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A System of Singularly Perturbed Parabolic Equations with a Power Boundary Layer

Written By

Asan Omuraliev and Peil Esengul Kyzy

Reviewed: 03 July 2022 Published: 19 October 2022

DOI: 10.5772/intechopen.106239

From the Edited Volume

Boundary Layer Flows - Modelling, Computation, and Applications of Laminar, Turbulent Incompressible and Compressible Flows

Edited by Vallampati Ramachandra Prasad, Valter Silva and João Cardoso

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Abstract

The work is devoted to the construction of the asymptotics of the solution of a singularly perturbed system of equations of parabolic type, in the case when the limit equation has a regular singularity as the small parameter tends to zero. The developed algorithm allows construction of the asymptotics of solutions containing power, parabolic, and angular boundary layer functions.

Keywords

  • singularly perturbed problems
  • power boundary layer
  • parabolic boundary layer
  • angular boundary-layer functions
  • regularized asymptotics

1. Introduction

1.1 A singularly perturbed system of parabolic equations

This article studies the problem

Lεuxtεε+ttuε2axx2uBtu=fxt,xtΩ,
ut=0=ux=0=ux=1=0,E1

where ε>0 is a small parameter, Ω={xt:x01,t(0,T]},u=u1u2un. Suppose the following conditions:

0<axC01,BtC0TCn2,fxtCΩ¯Cn;

The eigenvalues λit of the matrix Bt satisfy the conditions: Btψit=λitψit,i=1,2,,n,λit<0,λjtλit,ij,t0T ( is the real part of the complex number).

Singularly perturbed parabolic problems in various statements are devoted to Ref. [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23]. In Ref. [6, 9, 10, 11, 12, 13, 14, 15, 16, 17], by the method of boundary layer functions, various boundary value problems for parabolic equations with a small parameter are studied. The regularization asymptotic for solving parabolic problems [7, 8, 18, 19, 20, 21, 22, 23] with different occurrences of a small parameter in the equations and limit operators has different structures using the regularization method for singularly perturbed problems.

The papers [16, 23] are devoted to systems of singularly perturbed parabolic equations. In [16], the method of boundary functions, and in Ref. [23], the method of regularization. For singularly perturbed problems, equations are studied when the spatial derivative is preceded by a scalar function, and [24] is a matrix. In the latter case, the structure of the solution asymptotic is greatly multiplied. The order of the equation does not go down.

Problems with power boundary layers, that is, problems where, when a small parameter tends to zero, it acquires a regular feature were studied in Ref. [9, 10, 25, 26]. Ordinary differential equations with a power boundary layer are studied in Ref. [8, 9, 10, 26]. For equations of parabolic type, when a small parameter does not enter the factors of the spatial derivative, the asymptotic of the power boundary layer is constructed. In contrast to Ref. [9], in our equation, there is a small parameter in front of the spatial derivative and we will improve the algorithm for constructing the asymptotic.

1.2 Regularization of problems

We introduce regularizing variables

ξl=φlxε3,φlx=1l1l1xdsas,ηi=λi0ln1+t/ε,τ=ε1ln1+t/εE2

along with the independent variables xt of the function u˜Mε, M=xtξτη, ξ=ξ1ξ2, η=η1η2ηn such that

uMεζ=γxtεuxtε,
ζ=ξ,τη,γx,tε=φ1xε3,φ2xε3,λ10ln1+t/ε,,λn0ln1+t/εε1ln1+t/ε.

Hence, on the basis of (2), we find

tutu+i=1nλi0t+εηiu+1ε1t+ετuζ=γx,tε,xuxu+l=12φlxε3ξluζ=γx,tε,
x2ux2u+l=12φlxε32ξl2u+1ε3Dξ,luζ=γxtε,Dξ,l2φlxξlx2+φlxξl,

then, according to (1), we set the extended problem

Lξε1T0u+T1uεLξu+εtuε2Lxu=fxt,MQ,
ut=τ=ηi=0=ux=0,ξ1=0=ux=1,ξ2=0=0,Q=Ω×03.E3

The notation is entered here

T0τΔξ,T1i=1nλi0ηi+ttBt,Δξl=12ξl2,Lξl=12axDξ,l,Lxaxx2.

