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Research article Special Issues

Existence of generalized solutions for Keller-Segel-Navier-Stokes equations with degradation in dimension three

  • Received: 31 December 2020 Accepted: 06 July 2021 Published: 12 October 2021
  • We construct generalized solutions for the Keller-Segel system with a degradation source coupled to Navier Stokes equations in three dimensions, in case that the power of degradation is smaller than quadratic. Furthermore, if the logistic type source is purely damping with no growing effect, we prove that solutions converge to zero in some norms and provide upper bounds of convergence rates in time.

    Citation: Kyungkeun Kang, Dongkwang Kim. Existence of generalized solutions for Keller-Segel-Navier-Stokes equations with degradation in dimension three[J]. Mathematics in Engineering, 2022, 4(5): 1-25. doi: 10.3934/mine.2022041

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  • We construct generalized solutions for the Keller-Segel system with a degradation source coupled to Navier Stokes equations in three dimensions, in case that the power of degradation is smaller than quadratic. Furthermore, if the logistic type source is purely damping with no growing effect, we prove that solutions converge to zero in some norms and provide upper bounds of convergence rates in time.



    In 1940, Ulam [24] posed the stability problem concerning group homomorphisms. For Banach spaces, the problem was solved by Hyers [7] in the case of approximate additive mappings. And then Hyers' result was extended by Aoki [1] and Rassias [18] for additive mappings and linear mappings, respectively. In 1994, another further generalization, the so-called generalized Hyer-Ulam stability, was obtained by Gavruta [6]. Later, the stability of several functional equations has been extensively discussed by many mathematicians and there are many interesting results concerning this problem (see [2,8,9,10,19,20] and references therein); also, some stability results of different functional equations and inequalities were studied and generalized [5,11,12,15,16,17,26] in various matrix normed spaces like matrix fuzzy normed spaces, matrix paranormed spaces and matrix non-Archimedean random normed spaces.

    In 2017, Wang and Xu [25] introduced the following functional equation

    2k[f(x+ky)+f(kx+y)]=k(1s+k+ks+2k2)f(x+y)+k(1s3k+ks+2k2)f(xy)+2kf(kx)+2k(s+kks2k2)f(x)+2(1ks)f(ky)+2ksf(y) (1.1)

    where s is a parameter, k>1 and s12k. It is easy to verify that f(x)=ax+bx2(xR) satisfies the functional Eq (1.1), where a,b are arbitrary constants. They considered the general solution of the functional Eq (1.1), and then determined the generalized Hyers-Ulam stability of the functional Eq (1.1) in quasi-Banach spaces by applying the direct method.

    The main purpose of this paper is to employ the direct and fixed point methods to establish the Hyers-Ulam stability of the functional Eq (1.1) in matrix intuitionistic fuzzy normed spaces. The paper is organized as follows: In Sections 1 and 2, we present a brief introduction and introduce related basic definitions and preliminary results, respectively. In Section 3, we prove the Hyers-Ulam stability of the functional Eq (1.1) in matrix intuitionistic fuzzy normed spaces by applying the direct method. In Section 4, we prove the Hyers-Ulam stability of the functional Eq (1.1) in matrix intuitionistic fuzzy normed spaces by applying the fixed point method. Our results may be viewed as a continuation of the previous contribution of the authors in the setting of fuzzy stability (see [14,17]).

    For the sake of completeness, in this section, we present some basic definitions and preliminary results, which will be useful to investigate the Hyers-Ulam stability results in matrix intuitionistic fuzzy normed spaces. The notions of continuous t-norm and continuous t-conorm can be found in [14,22]. Using these, an intuitionistic fuzzy normed space (for short, IFNS) is defined as follows:

    Definition 2.1. ([14,21]) The five-tuple (X,μ,ν,,) is said to be an IFNS if X is a vector space, is a continuous t-norm, is a continuous t-conorm, and μ,ν are fuzzy sets on X×(0,) satisfy the following conditions. For every x,yX and s,t>0,

    (i) μ(x,t)+ν(x,t)1;

    (ii) μ(x,t)>0, (iii) μ(x,t)=1 if and only if x=0;

    (iii) μ(αx,t)=μ(x,t|α|) for each α0, (v) μ(x,t)μ(y,s)μ(x+y,t+s);

    (iv) μ(x,):(0,)[0,1] is continuous;

    (v) limtμ(x,t)=1 and limt0μ(x,t)=0;

    (vi) ν(x,t)<1, (ix) ν(x,t)=0 if and only if x=0;

    (vii) ν(αx,t)=ν(x,t|α|) for each α0, (xi) ν(x,t)ν(y,s)ν(x+y,t+s);

    (xiii) ν(x,):(0,)[0,1] is continuous;

    (ix) limtν(x,t)=0 and limt0ν(x,t)=1.

    In this case, (μ,ν) is called an intuitionistic fuzzy norm.

    The following concepts of convergence and Cauchy sequences are considered in [14,21]:

    Let (X,μ,ν,,) be an IFNS. Then, a sequence {xk} is said to be intuitionistic fuzzy convergent to xX if for every ε>0 and t>0, there exists k0N such that

    μ(xkx,t)>1ε

    and

    ν(xkx,t)<ε

    for all kk0. In this case we write

    (μ,ν)limxk=x.

    The sequence {xk} is said to be an intuitionistic fuzzy Cauchy sequence if for every ε>0 and t>0, there exists k0N such that

    μ(xkx,t)>1ε

    and

    ν(xkx,t)<ε

    for all k,k0. (X,μ,ν,,) is said to be complete if every intuitionistic fuzzy Cauchy sequence in (X,μ,ν,,) is intuitionistic fuzzy convergent in (X,μ,ν,,).

    Following [11,12], we will also use the following notations: The set of all m×n-matrices in X will be denoted by Mm,n(X). When m=n, the matrix Mm,n(X) will be written as Mn(X). The symbols ejM1,n(C) will denote the row vector whose jth component is 1 and the other components are 0. Similarly, EijMn(C) will denote the n×n matrix whose (i,j)-component is 1 and the other components are 0. The n×n matrix whose (i,j)-component is x and the other components are 0 will be denoted by EijxMn(X).

