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Research article

Approximate solutions of Atangana-Baleanu variable order fractional problems

  • The main aim of this paper is to propose a new approach for Atangana-Baleanu variable order fractional problems. We introduce a new reproducing kernel function with polynomial form. The advantage is that its fractional derivatives can be calculated explicitly. Based on this kernel function, a new collocation technique is developed for variable order fractional problems in the Atangana-Baleanu fractional sense. To show the accuracy and effectiveness of our approach, we provide three numerical experiments.

    Citation: Xiuying Li, Yang Gao, Boying Wu. Approximate solutions of Atangana-Baleanu variable order fractional problems[J]. AIMS Mathematics, 2020, 5(3): 2285-2294. doi: 10.3934/math.2020151

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  • The main aim of this paper is to propose a new approach for Atangana-Baleanu variable order fractional problems. We introduce a new reproducing kernel function with polynomial form. The advantage is that its fractional derivatives can be calculated explicitly. Based on this kernel function, a new collocation technique is developed for variable order fractional problems in the Atangana-Baleanu fractional sense. To show the accuracy and effectiveness of our approach, we provide three numerical experiments.


    A topological space can be presumed as an axiomatization of the notion of a point's closeness to a set. When a point is a member of the closure of a set, it is said to be close to the set. According to the theory of metric spaces, which is an axiomatization of the idea of a pair of points being close to one another in a metric space, the distance between any two points is assessed by a real number, and its basic properties are outlined by a set of axioms. Nevertheless, the class of metric spaces is inextricably linked with the fascinating class of metrizable spaces, which is a class of topological spaces and plays a significant role in applications of modern and general topology, as well as in the development of proper topological structures and relations. We place a high priority on metrizable spaces because they are utilized in numerous interesting topological spaces in multiple mathematical disciplines. Numerous researchers have been working on its extension, generalization or improvement because of its wide range of applications in numerous fields of mathematics. Topological spaces and metric spaces are both extensively used topics. As a special case of topological spaces, metric spaces are actually of interest, and the suggested axioms of certain spaces are geometrically meaningful. This makes metrizability a fascinating topic for topological spaces. Unsurprisingly, some spaces are not metrizable. As a result, researchers try to build more general and metrizable functions. Metric spaces are a unique type of topological space. In metric spaces, sequences are used to characterize topological properties. Sequences are completely inadequate for such convenience in topological spaces. Seeking classes that are largely independent of topological spaces and metric spaces is simple, and with members, sequences play an important role in assessing their topological properties.

    A variety of spaces have recently been built, as well as some new types of modified metric spaces. Weakening the axioms of certain modified metric spaces or of metric spaces, in general, is the crucial step in creating these spaces. It is frequently not stated what the topological characteristics of new modified metric spaces are, and it is frequently not taken into account how these modified metric spaces relate to previously modified metric spaces in terms of fixed point theorems. In this article, extended suprametric spaces are introduced, and the contraction principle is established using elementary properties of the greatest lower bound instead of the usual iteration procedure. Thereafter, some topological results and the Stone-type theorem are derived in terms of suprametric spaces. Also, we have shown that every suprametric space is metrizable.

    The metric space has been generalized in numerous research works to more abstract spaces, including the b-metric spaces of Bakhtin[1] and Bourbaki[2], the partial metric spaces of Matthews[3] and the rectangle metric spaces of Branciari[4]. The b-metric was developed due to the theories of Bakhtin[1] and Bourbaki[2]. Czerwik[5] established an axiom that was weaker than the triangular inequality and specifically defined a b-metric space in order to develop the Banach contraction result. Numerous authors have generalized the b-metric space as a result of being inspired by its idea and have produced a variety of fixed-point results (see, for example, [6,7,8,9]).

    Definition 2.1. Let χ be a non-empty set.

    (a) A suprametric on the set χ is a function D:χ×χR+ which satisfies the following conditions:

    (i) ( x,yχ) D(x,y)=0  x=y;

    (ii) ( x,yχ) D(x,y)=D(y,x);

    (iii) (ζR+)( x,y,zχ) D(x,z)D(x,y)+D(y,z)+ζD(x,y)D(y,z).

    (b) An extended suprametric on the set χ is a function D:χ×χR+ which satisfies conditions (i) and (ii) of item (a) and the following condition:

    (iv) there exists a function γ:χ×χ[1,+) such that :

    ( x,y,zχ) D(x,z)D(x,y)+D(y,z)+γ(x,z)D(x,y)D(y,z).

    If D is a suprametric (respectively, an extended suprametric) on χ, then the ordered pair (χ,D) is called a suprametric space[10] (respectively, an extended suprametric space).

