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Research article

Geotechnical characterization of Halsen-Stjørdal silt, Norway

  • Received: 13 August 2020 Accepted: 16 September 2020 Published: 21 September 2020
  • The evaluation of geotechnical parameters for design problems in silty soils is complicated due to partially drained conditions and irregular soil structure, including small layers and pockets of both coarser and finer material. Many established methods to define soil parameters in clay and sand exist but little guidance is given to practicing engineers on how to interpret soil parameters for silty materials. This paper presents the results of an extensive laboratory and field test program which was carried out at a silt testing site Halsen-Stjø rdal in Norway. The main objective is to broaden the database of the engineering behaviour of silts and to gain a better understanding of the behaviour of these soils. Cone penetration tests (CPTU) were performed and shear wave velocity measurements close to the site were used to supplement the CPTU results, confirming the coarse, silty nature of the deposit. In addition, several samples were taken using thin walled 54 mm steel sample tubes and examined in the laboratory by means of index, oedometer and triaxial tests. Recently developed methods to determine sample quality in intermediate low plastic soils were adopted and showed promising results. The interpretation of the oedometer tests was challenging due to the shapes of the curves. The results did not identify the yield or preconsolidation stress clearly partly due to the nature of the silt and partly due to sample disturbance. Triaxial test results on the silt showed a strong dilative behaviour developing negative pore pressures with increasing axial strain. The shape of the stress paths revealed no unique undrained shear strength of the silt. Although many researchers doubt the use of undrained shear strength (su) for partially drained materials, this parameter is still frequently used. Several methods were applied to determine values of an apparent su in the silt in order to provide an overview over the range of strength values. The results from this study contribute to the existing database and increase the understanding of silty soils.

    Citation: Annika Bihs, Mike Long, Steinar Nordal. Geotechnical characterization of Halsen-Stjørdal silt, Norway[J]. AIMS Geosciences, 2020, 6(3): 355-377. doi: 10.3934/geosci.2020020

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  • The evaluation of geotechnical parameters for design problems in silty soils is complicated due to partially drained conditions and irregular soil structure, including small layers and pockets of both coarser and finer material. Many established methods to define soil parameters in clay and sand exist but little guidance is given to practicing engineers on how to interpret soil parameters for silty materials. This paper presents the results of an extensive laboratory and field test program which was carried out at a silt testing site Halsen-Stjø rdal in Norway. The main objective is to broaden the database of the engineering behaviour of silts and to gain a better understanding of the behaviour of these soils. Cone penetration tests (CPTU) were performed and shear wave velocity measurements close to the site were used to supplement the CPTU results, confirming the coarse, silty nature of the deposit. In addition, several samples were taken using thin walled 54 mm steel sample tubes and examined in the laboratory by means of index, oedometer and triaxial tests. Recently developed methods to determine sample quality in intermediate low plastic soils were adopted and showed promising results. The interpretation of the oedometer tests was challenging due to the shapes of the curves. The results did not identify the yield or preconsolidation stress clearly partly due to the nature of the silt and partly due to sample disturbance. Triaxial test results on the silt showed a strong dilative behaviour developing negative pore pressures with increasing axial strain. The shape of the stress paths revealed no unique undrained shear strength of the silt. Although many researchers doubt the use of undrained shear strength (su) for partially drained materials, this parameter is still frequently used. Several methods were applied to determine values of an apparent su in the silt in order to provide an overview over the range of strength values. The results from this study contribute to the existing database and increase the understanding of silty soils.


    Let S denote the class of univalent functions which are analytic in the open unit disk D={zC:|z|<1} of the form

    f(z)=z+n=2anzn(zD). (1.1)

    Let P represent a class of analytic functions within the unit disk D of the form

    p(z)=1+n=1cnzn(zD) (1.2)

    and satisfy the condition of (p(z))>0. It is easy to know from the conclusion of [1], for p(z)P, there exists a Schwarz function w(z), making

    p(z)Pp(z)=1+w(z)1w(z).

    In 1976, Noonan and Thomas [2] defined the qth Hankel determinant for a function fS of form (1.1) as

    Hq,n(f)=|anan+1an+q1an+1an+2an+qan+q1an+qan+2q2|,

    where a1=1,n1,q1. In particular, we have

    H2,1(f)=a3a22,
    H2,2(f)=a2a4a23,
    H3,1(f)=a3(a2a4a23)a4(a4a2a3)+a5(a3a22)

    and

    H4,1(f)=a7H3,1(f)a6δ1+a5δ2a4δ3,

    where

    δ1=a3(a2a5a3a4)a4(a5a2a4)+a6(a3a22),
    δ2=a3(a3a5a24)a5(a5a2a4)+a6(a4a2a3),
    δ3=a4(a3a5a24)a5(a2a5a3a4)+a6(a2a4a23).

    Next, we recall the definition of subordination. We assume that f1 and f2 are two analytic functions in D. Then, we say that the function f1 is subordinate to the function f2, as we write f1(z)f2(z), for all zD. Then, there exists a Schwarz function w(z) with w(0)=0 and |w(z)|<1 to satisfy

    f1(z)=f2(w(z)).

