In this paper, the stability and bifurcation of a two–dimensional p53 gene regulatory network without and with time delay are taken into account by rigorous theoretical analyses and numerical simulations. In the absence of time delay, the existence and local stability of the positive equilibrium are considered through the Descartes' rule of signs, the determinant and trace of the Jacobian matrix, respectively. Then, the conditions for the occurrence of codimension–1 saddle–node and Hopf bifurcation are obtained with the help of Sotomayor's theorem and the Hopf bifurcation theorem, respectively, and the stability of the limit cycle induced by hopf bifurcation is analyzed through the calculation of the first Lyapunov number. Furthermore, codimension-2 Bogdanov–Takens bifurcation is investigated by calculating a universal unfolding near the cusp. In the presence of time delay, we prove that time delay can destabilize a stable equilibrium. All theoretical analyses are supported by numerical simulations. These results will expand our understanding of the complex dynamics of p53 and provide several potential biological applications.
Citation: Xin Du, Quansheng Liu, Yuanhong Bi. Bifurcation analysis of a two–dimensional p53 gene regulatory network without and with time delay[J]. Electronic Research Archive, 2024, 32(1): 293-316. doi: 10.3934/era.2024014
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In this paper, the stability and bifurcation of a two–dimensional p53 gene regulatory network without and with time delay are taken into account by rigorous theoretical analyses and numerical simulations. In the absence of time delay, the existence and local stability of the positive equilibrium are considered through the Descartes' rule of signs, the determinant and trace of the Jacobian matrix, respectively. Then, the conditions for the occurrence of codimension–1 saddle–node and Hopf bifurcation are obtained with the help of Sotomayor's theorem and the Hopf bifurcation theorem, respectively, and the stability of the limit cycle induced by hopf bifurcation is analyzed through the calculation of the first Lyapunov number. Furthermore, codimension-2 Bogdanov–Takens bifurcation is investigated by calculating a universal unfolding near the cusp. In the presence of time delay, we prove that time delay can destabilize a stable equilibrium. All theoretical analyses are supported by numerical simulations. These results will expand our understanding of the complex dynamics of p53 and provide several potential biological applications.
Let Σ(a,k) be the class of meromorphic functions f of the form
f(z)=az+∞∑n=kanzn(a>0,k≥2,k∈N), | (1.1) |
which is analytic in the punctured open unit disk U∗={z∈C:0<|z|<1}=U∖{0}.
In 2009, El-Ashwah [1] introduced the subclass ΣS∗k(a,β) of generalized meromorphically starlike of order β as follows,
ΣS∗k(a,β)={f(z)∈Σ(a,k):−Rezf′(z)f(z)>β,β∈[0,1)}. |
An analytic function g:U={z:|z|<1}→C is subordinate to an analytic function h:U→C, if there is a function ω satisfying ω(0)=0and|ω(z)|<1(z∈U), such that g(z)=h(ω(z))(z∈U). Note that g(z)≺h(z). Especially, if h is univalent in U, then the following conclusion is true (see [2]):
g(z)≺h(z)⟺g(0)=h(0)andg(U)⊂h(U). |
Using the subordinate relationship, we introduce the following subclass of generalized meromorphic Janowski functions (see [3,5,6]).
Definition 1. Let f(z)∈Σ(a,k),a>0,A,B∈R,|A|≤1,|B|≤1 and A≠B. The function f∈ΣS∗k(a,A,B) if and only if
−zf′(z)f(z)≺1+Az1+Bz(z∈U∗). |
It is obvious that
ΣS∗k(a,A,B)⊂ΣS∗k(a,1−A1−B)(−1≤B<A≤1). |
According to the subordination relationship, the function f(z)∈ΣS∗k(a,A,B) if and only if there exists an analytic function ω(z) in the open unit dick U satisfying ω(0)=0 and |ω(z)|<1, such that
−zf′(z)f(z)=1+Aw(z)1+Bw(z)(z∈U∗), |
which is equivalent to
|zf′(z)+f(z)Af(z)+Bzf′(z)|<1(z∈U∗). | (1.2) |
Let T(a,k) be the subclass of Σ(a,k), the function f belonging to T(a,k) of the following form
f(z)=az−∞∑n=k|an|zn. | (1.3) |
Let
¯ΣS∗k(a,A,B)=T(a,k)∩ΣS∗k(a,A,B). |
Next, we introduce the generalized λ-Hadamard product of the class T(a,k).
