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Research article

Convergence properties of a family of inexact Levenberg-Marquardt methods

  • Received: 17 April 2023 Revised: 13 May 2023 Accepted: 18 May 2023 Published: 02 June 2023
  • MSC : 90C33, 65K05

  • We present a family of inexact Levenberg-Marquardt (LM) methods for the nonlinear equations which takes more general LM parameters and perturbation vectors. We derive an explicit formula of the convergence order of these inexact LM methods under the H¨oderian local error bound condition and the H¨oderian continuity of the Jacobian. Moreover, we develop a family of inexact LM methods with a nonmonotone line search and prove that it is globally convergent. Numerical results for solving the linear complementarity problem are reported.

    Citation: Luyao Zhao, Jingyong Tang. Convergence properties of a family of inexact Levenberg-Marquardt methods[J]. AIMS Mathematics, 2023, 8(8): 18649-18664. doi: 10.3934/math.2023950

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  • We present a family of inexact Levenberg-Marquardt (LM) methods for the nonlinear equations which takes more general LM parameters and perturbation vectors. We derive an explicit formula of the convergence order of these inexact LM methods under the H¨oderian local error bound condition and the H¨oderian continuity of the Jacobian. Moreover, we develop a family of inexact LM methods with a nonmonotone line search and prove that it is globally convergent. Numerical results for solving the linear complementarity problem are reported.



    Fractional calculus has been concerned with integration and differentiation of fractional (non-integer) order of the function. Riemann and Liouville defined the concept of fractional order intgro-differential equations [1]. Fractional calculus has developed an extensive attraction in current years in applied mathematics such as physics, medical, biology and engineering [2,3,4,5,6,7,8]. Whenever dealing with the fractional integro-differential equation many authors consider the terms Caputo fractional derivative, Riemann-Liouville and Grunwald-Letnikvo [9,10,11,12,13]. The subject fractional calculus has many applications in widespread and diverse field of science and engineering such as fractional dynamics in the trajectory control of redundant manipulators, viscoelasticity, electrochemistry, fluid mechanics, optics and signals processing etc.

    Fractional integro-differential equations having some uncertainties in the form of boundary conditions, initial conditions and so on [14,15,16]. To resolve these type of uncertainties mathematicians introduced some concepts fuzzy set theory is one of them.

    Zadeh introduced the concept of fuzzy set theory [17,18,19,20]. Later on Prade and Dubois [21,22], Nahmias [23], Tanaka and Mizumoto [24]. All of them experienced that the fuzzy number as a location of r-cut 0r1.

    Many authors investigated some numerical techniques related to these problem which include the existence of the solution for discontinuous [25], reproducing kernel algorithm [26], integro-differential under generalized Caputo differentiability [27], A domain decomposition method [28], fractional differential transform method [29], Jacobi polynomial operational matrix [30], global solutions for nonlinear fuzzy equations [31], radioactivity decay model [32], Caputo-Katugampola fractional derivative approach [33], two-dimensional legendre wavelet method [34], fuzzy Laplace transform [35], fuzzy sumudu transform [36]. Further we can see [37,38,39,40]

    Optimal Homotopy Asymptotic Method (OHAM) is one of the powerful techniques introduced by Marinca at al. [41,42,43] for approximate solution of differential equations. OHAM attracted an enormous importance in solving various problems in different field of science. Iqbal et al. applied this technique to Klein-Gordon equations and singular Lane-Emden type equation [44]. Sheikholeslami et al. used the proposed method for investigation of the laminar viscous flow and magneto hydrodynamic flow in a permeable channel [45]. Hashmi et al. obtained the solution of nonlinear Fredholm integral equations using OHAM [46]. Nawaz at al. applied the proposed method for solution of fractional order integro-differential equations [47], fractional order partial differential equations [48] and three-dimensional integral equations [49].

    Aim of our study is to extend OHAM for solution of system of fuzzy Volterra integro differential equation of fractional order of the following form

    Dαxu(x)=h(x)+xak(x,t)u(t)dt,0α1,x[0,1], (1.1)

    with the given initial condition

    uk(0),u0k(x),k=1,2,3,....,η1,η1<α<η,ηN,

    Where Dαx represents the fuzzy fractional derivative in Caputo sense for fractional order of α with respect to x, h:[a,b]RF is fuzzy valued function, k(x,t) is arbitrary kernel u0(x)RF is an unknown solution. RF represent set of all fuzzy valued function on real line.

    The remaining paper is structured as follows: A brief overview on some elementary concept, notations and definitions of fuzzy calculus and fuzzy fractional calculus are discussed in section 2. Analysis of the technique is presented in section 3. Proposed method is applied to solve fuzzy fractional order Volterra integro-differential equations in section 4. Result and discussion of the paper is given in section 5 and section 6 is the conclusion of the paper.

    In literature there exist various definitions of fuzzy calculus and fuzzy fractional calculus [50]. Some elementary concept, notations and definitions of fuzzy calculus and fuzzy fractional calculus related to this study are provided in this section.

    Definition 2.1. The Riemann-Liouville fractional integral operator Iαx of order α is [50]:

    Iαxu(x)={1Γ(α)x0(xt)α1u(t)dt=0,α>0,u(x),α=0. (2.1)

    Definition 2.2. Caputo partial fractional Derivative operator Dαx of order α with respect to x is defined as follow [50]:

    Dαxu(x)={1Γ(ηα)x0(xt)ηα1u(n)(t)dt=0,η1<αη,dηu(x)dxη,α=ηN. (2.2)

    which clearly shows that

    DαxIαxu(x)=u(x) (2.3)

    Definition 2.3. A fuzzy number σ is a mapping σ:R[0,1], satisfy the following property:

    a. σ is normal that is, x0R with u(x0)=1 [51,52].

    b. σ is a convex fuzzy set that is, u(λx+(1λ)y)min{u(x),u(y)} for all x,yR, λ[0,1].

    c. σ is upper semi-continuous in R.

    d. ¯{xR:u(x)>0} is compact.

