Research article

On-line estimation of physiological states for monitoring and control of bioprocesses

  • An approach for monitoring of main physiological states of a class processes is proposed. This class is characterized by production and consumption of intermediate metabolite related to target product. The balance between these two phenomena is considered as key parameter for recognizing the process physiological states. A general structure of cascade software sensor of the key parameter is derived and applied for process monitoring and control. Two type processes are considered as case study. The first one is mono culture for simultaneous saccharification and fermentation of starch to ethanol by Saccharomyces cerevisiae and the second one is mixed culture for biopolymer production by L. delbrulckii and R. Eutropha. The good properties of the proposed monitoring and control schemes are demonstrated by simulation investigations.

    Citation: Velislava N Lyubenova, Maya N Ignatova. On-line estimation of physiological states for monitoring and control of bioprocesses[J]. AIMS Bioengineering, 2017, 4(1): 93-112. doi: 10.3934/bioeng.2017.1.93

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  • An approach for monitoring of main physiological states of a class processes is proposed. This class is characterized by production and consumption of intermediate metabolite related to target product. The balance between these two phenomena is considered as key parameter for recognizing the process physiological states. A general structure of cascade software sensor of the key parameter is derived and applied for process monitoring and control. Two type processes are considered as case study. The first one is mono culture for simultaneous saccharification and fermentation of starch to ethanol by Saccharomyces cerevisiae and the second one is mixed culture for biopolymer production by L. delbrulckii and R. Eutropha. The good properties of the proposed monitoring and control schemes are demonstrated by simulation investigations.


    Let H(U) be the class of analytic functions in the open unit disc U={zC:|z|<1} and let H[a,υ] be the subclass of H(U) including form-specific functions

    f(z)=a+aυzυ+aυ+1zυ+1+(aC),

    we denote by H=H[1,1].

    Also, A(p) should denote the class of multivalent analytic functions in U, with the power series expansion of the type:

    f(z)=zp+υ=p+1aυzυ(pN={1,2,3,..}). (1.1)

    Upon differentiating j-times for each one of the (1.1) we obtain:

    f(j)(z)=δ(p,j)zpj+υ=p+1δ(υ,j)aυzυjzU,δ(p,j)=p!(pj)!       (pN, jN0=N{0}, pj). (1.2)

    Numerous mathematicians, for instance, have looked at higher order derivatives of multivalent functions (see [1,3,6,9,16,27,28,31]).

    For f,H, the function f is subordinate to or the function is said to be superordinate to f in U and we write f(z)(z), if there exists a Schwarz function ω in U with ω(0)=0 and |ω(z)|<1, such that f(z)=(ω(z)), zU. If is univalent in U, then f(z)(z) iff f(0)=(0) and f(U)(U). (see [7,21]).

    In the concepts and common uses of fractional calculus (see, for example, [14,15] see also [2]; the Riemann-Liouville fractional integral operator of order αC ((α)>0) is one of the most widely used operators (see [29]) given by:

    (Iα0+f)(x)=1Γ(α)x0(xμ)α1f(μ)dμ(x>0;(α)>0) (1.3)

    applying the well-known (Euler's) Gamma function Γ(α). The Erd élyi-Kober fractional integral operator of order αC((α)>0) is an interesting alternative to the Riemann-Liouville operator Iα0+, defined by:

    (Iα0+;σ,ηf)(x)=σxσ(α+η)Γ(α)x0μσ(η+1)1(xσμσ)α1f(μ)dμ (1.4)
    (x>0;(α)>0),

    which corresponds essentially to (1.3) when σ1=η=0, since

    (Iα0+;1,0f)(x)=xα(Iα0+f)(x)(x>0;(α)>0).

    Mainly motivated by the special case of the definition (1.4) when x=σ=1, η=ν1 and α=ρν, here, we take a look at the integral operator p(ν,ρ,μ) with fA(p) by (see [11])

    p(ν,ρ;)f(z)=Γ(ρ+p)Γ(ν+p)Γ(ρν)10μν1(1μ)ρν1f(zμ)dμ
    (>0;ν,ρR;ρ>ν>p;pN).

