Research article Special Issues

Individual-based modeling of COVID-19 transmission in college communities

  • #Joint last authors
  • Received: 30 June 2022 Revised: 23 August 2022 Accepted: 01 September 2022 Published: 20 September 2022
  • The ongoing COVID-19 pandemic has created major public health and socio-economic challenges across the United States. Among them are challenges to the educational system where college administrators are struggling with the questions of how to mitigate the risk and spread of diseases on their college campus. To help address this challenge, we developed a flexible computational framework to model the spread and control of COVID-19 on a residential college campus. The modeling framework accounts for heterogeneity in social interactions, activities, environmental and behavioral risk factors, disease progression, and control interventions. The contribution of mitigation strategies to disease transmission was explored without and with interventions such as vaccination, quarantine of symptomatic cases, and testing. We show that even with high vaccination coverage (90%) college campuses may still experience sizable outbreaks. The size of the outbreaks varies with the underlying environmental and socio-behavioral risk factors. Complementing vaccination with quarantine and mass testing was shown to be paramount for preventing or mitigating outbreaks. Though our quantitative results are likely provisional on our model assumptions, sensitivity analysis confirms the robustness of their qualitative nature.

    Citation: Durward Cator, Qimin Huang, Anirban Mondal, Martial Ndeffo-Mbah, David Gurarie. Individual-based modeling of COVID-19 transmission in college communities[J]. Mathematical Biosciences and Engineering, 2022, 19(12): 13861-13877. doi: 10.3934/mbe.2022646

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  • The ongoing COVID-19 pandemic has created major public health and socio-economic challenges across the United States. Among them are challenges to the educational system where college administrators are struggling with the questions of how to mitigate the risk and spread of diseases on their college campus. To help address this challenge, we developed a flexible computational framework to model the spread and control of COVID-19 on a residential college campus. The modeling framework accounts for heterogeneity in social interactions, activities, environmental and behavioral risk factors, disease progression, and control interventions. The contribution of mitigation strategies to disease transmission was explored without and with interventions such as vaccination, quarantine of symptomatic cases, and testing. We show that even with high vaccination coverage (90%) college campuses may still experience sizable outbreaks. The size of the outbreaks varies with the underlying environmental and socio-behavioral risk factors. Complementing vaccination with quarantine and mass testing was shown to be paramount for preventing or mitigating outbreaks. Though our quantitative results are likely provisional on our model assumptions, sensitivity analysis confirms the robustness of their qualitative nature.



    The class of normalized analytic functions in the open unit disc Δ={zC:|z|<1} denoted by Ω consists of the functions f of the form

    f(z)=z+n=2anzn, (1.1)

    where f(0)1=f(0)=0. Let (z)Ω defined by

    (z)=z+n=2bnzn. (1.2)

    Then the Hadamard product, also known as the convolution of two function f(z) and (z) denoted by f is defined as

    (f)(z)=f(z)(z)=z+n=2anbnzn,zΔ.

    Moreover, f(z)(z), if there exist a Schwartz function χ(z) in A, satisfying the conditions χ(0)=0 and |χ(z)|<1, such that f(z)=(χ(z)). The symbol is used to denote subordination.

    Let S denote the subclass of Ω of univalent functions in Δ. Let P,C,S and K represent the subclasses of S known as the classes of Caratheodory functions, convex funtions, starlike functions, and close-to-convex functions, respectively.

    The concept of bounded rotations was introduced by Brannan in [7]. A lot of quality work on the generalization of this concept has already been done. Working in the same manner, we have defined the following new classes.

    Definition 1.1. Let

    ν(z)=1+n=1pnzn (1.3)

    be analytic in Δ such that ν(0)=1. Then for m2, ν(z)Pm((z)), if and only if

    ν(z)=(m4+12)ν1(z)(m412)ν2(z), (1.4)

    where (z) is a convex univalent function in Δ and νi(z)(z) for i=1,2.

    Particularly, for m=2, we get the class P((z)).

    Definition 1.2. Let f(z) and (z) be two analytic functions as defined in (1.1) and (1.2) such that (f)(z)0. Let (z) be a convex univalent function. Then for m2, fVm[(z);(z)] if and only if

    (z(f))(f)Pm((z)),zΔ. (1.5)

    Particularly, for m=2, we will get the class C[(z);(z)]. So, a function fC[(z);(z)] if and only if

    (z(f))(f)(z),zΔ.

    Definition 1.3. Let f(z) and (z) be the functions defined in (1.1) and (1.2), then f(z)Rm[(z);(z)] if and only if

    z(f)(f)Pm((z)),zΔ. (1.6)

    Particularly, for m=2, we get the class SΛ[(z);(z)], i.e., fSΛ[(z);(z)] if and only if

    z(f)(f)(z),zΔ.

    From (1.5) and (1.6) it can be easily noted that fVm[(z);(z)] if and only if zf(z)Rm[(z);(z)]. For m=2, this relation will hold for the classes C[(z);(z)] and SΛ[(z);(z)].

    Definition 1.4. Let f(z) and (z) be analytic function as defined in (1.1) and (1.2) and m2. Let (z) be the convex univalent function. Then, fTm[(z);(z)] if and only if there exists a function ψ(z)SΛ[(z);(z)] such that

    z(f)ψPm((z)),zΔ. (1.7)

    It is interesting to note that the particular cases of our newly defined classes will give us some well-known classes already discussed in the literature. Some of these special cases have been elaborated below.