Definition 2. Let a>0,p>0,q>0,λ≥0,k∈N,k≥2,fi(z)=az−∑∞n=k|an,j|zn∈T(a,k)(i=1,2). The generalized λ-Hadamard product (f1˜△f2)λ(p,q,a;z) of the function f1 and f2 is defined by
(f1˜Δf2)λ(p,q,a;z)=(1−λ)(f1Δf2)(p,q,a;z)−λz(f1Δf2)′(p,q,a;z), | (1.4) |
where
(f1Δf2)(p,q,a;z)=a2z−∞∑n=k|an,1|p|an,2|qzn. | (1.5) |
From (1.4) and (1.5), we get
(f1˜Δf2)λ(p,q,a;z)=a2z−∞∑n=k[1−(n+1)λ]|an,1|p|an,2|qzn. |
Selecting different parameters a,λ,p and q, we can obtain the following three special convolutions:
(ⅰ) For λ=0, (f1˜Δf2)0(p,q,a;z) is the generalized Hadamard product:
(f1˜Δf2)0(p,q,a;z)=:(f1Δf2)(p,q,a;z)=a2z−∞∑n=k|an,1|p|an,2|qzn. |
In particular, for a=1, the Hadamard product (f1Δf2)(p,q,1;z)=1z−∑∞n=k|an,1|p|an,2|qzn was studied by Janowski [3] and Tang et al. [6].
(ⅱ) For λ=0,p=q=1, (f1˜Δf2)0(1,1,a;z) is the general Hadamard product:
(f1˜Δf2)0(1,1,a;z)=:(f1∗f2)(a;z)=a2z−∞∑n=k|an,1||an,2|zn. |
In particular, for a=1, (f1∗f2)(1;z) is the well-known Hadamard product (see [1,4,7]):
(f1∗f2)(1;z)=(f1∗f2)(z)=1z−∞∑n=k|an,1||an,2|zn. |
For a=1,k=1, f1(z)=z−1+∑∞n=1(a)n(c)nzn,f2(z)=az−∑∞n=1|an,2|zn, (a)n=a(a+1)⋯(a+n−1)(n∈N), (f1∗f2)(1;z)=1z−∑∞n=1(a)n(c)n|an,2|zn was studied by Liu et al. [8] (see also [9]).
(ⅲ) For p=q=1, (f1˜Δf2)λ(1,1,a;z)=(f1¯∗f2)λ(z) is λ−Hadamard product:
(f1¯∗f2)λ(z)=(1−λ)(f1∗f2)+λz(f1∗f2)′(z)=a2z−∞∑n=k[1−(n+1)λ]|an,1||an,2|zn. |
In 1996, Choi et al. [4] studied the closure properties of the generalized λ-Hadamard product for univalent functions. In 2001, Liu et al. [8] investigated two new classes of meromorphically multivalent functions by using a linear operator. Based on this, in this paper, we mainly consider the closure properties of the generalized λ-Hadamard product for meromorphic functions. Firstly, we obtain the necessary and sufficient conditions of the class ¯ΣS∗k(a,A,B) by using subordination relationship. Secondly, we discuss the closure properties of the general Hadamard product, the generalized Hadamard product and λ-Hadamard product for functions of the class ¯ΣS∗k(a,A,B). Finally, we discuss the closure properties of the generalized λ-Hadamard product and generalize the known conclusions.
Unless otherwise noted, the parameters a,k,A and B satisfy the condition: a>0,k∈N,k≥2,A,B∈R,|A|≤1,|B|≤1 and A≠B.
