary layer

1. Asymptotics of the solution of the parabolic problem with a stationary phase and an additive-free member

1.1 Introduction

Singularly perturbed problems with rapidly oscillating free terms were studied in [1, 2, 3]. Ordinary differential equations with a rapidly oscillating free term whose phase does not have stationary points are studied in [1]. Using the regularization method for singularly perturbed problems [4], differential equations of parabolic type with a small parameter were studied in [2, 3] when fast-oscillating functions as a free member. The asymptotic solutions constructed in [1, 2, 3] contain a boundary layer function having a rapidly oscillating character of change. In addition to such a boundary layer function, ordinary differential equations contain an exponential [1], and parabolic equations - parabolic [2, 3] and angular boundary layer [2, 5] functions. If the phase of the free term has stationary points, then boundary layers arise additionally, having a power character of change. In this case, the asymptotic solution consists of regular and boundary layer terms. The boundary layer members are parabolic, power, rapidly oscillating boundary layer functions, and their products, which are called angular boundary layer functions [4]. In this chapter we used the methods of [4, 5].

1.2 Statement of the problem

In this chapter we study the following problem:

Lεuxtεtuε2axx2ubxtu=k=1Nfкxtexpiθкtε,xtΩ,E1
uxtεt=0=uxtεx=0=uxtεx=1=0

where ε>0is a small parameter and Ω = {(x, t): x01,t0T}.

The problem is solved under the following assumptions:

  1. ax>0,axС01,bxt,fxtСΩ¯.

  2. x01functionax>0.

  3. θktt=0=0is the phase function.

1.3 Regularization of the problem

For the regularization of problem (Eq. (1)), we introduce regularizing independent variables using methods [5, 6]:

η=tε2,rk=iθktθk0ε,ξν=φνxε,i=1,ζv=φνxε2,φνx=1ν1ν1xdsas,ν=1,2,σk=0texpiθкsθк0εdspкtε,l=0,r¯,j=0,kl1¯E2

Instead of the desired function uxtε, we will study the extended function

uˇMε,M=xtrησξζ,σ=σ1σ2σN,r=r1r2rN,ξ=ξ1ξ2,ζ=ζ1ζ2

such that its restriction by regularizing variables coincides with the desired solution:

uˇMεγ=pxtεuxtεγ=rσηξζE3

Taking into account (Eqs. (21)) and ((3)), we find the derivatives

On the basis of (Eqs. (1)(4)) for the extended function uˇMε, we set the problem:

tuxtε(tuˇMε+1ε2ηuˇMε+k=1N[iθktεrkuˇMε+exprkσkuˇMε)γ=pxtε,xuxtεxuˇ(Mε+ν=12φνxεξνuˇMε+φνxε2ζνuˇMεγ=pxtε,E4
x2uxtεx2uˇ(Mε+ν=12φνxε2ξν2uˇMε+φνxε22ζν2uˇMε+1εDξ,vuˇMε+1ε2Dζ,vuˇMεγ=pxtε,
Dξ,v2φνxx,ξν2+φνxξν,Dζ,v2φνxx,ζν2+φνxζν.
LεuˇMε1ε2T0uˇMε+k=1NiθktεrkuˇMε+T1uˇMε=k=1Nfкxtexprk+iθк0ε+LζuˇMε+εLξuˇMε+ε2LxuˇMε     uˇMεt=rk=η=0=uˇMεx=0,ξ1=ζ1=0=uˇMεx=1,ξ2=ζ2=0=0,T1ην=12ζv2,T2tν=12ξv2bxt+k=1Nexprkσk,Lξaxv=12Dξ,v,Lζaxv=12Dζ,v,Lxaxx2.E5

The problem (Eq. (5)) is regular in ε as ε → 0:

LεuˇMεq=qxtεLεuˇxtε.E6

1.4 Solution of iterative problems

The solution of problem (Eq. (5)) will be determined in the form of a series:

uˇMε=v=0εvuvM.E7

For the coefficients of this series, we obtain the following iterative problems:

