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Representation of solution of initial value problem for fuzzy linear multi-term fractional differential equation with continuous variable coefficient

  • We consider the representation of solutions of the initial value problems of fuzzy linear multi-term in-homogeneous fractional differential equations with continuous variable coefficients.

    Citation: Huichol Choi, Kinam Sin, Sunae Pak, Kyongjin Sok, Sungryol So. Representation of solution of initial value problem for fuzzy linear multi-term fractional differential equation with continuous variable coefficient[J]. AIMS Mathematics, 2019, 4(3): 613-625. doi: 10.3934/math.2019.3.613

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  • We consider the representation of solutions of the initial value problems of fuzzy linear multi-term in-homogeneous fractional differential equations with continuous variable coefficients.


    Nowadays the fractional differential equations (FDEs) are powerful tools representing many problems in various fields such as viscoelasticity, control engineering, diffusion processes, signal processing and electromagnetism.

    Recently many researchers are interested in research on fuzzy fractional differential equations(FFDEs). In one of the earliest papers, Agarwal et al. [1] considered a fractional differential equation of order α with uncertainty and introduced the concept of solution for proposed equations. Arshad and Lupulescu [2] presented the existence and uniqueness of solution for fractional differential equations with uncertainty using the concept of solution in means of [1]. Next, Khastan et al.[3] proved the existence result of solution for nonlinear fuzzy fractional differential equations with the Riemann-Liouville derivative by Schauder fixed-point theorem. Allahviranloo et al. [4] proved that uncertain fractional differential equation is equivalent to integral equation under Riemann-Liouville H-differentiability and obtained the explicit solution in case of the fuzzy linear fractional differential equations. Salahshour et al. [5] studied the analytical methods to solve the FFDE under the fuzzy Caputo fractional differentiability.

    Abdollahi et al. introduced the linear fuzzy Caputo-Fabrizio fractional differential equation with initial value problems and presented the general form of their solutions under generalized Hukuhara differentiability in [6] and H. V. Ngo et al. proved that the fractional fuzzy differential equation is not equal to the fractional fuzzy integral equation in general in [7]. Salahshour et al. [8] introduced that the solutions of FFDEs of order 0<β<1 are obtained by fuzzy Laplace transforms under Riemann-Liouville H-differentiability. Chehlabi and Allahviranloo [9] investigated the concreted solutions for fuzzy linear fractional differential equations of order 0<α<1 under Riemann-Liouville H-differentiability by using fractional hyperbolic functions and extended the results obtained in [8]. Also Arqub et al. introduced the exact and the numerical solutions of various fuzzy differential equations based on the reproducing kernel Hilbert space method under strongly generalized differentiability [10,11,12].

    Generally, the most of the FFDEs have no known exact solution. Thus the researches of approximate and numerical solutions of the FFDEs are important. Mazandarani and Vahidian Kamyad [13] obtained a fuzzy approximate solution to solve FFDEs with the Caputo-type fuzzy fractional derivative based on Hukuhara difference and strongly generalized fuzzy differentiability by using modified fractional Euler method. Also the operational matrix methods based on orthogonal polynomials have been proposed to solve FFDEs [14,15,16,17,18].

    Based on above considerations, we are going to obtain the representation of solutions of the initial value problems of fuzzy linear multi-term in-homogeneous fractional differential equations with continuous variable coefficients.

    The paper is organized as follows. In Section 2, we introduced some definitions and properties for fuzzy calculus and proposed basic results of our paper. The representation of solution of proposed problem is described in Section 3. The examples are presented to illustrate our result in Section 4. Finally, the conclusion is summarized in Section 5.

    We denote the set of all fuzzy numbers on R by RF. A fuzzy number is a mapping u:R[0,1] with the following properties:

    (ⅰ) u is normal, i.e., x0Ru(x0)=1,

    (ⅱ) u is a convex fuzzy subset, i.e., u(λx+(1λ)y)min{u(x),u(y)},x,yRλ[0,1],

    (ⅲ) u is upper semi-continuous on R,

    (ⅳ) The set ¯supp(u) is compact in R (where supp(u):={xR|u(x)>0}).

