On One Evolution Equation of Parabolic Type with ...

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Research Article On One Evolution Equation of Parabolic Type with Fractional Differentiation Operator in S Spaces V. V. Gorodetskiy , R. S. Kolisnyk, and N. M. Shevchuk Chernivtsi National University, Chernivtsi, Ukraine Correspondence should be addressed to V. V. Gorodetskiy; [email protected] Received 25 February 2020; Revised 31 May 2020; Accepted 10 June 2020; Published 18 July 2020 Academic Editor: Mayer Humi Copyright © 2020 V. V. Gorodetskiy et al. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. In the paper, we investigate a nonlocal multipoint by a time problem for the evolution equation with the operator A �(I Δ) ω/2 , Δ �(d 2 /dx 2 ), and ω [1; 2) is a fixed parameter. e operator A is treated as a pseudodifferential operator in a certain space of type S. e solvability of this problem is proved. e representation of the solution is given in the form of a convolution of the fundamental solution with the initial function which is an element of the space of generalized functions of ultradistribution type. e properties of the fundamental solution are investigated. e behavior of the solution at t +(solution stabilization) in the spaces of generalized functions of type Sand the uniform stabilization of the solution to zero on R are studied. 1. Introduction A rather wide class of differential equations with partial derivatives forms linear parabolic and B-parabolic equa- tions, and the theory of which originates from the study of the heat conduction equation. e classical theory of the Cauchy problem and boundary-value problems for such equations and systems of equations is constructed in the works of I.G. Petrovskiy, S.D. Eidelman, S.D. Ivasyshen, M.I. Matiychuk, M.V. Zhitarashu, A. Friedman, S. Teklind, V.O. Solonnikov, and V.V. Krekhivskiy et al. e Cauchy problem with initial data from the spaces of generalized functions of the type of distributions and ultradistributions was studied by G.E. Shilov, B.L. Gurevich, M.L. Gorbachuk, V.I. Gorbachuk, O.I. Kashpirovskiy, Ya.I. Zhytomyrskiy, S.D. Ivasyshen, V.V. Gorodetskiy, and V.A. Litovchenko et al. A formal extension of the class of parabolic type equations is the set of evolution equations with the pseudodifferential operator (PDO), which can be represented as A J 1 σx [a(t, x; σ )J xσ ], x, σ { } R n , t > 0, where a is a function (symbol) that satisfies certain conditions and J(J 1 ) is the direct (inverse) Fourier or Bessel transform. e PDO includes differential operators, fractional differentiation and integration operators, convolution operators, and the Bessel operator B ] �(d 2 /dx 2 )+(2] + 1)x 1 (d/dx), ] > −(1/2), which contains the expression (1/x) in its structure and is formally represented as B ] F 1 B ] [− σ 2 F B ] ], where F B ] is the Bessel integral transformation. If A is a nonnegative self-adjoint operator in a Hilbert space H, then it is known [1] that a continuous on [0,T) function u(t) is continuously differentiable solution of the operator differential equation u(t)+ Au(t)� 0, t (0,T), which refers to abstract equations of parabolic type, if and only if it is given as u(t)� e tA f, f u(0) H. It turns out [1] that all continuously differentiable functions within the interval (0,T) solutions of this equation are described by the same formula where f is an element of the wider than H space H a conjugate to the space H a of analytic vectors of the operator A; the role of A is played by the extension A of the operator A to the space H a , and the boundary value of u(t) at the point 0 exists in the space H a . If A �(I Δ) 1/2 , Δ �(d 2 /dx 2 ), then A is a nonnegative self-adjoint operator in H L 2 (R), since (id/dx) is a self- adjoint in L 2 (R) operator with the domain D(id/dx)� φ L 2 (R): φ L 2 (R) . If E λ , λ R, is the spectral function of the operator (id/dx), then, due to the basic spectral theorem for self-adjoint operators, Hindawi International Journal of Differential Equations Volume 2020, Article ID 1673741, 11 pages https://doi.org/10.1155/2020/1673741

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Research ArticleOn One Evolution Equation of Parabolic Type with FractionalDifferentiation Operator in S Spaces

V V Gorodetskiy R S Kolisnyk and N M Shevchuk

Chernivtsi National University Chernivtsi Ukraine

Correspondence should be addressed to V V Gorodetskiy vgorodetskiychnueduua

Received 25 February 2020 Revised 31 May 2020 Accepted 10 June 2020 Published 18 July 2020

Academic Editor Mayer Humi

Copyright copy 2020 V V Gorodetskiy et al )is is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited

In the paper we investigate a nonlocal multipoint by a time problem for the evolution equation with the operator A (I minus Δ)ω2Δ (d2dx2) and ω isin [1 minus 2) is a fixed parameter )e operator A is treated as a pseudodifferential operator in a certain space oftype S )e solvability of this problem is proved )e representation of the solution is given in the form of a convolution of thefundamental solution with the initial function which is an element of the space of generalized functions of ultradistribution type)e properties of the fundamental solution are investigated )e behavior of the solution at t⟶ +infin (solution stabilization) inthe spaces of generalized functions of type Sprime and the uniform stabilization of the solution to zero on R are studied

1 Introduction

A rather wide class of differential equations with partialderivatives forms linear parabolic and B-parabolic equa-tions and the theory of which originates from the study ofthe heat conduction equation )e classical theory of theCauchy problem and boundary-value problems for suchequations and systems of equations is constructed in theworks of IG Petrovskiy SD Eidelman SD IvasyshenMI Matiychuk MV Zhitarashu A Friedman S TeklindVO Solonnikov and VV Krekhivskiy et al )e Cauchyproblem with initial data from the spaces of generalizedfunctions of the type of distributions and ultradistributionswas studied by GE Shilov BL Gurevich ML GorbachukVI Gorbachuk OI Kashpirovskiy YaI ZhytomyrskiySD Ivasyshen VV Gorodetskiy and VA Litovchenkoet al

A formal extension of the class of parabolic type equationsis the set of evolution equations with the pseudodifferentialoperator (PDO) which can be represented asA Jminus 1

σ⟶x[a(t x σ)Jx⟶σ] x σ sub Rn tgt 0 where a is afunction (symbol) that satisfies certain conditions and J(Jminus 1)

is the direct (inverse) Fourier or Bessel transform )e PDOincludes differential operators fractional differentiation and

integration operators convolution operators and the Besseloperator B] (d2dx2) + (2] + 1)xminus 1(ddx) ]gt minus (12)which contains the expression (1x) in its structure and isformally represented as B] Fminus 1

B][minus σ2FB]

] where FB]is the

Bessel integral transformationIf A is a nonnegative self-adjoint operator in a Hilbert

space H then it is known [1] that a continuous on [0 T)

function u(t) is continuously differentiable solution of theoperator differential equation uprime(t) + Au(t) 0 t isin (0 T)which refers to abstract equations of parabolic type if andonly if it is given as u(t) eminus tAf f u(0) isin H It turns out[1] that all continuously differentiable functions within theinterval (0 T) solutions of this equation are described by thesame formula where f is an element of the wider than H

space Haprime conjugate to the space Ha of analytic vectors of the

operator A the role of A is played by the extension 1113954A of theoperator A to the space Ha

prime and the boundary value of u(t)

at the point 0 exists in the space Haprime

If A (I minus Δ)12 Δ (d2dx2) then A is a nonnegativeself-adjoint operator in H L2(R) since (iddx) is a self-adjoint in L2(R) operator with the domainD(iddx) φ isin L2(R) existφprime isin L2(R)1113864 1113865 If Eλ λ isin R is thespectral function of the operator (iddx) then due to thebasic spectral theorem for self-adjoint operators

HindawiInternational Journal of Differential EquationsVolume 2020 Article ID 1673741 11 pageshttpsdoiorg10115520201673741

Aφ I + iddx

1113888 1113889

2⎛⎝ ⎞⎠

12

φ 1113946+infin

minus infin1 + λ21113872 1113873

12dEλφ (1)

It is known (see for example [2]) that

Eλφ(t) 12π

1113946λ

minus infin1113946

+infin

minus infinφ(τ)e

iστdτ1113966 1113967eminus itσdσ (2)

It follows from this that dEλφ(t) (12π)

F[φ](λ)eminus itλdλ So

Aφ 12π

1113946+infin

minus infin1 + λ21113872 1113873

12F[φ](λ)e

minus itλdλ Fminus 1 1 + λ21113872 1113873

12F[φ]1113876 1113877

(3)

Following [3] we call the operator A the Bessel operatorof fractional differentiation )erefore A can be understoodas a pseudodifferential operator constructed on the functionsymbol (1 + λ2)12 λ isin R )is allows us to interpret thefunction eminus tAf that is a solution of the correspondingCauchy problem as a convolution of the form G(t middot)lowastf

[4 5] where G(t middot) Fminus 1[eminus (1+λ2)12]In this paper we give a similar depiction of the solution

of a nonlocal multipoint time problem for the equation(zu(t x)zt) + Au(t x) 0 (t x) isin (0 +infin) times R when theinitial condition u|t0 f is replaced by the condition1113936

mk0 αkBku(t middot)|ttk

where t0 0 t1 tm1113864 1113865 sub (0 +infin)α0 α1 αm1113864 1113865 sub R and m isin N are fixed and B1 Bm arepseudodifferential operators constructed of smooth symbols(if α0 1 α1 middot middot middot αm 0 B0 I then obviously we havea Cauchy problem) )is condition is interpreted in theclassical or weak sense as if f is a generalized function(generalized element of the operator A) of ultradistributiontype Properties of the fundamental solution of the specifiedmultipoint problem are investigated )e behavior of thesolution at t⟶ +infin (solution stabilization) in the spacesof generalized functions of type Sprime and the uniform stabi-lization of the solution to zero on R are studied

Note that the nonlocal multipoint time problem relatesto nonlocal problems for operator differential equations andpartial differential equations Such problems arise whenmodeling many processes and problems of practice withboundary-value problems for partial differential equationswhen describing correct problems for a particular operatorand constructing a general theory of boundary-valueproblems (see for example [6ndash11])

2 Spaces of Test and Generalized Functions

Gelfand and Shilov introduced in [12] a series of spaceswhich they called the spaces S )ey consist of infinitelydifferentiable on R functions which satisfy certain condi-tions on the decrease at infinity and the growth of deriva-tives )ese conditions are given by the inequalities|xkφ(m)(x)| le ckm x isin R k m sub Z+ where ckm1113864 1113865 is somedouble sequence of positive numbers If there are no re-strictions on elements of the sequence ckm1113864 1113865 then obviouslywe have L Schwartzrsquos space S equiv S(R) of quickly descendingat infinity functions However if the numbers ckm satisfycertain conditions then the corresponding specific spaces

are contained in S and they are called the spaces of S type Letus define some of them

For any α βgt 0 let us put

Sβα(R) equiv S

βα ≔ φ isin S |existcgt 0existAgt 0existBgt 0forall m n sub Z+1113864

forallx isin R xmφ(n)

(x)11138681113868111386811138681113868

11138681113868111386811138681113868 le cAm

Bnm

mαn

nβ1113883

(4)

)e introduced S spaces can also be described as in [12]Sβα consists of those infinitely differentiable on R func-

tions φ(x) that satisfy the following inequalities

φ(n)(x)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cBnn

nβ exp minus a|x|1α

1113872 1113873 n isin Z+ x isin R (5)

with some positive constants c B and a dependent only onthe function φ

If 0lt βlt 1 and αge 1 minus β then Sβα consists of only those

φ isin Cinfin(R) which admit an analytic extension into thecomplex plane C such that|φ(x + iy)| le c exp minus a|x|

1α+ b|y|

1(1minus β)1113872 1113873 cgt 0 agt 0 bgt 0

(6)

)e space S1α consists of functions φ isin Cinfin(R) which canbe analytically extended into some band |Imz|lt δ (depen-dent on φ) of the complex plane so that the estimate

|φ(x + iy)| le c exp minus a|x|1α

1113872 1113873 c agt 0 forally isin R (7)

is carried out)e spaces S

βα α βgt 0 are nontrivial if α + βge 1 and they

form dense sets in L2(R))e topological structure in S

βα is defined as follows )e

symbol SβB

αA A Bgt 0 denotes the set of functions φ isin Sβα that

satisfy the condition

forallAgtAforallBgtB xkφ(m)

(x)11138681113868111386811138681113868

11138681113868111386811138681113868 le cAkB

mk

kαm

mβ k m sub Z+ x isin R

(8)

)is set is transformed into a complete countably normedspace if the norms in it are defined by means of relations

φδρ supxkm

xkφ(m)(x)1113868111386811138681113868

1113868111386811138681113868

(A + δ)k(B + ρ)mkkαmmβ

δ ρ1113864 1113865 sub 11213 1113882 1113883

(9)

)e specified norm system is sometimes replaced by anequivalent norm system

φδρprime supxm

exp a(1 minus δ)|x|1α1113966 1113967 middot φ(m)(x)1113868111386811138681113868

1113868111386811138681113868

(B + ρ)mmmβ

δ ρ1113864 1113865 sub1213 1113882 1113883

a α

eA1α

(10)

If A1 ltA2 B1 ltB2 then SβB1αA1

is continuously embeddedinto S

βB2αA2

and Sβα cupABgt0S

βB

αA that is Sβα is endowed by the

inductive limit topology of the spaces SβB

αA [12] )erefore

2 International Journal of Differential Equations

the convergence of a sequence φ] ]ge 11113864 1113865 sub Sβα to zero in the

space Sβα is the convergence in the topology of some space

SβB

αA to which all the functions φ] belong In other words (see[12]) φ]⟶ 0 in S

βα as ]⟶infin if and only if for every

n isin Z+ the sequence φ(n)] ]ge 1 converges to zero uniformly

on an arbitrary segment [a b] sub R and for some c a Bgt 0independent of ] and the inequality

φ(n)] (x)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cBnn

nβ exp minus a|x|1α

1113966 1113967 x isin R n isin Z+ (11)

holdsIn S

βα the continuous translation operation Tx

φ(ξ)⟶ φ(ξ + x) is defined )is operation is also differ-entiable (even infinitely differentiable [12]) in the sense thatthe limit relation (φ(x + h) minus φ(x))hminus 1⟶ φprime(x) h⟶ 0is true for every function φ isin S

βα in the sense of convergence

in the Sβα-topology In S

βα the continuous differentiation

operator is also defined )e spaces of type S are perfect [12](that is the spaces and all bounded sets of which arecompact) and closely related to the Fourier transformnamely the formula F[S

βα] Sαβ α βgt 0 is correct where

F Sβα1113960 1113961 ≔ ψ ψ(σ) 1113946

Rφ(x)e

iσxdx φ isin Sβα1113882 1113883 (12)

Moreover the operator F Sβα⟶ Sαβ is continuousLet (S

βα)prime denote the space of all linear continuous

functionals on Sβα with weak convergence Since the trans-

lation operator Tx is defined in the space Sβα of test functions

the convolution of a generalized function f isin (Sβα)prime with the

test function φ isin Sβα is given by the following formula

(flowastφ) langf Tminus xφ(middot)rang equiv langf φ(x minus middot)rang φ(ξ) φ(minus ξ)

(13)

It follows from the infinite differentiability property ofthe argument translation operation in S

βα that the convo-

lution flowastφ is a usual infinitely differentiable function onRWe define the Fourier transform of a generalized

function f isin (Sβα)prime by the relation

langF[f] φrang langf F[φ]rang forallφ isin Sαβ (14)

where the operator F (Sβα)prime ⟶ (Sαβ)prime is continuous

Let f isin (Sβα)prime If flowastφ isin S

βα forallφ isin S

βα and the conver-

gence φ]⟶ 0 in the Sβα-topology as ]⟶infin implies that

flowastφ]⟶ 0 in the Sβα-topology as ]⟶infin then the

function f is called a convolutor in Sβα If f isin (S

βα)prime is a

convolutor in Sβα then for an arbitrary function φ isin S

βα the

formula F[flowastφ] F[f]F[φ] is valid where F[f] is amultiplier in S

βα [12]

Recall that the function g isin Cinfin(R) is called a multiplierin S

βα if gφ isin S

βα for an arbitrary function φ isin S

βα and the

mapping φ⟶ gφ is continuous in the space Sβα

3 Nonlocal Multipoint by Time Problem

Consider the function a(σ) (1 + σ2)ω2 σ isin R whereω isin [1 2) is a fixed parameter Obviously the function a(σ)

satisfies the inequality

a(σ) le cε exp ε|σ|ω

1113864 1113865 σ isin R (15)

where cε 2ω2 max 1 (1ε) for an arbitrary εgt 0 Usingdirect calculations and the Stirling formula we find that

Dmσ a(σ)

11138681113868111386811138681113868111386811138681113868 le c0B

m0 m le c1B

m1 m

m m isin N σ isin R (16)

where B0 B0(ω)gt 0 B1 B1(ω)gt 0 and c0 c1 gt 0 It fol-lows from (15) and (16) that a(σ) is a multiplier in S11ωIndeed let φ isin S11ω that is the function φ and its derivativessatisfy the inequality

Dkσφ(σ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cAkk

k exp minus a|σ|ω

1113864 1113865 k isin Z+ σ isin R (17)

with some positive constants c A and a )en using theLeibniz formula for differentiating the product of twofunctions as well as inequalities (15)ndash(17) we find that

(a(σ)φ(σ))(k)

11138681113868111386811138681113868

11138681113868111386811138681113868 le 1113944

k

l0C

lk a

(l)(σ)

11138681113868111386811138681113868

11138681113868111386811138681113868 φ(kminus l)

(σ)11138681113868111386811138681113868

11138681113868111386811138681113868 a(σ)φ(k)(σ)

11138681113868111386811138681113868

11138681113868111386811138681113868+

+ 1113944

k

l1C

lk a

(l)(σ)

11138681113868111386811138681113868

11138681113868111386811138681113868 middot φ(kminus l)(σ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cεcAkk

k exp minus a|σ|ω

1113864 1113865

+ c1c 1113944k

l1C

lkB

l1l

lA

kminus l(k minus l)

kminus l exp minus a|σ|ω

1113864 1113865

(18)

Since εgt 0 is arbitrary we can put ε (a2) )en

(a(σ)φ(σ))(k)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cεcAkk

k exp minusa

2|σ|

ω1113882 1113883

+ c1c 1113944k

l1C

lkB

l1l

lA

kminus l(k minus l)

(kminus l)

times exp minusa

2|σ|

ω1113882 1113883 le c2B

k2k

k exp minusa

2|σ|

ω1113882 1113883

σ isin R k isin Z+

(19)

where c2 c(cε + c1) and B2 2max A B11113864 1113865 It follows from(19) that a middot φ is an element of S11ω

)e operation of multiplying by a(σ) is continuous inthe space S11ω In fact let φn nge 11113864 1113865 be a sequence offunctions from S11ω convergent to 0 in this space)is meansthat φn nge 11113864 1113865 sub S

1B01ωA0

with some constants A0 B0 gt 0 and

φn

δρ sup

exp a(1 minus δ)|σ|ω middot φ(k)n (σ)

11138681113868111386811138681113868111386811138681113868

B0 + ρ( 1113857kkk

⟶ 0

n⟶infin

δ ρ1113864 1113865 sub1213 1113882 1113883

a 1

ωeAω0

(20)

In other words for an arbitrary 1113957εgt 0 there exists anumber n0 n0(1113957ε) such that for nge n0

φ(k)n (σ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le 1113957ε B0 + ρ( 1113857kk

k exp minus a(1 minus δ)|σ|ω

1113864 1113865 σ isin R k isin Z+

(21)

International Journal of Differential Equations 3

Using inequalities (15) and (16) (when putting in (15)ε (a2)(1 minus δ)) we get

a(σ)φn(σ)( 1113857(k)

11138681113868111386811138681113868

11138681113868111386811138681113868 le c31113957ε(1113957B + ρ)k exp minus

a

2(1 minus δ)|σ|

ω1113882 1113883

k isin Z+ nge n0

(22)

where c3 cε + c1 and 1113957B 2max B0 B11113864 1113865 It follows from thelast inequality that aφnδρ le c31113957ε forallnge n0 that is the se-

quence aφn nge 11113864 1113865 converges to zero in the space S11113957B1ω1113957A

where 1113957A 21ωA0 and 1113957B 2max B0 B11113864 1113865 )is means thatthe sequence aφn nge 11113864 1113865 converges to zero in the space S11ωwhich is what we needed to prove

Remark 1 It follows from the proven property that thefunction a(σ) (1 + σ2)ω2 σ isin R is also a multiplier inevery space S

β1ω where βgt 1 )erefore in the space S1ωβ

βge 1 defined as a continuous linear pseudodifferentialoperator A constructed by the function a(σ)

Aφ Fminus 1

[a(σ)F[φ]] forallφ isin S1ωβ (23)

where Aφ (I minus Δ)ω2φ Δ (d2dx2)Let us consider the evolution equation with the operator

A (the Bessel fractional differentiation operator)zu

zt+ Au 0 (t x) isin (0 +infin) times R equiv Ω (24)

By a solution of equation (24) we mean the functionu(t x) (t x) isin Ω such that (1) u(t middot) isin C1(0 +infin) forevery x isin R (2) u(middot x) isin S1ωβ for every t isin (0 +infin) (3)u(t x) is continuous at every point (0 x) of the boundaryΓΩ 0 times R of the region Ω (4) existMR⟶ [0infin) forallt isin (0 +infin) |ut(t x)| le M(x)intRM(x)dxltinfin and (5) u(t x) (t x) isin Ω satisfiesequation (24)

For equation (24) we consider the nonlocal multipointby time problem of finding a solution of equation (24) thatsatisfies the condition

μu(0 x) minus μ1B1u t1 x( 1113857 minus middot middot middot minus μmBmu tm x( 1113857 f(x)

x isin R f isin S1ωβ

(25)

where u(0 x) ≔ limt⟶+0u(t x) x isin R m isin Nμ μ1 μm1113864 1113865 sub (0infin) and t1 tm1113864 1113865 sub (0infin) are fixednumbers 0lt t1 lt t2 lt middot middot middot lt tm lt +infin μgt 1113936

mk1 μk and

B1 Bm are pseudodifferential operators in S1ωβ con-structed by functions (characters) gk R⟶ (0 +infin) re-spectively Bk Fminus 1[gk(σ)F] k isin 1 m )e functionsgk isin Cinfin(R) and k isin 1 m satisfy the conditions

forallεgt 0forallσ isin R gk(σ) le exp ε|σ|ω

1113864 1113865

existLk gt 0foralls isin N Dsσgk(σ)

11138681113868111386811138681113868111386811138681113868 le L

sks

s

(26)

Note that the above properties of the functions gk implythat gk k isin 1 m is a multiplier in S1ωβ

We are looking for the solution of problems (24) and(25) via the Fourier transform Due to condition (19)

F ut1113858 1113859 1113946R

ut(t x)eiσxdx

z

zt1113946R

u(t x)eiσxdx

z

ztF[u]

(27)

We introduce the notation F[u(t x)] v(t σ) Giventhe form of the operators A B1 Bm we get

F[Au(t x)] F Fminus 1

[a(σ)F[u]]1113960 1113961 a(σ)F[u(t x)] a(σ)v(t σ)

F Bku(t x)1113858 1113859 gk(σ)F[u(t x)] gk(σ)v(t σ) k isin 1 m

(28)

So for the function v Ω⟶ R we arrive at a problemwith parameter σ

dv(t σ)

dt+ a(σ)v(t σ) 0 (t σ) isin Ω (29)

μv(0 σ) minus 1113944

m

k1μkgk(σ)v tk σ( 1113857 1113957f(σ) σ isin R (30)

where 1113957f(σ) F[f](σ) )e general solution of equation(29) has the form

v(t σ) c exp minus ta(σ) (t σ) isin Ω (31)

where c c(σ) is determined by condition (30) Substituting(30) into (31) we find that

c 1113957f(σ) μ minus 1113944m

k1μkgk(σ)exp minus tka(σ)1113864 1113865⎛⎝ ⎞⎠

minus 1

σ isin R

(32)

Now put G(t x) ≔ Fminus 1[Q(t σ)] andQ(t σ) Q1(t σ)Q2(σ) and thus

Q1(t σ) exp minus ta(σ)

Q2(σ) μ minus 1113944m

k1μkgk(σ)Q1 tk σ( 1113857⎛⎝ ⎞⎠

minus 1

(33)

)en thinking formally we come to the relation

u(t x) 1113946R

G(t x minus ξ)f(ξ)dξ G(t x)lowastf(x) (34)

Indeed

u(t x) (2π)minus 1

1113946R

Q(t σ) 1113946R

f(ξ)eiσξdξ1113874 1113875e

minus iσxdσ

1113946R

(2π)minus 1

1113946R

Q(t σ)eminus iσ(xminus ξ)dσ1113874 1113875f(ξ)dξ

1113946R

G(t x minus ξ)f(ξ)dξ G(t x)lowastf(x) (t x) isin Ω

(35)

)e correctness of the transformations performed theconvergence of the corresponding integrals and conse-quently the correctness of formula (35) follow from theproperties of the function G which are given below )eproperties of G are determined by the properties of Q

4 International Journal of Differential Equations

because G Fminus 1[Q] So let us first examine the properties ofthe function Q(t σ) as a function of the variable σ

Lemma 1 For derivatives of Q1(t σ) (t σ) isin Ω theestimates

DsσQ1(t σ)

11138681113868111386811138681113868111386811138681113868 le cA

stcs

ss exp minus t|σ|

ω1113864 1113865 (36)

are valid where c 0 if 0lt t le 1 and c 1 if tgt 1 and theconstants cgt 1 and Agt 0 do not depend on t

Proof To prove the statement we use the Faa di Brunoformula for differentiation of a complex function

DsσF(g(σ)) 1113944

s

p1

dp

dgpF(g) 1113944

s

p1 pl

ddσ

g(σ)1113888 1113889

p1

1l

dl

dσlg(σ)1113888 1113889

pl

s isin N

(37)

where the sum sign is applied to all the solutions in positiveintegers of the equation p1 + 2p2 + middot middot middot + lpl sp1 + middot middot middot + pl m Put F eg and g minus ta(σ) )en

Dsσe

minus ta(σ) e

minus ta(σ)1113944

s

p11113944

s

p1 plΛ (38)

where the symbol Λ denotes the expression and

Λ ≔ddσ

(minus ta(σ))1113888 1113889

p1

middot12

d2

dσ2(minus ta(σ))1113888 1113889

p2

1l

dl

dσl(minus ta(σ))1113888 1113889

pl

(39)

Given estimate (16) is fulfilled we find that

|Λ| le cp1+middotmiddotmiddot+pl

0 Bp1+2p2+middotmiddotmiddot+lpl

0 tp1+middotmiddotmiddot+pl le 1113957c

s0t

pB

s0 1113957c0 max 1 c01113864 1113865

(40)

Using (40) and the Stirling formula we arrive at theinequalities

DsσQ1(t σ)

11138681113868111386811138681113868111386811138681113868 le 1113957c

s0B

s0t

css exp minus ta(σ)

le cAstcs

ss exp minus t|σ|

ω1113864 1113865 σ isin R

(41)

where c 0 if 0lt t le 1 and c 1 if tgt 1 and the values cgt 1and Agt 0 do not depend on t

Lemma is proved

Remark 2 It follows from estimate (41) that Q1(t middot) isin S11ωfor every tgt 0

Lemma 2 -e function Q2 is a multiplier in S21ω

Proof To prove the assertion let us estimate the derivativesof Q2 For this we use formula (37) in which we put F φminus 1

and φ R where

R(σ) μ minus 1113944m

k1μkgk(σ)exp minus tka(σ)1113864 1113865 (42)

)en Q2 F(φ) Rminus 1 and

DsσQ2(σ)

11138681113868111386811138681113868111386811138681113868 1113944

s

p1

dp

dRpR

minus 11113944

s

p1 pl

ddσ

R(σ)1113888 1113889

p11113868111386811138681113868111386811138681113868111386811138681113868

1l

dl

dσlR(σ)1113888 1113889

pl

s isin N

(43)

Given the properties of g1 gm and inequality (41)we find that

1l

dj

dσjR(σ)

11138681113868111386811138681113868111386811138681113868

11138681113868111386811138681113868111386811138681113868le

1j

1113944

m

k1μk 1113944

j

i0C

ij D

iσgk(σ)

11138681113868111386811138681113868

11138681113868111386811138681113868 middot Djminus iσ e

minus tka(σ)11138681113868111386811138681113868

11138681113868111386811138681113868

le1j

1113944

m

k1μk 1113944

i

j0C

ijL

iki1113957c

jminus i0 t

c(jminus i)

k (j minus i)eminus tka(σ)

le 1113944m

k1μk 1113944

j

i0C

ijL

ik1113957c

jminus i0 t

c(jminus i)

(44)

where we took into account that i(j minus i) le j)Let

1113957L max L1 Lm1113864 1113865

L0 2max 1113957L 1113957c0T1113864 1113865(45)

then

1j

dj

dσjR(σ)

11138681113868111386811138681113868111386811138681113868

11138681113868111386811138681113868111386811138681113868le α0L

j0 α0 1113944

m

k1μk j isin 1 l

ddσ

R(σ)1113888 1113889

p1111386811138681113868111386811138681113868111386811138681113868

111386811138681113868111386811138681113868111386811138681113868

1l

dl

dσlR(σ)1113888 1113889

pl111386811138681113868111386811138681113868111386811138681113868

111386811138681113868111386811138681113868111386811138681113868le α0L0( 1113857

p α0L201113872 1113873

p2 α0L

l01113872 1113873

pl

αp1+middotmiddotmiddot+pl

0 Lp1+2p2+middotmiddotmiddot+lpl

0 αp0L

s0

(46)

In addition (dpdRp)Rminus 1 (minus 1)ppRminus (p+1) and

Rminus 1

(σ) Q2(σ) μ minus 1113944m

k1μkgk(σ)exp minus tka(σ)1113864 1113865⎛⎝ ⎞⎠

minus 1

le μ minus 1113944m

k1μk exp minus t1|σ|

ω+ ε|σ|

ω1113864 1113865⎛⎝ le μ minus 1113944

m

k1μk

⎛⎝ ⎞⎠

minus 1

equiv β0 gt 0 ε t1

(47)

Since by assumption μgt 1113936mk1 μk (assuming the prop-

erties of g1 gm and 0lt t1 lt t2 lt middot middot middot lt tm) So

International Journal of Differential Equations 5

dp

dRpR

minus 111138681113868111386811138681113868111386811138681113868

11138681113868111386811138681113868111386811138681113868le βp+1

0 p

DsσQ2(σ)

11138681113868111386811138681113868111386811138681113868 le s 1113944

s

p1βp+10 pαp

0Ls0

le β0Ls0(s)

21113944

s

p1βp0 le β0L

s0s middot (s)

2βs1 le β2β

s3s

2s s isin N

(48)

It follows from the last inequality and boundedness ofthe function Q2(σ) on R that Q2 is a multiplier in S21ω

Corollary 1 For every tgt 0 the functionQ(t σ) Q1(t σ)Q2(σ) σ isin R is an element of the spaceS21ω and the estimates

DsσQ(t σ)

11138681113868111386811138681113868111386811138681113868 le 1113957c1113957A

stcs

s2s exp minus t|σ|

ω1113864 1113865 s isin Z+ (t σ) isin Ω

(49)

are valid where the constants 1113957c and 1113957Agt 0 do not depend on t

Taking into account the properties of the Fouriertransform (direct and inverse) and the formulaFminus 1[S21ω] S1ω2 we get that G(t middot) isin S1ω2 for every tgt 0 Weremove in the estimates of derivatives of the function G (inthe variable x) the dependence on t assuming tgt 1 To dothis we use the relations

xkD

sxF[φ](x) i

k+sF σsφ(σ)( 1113857

(k)1113876 1113877

ik+s

1113946R

σsφ(σ)( 1113857(k)

eiσxdσ k s sub Z+ φ isin S

21ω

(50)

So

xkD

sxG(t x) (2π)

minus 1ik+s

(minus 1)s1113946R

σsQ(t minus σ)( 1113857

(k)e

iσxdσ

(51)

Applying the Leibniz formula for differentiating theproduct of two functions and estimating the derivatives ofthe function Q(t σ) we find that

σsQ(t minus σ)( 1113857

(k)11138681113868111386811138681113868

11138681113868111386811138681113868 1113944k

p0C

p

k σs( 1113857

(p)Q

(kminus p)(t minus σ)

1113868111386811138681113868111386811138681113868111386811138681113868

1113868111386811138681113868111386811138681113868111386811138681113868

le σsQ

(k)(t minus σ)

11138681113868111386811138681113868

11138681113868111386811138681113868 + ks σsminus 1Q

(kminus 1)(t minus σ)

11138681113868111386811138681113868

11138681113868111386811138681113868

+k(k minus 1)

2s(s minus 1) σsminus 2

Q(kminus 2)

(t minus σ)11138681113868111386811138681113868

11138681113868111386811138681113868

+ middot middot middot le 1113957c 1113957Aktck 1113957B

stminus (sω)

mks + ks1113957Akminus 1

tc(kminus 1)

1113876

middot 1113957Bsminus 1

tminus ((sminus 1)ω)

mkminus 1sminus 1

+12

k(k minus 1)s(s minus 1)1113957Akminus 2

tc(kminus 2) 1113957B

sminus 2tminus ((sminus 2)ω)

middot mkminus 2sminus 2 + 1113961eminus (t2)|σ|ω

(52)

where mks k2kssω Taking into account the results in (see[12] p 236ndash243) we find that this double sequence satisfiesthe inequality

ksmkminus 1sminus 1

mks

le 1113957c(k + s) 1113957cgt 0 (53)

Bearing in mind the last inequality and also that tgt 1 weobtain

σsQ(t minus σ)( 1113857

(k)11138681113868111386811138681113868

11138681113868111386811138681113868 le 1113957c1113957Aktk 1113957B

smks 1 +

ks

1113957A1113957B

mkminus 1sminus 1

mks

+12

k(k minus 1)s(s minus 1)1

1113957A21113957B

2mkminus 2sminus 2

mks

+ middot middot middot1113888 1113889eminus (t2)|σ|ω le 1113957c1113957A

ktk 1113957B

smks

times 1 +ks

1113957A1113957B

mkminus 1sminus 1

mks

+12

11113957A21113957B

2 ksmkminus 1sminus 1

mks

(k minus 1)(s minus 1)mkminus 2sminus 2

mkminus 1sminus 1+ middot middot middot1113888 1113889e

minus (t2)|σ|ω

le 1113957c1113957Aktk 1113957B

smks 1 +

1113957c

1113957A1113957B(k + s) +

12

1113957c2

(1113957A1113957B)2(k + s)

2+ middot middot middot1113888 1113889e

minus (t2)|σ|ω

le c1AktkB

smkse

minus (t2)|σ|ω

A 1113957A exp1113957c

1113957A1113957B1113882 1113883

B 1113957B exp1113957c

1113957A1113957B1113882 1113883

(54)

6 International Journal of Differential Equations

So

xkD

sxG(t x)

11138681113868111386811138681113868

11138681113868111386811138681113868 le (2π)minus 1

c1AktkB

smks1113946

Re

minus (t2)|σ|ωdσ

c2AktkB

stminus (1ω)

k2k

ssω

k s sub Z+

(55)

)en

DsxG(t x)

11138681113868111386811138681113868111386811138681113868 le c2B

stminus (1ω)

ssω inf

k

Akk2k

tminus 1|x|( )k

le c3Bstminus (1ω)

ssω exp minus a0t

minus (12)|x|

121113966 1113967

(56)

where the constants c3 B and a0 gt 0 do not depend on there we used the well-known inequality (see [12] p 204)

eminus (αe)|ξ|1α le inf

k

Lkkkα

|ξ|kle ce

minus (αe)|ξ|1α c e

αe2 (57)

in which α 2 and L 1113957A )us such a statement is correct

Lemma 3 -e derivatives (by variable x) of the functionG(t x) for tgt 1 satisfy the inequality

DsxG(t x)

11138681113868111386811138681113868111386811138681113868 le c3B

ss

sωtminus (1ω) exp minus a0t

minus (12)|x|

121113966 1113967 s isin Z+

(58)

where the constants c3 B and a0 gt 0 do not depend on t

Remark 3 In what follows we assume that in condition(25) β 2

Here are a few more properties of the function G(t x)

Lemma 4 -e function G(t middot) t isin (0 +infin) as an abstractfunction of t with values in the space S1ω2 is differentiable in t

Proof It follows from the continuity of the Fourier trans-form (direct and inverse) that to prove the statement itsuffices to establish that the function F[G(t middot)] Q(t middot) as afunction of a parameter t with values in the space S21ω isdifferentiable in t In other words it is necessary to provethat the boundary value relation

ΦΔt(σ) ≔1Δt

[Q(t + Δt middot) minus Q(t middot)]⟶z

ztQ(t middot) Δt⟶ 0

(59)

is performed in the sense as follows

(1) DsσΦΔt(σ) ⟶

Δt⟶ 0Ds

σ(minus a(σ)Q(t σ)) s isin Z+ uni-formly on each segment [a b] sub R

(2) |DsσΦΔt(σ)| le cB

ss2s exp minus a|σ|ω s isin Z+ where the

constants c B and agt 0 do not depend on Δt if Δt issmall enough

)e function Q(t σ) (t σ) isin Ω is differentiable in t inthe usual sense Due to the Lagrange theorem on finiteincrements

ΦΔt(σ) minus a(σ)Q(t + θΔt σ) (60)

So

DsσΦΔt(σ) minus 1113944

s

l0C

lsD

lσa(σ)D

sminus lσ Q(t +θΔtσ) (61)

Dsσ ΦΔt(σ) minus

z

ztQ(tσ)1113888 1113889 minus 1113944

s

l0C

lsD

lσa(σ) D

sminus lσ Q(t +θΔtσ)1113960

minus Dsminus lσ Q(tσ)1113961

(62)

Since

Dsminus lσ Q(t + θΔt σ) minus D

sminus lσ Q(t σ) D

sminus l+1σ Q t + θ1Δt σ( 1113857

0lt θ1 lt 1

(63)

it follows from this and estimate (49) that

Dsminus l+1σ Q t + θ1Δt σ( 1113857θΔt⟶ 0 Δt⟶ 0 (64)

uniformly on an arbitrary segment [a b] sub R )en

DsσΦΔt(σ)⟶ D

z

ztQ(t σ)1113888 1113889 Δt⟶ 0 (65)

uniformly on each segment [a b] sub R too )ereforecondition 1 is satisfied

Taking into account (61) estimates (15) (16) and (49)for a(σ) Q(t σ) and their derivatives we find that

DsσΦΔt(σ)

11138681113868111386811138681113868111386811138681113868 le c1cε 1113944

s

l0C

lsB

l1l

l 1113957Asminus l

tc(sminus l)

(s minus l)2(sminus l)

exp minus (t + θΔt)|σ|ω

+ ε|σ|ω

1113864 1113865

(66)

where εgt 0 is arbitrary fixed Take ε (t2) and then

DsσΦΔt(σ)

11138681113868111386811138681113868111386811138681113868 le cB

ss2s exp minus a|σ|

ω1113864 1113865 s isin Z+ (67)

where B 2max B11113957Atc1113966 1113967 a (t2) and all the constants

do not depend on Δt

Corollary 2 -e formulaz

zt(flowastG(t middot)) f lowast

zG(t middot)

zt forallf isin S

1ω21113872 1113873prime tgt 0

(68)

is correct

Proof According to the definition of a convolution of ageneralized function with a test one we have

flowastG(t middot) langfξ Tminus xG(t ξ)rang

G(t ξ) G(t minus ξ)(69)

)enz

zt(flowastG(t middot)) lim

Δt⟶0

1Δt

[(flowastG(t + Δt middot)) minus (flowastG(t middot))]

limΔt⟶0langfξ

1Δt

Tminus xG(t + Δt ξ) minus Tminus x

G(t ξ)1113960 1113961rang

(70)

International Journal of Differential Equations 7

Due to Lemma 4 the relation1Δt

Tminus xG(t + Δt ξ) minus Tminus x

G(t ξ)1113960 1113961⟶z

ztTminus x

G(t ξ) Δt⟶ 0

(71)

is performed in the sense of S1ω2 topology So

z

zt(flowastG(t middot)) langfξ lim

Δt⟶0

1Δt

Tminus xG(t + Δt ξ) minus Tminus x

G(t ξ)1113960 1113961rang

langfξ z

ztTminus x

G(t ξ)rang langfξ Tminus x

z

ztG(t ξ)rang

flowastz

ztG(t middot)

(72)

