Research article

Some positive results for exponential-kernel difference operators of Riemann-Liouville type

  • Received: 18 November 2023 Revised: 05 February 2024 Accepted: 18 February 2024 Published: 08 April 2024
  • We established positivity of f obtained from a systematic computation of a composition of sequential fractional differences of the function f that satisfy certain conditions in a negative lower bound setup. First, we considered the different order sequential fractional differences in which we need a complicated condition. Next, we equalled the order of fractional differences and we saw that a simpler condition will be needed. We illustrated our positivity results for an increasing function of the rising type.

    Citation: Pshtiwan Othman Mohammed. Some positive results for exponential-kernel difference operators of Riemann-Liouville type[J]. Mathematical Modelling and Control, 2024, 4(1): 133-140. doi: 10.3934/mmc.2024012

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  • We established positivity of f obtained from a systematic computation of a composition of sequential fractional differences of the function f that satisfy certain conditions in a negative lower bound setup. First, we considered the different order sequential fractional differences in which we need a complicated condition. Next, we equalled the order of fractional differences and we saw that a simpler condition will be needed. We illustrated our positivity results for an increasing function of the rising type.



    Discrete operators and fractional differences/sums are important in many fields of science, including applied science, mathematics, engineering sciences, and physics, as well as some related research fields such as quantum mechanics, number theory, fluid dynamics, mathematical physics, and ordinary/partial differential equation (see, e.g., [1,2,3,4,5,6,7]). Recently, Wu et al. [8] studied the inverse problem model and concept of inverse-time fractional chaotic maps with some application involving Riemann-Liouville fractional difference operators; Wu et al. [9] studied Liouville-Caputo fractional difference operators with some more definitions of fractional differences and their applications to fractional maps were compared; Abdeljawad and Baleanu [10] defined Caputo-Fabrizio fractional difference operators and they studied the integration by parts and Euler-Lagrange equations on these operators; and Mohammed et al. [11] examined sharpness results analytically and numerically for those operators involving Atangana-Baleanu fractional differences.

    The positivity analysis, which is derived from discrete fractional operators, is one of the significant models in the context of discrete fractional calculus. This analysis is also known as monotonicity analysis, and it represents the positivity of f. The positivity analyses have been frequently utilized to check if a function is increasing or decreasing. Some of the featured applications of positivity analysis are related to Riemann-Liouville fractional difference type [12,13], Liouville-Caputo fractional difference type [14,15], Caputo-Fabrizio fractional difference type [16,17,18], and Atangana-Baleanu fractional difference type [19,20]. The published articles [21,22,23,24] are also important from a sequential aspect to understanding the behavior of monotonicity and positivity analyses in a composition of two discrete fractional operators.

    In the present study, we consider the sequential fractional difference operator

    (CFa+1αrlCFaβrlf)(τ) (1.1)

    defined on the pair set

    D1:={(α,β);0<α,β<1and1α+β<2forβα}, (1.2)

    or, in the case when α=β, on the set

    D2:={(α,β);0<α,β<1and1α+β<2forβ=α} (1.3)

    for α,βR. We will analyze (1.1) to produce the positivity of (f)(τ) under certain conditions. Incidentally, we can say that this article is the extension of the Liouville-Caputo work [17], but here in the sense of Riemann-Liouville operators.

    The subsequent sections of the article are organized as follows: In Section 2, we recall the definition of the Caputo-Fabrizio that occurred in (1.1) and we formulated the main lemmas. Moving on to Section 3, we delve into the topics of positivity and monotonicity in analysis for sequential operator Eq (1.1). We end with some concluding remarks about the main results in Section 4.

    The discrete operator that appeared in (1.1) is the fractional difference operator with exponential kernel or, briefly, is the Caputo-Fabrizio fractional difference operator in the sense of Riemann-Liouville. It is defined by the following summation formula (see [1,4]):

    (CFaαrlf)(τ)=B(α)τ[τs=a+1f(s)(1α)τs], (2.1)

    where τNa+1, aR, α[0,1), and B(α)>0 is a normalization constant. It is also worth mentioning that (f)(τ) is the difference operator of f and is given by f(τ)f(τ1).

    Lemma 2.1. If (α,β)D1, then

    σ1(i):=1αβ[β(1β)iα(1α)i]0 (2.2)

    is nonnegative, and

    σ2(i):=1αβ[(1β)i(1α)i] (2.3)

    is positive for each iN1.