    Now, we consider the following class S(g) as follows:

    S(g)={fS:zf(z)f(z)g(z)}, (1.3)

    where g is an analytic univalent function with positive real part in D, and g maps D onto a region starlike with respect to g(0)=1, g(0)>0, and is symmetric about the real axis. The class S(g) was introduced by Ma and Minda [3]. If we vary the function g on the right side of (1.3), we will obtain different results. In recent years, many researchers have also conducted a lot of research on this and obtained a series of conclusions. Some of them are as follows:

    (1) For g=21+ez, which was defined in [4].

    (2) For g=1+z, it has been further studied in [5].

    (3) For g=1+43z+23z2, it was introduced in [6] and further investigated in [7].

    (4) For g=ez, it was defined and studied in [8].

    (5) For g=z+1+z2, the class is denoted by Sl, and it was further studied in [9].

    (6) For g=1+sinh1z, the class Sp=S(1+sinh1z) was studied by Kumar and Arora [10].

    (7) For g=coshz, the class Scosh=S(g(z)) was introduced by Alotaibi et al. [11].

    The Fekete-Szegö inequality is one of the inequalities for the coefficients of univalent analytic functions found by Fekete and Szegö. The Fekete-Szegö inequality of various analytic functions has been studied by many researchers in the last few decades, for example, Huo Tang defined certain class of analytic functions related to the sine function (see [12])

    f(ζ)θ(ζf(ζ)f(ζ))1θ1+sin(ζ);(fS,0θ1)

    and investigated the upper bound of the second Hankel determinant and the Fekete-Szegö inequality for functions in this class. Many papers have been devoted to researching the Fekete-Szegö inequality for various sub-class functions (see [13,14]). Therefore, the study of the Fekete-Szegö inequality for different analytic functions is valuable and of great significance.

    In recent years, many papers have been devoted to finding the upper bounds of Hankel determinants for various sub-classes of analytic functions as well. For the basics and preliminaries, the readers are advised to see the academic achievements in [15,16,17,18]. Guangadharan studied a class of bounded turning functions related to the three leaf function in [19]. From this, it can be seen that the research on Hankel determinants of various analytic functions has become popular. Therefore, it is an interesting and hot topic to investigate the Hankel determinants for various classes of analytic functions. In addition, it is worth mentioning that a class of star like functions associated with the modified sigmoid function was defined by Goel and Kumar [20],

    SSG={fS:zf(z)f(z)21+ez}.

    Apart from the above, the coefficient bounds for certain analytic functions have been studied by many researchers, see [21,22,23,24,25]. Further, many star like functions have been defined and studied as well, see [26,27,28,29]. Not long ago, another class of analytic functions associated with the modified sigmoid function was defined and studied by Muhammad Ghaffar Khan [4],

    RSG={fS:f(z)21+ez}.

    It is well known that for each univalent function fS, there is an inverse function f1(w) which can be defined in (|w|<r;r14), where

    f1(w)=wa2w2+(2a22a3)w3(5a325a2a3+a4)w4+.

    A function fS is said to be bi-univalent in D if there exists a function gS such that g(z) is a univalent extension of f1 to D. Brannan [30] studied classes of bi-univalent functions and obtained estimates for their initial coefficients. Many classes of bi-univalent funtions were introduced and further studied in the past few years. Inspired by all the aforementioned works, in this paper we investigate another certain class of analytic functions H(λ,ψ), which are related to the modified sigmoid function, and discuss the upper bound of the fourth-order Hankel determinant in special cases, here we use another method to obtain improved results compared to [20]. And we also obtain the upper bound of third-order Hankel determinant of its inverse function. Furthermore, we discuss the Fekete-Szegő inequality for functions in this class when λ[0,1] and ψ=21+ez. Finally, we estimate the upper bounds of the initial coefficients for functions in this class when λ[0,1], ψ(0)=1, and ψ(0)>0, where its inverse function f1 also belongs to this class.

    Definition 1.1. Assume that fS, 0λ1, (f(z))1λ and (2zf(z)f(z)f(z))λ are analytic in D with f(z)0, and f(z)f(z) for all zD{0}. Furthermore, (f(z))λ = 1 at z=0, ψ(z) is a univalent and analytic function. Then, f(z) is said to be in the class H(λ,ψ) if the following condition is satisfied:

    (1λ)(f(z))1λ+λ(2zf(z)f(z)f(z))λψ(z).

    For convenience, we denote

    H(λ)=H(λ,21+ez).

    Remark 1.1. For any λ[0,1], we have that f(z)=zH(λ) always holds.

    Below we will evaluate bounds of the first six initial coefficients and non-sharp bound of the third Hankel determinant for functions belonging to H(1).

    Theorem 2.1. Let fH(1) and be of form (1.1). Then,

    |a2|14, (2.1)
    |a3|14, (2.2)
    |a4|18, (2.3)
    |a5|18, (2.4)
    |a6|731576, (2.5)
    |a7|388937241920. (2.6)

    The first four inequalities are sharp.