)e statement is provedSince

F[G(t middot)] Q(t σ) Q1(t σ)Q2(σ) 1113946R

G(t x)eiσxdx tgt 0

(73)

we get the formula

1113946R

G(t x)dx Q1(t 0)Q2(0) λ0eminus t

λ0 μ minus 1113944m

k1μkgk(0)e

minus tk⎛⎝ ⎞⎠

minus 1

(74)

where a(0) 1 is taken into account

Lemma 5 In (S1ω2 )prime the following relations are correct

(1) G(t middot)⟶ Fminus 1

Q21113858 1113859 t⟶ + 0

(2) μG(t middot) minus 1113944m

k1μkBkG tk middot( 1113857⟶ δ t⟶ + 0

(75)

where δ is the Dirac delta function

Proof (1) Taking into account the continuity property of theFourier transform (direct and inverse) in spaces of Sprime type itis sufficient to establish that

F[G(t middot)] Q1(t middot)Q2(middot)⟶ Q2(middot) t⟶ + 0 (76)

in the space (S21ω)prime To do this we take an arbitrary functionφ isin S21ω and using the fact thatQ2 is a multiplier in the spaceS21ω and the Lebesgue theorem on the limit passage underthe Lebesgue integral sign we find that

langQ1(t middot)Q2(middot)φrang langQ1(t middot) Q2(middot)φ(middot)rang

1113946R

Q1(t σ)Q2(σ)φ(σ)dσ ⟶t⟶+0

1113946R

Q2(σ)φ(σ)dσ

lang1 Q2(middot)φ(middot)rang langQ2φrang

(77)

)is leads us to statement (1) of Lemma 5(2) Given statement (1) and the form of the operators

B1 Bm we find that

μG(t middot) minus 1113944m

l1μlBlG tl middot( 1113857 ⟶

t⟶+0μF

minus 1Q21113858 1113859 minus 1113944

m

l1μlBlG tl middot( 1113857

μFminus 1

Q21113858 1113859 minus 1113944m

l1μlF

minus 1gl(middot)Q1 tl middot( 1113857Q2(middot)1113858 1113859

Fminus 1 μQ2(middot) minus 1113944

m

l1μlgl(middot⎡⎣ ⎞⎠Q1 tl middot( 1113857Q2(middot)]

Fminus 1 μ minus 1113944

m

l1μlgl(middot)Q1 tl middot( 1113857Q2(middot)⎡⎣ ⎤⎦

Fminus 1 μ minus 1113944

m

l1μlgl(middot)Q1 tl middot( 1113857⎛⎝ ⎞⎠⎡⎢⎢⎣

middot μ minus 1113944m

l1μlgl(middot)Q1 tl middot( 1113857⎛⎝ ⎞⎠

minus 1⎤⎥⎥⎥⎥⎦ F

minus 1[1] δ

(78)

)erefore relation (75) is fulfilled in the space (S1ω2 )primeLemma 5 is proved

Remark 4 If μ 1 μ1 middot middot middot μm 0 then problems (24)and (25) degenerate into the Cauchy problem for equation(24) In this case Q2(σ) 1 forallσ isin R G(t x) Fminus 1[eminus ta(σ)]and G(t middot)⟶ Fminus 1[1] δ for t⟶ +0 in the space (S1ω1 )prime

Corollary 3 Let

ω(t x) flowastG(t x) f isin S1ω2lowast1113872 1113873prime (t x) isin Ω (79)

where (S1ω2lowast )prime is a class of convolutors in S1ω2 -en in (S1ω2 )primethe following boundary relation holds

μω(t middot) minus 1113944m

k1μkBkω tk middot( 1113857⟶ f t⟶ + 0 (80)

Proof Let us prove that the relation

F μω(t middot) minus 1113944

m

k1μkBkω tk middot( 1113857⎡⎣ ⎤⎦⟶ F[f] t⟶ + 0

(81)

takes place in the space (S21ω)prime Since f isin (S1ω2lowast )prime andG(t middot) isin S1ω2 for every tgt 0 we have

F[ω(t middot)] F[flowastG(t middot)] F[f]F[G(t middot)] F[f]Q(t middot)

(82)

So we need to prove that

F[f] μQ(t middot) minus 1113944m

k1μkgk(middot)Q tk middot( 1113857⎛⎝ ⎞⎠⟶ F[f] (83)

in (S21ω)prime when t⟶ + 0 SinceQ(t middot) Q1(t middot)Q2(middot)⟶ Q2(middot) for t⟶ + 0 in the space(S21ω)prime (see the proof of assertion 1 and Lemma 5) thecorrelation

8 International Journal of Differential Equations

μQ(t middot) minus 1113944m

k1μkgk(middot)Q tk middot( 1113857 ⟶

t⟶+0μQ2(middot) minus 1113944

m

k1μkgk(middot)Q1 tk middot( 1113857Q2(middot)

μ minus 1113944m

k1μkgk(middot)Q1 tk middot( 1113857⎛⎝ ⎞⎠Q2(middot) 1

(84)

is realized in the space (S21ω)prime )us relation (81) and (80)are fulfilled in the corresponding spaces )e statement isproved

)e function G(t middot) satisfies equation (24) as tgt 0Indeed

z

ztG(t x)

z

ztF

minus 1[Q(t σ)] F

minus 1 z

ztQ(t σ)1113890 1113891

Fminus 1

[a(σ)Q(t σ)]

AG(t x) Fminus 1

a(σ)Fminus 1

[G(t middot)]1113960 1113961 Fminus 1

[a(σ)Q(t σ)]

(85)

So

zG(t x)

zt+ AG(t x) 0 (t x) isin Ω (86)

which is what had to be provedIt follows from Corollary 3 that the nonlocal multipoint

by the time problem for equation (24) can be formulated inthe following way find a function u(t x) (t x) isin Ω thatsatisfies equation (24) and the condition

μ limt⟶+0

u(t middot) minus 1113944

m

k1μkBku tk middot( 1113857 f f isin S

1ω2lowast1113872 1113873prime (87)

where boundary relation (87) is considered in the space(S1ω2 )prime and the constraints on the parameters μ μ1 μmt1 tm are the same as in case of problems (24) and(25))

Theorem 1 By nonlocal multipoint by time problem (24)(87) is solvable and its solution is given by the formula

u(t x) flowastG(t x) (t x) isin Ω (88)

where u(t middot) isin S1ω2 for every tgt 0

Proof Make sure that the function u(t x) (t x) isin Ω sat-isfies equation (24) Indeed (see Corollary 2)

zu(t x)

zt

z

zt(flowastG(t middot)) flowast

zG(t middot)

zt

Au(t x) Fminus 1

[a(σ)F[f lowastG(t middot)]]

(89)

Since f is a convolutor in S1ω2

F[flowastG(t middot)] F[f]F[G(t middot)] F[f]Q(t middot) (90)

So

Au(t x) Fminus 1

[a(σ)Q(t σ)F[f](σ)] minus Fz

ztQ(t middot)F[f]1113890 1113891

minus F Fz

ztG(t middot)1113890 1113891 middot F[f]1113890 1113891

minus Fminus 1

F flowastzG(t middot)

zt1113890 11138911113890 1113891 minus flowast

zG(t middot)

zt

(91)

It follows from this that the function u(t x) (t x) isin Ωsatisfies equation (24) From Corollary 3 we get that u(t x)

satisfies condition (87) in the defined sense )e theorem isproved

Remark 5 If in condition (87) B1 middot middot middot Bm I (I is theidentity operator) then we can prove that problems (24) and(87) are well posed and its solution is given by the formula

u(t x) flowastG(t x) f isin S1ω1lowast1113872 1113873prime (t x) isin Ω (92)

where G(t middot) Fminus 1[Q(t middot)] isin S1ω1 for every tgt 0 and

Q(t σ) eminus ta(σ) μ minus 1113944

m

k1μke

minus tka(σ)⎛⎝ ⎞⎠

minus 1

(93)

Theorem 2 Suppose u(t x) (t x) isin Ω is the solution ofproblems (24) and (87) -en u(t middot)⟶ 0 in the space(S1ω2 )prime as t⟶ +infin

Proof Recall that the solution of problems (24) and (87) isgiven by the formula

u(t x) flowastG(t x) langfξ G(t x minus ξ)rang

f isin S1ω2lowast1113872 1113873prime (t x) isin Ω

(94)

Let us prove that langu(t middot)φrang⟶ 0 as t⟶ +infin for anarbitrary function φ isin S1ω2 Put

Ψt(ξ) 1113946+infin

minus infinG(t x minus ξ)φ(x)dx

ΨtR(ξ) 1113946R

minus RG(t x minus ξ)φ(x)dx Rgt 0 tgt 1

(95)

In these notations we prove that (a) for every tgt 1 andRgt 0 the function ΨtR(ξ) belongs to the space S1ω2 andΨtR(ξ)⟶Ψt(ξ) in the space S1ω2 as R⟶infin (b)Ψt(ξ) isin S1ω2 for every tgt 1 From this we get

langu(t middot)φrang 1113946+infin

minus infinlangfξ G(t x minus ξ)rangφ(x)dx

langfξ 1113946+infin

minus infinG(t x)φ(x + ξ)dxrang

langfξ 1113946+infin

minus infinG(t minus y)φ(minus (y minus ξ))dyrang

langfξ 1113946+infin

minus infinG(t minus y)φ(y minus ξ)dyrang φ(z) φ(minus z)

(96)

International Journal of Differential Equations 9

where u(t middot) is interpreted as a regular generalized functionfrom (S1ω2 )prime for every tgt 0

So let us set property (a) For fixed k m sub Z+ we have

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le 1113946+R

minus Rξkφ(x)D

mξ G(t x minus ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868dx

le 1113946+infin

minus infinξkφ(ξ + η)D

mη G(t η)

11138681113868111386811138681113868

11138681113868111386811138681113868dη

(97)

Since φ isin S1ω2 the inequality

ξkD

mξ φ(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cLkM

mk2k

mmω

k m sub Z+ (98)

with some c L Mgt 0 is valid It follows from this that forevery η isin R

supξisinR

ξkφ(ξ)11138681113868111386811138681113868

11138681113868111386811138681113868 supyisinR

(y minus η)kφ(η)

11138681113868111386811138681113868

11138681113868111386811138681113868 supyisinR

1113944

k

l0C

lky

l(minus η)

kminus lφ(y)

1113868111386811138681113868111386811138681113868111386811138681113868

1113868111386811138681113868111386811138681113868111386811138681113868

le 1113944k

l0C

lk|η|

kminus l supyisinR

ykφ(y)

11138681113868111386811138681113868

11138681113868111386811138681113868 le c 1113944k

l0C

lkL

ll2l

|η|kminus l

(99)

Next we will use estimate (58) )en

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le c 1113944k

l0C

lkL

ll2l

1113946+infin

minus infin|η|

kminus lD

mη G(t η)

11138681113868111386811138681113868

11138681113868111386811138681113868dη

le cc3Bm

mmω

tminus (1ω)

1113944

k

l0C

lkL

ll2l

middot 1113946+infin

minus infin|η|

kminus l exp minus a0tminus (12)

|η|12

1113966 1113967dη

(100)

By the direct calculations we find that

1113946+infin

minus infin|η|

kminus l exp minus a0tminus (12)

|η|12

1113966 1113967dη le c41113957L

kk2(kminus l)

tkminus l+1

c41113957Lgt 0

(101)

So

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cc3c4Bm

t1minus (1ω)

mmω

1113944

k

l0C

lkL

l1113957Lkminus l

tkminus l

l2l

k2(kminus l)

le cBm

Lkk2k

mmω

(102)

where c cc3c4t1minus (1ω) and L 2max L 1113957Lt1113864 1113865 )erefore

ΨtR(ξ) isin S1ω2 for every tgt 1 and an arbitrary Rgt 0 Fur-thermore we make sure thatΨtR(ξ)⟶Ψt(ξ) as R⟶infin

with all its derivatives uniformly in ξ on each segment[a b] sub R In addition the set of functions ξk

Dmξ ΨtR(ξ)1113966 1113967

k m sub Z+ is uniformly bounded (by R) in the space S1ω2(this property follows from estimate (102) in which c B andLgt 0 do not depend on R) )is means that condition (a) isfulfilled

Condition (b) follows from (a) because in a perfectspace every bounded set is compact

Using properties (a) and (b) we obtain the equality

langu(t middot)φrang 1113946+infin

minus infinG(t minus y)(flowastφ)(y)dy (103)

Since f isin (S1ω2 )prime is a convolutor in S1ω2 we haveflowast φ isin S1ω2 )en given estimate (58) (at s 0) we find that

|langu(t middot)φrang| le 1113946+infin

minus infin|G(t minus y)||(flowastφ)(y)|dy

lec

t1ω1113946

+infin

minus infin|(flowastφ)(y)|dy le

c0

t1ω⟶ 0 t⟶ +infin

(104)

for an arbitrary function φ isin S1ω2 ie u(t middot)⟶ 0 in thespace (S1ω2 )prime as t⟶ +infin )e theorem is proved

If the generalized function f in (87) is finite (ie itssupport suppf is a finite set in R) then we can say onuniform tending a solution u(t x) of problems (24) and (87)to zero onR as t⟶ +infin Note that every finite generalizedfunction is a convolutor in S spaces )is property followsfrom the general result concerning the theory of perfectspaces (see [12] p 137) if Φ is a perfect space with dif-ferential translation operation then every finite functional isa convolutor inΦ Finite functionals form a fairly wide classIn particular every bounded set F sub R is a support of somegeneralized function [12]

Theorem 3 Let u(t x) be a solution of problems (24) and(87) with the boundary function f in condition (87) which isan element of (S

β2)prime sub (S1ω2 )prime βgt 1 with finite support in R

-en u(t x)⟶ 0 uniformly on R as t⟶ +infin

Here is a scheme for proving this statement Letsuppf sub [a1 b1] sub [a2 b2] sub R Consider the functionφ isin S

β2 βgt 1 such that φ(x) 1 x isin [a1 b1] and

suppφ sub [a2 b2] )is function exists when βgt 1 because Sβ2

contains finite functions [12] Consider the function

u(t x) langfξ φ(ξ)G(t x minus ξ)rang +langfξ c(ξ)G(t x minus ξ)rang(105)

where c 1 minus φ Since supp(c(ξ)G(t x minus ξ)) cap suppφ emptywe have

u(t x) tminus (1ω)langfξ t

1ωφ(ξ)G(t x minus ξ)rang (106)

To prove the formulated above statement it remains toestablish that the set of functions Φtx(ξ) t1ωφ(ξ)G(t x minus

ξ) is bounded in the space Sβ2 βgt 1 for large t and x isin R

For example if in condition (87) f δ then δ is aconvolutor in the space S supp δ 0 andu(t x) δ lowastG(t x) G(t x) It follows directly from es-timate (58) that G(t x)⟶ 0 uniformly onR as t⟶ +infin

4 Conclusion

In this paper the solvability of a nonlocal multipoint by thetime problem for the evolutionary equation(zuzt) + (I minus (d2dx2))ω2u 0 (t x) isin (0 +infin) times R isproved herewith the operator (I minus (d2dx2))ω2 is inter-preted as a pseudodifferential operator in the space S1ω2

10 International Journal of Differential Equations

constructed by function (1 + σ2)ω2 σ isin R while the non-local multipoint by time condition also contains pseudo-differential operators constructed on smooth symbols )erepresentation of the solution is given in the form of aconvolution of the fundamental solution with the initialfunction which is an element of the space of generalizedfunctions of the ultradistribution type (the coagulator inspace S1ω2 ) )e behavior of the solution u(t x) t⟶ +infinin the space of generalized functions (S1ω2 )prime is investigated)e conditions for the initial generalized function are foundunder which the solution is uniformly stabilized to zero onR )e method of research of a nonlocal multipoint by timeproblem offered in this paper allows to interpret differential-operator equations of a form (zuzt) + φ(i zzx)u 0 as anevolutionary equation with a pseudodifferential operatorφ(i zzx) Fminus 1[φ middot F] constructed by the function φ actingin certain countable-normalized space of infinite-differentialfunctions (the choice of space depends on the properties ofthe function which is the symbol of the operator φ(i zzx))

Data Availability

No data were used to support the findings of this study

Disclosure

)is study was conducted in the framework of scientificactivity at Yuriy Fedkovych Chernivtsi National University

Conflicts of Interest

)e authors declare that they have no conflicts of interest

References

[1] V I Gorbachuk and M L Gorbachuk Boundary-ValueProblems for Differential-Operator Equations NaukovaDumka Kiev Ukraine 1984

[2] V V Gorodetskiy N I Nagnibida and P P NastasievMethods for Solving Problems in Functional Analysis VyshkaSchool Vyshka Ukraine 1990

[3] S G Samko A A Kilbas and O I Marichev Integrals andFractional Derivatives and Some of -eir Applications Scienceand Technology Minsk Belarus 1987

[4] I M Gelfand and G E Shilov Some Questions of the-eory ofDifferential Equations Fizmatgiz Moscow Russia 1958

[5] V V Gorodetskiy and O M Lenyuk About the FractionalDifferentiation in Spaces of Sprime type NAS of Ukraine KyivUkraine 1998

[6] A A Dezin General Questions of the -eory of BoundaryValue Problems Nauka Moscow Russia 1980

[7] A M Nakhushev Equations of Mathematical Biology HigherSchool Moscow Russia 1995

[8] I A Belavin S P Kapitsa and S P Kurdyumov ldquoAmathematical model of global demographic processes withconsidering of the spatial distributionrdquo Zh CalculationMathematics Physics vol 38 no 6 pp 885ndash902 1998

[9] V V Gorodetskiy and O V Martynyuk-e Cauchy Problemand Nonlocal Problems for First-Order Evolution EquationsEby Time Variable Vidavnichy dim rdquoRODOVIDrdquo ChernivtsiUkraine 2015

[10] V V Gorodetskiy and V I Myronyk ldquoTwo-point problem forone class of evolution equationsrdquo Differents Equationsvol 46 no 3 pp 349ndash363 2010

[11] V V Gorodetskiy and V I Myronyk ldquoTwo-point problem forone class of evolution equationsrdquo Differents Equationsvol 46 no 4 pp 520ndash526 2010

[12] I M Gelfand and G E Shilov Spaces of Basic and GeneralizedFunctions Fizmatgiz Moscow Russia 1958

International Journal of Differential Equations 11

Page 2: On One Evolution Equation of Parabolic Type with ...

Aφ I + iddx

1113888 1113889

2⎛⎝ ⎞⎠

12

φ 1113946+infin

minus infin1 + λ21113872 1113873

12dEλφ (1)

It is known (see for example [2]) that

Eλφ(t) 12π

1113946λ

minus infin1113946

+infin

minus infinφ(τ)e

iστdτ1113966 1113967eminus itσdσ (2)

It follows from this that dEλφ(t) (12π)

F[φ](λ)eminus itλdλ So

Aφ 12π

1113946+infin

minus infin1 + λ21113872 1113873

12F[φ](λ)e

minus itλdλ Fminus 1 1 + λ21113872 1113873

12F[φ]1113876 1113877

(3)

Following [3] we call the operator A the Bessel operatorof fractional differentiation )erefore A can be understoodas a pseudodifferential operator constructed on the functionsymbol (1 + λ2)12 λ isin R )is allows us to interpret thefunction eminus tAf that is a solution of the correspondingCauchy problem as a convolution of the form G(t middot)lowastf

[4 5] where G(t middot) Fminus 1[eminus (1+λ2)12]In this paper we give a similar depiction of the solution

of a nonlocal multipoint time problem for the equation(zu(t x)zt) + Au(t x) 0 (t x) isin (0 +infin) times R when theinitial condition u|t0 f is replaced by the condition1113936

mk0 αkBku(t middot)|ttk

where t0 0 t1 tm1113864 1113865 sub (0 +infin)α0 α1 αm1113864 1113865 sub R and m isin N are fixed and B1 Bm arepseudodifferential operators constructed of smooth symbols(if α0 1 α1 middot middot middot αm 0 B0 I then obviously we havea Cauchy problem) )is condition is interpreted in theclassical or weak sense as if f is a generalized function(generalized element of the operator A) of ultradistributiontype Properties of the fundamental solution of the specifiedmultipoint problem are investigated )e behavior of thesolution at t⟶ +infin (solution stabilization) in the spacesof generalized functions of type Sprime and the uniform stabi-lization of the solution to zero on R are studied

Note that the nonlocal multipoint time problem relatesto nonlocal problems for operator differential equations andpartial differential equations Such problems arise whenmodeling many processes and problems of practice withboundary-value problems for partial differential equationswhen describing correct problems for a particular operatorand constructing a general theory of boundary-valueproblems (see for example [6ndash11])

2 Spaces of Test and Generalized Functions

Gelfand and Shilov introduced in [12] a series of spaceswhich they called the spaces S )ey consist of infinitelydifferentiable on R functions which satisfy certain condi-tions on the decrease at infinity and the growth of deriva-tives )ese conditions are given by the inequalities|xkφ(m)(x)| le ckm x isin R k m sub Z+ where ckm1113864 1113865 is somedouble sequence of positive numbers If there are no re-strictions on elements of the sequence ckm1113864 1113865 then obviouslywe have L Schwartzrsquos space S equiv S(R) of quickly descendingat infinity functions However if the numbers ckm satisfycertain conditions then the corresponding specific spaces

are contained in S and they are called the spaces of S type Letus define some of them

For any α βgt 0 let us put

Sβα(R) equiv S

βα ≔ φ isin S |existcgt 0existAgt 0existBgt 0forall m n sub Z+1113864

forallx isin R xmφ(n)

(x)11138681113868111386811138681113868

11138681113868111386811138681113868 le cAm

Bnm

mαn

nβ1113883

(4)

)e introduced S spaces can also be described as in [12]Sβα consists of those infinitely differentiable on R func-

tions φ(x) that satisfy the following inequalities

φ(n)(x)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cBnn

nβ exp minus a|x|1α

1113872 1113873 n isin Z+ x isin R (5)

with some positive constants c B and a dependent only onthe function φ

If 0lt βlt 1 and αge 1 minus β then Sβα consists of only those

φ isin Cinfin(R) which admit an analytic extension into thecomplex plane C such that|φ(x + iy)| le c exp minus a|x|

1α+ b|y|

1(1minus β)1113872 1113873 cgt 0 agt 0 bgt 0

(6)

)e space S1α consists of functions φ isin Cinfin(R) which canbe analytically extended into some band |Imz|lt δ (depen-dent on φ) of the complex plane so that the estimate

|φ(x + iy)| le c exp minus a|x|1α

1113872 1113873 c agt 0 forally isin R (7)

is carried out)e spaces S

βα α βgt 0 are nontrivial if α + βge 1 and they

form dense sets in L2(R))e topological structure in S

βα is defined as follows )e

symbol SβB

αA A Bgt 0 denotes the set of functions φ isin Sβα that

satisfy the condition

forallAgtAforallBgtB xkφ(m)

(x)11138681113868111386811138681113868

11138681113868111386811138681113868 le cAkB

mk

kαm

mβ k m sub Z+ x isin R

(8)

)is set is transformed into a complete countably normedspace if the norms in it are defined by means of relations

φδρ supxkm

xkφ(m)(x)1113868111386811138681113868

1113868111386811138681113868

(A + δ)k(B + ρ)mkkαmmβ

δ ρ1113864 1113865 sub 11213 1113882 1113883

(9)

)e specified norm system is sometimes replaced by anequivalent norm system

φδρprime supxm

exp a(1 minus δ)|x|1α1113966 1113967 middot φ(m)(x)1113868111386811138681113868

1113868111386811138681113868

(B + ρ)mmmβ

δ ρ1113864 1113865 sub1213 1113882 1113883

a α

eA1α

(10)

If A1 ltA2 B1 ltB2 then SβB1αA1

is continuously embeddedinto S

βB2αA2

and Sβα cupABgt0S

βB

αA that is Sβα is endowed by the

inductive limit topology of the spaces SβB

αA [12] )erefore

2 International Journal of Differential Equations

the convergence of a sequence φ] ]ge 11113864 1113865 sub Sβα to zero in the

space Sβα is the convergence in the topology of some space

SβB

αA to which all the functions φ] belong In other words (see[12]) φ]⟶ 0 in S

βα as ]⟶infin if and only if for every

n isin Z+ the sequence φ(n)] ]ge 1 converges to zero uniformly

on an arbitrary segment [a b] sub R and for some c a Bgt 0independent of ] and the inequality

φ(n)] (x)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cBnn

nβ exp minus a|x|1α

1113966 1113967 x isin R n isin Z+ (11)

holdsIn S

βα the continuous translation operation Tx

φ(ξ)⟶ φ(ξ + x) is defined )is operation is also differ-entiable (even infinitely differentiable [12]) in the sense thatthe limit relation (φ(x + h) minus φ(x))hminus 1⟶ φprime(x) h⟶ 0is true for every function φ isin S

βα in the sense of convergence

in the Sβα-topology In S

βα the continuous differentiation

operator is also defined )e spaces of type S are perfect [12](that is the spaces and all bounded sets of which arecompact) and closely related to the Fourier transformnamely the formula F[S

βα] Sαβ α βgt 0 is correct where

F Sβα1113960 1113961 ≔ ψ ψ(σ) 1113946

Rφ(x)e

iσxdx φ isin Sβα1113882 1113883 (12)

Moreover the operator F Sβα⟶ Sαβ is continuousLet (S

βα)prime denote the space of all linear continuous

functionals on Sβα with weak convergence Since the trans-

lation operator Tx is defined in the space Sβα of test functions

the convolution of a generalized function f isin (Sβα)prime with the

test function φ isin Sβα is given by the following formula

(flowastφ) langf Tminus xφ(middot)rang equiv langf φ(x minus middot)rang φ(ξ) φ(minus ξ)

(13)

It follows from the infinite differentiability property ofthe argument translation operation in S

βα that the convo-

lution flowastφ is a usual infinitely differentiable function onRWe define the Fourier transform of a generalized

function f isin (Sβα)prime by the relation

langF[f] φrang langf F[φ]rang forallφ isin Sαβ (14)

where the operator F (Sβα)prime ⟶ (Sαβ)prime is continuous

Let f isin (Sβα)prime If flowastφ isin S

βα forallφ isin S

βα and the conver-

gence φ]⟶ 0 in the Sβα-topology as ]⟶infin implies that

flowastφ]⟶ 0 in the Sβα-topology as ]⟶infin then the

function f is called a convolutor in Sβα If f isin (S

βα)prime is a

convolutor in Sβα then for an arbitrary function φ isin S

βα the

formula F[flowastφ] F[f]F[φ] is valid where F[f] is amultiplier in S

βα [12]

Recall that the function g isin Cinfin(R) is called a multiplierin S

βα if gφ isin S

βα for an arbitrary function φ isin S

βα and the

mapping φ⟶ gφ is continuous in the space Sβα

3 Nonlocal Multipoint by Time Problem

Consider the function a(σ) (1 + σ2)ω2 σ isin R whereω isin [1 2) is a fixed parameter Obviously the function a(σ)

satisfies the inequality

a(σ) le cε exp ε|σ|ω

1113864 1113865 σ isin R (15)

where cε 2ω2 max 1 (1ε) for an arbitrary εgt 0 Usingdirect calculations and the Stirling formula we find that

Dmσ a(σ)

11138681113868111386811138681113868111386811138681113868 le c0B

m0 m le c1B

m1 m

m m isin N σ isin R (16)

where B0 B0(ω)gt 0 B1 B1(ω)gt 0 and c0 c1 gt 0 It fol-lows from (15) and (16) that a(σ) is a multiplier in S11ωIndeed let φ isin S11ω that is the function φ and its derivativessatisfy the inequality

Dkσφ(σ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cAkk

k exp minus a|σ|ω

1113864 1113865 k isin Z+ σ isin R (17)

with some positive constants c A and a )en using theLeibniz formula for differentiating the product of twofunctions as well as inequalities (15)ndash(17) we find that

(a(σ)φ(σ))(k)

11138681113868111386811138681113868

11138681113868111386811138681113868 le 1113944

k

l0C

lk a

(l)(σ)

11138681113868111386811138681113868

11138681113868111386811138681113868 φ(kminus l)

(σ)11138681113868111386811138681113868

11138681113868111386811138681113868 a(σ)φ(k)(σ)

11138681113868111386811138681113868

11138681113868111386811138681113868+

+ 1113944

k

l1C

lk a

(l)(σ)

11138681113868111386811138681113868

11138681113868111386811138681113868 middot φ(kminus l)(σ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cεcAkk

k exp minus a|σ|ω

1113864 1113865

+ c1c 1113944k

l1C

lkB

l1l

lA

kminus l(k minus l)

kminus l exp minus a|σ|ω

1113864 1113865

(18)

Since εgt 0 is arbitrary we can put ε (a2) )en

(a(σ)φ(σ))(k)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cεcAkk

k exp minusa

2|σ|

ω1113882 1113883

+ c1c 1113944k

l1C

lkB

l1l

lA

kminus l(k minus l)

(kminus l)

times exp minusa

2|σ|

ω1113882 1113883 le c2B

k2k

k exp minusa

2|σ|

ω1113882 1113883

σ isin R k isin Z+

(19)

where c2 c(cε + c1) and B2 2max A B11113864 1113865 It follows from(19) that a middot φ is an element of S11ω

)e operation of multiplying by a(σ) is continuous inthe space S11ω In fact let φn nge 11113864 1113865 be a sequence offunctions from S11ω convergent to 0 in this space)is meansthat φn nge 11113864 1113865 sub S

1B01ωA0

with some constants A0 B0 gt 0 and

φn

δρ sup

exp a(1 minus δ)|σ|ω middot φ(k)n (σ)

11138681113868111386811138681113868111386811138681113868

B0 + ρ( 1113857kkk

⟶ 0

n⟶infin

δ ρ1113864 1113865 sub1213 1113882 1113883

a 1

ωeAω0

(20)

In other words for an arbitrary 1113957εgt 0 there exists anumber n0 n0(1113957ε) such that for nge n0

φ(k)n (σ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le 1113957ε B0 + ρ( 1113857kk

k exp minus a(1 minus δ)|σ|ω

1113864 1113865 σ isin R k isin Z+

(21)

International Journal of Differential Equations 3

Using inequalities (15) and (16) (when putting in (15)ε (a2)(1 minus δ)) we get

a(σ)φn(σ)( 1113857(k)

11138681113868111386811138681113868

11138681113868111386811138681113868 le c31113957ε(1113957B + ρ)k exp minus

a

2(1 minus δ)|σ|

ω1113882 1113883

k isin Z+ nge n0

(22)

where c3 cε + c1 and 1113957B 2max B0 B11113864 1113865 It follows from thelast inequality that aφnδρ le c31113957ε forallnge n0 that is the se-

quence aφn nge 11113864 1113865 converges to zero in the space S11113957B1ω1113957A

where 1113957A 21ωA0 and 1113957B 2max B0 B11113864 1113865 )is means thatthe sequence aφn nge 11113864 1113865 converges to zero in the space S11ωwhich is what we needed to prove

Remark 1 It follows from the proven property that thefunction a(σ) (1 + σ2)ω2 σ isin R is also a multiplier inevery space S

β1ω where βgt 1 )erefore in the space S1ωβ

βge 1 defined as a continuous linear pseudodifferentialoperator A constructed by the function a(σ)

Aφ Fminus 1

[a(σ)F[φ]] forallφ isin S1ωβ (23)

where Aφ (I minus Δ)ω2φ Δ (d2dx2)Let us consider the evolution equation with the operator

A (the Bessel fractional differentiation operator)zu

zt+ Au 0 (t x) isin (0 +infin) times R equiv Ω (24)

By a solution of equation (24) we mean the functionu(t x) (t x) isin Ω such that (1) u(t middot) isin C1(0 +infin) forevery x isin R (2) u(middot x) isin S1ωβ for every t isin (0 +infin) (3)u(t x) is continuous at every point (0 x) of the boundaryΓΩ 0 times R of the region Ω (4) existMR⟶ [0infin) forallt isin (0 +infin) |ut(t x)| le M(x)intRM(x)dxltinfin and (5) u(t x) (t x) isin Ω satisfiesequation (24)

For equation (24) we consider the nonlocal multipointby time problem of finding a solution of equation (24) thatsatisfies the condition

μu(0 x) minus μ1B1u t1 x( 1113857 minus middot middot middot minus μmBmu tm x( 1113857 f(x)

x isin R f isin S1ωβ

(25)

where u(0 x) ≔ limt⟶+0u(t x) x isin R m isin Nμ μ1 μm1113864 1113865 sub (0infin) and t1 tm1113864 1113865 sub (0infin) are fixednumbers 0lt t1 lt t2 lt middot middot middot lt tm lt +infin μgt 1113936

mk1 μk and

B1 Bm are pseudodifferential operators in S1ωβ con-structed by functions (characters) gk R⟶ (0 +infin) re-spectively Bk Fminus 1[gk(σ)F] k isin 1 m )e functionsgk isin Cinfin(R) and k isin 1 m satisfy the conditions

forallεgt 0forallσ isin R gk(σ) le exp ε|σ|ω

1113864 1113865

existLk gt 0foralls isin N Dsσgk(σ)

11138681113868111386811138681113868111386811138681113868 le L

sks

s

(26)

Note that the above properties of the functions gk implythat gk k isin 1 m is a multiplier in S1ωβ

We are looking for the solution of problems (24) and(25) via the Fourier transform Due to condition (19)

F ut1113858 1113859 1113946R

ut(t x)eiσxdx

z

zt1113946R

u(t x)eiσxdx

z

ztF[u]

(27)

We introduce the notation F[u(t x)] v(t σ) Giventhe form of the operators A B1 Bm we get

F[Au(t x)] F Fminus 1

[a(σ)F[u]]1113960 1113961 a(σ)F[u(t x)] a(σ)v(t σ)

F Bku(t x)1113858 1113859 gk(σ)F[u(t x)] gk(σ)v(t σ) k isin 1 m

(28)

So for the function v Ω⟶ R we arrive at a problemwith parameter σ

dv(t σ)

dt+ a(σ)v(t σ) 0 (t σ) isin Ω (29)

μv(0 σ) minus 1113944

m

k1μkgk(σ)v tk σ( 1113857 1113957f(σ) σ isin R (30)

where 1113957f(σ) F[f](σ) )e general solution of equation(29) has the form

v(t σ) c exp minus ta(σ) (t σ) isin Ω (31)

where c c(σ) is determined by condition (30) Substituting(30) into (31) we find that

c 1113957f(σ) μ minus 1113944m

k1μkgk(σ)exp minus tka(σ)1113864 1113865⎛⎝ ⎞⎠

minus 1

σ isin R

(32)

Now put G(t x) ≔ Fminus 1[Q(t σ)] andQ(t σ) Q1(t σ)Q2(σ) and thus

Q1(t σ) exp minus ta(σ)

Q2(σ) μ minus 1113944m

k1μkgk(σ)Q1 tk σ( 1113857⎛⎝ ⎞⎠

minus 1

(33)

)en thinking formally we come to the relation

u(t x) 1113946R

G(t x minus ξ)f(ξ)dξ G(t x)lowastf(x) (34)

Indeed

u(t x) (2π)minus 1

1113946R

Q(t σ) 1113946R

f(ξ)eiσξdξ1113874 1113875e

minus iσxdσ

1113946R

(2π)minus 1

1113946R

Q(t σ)eminus iσ(xminus ξ)dσ1113874 1113875f(ξ)dξ

1113946R

G(t x minus ξ)f(ξ)dξ G(t x)lowastf(x) (t x) isin Ω

(35)

)e correctness of the transformations performed theconvergence of the corresponding integrals and conse-quently the correctness of formula (35) follow from theproperties of the function G which are given below )eproperties of G are determined by the properties of Q

4 International Journal of Differential Equations

because G Fminus 1[Q] So let us first examine the properties ofthe function Q(t σ) as a function of the variable σ

Lemma 1 For derivatives of Q1(t σ) (t σ) isin Ω theestimates

DsσQ1(t σ)

11138681113868111386811138681113868111386811138681113868 le cA

stcs

ss exp minus t|σ|

ω1113864 1113865 (36)

are valid where c 0 if 0lt t le 1 and c 1 if tgt 1 and theconstants cgt 1 and Agt 0 do not depend on t

Proof To prove the statement we use the Faa di Brunoformula for differentiation of a complex function

DsσF(g(σ)) 1113944

s

p1

dp

dgpF(g) 1113944

s

p1 pl

ddσ

g(σ)1113888 1113889

p1

1l

dl

dσlg(σ)1113888 1113889

pl

s isin N

(37)

where the sum sign is applied to all the solutions in positiveintegers of the equation p1 + 2p2 + middot middot middot + lpl sp1 + middot middot middot + pl m Put F eg and g minus ta(σ) )en

Dsσe

minus ta(σ) e

minus ta(σ)1113944

s

p11113944

s

p1 plΛ (38)

where the symbol Λ denotes the expression and

Λ ≔ddσ

(minus ta(σ))1113888 1113889

p1

middot12

d2

dσ2(minus ta(σ))1113888 1113889

p2

1l

dl

dσl(minus ta(σ))1113888 1113889

pl

(39)

Given estimate (16) is fulfilled we find that

|Λ| le cp1+middotmiddotmiddot+pl

0 Bp1+2p2+middotmiddotmiddot+lpl

0 tp1+middotmiddotmiddot+pl le 1113957c

s0t

pB

s0 1113957c0 max 1 c01113864 1113865

(40)

Using (40) and the Stirling formula we arrive at theinequalities

DsσQ1(t σ)

11138681113868111386811138681113868111386811138681113868 le 1113957c

s0B

s0t

css exp minus ta(σ)

le cAstcs

ss exp minus t|σ|

ω1113864 1113865 σ isin R

(41)

where c 0 if 0lt t le 1 and c 1 if tgt 1 and the values cgt 1and Agt 0 do not depend on t

Lemma is proved

Remark 2 It follows from estimate (41) that Q1(t middot) isin S11ωfor every tgt 0

Lemma 2 -e function Q2 is a multiplier in S21ω

Proof To prove the assertion let us estimate the derivativesof Q2 For this we use formula (37) in which we put F φminus 1

and φ R where

R(σ) μ minus 1113944m

k1μkgk(σ)exp minus tka(σ)1113864 1113865 (42)

)en Q2 F(φ) Rminus 1 and

DsσQ2(σ)

11138681113868111386811138681113868111386811138681113868 1113944

s

p1

dp

dRpR

minus 11113944

s

p1 pl

ddσ

R(σ)1113888 1113889

p11113868111386811138681113868111386811138681113868111386811138681113868

1l

dl

dσlR(σ)1113888 1113889

pl

s isin N

(43)

Given the properties of g1 gm and inequality (41)we find that

1l

dj

dσjR(σ)

11138681113868111386811138681113868111386811138681113868

11138681113868111386811138681113868111386811138681113868le

1j

1113944

m

k1μk 1113944

j

i0C

ij D

iσgk(σ)

11138681113868111386811138681113868

11138681113868111386811138681113868 middot Djminus iσ e

minus tka(σ)11138681113868111386811138681113868

11138681113868111386811138681113868

le1j

1113944

m

k1μk 1113944

i

j0C

ijL

iki1113957c

jminus i0 t

c(jminus i)

k (j minus i)eminus tka(σ)

le 1113944m

k1μk 1113944

j

i0C

ijL

ik1113957c

jminus i0 t

c(jminus i)

(44)

where we took into account that i(j minus i) le j)Let

1113957L max L1 Lm1113864 1113865

L0 2max 1113957L 1113957c0T1113864 1113865(45)

then

1j

dj

dσjR(σ)

11138681113868111386811138681113868111386811138681113868

11138681113868111386811138681113868111386811138681113868le α0L

j0 α0 1113944

m

k1μk j isin 1 l

ddσ

R(σ)1113888 1113889

p1111386811138681113868111386811138681113868111386811138681113868

111386811138681113868111386811138681113868111386811138681113868

1l

dl

dσlR(σ)1113888 1113889

pl111386811138681113868111386811138681113868111386811138681113868

111386811138681113868111386811138681113868111386811138681113868le α0L0( 1113857

p α0L201113872 1113873

p2 α0L

l01113872 1113873

pl

αp1+middotmiddotmiddot+pl

0 Lp1+2p2+middotmiddotmiddot+lpl

0 αp0L

s0

(46)

In addition (dpdRp)Rminus 1 (minus 1)ppRminus (p+1) and

Rminus 1

(σ) Q2(σ) μ minus 1113944m

k1μkgk(σ)exp minus tka(σ)1113864 1113865⎛⎝ ⎞⎠

minus 1

le μ minus 1113944m

k1μk exp minus t1|σ|

ω+ ε|σ|

ω1113864 1113865⎛⎝ le μ minus 1113944

m

k1μk

⎛⎝ ⎞⎠

minus 1

equiv β0 gt 0 ε t1

(47)

