A new analytic solution of complex Langevin differential equations

Abstract

Purpose

In this study, the authors introduce a solvability of special type of Langevin differential equations (LDEs) in virtue of geometric function theory. The analytic solutions of the LDEs are considered by utilizing the Caratheodory functions joining the subordination concept. A class of Caratheodory functions involving special functions gives the upper bound solution.

Design/methodology/approach

The methodology is based on the geometric function theory.

Findings

The authors present a new analytic function for a class of complex LDEs.

Originality/value

The authors introduced a new class of complex differential equation, presented a new technique to indicate the analytic solution and used some special functions.

Citation

Ibrahim, R.W. (2023), "A new analytic solution of complex Langevin differential equations", Arab Journal of Mathematical Sciences, Vol. 29 No. 1, pp. 83-99. https://doi.org/10.1108/AJMS-04-2021-0085

1. Introduction

Langevin differential equation (LDE) is one of the most important differential equation in mathematical sciences, including fluid, Brownian motion, thermal and wavelet studies. It investigated wildly in view of various types of geometric, stochastic and analysis studies (see for example references [1–5]). An arbitrary model of LDEs is studied in [6–8] including analytic solutions. The existence and stability of a class of LDEs with two Hilfer-Katugampola fractional derivatives is investigated in [9]. Moreover, the existence of LDE is illustrated suggesting different types of geometry [10, 11].

LDEs of a complex variable are applied to simulate special types of polymer and nanomaterials, including the conduct of the polymers [12]. Based on this priority of LDEs of a complex variable, we aim to study this class analytically. The technique of the geometric function theory is used recently by Ibrahim and Baleanu [13] to determine the fractal solution. They utilized different notions such as the subordination and super-ordination, majorization, Caratheodory functions, convex functions and special functions (see [14–16]).

Here, we discuss the upper bound solution of LDEs of a complex variable in feature of geometric function theory. We illustrated a list of conditions that implies a univalent result in ∪ (the unit disk). The mechanism of our proof is considered utilizing the Caratheodory functions joining the subordination concept. A class of Caratheodory functions involving special functions gives the upper bound solution.

2. Complex Langevin differential equations

The LDE of a complex variable can be realized by the next formula [17].

(2.1)χ″(z)+αχ′(z)=F(χ(z)),z∈C,

where α > 0 indicates the oscillation coefficient, and F is the noise factor. To study the geometric properties of Eq.(2.1), we consider z∈∪={z∈C:|z|<1}, and χ(z) is a normalized function satisfying the expansion χ(z)=z+∑n=2∞χnzn. Rearrange Eqn (2.1) with complex coefficient, then the homogeneous formula is given by

(2.2)Ψ(z)≔ς(z)z2χ″(z)χ(z)+zχ′(z)χ(z),z∈∪,

where ς(z) is analytic function in ∪. It is clear that Ψ(0) = 1, for all ς(z) ∈ ∪ (see the following example).

Example 2.1.

Let data given by

1. χ(z) = z/(1 − z), ς(z) = z, then we have Ψ(z) = 1 + z + 3z² + 5z³ + 7z⁴ + 9z⁵ + O(z⁶);

2. χ(z) = z/(1 − z)², ς(z) = z, then we get Ψ(z) = 1 + 2z + 6z² + 12z³ + 18z⁴ + 24z⁵ + O(z⁶);

3. ς(z) = 1 − z and χ(z) = z/(1 − z), then we obtain Ψ(z) = 1 + 3z + 3z² + 3z³ + 3z⁴ + 3z⁵ + O(z⁶)

4. ς(z) = 1 and χ(z) = z/(1 − z), then we obtain Ψ(z) = 1 + 3z + 5z² + 7z³ + 9z⁴ + 11z⁵ + O(z⁶).

We denote by P(A,B), the class of functions

ρ(z)=1+Aw(z)1+Bw(z)≺1+Az1+Bz,

where w satisfies w(0) = 0 and |w(z)| < 1; and −1 ≤ B < A ≤ 1, then P(A,B)⊂P(1−A1−B) is the Janowski class. Next, we define a class of analytic functions.

Definition 2.2.

The function χ(z)=z+∑n=2∞χnzn,z∈∪ is in M_ς(ρ) if and only if

(2.3) Ψ(z)=ς(z)z2χ″(z)χ(z)+zχ′(z)χ(z)≺ρ(z).

z∈∪,ρ(0)=1,ρ′(0)>1,ς∈∪,ς(z)∈∪.