    Example 2.2. Assume that χ is a set of natural numbers. Define D:χ×χR+ by D(x,y)=(xy)2 and γ:χ×χR by γ(x,y)=ex+y. Then (χ,D) is an extended suprametric space.

    Proof. Clearly, conditions (i) and (ii) of item (a) holds for all x,yχ.

    For any x,y,zχ, consider

    D(x,z)=(xz)2=(xy+yz)2=(xy)2+(yz)2+2(xy)(yz)(xy)2+(yz)2+2(xy)2(yz)2<(xy)2+(yz)2+ex+z(xy)2(yz)2.

    Therefore, D(x,z)D(x,y)+D(y,z)+γ(x,z)D(x,y)D(y,z) for all x,y,zχ. Hence (χ,D) is an extended suprametric space.

    Remark 2.3. If γ(x,y)=ζ for ζ1 then we obtain the definition of a suprametric space.

    Definition 2.4. A sequence {xn} in χ is said to be a convergent sequence in an extended suprametric space (χ,D) if for every ε>0 there is M=M(ε)N such that D(xn,x)<ε, for all nM and xχ. In this case, we write it as limnxn=x.

    Definition 2.5. A sequence {xn} in χ is said to be a Cauchy sequence in an extended suprametric space (χ,D) for every ε>0 there is M=M(ε)N such that D(xm,xn)<ε, for all m,nM.

    Definition 2.6. An extended suprametric space (χ,D) is complete if and only if every Cauchy sequence in χ is convergent.

    Remark 2.7. Assume that (χ,D) is an extended suprametric space. If D is continuous, then every convergent sequence has a unique limit.

    Remark 2.8. Whenever a sequence {xn}nN is a Cauchy sequence in a complete extended suprametric space then there exists xχ such that limnD(xn,x)=0 and that every subsequence {xn()}N converges to x.

    Let K represent a non-empty collection of positive real numbers that is bounded below. Then c is an infimum of K and any number in K that exceeds c can't be the lower bound of K according to the infimum property of R (see [11]).

    A straightforward result of the infimum's properties is as follows:

    Lemma 2.9. Let K={a/a is a nonnegative real number} be a nonempty set with zero as its greatest lower bound (shortly, glb). Then limngn=0, where the sequence {gn}n=1 exists in K.

    Theorem 2.10. Assume that (χ,D) is a complete extended suprametric space such that D is continuous and Z:χχ be a mapping. Consider that there exists θ[0,1) such that

    D(Zx,Zy)θD(x,y),  for all x,yχ. (2.1)

    Then, Z has a unique fixed point, and for every x0χ the iterative sequence defined by xn=Zxn1, nN converges to this fixed point.

    Proof. Define the sequence xn by xn=Zxn1 for all nN for some arbitrary x0χ. Now, we demonstrate that the fixed point's existence can be effectively established by utilizing only the basic properties of an extended suprametric space and an infimum, omitting the usual iteration process.

    Case-1: For θ=0.

    This case is trivial since for θ=0, Z is a constant map, so Z has a fixed point.

    Case-2: For 0<θ<1.

    We define D={D(x,Zx)/xχ} and put p=infD. Suppose that p>0. Since 0<θ<1, we have pθ>p, so there exists xpD such that D(xp,Zxp)<pθ. Then

    D(Zxp,Z2xp)θD(xp,Zxp)<p, a contradiction.

    The contradiction obtained shows that p=0. Now, since p=0, there exists a sequence {xn} of members of χ such that limnD(xn,Zxn)=0. Thus, for all ε>0, there exists N such that for all n, as a result

    D(xn,xn+1)<ε. (2.2)

    Now we will demonstrate that the sequence {xn} is Cauchy.

    Using the existing assumptions and (2.2), and for large enough integers a,b such that b>a>, consider

    D(xa,xb)D(xa,xa+1)+D(xa+1,xb)+γ(xa,xb)D(xa,xa+1)D(xa+1,xb)θaD(x,x+1)+D(xa+1,xb)+γ(xa,xb)θaD(x,x+1)D(xa+1,xb)θaε+D(xa+1,xb)+γ(xa,xb)θaεD(xa+1,xb)θaε+[1+γ(xa,xb)εθa]D(xa+1,xb). (2.3)

    Similarly,

    D(xa+1,xb)D(xa+1,xa+2)+D(xa+2,xb)+γ(xa+1,xb)D(xa+1,xa+2)D(xa+2,xb)θa+1D(x+1,x+2)+D(xa+2,xb)+γ(xa+1,xb)θa+1D(x+1,x+2)D(xa+2,xb)θa+1ε+(1+γ(xa+1,xb)εθa+1)D(xa+2,xb). (2.4)