Since by assumption μgt 1113936mk1 μk (assuming the prop-

erties of g1 gm and 0lt t1 lt t2 lt middot middot middot lt tm) So

International Journal of Differential Equations 5

dp

dRpR

minus 111138681113868111386811138681113868111386811138681113868

11138681113868111386811138681113868111386811138681113868le βp+1

0 p

DsσQ2(σ)

11138681113868111386811138681113868111386811138681113868 le s 1113944

s

p1βp+10 pαp

0Ls0

le β0Ls0(s)

21113944

s

p1βp0 le β0L

s0s middot (s)

2βs1 le β2β

s3s

2s s isin N

(48)

It follows from the last inequality and boundedness ofthe function Q2(σ) on R that Q2 is a multiplier in S21ω

Corollary 1 For every tgt 0 the functionQ(t σ) Q1(t σ)Q2(σ) σ isin R is an element of the spaceS21ω and the estimates

DsσQ(t σ)

11138681113868111386811138681113868111386811138681113868 le 1113957c1113957A

stcs

s2s exp minus t|σ|

ω1113864 1113865 s isin Z+ (t σ) isin Ω

(49)

are valid where the constants 1113957c and 1113957Agt 0 do not depend on t

Taking into account the properties of the Fouriertransform (direct and inverse) and the formulaFminus 1[S21ω] S1ω2 we get that G(t middot) isin S1ω2 for every tgt 0 Weremove in the estimates of derivatives of the function G (inthe variable x) the dependence on t assuming tgt 1 To dothis we use the relations

xkD

sxF[φ](x) i

k+sF σsφ(σ)( 1113857

(k)1113876 1113877

ik+s

1113946R

σsφ(σ)( 1113857(k)

eiσxdσ k s sub Z+ φ isin S

21ω

(50)

So

xkD

sxG(t x) (2π)

minus 1ik+s

(minus 1)s1113946R

σsQ(t minus σ)( 1113857

(k)e

iσxdσ

(51)

Applying the Leibniz formula for differentiating theproduct of two functions and estimating the derivatives ofthe function Q(t σ) we find that

σsQ(t minus σ)( 1113857

(k)11138681113868111386811138681113868

11138681113868111386811138681113868 1113944k

p0C

p

k σs( 1113857

(p)Q

(kminus p)(t minus σ)

1113868111386811138681113868111386811138681113868111386811138681113868

1113868111386811138681113868111386811138681113868111386811138681113868

le σsQ

(k)(t minus σ)

11138681113868111386811138681113868

11138681113868111386811138681113868 + ks σsminus 1Q

(kminus 1)(t minus σ)

11138681113868111386811138681113868

11138681113868111386811138681113868

+k(k minus 1)

2s(s minus 1) σsminus 2

Q(kminus 2)

(t minus σ)11138681113868111386811138681113868

11138681113868111386811138681113868

+ middot middot middot le 1113957c 1113957Aktck 1113957B

stminus (sω)

mks + ks1113957Akminus 1

tc(kminus 1)

1113876

middot 1113957Bsminus 1

tminus ((sminus 1)ω)

mkminus 1sminus 1

+12

k(k minus 1)s(s minus 1)1113957Akminus 2

tc(kminus 2) 1113957B

sminus 2tminus ((sminus 2)ω)

middot mkminus 2sminus 2 + 1113961eminus (t2)|σ|ω

(52)

where mks k2kssω Taking into account the results in (see[12] p 236ndash243) we find that this double sequence satisfiesthe inequality

ksmkminus 1sminus 1

mks

le 1113957c(k + s) 1113957cgt 0 (53)

Bearing in mind the last inequality and also that tgt 1 weobtain

σsQ(t minus σ)( 1113857

(k)11138681113868111386811138681113868

11138681113868111386811138681113868 le 1113957c1113957Aktk 1113957B

smks 1 +

ks

1113957A1113957B

mkminus 1sminus 1

mks

+12

k(k minus 1)s(s minus 1)1

1113957A21113957B

2mkminus 2sminus 2

mks

+ middot middot middot1113888 1113889eminus (t2)|σ|ω le 1113957c1113957A

ktk 1113957B

smks

times 1 +ks

1113957A1113957B

mkminus 1sminus 1

mks

+12

11113957A21113957B

2 ksmkminus 1sminus 1

mks

(k minus 1)(s minus 1)mkminus 2sminus 2

mkminus 1sminus 1+ middot middot middot1113888 1113889e

minus (t2)|σ|ω

le 1113957c1113957Aktk 1113957B

smks 1 +

1113957c

1113957A1113957B(k + s) +

12

1113957c2

(1113957A1113957B)2(k + s)

2+ middot middot middot1113888 1113889e

minus (t2)|σ|ω

le c1AktkB

smkse

minus (t2)|σ|ω

A 1113957A exp1113957c

1113957A1113957B1113882 1113883

B 1113957B exp1113957c

1113957A1113957B1113882 1113883

(54)

6 International Journal of Differential Equations

So

xkD

sxG(t x)

11138681113868111386811138681113868

11138681113868111386811138681113868 le (2π)minus 1

c1AktkB

smks1113946

Re

minus (t2)|σ|ωdσ

c2AktkB

stminus (1ω)

k2k

ssω

k s sub Z+

(55)

)en

DsxG(t x)

11138681113868111386811138681113868111386811138681113868 le c2B

stminus (1ω)

ssω inf

k

Akk2k

tminus 1|x|( )k

le c3Bstminus (1ω)

ssω exp minus a0t

minus (12)|x|

121113966 1113967

(56)

where the constants c3 B and a0 gt 0 do not depend on there we used the well-known inequality (see [12] p 204)

eminus (αe)|ξ|1α le inf

k

Lkkkα

|ξ|kle ce

minus (αe)|ξ|1α c e

αe2 (57)

in which α 2 and L 1113957A )us such a statement is correct

Lemma 3 -e derivatives (by variable x) of the functionG(t x) for tgt 1 satisfy the inequality

DsxG(t x)

11138681113868111386811138681113868111386811138681113868 le c3B

ss

sωtminus (1ω) exp minus a0t

minus (12)|x|

121113966 1113967 s isin Z+

(58)

where the constants c3 B and a0 gt 0 do not depend on t

Remark 3 In what follows we assume that in condition(25) β 2

Here are a few more properties of the function G(t x)

Lemma 4 -e function G(t middot) t isin (0 +infin) as an abstractfunction of t with values in the space S1ω2 is differentiable in t

Proof It follows from the continuity of the Fourier trans-form (direct and inverse) that to prove the statement itsuffices to establish that the function F[G(t middot)] Q(t middot) as afunction of a parameter t with values in the space S21ω isdifferentiable in t In other words it is necessary to provethat the boundary value relation

ΦΔt(σ) ≔1Δt

[Q(t + Δt middot) minus Q(t middot)]⟶z

ztQ(t middot) Δt⟶ 0

(59)

is performed in the sense as follows

(1) DsσΦΔt(σ) ⟶

Δt⟶ 0Ds

σ(minus a(σ)Q(t σ)) s isin Z+ uni-formly on each segment [a b] sub R

(2) |DsσΦΔt(σ)| le cB

ss2s exp minus a|σ|ω s isin Z+ where the

constants c B and agt 0 do not depend on Δt if Δt issmall enough

)e function Q(t σ) (t σ) isin Ω is differentiable in t inthe usual sense Due to the Lagrange theorem on finiteincrements

ΦΔt(σ) minus a(σ)Q(t + θΔt σ) (60)

So

DsσΦΔt(σ) minus 1113944

s

l0C

lsD

lσa(σ)D

sminus lσ Q(t +θΔtσ) (61)

Dsσ ΦΔt(σ) minus

z

ztQ(tσ)1113888 1113889 minus 1113944

s

l0C

lsD

lσa(σ) D

sminus lσ Q(t +θΔtσ)1113960

minus Dsminus lσ Q(tσ)1113961

(62)

Since

Dsminus lσ Q(t + θΔt σ) minus D

sminus lσ Q(t σ) D

sminus l+1σ Q t + θ1Δt σ( 1113857

0lt θ1 lt 1

(63)

it follows from this and estimate (49) that

Dsminus l+1σ Q t + θ1Δt σ( 1113857θΔt⟶ 0 Δt⟶ 0 (64)

uniformly on an arbitrary segment [a b] sub R )en

DsσΦΔt(σ)⟶ D

z

ztQ(t σ)1113888 1113889 Δt⟶ 0 (65)

uniformly on each segment [a b] sub R too )ereforecondition 1 is satisfied

Taking into account (61) estimates (15) (16) and (49)for a(σ) Q(t σ) and their derivatives we find that

DsσΦΔt(σ)

11138681113868111386811138681113868111386811138681113868 le c1cε 1113944

s

l0C

lsB

l1l

l 1113957Asminus l

tc(sminus l)

(s minus l)2(sminus l)

exp minus (t + θΔt)|σ|ω

+ ε|σ|ω

1113864 1113865

(66)

where εgt 0 is arbitrary fixed Take ε (t2) and then

DsσΦΔt(σ)

11138681113868111386811138681113868111386811138681113868 le cB

ss2s exp minus a|σ|

ω1113864 1113865 s isin Z+ (67)

where B 2max B11113957Atc1113966 1113967 a (t2) and all the constants

do not depend on Δt

Corollary 2 -e formulaz

zt(flowastG(t middot)) f lowast

zG(t middot)

zt forallf isin S

1ω21113872 1113873prime tgt 0

(68)

is correct

Proof According to the definition of a convolution of ageneralized function with a test one we have

flowastG(t middot) langfξ Tminus xG(t ξ)rang

G(t ξ) G(t minus ξ)(69)

)enz

zt(flowastG(t middot)) lim

Δt⟶0

1Δt

[(flowastG(t + Δt middot)) minus (flowastG(t middot))]

limΔt⟶0langfξ

1Δt

Tminus xG(t + Δt ξ) minus Tminus x

G(t ξ)1113960 1113961rang

(70)

International Journal of Differential Equations 7

Due to Lemma 4 the relation1Δt

Tminus xG(t + Δt ξ) minus Tminus x

G(t ξ)1113960 1113961⟶z

ztTminus x

G(t ξ) Δt⟶ 0

(71)

is performed in the sense of S1ω2 topology So

z

zt(flowastG(t middot)) langfξ lim

Δt⟶0

1Δt

Tminus xG(t + Δt ξ) minus Tminus x

G(t ξ)1113960 1113961rang

langfξ z

ztTminus x

G(t ξ)rang langfξ Tminus x

z

ztG(t ξ)rang

flowastz

ztG(t middot)

(72)

)e statement is provedSince

F[G(t middot)] Q(t σ) Q1(t σ)Q2(σ) 1113946R

G(t x)eiσxdx tgt 0

(73)

we get the formula

1113946R

G(t x)dx Q1(t 0)Q2(0) λ0eminus t

λ0 μ minus 1113944m

k1μkgk(0)e

minus tk⎛⎝ ⎞⎠

minus 1

(74)

where a(0) 1 is taken into account

Lemma 5 In (S1ω2 )prime the following relations are correct

(1) G(t middot)⟶ Fminus 1

Q21113858 1113859 t⟶ + 0

(2) μG(t middot) minus 1113944m

k1μkBkG tk middot( 1113857⟶ δ t⟶ + 0

(75)

where δ is the Dirac delta function

Proof (1) Taking into account the continuity property of theFourier transform (direct and inverse) in spaces of Sprime type itis sufficient to establish that

F[G(t middot)] Q1(t middot)Q2(middot)⟶ Q2(middot) t⟶ + 0 (76)

in the space (S21ω)prime To do this we take an arbitrary functionφ isin S21ω and using the fact thatQ2 is a multiplier in the spaceS21ω and the Lebesgue theorem on the limit passage underthe Lebesgue integral sign we find that

langQ1(t middot)Q2(middot)φrang langQ1(t middot) Q2(middot)φ(middot)rang

1113946R

Q1(t σ)Q2(σ)φ(σ)dσ ⟶t⟶+0

1113946R

Q2(σ)φ(σ)dσ

lang1 Q2(middot)φ(middot)rang langQ2φrang

(77)

)is leads us to statement (1) of Lemma 5(2) Given statement (1) and the form of the operators

B1 Bm we find that

μG(t middot) minus 1113944m

l1μlBlG tl middot( 1113857 ⟶

t⟶+0μF

minus 1Q21113858 1113859 minus 1113944

m

l1μlBlG tl middot( 1113857

μFminus 1

Q21113858 1113859 minus 1113944m

l1μlF

minus 1gl(middot)Q1 tl middot( 1113857Q2(middot)1113858 1113859

Fminus 1 μQ2(middot) minus 1113944

m

l1μlgl(middot⎡⎣ ⎞⎠Q1 tl middot( 1113857Q2(middot)]

Fminus 1 μ minus 1113944

m

l1μlgl(middot)Q1 tl middot( 1113857Q2(middot)⎡⎣ ⎤⎦

Fminus 1 μ minus 1113944

m

l1μlgl(middot)Q1 tl middot( 1113857⎛⎝ ⎞⎠⎡⎢⎢⎣

middot μ minus 1113944m

l1μlgl(middot)Q1 tl middot( 1113857⎛⎝ ⎞⎠

minus 1⎤⎥⎥⎥⎥⎦ F

minus 1[1] δ

(78)

)erefore relation (75) is fulfilled in the space (S1ω2 )primeLemma 5 is proved

Remark 4 If μ 1 μ1 middot middot middot μm 0 then problems (24)and (25) degenerate into the Cauchy problem for equation(24) In this case Q2(σ) 1 forallσ isin R G(t x) Fminus 1[eminus ta(σ)]and G(t middot)⟶ Fminus 1[1] δ for t⟶ +0 in the space (S1ω1 )prime

Corollary 3 Let

ω(t x) flowastG(t x) f isin S1ω2lowast1113872 1113873prime (t x) isin Ω (79)

where (S1ω2lowast )prime is a class of convolutors in S1ω2 -en in (S1ω2 )primethe following boundary relation holds

μω(t middot) minus 1113944m

k1μkBkω tk middot( 1113857⟶ f t⟶ + 0 (80)

Proof Let us prove that the relation

F μω(t middot) minus 1113944

m

k1μkBkω tk middot( 1113857⎡⎣ ⎤⎦⟶ F[f] t⟶ + 0

(81)

takes place in the space (S21ω)prime Since f isin (S1ω2lowast )prime andG(t middot) isin S1ω2 for every tgt 0 we have

F[ω(t middot)] F[flowastG(t middot)] F[f]F[G(t middot)] F[f]Q(t middot)

(82)

So we need to prove that

F[f] μQ(t middot) minus 1113944m

k1μkgk(middot)Q tk middot( 1113857⎛⎝ ⎞⎠⟶ F[f] (83)

in (S21ω)prime when t⟶ + 0 SinceQ(t middot) Q1(t middot)Q2(middot)⟶ Q2(middot) for t⟶ + 0 in the space(S21ω)prime (see the proof of assertion 1 and Lemma 5) thecorrelation

8 International Journal of Differential Equations

μQ(t middot) minus 1113944m

k1μkgk(middot)Q tk middot( 1113857 ⟶

t⟶+0μQ2(middot) minus 1113944

m

k1μkgk(middot)Q1 tk middot( 1113857Q2(middot)

μ minus 1113944m

k1μkgk(middot)Q1 tk middot( 1113857⎛⎝ ⎞⎠Q2(middot) 1

(84)

is realized in the space (S21ω)prime )us relation (81) and (80)are fulfilled in the corresponding spaces )e statement isproved

)e function G(t middot) satisfies equation (24) as tgt 0Indeed

z

ztG(t x)

z

ztF

minus 1[Q(t σ)] F

minus 1 z

ztQ(t σ)1113890 1113891

Fminus 1

[a(σ)Q(t σ)]

AG(t x) Fminus 1

a(σ)Fminus 1

[G(t middot)]1113960 1113961 Fminus 1

[a(σ)Q(t σ)]

(85)

So

zG(t x)

zt+ AG(t x) 0 (t x) isin Ω (86)

which is what had to be provedIt follows from Corollary 3 that the nonlocal multipoint

by the time problem for equation (24) can be formulated inthe following way find a function u(t x) (t x) isin Ω thatsatisfies equation (24) and the condition

μ limt⟶+0

u(t middot) minus 1113944

m

k1μkBku tk middot( 1113857 f f isin S

1ω2lowast1113872 1113873prime (87)

where boundary relation (87) is considered in the space(S1ω2 )prime and the constraints on the parameters μ μ1 μmt1 tm are the same as in case of problems (24) and(25))

Theorem 1 By nonlocal multipoint by time problem (24)(87) is solvable and its solution is given by the formula

u(t x) flowastG(t x) (t x) isin Ω (88)

where u(t middot) isin S1ω2 for every tgt 0

Proof Make sure that the function u(t x) (t x) isin Ω sat-isfies equation (24) Indeed (see Corollary 2)

zu(t x)

zt

z

zt(flowastG(t middot)) flowast

zG(t middot)

zt

Au(t x) Fminus 1

[a(σ)F[f lowastG(t middot)]]

(89)

Since f is a convolutor in S1ω2

F[flowastG(t middot)] F[f]F[G(t middot)] F[f]Q(t middot) (90)

So

Au(t x) Fminus 1

[a(σ)Q(t σ)F[f](σ)] minus Fz

ztQ(t middot)F[f]1113890 1113891

minus F Fz

ztG(t middot)1113890 1113891 middot F[f]1113890 1113891

minus Fminus 1

F flowastzG(t middot)

zt1113890 11138911113890 1113891 minus flowast

zG(t middot)

zt

(91)

It follows from this that the function u(t x) (t x) isin Ωsatisfies equation (24) From Corollary 3 we get that u(t x)

satisfies condition (87) in the defined sense )e theorem isproved

Remark 5 If in condition (87) B1 middot middot middot Bm I (I is theidentity operator) then we can prove that problems (24) and(87) are well posed and its solution is given by the formula

u(t x) flowastG(t x) f isin S1ω1lowast1113872 1113873prime (t x) isin Ω (92)

where G(t middot) Fminus 1[Q(t middot)] isin S1ω1 for every tgt 0 and

Q(t σ) eminus ta(σ) μ minus 1113944

m

k1μke

minus tka(σ)⎛⎝ ⎞⎠

minus 1

(93)

Theorem 2 Suppose u(t x) (t x) isin Ω is the solution ofproblems (24) and (87) -en u(t middot)⟶ 0 in the space(S1ω2 )prime as t⟶ +infin

Proof Recall that the solution of problems (24) and (87) isgiven by the formula

u(t x) flowastG(t x) langfξ G(t x minus ξ)rang

f isin S1ω2lowast1113872 1113873prime (t x) isin Ω

(94)

Let us prove that langu(t middot)φrang⟶ 0 as t⟶ +infin for anarbitrary function φ isin S1ω2 Put

Ψt(ξ) 1113946+infin

minus infinG(t x minus ξ)φ(x)dx

ΨtR(ξ) 1113946R

minus RG(t x minus ξ)φ(x)dx Rgt 0 tgt 1

(95)

In these notations we prove that (a) for every tgt 1 andRgt 0 the function ΨtR(ξ) belongs to the space S1ω2 andΨtR(ξ)⟶Ψt(ξ) in the space S1ω2 as R⟶infin (b)Ψt(ξ) isin S1ω2 for every tgt 1 From this we get

langu(t middot)φrang 1113946+infin

minus infinlangfξ G(t x minus ξ)rangφ(x)dx

langfξ 1113946+infin

minus infinG(t x)φ(x + ξ)dxrang

langfξ 1113946+infin

minus infinG(t minus y)φ(minus (y minus ξ))dyrang

langfξ 1113946+infin

minus infinG(t minus y)φ(y minus ξ)dyrang φ(z) φ(minus z)

(96)

International Journal of Differential Equations 9

where u(t middot) is interpreted as a regular generalized functionfrom (S1ω2 )prime for every tgt 0

So let us set property (a) For fixed k m sub Z+ we have

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le 1113946+R

minus Rξkφ(x)D

mξ G(t x minus ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868dx

le 1113946+infin

minus infinξkφ(ξ + η)D

mη G(t η)

11138681113868111386811138681113868

11138681113868111386811138681113868dη

(97)

Since φ isin S1ω2 the inequality

ξkD

mξ φ(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cLkM

mk2k

mmω

k m sub Z+ (98)

with some c L Mgt 0 is valid It follows from this that forevery η isin R

supξisinR

ξkφ(ξ)11138681113868111386811138681113868

11138681113868111386811138681113868 supyisinR

(y minus η)kφ(η)

11138681113868111386811138681113868

11138681113868111386811138681113868 supyisinR

1113944

k

l0C

lky

l(minus η)

kminus lφ(y)

1113868111386811138681113868111386811138681113868111386811138681113868

1113868111386811138681113868111386811138681113868111386811138681113868

le 1113944k

l0C

lk|η|

kminus l supyisinR

ykφ(y)

11138681113868111386811138681113868

11138681113868111386811138681113868 le c 1113944k

l0C

lkL

ll2l

|η|kminus l

(99)

Next we will use estimate (58) )en

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le c 1113944k

l0C

lkL

ll2l

1113946+infin

minus infin|η|

kminus lD

mη G(t η)

11138681113868111386811138681113868

11138681113868111386811138681113868dη

le cc3Bm

mmω

tminus (1ω)

1113944

k

l0C

lkL

ll2l

middot 1113946+infin

minus infin|η|

kminus l exp minus a0tminus (12)

|η|12

1113966 1113967dη

(100)

By the direct calculations we find that

1113946+infin

minus infin|η|

kminus l exp minus a0tminus (12)

|η|12

1113966 1113967dη le c41113957L

kk2(kminus l)

tkminus l+1

c41113957Lgt 0

(101)

So

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cc3c4Bm

t1minus (1ω)

mmω

1113944

k

l0C

lkL

l1113957Lkminus l

tkminus l

l2l

k2(kminus l)

le cBm

Lkk2k

mmω

(102)

where c cc3c4t1minus (1ω) and L 2max L 1113957Lt1113864 1113865 )erefore

ΨtR(ξ) isin S1ω2 for every tgt 1 and an arbitrary Rgt 0 Fur-thermore we make sure thatΨtR(ξ)⟶Ψt(ξ) as R⟶infin

with all its derivatives uniformly in ξ on each segment[a b] sub R In addition the set of functions ξk

Dmξ ΨtR(ξ)1113966 1113967

k m sub Z+ is uniformly bounded (by R) in the space S1ω2(this property follows from estimate (102) in which c B andLgt 0 do not depend on R) )is means that condition (a) isfulfilled

Condition (b) follows from (a) because in a perfectspace every bounded set is compact

Using properties (a) and (b) we obtain the equality

langu(t middot)φrang 1113946+infin

minus infinG(t minus y)(flowastφ)(y)dy (103)

Since f isin (S1ω2 )prime is a convolutor in S1ω2 we haveflowast φ isin S1ω2 )en given estimate (58) (at s 0) we find that

|langu(t middot)φrang| le 1113946+infin

minus infin|G(t minus y)||(flowastφ)(y)|dy

lec

t1ω1113946

+infin

minus infin|(flowastφ)(y)|dy le

c0

t1ω⟶ 0 t⟶ +infin

(104)

for an arbitrary function φ isin S1ω2 ie u(t middot)⟶ 0 in thespace (S1ω2 )prime as t⟶ +infin )e theorem is proved

If the generalized function f in (87) is finite (ie itssupport suppf is a finite set in R) then we can say onuniform tending a solution u(t x) of problems (24) and (87)to zero onR as t⟶ +infin Note that every finite generalizedfunction is a convolutor in S spaces )is property followsfrom the general result concerning the theory of perfectspaces (see [12] p 137) if Φ is a perfect space with dif-ferential translation operation then every finite functional isa convolutor inΦ Finite functionals form a fairly wide classIn particular every bounded set F sub R is a support of somegeneralized function [12]

Theorem 3 Let u(t x) be a solution of problems (24) and(87) with the boundary function f in condition (87) which isan element of (S

β2)prime sub (S1ω2 )prime βgt 1 with finite support in R

-en u(t x)⟶ 0 uniformly on R as t⟶ +infin

Here is a scheme for proving this statement Letsuppf sub [a1 b1] sub [a2 b2] sub R Consider the functionφ isin S

β2 βgt 1 such that φ(x) 1 x isin [a1 b1] and

suppφ sub [a2 b2] )is function exists when βgt 1 because Sβ2

contains finite functions [12] Consider the function

u(t x) langfξ φ(ξ)G(t x minus ξ)rang +langfξ c(ξ)G(t x minus ξ)rang(105)

where c 1 minus φ Since supp(c(ξ)G(t x minus ξ)) cap suppφ emptywe have

u(t x) tminus (1ω)langfξ t

1ωφ(ξ)G(t x minus ξ)rang (106)

To prove the formulated above statement it remains toestablish that the set of functions Φtx(ξ) t1ωφ(ξ)G(t x minus

ξ) is bounded in the space Sβ2 βgt 1 for large t and x isin R

For example if in condition (87) f δ then δ is aconvolutor in the space S supp δ 0 andu(t x) δ lowastG(t x) G(t x) It follows directly from es-timate (58) that G(t x)⟶ 0 uniformly onR as t⟶ +infin

4 Conclusion

In this paper the solvability of a nonlocal multipoint by thetime problem for the evolutionary equation(zuzt) + (I minus (d2dx2))ω2u 0 (t x) isin (0 +infin) times R isproved herewith the operator (I minus (d2dx2))ω2 is inter-preted as a pseudodifferential operator in the space S1ω2

10 International Journal of Differential Equations

constructed by function (1 + σ2)ω2 σ isin R while the non-local multipoint by time condition also contains pseudo-differential operators constructed on smooth symbols )erepresentation of the solution is given in the form of aconvolution of the fundamental solution with the initialfunction which is an element of the space of generalizedfunctions of the ultradistribution type (the coagulator inspace S1ω2 ) )e behavior of the solution u(t x) t⟶ +infinin the space of generalized functions (S1ω2 )prime is investigated)e conditions for the initial generalized function are foundunder which the solution is uniformly stabilized to zero onR )e method of research of a nonlocal multipoint by timeproblem offered in this paper allows to interpret differential-operator equations of a form (zuzt) + φ(i zzx)u 0 as anevolutionary equation with a pseudodifferential operatorφ(i zzx) Fminus 1[φ middot F] constructed by the function φ actingin certain countable-normalized space of infinite-differentialfunctions (the choice of space depends on the properties ofthe function which is the symbol of the operator φ(i zzx))

Data Availability

No data were used to support the findings of this study

Disclosure

)is study was conducted in the framework of scientificactivity at Yuriy Fedkovych Chernivtsi National University

Conflicts of Interest

)e authors declare that they have no conflicts of interest

References

[1] V I Gorbachuk and M L Gorbachuk Boundary-ValueProblems for Differential-Operator Equations NaukovaDumka Kiev Ukraine 1984

[2] V V Gorodetskiy N I Nagnibida and P P NastasievMethods for Solving Problems in Functional Analysis VyshkaSchool Vyshka Ukraine 1990

[3] S G Samko A A Kilbas and O I Marichev Integrals andFractional Derivatives and Some of -eir Applications Scienceand Technology Minsk Belarus 1987

[4] I M Gelfand and G E Shilov Some Questions of the-eory ofDifferential Equations Fizmatgiz Moscow Russia 1958

[5] V V Gorodetskiy and O M Lenyuk About the FractionalDifferentiation in Spaces of Sprime type NAS of Ukraine KyivUkraine 1998

[6] A A Dezin General Questions of the -eory of BoundaryValue Problems Nauka Moscow Russia 1980

[7] A M Nakhushev Equations of Mathematical Biology HigherSchool Moscow Russia 1995

[8] I A Belavin S P Kapitsa and S P Kurdyumov ldquoAmathematical model of global demographic processes withconsidering of the spatial distributionrdquo Zh CalculationMathematics Physics vol 38 no 6 pp 885ndash902 1998

[9] V V Gorodetskiy and O V Martynyuk-e Cauchy Problemand Nonlocal Problems for First-Order Evolution EquationsEby Time Variable Vidavnichy dim rdquoRODOVIDrdquo ChernivtsiUkraine 2015

[10] V V Gorodetskiy and V I Myronyk ldquoTwo-point problem forone class of evolution equationsrdquo Differents Equationsvol 46 no 3 pp 349ndash363 2010

[11] V V Gorodetskiy and V I Myronyk ldquoTwo-point problem forone class of evolution equationsrdquo Differents Equationsvol 46 no 4 pp 520ndash526 2010

[12] I M Gelfand and G E Shilov Spaces of Basic and GeneralizedFunctions Fizmatgiz Moscow Russia 1958

International Journal of Differential Equations 11

Page 3: On One Evolution Equation of Parabolic Type with ...

the convergence of a sequence φ] ]ge 11113864 1113865 sub Sβα to zero in the

space Sβα is the convergence in the topology of some space

SβB

αA to which all the functions φ] belong In other words (see[12]) φ]⟶ 0 in S

βα as ]⟶infin if and only if for every

n isin Z+ the sequence φ(n)] ]ge 1 converges to zero uniformly

on an arbitrary segment [a b] sub R and for some c a Bgt 0independent of ] and the inequality

φ(n)] (x)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cBnn

nβ exp minus a|x|1α

1113966 1113967 x isin R n isin Z+ (11)

holdsIn S

βα the continuous translation operation Tx

φ(ξ)⟶ φ(ξ + x) is defined )is operation is also differ-entiable (even infinitely differentiable [12]) in the sense thatthe limit relation (φ(x + h) minus φ(x))hminus 1⟶ φprime(x) h⟶ 0is true for every function φ isin S

βα in the sense of convergence

in the Sβα-topology In S

βα the continuous differentiation

operator is also defined )e spaces of type S are perfect [12](that is the spaces and all bounded sets of which arecompact) and closely related to the Fourier transformnamely the formula F[S

βα] Sαβ α βgt 0 is correct where

F Sβα1113960 1113961 ≔ ψ ψ(σ) 1113946

Rφ(x)e

iσxdx φ isin Sβα1113882 1113883 (12)

Moreover the operator F Sβα⟶ Sαβ is continuousLet (S

βα)prime denote the space of all linear continuous

functionals on Sβα with weak convergence Since the trans-

lation operator Tx is defined in the space Sβα of test functions

the convolution of a generalized function f isin (Sβα)prime with the

test function φ isin Sβα is given by the following formula

(flowastφ) langf Tminus xφ(middot)rang equiv langf φ(x minus middot)rang φ(ξ) φ(minus ξ)

(13)

It follows from the infinite differentiability property ofthe argument translation operation in S

βα that the convo-

lution flowastφ is a usual infinitely differentiable function onRWe define the Fourier transform of a generalized

function f isin (Sβα)prime by the relation

langF[f] φrang langf F[φ]rang forallφ isin Sαβ (14)

where the operator F (Sβα)prime ⟶ (Sαβ)prime is continuous

Let f isin (Sβα)prime If flowastφ isin S

βα forallφ isin S

βα and the conver-

gence φ]⟶ 0 in the Sβα-topology as ]⟶infin implies that

flowastφ]⟶ 0 in the Sβα-topology as ]⟶infin then the

function f is called a convolutor in Sβα If f isin (S

βα)prime is a

convolutor in Sβα then for an arbitrary function φ isin S

βα the

formula F[flowastφ] F[f]F[φ] is valid where F[f] is amultiplier in S

βα [12]

Recall that the function g isin Cinfin(R) is called a multiplierin S

βα if gφ isin S

βα for an arbitrary function φ isin S

βα and the

mapping φ⟶ gφ is continuous in the space Sβα

3 Nonlocal Multipoint by Time Problem

Consider the function a(σ) (1 + σ2)ω2 σ isin R whereω isin [1 2) is a fixed parameter Obviously the function a(σ)

satisfies the inequality

a(σ) le cε exp ε|σ|ω

1113864 1113865 σ isin R (15)

where cε 2ω2 max 1 (1ε) for an arbitrary εgt 0 Usingdirect calculations and the Stirling formula we find that

Dmσ a(σ)

11138681113868111386811138681113868111386811138681113868 le c0B

m0 m le c1B

m1 m

m m isin N σ isin R (16)

where B0 B0(ω)gt 0 B1 B1(ω)gt 0 and c0 c1 gt 0 It fol-lows from (15) and (16) that a(σ) is a multiplier in S11ωIndeed let φ isin S11ω that is the function φ and its derivativessatisfy the inequality

Dkσφ(σ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cAkk

k exp minus a|σ|ω

1113864 1113865 k isin Z+ σ isin R (17)

with some positive constants c A and a )en using theLeibniz formula for differentiating the product of twofunctions as well as inequalities (15)ndash(17) we find that

(a(σ)φ(σ))(k)

11138681113868111386811138681113868

11138681113868111386811138681113868 le 1113944

k

l0C

lk a

(l)(σ)

11138681113868111386811138681113868

11138681113868111386811138681113868 φ(kminus l)

(σ)11138681113868111386811138681113868

11138681113868111386811138681113868 a(σ)φ(k)(σ)

11138681113868111386811138681113868

11138681113868111386811138681113868+

+ 1113944

k

l1C

lk a

(l)(σ)

11138681113868111386811138681113868

11138681113868111386811138681113868 middot φ(kminus l)(σ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cεcAkk

k exp minus a|σ|ω

1113864 1113865

+ c1c 1113944k

l1C

lkB

l1l

lA

kminus l(k minus l)

kminus l exp minus a|σ|ω

1113864 1113865

(18)

Since εgt 0 is arbitrary we can put ε (a2) )en

(a(σ)φ(σ))(k)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cεcAkk

k exp minusa

2|σ|

ω1113882 1113883

+ c1c 1113944k

l1C

lkB

l1l

lA

kminus l(k minus l)

(kminus l)

times exp minusa

2|σ|

ω1113882 1113883 le c2B

k2k

k exp minusa

2|σ|

ω1113882 1113883

σ isin R k isin Z+

(19)

where c2 c(cε + c1) and B2 2max A B11113864 1113865 It follows from(19) that a middot φ is an element of S11ω

)e operation of multiplying by a(σ) is continuous inthe space S11ω In fact let φn nge 11113864 1113865 be a sequence offunctions from S11ω convergent to 0 in this space)is meansthat φn nge 11113864 1113865 sub S

1B01ωA0

with some constants A0 B0 gt 0 and

φn

δρ sup

exp a(1 minus δ)|σ|ω middot φ(k)n (σ)

11138681113868111386811138681113868111386811138681113868

B0 + ρ( 1113857kkk

⟶ 0

n⟶infin

δ ρ1113864 1113865 sub1213 1113882 1113883

a 1

ωeAω0

(20)

In other words for an arbitrary 1113957εgt 0 there exists anumber n0 n0(1113957ε) such that for nge n0

φ(k)n (σ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le 1113957ε B0 + ρ( 1113857kk

k exp minus a(1 minus δ)|σ|ω

1113864 1113865 σ isin R k isin Z+

(21)

International Journal of Differential Equations 3

Using inequalities (15) and (16) (when putting in (15)ε (a2)(1 minus δ)) we get

a(σ)φn(σ)( 1113857(k)

11138681113868111386811138681113868

11138681113868111386811138681113868 le c31113957ε(1113957B + ρ)k exp minus

a

2(1 minus δ)|σ|

ω1113882 1113883

k isin Z+ nge n0

(22)

where c3 cε + c1 and 1113957B 2max B0 B11113864 1113865 It follows from thelast inequality that aφnδρ le c31113957ε forallnge n0 that is the se-

quence aφn nge 11113864 1113865 converges to zero in the space S11113957B1ω1113957A

where 1113957A 21ωA0 and 1113957B 2max B0 B11113864 1113865 )is means thatthe sequence aφn nge 11113864 1113865 converges to zero in the space S11ωwhich is what we needed to prove

Remark 1 It follows from the proven property that thefunction a(σ) (1 + σ2)ω2 σ isin R is also a multiplier inevery space S

β1ω where βgt 1 )erefore in the space S1ωβ

βge 1 defined as a continuous linear pseudodifferentialoperator A constructed by the function a(σ)

Aφ Fminus 1

[a(σ)F[φ]] forallφ isin S1ωβ (23)

where Aφ (I minus Δ)ω2φ Δ (d2dx2)Let us consider the evolution equation with the operator

A (the Bessel fractional differentiation operator)zu

zt+ Au 0 (t x) isin (0 +infin) times R equiv Ω (24)

By a solution of equation (24) we mean the functionu(t x) (t x) isin Ω such that (1) u(t middot) isin C1(0 +infin) forevery x isin R (2) u(middot x) isin S1ωβ for every t isin (0 +infin) (3)u(t x) is continuous at every point (0 x) of the boundaryΓΩ 0 times R of the region Ω (4) existMR⟶ [0infin) forallt isin (0 +infin) |ut(t x)| le M(x)intRM(x)dxltinfin and (5) u(t x) (t x) isin Ω satisfiesequation (24)

For equation (24) we consider the nonlocal multipointby time problem of finding a solution of equation (24) thatsatisfies the condition

μu(0 x) minus μ1B1u t1 x( 1113857 minus middot middot middot minus μmBmu tm x( 1113857 f(x)

x isin R f isin S1ωβ

(25)

where u(0 x) ≔ limt⟶+0u(t x) x isin R m isin Nμ μ1 μm1113864 1113865 sub (0infin) and t1 tm1113864 1113865 sub (0infin) are fixednumbers 0lt t1 lt t2 lt middot middot middot lt tm lt +infin μgt 1113936

mk1 μk and

B1 Bm are pseudodifferential operators in S1ωβ con-structed by functions (characters) gk R⟶ (0 +infin) re-spectively Bk Fminus 1[gk(σ)F] k isin 1 m )e functionsgk isin Cinfin(R) and k isin 1 m satisfy the conditions

forallεgt 0forallσ isin R gk(σ) le exp ε|σ|ω

1113864 1113865

existLk gt 0foralls isin N Dsσgk(σ)

11138681113868111386811138681113868111386811138681113868 le L

sks

s

(26)

Note that the above properties of the functions gk implythat gk k isin 1 m is a multiplier in S1ωβ

We are looking for the solution of problems (24) and(25) via the Fourier transform Due to condition (19)

F ut1113858 1113859 1113946R

ut(t x)eiσxdx

z

zt1113946R

u(t x)eiσxdx

z

ztF[u]

(27)

We introduce the notation F[u(t x)] v(t σ) Giventhe form of the operators A B1 Bm we get

F[Au(t x)] F Fminus 1

[a(σ)F[u]]1113960 1113961 a(σ)F[u(t x)] a(σ)v(t σ)

F Bku(t x)1113858 1113859 gk(σ)F[u(t x)] gk(σ)v(t σ) k isin 1 m

(28)

So for the function v Ω⟶ R we arrive at a problemwith parameter σ

dv(t σ)

dt+ a(σ)v(t σ) 0 (t σ) isin Ω (29)

μv(0 σ) minus 1113944

m

k1μkgk(σ)v tk σ( 1113857 1113957f(σ) σ isin R (30)

where 1113957f(σ) F[f](σ) )e general solution of equation(29) has the form

v(t σ) c exp minus ta(σ) (t σ) isin Ω (31)

where c c(σ) is determined by condition (30) Substituting(30) into (31) we find that

c 1113957f(σ) μ minus 1113944m

k1μkgk(σ)exp minus tka(σ)1113864 1113865⎛⎝ ⎞⎠

minus 1

σ isin R

(32)

Now put G(t x) ≔ Fminus 1[Q(t σ)] andQ(t σ) Q1(t σ)Q2(σ) and thus

Q1(t σ) exp minus ta(σ)

Q2(σ) μ minus 1113944m

k1μkgk(σ)Q1 tk σ( 1113857⎛⎝ ⎞⎠

minus 1

(33)

)en thinking formally we come to the relation

u(t x) 1113946R

G(t x minus ξ)f(ξ)dξ G(t x)lowastf(x) (34)

Indeed

u(t x) (2π)minus 1

1113946R

Q(t σ) 1113946R

f(ξ)eiσξdξ1113874 1113875e

minus iσxdσ

1113946R

(2π)minus 1

1113946R

Q(t σ)eminus iσ(xminus ξ)dσ1113874 1113875f(ξ)dξ

1113946R

G(t x minus ξ)f(ξ)dξ G(t x)lowastf(x) (t x) isin Ω

(35)

)e correctness of the transformations performed theconvergence of the corresponding integrals and conse-quently the correctness of formula (35) follow from theproperties of the function G which are given below )eproperties of G are determined by the properties of Q

4 International Journal of Differential Equations

because G Fminus 1[Q] So let us first examine the properties ofthe function Q(t σ) as a function of the variable σ

Lemma 1 For derivatives of Q1(t σ) (t σ) isin Ω theestimates

DsσQ1(t σ)

11138681113868111386811138681113868111386811138681113868 le cA

stcs

ss exp minus t|σ|

ω1113864 1113865 (36)

are valid where c 0 if 0lt t le 1 and c 1 if tgt 1 and theconstants cgt 1 and Agt 0 do not depend on t