    Special Cases: Let (z) be the identity function defined as z1z denoted by I i.e., f=fI=f. Then

    (1) For (z)=1+z1z we have Pm((z))=Pm,Rm[(z);(z)]=Rm introduced by Pinchuk [23] and the class Vm[(z);(z)]=Vm defined by Paatero [21]. For m=2, we will get the well-known classes of convex functions C and the starlike functions SΛ.

    (2) Taking (z)=1+(12δ)z1z, we get the classes Pm(δ),Rm(δ) and Vm(δ) presented in [22]. For m=2, we will get the classes C(δ) and SΛ(δ).

    (3) Letting (z)=1+Az1+Bz, with 1B<A1 introduced by Janowski in [12], the classes Pm[A,B],Rm[A,B] and Vm[A,B] defined by Noor [16,17] can be obtained. Moreover, the classes C[A,B] and SΛ[A,B] introduced by [12] can be derived by choosing m=2.

    A significant work has already been done by considering (z) to be different linear operators and (z) to be any convex univalent function. For the details see ([4,9,18,19,24]).

    The importance of Mittag-Leffler functions have tremendously been increased in the last four decades due to its vast applications in the field of science and technology. A number of geometric properties of Mittag-Leffler function have been discussed by many researchers working in the field of Geometric function theory. For some recent and detailed study on the Geometric properties of Mittag-Leffler functions see ([2,3,31]).

    Special function theory has a vital role in both pure and applied mathematics. Mittag-Leffler functions have massive contribution in the theory of special functions, they are used to investigate certain generalization problems. For details see [11,26]

    There are numerous applications of Mittag-Leffler functions in the analysis of the fractional generalization of the kinetic equation, fluid flow problems, electric networks, probability, and statistical distribution theory. The use of Mittag-Leffler functions in the fractional order integral equations and differential equations attracted many researchers. Due to its connection and applications in fractional calculus, the significance of Mittag-Leffler functions has been amplified. To get a look into the applications of Mittag-Leffler functions in the field of fractional calculus, (see [5,27,28,29,30]).

    Here, in this article we will use the operator Hγ,κλ,η:ΩΩ, introduced by Attiya [1], defined as

    Hγ,κλ,η(f)=μγ,κλ,ηf(z),zΔ, (1.8)

    where η,γC, (λ)>max{0,(k)1} and (k)>0. Also, (λ)=0 when (k)=1;η0. Here, μγ,κλ,η is the generalized Mittag-Leffler function, defined in [25]. The generalized Mittag-Leffler function has the following representation.

    μγ,κλ,η=z+n=2Γ(γ+nκ)Γ(λ+η)Γ(γ+κ)Γ(η+λn)n!zn.

    So, the operator defined in (1.8) can be rewritten as:

    Hγ,κλ,η(f)(z)=z+n=2Γ(γ+nκ)Γ(λ+η)Γ(γ+κ)Γ(η+λn)n!anzn,zΔ. (1.9)

    Attiya [1] presented the properties of the aforesaid operator as follows:

    z(Hγ,κλ,η(f(z)))=(γ+κκ)(Hγ+1,κλ,η(f(z)))(γκ)(Hγ,κλ,η(f(z))), (1.10)

    and

    z(Hγ,κλ,η+1(f(z)))=(λ+ηλ)(Hγ,κλ,η(f(z)))(ηλ)(Hγ,κλ,η+1(f(z))). (1.11)

    However, as essential as real-world phenomena are, discovering a solution for the commensurate scheme and acquiring fundamentals with reverence to design variables is challenging and time-consuming. Among the most pragmatically computed classes, we considered the new and novel class which is very useful for efficiently handling complex subordination problems. Here, we propose a suitably modified scheme in order to compute the Janowski type function of the form (z)=(1+Az1+Bz)β, where 0<β1 and 1B<A1, which is known as the strongly Janowski type function. Moreover, for (z), we will use the function defined in (1.9). So, the classes defined in Definition 1.1–1.4 will give us the following novel classes.

    Definition 1.5. A function ν(z) as defined in Eq (1.3) is said to be in the class P(m,β)[A,B] if and only if for m2 there exist two analytic functions ν1(z) and ν2(z) in Δ, such that

    ν(z)=(m4+12)ν1(z)(m412)ν2(z),

    where νi(z)(1+Az1+Bz)β for i=1,2. For m=2, we get the class of strongly Janowki type functions Pβ[A,B].

    Moreover,

    V(m,β)[A,B;γ,η]={fΩ:(z(Hγ,κλ,ηf(z)))(Hγ,κλ,ηf(z))P(m,β)[A,B]},
    R(m,β)[A,B;γ,η]={fΩ:z(Hγ,κλ,ηf(z))Hγ,κλ,ηf(z)P(m,β)[A,B]},
    Cβ[A,B,γ,η]={fΩ:(z(Hγ,κλ,ηf(z)))(Hγ,κλ,ηf(z))Pβ[A,B]},
    SΛβ[A,B,γ,η]={fΩ:z(Hγ,κλ,ηf(z))Hγ,κλ,ηf(z)Pβ[A,B]},
    T(m,β)[A,B;γ,η]={fΩ:z(Hγ,κλ,ηf(z))Hγ,κλ,ηψ(z)P(m,β)[A,B],whereψ(z)SΛβ[A,B,γ,η]},

    where η,γC, (λ)>max{0,(k)1} and (k)>0. Also, (λ)=0 when (k)=1;η0. It can easily be noted that there exists Alexander relation between the classes V(m,β)[A,B;γ,η] and R(m,β)[A,B;γ,η], i.e.,

    fV(m,β)[A,B;γ,η]zfR(m,β)[A,B;γ,η]. (1.12)

    Throughout this investigation, 1B<A1, m2 and 0<β1 unless otherwise stated.

