Research ArticleSolving Poisson Equation by Distributional HK-IntegralProspects and Limitations

Amila J Maldeniya 1 Naleen C Ganegoda 2 Kaushika De Silva2

and Sanath K Boralugoda3

1Department of Mathematics General Sir John Kotelawala Defence University Kandawala Road Ratmalana Sri Lanka2Department of Mathematics University of Sri Jayewardenepura Gangodawila Nugegoda Sri Lanka3Department of Mathematics University of Colombo Cumarathunga Munidasa Mawatha Colombo Sri Lanka

Correspondence should be addressed to Amila J Maldeniya amilamkduaclk

Received 11 February 2021 Revised 14 May 2021 Accepted 24 June 2021 Published 12 July 2021

Academic Editor Raul E Curto

Copyright copy 2021 Amila J Maldeniya et al ampis 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 this paper we present some properties of integrable distributions which are continuous linear functional on the space of testfunction D(R2) Here it uses two-dimensional HenstockndashKurzweil integral We discuss integrable distributional solution forPoissonrsquos equation in the upper half space R3

+ with Dirichlet boundary condition

1 Introduction

In this paper we find an integrable distributional solutionfor Poissonrsquos equation in the upper half space Poissonrsquosequation is a second-order elliptic partial differentialequation which has many applications in physics and en-gineering ampere are three types of boundary conditions forPoissonrsquos equation Dirichlet condition Neumann condi-tions and Robin conditions ampis article is mainly con-cerning on Poissonrsquos equation with Dirichlet boundarycondition (1)

u f inΩ sub Rn

u g on zΩ 1113896 (1)

Partial differential equations (PDEs) are more difficult tosolve than ordinary differential equations (ODEs) ampere-fore numerical approximations are widely used in appli-cation ampere are standard numerical methods such as finitedifference and finite element etc to solve (1) In numericalapproaches to solving PDEs particularly in finite differencemethods we may face two ways of accuracy loss because ofdiscretization of domain and approximating partial deriv-atives by difference formulas Finite element methods will

improve the accuracy of problems with irregular boundariesbut still abide by approximations as one has to do mini-mizing functionals with smaller class of functions [1] Alsoartificial neural network methods can be used for approx-imate nonsmooth solutions See [2] for application of ar-tificial neural network methods for Poissonrsquos equation withboundary value problems in domain R2 ampis method trainsdata to optimize the algorithm which gives numerical so-lutions to partial differential equations Convergent prop-erties and accuracy of solution are not discussed in the givenmethod of [2] Finite element method can be used to solveelliptic problems with HenstockndashKurzweil integrablefunctions Such an approximation is used in [3] for ODEswhere existence and uniqueness of the solution are notdiscussed ampe proposed analytical approach in our paperwill direct to a solution where if its explicit form is notavailable one has to only do a numerical integration

ampere are several methods to find analytical solution forDirichlet problem such as Greenrsquos function Dirichletrsquosprinciple layer potentials energy methods and Perronrsquosmethod In 1850 [4] George Green gives theoretical ap-proach for Dirichlet problem which is now called as Greenrsquosfunction ampis assumes that Greenrsquos function exists for anydomain which is not true However this idea influences

HindawiInternational Journal of Mathematics and Mathematical SciencesVolume 2021 Article ID 5511283 9 pageshttpsdoiorg10115520215511283

modern techniques and distributional theory Using Greenrsquosfunction one can obtain a general representation solutionfor Poissonrsquos equation

General representation solution depends on the domainΩ and nature of the functions f and g For Lebesgue in-tegration it is required that the conditions g isin C(zΩ) andf isin C(Ω) get a unique analytical solutionu isin C2(Ω)capC(Ω) for (1) where Ω needs to be open andbounded in Rn ([5] pg 28) In classical solution it is worthto use HenstockndashKurzweil integral (HK-integral) instead ofthe Lebesgue integral

HK-integral is more advanced and it includes theLebesgue and Riemann integral ampis integral was first in-troduced by Henstock and Kurzweil in 1957 It has manyadvantages such as convergence theorems integration onunbounded intervalsfunctions Fubinirsquos theorem andfundamental theorem of calculus with full generality fordetails see [6ndash9] If HK is the space of HK-integrablefunctions then it has proper inclusion L1 ⊊ HK For anexample a highly oscillating function f(x) (1x)

sin(1x3) is neither Lebesgue nor Riemann integrable But f

is HK-integrable ampere are only few articles which use HK-integration to find analytical solution for Dirichlet problem[10 11] Talvila uses HK-integration to find solution forDirichlet problem in the upper half plane see example in[10] Also the same author obtains useful results of Poissonkernel in the unit disk via HK-integral with application intoDirichlet problem [11] However both articles [10 11] proveits results in R2 and do not extend into R3 It is an obviousfact that properties of HK-integration are not easily ex-tended for higher dimension Even the HK-integration issuperior to Lebesgue integration there are a few drawbacksMost significantly there is no known natural topology forHK so that it is not a Banach spaceamperefore there are notmuch functional analysis tools to work with the problemHowever there is a natural seminorm on HK defined byAlexiewicz [12]

Looking at an alternative we moved into distributiontheory specially the space of integrable distribution AC

which is isomorphic to completion of the spaceHK whichis presented in Section 2 It seems the concept of integrabledistribution is first introduced by Mikusinski and Ostas-zewski [13] and further developed by several authors[14 15] In recent years integrable distributional solution fordifferential equations has been used by many authors[16ndash19] We define derivative (in distributional sense) ofAC

and the integral on it In Section 3 the space of Har-dyndashKrause variation functions is identified as a multiplier forHK Here it uses HenstockndashStieltjes integral and integra-tion by parts defined on AC In [15] example 144 Talviladiscussed the Dirichlet boundary condition of Laplaceequation in the half space ampere it uses Poisson Kernel andproof of integrable distributional solution for the Laplace

equation is not givenWe extend the proof of this problem inSection 4 and obtain integrable distributional solution for (1)when f 0

2 Integrable Distributions

Distribution theory was frequently used with the advent ofLaurent Schwartz [20] in 1950 Distributions or generalizedfunctions are an important functional analysis tool inmodern analysis especially in the field of PDErsquos Whenfunctions are nonsmooth then distribution theory allows toperform operations translation differentiation convolutionetc Here we use integrable distributions which are a specialclass of distributions that will be important for applicationsof distribution theory to partial differential equations fordetails see [13 14 21] ampere are several articles on solvingordinary and partial differential equations using integrabledistributions [16 17 19] Integrable distribution has used tofind a distributional solution for Poissonrsquos equation withDirichlet boundary condition for the upper half plane ampisarticle extends the distributional solution for Poissonrsquosequation for the 3-dimensional upper half space

Let the space of test functions isD(R2) ϕ R2⟶ R|ϕ isin Cinfin1113864 and ϕ has compact sup-port ampen D(R2) forms a normed space under usualpointwise operations ampe dual spaceDprime(R) is called as thespace of distributions ampat is the distributions are thecontinuous linear functionals on D(R2) For a given dis-tribution T denotes its action on test function ϕ isin D(R2) bylangT ϕrang For an example if f is a locally HK-integrablefunction then its corresponding distribution is definedlangTfϕrang 1113938

infinminus infin 1113938infinminus infin fϕ Hereafter all integral will be HK-

integral unless stated otherwise Next consider derivative ofa distribution sometimes called as distributional derivativeor weak derivative If F is continuously differentiable then1113938infinminus infin Fprimeϕ minus 1113938

infinminus infin Fϕprime for any test function ϕ ampe compact

support of ϕ was used in this integration by parts With thissuggestion we define the distributional derivative of adistribution F isin Dprime(R) by langzαF ϕrang (minus 1)|α|langF zαϕrang forany ϕ isin D(R2) amperefore in particularlangz12F ϕrang langF z12ϕrang where z12 z2zy zx Now considera specific type of distributions called as integrable distri-butions ampis definition seems to have been first introducedby Mikusinski and Ostaszewski [13] After that it was de-veloped in detail in the plane by Ang et al [14] and onR andR

2 by Talvila [15 21]For the extended real numbers R consider C0(R

2) be

the space of continuous functions so that bothlimx⟶minus infinF(x y) and limy⟶minus infinF(x y) exit for any

F isin C(R2) ampen define BC(R 2) F isin C0(R

2)1113882 |F(minus infin

y) F(x minus infin) 0forallx y isin R ampen BC(R2) forms a

Banach space under the uniform norm middotinfin Next define the

2 International Journal of Mathematics and Mathematical Sciences

integrable distribution as the derivative of function inBC(R

2)

Definition 1 Let f isin Dprime(R2) be distribution ampe space ofintegrable distributions is defined and denoted by

AC R2

1113874 1113875 f isin Dprime R21113872 1113873|f z12F F isinBC R

21113874 11138751113882 1113883 (2)

Here f z12F be the sense of distributional derivativeie langf ϕrang langz12F ϕrang langF z12ϕrang for ϕ isin D(R2) ampefunction F is called as a primitive of f Generally primitiveis not unique and differed by a constant However with ourchoice F(minus infin y) F(x minus infin) 0 primitive is unique forany given f isin Dprime(R2)

Theorem 1 Any given f isin AC(R2) has a unique primitive

in BC(R2)

Proof Let F1 and F2 are primitives of f isin AC(R2) ampen

f z12F1 z12F2 so that z12(F1 minus F2) 0 ampis impliesF1 minus F2 X(x) + Y(y) for some X Y isin C(R

2) Since

(F1 minus F2) isinBC(R2) for a fixed y0

0 limx⟶minus infin

F1 minus F2( 1113857 limx⟶minus infin

X(x) + limx⟶minus infin

Y y0( 1113857 (3)

amperefore limx⟶minus infinX(x) minus Y(y0) for any x isin R ampisimplies that X(x) is a constant function For fixed x0 lettingy⟶ minus infin gives Y(y) is a constant function for the sameargument ampose constant functions sums into 0 and henceF1 F2

ampe uppercase letter is denoted the primitive of anyintegrable distribution f If f isin AC(R

2) with the primitive

F isinBC(R2) then for any test function ϕ isin D(R2)

langf ϕrang langz12F ϕrang langF z12ϕrang 1113946infin

minus infin1113946infin

minus infinFz12ϕ (4)

where it uses HK-integral and here onward the same unlessotherwise stated Since the linearity of derivatives AC(R

2)

forms a vector space with usual operations of functionals Toget Banach space structure of AC(R

2) it requires to define

the suitable norm Using the unique primitive off isin AC(R

2) we define the Alexiewicz norm as

f sup 1113946x

minus infin1113946

y

minus infinf

1113868111386811138681113868111386811138681113868

1113868111386811138681113868111386811138681113868 (x y) isin R2

1113882 1113883 supR2

|F| Finfin

(5)

Note that the space HK with the Alexiewicz norm is anormed space but not a Banach space [12 22] Absence ofBanach space structure ofHK is main disadvantage on it Itis one of the reasons to move into AC(R

2)

Theorem 2 AC(R2) is a Banach space with the Alexiewicz

norm middot

Proof First prove middot is a norm

(i) If f equiv 0 then clearly f 0 If f 0 and thenFinfin 0 ampis implies F equiv 0 and hencef z12F equiv 0

(ii) For any λ isin R and f isin AC(R2)

λf sup 1113946x

minus infin1113946

y

minus infinλf

1113868111386811138681113868111386811138681113868

1113868111386811138681113868111386811138681113868 (x y) isin R2

1113882 1113883

|λ|sup 1113946x

minus infin1113946

y

minus infinf

1113868111386811138681113868111386811138681113868

1113868111386811138681113868111386811138681113868 (x y) isin R2

1113882 1113883 |λ|f

(6)

(iii) If f g isin AC(R2) the primitive of (f + g) is

(F + G) ampen

f + g F + Ginfin le Finfin +Ginfin f +g (7)

amperefore AC(R2) is a normed space To show com-

pleteness let any fn1113864 1113865subeAC(R2) be Cauchy sequence in middot

ampen Fn1113864 1113865subeBC(R2) is a Cauchy sequence in middotinfin because

fn minus fm Fn minus Fminfin ButBC(R2) is a Banach space so

that there exists a F isinBC(R2) and Fn⟶ F in middotinfin Let

f z12F ampen f isin AC(R2) and fn minus f Fn minus

Finfin⟶ 0 ampis shows fn⟶ f isin AC(R2) in middot and

hence the lemma

Alexiewicz norm dose not induce an inner product inAC(R

2) because the norm does not satisfy the parallelogram

law ampereforeAC(R2) is not a Hilbert space under the norm

middot However the space AC(R2) with middot is isomorphic to

BC(R2) with middotinfin ampe mapping f⟶ F gives a linear

isomorphism between the spaces Uniqueness of primitivegives the mapping is injective and by the definition it issurjective mapping Furthermore the mapping preserves normsince f Finfin ampus the space AC(R

2) is identified with

the space of continuous functions vanish at infinity

Proposition 1 (AC(R2) middot) is isometrically isomorphic to

(BC(R2) middotinfin)

Next we define the integral of an integrable distributionas it uses the primitive F isinBC(R

2) of f isin AC(R

2)

Definition 2 (see [15] Definition 43) Let f isin AC(R2) with

the primitive F isinBC(R2) We define integral of f on I

[a b] times [c d]subeR2 by

1113946If 1113946

b

a1113946

d

cf F(a c) + F(b d) minus F(a d) minus F(b c)

(8)

ampis integral is uniquely defined by uniqueness of theprimitive ampeorem 1 Also it is very general and includesthe Riemann integral Lebesgue integral and HK-integralNorm preservation of Proposition 1 gives that the corre-sponding integral values agree on each integral as presentedin ampeorem 3

International Journal of Mathematics and Mathematical Sciences 3

Theorem 3 For any test function ϕ isin D(R2) if f isin L1(R2)

is identity with the corresponding distributionf ϕ⟼ (L)1113938

R2fϕ (or with HK-integral) then

(i) f isin AC(R2)

(ii) AC(R2) is the completion of L1(R2) (or HK(R2))

with respect to the norm

f sup (L) 1113946x

minus infin1113946

y

minus infinf

1113868111386811138681113868111386811138681113868

1113868111386811138681113868111386811138681113868 (x y) isin R2

1113882 1113883 (9)

(or in the sense of HK-integral)(iii) 1113938

R2f (L)1113938R2f(or (HK)1113938

R2f)

Proof

(i) Let f isin L1(R2) For any (x y) isin R2 we defineFL(x y) (L) 1113938

x

minus infin 1113938y

minus infin f ampen FL is a primitiveof f in the Lebesgue integral (or HK-integral)ampen FL isin C0(R2) since the primitive of Lebesgueintegral (or HK-integral) is continuous ClearlyFL(minus infin y) FL(x minus infin) 0 for any (x y) isin R2amperefore f isin AC(R

2)

(ii) From (i) L1(R2) sub AC(R2) Let any f isin AC(R

2)

ampen it needs to show f isin L1(R2) We prove this byshowing f is a limit point of L1(R2) Sincef isin AC(R

2) there is F isinBC(R

2) Now choose a

sequence Fn1113864 1113865 sub C2(R2) so that Fn minus Finfin⟶ 0

Since Fn converges to F uniformly andF(minus infin y) F(x minus infin) 0 we can assume thatFn(minus infin y) Fn(x minus infin) 0 for each n Now takefn zFn in the distributional sense ampenfn zFn isin C(R2) sub L1(R2) Moreover fn minus f

Fn minus Finfin⟶ 0 Hence f is a limit point ofL1(R2)

(iii) Any Lebesgue integrable function is HK-integrablewith the same value ampus the result

Note that if f isin L1(R2) then its Alexiewicz norm and L1

norm are not equivalent for an example see [15] (pg 9) ampenorms are equivalent if fge 0 for almost everywhere Fromthe definition of integral of f isin AC(R

2) and its unique

primitive we obtain following fundamental theorem ofcalculus

Theorem 4 (fundamental theorem of calculus)

(i) If F isin C(R2) then 1113938x

minus infin 1113938y

minus infin zF F(x y)

+ F(infininfin) minus F(minus infin y) minus F(x minus infin)

(ii) Let f isin AC(R2) and F(x y) 1113938

x

minus infin 1113938y

minus infin f 8enF isinBC(R

2) and f zF

Proof See [15]

3 Multiplier for Integrable Distributions

We will turn now to our discussion on multipliers forAC(R

2) A multiplier for a class of functionsF is a function

g such that fg isinF for each f isin F For an example inLebesgue integral g isin Linfin(Rn) if and only if fg isin L1(Rn)

for each f isin L1(Rn) ie the space Linfin(Rn) is multipliers forL1(Rn) For HK-integral in one-dimensional case 2 space ofbounded variation functions BV is multipliers for HK

[9] (ampeorem 615) In multidimension case there is nounique way to extend the notion of variation to functionAdams and Clarkson [23] give six such extensions butmostly recognized by Vitali variation and HardyndashKrausevariation To establish integration by parts formula andmultipliers for AC(R

2) we begin with HardyndashKrause

variationTwo intervals in R

2 are nonoverlapping if their inter-section is of Lebesgue measure zero A division of R2 is afinite collection of nonoverlapping intervals whose union isR

2 If φ R2⟶ R then its total variation (in the sense of

Vitali) is given by

V12φ supD

1113944i

φ ai ci( 1113857 + φ bi di( 1113857 minus φ ai di( 1113857 minus φ bi ci( 11138571113868111386811138681113868

1113868111386811138681113868

(10)

where the supremum is taken over all divisions D of R2 andinterval Ii [ai bi] times [ci di] isin D

Definition 3 (HardyndashKrause bounded variation) A functionφ R

2⟶ R is said to be HardyndashKrause bounded variationif the following conditions are satisfied

(i) V12φ is finite

(ii) ampe function x⟼φ(x y0) is bounded variation onR for some y0 isin R

(iii) ampe function y⟼φ(x0 y) is bounded variationon R for some x0 isin R

ampe set of functions in HardyndashKrause variation isdenoted by BVHK(R

2) As in [15] BVHK(R

2) forms a

Banach space with norm defined byφBV φinfin + V1φinfin + V2φinfin + V12φ Here V1φ andV2φ are bounded variation to the functions of (ii) and (iii) inDefinition 3 For any multidimensional case any function of

4 International Journal of Mathematics and Mathematical Sciences

HardyndashKrause variation is a multiplier for HK-integrablefunctions see [9] amperefore define following integration byparts as a two-dimensional case which is presented in ([9]ampeorem 659)

Definition 4 (integration by parts) Let f isin AC(R2) with its

primitive F isinBC(R2) If g isinBVHK(R

2) then define

integration by parts on [a b] times [c d]subeR2

1113946b

a1113946

d

cfg F(a c)g(a c) + F(b d)g(b d) minus F(a d)g(a d) minus F(b c)g(b c)

minus 1113946b

aF(x d)d1g(x d) + 1113946

b

aF(x c)d1g(x c)

minus 1113946d

cF(b y)d2g(b y) + 1113946

d

cF(a y)d2g(a y)