    From (2.3) and (2.4), we get

    D(xa,xb)εθa+[1+γ(xa,xb)εθa]D(xa+1,xb)εθa+[1+γ(xa,xb)εθa]{εθa+1+[1+γ(xa+1,xb)θa+1]D(xa+2,xb)}εθa+[1+γ(xa,xb)εθa]{εθa+1+D(xa+2,xb)+εθa+1γ(xa+1,xb)D(xa+2,xb)}εθa+εθa+1+D(xa+2,xb)+εθa+1γ(xa+1,xb)D(xa+2,xb)    +ε2θaθa+1γ(xa,xb)+εθaγ(xa,xb)D(xa+2,xb)+ε2θaθa+1γ(xa,xb)γ(xa+1,xb)D(xa+2,xb)εθa+εθa+1[1+γ(xa,xb)εθa]    +[1+γ(xa+1,xb)εθa+1+γ(xa,xb)εθa+γ(xa,xb)γ(xa+1,xb)ε2θaθa+1]D(xa+2,xb)εθa+εθa+1[1+γ(xa,xb)εθa][(1+εθaγ(xa,xb))(1+εθa+1γ(xa+1,xb))]D(xa+2,xb).

    Performing this process repeatedly and using (2.2) in each term of the sum until we reach continuing this process and using (2.2) in every term of the sum, until we obtain

    D(xa,xb)εθaba1i=0θii1η=0[1+εγ(xa+η,xb)θa+η]

    since θ[0,1), it follows that

    D(xa,xb)εθaba1i=0θii1η=0[1+εγ(xa+η,xb)θη]. (2.5)

    Let Ui=θii1η=0[1+εγ(xa+η,xb)θη].

    By Ratio test i=0Ui is converges, since limi|Ui+1Ui|<1 as θ[0,1).

    Hence from (2.5), we can deduce that, D(xa,xb) tends to zero as a,b tend to infinity, that suggests the sequence {xn} is Cauchy. Therefore by completeness of χ, as a result of this {xn} converges to some xχ(say).

    We will now prove that x is a fixed point of Z.

    By utilizing conditions (2) and (3) of Remark 2.7 and (2.1), we get

    D(Zxn(),Zx)θD(xn(),x).

    Therefore as , thus, we conclude x=Zx. Therefore, our assertion is true.

    To prove uniqueness, let us assume xa and xb are two fixed points of Z, additionally from (2.5), we get,

    D(xa,xb)=D(Zxa,Zxb)θD(xa,xb)<D(xa,xb),  a contradiction.

    Hence xa=xb. This completes the proof of the theorem.

    From the outset of General Topology, metrization has been and continues to be one of its most crucial fields. In the literature, there are numerous metrization theorems (see [12,13,14,15,16,17,18]). The metrizability hypotheses vary greatly from one metrization theorem to another, even though the thesis is always the same. Furthermore, not only are the proofs very dissimilar, but it is also difficult to draw any conclusions about one metrization theorem from another. The Stone-type theorem is derived in terms of suprametric spaces in this section. In addition, we have shown that every suprametric space is metrizable.

    Assume that (χ,D) is a suprametric space, x0χ and ϱ a positive number, the set B(x0,ϱ)={xχ/D(x0,x)<ϱ} is called the open ball with centre x0χ and radius ϱ>0 or, briefly the ϱ-ball about x0 moreover, B[x0,ϱ]={xχ/D(x0,x)ϱ} is the closed ball. For a set Aχ and a positive number ϱ, by the ϱ-ball about A we mean the set B(A,ϱ)=xAB(x,ϱ); let us note that xB(x,ϱ) so that AB(A,ϱ). Let us also observe that if x1B(x0,ϱ), then B(x0,ϱ1)B(x0,ϱ) for ϱ1=ϱD(x0,x1)1+ζD(x0,x1)>0.

    A subset Y of χ is said to be open if for any point yY, there is ϱ>0 such that B(y,ϱ)Y. Let s={Yχ/ xY, there exists ϱ>0 such that B(x,ϱ)Y}. One can easily see that s is a topology on χ.

    For every xχ we define a collection of families of subsets M(x) of χ that have the following characteristics:

    (Bp1) For every xχ,M(x) and for every QM(x), xQ.

    (Bp2) If xQM(y) then there is a RM(x) such that RQ.

    (Bp3) For any Q1,Q2M(x) there exists a QM(x) such that QQ1Q2;

    where M(x)={B(x,ϱ)/ϱ>0}.

    Let be the family of all subsets of χ that are unions of subfamilies of M(x). That is, let Q iff Q=B0 for a subfamily B0 of M(x). Clearly M(x) is a base for the space (χ,) and is the topology generated by the base M(x). Since =B0 for B0=, χ=B0 for B0=M(x).