    Let (X,) be a normed space. Note that (X,{n}) is a matrix normed space if and only if (Mn(X),n) is a normed space for each positive integer n and

    AxBkABxn

    holds for AMk,n, x=[xij]Mn(X) and BMn,k, and that (X,{n}) is a matrix Banach space if and only if X is a Banach space and (X,{n}) is a matrix normed space.

    Following [23], we introduce the concept of a matrix intuitionistic fuzzy normed space as follows:

    Definition 2.2. ([23]) Let (X,μ,ν,,) be an intuitionistic fuzzy normed space, and the symbol θ for a rectangular matrix of zero elements over X. Then:

    (1) (X,{μn},{νn},,) is called a matrix intuitionistic fuzzy normed space (briefly, MIFNS) if for each positive integer n, (Mn(X),μn,νn,,) is an intuitionistic fuzzy normed space, μn and νn satisfy the following conditions:

    (i) μn+m(θ+x,t)=μn(x,t),νn+m(θ+x,t)=νn(x,t) for all t>0, x=[xij]Mn(X), θMn(X);

    (ii) μk(AxB,t)μn(x,tAB), νk(AxB,t)νn(x,tAB) for all t>0, AMk,n(R), x=[xij]Mn(X) and BMn,k(R) with AB0.

    (2) (X,{μn},{νn},,) is called a matrix intuitionistic fuzzy Banach space if (X,μ,ν,,) is an intuitionistic fuzzy Banach space and (X,{μn},{νn},,) is a matrix intuitionistic fuzzy normed space.

    The following Lemma 2.3 was found in [23].

    Lemma 2.3. ([23]) Let (X,{μn},{νn},,) be a matrix intuitionistic fuzzy normed space. Then,

    (1) μn(Eklx,t)=μ(x,t), νn(Eklx,t)=ν(x,t) for all t>0 and xX.

    (2) For all [xij]Mn(X) and t=ni,j=1tij>0,

    μ(xkl,t)μn([xij],t)min{μ(xij,tij):i,j=1,2,,n},μ(xkl,t)μn([xij],t)min{μ(xij,tn2):i,j=1,2,,n},

    and

    ν(xkl,t)νn([xij],t)max{ν(xij,tij):i,j=1,2,,n},ν(xkl,t)νn([xij],t)max{ν(xij,tn2):i,j=1,2,,n}.

    (3) limmxm=x if and only if limmxijm=xij for xm=[xijm],x=[xij]Mn(X).

    For explicit later use, we also recall the following Lemma 2.4 is due to Diaz and Margolis [4], which will play an important role in proving our stability results in this paper.

    Lemma 2.4. (The fixed point alternative theorem [4]) Let (E,d) be a complete generalized metric space and J: EE be a strictly contractive mapping with Lipschitz constant L<1. Then for each fixed element xE, either

    d(Jnx,Jn+1x)=,n0,

    or

    d(Jnx,Jn+1x)<,nn0,

    for some natural number n0. Moreover, if the second alternative holds then:

    (i) The sequence {Jnx} is convergent to a fixed point y of J.

    (ii)y is the unique fixed point of J in the set E:={yEd(Jn0x,y)<+} and d(y,y)11Ld(y,Jy),x,yE.

    From now on, let (X,{μn},{νn},,) be a matrix intuitionistic fuzzy normed space and (Y,{μn},{νn},,) be a matrix intuitionistic fuzzy Banach space. In this section, we will prove the Hyers-Ulam stability of the functional Eq (1.1) in matrix intuitionistic fuzzy normed spaces by using the direct method. For the sake of convenience, given mapping f: XY, we define the difference operators Df: X2Y and Dfn: Mn(X2)Mn(Y) of the functional Eq (1.1) by

    Df(a,b):=2k[f(a+kb)+f(ka+b)]k(1s+k+ks+2k2)f(a+b)k(1s3k+ks+2k2)f(ab)2kf(ka)2k(s+kks2k2)f(a)2(1ks)f(kb)2ksf(b),Dfn([xij],[yij]):=2k[fn([xij]+k[yij])+fn(k[xij]+[yij])]k(1s+k+ks+2k2)fn([xij]+[yij])k(1s3k+ks+2k2)fn([xij][yij])2kfn(k[xij])2k(s+kks2k2)fn([xij])2(1ks)fn(k[yij])2ksfn([yij])

    for all a,bX and all x=[xij],y=[yij]Mn(X).

    We start with the following lemmas which will be used in this paper.

    Lemma 3.1. ([25]) Let V and W be real vector spaces. If an odd mapping f: VW satisfies the functional Eq (1.1), then f is additive.

    Lemma 3.2. ([25]) Let V and W be real vector spaces. If an even mapping f: VW satisfies the functional Eq (1.1), then f is quadratic.

    Theorem 3.3. Let φo: X2[0,) be a function such that for some real number α with 0<α<k,

    φo(ka,kb)=αφo(a,b) (3.1)

    for all a,bX. Suppose that an odd mapping f: XY satisfies the inequality

    {μn(Dfn([xij],[yij]),t)tt+ni,j=1φo(xij,yij),νn(Dfn([xij],[yij]),t)ni,j=1φo(xij,yij)t+ni,j=1φo(xij,yij) (3.2)

    for all x=[xij],y=[yij]Mn(X) and all t>0. Then there exists a unique additive mapping A: XY such that

    {μn(fn([xij])An([xij]),t)(kα)(2k+s1)t(kα)(2k+s1)t+n2ni,j=1φo(0,xij),νn(fn([xij])An([xij]),t)n2ni,j=1φo(0,xij)(kα)(2k+s1)t+n2ni,j=1φo(0,xij) (3.3)

    for all x=[xij]Mn(X) and all t>0.