    Evaluating (Euler's) Gamma function by using the Eulerian Beta-function integral as following:

    B(α,β):={10μα1(1μ)β1dμ(min{(α),(β)}>0)Γ(α)Γ(β)Γ(α+β)(α,βCZ0),

    we readily find that

    p(ν,ρ;)f(z)={zp+Γ(ρ+p)Γ(ν+p)υ=p+1Γ(ν+υ)Γ(ρ+υ)aυzυ(ρ>ν)f(z)(ρ=ν). (1.5)

    It is readily to obtain from (1.5) that

    z(p(ν,ρ;)f(z))=(ν+p)(p(ν+1,ρ;)f(z))ν(p(ν,ρ;)f(z)). (1.6)

    The integral operator p(ν,ρ;)f(z) should be noted as a generalization of several other integral operators previously discussed for example,

    (ⅰ) If we set p=1, we get ˜I(ν,ρ;)f(z) defined by Ŕaina and Sharma ([22] with m=0);

    (ⅱ) If we set ν=β,ρ=β+1 and  =1, we obtain βpf(z)(β>p) it was presented by Saitoh et al.[24];

    (ⅲ) If we set ν=β,ρ=α+βδ+1, =1, we obtain α,δβ,pf(z)(δ>0; αδ1; β>p) it was presented by Aouf et al. [4];

    (ⅳ) If we put ν=β,ρ=α+β, =1, we get Qαβ,pf(z)(α0;β>p) it was investigated by Liu and Owa [18];

    (ⅴ) If we put p=1, ν=β,ρ=α+β, =1, we obtain αβf(z)(α0;β>1) it was introduced by Jung et al. [13];

    (ⅵ) If we put p=1, ν=α1, ρ=β1, =1, we obtain L(α,β)f(z)(α,βCZ0,Z0={0,1,2,...}) which was defined by Carlson and Shaffer [8];

    (ⅶ) If we put p=1, ν=ν1, ρ=j, =1 we obtain Iν,jf(z)(ν>0;j1) it was investigated by Choi et al. [10];

    (ⅷ) If we put p=1, ν=α,ρ=0, =1, we obtain Dαf(z)(α>1) which was defined by Ruscheweyh [23];

    (ⅸ) If we put p=1, ν=1, ρ=m, =1, we obtain Imf(z)(mN0) which was introduced by Noor [21];

    (ⅹ) If we set p=1, ν=β,ρ=β+1, =1 we obtain βf(z) which was studied by Bernadi [5];

    (ⅹⅰ) If we set p=1, ν=1, ρ=2, =1 we get f(z) which was defined by Libera [17].

    We state various definition and lemmas which are essential to obtain our results.

    Definition 1. ([20], Definition 2, p.817) We denote by Q the set of the functions f that are holomorphic and univalent on ¯UE(f), where

    E(f)={ζ:ζU  and  limzζf(z)=},

    and satisfy f(ζ)0 for ζUE(f).

    Lemma 1. ([12]; see also ([19], Theorem 3.1.6, p.71)) Assume that h(z) is convex (univalent) function in U with h(0)=1, and let φ(z)H, is analytic in U. If

    φ(z)+1γzφ(z)h(z)(zU),

    where γ0 and Re(γ)0. Then

    φ(z)Ψ(z)=γzγz0tγ1h(t)dth(z)(zU),

    and Ψ(z) is the best dominant.

    Lemma 2. ([26]; Lemma 2.2, p.3) Suppose that q is convex function in U and let  ψC with ϰC=C{0} with

    Re(1+zq(z)q(z))>max{0;Reψϰ},zU.

    If λ(z) is analytic in U, and

    ψλ(z)+ϰzλ(z)ψq(z)+ϰzq(z),

    therefore λ(z)q(z), and q is the best dominant.