    Definition 2.4. Parametric form of fuzzy number σ represented by an order pair (σ_,ˉσ) of the function (σ_(r),ˉσ(r)), satisfies the following conditions [52,53]:

    a. σ_(r) is bounded monotonic increasing left continuous r[0,1].

    b. ˉσ(r) is bounded monotonic decreasing left continuous r[0,1].

    c. σ_(r)ˉσ(r)r[0,1].

    Definition 2.5. Addition and scalar multiplication of fuzzy number is given as:

    a. (σ1σ2)=(σ_1(r)+σ_2(r),ˉσ1(r)+ˉσ2(r))

    b. (kσ)={(σ_(r),ˉσ(r)),k0,(σ_(r),ˉσ(r)),k<0.

    Definition 2.6. A fuzzy real valued function σ1,σ2:[a,b]R, then in [54]:

    DU(σ1,σ2)=sup{D(σ1(x),σ2(x))|x[a,b]}.

    Definition 2.7. Assume u:[a,b]RF. For every partition P={σ0,σ1,σ2,σ3,....,σn} and arbitrary i:σi1iσi, 2in consider

    Rp=nΣi=2u(j)(σiσi1). The definite integral of u(x) over [α,β] is

    βαu(x)dx=limRρ,

    which show existence of limit in metric [55].

    Definite integral exist if u(x) is continuous in metric D [51]:

    (βαu(x)dx_)=βαu_(x)dx,(¯βαu(x)dx)=βα¯u(x)dxt.

    By considering definition 2.4. as discussed in section 2, Eq (1.1) becomes:

    {Dαxu(x,r)h(x,r)xak(x,t)u(t,r)dt=0,Dαxˉu(x,r)h(x,r)xak(x,t)ˉu(t,r)dt=0,0α1,0r1,x[0,1], (3.1)

    with the given initial condition

    [uk(0)]r,(u0k(x,r),ˉu0k(x,r)),k=1,2,3,....,η1,η1<α<η,ηN, (3.2)

    The homotopy of OHAM [41,42,43], constructed as follow:

    {(1ρ)(αυ(x,r;ρ)tαh(x,r))=H(ρ)(αυ(x,r;ρ)tαh(x,r)δ(υ,r)),(1ρ)(αˉυ(x,r;ρ)tαˉh(x,r))=H(ρ)(αˉυ(x,r;ρ)tαˉh(x,r)ˉδ(ˉυ,r)). (3.3)

    where ρ[0,1], H(ρ)=m1cmρm for all ρ0 is an auxiliary function, if ρ=0 then H(0)=0 where

    {υ(x,r,0)=u0(x,r)υ(x,r;1)=u(x,r),ˉυ(x,r,0)=ˉu0(x,r)ˉυ(x,r;1)=ˉu(x,r).

    and cm represent auxiliary constants. Using Taylor's series to expand υ(x,r;ρ) about ρ we get

    {υ(x,r;ρ)=u0(x,r)+m1um(x,r)ρm,ˉυ(x,r;ρ)=ˉu0(x,r)+m1ˉum(x,r)ρm. (3.4)

    Inserting Eq (3.4) into Eq (3.3) we get series of the problems by comparing the like power of ρ given as follow:

    ρ0:{u0(x,r)h(x,r)=0,ˉu0(x,r)ˉh(x,r)=0. (3.5)
    ρ1:{u1(x,r)+c1δ(u0)+(1+c1)+u0(x,r)=0,ˉu1(x,r)+c1δ(ˉu0)+(1+c1)+ˉu0(x,r)=0. (3.6)
    ρ2:{u2(x,r)+c1δ(u1)+c2δ(u0)+c2(h(x,r)u0(x,r))(1+c1)u1(x,r)=0,ˉu2(x,r)+c1δ(ˉu1)+c2δ(ˉu0)+c2(ˉh(x,r)ˉu0(x,r))(1+c1)ˉu1(x,r)=0. (3.7)
    ρn:{un(x,r)+c1δ(un)+c2δ(un1)+c3(h+δ(u0))...c2un1(x,r)(1+c1)un(x,r)=0,ˉun(x,r)+c1δ(ˉun)+c2δ(ˉun1)+c3(h+δ(ˉu0))...c2ˉun1(x,r)(1+c1)ˉun(x,r)=0. (3.8)

    For calculating the constants c1,c2,c3..., mth order optimum solution becomes

    {um(x,r,cl)=u0(x,r)+mk=1uk(x,r,cl),l=1,2,3,...m,ˉum(x,r,cl)=ˉu0(x,r)+mk=1ˉuk(x,r,cl),l=1,2,3,...m. (3.9)

    Putting Eq (3.9) into Eq (3.1), we can found our residual given as follow:

    {R(x,r;cl)=um(x,r;cl)h(x,r)δ(u),l=1,2,...ˉR(x,r;cl)=ˉum(x,r;cl)ˉh(x,r)δ(ˉu),l=1,2,... (3.10)

    If R(x,r;cl)=0, then um(x,r;cl)&ˉum(x,r;cl) will be the exact solutions.

    Optimum solution contains some auxiliary constants; the optimal values of these constants are obtained through various techniques. In the present work, we have used the least square method [56,57]. The method of least squares is a powerful technique for obtaining the values of auxiliary constants. By putting the optimal values of these constants in Eq (8), we obtain the OHAM solution.

    Problem 4.1. Consider system of fuzzy fractional order Volterra integro-differential equation as [58]:

    {Dαxu_(x,r)=(r1)+x0u_(t,r)dtDαxˉu(x,r)=(1r)+x0ˉu(t,r)dt,0<α1,x[0,1], (4.1)

    subject to the fuzzy initial condition [u(0)]r=[r1,1r], and for α=1 fuzzy fractional order Volterra integro-differential equations the exact solution is [u(x)]r=[r1,1r]Sinh(x) and 0r1.