T1u0M=0,T1u1M=ik=1Nθktrku0M,T1u2M==ik=1Nθktrku1MT2u0M+k=1Nfкxtexprk+iθк0ε+Lζu0M,T1uvM=ik=1Nθktrkuv1MT2uv2M+Lζuv2+Lξuv3M+Lxuv4M.E8

The solution of this problem contains parabolic boundary layer functions; internal power boundary layer functions which are connected with a rapidly oscillating free term in a phase which are vanished at t=tl,l=0,1,,nin addition; and the asymptotic also contain angular boundary layer functions. We introduce a class of functions in which the iterative problems will be solved:

G0CΩ¯,G1=uM:uM=l=12G0erfcξl2t,

G2=uM:uM=k=1NG0exprk,
G3=uM:uM=k=1Nl=12YklNlexprkYklNl<cexpςl28η,
G4=uM:uM=k=1NG0l=12G0erfcξl2tσk,Nl=xtης1ς2.

From these spaces we construct a new space:

G=l=04Gl.

The element uMϵGhas the form:

uM=vxt+l=12wlxterfcξl2t+k=1Nckxt+l=12YklNlexprk+k=1Nzkxt+l=12qklxterfcξl22tσk.E9

1.5 Solvability of intermediate tasks

The iterative problems (Eq. (9)) in general form will be written:

T1uM=HM.E10

Theorem 1. Suppose that the conditions (1)(3) and HMϵG3are satisfied. Then, equation (Eq. (10)) is solvable in G.

Proof. Let the free term HMϵG3be representable in the form:

HM=k=1Nl=12HklNl,HklNl<cexpςl28η.

Then, by directly substituting function uMϵGfrom (Eq. (9)) in (Eq. (10)), we see that this function is a solution if and only if the function YklNlwill be a solution of equation:

ηYklNl=ςl2YklNl+HklNl,l=1,2,k=1,2,,N.E11

With the corresponding boundary conditions, this equation has a solution which have the estimate:

YklNl<cexpςl28η.

The theorem is proven.

Theorem 2. Suppose that the conditions of Theorem 1 are satisfied. Then, under additional conditions:

  1. uMt=η=0=0uMx=l1,ξl=0,ςl=0=0,l=1,2.

  • LςuM=0,LξuM=0.

  • ik=1NθktrkuvM+T2uv1M+hMG3.

  • Eq. (10) is uniquely solvable.

    Proof. By Theorem 1 equation (Eq. (10)) has a solution that is representable in the form (Eq. (9)). With satisfying condition (1), we obtain

    vxtt=0=k=1Nckx0wlxtt=0=w¯lx,E12

    YklNlt=η=0=0qklxtt=0=q¯klx,dklxtt=0=d¯klx,

    wlxtx=l1=ckl1tqklxtx=l1=zkl1t,l=1,2.

    Due to the fact that the function erfcθ2tis zero at θ=0, the values for wlxtt=0qklxtt=0are chosen arbitrarily.

    We calculate

    ik=1NθktrkuvM+T2uv1M+hM=ik=1Nθktck,vxt+l=12Yk,vlNlexprk+tvv1xtbxtvv1(xt)+l=12twv1lxtbxtwv1lxterfcξl2t+k=1Ntck,v1xtbxtck,v1xt+l=12tYk,v1lNlbxtYk,v1lNlexpτk+k=1Ntzk,v1xtxtzk,v1xt+l=12tqk,v1lxtbxtqk,v1lxterfcξl2tσk+k=1Nzk,v1xt+l=12qk,v1lxterfcξl2texpτk+h0xt+l=12h1lxterfcξl2t+k=1Nh2kxt+l=12h2l,kxtexpτk+k=1Nh3kxt+l=12h3l,kxterfcξl2tσk.E13

