# Analytic Normalized Solutions of 2D Fractional Saint-Venant Equations of a Complex Variable

Authors: Najla M. Alarifi; Rabha W. Ibrahim
Journal: Journal of Function Spaces
 (2021)

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## Abstract

Saint-Venant equations describe the flow below a pressure surface in a fluid. We aim to generalize this class of equations using fractional calculus of a complex variable. We deal with a fractional integral operator type Prabhakar operator in the open unit disk. We formulate the extended operator in a linear convolution operator with a normalized function to study some important geometric behaviors. A class of integral inequalities is investigated involving special functions. The upper bound of the suggested operator is computed by using the Fox-Wright function, for a class of convex functions and univalent functions. Moreover, as an application, we determine the upper bound of the generalized fractional 2-dimensional Saint-Venant equations (2D-SVE) of diffusive wave including the difference of bed slope.


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## Body

## 1. Introduction

Newly, fractional calculus has expanded considerable attention primarily appreciations to the growing occurrence of investigation mechanisms in the life sciences, allowing for simulations found by fractional operators [1] including differential and integral formulas. Further, the mathematical investigation of fractional calculus has advanced, chief to connections with other mathematical areas such as probability theory, mathematical physics [2], and mathematical biology [3–7] and the investigation of stochastic processes in real cases. In addition, it appears in studies of complex analysis. Now the literature, several different definitions of fractional integrals and derivatives are presented. Some of them such as the Riemann-Liouville integral, the Caputo, and the Riemann-Liouville differential operators are extensively employed in mathematics and physics and actually utilized in applied structures, modeling systems in real cases. While, in complex analysis, especially the theory of geometric functions, the researchers are focusing on Srivastava-Owa integral and differential operators [8], Tremblay differential operator, and the most recent fractional operator in [9, 10]. A new investigation of the complex ABC-fractional operator is presented to formulate different classes of analytic functions [11]. Some definitions such as the Hilfer and Prabhakar results [12] (differential and integral operators) are essentially the theme of mathematical study.Our study is aimed to extend the Prabhakar operator [13] to the open unit disk utilizing the class of normalized analytic functions. We formulate this operates in a linear convolution operator to study some important geometric behaviors. A class of integral inequalities is investigated involving special functions. The upper bound of the suggested operator is computed by using the Fox-Wright function, for a class of convex functions and univalent functions, and other studies are illustrated in the sequel.


