Research article Special Issues

Shape reconstruction of acoustic obstacle with linear sampling method and neural network

  • Received: 03 February 2024 Revised: 29 March 2024 Accepted: 07 April 2024 Published: 12 April 2024
  • MSC : 65B99, 68T07

  • We consider the inverse scattering problem of reconstructing the boundary of an obstacle by using far-field data. With the plane wave as the incident wave, a priori information of the impenetrable obstacle can be obtained via the linear sampling method. We have constructed the shape parameter inversion model based on a neural network to reconstruct the obstacle. Numerical experimental results demonstrate that the model proposed in this paper is robust and performs well with a small number of observation directions.

    Citation: Bowen Tang, Xiaoying Yang, Lin Su. Shape reconstruction of acoustic obstacle with linear sampling method and neural network[J]. AIMS Mathematics, 2024, 9(6): 13607-13623. doi: 10.3934/math.2024664

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  • We consider the inverse scattering problem of reconstructing the boundary of an obstacle by using far-field data. With the plane wave as the incident wave, a priori information of the impenetrable obstacle can be obtained via the linear sampling method. We have constructed the shape parameter inversion model based on a neural network to reconstruct the obstacle. Numerical experimental results demonstrate that the model proposed in this paper is robust and performs well with a small number of observation directions.



    In the literature, special functions have a great importance in a variety of fields of mathematics, such as mathematical physics, mathematical biology, fluid mechanics, geometry, combinatory and statistics. Due of the essential position of special functions in mathematics, they continue to play an essential role in the subject as well as in the geometric function theory. For geometric behavior of some other special functions, one can refer to [1,2,3,4,5,6,7,8,9,10,11,12]. An interesting way to discuss the geometric properties of special functions is by the means of some criteria due to Ozaki, Fejér and MacGregor. One of the important special functions is the Mathieu series that appeared in the nineteenth century in the monograph [13] defined on R by

    S(r)=n12n(n2+r2)2. (1.1)

    Surprisingly, the Mathieu series is considered in a variety of fields of mathematical physics, namely, in the elasticity of solid bodies [13]. For more applications regarding the Mathieu series, we refer the interested reader to [14, p. 258, Eq (54)]. The functions bear the name of the mathematician Émile Leonard Mathieu (1835–1890). Recently, a more general family of the Mathieu series was studied by Diananda [15] in the following form:

    Sμ(r)=n12n(n2+r2)μ+1(μ>0,rR). (1.2)

    In 2020, Gerhold et al. [16], considered a new Mathieu type power series, defined by

    Sα,β,μ(r;z)=k=0(k!)αzk((k!)β+r2)μ+1, (1.3)

    where α,μ0,β,r>0 and |z|1, such that α<β(μ+1).

    In [17], Bansal and Sokól have determined sufficient conditions imposed on the parameters such that the normalized form of the function S(r,z) belong to a certain class of univalent functions, such as starlike and close-to-convex. In [18], the authors presented some generalizations of the results of Bansal and Sokól by using the same technique. In addition, Gerhold et al. [18, Theorems 5 and 6] has established some sufficient conditions imposed on the parameter of the normalized form of the function S1,2,μ(r;z) defined by

    Qμ(r;z):=z+n=2n!(r2+1)μ+1((n!)2+r2)μ+1zn, (1.4)

    to be starlike and close-to-convex in the open unit disk. The main focus of the present paper is to extend and improve some results from [18] by using a completely different method. More precisely, in this paper we present some sufficient conditions, such as the normalized form of the function S1,β,μ(r;z) defined by

    Qμ,β(r;z)=z+n=2n!(r2+1)μ+1zn((n!)β+r2)μ+1, (1.5)

    satisfying several geometric properties such as starlikeness, convexity and close-to-convexity.

    We denoted by H the class of all analytic functions inside the unit disk

    D={z:zCand|z|<1}.

    Assume that A denoted the collection of all functions fH, satisfying the normalization f(0)=f(0)1=0 such that

    f(z)=z+k=2akzk,(zD).

    A function fA is said to be a starlike function (with respect to the origin zero) in D, if f is univalent in D and f(D) is a starlike domain with respect to zero in C. This class of starlike functions is denoted by S. The analytic characterization of S is given [19] below:

    (zf(z)f(z))>0(zD).

