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Single-cell data-driven mathematical model reveals possible molecular mechanisms of embryonic stem-cell differentiation

  • Received: 30 January 2019 Accepted: 10 June 2019 Published: 24 June 2019
  • Embryonic development is widely studied due to its application in disease treatment. The published literature demonstrated that Krüppel-like factor 8(KLF8) plays an important role in modulating mesendoderm to definitive endoderm (DE) differentiation. However, it is not clear how KLF8 interacts with other key genes and affects the differentiation process. To qualitatively and quantitatively explore the molecular mechanisms of KLF8 during the differentiation of human embryonic stem cells (hESCs) in detail, we developed a mathematical model to describe the dynamics between KLF8 and two other significant genes, E-cadherin(CDH1) and Zinc-finger E-box-binding homeobox1(ZEB1). Based on the single-cell RNA-seq data, the model structure and parameters were obtained using particle swarm optimization (PSO). The bifurcation analysis and simulation results reveal that the system can exhibit a complex tristable transition, which corresponds to the three states of embryonic development at the single-cell level. We further predict that the novel important gene KLF8 promotes the formation of DE cells by reciprocal inhibition between CDH1 and KLF8 and promotion of the expression of ZEB1. These results may help to shed light on the biological mechanism in the differentiation process of hESCs.

    Citation: Xiao Tu, Qinran Zhang, Wei Zhang, Xiufen Zou. Single-cell data-driven mathematical model reveals possible molecular mechanisms of embryonic stem-cell differentiation[J]. Mathematical Biosciences and Engineering, 2019, 16(5): 5877-5896. doi: 10.3934/mbe.2019294

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  • Embryonic development is widely studied due to its application in disease treatment. The published literature demonstrated that Krüppel-like factor 8(KLF8) plays an important role in modulating mesendoderm to definitive endoderm (DE) differentiation. However, it is not clear how KLF8 interacts with other key genes and affects the differentiation process. To qualitatively and quantitatively explore the molecular mechanisms of KLF8 during the differentiation of human embryonic stem cells (hESCs) in detail, we developed a mathematical model to describe the dynamics between KLF8 and two other significant genes, E-cadherin(CDH1) and Zinc-finger E-box-binding homeobox1(ZEB1). Based on the single-cell RNA-seq data, the model structure and parameters were obtained using particle swarm optimization (PSO). The bifurcation analysis and simulation results reveal that the system can exhibit a complex tristable transition, which corresponds to the three states of embryonic development at the single-cell level. We further predict that the novel important gene KLF8 promotes the formation of DE cells by reciprocal inhibition between CDH1 and KLF8 and promotion of the expression of ZEB1. These results may help to shed light on the biological mechanism in the differentiation process of hESCs.


    In 1940, Ulam [25] proposed the following question concerning the stability of group homomorphisms: Under what condition does there is an additive mapping near an approximately additive mapping between a group and a metric group? In the next year, Hyers [7,8] answered the problem of Ulam under the assumption that the groups are Banach spaces. A generalized version of the theorem of Hyers for approximately linear mappings was given by Rassias [22]. Since then, the stability problems of various functional equation have been extensively investigated by a number of authors (see [3,6,13,14,16,17,22,24,26,27,28]). By regarding a large influence of Ulam, Hyers and Rassias on the investigation of stability problems of functional equations the stability phenomenon that was introduced and proved by Rassias [22] in the year 1978 is called the Hyers-Ulam-Rassias stability.

    Consider the functional equation

    f(x+y)+f(xy)=2f(x)+2f(y). (1.1)

    The quadratic function f(x)=cx2 is a solution of this functional Equation (1.1), and so one usually is said the above functional equation to be quadratic [5,10,11,12]. The Hyers-Ulam stability problem of the quadratic functional equation was first proved by Skof [24] for functions between a normed space and a Banach space. Afterwards, the result was extended by Cholewa [2] and Czerwik [4].

    Now, we consider the following functional equation:

    f(2x+y)+f(2xy)=4f(x+y)+4f(xy)+24f(x)6f(y). (1.2)

    It is easy to see that the function f(x)=cx4 satisfies the functional equation (1.2). Hence, it is natural that Eq (1.2) is called a quartic functional equation and every solution of the quartic functional equation is said to be a quartic mapping (see [15,19]).