## 2. Complex Prabhakar Operator (CPO)

The Prabhakar integral operator is defined for analytic functionϕz∈H0,1=ϕ∈∪:ϕ1z+ϕ2z2+⋯ by the formula [14–20](1)Pα,βγ,ωϕz=∫0zz−ζβ−1Ξα,βγωz−ζαϕζdζ=ϕ·ρα,βγ,ωz,α,β,γ,ω∈ℂ,z∈∪≔z∈ℂ:z<1,Rα,Rβ>0, where, [15](2)ρα,βγ,ωz≔zβ−1Ξα,βγωzα,Ξα,βγχ=∑n=0∞Γγ+nΓγΓαn+βχnn!.For example, letϕz=zε−1 then in view of [21] Corollary 2.3, we have(3)Pα,βγ,ωzε−1=∫0zz−ζβ−1Ξα,βγωz−ζαζε−1dζ=Γεzβ+ε−1Ξα,β+εγωzα.The Prabhakar derivative can be computed by the formula [13](4)Dα,βγ,ωϕz=dkdzkPα,k−β−γ,ωϕz,z∈∪.To study the geometric indications of CPO, we introduce the following class of analytic function: a normalized analytic functionϕz∈∧,z∈∪ achieving the power series(5)ϕz=z+∑n=2∞ϕnzn,z∈∪,Two analytic functionsf,g are called convoluted, denoting by f∗g if and only if(6)f∗gz=∑n=0∞anzn∗∑n=0∞gnzn=∑n=0∞angnzn.Definition 1.
Define a new functionΩ:∪⟶∪, such that(7)Ωα,βγ,ωz≔Γα+βγω1/αz1−βρα,βγ,ωz1α=Γα+βγω1/αΞα,βγω1αz1αα=Γα+βγω1/α∑n=0∞Γγ+nΓγΓαn+βω1/αznn!=Γα+βγω1/α∑n=0∞Γγ+nωn/αΓγΓαn+βznn!,α,β,γ,ω∈ℂ,z∈∪,Rα,Rβ>0.
Note that,(8)Ω1,11,1z=∑n=0∞znn!=ez,where ez is a transcendental function. Utilizing the functional Ωα,βγ,ω, we define the modified complex linear Prabhakar operator(9)ℙα,βγ,ωϕz≔Ωα,βγ,ω∗ϕz,ϕ∈∧.We have the following result.Proposition 2.
Ifϕ∈∧, then ℙα,βγ,ωϕz=Ωα,βγ,ω∗ϕ∈∧, where ∗ indicates the convolution product.Proof.
Letϕ∈∧, then we have(10)Ωα,βγ,ω∗ϕz=Ωα,βγ,ωz∗ϕz=Γα+βγω1/α∑n=0∞Γγ+nωn/αΓγΓαn+βznn!∗z+∑n=2∞ϕnzn=Γα+βΓβγω1/α+Γα+βγω1/αγω1/αΓα+βz+⋯∗0+z+ϕ2z2+⋯=z+∑n=2∞Γα+βγω1/αΓγ+nωn/αΓγΓαn+βϕnn!zn≔z+∑n=2∞Φnzn,where(11)Φn=Γα+βγω1/αΓγ+nωn/αΓγΓαn+βϕnn!=Γα+βΓγ+1ω1/αΓγ+nωn/αΓαn+βϕnn!.
Hence,ℙα,βγ,ωϕz=Ωα,βγ,ω∗ϕz∈∧.We need the following concepts to study theℙα,βγ,ωϕz geometrically.Definition 3.
A functionϕ∈∧ is indicated to be starlike including the origin of the linear slice contains the origin to all further point of ϕ in ϕz∈∪. A univalent function (one-one) ϕ is indicated to be convex in ∪ if the linear slice relating every two points of ϕz lies completely in ϕz∈∪. We indicate these classes by S∗ and C for starlike and convex consistently. Consider the class P includes all mappings ℘ smooth in ∪ with a positive real part in ∪ realizing ℘0=1,℘′0>0. Precisely, ϕ∈S∗⇔zϕ′z/ϕz∈P and ϕ∈C⇔1+zϕ′′z/ϕ′z∈P. Regularly, Rzϕ′z/ϕz>0 for the starlikeness and R1+zϕ′′z/ϕ′z>0 for the convexity.Two analytic functionsf1,f2 in U are called subordinated [22] denoting by f1≺f2 or f1z≺f2z,z∈∪, if there occurs an analytic function ω,∣ω∣≤∣z∣<1 satisfying(12)f1z=f2ωz,z∈∪.Next, lemma can be located in [22], P138-140.Lemma 4.
Leth∈∧.Then,(a)
hz+azh′z≺1+az+az2⇒hz≺z, when a∈0,1/3(b)
zh′z1+hz+ah2z≺z+1+az2⇒hz≺z, when ∣1+a∣≤1/4(c)
zh′z−hzeahz+ehz≺ez⇒hz≺z, when ∣a−1∣≤π/2(d)
zh′z1+ahz+hz≺2z+az2⇒hz≺z, when ∣a∣≤1/2(e)
zh′zeahz+hz≺z1+azeaz⇒hz≺z, when ∣a∣≤1(f)
hz+zh′z/1+ahz≺z⇒hz≺z, when ∣a∣≤1and the solution is sharp.Definition 5.
The Fox-Wright functionpWqz (the extension function of hypergeometric function) is formulated by(13)pWqx1,X1x2,X2⋯xp,Xpy1,Y1y2,Y2⋯yq,Yq;z=∑n=0∞Γx1+X1n⋯Γxp+XpnΓy1+Y1n⋯Γyq+Yqnznn!.
And it normalized by(14)pWq∗x1,X1x2,X2⋯xp,Xpy1,Y1y2,Y2⋯yq,Yq;z=Γy1⋯ΓyqΓx1⋯Γxp∑n=0∞Γx1+X1n⋯Γxp+XpnΓy1+Y1n⋯Γyq+Yqnznn!.
Note that the series is converged when(15)⊳≔∑j=1qYj−∑i=1pXi≥−1.
Moreover, it converges for all finite valuesz to the entire function provided ⊳<−1. In addition, at the boundary ∣z∣=1, it has the convergence value (see [23])(16)R⊲≔R∑j=1qyj−∑i=1pxi+p−q−12>0.
The significance of the Fox-Wright function arises regularly from its part in fractional calculus (see [1]). Further fascinating applications correspondingly occur. Wright’s original attentiveness in this function was connected to the asymptotic theory of partitions [24]. The formula ⊲ is generated in [23] by adding a positive parameter θ>0 as follows:(17)⊲θ≔∑j=1qyj+θ−∑i=1pxi+p−q−12.Based on this generalization, the authors in [24] introduced the following lemma.Lemma 6.
Assume that⊳=−1,θ>0 and R⊲θ>0. Then,(18)pWqz=Γθ∫01Hρ−1tdtt1−tzρ−1θ,z<ρ,where(19)H≔12πi∫∪Πς,θz−ς/Γς+θdς<∞,is the delta-neutral H function and Π indicates the Fox-Wright coefficients.Proposition 7.
Letϕ∈C (convex in the open unit disk), then(20)∣Ωα,βγ,ω∗ϕ′z∣≤1W1∗1+γ,1α+β,α;ω1/αr,α>0,β>0,∣z∣≔r<1,γ,ω∈ℝ+,Proof.