    Proof. When n=1, (3.2) is equivalent to

    μ(Df(a,b),t)tt+φo(a,b)andν(Df(a,b),t)φo(a,b)t+φo(a,b) (3.4)

    for all a,bX and all t>0. Putting a=0 in (3.4), we have

    {μ(2(2k+s1)f(kb)2(2k+s1)kf(b),t)tt+φo(0,b),ν(2(2k+s1)f(kb)2(2k+s1)kf(b),t)φo(0,b)t+φo(0,b) (3.5)

    for all bX and all t>0. Replacing a by kpa in (3.5) and using (3.1), we get

    {μ(f(kp+1a)kp+1f(kpa)kp,t2k(2k+s1)kp)tt+αpφo(0,a),ν(f(kp+1a)kp+1f(kpa)kp,t2k(2k+s1)kp)αpφo(0,a)t+αpφo(0,a) (3.6)

    for all aX and all t>0. It follows from (3.6) that

    {μ(f(kp+1a)kp+1f(kpa)kp,αpt2k(2k+s1)kp)tt+φo(0,a),ν(f(kp+1a)kp+1f(kpa)kp,αpt2k(2k+s1)kp)φo(0,a)t+φo(0,a) (3.7)

    for all aX and all t>0. It follows from

    f(kpa)kpf(a)=p1=0(f(k+1a)k+1f(ka)k)

    and (3.7) that

    {μ(f(kpa)kpf(a),p1=0αt2k(2k+s1)k)p1=0μ(f(k+1a)k+1f(ka)k,αt2k(2k+s1)k)tt+φo(0,a),ν(f(kpa)kpf(a),p1=0αt2k(2k+s1)k)p1=0ν(f(k+1a)k+1f(ka)k,αt2k(2k+s1)k)φo(0,a)t+φo(0,a) (3.8)

    for all aX and all t>0, where

    pj=0aj=a1a2ap,   pj=0aj=a1a2ap.

    By replacing a with kqa in (3.8), we have

    {μ(f(kp+qa)kp+qf(kqa)kq,p1=0αt2k(2k+s1)k+q)tt+αqφo(0,a),ν(f(kp+qa)kp+qf(kqa)kq,p1=0αt2k(2k+s1)k+q)αqφo(0,a)t+αqφo(0,a) (3.9)

    for all aX, t>0, p>0 and q>0. Thus

    {μ(f(kp+qa)kp+qf(kqa)kq,p+q1=qαt2k(2k+s1)k)tt+φo(0,a),ν(f(kp+qa)kp+qf(kqa)kq,p+q1=qαt2k(2k+s1)k)φo(0,a)t+φo(0,a) (3.10)

    for all aX, t>0, p>0 and q>0. Hence

    {μ(f(kp+qa)kp+qf(kqa)kq,t)tt+p+q1=qα2k(2k+s1)kφo(0,a),ν(f(kp+qa)kp+qf(kqa)kq,t)p+q1=qα2k(2k+s1)kφo(0,a)t+p+q1=qα2k(2k+s1)kφo(0,a) (3.11)

    for all aX, t>0, p>0 and q>0. Since 0<α<k and

    =0α2k(2k+s1)k<,

    the Cauchy criterion for convergence in IFNS shows that {f(kpa)kp} is a Cauchy sequence in (Y,μ,ν,,). Since (Y,μ,ν,,) is an intuitionistic fuzzy Banach space, this sequence converges to some point A(a)Y. So one can define the mapping A: XY such that

    A(a):=(μ,ν)limpf(kpa)kp.

    Moreover, if we put q=0 in (3.11), we get

    {μ(f(kpa)kpf(a),t)tt+p1=0α2k(2k+s1)kφo(0,a),ν(f(kpa)kpf(a),t)p1=0α2k(2k+s1)kφo(0,a)t+p1=0α2k(2k+s1)kφo(0,a) (3.12)

    for all aX, t>0 and p>0. Thus, we obtain

    {μ(f(a)A(a),t)μ(f(a)f(kpa)kp,t2)μ(f(kpa)kpA(a),t2)tt+p1=0αk(2k+s1)kφo(0,a),ν(f(a)A(a),t)ν(f(a)f(kpa)kp,t2)ν(f(kpa)kpA(a),t2)p1=0αk(2k+s1)kφo(0,a)t+p1=0αk(2k+s1)kφo(0,a) (3.13)

    for every aX, t>0 and large p. Taking the limit as p and using the definition of IFNS, we get

    {μ(f(a)A(a),t)(kα)(2k+s1)t(kα)(2k+s1)t+φo(0,a),ν(f(a)A(a),t)φo(0,a)(kα)(2k+s1)t+φo(0,a). (3.14)

    Replacing a and b by kpa and kpb in (3.4), respectively, and using (3.1), we obtain

    μ(1kpDf(kpa,kpb),t)tt+(αk)pφo(a,b)andν(1kpDf(kpa,kpb),t)(αk)pφo(a,b)t+(αk)pφo(a,b) (3.15)

    for all a,bX and all t>0. Letting p in (3.15), we obtain

    μ(DA(a,b),t)=1andν(DA(a,b),t)=0 (3.16)

    for all a,bX and all t>0. This means that A satisfies the functional Eq (1.1). Since f: XY is an odd mapping, and using the definition A, we have A(a)=A(a) for all aX. Thus by Lemma 3.1, the mapping A: XY is additive. To prove the uniqueness of A, let A: XY be another additive mapping satisfying (3.14). Let n=1. Then we have

    {μ(A(a)A(a),t)=μ(A(kpa)kpA(kpa)kp,t)μ(A(kpa)kpf(kpa)kp,t2)μ(f(kpa)kpA(kpa)kp,t2)(kα)(2k+s1)t(kα)(2k+s1)t+2(αk)pφo(0,a),ν(A(a)A(a),t)=ν(A(kpa)kpA(kpa)kp,t)ν(A(kpa)kpf(kpa)kp,t2)ν(f(kpa)kpA(kpa)kp,t2)2(αk)pφo(0,a)(kα)(2k+s1)t+2(αk)pφo(0,a) (3.17)

    for all aX, t>0 and p>0. Letting p in (3.17), we get

    μ(A(a)A(a),t)=1andν(A(a)A(a),t)=0

    for all aX and t>0. Hence we get A(a)=A(a) for all aX. Thus the mapping A: XY is a unique additive mapping.