    Lemma 3. ([20]; Theorem 8, p.822) Assume that q is convex univalent in U and suppose δC, with Re(δ)>0. If λH[q(0),1]Q and λ(z)+δzλ(z) is univalent in U, then

    q(z)+δzq(z)λ(z)+δzλ(z),

    implies

    q(z)λ(z)     (zU)

    and q is the best subordinant.

    For a,ϱ,c and c(cZ0) real or complex number the Gaussian hypergeometric function is given by

    2F1(a,ϱ;c;z)=1+aϱc.z1!+a(a+1)ϱ(ϱ+1)c(c+1).z22!+....

    The previous series totally converges for zU to a function analytical in U (see, for details, ([30], Chapter 14)) see also [19].

    Lemma 4. For a,ϱ and c (cZ0), real or complex parameters,

    10tϱ1(1t)cϱ1(1zt)xdt=Γ(ϱ)Γ(ca)Γ(c)2F1(a,ϱ;c;z)(Re(c)>Re(ϱ)>0); (2.1)
    2F1(a,ϱ;c;z)=2F1(ϱ,a;c;z); (2.2)
    2F1(a,ϱ;c;z)=(1z)a2F1(a,cϱ;c;zz1); (2.3)
    2F1(1,1;2;azaz+1)=(1+az)ln(1+az)az; (2.4)
    2F1(1,1;3;azaz+1)=2(1+az)az(1ln(1+az)az). (2.5)

    Throughout the sequel, we assume unless otherwise indicated 1D<C1, δ>0, >0, ν,ρR, ν>p, pN and (ρj)0. We shall now prove the subordination results stated below:

    Theorem 1. Let 0j<p, 0<r1 and for fA(p) assume that

    (p(ν,ρ;)f(z))(j)zpj0,    zU, (3.1)

    whenever δ(0,+)N. Let define the function Φj by

    Φj(z)=(1α)((p(ν,ρ;)f(z))(j)zpj)δ+α(p(ν+1,ρ;)f(z))(j)zpj((p(ν,ρ;)f(z))(j)zpj)δ1,

    such that the powers are all the principal ones, i.e., log1 = 0. Whether

    Φj(z)[p!(pj)!]δ(1+Cz1+Dz )r, (3.2)

    then

    ((p(ν,ρ;)f(z))(j)zpj)δ[p!(pj)!]δp(z), (3.3)

    where

    p(z)={(CD)ri0(r)ii!(CDC)i(1+Dz)i 2F1(i,1;1+δ(ν+p)α;Dz1+Dz)(D0);2F1(r,δ(ν+p)α;1+δ(ν+p)α;Cz)                                     (D=0),

    and [p!(pj)!]δp(z) is the best dominant of (3.3). Moreover, there are

    ((p(ν,ρ;)f(z))(j)zpj)δ>[p!(pj)!]δζ,     zU, (3.4)

    where ζ is given by:

    ζ={(CD)ri0(r)ii!(CDC)i(1D)i 2F1(i,1;1+δ(ν+p)α;DD1)(D0);2F1(r,δ(ν+p)α;1+δ(ν+p)α;C)                                     (D=0),

    then (3.4) is the best possible.

    Proof. Let

    ϕ(z)=((pj)!p!(p(ν,ρ;)f(z))(j)zpj)δ,   (zU). (3.5)

    It is observed that the function ϕ(z)H, which is analytic in U and ϕ(0)=1. Differentiating (3.5) with respect to z, applying the given equation, the hypothesis (3.2), and the knowing that

    z(p(ν,ρ;)f(z))(j+1)=(ν+p)(p(ν+1,ρ;)f(z))(j)(ν+j)(p(ν,ρ;)f(z))(j)   (0j<p), (3.6)

    we get

    ϕ(z)+zϕ(z)δ(ν+p)α(1+Cz1+Dz )r=q(z)     (zU).

    We can verify that the above equation q(z) is analytic and convex in U as following

    Re(1+zq(z)q(z))=1+(1r)(11+Cz)+(1+r)(11+Dz)>1+1r1+|C|+1+r1+|D|0   (zU).

