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Research article Special Issues

Bifurcation analysis of a two–dimensional p53 gene regulatory network without and with time delay

  • Received: 29 August 2023 Revised: 19 November 2023 Accepted: 21 November 2023 Published: 25 December 2023
  • 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,k2,kN), (1.1)

    which is analytic in the punctured open unit disk U={zC:0<|z|<1}=U{0}.

    In 2009, El-Ashwah [1] introduced the subclass ΣSk(a,β) of generalized meromorphically starlike of order β as follows,

    ΣSk(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:UC, if there is a function ω satisfying ω(0)=0and|ω(z)|<1(zU), such that g(z)=h(ω(z))(zU). 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,BR,|A|1,|B|1 and AB. The function fΣSk(a,A,B) if and only if

    zf(z)f(z)1+Az1+Bz(zU).

    It is obvious that

    ΣSk(a,A,B)ΣSk(a,1A1B)(1B<A1).

    According to the subordination relationship, the function f(z)ΣSk(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)(zU),

    which is equivalent to

    |zf(z)+f(z)Af(z)+Bzf(z)|<1(zU). (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)=azn=k|an|zn. (1.3)

    Let

    ¯ΣSk(a,A,B)=T(a,k)ΣSk(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,λ0,kN,k2,fi(z)=azn=k|an,j|znT(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)=(1λ)(f1Δf2)(p,q,a;z)λz(f1Δf2)(p,q,a;z), (1.4)

    where

    (f1Δf2)(p,q,a;z)=a2zn=k|an,1|p|an,2|qzn. (1.5)

    From (1.4) and (1.5), we get

    (f1˜Δf2)λ(p,q,a;z)=a2zn=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)=a2zn=k|an,1|p|an,2|qzn.

    In particular, for a=1, the Hadamard product (f1Δf2)(p,q,1;z)=1zn=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)=:(f1f2)(a;z)=a2zn=k|an,1||an,2|zn.

    In particular, for a=1, (f1f2)(1;z) is the well-known Hadamard product (see [1,4,7]):

    (f1f2)(1;z)=(f1f2)(z)=1zn=k|an,1||an,2|zn.

    For a=1,k=1, f1(z)=z1+n=1(a)n(c)nzn,f2(z)=azn=1|an,2|zn, (a)n=a(a+1)(a+n1)(nN), (f1f2)(1;z)=1zn=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 λHadamard product:

    (f1¯f2)λ(z)=(1λ)(f1f2)+λz(f1f2)(z)=a2zn=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 ¯ΣSk(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 ¯ΣSk(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,kN,k2,A,BR,|A|1,|B|1 and AB.

    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|AB|, (2.1)

    then f(z)ΣSk(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(AB)|+|n=k(A+nB)|an|zn|n=k(n+1)|an|a|AB|+n=k|A+nB||an|=n=k[(n+1)+|A+nB|]|an|a|AB|0.

    Using the principle of maximum modulus, we get

    |zf(z)+f(z)||Af(z)+Bzf(z)|<0(zU).

    Then we conclude that f(z)ΣSk(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 0B<A1. Then the function f(z)¯ΣSk(a,A,B) if and only if

    n=k[(n+1)+(A+nB)]|an|a(AB). (2.2)

    (ⅱ) If 1A<B0. Then the function f(z)¯ΣSk(a,A,B) if and only if

    n=k[(n+1)(A+nB)]|an|a(BA). (2.3)

    Proof. It appears from (1.3) that ¯ΣSk(a,A,B)ΣSk(a,A,B). In view of Lemma 1, we just need prove the necessity part of Lemma 2. Let f(z)¯ΣSk(a,A,B). By (1.2), we have

    |zf(z)+f(z)Af(z)+Bzf(z)|=|n=k(n+1)|an|zn1a(AB)n=k(A+nB)|an|zn1|<1.

    Since |Rez||z| holds true for all zC. If 0B<A1, then

    Re{n=k(1+n)|an|zn1a(AB)n=k(A+nB)|an|zn1}<1. (2.4)

    Let z=r1. 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)=aza|AB|(k+1)+|A+kB|zk,k2. (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 ¯ΣSk(a,A,B).