+ 1113946b

a1113946

d

cF(x y)d12g(x y)

(11)

Here dg(x y) is HenstockndashStieltjes integral of relevantvariables see in detail [24]ampis integration by parts formulainduces the space of HardyndashKrause bounded variationfunctions BVHK(R

2) as a multiplier for AC(R

2) ampis

multiplier is important to define convolution and obtainconvergence theorem

It needs to obtain continuity of the Alexiewicz normsince the Dirichlet boundary condition is taken on it For any(s t) isin R2 the translation of f isin AC(R

2)subeDprime(R

2) is de-fined by langτ(st)f ϕrang langf τ(minus sminus t)ϕrang for ϕ isin D(R2) whereτ(st)ϕ(x y) ϕ(x minus s y minus t) Translation is invariant andcontinuous under the Alexiewicz norm

Theorem 5 Let f isin AC(R2) and (s t) isin R2 8en (i)

τ(st)f isin AC(R2) (ii) f τ(st)f and (iii)

f minus τ(st)f⟶ 0 as (s t)⟶ (0 0) ie continuity of f inthe Alexiewicz norm

Proof Let F isinBC(R2) be the primitive of f isin AC(R

2)

ampen τ(st)F is a continuous function

(i) For any ϕ isin D(R2) applying change of variable inintegral

langτ(st)f ϕrang langf τ(minus sminus t)ϕrang

1113946R

1113946R

f(x y)ϕ(x + s y + t)dxdy

1113946R

1113946R

f(x minus s y minus t)ϕ(x y)dxdy

1113946R

1113946R

z12F(x minus s y minus t)ϕ(x y)dxdy

1113946R

1113946R

z12 τ(st)F(x y)1113872 1113873ϕ(x y)dxdy

(12)

amperefore τ(st)F isinBC(R2) is the primitive of

τ(st)f(ii) Clearly Finfin τ(st)Finfin for any (s t) isin R2

amperefore f τ(st)f(iii) If f isin AC(R

2) then (f minus τ(st)f) isin AC(R

2) so that

f minus τ(st)f

F minus τ(st)F

infin sup(xy)isinR2

|F(x y) minus F(x minus s y minus t)|⟶ 0(13)

when (s t)⟶ (0 0) because of uniform continuity ofF

4 Integrable Distributional Solution forPoisson Equation

Poissonrsquos kernel and its convolution are used to find aclassical solution to the Dirichlet problem In this section weuse convolution and convergence theorem on AC(R

2) to

find integrable distributional solution If g isinBVHK(R2)

then τ(minus sminus t)g isinBVHK(R2) From integration by parts the

convolution

flowastg(x y) 1113946R

1113946R

fτ(minus sminus t)gdsdt

1113946R

1113946R

f(x minus s y minus t)g(x y)dsdt

(14)

exists for f isin AC(R2) and g isinBVHK(R

2) Proposition 2

gives some basic properties of convolution which help withour problem

International Journal of Mathematics and Mathematical Sciences 5

Proposition 2 If f isin AC(R2) and g isinBVHK(R

2) then

(i) flowastg exists on R2 and (ii) flowastg glowastf

Proof See [15] ampeorem 141

Convergence theorems on Banach space are importantwhen solving partial differential equations ampere are severalconvergence theorems in AC such as strong convergence inAC weakstrong convergence in BV and weakstrongconvergence inD Details are discussed in the plane by AngSchmitt and Vy [14] and on R by Talvila [2] Now we moveinto the convergence theorem onAC(R

2) which will help to

interchange limit of integral

Proposition 3 Let φn be a sequence inBVHK(R2) so that

φnBVltinfin and limn⟶infinφn φ pointwise on R2 for a

function φ If g isin AC(R2) then g isinBVHK(R

2) and

limn⟶infin

1113946R

1113946R

gφn 1113946R

1113946R

gφ (15)

Proof See [15] Proposition 91

Let us move into the main problem As in [25] (pg 84)Poissonrsquos equation with Dirichlet boundary condition iseasily reduced to the case where either f 0 or g 0 in (1)amperefore consider integrable distributional solution for thefollowing problem

u 0 inR3+

u g on zR3+

⎧⎨

⎩ (16)

where R3+ (x y z) isin R3 zgt 01113864 1113865 For zgt 0 consider

Poissonrsquos kernel for the upper half space in 3-dimension

K(x y z) z x

2+ y

2+ z

21113872 1113873

minus (32)

2π (17)

Note that 1113938R

1113938R

Kdxdy 1 If g isin C(R2)cap Linfin(R2)then classical solution for the problem is given by the fol-lowing convolution

u(x y z) K(x y )lowastg(x y)

1113946R

1113946R

K(x minus s y minus t )g(s t)dtds(18)

ampis convolution gives integrable distributional solutionfor (16)

Theorem 6 Let g isin AC(R2) and define u(x y z) by (18)

8en

(i) For each x y isin R and zgt 0 the convolutionK(x y )lowastg(x y) u(x y z) exists andu isin AC(R

2)

(ii) u(x y z) 0 in R3+

(iii) limz⟶0+

u(x y z) minus g(x y) 0 for x y isin R

Proof

(i) Fix x y isin R and zgt 0 ampen the function(s t)⟼K(x minus s y minus t z) is of HardyndashKrausevariation on R2 From Proposition 2 Klowastg existson the upper half space Also u Klowastg isin AC(R

2) because any function of HardyndashK-

rause variation is a multiplier for HK-integrablefunctions

(ii) Fix x y isin R and zgt 0 To get ux let xn⟶ x

be a sequence so that xn nex For any (s t) isin R2define

φn(s t) K xn minus s y minus t z( 1113857 minus K(x minus s y minus t z)

xn minus x

(19)

ampen ux limn

1113938R

1113938R

g(s t)φn(s t)dsdt SinceK(x minus s y minus t z) of HardyndashKrause bounded vari-ation so does φn(s t) for each n Since g isin AC(R

2)

from Proposition 3

ux 1113946R

1113946R

limn⟶infin

g(s t)φn(s t)dsdt

1113946R

1113946R

g(s t)Kx(x minus s y minus t z)dsdt

(20)

Repeat the same argument to obtain uxx uy uyy uzand uzz From the linearity of integral

u(x y z) 1113946R

1113946R

g(s t)K(x minus s y minus t z)dsdt

(21)

But Poisson kernel is harmonic so thatu(x y z) 0 in R3

+

6 International Journal of Mathematics and Mathematical Sciences

(iii) Let any α β isin R For Poisson kernel1113938R

1113938R

Kdxdy 1 Using FubinindashTonelli theoremand interchanging iterated integral

1113946α

minus infin1113946β

minus infin(u minus g)dxdy 1113946

α

minus infin1113946β

minus infin[u(x y z) minus g(x y)]dxdy

1113946α

minus infin1113946β

minus infin[g(x y)lowastK(x y z) minus g(x y)]dxdy

1113946α

minus infin1113946β

minus infin1113946R

1113946R

K(s t z)[g(x minus s y minus t) minus g(x y)]dsdtdxdy

1113946α

minus infin1113946β

minus infin1113946R

1113946R

z

2π s2

+ t2

+ z2

1113872 111387332 [g(x minus s y minus t) minus g(x y)]dsdtdxdy

1113946α

minus infin1113946β

minus infin1113946R

1113946R

1

2π s2

+ t2

+ 11113872 111387332 [g(x minus zs y minus zt) minus g(x y)]dsdtdxdy

1113946R

1113946R

1

2π s2

+ t2

+ 11113872 111387332 1113946

α

minus infin1113946β

minus infin[g(x minus zs y minus zt) minus g(x y)]dxdydsdt

le1113946R

1113946R

K(s t 1)g(x minus zs y minus zt) minus g(x y)dsdt

le 2g

(22)

Since α and β are arbitrary

u(x y z) minus g(x y)le1113946R

1113946R

K(s t 1)g(x minus zs y minus zt) minus g(x y)dsdtle 2g (23)

To let z⟶ 0+ apply dominated convergence theorem(HK-integral) for the sequence of functions

gn(s t) K(s t 1) g x minus1n

s y minus1n

t1113874 1113875 minus g(x y)

(24)

ampen |gn(s t)|le 2gK(s t 1) isinHK(R2) since1113938R

1113938R

K(s t 1)dsdt 1 It gives

limn⟶infin

1113946R

1113946R

gn(s t)dsdt limn⟶infin1113946

1113946R

K(s t 1)g(x minus zs y minus zt) minus g(x y)dsdt

1113946R

1113946R

limn⟶infin

gn(s t)dsdt

1113946R

1113946R

K(s t 1) limn⟶infin

g x minus1n

s y minus1n

t1113874 1113875 minus g(x y)

d s dt

(25)

From part (iii) of ampeorem 5 we get limn⟶infing(x minus

(1n)s y minus (1n)t) minus g(x y) 0 so that limit of (23)limz⟶0+ u(x y z) minus g(x y) 0 and this completes thetheorem

International Journal of Mathematics and Mathematical Sciences 7

Next we give an example which will apply ampeorem 6

Example 1 Consider the problem (16) in which

g(x y)

sin x siny

xy xne 0 andyne 0

0 x 0 ory 0

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

(26)

ampen g is HK-integrable but not absolutely integrableHence g isin AC(R

2)∖L1(R2) From ampeorem 6

u(x y z) 1113946R

1113946R

K(x minus s y minus t )g(s t)dtds

1113946R

1113946R

z

2π(x minus s)

2+(y minus t)

2+ z

21113960 1113961

minus (32) sin s sin t

st1113874 1113875dtds

(27)

exists and gives solution for the example Note thatKg isinHK(R2) since K isinBVHK(R

2) is a multiplier for

HK(R2) However the solution does not exist with theLebesgue integral For contraposition suppose thatKg isin L1(R2) ampen it follows from Fubinirsquos theorem [9](ampeorem 255) that the function

h s⟼K(x minus s y minus t z)g(s t) (28)

belongs to L1(R1) for almost all t isin R ampat is

h(s) z sin t

2πt1113874 1113875

sin s

s (x minus s)2

+(y minus t)2

+ z2

1113960 111396132 isin L

1R

11113872 1113873

h0(s)sin s

s1113874 1113875

(29)

where h0(s) (z sin t2πt)(1[(x minus s)2 + (y minus t)2 + z2]32)ampis contradicts that h(s) notin L1(R1) To see that supposeh(s) isin L1[0 1] sub L1(R1) Since h0(s) isin Linfin[0 1] is multi-plier for L1[0 1] it follows that sin ss isin L1[0 1] ampis im-plies that sin ss absolutely integrable on [0 1] which is acontradiction Hence h(s) notin L1(R1)

Even HK-integral exists in (27) it may not be able toevaluate and get explicit function In that case numericalintegration can be applied Numerical integration should bedone with respect to s and t while fixing x y and z details in[26] ampis reduces solving the partial differential equationinto applying numerical integration for the double integral(27)

5 Discussion

In this work integrable distributional solution is obtainedfor Poisson equation with the Dirichlet boundary conditionin the upper half space Boundary conditions are taken in theAlexiewicz norm Lebesgue integral uses for the classicalsolution to Poisson equation Because of the proper inclu-sion L1 ⊊ HR we may use HK-integral to solve Poissonequation However there are some limitations due to the

absence of any natural topology ampat is HK is not aBanach space nevertheless there is a natural seminormcalled Alexiewicz ampis is one of the main reasons to moveinto the space of integrable distribution AC(R

2) which

contains HK and leads to a Banach space with Alexiewicznorm

Integrable distribution is defined as derivative (distri-butional sense) of function in BC(R

2) In this definition

any f isin AC(R2) identifies with the unique primitive

F isinBC(R2) ampis unique primitive allows to define a

Alexiewicz norm middot on AC(R2) and leads to a Banach

space Alexiewicz norm does not induce an inner product inAC(R

2) because it does not satisfy the parallelogram law

amperefore AC(R2) is not a Hilbert space under the norm

middot However the space (AC(R2) middot) is isometrically iso-

morphic to the space (BC(R2) middotinfin) Integral on AC(R

2)

is uniquely defined by uniqueness of the primitive AC(R2)

is the completion of L1(R2) (or HK(R2)) with respect tothe corresponding Alexiewicz norm

In one-dimensional HK-integral space of boundedvariation functions BV is multipliers for HK In mul-tidimension case there are several extensions to variation offunction To establish integration by parts formula andmultipliers for AC(R

2) we begin with HardyndashKrause

variation ampe space of HardyndashKrause bounded variationfunctions BVHK(R

2) as a multiplier for AC(R

2) ampis

multiplier is important to define convolution and obtainconvergence theorem BVHK(R

2) is used to define con-

volution and obtain convergence theorem on AC(R2)

ampere are several convergence theorems in AC such asstrong convergence in AC weakstrong convergence inBV and weakstrong convergence in D Convergencetheorem on AC(R

2) and dominated convergence theorem

in HK(R2) are used to obtain integrable distributionalsolution Finding of this article gives solution for Poissonequation for broader initial condition in the upper halfspace Dirichlet problem in R3 is considered and the so-lution has obtained by HK-integration where it is notpossible from Lebesgue integral Because of functionrsquoscomplexity it may not be able to have solution in explicit

8 International Journal of Mathematics and Mathematical Sciences

form It would help keep an analytical outlook in the so-lution which would be better than solving the partial dif-ferential equation by numerical methods such as finitedifference methods and finite element methods

ampere is potential to find integrable distributional so-lution for Poisson equation in other boundary conditionsNeumann and Robin But it needs to develop suitableconvergence theorems on AC and its multipliers BVHKAlso it is possible to consider Poisson equation in unit ballwith the relevant Poisson kernel Talvila in [15] gives someconstruction to develop integrable distribution on Rn in-tegration by parts and mean value theorem and definesAC(R

n) and BC(R

n) for multidimensional case Hence

integrable distributional solution is possible to obtain forPoisson equation with Dirichlet boundary condition in Rn

+

with suitable convergence theorem

Data Availability

No data were used to support this study

Conflicts of Interest

ampe authors declare that they have no conflicts of interest

References

[1] R L Burden and J D Faires Numerical Analysis ampomsonBrooksCole USA 8th edition 2005

[2] C Anitescu E Atroshchenko N Alajlan and T RabczukldquoArtificial neural network methods for the solution of secondorder boundary value problemsrdquo Computers Materials ampContinua vol 59 no 1 pp 345ndash359 2019

[3] D A Leon-Velasco M M Morın-Castillo J J Oliveros-Oliveros T Perez-Becerra and J A Escamilla-Reyna ldquoNu-merical solution of some differential equations withHenstockndash Kurzweil functionsrdquo Journal of Function Spacesvol 2019 Article ID 8948570 9 pages 2019

[4] G Green ldquoAn essay on the application of mathematicalanalysis to the theories of electricity and magnetismrdquo Journalfur die reine und angewandte Mathematik (Crelles Journal)vol 1850 no 39 pp 73ndash89 1850

[5] L C Evans Partial Differential Equations American Math-ematical Society Providence RI USA 1998

[6] R Bartle A Modern 8eory of Integration American Math-ematical Society RI Providence RI USA 2001

[7] R Henstock 8e General 8eory of Integration OxfordUniversity Press Oxford UK 1991

[8] C Swartz Introduction to Gauge Integrals World ScientificSingapore 2001

[9] T-Y LeeHenstockndashKurzweil Integration on Euclidean SpacesWorld Scientific Singapore 2011

[10] E Talvila ldquoContinuity in the Alexiewicz normrdquoMathematicaBohemica vol 131 no 2 pp 189ndash196 2006

[11] E Talvila ldquoEstimates of henstock-kurzweil Poisson integralsrdquoCanadian Mathematical Bulletin vol 48 no 1 pp 133ndash1462005

[12] A Alexiewicz ldquoLinear functionals on Denjoy-integrablefunctionsrdquo Colloquium Mathematicum vol 1 no 4pp 289ndash293 1948

[13] P Mikusinksi and K Ostaszewski ldquoEmbedding Henstockintegrable functions into the space of Schwartz distributionsrdquoReal Analysis Exchange vol 14 no -89 pp 24ndash29 1988

[14] D D Ang K Schmitt and L K Vy ldquoA multidimensionalanalogue of the denjoyndashperronndashhenstockndashkurzweil integralrdquo8e Bulletin of the Belgian Mathematical Society mdash SimonStevin vol 4 pp 355ndash371 1997

[15] E Talvila ldquoampe continuous primitive integral in the planerdquoReal Analysis Exchange vol 45 no 2 pp 283ndash326 2020

[16] S Heikkila ldquoOn nonlinear distributional and impulsiveCauchy problemsrdquo Nonlinear Analysis 8eory Methods ampApplications vol 75 no 2 pp 852ndash870 2012

[17] B Liang G Ye W Liu and H Wang ldquoOn second ordernonlinear boundary value problems and the distributionalHenstock-Kurzweil integralrdquo Boundary Value Problemvol 2015 2015

[18] A Sikorska-Nowak ldquoExistence theory for integrodifferentialequations and henstock-kurzweil integral in Banach spacesrdquoJournal of Applied Mathematics vol 2007 Article ID 3157212 pages 2007

[19] M Tvrdy ldquoLinear distributional differential equations of thesecond orderrdquo Math Bohemvol 119 no 4 pp 415ndash4361994

[20] L Schwartz 8eorie des Distributions Hermann ParisFrance 1950

[21] E Talvila ldquoampe distributional denjoy integralrdquo Real AnalysisExchange vol 33 no 1 pp 51ndash82 2008

[22] K Ostaszewski ldquoampe space of Henstock integrable functionsof two variablesrdquo International Journal of Mathematics andMathematical Sciences vol 11 no 1 pp 15ndash21 1988

[23] C R Adams and J A Clarkson ldquoOn definitions of boundedvariation for functions of two variablesrdquo Transactions of theAmerican Mathematical Society vol 35 no 4 pp 824ndash8541933

[24] B Bongiorno ldquoampe henstockndashkurzweil integralrdquo inHandbookof Measure 8eory E Pap Ed pp 589ndash615 North HollandDordrecht Netherlands 2002

[25] G B Folland Introduction to Partial Differential EquationsPrinceton University Press Princeton NY USA 1995

[26] S M Aljassas ldquoNumerical methods for evaluation of doubleintegrals with continuous integrandsrdquo International Journalof Pure and Applied Mathematics vol 119 no 10 2018

International Journal of Mathematics and Mathematical Sciences 9

integrable distribution as the derivative of function inBC(R

2)

Definition 1 Let f isin Dprime(R2) be distribution ampe space ofintegrable distributions is defined and denoted by

AC R2

1113874 1113875 f isin Dprime R21113872 1113873|f z12F F isinBC R

21113874 11138751113882 1113883 (2)

Here f z12F be the sense of distributional derivativeie langf ϕrang langz12F ϕrang langF z12ϕrang for ϕ isin D(R2) ampefunction F is called as a primitive of f Generally primitiveis not unique and differed by a constant However with ourchoice F(minus infin y) F(x minus infin) 0 primitive is unique forany given f isin Dprime(R2)