    If f(z) is a univalent function in D and f(D) is a convex domain in C, then fA is said to be a convex function in D. We denote this class of convex functions by K, which can also be described as follows:

    (1+zf(z)f(z))>0(zD).

    An analytic function f in A is called close-to-convex in the open unit disk D if there exists a function g(z), which is starlike in D such that

    (zf(z)g(z))>0,zD.

    It can be noted that every close-to-convex function in D is also univalent in D (see, for details, [19,20]).

    In order to show the main results, the following preliminary lemmas will be helpful. The first result is due to Ozaki (see also [21, Lemma 2.1]).

    Lemma 1.1. [22] Let

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

    be analytic in D. If

    12a2(n+1)an+10,

    or if

    12a2(n+1)an+12,

    then f is close-to-convex with respect to the function log(1z).

    Remark 1.2. We note that, as Ponnusamy and Vuorinen pointed out in [21], proceeding exactly as in the proof of Lemma 1.1, one can verify directly that if a function f:DC satisfies the hypothesis of the above lemma, then it is close-to-convex with respect to the convex function

    z1z.

    The next two lemmas are due to Fejér [23].

    Lemma 1.3. Suppose that a function f(z)=1+k=2akzk1, with ak0(k2) as analytic in D. If (ak)k1 is a convex decreasing sequence, i.e., ak2ak+1+ak+20 and akak+10 for all k1, then

    (f(z))>12(zD).

    Lemma 1.4. Suppose that a f(z)=z+k=2akzk, with ak0(k2) as analytic in D. If (kak)k1 and (kak(k+1)ak+1)k1 both are decreasing, then f is starlike in D.

    Lemma 1.5 ([24]). Assume that fA. If the following inequality

    |f(z)z1|<1,

    holds for all zD, then f is starlike in

    D12:={zCand|z|<12}.

    Lemma 1.6 ([25]). Assume that fA and satisfies

    |f(z)1|<1,

    for each zD, then f is convex in D12.

    Theorem 2.1. Let μ,β>0 and 0<r1 such that β1+2μ+1. In addition, if the following condition holds:

    H:(2β+12)μ+14,

    then the function Qμ,β(r;z) is close-to-convex in D with respect to the function log(1z).

    Proof. For the function Qμ,β(r;z), we have

    a1=1andak=k!(r2+1)μ+1((k!)β+r2)μ+1(k2).

    To prove the result, we need to show that the sequence {kak}k1 is decreasing under the given conditions. For k2 we have

    kak(k+1)ak+1=(r2+1)μ+1[kk!((k!)β+r2)μ+1(k+1)(k+1)!(((k+1)!)β+r2)μ+1]=k!(r2+1)μ+1[k((k!)β+r2)μ+1(k+1)2(((k+1)!)β+r2)μ+1]=k!(r2+1)μ+1Ak(β,μ,r)[((k!)β+r2)(((k+1)!)β+r2)]μ+1, (2.1)

    where

    Ak(β,μ,r)=k(((k+1)!)β+r2)μ+1(k+1)2((k!)β+r2)μ+1,k2.

    However, we have

    Ak(β,μ,r)=(k1μ+1((k+1)!)β+k1μ+1r2)μ+1((k+1)2μ+1(k!)β+(k+1)2μ+1r2)μ+1=exp((μ+1)log[k1μ+1((k+1)!)β+k1μ+1r2])exp((μ+1)log[(k+1)2μ+1(k!)β+(k+1)2μ+1r2])=j=0[logj(k1μ+1((k+1)!)β+k1μ+1r2)logj((k+1)2μ+1(k!)β+(k+1)2μ+1r2)](μ+1)jj!. (2.2)