    By Lemma 2.3 and (3.14), we get

    {μn(fn([xij])An([xij]),t)min{μ(f(xij)A(xij),tn2):i,j=1,,n} (kα)(2k+s1)t(kα)(2k+s1)t+n2ni,j=1φo(0,xij),νn(fn([xij])An([xij]),t)max{ν(f(xij)A(xij),tn2):i,j=1,,n} n2ni,j=1φo(0,xij)(kα)(2k+s1)t+n2ni,j=1φo(0,xij)

    for all x=[xij]Mn(X) and all t>0. Thus A: XY is a unique additive mapping satisfying (3.3), as desired. This completes the proof of the theorem.

    Theorem 3.4. Let φe: X2[0,) be a function such that for some real number α with 0<α<k2,

    φe(ka,kb)=αφe(a,b) (3.18)

    for all a,bX. Suppose that an even mapping f: XY with f(0)=0 satisfies the inequality

    {μn(Dfn([xij],[yij]),t)tt+ni,j=1φe(xij,yij),νn(Dfn([xij],[yij]),t)ni,j=1φe(xij,yij)t+ni,j=1φe(xij,yij) (3.19)

    for all x=[xij], y=[yij]Mn(X) and all t>0. Then there exists a unique quadratic mapping Q: XY such that

    {μn(fn([xij])Qn([xij]),t)(k2α)(2k+s1)t(k2α)(2k+s1)t+n2ni,j=1φe(0,xij),νn(fn([xij])Qn([xij]),t)n2ni,j=1φe(0,xij)(k2α)(2k+s1)t+n2ni,j=1φe(0,xij) (3.20)

    for all x=[xij]Mn(X) and all t>0.

    Proof. When n=1, (3.19) is equivalent to

    μ(Df(a,b),t)tt+φe(a,b)andν(Df(a,b),t)φe(a,b)t+φe(a,b) (3.21)

    for all a,bX and all t>0. Letting a=0 in (3.21), we obtain

    {μ(2(2k+s1)f(kb)2(2k+s1)k2f(b),t)tt+φe(0,b),ν(2(2k+s1)f(kb)2(2k+s1)k2f(b),t)φe(0,b)t+φe(0,b) (3.22)

    for all bX and all t>0. Replacing a by kpa in (3.22) and using (3.18), we get

    {μ(f(kp+1a)k2(p+1)f(kpa)k2p,t2k2(2k+s1)k2p)tt+αpφe(0,a),ν(f(kp+1a)k2(p+1)f(kpa)k2p,t2k2(2k+s1)k2p)αpφe(0,a)t+αpφe(0,a) (3.23)

    for all aX and all t>0. It follows from (3.23) that

    {μ(f(kp+1a)k2(p+1)f(kpa)k2p,αpt2k2(2k+s1)k2p)tt+φe(0,a),ν(f(kp+1a)k2(p+1)f(kpa)k2p,αpt2k2(2k+s1)k2p)φe(0,a)t+φe(0,a) (3.24)

    for all aX and all t>0. It follows from

    f(kpa)k2pf(a)=p1=0(f(k+1a)k2(+1)f(ka)k2)

    and (3.24) that

    {μ(f(kpa)k2pf(a),p1=0αt2k2(2k+s1)k2)p1=0μ(f(k+1a)k2(+1)f(ka)k2,αt2k2(2k+s1)k2)tt+φe(0,a),ν(f(kpa)k2pf(a),p1=0αt2k2(2k+s1)k2)p1=0ν(f(k+1a)k2(+1)f(ka)k2,αt2k2(2k+s1)k2)φe(0,a)t+φe(0,a) (3.25)

    for all aX and all t>0, where

    pj=0aj=a1a2ap,   pj=0aj=a1a2ap.

    By replacing a with kqa in (3.25), we have

    {μ(f(kp+qa)k2(p+q)f(kqa)k2q,p1=0αt2k2(2k+s1)k2(+q))tt+αqφe(0,a),ν(f(kp+qa)k2(p+q)f(kqa)k2q,p1=0αt2k2(2k+s1)k2(+q))αqφe(0,a)t+αqφe(0,a) (3.26)

    for all aX, t>0, p>0 and q>0. Thus

    {μ(f(kp+qa)k2(p+q)f(kqa)k2q,p+q1=qαt2k2(2k+s1)k2)tt+φe(0,a),ν(f(kp+qa)k2(p+q)f(kqa)k2q,p+q1=qαt2k2(2k+s1)k2)φe(0,a)t+φe(0,a) (3.27)

    for all aX, t>0, p>0 and q>0. Hence

    {μ(f(kp+qa)k2(p+q)f(kqa)k2q,t)tt+p+q1=qα2k2(2k+s1)k2φe(0,a),ν(f(kp+qa)k2(p+q)f(kqa)k2q,t)p+q1=qα2k2(2k+s1)k2φe(0,a)t+p+q1=qα2k2(2k+s1)k2φe(0,a) (3.28)

    for all aX, t>0, p>0 and q>0. Since 0<α<k2 and

    =0α2k2(2k+s1)k2<,

    the Cauchy criterion for convergence in IFNS shows that {f(kpa)k2p} is a Cauchy sequence in (Y,μ,ν,,). Since (Y,μ,ν,,) is an intuitionistic fuzzy Banach space, this sequence converges to some point Q(a)Y. So one can define the mapping Q: XY such that

    Q(a):=(μ,ν)limpf(kpa)k2p.