    The theory of random normed spaces (briefly, RN-spaces) is important as a generalization of deterministic result of normed spaces and also in the study of random operator equations. The notion of an RN-space corresponds to the situations when we do not know exactly the norm of the point and we know only probabilities of possible values of this norm. Random theory is a setting in which uncertainty arising from problems in various fields of science, can be modelled. It is a practical tool for handling situations where classical theories fail to explain. Random theory has many application in several fields, for example, population dynamics, computer programming, nonlinear dynamical system, nonlinear operators, statistical convergence and so forth.

    In 2008, Mihet and Radu [18] applied fixed point alternative method to prove the stability theorems of the Cauchy functional equation:

    f(x+y)f(x)f(y)=0

    in random normed spaces. In 2008, Najati and Moghimi [20] obtained a stability of the functional equation deriving from quadratic and additive function:

    f(2x+y)+f(2xy)+2f(x)f(x+y)f(xy)2f(2x)=0 (1.3)

    by using the direct method. After that, Jin and Lee [9] proved the stability of the above mentioned functional equation in random normed spaces.

    In 2011, Saadati et al. [21] proved the nonlinear stability of the quartic functional equation of the form

    16f(x+4y)+f(4xy)=306[9f(x+y3)+f(x+2y)]+136f(xy)1394f(x+y)+425f(y)1530f(x)

    in the setting of random normed spaces. Furthermore, the interdisciplinary relation among the theory of random spaces, the theory of non-Archimedean spaces, the fixed point theory, the theory of intuitionistic spaces and the theory of functional equations were also presented. Azadi Kenary [1] investigated the Ulam stability of the following nonlinear function equation

    f(f(x)f(y))+f(x)+f(y)=f(x+y)+f(xy),

    in random normed spaces.

    In this note, we investigate the general solution for the quartic functional equation of the form

    (3n+4)f(ni=1xi)+nj=1f(nxj+ni=1,ijxi)=(n2+2n+1)ni=1,ijkf(xi+xj+xk)12(3n32n213n8)ni=1,ijf(xi+xj)+12(n3+2n2+n)ni=1,ijf(xixj)+12(3n45n37n2+13n+12)ni=1f(xi) (1.4)

    (nN,n>4) and also investigate the Hyers-Ulam stability of the quartic functional equation in random normed spaces by using the direct approach and the fixed point approach.

    In this part, we make some notations and basic definitions used in this article.

    Definition 2.1. A function T:[0,1]×[0,1][0,1] is called a continuous triangular norm, if T satisfies the following condition

    (a) T is commutative and associative;

    (b) T is continuous;

    (c) T(a,1)=a for all a[0,1];

    (d) T(a,b)T(c,d) when ac and bd for all a,b,c,d[0,1].

    Typical examples of continuous tnorms are Tp(a,b)=ab,Tm(a,b)=min(a,b) and TL(a,b)=max(a+b1,0) (The Lukasiewicz tnorm). Recall [23] that if T is a tnorm and {xn} is a given sequence of numbers in [0,1], then Tni=1xn+i is defined recurrently by Ti=1xi=xi and Tni=1xi=T(Tn1i=1xi,xn) for n2,Ti=1xi is defined as Ti=1xn+i. It is known that, for the Lukasiewicz tnorm, the following implication holds:

    limn(TL)i=1xn+i=1n=1(1xn)<.

    Definition 2.2. A random normed space (briefly, RNspace) is a triple (X,μ,T), where X is a vector space. T is a continuous tnorm and μ is a mapping from X into D+ satisfying the following conditions:

    (RN1) μx(t)=ε0(t) for all t>0 if and only if x=0;

    (RN2) μαx(t)=μx(t|α|) for all xX, and alpha with α0;

    (RN3) μx+y(t+s)T(μx(t),μy(s)) for all x,yX and t,s0.

    Definition 2.3. Let (X,μ,T) be an RNspace.

    1). A sequence {xn} in X is said to be convergent to a point xX if, for any ε>0 and λ>0, there exists a positive integer N such that μxnx(ε)>1λ for all n>N.

    2). A sequence {xn} in X is called a Cauchy sequence if, for any ε>0andλ>0, there exists a positive integer N such that μxnxm(ε)>1λ for all nmN.