Proof To prove the statement we use the Faa di Brunoformula for differentiation of a complex function

DsσF(g(σ)) 1113944

s

p1

dp

dgpF(g) 1113944

s

p1 pl

ddσ

g(σ)1113888 1113889

p1

1l

dl

dσlg(σ)1113888 1113889

pl

s isin N

(37)

where the sum sign is applied to all the solutions in positiveintegers of the equation p1 + 2p2 + middot middot middot + lpl sp1 + middot middot middot + pl m Put F eg and g minus ta(σ) )en

Dsσe

minus ta(σ) e

minus ta(σ)1113944

s

p11113944

s

p1 plΛ (38)

where the symbol Λ denotes the expression and

Λ ≔ddσ

(minus ta(σ))1113888 1113889

p1

middot12

d2

dσ2(minus ta(σ))1113888 1113889

p2

1l

dl

dσl(minus ta(σ))1113888 1113889

pl

(39)

Given estimate (16) is fulfilled we find that

|Λ| le cp1+middotmiddotmiddot+pl

0 Bp1+2p2+middotmiddotmiddot+lpl

0 tp1+middotmiddotmiddot+pl le 1113957c

s0t

pB

s0 1113957c0 max 1 c01113864 1113865

(40)

Using (40) and the Stirling formula we arrive at theinequalities

DsσQ1(t σ)

11138681113868111386811138681113868111386811138681113868 le 1113957c

s0B

s0t

css exp minus ta(σ)

le cAstcs

ss exp minus t|σ|

ω1113864 1113865 σ isin R

(41)

where c 0 if 0lt t le 1 and c 1 if tgt 1 and the values cgt 1and Agt 0 do not depend on t

Lemma is proved

Remark 2 It follows from estimate (41) that Q1(t middot) isin S11ωfor every tgt 0

Lemma 2 -e function Q2 is a multiplier in S21ω

Proof To prove the assertion let us estimate the derivativesof Q2 For this we use formula (37) in which we put F φminus 1

and φ R where

R(σ) μ minus 1113944m

k1μkgk(σ)exp minus tka(σ)1113864 1113865 (42)

)en Q2 F(φ) Rminus 1 and

DsσQ2(σ)

11138681113868111386811138681113868111386811138681113868 1113944

s

p1

dp

dRpR

minus 11113944

s

p1 pl

ddσ

R(σ)1113888 1113889

p11113868111386811138681113868111386811138681113868111386811138681113868

1l

dl

dσlR(σ)1113888 1113889

pl

s isin N

(43)

Given the properties of g1 gm and inequality (41)we find that

1l

dj

dσjR(σ)

11138681113868111386811138681113868111386811138681113868

11138681113868111386811138681113868111386811138681113868le

1j

1113944

m

k1μk 1113944

j

i0C

ij D

iσgk(σ)

11138681113868111386811138681113868

11138681113868111386811138681113868 middot Djminus iσ e

minus tka(σ)11138681113868111386811138681113868

11138681113868111386811138681113868

le1j

1113944

m

k1μk 1113944

i

j0C

ijL

iki1113957c

jminus i0 t

c(jminus i)

k (j minus i)eminus tka(σ)

le 1113944m

k1μk 1113944

j

i0C

ijL

ik1113957c

jminus i0 t

c(jminus i)

(44)

where we took into account that i(j minus i) le j)Let

1113957L max L1 Lm1113864 1113865

L0 2max 1113957L 1113957c0T1113864 1113865(45)

then

1j

dj

dσjR(σ)

11138681113868111386811138681113868111386811138681113868

11138681113868111386811138681113868111386811138681113868le α0L

j0 α0 1113944

m

k1μk j isin 1 l

ddσ

R(σ)1113888 1113889

p1111386811138681113868111386811138681113868111386811138681113868

111386811138681113868111386811138681113868111386811138681113868

1l

dl

dσlR(σ)1113888 1113889

pl111386811138681113868111386811138681113868111386811138681113868

111386811138681113868111386811138681113868111386811138681113868le α0L0( 1113857

p α0L201113872 1113873

p2 α0L

l01113872 1113873

pl

αp1+middotmiddotmiddot+pl

0 Lp1+2p2+middotmiddotmiddot+lpl

0 αp0L

s0

(46)

In addition (dpdRp)Rminus 1 (minus 1)ppRminus (p+1) and

Rminus 1

(σ) Q2(σ) μ minus 1113944m

k1μkgk(σ)exp minus tka(σ)1113864 1113865⎛⎝ ⎞⎠

minus 1

le μ minus 1113944m

k1μk exp minus t1|σ|

ω+ ε|σ|

ω1113864 1113865⎛⎝ le μ minus 1113944

m

k1μk

⎛⎝ ⎞⎠

minus 1

equiv β0 gt 0 ε t1

(47)

Since by assumption μgt 1113936mk1 μk (assuming the prop-

erties of g1 gm and 0lt t1 lt t2 lt middot middot middot lt tm) So

International Journal of Differential Equations 5

dp

dRpR

minus 111138681113868111386811138681113868111386811138681113868

11138681113868111386811138681113868111386811138681113868le βp+1

0 p

DsσQ2(σ)

11138681113868111386811138681113868111386811138681113868 le s 1113944

s

p1βp+10 pαp

0Ls0

le β0Ls0(s)

21113944

s

p1βp0 le β0L

s0s middot (s)

2βs1 le β2β

s3s

2s s isin N

(48)

It follows from the last inequality and boundedness ofthe function Q2(σ) on R that Q2 is a multiplier in S21ω

Corollary 1 For every tgt 0 the functionQ(t σ) Q1(t σ)Q2(σ) σ isin R is an element of the spaceS21ω and the estimates

DsσQ(t σ)

11138681113868111386811138681113868111386811138681113868 le 1113957c1113957A

stcs

s2s exp minus t|σ|

ω1113864 1113865 s isin Z+ (t σ) isin Ω

(49)

are valid where the constants 1113957c and 1113957Agt 0 do not depend on t

Taking into account the properties of the Fouriertransform (direct and inverse) and the formulaFminus 1[S21ω] S1ω2 we get that G(t middot) isin S1ω2 for every tgt 0 Weremove in the estimates of derivatives of the function G (inthe variable x) the dependence on t assuming tgt 1 To dothis we use the relations

xkD

sxF[φ](x) i

k+sF σsφ(σ)( 1113857

(k)1113876 1113877

ik+s

1113946R

σsφ(σ)( 1113857(k)

eiσxdσ k s sub Z+ φ isin S

21ω

(50)

So

xkD

sxG(t x) (2π)

minus 1ik+s

(minus 1)s1113946R

σsQ(t minus σ)( 1113857

(k)e

iσxdσ

(51)

Applying the Leibniz formula for differentiating theproduct of two functions and estimating the derivatives ofthe function Q(t σ) we find that

σsQ(t minus σ)( 1113857

(k)11138681113868111386811138681113868

11138681113868111386811138681113868 1113944k

p0C

p

k σs( 1113857

(p)Q

(kminus p)(t minus σ)

1113868111386811138681113868111386811138681113868111386811138681113868

1113868111386811138681113868111386811138681113868111386811138681113868

le σsQ

(k)(t minus σ)

11138681113868111386811138681113868

11138681113868111386811138681113868 + ks σsminus 1Q

(kminus 1)(t minus σ)

11138681113868111386811138681113868

11138681113868111386811138681113868

+k(k minus 1)

2s(s minus 1) σsminus 2

Q(kminus 2)

(t minus σ)11138681113868111386811138681113868

11138681113868111386811138681113868

+ middot middot middot le 1113957c 1113957Aktck 1113957B

stminus (sω)

mks + ks1113957Akminus 1

tc(kminus 1)

1113876

middot 1113957Bsminus 1

tminus ((sminus 1)ω)

mkminus 1sminus 1

+12

k(k minus 1)s(s minus 1)1113957Akminus 2

tc(kminus 2) 1113957B

sminus 2tminus ((sminus 2)ω)

middot mkminus 2sminus 2 + 1113961eminus (t2)|σ|ω

(52)

where mks k2kssω Taking into account the results in (see[12] p 236ndash243) we find that this double sequence satisfiesthe inequality

ksmkminus 1sminus 1

mks

le 1113957c(k + s) 1113957cgt 0 (53)

Bearing in mind the last inequality and also that tgt 1 weobtain

σsQ(t minus σ)( 1113857

(k)11138681113868111386811138681113868

11138681113868111386811138681113868 le 1113957c1113957Aktk 1113957B

smks 1 +

ks

1113957A1113957B

mkminus 1sminus 1

mks

+12

k(k minus 1)s(s minus 1)1

1113957A21113957B

2mkminus 2sminus 2

mks

+ middot middot middot1113888 1113889eminus (t2)|σ|ω le 1113957c1113957A

ktk 1113957B

smks

times 1 +ks

1113957A1113957B

mkminus 1sminus 1

mks

+12

11113957A21113957B

2 ksmkminus 1sminus 1

mks

(k minus 1)(s minus 1)mkminus 2sminus 2

mkminus 1sminus 1+ middot middot middot1113888 1113889e

minus (t2)|σ|ω

le 1113957c1113957Aktk 1113957B

smks 1 +

1113957c

1113957A1113957B(k + s) +

12

1113957c2

(1113957A1113957B)2(k + s)

2+ middot middot middot1113888 1113889e

minus (t2)|σ|ω

le c1AktkB

smkse

minus (t2)|σ|ω

A 1113957A exp1113957c

1113957A1113957B1113882 1113883

B 1113957B exp1113957c

1113957A1113957B1113882 1113883

(54)

6 International Journal of Differential Equations

So

xkD

sxG(t x)

11138681113868111386811138681113868

11138681113868111386811138681113868 le (2π)minus 1

c1AktkB

smks1113946

Re

minus (t2)|σ|ωdσ

c2AktkB

stminus (1ω)

k2k

ssω

k s sub Z+

(55)

)en

DsxG(t x)

11138681113868111386811138681113868111386811138681113868 le c2B

stminus (1ω)

ssω inf

k

Akk2k

tminus 1|x|( )k

le c3Bstminus (1ω)

ssω exp minus a0t

minus (12)|x|

121113966 1113967

(56)

where the constants c3 B and a0 gt 0 do not depend on there we used the well-known inequality (see [12] p 204)

eminus (αe)|ξ|1α le inf

k

Lkkkα

|ξ|kle ce

minus (αe)|ξ|1α c e

αe2 (57)

in which α 2 and L 1113957A )us such a statement is correct

Lemma 3 -e derivatives (by variable x) of the functionG(t x) for tgt 1 satisfy the inequality

DsxG(t x)

11138681113868111386811138681113868111386811138681113868 le c3B

ss

sωtminus (1ω) exp minus a0t

minus (12)|x|

121113966 1113967 s isin Z+

(58)

where the constants c3 B and a0 gt 0 do not depend on t

Remark 3 In what follows we assume that in condition(25) β 2

Here are a few more properties of the function G(t x)

Lemma 4 -e function G(t middot) t isin (0 +infin) as an abstractfunction of t with values in the space S1ω2 is differentiable in t

Proof It follows from the continuity of the Fourier trans-form (direct and inverse) that to prove the statement itsuffices to establish that the function F[G(t middot)] Q(t middot) as afunction of a parameter t with values in the space S21ω isdifferentiable in t In other words it is necessary to provethat the boundary value relation

ΦΔt(σ) ≔1Δt

[Q(t + Δt middot) minus Q(t middot)]⟶z

ztQ(t middot) Δt⟶ 0

(59)

is performed in the sense as follows

(1) DsσΦΔt(σ) ⟶

Δt⟶ 0Ds

σ(minus a(σ)Q(t σ)) s isin Z+ uni-formly on each segment [a b] sub R

(2) |DsσΦΔt(σ)| le cB

ss2s exp minus a|σ|ω s isin Z+ where the

constants c B and agt 0 do not depend on Δt if Δt issmall enough

)e function Q(t σ) (t σ) isin Ω is differentiable in t inthe usual sense Due to the Lagrange theorem on finiteincrements

ΦΔt(σ) minus a(σ)Q(t + θΔt σ) (60)

So

DsσΦΔt(σ) minus 1113944

s

l0C

lsD

lσa(σ)D

sminus lσ Q(t +θΔtσ) (61)

Dsσ ΦΔt(σ) minus

z

ztQ(tσ)1113888 1113889 minus 1113944

s

l0C

lsD

lσa(σ) D

sminus lσ Q(t +θΔtσ)1113960

minus Dsminus lσ Q(tσ)1113961

(62)

Since

Dsminus lσ Q(t + θΔt σ) minus D

sminus lσ Q(t σ) D

sminus l+1σ Q t + θ1Δt σ( 1113857

0lt θ1 lt 1

(63)

it follows from this and estimate (49) that

Dsminus l+1σ Q t + θ1Δt σ( 1113857θΔt⟶ 0 Δt⟶ 0 (64)

uniformly on an arbitrary segment [a b] sub R )en

DsσΦΔt(σ)⟶ D

z

ztQ(t σ)1113888 1113889 Δt⟶ 0 (65)

uniformly on each segment [a b] sub R too )ereforecondition 1 is satisfied

Taking into account (61) estimates (15) (16) and (49)for a(σ) Q(t σ) and their derivatives we find that

DsσΦΔt(σ)

11138681113868111386811138681113868111386811138681113868 le c1cε 1113944

s

l0C

lsB

l1l

l 1113957Asminus l

tc(sminus l)

(s minus l)2(sminus l)

exp minus (t + θΔt)|σ|ω

+ ε|σ|ω

1113864 1113865

(66)

where εgt 0 is arbitrary fixed Take ε (t2) and then

DsσΦΔt(σ)

11138681113868111386811138681113868111386811138681113868 le cB

ss2s exp minus a|σ|

ω1113864 1113865 s isin Z+ (67)

where B 2max B11113957Atc1113966 1113967 a (t2) and all the constants

do not depend on Δt

Corollary 2 -e formulaz

zt(flowastG(t middot)) f lowast

zG(t middot)

zt forallf isin S

1ω21113872 1113873prime tgt 0

(68)

is correct

Proof According to the definition of a convolution of ageneralized function with a test one we have

flowastG(t middot) langfξ Tminus xG(t ξ)rang

G(t ξ) G(t minus ξ)(69)

)enz

zt(flowastG(t middot)) lim

Δt⟶0

1Δt

[(flowastG(t + Δt middot)) minus (flowastG(t middot))]

limΔt⟶0langfξ

1Δt

Tminus xG(t + Δt ξ) minus Tminus x

G(t ξ)1113960 1113961rang

(70)

International Journal of Differential Equations 7

Due to Lemma 4 the relation1Δt

Tminus xG(t + Δt ξ) minus Tminus x

G(t ξ)1113960 1113961⟶z

ztTminus x

G(t ξ) Δt⟶ 0

(71)

is performed in the sense of S1ω2 topology So

z

zt(flowastG(t middot)) langfξ lim

Δt⟶0

1Δt

Tminus xG(t + Δt ξ) minus Tminus x

G(t ξ)1113960 1113961rang

langfξ z

ztTminus x

G(t ξ)rang langfξ Tminus x

z

ztG(t ξ)rang

flowastz

ztG(t middot)

(72)

)e statement is provedSince

F[G(t middot)] Q(t σ) Q1(t σ)Q2(σ) 1113946R

G(t x)eiσxdx tgt 0

(73)

we get the formula

1113946R

G(t x)dx Q1(t 0)Q2(0) λ0eminus t

λ0 μ minus 1113944m

k1μkgk(0)e

minus tk⎛⎝ ⎞⎠

minus 1

(74)

where a(0) 1 is taken into account

Lemma 5 In (S1ω2 )prime the following relations are correct

(1) G(t middot)⟶ Fminus 1

Q21113858 1113859 t⟶ + 0

(2) μG(t middot) minus 1113944m

k1μkBkG tk middot( 1113857⟶ δ t⟶ + 0

(75)

where δ is the Dirac delta function

Proof (1) Taking into account the continuity property of theFourier transform (direct and inverse) in spaces of Sprime type itis sufficient to establish that

F[G(t middot)] Q1(t middot)Q2(middot)⟶ Q2(middot) t⟶ + 0 (76)

in the space (S21ω)prime To do this we take an arbitrary functionφ isin S21ω and using the fact thatQ2 is a multiplier in the spaceS21ω and the Lebesgue theorem on the limit passage underthe Lebesgue integral sign we find that

langQ1(t middot)Q2(middot)φrang langQ1(t middot) Q2(middot)φ(middot)rang

1113946R

Q1(t σ)Q2(σ)φ(σ)dσ ⟶t⟶+0

1113946R

Q2(σ)φ(σ)dσ

lang1 Q2(middot)φ(middot)rang langQ2φrang

(77)

)is leads us to statement (1) of Lemma 5(2) Given statement (1) and the form of the operators

B1 Bm we find that

μG(t middot) minus 1113944m

l1μlBlG tl middot( 1113857 ⟶

t⟶+0μF

minus 1Q21113858 1113859 minus 1113944

m

l1μlBlG tl middot( 1113857

μFminus 1

Q21113858 1113859 minus 1113944m

l1μlF

minus 1gl(middot)Q1 tl middot( 1113857Q2(middot)1113858 1113859

Fminus 1 μQ2(middot) minus 1113944

m

l1μlgl(middot⎡⎣ ⎞⎠Q1 tl middot( 1113857Q2(middot)]

Fminus 1 μ minus 1113944

m

l1μlgl(middot)Q1 tl middot( 1113857Q2(middot)⎡⎣ ⎤⎦

Fminus 1 μ minus 1113944

m

l1μlgl(middot)Q1 tl middot( 1113857⎛⎝ ⎞⎠⎡⎢⎢⎣

middot μ minus 1113944m

l1μlgl(middot)Q1 tl middot( 1113857⎛⎝ ⎞⎠

minus 1⎤⎥⎥⎥⎥⎦ F

minus 1[1] δ

(78)

)erefore relation (75) is fulfilled in the space (S1ω2 )primeLemma 5 is proved

Remark 4 If μ 1 μ1 middot middot middot μm 0 then problems (24)and (25) degenerate into the Cauchy problem for equation(24) In this case Q2(σ) 1 forallσ isin R G(t x) Fminus 1[eminus ta(σ)]and G(t middot)⟶ Fminus 1[1] δ for t⟶ +0 in the space (S1ω1 )prime

Corollary 3 Let

ω(t x) flowastG(t x) f isin S1ω2lowast1113872 1113873prime (t x) isin Ω (79)

where (S1ω2lowast )prime is a class of convolutors in S1ω2 -en in (S1ω2 )primethe following boundary relation holds

μω(t middot) minus 1113944m

k1μkBkω tk middot( 1113857⟶ f t⟶ + 0 (80)

Proof Let us prove that the relation

F μω(t middot) minus 1113944

m

k1μkBkω tk middot( 1113857⎡⎣ ⎤⎦⟶ F[f] t⟶ + 0

(81)

takes place in the space (S21ω)prime Since f isin (S1ω2lowast )prime andG(t middot) isin S1ω2 for every tgt 0 we have

F[ω(t middot)] F[flowastG(t middot)] F[f]F[G(t middot)] F[f]Q(t middot)

(82)

So we need to prove that

F[f] μQ(t middot) minus 1113944m

k1μkgk(middot)Q tk middot( 1113857⎛⎝ ⎞⎠⟶ F[f] (83)

in (S21ω)prime when t⟶ + 0 SinceQ(t middot) Q1(t middot)Q2(middot)⟶ Q2(middot) for t⟶ + 0 in the space(S21ω)prime (see the proof of assertion 1 and Lemma 5) thecorrelation

8 International Journal of Differential Equations

μQ(t middot) minus 1113944m

k1μkgk(middot)Q tk middot( 1113857 ⟶

t⟶+0μQ2(middot) minus 1113944

m

k1μkgk(middot)Q1 tk middot( 1113857Q2(middot)

μ minus 1113944m

k1μkgk(middot)Q1 tk middot( 1113857⎛⎝ ⎞⎠Q2(middot) 1

(84)

is realized in the space (S21ω)prime )us relation (81) and (80)are fulfilled in the corresponding spaces )e statement isproved

)e function G(t middot) satisfies equation (24) as tgt 0Indeed

z

ztG(t x)

z

ztF

minus 1[Q(t σ)] F

minus 1 z

ztQ(t σ)1113890 1113891

Fminus 1

[a(σ)Q(t σ)]

AG(t x) Fminus 1

a(σ)Fminus 1

[G(t middot)]1113960 1113961 Fminus 1

[a(σ)Q(t σ)]

(85)

So

zG(t x)

zt+ AG(t x) 0 (t x) isin Ω (86)

which is what had to be provedIt follows from Corollary 3 that the nonlocal multipoint

by the time problem for equation (24) can be formulated inthe following way find a function u(t x) (t x) isin Ω thatsatisfies equation (24) and the condition

μ limt⟶+0

u(t middot) minus 1113944

m

k1μkBku tk middot( 1113857 f f isin S

1ω2lowast1113872 1113873prime (87)

where boundary relation (87) is considered in the space(S1ω2 )prime and the constraints on the parameters μ μ1 μmt1 tm are the same as in case of problems (24) and(25))

Theorem 1 By nonlocal multipoint by time problem (24)(87) is solvable and its solution is given by the formula

u(t x) flowastG(t x) (t x) isin Ω (88)

where u(t middot) isin S1ω2 for every tgt 0

Proof Make sure that the function u(t x) (t x) isin Ω sat-isfies equation (24) Indeed (see Corollary 2)

zu(t x)

zt

z

zt(flowastG(t middot)) flowast

zG(t middot)

zt

Au(t x) Fminus 1

[a(σ)F[f lowastG(t middot)]]

(89)

Since f is a convolutor in S1ω2

F[flowastG(t middot)] F[f]F[G(t middot)] F[f]Q(t middot) (90)

So

Au(t x) Fminus 1

[a(σ)Q(t σ)F[f](σ)] minus Fz

ztQ(t middot)F[f]1113890 1113891

minus F Fz

ztG(t middot)1113890 1113891 middot F[f]1113890 1113891

minus Fminus 1

F flowastzG(t middot)

zt1113890 11138911113890 1113891 minus flowast

zG(t middot)

zt

(91)

It follows from this that the function u(t x) (t x) isin Ωsatisfies equation (24) From Corollary 3 we get that u(t x)

satisfies condition (87) in the defined sense )e theorem isproved

Remark 5 If in condition (87) B1 middot middot middot Bm I (I is theidentity operator) then we can prove that problems (24) and(87) are well posed and its solution is given by the formula

u(t x) flowastG(t x) f isin S1ω1lowast1113872 1113873prime (t x) isin Ω (92)

where G(t middot) Fminus 1[Q(t middot)] isin S1ω1 for every tgt 0 and

Q(t σ) eminus ta(σ) μ minus 1113944

m

k1μke

minus tka(σ)⎛⎝ ⎞⎠

minus 1

(93)

Theorem 2 Suppose u(t x) (t x) isin Ω is the solution ofproblems (24) and (87) -en u(t middot)⟶ 0 in the space(S1ω2 )prime as t⟶ +infin

Proof Recall that the solution of problems (24) and (87) isgiven by the formula

u(t x) flowastG(t x) langfξ G(t x minus ξ)rang

f isin S1ω2lowast1113872 1113873prime (t x) isin Ω

(94)

Let us prove that langu(t middot)φrang⟶ 0 as t⟶ +infin for anarbitrary function φ isin S1ω2 Put

Ψt(ξ) 1113946+infin

minus infinG(t x minus ξ)φ(x)dx

ΨtR(ξ) 1113946R

minus RG(t x minus ξ)φ(x)dx Rgt 0 tgt 1

(95)

In these notations we prove that (a) for every tgt 1 andRgt 0 the function ΨtR(ξ) belongs to the space S1ω2 andΨtR(ξ)⟶Ψt(ξ) in the space S1ω2 as R⟶infin (b)Ψt(ξ) isin S1ω2 for every tgt 1 From this we get

langu(t middot)φrang 1113946+infin

minus infinlangfξ G(t x minus ξ)rangφ(x)dx

langfξ 1113946+infin

minus infinG(t x)φ(x + ξ)dxrang

langfξ 1113946+infin

minus infinG(t minus y)φ(minus (y minus ξ))dyrang

langfξ 1113946+infin

minus infinG(t minus y)φ(y minus ξ)dyrang φ(z) φ(minus z)

(96)

International Journal of Differential Equations 9

where u(t middot) is interpreted as a regular generalized functionfrom (S1ω2 )prime for every tgt 0

So let us set property (a) For fixed k m sub Z+ we have

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le 1113946+R

minus Rξkφ(x)D

mξ G(t x minus ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868dx

le 1113946+infin

minus infinξkφ(ξ + η)D

mη G(t η)

11138681113868111386811138681113868

11138681113868111386811138681113868dη

(97)

Since φ isin S1ω2 the inequality

ξkD

mξ φ(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cLkM

mk2k

mmω

k m sub Z+ (98)

with some c L Mgt 0 is valid It follows from this that forevery η isin R

supξisinR

ξkφ(ξ)11138681113868111386811138681113868

11138681113868111386811138681113868 supyisinR

(y minus η)kφ(η)

11138681113868111386811138681113868

11138681113868111386811138681113868 supyisinR

1113944

k

l0C

lky

l(minus η)

kminus lφ(y)

1113868111386811138681113868111386811138681113868111386811138681113868

1113868111386811138681113868111386811138681113868111386811138681113868

le 1113944k

l0C

lk|η|

kminus l supyisinR

ykφ(y)

11138681113868111386811138681113868

11138681113868111386811138681113868 le c 1113944k

l0C

lkL

ll2l

|η|kminus l

(99)

Next we will use estimate (58) )en

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le c 1113944k

l0C

lkL

ll2l

1113946+infin

minus infin|η|

kminus lD

mη G(t η)

11138681113868111386811138681113868

11138681113868111386811138681113868dη

le cc3Bm

mmω

tminus (1ω)

1113944

k

l0C

lkL

ll2l

middot 1113946+infin

minus infin|η|

kminus l exp minus a0tminus (12)

|η|12

1113966 1113967dη

(100)

By the direct calculations we find that

1113946+infin

minus infin|η|

kminus l exp minus a0tminus (12)

|η|12

1113966 1113967dη le c41113957L

kk2(kminus l)

tkminus l+1

c41113957Lgt 0

(101)

So

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cc3c4Bm

t1minus (1ω)

mmω

1113944

k

l0C

lkL

l1113957Lkminus l

tkminus l

l2l

k2(kminus l)

le cBm

Lkk2k

mmω

(102)

where c cc3c4t1minus (1ω) and L 2max L 1113957Lt1113864 1113865 )erefore

ΨtR(ξ) isin S1ω2 for every tgt 1 and an arbitrary Rgt 0 Fur-thermore we make sure thatΨtR(ξ)⟶Ψt(ξ) as R⟶infin

with all its derivatives uniformly in ξ on each segment[a b] sub R In addition the set of functions ξk

Dmξ ΨtR(ξ)1113966 1113967

k m sub Z+ is uniformly bounded (by R) in the space S1ω2(this property follows from estimate (102) in which c B andLgt 0 do not depend on R) )is means that condition (a) isfulfilled

Condition (b) follows from (a) because in a perfectspace every bounded set is compact

Using properties (a) and (b) we obtain the equality

langu(t middot)φrang 1113946+infin

minus infinG(t minus y)(flowastφ)(y)dy (103)

Since f isin (S1ω2 )prime is a convolutor in S1ω2 we haveflowast φ isin S1ω2 )en given estimate (58) (at s 0) we find that

|langu(t middot)φrang| le 1113946+infin

minus infin|G(t minus y)||(flowastφ)(y)|dy

lec

t1ω1113946

+infin

minus infin|(flowastφ)(y)|dy le

c0

t1ω⟶ 0 t⟶ +infin

(104)

for an arbitrary function φ isin S1ω2 ie u(t middot)⟶ 0 in thespace (S1ω2 )prime as t⟶ +infin )e theorem is proved

If the generalized function f in (87) is finite (ie itssupport suppf is a finite set in R) then we can say onuniform tending a solution u(t x) of problems (24) and (87)to zero onR as t⟶ +infin Note that every finite generalizedfunction is a convolutor in S spaces )is property followsfrom the general result concerning the theory of perfectspaces (see [12] p 137) if Φ is a perfect space with dif-ferential translation operation then every finite functional isa convolutor inΦ Finite functionals form a fairly wide classIn particular every bounded set F sub R is a support of somegeneralized function [12]

Theorem 3 Let u(t x) be a solution of problems (24) and(87) with the boundary function f in condition (87) which isan element of (S

β2)prime sub (S1ω2 )prime βgt 1 with finite support in R

-en u(t x)⟶ 0 uniformly on R as t⟶ +infin

Here is a scheme for proving this statement Letsuppf sub [a1 b1] sub [a2 b2] sub R Consider the functionφ isin S

β2 βgt 1 such that φ(x) 1 x isin [a1 b1] and

suppφ sub [a2 b2] )is function exists when βgt 1 because Sβ2

contains finite functions [12] Consider the function

u(t x) langfξ φ(ξ)G(t x minus ξ)rang +langfξ c(ξ)G(t x minus ξ)rang(105)

where c 1 minus φ Since supp(c(ξ)G(t x minus ξ)) cap suppφ emptywe have

u(t x) tminus (1ω)langfξ t

1ωφ(ξ)G(t x minus ξ)rang (106)

To prove the formulated above statement it remains toestablish that the set of functions Φtx(ξ) t1ωφ(ξ)G(t x minus

ξ) is bounded in the space Sβ2 βgt 1 for large t and x isin R

For example if in condition (87) f δ then δ is aconvolutor in the space S supp δ 0 andu(t x) δ lowastG(t x) G(t x) It follows directly from es-timate (58) that G(t x)⟶ 0 uniformly onR as t⟶ +infin

4 Conclusion

In this paper the solvability of a nonlocal multipoint by thetime problem for the evolutionary equation(zuzt) + (I minus (d2dx2))ω2u 0 (t x) isin (0 +infin) times R isproved herewith the operator (I minus (d2dx2))ω2 is inter-preted as a pseudodifferential operator in the space S1ω2

10 International Journal of Differential Equations

constructed by function (1 + σ2)ω2 σ isin R while the non-local multipoint by time condition also contains pseudo-differential operators constructed on smooth symbols )erepresentation of the solution is given in the form of aconvolution of the fundamental solution with the initialfunction which is an element of the space of generalizedfunctions of the ultradistribution type (the coagulator inspace S1ω2 ) )e behavior of the solution u(t x) t⟶ +infinin the space of generalized functions (S1ω2 )prime is investigated)e conditions for the initial generalized function are foundunder which the solution is uniformly stabilized to zero onR )e method of research of a nonlocal multipoint by timeproblem offered in this paper allows to interpret differential-operator equations of a form (zuzt) + φ(i zzx)u 0 as anevolutionary equation with a pseudodifferential operatorφ(i zzx) Fminus 1[φ middot F] constructed by the function φ actingin certain countable-normalized space of infinite-differentialfunctions (the choice of space depends on the properties ofthe function which is the symbol of the operator φ(i zzx))

Data Availability

No data were used to support the findings of this study

Disclosure

)is study was conducted in the framework of scientificactivity at Yuriy Fedkovych Chernivtsi National University

Conflicts of Interest

)e authors declare that they have no conflicts of interest

References

[1] V I Gorbachuk and M L Gorbachuk Boundary-ValueProblems for Differential-Operator Equations NaukovaDumka Kiev Ukraine 1984

[2] V V Gorodetskiy N I Nagnibida and P P NastasievMethods for Solving Problems in Functional Analysis VyshkaSchool Vyshka Ukraine 1990

[3] S G Samko A A Kilbas and O I Marichev Integrals andFractional Derivatives and Some of -eir Applications Scienceand Technology Minsk Belarus 1987

[4] I M Gelfand and G E Shilov Some Questions of the-eory ofDifferential Equations Fizmatgiz Moscow Russia 1958

[5] V V Gorodetskiy and O M Lenyuk About the FractionalDifferentiation in Spaces of Sprime type NAS of Ukraine KyivUkraine 1998

[6] A A Dezin General Questions of the -eory of BoundaryValue Problems Nauka Moscow Russia 1980

[7] A M Nakhushev Equations of Mathematical Biology HigherSchool Moscow Russia 1995

[8] I A Belavin S P Kapitsa and S P Kurdyumov ldquoAmathematical model of global demographic processes withconsidering of the spatial distributionrdquo Zh CalculationMathematics Physics vol 38 no 6 pp 885ndash902 1998

[9] V V Gorodetskiy and O V Martynyuk-e Cauchy Problemand Nonlocal Problems for First-Order Evolution EquationsEby Time Variable Vidavnichy dim rdquoRODOVIDrdquo ChernivtsiUkraine 2015

[10] V V Gorodetskiy and V I Myronyk ldquoTwo-point problem forone class of evolution equationsrdquo Differents Equationsvol 46 no 3 pp 349ndash363 2010

[11] V V Gorodetskiy and V I Myronyk ldquoTwo-point problem forone class of evolution equationsrdquo Differents Equationsvol 46 no 4 pp 520ndash526 2010

[12] I M Gelfand and G E Shilov Spaces of Basic and GeneralizedFunctions Fizmatgiz Moscow Russia 1958

International Journal of Differential Equations 11

Page 4: On One Evolution Equation of Parabolic Type with ...

Using inequalities (15) and (16) (when putting in (15)ε (a2)(1 minus δ)) we get

a(σ)φn(σ)( 1113857(k)

11138681113868111386811138681113868

11138681113868111386811138681113868 le c31113957ε(1113957B + ρ)k exp minus

a

2(1 minus δ)|σ|

ω1113882 1113883

k isin Z+ nge n0

(22)

where c3 cε + c1 and 1113957B 2max B0 B11113864 1113865 It follows from thelast inequality that aφnδρ le c31113957ε forallnge n0 that is the se-

quence aφn nge 11113864 1113865 converges to zero in the space S11113957B1ω1113957A

where 1113957A 21ωA0 and 1113957B 2max B0 B11113864 1113865 )is means thatthe sequence aφn nge 11113864 1113865 converges to zero in the space S11ωwhich is what we needed to prove

Remark 1 It follows from the proven property that thefunction a(σ) (1 + σ2)ω2 σ isin R is also a multiplier inevery space S

β1ω where βgt 1 )erefore in the space S1ωβ

βge 1 defined as a continuous linear pseudodifferentialoperator A constructed by the function a(σ)

Aφ Fminus 1

[a(σ)F[φ]] forallφ isin S1ωβ (23)

where Aφ (I minus Δ)ω2φ Δ (d2dx2)Let us consider the evolution equation with the operator

A (the Bessel fractional differentiation operator)zu

zt+ Au 0 (t x) isin (0 +infin) times R equiv Ω (24)

By a solution of equation (24) we mean the functionu(t x) (t x) isin Ω such that (1) u(t middot) isin C1(0 +infin) forevery x isin R (2) u(middot x) isin S1ωβ for every t isin (0 +infin) (3)u(t x) is continuous at every point (0 x) of the boundaryΓΩ 0 times R of the region Ω (4) existMR⟶ [0infin) forallt isin (0 +infin) |ut(t x)| le M(x)intRM(x)dxltinfin and (5) u(t x) (t x) isin Ω satisfiesequation (24)

For equation (24) we consider the nonlocal multipointby time problem of finding a solution of equation (24) thatsatisfies the condition

μu(0 x) minus μ1B1u t1 x( 1113857 minus middot middot middot minus μmBmu tm x( 1113857 f(x)

x isin R f isin S1ωβ

(25)

where u(0 x) ≔ limt⟶+0u(t x) x isin R m isin Nμ μ1 μm1113864 1113865 sub (0infin) and t1 tm1113864 1113865 sub (0infin) are fixednumbers 0lt t1 lt t2 lt middot middot middot lt tm lt +infin μgt 1113936

mk1 μk and

B1 Bm are pseudodifferential operators in S1ωβ con-structed by functions (characters) gk R⟶ (0 +infin) re-spectively Bk Fminus 1[gk(σ)F] k isin 1 m )e functionsgk isin Cinfin(R) and k isin 1 m satisfy the conditions

forallεgt 0forallσ isin R gk(σ) le exp ε|σ|ω

1113864 1113865

existLk gt 0foralls isin N Dsσgk(σ)

11138681113868111386811138681113868111386811138681113868 le L

sks

s

(26)

Note that the above properties of the functions gk implythat gk k isin 1 m is a multiplier in S1ωβ

We are looking for the solution of problems (24) and(25) via the Fourier transform Due to condition (19)

F ut1113858 1113859 1113946R

ut(t x)eiσxdx

z

zt1113946R

u(t x)eiσxdx

z

ztF[u]

(27)

We introduce the notation F[u(t x)] v(t σ) Giventhe form of the operators A B1 Bm we get

F[Au(t x)] F Fminus 1

[a(σ)F[u]]1113960 1113961 a(σ)F[u(t x)] a(σ)v(t σ)

F Bku(t x)1113858 1113859 gk(σ)F[u(t x)] gk(σ)v(t σ) k isin 1 m

(28)

So for the function v Ω⟶ R we arrive at a problemwith parameter σ

dv(t σ)

dt+ a(σ)v(t σ) 0 (t σ) isin Ω (29)

μv(0 σ) minus 1113944

m

k1μkgk(σ)v tk σ( 1113857 1113957f(σ) σ isin R (30)

where 1113957f(σ) F[f](σ) )e general solution of equation(29) has the form

v(t σ) c exp minus ta(σ) (t σ) isin Ω (31)

where c c(σ) is determined by condition (30) Substituting(30) into (31) we find that

c 1113957f(σ) μ minus 1113944m

k1μkgk(σ)exp minus tka(σ)1113864 1113865⎛⎝ ⎞⎠

minus 1

σ isin R

(32)

Now put G(t x) ≔ Fminus 1[Q(t σ)] andQ(t σ) Q1(t σ)Q2(σ) and thus

Q1(t σ) exp minus ta(σ)

Q2(σ) μ minus 1113944m

k1μkgk(σ)Q1 tk σ( 1113857⎛⎝ ⎞⎠

minus 1

(33)

)en thinking formally we come to the relation

u(t x) 1113946R

G(t x minus ξ)f(ξ)dξ G(t x)lowastf(x) (34)

Indeed

u(t x) (2π)minus 1

1113946R

Q(t σ) 1113946R

f(ξ)eiσξdξ1113874 1113875e

minus iσxdσ

1113946R

(2π)minus 1

1113946R

Q(t σ)eminus iσ(xminus ξ)dσ1113874 1113875f(ξ)dξ

1113946R

G(t x minus ξ)f(ξ)dξ G(t x)lowastf(x) (t x) isin Ω

(35)

)e correctness of the transformations performed theconvergence of the corresponding integrals and conse-quently the correctness of formula (35) follow from theproperties of the function G which are given below )eproperties of G are determined by the properties of Q

4 International Journal of Differential Equations

because G Fminus 1[Q] So let us first examine the properties ofthe function Q(t σ) as a function of the variable σ

Lemma 1 For derivatives of Q1(t σ) (t σ) isin Ω theestimates

DsσQ1(t σ)

11138681113868111386811138681113868111386811138681113868 le cA

stcs

ss exp minus t|σ|

ω1113864 1113865 (36)

are valid where c 0 if 0lt t le 1 and c 1 if tgt 1 and theconstants cgt 1 and Agt 0 do not depend on t

Proof To prove the statement we use the Faa di Brunoformula for differentiation of a complex function

DsσF(g(σ)) 1113944

s

p1

dp

dgpF(g) 1113944

s

p1 pl

ddσ

g(σ)1113888 1113889

p1

1l

dl

dσlg(σ)1113888 1113889

pl

s isin N

(37)

where the sum sign is applied to all the solutions in positiveintegers of the equation p1 + 2p2 + middot middot middot + lpl sp1 + middot middot middot + pl m Put F eg and g minus ta(σ) )en

Dsσe

minus ta(σ) e

minus ta(σ)1113944

s

p11113944

s

p1 plΛ (38)

where the symbol Λ denotes the expression and

Λ ≔ddσ

(minus ta(σ))1113888 1113889

p1

middot12

d2

dσ2(minus ta(σ))1113888 1113889

p2

1l

dl

dσl(minus ta(σ))1113888 1113889

pl

(39)

Given estimate (16) is fulfilled we find that

|Λ| le cp1+middotmiddotmiddot+pl

0 Bp1+2p2+middotmiddotmiddot+lpl

0 tp1+middotmiddotmiddot+pl le 1113957c

s0t

pB

s0 1113957c0 max 1 c01113864 1113865

(40)

Using (40) and the Stirling formula we arrive at theinequalities

DsσQ1(t σ)