Theorem 1 Any given f isin AC(R2) has a unique primitive

in BC(R2)

Proof Let F1 and F2 are primitives of f isin AC(R2) ampen

f z12F1 z12F2 so that z12(F1 minus F2) 0 ampis impliesF1 minus F2 X(x) + Y(y) for some X Y isin C(R

2) Since

(F1 minus F2) isinBC(R2) for a fixed y0

0 limx⟶minus infin

F1 minus F2( 1113857 limx⟶minus infin

X(x) + limx⟶minus infin

Y y0( 1113857 (3)

amperefore limx⟶minus infinX(x) minus Y(y0) for any x isin R ampisimplies that X(x) is a constant function For fixed x0 lettingy⟶ minus infin gives Y(y) is a constant function for the sameargument ampose constant functions sums into 0 and henceF1 F2

ampe uppercase letter is denoted the primitive of anyintegrable distribution f If f isin AC(R

2) with the primitive

F isinBC(R2) then for any test function ϕ isin D(R2)

langf ϕrang langz12F ϕrang langF z12ϕrang 1113946infin

minus infin1113946infin

minus infinFz12ϕ (4)

where it uses HK-integral and here onward the same unlessotherwise stated Since the linearity of derivatives AC(R

2)

forms a vector space with usual operations of functionals Toget Banach space structure of AC(R

2) it requires to define

the suitable norm Using the unique primitive off isin AC(R

2) we define the Alexiewicz norm as

f sup 1113946x

minus infin1113946

y

minus infinf

1113868111386811138681113868111386811138681113868

1113868111386811138681113868111386811138681113868 (x y) isin R2

1113882 1113883 supR2

|F| Finfin

(5)

Note that the space HK with the Alexiewicz norm is anormed space but not a Banach space [12 22] Absence ofBanach space structure ofHK is main disadvantage on it Itis one of the reasons to move into AC(R

2)

Theorem 2 AC(R2) is a Banach space with the Alexiewicz

norm middot

Proof First prove middot is a norm

(i) If f equiv 0 then clearly f 0 If f 0 and thenFinfin 0 ampis implies F equiv 0 and hencef z12F equiv 0

(ii) For any λ isin R and f isin AC(R2)

λf sup 1113946x

minus infin1113946

y

minus infinλf

1113868111386811138681113868111386811138681113868

1113868111386811138681113868111386811138681113868 (x y) isin R2

1113882 1113883

|λ|sup 1113946x

minus infin1113946

y

minus infinf

1113868111386811138681113868111386811138681113868

1113868111386811138681113868111386811138681113868 (x y) isin R2

1113882 1113883 |λ|f

(6)

(iii) If f g isin AC(R2) the primitive of (f + g) is

(F + G) ampen

f + g F + Ginfin le Finfin +Ginfin f +g (7)

amperefore AC(R2) is a normed space To show com-

pleteness let any fn1113864 1113865subeAC(R2) be Cauchy sequence in middot

ampen Fn1113864 1113865subeBC(R2) is a Cauchy sequence in middotinfin because

fn minus fm Fn minus Fminfin ButBC(R2) is a Banach space so

that there exists a F isinBC(R2) and Fn⟶ F in middotinfin Let

f z12F ampen f isin AC(R2) and fn minus f Fn minus

Finfin⟶ 0 ampis shows fn⟶ f isin AC(R2) in middot and

hence the lemma

Alexiewicz norm dose not induce an inner product inAC(R

2) because the norm does not satisfy the parallelogram

law ampereforeAC(R2) is not a Hilbert space under the norm

middot However the space AC(R2) with middot is isomorphic to

BC(R2) with middotinfin ampe mapping f⟶ F gives a linear

isomorphism between the spaces Uniqueness of primitivegives the mapping is injective and by the definition it issurjective mapping Furthermore the mapping preserves normsince f Finfin ampus the space AC(R

2) is identified with

the space of continuous functions vanish at infinity

Proposition 1 (AC(R2) middot) is isometrically isomorphic to

(BC(R2) middotinfin)

Next we define the integral of an integrable distributionas it uses the primitive F isinBC(R

2) of f isin AC(R

2)

Definition 2 (see [15] Definition 43) Let f isin AC(R2) with

the primitive F isinBC(R2) We define integral of f on I

[a b] times [c d]subeR2 by

1113946If 1113946

b

a1113946

d

cf F(a c) + F(b d) minus F(a d) minus F(b c)

(8)

ampis integral is uniquely defined by uniqueness of theprimitive ampeorem 1 Also it is very general and includesthe Riemann integral Lebesgue integral and HK-integralNorm preservation of Proposition 1 gives that the corre-sponding integral values agree on each integral as presentedin ampeorem 3

International Journal of Mathematics and Mathematical Sciences 3

Theorem 3 For any test function ϕ isin D(R2) if f isin L1(R2)

is identity with the corresponding distributionf ϕ⟼ (L)1113938

R2fϕ (or with HK-integral) then

(i) f isin AC(R2)

(ii) AC(R2) is the completion of L1(R2) (or HK(R2))

with respect to the norm

f sup (L) 1113946x

minus infin1113946

y

minus infinf

1113868111386811138681113868111386811138681113868

1113868111386811138681113868111386811138681113868 (x y) isin R2

1113882 1113883 (9)

(or in the sense of HK-integral)(iii) 1113938

R2f (L)1113938R2f(or (HK)1113938

R2f)

Proof

(i) Let f isin L1(R2) For any (x y) isin R2 we defineFL(x y) (L) 1113938

x

minus infin 1113938y

minus infin f ampen FL is a primitiveof f in the Lebesgue integral (or HK-integral)ampen FL isin C0(R2) since the primitive of Lebesgueintegral (or HK-integral) is continuous ClearlyFL(minus infin y) FL(x minus infin) 0 for any (x y) isin R2amperefore f isin AC(R

2)

(ii) From (i) L1(R2) sub AC(R2) Let any f isin AC(R

2)

ampen it needs to show f isin L1(R2) We prove this byshowing f is a limit point of L1(R2) Sincef isin AC(R

2) there is F isinBC(R

2) Now choose a

sequence Fn1113864 1113865 sub C2(R2) so that Fn minus Finfin⟶ 0

Since Fn converges to F uniformly andF(minus infin y) F(x minus infin) 0 we can assume thatFn(minus infin y) Fn(x minus infin) 0 for each n Now takefn zFn in the distributional sense ampenfn zFn isin C(R2) sub L1(R2) Moreover fn minus f

Fn minus Finfin⟶ 0 Hence f is a limit point ofL1(R2)

(iii) Any Lebesgue integrable function is HK-integrablewith the same value ampus the result

Note that if f isin L1(R2) then its Alexiewicz norm and L1

norm are not equivalent for an example see [15] (pg 9) ampenorms are equivalent if fge 0 for almost everywhere Fromthe definition of integral of f isin AC(R

2) and its unique

primitive we obtain following fundamental theorem ofcalculus

Theorem 4 (fundamental theorem of calculus)

(i) If F isin C(R2) then 1113938x

minus infin 1113938y

minus infin zF F(x y)

+ F(infininfin) minus F(minus infin y) minus F(x minus infin)

(ii) Let f isin AC(R2) and F(x y) 1113938

x

minus infin 1113938y

minus infin f 8enF isinBC(R

2) and f zF

Proof See [15]

3 Multiplier for Integrable Distributions

We will turn now to our discussion on multipliers forAC(R

2) A multiplier for a class of functionsF is a function

g such that fg isinF for each f isin F For an example inLebesgue integral g isin Linfin(Rn) if and only if fg isin L1(Rn)

for each f isin L1(Rn) ie the space Linfin(Rn) is multipliers forL1(Rn) For HK-integral in one-dimensional case 2 space ofbounded variation functions BV is multipliers for HK

[9] (ampeorem 615) In multidimension case there is nounique way to extend the notion of variation to functionAdams and Clarkson [23] give six such extensions butmostly recognized by Vitali variation and HardyndashKrausevariation To establish integration by parts formula andmultipliers for AC(R

2) we begin with HardyndashKrause

variationTwo intervals in R

2 are nonoverlapping if their inter-section is of Lebesgue measure zero A division of R2 is afinite collection of nonoverlapping intervals whose union isR

2 If φ R2⟶ R then its total variation (in the sense of

Vitali) is given by

V12φ supD

1113944i

φ ai ci( 1113857 + φ bi di( 1113857 minus φ ai di( 1113857 minus φ bi ci( 11138571113868111386811138681113868

1113868111386811138681113868

(10)

where the supremum is taken over all divisions D of R2 andinterval Ii [ai bi] times [ci di] isin D

Definition 3 (HardyndashKrause bounded variation) A functionφ R

2⟶ R is said to be HardyndashKrause bounded variationif the following conditions are satisfied

(i) V12φ is finite

(ii) ampe function x⟼φ(x y0) is bounded variation onR for some y0 isin R

(iii) ampe function y⟼φ(x0 y) is bounded variationon R for some x0 isin R

ampe set of functions in HardyndashKrause variation isdenoted by BVHK(R

2) As in [15] BVHK(R

2) forms a

Banach space with norm defined byφBV φinfin + V1φinfin + V2φinfin + V12φ Here V1φ andV2φ are bounded variation to the functions of (ii) and (iii) inDefinition 3 For any multidimensional case any function of

4 International Journal of Mathematics and Mathematical Sciences

HardyndashKrause variation is a multiplier for HK-integrablefunctions see [9] amperefore define following integration byparts as a two-dimensional case which is presented in ([9]ampeorem 659)

Definition 4 (integration by parts) Let f isin AC(R2) with its

primitive F isinBC(R2) If g isinBVHK(R

2) then define

integration by parts on [a b] times [c d]subeR2

1113946b

a1113946

d

cfg F(a c)g(a c) + F(b d)g(b d) minus F(a d)g(a d) minus F(b c)g(b c)

minus 1113946b

aF(x d)d1g(x d) + 1113946

b

aF(x c)d1g(x c)

minus 1113946d

cF(b y)d2g(b y) + 1113946

d

cF(a y)d2g(a y)

+ 1113946b

a1113946

d

cF(x y)d12g(x y)

(11)

Here dg(x y) is HenstockndashStieltjes integral of relevantvariables see in detail [24]ampis integration by parts formulainduces the space of HardyndashKrause bounded variationfunctions BVHK(R

2) as a multiplier for AC(R

2) ampis

multiplier is important to define convolution and obtainconvergence theorem

It needs to obtain continuity of the Alexiewicz normsince the Dirichlet boundary condition is taken on it For any(s t) isin R2 the translation of f isin AC(R

2)subeDprime(R

2) is de-fined by langτ(st)f ϕrang langf τ(minus sminus t)ϕrang for ϕ isin D(R2) whereτ(st)ϕ(x y) ϕ(x minus s y minus t) Translation is invariant andcontinuous under the Alexiewicz norm

Theorem 5 Let f isin AC(R2) and (s t) isin R2 8en (i)

τ(st)f isin AC(R2) (ii) f τ(st)f and (iii)

f minus τ(st)f⟶ 0 as (s t)⟶ (0 0) ie continuity of f inthe Alexiewicz norm

Proof Let F isinBC(R2) be the primitive of f isin AC(R

2)

ampen τ(st)F is a continuous function

(i) For any ϕ isin D(R2) applying change of variable inintegral

langτ(st)f ϕrang langf τ(minus sminus t)ϕrang

1113946R

1113946R

f(x y)ϕ(x + s y + t)dxdy

1113946R

1113946R

f(x minus s y minus t)ϕ(x y)dxdy

1113946R

1113946R

z12F(x minus s y minus t)ϕ(x y)dxdy

1113946R

1113946R

z12 τ(st)F(x y)1113872 1113873ϕ(x y)dxdy

(12)

amperefore τ(st)F isinBC(R2) is the primitive of

τ(st)f(ii) Clearly Finfin τ(st)Finfin for any (s t) isin R2

amperefore f τ(st)f(iii) If f isin AC(R

2) then (f minus τ(st)f) isin AC(R

2) so that

f minus τ(st)f

F minus τ(st)F

infin sup(xy)isinR2

|F(x y) minus F(x minus s y minus t)|⟶ 0(13)

when (s t)⟶ (0 0) because of uniform continuity ofF

4 Integrable Distributional Solution forPoisson Equation

Poissonrsquos kernel and its convolution are used to find aclassical solution to the Dirichlet problem In this section weuse convolution and convergence theorem on AC(R

2) to

find integrable distributional solution If g isinBVHK(R2)

then τ(minus sminus t)g isinBVHK(R2) From integration by parts the

convolution

flowastg(x y) 1113946R

1113946R

fτ(minus sminus t)gdsdt

1113946R

1113946R

f(x minus s y minus t)g(x y)dsdt

(14)

exists for f isin AC(R2) and g isinBVHK(R

2) Proposition 2

gives some basic properties of convolution which help withour problem

International Journal of Mathematics and Mathematical Sciences 5

Proposition 2 If f isin AC(R2) and g isinBVHK(R

2) then

(i) flowastg exists on R2 and (ii) flowastg glowastf

Proof See [15] ampeorem 141

Convergence theorems on Banach space are importantwhen solving partial differential equations ampere are severalconvergence theorems in AC such as strong convergence inAC weakstrong convergence in BV and weakstrongconvergence inD Details are discussed in the plane by AngSchmitt and Vy [14] and on R by Talvila [2] Now we moveinto the convergence theorem onAC(R

2) which will help to

interchange limit of integral

Proposition 3 Let φn be a sequence inBVHK(R2) so that

φnBVltinfin and limn⟶infinφn φ pointwise on R2 for a

function φ If g isin AC(R2) then g isinBVHK(R

2) and

limn⟶infin

1113946R

1113946R

gφn 1113946R

1113946R

gφ (15)

Proof See [15] Proposition 91

Let us move into the main problem As in [25] (pg 84)Poissonrsquos equation with Dirichlet boundary condition iseasily reduced to the case where either f 0 or g 0 in (1)amperefore consider integrable distributional solution for thefollowing problem

u 0 inR3+

u g on zR3+

⎧⎨

⎩ (16)

where R3+ (x y z) isin R3 zgt 01113864 1113865 For zgt 0 consider

Poissonrsquos kernel for the upper half space in 3-dimension

K(x y z) z x

2+ y

2+ z

21113872 1113873

minus (32)

2π (17)

Note that 1113938R

1113938R

Kdxdy 1 If g isin C(R2)cap Linfin(R2)then classical solution for the problem is given by the fol-lowing convolution

u(x y z) K(x y )lowastg(x y)

1113946R

1113946R

K(x minus s y minus t )g(s t)dtds(18)

ampis convolution gives integrable distributional solutionfor (16)

Theorem 6 Let g isin AC(R2) and define u(x y z) by (18)

8en

(i) For each x y isin R and zgt 0 the convolutionK(x y )lowastg(x y) u(x y z) exists andu isin AC(R

2)

(ii) u(x y z) 0 in R3+

(iii) limz⟶0+

u(x y z) minus g(x y) 0 for x y isin R

Proof

(i) Fix x y isin R and zgt 0 ampen the function(s t)⟼K(x minus s y minus t z) is of HardyndashKrausevariation on R2 From Proposition 2 Klowastg existson the upper half space Also u Klowastg isin AC(R

2) because any function of HardyndashK-

rause variation is a multiplier for HK-integrablefunctions

(ii) Fix x y isin R and zgt 0 To get ux let xn⟶ x

be a sequence so that xn nex For any (s t) isin R2define

φn(s t) K xn minus s y minus t z( 1113857 minus K(x minus s y minus t z)

xn minus x

(19)

ampen ux limn

1113938R

1113938R

g(s t)φn(s t)dsdt SinceK(x minus s y minus t z) of HardyndashKrause bounded vari-ation so does φn(s t) for each n Since g isin AC(R

2)

from Proposition 3

ux 1113946R

1113946R

limn⟶infin

g(s t)φn(s t)dsdt

1113946R

1113946R

g(s t)Kx(x minus s y minus t z)dsdt

(20)

Repeat the same argument to obtain uxx uy uyy uzand uzz From the linearity of integral

u(x y z) 1113946R

1113946R

g(s t)K(x minus s y minus t z)dsdt

(21)

But Poisson kernel is harmonic so thatu(x y z) 0 in R3

+

6 International Journal of Mathematics and Mathematical Sciences

(iii) Let any α β isin R For Poisson kernel1113938R

1113938R

Kdxdy 1 Using FubinindashTonelli theoremand interchanging iterated integral

1113946α

minus infin1113946β

minus infin(u minus g)dxdy 1113946

α

minus infin1113946β

minus infin[u(x y z) minus g(x y)]dxdy

1113946α

minus infin1113946β

minus infin[g(x y)lowastK(x y z) minus g(x y)]dxdy

1113946α

minus infin1113946β

minus infin1113946R

1113946R

K(s t z)[g(x minus s y minus t) minus g(x y)]dsdtdxdy

1113946α

minus infin1113946β

minus infin1113946R

1113946R

z

2π s2

+ t2

+ z2

1113872 111387332 [g(x minus s y minus t) minus g(x y)]dsdtdxdy

1113946α

minus infin1113946β

minus infin1113946R

1113946R

1

2π s2

+ t2

+ 11113872 111387332 [g(x minus zs y minus zt) minus g(x y)]dsdtdxdy

1113946R

1113946R

1

2π s2

+ t2

+ 11113872 111387332 1113946

α

minus infin1113946β

minus infin[g(x minus zs y minus zt) minus g(x y)]dxdydsdt

le1113946R

1113946R

K(s t 1)g(x minus zs y minus zt) minus g(x y)dsdt

le 2g

(22)

Since α and β are arbitrary

u(x y z) minus g(x y)le1113946R

1113946R

K(s t 1)g(x minus zs y minus zt) minus g(x y)dsdtle 2g (23)

To let z⟶ 0+ apply dominated convergence theorem(HK-integral) for the sequence of functions

gn(s t) K(s t 1) g x minus1n

s y minus1n

t1113874 1113875 minus g(x y)

(24)

ampen |gn(s t)|le 2gK(s t 1) isinHK(R2) since1113938R

1113938R

K(s t 1)dsdt 1 It gives

limn⟶infin

1113946R

1113946R

gn(s t)dsdt limn⟶infin1113946

1113946R

K(s t 1)g(x minus zs y minus zt) minus g(x y)dsdt

1113946R

1113946R

limn⟶infin

gn(s t)dsdt

1113946R

1113946R

K(s t 1) limn⟶infin

g x minus1n

s y minus1n

t1113874 1113875 minus g(x y)

d s dt

(25)

From part (iii) of ampeorem 5 we get limn⟶infing(x minus

(1n)s y minus (1n)t) minus g(x y) 0 so that limit of (23)limz⟶0+ u(x y z) minus g(x y) 0 and this completes thetheorem