Sinceϕ∈C, then for each n≥2, we have ∣ϕn∣≤1. Then, a computation implies that(21)Ωα,βγ,ω∗ϕz≤Γα+βΓγ+1ω1/α∑n=1∞Γγ+nωn/αΓαn+β∣ϕn∣n!rn≤Γα+βΓγ+1ω1/α∑n=1∞Γγ+nωn/αΓαn+βrnn!=Γα+βΓγ+1ω1/α∑n=1∞Γγ+nΓαn+βω1/αrnn!=Γα+βΓγ+1ω1/α∑n=0∞Γγ+1+nΓαn+α+βω1/αrn+1n+1!=ω1/αrΓα+βΓγ+1ω1/α∑n=0∞Γ1+nΓ1+γ+nΓ2+nΓαn+α+βω1/αrnn!=rΓα+βΓγ+1∑n=0∞Γ1+nΓ1+γ+nΓ2+nΓαn+α+βω1/αrnn!=rΓα+βΓγ+12W21,11+γ,12,1α+β,α;ω1/αr=rΓ2Γα+βΓγ+1Γ12W21,11+γ,12,1α+β,α;ω1/αr=r2W2∗1,11+γ,12,1α+β,α;ω1/αr.
Now, for the derivative, we have(22)∣Ωα,βγ,ω∗ϕ′z∣≤∑n=1∞Γα+βγω1/αΓγ+nωn/αΓγΓαn+βn∣ϕn∣n!∣z∣∧n−1≤∑n=0∞Γα+βΓγ+1ω1/αΓγ+n+1ωn+1/αΓαn+α+βrnn!=∑n=0∞Γα+βΓγ+1Γγ+n+1ωn/αΓαn+α+βrnn!=Γα+βΓγ+11W11+γ,1α+β,α;ω1/αr=1W1∗1+γ,1α+β,α;ω1/αr.
This completes the proof.Remark 8.
(i)
It is clear that the above upper bound ofΩα,βγ,ω∗ϕz,∣z∣⟶1,ω=1 converges at(23)R⊲=R∑j=1qyj−∑i=1pxi+p−q−12>0⟶α+β−γ−1/2>0⟶α+β>γ+1/2.
Moreover, the upper bound ofΩα,βγ,ω∗ϕz converges, when(24)R⊲=R∑j=1qyj−∑i=1pxi+p−q−12>0⟶α+β−γ−3/2>0⟶α+β>γ+3/2.(ii)
For special case: by Proposition7 and Lemma 6, we have(25)Ω2,βγ,1∗ϕz≤rΓ2Γ2+βΓγ+1Γ12W21,11+γ,12,12+β,2;r≤rΓ2+βΓγ+1∫01Hr−1tdtt1−tzr−1,
provided that5/2+β>γ and θ=1.Proposition 9.
Letϕ be univalent in the open unit disk. Then,(26)∣Ωα,βγ,ω∗ϕz∣≤r1W1∗1+γ,1α+β,α;ω1/αr,∣Ωα,βγ,ω∗ϕz∣≤r1W1∗1+γ,1α+β,α;ω1/αr′α≠0,r<1,β,γ∈ℝ+.Proof.
Sinceϕ is univalent, then for each n≥2, we have ∣ϕn∣≤n. Then a calculation indicates that(27)∣Ωα,βγ,ω∗ϕz∣≤Γα+βΓγ+1ω1/α∑n=1∞Γγ+nωn/αΓαn+β∣ϕn∣n!rn≤Γα+βΓγ+1ω1/α∑n=1∞nΓγ+nωn/αΓαn+βrnn!=Γα+βΓγ+1ω1/α∑n=1∞Γγ+nΓαn+βω1/αrnn−1!=Γα+βΓγ+1ω1/α∑n=0∞Γγ+1+nΓαn+α+βω1/αrn+1n!=rΓα+βΓγ+1∑n=0∞Γ1+γ+nΓαn+α+βω1/αrnn!=rΓα+βΓγ+1W11+γ,1α+β,α;ω1/αr1=rW1∗11+γ,1α+β,α;ω1/αr.
Now, we return to the upper bound of the derivative(28)∣Ωα,βγ,ω∗ϕz∣≤∑n=1∞Γα+βγω1/αΓγ+nωn/αΓγΓαn+βnϕnn!∣zn−1∣≤∑n=1∞Γα+βγω1/αΓγ+nωn/αΓγΓαn+βn∣ϕn∣n!rn−1≤∑n=1∞Γα+βγω1/αΓγ+nωn/αΓγΓαn+βnn−1!rn−1=∑n=0∞Γα+βΓγ+1Γγ+n+1Γαn+β+αn+1n!ω1/αrn=Γα+βΓγ+1∑n=0∞Γ1+γ+nΓαn+α+βn+1ω1/αrnn!=Γα+βΓγ+1r1W11+γ,1α+β,α;ω1/αr′=r1W1∗1+γ,1α+β,α;ω1/αr′.
This completes the proof.More integral inequality results will consider in the following theorem.Theorem 10.
