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Original article
32 (
6
); 2790-2796
doi:
10.1016/j.jksus.2020.06.016

The HK-Sobolev space and applications to one-dimensional boundary value problems

Universidad Tecnológica de la Mixteca, División de Estudios de Postgrado, Km. 2.5 Carretera a Acatlima, 69000 Oaxaca, Mexico
Universidad Tecnológica de la Mixteca, Instituto de Física y Matemáticas, Km. 2.5 Carretera a Acatlima, 69000 Oaxaca, Mexico
Benemérita Universidad Autónoma de Puebla, Facultad de Ciencias Físico Matemáticas, Río Verde y Av. San Claudio, San Manuel, Puebla, Pue. 72570, Mexico

⁎Corresponding author. es21254@yahoo.com.mx (S. Sánchez-Perales) ssanchez@mixteco.utm.mx (S. Sánchez-Perales)

Disclaimer:
This article was originally published by Elsevier and was migrated to Scientific Scholar after the change of Publisher.
v denotes de usual derivative of v. From u1ACG([0,1]) it follows that u1 exists a.e on [0,1].

Abstract

In this paper, we introduce the HK-Sobolev space and establish a fundamental theorem of calculus and an integration by parts formula, then we give sufficient conditions for the existence and uniqueness of a solution to a variational problem associated with a Sturm–Liouville type equation involving Henstock-Kurzweil integrable functions as source terms.

Keywords

Henstock–Kurzweil integral
HK-Sobolev space
Weak derivative
Ordinary differential equations
1

1 Introduction

By way of introduction, let us begin with the motivation for this work.

A classical variational problem. The Sobolev space is given by

(1)
H1([0,1])=uL2([0,1]):gL2([0,1])suchthat01uφ'=-01gφ,φCc1(0,1),where Cc1(0,1) is the space of continuously differentiable functions defined on [0,1] with support in (0,1). The function g in (1) is named the weak derivative of u and is denoted by u̇. We setH01([0,1])={uH1([0,1]):u(0)=u(1)=0}.Consider the following variational problem:

Given fL2([0,1]), find uH01([0,1])such that

(2)
01u̇φ̇+01uφ=01fφfor all φH01([0,1]). This variational problem arises from considering the following boundary problem (Brezis, 2011):
(3)
-u¨+u=fa.e.on(0,1);u(0)=0,u(1)=0.

The space H01([0,1]) has certain properties that are useful in order to find a solution to variational problems (2): it is a Hilbert, separable and reflexive space.

What happens if f is not of square Lebesgue integrable? In several physical phenomena, highly oscillating or singular functions appear (Condon et al., 2009; Hamed and Cummins, 1991; Hong and Xu, 2001; Samoilov et al., 2005). The Lebesgue integral is not enough for some highly oscillating functions leading to the possibility that the integral on the right side of the Eq. (2) does not exist for this type of functions and so the variational problem (2) would not be well defined. One way to solve this problem is to change the type of integral to be considered, in this work we will use the Henstock–Kurzweil integral. Different authors have studied differential equations involving Henstock–Kurzweil integrable functions. In León-Velasco et al. (2019) the authors use the Finite Element Method (FEM) for finding numerical solutions of elliptic problems with Henstock–Kurzweil integrable functions. They use open quadratures and Lobatto quadratures to approximate numerically the integrals that appear in the FEM. In Liu et al. (2018) are given conditions to establish the existence of a solution to nonlinear second-order differential equations of type -D2x=f(t,x)+g(t,x)Du subject to the boundary conditions x(0)=βDx(0),Dx(1)+Dx(η)=0, where the derivatives are in the distributional sense, x,u are regulated functions and g is of bounded variation. In that paper the Henstock–Kurzweil–Stieltjes integral is used to transform the distributional differential equation into an integral equation, then the Leray–Schauder nonlinear alternative theorem is applied for finding a solution. In Sánchez-Perales and Mendoza-Torres (2020) the existence and uniqueness of the Shrödinger equation, -y+qy=f a.e. on [a,b] subject to arbitrary boundary values, is guaranteed for functions f,q Henstock–Kurzweil integrable. Properties of the inverse of the Shrödinger operator are established, then the authors give conditions so that the solution of the differential equation can be expressed as a Fourier type series.