    By follow the technique as discussed in section 3, we get series of problems and their solutions as:

    {Dxαu_0(x,r)+(1r)=0,Dxα¯u0(x,r)+(r1)=0. (4.2)
    {Dxαu_1(x,r)1+rc1+rc1+(x0u_0(t,r)dt)c1Dxαu_0(x,r)c1Dxαu_0(x,r)=0,Dxαˉu1(x,r)+1r+c1rc1+(x0ˉu0(t,r)dt)c1Dxαˉu0(x,r)c1Dxαˉu(x,r)=0. (4.3)
    {Dxαu_2(x,r)+(x0u_1(t,r)dt)c1c2+rc2+(x0u_0(t,r)dt)c2c2Dxαu_0(x,r)Dxαu_1(x,r)c1Dxαu_0(x,r)=0,Dxαˉu2(x,r)+(x0ˉu1(t,r)dt)c1+c2rc2+(x0ˉu0(t,r)dt)c2c2Dxαˉu0(x,r)Dxαˉu1(x,r)c1Dxαˉu0(x,r)=0. (4.4)
    {Dxαu_3(x,r)+(x0u_2(t,r)dt)c1+(x0u_1(t,r)dt)c2c3+rc3+(x0u_0(t,r)dt)c3c3Dxαu_0(x,r)c2Dxαu_1(x,r)Dxαu_2(x,r)c1Dxαu_2(x,r)=0,Dxαˉu3(x,r)+(x0ˉu2(t,r)dt)c1+(x0ˉu1(t,r)dt)c2+c3rc3+(x0ˉu0(t,r)dt)c3c3Dxαˉu0(x,r)c2Dxαˉu1(x,r)Dxαˉu2(x,r)c1Dxαˉu2(x,r)=0. (4.5)

    Their solutions are

    {u_0(x,r)=(1+r)xααΓ(α)ˉu0(x,r)=(1+r)xααΓ(α), (4.6)
    {u_1(x,r)=(1+r)x1+2αc1Γ(2+2α),ˉu1(x,r)=(1+r)x1+2αc1Γ(2+2α). (4.7)
    {u_2(x,r)=(1+r)x1+2α(x1+αc21Γ(3+3α)c1+c21+c2Γ(2+2α)),ˉu2(x,r)=(1+r)x1+α(x1+αc21Γ(3+3α)+c1+c21+c2Γ(2+2α)). (4.8)
    {u_3(x,r)=(1+r)x1+2α(x2+2αc31Γ(4+4α)+2x1+αc1(c1+c21+c2)Γ(3+3α)c1+2c21+c31+c2+2c1c2+c3Γ(2+2α)),ˉu3(x,r)=(1+r)x1+2α(x2+2αc31Γ(4+4α)2x1+αc1(c1+c21+c2)Γ(3+3α)+c1+2c21+c31+c2+2c1c2+c3Γ(2+2α)). (4.9)

    Adding (4.6), (4.7), (4.8) and (4.9), one can construct u_(x,r) & \bar u(x, r) :

    \left\{ \begin{array}{l} \underline u (x, r) = ( - 1 + r){x^\alpha }\left( \begin{gathered} \frac{1}{{\Gamma (1 + \alpha )}} - \frac{{{x^{3 + 3\alpha }}c_1^3}}{{\Gamma (4 + 4\alpha )}} + \frac{{{x^{2 + 2\alpha }}{c_1}({c_1}(3 + 2{c_1}) + 2{c_2})}}{{\Gamma (3 + 3\alpha )}} - \hfill \\ \frac{{{x^{1 + \alpha }}(2{c_2} + {c_1}(3 + {c_1}(3 + {c_1}) + 2{c_2}) + {c_3})}}{{\Gamma (2 + 2\alpha )}} \hfill \\ \end{gathered} \right), \hfill \\ \bar u(x, r) = ( - 1 + r){x^\alpha }\left( \begin{gathered} - \frac{1}{{\Gamma (1 + \alpha )}} + \frac{{{x^{3 + 3\alpha }}c_1^3}}{{\Gamma (4 + 4\alpha )}} - \frac{{{x^{2 + 2\alpha }}{c_1}({c_1}(3 + 2{c_1}) + 2{c_2})}}{{\Gamma (3 + 3\alpha )}} + \hfill \\ \frac{{{x^{1 + \alpha }}(2{c_2} + {c_1}(3 + {c_1}(3 + {c_1}) + 2{c_2}) + {c_3})}}{{\Gamma (2 + 2\alpha )}} \hfill \\ \end{gathered} \right). \hfill \\ \end{array} \right. (4.10)

    Values of {c_1}, \, \, {c_2} and {c_3} contain is in Eq (4.10)

    Substituting the values from Table 1 into Eq (4.10), the approximate solutions for \underline u (x, r) & \bar u(x, r) at different values of \alpha taking r = 0.75 respectively is as follow

    \alpha = 0.7
    \left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.2751369{x}^{0.7}-0.08602097{x}^{2.4}-0.25{x}^{2.4}(-0.00904995+0.0376716{x}^{1.7})\\ -0.25{x}^{2.4}(0.000425859-0.00198165{x}^{1.7}+0.0021734872{x}^{3.4}), \\ \overline{u}(x, r)\approx 0.2751369{x}^{0.7}+0.08602097{x}^{2.4}-0.25{x}^{2.4}(0.00904995-0.0376716{x}^{1.7})-\\ 0.25{x}^{2.4}(-0.000425859+0.00198165{x}^{1.7}-0.0021734872{x}^{3.4}).\end{array} \right. (4.11)
    \alpha = 0.8
    \left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.2684178{x}^{0.8}-0.0685589{x}^{2.6}-0.25{x}^{2.6}(-0.005391327+0.023297809{x}^{1.8})\\ -0.25{x}^{2.6}(0.000197513-0.0009160448{x}^{1.8}+0.001008410504{x}^{3.6}), \\ \overline{u}(x, r)\approx 0.2684178{x}^{0.8}+0.0685589{x}^{2.6}-0.25{x}^{2.6}(0.0053913-0.023297809{x}^{1.8})\\ -0.25{x}^{2.6}(-0.000197513+0.0009160448{x}^{1.8}-0.001008410504{x}^{3.6}).\end{array} \right. (4.12)
    \alpha = 0.9
    \left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.2599385{x}^{0.9}-0.0540269{x}^{2.8}-0.25{x}^{2.8}(-0.0031639+0.01418901{x}^{1.9})\\ -0.25{x}^{2.8}(0.00008989-0.0004154745{x}^{1.9}+0.000458881{x}^{3.8}), \\ \overline{u}(x, r)\approx 0.2599385{x}^{0.9}+0.0540269{x}^{2.8}-0.25{x}^{2.8}(0.0031639-0.014189097{x}^{1.9})\\ -0.25{x}^{2.8}(-0.00008989+0.0004154745{x}^{1.9}-0.000458881{x}^{3.8}).\end{array} \right. (4.13)
    \alpha = 1
    \left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.25x-0.042114377{x}^{3}-0.25{x}^{3}(-0.001829736+0.0085133796{x}^{2})\\ -0.25{x}^{3}(0.0000401535-0.0001849397{x}^{2}+0.00020487753{x}^{4}), \\ \overline{u}(x, r)\approx 0.25x+0.042114377{x}^{3}-0.25{x}^{3}(0.001829736-0.0085133796{x}^{2})\\ -0.25{x}^{3}(-0.0000401535+0.0001849397{x}^{2}-0.0002048775{x}^{4}).\end{array} \right. (4.14)
    Table 1.  at r = 0.75.
    \alpha {\underline c _1} & {\overline c _1} {\underline c _2} & {\overline c _2} {\underline c _3} & {\overline c _3}
    0.7 −1.0257850714449026 5.298291106236844×10−4 −3.040859671410477×10−5
    0.8 −1.0193406988378892 3.249294721058776×10−4 −1.5364294422415488×10−5
    0.9 −1.014446487354385 1.9694362983845834×10−4 −7.63055570435551×10−6
    1 −1.0107450504316333 1.1791102776455743×10−4 −3.779171763451589×10−6