In order to establish our results, we need the following lemmas.
Lemma 1. Let f be of the form (1.1). If
∞∑n=k[(n+1)+|A+nB|]|an|≤a|A−B|, | (2.1) |
then f(z)∈ΣS∗k(a,A,B).
Proof. Let |z|=1. By (2.1), we obtain
|zf′(z)+f(z)|−|Af(z)+Bzf′(z)|≤|∞∑n=k(n+1)|an|zn|−|az(A−B)|+|∞∑n=k(A+nB)|an|zn|≤∞∑n=k(n+1)|an|−a|A−B|+∞∑n=k|A+nB||an|=∞∑n=k[(n+1)+|A+nB|]|an|−a|A−B|≤0. |
Using the principle of maximum modulus, we get
|zf′(z)+f(z)|−|Af(z)+Bzf′(z)|<0(z∈U∗). |
Then we conclude that f(z)∈ΣS∗k(a,A,B). Therefore, we complete the proof of Lemma 1.
Lemma 2. Let the function f be of the form (1.3).
(ⅰ) If 0≤B<A≤1. Then the function f(z)∈¯ΣS∗k(a,A,B) if and only if
∞∑n=k[(n+1)+(A+nB)]|an|≤a(A−B). | (2.2) |
(ⅱ) If −1≤A<B≤0. Then the function f(z)∈¯ΣS∗k(a,A,B) if and only if
∞∑n=k[(n+1)−(A+nB)]|an|≤a(B−A). | (2.3) |
Proof. It appears from (1.3) that ¯ΣS∗k(a,A,B)⊂ΣS∗k(a,A,B). In view of Lemma 1, we just need prove the necessity part of Lemma 2. Let f(z)∈¯ΣS∗k(a,A,B). By (1.2), we have
|zf′(z)+f(z)Af(z)+Bzf′(z)|=|∑∞n=k(n+1)|an|zn−1a(A−B)−∑∞n=k(A+nB)|an|zn−1|<1. |
Since |Rez|≤|z| holds true for all z∈C. If 0≤B<A≤1, then
Re{∑∞n=k(1+n)|an|zn−1a(A−B)−∑∞n=k(A+nB)|an|zn−1}<1. | (2.4) |
Let z=r→1−. From (2.4), we can get (2.2). Using the same method, we can obtain the result of (ⅱ). The estimates of (ⅰ) and (ⅱ) are sharp for the function f given by
f(z)=az−a|A−B|(k+1)+|A+kB|zk,k≥2. | (2.5) |
So, we complete the proof of Lemma 2.
First of all, we discuss the closure properties of the general Hadamard product of the class ¯ΣS∗k(a,A,B).
Theorem 1. Let fi(z)=az−∑∞n=k|an|zn∈¯ΣS∗k(a,A,B)(i=1,2). If A and B satisfy the condition
0≤B<A≤1or−1≤A<B≤0, |
then 1a(f1∗f2)(a;z)∈¯ΣS∗k(a,A,B).
Proof. Let 0≤B<A≤1. If fi(z)∈¯ΣS∗k(a,A,B)(i=1,2), from (2.2), we obtain
∞∑n=k[(n+1)+(A+nB)]|an,1|≤a(A−B) | (3.1) |
and
∞∑n=k[(n+1)+(A+nB)]|an,2|≤a(A−B). | (3.2) |
To obtain 1a(f1∗f2)(a;z)∈¯ΣS∗k(a,A,B), we just prove
∞∑n=k1a[(n+1)+(A+nB)]|an,1||an,2|≤a(A−B). | (3.3) |
By using Cauchy-Schwarz inequality, from (3.1) and (3.2), we have
∞∑n=k[(n+1)+(A+nB)]√|an,1||an,2|≤a(A−B). | (3.4) |
According to (3.3) and (3.4), we only need to prove
√|an,1||an,2|≤a. |
From (3.4) and the condition 0≤B<A≤1, we have
√|an,1||an,2|≤a(A−B)(n+1)+(A+nB)≤a. |
Therefore, we conclude that 1a(f1∗f2)(a;z)∈¯ΣS∗k(a,A,B).