Note that the identity holds

Lεuζ=γxtεLεu,E4

solutions of problem (3) will be defined as

uMε=k=0εk2ukM,E5

then for the coefficients of this series we obtain the following iterative problems:

T0uν=0,ν=0,1,T0u2=fxtT1u0,T0uk=T1uk2+Lξuk3tuk4+Lxuk6,
ukt=τ=ηi=0=0,ukx=0,ξ1=0=ukx=1,ξ2=0=0,k3.E6

1.3 Solvability of iterative problems

We introduce the space of functions in which the iterative problems will be solved:

U1={u1N1:u1N1=<vxt+cxt+ΛPxeη,ψt>,
vxtCΩ¯Cn,PxC01Cn,cxtCΩ¯Cn2,ψtC0TCn},
U2={u2Nl:u2Nl=l=12<YNl+zNleη,ψt>,YNl<ceξl28τ,zNl<ceξl28τ},
YNl=colY1NlY2NlYnNl,zNl=zijNl,i,j=1,n¯,
Nl=xtξlτ,η=colη1η2ηn,ΛPx=diagP1xP2xPnx,
<cxt+ΛPxeη,ψt>=i,j=1ncijxteηjψit+i=1nPixeηiψit.
<YNl+zNleη,ψt>=i=1nYiNl+j=1nzijNleηjψit.

Here colη1η2ηn is the vector, diagP1xP2xPnx is a diagonal matrix. Element ukMU=U1U2 has an idea

ukM=<vkxt+ckxt+ΛPkxeη,ψt>+l=12<YkNl+zkNleη,ψt>.E7

We calculate the action of the operators included in the extended equation on the function ukMM, for which we first decompose ψit=j=1nαijtψjt, or we write by entering the notation

T0ukM=l=12<τYkNlξl2YkNl+τzkNlξl2zkNleη,ψt>,
T1ukM=<ttvkxt+tATtvkxtBtvkxt+ttckxtE8
+tATtckxtBtckxt+ckxtΛ0BtΛPkx+tATtΛPkx
+ΛPkxΛ0,ψt>+l=12<[ttYkNl+tATtYkNlBtYkNl
+ttzkNl+tATtzkNlBtzkNl+zkNlΛ0]eη,ψt>
=<D1vkxt+D2ckxt+ΛPkxeη+l=12<D1YkNl+D2zkNleη,ψt>,
D1tt+tATtBt,D2D1+Dλ,Dλ=i=1nλi0ηi,
Λ0=diagλ10λ20λn0,
At=αij,αij=ψitψjt;
Lξuk=<axl=12{2φlxxξl2+φlxξlYkNl+2φlxxξl2+φlxξlzkNleη,ψt}>,
LxukM=<Lxvkxt+l=12LxYkNl+Lxckxt+LxΛPkx+l=12LxzkNleη,ψt>.

Iterative equations are written in the form.

T0ukM=hkM.E9

Theorem 1: Let conditions (1), (2), and hkMU2 hold. Then Eq. (9) is solvable in U.

Proof: Let hkMU2:

hkM=l=12<hk,1Nl+hk,2Nleη,ψt>.E10

Function (7) will be the solution of Eq. (9) in U. If the functions YkNl,zkNl are the solution of the equations

τYkNlξl2YkNl=hk,1Nl,τzkNlξl2zkNl=hk,2Nl.E11

These equations are obtained by substituting (7) into Eq. (9), taking into account calculations (8) and representation (10). Eqs. (11), with appropriate boundary conditions, have solutions that satisfy the estimates [25]:

YkNl<ceξl28τ,zkNl<ceξl28τ.

Theorem 2: Let conditions (1), (2), and hk,3x0¯=0 hold. Then problem

<D2ckxt+ΛPkx,ψt>=<hk,3xt,ψt>E12
ckx0¯=Λvkx0ΛPkxΛckx0¯¯1,1=col111,E13
ciikxtt=0=vk,ix0Pikxj=1ijncijk(x0),

has a unique solution. Hereinafter, c¯ denotes a matrix with nonzero diagonal, and c¯¯ with zero diagonal elements, that is, c=c¯+c¯¯.

Proof: We write the relation (12) in the coordinate form

i,j=1nttcijkxt+μ=1nαμ,it(cμ,ik(xt)+Pμkx)+λj0λitcijk(xt)ψiteηj
+i=1nλi0λitPikxeηiψit=i,j=1nhijk,3xteηjψit.

Removing the degeneracy of this equation, assuming that

λj0λitcijk(xt)t=0=hijk,3x0,ij.E14

Equating the coefficients ψ, we get

ttciikxt+αiitciik(xt)=λitλi0Pikx+hiik,3xttμ=1nαμ,itPμkx,i=1,n¯,E15
ttcijkxt+μ=1iμnαμ,itcμ,jk(xt)+λj0λitcijkxt=hijkxt,ij,i,j=1,n¯.E16

Eq. (15), by virtue of condition hiikx0=0, under the initial condition (13), and Eq. (16) with condition (14) have one-to-one solutions.