     | Show Table
    DownLoad: CSV

    Substituting the values from Table 2 into Eq (4.10), the approximate solutions for \underline u (x, r) & \bar u(x, r) at different values of \alpha taking r = 0.5 respectively is as follow

    \alpha = 0.7
    \left\{ \begin{array}{l} \underline u (x, r) \approx - 0.5502737{x^{0.7}} - 0.172041940{x^{2.4}} - 0.5{x^{2.4}}( - 0.00904995 + 0.0376716{x^{1.7}}) \hfill \\ - 0.5{x^{2.4}}(0.0004258594 - 0.0019816476{x^{1.7}} + 0.0021734872{x^{3.4}}), \hfill \\ \bar u(x, r) \approx 0.5502737{x^{0.7}} + 0.172041940{x^{2.4}} - 0.5{x^{2.4}}(0.00904995 - 0.03767164{x^{1.7}}) \hfill \\ - 0.5{x^{2.4}}( - 0.0004258594 + 0.0019816476{x^{1.7}} - 0.0021734872{x^{3.4}}). \hfill \\ \end{array} \right. (4.15)
    \alpha = 0.8
    \left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.53683564{x}^{0.8}-0.137117858{x}^{2.6}-0.5{x}^{2.6}(-0.00539133+0.02329781{x}^{1.8})\\ -0.5{x}^{2.6}(0.00019751-0.0009160448{x}^{1.8}+0.0010084105{x}^{3.6}), \\ \overline{u}(x, r)\approx 0.53683564{x}^{0.8}+0.137117858{x}^{2.6}-0.5{x}^{2.6}(0.00539133-0.02329781{x}^{1.8})\\ -0.5{x}^{2.6}(-0.00019751+0.0009160448{x}^{1.8}-0.0010084105{x}^{3.6}).\end{array} \right. (4.16)
    \alpha = 0.9
    \left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.5198771{x}^{0.9}-0.10805378{x}^{2.8}-0.5{x}^{2.8}(-0.0031640+0.0141891{x}^{1.9})\\ -0.5{x}^{2.8}(0.0000898945-0.0004154745{x}^{1.9}+0.000458881{x}^{3.8}), \\ \overline{u}(x, r)\approx 0.5198771{x}^{0.9}+0.10805378{x}^{2.8}-0.5{x}^{2.8}(0.0031640-0.0141891{x}^{1.9})\\ -0.5{x}^{2.8}(-0.0000898945+0.0004154745{x}^{1.9}-0.000458881{x}^{3.8}).\end{array} \right. (4.17)
    \alpha = 1
    \left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.5x-0.0842288{x}^{3}-0.5{x}^{3}(-0.0018298+0.008513{x}^{2})-0.5{x}^{3}\\ (0.0000401612-0.00018494622{x}^{2}+0.0002048777{x}^{4}), \\ \overline{u}(x, r)\approx 0.5x+0.0842288{x}^{3}-0.5{x}^{3}(0.0018298-0.008513{x}^{2})-0.5{x}^{3}\\ (-0.000040162+0.00018494622{x}^{2}-0.0002048777{x}^{4}).\end{array} \right. (4.18)
    Table 2.  at r = 0.5.
    \alpha {\underline c _1} & {\overline c _1} {\underline c _2} & {\overline c _2} {\underline c _3} & {\overline c _3}
    0.7 −1.0257850714449026 5.298291106236844×10−4 −3.040859671410477×10−5
    0.8 −1.0193406988378892 3.249294721058776×10−4 −1.5364294422415488×10−5
    0.9
    1
    −1.014446487354385
    −1.0107453381292266
    1.9694362983845834×10−4
    1.1800167363027721×10−4
    −7.630555570435551×10−6
    −3.726389827252244×10−6

     | Show Table
    DownLoad: CSV

    Problem 4.2. Consider system of fuzzy fractional order Volterra integro-differential equation as [59]:

    D_x^\alpha u(x, r) + \int\limits_0^t {u(t, r)dt = 0} , \, \, \, \, 0 < \alpha \leqslant 1, \, \, x \in [0, 1], (4.19)

    subject to the fuzzy initial condition {\left[ {u(0)} \right]^r} = \left[ {r - 1, 1 - r} \right], and the exact solution is \underline u (x, r) = \left( {r - 1} \right){E_{\alpha + 1}}\left( { - {t^{\alpha + 1}}} \right), \bar u(x, r) = \left( {1 - r} \right){E_{\alpha + 1}}\left( { - {t^{\alpha + 1}}} \right),

    where {E_{\alpha + 1}} is a Mittag-Leffler function and 0 \leqslant r \leqslant 1.