International Journal of Mathematics and Mathematical Sciences 7

Next we give an example which will apply ampeorem 6

Example 1 Consider the problem (16) in which

g(x y)

sin x siny

xy xne 0 andyne 0

0 x 0 ory 0

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

(26)

ampen g is HK-integrable but not absolutely integrableHence g isin AC(R

2)∖L1(R2) From ampeorem 6

u(x y z) 1113946R

1113946R

K(x minus s y minus t )g(s t)dtds

1113946R

1113946R

z

2π(x minus s)

2+(y minus t)

2+ z

21113960 1113961

minus (32) sin s sin t

st1113874 1113875dtds

(27)

exists and gives solution for the example Note thatKg isinHK(R2) since K isinBVHK(R

2) is a multiplier for

HK(R2) However the solution does not exist with theLebesgue integral For contraposition suppose thatKg isin L1(R2) ampen it follows from Fubinirsquos theorem [9](ampeorem 255) that the function

h s⟼K(x minus s y minus t z)g(s t) (28)

belongs to L1(R1) for almost all t isin R ampat is

h(s) z sin t

2πt1113874 1113875

sin s

s (x minus s)2

+(y minus t)2

+ z2

1113960 111396132 isin L

1R

11113872 1113873

h0(s)sin s

s1113874 1113875

(29)

where h0(s) (z sin t2πt)(1[(x minus s)2 + (y minus t)2 + z2]32)ampis contradicts that h(s) notin L1(R1) To see that supposeh(s) isin L1[0 1] sub L1(R1) Since h0(s) isin Linfin[0 1] is multi-plier for L1[0 1] it follows that sin ss isin L1[0 1] ampis im-plies that sin ss absolutely integrable on [0 1] which is acontradiction Hence h(s) notin L1(R1)

Even HK-integral exists in (27) it may not be able toevaluate and get explicit function In that case numericalintegration can be applied Numerical integration should bedone with respect to s and t while fixing x y and z details in[26] ampis reduces solving the partial differential equationinto applying numerical integration for the double integral(27)

5 Discussion

In this work integrable distributional solution is obtainedfor Poisson equation with the Dirichlet boundary conditionin the upper half space Boundary conditions are taken in theAlexiewicz norm Lebesgue integral uses for the classicalsolution to Poisson equation Because of the proper inclu-sion L1 ⊊ HR we may use HK-integral to solve Poissonequation However there are some limitations due to the

absence of any natural topology ampat is HK is not aBanach space nevertheless there is a natural seminormcalled Alexiewicz ampis is one of the main reasons to moveinto the space of integrable distribution AC(R

2) which

contains HK and leads to a Banach space with Alexiewicznorm

Integrable distribution is defined as derivative (distri-butional sense) of function in BC(R

2) In this definition

any f isin AC(R2) identifies with the unique primitive

F isinBC(R2) ampis unique primitive allows to define a

Alexiewicz norm middot on AC(R2) and leads to a Banach

space Alexiewicz norm does not induce an inner product inAC(R

2) because it does not satisfy the parallelogram law

amperefore AC(R2) is not a Hilbert space under the norm

middot However the space (AC(R2) middot) is isometrically iso-

morphic to the space (BC(R2) middotinfin) Integral on AC(R

2)

is uniquely defined by uniqueness of the primitive AC(R2)

is the completion of L1(R2) (or HK(R2)) with respect tothe corresponding Alexiewicz norm

In one-dimensional HK-integral space of boundedvariation functions BV is multipliers for HK In mul-tidimension case there are several extensions to variation offunction To establish integration by parts formula andmultipliers for AC(R

2) we begin with HardyndashKrause

variation ampe space of HardyndashKrause bounded variationfunctions BVHK(R

2) as a multiplier for AC(R

2) ampis

multiplier is important to define convolution and obtainconvergence theorem BVHK(R

2) is used to define con-

volution and obtain convergence theorem on AC(R2)

ampere are several convergence theorems in AC such asstrong convergence in AC weakstrong convergence inBV and weakstrong convergence in D Convergencetheorem on AC(R

2) and dominated convergence theorem

in HK(R2) are used to obtain integrable distributionalsolution Finding of this article gives solution for Poissonequation for broader initial condition in the upper halfspace Dirichlet problem in R3 is considered and the so-lution has obtained by HK-integration where it is notpossible from Lebesgue integral Because of functionrsquoscomplexity it may not be able to have solution in explicit

8 International Journal of Mathematics and Mathematical Sciences

form It would help keep an analytical outlook in the so-lution which would be better than solving the partial dif-ferential equation by numerical methods such as finitedifference methods and finite element methods

ampere is potential to find integrable distributional so-lution for Poisson equation in other boundary conditionsNeumann and Robin But it needs to develop suitableconvergence theorems on AC and its multipliers BVHKAlso it is possible to consider Poisson equation in unit ballwith the relevant Poisson kernel Talvila in [15] gives someconstruction to develop integrable distribution on Rn in-tegration by parts and mean value theorem and definesAC(R

n) and BC(R

n) for multidimensional case Hence

integrable distributional solution is possible to obtain forPoisson equation with Dirichlet boundary condition in Rn

+

with suitable convergence theorem

Data Availability

No data were used to support this study

Conflicts of Interest

ampe authors declare that they have no conflicts of interest

References

[1] R L Burden and J D Faires Numerical Analysis ampomsonBrooksCole USA 8th edition 2005

[2] C Anitescu E Atroshchenko N Alajlan and T RabczukldquoArtificial neural network methods for the solution of secondorder boundary value problemsrdquo Computers Materials ampContinua vol 59 no 1 pp 345ndash359 2019

[3] D A Leon-Velasco M M Morın-Castillo J J Oliveros-Oliveros T Perez-Becerra and J A Escamilla-Reyna ldquoNu-merical solution of some differential equations withHenstockndash Kurzweil functionsrdquo Journal of Function Spacesvol 2019 Article ID 8948570 9 pages 2019

[4] G Green ldquoAn essay on the application of mathematicalanalysis to the theories of electricity and magnetismrdquo Journalfur die reine und angewandte Mathematik (Crelles Journal)vol 1850 no 39 pp 73ndash89 1850

[5] L C Evans Partial Differential Equations American Math-ematical Society Providence RI USA 1998

[6] R Bartle A Modern 8eory of Integration American Math-ematical Society RI Providence RI USA 2001

[7] R Henstock 8e General 8eory of Integration OxfordUniversity Press Oxford UK 1991

[8] C Swartz Introduction to Gauge Integrals World ScientificSingapore 2001

[9] T-Y LeeHenstockndashKurzweil Integration on Euclidean SpacesWorld Scientific Singapore 2011

[10] E Talvila ldquoContinuity in the Alexiewicz normrdquoMathematicaBohemica vol 131 no 2 pp 189ndash196 2006

[11] E Talvila ldquoEstimates of henstock-kurzweil Poisson integralsrdquoCanadian Mathematical Bulletin vol 48 no 1 pp 133ndash1462005

[12] A Alexiewicz ldquoLinear functionals on Denjoy-integrablefunctionsrdquo Colloquium Mathematicum vol 1 no 4pp 289ndash293 1948

[13] P Mikusinksi and K Ostaszewski ldquoEmbedding Henstockintegrable functions into the space of Schwartz distributionsrdquoReal Analysis Exchange vol 14 no -89 pp 24ndash29 1988

[14] D D Ang K Schmitt and L K Vy ldquoA multidimensionalanalogue of the denjoyndashperronndashhenstockndashkurzweil integralrdquo8e Bulletin of the Belgian Mathematical Society mdash SimonStevin vol 4 pp 355ndash371 1997

[15] E Talvila ldquoampe continuous primitive integral in the planerdquoReal Analysis Exchange vol 45 no 2 pp 283ndash326 2020

[16] S Heikkila ldquoOn nonlinear distributional and impulsiveCauchy problemsrdquo Nonlinear Analysis 8eory Methods ampApplications vol 75 no 2 pp 852ndash870 2012

[17] B Liang G Ye W Liu and H Wang ldquoOn second ordernonlinear boundary value problems and the distributionalHenstock-Kurzweil integralrdquo Boundary Value Problemvol 2015 2015

[18] A Sikorska-Nowak ldquoExistence theory for integrodifferentialequations and henstock-kurzweil integral in Banach spacesrdquoJournal of Applied Mathematics vol 2007 Article ID 3157212 pages 2007

[19] M Tvrdy ldquoLinear distributional differential equations of thesecond orderrdquo Math Bohemvol 119 no 4 pp 415ndash4361994

[20] L Schwartz 8eorie des Distributions Hermann ParisFrance 1950

[21] E Talvila ldquoampe distributional denjoy integralrdquo Real AnalysisExchange vol 33 no 1 pp 51ndash82 2008

[22] K Ostaszewski ldquoampe space of Henstock integrable functionsof two variablesrdquo International Journal of Mathematics andMathematical Sciences vol 11 no 1 pp 15ndash21 1988

[23] C R Adams and J A Clarkson ldquoOn definitions of boundedvariation for functions of two variablesrdquo Transactions of theAmerican Mathematical Society vol 35 no 4 pp 824ndash8541933

[24] B Bongiorno ldquoampe henstockndashkurzweil integralrdquo inHandbookof Measure 8eory E Pap Ed pp 589ndash615 North HollandDordrecht Netherlands 2002

[25] G B Folland Introduction to Partial Differential EquationsPrinceton University Press Princeton NY USA 1995

[26] S M Aljassas ldquoNumerical methods for evaluation of doubleintegrals with continuous integrandsrdquo International Journalof Pure and Applied Mathematics vol 119 no 10 2018

International Journal of Mathematics and Mathematical Sciences 9

Theorem 3 For any test function ϕ isin D(R2) if f isin L1(R2)

is identity with the corresponding distributionf ϕ⟼ (L)1113938

R2fϕ (or with HK-integral) then

(i) f isin AC(R2)

(ii) AC(R2) is the completion of L1(R2) (or HK(R2))

with respect to the norm

f sup (L) 1113946x

minus infin1113946

y

minus infinf

1113868111386811138681113868111386811138681113868

1113868111386811138681113868111386811138681113868 (x y) isin R2

1113882 1113883 (9)

(or in the sense of HK-integral)(iii) 1113938

R2f (L)1113938R2f(or (HK)1113938

R2f)

Proof

(i) Let f isin L1(R2) For any (x y) isin R2 we defineFL(x y) (L) 1113938

x

minus infin 1113938y

minus infin f ampen FL is a primitiveof f in the Lebesgue integral (or HK-integral)ampen FL isin C0(R2) since the primitive of Lebesgueintegral (or HK-integral) is continuous ClearlyFL(minus infin y) FL(x minus infin) 0 for any (x y) isin R2amperefore f isin AC(R

2)

(ii) From (i) L1(R2) sub AC(R2) Let any f isin AC(R

2)

ampen it needs to show f isin L1(R2) We prove this byshowing f is a limit point of L1(R2) Sincef isin AC(R

2) there is F isinBC(R

2) Now choose a

sequence Fn1113864 1113865 sub C2(R2) so that Fn minus Finfin⟶ 0

Since Fn converges to F uniformly andF(minus infin y) F(x minus infin) 0 we can assume thatFn(minus infin y) Fn(x minus infin) 0 for each n Now takefn zFn in the distributional sense ampenfn zFn isin C(R2) sub L1(R2) Moreover fn minus f

Fn minus Finfin⟶ 0 Hence f is a limit point ofL1(R2)

(iii) Any Lebesgue integrable function is HK-integrablewith the same value ampus the result

Note that if f isin L1(R2) then its Alexiewicz norm and L1

norm are not equivalent for an example see [15] (pg 9) ampenorms are equivalent if fge 0 for almost everywhere Fromthe definition of integral of f isin AC(R

2) and its unique

primitive we obtain following fundamental theorem ofcalculus

Theorem 4 (fundamental theorem of calculus)

(i) If F isin C(R2) then 1113938x

minus infin 1113938y

minus infin zF F(x y)

+ F(infininfin) minus F(minus infin y) minus F(x minus infin)

(ii) Let f isin AC(R2) and F(x y) 1113938

x

minus infin 1113938y

minus infin f 8enF isinBC(R

2) and f zF

Proof See [15]

3 Multiplier for Integrable Distributions

We will turn now to our discussion on multipliers forAC(R

2) A multiplier for a class of functionsF is a function

g such that fg isinF for each f isin F For an example inLebesgue integral g isin Linfin(Rn) if and only if fg isin L1(Rn)

for each f isin L1(Rn) ie the space Linfin(Rn) is multipliers forL1(Rn) For HK-integral in one-dimensional case 2 space ofbounded variation functions BV is multipliers for HK

[9] (ampeorem 615) In multidimension case there is nounique way to extend the notion of variation to functionAdams and Clarkson [23] give six such extensions butmostly recognized by Vitali variation and HardyndashKrausevariation To establish integration by parts formula andmultipliers for AC(R

2) we begin with HardyndashKrause

variationTwo intervals in R

2 are nonoverlapping if their inter-section is of Lebesgue measure zero A division of R2 is afinite collection of nonoverlapping intervals whose union isR

2 If φ R2⟶ R then its total variation (in the sense of

Vitali) is given by

V12φ supD

1113944i

φ ai ci( 1113857 + φ bi di( 1113857 minus φ ai di( 1113857 minus φ bi ci( 11138571113868111386811138681113868

1113868111386811138681113868

(10)

where the supremum is taken over all divisions D of R2 andinterval Ii [ai bi] times [ci di] isin D

Definition 3 (HardyndashKrause bounded variation) A functionφ R

2⟶ R is said to be HardyndashKrause bounded variationif the following conditions are satisfied

(i) V12φ is finite

(ii) ampe function x⟼φ(x y0) is bounded variation onR for some y0 isin R

(iii) ampe function y⟼φ(x0 y) is bounded variationon R for some x0 isin R

ampe set of functions in HardyndashKrause variation isdenoted by BVHK(R

2) As in [15] BVHK(R

2) forms a

Banach space with norm defined byφBV φinfin + V1φinfin + V2φinfin + V12φ Here V1φ andV2φ are bounded variation to the functions of (ii) and (iii) inDefinition 3 For any multidimensional case any function of

4 International Journal of Mathematics and Mathematical Sciences

HardyndashKrause variation is a multiplier for HK-integrablefunctions see [9] amperefore define following integration byparts as a two-dimensional case which is presented in ([9]ampeorem 659)

Definition 4 (integration by parts) Let f isin AC(R2) with its

primitive F isinBC(R2) If g isinBVHK(R

2) then define

integration by parts on [a b] times [c d]subeR2

1113946b

a1113946

d

cfg F(a c)g(a c) + F(b d)g(b d) minus F(a d)g(a d) minus F(b c)g(b c)

minus 1113946b

aF(x d)d1g(x d) + 1113946

b

aF(x c)d1g(x c)

minus 1113946d

cF(b y)d2g(b y) + 1113946

d

cF(a y)d2g(a y)

+ 1113946b

a1113946

d

cF(x y)d12g(x y)

(11)

Here dg(x y) is HenstockndashStieltjes integral of relevantvariables see in detail [24]ampis integration by parts formulainduces the space of HardyndashKrause bounded variationfunctions BVHK(R

2) as a multiplier for AC(R

2) ampis

multiplier is important to define convolution and obtainconvergence theorem

It needs to obtain continuity of the Alexiewicz normsince the Dirichlet boundary condition is taken on it For any(s t) isin R2 the translation of f isin AC(R

2)subeDprime(R

2) is de-fined by langτ(st)f ϕrang langf τ(minus sminus t)ϕrang for ϕ isin D(R2) whereτ(st)ϕ(x y) ϕ(x minus s y minus t) Translation is invariant andcontinuous under the Alexiewicz norm

Theorem 5 Let f isin AC(R2) and (s t) isin R2 8en (i)

τ(st)f isin AC(R2) (ii) f τ(st)f and (iii)

f minus τ(st)f⟶ 0 as (s t)⟶ (0 0) ie continuity of f inthe Alexiewicz norm

Proof Let F isinBC(R2) be the primitive of f isin AC(R

2)

ampen τ(st)F is a continuous function

(i) For any ϕ isin D(R2) applying change of variable inintegral

langτ(st)f ϕrang langf τ(minus sminus t)ϕrang

1113946R

1113946R

f(x y)ϕ(x + s y + t)dxdy

1113946R

1113946R

f(x minus s y minus t)ϕ(x y)dxdy

1113946R

1113946R

z12F(x minus s y minus t)ϕ(x y)dxdy

1113946R

1113946R

z12 τ(st)F(x y)1113872 1113873ϕ(x y)dxdy

(12)

amperefore τ(st)F isinBC(R2) is the primitive of

τ(st)f(ii) Clearly Finfin τ(st)Finfin for any (s t) isin R2

amperefore f τ(st)f(iii) If f isin AC(R

2) then (f minus τ(st)f) isin AC(R

2) so that

f minus τ(st)f

F minus τ(st)F

infin sup(xy)isinR2

|F(x y) minus F(x minus s y minus t)|⟶ 0(13)

when (s t)⟶ (0 0) because of uniform continuity ofF

4 Integrable Distributional Solution forPoisson Equation

Poissonrsquos kernel and its convolution are used to find aclassical solution to the Dirichlet problem In this section weuse convolution and convergence theorem on AC(R

2) to

find integrable distributional solution If g isinBVHK(R2)

then τ(minus sminus t)g isinBVHK(R2) From integration by parts the

convolution

flowastg(x y) 1113946R

1113946R

fτ(minus sminus t)gdsdt

1113946R

1113946R

f(x minus s y minus t)g(x y)dsdt

(14)

exists for f isin AC(R2) and g isinBVHK(R

2) Proposition 2

gives some basic properties of convolution which help withour problem

International Journal of Mathematics and Mathematical Sciences 5

Proposition 2 If f isin AC(R2) and g isinBVHK(R

2) then

(i) flowastg exists on R2 and (ii) flowastg glowastf

Proof See [15] ampeorem 141

Convergence theorems on Banach space are importantwhen solving partial differential equations ampere are severalconvergence theorems in AC such as strong convergence inAC weakstrong convergence in BV and weakstrongconvergence inD Details are discussed in the plane by AngSchmitt and Vy [14] and on R by Talvila [2] Now we moveinto the convergence theorem onAC(R

2) which will help to

interchange limit of integral

Proposition 3 Let φn be a sequence inBVHK(R2) so that

φnBVltinfin and limn⟶infinφn φ pointwise on R2 for a

function φ If g isin AC(R2) then g isinBVHK(R

2) and

limn⟶infin

1113946R

1113946R

gφn 1113946R

1113946R

gφ (15)

Proof See [15] Proposition 91

Let us move into the main problem As in [25] (pg 84)Poissonrsquos equation with Dirichlet boundary condition iseasily reduced to the case where either f 0 or g 0 in (1)amperefore consider integrable distributional solution for thefollowing problem

u 0 inR3+

u g on zR3+

⎧⎨

⎩ (16)

where R3+ (x y z) isin R3 zgt 01113864 1113865 For zgt 0 consider

Poissonrsquos kernel for the upper half space in 3-dimension

K(x y z) z x

2+ y

2+ z

21113872 1113873

minus (32)