    3). An RNspace (X,μ,T) is said to be complete, if every Cauchy sequence in X is convergent to a point in X.

    Throughout this paper, we use the following notation for a given mapping f:XY as

    Df(x1,,xn):=(3n+4)f(ni=1xi)+nj=1f(nxj+ni=1,ijxi)(n2+2n+1)ni=1,ijkf(xi+xj+xk)12(3n32n213n8)ni=1,ijf(xi+xj)+12(n3+2n2+n)ni=1,ijf(xixj)+12(3n45n37n2+13n+12)ni=1f(xi)

    for all x1,x2,,xnX.

    In this section we investigate the general solution of the n-variable quartic functional equation (1.4).

    Theorem 3.1. Let X and Y be real vector spaces. If a mapping f:XY satisfies the functional equation (1.4) for all x1,,xnX, then f:XY satisfies the functional equation (1.2) for all x,yX.

    Proof. Assume that f satisfies the functional equation (1.4). Putting x1=x2==xn=0 in (1.4), we get

    (3n+4)f(0)=(n2+2n+1){6n20+4(n4)(n5)2+(n4)(n5)(n6)6}f(0)12(3n32n213n8){4n10+(n29n+20)2}f(0)+12(n3+2n2+n){4n10+(n29n+20)2}f(0)n2(3n45n37n2+13n+12)f(0) (3.1)

    It follows from (3.1) that

    f(0)=0. (3.2)

    Replacing (x1,x2,,xn) by (x,0,,0(n1)times) in (1.4), we have

    (3n+4)f(x)+n4f(x)+(n1)f(x)=(n4n33n2+n+22)f(x)12(3n45n311n2+5n+8)f(x)+12(n4+n3+n2n)f(x)+12(3n45n37n2+13n+12)f(x) (3.3)

    for all xX. It follows from (3.3) that

    f(x)=f(x)

    for all xX. Setting x1=x2==xn=x in (1.4), we obtain

    (3n+4)f(x)+nf(x)=(n2+2n+1){6n20+4(27)(n4)(n5)2+(n4)(n5)(n6)6}f(x)12(3n32n213n8){4n10+(n29n+20)2}f(2x)+12(n3+2n2+n){4n10+(n29n+20)2}f(0)+n2(3n45n37n2+13n+12)f(x) (3.4)

    for all xX. It follows from (3.4) and (3.2) that

    f(2x)=16f(x)

    for all xX. Setting x1=x2==xn=x in (1.4), we get

    (3n+4)f(x)+nf(x)=(n2+2n+1){6n20+4(n29n+20)2+(n29n+20)(n6)6}f(3x)12(3n32n213n8){4n10+16(n29n+20)2}f(x)+12(n3+2n2+n){4n10+(n29n+20)2}f(0)+n2(3n45n37n2+13n+12)f(x) (3.5)

    for all xX. It follows from (3.5) and (3.2) that

    f(3x)=81f(x)

    for all xX. In general for any positive m, we get

    f(mx)=m4f(x)

    for all xX. Replacing (x1,x2,,xn) by (x,x,,x(n1)times,y) in (1.4), we get

    (3n+4)(f((n1)x+y)+(n1)f(2x+y))+f((n1)xny)=(n2+2n+1)(3n9+(n23n+2)2)f(2x+y)(n2+2n+1)(3n11+(3n227n+602)+((n4)(n5)(n6)6))f(3x)12(3n32n213n8){3n9+(n29n+20)2}f(2x)(n1)2(3n32n213n8)f(x+y)+12(n3+2n2+n)(n1)f(xy)+12(3n45n37n2+13n+12)(f(y)+(n1)f(x)) (3.6)

    for all x,yX. It follows from (3.6) that

    (3n+4)(f((n1)x+y)+(n1)f(2x+y))+f((n1)xny)=(n2+2n+1)(n23n+22)f(2x+y)+81(n2+2n+1)(n35n2+11n66)f(x)16(3n32n213n8)(n23n+24)f(x)(n1)2(3n32n213n8)f(x+y)+12(n3+2n2+n)(n1)f(xy)+12(3n45n37n2+13n+12)(f(y)+(n1)f(x)) (3.7)