    Moreover, if we put q=0 in (3.28), we get

    {μ(f(kpa)k2pf(a),t)tt+p1=0α2k2(2k+s1)k2φe(0,a),ν(f(kpa)k2pf(a),t)p1=0α2k2(2k+s1)k2φe(0,a)t+p1=0α2k2(2k+s1)k2φe(0,a) (3.29)

    for all aX, t>0 and p>0. Thus, we obtain

    {μ(f(a)Q(a),t)μ(f(a)f(kpa)k2p,t2)μ(f(kpa)k2pQ(a),t2)tt+p1=0αk2(2k+s1)k2φe(0,a),ν(f(a)Q(a),t)ν(f(a)f(kpa)k2p,t2)ν(f(kpa)k2pQ(a),t2)p1=0αk2(2k+s1)k2φe(0,a)t+p1=0αk2(2k+s1)k2φe(0,a) (3.30)

    for every aX, t>0 and large p. Taking the limit as p and using the definition of IFNS, we get

    {μ(f(a)Q(a),t)(k2α)(2k+s1)t(k2α)(2k+s1)t+φe(0,a),ν(f(a)Q(a),t)φe(0,a)(k2α)(2k+s1)t+φe(0,a). (3.31)

    Replacing a and b by kpa and kpb in (3.21), respectively, and using (3.18), we obtain

    μ(1k2pDf(kpa,kpb),t)tt+(αk2)pφe(a,b),ν(1k2pDf(kpa,kpb),t)(αk2)pφe(a,b)t+(αk2)pφe(a,b) (3.32)

    for all a,bX and all t>0. Letting p in (3.32), we obtain

    μ(DQ(a,b),t)=1andν(DQ(a,b),t)=0 (3.33)

    for all a,bX and all t>0. This means that Q satisfies the functional Eq (1.1). Since f: XY is an even mapping, and using the definition Q, we have Q(a)=Q(a) for all aX. Thus by Lemma 3.2, the mapping Q: XY is quadratic. To prove the uniqueness of Q, let Q: XY be another quadratic mapping satisfying (3.31). Let n=1. Then we have

    {μ(Q(a)Q(a),t)=μ(Q(kpa)k2pQ(kpa)k2p,t)  μ(Q(kpa)k2pf(kpa)k2p,t2)μ(f(kpa)k2pQ(kpa)k2p,t2)  (k2α)(2k+s1)t(k2α)(2k+s1)t+2(αk2)pφe(0,a),ν(Q(a)Q(a),t)=ν(Q(kpa)k2pQ(kpa)k2p,t)  ν(Q(kpa)k2pf(kpa)k2p,t2)ν(f(kpa)kpQ(kpa)k2p,t2)  2(αk2)pφe(0,a)(k2α)(2k+s1)t+2(αk2)pφe(0,a) (3.34)

    for all aX, t>0 and p>0. Letting p in (3.34), we get

    μ(Q(a)Q(a),t)=1andν(Q(a)Q(a),t)=0

    for all aX and t>0. Hence we get Q(a)=Q(a) for all aX. Thus the mapping Q: XY is a unique quadratic mapping.

    By Lemma 2.3 and (3.31), we get

    {μn(fn([xij])Qn([xij]),t)min{μ(f(xij)Q(xij),tn2):i,j=1,,n}(k2α)(2k+s1)t(k2α)(2k+s1)t+n2ni,j=1φe(0,xij),νn(fn([xij])Qn([xij]),t)max{ν(f(xij)Q(xij),tn2):i,j=1,,n}n2ni,j=1φe(0,xij)(k2α)(2k+s1)t+n2ni,j=1φe(0,xij)

    for all x=[xij]Mn(X) and all t>0. Thus Q: XY is a unique quadratic mapping satisfying (3.20), as desired. This completes the proof of the theorem.

    Theorem 3.5. Let φ: X2[0,) be a function such that for some real number α with 0<α<k,

    φ(ka,kb)=αφ(a,b) (3.35)

    for all a,bX. Suppose that a mapping f: XY with f(0)=0 satisfies the inequality

    {μn(Dfn([xij],[yij]),t)tt+ni,j=1φ(xij,yij),νn(Dfn([xij],[yij]),t)ni,j=1φ(xij,yij)t+ni,j=1φ(xij,yij) (3.36)

    for all x=[xij],y=[yij]Mn(X) and all t>0. Then there exist a unique quadratic mapping Q: XY and a unique additive mapping A: XY such that

    {μn(fn([xij])Qn([xij])An([xij]),t)(kα)(2k+s1)t(kα)(2k+s1)t+2n2ni,j=1˜φ(0,xij),νn(fn([xij])Qn([xij])An([xij]),t)2n2ni,j=1˜φ(0,xij)(kα)(2k+s1)t+2n2ni,j=1˜φ(0,xij) (3.37)

    for all x=[xij]Mn(X) and all t>0, ˜φ(a,b)=φ(a,b)+φ(a,b) for all a,bX.

    Proof. When n=1, (3.36) is equivalent to

    μ(Df(a,b),t)tt+φ(a,b)andν(Df(a,b),t)φ(a,b)t+φ(a,b) (3.38)

    for all a,bX and all t>0. Let

    fe(a)=f(a)+f(a)2

    for all all aX. Then fe(0)=0,fe(a)=fe(a). And we have

    {μ(Dfe(a,b),t)=μ(12Df(a,b)+12Df(a,b),t)=μ(Df(a,b)+Df(a,b),2t)μ(Df(a,b),t)μ(Df(a,b),t)min{μ(Df(a,b),t),μ(Df(a,b),t)}tt+˜φ(a,b),ν(Dfe(a,b),t)=ν(12Df(a,b)+12Df(a,b),t)=ν(Df(a,b)+Df(a,b),2t)ν(Df(a,b),t)ν(Df(a,b),t)max{ν(Df(a,b),t),ν(Df(a,b),t)}˜φ(a,b)t+˜φ(a,b) (3.39)