Considerϕ∈∧ and the operator Ωα,βγ,ω∗ϕz, where α≠0,z∈∪,β,γ∈ℝ+. If one of the following subordination inequalities hold(29)Ωα,βγ,ω∗ϕz+zΩα,βγ,ω∗ϕz1+σΩα,βγ,ω∗ϕz≺σz2+2z,σ≤12,Ωα,βγ,ω∗ϕz+zΩα,βγ,ω∗ϕzeσz≺1+σzeσzz,σ≤1,Ωα,βγ,ω∗ϕz+zΩα,βγ,ω∗ϕz1+σΩα,βγ,ω∗ϕz≺z1+σz1+σz,σ≤1,zΩα,βγ,ω∗ϕz−Ωα,βγ,ω∗ϕzeσΩα,βγ,ω∗ϕz+eσΩα,βγ,ω∗ϕz≺ez,σ−1≤π2,Ωα,βγ,ω∗ϕz+σzΩα,βγ,ω∗ϕz≺1+σz+σz2,0<σ≤13,Ωα,βγ,ω∗ϕz+zΩα,βγ,ω∗ϕz≺2z,Ωα,βγ,ω∗ϕz1+zΩα,βγ,ω∗ϕzΩα,βγ,ω∗ϕz≺2z,Ωα,βγ,ω∗ϕz1+zΩα,βγ,ω∗ϕz2≺2z,
thenΨz≔1/eσz∫0zΩα,βγ,ω∗ϕζeσζdζ∈H0,1 and(30)1eσz∫0zΩα,βγ,ω∗ϕζeσζdζ<12−∣σ∣,∣σ∣<2.Proof.
Suppose that the operatorΩα,βγ,ω∗ϕ achieves one of the subordination inequalities (a)-(h) then, in view of Proposition 2 and results in Lemma 4, we have(31)Ωα,βγ,ω∗ϕz≺z,z∈∪.
Consequently, we obtain the upper bound inequality(32)Ωα,βγ,ω∗ϕz<1,z∈∪.
The functioneσz achieves the real inequality(33)Rzeσz′eσz+2=Rσz+2>0,provided that ∣σ∣<2. Moreover, we have superior inequality(34)χ=sup∣z∣<1∣eσz∣∣zeσz′+2eσz∣=sup∣z∣<1∣eσz∣∣zσeσz+2eσz∣≤12−∣σ∣<∞,∣σ∣<2.
Hence, in view of [22] [Corollary 4.3a.2, P210], we conclude that(35)1eσz∫0zΩα,βγ,ω∗ϕζeσζdζ≤z22−∣σ∣,∣σ∣<2,which yields(36)1eσz∫0zΩα,βγ,ω∗ϕζeσζdζ<12−∣σ∣,σ<2,z∈∪,,
hence the proof.Now, we investigate another integral inequality involving the operatorΩα,βγ,ω∗ϕ where ϕ∈∧.Theorem 11.
Considerϕ∈∧ and the operator Ωα,βγ,ω∗ϕz, where α,β,γ,ω∈ℂ,z∈∪,Rα,Rβ>0. If one of the subordination inequalities in Theorem 10 holds, then(37)Gz≔1eσz∫0zΩα,βγ,ω∗ϕζgζζσ−1dζ∈H0,1,∣1eσz∫0zΩα,βγ,ω∗ϕζgζdζ∣<∣gz∣1−∣σ∣,∣σ∣<1,where g∈H1,1,g≠0.Proof.
Suppose that the operatorΩα,βγ,ω∗ϕ has one of the subordination inequalities (a)-(h) in Theorem 10 then, in view of Proposition 2, we have(38)Ωα,βγ,ω∗ϕz≺z,z∈∪,which implies that(39)∣Ωα,βγ,ω∗ϕz∣<1,z∈∪.
As in Theorem10, the function eσz admits the real inequality(40)Rzeσz′eσz+1=Rσz+1>0,provided that ∣σ∣<1. Moreover, we have superior inequality(41)ν=sup∣z∣<1gzzeσz′+1+σeσz=supz<1gzzσeσz+eσz=supz<1gz1−σ,σ<1<∞,
Thus, in view of [22] [Theorem 4.3a, P207], we indicate that then Gz∈H0,1 and(42)∣1eσz∫0zΩα,βγ,ω∗ϕζgζdζ∣<∣gz∣1−∣σ∣,∣σ∣<1,where g∈H1,1,g≠0, hence the proof.In addition, we have the following result by replacingeσz by zeσz:Theorem 12.