    Then RF is called the space of fuzzy numbers.

    We denote the r-level form of the fuzzy number uRF by [u]r:=[u1(r),u2(r)], 0r1.

    Let's u,vRF. The distance d:RF×RFR+ is defined by

    d(u,v):=supr[0,1]max{|u1(r)v1(r)|,|u2(r)v2(r)|}.

    We introduce following notations to understand our paper.

    C(I) is the set of all continuous real functions on I.

    CF(I) is the set of all continuous fuzzy-valued functions on I.

    ACF(I) is the set of all absolutely continuous fuzzy-valued functions on I.

    LF(I) is the space of all Lebesque integrable fuzzy-valued functions on I, where I=[0,L] and without losing generality, we promise that L=1.

    Definition 2.1. [4] Let x,yRF. If there exists zRF such that x=yz, then z is called the H-difference of x and y, and it is denoted by z=xHy.

    Definition 2.2. [8] Let f:IRF and fix x0(0,1).

    We say that f is (1)-differentiable at x0, if for all h>0 sufficiently near to 0, f(x0+h)Hf(x0), f(x0)Hf(x0h) and the relation

    limh0+f(x0+h)Hf(x0)h=limh0+f(x0)Hf(x0h)h

    exists. Then the limit is defined by D(1)f(x0).

    Definition 2.3. [4] Let fLF(I). The fuzzy Riemann-Liouville integral of fuzzy-valued function f is defined as follow

    Iβ0+f(x)=1Γ(β)x0(xs)β1f(s)ds,x(0,1],

    where 0<ν1.

    Definition 2.4. [4,7] Let f:IRF,0<β<1. We say that f is fuzzy Riemann-Liouville H-differentiable of order β if

    I1β0+f(x)=x0(xs)βΓ(1β)f(s)ds,x(0,1]

    is (1)-differentiable. Then fuzzy Riemann-Liouville H-derivative of order β of function f is denoted RLDβ0+f(x):=D(1)I1β0+f(x).

    Definition 2.5. Let f:IRF,0<β<1. We say that f is fuzzy Caputo-type differentiable of order β if H-difference f(x)Hf(0) exists and I1β0+(f(x)Hf(0))ACF(I) satisfies. Then fuzzy Caputo-type derivative of order β of function f is denoted by

    (cDβ0+f)(x):=RLDβ0+(f(x)Hf(0)),x(0,1].

    We consider initial value problem of fuzzy multi-term fractional differential equation as

    cDα0+y(x)=f(x,y(x),cDβ0+y(x)),xI,y(0)=y0,y0RF. (2.1)

    Definition 2.6. Let y:IRF. y is called the solution of Eq. (2.1) if y is to be cDα0+y(x),cDβ0+y(x)CF(I) and satisfies Eq. (2.1).

    Now let consider as follows

    cDα0+y(x)=z(x),xI,y(0)=y0,y0RF. (2.2)

    Theorem 2.1. Let f of Eq. (2.1) be a continuous function with respect to every variable. If y(x) is the solution of Eq. (2.1), the fuzzy-valued function z(x) which is constructed by z(x):=cDα0+y(x) is the solution in CF(I) of fuzzy integral equation as follows

    z(x)=f(x,y01Γ(α)x0z(s)(xs)1αds,Iαβ0+z(x)),xI. (2.3)

    Conversely if z(x) is the solution in CF(I) of fuzzy integral equation (2.3), y(x) which is constructed by

    y(x)=y01Γ(α)x0z(s)(xs)1αds

    is the solution of Eq. (2.1).

    Let use the following distance structure in CF(I)

    u,vCF(I),d(u,v):=maxtId(u(t),v(t)).