    In addition, for all k2, we have

    k1μ+1((k+1)!)β+k1μ+1r2(k+1)2μ+1(k!)β+(k+1)2μ+1r2=r2(k1μ+1(k+1)2μ+1)+k1μ+1((k+1)!)β(k+1)2μ+1(k!)β[k1μ+1(k+1)2μ+1+k1μ+1((k+1)!)β2]+[k1μ+1((k+1)!)β2(k+1)2μ+1(k!)β]=k1μ+1(1+((k+1)!)β2((k+1)2k)1μ+1)+(k!)β(k1μ+1(k+1)β2(k+1)2μ+1)k1μ+1(k+1)2μ+1(1+(k!)β(k+1)21k1μ+1)+(k!)β(k+1)2μ+1(k1μ+1(k+1)21)k1μ+1(k+1)2μ+1(1+(k!)βk1μ+11k1μ+1)+(k!)β(k+1)2μ+1(k1μ+11), (2.3)

    which is positive by our assumption. Having (2.1)–(2.3), we conclude that the sequence (kak)k2 is decreasing. Finally, we see that the condition (H) implies that a12a2, then the function Qμ,β(r;z) is close-to-convex in D with respect to the function log(1z) by Lemma 1.1.

    If we set β=32 in Theorem 2.1, we derive the following result as follows:

    Corollary 2.2. Let 0<r1. If μ3, then the function Qμ,32(r;z) is close-to-convex in D with respect to the function log(1z).

    Upon setting μ=2 in Theorem 2.1, we get the following result:

    Corollary 2.3. Let 0<r1. If β53, then the function Q2,β(r;z) is close-to-convex in D with respect to the function log(1z).

    Remark 2.4. In [18], it is established that the function Qμ,2(r;z)=:Qμ(r;z) is close-to-convex in D with respect to the function z1z for all 0<rμ. Moreover, in view of Remark 1.2, we conclude that the function Qμ,2(r;z) is close-to-convex in D with respect to the function log(1z) for all 0<rμ. However, in view of Corollaries 2.2 and 2.3, we deduce that Theorem 2.1 improves the corresponding result available in [18, Theorem 5] for 0<r1.

    Theorem 2.5. Assume that μ,β>0,0<r1 such that β1+1μ+1. In addition, if the condition (H) holds, then

    (Qμ,β(r;z)z)>12,

    for all zD.

    Proof. For k1, we get

    ((k!)β+r2)μ+1(((k+1)!)β+r2)μ+1(akak+1)(r2+1)μ+1=k!(((k+1)!)β+r2)μ+1(k+1)!((k!)β+r2)μ+1=[(k!)1μ+1((k+1)!)β+r2)]μ+1[((k+1)!)1μ+1((k!)β+r2)]μ+1. (2.4)

    Further, for all k1, we have

    (k!)1μ+1((k+1)!)β+r2)((k+1)!)1μ+1((k!)β+r2)=r2[(k!)1μ+1((k+1)!)1μ+1]+(k!)1μ+1((k+1)!)β((k+1)!)1μ+1(k!)β(k!)1μ+1((k+1)!)1μ+1+(k!)1μ+1((k+1)!)β((k+1)!)1μ+1(k!)β=(k!)1μ+1[1+(k!)β(k+1)β2(k+1)1μ+1]+(k!)β+1μ+1[(k+1)β2(k+1)1μ+1](k!)1μ+1[1+(k+1)1+1μ+12(k+1)1μ+1]+(k!)β+1μ+1[(k+1)1+1μ+12(k+1)1μ+1]=(k!)1μ+1[1+(k+1)1μ+1((k+1)21)]+(k!)β+1μ+1(k+1)1μ+1((k+1)21)>0. (2.5)

    Hence, in view of (2.4) and (2.5), we deduce that the sequence (ak)k1 is decreasing. Next, we prove that (ak)k1 is a convex decreasing sequence, then, for k2 we obtain

    ((k!)β+r2)μ+1(((k+1)!)β+r2)μ+1(ak2ak+1)(r2+1)μ+1=k!(((k+1)!)β+r2)μ+12(k+1)!((k!)β+r2)μ+1=[(k!)1μ+1((k+1)!)β+r2)]μ+1[(2(k+1)!)1μ+1((k!)β+r2)]μ+1. (2.6)