Now consider starlike function as follows:

ρe(z)=zez−1=1−z2+z212−z4720+…

and a convex function

ϱe(z)≔1/ρe(z)=1+z2+z26+z324+z4120+…

(see [18]-P415). We note here that the coefficients are approximating to the Bernoulli numbers such that

Reξz−1ξz≥12,0<ξ≤1.793….

Hence, Reξz−1ξz≥1/ρe(−1)=12.

Our design is generated by the Caratheodory functions, which are operated in [19]. In this situation, we establish the necessary conditions of the joining bounds of Ψ(z) consuming a Caratheodory function. Note that, when ς(z) is a constant, the class M_ς(ρ) reduces to the well-known class in [20].

2.1 Geometric properties

Some geometric properties are illustrated as follows:

Proposition 2.3.

Consider the functional Ψ(z) such that p(z)=zχ(z)′/χ(z). Then χ(z) is starlike in ∪, whenever ς(z)=z,R(z)>0 and

−(3)(R(z))<I(z)<(3)(R(z)).

Proof.

Suppose the functional

Ψ(z)=ς(z)z2χ′′(z)χ(z)+zχ′(z)χ(z).

Let p(z) = zχ(z)′/χ(z), then

(z)zχ(z)″χ(z)=z(zp′(z))+zp2(z)−zp(z)

yields that

Ψ(z)=z(zp′(z))+zp2(z)+(1−z)p(z),z∈∪.

By [18]-Example 2.4m, we have A(z) = z, B(z) = z, C(z) = 1 − z and D(z) = 0, where the assumptions imply that R(z)>0 we get the conclusion

R[A(z)zp′(z)+B(z)p2(z)+C(z)p(z)+D(z)]=R[z(zp′(z))+zp2(z)+(1−z)p(z)]>0⇒R(p(z))>0.

Corresponding to the above conclusion, we indicate that χ(z) is starlike. □

Proposition 2.3 can be generated for R(ς(z))>0 as follows:

Proposition 2.4.

Consider the functional Ψ(z) such that p(z)=zχ(z)′/χ(z). Then χ(z) is starlike in ∪, whenever R(ς(z))>0 and

[I(ς(z))]2<3(R(ς(z)))2.

Proposition 2.5.

(Integral existence result)

Consider the functional Ψ(z) = p(z).q(z), where p(z) = zχ(z)′/χ(z) and

q(z)=ς(z)1+zχ′′(z)χ′(z).

If ς(0) = 1 and the subordination

(α1+α2)+zq′(z)q(z)≺(α1+α2)1+z1−z+2z1−z2

holds such that α₁ + α₂ = β₁ + β₂ > 0, then the integral

V(z)≔β1+β2z(β2)p(z)∫0zζα1+α2−1q(ζ)dζ1/(β1)

satisfies the following conclusion

V(z)∈∧,V(z)/z≠0,Rβ1zV′(z)V(z)+zq′(z)q(z)+β2>0.

Proof.

Consider Ψ(z) with ς(z) = 1, then a computation implies that

Ψ(z)=zχ′(z)χ(z)1+zχ′′(z)χ′(z).

Since, p(0) = 1 and q(0) = 1 with Ψ(z) = p(z)q(z) ≠ 0 for some z₀ ∈ ∪. Then in view of [18]-Theorem 2.5c, we have the desired conclusion. □

Proposition 2.6.

Consider the functional Ψ(z) = p(z).q(z), where p(z) = zχ(z)′/χ(z) and

q(z)=1+zχ′′(z)χ′(z),ς(0)=1

If one of the following facts is indicated

1. R(zq′(z)q(z))+δ>0;

2. |I(zq′(z)q(z)+δ)|<1+2δ,δ>0;

3. |zq′(z)q(z)|<δ+1,δ>0,

then the integral

W(z)≔δzδ−1p(z)∫0zζδ−1q(ζ)dζ

satisfies the conclusion

W(z)∈∧,W(z)/z≠0,RzW′(z)W(z)+zq′(z)q(z)+1>0.

Proof.

A computation implies that

Ψ(z)=ς(z)1+zχ′′(z)χ′(z)zχ′(z)χ(z),ς(z)∈∪.