    We need the following lemmas to prove the above theorem:

    Lemma 2.1. [4] Let pP, then |cn|2.

    Lemma 2.2. [17] Let pP, then for all n,mN, if 0ζ1, there is |cm+nζcmc1|2. If ζ<0 or ζ>1, there is |cm+nζcmc1|2|2ζ1|.

    Lemma 2.3. [4] Let pP, then

    |αc31βc1c2+γc3|2|α|+2|β2α|+2|αβ+γ|,

    where α, β and γ are real numbers.

    Lemma 2.4. [4] Let α, β, γ, and ζ satisfy the inequalities 0<γ<1,0<β<1, and

    8β(1β)[(γζ2α)2+(γ(β+γ)ζ)2]+γ(1γ)(ζ2βγ)24γ2(1γ)2β(1β).

    If pP, then

    |αc41+βc22+2γc1c332ζc21c2c4|2.

    Proof. If fH(1), there exists a Schwarz function w(z) to satisfy

    2zf(z)f(z)f(z)=21+ew(z).

    Also, if pP, it can be written in terms of the Schwarz function w(z) as

    p(z)=1+c1z+c2z2+c3z3+=1+w(z)1w(z),

    or equivalently,

    w(z)=p(z)1p(z)+1=12c1z+(12c214c21)z2+(18c3112c1c2+12c3)z3+. (2.7)

    Now, we set

    2zf(z)f(z)f(z)=1+b1z+b2z2+b3z3+b4z4+b5z5+b6z6+=21+ew(z). (2.8)

    In addition,

    2zf(z)f(z)f(z)=1+2a2z+3a3z2+5a5z4+6a6z5+7a7z6+1+a3z2+a5z4+a7z6+. (2.9)

    Using (2.8) and (2.9), we can get

    b1=2a2, (2.10)
    b2=2a3, (2.11)
    b3=4a42a2a3, (2.12)
    b4=4a52a23, (2.13)
    b5=6a64a3a4+2a2a232a2a5, (2.14)
    b6=6a74a3a5+2a232a3a5. (2.15)

    Substituting (2.7) into the right side of (2.8), by simplifying, using (2.10)–(2.15), and comparing the coefficients on both sides of the equation, we can get

    a2=18c1, (2.16)
    a3=12(c24c218), (2.17)
    a4=14(124c31732c1c2+c34), (2.18)
    a5=116(116c41916c21c2+c1c3+38c22c4), (2.19)
    a6=16b5+23a3a413a2a23+13a2a5, (2.20)
    a7=16b6+a3(3a5a23)3+a3a53, (2.21)

    where

    b5=13072(5c61+122c41c2288c31c3432c21c22+528c21c4+1056c1c2c3768c1c5+176c32768c2c4384c23+768c6),
    b6=15160960(2537c7150400c51c2+204960c41c3+409920c31c22483840c4c311451520c21c2c3+887040c21c5483840c1c32+1774080c1c2c4+887040c1c231290240c1c6+887040c22c31290240c2c51290240c3c4+1290240c7).

    Applying Lemma 2.1, we have

    |a2|14.

    The above inequality is sharp with extremal function f(z)=z021+etdt.

    |a3|=18|c212c21|14.

    The above inequality is sharp for the function p(z)=(1+z2)/(1z2).

    Applying Lemma 2.3, we have

    |a4|14[2|124|+2|732112|+2|124732+14|]=18.

    The above inequality is sharp with extremal function f(z)=z021+et3dt.

    Applying Lemma 2.4, we have

    |a5|=|116(116c41916c21c3+c1c3+38c22c4)|18.

    The above inequality is sharp for the function p(z)=(1+z4)/(1z4).

    Applying the triangle inequality, we have

    |16b5|118432(122|c1|4|c25122c21|+1056|c1||c3||c2311c21|+528|c1|2|c4911c22|+768|c6c1c5|+768|c2||c48889c22|+384|c3|2).

    By applying Lemmas 2.1 and 2.2, we have

    |16b5|355288,

    and then applying the triangle inequality and (2.1)–(2.4), we have

    |23a3a413a2a23+13a2a5|23|a3||a4|+13|a2||a3|2+13|a2||a5|7192,

    and from (2.20) we can obtain

    |a6|355288+7192=731576.

    By applying triangle inequality, we have

    |16b6|130965760(20965760|c1|4|c3105427c1c2|+483840|c1|3|c46172c21|+1290240|c1||c61116c1c5|+1774080|c1||c2||c4911c1c3|+887040|c2|2|c3611c1c2|+1290240|c7c2c5|+1290240|c3||c41116c1c3|+2537|c1|7).

    Then, from (2.2) and Lemmas 2.1, 2.2, and 2.4, we have

    |16b6|381377241920,
    |a3(2a5a23)3|=124|a3||332c41+12c22+c1c31116c21c2c4|148.

    From (2.2) and (2.4), we have

    |a3a53|196.

    Then, applying the triangle inequality and (2.21), we can get

    |a7|381377241920+148+196=388937241920.

    This completes our proof.