2π (17)

Note that 1113938R

1113938R

Kdxdy 1 If g isin C(R2)cap Linfin(R2)then classical solution for the problem is given by the fol-lowing convolution

u(x y z) K(x y )lowastg(x y)

1113946R

1113946R

K(x minus s y minus t )g(s t)dtds(18)

ampis convolution gives integrable distributional solutionfor (16)

Theorem 6 Let g isin AC(R2) and define u(x y z) by (18)

8en

(i) For each x y isin R and zgt 0 the convolutionK(x y )lowastg(x y) u(x y z) exists andu isin AC(R

2)

(ii) u(x y z) 0 in R3+

(iii) limz⟶0+

u(x y z) minus g(x y) 0 for x y isin R

Proof

(i) Fix x y isin R and zgt 0 ampen the function(s t)⟼K(x minus s y minus t z) is of HardyndashKrausevariation on R2 From Proposition 2 Klowastg existson the upper half space Also u Klowastg isin AC(R

2) because any function of HardyndashK-

rause variation is a multiplier for HK-integrablefunctions

(ii) Fix x y isin R and zgt 0 To get ux let xn⟶ x

be a sequence so that xn nex For any (s t) isin R2define

φn(s t) K xn minus s y minus t z( 1113857 minus K(x minus s y minus t z)

xn minus x

(19)

ampen ux limn

1113938R

1113938R

g(s t)φn(s t)dsdt SinceK(x minus s y minus t z) of HardyndashKrause bounded vari-ation so does φn(s t) for each n Since g isin AC(R

2)

from Proposition 3

ux 1113946R

1113946R

limn⟶infin

g(s t)φn(s t)dsdt

1113946R

1113946R

g(s t)Kx(x minus s y minus t z)dsdt

(20)

Repeat the same argument to obtain uxx uy uyy uzand uzz From the linearity of integral

u(x y z) 1113946R

1113946R

g(s t)K(x minus s y minus t z)dsdt

(21)

But Poisson kernel is harmonic so thatu(x y z) 0 in R3

+

6 International Journal of Mathematics and Mathematical Sciences

(iii) Let any α β isin R For Poisson kernel1113938R

1113938R

Kdxdy 1 Using FubinindashTonelli theoremand interchanging iterated integral

1113946α

minus infin1113946β

minus infin(u minus g)dxdy 1113946

α

minus infin1113946β

minus infin[u(x y z) minus g(x y)]dxdy

1113946α

minus infin1113946β

minus infin[g(x y)lowastK(x y z) minus g(x y)]dxdy

1113946α

minus infin1113946β

minus infin1113946R

1113946R

K(s t z)[g(x minus s y minus t) minus g(x y)]dsdtdxdy

1113946α

minus infin1113946β

minus infin1113946R

1113946R

z

2π s2

+ t2

+ z2

1113872 111387332 [g(x minus s y minus t) minus g(x y)]dsdtdxdy

1113946α

minus infin1113946β

minus infin1113946R

1113946R

1

2π s2

+ t2

+ 11113872 111387332 [g(x minus zs y minus zt) minus g(x y)]dsdtdxdy

1113946R

1113946R

1

2π s2

+ t2

+ 11113872 111387332 1113946

α

minus infin1113946β

minus infin[g(x minus zs y minus zt) minus g(x y)]dxdydsdt

le1113946R

1113946R

K(s t 1)g(x minus zs y minus zt) minus g(x y)dsdt

le 2g

(22)

Since α and β are arbitrary

u(x y z) minus g(x y)le1113946R

1113946R

K(s t 1)g(x minus zs y minus zt) minus g(x y)dsdtle 2g (23)

To let z⟶ 0+ apply dominated convergence theorem(HK-integral) for the sequence of functions

gn(s t) K(s t 1) g x minus1n

s y minus1n

t1113874 1113875 minus g(x y)

(24)

ampen |gn(s t)|le 2gK(s t 1) isinHK(R2) since1113938R

1113938R

K(s t 1)dsdt 1 It gives

limn⟶infin

1113946R

1113946R

gn(s t)dsdt limn⟶infin1113946

1113946R

K(s t 1)g(x minus zs y minus zt) minus g(x y)dsdt

1113946R

1113946R

limn⟶infin

gn(s t)dsdt

1113946R

1113946R

K(s t 1) limn⟶infin

g x minus1n

s y minus1n

t1113874 1113875 minus g(x y)

d s dt

(25)

From part (iii) of ampeorem 5 we get limn⟶infing(x minus

(1n)s y minus (1n)t) minus g(x y) 0 so that limit of (23)limz⟶0+ u(x y z) minus g(x y) 0 and this completes thetheorem

International Journal of Mathematics and Mathematical Sciences 7

Next we give an example which will apply ampeorem 6

Example 1 Consider the problem (16) in which

g(x y)

sin x siny

xy xne 0 andyne 0

0 x 0 ory 0

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

(26)

ampen g is HK-integrable but not absolutely integrableHence g isin AC(R

2)∖L1(R2) From ampeorem 6

u(x y z) 1113946R

1113946R

K(x minus s y minus t )g(s t)dtds

1113946R

1113946R

z

2π(x minus s)

2+(y minus t)

2+ z

21113960 1113961

minus (32) sin s sin t

st1113874 1113875dtds

(27)

exists and gives solution for the example Note thatKg isinHK(R2) since K isinBVHK(R

2) is a multiplier for

HK(R2) However the solution does not exist with theLebesgue integral For contraposition suppose thatKg isin L1(R2) ampen it follows from Fubinirsquos theorem [9](ampeorem 255) that the function

h s⟼K(x minus s y minus t z)g(s t) (28)

belongs to L1(R1) for almost all t isin R ampat is

h(s) z sin t

2πt1113874 1113875

sin s

s (x minus s)2

+(y minus t)2

+ z2

1113960 111396132 isin L

1R

11113872 1113873

h0(s)sin s

s1113874 1113875

(29)

where h0(s) (z sin t2πt)(1[(x minus s)2 + (y minus t)2 + z2]32)ampis contradicts that h(s) notin L1(R1) To see that supposeh(s) isin L1[0 1] sub L1(R1) Since h0(s) isin Linfin[0 1] is multi-plier for L1[0 1] it follows that sin ss isin L1[0 1] ampis im-plies that sin ss absolutely integrable on [0 1] which is acontradiction Hence h(s) notin L1(R1)

Even HK-integral exists in (27) it may not be able toevaluate and get explicit function In that case numericalintegration can be applied Numerical integration should bedone with respect to s and t while fixing x y and z details in[26] ampis reduces solving the partial differential equationinto applying numerical integration for the double integral(27)

5 Discussion

In this work integrable distributional solution is obtainedfor Poisson equation with the Dirichlet boundary conditionin the upper half space Boundary conditions are taken in theAlexiewicz norm Lebesgue integral uses for the classicalsolution to Poisson equation Because of the proper inclu-sion L1 ⊊ HR we may use HK-integral to solve Poissonequation However there are some limitations due to the

absence of any natural topology ampat is HK is not aBanach space nevertheless there is a natural seminormcalled Alexiewicz ampis is one of the main reasons to moveinto the space of integrable distribution AC(R

2) which

contains HK and leads to a Banach space with Alexiewicznorm

Integrable distribution is defined as derivative (distri-butional sense) of function in BC(R

2) In this definition

any f isin AC(R2) identifies with the unique primitive

F isinBC(R2) ampis unique primitive allows to define a

Alexiewicz norm middot on AC(R2) and leads to a Banach

space Alexiewicz norm does not induce an inner product inAC(R

2) because it does not satisfy the parallelogram law

amperefore AC(R2) is not a Hilbert space under the norm

middot However the space (AC(R2) middot) is isometrically iso-

morphic to the space (BC(R2) middotinfin) Integral on AC(R

2)

is uniquely defined by uniqueness of the primitive AC(R2)

is the completion of L1(R2) (or HK(R2)) with respect tothe corresponding Alexiewicz norm

In one-dimensional HK-integral space of boundedvariation functions BV is multipliers for HK In mul-tidimension case there are several extensions to variation offunction To establish integration by parts formula andmultipliers for AC(R

2) we begin with HardyndashKrause

variation ampe space of HardyndashKrause bounded variationfunctions BVHK(R

2) as a multiplier for AC(R

2) ampis

multiplier is important to define convolution and obtainconvergence theorem BVHK(R

2) is used to define con-

volution and obtain convergence theorem on AC(R2)

ampere are several convergence theorems in AC such asstrong convergence in AC weakstrong convergence inBV and weakstrong convergence in D Convergencetheorem on AC(R

2) and dominated convergence theorem

in HK(R2) are used to obtain integrable distributionalsolution Finding of this article gives solution for Poissonequation for broader initial condition in the upper halfspace Dirichlet problem in R3 is considered and the so-lution has obtained by HK-integration where it is notpossible from Lebesgue integral Because of functionrsquoscomplexity it may not be able to have solution in explicit

8 International Journal of Mathematics and Mathematical Sciences

form It would help keep an analytical outlook in the so-lution which would be better than solving the partial dif-ferential equation by numerical methods such as finitedifference methods and finite element methods

ampere is potential to find integrable distributional so-lution for Poisson equation in other boundary conditionsNeumann and Robin But it needs to develop suitableconvergence theorems on AC and its multipliers BVHKAlso it is possible to consider Poisson equation in unit ballwith the relevant Poisson kernel Talvila in [15] gives someconstruction to develop integrable distribution on Rn in-tegration by parts and mean value theorem and definesAC(R

n) and BC(R

n) for multidimensional case Hence

integrable distributional solution is possible to obtain forPoisson equation with Dirichlet boundary condition in Rn

+

with suitable convergence theorem

Data Availability

No data were used to support this study

Conflicts of Interest

ampe authors declare that they have no conflicts of interest

References

[1] R L Burden and J D Faires Numerical Analysis ampomsonBrooksCole USA 8th edition 2005

[2] C Anitescu E Atroshchenko N Alajlan and T RabczukldquoArtificial neural network methods for the solution of secondorder boundary value problemsrdquo Computers Materials ampContinua vol 59 no 1 pp 345ndash359 2019

[3] D A Leon-Velasco M M Morın-Castillo J J Oliveros-Oliveros T Perez-Becerra and J A Escamilla-Reyna ldquoNu-merical solution of some differential equations withHenstockndash Kurzweil functionsrdquo Journal of Function Spacesvol 2019 Article ID 8948570 9 pages 2019

[4] G Green ldquoAn essay on the application of mathematicalanalysis to the theories of electricity and magnetismrdquo Journalfur die reine und angewandte Mathematik (Crelles Journal)vol 1850 no 39 pp 73ndash89 1850

[5] L C Evans Partial Differential Equations American Math-ematical Society Providence RI USA 1998

[6] R Bartle A Modern 8eory of Integration American Math-ematical Society RI Providence RI USA 2001

[7] R Henstock 8e General 8eory of Integration OxfordUniversity Press Oxford UK 1991

[8] C Swartz Introduction to Gauge Integrals World ScientificSingapore 2001

[9] T-Y LeeHenstockndashKurzweil Integration on Euclidean SpacesWorld Scientific Singapore 2011

[10] E Talvila ldquoContinuity in the Alexiewicz normrdquoMathematicaBohemica vol 131 no 2 pp 189ndash196 2006

[11] E Talvila ldquoEstimates of henstock-kurzweil Poisson integralsrdquoCanadian Mathematical Bulletin vol 48 no 1 pp 133ndash1462005

[12] A Alexiewicz ldquoLinear functionals on Denjoy-integrablefunctionsrdquo Colloquium Mathematicum vol 1 no 4pp 289ndash293 1948

[13] P Mikusinksi and K Ostaszewski ldquoEmbedding Henstockintegrable functions into the space of Schwartz distributionsrdquoReal Analysis Exchange vol 14 no -89 pp 24ndash29 1988

[14] D D Ang K Schmitt and L K Vy ldquoA multidimensionalanalogue of the denjoyndashperronndashhenstockndashkurzweil integralrdquo8e Bulletin of the Belgian Mathematical Society mdash SimonStevin vol 4 pp 355ndash371 1997

[15] E Talvila ldquoampe continuous primitive integral in the planerdquoReal Analysis Exchange vol 45 no 2 pp 283ndash326 2020

[16] S Heikkila ldquoOn nonlinear distributional and impulsiveCauchy problemsrdquo Nonlinear Analysis 8eory Methods ampApplications vol 75 no 2 pp 852ndash870 2012

[17] B Liang G Ye W Liu and H Wang ldquoOn second ordernonlinear boundary value problems and the distributionalHenstock-Kurzweil integralrdquo Boundary Value Problemvol 2015 2015

[18] A Sikorska-Nowak ldquoExistence theory for integrodifferentialequations and henstock-kurzweil integral in Banach spacesrdquoJournal of Applied Mathematics vol 2007 Article ID 3157212 pages 2007

[19] M Tvrdy ldquoLinear distributional differential equations of thesecond orderrdquo Math Bohemvol 119 no 4 pp 415ndash4361994

[20] L Schwartz 8eorie des Distributions Hermann ParisFrance 1950

[21] E Talvila ldquoampe distributional denjoy integralrdquo Real AnalysisExchange vol 33 no 1 pp 51ndash82 2008

[22] K Ostaszewski ldquoampe space of Henstock integrable functionsof two variablesrdquo International Journal of Mathematics andMathematical Sciences vol 11 no 1 pp 15ndash21 1988

[23] C R Adams and J A Clarkson ldquoOn definitions of boundedvariation for functions of two variablesrdquo Transactions of theAmerican Mathematical Society vol 35 no 4 pp 824ndash8541933

[24] B Bongiorno ldquoampe henstockndashkurzweil integralrdquo inHandbookof Measure 8eory E Pap Ed pp 589ndash615 North HollandDordrecht Netherlands 2002

[25] G B Folland Introduction to Partial Differential EquationsPrinceton University Press Princeton NY USA 1995

[26] S M Aljassas ldquoNumerical methods for evaluation of doubleintegrals with continuous integrandsrdquo International Journalof Pure and Applied Mathematics vol 119 no 10 2018

International Journal of Mathematics and Mathematical Sciences 9

HardyndashKrause variation is a multiplier for HK-integrablefunctions see [9] amperefore define following integration byparts as a two-dimensional case which is presented in ([9]ampeorem 659)

Definition 4 (integration by parts) Let f isin AC(R2) with its

primitive F isinBC(R2) If g isinBVHK(R

2) then define

integration by parts on [a b] times [c d]subeR2

1113946b

a1113946

d

cfg F(a c)g(a c) + F(b d)g(b d) minus F(a d)g(a d) minus F(b c)g(b c)

minus 1113946b

aF(x d)d1g(x d) + 1113946

b

aF(x c)d1g(x c)

minus 1113946d

cF(b y)d2g(b y) + 1113946

d

cF(a y)d2g(a y)

+ 1113946b

a1113946

d

cF(x y)d12g(x y)

(11)

Here dg(x y) is HenstockndashStieltjes integral of relevantvariables see in detail [24]ampis integration by parts formulainduces the space of HardyndashKrause bounded variationfunctions BVHK(R

2) as a multiplier for AC(R

2) ampis

multiplier is important to define convolution and obtainconvergence theorem

It needs to obtain continuity of the Alexiewicz normsince the Dirichlet boundary condition is taken on it For any(s t) isin R2 the translation of f isin AC(R

2)subeDprime(R

2) is de-fined by langτ(st)f ϕrang langf τ(minus sminus t)ϕrang for ϕ isin D(R2) whereτ(st)ϕ(x y) ϕ(x minus s y minus t) Translation is invariant andcontinuous under the Alexiewicz norm

Theorem 5 Let f isin AC(R2) and (s t) isin R2 8en (i)

τ(st)f isin AC(R2) (ii) f τ(st)f and (iii)

f minus τ(st)f⟶ 0 as (s t)⟶ (0 0) ie continuity of f inthe Alexiewicz norm

Proof Let F isinBC(R2) be the primitive of f isin AC(R

2)

ampen τ(st)F is a continuous function

(i) For any ϕ isin D(R2) applying change of variable inintegral

langτ(st)f ϕrang langf τ(minus sminus t)ϕrang

1113946R

1113946R

f(x y)ϕ(x + s y + t)dxdy

1113946R

1113946R

f(x minus s y minus t)ϕ(x y)dxdy

1113946R

1113946R

z12F(x minus s y minus t)ϕ(x y)dxdy

1113946R

1113946R

z12 τ(st)F(x y)1113872 1113873ϕ(x y)dxdy

(12)

amperefore τ(st)F isinBC(R2) is the primitive of

τ(st)f(ii) Clearly Finfin τ(st)Finfin for any (s t) isin R2

amperefore f τ(st)f(iii) If f isin AC(R

2) then (f minus τ(st)f) isin AC(R

2) so that

f minus τ(st)f

F minus τ(st)F

infin sup(xy)isinR2

|F(x y) minus F(x minus s y minus t)|⟶ 0(13)

when (s t)⟶ (0 0) because of uniform continuity ofF

4 Integrable Distributional Solution forPoisson Equation

Poissonrsquos kernel and its convolution are used to find aclassical solution to the Dirichlet problem In this section weuse convolution and convergence theorem on AC(R

2) to

find integrable distributional solution If g isinBVHK(R2)

then τ(minus sminus t)g isinBVHK(R2) From integration by parts the

convolution

flowastg(x y) 1113946R

1113946R

fτ(minus sminus t)gdsdt

1113946R

1113946R

f(x minus s y minus t)g(x y)dsdt

(14)

exists for f isin AC(R2) and g isinBVHK(R

2) Proposition 2

gives some basic properties of convolution which help withour problem

International Journal of Mathematics and Mathematical Sciences 5

Proposition 2 If f isin AC(R2) and g isinBVHK(R

2) then

(i) flowastg exists on R2 and (ii) flowastg glowastf

Proof See [15] ampeorem 141

Convergence theorems on Banach space are importantwhen solving partial differential equations ampere are severalconvergence theorems in AC such as strong convergence inAC weakstrong convergence in BV and weakstrongconvergence inD Details are discussed in the plane by AngSchmitt and Vy [14] and on R by Talvila [2] Now we moveinto the convergence theorem onAC(R

2) which will help to

interchange limit of integral

Proposition 3 Let φn be a sequence inBVHK(R2) so that

φnBVltinfin and limn⟶infinφn φ pointwise on R2 for a

function φ If g isin AC(R2) then g isinBVHK(R

2) and

limn⟶infin

1113946R

1113946R

gφn 1113946R

1113946R

gφ (15)