    (CF(I),d) is a complete metric space. For any k, let consider distance as

    u,vCF(I),dk(u,v):=maxtIektd(u(t),v(t)).

    Then the distance dk is equal to the distance d. Namely

    M,m>0;u,vCF(I),mdk(u,v)d(u,v)Mdk(u,v).

    Theorem 2.2. Assume that the function in Eq. (2.3) is continuous in its all variables and especially, for y1,y2,z1,z2RF and f satisfies the following condition

    d(f(x,y1,z1),f(x,y2,z2))L1d(y1,y2)+L2d(z1,z2).

    Then the fuzzy integral equation (2.3) has the unique solution.

    We consider the representation of solution of initial value problem for fuzzy linear multi-term fractional differential equation with continuous variable coefficient as

    cDα0+y(x)=a(x)cDβ0+y(x)b(x)y(x)g(x),xI,0<β<α<1,y(0)=y0,y0RF, (3.1)

    where a,bC(I),gCF(I).

    First we will obtain the representation of solution for corresponding fuzzy integral equation (2.3).

    If z(x):=cDα0+y(x), the following relations satisfy

    y(x)=y0Iα0+z(x).

    Then we have

    cDβ0+y(x)=cDβ0+Iα0+z(x)=Iαβ0+z(x).
    z(x)=a(x)Iαβ0+z(x)b(x)(y0Iα0+z(x))g(x)=a(x)Iαβ0+z(x)b(x)Iα0+z(x)g(x)b(x)y0. (3.2)

    Now let define the operator L by

    (Lz)(x):=a(x)Iαβ0+z(x)b(x)Iα0+z(x),xI.

    From a,bC(I), it is obvious that L:CF(I)CF(I). By using the operator L, the integral equation (3.2) can be expressed as

    z(x)=(Lz)(x)g(x)b(x)y0.

    Let ˆg(x):=g(x)b(x)y0. Then the integral equation (3.2) is denoted by

    z(x)=(Lz)(x)ˆg(x). (3.3)

    Also let denote the operator IHL as

    (IHL)z:=zHLz,

    where the operator I is identity operator. From Eq. (3.3), we have

    z(x)H(Lz)(x)=ˆg(x),(IHL)z(x)=ˆg(x). (3.4)

    Lemma 3.1. Let a,b,c,dRF. The following relations are satisfied

    (1) If (ab)Hb exists, then (ab)Hb=a.

    (2) If (ab)H(cd),(aHc) and (bHd) exist, then

    (ab)H(cd)=(aHc)(bHd).

    Proof. First we prove (1). Now let E:=(ab)Hb. Then we get

    Eb=ab.

    Also for r[0,1], we have [E]r=[Er,Er+],[a]r=[ar,ar+] and [b]r=[br,br+]. Therefore following relations holds

    [Eb]r=[ab]r,[Er+br,Er++br+]=[ar+br,ar++br+],{Er+br=ar+br,Er++br+=ar++br+,{Er=ar,Er+=ar+.

    Namely E=a.

    We prove (2). Let E:=(ab)H(cd), F:=(aHc)(bHd). As E(cd)=ab stands, for r[0,1], we get

    [Er+cr+dr,Er++cr++dr+]=[ar+br,ar++br+],{Er+cr+dr=ar+br,Er++cr++dr+=ar++br+.

    Therefore

    {Er=ar+brcrdr,Er+=ar++br+cr+dr+.

    In the one hand, let F:=(aHc)(bHd),G:=aHc. Because Gc=a stands, we have

    [Gr+cr,Gr++cr+]=[ar,ar+],{Gr=arcr,Gr+=ar+cr+,{Fr=arcr+brdr,Fr+=ar+cr++br+dr+.

    Therefore the proof is completed.

    If zCF(I) is the solution of the fuzzy integral equation (3.3) and D(IHL) is the domain of the operator IHL, we can think the following Lemma.