Remarks. In iterative problems, the condition hiikx0=0 will be provided by the choice of the function Pikx.

Theorem 3: Let conditions (1) and (2) be fulfilled. Then Eq. (9) has a unique solution satisfying the conditions: (a) ukMt=τ=η=0=0ukMx=l1,ξl=0=0; (b) T1ukM+hkMU2; (c) LξukM=0.

Proof: Let satisfy the function (7) to the boundary conditions (a):

<vkx0+ckx0+ΛPkx1+l=12YkNl+zkNl1t=τ=0,ψ0>=0,
<vkl1t+[ck(l1t)+ΛPkx]eη+l=12[YkNlx=l1,ξl=0
+zkNlx=l1,ξl=0eη],ψt>=0,

Hence, we define

ckx0¯=Λvkx0ΛPkxΛckx0¯¯1,
YkNlt=τ=0=0,zkNlt=τ=0=0,E17
zkNl|ξl=0=Wk,lxt,Wk,lxt|x=l1=ckl1tΛPkx,YkNl|ξl=0=dk,lxt,
dk,lxtx=l1=vkl1t.

Ensuring the solvability of Eq. (9) with the right side of

FkM=T1ukM+hkMU2,

based on the calculations (8) and the representation of

hkM=<hk,1xt+hk,2xteη+l=12hk,3Nl+hk,4Nleη,ψt>

assuming that

<D2ckxt+ΛPkxeη,ψt>=<hk,2xteη,ψt>,E18
<D1vkxthk,1xt,ψt>=0.E19

Eq. (18) under the initial condition (17), on the basis of Theorem 2, are uniquely solvable. Eqs. (19) are solved without an initial condition and have a bounded solution [27].

Eqs. (11) with boundary conditions (17) have solutions that can be represented as

YikNl=dik,lxterfcξl/2τ1/2+hik,3xtI1ξlτ,
zijkNl=Wijk,lxterfcξl/2τ1/2+hijk,4xtI2ξlτ,

where erfcx=2π1/20xet2dt is integral of the additional function and describes a parabolic boundary layer, dk,lxt,Wk,lxt are arbitrary functions that are chosen, like solving the equations

Dxdik,lxt=Dxhik,3xt,DxWijk,3xt=Dxhijk,4xt,Dx2φlxx+φlx,E20

with boundary conditions from (17). Eqs. (20) are obtained by satisfying condition c) and taking into account that the functions erfcξl/2τ1/2 and Ilξlτ have single estimates.

Thus, a unique solution to Eq. (9) is obtained that satisfies conditions (a)–(c).

1.4 Construction of solutions of iterative equations

For ν=0,1 the equations for uνM are homogeneous; therefore, the condition of Theorem 1 holds; therefore, the solution of these equations exists and can be represented in the form (7).

The following iterative equation, on the grounds (8), has a free term

F2M=fxtT1u0=fxt<D1v0xt+D2c0xt+ΛP0xeη
+l=12D1Y0Nl+D2z0Nleη,ψt>.

We decompose fxt by the system ψit and substitute it with the previous relation. Further, providing the conditions of Theorem 1, we set

D1v0xt=fxt,ttv0,ixt+μ=1nαtv0,μ(xt)λitv0ixt=fxtψi,
<D2c0xt+ΛP0x,ψt>=0.

The first system has a smooth solution, and the second system by Theorem 2 is solvable if

ttcii0+αiitcii0(xt)=λitλi0Pi0xtμ=1nαμ,itPμ0x,
cii0xtt=0=vi0x0Pi0x,E21
ttcij0+μ=1μinαμ,itcμi0(xt)+λj0λitcij0xt=0,
λj0λitcij0x0=0,ij,i,j=1,n¯.E22

Problem (21) is uniquely solvable, and problem (22) has a trivial solution.

For k=3, by Theorem 3, ensuring condition (c) for d0,lxt and W0,lxt, we obtain problem

Dxdi0,lxt=0,di0,lxtx=l1=v0,il1t,
DxWij0,lxt=0,Wij0,lxtx=l1=cij0l1t.

By this, we have determined the main term of the asymptotic. In addition, conditions (b) of Theorem 3 gives

D1v1,i=0,ttcii1+αiitcii1=λitλi0Pi1xtμ=1nαμ,itPμ1x,
cii1xtt=0=v1ix0Pi1x,
ttcij1+μ=1μinαμ,itcμj1(xt)+(λj0λitcij1xt=0,cij1xtt=0.