    Condition (3) of the theorem will be ensured, if we choose arbitrarily (Eq. (9)) as the solutions of the following equations:

    tvv1xtbxtvv1xt=h0xt,twv1lxtbxtwv1lxt=h1lxt,tYk,v1lNlbxtYk,v1lNl=h2l,kxt+qk,v1lxterfcςl2η,tzk,v1xtbxtzk,v1xt=h3kxt,tqk,v1lxtbxtqk,v1lxt=h3l,kxt,iθktck,vxt=zk,v1xttck,v1xtbxtck,v1xth2kxt.E14

    After this choice of arbitrariness, expression (Eq. (13)) is rewritten:

    ik=1NθktrkuvM+T2uv1M+hM=k=1Nl=12ktYk,vlNlexpτkG3

    In (Eq. (14)), transition was made from ξl/2tto variable ςl/2η. The function YklNlis defined as the solution of equation (Eq. (30)) under the boundary conditions from (Eq. (12)) in the form:

    YklNl=dklxterfcςl2η+12π0η0Hkl·ητexpςly24ητexpςl+y24ητdydτ.E15

    We substitute this function in the corresponding equation from (Eq. (14)); then with respect to dklxt, we obtain a differential equation, which is solving under the initial condition dklxtt=0=d¯klx, and we find

    dklxt=d¯klxtBxt+Pklxt,Bxt=exp0tbxsds,

    where Pklxtis known as the function.

    By substituting the obtained function into condition for dklxtx=l1from (Eq. (12)), we define the value of d¯klxx=l1. The obtained value is used as an initial condition for a differential equation with respect to d¯klx, which is obtained after substitution dklxtinto the first condition of (2). With that we ensure fulfillment of this condition and uniqueness of the function YklNl.The last equation from (Eq. (14)) due to the fact that θktk=0is solvable if

    zk,v1lx0=h2kx0tck,v1xtbxtck,v1xtt=0.

    The obtained ratio is used as the initial condition for the differential equation with respect to zk,v1lxtfrom (Eq. (14)).

    The equation with respect to vv1xtunder the initial condition from (12) determines this function uniquely. Equations with respect to wk,v1lxt,qk,v1lxtunder the corresponding condition from (Eq. (12)) have solutions representable in the form:

    wk,v1lxt=w¯k,v1lxBxt+H1,v1lxt,qk,v1lxt=q¯k,v1lxBxt+H2,v1lxtE16

    where H1,v1lxt,H2,v1lxt- are known functions.

    With substituting (Eq. (16)) into the conditions under x=l1from (Eq. (12)), we define values of w¯k,v1lxx=l1q¯k,v1lxx=l1. These conditions are used in solving differential equations which are obtained from the second condition of (Eq. (21)):

    Lξwk,v1lxterfcξl2t=0,Lξqk,v1lxterfcξl2t=0.

    Thus, function uMis determined uniquely. The theorem is proven.

    1.6 Solution of iterative problems

    Eq. (8) is homogeneous for k = 0; therefore, by Theorem 1, it has a solution in G, representable in the form:

    u0M=v0xt+l=12wlxterfcξl2t+k=1Nck,0xt+l=12Yk,0lNlerk+zk,0xt+l=12qk,0lxterfcξl2tσkE17

    If the function Yk,0lNlis the solution of the equation ηYk,0lNl=ςl2Yk,0lNlwhich is satisfying that

    Yk,0lNlt=η=0=0Yk,0lNlx=l1,ςl=0=ck,0l1t.

    from the last problem, we define

    Yk,0lNl=dk,0lxterfcςl2η,dk,0lxtx=l1=ck,0l1twheredk,0lxtt=0=d¯k,0lx.

    d¯k,0lxis the arbitrary function. In the next step, equation (Eq. (8)) for k = 1 takes the form:

    T1u1M=ik=1Nθktck,0xt+l=12Yk,0lNlerk.