Proof See [15] Proposition 91

Let us move into the main problem As in [25] (pg 84)Poissonrsquos equation with Dirichlet boundary condition iseasily reduced to the case where either f 0 or g 0 in (1)amperefore consider integrable distributional solution for thefollowing problem

u 0 inR3+

u g on zR3+

⎧⎨

⎩ (16)

where R3+ (x y z) isin R3 zgt 01113864 1113865 For zgt 0 consider

Poissonrsquos kernel for the upper half space in 3-dimension

K(x y z) z x

2+ y

2+ z

21113872 1113873

minus (32)

2π (17)

Note that 1113938R

1113938R

Kdxdy 1 If g isin C(R2)cap Linfin(R2)then classical solution for the problem is given by the fol-lowing convolution

u(x y z) K(x y )lowastg(x y)

1113946R

1113946R

K(x minus s y minus t )g(s t)dtds(18)

ampis convolution gives integrable distributional solutionfor (16)

Theorem 6 Let g isin AC(R2) and define u(x y z) by (18)

8en

(i) For each x y isin R and zgt 0 the convolutionK(x y )lowastg(x y) u(x y z) exists andu isin AC(R

2)

(ii) u(x y z) 0 in R3+

(iii) limz⟶0+

u(x y z) minus g(x y) 0 for x y isin R

Proof

(i) Fix x y isin R and zgt 0 ampen the function(s t)⟼K(x minus s y minus t z) is of HardyndashKrausevariation on R2 From Proposition 2 Klowastg existson the upper half space Also u Klowastg isin AC(R

2) because any function of HardyndashK-

rause variation is a multiplier for HK-integrablefunctions

(ii) Fix x y isin R and zgt 0 To get ux let xn⟶ x

be a sequence so that xn nex For any (s t) isin R2define

φn(s t) K xn minus s y minus t z( 1113857 minus K(x minus s y minus t z)

xn minus x

(19)

ampen ux limn

1113938R

1113938R

g(s t)φn(s t)dsdt SinceK(x minus s y minus t z) of HardyndashKrause bounded vari-ation so does φn(s t) for each n Since g isin AC(R

2)

from Proposition 3

ux 1113946R

1113946R

limn⟶infin

g(s t)φn(s t)dsdt

1113946R

1113946R

g(s t)Kx(x minus s y minus t z)dsdt

(20)

Repeat the same argument to obtain uxx uy uyy uzand uzz From the linearity of integral

u(x y z) 1113946R

1113946R

g(s t)K(x minus s y minus t z)dsdt

(21)

But Poisson kernel is harmonic so thatu(x y z) 0 in R3

+

6 International Journal of Mathematics and Mathematical Sciences

(iii) Let any α β isin R For Poisson kernel1113938R

1113938R

Kdxdy 1 Using FubinindashTonelli theoremand interchanging iterated integral

1113946α

minus infin1113946β

minus infin(u minus g)dxdy 1113946

α

minus infin1113946β

minus infin[u(x y z) minus g(x y)]dxdy

1113946α

minus infin1113946β

minus infin[g(x y)lowastK(x y z) minus g(x y)]dxdy

1113946α

minus infin1113946β

minus infin1113946R

1113946R

K(s t z)[g(x minus s y minus t) minus g(x y)]dsdtdxdy

1113946α

minus infin1113946β

minus infin1113946R

1113946R

z

2π s2

+ t2

+ z2

1113872 111387332 [g(x minus s y minus t) minus g(x y)]dsdtdxdy

1113946α

minus infin1113946β

minus infin1113946R

1113946R

1

2π s2

+ t2

+ 11113872 111387332 [g(x minus zs y minus zt) minus g(x y)]dsdtdxdy

1113946R

1113946R

1

2π s2

+ t2

+ 11113872 111387332 1113946

α

minus infin1113946β

minus infin[g(x minus zs y minus zt) minus g(x y)]dxdydsdt

le1113946R

1113946R

K(s t 1)g(x minus zs y minus zt) minus g(x y)dsdt

le 2g

(22)

Since α and β are arbitrary

u(x y z) minus g(x y)le1113946R

1113946R

K(s t 1)g(x minus zs y minus zt) minus g(x y)dsdtle 2g (23)

To let z⟶ 0+ apply dominated convergence theorem(HK-integral) for the sequence of functions

gn(s t) K(s t 1) g x minus1n

s y minus1n

t1113874 1113875 minus g(x y)

(24)

ampen |gn(s t)|le 2gK(s t 1) isinHK(R2) since1113938R

1113938R

K(s t 1)dsdt 1 It gives

limn⟶infin

1113946R

1113946R

gn(s t)dsdt limn⟶infin1113946

1113946R

K(s t 1)g(x minus zs y minus zt) minus g(x y)dsdt

1113946R

1113946R

limn⟶infin

gn(s t)dsdt

1113946R

1113946R

K(s t 1) limn⟶infin

g x minus1n

s y minus1n

t1113874 1113875 minus g(x y)

d s dt

(25)

From part (iii) of ampeorem 5 we get limn⟶infing(x minus

(1n)s y minus (1n)t) minus g(x y) 0 so that limit of (23)limz⟶0+ u(x y z) minus g(x y) 0 and this completes thetheorem

International Journal of Mathematics and Mathematical Sciences 7

Next we give an example which will apply ampeorem 6

Example 1 Consider the problem (16) in which

g(x y)

sin x siny

xy xne 0 andyne 0

0 x 0 ory 0

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

(26)

ampen g is HK-integrable but not absolutely integrableHence g isin AC(R

2)∖L1(R2) From ampeorem 6

u(x y z) 1113946R

1113946R

K(x minus s y minus t )g(s t)dtds

1113946R

1113946R

z

2π(x minus s)

2+(y minus t)

2+ z

21113960 1113961

minus (32) sin s sin t

st1113874 1113875dtds

(27)

exists and gives solution for the example Note thatKg isinHK(R2) since K isinBVHK(R

2) is a multiplier for

HK(R2) However the solution does not exist with theLebesgue integral For contraposition suppose thatKg isin L1(R2) ampen it follows from Fubinirsquos theorem [9](ampeorem 255) that the function

h s⟼K(x minus s y minus t z)g(s t) (28)

belongs to L1(R1) for almost all t isin R ampat is

h(s) z sin t

2πt1113874 1113875

sin s

s (x minus s)2

+(y minus t)2

+ z2

1113960 111396132 isin L

1R

11113872 1113873

h0(s)sin s

s1113874 1113875

(29)

where h0(s) (z sin t2πt)(1[(x minus s)2 + (y minus t)2 + z2]32)ampis contradicts that h(s) notin L1(R1) To see that supposeh(s) isin L1[0 1] sub L1(R1) Since h0(s) isin Linfin[0 1] is multi-plier for L1[0 1] it follows that sin ss isin L1[0 1] ampis im-plies that sin ss absolutely integrable on [0 1] which is acontradiction Hence h(s) notin L1(R1)

Even HK-integral exists in (27) it may not be able toevaluate and get explicit function In that case numericalintegration can be applied Numerical integration should bedone with respect to s and t while fixing x y and z details in[26] ampis reduces solving the partial differential equationinto applying numerical integration for the double integral(27)

5 Discussion

In this work integrable distributional solution is obtainedfor Poisson equation with the Dirichlet boundary conditionin the upper half space Boundary conditions are taken in theAlexiewicz norm Lebesgue integral uses for the classicalsolution to Poisson equation Because of the proper inclu-sion L1 ⊊ HR we may use HK-integral to solve Poissonequation However there are some limitations due to the

absence of any natural topology ampat is HK is not aBanach space nevertheless there is a natural seminormcalled Alexiewicz ampis is one of the main reasons to moveinto the space of integrable distribution AC(R

2) which

contains HK and leads to a Banach space with Alexiewicznorm

Integrable distribution is defined as derivative (distri-butional sense) of function in BC(R

2) In this definition

any f isin AC(R2) identifies with the unique primitive

F isinBC(R2) ampis unique primitive allows to define a

Alexiewicz norm middot on AC(R2) and leads to a Banach

space Alexiewicz norm does not induce an inner product inAC(R

2) because it does not satisfy the parallelogram law

amperefore AC(R2) is not a Hilbert space under the norm

middot However the space (AC(R2) middot) is isometrically iso-

morphic to the space (BC(R2) middotinfin) Integral on AC(R

2)

is uniquely defined by uniqueness of the primitive AC(R2)

is the completion of L1(R2) (or HK(R2)) with respect tothe corresponding Alexiewicz norm

In one-dimensional HK-integral space of boundedvariation functions BV is multipliers for HK In mul-tidimension case there are several extensions to variation offunction To establish integration by parts formula andmultipliers for AC(R

2) we begin with HardyndashKrause

variation ampe space of HardyndashKrause bounded variationfunctions BVHK(R

2) as a multiplier for AC(R

2) ampis

multiplier is important to define convolution and obtainconvergence theorem BVHK(R

2) is used to define con-

volution and obtain convergence theorem on AC(R2)

ampere are several convergence theorems in AC such asstrong convergence in AC weakstrong convergence inBV and weakstrong convergence in D Convergencetheorem on AC(R

2) and dominated convergence theorem

in HK(R2) are used to obtain integrable distributionalsolution Finding of this article gives solution for Poissonequation for broader initial condition in the upper halfspace Dirichlet problem in R3 is considered and the so-lution has obtained by HK-integration where it is notpossible from Lebesgue integral Because of functionrsquoscomplexity it may not be able to have solution in explicit

8 International Journal of Mathematics and Mathematical Sciences

form It would help keep an analytical outlook in the so-lution which would be better than solving the partial dif-ferential equation by numerical methods such as finitedifference methods and finite element methods

ampere is potential to find integrable distributional so-lution for Poisson equation in other boundary conditionsNeumann and Robin But it needs to develop suitableconvergence theorems on AC and its multipliers BVHKAlso it is possible to consider Poisson equation in unit ballwith the relevant Poisson kernel Talvila in [15] gives someconstruction to develop integrable distribution on Rn in-tegration by parts and mean value theorem and definesAC(R

n) and BC(R

n) for multidimensional case Hence

integrable distributional solution is possible to obtain forPoisson equation with Dirichlet boundary condition in Rn

+

with suitable convergence theorem

Data Availability

No data were used to support this study

Conflicts of Interest

ampe authors declare that they have no conflicts of interest

References

[1] R L Burden and J D Faires Numerical Analysis ampomsonBrooksCole USA 8th edition 2005

[2] C Anitescu E Atroshchenko N Alajlan and T RabczukldquoArtificial neural network methods for the solution of secondorder boundary value problemsrdquo Computers Materials ampContinua vol 59 no 1 pp 345ndash359 2019

[3] D A Leon-Velasco M M Morın-Castillo J J Oliveros-Oliveros T Perez-Becerra and J A Escamilla-Reyna ldquoNu-merical solution of some differential equations withHenstockndash Kurzweil functionsrdquo Journal of Function Spacesvol 2019 Article ID 8948570 9 pages 2019

[4] G Green ldquoAn essay on the application of mathematicalanalysis to the theories of electricity and magnetismrdquo Journalfur die reine und angewandte Mathematik (Crelles Journal)vol 1850 no 39 pp 73ndash89 1850

[5] L C Evans Partial Differential Equations American Math-ematical Society Providence RI USA 1998

[6] R Bartle A Modern 8eory of Integration American Math-ematical Society RI Providence RI USA 2001

[7] R Henstock 8e General 8eory of Integration OxfordUniversity Press Oxford UK 1991

[8] C Swartz Introduction to Gauge Integrals World ScientificSingapore 2001

[9] T-Y LeeHenstockndashKurzweil Integration on Euclidean SpacesWorld Scientific Singapore 2011

[10] E Talvila ldquoContinuity in the Alexiewicz normrdquoMathematicaBohemica vol 131 no 2 pp 189ndash196 2006

[11] E Talvila ldquoEstimates of henstock-kurzweil Poisson integralsrdquoCanadian Mathematical Bulletin vol 48 no 1 pp 133ndash1462005

[12] A Alexiewicz ldquoLinear functionals on Denjoy-integrablefunctionsrdquo Colloquium Mathematicum vol 1 no 4pp 289ndash293 1948

[13] P Mikusinksi and K Ostaszewski ldquoEmbedding Henstockintegrable functions into the space of Schwartz distributionsrdquoReal Analysis Exchange vol 14 no -89 pp 24ndash29 1988

[14] D D Ang K Schmitt and L K Vy ldquoA multidimensionalanalogue of the denjoyndashperronndashhenstockndashkurzweil integralrdquo8e Bulletin of the Belgian Mathematical Society mdash SimonStevin vol 4 pp 355ndash371 1997

[15] E Talvila ldquoampe continuous primitive integral in the planerdquoReal Analysis Exchange vol 45 no 2 pp 283ndash326 2020

[16] S Heikkila ldquoOn nonlinear distributional and impulsiveCauchy problemsrdquo Nonlinear Analysis 8eory Methods ampApplications vol 75 no 2 pp 852ndash870 2012

[17] B Liang G Ye W Liu and H Wang ldquoOn second ordernonlinear boundary value problems and the distributionalHenstock-Kurzweil integralrdquo Boundary Value Problemvol 2015 2015

[18] A Sikorska-Nowak ldquoExistence theory for integrodifferentialequations and henstock-kurzweil integral in Banach spacesrdquoJournal of Applied Mathematics vol 2007 Article ID 3157212 pages 2007

[19] M Tvrdy ldquoLinear distributional differential equations of thesecond orderrdquo Math Bohemvol 119 no 4 pp 415ndash4361994

[20] L Schwartz 8eorie des Distributions Hermann ParisFrance 1950

[21] E Talvila ldquoampe distributional denjoy integralrdquo Real AnalysisExchange vol 33 no 1 pp 51ndash82 2008

[22] K Ostaszewski ldquoampe space of Henstock integrable functionsof two variablesrdquo International Journal of Mathematics andMathematical Sciences vol 11 no 1 pp 15ndash21 1988

[23] C R Adams and J A Clarkson ldquoOn definitions of boundedvariation for functions of two variablesrdquo Transactions of theAmerican Mathematical Society vol 35 no 4 pp 824ndash8541933

[24] B Bongiorno ldquoampe henstockndashkurzweil integralrdquo inHandbookof Measure 8eory E Pap Ed pp 589ndash615 North HollandDordrecht Netherlands 2002

[25] G B Folland Introduction to Partial Differential EquationsPrinceton University Press Princeton NY USA 1995

[26] S M Aljassas ldquoNumerical methods for evaluation of doubleintegrals with continuous integrandsrdquo International Journalof Pure and Applied Mathematics vol 119 no 10 2018

International Journal of Mathematics and Mathematical Sciences 9

Proposition 2 If f isin AC(R2) and g isinBVHK(R

2) then

(i) flowastg exists on R2 and (ii) flowastg glowastf

Proof See [15] ampeorem 141

Convergence theorems on Banach space are importantwhen solving partial differential equations ampere are severalconvergence theorems in AC such as strong convergence inAC weakstrong convergence in BV and weakstrongconvergence inD Details are discussed in the plane by AngSchmitt and Vy [14] and on R by Talvila [2] Now we moveinto the convergence theorem onAC(R

2) which will help to

interchange limit of integral

Proposition 3 Let φn be a sequence inBVHK(R2) so that

φnBVltinfin and limn⟶infinφn φ pointwise on R2 for a

function φ If g isin AC(R2) then g isinBVHK(R

2) and

limn⟶infin

1113946R

1113946R

gφn 1113946R

1113946R

gφ (15)

Proof See [15] Proposition 91

Let us move into the main problem As in [25] (pg 84)Poissonrsquos equation with Dirichlet boundary condition iseasily reduced to the case where either f 0 or g 0 in (1)amperefore consider integrable distributional solution for thefollowing problem

u 0 inR3+

u g on zR3+

⎧⎨

⎩ (16)

where R3+ (x y z) isin R3 zgt 01113864 1113865 For zgt 0 consider

Poissonrsquos kernel for the upper half space in 3-dimension

K(x y z) z x

2+ y

2+ z

21113872 1113873

minus (32)

2π (17)

Note that 1113938R

1113938R

Kdxdy 1 If g isin C(R2)cap Linfin(R2)then classical solution for the problem is given by the fol-lowing convolution

u(x y z) K(x y )lowastg(x y)

1113946R

1113946R

K(x minus s y minus t )g(s t)dtds(18)

ampis convolution gives integrable distributional solutionfor (16)

Theorem 6 Let g isin AC(R2) and define u(x y z) by (18)

8en

(i) For each x y isin R and zgt 0 the convolutionK(x y )lowastg(x y) u(x y z) exists andu isin AC(R

2)

(ii) u(x y z) 0 in R3+

(iii) limz⟶0+

u(x y z) minus g(x y) 0 for x y isin R

Proof

(i) Fix x y isin R and zgt 0 ampen the function(s t)⟼K(x minus s y minus t z) is of HardyndashKrausevariation on R2 From Proposition 2 Klowastg existson the upper half space Also u Klowastg isin AC(R

2) because any function of HardyndashK-

rause variation is a multiplier for HK-integrablefunctions

(ii) Fix x y isin R and zgt 0 To get ux let xn⟶ x

be a sequence so that xn nex For any (s t) isin R2define

φn(s t) K xn minus s y minus t z( 1113857 minus K(x minus s y minus t z)

xn minus x

(19)

ampen ux limn

1113938R

1113938R

g(s t)φn(s t)dsdt SinceK(x minus s y minus t z) of HardyndashKrause bounded vari-ation so does φn(s t) for each n Since g isin AC(R

2)

from Proposition 3

ux 1113946R

1113946R

limn⟶infin

g(s t)φn(s t)dsdt

1113946R

1113946R

g(s t)Kx(x minus s y minus t z)dsdt

(20)

Repeat the same argument to obtain uxx uy uyy uzand uzz From the linearity of integral

u(x y z) 1113946R

1113946R

g(s t)K(x minus s y minus t z)dsdt

(21)

But Poisson kernel is harmonic so thatu(x y z) 0 in R3

+

6 International Journal of Mathematics and Mathematical Sciences

(iii) Let any α β isin R For Poisson kernel1113938R

1113938R

Kdxdy 1 Using FubinindashTonelli theoremand interchanging iterated integral

1113946α

minus infin1113946β

minus infin(u minus g)dxdy 1113946

α

minus infin1113946β

minus infin[u(x y z) minus g(x y)]dxdy

1113946α

minus infin1113946β

minus infin[g(x y)lowastK(x y z) minus g(x y)]dxdy

1113946α

minus infin1113946β

minus infin1113946R

1113946R

K(s t z)[g(x minus s y minus t) minus g(x y)]dsdtdxdy

1113946α

minus infin1113946β

minus infin1113946R

1113946R

z

2π s2

+ t2

+ z2

1113872 111387332 [g(x minus s y minus t) minus g(x y)]dsdtdxdy

1113946α

minus infin1113946β

minus infin1113946R

1113946R

1

2π s2

+ t2

+ 11113872 111387332 [g(x minus zs y minus zt) minus g(x y)]dsdtdxdy

1113946R

1113946R

1

2π s2

+ t2

+ 11113872 111387332 1113946

α

minus infin1113946β

minus infin[g(x minus zs y minus zt) minus g(x y)]dxdydsdt

le1113946R

1113946R

K(s t 1)g(x minus zs y minus zt) minus g(x y)dsdt

le 2g

(22)