From here we define v1ixt=0, below it will be shown that Pi1x=0, therefore cii1xt=0, and from the last equation we find cij1xt=0,ij.

In the next k=4 step free member

t,F4M=T1u2+Lξu1tu0.

Satisfying condition (c) of Theorem 3, we obtain the problems

Dxdi1,lxt=0,di1,lxtx=l1=v1,il1t=0,
DxWij1,lxt=0,Wij1,lxtx=l1=cij1l1t=0.

That is, di1,lxt=0, Wij1,lxt=0, and, therefore, u1M=0. Regarding Yi3Nl,Wij3Nl we obtain homogeneous equations, therefore

Yi3Nl=di3lxterfcξl/2τ1/2,Wij3Nl=Wij3,lxterfcξl/2τ1/2.

We calculate

F4M=<D1v2xt+D2c2xt+ΛP2xeη
+l=12Y2Nl+z2Nleη,ψt><tv0xt+ATtv0xt
+tc0xt+ATtc0(xt)+AtΛP0xeη
+l=12tY0Nl+ATtY0Nl+tz0Nl+ATtz0Nleη,ψt>

and ensuring the solvability in U of the iteration equation for k=4, we assume

<Dtv2xt,ψt>=<tv0xt+Atv0xt,ψt>,E23
D2c2xt+ΛP2x,ψt>=<tc0xt+Atc0xt+AtΛP0x,ψt>.

The solvability of the second system is ensured by the choice of P0x:

ΛP0x=A1ttc0xt+Atc0xt¯,Pi0x=αii1tcii0xt+k=1nαkitcki0xtt=0,

as well as

<c2xtΛ0Λtc2xt,ψt¯>|t=0=<tc0xt+Atc0xt+AtΛP0x,ψt¯¯>
cij2xtt=0=1λj0λi0tcij0xt+k=1nαkitckj0(xt)+αjitPj0xt=0,ij.E24

Under such assumptions, the second system of (23) is solvable. It is solved under the initial conditions (24) and the initial condition cii2xtt=0=v2ix0Pi2x, which is obtained from (17). This completely defines the main term of the asymptotic.

Further, repeating the process described above, we find all the coefficients of the partial sum (5), and the coefficients with odd indices are zero.

1.5 Estimates of the remainder term

Denote by RεnM=uMεuεnM, where uεnM=k=0nεku2kM. Substituting uMε=uεnM+RεnM into the extended problem (3), then, taking into account (6), with respect to RεnM, we obtain

LεRεnM=εn+1gεnM,RεnMt=τ=0=RεnMx=l1,ξl=0=0,l=1,2,

where the function gεnM is expressed through uε,kM. Producing constriction this problem, taking into account the identity (4), with respect to Rε,nxtεRεnMζ=γxtε we get the problem

LεRε,nxtε=εn+1gεnxtε,Rεnt=0=Rεnx=l1=0,l=1,2.

From the construction of solutions to iterative problems, it can be seen that the function gεnxtε is uniformly bounded in Ω¯. Applying the maximum principle [28], we can establish a uniform estimate in Ω¯

Rεnxtε<cεn+1.E25

Theorem 4: Let conditions (1) and (2) be fulfilled. Then the function uεnxtε is a uniform asymptotic solution of problem (1), that is, the estimate (25) holds.

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2. A singularly perturbed parabolic equation system

We consider the first boundary-value problem for a system of singularly perturbed parabolic equations

Lεuε+ttuε2Axx2uDtu=fxt,xtΩ,E26
ux0ε=hx,u0tε=u1tε=0,

where Ω=01×0T,ε>0 is a small parameter,

u=uxtε=colu1xtεu2xtεunxtε,AxC01n2,
DtC0Tn2,fxtCΩ¯n.

The work is a continuation of [29], where instead of the matrix-function Ax, there was a scalar function and an asymptotic of the solution was constructed containing two functions describing the boundary layers along x=0 and x=1. Lomov was the first to introduce the concept of a power-law boundary layer based on the study of the Lighthill equation and he based his method on it [29]. In this case, the asymptotics contains 2m parabolic boundary layer functions describing the boundary layers along x=0 and x=1.

The asymptotics of the solution to this problem, along with the parabolic boundary layer function (the parabolic boundary layer is described by the function), erfcφx2εt, also contain the power boundary layer function

Πεt=εε+tλ,λ>0

as well as their products, which describe the corner boundary layer in the vicinity of 00.