    According to Theorem 1, this equation is solvable in U, if ck,0xt=0; the function Yk,0lNlis the solution of the differential equation ηYk,0lNl=ςl2Yk,0lNl+Hk,0lNl, and its solution is representable in the form (Eq. (14)), where Hkl0=ktYk,0lNl.Satisfying condition (1)(3) of Theorem 1, we obtain (see (Eq. (14)))

    tv0bxtv0xt=0,tw0lxtbxtw0lxt=0,tdk,0lxtbxtdk,0lxt=qk,0lxt,tzk,0xtbxtzk,0xt=0,tqk,0lxtbxtqk,0lxt=0,iθktck,1xt=zk,0xt+fkxtexpiθк0ε,Lςdk,0lxterfcςl2η=0.E18

    When the equation is obtained with respect to dk,0lxtin the qk,0lxterfcξl2t, a transition ξl2t=ςl2ηoccurs.

    The initial conditions for equation (Eq. (18)) are determined from (Eq. (12)). Functions w0lxt,dk,0lxt,qk,0lxtare expressed through arbitrary functions w¯0lx,d¯k,0lx,q¯k,0lx. These arbitrary functions provide the condition:

    Lξukm=0,Lςukm=0,

    ensuring the solvability of the equation with respect to ck,1lxt.Suppose that

    Zk,0xtt=0=fkxtexpiθk0ε.

    This relation is used by the initial condition for determining Zk,0xtfrom the equation entering into (Eq. (18)).

    Further repeating this process, we can determine all the coefficients of ukmof the partial sum:

    uεnm=i=0nεiuim.

    In each iteration with respect to vixt,wilxt,dk,ilxt,zk,ixt,qk,ilxt, we obtain inhomogeneous equations.

    1.7 Assessment of the remainder term

    For the remainder term

    RεnxtεRεnmεγ=ρxtε=uxtεi=0nεiuimγ=ρxtε,

    taking into account (Eqs. (3) and (6)), we obtain the equation

    LεRεnxtε=εn+1gnxtε

    with homogeneous boundary conditions. Using the maximum principle, like work of [7], we get the estimate:

    Rεnxtε<cεn+1.E19

    Theorem 3. Suppose that conditions (1)(3) are satisfied. Then, the constructed solution is an asymptotic solution of problem (Eq. (1)), i.e., n=0,1,2,; the estimate is fair (Eq. (18)).

    2. Two-dimensional parabolic problem with a rapidly oscillating free term

    2.1 Introduction

    In the case when a small parameter is also included as a multiplier with a temporal derivative, the asymptotic of the solution acquires a complex structure.

    Different classes of singularly perturbed parabolic equations are studied in [2]. There, regularized asymptotics of the solution of these equations are constructed, when a small parameter is in front of the time derivative and with one spatial derivative. It is shown that the constructed asymptotic contains exponential, parabolic, and angular products of exponential and parabolic boundary layer functions. The equations are studied when the limiting equation has a regular singularity. Such equations have a power boundary layer. If a small parameter is entering as the multiplier for all spatial derivatives, then the asymptotic solution contains a multidimensional parabolic boundary layer function. When entering into the equation, as free terms of rapidly oscillating functions, then the asymptotic of the solution additionally contains fast-oscillating boundary layer functions. If it is additionally assumed that the phase of this free term has a stationary point, in addition to the rapidly oscillating boundary layer function that arises as a power boundary layer.

    This section is devoted to a two-dimensional equation of parabolic type.

    2.2 Statement of the problem

    Consider the problem:

    Lεuxtεtuε2aubxtu=fxtexptε,xtϵE,
    ut=0=0uΩ=0=0,E20

    where ε>0is the small parameter, x=x1x2, Ω=0<x1<1x0<x2<1,E=0<tT,al=12alxlxl2.

    The problem is solved under the following assumptions:

    1. xl01the function alxlС01,l=1,2.

    2. bxt,fxtСE.

    3. θ0=0.