    Theorem 1. Let fi(z)=azn=k|an|zn¯ΣSk(a,A,B)(i=1,2). If A and B satisfy the condition

    0B<A1or1A<B0,

    then 1a(f1f2)(a;z)¯ΣSk(a,A,B).

    Proof. Let 0B<A1. If fi(z)¯ΣSk(a,A,B)(i=1,2), from (2.2), we obtain

    n=k[(n+1)+(A+nB)]|an,1|a(AB) (3.1)

    and

    n=k[(n+1)+(A+nB)]|an,2|a(AB). (3.2)

    To obtain 1a(f1f2)(a;z)¯ΣSk(a,A,B), we just prove

    n=k1a[(n+1)+(A+nB)]|an,1||an,2|a(AB). (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(AB). (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 0B<A1, we have

    |an,1||an,2|a(AB)(n+1)+(A+nB)a.

    Therefore, we conclude that 1a(f1f2)(a;z)¯ΣSk(a,A,B).

    For 1A<B0, using the same method above, we get 1a(f1f2)(a;z)¯ΣSk(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)=azn=k|an|zn¯ΣSk(a,A,B)(i=1,2) and p>1. If anyone of the following conditions holds true:

    (ⅰ) For 0B<A1, ˆB and a satisfy

    {0ˆBaA[(k+1)+(A+kB)](AB)(k+1+A)a[(k+1)+(A+kB)]+(AB)k,0<a<1,ˆBaA(1+B)(AB)a(1+B)+(AB),a>1.

    (ⅱ) For 1A<B0, 1A<ˆB0 and a satisfies a>0. Then 1a(f1Δf2)(1p,p1p;a,z)¯ΣSk(a,A,ˆB).

    Proof. (ⅰ) For 0B<A1, by Lemma 2, we can get

    n=k(n+1)+(A+nB)a(AB)|an,i|1(i=1,2).

    We deduce that

    {n=k(n+1)+(A+nB)a(AB)|an,1|}1p1 (3.5)

    and

    {n=k(n+1)+(A+nB)a(AB)|an,2|}p1p1. (3.6)

    According to (1.5), we have

    (f1Δf2)(1p,p1p)=a2zn=k|an,1|1p|an,2|p1pzn.

    To prove 1a(f1f2)(1p,p1p,a;z)¯ΣSk(a,A,ˆB), we only need to show

    n=k(n+1)+(A+nˆB)a2(AˆB)|an,1|1p|an,2|p1p1. (3.7)

    Applying Hölder inequality, from (3.5) and (3.6), we obtain

    n=k(n+1)+(A+nB)a(AB)|an,1|1p|an,2|p1p1.

    Thus, (3.7) holds if

    (n+1)+(A+nˆB)a2(AˆB)(n+1)+(A+nB)a(AB)(nN,nk),

    that is,

    ˆB[a[(n+1)+(A+nB)]AB+n]aA[(n+1)+(A+nB)]AB(n+1+A). (3.8)

    Let

    F1(n)=a[(n+1)+(A+nB)]AB+n

    and

    G1(n)=aA[(n+1)+(A+nB)]AB(n+1+A).

    It is easy to see F1(n)>0 holds true for nk. To obtain (3.8), we only need to prove

    ˆBG1(n)F1(n)=aA[(n+1)+(A+nB)](AB)(n+1+A)a[(n+1)+(A+nB)]+(AB)n.

    Because G1(n)F1(n) is an increasing function with respect to n and the condition (i) holds true, so we have

    ˆBaA[(k+1)+(A+kB)](AB)(k+1+A)a[(k+1)+(A+kB)]+(AB)k=G(k)F(k)G(n)F(n).

    Therefore, we conclude that 1a(f1Δf2)(1p,p1p;a,z)¯ΣSk(a,A,ˆB).