    Using Lemma 1, there will be

    ϕ(z)p(z)=δ(ν+p)αzδ(ν+p)αz0tδ(ν+p)α1(1+Ct1+Dt)rdt.

    In order to calculate the integral, we define the integrand in the type

    tδ(ν+p)α1(1+Ct1+Dt)r=tδ(ν+p)α1(CD)r(1CDC+CDt)r,

    using Lemma 4 we obtain

    p(z)=(CD)ri0(r)ii!(CDC)i(1+Dz)i 2F1(i,1;1+δ(ν+p)α;Dz1+Dz)(D0).

    On the other hand if D=0 we have

    p(z)=2F1(r,δ(ν+p)α;1+δ(ν+p)α;Cz),

    where the identities (2.1)–(2.3), were used after changing the variable, respectively. This proof the inequality (3.3).

    Now, we'll verify it

    inf{p(z):|z|<1}=p(1). (3.7)

    Indeed, we have

    (1+Cz1+Dz )r(1Cσ1Dσ)r   (|z|<σ<1).

    Setting

    (s,z)=(1+Csz1+Dsz)r   (0s1; zU)

    and

    dv(s)=δ(ν+p)αsδ(ν+p)α1ds

    where dv(s) is a positive measure on the closed interval [0, 1], we get that

    p(z)=10(s,z)dv(s),

    so that

    p(z)10(1Csσ1Dsσ)rdv(s)=p(σ)   (|z|<σ<1).

    Now, taking σ1 we get the result (3.7). The inequality (3.4) is the best possible since [p!(pj)!]δp(z) is the best dominant of (3.3).

    If we choose j=1 and α=δ=1 in Theorem 1, we get:

    Corollary 1. Let 0<r1. If

    (p(ν+1,ρ;)f(z))zp1p(1+Cz1+Dz )r,

    then

    ((p(ν,ρ;)f(z))zp1)>pζ1,     zU, (3.8)

    where ζ1 is given by:

    ζ1={(CD)ri0(r)ii!(CDC)i(1D)i 2F1(i,1;1+(ν+p);DD1)(D0);2F1(r,(ν+p);1+(ν+p);C)                                     (D=0),

    then (3.8) is the best possible.

    If we choose ν=ρ=0 and  =1 in Theorem 1, we get:

    Corollary 2. Let 0j<p, 0<r1 and as fA(p) assume that

    f(j)(z)zpj0,    zU,

    whenever δ(0,+)N. Let define the function Φj by

    Φj(z)=[1α(1jp)](f(j)(z)zpj)δ+α(zf(j+1)(z)pf(j)(z))(f(j)(z)zpj)δ, (3.9)

    such that the powers are all the principal ones, i.e., log1 = 0. If

    Φj(z)[p!(pj)!]δ(1+Cz1+Dz )r,

    then

    (f(j)(z)zpj)δ[p!(pj)!]δp1(z), (3.10)

    where

    p1(z)={(CD)ri0(r)ii!(CDC)i(1+Dz)i 2F1(i,1;1+δpα;Dz1+Dz)(D0);2F1(r,δpα;1+δpα;Cz)                                     (D=0),

    and [p!(pj)!]δp1(z) is the best dominant of (3.10). Morover, there are

    (f(j)(z)zpj)δ>[p!(pj)!]δζ2,     zU, (3.11)

    where ζ2 is given by

    ζ2={(CD)ri0(r)ii!(CDC)i(1D)i 2F1(i,1;1+δpα;DD1)(D0);2F1(r,δpα;1+δpα;C)                                     (D=0),

    then (3.11) is the best possible.

    If we put δ=1 and  r=1 in Corollary 2, we get:

    Corollary 3. Let 0j<p, and for fA(p) say it

    f(j)(z)zpj0,    zU.

    Let define the function Φj by

    Φj(z)=[(1α(1jp)]f(j)(z)zpj+αf(j+1)(z)pzpj1.