    2.3 Regularization of the problem

    Following the method of regularization of singularly perturbed problems [1, 2], along with the independent variables xt, we introduce regularizing variables:

    μ=tε,ξl=1l1ε3l1x1dsa1s,ηl=φlx1ε2ξl+2=1l1ε3l1x2dsa2s,ηl+2=φl+2x2ε3σ=0teiθsθ0εds,τ2=iθtθ0ε,τ1=tε2,φlxr=1l1l1xrdsars,E21

    For extended function uMε,M=xtτξηsuch that

    uMεμ=ψxtεuxtε,χ=τξη,τ=τ1τ2,ξ=ξ1ξ2ξ3ξ4,η=η1η2η3η4,ψxtε=tε2tεiθtθ0εφxεφxε2,φx=φ1x1φ2x1φ3x2φ4x2uMεμ=ψxtεuxtε,χ=τξη,τ=τ1τ2,ξ=ξ1ξ2ξ3ξ4,η=η1η2η3η4,ψxtε=tε2tεiθtθ0εφxεφxε2,φx=φ1x1φ2x1φ3x2φ4x2.E22

    Find from (Eq. (22)) the derivatives based on

    tutu+1εμu+1ε2τ1u+iθtετ2u+expτ2σuχ=ψxtε,xruxru+l=2r12rφlxrε3ξlu+φlxrε2ζluχ=ψxtε,xr2uxr2u+l=2r12rφl2xrε3ξl2u+φl2xrε4ζl2u+l=2r12r2φlxrε3xrξl2u+φlxrε3ξlu+1ε2φlxrxrηl2u+φlxrˇηluχ=ψxtεE23

    Below, it is shown that the solution of the iterative problems does not contain terms depending on ξ1ξ2,ξ3ξ4,ζ1ζ2,ζ3ζ4,ξlζk,l,k=1,2.Therefore, to simplify recording, the mixed derivatives of these variables are omitted. Based on (Eq. (20)), (Eq. (22)), and (Eq. (23)) for extended function uMε, set the problem:

    Lεu1ε2T0u+1εiθtτ2u+1εT1u+DσuLηuεLξuε2Δau=fxtexpτ2+0ε,u|t=τ1=τ2=0=0,u|xl=r1,ξk=ηk=0=0,r=1,2,l=1,2,k=1,4¯T0τ1Δη,T1μ+Δξ,DσDt+expτ2σ,Dtt+bxt,Lηr=12l=2r12rarxrDx,ηr,l,Dx,ξr,l2φlxrxrξl2+φlxrηl,Δηk=14ηk2,E1=Eх010E24

    In this case, the identity is satisfied:

    Lεuχ=ψxtεLεuxtε.E25

    2.4 Solution of iterative problems

    For the solution of the extended function (Eq. (24)), we search in the form of series

    uMε=i=0εi2uiM.E26

    Then, for the coefficients of this series, we get the following problems:

    T0uvM=0,v=0,1,T0uq=iθtτ2uq2T1uq2,q=2,3.T0u4=fxtexpτ2+0εT1u2Dσu0+Lηu0,T0ui=iθtτ2ui2T1ui2Dσui4+Lηui4+Lξui5+Δaui8,uit=τ=0=0uixl=r1,ξk=ηk=0=0,l,r=1,2.k=1,4¯E27

    We introduce a class of functions:

    U0=V0N=cxt+F1N+F2Nexpτ2F1NU4F2NU5cxtCE¯,U1=V1M:V1M=vxt+F1M+F2MF1MU4F2MU5vxtCE¯,U2=V2M:V2M=zxt+F1M+F2MσF1MU4F2MU5zxtCE¯,U4=V4M:V1M=l=14YlNlYlNl<cexpηl28τ1,U5=V5M:V2M=r,l=14Yr+2,lNr+2,lYr+2,lNr+2,l<cexpηr,l28τ1ηr,l=ηr2+ηl2.