Since p(0) = 1 and q(0) = 1 with Ψ(z) = p(z)q(z) ≠ 0 for some z₀ ∈ ∪. Then in view of [18]-Corollary 2.5c.1, we get the desired conclusion. □

Next example shows the integral existence result of the convex Koebe function z/(1 − z). We confirm that the integral formula is also convex because it is majorized by z/(1 − z).

Example 2.7.

Let χ(z) = z/(1 − z) and ς(z) = 1, then we have p(z) = 1/(1 − z) and

q(z)=1+zχ″(z)χ′(z)=1+z1−z=1+2z2+2z3+2z4+2z5+O(z6).

Thus, we obtain

|zq′(z)q(z)|<δ+1⇔δ>0,δδ+2<R(z)<δ.

Then by letting δ = 1, we conclude that the integral existence result satisfies

W(z)=δzδ−1p(z)∫0zζδ−1q(ζ)dζ=(1−z)∫0z(1+ζ1−ζ)dζ=(z−1)(z+2⁡log(1−z))=z−z3/3−z4/6−z5/10−z6/15+O(z7)∈∧,

which achieves all indicated facts in Proposition 2.5. Since the coefficient bounds of W(z) are motorized by the coefficient bounds of χ(z), then we conclude that W(z) is convex. Moreover, the iteration of the integral existence theorem of a convex function (χ(z)) remains convex in the open unit disk (W(W…(W(z)))). As a conclusion, this example provides a chain of analytic convex solutions of LDEs in ∪. Next remark shows the important relation of W(z) with the function of nephroid plane curve. This leads to use W(z) as an official formula in a nephroid plane curve instead of using parametric functions. Moreover, Proposition 2.5 implies a positive real solution of LDEs; for example, by assuming β₁ = 1, β₂ = 0, we get RzW′(z)W(z)+zq′(z)q(z)>0 because W(z) and q(z) are starlike in ∪ satisfying RzW′(z)W(z)>0 and Rzq′(z)q(z)>0. As a comparison with recent methods, our method provides in spite of an analytic solution, the strategy of the existing integral formula involving the analytic solution is still analytic in ∪. Note that this solution is univalent in ∪. All recent techniques provided an analytic solution without geometric presentations. Our method describes the analytic solution and its integration geometrically.

Remark 2.8.

It is well known that the function ω(z) = 1 + z − z³/3 (see Figure 1) translates the unit circle onto a 2-cusped curve tilled nephroid satisfying ((w−1)2+v2−4/9)3−4/3v2=0. The functional ϖ(z)≔1 + W(z) can be expanded by (see Figure 2)

ϖ(z)=ω(z)−z4/6−z5/10−z6/15+O(z7).

We shall use ϖ(z) to define some interesting classes of analytic functions.

3. Computations

This section deals with some computational outcomes utilizing a sigmoid function. Note that a sigmoid function is bounded analytic in convex complex domain (see Figure 3).

Theorem 3.1.

Suppose that χ ∈ ∧ achieves the inequality

1+μzΨ′(z)[Ψ(z)]k≺21+e−z,k=0,1,2,

where Ψ(z)=ς(z)z2χ′′(z)χ(z)+zχ′(z)χ(z),z∈∪. Then

Ψ(z)≺ρe(z)=zez−1,z∈∪

when μ ≥ max μ_k.

• (1)maxμ0=max3572(e−1),(35(e−1))(72(e−2))≈1.1628.

• (2)maxμ1=max{−35(72(log(e−1)−1)),35(72⁡log(e−1))}≈1.1.

• (3)  maxμ2=max{35e72,3572(e−2)}≈1.321.

Proof.

Case [A]: assume that k=0⇒1+μzΨ′(z)≺21+e−z.

Formulate a function Xμ:∪→C by the structure

Xμ(z)=1+1μz2−z372+z51200+…,

where

∫0zeζ−1ζ(1+eζ)=

z2−z372+z51200−(17z7)282240+(31z9)6531840−(691z11)1756339200+O(z13)+constant.

It is clear that X_μ(z) is an analytic solution of

(3.1)1+μzXμ′(z)=21+e−z,z∈∪.

Consider the functional U(z)≔μzXμ′(z)=21+e−z−1=e−z−1e−z+1, which is starlike in ∪ [19]. This implies that for G(z)≔U(z)+1, we have

RzU′(z)U(z)=RzG′(z)U(z)>0.

Consequently, Miller-Mocanu Lemma [18] indicates that

1+μzΨ′(z)≺1+μzXμ′(z)⇒Ψ(z)≺Xμ(z).