For −1≤A<B≤0, using the same method above, we get 1a(f1∗f2)(a;z)∈¯ΣS∗k(a,A,B). Therefore, we complete the proof of Theorem 1.
Next, we prove the closure properties of the generalized Hadamard product.
Theorem 2. Let fi(z)=az−∑∞n=k|an|zn∈¯ΣS∗k(a,A,B)(i=1,2) and p>1. If anyone of the following conditions holds true:
(ⅰ) For 0≤B<A≤1, ˆB and a satisfy
{0≤ˆB≤aA[(k+1)+(A+kB)]−(A−B)(k+1+A)a[(k+1)+(A+kB)]+(A−B)k,0<a<1,ˆB≤aA(1+B)−(A−B)a(1+B)+(A−B),a>1. |
(ⅱ) For −1≤A<B≤0, −1≤A<ˆB≤0 and a satisfies a>0. Then 1a(f1Δf2)(1p,p−1p;a,z)∈¯ΣS∗k(a,A,ˆB).
Proof. (ⅰ) For 0≤B<A≤1, by Lemma 2, we can get
∞∑n=k(n+1)+(A+nB)a(A−B)|an,i|≤1(i=1,2). |
We deduce that
{∞∑n=k(n+1)+(A+nB)a(A−B)|an,1|}1p≤1 | (3.5) |
and
{∞∑n=k(n+1)+(A+nB)a(A−B)|an,2|}p−1p≤1. | (3.6) |
According to (1.5), we have
(f1Δf2)(1p,p−1p)=a2z−∞∑n=k|an,1|1p|an,2|p−1pzn. |
To prove 1a(f1∗f2)(1p,p−1p,a;z)∈¯ΣS∗k(a,A,ˆB), we only need to show
∞∑n=k(n+1)+(A+nˆB)a2(A−ˆB)|an,1|1p|an,2|p−1p≤1. | (3.7) |
Applying Hölder inequality, from (3.5) and (3.6), we obtain
∞∑n=k(n+1)+(A+nB)a(A−B)|an,1|1p|an,2|p−1p≤1. |
Thus, (3.7) holds if
(n+1)+(A+nˆB)a2(A−ˆB)≤(n+1)+(A+nB)a(A−B)(∀n∈N,n≥k), |
that is,
ˆB[a[(n+1)+(A+nB)]A−B+n]≤aA[(n+1)+(A+nB)]A−B−(n+1+A). | (3.8) |
Let
F1(n)=a[(n+1)+(A+nB)]A−B+n |
and
G1(n)=aA[(n+1)+(A+nB)]A−B−(n+1+A). |
It is easy to see F1(n)>0 holds true for n≥k. To obtain (3.8), we only need to prove
ˆB≤G1(n)F1(n)=aA[(n+1)+(A+nB)]−(A−B)(n+1+A)a[(n+1)+(A+nB)]+(A−B)n. |
Because G1(n)F1(n) is an increasing function with respect to n and the condition (i) holds true, so we have
ˆB≤aA[(k+1)+(A+kB)]−(A−B)(k+1+A)a[(k+1)+(A+kB)]+(A−B)k=G(k)F(k)≤G(n)F(n). |
Therefore, we conclude that 1a(f1Δf2)(1p,p−1p;a,z)∈¯ΣS∗k(a,A,ˆB).