11138681113868111386811138681113868111386811138681113868 le 1113957c

s0B

s0t

css exp minus ta(σ)

le cAstcs

ss exp minus t|σ|

ω1113864 1113865 σ isin R

(41)

where c 0 if 0lt t le 1 and c 1 if tgt 1 and the values cgt 1and Agt 0 do not depend on t

Lemma is proved

Remark 2 It follows from estimate (41) that Q1(t middot) isin S11ωfor every tgt 0

Lemma 2 -e function Q2 is a multiplier in S21ω

Proof To prove the assertion let us estimate the derivativesof Q2 For this we use formula (37) in which we put F φminus 1

and φ R where

R(σ) μ minus 1113944m

k1μkgk(σ)exp minus tka(σ)1113864 1113865 (42)

)en Q2 F(φ) Rminus 1 and

DsσQ2(σ)

11138681113868111386811138681113868111386811138681113868 1113944

s

p1

dp

dRpR

minus 11113944

s

p1 pl

ddσ

R(σ)1113888 1113889

p11113868111386811138681113868111386811138681113868111386811138681113868

1l

dl

dσlR(σ)1113888 1113889

pl

s isin N

(43)

Given the properties of g1 gm and inequality (41)we find that

1l

dj

dσjR(σ)

11138681113868111386811138681113868111386811138681113868

11138681113868111386811138681113868111386811138681113868le

1j

1113944

m

k1μk 1113944

j

i0C

ij D

iσgk(σ)

11138681113868111386811138681113868

11138681113868111386811138681113868 middot Djminus iσ e

minus tka(σ)11138681113868111386811138681113868

11138681113868111386811138681113868

le1j

1113944

m

k1μk 1113944

i

j0C

ijL

iki1113957c

jminus i0 t

c(jminus i)

k (j minus i)eminus tka(σ)

le 1113944m

k1μk 1113944

j

i0C

ijL

ik1113957c

jminus i0 t

c(jminus i)

(44)

where we took into account that i(j minus i) le j)Let

1113957L max L1 Lm1113864 1113865

L0 2max 1113957L 1113957c0T1113864 1113865(45)

then

1j

dj

dσjR(σ)

11138681113868111386811138681113868111386811138681113868

11138681113868111386811138681113868111386811138681113868le α0L

j0 α0 1113944

m

k1μk j isin 1 l

ddσ

R(σ)1113888 1113889

p1111386811138681113868111386811138681113868111386811138681113868

111386811138681113868111386811138681113868111386811138681113868

1l

dl

dσlR(σ)1113888 1113889

pl111386811138681113868111386811138681113868111386811138681113868

111386811138681113868111386811138681113868111386811138681113868le α0L0( 1113857

p α0L201113872 1113873

p2 α0L

l01113872 1113873

pl

αp1+middotmiddotmiddot+pl

0 Lp1+2p2+middotmiddotmiddot+lpl

0 αp0L

s0

(46)

In addition (dpdRp)Rminus 1 (minus 1)ppRminus (p+1) and

Rminus 1

(σ) Q2(σ) μ minus 1113944m

k1μkgk(σ)exp minus tka(σ)1113864 1113865⎛⎝ ⎞⎠

minus 1

le μ minus 1113944m

k1μk exp minus t1|σ|

ω+ ε|σ|

ω1113864 1113865⎛⎝ le μ minus 1113944

m

k1μk

⎛⎝ ⎞⎠

minus 1

equiv β0 gt 0 ε t1

(47)

Since by assumption μgt 1113936mk1 μk (assuming the prop-

erties of g1 gm and 0lt t1 lt t2 lt middot middot middot lt tm) So

International Journal of Differential Equations 5

dp

dRpR

minus 111138681113868111386811138681113868111386811138681113868

11138681113868111386811138681113868111386811138681113868le βp+1

0 p

DsσQ2(σ)

11138681113868111386811138681113868111386811138681113868 le s 1113944

s

p1βp+10 pαp

0Ls0

le β0Ls0(s)

21113944

s

p1βp0 le β0L

s0s middot (s)

2βs1 le β2β

s3s

2s s isin N

(48)

It follows from the last inequality and boundedness ofthe function Q2(σ) on R that Q2 is a multiplier in S21ω

Corollary 1 For every tgt 0 the functionQ(t σ) Q1(t σ)Q2(σ) σ isin R is an element of the spaceS21ω and the estimates

DsσQ(t σ)

11138681113868111386811138681113868111386811138681113868 le 1113957c1113957A

stcs

s2s exp minus t|σ|

ω1113864 1113865 s isin Z+ (t σ) isin Ω

(49)

are valid where the constants 1113957c and 1113957Agt 0 do not depend on t

Taking into account the properties of the Fouriertransform (direct and inverse) and the formulaFminus 1[S21ω] S1ω2 we get that G(t middot) isin S1ω2 for every tgt 0 Weremove in the estimates of derivatives of the function G (inthe variable x) the dependence on t assuming tgt 1 To dothis we use the relations

xkD

sxF[φ](x) i

k+sF σsφ(σ)( 1113857

(k)1113876 1113877

ik+s

1113946R

σsφ(σ)( 1113857(k)

eiσxdσ k s sub Z+ φ isin S

21ω

(50)

So

xkD

sxG(t x) (2π)

minus 1ik+s

(minus 1)s1113946R

σsQ(t minus σ)( 1113857

(k)e

iσxdσ

(51)

Applying the Leibniz formula for differentiating theproduct of two functions and estimating the derivatives ofthe function Q(t σ) we find that

σsQ(t minus σ)( 1113857

(k)11138681113868111386811138681113868

11138681113868111386811138681113868 1113944k

p0C

p

k σs( 1113857

(p)Q

(kminus p)(t minus σ)

1113868111386811138681113868111386811138681113868111386811138681113868

1113868111386811138681113868111386811138681113868111386811138681113868

le σsQ

(k)(t minus σ)

11138681113868111386811138681113868

11138681113868111386811138681113868 + ks σsminus 1Q

(kminus 1)(t minus σ)

11138681113868111386811138681113868

11138681113868111386811138681113868

+k(k minus 1)

2s(s minus 1) σsminus 2

Q(kminus 2)

(t minus σ)11138681113868111386811138681113868

11138681113868111386811138681113868

+ middot middot middot le 1113957c 1113957Aktck 1113957B

stminus (sω)

mks + ks1113957Akminus 1

tc(kminus 1)

1113876

middot 1113957Bsminus 1

tminus ((sminus 1)ω)

mkminus 1sminus 1

+12

k(k minus 1)s(s minus 1)1113957Akminus 2

tc(kminus 2) 1113957B

sminus 2tminus ((sminus 2)ω)

middot mkminus 2sminus 2 + 1113961eminus (t2)|σ|ω

(52)

where mks k2kssω Taking into account the results in (see[12] p 236ndash243) we find that this double sequence satisfiesthe inequality

ksmkminus 1sminus 1

mks

le 1113957c(k + s) 1113957cgt 0 (53)

Bearing in mind the last inequality and also that tgt 1 weobtain

σsQ(t minus σ)( 1113857

(k)11138681113868111386811138681113868

11138681113868111386811138681113868 le 1113957c1113957Aktk 1113957B

smks 1 +

ks

1113957A1113957B

mkminus 1sminus 1

mks

+12

k(k minus 1)s(s minus 1)1

1113957A21113957B

2mkminus 2sminus 2

mks

+ middot middot middot1113888 1113889eminus (t2)|σ|ω le 1113957c1113957A

ktk 1113957B

smks

times 1 +ks

1113957A1113957B

mkminus 1sminus 1

mks

+12

11113957A21113957B

2 ksmkminus 1sminus 1

mks

(k minus 1)(s minus 1)mkminus 2sminus 2

mkminus 1sminus 1+ middot middot middot1113888 1113889e

minus (t2)|σ|ω

le 1113957c1113957Aktk 1113957B

smks 1 +

1113957c

1113957A1113957B(k + s) +

12

1113957c2

(1113957A1113957B)2(k + s)

2+ middot middot middot1113888 1113889e

minus (t2)|σ|ω

le c1AktkB

smkse

minus (t2)|σ|ω

A 1113957A exp1113957c

1113957A1113957B1113882 1113883

B 1113957B exp1113957c

1113957A1113957B1113882 1113883

(54)

6 International Journal of Differential Equations

So

xkD

sxG(t x)

11138681113868111386811138681113868

11138681113868111386811138681113868 le (2π)minus 1

c1AktkB

smks1113946

Re

minus (t2)|σ|ωdσ

c2AktkB

stminus (1ω)

k2k

ssω

k s sub Z+

(55)

)en

DsxG(t x)

11138681113868111386811138681113868111386811138681113868 le c2B

stminus (1ω)

ssω inf

k

Akk2k

tminus 1|x|( )k

le c3Bstminus (1ω)

ssω exp minus a0t

minus (12)|x|

121113966 1113967

(56)

where the constants c3 B and a0 gt 0 do not depend on there we used the well-known inequality (see [12] p 204)

eminus (αe)|ξ|1α le inf

k

Lkkkα

|ξ|kle ce

minus (αe)|ξ|1α c e

αe2 (57)

in which α 2 and L 1113957A )us such a statement is correct

Lemma 3 -e derivatives (by variable x) of the functionG(t x) for tgt 1 satisfy the inequality

DsxG(t x)

11138681113868111386811138681113868111386811138681113868 le c3B

ss

sωtminus (1ω) exp minus a0t

minus (12)|x|

121113966 1113967 s isin Z+

(58)

where the constants c3 B and a0 gt 0 do not depend on t

Remark 3 In what follows we assume that in condition(25) β 2

Here are a few more properties of the function G(t x)

Lemma 4 -e function G(t middot) t isin (0 +infin) as an abstractfunction of t with values in the space S1ω2 is differentiable in t

Proof It follows from the continuity of the Fourier trans-form (direct and inverse) that to prove the statement itsuffices to establish that the function F[G(t middot)] Q(t middot) as afunction of a parameter t with values in the space S21ω isdifferentiable in t In other words it is necessary to provethat the boundary value relation

ΦΔt(σ) ≔1Δt

[Q(t + Δt middot) minus Q(t middot)]⟶z

ztQ(t middot) Δt⟶ 0

(59)

is performed in the sense as follows

(1) DsσΦΔt(σ) ⟶

Δt⟶ 0Ds

σ(minus a(σ)Q(t σ)) s isin Z+ uni-formly on each segment [a b] sub R

(2) |DsσΦΔt(σ)| le cB

ss2s exp minus a|σ|ω s isin Z+ where the

constants c B and agt 0 do not depend on Δt if Δt issmall enough

)e function Q(t σ) (t σ) isin Ω is differentiable in t inthe usual sense Due to the Lagrange theorem on finiteincrements

ΦΔt(σ) minus a(σ)Q(t + θΔt σ) (60)

So

DsσΦΔt(σ) minus 1113944

s

l0C

lsD

lσa(σ)D

sminus lσ Q(t +θΔtσ) (61)

Dsσ ΦΔt(σ) minus

z

ztQ(tσ)1113888 1113889 minus 1113944

s

l0C

lsD

lσa(σ) D

sminus lσ Q(t +θΔtσ)1113960

minus Dsminus lσ Q(tσ)1113961

(62)

Since

Dsminus lσ Q(t + θΔt σ) minus D

sminus lσ Q(t σ) D

sminus l+1σ Q t + θ1Δt σ( 1113857

0lt θ1 lt 1

(63)

it follows from this and estimate (49) that

Dsminus l+1σ Q t + θ1Δt σ( 1113857θΔt⟶ 0 Δt⟶ 0 (64)

uniformly on an arbitrary segment [a b] sub R )en

DsσΦΔt(σ)⟶ D

z

ztQ(t σ)1113888 1113889 Δt⟶ 0 (65)

uniformly on each segment [a b] sub R too )ereforecondition 1 is satisfied

Taking into account (61) estimates (15) (16) and (49)for a(σ) Q(t σ) and their derivatives we find that

DsσΦΔt(σ)

11138681113868111386811138681113868111386811138681113868 le c1cε 1113944

s

l0C

lsB

l1l

l 1113957Asminus l

tc(sminus l)

(s minus l)2(sminus l)

exp minus (t + θΔt)|σ|ω

+ ε|σ|ω

1113864 1113865

(66)

where εgt 0 is arbitrary fixed Take ε (t2) and then

DsσΦΔt(σ)

11138681113868111386811138681113868111386811138681113868 le cB

ss2s exp minus a|σ|

ω1113864 1113865 s isin Z+ (67)

where B 2max B11113957Atc1113966 1113967 a (t2) and all the constants

do not depend on Δt

Corollary 2 -e formulaz

zt(flowastG(t middot)) f lowast

zG(t middot)

zt forallf isin S

1ω21113872 1113873prime tgt 0

(68)

is correct

Proof According to the definition of a convolution of ageneralized function with a test one we have

flowastG(t middot) langfξ Tminus xG(t ξ)rang

G(t ξ) G(t minus ξ)(69)

)enz

zt(flowastG(t middot)) lim

Δt⟶0

1Δt

[(flowastG(t + Δt middot)) minus (flowastG(t middot))]

limΔt⟶0langfξ

1Δt

Tminus xG(t + Δt ξ) minus Tminus x

G(t ξ)1113960 1113961rang

(70)

International Journal of Differential Equations 7

Due to Lemma 4 the relation1Δt

Tminus xG(t + Δt ξ) minus Tminus x

G(t ξ)1113960 1113961⟶z

ztTminus x

G(t ξ) Δt⟶ 0

(71)

is performed in the sense of S1ω2 topology So

z

zt(flowastG(t middot)) langfξ lim

Δt⟶0

1Δt

Tminus xG(t + Δt ξ) minus Tminus x

G(t ξ)1113960 1113961rang

langfξ z

ztTminus x

G(t ξ)rang langfξ Tminus x

z

ztG(t ξ)rang

flowastz

ztG(t middot)

(72)

)e statement is provedSince

F[G(t middot)] Q(t σ) Q1(t σ)Q2(σ) 1113946R

G(t x)eiσxdx tgt 0

(73)

we get the formula

1113946R

G(t x)dx Q1(t 0)Q2(0) λ0eminus t

λ0 μ minus 1113944m

k1μkgk(0)e

minus tk⎛⎝ ⎞⎠

minus 1

(74)

where a(0) 1 is taken into account

Lemma 5 In (S1ω2 )prime the following relations are correct

(1) G(t middot)⟶ Fminus 1

Q21113858 1113859 t⟶ + 0

(2) μG(t middot) minus 1113944m

k1μkBkG tk middot( 1113857⟶ δ t⟶ + 0

(75)

where δ is the Dirac delta function

Proof (1) Taking into account the continuity property of theFourier transform (direct and inverse) in spaces of Sprime type itis sufficient to establish that

F[G(t middot)] Q1(t middot)Q2(middot)⟶ Q2(middot) t⟶ + 0 (76)

in the space (S21ω)prime To do this we take an arbitrary functionφ isin S21ω and using the fact thatQ2 is a multiplier in the spaceS21ω and the Lebesgue theorem on the limit passage underthe Lebesgue integral sign we find that

langQ1(t middot)Q2(middot)φrang langQ1(t middot) Q2(middot)φ(middot)rang

1113946R

Q1(t σ)Q2(σ)φ(σ)dσ ⟶t⟶+0

1113946R

Q2(σ)φ(σ)dσ

lang1 Q2(middot)φ(middot)rang langQ2φrang

(77)

)is leads us to statement (1) of Lemma 5(2) Given statement (1) and the form of the operators

B1 Bm we find that

μG(t middot) minus 1113944m

l1μlBlG tl middot( 1113857 ⟶

t⟶+0μF

minus 1Q21113858 1113859 minus 1113944

m

l1μlBlG tl middot( 1113857

μFminus 1

Q21113858 1113859 minus 1113944m

l1μlF

minus 1gl(middot)Q1 tl middot( 1113857Q2(middot)1113858 1113859

Fminus 1 μQ2(middot) minus 1113944

m

l1μlgl(middot⎡⎣ ⎞⎠Q1 tl middot( 1113857Q2(middot)]

Fminus 1 μ minus 1113944

m

l1μlgl(middot)Q1 tl middot( 1113857Q2(middot)⎡⎣ ⎤⎦

Fminus 1 μ minus 1113944

m

l1μlgl(middot)Q1 tl middot( 1113857⎛⎝ ⎞⎠⎡⎢⎢⎣

middot μ minus 1113944m

l1μlgl(middot)Q1 tl middot( 1113857⎛⎝ ⎞⎠

minus 1⎤⎥⎥⎥⎥⎦ F

minus 1[1] δ

(78)

)erefore relation (75) is fulfilled in the space (S1ω2 )primeLemma 5 is proved

Remark 4 If μ 1 μ1 middot middot middot μm 0 then problems (24)and (25) degenerate into the Cauchy problem for equation(24) In this case Q2(σ) 1 forallσ isin R G(t x) Fminus 1[eminus ta(σ)]and G(t middot)⟶ Fminus 1[1] δ for t⟶ +0 in the space (S1ω1 )prime

Corollary 3 Let

ω(t x) flowastG(t x) f isin S1ω2lowast1113872 1113873prime (t x) isin Ω (79)

where (S1ω2lowast )prime is a class of convolutors in S1ω2 -en in (S1ω2 )primethe following boundary relation holds

μω(t middot) minus 1113944m

k1μkBkω tk middot( 1113857⟶ f t⟶ + 0 (80)

Proof Let us prove that the relation

F μω(t middot) minus 1113944

m

k1μkBkω tk middot( 1113857⎡⎣ ⎤⎦⟶ F[f] t⟶ + 0

(81)

takes place in the space (S21ω)prime Since f isin (S1ω2lowast )prime andG(t middot) isin S1ω2 for every tgt 0 we have

F[ω(t middot)] F[flowastG(t middot)] F[f]F[G(t middot)] F[f]Q(t middot)

(82)

So we need to prove that

F[f] μQ(t middot) minus 1113944m

k1μkgk(middot)Q tk middot( 1113857⎛⎝ ⎞⎠⟶ F[f] (83)

in (S21ω)prime when t⟶ + 0 SinceQ(t middot) Q1(t middot)Q2(middot)⟶ Q2(middot) for t⟶ + 0 in the space(S21ω)prime (see the proof of assertion 1 and Lemma 5) thecorrelation

8 International Journal of Differential Equations

μQ(t middot) minus 1113944m

k1μkgk(middot)Q tk middot( 1113857 ⟶

t⟶+0μQ2(middot) minus 1113944

m

k1μkgk(middot)Q1 tk middot( 1113857Q2(middot)

μ minus 1113944m

k1μkgk(middot)Q1 tk middot( 1113857⎛⎝ ⎞⎠Q2(middot) 1

(84)

is realized in the space (S21ω)prime )us relation (81) and (80)are fulfilled in the corresponding spaces )e statement isproved

)e function G(t middot) satisfies equation (24) as tgt 0Indeed

z

ztG(t x)

z

ztF

minus 1[Q(t σ)] F

minus 1 z

ztQ(t σ)1113890 1113891

Fminus 1

[a(σ)Q(t σ)]

AG(t x) Fminus 1

a(σ)Fminus 1

[G(t middot)]1113960 1113961 Fminus 1

[a(σ)Q(t σ)]

(85)

So

zG(t x)

zt+ AG(t x) 0 (t x) isin Ω (86)

which is what had to be provedIt follows from Corollary 3 that the nonlocal multipoint

by the time problem for equation (24) can be formulated inthe following way find a function u(t x) (t x) isin Ω thatsatisfies equation (24) and the condition

μ limt⟶+0

u(t middot) minus 1113944

m

k1μkBku tk middot( 1113857 f f isin S

1ω2lowast1113872 1113873prime (87)

where boundary relation (87) is considered in the space(S1ω2 )prime and the constraints on the parameters μ μ1 μmt1 tm are the same as in case of problems (24) and(25))

Theorem 1 By nonlocal multipoint by time problem (24)(87) is solvable and its solution is given by the formula

u(t x) flowastG(t x) (t x) isin Ω (88)

where u(t middot) isin S1ω2 for every tgt 0

Proof Make sure that the function u(t x) (t x) isin Ω sat-isfies equation (24) Indeed (see Corollary 2)

zu(t x)

zt

z

zt(flowastG(t middot)) flowast

zG(t middot)

zt

Au(t x) Fminus 1

[a(σ)F[f lowastG(t middot)]]

(89)

Since f is a convolutor in S1ω2

F[flowastG(t middot)] F[f]F[G(t middot)] F[f]Q(t middot) (90)

So

Au(t x) Fminus 1

[a(σ)Q(t σ)F[f](σ)] minus Fz

ztQ(t middot)F[f]1113890 1113891

minus F Fz

ztG(t middot)1113890 1113891 middot F[f]1113890 1113891

minus Fminus 1

F flowastzG(t middot)

zt1113890 11138911113890 1113891 minus flowast

zG(t middot)

zt

(91)

It follows from this that the function u(t x) (t x) isin Ωsatisfies equation (24) From Corollary 3 we get that u(t x)

satisfies condition (87) in the defined sense )e theorem isproved

Remark 5 If in condition (87) B1 middot middot middot Bm I (I is theidentity operator) then we can prove that problems (24) and(87) are well posed and its solution is given by the formula

u(t x) flowastG(t x) f isin S1ω1lowast1113872 1113873prime (t x) isin Ω (92)

where G(t middot) Fminus 1[Q(t middot)] isin S1ω1 for every tgt 0 and

Q(t σ) eminus ta(σ) μ minus 1113944

m

k1μke

minus tka(σ)⎛⎝ ⎞⎠

minus 1

(93)

Theorem 2 Suppose u(t x) (t x) isin Ω is the solution ofproblems (24) and (87) -en u(t middot)⟶ 0 in the space(S1ω2 )prime as t⟶ +infin

Proof Recall that the solution of problems (24) and (87) isgiven by the formula

u(t x) flowastG(t x) langfξ G(t x minus ξ)rang

f isin S1ω2lowast1113872 1113873prime (t x) isin Ω

(94)

Let us prove that langu(t middot)φrang⟶ 0 as t⟶ +infin for anarbitrary function φ isin S1ω2 Put

Ψt(ξ) 1113946+infin

minus infinG(t x minus ξ)φ(x)dx

ΨtR(ξ) 1113946R

minus RG(t x minus ξ)φ(x)dx Rgt 0 tgt 1

(95)

In these notations we prove that (a) for every tgt 1 andRgt 0 the function ΨtR(ξ) belongs to the space S1ω2 andΨtR(ξ)⟶Ψt(ξ) in the space S1ω2 as R⟶infin (b)Ψt(ξ) isin S1ω2 for every tgt 1 From this we get

langu(t middot)φrang 1113946+infin

minus infinlangfξ G(t x minus ξ)rangφ(x)dx

langfξ 1113946+infin

minus infinG(t x)φ(x + ξ)dxrang

langfξ 1113946+infin

minus infinG(t minus y)φ(minus (y minus ξ))dyrang

langfξ 1113946+infin

minus infinG(t minus y)φ(y minus ξ)dyrang φ(z) φ(minus z)

(96)

International Journal of Differential Equations 9

where u(t middot) is interpreted as a regular generalized functionfrom (S1ω2 )prime for every tgt 0

So let us set property (a) For fixed k m sub Z+ we have

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le 1113946+R

minus Rξkφ(x)D

mξ G(t x minus ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868dx

le 1113946+infin

minus infinξkφ(ξ + η)D

mη G(t η)

11138681113868111386811138681113868

11138681113868111386811138681113868dη

(97)

Since φ isin S1ω2 the inequality

ξkD

mξ φ(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cLkM

mk2k

mmω

k m sub Z+ (98)

with some c L Mgt 0 is valid It follows from this that forevery η isin R

supξisinR

ξkφ(ξ)11138681113868111386811138681113868

11138681113868111386811138681113868 supyisinR

(y minus η)kφ(η)

11138681113868111386811138681113868

11138681113868111386811138681113868 supyisinR

1113944

k

l0C

lky

l(minus η)

kminus lφ(y)

1113868111386811138681113868111386811138681113868111386811138681113868

1113868111386811138681113868111386811138681113868111386811138681113868

le 1113944k

l0C

lk|η|

kminus l supyisinR

ykφ(y)

11138681113868111386811138681113868

11138681113868111386811138681113868 le c 1113944k

l0C

lkL

ll2l

|η|kminus l

(99)

Next we will use estimate (58) )en

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le c 1113944k

l0C

lkL

ll2l

1113946+infin

minus infin|η|

kminus lD

mη G(t η)

11138681113868111386811138681113868

11138681113868111386811138681113868dη

le cc3Bm

mmω

tminus (1ω)

1113944

k

l0C

lkL

ll2l

middot 1113946+infin

minus infin|η|

kminus l exp minus a0tminus (12)

|η|12

1113966 1113967dη

(100)

By the direct calculations we find that

1113946+infin

minus infin|η|

kminus l exp minus a0tminus (12)

|η|12

1113966 1113967dη le c41113957L

kk2(kminus l)

tkminus l+1

c41113957Lgt 0

(101)

So

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cc3c4Bm

t1minus (1ω)

mmω

1113944

k

l0C

lkL

l1113957Lkminus l

tkminus l

l2l

k2(kminus l)

le cBm

Lkk2k

mmω

(102)

where c cc3c4t1minus (1ω) and L 2max L 1113957Lt1113864 1113865 )erefore

ΨtR(ξ) isin S1ω2 for every tgt 1 and an arbitrary Rgt 0 Fur-thermore we make sure thatΨtR(ξ)⟶Ψt(ξ) as R⟶infin

with all its derivatives uniformly in ξ on each segment[a b] sub R In addition the set of functions ξk

Dmξ ΨtR(ξ)1113966 1113967

k m sub Z+ is uniformly bounded (by R) in the space S1ω2(this property follows from estimate (102) in which c B andLgt 0 do not depend on R) )is means that condition (a) isfulfilled

Condition (b) follows from (a) because in a perfectspace every bounded set is compact

Using properties (a) and (b) we obtain the equality

langu(t middot)φrang 1113946+infin

minus infinG(t minus y)(flowastφ)(y)dy (103)

Since f isin (S1ω2 )prime is a convolutor in S1ω2 we haveflowast φ isin S1ω2 )en given estimate (58) (at s 0) we find that

|langu(t middot)φrang| le 1113946+infin

minus infin|G(t minus y)||(flowastφ)(y)|dy

lec

t1ω1113946

+infin

minus infin|(flowastφ)(y)|dy le

c0

t1ω⟶ 0 t⟶ +infin

(104)

for an arbitrary function φ isin S1ω2 ie u(t middot)⟶ 0 in thespace (S1ω2 )prime as t⟶ +infin )e theorem is proved

If the generalized function f in (87) is finite (ie itssupport suppf is a finite set in R) then we can say onuniform tending a solution u(t x) of problems (24) and (87)to zero onR as t⟶ +infin Note that every finite generalizedfunction is a convolutor in S spaces )is property followsfrom the general result concerning the theory of perfectspaces (see [12] p 137) if Φ is a perfect space with dif-ferential translation operation then every finite functional isa convolutor inΦ Finite functionals form a fairly wide classIn particular every bounded set F sub R is a support of somegeneralized function [12]

Theorem 3 Let u(t x) be a solution of problems (24) and(87) with the boundary function f in condition (87) which isan element of (S

β2)prime sub (S1ω2 )prime βgt 1 with finite support in R

-en u(t x)⟶ 0 uniformly on R as t⟶ +infin

Here is a scheme for proving this statement Letsuppf sub [a1 b1] sub [a2 b2] sub R Consider the functionφ isin S

β2 βgt 1 such that φ(x) 1 x isin [a1 b1] and

suppφ sub [a2 b2] )is function exists when βgt 1 because Sβ2

contains finite functions [12] Consider the function

u(t x) langfξ φ(ξ)G(t x minus ξ)rang +langfξ c(ξ)G(t x minus ξ)rang(105)

where c 1 minus φ Since supp(c(ξ)G(t x minus ξ)) cap suppφ emptywe have

u(t x) tminus (1ω)langfξ t

1ωφ(ξ)G(t x minus ξ)rang (106)

To prove the formulated above statement it remains toestablish that the set of functions Φtx(ξ) t1ωφ(ξ)G(t x minus

ξ) is bounded in the space Sβ2 βgt 1 for large t and x isin R

For example if in condition (87) f δ then δ is aconvolutor in the space S supp δ 0 andu(t x) δ lowastG(t x) G(t x) It follows directly from es-timate (58) that G(t x)⟶ 0 uniformly onR as t⟶ +infin

4 Conclusion

In this paper the solvability of a nonlocal multipoint by thetime problem for the evolutionary equation(zuzt) + (I minus (d2dx2))ω2u 0 (t x) isin (0 +infin) times R isproved herewith the operator (I minus (d2dx2))ω2 is inter-preted as a pseudodifferential operator in the space S1ω2

10 International Journal of Differential Equations

constructed by function (1 + σ2)ω2 σ isin R while the non-local multipoint by time condition also contains pseudo-differential operators constructed on smooth symbols )erepresentation of the solution is given in the form of aconvolution of the fundamental solution with the initialfunction which is an element of the space of generalizedfunctions of the ultradistribution type (the coagulator inspace S1ω2 ) )e behavior of the solution u(t x) t⟶ +infinin the space of generalized functions (S1ω2 )prime is investigated)e conditions for the initial generalized function are foundunder which the solution is uniformly stabilized to zero onR )e method of research of a nonlocal multipoint by timeproblem offered in this paper allows to interpret differential-operator equations of a form (zuzt) + φ(i zzx)u 0 as anevolutionary equation with a pseudodifferential operatorφ(i zzx) Fminus 1[φ middot F] constructed by the function φ actingin certain countable-normalized space of infinite-differentialfunctions (the choice of space depends on the properties ofthe function which is the symbol of the operator φ(i zzx))

Data Availability

No data were used to support the findings of this study

Disclosure

)is study was conducted in the framework of scientificactivity at Yuriy Fedkovych Chernivtsi National University

Conflicts of Interest

)e authors declare that they have no conflicts of interest

References

[1] V I Gorbachuk and M L Gorbachuk Boundary-ValueProblems for Differential-Operator Equations NaukovaDumka Kiev Ukraine 1984

[2] V V Gorodetskiy N I Nagnibida and P P NastasievMethods for Solving Problems in Functional Analysis VyshkaSchool Vyshka Ukraine 1990

[3] S G Samko A A Kilbas and O I Marichev Integrals andFractional Derivatives and Some of -eir Applications Scienceand Technology Minsk Belarus 1987

[4] I M Gelfand and G E Shilov Some Questions of the-eory ofDifferential Equations Fizmatgiz Moscow Russia 1958

[5] V V Gorodetskiy and O M Lenyuk About the FractionalDifferentiation in Spaces of Sprime type NAS of Ukraine KyivUkraine 1998

[6] A A Dezin General Questions of the -eory of BoundaryValue Problems Nauka Moscow Russia 1980

[7] A M Nakhushev Equations of Mathematical Biology HigherSchool Moscow Russia 1995

[8] I A Belavin S P Kapitsa and S P Kurdyumov ldquoAmathematical model of global demographic processes withconsidering of the spatial distributionrdquo Zh CalculationMathematics Physics vol 38 no 6 pp 885ndash902 1998

[9] V V Gorodetskiy and O V Martynyuk-e Cauchy Problemand Nonlocal Problems for First-Order Evolution EquationsEby Time Variable Vidavnichy dim rdquoRODOVIDrdquo ChernivtsiUkraine 2015

[10] V V Gorodetskiy and V I Myronyk ldquoTwo-point problem forone class of evolution equationsrdquo Differents Equationsvol 46 no 3 pp 349ndash363 2010

[11] V V Gorodetskiy and V I Myronyk ldquoTwo-point problem forone class of evolution equationsrdquo Differents Equationsvol 46 no 4 pp 520ndash526 2010

[12] I M Gelfand and G E Shilov Spaces of Basic and GeneralizedFunctions Fizmatgiz Moscow Russia 1958

International Journal of Differential Equations 11

Page 5: On One Evolution Equation of Parabolic Type with ...

because G Fminus 1[Q] So let us first examine the properties ofthe function Q(t σ) as a function of the variable σ

Lemma 1 For derivatives of Q1(t σ) (t σ) isin Ω theestimates

DsσQ1(t σ)

11138681113868111386811138681113868111386811138681113868 le cA

stcs

ss exp minus t|σ|

ω1113864 1113865 (36)

are valid where c 0 if 0lt t le 1 and c 1 if tgt 1 and theconstants cgt 1 and Agt 0 do not depend on t

Proof To prove the statement we use the Faa di Brunoformula for differentiation of a complex function

DsσF(g(σ)) 1113944

s

p1

dp

dgpF(g) 1113944

s

p1 pl

ddσ

g(σ)1113888 1113889

p1

1l

dl

dσlg(σ)1113888 1113889

pl

s isin N

(37)

where the sum sign is applied to all the solutions in positiveintegers of the equation p1 + 2p2 + middot middot middot + lpl sp1 + middot middot middot + pl m Put F eg and g minus ta(σ) )en

Dsσe

minus ta(σ) e

minus ta(σ)1113944

s

p11113944

s

p1 plΛ (38)

where the symbol Λ denotes the expression and

Λ ≔ddσ

(minus ta(σ))1113888 1113889

p1

middot12

d2

dσ2(minus ta(σ))1113888 1113889

p2

1l

dl

dσl(minus ta(σ))1113888 1113889

pl

(39)

Given estimate (16) is fulfilled we find that

|Λ| le cp1+middotmiddotmiddot+pl

0 Bp1+2p2+middotmiddotmiddot+lpl

0 tp1+middotmiddotmiddot+pl le 1113957c

s0t

pB

s0 1113957c0 max 1 c01113864 1113865

(40)

Using (40) and the Stirling formula we arrive at theinequalities

DsσQ1(t σ)

11138681113868111386811138681113868111386811138681113868 le 1113957c

s0B

s0t

css exp minus ta(σ)

le cAstcs

ss exp minus t|σ|

ω1113864 1113865 σ isin R

(41)

where c 0 if 0lt t le 1 and c 1 if tgt 1 and the values cgt 1and Agt 0 do not depend on t

Lemma is proved

Remark 2 It follows from estimate (41) that Q1(t middot) isin S11ωfor every tgt 0

Lemma 2 -e function Q2 is a multiplier in S21ω

Proof To prove the assertion let us estimate the derivativesof Q2 For this we use formula (37) in which we put F φminus 1

and φ R where

R(σ) μ minus 1113944m

k1μkgk(σ)exp minus tka(σ)1113864 1113865 (42)

)en Q2 F(φ) Rminus 1 and

DsσQ2(σ)

11138681113868111386811138681113868111386811138681113868 1113944

s

p1

dp

dRpR

minus 11113944

s

p1 pl

ddσ

R(σ)1113888 1113889

p11113868111386811138681113868111386811138681113868111386811138681113868

1l

dl

dσlR(σ)1113888 1113889

pl

s isin N

(43)

Given the properties of g1 gm and inequality (41)we find that

1l

dj

dσjR(σ)

11138681113868111386811138681113868111386811138681113868

11138681113868111386811138681113868111386811138681113868le

1j

1113944

m

k1μk 1113944

j

i0C

ij D

iσgk(σ)

11138681113868111386811138681113868

11138681113868111386811138681113868 middot Djminus iσ e

minus tka(σ)11138681113868111386811138681113868

11138681113868111386811138681113868

le1j

1113944

m

k1μk 1113944

i

j0C

ijL

iki1113957c

jminus i0 t

c(jminus i)

k (j minus i)eminus tka(σ)

le 1113944m

k1μk 1113944

j

i0C

ijL

ik1113957c

jminus i0 t

c(jminus i)

(44)

where we took into account that i(j minus i) le j)Let

1113957L max L1 Lm1113864 1113865

L0 2max 1113957L 1113957c0T1113864 1113865(45)

then

1j

dj

dσjR(σ)

11138681113868111386811138681113868111386811138681113868

11138681113868111386811138681113868111386811138681113868le α0L

j0 α0 1113944

m

k1μk j isin 1 l

ddσ

R(σ)1113888 1113889

p1111386811138681113868111386811138681113868111386811138681113868

111386811138681113868111386811138681113868111386811138681113868

1l

dl

dσlR(σ)1113888 1113889

pl111386811138681113868111386811138681113868111386811138681113868

111386811138681113868111386811138681113868111386811138681113868le α0L0( 1113857

p α0L201113872 1113873

p2 α0L

l01113872 1113873

pl

αp1+middotmiddotmiddot+pl

0 Lp1+2p2+middotmiddotmiddot+lpl

0 αp0L

s0

(46)

In addition (dpdRp)Rminus 1 (minus 1)ppRminus (p+1) and

Rminus 1

(σ) Q2(σ) μ minus 1113944m

k1μkgk(σ)exp minus tka(σ)1113864 1113865⎛⎝ ⎞⎠

minus 1

le μ minus 1113944m

k1μk exp minus t1|σ|

ω+ ε|σ|

ω1113864 1113865⎛⎝ le μ minus 1113944

m

k1μk

⎛⎝ ⎞⎠

minus 1

equiv β0 gt 0 ε t1

(47)

Since by assumption μgt 1113936mk1 μk (assuming the prop-

erties of g1 gm and 0lt t1 lt t2 lt middot middot middot lt tm) So

International Journal of Differential Equations 5

dp

dRpR

minus 111138681113868111386811138681113868111386811138681113868

11138681113868111386811138681113868111386811138681113868le βp+1

0 p

DsσQ2(σ)

11138681113868111386811138681113868111386811138681113868 le s 1113944

s

p1βp+10 pαp

0Ls0

le β0Ls0(s)

21113944

s

p1βp0 le β0L

s0s middot (s)

2βs1 le β2β

s3s

2s s isin N

(48)

It follows from the last inequality and boundedness ofthe function Q2(σ) on R that Q2 is a multiplier in S21ω

Corollary 1 For every tgt 0 the functionQ(t σ) Q1(t σ)Q2(σ) σ isin R is an element of the spaceS21ω and the estimates

DsσQ(t σ)

11138681113868111386811138681113868111386811138681113868 le 1113957c1113957A

stcs

s2s exp minus t|σ|

ω1113864 1113865 s isin Z+ (t σ) isin Ω

(49)

are valid where the constants 1113957c and 1113957Agt 0 do not depend on t

Taking into account the properties of the Fouriertransform (direct and inverse) and the formulaFminus 1[S21ω] S1ω2 we get that G(t middot) isin S1ω2 for every tgt 0 Weremove in the estimates of derivatives of the function G (inthe variable x) the dependence on t assuming tgt 1 To dothis we use the relations

xkD

sxF[φ](x) i

k+sF σsφ(σ)( 1113857

(k)1113876 1113877

ik+s

1113946R

σsφ(σ)( 1113857(k)

eiσxdσ k s sub Z+ φ isin S

21ω

(50)

So

xkD

sxG(t x) (2π)

minus 1ik+s

(minus 1)s1113946R

σsQ(t minus σ)( 1113857

(k)e

iσxdσ

(51)

Applying the Leibniz formula for differentiating theproduct of two functions and estimating the derivatives ofthe function Q(t σ) we find that

σsQ(t minus σ)( 1113857

(k)11138681113868111386811138681113868

11138681113868111386811138681113868 1113944k

p0C

p

k σs( 1113857

(p)Q

(kminus p)(t minus σ)

1113868111386811138681113868111386811138681113868111386811138681113868

1113868111386811138681113868111386811138681113868111386811138681113868

le σsQ

(k)(t minus σ)

11138681113868111386811138681113868

11138681113868111386811138681113868 + ks σsminus 1Q

(kminus 1)(t minus σ)

11138681113868111386811138681113868

11138681113868111386811138681113868

+k(k minus 1)

2s(s minus 1) σsminus 2

Q(kminus 2)

(t minus σ)11138681113868111386811138681113868

11138681113868111386811138681113868

+ middot middot middot le 1113957c 1113957Aktck 1113957B

stminus (sω)

mks + ks1113957Akminus 1

tc(kminus 1)

1113876

middot 1113957Bsminus 1

tminus ((sminus 1)ω)

mkminus 1sminus 1

+12

k(k minus 1)s(s minus 1)1113957Akminus 2

tc(kminus 2) 1113957B

sminus 2tminus ((sminus 2)ω)

middot mkminus 2sminus 2 + 1113961eminus (t2)|σ|ω

(52)

where mks k2kssω Taking into account the results in (see[12] p 236ndash243) we find that this double sequence satisfiesthe inequality

ksmkminus 1sminus 1

mks

le 1113957c(k + s) 1113957cgt 0 (53)