    Proof. By using the mathematical induction process, first for i=1, we have

    σ1(1):=1αβ[β(1β)α(1α)]=1+α+β0,

    since α+β1. Now, we suppose that

    σ1(ȷ)=1αβ[β(1β)ȷα(1α)ȷ]0, (2.4)

    for some ȷN1, then we have to prove that σ1(ȷ+1)0. Here, two cases arise: the case when α>β leads to

    σ1(ȷ+1)=1αβ>0[β(1β)ȷ+1α(1α)ȷ+1],

    which remains to prove that

    [β(1β)ȷ+1α(1α)ȷ+1]0.

    It is clear that

    β(1β)ȷ+1by(2.4)α(1α)ȷ(1β)α(1α)ȷ+1,

    where the fact 1β>1α>0 has been used and thus σ1(ȷ+1)0 for α>β. By the same process, we can prove that σ1(ȷ+1)0 for α<β. Hence, σ1(i)0 for each (α,β)D1 and iN1. Consequently, the first part of the lemma is proved.

    To prove the inequality (2.3): for each i1, we have

    1αβ[(1β)i(1α)i]>1αβ>0[(1β)i(1β)i]=0=0,

    for α>β ( 1β>1α>0), and

    1αβ[(1β)i(1α)i]=1βα[(1α)i(1β)i]>1βα>0[(1β)i(1β)i]=0=0,

    for β>α (hence, 1α>1β>0). Thus, the proof is completed.

    Lemma 2.2. If α[12,1), then

    σ3(i):=(1α)i1[αi(1α)] (2.5)

    is nonnegative for each iN1.

    Proof. Applying induction on (2.5), we have for i=1, σ3(1)=2α1, and it is clear that is nonnegative as α[12,1). Suppose that σ3(ȷ)0; that is,

    (1α)ȷ1[αȷ(1α)]0, (2.6)

    for some ȷN1, then we have to prove that σ3(ȷ+1)0. It is clear that

    (1α)ȷ[α(ȷ+1)(1α)]=(1α)>0(1α)ȷ1[αȷ(1α)]0by claim (2.6)+α(1α)ȷ>00,

    and this tells us σ3(ȷ+1)0. Thus, we see that (2.5) is true for each iN1. The proof is done.

    Lemma 2.3. If f is a function f: NaR, then we have

    (CFaαrlf)(τ)=B(α)[(f)(τ)αf(a+1)(1α)τ2aατ1s=a+2(f)(s)(1α)τs1],

    for α(0,1) and τNa+2.

    Proof. Considering the definition of (2.1), we have

    (CFaαrlf)(τ)=B(α)[τs=a+1f(s)(1α)τsτ1s=a+1f(s)(1α)τs1]=B(α)[f(τ)ατ1s=a+1f(s)(1α)τs1].

    It follows from this that

    (CFaαrlf)(τ)=B(α)[(f)(τ)ατ1s=a+1f(s)(1α)τs1+ατ2s=a+1f(s)(1α)τs2]=B(α)[(f)(τ)αf(a+1)(1α)τ2aατ1s=a+2f(s)(1α)τs1+ατ1s=a+2f(s1)(1α)τs1]=B(α)[(f)(τ)αf(a+1)(1α)τ2aατ1s=a+2(f)(s)(1α)τs1], (2.7)

    for each τNa+2. This completes the proof.

    In this section, we give some positivity results on the sequential fractional difference (1.1) when there is a negative lower bound in the right side of the inequality in Theorem 3.1. These results are based on the set D1 as plotted in Figure 1.

    Figure 1.  The region of D1.

    Theorem 3.1. If

    ξ0andNȷa:={a,a+1,,ȷ}

    and the function f: NaR satisfies

    (i) (f)(a+2)0;

    (ii)

    (CFa+1αrlCFaβrlf)(τ)ξB(α)B(β){f(a+2)βf(a+1)};

    (iii)

    β(1β)τ2aα(1α)τ2aαβξ;

    for each (α,β)D1 and τNȷa+3, for some ȷNa+3, then, (f)(τ)0 for every τNȷa+2.