    Moreover, we get

    (k!)1μ+1(((k+1)!)β+r2)(2(k+1)!)1μ+1((k!)β+r2)=r2[(k!)1μ+1(2(k+1)!)1μ+1]+(k!)1μ+1((k+1)!)β(2(k+1)!)1μ+1(k!)β>[(k!)1μ+1(2(k+1)!)1μ+1]+(k!)1μ+1((k+1)!)β3+2(k!)β+1μ+13[(k+1)β3.2μμ+1(k+1)1μ+1](k!)1μ+1[1+((k+1)!)1+1μ+13(2(k+1))1μ+1]+2(k!)β+1μ+13[(k+1)1+1μ+13.2μμ+1(k+1)1μ+1]=(k!)1μ+1[1+(k+1)1μ+1{(k+1)(k!)1+1μ+1321μ+1}]+2(k!)β+1μ+1(k+1)1μ+13[(k+1)3.2μμ+1]>2[12μμ+1](k!)β+1μ+1(k+1)1μ+1>0. (2.7)

    Keeping (2.6) and (2.7) in mind, we have ak2ak+1>0 for all k2. In addition, the condition (H) implies a12a20. This in turn implies that the sequence (ak)k1 is convex. Finally, by Lemma 1.3, we obtain the desired result.

    Taking β=32 in Theorem 2.5, we derive the following result:

    Corollary 2.6. Assume that r(0,1]. If μlog(4)log(232+1)log(2)11.14, then

    (Qμ,32(r;z)z)>12(zD).

    Setting μ=1 in Theorem 2.5, we established the following result which reads as follows:

    Corollary 2.7. Let 0<r1. If βlog(3)log(2), then

    (Q1,β(r;z)z)>12(zD).

    Remark 2.8. The result obtained in the above theorem has been derived from [18, Theorem 6] for β=2,μ>0 and 0<r<μ. Hence, in view of Corollaries 2.2 and 2.6, we deduce that Theorem 2.5 improves the corresponding result given in [18, Theorem 6] for 0<r1.

    Theorem 2.9. Assume that min(μ,β)>0,0<r1 such that β1+3μ+1, then the function Qμ,β(r;z) is starlike in D.

    Proof. We see in the proof of Theorem 2.1 that the sequence (kak)k1 is decreasing. Hence, with the aid of Lemma 1.4 to show that the function Qμ,β(r;z) is starlike in D, it suffices to prove that the sequence (kak(k+1)ak+1)k1 is decreasing. We have

    kak2(k+1)ak+1=k!(r2+1)μ+1Bk(β,μ,r)[((k!)β+r2)((k+1)!)β+r2)]μ+1, (2.8)

    where

    Bk(β,μ,r)=k(((k+1)!)β+r2)μ+12(k+1)2((k!)β+r2)μ+1,k1.

    For k2, we have

    k1μ+1(((k+1)!)β+r2)(2(k+1)2)1μ+1((k!)β+r2)k1μ+1(2(k+1)2)1μ+1+k1μ+1((k+1)!)β2+[k1μ+1((k+1)!)β2(2(k+1)2)1μ+1(k!)β]=k1μ+1+k1μ+1((k+1)!)β2(2(k+1)2)1μ+1+(k!)β(k1μ+1(k+1)β2(2(k+1)2)1μ+1)k1μ+1+(k+1)2μ+1(k1μ+1(k!)β(k+1)221μ+1)+(k!)β(k+1)2μ+1(k1μ+1(k+1)221μ+1)k1μ+1+(k+1)2μ+1(k1μ+1(k!)β21μ+1)+(k!)β(k+1)2μ+1(k1μ+121μ+1)>0, (2.9)

    which in turn implies that

    Bk(β,μ,r)>0,

    for all k2, and consequently, the sequence (kak(k+1)ak+1)k2 is decreasing. Further, a simple computation gives

    a14a2+3a3(1+r2)μ+1=1(1+r2)μ+18(2β+r2)μ+1+18(6β+r2)μ+112μ+182β(μ+1)+18(6β+r2)μ+1=2β(μ+1)2μ+42(β+1)(μ+1)+18(6β+r2)μ+12μ+42μ+42(β+1)(μ+1)+18(6β+r2)μ+1>0.

    Therefore, (kak(k+1)ak+1)k1 is decreasing, which leads us to the asserted result.