Construction of the asymptotic solution of a singularly perturbed system of parabolic equations is devoted to works [8, 9, 10] and [30]. In Ref. [8], a regularized asymptotic is constructed in the case when the matrix of coefficients for the desired function has zero multiple eigenvalues. A similar problem was studied in [9] and an asymptotic of the boundary layer type was constructed. The method of boundary functions in [10] studied the bisingular problem for systems of parabolic equations, which is characterized by the presence of non-smoothness of the asymptotic terms and a singular dependence on a small parameter. In Ref. [30] and [26], various problems for split systems of two equations of parabolic type were studied, and asymptotics of the boundary layer type were constructed. The problems of differential equations of parabolic type with a small parameter were studied in Ref. [24, 31, 32].

2.1 Statement of the problem

We consider the first boundary-value problem (26). The problem is solved under the following assumptions:

  1. For n-dimensional vector functions fxt and hx, the inclusions

    fxtCΩ¯n,hxC01n,

    are fulfilled for n×n-matrix-valued functions Dt and Ax-inclusions

    DtC0Tn×n,AxC01n×n;

  2. The real parts of all roots λix,i=1,n¯, of the equation detAxλE=0 are positive and λixλjx for all x01, ij,i,j=1,n¯;

  3. The real parts of the eigenvalues βjt,j=1,n¯ of the matrix Dt are negative, that is, Reβj0<0 and βi0βjtt0T,ij,i,j=1,n¯;

  4. Completed the conditions of approval of the initial and boundary conditions h0=h1=0.

2.2 Regularization of the problem

Following [29], p. 316; 30, P. 18], we introduce regularizing variables

ξi,l=φi,lxε3;φi,lx=1l1l1xdsλis,l=1,2,i=1,n¯,E27
τ=1εlnt+εε,μj=βj0lnt+εεKjtε,j=1,n¯,

and an extended function such that

uMεξ=φx/εuxtε,M=xtξτμ,ξ=ξ1ξ2,
ξl=ξ1,lξ2,lξn,l,φx=φ1xφ2x,E28
φlx=φ1,lφ2,lxφn,lx,l=1,2,μ=μ1μ2μn.

Based on (27), we find the derivatives from (28):

tutu+1εt+ετu+j=1nβj0t+εμjuθ=χxtξτμ,
xuxu+l=12i=1n1ε3φi,lxξi,luθ=χxtξτμ,
x2ux2u+l=12i=1n1ε3φi,lx2ξi,l2u+1ε3Li,lξuθ=χxtξτμ,
Li,lξu=2φi,lxxξi,l2u+φi,lxξi,lu,χxtε=φxε31εlnt+εεK(tε),
Ktε=K1tεK2(tε)Kn(tε),θ=τμξ.

According to these calculations, as well as (26) and (28), we pose the following extended problem

Lεuεtu+1εT0u+T1uεLξuttuε2Lxu=fxt,E29
ut=0=0,ux=0,ξi,1=0=ux=1,ξi,2=0=0,i=1,n¯,
T0u=τul=12i=1nφi,lx2Axξi,l2u,T1u=ttu+j=1βj0μjuDtu,
Lξu=l=12i=1nAxLi,lξu,Lxu=Axx2u.

In this case, the identity holds

LεuMεθ=χxtξτμLεuxtε.E30

The solution to the extended problem (29) will be defined as a series

uMε=k=0εk/2ukM.E31

Substituting (31) into problem (29) and equating the coefficients for the same powers of P, we obtain the following equations:

T0u0=0,T0u1=0,T0u2=fxtT1u0,T0u3=Lξu0T1u1,E32
T0uk=Lξuk3+Lxuk6tuk4T1uk2.

The initial and boundary conditions for them are set in the form

ukMt=μ=τ=0=0,
ukMx=l1,ξi,l=0=0,k0,i=1,n¯,l=1,2.