Henstock–Kurzweil–Sobolev space. Around the 1960s, R. Henstock and J. Kurzweil, independently, define a Riemann-type integral, known as Henstock–Kurzweil integral, which is equivalent to Denjoy and Perron integrals. This integral is more general than the Lebesgue integral. In this work we introduce, using the Henstock–Kurzweil integral instead of the Lebesgue integral, a space analogous to H1([0,1]), which we will call the Henstock–Kurzweil–Sobolev space and denote it by WHK. Since the product of two Henstock–Kurzweil integrable functions is not necessarily an integrable function, the WHKspace is not provided with a natural internal product. Thus we cannot apply classical theorems, such as Lax–Milgram’s, to guarantee the existence and uniqueness of the solution to variational problems such as (2). In this paper, we will use Fredholm’s alternative for compact operators and the properties of the Henstock–Kurzweil–Sobolev space to solve such problems.

2

2 Preliminaries

The symbol R denotes the set of real numbers, C stands for the complex numbers and [a,b]R is a closed finite interval. A tagged partition {([ti-1,ti],ξi):i=1,,n} of [a,b] is a finite collection of non-overlapping intervals [ti-1,ti] such that [a,b]=i=1n[ti-1,ti], and ξi[ti-1,ti] for all i=1,,n. A function δ:[a,b]R is a gauge on [a,b] if δ(t)>0 for every t[a,b]. Given a gauge δ on [a,b] and a tagged partition P={([ti-1,ti],ξi):i=1,,n} of [a,b],P is δ-fine if[ti-1,ti](ξi-δ(ξi),ξi+δ(ξi)),i=1,,n.

A function f:[a,b]C is Henstock–Kurzweil integrable (HK-integrable) on [a,b] if there exists a number I such that for every >0 there exists a gauge δ on [a,b] such that for each δ-fine tagged partition {([ti-1,ti],ξi):i=1,,n} of [a,b],i=1nf(ξi)(ti-ti-1)-I<.

The number I is called the integral of f over [a,b] and it is denoted by abf. The space of Henstock–Kurzweil integrable functions is denoted by HK([a,b]). The Alexiewicz semi-norm of a function fHK([a,b]) is defined byfA=supt[a,b]atf.

A function φ:[a,b]C is of bounded variation on [a,b] (φBV([a,b])) ifV[a,b]φsupi=1n|φ(di)-φ(ci)|<,where the supremum is taken over all finite collections {[ci,di]:i=1,,n} of non-overlapping intervals of [a,b].

Theorem 2.1

(Talvila, 1999, Lemma 24) If fHK([a,b]) and gBV([a,b]), then fgHK([a,b]) andabfginft[a,b]|g(t)|abf(t)dt+fAV[a,b]g.

Theorem 2.2

(Sargent, 1948, Theorem D) Let f,φ be functions such that f is of real values and fHK([a,b]). Then, fφHK([a,b]) if and only if there exists φ1BV([a,b]) such that φ=φ1 a.e. on [a,b].

The next Fubini’s Theorem is a direct consequence of Talvila (2002, Lemma 25).

Theorem 2.3

If fHK([a,b]) and hBV([a,b]), then for any subintervals A,B of[a,b], we haveABf(t)h(x)dtdx=BAf(t)h(x)dxdt.

A function F:[a,b]C is absolutely continuous (respectively, absolutely continuous in the restricted sense) on a set E[a,b], if for each >0 there exists δ>0 such that i=1s|F(di)-F(ci)|< (respectively, i=1ssup{|F(x)-F(y)|:x,y[ci,di]}<) whenever {[ci,di]}i=1s is a collection of non-overlapping intervals with endpoints in E and such that i=1s(di-ci)<δ. The space of absolutely continuous functions on E is denoted by AC(E), and the space of absolutely continuous functions in the restricted sense on E is denoted by AC(E).

The function F is generalized absolutely continuous in the restricted sense on [a,b] (FACG([a,b])), if F is continuous on [a,b] and there exists a countable collection (En)n=1 of subsets of [a,b] such that [a,b]=i=1En and FAC(En) for all nN. This concept leads to a very strong version of the fundamental theorem of calculus:

Theorem 2.4

[Fundamental theorem of calculus] (Gordon, 1994) Let f,F:[a,b]C be functions and let c[a,b].

  1. If fHK([a,b]) and F(x)=cxf for all x[a,b], then FACG([a,b]) and F=f almost everywhere on [a,b]. In particular, if f is continuous at x[a,b], then F(x)=f(x).