    In the next Theorem we present another set of sufficient conditions to be imposed on the parameters so that the function Qμ,β(r;z) is starlike in D.

    Theorem 2.10. Let the parameters be the same as in Theorem 2.1. In addition, if the following conditions

    H:(2β+12)μ+18(e2),

    hold true, then the function Qμ,β(r;z) is starlike in D.

    Proof. First of all, we need to prove that the sequences (uk)k2 and (vk)k2 defined by

    uk=(k!)2(r2+1)μ+1((k!)β+r2)μ+1andvk=(k1)(k!)2(r2+1)μ+1((k!)β+r2)μ+1,

    are decreasing. Indeed, we have

    ((k!)β+r2)μ+1(((k+1)!)β+r2)μ+1(ukuk+1)(k!)2(r2+1)μ+1=(((k+1)!)β+r2)μ+1(k+1)2((k!)β+r2)μ+1. (2.10)

    In addition, for any k2, we have

    ((k+1)!)β+r2(k+1)2μ+1((k!)β+r2)=r2(1(k+1)2μ+1)+((k+1)!)β(k+1)2μ+1(k!)β1(k+1)2μ+1+((k+1)!)β(k+1)2μ+1(k!)β=1+(((k+1)!)β2(k+1)2μ+1)+(((k+1)!)β2(k+1)2μ+1(k!)β)1+((k!)β(k+1)1+2μ+12(k+1)2μ+1)+(k!)β((k+1)1+2μ+12(k+1)2μ+1)=1+(k+1)2μ+1((k!)β(k+1)21)+(k!)β(k+1)2μ+1(k+121)>0. (2.11)

    According to (2.10) and (2.11) we conclude that the sequence (uk)k2 is decreasing. Also, for k2, we have

    ((k!)β+r2)μ+1(((k+1)!)β+r2)μ+1(vkvk+1)(k!)2(r2+1)μ+1=(k1)(((k+1)!)β+r2)μ+1k(k+1)2((k!)β+r2)μ+1. (2.12)

    Moreover, for all k2, we find

    (k1)1μ+1(((k+1)!)β+r2)(k(k+1)2)1μ+1((k!)β+r2)=r2((k1)1μ+1(k(k+1)2)1μ+1)+(k1)1μ+1((k+1)!)β(k(k+1)2)1μ+1(k!)β(k1)1μ+1(k(k+1)2)1μ+1+(k1)1μ+1((k+1)!)β3+2(k1)1μ+1((k+1)!)β3(k(k+1)2)1μ+1(k!)β(k1)1μ+1+(k+1)2μ+1((k1)1μ+1(k!)1+2μ+1(k+1)3k1μ+1)+(k!)β(k+1)2μ+1(2(k1)1μ+1(k+1)3k1μ+1)(k1)1μ+1+(k+1)2μ+1((k1)1μ+1(k!)1+2μ+1k1μ+1)+(k!)β(k+1)2μ+1(2(k1)1μ+1k1μ+1). (2.13)

    Since the sequence (k/(k1))n2 is decreasing, we deduce that kk12 for all k2 and consequently,

    (kk1)1μ+121μ+12(k2,μ>0).

    Hence, in view of the above inequality combined with (2.13) and (2.12), we conclude that the sequence (vk)k2 is decreasing. Now, we set

    ˜Qμ,β(r;z):=z[Qμ,β(r;z)]Qμ,β(r;z),zD.

    We see that the function ˜Qμ,β(r;z) is analytic in D and satisfies ˜Qμ,β(r;0)=1. Hence, to derive the desired result, it suffices to prove that, for any zD, we have

    (˜Qμ,β(r;z))>0.

    For this goal in view, it suffices to show that

    |˜Qμ,β(r;z)1|<1(zD).