    Lemma 2.1. ([13]) Let ν(z) as defined in (1.3) be in P(m,β)[A,B]. Then ν(z)Pm(ϱ), where 0ϱ=(1A1B)β<1.

    Lemma 2.2. ([8]) Let (z) be convex univalent in Δ with h(0)=1 and (ζ(z)+α)>0(ζC). Let p(z) be analytic in Δ with p(0)=1, which satisfy the following subordination relation

    p(z)+zp(z)ζp(z)+α(z),

    then

    p(z)(z).

    Lemma 2.3. ([10]) Let (z)P. Then for |z|<r, 1r1+r((z)) |(z)|1+r1r, and |h(z)|2r(z)1r2.

    Theorem 3.1. Let ϱ=(1A1B)β. Then for (γκ)>ϱ,

    R(m,β)[A,B,γ+1,η]R(m,β)[A,B,γ,η].

    Proof. Let f(z)R(m,β)[A,B,γ+1,η]. Set

    φ(z)=z(Hγ+1,κλ,ηf(z))Hγ+1,κλ,ηf(z), (3.1)

    then φ(z)P(m,β)[A,B]. Now, Assume that

    ψ(z)=z(Hγ,κλ,ηf(z))Hγ,κλ,ηf(z). (3.2)

    Plugging (1.10) in (3.2), we get

    ψ(z)=(γ+κκ)(Hγ+1,κλ,ηf(z))(γκ)(Hγ,κλ,ηf(z))Hγ,κλ,ηf(z).

    It follows that

    Hγ,κλ,ηf(z)(κγ+κ)(ψ(z)+γκ)=Hγ+1,κλ,ηf(z).

    After performing logarithmic differentiation and simple computation, we get

    ψ(z)+zψ(z)ψ(z)+γκ=φ(z). (3.3)

    Now, for m2, consider

    ψ(z)=(m4+12)ψ1(z)(m412)ψ2(z). (3.4)

    Combining (3.3) and (3.4) with the similar technique as used in Theorem 3.1 of [20], we get

    φ(z)=(m4+12)φ1(z)(m412)φ2(z),

    where

    φi(z)=ψi(z)+zψi(z)ψi(z)+γκ,

    for i=1,2. Since φ(z)P(m,β)[A,B], therefore

    φi(z)=ψi(z)+zψi(z)ψi(z)+γκ(1+Az1+Bz)β,

    for i=1,2. By using Lemma 2.1 and the condition (γκ)>ϱ, we have

    (γκ+(1+Az1+Bz)β)>0,

    where ϱ=(1A1B)β. Hence, in view of Lemma 2.2, we have

    ψi(z)(1+Az1+Bz)β,

    for i = 1, 2. This implies ψ(z)P(m,β)[A,B], so

    f(z)R(m,β)[A,B,γ,η],

    which is required to prove.

    Theorem 3.2. If (λη)>ϱ, where ϱ=(1A1B)β, then

    R(m,β)[A,B,γ,η]R(m,β)[A,B,γ,η+1].

    Proof. Let f(z)R(m,β)[A,B,γ,η]. Taking

    φ(z)=z(Hγ,κλ,ηf(z))Hγ,κλ,ηf(z), (3.5)

    we have φ(z)P(m,β)[A,B]. Now, suppose that

    ψ(z)=z(Hγ,κλ,η+1f(z))Hγ,κλ,η+1f(z). (3.6)

    Applying the relation (1.11) in the Eq (3.6), we have

    ψ(z)=(λ+ηλ)(Hγ,κλ,ηf(z))(ηλ)(Hγ,κλ,η+1f(z))Hγ,κλ,η+1f(z).

    arrives at

    Hγ,κλ,η+1f(z)(λη+λ)(ψ(z)+ηλ)=Hγ,κλ,ηf(z).

    So by the logarithmic differentiation and simple computation we get,

    ψ(z)+zψ(z)ψ(z)+ηλ=φ(z). (3.7)

    Therefore, for m2, take

    ψ(z)=(m4+12)ψ1(z)(m412)ψ2(z). (3.8)

    Combining Eqs (3.6) and (3.7) using the similar technique as in Theorem 3.1 of [20], we get

    φ(z)=(m4+12)φ1(z)(m412)φ2(z),

    where

    φi(z)=ψi(z)+zψi(z)ψi(z)+ηλ,

    for i=1,2. Since φ(z)P(m,β)[A,B], therefore

    φi(z)=ψi(z)+zψi(z)ψi(z)+ηλ(1+Az1+Bz)β,

    for i=1,2. Applying Lemma 2.1 and the condition (ηλ)>ϱ, we get

    (ηλ+(1+Az1+Bz)β)>0,

    where ϱ=(1A1B)β. Hence, by Lemma 2.2, we have

    ψi(z)(1+Az1+Bz)β,

    for i = 1, 2. This implies ψ(z)P(m,β)[A,B], so

    f(z)R(m,β)[A,B,γ,η+1],

    which completes the proof.

    Corollary 3.1. For m=2, if (γκ)>ϱ, where ϱ=(1A1B)β. Then

    SΛβ[A,B,γ+1,η]SΛβ[A,B,γ,η].