Bearing in mind the last inequality and also that tgt 1 weobtain

σsQ(t minus σ)( 1113857

(k)11138681113868111386811138681113868

11138681113868111386811138681113868 le 1113957c1113957Aktk 1113957B

smks 1 +

ks

1113957A1113957B

mkminus 1sminus 1

mks

+12

k(k minus 1)s(s minus 1)1

1113957A21113957B

2mkminus 2sminus 2

mks

+ middot middot middot1113888 1113889eminus (t2)|σ|ω le 1113957c1113957A

ktk 1113957B

smks

times 1 +ks

1113957A1113957B

mkminus 1sminus 1

mks

+12

11113957A21113957B

2 ksmkminus 1sminus 1

mks

(k minus 1)(s minus 1)mkminus 2sminus 2

mkminus 1sminus 1+ middot middot middot1113888 1113889e

minus (t2)|σ|ω

le 1113957c1113957Aktk 1113957B

smks 1 +

1113957c

1113957A1113957B(k + s) +

12

1113957c2

(1113957A1113957B)2(k + s)

2+ middot middot middot1113888 1113889e

minus (t2)|σ|ω

le c1AktkB

smkse

minus (t2)|σ|ω

A 1113957A exp1113957c

1113957A1113957B1113882 1113883

B 1113957B exp1113957c

1113957A1113957B1113882 1113883

(54)

6 International Journal of Differential Equations

So

xkD

sxG(t x)

11138681113868111386811138681113868

11138681113868111386811138681113868 le (2π)minus 1

c1AktkB

smks1113946

Re

minus (t2)|σ|ωdσ

c2AktkB

stminus (1ω)

k2k

ssω

k s sub Z+

(55)

)en

DsxG(t x)

11138681113868111386811138681113868111386811138681113868 le c2B

stminus (1ω)

ssω inf

k

Akk2k

tminus 1|x|( )k

le c3Bstminus (1ω)

ssω exp minus a0t

minus (12)|x|

121113966 1113967

(56)

where the constants c3 B and a0 gt 0 do not depend on there we used the well-known inequality (see [12] p 204)

eminus (αe)|ξ|1α le inf

k

Lkkkα

|ξ|kle ce

minus (αe)|ξ|1α c e

αe2 (57)

in which α 2 and L 1113957A )us such a statement is correct

Lemma 3 -e derivatives (by variable x) of the functionG(t x) for tgt 1 satisfy the inequality

DsxG(t x)

11138681113868111386811138681113868111386811138681113868 le c3B

ss

sωtminus (1ω) exp minus a0t

minus (12)|x|

121113966 1113967 s isin Z+

(58)

where the constants c3 B and a0 gt 0 do not depend on t

Remark 3 In what follows we assume that in condition(25) β 2

Here are a few more properties of the function G(t x)

Lemma 4 -e function G(t middot) t isin (0 +infin) as an abstractfunction of t with values in the space S1ω2 is differentiable in t

Proof It follows from the continuity of the Fourier trans-form (direct and inverse) that to prove the statement itsuffices to establish that the function F[G(t middot)] Q(t middot) as afunction of a parameter t with values in the space S21ω isdifferentiable in t In other words it is necessary to provethat the boundary value relation

ΦΔt(σ) ≔1Δt

[Q(t + Δt middot) minus Q(t middot)]⟶z

ztQ(t middot) Δt⟶ 0

(59)

is performed in the sense as follows

(1) DsσΦΔt(σ) ⟶

Δt⟶ 0Ds

σ(minus a(σ)Q(t σ)) s isin Z+ uni-formly on each segment [a b] sub R

(2) |DsσΦΔt(σ)| le cB

ss2s exp minus a|σ|ω s isin Z+ where the

constants c B and agt 0 do not depend on Δt if Δt issmall enough

)e function Q(t σ) (t σ) isin Ω is differentiable in t inthe usual sense Due to the Lagrange theorem on finiteincrements

ΦΔt(σ) minus a(σ)Q(t + θΔt σ) (60)

So

DsσΦΔt(σ) minus 1113944

s

l0C

lsD

lσa(σ)D

sminus lσ Q(t +θΔtσ) (61)

Dsσ ΦΔt(σ) minus

z

ztQ(tσ)1113888 1113889 minus 1113944

s

l0C

lsD

lσa(σ) D

sminus lσ Q(t +θΔtσ)1113960

minus Dsminus lσ Q(tσ)1113961

(62)

Since

Dsminus lσ Q(t + θΔt σ) minus D

sminus lσ Q(t σ) D

sminus l+1σ Q t + θ1Δt σ( 1113857

0lt θ1 lt 1

(63)

it follows from this and estimate (49) that

Dsminus l+1σ Q t + θ1Δt σ( 1113857θΔt⟶ 0 Δt⟶ 0 (64)

uniformly on an arbitrary segment [a b] sub R )en

DsσΦΔt(σ)⟶ D

z

ztQ(t σ)1113888 1113889 Δt⟶ 0 (65)

uniformly on each segment [a b] sub R too )ereforecondition 1 is satisfied

Taking into account (61) estimates (15) (16) and (49)for a(σ) Q(t σ) and their derivatives we find that

DsσΦΔt(σ)

11138681113868111386811138681113868111386811138681113868 le c1cε 1113944

s

l0C

lsB

l1l

l 1113957Asminus l

tc(sminus l)

(s minus l)2(sminus l)

exp minus (t + θΔt)|σ|ω

+ ε|σ|ω

1113864 1113865

(66)

where εgt 0 is arbitrary fixed Take ε (t2) and then

DsσΦΔt(σ)

11138681113868111386811138681113868111386811138681113868 le cB

ss2s exp minus a|σ|

ω1113864 1113865 s isin Z+ (67)

where B 2max B11113957Atc1113966 1113967 a (t2) and all the constants

do not depend on Δt

Corollary 2 -e formulaz

zt(flowastG(t middot)) f lowast

zG(t middot)

zt forallf isin S

1ω21113872 1113873prime tgt 0

(68)

is correct

Proof According to the definition of a convolution of ageneralized function with a test one we have

flowastG(t middot) langfξ Tminus xG(t ξ)rang

G(t ξ) G(t minus ξ)(69)

)enz

zt(flowastG(t middot)) lim

Δt⟶0

1Δt

[(flowastG(t + Δt middot)) minus (flowastG(t middot))]

limΔt⟶0langfξ

1Δt

Tminus xG(t + Δt ξ) minus Tminus x

G(t ξ)1113960 1113961rang

(70)

International Journal of Differential Equations 7

Due to Lemma 4 the relation1Δt

Tminus xG(t + Δt ξ) minus Tminus x

G(t ξ)1113960 1113961⟶z

ztTminus x

G(t ξ) Δt⟶ 0

(71)

is performed in the sense of S1ω2 topology So

z

zt(flowastG(t middot)) langfξ lim

Δt⟶0

1Δt

Tminus xG(t + Δt ξ) minus Tminus x

G(t ξ)1113960 1113961rang

langfξ z

ztTminus x

G(t ξ)rang langfξ Tminus x

z

ztG(t ξ)rang

flowastz

ztG(t middot)

(72)

)e statement is provedSince

F[G(t middot)] Q(t σ) Q1(t σ)Q2(σ) 1113946R

G(t x)eiσxdx tgt 0

(73)

we get the formula

1113946R

G(t x)dx Q1(t 0)Q2(0) λ0eminus t

λ0 μ minus 1113944m

k1μkgk(0)e

minus tk⎛⎝ ⎞⎠

minus 1

(74)

where a(0) 1 is taken into account

Lemma 5 In (S1ω2 )prime the following relations are correct

(1) G(t middot)⟶ Fminus 1

Q21113858 1113859 t⟶ + 0

(2) μG(t middot) minus 1113944m

k1μkBkG tk middot( 1113857⟶ δ t⟶ + 0

(75)

where δ is the Dirac delta function

Proof (1) Taking into account the continuity property of theFourier transform (direct and inverse) in spaces of Sprime type itis sufficient to establish that

F[G(t middot)] Q1(t middot)Q2(middot)⟶ Q2(middot) t⟶ + 0 (76)

in the space (S21ω)prime To do this we take an arbitrary functionφ isin S21ω and using the fact thatQ2 is a multiplier in the spaceS21ω and the Lebesgue theorem on the limit passage underthe Lebesgue integral sign we find that

langQ1(t middot)Q2(middot)φrang langQ1(t middot) Q2(middot)φ(middot)rang

1113946R

Q1(t σ)Q2(σ)φ(σ)dσ ⟶t⟶+0

1113946R

Q2(σ)φ(σ)dσ

lang1 Q2(middot)φ(middot)rang langQ2φrang

(77)

)is leads us to statement (1) of Lemma 5(2) Given statement (1) and the form of the operators

B1 Bm we find that

μG(t middot) minus 1113944m

l1μlBlG tl middot( 1113857 ⟶

t⟶+0μF

minus 1Q21113858 1113859 minus 1113944

m

l1μlBlG tl middot( 1113857

μFminus 1

Q21113858 1113859 minus 1113944m

l1μlF

minus 1gl(middot)Q1 tl middot( 1113857Q2(middot)1113858 1113859

Fminus 1 μQ2(middot) minus 1113944

m

l1μlgl(middot⎡⎣ ⎞⎠Q1 tl middot( 1113857Q2(middot)]

Fminus 1 μ minus 1113944

m

l1μlgl(middot)Q1 tl middot( 1113857Q2(middot)⎡⎣ ⎤⎦

Fminus 1 μ minus 1113944

m

l1μlgl(middot)Q1 tl middot( 1113857⎛⎝ ⎞⎠⎡⎢⎢⎣

middot μ minus 1113944m

l1μlgl(middot)Q1 tl middot( 1113857⎛⎝ ⎞⎠

minus 1⎤⎥⎥⎥⎥⎦ F

minus 1[1] δ

(78)

)erefore relation (75) is fulfilled in the space (S1ω2 )primeLemma 5 is proved

Remark 4 If μ 1 μ1 middot middot middot μm 0 then problems (24)and (25) degenerate into the Cauchy problem for equation(24) In this case Q2(σ) 1 forallσ isin R G(t x) Fminus 1[eminus ta(σ)]and G(t middot)⟶ Fminus 1[1] δ for t⟶ +0 in the space (S1ω1 )prime

Corollary 3 Let

ω(t x) flowastG(t x) f isin S1ω2lowast1113872 1113873prime (t x) isin Ω (79)

where (S1ω2lowast )prime is a class of convolutors in S1ω2 -en in (S1ω2 )primethe following boundary relation holds

μω(t middot) minus 1113944m

k1μkBkω tk middot( 1113857⟶ f t⟶ + 0 (80)

Proof Let us prove that the relation

F μω(t middot) minus 1113944

m

k1μkBkω tk middot( 1113857⎡⎣ ⎤⎦⟶ F[f] t⟶ + 0

(81)

takes place in the space (S21ω)prime Since f isin (S1ω2lowast )prime andG(t middot) isin S1ω2 for every tgt 0 we have

F[ω(t middot)] F[flowastG(t middot)] F[f]F[G(t middot)] F[f]Q(t middot)

(82)

So we need to prove that

F[f] μQ(t middot) minus 1113944m

k1μkgk(middot)Q tk middot( 1113857⎛⎝ ⎞⎠⟶ F[f] (83)

in (S21ω)prime when t⟶ + 0 SinceQ(t middot) Q1(t middot)Q2(middot)⟶ Q2(middot) for t⟶ + 0 in the space(S21ω)prime (see the proof of assertion 1 and Lemma 5) thecorrelation

8 International Journal of Differential Equations

μQ(t middot) minus 1113944m

k1μkgk(middot)Q tk middot( 1113857 ⟶

t⟶+0μQ2(middot) minus 1113944

m

k1μkgk(middot)Q1 tk middot( 1113857Q2(middot)

μ minus 1113944m

k1μkgk(middot)Q1 tk middot( 1113857⎛⎝ ⎞⎠Q2(middot) 1

(84)

is realized in the space (S21ω)prime )us relation (81) and (80)are fulfilled in the corresponding spaces )e statement isproved

)e function G(t middot) satisfies equation (24) as tgt 0Indeed

z

ztG(t x)

z

ztF

minus 1[Q(t σ)] F

minus 1 z

ztQ(t σ)1113890 1113891

Fminus 1

[a(σ)Q(t σ)]

AG(t x) Fminus 1

a(σ)Fminus 1

[G(t middot)]1113960 1113961 Fminus 1

[a(σ)Q(t σ)]

(85)

So

zG(t x)

zt+ AG(t x) 0 (t x) isin Ω (86)

which is what had to be provedIt follows from Corollary 3 that the nonlocal multipoint

by the time problem for equation (24) can be formulated inthe following way find a function u(t x) (t x) isin Ω thatsatisfies equation (24) and the condition

μ limt⟶+0

u(t middot) minus 1113944

m

k1μkBku tk middot( 1113857 f f isin S

1ω2lowast1113872 1113873prime (87)

where boundary relation (87) is considered in the space(S1ω2 )prime and the constraints on the parameters μ μ1 μmt1 tm are the same as in case of problems (24) and(25))

Theorem 1 By nonlocal multipoint by time problem (24)(87) is solvable and its solution is given by the formula

u(t x) flowastG(t x) (t x) isin Ω (88)

where u(t middot) isin S1ω2 for every tgt 0

Proof Make sure that the function u(t x) (t x) isin Ω sat-isfies equation (24) Indeed (see Corollary 2)

zu(t x)

zt

z

zt(flowastG(t middot)) flowast

zG(t middot)

zt

Au(t x) Fminus 1

[a(σ)F[f lowastG(t middot)]]

(89)

Since f is a convolutor in S1ω2

F[flowastG(t middot)] F[f]F[G(t middot)] F[f]Q(t middot) (90)

So

Au(t x) Fminus 1

[a(σ)Q(t σ)F[f](σ)] minus Fz

ztQ(t middot)F[f]1113890 1113891

minus F Fz

ztG(t middot)1113890 1113891 middot F[f]1113890 1113891

minus Fminus 1

F flowastzG(t middot)

zt1113890 11138911113890 1113891 minus flowast

zG(t middot)

zt

(91)

It follows from this that the function u(t x) (t x) isin Ωsatisfies equation (24) From Corollary 3 we get that u(t x)

satisfies condition (87) in the defined sense )e theorem isproved

Remark 5 If in condition (87) B1 middot middot middot Bm I (I is theidentity operator) then we can prove that problems (24) and(87) are well posed and its solution is given by the formula

u(t x) flowastG(t x) f isin S1ω1lowast1113872 1113873prime (t x) isin Ω (92)

where G(t middot) Fminus 1[Q(t middot)] isin S1ω1 for every tgt 0 and

Q(t σ) eminus ta(σ) μ minus 1113944

m

k1μke

minus tka(σ)⎛⎝ ⎞⎠

minus 1

(93)

Theorem 2 Suppose u(t x) (t x) isin Ω is the solution ofproblems (24) and (87) -en u(t middot)⟶ 0 in the space(S1ω2 )prime as t⟶ +infin

Proof Recall that the solution of problems (24) and (87) isgiven by the formula

u(t x) flowastG(t x) langfξ G(t x minus ξ)rang

f isin S1ω2lowast1113872 1113873prime (t x) isin Ω

(94)

Let us prove that langu(t middot)φrang⟶ 0 as t⟶ +infin for anarbitrary function φ isin S1ω2 Put

Ψt(ξ) 1113946+infin

minus infinG(t x minus ξ)φ(x)dx

ΨtR(ξ) 1113946R

minus RG(t x minus ξ)φ(x)dx Rgt 0 tgt 1

(95)

In these notations we prove that (a) for every tgt 1 andRgt 0 the function ΨtR(ξ) belongs to the space S1ω2 andΨtR(ξ)⟶Ψt(ξ) in the space S1ω2 as R⟶infin (b)Ψt(ξ) isin S1ω2 for every tgt 1 From this we get

langu(t middot)φrang 1113946+infin

minus infinlangfξ G(t x minus ξ)rangφ(x)dx

langfξ 1113946+infin

minus infinG(t x)φ(x + ξ)dxrang

langfξ 1113946+infin

minus infinG(t minus y)φ(minus (y minus ξ))dyrang

langfξ 1113946+infin

minus infinG(t minus y)φ(y minus ξ)dyrang φ(z) φ(minus z)

(96)

International Journal of Differential Equations 9

where u(t middot) is interpreted as a regular generalized functionfrom (S1ω2 )prime for every tgt 0

So let us set property (a) For fixed k m sub Z+ we have

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le 1113946+R

minus Rξkφ(x)D

mξ G(t x minus ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868dx

le 1113946+infin

minus infinξkφ(ξ + η)D

mη G(t η)

11138681113868111386811138681113868

11138681113868111386811138681113868dη

(97)

Since φ isin S1ω2 the inequality

ξkD

mξ φ(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cLkM

mk2k

mmω

k m sub Z+ (98)

with some c L Mgt 0 is valid It follows from this that forevery η isin R

supξisinR

ξkφ(ξ)11138681113868111386811138681113868

11138681113868111386811138681113868 supyisinR

(y minus η)kφ(η)

11138681113868111386811138681113868

11138681113868111386811138681113868 supyisinR

1113944

k

l0C

lky

l(minus η)

kminus lφ(y)

1113868111386811138681113868111386811138681113868111386811138681113868

1113868111386811138681113868111386811138681113868111386811138681113868

le 1113944k

l0C

lk|η|

kminus l supyisinR

ykφ(y)

11138681113868111386811138681113868

11138681113868111386811138681113868 le c 1113944k

l0C

lkL

ll2l

|η|kminus l

(99)

Next we will use estimate (58) )en

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le c 1113944k

l0C

lkL

ll2l

1113946+infin

minus infin|η|

kminus lD

mη G(t η)

11138681113868111386811138681113868

11138681113868111386811138681113868dη

le cc3Bm

mmω

tminus (1ω)

1113944

k

l0C

lkL

ll2l

middot 1113946+infin

minus infin|η|

kminus l exp minus a0tminus (12)

|η|12

1113966 1113967dη

(100)

By the direct calculations we find that

1113946+infin

minus infin|η|

kminus l exp minus a0tminus (12)

|η|12

1113966 1113967dη le c41113957L

kk2(kminus l)

tkminus l+1

c41113957Lgt 0

(101)

So

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cc3c4Bm

t1minus (1ω)

mmω

1113944

k

l0C

lkL

l1113957Lkminus l

tkminus l

l2l

k2(kminus l)

le cBm

Lkk2k

mmω

(102)

where c cc3c4t1minus (1ω) and L 2max L 1113957Lt1113864 1113865 )erefore

ΨtR(ξ) isin S1ω2 for every tgt 1 and an arbitrary Rgt 0 Fur-thermore we make sure thatΨtR(ξ)⟶Ψt(ξ) as R⟶infin

with all its derivatives uniformly in ξ on each segment[a b] sub R In addition the set of functions ξk

Dmξ ΨtR(ξ)1113966 1113967

k m sub Z+ is uniformly bounded (by R) in the space S1ω2(this property follows from estimate (102) in which c B andLgt 0 do not depend on R) )is means that condition (a) isfulfilled

Condition (b) follows from (a) because in a perfectspace every bounded set is compact

Using properties (a) and (b) we obtain the equality

langu(t middot)φrang 1113946+infin

minus infinG(t minus y)(flowastφ)(y)dy (103)

Since f isin (S1ω2 )prime is a convolutor in S1ω2 we haveflowast φ isin S1ω2 )en given estimate (58) (at s 0) we find that

|langu(t middot)φrang| le 1113946+infin

minus infin|G(t minus y)||(flowastφ)(y)|dy

lec

t1ω1113946

+infin

minus infin|(flowastφ)(y)|dy le

c0

t1ω⟶ 0 t⟶ +infin

(104)

for an arbitrary function φ isin S1ω2 ie u(t middot)⟶ 0 in thespace (S1ω2 )prime as t⟶ +infin )e theorem is proved

If the generalized function f in (87) is finite (ie itssupport suppf is a finite set in R) then we can say onuniform tending a solution u(t x) of problems (24) and (87)to zero onR as t⟶ +infin Note that every finite generalizedfunction is a convolutor in S spaces )is property followsfrom the general result concerning the theory of perfectspaces (see [12] p 137) if Φ is a perfect space with dif-ferential translation operation then every finite functional isa convolutor inΦ Finite functionals form a fairly wide classIn particular every bounded set F sub R is a support of somegeneralized function [12]

Theorem 3 Let u(t x) be a solution of problems (24) and(87) with the boundary function f in condition (87) which isan element of (S

β2)prime sub (S1ω2 )prime βgt 1 with finite support in R

-en u(t x)⟶ 0 uniformly on R as t⟶ +infin

Here is a scheme for proving this statement Letsuppf sub [a1 b1] sub [a2 b2] sub R Consider the functionφ isin S

β2 βgt 1 such that φ(x) 1 x isin [a1 b1] and

suppφ sub [a2 b2] )is function exists when βgt 1 because Sβ2

contains finite functions [12] Consider the function

u(t x) langfξ φ(ξ)G(t x minus ξ)rang +langfξ c(ξ)G(t x minus ξ)rang(105)

where c 1 minus φ Since supp(c(ξ)G(t x minus ξ)) cap suppφ emptywe have

u(t x) tminus (1ω)langfξ t

1ωφ(ξ)G(t x minus ξ)rang (106)

To prove the formulated above statement it remains toestablish that the set of functions Φtx(ξ) t1ωφ(ξ)G(t x minus

ξ) is bounded in the space Sβ2 βgt 1 for large t and x isin R

For example if in condition (87) f δ then δ is aconvolutor in the space S supp δ 0 andu(t x) δ lowastG(t x) G(t x) It follows directly from es-timate (58) that G(t x)⟶ 0 uniformly onR as t⟶ +infin

4 Conclusion

In this paper the solvability of a nonlocal multipoint by thetime problem for the evolutionary equation(zuzt) + (I minus (d2dx2))ω2u 0 (t x) isin (0 +infin) times R isproved herewith the operator (I minus (d2dx2))ω2 is inter-preted as a pseudodifferential operator in the space S1ω2

10 International Journal of Differential Equations

constructed by function (1 + σ2)ω2 σ isin R while the non-local multipoint by time condition also contains pseudo-differential operators constructed on smooth symbols )erepresentation of the solution is given in the form of aconvolution of the fundamental solution with the initialfunction which is an element of the space of generalizedfunctions of the ultradistribution type (the coagulator inspace S1ω2 ) )e behavior of the solution u(t x) t⟶ +infinin the space of generalized functions (S1ω2 )prime is investigated)e conditions for the initial generalized function are foundunder which the solution is uniformly stabilized to zero onR )e method of research of a nonlocal multipoint by timeproblem offered in this paper allows to interpret differential-operator equations of a form (zuzt) + φ(i zzx)u 0 as anevolutionary equation with a pseudodifferential operatorφ(i zzx) Fminus 1[φ middot F] constructed by the function φ actingin certain countable-normalized space of infinite-differentialfunctions (the choice of space depends on the properties ofthe function which is the symbol of the operator φ(i zzx))

Data Availability

No data were used to support the findings of this study

Disclosure

)is study was conducted in the framework of scientificactivity at Yuriy Fedkovych Chernivtsi National University

Conflicts of Interest

)e authors declare that they have no conflicts of interest

References

[1] V I Gorbachuk and M L Gorbachuk Boundary-ValueProblems for Differential-Operator Equations NaukovaDumka Kiev Ukraine 1984

[2] V V Gorodetskiy N I Nagnibida and P P NastasievMethods for Solving Problems in Functional Analysis VyshkaSchool Vyshka Ukraine 1990

[3] S G Samko A A Kilbas and O I Marichev Integrals andFractional Derivatives and Some of -eir Applications Scienceand Technology Minsk Belarus 1987

[4] I M Gelfand and G E Shilov Some Questions of the-eory ofDifferential Equations Fizmatgiz Moscow Russia 1958

[5] V V Gorodetskiy and O M Lenyuk About the FractionalDifferentiation in Spaces of Sprime type NAS of Ukraine KyivUkraine 1998

[6] A A Dezin General Questions of the -eory of BoundaryValue Problems Nauka Moscow Russia 1980

[7] A M Nakhushev Equations of Mathematical Biology HigherSchool Moscow Russia 1995

[8] I A Belavin S P Kapitsa and S P Kurdyumov ldquoAmathematical model of global demographic processes withconsidering of the spatial distributionrdquo Zh CalculationMathematics Physics vol 38 no 6 pp 885ndash902 1998

[9] V V Gorodetskiy and O V Martynyuk-e Cauchy Problemand Nonlocal Problems for First-Order Evolution EquationsEby Time Variable Vidavnichy dim rdquoRODOVIDrdquo ChernivtsiUkraine 2015

[10] V V Gorodetskiy and V I Myronyk ldquoTwo-point problem forone class of evolution equationsrdquo Differents Equationsvol 46 no 3 pp 349ndash363 2010

[11] V V Gorodetskiy and V I Myronyk ldquoTwo-point problem forone class of evolution equationsrdquo Differents Equationsvol 46 no 4 pp 520ndash526 2010

[12] I M Gelfand and G E Shilov Spaces of Basic and GeneralizedFunctions Fizmatgiz Moscow Russia 1958

International Journal of Differential Equations 11

Page 6: On One Evolution Equation of Parabolic Type with ...

dp

dRpR

minus 111138681113868111386811138681113868111386811138681113868

11138681113868111386811138681113868111386811138681113868le βp+1

0 p

DsσQ2(σ)

11138681113868111386811138681113868111386811138681113868 le s 1113944

s

p1βp+10 pαp

0Ls0

le β0Ls0(s)

21113944

s

p1βp0 le β0L

s0s middot (s)

2βs1 le β2β

s3s

2s s isin N

(48)

It follows from the last inequality and boundedness ofthe function Q2(σ) on R that Q2 is a multiplier in S21ω

Corollary 1 For every tgt 0 the functionQ(t σ) Q1(t σ)Q2(σ) σ isin R is an element of the spaceS21ω and the estimates

DsσQ(t σ)

11138681113868111386811138681113868111386811138681113868 le 1113957c1113957A

stcs

s2s exp minus t|σ|

ω1113864 1113865 s isin Z+ (t σ) isin Ω

(49)

are valid where the constants 1113957c and 1113957Agt 0 do not depend on t

Taking into account the properties of the Fouriertransform (direct and inverse) and the formulaFminus 1[S21ω] S1ω2 we get that G(t middot) isin S1ω2 for every tgt 0 Weremove in the estimates of derivatives of the function G (inthe variable x) the dependence on t assuming tgt 1 To dothis we use the relations

xkD

sxF[φ](x) i

k+sF σsφ(σ)( 1113857

(k)1113876 1113877

ik+s

1113946R

σsφ(σ)( 1113857(k)

eiσxdσ k s sub Z+ φ isin S

21ω

(50)

So

xkD

sxG(t x) (2π)

minus 1ik+s

(minus 1)s1113946R

σsQ(t minus σ)( 1113857

(k)e

iσxdσ

(51)

Applying the Leibniz formula for differentiating theproduct of two functions and estimating the derivatives ofthe function Q(t σ) we find that

σsQ(t minus σ)( 1113857

(k)11138681113868111386811138681113868

11138681113868111386811138681113868 1113944k

p0C

p

k σs( 1113857

(p)Q

(kminus p)(t minus σ)

1113868111386811138681113868111386811138681113868111386811138681113868

1113868111386811138681113868111386811138681113868111386811138681113868

le σsQ

(k)(t minus σ)

11138681113868111386811138681113868

11138681113868111386811138681113868 + ks σsminus 1Q

(kminus 1)(t minus σ)

11138681113868111386811138681113868

11138681113868111386811138681113868

+k(k minus 1)

2s(s minus 1) σsminus 2

Q(kminus 2)

(t minus σ)11138681113868111386811138681113868

11138681113868111386811138681113868

+ middot middot middot le 1113957c 1113957Aktck 1113957B

stminus (sω)

mks + ks1113957Akminus 1

tc(kminus 1)

1113876

middot 1113957Bsminus 1

tminus ((sminus 1)ω)

mkminus 1sminus 1

+12

k(k minus 1)s(s minus 1)1113957Akminus 2

tc(kminus 2) 1113957B

sminus 2tminus ((sminus 2)ω)

middot mkminus 2sminus 2 + 1113961eminus (t2)|σ|ω

(52)

where mks k2kssω Taking into account the results in (see[12] p 236ndash243) we find that this double sequence satisfiesthe inequality

ksmkminus 1sminus 1

mks

le 1113957c(k + s) 1113957cgt 0 (53)

Bearing in mind the last inequality and also that tgt 1 weobtain

σsQ(t minus σ)( 1113857

(k)11138681113868111386811138681113868

11138681113868111386811138681113868 le 1113957c1113957Aktk 1113957B

smks 1 +

ks

1113957A1113957B

mkminus 1sminus 1

mks

+12

k(k minus 1)s(s minus 1)1

1113957A21113957B

2mkminus 2sminus 2

mks

+ middot middot middot1113888 1113889eminus (t2)|σ|ω le 1113957c1113957A

ktk 1113957B

smks

times 1 +ks

1113957A1113957B

mkminus 1sminus 1

mks

+12

11113957A21113957B

2 ksmkminus 1sminus 1

mks

(k minus 1)(s minus 1)mkminus 2sminus 2

mkminus 1sminus 1+ middot middot middot1113888 1113889e

minus (t2)|σ|ω

le 1113957c1113957Aktk 1113957B

smks 1 +

1113957c

1113957A1113957B(k + s) +

12

1113957c2

(1113957A1113957B)2(k + s)

2+ middot middot middot1113888 1113889e

minus (t2)|σ|ω

le c1AktkB

smkse

minus (t2)|σ|ω

A 1113957A exp1113957c

1113957A1113957B1113882 1113883

B 1113957B exp1113957c

1113957A1113957B1113882 1113883

(54)

6 International Journal of Differential Equations

So

xkD

sxG(t x)

11138681113868111386811138681113868

11138681113868111386811138681113868 le (2π)minus 1

c1AktkB

smks1113946

Re

minus (t2)|σ|ωdσ

c2AktkB

stminus (1ω)

k2k

ssω

k s sub Z+

(55)

)en

DsxG(t x)

11138681113868111386811138681113868111386811138681113868 le c2B

stminus (1ω)

ssω inf

k

Akk2k

tminus 1|x|( )k

le c3Bstminus (1ω)

ssω exp minus a0t

minus (12)|x|

121113966 1113967

(56)

where the constants c3 B and a0 gt 0 do not depend on there we used the well-known inequality (see [12] p 204)

eminus (αe)|ξ|1α le inf

k

Lkkkα

|ξ|kle ce

minus (αe)|ξ|1α c e

αe2 (57)

in which α 2 and L 1113957A )us such a statement is correct

Lemma 3 -e derivatives (by variable x) of the functionG(t x) for tgt 1 satisfy the inequality

DsxG(t x)

11138681113868111386811138681113868111386811138681113868 le c3B

ss

sωtminus (1ω) exp minus a0t

minus (12)|x|

121113966 1113967 s isin Z+

(58)

where the constants c3 B and a0 gt 0 do not depend on t

Remark 3 In what follows we assume that in condition(25) β 2

Here are a few more properties of the function G(t x)

Lemma 4 -e function G(t middot) t isin (0 +infin) as an abstractfunction of t with values in the space S1ω2 is differentiable in t

Proof It follows from the continuity of the Fourier trans-form (direct and inverse) that to prove the statement itsuffices to establish that the function F[G(t middot)] Q(t middot) as afunction of a parameter t with values in the space S21ω isdifferentiable in t In other words it is necessary to provethat the boundary value relation

ΦΔt(σ) ≔1Δt

[Q(t + Δt middot) minus Q(t middot)]⟶z

ztQ(t middot) Δt⟶ 0

(59)

is performed in the sense as follows

(1) DsσΦΔt(σ) ⟶

Δt⟶ 0Ds

σ(minus a(σ)Q(t σ)) s isin Z+ uni-formly on each segment [a b] sub R

(2) |DsσΦΔt(σ)| le cB

ss2s exp minus a|σ|ω s isin Z+ where the

constants c B and agt 0 do not depend on Δt if Δt issmall enough

)e function Q(t σ) (t σ) isin Ω is differentiable in t inthe usual sense Due to the Lagrange theorem on finiteincrements

ΦΔt(σ) minus a(σ)Q(t + θΔt σ) (60)

So

DsσΦΔt(σ) minus 1113944

s

l0C

lsD

lσa(σ)D

sminus lσ Q(t +θΔtσ) (61)

Dsσ ΦΔt(σ) minus

z

ztQ(tσ)1113888 1113889 minus 1113944

s

l0C

lsD

lσa(σ) D

sminus lσ Q(t +θΔtσ)1113960

minus Dsminus lσ Q(tσ)1113961

(62)

Since

Dsminus lσ Q(t + θΔt σ) minus D

sminus lσ Q(t σ) D

sminus l+1σ Q t + θ1Δt σ( 1113857

0lt θ1 lt 1

(63)

it follows from this and estimate (49) that

Dsminus l+1σ Q t + θ1Δt σ( 1113857θΔt⟶ 0 Δt⟶ 0 (64)

uniformly on an arbitrary segment [a b] sub R )en

DsσΦΔt(σ)⟶ D

z

ztQ(t σ)1113888 1113889 Δt⟶ 0 (65)

uniformly on each segment [a b] sub R too )ereforecondition 1 is satisfied

Taking into account (61) estimates (15) (16) and (49)for a(σ) Q(t σ) and their derivatives we find that

DsσΦΔt(σ)

11138681113868111386811138681113868111386811138681113868 le c1cε 1113944

s

l0C

lsB

l1l

l 1113957Asminus l

tc(sminus l)

(s minus l)2(sminus l)

exp minus (t + θΔt)|σ|ω

+ ε|σ|ω

1113864 1113865

(66)

where εgt 0 is arbitrary fixed Take ε (t2) and then

DsσΦΔt(σ)

11138681113868111386811138681113868111386811138681113868 le cB

ss2s exp minus a|σ|

ω1113864 1113865 s isin Z+ (67)

where B 2max B11113957Atc1113966 1113967 a (t2) and all the constants

do not depend on Δt

Corollary 2 -e formulaz

zt(flowastG(t middot)) f lowast

zG(t middot)

zt forallf isin S

1ω21113872 1113873prime tgt 0

(68)

is correct

Proof According to the definition of a convolution of ageneralized function with a test one we have

flowastG(t middot) langfξ Tminus xG(t ξ)rang

G(t ξ) G(t minus ξ)(69)

)enz

zt(flowastG(t middot)) lim

Δt⟶0

1Δt

[(flowastG(t + Δt middot)) minus (flowastG(t middot))]

limΔt⟶0langfξ

1Δt

Tminus xG(t + Δt ξ) minus Tminus x

G(t ξ)1113960 1113961rang

(70)

International Journal of Differential Equations 7

Due to Lemma 4 the relation1Δt

Tminus xG(t + Δt ξ) minus Tminus x

G(t ξ)1113960 1113961⟶z

ztTminus x

G(t ξ) Δt⟶ 0

(71)

is performed in the sense of S1ω2 topology So

z

zt(flowastG(t middot)) langfξ lim

Δt⟶0

1Δt

Tminus xG(t + Δt ξ) minus Tminus x

G(t ξ)1113960 1113961rang

langfξ z

ztTminus x

G(t ξ)rang langfξ Tminus x

z

ztG(t ξ)rang

flowastz

ztG(t middot)

(72)

)e statement is provedSince

F[G(t middot)] Q(t σ) Q1(t σ)Q2(σ) 1113946R

G(t x)eiσxdx tgt 0

(73)

we get the formula

1113946R

G(t x)dx Q1(t 0)Q2(0) λ0eminus t

λ0 μ minus 1113944m

k1μkgk(0)e

minus tk⎛⎝ ⎞⎠

minus 1

(74)

where a(0) 1 is taken into account

Lemma 5 In (S1ω2 )prime the following relations are correct

(1) G(t middot)⟶ Fminus 1

Q21113858 1113859 t⟶ + 0

(2) μG(t middot) minus 1113944m

k1μkBkG tk middot( 1113857⟶ δ t⟶ + 0

(75)

where δ is the Dirac delta function

Proof (1) Taking into account the continuity property of theFourier transform (direct and inverse) in spaces of Sprime type itis sufficient to establish that

F[G(t middot)] Q1(t middot)Q2(middot)⟶ Q2(middot) t⟶ + 0 (76)

in the space (S21ω)prime To do this we take an arbitrary functionφ isin S21ω and using the fact thatQ2 is a multiplier in the spaceS21ω and the Lebesgue theorem on the limit passage underthe Lebesgue integral sign we find that

langQ1(t middot)Q2(middot)φrang langQ1(t middot) Q2(middot)φ(middot)rang

1113946R

Q1(t σ)Q2(σ)φ(σ)dσ ⟶t⟶+0

1113946R

Q2(σ)φ(σ)dσ

lang1 Q2(middot)φ(middot)rang langQ2φrang

(77)

)is leads us to statement (1) of Lemma 5(2) Given statement (1) and the form of the operators

B1 Bm we find that

μG(t middot) minus 1113944m

l1μlBlG tl middot( 1113857 ⟶

t⟶+0μF

minus 1Q21113858 1113859 minus 1113944

m

l1μlBlG tl middot( 1113857

μFminus 1

Q21113858 1113859 minus 1113944m

l1μlF

minus 1gl(middot)Q1 tl middot( 1113857Q2(middot)1113858 1113859

Fminus 1 μQ2(middot) minus 1113944

m

l1μlgl(middot⎡⎣ ⎞⎠Q1 tl middot( 1113857Q2(middot)]

Fminus 1 μ minus 1113944

m

l1μlgl(middot)Q1 tl middot( 1113857Q2(middot)⎡⎣ ⎤⎦

Fminus 1 μ minus 1113944

m

l1μlgl(middot)Q1 tl middot( 1113857⎛⎝ ⎞⎠⎡⎢⎢⎣

middot μ minus 1113944m

l1μlgl(middot)Q1 tl middot( 1113857⎛⎝ ⎞⎠

minus 1⎤⎥⎥⎥⎥⎦ F

minus 1[1] δ

(78)

)erefore relation (75) is fulfilled in the space (S1ω2 )primeLemma 5 is proved

Remark 4 If μ 1 μ1 middot middot middot μm 0 then problems (24)and (25) degenerate into the Cauchy problem for equation(24) In this case Q2(σ) 1 forallσ isin R G(t x) Fminus 1[eminus ta(σ)]and G(t middot)⟶ Fminus 1[1] δ for t⟶ +0 in the space (S1ω1 )prime

Corollary 3 Let

ω(t x) flowastG(t x) f isin S1ω2lowast1113872 1113873prime (t x) isin Ω (79)

where (S1ω2lowast )prime is a class of convolutors in S1ω2 -en in (S1ω2 )primethe following boundary relation holds

μω(t middot) minus 1113944m

k1μkBkω tk middot( 1113857⟶ f t⟶ + 0 (80)

Proof Let us prove that the relation

F μω(t middot) minus 1113944

m

k1μkBkω tk middot( 1113857⎡⎣ ⎤⎦⟶ F[f] t⟶ + 0

(81)

takes place in the space (S21ω)prime Since f isin (S1ω2lowast )prime andG(t middot) isin S1ω2 for every tgt 0 we have

F[ω(t middot)] F[flowastG(t middot)] F[f]F[G(t middot)] F[f]Q(t middot)

(82)

So we need to prove that

F[f] μQ(t middot) minus 1113944m

k1μkgk(middot)Q tk middot( 1113857⎛⎝ ⎞⎠⟶ F[f] (83)

in (S21ω)prime when t⟶ + 0 SinceQ(t middot) Q1(t middot)Q2(middot)⟶ Q2(middot) for t⟶ + 0 in the space(S21ω)prime (see the proof of assertion 1 and Lemma 5) thecorrelation

8 International Journal of Differential Equations

μQ(t middot) minus 1113944m

k1μkgk(middot)Q tk middot( 1113857 ⟶

t⟶+0μQ2(middot) minus 1113944

m

k1μkgk(middot)Q1 tk middot( 1113857Q2(middot)

μ minus 1113944m

k1μkgk(middot)Q1 tk middot( 1113857⎛⎝ ⎞⎠Q2(middot) 1

(84)

is realized in the space (S21ω)prime )us relation (81) and (80)are fulfilled in the corresponding spaces )e statement isproved

)e function G(t middot) satisfies equation (24) as tgt 0Indeed

z

ztG(t x)

z

ztF

minus 1[Q(t σ)] F

minus 1 z

ztQ(t σ)1113890 1113891

Fminus 1

[a(σ)Q(t σ)]

AG(t x) Fminus 1

a(σ)Fminus 1

[G(t middot)]1113960 1113961 Fminus 1

[a(σ)Q(t σ)]

(85)

So

zG(t x)

zt+ AG(t x) 0 (t x) isin Ω (86)

which is what had to be provedIt follows from Corollary 3 that the nonlocal multipoint

by the time problem for equation (24) can be formulated inthe following way find a function u(t x) (t x) isin Ω thatsatisfies equation (24) and the condition