    For all zD, we get

    |[Qμ,β(r;z)]Qμ,β(r;z)z|<k=2(k1)k!(r2+1)μ+1((k!)β+r2)μ+1=k=2vkk!v2(e2). (2.14)

    In addition, in view of the inequality:

    |a+b|||a||b||,

    we obtain

    |Qμ,β(r;z)z|>1k=2(k!)(r2+1)μ+1((k!)β+r2)μ+1=1k=2ukk!1u2(e2). (2.15)

    By using (2.14) and (2.15), for zD, we get

    |˜Qμ,β(r;z)1|=|[Qμ,β(r;z)]Qμ,β(r;z)z||Qμ,β(r;z)z|<v2(e2)1u2(e2). (2.16)

    Furthermore, by using the fact that the function rχμ,β(r)=(r2+1r2+2β)μ+1 is strictly increasing on (0,1], and with the aid of condition (H), we obtain

    (v2+u2)(e2)=8(e2)(r2+1)μ+1(2β+r2)μ+1<8(e2)(22β+1)μ+11. (2.17)

    Finally, by combining (2.16) and (2.17), we derived the desired results.

    By setting β=2 in Theorem 2.10, we obtain the following corollary:

    Corollary 2.11. If 0<r1 and μ1, then the function Qμ(r;z) defined in (1.4) is starlike in D.

    Taking β=32 in Theorem 2.10, we obtain:

    Corollary 2.12. Under the assumptions of Corollary 2.2, the function Qμ,32(r;z) is starlike in D.

    Setting in Theorem 2.10 the values μ=2, we compute the following corollary:

    Corollary 2.13. Suppose that all hypotheses of Corollary 2.3 hold, then the function Q2,β(r;z) is starlike in D.

    Example 2.14. The functions Q3,32(1/2;z) and Q2,53(1/2;z) are starlike in D.

    Figure 1 illustrates the mappings of the above examples in D.

    Figure 1.  Mappings of Qμ,β(r;z) over D.

    Theorem 2.15. Let μ,β>0 and 0<r1 such that β1+3μ+1. If the following condition

    H:(2β+12)μ+116(e2),

    holds true, then the function Qμ,β(r;z) is convex in D.

    Proof. We define the sequences (xk)k2 and (yk)k2 by

    xk=k(k!)2(r2+1)μ+1((k!)β+r2)μ+1andyk=k(k1)(k!)2(r2+1)μ+1((k!)β+r2)μ+1.

    Let k2, then

    ((k!)β+r2)μ+1(((k+1)!)β+r2)μ+1(xkxk+1)(k!)2(r2+1)μ+1=k(((k+1)!)β+r2)μ+1(k+1)3((k!)β+r2)μ+1. (2.18)

    However, we have

    k1μ+1(((k+1)!)β+r2)(k+1)3μ+1((k!)β+r2)k1μ+1(k+1)3μ+1+k1μ+1((k+1)!)β(k+1)3μ+1(k!)β=k1μ+1+(k1μ+1(k!)β(k+1)β2(k+1)3μ+1)+(k1μ+1(k!)β(k+1)β2(k+1)3μ+1(k!)β)k1μ+1+(k+1)3μ+1(k1μ+1(k!)β(k+1)21)+(k!)β(k+1)3μ+1(k1μ+1(k+1)21)>0. (2.19)

    Hence, in view of (2.18) and (2.19), we get that (xk)k2 is decreasing. Also, we have

    ((k!)β+r2)μ+1(((k+1)!)β+r2)μ+1(ykyk+1)k(k!)2(r2+1)μ+1=(k1)(((k+1)!)β+r2)μ+1(k+1)3((k!)β+r2)μ+1. (2.20)

    Moreover, for k2, we find that

    (k1)1μ+1(((k+1)!)β+r2)(k+1)3μ+1((k!)β+r2)(k1)1μ+1+((k1)1μ+1(k!)β(k+1)β2(k+1)3μ+1)+(k!)β((k1)1μ+1(n+1)β2(n+1)3μ+1)(k1)1μ+1+(k+1)3μ+1((k1)1μ+1(k!)β(k+1)21)+(k!)β(k+1)3μ+1((k1)1μ+1(k+1)21)>0. (2.21)

    Having (2.20) and (2.21) in mind, we deduce that the sequence (yk)k2 is decreasing. To show that the function Qμ,β(r;z) is convex in D, it suffices to establish that the function

    ˆQμ,β(r;z):=z[Qμ,β(r;z)],

    is starlike in D. For this objective in view, it suffices to find that

    |z[ˆQμ,β(r;z)]ˆQμ,β(r;z)1|<1(zD).