    Proof. Let

    h(τ):=(CFaβrlf)(τ),

    then (2.1) enables us to write

    (CFa+1αrlCFaβrlf)(τ)=(CFa+1αrlh)(τ)=B(α)[τs=a+2h(s)(1α)τsτ1s=a+2h(s)(1α)τ1s]=B(α)[(1α)τa2h(a+2)+τs=a+3(h)(s)(1α)τs]. (3.1)

    It follows from Lemma 2.3 that

    (CFa+1αrlCFaβrlf)(τ)=B(α)B(β)[(1α)τa2{f(a+2)βf(a+1)}+(f)(τ)+τ1s=a+3(f)(s)(1α)τsβf(a+1)τs=a+3(1β)sa2(1α)τsβτs=a+3s1r=a+2(f)(r)(1β)sr1(1α)τs]:=B(α)B(β)[(1α)τa2{f(a+2)βf(a+1)}+A1(τ)A2(τ)f(a+1)A3(τ)], (3.2)

    where the following is used,

    h(a+2)=(CFaβrlf)(a+2)=f(a+2)βf(a+1).

    Compute A1(τ), A2(τ), and A3(τ) to get

    A1(τ):=τs=a+3(f)(s)(1α)τs=(f)(τ)+τ1s=a+3(f)(s)(1α)τs, (3.3)
    A2(τ):=βτs=a+3(1β)sa2(1α)τs=β(1β)(1α)τa3τa3κ=0(1β1α)κ=β(1β)(1α)τa31(1β1α)τa211β1α=β(1β)((1β)τa2(1α)τa2αβ), (3.4)

    and

    A3(τ):=βτs=a+3s1r=a+2(f)(r)(1β)sr1(1α)τs=βτ1r=a+2[(f)(r)(1α)τ(1β)r+1τs=r+1(1β1α)s]=βτ1r=a+2(f)(r)(1α)τ1r1(1β1α)τr11β1α=βτ1r=a+2(f)(r)((1β)τr(1α)τrαβ)=β((1β)τa2(1α)τa2αβ)(f)(a+2)+βτ1r=a+3(f)(r)((1β)τr(1α)τrαβ). (3.5)

    By making use of (3.3)–(3.5) and (ⅱ) in (3.2), together with the fact that 0<B(α)B(β), we obtain

    (f)(τ)(1α)τa2{f(a+2)βf(a+1)}ξ{f(a+2)βf(a+1)}+β((1β)τa2(1α)τa2αβ)(f)(a+2)+β(1β)((1β)τa2(1α)τa2αβ)f(a+1)τ1s=a+3(f)(s)(1α)τs+βτ1r=a+3(f)(r)((1β)τr(1α)τrαβ)=(β(1β)τa2α(1α)τa2αβξ)0by(iii)×{f(a+2)βf(a+1)}0by(i)+τ1s=a+3(f)(s)1αβ[β(1β)τsα(1α)τs]0by(2.2), (3.6)

    for τNȷa+2. Now, we know from (ⅰ) that (f)(a+2)0, which implies that

    f(a+2)f(a+1)>βf(a+1).

    By substituting τ=a+3 into (3.6), we get

    (f)(a+3)(β(1β)α(1α)αβ) {f(a+2)βf(a+1)}+a+2s=a+3()=0(β+α1){f(a+2)βf(a+1)}0.

    Repeating this action together with the help of (2.2), we reach that (f)(τ)0 for each τNȷa+2, as desired.

    To confirm the validity of the above theorem, we consider the following example.

    Example 3.1. Suppose that f is a function f: N0R defined by

    f(τ)=τ¯α+β.

    First of all, for α=0.5 and β=0.55, we see that

    1<α+β<21<1.05<2,

    which verifies that (α,β)D1.

    Now, by using (3.2) with ξ=0.001, τ=a+3, and a=0, we get

    (CF1αrlCF0βrlf)(2)=B(α)B(β)[(1α){f(2)βf(1)}+3s=3(f)(s)(1α)3sβf(1)3s=3(1β)s2(1α)3sβ3s=3s1r=2(f)(r)(1β)sr1(1α)3s]=1.0638B(α)B(β)0.0015B(α)B(β).

    Hence, the second condition of Theorem 3.1 is satisfied for τ=a+3. Furthermore, the first condition of Theorem 3.1

    (f)(2)=1.07330

    holds. The last condition

    β(1β)α(1α)αβ=0.050.001

    is satisfied. Hence, τα+β_ is increasing at τ=a+3, according to Theorem 3.1. Moreover, τα+β_ is an increasing function on N1, as its plot has been shown in Figure 2.

    Figure 2.  Graph of τ¯α+β for α=0.5 and β=0.55.

    Our second result is based on the set D2, which is plotted clearly in Figure 3.