    From these classes we will construct a new one, as a direct sum:

    U=U0U1U2.

    Any item uMUisrepresentable in the form:

    uM=vxt+cxtexpτ2+zxtσ+l=14YlNl+r,l=12Yr+2,lNr+2,lexpτ2+l=14wlxterfcξl2μ+l,r=12wr+2,lMr+2,l+l=14qlxterfcξl2μ+l,r=12zr+2,lMr+2,lσ,Nl=xtτ1ηl,Nr+2,l=xtτ1ηlηr+2,Ml=xtμξl,Mr+2,l=xtμξlξr+2.E28

    Let’s satisfy this function to the boundary conditions:

    vx0=cx0YlNlt=τ1=0=0E29
    Yr+2,lNr+2,lt=τ1=0=0wlt=0=w¯lx,
    qlt=0=q¯lx,
    wr+2,lMr+2,lt=μ=0=0,
    zr+2,lMr+2,lt=μ=0=0,
    wlxtx1=l1=vl1x2t,
    qlxtx1=l1=zl1x2t,

    Ylx1=l1,ηl=0=cl1x2t,Yr+2,lx1=l1,ηl=0=Yr+2,lNr+2,lx1=l1,

    wr+2,lx1=l1,ξl=0=wr+2l1x2terfcξr+22t,
    zr+2,lx1=l1,ξl=0=qr+2l1x2terfcξr+22t,
    wlxtxr=l1=vxtxr=l1,
    qlxtxr=l1=zxtxr=l1,
    Yr+2x2=l1,ηr+2=0=cx1l1t,
    Yr+2,lx2=l1,ηr+2=0=Ylx2=l1,
    wr+2,lx2=l1,ξr+2=0=wlx2=l1erfcξl2t,
    zr+2,lx2=l1,ξr+2=0=qlx2=l1erfcξl2t,l,r=1,2.

    We compute the action of the operators T0,T1,Lη,Lξon function uMU, and we have

    T1uM=r,l=12μwr+2,lξwr+2,l+σμzr+2,lξzr+2,l,
    Lηu=r=12l=2r12rDx,ηr,lYlNl+v=12r,l=12Dx,ηv,lYr+2,lNr+2,l,Lξu=r=12l=2r12rDx,ξr,lwlxterfcξl2μ+v=12r,l=12Dx,ξv,lwr+2,lMr+2,l+σr=12l=2r12rDx,ξr,lqlxterfcξl2μ+v=12r,l=12Dx,ξv,lzr+2,lMr+2,l,DσuM=Dtvxt+l=14Dtwlxterfcξl2μ+r,l=12Dtwr+2,lMr+2,l+Dtcxt+l=14DtYlNl+r,l=12DtYr+2,lNr+2,lexpτ2+σDtzxt+l=14Dtqlxterfcξl2μ+r,l=12Dtzr+2,lMr+2,l+zxt+l=14qlxterfcξl2μ+r,l=12zr+2,lMr+2,lexpτ2

    We write iterative equation (8) in the form:

    T0uM=HM.E31

    Theorem 1. Let be HMU4U5and condition (1) is satisfied. Then, Eq. (31) is solvable in U, if the equations are solvable:

    T0YlNl=H1Nl,l=1,4¯,T0Yr+2,lNr+2,l=H2Nr+2,l,r,l=1,2.

    Theorem 2. Let beH1NlU4. Then, the problem τ1YlNl=ηYlNl+H1Nl,YlNlτ1=0=0YlNlηl=0=dlxt,l=1,4¯(Eq. (32)) has a solutionYlNlU4.

    Theorem 3. Let be H2Nr+2,lU5,YlNlU4,and then the problem τ1Yr+2,lNr+2,l=ηYr+2,lNr+2,l+H2Nr+2,l,Yr+2,lNr+2,lηl=0=Yr+2Nr+2Yr+2,lNr+2,lηr+2=0=YlNl,r,l=1,2has a solution Yr+2,lNr+2,lU5.