    Lemma 3.2. If zD(L) is the solution of the integral equation (3.3) and

    k{0,},LkˆgD(L). (3.5)

    Then the following relations are satisfied

    (1) kN,LkzD(IHL),

    (2) nk=0LkzD(IHL),

    where D(L) is the domain of the operator L.

    Proof. First we prove (1). Because zD(L) is the solution of the integral equation (3.3), the following relations are satisfied

    z(x)=(Lz)(x)ˆg(x),Lz(x)=(L2z)(x)Lˆg(x),Lz(x)H(L2z)(x)=Lˆg(x),(IHL)Lz(x)=Lˆg(x).

    Consequently we have LzD(IHL). Similarity we can prove

    kN,LkzD(IHL).

    Next let prove (2). By the assumption of z, we get

    z(x)Lz(x)=(Lz)(x)ˆg(x)(L2z)(x)Lˆg(x).

    Also the following relation is satisfied

    z(x)Lz(x)=(Lz)(x)(L2z)(x)ˆg(x)Lˆg(x)=L(z(x)Lz(x))ˆg(x)Lˆg(x).

    Therefore we have

    (z(x)Lz(x))HL(z(x)Lz(x))=ˆg(x)Lˆg(x),
    (IHL)(z(x)Lz(x))=ˆg(x)Lˆg(x),
    z(x)Lz(x)D(IHL).

    In the same way, we can prove nk=0LkzD(IHL).

    For zD(IHL), we can know that

    (IHL)nk=0Lkz(x)=nk=0Lkz(x)HLnk=0Lkz(x)=nk=0Lkz(x)Hnk=0Lk+1z(x)
    =(z(x)Lz(x)Lnz(x))H(Lz(x)Ln+1z(x)).

    where the symbol means that the operator (IHL) is applied to nk=0Lkz(x).

    Because zD(IHL), z(x)HLz(x) exists. Also applying the operator L to the both of Eq. (3.4), we get

    L(IHL)z(x)=Lˆg(x).

    Similarity, we can obtain

    Ln(z(x)HLz(x))=Lnz(x)HLn+1z(x).

    On the other hand, from (2) of Lemma 3.1, we get

    z(x)=(z(x)H0)(Lz(x)HLz(x))=(z(x)Lz(x))H(0Lz(x))=(z(x)Lz(x))H(Lz(x)0)=(IHL)z(x)Lz(x)=(IHL)z(x)(IHL)Lz(x)L2z(x)=(IHL)(z(x)Lz(x))L2z(x)=(IHL)nk=0Lkz(x)Ln+1z(x).

    Namely

    (IHL)nk=0Lkz(x)=z(x)HLn+1z(x). (3.6)

    Let estimate dk(ˆ0,Lz) for k>0. Since

    dk(ˆ0,Lz)=maxtIektd(ˆ0,Lz(t))=maxtIektd(ˆ0,a(t)Iαβ0+z(t)b(t)Iα0+z(t))maxtIekt{d(ˆ0,a(t)Iαβ0+z(t))+d(ˆ0,b(t)Iα0+z(t))}=maxtIekt{|a(t)|d(ˆ0,Iαβ0+z(t))+|b(t)|d(ˆ0,Iα0+z(t))},
    d(ˆ0,Iα0+z(t))Iα0+d(ˆ0,z(t))ektkαdk(ˆ0,z)

    and

    d(ˆ0,Iαβ0+z(t))ektkαβdk(ˆ0,z),

    we have

    dk(ˆ0,Lz)maxtIekt{|a(t)|d(ˆ0,Iαβ0+z(t))+|b(t)|d(ˆ0,Iα0+z(t))}maxtIekt{|a(t)|ektkαβdk(ˆ0,z)+|b(t)|ektkαdk(ˆ0,z)}={||a(t)||C(I)1kαβ+||b(t)||C(I)1kα}dk(ˆ0,z).