    Theorem 2.2. If f of the form (1.1) belongs to H(1), then

    |a3a22|14.

    The result is sharp for the function p(z)=(1+z2)/(1z2).

    Proof. Using (2.16), (2.17), and Lemma 2.2, we have

    |a3a22|=18|c258c21|14.

    Theorem 2.3. If f of the form (1.1) belongs to H(1), then

    |a2a3a4|18.

    The result is sharp with the extremal function f(z)=z021+et3dt.

    Proof. Using (2.16)–(2.18), we can get

    |a2a3a4|=|7384c319128c1c2+116c3|.

    Applying Lemma 2.3,

    |7384c319128c1c2+116c3|2|7384|+2|91287192|+2|73849128+116|=18.

    Theorem 2.4. If f of the form (1.1) belongs to H(1), then

    |a2a4a23|116.

    The result is sharp with the extremal function f(z)=z021+et3dt.

    Proof. Using (2.16)–(2.18), we can get

    |a2a4a23|=164|16c41+916c21c2+12c1c3c22|.

    Now, in order to get the desired bound, we shall prove that

    |16c41+916c21c2+12c1c3c22|4. (2.22)

    Next we will use the following Lemma:

    Lemma 2.5. [17] Let pP. Then, there exists some x, y with |x|1,|y|1 such that

    2c2=c21+x(4c21),
    4c3=c31+2c1x(4c21)(4c21)c1x2+2(4c21)(1|x|2)y.

    Using the invariant property under rotation, we can assume that c=c1[0,2], and then from Lemma 2.5, substituting the expression for c2, c3 and simplifying, we can obtain

    16c41+916c21c2+12c1c3c22=196c4+132c2(4c2)x14(4c2)(412c2)x2+14c(4c2)(1|x|2)y.

    If c=0, there is

    |16c41+916c21c2+12c1c3c22|=4|x|24.

    If c=2, there is

    |16c41+916c21c2+12c1c3c22|=16.

    Next, we will discuss the case of c(0,2). At this time,

    16c41+916c21c2+12c1c3c22=14c(4c2)[px2+qx+t+(1|x|2)y],

    where

    p=c282c,q=c8,t=c324(4c2),

    and then we denote

    I=14c(4c2)[px2+qx+t+(1|x|2)y],

    where p<0,q>0, and t<0 always holds due to the fact that c(0,2). Then, by using the triangle inequality, we have

    |I|14c(4c2)(1|x|2+|p||x|2+|q||x|+|t|)=14c(4c2)[(p+1)|x|2+q|x|t+1].

    Since q2(p+1)<0 always holds, we can obtain

    |I|14c(4c2)(p+qt)=548c4118c2+4=f(c).

    By computation, it can be revealed that

    f(c)<max{f(0),f(2)}=4.

    In summary, |I|4, that is, (2.22) holds, which evidently yields

    |16c41+916c21c2+12c1c3c22|4.

    This completes the proof.

    Theorem 2.5. If f of the form (1.1) belongs to H(1), then

    |H3,1(f)|116.

    Proof.

    H3,1(f)=a3(a2a4a23)a4(a4a2a3)+a5(a3a22).

    By using the triangle inequality, we have

    |H3,1(f)||a3||a2a4a23|+|a4||a4a2a3|+|a5||a3a22|.

    According to Theorem 2.1, we have

    |a3|14,|a4|18,|a5|18.

    According to Theorems 2.2–2.4, we have

    |a2a4a23|116,|a4a2a3|18,|a3a22|14.

    Therefore,

    |H3,1(f)|116.

    Below we will evaluate the non-sharp bound of the fourth determinant for functions belonging to H(1).

    Theorem 3.1. If f of the form (1.1) belongs to H(1), then

    |a2a5a3a4|116.

    Proof. Using (2.16)–(2.19), we can get

    |a2a5a3a4|=|16144c5113072c31c21256c21c3+1256c1c22+1128c1c41128c2c3|.

    Then, by applying the triangle inequality, we have

    |a2a5a3a4|13072|c31(c2c212)|+1128|c2(c3c1c22)|+1128|c1(c4c1c32)|.

    We denote |c1|=c, and from Lemmas 2.1 and 2.2 we can obtain

    |c31(c2c212)|c3(212c2),1128|c2(c3c1c22)|132,1128|c1(c4c1c32)|c64.

    Thus,

    |a2a5a3a4|13072c3(2c22)+c64+132=G(c),
    G(c)=5c46144+c2512+1640c[0,2].

    Therefore,

    G(c)G(2)=116.

    This completes the proof.

    Theorem 3.2. If f of the form (1.1) belongs to H(1), then

    |a5a2a4|18.

    Proof. By using (2.16), (2.18), and (2.19), we have

    |a5a2a4|=116|112c414364c21c2+98c1c3+38c22c4|.

    By applying Lemma 2.4, we can get the sharp result for the function p(z)=(1+z4)/(1z4).

    Theorem 3.3. If f of the form (1.1) belongs to H(1), then

    |a3a5a24|689+14439216.