To end this organization, we aim to show that X_μ(z) ≺ ρ_e(z). Obviously, X_μ(z) increases in (−1, 1) for some μ that is fulfilling

1+(−1+2ee−1+log(2)+log(e−1)−log(3e−1))μ≈1+−2.8μ≤Xμ(−1)≤Xμ(1)≈1+(1+e)(e−1)+2⁡tanh(−1)(1+e)(3−5e)/μ≈1+1.4306μ.

Since the function ρ_e(z) fulfills the relation

(e−1)−1≤R(ρe(z))≈1−cos(ϑ)2+∑n=1∞β2n⁡cos(2nϑ)(2n)!≤e(e−1)−1,

then consequently, we arrive at the inequality

(e−1)−1≤Xμ(−1)≤Xμ(1)≤e(e−1)−1

whenever μ satisfies

μ≥maxμ0=max3572(e−1),(35(e−1))(72(e−2))≈1.1628.

Consequently, we obtain

Xμ(z)≺zez−1⇒Ψ(z)≺zez−1.

Case [B]: consume the case k=1⇒1+μzΨ′(z)Ψ(z)≺21+e−z.

Formulate a function Yμ:∪→C by the equation

Yμ(z)=exp1μz2−z372+z51200+….

Clearly, we have a solution Y_μ(z) (Y_μ(0) = 1) of the differential equation

(3.2)1+μzYμ′(z)Yμ(z)=21+e−z,z∈∪.

Consider the starlike function G(z)=2/(1+e−z)−1 then the functional H(z)=G(z)+1 implies RzH′(z)G(z)=RzG′(z)G(z)>0. Again Miller-Mocanu Lemma gives

1+μzΨ′(z)Ψ(z)≺1+μzYμ′(z)Yμ(z)⇒Ψ(z)≺Yμ(z).

Proceeding, we have

(e−1)−1≤Yμ(−1)≤Yμ(1)≤e(e−1)−1

if μ when

μ≥maxμ1=max{−35(72(log(e−1)−1)),35(72⁡log(e−1))}≈1.1.

This implies

Yμ(z)≺zez−1⇒Ψ(z)≺zez−1,z∈∪.

Case [C]: assume that k=2⇒1+μzΨ′(z)Ψ2(z)≺21+e−z.

The function

Dμ(z)=1−1μz2−z372+z51200+…−1.

is a solution for the differential equation

(3.3)1+μzDμ′(z)Dμ2(z)=21+e−z.

As a conclusion, Miller-Mocanu Lemma yields

1+μzΨ′(z)Ψ2(z)≺1+μzDμ′(z)Dμ2(z)⇒Ψ(z)≺Dμ(z).

Accordingly, we have

(e−1)−1≤Dμ(−1)≤Dμ(1)≤e(e−1)−1

if μ₂ recognizes the upper and lower bounds

μ≥maxμ2=max{35e72,3572(e−2)}≈1.321.

This indicates the relation

Dμ(z)≺zez−1⇒Ψ(z)≺zez−1,z∈∪.

□

Theorem 3.1 can be extended to functions in P. We omit the proof.

Theorem 3.2.

Let p∈P achieving the inequality

1+μzp′(z)[p(z)]k≺21+e−z,k=0,1,2,μ≥1.321.

Then

p(z)≺ρe(z)=zez−1,z∈∪.

We deal with the function ϱe(z)=ez−1z, which is convex univalent.

Theorem 3.3.

Consider the hypotheses of Theorem 3.1. Then

Ψ(z)≺ϱe(z).

when υ ≥ max υ_k.

• (1)      maxυ0=max{35(72(e−2)),(35e)72}≈1.321

• (2)     maxυ1=max{35(72⁡log(e−1)),−35(72(log(e−1)−1))}≈1.12.

• (3)    maxυ2=max{(35(e−1))(72(e−2)),3572(e−1)}≈1.162.

Proof.

Clearly, we have (e−1)/e≤R(ϱe(z))≤e−1. Consequently, we obtain (e − 1)/e ≤ X_υ(−1) ≤ X_υ(1) ≤ e − 1 whenever υ satisfies

υ≥maxυ0=maxυ0=max{35(72(e−2)),(35e)72}≈1.321.

This implies the relation

Xυ(z)≺ez−1z⇒Ψ(z)≺ez−1z.