    Moreover, if (λη)>ϱ, then

    SΛβ[A,B,γ,η]SΛβ[A,B,γ,η+1].

    Theorem 3.3. Let ϱ=(1A1B)β. Then for (γκ)>ϱ,

    V(m,β)[A,B,γ+1,η]V(m,β)[A,B,γ,η].

    Proof. By means of theorem 3.1 and Alexander relation defined in (1.12), we get

    fV(m,β)[A,B,γ+1,η]zfR(m,β)[A,B,γ+1,η]zfR(m,β)[A,B,γ,η]fV(m,β)[A,B,γ,η].

    Hence the result.

    Analogously, we can prove the following theorem.

    Theorem 3.4. If (λη)>ϱ, where ϱ=(1A1B)β, then

    V(m,β)[A,B,γ,η]V(m,β)[A,B,γ,η+1].

    Corollary 3.2. For m=2, if (γκ)>ϱ, where ϱ=(1A1B)β. Then

    Cβ[A,B,γ+1,η]Cβ[A,B,γ,η].

    Moreover, if (λη)>ϱ, then

    Cβ[A,B,γ,η]Cβ[A,B,γ,η+1].

    Theorem 3.5. Let ϱ=(1A1B)β, and (γκ)>ϱ. Then

    T(m,β)[A,B;γ+1,η]T(m,β)[A,B;γ,η].

    Proof. Let f(z)T(m,β)[A,B,γ+1,η]. Then there exist ψ(z)SΛβ[A,B,γ+1,η] such that

    φ(z)=z(Hγ+1,κλ,ηf(z))Hγ+1,κλ,ηψ(z)P(m,β)[A,B]. (3.9)

    Now consider

    ϕ(z)=z(Hγ,κλ,ηf(z))Hγ,κλ,ηψ(z). (3.10)

    Since ψ(z)SΛβ[A,B,γ+1,η] and (γκ)>ϱ, therefore by Corollary 3.3, ψ(z)SΛβ[A,B,γ,η]. So

    q(z)=z(Hγ,κλ,ηψ(z))Hγ,κλ,ηψ(z)Pβ[A,B]. (3.11)

    By doing some simple calculations on (3.11), we get

    (κq(z)+γ)Hγ,κλ,ηψ(z)=(γ+κ)Hγ+1,κλ,ηψ(z). (3.12)

    Now applying the relation (1.10) on (3.10), we get

    ϕ(z)Hγ,κλ,ηψ(z)=γ+κκHγ+1,κλ,ηf(z)γκHγ,κλ,ηf(z). (3.13)

    Differentiating both sides of (3.13), we have

    ϕ(z)(Hγ,κλ,ηψ(z))+ϕ(z)Hγ,κλ,ηψ(z)=γ+κκ(Hγ+1,κλ,ηf(z))γκ(Hγ,κλ,ηf(z)).

    By using (3.12) and with some simple computations, we get

    ϕ(z)+zϕ(z)q(z)+γκ=φ(z)P(m,β)[A,B], (3.14)

    with (q(z)+γκ)>0, since q(z)Pβ[A,B], so by Lemma 2.1, (q(z)>ϱ and (γκ)>ϱ. Now consider

    ϕ(z)=(m4+12)ϕ1(z)(m412)ϕ2(z). (3.15)

    Combining (3.14) and (3.15) with the similar technique as used in Theorem 3.1 of [20], we get

    φ(z)=(m4+12)φ1(z)(m412)φ2(z), (3.16)

    where

    φi(z)=ϕ(z)+zϕzq(z)+γκ,

    for i=1,2. Since φ(z)P(m,β)[A,B], therefore

    φi(z)(1+Az1+Bz)β,i=1,2.

    Using the fact of Lemma 2.2, we can say that

    ϕi(z)(1+Az1+Bz)β,i=1,2.

    So, ϕ(z)P(m,β)[A,B]. Hence we get the required result.

    Theorem 3.6. If (λη)>ϱ, where ϱ=(1A1B)β, then

    T(m,β)[A,B,γ,η]T(m,β)[A,B,γ,η+1].

    Let f(z)T(m,β)[A,B,γ,η]. Then there exist ψ(z)SΛβ[A,B,γ,η] such that

    φ(z)=z(Hγ,κλ,ηf(z))Hγ,κλ,ηψ(z)P(m,β)[A,B]. (3.17)

    Taking

    ϕ(z)=z(Hγ,κλ,η+1f(z))Hγ,κλ,η+1ψ(z). (3.18)

    As we know that, ψ(z)SΛβ[A,B,γ,η] and (ηλ)>ϱ, therefore by Corollary 3.3, ψ(z)SΛβ[A,B,γ,η+1]. So

    q(z)=z(Hγ,κλ,η+1ψ(z))Hγ,κλ,η+1ψ(z)Pβ[A,B]. (3.19)

    By doing some simple calculations on (3.19) with the help of (1.11), we get

    (λq(z)+η)Hγ,κλ,η+1ψ(z)=(η+λ)Hγ,κλ,ηψ(z). (3.20)

    Now, applying the relation (1.11) on (3.18), we get

    ϕ(z)Hγ,κλ,η+1ψ(z)=η+λλHγ,κλ,ηf(z)ηλHγ,κλ,η+1f(z). (3.21)

    Differentiating both sides of Eq (3.21), we have

    ϕ(z)(Hγ,κλ,η+1ψ(z))+ϕ(z)Hγ,κλ,η+1ψ(z)=η+λλ(Hγ,κλ,ηf(z))ηλ(Hγ,κλ,η+1f(z)),

    some simple calculations along with using (3.20) give us

    ϕ(z)+zϕ(z)q(z)+ηλ=φ(z)P(m,β)[A,B], (3.22)

    with (q(z)+ηλ)>0. Since q(z)Pβ[A,B], so applying Lemma 2.1, we have (q(z)>ϱ and (ηλ)>ϱ.