    If

    Φj(z)p!(pj)!1+Cz1+Dz,

    then

    f(j)(z)zpjp!(pj)!p2(z), (3.12)

    where

    p2(z)={CD+(1CD)(1+Dz)1 2F1(1,1;1+pα;Dz1+Dz)(D0);1+pp+αCz,                                                    (D=0),

    and p!(pj)!p2(z) is the best dominant of (3.12). Morover there will be

    (f(j)(z)zpj)>p!(pj)!ζ3,     zU, (3.13)

    where ζ3 is given by:

    ζ3={CD+(1CD)(1D)1 2F1(1,1;1+pα;DD1)(D0);1pp+αC,                                                    (D=0),

    then (3.13) is the best possible.

    For C=1,D=1 and j=1 Corollary 3, leads to the next example:

    Example 1. (i) For fA(p) suppose that

    f(z)zp10,    zU.

    Let define the function Φj by

    Φj(z)=[1(ααp)]f(z)zp1+αf(z)pzp2p1+z1z,

    then

    f(z)zp1p1+z1z, (3.14)

    and

    (f(z)zp1)>pζ4,     zU, (3.15)

    where ζ4 is given by:

    ζ4=1+ 2F1(1,1;p+αα;12),

    then (3.15) is the best possible.

    (ii) For p=α=1, (i) leads to:

    For fA suppose that

    f(z)0,    zU.

    Let define the function Φj by

    Φj(z)=f(z)+zf(z)1+z1z,

    then

    (f(z))>1+2ln2,     zU.

    So the estimate is best possible.

    Theorem 2. Let 0j<p, 0<r1 as for fA(p). Assume that Fα is defined by

    Fα(z)=α(ν+p)(p(ν+1,ρ;)f(z))+(1αα(ν))(p(ν,ρ;)f(z)).  (3.16)

    If

    F(j)α(z)zpj(1α+αp)p!(pj)!(1+Cz1+Dz )r, (3.17)

    then

    (p(ν,ρ;)f(z))(j)zpjp!(pj)!p(z), (3.18)

    where

    p(z)={(CD)ri0(r)ii!(CDC)i(1+Dz)i 2F1(i,1;1+(1α+αp)α;Dz1+Dz)(D0);2F1(r,(1α+αp)α;1+(1α+αp)α;Cz)                                     (D=0),

    and p!(pj)!p(z) is the best dominant of (3.18). Moreover, there will be

    ((p(ν,ρ;)f(z))(j)zpj)>p!(pj)!η,  zU, (3.19)

    where η is given by:

    η={(CD)ri0(r)ii!(CDC)i(1+D)i 2F1(i,1;1+(1α+αp)α;DD1)(D0);2F1(r,(1α+αp)α;1+(1α+αp)α;C)                                     (D=0),

    then (3.19) is the best possible.

    Proof. By using the definition (3.16) and the inequality (3.6), we have

    F(j)α(z)=αz(p(ν,ρ;)f(z))(j+1)+(1α+αj)(p(ν,ρ;)f(z))(j),  (3.20)

    for 0j<p. Putting

    ϕ(z)=(pj)!p!(p(ν,ρ;)f(z))(j)zpj,   (zU), (3.21)

    we have that ϕH. Differentiating (3.21), and using (3.17), (3.20), we get

    ϕ(z)+zϕ(z)(1α+αp)α(1+Cz1+Dz )r     (zU).

    Following the techniques of Theorem 1, we can obtain the remaining part of the proof.

    If we choose j=1 and r=1 in Theorem 2, we get:

    Corollary 4. For fA(p) let the function Fα define by 3.16. If

    Fα(z)zp1p(1α+αp)1+Cz1+Dz ,

    then

    ((p(ν,ρ;)f(z))zp1)>pη1,  zU, (3.22)

    where η1 is given by:

    η1={CD+(1CD)(1D)1 2F1(1,1;1+1α+αpα;DD1)(D0);11α+αp1+αpC                                                         (D=0),

    then (3.22) is the best possible.

    Example 2. If we choose p=C=α=1 and D=1 in Corollary 4, we obtain:

    For

    F(z)=(ν+1)((ν+1,ρ;)f(z))(ν)((ν,ρ;)f(z)).