    for all x,yX. Replacing y by y in (3.7), we get

    (3n+4)(f((n1)xy)+(n1)f(2x+y))+f(()x+y)=(n2+2n+1)(n23n+22)f(2xy)+81(n2+2n+1)(n35n2+11n66)f(x)16(3n32n213n8)(n23n+24)f(x)(n1)2(3n32n213n8)f(xy)+12(n3+2n2+n)(n1)f(x+y)+12(3n45n37n2+13n+12)(f(y)+(n1)f(x)) (3.8)

    for all x,yX. Adding (3.7) and (3.8), we get

    (3n+4)((n1)2{f(x+y)+f(xy)}+2{f((n1)x)(n1)2f(x)}+2{f(y)(n1)2f(y)})+(n1)(f(2x+y)+f(2x+y))+(f((n1)xny)+f((n1)x+ny))=(n2+2n+1)(n23n+22)(f(2xy)+f(2x+y))+162(n2+2n+1)(n35n2+11n66)f(x)32(3n32n213n8)(n23n+24)f(x)(n1)2(3n32n213n8)(f(xy)+f(x+y))+12(n3+2n2+n)(n1)(f(x+y)+f(xy))+(3n45n37n2+13n+12)(f(y)+(n1)f(x)) (3.9)

    for all x,yX. It follows from (3.9) that

    (3n+4)((n1)2{f(x+y)+f(xy)}+2{f((n1)x)(n1)2f(x)}+2{f(y)(n1)2f(y)})+(n1)(f(2xy)+f(2x+y))+n2(n1)2(f(x+y)+f(xy))+2{f((n1)x)n2(n1)2f(x)+n4f(y)n2(n1)2f(y)}=(n2+2n+1)(n23n+22)(f(2xy)+f(2x+y))+162(n2+2n+1)(n35n2+11n66)f(x)32(3n32n213n8)(n23n+24)f(x)(n1)2(3n32n213n8)(f(xy)+f(x+y))+12(n3+2n2+n)(n1)(f(x+y)+f(xy))+(3n45n37n2+13n+12)(f(y)+(n1)f(x)) (3.10)

    for all x,yX. It follows from (3.10) that

    2f(2x+y)2f(2xy)=8f(x+y)8f(xy)48f(x)+12f(y) (3.11)

    for all x,yX. From (3.11), we get

    (2x+y)+f(2xy)=4f(x+y)+4f(xy)+24f(x)6f(y)

    for all x,yX. Thus the mapping f:XY is quartic.

    In this section, the Ulam-Hyers stability of the quartic functional equation (1.4) in RNspace is provided. Throughout this part, let X be a linear space and (Y,μ,T) be a complete RNspace.

    Theorem 4.1. Let j=±1, f:XY be a mapping for which there exists a mapping η:XnD+ satisfying

    limkTi=0(η2(k+i)x1,2(k+i)x2,...,2(k+i)xn(24(k+i+1)jt))=1=limkη2kjx1,2kjx2,..,2kjxn(24kjt)

    such that f(0)=0 and

    μDf(x1,x2,...,xn)(t)η(x1,x2,...,xn)(t) (4.1)

    for all x1,x2,...,xnX and all t>0. Then there exists a unique quartic mapping Q:XY satisfying the functional equation (1.4) and

    μQ(x)f(x)(t)Ti=0(η2(i+1)jx,2(i+1)jx,,...,2(i+1)jx(14(3n55n411n3+5n2+8n)24(i+1)jt))

    for all xX and all t>0. The mapping Q(x) is defined by

    μC(x)(t)=limkμf(2kjx)24kj(t) (4.2)

    for all xX and all t>0.

    Proof. Assume j=1. Setting (x1,x2,...,xn) by (x,x,...,xntimes) in (4.1), we obtain

    μ(3n55n411n3+5n2+8n)4f(2x)4(3n55n411n3+5n2+8n)f(x)(t)η(x,x,...,xntimes)(t) (4.3)

    for all xX and all t>0. It follows from (4.2) and (RN2) that

    μf(2x)16f(x)(t)η(x,x,...,xntimes)(24(3n55n411n3+5n2+8n)t)

    for all xX and all t>0. Replacing x by 2kx in (4.3), we have

    μf(2k+1x)24(k+1)f(2kx)24k(t)η(2kx,2kx,...,2kxntimes)(24k(3n55n411n3+5n2+8n)16t)η(x,x,...,xntimes)(24k(3n55n411n3+5n2+8n)16αkt) (4.4)

    for all xX and all t>0. It follows from f(2nx)24nf(x)=n1k=0f(2k+1x)24(k+1)f(2kx)24k and (4.4) that