    (ⅱ) It is similar to the proof of (ⅰ). For 1A<B0, we only need to show

    (n+1)(A+nˆB)a2(ˆBA)(n+1)(A+nB)a(BA)(nN,nk),

    that is,

    ˆB[a[(n+1)(A+nB)]BA+n]aA[(n+1)(A+nB)]BA+(n+1A). (3.9)

    Let

    F2(n)=a[(n+1)(A+nB)]BA+n

    and

    G2(n)=aA[(n+1)(A+nB)]BA+(n+1A).

    It is easy to see F2(n)>0 holds true for nk. To make (3.9) establish, we only need to prove

    ˆBG2(n)F2(n)=aA[(n+1)(A+nB)]+(BA)(n+1A)a[(n+1)+(A+nB)]+(BA)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)](BA)(k+1A)a[(k+1)(A+kB)]+(BA)k,

    then we have

    ˆBaA[(k+1)(A+kB)](BA)(k+1A)a[(k+1)(A+kB)]+(BA)k=G2(k)F2(k)G2(n)F2(n).

    It can be concluded that 1a(f1Δf2)(1p,p1p;a,z)¯ΣSk(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(1B)+(BA)a(1B)+(BA).

    Since A>aA(1B)+(BA)a(1B)+(BA), we have

    ˆB>A>aA(1B)+(BA)a(1B)+(BA)=limn+G2(n)F2(n)G2(n)F2(n).

    Thus, we conclude that 1a(f1Δf2)(1p,p1p;a,z)¯ΣSk(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)=azn=k|an,i|zn¯ΣSk(a,1,0)(i=1,2) and p>1. If ˆB and a satisfy

    {0ˆB(a1)(k+2)a(k+2)+k,0<a<1,ˆBa1a+1,a>1.

    Then 1a(f1f2)(1p,p1p;a,z)¯ΣSk(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)=azn=k|an,i|zn¯ΣSk(a,A,B)(i=1,2). If A,B and λ satisfy

    {(AB)(k+1AnB)(AB)(k+1)<λ<1k+1,0B<A1,λ<1k+1,1A<B0.

    Then 1a(f1¯f2)λ(z)¯ΣSk(a,A,B).

    Proof. Let fi(z)¯ΣSk(a,A,B)(i=1,2). If 0B<A1, from Lemma 2, we obtain

    n=k((n+1)+(A+nB))|an,1|a(AB), (3.11)

    and

    n=k((n+1)+(A+nB))|an,2|a(AB). (3.12)

    By Definition 2, we have

    1a(f1ˉf2)λ(z)=azn=k1a(1(n+1)λ|an,1||an,2|zn.

    To show 1a(f1ˉf2)λ(z)¯ΣSk(a,A,B), we just prove

    n=k1a(1(n+1)λ)((n+1)+(A+nB))|an,1||an,2|a(AB). (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(AB). (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(AB)(n+1)+(A+nB), we only need to prove

    AB(n+1)+(A+nB)1(n+1)λ,nk. (3.15)

    Let

    M(n)=(n+1)+(A+nB)1(n+1)λ.

    Clearly, M(n) is an increasing function for nk. This implies that (3.15) holds true if

    k+1+A+nB1(k+1)λAB.

    For λ<1k+1, the above formula can be expressed as λ(AB)(k+1AnB)(AB)(k+1).

    To summarize, we conclude that 1a(f1¯f2)λ(z)¯ΣSk(a,A,B) for (AB)(k+1AnB)(AB)(k+1)<λ<1k+1. Using the same method, we can get 1a(f1¯f2)λ(z)¯ΣSk(a,A,B) for

    λ<min{(BA)(k+1AnB)(BA)(k+1),1k+1}=1k+1.

    Thus, we complete the proof of Theorem 3.

    Theorem 4. Let fi(z)=azn=k|an|zn¯ΣSk(a,A,B)(i=1,2) and p>1.

    (1) For 0B<A1,0ˆB<A. If anyone of the following conditions holds true.