    If

    F(z)1+z1z,

    then

    (((ν,ρ;)f(z)))>1+2ln2,  zU,

    the result is the best possible.

    Theorem 3. Let 0j<p, 0<r1 as for θ>p assume that Jp,θ:A(p)A(p) defined by

    Jp,θ(f)(z)=p+θzθz0tθ1f(t)dt,    zU. (3.23)

    If

    (p(ν,ρ;)f(z))(j)zpjp!(pj)!(1+Cz1+Dz )r, (3.24)

    then

    (p(ν,ρ;)Jp,θ(f)(z))(j)zpjp!(pj)!p(z), (3.25)

    where

    p(z)={(CD)ri0(r)ii!(CDC)i(1+Dz)i 2F1(i,1;1+θ+p;Dz1+Dz)(D0);2F1(r,θ+p;1+θ+p;Cz)                                     (D=0),

    and p!(pj)!p(z) is the best dominant of (3.25). Moreover, there will be

    ((p(ν,ρ;)Jp,θ(f)(z))(j)zpj)>p!(pj)!β,      zU, (3.26)

    where β is given by:

    β={(CD)ri0(r)ii!(CDC)i(1+D)i 2F1(i,1;1+θ+p;DD1)(D0);2F1(r,θ+p;1+θ+p;C)                                     (D=0),

    then (3.26) is the best possible.

    Proof. Suppose

    ϕ(z)=(pj)!p!(p(ν,ρ;)Jp,θ(f)(z))(j)zpj,   (zU),

    we have that ϕH. Differentiating the above definition, by using (3.24) and

    z(p(ν,ρ;)Jp,θ(f)(z))(j+1)=(θ+p)(p(ν,ρ;)f(z))(j)(θ+j)(p(ν,ρ;)Jp,θ(f)(z))(j)   (0j<p),

    we get

    ϕ(z)+zϕ(z)θ+p(1+Cz1+Dz )r.

    Now, we obtain (3.25) and the inequality (3.26) follow by using the same techniques in Theorem 1.

    If we set j=1 and r=1 in Theorem 3, we get:

    Corollary 5. For θ>p, let the operator Jp,θ:A(p)A(p) defined by (3.25). If

    (p(ν,ρ;)f(z))zp1p1+Cz1+Dz ,

    then

    ((p(ν,ρ;)Jp,θ(f)(z))zp1)>pβ1,     zU, (3.27)

    where β1 is given by:

    β1={CD+(1CD)(1D)1 2F1(1,1;1+θ+p;DD1)(D0);1θ+p1+θ+pC                                                   (D=0),

    then (3.27) is the best possible.

    Example 3. If we choose p=C=θ=1 and D=1 in Corollary 5, we get:

    If

    ((ν,ρ;)f(z))1+z1z,

    then

    (((ν,ρ;)J1,1(f)(z)))>1+4(1ln2),

    the result is the best possible.

    Theorem 4. Let q is univalent function in U, such that q satisfies

    Re(1+zq(z)q(z))>max{0;δ(ν+p)α},  zU. (3.28)

    Let 0j<p, 0<r1 and for fA(p) assume that

    (p(ν,ρ;)f(z))(j)zpj0,    zU,

    whenever δ(0,+)N. Let the function Φj defined by (3.1), and assume that it satisfies:

    [(pj)!p!]δΦj(z)q(z)+αδ(ν+p)zq(z). (3.29)

    Then,

    ((pj)!p!(p(ν,ρ;)f(z))(j)zpj)δq(z), (3.30)

    and q(z) is the best dominant of (3.30).

    Proof. Let ϕ(z) is defined by (3.5), from Theorem 1 we get

    [(pj)!p!]δΦj(z)=ϕ(z)+αδ(ν+p)zϕ(z). (3.31)

    Combining (3.29) and (3.31) we find that

    ϕ(z)+αδ(ν+p)zϕ(z)q(z)+αδ(ν+p)zq(z). (3.32)

    The proof of Theorem 4 follows by using Lemma 2 and (3.32).