    Take Q1,Q2, which yield Q1=sSQs and Q2=tTQt, where Qs,QtM(x) for sS and tT. Q1Q2=sS,tTQsQt, since for every xQsQt there exists a Q(x)M(x) such that xQ(x)QsQt, that suggests QsQt=B0 for B0={Q(x)/xQsQt}. As a result, we construct a topology on the set χ is called the topology induced by the metric D. One can easily check that and s coincides. Clearly the family of all open balls is a base for (χ,). The family of all 1i balls about x0, where i=1,2,3... is a base for (χ,) at the point x0, that would suggest the space (χ,) is first countable.

    The topology induced by the suprametric D on a set χ was introduced by Maher Berzig[10] in Proposition 1.2.

    The topology is Hausdorff space according to [10], since for every pair x,y of distinct points of χ, it follows from the (iii) of Definition 2.1 that B(x,ϱ2) and B(y,ϱ2+ζϱ) are disjoint neighbourhoods of x and y, where ϱ=D(x,y)>0. Also note that every suprametric space is continuous[10].

    Example 3.1. We define a suprametric on set of all infinite sequence {pi} of real numbers satisfying the condition i=0p2i< (say H).

    That is, H={set of all infinite sequence {pi}of real numbers/i=1p2i<} which is called Hilbert space.

    Let us define,

    D(p,q)=i=0(piqi)2 for p=pi,q=qi.

    We will prove that D is a suprametric. First of all to prove that D is well-defined. That is, the series in the definition of D is convergent.

    Let us note that for every pair of points p=pi,q=qi in H and any positive integer , we have

    i=1(piqi)2=(i=1p2i+i=1q2i)2.

    Thus the series in the notion of D is convergent and D(p,q) is well-defined. Now we will prove that D is a suprametric. Clearly, D satisfies (D1),(D2) of suprametric definition. We shall show that condition (iii) of Definition 2.1 is also satisfied.

    Let p=pi,q=qi and ϱ=ϱi be any points of H and let

    p={p1,p2,....,p,0,0,....};
    q={q1,q2,....,q,0,0,....};
    ϱ={ϱ1,ϱ2,....,ϱ,0,0,....};

    and ui=piqi, vi=qiϱi, wi=piϱi.

    By Cauchy inequality we have

    [D(p,ϱ)]2=i=1w2i=i=1(ui+vi)2=i=1u2i+2i=1uivi+i=1v2ii=1u2i+2i=1u2ii=1v2i+i=1v2ii=1u2i+2i=1u2ii=1v2i+i=1v2i+2ζi=1u2ii=1v2i       +2ζi=1u2ii=1v2i+ζ2i=1u2ii=1v2i=(i=1u2i+i=1v2i+ζi=1u2ii=1v2i)2=(i=1u2i+i=1v2i+ζi=1u2ii=1v2i)2=(i=1(piqi)2+i=1(qiϱi)2+ζi=1(piqi)2i=1(qiϱi)2)2.

    For =1,2,...., we have

    D(p,ϱ)D(p,q)+D(q,ϱ)+ζD(p,q)D(q,ϱ)

    which implies,

    D(p,ϱ)D(p,q)+D(q,ϱ)+ζD(p,q)D(q,ϱ).

    Hence D is a suprametric on H.

    Definition 3.2. [19] Suppose H={Hq:qS} be a family of subsets of topological space χ.

    ● If for all xχ there will be a neighbourhood Qx of x, so that the family {qS: QxHq} is finite then H is known as locally finite.

    ● If for all xχ, there will be a neighbourhood Qx of x, so that the family {pS:HpQx} will have at most one element then H is known as discrete. It is obvious that any finite family is locally finite.

    ● For every locally finite Hi, if H=i=NHi, then the family H is called σ-locally finite.

    ● For every Hi is discrete, if H=i=NHi then the family H is called σ-discrete

    ● If pSHp=χ, then the family H is called a cover of χ.

    ● If for all iI there will be pS, so that BiAp, then a cover B of subsets of χ is known as a refinement of the cover H, where B={Bi/iI}.

    Definition 3.3. Let O and L are two disjoint non-empty closed subsets of χ. If OQ and LR for two non-empty disjoint open sets Q and R in the topological space (χ,). Then χ is said to be a regular space.

    Definition 3.4. Assume that (χ,D) is a suprametric space and χ is a super set of W. Accordingly,

    (1) W is said to be an open set whenever W.

    (2) W is said to be a closed set whenever χW.

    (3) xχ is called to be a limit point of W whenever there is ϱ>0 such that (W(x,ϱ){x})W having an infinite number of points of W.

    (4) Denoted by W, the collection of all limit points of W is known as the derived set of W.