    Assume that

    ϕ(z)=(m4+12)ϕ1(z)(m412)ϕ2(z). (3.23)

    Combining (3.22) and (3.23), along with using the similar technique as in Theorem 3.1 of [20], we get

    φ(z)=(m4+12)φ1(z)(m412)φ2(z), (3.24)

    where

    φi(z)=ϕ(z)+zϕzq(z)+ηλ,

    for i=1,2. Since φ(z)P(m,β)[A,B], therefore

    φi(z)(1+Az1+Bz)β,i=1,2.

    Applying the fact of Lemma 2.2, we have

    ϕi(z)(1+Az1+Bz)β,i=1,2.

    So ϕ(z)P(m,β)[A,B]. Which gives us the required result.

    Corollary 3.3. If ϱ>min{(γκ),(λη)}, where ϱ=(1A1B)β, then we have the following inclusion relations:

    (i) R(m,β)[A,B,γ+1,η]R(m,β)[A,B,γ,η]R(m,β)[A,B,γ,η+1].

    (ii)V(m,β)[A,B,γ+1,η]V(m,β)[A,B,γ,η]V(m,β)[A,B,γ,η+1].

    (iii)T(m,β)[A,B,γ+1,η]T(m,β)[A,B,γ,η]T(m,β)[A,B,γ,η+1].

    Now, we will discuss some radius results for our defined classes.

    Theorem 3.7. Let ϱ=(1A1B)β, and (γκ)>ϱ. Then

    R(m,β)[A,B,γ,η]R(m,β)[ϱ,γ+1,η]

    whenever

    |z|<ro=1ϱ2ϱ+32ϱ,where0ϱ<1.

    Proof. Let f(z)R(m,β)[A,B,γ,η]. Then

    ψ(z)=z(Hγ,κλ,ηf(z))Hγ,κλ,ηf(z)P(m,β)[A,B]. (3.25)

    In view of Lemma 2.1 P(m,β)[A,B]Pm(ϱ), for ϱ=(1A1B)β, therefore ψ(z)Pm(ϱ). So by the Definition of Pm(ϱ) given in [22], there exist two functions ψ1(z),ψ2(z)P(ϱ) such that

    ψ(z)=(m4+12)ψ1(z)(m412)ψ2(z), (3.26)

    with m2 and (ψi(z))>ϱ,i=1,2. We can write

    ψi(z)=(1ϱ)hi(z)+ϱ, (3.27)

    where hi(z)P and (hi(z)>0, for i=1,2. Now, let

    ϕ(z)=z(Hγ+1,κλ,ηf(z))Hγ+1,κλ,ηf(z). (3.28)

    We have to check when ϕ(z)Pm(ϱ). Using relation (1.10) in (3.25), we get

    ψ(z)Hγ+1,κλ,ηf(z)=(γ+κκ)(Hγ+1,κλ,η(f(z)))(γκ)(Hγ,κλ,η(f(z))).

    So, by simple calculation and logarithmic differentiation, we get

    ψ(z)+zψzψ(z)+γκ=ϕ(z). (3.29)

    Now, consider

    ϕ(z)=(m4+12)ϕ1(z)(m412)ϕ2(z),

    where

    ϕi(z)=ψi(z)+zψizψi(z)+γκ,i=1,2.

    To derive the condition for ϕi(z) to be in P(ϱ), consider

    (ϕi(z)ϱ)=(ψi(z)+zψizψi(z)+γκϱ).

    In view of (3.27), we have

    (ϕi(z)ϱ)=((1ϱ)hi(z)+ϱ+z(1ϱ)hi(z)γκ+ϱ+(1ϱ)hi(z)ϱ)(1ϱ)(hi(z))(1ϱ)|zhi(z)|(γκ+ϱ)+(1ϱ)(hi(z)). (3.30)

    We have, (γκ+ϱ)>0 since (γκ)>ϱ. Since hi(z)P, hence by using Lemma 2.3 in inequality (3.30), we have

    (ϕi(z)ϱ)(1ϱ)(hi(z))1ϱ2r1r2(hi(z))(1ϱ)(1r1+r)=(1ϱ)(hi(z))[(1r)2(1ϱ)2r(1r)2(1ϱ)](1ϱ)(1r1+r)[(1r)2(1ϱ)2r(1r)2(1ϱ)]=r2(1ϱ)2r(2ϱ)+(1ϱ)1r2. (3.31)

    Since 1r2>0, letting T(r)=r2(1ϱ)2r(2ϱ)+(1ϱ). It is easy to note that T(0)>0 and T(1)<0. Hence, there is a root of T(r) between 0 and 1. Let ro be the root then by simple calculations, we get

    ro=1ϱ2ϱ+32ϱ.

    Hence ϕ(z)Pm(ϱ) for |z|<ro. Thus for this radius ro the function f(z) belongs to the class R(m,β)[ϱ,γ+1,η], which is required to prove.