Considerϕ∈∧ and the operator Ωα,βγ,ω∗ϕz, where α≠0,z∈∪β,γ∈ℝ+. If one of the subordination inequalities in Theorem 10 holds, then(43)Lz≔1zκeσz∫0zΩα,βγ,ω∗ϕζ1+σζeσζζκ−1dζ∈H0,1,∣1zκeσz∫0zΩα,βγ,ω∗ϕζ1+σζeσζζκ−1dζ∣<1+∣σ∣1−∣σ∣+κ,κ>0,∣σ∣≤1.Proof.
Suppose that the operatorΩα,βγ,ω∗ϕ has one of the subordination inequalities (a)-(h) then, in view of Proposition 2 and results in [22], P138-140, we have(44)Ωα,βγ,ω∗ϕz≺z,z∈∪,which leads to(45)∣Ωα,βγ,ω∗ϕz∣<1,z∈∪.
The functionzeσz admits the following properties(46)zeσz∈∧,zeσz′zeσzz=e2σz+σze2σz≠0,∣σ∣≤1;Rzzeσz′zeσz+κ=R1+σz+κ>0.
Moreover, we have superior inequality(47)S=sup∣z∣<1∣zzeσz′∣∣zzeσz′+κzeσz∣=sup∣z∣<11+σz1+σz+κ=sup∣z∣<11+∣σ∣1−∣σ∣+κ,∣σ∣<1,κ>0<∞.
Thus, in view of [22]-[Corollary 4.3a.1, P208], Lz∈H0,1 and(48)∣1zκeσz∫0zΩα,βγ,ω∗ϕζ1+σζeσζζκ−1dζ∣<1+∣σ∣1−∣σ∣+κ,κ>0,σ∣≤1,z∈∪,ϕ∈∧.
This completes the proof.Theorem 13.
Consider thatϕ is convex univalent function in the open unit disk and the operator Ωα,βγ,ω∗ϕz, where α≠0,z∈∪,β,γ∈ℝ+. Then,(49)Ψz∈H0,1,
with(50)∣1eσz∫0zΩα,βγ,ω∗ϕζeσζdζ∣<2W2∗1,11+γ,12,1α+β,α;ω1/α2−∣σ∣,∣σ∣<2.
Gz∈H0,1 with(51)∣1eσz∫0zΩα,βγ,ω∗ϕζgζdζ∣<∣gz∣2W2∗1,11+γ,12,1α+β,α;ω1/α1−∣σ∣,∣σ∣<1,
whereg∈H1,1,g≠0.
Lz∈H0,1 with(52)∣1zκeσz∫0zΩα,βγ,ω∗ϕζ1+σζeσζζκ−1dζ∣<1+∣σ∣2W2∗1,11+γ,12,1α+β,α;ω1/α1+∣σ∣+κ,κ>0,∣σ∣≤1,z∈∪.Proof.
Letϕ be convex univalent in ∪. Then, in view of Proposition 7, we have(53)∣Ωα,βγ,ω∗ϕz∣<r2W2∗1,11+γ,12,1α+β,α;ω1/αr.
Consequently, by assumingr⟶1, we obtain(54)∣Ωα,βγ,ω∗ϕz∣<2W2∗1,11+γ,12,1α+β,α;ω1/α.
By the proof of Theorem10, we conclude that (A). Similarly, by using the proof in Theorems 11 and 12, we have (B) and (C), respectively. This ends the proof.Theorem 14.
Consider thatϕ is univalent function in the open unit disk and the operator Ωα,βγ,ω∗ϕz, where α≠0,z∈∪,β,γ∈ℝ+. Then
Ψz∈H0,1 with(55)1eσz∫0zΩα,βγ,ω∗ϕζeσζdζ<1W1∗1+γ,1α+β,α;ω1/α2−∣σ∣,∣σ∣<2.
Gz∈H0,1 with(56)1eσz∫0zΩα,βγ,ω∗ϕζgζdζ<∣gz∣1W1∗1+γ,1α+β,α;ω1/α1−∣σ∣,∣σ∣<1,where g∈H1,1,g≠0.
Lz∈H0,1 with(57)1zκeσz∫0zΩα,βγ,ω∗ϕζ1+σζeσζζκ−1dζ<1+σ1W1∗1+γ,1α+β,α;ω1/α1+σ+κ,κ>0,σ≤1,z∈∪.Proof.
Letϕ be convex univalent in ∪. Then, in view of Proposition 9, we have(58)Ωα,βγ,ω∗ϕz<r1W1∗1+γ,1α+β,α;ω1/αr.
Consequently, by assumingr⟶1, we have(59)Ωα,βγ,ω∗ϕz<1W1∗1+γ,1α+β,α;ω1/α.