    Proposition 3.5. Closed balls are closed set in a suprametric space (χ,D).

    Proof. Take xχ, ϱ>0, and the closed ball B[x,ϱ]. In order to show that B[x,ϱ] is closed, then it suffices to depict χB[x,ϱ]=F(fix) is open. Let yF. Thus D(x,y)=ϱ(say)>ϱ. We must now find some s>0 such that B(y,s)F.

    Choose s>0 such that s<ϱϱ1+ζϱ. Let aB(y,s), thus D(a,y)<s. Besides xχ,yF and aB(y,s), we now have D(x,y)D(x,a)+D(a,y)+ζD(x,a)D(a,y).

    D(x,y)D(x,y)D(a,y)1+ζD(a,y)>ϱs1+ζs>ϱ(ϱϱ1+ζϱ)1+ζ(ϱϱ1+ζϱ)=ϱ(ϱζ+1)1+ζϱ=ϱ.

    Thus D(x,a)>ϱ if aB(y,s), having 0<s<ϱϱ1+ζϱ. Which yields F is an open set. Consequently B[x,ϱ] is a closed set.

    Theorem 3.6. Assume that (χ,D) is a suprametric space. Whenever E be a closed subset of χ and xχE then there are two disjoint open sets Q and R containing E and x.

    Proof. As xχE, E is closed. Which yields, D(x,a)>0 for all aE.

    Let 2ϱ=inf{D(x,a)/aE}, where ϱ>0. Let us assume the open ball B(x,ϱ2)=R (fix) and the open set Q=aEB(a,3ϱ2+ζϱ). Therefore EQ.

    We will prove that QR=. Assume that there is ξQR. Thus for any aE,

    D(x,a)D(x,ξ)+D(ξ,a)+ζD(x,ξ)D(ξ,a)<ϱ2+3ϱ2+ζϱ+ζϱ23ϱ2+ζϱ=2ϱ.

    Our assumption is contradicted by this, so Q and R are two disjoint non-empty open sets in χ, each containing E and x.

    Theorem 3.7. Let O and L be two disjoint non-empty closed subsets of a suprametric space (χ,D). Then OQ and LR for two disjoint open sets R and Q in χ.

    Proof. Let aO, bL and ab which implies D(a,b)>0.

    Let 2ϱ=glb{D(a,b)/aO, bL}. Let R=bLB(b,ϱ2) which consists L, where R is an open set. Now for aO, bL and ϱ>0.

    Consider Q=aOB(a,3ϱ2+ζϱ). Hence Q is open and OQ.

    Now we will prove that Q and R are disjoint. If this is not the case, there exist ξQR. Whereupon for every aO and bL, D(a,ξ)<3ϱ2+ζϱ,D(b,ξ)<ϱ2.

    So for aO,bL and ξQR,

    D(a,b)D(a,ξ)+D(ξ,b)+ζD(a,ξ)D(ξ,b)<3ϱ2+ζϱ+ϱ2+ζ3ϱ2+ζϱϱ2=3ϱ2+ζϱ(1+ζϱ2)+ϱ2=2ϱ, a contradiction.

    Our assumption is contradicted by this, so QR=. This completes the proof.

    Definition 3.8. A topological space χ is called suprametrizable if there exists a suprametric D on χ which induces the topology of χ.

    Theorem 3.9. [13] (The Stone Theorem) A metrizable space has an open refinement for each open cover that is both σ-discrete and locally finite.

    Theorem 3.10. [14] (The Bing Metrization Theorem) A topological space is metrizable if and only if it is regular and has a σ-discrete base.

    Theorem 3.11. [15] (Collins-Roscoe Metrization Theorem) Let χ be a T1-space such that, for every xχ, there exists a countable neighborhood base B(x)={W(n,x):nN} at x satisfying the following conditions:

    (1) (nN) W(n+1,x)W(n,x);

    (2) (xχ)(nN)(rN)(nr( xW(r,x)))(xW(n,y)W(n,y)W(n,y)).

    Then χ is metrizable.

    Theorem 3.12. (Stone-type Theorem) Assume (χ,D) is a suprametric space. Then, for every open cover of χ, there is an open refinement that is both locally finite and σ-discrete.

    Proof. Let W={W}S be an open cover of χ. Take a suprametric D on the space χ and < is a well-order relation on S. Let Xi={R,i}Sχ

    R,i=δCB(δ,12i), (3.1)

    where the union is taken over all points δχ gratifying the below mentioned assertions.

    (c1) In order for δW to exist, must be the smallest element in S.

    (c2) δRp,η for η<i and pS.

    (c3) B(δ,32i(1+ζ2i)+ζ223i)W.