    for all aX and all t>0. Let

    fo(a)=f(a)f(a)2

    for all all aX. Then f0(0)=0,fo(a)=fo(a). And we obtain

    {μ(Dfo(a,b),t)=μ(12Df(a,b)12Df(a,b),t)=μ(Df(a,b)Df(a,b),2t)μ(Df(a,b),t)μ(Df(a,b),t)=min{μ(Df(a,b),t),μ(Df(a,b),t)}tt+˜φ(a,b),ν(Dfo(a,b),t)=ν(12Df(a,b)12Df(a,b),t)=ν(Df(a,b)Df(a,b),2t)ν(Df(a,b),t)ν(Df(a,b),t)=max{ν(Df(a,b),t),ν(Df(a,b),t)}˜φ(a,b)t+˜φ(a,b) (3.40)

    for all aX and all t>0. It follows that the definition of ˜φ that ˜φ(ka,kb)=α˜φ(a,b) for all a,bX. It is easy to check that the condition of Theorems 3.3 and 3.4 are satisfying. Then applying the proofs of Theorems 3.3 and 3.4, we know that there exists a unique quadratic mapping Q: XY and a unique additive mapping A: XY satisfying

    {μ(fe(a)Q(a),t)(k2α)(2k+s1)t(k2α)(2k+s1)t+˜φ(0,a),ν(fe(a)Q(a),t)˜φ(0,a)(k2α)(2k+s1)t+˜φ(0,a) (3.41)

    and

    {μ(fo(a)A(a),t)(kα)(2k+s1)t(kα)(2k+s1)t+˜φ(0,a),ν(fo(a)A(a),t)˜φ(0,a)(kα)(2k+s1)t+˜φ(0,a) (3.42)

    for all aX and all t>0. Therefore

    {μ(f(a)Q(a)A(a),t)=μ(fe(a)Q(a)+fo(a)A(a),t)μ(fe(a)Q(a),t2)μ(fo(a)A(a),t2)=min{μ(fe(a)Q(a),t2),μ(fo(a)A(a),t2)}min{(k2α)(2k+s1)t(k2α)(2k+s1)t+2˜φ(0,a),(kα)(2k+s1)t(kα)(2k+s1)t+2˜φ(0,a)}=(kα)(2k+s1)t(kα)(2k+s1)t+2˜φ(0,a),ν(f(a)Q(a)A(a),t)=ν(fe(a)Q(a)+fo(a)A(a),t)ν(fe(a)Q(a),t2)ν(fo(a)A(a),t2)=max{ν(fe(a)Q(a),t2),ν(fo(a)A(a),t2)}max{2˜φ(0,a)(k2α)(2k+s1)t+2˜φ(0,a),2˜φ(0,a)(kα)(2k+s1)t+2˜φ(0,a)}=2˜φ(0,a)(kα)(2k+s1)t+2˜φ(0,a). (3.43)

    By Lemma 2.3 and (3.43), we have

    {μn(fn([xij])Qn([xij])An([xij]),t)min{μ(f(xij)Q(xij)A(xij),tn2):i,j=1,,n}(kα)(2k+s1)t(kα)(2k+s1)t+2n2ni,j=1˜φ(0,xij),νn(fn([xij])Qn([xij])An([xij]),t)max{ν(f(xij)Q(xij)A(xij),tn2):i,j=1,,n}2n2ni,j=1˜φ(0,xij)(kα)(2k+s1)t+2n2ni,j=1˜φ(0,xij)

    for all x=[xij]Mn(X) and all t>0. Thus Q: XY is a unique quadratic mapping and a unique additive mapping A: XY satisfying (3.37), as desired. This completes the proof of the theorem.

    Corollary 3.6. Let r,θ be positive real numbers with r<1. Suppose that a mapping f: XY with f(0)=0 satisfies the inequality

    {μn(Dfn([xij],[yij]),t)tt+ni,j=1θ(xijr+yijr),νn(Dfn([xij],[yij]),t)ni,j=1θ(xijr+yijr)t+ni,j=1θ(xijr+yijr) (3.44)

    for all x=[xij],y=[yij]Mn(X) and all t>0. Then there exist a unique quadratic mapping Q: XY and a unique additive mapping A: XY such that

    {μn(fn([xij])Qn([xij])An([xij]),t)(kkr)(2k+s1)t(kkr)(2k+s1)t+4n2ni,j=1θxijr,νn(fn([xij])Qn([xij])An([xij]),t)4n2ni,j=1θxijr(kkr)(2k+s1)t+4n2ni,j=1θxijr (3.45)

    for all x=[xij]Mn(X) and all t>0.

    Proof. The proof follows from Theorem 3.5 by taking φ(a,b)=θ(ar+br) for all a,bX, we obtain the desired result.

    In this section, we will prove the Hyers-Ulam stability of the functional Eq (1.1) in matrix intuitionistic fuzzy normed spaces by applying the fixed point method.

    Theorem 4.1. Let φo: X2[0,) be a function such that for some real number ρ with 0<ρ<1 and

    φo(a,b)=ρkφo(ka,kb) (4.1)

    for all a,bX. Suppose that an odd mapping f: XY satisfies the inequality

    {μn(Dfn([xij],[yij]),t)tt+ni,j=1φo(xij,yij),νn(Dfn([xij],[yij]),t)ni,j=1φo(xij,yij)t+ni,j=1φo(xij,yij) (4.2)

    for all x=[xij],y=[yij]Mn(X) and all t>0. Then there exists a unique additive mapping A: XY such that

    {μn(fn([xij])An([xij]),t)2k(2k+s1)(1ρ)t2k(2k+s1)(1ρ)t+ρn2ni,j=1φo(0,xij),νn(fn([xij])An([xij]),t)ρn2ni,j=1φo(0,xij)2k(2k+s1)(1ρ)t+ρn2ni,j=1φo(0,xij) (4.3)

    for all x=[xij]Mn(X) and all t>0.