    μf(2nx)24nf(x)(tn1k=0αk24k(3n55n411n3+5n2+8n)16)Tn1k=0(ηx,x,0,...,0(t))=η(x,x,...,xntimes)(t),
    μf(2nx)24nf(x)(t)η(x,x,...,xntimes)(tn1k=0αk24k(3n55n411n3+5n2+8n)24) (4.5)

    for all xX and all t>0. Replacing x by 2mx in (4.5), we get

    μf(2n+mx)24(n+m)f(2mx)24m(t)η(x,x,...,xntimes)(tn+mk=mαk24k(3n55n411n3+5n2+8n)16). (4.6)

    Since η(x,x,...,xntimes)(tn+mk=mαk24k(3n55n411n3+5n2+8n)16)1 as m,n, {f(2nx)24n} is a Cauchy sequence in (Y,μ,T). Since (Y,μ,T) is a complete RN-space, this sequence converges to some point C(x)Y. Fix xX and put m=0 in (4.6). Then we have

    μf(2nx)24nf(x)(t)η(x,x,...,xntimes)(tn1k=0αk24k(3n55n411n3+5n2+8n)16)

    and so, for every δ>0, we get

    μC(x)f(x)(t+δ)T(μQ(x)f(2nx)24n(δ),μf(2nx)24nf(x)(t))T(μQ(x)f(2nx)24n(δ),η(x,x,...,xntimes)(tn1k=0αk24k(3n55n411n3+5n2+8n)16)). (4.7)

    Taking the limit as n and using (4.7), we have

    μC(x)f(x)(t+δ)η(x,x,...,xntimes)((3n55n411n3+5n2+8n)(24α)t). (4.8)

    Since δ is arbitrary, by taking δ0 in (4.8), we have

    μQ(x)f(x)(t)η(x,x,...,xntimes)((3n55n411n3+5n2+8n)(24α)t). (4.9)

    Replacing (x1,x2,...,xn) by (2nx1,2nx2,...,2nxn) in (4.1), respectively, we obtain

    μDf(2nx1,2nx2,...,2nxn)(t)η2nx1,2nx2,...,2nxn(24nt)

    for all x1,x2,...,xnX and for all t>0. Since

    limkTi=0(η2k+ix1,2k+ix2,...,2k+ixn(24(k+i+1)jt))=1,

    we conclude that Q fulfils (1.1). To prove the uniqueness of the quartic mapping Q, assume that there exists another quartic mapping D from X to Y, which satisfies (4.9). Fix xX. Clearly, Q(2nx)=24nQ(x) and D(2nx)=24nD(x) for all xX. It follows from (4.9) that

    μQ(x)D(x)(t)=limnμQ(2nx)24nD(2nx)24n(t)μQ(2nx24n(t)min{μQ(2nx)24nf(2nx)24n(t2),μD(2nx)24nf(2nx)24n(t2)}
    η(2nx,2nx,...,2nxntimes)(24n(3n55n411n3+5n2+8n)(24α)t)η(x,x,...,xntimes)(24n(3n55n411n3+5n2+8n)(24α)tαn).

    Since limn(24n(3n55n411n3+5n2+8n)(24α)tαn)=, we get

    limnηx,x,0,...,0(24n(3n55n411n3+5n2+8n)(24α)tαn)=1.

    Therefore, it follows that μQ(x)D(x)(t)=1 for all t>0 and so Q(x)=D(x).

    This completes the proof.

    The following corollary is an immediate consequence of Theorem 4.1, concerning the stability of (1.4).