2.3 Solvability of iterative problems

Each of the problems (32) has innumerable solutions, therefore, we single out a class of functions in which these problems were uniquely solvable. We introduce the following function classes:

U1=Vxt:V(xt)=i=1nvi(xt)ψitvi(xt)CΩ¯,
U2=YN:YN=l=12i=1nyilNilbixyilNil<cexpξi,l8τ,
U3=Cxt:C(xt)=i=1nj=1ncijxtexpμj+pixψitcij(xt)CΩ¯,
U4=ZN:ZN=l=12i,j=1nzi,jlNilbixexpμjzi,jlNil<cexpξi,l28τ,

where Nil=xtξi,lτμi,i=1,2,,n,l=1,2. From these classes of functions we construct a new class as a direct sum: U=U1U2U3U4. The function ukMU is representable in vector form

ukM=ΨtVkxt+Ck(xt)expμ
+l=12BxYk,lNl+Zk,lNlexpμ,Ckxt=C1kxt+ΛPx,
C1kxt=cijxt,Vkxt=colvk1vk2vkn,
Yk,lNl=coly1k,lN1ly2k,lN2lynk,lNnl,Zk,lNl=zijk,lNil,
Ψt=ψ1tψ2tψnt,Bx=b1xb2xbnx,
expμ=colexpμ1expμ2expμn

or in coordinate form

ukM=i=1nvk,ixtψit+l=12i=1nyik,lNilbixE33
+i,j=1nci,jkxtψit+l=12zi,jk,lNilbixexpμj+i=1npikxψitexpμi.

The vector-functions bix,ψit included in these classes are eigenfunctions of the matrices Ax and Dt, respectively

Axbix=λixbix,Dtψit=βitψit,i=1,n¯.E34

Moreover, according to condition (1), they are smooth in their arguments.

Along with the eigenvectors bix and ψit, the eigenvectors bix,ψit,i=1,n¯ of the conjugate matrices Ax,Dt will be used

Axbix=λ¯ixbix,Dtψit=β¯itψit

and they are selected biorthogonal

bixbjx=δi,j,ψitψjt=δi,j,i,j=1,n¯,

where δi,j is the Kronecker symbol.

We calculate the action of the operators T0,T1,Lξ,Lx on the function ukMε from (34), taking into account relations (35) and

φi,l'2x=1λix,i=1,n¯

we have

T0ukMi=1nl=12τyik,lNilξi,l2yik,lNilE35
+j=1nτzi,jk,lNilξi,l2zi,jk,lNilexpμjbix;

or vector form

T0ukMl=12BxτYk,lNlξl2Yk,lNl+τZk,lNlξl2Zk,lNlexpμ,

Yk,lNl is n-vector, Zk,lNl is n×n-matrix. Here Bx is a matrix function n×n whose columns are the eigenvectors bix of the matrix Ax. We calculate

T1uk=ttuk+j=1nβj0μjukDtuk
=ti=1ntvk,i+r=1nαr,itvk,r(xt)ψit+l=12tyik,lNilbix
+j=1ntcijkxt+r=1nαritcr,jk(xt)+αjitpjkxψitE36
+l=12zijk,lNilbixexpμj
+i,j=1nβj0cijkxtψitexpμj+i=1nβi0pikxψitexpμi
+l=12i,j=1nβj0zi,jk,lxtbixexpμj
i=1nβitvki(xt)ψit+l=12r=1nγr,i(xt)yik,lNilbix
i,j=1nβjtci,jkxtψitexpμji=1nβitpikxψitexpμi
l=12i,j=1nr=1nγr,ixtzr,jk,lNilbixexpμj.

Here αi,r=ψitψrt,γi,rxt=Dtbixbrx.

It will be shown below that the scalar functions yik,lNl and zi,jk,lNl are representable in the form

yik,lNil=dik,lxtyik,lξlτ,zi,jk,lNil=ωi,jk,lxtzi,jk,lξlτ.

Given these representations, we calculate

LξukM=l=12Axi=1n2φixbixdik,l(xt)xE37
+φixbixdik,l(xt)ξi,lyik,lξi,lτ
+j=1n2φixbixωi,jk,l(xt)x+φixbixωi,jk,l(xt)ξi,lzi,jk,lξi,lτexpμj,
LxukM=Axx2Vkxt+l=12x2BxYl,kNl
+x2ΨtCk,0(xt)expμ+l=12x2BxZl,kNlexpμ.

Satisfy the function (34) of the boundary conditions from (29)

yik,lNilt=τ=μ=0=0zi,jk,lNilt=τ=μ=0=0,E38
ciikx0=vk,ix0pikxijcijkx0,
yik,lNilξi,l=0=dik,l(xt)dik,l(xt)bixx=l1=vil1tψit,
zi,jk,lNilξi,l=0=ωi,jk,l(xt)ωi,jk,l(xt)bixx=l1=ci,jl1t+pil1ψit.

In general form, the iterative Eqs. (32) are written as

T0ukM=hkM.E39

Theorem 1: Let hkMU and conditions (2) and (3) on hold. Then Eq. (40) has a solution ukMU, if the equations are solvable

τyik,lNilξi,l2yik,lNil=hik,1Nilh¯ik,1xth¯¯ik,1ξi,1τ,E40
τzi,jk,lNilξi,l2zi,jk,lNil=hijk,2Nilh¯ijk,2xth¯¯ik,2ξi,2τ.