    Proof. When n=1, similar to the proof of Theorem 3.3, we have

    {μ(2(2k+s1)f(ka)2(2k+s1)kf(a),t)tt+φo(0,a),ν(2(2k+s1)f(ka)2(2k+s1)kf(a),t)φo(0,a)t+φo(0,a) (4.4)

    for all aX and all t>0.

    Let S1={g1:XY}, and introduce a generalized metric d1 on S1 as follows:

    d1(g1,h1):=inf{λR+|{μ(g1(a)h1(a),λt)tt+φo(0,a),ν(g1(a)h1(a),λt)φo(0,a)t+φo(0,a),aX,t>0}.

    It is easy to prove that (S1,d1) is a complete generalized metric space ([3,13]). Now, we define the mapping J1: S1S1 by

    J1g1(a):=kg1(ak),for allg1S1andaX. (4.5)

    Let g1,h1S1 and let λR+ be an arbitrary constant with d1(g1,h1)λ. From the definition of d1, we get

    {μ(g1(a)h1(a),λt)tt+φo(0,a),ν(g1(a)h1(a),λt)φo(0,a)t+φo(0,a)

    for all aX and t>0. Therefore, using (4.1), we get

    {μ(J1g1(a)J1h1(a),λρt)=μ(kg1(ak)kh1(ak),λρt)=μ(g1(ak)h1(ak),λρtk)ρktρkt+ρkφo(0,a)=tt+φo(0,a),ν(J1g1(a)J1h1(a),λρt)=ν(kg1(ak)kh1(ak),λρt)=ν(g1(ak)h1(ak),λρtk)ρkφo(0,a)ρkt+ρkφo(0,a)=φo(0,a)t+φo(0,a) (4.6)

    for some ρ<1, all aX and all t>0. Hence, it holds that d1(J1g1,J1h1)λρ, that is, d1(J1g1,J1h1)ρd1(g1,h1) for all g1,h1S1.

    Furthermore, by (4.1) and (4.4), we obtain the inequality

    d(f,J1f)ρ2k(2k+s1).

    It follows from Lemma 2.4 that the sequence Jp1f converges to a fixed point A of J1, that is, for all aX and all t>0,

    A:XY,A(a):=(μ,ν)limpkpf(akp) (4.7)

    and

    A(ka)=kA(a). (4.8)

    Meanwhile, A is the unique fixed point of J1 in the set

    S1={g1S1:d1(f,g1)<}.

    Thus, there exists a λR+ such that

    {μ(f(a)A(a),λt)tt+φo(0,a),ν(f(a)A(a),λt)φo(0,a)t+φo(0,a)

    for all aX and all t>0. Also,

    d1(f,A)11ρd(f,J1f)ρ2k(1ρ)(2k+s1).

    This means that the following inequality

    {μ(f(a)A(a),t)2k(2k+s1)(1ρ)t2k(2k+s1)(1ρ)t+ρφo(0,a),ν(f(a)A(a),t)ρφo(0,a)2k(2k+s1)(1ρ)t+ρφo(0,a) (4.9)

    holds for all aX and all t>0. It follows from (3.4) and (4.1) that

    μ(kpDf(akp,bkp),t)tt+ρpφo(a,b),ν(kpDf(akp,bkp),t)ρpφo(a,b)t+ρpφo(a,b) (4.10)

    for all a,bX and all t>0. Letting p in (4.10), we obtain

    μ(DA(a,b),t)=1andν(DA(a,b),t)=0 (4.11)

    for all a,bX and all t>0. This means that A satisfies the functional Eq (1.1). Since f: XY is an odd mapping, and using the definition A, we have A(a)=A(a) for all aX. Thus by Lemma 3.1, the mapping A: XY is additive.

    By Lemma 2.3 and (4.9), we get

    {μn(fn([xij])An([xij]),t)min{μ(f(xij)A(xij),tn2):i,j=1,,n}2k(2k+s1)(1ρ)t2k(2k+s1)(1ρ)t+ρn2ni,j=1φo(0,xij),νn(fn([xij])An([xij]),t)max{ν(f(xij)A(xij),tn2):i,j=1,,n}ρn2ni,j=1φo(0,xij)2k(2k+s1)(1ρ)t+ρn2ni,j=1φo(0,xij)

    for all x=[xij]Mn(X) and all t>0. Thus A: XY is a unique additive mapping satisfying (4.3), as desired. This completes the proof of the theorem.

    Theorem 4.2. Let φe: X2[0,) be a function such that for some real number ρ with 0<ρ<1 and

    φe(a,b)=ρk2φe(ka,kb) (4.12)

    for all a,bX. Suppose that an even mapping f: XY satisfies the inequality

    {μn(Dfn([xij],[yij]),t)tt+ni,j=1φe(xij,yij),νn(Dfn([xij],[yij]),t)ni,j=1φe(xij,yij)t+ni,j=1φe(xij,yij) (4.13)

    for all x=[xij],y=[yij]Mn(X) and all t>0. Then there exists a unique quadratic mapping Q: XY such that

    {μn(fn([xij])Qn([xij]),t)2k2(2k+s1)(1ρ)t2k2(2k+s1)(1ρ)t+ρn2ni,j=1φe(0,xij),νn(fn([xij])Qn([xij]),t)ρn2ni,j=1φe(0,xij)2k2(2k+s1)(1ρ)t+ρn2ni,j=1φe(0,xij) (4.14)

    for all x=[xij]Mn(X) and all t>0.

    Proof. When n=1, similar to the proof of Theorem 3.4, we obtain

    {μ(2(2k+s1)f(ka)2(2k+s1)k2f(a),t)tt+φe(0,a),ν(2(2k+s1)f(ka)2(2k+s1)k2f(a),t)φe(0,a)t+φe(0,a) (4.15)

    for all aX and all t>0.