    Corollary 4.2. Let Ω and be nonnegative real numbers. Let f:XY satisfy the inequality

    μDf(x1,x2,...,xn)(t){ηΩ(t),ηΩni=1xi(t)ηΩ(ni=1xin)(t),ηΩ(ni=1xi+ni=1xin)(t)

    for all x1,x2,x3,...,xnX and all t>0, then there exists a unique quartic mapping Q:XY such that

    μf(x)Q(x)(t){ηΩ|60|(3n55n411n3+5n2+8n)(t)ηΩx|4|(3n55n411n3+5n2+8n)|242|(t),4ηΩxn|4|(3n55n411n3+5n2+8n)|242n|(t),4nη(n+1)Ωxn|4|(3n55n411n3+5n2+8n)|242n|(t),4n

    for all xX and all t>0.

    In this section, we prove the Ulam-Hyers stability of the functional equation (1.4) in random normed spaces by the using fixed point method.

    Theorem 5.1. Let f:XY be a mapping for which there exists a mapping η:XnD+ with the condition

    limkηδkix1,δkix2,...,δkixn(δkit)=1

    for all x1,x2,x3,...,xnX and all 0, where

    δi={2i=0;12i=1;

    and satisfy the functional inequality

    μDf(x1,x2,...,xn)(t)ηx1,x2,...,xn(t)

    for all x1,x2,x3,...,xnX and all 0. If there exists L=L(i) such that the function xβ(x,t)=η(x2,x2,...,x2ntimes)((3n55n411n3+5n2+8n)t) has the property that

    β(x,t)L1δ4iβ(δix,t) (5.1)

    for all xX and t>0, then there exists a unique quartic mapping Q:XY satisfying the functional equation (1.4) and

    μQ(x)f(x)(L1i1Lt)β(x,t)

    for all xX and t>0.

    Proof. Let d be a general metric on Ω such that

    d(p,q)=inf{k(0,)/μ(p(x)q(x))(kt)β(x,t),xX,t>0}.

    It is easy to see that (Ω,d) is complete. Define T:ΩΩ by Tp(x)=1δ4ip(δix) for all xX. Now assume that for p,qΩ, we have d(p,q)K. Then

    μ(p(x)q(x))(kt)β(x,t)μ(p(x)q(x))(Ktδ4i)β(x,t)d(Tp(x),Tq(x))KLd(Tp,Tq)Ld(p,q) (5.2)

    for all p,qΩ. Therefore, T is a strictly contractive mapping on Ω with Lipschitz constant L. It follows from (5.2) that

    μ(3n55n411n3+5n2+8n)4f(2x)4(n3n55n411n3+5n2+8n)f(x)(t)η(x,x,...,xntimes)(t) (5.3)

    for all xX. It follows from (5.3) that

    μf(2x)16f(x)(t)η(x,x,...,xntimes)((3n55n411n3+5n2+8n)16t) (5.4)

    for all xX. By using (5.1) for the case i=0, it reduce to

    μf(2x)16f(x)(t)Lβ(x,t)

    for all xX. Hence we obtain

    d(μTf,f)L=L1i< (5.5)

    for all xX. Replacing x by x2 in (5.4), we get

    μf(x)16f(x2)(t)η(x2,x2,...,x2ntimes)((3n55n411n3+5n2+8n)16t)

    for all xX. By using (5.1) for the case i=1, it reduce to

    μ16f(x2)f(x)(t)β(x,t)μTf(x)f(x)(t)β(x,t)

    for all xX. Hence we get

    d(μTf,f)L=L1i< (5.6)

    for all xX. From (5.5) and (5.6), we can conclude

    d(μTf,f)L=L1i<

    for all xX.

    The remaining proof is similar to the proof of Theorem 4.1. So Q is a unique fixed point of T in the set such that

    μ(f(x)Q(x))(L1i1Lt)β(x,t)

    for all xX and t>0. This completes the proof of the theorem.

    From Theorem 4.1, we obtain the following corollary concerning the stability for the functional equation (1.4).

    Corollary 5.2. Suppose that a mapping f:XY satisfies the inequality

    μDf(x1,x2,...,xn)(t){ηΩ(t),ηΩni=1xi(t)ηΩ(ni=1xi)(t),ηΩ(ni=1xi+ni=1xin)(t),

    for all x1,x2,...,xnX and all t>0, where Ω, are constants with Ω>0. Then there exists a unique quartic mapping Q:XY such that

    μf(x)Q(x)(t){ηΩ|60|(3n55n411n3+5n2+8n)(t)ηΩx|4|(3n55n411n3+5n2+8n)|242|(t),4ηΩxn|4|(3n55n411n3+5n2+8n)|242n|(t),4nη(n+1)Ωxn|4|(3n55n411n3+5n2+8n)|242n|(t),4n

    for all xX and all t>0.