    Theorem 3.8. Let ϱ=(1A1B)β, and (λη)>ϱ. Then

    R(m,β)[A,B,γ,η+1]R(m,β)[ϱ,γ,η],

    whenever

    |z|<ro=1ϱ2ϱ+32ϱ,where0ϱ<1.

    Proof. Let f(z)R(m,β)[A,B,γ,η+1]. Then

    ψ(z)=z(Hγ,κλ,η+1f(z))Hγ,κλ,η+1f(z)P(m,β)[A,B]. (3.32)

    By applying of Lemma 2.1, we get P(m,β)[A,B]Pm(ϱ), for ϱ=(1A1B)β, therefore ψ(z)Pm(ϱ). Hence, the Definition of Pm(ϱ) given in [22], there exist two functions ψ1(z),ψ2(z)P(ϱ) such that

    ψ(z)=(m4+12)ψ1(z)(m412)ψ2(z), (3.33)

    with m2 and (ψi(z))>ϱ,i=1,2. We can say that

    ψi(z)=(1ϱ)hi(z)+ϱ, (3.34)

    where hi(z)P and (hi(z)>0, for i=1,2. Now, assume

    ϕ(z)=z(Hγ,κλ,ηf(z))Hγ,κλ,ηf(z). (3.35)

    Here, We have to obtain the condition for which ϕ(z)Pm(ϱ). Using relation (1.11) in (3.51), we get

    ψ(z)Hγ,κλ,ηf(z)=(η+λλ)(Hγ,κλ,η(f(z)))(ηλ)(Hγ,κλ,η+1(f(z))).

    Thus, by simple calculation and logarithmic differentiation, we have

    ψ(z)+zψzψ(z)+ηλ=ϕ(z). (3.36)

    Now, consider

    ϕ(z)=(m4+12)ϕ1(z)(m412)ϕ2(z),

    where

    ϕi(z)=ψi(z)+zψizψi(z)+ηλ,i=1,2.

    To derive the condition for ϕi(z) to be in P(ϱ), consider

    (ϕi(z)ϱ)=(ψi(z)+zψizψi(z)+ηλϱ).

    In view of (3.34), we have

    (ϕi(z)ϱ)=((1ϱ)hi(z)+ϱ+z(1ϱ)hi(z)ηλ+ϱ+(1ϱ)hi(z)ϱ)(1ϱ)(hi(z))(1ϱ)|zhi(z)|(ηλ+ϱ)+(1ϱ)(hi(z)). (3.37)

    Here, (ηλ+ϱ)>0 since (ηλ)>ϱ. We know that hi(z)P, therefore by using Lemma 2.3 in inequality (3.37), we have

    (ϕi(z)ϱ)(1ϱ)(hi(z))1ϱ2r1r2(hi(z))(1ϱ)(1r1+r)=(1ϱ)(hi(z))[(1r)2(1ϱ)2r(1r)2(1ϱ)](1ϱ)(1r1+r)[(1r)2(1ϱ)2r(1r)2(1ϱ)]=r2(1ϱ)2r(2ϱ)+(1ϱ)1r2. (3.38)

    Since 1r2>0, letting T(r)=r2(1ϱ)2r(2ϱ)+(1ϱ). It can easily be seen that T(0)>0 and T(1)<0. Hence, there is a root of T(r) between 0 and 1. Let ro be the root then by simple calculations, we get

    ro=1ϱ2ϱ+32ϱ.

    Hence ϕ(z)Pm(ϱ) for |z|<ro. Thus for this radius ro the function f(z) belongs to the class R(m,β)[ϱ,γ,η], which is required to prove.

    Corollary 3.4. Let ϱ=(1A1B)β. Then, for m=2, and |z|<ro=1ϱ2ϱ+32ϱ,

    (i) If (γκ)>ϱ, then SΛβ[A,B,γ,η]SΛβ[ϱ,γ+1,η].

    (ii) If(λη)>ϱ, then SΛβ[A,B,γ,η+1]SΛβ[ϱ,γ,η].

    Theorem 3.9. Let ϱ=(1A1B)β. Then for |z|<ro=1ϱ2ϱ+32ϱ, we have

    (1)V(m,β)[A,B,γ,η]V(m,β)[ϱ,γ+1,η], if (γκ)>ϱ.

    (2)V(m,β)[A,B,γ,η+1]V(m,β)[ϱ,γ,η], if (λη)>ϱ.

    Proof. The above results can easily be proved by using Theorem 3.10, Theorem 3.11 and the Alexander relation defined in (1.12).

    Theorem 3.10. Let ϱ=(1A1B)β, and (γκ)>ϱ. Then

    T(m,β)[A,B,γ,η]T(m,β)[ϱ,γ+1,η],

    whenever

    |z|<ro=1ϱ2ϱ+32ϱ,where0ϱ<1.

    Proof. Let fT(m,β)[A,B,γ,η], then there exist ψ(z)SΛβ[A,B,γ,η] such that

    φ(z)=z(Hγ,κλ,ηf(z))Hγ,κλ,ηψ(z)P(m,β)[A,B]. (3.39)

    Since by Lemma 2.1 we know that P(m,β)[A,B]Pm(ϱ), where ϱ=(1A1B)β, therefore φ(z)Pm(ϱ). So by using the Definition of Pm(ϱ) defined in [22], there exist two functions φ1(z) and φ2(z) such that

    φ(z)=(m4+12)φ1(z)(m412)φ2(z), (3.40)

    where φi(z)P(ϱ),i=1,2. We can write

    φi(z)=ϱ+(1ϱ)hi(z), (3.41)

    where hi(z)P. Now, let

    ϕ(z)=z(Hγ+1,κλ,ηf(z))Hγ+1,κλ,ηψ(z).