  2. FACG([a,b]) if and only if F exists almost everywhere on [a,b] and cxF=F(x)-F(c) for all x[a,b].

Theorem 2.5

[Integration by parts formula] (Sánchez-Perales and Mendoza-Torres, 2020, Corollary 2.4) If uACG([a,b]) and vAC([a,b]), then uvHK([a,b]),uvL([a,b]) andabu(t)v(t)dt=u(b)v(b)-u(a)v(a)-abu(t)v(t)dt.

3

3 The HK-Sobolev space

In what follows from this document, we have presented the results on the interval [0,1] without loss of generality, since they can be generalized to any compact interval. Let CP2([0,1]) the space of all functions φC([0,1]) for which there exists {[ti-1,ti]}i=1n a partition of [0,1] such that φC2((ti-1,ti)) for all i=1,,n; and φ(k)(t0+),φ(k)(t1-),φ(k)(t1+),,φ(k)(tn-1-),φ(k)(tn-1+),φ(k)(tn-) exist for all k=1,2. We setV={φCP2([0,1]):φ(0)=φ(1)=0}.

It is clear that if φCP2([0,1]), then φ and φ belong to AC([0,1])(BV([0,1])). The next theorem is proved in a similar way to Hestenes (1966, Lemma 15.2, p. 51) with some modifications.

Lemma 3.1

[Fundamental lemma of calculus of variations] Let f,gHK([0,1]) with g continuous on the right at t=0. Then01[f(t)φ(t)+g(t)φ(t)]dt=0for every φV, if and only if,g(t)=0tf(s)ds+g(0),for almost all t[0,1].

Proof

We assume that f and g are HK-integrable on [0,1], then the function G(t)0tg is differentiable except for a set K of measure zero. Let t̂(0,1)K, choose such that 0<<t̂/2, and define the functionφ(t)=t,if0t,1,iftt̂-,t̂-t,ift̂-tt̂,0,ift̂t1.

Then φV and so

(4)
0=10f(t)tdt+10g+t̂-f+1t̂-t̂f(t)(t̂-t)dt-1t̂-t̂g.

Since g is continuous on the right at 0, it follows thatlim0+10g=g(0).

From Bartle (2001, Theorem 12.5),lim010f(t)tdt=0=lim01t̂-t̂f(t)(t̂-t)dt.

Thus the right side of (4) tends to g(0)+0t̂f-G(t̂) as 0. Therefore, g(0)+0t̂f-G(t̂)=0, i.e.g(t̂)=0t̂f+g(0).

Conversely, suppose thatg(t)=0tf(s)ds+g(0),for almost all t[0,1]. We set H(t)=0tf(s)ds+g(0), then H=g and H=f a.e. on [0,1], hence by Theorem 2.5,01[fφ+gφ]=01[Hφ+Hφ]=H(1)φ(1)-H(0)φ(0)=0.

Corollary 3.2

Let fHK[0,1]. If01fφ=0,for every φV, then f=0 a.e on [0,1].

Proof

Taking g=0 in Lemma 3.1, we obtain that0tf=0,for almost all t[0,1]. We set F(t)=0tf, by continuity of F and the zero function, we have that f=0 a.e. on [0,1].

Corollary 3.3

Let fHK[0,1]. If01fφ=0,for every φV, then there exists a constant C such that f=C a.e on [0,1].

Proof

Consider the functionψ(t)=4t,if0t12;-4t+4,if12t1.

Let φV and definez(t)=0tφ-01φψ.

Then zCP2([0,1]) and z(0)=z(1)=0, hence zV. Thus by hypothesis and Fubini’s Theorem, we have that0=01fz=01f(t)φ(t)-01φ(x)dxψ(t)dt=01f(t)φ(t)dt-0101f(t)φ(x)ψ(t)dxdt=01f(x)φ(x)dx-0101f(t)φ(x)ψ(t)dtdx=01f(x)-01f(t)ψ(t)dtφ(x)dx.

Therefore, by Corollary 3.2, f-01fψ=0 a.e. on [0,1].

Theorem 3.4

Let gHK([0,1]). For a fixed y0[0,1] definev(x)=y0xg(t)dt,x[0,1].

Then vC[0,1] and01vφ=-01gφ,for all φV.