    Proof. Set

    μDf(x1,x2,...,xn)(t){ηΩ(t),ηΩni=1xi(t),ηΩ(ni=1xi)(t),ηΩ(ni=1xi+ni=1xin)(t)

    for all x1,x2,...,xnX and all t>0. Then

    η(δkix1,δkix2,...,δkixn)(δ4kit)={ηΩδ4ki(t),ηΩni=1xiδ(4n)ki(t),ηΩ(ni=1xiδ(4n)ki)(t),ηΩ(ni=1xiδ(4n)ki+ni=1xin)(t),
    ={  1  as  k  1  as  k  1  as  k  1  as  k.

    But we have β(x,t)=η(x2,x2,...,x2ntimes)(14(3n55n411n3+5n2+8n)t) has the property L1δ4iβ(δix,t) for all xX and t>0. Now

    β(x,t)={η4Ω3n55n411n3+5n2+8n(t)η4Ωx24(3n55n411n3+5n2+8n)(t)η4Ωxn24(3n55n411n3+5n2+8n)(t)η4Ωxns2n(3n55n411n3+5n2+8n)(t),
    L1δ4iβ(δix,t)={ηδ4iβ(x)(t)ηδ4iβ(x)(t)ηδ4iβ(x)(t)ηδn4iβ(x)(t).

    By (4.1), we prove the following eight cases:

    L=24  ifi=0  and  L=24  if  i=1L=24  for  <4  ifi=0  and  L=24  for  >4  if  i=1L=2n4  for  <4n  ifi=0  and  L=24n  for  >4n  if  i=1L=2n4  for  <4n  ifi=0  and  L=24n  for  >4n  if  i=1

    Case 1: L=24 if i=0

    μf(x)Q(x)(t)L1δ4iβ(δix,t)(t)η(Ω60(3n55n411n3+5n2+8n))(t).

    Case 2: L=24 if i=1

    μf(x)Q(x)(t)L1δ4iβ(δix,t)(t)η(Ω60(3n55n411n3+5n2+8n))(t).

    Case 3: L=24 for <4 if i=0

    μf(x)Q(x)(t)L1δ4iβ(δix,t)(t)η(Ωnx4(3n55n411n3+5n2+8n)(242))(t).

    Case 4: L=24 for >4 if i=1

    μf(x)Q(x)(t)L1δ4iβ(δix,t)(t)η(nΩxs4(3n55n411n3+5n2+8n)(224))(t).

    Case 5: L=2n4 for <4n if i=0

    μf(x)Q(x)(t)L1δ4iβ(δix,t)(t)η(Ωxn4(3n55n411n3+5n2+8n)(242n))(t).

    Case 6: L=24n for >4n if i=1

    μf(x)Q(x)(t)L1δ4iβ(δix,t)(t)η(Ωxn4(3n55n411n3+5n2+8n)(2n24))(t).

    Case 7: L=2n4 for <4n if i=0

    μf(x)Q(x)(t)L1δ4iβ(δix,t)(t)η((n+1)xn4(3n55n411n3+5n2+8n)(242n))(t).

    Case 8 : L=24n for >4n if i=1

    μf(x)Q(x)(t)L1δ4iβ(δix,t)(t)η((n+1)Ωxn4(3n55n411n3+5n2+8n)(2n24))(t).

    Hence the proof is complete.

    In this note we investigated the general solution for the quartic functional equation (1.4) and also investigated the Hyers-Ulam stability of the quartic functional equation (1.4) in random normed space using the direct approach and the fixed point approach. This work can be applied to study the stability in various spaces such as intuitionistic random normed spaces, quasi-Banach spaces and fuzzy normed spaces. Moreover, the results can be applied to investigate quartic homomorphisms and quratic derivations in Banach algebras, random normed algebras, fuzzy Banach algebras and C-ternary algebras.

    This work was supported by Incheon National University Research Grant 2020-2021. We would like to express our sincere gratitude to the anonymous referee for his/her helpful comments that will help to improve the quality of the manuscript.

    The authors declare that they have no competing interests.



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