    Since ψ(z)SΛβ[A,B,γ,η], therefore

    q(z)=z(Hγ,κλ,ηψ(z))Hγ,κλ,ηψ(z)Pβ[A,B], (3.42)

    then by using relation (1.10) and doing some simple computation on Eq (3.42), we have

    (κq(z)+γ)Hγ,κλ,ηψ(z)=(γ+κ)Hγ+1,κλ,ηψ(z). (3.43)

    Now, using relation (1.10) in (3.39), we get

    φ(z)=(γ+κκ)(Hγ+1,κλ,ηf(z))(γκ)(Hγ,κλ,ηf(z))Hγ,κλ,ηψ(z). (3.44)

    By some simple calculations along with differentiation of both sides of (3.44) and then applying (3.43) we get the following relation

    φ(z)+zφ(z)q(z)+(γκ)=ϕ(z).

    Let us consider

    ϕ(z)=(m4+12)ϕ1(z)(m412)ϕ2(z),

    where

    ϕi(z)=φi(z)+zφi(z)q(z)+(γκ),

    i=1,2. Since q(z)Pβ[A,B]P(ϱ). Therefore, we can write

    q(z)=ϱ+(1ϱ)qo(z), (3.45)

    where qo(z)P. We have to check when ϕi(z)Pm(ϱ). For this consider

    (ϕi(z)ϱ)=(φi(z)+zφi(z)q(z)+(γκ)ϱ).

    Using (3.41) and (3.45), we have

    (ϕi(z)ϱ)=(ϱ+(1ϱ)hi(z)+(1ϱ)zhi(z)ϱ+(1ϱ)qo(z)+(γκ)ϱ),

    where hi(z),qo(z)P.

    (ϕi(z)ϱ)=(1ϱ)(hi(z))(1ϱ)|zhi(z)|(ϱ+γκ)+(1ϱ)qo(z).

    Since (γκ)>ϱ, so (ϱ+γκ)>0. Now by using the distortion results of Lemma 2.3, we have

    (ϕi(z)ϱ)=((1ϱ)hi(z)+ϱ+z(1ϱ)hi(z)γκ+ϱ+(1ϱ)hi(z)ϱ)(1ϱ)(hi(z))(1ϱ)|zhi(z)|(γκ+ϱ)+(1ϱ)(hi(z)). (3.46)

    Since hi(z)P, so (hi(z))>0 and (γκ+ϱ)>0 for (γκ)>ϱ. Hence, by using Lemma 2.3 in inequality (3.46), we have

    (ϕi(z)ϱ)(1ϱ)(hi(z))1ϱ2r1r2(hi(z))(1ϱ)(1r1+r)r2(1ϱ)2r(2ϱ)+(1ϱ)1r2.

    Since 1r2>0, taking T(r)=r2(1ϱ)2r(2ϱ)+(1ϱ). Let ro be the root then by simple calculations, we get

    ro=1ϱ2ϱ+32ϱ.

    Hence ϕ(z)Pm(ϱ) for |z|<ro. Thus for this radius ro the function f(z) belongs to the class T(m,β)[ϱ,γ+1,η], which is required to prove.

    Using the analogous approach used in Theorem 3.14, one can easily prove the following theorem.

    Theorem 3.11. Let ϱ=(1A1B)β, and (ηλ)>ϱ. Then

    T(m,β)[A,B,γ,η+1]T(m,β)[ϱ,γ,η]

    whenever

    |z|<ro=1ϱ2ϱ+32ϱ,where0ϱ<1.

    Integral Preserving Property: Here, we will discuss some integral preserving properties of our aforementioned classes. The generalized Libera integral operator Iσ introduced and discussed in [6,14] is defined by:

    Iσ(f)(z)=σ+1zσz0tσ1f(t)dt, (3.47)

    where f(z)A and σ>1.

    Theorem 3.12. Let σ>ϱ, where ϱ=(1A1B)β. If fR(m,β)[A,B,γ,η] then Iσ(f)R(m,β)[A,B,γ,η].

    Proof. Let fR(m,β)[A,B,γ,η], and set

    ψ(z)=z(Hγ,κλ,ηIσ(f)(z))Hγ,κλ,ηIσ(f)(z), (3.48)

    where ψ(z) is analytic and ψ(0)=1. From definition of Hγ,κλ,η(f) given by [1] and using Eq (3.47), we have

    z(Hγ,κλ,ηIσ(f)(z))=(σ+1)Hγ,κλ,ηf(z)σHγ,κλ,ηIσ(f)(z). (3.49)

    Then by using Eqs (3.48) and (3.49), we have

    (σ+1)Hγ,κλ,ηf(z)Hγ,κλ,ηIσ(f)(z)=ψ(z)+σ.

    Logarithmic differentiation and simple computation results in

    ϕ(z)=ψ(z)+zψ(z)ψ(z)+σ=z(Hγ,κλ,ηf(z))Hγ,κλ,ηf(z)P(m,β)[A,B], (3.50)

    with (ψ(z)+σ)>0, since (σ)>ϱ. Now, consider

    ψ(z)=(m4+12)ψ1(z)(m412)ψ2(z). (3.51)

    Combining (3.50) and (3.51), we get

    ϕ(z)=(m4+12)ϕ1(z)(m412)ϕ2(z),

    where ϕi(z)=ψi(z)+zψi(z)ψi(z)+σ, i=1,2. Since ϕ(z)P(m,β)[A,B], therefore

    ϕi(z)(1+Az1+Bz)β,

    which implies

    ψi(z)+zψi(z)ψi(z)+σ(1+Az1+Bz)βi=1,2.