Proof

It is clear that v is continuous. By Fubini’s Theorem it follows that01vφ'=01y0xg(t)φ'(x)dtdx=-0y0xy0g(t)φ'(x)dtdx+y01y0xg(t)φ'(x)dtdx=-0y00tg(t)φ'(x)dxdt+y01t1g(t)φ'(x)dxdt=-0y0g(t)(φ(t)-φ(0))dt+y01g(t)(φ(1)-φ(t))dt=-01g(t)φ(t)dt.

Definition 1

The HK-Sobolev space WHK is defined to beWHK=uHK([0,1]):gHK([0,1])suchthat01uφ=-01gφ,φV.

For uWHK we define the weak derivative of u, denoted by u̇, asu̇=g,where g is the function given in Definition 1. Observe, by Corollary 3.2, u̇ is well defined. Also, from Theorem 3.4 we have that for every fHK([0,1]), there exists a continuous function v defined on [0,1] such that v̇=f, that is, each HK-integrable function is the weak derivative of a some continuous function.

Proposition 3.5

If u=u1 a.e on [0,1] and u1 belongs to ACG([0,1]), then uWHK and u̇=u1.1

Proof

Let u,u1 be functions such that u1ACG([0,1]) and u=u1 a.e. on [0,1]. Then u1 exists a.e on [0,1], and by Theorem 2.5,01u1φ=u1φ|01-01u1φfor all φV. Therefore01uφ=01u1φ=-01u1φfor all φV. Consequently, uWHK and u̇=u1. □

Remark 1

As an immediate consequence of the previous proposition, we have that if u is a continuous function on I such that u exists except on a countable set, then uWHK and u̇=u.

Theorem 3.6

[Fundamental theorem of calculus] Let uWHK. Then there exists a function ũC([0,1]) such thatu=ũa.e.on[0,1]andũ(d)-ũ(c)=cdu̇,forallc,d[0,1].

Proof

Write û(x)=0xu̇. By Theorem 3.4 we have01ûφ=-01u̇φ,for all φV. On the other hand, as uWHK, then-01u̇φ=01uφ.

Therefore01(u-û)φ=0,for all φV. Consequently by Corollary 3.3, there exists a constant C such that u-û=C a.e. on [0,1]. The function ũ=û+C is the desired and satisfies the second part of the theorem. □

Remark 2

  1. Every function uWHK admits one (and only one) continuous representative ũ on [0,1]. Therefore uL2([0,1]). Moreover, ũACG([0,1]) and by Proposition 3.5, u̇=ũ.

  2. If vWHK and there exists wC([0,1]) such that w=v̇ a.e on [0,1], then by Theorem 3.6, ṽC1([0,1]), i.e. v has a continuously differentiable representative on [0,1]. Therefore by Theorem 2.2, gvHK([0,1]) for all real valued function gHK([0,1]).

Corollary 3.7

[Integration by parts formula] If  u,vWHK, then uvWHK and(uv)̇=u̇v+uv̇.

Also, if u̇vHK([0,1]) and u(0+)=u(0),v(0+)=v(0),u(1-)=u(1) and v(1-)=v(1), then

(5)
01u̇v=uv|01-01uv̇.

Proof

Let u,vWHK. Then by Remark 2, ũ,ṽACG([0,1]),u̇=ũ and v̇=ṽ. From Sánchez-Perales and Mendoza-Torres (2020, Proposition 2.5), ũṽACG([0,1]). Thus by Proposition 3.5, uvWHK and

(6)
(uv)̇=(ũṽ)=ṽṽ+ũṽ=u̇v+uv̇.

Integrating (6) we obtain (5). □

4

4 Existence and uniqueness of a solution of a boundary value problem

Define the spacesWHK2={uWHK:u̇WHK}andWHK0={uWHK:u(0)=u(0+)=u(1)=u(1-)=0}.

Let fHK([0,1]) and let q,ρ be real valued functions such that qL2([0,1]),ρACG([0,1])BV([0,1]) and |ρ(x)|α for all x[0,1] and some α>0. Consider the following problems:

  • I. Find uWHK2 that satisfies

    (7)
    -[ρu̇]̇+qu=fa.e.on(0,1);u(0)=u(0+)=0,u(1)=u(1-)=0.

  • II. Find uWHK0 that satisfies

    (8)
    01ρu̇φ+01quφ=01fφfor all φV.