Proof: Let hkM=l=12i=1nhik,1Nil+j=1nhijk,2NilexpμjbixU. Pose (34) into Eq. (40), then, on the basis of calculations (38), with respect yik,lNil,zijk,lNil, we obtain Eqs. (41). These equations, under appropriate boundary value conditions

yik,lNilt=τ=μ=0=0yik,lNilξi,l=0=dik,lxt,
zi,jk,lNilt=τ=μ=0=0zi,jk,lNilξi,l=0=ωi,jk,lxt.

Solutions are represented in the form

yik,lNil=dik,lxterfcξi,l2τ+h¯ik,1xtI1ξi,lτ,
zi,jk,lNil=ωi,jk,lxterfcξi,l2τ+h¯ijk,2xtI2ξi,lτ,
Irξi,lτ=12π0τ0h¯¯ik,rηsτsexpξi,lη24τs
expξi,l+η24τsdηds,r=1,2,

where h¯ik,rxt,h¯ik,rηs are known functions. Evaluation of the integral

Irξi,lτcexpξi,l28τ.

Theorem 2: Let conditions (1)(4) be satisfied, then Eq. (32) under additional conditions

  1. ukt=μ=τ=0=0ukx=l1,ξi,l=0=0,l=1,2;

  2. T1uktuk2+Lxuk4U2U4;

  3. Lξuk=0,

has a unique solutionin U.

Proof: Satisfying the function ukMU with the boundary conditions from (26) we obtain (39). Based on calculations (36–38) we have

T1uktuk2+Lxuk4=ti=1ntvk,ixtψit
+r=1nαritvk,rxtψit+r=1nαritprkxψitexpμr
+j=1ntcijk+r=1nαr,itcr,jk(xt)ψitexpμj
+l=12tyik,lNl+j=1ntzijkNlexpμjbix
i,j=1nψitci,jkxtβj0βjtexpμj
i=1nψitpikxβi0βitexpμi
+l=12i=1nr=1nγirxtbrxyik,lNl+j=1nr=1nγi,r(xt)brxzi,jk,lNlexpμj
i=1nr=1nαr,itprk2xexpμrψit
i=1ntvk2,i+r=1nαr,itvk2,r+j=1ntci,jk2+r=1nαr,itcr,jk2(xt)expμjψit
l=12i=1ntyik,lNl+j=1ntzi,jk,lNlexpμjbix
+Axi=1nx2vk4,i+j=1nx2ci,jk4xt+pik4xexpμjψit
+Axl=12i=1nx2yik,lNl+j=1nzi,jk,lNlexpμjbix.

Providing the condition of Theorem 1, we set

ttVk+ATtVk=tVk2LxVk4xt,E41
ttCk+ATtCkxt+ΛPkx+Ckxt+ΛPkxΛβ0
ΛβtCkxt+ΛPkxE42
=tCk2ATtCk2+ΛPk2x+LxCk4+ΛPk4x,

Eq. (43) is solved without an initial condition and uniquely determines the function Vkxt [27].

Providing the solvability of Eq. (44) we set

CkxtΛβ0ΛβtCkxt¯¯t=0
=tCk2ATtCk2xtATtΛPk2+LxCk4¯¯t=0
ATtΛPk2x¯=tCk2ATtCk2LxCk4xt+ΛPk4x¯

or in coordinate form

cijkxtt=0=1βj0βittcijk2+rjαritcrjk2(xt)qij(xt)t=0,
pik2x=1αiittciik2+αiitciik2qii(xt)t=0,

where qijxt is known function included in Axx2Ck4xt+ΛPk4x.

On the basis of (38), condition (3), Theorem 2 is ensured if arbitrary functions dil,kxtbix,.

ωijk,lxtbix are solutions to the problems

2φi,lxdil,kxtbixx+φi,lxdil,kxtbix=0,
dil,kxtbixx=l1=vkl1tψit,i=1,n¯,
2φi,lxωi,jl,kxtbixx+φi,lxωi,jl,kxtbix=0,
ωi,jl,kxtbixx=l1=cijkl1t+pil1ψjt.

Thus, arbitrary functions dil,kxt,ωijk,lxt,vkixt,cijk,lxt included in (34) are uniquely determined.