    Therefore, using Lemma 2.2 we get

    ψi(z)(1+Az1+Bz)β,

    or ψ(z)P(m,β)[A,B]. Hence the result.

    Corollary 3.5. Let σ>ϱ. Then for m=2, if fSΛβ[A,B,γ,η] then Iσ(f)SΛβ[A,B,γ,η], where ϱ=(1A1B)β.

    Theorem 3.13. Let σ>ϱ, where ϱ=(1A1B)β. If fV(m,β)[A,B,γ,η] then Iσ(f)V(m,β)[A,B,γ,η].

    Proof. Let fV(m,β)[A,B,γ,η]. Then by using relation (1.12), we have

    zf(z)R(m,β)[A,B,γ,η],

    so by using Theorem 3.16, we can say that

    Iσ(zf(z))R(m,β)[A,B,γ,η],

    equivalently

    z(Iσ(f(z)))R(m,β)[A,B,γ,η],

    so again by using the relation (1.12), we get

    Iσ(f)V(m,β)[A,B,γ,η].

    Theorem 3.14. Let σ>ϱ, where ϱ=(1A1B)β. If fT(m,β)[A,B,γ,η] then Iσ(f)T(m,β)[A,B,γ,η].

    Proof. Let fT(m,β)[A,B,γ,η]. Then there exists ψ(z)SΛβ[A,B,γ,η], such that

    φ(z)=z(Hγ,κλ,ηf(z))(Hγ,κλ,ηψ(z)P(m,β)[A,B]. (3.52)

    Consider

    ϕ(z)=z(Hγ,κλ,ηIσ(f)(z))Hγ,κλ,ηIσ(ψ)(z). (3.53)

    Since ψ(z)SΛβ[A,B,γ,η], then by Corollary 3.17, Iσ(ψ)(z)SΛβ[A,B,γ,η]. Therefore

    q(z)=z(Hγ,κλ,ηIσ(ψ)(z))Hγ,κλ,ηIσ(ψ)(z)Pβ[A,B]. (3.54)

    By using (3.47) and Definition of Hγ,κλ,η, we get

    q(z)Hγ,κλ,ηIσ(ψ)(z)=(σ+1)Hγ,κλ,η(ψ)(z)σHγ,κλ,ηIσ(ψ)(z),

    or we can write it as

    Hγ,κλ,ηIσ(ψ)(z)=σ+1q(z)+σHγ,κλ,η(ψ)(z). (3.55)

    Now using the relation (3.47) and the Definition of Hγ,κλ,η, in (3.53), we have

    ϕ(z)Hγ,κλ,ηIσ(ψ)(z)=(σ+1)Hγ,κλ,η(f)(z)σHγ,κλ,ηIσ(f)(z). (3.56)

    Differentiating both sides of (3.56), we have

    ϕ(z)Hγ,κλ,ηIσ(ψ)(z)+ϕ(z)(Hγ,κλ,ηIσ(ψ)(z))=(σ+1)(Hγ,κλ,η(f)(z))σ(Hγ,κλ,ηIσ(f)(z)),

    then by simple computations and using (3.53)–(3.55), we get

    ϕ(z)+zϕ(z)q(z)+σ=φ(z), (3.57)

    with (σ)>ϱ, so (q(z)+σ)>0, since q(z)Pβ[A,B]P(ϱ). Consider

    ϕ(z)=(m4+12)ϕ1(z)(m412)ϕ2(z), (3.58)

    Combining Eqs (3.57) and (3.58), we have

    φ(z)=(m4+12)φ1(z)(m412)φ2(z), (3.59)

    where φi(z)=ϕi(z)+zϕi(z)q(z)+σ, i=1,2.

    Since φ(z)P(m,β)[A,B], thus we have

    φi(z)(1+Az1+Bz)β,

    then

    ϕi(z)+zϕi(z)q(z)+σ(1+Az1+Bz)β,i=1,2.

    Since (q(z)+σ)>0, therefore using Lemma 2.2 we get

    ϕi(z)(1+Az1+Bz)β,i=1,2,

    thus ϕ(z)P(m,β)[A,B]. Hence the result.

    Due to their vast applications, Mittag-Leffler functions have captured the interest of a number of researchers working in different fields of science. The present investigation may help researchers comprehend some stimulating consequences of the special functions. In the present article, we have used generalized Mittag-Leffler functions to define some novel classes related to bounded boundary and bounded radius rotations. Several inclusion relations and radius results for these classes have been discussed. Moreover, it has been proved that these classes are preserved under the generalized Libera integral operator. Finally, we can see that the projected solution procedure is highly efficient in solving inclusion problems describing the harmonic analysis. It is hoped that our investigation and discussion will be helpful in cultivating new ideas and applications in different fields of science, particularly in mathematics.

    Δ Open Unit Disc.

    Ω Class of normalized analytic functions.

    Real part of complex number.

    Γ Gamma function.

    χ(z) Schwartz function.

    The authors declare that they have no competing interests.

    The authors would like to thank the Rector of COMSATS Univeristy Islamabad, Pakistan for providing excellent research oriented environment. The author Thabet Abdeljawad would like to thank Prince Sultan University for the support through TAS research Lab.



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