By the proof of Theorem10, we conclude that (A). Similarly, by using the proof in Theorems 11 and 12, we have (B) and (C), respectively. This ends the proof.In the next result, we discuss the starlikeness of the operatorΩα,βγ,ω∗ϕ.Theorem 15.
Consider the operatorΩα,βγ,ω∗ϕ,ϕ∈∧.(A)
IfΩα,βγ,ω∗ϕz/eσzσz+1−1<0.04, where ∣σ∣<1/213−3≈0.3027 then Ωα,βγ,ω∗ϕ∈S∗(B)
Ifzϕ′z/ϕz−1<0.374 and supΩα,βγ,ω∗ϕz/Ωα,βγ,ω∗ϕz≡σ,σ<1 then(60)∫0zΩα,βγ,ω∗ϕzϕ′zdz∈S∗.(C)
If∣Ωα,βγ,ω∗ϕz∣<2/5 then Ωα,βγ,ω∗ϕz∈S∗Proof.
For part (A), assume thatgz=zeσz then(61)sup∣g′′zg′z∣=sup∣2σ+zσ21+zσ∣.
Consequently, we have(62)sup∣2σ+zσ21+zσ∣≤2∣σ∣+σ21−∣σ∣<1,∣σ∣<1.
The value∣σ∣<1/213−3≈0.3027 implies that 2∣σ∣+σ2/1−∣σ∣≈0.98<1. Since(63)Ωα,βγ,ω∗ϕzeσzσz+1−1<0.04,then in view of [22]-[Theorem 5.5c, P296], we conclude that Ωα,βγ,ω∗ϕ∈S∗.
For part (B), since(64)zϕ′zϕz−1<0.374,where the number 0.374 is a solution of the equation 1+xex=2 then by [22] [Theorem 5.5g, P299], we have ∣ϕ′z−1∣<1. Moreover, in terms of ∣σ∣, we have(65)ϕ′z−1<2−∣σ∣22−2∣σ∣+1−∣σ∣2−∣σ∣2+1<1,∣σ∣<1,lim∣σ∣⟶12−∣σ∣22−2∣σ∣+1−∣σ∣2−∣σ∣2+1≈0.by [22] [Theorem 5.5d, P298] then(66)∫0zΩα,βγ,ω∗ϕzϕ′zdz∈S∗.
The last part immediately comes from [22] [Corollary5.5.a,P294]. This ends the proof.Theorem 16.
Consider the operatorΩα,βγ,ω∗ϕ,ϕ∈∧.(A)
If the following inequality holds(67)zΩα,βγ,ω∗ϕzΩα,βγ,ω∗ϕz≺ez−1,z∈∪,then(68)Ωα,βγ,ω∗ϕz≺exp∫0zeζ−1ζ−1dζ,z∈∪,(B)
If the subordination occurs(69)zΩα,βγ,ω∗ϕzΩα,βγ,ω∗ϕz−1≺ez−1,z∈∪,then(70)Ωα,βγ,ω∗ϕzz≺exp∫0zeζ−1ζ−1dζ,z∈∪,(C)
If the next relation exists(71)zΩα,βγ,ω∗ϕz≺ez−1,z∈∪,then(72)Ωα,βγ,ω∗ϕz≺∫0zeζ−1ζ−1dζ.Proof.
It is clear that the function(73)fz=ez−1=z+z22+z36+z424+z5120+Oz6,satisfies f0=0 and it is convex in the open unit disk. Consequently, it is starlike. By Proposition 2, the operator Ωα,βγ,ω∗ϕz∈∧ and hence Ωα,βγ,ω∗ϕ0=1 that is Ωα,βγ,ω∗ϕz∈H1,1. Similarly for the function(74)Ωα,βγ,ω∗ϕzz∈H1,1.
Thus, in view of [22]-[Corollary 3.1d.1, P76], we have the desire results.
For the last part (C),zΩα,βγ,ω∗ϕz∈H0,1; thus, in virtue of [22]-[Theorem 3.1d, P76], where a=0, we conclude the last subordination.