    If there is a positive number k satisfying

    w:=||a(t)||C(I)1kαβ+||b(t)||C(I)1kα<1,

    then we have

    dk(ˆ0,Lz)wdk(ˆ0,z).

    Therefore

    dk(ˆ0,L2z)wdk(ˆ0,Lz)w2dk(ˆ0,z),dk(ˆ0,Lnz)wndk(ˆ0,z). (3.7)

    Now let denote as

    S(x):=z(x)HLn+1z(x).

    Then we get

    S(x)Ln+1z(x)=z(x).

    Therefore

    d(z(x)HLn+1z(x),z(x))=d(S(x),S(x)Ln+1z(x))=d(ˆ0,Ln+1z(x)).

    From Eq. (3.7), we have

    dk(zHLn+1z,z)=dk(ˆ0,Ln+1z)wn+1dk(ˆ0,z).

    Namely

    limndk(zHLn+1z,z)=0.

    Consequently from Eq. (3.6), we hold

    (IHL)k=0Lkz(x)=z(x).

    Similarity from Eq. (3.4), Since

    k=0Lk(IHL)z(x)=k=0Lkˆg(x),

    we get

    z(x)=k=0Lkˆg(x). (3.8)

    The following Theorem gives the representation of solution for the proposed problem (3.1).

    Theorem 3.1. If k{0,1,},LkˆgD(L) holds, the initial value problem (3.1) has the representation of the solution as follows

    y(x)=y0Iα0+g(x)Iα0+(b(x)y0)k=1Iα0+(a(x)Iαβ0+b(x)Iα0+)kˆg(x).

    Proof. Eq. (3.8) can be rewritten as

    z(x)=ˆg(x)k=1Lkˆg(x)=ˆg(x)k=1(a(x)Iαβ0+b(x)Iα0+)kˆg(x).

    Therefore we obtain

    y(x)=y0Iα0+z(x)=y0Iα0+(ˆg(x)k=1(a(x)Iαβ0+b(x)Iα0+)kˆg(x))=y0Iα0+ˆg(x)k=1Iα0+(a(x)Iαβ0+b(x)Iα0+)kˆg(x)=y0Iα0+(g(x)b(x)y0)k=1Iα0+(a(x)Iαβ0+b(x)Iα0+)kˆg(x)=y0Iα0+g(x)Iα0+(b(x)y0)k=1Iα0+(a(x)Iαβ0+b(x)Iα0+)kˆg(x).

    The proof is completed.

    We consider the analytical representation of solution for the fuzzy fractional differential equation as following

    cDα0+y(x)=λy(x),xI,y(0)=y0,y0RF. (4.1)

    By using Theorem 3.3, we get ˆg(x):=λy0, (Lz)(x):=λIα0+z(x),xI. Therefore the conditions of Theorem 3.1 are satisfied.

    Corollary 4.1. The initial value problem (4.1) have the solution as follows

    in case of λ>0,

    y(x)=Eα(λxα)y0,

    in case of λ<0,

    y(x)=E2α,1(λ2x2α)y0λxαE2α,α+1(λ2x2α)y0.

    Proof. In case of λ>0, we get

    y(x)=y0Iα0+(λy0)k=1Iα0+(λIα0+)kλy0=y0Iα0+(λy0)k=1λk+1(Iα0+)k+1y0=y0λxαΓ(α+1)y0k=1λk+1x(k+1)αΓ((k+1)α+1)y0=y0k=0λk+1x(k+1)αΓ((k+1)α+1)y0=k=0(λxα)kΓ(kα+1)y0=Eα(λxα)y0.