    Proof. Using (2.17)–(2.19), we can obtain

    |a3a5a24|=|536864c611912288c41c2+1384c31c3+4716384c21c221256c21c411024c1c2c331024c32+1128c2c41256c23|.

    By applying the triangle inequality, we get

    |a3a5a24|536864|c1|6+149152|c2||76c41141c21c2+48c1c3+144c22|+1384|c3||c3132c3|+1128|c4||c212c21|.

    In order to get the desired bound, we shall prove that

    |76c41141c21c2+48c1c3+144c22|856.

    Using the invariant property under rotation, we can assume that c=c1[0,2], and then from Lemma 2.5, substituting the expression for c2, c3 and simplifying, we have

    76c41141c21c2+48c1c3+144c22=1072c4+512c2(4c2)x+48(4c2)(3c2)x2+24c(4c2)(1|x|2)y.

    If c=0,

    |76c41141c21c2+48c1c3+144c22|=576. (3.1)

    If c=2,

    |76c41141c21c2+48c1c3+144c22|=856. (3.2)

    If c0 and c2, we have

    76c41141c21c2+48c1c3+144c22=24c(4c2)[p+qx+tx2+(1|x|2)y],

    where

    p=107c348(4c2),q=17c16,t=62c2c.

    p,q>0 always holds due to the fact that c(0,2). Then, by using the triangle inequality, we have

    |76c41141c21c2+48c1c3+144c22|24c(4c2)(1|x|2+p+q|x|+|t||x|2).

    We denote

    I=24c(4c2)(1|x|2+p+q|x|+|t||x|2).

    For suitability, we divide the calculation in five cases:

    Case (I). t0 if and only if 3=c1c<2. At this time,

    I=24c(4c2)[(1+t)|x|2+q|x|+p+1],

    and when c1c<c2, there is q2(1+t)<1, where c2=16+824781. Then, we have

    I24c(4c2)4(1+t)pq24(1+t)+24c(4c2)=107c42+867128c42+c3+2c+24c(4c2)=f1(c).

    By computation, it can be revealed that

    f1(c)<856,c[c1,c2). (3.3)

    Case (II). For c[c2,2), there is q2(1+t)1, and we then have

    I24c(4c2)(p+qt)=20c4+438c2576=f2(c).

    By computation, we have

    f2(c)<856,c[c2,2). (3.4)

    Case (III). For c1>c>c3=16+814547, we have

    I=24c(4c2)[(t1)|x|2+q|x|+p+1],

    where q2(1t)<1, and we then obtain

    I24c(4c2)4(t1)pq24(t1)+24c(4c2)=107c42867128c42c32c+24c(4c2)=f3(c).

    Now, computation reveals that

    f3(c)<856,c(c3,c1). (3.5)

    Case (IV). For c3c>c4, where c4=32, there is q2(1t)>1. We can get

    I24c(4c2)(t+p+q)=76c4234c2+576=f4(c).

    A computation shows that

    f4(c)<856,c(c4,c3]. (3.6)

    Case (V). For the case of c(0,c4], we have

    I=24c(4c2)[(t1)|x|2+q|x|+p+1],

    where t1>0 holds for c(0,32). Thus, we have

    Imax{24c(4c2)(p+1),24c(4c2)(p+q+t)},

    or, equivalently,

    Imax{107c42+24c(4c2),76c4234c2+576}.

    Now, we denote

    g1(c)=107c42+24c(4c2),g2(c)=76c4234c2+576
    g1(c)=314c372c2+96,
    g1(c)=144c(471c721).

    g1(c) attains its minimum at c0=72471, g1(c0)>0, which evidently yields that g1(c)>0 holds for c(0,32). Therefore, g1(c)<g1(32)=333.84375. On the other hand,

    g2(c)<max{g2(0),g2(32)}=576.

    Thus,

    I<576,c(0,c4]. (3.7)

    From (3.11)–(3.17), we conclude that I856, which implies

    |76c41141c21c2+48c1c3+144c22|856. (3.8)

    Next, we will use the following lemma:

    Lemma 3.1. Let p(z)=1+c1z+c2z2+c3z3+P. Then, for any real number μ,

    |μc3c31|{2|μ4|(μ43),2μμμ1(43<μ).

    By Lemmas 2.1, 2.2, and 3.1, we can obtain

    |536864c61|5576,149152|c2||76c41141c21c2+48c1c3+144c22|1073072. (3.9)
    1384|c3||c3132c3|364,1128|c4||c212c21|132. (3.10)

    From (3.9) and (3.10), we have

    |a3a5a24|689+14439216.

    This completes our proof.

    Theorem 3.4. If f of the form (1.1) belongs to H(1), then

    |H4,1(f)|215139562371589120+334096.

    Proof. We can write H4,1(f) as

    H4,1(f)=a7H3,1(f)a6δ1+a5δ2a4δ3,

    where

    δ1=a3(a2a5a3a4)a4(a5a2a4)+a6(a3a22),
    δ2=a3(a3a5a24)a5(a5a2a4)+a6(a4a2a3),
    δ3=a4(a3a5a24)a5(a2a5a3a4)+a6(a2a4a23).