    By follow the technique as discussed in section 3, we get series of problems and their solutions as:

    \left\{ \begin{array}{l} {D_x}^\alpha {\underline u _0}(x, r) = 0, \hfill \\ {D_x}^\alpha {{\bar u}_0}(x, r) = 0. \hfill \\ \end{array} \right. (4.20)
    \left\{ \begin{array}{l} {D_x}^\alpha {\underline u _1}(x, r) + \left( { - \mathop \smallint \limits_0^x {{\underline u }_0}(t, r)dt} \right){c_1} - {D_x}^\alpha {\underline u _0}(x, r) - {c_1}{D_x}^\alpha {\underline u _0}(x, r) = 0, \hfill \\ {D_x}^\alpha {{\bar u}_1}(x, r) + \left( { - \mathop \smallint \limits_0^x {{\bar u}_0}(t, r)dt} \right){c_1} - {D_x}^\alpha {{\bar u}_0}(x, r) - {c_1}{D_x}^\alpha \bar u(x, r) = 0. \hfill \\ \end{array} \right. (4.21)
    \left\{ \begin{array}{l} {D_x}^\alpha {\underline u _2}(x, r) + \left( {\mathop \smallint \limits_0^x {{\underline u }_1}(t, r)dt} \right){c_1} - \left( {\mathop \smallint \limits_0^x {{\underline u }_0}(t, r)dt} \right){c_2} - {c_2}{D_x}^\alpha {\underline u _0}(x, r) - {D_x}^\alpha {\underline u _1}(x, r) - {c_1}{D_x}^\alpha {\underline u _0}(x, r) = 0, \hfill \\ {D_x}^\alpha {{\bar u}_2}(x, r) + \left( { - \mathop \smallint \limits_0^x {{\bar u}_1}(t, r)dt} \right){c_1} - \left( {\mathop \smallint \limits_0^x {{\bar u}_0}(t, r)dt} \right){c_2} - {c_2}{D_x}^\alpha {{\bar u}_0}(x, r) - {D_x}^\alpha {{\bar u}_1}(x, r) - {c_1}{D_x}^\alpha {{\bar u}_0}(x, r) = 0. \hfill \\ \end{array} \right. (4.22)
    \left\{ \begin{array}{l} {D_x}^\alpha {\underline u _3}(x, r) - \left( {\mathop \smallint \limits_0^x {{\underline u }_2}(t, r)dt} \right){c_1} - \left( {\mathop \smallint \limits_0^x {{\underline u }_1}(t, r)dt} \right){c_2} - \left( {\mathop \smallint \limits_0^x {{\underline u }_0}(t, r)dt} \right){c_3} - {c_3}{D_x}^\alpha {\underline u _0}(x, r) \hfill \\ - {c_2}{D_x}^\alpha {\underline u _1}(x, r) - {D_x}^\alpha {\underline u _2}(x, r) - {c_1}{D_x}^\alpha {\underline u _2}(x, r) = 0, \hfill \\ {D_x}^\alpha {{\bar u}_3}(x, r) - \left( {\mathop \smallint \limits_0^x {{\bar u}_2}(t, r)dt} \right){c_1} - \left( {\mathop \smallint \limits_0^x {{\bar u}_1}(t, r)dt} \right){c_2} - \left( {\mathop \smallint \limits_0^x {{\bar u}_0}(t, r)dt} \right){c_3} - {c_3}{D_x}^\alpha {{\bar u}_0}(x, r) \hfill \\ - {c_2}{D_x}^\alpha {{\bar u}_1}(x, r) - {D_x}^\alpha {{\bar u}_2}(x, r) - {c_1}{D_x}^\alpha {{\bar u}_2}(x, r) = 0. \hfill \\ \end{array} \right. (4.23)

    And their solutions are

    \left\{ \begin{array}{l} {\underline u _0}(x, r) = r - 1, \hfill \\ {{\bar u}_0}(x, r) = 1 - r. \hfill \\ \end{array} \right. (4.24)
    \left\{ \begin{array}{l} {\underline u _1}(x, \underline r ) = \frac{{( - 1 + r){x^{1 + \alpha }}{c_1}}}{{(\alpha + {\alpha ^2})\Gamma (\alpha )}}, \hfill \\ {\overline u _1}(x, \overline r ) = - \frac{{( - 1 + r){x^{1 + \alpha }}{c_1}}}{{\alpha (1 + \alpha )\Gamma (\alpha )}}, \hfill \\ \end{array} \right. (4.25)
    \left\{ \begin{array}{l} {\underline u _2}(x, r) = ( - 1 + r){x^{1 + \alpha }}\left( {\frac{{{x^{1 + \alpha }}c_1^2}}{{\Gamma (3 + 2\alpha )}} + \frac{{{c_1} + c_1^2 + {c_2}}}{{\Gamma (2 + \alpha )}}} \right), \hfill \\ {{\bar u}_2}(x, r) = ( - 1 + r){x^{1 + \alpha }}\left( { - \frac{{{x^{1 + \alpha }}c_1^2}}{{\Gamma (3 + 2\alpha )}} - \frac{{{c_1} + c_1^2 + {c_2}}}{{\Gamma (2 + \alpha )}}} \right). \hfill \\ \end{array} \right. (4.26)
    \left\{ \begin{array}{l} {\underline u _3}(x, r) = \frac{{( - 1 + r){x^{1 + \alpha }}\left( {{c_2} + {c_1}\left( {1 + \frac{{{x^{2 + 2\alpha }}c_1^2}}{{\Gamma (4 + 2\alpha )}} + {c_1}(2 + {c_1}) + 2{c_2} + \frac{{2{x^{1 + \alpha }}({c_1} + c_1^2 + {c_2})}}{{\Gamma (3 + 3\alpha )}}} \right) + {c_3}} \right)}}{{\Gamma (1 + \alpha )}}, \hfill \\ {{\bar u}_3}(x, r) = ( - 1 + r){x^{1 + \alpha }}\left( { - \frac{{{x^{2 + 2\alpha }}c_1^3}}{{\Gamma (4 + 3\alpha )}} - \frac{{2{x^{1 + \alpha }}{c_1}({c_1} + c_1^2 + {c_2})}}{{\Gamma (3 + 2\alpha )}} - \frac{{{c_2} + {c_1}({{(1 + {c_1})}^2} + 2{c_2}) + {c_3}}}{{\Gamma (2 + \alpha )}}} \right), \hfill \\ \end{array} \right. (4.27)