    The proof of these theorems is given in [2].

    2.5 The decision of the iterative problems

    Eq. (27) under v=0,1is homogeneous. By Theorem 1, it has a solution representable in the form u0MUif functions YlNland Yr+2,lNr+2,l–are solutions of the following equations:

    T0YvlNl=0,T0Yvr+2,lNr+2,l=0.

    Based on the boundary conditions from (Eq. (29)), the solution is written:

    YvlNl=dvlxterfcηl2τ1,l=1,2,3,4.E32
    Yvr+2,lNr+2,l=0τ10YvlξGNlξητ1τξ=0dηdτ0t0Yvr+2ηGNr+2,lξητ1τη=0dξdτ,

    wheredlxt– is arbitrary function such as

    dvpxtt=0=d¯vpxdvlxtx1=l1=cvl1x2t,Gηlηr+2,lξητ1=14πτ1expηlξ24τ1expηl+ξ24τ1expηr+2η24τ1expηr+2+η24τ1.E33

    Due to the fact that the function dvlxtпри t=τ1=0multiplied by the function becomes as d0lxtt=0=d¯0lx, an arbitrary function is accepted, and its values under x1=l1are determined from the second relation. According to Theorems 2 and 3, the functions found by the formula (Eq. (33)) satisfy the estimates:

    YvlNl<cexpηl28τ1,Yvr+2,lNr+2,l<cexpηr+22+ηl28τ1,r,l=1,2.E34

    Free member of equation (Eq. (27)) under v=2,3has a form

    Fv2MT1uv2M+iθtσuv2M=iθtcv2xt+l=14Yv2lNl+r,l=12Yv2r+2,lNr+2,lexpτ2+l,r=12μwv2r+2,lξwv2r+2,l+σμzv2r+2,lξzv2r+2,l,

    so that equation (Eq. (27)), under v=2,3, has a solution in U; we set

    cv2xt=0,T1wv2r+2,l=0,T1zv2r+2,l=0.

    Solutions of the last equations under the boundary conditions from (Eq. (29)) have a form (Eq. (33)) for which estimates of the form (Eq. (35) are fair. Eq. (27), i=4, has a free term:

    F4M=iθtτ2T1u2+fxtexp0εDσu0+Lηu0=iθtc2xt+l=14Y2lNl+r,l=12Y2r+2,lNr+2,lexpτ2l,r=12T0w2r+2,lMr+2,l+σT0z2r+2,lDtv0xtl=14Dtw0lxterfcξl2μl,r=12Dtw0r+2,lxtexpτ2tc0xt+l=14tY0l+l,r=12DtY0r+2,l
    σDtz0xt+l=14Dtq0lxterfcξl2μ+r,l=12Dtz0r+2,lMr+2,lz0xt+l=14q0lxterfcξl2μ+r,l=12z0r+2,lMr+2,lexpτ2+r=12 l=2r12rDx,ξr,lw0pxterfcξl2μ+v=12 r,l=12Dx,ηv,lY0r+2,lNr+2,l.

    By providing F4MU4U5with regard to cvxt=0,v=0,1,we set

    iθtc2xt+fxtexp0εz0xt=0,Dtv0xt=0,Dtz0xt=0,DtY0lNl,T0w2r+2,l=0,T0z2r+2,l=0,Dtw0l=0,Dtw0r+2,l=0,DtY0r+2,l=0,Dtq0lxt=0,Dtz0r+2,lxt=0,Dx,ξr,lw0lxt=0,Dx,ηv,lY0r+2,l=0,Dx,ηr,lY0l=0,E35

    then

    F4M=iθtl=14Y2lNl+r,l=12Y2r+2,lNr+2,lexpτ2l=14q0lxterfcηl2τ2+r,l=12z0r+2,lNr+2,lexpτ2.

    In the last bracket, the transition is from the variables ξl2μto the variables ηl2τ2.