(ⅱ) It is similar to the proof of (ⅰ). For −1≤A<B≤0, we only need to show
(n+1)−(A+nˆB)a2(ˆB−A)≤(n+1)−(A+nB)a(B−A)(∀n∈N,n≥k), |
that is,
ˆB[a[(n+1)−(A+nB)]B−A+n]≥aA[(n+1)−(A+nB)]B−A+(n+1−A). | (3.9) |
Let
F2(n)=a[(n+1)−(A+nB)]B−A+n |
and
G2(n)=aA[(n+1)−(A+nB)]B−A+(n+1−A). |
It is easy to see F2(n)>0 holds true for n≥k. To make (3.9) establish, we only need to prove
ˆB≥G2(n)F2(n)=aA[(n+1)−(A+nB)]+(B−A)(n+1−A)a[(n+1)+(A+nB)]+(B−A)n. | (3.10) |
It is clear that G2(n)F2(n) is an decreasing function with respect to n for 0<a<1 and
A>aA[(k+1)−(A+kB)]−(B−A)(k+1−A)a[(k+1)−(A+kB)]+(B−A)k, |
then we have
ˆB≥aA[(k+1)−(A+kB)]−(B−A)(k+1−A)a[(k+1)−(A+kB)]+(B−A)k=G2(k)F2(k)≥G2(n)F2(n). |
It can be concluded that 1a(f1Δf2)(1p,p−1p;a,z)∈¯ΣS∗k(a,A,ˆB). Clearly, G2(n)F2(n) is an increasing function with respect to n for a>1 and
limn→+∞G2(n)F2(n)=aA(1−B)+(B−A)a(1−B)+(B−A). |
Since A>aA(1−B)+(B−A)a(1−B)+(B−A), we have
ˆB>A>aA(1−B)+(B−A)a(1−B)+(B−A)=limn→+∞G2(n)F2(n)≥G2(n)F2(n). |
Thus, we conclude that 1a(f1Δf2)(1p,p−1p;a,z)∈¯ΣS∗k(a,A,ˆB). The proof of Theorem 2 is completed.
Putting A=1,B=0 in Theorem 2, we obtain the following corollary.
Corollary 1. Let fi(z)=az−∑∞n=k|an,i|zn∈¯ΣS∗k(a,1,0)(i=1,2) and p>1. If ˆB and a satisfy
{0≤ˆB≤(a−1)(k+2)a(k+2)+k,0<a<1,ˆB≤a−1a+1,a>1. |
Then 1a(f1△f2)(1p,p−1p;a,z)∈¯ΣS∗k(a,A,ˆB). Finally, we consider the closure properties of λ-Hadamard product (f1¯∗f2)λ(z) and the generalized λ-Hadamard product (f1˜Δf2)λ(p.q,a;z) as follows.
Theorem 3. Let fi(z)=az−∑∞n=k|an,i|zn∈¯ΣS∗k(a,A,B)(i=1,2). If A,B and λ satisfy
{(A−B)−(k+1−A−nB)(A−B)(k+1)<λ<1k+1,0≤B<A≤1,λ<1k+1,−1≤A<B≤0. |
Then 1a(f1¯∗f2)λ(z)∈¯ΣS∗k(a,A,B).
Proof. Let fi(z)∈¯ΣS∗k(a,A,B)(i=1,2). If 0≤B<A≤1, from Lemma 2, we obtain
∞∑n=k((n+1)+(A+nB))|an,1|≤a(A−B), | (3.11) |
and
∞∑n=k((n+1)+(A+nB))|an,2|≤a(A−B). | (3.12) |
By Definition 2, we have
1a(f1ˉ∗f2)λ(z)=az−∞∑n=k1a(1−(n+1)λ|an,1||an,2|zn. |
To show 1a(f1ˉ∗f2)λ(z)∈¯ΣS∗k(a,A,B), we just prove
∞∑n=k1a(1−(n+1)λ)((n+1)+(A+nB))|an,1||an,2|≤a(A−B). | (3.13) |
Applying Cauchy-Schwarz inequality to (3.11) and (3.12), we get
∞∑n=k((n+1)+(A+nB))√|an,1||an,2|≤a(A−B). | (3.14) |
According to (3.13) and (3.14), we only need to prove
√|an,1||an,2|≤a1−(n+1)λ. |
Since √|an,1||an,2|≤a(A−B)(n+1)+(A+nB), we only need to prove
A−B≤(n+1)+(A+nB)1−(n+1)λ,n≥k. | (3.15) |
Let
M(n)=(n+1)+(A+nB)1−(n+1)λ. |
Clearly, M(n) is an increasing function for n≥k. This implies that (3.15) holds true if
k+1+A+nB1−(k+1)λ≥A−B. |
For λ<1k+1, the above formula can be expressed as λ≥(A−B)−(k+1−A−nB)(A−B)(k+1).