μ limt⟶+0

u(t middot) minus 1113944

m

k1μkBku tk middot( 1113857 f f isin S

1ω2lowast1113872 1113873prime (87)

where boundary relation (87) is considered in the space(S1ω2 )prime and the constraints on the parameters μ μ1 μmt1 tm are the same as in case of problems (24) and(25))

Theorem 1 By nonlocal multipoint by time problem (24)(87) is solvable and its solution is given by the formula

u(t x) flowastG(t x) (t x) isin Ω (88)

where u(t middot) isin S1ω2 for every tgt 0

Proof Make sure that the function u(t x) (t x) isin Ω sat-isfies equation (24) Indeed (see Corollary 2)

zu(t x)

zt

z

zt(flowastG(t middot)) flowast

zG(t middot)

zt

Au(t x) Fminus 1

[a(σ)F[f lowastG(t middot)]]

(89)

Since f is a convolutor in S1ω2

F[flowastG(t middot)] F[f]F[G(t middot)] F[f]Q(t middot) (90)

So

Au(t x) Fminus 1

[a(σ)Q(t σ)F[f](σ)] minus Fz

ztQ(t middot)F[f]1113890 1113891

minus F Fz

ztG(t middot)1113890 1113891 middot F[f]1113890 1113891

minus Fminus 1

F flowastzG(t middot)

zt1113890 11138911113890 1113891 minus flowast

zG(t middot)

zt

(91)

It follows from this that the function u(t x) (t x) isin Ωsatisfies equation (24) From Corollary 3 we get that u(t x)

satisfies condition (87) in the defined sense )e theorem isproved

Remark 5 If in condition (87) B1 middot middot middot Bm I (I is theidentity operator) then we can prove that problems (24) and(87) are well posed and its solution is given by the formula

u(t x) flowastG(t x) f isin S1ω1lowast1113872 1113873prime (t x) isin Ω (92)

where G(t middot) Fminus 1[Q(t middot)] isin S1ω1 for every tgt 0 and

Q(t σ) eminus ta(σ) μ minus 1113944

m

k1μke

minus tka(σ)⎛⎝ ⎞⎠

minus 1

(93)

Theorem 2 Suppose u(t x) (t x) isin Ω is the solution ofproblems (24) and (87) -en u(t middot)⟶ 0 in the space(S1ω2 )prime as t⟶ +infin

Proof Recall that the solution of problems (24) and (87) isgiven by the formula

u(t x) flowastG(t x) langfξ G(t x minus ξ)rang

f isin S1ω2lowast1113872 1113873prime (t x) isin Ω

(94)

Let us prove that langu(t middot)φrang⟶ 0 as t⟶ +infin for anarbitrary function φ isin S1ω2 Put

Ψt(ξ) 1113946+infin

minus infinG(t x minus ξ)φ(x)dx

ΨtR(ξ) 1113946R

minus RG(t x minus ξ)φ(x)dx Rgt 0 tgt 1

(95)

In these notations we prove that (a) for every tgt 1 andRgt 0 the function ΨtR(ξ) belongs to the space S1ω2 andΨtR(ξ)⟶Ψt(ξ) in the space S1ω2 as R⟶infin (b)Ψt(ξ) isin S1ω2 for every tgt 1 From this we get

langu(t middot)φrang 1113946+infin

minus infinlangfξ G(t x minus ξ)rangφ(x)dx

langfξ 1113946+infin

minus infinG(t x)φ(x + ξ)dxrang

langfξ 1113946+infin

minus infinG(t minus y)φ(minus (y minus ξ))dyrang

langfξ 1113946+infin

minus infinG(t minus y)φ(y minus ξ)dyrang φ(z) φ(minus z)

(96)

International Journal of Differential Equations 9

where u(t middot) is interpreted as a regular generalized functionfrom (S1ω2 )prime for every tgt 0

So let us set property (a) For fixed k m sub Z+ we have

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le 1113946+R

minus Rξkφ(x)D

mξ G(t x minus ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868dx

le 1113946+infin

minus infinξkφ(ξ + η)D

mη G(t η)

11138681113868111386811138681113868

11138681113868111386811138681113868dη

(97)

Since φ isin S1ω2 the inequality

ξkD

mξ φ(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cLkM

mk2k

mmω

k m sub Z+ (98)

with some c L Mgt 0 is valid It follows from this that forevery η isin R

supξisinR

ξkφ(ξ)11138681113868111386811138681113868

11138681113868111386811138681113868 supyisinR

(y minus η)kφ(η)

11138681113868111386811138681113868

11138681113868111386811138681113868 supyisinR

1113944

k

l0C

lky

l(minus η)

kminus lφ(y)

1113868111386811138681113868111386811138681113868111386811138681113868

1113868111386811138681113868111386811138681113868111386811138681113868

le 1113944k

l0C

lk|η|

kminus l supyisinR

ykφ(y)

11138681113868111386811138681113868

11138681113868111386811138681113868 le c 1113944k

l0C

lkL

ll2l

|η|kminus l

(99)

Next we will use estimate (58) )en

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le c 1113944k

l0C

lkL

ll2l

1113946+infin

minus infin|η|

kminus lD

mη G(t η)

11138681113868111386811138681113868

11138681113868111386811138681113868dη

le cc3Bm

mmω

tminus (1ω)

1113944

k

l0C

lkL

ll2l

middot 1113946+infin

minus infin|η|

kminus l exp minus a0tminus (12)

|η|12

1113966 1113967dη

(100)

By the direct calculations we find that

1113946+infin

minus infin|η|

kminus l exp minus a0tminus (12)

|η|12

1113966 1113967dη le c41113957L

kk2(kminus l)

tkminus l+1

c41113957Lgt 0

(101)

So

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cc3c4Bm

t1minus (1ω)

mmω

1113944

k

l0C

lkL

l1113957Lkminus l

tkminus l

l2l

k2(kminus l)

le cBm

Lkk2k

mmω

(102)

where c cc3c4t1minus (1ω) and L 2max L 1113957Lt1113864 1113865 )erefore

ΨtR(ξ) isin S1ω2 for every tgt 1 and an arbitrary Rgt 0 Fur-thermore we make sure thatΨtR(ξ)⟶Ψt(ξ) as R⟶infin

with all its derivatives uniformly in ξ on each segment[a b] sub R In addition the set of functions ξk

Dmξ ΨtR(ξ)1113966 1113967

k m sub Z+ is uniformly bounded (by R) in the space S1ω2(this property follows from estimate (102) in which c B andLgt 0 do not depend on R) )is means that condition (a) isfulfilled

Condition (b) follows from (a) because in a perfectspace every bounded set is compact

Using properties (a) and (b) we obtain the equality

langu(t middot)φrang 1113946+infin

minus infinG(t minus y)(flowastφ)(y)dy (103)

Since f isin (S1ω2 )prime is a convolutor in S1ω2 we haveflowast φ isin S1ω2 )en given estimate (58) (at s 0) we find that

|langu(t middot)φrang| le 1113946+infin

minus infin|G(t minus y)||(flowastφ)(y)|dy

lec

t1ω1113946

+infin

minus infin|(flowastφ)(y)|dy le

c0

t1ω⟶ 0 t⟶ +infin

(104)

for an arbitrary function φ isin S1ω2 ie u(t middot)⟶ 0 in thespace (S1ω2 )prime as t⟶ +infin )e theorem is proved

If the generalized function f in (87) is finite (ie itssupport suppf is a finite set in R) then we can say onuniform tending a solution u(t x) of problems (24) and (87)to zero onR as t⟶ +infin Note that every finite generalizedfunction is a convolutor in S spaces )is property followsfrom the general result concerning the theory of perfectspaces (see [12] p 137) if Φ is a perfect space with dif-ferential translation operation then every finite functional isa convolutor inΦ Finite functionals form a fairly wide classIn particular every bounded set F sub R is a support of somegeneralized function [12]

Theorem 3 Let u(t x) be a solution of problems (24) and(87) with the boundary function f in condition (87) which isan element of (S

β2)prime sub (S1ω2 )prime βgt 1 with finite support in R

-en u(t x)⟶ 0 uniformly on R as t⟶ +infin

Here is a scheme for proving this statement Letsuppf sub [a1 b1] sub [a2 b2] sub R Consider the functionφ isin S

β2 βgt 1 such that φ(x) 1 x isin [a1 b1] and

suppφ sub [a2 b2] )is function exists when βgt 1 because Sβ2

contains finite functions [12] Consider the function

u(t x) langfξ φ(ξ)G(t x minus ξ)rang +langfξ c(ξ)G(t x minus ξ)rang(105)

where c 1 minus φ Since supp(c(ξ)G(t x minus ξ)) cap suppφ emptywe have

u(t x) tminus (1ω)langfξ t

1ωφ(ξ)G(t x minus ξ)rang (106)

To prove the formulated above statement it remains toestablish that the set of functions Φtx(ξ) t1ωφ(ξ)G(t x minus

ξ) is bounded in the space Sβ2 βgt 1 for large t and x isin R

For example if in condition (87) f δ then δ is aconvolutor in the space S supp δ 0 andu(t x) δ lowastG(t x) G(t x) It follows directly from es-timate (58) that G(t x)⟶ 0 uniformly onR as t⟶ +infin

4 Conclusion

In this paper the solvability of a nonlocal multipoint by thetime problem for the evolutionary equation(zuzt) + (I minus (d2dx2))ω2u 0 (t x) isin (0 +infin) times R isproved herewith the operator (I minus (d2dx2))ω2 is inter-preted as a pseudodifferential operator in the space S1ω2

10 International Journal of Differential Equations

constructed by function (1 + σ2)ω2 σ isin R while the non-local multipoint by time condition also contains pseudo-differential operators constructed on smooth symbols )erepresentation of the solution is given in the form of aconvolution of the fundamental solution with the initialfunction which is an element of the space of generalizedfunctions of the ultradistribution type (the coagulator inspace S1ω2 ) )e behavior of the solution u(t x) t⟶ +infinin the space of generalized functions (S1ω2 )prime is investigated)e conditions for the initial generalized function are foundunder which the solution is uniformly stabilized to zero onR )e method of research of a nonlocal multipoint by timeproblem offered in this paper allows to interpret differential-operator equations of a form (zuzt) + φ(i zzx)u 0 as anevolutionary equation with a pseudodifferential operatorφ(i zzx) Fminus 1[φ middot F] constructed by the function φ actingin certain countable-normalized space of infinite-differentialfunctions (the choice of space depends on the properties ofthe function which is the symbol of the operator φ(i zzx))

Data Availability

No data were used to support the findings of this study

Disclosure

)is study was conducted in the framework of scientificactivity at Yuriy Fedkovych Chernivtsi National University

Conflicts of Interest

)e authors declare that they have no conflicts of interest

References

[1] V I Gorbachuk and M L Gorbachuk Boundary-ValueProblems for Differential-Operator Equations NaukovaDumka Kiev Ukraine 1984

[2] V V Gorodetskiy N I Nagnibida and P P NastasievMethods for Solving Problems in Functional Analysis VyshkaSchool Vyshka Ukraine 1990

[3] S G Samko A A Kilbas and O I Marichev Integrals andFractional Derivatives and Some of -eir Applications Scienceand Technology Minsk Belarus 1987

[4] I M Gelfand and G E Shilov Some Questions of the-eory ofDifferential Equations Fizmatgiz Moscow Russia 1958

[5] V V Gorodetskiy and O M Lenyuk About the FractionalDifferentiation in Spaces of Sprime type NAS of Ukraine KyivUkraine 1998

[6] A A Dezin General Questions of the -eory of BoundaryValue Problems Nauka Moscow Russia 1980

[7] A M Nakhushev Equations of Mathematical Biology HigherSchool Moscow Russia 1995

[8] I A Belavin S P Kapitsa and S P Kurdyumov ldquoAmathematical model of global demographic processes withconsidering of the spatial distributionrdquo Zh CalculationMathematics Physics vol 38 no 6 pp 885ndash902 1998

[9] V V Gorodetskiy and O V Martynyuk-e Cauchy Problemand Nonlocal Problems for First-Order Evolution EquationsEby Time Variable Vidavnichy dim rdquoRODOVIDrdquo ChernivtsiUkraine 2015

[10] V V Gorodetskiy and V I Myronyk ldquoTwo-point problem forone class of evolution equationsrdquo Differents Equationsvol 46 no 3 pp 349ndash363 2010

[11] V V Gorodetskiy and V I Myronyk ldquoTwo-point problem forone class of evolution equationsrdquo Differents Equationsvol 46 no 4 pp 520ndash526 2010

[12] I M Gelfand and G E Shilov Spaces of Basic and GeneralizedFunctions Fizmatgiz Moscow Russia 1958

International Journal of Differential Equations 11

Page 7: On One Evolution Equation of Parabolic Type with ...

So

xkD

sxG(t x)

11138681113868111386811138681113868

11138681113868111386811138681113868 le (2π)minus 1

c1AktkB

smks1113946

Re

minus (t2)|σ|ωdσ

c2AktkB

stminus (1ω)

k2k

ssω

k s sub Z+

(55)

)en

DsxG(t x)

11138681113868111386811138681113868111386811138681113868 le c2B

stminus (1ω)

ssω inf

k

Akk2k

tminus 1|x|( )k

le c3Bstminus (1ω)

ssω exp minus a0t

minus (12)|x|

121113966 1113967

(56)

where the constants c3 B and a0 gt 0 do not depend on there we used the well-known inequality (see [12] p 204)

eminus (αe)|ξ|1α le inf

k

Lkkkα

|ξ|kle ce

minus (αe)|ξ|1α c e

αe2 (57)

in which α 2 and L 1113957A )us such a statement is correct

Lemma 3 -e derivatives (by variable x) of the functionG(t x) for tgt 1 satisfy the inequality

DsxG(t x)

11138681113868111386811138681113868111386811138681113868 le c3B

ss

sωtminus (1ω) exp minus a0t

minus (12)|x|

121113966 1113967 s isin Z+

(58)

where the constants c3 B and a0 gt 0 do not depend on t

Remark 3 In what follows we assume that in condition(25) β 2

Here are a few more properties of the function G(t x)

Lemma 4 -e function G(t middot) t isin (0 +infin) as an abstractfunction of t with values in the space S1ω2 is differentiable in t

Proof It follows from the continuity of the Fourier trans-form (direct and inverse) that to prove the statement itsuffices to establish that the function F[G(t middot)] Q(t middot) as afunction of a parameter t with values in the space S21ω isdifferentiable in t In other words it is necessary to provethat the boundary value relation

ΦΔt(σ) ≔1Δt

[Q(t + Δt middot) minus Q(t middot)]⟶z

ztQ(t middot) Δt⟶ 0

(59)

is performed in the sense as follows

(1) DsσΦΔt(σ) ⟶

Δt⟶ 0Ds

σ(minus a(σ)Q(t σ)) s isin Z+ uni-formly on each segment [a b] sub R

(2) |DsσΦΔt(σ)| le cB

ss2s exp minus a|σ|ω s isin Z+ where the

constants c B and agt 0 do not depend on Δt if Δt issmall enough

)e function Q(t σ) (t σ) isin Ω is differentiable in t inthe usual sense Due to the Lagrange theorem on finiteincrements

ΦΔt(σ) minus a(σ)Q(t + θΔt σ) (60)

So

DsσΦΔt(σ) minus 1113944

s

l0C

lsD

lσa(σ)D

sminus lσ Q(t +θΔtσ) (61)

Dsσ ΦΔt(σ) minus

z

ztQ(tσ)1113888 1113889 minus 1113944

s

l0C

lsD

lσa(σ) D

sminus lσ Q(t +θΔtσ)1113960

minus Dsminus lσ Q(tσ)1113961

(62)

Since

Dsminus lσ Q(t + θΔt σ) minus D

sminus lσ Q(t σ) D

sminus l+1σ Q t + θ1Δt σ( 1113857

0lt θ1 lt 1

(63)

it follows from this and estimate (49) that

Dsminus l+1σ Q t + θ1Δt σ( 1113857θΔt⟶ 0 Δt⟶ 0 (64)

uniformly on an arbitrary segment [a b] sub R )en

DsσΦΔt(σ)⟶ D

z

ztQ(t σ)1113888 1113889 Δt⟶ 0 (65)

uniformly on each segment [a b] sub R too )ereforecondition 1 is satisfied

Taking into account (61) estimates (15) (16) and (49)for a(σ) Q(t σ) and their derivatives we find that

DsσΦΔt(σ)

11138681113868111386811138681113868111386811138681113868 le c1cε 1113944

s

l0C

lsB

l1l

l 1113957Asminus l

tc(sminus l)

(s minus l)2(sminus l)

exp minus (t + θΔt)|σ|ω

+ ε|σ|ω

1113864 1113865

(66)

where εgt 0 is arbitrary fixed Take ε (t2) and then

DsσΦΔt(σ)

11138681113868111386811138681113868111386811138681113868 le cB

ss2s exp minus a|σ|

ω1113864 1113865 s isin Z+ (67)

where B 2max B11113957Atc1113966 1113967 a (t2) and all the constants

do not depend on Δt

Corollary 2 -e formulaz

zt(flowastG(t middot)) f lowast

zG(t middot)

zt forallf isin S

1ω21113872 1113873prime tgt 0

(68)

is correct

Proof According to the definition of a convolution of ageneralized function with a test one we have

flowastG(t middot) langfξ Tminus xG(t ξ)rang

G(t ξ) G(t minus ξ)(69)

)enz

zt(flowastG(t middot)) lim

Δt⟶0

1Δt

[(flowastG(t + Δt middot)) minus (flowastG(t middot))]

limΔt⟶0langfξ

1Δt

Tminus xG(t + Δt ξ) minus Tminus x

G(t ξ)1113960 1113961rang

(70)

International Journal of Differential Equations 7

Due to Lemma 4 the relation1Δt

Tminus xG(t + Δt ξ) minus Tminus x

G(t ξ)1113960 1113961⟶z

ztTminus x

G(t ξ) Δt⟶ 0

(71)

is performed in the sense of S1ω2 topology So

z

zt(flowastG(t middot)) langfξ lim

Δt⟶0

1Δt

Tminus xG(t + Δt ξ) minus Tminus x

G(t ξ)1113960 1113961rang

langfξ z

ztTminus x

G(t ξ)rang langfξ Tminus x

z

ztG(t ξ)rang

flowastz

ztG(t middot)

(72)

)e statement is provedSince

F[G(t middot)] Q(t σ) Q1(t σ)Q2(σ) 1113946R

G(t x)eiσxdx tgt 0

(73)

we get the formula

1113946R

G(t x)dx Q1(t 0)Q2(0) λ0eminus t

λ0 μ minus 1113944m

k1μkgk(0)e

minus tk⎛⎝ ⎞⎠

minus 1

(74)

where a(0) 1 is taken into account

Lemma 5 In (S1ω2 )prime the following relations are correct

(1) G(t middot)⟶ Fminus 1

Q21113858 1113859 t⟶ + 0

(2) μG(t middot) minus 1113944m

k1μkBkG tk middot( 1113857⟶ δ t⟶ + 0

(75)

where δ is the Dirac delta function

Proof (1) Taking into account the continuity property of theFourier transform (direct and inverse) in spaces of Sprime type itis sufficient to establish that

F[G(t middot)] Q1(t middot)Q2(middot)⟶ Q2(middot) t⟶ + 0 (76)

in the space (S21ω)prime To do this we take an arbitrary functionφ isin S21ω and using the fact thatQ2 is a multiplier in the spaceS21ω and the Lebesgue theorem on the limit passage underthe Lebesgue integral sign we find that

langQ1(t middot)Q2(middot)φrang langQ1(t middot) Q2(middot)φ(middot)rang

1113946R

Q1(t σ)Q2(σ)φ(σ)dσ ⟶t⟶+0

1113946R

Q2(σ)φ(σ)dσ

lang1 Q2(middot)φ(middot)rang langQ2φrang

(77)

)is leads us to statement (1) of Lemma 5(2) Given statement (1) and the form of the operators

B1 Bm we find that

μG(t middot) minus 1113944m

l1μlBlG tl middot( 1113857 ⟶

t⟶+0μF

minus 1Q21113858 1113859 minus 1113944

m

l1μlBlG tl middot( 1113857

μFminus 1

Q21113858 1113859 minus 1113944m

l1μlF

minus 1gl(middot)Q1 tl middot( 1113857Q2(middot)1113858 1113859

Fminus 1 μQ2(middot) minus 1113944

m

l1μlgl(middot⎡⎣ ⎞⎠Q1 tl middot( 1113857Q2(middot)]

Fminus 1 μ minus 1113944

m

l1μlgl(middot)Q1 tl middot( 1113857Q2(middot)⎡⎣ ⎤⎦

Fminus 1 μ minus 1113944

m

l1μlgl(middot)Q1 tl middot( 1113857⎛⎝ ⎞⎠⎡⎢⎢⎣

middot μ minus 1113944m

l1μlgl(middot)Q1 tl middot( 1113857⎛⎝ ⎞⎠

minus 1⎤⎥⎥⎥⎥⎦ F

minus 1[1] δ

(78)

)erefore relation (75) is fulfilled in the space (S1ω2 )primeLemma 5 is proved

Remark 4 If μ 1 μ1 middot middot middot μm 0 then problems (24)and (25) degenerate into the Cauchy problem for equation(24) In this case Q2(σ) 1 forallσ isin R G(t x) Fminus 1[eminus ta(σ)]and G(t middot)⟶ Fminus 1[1] δ for t⟶ +0 in the space (S1ω1 )prime

Corollary 3 Let

ω(t x) flowastG(t x) f isin S1ω2lowast1113872 1113873prime (t x) isin Ω (79)

where (S1ω2lowast )prime is a class of convolutors in S1ω2 -en in (S1ω2 )primethe following boundary relation holds

μω(t middot) minus 1113944m

k1μkBkω tk middot( 1113857⟶ f t⟶ + 0 (80)

Proof Let us prove that the relation

F μω(t middot) minus 1113944

m

k1μkBkω tk middot( 1113857⎡⎣ ⎤⎦⟶ F[f] t⟶ + 0

(81)

takes place in the space (S21ω)prime Since f isin (S1ω2lowast )prime andG(t middot) isin S1ω2 for every tgt 0 we have

F[ω(t middot)] F[flowastG(t middot)] F[f]F[G(t middot)] F[f]Q(t middot)

(82)

So we need to prove that

F[f] μQ(t middot) minus 1113944m

k1μkgk(middot)Q tk middot( 1113857⎛⎝ ⎞⎠⟶ F[f] (83)

in (S21ω)prime when t⟶ + 0 SinceQ(t middot) Q1(t middot)Q2(middot)⟶ Q2(middot) for t⟶ + 0 in the space(S21ω)prime (see the proof of assertion 1 and Lemma 5) thecorrelation

8 International Journal of Differential Equations

μQ(t middot) minus 1113944m

k1μkgk(middot)Q tk middot( 1113857 ⟶

t⟶+0μQ2(middot) minus 1113944

m

k1μkgk(middot)Q1 tk middot( 1113857Q2(middot)

μ minus 1113944m

k1μkgk(middot)Q1 tk middot( 1113857⎛⎝ ⎞⎠Q2(middot) 1

(84)

is realized in the space (S21ω)prime )us relation (81) and (80)are fulfilled in the corresponding spaces )e statement isproved

)e function G(t middot) satisfies equation (24) as tgt 0Indeed

z

ztG(t x)

z

ztF

minus 1[Q(t σ)] F

minus 1 z

ztQ(t σ)1113890 1113891

Fminus 1

[a(σ)Q(t σ)]

AG(t x) Fminus 1

a(σ)Fminus 1

[G(t middot)]1113960 1113961 Fminus 1

[a(σ)Q(t σ)]

(85)

So

zG(t x)

zt+ AG(t x) 0 (t x) isin Ω (86)

which is what had to be provedIt follows from Corollary 3 that the nonlocal multipoint

by the time problem for equation (24) can be formulated inthe following way find a function u(t x) (t x) isin Ω thatsatisfies equation (24) and the condition

μ limt⟶+0

u(t middot) minus 1113944

m

k1μkBku tk middot( 1113857 f f isin S

1ω2lowast1113872 1113873prime (87)

where boundary relation (87) is considered in the space(S1ω2 )prime and the constraints on the parameters μ μ1 μmt1 tm are the same as in case of problems (24) and(25))

Theorem 1 By nonlocal multipoint by time problem (24)(87) is solvable and its solution is given by the formula

u(t x) flowastG(t x) (t x) isin Ω (88)

where u(t middot) isin S1ω2 for every tgt 0

Proof Make sure that the function u(t x) (t x) isin Ω sat-isfies equation (24) Indeed (see Corollary 2)

zu(t x)

zt

z

zt(flowastG(t middot)) flowast

zG(t middot)

zt

Au(t x) Fminus 1

[a(σ)F[f lowastG(t middot)]]

(89)

Since f is a convolutor in S1ω2

F[flowastG(t middot)] F[f]F[G(t middot)] F[f]Q(t middot) (90)

So

Au(t x) Fminus 1

[a(σ)Q(t σ)F[f](σ)] minus Fz

ztQ(t middot)F[f]1113890 1113891

minus F Fz

ztG(t middot)1113890 1113891 middot F[f]1113890 1113891

minus Fminus 1

F flowastzG(t middot)

zt1113890 11138911113890 1113891 minus flowast

zG(t middot)

zt

(91)

It follows from this that the function u(t x) (t x) isin Ωsatisfies equation (24) From Corollary 3 we get that u(t x)

satisfies condition (87) in the defined sense )e theorem isproved

Remark 5 If in condition (87) B1 middot middot middot Bm I (I is theidentity operator) then we can prove that problems (24) and(87) are well posed and its solution is given by the formula

u(t x) flowastG(t x) f isin S1ω1lowast1113872 1113873prime (t x) isin Ω (92)

where G(t middot) Fminus 1[Q(t middot)] isin S1ω1 for every tgt 0 and

Q(t σ) eminus ta(σ) μ minus 1113944

m

k1μke

minus tka(σ)⎛⎝ ⎞⎠

minus 1

(93)

Theorem 2 Suppose u(t x) (t x) isin Ω is the solution ofproblems (24) and (87) -en u(t middot)⟶ 0 in the space(S1ω2 )prime as t⟶ +infin

Proof Recall that the solution of problems (24) and (87) isgiven by the formula

u(t x) flowastG(t x) langfξ G(t x minus ξ)rang

f isin S1ω2lowast1113872 1113873prime (t x) isin Ω

(94)

Let us prove that langu(t middot)φrang⟶ 0 as t⟶ +infin for anarbitrary function φ isin S1ω2 Put

Ψt(ξ) 1113946+infin

minus infinG(t x minus ξ)φ(x)dx

ΨtR(ξ) 1113946R

minus RG(t x minus ξ)φ(x)dx Rgt 0 tgt 1

(95)

In these notations we prove that (a) for every tgt 1 andRgt 0 the function ΨtR(ξ) belongs to the space S1ω2 andΨtR(ξ)⟶Ψt(ξ) in the space S1ω2 as R⟶infin (b)Ψt(ξ) isin S1ω2 for every tgt 1 From this we get

langu(t middot)φrang 1113946+infin

minus infinlangfξ G(t x minus ξ)rangφ(x)dx

langfξ 1113946+infin

minus infinG(t x)φ(x + ξ)dxrang

langfξ 1113946+infin

minus infinG(t minus y)φ(minus (y minus ξ))dyrang

langfξ 1113946+infin

minus infinG(t minus y)φ(y minus ξ)dyrang φ(z) φ(minus z)

(96)

International Journal of Differential Equations 9

where u(t middot) is interpreted as a regular generalized functionfrom (S1ω2 )prime for every tgt 0

So let us set property (a) For fixed k m sub Z+ we have

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le 1113946+R

minus Rξkφ(x)D

mξ G(t x minus ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868dx

le 1113946+infin

minus infinξkφ(ξ + η)D

mη G(t η)

11138681113868111386811138681113868

11138681113868111386811138681113868dη

(97)

Since φ isin S1ω2 the inequality

ξkD

mξ φ(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cLkM

mk2k

mmω

k m sub Z+ (98)

with some c L Mgt 0 is valid It follows from this that forevery η isin R

supξisinR

ξkφ(ξ)11138681113868111386811138681113868

11138681113868111386811138681113868 supyisinR

(y minus η)kφ(η)

11138681113868111386811138681113868

11138681113868111386811138681113868 supyisinR

1113944

k

l0C

lky

l(minus η)

kminus lφ(y)

1113868111386811138681113868111386811138681113868111386811138681113868

1113868111386811138681113868111386811138681113868111386811138681113868

le 1113944k

l0C

lk|η|

kminus l supyisinR

ykφ(y)

11138681113868111386811138681113868

11138681113868111386811138681113868 le c 1113944k

l0C

lkL

ll2l

|η|kminus l

(99)

Next we will use estimate (58) )en

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le c 1113944k

l0C

lkL

ll2l

1113946+infin

minus infin|η|

kminus lD

mη G(t η)

11138681113868111386811138681113868

11138681113868111386811138681113868dη

le cc3Bm

mmω

tminus (1ω)

1113944

k

l0C

lkL

ll2l

middot 1113946+infin

minus infin|η|

kminus l exp minus a0tminus (12)

|η|12

1113966 1113967dη

(100)

By the direct calculations we find that

1113946+infin

minus infin|η|

kminus l exp minus a0tminus (12)

|η|12

1113966 1113967dη le c41113957L

kk2(kminus l)

tkminus l+1

c41113957Lgt 0

(101)

So

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cc3c4Bm

t1minus (1ω)

mmω

1113944

k

l0C

lkL

l1113957Lkminus l

tkminus l

l2l

k2(kminus l)

le cBm

Lkk2k

mmω

(102)

where c cc3c4t1minus (1ω) and L 2max L 1113957Lt1113864 1113865 )erefore

ΨtR(ξ) isin S1ω2 for every tgt 1 and an arbitrary Rgt 0 Fur-thermore we make sure thatΨtR(ξ)⟶Ψt(ξ) as R⟶infin

with all its derivatives uniformly in ξ on each segment[a b] sub R In addition the set of functions ξk

Dmξ ΨtR(ξ)1113966 1113967

k m sub Z+ is uniformly bounded (by R) in the space S1ω2(this property follows from estimate (102) in which c B andLgt 0 do not depend on R) )is means that condition (a) isfulfilled

Condition (b) follows from (a) because in a perfectspace every bounded set is compact

Using properties (a) and (b) we obtain the equality

langu(t middot)φrang 1113946+infin

minus infinG(t minus y)(flowastφ)(y)dy (103)

Since f isin (S1ω2 )prime is a convolutor in S1ω2 we haveflowast φ isin S1ω2 )en given estimate (58) (at s 0) we find that

|langu(t middot)φrang| le 1113946+infin

minus infin|G(t minus y)||(flowastφ)(y)|dy

lec

t1ω1113946

+infin

minus infin|(flowastφ)(y)|dy le

c0

t1ω⟶ 0 t⟶ +infin

(104)

for an arbitrary function φ isin S1ω2 ie u(t middot)⟶ 0 in thespace (S1ω2 )prime as t⟶ +infin )e theorem is proved

If the generalized function f in (87) is finite (ie itssupport suppf is a finite set in R) then we can say onuniform tending a solution u(t x) of problems (24) and (87)to zero onR as t⟶ +infin Note that every finite generalizedfunction is a convolutor in S spaces )is property followsfrom the general result concerning the theory of perfectspaces (see [12] p 137) if Φ is a perfect space with dif-ferential translation operation then every finite functional isa convolutor inΦ Finite functionals form a fairly wide classIn particular every bounded set F sub R is a support of somegeneralized function [12]

Theorem 3 Let u(t x) be a solution of problems (24) and(87) with the boundary function f in condition (87) which isan element of (S

β2)prime sub (S1ω2 )prime βgt 1 with finite support in R

-en u(t x)⟶ 0 uniformly on R as t⟶ +infin

Here is a scheme for proving this statement Letsuppf sub [a1 b1] sub [a2 b2] sub R Consider the functionφ isin S

β2 βgt 1 such that φ(x) 1 x isin [a1 b1] and

suppφ sub [a2 b2] )is function exists when βgt 1 because Sβ2

contains finite functions [12] Consider the function

u(t x) langfξ φ(ξ)G(t x minus ξ)rang +langfξ c(ξ)G(t x minus ξ)rang(105)

where c 1 minus φ Since supp(c(ξ)G(t x minus ξ)) cap suppφ emptywe have

u(t x) tminus (1ω)langfξ t

1ωφ(ξ)G(t x minus ξ)rang (106)

To prove the formulated above statement it remains toestablish that the set of functions Φtx(ξ) t1ωφ(ξ)G(t x minus

ξ) is bounded in the space Sβ2 βgt 1 for large t and x isin R

For example if in condition (87) f δ then δ is aconvolutor in the space S supp δ 0 andu(t x) δ lowastG(t x) G(t x) It follows directly from es-timate (58) that G(t x)⟶ 0 uniformly onR as t⟶ +infin

4 Conclusion

In this paper the solvability of a nonlocal multipoint by thetime problem for the evolutionary equation(zuzt) + (I minus (d2dx2))ω2u 0 (t x) isin (0 +infin) times R isproved herewith the operator (I minus (d2dx2))ω2 is inter-preted as a pseudodifferential operator in the space S1ω2

10 International Journal of Differential Equations

constructed by function (1 + σ2)ω2 σ isin R while the non-local multipoint by time condition also contains pseudo-differential operators constructed on smooth symbols )erepresentation of the solution is given in the form of aconvolution of the fundamental solution with the initialfunction which is an element of the space of generalizedfunctions of the ultradistribution type (the coagulator inspace S1ω2 ) )e behavior of the solution u(t x) t⟶ +infinin the space of generalized functions (S1ω2 )prime is investigated)e conditions for the initial generalized function are foundunder which the solution is uniformly stabilized to zero onR )e method of research of a nonlocal multipoint by timeproblem offered in this paper allows to interpret differential-operator equations of a form (zuzt) + φ(i zzx)u 0 as anevolutionary equation with a pseudodifferential operatorφ(i zzx) Fminus 1[φ middot F] constructed by the function φ actingin certain countable-normalized space of infinite-differentialfunctions (the choice of space depends on the properties ofthe function which is the symbol of the operator φ(i zzx))

Data Availability

No data were used to support the findings of this study

Disclosure

)is study was conducted in the framework of scientificactivity at Yuriy Fedkovych Chernivtsi National University

Conflicts of Interest

)e authors declare that they have no conflicts of interest

References

[1] V I Gorbachuk and M L Gorbachuk Boundary-ValueProblems for Differential-Operator Equations NaukovaDumka Kiev Ukraine 1984

[2] V V Gorodetskiy N I Nagnibida and P P NastasievMethods for Solving Problems in Functional Analysis VyshkaSchool Vyshka Ukraine 1990

[3] S G Samko A A Kilbas and O I Marichev Integrals andFractional Derivatives and Some of -eir Applications Scienceand Technology Minsk Belarus 1987

[4] I M Gelfand and G E Shilov Some Questions of the-eory ofDifferential Equations Fizmatgiz Moscow Russia 1958

[5] V V Gorodetskiy and O M Lenyuk About the FractionalDifferentiation in Spaces of Sprime type NAS of Ukraine KyivUkraine 1998

[6] A A Dezin General Questions of the -eory of BoundaryValue Problems Nauka Moscow Russia 1980

[7] A M Nakhushev Equations of Mathematical Biology HigherSchool Moscow Russia 1995

[8] I A Belavin S P Kapitsa and S P Kurdyumov ldquoAmathematical model of global demographic processes withconsidering of the spatial distributionrdquo Zh CalculationMathematics Physics vol 38 no 6 pp 885ndash902 1998

[9] V V Gorodetskiy and O V Martynyuk-e Cauchy Problemand Nonlocal Problems for First-Order Evolution EquationsEby Time Variable Vidavnichy dim rdquoRODOVIDrdquo ChernivtsiUkraine 2015

[10] V V Gorodetskiy and V I Myronyk ldquoTwo-point problem forone class of evolution equationsrdquo Differents Equationsvol 46 no 3 pp 349ndash363 2010

[11] V V Gorodetskiy and V I Myronyk ldquoTwo-point problem forone class of evolution equationsrdquo Differents Equationsvol 46 no 4 pp 520ndash526 2010

[12] I M Gelfand and G E Shilov Spaces of Basic and GeneralizedFunctions Fizmatgiz Moscow Russia 1958

International Journal of Differential Equations 11

Page 8: On One Evolution Equation of Parabolic Type with ...