The boundary value problem (I) and the variational problem (II) are equivalent. Indeed, suppose that uWHK2 is a solution of the boundary value problem (I). Multiplying both sides of the differential equation in (7) by φV and integrating it from 0 to 1, we obtain that-01[ρu̇]̇φ+01quφ=01fφ.

Therefore01ρu̇φ+01quφ=01fφfor all φV. Conversely, suppose that uWHK0 satisfies (8) for all φV. Then,01ρu̇φ=-01(qu-f)φfor all φV. Thus ρu̇WHK. Since 1ρACG([0,1]), it follows by Proposition 3.5 and Corollary 3.7 that u̇WHK, consequently uWHK2. On the other hand, since ρu̇WHK we have that01ρu̇φ=-01(ρu̇)̇φfor all φV. Consequently, from (8),01[-(ρu̇)̇+qu-f]φ=0for all φV. Hence, by Corollary 3.2, -(ρu̇)̇+qu=f a.e. on [0,1].

To find a solution to the boundary value problem (I), we demonstrate the existence and uniqueness of the variational problem (II). Define B on WHK0×V and lu,lf on V byB(u,φ)=01ρu̇φ,lu(φ)=01quφ,lf(φ)=01fφ.

It is clear that lu and lf are linear operators and B is a bilinear operator. The variational problem (8) is equivalent to find uWHK0 such that

(9)
B(u,φ)+lu(φ)=lf(φ)for all φV.

Affirmation 1. There exists an operator A:WHK0WHK0 such that lu(φ)=B(A(u),φ) for all uWHK0 and φV.

Proof

Define the functionshu(t)=-1ρ(t)0tquandzu(t)=0thu-αu1ρ(t),where αu=01hu011ρ. Observe zuWHK and zu(0)=zu(0+)=zu(1)=zu(1-)=0. Define the operator A:WHK0WHK0 by

(10)
A(u)=zu.

Then A is a linear operator, and for each φV,B(A(u),φ)=01ρzu̇φ'=01ρhu-αu1ρφ'=01(ρhu-αu)φ'=010t-qu-αuφ'(t)dt=01quφ=lu(φ).

Affirmation 2. There exists a function vfWHK0 for which lf(φ)=B(vf,φ) for all φV.

Proof

Let F be the primitive of f, observe that Ḟ=f. Then vf defined byvf(t)=0t1ρ(F-βu),where βu=01Fρ011ρ, satisfies that vfWHK0 andB(vf,φ)=01ρvḟφ=01(F-βu)φ=01fφ=lf(φ),for all φV. 

Therefore, the problem (9) is equivalent to find uWHK0 such thatB(u,φ)+B(A(u),φ)=B(vf,φ),for all φV, orB(u+A(u)-vf,φ)=0,for all φV.

Remember that an operator T:XY between two normed spaces is compact if and only if for any bounded sequence (xn) in X, the sequence (Txn) contains a converging subsequence. The following is the Arzelá-Ascoli Theorem.

Theorem 4.1

A subset H ofC([0,1]) is relatively compact on C([0,1]) if and only if:

  • (i) H is pointwise bounded, i.e. for every x[0,1],supvH|v(x)|<.

  • (ii) H is equicontinuous, i.e. for every >0 there exists δ>0 such that|v(x)-v(y)|<,for all x,y[0,1] with |x-y|<δ, and for each vH.

Affirmation 3. The operator A:(WHK0,·)(WHK0,·) is compact, where u=u2+u̇A.

Proof

Let ΩWHK0 be bounded. Then there exists M>0 such that uM for all uΩ. LetH=hu-αu1ρ:uΩ.

We shall use the Arzelá-Ascoli theorem to prove that H is relatively compact in C([0,1]) with the uniform norm.

  • (i) First observe that if uΩ, then|αu|=01hu011ρ011|ρ(t)|0tqudt011ρq2u2α011ρq2Mα011ρ.

    Let x[0,1]. Then for every uΩ,hu(x)-αu1ρ(x)1|ρ(x)|0x|qu|+|αu|1αq2u2+|αu|q2Mα1+1α011ρ.

  • (ii) We set Q(t)=0t|q|2. Let >0, since Q and 1ρ are continuous on [0,1], there exists δ>0 such that if |y-x|<δ, then1ρ(y)-1ρ(x)<2q2M1+1α011ρand|Q(y)-Q(x)|<α2M.