    The set R,i is open, as it is given in Proposition 1.1 of [10]. Note that ζ0 by Definition 1.1 of [10]. Since B(δ,12i)B(δ,32i(1+ζ2i)+ζ223i), from condition (c3), we get R,iW. Let x be a point of χ, take the smallest element S such that xW and a natural number i such that

    B(x,32i(1+ζ2i)+ζ223i)W.

    It implies that xC if and only if xRp,η for all η<i and all pS. In this case xR,i. Then we either have xRp,η for some η<i and some pS or xR,i. Hence the union X=i=1Ri is an open refinement of the cover {W}S.

    We shall prove that for every i. If x1R1,i,x2R2,i and 12 then

    D(x1,x2)>12i, (3.2)

    and this will show that the families Ri are discrete, because every 12i+1 ball meets at most one member of Ri.

    Let x1R1,i and x2R2,i iN having 12. Let us assume that 1<2. By the definition of R1,i and R2,i, there exist points δ1,δ2 satisfying conditions (c1)(c3) such that xB(δ,12i)R,i for =1,2.

    From condition (c3), as a result of this, B(δ1,32i(1+ζ2i)+ζ223i)W1 and from (c1) we see that δ2W1, so that D(δ1,δ2)32i(1+ζ2i)+ζ223i.

    Hence

    D(x1,x2)D(δ1,δ2)D(δ1,x1)D(x2,δ2)ζD(δ1,x1)D(x2,δ2)1+ζD(x2,δ2)+ζD(δ1,x1)+ζ2D(δ1,x1)D(δ2,x2)32i(1+ζ2i)+ζ223i12i12iζ22i1+ζ2i+ζ2i+ζ222i32i+3ζ22i+ζ223i22iζ22i1+2ζ2i+ζ222i=12i+2ζ22i+ζ223i(1+ζ2i)2=12i(1+ζ2i)2(1+ζ2i)2=12i.

    Hence D(x1,x2)12i.

    If there exists xχ such that x1,x2B(x,1+1+ζ2iζ) then we have

    12iD(x1,x2)D(x1,x)+D(x,x2)+ζD(x1,x)D(x,x2)1+1+ζ2iζ+1+1+ζ2iζ+ζ(1±1+ζ2iζ)2=2(1+1+ζ2iζ)+(1+1+ζ2i)2ζ=12i.

    Since this goes against our assumption, so radius 1+1+ζ2iζ of every ball coincides at only one element of Ri. This can be written as X=iNRi is σ-discrete.

    Assume that iN, additionally for each pS, iη+ and δC which gives δRp,η. Here whenever B(x,12)Rp,η then δB(x,12), which yields that D(x,δ)12. Note that η++1 and i+1 then 12η+12+1 and 12i12+1.

    Let ψ=max{12η+,12i} and ψ<1+1+ζ2ζ12+1.

    Suppose to the contrary that there exists yB(x,12η+)B(δ,12i) then

    D(x,δ)D(x,y)+D(y,δ)+ζD(x,δ)D(y,δ)12η++12i+ζ(12η+)(12i)1+1+ζ2ζ+1+1+ζ2ζ+ζ(1+1+ζ2ζ)212

    which concludes that B(x,12η+)B(δ,12i)=.

    This implies B(x,12η+)R,i for iη+ and S with B(x,12)Rp,η. Let xχ, as such R is refinement of W, there exist l,η and p so D(x,12l)Rp,η and thus there is ,η and p so D(x,12)Rp,η. Thus the ball B(x,12η+) fulfils at most η+1 members of R. Which concludes, R is σ-locally finite as Xi is locally finite.

    Corollary 3.13. Assume that (χ,D) is a suprametric space. Thus, there exists σ-discrete base for χ.

    Proof. Let Fi={B(x,1i):xχ} for all iN. Which yields χ has an open cover Fi. Making use of Theorem 3.12, Fi has an open σ-discrete refinement Bi. We get our claim that χ has a σ-discrete base iNBi, as one can easily prove that iNBi is a base of χ.

    Theorem 3.14. Every suprametric space (χ,D) is metrizable.

    Proof. Now, we present two proofs of the metrizability of suprametric spaces using two different approaches. In the first approach, Stone's theorem is used, while in the second, the Collins-Roscoe metrization theorem is used.

    Approach-Ⅰ.

    One can easily conclude that χ is regular space with σ-discrete base from Corollary 3.13 and Theorem 3.6. Thus χ is metrizable by making use of Theorem 3.10.

    Approach-Ⅱ.

    Now, assume that D is a suprametric on a set χ, and ζR+ is a fixed constant such that:

    (x,y,zχ) D(x,z)D(x,y)+D(y,z)+ζD(x,y)D(y,z).