To summarize, we conclude that 1a(f1¯∗f2)λ(z)∈¯ΣS∗k(a,A,B) for (A−B)−(k+1−A−nB)(A−B)(k+1)<λ<1k+1. Using the same method, we can get 1a(f1¯∗f2)λ(z)∈¯ΣS∗k(a,A,B) for
λ<min{(B−A)−(k+1−A−nB)(B−A)(k+1),1k+1}=1k+1. |
Thus, we complete the proof of Theorem 3.
Theorem 4. Let fi(z)=az−∑∞n=k|an|zn∈¯ΣS∗k(a,A,B)(i=1,2) and p>1.
(1) For 0≤B<A≤1,0≤ˆB<A. If anyone of the following conditions holds true.
(ⅰ) a<1,λ<min{12(n+1),a(1+B)+(A−B)(2n+1)(A−B),k(A−B)+a(k+1+A+kB)k(k+1)(A−B)},
(ⅱ) a>1,12(k+1)<λ<min{a(1+B)+(A−B)(2n+1)(A−B),k(A−B)+a(k+1+A+kB)k(k+1)(A−B)},
(ⅲ) a<1,max{a(1+B)+(A−B)(2k+1)(A−B),k(A−B)+a(k+1+A+kB)k(k+1)(A−B)}<λ<12(n+1),
(ⅳ) a>1,max{12(k+1),a(1+B)+(A−B)(2k+1)(A−B),k(A−B)+a(k+1+A+kB)k(k+1)(A−B)}<λ.
Then the generalized λ-Hadamard product: 1a(f1˜Δf2)λ(1p,p−1p,a;z)∈¯ΣS∗k(a,A,B).
(2) For −1≤A<B≤0. If anyone of the following conditions holds true.
(ⅴ) a<1,λ<min{12(n+1),(B−A)+a(1−B)(2k−1)(B−A),k(B−A)+a(k+1−A−kB)k(k+1)(B−A)},
max(A,aA[(k+1)−(A+kB)]+[1−(k+1)λ](k+1+A)(B−A)a[(k+1)−(A+kB)]−k[1−(k+1)λ](B−A))≤ˆB≤0,
(ⅵ) a>1,12(k+1)<λ<min{(B−A)+a(1−B)(2k−1)(B−A),k(B−A)+a(k+1−A−kB)k(k+1)(B−A)},
max(A,aA[(k+1)−(A+kB)]+[1−(k+1)λ](k+1+A)(B−A)a[(k+1)−(A+kB)]−k[1−(k+1)λ](B−A))≤ˆB≤0,
(ⅶ) a<1,max{(B−A)+a(1−B)(2k−1)(B−A),k(B−A)+a(k+1−A−kB)k(k+1)(B−A)}<λ<12(n+1),A≤ˆB<0,
(ⅷ) a>1,max{12(k+1),(B−A)+a(1−B)(2k−1)(B−A),k(B−A)+a(k+1−A−kB)k(k+1)(B−A)}<λ,A≤ˆB<0.
Then the generalized λ-Hadamard product: 1a(f1˜Δf2)λ(1p,p−1p,a;z)∈¯ΣS∗k(a,A,B).