    Adding (4.24), (4.25), (4.26) and (4.27), one can construct \underline u (x, r) & \bar u(x, r) :

    \left\{ \begin{array}{l} \underline u (x, r) = - 1 + r + ( - 1 + r){x^{1 + \alpha }}\left( \begin{gathered} \frac{{{x^{1 + \alpha }}c_1^2}}{{\Gamma (3 + 2\alpha )}} + \frac{{{x^{2 + 2\alpha }}c_1^3}}{{\Gamma (1 + \alpha )\Gamma (4 + 2\alpha )}} + \frac{{2{x^{1 + \alpha }}{c_1}({c_1} + c_1^2 + {c_2})}}{{\Gamma (1 + \alpha )\Gamma (3 + \alpha )}} \hfill \\ + \frac{{{c_1}(2 + {c_1}) + {c_2}}}{{\Gamma (2 + \alpha )}} + \frac{{{c_2} + {c_1}({{(1 + {c_1})}^2} + 2{c_2}) + {c_3}}}{{\Gamma (1 + \alpha )}} \hfill \\ \end{gathered} \right), \hfill \\ \bar u(x, r) = 1 - r + ( - 1 + r){x^{1 + \alpha }}\left( \begin{gathered} - \frac{{{x^{2 + 2\alpha }}c_1^3}}{{\Gamma (4 + 3\alpha )}} - \frac{{{x^{1 + \alpha }}{c_1}({c_1}(3 + 2{c_1}) + 2{c_2})}}{{\Gamma (3 + 2\alpha )}} - \hfill \\ \frac{{2{c_2} + {c_1}(3 + {c_1}(3 + {c_1}) + 2{c_2}) + {c_3}}}{{\Gamma (2 + \alpha )}} \hfill \\ \end{gathered} \right). \hfill \\ \end{array} \right. (4.28)

    Values of {c_1}, \, \, {c_2} and {c_3} contain in Eq (4.28)

    Substituting the values from Tables 3 and 4 into Eq (4.28), the approximate solutions for \underline u (x, r) & \bar u(x, r) at different values of \alpha taking r = 0.5 is as follow

    \alpha = 0.2
    \left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.5-0.5{x}^{1.2}(-0.905948+0.325139{x}^{1.2}-0.055807{x}^{2.4}), \\ \overline{u}(x, r)\approx 0.5-0.5{x}^{1.2}(0.905948-0.325139{x}^{1.2}+0.055807{x}^{2.4}).\end{array} \right. (4.29)
    \alpha = 0.4
    \left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.5-0.5{x}^{1.4}(-0.804545+0.210093{x}^{1.4}-0.025570{x}^{2.8}), \\ \overline{u}(x, r)\approx 0.5-0.5{x}^{1.4}(0.80454545-0.210093{x}^{1.4}+0.025570{x}^{2.8}).\end{array} \right. (4.30)
    \alpha = 0.6
    \left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.5-0.5{x}^{1.6}(-0.699349+0.128177{x}^{1.6}-0.010450{x}^{3.2}), \\ \overline{u}(x, r)\approx 0.5-0.5{x}^{1.6}(0.699349-0.128177{x}^{1.6}+0.010450{x}^{3.2}).\end{array} \right. (4.31)
    \alpha = 0.8
    \left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.5-0.5{x}^{1.8}(-0.596450+0.074561{x}^{1.8}-0.0038863{x}^{3.6}), \\ \overline{u}(x, r)\approx 0.5-0.5{x}^{1.8}(0.5964503-0.074561{x}^{1.8}+0.0038863{x}^{3.6}).\end{array} \right. (4.32)
    \alpha = 1
    \left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.5-0.5{x}^{2}(-0.499992+0.04163089{x}^{2}-0.001336253{x}^{4}), \\ \overline{u}(x, r)\approx 0.5-0.5{x}^{2}(0.4999914-0.04163076{x}^{2}+0.001336247{x}^{4}).\end{array} \right. (4.33)
    Table 3.  at r = 0.5.
    \alpha {\underline c _1} {\underline c _2} {\underline c _3}
    0.2 −0.8038238618267683 7.6178003377104005×10−3 −1.334733828882206×10−3
    0.4 −0.739676946061329 0.02530379950927192 −3.78307542584476×10−3
    0.6 −0.6725325561865596 0.04771922184372877 −8.140310215705777×10−3
    0.8 −0.6062340661192892 0.06889015391501904 −0.015746150088743013
    1.0 −0.5432795308983783 0.08615024033359142 −0.026582381449582644

     | Show Table
    DownLoad: CSV
    Table 4.  at r = 0.5.
    \alpha {\overline c _1} {\overline c _2} {\overline c _3}
    0.2 −0.9072542694138958 3.5727393445527333×10−3 3.637119200533134×10−4
    0.4 −0.9409187563211361 1.9803045412863643×10−3 1.7807647588483674×10−4
    0.6 −0.9636043521097131 9.808463578935658×10−4 7.369503633679291×10−5
    0.8 −0.9781782948007163 4.4492367269725435×10−4 2.6729929232143615×10−5
    1.0 −0.9872029432879605 1.896941835601795×10−4 1.021455544474314×10−5

     | Show Table
    DownLoad: CSV

    Substituting the values from Tables 5 and 6 into Eq (4.28), the approximate solutions for \underline u (x, r) & \bar u(x, r) at different values of r taking \alpha = 0.5 is as follow