Since α and β are arbitrary

u(x y z) minus g(x y)le1113946R

1113946R

K(s t 1)g(x minus zs y minus zt) minus g(x y)dsdtle 2g (23)

To let z⟶ 0+ apply dominated convergence theorem(HK-integral) for the sequence of functions

gn(s t) K(s t 1) g x minus1n

s y minus1n

t1113874 1113875 minus g(x y)

(24)

ampen |gn(s t)|le 2gK(s t 1) isinHK(R2) since1113938R

1113938R

K(s t 1)dsdt 1 It gives

limn⟶infin

1113946R

1113946R

gn(s t)dsdt limn⟶infin1113946

1113946R

K(s t 1)g(x minus zs y minus zt) minus g(x y)dsdt

1113946R

1113946R

limn⟶infin

gn(s t)dsdt

1113946R

1113946R

K(s t 1) limn⟶infin

g x minus1n

s y minus1n

t1113874 1113875 minus g(x y)

d s dt

(25)

From part (iii) of ampeorem 5 we get limn⟶infing(x minus

(1n)s y minus (1n)t) minus g(x y) 0 so that limit of (23)limz⟶0+ u(x y z) minus g(x y) 0 and this completes thetheorem

International Journal of Mathematics and Mathematical Sciences 7

Next we give an example which will apply ampeorem 6

Example 1 Consider the problem (16) in which

g(x y)

sin x siny

xy xne 0 andyne 0

0 x 0 ory 0

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

(26)

ampen g is HK-integrable but not absolutely integrableHence g isin AC(R

2)∖L1(R2) From ampeorem 6

u(x y z) 1113946R

1113946R

K(x minus s y minus t )g(s t)dtds

1113946R

1113946R

z

2π(x minus s)

2+(y minus t)

2+ z

21113960 1113961

minus (32) sin s sin t

st1113874 1113875dtds

(27)

exists and gives solution for the example Note thatKg isinHK(R2) since K isinBVHK(R

2) is a multiplier for

HK(R2) However the solution does not exist with theLebesgue integral For contraposition suppose thatKg isin L1(R2) ampen it follows from Fubinirsquos theorem [9](ampeorem 255) that the function

h s⟼K(x minus s y minus t z)g(s t) (28)

belongs to L1(R1) for almost all t isin R ampat is

h(s) z sin t

2πt1113874 1113875

sin s

s (x minus s)2

+(y minus t)2

+ z2

1113960 111396132 isin L

1R

11113872 1113873

h0(s)sin s

s1113874 1113875

(29)

where h0(s) (z sin t2πt)(1[(x minus s)2 + (y minus t)2 + z2]32)ampis contradicts that h(s) notin L1(R1) To see that supposeh(s) isin L1[0 1] sub L1(R1) Since h0(s) isin Linfin[0 1] is multi-plier for L1[0 1] it follows that sin ss isin L1[0 1] ampis im-plies that sin ss absolutely integrable on [0 1] which is acontradiction Hence h(s) notin L1(R1)

Even HK-integral exists in (27) it may not be able toevaluate and get explicit function In that case numericalintegration can be applied Numerical integration should bedone with respect to s and t while fixing x y and z details in[26] ampis reduces solving the partial differential equationinto applying numerical integration for the double integral(27)

5 Discussion

In this work integrable distributional solution is obtainedfor Poisson equation with the Dirichlet boundary conditionin the upper half space Boundary conditions are taken in theAlexiewicz norm Lebesgue integral uses for the classicalsolution to Poisson equation Because of the proper inclu-sion L1 ⊊ HR we may use HK-integral to solve Poissonequation However there are some limitations due to the

absence of any natural topology ampat is HK is not aBanach space nevertheless there is a natural seminormcalled Alexiewicz ampis is one of the main reasons to moveinto the space of integrable distribution AC(R

2) which

contains HK and leads to a Banach space with Alexiewicznorm

Integrable distribution is defined as derivative (distri-butional sense) of function in BC(R

2) In this definition

any f isin AC(R2) identifies with the unique primitive

F isinBC(R2) ampis unique primitive allows to define a

Alexiewicz norm middot on AC(R2) and leads to a Banach

space Alexiewicz norm does not induce an inner product inAC(R

2) because it does not satisfy the parallelogram law

amperefore AC(R2) is not a Hilbert space under the norm

middot However the space (AC(R2) middot) is isometrically iso-

morphic to the space (BC(R2) middotinfin) Integral on AC(R

2)

is uniquely defined by uniqueness of the primitive AC(R2)

is the completion of L1(R2) (or HK(R2)) with respect tothe corresponding Alexiewicz norm

In one-dimensional HK-integral space of boundedvariation functions BV is multipliers for HK In mul-tidimension case there are several extensions to variation offunction To establish integration by parts formula andmultipliers for AC(R

2) we begin with HardyndashKrause

variation ampe space of HardyndashKrause bounded variationfunctions BVHK(R

2) as a multiplier for AC(R

2) ampis

multiplier is important to define convolution and obtainconvergence theorem BVHK(R

2) is used to define con-

volution and obtain convergence theorem on AC(R2)

ampere are several convergence theorems in AC such asstrong convergence in AC weakstrong convergence inBV and weakstrong convergence in D Convergencetheorem on AC(R

2) and dominated convergence theorem

in HK(R2) are used to obtain integrable distributionalsolution Finding of this article gives solution for Poissonequation for broader initial condition in the upper halfspace Dirichlet problem in R3 is considered and the so-lution has obtained by HK-integration where it is notpossible from Lebesgue integral Because of functionrsquoscomplexity it may not be able to have solution in explicit

8 International Journal of Mathematics and Mathematical Sciences

form It would help keep an analytical outlook in the so-lution which would be better than solving the partial dif-ferential equation by numerical methods such as finitedifference methods and finite element methods

ampere is potential to find integrable distributional so-lution for Poisson equation in other boundary conditionsNeumann and Robin But it needs to develop suitableconvergence theorems on AC and its multipliers BVHKAlso it is possible to consider Poisson equation in unit ballwith the relevant Poisson kernel Talvila in [15] gives someconstruction to develop integrable distribution on Rn in-tegration by parts and mean value theorem and definesAC(R

n) and BC(R

n) for multidimensional case Hence

integrable distributional solution is possible to obtain forPoisson equation with Dirichlet boundary condition in Rn

+

with suitable convergence theorem

Data Availability

No data were used to support this study

Conflicts of Interest

ampe authors declare that they have no conflicts of interest

References

[1] R L Burden and J D Faires Numerical Analysis ampomsonBrooksCole USA 8th edition 2005

[2] C Anitescu E Atroshchenko N Alajlan and T RabczukldquoArtificial neural network methods for the solution of secondorder boundary value problemsrdquo Computers Materials ampContinua vol 59 no 1 pp 345ndash359 2019

[3] D A Leon-Velasco M M Morın-Castillo J J Oliveros-Oliveros T Perez-Becerra and J A Escamilla-Reyna ldquoNu-merical solution of some differential equations withHenstockndash Kurzweil functionsrdquo Journal of Function Spacesvol 2019 Article ID 8948570 9 pages 2019

[4] G Green ldquoAn essay on the application of mathematicalanalysis to the theories of electricity and magnetismrdquo Journalfur die reine und angewandte Mathematik (Crelles Journal)vol 1850 no 39 pp 73ndash89 1850

[5] L C Evans Partial Differential Equations American Math-ematical Society Providence RI USA 1998

[6] R Bartle A Modern 8eory of Integration American Math-ematical Society RI Providence RI USA 2001

[7] R Henstock 8e General 8eory of Integration OxfordUniversity Press Oxford UK 1991

[8] C Swartz Introduction to Gauge Integrals World ScientificSingapore 2001

[9] T-Y LeeHenstockndashKurzweil Integration on Euclidean SpacesWorld Scientific Singapore 2011

[10] E Talvila ldquoContinuity in the Alexiewicz normrdquoMathematicaBohemica vol 131 no 2 pp 189ndash196 2006

[11] E Talvila ldquoEstimates of henstock-kurzweil Poisson integralsrdquoCanadian Mathematical Bulletin vol 48 no 1 pp 133ndash1462005

[12] A Alexiewicz ldquoLinear functionals on Denjoy-integrablefunctionsrdquo Colloquium Mathematicum vol 1 no 4pp 289ndash293 1948

[13] P Mikusinksi and K Ostaszewski ldquoEmbedding Henstockintegrable functions into the space of Schwartz distributionsrdquoReal Analysis Exchange vol 14 no -89 pp 24ndash29 1988

[14] D D Ang K Schmitt and L K Vy ldquoA multidimensionalanalogue of the denjoyndashperronndashhenstockndashkurzweil integralrdquo8e Bulletin of the Belgian Mathematical Society mdash SimonStevin vol 4 pp 355ndash371 1997

[15] E Talvila ldquoampe continuous primitive integral in the planerdquoReal Analysis Exchange vol 45 no 2 pp 283ndash326 2020

[16] S Heikkila ldquoOn nonlinear distributional and impulsiveCauchy problemsrdquo Nonlinear Analysis 8eory Methods ampApplications vol 75 no 2 pp 852ndash870 2012

[17] B Liang G Ye W Liu and H Wang ldquoOn second ordernonlinear boundary value problems and the distributionalHenstock-Kurzweil integralrdquo Boundary Value Problemvol 2015 2015

[18] A Sikorska-Nowak ldquoExistence theory for integrodifferentialequations and henstock-kurzweil integral in Banach spacesrdquoJournal of Applied Mathematics vol 2007 Article ID 3157212 pages 2007

[19] M Tvrdy ldquoLinear distributional differential equations of thesecond orderrdquo Math Bohemvol 119 no 4 pp 415ndash4361994

[20] L Schwartz 8eorie des Distributions Hermann ParisFrance 1950

[21] E Talvila ldquoampe distributional denjoy integralrdquo Real AnalysisExchange vol 33 no 1 pp 51ndash82 2008

[22] K Ostaszewski ldquoampe space of Henstock integrable functionsof two variablesrdquo International Journal of Mathematics andMathematical Sciences vol 11 no 1 pp 15ndash21 1988

[23] C R Adams and J A Clarkson ldquoOn definitions of boundedvariation for functions of two variablesrdquo Transactions of theAmerican Mathematical Society vol 35 no 4 pp 824ndash8541933

[24] B Bongiorno ldquoampe henstockndashkurzweil integralrdquo inHandbookof Measure 8eory E Pap Ed pp 589ndash615 North HollandDordrecht Netherlands 2002

[25] G B Folland Introduction to Partial Differential EquationsPrinceton University Press Princeton NY USA 1995

[26] S M Aljassas ldquoNumerical methods for evaluation of doubleintegrals with continuous integrandsrdquo International Journalof Pure and Applied Mathematics vol 119 no 10 2018

International Journal of Mathematics and Mathematical Sciences 9

(iii) Let any α β isin R For Poisson kernel1113938R

1113938R

Kdxdy 1 Using FubinindashTonelli theoremand interchanging iterated integral

1113946α

minus infin1113946β

minus infin(u minus g)dxdy 1113946

α

minus infin1113946β

minus infin[u(x y z) minus g(x y)]dxdy

1113946α

minus infin1113946β

minus infin[g(x y)lowastK(x y z) minus g(x y)]dxdy

1113946α

minus infin1113946β

minus infin1113946R

1113946R

K(s t z)[g(x minus s y minus t) minus g(x y)]dsdtdxdy

1113946α

minus infin1113946β

minus infin1113946R

1113946R

z

2π s2

+ t2

+ z2

1113872 111387332 [g(x minus s y minus t) minus g(x y)]dsdtdxdy

1113946α

minus infin1113946β

minus infin1113946R

1113946R

1

2π s2

+ t2

+ 11113872 111387332 [g(x minus zs y minus zt) minus g(x y)]dsdtdxdy

1113946R

1113946R

1

2π s2

+ t2

+ 11113872 111387332 1113946

α

minus infin1113946β

minus infin[g(x minus zs y minus zt) minus g(x y)]dxdydsdt

le1113946R

1113946R

K(s t 1)g(x minus zs y minus zt) minus g(x y)dsdt

le 2g

(22)

Since α and β are arbitrary

u(x y z) minus g(x y)le1113946R

1113946R

K(s t 1)g(x minus zs y minus zt) minus g(x y)dsdtle 2g (23)

To let z⟶ 0+ apply dominated convergence theorem(HK-integral) for the sequence of functions

gn(s t) K(s t 1) g x minus1n

s y minus1n

t1113874 1113875 minus g(x y)

(24)

ampen |gn(s t)|le 2gK(s t 1) isinHK(R2) since1113938R

1113938R

K(s t 1)dsdt 1 It gives

limn⟶infin

1113946R

1113946R

gn(s t)dsdt limn⟶infin1113946

1113946R

K(s t 1)g(x minus zs y minus zt) minus g(x y)dsdt

1113946R

1113946R

limn⟶infin

gn(s t)dsdt

1113946R

1113946R

K(s t 1) limn⟶infin

g x minus1n

s y minus1n

t1113874 1113875 minus g(x y)

d s dt

(25)

From part (iii) of ampeorem 5 we get limn⟶infing(x minus

(1n)s y minus (1n)t) minus g(x y) 0 so that limit of (23)limz⟶0+ u(x y z) minus g(x y) 0 and this completes thetheorem

International Journal of Mathematics and Mathematical Sciences 7

Next we give an example which will apply ampeorem 6

Example 1 Consider the problem (16) in which

g(x y)

sin x siny

xy xne 0 andyne 0

0 x 0 ory 0

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

(26)

ampen g is HK-integrable but not absolutely integrableHence g isin AC(R

2)∖L1(R2) From ampeorem 6

u(x y z) 1113946R

1113946R

K(x minus s y minus t )g(s t)dtds

1113946R

1113946R

z

2π(x minus s)

2+(y minus t)

2+ z

21113960 1113961

minus (32) sin s sin t

st1113874 1113875dtds

(27)

exists and gives solution for the example Note thatKg isinHK(R2) since K isinBVHK(R

2) is a multiplier for

HK(R2) However the solution does not exist with theLebesgue integral For contraposition suppose thatKg isin L1(R2) ampen it follows from Fubinirsquos theorem [9](ampeorem 255) that the function

h s⟼K(x minus s y minus t z)g(s t) (28)

belongs to L1(R1) for almost all t isin R ampat is

h(s) z sin t

2πt1113874 1113875

sin s

s (x minus s)2

+(y minus t)2

+ z2

1113960 111396132 isin L

1R

11113872 1113873

h0(s)sin s

s1113874 1113875

(29)

where h0(s) (z sin t2πt)(1[(x minus s)2 + (y minus t)2 + z2]32)ampis contradicts that h(s) notin L1(R1) To see that supposeh(s) isin L1[0 1] sub L1(R1) Since h0(s) isin Linfin[0 1] is multi-plier for L1[0 1] it follows that sin ss isin L1[0 1] ampis im-plies that sin ss absolutely integrable on [0 1] which is acontradiction Hence h(s) notin L1(R1)

Even HK-integral exists in (27) it may not be able toevaluate and get explicit function In that case numericalintegration can be applied Numerical integration should bedone with respect to s and t while fixing x y and z details in[26] ampis reduces solving the partial differential equationinto applying numerical integration for the double integral(27)

5 Discussion

In this work integrable distributional solution is obtainedfor Poisson equation with the Dirichlet boundary conditionin the upper half space Boundary conditions are taken in theAlexiewicz norm Lebesgue integral uses for the classicalsolution to Poisson equation Because of the proper inclu-sion L1 ⊊ HR we may use HK-integral to solve Poissonequation However there are some limitations due to the

absence of any natural topology ampat is HK is not aBanach space nevertheless there is a natural seminormcalled Alexiewicz ampis is one of the main reasons to moveinto the space of integrable distribution AC(R

2) which

contains HK and leads to a Banach space with Alexiewicznorm

Integrable distribution is defined as derivative (distri-butional sense) of function in BC(R

2) In this definition

any f isin AC(R2) identifies with the unique primitive

F isinBC(R2) ampis unique primitive allows to define a

Alexiewicz norm middot on AC(R2) and leads to a Banach

space Alexiewicz norm does not induce an inner product inAC(R

2) because it does not satisfy the parallelogram law

amperefore AC(R2) is not a Hilbert space under the norm

middot However the space (AC(R2) middot) is isometrically iso-

morphic to the space (BC(R2) middotinfin) Integral on AC(R

2)

is uniquely defined by uniqueness of the primitive AC(R2)

is the completion of L1(R2) (or HK(R2)) with respect tothe corresponding Alexiewicz norm

In one-dimensional HK-integral space of boundedvariation functions BV is multipliers for HK In mul-tidimension case there are several extensions to variation offunction To establish integration by parts formula andmultipliers for AC(R

2) we begin with HardyndashKrause

variation ampe space of HardyndashKrause bounded variationfunctions BVHK(R

2) as a multiplier for AC(R

2) ampis

multiplier is important to define convolution and obtainconvergence theorem BVHK(R

2) is used to define con-

volution and obtain convergence theorem on AC(R2)

ampere are several convergence theorems in AC such asstrong convergence in AC weakstrong convergence inBV and weakstrong convergence in D Convergencetheorem on AC(R

2) and dominated convergence theorem

in HK(R2) are used to obtain integrable distributionalsolution Finding of this article gives solution for Poissonequation for broader initial condition in the upper halfspace Dirichlet problem in R3 is considered and the so-lution has obtained by HK-integration where it is notpossible from Lebesgue integral Because of functionrsquoscomplexity it may not be able to have solution in explicit

8 International Journal of Mathematics and Mathematical Sciences

form It would help keep an analytical outlook in the so-lution which would be better than solving the partial dif-ferential equation by numerical methods such as finitedifference methods and finite element methods

ampere is potential to find integrable distributional so-lution for Poisson equation in other boundary conditionsNeumann and Robin But it needs to develop suitableconvergence theorems on AC and its multipliers BVHKAlso it is possible to consider Poisson equation in unit ballwith the relevant Poisson kernel Talvila in [15] gives someconstruction to develop integrable distribution on Rn in-tegration by parts and mean value theorem and definesAC(R

n) and BC(R

n) for multidimensional case Hence

integrable distributional solution is possible to obtain forPoisson equation with Dirichlet boundary condition in Rn

+

with suitable convergence theorem

Data Availability

No data were used to support this study

Conflicts of Interest

ampe authors declare that they have no conflicts of interest

References

[1] R L Burden and J D Faires Numerical Analysis ampomsonBrooksCole USA 8th edition 2005