Proof. Let fi(z)∈¯ΣS∗k(a,A,B)(i=1,2). The method for the proof of Theorem 4 is similar to that of Theorem 2. If −1≤A<B≤0, we just need to prove
[1−(n+1)λ][(n+1)−(A+nˆB)]a2(A−ˆB)≤(n+1)−(A−nB)a(B−A), | (3.16) |
that is
ˆB[a[(n+1)−(A+nB)]B−A+n(1−(n+1)λ)]≤aA[(n+1)−(A+nB)]B−A+[1−(n+1)λ](n+1−A). |
Let
P(n)=a[(n+1)−(A+nB)]B−A+n[1−(n+1)λ] |
and
Q(n)=aA[(n+1)+(A+nB)]B−A+[1−(n+1)λ](n+1−A). |
The following discussion can be divided into two cases:
(a) It is easy to see that P(n)>0 if λ<(B−A)+a(1−B)(B−A)(2n−1). To prove that (3.16) holds true, we need only to prove the following inequality
ˆB≤Q(n)P(n)=aA[(n+1)−(A+nB)]B−A+[1−(n+1)λ](n+1−A)a[(n+1)−(A+nB)]B−A+n[1−(n+1)λ]. | (3.17) |
Clearly, Q(n)P(n) is an increasing function with respect to n for a<1,λ<12(n+1) or a>1,λ>12(k+1). If λ<k(B−A)+a(k+1−A−kB)k(k+1)(B−A), then we have
ˆB≤aA[(k+1)−(A+kB)]+[1−(k+1)λ](k+1−A)(B−A)a[(k+1)−(A+kB)]+k[1−(k+1)λ](B−A)=Q(k)P(k)≤Q(n)P(n). |
Therefore, 1a(f1˜△f2)λ(1p,p−1p,a,z)∈¯ΣS∗k(a,A,B) if the conditions (v) or (vi) are satisfied.
(b) It is easy to see that P(n)<0 if λ>(B−A)+a(1−B)(B−A)(2n−1) and P(k)=a[(k+1)−(A+kB)]B−A+k[1−(k+1)λ]>0. To prove that (3.16) holds true, we need only to prove the following inequality
ˆB≤Q(n)P(n)=aA[(n+1)−(A+nB)]A−B−[1−(n+1)λ])(n+1−A)a[(n+1)−(A+nB)]B−A+n[1−(n+1)λ]. | (3.18) |
Clearly, Q(n)P(n) is a decreasing function with respect to n for a<1,λ<12(n+1) or a>1,λ>12(k+1).
ˆB≤limn→+∞Q(n)P(n)≤Q(n)P(n). |
Therefore, 1a(f1˜△f2)λ(1p,p−1p,a,z)∈¯ΣS∗k(a,A,B) if the conditions (vii) or (viii) are satisfied.
To summarize, we conclude that the conclusion (2) is established. Using the same method to that in the proof of (2), the conclusion (1) holds true. Thus, we complete the proof of Theorem 4.
Let A=1 and B=0 in Theorem 4, we have the following corollary.
Corollary 2. Let fi(z)∈¯ΣS∗k(a,1,0)(i=1,2) and p>1. If anyone of the following conditions is satisfied.
(ⅰ) a<1,λ<min{12(n+1),a+1(2n+1),k+a(k+2)k(k+1)},
(ⅱ) a>1,12(k+1)<λ<min{a+1(2n+1),k+a(k+2)k(k+1)},
(ⅲ) a<1,max{a+1(2k+1),k+a(k+2)k(k+1)}<λ<12(n+1),
(ⅳ) a>1,max{12(k+1),a+1(2k+1),k+a(k+2)k(k+1)}<λ.
Then the generalized λ-Hadamard product (f1˜Δf2)λ(1p,p−1p,a;z)∈ ¯ΣS∗k(a,1−2β,ˆB).
This work is supported by the Natural Science Foundation of the People's Republic of China under Grant 11561001, the Natural Science Foundation of Inner Mongolia of the People's Republic of China under Grants 2018MS01026, 2019MS01023 and 2020MS01011, the Program for Young Talents of Science and Technology in Universities of Inner Mongolia Autonomous Region under Grant NJYT-18-A14 and the Higher-School Science Foundation of Inner Mongolia of the People's Republic of China under Grants NJZY20198 and NJZY20200.
All authors declare no conflict of interest in this paper.
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