    By applying Theorems 2.1–2.5, 3.1–3.3, and the triangle inequality, we have

    |H4,1(f)||a7||H3,1(f)|+|a6||δ1|+|a5||δ2|+|a4||δ3|, (3.11)
    |a7||H3,1|388937241920×116=3889373870720, (3.12)
    |δ1||a3||a2a5a3a4|+|a4||a5a2a4|+|a6||a3a22|8032304, (3.13)
    |δ2||a3||a3a5a24|+|a5||a5a2a4|+|a6||a4a2a3|237112288+3256, (3.14)
    |δ3||a4||a3a5a24|+|a5||a2a5a3a4|+|a6||a2a4a23|2371+48324576. (3.15)

    Thus, from (3.11)–(3.15), we obtain

    |H4,1(f)|215139562371589120+334096.

    Theorem 4.1. If the function fH(1) given by (1.1) and f1(w)=w+n=2dnwn is the analytic continuation to D of the inverse function of f with |w|<r0, where r014 is the radius of the Koebe domain, then

    |d2|14, (4.1)
    |d3|14, (4.2)
    |d4|65384, (4.3)
    |d5|167256. (4.4)

    The first three inequalities are sharp.

    Proof. If

    f1(w)=w+n=2dnwn

    is the inverse function of f, it can be seen that

    f1(f(z))=f(f1(z))=z.

    Equivalently,

    n=1dn(z+k=2dkwk)n=z(d1=1). (4.5)

    By comparing the coefficients on both sides of (4.5), we can obtain

    d2=a2, (4.6)
    d3=2a22a3, (4.7)
    d4=(5a325a2a3+a4), (4.8)
    d5=14a4221a22a3+6a2a4+3a23a5. (4.9)

    Applying Lemmas 2.1 and 2.2, (2.16), and (2.17), we have

    |d2|=|a2|=|18c1|14,
    |d3|=18|c234c21|14.

    Applying Lemma 2.3 and (2.16)–(2.18), we have

    |d4|=|911536c3117128c1c2+116c3|2|911536|+2|1712891768|+2|91153617128+116|=65384,
    |d5|=|972048c4121128c21c2+764c1c3+9128c22116c4|.

    By applying the triangle inequality, we can get

    |d5||972048c4121128c21c2+764c1c3|+|9128c22116c4|.

    Using Lemmas 2.1 and 2.3, we have

    |972048c4121128c21c2+764c1c3|=|c1||972048c3121128c1c2+764c3|127256.

    Using Lemma 2.2, we obtain

    |9128c22116c4|=116|c498c22|532.

    Therefore,

    |d5|127256+532=167256.

    This completes the proof.

    Theorem 4.2. If the function fH(1) given by (1.1) and f1(w)=w+n=2dnwn is the analytic continuation to D of the inverse function of f with |w|<r0, where r014 is the radius of the Koebe domain, then

    |H3,1(f1)|34171147456.

    Proof. From Theorem 3.2, we have

    |d3d22|=|a22a3|14. (4.10)

    Applying (4.6)–(4.8) and (2.16)–(2.18),

    |d2d3d4|=|3a324a2a3+a4|=|731536c3115128c1c2+116c3|.

    Using Lemma 2.3,

    |d2d3d4|2|731536|+2|1512873768|+2|73153615128+116|=59384. (4.11)

    Applying (4.6)–(4.8) and (2.16)–(2.18),

    |d2d4d23|=|a42a22a3+a2a4a23|=164|17192c41+c22716c21c212c1c3|.

    We denote |c1|=c[0,2],|x|=t[0,1], and referring to Lemma 2.5, we have

    |c2|c2+t(4c2)2,|c3|c34+c(4c2)t2+(4c2)ct24+(4c2)(1t2)2.

    Using the triangle inequality, we have

    |17192c41+c22716c21c212c1c3|131192c4+(4c2)c4+(c+2)(c+4)(c2)28t2+31c2(4c2)32t=F(c,t).
    Ft=(c+2)(c+4)(c2)24t+31c2(4c2)32>0.

    Therefore,

    F(c,t)F(c,1)=31192c4+198c2+4=G(c),
    G(c)=194c(131228c2)0.

    This leads to

    G(c)G(2)=13112,
    |d2d4d23|13112164=131768. (4.12)

    Applying (4.2)–(4.4), (4.10)–(4.12), and the triangle inequality, we have

    |H3,1(f)|=|d3(d2d4d23)d4(d4d2d3)+d5(d3d22)||d3||d2d4d23|+|d4||d4d2d3|+|d5||d3d22|34171147456,

    which completes the proof.

    Theorem 5.1. If fH(λ) and is of the form (1.1), then

    |a2|18λ28λ+4,
    |a3|110λ212λ+6,
    |a3νa22|{ν[3(1λ)2+2λ2]2(1λ)λ16[3(1λ)2+2λ2][(1λ)2+λ2]2,νt1,12[3(1λ)2+2λ2],t2<ν<t1,2(1λ)λν[3(1λ)2+2λ2]16[3(1λ)2+2λ2][(1λ)2+λ2]2,νt2, (5.1)

    where

    t1=8[(1λ)2+λ2]2+2λ(1λ)3(1λ)2+2λ2,t2=2λ(1λ)8[(1λ)2+λ2]23(1λ)2+2λ2.