    Let S2:={g2:XY}, and introduce a generalized metric d2 on S2 as follows:

    d2(g2,h2):=inf{λR+|{μ(g2(a)h2(a),λt)tt+φe(0,a),ν(g2(a)h2(a),λt)φe(0,a)t+φe(0,a),aX,t>0}.

    It is easy to prove that (S2,d2) is a complete generalized metric space ([3,13]). Now, we define the mapping J2: S2S2 by

    J2g2(a):=k2g2(ak),for allg2S2andaX. (4.16)

    Let g2,h2S2 and let λR+ be an arbitrary constant with d2(g2,h2)λ. From the definition of d2, we get

    {μ(g2(a)h2(a),λt)tt+φe(0,a),ν(g2(a)h2(a),λt)φe(0,a)t+φe(0,a)

    for all aX and t>0. Therefore, using (4.12), we get

    {μ(J2g2(a)J2h2(a),λρt)=μ(k2g2(ak)k2h2(ak),λρt)=μ(g2(ak)h2(ak),λρtk2)ρk2tρk2t+ρk2φe(0,a)=tt+φe(0,a),ν(J2g2(a)J2h2(a),λρt)=ν(k2g2(ak)k2h2(ak),λρt)=ν(g2(ak)h2(ak),λρtk2)ρk2φe(0,a)ρk2t+ρk2φe(0,a)=φe(0,a)t+φe(0,a) (4.17)

    for some ρ<1, all aX and all t>0. Hence, it holds that d2(J2g2,J2h2)λρ, that is, d2(J2g2,J2h2)ρd2(g2,h2) for all g2,h2S2.

    Furthermore, by (4.12) and (4.15), we obtain the inequality

    d(f,J2f)ρ2k2(2k+s1).

    It follows from Lemma 2.4 that the sequence Jp2f converges to a fixed point Q of J2, that is, for all aX and all t>0,

    Q:XY,Q(a):=(μ,ν)limpk2pf(akp) (4.18)

    and

    Q(ka)=k2Q(a). (4.19)

    Meanwhile, Q is the unique fixed point of J2 in the set

    S2={g2S2:d2(f,g2)<}.

    Thus there exists a λR+ such that

    {μ(f(a)Q(a),λt)tt+φe(0,a),ν(f(a)Q(a),λt)φe(0,a)t+φe(0,a)

    for all aX and all t>0. Also,

    d2(f,Q)11ρd(f,J2f)ρ2k2(1ρ)(2k+s1).

    This means that the following inequality

    {μ(f(a)Q(a),t)2k2(2k+s1)(1ρ)t2k2(2k+s1)(1ρ)t+ρφe(0,a),ν(f(a)Q(a),t)ρφe(0,a)2k2(2k+s1)(1ρ)t+ρφe(0,a) (4.20)

    holds for all aX and all t>0. The rest of the proof is similar to the proof of Theorem 4.1. This completes the proof of the theorem.

    Theorem 4.3. Let φ: X2[0,) be a function such that for some real number ρ with 0<ρ<k,

    φ(a,b)=ρk2φ(ka,kb) (4.21)

    for all a,bX. Suppose that a mapping f: XY with f(0)=0 satisfies the inequality

    {μn(Dfn([xij],[yij]),t)tt+ni,j=1φ(xij,yij),νn(Dfn([xij],[yij]),t)ni,j=1φ(xij,yij)t+ni,j=1φ(xij,yij) (4.22)

    for all x=[xij],y=[yij]Mn(X) and all t>0. Then there exist a unique quadratic mapping Q: XY and a unique additive mapping A: XY such that

    {μn(fn([xij])Qn([xij])An([xij]),t)k(2k+s1)(1ρ)tk(2k+s1)(1ρ)t+ρn2ni,j=1˜φ(0,xij),νn(fn([xij])Qn([xij])An([xij]),t)ρn2ni,j=1˜φ(0,xij)k(2k+s1)(1ρ)t+ρn2ni,j=1˜φ(0,xij) (4.23)

    for all x=[xij]Mn(X) and all t>0, ˜φ(a,b)=φ(a,b)+φ(a,b) for all a,bX.

    Proof. The proof follows from Theorems 4.1 and 4.2, and a method similar to Theorem 3.5. This completes the proof of the theorem.

    Corollary 4.4. Let r,θ be positive real numbers with r>2. Suppose that a mapping f: XY with f(0)=0 satisfies the inequality

    {μn(Dfn([xij],[yij]),t)tt+ni,j=1θ(xijr+yijr),νn(Dfn([xij],[yij]),t)ni,j=1θ(xijr+yijr)t+ni,j=1θ(xijr+yijr) (4.24)

    for all x=[xij],y=[yij]Mn(X) and all t>0. Then there exist a unique quadratic mapping Q: XY and a unique additive mapping A: XY such that

    {μn(fn([xij])Qn([xij])An([xij]),t)(2k+s1)(krk2)t(2k+s1)(krk2)t+2kn2ni,j=1θxijr,νn(fn([xij])Qn([xij])An([xij]),t)2kn2ni,j=1θxijr(2k+s1)(krk2)t+2kn2ni,j=1θxijr (4.25)

    for all x=[xij]Mn(X) and all t>0.

    Proof. Taking φ(a,b)=θ(ar+br) for all a,bX and ρ=k2r in Theorem 4.3, we get the desired result.

    We use the direct and fixed point methods to investigate the Hyers-Ulam stability of the functional Eq (1.1) in the framework of matrix intuitionistic fuzzy normed spaces. We therefore provide a link two various discipline: matrix intuitionistic fuzzy normed spaces and functional equations. We generalized the Hyers-Ulam stability results of the functional Eq (1.1) from quasi-Banach spaces to matrix intuitionistic fuzzy normed spaces. These circumstances can be applied to other significant functional equations.

    The author declare he has not used Artificial Intelligence (AI) tools in the creation of this article.

    The author is grateful to the referees for their helpful comments and suggestions that help to improve the quality of the manuscript.

    The author declares no conflict of interest in this paper.



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