    Figure 3.  The region of D2.

    Let us start stating and proving our last theorem.

    Theorem 3.2. If 0ξ and α[12,1) and the function f: NaR satisfies

    (i) (f)(a+2)0;

    (ii)

    (CFa+1αrlCFaαrlf)(τ)ξB2(α){f(a+2)αf(a+1)};

    (iii)

    ξ(1α)τa2[α(τa2)(1α)];

    for each (α,β)D2 and τNȷa+3, for some ȷNa+3, then, (f)(τ)0 for every τNȷa+2.

    Proof. We know from (3.1) that when β=α,

    (CFa+1αrlCFaβrlf)(τ)=B2(α)[(1α)τa2{f(a+2)αf(a+1)}+τs=a+3(f)(s)(1α)τsαf(a+1)τs=a+3(1α)τa2ατs=a+3s1r=a+2(f)(r)(1α)τr1]:=B2(α)[(1α)τa2{f(a+2)αf(a+1)}+B1(τ)f(a+1)B2(τ)B3(τ)]. (3.7)

    Compute B1(τ), B2(τ), and B3(τ) to get

    B1(τ):=τs=a+3(f)(s)(1α)τs=(f)(τ)+τ1s=a+3(f)(s)(1α)τs, (3.8)
    B2(τ):=ατs=a+3(1α)τa2=α(1α)τa2τa2κ=1(1)=α(τa2)(1α)τa2 (3.9)

    and

    B3(τ):=ατs=a+3s1r=a+2(f)(r)(1α)τr1=ατ1r=a+2(f)(r)(1α)τr1τs=r+1(1)=α(τa2)(1α)τa3(f)(a+2)+ατ1r=a+3(f)(r)(τr)(1α)τr1. (3.10)

    By making use of (3.8)–(3.10) and using (ⅱ) in (3.7), together with the fact that 0<B(α)B(β), we get

    (f)(τ)(1α)τa2{f(a+2)αf(a+1)}ξ{f(a+2)αf(a+1)}+α(τa2)(1α)τa3(f)(a+2)+α(τa2)(1α)τa2f(a+1)τ1s=a+3(f)(s)(1α)τs+ατ1r=a+3(f)(r)(τr)(1α)τr1=((1α)τa2[α(τa2)(1α)]ξ)0by(iii)×{f(a+2)βf(a+1)}0by(i)+τ1s=a+3(f)(s)(1α)τs1[α(τs)(1α)]0by(2.5). (3.11)

    By considering (ⅰ) and the last positive inequality (3.11), we can deduce that (f)(τ)0 for each τNȷa+2, as desired.

    We consider the following example in order to see the validity of the above theorem.

    Example 3.2. We consider the same function defined in Example 3.1, then for choosing α=β=0.6, we have

    1<α+β<21<1.2<2,

    which verifies that (α,β)D2. Moreover,

    (f)(2)=0.87130,

    and

    (CF1αrlCF0αrlf)(2)=B2(α)[(1α){f(2)αf(1)}+3s=3(f)(s)(1α)3sαf(a+1)τs=3(1)α3s=3s1r=2(f)(r)(1α)2r]=1.2951B2(α)0.0018B2(α).

    In addition,

    (1α)τa2[α(τa2)(1α)]=0.080.001.

    Thus, all the conditions of Theorem 3.2 are satisfied. Therefore, the increase of τ¯α+β is proved at τ=a+3. For more clarification, see below Figure 4 as τ¯α+β is increasing on N1.

    Figure 4.  Graph of τ¯α+β for α=β=0.4.

    In this study, we have considered analyzing (CFa+1αrlCFaβrlf)(τ) on the set D1 in which βα, and (CFa+1αrlCFaαrlf)(τ) on the set D2 in which β=α. The positivity (f)(τ) has been examined from analyzing these sequential fractional differences on a finite time set Nȷa+2 for some ȷNa+3 in both cases when (α,β)D1 or (α,β)D2. In the first case, we have used a complex condition as appeared in Theorem 3.1 (ⅲ). However, a simpler condition has been applied in the second case as stated in Theorem 3.2 (ⅲ). In the end, we have provided an increasing function to support our claims and results, and it has been shown that the positivity of our main theorems is accurate and can be obtained under the certain conditions.

    The author declares he has not used Artificial Intelligence (AI) tools in the creation of this article.

    Many thanks for the academic editor and the anonymous referees.

    The author declares no conflict interest.



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