    The result is sharp for the function p(z)=(1+z2)/(1z2).

    Proof.

    f(z)=1+2a2z+3a3z2+4a4z3+5a5z4+,
    (1λ)(f(z))1λ=(1λ)+2(1λ)2a2z+(3(1λ)2a32(1λ)2λa22)z2+, (5.2)
    2zf(z)f(z)f(z)=1+2a2z+2a3z2+(4a42a2a3)z3+,
    λ(2zf(z)f(z)f(z))λ=λ+2λ2a2z+(2λ2a32λ2(1λ)a22)z2+. (5.3)

    In addition,

    (1λ)(f(z))1λ+λ(2zf(z)f(z)f(z))λ=21+ew(z)=1+c14z+(c24c218)z2+, (5.4)

    where

    w(z)=p(z)1p(z)+1=12c1z+(12c214c21)z2+(18c3112c2c1+12c3)z3+.
    p(z)=1+c1z+c2z2+c3z3+P.

    Substituting (5.2) and (5.3) into (5.4) and comparing the coefficients on both sides of (5.4), we can obtain

    [2(1λ)2+2λ2]a2=c14, (5.5)
    [3(1λ)2+2λ2]a3[2(1λ)2λ+2λ2(1λ)]a22=c24c218. (5.6)

    From (5.5) and (5.6), we have

    a2=c18[(1λ)2+λ2], (5.7)
    a3=13(1λ)2+2λ2A, (5.8)

    where

    A={(1λ)λ32[(1λ)2+λ2]218}c21+c24.

    Hence,

    |a3|=14[3(1λ)2+2λ2]|c2(12(1λ)λ8[(1λ)2+λ2]2)c21|,

    and, since

    |2(12(1λ)λ8[(1λ)2+λ2]2)1|=(1λ)λ4[(1λ)2+λ2]2<1λ[0,1],

    we can apply Lemma 2.2 to get

    |a3|110λ212λ+6.

    From (5.7) and (5.8), we can get

    |a3νa22|=14[3(1λ)2+2λ2]|c2{ν[3(1λ)2+2λ2]16[(1λ)2+λ2]2(1λ)λ8[(1λ)2+λ2]2+12}c21|.

    Applying Lemma 2.2, we can obtain

    |a3νa22|12[3(1λ)2+2λ2]max{1,|ν[3(1λ)2+2λ2]2(1λ)λ8[(1λ)2+λ2]2|}.

    Then, we get (5.1), which completes the proof.

    Corollary 5.1. If fH(12) and is of the form (1.1), then

    |a3νa22|{5ν220,ν2,25,65<ν<2,25ν20,ν65. (5.9)

    Now, we assume that ψ(z) is an analytic and univalent function with positive real part in D, and ψ(z) satisfies the condition of ψ(0)=1 and ψ(0)>0. It is easy to know that ψ(z) has a series expansion of the form

    ψ(z)=1+A1z+A2z2+A3z3+.

    Next, we are going to estimate the upper bounds of the initial coefficients for f, where f and f1 belong to H(λ,ψ). Since ψ(0)>0, we have A1>0.

    Remark 6.1. For ψ(z)=1+z, f(z)=z, we have that f(z) and f1(z) belong to H(λ,ψ) always holds.

    Theorem 6.1. If f, g belong to H(λ,ψ) and are of the form (1.1), where g is the inverse function of f, then we have

    |a2|min{A12[λ2+(1λ)2],A1+|A2A1|7λ28λ+3},
    |a3|min{A12λ2+3(1λ)2+A214[(1λ)2+λ2]2,A1+|A2A1|7λ28λ+3}.

    Proof. Since f,gH(λ,ψ), there exists two analytic functions u,v:DD, where u(0)=v(0)=0, such that

    (1λ)(f(z))1λ+λ(2zf(z)f(z)f(z))λ=ψ(u(z)), (6.1)
    (1λ)(g(z))1λ+λ(2zg(z)g(z)g(z))λ=ψ(v(z)). (6.2)

    Let us define the functions p and q by

    p(z)=1+p1z+p2z2+=1+u(z)1u(z),
    q(z)=1+q1z+q2z2+=1+v(z)1v(z).

    Or, equivalently,

    u(z)=p(z)1p(z)+1=12p1z+(p22p214)z2+,
    v(z)=q(z)1q(z)+1=12q1z+(q22q214)z2+.