    (ⅰ) a<1,λ<min{12(n+1),a(1+B)+(AB)(2n+1)(AB),k(AB)+a(k+1+A+kB)k(k+1)(AB)},

    (ⅱ) a>1,12(k+1)<λ<min{a(1+B)+(AB)(2n+1)(AB),k(AB)+a(k+1+A+kB)k(k+1)(AB)},

    (ⅲ) a<1,max{a(1+B)+(AB)(2k+1)(AB),k(AB)+a(k+1+A+kB)k(k+1)(AB)}<λ<12(n+1),

    (ⅳ) a>1,max{12(k+1),a(1+B)+(AB)(2k+1)(AB),k(AB)+a(k+1+A+kB)k(k+1)(AB)}<λ.

    Then the generalized λ-Hadamard product: 1a(f1˜Δf2)λ(1p,p1p,a;z)¯ΣSk(a,A,B).

    (2) For 1A<B0. If anyone of the following conditions holds true.

    (ⅴ) a<1,λ<min{12(n+1),(BA)+a(1B)(2k1)(BA),k(BA)+a(k+1AkB)k(k+1)(BA)},

    max(A,aA[(k+1)(A+kB)]+[1(k+1)λ](k+1+A)(BA)a[(k+1)(A+kB)]k[1(k+1)λ](BA))ˆB0,

    (ⅵ) a>1,12(k+1)<λ<min{(BA)+a(1B)(2k1)(BA),k(BA)+a(k+1AkB)k(k+1)(BA)},

    max(A,aA[(k+1)(A+kB)]+[1(k+1)λ](k+1+A)(BA)a[(k+1)(A+kB)]k[1(k+1)λ](BA))ˆB0,

    (ⅶ) a<1,max{(BA)+a(1B)(2k1)(BA),k(BA)+a(k+1AkB)k(k+1)(BA)}<λ<12(n+1),AˆB<0,

    (ⅷ) a>1,max{12(k+1),(BA)+a(1B)(2k1)(BA),k(BA)+a(k+1AkB)k(k+1)(BA)}<λ,AˆB<0.

    Then the generalized λ-Hadamard product: 1a(f1˜Δf2)λ(1p,p1p,a;z)¯ΣSk(a,A,B).

    Proof. Let fi(z)¯ΣSk(a,A,B)(i=1,2). The method for the proof of Theorem 4 is similar to that of Theorem 2. If 1A<B0, we just need to prove

    [1(n+1)λ][(n+1)(A+nˆB)]a2(AˆB)(n+1)(AnB)a(BA), (3.16)

    that is

    ˆB[a[(n+1)(A+nB)]BA+n(1(n+1)λ)]aA[(n+1)(A+nB)]BA+[1(n+1)λ](n+1A).

    Let

    P(n)=a[(n+1)(A+nB)]BA+n[1(n+1)λ]

    and

    Q(n)=aA[(n+1)+(A+nB)]BA+[1(n+1)λ](n+1A).

    The following discussion can be divided into two cases:

    (a) It is easy to see that P(n)>0 if λ<(BA)+a(1B)(BA)(2n1). To prove that (3.16) holds true, we need only to prove the following inequality

    ˆBQ(n)P(n)=aA[(n+1)(A+nB)]BA+[1(n+1)λ](n+1A)a[(n+1)(A+nB)]BA+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(BA)+a(k+1AkB)k(k+1)(BA), then we have

    ˆBaA[(k+1)(A+kB)]+[1(k+1)λ](k+1A)(BA)a[(k+1)(A+kB)]+k[1(k+1)λ](BA)=Q(k)P(k)Q(n)P(n).

    Therefore, 1a(f1˜f2)λ(1p,p1p,a,z)¯ΣSk(a,A,B) if the conditions (v) or (vi) are satisfied.

    (b) It is easy to see that P(n)<0 if λ>(BA)+a(1B)(BA)(2n1) and P(k)=a[(k+1)(A+kB)]BA+k[1(k+1)λ]>0. To prove that (3.16) holds true, we need only to prove the following inequality

    ˆBQ(n)P(n)=aA[(n+1)(A+nB)]AB[1(n+1)λ])(n+1A)a[(n+1)(A+nB)]BA+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).

    ˆBlimn+Q(n)P(n)Q(n)P(n).

    Therefore, 1a(f1˜f2)λ(1p,p1p,a,z)¯ΣSk(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)¯ΣSk(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,p1p,a;z) ¯ΣSk(a,12β,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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