    Taking q(z)=(1+Cz1+Dz)r in Theorem 4, we obtain:

    Corollary 6. Suppose that

    Re(1Dz1+Dz+(r1)(CD)z(1+Dz)(1+Cz))>max{0;δ(ν+p)α},  zU.

    Let 0j<p, 0<r1 and for fA(p) satisfies

    (p(ν,ρ;)f(z))(j)zpj0,    zU,

    whenever δ(0,+)N. Let the function Φj defined by (3.1), satisfies:

    [(pj)!p!]δΦj(z)(1+Cz1+Dz )r+αδ(ν+p)(1+Cz1+Dz )rr(CD)z(1+Dz)(1+Cz).

    Then,

    ((pj)!p!(p(ν,ρ;)f(z))(j)zpj)δ(1+Cz1+Dz )r, (3.33)

    so (1+Cz1+Dz)r is the best dominant of (3.33).

    Taking q(z)=1+Cz1+Dz in Theorem 4, we get:

    Corollary 7. Suppose that

    Re(1Dz1+Dz)>max{0;δ(ν+p)α},  zU.

    Let 0j<p, 0<r1 and for fA(p) satisfies

    (p(ν,ρ;)f(z))(j)zpj0,    zU,

    whenever δ(0,+)N. Let the function Φj defined by (3.1), satisfies:

    [(pj)!p!]δΦj(z)1+Cz1+Dz +αδ(ν+p)(CD)z(1+Dz)2.

    Then,

    ((pj)!p!(p(ν,ρ;)f(z))(j)zpj)δ1+Cz1+Dz , (3.34)

    so 1+Cz1+Dz is the best dominant of (3.34).

    If we put ν=ρ=0 and  =1 in Theorem 4, we get:

    Corollary 8. Let q is univalent function in U, such that q satisfies

    Re(1+zq(z)q(z))>max{0;δpα},  zU.

    For fA(p) satisfies

    f(j)(z)zpj0,    zU.

    Let the function Φj defined by (3.9), satisfies:

    [(pj)!p!]δΦj(z)q(z)+αδpzq(z). (3.35)

    Then,

    ((pj)!p!f(j)(z)zpj)δq(z), (3.36)

    so q(z) is the best dominant of (3.36).

    Taking C=1 and D=1 in Corollaries 6 and 7 we get:

    Example 4. (i) For fA(p) assume that

    (p(ν,ρ;)f(z))(j)zpj0,    zU.

    Let the function Φj defined by (3.1), and assume that it satisfies:

    [(pj)!p!]δΦj(z)(1+z1z)r+αδ(ν+p)(1+z1z)r2rz1z2.

    Then,

    ((pj)!p!(p(ν,ρ;)f(z))(j)zpj)δ(1+z1z)r, (3.37)

    so (1+z1z)r is the best dominant of (3.37).

    (ii) For fA(p) say it

    (p(ν,ρ;)f(z))(j)zpj0,    zU.

    Let the function Φj defined by (3.1), and assume that it satisfies:

    [(pj)!p!]δΦj(z)1+z1z+αδ(ν+p)2z1z2.

    Then,

    ((pj)!p!(p(ν,ρ;)f(z))(j)zpj)δ1+z1z, (3.38)

    so 1+z1z is the best dominant of (3.38).

    If we put p=C=α=δ=1, D=1 and j=0 in Corollary 8 we get:

    Example 5. For fA suppose that

    f(z)z0,    zU,

    and

    f(z)(1+z1z)r+(1+z1z)r2rz1z2.

    Then,

    f(z)z(1+z1z)r, (3.39)

    and (1+z1z)r is the best dominant of (3.39).

    Remark 1. For  ν=ρ=0, =p=r=1 and j=0 in Theorem 4, we get the results investigated by Shanmugam et al. ([25], Theorem 3.1).