    Let x,y[0,1] be such that |y-x|<δ. Suppose that x<y. Thenhu(y)-αu1ρ(y)-hu(x)-αu1ρ(x)|hu(y)-hu(x)|+|αu|1ρ(y)-1ρ(x)1ρ(y)-1ρ(x)(q2M+|αu|)+1αxy|qu|1ρ(y)-1ρ(x)q2M1+1α011ρ+MαQ(y)-Q(x)<.

Therefore, by the Arzelá-Ascoli theorem, H is relatively compact in C([0,1]) with the uniform norm. Now, let (un) be a bounded sequence in WHK0, then for the above, there exists a subsequence (unk) of (un) and gC([0,1]) such that hunk-αunk1ρg uniformly. This implies thatzunk(t)=0thunk-αunk1ρ0tguniformly for all t[0,1]. We set z(t)=0tg Then zACG([0,1]) and z(0)=z(0+)=z(1)=z(1-)=0. Therefore, zWHK0. It is clear that zunk-z20 and zunk̇-żA=hunk-αunk1ρ-gA0. ThusA(unk)-z0.Therefore A is a compact operator. 

We shall use the Fredholm’s Alternative Theorem:

Theorem 4.2

(Kress, 1989, Theorem 3.4) If A:XX is a compact operator on a normed space X. The equation (A+I)u=f has a unique solution for all fX if and only if the homogeneous equation (A+I)u=0 has only the trivial solution.

Affirmation 4. Let A:WHK0WHK0 be the operator defined in (10). If ρ>0 and q>0 then the equation (A+I)u=0 has only the trivial solution in WHK0.

Proof

Let uWHK0 be such that (A+I)u=0 a.e. on [0,1]. Then zu=-u hence by Proposition 3.5 u̇=-hu+αu1ρ. This implies thatρ(t)u̇(t)=0tqu+αufor almost all t[0,1]. Again, by Proposition 3.5,

(11)
(ρu̇)̇=qu.

Denote by u the conjugate of u, it is clear that uWHK0. Then by Corollary 3.7,01quu=01(ρu̇)̇u=ρu̇̃u01-01ρu̇(u)̇=-01ρu̇(u̇).

Thus01q|u|2+01ρ|u̇|2=0which implies that u=0 a.e. on [0,1]. 

Affirmation 5. If ρ>0 and q>0, then the uniqueness of the variational problem (8) holds.

Proof

Suppose that there exist u,vWHK0 such that satisfy the variational problem (8). Then01ρ(u̇-v̇)φ=-01q(u-v)φfor all vV. Therefore

(12)
[ρ(u̇-v̇)]̇=q(u-v).

Thus in similar way to Affirmation 4, we have that01q|u-v|2+01ρ|u̇-v̇|2=0.

Therefore u=v a.e. on [0,1]. 

Remark 3

From (11) and (12) it follows that the conclusions of Affirmation 4 and Affirmation 5 are also fulfilled if we assume that the homogeneous problem-[ρu̇]̇+qu=0a.e.on(0,1);u(0)=u(0+)=0,u(1)=u(1-)=0has only the trivial solution.

Finally, we present the existence and uniqueness theorem.

Theorem 4.3

Let fHK([0,1]) and let q,ρ be real valued functions such that qL2([0,1]),ρACG([0,1])BV([0,1]) and |ρ(x)|αfor all x[0,1] and some α>0. If one of the following conditions holds:

  • (i) ρ>0 and q>0,

  • (ii) the homogeneous problem-[ρu̇]̇+qu=0a.e.on(0,1);u(0)=u(0+)=0,u(1)=u(1-)=0has only the trivial solution,

then there exists a unique uWHK2 such that-[ρu̇]̇+qu=fa.e.on(0,1);u(0)=u(0+)=0,u(1)=u(1-)=0.

Example 1

The unique solution of the problem-[ρu̇]̇+qu=fa.e.on(0,1);u(0)=u(0+)=0,u(1)=u(1-)=0;where ρ(x)=x+1,q(x)=1 and f(x)=3x12+2-12x+x-x2, is given byu(x)=x(x-1).

In this case fHK([0,1]).

Acknowledgement

Authors acknowledge Mexican National Council for Science and Technology, Technological University of the Mixteca, and PRODEP-SEP for the support.

Declaration of Competing Interest

The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.

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