In the same manner, we get

υ≥maxυ1=max{35(72⁡log(e−1)),−35(72(log(e−1)−1))}≈1.12

Consequently, we obtain

Yυ(z)≺ez−1z⇒Ψ(z)≺ez−1z.

Finally, we have

υ≥maxυ2=max{(35(e−1))(72(e−2)),3572(e−1)}≈1.162.

This implies that the result

Dυ(z)≺ez−1z⇒Ψ(z)≺ez−1z.

□

Theorem 3.3 can be generalized by utilizing p∈P. The proof is similar to the above proof.

Theorem 3.4.

Suppose that p∈P satisfies

1+μzp′(z)[p(z)]k≺21+e−z,k=0,1,2,μ≥1.321.

Then

p(z)≺ϱe(z)=ez−1z,z∈∪.

Next result indicates the upper bound:

J(z)=1+Az1+Bz,(−1≤B<A≤1),

(bi-linear transformation) which is starlike function with positive real part.

Theorem 3.5.

Consider one of the following inequalities

1. 1+ℓzΨ′(z)≺z+1,ℓ≥max{ℓ0,ℓ1}, where

   ℓ0=2(0.22599B+0.22599)(A−B),B+1≠0,A−B≠0;

   and

   ℓ1=2((B−1)(log(2)−1))(A−B),B−1≠0,A−B≠0.

2. 1+ℓzΨ′(z)Ψ(z)≺z+1,ℓ≥max{ℓ2,ℓ3}, where

   ℓ2=2(i(−1+(2)+log(2)−log(1+(2))))2πn−i⁡logB+1A+1,

   B+1≠0,A+1≠0,logB+1A+1+2πni≠0

   and

   ℓ3=−2(i(log(2)−1))2πn−i⁡logA−1B−1;logA−1B−1+2iπn≠0,A≠1,B≠1;

3. 1+ℓzΨ′(z)Ψ2(z)≺z+1,ℓ≥max{ℓ4,ℓ5}, where

   ℓ4=2(0.225987A+0.225987)(A−B),B+1≠0,A≠B;

   ℓ5=2((A−1)(log(2)−1))(A−B),B−1≠0A≠B.

Then Ψ(z)≺1+Az1+Bz,(−1≤B<A≤1).

Proof.

Case [A]: Let k=0⇒1+ℓzΨ′(z)≺z+1.

Define a function Fℓ:∪→C admitting the structure

Fℓ(z)=1+2ℓz+1−log(1+z+1)−1+log(2).

It is clear that F_ℓ(z) is analytic in ∪ satisfying F_ℓ(0) = 1, and it is a solution of the differential equation

(3.4)1+ℓzFℓ′(z)=z+1,z∈∪.

Therefore, this yields U(z)≔ℓzFℓ′(z)=z+1−1 is starlike in ∪. So in view of Miller-Mocanu Lemma, we get

1+μzΨ′(z)≺1+ℓzFℓ′(z)⇒Ψ(z)≺Fℓ(z).

To end this argument, we must show that F_ℓ(z) ≺ J(z). Evidently, F_ℓ(z) increases in (−1, 1), such that F_ℓ(−1) ≤ F_ℓ(1). Since

1−A1−B≤Fℓ(−1)≤Fℓ(1)≤1+A1+B

whenever ℓ ≥ max{ℓ₀, ℓ₁} where

ℓ0=2(0.22599B−0.22599)(A−B),B+1≠0,A−B≠0;

and

ℓ1=2((B−1)(log(2)−1))(A−B),B−1≠0,A−B≠0.

Consequently, we obtain

Fℓ(z)≺J(z)⇒Ψ(z)≺J(z),z∈∪.

Case [B]: assume that k=1⇒1+ℓzΨ′(z)Ψ(z)≺z+1.

The function

Sℓ(z)=exp2ℓz+1−log(1+z+1)−1+log(2).

is a solution of the differential equation

(3.5)1+ℓzSℓ′(z)Sℓ(z)=z+1,z∈∪.

Then again, in virtue of the Miller-Mocanu Lemma, we arrive at

1+μzΨ′(z)Ψ(z)≺1+ℓzSℓ′(z)Sℓ(z)⇒Ψ(z)≺Sℓ(z).