Iterative Eq. (32) for k=0,1 is homogeneous; therefore, by Theorem 1, it has a solution ukMU if the functions yil,kNil,zi,jl,kNil are solutions of the equations

τyik,lNil=ξi,l2yik,lNil,τzi,jk,lNil=ξi,l2zi,jk,lNil

for boundary value conditions

yik,lNilt=τ=0=0yik,lNilξi,l=0=dik,lxt,
zi,jk,lNilt=τ=0=0zi,jk,lNilξi,l=0=ωi,jk,lxt.

From this problem, we find

yi0,lNil=di0,lxterfcξi,l2τ,zi,j0,lNil=ωi,j0,lxterfcξi,l2τ.

The functions di0,lxt,ωi,j0,lxt are determined from problems (44) which ensure that condition Lξu0=0 is satisfied. Using calculations (37), the free term of iterative Eq. (32) at k=2 is written as F2M=T1u0M+fxt by Theorem 1, an equation with such a free term is solvable in U, if

ti=1ntv0,ixt+r=1nαrixv0,r(xt)βitv0,i(xt)=fxt
ti,j=1ntcij0xt+r=1nαritcrj0(xt)+αjitpj0x
+i,j=1nβj0βitcij0xt+i=1nβi0βitpi0x=0.E43

From (47) we uniquely determine ci,j0xt=0,ij and the function ci,i0xt is determined from the equation

ttcii0xt+αiitcii0(xt)+βi0βitcii0xt+βi0βitpi0x=0

under the initial condition

cii0x0=v0,ix0pi0x.

The first equation from (47), by virtue of condition (2), has a solution satisfying the condition [27] v0x0<. We calculate the free term of Eq. (32) at k=3:

F3M=T1u1,

which has the same view as T1u0. Providing the solvability of equation T0u3=T1u1 in U, with respect to cij1xt,v1ixt we obtain Eqs. (47).

In the next step, k=4 the free term has the view

F4M=T1u2tu0+Lξu1.

The functions di1,lxt,ωi,j1,lxt entering the u1M provide the condition Lξu1=0. Providing the solvability of the iterative equation at k=4, we set

ttv2ixt+k=1nαkixv2,k(xt)βitv2,ixt=tv0,ixt.

For cij2xt we obtain the same equation of the form (47), but with the right-hand side tci,j0xt+k=1nαk,ixck,j0xt+αj,ixpj0x. Taking off the degeneracy of this equation as t=0, we set pi0x=1αiittcii0+αiixcii0t=0. Further repeating the described process, using Theorems 1 and 2, sequentially determining ukM,k=0,1,,n, we construct a partial sum

uεnM=k=0nεk/2ukM.E44

2.4 Estimate for the remainder

For the remainder term

RεnM=uMεuεnM=uMεk=0n+2εk/2ukM+l=12εn+l2un+l

we get the problem

LεRεnM=εn+12gεnM,
RεnMt=τ=μ=0=Rεnx=l1,ξ=0=0,l=1,2,

where

gM=l=01T1un+1+lεl/2l=03εl/2tun1+l+l=05εl/2Lxun3+ll=01εl/2Lεun+1+l.

From the form (34) of the function U, based on conditions (1)(3) and the form of regularizing variables from (27), we conclude that

gM<C.

In the equation for R, we make the restriction by means of regularizing functions, then, based on (30), we obtain the problem

LεRεnxtε=εn+1/2gxtε,
Rεnt=0=Rεnx=0=Rεnx=1=0.

We divide both sides of this equation by t+ε, while the properties of the matrices A are preserved. Therefore, using Theorem 5.5 of [33], we obtain the estimate

Rεnxtε<cεn+1/2.

Passing to Euclidean norms, like [28], or the same estimate can be obtained using the maximum principle [33], p. 23.

By narrowing this problem by means of regularizing functions. Following [33] and [28], passing to Euclidean norms, we obtain a problem that is limited by the maximum principle

Rεnxtε<cεn+12.E45

Theorem 3: Let conditions (1)(4) be satisfied. Then the restriction of the constructed solution (44) is an asymptotic solution to the problem (26), that is, n=0,1, the estimate (45) holds at ε0.

Proof: Let us rewrite the problem (31)

tuε21t+εAxx2u1t+εDtu=1t+εfxt

here, the expression t+ε does not affect sufficiently small ε the properties of the matrices Ax,Dt for which the conditions of the maximum principle theorem are valid [28], p. 20]. Therefore, on the basis of this theorem, it is easy to establish an estimate (45).

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Written By

Asan Omuraliev and Peil Esengul Kyzy

Reviewed: 03 July 2022 Published: 19 October 2022