    Let D be the topology induced by D. It is trivial that (χ,D) is a T1-space because if x,yχ and xy, then D(x,y)>0 (the author of [10] noticed that (χ,D) is even a Hausdorff space). For every xχ and nN, we put W(n,x)={yχ:D(x,y)<12n}. Clearly, for every xχ, the family {W(n,x):nN} is a local neighborhood base at x in (χ,D). Of course, for every xχ and nN, W(n+1,x)W(n,x). Let us fix xχ and nN. We can fix rN such that nr and 12r1+ζ22r<12n. Consider any yW(r,n). Then xW(r,x)W(n,y). To show that W(r,y)W(n,x), we consider any zW(r,y). Then D(x,z)D(x,y)+D(y,z)+ζD(x,y)D(y,z)<12r+12r+ζ122r<12r. This implies that zW(n,x), so W(r,y)W(n,x). It follows from the Collins-Roscoe Metrization Theorem 3.11 that (χ,D) is metrizable.

    Remark 3.15. It is well known that Stone's theorem is unprovable in ZF (see Theorem 2 of [20]). Additionally, in ZF, it is unprovable to demonstrate that any metric space has a σ-locally finite basis. (see From 232 in the book: Howard, Paul; Rubin, Jean E. (1998). Consequences of the axiom of choice. Mathematical Surveys and Monographs. Vol. 59. Providence, Rhode Island: American Mathematical Society. ISBN 9780821809778. [21]). Therefore Theorem 3.12 and Corollary 3.13 of the manuscript are both unprovable in ZF. As a result, this manuscript's Theorem 3.12, which involves the axiom of choice, leads to the conclusion that any suprametrizable space is metrizable. Furthermore, the Collins-Roscoe Metrization Theorem, which is an alternate and more straightforward argument, is used to demonstrate that suprametrizable space is metrizable.

    Remark 3.16. Theorem 3.12 and Corollary 3.13 are straightforward from the perspective of Approach-Ⅱ. Furthermore, any suprametrizable space is normal, which follows directly from Approach-Ⅱ. Therefore, according to Approach-Ⅱ, Theorems 3.6 and 3.7 are not necessary in order to establish that X is metrizable.

    We will examine the existence and uniqueness of a random solution to a stochastic integral equation of the following type in this section:

    z(t,λ)=t0φ(τ,z(τ,λ))dτ+t0ψ(τ,z(τ,λ))dδ(τ), (4.1)

    where t[0,1]. The subsequent integral, defined in regard to scalar Brownian motion procedures {δ(t)}, where t[0,1], is an Ito-type stochastic integral. The primary integral is a Lebesgue integral. Keep in mind that C([0,1],L2(ϕ,S,K))Cc(R+,L2(ϕ,S,K)). The operators will be defined as U and U from C([0,1],L2(ϕ,S,K)) into C([0,1],L2(ϕ,S,K)) by

    Uz(t,λ)=t0z(t,λ)dτ (4.2)

    and

    Uz(t,λ)=t0z(t,λ)dδ(τ). (4.3)

    Here z(t,λ)C([0,1],L2(ϕ,S,K)).

    Lemma 4.1. [22] U and U are continuous operators from C([0,1],L2(ϕ,S,K)) into C([0,1],L2(ϕ,S,K)). These operators are characterized by (4.2) and (4.3), accordingly.

    Let T be a linear operator, and let A and B be a pair of Banach spaces. The preceding lemma, which is relevant to the examination of this section, is given. It is employed in the main theorem.

    Lemma 4.2. [22] Let T be a continuous operator from Cc(R+,L2(ϕ,S,K)) into itself. If A and B are Banach spaces stronger than Cc and the pair (A,B) is admissible with respect to T, then T is a continuous operator from A to B.

    Definition 4.3. [22] The pair of spaces (A,B) will be called admissible with respect to the operator T:Cc(R+,L2(ϕ,S,K))Cc(R+,L2(ϕ,S,K)) if and only if T(B)A.

    Definition 4.4. [22] By stating that the Banach space B is stronger than the space Cc(R+,L2(ϕ,S,K))we mean that every convergent sequence in B, with respect to its norm, will also converge in Cc.

    The aforementioned theorem identifies the necessary conditions for a second-order stochastic process, a unique random solution to Eq (4.1), to exist.

    Theorem 4.5. Let χ=C([0,1],L2(ϕ,S,K)) be the space of all continuous and bounded functions on [0,1] with values in L2(ϕ,S,K). Note that χ is extended suprametric space by considering D(z(t,λ),ˆz(t,λ))=sup|z(t,λ)ˆz(t,λ)|2 with γ(z(t,λ),ˆz(t,λ))=ez(t,λ)+ˆz(t,λ), where γ:χ×χ[1,).

    Considering the aforementioned assumptions, take into account the stochastic integral equation (4.1).