Thus, we obtain

1−A1−B≤Sℓ(−1)≤Sℓ(1)≤1+A1+B

whenever ℓ ≥ max{ℓ₂, ℓ₃} where

ℓ2=(2i(−1+(2)+log(2)−log(1+(2))))2πn−i⁡logB+1A+1,

B+1≠0,A+1≠0,logB+1A+1+2πni≠0

and

ℓ3=−2(i(log(2)−1))2πn−i⁡logA−1B−1;logA−1B−1+2iπn≠0,A≠1,B≠1.

This indicates the relations

Sℓ(z)≺J(z)⇒Ψ(z)≺J(z),z∈∪.

Case[C]: suppose that k=2⇒1+ℓzΨ′(z)Ψ2(z)≺z+1.

The function

Qℓ(z)=1−2ℓz+1−log(1+z+1)−1+log(2)−1

is a solution of the differential equation

(3.6)1+μzQℓ′(z)Qℓ(z)=z+1,z∈∪.

Clearly, Miller-Mocanu Lemma implies

1+ℓzΨ′(z)Ψ2(z)≺1+ℓzQℓ′(z)Qℓ2(z)⇒Ψ(z)≺Qℓ(z).

Accordingly, we have

1−A1−B≤Qℓ(−1)≤Qℓ(1)≤1+A1+B

if ℓ₂ recognizes the upper and lower bounds

ℓ4=2(0.225987A+0.225987)(A−B),B+1≠0,A≠B;

ℓ5=2((A−1)(log(2)−1))(A−B),B−1≠0A≠B.

This brings that

Qℓ(z)≺J(z)⇒Ψ(z)≺J(z).

□

Note that, in Theorem 3.5, we can replace Ψ(z) by the general function p(z)∈P to get p(z) ≺ J(z). We advance to extant the upper bound result of Eq. (2.3) by the singular function λ(z) = 1 + sin(z), z ∈ ∪, where it is with positive real part. The proof is quite similar to Theorem 3.5; therefore, we omit it.

Theorem 3.6.

Consider one of the following inequalities

1. 1+τzΨ′(z)≺z+1,τ≥max{τ0,τ1}, where

   τ0=2(−1+(2)+log(2)−log(1+(2)))csc(1);

   and

   τ1=2(log(2)−1)(−csc(1)).

2. 1+τzΨ′(z)Ψ(z)≺z+1,τ≥max{τ2,τ3}, where

   τ2=2(−1+(2)+log(2)−log(1+(2)))log(1+sin(1)),

   and

   τ3=2(log(2)−1)log(1−sin(1)).

3. 1+τzΨ′(z)Ψ2(z)≺z+1,τ≥max{τ4,τ5}, where

   τ4=2(−1+(2)+log(2)−log(1+(2)))(1+csc(1));

   τ5=2(log(2)−1)(−(csc(1)−1)).

Then Ψ(z) ≺ 1 + sin(z), z ∈ ∪ .

By using the technique of Theorem 3.5, we have the following result using ϖ(z).

Theorem 3.7.

Consider one of the following inequalities

1. 1+ℓzΨ′(z)≺z+1,ℓ≥max{ℓ0,ℓ1}={−104(log(2)−1),21(log(27)−6)(log(8)−4)}≈2.816;

2. 1+ℓzΨ′(z)Ψ(z)≺z+1,ℓ≥max{ℓ2,ℓ3}={2(1−log(2))log(5),2(log(3)−2)(−1−log(2)+log(3))}≈3;

3. 1+ℓzΨ′(z)Ψ2(z)≺z+1,ℓ≥max{ℓ4,ℓ5}={2−(4(log(3)−2))(log(16)−3),2(2−log(3))(log(2)−5)}≈−0.4.

Then

Ψ(z)≺ϖ(z)=1+z−z3/3−z4/6−z5/10−z6/15+O(z7),z∈∪.

In Theorems 3.6 and 3.7, one can replace Ψ(z) by p(z) to get more general results p(z) ≺ 1 + sin(z) and p(z) ≺ ϖ(z) respectively.

4. Conclusion

From above, we conclude that LDEs can be recognized in terms of a complex variable z ∈ ∪. We illustrated a list of sufficient conditions for the existence of holomorphic univalent solutions. Our next study will be considered for a generalized class of analytic functions in the open unit disk.

Figures

Figure 1

The plot of ω(z)

Figure 2

The plot of ϖ(z)

Figure 3

The plot of sigmoid function s(z)=21+e−z, where its max is at 2e1+e, and its min is at 21+e