    r = 0
    \left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -1-{x}^{1.5}(-0.75199032254+0.165168588631{x}^{1.5}-0.016559344247{x}^{3.}), \\ \overline{u}(x, r)\approx 1-{x}^{1.5}(0.75199032256-0.165168588653{x}^{1.5}+0.016559344262{x}^{3.}).\end{array} \right. (4.34)
    r = 0.2
    \left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.8-0.8{x}^{1.5}(-0.751990322550+0.165168588{x}^{1.5}-0.01655934425{x}^{3.}), \\ \overline{u}(x, r)\approx 0.8-0.8{x}^{1.5}(0.751990322554-0.165168588{x}^{1.5}+0.01655934426{x}^{3.}). \end{array} \right. (4.35)
    r = 0.4
    \left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.6-0.6{x}^{1.5}(-0.7519903225+0.1651685886{x}^{1.5}-0.016559344250{x}^{3.}), \\ \overline{u}(x, r)\approx 0.6-0.6{x}^{1.5}(0.7519903225-0.1651685886{x}^{1.5}+0.0165593442496{x}^{3.}).\end{array} \right. (4.36)
    r = 0.6
    \left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.4-0.4{x}^{1.5}(-0.751990322550+0.16516858863{x}^{1.5}-0.01655934425{x}^{3.}), \\ \overline{u}(x, r)\approx 0.4-0.4{x}^{1.5}(0.751990322554-0.16516858865{x}^{1.5}+0.01655934426{x}^{3.}).\end{array} \right. (4.37)
    r = 0.8
    \left\{ \begin{array}{l} \underset{\_}{u}(x, r)\approx -0.1999910-0.1999910{x}^{1.5}(-0.7519903+0.16516859{x}^{1.5}-0.0165593{x}^{3.}), \\ \overline{u}(x, r)\approx 0.19999910-0.19999910{x}^{1.5}(0.7519903-0.16516859{x}^{1.5}+0.0165593{x}^{3.}).\end{array} \right. (4.38)
    Table 5.  at \alpha = 0.5.
    r {\underline c _1} {\underline c _2} {\underline c _3}
    0 - 0.7062087686601037 0.03638991875243272 - 5.6065011933001474 \times {10^{ - 3}}
    0.2 - 0.7062087687083373 0.03638991875755982 - 5.606501189519195 \times {10^{ - 3}}
    0.4 - 0.7062087686911187 0.03638991875566347 - 5.606501190876118 \times {10^{ - 3}}
    0.6 - 0.7062087687083373 0.03638991875755982 - 5.606501189519195 \times {10^{ - 3}}
    0.8 - 0.7062087686312865 0.036389918749298394 - 5.606501195566148 \times {10^{ - 3}}

     | Show Table
    DownLoad: CSV
    Table 6.  at \alpha = 0.5.
    r {\overline c _1} {\overline c _2} {\overline c _3}
    0 - 0.9534544876637709 1.4110239362094313 \times {10^{ - 3}} 1.1669890178958486 \times {10^{ - 4}}
    0.2 - 0.9534544876175205 1.4110239407705756 \times {10^{ - 3}} 1.1669890245071123 \times {10^{ - 4}}
    0.4 - 0.9534544874104965 1.4110239614371703 \times {10^{ - 3}} 1.1669890547775563 \times {10^{ - 4}}
    0.6 - 0.9534544876175205 1.4110239407705756 \times {10^{ - 3}} 1.1669890245071123 \times {10^{ - 4}}
    0.8 - 0.9534544876289686 1.4110239398903034 \times {10^{ - 3}} 1.1669890235330204 \times {10^{ - 4}}

     | Show Table
    DownLoad: CSV

    Tables 16 show the values of auxiliary constant at different values of r & \alpha for both lower and upper solution of OHAM for the solved problems. Tables 7 and 8 show the comparison of absolute error of 3rd order OHAM with Fractional Residual Power Series (FRPS) Method for 5-approximated solution and k = 5 for both lower and upper solutions of OHAM at different value of \alpha for problem 1. Comparison of absolute error of 3rd orders OHAM for both lower and upper solution of OHAM are shown in Tables 9 and 10. Numerical result show that OHAM provide more accuracy as compared to the other method and as \alpha \to 1 the approximate solution become very close to the exact solution. Graphical representation confirmed the convergence of fractional order solution towards the integer order solution. In Figure 1 graphical representation of OHAM at \alpha = 0.7, \, \, 0.8, \, \, 0.9\, , \, \, 1, \, \, r = 0.75 and \alpha = 0.7, \, \, 0.8, \, \, 0.9\, , \, \, 1, \, \, r = 0.50 are discussed for both \underline u (x, r) & \bar u(x, r) for problem 1. Figures 2 and 3 show the comparison of OHAM with the exact solution at different values of and taking r = 0.75 & r = 0.5 respectively for problem 1. Figure 4 represent the comparison of OHAM at \alpha = 0.2, \, 0.4, \, \, 0.6, \, \, 0.8, \, \, 1, \, \, r = 0.5 and r = 0, \, \, 0.2, \, 0.4, \, \, 0.6, \, \, 0.8, \, \, \alpha \, = 0.5 for both \underline u (x, r) and \bar u(x, r) for problem 2. Figure 5 shows the comparison of OHAM with the exact solution at different values of and r = 0.5 while Figure 6 shows the comparison of OHAM with the exact solution at different values of r and = 0.5 for problem 2.

    Table 7.  Comparison of Absolute Error (Abs Err.) of 3rd order OHAM for \underline u (x, r) and Fractional Residual Power Series (FRPS) [54] Method for 5-approximated solution and k = 5 for problem 1.
    r x FRPS [58]
    \alpha = 0.7
    OHAM FRPS [58]
    \alpha = 0.8
    OHAM FRPS [58]
    \alpha = 0.9
    OHAM FRPS [58]
    \alpha = 1
    OHAM
    0.75 0.2 0.042797 0.040621 0.025514 0.024764 0.011512 0.011321 6.35273×10−10 2.14676×10−9
    0.4 0.059664 0.051698 0.035840 0.032584 0.016392 0.015405 8.14507×10−8 1.00832×10−8
    0.6 0.075769 0.058997 0.045171 0.037635 0.020545 0.018031 1.39554×10−6 1.11336×10−8
    0.8 0.094364 0.066136 0.055863 0.04232 0.025182 0.020362 1.04955×10−5 6.21567×10−9
    0.50 0.2 0.085595 0.081241 0.051027 0.049528 0.011321 0.022643 1.27055×10−9 4.25946×10−9
    0.4 0.119328 0.103396 0.071680 0.065167 0.015405 0.030811 1.62901×10−7 1.98877×10−8
    0.6 0.151537 0.117994 0.090342 0.075269 0.018031 0.036062 2.79107×10−6 2.12923×10−8
    0.8 0.188728 0.132271 0.111723 0.084640 0.020362 0.040725 2.09911×10−5 1.00120×10−8