    For all zD and since (yk)k2 is decreasing, we get

    |[ˆQμ,β(r;z)]ˆQμ,β(r;z)z|<k=2k(k1)k!(r2+1)μ+1((k!)β+r2)μ+1=k=2ykk!y2(e2). (2.22)

    Further, for any zD, we obtain

    |ˆQμ,β(r;z)z|>1k=2k(k!)(r2+1)μ+1((k!)β+r2)μ+1=1k=2xkk!1x2(e2). (2.23)

    Keeping (2.22) and (2.23) in mind, for zD, we get

    |z[ˆQμ,β(r;z)]ˆQμ,β(r;z)1|=|[ˆQμ,β(r;z)]ˆQμ,β(r;z)z||ˆQμ,β(r;z)z|<y2(e2)1x2(e2)=8(e2)(r2+1)μ+1(2β+r2)μ+18(e2)(r2+1)μ+1. (2.24)

    Again, by using the fact that the function rχμ,β(r) is increasing on (0,1] and with the aid of hypothesis (H) we obtain that

    8(e2)(r2+1)μ+1(2β+r2)μ+18(e2)(r2+1)μ+1<1. (2.25)

    Finally, by combining the above inequality and (2.24), we obtain the desired result asserted by Theorem 2.15.

    Taking β=2 in Theorem 2.15, in view of (1.4), the following result holds true:

    Corollary 2.16. Let 0<r1. If μ2, then the function Qμ(r;z) is convex in D.

    If we set μ=1 in Theorem 2.15, in view of (1.5), we derive the following result:

    Corollary 2.17. Let 0<r1. If βlog(8e21)log(2), then the function Q1,β(r;z) is convex in D.

    Example 2.18. The functions Q2(r;z) and Q1,83(r;z) are convex in D.

    Figure 2 gives the mappings of the above presented examples in D.

    Figure 2.  Mappings of Qμ,β(r;z) over D.

    Theorem 2.19. Let the parameters be the same as in Theorem 2.1, then the function Qμ,β(r;z) is starlike in D12.

    Proof. For any zD we get

    |Qμ,β(r;z)z1|<k=2k!(r2+1)μ+1(k!)β+r2)μ+1=k=2ckk!, (2.26)

    where

    ck:=(k!)2(r2+1)μ+1((k!)β+r2)μ+1,k2.

    Straightforward calculation gives

    [((k!)β+r2)(((k+1)!)β+r2)]μ+1(ckck+1)(k!)2(r2+1)μ+1=(((k+1)!)β+r2)μ+1((k+1)2μ+1((k!)β+r2))μ+1. (2.27)

    Furthermore, for k2, we get

    ((k+1)!)β+r2(n+1)2μ+1((k!)β+r2)=r2(1(k+1)2μ+1)+((k+1)!)β(k!)β(k+1)2μ+1(1+((k+1)!)β2(k+1)2μ+1)+(k!)β((k+1)β2(k+1)2μ+1)(1+(k!)β(k+1)1+2μ+12(k+1)2μ+1)+(k!)β(k+1)2μ+1(k1)2(1+(k+1)2μ+1((k!)β(k+1)2)2)+(k!)β(k+1)2μ+1(k1)2>0. (2.28)

    Thus, the sequence (ck)k2 is decreasing. However, in view of (2.26), for zD we obtain

    |Qμ,β(r;z)z1|<k=2k!(r2+1)μ+1((k!)β+r2)μ+1=k=2c2k!=c2(e2)=4(e2)(r2+1)μ+1(2β+r2)μ+1. (2.29)

    According to the monotony property of the function rχβ,μ(r) on (0,1) we get

    χβ,μ(r)<14. (2.30)

    Hence, in view (2.29) and (2.30) we find for all zD that

    |Qμ,β(r;z)z1|<(e2)<1.

    With the help of Lemma 1.5, we deduce that the function Qμ,β(r;z) is starlike in D12.

    Corollary 2.20. Assume that all conditions of Corollary 2.2 are satisfied, then the function Qμ,32(r;z) is starlike in D12.

    Corollary 2.21. Suppose that all hypotheses of Corollary 2.3 hold, then the function Q2,β(r;z) is starlike in D12.