    (1) A and B are subsets in C([0,1],L2(ϕ,S,K)) which are stronger than C([0,1],L2(ϕ,S,K)) such that (A,B) is admissible with respect to the operators U and U;

    (2) z(t,λ)φ(τ,z(τ,λ)) is an operator on S={z(t,λ)/z(t,λ)B and |z(t,λ)|κ} with values in A satisfying

    |φ(τ,z(τ,λ))φ(τ,ˆz(τ,λ))|Aη1|z(τ,λ)ˆz(τ,λ)|B;

    (3) z(t,λ)ψ(τ,z(τ,λ)) is an operator on S into A satisfying

    |ψ(τ,z(τ,λ))ψ(τ,ˆz(τ,λ))|Aη2|z(τ,λ)ˆz(τ,λ)|B;

    (4) (1η1+2η2)2<1 and |φ(t,0)|A+|ψ(t,0)|Ae(11η12η2);

    (5) Lemma 4.2 holds in view of U and U, i.e.,

    |(Uz)(t,λ)|B1|z(t,λ)|A and
    |(Uz)(t,λ)|B2|z(t,λ)|A,

    where 1,2<1.

    Then Eq (4.1) has a unique random solution.

    Proof. We split the proof into four steps.

    Step-1. Constructing an operator M:SB to which we can apply our Theorem 2.10, for z(t,λ)S,

    Mz(t,λ)=t0φ(τ,z(τ,λ))dτ+t0ψ(τ,z(τ,λ))dδ(τ). (4.4)

    The set S is a closed subset of χ=C([0,1],L2(ϕ,S,K)) endowed with suprametric D. So (χ,D) is complete.

    Step-2. We will show that M is a contraction operator on χ.

    Let z(t,λ),ˆz(t,λ)χ.

    Consider,

    |Mz(t,λ)Mˆz(t,λ)|2B=[|t0[φ(τ,z(τ,λ))φ(τ,ˆz(τ,λ))]Bdτ   +t0[ψ(τ,z(τ,λ))ψ(τ,ˆz(τ,λ))]dδ(τ)|]2B[t0|φ(τ,z(τ,λ))φ(τ,ˆz(τ,λ))|Adτ   +t0|ψ(τ,z(τ,λ))ψ(τ,ˆz(τ,λ))|dδ(τ)]2A[1|φ(t,z(t,λ))φ(t,ˆz(t,λ))|+2|ψ(t,z(t,λ))ψ(t,ˆz(t,λ))|]2[1η1|z(t,λ)ˆz(t,λ)|B+2η2|z(t,λ)ˆz(t,λ)|B]2[(1η1+2η2)|z(t,λ)ˆz(t,λ)|B]2(1η1+2η2)2|z(t,λ)ˆz(t,λ)|2B. (4.5)

    Step-3. Observe that M:χχ, for z(t,λ)χ, we need to show that (Mz)χ.

    For any element z(t,λ)χ, we have

    |(Mz)(t,λ)|B=|t0φ(τ,z(τ,λ))dτ+t0ψ(τ,z(τ,λ))dδ(τ)||t0φ(τ,z(τ,λ))dτ|B+|t0ψ(τ,z(τ,λ))dδ(τ)|B1|φ(t,z(t,λ))|A+2|ψ(t,z(t,λ))|A1η1|z(t,λ)|B+2η2|z(t,λ)|B+1|φ(t,0)|A+2|ψ(t,0)|A.

    Since z(t,λ)S, it follows that

    |(Mz)(t,λ)|B|κ(1η1+2η2)+|φ(t,0)|A+|ψ(t,0)|κ.

    Thus (Mz)S.

    Step-4. From Eq (4.5), |Mz(t,λ)Mˆz(t,λ)|2(1η1+2η2)2|z(t,λ)ˆz(t,λ)|2.

    Taking supremum on both sides, we get D(Mz(t,λ),Mˆz(t,λ))θD(z(t,λ),ˆz(t,λ)) where θ=(1η1+2η2)2<1.

    Thus, all the conditions of Theorem 2.10 satisfied. Thus, the existence and uniqueness of a random solution of Eq (4.1) follow from the Theorem 2.10.

    Open Question: Let (χ,D) be a extended suprametric space. If D is continuous in one variable, then χ is metrizable?

    In this article, we focus on the extended suprametric space which opens new rooms for researchers. We consider that this new structure shall lead to the help of the solutions of certain differential equations and hence, produce new applications. In addition, we foresee that it shall allow us to achieve more refined results in existing applications.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    Authors show their gratitude to those reviewers because their comments improved the paper, and therefore they deserve the acknowledgment.

    The authors declare that they do not have any conflict interests.



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