    Theorem 5. Let 0j<p, and for fA(p) assume that

    (p(ν,ρ;)f(z))(j)zpj0,    zU,

    whenever δ(0,+)N. Suppose that

    ((pj)!p!(p(ν,ρ;)f(z))(j)zpj)δHQ

    such that [(pj)!p!]δΦj(z) is univalent in U, where the function Φj is defined by (3.1). If q is convex (univalent) function in U, and

    q(z)+αδ(ν+p)zq(z)[(pj)!p!]δΦj(z),

    then

    q(z)((pj)!p!(p(ν,ρ;)f(z))(j)zpj)δ, (3.40)

    so q(z) is the best subordinate of (3.40).

    Proof. Let ϕ is defined by (3.5), from (3.31) we get

    q(z)+αδ(ν+p)zq(z)[(pj)!p!]δΦj(z)=ϕ(z)+αδ(ν+p)zϕ(z).

    The proof of Theorem 5 followes by an application of Lemma 3.

    Taking q(z)=(1+Cz1+Dz)r in Theorem 5, we get:

    Corollary 9. Let 0j<p, 0<r1 and for fA(p) assume that

    (p(ν,ρ;)f(z))(j)zpj0,    zU.

    Suppose that

    ((pj)!p!(p(ν,ρ;)f(z))(j)zpj)δHQ

    such that [(pj)!p!]δΦj(z) is univalent in U, where the function Φj is defined by (3.1). If

    (1+Cz1+Dz )r+αδ(ν+p)(1+Cz1+Dz )rr(CD)z(1+Dz)(1+Cz)[(pj)!p!]δΦj(z),

    then

    (1+Cz1+Dz )r((pj)!p!(p(ν,ρ;)f(z))(j)zpj)δ, (3.41)

    so (1+Cz1+Dz)r is the best dominant of (3.41).

    Taking q(z)=1+Cz1+Dz and r=1 in Theorem 5, we get:

    Corollary 10. Let 0j<p, and for fA(p) assume that

    (p(ν,ρ;)f(z))(j)zpj0,    zU,

    whenever δ(0,+)N. Assume that

    ((pj)!p!(p(ν,ρ;)f(z))(j)zpj)δHQ

    such that [(pj)!p!]δΦj(z) is univalent in U, where the function Φj is defined by (3.1). If

    1+Cz1+Dz +αδ(ν+p)(CD)z(1+Dz)2[(pj)!p!]δΦj(z),

    then

    1+Cz1+Dz ((pj)!p!(p(ν,ρ;)f(z))(j)zpj)δ, (3.42)

    so 1+Cz1+Dz is the best dominant of (3.42).

    Combining results of Theorems 4 and 5, we have

    Theorem 6. Let 0j<p, and for fA(p) assume that

    (p(ν,ρ;)f(z))(j)zpj0,    zU.

    Suppose that

    ((pj)!p!(p(ν,ρ;)f(z))(j)zpj)δH[q(0),1]Q

    such that [(pj)!p!]δΦj(z) is univalent in U, where the function Φj is defined by (3.1). Let q1 is convex (univalent) function in U, and assume that q2 is convex in U, that q2 satisfies (3.28). If

    q1(z)+αδ(ν+p)zq1(z)[(pj)!p!]δΦj(z)q2(z)+αδ(ν+p)zq2(z),

    then

    q1(z)((pj)!p!(p(ν,ρ;)f(z))(j)zpj)δq2(z)

    and q1(z) and q2(z) are respectively the best subordinate and best dominant of the above subordination.

    We used the application of higher order derivatives to obtained a number of interesting results concerning differential subordination and superordination relations for the operator p(ν,ρ;)f(z) of multivalent functions analytic in U, the differential subordination outcomes are followed by some special cases and counters examples. Differential sandwich-type results have been obtained. Our results we obtained are new and could help the mathematicians in the field of Geometric Function Theory to solve other special results in this field.

    This research has been funded by Deputy for Research & innovation, Ministry of Education through initiative of institutional funding at university of Ha'il, Saudi Arabia through project number IFP-22155.

    The authors declare no conflict of interest.

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