Due to Lemma 4 the relation1Δt

Tminus xG(t + Δt ξ) minus Tminus x

G(t ξ)1113960 1113961⟶z

ztTminus x

G(t ξ) Δt⟶ 0

(71)

is performed in the sense of S1ω2 topology So

z

zt(flowastG(t middot)) langfξ lim

Δt⟶0

1Δt

Tminus xG(t + Δt ξ) minus Tminus x

G(t ξ)1113960 1113961rang

langfξ z

ztTminus x

G(t ξ)rang langfξ Tminus x

z

ztG(t ξ)rang

flowastz

ztG(t middot)

(72)

)e statement is provedSince

F[G(t middot)] Q(t σ) Q1(t σ)Q2(σ) 1113946R

G(t x)eiσxdx tgt 0

(73)

we get the formula

1113946R

G(t x)dx Q1(t 0)Q2(0) λ0eminus t

λ0 μ minus 1113944m

k1μkgk(0)e

minus tk⎛⎝ ⎞⎠

minus 1

(74)

where a(0) 1 is taken into account

Lemma 5 In (S1ω2 )prime the following relations are correct

(1) G(t middot)⟶ Fminus 1

Q21113858 1113859 t⟶ + 0

(2) μG(t middot) minus 1113944m

k1μkBkG tk middot( 1113857⟶ δ t⟶ + 0

(75)

where δ is the Dirac delta function

Proof (1) Taking into account the continuity property of theFourier transform (direct and inverse) in spaces of Sprime type itis sufficient to establish that

F[G(t middot)] Q1(t middot)Q2(middot)⟶ Q2(middot) t⟶ + 0 (76)

in the space (S21ω)prime To do this we take an arbitrary functionφ isin S21ω and using the fact thatQ2 is a multiplier in the spaceS21ω and the Lebesgue theorem on the limit passage underthe Lebesgue integral sign we find that

langQ1(t middot)Q2(middot)φrang langQ1(t middot) Q2(middot)φ(middot)rang

1113946R

Q1(t σ)Q2(σ)φ(σ)dσ ⟶t⟶+0

1113946R

Q2(σ)φ(σ)dσ

lang1 Q2(middot)φ(middot)rang langQ2φrang

(77)

)is leads us to statement (1) of Lemma 5(2) Given statement (1) and the form of the operators

B1 Bm we find that

μG(t middot) minus 1113944m

l1μlBlG tl middot( 1113857 ⟶

t⟶+0μF

minus 1Q21113858 1113859 minus 1113944

m

l1μlBlG tl middot( 1113857

μFminus 1

Q21113858 1113859 minus 1113944m

l1μlF

minus 1gl(middot)Q1 tl middot( 1113857Q2(middot)1113858 1113859

Fminus 1 μQ2(middot) minus 1113944

m

l1μlgl(middot⎡⎣ ⎞⎠Q1 tl middot( 1113857Q2(middot)]

Fminus 1 μ minus 1113944

m

l1μlgl(middot)Q1 tl middot( 1113857Q2(middot)⎡⎣ ⎤⎦

Fminus 1 μ minus 1113944

m

l1μlgl(middot)Q1 tl middot( 1113857⎛⎝ ⎞⎠⎡⎢⎢⎣

middot μ minus 1113944m

l1μlgl(middot)Q1 tl middot( 1113857⎛⎝ ⎞⎠

minus 1⎤⎥⎥⎥⎥⎦ F

minus 1[1] δ

(78)

)erefore relation (75) is fulfilled in the space (S1ω2 )primeLemma 5 is proved

Remark 4 If μ 1 μ1 middot middot middot μm 0 then problems (24)and (25) degenerate into the Cauchy problem for equation(24) In this case Q2(σ) 1 forallσ isin R G(t x) Fminus 1[eminus ta(σ)]and G(t middot)⟶ Fminus 1[1] δ for t⟶ +0 in the space (S1ω1 )prime

Corollary 3 Let

ω(t x) flowastG(t x) f isin S1ω2lowast1113872 1113873prime (t x) isin Ω (79)

where (S1ω2lowast )prime is a class of convolutors in S1ω2 -en in (S1ω2 )primethe following boundary relation holds

μω(t middot) minus 1113944m

k1μkBkω tk middot( 1113857⟶ f t⟶ + 0 (80)

Proof Let us prove that the relation

F μω(t middot) minus 1113944

m

k1μkBkω tk middot( 1113857⎡⎣ ⎤⎦⟶ F[f] t⟶ + 0

(81)

takes place in the space (S21ω)prime Since f isin (S1ω2lowast )prime andG(t middot) isin S1ω2 for every tgt 0 we have

F[ω(t middot)] F[flowastG(t middot)] F[f]F[G(t middot)] F[f]Q(t middot)

(82)

So we need to prove that

F[f] μQ(t middot) minus 1113944m

k1μkgk(middot)Q tk middot( 1113857⎛⎝ ⎞⎠⟶ F[f] (83)

in (S21ω)prime when t⟶ + 0 SinceQ(t middot) Q1(t middot)Q2(middot)⟶ Q2(middot) for t⟶ + 0 in the space(S21ω)prime (see the proof of assertion 1 and Lemma 5) thecorrelation

8 International Journal of Differential Equations

μQ(t middot) minus 1113944m

k1μkgk(middot)Q tk middot( 1113857 ⟶

t⟶+0μQ2(middot) minus 1113944

m

k1μkgk(middot)Q1 tk middot( 1113857Q2(middot)

μ minus 1113944m

k1μkgk(middot)Q1 tk middot( 1113857⎛⎝ ⎞⎠Q2(middot) 1

(84)

is realized in the space (S21ω)prime )us relation (81) and (80)are fulfilled in the corresponding spaces )e statement isproved

)e function G(t middot) satisfies equation (24) as tgt 0Indeed

z

ztG(t x)

z

ztF

minus 1[Q(t σ)] F

minus 1 z

ztQ(t σ)1113890 1113891

Fminus 1

[a(σ)Q(t σ)]

AG(t x) Fminus 1

a(σ)Fminus 1

[G(t middot)]1113960 1113961 Fminus 1

[a(σ)Q(t σ)]

(85)

So

zG(t x)

zt+ AG(t x) 0 (t x) isin Ω (86)

which is what had to be provedIt follows from Corollary 3 that the nonlocal multipoint

by the time problem for equation (24) can be formulated inthe following way find a function u(t x) (t x) isin Ω thatsatisfies equation (24) and the condition

μ limt⟶+0

u(t middot) minus 1113944

m

k1μkBku tk middot( 1113857 f f isin S

1ω2lowast1113872 1113873prime (87)

where boundary relation (87) is considered in the space(S1ω2 )prime and the constraints on the parameters μ μ1 μmt1 tm are the same as in case of problems (24) and(25))

Theorem 1 By nonlocal multipoint by time problem (24)(87) is solvable and its solution is given by the formula

u(t x) flowastG(t x) (t x) isin Ω (88)

where u(t middot) isin S1ω2 for every tgt 0

Proof Make sure that the function u(t x) (t x) isin Ω sat-isfies equation (24) Indeed (see Corollary 2)

zu(t x)

zt

z

zt(flowastG(t middot)) flowast

zG(t middot)

zt

Au(t x) Fminus 1

[a(σ)F[f lowastG(t middot)]]

(89)

Since f is a convolutor in S1ω2

F[flowastG(t middot)] F[f]F[G(t middot)] F[f]Q(t middot) (90)

So

Au(t x) Fminus 1

[a(σ)Q(t σ)F[f](σ)] minus Fz

ztQ(t middot)F[f]1113890 1113891

minus F Fz

ztG(t middot)1113890 1113891 middot F[f]1113890 1113891

minus Fminus 1

F flowastzG(t middot)

zt1113890 11138911113890 1113891 minus flowast

zG(t middot)

zt

(91)

It follows from this that the function u(t x) (t x) isin Ωsatisfies equation (24) From Corollary 3 we get that u(t x)

satisfies condition (87) in the defined sense )e theorem isproved

Remark 5 If in condition (87) B1 middot middot middot Bm I (I is theidentity operator) then we can prove that problems (24) and(87) are well posed and its solution is given by the formula

u(t x) flowastG(t x) f isin S1ω1lowast1113872 1113873prime (t x) isin Ω (92)

where G(t middot) Fminus 1[Q(t middot)] isin S1ω1 for every tgt 0 and

Q(t σ) eminus ta(σ) μ minus 1113944

m

k1μke

minus tka(σ)⎛⎝ ⎞⎠

minus 1

(93)

Theorem 2 Suppose u(t x) (t x) isin Ω is the solution ofproblems (24) and (87) -en u(t middot)⟶ 0 in the space(S1ω2 )prime as t⟶ +infin

Proof Recall that the solution of problems (24) and (87) isgiven by the formula

u(t x) flowastG(t x) langfξ G(t x minus ξ)rang

f isin S1ω2lowast1113872 1113873prime (t x) isin Ω

(94)

Let us prove that langu(t middot)φrang⟶ 0 as t⟶ +infin for anarbitrary function φ isin S1ω2 Put

Ψt(ξ) 1113946+infin

minus infinG(t x minus ξ)φ(x)dx

ΨtR(ξ) 1113946R

minus RG(t x minus ξ)φ(x)dx Rgt 0 tgt 1

(95)

In these notations we prove that (a) for every tgt 1 andRgt 0 the function ΨtR(ξ) belongs to the space S1ω2 andΨtR(ξ)⟶Ψt(ξ) in the space S1ω2 as R⟶infin (b)Ψt(ξ) isin S1ω2 for every tgt 1 From this we get

langu(t middot)φrang 1113946+infin

minus infinlangfξ G(t x minus ξ)rangφ(x)dx

langfξ 1113946+infin

minus infinG(t x)φ(x + ξ)dxrang

langfξ 1113946+infin

minus infinG(t minus y)φ(minus (y minus ξ))dyrang

langfξ 1113946+infin

minus infinG(t minus y)φ(y minus ξ)dyrang φ(z) φ(minus z)

(96)

International Journal of Differential Equations 9

where u(t middot) is interpreted as a regular generalized functionfrom (S1ω2 )prime for every tgt 0

So let us set property (a) For fixed k m sub Z+ we have

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le 1113946+R

minus Rξkφ(x)D

mξ G(t x minus ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868dx

le 1113946+infin

minus infinξkφ(ξ + η)D

mη G(t η)

11138681113868111386811138681113868

11138681113868111386811138681113868dη

(97)

Since φ isin S1ω2 the inequality

ξkD

mξ φ(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cLkM

mk2k

mmω

k m sub Z+ (98)

with some c L Mgt 0 is valid It follows from this that forevery η isin R

supξisinR

ξkφ(ξ)11138681113868111386811138681113868

11138681113868111386811138681113868 supyisinR

(y minus η)kφ(η)

11138681113868111386811138681113868

11138681113868111386811138681113868 supyisinR

1113944

k

l0C

lky

l(minus η)

kminus lφ(y)

1113868111386811138681113868111386811138681113868111386811138681113868

1113868111386811138681113868111386811138681113868111386811138681113868

le 1113944k

l0C

lk|η|

kminus l supyisinR

ykφ(y)

11138681113868111386811138681113868

11138681113868111386811138681113868 le c 1113944k

l0C

lkL

ll2l

|η|kminus l

(99)

Next we will use estimate (58) )en

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le c 1113944k

l0C

lkL

ll2l

1113946+infin

minus infin|η|

kminus lD

mη G(t η)

11138681113868111386811138681113868

11138681113868111386811138681113868dη

le cc3Bm

mmω

tminus (1ω)

1113944

k

l0C

lkL

ll2l

middot 1113946+infin

minus infin|η|

kminus l exp minus a0tminus (12)

|η|12

1113966 1113967dη

(100)

By the direct calculations we find that

1113946+infin

minus infin|η|

kminus l exp minus a0tminus (12)

|η|12

1113966 1113967dη le c41113957L

kk2(kminus l)

tkminus l+1

c41113957Lgt 0

(101)

So

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cc3c4Bm

t1minus (1ω)

mmω

1113944

k

l0C

lkL

l1113957Lkminus l

tkminus l

l2l

k2(kminus l)

le cBm

Lkk2k

mmω

(102)

where c cc3c4t1minus (1ω) and L 2max L 1113957Lt1113864 1113865 )erefore

ΨtR(ξ) isin S1ω2 for every tgt 1 and an arbitrary Rgt 0 Fur-thermore we make sure thatΨtR(ξ)⟶Ψt(ξ) as R⟶infin

with all its derivatives uniformly in ξ on each segment[a b] sub R In addition the set of functions ξk

Dmξ ΨtR(ξ)1113966 1113967

k m sub Z+ is uniformly bounded (by R) in the space S1ω2(this property follows from estimate (102) in which c B andLgt 0 do not depend on R) )is means that condition (a) isfulfilled

Condition (b) follows from (a) because in a perfectspace every bounded set is compact

Using properties (a) and (b) we obtain the equality

langu(t middot)φrang 1113946+infin

minus infinG(t minus y)(flowastφ)(y)dy (103)

Since f isin (S1ω2 )prime is a convolutor in S1ω2 we haveflowast φ isin S1ω2 )en given estimate (58) (at s 0) we find that

|langu(t middot)φrang| le 1113946+infin

minus infin|G(t minus y)||(flowastφ)(y)|dy

lec

t1ω1113946

+infin

minus infin|(flowastφ)(y)|dy le

c0

t1ω⟶ 0 t⟶ +infin

(104)

for an arbitrary function φ isin S1ω2 ie u(t middot)⟶ 0 in thespace (S1ω2 )prime as t⟶ +infin )e theorem is proved

If the generalized function f in (87) is finite (ie itssupport suppf is a finite set in R) then we can say onuniform tending a solution u(t x) of problems (24) and (87)to zero onR as t⟶ +infin Note that every finite generalizedfunction is a convolutor in S spaces )is property followsfrom the general result concerning the theory of perfectspaces (see [12] p 137) if Φ is a perfect space with dif-ferential translation operation then every finite functional isa convolutor inΦ Finite functionals form a fairly wide classIn particular every bounded set F sub R is a support of somegeneralized function [12]

Theorem 3 Let u(t x) be a solution of problems (24) and(87) with the boundary function f in condition (87) which isan element of (S

β2)prime sub (S1ω2 )prime βgt 1 with finite support in R

-en u(t x)⟶ 0 uniformly on R as t⟶ +infin

Here is a scheme for proving this statement Letsuppf sub [a1 b1] sub [a2 b2] sub R Consider the functionφ isin S

β2 βgt 1 such that φ(x) 1 x isin [a1 b1] and

suppφ sub [a2 b2] )is function exists when βgt 1 because Sβ2

contains finite functions [12] Consider the function

u(t x) langfξ φ(ξ)G(t x minus ξ)rang +langfξ c(ξ)G(t x minus ξ)rang(105)

where c 1 minus φ Since supp(c(ξ)G(t x minus ξ)) cap suppφ emptywe have

u(t x) tminus (1ω)langfξ t

1ωφ(ξ)G(t x minus ξ)rang (106)

To prove the formulated above statement it remains toestablish that the set of functions Φtx(ξ) t1ωφ(ξ)G(t x minus

ξ) is bounded in the space Sβ2 βgt 1 for large t and x isin R

For example if in condition (87) f δ then δ is aconvolutor in the space S supp δ 0 andu(t x) δ lowastG(t x) G(t x) It follows directly from es-timate (58) that G(t x)⟶ 0 uniformly onR as t⟶ +infin

4 Conclusion

In this paper the solvability of a nonlocal multipoint by thetime problem for the evolutionary equation(zuzt) + (I minus (d2dx2))ω2u 0 (t x) isin (0 +infin) times R isproved herewith the operator (I minus (d2dx2))ω2 is inter-preted as a pseudodifferential operator in the space S1ω2

10 International Journal of Differential Equations

constructed by function (1 + σ2)ω2 σ isin R while the non-local multipoint by time condition also contains pseudo-differential operators constructed on smooth symbols )erepresentation of the solution is given in the form of aconvolution of the fundamental solution with the initialfunction which is an element of the space of generalizedfunctions of the ultradistribution type (the coagulator inspace S1ω2 ) )e behavior of the solution u(t x) t⟶ +infinin the space of generalized functions (S1ω2 )prime is investigated)e conditions for the initial generalized function are foundunder which the solution is uniformly stabilized to zero onR )e method of research of a nonlocal multipoint by timeproblem offered in this paper allows to interpret differential-operator equations of a form (zuzt) + φ(i zzx)u 0 as anevolutionary equation with a pseudodifferential operatorφ(i zzx) Fminus 1[φ middot F] constructed by the function φ actingin certain countable-normalized space of infinite-differentialfunctions (the choice of space depends on the properties ofthe function which is the symbol of the operator φ(i zzx))

Data Availability

No data were used to support the findings of this study

Disclosure

)is study was conducted in the framework of scientificactivity at Yuriy Fedkovych Chernivtsi National University

Conflicts of Interest

)e authors declare that they have no conflicts of interest

References

[1] V I Gorbachuk and M L Gorbachuk Boundary-ValueProblems for Differential-Operator Equations NaukovaDumka Kiev Ukraine 1984

[2] V V Gorodetskiy N I Nagnibida and P P NastasievMethods for Solving Problems in Functional Analysis VyshkaSchool Vyshka Ukraine 1990

[3] S G Samko A A Kilbas and O I Marichev Integrals andFractional Derivatives and Some of -eir Applications Scienceand Technology Minsk Belarus 1987

[4] I M Gelfand and G E Shilov Some Questions of the-eory ofDifferential Equations Fizmatgiz Moscow Russia 1958

[5] V V Gorodetskiy and O M Lenyuk About the FractionalDifferentiation in Spaces of Sprime type NAS of Ukraine KyivUkraine 1998

[6] A A Dezin General Questions of the -eory of BoundaryValue Problems Nauka Moscow Russia 1980

[7] A M Nakhushev Equations of Mathematical Biology HigherSchool Moscow Russia 1995

[8] I A Belavin S P Kapitsa and S P Kurdyumov ldquoAmathematical model of global demographic processes withconsidering of the spatial distributionrdquo Zh CalculationMathematics Physics vol 38 no 6 pp 885ndash902 1998

[9] V V Gorodetskiy and O V Martynyuk-e Cauchy Problemand Nonlocal Problems for First-Order Evolution EquationsEby Time Variable Vidavnichy dim rdquoRODOVIDrdquo ChernivtsiUkraine 2015

[10] V V Gorodetskiy and V I Myronyk ldquoTwo-point problem forone class of evolution equationsrdquo Differents Equationsvol 46 no 3 pp 349ndash363 2010

[11] V V Gorodetskiy and V I Myronyk ldquoTwo-point problem forone class of evolution equationsrdquo Differents Equationsvol 46 no 4 pp 520ndash526 2010

[12] I M Gelfand and G E Shilov Spaces of Basic and GeneralizedFunctions Fizmatgiz Moscow Russia 1958

International Journal of Differential Equations 11

Page 9: On One Evolution Equation of Parabolic Type with ...

μQ(t middot) minus 1113944m

k1μkgk(middot)Q tk middot( 1113857 ⟶

t⟶+0μQ2(middot) minus 1113944

m

k1μkgk(middot)Q1 tk middot( 1113857Q2(middot)

μ minus 1113944m

k1μkgk(middot)Q1 tk middot( 1113857⎛⎝ ⎞⎠Q2(middot) 1

(84)

is realized in the space (S21ω)prime )us relation (81) and (80)are fulfilled in the corresponding spaces )e statement isproved

)e function G(t middot) satisfies equation (24) as tgt 0Indeed

z

ztG(t x)

z

ztF

minus 1[Q(t σ)] F

minus 1 z

ztQ(t σ)1113890 1113891

Fminus 1

[a(σ)Q(t σ)]

AG(t x) Fminus 1

a(σ)Fminus 1

[G(t middot)]1113960 1113961 Fminus 1

[a(σ)Q(t σ)]

(85)

So

zG(t x)

zt+ AG(t x) 0 (t x) isin Ω (86)

which is what had to be provedIt follows from Corollary 3 that the nonlocal multipoint

by the time problem for equation (24) can be formulated inthe following way find a function u(t x) (t x) isin Ω thatsatisfies equation (24) and the condition

μ limt⟶+0

u(t middot) minus 1113944

m

k1μkBku tk middot( 1113857 f f isin S

1ω2lowast1113872 1113873prime (87)

where boundary relation (87) is considered in the space(S1ω2 )prime and the constraints on the parameters μ μ1 μmt1 tm are the same as in case of problems (24) and(25))

Theorem 1 By nonlocal multipoint by time problem (24)(87) is solvable and its solution is given by the formula

u(t x) flowastG(t x) (t x) isin Ω (88)

where u(t middot) isin S1ω2 for every tgt 0

Proof Make sure that the function u(t x) (t x) isin Ω sat-isfies equation (24) Indeed (see Corollary 2)

zu(t x)

zt

z

zt(flowastG(t middot)) flowast

zG(t middot)

zt

Au(t x) Fminus 1

[a(σ)F[f lowastG(t middot)]]

(89)

Since f is a convolutor in S1ω2

F[flowastG(t middot)] F[f]F[G(t middot)] F[f]Q(t middot) (90)

So

Au(t x) Fminus 1

[a(σ)Q(t σ)F[f](σ)] minus Fz

ztQ(t middot)F[f]1113890 1113891

minus F Fz

ztG(t middot)1113890 1113891 middot F[f]1113890 1113891

minus Fminus 1

F flowastzG(t middot)

zt1113890 11138911113890 1113891 minus flowast

zG(t middot)

zt

(91)

It follows from this that the function u(t x) (t x) isin Ωsatisfies equation (24) From Corollary 3 we get that u(t x)

satisfies condition (87) in the defined sense )e theorem isproved

Remark 5 If in condition (87) B1 middot middot middot Bm I (I is theidentity operator) then we can prove that problems (24) and(87) are well posed and its solution is given by the formula

u(t x) flowastG(t x) f isin S1ω1lowast1113872 1113873prime (t x) isin Ω (92)

where G(t middot) Fminus 1[Q(t middot)] isin S1ω1 for every tgt 0 and

Q(t σ) eminus ta(σ) μ minus 1113944

m

k1μke

minus tka(σ)⎛⎝ ⎞⎠

minus 1

(93)

Theorem 2 Suppose u(t x) (t x) isin Ω is the solution ofproblems (24) and (87) -en u(t middot)⟶ 0 in the space(S1ω2 )prime as t⟶ +infin

Proof Recall that the solution of problems (24) and (87) isgiven by the formula

u(t x) flowastG(t x) langfξ G(t x minus ξ)rang

f isin S1ω2lowast1113872 1113873prime (t x) isin Ω

(94)

Let us prove that langu(t middot)φrang⟶ 0 as t⟶ +infin for anarbitrary function φ isin S1ω2 Put

Ψt(ξ) 1113946+infin

minus infinG(t x minus ξ)φ(x)dx

ΨtR(ξ) 1113946R

minus RG(t x minus ξ)φ(x)dx Rgt 0 tgt 1

(95)

In these notations we prove that (a) for every tgt 1 andRgt 0 the function ΨtR(ξ) belongs to the space S1ω2 andΨtR(ξ)⟶Ψt(ξ) in the space S1ω2 as R⟶infin (b)Ψt(ξ) isin S1ω2 for every tgt 1 From this we get

langu(t middot)φrang 1113946+infin

minus infinlangfξ G(t x minus ξ)rangφ(x)dx

langfξ 1113946+infin

minus infinG(t x)φ(x + ξ)dxrang

langfξ 1113946+infin

minus infinG(t minus y)φ(minus (y minus ξ))dyrang

langfξ 1113946+infin

minus infinG(t minus y)φ(y minus ξ)dyrang φ(z) φ(minus z)

(96)

International Journal of Differential Equations 9

where u(t middot) is interpreted as a regular generalized functionfrom (S1ω2 )prime for every tgt 0

So let us set property (a) For fixed k m sub Z+ we have

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le 1113946+R

minus Rξkφ(x)D

mξ G(t x minus ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868dx

le 1113946+infin

minus infinξkφ(ξ + η)D

mη G(t η)

11138681113868111386811138681113868

11138681113868111386811138681113868dη

(97)

Since φ isin S1ω2 the inequality

ξkD

mξ φ(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cLkM

mk2k

mmω

k m sub Z+ (98)

with some c L Mgt 0 is valid It follows from this that forevery η isin R

supξisinR

ξkφ(ξ)11138681113868111386811138681113868

11138681113868111386811138681113868 supyisinR

(y minus η)kφ(η)

11138681113868111386811138681113868

11138681113868111386811138681113868 supyisinR

1113944

k

l0C

lky

l(minus η)

kminus lφ(y)

1113868111386811138681113868111386811138681113868111386811138681113868

1113868111386811138681113868111386811138681113868111386811138681113868

le 1113944k

l0C

lk|η|

kminus l supyisinR

ykφ(y)

11138681113868111386811138681113868

11138681113868111386811138681113868 le c 1113944k

l0C

lkL

ll2l

|η|kminus l

(99)

Next we will use estimate (58) )en

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le c 1113944k

l0C

lkL

ll2l

1113946+infin

minus infin|η|

kminus lD

mη G(t η)

11138681113868111386811138681113868

11138681113868111386811138681113868dη

le cc3Bm

mmω

tminus (1ω)

1113944

k

l0C

lkL

ll2l

middot 1113946+infin

minus infin|η|

kminus l exp minus a0tminus (12)

|η|12

1113966 1113967dη

(100)

By the direct calculations we find that

1113946+infin

minus infin|η|

kminus l exp minus a0tminus (12)

|η|12

1113966 1113967dη le c41113957L

kk2(kminus l)

tkminus l+1

c41113957Lgt 0

(101)

So

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cc3c4Bm

t1minus (1ω)

mmω

1113944

k

l0C

lkL

l1113957Lkminus l

tkminus l

l2l

k2(kminus l)

le cBm

Lkk2k

mmω

(102)

where c cc3c4t1minus (1ω) and L 2max L 1113957Lt1113864 1113865 )erefore

ΨtR(ξ) isin S1ω2 for every tgt 1 and an arbitrary Rgt 0 Fur-thermore we make sure thatΨtR(ξ)⟶Ψt(ξ) as R⟶infin

with all its derivatives uniformly in ξ on each segment[a b] sub R In addition the set of functions ξk

Dmξ ΨtR(ξ)1113966 1113967

k m sub Z+ is uniformly bounded (by R) in the space S1ω2(this property follows from estimate (102) in which c B andLgt 0 do not depend on R) )is means that condition (a) isfulfilled

Condition (b) follows from (a) because in a perfectspace every bounded set is compact

Using properties (a) and (b) we obtain the equality

langu(t middot)φrang 1113946+infin

minus infinG(t minus y)(flowastφ)(y)dy (103)

Since f isin (S1ω2 )prime is a convolutor in S1ω2 we haveflowast φ isin S1ω2 )en given estimate (58) (at s 0) we find that

|langu(t middot)φrang| le 1113946+infin

minus infin|G(t minus y)||(flowastφ)(y)|dy

lec

t1ω1113946

+infin

minus infin|(flowastφ)(y)|dy le

c0

t1ω⟶ 0 t⟶ +infin

(104)

for an arbitrary function φ isin S1ω2 ie u(t middot)⟶ 0 in thespace (S1ω2 )prime as t⟶ +infin )e theorem is proved

If the generalized function f in (87) is finite (ie itssupport suppf is a finite set in R) then we can say onuniform tending a solution u(t x) of problems (24) and (87)to zero onR as t⟶ +infin Note that every finite generalizedfunction is a convolutor in S spaces )is property followsfrom the general result concerning the theory of perfectspaces (see [12] p 137) if Φ is a perfect space with dif-ferential translation operation then every finite functional isa convolutor inΦ Finite functionals form a fairly wide classIn particular every bounded set F sub R is a support of somegeneralized function [12]

Theorem 3 Let u(t x) be a solution of problems (24) and(87) with the boundary function f in condition (87) which isan element of (S

β2)prime sub (S1ω2 )prime βgt 1 with finite support in R

-en u(t x)⟶ 0 uniformly on R as t⟶ +infin

Here is a scheme for proving this statement Letsuppf sub [a1 b1] sub [a2 b2] sub R Consider the functionφ isin S

β2 βgt 1 such that φ(x) 1 x isin [a1 b1] and

suppφ sub [a2 b2] )is function exists when βgt 1 because Sβ2

contains finite functions [12] Consider the function

u(t x) langfξ φ(ξ)G(t x minus ξ)rang +langfξ c(ξ)G(t x minus ξ)rang(105)

where c 1 minus φ Since supp(c(ξ)G(t x minus ξ)) cap suppφ emptywe have

u(t x) tminus (1ω)langfξ t

1ωφ(ξ)G(t x minus ξ)rang (106)

To prove the formulated above statement it remains toestablish that the set of functions Φtx(ξ) t1ωφ(ξ)G(t x minus

ξ) is bounded in the space Sβ2 βgt 1 for large t and x isin R

For example if in condition (87) f δ then δ is aconvolutor in the space S supp δ 0 andu(t x) δ lowastG(t x) G(t x) It follows directly from es-timate (58) that G(t x)⟶ 0 uniformly onR as t⟶ +infin

4 Conclusion

In this paper the solvability of a nonlocal multipoint by thetime problem for the evolutionary equation(zuzt) + (I minus (d2dx2))ω2u 0 (t x) isin (0 +infin) times R isproved herewith the operator (I minus (d2dx2))ω2 is inter-preted as a pseudodifferential operator in the space S1ω2

10 International Journal of Differential Equations

constructed by function (1 + σ2)ω2 σ isin R while the non-local multipoint by time condition also contains pseudo-differential operators constructed on smooth symbols )erepresentation of the solution is given in the form of aconvolution of the fundamental solution with the initialfunction which is an element of the space of generalizedfunctions of the ultradistribution type (the coagulator inspace S1ω2 ) )e behavior of the solution u(t x) t⟶ +infinin the space of generalized functions (S1ω2 )prime is investigated)e conditions for the initial generalized function are foundunder which the solution is uniformly stabilized to zero onR )e method of research of a nonlocal multipoint by timeproblem offered in this paper allows to interpret differential-operator equations of a form (zuzt) + φ(i zzx)u 0 as anevolutionary equation with a pseudodifferential operatorφ(i zzx) Fminus 1[φ middot F] constructed by the function φ actingin certain countable-normalized space of infinite-differentialfunctions (the choice of space depends on the properties ofthe function which is the symbol of the operator φ(i zzx))

Data Availability

No data were used to support the findings of this study

Disclosure

)is study was conducted in the framework of scientificactivity at Yuriy Fedkovych Chernivtsi National University

Conflicts of Interest

)e authors declare that they have no conflicts of interest

References

[1] V I Gorbachuk and M L Gorbachuk Boundary-ValueProblems for Differential-Operator Equations NaukovaDumka Kiev Ukraine 1984

[2] V V Gorodetskiy N I Nagnibida and P P NastasievMethods for Solving Problems in Functional Analysis VyshkaSchool Vyshka Ukraine 1990

[3] S G Samko A A Kilbas and O I Marichev Integrals andFractional Derivatives and Some of -eir Applications Scienceand Technology Minsk Belarus 1987

[4] I M Gelfand and G E Shilov Some Questions of the-eory ofDifferential Equations Fizmatgiz Moscow Russia 1958

[5] V V Gorodetskiy and O M Lenyuk About the FractionalDifferentiation in Spaces of Sprime type NAS of Ukraine KyivUkraine 1998

[6] A A Dezin General Questions of the -eory of BoundaryValue Problems Nauka Moscow Russia 1980

[7] A M Nakhushev Equations of Mathematical Biology HigherSchool Moscow Russia 1995

[8] I A Belavin S P Kapitsa and S P Kurdyumov ldquoAmathematical model of global demographic processes withconsidering of the spatial distributionrdquo Zh CalculationMathematics Physics vol 38 no 6 pp 885ndash902 1998

[9] V V Gorodetskiy and O V Martynyuk-e Cauchy Problemand Nonlocal Problems for First-Order Evolution EquationsEby Time Variable Vidavnichy dim rdquoRODOVIDrdquo ChernivtsiUkraine 2015

[10] V V Gorodetskiy and V I Myronyk ldquoTwo-point problem forone class of evolution equationsrdquo Differents Equationsvol 46 no 3 pp 349ndash363 2010

[11] V V Gorodetskiy and V I Myronyk ldquoTwo-point problem forone class of evolution equationsrdquo Differents Equationsvol 46 no 4 pp 520ndash526 2010

[12] I M Gelfand and G E Shilov Spaces of Basic and GeneralizedFunctions Fizmatgiz Moscow Russia 1958

International Journal of Differential Equations 11

Page 10: On One Evolution Equation of Parabolic Type with ...

where u(t middot) is interpreted as a regular generalized functionfrom (S1ω2 )prime for every tgt 0

So let us set property (a) For fixed k m sub Z+ we have

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le 1113946+R

minus Rξkφ(x)D

mξ G(t x minus ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868dx

le 1113946+infin

minus infinξkφ(ξ + η)D

mη G(t η)

11138681113868111386811138681113868

11138681113868111386811138681113868dη

(97)

Since φ isin S1ω2 the inequality

ξkD

mξ φ(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cLkM

mk2k

mmω

k m sub Z+ (98)

with some c L Mgt 0 is valid It follows from this that forevery η isin R

supξisinR

ξkφ(ξ)11138681113868111386811138681113868

11138681113868111386811138681113868 supyisinR

(y minus η)kφ(η)

11138681113868111386811138681113868

11138681113868111386811138681113868 supyisinR

1113944

k

l0C

lky

l(minus η)

kminus lφ(y)

1113868111386811138681113868111386811138681113868111386811138681113868

1113868111386811138681113868111386811138681113868111386811138681113868

le 1113944k

l0C

lk|η|

kminus l supyisinR

ykφ(y)

11138681113868111386811138681113868

11138681113868111386811138681113868 le c 1113944k

l0C

lkL

ll2l

|η|kminus l

(99)

Next we will use estimate (58) )en

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le c 1113944k

l0C

lkL

ll2l

1113946+infin

minus infin|η|

kminus lD

mη G(t η)

11138681113868111386811138681113868

11138681113868111386811138681113868dη

le cc3Bm

mmω

tminus (1ω)

1113944

k

l0C

lkL

ll2l

middot 1113946+infin

minus infin|η|

kminus l exp minus a0tminus (12)

|η|12

1113966 1113967dη

(100)

By the direct calculations we find that

1113946+infin

minus infin|η|

kminus l exp minus a0tminus (12)

|η|12

1113966 1113967dη le c41113957L

kk2(kminus l)

tkminus l+1

c41113957Lgt 0

(101)

So

ξkD

mξ ΨtR(ξ)

11138681113868111386811138681113868

11138681113868111386811138681113868 le cc3c4Bm

t1minus (1ω)

mmω

1113944

k

l0C

lkL

l1113957Lkminus l

tkminus l

l2l

k2(kminus l)

le cBm

Lkk2k

mmω

(102)

where c cc3c4t1minus (1ω) and L 2max L 1113957Lt1113864 1113865 )erefore

ΨtR(ξ) isin S1ω2 for every tgt 1 and an arbitrary Rgt 0 Fur-thermore we make sure thatΨtR(ξ)⟶Ψt(ξ) as R⟶infin

with all its derivatives uniformly in ξ on each segment[a b] sub R In addition the set of functions ξk

Dmξ ΨtR(ξ)1113966 1113967

k m sub Z+ is uniformly bounded (by R) in the space S1ω2(this property follows from estimate (102) in which c B andLgt 0 do not depend on R) )is means that condition (a) isfulfilled

Condition (b) follows from (a) because in a perfectspace every bounded set is compact

Using properties (a) and (b) we obtain the equality

langu(t middot)φrang 1113946+infin

minus infinG(t minus y)(flowastφ)(y)dy (103)

Since f isin (S1ω2 )prime is a convolutor in S1ω2 we haveflowast φ isin S1ω2 )en given estimate (58) (at s 0) we find that

|langu(t middot)φrang| le 1113946+infin

minus infin|G(t minus y)||(flowastφ)(y)|dy

lec

t1ω1113946

+infin

minus infin|(flowastφ)(y)|dy le

c0

t1ω⟶ 0 t⟶ +infin

(104)

for an arbitrary function φ isin S1ω2 ie u(t middot)⟶ 0 in thespace (S1ω2 )prime as t⟶ +infin )e theorem is proved

If the generalized function f in (87) is finite (ie itssupport suppf is a finite set in R) then we can say onuniform tending a solution u(t x) of problems (24) and (87)to zero onR as t⟶ +infin Note that every finite generalizedfunction is a convolutor in S spaces )is property followsfrom the general result concerning the theory of perfectspaces (see [12] p 137) if Φ is a perfect space with dif-ferential translation operation then every finite functional isa convolutor inΦ Finite functionals form a fairly wide classIn particular every bounded set F sub R is a support of somegeneralized function [12]

Theorem 3 Let u(t x) be a solution of problems (24) and(87) with the boundary function f in condition (87) which isan element of (S

β2)prime sub (S1ω2 )prime βgt 1 with finite support in R

-en u(t x)⟶ 0 uniformly on R as t⟶ +infin

Here is a scheme for proving this statement Letsuppf sub [a1 b1] sub [a2 b2] sub R Consider the functionφ isin S

β2 βgt 1 such that φ(x) 1 x isin [a1 b1] and

suppφ sub [a2 b2] )is function exists when βgt 1 because Sβ2

contains finite functions [12] Consider the function

u(t x) langfξ φ(ξ)G(t x minus ξ)rang +langfξ c(ξ)G(t x minus ξ)rang(105)

where c 1 minus φ Since supp(c(ξ)G(t x minus ξ)) cap suppφ emptywe have

u(t x) tminus (1ω)langfξ t

1ωφ(ξ)G(t x minus ξ)rang (106)

To prove the formulated above statement it remains toestablish that the set of functions Φtx(ξ) t1ωφ(ξ)G(t x minus

ξ) is bounded in the space Sβ2 βgt 1 for large t and x isin R

For example if in condition (87) f δ then δ is aconvolutor in the space S supp δ 0 andu(t x) δ lowastG(t x) G(t x) It follows directly from es-timate (58) that G(t x)⟶ 0 uniformly onR as t⟶ +infin

4 Conclusion

In this paper the solvability of a nonlocal multipoint by thetime problem for the evolutionary equation(zuzt) + (I minus (d2dx2))ω2u 0 (t x) isin (0 +infin) times R isproved herewith the operator (I minus (d2dx2))ω2 is inter-preted as a pseudodifferential operator in the space S1ω2

10 International Journal of Differential Equations

constructed by function (1 + σ2)ω2 σ isin R while the non-local multipoint by time condition also contains pseudo-differential operators constructed on smooth symbols )erepresentation of the solution is given in the form of aconvolution of the fundamental solution with the initialfunction which is an element of the space of generalizedfunctions of the ultradistribution type (the coagulator inspace S1ω2 ) )e behavior of the solution u(t x) t⟶ +infinin the space of generalized functions (S1ω2 )prime is investigated)e conditions for the initial generalized function are foundunder which the solution is uniformly stabilized to zero onR )e method of research of a nonlocal multipoint by timeproblem offered in this paper allows to interpret differential-operator equations of a form (zuzt) + φ(i zzx)u 0 as anevolutionary equation with a pseudodifferential operatorφ(i zzx) Fminus 1[φ middot F] constructed by the function φ actingin certain countable-normalized space of infinite-differentialfunctions (the choice of space depends on the properties ofthe function which is the symbol of the operator φ(i zzx))

Data Availability

No data were used to support the findings of this study

Disclosure

)is study was conducted in the framework of scientificactivity at Yuriy Fedkovych Chernivtsi National University

Conflicts of Interest

)e authors declare that they have no conflicts of interest

References

[1] V I Gorbachuk and M L Gorbachuk Boundary-ValueProblems for Differential-Operator Equations NaukovaDumka Kiev Ukraine 1984

[2] V V Gorodetskiy N I Nagnibida and P P NastasievMethods for Solving Problems in Functional Analysis VyshkaSchool Vyshka Ukraine 1990

[3] S G Samko A A Kilbas and O I Marichev Integrals andFractional Derivatives and Some of -eir Applications Scienceand Technology Minsk Belarus 1987

[4] I M Gelfand and G E Shilov Some Questions of the-eory ofDifferential Equations Fizmatgiz Moscow Russia 1958

[5] V V Gorodetskiy and O M Lenyuk About the FractionalDifferentiation in Spaces of Sprime type NAS of Ukraine KyivUkraine 1998

[6] A A Dezin General Questions of the -eory of BoundaryValue Problems Nauka Moscow Russia 1980

[7] A M Nakhushev Equations of Mathematical Biology HigherSchool Moscow Russia 1995

[8] I A Belavin S P Kapitsa and S P Kurdyumov ldquoAmathematical model of global demographic processes withconsidering of the spatial distributionrdquo Zh CalculationMathematics Physics vol 38 no 6 pp 885ndash902 1998

[9] V V Gorodetskiy and O V Martynyuk-e Cauchy Problemand Nonlocal Problems for First-Order Evolution EquationsEby Time Variable Vidavnichy dim rdquoRODOVIDrdquo ChernivtsiUkraine 2015

[10] V V Gorodetskiy and V I Myronyk ldquoTwo-point problem forone class of evolution equationsrdquo Differents Equationsvol 46 no 3 pp 349ndash363 2010

[11] V V Gorodetskiy and V I Myronyk ldquoTwo-point problem forone class of evolution equationsrdquo Differents Equationsvol 46 no 4 pp 520ndash526 2010

[12] I M Gelfand and G E Shilov Spaces of Basic and GeneralizedFunctions Fizmatgiz Moscow Russia 1958

International Journal of Differential Equations 11

Page 11: On One Evolution Equation of Parabolic Type with ...

constructed by function (1 + σ2)ω2 σ isin R while the non-local multipoint by time condition also contains pseudo-differential operators constructed on smooth symbols )erepresentation of the solution is given in the form of aconvolution of the fundamental solution with the initialfunction which is an element of the space of generalizedfunctions of the ultradistribution type (the coagulator inspace S1ω2 ) )e behavior of the solution u(t x) t⟶ +infinin the space of generalized functions (S1ω2 )prime is investigated)e conditions for the initial generalized function are foundunder which the solution is uniformly stabilized to zero onR )e method of research of a nonlocal multipoint by timeproblem offered in this paper allows to interpret differential-operator equations of a form (zuzt) + φ(i zzx)u 0 as anevolutionary equation with a pseudodifferential operatorφ(i zzx) Fminus 1[φ middot F] constructed by the function φ actingin certain countable-normalized space of infinite-differentialfunctions (the choice of space depends on the properties ofthe function which is the symbol of the operator φ(i zzx))

Data Availability

No data were used to support the findings of this study

Disclosure

)is study was conducted in the framework of scientificactivity at Yuriy Fedkovych Chernivtsi National University

Conflicts of Interest

)e authors declare that they have no conflicts of interest

References

[1] V I Gorbachuk and M L Gorbachuk Boundary-ValueProblems for Differential-Operator Equations NaukovaDumka Kiev Ukraine 1984

[2] V V Gorodetskiy N I Nagnibida and P P NastasievMethods for Solving Problems in Functional Analysis VyshkaSchool Vyshka Ukraine 1990

[3] S G Samko A A Kilbas and O I Marichev Integrals andFractional Derivatives and Some of -eir Applications Scienceand Technology Minsk Belarus 1987

[4] I M Gelfand and G E Shilov Some Questions of the-eory ofDifferential Equations Fizmatgiz Moscow Russia 1958

[5] V V Gorodetskiy and O M Lenyuk About the FractionalDifferentiation in Spaces of Sprime type NAS of Ukraine KyivUkraine 1998

[6] A A Dezin General Questions of the -eory of BoundaryValue Problems Nauka Moscow Russia 1980

[7] A M Nakhushev Equations of Mathematical Biology HigherSchool Moscow Russia 1995

[8] I A Belavin S P Kapitsa and S P Kurdyumov ldquoAmathematical model of global demographic processes withconsidering of the spatial distributionrdquo Zh CalculationMathematics Physics vol 38 no 6 pp 885ndash902 1998

[9] V V Gorodetskiy and O V Martynyuk-e Cauchy Problemand Nonlocal Problems for First-Order Evolution EquationsEby Time Variable Vidavnichy dim rdquoRODOVIDrdquo ChernivtsiUkraine 2015

[10] V V Gorodetskiy and V I Myronyk ldquoTwo-point problem forone class of evolution equationsrdquo Differents Equationsvol 46 no 3 pp 349ndash363 2010

[11] V V Gorodetskiy and V I Myronyk ldquoTwo-point problem forone class of evolution equationsrdquo Differents Equationsvol 46 no 4 pp 520ndash526 2010

[12] I M Gelfand and G E Shilov Spaces of Basic and GeneralizedFunctions Fizmatgiz Moscow Russia 1958

International Journal of Differential Equations 11