    Substituting the value Y0lNl=d0lxterfcηl2τ1into equation DtY0lNl=0,with respect to d0lxt,we get the equation Dtd0lxt=0,which is solved under an arbitrary initial condition d0lxtt=0=d¯0lx. This arbitrary function provides the condition LηY0l=0;therefore, Dx,ηY0l=0.The initial condition for this equation is determined from the relation:

    d0lxtx1=l1=c0l1x2td0l+2xtx2=l1=c0x1l1t,

    which comes out from (Eq. (29)) and (Eq. (33)). The functionY0r+2,lNr+2,lexpresses through Y0lNltherefore provided that

    DtY0r+2,l=0,Dx,ηv,lY0r+2,l=0.

    The same is true for functions w0r+2,lMr+2,l,z0r+2,lMr+2,l; in other words, the following relations hold: Dtw0r+2,l=0,Dtz0r+2,l=0,Dx,ξv,lw0r+2,l=0,Dx,ξv,lz0r+2,l=0.

    Solutions of equations with respect w0r+2,l,z0r+2,lunder appropriate boundary conditions from (Eq. (29) are representable as (Eq. (33)), and they are expressed through w2lxt,q2lxt.The first equation (Eq. (36)) is solvable, ifz0xtt=0=fx0exp0ε.This ratio is used by the initial condition for the equation

    Dtz0xt=0. The remaining equations from (Eq. (36)) are solvable under the initial conditions from (Eq. (29)).

    Thus, the main term of the asymptotics is uniquely determined. As can be seen from the representation (Eq. (28)) and the estimates (Eq. (35)), we note that the asymptotics of the solution have a complex structure. In addition to regular members, it contains various boundary layer functions. Parabolic boundary layer functions have an estimate:

    YlNl<cexpηl28τ1,wlxterfcξl2μ<cexpξl28μ.

    Multidimensional and angular parabolic boundary layer functions have an estimate:

    Yr+2,lNr+2,l<cexpηr+22+ηl28τ1
    wr+2,lMr+2,l<cexpξr+22+ξl28μ.

    The boundary layer functions with rapidly oscillating exponential and power type of change:

    cxtexpτ2,σ=0teiθsθ0εds.

    In addition, the asymptotic contains the product of the abovementioned boundary layer functions.

    Repeating the above process, we construct a partial sum:

    uεnM=i=0nεi2uiM.E36

    2.6 Assessment of remainder term

    Substituting the function uMε=uεnM+εn+12RεnMinto problem (Eq. (24)), then taking into account the iterative tasks of (Eq. (27)) and (Eq. (29)), we obtain the following problem for the remainder term RεnM:

    LεRεnM=gnMε,RεnMt=0=RεnMxl=r1,ξr=0,ηk=0,r=1,2;k=1,4¯,E37

    where gnMε=iθtτ2un1ε12iθtτ2unMT1un1Mε12T1unMDσLηk=03εk2un3+kM+Lηk=05εk2un5+kM+Δak=07εk2un7+kM.

    We put in both parts (Eq. (38))χ=ψxtεconsidering (Eq. (25)), with respect to

    LεRεnxtε=gεnxtε,Rεnt=0=0RεnΩ=0.

    By virtue of the above constructions, the function is gεnxtε<c,xt;

    therefore, applying the maximum principle, an estimate is established:

    Rεnxtε<c.

    Thus, we have proven the following:

    Theorem 4. Suppose that the conditions (1)(3) are satisfied. Then, using the above method for solving uxtεof the problem (Eq. (20)), a regularized series (Eq. (26)) such that n=0,1,2,can be constructed, and for small enough ε>0,inequality is fair:

    uxtεuεnxtε=Rεnxtε<n+12,

    where cis independent of ε.

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    Asan Omuraliev and Ella Abylaeva (September 27th 2019). Singularly Perturbed Parabolic Problems [Online First], IntechOpen, DOI: 10.5772/intechopen.84339. Available from:

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