[2] C Anitescu E Atroshchenko N Alajlan and T RabczukldquoArtificial neural network methods for the solution of secondorder boundary value problemsrdquo Computers Materials ampContinua vol 59 no 1 pp 345ndash359 2019

[3] D A Leon-Velasco M M Morın-Castillo J J Oliveros-Oliveros T Perez-Becerra and J A Escamilla-Reyna ldquoNu-merical solution of some differential equations withHenstockndash Kurzweil functionsrdquo Journal of Function Spacesvol 2019 Article ID 8948570 9 pages 2019

[4] G Green ldquoAn essay on the application of mathematicalanalysis to the theories of electricity and magnetismrdquo Journalfur die reine und angewandte Mathematik (Crelles Journal)vol 1850 no 39 pp 73ndash89 1850

[5] L C Evans Partial Differential Equations American Math-ematical Society Providence RI USA 1998

[6] R Bartle A Modern 8eory of Integration American Math-ematical Society RI Providence RI USA 2001

[7] R Henstock 8e General 8eory of Integration OxfordUniversity Press Oxford UK 1991

[8] C Swartz Introduction to Gauge Integrals World ScientificSingapore 2001

[9] T-Y LeeHenstockndashKurzweil Integration on Euclidean SpacesWorld Scientific Singapore 2011

[10] E Talvila ldquoContinuity in the Alexiewicz normrdquoMathematicaBohemica vol 131 no 2 pp 189ndash196 2006

[11] E Talvila ldquoEstimates of henstock-kurzweil Poisson integralsrdquoCanadian Mathematical Bulletin vol 48 no 1 pp 133ndash1462005

[12] A Alexiewicz ldquoLinear functionals on Denjoy-integrablefunctionsrdquo Colloquium Mathematicum vol 1 no 4pp 289ndash293 1948

[13] P Mikusinksi and K Ostaszewski ldquoEmbedding Henstockintegrable functions into the space of Schwartz distributionsrdquoReal Analysis Exchange vol 14 no -89 pp 24ndash29 1988

[14] D D Ang K Schmitt and L K Vy ldquoA multidimensionalanalogue of the denjoyndashperronndashhenstockndashkurzweil integralrdquo8e Bulletin of the Belgian Mathematical Society mdash SimonStevin vol 4 pp 355ndash371 1997

[15] E Talvila ldquoampe continuous primitive integral in the planerdquoReal Analysis Exchange vol 45 no 2 pp 283ndash326 2020

[16] S Heikkila ldquoOn nonlinear distributional and impulsiveCauchy problemsrdquo Nonlinear Analysis 8eory Methods ampApplications vol 75 no 2 pp 852ndash870 2012

[17] B Liang G Ye W Liu and H Wang ldquoOn second ordernonlinear boundary value problems and the distributionalHenstock-Kurzweil integralrdquo Boundary Value Problemvol 2015 2015

[18] A Sikorska-Nowak ldquoExistence theory for integrodifferentialequations and henstock-kurzweil integral in Banach spacesrdquoJournal of Applied Mathematics vol 2007 Article ID 3157212 pages 2007

[19] M Tvrdy ldquoLinear distributional differential equations of thesecond orderrdquo Math Bohemvol 119 no 4 pp 415ndash4361994

[20] L Schwartz 8eorie des Distributions Hermann ParisFrance 1950

[21] E Talvila ldquoampe distributional denjoy integralrdquo Real AnalysisExchange vol 33 no 1 pp 51ndash82 2008

[22] K Ostaszewski ldquoampe space of Henstock integrable functionsof two variablesrdquo International Journal of Mathematics andMathematical Sciences vol 11 no 1 pp 15ndash21 1988

[23] C R Adams and J A Clarkson ldquoOn definitions of boundedvariation for functions of two variablesrdquo Transactions of theAmerican Mathematical Society vol 35 no 4 pp 824ndash8541933

[24] B Bongiorno ldquoampe henstockndashkurzweil integralrdquo inHandbookof Measure 8eory E Pap Ed pp 589ndash615 North HollandDordrecht Netherlands 2002

[25] G B Folland Introduction to Partial Differential EquationsPrinceton University Press Princeton NY USA 1995

[26] S M Aljassas ldquoNumerical methods for evaluation of doubleintegrals with continuous integrandsrdquo International Journalof Pure and Applied Mathematics vol 119 no 10 2018

International Journal of Mathematics and Mathematical Sciences 9

Next we give an example which will apply ampeorem 6

Example 1 Consider the problem (16) in which

g(x y)

sin x siny

xy xne 0 andyne 0

0 x 0 ory 0

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

(26)

ampen g is HK-integrable but not absolutely integrableHence g isin AC(R

2)∖L1(R2) From ampeorem 6

u(x y z) 1113946R

1113946R

K(x minus s y minus t )g(s t)dtds

1113946R

1113946R

z

2π(x minus s)

2+(y minus t)

2+ z

21113960 1113961

minus (32) sin s sin t

st1113874 1113875dtds

(27)

exists and gives solution for the example Note thatKg isinHK(R2) since K isinBVHK(R

2) is a multiplier for

HK(R2) However the solution does not exist with theLebesgue integral For contraposition suppose thatKg isin L1(R2) ampen it follows from Fubinirsquos theorem [9](ampeorem 255) that the function

h s⟼K(x minus s y minus t z)g(s t) (28)

belongs to L1(R1) for almost all t isin R ampat is

h(s) z sin t

2πt1113874 1113875

sin s

s (x minus s)2

+(y minus t)2

+ z2

1113960 111396132 isin L

1R

11113872 1113873

h0(s)sin s

s1113874 1113875

(29)

where h0(s) (z sin t2πt)(1[(x minus s)2 + (y minus t)2 + z2]32)ampis contradicts that h(s) notin L1(R1) To see that supposeh(s) isin L1[0 1] sub L1(R1) Since h0(s) isin Linfin[0 1] is multi-plier for L1[0 1] it follows that sin ss isin L1[0 1] ampis im-plies that sin ss absolutely integrable on [0 1] which is acontradiction Hence h(s) notin L1(R1)

Even HK-integral exists in (27) it may not be able toevaluate and get explicit function In that case numericalintegration can be applied Numerical integration should bedone with respect to s and t while fixing x y and z details in[26] ampis reduces solving the partial differential equationinto applying numerical integration for the double integral(27)

5 Discussion

In this work integrable distributional solution is obtainedfor Poisson equation with the Dirichlet boundary conditionin the upper half space Boundary conditions are taken in theAlexiewicz norm Lebesgue integral uses for the classicalsolution to Poisson equation Because of the proper inclu-sion L1 ⊊ HR we may use HK-integral to solve Poissonequation However there are some limitations due to the

absence of any natural topology ampat is HK is not aBanach space nevertheless there is a natural seminormcalled Alexiewicz ampis is one of the main reasons to moveinto the space of integrable distribution AC(R

2) which

contains HK and leads to a Banach space with Alexiewicznorm

Integrable distribution is defined as derivative (distri-butional sense) of function in BC(R

2) In this definition

any f isin AC(R2) identifies with the unique primitive

F isinBC(R2) ampis unique primitive allows to define a

Alexiewicz norm middot on AC(R2) and leads to a Banach

space Alexiewicz norm does not induce an inner product inAC(R

2) because it does not satisfy the parallelogram law

amperefore AC(R2) is not a Hilbert space under the norm

middot However the space (AC(R2) middot) is isometrically iso-

morphic to the space (BC(R2) middotinfin) Integral on AC(R

2)

is uniquely defined by uniqueness of the primitive AC(R2)

is the completion of L1(R2) (or HK(R2)) with respect tothe corresponding Alexiewicz norm

In one-dimensional HK-integral space of boundedvariation functions BV is multipliers for HK In mul-tidimension case there are several extensions to variation offunction To establish integration by parts formula andmultipliers for AC(R

2) we begin with HardyndashKrause

variation ampe space of HardyndashKrause bounded variationfunctions BVHK(R

2) as a multiplier for AC(R

2) ampis

multiplier is important to define convolution and obtainconvergence theorem BVHK(R

2) is used to define con-

volution and obtain convergence theorem on AC(R2)

ampere are several convergence theorems in AC such asstrong convergence in AC weakstrong convergence inBV and weakstrong convergence in D Convergencetheorem on AC(R

2) and dominated convergence theorem

in HK(R2) are used to obtain integrable distributionalsolution Finding of this article gives solution for Poissonequation for broader initial condition in the upper halfspace Dirichlet problem in R3 is considered and the so-lution has obtained by HK-integration where it is notpossible from Lebesgue integral Because of functionrsquoscomplexity it may not be able to have solution in explicit

8 International Journal of Mathematics and Mathematical Sciences

form It would help keep an analytical outlook in the so-lution which would be better than solving the partial dif-ferential equation by numerical methods such as finitedifference methods and finite element methods

ampere is potential to find integrable distributional so-lution for Poisson equation in other boundary conditionsNeumann and Robin But it needs to develop suitableconvergence theorems on AC and its multipliers BVHKAlso it is possible to consider Poisson equation in unit ballwith the relevant Poisson kernel Talvila in [15] gives someconstruction to develop integrable distribution on Rn in-tegration by parts and mean value theorem and definesAC(R

n) and BC(R

n) for multidimensional case Hence

integrable distributional solution is possible to obtain forPoisson equation with Dirichlet boundary condition in Rn

+

with suitable convergence theorem

Data Availability

No data were used to support this study

Conflicts of Interest

ampe authors declare that they have no conflicts of interest

References

[1] R L Burden and J D Faires Numerical Analysis ampomsonBrooksCole USA 8th edition 2005

[2] C Anitescu E Atroshchenko N Alajlan and T RabczukldquoArtificial neural network methods for the solution of secondorder boundary value problemsrdquo Computers Materials ampContinua vol 59 no 1 pp 345ndash359 2019

[3] D A Leon-Velasco M M Morın-Castillo J J Oliveros-Oliveros T Perez-Becerra and J A Escamilla-Reyna ldquoNu-merical solution of some differential equations withHenstockndash Kurzweil functionsrdquo Journal of Function Spacesvol 2019 Article ID 8948570 9 pages 2019

[4] G Green ldquoAn essay on the application of mathematicalanalysis to the theories of electricity and magnetismrdquo Journalfur die reine und angewandte Mathematik (Crelles Journal)vol 1850 no 39 pp 73ndash89 1850

[5] L C Evans Partial Differential Equations American Math-ematical Society Providence RI USA 1998

[6] R Bartle A Modern 8eory of Integration American Math-ematical Society RI Providence RI USA 2001

[7] R Henstock 8e General 8eory of Integration OxfordUniversity Press Oxford UK 1991

[8] C Swartz Introduction to Gauge Integrals World ScientificSingapore 2001

[9] T-Y LeeHenstockndashKurzweil Integration on Euclidean SpacesWorld Scientific Singapore 2011

[10] E Talvila ldquoContinuity in the Alexiewicz normrdquoMathematicaBohemica vol 131 no 2 pp 189ndash196 2006

[11] E Talvila ldquoEstimates of henstock-kurzweil Poisson integralsrdquoCanadian Mathematical Bulletin vol 48 no 1 pp 133ndash1462005

[12] A Alexiewicz ldquoLinear functionals on Denjoy-integrablefunctionsrdquo Colloquium Mathematicum vol 1 no 4pp 289ndash293 1948

[13] P Mikusinksi and K Ostaszewski ldquoEmbedding Henstockintegrable functions into the space of Schwartz distributionsrdquoReal Analysis Exchange vol 14 no -89 pp 24ndash29 1988

[14] D D Ang K Schmitt and L K Vy ldquoA multidimensionalanalogue of the denjoyndashperronndashhenstockndashkurzweil integralrdquo8e Bulletin of the Belgian Mathematical Society mdash SimonStevin vol 4 pp 355ndash371 1997

[15] E Talvila ldquoampe continuous primitive integral in the planerdquoReal Analysis Exchange vol 45 no 2 pp 283ndash326 2020

[16] S Heikkila ldquoOn nonlinear distributional and impulsiveCauchy problemsrdquo Nonlinear Analysis 8eory Methods ampApplications vol 75 no 2 pp 852ndash870 2012

[17] B Liang G Ye W Liu and H Wang ldquoOn second ordernonlinear boundary value problems and the distributionalHenstock-Kurzweil integralrdquo Boundary Value Problemvol 2015 2015

[18] A Sikorska-Nowak ldquoExistence theory for integrodifferentialequations and henstock-kurzweil integral in Banach spacesrdquoJournal of Applied Mathematics vol 2007 Article ID 3157212 pages 2007

[19] M Tvrdy ldquoLinear distributional differential equations of thesecond orderrdquo Math Bohemvol 119 no 4 pp 415ndash4361994

[20] L Schwartz 8eorie des Distributions Hermann ParisFrance 1950

[21] E Talvila ldquoampe distributional denjoy integralrdquo Real AnalysisExchange vol 33 no 1 pp 51ndash82 2008

[22] K Ostaszewski ldquoampe space of Henstock integrable functionsof two variablesrdquo International Journal of Mathematics andMathematical Sciences vol 11 no 1 pp 15ndash21 1988

[23] C R Adams and J A Clarkson ldquoOn definitions of boundedvariation for functions of two variablesrdquo Transactions of theAmerican Mathematical Society vol 35 no 4 pp 824ndash8541933

[24] B Bongiorno ldquoampe henstockndashkurzweil integralrdquo inHandbookof Measure 8eory E Pap Ed pp 589ndash615 North HollandDordrecht Netherlands 2002

[25] G B Folland Introduction to Partial Differential EquationsPrinceton University Press Princeton NY USA 1995

[26] S M Aljassas ldquoNumerical methods for evaluation of doubleintegrals with continuous integrandsrdquo International Journalof Pure and Applied Mathematics vol 119 no 10 2018

International Journal of Mathematics and Mathematical Sciences 9

form It would help keep an analytical outlook in the so-lution which would be better than solving the partial dif-ferential equation by numerical methods such as finitedifference methods and finite element methods

ampere is potential to find integrable distributional so-lution for Poisson equation in other boundary conditionsNeumann and Robin But it needs to develop suitableconvergence theorems on AC and its multipliers BVHKAlso it is possible to consider Poisson equation in unit ballwith the relevant Poisson kernel Talvila in [15] gives someconstruction to develop integrable distribution on Rn in-tegration by parts and mean value theorem and definesAC(R

n) and BC(R

n) for multidimensional case Hence

integrable distributional solution is possible to obtain forPoisson equation with Dirichlet boundary condition in Rn

+

with suitable convergence theorem

Data Availability

No data were used to support this study

Conflicts of Interest

ampe authors declare that they have no conflicts of interest

References

[1] R L Burden and J D Faires Numerical Analysis ampomsonBrooksCole USA 8th edition 2005

[2] C Anitescu E Atroshchenko N Alajlan and T RabczukldquoArtificial neural network methods for the solution of secondorder boundary value problemsrdquo Computers Materials ampContinua vol 59 no 1 pp 345ndash359 2019

[3] D A Leon-Velasco M M Morın-Castillo J J Oliveros-Oliveros T Perez-Becerra and J A Escamilla-Reyna ldquoNu-merical solution of some differential equations withHenstockndash Kurzweil functionsrdquo Journal of Function Spacesvol 2019 Article ID 8948570 9 pages 2019

[4] G Green ldquoAn essay on the application of mathematicalanalysis to the theories of electricity and magnetismrdquo Journalfur die reine und angewandte Mathematik (Crelles Journal)vol 1850 no 39 pp 73ndash89 1850

[5] L C Evans Partial Differential Equations American Math-ematical Society Providence RI USA 1998

[6] R Bartle A Modern 8eory of Integration American Math-ematical Society RI Providence RI USA 2001

[7] R Henstock 8e General 8eory of Integration OxfordUniversity Press Oxford UK 1991

[8] C Swartz Introduction to Gauge Integrals World ScientificSingapore 2001

[9] T-Y LeeHenstockndashKurzweil Integration on Euclidean SpacesWorld Scientific Singapore 2011

[10] E Talvila ldquoContinuity in the Alexiewicz normrdquoMathematicaBohemica vol 131 no 2 pp 189ndash196 2006

[11] E Talvila ldquoEstimates of henstock-kurzweil Poisson integralsrdquoCanadian Mathematical Bulletin vol 48 no 1 pp 133ndash1462005

[12] A Alexiewicz ldquoLinear functionals on Denjoy-integrablefunctionsrdquo Colloquium Mathematicum vol 1 no 4pp 289ndash293 1948

[13] P Mikusinksi and K Ostaszewski ldquoEmbedding Henstockintegrable functions into the space of Schwartz distributionsrdquoReal Analysis Exchange vol 14 no -89 pp 24ndash29 1988

[14] D D Ang K Schmitt and L K Vy ldquoA multidimensionalanalogue of the denjoyndashperronndashhenstockndashkurzweil integralrdquo8e Bulletin of the Belgian Mathematical Society mdash SimonStevin vol 4 pp 355ndash371 1997

[15] E Talvila ldquoampe continuous primitive integral in the planerdquoReal Analysis Exchange vol 45 no 2 pp 283ndash326 2020

[16] S Heikkila ldquoOn nonlinear distributional and impulsiveCauchy problemsrdquo Nonlinear Analysis 8eory Methods ampApplications vol 75 no 2 pp 852ndash870 2012

[17] B Liang G Ye W Liu and H Wang ldquoOn second ordernonlinear boundary value problems and the distributionalHenstock-Kurzweil integralrdquo Boundary Value Problemvol 2015 2015

[18] A Sikorska-Nowak ldquoExistence theory for integrodifferentialequations and henstock-kurzweil integral in Banach spacesrdquoJournal of Applied Mathematics vol 2007 Article ID 3157212 pages 2007

[19] M Tvrdy ldquoLinear distributional differential equations of thesecond orderrdquo Math Bohemvol 119 no 4 pp 415ndash4361994

[20] L Schwartz 8eorie des Distributions Hermann ParisFrance 1950

[21] E Talvila ldquoampe distributional denjoy integralrdquo Real AnalysisExchange vol 33 no 1 pp 51ndash82 2008

[22] K Ostaszewski ldquoampe space of Henstock integrable functionsof two variablesrdquo International Journal of Mathematics andMathematical Sciences vol 11 no 1 pp 15ndash21 1988

[23] C R Adams and J A Clarkson ldquoOn definitions of boundedvariation for functions of two variablesrdquo Transactions of theAmerican Mathematical Society vol 35 no 4 pp 824ndash8541933

[24] B Bongiorno ldquoampe henstockndashkurzweil integralrdquo inHandbookof Measure 8eory E Pap Ed pp 589ndash615 North HollandDordrecht Netherlands 2002

[25] G B Folland Introduction to Partial Differential EquationsPrinceton University Press Princeton NY USA 1995

[26] S M Aljassas ldquoNumerical methods for evaluation of doubleintegrals with continuous integrandsrdquo International Journalof Pure and Applied Mathematics vol 119 no 10 2018

International Journal of Mathematics and Mathematical Sciences 9