     | Show Table
    DownLoad: CSV
    Table 8.  Comparison of Absolute Error (Abs Err.) of 3rd order OHAM for \bar u(x, r) and Fractional Residual Power Series (FRPS) [58] Method for 5-approximated solution and k = 5 for problem 1.
    r x FRPS [58]
    \alpha = 0.7
    OHAM FRPS [58]
    \alpha = 0.8
    OHAM FRPS [58]
    \alpha = 0.9
    OHAM FRPS [58]
    \alpha = 1
    OHAM
    0.75 0.2 0.085595 0.081242 0.025514 0.024764 0.011512 0.011321 6.35273×10−10 2.14676×10−9
    0.4 0.119328 0.103396 0.035840 0.032584 0.016392 0.015405 8.14507×10−8 1.00832×10−8
    0.6 0.151537 0.117994 0.045171 0.037635 0.020545 0.018031 1.39554×10−6 1.11336×10−8
    0.8 0.188728 0.132271 0.055862 0.04232 0.025182 0.020362 1.04955×10−5 6.21567×10−9
    0.50 0.2 0.042797 0.040621 0.051027 0.049528 0.023025 0.022643 1.27055×10−9 4.25946×10−9
    0.4 0.059664 0.051698 0.071680 0.065167 0.032784 0.030811 1.62901×10−7 1.98877×10−8
    0.6 0.075769 0.058997 0.090342 0.075269 0.041090 0.036062 2.79107×10−6 2.12923×10−8
    0.8 0.094364 0.066136 0.111723 0.084640 0.050364 0.040725 2.09911×10−5 1.0012×10−8

     | Show Table
    DownLoad: CSV
    Table 9.  Comparison of Absolute Error (Abs Err.) of 3rd order OHAM for \underline u (x, \underline r ) & \bar u(x, r) at different values of \alpha taking r = 0.5 for problem 2.
    x \underline u (x, r)
    \alpha = 0.4
    \bar u(x, r) \underline u (x, r)
    \alpha = 0.6
    \bar u(x, r) \underline u (x, r)
    \alpha = 0.8
    \bar u(x, r) \underline u (x, r)
    \alpha = 1
    \bar u(x, r)
    0.2 1.2736×10−5 1.2736×10−5 3.2527×10−6 3.2527×10−6 6.93075×10−7 6.93076×10−7 1.29476×10−7 1.44585×10−7
    0.4 2.3337×10−6 2.3337×10−6 2.4426×10−6 2.4426×10−6 9.50573×10−7 9.50573×10−7 2.67538×10−7 3.26771×10−7
    0.6 9.9993×10-6 9.9993×10−6 1.8451×10−6 1.8451×10−6 3.8544×10−8 3.85438×10−8 1.10198×10−7 2.39031×10−7
    0.8 4.1251×10−6 4.1251×10−6 1.3416×10−7 1.3416×10−7 6.55017×10−8 6.55017×10−8 9.64628×10−9 2.27883×10−7
    1.0 1.312×10−5 1.312×10−5 1.8373×10−6 1.8373×10−6 5.49341×10−8 5.4934×10−8 7.7356×10−8 3.9735×10−7

     | Show Table
    DownLoad: CSV
    Table 10.  Comparison of Absolute Error (Abs Err.) of 3rd order OHAM for \underline u (x, r) & \bar u(x, r) at different values of r taking \alpha = 0.5 for problem 2.
    x \underline u (x, r)
    r = 0.4
    \bar u(x, r) \underline u (x, r)
    r = 0.6
    \bar u(x, r) \underline u (x, r)
    r = 0.8
    \bar u(x, r) \underline u (x, r)
    r = 1
    \bar u(x, r)
    0. 0. 0. 0. 0. 0. 0. 0. 0.
    0.2 1.0578×10−5 1.0578×10−5 7.9338×10−6 7.9338×10−6 5.28921×10−6 5.28921×10−6 2.64461×10−6 2.64461×10−6
    0.4 4.8940×10−6 4.8940×10−6 3.6705×10−6 3.6705×10−6 2.44698×10−6 2.44698×10−6 1.22349×10−6 1.22349×10−6
    0.6 7.4825×10−6 7.4825×10−6 5.6119×10−6 5.6119×10−6 3.74125×10−6 3.74125×10−6 1.87062×10−6 1.87062×10−6
    1.8 1.8038×10−6 1.8038×10−6 1.35291×10−6 1.35291×10−6 9.01939×10−7 9.01939×10−7 4.5097×10−7 4.5097×10−7

     | Show Table
    DownLoad: CSV
    Figure 1.  Solution plot of OHAM for \underline u (x, r) & \bar u(x, r) at different values of r & \alpha for problem 1.
    Figure 2.  Solution plot of OHAM and Exact for \underline u (x, r) & \bar u(x, r) at different values of \alpha taking r = 0.75 for problem 1.
    Figure 3.  Solution plot of OHAM and Exact for \underline u (x, r) & \bar u(x, r) at different values of \alpha taking r = 0.50 for problem 1.
    Figure 4.  Solution plot of OHAM for \underline u (x, r) & \bar u(x, r) at different values of r & \alpha for problem 2.
    Figure 5.  Solution plot of OHAM and Exact for \underline u (x, r) & \bar u(x, r) at different values of \alpha taking r = 0.5 for problem 2.
    Figure 6.  Solution plot of OHAM and Exact for \underline u (x, r) & \bar u(x, r) at different values of r taking \alpha = 0.5 for problem 2.

    In the research paper, a powerful technique known as Optimal Homotopy Asymptotic Method (OHAM) has been extended to the solution of system of fuzzy integro differential equations of fractional order. The obtained results are quite interesting and are in good agreement with the exact solution. Two numerical equations are taken as test examples which show the behavior and reliability of the proposed method. The extension of OHAM to system of fuzzy integro differential equations of fractional order is more accurate and as a result this technique will more appealing for the researchers for finding out optimum solutions of system of fuzzy integro differential equations of fractional order.

    The authors declare no conflict of interest.



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