    In addition,

    ψ(u(z))=1+12A1p1z+(A1(12p214p21)+14A2p21)z2+, (6.3)
    ψ(v(z))=1+12A1q1z+(A1(12q214q21)+14A2q21)z2+. (6.4)

    From (5.2), (5.3), (6.1), and (6.3), we have

    2[λ2+(1λ)2]a2=12A1p1, (6.5)
    [3(1λ)2+2λ2]a32λ(1λ)a22=A1(12p214p21)+14A2p21. (6.6)

    Since

    g(z)=za2z2+(2a22a3)z3(5a325a2a3+a4)z4+,

    we can obtain

    (1λ)(g(z))1λ+λ(2zg(z)g(z)g(z))λ=12[(1λ)2+λ2]a2z+[(3(1λ)2+2λ2)(2a22a3)2λ(1λ)a22]z2. (6.7)

    By using (6.2), (6.4), and (6.7), we have

    2[(1λ)2+λ2]a2=12A1q1, (6.8)
    [3(1λ)2+2λ2](2a22a3)2λ(1λ)a22=A1(12q214q21)+14A2q21. (6.9)

    From (6.5) and (6.8), we can get

    p1=q1, (6.10)

    and

    a22=A21(p21+q21)32[λ2+(1λ)2]2. (6.11)

    Since |pi|2,|qi|2(iN+), we obtain

    |a2|A12[λ2+(1λ)2]. (6.12)

    By adding (6.6) to (6.9), we can get

    a22=2A1(p2+q2)+(A2A1)(p21+q21)8(7λ28λ+3). (6.13)

    Since |pi|2and|qi|2(iN+), we can get

    |a2|A1+|A2A1|7λ28λ+3. (6.14)

    From (6.12) and (6.14), we can obtain the conclusion

    |a2|min{A12[λ2+(1λ)2],A1+|A2A1|7λ28λ+3}.

    By subtracting (6.6) from (6.9) and using (6.10), we have

    a3=A1(p2q2)4[2λ2+3(1λ)2]2+a22. (6.15)

    Using (6.10) and (6.11) in (6.15), we can obtain

    a3=A1(p2q2)4[3(1λ)2+2λ2]+A21p2116[(1λ)2+λ2]2.

    Therefore,

    |a3|A12λ2+3(1λ)2+A214[(1λ)2+λ2]2. (6.16)

    On the other hand, by using (6.10) and (6.13) in (6.15), we can obtain

    a3=A1(p2q2)4[3(1λ)2+2λ2]+A1(p2+q2)+(A2A1)p214[3(1λ)2+2λ2]8λ(1λ),

    or, equivalently,

    a3=(A1p2(14(5λ26λ+3)+14(7λ28λ+3))+A1q2(14(7λ28λ+3)14(5λ26λ+3))+(A2A1)p214(7λ28λ+3)).

    Using the triangle inequality and Lemma 2.2, we can obtain

    |a3|A17λ28λ+3+|A2A1|7λ28λ+3. (6.17)

    From (6.16) and (6.17), we have

    |a3|min{A12λ2+3(1λ)2+A214[(1λ)2+λ2]2,A1+|A2A1|7λ28λ+3},

    which completes the proof.

    Corollary 6.1. If f satisfies the condition of Theorem 6.1 and we let ψ(z)=1+43z+23z2, then

    |a2|min{23[λ2+(1λ)2],27λ28λ+3},
    |a3|min{43[2λ2+3(1λ)2]+49[(1λ)2+λ2]2,27λ28λ+3}.

    Corollary 6.2. If f satisfies the condition of Theorem 6.1 and we let fH(0,ψ), then

    |a2|min{A12,A1+|A2A1|3},
    |a3|min{A13+A214,A1+|A2A1|3}.

    Let fH(1,ψ), then

    |a2|min{A12,A1+|A2A1|2},
    |a3|min{A12+A214,A1+|A2A1|2}.

    In the present work, we defined new subclasses of analytic functions associated with the modified sigmoid function. Then, we mainly get upper bounds of the third-order Hankel determinant and fourth-order Hankel determinant in certain conditions. We also get the upper bound of the third-order Hankel determinant of its inverse function in the specific conditions mentioned above. Next, we investigated the upper bound of the Fekete-Szegö inequality for the analytic functions in the class H(λ). Finally, we estimated the upper bounds of the initial coefficients for the analytic functions in the class H(λ,ψ), where f1(z) also belongs to H(λ,ψ). The purpose of our study is to stimulate the interest of scholars in the field and to further stimulate their research in this kind of subject. In fact, this kind of problem plays a very important role in many other problems of mathematical analysis.

    We will further investigate the upper bounds of the third, fourth, and fifth-order Hankel determinants of functions belonging to H(λ) or H(λ,ψ) (0λ1). We can also research the upper bounds of the third or fourth Hankel determinant of a class of functions defined in [12]. Recently, the problems of the quantum calculus happens to provide another popular and interesting direction for researchers in complex analysis, which is evidenced by the recently-published review article by Srivastava [31]. Hence, the quantum extension of the results shown in this paper is quite worthwhile to further research. Apart from the above, we are motivated to explore how to get the upper bound of the Hankel determinant of certain analytic functions by other methods, from which we may get more precise or sharp upper bounds.

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

    This study is supported by the Guangdong Provincial Natural Science Foundation's general program. Fund number: 2021A1515010374.

    Also, the authors would like to thank the anonymous referee for the very thorough reading and contributions to improve our presentation of the paper.

    The authors declare that they have no competing interests.



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