### 2.1. Fractional Saint-Venant Equations

By using the fractional calculus of the construction, we formulate the fractional 2D-Saint-Venant equations utilizing the functional convolution operatorΩα,βγ,ω∗ϕ,ϕ∈Λ, and z∈∪.Example 1.
We investigate the upper bound of the 2-dimensional Saint-Venant equations (2D-SVE) of diffusive wave (this equation has measured the level of the water). This equation simply presents the formula(75)dΘzdz−Δz=0,
whereΘ is the height deviation of the horizontal pressure surface at two-dimensional position z=x+iy and Δς represents the difference of bed slope. By using the convolution operator, we generalize 2D-SVE into the form(76)Ωα,βγ,ω∗ϕz−Δz=0,z∈∪.
Multiplying both sides of Eq. (76) by z and let(77)Δz=ez−1z=1+z2+z26+z324+z4120+z5720+Oz6,we have(78)zΩα,βγ,ω∗ϕz=ez−1,z∈∪.
Thus, in view of Theorem16-(C), we conclude the upper solution of Eq. (78) is given by (see Figure 1, second row)(79)Ωα,βγ,ω∗ϕz≺∫0zeζ−1ζ−1dζ.
Hence, we obtain(80)Ωα,βγ,ω∗ϕz=c+Eiz−logz=c+0.577+z+z24+z318+z496+z5600+z64320+Oz7−iπargz+π2π,where c is a constant and(81)Eiz=0.577+lnz+z+⋯,indicates the exponential integral. Assuming that(82)c=iπargz+π2π−0.577,we get(83)Ωα,βγ,ω∗ϕz=z+z24+z318+z496+z5600+z64320+Oz7∈∧.
By the convexity ofez (see [22]-P139), we confirm that the solution is normalized analytic convex in ∪. Note that the term Ωα,βγ,ω∗ϕz is called the convective acceleration term. Figure 1 shows the behavior of solutions of 2D-SVE of diffusive wave.Figure 1
Saint-Venant equations (2D-SVE) of diffusive wave.
(a)
The first row represents the functionfz=ez−1 and the second row indicates the solution (79) of the extended 2D-SVE(b)
The solutionfz≔Ωα,βγ,ω∗ϕz


## 2.1. Fractional Saint-Venant Equations

By using the fractional calculus of the construction, we formulate the fractional 2D-Saint-Venant equations utilizing the functional convolution operatorΩα,βγ,ω∗ϕ,ϕ∈Λ, and z∈∪.Example 1.
We investigate the upper bound of the 2-dimensional Saint-Venant equations (2D-SVE) of diffusive wave (this equation has measured the level of the water). This equation simply presents the formula(75)dΘzdz−Δz=0,
whereΘ is the height deviation of the horizontal pressure surface at two-dimensional position z=x+iy and Δς represents the difference of bed slope. By using the convolution operator, we generalize 2D-SVE into the form(76)Ωα,βγ,ω∗ϕz−Δz=0,z∈∪.
Multiplying both sides of Eq. (76) by z and let(77)Δz=ez−1z=1+z2+z26+z324+z4120+z5720+Oz6,we have(78)zΩα,βγ,ω∗ϕz=ez−1,z∈∪.
Thus, in view of Theorem16-(C), we conclude the upper solution of Eq. (78) is given by (see Figure 1, second row)(79)Ωα,βγ,ω∗ϕz≺∫0zeζ−1ζ−1dζ.
Hence, we obtain(80)Ωα,βγ,ω∗ϕz=c+Eiz−logz=c+0.577+z+z24+z318+z496+z5600+z64320+Oz7−iπargz+π2π,where c is a constant and(81)Eiz=0.577+lnz+z+⋯,indicates the exponential integral. Assuming that(82)c=iπargz+π2π−0.577,we get(83)Ωα,βγ,ω∗ϕz=z+z24+z318+z496+z5600+z64320+Oz7∈∧.
By the convexity ofez (see [22]-P139), we confirm that the solution is normalized analytic convex in ∪. Note that the term Ωα,βγ,ω∗ϕz is called the convective acceleration term. Figure 1 shows the behavior of solutions of 2D-SVE of diffusive wave.Figure 1
Saint-Venant equations (2D-SVE) of diffusive wave.
(a)
The first row represents the functionfz=ez−1 and the second row indicates the solution (79) of the extended 2D-SVE(b)
The solutionfz≔Ωα,βγ,ω∗ϕz


## 3. Conclusion

From above, we have extended the Prabhakar operator in the open unit disk. We formulated it in a linear convolution operator with a normalized function. A class of integral inequalities is investigated involving special functions. The upper bound of the suggested operator is computed by using the Fox-Wright function, for a class of convex functions and univalent functions. Moreover, we applied the operator to generalize the 2D-SVE. A solution of the extended 2D-SVE is computed by using recent result (Theorem16).For future woks, one can consider extra studies in the geometric function theory by considering the operator in different classes of analytic functions, such as normalized functions, harmonic functions, and meromorphic functions.



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