    If we set β=2 in the above Theorem, in view of (1.4), the following result is true:

    Corollary 2.22. Let 0<r1 If μ1, then the function Qμ(r;z) is starlike in D12.

    Example 2.23. The functions Q3,32(1/2;z),Q1(1;z) and Q2,53(1/2;z) are starlike in D12.

    In Figure 3, we give the mappings of the above presented examples in D.

    Figure 3.  Mappings of Qμ,β(r;z) over D12.

    Theorem 2.24. Let β,μ>0 and 0<r<1. If β1+3μ+1, then the function Qμ,β(r;z) is convex in D12.

    Proof. For all zD, it follows that

    |Qμ,β(r;z)1|<k=2kk!(r2+1)μ+1((k!)β+r2)μ+1=k=2dkk(k1), (2.31)

    where

    dk:=k2(k1)k!(r2+1)μ+1(k!)β+r2)μ+1,k2.

    For all k2, we get

    ((k!)β+r2)μ+1(((k+1)!)β+r2)μ+1(dkdk+1)kk!(1+r2)μ+1=((k(k1))1μ+1[((k+1)!)β+r2])μ+1(((k+1))3μ+1[(k!)β+r2])μ+1. (2.32)

    However, for all k2 and under the conditions imposed on the parameters, we have

    (k(k1))1μ+1[((k+1)!)β+r2]((k+1))3μ+1[(k!)β+r2](k(k1))1μ+1((k+1))3μ+1+(k(k1))1μ+1((k+1)!)β(k+1)3μ+1(k!)β=(k(k1))1μ+1+((k(k1))1μ+1(k!)β(k+1)β2(k+1)3μ+1)+(k!)β((k(k1))1μ+1(k+1)β2(k+1)3μ+1)(k(k1))1μ+1+(n+1)3μ+1((k(k1))1μ+1(k!)β(k+1)21)+(k!)β(k+1)3μ+1((k(k1))1μ+1(k+1)21)(k(k1))1μ+1+(k+1)3μ+1((k(k1))1μ+1(k!)β1)+(k!)β(k+1)3μ+1((k(k1))1μ+11)>0. (2.33)

    Hence, in view of (2.32) and (2.33) we conclude that the sequence (dk)k2 is decreasing. Therefore, by (2.31), we conclude

    |Qμ,β(r;z)1|<k2d2k(k1)=d2. (2.34)

    Moreover, since β1+3μ+1 and r(0,1], we get

    (r2+1r2+2β)μ+118,

    and consequently, for all zD, we obtain

    |Qμ,β(r;z)1|<1. (2.35)

    Finally, with the means of Lemma 1.6, we conclude that the function Qμ,β(r;z) is convex in D12.

    If we take β=2 in Theorem 2.15, in view of (1.4), the following result holds true:

    Corollary 2.25. Let 0<r1. If μ2, then the function Qμ(r;z) is convex in D12.

    If we let μ=1 in Theorem 2.15, in view of (1.5), we derive the following result:

    Corollary 2.26. Let 0<r1. If β52, then the function Q1,β(r;z) is convex in D12.

    Example 2.27. The functions Q2(r;z) and Q1,52(r;z) are convex in D12.

    In Figure 4, we present the mappings of these examples in D.

    Figure 4.  Mappings of Qμ,β(r;z) over D12.

    Remark 2.28. The geometric properties of the function Qμ(r;z) derived in Corollaries 2.16, 2.22 and 2.25 are new.

    In our present paper, we have derived sufficient conditions such that a class of functions associated to the generalized Mathieu type power series are to be starlike, close-to-convex and convex in the unit disk D. The various results, which we have established in this paper, are believed to be new, and their importance is illustrated by several interesting corollaries and examples. Furthermore, we are confident that our paper will inspire further investigation in this field and pave the way for some developments in the study of geometric functions theory involving certain classes of functions related to the Mathieu type powers series.

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

    The authors extend their appreciation to the Deanship of Scientific Research at Northern Border University, Arar, KSA for funding this research work through the project number "NBU-FFR-2023-0093".

    The authors declare that they have no conflicts of interest.



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