    Also in case of λ<0, we have

    y(x)=y0Iα0+(λy0)k=1Iα0+(λIα0+)k(λy0)=y0Iα0+(λy0)k=1λk+1(Iα0+)k+1y0=y0λxαΓ(α+1)y0(k=1λ2k(Iα0+)2k1)y0(k=1λ2k+1(Iα0+)2k+11)y0=y0(k=1λ2kx2kαΓ(2kα+1))y0λxαΓ(α+1)y0(k=1λ2k+1x(2k+1)αΓ((2k+1)α+1))y0=y0(k=1λ2kx2kαΓ(2kα+1))y0(k=0λ2k+1x(2k+1)αΓ((2k+1)α+1))y0=(k=0(λxα)2kΓ(2kα+1))y0λxα(k=0(λxα)2kΓ(2kα+α+1))y0=E2α,1(λ2x2α)y0λxαE2α,α+1(λ2x2α)y0.

    Next we consider the representation of solution for initial value problem of the non-homogeneous fuzzy fractional differential equation with constant coefficient as following

    cDα0+y(x)=λy(x)+g(x),xI,y(0)=y0,y0RF. (4.2)

    Let's denote ˆg(x):=g(x)λy0, (Lz)(x):=λIα0+z(x),xI in order to use Theorem 3.1.

    Corollary 4.2. If k{0,1,},LkgD(L), then initial value problem (4.2) have the solution as follows

    in case of λ>0,

    y(t)=Eα(λxα)y0x0(xt)α1g(t)Eα,α(λ(xt)α)dt,

    in case of λ<0,

    y(x)=E2α,1(λ2x2α)y0λxαE2α,α+1(λ2x2α)y0x0(xt)α1E2α,α(λ2(xt)2α)g(t)dtλx0(xt)2α1E2α,2α(λ2(xt)2α)g(t)dt.

    Proof. In case of λ>0, we get

    y(x)=y0Iα0+g(x)Iα0+(λy0)k=1Iα0+(λIα0+)k(g(x)λy0)=y0Iα0+(λy0)k=1Iα0+(λIα0+)kλy0Iα0+g(x)k=1Iα0+(λIα0+)kg(x)=Eα(λxα)y0k=0λkI(k+1)α0+g(x)=Eα(λxα)y0k=0λk1Γ((k+1)α)x0g(t)(xt)1(k+1)αdt=Eα(λxα)y0x0(xt)α1Eα,α(λ(xt)α)g(t)dt.

    And in case of λ<0, we obtain

    y(x)=y0Iα0+g(x)Iα0+(λy0)k=1Iα0+(λIα0+)k(g(x)λy0)=y0k=0λ2kI(2k+1)α0+g(x)k=0λ2k+1I(2k+1)α0+y0k=1λ2k1I2kα0+g(x)k=1λ2k1I2kα0+λy0=k=0λ2kI2kα0+y0k=0λ2k+1I(2k+1)α0+y0k=0λ2kI(2k+1)α0+g(x)k=1λ2k1I2kα0+g(x)=(k=0(λ2x2α)kΓ(2kα+1))y0λxα(k=0(λ2x2α)kΓ(2kα+α+1))y0k=0λ2k1Γ((2k+1)α)x0g(t)(xt)1(2k+1)αdtk=1λ2k11Γ(2kα)x0g(t)(xt)12kαdt=E2α,1(λ2x2α)y0λxαE2α,α+1(λ2x2α)y0x0(xt)α1E2α,α(λ2(xt)2α)g(t)dtx0(xt)α1(k=0λ2k+1(xt)(2k+1)αΓ((2k+1)α+α))g(t)dt=E2α,1(λ2x2α)y0λxαE2α,α+1(λ2x2α)y0x0(xt)α1E2α,α(λ2(xt)2α)g(t)dtλx0(xt)2α1E2α,2α(λ2(xt)2α)g(t)dt.

    In this manuscript, we have studied the representation of the solution for initial value problem for fuzzy linear multi-term fractional differential equations with continuous variable coefficients. We obtained the representation of solutions for proposed problem by using the representation of solution of the corresponding fuzzy integral equations.

    The authors thank the referees for careful reading and helpful suggestions on the improvement of the manuscript.

    The authors declare no conflict of interest.



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